An Illustrated Introduction to Topology and Homotopy 9781482220810, 1482220814

Table of contents :
Front Cover
Table of Contents
Preface
Acknowledgment
Part I: Topology
Chapter 1. Sets, Numbers, and Cardinals
Chapter 2. Metric Spaces: Definition, Examples, and Basics
Chapter 3. Topological Spaces: Definition and Examples
Chapter 4. Subspaces, Quotient Spaces, Manifolds, and CW-Complexes
Chapter 5. Products of Spaces
Chapter 6. Connected Spaces and Path Connected Spaces
Chapter 7. Compactness and Related Matters
Chapter 8. Separation Properties
Chapter 9. Urysohn, Tietze, and Stone–Čech
Part 2: Homotopy
Chapter 10. Isotopy and Homotopy
Chapter 11.The Fundamental Group of a Circle and Applications
Chapter 12. Combinatorial Group Theory
Chapter 13. Seifert–van Kampen Theorem and Applications
Chapter 14. On Classifying Manifolds and Related Topics
Chapter 15. Covering Spaces, Part 1
Chapter 16. Covering Spaces, Part 2
Chapter 17. Applications in Group Theory
Bibliography
List of Symbols

Citation preview

Mathematics

TOPOLOGY and HOMOTOPY

The first part of the text covers basic topology, ranging from metric spaces and the axioms of topology through subspaces, product spaces, connectedness, compactness, and separation axioms to Urysohn’s lemma, Tietze’s theorems, ˇ and Stone-Cech compactification. Focusing on homotopy, the second part starts with the notions of ambient isotopy, homotopy, and the fundamental group. The book then covers basic combinatorial group theory, the Seifert-van Kampen theorem, knots, and low-dimensional manifolds. The last three chapters discuss the theory of covering spaces, the Borsuk-Ulam theorem, and applications in group theory, including various subgroup theorems. Requiring only some familiarity with group theory, the text includes a large number of figures as well as various examples that show how the theory can be applied. Each section starts with brief historical notes that trace the growth of the subject and ends with a set of exercises.

K12146

KALAJDZIEVSKI

Features • Provides a comprehensive treatment of basic topology and homotopy that uses a combination of rigorous proofs and extensive visualization • Gives full details of most proofs • Contains nearly 600 figures that help clarify difficult concepts • Presents historical notes that outline the growth of the subject • Includes about 750 exercises, many of which are relatively new

AN ILLUSTRATED INTRODUCTION TO

An Illustrated Introduction to Topology and Homotopy explores the beauty of topology and homotopy theory in a direct and engaging manner while illustrating the power of the theory through many, often surprising, applications. This selfcontained book takes a visual and rigorous approach that incorporates both extensive illustrations and full proofs.

TOPOLOGY and HOMOTOPY

AN ILLUSTRATED INTRODUCTION TO

AN ILLUSTRATED INTRODUCTION TO

TOPOLOGY and

HOMOTOPY

SASHO KALAJDZIEVSKI

w w w. c rc p r e s s . c o m

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AN ILLUSTRATED INTRODUCTION TO

TOPOLOGY and

HOMOTOPY

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AN ILLUSTRATED INTRODUCTION TO

TOPOLOGY and

HOMOTOPY

SASHO KALAJDZIEVSKI University of Manitoba Winnipeg, Canada

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CRC Press Taylor & Francis Group 6000 Broken Sound Parkway NW, Suite 300 Boca Raton, FL 33487-2742 © 2015 by Taylor & Francis Group, LLC CRC Press is an imprint of Taylor & Francis Group, an Informa business No claim to original U.S. Government works Version Date: 20141014 International Standard Book Number-13: 978-1-4822-2081-0 (eBook - PDF) This book contains information obtained from authentic and highly regarded sources. Reasonable efforts have been made to publish reliable data and information, but the author and publisher cannot assume responsibility for the validity of all materials or the consequences of their use. The authors and publishers have attempted to trace the copyright holders of all material reproduced in this publication and apologize to copyright holders if permission to publish in this form has not been obtained. If any copyright material has not been acknowledged please write and let us know so we may rectify in any future reprint. Except as permitted under U.S. Copyright Law, no part of this book may be reprinted, reproduced, transmitted, or utilized in any form by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying, microfilming, and recording, or in any information storage or retrieval system, without written permission from the publishers. For permission to photocopy or use material electronically from this work, please access www.copyright.com (http:// www.copyright.com/) or contact the Copyright Clearance Center, Inc. (CCC), 222 Rosewood Drive, Danvers, MA 01923, 978-750-8400. CCC is a not-for-profit organization that provides licenses and registration for a variety of users. For organizations that have been granted a photocopy license by the CCC, a separate system of payment has been arranged. Trademark Notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation without intent to infringe. Visit the Taylor & Francis Web site at http://www.taylorandfrancis.com and the CRC Press Web site at http://www.crcpress.com

To Timjan, Damjan, Darja and Nina

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Table of Contents Preface

xi

Acknowledgment

xv

Part I Topology Chapter 1 n Sets, Numbers, and Cardinals

3

1.1 Sets and Numbers

3

1.2 Sets and Cardinal Numbers

9

1.3 Axiom of Choice and Equivalent Statements

12

Chapter 2 n Metric Spaces: Definition, Examples, and Basics

17

2.1 Metric Spaces: Definition and Examples

17

2.2 Metric Spaces: Basics

23

Chapter 3 n Topological Spaces: Definition and Examples

33

3.1 The Definition and Some Simple Examples

33

3.2 Some Basic Notions

39

3.3 Bases

48

3.4 Dense and Nowhere Dense Sets

57

3.5 Continuous Mappings

61

Chapter 4 n Subspaces, Quotient Spaces, Manifolds, and CW-Complexes

71

4.1 Subspaces

71

4.2 Quotient Spaces

75

4.3 The Gluing Lemma, Topological Sums, and Some Special Quotient Spaces

81

4.4 Manifolds and CW-Complexes

90

vii

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viii   ◾    Table of Contents

Chapter 5 n Products of Spaces

101

5.1 Finite Products of Spaces

101

5.2 Infinite Products of Spaces

110

5.3 Box Topology

117

Chapter 6 n Connected Spaces and Path Connected Spaces

121

6.1 Connected Spaces: Definition and Basic Facts

121

6.2 Properties of Connected Spaces

128

6.3 Path Connected Spaces

133

6.4 Path Connected Spaces: More Properties and Related Matters

139

6.5 Locally Connected and Locally Path Connected Spaces 147

Chapter 7 n Compactness and Related Matters

153

7.1 Compact Spaces: Definition

153

7.2 Properties of Compact Spaces

159

7.3 Compact, Lindelöf, and Countably Compact Spaces

165

7.4 Bolzano, Weierstrass, and Lebesgue

170

7.5 Compactification

173

7.6 Infinite Products of Spaces and Tychonoff Theorem

179

Chapter 8 n Separation Properties

185

8.1 The Hierarchy of Separation Properties

185

8.2 Regular Spaces and Normal Spaces

191

8.3 Normal Spaces and Subspaces

195

Chapter 9 n Urysohn, Tietze, and Stone–Čech

199

9.1 THE Urysohn Lemma

199

9.2 The Tietze Extension Theorem

205

9.3 Stone–Čech Compactification

209

Part 2

Homotopy

Chapter 10 n Isotopy and Homotopy

217

10.1 Isotopy and Ambient Isotopy

218

10.2 Homotopy

228

10.3 Homotopy and Paths

232

10.4 The Fundamental Group of a Space

238

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Table of Contents   ◾    ix

Chapter 11 n The Fundamental Group of a Circle and Applications

247

11.1 The Fundamental Group of a Circle

247

11.2 Brouwer Fixed Point Theorem and The Fundamental Theorem of Algebra

255

11.3 The Jordan Curve Theorem

261

Chapter 12 n Combinatorial Group Theory

269

12.1 Group Presentations

269

12.2 Free Groups, Tietze, Dehn

275

12.3 Free Products and Free Products with Amalgamation

283

Chapter 13 n Seifert–van Kampen Theorem and Applications

291

13.1 Seifert–van Kampen Theorem

291

13.2 Seifert–van Kampen Theorem: Examples

299

13.3 Seifert–van Kampen Theorem and Knots

310

13.4 Torus Knots and Alexander’s Horned Sphere

319

13.5 Links

327

Chapter 14 n On Classifying Manifolds and Related Topics

333

14.1 1-Manifolds

333

14.2 Compact 2-Manifolds: Preliminary Results

336

14.3 Compact 2-Manifolds: Classification

343

14.4 Regarding Classification of CW-Complexes and Higher Dimensional Manifolds

356

14.5 Higher Homotopy Groups: A Brief Overview

365

Chapter 15 n Covering Spaces, Part 1

371

15.1 Covering Spaces: Definition, Examples, and Preliminaries 371 15.2 Lifts of Paths

375

15.3 Lifts of Mappings

379

15.4 Covering Spaces and Homotopy

388

Chapter 16 n Covering Spaces, Part 2

397

16.1 Covering Spaces and Sheets

397

16.2 Covering Transformations

400

16.3 Covering Spaces and Groups Acting Properly Discontinuously

405

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x   ◾    Table of Contents

16.4 Covering Spaces: Existence

415

16.5 The Borsuk–Ulam Theorem

421

Chapter 17 n Applications in Group Theory

429

17.1 Cayley Graphs and Covering Spaces

429

17.2 Topographs and Presentations

436

17.3 Subgroups of Free Groups

442

17.4 Two Subgroup Theorems

448

Bibliography

455

List of Symbols

459

Index

463

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Preface Chronology. In the year 2000 I was given two courses to teach: analysis and topology. The Analysis course was assigned to me due to an emergency redistribution; Topology was thrown in as an award for accepting to teach Analysis on such short notice. Algebraic topology, which covered only homotopy theory in its curriculum, was (re)introduced into the undergraduate-graduate program of the Mathematics Department (University of Manitoba) in 2004. I have taught these two courses almost continuously ever since. This book is primarily an upshot of my teaching lifestyle over these years. The level of the book, and the target audience. The book has two clearly differentiated parts: Part 1 is topology and Part 2 is homotopy. As indicated by the title of the book, we do not have any pretense to go very deeply into these subjects. On the other hand, neither could the content be described as too breezy, for we include most of the important theorems of the basic theory. The material covered in this book is on the fuzzy boundary between the undergraduate and the graduate level. One may tentatively state that Part 1 is mostly at the advanced undergraduate level, Part 2 is mostly at the early graduate level. Indeed, the Algebraic topology (Homotopy theory) course that I have been teaching has always been cross-listed, and has almost always been taken by both graduate and advanced undergraduate students. It is my hope that this book will bring this beautiful theory closer to the undergraduate curriculum.

Content of the Book Part 1. Part 1 begins with a brief introductory chapter on sets and (cardinal) numbers, followed by an equally short chapter about metric spaces. Metric spaces, and metrizable spaces, are discussed (along the way) in the rest of Part 1. The proper topology theory starts with the axioms given at the beginning of Chapter 3. The development of the theory from that point on follows the classical pattern. The basic notions are introduced in Chapter 3, followed by increasingly more complex ideas and properties (Chapters 4–7). The theory given in Part 1 culminates with Tychonoff’s theorem (7.6) and a few other significant results (Chapters 8 and 9). Various applications are discussed as they emerge from the theory. The connection between Part 1 and Part 2. The basic concepts (as are continuity, bases, connectedness, and compactness) from Part 1 are used extensively in Part 2. There are a number of simple results (for example, that the components of open sets in locally path xi

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xii   ◾    Preface

connected spaces are open), and more significant theorems (for example, Tietze’s theorem on extending functions) that are invoked in Part 2. However, the intellectual flavors of the two parts of this book are rather different, primarily because of the extensive link in Part 2 between topology theory and group theory. Part 2. The central theory in this part begins with the concept of homotopy. The introductory discussion of homotopy covers most of the first two chapters in Part 2; this is where we see the introduction of group theory through the notion of the fundamental group of a space. This initial part peaks with the proof that the fundamental group of a circle is infinite cyclic. It is rather amazing that this one, seemingly obvious fact yields such a diverse array of applications, as are, among others, the Brouwer fixed point theorem, the fundamental theorem of algebra, and the Jordan curve theorem. In order to access the core theory of Part 2 one needs combinatorial group theory. Group presentations and the underlying basic theory are introduced in Chapter 12. The statement and the proof of the crucial Seifert–van Kampen theorem given in Chapter 13 rely on Chapter 12. The rest of Chapter 13 is devoted to applications of this theorem. The classification theorem for compact connected 2-manifolds is proven in Chapter 14. The proof is incomplete: due to technical reasons, and rather reluctantly, I have decided against my initial plan to include a proof of Radó’s theorem on triangulability of 2-manifolds in an appendix of the book. This chapter ends with an overview of higherdimensional manifolds and higher homotopy groups. Two chapters are devoted to covering spaces and their immediate applications. The interaction between groups and topological spaces is the most interesting in this last part of the book (Chapters 15–17). It is remarkable (but certainly not unique) that, through the theory of covering spaces, one needs groups to prove such a glaringly topological statement as is the Borsuk–Ulam theorem. Applications of covering spaces theory in group theory are given in the last chapter (17). This part reflects my own mathematical affinities.

Remarks Regarding Using This Book as a Textbook Topology (Part 1). The material in Part 1 roughly fits a 1.5-semester course. Depending on the audience, it could easily fit into a full year course. Typically, during these years of teaching topology as a one-semester course, I cover the following in my lectures: the entire Chapter 3, Sections 4.1 and 4.2, small parts of 4.3 and 4.4, 5.1, 5.2 and the definition of Box topology in 5.3, Chapter 6, 7.1, 7.2 and large parts of 7.3, 7.4 and 7.5, and, finally, 8.1 and 8.2. For the remaining teaching time, if any, I choose one, or at most two of 7.6 (Tychonoff), 8.3 (Tychonoff plank), 9.1 (Urysohn), 9.2 (Tietze) and 9.3 (Stone–Čech). Homotopy (Part 2). Part 2 also has more material than what would fit in a reasonably paced one-semester course. Here is one recommendation for a one-semester course based on my personal experience: 10.1 (briefly), 10.2–10.4, 11.1 and 11.2, 11.3 (briefly or time permitting), 12.1–12.3, 13.1 and 13.2, then 13.3 (through a couple of examples and a short overview) 13.4 (one example of a torus link; I usually have time only to mention Alexander’s horned

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Preface   ◾    xiii

sphere), 13.5 (one example); 14.2 and 14.3 (I never bother much with 14.1 and 1-manifolds), 14.4 (overview) and 14.5 (overview); the entire sections 15.1–15.3, and parts of 15.4; 16.1– 16.3 (almost entirely); I rarely have enough time to prove the existence theorem in 16.4; 16.5 is usually done completely. By the end, there may be time to show one application in group theory. I often choose the claim that subgroups of free groups are free (17.3). Parts 1 and 2. A year-long Topology-Homotopy course is also an option, but I have never done it. The various kinds of beauty of the subject. One does not have to be over-enthusiastic to declare that the mathematical subjects we cover here are of exquisite and diverse beauty. This aesthetic richness does not end within the intellectual realm. There are about 500 figures or illustrations in the book. With that we justify the word ‘illustrated’ in the title. The inclusion of so many figures is partly a consequence of my personal affinity for visualization, partly in order to make some of the arguments, ideas and concepts more evident, and finally, to hint at that “other beauty” of the theory we cover. Most often the first two motives dominate. In a rather small number of instances (say, in case of fractals, Section 2.2) some graphics are included mainly because of their perceived aesthetic merit. Exercises in the book. There are a total of 756 problems in this book, 435 in Part 1, and 321 in Part 2. Most of the exercises came my way as I was writing the theory, so some of them should be new. The exercises are rather roughly ordered according to their level of difficulty in each section. Occasionally the exercises at the end of a section follow the chronology of the topics of that section, in which case easier exercises may appear later. Historical notes in the book. There is also one more somewhat unorthodox feature in this book: I preface almost every section with a brief historical remark. The aim of these remarks is to indicate the chronology of the development of the theory, and also to mention the names of some remarkable mathematicians who contributed in important ways. These encapsulated historical remarks are very laconic, but my hope is that they serve the indicated purpose. Regarding the Style. I have not chosen the style of this book; it chose me. It manifests, for example, through my fondness of pictures, and through my resentment of overly condensed proofs, sometimes called “short proofs”, where it may be hard to easily separate obvious implicit claims from the gaps in the argument of various sizes. Consequently, I have tried to avoid short proofs of such kind. I can see both the advantages and the disadvantages, but I would rather say a few words about the former. I try, most of the time, to give complete proofs of the claims we set out to prove completely. Along the way, we will inevitable encounter obvious, repetitive, or easy arguments. Sometimes the latter are relegated to the set of exercises. Most importantly, when we skip a step during an argument, we will announce it explicitly most of the time. This slightly—but not steeply—diminishes in frequency along the way, after some of the often-mentioned results become sufficiently folklorized. Technical Aspects. Most of the statements in this book are labeled, and prefixed in the standard way. Thus we usually use the word “theorem” as an attribute to a more difficult,

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xiv   ◾    Preface

or (mathematically or historically) more important statement. Most of the time we prefix the statements with the neutral term “proposition”. The proofs of the labeled claims will alternatively end with filled (black) squares ◼. The examples we cover more thoroughly will also be clearly separated from the rest of the text, and the end of the related discussion will usually be indicated with the white square ☐. I used PovRay to make most of the 3D graphics, sometime in combination with Mathematica; exceptionally I used EI Animator. For 2D graphics and for processing 3D graphics I worked with Adobe Illustrator. Labels within the graphics were mostly done with MathType. Images from Art Explosion 525 000 were used in Figures 1.2.2, 1.3.1, 2.1.9, 4.3.2, 4.3.3, and 6.5.2. I will maintain a web page dedicated to this book: http://home.cc.umanitoba.ca/~sasho/sk/topology_homotopy.html The web page contains the figures from the book, as well as additional graphics that did not make the final cut. One advantage is that many of the figures, rendered gray in the book because of cost-related issues, could now be viewed in their original color versions. Posting the figures here also makes them available for usage in lectures. I will eventually add short animations. In the web page I also include various, ostensibly non-mathematical tidbits (say, an insert of Calvino’s All At One Point).

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Acknowledgment The very first course I taught as a teaching assistant was topology, under the mentorship of Smilka Zdravkovska. My own approach in teaching topology during the last decade or so was also influenced by Murray Bell. I am indebted to both of them. Thanks to Chris Camfield for a review of the project in its early stages, to Yemon Choi for pointing to me an application of Tychonoff’s theorem in Ramsey Theory, and to David Gunderson for sharing his library and for drawing my attention to a solution of a combinatorial problem. Many thanks to Adam Clay for pointing some inaccuracies as well for suggesting a few exercises. I am grateful to the following people for helping me with the mathematical content of the book. Grant Woods read a large portion of Part 1 of the book; his comments and suggestion were most valuable to me. Jaydeep Chipalkatti was very kind to read Part 2 of the book: his lucid remarks impacted the treatment of the whole book. I am very grateful to Allan Edmonds for his extensive review of both Part 1 and Part 2 of the book. Moreover, he was extremely gracious to allow me to be in contact with him during the completion of the project, replying thoroughly and promptly to many questions and doubts that I had. Special thanks to Nina Zorboska, who was a discreet contributor of many ideas and exercises. My in-home editor Darja Kalajdzievska was invaluable to me during this project. Thank you very much Dar! Last, but not least, I thank my many students, whose collective input made the most profound impact on this project. Thanks to James Requeima, Clint Enns, Trevor Wares, Adam Bookatz, Iian Smythe, Max Bennett, Blake Madill, Andrii Arman, Mykhailo Akhtariiev, and special thanks to David Gabrielson (who was kind to give me Latex-ed notes of my early topology lectures), Karen Johannson (for sharing with me her carefully handwritten notes of my early algebraic topology lectures), Ivan Levinski (for pointing to me some results I was not aware of), Xiangui Zhao (who found a counterexample to a false claim), Danylo Radchenko (for a few excellent proofs and exercises), Ievgen Bilokopytov (who profoundly influenced one section of this book), Angel Daniel Barría Comicheo (for his commentaries on some exercises), and Damjan Kalajdzievski (for his

xv

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xvi   ◾    Acknowledgment

contributions and for his illuminating and uncompromising critique of “visual proofs” in general). Thanks to Timana too. Bob Stern, Taylor & Francis editor, was very supportive and helpful from the start to almost the end of this project. Also many thanks to Suzanne Lassandro for her help and patience during the proofreading stages of the project. Jos Leys is the author of the artwork in Figures 14.3.24 and 16.3.2. He was also very generous to share with me some of his PovRay codes, and some of the tricks of his trade. I’ve made use of a few crop circles. Thanks to their creators.

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I Topology

1

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Chapter

1

Sets, Numbers, and Cardinals

W

e start with a brief review of some basic notions of set theory. This will be done in a mostly informal manner, for a precise axiomatic set-theoretical approach requires much more than what could be covered in one preliminary chapter. However, it should be understood that these set-theoretical notions lie in the deeper basis of topology or, for that matter, of virtually every mathematical theory. A good introductory book on the subjects in this chapter is [29].

1.1  Sets and Numbers A brief historical note: The notions of sets and numbers are ancient. The first rigorous definition of the set of real numbers (1871) is attributed to Georg Cantor (1845–1918).

The term “set” will be occasionally replaced by “family,” “class,” or “collection,” especially when there is a hierarchy of sets under inclusion, and for the purpose of avoiding the selfreferential, paradox-generating notions such as “sets of all sets.” The reader may assume that these four terms are interchangeable throughout this book. We will use the standard notation of basic set theory. Thus, ∅ (the empty set), ⊆ (subset),  (proper subset), ∈ (belongs to),  ∪ (union), ∩ (intersection), and Ac (the complement X \ A of A in X) all have the usual meaning. The only exception is ⊂, which will take the role of ⊆ in this book (thus, making the latter unnecessary). This is justified by the fact that we will rarely be encountering . In the few cases where we will want to emphasize that a set A is a proper subset of a set B we will indicate it.

3

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4   ◾    Sets, Numbers, and Cardinals

Basic results in set theory, such as de Morgan’s laws, are assumed to be known to the reader. We will consistently use the following notation:  = {0,1,2,…} is the set of natural numbers.  = {…, −2, −1,0,1,2,…} is the set of integers. + = {1,2,…} is the set of positive integers, or the set of positive natural numbers. Given two sets A and B, their set product is the set A × B = {(a, b): a ∈ A, b ∈B} . Instead write  A2 . The set product of the sets X1 ,  X 2 ,… , Xn , denoted of A × A, we will sometimes n X1 × X 2 ×  × Xn , or Π Xi , is the set {( x1 , x 2 , … , xn ) :  xi ∈ Xi , i = 1,2, … , n}. As in the case i =1 n for two sets, if X1 = X 2 = … = Xn = X , then we will sometimes use X n as short for Π Xi . i =1

A mapping f : X → Y is a subset S of X × Y such that for every x ∈ X, there is exactly one y ∈Y, such that ( x , y ) ∈S. We will use the standard notation and write y = f ( x ) instead of ( x , y ) ∈S. Surjection (onto mapping), injection (one-to-one mapping), and bijection will have the usual meaning. Occasionally we will use the word “map” instead of “mapping.” If Y is the set of real numbers (to be defined further below), then we will say function, instead of mapping. A relation R on the set A is a subset of A × A. Instead of ( x , y ) ∈R we will more often write x R y. We list the standard terminology regarding a relation R on a set X: • reflexive, if for every x ∈ X , ( x , x ) ∈R, • irreflexive, if for every x ∈ X , ( x , x ) ∉R, • symmetric, if for every x , y ∈ X , if ( x , y ) ∈R then ( y , x ) ∈R, • antisymmetric, if for every x , y ∈ X, if ( x , y ) ∈R and x ≠ y then ( y , x ) ∉R, • transitive, if for every x , y , z ∈ X , if ( x , y ) ∈R and ( y , z ) ∈R, then ( x , y ) ∈R. An order of a set X is a relation that is antisymmetric and transitive; the set X is then ordered by such a relation. Order relations will usually be denoted by ≤. A strict order, denoted 0 there is an integer N such that if n, m > N , then an − am < ε . We define an equivalence relation on the set of all Cauchy sequences: two Cauchy sequences a1 , a2 ,…, an ,… and b1 , b2 ,…, bn ,… are equivalent if for every rational number ε > 0 there is an integer N such that if n > N , then an − bn < ε . It is straightforward to show that this is indeed an equivalence relation. The set of real numbers, denoted by  from now on, is the set of equivalence classes of Cauchy sequences of rational numbers (under the just defined equivalence). Each real number can be thus represented by any Cauchy sequence in the corresponding equivalence class. Integers, and more generally, rational numbers can be represented by constant Cauchy sequences. For example, a Cauchy sequence representative of the integer 2 is 2,2,…,2,… . Loosely speaking, a Cauchy sequence of rational numbers converges toward the real number it defines, where we are using the intuitive meaning of the term “converge” (and where we will evade explaining any meaning of the word “intuitive”). The skeleton of the construction of the basic structure of  is outlined in the next list of steps where we denote by [an ] the real number (equivalence class) defined by the Cauchy sequence (an ) = a1 , a2 ,…, an ,… . • Define addition, multiplication, and subtraction in a natural way. For example [an ] + [bn ] = [an + bn ] . • Define a real number [an ] to be positive if there is a positive rational number q such that, from some point on, the members of the Cauchy sequence (an ) are all larger than q. • Define an order < as follows: [an ] < [bn ] if [bn ] − [an ] is positive.

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1.1  Sets and Numbers   ◾    7  

The last property gives us the usual, linear order of the set of real numbers. We will say a few more things about this order in the next section. Each positive real number r = [an ], as well as r = 0 , can be represented in a unique way as an infinite decimal (sequence) n . d1d2 …dn … where n ∈ , and 0 ≤ di ≤ 9 for all i ∈+ , such that no right-hand side portion of that infinite sequence consists only of 9s. Here is the idea of how that is done (without going into details): Start with a real number r. There exists the largest integer n such that n ≤ r . Next, consider r − n: by construction it is a number in the interval [0,1), and so 0 ≤ 10(r − n) < 10 . Let d1 be the largest integer between 0 and 9 such that d1 ≤ 10(r − n). So we have 0 ≤ 10(r − n) − d1 < 1, and thereby 0 ≤ 10(10(r − n) − d1 ) < 10. Now find the largest integer d2 between 0 and 9 such that d2 ≤ 10(10(r − n) − d1 ) . Keep going! This procedure gives the representation of positive real numbers as (what we call) decimal numbers. For example, the decimal expression of the real number [9/10, 99/100, 999/1000, …] is 1.000…. The set of positive real numbers will be denoted by  + . Once we have the positive real numbers represented as decimals, it is easy to do the same to the negative numbers. This gives the well-known decimal representations of real numbers. In this context, rational numbers are real numbers that are expressible as either terminating decimals (the decimal digits are all 0 from some point on, as in 3.24300000…) or as periodic decimals (where the digits of the decimal from one point on are made of a repeating finite sequence of digits (for example, 1.23245454545(45)…). Irrational numbers are the real numbers that are not rational. We will often invoke the standard visualization of  as a line equipped with distance to a fixed point on that line. As a consequence,  2 is a plane with a coordinate system, etc.

Illustration 1.3  A part of  3 (a solid cube) cutting out a planar area (a part of  2 ) and a line

segment (a part of ), all of them embedded in a space called Old Mills Beach. And yes, there will be unnecessary illustrations in this book!

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8   ◾    Sets, Numbers, and Cardinals

In what follows in this section we draw attention to the Least Upper Bound Property, and its twin, the Greatest Lower Bound Property, since they are behind virtually every topological property of the set . The set , endowed with increasingly richer structure as we go, will be the blueprint model in the development of our theory. The Least Upper Bound Property is sometimes given as an axiom in the formal theory of real numbers, and that is how we should take it. A subset A of an ordered set X is bounded from above if there is an x ∈ X , such that for every a ∈ A, we have a ≤ x ; if there is a y ∈ X , such that for every a ∈ A, y ≤ a, then we say that A is bounded from below. The Least Upper Bound Property. Every subset of  that is bounded from above has the least upper bound. Explicitly, this means that for every subset A of , if there is a real number r such that a ≤ r for every a in A, then there is a number c in  such that c is the smallest of all such numbers r. The number c in the statement of the Least Upper Bound Property is called the supremum of the set A and will be denoted by sup(A). Here is the Least Upper Bound Property’s twin: The Greatest Lower Bound Property. Every subset of  that is bounded from below has the greatest lower bound. This greatest lower bound is called the infimum of the associated set A, and it is denoted by inf (A). Exercises 1. Find a set A such that for every a ∈ A, we have a ⊂ A. 2. Given a set X, show that the relation ⊂ is an order of the set of all subsets of X. For which sets of X is this order linear? 3. Describe a linear order on (a) the set  2 , and (b) the set  2. 4. Show that if ~ is an equivalence relation on a set X, then every two equivalence classes are either disjoint or equal. 5. Which of the following relations are equivalence relations on the set ? If a relation is an equivalence relation, describe the equivalence classes.

(a) x ~ y if x − y is rational.



(b) x ~ y if x − y is irrational.



(c) x ~ y if x − y is divisible by 5 (that is, if

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x− y 5

is an integer).

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1.2  Sets and Cardinal Numbers   ◾    9  

6. Is there a relation R on the set , such that R is different from the relation =, and such that R is both an order and an equivalence relation? 7. Let X be a nonempty set and let f : X → Y be any mapping. Show that “u ~ v if and only if f (u) = f (v )” defines an equivalence relation on X.

1.2  Sets and Cardinal Numbers A brief historical note: The first proof that the set of real numbers has more elements than the set of integers was given by Georg Cantor (1874). The claim that there is no set with cardinality strictly between ℵ0 and 2ℵ0 is known as the Continuum Hypothesis.

Two sets A and B have the same cardinality if there is a bijection (one-to-one and onto mapping) from one set onto the other set. The cardinality of a set A, denoted by A , is the class of all sets of the same cardinality as the set A. A set A is finite if there is an n ∈+ such that A has the same cardinality as the set {1,2,…, n}, or if it is the empty set. In the former case we denote A = n , and we will say that A has n elements; in the case when A = ∅, we write A = 0 . An infinite set is any set that is not finite, and when A is infinite we will write A = ∞ . A set A is countably infinite if it is of the same cardinality as the set  . In that case we will write A =ℵ0 (aleph zero). A set is countable if it is either finite or countably infinite. An uncountable set is a set that is not countable. Example 1 The set  is countably infinite because the mapping that sends n ∈  to the element that appears at position number (n + 1) in the infinite list (specified by the following ☐ pattern) 0, −1,1, − 2, 2, − 3, 3,… is a bijection from  onto . As this example indicates, in order to show that a set A is countably infinite, it suffices to exhibit a well-defined sequence a0 , a1 , a2 ,…, an ,… that has no repetitions and consists of all elements of A (in which case, obviously, n  an is a bijection from  onto A). It follows that every infinite subset of a countably infinite set is also countably infinite. We define a linear order on the class of all cardinal numbers as follows: A ≤ B if there exists a one-to-one mapping from A into B. It is straightforward to show that this relation is well defined, reflexive and transitive. Proving that it is antisymmetric is not so elementary and we will skip it. Theorem 1. (Cantor–Bernstein1) If A ≤ B and B ≤ A , then A = B . Our definition of the relation ≤ among cardinals was given in terms of one-to-one mappings. Onto mappings could have been used to the same effect: Proposition 2. A ≤ B if and only if there is an onto mapping g : B → A. 1

Also known as Schröder-Bernstein theorem.

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10   ◾    Sets, Numbers, and Cardinals

Proof. (Warning: in the second part of the following argument we use a seemingly harmless claim, which we will identify following the proof.) ⇒ Suppose A ≤ B , so that there is a one-to-one mapping f : A → B. The mapping f −1 : f ( A) → A is then well defined. Define  g : B → A to be f −1 over f (A) and arbitrary over B \ f ( A). Then g is obviously onto. ⇐ Suppose there is an onto mapping g : B → A. We will show that there is a one-to-one mapping f : A → B, thus establishing that A ≤ B . Define f  in terms of g as follows: for every a ∈ A , choose precisely one element ba from the subset g −1 (a) of B, and set f (a) = ba. Since we are assuming that g is onto, we have that g −1 (a) ≠ ∅ for every a ∈ A , and so this definition of f makes sense. It is evident that f is one-to-one, and so A ≤ B . We have deduced what we claimed—by closing our eyes for a moment and claiming the obvious—that we can choose one element from each nonempty set g −1 (a). This claim is the Axiom of Choice. We will come back to it in the next section. For the time being we merely note that by using it we are acknowledging our acceptance of it. In Illustration 1.4, we use Proposition 2 to exhibit the standard justification of the claim that the set of positive rational numbers has cardinality ≤ than ℵ0 . Since the opposite inequality is obvious, it follows that this set of positive rational numbers is countably infinite. One consequence (via a short argument) is that the set  of all rational numbers is also countably infinite (i.e., has cardinality ℵ0 ). 1

1

1

1

1

2

3

4

2

2

2

1

2

3

3

3

1

2

4 1

Illustration 1.4  Following the arrows we

en­cou­nter all positive rational numbers + (with rep­etition), listed according to the sum of the numbers in the numerator and denominator. This defines a mapping from the set  onto + showing that + ≤  .

The power set of a set A is the set of all subsets of A; it will be denoted by P ( A). The following simple but important proposition establishes a way of producing large cardinals. Proposition 3. For every set A, we have A < P( A) . Proof. Since a  {a} is a one-to-one mapping A → P ( A), it follows that A ≤ P( A) . Suppose  | A| = |P ( A)|, i.e., suppose there is a bijection f : A → P ( A). Consider the subset B = {b ∈ A : b ∉ f (b)} of A. Since f is a bijection, there exists a unique x ∈ A such that f ( x ) = B. Is x ∈B? If yes, then it follows by the definition of x that x ∉ f ( x ) = B; this is obviously not possible. On the other hand, if x ∉B, then the definition of B implies that x ∈B, again yielding a contradiction.

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1.2  Sets and Cardinal Numbers   ◾    11  

It is easy to verify by induction that for every n ∈ , if A = n then |P ( A)| = 2n . By analogy, if A = m (infinite cardinals included), then |P ( A)| will be denoted by 2m . For example, P ( ) = 2ℵ0 . Since P ( ) = (0,1), and since (0,1) =  (Exercise 4), it follows that  = 2ℵ0. One way to verify that P ( ) = (0,1) is to express the real numbers in the interval (0,1) in binary notation and associate to every subset A of  the real number in (0,1) whose n-th decimal is 1 only if n ∈ A . The following proposition is a short summary of the results on cardinals that we will invoke later on. Proposition 4. (a) If J is countable and if each A j , j ∈ J , is countable, then so is ∪ A j . j∈J

(b) If for every i ∈{1, 2, …, n} the set Xi is countable, then so is the set product X1 × X 2 ×  × X n . Proof. Exercise 5. Exercises 1. Let X be an infinite set. Show that for every finite subset A of X, X \ A = X . Show that there is subset B of X such that B =ℵ0 and such that X \ B = X . The Hilbert Hotel, with countably many rooms, is a popular rendering of this exercise in the case where X =ℵ0; see Illustration 1.5 for a special subcase.



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Illustration 1.5

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12   ◾    Sets, Numbers, and Cardinals

2. Show that if A and B are finite sets and if S is the set of all mappings A → B , then A S=B . 3. Let A = A1 , B = B1 , let S be the set of all mappings A → B , and let S1 be the set of all mappings A1 → B1. Show that S1 = S . 4. Show that  = (0,1) . 5. Prove Proposition 4. 6. Is the set of all mappings  → {a, b} countable or uncountable? Justify! 7. Prove that if A = B , then P ( A) = P ( B) . 8. If f : X → Y , define f # : P (Y ) → P ( X ) by f # ( A) = f −1 ( A). Prove that f is one-to-one if and only if f # is onto, and that f is onto if and only if f # is one-to-one. 9. Let A = n, let B = m (where n and m are any cardinal numbers). Define m + n to be A1 ∪ B1 , where A1 and B2 are any two disjoint copies of A and B, respectively. Show that this operation is well defined (i.e., show it does not depend on the choice of A1 and B1 ). Show that 2 + 3 = 5 . 10. Let A = n, let B = m (where n and m are any cardinal numbers). Define m ⋅ n to be A× B . (a) Show that 2 ⋅ 3 = 6. (b) Show that ℵ0 ⋅ℵ0 =ℵ0 . 11. (a)  Prove that if n ≥ m, then 2n ≥ 2m . (b) Prove that if 2n ≥ℵ0 , then 2n ≥ 2ℵ0 .

1.3 Axiom of Choice and Equivalent Statements A brief historical note: The Axiom of Choice was first explicitly formulated in 1904 by Ernst Zermelo (1871–1953).

We accept the following axiom and, consequently, all of its consequences (a few of which we will mention further below). Axiom of Choice. For every family { Ai : i ∈ I } of pairwise disjoint sets there exists a mapping f : I → ∪ Ai such that f (i ) ∈ Ai . i∈I

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1.3 Axiom of Choice and Equivalent Statements   ◾    13  

Illustration 1.6  If each Ai is a (possibly infinite) set of eggs in a basket, and if each of the

(possibly infinitely many) baskets contains eggs, then we use f to choose exactly one egg  f (i) from each of the baskets.

Informally speaking, we assert in this tame-looking axiom that we can choose, by means of f, exactly one element from each of the sets Ai (see Illustration 1.6). The Pandora’s box of unexpected, and often counter-intuitive, consequences is opened by the stipulation of the existence of the “choice-function” f, without specifying exactly how that mapping is to be constructed. We have already defined the set product of finitely many sets. More generally, the set product of the sets Ai , i ∈ I , denoted ∏ Ai , is the set of all mappings I → ∪ Ai . Very infori∈I

i ∈I

mally, this is the set of all | I | -tuples of elements in ∪ Ai , one element per each set Ai . i∈I

We will usually denote a generic element of ∏ Ai by (ai )i ∈I . When I is finite (with n i ∈I elements) or countable infinite we will sometimes denote the elements of the respective set-products by (a1 , a2 ,… , an ) and (a1 , a2 ,… , an ,…). The Axiom of Choice is equivalent to the following statement (Exercise 1). Axiom for Products. Let { Ai :  i ∈ I } be a collection of sets. If each Ai is nonempty, then so is ∏ Ai . i ∈I

Some other equivalent versions of the Axiom of Choice are not so simple to deduce. In order to state them we need a few preliminaries.

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14   ◾    Sets, Numbers, and Cardinals

The usual order of  was mentioned in Section 1.1; it is a linear order. The restriction of that linear order to  , , and  yields the usual linear order of these sets. Let A be an ordered set. An element c is a maximal element in A if for every x ∈ A, if c ≤ x or x ≤ c , then x ≤ c . In other words, c is maximal if x ≤ c for every x ∈ A that is comparable via ≤ to c. Maximal elements need not be unique. For example, A = {a, b, c} with an order < defined by a < b, a < c , has two maximal elements, b and c. A chain in A is any subset B of A that inherits a linear order from the order of A. For example, with the partial order on A = {a, b, c} as above, the set {a, b} is a chain in A. An element u ∈ A is an upper bound of subset B of A if b ≤ u for every b ∈B . Considering again the set A = {a, b, c} with the given order, we notice that every element of A is an upper bound for {a}, and that {b, c} has no upper bound. A lower bound of B is defined analogously. We now state two versions of Zorn’s Lemma (the first one without a proof). We will invoke the second, stronger version several times later on. Zorn’s Lemma 1. If every chain in an ordered nonempty set A has an upper bound in A, then A contains a maximal element. Zorn’s Lemma 2. If every chain in an ordered nonempty set A has an upper bound in A, then for every a ∈ A there is a maximal element m ∈ A such that a ≤ m. Proof. (Based on Zorn’s Lemma 1.) Take any a ∈ A and consider the set Y = { y ∈ A : a ≤ y} . Clearly a ∈Y , so Y ≠ ∅ . By assumption, every chain of elements in Y (being at the same time a chain of elements in A) has an upper bound in A. It follows from the definition of Y that this upper bound is also in Y. By Zorn’s Lemma 1 (applied to the set Y), Y contains a maximal element m. Again invoking the definition of Y, it follows that the element m is also maximal in A, and, since a ≤ m , our search for a maximal element in A comparable to a is over. The following example illustrates a typical usage of Zorn’s Lemma. Example 1 Let X be a set with X ≥ 2 and ordered with ≤. Fix an element x ∈ X , and consider the collection P of all subsets A of X such that A is linearly ordered by ≤ and such that x ∉ A. The relation ⊂ (“subset of”) is an order of the collection P . We show that there exists Y ∈ P , such that Y is maximal with respect to ⊂. Since X ≥ 2 , we have that P ≠ ∅ . Let Q be any chain in P (with respect to ⊂): this Z . It means that for every A, B ∈ Q , either A ⊂ B or B ⊂ A. Consider the set C = Z∪ ∈Q is obvious that x ∉C. It is very easy to show that C is a chain with respect to ≤. (Prove it! That is, show that for every a, b ∈C, either a ≤ b or b ≤ a.) Finally, it is also clear that Z ⊂ C for every Z ∈Q . This shows that C is an upper bound for Q . The conclusion we ☐ wanted now follows from Zorn’s Lemma 1. Our acceptance of Zorn’s Lemma is justified by the following.

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1.3 Axiom of Choice and Equivalent Statements   ◾    15  

Theorem 1. The Axiom of Choice and Zorn’s Lemma are equivalent statements (in a set theory with standard basic axioms). Suppose A is ordered by ≤, and let B be a subset of A. An element b ∈B is the least element for B if for every x ∈B, b ≤ x. The definite article in the preceding sentence (i.e., the uniqueness of the least element, if it exists) is easy to justify. The greatest element c ∈B is such that x ≤ c for every x ∈B . A set A is well ordered if it is linearly ordered (by ≤) and if every nonempty subset of A has the least element. In that case we say that ≤ is a well ordering of A. For example, the usual order of  makes it well ordered. However, the usual linear order of  does not make it well ordered, since, for example,  itself does not have the least element. It is easy to make  well ordered: for example, 0 ≤ −1 ≤ 1 ≤ −2 ≤ 2 ≤ −3 ≤ 3 ≤ … will do. The problem of explicitly well ordering  turned out to be deeper and more difficult. The existence part is established by Theorem 2 below, but only because we have accepted the Axiom of Choice. Well Ordering Principle. Every nonempty set has a well ordering. Theorem 2. The Axiom of Choice and the Well Ordering Principle are equivalent statements (in a set theory with standard basic axioms). Consequently, in the set theory based on the Axiom of Choice and the other standard axioms, the set  of all real numbers allows well ordering. (Observe that the usual linear order of  is not a well ordering.) The following example is important, since it is a part of a structure that we will explore later on. Example 2 Let < be a well ordering of  ; we know nothing about this ordering beyond its bare existence (based on the Well Ordering Principle). Keep in mind that this is not the usual linear order of  . Let ∗ be a well ordered copy of + disjoint from  ; for  example, we may take ∗ = {1 = (1,0),  2 = (2,0),  3 = (3,0),…} with 1 < 2 < 3 0), denoted B(a, r), is the set {x ∈ X : d(a, x ) < r }. The associated closed ball is defined to be {x ∈ X : d(a, x ) ≤ r } . A subset U of X is open if for every element x ∈ U there is a ball B(x, r) contained in U. Closed subsets of X are the complements in X of open subsets of X.

Open

Closed

Illustration 2.7  Open sets will often be shown with dashed bounding lines, closed sets will be with full lines.

The following property will later be used to generalize the notion of metric spaces to the more general concept of topological spaces. Proposition 1. The following are true for every metric space X: (i) The sets ∅ and X are open. (ii) Unions of open sets are open. (iii) Intersections of finitely many open sets are open. Proof. The sets X and ∅ are plainly open, as are unions of open sets. We confirm (iii): assume U1 ,U 2 ,…,U n are open sets and consider their intersection

U1

U2

n

V = ∩U i . Choose a point x ∈ V. Then, x ∈U i i =1

for i =1, 2, … , n . Since the sets U1 ,U 2 ,…,U n are open, there are balls B( x , r1 ), B( x , r2 ),…, B( x , rn ) contained in these sets, respectively. Choose

r = min{r1 , r2 ,…, rn } . Then B( x , r ) ⊂ V and so V is ◼ open too (Illustration 2.8).

Illustration 2.8  The smaller of the two disks corresponding to U1 and U2 tells us that the intersection U1 ∩ U 2 is open.

The interconnectedness and the mutual relationship of the open sets in a metric space, and, later, in a topological space, are what constitutes the inner (topological) structure of the space. Generally speaking, our primary object of study will be this deeper anatomy of the spaces, as opposed to the exterior, geometrical considerations, which in this theory will play a marginal role. Two metrics d1 and d2 on a set X are said to be equivalent if they generate the same classes of open sets. The two metric spaces defined in Examples 3 and 4 are equivalent (Exercise 2), but neither is equivalent to the discrete metric space. On the other hand, every metric space over a finite set X is equivalent to the discrete metric space (Exercise 3). As a

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22   ◾    Metric Spaces: Definition, Examples, and Basics

consequence, finite metric spaces are not interesting from the structural point of view; we have seen them in Illustrations 2.4 and 2.5 and we will not mention them again. Proposition 2. Let X be equipped with two metrics, d1 and d2 . Assume that for every ball B(a,r) in any one of the metric spaces ( X , d1 ), ( X , d2 ) , there is a ball B(a, r′ ) in the other metric space such that B(a, r ′ ) ⊂ B(a, r ). Then, ( X , d1 ) and ( X , d2 ) are equivalent. Proof. Straightforward. In the following example we describe a metric space over  2 that is not equivalent to the Euclidean metric space. Example 5: The Post Office Metric We define a metric d p over  2 as follows: we fix a point p in  2 and set d p ( x , y ) = d( x , p) + d( y , p) when x ≠ y, and d p ( x , x ) = 0, where d is the Euclidean metric (Illustration 2.9). It is easy to check that this is indeed a metric. Here, every set A that does not contain the point p is open, while only some of the sets containing p are open (Exercise 5). It is evident that this space is not equivalent to the Euclidean ☐ metric space over  2 .

A

B

Illustration 2.9  The post office metric:

The shortest path leads through a fixed point (the post office, represented by a mailbox).

Exercises 1. Show that for every metric space X and for every x ∈ X the singleton {x } is closed. 2. Prove that the Euclidean metric space and the city metric over  2 are equivalent. 3. Show that every finite metric space is equivalent to the discrete space over the same set. 4. Show that for every two distinct points x and y in a metric space X, there exist disjoint open sets U and V such that x ∈ U and y ∈ V. 5. Consider  2 with the topology induced by the post office metric with respect to a fixed point p.

(a) Show that if p ∉ A ⊂  2 then A is open.



(b) Show that if p ∈ A ⊂  2 then A is open if and only if there is a Euclidean ball B(p, r) (that is, a ball with respect to the Euclidean metric) that is contained in A.

6. Suppose d1 and d2 are two metrics over a set X. Show that d3 defined by d3 ( x , y ) = min{d1 ( x , y ), d2 ( x , y )} is also a metric over X.

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2.2  Metric Spaces: Basics   ◾    23  



7. Is the intersection of countably many open sets necessarily open? Give a proof or a counterexample. What happens when you try to generalize the proof of Proposition 1 (iii)?

8. Let F be a closed subset of a metric space (X, d). For every point x ∈ X, define d( x , F ) = inf{d( x , y ) : y ∈ F }. Show that x ∈ X \ F if and only if d( x , F ) > 0 . 9. Let A and B be two subsets of a metric space (X, d). Define d( A, B) = inf{d ( a, b ) : a ∈ A, b ∈B}. Find a metric space X and two disjoint closed subsets A and B of X such that d( A, B) = 0. 10. Show that if F and G are two disjoint closed sets in a metric space X, then there are two disjoint open sets U and V such that F ⊂ U and G ⊂ U . [Hint: Exercise 8.] 11. Consider a set R of all open rectangles in  2 above the x-axis, bounded by edges of slopes 1 and −1, with one corner at the x-axis, and such that every point (x, 0) on the x-axis is a corner of exactly one member Rx of R (see Illustration 2.10). Show that there is a rational number q and an irrational number z such that Rq ∩ Rz ≠ ∅. (We will refer to this exercise in Section 8.2.)

Ra

1

Ra

2

a2

a1

Illustration 2.10  We show a few rectangles

from the set R.

2.2  Metric Spaces: Basics A brief historical note: The French mathematician Augustin-Louis Cauchy (1789–1857) is considered to have been one of the founders of Analysis.

Let (X,  d ) be a metric space, and let A be a nonempty subset of X. It is very easy to verify that the restriction d A × A of d to A × A is a metric on A. We say that ( A, d A × A ) is a metric subspace of (X,  d). We will prefer the less accurate but simpler (A,  d) instead of ( A, d A × A ). When d is understood, then we say that A is a (metric) subspace of X. The following proposition establishes the principal link between the open sets in X and the open sets in A.

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24   ◾    Metric Spaces: Definition, Examples, and Basics

Proposition 1. Let (A, d) be a metric subspace of (X, d). For every V ⊂ A,V is open in A if and only if there is an open subset U of X such that V = A ∩U . We will eventually prove Proposition 1 in a more general setting of topological spaces; at this stage we leave it as an exercise (Exercise 3). The relationship between the closed subsets of A and X is similar: F is closed in A if and only if there is a closed subset G of X such that F = A ∩ G (Exercise 4). A sequence ( xn ) of elements in a metric space X converges to x, denoted lim xn = x , if n→∞ for every ε > 0, there is an integer N such that for every n > N , xn ∈ B( x , ε ).

ε xN

x xN+2 xN+1

x2 x3



x1

Illustration 2.11

The following theorem will be needed. Theorem 2 (Cantor’s Nested Intervals Theorem). Let {[an , bn ] : n = 1,2,…} be a set of closed ∞ ∈{1,2,…}, [an+1 , bn+1 ] ⊂ [an , bn ]. Then, ∩[an , bn ] ≠ ∅. In intervals in  such that for every n ∞ n=1 addition, if lim(bn − an ) = 0 , then ∩[an , bn ] contains exactly one element. n→∞

n=1

Proof. Our assumptions imply that for every n ∈ {1, 2, …}, an ≤ b1 and so the set {an  : n = 1,2,…} is bounded from above. By the Least Upper Bound Property, there is the least upper bound; denote it by a. Notice that a ≤ bn for every n ∈+ , for if a > bm , then bm would be an upper bound for {an  : n = 1,2,…} that is smaller than a. Similarly, the set {bn  : n = 1,2,…} has the largest lower bound b, for which we have an ≤ b for every n ∈+. This set∞of inequalities, together with a ≤ bn for every n ∈+ , implies that a ≤ b , and that [a, b] = ∩ [an , bn ]. This completes the proof of the first part of the theorem. n=1 Since an +1 ≥ an and bn +1 ≤ bn for all n =1, 2, … , it follows easily that lim an = a and n→∞

lim bn = b. If we further assume that lim(bn − an ) = 0, then 0 = lim(bn − an ) = lim bn − lim an =

n→∞

n→∞ ∞

b − a, so that a = b and hence [a, b] = ∩ [an , bn ] is a singleton.

n→∞

n→∞

n→∞

n =1

A sequence ( xn ) in a metric space (X, d) is a Cauchy sequence if for every ε > 0, there is an integer N such that if n, m ≥ N , then d( xm , xn ) < ε. It is obvious that every convergent sequence is a Cauchy sequence. The converse is not always true. For example ( n1 ) is a Cauchy sequence in the interval (0, 1) (considered as a

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2.2  Metric Spaces: Basics   ◾    25  

metric subspace of ), but it does not converge there. Metric spaces in which every Cauchy sequence converges are called complete. Example 1:  Is Complete We show in this example that  is a complete metric space. There will be three small gaps in the argument: one will be called “Exercise,” and we will alert the reader of the other two by following them with “Why?” These three exercises deserve some thought. ∞ Let ( xn ) be a Cauchy sequence in . Then so is ( xn )n = k , k ≥1. Every Cauchy sequence is (obviously) bounded. Denote the least upper bound of ( xn )n∞= k by bk . Then {bk } is a nonincreasing sequence (i.e., bn +1 ≤ bn for all n), and it is bounded from below. Hence, it converges (Exercise 2); denote lim bk = b. Symmetrically, if the greatk→∞ est lower bound of ( xn )n∞= k is denoted by ck , then (ck ) is nondecreasing and bounded from above, and so it converges; denote lim ck = c. Observe that ci ≤ b j all the time, k→∞ so that c ≤ b. Suppose c < b. The interval (c, b) must have members of ( xn ) , or else the sequence fails to be Cauchy. (Why?) On the other hand, it follows from our construction of c and b that x k ≥ b or x k ≤ c for every k. (Why?) This clearly implies that x k ∉(c , b) for every k. Our assumption c < b led us to a contradiction. Hence, b = c. It is now easy to ☐ confirm that ( xn ) converges to b = c. Let ( X , d1 ) and (Y , d2 ) be metric spaces. We say that a mapping f : X → Y is conti­nuous at a point x ∈ X if for every B( f ( x ), ε) in Y, there is B( x , δ) in X such that f ( B( x , δ )) ⊂ B( f ( x ), ε). The reader should notice that in the case where X = Y =  and d1 = d2 = the Euclidean metric, the above definition of continuity reduces to the standard “ε −δ definition.” Proposition 3. Let ( X , d1 ) and (Y , d2 ) be metric spaces. A mapping f : X → Y is continuous at x ∈ X if and only if for every sequence ( xn ) converging to a in X, the sequence ( f ( xn )) converges to f (a) in Y. X xn

x2

x1

a

f

f (x1)

Y

f (x2) f (a) f (xn)

Illustration 2.12  X is a metric space; f is continuous if and only if images of sequences con-

verging to a are sequences converging to f (a).

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26   ◾    Metric Spaces: Definition, Examples, and Basics

Proof. ⇒ Suppose ( xn ) converges to a in X, and choose any B( f ( x ), ε), ε > 0. Then, by assumption, there is B(a, δ) such that f ( B(a, δ)) ⊂ B( f (a), ε). Since lim xn = a there n→∞ exists N such that for every n > N , xn ∈ B(a, δ ). It is then evident that for every n > N , f ( xn ) ∈ f ( B(a, δ )) ⊂ B( f (a), ε ), showing that lim f ( xn ) = f (a). n→∞

⇐ Suppose f : X → Y is not continuous at a. Then there is B( f (a), ε) such that for every B(a, δ), we have f ( B(a, δ )) ⊂/ B( f (a), ε). Choose xn in B ( a, n1 ) such that f ( xn ) ∉ B( f (a), ε), n =1, 2, … . Then lim xn = a , and lim f ( xn ) ≠ f (a). n→∞

n→∞

Remark: In the proof of Proposition 3 we have used the symbols ⇒ and ⇐. We will do this consistently throughout this book. If a statement is of type “P if and only if Q,” then ⇒ indicates that we are focusing on “P implies Q,” while ⇐ means that we are dealing with “Q implies P.” A mapping f : X → Y between two metric spaces is continuous if it is continuous at every point of X. If, in addition, f is a bijection with a continuous inverse f −1 : Y → X , then we say that f is a homeomorphism, and that the metric spaces X and Y are homeomorphic. If X and Y are homeomorphic then we write X ≅ Y . The following proposition will metamorphose into a definition in Chapter 3. Proposition 4. Let ( X , d1 ) and (Y , d2 ) be metric spaces. A mapping f : X → Y is continuous if and only if for every open subset U of Y, f −1 (U ) is open in X. Proof. ⇒ Let f be continuous, and let U be an open subset of Y. Showing that f −1 (U ) is open in X is equivalent to showing that every x ∈ f −1 (U ) is the center of a ball that is contained in f −1 (U ) . Consider f ( x ) ∈ U . Since U is open, there is a ball B( f ( x ), ε) such that B( f ( x ), ε) ⊂ U . Since f is continuous, it is continuous at x, which in turn implies that there is a ball B( x , δ) such that f ( B( x , δ)) ⊂ B( f ( x ), ε). It is then evident that B( x , δ ) ⊂ f −1 (U ) . ⇐ Assuming f −1 (U ) is open in X for every open subset U of Y, we need to show that f is continuous at every point x ∈ X . Start with a ball B( f ( x ), ε). Then, since B( f ( x ), ε) is open in Y, f −1 ( B( f ( x ), ε )) is open in X. Choose a ball B centered at x and such that B ⊂ f −1 ( B( f ( x ), ε )). Then, f ( B) ⊂ f ( f −1 ( B( f ( x ), ε))) = B( f ( x ), ε). Example 2 Let A be a subset of a set X. The inclusion in : A → X is defined by in(a) = a for every a ∈ A. If X is a metric space, then A becomes a metric subspace of X. In this case the inclusion in is necessarily continuous (Exercise 6). If f : X → Y is a continuous mapping ☐ between two metric spaces, then the restriction f A : A → Y is also continuous. n

Let ( Xi , di ), i = 1, 2, …, n, be metric spaces and consider the set product X = ∏ Xi . We i =1

notice two obvious types of mappings: the projections pi : X → Xi , i ∈{1, 2, …, n}, defined

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2.2  Metric Spaces: Basics   ◾    27  

by pi ( x1 , x 2 ,…, xn ) = xi , and the coordinate mappings cia : Xi → X (i ∈ {1, 2, … , n}) defined by cia ( xi ) = (a1 , a2 ,…, ai−1 , xi , ai+1 ,…, an ), where a = (a1 , a2 ,…, an ) is a fixed element of X. n

We want a metric over X = ∏ Xi such that all projections pi and all coordinate mappings i =1

cia , for all choices of a, are continuous. One natural choice is the product metric d p , defined as follows: for every ( x1 , x 2 ,…, xn ), ( y1 , y 2 ,…, yn ) ∈ X , d p (( x1 , x 2 ,…, xn ), ( y1 , y 2 ,…, yn )) =

(d (x , y ) 1

1

1

2

)

1

+ d2 ( x 2 , y 2 )2 +  + dn ( xn , yn )2 2 .

The product metric is a generalization of the Euclidean metric for n. Confirming that d p is indeed a metric follows along the lines of the argument given in Example 3 of 2.1. As was the case there, it is the triangle inequality that needs primary attention: n

∑d (x , z ) i



i

i

i =1

2

n

≤

n

∑d (x , y ) + ∑d ( y , z ) . i

i

i

2

i

i =1

i

2

i

i =1

This is sometimes called the Cauchy–Schwartz inequality. Simple manipulation reduces n

n

i =1

i =1

it to proving the following: Σ di ( xi , yi )di ( yi , zi ) ≤ Σ di ( xi , yi )2

n

Σ di ( yi , zi )2 , which we

i =1

recognize as the more general version of the Cauchy inequality. The proof of the Cauchy inequality provided in 2.1 needs only a slight modification to justify the inequality shown here. We leave the details to the reader (Exercise 11). The space ( X , d p ) is called the product metric space. That it works as advertised is left as an exercise (Exercise 9). n

If ( Xi , di ) = (Y , d∗ ) for all i = 1, 2, …, n, then we will abbreviate ∏ Xi to Y n . i =1

The product metric d p is certainly not the only metric over X making the projections and the coordinate mappings continuous. For example, dm (( x1 , x 2 ,…, xn ), ( y1 , y 2 ,…, yn )) = max{d1 ( x1 , y1 ), d2 ( x 2 , y 2 ),…, dn ( xn , yn )} is another such metric over X. In fact, these two metrics over X are equivalent (Exercise 12). So, from the point of view of the structure of the set of open subsets of X, it really does not matter which of the two we choose to work with. Example 3: Continuity of the Metric As usual, let (X, d) denote a metric space. Consider the product metric space ( X 2 , d p ) and the metric d : X 2 →  . We will now show that d is continuous. Suppose (( xi , yi )) is a sequence converging to a point (a, b) ∈ X 2 . This implies that lim xi = a and that lim yi = b (Exercise 14). So, for every ε > 0, there is an integer N, i →∞ i →∞ such that for every i > N , d( xi , a) < 2ε and d( yi , b) < 2ε . Hence, by the triangle inequality, d( xi , yi ) ≤ d( xi , a) + d(a, yi ) ≤ d( xi , a) + d(a, b) + d(b, yi ) ≤ d(a, b) + ε, for all i > N . Similarly, d(a, b) ≤ d(a, xi ) + d( xi , yi ) + d( yi , b) ≤ d( xi , yi ) + ε, for all i > N . It follows that, for all i > N , d(a, b) − ε ≤ d( xi , yi ) ≤ d(a, b) + ε. All this tells us that the sequence (d( xi , yi )) converges to the number d(a,b).

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28   ◾    Metric Spaces: Definition, Examples, and Basics

It follows from Proposition 3 that d, considered as a mapping X 2 → , where X 2 is ☐ equi­pped with the product metric d p , is always continuous. We now strengthen the definition of continuity. Let ( X , d1 ) and (Y , d2 ) be metric spaces. A mapping f : X → Y is uniformly continuous if for every ε > 0, there is a δ > 0 such that for every x1 , x 2 ∈ X , if d1 ( x1 , x 2 ) < δ, then d2 ( f ( x1 ), f ( x 2 )) < ε. It is obvious (by temporarily fixing one of x1 , x 2) that uniform continuity implies continuity at every given point; thus uniform continuity implies continuity. A standard counterexample showing that the converse fails is the function f ( x ) = x1 from the interval (0,1) (considered as a metric subspace of ) into . The conditions under which continuity implies uniform continuity will be studied later. Continuous mappings do not always carry a Cauchy sequence into a Cauchy sequence (Exercise 7). However, uniformly continuous mappings do! Proposition 5. Let ( X1 , d1 ) and ( X 2 , d2 ) be two metric spaces, let ( xn ) be a Cauchy sequence in X1, and let f : X1 → X 2 be a uniformly continuous mapping. Then ( f ( xn )) is a Cauchy sequence in X 2 . Proof. Take any ε > 0. Since f is uniformly continuous, there is a δ > 0 such that if d1 ( y1 , z1 ) < δ, then d2 ( f ( y1 ), f (z1 )) < ε. Since ( xn ) is a Cauchy sequence, there exists an integer N such that if n, m > N , then d1 ( xn , xm ) < δ. Hence, for the same n and m, we have d2 ( f ( xn ), f ( xm )) < ε, establishing that ( f ( xn )) is a Cauchy sequence. Given two metric spaces ( X1 , d1 ) and ( X 2 , d2 ) a mapping f : X1 → X 2 is a dilation if there is a positive number α such that for every x1 , x 2 ∈ X1 we have d2 ( f ( x1 ), f ( x 2 )) = αd1 ( x1 , x 2 ). A dilation for which α =1 is called an isometry. Every dilation, hence every isometry, is one-to-one and uniformly continuous. We leave the justification as an exercise (Exercise 19). A dilation with 0 < α < 1 is called a contraction by a fixed factor α. If there is α ∈(0, 1) such that d2 ( f ( x1 ), f ( x 2 )) ≤ αd1 ( x1 , x 2 ) for every x1 , x 2 ∈ X1 , then we say that f is a contraction. It is very easy to see that contractions must be continuous. A mapping f : X → X has a fixed point if there is an element x ∈ X such that f ( x ) = x . Theorem 6 (Banach fixed point theorem). Let X be a complete metric space. Every contraction X → X has a unique fixed point. Proof. Let f : X → X be a contraction. So, there is α ∈ (0, 1) such that d( f (u), f (v )) ≤ αd(u, v ) for every u, v ∈ X . The sequence x ,  f ( x ),  f 2 ( x ), … is Cauchy, since d( f m ( x ), f m+k ( x )) ≤ α md( x , f k ( x )) ≤ α m (d( x , f ( x )) + d( f ( x ), f 2 ( x )) +  + d( f k−1 ( x ), f k ( x ))) ≤ ≤ α m (d( x , f ( x )) + αd( x , f ( x )) + α 2d( x , f ( x )) +  + α k−1d( x , f ( x ))) =



= α md( x , f ( x ))(1 + α +  + α k−1 ) = α md( x , f ( x ))

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1 − αk 1 ≤ α md( x , f ( x )) , 1− α 1− α

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2.2  Metric Spaces: Basics   ◾    29  

so that d( f m ( x ), f m+ k ( x )) can be made arbitrarily small > 0 by choosing m sufficiently large, for every k ≥ 0. Since we have assumed that the space X is complete, it follows that x ,  f ( x ),  f 2 ( x ), … converges to some z ∈ X . Since applying f to the sequence x ,  f ( x ),  f 2 ( x ), … produces the sequence   f ( x ),  f 2 ( x ),  f 2 ( x ),…, converging to the same point, and since f is continuous, it follows that f (z ) = z . The uniqueness of the fixed point is immediate, for if there is an element z1 ∈ X such that f (z1 ) = z1, then d(z , z1 ) = d( f (z ), f (z1 )) ≤ αd(z , z1 ), which, since 0 < α < 1, is true only if d(z , z1 ) = 0 . Hence. z = z1 . Example 4: Fractals We briefly and tangentially mention fractals; we will not see them further in this book. A metric space (X, d) is a fractal if there is a contraction f : X → X by a fixed factor (called the stretching factor), such that f ( X ) X . Intervals in  (considered as metric subspaces of ), except the degenerate onepoint intervals and except  = (−∞, ∞) itself, are simple and not very interesting fractals. The only not entirely obvious part of this claim is that  is not a fractal (Exercise 22). Notice that the interval (0,1) is a fractal and  is not, even though they are homeomorphic. Hence, being a fractal is a geometric property per se, not recognized by the structure of the open subsets of the metric space. A couple of nontrivial fractal metric subspaces—one of  2 , the other of  3—are shown in Illustrations 2.13 and 2.14, respectively; we justify their inclusion on the merits of their visual beauty only.

Illustration 2.13  Pentaflake.

Illustration 2.14  Menger sponge.

We will encounter more properties of metric spaces as we go.

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☐

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30   ◾    Metric Spaces: Definition, Examples, and Basics

Exercises 1. Show that if lim xn = a and if lim xn = b in a metric space, then a = b. n→∞

n→∞

2. Let (an ) be a nondecreasing sequence in  that is bounded from above. Show that (an ) converges. 3. Prove Proposition 1. 4. Let (A, d) be a metric subspace of (X, d). Prove that F is closed in A if and only if there is a closed subset G of X such that F = A ∩ G . 5. Show that a subset A of a metric space X is closed if and only if every sequence (an ) of elements in A that converges in X also converges in A. 6. Let A be a subset of a metric space X. (a) Consider A as a metric subspace of X. Show that the inclusion in : A → X is continuous. (b) Let d1 be a metric over A, and let d be the metric over X. Is the following true: if the inclusion in : A → X is continuous, then A is a metric subspace of X?

7. Show that a continuous image of a Cauchy sequence need not be a Cauchy sequence.

8. Show that if ( xn ) is a Cauchy sequence that has a convergent subsequence, then ( xn ) is also convergent. n

9. Let ( Xi , di ), i =1, 2, … , n, be metric spaces, and denote their product metric space ∏ Xi i =1

by X. Show that every projection pi : X → Xi and every coordinate mapping cia : Xi → X is continuous. n

10. Let ( Xi , di ) , i =1, 2, … , n, be metric spaces, and consider the set product X = ∏ Xi . i =1

Define  d # (( x1 , x 2 ,…, xn ), ( y1 , y 2 ,…, yn )) = d1 ( x1 , y1 ) + d2 ( x 2 , y 2 ) +  + dn ( xn , yn ) , for every ( x1 , x 2 ,…, xn ), ( y1 , y 2 ,…, yn ) ∈ X . Show that d # is a metric on the set X, and show that it is equivalent to the product metric over X. 11. Let ( Xi , di ), i =1, 2, … , n, be metric spaces. Prove that n

n

∑ d (x , y )d ( y , z ) ≤ ∑ d (x , y ) ∑ d ( y , z ) . i



n

i

i

i =1

i

i

i

i

i =1

i

i

2

i

i

i

2

i =1

12. Show that the product metric d p and the metric dm as defined in this section (following Example 2) are equivalent. 13. Show that n × m ≅ n+m. 14. Let ( X1 , d1 ) and ( X 2 , d2 ) be metric spaces and let (( xi , yi )) be a sequence in the product metric space X1 × X 2 . Show that lim( xi , yi ) = (a, b) in X1 × X 2 if and only if lim xi = a i →∞ i →∞ in X1 and lim yi = b in X 2 . i →∞

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2.2  Metric Spaces: Basics   ◾    31  

15. (a) Find a set {(an , bn ) : n = 1,2,…} of open intervals in  such that for every ∞ n ∈{1,2,…}, (an+1 , bn+1 ) ⊂ (an , bn ), and such that ∩ (an , bn ) = ∅ . n=1

(b) Let {(an , bn ) : n = 1,2,…} be a set of open ∞intervals in  such that for every n ∈{1,2,…}, an < an+1 < bn+1 < bn . Show that ∩ (an , bn ) ≠ ∅ . n=1

16. Let f :[0, 1] → [0, 1] be a continuous function. Show that the function g :[0, 1] → [0, 1] defined by g (t ) = sup{ f ( x ): x ∈[0, t ]} is continuous and nondecreasing. Draw a sketch comparing the graphs of f and g if f ( x ) = − sin( 2πx ). 17. Let S be a set, and let (X, d) be a metric space. We say that a sequence ( f j ) of functions f j : S → X , j = 1, 2, 3, …, is a Cauchy sequence of functions if for every ε > 0 there exists an integer N such that for every n , m > N , and for every x ∈S , we have d( fn ( x ), fm ( x )) < ε. The same sequence of functions converges uniformly to a function f : S → X if for every ε > 0 there exists an integer N such that for every n > N and for every x ∈S , we have d ( fn ( x ), f ( x )) < ε. Show that if X is complete, then for every Cauchy sequence ( f j ) of functions f j : S → X there exists a function f : S → X such that ( f j ) converges uniformly to f. 18. Let ( X , d1 ) and ( X , d2 ) be two metric spaces, let X = A ∪ B and let A ∩ B = C ≠ ∅. Assume also that d1 (c1 , c2 ) = d2 (c1 , c2 ) for every c1 , c2 ∈C . Define d : X × X →  as follows:  d1 ( x , y )                                   if x , y ∈ A  d2 ( x , y )                                   if x , y ∈ B d( x , y ) =   inf{d ( x , z ) + d ( y , z ) :  z ∈C}  otherwise. 1 2  Show that d is a metric. 19. Show that every dilation is one-to-one and uniformly continuous. 20. Find an example of homeomorphic metric spaces which have no dilations from one onto the other. 21. Let ( X1 , d1 ) and ( X 2 , d2 ) be metric spaces. Show that if f : X1 → X 2 is a dilation, then f −1 : f ( X1 ) → X1 is continuous. 22. Show that  is not a fractal. 23. Let f :  2 →  2 be a dilation. Prove the following: (a) f maps lines to lines. (b) f sends angles (pairs of rays emanating from a single point) to angles of equal size. (c) f maps circles to circles. (d)  2 is not a fractal. [Note: Justifying that n is not a fractal for all, n ≥ 3 can be done in steps similar to the ones indicated in this exercise.]

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Chapter

3

Topological Spaces: Definition and Examples

O

ur short theory of metric spaces will now be used as a springboard to launch us into the study of the theory of topological spaces.

3.1  The Definition and Some Simple Examples A brief historical note: Historically, topology evolved from analysis. Indeed, in its early stages it was called Analysis Situs (positional analysis; “Analysis Situs” is the title of the seminal paper by Henri Poincaré, published 1895). The term topology was coined in 1914 by Felix Hausdorff (1868–1942), abstract spaces with topological structures were introduced in 1906 by Maurice Fréchet (1878–1973), and the modern, axiomatic setup of topology—the one we will deal with in this book—was introduced in 1922 by Kazimierz Kuratowski (1896–1980).

Proposition 1, Section 2.1 does not explicitly involve the notion of metric, hence we can abstract the following generalization from it. A topological space is a set X and a set τ of subsets of X satisfying the following axioms: (i) ∅ and X are in τ. (ii) If U i , i ∈ I are all in τ, then so is ∪U i . i∈I

n

(iii) If U1 , U 2 ,… ,U n are all in τ, then so is ∩U i . i =1

It follows from (i) that τ is never empty. In the case when the set X is empty τ must be {∅}—the most boring topological space one can imagine. We do not want to deal with it and so:

We will assume now and forever that the set X in this definition is not empty. 33

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34   ◾    Topological Spaces: Definition and Examples

The collection τ is called a topology over X. Formally, a topological space is the pair (X, τ), where the topology τ satisfies the above axioms. We will sacrifice formality for brevity and, when the topology τ is understood from the context, we will often refer to X itself as a topological space. The sets contained in τ are called open sets. Thus, according to axiom (i), the sets ∅ and X are always open; axiom (ii) stipulates that unions of open sets must be open, while in axiom (iii) we request that finite intersections of open sets be open. Example 1: Metric Spaces Before we see more exotic spaces, we back up for a moment to our starting models: the metric spaces. Proposition 1 in 2.1, as announced, makes a topological space out of every metric space. Importantly, the Euclidean metric space n now becomes the ☐ Euclidean topological space (or the usual topological space). We reiterate the following convention (stated for the first time in 2.1): Unless otherwise stated, n stands for the Euclidean topological space.



We will remind the reader of this a few more times. In special cases,  is the Euclidean line,  2 is the Euclidean plane. At times we will use the all-encompassing term space even when n = 1 or n = 2. Example 2: The Indiscrete Topology If ∅ and X happen to be the only members of τ, then we get one of the two extreme topologies over a set X: the indiscrete topology. Since in this case the sets U i mentioned in the axioms can only be ∅ or X, all of the axioms are satisfied. The indiscrete topological space is the simplest, and in a way, the least interesting ☐ of all spaces. Example 3: The Discrete Topology Suppose now that τ is the set P (X) of all subsets of a set X. It is again plain that the three axioms are satisfied so that τ is a topology over X. This is the other most extreme example of a topology, called the discrete topology over X. So, not surpris☐ ingly, this space possesses some extreme properties. Example 4: A Finite Topological Space Take X = {a , 1, $} and τ = {∅ , {a , 1, $}, {a}, {a , 1}}. Since ∅ and X are given to be in τ, the first axiom is satisfied. The union of any members of τ is the largest of them according to the linear order ∅ ⊂ {a} ⊂ {a , 1} ⊂ {a , 1, $}, and so it is also in τ. This shows that the second axiom is fulfilled. Similarly, any (finite) intersection of members

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3.1  The Definition and Some Simple Examples   ◾    35  

of τ is the smallest of them with respect to ∅ ⊂ {a} ⊂ {a , 1} ⊂ {a , 1, $}, and so it is in τ. ☐ Consequently the last axiom holds too and thereby τ is a topology. Finite topological spaces will rarely be encountered in this book. Example 5: Finite Complement Topology or Co-finite Topology With X being any (nonempty) set, define open sets (elements of τ) to be complements in X of finite sets, as well as the empty set. We get the finite complement topology, or, more succinctly, the co-finite topology (Illustration 3.1).

Illustration 3.1  An open set in the finite complement topology over a plane: they are obtained

by removing finitely many points in the plane (indicated by white disks in the picture).

This time we will prove that τ is indeed a topology. (i) The set ∅ is open since we explicitly included it in τ, while X is open since it is the complement of the finite set ∅. (ii) Suppose {U i : i ∈ I } is a nonempty family of open sets. We want to show that ∪U i is also open. Since U i , i ∈ I , are open, each U i is a complement of some i∈I

finite Vi . Then we have ∪U i

assumption

=

i∈I

∪ (Vi )

c

de Morgan law

=

i∈I

( ∩V ) . Since ∩V i∈I

i

c

i∈I

i

is a

subset of each Vi it must be finite too, and so ∪U i is a complement of a finite i∈I set. Thereby, it is open. n (iii) Suppose that U1 , U 2 ,… ,U n are open. We show that ∩U i is open. By i∈I

assumption, for every i = 1, 2, …, n , U i = (Vi )c for some finite set Vi . So n

∩U i

i =1

assumption n

=

∩(Vi )c

i =1

n

de Morgan law

=

(∪V ) . Since all V , V ,…,V are finite, so is their c

n

i =1

i

1

2

n

union. Thereby, ∩U i is a complement of a finite set, and so it is indeed open. ☐ i∈I

Observe that infinite intersections of open sets in the cofinite topology are not necessarily open.

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36   ◾    Topological Spaces: Definition and Examples

Example 6: Countable Complement, or Co-countable Topology Let X be any (always nonempty) set, and let τ be the collection of all complements in X of (finite or infinite) countable sets. In addition we take that the empty set is in τ. Proving that τ is a topology is left as an exercise. This is the countable complement topology, sometimes also called the co-countable topology (Illustration 3.2).

Illustration 3.2  The countable complement topology over a plane: the open sets are obtained

by removing countably many points in the plane (indicated by white disks in the picture).

Note that justifying one of the three axioms requires Proposition 4, Section 1.2. ☐ Example 7: Nested Topology

{

∞

For any (nonempty) space X, let τ = ∅, X , U1 ,U 2 ,…,U n ,… , ∪U i i =1

}

be a family of

subsets of X such that U1 ⊂ U 2 ⊂  ⊂ U n ⊂ U n +1 ⊂  . (Recall again that ⊂ includes =, and so τ could be finite.) Then, τ is a topology over X, called a nested topology over X. The finite topological space given in Example 4 is an example of a nested topology. Here is one more example: X =  = {…, − 2, − 1, 0, 1, 2, …}, the set of integers, τ = {∅ , {−1, 1}, {−2, − 1, 1, 2}, {−3, − 2, − 1, 1, 2, 3}, … , X }. It is easy to show (we leave it to the reader) that nested topologies τ satisfy our axioms of topological spaces. ☐ One more example of a nested topology is shown in Illustrations 3.3 and 3.4.

Illustration 3.3

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3.1  The Definition and Some Simple Examples   ◾    37  

   Illustration 3.4

The set X (Illustration 3.4) consists of all of the points in the set of infinitely many pentagonal stars positioned as shown in the picture. A topology τ over X consists of the empty set, the set X itself, and the members of the infinite sequence of finitely many concentric stars as indicated in Illustration 3.3 (only the first three members of that sequence are shown). The topological structure of the set X equipped with τ is obviously different from what one might expect just by viewing the object as a metric subspace of  2 . Example 8: Order Topology Let X be equipped with a linear order 0}) is not open ☐ there. Is it closed in  2 ? Example 4 The set { 1n ∈  : n = 1,2,3,…} is not closed in  since its complement is not open. The complement is not open since the point 0 is not an interior point for the complement; that is so since any open ball around 0 must contain elements of the set { 1n ∈  : n = 1, 2, 3, …} . ☐ The point 0 is an accumulation point for the set { 1n ∈  : n = 1,2,3,…}. We define the concept properly in the next two paragraphs. An open neighborhood of a point x in a space X is an open subset U of X that contains x. A point x is an accumulation point (or a limit point) for the subset A of a space X if every open neighborhood of x contains other elements of A. That is, for every open neighborhood Ux of x, we have U x ∩ ( A\{x }) ≠ ∅. The set of all accumulation points for a subset A of a space X will be denoted by A′. y x

A z

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Illustration 3.7  A subset A of  2: the points x,

y, and z are in A′.

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42   ◾    Topological Spaces: Definition and Examples

Note that we do not stipulate that the accumulation point for a set A must be in A. Indeed, A′ \ A ≠ ∅ in Example 4. In that example, the set A failed to be closed since it did not contain its only accumulation point. In the next theorem we link these two notions explicitly. Theorem 4. A subset A of a space X is closed if and only if it contains all of its accumulation points. Proof. ⇒ Let A be a closed subset of X and suppose it does not contain an accumulation point x. Then x ∈ Ac , and so Ac is an open neighborhood of x that is disjoint from A. Hence x is not accumulation point for A, and we have a contradiction. ⇐ Suppose A contains all of its accumulation points and take a point x in Ac . Since x is not an accumulation point for A there is an open neighborhood of x that does not contain any points from A. That makes x an interior point for Ac . So, every point of Ac is an interior point, and so Ac is open. That makes A closed. Example 5 If  is equipped with the usual topology, then ′ =  . This is true since every open set in  (with the usual topology) contains as a subset an open interval, and every open interval contains a rational number (in fact, infinitely many of them). If, on the other hand,  has the countable-complement topology, then ′ = ∅ since for every x in  , the set {x } ∪ ( \ ) is an open neighborhood of x that has an empty intersec☐ tion with  \{x }. A subset of n is bounded if it is contained in some ball B(a, r), where we remind the reader that, unless otherwise stated, we are assuming that n comes with the usual, Euclidean (metric) topology. The following theorem, as well as Proposition 6, will be generalized in Section 7.4. Theorem 5 (Bolzano–Weierstrass). Every bounded and infinite subset A of  has an accumulation point. Proof. Without loss of generality we can assume that A is a subset of [0, 1]. Subdivide [0, 1] into [0, 1 2] and [ 1 2 ,1]. Since A is infinite, at least one of these two intervals contains infinitely many points. Subdivide that interval into two closed subintervals of equal length, and repeat the argument in the previous sentence. Iterate this procedure to get an infinite sequence of subintervals [0,1] ⊃ I1 ⊃ I 2 ⊃ I 3 ⊃ … . It follows from the Cantor’s Nested ∞ ∞ Intervals Theorem (Theorem 2, Section 2.2) that ∩ I j ≠ ∅ . Take a ∈∩ I j ≠ ∅ and an open j =1

j =1

ball (a − ε , a + ε ) around a. Since, by construction, the lengths of the intervals I j tend to 0 as j tends to infinity, and since a belongs to each of these intervals I j , there is a large enough j such that a ∈ I j ⊂ (a − ε , a + ε ). By construction, each I j contains infinitely many points

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3.2  Some Basic Notions   ◾    43  

of A, and thus, so does (a − ε , a + ε ). So, the set (a − ε , a + ε ) contains a point from A other than a, implying that a ∈ A′. A similar argument justifies the following. Proposition 6. Every infinite and bounded subset of n has an accumulation point. The closure of a subset A of a space X, denoted A, is the smallest closed subset of X such that A ⊂ A. Theorem 7. (a) If A ⊂ B then A′⊂ B′. (b) For every subset A of a space X we have that A = A ∪ A′. Proof. (a) This is left as an exercise (Exercise 7). (b) First we show that A ∪ A′ is closed. According to Theorem 4, it suffices to show that A ∪ A′ contains all of its accumulation points. Suppose otherwise; that is, suppose that x is an accumulation point for A ∪ A′ and x ∉ A ∪ A′. If every open neighborhood of x contains points from A, then x ∈ A′ and so x ∈( A ∪ A′ ) ; contradiction. So, there is an open neighborhood U of x that contains points of U x A′ and does not contain points of A. Let y ∈U ∩ A′ . Then (see Illustration 3.8) U would be an open neighborhood of y ∈ A′ that y does not contain any points of A, arriving again at a contradiction. We have exhausted all possibilities and we can conclude that Illustration 3.8 A ∪ A′ contains all of its accumulation points. So, it is closed. Since A is the smallest closed set containing A, we have that A ⊂ A ∪ A′ . If the inclusion is proper then there must be some x ∈ A′ such that x ∉ A . Since A is closed, Theorem 4 implies that ( A)′ ⊂ A . So x ∉( A)′ , and thereby A′ ⊂ ( A)′ . The last statement contradicts part (a). A sequence x1 , x 2 ,… , xn ,… of points in a topological space X converges to a point x if for every open neighborhood U of x, there is a positive integer N such that xn ∈U for every n > N . This definition is compatible with the definition of the same concept given for metric spaces. We will, as before, use the shorter notation ( xn ) for a sequence x1 , x 2 ,… , xn ,…. If the sequence ( xn ) converges to x we will write either lim xn = x or ( xn ) → x . So, ( xn ) → x if n→∞ for any open neighborhood of x, from some point on, all of the members of the sequence are in that neighborhood. Proposition 8. Suppose X is a space and x1 , x 2 ,…, xn ,…∈ A ⊂ X . If ( xn ) → x then x ∈ A .

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44   ◾    Topological Spaces: Definition and Examples

Proof. If x = xm for some m, then x ∈ A and so x ∈ A . Otherwise, it is obvious that x ∈ A′, and by Theorem 7(b) ◼ it follows that x ∈ A . The converse of Proposition 8 is true if X is a metric space, but not true in general (Exercise 9). A point x in a subset A of a space X is a boundary point for A if for every open neighborhood Ux of x, we have that Ux ∩ A ≠ ∅ ≠ Ux ∩ Ac (Illustration 3.9). The set of all boundary points of A will be denoted by ∂A, and will be called the boundary of A.

y

A z

Illustration 3.9  A subset A of

 2: the points y and z are in ∂A; compare with Illustration 3.7.

Example 6 For subsets A of the Euclidean space n the notion of boundary is compatible with the usual meaning of the word, at least when the sets A are simple enough. For example, if A = {( x , y ) :  x 2 + y 2 ≤ 1} (the closed unit disk), then ∂ A = {( x , y ) :  x 2 + y 2 = 1} ☐ (the unit circle). Our intuition and expectations should be restrained when dealing with more exotic spaces or sets. For example, in Exercise 15 the boundary of a set is the “interior” (the dictionary meaning) of that set. Theorem 9. For every subset A of a space X the following is true. (a) ∂A is a closed set. (b) A = int A ∪ ∂ A and int A ∩ ∂A = ∅ (and so ∂ A = A \ int A = A ∩ (int A)c ). (c) A is open and closed (i.e., a clopen) if and only if ∂A = ∅. Proof. (a) Every point in (∂ A)c is either in A\ ∂ A or in Ac \ ∂ A . Suppose x ∈ A\ ∂ A. Then there must be an open neighborhood of x that is entirely in A\ ∂ A because otherwise every open neighborhood must contain points from ∂A or from Ac \ ∂ A , which would force x to be in ∂A. (Why is it true that if every open neighborhood of x contains points from ∂A then x is in ∂A?) This makes every point of A\ ∂ A an interior point. So A\ ∂ A is open. By symmetry, so is Ac \ ∂ A . So (∂ A)c is open. So ∂A is closed. (b) It is obvious that intA ∩ ∂A = ∅ (just compare the definitions of interior and boundary points). It is also obvious that every point of A is either an interior point or a boundary point. So, A ⊂ int A ∪ ∂ A. If x is an accumulation point for A that is not in A, then the definitions of accumulation points and boundary points imply that x ∈∂A . If x is an accumulation

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3.2  Some Basic Notions   ◾    45  

point for ∂A then x is in ∂A; this again follows from the definition of boundary points. We have thus shown that int A ∪ ∂A contains all of its accumulation points and so it is closed. Since A is the smallest closed set containing A, it follows that A ⊂ int A ∪ ∂ A. Take any point x ∈intA ∪ ∂ A . If x ∈ A , then x ∈ A . If x ∉ A , then x ∈∂A . It follows directly from the definition of ∂A that in this case x ∈ A′. Hence, in all cases x ∈intA ∪ ∂ A implies x ∈ A ∪ A′ = A . So int A ∪ ∂ A ⊂ A. (c) It follows from part (b) that intA = A if and only if ∂A = ∅ . Since intA ⊂ A ⊂ A we have that int A = A = A if and only if ∂A = ∅, so that A is both open and closed if and only if ∂A = ∅. We end this section with a notion that will be encountered frequently as we proceed. A space X is Hausdorff if for every distinct point x , y ∈ X , there are disjoint open neighborhoods U and V of x and y, respectively (see Illustration 3.10). X

U x

y

V Illustration 3.10  In a Hausdorff space every

two points can be separated with two disjoint open sets.

Every metric space is Hausdorff (Exercise 4, Section 2.1). The indiscrete space with more than two elements is clearly not Hausdorff. Exercises 1. Consider Illustration 3.5 and its caption. Find a topology over  2 such that b is an interior point for S and at the same time a is not an interior point for S. 2. Consider the square A = {( x , y ) :  x − 2 < 1,  y − 2 ≤ 1} as a subset of the set  2 . Find intA if  2 is equipped with (i) the discrete topology, (ii) the usual topology, (iii) the topology induced by the post office metric (Section 2.3) with (0, 0) being the special point (the mailbox). 3. Show that unions of finitely many closed sets are closed. Give an example of a family of closed sets in  (usual topology), such that their union is not closed. 4. Find A′ if A is as in Exercise 2.

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46   ◾    Topological Spaces: Definition and Examples

5. Find ′ and ( \ )′ if  is equipped with (i) discrete topology, (ii) indiscrete topology, (iii) finite complement topology. 6. (a) Give an example of a nonempty subset A of  2 (usual topology) such that A′ = ∅ . (b) Give an example of a countable subset A of  2 (usual topology) such that A′ =  2 . (c) Show that there is no nonempty and bounded subset A of  2 (usual topology) such that A′ = A and ∂A = ∅ . 7. Let A and B be subsets of a space X and assume A ⊂ B in parts (a), (b), and (c). Prove the following. (a) A′ ⊂ B′ . (b) int( A ) ⊂ int( B ). (c) A ⊂ B . (d)

( A ∪ B )′ = A′ ∪ B′.

8. Let A be a subset of a metric space X. Show that ( A′ )′ ⊂ A′ . Give an example of a subset A of  2 (always usual topology unless otherwise stated!) for which the inclusion is proper, and give an example of A ⊂  2 for which ( A′ )′ = A′ . 9. Let X be a metric space and let A be a subset of X. Show that if x ∈ A′ then there is a sequence ( xn ) of elements in A that are distinct from x such that ( xn ) → x . Show that that is not necessarily true in the case where X is not a metric space. 10. Show that the following is true for every subset A of a space X. (a) int(int A) = int A. (b) ∂( ∂A ) ⊂ ∂A . (c) ( A) = A . 11. Consider the space  and let  1 and  2 be two disjoint copies of that space. Define a topology over  1 ∪  2 by declaring open sets to be all of the sets of type A1 ∪ A2 where A1 and A2 are copies in  1 and  2 , respectively, of a single open set A in . (a) Show that this defines a topology over  1 ∪  2. (b) Find a subset A of  1 ∪  2 (with the above topology) such that ∂(∂A) is a proper subset of ∂A.

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3.2  Some Basic Notions   ◾    47  

12. In Illustration 3.11 we depict a set F of points in  2 (only the black dots count). Find a topology over  2 such that int F = F and F ′ = F . Find another topology over  2 such that int F = F and F′ = ∅ . 13. Find a topology over  2 such that the boundary of the square A = {( x , y ) :  x ≤ 1,  y ≤ 1} is {( x , y ) ;  x < 1,  y < 1}. 14. Which of the following topologies over  is Hausdorff?

Illustration 3.11

(a) Co-finite topology. (b) Co-countable topology. 15. Define a topology over  2 as follows. Open sets are the empty sets, {( x , y ) ∈ 2 : x 2 + y 2 = 1} , and all sets of type A ∪ B where A is any subset of the open disk {( x , y ) ∈ 2  :  x 2 + y 2 < 1} and B is any subset of the complement of that open disk that contains a ring {( x , y ) ∈ 2  : 1 ≤ x 2 + y 2 ≤ r } for some r >1 (see Illustration 3.12). Call it the In–Out topology over 2.

B

A

Illustration 3.12

(a) Prove that the In–Out topology over  2 is indeed a topology. (b) Show that in the In–Out topology the boundary of the closed unit disk D 2 = {( x , y ) ∈ 2  :  x 2 + y 2 ≤ 1} is the open unit disk {( x , y ) ∈ 2  : x 2 + y 2 < 1}. What is the interior of D 2 ? 16. Consider  2 with the post office topology with (0, 0) being the designated point (the post office). For A = {( x , y ) ∈ 2  : ( x − 1)2 + y 2 < 1} find intA, A, A′, and ∂A. 17. Show that for every two subsets A and B of a space X the following are true. (a) int( A ∩ B ) = (int A ) ∩ (int B ). (b) (int A ) ∪ (int B ) ⊂ int( A ∪ B ) (and give an example where the inclusion is proper). (c) A ∪ B = A ∪ B . (d) A ∩ B ⊂ A ∩ B . 18. Suppose X is a metric space and A is a subset of X. Show that if x ∈ A , then there is a sequence ( xn ) of elements in A converging to x.

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48   ◾    Topological Spaces: Definition and Examples

19. Show that every sequence of infinitely many elements in a bounded subset of  contains a convergent subsequence. 20. (a) Suppose a space X is a Hausdorff space. Show that if a is an accumulation point for a subset A of X, then every open neighborhood of a contains infinitely many points of A. (b) Show that every metric space is Hausdorff. 21. (a)  Find three distinct open subsets A, B, and C of  2 such that ∂A = ∂B = ∂C . (b) Find three pairwise disjoint subsets A, B, and C of  2 such that A ∪ B ∪ C =  2 and such that ∂A = ∂B = ∂C . (c) Find four pairwise disjoint subsets A, B, C, and D of  2 such that A ∪ B ∪ C ∪ D =  2 and such that ∂A = ∂B = ∂C = ∂D . (d) Find infinitely many pairwise disjoint subsets of  2 that share a common boundary. 22. Denote the usual (Euclidean) topology over  by τ , and the set of irrational numbers by J. Define σ = {U ∪ T : U ∈ τ, T ⊂ J }. (a) Prove that σ is a topology over  and that τ ⊂ σ. (b) A point x in a space X is isolated if {x} is open in X. Find all isolated points in  equipped with σ.

3.3  Bases A brief historical note: The topological space introduced in Example 6 was first described by the American mathematician Robert Sorgenfrey (1915–1995); the spaces given in Examples 7 and 8 were found independently by Robert Lee Moore (1882–1974) and Viktor Niemytzki (also spelled Nemytskii) (1900–1967).

A basis for a topological space X is a family B of open sets such that every open subset of X is a union of some members of B. Sometimes such a set B is called an open basis for X: this finesse in the terminology emphasizes the presence of a topology over X before the basis is identified (compare with Theorem 2 further below). The empty set ∅, which is open in every space, should be considered as the empty union of the elements of the basis. Proposition 1. A collection B of open sets in a space X is a basis for the space X if and only if for every open subset U of X, and every x ∈U , there is a B ∈B such that x ∈B ⊂ U .

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3.3  Bases   ◾    49   B1

U

x1 B2 x2

B3

x3

Illustration 3.13  Every point xi in U is in

some Bi ∈B , such that Bi ⊂ U .

Proof. ⇒ If B is a basis for the space X, then every open subset U in X is a union of members of B , and so for every x ∈U, there is a B ∈B such that x ∈B ⊂ U . ⇐ Let B be a collection of open sets such that, for every open set U, and for every x ∈U , there is Bx ∈B such that x ∈ Bx ⊂ U . Then U = ∪ Bx . x ∈U

Example 1 The set of all open sets is surely a basis for the underlying topological space. This is the most uninteresting example of a basis. The point of introducing the concept of a basis is precisely in getting smaller families of open sets, used as building blocks to ☐ assemble the whole topology. Example 2 The set of open intervals is a basis for the usual topology over . The set of all open balls B(a, r), a ∈n , r ∈ + , is a basis for the usual topology over n . More generally, the set of open balls B(a, r) in a metric space X , a ∈ X , r ∈ + , is a basis for the metric space topology over X. All of these three claims follow directly from the ☐ definitions. Example 3 The set {X} is a basis for the indiscrete topology over X. A minimal (under inclusion) ☐ basis for the discrete topology consists of all one-element sets (singletons). Now we turn to a related problem: if X is merely a set, and if B is a collection of subsets X, when is B a basis for some topology over the set X? One answer is: the set of unions of elements in B must satisfy the axioms of topological spaces. Before we state a better criterion, we take a look at a couple of negative examples. Example 4 The set D = {{1}, {2}} is not a basis for a topology over the set X = {1, 2, 3}. This is true since X is must be an open set in any topology over X, and since it is not a union of ☐ the elements of D.

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50   ◾    Topological Spaces: Definition and Examples

Example 5 The set of intervals D = {(n , n + 2): n ∈} is not a basis for a topology over . The set S of all unions of elements of D does not satisfy all of the axioms of topological spaces: the intervals (1,3) and (2,4) are obviously in S (they are in D ), but their inter☐ section is not (and so the last axiom of topological spaces fails). Theorem 2. A collection B of subsets of a set X is a basis for some topology over X if and only if the following two conditions are satisfied. (1) ∪ B = X (that is, X is the union of the elements of B). B∈B

(2) For every B1 and B2 in B and for every x ∈B1 ∩ B2 , there is B ∈B such that x ∈B ⊂ B1 ∩ B2 (see Illustration 3.12).

B1

B x

B2 Illustration 3.14

Proof. ⇒ Suppose B is a basis for a topology τ over X. This means that the set of unions of elements of B satisfies the axioms for topological spaces. In particular the first axiom is satisfied, and so ∪ B = X . B∈B Since the elements of B are clearly open in the topology τ, so are their finite intersections. Consequently, for every B1 and B2 in B , B1 ∩ B2 must also be a union of elements of B . So, every element in that intersection must be in some member of B that comprises that union. So (2) is true too. ⇐ Suppose (1) and (2) are true. We need to show that the set τ of all unions of elements of B is a topology over X. So, we need to check the axioms for topological spaces. The empty set is the empty union of elements of B , while it follows from (1) that X is also a union of elements of B. It is plain that unions of elements of τ are also in τ. It remains to be shown that finite intersections of elements of τ are also in τ. We will do that for intersections of pairs of elements of τ; the rest follows by induction. Take any  U1 and U 2 in τ; so U1 = ∪ Bi and i∈I

U 2 = ∪ B j , where all Bi ’s and B j ’s are members of B. Since A ∩ ( ∪ B j ) = ∪ ( A ∩ B j ) for every j∈J j ∈J j ∈J   sets A, B j , j ∈ J , we have, U1 ∩ U 2 = ∪ Bi ∩  ∪ B j  = ∪ ( Bi ∩ B j ). So, it suffices to show that  j∈J  i ∈I i∈I

( )

j∈J

each Bi ∩ B j is a union of elements of B. Assumption (2) implies that for every x ∈ Bi ∩ B j there is a member Bx of B such that x ∈ Bx ⊂ Bi ∩ B j . This implies that Bi ∩ B j = ∪ Bx . x ∈Bi ∩B j

According to the last theorem, if B satisfies (1) and (2) in that theorem, then the family τ of all unions of elements of B is a topology over the underlying set X. The topology τ is called the topology generated by the basis B.

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3.3  Bases   ◾    51  

Remark. Let B be a collection of open subsets of a space X satisfying the conditions (1) and (2) of Theorem 2. Then B is not necessarily a basis for the given topology over X. Prove this claim through a simple example. We say that a topology τ1 is weaker or coarser than a topology τ 2 over the same set X if τ1 ⊂ τ 2 . At the same time we say that τ2 is stronger or finer than τ1 . The topology generated by a basis B is the weakest topology that contains the elements of B (Exercise 7). Defining a topology by specifying a basis that generates that topology is an efficient approach that is often used when introducing new topological spaces. The next few examples are important and will be invoked many times. Example 6: Sorgenfrey Line Consider the set of all half-open intervals of type [a, b). The union of all such halfopen intervals is , and the intersection of any two non-disjoint intervals of that type is an interval of the same type. Consequently the set of all half-open intervals of that type is a basis for a topology over . This topology is called the right half-open interval topology over . The left half-open interval topology over  is defined similarly, using intervals of type (a, b] as a basis. Since these two topologies have essentially the same properties, we will deal only with the right half-open interval topology, also ☐ called the Sorgenfrey line. We will use this latter terminology from now on. Example 7: Half-Disk Topology Consider the upper closed half plane  2up = {( x , y ) :  y ≥ 0} . Let B consist of all open balls B( x , r ) that are entirely in  2up (as is the set A in Illustration 3.15) and of sets of type B( x , r )up ∪ {x} where B( x , r )up is an open half ball {( x , y ) : ( x − a)2 + y 2 < r ,  y > 0} centered at the point x = (a,0) on the x-axis (as is the set B in Illustration 3.15). A B Illustration 3.15

It is very easy to see that B satisfies the requirements of Theorem 2. The topology ☐ generated by B is called half-disk topology. Example 8: Tangent Disk Topology The underlying set is again the upper closed half plane  2up = {( x , y ) :  y ≥ 0} . Let B consist of all open balls B( x , r ) that are entirely in  2up and are not tangential to the x-axis (as is the set A in Illustration 3.16), and of sets of type B(a , r ) ∪ {x} where B(a , r ) is an open ball touching the x-axis at the point x (depicted as the set B in Illustration 3.16).

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52   ◾    Topological Spaces: Definition and Examples A B x

Illustration 3.16

It is again easy to show that B is indeed a basis (Exercise 3). The topology over  2up generated by B is called the tangent disk topology (introduced independently by ☐ Niemytzki and Moore). Example 9: Order Topology, Again Let X be a set equipped with a linear order 1  It is also easy to see that f is closed. It follows from Proposition 2 that  ~ is homeof morphic to . As we saw in Example 1, the space  ~ is obtained by shrinking the f ☐ closed interval [0,1] in  to a point. Example 5

Suppose now that X is a space, Y is just a set, and f : X → Y is a mapping. The space X ~ f is called the quotient space determined by f (Illustration 4.10). f

X

Y

Illustration 4.10  The quotient space X ~ is determined by the mapping f. f

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4.2  Quotient Spaces   ◾    79  

Recall that specifying a subset U of Y to be open if and only if f −1 (U ) is open in X, yields the topology over Y induced by f (Exercise 12, Section 3.1). The mapping f : X → Y then becomes continuous. Proposition 2 implies that the mapping g : X ~ → Y defined by f

g ([x ]) = f ( x ) is well defined and continuous. In this case, as we see further below, it is automatically a homeomorphism onto f(X). We show that g is open; the rest is obvious. Let V be an open subset of X ~ . That means that ∪ [v] is open in X. We want to show f [v ] ∈V   −1 that g(V) is open in Y, or, equivalently, that f ( g (V )) is open in X. Now, g (V ) = f  ∪ [v] , so  [v ] ∈V    −1 −1   that f ( g (V )) = f  f ∪ [v]  . It follows from the definition of [v] that f ([v]) = { f (v )},   [v ]∈V      and this in turn implies that f −1  f  ∪ [v]   = ∪ [v]. So, g(V) is indeed open. In light   [v ] ∈V   [v ]∈V of Proposition 2, we proved the following: Proposition 3. Given any space X and any set mapping f : X → Y such that f is onto, the quotient space X ~ is homeomorphic (via the mapping g) to the space Y induced f by f. Exercises

1. Give a sufficient and necessary criterion for a quotient map to be a homeomorphism. 2. Suppose X =  2. Describe (visualize) the space X ~ if ~ is the smallest equivalence relation satisfying the following conditions.

(a) ( x , y ) ~ ( x ′ , y ′ ) if and only if x = x ′ − 1.



(b) ( x , y ) ~ ( x ′ , y ′ ) if and only if x = x ′ − 1 and y = y ′ − 1.



(c) ( x , y ) ~ ( x ′ , y ′ ) if and only if x = x ′ − 1 or y = y ′ − 1.

3. Find an equivalence relation ~ on  2 such that

2

(a)  ~ is (homeomorphic to) a circle (equipped with the usual topology). 2 (b)  ~ is (homeomorphic to) a sphere (equipped with the usual topology).

{

}

4. Denote, as usual, S 2 = ( x , y , z ) ∈ 3  :  x 2 + y 2 + z 2 = 1

{

}

(the unit sphere), and

D 3 = ( x , y , z ) ∈ 3  :  x 2 + y 2 + z 2 ≤ 1 (the closed unit ball).

K12146_C004.indd 79

2

(a) Define f : S → S by f ( x , y , z ) = ( x , y , − z ). Describe the space S ~ . f 3 3 3 2 (b) Define g : D → D to be f over S and the identity elsewhere. Describe D ~ . 2

2

g

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80   ◾    Subspaces, Quotient Spaces, Manifolds, and CW-Complexes

5. Show that the quotient mapping q : X → X ~ is closed if and only if for every closed A ⊂ X, the set ∪ [x ] is closed in X. Show that the same is true if “closed” is [ x ]∩ A≠∅

replaced by “open.”

6. Show that a quotient mapping q : X → X ~ need not be open. 7. Let ~ be an equivalence relation on a space X, and let f : X → Y be a continuous mapping such that f ( x1 ) = f ( x 2 ) whenever x1 ~ x 2 . Show that the mapping g : X ~ → Y defined by g ([x ]) = f ( x ) is well defined and continuous. 8. (a) Define an equivalence relation ~ on the circle S1 = {( x , y ) :  x 2 + y 2 = 1} by 1 ( x , y ) ~ (− x , − y ). Show through pictures that S ~ is homeomorphic to S1. (b) Define an equivalence relation ~ on the sphere S 2 = {( x , y , z ) :  x 2 + y 2 + z 2 = 1} by 2 ( x , y , z ) ~ (− x , − y , − z ). Show through pictures that S ~ is homeomorphic to S 2. (c) Define an equivalence relation ~ on the disk D 2 = {( x , y ) :  x 2 + y 2 ≤ 1} by ( x , y ) ~ (− x , − y ) if x 2 + y 2 = 1 and (x, y) is only equivalent to itself in other cases. 2 Show through pictures that D ~ is homeomorphic to S 2 . 2 = {( x , y ) :  y ≥ 0} equipped with the tangent-disk 9. Consider the upper half-plane up 2 topology. Partition up into equivalence classes, one of which consists of all elements of the x-axis, while the others are singletons; denote the resulting equivalence relation by ~. ∞

2 ? Justify your answer. (a) Is the sequence  (−1)n , 1  convergent in up   n n=1

2 (b) Find a countable local basis of [(0,0)] ∈ up ~ . (Remark: [(0,0)] is the equivalence class of (0,0).) 10. Define an equivalence relation ~ on  as follows: x ~ y if and only if either x = y or x , y ∈[0,1]. Show directly that  ~ is homeomorphic to . 11. Define an equivalence relation ~ on  as follows: x ~ y if and only if either x = y or x , y ∈[0,1). Show that  ~ is not homeomorphic to  . Show that the same con­clusion holds if [0,1) is replaced by (0,1] or by (0,1). 12. Define ~ on  as follows: x ~ y and x ≠ y if and only if one of x, y is in (0,1] and the other is larger by exactly 1. Show that  ~ is not homeomorphic to  . 13. Let ~ be the smallest relation on  defined by x ~ y if x and y are integers. Then the quotient space  ~ is a wedge of countably many circles (countably many circles with one common point). (a) Show that  ~ is not embeddable in  2 .

(b) Show that  ~ is not first countable (hence, neither is it second countable).

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14. Identify all rational numbers in  (x ~ y and x ≠ y only if x and y are rational numbers). Describe the open subsets in  ~. 15. Define ~ on  as follows: x ~ y if x − y ∈. Confirm that this is an equivalence relation and describe the open sets in the quotient space. 16. Let X be a space, and let G be a subgroup of the group of all homeomorphisms X → X (where the operation in the group is the composition of homeomorphisms). Define a relation ~G on X as follows: x1 ~G x 2 if there is f ∈G such that f ( x1 ) = x 2 .

(a) Show that ~G is an equivalence relation.



(b) Denote the quotient mapping by q : X → X ~ . Show that q is open. G

4.3 The Gluing Lemma, Topological Sums, and Some Special Quotient Spaces A brief historical note: The notion of sums of general topological spaces seems to appear for the first time in the first edition (1940) of the book General Topology by Nicolas Bourbaki (a collective pseudonym for a group of mathematicians).

The following useful lemma will be encountered many times. Lemma 1. (The Gluing Lemma) Let A and B be two closed subsets of a space X, let X = A ∪ B, and let f : A → Y , g : B → Y be continuous mappings such that f ( x ) = g ( x ) for  f ( x ) if x ∈A every x ∈ A ∩ B. Then the mapping h( x ) =  is a continuous mapping g ( x ) if x ∈B   X →Y. Proof. Let V be a closed subset of Y. We show that h is continuous by proving that h −1 (V ) is closed (Proposition 4, Section 3.5). It is easy to see that h −1 (V ) = f −1 (V ) ∪ g −1 (V ). Since f is continuous, f −1 (V ) is a closed subset of A. Proposition 2, Section 4.1 implies that f −1 (V ) = F ∩ A for some closed subset F of X. Since A is closed in X it follows that f −1 (V ) = F ∩ A is closed in X. By symmetry, −1 −1 −1 g −1 (V ) is also closed in X. Hence, h (V ) = f (V ) ∪ g (V ) is closed in X , too.

Lemma 1 holds if we replace “closed” by “open” (Exercise 1). The mapping h : A ∪ B → Y defined in the last proposition is called the union of the mappings f and g. We now turn to other methods of constructing new spaces from old spaces.

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Let Xi , i ∈I be a collection of pairwise disjoint (nonempty) spaces. The family B of subsets of ∪ Xi that are open in some Xi is a basis for a topology over ∪ Xi. The space over the set i∈I i∈I ∪ Xi generated by B is called the discrete sum (or the topological sum) of the spaces Xi , i ∈I, i∈I and is usually denoted by ⊕ Xi . For the discrete sum of two spaces X and Y we will use X ⊕ Y . i∈I

Proposition 2. A subset U of ∪ Xi is open in ⊕ Xi if and only if U ∩ Xi is open in Xi for i∈I i∈I every i ∈I . Proof. Exercise 7. It follows from the gluing lemma that if fi : Xi → Y is continuous for every i ∈I, then so is the mapping f : ⊕ Xi → Y defined by f ( x ) = fi ( x ) if x ∈ Xi . i∈I

Example 1 Let X1 = {a, b} be equipped with the discrete topology, and let X 2 = {c , d} be equipped with the indiscrete topology. The open subsets of the discrete sum X1 ⊕ X 2 are of type U1 ∪ U 2 , where U1 is open in X1 and U 2 is open in X 2. We list them all: {∅,{a},{b},{a, b}, {c , d},{a, c , d},{b, c , d},{a, b, c , d}}. ☐ The discrete sum topology over ∪ Xi is induced by the inclusions Xi → ⊕ Xi , i.e., it is the i∈I

i∈I

finest topology making all the inclusions Xi → ⊕ Xi continuous (Exercise 8). i ∈I

Example 2 Take pairwise disjoint copies Xi , i ∈I, of the circle S1 = {( x , y ) :  x 2 + y 2 = 1} (viewed as a subspace of  2). The space ⊕ Xi can be viewed as I -many separate, unrelated circles, as i ∈I shown in Illustration 4.11. (This space is homeomorphic to soon-to-be-defined product ☐ space I × S1, where I is equipped with the discrete topology; Exercise 3, Section 5.1.)

Illustration 4.11

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4.3 The Gluing Lemma, Topological Sums, and Some Special Quotient Spaces   ◾    83  

The more general case, when the spaces in the collection A = { Xi  : i ∈ I } are not necessarily disjoint, can be approached through the criterion given in Proposition 2: the set T ( A ) = {U ⊂ ∪ Xi  : U ∩ Xi is open in Xi for every i ∈I } is a topology over ∪ Xi . It is tradii∈I

i∈I

tionally called the weak topology over ∪ Xi determined by A . As was the case when the i∈I sets Xi were pairwise disjoint, it is again true that this topology is the finest topology making the inclusions Xi → ∪ Xi continuous. i∈I

Example 3 Let X1 = {a, b} be with the discrete topology and let X 2 = {b, c} be with the indiscrete topology. The weak topology over {a, b, c} determined by A = { X1 , X 2 } is {∅,{a},{b, c},{a, b, c}}. ☐ Let X and Y be two spaces, and suppose X ∩ Y = A. Choose disjoint copies X ′ and Y ′ of X and Y respectively, and for each a ∈ A denote by a1 and a2 the corresponding elements in X ′ and Y ′, respectively. Partition X ′ ∪ Y ′ into {a1 , a2 } for every a ∈ A, and into singletons for other elements of X ′ ∪ Y ′ . This defines an equivalence ~ over X ′ ∪ Y ′ . It is straightforward to show that X ′ ∪ Y ′ ~ and the weak topology over X ∪ Y determined by { X ,Y } define homeomorphic spaces. (Exercise: prove it!) We now pay more attention to spaces obtained in a similar way as X ′ ∪ Y ′ ~ . Let A be a nonempty subset of a space X, let Y be any space disjoint from X, and let f : A → Y be any mapping. Denote B = f ( A), and partition X ∪ Y into classes of type { y} ∪ f −1 ({ y}) when y ∈B, and singletons otherwise. This defines an equivalence relation ~ over X ∪ Y . The quotient space X ⊕ Y ~ is called the space obtained from X and Y by identifying (or by gluing) the subspaces A and B along f; we will denote it by X ∪ f Y . When the rest is clear from the context, we will simply say that the new space is obtained by identifying (or gluing) A and B. In a special case when both X and Y are disjoint subspaces of a fixed space Z, and when A coincides with X, then the above partition of X ∪ Y , extended by the singletons {{z } : z ∈Z \ ( X ∪ Y )} , determines an equivalence relation ~ on Z. In this case the space Z ~ will again be referred to as being obtained from Z by identifying (gluing) the subspaces A and B (along f). We will denote it by Z f . Summarizing the somewhat superficial difference between the above two definitions: X ∪ f Y is obtained by identifying subspaces of two disjoint spaces X and Y, while in the case of Z f we identify (glue) two disjoint subspaces of a single space Z. Example 4 The space X is a truncated ball {( x , y , z ) :  x 2 + y 2 + z 2 ≤ 1,  z ≤ 3 4} while Y is the truncated part {( x , y , z ) :  x 2 + y 2 + (z − 1)2 ≤ 1,  z ≥ 7 4}. The mapping f sends every point in the disk {( x , y , z ) :  x 2 + y 2 + z 2 ≤ 1,  z = 3 4} to the point in the disk {( x , y , z ) :  x 2 + y 2 + (z − 1)2 ≤ 1,  z = 7 4} with the same first two coordinates. The space X ∪ f Y is a whole ball. Illustrations 4.12 and 4.13 repeat the whole story.

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84   ◾    Subspaces, Quotient Spaces, Manifolds, and CW-Complexes

Illustration 4.12  The space X is the lower truncated ball, Y is the upper truncated ball. The

map f is indicated by the arrow.

Illustration 4.13  After gluing the two disks we get the space X ∪ f Y : a whole ball.

☐

Example 5 Denote X = [0,1] × [1,3], A = [0,1] × [1,2], Y = [0,2] × {0} (both subspaces of  2 ; see Illustration 4.14), and let f : A → Y be the projection f ( x , y ) = ( x ,0 ). A copy of the space X ∪ f Y is shown in Illustration 4.15. 3

X

2

A 1

f 1

Y

2

Illustration 4.14

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1

X ∪f Y 2

Illustration 4.15  The space A gets flattened into a line segment within X ∪ f Y .

☐

Example 6 Let X =[0,1] and Y =[2,3]. If A = {0,1} ⊂ X and g : A → Y is the embedding defined by

g (0) = 2, g (1) = 3, then X ∪ g Y is homeomorphic to a circle (Illustration 4.16).

0

2



[0, 1]

0

[0, 1]

1

1

3

[2, 3]

2



[2, 3]

Illustration 4.16

3

Illustration 4.17

2 if x < 1 If A = X and f is defined by f ( x ) =  (Illustration 4.17), then the topological 3 if x = 1 ☐ structure of X ∪ f Y is different from any subspace of n (Exercise 15). inclusion

Neither of the compositions X   →   X ⊕ Y

quotient map

→

inclusion

X ∪ f Y and Y   →   X ⊕ Y

quotient map

→

X ∪ f Y in the last example (Illustration 4.17) is an embedding. In that sense, the structure of the space X is very easy to destroy: if f is not one-to-one, then the first of the two compositions is certainly not an embedding. However, the space Y is much more resilient, and, as we see inclusion

in Proposition 3, the composition Y   →   X ⊕ Y even when f is not one-to-one.

quotient map

→

X ∪ f Y could be an embedding

Proposition 3. Let X and Y be two spaces and let A be a closed subspace of X. If f : A → Y inclusion

is continuous, then the composition Y   →   X ⊕ Y

quotient map

→

X ∪ f Y is an embedding.

Remark. Recall that X ∪ f Y is a quotient space of X ⊕ Y , so that we may talk about the quotient map X ⊕ Y → X ∪ f Y . Also recall that the inclusion Y → X ⊕ Y (sending each y ∈Y to y ∈ X ⊕ Y ) is an embedding, and consequently we can assume that Y is a

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86   ◾    Subspaces, Quotient Spaces, Manifolds, and CW-Complexes

subspace of X ⊕ Y . Thereby it suffices to show that the restriction of X ⊕ Y → X ∪ f Y to Y is an embedding. Proof. We want to show that the mapping g : Y → X ∪ f Y defined by g ( y ) = [ y] is an embedding. The equivalence class [y] of y ∈Y is { y} ∪ f −1 ( y ). If y1 ≠ y 2 , then ({ y1 } ∪ f −1 ( y1 )) ∩ ({ y 2 } ∪ f −1 ( y 2 )) = ∅ , and so [ y1 ] ≠ [ y 2 ]. Hence, g is one-to-one. The mapping g is continuous since it is a restriction of a continuous map. We complete the proof by showing that g : Y → g (Y ) is a closed mapping. Let B be a closed subset of Y. Then g ( B) = {[ y] ∈ X ∪ f Y  :  y ∈B} = {{ y} ∪ f −1 ( y )  :  y ∈B}. It follows from Proposition 1 in 4.2 that {{ y} ∪ f −1 ( y )  :  y ∈B} is closed in X ∪ f Y if and only if ∪ ({ y} ∪ f −1 ( y )) is closed in X ⊕ Y . Notice that ∪ ({ y} ∪ f −1 ( y ))  = B ∪ f −1 ( B) and that y ∈B −1

y ∈B

f ( B) is closed in A since f is continuous. Since A is equipped with the subspace topology of X, we have that f −1 ( B) = C ∩ A for some closed subset C of X. So, B ∪ f −1 ( B) = B ∪ (C ∩ A), and this is closed in X ⊕ Y since B is closed in Y, and since C ∩ A is closed in X (we have just used our assumption that A is closed in X). So ∪ ({ y} ∪ f −1 ( y )) is closed in X ⊕ Y , and y ∈B

so g(B) is closed in X ∪ f Y . Consequently, g(B) is closed in g(Y). Under the assumptions of Proposition 3, in addition to Y “surviving” within X ∪ f Y , the subspace X \ A of X also remains intact within X ∪ f Y (Exercise 16). Because of this proposition, the notion of attaching a space to a space via a mapping is sometimes defined under the additional assumptions that f is continuous and A is closed in X. Indeed, as will be evident, most of the time we will be attaching spaces under such conditions. Example 7: Sphere Obtained by Identifying Subspaces of a Disk Start with a closed disk D, and subdivide its boundary into two closed arcs as in the first frame of Illustration 4.18 (where both arcs are labeled by a). Then identify these two segments via the mapping f that projects the points on the top segment vertically down to the points of the bottom segment. We get a space denoted D f . The sequence of frames shown in Illustration 4.18 animates the construction of D f . As we see, the resulting quotient space D f is a sphere. Alternatively, we can introduce a new closed interval Y disjoint from D, directly below the horizontal diagonal of D, then project the boundary S1 of D vertically onto Y. Denoting this projection by g, the resulting space is D ∪ g Y . It is clear that this space is homeomorphic to D f (Illustration 4.19).

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4.3 The Gluing Lemma, Topological Sums, and Some Special Quotient Spaces   ◾    87   a a

a

a

a

a a

a

a

Illustration 4.18  From top left to bottom right: Start with a disk and identify the boundary points according to the arrows (i.e., according to the mapping f  ).

D

g Y

Illustration 4.19  Y = [−1, −1] × {−2} is a subspace of  2 , and the mapping f : S1 → Y is the projection defined by f ( x , y ) = ( x , −2). ☐

Exercises 1. (This exercise generalizes the gluing lemma.) Let X and Y be spaces, and let f : X → Y be a mapping. (a) Let X = ∪U i , where each U i is open in X, and suppose for every i ∈I, the restriction f

i∈I

Ui

is continuous. Prove that f is continuous.

n

(b) Let X = ∪ Fi , where each Fi is closed in X, and suppose for every i ∈{1,2,…, n}, the i =1

restriction f

Fi

is continuous. Prove that f is continuous.

(c) Does (b) remain true if we replace {Fi  : i = 1,2,…, n} by an infinite collection of closed subsets of X?

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88   ◾    Subspaces, Quotient Spaces, Manifolds, and CW-Complexes n

2. Let { Ai  : i = 1,2,3,…, n} be a family of closed subsets of X such that ∪ Ai = X and such i =1

that Ai ∩ A j = ∅ for j ≠ i + 1, i = 1, 2, 3, …, n − 1, and let { fi : Ai → Y : i = 1, 2, 3, …, n} be a set of continuous mappings such that fi and fi+1 agree on Ai ∩ Ai+1 , i = 1, 2, 3, …, n − 1. Prove that the mapping g : X → Y defined by g ( x ) = fi ( x ) if x ∈ Ai is well defined and continuous. 3. (a) Let { Ai  : i = 1,2,3,…} be a family of closed subsets of a metric space X such ∞  ′ that A j ∩  ∪ Ai  = ∅ , j = 1, 2, 3, …, such that ∪ Ai = X , and such that i =1  i− j ≥ 2  Ai ∩ A j = ∅ for | i − j| ≥ 2 . Let { fi : Ai → Y : i = 1,2,3,…} be a set of continuous mappings such that fi and fi+1 agree on Ai ∩ Ai +1 , i = 1, 2, 3, … . Then the mapping g : X → Y defined by g ( x ) = fi ( x ) if x ∈ Ai is well defined and continuous. [Hint: Show that inverse images of closed sets are closed.]  ′ (b) Show that the condition A j ∩  ∪ Ai  = ∅ , j = 1, 2, 3, … is necessary.  i− j ≥ 2  4. Show that if we omit the requirement that both A and B must be closed or open in X in the statement of the gluing lemma, then the union of f and g need not be continuous. 5. Suppose X1 = {0, a} (a is not a number) is equipped with the indiscrete topology, and suppose X 2 = . Describe the open subsets of the discrete sum X1 ⊕ X 2. 6. Show that if fi : Xi → Y is an embedding for every i ∈I, such that fi ( Xi ) ∩ f j ( X j ) = ∅ for every i ≠ j, then so is the mapping f : ⊕ Xi → Y defined by f ( x ) = fi ( x ) if x ∈ Xi .

i∈I

7. Prove Proposition 2.

8. Show that the discrete sum topology over the disjoint union ∪ Xi of the spaces Xi , i∈I

i ∈I, is the finest topology such that all of the inclusions ini : Xi → ∪ Xi (defined by i∈I

ini ( x ) = x ) are continuous. Show that the inclusions ini : Xi → ⊕ Xi are embeddings. i∈I

9. Let A = { Xi  : i ∈I } be a collection of (not necessarily disjoint) spaces and consider the

{

}

family S = U ⊂ ∪ Xi : U ⊂ Xi for some i ∈I . Show that S is a subbasis for a topology i∈I

open

τ over ∪ Xi , and show that τ is finer than the weak topology over ∪ Xi determined by i∈I

i∈I

A. Show that the inclusions ini : Xi → ∪ Xi need not be continuous with respect to τ. i∈I

10. Find two disjoint subsets A and B of the unit circle S1 and a bijection f : A → B such that S1f is homeomorphic to S1. Do the same for the unit sphere S 2 in place of S1.

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11. Let X be the closed unit ball {( x , y , z ) ∈ 3  :  x 2 + y 2 + z 2 ≤ 1}. Denote A = {( x , y , z ) ∈ X  : x 2 + y 2 + z 2 = 1,  z > 0} and B = {( x , y , z ) ∈ X  :  x 2 + y 2 + z 2 = 1,  z < 0}. Define f : A → B by f ( x , y , z ) = ( x , y , − z ). Show that the space X f , obtained by identifying A and B along f, is homeomorphic to the three-sphere S 3 = {( x , y , z , u) ∈ 4  :  x 2 + y 2 + z 2 + u 2 = 1}. (Compare with Example 7.) 12. Let X and Y be two disjoint copies of the solid torus S1 × D 2 (as a product of two metric spaces), and denote by A and B the subspaces of X and Y corresponding to the subspace S1 × S1 of S1 × D 2. Define a mapping f : A → B by f ( x , y ) = ( y , x ) for every ( x , y ) ∈ A ≅ S1 × S1. Show visually and algebraically that X ∪ f Y is homeomorphic to the three-sphere S 3 = {( x , y , z , u) ∈  4  :  x 2 + y 2 + z 2 + u 2 = 1}. 13. Let f : A1 → A2 and g : B1 → B2 be homeomorphisms. Let C be a subspace of A1, let D be a subspace of B1 and let h : C → D be a homeomorphism. Show that A1 ∪h B1 is B obtained by identifying A2 and B2 homeomorphic to the space A2 ∪ g h f −1 ( D ) ( f (C ) ) 2 −1 along ( g D )  h  f f (C ) .

(

)

14. Let X ∪ f Y be the space obtained by identifying along a mapping f : A → Y , A ⊂ X . For z ∈ X ∪ Y denote its equivalence class in X ∪ f Y by [z]. (a) Show that if f is an embedding, then the mapping k : X → X ∪ f Y defined by k( x ) = [x ] is also an embedding. (b) Show that if A is a closed subset of X and f is a closed mapping then k(X), k defined as in part (a), is a closed subset of X ∪ f Y . 15. Let A = X =[0,1], let Y =[2, 3] and let f be the mapping A → Y defined by 2 if x < 1 . Show that the space X ∪ f Y is not embeddable in any n. f (x ) =  3 if = 1 x 

16. Let X and Y be two spaces, let A be a closed subspace of X, and let f : A → Y be con-

tinuous. Show that the restriction of the quotient map X ⊕ Y → X ∪ f Y to X \ A is an embedding. 17. Let X and Y be two spaces, let A be an open subspace of X, and let f : A → Y be continuous. Show that the restriction of the quotient map X ⊕ Y → X ∪ f Y to Y is an embedding. 18 (a) Find two disjoint subspaces A and B of , find (nonempty) A1 ⊂ A, B1 ⊂ B, and find a homeomorphism f : A1 → B1, such that the space ( A ∪ B ) f is homeomorphic to . (b) Consider f :[0,1) → [1, 2) defined by f ( x ) = x + 1. Show that the space  f is not homeomorphic to .

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4.4  Manifolds and CW-Complexes A brief historical note: Low dimensional manifolds (line, plane, circle, sphere, etc.) have been mathematical folklore for a long time. The study of higher dimensional manifolds was initiated by Henri Poincaré (1854–1912). The Invariance of Domain theorem (Theorem 1) and Corollary 2 were first proved in 1912 by Luitzen Egbertus Jan Brouwer (1881–1966), whom we will encounter many times. CW-complexes were introduced in 1940 by John Henry Constantine Whitehead (1904–1960).

In this section we introduce manifolds and CW-complexes. Further analysis will be provided later on. A space X is an n-manifold (n ∈+ ) if the following conditions are satisfied. (i) X is Hausdorff. (ii) X is second countable. (iii) For every x ∈ X there exists an open neighborhood U of x such that U is homeomorphic to the open n-ball Bn = {( x1 , x 2 ,…, xn ) ∈n :  x12 + x 22 +  + xn2 < 1}. The positive integer n in this definition is called the dimension of the manifold X. It is well defined since n ≅ m only if n = m. This is a consequence (Corollary 2) of the following theorem that we will not prove. Theorem 1. (Invariance of Domain) If U is an open subspace of n and if f : U → n is an embedding, then f(U) is open in n. In the old terminology, we assert in Theorem 1 that domains (as open sets in n used to be called) are sent to domains via embeddings, explaining the origin of the name “Invariance of Domain.” It is one of these innocent-looking theorems that appear deceptively easy, yet no elementary proof is known. Corollary 2. n ≅ m if and only if n = m. Proof. The only nontrivial part of the claim is ⇒. Suppose n < m, and suppose there is a homeomorphism g : n → m. Take a nonempty open subset U of m . Then g −1 (U ) is a nonempty open subset of n. Fix an+1 , an+2 ,…, am ∈ and let c : n → m be defined by c( x1 , x 2 ,…, xn ) = ( x1 , x 2 ,…, xn , an+1 ,…, am ). Since c is an embedding and g is a homeomor­phism, the composition c  g −1 : m → m is also an embedding. Theorem 1 implies that c  g −1 (U ) is open in m . This means that c sends the nonempty open subset V = g −1 (U ) of n onto a nonempty open subset of m . But this is not possible since c(V ) ⊂ n × {an+1 } × {an+2 } × × {am }, and no subset of n × {an +1 } × {an + 2 } ×  × {am } has any interior points.

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The condition (iii) in the definition of n-manifolds is crucial. The other two conditions are included in order to avoid pathological manifolds; see Exercise 1. In the examples covered in this section we will focus only on part (iii): the first two conditions will be evident in all cases and so we will not pay attention to them. In Chapter 14, where manifolds will be covered in more detail, we will make full use of all of these conditions. Example 1 The Euclidean space n is an n-manifold. This is trivial.

☐

Example 2 The circle S1 = {( x , y ) ∈ 2 :  x 2 + y 2 = 1} is a 1-manifold. We will see later that, in a ☐ sense, S1 is the only 1-manifold other than 1. The family of 2-manifolds is much richer. In the next example we show only a few of infinitely many fundamentally different 2-manifolds. Example 3 In Illustration 4.20, we show three 2-manifolds.

Illustration 4.20  From left to right: a simple sphere, two unbounded cylindrical surfaces capped with half-spheres and joined with three short tubes, and a “dog-bone” (surface only). ☐

Example 4: Torus Start with a closed filled rectangle Z (as a subspace of  2 ) bounded by two vertical and two horizontal edges. Then identify the two vertical edges (labeled by a in Illustration 4.21) along the function f that projects every point on the left-hand side edge horizontally to a point on the right-hand side edge. The resulting space Z f is a cylindrical surface; the two horizontal edges in the starting square became two circles, both denoted by b (Illustration 4.21).

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92   ◾    Subspaces, Quotient Spaces, Manifolds, and CW-Complexes b b

a

a a

b

b

Illustration 4.21  We get a cylinder by identifying the edges labeled a along the horizontal

projection.

Proceeding further, we identify these two circles along the function that projects each point on the top circle vertically onto the point on the bottom circle. We animate the process in Illustration 4.22. b b a

b

a

Illustration 4.22  We identify the bounding circles along the vertical projection.

The resulting quotient space is a torus. The last frame of the visualization shown in Illustration 4.21 represents the torus faithfully in the sense that if we consider it as a subspace of  3, it is homeomorphic to the torus as the quotient space of the rectangle Z. In Illustration 4.23 we summarize the above procedure: we get a torus from a closed rectangle by identifying two pairs of edges as indicated. Identifying edges in filled closed polygons will be our main route for introducing 2-manifolds later on. b a

a

b

a

b

Illustration 4.23  A torus as a quotient space of a rectangle.

The torus is an important 2-manifold, and we will encounter it again.

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Example 5: Torus via a Bouquet of Two Circles Observe that the four edges of the starting rectangle Z in Example 4 (Illustration 4.23) became two circles in the torus, one labeled by a, the other by b, sharing a point v (as in Illustration 4.24). b

v

a

Illustration 4.24

Denote the subspace of the rectangle Z consisting of the four boundary edges by A, and denote the space shown in Illustration 4.24 by B. Define f : A → B by winding each of the four edges once around the circle in B with the same label, starting at v, and in the direction of matching arrows. Since ~ f yields precisely the same partition ☐ of Z as in Example 4, the space Z ∪ f B is again a torus. Example 6: Klein Bottle The Klein bottle is the quotient space of the square shown in Illustration 4.25 obtained by identifying the edges with the same labels so that the arrows match. b

a

a

b

Illustration 4.25  Klein bottle from a square.

The only difference here is that the arrows of the horizontal edges point in opposite directions. Formally, that part of the identifying mapping is defined by first rotating one of the two horizontal edges around the midpoint through 180°, then projecting vertically to the other edge. An animation of this step and the resulting quotient space is shown in Illustration 4.26.

Illustration 4.26  Identifying the a-edges of the square gives a cylindrical surface, as for the torus. However, this time the bounding circles are identified differently, corresponding to the orientation of the b-edges.

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The Klein bottle is not embeddable in 3; this claim is beyond the scope of this book. So, the animation shown in Illustration 4.26 has an unavoidable deficiency: we must put up with self-intersections of the subspaces of 3 representing the Klein bottle. Nevertheless, it is a beautiful object that deserves even an imperfect visualization. ☐ In the next example we describe and depict a 3-manifold. Example 7: A 3-Manifold The interesting 3-manifolds, with the possible exception of  3, are not embeddable in  3. That notwithstanding, all of them can be described as quotients of spaces that are embeddable in  3. That is what we will do in this example. Another formulation of the space in this example will be provided in the next section (Exercise 17, Section 5.1). A solid torus T (defined in Exercise 12, Section 4.3) is the closed subspace of  3 bounded and enclosed by a torus T 2 . For convenience we will assume that the largest (equatorial) circle of T 2 is in the plane z = 0. Split T 2 into two closed sets: F = T 2 ∩ {( x , y , z ): z ≥ 0} and G = T 2 ∩ {( x , y , z ) : z ≤ 0}. Project the points of F vertically onto the points of G, then identify these two sets along that projection. In plain terms, we identify every point of the upper half-torus with the point straight below it on the lower half-torus, leaving the equatorial circles untouched. The resulting space is a 3-manifold. A visual justification is given in Illustration 4.27 and in the next paragraph.

A B

A Illustration 4.27  Every point in the quotient space has a neighborhood homeomorphic to the

open 3-disk.

Illustration 4.27 depicts the solid torus T, the boundary of which is to be identified as indicated above to get the space T ~. The points in the interior of T are unaffected in the sense that small open 3-disks around such point give rise to open neighborhoods of the corresponding points in T ~. One such point is B in Illustration 4.27. Open 3-disks also become neighborhoods in T ~ of what used to be the boundary points of T. We illustrate that claim for the two points labeled A in Illustration 4.27 (becoming one point after identification): the two half-disks in T as shown will fit into one open 3-disk in T ~ around the single point we get by identifying the pair of points A. A similar

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argument works for the points on the two equatorial circles. Hence, every point in T ~ has a neighborhood homeomorphic to an open 3-disk, and so T ~ is a 3-manifold. ☐ Recall that we have chosen not to pay attention to Hausdorff-ness and second countability in our examples. The readers are encouraged to devote a few moments and check that these two properties are satisfied in all of the above examples. We turn our attention to CW-complexes. A 0-cell is a point; a 1-cell is the interval [−1,1], and, in general, an n-cell is the closed unit n-ball D n = {( x1 , x 2 ,… , xn ) ∈n :  x12 + x 22 +  + xn2 ≤ 1}. All of these together will be called cells. A CW-complex X is defined recursively as follows. We start with any discrete space; it constitutes the 0-skeleton X 0 of X, and it can be viewed as a discrete sum of 0-cells. Then we attach copies of 1-cells via any mapping that sends the boundary {−1,1} of each 1-cell into X 0 . Note that, since {−1,1} is a discrete subspace of [−1,1], such mappings are always continuous. The resulting space is the 1-skeleton of X and is denoted by X 1. Assuming we have the n-skeleton X n, n ∈, the (n + 1)-skeleton is obtained by attaching copies of D n+1 via continuous mappings sending ∂(D n +1 ) = S n into X n. If this procedure terminates, then X = X n for some smallest n ∈, and the finite number n is the dimension of the ∞ CW-complex X. Otherwise, X = ∪ X n, equipped with the weak1 topology determined by n=0 { Xi  : i ∈I } (that is, U ⊂ X if and only if U ∩ X n is open for every n ∈). In this case we open

take that the dimension of the CW-complex X is infinity. Remark. X n as the n-skeleton and X n as an n-fold product of the space X with itself is an overlapping notation. It should be clear from the context which of the two meanings applies. A process of building a simple CW-complex of dimension 2 is shown in Illustration 4.28.

Illustration 4.28  From left to right: X 0 , X 1 , and X 2 . ∞

It follows from Proposition 3, Section 4.3 that each skeleton X n of a CW-complex X = ∪ X n n=0 is embeddable in X, and so it could be considered a subspace of X. 1

The word “weak” is the source of W in CW; C stands for “closure finite.”

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If D n is attached to the (n − 1) -skeleton of X via a continuous map f : ∂(D n ) → X n −1, then   map → X n −1 ⊕ D n quotient  → X n −1 ∪ f D n will the image of D n under the mapping D n inclusion be called an n-cell in X. Since X n survives intact in X, the terminology makes sense. A CW-complex is finite if it has finitely many cells. Otherwise, it is infinite. We will mostly deal with finite complexes. A subcomplex Y of a CW-complex X is a subspace of X obtained in such a way that for every n ≥ 0 the (n + 1)-st skeleton of Y is obtained from the n-th skeleton of Y by attaching a subset of the (n + 1)-cells used to construct X in the same way they have been attached in X. For example, every n-skeleton of X is a subcomplex of X. Example 8: n as a CW-Complex The Euclidean space  is an infinite CW-complex of dimension 1. It is homeomorphic to the complex we get by attaching copies of 1-cells between each two consecutive integers. A similar construction shows that n is a CW-complex of dimension n (Exercise 7). With respect to this construction  is a subcomplex of  2 . Exercise 9 ☐ tells us that this is not always true. A CW-complex X of dimension 1 is also called a graph. In this setting the points in the 0-skeleton are called the vertices of X, and the images of the 1-cells under the quotient mapping are called the edges of the graph X. A special graph is introduced in the next example. Example 9: Bouquets of Circles (Roses) A bouquet of circles (or a rose) is a graph consisting of only one vertex and a nonempty set of edges. The edges (circles) in the bouquet are sometimes called petals. We will mostly be concerned with bouquets with finitely many petals. (Note that, in Illustration 4.29, when there are infinitely many petals, one should not assume from the figure that the topology is inherited from  3; we will come back to this point again in 13.2.)



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Illustration 4.29  A bouquet of circles.

☐

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4.4  Manifolds and CW-Complexes   ◾    97  

Observe that it follows from Example 5 that a torus is a CW-complex of dimension 2 obtained from a bouquet of two circles by attaching one 2-cell. We will eventually confirm that every 2-manifold is a CW-complex of dimension 2. Exercises 1. (a) Find a space X such that the conditions (i) and (iii) in the definition of the n-manifold are satisfied, and the condition (ii) fails. (b) Find a space X such that the conditions (ii) and (iii) in the definition of the n-manifold are satisfied, and the condition (i) fails. 2. Identify the edges of the filled polygon according to the labels and arrows shown in Illustration 4.30, as explained in Examples 4 and 6; for example, the a-edges are identified along the obvious (linear) homeomorphism that follows the arrows. Describe (or visualize) the resulting 2-manifold. a c

b

c

a b

   Illustration 4.30

3. A Möbius band is the quotient space obtained from a filled rectangle by identifying the two edges labeled a according to the arrows (top figure, Illustration 4.31). Start with two copies of a Möbius band and identify the boundaries according to the labels and arrows in the second row of figures in Illustration 4.31. Which 2-manifold is the resulting quotient space?

a

a

b a1

b

a1

a2

c

a2

c

Illustration 4.31

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98   ◾    Subspaces, Quotient Spaces, Manifolds, and CW-Complexes

4. (a) Define an equivalence relation ~ on the three-dimensional unit ball D 3 = {( x , y , z ) :  x 2 + y 2 + z 2 ≤ 1} by partitioning it into classes consisting of singletons or of sets of type {( x , y , z ),( x , y , − z )} whenever  x 2 + y 2 + z 2 = 1. Visualize 3 a homeomorphism between D and the three-sphere S 3 . [Hint: Slice D 3 with ~ planes and see what happens to these slices after the identification.] (b) Let f : D 3 → X and g : D 3 → Y be two homeomorphism (so that X and Y are two copies of D 3). Identify the points f ( x , y , z ) and g ( x , y , z ) for each ( x , y , z ) ∈S 2. What is the resulting quotient space? 5. Let O be a subset of  3. Denote the rotation around the x-axis through α radians (α ∈ ) by fα , and denote the rotation around the z-axis through γ radians ( γ ∈ ) by hγ . The configuration space C(O) of O with respect to the rotations fα and hγ (in that 2 order) is the quotient space  ~ of  2 , where ~ is defined as follows: (a, b) ~ (c , d ) if hb ( fa (O )) = hd ( fc (O )). Describe the configuration space C(O) if (a) O = {(0, 0, z ): − 1 ≤ z ≤ 1}.

1 1 (b) O = {(0, 0, z ):  −1 ≤ z ≤ 1} ∪ {(0, y ,1) : 0 ≤ y ≤ } ∪ {(0, y , −1) : 0 ≤ y ≤ } . (See Illus­t­ 2 2 ration 4.32.) z

y

x

   Illustration 4.32

6. (a) Let G be the group of symmetries of  2 consisting of all compositions of translations along integer multiples of the vectors u = (0,1) and v = (1,0). Partition  2 into sets of type { g ( x ): g ∈G}, x ∈ 2 (these are the orbits of the elements of  2 under the action of G), and let ~ be the corresponding equivalence relation. Describe the quotient space  2 . ~ (b) Same question, but G generated by a translation along u and the glide reflection that is the composition of the translation along v followed by the reflection with respect to the x-axis.

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7. Show that n is a CW-complex of dimension n.

8. Let X be the CW-complex obtained by attaching the closed disk via the mapping that sends the boundary of the disk D to the only edge of the graph Γ as indicated by the labels (see Illustration 4.33). Let Y be the space obtained from two open cones sharing a slant height by identifying the two base circles labeled c, as indicated in Illustration 4.34. Explain why X is homeomorphic to Y. Y

a Γ a

a



a

D

a

  

c

c

   

Illustration 4.33

Illustration 4.34

9. Find a CW complex Y of dimension 2 such that Y is homeomorphic to  2 and such that no subcomplex of Y is homeomorphic to . 10. Show that every finite CW-complex is Hausdorff. 11. Show that the k-skeleton X k of a finite CW-complex X is closed in X. 2

∞ 1 1 12. Consider the subspace X = ∪ Ck of  2 , where Ck = {( x , y ) ∈ :  x −  + y 2 = 2 }, k=1   k k k ∈ (see Illustration 4.35). Why is X not a CW-complex?

1.0

0.5

0.5

1.0

1.5

2.0

–0.5

–1.0

   Illustration 4.35

13. Show that the Klein bottle is a CW-complex of dimension 2.

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Chapter

5

Products of Spaces

T

here are a couple of straightforward and reasonable ways to extend the notion of products of metric spaces (Section 2.2) in the setting of topological spaces. The pre­ ference for one of these two choices will be clearer after the next chapter.

5.1  Finite Products of Spaces A brief historical note: Fréchet was the first to study finite products of abstract (topological) spaces (1910).

The product space of two spaces X and Y is the set X × Y, together with the topology gener­ ated by all of the sets of type U × V , where U is open in X and V is open in Y. This topology over X × Y is called the product topology. It is easy to verify that the subsets of the set X × Y of type U × V , where U ⊂ X and open V ⊂ Y , indeed constitute a basis for a topology over X × Y . That particular basis will be open

called the standard basis for X × Y We will denote both the product space and the product set by X × Y, and, unless otherwise stated, we will always assume that X × Y denotes the product space. This definition is natural in the sense of the following proposition. Proposition 1. The product topology over X × Y is induced by the projections p1 : X × Y → X and p2 : X × Y → Y . Proof. Denote the product topology by τ1 , and denote the topology induced by { p1 , p2 } by τ 2 . By Proposition 6 in 3.5, τ 2 is generated by the subbasis

{

}{

}

S = p1−1 (U ) : U ⊂ X ∪ p2−1 (V ) : V ⊂ Y . Since p1−1 (U ) = U × Y and p2−1 (V ) = X × V , we open

open

101

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have that p1−1 (U ) ∩ p2−1 (V ) = U × V , and so τ1 ⊂ τ 2 . On the other hand it is clear that S ⊂ τ1, and so τ 2 ⊂ τ1. We will now provide an alternative interpretation of product spaces, more suitable for visualization. Start with two spaces, X and Y. For every x ∈ X choose a copy Yx of Y, such that if x1 ≠ x 2 , then Yx1 ∩ Yx 2 = ∅, and let φ x : Y → Yx be the associated homeomorphism. Consider the set ∪ Yx . You may think of it as being obtained by replacing each point x of x ∈X

X with a copy Yx of the space Y. Introduce a topology over that set as follows: it is generated by the basis consisting of all subsets of the set ∪ Yx of type ∪ Vx , where U is an open x ∈X

x ∈U

subset of Y, and where for every x ∈U , Vx = φ x (V ) for some fixed open subset V of Y. It is fairly easy to show that these types of sets indeed constitute a basis. We get a topological space over ∪ Yx that we temporarily denote by P(X,Y). That temporary notation will be all x ∈X but extinguished by the following proposition. Proposition 2. The product space X × Y and the space P(X,Y) are homeomorphic. Proof. Define f : X × Y → P ( X , Y ) by f ( x , y ) = φ x ( y ). It is easy to see that f is a bijec­

( )

tion. The continuity of f follows from the observation that f −1 ∪ Vx = U × V , and from x ∈U

Proposition  3(c) of Section 3.5. Similarly, since ∪ Vx = f (U × V ), the same proposition implies that f −1 is continuous, and so f is open.

x ∈U

Corollary 3. For each fixed y ∈Y , the subspace {φ x ( y ) :  x ∈ X } of P(X, Y) is homeomor­ phic to X. Proof. The restriction of f to X × { y} is a homeomorphism onto {φ x ( y ) :  x ∈ X } . On the other hand it is obvious that X ≅ X × { y}. In the view of Proposition 2 we may regard P(X, Y) as being the same as X × Y . So, in order to visualize P(X, Y) or X × Y we need to replace each point of X with a copy Yx of the space Y. Corollary 3 then asserts that for each y ∈Y the sets {φ x ( y ) :  x ∈ X } could be viewed as a family of pairwise disjoint copies of X. Example 1 Let X and Y both be homeomorphic copies of the circle S1 = {( x , y ) :  x 2 + y 2 = 1} (viewed as a subspace of  2). Then X × Y, or P(X,Y), is depicted in Illustrations 5.1 and 5.2.

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5.1  Finite Products of Spaces   ◾    103   Illustration 5.1  This is an illustration of the

space X × Y = S1 × S1. It is, of course, incom­ plete, mainly because only finitely many points of X have been replaced by copies of the space  Y. A better illustration is given in Illustration 5.2. However, in more complicated cases, when X  ×  Y is not embeddable in our visualizing space  3, the complete visualiza­ tion as in Illustration 5.2 will often obscure the view of the main features of the space, and so the restricted visualization as in Illustration 5.1 will sometimes be preferable.

Illustration 5.2  This is a better view of S1 × S1.

U Vx

The  spaces Yx are assembled continuously within the product space X × Y . The topo­ logy is determined by the standard basis, and a basis element U × V , where Vx (a copy of V in Yx ), is shown in this figure. We get a standard depiction of the space S1 × S1 . It is fairly faithful, since S1 × S1 is embeddable in  3 . Note, however, that even  3 is not good enough to properly depict the symmetry that exists in X × Y between X and Y.



☐

Example 2 Suppose now that Y is the circle S1 = {( x , y ) :  x 2 + y 2 = 1} as a subspace of  2 , and suppose X is the same circle, but with the discrete topology. Then the structure of the space X × Y is entirely different from the space X × Y in Example 1. See Illustration 5.3 and its caption for further explanation.

Illustration 5.3  The topology of this space is

U Vx



K12146_C005.indd 103

different from the space shown in illustration. One member of the standard basis of X × Y is U × V, where Vx is a copy of an open arc V in some Yx , while U is just the point x. There is really nothing that holds the circles Yx together in this space; X × Y is now just a bunch of circles, one per each point in X. See also Exercise 3 below. ☐

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The product space T 2 = S1 × S1 , where both copies of S1 are with the usual topology (inher­ ited from  2), is homeomorphic to the torus (as defined in Section 4.41). According to the following proposition, higher dimensional manifolds can easily be generated through products of spaces. Proposition 4. Let X be an n-manifold, and let Y be an m-manifold. Then X × Y is an n + m -manifold. Proof. Take a point ( x , y ) ∈ X × Y . Since X is an n-manifold, there is an open neighborhood U of x such that U ≅ n . Similarly, there is an open neighborhood V of y, such that V ≅ m. Then U × V ≅ n × m (Exercise 1), and n × m ≅ n +m (Exercise 13, Section 2.2). Example 3 By Proposition 4, the space T 2 × S 2 , where T 2 is a torus and S 2 is the unit sphere is a 4-manifold. It is embeddable (and visualizable) in  6 (Exercise 5). Nevertheless, Proposition 2 allows us to depict the main features of that space in Illustration 5.4; the caption explains some of the limitations of that illustration.

Illustration 5.4  All of the points of the torus are replaced by disjoint copies of the sphere—for obvious, technical reasons, our illustration only shows finitely many spheres. But that is just one of the limitations of our rendering of T 2 × S 2 . We also show three, semi-transparent, concentric tori in this illustration; each one of the tori is made of points on the spheres, one point per a copy of the sphere S 2 , and each of these points correspond to a fixed point of the original space S 2 . For example, the north poles of all spheres together make one torus.

According to Corollary 3, a properly depicted T 2 × S 2 should show infinitely many copies of T 2 , too. It is difficult to show more than one, since the largest will obscure the view of the smaller tori. We use three transparent tori in Illustration 5.4 to stress 1

As in the case of S 2 or  2, the superscript in T 2 is to indicate that we deal with a two-dimensional manifold.

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5.1  Finite Products of Spaces   ◾    105  

that point. We are reaching the limits of visual depiction of manifolds as a way of ☐ gaining intuitive understanding. Example 4 In Illustration 5.5 we depict T 2 × K , where T 2 is a torus and K is a Klein bottle. In Illustration 5.6 we zoom in and slice to show the texture of the space, consisting of small copies of a Klein bottle, one per each point of the torus.

Illustration 5.5  T  × K.

 Illustration 5.6  We zoom in and cut to see

better.

This is a standard visualization procedure used to model the space we sometimes call the universe. For example, the Calabi–Yau model in string theory is the ten-dimensional manifold  4 × CY , where  4 accounts for  3 (the three “spatial dimensions”) and for linear time as the fourth dimension, and where each point of  4 is replaced by a small (and relatively complicated) 6-manifold CY (Celabi–Yau space) according to ☐ our procedure P. We enumerate a few basic properties of products of spaces. Proposition 5. The mapping f : X × Y → Y × X defined by f ( x , y ) = ( y , x ) is a home­ omorphism. Proof. It is clear that f is a bijection. Since f −1 (U × V ) = V × U for every set U × V in the standard basis for X × Y ,  f  is continuous. It follows by symmetry that f is open. Example 5 In the case of relatively simple topological spaces we can literally see that X × Y ≅ Y × X . For example, in Illustrations 5.7 and 5.8 we show what we get when the space X is a copy of the circle S1 and the space Y is the interval [0,1] (depicted as a vertical line segment).

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106   ◾    Products of Spaces

Illustration 5.7  In the space S1 × [0,1] we replace 1

each point of S by a (vertical) line segment …

Illustration 5.8  … while in the space [0,1] × S1

we replace each point of [0,1] by a circle.

We can see in Illustrations 5.7 and 5.8 that X × Y is homeomorphic to Y × X . Both illustrations, as we have indicated several times by now, are incomplete skeletons ☐ of the full objects. Both X × Y and Y × X are cylinders. Example 6 In the case of subspaces of n , n ≥ 4, visually justifying that X × Y ≅ Y × X is less convincing. For example, the spaces T 2 × S 2 and S 2 × T 2 , as depicted in Illustrations 5.4 and 5.9, are homeomorphic.

Illustration 5.9  S 2 × T 2 .

The space T 2 × S 2 is shown in Illustration 5.4: every point of the torus is replaced by a (small) sphere. In Illustration 5.9, we show the space S 2 × T 2 where each point on the sphere is replaced by a small torus. The difference between the two spaces is geo­ metrical rather than topological: in Illustration 5.4 the tori are large and the spheres ☐ are small, while in Illustration 5.9 it is the other way around. The product space of the spaces X, Y, and Z (in that order) is defined to be ( X × Y ) × Z . The product space of the spaces Xi , i = 1,2,… , n , (in that order) is (…(( X1 × X 2 ) × X 3 ) × … × Xn ).

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Proposition 6. The collection of sets of type (U × V ) × W = {(( x , y ), z ): x ∈U , y ∈V , z ∈W }, where U, V, and W range through the open sets in X, Y, and Z, respectively, is a basis for the product space ( X × Y ) × Z . Proof. If B1 and B 2 are bases for the spaces X and Y, respectively, then the collection of sets {U × V  : U ∈B1 , V ∈B 2 } is a basis for the product space X × Y (Exercise 6). That, together with the fact that the collection of open sets U × V = {( x , y ): x ∈U , y ∈V }, U is open in X and V is open in Y, is a basis for X × Y , implies the statement in the proposition. The following simple proposition will allow us to further simplify the notation. Proposition 7. For any three spaces X, Y, and Z, the mapping h : (( x , y ), z )  ( x ,( y , z )) defines a homeomorphism from ( X × Y ) × Z onto X × (Y × Z ). Proof. The mapping h is a bijection. The argument given in Proposition 6 needs only a minor modification in order to conclude that the collection U × (V × W ) = {( x , ( y , z )): x ∈U , y ∈V , z ∈W }, U, V, and W range through the open sets in X, Y, and Z, respectively, is a basis for X × (Y × Z ). Since h −1 (U × (V × W )) = ((U × V ) × W ), it follows that h is continuous. By symmetry, h is open. Proposition 7 (together with a straightforward induction) allows us to disregard the paren­ theses. Hence we will denote the product space of X, Y, and Z by X × Y × Z, and wen will denote the product space of Xi , i = 1,2,…n, by X1 × X 2 × … × Xn , or, more succinctly, by ∏ Xi . Example 7

i =1

The two images in Illustration 5.10 depict the space (S 2 × T 2 ) × S 2 , where, as before, S 2 is the unit sphere and T 2 is a torus.

Illustration 5.10  The space (S 2 × T 2 ) × S 2 is obtained by first replacing each point in S 2 by a

copy of T, and then replacing each point in each copy of T 2 by a copy of S 2 . The right-handside illustration is a magnification of a part of the left-hand-side picture: we see the small spheres distributed along the corresponding tori. We repeat again that there are uncountably many small tori and small spheres within the space ( X × Y ) × X and that we can only depict ☐ finitely many of them. The space ( X × Y ) × X is a 6-manifold.

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Almost all of the properties of topological spaces that we have introduced so far are pre­ served in (finite) product spaces. n

Proposition 8. (a) If Xi , i = 1,2,… , n , are first countable, then so is ∏ Xi . i =1

n

(b) If Xi , i = 1,2,… , n , are second countable, then so is ∏ Xi . n

i =1

(c) If Xi , i = 1,2,… , n , are separable, then so is ∏ Xi . i =1

n

(d) If Xi , i = 1,2,… , n , are metrizable, then so is ∏ Xi . i =1

n

Proof. (a) Take a point x = ( x1 , x 2 ,… , xn ) in ∏ Xi . There is a countable local basis Bi for xi , i =1

n

i = 1, 2, …, n. Consider the collection B x of sets of type ∏ Bi , where Bi ∈Bi . Proposition 4 i =1

in 1.2 implies that B x is countable. We show it is a local basis at x, thus justifying (a). Take n

n

i =1

i =1

an open subset U of ∏ Xi . Then there is a standard basis set ∏ U i containing x and such n

that ∏ U i ⊂ U . Since each U i is an open set containing xi and since Bi is a local basis at xi , n

i =1

n

it follows that there is an element Bi of Bi such that xi ∈ Bi ⊂ U i . But then x ∈∏ Bi ⊂ ∏ U i , n i =1 i =1 and so x ∈∏ Bi ⊂ U . i =1

We leave the proofs of (b) and (c) as exercises. (d) Suppose Xi , i = 1,2,… , n are metrizable. So, there is a metric di for Xi such that the metric space generated by di is the same as the original topological space. Consider the pro­ n duct metric over ∏ Xi (as defined in 2.2: for every x = ( x1 , x 2 ,… , xn ) and y = ( y1 , y 2 ,… , yn ) n i =1 in ∏ Xi , we have d( x , y ) = d1 ( x1 , y1 )2 + d2 ( x 2 , y 2 )2 +  + dn ( xn , yn )2 ). Then it is straight­ i =1 forward to check that the topology generated by that metric is the same as the product n topology over ∏ Xi (Exercise 10). i =1

Exercises 1. Show that if X1 ≅ X 2 and Y1 ≅ Y2 , then X1 × Y1 ≅ X 2 × Y2 . 2. T = {{n, n + 1, n + 2, …} : n ∈+ } ∪ {∅} is a topology over + (Exercise 4, Section 3.1), and Ω = {{…, −n − 2, −n − 1, −n, n, n + 1, n + 2, …} : n ∈+ } ∪ {∅} is a topology over \{0}. Show that (  \ {0}, Ω ) is homeomorphic to the product space X = + × {0,1}, where + is equipped with T , and {0,1} is with the indiscrete topology. Show that  equipped with the topology Ω1 = {{…, −n − 2, −n − 1, −n, n, n + 1, n + 2,…} : n ∈} ∪ {∅} is not homeomorphic to X. 3. Show that if X is a discrete space, then the product space X × Y is homeomorphic to the discrete sum ⊕ Yx , where Yx , x ∈ X , are pairwise disjoint copies of the space Y. x ∈X

4. (a) Prove that projections are open. (b) Show that projections are not always closed. 5. Let A be a subspace of X and let B be a subspace of Y. Show that A × B with the pro­ duct space topology is homeomorphic to A × B considered as a subspace of X × Y.

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6. Let B1 and B 2 be bases for the spaces X and Y, respectively. Show that the collection of sets {U × V  : U ∈B1 , V ∈B 2 } is a basis for the product space X × Y .

7. Visualize the space X × Y if (a) X is the number 8 (viewed as a subspace of  2 ) and Y is the (unit) circle S1. (b) The spaces X and Y are copies of the number 8 (viewed as a subspace of  2 ).

8. Visualize the space X × Y × X where (a) X is the unit interval [0,1] and Y is the (unit) circle S1. (b) X is the circle S1 and Y is a torus.

n

9. (a)  Show that if each Xi , i = 1,2,… , n , is separable, then so is ∏ Xi . i =1

n

(b) Show that if each Xi , i = 1,2,… , n , is second countable, then so is ∏ Xi . i =1

10. Show that if Xi , i = 1,2,… , n , are metric spaces, then the product space and the prod­ n

uct metric space over ∏ Xi are homeomorphic. i =1

11. Let f : X → Y be continuous, and let Z be the subspace of X × Y defined as follows: Z = {( x , y ) ∈ X × Y  :  y = f ( x )} . (a) Show that Z is homeomorphic to X. (b) Show that if Y is Hausdorff, then Z is closed in X × Y . 12. Let f : Y → X and g : Z → X be homeomorphisms, let p1 : Y × Z → Y , p2 : Y × Z → Z be the projections, and denote W = {( y , z ) ∈Y × Z : f ( y ) = g ( z )} . Show that the restrictions pi

W

, i = 1, 2 are both homeomorphisms onto Y and Z respectively. n

n

i =1

i =1

13. Let Ai be a subset of the space Xi , i = 1, 2, …, n (so that ∏   Ai is a subset of   ∏ Xi ). n

n

i =1

i =1

Show that ∏ Ai   =   ∏ Ai . (Hint: Do it for n = 2 and then use induction.) n

14. Let Xi , i = 1, 2, …, n, Y be spaces, and let f : Y → ∏ Xi be a mapping. Show that f is i =1

continuous if and only if the compositions p j  f are continuous for every projection n map p j : ∏ Xi → X j j = 1, 2, …, n. i =1

15. Show that if Y is Hausdorff and f : X → Y is continuous, then {( x1 , x 2 ) ∈ X × X : f ( x1 ) = f ( x 2 )} is a closed subset of X × X . 16. Show that if Z is Hausdorff and f : X → Z , g : Y → Z are continuous, then {( x , y ) ∈ X × Y : f ( x ) = g ( y )} is a closed subset of X × Y . 17. Describe the 3-manifold introduced in Example 7, Section 4.4, as a product of known topological spaces. 18. Starting with the space X = {( x , y , z ) ∈ 3 : 1 ≤ x 2 + y 2 + z 2 ≤ 4} identify the bound­ ing spheres along the mapping f that sends a point (x, y, z), x 2 + y 2 + z 2 = 1 , to the

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110   ◾    Products of Spaces

point (2x, 2y, 2z). We get the quotient space X f . Explain why X f is a 3-manifold and describe it as a product of known spaces. 19. (a) Find a subset U of  2 such that the intersection of U with any horizontal or vertical line is open in that line (considered as a subspace of  2 ), yet U is not open in  2 . (b) Show that the usual topology over X × Y is, in general, strictly weaker than the topology induced by all coordinate mappings c1z0 : X → X × Y , c2z0 : Y → X × Y , where z 0 = ( x 0 , y0 ) ranges through X × Y , and where we define c1z0 ( x ) = ( x , y0 ), and c2z0 ( y ) = ( x 0 , y ) .

5.2 Infinite Products of Spaces A brief historical note: Infinite products of spaces were first introduced in 1930 by Andrey Tychonoff (also spelled Tikhonov; 1906–1993). The product space topology is sometimes called Tychonoff topology.

In this section we introduce a topology over the set-product ∏ Xi , where each Xi is a (noni∈I empty) topological space, and I is any non-empty set, thus generalizing the finite case. ⋅

Recall that, formally, the elements of the set ∏ Xi are mappings f : I → ∪ Xi , such that i∈I i∈I f (i ) ∈ Xi for every i ∈I (the dot over the union symbol indicates that the sets Xi , i ∈I are assumed to be pair-wise disjoint). We will more often use a simpler notation: instead of ⋅

f : I → ∪ Xi , such that f (i ) ∈ Xi for every i ∈I , we will use x = ( xi )i ∈I , where xi = f (i ) . In i∈I case I is countable, instead of x = ( xi )i ∈I we may write x = ( x1 , x 2 ,…, xn ,…), or, occasion­ ally, x = ( xn ) The elements xi are called the coordinates or components of x. As noted earlier, we accept the validity of the Axiom of Choice, and so we take that ∏ Xi ≠ ∅ when all Xi ≠ ∅ and I ≠ ∅. i ∈I

The projections p j : ∏ Xi → X j are defined by p j (( xi )i ∈I ) = x j . The product topology i ∈I

over the set ∏ Xi is induced by the projections p j : ∏ Xi → X j , j ∈I ; this means that it i ∈I

i ∈I

is the weakest topology making the projections continuous. The product space will be denoted by the same symbol ∏ Xi . When the index set I is the finite set {1, 2, …, n}, we n

i ∈I

will use the notation ∏ Xi , or X1 × X 2 ×  × Xn , as introduced in the preceding section. ∞

i =1

When I is countably infinite, then we will denote the product space by ∏ Xi . When all n

i =1

Xi , i = 1, 2, …, n are copies of a single space X, then instead of ∏ Xi we will more often i =1 write X n . Similarly, if all of the spaces Xi , i = 1, 2, …, n, …, are copies of a single space ∞ X, then instead of ∏ Xi we will often write X ∞ . The countably infinite interpretation i =1

of the symbol ∞ in this context will be clear from the setup.

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5.2 Infinite Products of Spaces   ◾    111  

The following theorem describes the product topology more explicitly. Theorem 1. The product topology over ∏ Xi is generated by all sets of type ∏ U i where for i ∈I i ∈I every i ∈I , U i is an open subset of Xi , and where U i is Xi for all but finitely many i ∈I . (In other words, the collection B of all sets of type ∏ U i , U i is open in Xi , and for all i ∈I but finitely many i ∈I , U i = Xi , is a basis for the product topology.) Proof. Showing that B is a basis for some topology τ over ∏ Xi is straightforward i ∈I (Exercise 18 in Section 3.3). It is evident that the projections are continuous with respect to the topology τ , so that the product topology is a subset of τ. It remains to show that the elements of B are open in the product topology over ∏ Xi . Since, by assumption, each i ∈I

projection p j is continuous, for every open subset V of X j , the set p −j 1 (V ) is open in the product space ∏ Xi . Since p −j 1 (V ) = ∏ U i where U j = V and where U i = Xi for i ≠ j , and i ∈I

i ∈I

since all finite intersections of sets of such type constitute the collection B , it follows that the elements of B are open in ∏ Xi . i ∈I

As in the case of finite product spaces, the basis B will be called the standard basis for the product space ∏ Xi . i ∈I

Example 1 Suppose for each i ∈ , Xi is a copy of the discrete space over the set {0, 2}. The set ∏ Xi has the same cardinality as the set  . The nonempty open sets in the product

i ∈

space ∏ Xi are unions of sets of type ∏ U i where in all but finitely many cases U i is i ∈

i ∈

Xi and in the remaining cases U i is a singleton. We will see this space again in this section.

☐

Note that the collection of sets of type ∏ U i , U i is open in Xi , is also a basis for a topol­ i ∈I ogy over the set ∏ Xi : if I is infinite, then this last topology, called the box topology, is dif­ i ∈I

ferent from the product topology. Even though one may argue that this generating basis is more natural in some way than the standard basis, the box topology fails to satisfy many properties that the product topology fulfills. The main culprit for the shortcom­ ings of the box topology is the overabundance of open sets. We will say more about the box topology in the next section. In the next example we will introduce the Hilbert metric space H. We will prove in Section 7.6 that (the countable) product  ∞ is homeomorphic to a subspace of H. This implies that  ∞ is metrizable. In fact, it is true that countable products of metric spaces are always metrizable (we will not prove that). However, as we will show in the Example 3, this fails in case of uncountable products of metric spaces.

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112   ◾    Products of Spaces

Example 2: The Hilbert Space Let H be the set of all ( xn ) ∈ ∞ for which the sequence

(

x12 + x 22 +  + xn2

)

∞

n=1

con­

verges in . We will denote the point of con­vergence of a convergent sequence

(

x12 + x 22 +  + xn2

)

∞

n=1

∞

Σ xi2 .

by

i =1

If {xn } and { yn } are in H, then so is {xn − yn } . In order to justify that claim, we need to

( y + y +  + y ) converge, then so does ( (x − y ) + (x − y ) +  + (x − y ) ) . The latter sequence converges if and (

show that if 1

x12 + x 22 +  + xn2

2

1

2

)

n=1

2

2

∞

n

(

2 1

and n

2

2 2

2 n

∞

n=1 2 2

only if the sequence ( x1 − y1 ) + ( x 2 − y ) + … + ( xn − yn )2 2

∞

n=1

)

∞

converges, and since

n =1

the last sequence is an increasing sequence of positive numbers, it suffices to show that (( x1 − y1 )2 + ( x 2 − y 2 )2 +  + ( xn − yn )2 )n∞=1 is bounded from above. We compute: n

n

i =1

i =1 n

n

n

n

n

n

n

i =1

i =1

i =1

i =1

i =1

n

n

i =1

i =1 ∞

Σ ( xi − yi )2 = Σ xi2 − 2 Σ xi yi + Σ yi2 = Σ xi2 − 2 Σ xi yi + Σ yi2 ≤ Σ xi2 + 2 Σ xi yi + Σ yi2 . i =1

The numbers Σ x i =1

2 i

n

∞

and Σ y are not larger than the numbers Σ x and Σ yi2 , i =1

2 i

i =1

2 i

respectively (where the latter two series converge since, by assumption, ∞

∞

i =1

2 i

Σ x and

i =1

Σ yi2 converge). By the Cauchy–Schwarz inequality, for every n we have that

i =1 n

n

Σ xi yi ≤ Σ xi2

i =1

n

Σ yi2 . Hence we have that

i =1

i =1

∞

∞

∞

i =1

i =1

i =1

n

n

n

∞

i =1

i =1

i =1

i =1

Σ xi2 + 2 Σ xi yi + Σ yi2 ≤ Σ xi2 +

2 Σ xi2 ⋅ Σ yi2 + Σ yi2 , and since the expression to the right defines a (finite) num­

ber, we have justified our claim.

∞

We can now define a metric d over H by d( x , y ) = Σ ( xi − yi )2 for every x = ( xn ) i =1 and y = ( yn ) in H. An argument similar to the one given for the Euclidean metric over  ∞ shows that this is indeed a metric. The metric space generated by this metric is the Hilbert space. Notice that the metric that defines the Hilbert space is a straightforward extension of the usual metric over n . The set H can thus be viewed as consisting of all points ☐ in  ∞ at a finite distance from the origin. Example 3 Let X be the discrete space over {0,1}. Then the product space ∏ X is not metrizable. i ∈

We will show this by exhibiting a subset A of ∏ X and a point x ∈ ∏ X such that i ∈ i ∈ x ∈ A and such that no sequence of elements in A converges to x. That is not possible in metric spaces (Exercise 20, Section 3.2).

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Let A be the subset of ∏ X consisting of the elements of the type ( xi )i ∈ where i ∈

x j = 0 for every j in a finite subset J ⊂  , and where xi = 1 for i ∈  \ J. Denote x = ( 0 )i ∈ (each coordinate is 0). Every element in the standard basis for ∏ X con­ i ∈

taining x must be of the type ∏ U i where finitely many U i are {0} while the rest of i ∈ them are equal to X. Such an open set ∏ U i will contain the elements of A that have i ∈ 0’s at the same coordinates where U i ’s are {0}. It follows that every open set contain­ ing x must contain points from A. So, x is an accumulation point for A, and thus x ∈A . Let s be a sequence of elements in A. We show that s does not converge to x. Each element of A has 0 in finitely many coordinates, and since every sequence has count­ ably many members, the cardinality {k ∈  :  x k = 0 for some member ( xi )i∈ of s} is countable (as a countable union of finite sets). So, there exists a number j ∈ such that x j = 1 for every member ( xi )i ∈ of the sequence s. Consider the following open neighborhood of x: ∏ U i , where U j = {0} and U i = X if i ≠ j . It does not i ∈

contain any member of s, and so s does not converge to x. It follows from this example that ∏  is not metrizable, too (Exercise 12). i ∈

☐

∞

In the case where the index set I is countably infinite, the product space ∏ Xi can also i =1

be described by means of the procedure P that we have utilized to introduce the finite products: we start with the space X1 , then we replace each of its points by disjoint copies of X 2 , then we replace each point in each copy of X 2 by disjoint copies of X 3 ; continue ∞ this procedure recursively. In this way a point ( x1 , x 2 ,..., xi ,...) ∈∏ Xi can be viewed as cor­ i =1 responding to a sequence of such steps. The topology is the same: a basic neighborhood of the point ( x1 , x 2 ,…, xi ,…) consists of a sequence of open sets, all but finitely many of them being the whole spaces Xi . The procedure P allows us again to provide crude visualizations of products of countably many spaces. Example 4 Suppose the spaces X 2n+1, n = 0, 1, 2,… are (disjoint) copies of a torus, and suppose the spaces X 2n, n = 1, 2,… are (disjoint) copies of a sphere. A very rudimentary illus­ ∞ ∞ tration of the space ∏ Xi is provided in Illustration 5.11. In order to get ∏ Xi we start i =1 i =1 with a torus X1 , then replace each of its points by a copy of a sphere (only 4 such points are shown), then we replace each point of each copy of the sphere by a copy of the torus (only 16 such points are shown) and iterate countably many times. In the picture we show only 5 steps of that iteration.

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114   ◾    Products of Spaces

Illustration 5.11  An oversimplified illustration of the space T 2 × S 2 × T 2 × S 2 × T 2 × S 2 × … . ☐

The product space depicted in Illustration 5.11 is (homeomorphic to) a fractal (Exercise 16). Example 5: The Cantor Set The Cantor set we are about to construct is a subset of the closed interval [0,1] (with the usual topology). We delete from [0,1] the open middle interval ( 13 , 23 ) to get C1 (the first step in the construction; see Illustration 5.12). Then we delete the middle intervals ( 19 , 92 ) and ( 79 , 89 ) from the remaining two intervals to get C2 . Iterate: We get Cn+1 from Cn by deleting the open middle thirds of the closed intervals whose union constitutes Cn . The Cantor set C is the set we get after performing this procedure ∞ infinitely many times; that is, C = ∩ Cn (with the subspace topology). n=1

0

0

1 9

2 9

1 3

2 3

1 3

2 3

1

7 9

8 9

1

C1

C2

C3

Illustration 5.12  The first three steps in the construction of the Cantor set C.

It appears at first glance that after removing so many intervals not much is left in C. However, C is still uncountable. Moreover, C is homeomorphic to the space ∏ Xi i ∈ given in Example 1 (the product of uncountably many two-element discrete spaces).

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5.2 Infinite Products of Spaces   ◾    115  

It is fairly simple to provide a justification for both of these two claims. Write the numbers in the interval [0,1] as decimals 0. x1 x 2 … xn … using ternary base (using digits 0,1, and 2; note that 1 = 0.222… ). The digits we use are the numerators in the sum x31 + x322 +  + 3xnn + . When we remove the first middle subinterval to get C1 we are removing the set of real numbers for which x1 = 1 (i.e., the numbers that can be writ­ ten as 0.1∗∗∗ in ternary base, where ∗∗∗ substitutes the rest of the decimal number), except the number 0.1. Since 0.1 = 0.0222… , this number allows decimal representa­ tion without using the digit 1. Similarly, the step from C1 to C2 removes all decimals for which x 2 = 1 , except the number 0.01 and 0.21. These two can also be represented without using the digit 1 as follows : 0.01 = 0.00222…, and 0.21 = 0.20222… . We see by induction that the numbers of the set C are precisely those that can be written without using the digit 1 in decimal notation in ternary basis. It is now clear that C is uncountable: the correspondence that associates to the number 0.  x1 x 2 … xn … in C the subset {n ∈+:  xn = 2} of the set of positive integers is a bijection from C onto the power set P (+ ), and, since by Proposition 3 in 1.2 P (+ ) is uncountable, so is C. There is now a very straightforward candidate for a homeomorphism f : C → ∏ Xi , where ∏ Xi is as in Example 1: writing the elements of C as ternary i ∈

i ∈

decimals as in the previous paragraph, we set f (0.  x1 x 2 … xn …) = ( xi )i∞=1 . We leave the justification that f is indeed a homeomorphism as an exercise (Exercise 13). We digress to note that the Cantor set gives rise to some exotic but important spaces. In Illustration 5.13, we show a Cantor interval, an interesting homeomorphic copy of the closed interval [0,1]. We will use this idea in the construction of the Alexander’s Horned Sphere (to be seen in Section 13.4). CS 1 1

32 16

Illustration 5.13  The Cantor interval. The dotted lines are only to indicate the stages of the construction of the Cantor set. In each stage of this recursive construction we shrink slightly so that no line in the Cantor interval is vertical. The vertical projection is then a homeomor­ phism onto the base interval. ☐

Let f : Y → ∏ Xi be a mapping. For every fixed j ∈I, the component (or coordinate) mapi ∈I

ping f j : Y → X j of f is defined by f j ( y ) = p j ( f ( y )), where p j : ∏ Xi → X j is the projection. We end this section with a useful property.

i ∈I

Proposition 2. A mapping f : Y → ∏ Xi is continuous if and only if the component map­ i ∈I

ping f j is continuous for every j ∈I .

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116   ◾    Products of Spaces

Proof. ⇒ If f is continuous, then, since each component f j is a composition of two continuous functions, it is also continuous. ⇐ Suppose each component f j is continuous. In order to show that f is continuous it suf­ fices to show that the inverse image of every member of the standard subbasis for ∏ Xi is i ∈I

open in Y (Exercise 12, Section 3.5). A member U of the standard subbasis of ∏ Xi is of i ∈I

the type ∏ U i , where U j0 is open in X j0 for some fixed j0 ∈ I , and U i = Xi otherwise. It i ∈I is easy to check that f −1 ∏U i  = f −j01(U j0 ). Since f j0 is continuous by assumption, the set i∈I  −1 f j0 (U j0 ) is open in Y, and thereby, so is f −1 ∏ U i  . i ∈I  Exercises 1. Show that each projection p j : ∏ Xi → X j is open. i ∈I

2. Show that for every element z 0 = ( xi )i ∈I the coordinate mapping c zj 0 : X j → ∏ Xi i ∈I defined by c zj 0 ( x ) = ( yi )i ∈I , where yi = xi for i ≠ j , and y j = x , is an embedding. 3. Show that it is not true that if each of the spaces Xi is separable (first countable, second countable), then the space ∏ Xi must be separable (first countable, second i ∈I countable, respectively). 4. (a) Let Y j be a closed subset of X j , for every j ∈ J . Show that ∏ Y j is a closed subset j ∈J of ∏ X j . j ∈J

(b) Show that if Xi , i ∈I , are topological spaces and Ai ⊂ Xi are nonempty subsets, then ∏ Ai = ∏ Ai  , where the closure to the right is taken in ∏ Xi . i ∈I  i ∈I i ∈I (c) In the setting of part (b), is it necessarily true that int ∏ Ai  = ∏ (int Ai )? Justify i ∈I  i ∈I your answer. 5. Show that if I is infinite, and if for every i ∈I , Xi is homeomorphic to a fixed space X, then for any fixed j ∈I , the spaces ∏ Xi and ∏ Xi are homeomorphic. i ∈I

i ∈I i≠ j

6. (a) Show that if Xi is homeomorphic to Yi , i ∈I , then ∏ Xi is homeomorphic to i ∈I ∏ Yi . i ∈I

(b) Show that if Xi is a subspace of Yi , i ∈I , then ∏ Xi is a subspace of ∏ Yi .

i ∈I

i ∈I

7. Show that if I is infinite then the set ∏ U i , where each U i is a nonempty proper open i ∈I subset of Xi , is not open in ∏ Xi . i ∈I

8. Show that if each Xi , i ∈I , is Hausdorff, then so is ∏ Xi . i ∈I

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5.3  Box Topology   ◾    117  

9. (a)  Let X be any space over the set of natural numbers. Show that X ∞ is separable. ∞

(b) Show that if Xi , i = 1, 2, …, n , …, are spaces such that Xi =ℵ0 , then ∏ Xi is i =1 separable. 10. Show that if Di is dense in Xi , i ∈ I , then the set ∏ Di is dense in the space ∏ Xi . i ∈I

i ∈I

11. Let ( xi )i ∈I be a fixed element in ∏ Xi . For every finite subset T of I, denote C(T ) = ∏ Ai , i ∈I

i ∈I

where Ai = {xi } if i ∉T , and Ai = Xi if i ∈T . Show that the set Y = ∪ C(T ) is dense T ⊂ I

in ∏ Xi .

finite

i ∈I

12. Show that the product space ∏  is not metrizable. (Hint: Use Example 3.) i∈

13. Show that the mapping f (0.  x1 x 2  xn ) = ( xi )i∞=1 from the Cantor set C to product space ∏ Xi (with each Xi a copy of the discrete space over {0, 2}, as in Example 5) is i ∈I

indeed a homeomorphism. 14. Show that if X is second countable, then so is X ∞ . 15. (a) Show that the Cantor set C, considered as a subspace of , is a fractal (see Example 4, Section 2.2). (b) Visualize the product spaces C × C and C × C × C . (c) Show that if X is a fractal then the product space X n is a fractal. 16. Consider the space X = T 2 × S 2 × T 2 × S 2 × T 2 × S 2 × … , as depicted in Illustration 5.11, as a subspace of the metric space  ∞ (which you may assume is homeomorphic to the Hilbert metric space H). Show that X is (homeomorphic to) a fractal.

5.3  Box Topology A brief historical note: The box topology was first introduced in 1923 by Heinrich Tietze (1880–1964).

In this section we will briefly discuss the box topology mentioned in Section 5.2 (following Example 1 there). Given a set of spaces Xi , i ∈I , the box space is the space over the set product ∏ Xi i ∈I

generated by the basis consisting of all sets of type ∏ U i , where each U i is open in Xi . i ∈I

That these sets indeed make a basis for a topology is left as an exercise (Exercise 1). We Box

denote the box space by ∏ Xi . The basis we have used to define the box space will be Box

i ∈I

called the standard basis for the box space ∏ Xi . The corresponding topology is called the box topology over the set product ∏ Xi .

i ∈I

i ∈I

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118   ◾    Products of Spaces Box

If the index set I is finite, then the box space ∏ Xi coincides with the product space i ∈I

∏ Xi . Otherwise the box topology is strictly finer than the product topology. For example,

i ∈I

if each of the spaces Xi , i ∈+ is equipped with the discrete topology over {0,1}, then Box

{(0, 0, 0, …, 0, …)} is an open subset of ∏ Xi (it is an element of the standard basis), but it is i ∈I not open in the product space ∏ Xi . i ∈I

The main problem with the box space is that it has too many open sets. As a conse­ quence, some important properties which are preserved in the product space are in general lost in the box spaces. For example, the analogue of Proposition 2, Section 5.2 is no longer true (Exercise 2). Another example is given in the next proposition. ∞

Proposition 1. The box topology over the set ∏  is not first countable. i =1

∞

Proof. Take a point x = ( x1 , x 2 ,…, xn ,…) in ∏  and suppose B x = {B j :  j = 1,2,…} is a count­ i =1

able local basis at x. Then there is a countable local basis at x consisting of sets in the standard basis (with respect to the box topology). Hence, we may assume that the elements of B x are ∞

all from the standard basis of the box space, so that each of them is of the form B j = ∏ U j , i , i =1 the usual topology). Let Vi be a proper where each U j , i is nonempty and open in  (with ∞ open subset of U i ,i containing xi . Then the set ∏ Vi is an open neighborhood of x. Hence ∞

i =1

there is a member of B x that is a subset of ∏ Vi . However our construction prevents that from i =1

happening (see Illustration 5.14 and its caption for further explanation). So, B x could not be a local basis at x. U1,1

U1,2

U1,3

U1,n

U2,1

U2,2

U2,3

U2,n

U3,1

U3,2

U3,3

U3,n

Un,1

Un,2

Un,3

Un,n

Illustration 5.14  We use the standard diag­

onal argument: by shrinking the indicated open subsets of  (shown in frames in the array to the left) to proper subsets Vi , we get ∞ a new open neighborhood ∏ Vi of x such i =1

that no listed neighborhood of x is a subset ∞ of ∏ Vi . i =1

∞

Corollary 2. The box space over ∏  is not second countable and it is not metrizable. i =1

Proof. The first part follows from Proposition 1 here and Exercise 10 in 3.3. The second part is a consequence of Proposition 1 here and Exercise 13, Section 3.3. The main reason why the product topology is preferable to the box topology will be seen in Chapter 7: the former preserves compactness, the latter does not.

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5.3  Box Topology   ◾    119  

Exercises 1. For a collection of spaces Xi , i ∈I , show that the set of all sets of type ∏ U i , U i is i ∈I open in Xi , is a basis for a topology over the set ∏ Xi . i ∈I

Box

2. Find a mapping f :  → ∏  such that each coordinate function fi , i = 1, 2,… is i∈+

continuous, but f is not continuous. (Compare with Proposition 2, Section 5.2.) Box

Box

i ∈I

i ∈I

3. Show that ∏ Ai = ∏ Ai . Box

4. Show that if each Xi , i ∈I , is Hausdorff, then so is ∏ Xi . i ∈I

5. Consider the set X = ∏  of all functions   → ; let B be the set of all bounded i ∈

functions in X; so f ∈B if there are numbers u , v ∈, such that u ≤ f ( x ) ≤ v for every x ∈ . (a) Show that the sets of type ∏ ( f (i ) − ε , f (i ) + ε) , f ∈B ,  ε > 0, comprise a basis for i ∈

a topology over B. [Here ( f (i ) −  ε , f (i )  +   ε) denotes an interval; two basis sets are shown in Illustration 5.15.] g + ε1 g

g – ε1 f + ε2 f f – ε2

   Illustration 5.15  Two sets from our basis of bounded real valued functions. What is

their intersection? Careful!

(b) Define ρ : B × B →  by ρ( f , g ) = sup{ f (t ) − g (t ) : t ∈} . Show that ρ is a metric. (c) Show that the topology T generated by the basis in part (a) is the same as the topology of the metric space (B, ρ). (d) Show that the topology T is strictly coarser than topology over B considered as a subspace of X = ∏  equipped with the box topology. i∈

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Chapter

6

Connected Spaces and Path Connected Spaces

6.1  Connected Spaces: Definition and Basic Facts A brief historical note: The notion of connectedness of some subsets of  2 was introduced in 1893 by Camille Jordan (1838–1922). A systematic study of connected topological spaces started in 1914 with a book by Felix Hausdorff.

The definition of connectedness is fairly simple. A space X is disconnected if the set X is a union of two disjoint, nonempty, open subsets A and B of X. Otherwise we say that X is connected. The set {A, B} is called a separation of the space X (Illustration 6.1). X

A

B

Illustration 6.1  A space X is disconnected if it is a union of two disjoint, nonempty, open subsets A and B.

121

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122   ◾    Connected Spaces and Path Connected Spaces

Example 1 Suppose X = {a, b} is with the discrete topology. Then X is disconnected, since X = {a} ∪ {b}, where {a} and {b} are two disjoint, nonempty, open subsets of X. If the same set X is equipped with the indiscrete topology over the same set, then it is connected. The ☐ indiscrete space over any set is always connected. Example 2 The Euclidean space  is connected. We will show that  is not a union of two disjoint, nonempty subsets. Suppose otherwise:  = A ∪ B, where A and B are two disjoint, nonempty, open subsets of . Since A ≠ ∅, there is an element a in A. Since B ≠ ∅, there is an element b ∉ A such either a < b or b > a. These two cases are symmetric, so we consider one of them, say a < b. Look at the set S = {x ∈ A : x < b} : it is nonempty (since a ∈S ), and it is bounded from above by b. By the Least Upper Bound Property, there exists the least upper bound for S. This means that there is an element u ∈ such that x < u for every x ∈S , and such that u is the smallest element with that property. Where is this least upper bound u? Since  = A ∪ B, it must be either in A or in B. We will show that in both cases we get a contradiction and so  cannot be disconnected. A B a

u

b

Illustration 6.2  Where is u: in A or in B?

Suppose u ∈A. Since A is open, u must be an interior point. So, there is an interval (u − r , u + r ) such that (u − r , u + r ) ⊂ A. Since b is not in A, it is not in (u − r , u + r ). Since u ≤ b (u being the least upper bound), it follows that b must be larger than all r r of the elements in (u − r , u + r ). In particular, u + < b . But, that would put u + in 2 2 r S, and since u < u + , u is not an upper bound for S. Contradiction. 2 Suppose u ∈B. Because B is open, there is an interval (u − r , u + r ) such that (u − r , u + r )⊂ B. Virtually the same argument as in the preceding paragraph implies that r u − is a smaller upper bound for S, contradicting the choice of u. ☐ 2 Since the last example is important for us, we promote its conclusion to a proposition. Proposition 1. The Euclidean space  is connected. Suppose now that we have an order topology over a set X (Example 8, Section 3.1) and suppose the space X satisfies the Least Upper Bound Property with respect to that linear order on X. Would a slight modification of the proof of Proposition 1 be sufficient to show that X is connected? The word “slight” refers primarily to the use of the interval

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6.1  Connected Spaces: Definition and Basic Facts   ◾    123  

(u − r , u + r ) around u in  : in the setting of the ordered topology, it should be replaced with an open interval (c, d) containing the point u. We see a problem: in the above proof of Proposition 1 we have found an element in (u − r , u + r ) that is between u and u + r (the number u + 2r ). There need not be an element in an interval (x, y) in the linearly ordered space X, and any modification of the above proof to the more general setting must account for that difference. It is easy to see that this would be sufficient: if a linear order in X is such that for every x < y in X, there is an element z ∈ X such that x < z < y , then a minor modification of the above proof of Proposition 1 will work for ordered spaces (with the two extra conditions). We summarize this in the following definition and in Proposition 1ʹ. An ordered space X is a linear continuum if the linear order satisfies the following two properties. (i) (The Least Upper Bound Property) If A is a subset of X that is bounded from above, then it has the least upper bound in X. (ii) For every x , y ∈ X such that x < y , there is an element z ∈ X such that x < z < y . Proposition 1′. Every linear continuum is connected. Connected subsets of  are easy to classify. Proposition 2. A subset A of  is connected if and only if it is an interval. Proof. ⇒ Suppose A is connected. Suppose the largest lower bound x for A and the least upper bound y for A exist. If x = y, then A = [x , x ], a singleton. Suppose x < y. There is no element c strictly between x and y that is out of A, since otherwise { A ∩ (−∞, c ), A ∩ (c , ∞)} would be a separation of A. It follows that A is ([x , y]), where ([ and ]) mean “open or closed” to the left and right, respectively. If the largest lower bound x for A does not exist (but y exists) then the set A is the interval (−∞, y]) . Similarly, if the least upper bound for A does not exist, while x exists, then A is ([x , ∞). Otherwise A = (−∞, ∞), and we are back to Proposition 1. ⇐ Repeat the argument given in Proposition 1. Example 3 Recall that the Sorgenfrey (half-open) topology over  is generated by all halfopen intervals of type [a, b). This topology is a refinement of the Euclidean topology (Exercise 2 in 3.3), and the additional open sets cause its disconnectedness. For exa­mple,  = (−∞,1) ∪ [1, ∞), where the two intervals are open in the Sorgenfrey topo­logy. If  one is to visualize the Euclidean  as a smooth, unbounded rod,

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124   ◾    Connected Spaces and Path Connected Spaces

then the Sorgenfrey space  is the same rod with infinitely many cracks in it (see Illustration 6.3).

Illustration 6.3  In order to depict the structure of the (infinitely thin) Sorgenfrey line  we

fatten it up to the space  × S1 (that shares many properties related to connectedness with the Sorgenfrey line ). Roughly speaking, there are natural cracks in the fabric of the space  × S1 (left-hand-side picture), and we can split the space into two, disjoint subspaces without damaging the “connecting fiber” (the right-hand-side picture). What we mean by “splitting ☐ the space without damaging the fiber” is clarified by Proposition 3.

Proposition 3. Let A and B be open subspaces of a space X. {A, B} is a separation of the space X if and only if X = A ⊕ B. Proof. ⇒ Assume {A,B} is a separation for X. We prove that the topology of X is the same as the topology of the discrete sum A ⊕ B. If U is open in X, then U ∩ A is open in A and U ∩ B is open in B, so that U is open in A ⊕ B. Conversely, if U is open in A ⊕ B, then, by Proposition 2 in 4.3, U ∩ A is open in A and U ∩ B is open in B, and since A and B are open in X, it follows that (U ∩ A ) ∪ (U ∩ B ) = U is open in X. ⇐ Suppose X = A ⊕ B , where A and B are equipped with the subspace topology. Then both A and B are open in X , X = A ∪ B , and A and B are nonempty, so that { A, B} is a separation of X. The following is a simple criterion for connectedness. Theorem 4. A space X is connected if and only if the only open and closed sets (clopens) are X and ∅. Proof. ⇒ Suppose A is a proper nonempty subset of X that is both open and closed. Then Ac is also open (since A is closed), and it is nonempty and not equal to X (since A is such). So, X = A ∪ Ac and X is disconnected. ⇐ Suppose X is disconnected. So, X = A ∪ B for some nonempty, disjoint, open subsets ⋅ A and B. Since A and B are nonempty and disjoint, they are not X or ∅. Since X = A ∪ B,

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6.1  Connected Spaces: Definition and Basic Facts   ◾    125  

the sets A and B are mutual complements of each other and so they are both open and closed. A subset A of a space X is connected if it is connected as a subspace of X. So the attribute “connected” applies to subsets and subspaces in the same way. The following observation will be used without explicitly referring to it: if A ⊂ X ⊂ Y and Y is a space, then A is a connected subset of Y if and only if A is a connected subset of the space X considered as a subspace of Y. We will refer to the next theorem quite a few times later on. Theorem 5. If A is a connected subset of a space X, then so is A. Proof. Suppose A is disconnected, so that A = B ∪ C for some nonempty, disjoint, open subsets of A (considered as a subspace of X). It does not matter whether we consider A as a subspace of X or as a subspace of A (see Exercise 1, Section 4.1) and so we will consider A as a subspace of A. Suppose A ⊂ B. Then C ⊂ A \ A, and so C is disjoint from A. This is not possible since C is a non-empty open set and since the only points of A \ A are the accumulation points of A. So, A is not a subset of B. By symmetry, A is not a subset of C. Consequently, A ∩ B and A ∩ C are two nonempty and disjoint subsets of A. They are open in A (since B and C are open in A). So, A is disconnected. The following proposition allows us to build larger connected spaces by taking unions of connected subspaces. Proposition 6. (a)  Let Ai , i ∈I , be connected subsets of a space X, and suppose ∩ Ai ≠ ∅. Then ∪ Ai i∈I

i∈I

A4

is a connected subset of X (see Illustration 6.4).

(b)  Let A0 and Ai , i ∈I , be connected subsets of a space X and suppose A0 ∩ Ai ≠ ∅ for every i ∈I .

(

)

Then, A0 ∪ ∪ Ai is a connected subset of X. i∈I

A5

A1

A2 A3

∩ Ai

i∈I Proof.  (a)  Suppose ∪ Ai is disconnected. By i∈I Theorem 4, there is a nonempty, proper subset B of ∪ Ai such that B is a clopen in ∪ Ai . Since B ≠ ∅ , Illustration 6.4  The intersection of the conn­

i∈I

i∈I

ected sets A (shaded part) is not empty.

i there is some j ∈I , such that B ∩ A j ≠ ∅. The only clopens in Aj are ∅ and Aj ; hence, B ∩ A j = A j. So Aj ⊂ B.We proved that if B ∩ Ai ≠ ∅ then Ai ⊂ B . It follows that B = ∪ Ai for some proper subset J of I.

i∈J

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126   ◾    Connected Spaces and Path Connected Spaces

Since ∅ ≠ ∩ Ai ⊂ A j ⊂ B , it follows that Ai ∩ B ≠ ∅ i ∈I

A0

A1

for every i ∈ I . Hence Ai ⊂ B for every i ∈ I . Consequently ∪ Ai ⊂ B , and so ∪ Ai = B . We proved that the only noni ∈I

A2

i ∈I

empty clopen in ∪ Ai is ∪ Ai itself, thus establishing its i ∈I i ∈I connectedness. (b)  Illustration 6.5 depicts the setting of this part of the propo­sition. An argument that is very similar to the one provided in part (a) suffices. We leave the details as ◼ an exercise (Exercise 2).

A3 A4

A5 A6

Illustration 6.5

Example 4

(

∞

)

Consider the subspace X = ∪ {( x , 1 n) :  x ∈} ∪ {( x ,0) :  x ∈} ∪ {(0, y ) :  y ∈} of 2

n=1

 (Illustration 6.6). The vertical line and all horizontal lines are connected (being

copies of ). Since each of the horizontal lines intersects with the vertical line, it ☐ follows from Proposition 6(b) that this space is connected.

Illustration 6.6  This space is connected; the dots at the ends of the lines are there only to

indicate that the lines are unbounded.

Example 5 Suppose now that we delete the point (0,0) from the space X described in Example 4. The line X 0 = {( x ,0) :  x ∈} does not intersect the vertical line any more and so Proposition 6 does not apply. Is the set X \ {(0,0)} still connected? Perhaps, somewhat surprisingly, the answer is affirmative! The closure of X \ X 0 in X \ {(0,0)} is X \ {(0,0)}. This is so since all of the points in X 0 \ {(0,0)} are accumulation points in X \ {(0,0)} for the set X \ X 0. Since X\X0 is connected (Proposition 6(b)), it follows from Theorem 5 that X \ X 0 (closure taken in X \ {(0,0)}) is connected, and since X \ X 0 is X \ {(0,0)}, it follows that X \ {(0,0)} is connected. In fact, we can delete as many points from X0 as we wish; a similar argument applies and the remaining set will still be connected (Illustration 6.7).

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6.1  Connected Spaces: Definition and Basic Facts   ◾    127  



Illustration 6.7  We have deleted some points from the x-axis.

☐

Remark. Example 5 suggests that our definition of connectedness might not match the intuitive (dictionary) meaning of that word: even though the set X \ {(0,0)} appears to be visually disconnected (Illustration 6.7), there are internal links that make it connected. (A better match for the dictionary meaning of connectedness might be the path connectedness. We will introduce this later in this chapter.) A component C in a space X is a connected subset of X such that if C ⊂ D and D is connected, then C = D . Components are maximal (under inclusion) connected subsets. The component containing an element x in X is the union of all connected subsets of X containing x (Exercise 5). This union is connected by Proposition 6(a). Proposition 7. Components are closed. Proof. Suppose A is a component in X. Suppose A is not closed. Then A properly contains A and, by Theorem 5, it is connected. That contradicts the assumption that A is a component. Components need not be open (Exercise 8). Proposition 8. A space is connected if and only if it has only one component. Proof. Easy, and we leave it as an exercise. Observe that the class of all components of a space is a partition of the space. Exercises 1. Show that X is disconnected if and only if there are two nonempty open subsets A and B of X, such that X = A ∪ B and A ∩ B = ∅ = A ∩ B. 2. Prove Proposition 6(b).

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128   ◾    Connected Spaces and Path Connected Spaces

3. Let Ai , i = 1,2, … , be a collection of connected subsets of a space X such that ∞

Ai ∩ Ai+1 ≠ ∅ for every i = 1,2, …. Show that ∪ Ai is connected.

(

)

i =1

4. Show that the set { x ,sin 1 x  :  x ∈ + } ∪ {(0, y ) :  −1 ≤ y ≤ 1} is connected. 5. Show that for every x in a space X, the component of X containing x is the union of all connected subsets of X containing x. 6. A space is totally disconnected if its components are singletons. Show that  is a union of two totally disconnected subspaces.

7. Find three distinct connected subsets A, B, and C, of  2 such that ∂ A = ∂ B = ∂C .

8. (a) Show that if X has finitely many components, then each of them is both open and closed. (b) Show that, in general, components need not be open. 9. Describe the components of the space X. (a) X is the subspace of  consisting of all irrational numbers. (b) X is the Sorgenfrey line. 10. Show that  2up = {( x , y ) :  y ≥ 0} with the tangent disk topology is connected. 11. Find a subspace of  2 with uncountably many components, all of them uncountable. 12. Show that if there is a dense connected subset of a space X, then X is connected. 13. Show that every countable subset A of a metric space such that | A| > 1 is disconnected. 14. (a) Find a subspace X of  2 such that for every points u, v ∈ X , the set X \ {u, v} is connected, and there exist points x , y , z ∈ X, such that the set X \ {x , y , z} is disconnected. (b) Find a subspace X of   2 such that for every n ∈+ and every points x1 , x 2 ,…, xn ∈ X , the set X \ {x1 , x 2 ,…, xn } is connected, and there exist points y1 , y 2 ,…, yn , yn+1 ∈ X , such that X \ { y1 , y 2 ,…, yn , yn+1 } is disconnected. 15. Show that no closed interval in  is a union of m many pairwise disjoint closed sets, where 2 ≤ m ≤ℵ0.

6.2 Properties of Connected Spaces A brief historical note: The finite case of Theorem 5 was proven by David van Dantzig (1900–1959) in 1930.

We start with a property that is very easy to prove, but, nonetheless, is very useful.

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6.2 Properties of Connected Spaces   ◾    129  

Proposition 1. If X is connected and f : X → Y is continuous then f ( X ) is a connected subset of Y. Proof. Suppose f ( X ) is disconnected and A, B is a separation of f ( X ). Then, f −1 ( A), f −1 ( B) is a separation of X: these two sets are disjoint (since A and B are disjoint), their union is X, they are nonempty (since A and B are nonempty) and, finally, f −1 ( A) and f −1 ( B) are open since f is continuous and since A and B are open. Corollary 2. Connectedness is a topological property. Corollary 3. (The Intermediate Value Theorem) If f : [a, b] →  is a continuous function and x is between f (a) and f (b) , then there is a number c ∈ [a, b] such that f (c ) = x . Proof. Let x be between f (a) and f (b), and suppose that there is no c ∈ [a, b] such that f (c ) = x . Then {(−∞, x ) ∩ f ([a, b]),( x , ∞) ∩ f ([a, b])} is a separation of f ([a, b]) contradicting Proposition 1 and the fact that [a, b] is a connected subset of . We illustrate the usage of Proposition 1 with a couple of examples. Example 1 The subspace X = {( x , y ) :  x 2 + y 2 = 1} ∪ {( x , y ) :  x 2 + ( y − 2)2 = 1} of  2 (a bouquet of two circles) is not homeomorphic to the subspace S1 = {( x , y ) :  x 2 + y 2 = 1} of  2 . Suppose otherwise and let f : S1 → X be a homeomorphism. X S1

f

(0, 1)

P

  Illustration 6.8

So, there is a point P of S1 such that f ( P ) = (0,1). The space S1 \ {P} is connected (since it is homeomorphic to  via a stereographic projection), and the restriction of f to S1 \ {P} is a homeomorphism from S1 \ {P} onto X \ {(0,1)}. By Proposition 1, we must have that f (S1 \ {P}) = X \ {(0,1)} is also connected, which is ☐ clearly false. So, we have a contradiction. Example 2 The semi-open interval (0,1] is not homeomorphic to the open interval (0,1) (both as subspaces of ), since (0,1]\ {1} is connected while (0,1)\ { p} is disconnected for every ☐ p ∈(0,1).

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130   ◾    Connected Spaces and Path Connected Spaces

Example 3 There does not exist a continuous one-to-one mapping from the closed disk D2 onto the circle S1, and so D2 is not homeomorphic to a circle. Otherwise, if f : D 2 → S1 is continuous, one-to-one, and onto, then f D 2 \{ x , y } , where x and y are any two distinct points in D2, is also continuous, one-to-one, and onto S1 \ { f ( x ), f ( y )}. But this could not be ☐ true, since D 2 \ {x , y} is connected, while f (D 2 \ {x , y}) = S1 \ { f ( x ), f ( y )} is not. Continuous mappings can be used to give yet another criterion for connectedness. Proposition 4. A space X is connected if and only if there is no continuous, onto mapping f : X → {0,1},  where {0,1} is equipped with the discrete topology. Proof. ⇒ Suppose there is a continuous, onto mapping f : X → {0,1}. Then f −1 (0) is both open and closed (since {0} is both open and closed and since f is continuous), nonempty (since f is onto), and not equal to X (again because f is onto, so that some points are mapped to 1). It follows by Theorem 4, Section 6.1, that X is disconnected. ⇐ Suppose X is disconnected and A, B is a separation of X. Then the mapping 0  if x ∈ A f : X → {0,1} defined by f ( x ) =  is continuous and onto.  1  if x ∈B Connectedness is preserved in product spaces; this will require slightly more effort to justify. Theorem 5. Suppose Xi , i ∈I are connected spaces. Then ∏ Xi is also connected. i∈I

Proof. In order to (literally) see what happens we will first prove the case of the product of two spaces. Then we will consider arbitrarily many spaces. Suppose X and Y are connected and consider the product space X × Y . Obviously, X × Y = ∪ X × { y}. Each of the subspaces X × { y} is homeomorphic to X. It follows from y ∈Y

Corollary 2 that the spaces X × { y}, y ∈Y , are connected. By a symmetric argument, for any fixed x 0 , the space {x 0 } × Y is connected. Observe now that for every y ∈Y , we have ({x 0 } × Y ) ∩ ( X × { y}) = {( x 0 , y )} ≠ ∅ . So, by Proposition 6(b) in 6.1, the space X × Y = ∪ X × { y} is also connected. Illustration 6.9 summarizes the argument. y ∈Y

X×Y

X × {yk }

X × {yi} X × {yj}

{x0} × Y

Illustration 6.9

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6.2 Properties of Connected Spaces   ◾    131  

A straightforward induction shows that finite products of connected spaces are connected. Now, consider the product ∏ Xi of connected spaces, where the set I is infinite. Fix an elei∈I ment x = ( xi )i ∈I in ∏ Xi . For every finite subset T of I, denote C(T ) = ∏ Ai , where Ai = {xi } i∈I

i∈I

if i ∉T , and Ai = Xi if i ∈T . Then C(T) is homeomorphic to ∏ Xi , and, by the first part i ∈T

of the proof, it is connected. Since x ∈ ∩ C(T ) , it follows by Proposition 6(a) in 6.1 that T ⊂ I finite

Y = ∪ C(T ) is connected. The set Y is dense in ∏ Xi (Exercise 11, Section 5.2). It follows T ⊂ I

i∈I

finite

from Theorem 5 in 6.1 (or Exercise 12 in 6.1) that ∏ Xi is connected. i∈I

Example 4 An immediate consequence of this theorem and of Proposition 1 in 6.1 is that the product space ∏  for any index set I is connected. In particular, the Euclidean i∈I

spaces n and the Hilbert space are connected.

☐

The analogue of Theorem 5 for the box topology fails. Example 5: Box Products of Connected Spaces Need Not Be Connected Consider the set product ∏  and recall the formal definition of the set products: i ∈

every element of ∏  is a function  → . Some of these functions are bounded i ∈

(their graphs are between two horizontal lines), and some are unbounded. Let B be the subset of ∏  containing all bounded functions and let U be the subset of ∏  i ∈

i ∈

containing all unbounded functions. Obviously, B ∩ U = ∅ and B ∪ U = ∏  . We i∈

will now (literally) see that both B and U are open in the box topology over ∏  , i ∈

box

so that the box space ∏  is disconnected. i ∈

f

f

    

    Illustration 6.10  A bounded function.   Illustration 6.11  An unbounded function.

If f is a function in B, then ∏ ( f (i ) − ε, f (i ) + ε) is an open neighborhood of f in the box space box

i∈

∏  (because each of the intervals ( f (i ) − ε, f (i ) + ε) is open in ). Since f is bounded it

i ∈

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132   ◾    Connected Spaces and Path Connected Spaces

follows that every function in ∏ ( f (i ) − ε, f (i ) + ε) is bounded (see Illustration 6.12, left). i∈

So, every point of B is an interior point, and so B is open. Similar argument (see Illustration 6.12, right) shows that U is open.

f–ε

f–ε f

f

f+ε

f+ε

Illustration 6.12  Every function in the open neighborhood (in the box topology) of a bounded

(unbounded) function is a bounded (respectively, unbounded) function. The neighborhoods of the functions are shaded. (Warning: Do not misinterpret the above illustration. The elements in the shaded neighborhoods are the functions with graphs within the shaded neighborhood, ☐ not merely points in the shaded part of  2.)

Exercises 1. Show that in each case below X is not homeomorphic to Y. (a) X is the subspace of  made of all irrational numbers, Y is . (b) X is , Y is  2. (c) X is the letter X, Y is the letter Y (both considered as subspaces of  2). (d) X=

Y=

(e) X=

Y=

(f) X=

Y= Illustration 6.13

2. Let X be obtained from  ×[0,1] and  ×[2,3] by gluing the pairs of points (z, 1) and ( z ,2), z ∈. Show that X is not homeomorphic to a (closed or open) disk in  2 .

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6.3 Path Connected Spaces   ◾    133  

3. Let A be a component of X and let f : X → Y be continuous. Show through a counter­ example that f (A) need not be a component of the subspace f (X) of Y. 4. Let { Xi  : i ∈I } be a class of pairwise disjoint spaces such that ∏ Xi is a connected space

{

i∈I

and let ( xi ) ∈ ∏ Xi . The partition {{ xi  : i ∈I }} ∪ {x } :  x ∈∪ Xi ,  x ≠ xi for every i ∈I i∈I

i∈I

}

of ∪ Xi defines an equivalence relation ~ on ∪ Xi . Show that the quotient space i∈I

⊕ Xi

i∈I

~ is connected. 5. Let X and Y be spaces, and let A ⊂ X , B ⊂ Y . Show that A is a component in X and B is a component in Y if and only if A × B is a component in the product space X × Y . i∈I

6. Let X be connected and consider the subset D = {( x , x ) :  x ∈ X } of X × X. Show that D is connected.

7. Use properties of connectedness to show that there is no continuous mapping  →  that sends the rational numbers to irrational numbers and irrational numbers to rational numbers. (Another solution was given in Example 4, Section 3.5.)

8. Show that if A is a countable subset of  2 , then  2 \A is connected. 9. Show that if f :  2 →  is continuous and if both 1 and –1 are images under f of some points in  2 , then f −1 (0) is uncountable. 10. Show that if f :[0,1] → [0,1] is continuous and onto then f has a fixed point. 11. An involution f : X → X is a continuous map such that f  °  f = identity. (So, involutions are self-inverse homeomorphisms.) Show that if f :  →  is an involution, then f must have a fixed point. 12. Let A = {0} ∪ [1,2] ∪ {3} and B = [0,1] ∪ {2} ∪ {3} (as subspaces of  ). Show that A and B are homeomorphic but there is no homeomorphism from  onto  taking A onto B. 13. (a) Show that if f : S1 → S1 is antipode-preserving, then f is onto. (A mapping f : S1 → S1 is antipode preserving if f ( x ) = − f (− x ) for every x ∈S1.) (b) Show that if f : S1 → S1 is an embedding than it is a homeomorphism. 14. Let U be a subset of  2 with a nonempty interior. Show that there is no embedding f : U → S1.

6.3 Path Connected Spaces We start by introducing the main notions of this section; we will encounter them frequently, especially in Part 2 of this book. A path in a space X is a continuous mapping α :[0,1] → X . The image α([0,1]) will sometimes be called an arc in X. The point α(0) is called the initial point of the path α, and the

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134   ◾    Connected Spaces and Path Connected Spaces

point α(1) is the terminal point α. We also say that α starts at α(0) and ends at α(1), or, more succinctly, that α is a path from α(0) to α(1). The points α(0) and α(1) are jointly called the endpoints of the path α. A space X is path connected if for every two points in X there is a path starting at one of these two points and ending at the other. X α(0) α

0

α(1)

1

Illustration 6.14  A path in X from α(0) to α(1).

The main reason we choose the closed unit interval [0, 1] to be the domain of every path α is convenience. Occasionally we will encounter continuous mappings λ :[a, b] → X . The path α λ :[0, 1] → X associated to the mapping λ is defined by α λ (t ) = λ ((b − a)t + a ) . From now on, and unless otherwise stated, we will use the letter I to designate the unit interval [0,1] (as a subspace of ). We will repeat this convention several times as we go. Since all closed intervals are homeomorphic, the existence of a path in a space X starting at x and ending at y is equivalent to the existence of a continuous map g from any closed interval [a, b], a < b into X such that g (a) = x and g (b) = y . Example 1 The Euclidean space  is path connected. Given any a, b ∈ the mapping α :[0,1] →  , defined by g (t ) = a + t (b − a) is a path from a to b. Similar argument gives that n is ☐ path connected for every n ∈+. Example 2: A Space That Is Connected but Not Path Connected Illustration 6.15 depicts a space called the topological comb: it consists of the vertical line segment {( x , y ) : x = 0, 0 ≤ y ≤ 1} and the horizontal line segments {( x , y ) : y = 1n ,  0 ≤ x ≤ 1}, n = 1, 2, 3,…, and {( x , y ) : y = 0,0 ≤ x ≤ 1}, considered as a subspace of  2 (Illustration 6.15). Illustration 6.16 is the same space, with the point A at the lowest horizontal segment deleted (we denote the resulting space by X). Its topological structure is in many ways similar to the structure of the space described in Example 5, Section 6.1. Specifically, the argument used there to show that the space in that example is connected could be used almost without any change to show that the space X is also connected. Is the space X path connected? Our intuition suggests a negative

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6.3 Path Connected Spaces   ◾    135  

answer: there should be no way to continuously bridge the gap at the point A with a path starting at B and ending at C. We will, however, resist the temptation to use obviousness as an argument in this case and we will prove that the space X is indeed not path connected. Obvious claims are the ultimate test of a theory; if they are provable then we can say that the theory indeed performs the way we want it to perform.

Illustration 6.15  A topological comb.

We will show that there is no path in X starting at B and ending at C. Suppose A = (a,0), a > 0, and consider the points a, 1 n , n = 1,2,3,… (above A; see Illustration 6.16). If a point a, 1 m is not in f ([0,1]), then no point to the right of it (along the horizontal segment) is in f ([0,1]), for otherwise the pair of open sets f −1 (U ),  f −1 (V ) , U, and V, as in Illustration 6.17, would be a separation of [0, 1].

(

(

)

)

X

B

A

C

Illustration 6.16  A topological comb with one point deleted.

X V U

1/m B

A

C

Illustration 6.17

(

)

Suppose that only finitely many points a, 1 n are in f ([0,1]). Let m be the largest integer such that a, 1 m is in f ([0,1]); take any m if no point a, 1 n is in f ([0,1]). −1

(

−1

)

(

)

Then, f (U ),  f (V ), where U and V are as in Illustration 6.18, would be a separation of

[0,1]. Note that both f −1 (U ) and f −1 (V ) are not empty since the former contains the point

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136   ◾    Connected Spaces and Path Connected Spaces

f −1 (C ) while the latter contains the point f −1 ( B). Also note that [0,1] = f −1 (U ) ∪ f −1 (V ) (even though U ∪ V is not X). X V 1/m

C

A

B

U

Illustration 6.18

(

)

Suppose now that infinitely many points a, 1 n are in f ([0,1]). Denote S = a, 1 n : n = 1,2,… . So the set f −1 (S ) is an infinite set in the closed (and bounded) interval [0,1]. By the Bolzano–Weierstrass Theorem (Theorem 5, Section 3.2), there must be an accumulation point P for f −1 (S ) and, since [0,1] is closed, that accumulation point is in [0,1]. Since f is continuous, f (P) must be an accumulation point for f f −1 (S ) (this is an easy exercise). However, f f −1 (S ) is a subset of the set a, 1 n  : n = 1,2,… ,

{( )

}

(

(

{( )

)

)

}

and the latter set has no accumulation points in X (precisely because A is missing). We have a contradiction. ☐ The last example shows that, in general, connectedness does not imply path connectedness. However, the converse is true. Proposition 1. If a space X is path connected, then it is connected. Proof. Suppose X is disconnected and suppose that A, B is a separation of X. Since both A and B are not empty, there are points P and Q in A and B, respectively. X A B α 0

α–1(A) α–1(B)

1

Illustration 6.19  A separation of X would generate a separation of the interval [0,1].

There is no path α starting at P and ending at Q since otherwise α −1 ( A), α −1 ( B) would be a separation of the interval [0,1].

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6.3 Path Connected Spaces   ◾    137  

Next, we introduce an algebraic structure over the set of paths. Let α :[0,1] → X and β :[0,1] → X . be two paths in X such that α(1) = β(0). It is clear that concatenating these two paths as in Illustration 6.20 gives another path in the space. We make it formal as follows: consider the mapping γ :[0,1] → X defined by  α(2t ) γ (t ) =  β(2t − 1) 

if if

β α 0

1

Illustration 6.20

0 ≤ t ≤ 12 . 1 ≤ t ≤1 2

We call it the product of the paths α and β, and denote it by αβ. Proposition 2. The product of two paths is a path. Proof. We only need to show that αβ is continuous. This follows from the gluing lemma with the roles of f, g, A, and B in that lemma taken by α, β, 0, 1 2  , and  1 2 ,1, respectively. The algebraic structure we have introduced so far (a partial groupoid) is rather weak. We will expand it further in Part 2 of this book. A subset A of a space X is a path connected subset if A is path connected as a subspace of X. Proposition 3. Let Ai , i = 1,2, …, be path connected subsets of a space X such that n Ai ∩ Ai+1 ≠ ∅ for every i = 1,2, … . Then for every n = 1,2, …, ∪ Ai is path connected. i =1

Proof. We use induction, with the first step being our assumption that A1 is path conk +1

k

nected. Assume ∪ Ai is path connected, and let P and Q be two points in ∪ Ai . The only i =1

i =1

k

nontrivial case is when P ∈∪ Ai and Q ∈ Ak +1. i =1

P

R

Q

Illustration 6.21 k

Since, by assumption, Ak ∩ Ak+1 ≠ ∅ there is a point R in Ak ∩ Ak+1. Since ∪ Ai is path k i =1 connected, there is a path α in ∪ Ai from P to R. Since Ak+1 is path connected, there i =1

k +1

is a path β in Ak+1 from R to Q. The product path αβ is then a path in ∪ Ai from P i =1 to Q. In the case of Euclidean spaces n, connectedness plus a bit more implies path connectedness.

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138   ◾    Connected Spaces and Path Connected Spaces

Proposition 4. Every connected and open subset of n is path connected. Proof. Suppose A is a connected and open subset of n and let P and Q be two points in A. Consider the collection O of all open balls W such that W ⊂ A. Since A is open, A = ∪ W . W ∈O

Construct a sequence of open sets as follows: U1 is an open ball in O containing P, U2 is the union of all open balls in O having nonempty intersections with U1, and, inductively, Un + 1 is ∞

the union of all open balls in O having nonempty intersections with Un. Denote VP = ∪U i . i =1

Assume that there is an open ball in O not used in the construction of VP . Then the union U of all such balls is a nonempty open set disjoint from VP , and A = U ∪ VP . That contradicts the connectedness of A. So, Q ∈VP , and so there is a sequence of balls Bi such that Bi ∩ Bi+1 ≠ ∅, P ∈ B1, and Q ∈ Bm for some m. Since every ball in n is path connected, the claim of this proposition now follows from Proposition 3. Exercises 1. Let Z be a space and let X ⊂ Y ⊂ Z. Show that X is path connected as a subset of Z if and only if it is path connected as a subset of the subspace Y of Z. 2. Show that every convex subset of n is path connected. 2

1

–2

–1

1

2

–1

–2

   Illustration 6.22

3. In Illustration 6.22 we show the subspace X of n consisting of the points on the spiral r = 1 + 10 ϕ in polar coordinates, ϕ ∈[2π , ∞), together with the points on the circle r = 1. Show that X is connected but not path connected. 4. True or false: If A is an open path connected subset of n then A is also path connected. 5. (a) Let Ai , i ∈I , be path connected subsets of a space X and suppose ∩ Ai ≠ ∅. Show i∈I that ∪ Ai is a path-connected subset of X. i∈I

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6.4 Path Connected Spaces: More Properties and Related Matters   ◾    139  

(b) Let A0 and Ai , i ∈I , be path connected subsets of a space X and suppose A0 ∩ Ai ≠ ∅

( )

for every i ∈I . Show that A0 ∪ ∪ Ai is path connected. i∈I

(c) Let Ai , i = 1, 2, …, be path-connected subsets of a space X such that Ai ∩ Ai+1 ≠ ∅ ∞

for every i = 1, 2, …. Show that ∪ Ai is path connected. i =1

6. Find a path-connected space that is not second countable. 7. An ambient path in a space X starting from P ∈ X and ending at Q ∈ X is a mapping h: X × I → X such that the restriction h X ×{t } is a homeomorphism for every t in I, such that h X ×{0} : ( x ,0)  x for every x ∈ X, and such that h X ×{1} sends P to Q. A space is ambient path connected if for every two points P and Q in X there is an ambient path in X from P to Q. (a) Show that every ambient-path-connected space is path connected. (b) Show that the (path-connected) subspace of  2 depicted in Illustration 6.23 below is not ambient path connected.

   Illustration 6.23

8. Let X be the subspace of  2 consisting of two intersecting line segments, such that the intersection point is not an end point of any of the segments. Show that there is no ambient path from the intersection point to any other point in X.

6.4 Path Connected Spaces: More Properties and Related Matters A path component in a space X is a maximal (under inclusion) path connected subset of X. The path component of X containing a point x is the set of all points y in X for which there is a path starting at x and ending at y.

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140   ◾    Connected Spaces and Path Connected Spaces

Example 1

X

There are two path components in the topological comb, with the point A = (a,0), 0 ≤ a < 1, deleted: the line segment {(t ,0) : a < t ≤ 1} is one path component, the rest of X is another path component. ☐ A

The following is an observation.

Illustration 6.24

Proposition 1. A space is path connected if and only if it has one path component. Proof. Obvious. Example 2: A Connected Space with Infinitely Many Path Components Let Xi , i ∈ , be a copy of the interval [0, ∞) (thus we have  copies of [0, ∞)). Assume these copies of [0, ∞) are pair-wise disjoint. Define a linear order on Y = ∪ Xi as follows: the order within each single copy of [0, ∞) is the usual linear i∈

order of real numbers, while for every i < j (usual linear order of real numbers), each element of Xi is less than each element of Xj. The linear order we get in such a way generates the order topology over Y. It is easy to check that Y is a linear continuum. Consequently, it is a connected space (Proposition 1′, Section 6.1). Xi

Xj

Illustration 6.25  Each Xi is a copy of [0, ∞ ). The copies of Xi with larger index (with respect to

the usual order in ) are to the right and contain larger elements in Y.

We will now see that Y is not path connected. Take any distinct i , j ∈ with i < j and suppose x ∈ Xi and y ∈ X j . Suppose there is a path α :[0,1] → Y from x to y. Then every point z between x and y (in the linear order of Y) is in α([0,1]) for otherwise α −1 ({w ∈Y  : w < z }) and α −1 ({w ∈Y  : w > z }) would be a separation of [0, 1]. Choose any open interval in each X k , i < k < j ; for example, choose a copy Z k in X k, i < k < j , of the open interval (0,1). Since α is continuous, each α −1 ( Z k ) is open in [0, 1], and since the elements of Z k are between x and y, the open sets α −1 ( Z k ) are not empty. So, each of these sets α −1 ( Z k ), i < k < j , contains (as a subset) an open interval in [0, 1]. Choose one such interval Uk for each α −1 ( Z k ). Since the sets Z k, i < k < j , are pairwise disjoint, so are the sets α −1 ( Z k ), i < k < j , and so are the intervals U k, i < k < j. Define a mapping f : (i , j ) → [0,1] (where (i, j) denotes the open interval in ) as follows: for every k ∈(i , j ), choose any rational number qk in U k and set f (k ) = qk . Since U k ’s are intervals, they contain rational numbers, and so f is well defined. Since the intervals U k , i < k < j , are pairwise disjoint, the mapping f is one-to-one.

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6.4 Path Connected Spaces: More Properties and Related Matters   ◾    141  

(

0

)

Xk

Xl

( )

1

1

0 α–1 0

(

qk

)

(

ql

)

1

Illustration 6.26  f sends k to the rational number qk , and it sends l to the rational number ql.

So, summarizing, we got a one-to-one mapping f from the uncountable set of real numbers in the interval (i , j ) into a subset of the countable set of rational numbers in [0,1]. That is not possible (see Section 1.2). So, Y is not path connected. In fact, we proved more: we showed that if a subset A of Y intersects (non-emptily) both Xi and X j , i ≠ j , then it is not path connected. Consequently, the path components of Y are the sets Xi , i ∈ , each of them a copy of [0, ∞). So, Y has uncountably many ☐ path components and only one component. The next proposition tells us that path connectedness is preserved under continuous maps (and so it is a topological property). Proposition 2. If a space X is path connected and f : X → Y is continuous, then f ( X ) is a path connected subset of Y. Proof. Suppose P and Q are two points in f ( X ). Let p, q ∈ X be such that f ( p) = P and f (q ) = Q .

X α

p

Y

q f

P

Q

f(X)

Illustration 6.27

Since X is path connected, there is a path α in X starting at p and ending at q. The composition f α is then a path in f ( X ) starting at P and ending at Q. Example 3 Every quotient space of a path connected space is path connected since quotient maps are continuous and surjective. ☐ Example 4 Suppose X is path connected and consider the diagonal D = {( x ) j ∈ J  :  x ∈ X } as a subspace of ∏ X . Then, D is path connected, too. It is clear that the mapping f : X → D j∈ J

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142   ◾    Connected Spaces and Path Connected Spaces

defined by f ( x ) = ( x ) j ∈ J is onto. In order to see that it is continuous take a standard basis neighborhood ∏ U j of ( x ) j ∈ J in ∏ X ; so all but finitely many U ’s are j∈ J

j∈ J

the entire space X, and the remaining {U j1 ,U j2 ,…,U jk } are open neighborhoods of x in X. It is visible that

k

f −1 ( ∏ U j ) = ∩ U j = ∩U ji , an open set. So, f is conj ∈J

j ∈J

i =1

tinuous. Since D = f ( X ) Proposition 2 applies, and D is path connected.

☐

Proposition 3. If X j , j ∈J are path-connected spaces, then so is ∏ X j . j∈ J

Proof. The idea of the proof is outlined in Illustration 6.28, where we take J = {1,2}, and where X1 and X2 are copies of [0, ∞).

]

X2

[

(α,β)(t) = (α(t),β(t))

[

β

]

(α,β) X1 α

Illustration 6.28

Formally, suppose  X j , j ∈ J  are path-connected spaces, and let  P , Q ∈ ∏ X j . Then for every j ∈J

fixed i ∈ J there is a path α i from pi ( P ) to pi (Q ) in Xi , where, as usual, pi denotes the projection ∏ X j → Xi . Consider now the mapping ∏ α j :[0,1] → ∏ X j defined by

( )

j ∈J

j∈ J

j∈ J

∏ α j (t ) = ∏ α j (t ) . By Proposition 2 in Section 5.2, ∏ α j is a (continuous) path, and it j ∈J

j ∈J

starts at P and ends at Q.

j∈ J

In the rest of this section we explore the notion of travel in a space through some natural generalizations of path connectedness. We will not use these generalizations in the sequel. However, we will come close to them in Section 10.1 where we introduce ambient isotopy. In a path-connected space X a point can travel between any two positions in the space. If path connectedness emulates movement of points, then we need a suitable generalization to correspond to movement of other objects in a space. The next definition does that to a certain degree: we allow the starting object to be deformed along a path as long as it does not change up to a homeomorphism.

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6.4 Path Connected Spaces: More Properties and Related Matters   ◾    143  

Let X be a space and S a subspace of X. We say that X is S-connected if for every two embeddings f : S → X and g : S → X there is a continuous map F : S × [0,1] → S (where S ×[0,1] is equipped with the product topology) such that F (s ,0) = f (s ), F (s ,1) = g (s ) for every s in S, and such that F (s , t0 ): S × {t0 } → X is an embedding for every t0 ∈ I. As in the definition of path connectedness, the usage of the interval I = [0,1] in the last definition is for the purpose of standardizing the domain and is not essential: we could replace it by any other closed intervals without affecting the intended meaning of the definition. We may view I in this context as denoting time, so that traveling from position F (s ,0) = f (s ), s ∈S to position F (s ,1) = g (s ), s ∈S is always executed in one unit of time. The definition of S-connectedness generalizes the notion of path connectedness: if S is a one-point space, then being S-connected is the equivalent to being path connected. Example 5 In Illustration 6.29 we depict a hollow box B with a point-thin slit at the top face. Keeping the notation used above, we take X =  3 \ B. X

Y

Z

Illustration 6.29

The space Y is a stickman with a flat head, and Z is a stickman with a spherical head, both as subspaces of X. The space X is Y-connected, but not Z-connected. The first claim is clear, and the second is equivalent to the statement that there is no subspace  V2 ∪  V3 , V1 is inside B, V2 is in the slit (i.e., V2 is homeomorV of X such that V = V1 ∪ phic to a subspace of a line segment), and V3 is outside B, and such that V ≅ S 2 . This ☐ merits attention (Exercise 10). Example 6 Let X be a cylinder (see Illustration 6.30). Then X is D2-connected (where we recall that D 2 = {( x , y ) ∈  2 : x 2 + y 2 ≤ 1}). However, X is not S 1-connected ( S1 = {( x , y ) ∈ 2 : x 2 + y 2 = 1} ). There is no way to travel from the homeomorphic image S11 of the circle S1 to the homeomorphic image S21 (Illustration 6.30). A simple formal justification of this statement will come out of the theory we will develop in Part 2.

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144   ◾    Connected Spaces and Path Connected Spaces

D2 D1 S21

S11

Illustration 6.30  It is not possible to travel from position S11 to position S21 as shown in this picture.

This is true even if we drop the requirement that at each moment of time the image of S1 on the cylinder is homeomorphic to the original. ☐

S-connectedness is a topological property but it is not necessarily preserved under continuous mappings. Finally, we will model another type of movement, whereby we deform the whole space X in order for a given portion of it to reach a predetermined position. Let X be a space and let S be a subspace of X. We say that X is isotopically S­- connected, if for every two embeddings f : S → X and g : S → X there is a continuous mapping F : X × I → X such that F X ×{t } is a homeomorphism for every t ∈I, such that F ( x , 0) = x for every x ∈ X , and such that F ( f (s ),1) = g (s ) for every s ∈S. (The mapping F here is called an ambient isotopy within X from f (S ) to g (S ); we will encounter ambient isotopies in Part 2 of this book.) Proposition 4. If a space X is isotopically S­-connected, then it is S-connected. Proof. Exercise 13. Example 7 The double cone in Illustration 6.31 is neither S1-connected nor isotopically S1-connected. These claims constitute Exercise 14. The space X depicted in Illustration 6.32 (obtained by deleting two open disks on two spheres and then joining the boundaries of these disks with a cylindrical surface) is both S1-connected and isotopically S1-connected. Can you visualize that claim? Finally, the space Y depicted in Illustration 6.33 (obtained by joining two solid balls with a cylindrical surface) is S1-connected but not isotopically S1-connected (Exercise 15).

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6.4 Path Connected Spaces: More Properties and Related Matters   ◾    145  

Illustration 6.31  A double-cone.



Illustration 6.32  The space X.

Illustration 6.33  The space Y.

☐

Exercises 1. Describe the path components in  with the half open interval topology. 2. (a)  Find a connected subset A of  2 such that ∂A has three path components. (b) Find a connected subset A of  2 such that ∂A has infinitely many path components. 3. Show that path components of open subsets of  2 are open. 4. Show that each path component is contained in a component. 5. Consider the set  2 equipped with the post office metric (Section 2.3), where we assume that the designated point (the post office) is the origin. Denote the metric space generated by that metric by X.

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146   ◾    Connected Spaces and Path Connected Spaces

(a) Which of the following three mappings from [0,1] into X is a path in X from the point (1,0) to the point (0,1)?   (1 − 2t ,0) α(t ) =   (0,2t − 1) 

β(t) = (1, 0) + t(−1, 1)



 (1,0)  γ (t ) =  (0,0)  (0,1) 

if if if

0 ≤ t  ≤ 1 2

if

1 ≤ t  ≤ 1 2

if

t=0 0  a . Repeat the procedure from the previous paragraph, with the role of b1 now taken by b2 : we get b3 = inf {c j  :  j ∈ J , b2 ∈(c j , d j )} , and a subset of U consisting of at most two intervals covering [b3 , b2 ]. As in the first step, none of the intervals in U containing b3 also contains b2 . If b3 ≤ a , then the union of the two finite covers of [b3 , b2 ] and [b2 , b1 ] is a finite cover of A. Otherwise b3 > a and we iterated the procedure. In step k we get a subset of U consisting of at most two intervals, covering [bk +1 , bk ], where, as in the first step, none of the intervals in U containing bk+1 contain bk . If after k iterations we have bk +1 ≤ a for the first time, then the union of the finite covers of [bi , bi−1 ], i = 2, 3, … , k + 1 is a finite subcover of U that we were looking for. Otherwise bk > a for every k ∈+ . Thus, we have a decreasing sequence (b j ) of elements in A. Since A is bounded this sequence converges to some b. Since A is closed, b ∈ A. Hence, there is an interval in U ∈U containing b. It follows that, from some point on, all the members of the sequence (b j ) are in U. In particular, there will be two consecutive bk , bk +1 ∈U . This is a contradiction (see the last sentence two paragraphs earlier). So, the preceding cases apply, and we have established that A is compact. Finally, we eliminate the two initial restrictions on the cover and on the set A. Let U be any open cover. We replace it with the family of all intervals that are subsets of the open sets in U , then we use the preceding argument to find a finite subcover of that cover made of intervals, and, at the end, we replace each interval of that finite subcover with one element of U that contains it as a subset. The second restriction was that A was a closed interval. Suppose now that A is any closed set. Then, since it is bounded, it is a subset of an interval [c, d]. Hence, U ∪{ \ A} is an open cover of [c, d]. By the above argument, there is a finite subclass of U ∪{ \ A} that covers [c, d]; we simply eliminate {\ A} from this subclass to get a finite subcover of U that covers A. ⇐ Suppose now that every open cover of a subset A of  has a finite subcover. First we show that A is bounded. Otherwise there is a sequence ai , i = 1,2, … , of elements of A such that for every positive integer n there is an element of that sequence out of the interval (−n, n). Then {(−n, n) : n ∈+ } would be an open cover of A that does not have a finite subcover. Now we show that A is closed. Suppose otherwise. Then there is an accumulation point for A that is not contained in A. Denote it by x. Take any a, a < x , and any b, x < b, and consider the following cover of A made of open intervals in  (where ε is any positive number).

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1 1     {(−∞, a)} ∪  a − ε, x −  : n = 1,2, … ∪  x + , b + ε  : n = 1,2, … ∪ {(b, ∞)}.     n n    

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7.1  Compact Spaces: Definition   ◾    157  

Extracting a finite subcover

{(−∞, a)} ∪  a − ε, x − 



1 1      : n = 1,2, …, p  ∪  x + , b + ε  : m = 1,2, …, q  ∪ {(b, ∞)} n m   

(

)

of this cover is not possible, since otherwise the interval x − p1+1 , x + q1+1 would contain no elements of A, contradicting the assumption that x is an accumulation point for A. The Heine–Borel theorem holds for n in place of  (Theorem 6, Section 7.2). Theorem 3 cannot be generalized to all metric spaces. For example, the Hilbert space H is a metric space having closed and bounded subsets that are not compact [Exercise 10, Section 7.4]. Generalizations often offer views from afar that reveal properties that are hard to see from within a specific model of a theory. Compare, for example, the proof of Proposition 1 and the proof of the following generalization; the advantages of the latter are obvious. Proposition 4. Compactness is preserved under continuous mappings. That is, if A is a compact subspace of a space X and f : X → Y is continuous, then f (A) is a compact subspace of Y. Proof. Suppose U = {U i  : i ∈ J } is an open cover of f (A). Then, since f is continuous, { f −1 (U i ) : i ∈ J } is an open cover of A. Since A is compact, that cover possesses a finite subn

(

n

)

cover { f −1 (U i ) : i = 1,2, … , n} of A. Hence, A ⊂ ∪ f −1 (U i ) . Then f ( A) ⊂ f ∪ f −1 (U i ) = n

−1

i =1

n

∪ f ( f (U i )) = ∪U i , and so {U i  : i = 1,2, … , n} is a finite subcover of f (A).

i =1

i =1

i =1

As a consequence, compactness is a topological property (i.e., it is preserved under homeomorphisms). We finish this section by extracting an easy alternative criterion for compactness. A family F = {Fi  : i ∈I } of subsets of X has the finite intersection property if every finite intersection of elements of F has a nonempty intersection. Proposition 5. A space X is compact if and only if every family of closed subsets of X with the finite intersection property has a nonempty intersection. Proof. ⇒ Let X be compact and let F = {Fi  : i ∈I } be a family of closed subsets of X with the finite intersection property. We need to show that ∩ Fi ≠ ∅ . Otherwise ∩ Fi = ∅ , and so, i∈I

i∈I

by de Morgan laws, ∪ Fic = X , so that the family {Fjc  : i ∈I } of open sets covers X. Since X is i∈I

n

compact, there is a finite subcover {F1c , F2c ,… , Fnc } , so that ∪ Fi c = X . Applying de Morgan i =1 n laws again, we have ∩ Fi = ∅ , contradicting our assumption that F has the finite interseci =1 tion property.

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158   ◾    Compactness and Related Matters

⇐ Suppose every family of closed subsets of X with the finite intersection property has a nonempty intersection. Let {U i  : i ∈I } be an open cover for X, so that ∪U i = X . Then, by de i∈I

Morgan laws, ∩U ic = ∅ . Since all U ic are closed sets, our assumption implies that for some i∈I

finite subset J of I, ∩U ic = ∅ . Applying de Morgan laws again we get ∪U i = X . i∈J

i∈J

Here is a variant of the last proposition. Proposition 6. A space X is compact if and only if for every family { Ai  : i ∈ I } of subsets of X satisfying the finite intersection property, we have ∩ Ai ≠ ∅ . i∈I

Proof. Exercise 16. Exercises 1. When is a subspace of a discrete space X compact? What if X is indiscrete? 2. Let ( X ,T1 ) and ( X ,T 2 ) be topological spaces such that T1 ⊂ T 2 . Show that if ( X ,T 2 ) is compact, then so is ( X ,T1 ) . 3. Prove Proposition 2. 4. Show that the Cantor set is compact. 5. Show that the Sorgenfrey line is not compact. 6. Show that a quotient space of a compact space is compact.

7. (a)  Show that the interval [0, ∞) is not compact. (b) Show that [0, ∞) is not homeomorphic to the circle.

8. Show that if K is a compact subspace of  2 , then  2 \ K has exactly one unbounded path component. 9. Prove that if X is a compact space, then the diagonal {( x , x ) : x ∈ X } is a compact subspace of X × X . 10. Show that a space X is compact if and only if for every basis B of the space X, every cover of X consisting of elements of B has a finite subcover. 11. Show that if a set X is well ordered, and if there exists the largest element m in X, then X with the order topology is a compact space. 12. Show that if X is a compact metric space then every sequence has a convergent subsequence. 13. Show that every compact metric space is complete. 14. Show that the topological sum ⊕ Xi of compact spaces is compact if and only if the i ∈I index set I is finite.

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7.2 Properties of Compact Spaces   ◾    159  

15. (a) Show that there is no homeomorphism f :  2 →  2 such that f ( A) = B (see Illustration 7.3).

A

B

Illustration 7.3  The subspace A of  2 is shown to the left (all black points); B is to the right.

(b) Describe (visually or otherwise) a homeomorphism g :  2 →  2 such g ( A) = B (see Illustration 7.4).

A

B

   Illustration 7.4

16. Prove Proposition 6: A space X is compact if and only if for every family { Ai  : i ∈I } of subsets of X satisfying the finite intersection property, we have ∩ Ai ≠ ∅ . i∈I

17. Show that if { A j : j ∈1,2,…} is a family of compact nonempty subsets of a Hausdorff ∞ space X such that A1 ⊃ A2 ⊃  ⊃ A j ⊃ , then ∩ A j ≠ ∅ . Show that if “Hausdorff” is omitted than the new statement can fail.

j =1

18. (a)  Prove that compact subspaces of a metric space are closed and bounded. (b) Show that in the Hilbert space (Example 2, Section 5.2) there are closed and bounded sets that are not compact. 19. Show that every finite CW-complex is compact.

7.2 Properties of Compact Spaces A brief historical note: Proposition 2 is due to Vietoris (1921).

Compactness is obviously not hereditary; for example, no nonempty, open, proper subspace of the closed interval [0,1] (with the usual topology) is compact. In the next proposition we postulate an extra condition for heredity of compactness.

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160   ◾    Compactness and Related Matters

Proposition 1. Every closed subspace of a compact space is compact. Proof. Let X be a compact space and A be a closed subset of X. Note that Proposition 2, Section 7.1 allows us the freedom to choose a cover made of open subsets of A or a cover made of open subsets of X. We choose the latter. The proof is contained in Illustrations 7.5–7.8 and the associated captions.

A

X

Illustration 7.5  Let U = {U i  : i ∈I } be an open

cover of A.

A

X

Illustration 7.7  Since X is compact, there is a

finite subcover U ∗ of U ∪ { Ac }.

A

X

Illustration 7.6  Then U ∪ { Ac } is an open

cover of X.

A

X

Illustration 7.8  Then U ∗ \{ Ac } is a finite sub-

cover of U covering A.

Example 1 That the Sorgenfrey line S (generated by all intervals of type (a, b]) is not compact is an easy exercise (Exercise 5, Section 7.1). It is equally easy to show that any open interval [a, b) (considered as a subspace of S ) also fails to be compact. Since these intervals are also closed in the Sorgenfrey topology, it follows from Proposition 1 that any subspace of S that contains some [a, b) as a subset cannot be compact. In particular, closed intervals [a, b] are not compact in S . ☐

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7.2 Properties of Compact Spaces   ◾    161  

Compact subspaces of a compact space X need not be closed in X (Exercise 3). For the compact subspaces to be closed in X it is sufficient to assume that the space X is Hausdorff. We have defined that notion earlier; it is now time to raise its prominence. Proposition 2. Every compact subspace of a Hausdorff space is closed. Proof. Let X be a Hausdorff space and let A be a compact subspace of X. We show that Ac is open. It suffices to show that every point x ∈ Ac is interior for Ac . The proof is given in Illustrations 7.9–7.12 and in the captions.

X A

y

X

Uxy

Vy

A

x

Illustration 7.9  Since X is Hausdorff, for x and

for every y ∈ A there are two disjoint open neighborhoods U xy and Vy of x and y, respectively.

Illustration 7.10  Repeat that for every y ∈ A.

We get an open cover {Vy  :  y ∈ A} for A and a class of open neighborhoods {U xy  :  y ∈ A} of x.

X

X

A

A

x

Illustration 7.11  Since A is compact, there is

a finite subcover {V1 ,V2 ,…,Vn } of {Vy : y ∈ A} for A. Choose the associated neighborhoods {U1 ,U 2 ,…,U n } of x.

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x

x

n

Illustration 7.12  The finite intersection ∩ U i is i =1

then an open neighborhood of x having empty n

n

i =1

i =1

inter­section with ∪ Vi . So, ∩ U i is entirely with­in Ac.

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The following easy consequence will be used later on. Corollary 3. Let X be a compact space, let Y be Hausdorff, and let f : X → Y be continuous and bijective. Then f is a homeomorphism. Proof. Exercise 7. We now turn our attention to product spaces: Is the product of a collection of compact spaces a compact space? The affirmative answer to that question—which we will establish in due course—is the primary reason the product topology is regarded to be “more natural” or “better behaved” than the box topology. The case of finite products of spaces is much more accessible; the general case is postponed for a while. Theorem 4. A finite product of compact spaces is a compact space. Proof. We prove the claim for two spaces; the rest follows by induction. So, given two compact spaces X and Y, we need to prove that X × Y is also compact. Start with any open cover U of X × Y . In Illustrations 7.13–7.17, the space X is represented by a horizontal ellipse, while Y is depicted as a vertical line segment. The first part of the proof is given in the captions of these illustrations.

x0 × Y

X×Y

x0

Illustration 7.13  For any point x 0 of X the

subspace x 0 × Y of X × Y is compact (being Illustration 7.14  So, we can extract a finite subcover of U  for the space x 0 × Y . homeomorphic to the space Y).

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7.2 Properties of Compact Spaces   ◾    163  

Illustration 7.15  Remove X × Y to see more

clearly the “skewer” of open sets surrounding x0 × Y.

Illustration 7.16  Project these open sets over

X to get finitely many open sets in X. Take any open subset V of their intersection containing x 0 (we may take the intersection itself).

Illustration 7.17  The open tube V × Y con-

tains (as a subset) x0 × Y (but is not necessarily a member of U ).

Repeating the above procedure for every element x 0 of X we get a collection C of open neighborhoods V covering X and a collection of tubes V × Y covering all X × Y . Since X is compact, there is a finite subset of C covering X. The associated finite set of tubes will therefore cover all of X × Y . Now replace each of these finitely many tubes with the associated “skewer” (see Illustrations 7.15 and 7.16) consisting of finitely many open subsets of U whose union contains the tube as a subset. We get a finite subcover of U covering X × Y.

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Denote the unit closed interval [0,1] by I. The Euclidean unit n-cube is

{

}

I n = ( x1 , x 2 ,… , xn ) ∈n : 0 ≤ x1 ≤ 1, 0 ≤ x 2 ≤ 1,… , 0 ≤ xn ≤ 1 .

Corollary 5. The Euclidean cube I n is compact. Proof. The unit interval I is compact, since it is bounded and closed. The claim follows from Theorem 4. It is now easy to generalize Theorem 3, Section 7.1. Theorem 6 (Heine–Borel). A subset A of the space n is compact if and only if it is closed and bounded. Proof. ⇒ Suppose A is a compact subspace of n. Since n is Hausdorff, it follows from Proposition 2 that A is closed. The set A could not be unbounded since otherwise, for every x ∈n , the open cover consisting of the open balls B( x , m), m = 1, 2, … would not have a finite subcover (and A would not be compact). ⇐ Suppose A is a closed and bounded subset of n. Since A is bounded, there is a large cube C = [a1 , b1 ] × [a2 , b2 ] ×  × [an , bn ] in n containing A as a subset. Since I n is compact (Corollary 5) and since any two cubes in n are homeomorphic, it follows that C is compact. So, A is a closed subset of the compact space C. It follows from Proposition 1 that A is compact. Exercises 1. Show that a finite union of compact subspaces of a space X is a compact subspace of X. 2. Recall that for two subsets A and B of a metric space (X, d), d(A, B) stands for inf {d(a, b) : a ∈ A, b ∈ B}. Is there a metric space X and two disjoint closed subsets A and B of X, at least one of them compact, such that d( A, B) = 0? (Compare to Exercise 9, Section 2.1.) 3. Show that a compact subspace of a compact space X need not be closed in X. 4. Let X be a Hausdorff space.n Prove that for every collection {F1 , F2 ,… , Fn } of closed subsets of X, the subspace ∪ Fj of X is compact if and only if Fj is compact for every j =1 j = 1, 2, … , n. 5. Show that the closure of a compact subspace of a space X need not be compact. (Hint: Consider the Particular-point topology over an infinite set X, defined by stipulating that open sets are only the sets containing a fixed point of X.) What if X is Hausdorff? 6. Let X be a compact space, let Y  be a Hausdorff space, and let f : X → Y be continuous mapping. Show that for every A ⊂ X , f ( A) = f ( A) .

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7.3  Compact, Lindelöf, and Countably Compact Spaces   ◾    165  



7. Let X be a compact space, let Y be Hausdorff, and let f : X → Y be a bijection.

(a) Show that if f is continuous that it is open.



(b) Show that if f is open then it is continuous.

8. (a)  Show that if X is Hausdorff then intersections of compact subsets are compact.

(b) Find a space X and two compact subspaces of X such that their intersection is not compact.

9. (a) Show that if A is a compact subspace of a Hausdorff space X and if x ∈ X \ A, then there are two disjoint, open subsets U and V of X such that A ⊂ U and x ∈V .

(b) Show that for every two disjoint compact subspaces A and B of a Hausdorff space X, then there are two disjoint, open sets U and V such that A ⊂ U and B ⊂ V .

10. Show that there does not exist a class of infinitely many pairwise disjoint open subsets of a closed interval [a, b] ⊂  , a < b. 11. Denote X = {( x ,0) : 0 ≤ x ≤ 1} ∪ ∪ ({s} × [0,1]) , where J is a totally disconnected subset s ∈J

of the interval [0,1] (see the following illustration). Show that if f :[0, 1] → X is a continuous mapping, then f ([0,1]) ∩ ∪ ({s} × {1}) is finite. s ∈J

   Illustration 7.18

7.3  Compact, Lindelöf, and Countably Compact Spaces A brief historical note: Lindelöf spaces are named after the Finnish topologist Ernst Lindelöf (1870–1946), who in 1903 proved that open subsets of n are (what was later called) Lindelöf spaces.

We have discovered in Proposition 4, Section 7.1, that compactness is preserved under continuous mappings. Other relationships between compactness and continuity that we note in this section are consequences of that proposition. Proposition 1. Every continuous mapping from a compact space into a Hausdorff space is closed.

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Proof. Suppose X is compact, Y is Hausdorff, and f : X → Y is continuous. Choose a closed subset F of X. Then, by Proposition 1, Section 7.2, F is compact. So, by Proposition 4, Section 7.1, f (F) is also compact. Finally, since Y is Hausdorff, it follows from Proposition 2, Section 7.2 that f (F) is closed. Recall that if f : X → Y is a mapping, then ~ f is the equivalence relation defined by x ~ f y if f ( x ) = f ( y ). Proposition 2. If f : X → Y is a continuous mapping from a compact space X onto a Hausdorff space Y, then X ~ is homeomorphic to Y. f

Proof. This is a consequence of Proposition 1 here and Proposition 2, Section 4.2. Example 1 Consider the mapping f : S 2 →  6 defined by f ( x , y , z ) = ( x 2 , y 2 , z 2 , xy , xz , yz ) . Since the component functions are continuous, so is f (Exercise 14, Section 5.1). By 2 Proposition 2, f (S 2 ) is homeomorphic to S ~ . In this case, for every x , y ∈S 2 such f

that x ≠ y , we have f ( x ) = f ( y ) if and only if x and y are two antipodal points in S 2 (i.e., if and only if x = − y ). So, f (S 2 ) is homeomorphic to the quotient space obtained from S 2 by identifying all pairs of antipodal points. This quotient space is the projective plane P 2 ; it will be important to us later on. We visualize it in Illustrations 7.19 and 7.20.

–x

x

Illustration 7.20  The space we get is called Illustration 7.19  We identify each pair of

antipodal points x and −x on the sphere S 2 .

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the projective plane. It is not embeddable in  3 , so we cannot see it. ☐

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The real-valued functions are important to us since they link general spaces with the wellstudied blueprint model . Proposition 3. If X is compact and f : X →  is continuous, then there are points a, b ∈ X such that f ( X ) ⊂ [ f (a), f (b)]. Proof. Suppose X is compact. Then f (X) is also compact. By Theorem 3, Section 7.1, f (X) is a closed and bounded subset of . Since f (X) is bounded it has the least upper bound and the greatest lower bound. Since f (X) is closed, these two are elements of f (X). As a consequence every continuous mapping f :[c , d] →  attains its absolute maximum and absolute minimum (i.e., we got the Extreme Value Theorem from basic calculus). Example 2 Let (X, d) be any compact metric space, let A be a closed subset of X and let b ∈ X \ A (see Illustration 7.21). Define a mapping f : A →  by f ( x ) = d( x , b) for every x in A. Then, by Example 3, Section 2.2, and by simple considerations, f is continuous at every point x of A. Since A is a closed set in a compact space X, it is itself compact. By Proposition 3, the mapping f attains the minimal value at some point amin ∈ A , and it attains the maximal value at some point amax ∈ A . Since b ∉ A, we have d(amin , b) > 0 and d(amax , b) > 0. amax A amin

b

Illustration 7.21  There is a point in the closed set A that is the closest to b and there is a point in A that is the farthest from b. ☐

X

The following proposition concerning compact metric spaces is important in basic analysis: Proposition 4. Let ( X , d1 ) be a compact metric space, and let (Y , d2 ) be a metric space. A mapping f : X → Y is continuous if and only if it is uniformly continuous. Proof. We noticed in Section 2.2 that ⇐ holds. For the other implication assume that f : X → Y is continuous, and choose any ε > 0. The continuity of f implies that for every x ∈ X , there is a δ x > 0, such that if x ′ ∈ X is such that d1 ( x , x ′ ) < δ x , then d2 ( f ( x ), f ( x ′ )) < 2ε .

{( )

}

The family of balls B x , δ2x  :  x ∈ X is an open cover of X. Since X is compact, there exists

{( )

}

a finite subcover B x j , δ2x j  :  j = 1,2,…, n . Choose δ = min

{

δx j 2

}

 :  j = 1,2,…, n .

We will complete the proof by showing that if u, v ∈ X are such that d1 (u, v ) < δ , then δ d2 ( f (u), f (v )) < ε . So, take u, v ∈ X , such that d1 (u, v ) < δ. Some B x k , x2k contains u,

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(

)

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168   ◾    Compactness and Related Matters δ

and we have: d( x k , v ) ≤ d1 ( x k , u) + d1 (u, v ) < δ x2k + x2k = δ x k. So, d( x k , v ) < δ x k and, by symmetry, d( x k , u) < δ x k . Hence, by assumption, d2 ( f ( x k ), f (v )) < 2ε and d2 ( f ( x k ), f (u)) < 2ε . Consequently, d2 ( f (u), f (v )) ≤ d2 ( f (u), f ( x k )) + d2 ( f ( x k ), f (v )) < 2ε + 2ε = ε . The following two concepts are weaker than the notion of compactness. A space X is Lindelöf if every open cover of X has a countable subcover. A space X is countably compact if every countable open cover of X has a finite subcover. Obviously a space X is compact if and only if it is both Lindelöf and countably compact. Example 3 We mentioned in Example 1, Section 7.2, that the space X =  equipped with the Sorgenfrey line topology S is not compact. However, this space is Lindelöf. In this example we justify this claim. Start with any open cover O of X and consider the set A of all intervals of the type (a, b) such that (a, b) ⊂ O for some O ∈O . Obviously A is a cover of the subspace X • = ∪ (a, b) of X. ( a ,b )∈A

For every x in X • choose (a,b) that contains it, then find two rational numbers, q1x between a and x and q2x between x and b, so that x ∈(q1x , q2x ) ⊂ (a, b). There are clearly countably many such intervals (q1x , q2x ), and it follows from the definition of A that they cover X •. For each of them choose one (a, b) that contains it as a subset to get a countable subcover B of A. Now, for each of the members of B choose an open set from O that contains it as a subset (which we can do by the definition of A ) to get a countable subset C of O that covers X •. To complete our justification it suffices to find a countable subcover D of O that covers X \ X • , in which case C ∪ D will then be a countable subcover of O that covers all of X. Since the intervals of the type [a, b) make a basis for the Sorgenfrey topology, for every x in X \ X • and every Ox ∈O that contains it, there is [ax , bx ) such that x ∈[ax , bx ) ⊂ Ox . Then, x ∈[x , bx ) ⊂ Ox . For every x1 , x 2 ∈ X \ X ∗, x1 < x 2 , we have that [x1 , bx1 ) ∩[x 2 , bx 2 ) = ∅, for otherwise ( x1 , bx1 ) would contain x 2 , and since ( x1 , bx1 ) ⊂ Ox1 ∈O , this would imply that x 2 is in X • . So, all of these intervals [x , bx ) are pairwise disjoint as x ranges through X \ X ∗. Since each of them contains a rational number, there are countably many of them. So, there are countably many elements in X \ X • . Now choose one open set from O for each element of X \ X • to get ☐ a countable subcover of O that covers all of X \ X ∗ . That is the desired D . Proposition 5. (Lindelöf Lemma) Every second countable space is Lindelöf. Proof. Let O be an open cover of a second countable space X, and let B be a countable basis for X. Each member of O is a union of elements of B . Replace each member of O by the

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elements of B comprising that union. We get a new cover of X, and since each member of that cover is a member of B , this is a countable cover. Now replace each member of that new cover by any element of O containing it as a subset. We get a countable subcover of O . Example 4 The Euclidean line  is Lindelöf. This is now a consequence of Proposition 5 and the ☐ fact that  is second countable. Observe that the argument in Example 4 is not transferable to the case when  is equipped with the Sorgenfrey topology, since that topology is not second countable (Example 15, Section 3.3). Exercises 1. Let f :[a, b] →  be a continuous function. Show that f ([a, b]) is a closed and bounded interval. 2. Show that among all Hausdorff topologies over a fixed set X, compact topologies are minimal. In other words, show that if ( X ,T1 ) is a Hausdorff space, ( X ,T 2 ) is a compact space and T1 ⊂ T 2 , then T1 = T 2 . 3. (a)  Prove that a space X is countably compact if and only if for every countable collection {Fj :  j = 1,2,…} of closed subsets of X that has the finite intersection ∞ property, ∩ Fj ≠ ∅. j =1

(b) Let Y be a countably compact subspace of a space X, and let {Fj  :  j = 1,2,3,…} be a collection of closed subsets of X such that F1 ⊃ F2 ⊃ F3 ⊃  . Show that ∞ ∩ Fj ≠ ∅.



j =1

4. Show that if f : X → Y is continuous and onto and X is Lindelöf, then so is Y. 5. (a)  Show that closed subspaces of Lindelöf spaces are Lindelöf. (b) Let X =  be equipped with the Sorgenfrey line topology. Show that X × X is not Lindelöf. (Hence products of Lindelöf spaces need not be Lindelöf.)



6. Show that if X is compact and Y is Lindelöf, then X × Y is Lindelöf.

7. Show that the Lindelöf property is not hereditary.

8. Define a sequence ( f j ) ( j ∈+ ) of functions I → I × I through the frames shown in the Illustration 7.22.

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Illustration 7.22  From left to right: f1, f 2 , and f3 ; iterate to get each member of the sequence ( f j ).

The sequence ( f j ) is obviously a Cauchy sequence of functions (“obvious” in this sentence means “as seen by the naked eye”—the visually introduced functions f j do not allow any rigor). According to Exercise 17, Section 2.2, it converges uniformly to a continuous function f : I → I × I ; f is the space filling Hilbert curve. Does it follow from Corollary 3, Section 7.2, that I and I × I are homeomorphic? Justify!

7.4  Bolzano, Weierstrass, and Lebesgue A brief historical note: Bolzano proved in 1817 that every bounded sequence in n has a convergent subsequence. This is a special case of Corollary 2 below. Bolzano’s result was later generalized by Weierstrass. The notion of a Lebesgue number stems from Lebesgue’s work on measure theory, starting with his dissertation in 1902.

We continue investigating properties of compact spaces, or of spaces that are close to being compact spaces. A space X is a Bolzano–Weierstrass space, or a BW-space, if every infinite subset of X has an accumulation point in X (Illustration 7.23).

Illustration 7.23  BW-space: infinite sets have accumulation points.

Proposition 1. Suppose X is a Hausdorff space. Then, X is a BW-space if and only if it is countably compact.

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Proof. ⇒ Assume X is a BW-space and let {Fi : i = 1,2,3,…} be a family of closed subsets of X ∞

satisfying the finite intersection property. It suffices to show that ∩ Fi ≠ ∅ (Exercise 3, i =1

n

Section 7.3). Take an element xn ∈∩ Fi , n ∈+ . If the sequence ( xn ) stabilizes at x, then i =1

this x is in each Fi , and so it is in the intersection. Otherwise ( xn ) has infinitely many elements. Since X is a BW-space there is an accumulation point x for ( xn ). Every open set U around x contains infinitely many elements from ( xn ) (Exercise 20, Section 3.2) and so it intersects every Fi . Consequently, x is an accumulation point for each Fi . Since each Fi is ∞ ∞ closed, x ∈Fi for every i = 1,2,3,…. Hence, x ∈∩ Fi and we proved that ∩ Fi ≠ ∅ . i =1

i =1

⇐ Suppose X is countably compact, and let B be an infinite set. Then, B has a countably infinite subset A = {a1 , a2 ,…}. If A does not have an accumulation point in X, then neither does any of the sets Ak = {ak , ak +1 ,…} , k = 2, 3, …. Consequently, each Ak is closed. However these sets clearly satisfy the finite intersection property, and it is equally obvious ∞ that ∩ Ai = ∅ , contradicting the assumption that X is countably compact (and Exercise 3, i =1

Section 7.3). We did not need the assumption that X is Hausdorff to prove that countably compact implies BW. Since every compact space is countably compact, the following is then an immediate consequence. Corollary 2. Every compact space is a BW-space. We will see in a couple of propositions that for metric spaces the converse of Corollary 2 is also true. Proposition 3. For every open cover O of a metric BW-space (X, d) there is a positive number ε such that, for every x ∈ X , the open ball B(x, ε) is a subset of some member of O . Proof. Let X be a metric BW-space, and let O be an open cover of X. Assume that the conclusion of the proposition fails. This means that, however small a number δ we choose, there is a point x ∈ X , such that B( x , δ ) is not a subset of a member of O. In particular, for every δ = n1 , n ∈+ , there is a point xn ∈ X such that B ( xn , n1 ) is not a subset of a member of O. The set {xn  : n ∈+ } cannot be finite; otherwise, there is a number m ∈+ such that xn = xm for every n ∈{n1 , n2 ,… , nk ,…}, m < n1 < n2 < … < nk < … , which leads to the statement that B ( xm , n1 ) is not a subset of a member of O for every n ∈{n1 , n2 ,…, nk ,…}. This is not possible since xm must be in some open set in O , and since B ( xm , n1 ) : n ∈+ is a local basis at xm . So {xn : n ∈+ } is infinite. The assumption that X is a BW-space implies that {xn  : n ∈+ } has an accumu­lation point x ∈ X . Since O is a cover, there exists a U ∈O such that x ∈U . Since U is open, there

{

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is a ball B(x, r) that is a subset of U. Look at the ball B ( x , 2r ) (see Illustration 7.24). Since X is a metric space, B ( x , 2r ) must contain infinitely many members of the set {xn  : n ∈+ }. Choose any s such that 1s < 2r . Then (see Illustration 7.24), B ( x s , 1s ) ⊂ B( x , r ) ⊂ U ∈O , and ◼ we have a contradiction. The number λ = 2ε, where ε is as in Proposition 3, is called a Lebesgue number for the covering O of the compact space X. Explicitly, λ is a Lebesgue number for O if every subset of X with a diameter less than λ is contained within some member of O . We will make use of the existence of the Lebesgue number several times in Part 2 of this book.

r x

xs

U

Illustration 7.24  The radius

1 s

of the ball centered at x s is small enough so that this ball stays within B(x, r).

Proposition 4. A metric space is compact if and only if it is a BW-space. Proof. ⇒ This was settled in Corollary 2. ⇐ Let X be a BW-space, and let U be an open cover of X. By Proposition 3 there exists a Lebesgue number λ associated to U : every open ball B(x, λ) is a subset of some member U x of U . In the next two paragraphs we will find a finite subset {x1 , x 2 ,… , xm } of X such that { B( xi , λ ) : i = 1,2,…, m} is a cover of X. Then {U xi : i = 1,2,… , m} is a finite subcover of U , and so X is compact. Choose any x1 ∈ X . If there are no points in X out of B( x1 , λ ) then we can take m = 1 and we are done. Otherwise there is an x 2 ∉ B( x1 , λ ) ; so x1 and x 2 are at distance > λ . Suppose we have chosen {x1 , x 2 ,… , xn } such that for every j ≠ k , d( x j , x k ) > λ . If n

B( xi , λ ). {B(xi , λ ) : i = 1,2,…, n} is a cover for X, we are done. Otherwise there is an xn+1 ∉∪ i =1

Suppose this procedure never ends. Then we have an infinite sequence ( xi ) such that, for every j ≠ k , d( x j , x k ) > λ . Since X is a BW-space, there must be an accumulation point x for ( xi ) . Consider the ball B ( x , λ 2 ) : it contains infinitely many members of ( xi ). Any two of them are at distance less than λ, yielding a contradiction. Hence, the procedure must end after finitely many steps, and we have the desired finite cover of open balls. Corollary 5. A metric space is compact if and only if it is countably compact. Proof. That compact implies countably compact is clear. For the converse, suppose X is a countably compact metric space. Since metric spaces are Hausdorff, it follows from Proposition 1 that X is a BW-space. Proposition 4 now implies that X is compact.

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Exercises 1. Show that every closed subspace of a BW-space is a BW-space. 2. (a)  Check that the collection B = {{−n, n} : n ∈} is a basis for a topology T over . (b) Show that ( ,T ) is not a BW-space.



3. Let X be an uncountable space equipped with the co-countable topology, and let Y = [0, 1] be with the indiscrete topology. Show that X × Y is a noncompact BW-space. 4. Show that every uncountable space X equipped with co-countable topology is not a BW-space. 5. Let X be a BW-space, and let ( Fi ) be a sequence of closed, nonempty subsets of X such ∞ that Fi +1 ⊂ Fi for every i ∈+ . Show that ∩ Fi ≠ ∅ . i =1

6. Show that every compact metric space is complete. 7. A space X is sequentially compact if every sequence has a convergent subsequence.



(a) Show that if X is sequentially compact then it is a BW-space.



(b) Show that if X is first countable and a BW-space then X is sequentially compact.

8. A mapping p : X → Y is perfect if it is continuous, closed, surjective, and if for every y ∈Y , p −1 ( y ) is a compact subset of X. Prove that if p : X → Y is a perfect mapping and Y is compact, then X is compact. 9. Show that every compact subspace of the Sorgenfrey line is countable. [Hint: You may want to start with the identity map from the Sorgenfrey line onto the Euclidean line .] 10. Let H be the Hilbert metric space. Show that the unit closed ball ∞   D = ( x j )∞j =1 ∈ ∞ : ∑ xi2 ≤ 1 in H is not compact (despite being closed and i =1   bounded). [Hint: Show that the set ( x j )∞j =1 ∈ ∞ : x j0 = 1, and x j = 0 for j ≠ j0 does not have an accumulation point.]

{

}

7.5  Compactification A brief historical note: In 1913 Constantin Carathéodory (1873–1950) was the first to study compactifications of spaces (he worked on open subsets of  2).

We saw in Sections 7.1–7.4 that compact spaces have important properties. Hence, given a space X it makes sense to try to compactify it by embedding it into a compact space.

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Achieving just that is rather simple; in fact we can do more without much effort. As the following example shows we can easily extend X to a compact space Y such that X is dense in Y. Example 1 Start with a space X and choose a point p not in X. Define a topology over Y = X ∪ { p} as follows: open sets in Y are Y and the open sets in X. Then Y is a compact space: any cover of Y must contain Y as a member (since that is the only open set containing p), and so {Y} would be a finite subcover. It is also clear that X (with the original topology) ☐ is a dense subspace of Y. The plainness of this example hints at the necessity of upgrading the properties of the target space Y. Observe that Y in Example 1 is never Hausdorff: recall that we have assumed that our spaces are not empty, and so X has a point that cannot be separated from p with two disjoint open sets. We do not want such an ugly compact extension Y of X. However, it is a good start. A compact space Y is a compactification of a space X if the following two conditions are satisfied: (i) There is an embedding e : X → Y . (ii) e(X) is dense in Y. We are looking for a procedure that will associate Hausdorff compactifications under some reasonable conditions. Before we present one, let’s look at our blueprint model again. Example 2 NP The space X is the circle without the point NP (the north pole). Consider the homeomorphism from X onto  P4 P1 defined by the stereographic projection f (Illustration 7.25). If we simply P2 add the north pole to X we clearly get P3 a compactification S1 = X ∪ {NP} f (P2) f (P3) f (P4) of X. We want to add a new point f (P1) to  such that that the extension Illustration 7.25 f : S1 →  ∪ {the new point} of f is also a homeomorphism (so that  ∪ {the new point} also becomes a compact space). Since we may think of the mapping f as sending the north pole to “infinity,” we will denote that new point by ∞ and we will call it an infinity point. The set  ∪ {∞} will be denoted by  ∞ . As we have pointed out, we want it to be homeomorphic to the circle via f . So, the set of open subsets of  ∞ should be { f (U ) : U  open in S1 }.

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There are two types of open sets in S1 : open sets containing NP and open sets not containing NP. In the latter case, f operates the same way as f, so that the open sets in  are also open in  ∞ . In the former case, as is easy to see, the complement of an open set containing NP is mapped via f onto the complement in  of a closed and bounded subset of  , so that f maps such sets into complements in  ∞ of closed and bounded subsets of  . Consequently, open subsets  ∞ containing the infinity ☐ point are complements in  ∞ of compact subsets of  . We now promote Example 2 into a general recipe for obtaining a special compactification. Suppose X is noncompact, and ∞ a point not in X. The one-point compactification of X, denoted X ∞ , is the space over X ∪ {∞} with topology specified as follows: (a) ∅ and X ∪ {∞} are open sets. (b) Every open subset of X is open in X ∪ {∞}. (c) A subset A of X ∪ {∞} containing the point ∞ is open if and only if X \ A is a closed and compact subset of X. If X is Hausdorff, then stipulating that X \ A is closed is redundant, since compact subsets of Hausdorff space are closed. Example 3 The one-point compactification of  2 is homeomorphic to the sphere S 2 , and the stereographic projection (see Example 7, Section 3.5) extended by sending the north pole of the sphere to the ∞-point is a homeomorphism from S 2 onto  2∞ . As emphasized in Illustration 7.26 the open sets in S 2 containing the North Pole correspond to complements in  2∞ of compact (and closed) subsets of  2 . U

S2 \ U

f (S 2 \ U) f (U)

Illustration 7.26  The dark-shaded open set U in S 2 contains the North Pole. The correspond-

ing open set in  2∞ of matching shade is the complement of the compact set f (S 2 \ U ), where f is the stereographic projection.

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The one-point compactification of an open disk in  2 is also homeomorphic to S 2 (see ☐ Exercise 2). The following proposition justifies the terminology used in the last definition. Proposition 1. For every noncompact space X, the one-point compactification X ∞ is a compactification of X. Proof. We leave the proof that X ∞ is indeed a topological space as an exercise (Exercise 1). X is a subspace of X ∞ (i.e., the inclusion X → X ∞ is an embedding): We check that a subset U of X is open in X if and only if it is V ∩ X for some open subset V of X ∞ . It follows from the definition of the topology over X ∞ that if U is open in X then it is open in X ∞ , and so we may take V = U . Conversely, we show that for every open subset V of X ∞ , V ∩ X is open in X. Open subsets V of X ∞ are either open subsets of X (and so their intersection with X is open in X), or they are such that X \V is closed (and compact) in X (and so X \( X \ V ) = X ∩ V is open in X ). X is dense in X ∞ : The set {∞} is the only nonempty subset of X ∞ that does not intersect X. Since X is not compact, {∞} is not open in X ∞ . It follows that X is dense in X ∞ . X ∞ is compact: Let O be an open cover of X ∞ . So, some member O of O contains the point ∞. By assumption O c is compact in X. By Proposition 2, Section 7.1, O c is compact in X ∞ . So, the open cover O of O c has a finite subcover O ′ . Then, O ′ ∪ {O} is a finite subcover of O covering X ∞ . In order to achieve Hausdorff-ness of the one-point compactification we need to pay attention to a local version of compactness. A space X is locally compact at a point x ∈ X , if there is an open neighborhood U of x such that the closure U is compact. A space X is locally compact if it is locally compact at every point of X. Obviously every compact space is locally compact. (Why?) The converse is obviously not true. (Why? Exercise 9.) Proposition 2. If a space X is locally compact at a point x then there is a compact set K and an open set U such that x ∈U ⊂ K . For Hausdorff spaces X the converse is also true.

x

U K

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Illustration 6.27  Locally compact: every point x within a compact set K possesses an open neighborhood U that is a subset of K.

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7.5  Compactification   ◾    177  

Proof. ⇒ Suppose that there is an open neighborhood of x such that its closure is compact. Take K to be that closure. ⇐ Let K be a compact set containing x and let U be open in X such that x ∈U ⊂ K . Since K is compact in X and since X is Hausdorff, K must be closed. So the closure U of U is a subset of K. U is compact since it is a closed subset of the compact space K. Example 4 The space  is locally compact. However, the subspace  is not. Suppose other­ wise. Take any point x in . Then there is an open subset U of  and a compact subspace F of  such that x ∈U ⊂ F . Since U is open in  , there must be an open interval (a, b) in  such that x ∈(a, b) ⊂ U . Take any two rational numbers p < r in (a, b) and consider the closed interval [p, q] (with only rational numbers in it). Since it is closed in  and since F contains [p, q], it follows that [p, q] is closed in F. But F is compact, and closed subsets of compact spaces are compact, so that [p, q] is compact. We now show that this is not true: take any irrational point y in [p, q] and consider all closed intervals [ y − ε , y + ε] that are completely within [p, q] (so, we should take ε to be small enough). This set clearly satisfies the finite intersection property. But the intersection of all of them is empty (keep in mind that y is irrational). So, [p, q] is not compact and we have arrived at a contradiction. ☐ Consequently,  is not locally compact. We now prove that in order to obtain a compactification that is a Hausdorff space, the starting space X must be Hausdorff and locally compact. Proposition 3. A space X is locally compact and Hausdorff if and only if the one-point compactification X ∞ is Hausdorff. Proof. ⇒ Suppose X is Hausdorff and locally compact. Take any two points x and y in X ∞ . The only interesting case is when one of them is ∞; so, assume y = ∞. Since X is locally compact there is an open neighborhood U of x such that U is a compact subset of X. Then U and U c ∪ {∞} are two disjoint and open (in X ∞ ) neighborhoods of x and ∞, respectively. ⇐ Suppose X ∞ is Hausdorff. Since being Hausdorff is hereditory, it follows that X is Hausdorff, too. We check that it is locally compact. Let x ∈ X . Since X ∞ is Hausdorff, we can separate x and ∞ with two open sets, U x and U ∞ , respectively. Then X \ U ∞ is compact in X (by the definition of open sets around ∞), and it contains U x . The conclusion now follows from Proposition 2. Example 5 The one-point compactification  ∞ of  is not Hausdorff since  is not locally ☐ compact (Example 4). In Chapter 8 we will see another important compactification procedure: the Čech–Stone compactification.

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Exercises 1. Show that the one-point compactification X ∞ of a space X is a topological space. 2. Show that if X and Y are homeomorphic, then so are their one-point compacti­fi­cations. 3. Let X be the subspace of the unit circle S1 consisting of the points in S1 with rational polar angles (in degrees), and let Y = X \ {(0, 1)}. Is X homeomorphic to the onepoint compactification of Y? 4. (a)  What familiar space is (homeomorphic to) the one-point compactification of the interval [0,1)? (b) Visualize the one-point compactification of the union of two intervals (0,1) ∪ (1, 2). (c) Visualize the one-point compactification of the quotient space of (0,2) ∪ (3,5) obtained by identifying the points 1 and 4. (d) Visualize and describe the structure of the one-point compactification of the subspace of  obtained from  by deleting all integers. 5. Find the one-point compactification of a discrete space X with countably many elements. Identify a subspace of  which is isomorphic to the one-point compactification of  . 6. Denote the set of nonnegative real numbers by  ≥0 , and its one-point compactification by  ≥∞0. Attach the line segment {x } × [0, 1] at every x ∈ ∪ {∞} by identifying x with {x } × {0}. Is the resulting space homeomorphic to the topological comb? Justify your answer.

7. Show that if X is a subspace of a compact space Y, then there is a compact subspace Z of Y such that X ⊂ Z and such that X is dense in Z.

8. (a)  Show that if X is a closed subspace of Y, and the point ∞ is not in Y then X ∞ is a subspace of Y∞ . (b) Show that if the assumption that X is closed in Y is dropped then the conclusion fails. 9. Show that a locally compact space need not be compact. 10. (a)  Show that each compact space is locally compact. (b) Show that finite products of locally compact spaces are locally compact. 11. (a)  S how that a closed subset of a locally compact space is locally compact. [Hint: You might need Exercise 16, Section 4.1.] (b) Show that an open subset of a locally compact space need not be locally compact.

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7.6 Infinite Products of Spaces and Tychonoff Theorem   ◾    179  

7.6 Infinite Products of Spaces and Tychonoff Theorem A brief historical note: Tychonoff proved in 1930 that for every n ∈+ , the cube [0,1]n is compact. The statement of Theorem 2 appears for the first time in his paper published in 1936.

Our goal in this section is to show that (any) product of compact spaces is compact (Tychonoff’s theorem). The following theorem, which will be used in the proof of Tychonoff’s theorem, is interesting in its own right. Theorem 1. (The Alexander Subbasis Theorem) A space X is compact if and only if there is a subbasis S for X such that every cover of X by (some of the) elements in S has a finite subcover. Proof. ⇒ If X is compact, then we can take S to be the whole topology over X. ⇐ Suppose there is a subbasis S for X such that every cover of X by elements in S has a finite subcover. Suppose X is not compact. Then, there is a collection O of open sets covering X such that no finite subcollection of O covers X. Let U be the family of all open covers O ′ of X such that O ⊂ O ′ and such that O ′ possesses no finite subcovers of X. Since O ∈U ,U is not empty. Suppose now that K is a chain of elements of U (ordered under inclusion); so, each element C of K is a cover of X that has no finite subcover, and such that O ⊂ C . Then the union of the elements of K is a collection of open sets that certainly covers X. We now show that it is also in U . Assume that union has a finite subcover {V1 , V2 ,… , Vm } . Then each Vk is in some member of Cik ∈K . Since the members of K are ordered under inclusion, the set {V1 , V2 ,… , Vm } must be a subset of one of Ci1 , Ci2 ,… , Cik , which would thus have a finite subcover. We have a contradiction. Consequently, the union of the elements of K does not have a finite subcover, and so it is in U . So, each chain of elements of U is bounded from above by a member of U . By Zorn’s lemma, there exists an M ∈U such that O ⊂ M , and such that M is a maximal (under inclusion) element of U . Suppose now that M ∈M , and x ∈ M . Since M is an open set and since S is a subbasis, there are elements S1 , S2 ,… , Sn of S such that x ∈S1 ∩ S2 ∩ … ∩ Sn ⊂ M . Suppose that each of S1 , S2 ,… , Sn is not a member of M. Then M ∪{Sk } is a proper overset of M for every k = 1, 2, …, n. Since M is a maximal cover that has no finite subcover, it follows that M ∪{Sk } must have a finite subcover. So, for every such k, there is a finite subcollection M k of M such that M k ∪{Sk } is a finite cover of X. We claim that M1 ∪ M 2 ∪ … ∪ M k ∪ … ∪ M n ∪ {S1 ∩ S2 ∩ … ∩ Sk ∩ … ∩ Sn } covers X. This is true since any element that is not in Sk must be in the union of the elements of M k (since M k ∪{Sk } covers X) and so the elements out of S1 ∩ S2 ∩ … ∩ Sn are in the union of the sets in M1 ∪ M 2 ∪ … ∪ M k ∪ … ∪ M n . Since M1 ∪ M 2 ∪ … ∪ M k ∪ … ∪ M n ∪ {S1 ∩ S2 ∩ … ∩ Sk ∩ … ∩ Sn } covers X and since S1 ∩ S2 ∩… ∩ Sn ⊂ M , it follows that M1 ∪ M 2 ∪ … ∪ M k ∪ … ∪ M n ∪{ M } covers X, too. We

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180   ◾    Compactness and Related Matters

get a finite subcover of M , contradicting our assumption. So, at least one of S1 , S2 ,… , Sn is a member of M . We have shown that every element in X is contained in a set Sk ∈M ∩ S . So, M ∩ S is an open cover of X. Since it is a part of our subbasis, it follows that it has a finite subcover. That finite subcover is also a finite subcover of M . This is a contradiction. So, X is compact. Theorem 2. (Tychonoff Theorem) If X j , j ∈ J , is a family of compact spaces, then ∏ X j j ∈J is also compact. Proof. Let X j , j ∈ J , be compact spaces, and denote X = ∏ X j . According to the j ∈J

Alexander Subbasis Theorem, in order to show that X is compact it suffices to choose one subbasis and extract a finite cover out of any cover of X made of some of the elements of that subbasis. Not surprisingly we choose the standard subbasis S , consisting of the sets of type p −j 1 (U j ), where U j is open in X j and p j : X → X j is the projection. So we start with a cover O of X consisting of such sets. Assume that O does not allow extraction of a finite subcover. This will lead to contradiction. Consider the set O j = {U j ⊂ X j  :  p −j 1 (U j ) ∈O } . It follows from the definition of O that the sets of O j are all open in X j . Assume O j covers X j for some j ∈ J . Then, since X j is compact, we can extract a finite subcover O jfin . It is then clear that p −j 1 (U j ) : U j ∈O jfin covers X, and that it is a finite subcover of O . This contradicts the assumption that O does not allow extraction of a finite subcover. Assume O j does not cover X j for every j ∈ J . So, there is an x j in each X j that is not covered by O j . This implies that x = ( x j ) j ∈J is an element in X = ∏ X j out of the cover O , j ∈J contradicting the assumption that O covers X.

{

}

The Tychonoff theorem is one of the primary reasons why the product topology is in a way superior to the boxed product topology; the boxed product topology of compact spaces need not be compact (Exercise 2). As ∞the first application of this theorem we will now show that the product space ∞  = ∏  is metrizable. i =1

Example 1:  ∞ Is Metrizable Since subspaces of metric spaces are metric spaces (through the inherited metric), it suffices to show that  ∞ can be embedded into a metric space. Since  is homeomorphic to ∞the interval (0,1), it follows that  ∞ can be embedded into the product space I ∞ = ∏[0,1], called the Hilbert cube. So, it suffices to embed I ∞ into a metric i =1 space. Recall the Hilbert space H, introduced in Example 2, Section 5.2, consists of all ∞ converges in  . We sequences ( xn ) for which the sequence x12 + x 22 +  + xn2 ∞

2

(

)

showed there that d( x , y ) = ∑ ( xi − yi ) is a metric over H.

n=1

i =1

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7.6 Infinite Products of Spaces and Tychonoff Theorem   ◾    181  

x  x  Define f : I ∞ → H by f ( x1 , x 2 ,… , xn ,…) =  x1 , 2 ,… , n ,… . This is well defined   n 2 ∞  2 x2 2 xn 2  since  x1 + 2 +  + n  converges when xi ∈[0,1], i = 1, 2, …. It is clear   n=1 ∞ that f ( I ∞ ) = ∏ [ 0, 1i ] and that f is one-to-one. Since f is continuous (next paragraph), i =1 since H is Hausdorff, and since I ∞ is compact (Tychonoff’s theorem), it follows from Corollary 3, Section 7.2 that f is a homeomorphism from I ∞ onto a subspace of H, and so I ∞ is metrizable. It remains to be shown that f is continuous. Choose any open set U in f ( I ∞ ) and any point a = ann in U. Notice that f (a1 , a2 ,… , an ,…) = a . Since U is open, there is an open ball B(a , r ) contained in U. It consists of all z1 , z22 ,…, znn ,… , zi ∈[0,1] such that

( )

( )

( )

∞

∑

i =1

(

ai i

− zii

∞

)

2

(

∞

< r . Choose m large enough such that

∑

i =m

)

< 2r . This can be done

1 i2

since ∑ i12 converges. Choose intervals (ai − εi , ai + εi ) , i = 1, 2, … , m − 1, small enough i =1

such that

m−1

∑

i =1

(

ai i

− zii

)

2

< 2r for all z i ∈(ai − εi , ai + εi ) , i = 1, 2, … , m − 1. This is doable

since the last sum is finite. Then for every ( z1 , z 2 ,… , zn ,…) ∈(a1 − ε1 , a1 + ε1 ) × ∞

× (am−1 − εm−1 , am−1 + εm−1 ) × I × I  we have  ∑ < 2r + 2r = r .

We

proved

that

the

open

i =1

(

ai i

− zii

)

2

≤

neighborhood

m −1

∑

i =1

(

ai i

− zii

)

2

∞

+∑

i =m

1 i2

(a1 − ε1 , a1 + ε1 ) ×

× (am−1 − εm−1 , am−1 + εm−1 ) × I × I  of (a1 , a2 ,… , an ,…) is sent into B(a , r ) ⊂ U via f, thus establishing that f is continuous at (a1 , a2 ,… , an ,…) . ☐ The versatility of Tychonoff’s theorem is illustrated in the next example, where we enter the territory of pure combinatorics. Example 2. Ramsey Theorem (Finite Version) A k-coloring of a set A by a set of colors {c1 , c2 ,…, ck } is a mapping c : A → {c1 , c2 ,…, ck }. A subset B of A is monochromatic if all elements of B are colored with one color. Precisely, B is monochromatic if there exists j ∈{1, 2, … , k}, such that c( B) = {c j }. For a set A and a fixed integer r, 0 ≤ r ≤ A , we will denote the set of all subsets C of A such that C = r by A(r ) . In this example we discuss colorings of A(r ) . Our starting point is the infinite version of Ramsey’s theorem. Theorem 3. (Ramsey Theorem; infinite case) For every k-coloring of (r ) , there is an infinite subset M of  such that the subset M (r ) of (r ) is monochromatic. It is not difficult to prove Theorem 3 using induction; see, for example [5], ◼ page 157. In the finite version of Ramsey’s theorem we consider k-colorings of sets {1,2,… , n}(r ) . Given n, k, r, and m, there may be a coloring of {1,2,…, n}(r ) such that for no subset M of {1, 2, … , n} with M = m , the subset M (r ) of {1,2,… , n}(r ) is monochromatic. One

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182   ◾    Compactness and Related Matters 1

such example, for k = 2, n = 5, m = 3 and r = 2, is given in Illustration 7.28: the elements of {1,2,3,4,5}( 2) are represented by “colored” edges, one set of edges colored with a fill line, and the rest of the edges colored with a dashed 5 line. It is easy to see that for no three-element subset M of {1, 2, 3, 4, 5}, M ( 2) is monochromatic. In terms of the colored edges, no subgraph of the graph shown in Illustration 7.28 that is spanned by three vertices, has all the edges in 4 3 one color (where “spanned” means that we include all of Illustration 7.28 the edges joining the chosen vertices). The finite version of Ramsey’s theorem tells that, for fixed k, r, and m, there is a number n which precludes such a case. Theorem 4. (Ramsey Theorem; finite case): For every k, r, and m, there is an n such that, for every k-coloring of {1,2,… , n}(r ) , there is a subset M of {1,2,…,n} such that M = m , and such that M (r ) is a monochromatic subset of {1,2,… , n}(r ).

1

2

2

6

3

For example, when k = 2 (two colors), r = 2 (we are coloring subsets of size 2) and m = 3, we can take n = 6 : however we color the edges of the graph Γ shown in Illustration 7.29, there must be a 5 4 three-element subset M of {1, 2, 3, 4, 5, 6} such that Illustration 7.29 the set of edges of the subgraph Ζ of Γ spanned by the vertices in M is monochromatic. In the coloring we show in Illustration 7.29 we can choose Ζ to be with the set of vertices being either {2, 3, 5} or {1, 2, 4}. Proof (based on [5]). In the first two paragraphs we introduce the notation and make some preliminary observations. The actual proof of the theorem is in the last paragraph. A k-coloring of (r ) is a mapping (r )  {c1 , c2 ,… , ck }, which in turn is an element of X = ∏ Xi , where Xi = {c1 , c2 ,… , ck } for every i ∈(r ). Equip {c1 , c2 ,… , ck } with the i ∈( r )

discrete topology. Since it is a finite space, it is compact. By Tychonoff ’s theorem, so is the space X. A ≥ r . The set Let A be a finite subset of , such that (r ) (r )  C A = {ϕ ∈ X  :  A is a monochromatic subset of with respect to the coloring ϕ} is       the set   ∏ {c1 } × ∏ Xi  ∪ ∪   ∏ {ck } × ∏ Xi   , where all A(r )-coordinates  i ∈A( r )  i ∈( r )  i ∈A( r )  i ∈( r )     i ∉A( r ) i ∉A( r ) are equal. Clearly, each C A is open in X. Suppose the statement of Theorem 4 fails. This means that there are k , r , m ∈+ such that for every n ∈+ , n ≥ r , there exists a k-coloring of {1,2,… , n}(r ) such that for every M ⊂ {1, 2, … , n}, M = m , the set M (r ) is

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7.6 Infinite Products of Spaces and Tychonoff Theorem   ◾    183  

not monochromatic. It  follows that (for such k , r , m ∈+ ), for every finite subset F of  with at least m elements, F (r ) is not monochromatic. Consequently, for every A1 , A2 ,…, Ap such that A1 = A2 =  = Ap = m , the set C A1 ∪ C A2 ∪… ∪ C Ap cannot be the entire space X. Since X  is compact, it follows that ∪ C A ≠ X , and hence ∩ (C A )c ≠ ∅ . Notice that (C A )c is the set |A|=m

|A|=m

{ϕ ∈ X : A(r ) is not a monochromatic subset of (r ), with respect to the coloring ϕ}. Then,  for every coloring ψ ∈ ∩ (C A )c ≠ ∅ of (r ) , there is no subset B of  |A|=m

such that | B| = m and such that B(r ) is monochromatic. This contradicts Theorem 3. ☐ Exercises 1. Prove the converse of the Tychonoff theorem: If ∏ X j is compact, then so is every X j , i ∈J j ∈J. Box

2. Find compact spaces X j , j ∈ J , such that ∏ Xi is not compact. i ∈I

3. Show that an infinite and countable discrete space X is never equal to the product space ∏ Y j , where each Y j is not homeomorphic to X. j ∈J

4. Use the Tychonoff theorem to prove that the Cantor set is compact. 5. Consider the Tychonoff cube X = ∏ I j , where each I j is a copy of I = [0, 1]. Let H j ∈I

be the subspace of X consisting of all nondecreasing functions I → I (H is the Helly space). Show that H is compact. [Hint: Exhibit H as f (X) for a continuous f : X → X .] 6. Consider the space ∏ X j , where each X j is a fixed compact space X. Prove that the j ∈J

diagonal D = {( x ) j ∈J :  x ∈ X } is compact in ∏ X j . j ∈J

7. Let Xi , i ∈ I , be nonempty sets, let yi ∉ Xi for every i ∈ I , and define topologies over Yi = Xi ∪ { yi } , i ∈ I , by declaring open sets to be {∅, Xi , { yi }, Yi } . Denote Y = ∏ Yi , and let pi : Y → Yi , be the projections. i ∈I

{

}

(a) Use the Tychonoff theorem to show that C = pi−1 ( yi ) : i ∈I is not a cover of Y. (b) Deduce the Axiom of Choice; that is, show that there exists f : I → ∪ Xi , such i∈I that f (i ) ∈ Xi for every i ∈ I .

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Chapter

8

Separation Properties

I

n a metric space every two disjoint closed subsets are respective subsets of two disjoint open sets (Exercise 10, Section 2.1). Using alternative terminology, in metric spaces every two closed subsets can be separated by two disjoint open subsets. Spaces satisfying that property (with some additional constraints to be discussed later) are called normal spaces. As we will see, not every normal space is metrizable. Nevertheless, normal spaces have fine enough structure and possess important properties. For example, we will prove later that any continuous map into  defined on a closed subset of a normal space can be extended over the whole space (Tietze extension theorem). In this chapter we will consider spaces in which various types of disjoint sets can be separated (in some way) by open sets. We will start from rough topologies and introduce more and more refining properties leading to near-metric spaces.

8.1  The Hierarchy of Separation Properties A brief historical note: Andrey Kolmogorov (1903–1987) introduced T0 -spaces around 1930; T1 spaces were first defined in 1907 by Frigyes Riesz (1880–1956). These classes of spaces are sometimes—but lately rather rarely—named after their inventors. On the other hand, T2 -spaces are commonly known as Hausdorff spaces; Hausdorff introduced them in 1914. Regular spaces were defined by Vietoris in 1921; normal spaces were introduced by Tietze in 1923, and independently by Alexandroff and Urysohn in 1929.

We start by listing the (tentatively called) main separation properties. The first one will be used infrequently. A space X is said to be a T0 -space if for every x , y ∈ X there are open sets U and V such that x ∈U , y ∈V and such that y ∉U or x ∉V .

185

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186   ◾    Separation Properties

U y

V

Illustration 8.1  T0 -space: x ∈U , y ∈V ; how-

ever, at least one of U and V does not contain y or x. In this illustration we have x ∉V .

x

The discrete topology over any X will be a universal—albeit, rather uninteresting— example of a space possessing this, and every other separation property in our emerging list. A less trivial example follows. Example 1 Let X = {a , b} and T = {∅ , {a}, X }. Then, as is very easy to see, this is a simple T0 -space. ☐ A space X is said to be a T1-space if for every x , y ∈ X there are open sets U and V such that x ∈U , y ∈V and such that y ∉U and x ∉V . U V x

y

Illustration 8.2  T1 -space: x ∈U , y ∈V , x ∉V , y ∉U . Note

that U and V need not be disjoint.

It is obvious that every T1-space is a T0 -space. On the other hand, the T0-space described in Example 1 is not a T1-space. Consequently, the class of T1-spaces is a proper subclass of the class of T0 -spaces. Example 2 Let X be an infinite set equipped with the co-finite topology. Then X is a T1-space: for every x , y ∈ X , the sets U = X \{ y } and V = X \{ x } perform as in the above definition ☐ of a T1-space. T1-spaces are important because of the following simple result. Proposition 1. A space X is a T1-space if and only if each singleton in X is closed. Proof. ⇒ Suppose X is a T1-space and take any x ∈ X . Then it follows straight from the definition of T1-spaces that every point in { x }c is an interior point for { x }c , so that { x }c is open, and thus {x} is closed. ⇐ Assume all singletons are closed and take any two distinct elements x , y ∈ X . Then, { x }c and { y }c are open, x ∈{ y}c , y ∈{x }c , and y ∉{ y}c and x ∉{x }c .

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8.1  The Hierarchy of Separation Properties   ◾    187  

Next, we meet an old acquaintance of ours, under a different name. A space X is said to be a T2 -space if for every x , y ∈ X there are disjoint open sets U and V such that x ∈U , y ∈V . What we are now calling a T2 -space was earlier called a Hausdorff space. We will mostly stick with the old terminology except where it is convenient to talk about Ti -spaces in general, for various choices of the index i. U

V x

Illustration 8.3  T2 -space: x ∈U , y ∈V , and U and V are disjoint.

y

It is plain that every T2 -space is a T1 -space. That the converse is not true is shown in Example 2 above. Hence, the class of T2 -spaces is a proper subclass of the class of T1 -spaces. (For another T1 -space that fails to be a T2 -space see Exercise 5.) The next example will be of a T2 (Hausdorff) space which is not a “ T3 ”-space (the definition follows soon). Example 3 Consider the half-disk topology over  2up = {( x , y ) :  y ≥ 0} (Example 7, Section 3.3). It ☐ is clear that it makes  2up a Hausdorff space. It is convenient at this stage to introduce the following terminology: Given two disjoint subsets A and B of a space X, we will say that two disjoint open sets U and V separate A and B, respectively, if A ⊂ U and B ⊂ V . This terminology is compatible with the one used in relation to connectedness of spaces (Exercise 3). A space X is a regular space if for every x ∈ X and every closed subset F of X, there are disjoint open sets U and V such that x ∈U and F ⊂ V . It is a T3-space if it is regular and a T1-space.1 V

U x

F

Illustration 8.4  T3 -space or regular space:

separating points and closed sets with disjoint open sets.

1

There is not much consistency in the literature regarding this terminology; sometimes “regular = T3,” sometimes “regular = T3 + T1”; in this book we will stick with “regular + T1 = T3.”

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188   ◾    Separation Properties

Regular spaces allow us to separate every point and every closed set not containing that point with two disjoint open sets. The stipulation that X be T1 is needed to establish the following proposition. Proposition 2. Every T3 -space is a T2 -space. Proof. Left as an easy exercise (Exercise 6). As announced, Example 3 furnishes a space that is T2 but not T3 . It is not T3 since a point (a, 0) on the x-axis and the closed set {( x ,0) ∈ 2 :  x ≠ a} cannot be separated by two disjoint open sets. This, together with Proposition 2, implies that the class of T3 -spaces is a proper subclass of the class of T2 -spaces. The next example is of a T3 -space. Example 4 The Sorgenfrey topology over  makes it a T3 -space. This claim is easy to justify. First note that this topology makes  a T1-space. For the rest, start by taking a closed subset F of  and a point a out of F. There must be an open set (open in the Sorgenfrey topology, of course) [a, b) that is disjoint from F, for otherwise we would have a ∈ F ′, and so a ∈ F = F. The complement [a, b)c is ( −∞ , a ) ∪[b , ∞ ), and it is open in the Sorgenfrey topology. Then U =[a , b ) and V = ( −∞ , a ) ∪[b , ∞ ) is a separation of a and ☐ F by two disjoint open sets. At the top of our hierarchy of Ti -spaces are the T4 -spaces. A space X is said to be a normal space if for every two disjoint closed subsets F and G of X, there are disjoint open sets U and V such that F ⊂ U and G ⊂ V . The space X is a T4 -space, if in addition to being normal, it is a T1 -space. V

U F

G

Illustration 8.5  T4 -space or normal space:

separating disjoint closed sets with disjoint open sets.

The stipulation that T4 -spaces are T1 makes it obvious that every T4 -space is a T3 -space. As we saw in Example 4, the Sorgenfrey topology makes  a T3 -space. We will prove later that it is also a T4 -space. For the time being we will not show an example of a T3 -space that is not a T4 -space; for that we need to work harder (Example 2, Section 8.2). Illustration 8.6 depicts the hierarchy of the various types of spaces introduced in this section; the question marks indicate instances where strict inequality has not yet been established.

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8.1  The Hierarchy of Separation Properties   ◾    189  

T0 T1 T2

T3 ? T4 ?

metric

Illustration 8.6

The next two theorems establish the fact that Ti -properties, i = 0, 1, 2, 3, are hereditary and preserved under products of spaces. Observe that T4 spaces are excluded; we will justify that exclusion in the next section. Theorem 3. Subspaces of Ti -spaces are Ti -spaces, i = 0, 1, 2, 3. Proof. We leave it as an exercise (Exercise 7). Theorem 4. If X j , j ∈ J are Ti -spaces (i ∈{0, 1, 2, 3}), then so is ∏ X j . j ∈J

Proof. We deal only with i = 3; the rest is left as an exercise (Exercise 8). Suppose X j , j ∈ J are all regular spaces, and consider X = ∏ X j . It follows from Exercise 8 j∈J

that, since each of X j is T1 ( j ∈ J ), the same is true for X. Let x ∈X , and let F be a closed

subset of X not containing x. Then F c is an open set containing x, and hence there is a standard basis element U for X such that x ∈U ⊂ F c . The idea of the rest of the proof is outlined in Illustration 8.7 and its caption. V2

V1

x

F

U

Illustration 8.7  We will separate x and U c with

the disjoint open sets V1 and V2 , thus achieving a separation of x and F by the same open sets.

Uc

The open set U is of the type U = ∏ U j where all but finitely many U j coincide with the j ∈J

respective spaces X j . Since F ∩ U = ∅, U is not all of X, and so there is j0 ∈ J such that U jo ≠ X j0 . Since X j0 is regular, we can separate the j0 -th coordinate x j0 of x and U cj0 with two disjoint open sets W1 and W2 , respectively. Denote, as usual, the projection

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190   ◾    Separation Properties

X → X j0 by p j0 . Then V1 = p −j01 (W1 ) and V2 = p −j01 (W2 ) are two disjoint open sets separating x and U c ( x ∈V1 , U c ⊂ V2 ), and since F ⊂ U c , these two open sets separate x and F, respectively. As hinted by the last two theorems, T4 -spaces are exceptional compared to the other Ti -spaces, i = 0,1, 2, 3. We will examine them more carefully in the following sections. Exercises 1. Show that X is a T0 space if and only if for every distinct x , y ∈ X , {x } ≠ { y}. 2. Show that the number of T0 -topologies over a set with n elements is at least 2n . 3. Let A and B be two disjoint subsets of a space X. Show that there are U and V that separate A and B if and only if {A, B} is a separation of the subspace A ∪ B of X. 4. Show that every finite T1 space is discrete. 5. Show that the Zariski topology over n is T1 but not T2 . 6. Prove Proposition 2: Show that every T3 -space is Hausdorff.

7. Prove Theorem 3: Subspaces of Ti -spaces are Ti -spaces, i = 0, 1, 2, 3.

8. Prove Theorem 4: Show that if X j , j ∈ J are Ti-spaces (i ∈{0 , 1, 2}), then so is ∏ X j . j ∈J

9. (a)  Find a normal space X and a quotient space Y = X ~ such that Y is not Hausdorff. (b) Find a normal space X and a quotient space Y = X ~ such that Y is not a T0 -space. 10. Show that every subspace of a space X is normal if and only if every open subspace of X is normal. 11. (a) Show that if for every two distinct elements x , y ∈ X there is a continuous function f : X →  such that f ( x ) = 0 and f ( y ) = 1 then the space X is Hausdorff. (b) Show if X is a T1 -space and if for every x ∈ X and every closed subset F of X not containing x there is a continuous function f : X →  such that f ( x ) = 0 and f ( F ) = {1}, then X is a T3 -space. (c) Show that if X is a T1-space and if for every two disjoint closed subsets F and G of X there is a continuous function f : X →  such that f ( F ) = {0} and f (G ) = {1} then X is a T4 -space. [Note: The converses of (a) and (b) are not true. The converse of (c) is a major theorem—Urysohn’s lemma—and will be proven later.]

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8.2 Regular Spaces and Normal Spaces   ◾    191  

8.2 Regular Spaces and Normal Spaces A brief historical note: The name T3 1 -spaces for completely regular spaces, introduced by 2

Tychonoff in 1930, was a joke, that somehow became semi-standardized terminology.

The following useful theorem gives us two new criteria for normality. Theorem 1. The following statements are equivalent. (i) X is a normal space. (ii) For every closed subset F of X and every open subset U of X such that F ⊂ U , there is an open subset W of X such that F ⊂ W ⊂ W ⊂ U . (iii) For every two disjoint closed subsets F and G of X there are two open sets U and V such that F ⊂ U and G ⊂ V , and such that U ∩ V = ∅ .

U

X

F W W

Illustration 8.8  Part (ii) of Theorem 1.

Proof. (i) ⇒ (ii) Assume X is a normal space, and choose a closed subset F of an open set U. Then F and U c are disjoint closed sets. By the normality of X there are two disjoint open sets V and W such that F ⊂ W and U c ⊂ V (see Illustration 8.9). W

V

F

U

Uc

Illustration 8.9

Since W ∩ V = ∅, we have that W ⊂ V c . Hence, W ⊂ V c . Since V c is closed V c = V c , and so F ⊂ W ⊂ W ⊂ V c ⊂ U , where V c ⊂ U since U c ⊂ V .

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192   ◾    Separation Properties

(ii) ⇒ (iii) Take any two disjoint closed sets F and G. By (ii) there is an open set U such that F ⊂ U ⊂ U ⊂ G c . The inclusion U ⊂ G c implies G ⊂ (U )c , and we can use (ii) again to claim existence of an open set V such that G ⊂ V ⊂ V ⊂ (U )c . Since V ⊂ (U )c , it follows that U and V are disjoint, and so the open sets U and V we have found satisfy the requirements of (iii). (iii) ⇒ (i) is obvious. An analogous statement is true for regular spaces. In the next theorem we make this more explicit. Theorem 2. A space is regular if and only if for every x ∈ X , and for every open subset U of X containing x, there is an open subset W of X such that x ∈W ⊂ W ⊂ U . Proof. The proof of this theorem is similar to the corresponding part of the proof of Theorem 1; so we leave it as an exercise (Exercise 2). In the next theorem we establish that the difference between being regular and normal is not more than being Lindelöf. Theorem 3. If a space is regular and Lindelöf, then it is normal. Proof. Start with two disjoint closed subsets F and G of a regular and Lindelöf space X. Since X is regular, for every point x ∈G there are two disjoint open sets Vx and U x such that x ∈Vx and F ⊂ U x . The set V = {Vx :  x ∈G} is an open cover of G, and so V ∪{G c } is an open cover of X. The space X being Lindelöf, we can extract a countable open cover of X from V , which, after throwing out G c , becomes a countable open cover {Vx1 ,Vx 2 ,…,Vxn ,…} of G. Easing the notation by relabeling each Vxi by Vi , we get a countable open cover {V1 , V2 ,… , Vn ,…} of G.

Since Vx ⊂ (U x )c , we have that Vx ⊂ (U x )c = (U x )c ⊂ F c , and so the closures of the members of the cover V stay away from F. Consequently, the countable cover {V1 , V2 ,… , Vn ,…} of G has the property that Vi ∩ F = ∅ for every i = 1, 2, …. Symmetric argument yields an open cover {U1 ,U 2 ,… ,U n ,…} of F such that U i ∩ G = ∅ for every i = 1, 2, …. ∞ ∞ The problem we need to fix is that ∪ Vi and ∪U i need not be disjoint. We will do it by i =1

i =1

using countability to recursively modify the two families of open sets. Define V 1• = V1 \ U1 , Vn• = Vn \ (U1 ∪ U 2 ∪ … ∪ U n ), n = 2, 3, …. Note that since V •n = Vn ∩   (U1 )c ∩ (U 2 )c ∩ ... ∩ (U n )c all these sets are open. Since U i ∩ G = ∅ for every i = 1, 2, …, the family of open sets V1• , V2• ,… , Vn• ,… is still a cover of G. Symmetrically, U 1• = U1 \ V1 , U n• = U n \ (V1 ∪ V2 ∪ … ∪ Vn ), n = 2, 3,… defines an open cover of F. It follows ∞ ∞ from our construction that U n• ∩ V •m = ∅ for every n , m ∈{1, 2, …}, so ∪ V i• and ∪U •i are i =1 ∞ i =1 disjoint. Summarizing, we have found what we wanted: a pair of disjoint open sets ∪ V •i i =1 ∞ ∞ ∞ and ∪U •i such that G ⊂ ∪V •i and F ⊂ ∪U •i .

(

i =1

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)

i =1

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8.2 Regular Spaces and Normal Spaces   ◾    193  

Example 1 We know that the Sorgenfrey line is a regular space (Example 4, Section 8.1) and that it is Lindelöf (Example 3 in Section 7.3). Theorem 3 thus implies that the Sorgenfrey line is ☐ normal. Corollary 4. Every second countable and regular space is normal. Proof. This is a consequence of Theorem 3 and Lindelöf’s lemma (Proposition 5, Section 7.3). We are ready for the promised regular space that fails to be normal. Example 2. A Regular, but Not Normal Space The Sorgenfrey plane is the product space S × S of two Sorgenfrey lines. The rectangles of type [a , b ) × [c , d ) constitute a basis for S × S . Since S is regular (Example 4, Section 8.1), it follows from Theorem 4, Section 8.1, that S × S is also regular. However, the Sorgenfrey plane is not normal: take the line y = − x (or, for that matter, any line of negative slope) and consider it as a subspace E of S. Let Q be the set of all points in E of rational distance from (0,0), and let P consist of points in E of irrational distance from (0, 0). Then both Q and P are closed subsets of S × S (see Illustration 8.10). S×S

E

y x

P

z

Illustration 8.10  The three generic cases

of points x, y, and z not in Q showing that Q c is open (and that Q is closed).

These two closed sets cannot be separated with two disjoint open sets (Exercise 11, ☐ Section 2.1). The last example confirms that the class of T4 -spaces is a proper subclass of the class of T3 -spaces. There is one more important class of spaces squeezed between these two; it is sometimes called the class of T3 1 -spaces. The related notion is that of completely regular 2 spaces. A space X is completely regular if for every x ∈ X and every closed subset F of X that does not contain x, there exists a continuous function f : X → [0, 1] such that f ( x ) = 0 and f ( F ) = {1}. If in addition X is T1 , then we say that it is a T3 1 -space.1 2

1

A completely regular Hausdorff space is called a Tychonoff space; the terminology (completely regular, T3 1 , Tychonoff) 2 varies from author to author.

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Every completely regular space is regular (Exercise 11, Section 8.1). A direct consequence of Urysohn’s lemma (to be stated and proven in Section 9.1) is that every normal T1-space is completely regular. We will show in 8.3 that the class of normal T1-space is a proper subclass of the class of the class of completely regular spaces. For that we need the following simple result. Proposition 5. Every subspace of a completely regular space is completely regular. Proof. Let X be a completely regular space, and let Y be a subspace of X. Consider a closed subset F of Y and a point x ∈Y \ F . Then F = G ∩ Y for some closed subset G of X not containing x. By assumption there exists a continuous mapping f : X → [0, 1] such that f ( x ) = 0 and f (G ) = {1}. The restriction of f to Y confirms that Y is completely regular. There are regular spaces that fail to be completely regular. We will not provide such examples here (see, for example, [71]). Exercises 1. (a)  Let X and Y be normal spaces. Show that X ⊕ Y is a normal space. (b) Let X and Y be completely regular spaces. Show that X ⊕ Y is a completely regular space. 2. Prove Theorem 2; that is, show that a T1-space is regular if and only if for every x ∈ X and every open subset U of X containing x, there is an open subset W of X such that x ∈W ⊂ W ⊂ U . 3. Show that every closed subspace of a normal space is normal. 4. Use Theorem 3 to show that a product of two Lindelöf spaces need not be Lindelöf. 5. Show that if X and Y are completely regular spaces, then so is X × Y . 6. Let X be a Hausdorff space. Prove that for every compact subset K of X and every point x ∈ X \ K , there are open disjoint subsets U and V such that K ⊂ U and x ∈V .

7. Show that every locally compact Hausdorff space is regular.

8. Let A be a compact subspace of a T3 -space X. Show that for every closed set F such that A ∩ F = ∅, there exist open sets U , V ⊂ X such that A ⊂ U , F ⊂ V , and U ∩ V = ∅. 9. Let X be a normal separable space, and let E be a subset of X such that E ≥ 2ℵ0. Assume that E has no accumulation points. (a) Show that for every Y ⊂ E , Y and E \Y are disjoint and closed subsets of X. (b) Define a mapping f : P ( E ) → P ( D ) as follows: for every Y ⊂ E choose two disjoint open sets UY and VE \Y such that Y ⊂ UY and E \ Y ⊂ VE\Y , and set f (Y ) = UY . Show that f is well defined and one-to-one. (c) Deduce from (a) and (b) that if a subset E of a normal and separable space, satisfies E ≥ 2ℵ0 , then it must have an accumulation point. 10. Show that any countable, completely regular, Hausdorff space with at least two elements is disconnected.

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8.3 Normal Spaces and Subspaces   ◾    195  

8.3 Normal Spaces and Subspaces A brief historical note: The space in Example 1 (the Tychonoff plank) was constructed in 1930, not surprisingly, by Tychonoff.

If we weaken the requirement of regularity to Hausdorff-ness and strengthen the requirement of Lindelöf-ness to compactness in Theorem 3, Section 8.2, then the conclusion is the same: Theorem 1. Every compact Hausdorff space is normal. Proof. In view of Theorem 3 in 8.2 it suffices to show that compact Hausdorff spaces are regular. Let F be a closed subset of a compact Hausdorff space X, and let a be a point in X \ F. Since X is Hausdorff, for every x ∈F there are open disjoint neighborhoods U x and U ax around x and a, respectively. The class {U x :  x ∈ F } is an open cover of F. Since closed subsets of a compact space are compact, F is compact, and hence there is a finite subcover n n {U x1 ,U x 2 ,… ,U xn } of {U x :  x ∈ F } . Then ∩U xi and ∪U axi is a separation of F and a by two i =1 i =1 open sets. The examples that follow will give us a nonhereditary normal space, showing, as promised, that Theorem 3, Section 8.1 fails for normal spaces. Example 1 Recall the well-ordered set TP defined in Example 2, Section 1.3: briefly, it is the closed interval [1, ω], containing a copy of + , followed (with respect to the order in TP) by a left segment of a well ordered copy of the set of real numbers bounded on the right by the least real number ω that is preceded by uncountably many elements. The order topology (Example 9, Section 3.3) makes it a topological space. Explicitly, this means that the intervals of type [1, b), (a, b) and (a, ω] constitute a basis for TP; in this section we will call this basis a standard basis for TP. We will now show that TP is a compact Hausdorff space. Hence, by Theorem 1, TP is a normal space. Hausdorff-ness: Consider any a, b ∈[1, ω]; we may suppose a < b. If there is no x ∈[1, ω] such that a < x < b (i.e., if a is the immediate predecessor of b), then the sets [1, b) and (a, ω] are two disjoint (basic) open sets separating a and b. If there is an x ∈[1, ω] such that a < x < b , then [1, x ) and ( x , ω] are two disjoint (basic) open sets separating a and b. Compactness: First we show that TP is a Bolzano–Weirstrass space. It is easy to see that ω is an accumulation point of every uncountable subset of TP (Exercise 3). Thereby it suffices to show that every infinite countable set has an accumulation point. Let A be an infinite countable set in the space TP, and let x be the least element of

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196   ◾    Separation Properties

TP that is preceded by infinitely many elements of A. We show that x ∈ A′. Take any standard basis neighborhood U of x in TP. If U = [1, b) for some b, then it is obvious that U ∩ A ≠ ∅. Consider the case when U = (a , b ) or U = (a , ω] for some a and b. There must be an element y < x such that y ∈U , or else a would be an immediate predecessor of x larger than infinitely many elements of A, contradicting the choice of x. If U ∩ A = ∅ then y would be an upper bound of infinitely many elements of A smaller than x, again contradicting the choice of x. So, U ∩ A ≠ ∅ for every open set U containing x, showing that x ∈ A′ , and establishing that TP is a Bolzano–Weirstrass space. This result, together with Proposition 1, Section 7.4, implies that TP is countably compact. In order to show compactness it suffices to extract a countable subcover from any cover O consisting of standard basis sets for TP = [1, ω]. Since ω is covered, some (a,ω] is in O. Then [1, a] = TP \ (a, ω] is countable, or else a would have uncountably many predecessors, contradicting the choice of ω. For every x ∈[1, a] choose Ox ∈O containing it. Then {Ox ∈O :  x ∈[1, a]} ∪ {(a, ω]} is a count☐ able subcover of O . We are ready to give the example we promised. Example 2. Tychonoff Plank; A Normal Space with a Non-normal Subspace) Let ω 0 be the smallest element in TP larger than all natural numbers (so that [1, ω 0 ] = {1, 2 , 3,… , n ,… , ω 0 }). Since [1, ω 0 ] is a closed subspace of the compact space TP = [1, ω], it is also compact. It is clear that [1, ω 0 ] is Hausdorff. The product space X = [1, ω] × [1, ω 0 ] is called the Tychonoff plank (see Illustration 8.11). Since a product of Hausdorff and compact spaces is Hausdorff and compact, the space X is Hausdorff and compact. By Theorem 1, X is normal. However, the subspace Y = X \ {(ω , ω 0 )}—the deleted Tychonoff plank—is not normal. {ω} × [1,ω0) ω0

(ω,ω0)

n 3 2 1

[1,ω] × {ω0}

an

1 2 3 4

ω0

a3 a2 a1 x ω

Illustration 8.11  Tychonoff plank; the points ai and x will emerge further in our argument.

The sets A1 = {ω} × [1, ω 0 ) and B1 = [1, ω] × {ω 0 } (see Illustration 8.11) are closed in X, and so A = A1\{(ω , ω 0 )} and B = B1\{(ω , ω 0 )} are closed in Y. We will show that A and B cannot be separated with two disjoint open sets.

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8.3 Normal Spaces and Subspaces   ◾    197  

Let U be an open subset of Y such that A ⊂ U (see Illustration 8.12). Since (ω , n ) ∈ A ⊂ U , we have that U ∩ ([1, ω] × {n}) ≠ ∅ , for every n ∈{1, 2 ,…} . Denote the least element of U ∩ ([1, ω] × {n}) by an , and the image of an under the projection [1, ω] × {n} → [1, ω] by bn . The set Z = {bn : n = 1,2,…} is a countable set. Observe that none of bn is ω, since otherwise U ∩ ([1, ω] × {n}) consists only of the pair (ω , n ) , contradicting the fact that U ∩ ([1, ω] × {n}) is open in [1, ω] × {n}. Since ω > bn for every n ∈+ , the set of all elements in [1, ω] larger than all elements in Z is not empty, and thus has the least element x. We note that x ≠ ω since ∞ x is preceded by the elements in the countable set ∪[1, bi ], while ω has uncountably i =1 many predecessors. We proved that there exists an x ∈[1, ω ) such that ( x , ω ) × n ⊂ U for every n (where, to avoid misunderstanding, we point out that here (x, ω) denotes an open interval). This means that ( x , ω ) × [1, ω 0 ) is a subset of U (as indicated in Illustration 8.12). Our argument will be complete once we show that ( x , ω ) × [1, ω 0 ) has a nonempty intersection with every open subset V of Y such that B ⊂ V . {y} × {u} ω0

V

n 3 2 1



U

1 2 3 4

ω0

x ω

Illustration 8.12

Let V be an open subset of Y such that B ⊂ V . Choose an element y in the interval (x, ω). The set ({ y} × [1, ω 0 ]) ∩ V is open in { y} × [1, ω 0 ] and by construction it contains the point { y} × {ω 0 } . So, there is an open interval ( z , ω 0 ] in [1, ω 0 ] such that { y} × {ω 0 } ∈{ y} × ( z , ω 0 ] ⊂ ({ y} × [1, ω 0 ]) ∩ V . Finally, notice that for every u ∈( z , ω 0 ], the point { y } × {u} is in both V and  U, the latter being true since ☐ y > x . We proved that U ∩ V ≠ ∅ and that Y is not normal. We have established the truth of the following claim. Proposition 2. There exists a normal space with a non-normal subspace. Remark: The whole construction of Tychonoff’s plank and the subsequent examples and counterexamples are all based on a well ordering of . The existence of the latter is a consequence of the Axiom of Choice. However, there is no known (and relatively simple) explicit well ordering of , and so our constructs are existential only.

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198   ◾    Separation Properties

Exercises 1. Give an example of a compact space that is not normal. 2. Let X be a compact and Hausdorff space and ~ an equivalence relation over X. Show that the quotient mapping q : X → X~ is closed if and only if the set {( x , y ): x ~ y } is a closed subset of X × X . 3. Show that ω is an accumulation point for TP, and that there is no sequence in TP \{ω} converging to ω. [Hint for the second part: The argument is essentially the same as the one provided for the set Z in Example 2.] 4. Show that the point (ω, ω0 ) in the Tychonoff plank is not an intersection of countably many open sets. [Hint: Exercise 3.] 5. A space X is perfectly normal if it is normal and each closed subset is a countable intersection of open sets. (a) Show that being perfectly normal is hereditary. (b) Show that the Tychonoff plank is not perfectly normal.

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Chapter

9

Urysohn, Tietze, and Stone–Čech

I

n this chapter we focus on a few important theorems.

9.1  The Urysohn Lemma A brief historical note: Pavel (Paul) Urysohn and Pavel (Paul) Alexandroff were mentioned together several times earlier in the book. They were together one summer day in 1924 when they decided to take a swim in the rough waters of the Atlantic Ocean, off the coast of Brittany, France. Urysohn drowned. He was only 26.

{

}

Recall that D = m2i  : m ∈, i ∈ is the set of dyadic numbers. According to Exercise 8, Section 3.4, this set is dense in . It is then easy to show that D ∩[0, 1] is dense in [0,1]. Theorem 1. (The Urysohn Lemma) Let X be a T1 -space. Then X is normal if and only if for every two disjoint closed sets A and B there is a continuous mapping f : X → [0, 1] such that f ( A) = {0} and f ( B) = {1}. Proof. ⇐ Take any two disjoint closed subsets A and B of X. By assumption there is a continuous f : X → [0,1] with f ( A) = {0} and f ( B) = {1}. Let a and b be any two numbers such that 0 < a < b < 1. Then, f −1 ([0, a)) and f −1 ((b,1]) are two disjoint open sets containing A and B, respectively. ⇒ Suppose that X is normal, and let A and B be two disjoint closed subsets of X. Then A ⊂ Bc , and since Bc is open, we can use Theorem 1, Section 8.2 to conclude that there is an open set U 1 2 satisfying A ⊂ U 1 2 ⊂ U 1 2 ⊂ Bc . (The reason we use dyadic numbers as subsripts will become clearer as we go.) Denote A = U 0 and Bc = U1, and define f1 : X → [0,1] as follows: f1 ( x ) = inf{t  ∈{0, 1 2 ,1} :  x ∈U t } if x ∈U1, and f1 ( x ) = 1 if x ∈ B. This means that 199

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200   ◾    Urysohn, Tietze, and Stone–Čech

f1 sends the points of A to 0, sends the points of U 1 2 \ A to 1 2 , and sends the rest of the points of X to 1. The graph of f1 is shown below, and the caption explains our intention.

B B

X

U½ \ A

X

A

A

Illustration 9.1  We start with the closed subsets A and B of X and with a function shown to the

left (sending A to 0 and B to 1). Then we build a rough stairway from A to B via the function f1, the graph of which is shown to the right.

Apply Theorem 1, Section 8.2 to U 0 ⊂ U 1 2 and to U 1 2 ⊂ U1 to establish the existence of open sets U 1 4 and U 3 4 such that U 0 ⊂ U 1 4 ⊂ U 1 4 ⊂ U 2 4 and U 2 4 ⊂ U 3 4 ⊂ U 3 4 ⊂ U1 (where we have written the index 1 2 as 2 4 ). We get: U0 ⊂ U 1 4 ⊂ U 1 4 ⊂ U2 4 ⊂ U 2 4 ⊂ U 2 4 ⊂ U 3 4 ⊂ U1 . Now define f 2 ( x ) = inf{t  ∈{ i 22  : i = 0,1,2,3,4} :  x ∈U t } if x ∈U1, and f1 ( x ) = 1 if x ∈ B. The graph of f 2 is shown in Illustration 9.2. Proceeding inductively, in the k­-th step we get U 0 ⊂ U 1 k ⊂ U 1 ⊂ U 2 k ⊂ U 2 ⊂  2

2k

2

2k

⊂ U(2k −1) ⊂ U(2k −1) ⊂ U1 and f k ( x ) = inf{t ∈{ i 2k  : i = 0,1,…,2k } :  x ∈U t } if x ∈U1, f k ( x ) = 1 2k

if x ∈ B.

2k

B

B U³⁄ ₄ \U²⁄ ₄

A

U¹⁄ ₄ \ A

U²⁄ ₄ \U¹⁄ ₄

A

U¹⁄ ₈ \ A

U²⁄ ₈ \U¹⁄ ₈

U⅜ \U²⁄ ₈

U⁷⁄ ₈ \U⁶⁄ ₈ U⁶⁄ ₈ \U⁵⁄ ₈ U⁵⁄ ₈ \U⁴⁄ ₈ U⁴⁄ ₈ \U⅜

Illustration 9.2  We show the graphs of the functions f2 (left) and f3 (right); our stairway becomes

more refined.

At this point we observe the following: (i) If U t1 and U t2 are two distinct open sets we get in this procedure and if U t1 ⊂ U t2 , then t1 < t 2 . (ii) For each dyadic number t in the interval [0,1], there is an open set U t introduced in some step of the above procedure. Define f : X → [0, 1] by f ( x ) = lim k→∞ f k ( x ) . We will now show that f is the promised smooth transition from the ­A-plateau at altitude 0 to the ­B-plateau at altitude 1 (see Illustration 9.3).

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9.1  The Urysohn Lemma   ◾    201  

B

A

Illustration 9.3  The function f provides a continuous transition from A (at altitude 0) to B

(at altitude1).

It follows from (i) that for every x ∈ X , the sequence ( f k ( x ))∞k =1 is nonincreasing. Since it is bounded from below it must converge. So the mapping f is well defined. Since f k (a) = 0 for every a ∈ A and for every k =1, 2, … , it follows that f ( A) = {0}. Similarly, since f k (b) = 1 for every b ∈B, and for every k =1, 2, … , it follows that f ( B) = {1}. It remains to be shown that f is continuous. First, observe (directly from the definitions of f k and f ) that f ( x ) = inf{t ∈D  :  x ∈U t } if x ∈U1, and f ( x ) = 1 if x ∈ B. The continuity of f follows from the following two properties, and from Exercise 12, Section 3.5 (since the intervals of types [0, a) and (b,1] form a subbasis for the topology over [0,1]): (1) f −1 ([0, a)) = ∪ U t t b

(1) Take any x ∈ f −1 ([0, a)) . This means that f ( x ) ∈[0, a) . Since D is dense, there is an element t0 ∈D with f ( x ) < t0 < a. If x ∉U t0 then inf{t ∈D  :  x ∈U t } ≥ t0 , and so f ( x ) ≥ t0 . So, it must be that x ∈U t0 . Hence, x ∈∪ U t . We have proven that t t0 . Recalling that to > b we conclude that f ( x ) ∈ (b, 1], i.e., that x ∈ f −1 ((b,1]) . We have proven that ∪(U t )c ⊂ f −1 ((b,1]) and so we have established that (2) is true. t >b

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202   ◾    Urysohn, Tietze, and Stone–Čech

Here is a simple consequence. Corollary 2. There is a completely regular space that is not normal. Proof. Since the Tychonoff’s plank TP is normal, the Urysohn lemma (applied to a singleton— which is a closed set since TP is T1—and to a disjoint closed set) implies that it is completely regular. Consider the subspace Y of TP defined in Example 2, Section 8.3. We showed there that Y is not normal. On the other hand, by Proposition 5, Section 8.2, Y is completely regular. The following result, based on the Urysohn lemma, shows that small regular T1 spaces, those that have countable bases, are metrizable. Theorem 3. (Urysohn)1 Every second countable T3 space is metrizable. Proof. First we give a short synopsis of the proof. Start with a countable basis for X, then extract pairs ( B∗ , B) of basis sets such that B∗ ⊂ B. By the Urysohn lemma, each such pair gives rise to a mapping onto {0,1} separating B∗ and Bc . Countably many such mappings will give rise to an embedding from X into the Hilbert spaces H, and since H is a metric space, so is X. Let B = {Bi : i ∈+ } be a countable basis for a regular space X. We may suppose that ∅ ∉B , so that there exists an element xi ∈ Bi for each i ∈+ . By Theorem 2, Section 8.2, for every xi and Bi , there exists an open set U i such that xi ∈U i ⊂ U i ⊂ Bi . Since B is a basis, for each such U i there exists a set Bx∗i ∈B such that xi ∈ Bx∗i ⊂ U i , and so, since Bx∗i ⊂ U i , we have established that for every Bi ∈B , and for every xi ∈ Bi , there exists some Bx∗i ∈B such that xi ∈ Bx∗i ⊂ Bx∗i ⊂ Bi . The set of all such pairs Bx∗i , Bi is countable, and so, after reindexing 2 if necessary, we can denote the elements of that set by Bi∗ , Bi , i ∈+ . By Corollary 4, Section 8.2, the space X is normal. By the Urysohn lemma, the mapping that sends Bi∗ to {0} and Bic to {1} can be extended to a continuous mapping fi : X → [0,1], i ∈+ . Choose an arbitrary x ∈ X and consider the

(

(

)

f 2 ( x ) f3 ( x ) 2 3 fn ( x ) 2 n n =1 ∞ 2 fn ( x ) n n =1

sequence of real numbers f1 ( x ),

,

( ) ( ( f ( x )) + ( ) + ( ) +  + ( ) )

( )

fn ( x ) 2 n

∞

≤ n12 and so the series ∑ f2 ( x ) 2 2

2

1

f3 ( x ) 3

,…,

)

fn ( x ) n

)

f3 ( x ) 2 3

,… ,

fn ( x ) n

,… . Since 0 ≤ fi ( x ) ≤ 1 , we have

converges. This means that the sequence converges, establishing that

( f (x ), 1

f2 ( x ) 2

,

,… is an element of the Hilbert space H (Example 2, Section 5.2).

(

f (x )

f (x )

f (x )

)

So far we have shown that F : x  f1 ( x ), 2 2 , 3 3 ,… , nn ,… is a mapping from X into H. We will show that it is an embedding; that will make X homeomorphic to a subspace of H. Since H is a metric space, so is every subspace of H, and so we would thus establish that X is metrizable. 1 2

Theorem 3 is sometimes called the Urysohn–Tychonoff Metrization Theorem. In order to unburden the notation, we are allowing some ambiguity here: Bi as a member of B , and Bi as the second coordinate of Bi∗ , Bi may be different sets; which of these we are using should be clear from the context.

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(

)

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9.1  The Urysohn Lemma   ◾    203  

Step 1. (F is continuous.) Choose an open neighborhood V of some F (x). Since H is a metric space, there is an ε > 0 such that the open ball B = {y ∈H : d( F ( x ), y ) < ε} is a subset of V. We are searching for an open neighborhood U of x such that F (U ) ⊂ B. ∞

Find N such that ∑

i = N +1

1 i2

∞

2

< ε4 ; this can be done since the series ∑ i12 converges. Since i =1

each of f1 , f 2 ,…, f N is a continuous mapping X →[0, 1], for each i ∈ {1, 2, … , N } there is an open neighborhood U i of x such that fi (U i ) is a subset of the interval N ( fi ( x ) − 2εN , fi ( x ) + 2εN ) ∩[0,1] , that is open in [0, 1] ⊂ . Denote U = ∩U i ; then, for every i =1 y ∈ U , and for every i ∈ {1, 2, … , N }, we have fi ( y ) − fi ( x ) < 2εN . We prove that F (U ) ⊂ B by establishing that d( F ( x ), F ( y )) < ε for every y ∈ U . Deno­t ing F ( x ) =

( )

fi ( x ) ∞ i i =1

∞

can thus be written as follows:

∑

i =1

this inequa­l ity, we have: ∞

∑ i =1

≤

∑ i =1 N

=

N

2

∑

 fi ( x ) fi ( y )  −   ≤ i i 

N

∑ i =1

i =1

2

 fi ( x ) fi ( y )  −   + i i  fi ( x ) − fi ( y ) + i

( )

fi ( y ) ∞ i i =1

and F ( y ) =

∞

∑

(

fi ( x ) i

−

i = N +1

+

∞

∑

≤

ε2 4

, hence

( fi ( x )− fi ( y ))

i = N +1

i2

F is continuous.

2

∞

i = N +1

2

∑

 fi ( x ) fi ( y )  −   ≤ i i i = N +1 2

∑

 fi ( x ) fi ( y )  −   = i i  i = N +1

( f i ( x ) − f i ( y ))2 i2

i = N +1

∑

∞

∞

ε 2N

≤

∑ f (x ) − f ( y ) + ∑ i

i

( f i ( x ) − f i ( y ))2 . i2

i = N +1 N

for every i ∈ {1, 2, … , N }, so that ∑ fi ( x ) − fi ( y )

( fi ( x )− fi ( y ))2 i2

≤ 2ε + 2ε = ε , and so

∞

N

i =1

i =1

fi (z ) ∈[0,1] for every z ∈X , we have that

( fi ( x ) − fi ( y ))2 ≤ 1, i ∈ {N , N + 1, …}, and so 1 i2

< ε . Focusing on the left-hand side of

2

≤ N 2εN = 2ε . On the other hand, since ∞

)

fi ( y ) 2 i

 fi ( x ) fi ( y )  −   + i i 

By assumption fi ( y ) − fi ( x )
−0.25} and U 2 = {( x , y ) ∈ 2:  x 2 + y 2 = 1,  y < 0.25}. Then {U1 ,U 2 } is an open cover of S1 (Illustration 11.4). Observe that both U1 and U 2 are simply connected. U1 S1 y

x

U2

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Illustration 11.4  x = (1,0); y = (−1,0).

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250   ◾    The Fundamental Group of a Circle and Applications

Let β be a loop at x in S1, giving rise to the element [ β] ∈π1 (S1, x ). Let β1β 2 …βm be a Lebesgue subdivision of the path β relative to the cover {U1 ,U 2 } . This means that for every i = 1,2,…, m, βi ([ti −1 , ti ]) ⊂ U j for some j ∈{1,2}. If both βi ([ti −1 , ti ]) and βi +1 ([ti , ti +1 ]) are contained in a single U j , then we can replace βi and βi+1 by their product in the Lebesque subdivision. Repeating this procedure we can guarantee that no two βi ([ti −1 , ti ]) and βi +1 ([ti , ti +1 ]) are in a single U j . After relabeling, we have a Lebesgue subdivision β1β 2 …βm of β satisfying that property. Denote y = (−1,0). There are two path components of U1 ∩ U 2 : one contains x, the other y. We denote the former by U x , and the latter by U y . The modification of the original Lebesgue subdivision of β that we have performed in the preceding paragraph guarantees that the endpoints βi (ti −1 ) and βi (ti ) of each βi are either in U x or in U y . If βi (ti ) ∈U z , z ∈{x , y}, then we choose a path γ i in U z from βi (ti ) to z; this could be done since U z is path connected. Since γ i γ i−1 is homotopic to the constant path at βi (ti ) , it follows from the associativity of the operation of multiplication of homotopy classes of paths that [β1β 2β3 …βm−1βm ] = [β1γ 1 ][γ 1−1β 2 γ 2 ][γ 2−1β3 γ 3 ]…[γ m−1−2βm−1γ m−1 ][γ m−1−1βm ], where each path in square parenthesis leads from x or y to x or y (it may be a loop) within one single U j , j ∈{1,2}. If some γ i−−11βi γ i is a loop at z ∈{x , y}, then, since U1 and U 2 are simply connected, it is homotopic to the constant map at z, and so [γ i−−11βi γ i ] could be omitted without affecting [β]. So, we may suppose that every γ i−−11βi γ i is a path from one of x, y to the other. Then, if necessary, we repeat the same argument as above to ensure again that no two consecutive γ i−−11βi γ i and γ i−1βi +1 γ i +1 are paths in a single U j . We relabel again (if needed) to end up with [β] = [β1γ 1 ][γ 1−1β 2 γ 2 ][γ 2−1β3 γ 3 ]…[γ m−1−2βm−1γ m−1 ][γ m−1−1βm ] where the paths in the square parentheses join x and y (in some order), and each two consecutives paths are in different U1 and U 2 . It is easy to confirm that all of the above changes of the original Lebesgue subdivision β1β 2 …βm of β had no effect whatsoever on the fact that the first path of the product starts at x, and that the last path of the product ends at the same point, i.e., that we are dealing with a loop at x. We are almost done: β1 γ 1 is a path from x to y in U j ( j ∈{1,2}). Suppose it is in U1 . Since α1−1 is a path from y to x in U1, and since U1 is simply connected, it follows that [β1 γ 1 ] = [α1 ] (Exercise 4 in 10.4). Hence [β] = [α1 ][γ 1−1β 2 γ 2 ][γ 2−1β3 γ 3 ]…[γ m−1−2βm−1 γ m−1 ][γ m−1−1βm ]. The next path γ 1−1β 2 γ 2 must start at y and end at x, and it must be in U 2 . Arguing similarly as for β1 γ 1 we conclude that [γ 1−1β 2 γ 2 ] = [α 2 ]. So, [β] = [α1 ][α 2 ][γ 2 −1β3 γ 3 ]…[γ −1m−2βm−1γ m−1 ] [γ −1m−1βm ]. Repeating this argument m many times (and using induction), we conclude that [β] = [α1 ][α 2 ][α1 ][α 2 ]…[α1 ][α 2 ]; the last homotopy class in this product must be [α 2 ] since γ m−1−1βm ends at x. At the same time we observe that m is even. Since [α1 ][α 2 ] = [α] m it follows that [β] = [α] 2 . only slight modifications of the If in the initial step β1 γ 1 is a path from x to y in U 2 , then −m above argument are needed to conclude that [β] = [α] 2 . Summarizing: we showed in this part that every [β] ∈π1 (S1 , x ) is a power of [α] or its inverse [α]−1 , confirming that π1 (S1 , x ) is a cyclic group generated by [α] and completing the proof of Part 1.

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11.1  The Fundamental Group of a Circle   ◾    251  

Proof of Part 2: Here is a sketch of the basic idea of the rest of the proof: We associate with every loop at x = (1,0) in S1 a path in  that we call a lift of the loop, and then we show that two loops in S1 are homotopic if and only if their lifts starting at a fixed point must end at the same point. At the end we merely notice that, for every m ≠ n, the lifts of α n and α m starting n m at 0 ∈ end at different points, so that [α ] and [α ] must be different. It follows that π1 (S1 , x ) is an infinite cyclic group. It suffices to show that [α]j ≠ [α]k for j ≠ k. The proof we will give is a special case of the theory of covering spaces that we will develop in a few chapters. Consider the continuous mapping p :  → S1 defined by p(s ) = (cos2π , sin 2πs ) and depicted in Illustration 11.5. 





p–1(U1)

Illustration 11.5

p

p (1,0)

0

0

0

S1

1 p–1(U2)

1

1

p

2

2

2

U1 S1

Illustration 11.6

(1,0)

S1

(1,0) U2

Illustration 11.7

If U j ∈{U1 ,U 2 } , where {U1 ,U 2 } is the cover of S1 as in the proof of Part 1, then p −1 (U j ) ( j ∈{1,2}) is a disjoint union of its path components, each of which is homeomorphic to U j (Illustrations 11.6 and 11.7), and each containing precisely one integer. Denote the component of p −1 (U j ) containing the integer a by Vj ,a , and the restriction of p to Vj ,a by p j ,a ; note that each p j , a is a homeomorphism. Observe also that if C and D are path components of p −1 (U1 ) and p −1 (U 2 ), respectively, then C ∩ D is either empty or it is path connected. Given a continuous mapping f : I × [0, c] → S1 , 0 ≤ c , such that f (0, s ) = f (1, s ) = x for every s ∈[0, c], we will construct a continuous mapping f : I × [0, c] →  such that p  f = f . We call f the lift of f; this lift will satisfy additional properties, which we will identify along the way. Before we proceed, we point out that we are primarily concerned with two cases: c = 0, in which case f is a loop at x, and c = 1 in which case f is a homotopy relative to {0,1} from the loop f I ×{0} at x to the loop f I ×{1} at x, where we identify I with I × {0} and I × {1} via the evident homeomorphisms.

Let { fi ,k : i ∈{0,1, … , n − 1},  k ∈{0,1, … , m − 1}} be a Lebesgue subdivision of the mapping f (relative to the cover {U1 ,U 2 }) of S1. The domain of each fi .k is the rectangle Ri ,k as in Illustration 11.8.

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252   ◾    The Fundamental Group of a Circle and Applications

fi ,k : Ri ,k →  of each fi ,k , so that these lifts agree over the intersecWe construct a lift  tions of their domains. By the gluing lemma, the mapping f : I × [0, c] →  defined to be  fi ,k over Ri ,k (i ∈{0,1,…, n − 1}, k ∈{0,1,…, m − 1}), would then be a continuous. In the first step we construct  f0,0 as follows: since f0,0 ( R0,0 ) ⊂ U j for some j ∈{1,2}, −1  f0,0 ( R0,0 ) is a subset of the component of p −1 (U j ) define f0,0 = p j ,0  f0,0 . Observe that  containing the integer 0. Proceed recursively, and assume that we have constructed, row by row, from the bottom to the top and from left to right, the lifts corresponding to all shaded rectangles in Illustration 11.8 agreeing over their intersections. The union of these mappings is a continuous mapping  g  defined over these shaded rectangles. We now pay attention to Ri ,k . We lift fi ,k : Ri ,k → S1 to a continuous  fi ,k : Ri ,k →  as follows: By construction fi ,k ( Ri ,k ) is a subset of some U j ( j ∈{1,2}). Let us assume, to simplify the indexing, that j = 1; the case when j = 2 is symmetric. So, fi ,k ( Ri ,k ) ⊂ U1 . We will distinguish two cases: (1) both fi −1,k ( Ri −1,k ) and fi ,k −1 ( Ri ,k −1 ) are subsets of U1 , and (2) at least one of them is a subset of U 2 . Case (1): Since Ri −1,k ∩ Ri ,k −1 ≠ ∅ , it follows that  fi−1,k ( Ri−1,k ) ∩  fi ,k−1 ( Ri ,k−1 ) is also non  empty. Hence the sets fi−1,k ( Ri−1,k ) and fi ,k−1 ( Ri ,k−1 ) are contained in the same path component of p −1 (U1 ) . Suppose that path component contains the integer a. Then define  fi ,k = p1,−1a  fi ,k . Since the same p1,−1a has been used to construct all three of  f i ,k ,  fi ,k−1 , it follows that fi−1,k and 

they agree over the intersections of their domains, and so the extension of g by the mapping  fi , k is also continuous.  a+1 R0, m–1 R1, m–1

R0, k

R1, k

R0, k–1 R1, k–1

Ri–1, m–1 Ri, m–1

Ri–1, k

Ri, k

Ri–1, k–1 Ri, k–1

Rn–1, k

R1, 1

Ri–1, 1

Ri, 1

Rn–1, 1

R0, 0

R1, 0

Ri–1, 0

Ri, 0

Rn–1, 0

a a−1

fi–1,k p

Rn–1, k–1

R0, 1

Illustration 11.8

fi,k

Rn–1, m–1

Ri−1,k

fi,k

Ri,k

Ri−1,k−1 Ri,k−1

fi−1,k

x = (0,1) S1

Illustration 11.9

Case (2): We may suppose that fi −1,k ( Ri −1,k ) ⊂ U 2 . By the recursive step, there is an integer a such that  fi−1,k = p2,−1a   fi−1,k . Pay attention to the vertical edge E common to Ri −1,k and Ri ,k : since fi ,k ( Ri ,k ) ⊂ U1 , it follows that fi −1,k ( E ) ⊂ U1 ∩ U 2 . Consequently  fi−1,k ( E ) is a subset of the intersection of a path component of p −1 (U1 ) with a path component of p −1 (U 2 ).

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11.1  The Fundamental Group of a Circle   ◾    253  

There are two components of p −1 (U1 ) intersecting p2,−1a (U 2 ); one contains the integer a, the other contains the integer a −1. In both cases the intersection of the component of p −1 (U1 ) and p2,−1a (U 2 ) contains  fi−1,k ( E ). Let b ∈{a − 1, a} be the (only) integer in  fi−1,k ( E ). 1 − (In Illustration 11.9 we depict the case when b = a.) Define  fi ,k = p1,b   fi ,k . It follows imme  diately that fi−1,k and fi ,k agree over E (the intersection of their domains). Since Ri −1,k ∩ Ri ,k −1 ≠ ∅ , and because g  (introduced just before Case 1) is well defined fi−1,k and  fi ,k−1 agree over their (that is, since  fi ,k−1 = intersection), we have that either  2 p2,−1b   fi ,k −1, or  fi ,k −1 = p1,−1b   fi ,k −1 (depending on whether f ( Ri ,k −1 ) is in U2 or in U1, respecfi ,k−1 tively). It is evident that in both cases  1  and fi ,k agree over the intersection of their domains. α2 0 p We have shown that the extension of g by fi ,k is continuous. the mapping  We have constructed a continuous mapα2 ping f : I × [0, c] →  for every c ≤ 0. When c = 0, the mapping f is a loop at x, and its 1 lift f is a path starting at 0. In Illustration 0 S1 2 11.10 we show the lift of the loop α , where x = (1,0) α is the generator of π1 (S1 , x ) introduced Illustration 11.10 in the first part of the proof of the theorem. It is clear from our construction that the lift of α i must end at i. So, if i ≠ k, then the lifts of the loops α i and α k are paths that end at different integers. The proof of our theorem will be completed once we make the following observations. If c = 1, then, as we have observed above, f is a homotopy relative to {0,1}. We claim that in that case f is also a homotopy relative to {0,1}. This has already been confirmed, except for the claim that the homotopy is relative to {0,1}. Note that since (by definition) f sends {0} × I to x, our construction guarantees that f sends the same set to 0, and since f sends {1} × I to x, f must send the same set to the terminal point of the lift of the path f I ×{0} . (In particular, the end-points of the lifts of the paths f I ×{0} and f I ×{1} must be the same.) So, two homotopic loops at x in S1 are lifted to homotopic paths in  starting at 0 and ending at the same point. Consequently, if j ≠ k , then [α]j ≠ [α]k , proving that [α] generates an infinite cyclic group. Hence π1 (S1 , x ) is an infinite cyclic group. Exercises 1. (a)  Find the fundamental group of a cylinder. (b) Find π1 (T 2 , x ) , where T 2 is a torus. (c) Find π1 ( 2 \{(0,0)}, x ).

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254   ◾    The Fundamental Group of a Circle and Applications

2. Show that the unit open disk B2 = {( x , y ) ∈ 2 : x 2 + y 2 < 1} is not homeomorphic to B2 ∪ Y , where Y is any non-empty subset of S1 . 3. Denote X = S 2 \{(0,0, −1),(0,0,1)}. Define the homeomorphism f : X → X by f ( x , y , z ) = (− x , y , z ). Show that f is not ambient isotopic within X to the identity mapping id X . [Hint: π1 ( X ) =  .] 4. (a) Show that the equatorial circle C in the torus T 2 (shown in Illustration 11.11) is a retract of T 2. (b) Show that C is not a deformation retract of T 2. T2

C

Illustration 11.11  

5. True or false: if A and B are path connected and simply connected subspaces of a space X, and if A ∩ B is nonempty, path connected, and simply connected, then so is A ∪ B. 6. Prove that every bouquet of n circles is a 1-skeleton of a simply connected 2-complex. 7. Let C be a thick cylinder {( x , y ) ∈ 2 : 0.5 ≤ x 2 + y 2 ≤ 1} × I , and let A and B be the two spirally subspaces of C shown in Illustration 11.12. (A winds exactly once around the outer surface of C; B winds exactly twice.) Show that there is no homeomorphism f : C → C such that f ( A) = B and such that f ( x ) = x  (x is shown in the Illustration 11.12).

C A

x

C

B

x

Illustration 11.12  To avoid clutter, we show C twice, first with A, then with B.

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11.2  Brouwer Fixed Point Theorem and The Fundamental Theorem of Algebra   ◾    255  

8. Show that the sphere S 2 is not homeomorphic to the disk D 2 . 9. The space X depicted in Illustration 11.13 is constructed as follows: starting with two concentric spheres, remove four open disk-like subspaces, D11 and D12 from the first and D21 and D22 from the second sphere; join the circular boundaries of ∂(D11 ) and ∂(D21 ) , and of ∂(D12 ) and ∂(D22 ) with two tubes, respectively. Find the fundamental group of X.

Illustration 11.13

11.2 Brouwer Fixed Point Theorem and The Fundamental Theorem of Algebra A brief historical note: Jan Brouwer proved the fixed point theorem in 1911. During his early studies his main interests were split between mathematics and philosophy, so much so that his Ph.D. advisor refused an early version of Brouwer’s dissertation (eventually defended 1907) because it incorporated substantial philosophical components. In 1806 (republished in 1813) Jean-Robert Argand (1768–1822) produced the first rigorous proof of the fundamental theorem of algebra.

Let A be a retract of a space X, and let a ∈ A. This means that there is a continuous r : ( X , a) → ( A, a) such that r  i = id A , where i : A → X is the inclusion. Hence, (r  i )∗ = (id A )∗, and so, since (r  i )∗ = r∗  i∗, and since (id A )∗ = idπ1 ( A ,a ) , we have r∗  i∗ = idπ1 ( A ,a ). In particular, r∗  i∗ must be a bijection. It is easy to verify that if f  g is a bijection, then g must be one-to-one, and f must be onto. We conclude from this that i∗ must be a monomorphism (= one-to-one homomorphism) and that r∗ must be an epimorphism (= onto homomorphism). We have proven the following initial result. Proposition 1. If r : ( X , a) → ( A, a) is a retraction then r∗ : π1 ( X , a) → π1 ( A, a) is an epimorphism, and the inclusion i : A → X induces a monomorphism i∗ : π1 ( A, a) → π1 ( X , a) . Example 1 The circle S1 is not a retract of the disk D 2 . Otherwise, it would follow from Proposition 1 that there would be an epimorphism from the trivial group π1 (D 2 , a) onto the cyclic group π1 (S1 , a) , which is clearly impossible. More generally, the same argument shows that if a subspace A of a simply connected space X is not simply con☐ nected, then A is not a retract of X. In the following examples we will see that π1 ( A, a) ≅ π1 ( X , a) is not a guarantee that a subspace A of X is a retract of X.

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256   ◾    The Fundamental Group of a Circle and Applications

Example 2 We have introduced the Möbius band in Exercise 3 of 4.4; it is the quotient space obtained from a filled rectangle by identifying two opposite edges (see Illustration 4.31 in 4.4). Illustration 11.14(a) shows a Möbius band, denoted M. The middle circle C of M is a deformation retract of M: the deformation retraction is indicated by the arrows in Illustration 11.14a. It follows from Corollary 8 of 10.4 that π1 ( M , x ) is infinite cyclic; it is generated by the homotopy class of the loop α shown in Illustration 11.14b.

B

M

α C

β x



Illustration 11.14a

Illustration 11.14b

The boundary B of M (see Illustration 11.14a) is obviously homeomorphic to S1, so its fundamental group is also infinite cyclic. We will now show that the boundary B of M is not a retract of M. Assume otherwise, i.e., assume that there is a retraction r : M → B. By Proposition 1, r∗ : π1 ( M , x ) → π1 ( B, x ) is an epimorphism from an infinite cyclic group to another infinite cyclic group. Since every such epimorphism sends a generator to a generator, it follows easily that r∗ must be an isomorphism. That, together with r∗  i∗ = idπ1 ( B , x ), implies that i∗ is also an isomorphism (where, as before, i : B → M is the inclusion). However, one of the two generators of π1 ( B, x ) is the homotopy class of the loop β (see Illustration 11.14b), and, since β  α 2 , we have that i∗ ([β]) = [α]2 . Hence, i∗ is not onto, and so not an ☐ isomorphism. We have a contradiction. Recall that D n = {( x1 , x 2 ,…, xn ) ∈n :  x12 + x 22 +  + xn2 ≤ 1} is the closed n-disk. Theorem 2. (Brouwer fixed-point theorem) Every continuous mapping D n → D n has a fixed point. The case n = 1 is Exercise 10 in 6.2. The case n = 2 is within the scope of our theory so far and will be proven below. We will not prove the statement for n > 2; an explanation will be provided in Section 14.5 when we touch on higher homotopy groups. Proof (n = 2). Suppose there is a continuous f : D 2 → D 2 such that f ( x ) ≠ x for every x ∈D 2 . This implies that there is a unique semi-line lx in  2 starting at f ( x ) and passing through x (Illustration 11.15).

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11.2  Brouwer Fixed Point Theorem and The Fundamental Theorem of Algebra   ◾    257  

S1

g(x) x

f(x)

D2

Illustration 11.15

Define a mapping g : D 2 → S1 as follows: g sends every x ∈D 2 to the intersection of lx and the circle S1 (Illustration 11.15). It is obvious that g is continuous, and that it fixes every x ∈S1, hence, it is a retraction D 2 → S1 . By Example 1, that is not possible, hence the initial hypothesis in this proof fails. Readers who do not like visual arguments are invited to provide an algebraic justification (Exercise 2). Since “having a fixed point” is a topological property (Exercise 17 in 3.5), we could replace D n in the statement of the Brouwer fixed point theorem with any homeomorphic space. If every continuous f : X → X has a fixed point then we say that X has the fixed-point property. Here are a couple of nonmathematical versions of instances of Theorem 2: (a) (Crumpling Map Theorem) Crumpling a map and throwing it over a copy of the same map will always have two corresponding points of the maps on the same vertical line (case n = 2). (b) (Cup of Coffee Theorem) Stirring a cup of coffee will always leave a point unmoved (case n = 3). The following example provides a method for establishing existence of solutions of systems of n equations with n unknowns. Our incomplete proof of the Brouwer fixed point theorem provides justification only when n = 2. Example 3 We want to establish that a system of n equations with n unknowns is consistent (i.e., that a solution exists). h1 ( x1 , x 2 ,… , xn ) = 0 h2 ( x1 , x 2 ,… , xn ) = 0                               

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hn ( x1 , x 2 ,… , xn ) = 0

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258   ◾    The Fundamental Group of a Circle and Applications

Suppose that each of h1 , h2 ,… , hn is a continuous function; then the function f : n → n defined by f ( x ) = (h1 ( x ) + x1 , h2 ( x ) + x 2 ,… , hn ( x ) + xn ) is also continuous. Suppose we can identify any subset A of n such that A ≅ D n, and such that f ( A) ⊂ A. Then, by the Brouwer fixed point theorem, there is an element a = (a1 , a2 ,… , an ) ∈n such that f (a ) = a . This a is ☐ then a solution of the original system, and our goal would thus be accomplished. Example 4 has a similar flavor. Example 4 Let us call a subspace C of  3 a conic subspace if it consists of all of the points on a set of rays originating at (0, 0, 0), and if the intersection of all rays in C and the sphere S 2 is homeomorphic to the closed disk D 2 (Illustration 11.16). For example, the first octant of  3, consisting of all points (x, y, z) such that x ≥ 0, y ≥ 0 and z ≥ 0 , is a conic subspace.

C

~ = D2

Illustration 11.16  In this figure we show the

boundary of a conic space C and the intersection of C with a sphere.

Let f :  3 →  3 be continuous and such that f (C \{(0,0,0)} ⊂ C \ {(0,0,0)} for some conic subspace C. Then, we claim that there is a point x 0 ∈C such that (0, 0, 0), x 0 and  f ( x 0 ) are collinear. This means that we claim that there is an element x 0 ∈C and a number k ∈ + such that f ( x 0 ) = kx 0 . We justify it as follows. Denote the restriction of f to D = S 2 ∩ C by g. Denote the coordinate functions of g by g 1 , g 2 , and g 3 , and define g by g ( x , y , z ) = g 1 ( x , y , z )2 + g 2 ( x , y , z )2 + g 3 ( x , y , z )2 . Observe g maps D into itself. Since D is homeomorphic to D 2 , there must be a fixed that g g point x 0 for . That gives g ( x 0 ) = g ( x 0 ) x 0 , and so f ( x 0 ) = f ( x 0 ) x 0 , that is, g f ( x 0 ) = kx 0 for k = f ( x 0 ) . Here is a seemingly unrelated simple consequence: every 3 × 3 matrix A with positive real entries has a positive eigenvalue. Every such matrix maps the first octant without the origin into itself, and the above argument establishes existence of a vector x such that Ax = A( x ) x . Hence, A( x ) is a positive eigenvalue for A, as asserted. ☐

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11.2  Brouwer Fixed Point Theorem and The Fundamental Theorem of Algebra   ◾    259  

We are still at the early stage of the development of our theory; yet what we have so far is sufficient to deduce a major theorem in algebra. Theorem 3. (The Fundamental Theorem of Algebra) Every polynomial a0 + a1 z + a2 z 2 +  + an z n with complex coefficients and of degree ≥1 has a complex root. Proof. Let an ≠ 0 , so that the degree of p( z ) = a0 + a1z + a2 z 2 +  + an z n is n. For what follows, there is no loss in generality to take that an = 1 . Assume p( z ) = 0 has no complex solution. Hence, p maps  2 into  2 \{(0,0)}. Consider the restriction of p to the circle rS1 = {r (cos2πt ,sin2πt ) : 0 ≤ t ≤ 1} of radius r ≥ 1, and let α : I → rS1 be the loop defined by α(t ) = r (cos2πt , sin2πt ). Then β = p  α is a loop in  2 \{(0,0)} (Illustration 11.17).

I

α

S1

p

zn

Illustration 11.17

Since the restriction of p to the closed disk bounded by rS1 is an extension of p rSit , it follows that the loop β is homotopic to the trivial loop (Exercise 15, Section 10.4). On the other hand, the loop δ we get by composing α with the mapping z n winds n times around the circle r n S1 (Illustration 11.17), and since we have assumed that n ≥ 1, that loop defines a non-trivial element in the fundamental group of r n S1 . Since r n S1 is a deformation retract of  2 \{(0,0)} , it follows that δ defines a non-trivial element in the fundamental group of  2 \{(0,0)} . We will get a contradiction by showing that a conjugate of β and δ are homotopic in  2 \{(0,0)} . We restrict p to rS1, and rearrange slightly: p( z ) = a0 + a1 z + a2 z 2 +  + an−1 z n−1 + z n = 1 z n (1 + q( z )) where q( z ) = n ( a0 + a1z + a2 z 2 +  + an−1z n−1 ). We claim that for sufficiently z large r, the mapping F : rS1 × I →  2 \{(0,0)} defined by F (z , t ) = z n (1 + tq(z )) is a homotopy from z n to p(z) (both restricted to rS1). The only non-obvious part of this claim is that the range of F avoids the origin (0, 0), and we handle it as follows: ( ∗)   a1 an −1  1  a0 1 + tq( z ) ≥ 1 − tq( z ) ≥ 1 − q( z ) ≥ 1 −  n + n −1 +  + ≥ 1 − ( a0 + a1 +  + an −1 ) , z z z  r

and since (∗) could be made as small as we wish by taking sufficiently large r, it follows that for such r we have 1 + tq(z ) > 0.

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260   ◾    The Fundamental Group of a Circle and Applications

We have proved that the loops β and δ are homotopic in  2 \{(0,0)}. It follows from Proposition 4 in 10.3 that a conjugate γβγ −1 of β, and δ, are homotopic relative to {0,1}, and so they define the same element in the fundamental group of  2 \{(0,0)}. We have a contradiction. Exercises 1. Let x 0 ∈S 2 and let B be the (3-dimensional, open) ball B( x 0 , ), where  ∈(0,2). Show that there is no continuous one-to-one mapping S 2 → S 2 such that f ( x ) = x for every x ∈S 2 ∩ B , and f ( x ) ≠ x for every x ∈S 2 \ B . 2. Provide an algebraic proof of the Brouwer fixed-point theorem (Theorem 2) for n = 2. [Hint: Find algebraically the intersection of the lines through x and f ( x ) and the circle S1.] 3. Let X be any nonempty subset of S1 . Find a continuous mapping f : D 2 \ X → D 2 \ X such that f does not have any fixed points. 4. Prove that the torus S1 × S1 is not a retract of the solid torus S1 × D 2. 5. Prove that the boundary B of the triply-twisted band T shown in Illustration 11.18 is not a retract of T.

T

B

   Illustration 11.18

6. Let A be a subspace of X, let a ∈ A. Show that if π1 ( A, a) > π1 ( X , a) then A is not a retract of X. 7. Let A be a retract of a space X, and let X have the fixed-point property. Show that A has the fixed-point property. 8. Show that the subspace of  2 depicted in Illustration 11.19 has the fixed-point property. Illustration 11.19  The space is not homeomorphic to a disk: each of the three objects juggled by the silhouette is touched at a single point.

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11.3  The Jordan Curve Theorem   ◾    261  

9. Show that every tree with finitely many vertices has the fixed-point property. 10. Let X be the space obtained from a sphere by identifying two distinct points. Prove that X is not simply connected. [Note: Prove it!] 11. Let f : D 3 → D 2 , f ( x , y , z ) = ( f1 ( x , y , z ), f 2 ( x , y , z )) be continuous and such that ( f1 ( x , y , z ), f 2 ( x , y , z )) ≤ 1 − z 2 for every ( x , y , z ) ∈D 3 . Prove that there are uncountably many points ( x , y , z ) ∈D 3 such that f ( x , y , z ) = ( x , y ). 12. An infinitely tall cup {( x , y , z ) :  x 2 + y 2 = 1,  z ≥ 0} ∪ D 2 is filled with coffee. Show that stirring the coffee always leaves uncountably many points such that for each of them its original position and its position after the stirring is on the same vertical line. [For those who do not like the often loose jargon of recreational mathematics, you are asked to prove that for every continuous mapping f : C → C , C = {( x , y , z ) :  x 2 + y 2 ≤ 1,  z ≥ 0}, there are uncountably many points x ∈C such that x and f ( x ) are on the same vertical line.] 13. Let f be a continuous mapping T → T , where T = S1 × D 2 is a solid torus. Show that there are uncountably many points x ∈T such that x and f ( x ) are on the same latitudinal circle S1 × { y} , for some y ∈D 2 . 14. Show that the following system is consistent. sin( xy + 1) − x = 0 cos( x + 2 y 2 + 4) − y = 0



15. Show that the following system is consistent. e xyz = cos( x + y + z ) = sin( x + y + z ) =



1 e

2 xyz

e

2 xyz

e

2 xyz

+1

1 +1

1 +1

x y z

11.3  The Jordan Curve Theorem A brief historical note: Bernard Bolzano was the first to state and try to prove Theorem 1 (1808–1813). Jordan’s proof in 1887 contained some gaps, and the first rigorous proof is usually credited to Oswald Veblen (1880–1960), who published it in 1905.

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262   ◾    The Fundamental Group of a Circle and Applications

The main goal in this section is to give a proof of the Jordan curve theorem and to present (without proofs) some important related results. We start with a couple of notions and the statement of the main theorem. A Jordan curve in  2 is the image f (S1 ) of the unit circle S1 under an embedding f : S1 →  2 . A simple arc in  2 is the image f ([a, b]) under an embedding f :[a, b] →  2 . There is no difference between path components and components of open subsets of  2 (Exercise 3 in 6.5). Hence, when we deal with open subsets of  2 (which is what we will do in this section), we will use the word component to also mean path component. Theorem 1. (Jordan Curve Theorem) For every Jordan curve C in  2 , the subspace  2 \ C has two components, one bounded, the other unbounded. The proof we give is based on Maehara [46]. Another proof can be found in Moise’s textbook [55]. The following result is interesting in its own right. Lemma 2. Denote the square {( x , y ) ∈ 2 : x ≤ 1 and y ≤ 1} by S, let f :[−1,1] → S be a continuous function satisfying f (−1) ∈{( x , −1) : x ≤ 1} and f (1) ∈{( x ,1) : x ≤ 1} , and let g :[−1,1] → S be a continuous function satisfying g (−1) ∈{(−1, y ) : y ≤ 1} and g (1) ∈{(1, y ) : y ≤ 1} . Then f ([−1,1]) ∩ g ([−1,1]) ≠ ∅. In other words, under the assumptions of the lemma we assert that the curves defined by f and g must intersect (see Illustration 11.20). y f (1) g (–1) x g (1)

f (–1)

Illustration 11.20  The curves f and g intersect.

Proof. Let f1 and f 2 be the coordinate functions of f, so that f (t ) = ( f1 (t ), f 2 (t )), t ∈[−1,1]. Similarly, let g (t ) = ( g 1 (t ), g 2 (t )), t ∈[−1,1]. We want to prove that there are s, t ∈[−1, 1] such that f (s ) = g (t ). Assume otherwise. So, for every s , t ∈[−1,1], either f1 (s ) ≠ g 1 (t ) or f 2 (s ) ≠ g 2 (t ). Hence, the function H (s , t ) = max f1 (s ) − g 1 (t ) , f 2 (s ) − g 2 (t ) is such  f (t ) − g 1 (s ) g 2 (t ) − f 2 (s )  that H (s , t ) > 0 all the time. Consequently K (s , t ) =  1 is a well,  H (s , t ) H (s , t )  f (t ) − g 1 (s ) g 2 (t ) − f 2 (s ) , ≤ 1, it follows defined function for every s , t ∈[−1,1]. Since −1 ≤ 1 H (s , t ) H (s , t ) that K (S ) ⊂ S. Since the square S is homeomorphic to D 2 , it follows from the Brouwer’s fixed

{

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11.3  The Jordan Curve Theorem   ◾    263  

point theorem that K must have a fixed point. Hence, there is a pair (s0 , t0 ) ∈S such that f (t ) − g 1 (s ) g 2 (t ) − f 2 (s ) K (s0 , t0 ) = (s0 , t0 ). Notice that, for every s , t ∈[−1,1], at least one of 1 , H (s , t ) H (s , t ) is in the set {−1,1}. Hence, either s0 = 1 or t0 = 1. By symmetry we may assume that s0 = 1. f (t ) − g 1 (1) = 1, If s0 = 1 , then K (1, t0 ) = (1, t0 ). This and the definition of K imply that 1 0 H (1, t0 ) which is not possible since f (t0 ) − g 1 (1) = f (t0 ) − 1 ≤ 0 and (as we have noticed earlier) f (t ) − g 1 (−1) = −1 which is also H (s , t ) > 0. If s0 = −1 , then K (−1, t0 ) = (−1, t0 ) and so 1 0 H (1, t0 ) impossible since f1 (t0 ) − g 1 (−1) ≥ 0 . Lemma 3. Let C be a Jordan curve in  2 . For every component U of  2 \ C , we have ∂(U ) = C . Proof. Since C is compact, it is closed in  2 , and so  2 \ C is open. It follows from Proposition 2 in 6.5 that the components of  2 \ C are also open. Consequently, for every component U of  2 \ C , ∂(U ) ⊂ C . Suppose there is a component U of  2 \ C such that ∂(U )  C . Hence, there is a point in c ∈C \ ∂(U ). There is an open neighborhood of c avoiding ∂(U ), or else c ∈∂(∂(U )), which could not be, since ∂(∂(U )) ⊂ ∂(U ) (Exercise 10(b) in 3.2). It follows that ∂(U ) is a subset of a closed arc A in C (where a closed arc in C is the image of a closed arc in S1 under the embedding sending S1 onto C). Let D be any closed disk such that C ⊂ D . Since the closed simple arc A is homeomorphic to any closed interval on  , the Tietze Extension Theorem (Theorem 1 in 9.2; see also Exercise 1 in Section 9.2) implies that the identity mapping A → A can be extended to a continuous g : D → A. There are two cases: (i) U is bounded, and (ii) U is unbounded.  g ( x ) if x ∈U . Observe In case (i) we define a mapping h : D → D as follows: h( x ) =  c  x if x ∈U ∩ D that U ⊂ D or else there will be points in ∂U \ D . Since U and U c are closed, and since g is the identity over ∂(U ) = U ∩ U c , it follows from the gluing lemma that h is continuous. Since U ⊂ D, it follows that h ∂ D = id ∂ D . Hence, ∂D is a retract of D, contradicting Example 1 in 11.2.  x if x ∈U . Case (ii) requires a similar argument: define h : D → D by h( x ) =  c  g ( x ) if x ∈U ∩ D  This time U c must be a subset of D, and the same argument as in part (i) brings us to the conclusion that ∂D is a retract of D, again achieving a contradiction. Hence, it is not possible that ∂(U )  C . Thereby ∂(U ) = C . We are ready for the proof of the main theorem. Proof of Theorem 1. Step 1. In the first part of the proof we will find a point in a bounded component of  2 \C. (That this point is indeed in a bounded component of  2 \C will be shown in Step 2.) Since C is closed, there are two points a, b ∈C such that d(a, b) is maximal. We may assume that d(a, b) = 2, and that a = (−1,0), b = (1,0). Hence, C lies within the rectangle R = {( x , y ) ∈ 2 : −1 ≤ x ≤ 1, − 1.1 ≤ y ≤ 1.1} , and a and b are the only points in ∂ R ∩ C .

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264   ◾    The Fundamental Group of a Circle and Applications

The points a and b split C into two simple curves Cn and Cs (sharing the endpoints a and b). By Lemma 2 both of these two simple curves intersect the vertical line l = {( x , y ) ∈ 2 : x = 0, −1.1 ≤ y ≤ 1.1}. Identify the point cn in C ∩ l that is closest to the point p = (0,1.1), and let Cn be the one of the two curves Cn and Cs that contains cn . Let dn be the point in Cn ∩ l farthest from p; it could happen that dn = cn . We illustrate all this in Illustration 11.21. R

p cn Cn dn b

a

Cs l

Illustration 11.21

From now on, for convenience, we will use the following notation. For distinct x , y ∈R we will denote the path from x to y along the line segment between x and y by xy ; we will use the same notation for the image of that path (an arc in  2 ). If f is a path in  2 and if f (s ) = x and , f (t ) = y for some s , t ∈I , then we will denote the restriction f [s , t ] , as well as its image by xy where f will either be clear from the context or will be mentioned explicitly. Denote q = (0, −1.1). It has to be that dn q ∩ Cs ≠ ∅, for otherwise the product of paths pcn c n dn dn q and the path Cs do not intersect, contradicting Lemma 2. Let cs ∈Cs ∩ dn q be the closest to q, and ds ∈Cs ∩ dn q farthest from q. Finally, let z be the point in the middle of the line segment dn ds . See Illustration 11.22. R

p cn

Cn dn z

a

ds Cs

b

ds cs l q

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Illustration 11.22

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11.3  The Jordan Curve Theorem   ◾    265  

Step 2. Let U be the component of  2 \C containing z. In this step we show that U is bounded. Suppose U is the only unbounded component of  2 \C (see Exercise 1). Then there is a path λ in U leading from z to a point outside R. Since ∂ R splits  2 into two components, there must be a point w ∈∂ R where λ meets ∂ R for the first time. Since a, b ∈C, w is neither a nor b. By symmetry, we may assume that w is on the vertical line segment containing b, and below b (see Illustration 11.23). R

p cn pcn

Cn

cndn dn

a

dnz

z ds Cs

b

zw

w

cs l

λ

δ

q

Illustration 11.23

Then the path δ from w vertically down to the corner (1,−1.1), and then horizontally to q    does not touch a and b. Look at the path pcn c n dn dn z zw δ , where cn dn is along Cn and zw is along λ: it does not intersect Cs , contradicting Lemma 2. Step 3. We show that U is the only bounded component of  2 \ C , and the theorem will be proven. Let W be another bounded component of  2 \ C . It follows from the argument given in  Step 2 that W ⊂ R \ C . The path α = pcn c n dn dn ds ds c s c s q (Illustration 11.24) does not intersect W since each of the indicated sub-paths is either in U, or in the unbounded component of  2 \C , or it is a part of C. R

p cn

Cn dn a

a1

z

b1 b

ds

Ua Cs

Ub

cs l

α β

q



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Illustration 11.24

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266   ◾    The Fundamental Group of a Circle and Applications

Since a and b are not points on α, they have open neighborhoods U a and U b , respectively, avoiding α. By Lemma 3, we have that a, b ∈∂W . Hence, there are a1 ∈W ∩ U a and b1 ∈W ∩ U b . Let β be any path in W from a1 to b1 . Then aa1 β b1b is path from a to b which does not intersect α (Illustration 11.24). This contradicts Lemma 2, and the proof is completed. The following is the version of Jordan curve theorem for S 2 in place of  2. Corollary 4. Let f : S1 → S 2 be an embedding. Then S 2 \ f (S1 ) has two components, and each of these two components has f (S1 ) as its boundary. Proof. Since f (S1 ) ≠ S 2 , there is x ∈S 2 \ f (S1 ) . We may assume that x is the north pole of the sphere. The stereographic projection s sends S 2 \{x } homeomorphically onto  2 , and so s( f (S1 )) is a Jordan curve. By Theorem 1, its complement in  2 has two components, one bounded, the other unbounded. By Lemma 2, the boundary of each of them is s( f (S1 )) . Denote that unbounded component by U and the bounded one by V. Then s −1 (U ) ∪ {x } and s −1 (V ) are the components of S 2 \ f (S1 ) , and f (S1 ) is the boundary of each of them. The Jordan curve theorem has been generalized in many directions. Theorem 5 is the higher dimensional version; it was proven by Brouwer in 1912. Its proof is beyond the scope of this book. Theorem 5. (Jordan–Brouwer Separation Theorem) If f : S n → n +1 is an embedding, then n +1 \ f (S n ) has two components, and f (S n ) is the boundary of each of these two components. A consequence is the Invariance of Domain Theorem, stated and used in Section 4.4. For a proof see, for example, [31], page 172. Another important generalization of the Jordan curve theorem is the following result. For a proof, see, for example, [73]. Theorem 6. (Jordan-Schönflies Theorem) Every embedding f : S1 →  2 can be extended to a homeomorphism ϕ :  2 →  2 . Theorem 6 does not allow a straightforward generalization to higher dimensions. Specifically, there exist embeddings f : S 2 →  3 that do not have extensions ϕ :  3 →  3 . We will exhibit two of them in Section 13.4.

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11.3  The Jordan Curve Theorem   ◾    267  

Exercises 1. Show that if C is a Jordan curve, then  2 \ C has exactly one unbounded component. 2. Show that if f :  →  2 is an embedding such that f () \ D has at least two components for every disk D intersecting f (), then  2 \ f ( ) has two components. 3. Prove that the Jordan–Schoenflies theorem (Theorem 6) implies the Jordan curve theorem (Theorem 1). 4. Let X = S 2 \{(0,0,1),(0,0, −1)}, where S 2 is the unit sphere. Show that the mapping f : X → X defined by f ( x , y , z ) = (− x , − y , − z ) is not ambient isotopic within X to id X . [Hint: transfer the statement of the problem from X to  2 \{(0,0)} by means of the stereographic projection, then use the Jordan curve theorem.] 5. [This problem is a commentary on the Invariance of Domain Theorem, but that theorem is not needed to solve the problem.] (a) Find a subset A of  2 , not closed in  2 , and find a continuous, bijective f : A→  2 such that f is not open. (b) Show that if K is a compact subset of  2, and if f : K→  2 is continuous and bijective, then f is open. (c) Find an example of a continuous and bijective mapping f : K→  2 over closed subspace K of  2 , such that f is not open. 6. (a) Let A, B, C and D be points on S1 in clockwise (or counterclockwise) order. Show that if α is a path from A to C in D 2, and if β is a path from B to D in D 2, then α( I ) ∩β( I ) ≠ ∅. (b) Use the results given in this section to show that the Möbius band is not embeddable in  2 .

7. Show that if f :[0,1] →  2 is an embedding, then  2 \ f ([0,1]) is connected. (That is, show that arcs do not separate  2 ).

8. Prove that the open Möbius band (= Möbius band without the points on the boundary) is not homeomorphic to the open cylinder S1 × (0,1).

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Chapter

12

Combinatorial Group Theory

I

n order to strengthen the π1 ( X , x )-connection between groups and spaces we need more knowledge of group theory. A natural group theoretical setup in this context is combinatorial group theory, and its branch, geometric group theory. In this chapter we give a brief introduction to this area of group theory. In addition to the basic group theoretical terminology mentioned at the beginning of 10.4, the reader is expected to be familiar with the following notions: subgroup, normal subgroup (N is a normal subgroup of G, denoted N  G, if gNg −1 = N for every g ∈ G ), index of a subgroup in a group, coset, and quotient group or factor group. Other basic terms will be encountered (and defined) along the way. There will not be many illustrations in the chapter, since visualizing this part requires an approach that would take us too far away from the mainstream exposition.

12.1  Group Presentations A brief historical note: The concept of defining groups through group presentation is due to Walther von Dyck (1856–1934); the corresponding paper was published in 1882.

For any nonempty set A = {a, b,…}, we will denote the set of formal inverses of the elements of A by A−1 = {a −1 , b −1 ,…}, and A ± will stand for A ∪ A−1. A word on A ± is a finite sequence of elements of A ± . For every x ∈ A ± we denote the word xxx…x, n-times, by x n , and we denote the word ( x −1 )n by x − n. For example, b −2a3bcc −4 denotes a word over the set A ± , where A = {a, b, c}. The number of appearances of elements of A ± in a word W is called the length of the word W, denoted W . For example, b−2a3bcc −4 = 11 . The empty word is the  word of length 0. If W = x1x 2 … xm−1xm , we denote W −1 = xm−1x m−1−1 … x 2−1x1−1, where we write ( x −1 )−1 = x for every x ∈ A. For every two words U and V, UV is the word we get by concatenating the word V after the word U. 269

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Given any set R = {P , Q ,…} of words on A ± , we now proceed to define a group which we will denote by A  ; R , or by a, b,…;  P , Q ,… (where, despite what the notation may suggest, we do not put any constraints on the cardinalities of the sets {a, b,…} and {P , Q ,…}). Step 1. Define an equivalence relation ~ on the set of all words on A ± with respect to R as follows: U ~ V if there is a sequence of words W1 = U ,W2 ,…,Wk−1 ,Wk = V such that for every j = 1,2,… , k − 1, Wj+1 is obtained from Wj by applying one of the following operations:

(a) Insert or delete xx −1, x ∈ A ±, somewhere in the word Wj ;



(b) Insert or delete a word X, X ∈{P , Q ,…} ∪ {P −1 , Q −1 ,…} somewhere in the word Wj .

It is straightforward to show that in this way we indeed get an equivalence relation. Denote the equivalence class of the word U on A ± by [U]. The set of all such equivalence classes will be denoted by 〈 A  ; R 〉. Step 2. Define an operation over the set 〈 A  ; R 〉 by [U ][V] = [UV]. This turns 〈 A  ; R 〉 into a group. We outline an argument justifying this claim; the reader is invited to supply the details (Exercise 1).

(i) The operation is well defined; this means that if U1 ~ U 2 and V1 ~ V2 , then [U1 ][V1 ] = [U 2 ][V2 ]. One way to prove this is by induction on the total number of steps of types (a) and (b) (as defined in Step 1) needed to make U1 ~ U 2 and V1 ~ V2 .



(ii) Associativity, or ([U ][V ])[W ] = [U ]([V ][W ]) : use induction on W . (iii) Denote the equivalence class of the empty word by []. For every word U, [U ][] = [][U ] = [U ], and so [] plays the role of the identity, which we will usually denote by 1.



(iv) [U ][U −1 ] = [U −1 ][U ] = [ ] for every word U. This is easy.

With this we have established that 〈 A  ; R 〉 = 〈a, b, … ;  P , Q ,…〉 , together with the operation just defined, is a group. It is generated by A = {a, b,…}. It is obvious that [P] = 1,[Q] = 1,… in 〈 A  ; R 〉. If U and V are two words on A ± such that [U ] = [V ] is satisfied in 〈 A  ; R 〉, then we will say that [U ] = [V ] is a relation in 〈 A  ; R 〉, and that it is a consequence of the relations [P] = 1,[Q] = 1, … . The relations [P] = 1,[Q] = 1, … are called the defining relations for the group. If [W ] = 1 is a relation in 〈 A  ; R 〉, then 〈 A  ; R 〉 and 〈 A  ; R ∪ {W }〉 are the same group (Exercise 2). In order to ease the notation, and in the context of the group 〈a, b, … ;  P , Q ,…〉 , we will often write U for [U ]. Thus, for example, we will often write P = 1, Q = 1,…, instead of [P] = 1,[Q] = 1,… . Along these lines, we will sometimes write 〈a, b, … ;  P = 1, Q = 1,…〉 instead of 〈a, b, … ;  P , Q , …〉 . Example 1 Consider the group 〈a ; a 2 〉 : it follows by induction on the length of words that every word on {a}± is equivalent to either 1 or a. That 1 and a are distinct in the group ­follows

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from the observation that the operations (a) and (b) of Step 1 do not change the parity ☐ of the length of the word. We conclude that 〈a ; a 2 〉 is a cyclic group of order 2. In the preceding example we needed a paragraph to state the “obvious”: that 〈a ; a 2 〉 is cyclic of order 2. To increase the efficiency of our theory we develop it a little further. Let G be a group, and let f : A → G be any mapping. Define a mapping, also denoted by f, from the set of all words on A ± into G as follows:   (1)  f ( x1α1 x 2αn … xnαn ) = ( f ( x1 ))

α1

( f (x 2 ))α … ( f (xn ))α , xi ∈ A,  αi ∈{1, −1} . n

2

We would like to extend it further and define a homomorphism f : 〈 A  ; R 〉 → G by   (2)  f ([x1α1 x 2αn … xnαn ]) = ( f ( x1 ))

α1

( f (x 2 ))α … ( f (xn ))α , xi ∈ A,  αi ∈{1, −1} . 2

n

However, this last mapping f need not be well defined. The question of when the last f is a well-defined homomorphism is settled by the following proposition. Proposition 1. For every group G, the extension f : 〈a, b,… ;  P , Q ,…〉 → G of f : A → G , as defined in (2), is a well-defined homomorphism if and only if f ( X ) = 1 in G for every X ∈{P , Q ,…}.

Proof. ⇒ This is clear, since the identity element is mapped to the identity element by any homomorphisms. ⇐ Suppose [U ] = [V ] in 〈a, b,… ;  P , Q ,…〉 . This means that U ~ V with respect to {P , Q ,…}, which in turn means that we can get V from U by means of the operations (a) and (b) as in Step 1 above. It is easy to see that if one such operation is required to get U ~ V, then  f (U ) = f (V ), hence f ([U ]) = f ([V ]). For example, if U = U1U 2 and V = U1PU 2 , then f (U ) = f (U1 ) f (U 2 ) and f (V ) = f (U1 ) f ( P ) f (U 2 ), and so, since we have assumed in this part of the proof that f ( P ) = 1, we have that f (V ) = f (U ) in G. Induction on the number of operations (a) and (b) needed to get V from U implies that f (U ) = f (V ). This shows f is well defined. That f is a homomorphism is an immediate consequence of its definition. Remark. We have used the same f with three different meanings: a mapping f : A → G , a mapping f : A ± → G defined by (1), and the homeomorphism defined f : 〈 A  ; R 〉 → G by (2). We will almost always deal with the first and the last meanings. Occasionally, we will want to explicitly distinguish them. Otherwise, we will consistently denote them by the same letter; the context will resolve any ambiguity. When the extension 〈 A  ; R 〉 → G of f : A → G , is an isomorphism, then we say that 〈 A  ; R 〉 is a presentation of the group G with respect to f : A → G . Alternatively, we say that G is presented by 〈 A  ; R 〉 with respect to f. If the set of generators and defining relations is finite, then we say that F is finitely presented by 〈 A  ; R 〉 (with respect to f ). Every group G can be presented by 〈{x  :  x ∈ G} ;  {xyz −1  :  x , y , z ∈ G and xy = z in G  }〉 (with respect to the identity mapping G → G ). One of the primary objectives in this

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context is to find presentations with small (possibly minimal) sets of generators and defining relations. We will be particularly interested in finitely presented groups. Example 2 Recall the cycle notation for permutations: ( x1   x 2 … xn ) denotes the permutation sending xi to xi+1, for i = 1,2,…, n − 1, sending xn to x1 and fixing the other elements of the underlying set. In the context of group presentations it is convenient to make an exception to our longstanding convention and accept that compositions (products) p1 p2 of permutations are from left to right. It is easy to confirm that the group G of permutations of the set {1,2,3} is generated by (1 2  3) and (1 2). We claim that 〈a, b ; a3 , b 2 , aba −2b〉 is a presentation of G with respect to f : a  (1 2 3), b  (1 2). Computation shows that (1 2 3)3 = id , (1 2)2 = id and that (1 2 3)(1 2) = (1 2)(1 2 3)2 confirming that f (a3 ) , f (b 2 ) , and f (aba −2b) are all the identity in G. Proposition 1 implies that the extension f : 〈a, b ; a3 , b 2 , aba −2b〉 → G is a well defined epimorphism. It remains to be shown that it is one-to-one. The identity aba −2b = 1 implies that ab = b −1a 2 . It follows that every word in the presentation is equivalent to some of  {anbm : n, m ∈}. Further, the identities a3 = 1 and b 2 = 1 allow us to further reduce the class of all equivalence classes of words in the presentation to {a0 = 1, a, a 2 , b, ab, a 2b}. So, 〈a, b ; a3 , b 2 , aba −2b〉 has no more than 6 elements. Since we have already established that f : 〈a, b ; a3 , b 2 , aba −2b〉 → G is onto, and since G ☐ has 6 elements, it follows that f must also be one-to-one. We recapitulate: in order to establish that 〈 A  ; R 〉 is a presentation of a group G with respect to f : A → G we need to (i) show that f : 〈 A  ; R 〉 → G is a well-defined homomorphism, (ii) show that f is onto, and (iii) show that f is one-to-one. Part (i) needs Proposition 1 and (ii) is equivalent to the claim that f(A) generates G. It is part (iii) that often needs more attention. Recall the following fact in basic group theory: a group homomorphism f : H → K is one-toone if and only if Ker ( f ) = {1}, where Ker( f ) = {x ∈H : f ( x ) = 1 in K } is the kernel of f. Here is a simple procedure for establishing that a homomorphism f : 〈 A  ; R 〉 → G is one-to-one: find a set of words representing all elements of A  ; R and such that for only one element W of that set of words, f (W ) = 1. Then Ker ( f ) = {1} and so f must be one-to-one. The case when G is finite is simpler: if (i) and (ii) are checked, then, since an onto mapping between two finite sets of equal cardinality must also be one to one, the only thing we need to do to confirm (iii) is to find G many words representing all elements of A  ; R . This is what we did in Example 2. An example where G is infinite follows.

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Example 3 A symmetry of a subspace A of n is an isometry of n mapping the subspace A onto itself. Consider the group G of all symmetries (operation is composition) of the Frieze pattern F shown in Illustration 12.1, regarded as a subspace of  2. A–2

A–1

B–1

A0

B0

A1

A2

B2

B1

v

Illustration 12.1  The Frieze pattern F.

It is easy to identify the elements of the group G of the symmetries of F, and then to show that the group G is generated by the translation t along the vector v (shown in Illustration 12.1) and the rotation r around A0 through 180°. For example, the rotation around B1 through 180° is r  t −1. We show that 〈 x , y ;  xyx −1 y = 1,  x 2 = 1〉 is a presentation of G under the mapping x → r , y → t . Checking that rtr −1t = 1 and r 2 = 1 are satisfied in G is straightforward. It remains to be shown that the extension f : 〈 x , y ;  xyx −1 y = 1,  x 2 = 1〉 → G of x  r,  y  t is one-to-one. The relation xyx −1 y = 1  can be written as xy = y −1x. This, together with x 2 = 1 implies that every word in 〈 x , y ;  xyx −1 y = 1,  x 2 = 1〉 is equivalent to some of y n x ε , n ∈, ε ∈{0,1}. The only such word that is sent to 1 in G is y 0 x 0 = 1, and so f ☐ is one-to-one. Example 4 Let G be the subgroup of the group of symmetries of the plane generated by the translations t u along the vector u = (1,0), and t v along the vector v = (0,1) . We claim that P = a, b  ; aba−1b−1 = 1 is a presentation of G with respect to a  t u , b  t v . It is easy to check that the extension f : P → G is a well-defined epimorphism. To prove that f  is one-to-one, first observe that aba −1b −1 = 1 tells us that ab = ba , that is, that the group P is commutative. It follows that the set {anbm : n, m ∈} contains all of the representatives of the equivalence classes of the words in P . Of all the elements of that set only 1 = a0b0 is mapped to 1 ∈ G , and so f is indeed one-to-one. ☐ There is something deeper going on in Example 4 than what may appear at first sight. We showed in this example that the group G was isomorphic to  ⊕ , a free commutative group on two generators. We already know that the fundamental group of the torus is isomorphic to  ⊕ . On the other hand, the quotient space of the plane  2 obtained by identifying all orbits { g ( x ) : g ∈ G} of the elements x ∈ 2 under the action of G is precisely the torus (see Exercise 6 in 4.4). As we will see in Example 3, 16.3, this is not a coincidence.

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Exercises 1. Complete the details of Step 2, showing that 〈a, b,… ;  P , Q ,…〉 is indeed a group. 2. (a) Show that if [U ] = 1 is a relation in a, b,…;  P , Q ,… , then the groups 〈a, b,… ; P , Q ,…〉 and 〈a, b,… ;  P , Q ,… , U 〉 are identical. (b) Let U be any word on {a, b,…}± and let x ∉{a, b,…}± . Show that the groups 〈a, b,… ;  P , Q ,…〉 and 〈a, b,… , x ;  P , Q ,… , Ux −1 〉 are isomorphic. 3. Let G be the group of symmetries of the Frieze pattern shown in Illustration 12.2. First, guess a presentation for G, then prove your guess.

   Illustration 12.2 

4. Show that the group of symmetries G of the Frieze pattern F shown in Illustration 12.1, Example 3, is generated by the translation t and the rotation r ′ centered at B0 and through 180°, find a presentation of G on the generating set {t , r ′}, and prove that it is indeed a presentation of G. 5. Let G be the subgroup of the group of all symmetries of  2 generated by the translation t v along the vector v = (0,1) and the composition g of the translation along the vector u = (1,0) followed by the reflection with respect to the x-axis. Guess a presentation for G, then justify your guess.  ε k  6. ([47]) Let G be the group of matrices of the type   over the ring of integers  0 1  modulo n, where ε ∈{−1,1}, k ∈{0,1,… , n − 1} (the operation is multiplication of matrices). Show that 〈a, b ; an , b 2 , abab −1 〉 l is a presentation of G under the mapping  1 1   −1 0  a   , b   . 1   0 1   0 7. The group G of all symmetries of the subspace of  2 depicted in Illustration 12.3 is generated by the translation tb along the vector b and the glide reflection g that is the composition of the translation f = t a 2 along the vector a 2 followed by the reflection with respect to the vertical line l. Show that the group G has a presentation 〈a, b  ; aba−1b〉 under the mapping a  g, b  f.

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a

b

Illustration 12.3

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8. Show that the group 〈 x , y ; yx −1 y −1 x 2 , y −1 x −1 yxyx 〉 is trivial. 9. Show that the group 〈a, b, c  ; a3, b 2 , c 3, aba 2b, a 2c 2ac , bcbc −1 〉 has exactly 18 elements. (Note: it is easy to show that the group has ≤ 18 elements; showing that the number of elements is exactly 18 requires a strategy.) 10. Show that 〈a, b  ; a 4 , b 2 , (ab)2 〉 is a presentation for the group of symmetries of a square. 11. Show that the groups 〈a, b, c  ; ababab 2 〉 and 〈 x , y  ; ∅〉 are isomorphic. [Hint: consider a  x 4 , b  x −3 , c  y.] 12. Show that if N is a normal subgroup of G = 〈a, b, … ;  P , Q , …〉 generated by {x j : j ∈ J }, then 〈a, b, … ;  P , Q , … , {x j  :  j ∈ J }〉 is a presentation of the group G N .

12.2  Free Groups, Tietze, Dehn A brief historical note: The group-theoretical work of both Heinrich Tietze and Max Dehn was inspired by Poincaré’s introduction of the fundamental group of a space. Tietze introduced (what were later called) Tietze transformations in a paper published in 1908, and Dehn stated his three fundamental problems in 1911, in a paper dedicated entirely to group presentations.

A special and important class of groups is the class of free groups. A group F is free if it is isomorphic to a group of type 〈a, b,… ; ∅〉, that is, if it could be presented with the empty set of defining relations. In this setting, the set {a, b,…} is called the set of free generators of F, or we say that F is freely generated by that set. The free group that is freely generated by the empty set is the trivial group. The simplest nontrivial free group is the infinite cyclic group 〈a  ; ∅〉. If a free group is freely generated by the sets A and B, then A = B (Exercise 13). The cardinality {a, b,… } of the set of free generators is the rank of F. It follows directly from the definition of presentations that free groups satisfy only the relations satisfied in all groups, namely those that are consequences of the relations of the type xx −1 = 1, x ∈{a, b,… }± . Denote G = 〈a, b,… ;  P , Q , …〉 , and F = 〈a, b,… ; ∅〉. Proposition 1 in 12.1 implies that the identity mapping over {a, b,…} can be extended to a homomorphism f : F → G . Since f is onto, it follows from the first isomorphism theorem for groups that the group G is isomorphic to the factor group F Ker ( f ). So, every group is a quotient group of the free group on the same generating set. The normal subgroup Ker(f) of F is the normal closure of the set {P , Q ,…} of elements in F (Exercise 1), where the normal closure NC(A) of a subset A of a group H is the smallest normal subgroup of H containing A as a subset. The elements of NC(A) are products of conjugates in H of the elements of A (the last claim is an easy exercise). Hence we have the following:

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Remark. If G = 〈a, b,… ; P , Q ,…〉, and if F = 〈a, b,… ; ∅〉, then G ≅ F / NC({P , Q ,…}). A group presented with a nonempty set of defining relations may be free. So, it makes sense to ask the following questions: if we know that a presentation defines a free group, is there a procedure that will transform any presentation of that group into a relations-free presentation? More generally, if two presentations define the same group, is there a procedure to change one of them into the other without altering (up to isomorphism) the group they define in each step. The affirmative answer was established by Tietze in 1908 in the following theorem. Theorem 1. If 〈a1 , a2 ,… ;  P1 , P2 , …〉 and 〈b1 , b2 ,… ; Q1 , Q2 , …〉 are two presentations of a single group, then the presentation 〈b1 , b2 ,… ; Q1 , Q2 , …〉 can be obtained from the presentation 〈a1 , a2 ,… ;  P1 , P2 , …〉 by applying finitely many steps of the following type, where each step yields a presentation of the same group. (T1) If Y1 ,Y2 ,… are consequences of X1 , X 2 ,…, then adjoin Y1 ,Y2 ,… in the set of defining relations of 〈 x1 , x 2 ,… ;  X1 , X 2 , …〉 to get 〈 x1 , x 2 ,… ;  X1 , X 2 , … , Y1 , Y2 ,… 〉 . (T2) [Reverse (T1)] If Y1 ,Y2 ,… are consequences of X1 , X 2 ,…, then delete Y1 ,Y2 ,… from the set of defining relations of 〈 x1 , x 2 ,… ;  X1 , X 2 , … , Y1 ,Y2 ,…〉 to get 〈 x1 , x 2 ,… ;  X1 , X 2 , …〉 . (T3) If W1 ( x1 , x 2 ,…), W2 ( x1 , x 2 ,…), … are words on {x1 , x 2 ,…}± , and y1 , y 2 ,… are new symbols, then change 〈 x1 , x 2 ,… ; X1 , X 2 ,…〉 to 〈x1 , x 2 ,… , y1 , y 2 ,… ; X1 , X 2 , … ,   y1 = W1 ( x1 , x 2 ,…), y 2 = W2 ( x1 , x 2 ,…) ,…〉. (T4) [Reverse (T3)] Change to 〈 x1 , x 2 ,… ,  y1 , y 2 ,… ;  X1 , X 2 , … ,  y1 = W1 ( x1 , x 2 ,…), y 2 = W2 ( x1 , x 2 ,…) ,…〉 to 〈 x1 , x 2 ,… ;  X1 , X 2 , … 〉. Remark. We do not assume in the statement of the theorem and in the proof below that the sets of generators and defining relations are countable, as the notation might suggest. Proof. (In a nutshell, we use (T1) and (T3) to change 〈a1 , a2 ,… ;  P1 , P2 , … 〉 to a presentation that is symmetric with respect to the two generating sets, then we apply (T2) and (T4) to arrive at 〈b1 , b2 ,… ; Q1 , Q2 , …〉 .) Suppose 〈a1 , a2 ,… ;  P1 , P2 , … 〉 is a presentation of G with respect to a1  g 1 , a2  g 2 ,…, and suppose 〈b1 , b2 ,… ; Q1 , Q2 , …〉 is a presentation of the same group with respect to b1  h1 , b2  h2 ,…. Since the set { g 1 , g 2 ,…} generates G, each of h1 , h2 ,… can be written as a word on { g 1 , g 2 ,…}± . Suppose h1 = W1 ( g 1 , g 2 ,…), h2 = W2 ( g 1 , g 2 ,…), …. It is easy to check (see Exercise 3) that 〈a1 , a2 ,… , b1 , b2 ,… ;  P1 , P2 , … , b1 = W1 (a1 , a2 ,…), b2 = W2 (a1 , a2 ,…),…  〉 is also a presentation of G under the mapping a1  g 1 , a2  g 2 ,…, b1  h1 , b2  h2 ,…. We have used (T3). Each of Q1, Q2,… is a word on {b1 , b2 ,…}± ; to emphasize that, we write Qi = Qi (b1 , b2 ,…). Since 〈b1 , b2 ,… ; Q1 (b1 , b2 ,…), Q2 (b1 , b2 ,…), …〉 is a presentation of G with respect to b1  h1 , b2  h2 ,… , it follows that Q1 (h1 , h2 ,…) = 1, Q2 (h1 , h2 ,…) = 1,… in G. Consequently, adjoining Q1 , Q2 ,… to the set of defining relations of 〈a1 , a2 ,… , b1 , b2 ,… ;  P1 , P2 , … ,  b1 = W1 (a1 , a2 ,…), b2 = W2 (a1 , a2 ,…),…〉 , and geting 〈a1 , a2 ,… , b1 , b2 ,… ;  P1 , P2 , … , Q1 , Q2 ,…,

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b1 = W1 (a1 , a2 ,…), b2 = W2 (a1 , a2 ,…),…〉 as the result, again yields a presentation of G under a1  g 1 , a2  g 2 ,… , b1  h1 , b2  h2 ,… . We have used (T1) in this step. Since h1 , h2 ,… also generate G, we can write g 1 = U1 (h1 , h2 ,…), g 2 = U 2 (h1 , h2 ,…), … for some words U1 ,U 2 ,…. Consequently, a1 = U1 (b1 , b2 ,…), a2 = U 2 (b1 , b2 ,…), … are true in 〈a1 , a2 ,…, b1 , b2 ,…;  P1 , P2 , …, Q1 , Q2 ,…, b1 = W1 (a1 , a2 ,…), b2 = W2 (a1 , a2 ,…),…〉 and adding them as relations will not affect the group. Thus, we get the following presentation of the same group with respect to the same mapping a1  g 1 , a2  g 2 ,… , b1  h1 , b2  h2 ,…: 〈a1 , a2 ,…, b1 , b2 ,…;  P1 , P2 , …, Q1 , Q2 , …, b1 = W1 (a1 , a2 ,…), b2 = W2 (a1 , a2 ,… ),… , a1 = U1 (b1 , b2 ,… ), a2 = U 2 (b1 , b2 ,…),… 〉 We have used (T3) again. In the last presentation there is symmetry between {a1 , a2 ,…} and {b1 , b2 ,…}; so tracing the steps backwards (thus employing (T2) and (T4)), and interchanging the roles of {a1 , a2 ,…} and {b1 , b2 ,…} in the process, leads us to 〈b1 , b2 ,…; Q1 , Q2 , …〉 as a presentation of G with respect to b1  h1 , b2  h2 ,… . The transformations (T1)–(T4) are called Tietze transformations. Theorem 1 can be rephrased as follows. Theorem 2.  The presentations 〈a1 , a2 ,… ;  P1 , P2 , … 〉 and 〈b1 , b2 ,… ; Q1 , Q2 , …〉 define isomorphic groups if and only if one of them can be obtained from the other by means of Tietze transformations. Proof.  ⇒ Let f : 〈a1 , a2 ,… ;  P1 , P2 , … 〉 → 〈b1 , b2 ,… ; Q1 , Q2 , … 〉 be an isomorphism. Then 〈a1 , a2 ,…; P1 , P2 , …〉 is a presentation of the group 〈b1 , b2 ,… ; Q1 , Q2 , …〉 with respect to f. On the other hand 〈b1 , b2 ,… ; Q1 , Q2 , …〉 is certainly a presentation of itself (eith respect to the identity mapping). The result now follows from Theorem 1. ⇐ This is an immediate consequence of the fact that Tietze transformations do not affect the isomorphism class of the group, as noted in the proof of Theorem 1. We will say that two presentations are equivalent if one can be obtained from the other by means of Tietze transformations. Corollary 3. 〈a1 , a2 ,… ;  P1 , P2 , … 〉 and 〈b1 , b2 ,… ; Q1 , Q2 , …〉 are presentations of a single group if and only if they are equivalent. Proof. ⇒ is established in Theorem 1. ⇐ By Theorem 2, equivalent presentations define isomorphic groups. On the other hand, it is indicated in the proof of Theorem 2 that two isomorphic presentations are presentations of a single group. Example 1 In a few chapters the presentations 〈a, b, d ; ab = ba, d 2 = a, da = ad , db = b −1d 〉 and 〈b, d ; dbd −1b〉 will arise as the fundamental group of a single space, using two different methods. By Theorem 1 there must be a sequence of Tietze transformations

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changing one to the other. The proof of Theorem 1 actually establishes a generic— but usually rather inefficient—sequence of Tietze transformations. Here is a shorter sequence of Tietze transformations, leading from one to the other presentation in this example: 4) 〈a, b, d ; ab = ba, d 2 = a, da = ad , db = b −1d 〉 (T →       2) 2 2 −1 2) 〈d , b ; d 2b = bd 2 , dd 2 = d 2d , db = b −1d 〉 (T 〈 d → , b ; d b = bd , db = b d 〉 (T →

〈d , b ; db = b −1d 〉 = 〈d , b ; dbd −1b〉

In the last step above we have eliminated d 2b = bd 2 ; we can do that since it is a consequence of db = b −1d, as follows: d 2b = d(db) = d(b −1d ) = (db −1 )d = (bd )d = bd 2 ☐ (and note that bd = db −1 is a consequence of db = b −1d ). Example 2 Consider the group 〈a, b, c ; a3 , b 2 , c 3 , aba 2c , acac 2 〉 . We use Tietze transformations to show that this group is (isomorphic to) the trivial group. First we use the third relation to get c −1 = aba 2, and then c = a −2b −1a −1 = aba 2. Hence c = c −1 , which yields c 2 = 1. This together with c 3 = 1 implies that c = 1. We can then eliminate c using (T1) and (T2) to get 〈a, b, c ; a3 , b 2 , c 3 , aba 2c , acac 2 〉 → 〈a, b ; a3 , b 2 , aba 2 , a 2 〉 . Since a3 = 1 = a 2 implies that a = 1, we can remove a in the same way to get 〈a, b ; a3 , b 2 , aba 2 , a 2 〉 → 〈b ; b 2 , b〉 , ☐ and this is clearly the trivial group. In Example 2 we managed to show that the presentation defines the trivial group. Could determining if any finitely presented group is trivial be done algorithmically? The answer is an emphatic “No!” More importantly, we have arrived at Dehn’s fundamental problems1. We first state these problems in terms of fundamental groups. Recall that two elements x, and y in a group G are conjugate if there is an element a ∈ G such that x = aya −1. (D1)∗ Is there an algorithm to decide if, in a connected space with a known presentation of its fundamental group, a loop is homotopic to the trivial loop? (D2)∗ Is there an algorithm to decide if, in a connected space with a known presentation of its fundamental group, the homotopy class of a loop at the base point is conjugate to the homotopy class of another loop at the base point? (D3)∗ Is there an algorithm to decide if two connected spaces with known presentations of their respective fundamental groups have isomorphic fundamental groups? A special case of (D3)∗ is the following: Is there an algorithm to decide if a space is simply connected? The general, group-theoretical versions of the same three problems follow:

1

Max Dehn is widely regarded as the founder of combinatorial/geometric group theory.

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(D1) (The Word Problem for Groups) Is there an algorithm to decide if in a given presentation a word determines the trivial element? (D2) (The Conjugacy Problem for Groups) Is there an algorithm to decide if in a given presentation two words are conjugate? (D3) (The Isomorphism Problem for Groups) Is there an algorithm to decide if two presentations determine isomorphic groups? There is a specific and finite presentation for which the word problem (D1) is unsolvable (Novikov 1954, Boone 1959)1; see, for example, [66], page 362. This answers (D1) in the negative, and as a consequence, (D2) and (D3) as well (Exercise 8). The Word Problem is solvable for some specific classes of groups. For example, that is the case with the class of groups presentable with not more than one defining relation (Wilhelm Magnus, 1930; see, for example, [45]). This is relevant, since, as we will see, the fundamental groups of compact, connected 2-manifolds are presentable with one defining relation. Example 3 We will solve the word problem for the group 〈a, b ; a 2 , b 2 〉. This example will be eclipsed by Corollary 4 in 12.3. It is easy to see that every word W on {a, b}± is equivalent to a word of type U = a εbaba…babδ , where ε, δ ∈ {0,1}. This class of words includes the empty word. The set of all such words of length 1 is {a,b}; the set of all such words of length 2 is {ab,ba}, etc. The change from W to U can be done effectively by repeatedly reducing all even exponents of a and b to empty words (a consequence of {a 2 = 1, b 2 = 1} in the group). Now we show that the only word of type U that is equivalent to the identity in the group is the empty word. Let P(x) be the ring of all polynomials on the variable x with coefficients in the field  2 = {0,1}. Let G be the group of all 2 × 2 matrices with entries   1 0   1 x   in P(x) and generated by    ,     (the operation is multiplication   x 1   0 1    1 0  of matrices). Consider the mapping f : {a, b} → G defined by f (a) =   and x 1    1 x  f (b) =   . Denote these two matrices by A and B, respectively. Our goal is  0 1  to show that none of ( AB)m, ( AB)mA (m ∈+ ) is the identity matrix. 1

The chain of results leading to the final solution is interesting. Here is a brief summary: In 1936 Turing produced the notion of (what is now called) a Turing machine. Roughly, a Turing machine consists of a string of squares containing 0 or 1, and a finite program determining what should be done when a particular square is being observed, the options being: (a) leave or change the entry of the observed square and (b) move forward, move backward, or stay on the same square of the string of squares. Since Turing machines (i.e., the corresponding finite programs) can be effectively enumerated, the following problem makes sense (The Halting Problem): Is there a Turing machine such that if given a number n and the string consisting only of 0-s, it will determine if the Turing machine numbered n will stop or run forever. The standard diagonal argument (similar to the one used to prove that the set  is not countable) shows that the answer to the Halting Problem is negative. In 1947 Emil Post (and, independently, A. A. Markov, same year) proved that the Word Problem for semigroups is unsolvable; the basic idea was to associate semigroups to Turing machines and to rely on the unsolvability of the Halting Problem. Novikov’s and Boone’s results for groups is based on Post’s solution for semigroups.

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 1 0   1 0  2 2 2 Since, as is easy to check, ( f (a)) = A2 =   and ( f (b)) = B =  ,  0 1   0 1  it follows from Proposition 1 in 12.1 that f is extendible to a homomorphism f : 〈a, b ; a 2 , b 2 〉 → G . This homomorphism is onto, since { A,  B} generate G.  1  x We compute AB =   , and use induction to show that the second entry 2 x + 1   x of the second row in ( AB)m is a polynomial of degree 2m, and that this is the largest

degree of all four polynomial entries of ( AB)m, m ∈+. The case m = 1 is obvious.  P11  P12  satisfies the inductive assumpInductively, assuming ( AB)k =  2k  P21 Q + x  tions (that is, assuming all P11 , P12 , P21, and Q have degrees less than 2k), we have  P11 ( AB)k+1 =   P21

P12 Q+x

2k

 P11 + xP12 =  p21 + xQ + x 2 k+1 

  

 1   x

x x +1 2

  

xP11 + ( x 2 + 1)P12 xP21 + Q + x 2 k + x 2Q + x 2 k+2

 .  

It is now easy to confirm that, since the degrees of P11 , P12 , P21, and Q are smaller than 2k, the degree of xP21 + Q + x 2 k + x 2Q + x 2 k + 2 is indeed 2k + 2, and that the other three polynomial entries of the last matrix have degrees smaller than 2k + 2. Similar argument shows that the entry in the second row, first column of ( AB)m A is a polynomial of degree 2m + 1, and it is the largest degree of all polynomial entries of that matrix. We have proven what we were aiming at: none of ( AB)m, ( AB)m A (m ∈+) is the identity matrix. Consequently, no word of type (ab)m, (ab)m a (m ∈+) is equivalent to the empty word in the presentation. By symmetry, no word of the type (ba)m , (ba)m b (m ∈+) is equivalent in the presentation to the empty word. We proved that no word U = a εbaba…babδ defines the identity in the presentation unless U is the empty word. ± Summarizing, here is an algorithm to check if a word W on {a, b} is the same as the identity in the presentation: change each appearance of a −1 and b −1 to a and b, respectively (using a 2 = 1 and b 2 = 1, respectively), then reduce W using {a 2 = 1, b 2 = 1} and decreasing the length of the word in each step. Stop when no further reductions are possible, and ☐ check if the word obtained is the empty word. W =1 if and only if that happens. The group 〈a, b ; a 2 , b 2 〉 is the free product of two cyclic groups of order 2: we will define and briefly cover such groups in the next section, where we will also show that, under some natural hypotheses, the word problem of such groups is solvable.

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The isomorphism problem for group presentations is, in general, more difficult than the word problem. For example, at the time of writing this, the isomorphism problem for one-relation groups is still open. That is, it is not known if there is an algorithm to determine if any two presentations, each with one defining relation, are isomorphic. Such algorithms exist for many classes of one-relation presentations, including for almost all that we will encounter here. Tietze’s transformations is one path to follow in order to confirm that two presentations are isomorphic. As Examples 4 and 5 illustrate, these transformations also come in handy when our goal is to show that two (groups given by their) presentations are not isomorphic. Example 4 We show that the free group F2 = 〈a, b  ; ∅〉 of rank 2 is not isomorphic to the free group F3 = 〈 x , y , z  ; ∅〉 of rank 3. Suppose otherwise, and let f : F3 → F2 be an isomorphism. Then f (z ) is a nonidentity element of F2 , and it is easy to deduce directly from the definition of presentations that G = 〈a, b  ;  f (z ) = 1〉 is not freely generated by {a,b}. That G is not freely generated by any set of two generators is an exercise left to the reader. However, the group H = 〈 x , y , z  ;  z = 1〉 is a free group on two generators. Consequently G and H are not isomorphic. On the other hand, it follows from the remark at beginning of this section that G ≅ F2 NC ({ f ( z )}) and H ≅ F3 NC ({ z }). Since f ( NC({z }) = NC({ f (z )}), a simple group-theoretical exercise implies that G ≅ H , giving a clear contradiction. (The exercise we referred to is the following: If f : G1 → G2 is an isomorphism, and if N ☐ is a normal subgroup of G1, then f ( N ) is normal in G2 and G1 N ≅ G2 f ( N )). Similar argument can be used to show that, in general, the rank of a free group is preserved under isomorphisms. For tackling more complicated presentations we need to develop a mini-machine. Let S = { X1 , X 2 ,… , Xn } be a set of symbols, and let W ( X1 , X 2 ,… , Xn ) be any word over the set S ±. For any group G we will denote the set {W ( g 1 , g 2 ,… , g n ) :  g 1 , g 2 ,… , g n ∈G} by {W ( X1 , X 2 ,… , Xn )}G . For example, if S = { X , Y } and if W ( X , Y ) = XYX −1Y −1, then { XYX −1Y −1 }G = { ghg −1h −1  :  g , h ∈G}. Proposition 4. If G = 〈a1 , a2 ,… ;  P1 , P2 , …〉 and H = 〈b1 , b2 ,… ; Q1 , Q2 , …〉 are two isomorphic groups, then for every word W ( X1 , X 2 ,… , Xn ), the groups G1 = 〈a1 , a2 ,… ;  P1 , P2 , … , {W ( X1 , X 2 ,…, Xn )}G 〉 and H1 = 〈b1 , b2 ,… ; Q1 , Q2 , … ,{W ( X1 , X 2 ,… , Xn )}H 〉 are also isomorphic. Proof. Let f : G → H be an isomorphism. It is easy to see that f ({W ( X1 , X 2 ,…, Xn )}G ) = {W ( X1 , X 2 ,… , Xn )}H . It is very easy to show that the subgroup of any group G generated by the elements {W ( X1 , X 2 ,…, Xn )}G is normal in G. The claim now follows from the group-theoretical exercise mentioned at the end of Example 4, and from Exercise 2. Remark. From now on the capital letters X ,Y , X1 , X 2 ,… within words appearing as defining relations in a presentation have the same meaning as in the setup of Proposition 4.

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Example 5 Here is an alternative way to show that the free groups F2 and F3 discussed in Example 4 are not isomorphic. Assume they are isomorphic. Then, by Proposition 4, so are the groups H = 〈a, b  ; { XYX −1Y −1 }H 〉 and G = 〈 x , y , z  ; { XYX −1Y −1 } G 〉 . Each relation in { XYX −1Y −1 }H is a consequence of aba −1b −1; hence, by Tietze, H = 〈a, b  ; aba−1b−1 〉 . Similarly G = 〈 x , y , z  ;  xyx −1 y −1 , xzx −1z −1 , yzy −1z −1 〉. It is then easy to see that H is a presentation of the group  ⊕ , and G is a presentation of  ⊕  ⊕  . These two groups are not isomorphic (by a simple exercise), and we have a contradiction. ☐ The last step in Example 5 is also a consequence of the following basic theorem of group theory, which will come handy at times later on. A proof can be found in any textbook on basic algebra. (In the statement below  m stands for the finite group of integers mod(m) under addition.) Theorem 5. Every finitely generated abelian group G is isomorphic to  m1 ⊕  m2 ⊕… ⊕  mk ⊕ ⊕ ⊕ … ⊕  where m1 , m2 ,…, mk are positive integers such that m1 > 1 and m1 | m2  | … | mk . Moreover, the sequence m1 , m2 ,…, mk (hence, the number k) and the number of infinite cyclic groups that appear as factors in this direct sum are uniquely determined by the group G. The group  ⊕  ⊕ … ⊕ , n copies of  , is called free abelian group of rank n. The following example will be needed in Chapter 13, where we consider the problem of classification of certain 2-manifolds. Example 6 Denote Gm = 〈a1 , b1 , a2 , b2 ,… , am , bm  ; a1b1a1−1b1−1a2b2a2−1b2−1 … ambmam−1bm−1 〉 , m ≥1, and Hn = 〈d1 , d2 ,… , dn  ; d12d22 … dn2 〉 , n ≥1. We show that for every m and n the group Gm is not isomorphic to the group Hn . Suppose otherwise. Proposition 4 then imp­ lies that Gm∗ = 〈a1 , b1 , a2 , b2 ,… , am , bm  ; a1b1a1−1b1−1a2b2a2−1b2−1 … ambmam−1bm−1 , { XYX −1Y −1 }〉 is iso­morphic to Hn∗ = 〈d1 , d2 ,… , dn  ; d12d22 … dn2 , { XYX −1Y −1 }〉 . The group Gm∗ is a free abelian group of rank 2m (i.e., Gm∗ ≅  ⊕  ⊕  … ⊕, 2m times). Turning our attention to Hn , introducing c = d1d2 … dn by means of Tietze, changes the given presentation of Hn to Hn∗ = 〈c , d1 , d2 ,… , dn  ; d12d22 … dn2 , c = d1d2 … dn , { XYX −1Y −1 }〉 . Using Tietze transformations two more times (in one of which we eliminate d1), we get Hn∗ = 〈c , d2 ,… , dn  ; c 2 , { XYX −1Y −1 }〉 . We see that Hn∗ ≅  2 ⊕  ⊕ … ⊕  (n − 1 appearances of the group  ). It now follows from Theorem 5 that Gm∗ and Hn∗ are ☐ not isomorphic. Contradiction! Exercises 1. Prove that if f : 〈a, b,… ; ∅〉 → 〈a, b,… ;  P , Q , …〉 is the extension of the identity map {a, b,…} → {a, b,…}, then Ker ( f ) = NC({P , Q ,…}). 2. Show that if H is a normal subgroup of G = 〈a, b,… ;  P , Q , …〉 generated by {h j  :  j ∈ J }, then 〈a, b,… ;  P , Q , … , {h j :  j ∈ J }〉 is a presentation for G H .

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3. Let H1 = 〈a1 , a2 ,… ; P1 , P2 ,…〉 be a presentation of a group G with respect to f1 : a1  g 1 , a2  g 2 ,… . Choose any h ∈G. Since { g 1 , g 2 ,…} generates G, h = W ( g 1 , g 2 ,…) for some word W on { g 1 , g 2 ,…}± . Choose any b ∉{a1 , a2 ,…}± . Show that H 2 = 〈a1 , a2 ,… , b ; P1 , P2 ,… , b = W (a1 , a2 ,…)〉 is a presentation of the same group G with respect to f 2 : a1  g 1 , a2  g 2 ,…, b  h. 4. Use Tietze transformations (and Theorem 2) to show that 〈a, b, c ; b3 , (bc )3 〉 and 〈 x , y , z ;  x 3 ,  y 3 〉 are isomorphic. 5. Use Tietze transformations to transform the group 〈a, b ; a 4 , b7 , ab3 = ba3 〉 to 〈d ; d 2 〉. Conclude that the order of the former is also 2. (Hint: d = ab.) 6. The groups 〈a, b, c ; a 2 , b3 , c11 , aba 2b 2 , acac 3 〉 and 〈 x ;  x 3 〉 are isomorphic. Prove it by means of Tietze transformations.

7. Prove that the groups < a, b ; ab = ba > and < x , y , z ; xy = yx , xyz = zxy > are not isomorphic.

8. Starting from the fact that the Word Problem for Groups is unsolvable, prove that the Conjugacy Problem and the Isomorphism Problem are also unsolvable. 9. Solve the word problem for the following group 〈a, b  ; a3 , b3 〉 . (Specifically, determine in finitely many steps if a word on the generators of the group defines the identity in the group.) [Hint: Modify the argument in Example 3.] 10. Show that if n and m are positive, distinct integers, Fn is a free group of rank n, and Fm is a free group of rank m, then Fn ≅ Fm . 11. Show that 〈a, b  ; an = bn+1 ,  XYX −1Y −1 〉 is an infinite cyclic group for every n ≥ 0. 12. Use Tietze transformations to change 〈a, b ; a 2b 2 〉 to 〈 x , y  ;  xyxy −1 〉 . 13. Generalize Exercise 10: prove that if F is isomorphic to 〈a, b,… ; ∅〉 and to 〈 x , y ,… ; ∅〉 , then {a, b,…} = {x , y ,…} . [Hint: add X 2 and XYX −1Y −1 to the relations in both presentations.]

12.3  Free Products and Free Products with Amalgamation A brief historical note: The first results on free products of groups, including Proposition 3, were obtained by Emil Artin and Otto Schreier (1901–1929) in 1926 or earlier. Free products with amalgamation were introduced by Schreier in 1927.

We start with an important definition. Let {a, b,…}± and {x , y ,…}± be disjoint sets. The free product of the groups G1 = 〈a, b,…; P , Q , …〉 and G2 = 〈 x , y ,…;  R , S , …〉 , denoted G1 ∗ G2, is the group G1 = 〈a, b,…,  x , y ,… ;

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P , Q , … ,  R , S , …〉 . More generally, if A ±j , j ∈ J , are pairwise disjoint sets, then the free product of the groups G j = 〈 A j  ; R j 〉, j ∈ J , is the group ∗ G j = ∪ A j ;  ∪ R j . The groups j∈J j∈J j∈J G j are factors of ∗ G j . j ∈J

If the index set J is clear from the context, then we will sometimes write ∗G j instead of ∗ G j . j ∈J

Example 1 If G1 = 〈u, v ; u 2 , uv 2 = vu 2 〉 and if G2 = 〈 x , y ; xy = yx 〉 , then G1 ∗ G2 = 〈u, v , x , y ; u 2 , uv 2 = vu 2 , xy = yx 〉. The group 〈a, b ; a 2 , b 2 〉 , introduced in Example 3, 12.2, is the free product 〈a ; a 2 〉∗〈b ; b 2 〉 . Every free group 〈 A  ; ∅〉 is the free product ∗ 〈a  ; ∅〉 of A -many infinite cyclic a∈ A groups. ☐ In the next proposition we assert that the groups G j survive intact within ∗ G j . j ∈J

Proposition 1. Let G be the free product of the groups G j = 〈 A j  ; R j 〉 , j ∈ J . For every fixed k ∈ J , the mapping ik : Gk → ∗ G j , determined by ik (a) = a for every a ∈ Ak , is a well j ∈J defined monomorphism of groups. Proof. Proposition 1 in 12.1 implies that ik is a well-defined homomorphism. The same argument implies that the mappings pk : ∗ G j → Gk , determined by pk (a) = a for every j∈ J

a ∈ Ak, and by pk ( x ) = 1 for every x in the set ∪ A j of the remaining generators of ∗ G j , is j∈J j≠k

j∈J

also a well-defined homomorphism. Since pk  ik = idGk it follows that ik is one-to-one. In view of Proposition 1 we may consider G j’s to be subgroups of their free product ∗ G j , j ∈J and i j -s to be inclusions. The following statement links free products with a universal property of homomorphisms. Free products are sometimes defined in terms of this universal property. Proposition 2. Let G be a group, let G j , j ∈ J , be a set of subgroups of G such that Gi ∩ Gk = {1} for every i , k ∈ J , i ≠ k. Further, let i j : G j → G denote the inclusions, and let G be generated by ∪ G j. Then G is the free product of G j , j ∈ J , if and only if for every group H and every set j∈ J

of homomorphism f j : G j → H, there is a unique homomorphism f : G → H making the following diagram commutative for every j ∈ J . ij

Gj

f

fj H

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G

    Illustration 12.4

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12.3  Free Products and Free Products with Amalgamation   ◾    285  

Proof. ⇒ Let G j = 〈 A j  ; R j 〉 and G = ∗ G j . Given the homomorphisms f j : G j → H , j ∈J

j ∈ J , let f : G → H be determined by f ( x ) = f j ( x ) for every x ∈ A j , and for every j ∈ J. Then it is visible that f is a well-defined homomorphism and that the above diagram commutes. The commutativity of the diagram implies that f ( x ) must be f j ( x ) for every x ∈ A j , every j ∈ J ; hence f is unique. ⇐ Let G j, j ∈ J , be subgroups of G as in the statement of the proposition, and let i j : G j → G be the inclusions. Suppose that for every H and homomorphisms f j there is a unique homomorphism f such that the above diagram commutes for every j ∈ J . In particular this is true for H = ∗ G j , where each f j is the inclusion established in Proposition 1 (see j ∈J Illustration 12.5). ij

Gj

G f

fj ∗ Gj

j J



Illustration 12.5

The proof will be completed once we establish that the homomorphism f is an isomorphism. It is onto since each generator g k ∈Gk ⊂ ∗ G j is f (ik ( g k )). Finally, f is one-to one, since j ∈J g  f = idG , where g : ∗ G j → G is the homomorphism defined over the set ∪ A j of generators j∈ J

j∈J

of ∗ G j by g ( x ) = i j ( x ) for every x ∈ A j . j ∈J

A sequence g 1 ,  g 2 , … ,  g n of elements in ∗ G j is reduced if either n = 0 (and the j ∈J

sequence is empty) or n > 0, each g k is a non-identity element in some G jk , and no two consecutive g k ,  g k+1 are in the same group G j . It is obvious that every element of ∗ G j j ∈J can be written as g 1 g 2 … g n where g 1 ,  g 2 , … ,  g n is a reduced sequence; the identity element of the group is obtained when n = 0. More is true: Proposition 3. For each element x of G = ∗ G j there is a unique n and a unique reduced j ∈J

sequence g 1 ,  g 2 , … ,  g n of elements ∪ G j such that x = g 1 g 2 … g n in G. j∈J

Proof 1. For every j ∈ J and every a ∈G j , a ≠1, we associate a permutation pa of the set of all reduced sequences as follows:  (a, g 1 , g 2 , … ,  g n )  if a ∈G j1 ,  g 1 ∈G j2 ,  j1 ≠ j2  pa ( g 1 , g 2 , … ,  g n ) =  (ag1 , g 2 , … ,  g n )  if a ∈G j ,  g 1 ∈G j , ag1 ≠ 1  ( g 2 , … ,  g n )   if a ∈G j ,  g 1 ∈G j , ag 1 = 1. 

1

Artin and van der Warden; see [45], page 176.

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(In the last case, ( g 2 , … ,  g n ) is the empty sequence if n =1.) We set pa to be the identity permutation if a =1 in G j . The mapping pa is indeed a permutation for every a ∈∪ G j , j∈J since pa−1 is its inverse. It is easy to confirm that for every a, b ∈G j , pab = pa  pb , so that for every j ∈ J , ϕ j : a  pa is a homomorphism from G j into the group P of all permutations of the set of reduced sequences of elements in G. By Proposition 2, we have a homomorphism ϕ : G → P making the corresponding diagrams (with ϕ j-s in place of f j -s) commutative. It follows from the construction of φ that if g 1 ,  g 2 , … ,  g n is a reduced sequence, then ϕ( g 1 g 2  …  g n ) = ϕ j1 ( g 1 )  ϕ j2 ( g 2 ) … ϕ jn ( g n ) =   pg1  pg 2  …  pgn. Observe that the last composition sends the empty sequence to ( g 1 , g 2 , … ,  g n ). Consequently, products g 1 g 2 … g n corresponding to different reduced sequences are sent by the homomorphism φ to different elements of the group P, and hence, since φ is well defined, two different reduced sequences give rise to different elements in G. Corollary 4. If the word problem is solvable for each G j , j ∈ J , then it is solvable for ∗ G j . j ∈J

Proof. The obvious algorithm (Exercise 1) expresses each x ∈ ∗ G j as g 1 g 2  …  g n , where j ∈J

( g 1 , g 2 , … ,  g n ) is reduced. If n ≥ 2, it follows from Proposition 3 that x ≠1 in ∗ G j . If n =1, j ∈J we can use the solvability of the factor groups to check if x = 1. For example, since 〈a ;  a 2 〉 is finite, it follows from Corollary 4 that the group 〈a, b; a 2 , b 2 〉 has a solvable word problem. Compare this with the work done in Example 3, 12.2. The following proposition is, in a way, a converse of Proposition 3. Proposition 5. Let G be a group and let {G j : j ∈ J } be a family of subgroups of G such that ∪ G j generates G, and such that G j ∩ Gk = {1} whenever j ≠ k. Suppose further that j∈J for every two sequences g 1 ,  g 2 , … ,  g n and h1 , h2 , … , hm which are reduced with respect to {G j : j ∈ J } , g 1 g 2  g n = h1h2 hm in G implies that the two sequences are equal. Then G is isomorphic to the free product ∗ G j . j∈ J

Proof. Consider the mapping φ : ∗ G j → G determined by φ( g ) = g for every g ∈∪ G j . j ∈J

j∈J

It follows plainly that ϕ is well defined; it is onto since ∪ G j generates G. In order to show j∈J it is one-to-one, assume φ( g ) = φ(h), for some g , h ∈ ∗ G j. By Proposition 3 we can write j ∈J g = g 1 g 2  g n and h = h1h2 hm where the sequences g 1 ,  g 2 , … ,  g n and h 1 , h2 , … , hm are reduced  in ∗ G j . According to our definition of ϕ, φ( g ) = g 1 g 2  g n and φ(h) = h1h2 hm, j ∈J

viewed as products of the generators ∪ G j of G. Our assumptions in the statement of this j∈J

proposition, together with φ( g ) = φ(h), force m = n and g k = hk in G jk . This implies that g = h in ∗ G j . j ∈J

In the last part of our brief encounter with combinatorial group theory we will generalize the concept of free product.

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12.3  Free Products and Free Products with Amalgamation   ◾    287  

Let G1 = 〈a, b,… ;  P , Q , …〉 and G2 = 〈 x , y ,… ;  R , S , …〉 be two groups such that {a, b,…}± and {x , y ,…}± are disjoint, let H be a group and let φ1 : H → G1 and φ2 : H → G2 be homomorphisms. The free product of the groups G1 and G2, amalgamating the subgroups φ1 ( H ) and φ2 ( H ) , denoted G1 ∗ G2 , is the group 〈a, b, … ,  x , y , … ;  P , Q , … ,  R , S , … , H {φ1 (h) = φ2 (h) : h ∈ H }〉 . Before we generalize to amalgamated products of more than two groups, we pause for a moment to explain in plain words what we are doing. In the amalgamated product of two groups G1 and G2 we first take G1 ∗ G2, then we identify, element by element, the two images φ1 ( H ) and φ2 ( H ) . Let G j = 〈 A j  ; R j 〉, j ∈ J , be groups with pairwise disjoint sets A ±j , let {H jk : j , k ∈ J } be a class of groups, and let φ jjk : H jk → G j , φkjk : H jk → Gk be homomorphisms for each j , k ∈ J , j ≠ k. The free product of the groups G j , j ∈ J , amalgamating the subgroups φ jjk( H jk ), φ kjk ( H jk ), j , k ∈ J , denoted ∗ G j, is the group ∪ A j ;  ∪ R j ∪ ∪ {φ jjk (h) = H jk

j ∈J

j ∈J

j , k ∈J

φkjk (h) : h ∈ H jk } . In this context the groups G j are again called factors, and the groups φ jjk( H jk ) , φ kjk ( H jk ) are amalgamated subgroups. If the groups H jk are all equal to a group H, then we will write ∗ G j instead of ∗ G j. In H H jk order to avoid even more cluttered notation, we have not included the homomorphisms φ jjk: H jk → G j , φkjk : H jk → Gk in the notation ∗ G j . Almost all of the time φ jjk ’s and φkjk ’s H jk will be clear from the context. If B jk is a generating set for the group H jk , j , k ∈ J , then we can reduce the set of relations {φ jjk(h) = φ kjk (h) : h ∈ H jk } in the last presentation given above to its subset {φ jjk (b) = φ kjk (b) : b ∈ B jk }, without affecting the group defined by the presentation. This is true by Tietze’s Theorem 1 in 12.2, since the assumption that φ jjk and φkjk are homomorphisms implies that if φ jjk(b) = φ kjk (b) holds for every b ∈ B jk , then φ jjk(h) = φ kjk (h) for every h ∈H jk . Summarizing, if B jk is a set of generators for H jk , j , k ∈ J , then we can write more efficiently: ∗ G j = ∪ A j  ;  ∪ R j ∪ ∪ {φ jjk (b) = φkjk (b) : b ∈ B jk } . H jk

j ∈J

j ∈J

j , k ∈J

If the groups H jk are trivial, then, obviously, ∗ G j = ∗G j , so that free product with H jk

amalgamation indeed generalizes the concept of a free product. Similarly, if each φ jjk , φkjk is trivial (sending each element to the identity element of the corresponding G j ), then again ∗ G j = ∗G j . H jk

Example 2 Let G1 = 〈a  ; a3 〉 , G2 = 〈b  ; b 4 〉 , H = 〈c  ; ∅〉 , φ1 (c ) = a 2 , φ2 (c ) = b3. Then G1 ∗ G2 = 〈a, b  ;  H a3 , b 4 , a 2 = b3 〉 . Since 1 = a6 = (a 2 )3 = (b3 )3 = b9 = b8b = b , the relation b =1 is a consequence of the relations in the presentation. This easily implies that a =1, and so G1 ∗ G2 is H ☐ the trivial group.

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Example 3 Let G1 = 〈a1 , a2 ; a1a2 = a2a1 〉, G2 = 〈b1 , b2 ; b1b2 = b2b1 〉, G3 = 〈c1 , c2 ; c1c2 = c2c1 〉, H = 2 〈h  ; ∅〉, φ112 (h) = a2 , φ12 (h) = b1 , φ 223 (h) = b2 , φ 323 (h) = c1 , φ331 (h) = c2 φ131 (h) = a1. Then ∗ G j = 〈a1 , a2 , b1 , b2 , c1 , c2  ; a1a2 = a2a1 , b1b2 = b2b1 , c1c2 = c2c1 , a2 = b1 , b2 = c1 , c2 = a1 〉. H

We eliminate b1 , b2 and c2 using Tietze transformations to get: H∗ G j = 〈a1 , a2 , c2  ; a1a2 = a2a1 , a2c1 = c1a2 , c1a1 = a1c1 〉 , which is isomorphic to   ⊕  ⊕  . ☐ We say that G1 ∗ G2 is a proper free product of the groups G1 and G2 amalgamating φ1 ( H ) H and φ2 ( H ) if φ1 ( H ) and φ2 ( H ) are proper nontrivial subgroups of G1 and G2 , respectively. Example 3 implies that a proper free product with amalgamation can be a nontrivial abelian group. (Compare this claim with Exercises 3(a) and 10.) Remark. In the remainder of this section we will discuss a few properties of the free products with amalgamation. We will not invoke or need these properties in the rest of the theory that we will develop. We see from Example 2 that in free products with amalgamation both factor groups could be annihilated. The culprits for this “misbehavior” are the homomorphisms φ1 and φ2 : neither is one-to-one in this example. When the homomorphisms φ1 and φ2 are not one-to-one, then the group G1 ∗ G2 could H behave rather pathologically. For example, it is easy to construct a finitely presented G1 ∗ G2 H

with unsolvable word problem, while at the same time the associated presentations of G1 and G2 have solvable word problems: Let N = 〈 A  ; R 〉 be a finitely presented group with unsolvable word problem; let G1 = 〈 A ; ∅〉, G2 = 〈z ; ∅〉 and H = 〈{hr : r ∈ R } ; ∅〉. Define φ1 : H → G1 by φ1 (hr ) = r , and set φ2 : H → G2 to be the trivial homomorphism. Both G1 and G2 are free, and so the word problem in these two groups is solvable (Corollary 4). However, G1 ∗ G2 = 〈 A ∪ {z } ; R 〉 H does not have a solvable word problem, because N does not have a solvable word problem. If, however, the homomorphisms φ j , j ∈ {1,2}, are all monomorphisms, then the word problem in G1 ∗ G2 is solvable if the associated presentations of G1 and G2 have solvable word H problems. Exercise 11 contains the steps needed to prove that claim. Stipulating that all φ jjk’s and φkjk’s (as in the general definition) are monomorphisms does not guarantee that the homomorphisms G j → ∗ G j, j ∈ J , induced by the inclusion of H jk

generators are also monomorphisms (Exercise 12). However, in the case of the free product with amalgamation G1 ∗ G2 involving only two groups, the corresponding statement is H true, and the groups G1 and G2 survive within G1 ∗ G2 . We give the formal statement below, H but we do not provide a proof (which can be found in, say, [47], starting on page 199). Proposition 6. If ϕ1 : H → G1 and ϕ 2 : H → G2 are both monomorphisms, then the homomorphisms in1 : G1 → G1 ∗ G2 and in2 : G2 → G1 ∗ G2 induced by the respective inclusions of H H the generators are both monomorphisms. The following proposition generalizes Proposition 2. It is not in the mainstream of our theory and we will not use it. We give it here since it is sometimes used as a route to the Seifert-van Kampen theorem (next chapter) bypassing group theory.

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12.3  Free Products and Free Products with Amalgamation   ◾    289  

Proposition 7. Let G j , j ∈ J , be subgroups of a group G such that G j ∩ Gk = {1} when j ≠ k , and such that ∪ G j generates G. Further, let {H jk : j , k ∈ J } be a set of groups, and let j∈J

k φ : H jk → G j , φ jk : H jk → Gk , j , k ∈ J , be group homomorphisms. The group G is the free product of the groups G j, j ∈ J , amalgamating the subgroups φ jjk( H jk ), φ kjk ( H jk ), j , k ∈ J , if and only if for every group K and every set {ψ j  :  j ∈ J } of homomorphisms making the diagram in Illustration 12.6 commutative for every j , k ∈ J , there is a unique homomorphism f : G → K such that, for every j , k ∈ J the diagram in Illustration 12.7 commutes (where i j , ik are the inclusions). j jk

φjkj

Gj

ψj

Hjk

K k φjk





Hjk φ j jk k φjk

ψk

Gk



Illustration 12.6

Gj

Gk

G

ij ψj ψk

ik

f

K

Illustration 12.7

Proof. Exercise 13. Exercises 1. Use induction to give a precise proof of the claim that every x ∈ ∗ G j is equal to some j ∈J g 1 g 2  …  g n , where ( g 1 , g 2 , … ,  g n ) is a reduced sequence. 2. Let G be the free product of at least two non-trivial groups. Show that the center Z (G ) = {x ∈ G : xy = yx for every  y ∈ G} of G must be trivial. 3. (a) Show that no non-trivial abelian group is the free product of two non-trivial groups. (b) Show that the group 〈a, b ; W 〉 , where W is any cyclically reduced word on both a and b not equal to 1 in 〈a, b ; ∅〉 , is not a free product of two non-trivial factors. 4. Show that 〈b, c  ; b3 , cb 2cb 2 〉 is a free product of two nontrivial groups. 5. Show that the group 〈a  ; an 〉∗ 〈b  ; bm 〉 is isomorphic to the group 〈 x  ;  x p 〉∗ 〈 y  ;  y q 〉 if and only if {n, m} = { p, q}. 6. Let φ1 : H → G1 × H , φ2 : H → G2 × H   be group homomorphisms defined by, φ1 (h) = (1G1 , h), φ2 (h) = (1G2 , h) for every h ∈ H , respectively. Show that (G1 × H ) ∗(G2 × H ) ≅ (G1 ∗ G2 ) × H. H



7. The braid group on three strings has a presentation a, b  ; aba = bab . Show that it is a non-trivial free product with amalgamation. [Hint: start by introducing c = ab via Tietze transformations.]

8. Consider the group G1 ∗ G2 where the amalgamation is with respect to the homoH

morphisms φ j : H → G j , j ∈ {1,2}. Prove that if the mappings i j : G j → G1 ∗ G2 ,

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290   ◾     Combinatorial Group Theory

i ∈{1,2}, defined by i j ( x ) = x , for every x ∈G j , are monomorphisms, then both φ1 and φ2 are monomorphisms. 9. Let H, G1, and G2 be groups, let G3 be the trivial group, and let φ1 : H → G1 and φ2 : H → G2 be two homomorphisms. Recalling that NC(A), the normal closure of a subset A of a group G, is the subgroup of G generated by all conjugates of the elements in A, show that ( G1 NC (φ1 ( H )) ) ∗ ( G2 NC (φ2 ( H )) ) is isomorphic to the free product of the groups G j , j ∈ {1,2,3}, amalgamated along H.

10. Let G = G1 ∗ G2 be a proper free product of the groups G1 and G2 amalgamating φ1 ( H ) H and φ2 ( H ). Suppose that NC(φ1 ( H )) ≠ G1 and NC(φ2 ( H )) ≠ G2. Prove that the center Z (G ) = {x ∈ G : xy = yx  for every  y ∈ G} of G must be trivial. Deduce that no abelian group is a proper free product of two groups with amalgamation. 11. Let G1 = a, b,… ;  P , Q , … , G2 = 〈 x , y ,… ;  R , S , …〉 have solvable word problems, let H be a group, and let φ1 : H → G1 and φ2 : H → G2 be monomorphisms.

(i) Show that each element x ∈ G can be written as ha1a2 … an (n ≥ 0), where h ∈ φ1 ( H ), a j ∈(G1 \ φ1 ( H )) ∪ (G2 \ φ2 ( H )) for every j, where no two consecutive a j , a j+1 belong to the same G1, G2, and where n is minimal with the above properties.



(ii) Show that an application of one of the operations (a) and (b) of Step 1 in the definition of the equivalence relation on the set of words in presentations (described in Section 12.1) does not affect n.

(iii) Deduce that if n > 0 then the product ha1a2 … an (as in part (a)) is not equal to the identity in G = G1 ∗ G2 . H



(iv) Prove that the word problem for G is solvable. (Note: only this part requires the assumption that φ1 : H → G1 and φ2 : H → G2 are monomorphisms.)

12. Let  G1 = 〈a1 , a2  ; ∅〉, G2 = 〈b1 , b2  ; ∅〉, G3 = 〈c1 , c2  ; ∅〉 , and H = H12 = H13 = H 23 = 〈d1 , d2 ; ∅〉 . Find monomorphisms ϕijk : Hij → Gk , i , j ∈{1,2,3}, i < j , k ∈{i , j}, such that the corresponding free product with amalgamation ∗ G j satisfies the following: the H homomorphism G1 → ∗ G j determined by a1  a1, a2  a2 is not one-to-one. H

13. Prove Proposition 7. 14. Find two groups G1 and G2 such that G1 × G2 is not isomorphic to any G1 ∗ G2 . H

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Chapter

13

Seifert–van Kampen Theorem and Applications

I

n the previous chapter we laid the groundwork for further expansion of the homotopy theory. The Seifert–van Kampen theorem is a major step forward; it provides a way to express the fundamental group of a space in terms of the fundamental groups of open subspaces forming a cover. We will give a proof of one of the generalizations of the original statement of the theorem, then illustrate the power of the theorem through many examples.

13.1  Seifert–van Kampen Theorem A brief historical note: In 1932 Egbert van Kampen (1908–1942) published a proof of Theorem 1 below. At about the same time, and independently, Herbert Seifert (1907–1996) produced a proof of Theorem 2 when the space X belongs to a class of CW complexes.

In order to clearly see the basic idea of the Seifert–van Kampen theorem (which we will refer to as the SvK theorem), we first state it in its simplest version. It is the version that we will use most often in our applications. Theorem 1 (Seifert–van Kampen) Let {U1 ,U 2 } be an open cover of a path connected space X, such that U1 , U 2 and U1 ∩ U 2 are non-empty and path connected. Let x ∈U1 ∩ U 2 , and denote the inclusions U1 ∩ U 2 → U1 and U1 ∩ U 2 → U 2 by i1 and i2 , respectively. Then π1 ( X , x ) = π1 (U1 , x ) ∗ π1 (U 2 , x ), amalgamating the subgroups (i1 )∗ (π1 (U1 ∩ U 2 , x )) π1 (U1 ∩U 2 , x )

and (i2 )∗ (π1 (U1 ∩ U 2 , x )) .

291

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292   ◾    Seifert–van Kampen Theorem and Applications X U2

U1 x U1 U2

Illustration 13.1

With only slightly more effort we will prove a more general variant of the SvK theorem: Theorem 2. (Seifert–van Kampen). Let X be a path connected space, let x ∈ X , and let U = {U j  :  j ∈ J } be an open cover of X, such that each U j is path connected, and such that each intersection of four or fewer members of U is path connected and contains x. Denote the inclusion U j ∩ U k ⊂ U j by i jjk . Then π1 ( X , x ) is a free product of the groups π1 (U j , x ) ,

( )

( )

amalgamating the subgroups i jkj (π1 (U j ∩ U k , x )) and i kjk (π1 (U j ∩ U k , x )). ∗

∗

Remark 1. The source of the mysterious stipulation regarding the intersection of four or fewer members of U is the fact that in a rectangular net every vertex is a corner of four or fewer rectangles. The precise connection will be established in the proof. Remark 2. It follows from the definition of free products with amalgamation that Theorem 1 can be reformulated as follows: if π1 (U j , x ) is presented by <  Y j  ;  R j > , j ∈ J , and if Z jk is a generating set for π1 (U j ∩ U k , x ), j , k ∈ J , then the group π1 ( X , x ) is presented by

( )

( )

. j ∈J

j ∈J

∗

∗

(∗)

Remark 3. The proliferation of indexes in the statement of Theorem 2 was virtually forced upon us by the nature of the statement. Nevertheless, it is rather distracting and could make the proof that follows unnecessarily saturated. We will, therefore, sacrifice some of the precision, trusting that the domains and the ranges of the mappings we will encounter will be clear from the context. Thus, we will use i j : U j ∩ U k → U j instead of i jkj : U j ∩ U k → U j , and use ik : U j ∩ U k → U k instead of i kjk : U j ∩ U k → U k . Further, we will use i ∗j : π1 (U j ∩ U k , x ) → π1 (U j , x ) and ik∗ : π1 (U j ∩ U k , x ) → π1 (U k , x ) for the homomorphisms induced by i j and ik , respectively (in place of i jkj and i kjk , respectively). With this, ∗ ∗ and in the light of Remark 2, we can reformulate the Seifert–Van Kampen theorem as follows.

( )

( )

Theorem 2, formulation 2. (Seifert–van Kampen). Let X be a path connected space, let x ∈ X , and let U = {U j  :  j ∈ J } be an open cover of X, such that each U j is path connected, and such that each intersection of four or fewer members of U is path connected and contains x. Denote each inclusion U j ∩ U k ⊂ U j by i j , and the induced homomorphism by i ∗j : π1 (U j ∩ U k , x ) → π1 (U j , x ). Let π1 (U j , x ) be presented by <  Y j  ;  R j > , and let Z jk be a generating set for π1 (U j ∩ U k , x ). Then is a j ∈J j ∈J presentation for the group π1 ( X , x ).

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13.1  Seifert–van Kampen Theorem   ◾    293  

Proof. Let g : → π1 ( X , x ) be the extension j ∈J

j ∈J

of the following mapping defined over the generating set: the homotopy class in Y j of a loop α at x in U j is sent to the homotopy class of the same loop considered as a loop in X. We will show that g is a isomorphism by proving that it is well defined (Step 1), onto (Step 2), and one-to-one (Step 3). The first two steps will be easy, the third will require some work. Step 1 (g is well defined). By Proposition 1 in 12.1, we need to check that for every r ∈ R j , j ∈ J , g (r ) is the identity element in π1 ( X , x ), and that g (i ∗j ( z )) = g (ik∗ ( z ))  for every z ∈ Z jk ,  j , k ∈ J . Let r ∈ R j for some j ∈ J . This means that r is the homotopy class of a loop at x that is entirely in U j , and that this loop is homotopic to the constant loop at x by means of a homotopy ranging entirely within U j . This clearly implies that r is homotopic to the constant loop at x within X, and so g(r) is the trivial element in π1 ( X , x ) . The assumption that i ∗j ( z ) = ik∗ ( z )  means that the homotopy class in U j of a loop α at x in U j ∩ U k representing z is the same as the homotopy class in U k of the same loop. It follows plainly (from the fact that we deal with a single loop) that α defines a unique homotopy class in π1 ( X , x ) , so that g (i ∗j ( z )) = g (ik∗ ( z )) . Step 2 (g is onto). Let f  be a loop at x, so that [f] is an element of π1 ( X , x ). Let f = f1 f 2 … fn be the Lebesgue subdivision of the path f relative to the cover U . This means that each of f ([ti , ti+1 ]), i ∈{0,1, … , n − 1} ( t0 = 0 ), is in some U i ∈U . As in the first part of the proof of Theorem 1 in 11.1, if f ([ti , ti+1 ]) and f ([ti +1 , ti + 2 ]) are both in a single U i , then we join [ti , ti+1 ] and [ti +1 , ti + 2 ] into a single interval. We do that until there are no such cases. After relabeling the intervals we may assume that f ([ti , ti+1 ]) and f ([ti +1 , ti + 2 ]) are always in different U i ’s. This implies that for every i ∈{0,1, … , n − 1}, f (ti ) ∈U i −1 ∩ U i . Since U i −1 ∩ U i is a path connected set containing x, there is a path δi from f (ti ) to x within U i −1 ∩ U i . Since f = f1 f 2 … fn is homotopic to ( f1 δ1 )(δ1−1 f 2δ 2 )(δ 2 −1 f3 δ 3 ) … (δn −1 fn ) , and since each of the paths in parentheses is a loop at x entirely within an open set belonging to our cover, it follows that [ f ] = [( f1 δ1 )(δ1−1 f 2 δ 2 )(δ 2 −1 f3 δ 3 ) … (δn−1 fn )] = [ f1 δ1 ][δ1−1 f 2 δ 2 ][δ 2 −1 f3 δ 3 ] … [δn−1 fn ], where each of the homotopy classes to the righthand side of the equality is an element of π1 (U i , x ), some U i ∈U . This shows that the union of the sets π1 (U j , x ) , j ∈ J , generates π1 ( X , x ) , and so g is onto. Step 3. It remains to be shown that g is one-to-one, or, equivalently, that the kernel of g consists only of the identity element in the group defined by the presentation. This is the crucial part of the proof. The basic idea is to construct a step-by-step homotopy between a loop in X and the constant loop at x, so that in each step the range of the homotopy is confined within a single member of the cover U. Take a product [ f1 ][ f 2 ] … [ fn ] of elements in {Y j  :  j ∈ J }. Consider [ f1 ], [ f 2 ], … , [ fn ] to be elements in π1 ( X , x ) (via g), so that f = f1 f 2 … fn is a loop at x in X. Assuming this loop is homotopic to the trivial loop in X at x, we need to show that [ f1 ][ f 2 ] … [ fn ] is the identity element in the presentation, or, more explicitly, that it is equivalent to the empty word in the above presentation.

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294   ◾    Seifert–van Kampen Theorem and Applications

Since f = f1 f 2 … fn is homotopic to the constant loop at x, there is a homotopy F : I × I → X relative to {0,1} from f to the constant loop at x (so, {0} × I and {1} × I are mapped to x). Since I × I is compact, we can split it into a net so that each rectangle in this net is mapped into some U j in the cover of X. Further subdivision of this net will ensure that the endpoints of the domains of each of f1 , f 2 ,… , fn are on the vertical lines of the net (see Illustration 13.2). ...

...

...

...

...

...

...

...

...

...

h10

g21

...

g2k

R11

h11 . . .

R1k

g11

g1k

h1k

...

...

Rij hij

hi(j–1)

...

...

...

...

g(i+1)j gij

...

...

...

...

g2m

...

...

R1m g1m

......

f1

fn

h1m

Illustration 13.2  I × I subdivided into rectangles, each of which is mapped into some U j via F.

Focusing at I ×{0} we see (in Illustration 13.2) that f1 f 2 … fn = g 11 g 12 … g 1m, where g 1 j is the restriction of F to the j-th segment of I ×{0} in the net. More generally, the restriction of F to any side of any small rectangle in the net defines a path in X. We denote the restrictions of F to the edges of the rectangle Rij as follows: the restriction of F to the bottom edge is g ij , and to the top edge is g (i +1) j ; the restriction of F to the left edge is hi( j −1) , and to the right edge it is hij (see Illustration 13.2). The range of any of these paths is in one member of U . They need not be loops at x in X. We will replace them with loops in the next step.

fi

fn

Illustration 13.3  We add closed line segments at some vertices in the net.

Assuming that the rectangular net is in a horizontal plane, we attach a vertical closed line segment I v at most of the vertices v of the net, as shown in Illustration 13.3. A vertex v at the bottom of a vertical line segment is mapped by F to a point in the intersection of the open sets in U containing the images of the rectangles adjacent to this vertex. There are

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either two such adjacent rectangles (when the vertex is at the bottom or at the top edge of I × I ) or four of them (for the other vertices). So, bearing in mind that the associated open sets in U need not be distinct, F (v) is contained in one, two, three, or four open sets in U . Since we have assumed that the open sets in U are path connected, and that the intersections of two, three, or four open sets in U are path connected, and since each of these intersections contains the base point x, it follows that there is a path τ F (v ) from F (v) to x within the intersection of these open sets. We extend F on the vertical line segments I v as follows: identifying I v with I, the restriction of the extension of F on I v is τ F (v ) . Take this path to be the constant at x if it happens that F (v) coincides with x. Denote this extension of F by Fˆ . It is clear that Fˆ is continuous (gluing lemma). For convenience, we modify the notation for τ F (v ): if sij denotes the vertical line segment at the right bottom corner of the rectangle Rij , 1 ≤ j < m, then we denote the path Fˆ sij by λ ij . In Illustration 13.4 we show this notation pertaining to the bottom edge of I × I extended by the vertical segments. λ11 g11

λ1(m–2) λ1(m–1)

λ1k g1k

g1(m–1)

g1m

f

Illustration 13.4 −1 −1 g 12 λ12 λ12 … g 1(m−1)λ1(m−1)λ1−(1m−1) g 1m = Observe that f = f1 f 2 … fn = g 11 g 12 … g 1m  g 11 λ11 λ11 −1 ( g 11λ11 ) λ11 g 12 λ12 … λ1−(1m−2) g 1(m−1)λ1(m−1) λ1−(1m−1) g 1m , where the parentheses in the last step are only for emphasis. Each path in the parentheses is, by construction, a loop at x entirely within one member of the starting cover U of X. Focus for a moment on f1 : recall that it is a loop within some U j ∈U defining an element in the generator set Y j of the presentation <  Y j  ;  R j > of π1 (U j , x ) . As we see in Illustration 13.2, f1 = g 11 g 12 … g 1k , and so g 11 , g 12 ,… , g 1k are also paths in U j . We achi­eved above that f1 = g 11 g 12 … g 1k is homotopic to ( g11λ11 ) λ11−1 g12λ12 … λ1−(1k−2) g1(k−1)λ1(k−1) λ1−(1k−1) g1k , where the paths λij in this product are also in U j , and where the corresponding homotopy shrinks each λ1i λ1−i1 to a point within U j . Moreover, all the paths in the parentheses are loops in U j . All this implies −1 that f1 and ( g 11λ11 ) λ11 g 12 λ12 … λ1−(1k−2) g 1( k−1)λ1( k−1) λ1−(1k−1) g 1k define the same element −1 g 12 λ12 … λ1−(1k−2) g 1( k−1)λ1( k−1) λ1−(1k−1) g 1k  in π1 (U j , x ). Consequently, [ f1 ] = ( g 11λ11 ) λ11   in <  Yj  ;  R j >, hence this is also true in the presentation (∗) of π1 ( X , x ). Since −1 −1 −1 ( g 11λ11 ) λ11 g 12 λ12 … λ1−(1k−1) g 1k  =  g 11λ11   λ11 g 12 λ12  …  λ1( k−1) g 1k  , we showed that   −1 g 12 λ12  …  λ1−(1k−1) g 1k  in the presentation (∗).  f1  =  g 11λ11   λ11 Similar argument holds for f 2 , f3 ,… , fn ; so we can state that [ f ] = [ g 11λ11 ] −1  λ11 g 12 λ12 …  λ1−(1m− 2) g 1(m−1)λ1(m−1)   λ1−(1m−1) g 1m  holds in the presentation (∗). Note again that the homotopy classes to the right are associated to loops at x, each of them entirely within one member of U , so that they also represent elements in the presentation (∗).

(

) (

)(

(

) (

)(

(

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) (

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296   ◾    Seifert–van Kampen Theorem and Applications

Now it suffices to show that the product of the loops in the right-hand side of the last equality is equal to the identity element in the presentation (∗). We will prove it by changing that product in steps, each involving one small rectangle in our net. In the first step we consider the square we have denoted by R11, together with the two vertical line segments we have attached to it.

λ21

g21

h10

t = 0.4

t = 0.8

R11 t = 0.6

t = 0.2

h11

λ11

g11

Illustration 13.5

A simple homotopy (within an open set U j containing R11) between the loops h10 g 11λ11 and −1 ( g 21λ 21 )(λ 21 h11λ11 ) is indicated in Illustration 13.5. Recall that h10 is the constant loop at −1 h11λ11  in the x. Since this happens within a single U j , we have that [ g 11λ11 ] = [ g 21λ 21 ]  λ 21

...

...

...

...

...

...

−1 presentation of our theorem. Hence, we can change  g 11λ11   λ11 g 12 λ12  …  λ1−(1m−1) g 1m  to ( g 21λ 21 )(λ 21−1h11λ11 )  λ11−1 g 12 λ12  …  λ −11(m−1) g 1m  without changing the element in the presentation. This completes the first step of our induction.

Rij

R11

...

...

...

...

...

...

...

...

...

g11

R1k g1k

f1

R1m g1m

...... fn

Illustration 13.6

Assume we have established that [ f ] and the product of homotopy classes of the loops associated to the edges of the rectangles along the thick line segments shown in Illustration 13.6 (extended by the vertical segment which we do not show in this figure), define the same element in the presentation (∗). Roughly speaking, we now change the path along the two thick edges of Rij (left and bottom) to a path over the other two edges of Rij (top and right). The argument is similar to the one used in the first step of the induction.

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λ(i+1)(j–1)

λ(i+1)j

gi(j+1) t = 0.4

hi(j–1)

Rij

t = 0.2

λi(j–1)

t = 0.6

t = 0.8 hij

λij

gij

Illustration 13.7

(

)(

)

A homotopy between the (products of) loops λ(−i1+1)( j −1)hi( j −1) λ i( j −1) λ i−(1j −1) g ij λ ij and λ(−i1+1)( j −1) g i( j +1) λ(i +1) j λ(−i1+1) j hij λ ij is indicated in Illustration 13.7. Since the loops in parentheses are at x, and since these loops, as well as the homotopy shown in Illustration 13.7, are entirely within an open set U j containing Rij , we have that  λ −1  λ   λ −1 h λ  in the presentation h λ g λ  =  λ −1 g λ  (i +1)( j −1) i( j −1) i( j −1)   i( j −1) ij ij   (i+1)( j −1) i( j +1) (i+1) j   (i+1) j ij ij  (∗). Thus, we can extend our homotopy over the rectangle Rij without changing the element in the presentation (∗). It follows by induction that the element in (∗) defined by f is equal to the element in the presentation defined by a path that goes along the left edge of I × I , then the top edge, and then along the right edge. The last path is the constant loop at x; hence, f is the same as the trivial element in the presentation (∗), completing the proof of the theorem.

(

)(

)

Question: Where were the identities {i ∗j ( z ) = ik∗ ( z ) :  z ∈ Z jk ,  j , k ∈ J } in the presentation given in the statement of the theorem used in the above proof? Answer: Consider, for example, the loop z = λ (−i1+1)( j −1) g i ( j +1)λ (i+1) j used in the last step of the proof. There, we treat it is as a loop in the open set Uj containing the image of the rectangle Rij . When we deal with the rectangle R(i+1) j that is just above Rij , we will take this loop to represent an element in the fundamental group of the open set U k containing R(i+1) j . However z is in the intersection of Uj and Uk, and consequently, since i ∗j ( z ) and ik∗ ( z )  have been equated in the presentation, we can replace the homotopy class z in Uj (used when we deal with the rectangle Rij ) with the homotopy class of the same loop (used when we deal with the rectangle R(i+1) j ), without changing the element in the presentation. So, i ∗j ( z ) = ik∗ ( z )  was utilized when switching from the loops made of the edges of one rectangle to the loops made of the edges of the adjacent rectangles. We remind the reader that in what follows we will often refer to Theorem 2 as the SvK theorem. Corollary 3. Suppose the conditions of Theorem 2 are satisfied. If each U j ∩ U k is simply connected, then π1 ( X , x ) is the free product of the fundamental groups π1 (U j , x ), j ∈ J . Proof. This follows immediately from Theorem 2. The following statement is a slight generalization of Corollary 3.

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Corollary 4. Let the conditions of Theorem 2 be satisfied and let the inclusion U j ∩ U k → U j and U j ∩ U k → U k induce the trivial homomorphisms of the fundamental groups for every j , k ∈ J . Then π1 ( X , x ) is the free product of the fundamental groups π1 (U j , x ) , j ∈ J . Proof. Using the notation of Theorem 2, the assumptions imply that both i ∗j ( z ) and ik∗ ( z ) are the identity element, and so i ∗j ( z ) = ik∗ ( z ) can be removed by Tietze transformations, ending up with the free product of the groups π1 (U j , x ), j ∈ J . Corollary 5. Suppose the conditions of Theorem 2 are satisfied. If each of the groups π1 (U j , x ) , j ∈ J , is trivial, then so is π1 ( X , x ) . Proof. This is evident. Here is a specific application of the SvK theorem. We will see many more in the coming sections. Corollary 6. The sphere S n, n ≥ 2, is simply connected. Proof. Let A denote the north pole (0,0,…,0,1) of S n and let B denote the south pole (0, 0, … , 0, − 1) of S n . Take U1 = S n \{ A} and U 2 = S n \{B}. The north-pole and southpole stereographic projections show (respectively) that the open sets U1 and U 2 are both homeomorphic to n , and so they are both simply connected. Since U1 ∩ U 2 is path connected for n ≥ 2, the conditions of the SvK theorem are fulfilled and Corollary 5 applies. So, S n is also simply connected. Exercises 1. Show that there does not exist a cover of the torus T 2 = S1 × S1 consisting of two open subsets U and V such that both U and V are not simply connected and such that U ∩ V is simply connected. 2. Let U1 , U 2 be an open cover of a path connected space X, and let V1 , V2 be an open cover of a path connected space Y, such that both covers satisfy the conditions of the SvK theorem (Theorem 1). Show that U1 × V1 , U1 × V2 , U 2 × V1 , U 2 × V2 is an open cover of X × Y satisfying the conditions of Theorem 1. 3. Consider the statement of Theorem 2. Show through a counterexample that if the sets in the cover U = {U j  :  j ∈ J } are assumed to be closed (while keeping the other conditions of Theorem 2 unchanged), then the conclusion of Theorem 2 fails. 4. Cover a sphere with two open sets as indicated in Illustration 13.8 and its caption. Use the SvK theorem to find a presentation for the fundamental group of the sphere, then use Tietze transformations to reduce that presentation to the trivial one.

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Illustration 13.8 U is made of the red (dark gray) two quarters, the yellow (light) strips included, and V is made of the green (light gray) two quarters, the yellow (light) strips included. The intersection consists of the yellow (light) thin strips along two perpendicular large circles.

5. (a)  Show that for every x ∈ 3 , the space  3 \{x } is simply connected. (b) Show that  2 is not homeomorphic to  3. (c) Show that for every x1 , x 2 ,… , xn ∈ 3 , the space  3 \{x1 , x 2 ,… , xn } is simply connected. 6. Show that  3 \ D 3 is simply connected (recall that D 3 = {( x , y , z ) ∈  :  x 2 + y 2 + z 2 ≤ 1}).

k

7. (a) Denote Sn2 = {( x , y , z ) : ( x − 2n)2 + y 2 + z 2 = 1} , n =1, 2, … , k. Show that ∪ S n2 n=0 (considered as a subspace of  3 ) is simply connected. (b) Use the SvK theorem to find a presentation of the fundamental group of a necklace of n (touching) spheres (see Illustration 13.9).

   Illustration 13.9

13.2  Seifert–van Kampen Theorem: Examples A brief historical note: Herbert Seifert stated and proved his version of the SvK theorem in his dissertation in 1930, when he was only 23 years old. His dissertation was about compact 3-manifolds, and he needed the theorem to compute the fundamental groups of some spaces that emerged in his work.

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The Seifert–van Kampen Theorem substantially expands the class of spaces whose fundamental group can be relatively easily determined. Example 1 Consider the space X depicted in Illustration 13.10. It is obtained by drilling a Y-shaped tunnel within a ball (open or closed, it does not matter). We use the SvK theorem directly to compute the fundamental group of X (there is a somewhat shorter computation, as indicated in Exercise 4). The open sets U1 and U 2 are depicted in Illustrations 13.11 and 13.12. They form a cover of X. The intersection U1 ∩ U 2 is shown in Illustration 13.13. We chose the base point x for all of the spaces mentioned here to be in this intersection.

Illustration 13.10  The space X.



Illustration 13.12  The subspace U 2 .

Illustration 13.11  The subspace U1.

Illustration 13.13  The subspace U1 ∩ U 2 .

Since, as we see, U1 ∩ U 2 is simply connected, it follows from Corollary 3 in 13.1 that π1 ( X , x ) = π1 (U1 , x ) ∗π1 (U 2 , x ). The spaces U1 and U 2 are homeomorphic. Further, both

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of them are homotopically equivalent to the space we get by drilling a tube through a ball, which in turn is homotopically equivalent to a circle. So, π1 ( X , x ) =  ∗  . Specifically π1 ( X , x ) = [α],[β] ; ∅ , where α is a loop at x that winds once around the branch of the Y-shaped tunnel entirely in U1 , and β is a loop at x that winds once ☐ around the branch of the Y-shaped tunnel entirely in U 2 . Example 2 Consider the space X depicted in Illustration 13.14 (two tunnels) and Illustration 13.15 (n tunnels): it is obtained by drilling tunnels out of the hollow ball {( x , y , z ) ∈ 3  : 1.5 ≤ x 2 + y 2 + z 2 ≤ 2}. In the case when n = 2, we cover the space with the open sets U and V, where U encompasses a bit more than the right half of the space X, and V a bit more than the left half of X, so that the intersection U ∩ V is a washer-like space. Since every loop in U ∩ V is contractible in each of U and V, it follows that the inclusion U ∩ V → X induces the trivial homomorphism π1 (U ∩ V , x ) → π1 ( X , x ). Hence, Corollary 4 in 13.1 applies, and consequently, π1 ( X , x ) is the free product of the fundamental groups of U and V. Each of these two open sets is homotopically equivalent to a ball with a tunnel along (say) a diagonal drilled out of it, which in turn is homotopically equivalent to a circle.

Illustration 13.14  A hollow ball with two tunnels.

Illustration 13.15  A hollow ball with n tunnels.

Consequently, π1 (U , x ) = < a  ; ∅ > , π1 (V , x ) = < b  ; ∅ > , and π1 ( X , x ) = < a, b  ; ∅ > is a free group of rank 2. In the case of n tunnels (Illustration 13.15) the fundamental group of that space is free of rank n. This follows by simple induction, since Corollary 4 of 13.1 applies if we choose a cover with two open sets, one encompassing one tunnel, the other the ☐ remaining n −1 tunnels. The following is a simple but important example.

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Example 3: A Bouquet of Circles Recall that we have defined a bouquet X to be a 1-dimensional CW-complex, whose 0-skeleton consists of a single point (denoted by x below). We cover X with open sets as shown in the middle figure of Illustration 13.16, where we depict a typical open set in a cover {U j  :  j ∈ J } of X. We also show the intersection of any two distinct open sets in the cover (the right-most figure in Illustration 13.16).

Uj Uj ∩Uk x

x

x

Illustration 13.16  From left to right: the space X, one typical member of the open cover, and

the intersection of any two such open sets.

Clearly π1 (U j ∩ U k , x ) = 1 for every j , k ∈ J . Since each U j is homotopically equivalent to a circle, we have that π1 (U j , x ) = 〈a j  ; ∅〉 . It follows from Corollary 3 of 13.1 ☐ that π1 ( X , x ) = 〈{a j :  j ∈ J } ; ∅〉. In the following example we contrast the CW-complex in Example 3 with the corresponding subspace of  3. Example 4: A Bouquet of Circles—Subspace Topology Let X be the space consisting of infinitely many circles C j , j ∈ J , of radius 1, where C j is obtained by rotating the circle C = {(0, y , z ) ∈ 3 : ( y − 1)2 + z 2 = 1} around the z-axis through j radians (0 ≤ j < 2π ). Hence, X is a space of circles sharing a common point x = (0, 0, 0). We consider it as a subspace of  3 . If J is finite, then the argument in Example 3 applies without changes, since the subspace topology and the CW-complex topology over X coincide. However, if J is infinite, and if X is a closed subset of  3 , then the topologies are different, and, in particular, at least one of the sets U j described in Example 3 is not open in this space X; this claim follows from the fact that the space X is compact (hence, Bolzano–Weierstrass). Thereby, the argument used in Example 3 does not apply. Nevertheless, we show in this example that the fundamental group of X is also a free group. Denote the set of points in the circles C j antipodal to the origin by A, and let {Bk : k ∈ K } be the set of all path components of the set A. For each k ∈ K denote Dk = ∪ C j . Then X is the union of the spaces Dk , k ∈ K , where the intersection C j ∩Bk ≠∅

of any two Dk1 and Dk2 is the singleton {x = (0, 0, 0)} (see Illustration 13.17).

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Dj

Dk

Bj

Illustration 13.17  A set Bs could be an arc (as is Bk in the figure), or it could be a point ( B j in the figure), in which case the corresponding D j is a circle.

Bk

We prove that π1 ( X , x ) is a free group of rank K . We will use Propositions 3 and 5, Section 12.3, as well as the finite case of Example 3 above. Observe first that each Dk is homotopically equivalent to a circle, and so π1 (Dk , x ) = 〈α k ; ∅〉 , where α k is the homotopy class of a loop at x that winds once around Dk . Next, we show that every [α ] ∈π1 ( X , x ) can be expressed uniquely as a product β1β 2 …βn , where β1 , β 2 , … , βn is a sequence reduced with respect to {π1 (Dk , x ) : k ∈K } (the terminology comes from 12.3). Since the interval I is compact, there is a small open neighborhood U x of x such that α( I ) ∩ (Ck \ {U x }) ≠ ∅ only for finitely many k ∈{k1 , k2 ,… , km } ⊂ K . Following the method used in m Example 3, the fundamental group of the subspace ∪ Dks is freely generated s =1

by {ak1 , ak2 ,… , akm } . Consequently [α ] can be expressed uniquely in terms of m {ak1 , ak2 ,… , akm } , such that the corresponding sequence of elements in ∪ π1 (Dks , x ) s =1

is reduced (Proposition 3 in 12.3). This element [α ] does not allow other exprest sion in terms of the elements in some ∪ π1 (D pr , x ) such that the associated r =1

sequence is reduced with respect to {π1 (D pr , x ) : r = 1,2,…, t }, for otherwise these two products would correspond to different sequences reduced with respect to {π1 (Dks , x ): s = 1,2,…, m} ∪ {π1 (D pr , x ): r = 1,2,…, t }, contradicting Example 3 and Proposition 3 in 12.3. We have proven that [α] can be expressed uniquely as a product β1β 2 … βn , where β1 , β 2 , … , βn is a sequence reduced with respect to {π1 (Dk , x ) : k ∈ K } . It follows from Proposition 5 in 12.3 that π1 ( X , x ) is the free product of {π1 (Dk , x ) : k ∈ K } , which, in ☐ this case, implies that π1 ( X , x ) ≅ 〈{ak : k ∈ K } ; ∅〉 . ∞

{

}

Remark. The subspace ∪ ( x , y ) ∈ 2 : ( x − n1 ) + y 2 = n12 of  2 , called Hawaiian Earrings, n=1

2

also consists of circles sharing a common point. However, its fundamental group is not free. This is not an easy result (see, for example, [70]). The pathological behavior of the Hawaiian Earrings will also be noted in Example 3, Section 15.4 (see Illustration 15.28 there).

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304   ◾    Seifert–van Kampen Theorem and Applications

Example 5: Torus, Again We regard the torus T 2 as the quotient space described in Illustration 4.22, Section 4.4. We cover it with two open sets as shown in Illustration 13.18. b

a

Illustration 13.18  U1 is inside the dashed

a

rectangles, and U2 is outside the smallest (closed) rectangle (so, both U1 and U2 are open). The intersection is as indicated in the figure. The base point x is in the intersection, as required.

U1 U2

x

b

U1 is simply connected, and U1 ∩ U 2 is homotopically equivalent to a circle. Hence, π1 (U1 , x ) is trivial and π1 (U1 ∩ U 2 , x ) is infinite cyclic generated by the homotopy class of the loop α shown in Illustration 13.20. A bouquet of two circles (extended by a line segment to reach the base point, as shown in the right-most figure of Illustration 13.19) is a deformation retract of U 2 (Illustration 13.19). It follows by Example 3 that π1 (U 2 , x ) = 〈[γaγ −1 ],[γbγ −1 ] ; ∅〉 , where γ is the path shown at the right-most figure of Illustration 13.19. U2 b a

b a

x

a

b a

x

b

a

b a

x

b

a

a a=

x

b

γ x b

b

Illustration 13.19  A bouquet of two circles as a deformation retract of U2.

Denoting the inclusions by i1 : U1 ∩ U 2 → U1 and i2 : U1 ∩ U 2 → U 2 , we have that (i1 )∗ ([α]) must be the identity in π1 (U1 , x ) , while we see in Illustration 13.20 that (i2 )∗ ([α]) = [γbab −1a −1γ −1 ] = [γbγ −1 ][γaγ −1 ][γb −1γ −1 ][γa −1γ −1 ] . (Do not forget that the edges labeled by a and b are identified in the torus, forcing all four of the vertices of the large rectangle to be identified into a single point.) b

a

α

x b

b

b

a

a

a

x γ

α b

a

a

x γ

α b

Illustration 13.20  The loop α is homotopic within U2 to the loop ( γbγ −1 )( γaγ −1 )( γb −1 γ −1 )( γa −1 γ −1 ).

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13.2  Seifert–van Kampen Theorem: Examples   ◾    305  

Denoting for brevity x = [γaγ −1 ] and y = [γbγ −1 ], the SvK theorem gives (again) ☐ 〈 x , y  ;  yxy −1 x −1 = 1〉 as a presentation for π1 (T 2 , x ). Example 6: Graphs Recall that a graph is a CW-complex of dimension 1; it is obtained by identifying the endpoints of closed intervals (1-cells) to the points of a discrete space (made of 0-cells or isolated points). Recall that the 0-cells in a graph X are called vertices, while the 1-cells in X give rise to the edges in the graph. If the 1-cell I is attached to the 0-skeleton of X by identifying 0 and 1 with u and v, respectively, then we say that the edge in X that corresponds to that 1-cell has u and v as its end-vertices, and that it is an edge from u to v. It is, of course, possible that these two end-vertices coincide in X. We will denote the set of edges in a graph X by E(X), and the set of vertices in X by V(X). Since we are interested in fundamental groups, we will consider only path connected graphs. A tree is a path connected and simply connected graph. A maximal tree in a graph X is a tree in X that is maximal in X (i.e., for which there is no tree of X that properly contains it as a subcomplex.) Proposition 1. Every tree in a graph X is a subtree of a maximal tree in X. Proof. The set of all trees in X is partially ordered with respect to ≤, where T1 ≤ T2 means that the tree T1 is a subcomplex of the tree T2 . Intending to use Zorn’s Lemma (Section 1.3), we consider a chain of trees T1 ≤ T2 ≤ … ≤ Tn ≤ … in X. It is obvious that T = ∪+ Tj is a subcomplex of X of dimension 1. Since each Tj is path connected, it j∈

follows from Proposition 3 in 6.3 that T is also path connected. If it is not simply connected, then there is a loop α at a vertex in V(T) that is not homotopic to the trivial (constant) loop at that vertex. If α(I) does not contain the entire interior of a 1-cell in T, then α is homotopic to a loop that does not intersect the interior of that 1-cell. Hence, we may assume that if a 1-cell in T intersects α(I), then it is a subset of α(I). Consider the set S of all 1-cells in X that are subsets of α(I). Suppose S is infinite. Let U s be a 1-cell s ∈ S, possibly extended slightly at the two end-vertices to get an open set. Then the family {U s : s ∈S} is an open cover of α(I) containing no finite subcover. This is not possible since α(I) is compact. So, S must be finite. Since every 1-cell in S is in ∪+ Tj , it must be in Tk , for some k ∈+ . This j∈

forces α(I) to be in Tk . Since Tk is simply connected, α is homotopic to the trivial loop within Tk , and so, it is homotopic to the trivial loop in T, contradicting our initial assumption. Hence, T is a tree. We have proven that every chain of trees in X is bounded by a tree. The statement of the proposition now follows from Zorn’s Lemma (Section 1.3).

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Let u and v be two vertices in a graph X. A geodesic between u and v in the graph X is a minimal tree in X containing u and v. Every geodesic has finitely many vertices and is homeomorphic to I =[0, 1] (Exercise 6). Every maximal tree T in X must contain all of the vertices of X. Otherwise, if G is a geodesic joining a vertex out of T with a vertex in T, then T ∪ G will be a tree properly containing T (contradicting the maximality of T). Proposition 2. Let X be a path connected graph containing exactly one edge that does not belong to a maximal tree T in X. Then π1 ( X , x ) is an infinite cyclic group. Proof. Let e be the edge in X out of a maximal tree T, and let G be a geodesic in T joining the end-vertices of e. Choose two open sets U and V covering X as in Illustration 13.21 (the figure to the right) and its caption. T T

G

x G x

G x e e

Illustration 13.21  The graph X is shown to the left, with the geodesic G shown thicker. We

split it into two open sets: U is shown up right, V is down right. The intersection U ∩ V is G extended by a few whiskers made out of small parts of the adjacent edges.

Then T and G are deformation retracts of U and U ∩ V , respectively, and so both U and U ∩ V are simply connected. Observe that G ∪ e is a deformation retract of V. Since G is homeomorphic to a proper closed interval in  , it follows that G ∪ e is homeomorphic to S1, and so π1 (V , x ) is infinite cyclic. It now follows from the simple version of the SvK theorem described in Corollary 3 of 13.1 that π1 ( X , x ) is an infinite cyclic group. Proposition 3. For every path connected graph X, π1 ( X , x ) is a free group freely generated by a set of cardinality E( X ) − E(T ) , where T is any maximal tree in X. Proof. Choose an open cover of X as in Illustration 13.22 and its caption.

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13.2  Seifert–van Kampen Theorem: Examples   ◾    307   Uj

X x

T

T

x ei

ek ej

ej

Illustration 13.22  For every e j ∈ E( X )\ E(T ), the open set U j is the union of an open set slightly

larger than T and the edge e j (compare X and U j around the edge ek ).

For every j , k ∈ J , T is a deformation retract of U j ∩ U k , and so U j ∩ U k is simply connected. The result now follows from Proposition 2 (above) and Corollary 4 of 13.1. ☐

Exercises 1. Show that {( x , y , z , w ): x 2 + y 2 + z 2 + w 2 ≥ 1} is simply connected. 2. Find two homotopic continuous mappings f , g : ( X , x ) → (Y , y ) such that the induced homomorphisms f∗ : π1 ( X , x ) → π1 (Y , y ) and g ∗ : π1 ( X , x ) → π1 (Y , y ) are not equal. (Compare with Theorem 6 in 10.4.) 3. Find a presentation of the fundamental group of the punctured torus (S1 × S1 )\ {( x , y )}. 4. Visualize a homotopy equivalence between the space X in Example 1 (Illustration 13.10) and the bouquet of two circles. Conclude again that the fundamental group of X is a free group of rank 2. 5. Show that A = (S1 × {x }) ∪ ({x } × S1 ), x ∈S1 , is not a retract of X = S1 × S1 . 6. Show that every geodesic joining two vertices in a connected graph X has finitely many edges, and that it is isomorphic to the unit closed interval I. 7. Let X be a tree, and let T1 and T2 be trees in X such that T1 ∩ T2 ≠ ∅ . Show that T1 ∩ T2 is also a tree. 8. Use the SvK theorem to find the fundamental group of a sphere together with one of its diameters. 9. Find the fundamental group of the space made of two concentric spheres connected with (a) one tunnel (Illustration 13.23), (b) two tunnels (Illustration 13.24), (c) n tunnels along rays emanating from the common center of the spheres.

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308   ◾    Seifert–van Kampen Theorem and Applications

Illustration 13.23

Illustration 13.24

10. Use the SvK theorem to compute the fundamental group of the space in Illustration 13.25 (a tunnel drilled out of a solid ball).

Illustration 13.25

Illustration 13.26

11. Let X be the space consisting of a ball, with finitely many disjoint smaller balls removed from its interior and with finitely many disjoint tunnels connecting each pair of the smaller balls. Assume in addition that X could be projected in  2 such that both the set of small balls and the set of tunnels are pairwise disjoint in the projection. (This somewhat mysterious condition is given to avoid any knotting or linkage of the tunnels; without it the conclusion is false in general.) An illustration of one such X, with 3 small balls removed and 6 tunnels connecting them, is provided in Illustration 13.26. Show that π1 ( X , x ) is a free group of rank equal to the number of tunnels. 12. Let B be an open ball containing the object C shown in Illustration 13.27, and let X be B \C. Use SvK to compute the fundamental group of the space X.

   Illustration 13.27

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13.2  Seifert–van Kampen Theorem: Examples   ◾    309  

13. Let B be a ball containing the solid knotted object K shown in Illustration 13.28, and let X be B \ C. Compute the fundamental group of the space X. (Hint: you do not have to use the SvK theorem! But, if you do, then it is a good idea first to change the space via an ambient isotopy within  3.)

   Illustration 13.28

14. Find the fundamental group of the space obtained from the hollow solid ball {( x , y , z ) ∈ 3  : 1 ≤ x 2 + y 2 + z 2 ≤ 2} by drilling n many non-overlapping tunnels out of it along rays emanating from the origin. 15. Let X be the space consisting of any finite number n ≥ 3 of closed (open) half-disks sharing a common diameter (Illustration 13.29). Show that X is not homeomorphic to the disk D 2 (to B2 , respectively). [Hint: take out a point from the common diameter.]

   Illustration 13.29

16. (a) Find the fundamental group of the space depicted in Illustration 13.30 (a torus and a line segment joining two points on the surface of the torus). (b) Find the fundamental group of the space depicted in Illustration 13.31 (n tori touching a circle)

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310   ◾    Seifert–van Kampen Theorem and Applications T1

T2

T3 Tn

Illustration 13.30 Illustration 13.31

17. (a) Show that the fundamental group of the subspace X of  3 depicted in Illustration 13.32 is a free group (circles converging on both ends to two points, A and B, both of which are excluded from X). Find a set of free generators of π1 ( X ) . (b) Show that π1 ( 3 \({ A, B} ∪ X )) is also a free group and find a set of free generators. A

B

Illustration 13.32

13.3  Seifert–van Kampen Theorem and Knots A brief historical note: In 1905 Wilhelm Wirtinger (1865–1945) invented a method for computing the fundamental group of the complement of a tame knot in  3. Another early procedure for describing such groups was published in 1910 by Max Dehn.

If two knots K1 and K 2 are ambient isotopic within  3, then they are called equivalent knots. For such knots their  3 -complements K1c and K 2c are also ambient isotopic within  3 (Corollary 3 in 10.1). Since ambient isotopic sets are necessarily homeomorphic, it follows that K1c and K 2c have isomorphic fundamental groups. Consequently, if π1 ( K1c , x ) and π1 ( K 2c , x ) are not isomorphic, then K1 and K 2 are not ambient isotopic within  3 . (The converse is false; more about that later in this section.) The fundamental group π1 ( K c , x ) of the  3 -complement K c of a knot K is called the knot group of K. In this section we will investigate some knot groups. Let B = {( x , y , z ) ∈ 3  :  x 2 + y 2 + z 2 < a } be an open ball enclosing a knot K. Then the knot group of K is isomorphic to π1 ( B \ K , x ) (Exercise 1). Hence, in order to obtain a presentation of the knot group it suffices to investigate the complement of the knot in B. This is what we will do in this section1. 1

Usually the knots are enclosed in the three sphere S3 (viewed as the one-point compactification of  3 ) instead of the ball B. The advantage of S3 is that it is a fixed compact manifold containing all knots. For the purposes of what we cover in this book, there is no difference between the two. We choose the ball B since it is readily visualizable in  3.

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13.3  Seifert–van Kampen Theorem and Knots   ◾    311  

The following lemma tells us how to thread a loop around crossings of planar representations of knots. It will be needed throughout the rest of this section. Lemma 1. With the setup shown in Illustration 13.33, αβα −1  δ in B \ K, where B is an open ball enclosing the knot K. Proof. The proof we give is a six-frame animation, as shown in Illustrations 13.33 to 13.38.

δ

δ

δ

α

α β

αβ

β

Illustration 13.33  The setup.

Illustration 13.34  First we homotopically change αβ.

homotopy of αβ.

δ

δ

δ

Illustration 13.35  Continuing

αβα–1 αβ

α

did what we wanted to αβ. Now we show α again, and start deforming αβα–1.

Illustration

13.36  We

αβα–1

Illustration 13.37  Showing the

result after performing a simple homotopy.

Illustration 13.38  It is now evident that αβα −1  δ .

This lemma facilitates the usage of the SvK theorem, as will be illustrated in the following three examples in which we apply the SvK theorem directly on the complements of some knots. By the end of this section we will provide a general method for computing knot groups. Example 1: The Trefoil Knot Group Consider the trefoil knot (as a subspace of an open ball B of radius, say, 2, as shown in Illustration 13.39). We cover its complement X in B with two open sets, as shown in Illustrations 13.40 and 13.41, and the corresponding captions.

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312   ◾    Seifert–van Kampen Theorem and Applications

α1 x

α3

Illustration 13.39  The trefoil knot; the space

we consider is its complement X in B.

α2

Illustration 13.40  The open set U: it consists of

the points in X out of a closed ball of radius 1.

β2

β1

γ1 γ2 x

β6

x β5

β4

β3

γ3

Illustration 13.41  The open set V: it consists of

the points in X in an open ball of radius 1.5.

Illustration 13.42  The intersection U ∩ V

contains the base point x. We also depict (here and in Illustration 13.40 and 13.41) a few loops that we will need.

The group π1 (U , x ) is free on a1 = [α1 ], a2 = [α 2 ], and a3 = [α 3 ] (Example 2 in 13.2). The space V is homotopically equivalent to a bouquet of three circles, and so its fundamental group is also a free group; it is freely generated by c1 = [γ 1 ], c2 = [γ 2 ], and c3 = [γ 3 ]. Finally π1 (U ∩ V , x ) is generated by b1 = [β1 ], b2 = [β 2 ], b3 = [β3 ], b4 = [β 4 ], b5 = [β5 ], and b6 = [β6 ]. (In fact, π1 (U ∩ V , x ) is generated by any 5 of the 6 shown elements; see Exercise 16 in 10.4.) Denoting the inclusions by i : U ∩ V → U and j : U ∩ V → V , we observe (by staring at Illustration 13.40–13.42) the following: i∗ (b1 ) = a1 , i∗ (b2 ) = a2−1 , i∗ (b3 ) = a2 , i∗ (b4 ) = a3−1, i∗ (b5 ) = a3 , and i∗ (b6 ) = a1−1 . Three of the following six equalities need Lemma 1 as indicated; the other three are obvious: j∗ (b1 ) = c1 , j∗ (b2 ) = c1−1c3−1c1 (Lemma 1), j∗ (b3 ) = c2 , j∗ (b4 ) = c2−1c1−1c2 (Lemma 1), j∗ (b5 ) = c3 , and j∗ (b6 ) = c3−1c2−1c3 (Lemma 1). The SvK theorem then gives us the following presentation for the trefoil knot group: 〈a1 , a2 , a3 , c1 , c2 , c3  ; a1 = c1 , a2−1 = c1−1 c3−1 c1 , a2 = c2 , a3−1 = c2−1c1−1c2 , a3 = c3 , a1−1 = c3−1c2−1c3 〉. Tietze transformations can reduce the redundancies in this presentation (for example, we could immediately eliminate all of the ai’s), eventually ending up with 〈 x ,  y ;   xyx = yxy 〉, one of the standard presentations for this group (Exercise 5a). ☐

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13.3  Seifert–van Kampen Theorem and Knots   ◾    313  

Example 2: A Truncated Trefoil Knot Consider the space Y depicted in Illustration 13.43, obtained from the space X in Example 1 by removing a solid spherical cap from the enclosing ball.

γ2 γ1

x

Illustration 13.43  The space Y; the loops γ 1 and γ 2 , giving rise to the generators c1 and c2 of

its fundamental group are shown to the right.

With U and V the same as in Example 1, U ∩ Y and V ∩ Y make an open cover of Y, and their intersection is path connected (so that the SvK theorem applies). The analysis employed in Example 1 requires only slight modifications: compared to U , U ∩ Y is lacking one tunnel, and its fundamental group has one less generator ( α1 ) ; compared to U ∩ V , U ∩ V ∩ Y has a couple of tunnels missing, and hence has two fewer generators ( β1 and β6 ); V ∩ Y is the same as V. The computation of π1 (Y , x ) now follows on the footsteps of the one done in Example 1, the only difference being the exclusion of the relations i∗ (b1 ) = j∗ (b1 ) , and i∗ (b6 ) = j∗ (b6 ) (corresponding to β1 and β6 ), and the exclusion of the generator a1 (corresponding to the loop α1 ). Consequently the following is a presentation of π1 (Y , x ): a2 , a3 , c1 , c2 , c3  ;  a2−1 = c1−1c3−1c1 , a2 = c2 , a3−1 = c2−1c1−1c2 , a3 = c3 . This could be reduced to c1 , c2  ; c1c2c1−1 = c2−1c1c2 (Exercise 5c), where c1 and c2 are the homotopy classes of γ 1 and γ 2 , respectively, as shown in Illustration 13.43. We will ☐ use this result in Example 3 below. Example 3: The Square Knot Group The Square Knot is shown in Illustration 13.44; we are after its knot group.

Illustration 13.44  The square knot. Illustration 13.45  The square knot tunneled out of an open disk.

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314   ◾    Seifert–van Kampen Theorem and Applications

δ2

δ1

x

Illustration 13.46  The open set

U. We also show two loops.

γ1

λ1

γ2

x

Illustration 13.47  The open set V (and two loops).

λ2 x

Illustration 13.48  U ∩ V .

We put the square knot in an open ball B (Illustration 13.45) and denote its complement in the ball by X, so that the associated knot group is π1 ( X , x ). Choose two open sets, U and V, covering X as in Illustrations 13.46 and 13.47 ; notice that both U and V are the same as the space denoted by Y and shown in Illustration 13.43. It follows from Example 2, that c1 , c2  ; c1c2c1−1 = c2−1c1c2 is a presentation of π1 (V , x ) , and d1 , d2  ; d1d2d1−1 = d2−1d1d2 is a presentation of π1 (U , x ) , where ci is a homotopy class of the loop denoted γ i and shown in Illustration 13.47 (i = 1, 2), and where di is a homotopy class of the loop denoted δi and shown in Illustration 13.46 (i = 1, 2). The group π1 (U ∩ V , x ) is (freely) generated by the homotopy classes of the loops λ1 and λ 2 (Illustration 13.46). As usual we denote the inclusions by i : U ∩ V → U and j : U ∩ V → V . We observe that i∗ ([λ1 ]) = d1 , and that j∗ ([λ1 ]) = c1 , so that c1 = d1 will appear as one of the relations in the presentation stemming from the SvK theorem. We now analyze the relation i∗ ([λ 2 ]) = j∗ ([λ 2 ]) in the SvK presentation. A straightforward approach would be to express i∗ ([λ 2 ]) and j∗ ([λ 2 ]) in terms of the generators of π1 (U , x ) and π1 (V , x ), respectively. For example, threading loops through V by means of Lemma 1, it can be seen that λ 2 is homotopic to γ −2 1 γ 1−1 γ 2 γ 1 γ 2 , implying that j∗ ([λ 2 ]) = c2−1c1−1c2c1c2 . Symmetrically, i∗ ([λ 2 ]) = d2−1d1−1d2d1d2 . Hence, by SvK theorem, c1 , c2 , d1 , d2  ; d1d2d1−1 = d2−1d1d2 , c1c2c1−1 = c2−1c1c2 , c1 = d1 ,  −1 −1 c2 c1 c2c1c2 = d2−1d1−1d2d1d2 is a presentation for π1 ( X , x ) . With that our computation is completed. Remark. There is a shorter route that makes it evident that it is completely irrelevant what i∗ ([λ 2 ]) and j∗ ([λ 2 ]) are in terms of the generators of π1 (U , x ) and π1 (V , x ). The loop λ 2 is homotopic within V to the loop λ1−1 : stare at Illustrations 13.47 and 13.48 and simply widen λ 2 enough so that we can thread the whole knotted tunnel through it, emerging on the other side as λ1−1 . So, j∗ ([λ 2 ]) = c1−1 holds in the presentation for π1 (V , x ) . By symmetry, δ 2 is homotopic to δ1−1 within U, and so i∗ ([λ 2 ]) = d1−1 holds in the presentation of π1 (U , x ). On the other hand we showed above that c1 = d1 holds in the SvK presentation  of  π1 ( X , x ).

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13.3  Seifert–van Kampen Theorem and Knots   ◾    315  

Thereby i∗ ([λ 2 ]) = j∗ ([λ 2 ]) is a consequence of the other relations in the SvK presentation of π1 ( X , x ), and thus, by Tietze, it could be omitted. We conclude that c1 , c2 , d1 , d2  ; d1d2d1−1 = d2−1d1d2 , c1c2c1−1 = c2−1c1c2 , c1 = d1 is a presentation of π1 ( X , x ). Eliminating d1 (Tietze), we get c1 , c2 , d2  ; c1d2c1−1 = d2−1c1d2 , c1c2c1−1 = c2−1c1c2 , which in turn is equivalent to c1 , c2 , d2  ; d2c1d2 = c1d2c1 , c2c1c2 = c1c2c1 , or 〈a, b, c  ; cac = aca,  bab = aba〉, which is one of the standard presentations of the square knot group. ☐ Example 4: The Granny Knot Group The Granny Knot is depicted in Illustration 13.49. The knot group of the granny knot is the same as the knot group of the square knot (Exercise 7).



Illustration 13.49  The granny knot.

However, the square knot is not ambient isotopic within  3 to the granny knot. This is a consequence of the fact that the complements of these two knots in  3 are not ☐ homeomorphic [20]. We end this section with the promised general procedure for computing knot groups. A Method for Computing Knot Groups Using Knot Diagrams For practical reasons we usually represent knots in the drawing plane, rather than in  3 , where they belong. More precisely, our illustrations represent p(K), where K is the knot placed above the drawing plane  2 , and p :  3 →  2 is the vertical projection. In doing so, we introduce intersections: these are points x ∈ p( K ) such that | p −1 ( x )| ≥ 2. We consider only knots for which it is possible to choose the drawing plane such that | p −1 ( x )| is either 1 or 2 for every x, and such that when | p −1 ( x )| = 2 then the intersection is transverse. Briefly, transverse crossing is when the two parts of the knot in the diagram cross in the manner the x-axis and the y-axis do (but not similar to the way y = x 2 and the x-axes cross). Precise definition and technical details can be found in books on knot theory. If p(K) satisfies all of the above con­ ditions, then we call it a knot diagram. The points x in a diagram p(K) where | p −1 ( x )| = 2 are called crossings. For example, the diagram for the trefoil knot (Illustration 13.50) has three crossings. The method we are about to describe applies to knots allowing knot diagrams with finitely many crossings.   Let D be K \  ∪ p −1 ( x ) . Denote the components of D by C1 , C2 ,…, Cm . Each com x is a crossing  ponent Ci of D projects injectively via p onto a component of the diagram of the knot with

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316   ◾    Seifert–van Kampen Theorem and Applications

the crossing points deleted. For example, the standard diagram of the trefoil knot is such that D has 6 components (Illustration 13.51). x3

C2

C4 x1

x2

C1

x3 C5 C6

x1

α1 x2 α5 α3

Illustration 13.50  Three crossing points

in the standard diagram for the trefoil knot.

C3

Illustration 13.51  The 6 components of D here are C1 to C3. The loop α i winds once around Ci. To avoid clutter, only the loops α1 , α 3 and α 5 are shown.

Choose a loop α i at a base point and winding once around exactly one component Ci (see Illustration 13.51). Then the homotopy classes of these loops generate the knot group. The last claim can be justified by repeated usage of the SvK theorem; we indicate how to do this below.

αi3

αi2 αi4

αi1

Illustration 13.52  The neighborhood U. Illustration 13.53  The relevant (homotopy classes of) loops in U are α i1 , α i2 , α i3 and α i4 .

Choose an open ball U containing the inverse image p −1 ( x1 ) of a crossing x, such that U is small enough to avoid the inverse images of any other crossing points (Illustration 13.52). Then  3 \ K is covered by U, and by the complement V of a smaller ball U1 ⊂ U encompassing p −1 ( x1 ). The intersection U ∩ V is a spherical shell with four tunnels, similar to the space depicted in Illustration 13.42. The four tunnels give rise to four (homotopy classes of) loops α i1 , α i2 , α i3 , and α i4 . Careful usage of the SvK theorem, together with Lemma 1, introduces the following two relations: α i1 =  α i2 and α i4 =  α i1 α i3 α i−11

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(see Illustration 13.53). If A  ;  R is a presentation of π1 (V , x ), then SvK produces A  ;  R , α i1 =  α i2 , α i4 =  α i1 α i3 α i−11 as a presentation for the knot group π1 ( 3 \ K , x ). If there are no other crossings then π1 (V , x ) is a free group, and we end up with A  ; α i1 =  α i2 , α i4 =  α i1 α i3 α i−11 as the final presentation of the knot group. Otherwise we repeat the above procedure, working with V in place of  3 \ K . Each step will bring relations of the type α i =  α j or α i =  α j α k α −j 1 . After finitely many steps (corresponding to the number of crossings), and after eliminating one half of the generators using the relations of type α i =  α j , we end with an explicit presentation 〈 A  ;  {α is =  α js α ks α −js1 : s = 1,2,…, n}〉 of the knot group π1 ( 3 \ K , x ) . This is the Wirtinger presentation of the knot group; for more details see, for example, [72], page 144. The following example makes it clear what should be done to precisely identify the relations in the presentation of a knot group. Example 5: Trefoil Knot Group, Again The last procedure allows us to essentially read a presentation of the trefoil knot group (or any other knot group) directly from the associated knot diagram. In Illustration 13.54, we assume that the base point of the space is close to the location of the viewer’s eyes above the plane containing the figure, and that the generating loops are winding in the indicated direction (to make this orientation of the loops consistent, we have introduced an orientation along the knot). α 2 = α1 α4 = α2α5α–1 2 α5 α2

α4

α1

α6 α3 = α4 α6 = α4α1α 4–1

α 5 = α6 α2 = α6α3α –1 6 α3

Illustration 13.54

The six relations that immerge are shown in the figure. Consequently, we have the following presentation of the trefoil knot group: α1 , α 2 , α 3 , α 4 , α 5 , α 6  ; α 2 =  α1 , α 4 = α 2α 5α −2 1 , α 5 =  α 6 , α 2 = α 6α 3α 6−1 , α 3 =  α 4 , α 6 = α 4α1α 4−1 . Using Tietze transformations we eliminate α2, α4 and α6 to end up with the Wirtinger presentation 〈α1 , α 3 , α 5 ; α 3 = α1α 5α1−1 , α1 = α 5α 3α 5−1 , α 5 = α 3α1α 3−1 of the trefoil knot group. That what we get is equivalent to the earlier presentations for this knot group is left as an ☐ exercise (Exercise 8).

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Exercises 1. Let K be a knot and let a be a positive number such that K ⊂ B = {( x , y , z ) ∈ 3 :  x 2 + y 2 + z 2 < a} . Show that the knot group of K is isomorphic to π1 ( B \ K , x ). 2. View S 3 as the one-point compactification of  3 so that the latter can be considered as a subspace of the former (via the embedding equal to the inverse of the stereographic projection). Show that the knot group of a knot K (in  3 ) is isomorphic to π1 (S 3 \ K , x ). 3. Find the knot group of an unknot. 4. Prove that each generator in every Wirtinger presentation defines a non-trivial loop in the complement of the knot. 5. (a)  S how that the presentation for the trefoil knot group obtained in Example 1 can be reduced by means of Tietze transformations to 〈 x ,  y  ;   xyx = yxy 〉. [Hint, first eliminate ai ’s, then, in the next step, c1 ] (b) Show that 〈a, b  ;  a3 = b 2 〉 is also a presentation of the trefoil knot group. (c) Show that the presentation of the fundamental group of the space Y from Example  2 can be reduced by means of Tietze transformations to 〈 x ,  y  ;  xyx −1 = y −1 xy 〉 . 6. Consider the open cover of the complement of the trefoil knot, as shown in Illustration 13.55.

Illustration 13.55  The open sets U (left) and V (middle), and U ∩ V (right).

Use the SvK theorem on this cover, and again find the trefoil knot group, then employ Tietze transformations to change that presentation to 〈 x ,  y ;   xyx = yxy 〉 . 7. (a) Use the SvK theorem to find a presentation for the granny knot shown in Illustration 13.49. (b) Use Tietze transformations to show that a presentation solving (a) is equivalent to a, b, c  ; cac = aca, bab = aba , concluding that the granny knot and the square knot have the same knot groups. 8. Use Tietze transformations to confirm that the presentation of the trefoil knot group obtained in Example 5 is equivalent to 〈 x ,  y ;   xyx = yxy 〉. 9. Show that the knot group of a knot representable by a knot diagram with finitely many crossing points is infinite.

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10. Find a presentation of the knot group of the knot shown in Illustration 13.56.

    Illustration 13.56

11. Find a presentation of the knot group of the Figure-Eight Knot shown in Illustration 13.57.

    Illustration 13.57

12. The notion of a connected sum of two knots is explained in Illustration 13.58 and its caption. (For a precise definition see, for example, [12].) K1

K2

K1 # K2

Illustration 13.58  We cut both knots between two crossing points, then glue them as in

the right figure.

Show that if one of the knot groups of the knots K1, K 2 is not cyclic, then every connected sum of these two knots is not an unknot.

13.4 Torus Knots and Alexander’s Horned Sphere A brief historical note: James Waddell Alexander (1888–1971) presented his wild embedding of the sphere in  3, or the “horned sphere,” in an article published in 1924.

Torus knots are knots along the surface of a torus. We will classify them completely through their knot groups. The Fox–Artin wild arc, as we will see, is related to torus knots.

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It is one of the two examples mentioned in this section where we indicate that a sphere can be properly knotted in  3. The other example is the Alexander’s horned sphere. Torus Knots A torus knot of type (p,q), where p and q are relatively prime positive integers, is a knot on the surface of a torus that winds p times in the latitudinal direction and q times in the meridinal (or longitudinal) direction. Alternatively, the knot k p , q : I → S1 × S1 ⊂  3, k p ,q (0) = k p ,q (1), is a torus knot of type (p,q), for p and q relatively prime positive integers, kp , q 1 → S1 × S1 p → S1 ( p1 is the projection onto the first S1) winds p times around if I  k p,q 2 → S1 × S1 p → S1 ( p2 is the projection onto the second S1) winds q times S1, and I  1 around that S . There is yet a third equivalent definition: if we consider the torus to be the quotient space of  2 under the equivalence defined by ( x , y ) ~ ( x ′ , y ′ ) if both x − x ′ p and y − y ′ are integers (Exercise 2 in 4.2), then the line y = q x in the plane, p and q relatively prime, becomes a torus knot of type (p,q). An example of a torus knot is given in Illustrations 13.59 and 13.60. It is evident that the knot is of type (4,5).

Illustration 13.59  A torus knot of type (4,5).

Illustration 13.60  The same torus knot of type (4,5) is shown on the surface of a torus.

Our first goal is to find the knot group of a torus knot of type (p,q). The illustrations we show are of the (4,5) knot, but the computation we give below is for the fundamental group of any torus knot of type (p,q). First we replace the torus knot with a tubular torus knot—that is, with the product of the torus knot with a small open disk, small enough to avoid self intersections. (This change is now not only for esthetic reasons.) Since these two are isotopic within  3 (Exercise 10 in 10.2), so are their complements. Consequently, we can compute the knot group of the torus knot by computing the fundamental group of the complement of this tubular torus knot. Finally, we can substitute that underlying space  3 with any large open ball B containing the knot. Our target now becomes finding the fundamental group of the complement B \ K of the tubular (p,q)-knot K. We cover B \ K with two open sets, U and V. The open set U is the grooved solid torus (S1 × B2 ) \ K, where, we recall, B2 is the open unit disk (see Illustration 13.61). The tube around the torus knot cuts a tubular groove on the surface of U, and so U has the same

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fundamental group as the solid torus S1 × B2 . Hence π1 (U , x ) is infinite cyclic, freely generated by the homotopy class a of the loop that follows the middle circle S1 × {(0,0)} , winding once around the torus.

Illustration 13.61  The open

set U.

Illustration 13.62  The open

set V.

Illustration 13.63  The open

set U ∩ V .

The open set V is the complement in B \ K of ( B \ K ) ∩ (S1 × B∗2 ), where B∗2 is an open disk of radius slightly smaller than 1 (“slightly” means “less than one half of the diameter of the tube in the tubular knot we used above”). The open set V is shown in Illustration 13.62. The difference between V and the complement of the solid torus S1 × B∗2 in B is inconsequential: V has tubular carvings on its walls, while B \(S1 × B∗2 ) does not. Consequently their fundamental groups are the same. Since the fundamental group of B \(S1 × B∗2 ) is infinite cyclic (see Exercise 3 in 13.3), so is π1 (V , x ). A generator is the homotopy class b of any loop that winds once around the solid torus S1 × B∗2 (passing through the central hole of the torus). The intersection U ∩ V is shown in Illustration 13.63: it is a strip that winds p times in the latitudinal direction and q times in the meridinal direction. Since U ∩ V is clearly homotopically equivalent to a circle, the group π1 (U , x ) is freely generated by the homotopy class of a loop δ winding along the strip. It is evident that δ defines a p in π1 (U , x ), and that it defines bq in π1 (V , x ) . It follows from the SvK theorem that π1 ( B \ K , x ) = 〈a, b  ; a p = bq 〉. Summarizing: the knot group of a torus knot of type (p, q) is 〈a, b  ; a p = bq 〉. If ( p1 , q1 ) and ( p2 , q2 ) are two distinct pairs of relatively prime positive integers, such that p j < q j , j ∈ {1, 2}, then 〈a, b  ; a p1 = bq1 〉 and 〈a, b  ; a p2 = bq2 〉 are not isomorphic (Exercise  3). Consequently, torus knots of distinct types ( p, q), p < q, are not ambient isotopic within  3 . Alexander’s Horned Sphere According to the Jordan-Schönflies theorem (Theorem 6 in 11.3), every embedding f : S1 →  2 can be extended to a homeomorphism fˆ :  2 →  2 that sends the open unit disk B2 onto the bounded component of  2 \ f (S1 ), and sends  2 \ D 2 onto the unbounded component of  2 \ f (S1 ) . For a while people believed that the analogue statement should hold a dimension higher, with S 2 in place of S1 and with  3 in place of  2 . At one point Alexander announced that he had a proof. Upon discovering a gap in his argument he came

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up with a counter example, now called the Alexander’s horned sphere. It is depicted in Illustration 13.64.

CS 1

1

32

16

Illustration 13.64  Alexander’s horned sphere AHS: it is homeomorphic to the sphere. The precise template for the “horns” is shown to the right and it comes from Illustration 5.13 (the Cantor interval). In the figure to the left we widen up the “horns” to show the linkage more clearly.

We show that the unbounded component X of  3 \ AHS is not simply connected. Consequently,  3 \ D 3 (i.e., the unbounded component of  3 \ S 2 ), being simply connected, is not homeomorphic to X, hence the once-expected extension of the Jordan–Schönflies theorem in three dimensions fails. First observe that π1 ( X , x ) is generated by the homotopy classes of the loops in A = {a11 , a12 } ∪ {a21 , a22 , a23 , a24 } ∪ {a31 , a32 , a33 , a34 , a35 , a36 , a37 , a38 } ∪ … as shown in Illu­st­­ ration 13.65. We assume that they all wind clockwise. We use the SvK theorem to find an incomplete presentation of the fundamental group of AHS over the set A. Step 1. We split X into two open sets, U1 and U 0 : U 0 is bounded from above by a plane that is just above the plane ∑1 and U1 is bounded from below by a plane that is just below ∑1 , as shown in Illustration 13.65. Using the notation shown in Illustration 13.65 it is easy to compute that π1 (U 0 , x ) = a11 , a12  ; a12 = a11−1 , and that π1 (U 0 ∩ U1 , x ) = 〈a11 , a12  ; ∅〉 , where aij -s denote both the loops and their homotopy classes in the respective open sets. As for U1 , for the time being we merely state that π1 (U1 , x ) = 〈 A  ;  R 〉 for some set of relations R. The amalgamation

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Σ3

a31

a21

a22

a11

a23

a24

a12

Σ2

Σ1

Illustration 13.65

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produces equating a11 and a12 with the corresponding generators in π1 (U 0 , x ) and π1 (U1 , x ) which we have denoted by the same symbols. Hence, these relations could be ignored, and the SvK theorem gives π1 ( X , x ) = A  ; a12 = a11−1 , R . Step 2. Now we focus on U1 and cover it with two open sets U 2 and U 3 , where U 2 is bounded from above by a horizontal plane just above ∑ 2 , and U 3 is bounded from below by a horizontal plane just below ∑ 2 as in Illustration 13.65. We compute: π1 (U 2 , x ) = 〈a21 , a22 , a23 , a24 ; ∅〉, π1 (U 2 ∩ U 3 , x ) = 〈a21 , a22 , a23 , a24 ; ∅〉, and π1 (U 3 , x ) = 〈 A \ {a11 , a12 };  R1 〉 , where R1 is a set of relations not involving a11 and a12 . Using the SvK theorem, this gives π1 (U1 , x ) = 〈 A \ {a11 , a12 };  R1 〉 . Now we back up and modify the outcome of Step 1: since the last presentation of π1 (U1 , x ) does not use a11 and a12 , the amalgamation of π1 (U 0 ∩ U1 , x ) = 〈a11 , a12  ; ∅〉 in Step 1 requires that we express a11 and a12 in terms of A \ {a11 , a12 }. This is easy to achieve: it is obvious that a12 = a22a24 , and by Lemma 1 in 13.3 we have a11 = a21a22a23a22 −1 . So, Step 1 and 2 together produce the following: π1 ( X , x ) = A  ; a12 = a11−1 , a11 = a21a22a23a22 −1 , a12 = a22a24 ,  R1 where the set R1 does not have relations involving a11 and a12 . To see the iterative procedure more clearly we do one more step. Step 3. Here we take a look at U 3 . We split it into open sets U 4 and U 5 as suggested by Steps 1 and 2: U 4 is bounded from above by a horizontal plane just above ∑3 , and U 5 is bounded from below by a horizontal plane just below ∑3 as in Illustration 13.65. It is again easy to compute that π1 (U 4 , x ) = 〈a31 , a32 , a33 , a34 , a35 , a36 , a37 , a38 ; ∅〉, π1 (U 4 ∩ U 5 , x ) = 〈a31 , a32 , a33 , a34 , a35 , a36 , a37 , a38 ; ∅〉, and π1 (U 5 , x ) = 〈 A \ {a11 , a12 , a21 , a22 , a23 , a24 };  R2 〉, where R2 does not contain relations involving the generators that we have deleted from the set A. SvK gives π1 (U 3 , x ) = 〈 A \ {a11 , a12 , a21 , a22 , a23 , a24 };  R2 〉 . We again back up to Step 2 and use this presentation there; the only thing we need to do is to express {a11 , a12 , a21 , a22 , a23 , a24 } in terms of the other generators of π1 (U 3 , x ). Since a11 , a12 were already given in terms of a21 , a22 , a23 , a24 , we focus on these last four generators. Fortunately (and for our limited purposes) we do not need to worry about all of them: just focus on a21 (the first one), and −1 a24 (the last one). We compute (using Lemma 1 in 13.3) that a21 = a31a32a33a32 , and we easily see that a24 = a37a38 . The remaining two generators, a22 and a23 are clearly expressible in terms of a32 , a33 , a34 , a35 , a36 , a37 ; this little piece of information is sufficient for our −1 −1 , a22a23 , a12 = a22a24 , a21 = a31a32a33a32 purposes. We have π1 ( X , x ) = A ; a12 = a11−1 , a11 = a21a23 a24 = a37a38 , Q1 , R2 , where Q1 involves only middle generators—those not of type ai1 , and not of type ai 2i —and where R2 contains relations not involving generators encountered so far. Continuing this recursive construction and using induction, one gets the following: −1 −1 −1 π1 ( X , x ) = A  ; a12 = a11−1 , a11 = a21a22a23a22 ,  a21 = a31a32a33a32 , a31 =  a41a42a43a42 ,  −1 a41 =  a51a52a53a52 ,…a12 = a22a24 , a24 = a37 a38 , a38 = a4 15a4 16 ,… , Q ,

where Q has only relations not involving generators of types ai1 and ai 2i .

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Now add a relation of type y =1 for each generator y ∈ A not of type ai1 and not of type ai 2i . The resulting group os a quotient group of π1 ( X , x ) . After applying Tietze transformations to eliminate redundant relations, we change this new presentation to A  ; a12 = a11−1 , a11 = a21 ,  a21 = a31 , a31 =  a41 ,  a41 =  a51 ,… , a12 = a24 , a24 = a38 , a38 = a4 16 ,…,



By means of evident Tietze transformations we can further reduce this presentation to 〈a12  ; ∅〉. Since 〈a12  ; ∅〉 is obviously not the trivial group, and since it is a quotient group of π1 ( X , x ), it follows that π1 ( X , x ) ≠ 1, and we have established what we were after. A description of the fundamental group of the complement of the Alexander’s horned sphere in  3 can be found in, say, [31], page 171. Exercises 1. Explain why the trefoil knot is a (2,3) torus knot. Conclude that 〈 x , y  ;  x 2 = y 3 〉 is a presentation for the knot group of the trefoil knot. Use Tietze transformations to show that this presentation is equivalent to 〈a, b ; aba = bab〉 (see Example 1 in Section 13.3). 2. Show that 〈 x , y  ;  x 4 = y 5 〉 is not isomorphic to 〈a, b  ; a5 = b6 〉. Deduce that the (4,5)-knot (Illustration 13.66) is not ambient isotopic within  3 to the (5,6)- knot (Illustration 13.67).

A (4,5)-strip. Illustration 13.66

A (5,6)-strip. Illustration 13.67

3. Generalize Exercise 2 as follows. Let ( p1 , q1 ) and ( p2 , q2 ) be two distinct pairs of relatively prime positive integers such that p j < q j , j ∈ {1, 2}. Denote G = x 1 ,  y1  ;  x 1p1 = y 1q1 and H = x 2 ,  y 2  ;  x 2p2 = y 2q2 . (a) Show that the centralizer of G is the subgroup of G generated by x1p . (Recall that the centralizer of a group G is {x ∈ G : xy = yx  for every  y ∈ G}.) [Hint: The subgroup N of G generated by x1p is normal in G and G N has trivial center.] (b) Show that G and H are not isomorphic. Deduce that a torus knot of type ( p1 , q1 ) is not ambient isotopic within  3 to a torus knot of type ( p2 , q2 ) .

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4. Take a torus knot K of type (p,q), where p and q are relatively prime. Entangle the torus so it becomes a thick trefoil knot (as in Illustration 13.68). The torus knot K changes and it becomes a trefoil knot of type (p,q). Compute its knot group.

    Illustration 13.68

5. Embed the Möbius band in  3 in the usual way (see Illustration 14.33). Observe that its boundary forms the unknot. Confirm (with naked eye) that the boundary of the triple twisted band (embedded in  3 by rotating an edge of a rectangle by 3π radians, then identifying it with the opposite edge) is the trefoil knot. Draw knot diagrams for the boundaries of (2n + 1)-times twisted bands. Find presentations for their knot groups. 6. In Illustration 13.69, we show a subspace H of  3 ; it is a solid object consisting of all points within the sphere and the knotted part, together with the apparent point of convergence at the top of the figure. Find the fundamental group of  3 \ H.

1 1 1

1

16

8

4

2

    Illustration 13.69

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7.* The Fox-Artin Embedding of S 2 in  3 We describe another embedding f : S 2 →  3 that cannot be extended to a homeomorphism f :  3 →  3 such that the unbounded component  3\ D 3 of  3\ S 2 is sent to the unbounded component of  3\ f (S 2 ). It is based on the so-called Fox-Artin wild arc (see Illustration 13.70).

Illustration 13.70  Fox-Artin wild arc, or a special tubular copy of S2.

The Fox-Artin wild arc, gives rise to an embedding f of S 2 in  3 as follows: view the knot as a tube, the diameter of which tapers off to a point as we move away to the left or to the right from the middle part. We close that tube with the two accumulation points shown to the right and to the left of the knot, thus obtaining a sphere embedded in  3.   It is known that the unbounded component of  3\ f (S 2 ) is not simply connected ([21]). We will indicate how to justify this claim based on the method we have used to compute the knot groups of torus knots.   We exhibit the Fox-Artin embedding of S 2 as a kind of a “torus knot,” where the role of the torus is now played by a special connected sum of tori (as shown in the Illustrations 13.71 and 13.72).

Illustration 13.71  Exhibiting the space from the illustration above as a “torus knot”.

Illustration 13.72  Zooming in to see it more clearly.

Denote the unbounded component of  3\ f (S 2 ) by X. Cover X with two open sets in the same way as with the torus knots. One set, U, is the intersection of X and the chain of open solid tori, the boundary of which contains the central curve of the tube. Illustrations 13.71 and 13.72 apply to U too, except that we should view the tori in the chain as being solid, with the knotted sphere along the Fox-Artin wild arc canalling a ridge on the surface of the object. The other open set, V, is the complement of a slightly smaller concentric chain of closed solid tori. We point out again that U and V

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are subsets of X, and so the two accumulation points at the extreme right and left of the arc are neither in U nor in V. (a) Apply the SvK theorem to the cover {U, V} of X to get a presentation for π1(X). (b)  Show that π1(X) is not the trivial group. (c) Deduce that the embedding f : S 2 →  3 cannot be extended to a homeomorphism f :  3 →  3.

13.5 Links A brief historical note: As objects of mathematical studies, links appeared for the first time in a note (1833) by Carl Friedrich Gauss (1777–1855), and in relation to his work on elliptic integrals.

In this short section we will define links and show a few examples. A link in  3 is a disjoint union of two knots. The link group of a link is the fundamental group of the complement of the link in  3 (or the complement of the link in an open disk that encloses it). Two links L1 and L2 are equivalent if they are ambient isotopic within  3. Two disjoint knots K1 and K 2 are unlinked (or, K1 and K 2 form an unlink) if there is  K 2′ , so that the  K 2 and a link K1′ ∪ an ambient isotopy within  3 between the link K1 ∪ knots K1′ ,  K 2′ can be separated by a plane. Otherwise they are linked. The two unknots in Illustration 13.73 are unlinked.

Illustration 13.73  Two unlinked unknots!

It follows from the SvK theorem that if K1 and K 2 are unlinked, then the associated link group is the free product of the knot groups of K1 and K 2 . In particular, if K1 and K 2 are unknots forming an unlink, as in Illustration 13.73, then the link group is the free product  ∗  = 〈a, b ; ∅〉 . Example 1: The Hopf Link The Hopf link is depicted in Illustration 13.74. An easy application of the SvK theorem—for example, by employing a slight modification of the knot-diagram method described in Section 13.3—establishes that the link group for the Hopf link is

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 ⊕  = a, b; ab = ba (Exercise 1). Since the groups   ∗  and  ⊕  are not isomorphic, we have an algebraic justification of what is visually clear: that the Hopf link is not equivalent to the unlink.

Illustration 13.74  The Hopf link.



☐

Example 2: The Generalized Hopf Link Let 〈 A ∪ {α} ; P 〉 be a presentation of the knot group of a knot K1 , and let 〈 B ∪ {β} ; R 〉 be a presentation of the knot group of a knot K 2 ; the generators α and β—or rather the corresponding loops—are shown in Illustration 13.76 below; each of them is not trivial in the corresponding knot group (Exercise 4 in 13.3). We link the knots K1 and K 2 as in Illustration 13.75. Our goal is to find a presentation of the link group of this link in terms of the given presentations of the two knot groups.

K1

K2

Illustration 13.75  A generalized Hopf link.

In Illustration 13.76, we show the link again, together with four loops: α1, α, β1, and β. It is visible that β1 and β are homotopic in the complement of the shown link: we merely thread the knot K 2 through one of these two loops, as indicated by the dotted loop at the right of Illustration 13.76. Similarly, the loops α1 and α are homotopic in the complement of the link. So, we can write β1 = β, and α1 = α in the link group.

K1

α1 α

β1

K2

β

Illustration 13.76  The dotted loop indicates how to thread K 2 through loops homotopic to β1.

At this stage, in order to compute the link group, we can either use the general procedure for computing knot groups (as described at the end of Section 13.3),

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13.5 Links   ◾    329  

modified slightly for links, or use the SvK theorem. The former is faster, and we see that there are two relations introduced by the two crossings: α = βαβ −1 and β = αβα −1 . Both tell us that αβ = βα. Consequently, the link group has a presentation 〈 A ∪ {α} ∪ B ∪ {β}; P ∪ R , αβ = βα〉. It follows from the assumption that α and β are not equal to the trivial elements in the respective presentation, from Proposition 3 in 12.3, and from the fact that αβα −1β −1 = 1 in this group, that this presentation is not equivalent to the free product of the two knot groups. Consequently, the knots K1 ☐ and K 2 are necessarily linked in the generalized Hopf link. A knot K1 is weakly unlinked 1 from the knot K 2 if every loop in K1 is homotopically trivial in  3 \ K 2 . Clearly, if the knots in a link are unlinked, then each of the knots is weakly unlinked from the other. The following example shows that in linked knots one of the knots could be weakly unlinked from the other. Example 3: The Whitehead Link

Illustration 13.77 K2 K2

K2 K1 α



Illustration 13.78

In Illustration 13.77 we depict the Whitehead link. The unknot K1 is weakly unlinked from the unknot K 2 since any generating loop of π1 ( K1 ) is contractible in  3 \ K 2 (see Illustration 13.78; do not forget that self-intersection of a loop during a homotopy ☐ is allowed), and so every loop in K1 is contractible in  3 \ K 2 .

1

Alternative terminology: if a knot K1 is weakly unlinked from a link K 2, then it is said that the former is link-homotopic to the unknot.

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330   ◾    Seifert–van Kampen Theorem and Applications

Exercises 1. (a)  Find the link group of an unlink. (b) Show that the link group of the Hopf link is isomorphic to  ⊕ . 2. Compute the link group of the Whitehead link. (Perhaps the view of the Whitehead link shown in Illustration 13.79 is better suited for computational purposes.)

    Illustration 13.79

3. Compute the link group of the link shown in Illustration 13.80, then show that it is not isomorphic to a free group of rank 2.

    Illustration 13.80

4. Find the link group of the link shown in Illustration 13.81.

    Illustration 13.81

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13.5 Links   ◾    331  

5. Show that the unknots K1 and K 2 in Illustration 13.82 are linked, show that K1 is weakly unlinked from K 2 and show that K 2 is weakly unlinked from K1 . [Hint: for the first part: compute the link group and show that it is not isomorphic to the link group of the unlink.] K1 K2

    Illustration 13.82

    Illustration 13.83

6. Compute the fundamental group of the complement of the link depicted in Illustration 13.83. 7. Exhibit a link of three unknots K1 , K 2 , and K 3 such that for every i , j ∈{1,2,3}, i ≠ j , the knots K i , K j are not unlinked, and K i is weakly unlinked from the knot K j . 8. Prove or find a counterexample: for every embedding f : D 2 →  3 , if a knot K is linked with f (∂D 2 ) = f (S1 ), then f (D 2 ) ∩ K ≠ ∅ (see Illustration 13.84 below).

f (D2)

K

  

Illustration 13.84  If K is linked with f (S1 ), does it follow that K perforates f (D 2 )?

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Chapter

14

On Classifying Manifolds and Related Topics

A brief historical note: The notion (and the name) of manifold was introduced by Bernhard Riemann (1826–1866).

14.1  1-Manifolds It is relatively easy to classify all connected 1-manifolds, and to provide a justification that, up to homeomorphism, there are only two of them: S1 (Theorem 1) and  (Exercise 3). Theorem 1. Every compact, connected 1-manifold is homeomorphic to S1 . Proof (based on Gale, [23]). Let X be a connected, compact 1-manifold, and let U = {U j  :  j ∈ J } be an open cover of X such that for each i ∈ J there is a homeomorphism ϕ j : U j → (0,1). The set {ϕ j  :  j ∈ J } is called a chart for X; the existence of this set is a consequence of the definition of a 1-manifold. Step 1. Let U j , U k ∈ U be such that U j \ U k, U j ∩ U k and U k \ U j are not empty. Then for every component C of U j ∩ U k , ϕ j (C ) is one of (0, a), (b, 1), for some a, b ∈(0,1), and ϕ k (C ) is one of (0, a′ ), (b′ ,1), for some a′ , b′ ∈(0,1). Proof of Step 1. Since U j ∩ U k is open and locally path connected (easy), every component C is open in U j ∩ U k (Proposition 1 in 6.5), and hence, open in each of U j , U k . So, ϕ j (C ) and ϕ k (C ) are open subintervals of (0,1).

333

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334   ◾  On Classifying Manifolds and Related Topics

Assume ϕ j (C ) = (c , d ) , 0 < c < d < 1. The points ϕ −j 1 (c ) and ϕ −j 1 (d ) are obviously in U j, and none of them is in U k , or else one of [c, d), (c, d] would be a connected subset of U j ∩ U k larger than C. Since ϕ k (C ) is an open connected subinterval of (0,1), it has to be that ϕ k (C ) = (e , f ), for some e , f , 0 ≤ e < f ≤ 1. Since U k \ U j ≠ ∅ (assumption), we have that U j ∩ U k ≠ U k , and so either 0 < e or f < 1. We may assume the former (the latter case is symmetric). Then ϕ −k 1 (e ) ∈U k and, arguing as above, ϕ −k 1 (e ) ∉U j . ∞ 1 −1 The sequence ϕ −k1 (e + ) of elements in C converges to ϕ k (e ) ∈U k \ U j . Observe ∞ n n=1 1 also that, since ϕ j and ϕ k are homeomorphisms, the sequence ϕ j  ϕ −k1 (e + ) conn n=1 sists of pairwise distinct elements in ϕ j (C ) = (c , d ). So, by Corollary 2 in 7.4, this sequence must have an accumulation point x ∈[c , d], and hence, it has a subsequence ∞ 1 −1 {xn } of ϕ j  ϕ −k 1 (e + ) that converges to x. It follows that the sequence {ϕ j ( xn )}, n n =1 ∞ 1 which is a subsequence of ϕ −k1 (e + ) , converges to ϕ −j 1 ( x ) ∈U j . (At this point we n n=1 have used the claim that ϕ −j 1 ([c , d]) is well defined; as we saw above, this is a consequence of our assumption that 0 < c < d < 1 .) However, we already know that every subsequence ∞ 1 of ϕ −k 1 (e + ) converges to ϕ −k 1 (e ) ∈U k \ U j . So ϕ −k 1 (e ) and ϕ −j 1 ( x ) are two distinct n n =1 points of convergence of {ϕ −j 1 ( xn )}, contradicting the Hausdorff-ness of X Hence either

{

{

{

}

}

{

{

}

}

}

c = 0 or d = 1. This proves Step 1. It follows from Step 1 that for every U j , U k ∈U , U j ∩ U k has at most two com­p o­nents. Step 2. If there exist U j , U k ∈ U such that U j ∩ U k has two components, then X is a circle. Proof of Step 2. Let U j , U k ∈ U be such that U j ∩ U k has two components. It follows that X = U j ∪ U k (Exercise 2). Denote the components of U j ∩ U k by C and D. By Step 1, we have that {ϕ j (C ), ϕ j (D )} = {(0, a), (b,1)} and {ϕ k (C ), ϕ k (D )} = {(0, a′ ), (b′ ,1)} for some a, b, a′ , b′ ∈(0,1), a  n , is still not fully understood. For example, π 3 (S 2 ) = , π 4 (S 2 ) =  2 , π 6 (S 2 ) = 12 , π14 (S 2 ) =  84 ⊕  2 ⊕  .2

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370   ◾  On Classifying Manifolds and Related Topics

The fact that π n (S n ) =  , n ≥ 1, fills a gap which we mentioned in Section 11.2: Brouwer’s fixed-point theorem for higher dimensions. Since π n (D n , x ) is trivial for every n ≥ 2, the argument used to prove Brouwer’s theorem for n = 2 needs only minor modifications to justify the general claim as stated in Theorem 2 of 11.2. Exercises 1. Show that π n ( X , x ) with the operation + is indeed a group. Visualize the homotopy that confirms the associative law and the existence of inverses. 2. Prove Proposition 1. 3. Prove that the mapping φ, as defined in the proof of Proposition 3, is well-defined and bijective. 4. Show that any space homeomorphic to the unit ball D n has the fixed-point property. 5. The space X shown in Illustration 14.67 below consists of three closed disks sharing a common diameter. Show that X has the fixed-point property.



Illustration 14.67

6. Prove that for all spaces X and Y, and for every n ≥ 1, the groups π n ( X × Y ,( x , y )) and π n ( X , x ) × π n (Y , y ) are isomorphic.

7. Use the results stated in this section to show that S 4 and S 2 × S 2 are not homotopically equivalent (yet π1 (S 4 ) = π1 (S 2 × S 2 )).

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Chapter

15

Covering Spaces, Part 1

I

n this and the remaining chapters we will develop and apply the theory of covering spaces. In a nutshell (and rather informally), we unwind spaces to reveal some aspects of their structures, and, by extension, the structures of the associated fundamental groups.

15.1  Covering Spaces: Definition, Examples, and Preliminaries A brief historical note: The notion of a (universal) covering space is implicit in Poincaré’s paper published in 1883. He made it explicit in a paper published in 1907.

 , p), where A covering space, or, briefly, a cover (or a covering) of a space X is a pair ( X   is a path connected and locally path connected space, and p : X → X is a continuous X onto mapping satisfying the following condition: for every x ∈ X there is an open, path connected neighborhood U of x such that the restriction of the mapping p on every path component of p−1 (U ) is a homeomorphism onto U (Illustration 15.1).

~ X p–1(U)

p U

X

Illustration 15.1  Covering space and an elementary neighborhood.

371

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372   ◾    Covering Spaces, Part 1

The mapping p in this definition is the covering map. Occasionally, when p is understood,  itself is a covering space, or a cover of X. The space X in this context is the we will say that X base space. The neighborhood U in the definition of covering space is called an elementary neighborhood of x ∈ X .  is path connected, it follows that X is path connected Since p is onto, and since X  forces X to be locally (Proposition 2, in 6.4). Moreover, the local path connectedness of X path connected as well (Proposition 2 below).  , p) be a covering space of X. If U is an elementary neighborhood, then Proposition 1. Let ( X . every path component of p −1 (U ) is open in X Proof. This is a consequence of Proposition 2 in 6.5.  , p) be a covering space of X. Then X is locally path connected. Proposition 2. Let ( X Proof. Let V be an open neighborhood of x ∈ X , and let U be an elementary neighborhood for x. If C is a path component of p −1 (U ), then the restriction p C is a homeomorphism onto U. Denote C ′ = C ∩ p −1 (V ∩ U ), and y = p −1 ( x ) ∩ C (Illustration 15.2).

~ X C´

C

p x V

U

X

Illustration 15.2

 is locally path connected, there Then C ′ is an open (in C) neighborhood of y. Since X  exists an open (in X ) path connected neighborhood W of y such that y ∈W ⊂ C ′. Then (since p C is a homeomorphism) p(W) is an open (in U) path connected neighborhood of x, such that p(W ) ⊂ V ∩ U . Since V ∩ U is open in X, p(W) is open in X. Clearly, p(W ) ⊂ V , and so we have achieved our goal. In view of Proposition 2, only path connected and locally path connected spaces have covering spaces, according to our definition. Hence, unless otherwise stated, we will assume the following: The base spaces are path connected and locally path connected.

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15.1  Covering Spaces: Definition, Examples, and Preliminaries   ◾    373  

Example 1 (X, id) is a covering space of X. This is the trivial covering space. We have seen non-trivial covering spaces earlier, and a notable example was encountered in Part 2 of the proof of Theorem 1, Section 11.1: (,  p), where p :  → S1 is defined by p(s ) = (cos 2πs , sin 2πs ), is a covering space of the circle S1. See Illustration 11.5 in 11.1 for a visualization of this ☐ covering space. Note that the cover  of S1 is simply connected. Such covers will play an important role  , p) of a space in the theory we are building, and we introduce them formally: A cover ( X  X is a universal cover of the space X if the space X is simply connected. In a few sections we will prove that universal covers are unique up to certain special homeomorphisms. In light of that uniqueness we will, at times, refer to a simply connected cover as the universal cover. There are, of course, covers that are neither trivial nor universal. Example 2 ~ X p

X

Illustration 15.3

 and the base space X are both S1 (Illustration 15.3). For every k ∈, The cover space X k ≠ 0, p(cos 2 πs , sin 2πs ) = (cos 2 kπs , sin 2kπs ), s ∈[0, 1], defines a covering map. For exam = S1 twice around the base ple, when k = 2, the mapping p wraps the covering circle X 1 space X = S . This last covering space is different from the trivial cover of S1. In precisely ☐ what sense it is different will be explained later. Example 3 We show a few coverings of the bouquet of two circles. The associated covering maps are defined through the labels of the edges. b

v

a

Illustration 15.4  The base space X is a bouquet of two circles.

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374   ◾    Covering Spaces, Part 1 b

a

a

a

b a

a

b

a

a

b

b

a

a a

a b

b

b a

b

Illustration 15.5  Here is a universal cover of the bouquet X.

b b a

a

a

a

b b

Illustration 15.6  A compact cover of the bouquet X.

b a b

b a

a b

b a

a b

Illustration 15.7  A non-compact, non-simply connected cover of the bouquet X.

Covering spaces of bouquets of circles are much more than cute graphs: as we will see ☐ later they have some interesting applications in group theory. We end this section with the following property of covering maps. Proposition 3. Covering maps are always open.  , p) be a covering space of X, and let V be an open subset of X  . Choose points Proof. Let ( X −1 x ∈ p(V ) and y ∈p ( x ) ∩ V. Let U be an elementary neighborhood of x with respect to  , p), and let C be the path component of p −1 (U ) containing y. Since C ∩ V is open in (X C, and since p C is a homeomorphism onto U, p(C ∩ V ) is open in U. Since U is open in X, p(C ∩ V ) is open in X. Obviously x ∈ p(C ∩ V ) ⊂ p(V ), hence x is an interior point for p(V).

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15.2 Lifts of Paths   ◾    375  

Exercises  , p) be a covering space of X and let U be an elementary neighborhood of x ∈ X . 1. Let ( X Show that if V is a path connected open set such that x ∈V ⊂ U , then V is also an elementary neighborhood of x. 2. Describe a universal cover of X equipped with the indiscrete topology. When does X with the discrete topology possess a universal cover?  , p) be a covering of X. Prove that for every x ∈ X , p−1 ( x ) is a discrete subspace of X.  3. Let ( X 4. Prove that if (Y, p) is a covering space of a Hausdorff space X, and if Y is a BW-space, then for every x ∈ X , p−1 ( x ) is finite. Deduce the same conclusion if Y is assumed to be compact. 5. Find a universal cover of the space depicted in Illustration 14.65, Section 14.5 (a circle with four attached spheres). 6. Describe a universal covering space, and two more covering spaces of the space shown in Illustration 14.66, Section 14.5 (a sphere with two antipodal points identified).  be locally path connected. Show that if X 7. Suppose we do not require that the cover X   is locally path is locally path connected, and ( X , p) is a covering space of X, then X connected. 8. Find a universal cover of  2 \ B2, where B2 = {( x , y ): x 2 + y 2 < 1}.  , p) be a covering of X. 9. Let ( X  (a) Show that if X Hausdorff, then so is X. (b) Show that if for every x ∈ X there exists an open neighborhood U x such that  U x is homeomorphic to m , for a fixed positive integer m, then for every x ∈ X m there exists an open neighborhood Vx such that Vx is homeomorphic to  .  , p) be a covering of X. Prove that if X  is an n-manifold then X is an n-manifold. 10. Let ( X 1 , p1 ) is a covering space of a (path connected and locally path connected) 11. Suppose ( X 2 , p2 ) is a covering space of a (path connected and locally space X1, and suppose ( X 1 × X 2 ,( p1 , p2 )) is a covering space of X1 × X 2. path connected) space X 2. Show that ( X

15.2 Lifts of Paths A brief historical note: Propositions 1 and 2 were proven for 2-manifolds in 1913 by Hermann Weyl. The formal study of covering spaces and their connection with the fundamental groups, based on the path-lifting theorems stated in this section, was inaugurated by Seifert and his teacher William Threlfall (1888–1949) through their influential book A Textbook of Topology, published in 1934.

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376   ◾    Covering Spaces, Part 1

The lifting properties that we will encounter in this section are at the core of the theory of covering spaces.  , p) be a covering space of X. A lift f Let f : Y → X be a continuous mapping and let ( X  of f is a continuous mapping Y → X making the diagram in Illustration 15.8 commutative: ~ X ~ f Y

p

f

X

Illustration 15.8

We start by showing that paths can be lifted.  , p) be a covering space of X, and let α be a path starting at x ∈ X . For Theorem 1. Let ( X −1 every y ∈ p ( x ) there exists a unique lift of α starting at y. Proof. 1 and α 2 be two distinct lifts of α starting at y. Denote Step 1 (Uniqueness). Let α 1 (0) = α 2 (0) we have that z ≥ 0, and the 1 (s ) = α 2 (s ) for every s ≤ t }. Since α z = sup{t ∈ I  : α   assumption that α1 and α 2 are distinct implies that z 0, the path β 1λ is not a loop. By Corollary 8 in 15.3, β = β1β 2 is not in p∗ (π1 ( ,0)). Since p∗ (π1 ( ,0)) is the trivial group, this means that β is not contractible. We have a contradiction. If z ≠ (1, 0) then compose each of the paths β, β1 , and β 2 with a rotation of S1 that sends z to (0,1), and repeat the above argument for the resulting paths. The next three results are easy consequences of Theorem 1. We call them variants of Theorem 1, because the four statements are equivalent (Exercise 3). Theorem 2 (Borsuk–Ulam variant 2). Let f : S n → n be such that f ( x ) = − f (− x ) for every x ∈S n . Then there exists y ∈S n such that f ( y ) = (0, 0, … , 0). Proof. Restrict f on the closed upper hemisphere of S n and apply Theorem 1. Theorem 3 (Borsuk–Ulam variant 3). There is no continuous antipode-preserving map S n → S n−1 , n ≥ 1. Proof. Suppose there is such an antipode-preserving mapping f : S n → S n−1 . This means that for every x ∈S n , f ( x ) = − f (− x ) . Now, if in : S n−1 → n is the inclusion, then in  f : S n → n is antipode preserving and for every x ∈S n , in  f ( x ) = f ( x ) = 1 , which violates the conclusion of Theorem 2. Theorem 4. (Borsuk–Ulam variant 4) If f : S n → n is continuous, then there exists y ∈S n such that f ( y ) = f (− y ).

–y

y

–y

y

–y = y

Illustration 16.26  The Borsuk–Ulam Theorem.

Proof. Suppose there is no y ∈S n such that f ( y ) = f (− y ). Then the mapping h : S n → S n−1, defined by h( y ) = ff (( yy ))−− ff (( −− yy )) , is well-defined, continuous, and antipode preserving, contradicting Theorem 3. A consequence of Theorem 4 is that S n is not embeddable into n . Even though we have deduced Theorems 2–4 as stated, it is fair to emphasize that, since we have proven Theorem 1 only in the special case when n = 2, the same is true for the other three versions. Proving the general result is not very difficult (see, for example, [49]). However, since the theory of covering spaces cannot be generalized in a straightforward manner for higher homotopy groups (see Section 14.5), the argument in the proof of Theorem 1 presented here for n = 2 does not allow a simple modification for n > 2.

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424   ◾    Covering Spaces, Part 2

The Brouwer fixed point theorem, proven in Theorem 2 in 11.2, is an easy consequence of Theorem 1. Corollary 5 (Brouwer fixed point theorem). Every continuous mapping D n → D n has a fixed point. Proof (n = 2). Suppose otherwise. As in the first proof of this theorem (Theorem 2 in 11.2), this hypothesis implies the existence of a mapping f : D 2 → S1 that is the identity over ∂D 2 . If in : S1 →  2 is the inclusion, then g = in  f : D 2 →  2 clearly satisfies g ( x ) = − g (− x ) for every x ∈S1 . It is also evident that there is no y ∈D 2 such that g ( y ) = 0, contradicting Theorem 1. The general case (any n) is deducible mutatis mutandis from the full strength of Theorem 1. The following corollary of the Borsuk–Ulam theorem is stated in general terms. We will prove it only for n = 2, because we have proven only the case n = 2 of the Borsuk–Ulam theorem, and since this is the highest dimension that allows a visualization of the argument. However, deducing Corollary 6 in its full strength from Theorem 4 requires only a modification of the argument we provide (see, for example, [49]). Corollary 6. Given n objects in n with well-defined, positive hypervolumes, there exists a hyperplane in n that cuts all of them into halves (hypervolume-wise). Proof (n = 2). We deal with two planar objects with well-defined positive areas. We do not assume that the regions are bounded. Our goal is to show that there is a line halving these two objects. First, we associate a line la ,b ,c in the xy-plane to every point (a, b, c ) ∈S 2 \{(0,0,1),(0,0, −1)}: la ,b ,c is defined by ax + by = c. Since the points in (a, b, c ) ∈S 2 \{(0,0,1),(0,0, −1)} satisfy a 2 + b 2 ≠ 0 , the vector v = (a, b) is not the zero-vector. The line la ,b ,c is perpendicular to that vector, hence to the line x = at , y = bt . It intersects this line a2c+b2 units away from the origin (in the direction of v if c > 0, in the opposite direction to v if c < 0; see Illustration 16.27). Observe that every line in the plane is la ,b ,c for some (a, b, c ) ∈S 2 \{(0,0,1),(0,0, −1)} . y7

y6

y8

y5

y9 l1 = l7

y4

l2 = l8 l = l 3 9 x

y3 y1

l4

l5

y

l6

y2

Illustration 16.27  We show 9 points on a meridian circle. The line associated to the point yi is denoted in the figure by li . The closer a point yi is to a pole, the further away from the origin is the associated line. In the illustration we have chosen y1 and y7 , y 2 and y8, y3 and y9 to be antipodal in pairs. The lines associated to antipodal points coincide.

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16.5 The Borsuk–Ulam Theorem   ◾    425  

For every line of type la ,b ,c denote the half-plane ax + by ≥ c by l a+,b ,c . Denote the two planar objects in  2 (as in the statement of the corollary) by R1 and R2 , and define a mapping f : S 2 \ {(0,0,1),(0,0, −1)} →  2 as follows: f (a, b, c ) = Area( R1 ∩ l a+,b ,c ), Area( R2 ∩ l a+,b ,c ) ; see Illustration 16.28. It is obvious that f is continuous: small changes in (a,b,c) result in small changes in la ,b ,c and l a+,b ,c , which in turn result in small changes in Area( R1 ∩ l a+,b ,c ), Area( R2 ∩ l a+,b ,c ) . c Notice that as (a,b,c) approaches (0,0,1), the distance a2 +b2 between la ,b ,c and the origin tends to ∞. This means that the distance between the half plane l a+,b ,c and the origin tends to infinity too, and so Area( R1 ∩ l a+,b ,c ), Area( R2 ∩ l a+,b ,c ) tends to (0,0). It follows that defining f (0, 0,1) = (0, 0) continuously extends f over S 2 \{(0,0, −1)} (we have c used Proposition 3 in 2.2). Similarly, as (a,b,c) approach (0, 0, − 1), a2 +b2 tends to −∞, and the distance from the origin to the half-plane  2 \ l a+,b ,c tends to ∞. Consequently, Area( R1 ∩ l a+,b ,c ), Area( R2 ∩ l a+,b ,c ) tends to ( Area( R1 ), Area( R2 )), and extending f by f (0,0, −1) = ( Area( R1 ), Area( R2 )) gives a continuous mapping f : S 2 →  2.

(

)

(

)

(

(

)

)

y6

y7

y5

y8 y9 + l7 l + 1

+ l8 l2+

+ l4 l5+

y4 y3

+ l9 l3+

y2 y1

+ l6

Illustration 16.28  The half plane associated to the point yi is denoted in the figure by li+ . The

half planes associated to a pair of antipodal points are complementary, except for the common boundary line.

By the Borsuk–Ulam theorem (Theorem 4), there exists y = (a0 , b0 , c0 ) ∈S 2 such that f ( y ) = f (− y ). Since ( Area( R1 ), Area( R2 )) ≠ (0,0) , this point y is not in {(0, 0,1), (0, 0, − 1)}. Hence, Area R1 ∩ l a+0 ,b0 ,c0 , Area R2 ∩ la+0 ,b0 ,c0 = Area R1 ∩ l−+a0 ,− b0 ,− c0 , Area R2 ∩ l−+a0 ,− b0 ,− c0

( (

)

(

)) (

(

)

(

))

for some (a0 , b0 , c0 ) ∈S 2 \ {(0,0,1),(0,0, −1)}. Finally, notice that l a+,b ,c and l +− a ,− b ,− c are the two half planes determined by the line la ,b ,c , and so the line la0 ,b0 ,c0 halves both Area( R1 ) and Area( R2 ).

More formal statement and justification of this corollary (in terms of Borel measure in place of “hypervolume”) can be found in the book by Matoušek [49], where there is a wealth of other interesting results related to the Borsuk–Ulam theorem. Corollary 6 is sometimes called the Ham Sandwich Theorem, because of the following pretty instance when n = 3 : a sandwich made of bread, cheese, and ham can be cut once so that the bread, the cheese, and the ham are all halved volume-wise (Illustration 16.29).

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426   ◾    Covering Spaces, Part 2

Illustration 16.29  Cutting

Illustration 16.30  The objects that we halve

a sandwich fairly.

could be distant.

Corollary 7. If two of A, B, and C are closed subsets of S 2 and if S 2 = A ∪ B ∪ C , then at least one of these three sets contains a pair of antipodal points. Proof. We may assume that A and B are closed in S 2 . Define a mapping f : S 2 →  2 by f ( x ) = (d( x , A) − d(− x , A), d( x , B) − d(− x , B)), where d(x, D) is the distance from a point x ∈S 2 to the set D ⊂ S 2 . Since f is a continuous mapping, Theorem 4 implies that there exists an element y ∈S 2 such that f ( y ) = f (− y ). This means that (d( y , A) − d(− y , A), d( y , B) − d(− y , B)) = (d(− y , A) − d( y , A), d(− y , B) − d( y , B)), which easily yields d( y , A) = d(− y , A) and d( y , B) = d(− y , B). Hence, (using the assumption that A is closed) y ∈ A if and only if − y ∈ A; by symmetry we have that y ∈B if and only if − y ∈B. It follows that y ∈S 2 \ ( A ∪ B) if and only if − y ∈S 2 \( A ∪ B), and so both y and − y are in at least one of A, B, and C. Exercises 1. Is it true that for every continuous f : S1 × S1 →  2 there exists ( x , y ) ∈S1 × S1 such that f ( x , y ) = f (− x , − y )? 2. Show that for every continuous f : D 3 →  2 there are uncountably many x ∈D 3 such that f ( x ) = f (− x ). 3. Deduce Theorem 1 from every Theorem n, n = 2, 3, 4. 4. Give a direct proof of Theorem 3 for n = 2. (Hint: Suppose f : S 2 → S1 is an antipode preserving continuous mapping, and let ( , p) be the universal cover of S1 . Let f be a lift of f with respect to ( , p) . Find a loop α: I → S 2 → S1 such that f α is not a loop in , thus obtaining a contradiction.) 5. Prove that there does not exist a continuous one-to-one mapping  2 →  .

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16.5 The Borsuk–Ulam Theorem   ◾    427  

6. Show that for every two objects in  3 with well-defined volumes there exists a vertical plane simultaneously halving the volumes of these two objects. 7. Denote the area of a planar object S by A(S), and consider the unit disk D 2 . Given any r ∈ + , find a planar object B of positive area, such that if a line l divides D 2 into C1 and C2 satisfying A(C1 ) = rA(C2 ), then B does not intersect l. (This exercise shows that the ratio 1:1 in the statement of Corollary 6, n = 2, cannot be replaced by another number.) 8. Let X be a subspace of  3 having precisely one point on each ray ra ,b ,c defined by x = at , y = bt , z = ct , t ∈(0, ∞), a 2 + b 2 + c 2 = 1. Suppose in addition that f : (a, b, c )  X ∩ ra ,b ,c defines a homeomorphism S 2 → X . Show that there are two antipodal points in X equidistant from the origin. 9. Show that if A and B are two closed subsets of S 2 such that A ∪ B = S 2 then at least one of them contains a pair of antipodal points. Show that the analogous statement for 4 closed subsets of S 2 fails. Show that Corollary 7 fails if we do not assume that two of the sets A, B, and C are closed.

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Chapter

17

Applications in Group Theory

S

o far we have used group theory as a tool to investigate various properties of topological spaces. In this chapter we will reverse the roles, and use homotopy theory to discover and deduce pure group-theoretical results.

17.1  Cayley Graphs and Covering Spaces A brief historical note: In 1878 Arthur Cayley (1821–1895) introduced (what were later called) “Cayley graphs” through an example. In his paper he sketched the (Cayley) graph of the group A4, generated by a pair of elements different from the ones we use in Example 2.

Let G be a group and let A be a set of generators of G. We construct a graph as follows: The vertices are the elements of G. For every vertex g ∈G , and for every a ∈ A, we introduce an oriented edge starting at g and ending at ga. We label that edge by a, and call it an a-edge. The graph constructed in such a way is called the Cayley graph of the group G on the generating set A. It will be denoted by Γ(G ,A ), or by Γ G when A is clear, or by Γ when both G and A are understood from the context. We now show a few pictures of Cayley graphs. Example 1: The Cayley Graph of a Cyclic Group Consider the cyclic group G = 〈 x  ;  x 4 〉 of order 4, generated by x. We depict Γ (G ,{ x }) in Illustration 17.1.

429

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430   ◾    Applications in Group Theory 1

x

x

x

x3

x

x

x2

Illustration 17.1



☐

In order to avoid clutter, we will sometimes omit labeling the vertices of Cayley graphs. Recall that the symmetric group Sn on the set {1,2,…,n} is the group of all permutations (bijections) of the elements of this set, where the operation is composition. The 1 2 3 … n permutation sending k to ik, k = 1,2,…, n, is often denoted by  .  i1 i2 i3 … in 

A k-cycle in Sn is a permutation of the type j1  j2  j3 …  jk  j1 , for some distinct j1 ,  j2 ,  j3 … ,  jk ∈{1,2, … , n}, fixing the other elements of {1,2, … , n} . Such a k-cycle is usually denoted by ( j1   j2   j3 …  jk ). For example, the 3-cycle (1 2 3) in S4 is the permutation 1  2, 2  3, 3  1, 4  4. The group Sn is generated by the set {(i   j ) : i , j ∈{1,2, … , n}, i ≠ j} of all 2-cycles, also called transpositions. A permutation in Sn is odd if it is a composition of an odd number of 2-cycles; otherwise, it is even. Even permutations in Sn make a subgroup of Sn called the alternating group, commonly denoted by An. Example 2: The Cayley Graph of an Alternating Group In this example we will depict a Cayley graph of the alternating group A 4 . The group A 4 is generated by x = (1 2 3) and the product (composition) y = (1 2)(3 4) of two transposition. We point out again that in case of compositions of permutations the left mapping acts first. This is not of consequence for the product defining y (since the transpositions commute), but it matters in general. The Cayley graph Γ of A 4 on the generating set {x , y} is shown in Illustration 17.2. As we see, the vertices and the edges of Γ fit nicely along the vertices and the edges of a truncated tetrahedron. That this is not merely an aesthetic observation will be clearer later (Exercise 3 in 17.2).

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17.1  Cayley Graphs and Covering Spaces   ◾    431   x=

1 2 3 4 2 3 1 4

1 2 3 4 3 1 2 4

1 2 3 4 1 2 3 4

1 2 3 4 1 4 2 3

1 2 3 4 4 2 1 3 1 2 3 4 3 4 1 2

1 2 3 4 4 3 2 1

y= 1 2 3 4 2 4 3 1

1 2 3 4 1 3 4 2

1 2 3 4 2 1 4 3

1 2 3 4 4 1 3 2

1 2 3 4 3 2 4 1

Illustration 17.2  The edges corresponding to the generator x are depicted in full lines, those

corresponding to the generator y are shown in dashed lines. Recall that every oriented edge labeled z, where z ∈ {x, y}, joins a vertex g ∈ A4 with the vertex gz, where in the composition g acts first. Thus, for example, the oriented edge at the bottom of the graph joins the vertex  1 2 3 4   1 2 3 4   1 2 3 4  ☐  3 2 4 1  with the vertex  1 3 4 2  =  3 2 4 1  x .

Cayley graphs of infinite groups have infinitely many vertices and edges. Here are a couple of them. Example 3:  ⊕  and  ∗ The Cayley graph Γ ⊕ of  ⊕  on the generating set {(1,0),(0,1)} is partially shown in Illustration 17.3, and in Illustration 17.4 we show a part of the Cayley graph Γ ∗ of the free group  ∗  on the set of generators consisting of two copies of 1 (one for each ).

Illustration 17.3  The edges corresponding to (1,0) ∈ ⊕  are in dashed lines, and the edges

corresponding to (0,1) ∈ ⊕  are in full lines.

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432   ◾    Applications in Group Theory

Illustration 17.4

We recognize Γ ∗ as the universal cover of the bouquet of two circles. This is not a ☐ coincidence, as we explain in the next few paragraphs. Let Γ be the Cayley graph of a group G on a generating set A. Each edge of Γ is oriented, and it is labeled by an element in A. Let R be the bouquet of A-many circles, labeled bijectively by the elements in A, and with the only vertex denoted by v. Then the labels of the edges define a continuous mapping l : Γ → R (sending the interiors of the edges in Γ homeomorphically onto the interiors of the loops in R, and sending the vertices of Γ to v). It is evident that l is a covering map, hence (Γ , l ) is a covering space of R. We will call l the labeling map. Each g ∈G acts as a homeomorphism of Γ as follows: if e is an oriented edge of Γ labeled by a ∈ A and staring at g 1 ∈G (and ending at g 1a ∈G ), then g(e) is the edge also labeled by a, and starting at gg 1 and ending at gg 1a; Illustration 17.5. g1a

Γ a g1



g

Γ

gg1a a gg1

Illustration 17.5

It is clear that the group G, considered as a group of homeomorphisms of Γ with the action just described, acts properly discontinuously on Γ. Moreover, since each g ∈G (regarded as a homeomorphism of Γ) maps every a-edge onto an a-edge, it follows that l  g = l , and so each g becomes a covering transformation of (Γ , l ). Finally, observe that for every g 1 , g 2 ∈G (considered as vertices in Γ) there is an element g ∈G (considered as a covering transformation) such that gg 1 = g 2 : simply take g = g 2 g 1−1. This implies that every covering transformation of (Γ , l ) agrees at a vertex with one covering

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transformation in G. It follows from Proposition 2(a) in 16.2 that G is isomorphic to the group A(Γ , l ) of covering transformations of (Γ , l ) . (As before, the “A” in A(Γ , l ) stands for the first letter of the word automorphism, and it should not be confused with the generating set A.) By Proposition 1 in 16.3, ( Γ ,l ) is a regular cover of R. Proposition 3(b) in 16.3 then tells us that the covers (Γ , l ) and (Γ , q), where q is the quotient mapping Γ → Γ G , are the same covering spaces (up to a homomorphism that makes the diagram in Proposition 3, 16.3 com­mutative). By Corollary 5 in 16.2, the group G is isomorphic to the quotient group π1 ( R ) l∗ ( π1 ( Γ )). We have proven the following. Proposition 1. Let Γ be the Cayley graph of a group G on a generating set B, and let R be the bouquet of B circles labeled by the elements of B. Then (Γ , l ) , where l is the labeling map, is a regular covering of R, and G ≅ π1 ( R ) l∗ ( π1 ( Γ )) ≅ A( Γ , l ) . In other words, the Cayley graph of a group G generated by B is a covering of the bouquet of B circles corresponding to a normal subgroup N of the group F freely generated by B, such that G ≅ F N . The converse is also true: Proposition 2. If Γ is a graph and if (Γ , p) is a regular cover of the bouquet R of circles, then it is a Cayley graph of the group π1 ( R ) p∗ ( π1 ( Γ )). Proof. (Exercise 3) The assumption in Proposition 2 that Γ is a graph will be made redundant in Section 17.3 where we will show that coverings of graphs must be graphs. The following is an immediate consequence of Proposition 1 here, and Exercise 12 in Section 12.1. Corollary 3. Let Γ, G, B, R, and (Γ , l ) be as in Proposition 1. Then 〈 B  ; l∗ (π1 (Γ ))〉 is a presentation of the group G. The link from groups to Cayley graphs and covering spaces allows conversion of grouptheoretical statements and proofs into corresponding claims within the setup of the theory of covering spaces. This is one route of visualizing group theory, and in some cases it gives a better view on the intricacies and interconnectedness of various theorems. As an illustration of the wide scope of the theory of covering spaces, we prove the following classical theorem. The proof we give is somewhat longer than the standard group-theoretical proof, partly because we did not aim the theory at such statements, partly because we give the details. Theorem 4 (Cauchy). If G is a group of order n and if p is a prime dividing n, then there is an element in G of order p.

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434   ◾    Applications in Group Theory

Proof. Let G be a group of order pm, where p is a prime and m ∈+. Let A be a minimal (under inclusion) generating set for the group G. The case when G is a cyclic group is easy, and so we may assume that A > 2 . Consider the Cayley graph Γ of G on the generating set A. By Proposition 1, (Γ , l ), where l is the labeling map, is a covering space of the bouquet R with A circles (labeled by the elements of A). Since A is minimal generating set, deleting from Γ all edges labeled by a fixed element a ∈ A, disconnects Γ. If the order of a is a multiple of p, then we are finished. Suppose otherwise, and denote the graph we get by deleting all a-edges from Γ by Δ. The restriction of l on every component Δc of Δ is a covering mapping of the rose R without the a-labelled circle. Since Γ is regular, so is Δc, and each two components of Δ are homeomorphic (via a covering transformation). There are two cases. Case 1. The number of vertices of any (hence every) component Δc of Δ is divisible by p. Since each Δc is a Cayley graph of a subgroup L of G, and since | L | is the number of vertices of Δc, it follows that p divides L. Since K < G we may use induction on the number of elements of the underlying group to conclude that there is an element b in L ⊂ G such that b p = 1. Case 2. The prime p does not divide the number of vertices of the components of Δ; in this case, since the number of vertices in Δ is the same as the number of elements in G, the number p divides the number s of components in Δ. The subgroup H of G generated by A \ {a} acts properly discontinuously on Γ as a group of covering transformations (Proposition 1). Then Γ, together with the quotient map q : Γ → Γ H , is a regular covering space of Γ H . The action of H on Γ is such that the orbit of an edge consists only of edges with the same label (see Illustration 17.5). Consequently, labeling an edge e of Γ H with the label of q −1 (e ) is well defined. We get a graph Γ H with edges labeled by A, where each edge labeled by A \ {a} is a loop. The labeling map from Γ H onto the bouquet R is a covering map. By Exercise 4 (or, if shortcuts tempt us, by symmetry), Γ H with the labeling map is a regular cover of R. By Proposition 2, Γ H is the Cayley graph of a certain group, which we denote by K. The cardinality of K is s (the number of vertices in Γ H ). Since p divides s, and since s < G we can use our (implicit) induction to conclude that there is an element in K of order p. Since the elements in A \ {a} are all trivial in K, this element must be of type al for some l. So alp = 1 in K, which means that every path in Γ H labeled by the word alp is a loop α in Γ H . Lift α to a path in Γ starting at a vertex x1. If the end vertex of this path is x 2 , then x1 and x 2 are in the same component of Δ (so that q sends this path to a loop in Γ H; see Illustration 17.6). ∆1 x2

x1

∆lp

∆2

∆3

∆4

Illustration 17.6  The vertices x1 and x 2 must be in the same component ∆1 of Δ. We do not

assume the components ∆ i are different.

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17.1  Cayley Graphs and Covering Spaces   ◾    435   ∆1

x1 = xm ∆lp

∆2

∆3

∆4

Illustration 17.7  There are no earlier overlapping of vertices along the path, else we would

contradict the assumption that the order of a in K is lp.

Now lift α starting at x 2; if the end point is x3, then all x1 , x 2 , and x 3 are in the same component of Δ. Iterate. Since we are dealing with finite graph, at some point we must have xi = xi+m−1, hence x1 = xm, for some smallest m ∈+ (Illustration 17.7). It follows that (alp )m = 1 in G and that at ≠ 1 for t < l pm. So, alm has order p. Exercises 1. Sketch the Cayley graphs of the following groups. (a) 〈a, b, c ; ∅〉 (b) 〈a, b, c ; a 2 , b 2 , c 2 〉 (c) 〈a, b, c ; a 2 , b 2 , c 2 , aba1b−1 , aca−1c −1 , bcb−1c −1 〉 2. Let Γ be the Cayley graph of a group G on a generating set A, and let Γ′ be a disjoint copy of Γ. (a) Identify the corresponding pairs of vertices in Γ and Γ′. Do you get a Cayley graph of a group? If yes, describe the group in terms of G. (b) Attach two edges joining each v ∈Γ to the corresponding v ′ ∈Γ ′. Do you get a Cayley graph of a group? If yes, describe that group in terms of G and some other known groups. 3. Prove Proposition 2. 4. Let X be a bouquet of circles (labeled by the elements of a set A), and let (Y, p) be a regular covering space of X. Let H be a subgroup of the group of covering transformations of Y with respect to p, and let q : Y → Y H be the quotient map (so that (Y, q) is a regular cover of Y H ). Define r : Y H → X by r ([ y]) = p ( y ), where, as always, for y ∈Y ,[ y] is the equivalence classes of y in Y H . Show that r is well defined and that (Y H , r ) is a regular covering of X. (You may assume that covers of graphs are graphs.)

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436   ◾    Applications in Group Theory

5. Let Γ be the Cayley graph of a group G on a generating set A, so that (Γ, l), where l is the labeling map, is a regular covering of the bouquet X with A circles. An element of A(Γ , l ) is an a-sliding transformation if it moves a vertex v to the vertex va. (Note that v  va fully specifies the images of the other points in Γ; see Theorem 2 in 15.3.) Show that for every a ∈ A, the subgroup H of A(Γ , l ), generated by all a-sliding transformations is a normal subgroup of A(Γ , p). 6. Find Γ H if Γ is the Cayley graph of the alternating group A4 as shown in Illustration 17.2, and H is the subgroup of A (Γ , l ) generated by all y-sliding transformations (y as in Example 2). Show that H is a proper subgroup of A (Γ , l ) and use Exercise 5 to deduce that A4 is not a simple group. 7. Let X be a space and let (Y, p) be its universal cover. Let R be a subspace of X such that π1 ( R , v ) generates π1 ( X , v ) and such that R is homeomorphic to a bouquet of circles. Show that p −1 ( R ) is homeomorphic (as a topological space) to the Cayley graph of π1 ( X , v ) on the generating set π1 ( R , v ). 8. Let X be a bouquet of two circles. (a) Find a regular cover (Y, p) of X and a regular cover (Z, q) of Y such that ( Z , p  q) is not a regular cover of X. (b) Find groups G, H and K such that G  H , H  K , and G ⋪ K.

17.2  Topographs and Presentations A brief historical note: The origin of the idea of “reducing peaks” in Cayley graphs can be found in two papers (1936–1937) by John Henry Constantine Whitehead, where he established the existence of an algorithm deciding if a cyclic word in a free group can be carried to another via an automorphism of the free group.

In this section we describe a method for obtaining efficient group presentations. Let Γ be a graph. We will consistently denote its set of vertices by V (Γ). A topograph is a graph Γ together with a mapping t : V (Γ ) →  , which we call topography of Γ. For every v ∈V (Γ ), the image t(v) will be called the altitude of v. A peak in a topograph Γ is a subgraph of Γ with two edges e1 and e2, configured as in one of Illustrations 17.8, 17.9, and 17.10. v

e2

v

w

e1 u

e1

     u

u e2

e1

v e2

w

   

w

Illustration 17.8  t (u) < t (v ) = t (w ).  Illustration 17.9  t (u) < t (v ) > t (w ).  Illustration 17.10  t (u) = t (v ) > t (w ).

We do not assume that the vertices u, v, and w in these illustrations are all distinct. If the edges e1−1 (considered as the inverse of the path from u to v along e1) and e2

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17.2  Topographs and Presentations   ◾    437  

coincide  in  Illustration 17.9 we will say that the peak is trivial. Trivial peaks will not interest us. A peak from u to w in a topograph Γ (with the notation as in Illustrations 17.8–17.10) is reducible if there is a path in Γ from u to w such that t (z ) < t (v ) for every vertex z in that path other than u and w (Illustrations 17.11–17.13). Such a path is then called a reduction of the corresponding peak. v e1 v

e2

e1 u

e´1

w

u

e´k e´2

Illustration 17.11

e´1

e2

e´2

w e´k

Illustration 17.12  In case u = w,

k is 0, and the reducing path is     the constant at u.  

e1

u

v e2

e´1 e´2

e´k

Illustration 17.13

w

    

If every peak in Γ is reducible then we say that Γ is a reducible topograph. If Γ is the Cayley graph of a group G on a generating set A, and if it is equipped with a topography, then we say it is a (Cayley) topograph of G on A. Since in this case V (Γ ) = G the topography is a mapping G → . We will continue to denote the labeling mapping with respect to which Γ is a covering space of the associated bouquet R of labeled circles by l. For every path e1e2 … en in Γ the image l(e1 )l(e2 )  l(en ) is an element in the free group π1 ( R ). In particular, if e1e2 is a peak and if e1′e2′ … ek′ is its reduction, then l(e1′ )l(e2′ ) … l(ek′ )l(e2 )−1 l(e1 )−1 is called a reducing element in π1 ( R ). Recall that a subgraph Λ of a graph Γ is spanned by a set S of vertices of Γ if it consists of S and of all edges in Γ joining the vertices in S. Proposition 1. Let Γ be a reducible Cayley topograph of a group G on a generating set A, let the topography mapping t attain its minimum, and let Λ be the subgraph of Γ spanned by the set of all vertices of minimal altitude. Assume that Λ is connected and let R be the set of all reducing elements of the Cayley topograph Γ. Then 〈 A  ; l∗ (π1 ( Λ )), R 〉 is a presentation of G. The subgraph Λ in Proposition 1 will be called the floor of Γ. The group π1(Λ) in the statement of Proposition 1 may be replaced by any generating set of that group without altering the meaning of the statement. Before we prove Proposition 1 we illustrate its usage with the following basic example. Example 1: A Dihedral Group The group D8 of all symmetries of the square {( x , y ) ∈ 2  :  x + y ≤ 1} is generated by the rotation ρ through π2 (counterclockwise) and centered at (0,0), and by the reflection r with respect to x = 0. The Cayley graph of D8 on these two generators is shown in Illustration 17.14. We make it a topograph in the indicated natural way, by simply

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438   ◾    Applications in Group Theory

interpreting Illustration 17.14 as roughly depicting the edges of a cube with one face at altitude 0, and the other at altitude 1. 1 ρ

1

ρ

ρ

1 ρ

r

1

r

r

r

ρ

r

r 0

r

0

ρ

ρ

0

r ρ

0

Illustration 17.14

The relevant reducing elements come from the peaks shown in Illustrations 17.15 and 17.16 and from their reducing paths in Illustrations 17.16 and 17.17. v

1

e2

ρ

1

r e1 0

Illustration 17.15 1

v

v r e2

r e1

e2

1

ρ

r

1 r

e1 0

Illustration 17.16       

0

ρ

0

Illustration 17.17

Hence, R = {ρr ρ−1r −1 , r 2 }. It is evident from Illustration 17.14 that Λ is the bottom rectangular graph. Hence, π1 ( Λ ) is a cyclic group, and l∗ (π1 ( Λ )) is generated by ρ4. It ☐ follows from Proposition 1 that 〈r , ρ ; ρr ρ−1r −1 , r 2 , ρ4 〉 is a presentation of D8. Proof of Proposition 1. By Proposition 1 in 17.1 and by Exercise 12 in 12.1, 〈 A  ; l∗ (π1 (Γ , x ))〉 is a presentation of G, where, for convenience, we choose x ∈V ( Λ ). The proof will be complete if we establish that every W ∈l∗ (π1 (Γ , x )) is a consequence of l∗ (π1 ( Λ )) ∪ R . The lift δ of W starting at x is a loop. If e1e2 is a peak in δ, then we replace it with its reduction e1′e2′ … ek′ . This corresponds to changing W from Ul(e1 )l(e2 )V to W1 = Ul(e1′ )l(e2′ ) … l(ek′ )V . Since this change can be realized by inserting l(e1′ )l(e2′ ) … l(ek′ )l(e2 )−1 l(e1 )−1 immediately following the subword U in W, we have that W = W1 is a consequence of the relation l(e1′ )l(e2′ ) … l(ek′ )l(e2 )−1 l(e1 )−1 = 1, and

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17.2  Topographs and Presentations   ◾    439  

so a consequence of R . Since we have chosen x to be in V(Λ), iterating this procedure changes δ to a path δ′ in Λ, and changes W to W ′ = l(δ ′ ). By induction, W = W ′ is a consequence of R . Since δ′ is in Λ, l(δ ′ ) is in l∗ (π1 ( Λ )), hence, W ′ = 1 is a consequence of l∗ (π1 ( Λ )). We proved that W = 1 is a consequence of l∗ (π1 ( Λ )) ∪ R . Example 2: The Symmetric Group We noticed earlier that the symmetric group (also known as the group of permutations) Sn on the set {1,2, … , n} is generated by the set T = {(i   j ) : i , j ∈{1,2,… , n}, i ≠ j} of all transpositions. An inversion in an element α ∈Sn is an occurrence of a pair (α(i1 ), α(i2 )) such that i1 < i2 and α(i1 ) > α(i2 ) . We define a topography t on the Cayley graph Γ of Sn on the generating set T as follows: for every α ∈Sn , t (α ) is the number of inversions in α. For example, if α is the 3-cycle (2 1 3) ∈S3 , then t (α ) = 1. In Illustration 17.18 we show the topograph Γ when n = 3. 3

(1 2

)

)

(2 3

2

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Illustration 17.18  The topograph Γ of S3: only the bottom vertex is labeled (by the identity permutation); the numbers by the vertices are their altitudes. We indicate the transpositions corresponding to the edges. Notice that since each transposition is of order 2, there are pairs of edges connecting every pair of vertices.

Let a, b, c, and d be distinct elements of {1,2, … , n}. There are three types of peaks in Γ: type 1 peaks correspond to a word of type (a  b)(a  b), type 2 peaks correspond to (a  b) (c  d), and type 3 peaks correspond to (a  b)(b  c). The first type of peak allows the trivial (constant) reduction, and the associated reducing element is (a b)2 (accounting again for the fact that the transpositions are of order 2). Since (a b) (c d) = (c d) (a b), the type 2 peaks allow a reduction with the associated reducing element (a  b) (c  d) (a  b) (c  d). The third kind of peak requires more care, but the work is rather straightforward, and we leave the details to the reader (Exercise 1). The reducing elements stemming from this case are of type {α  (a  b) α  (α(a) α(b)) : α , (a, b) ∈T }. This set contains the reducing elements we got

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440   ◾    Applications in Group Theory

in the second case. Since the graph spanned by the set of vertices of minimal altitude contains only one vertex, it follows from Proposition 1 that the following is a pre­ ☐ sentation of Sn : 〈T  ; {α 2 = 1 : α ∈ T } ∪ {α  (a  b) α  (α(a) α(b)) : α, (a, b) ∈ T }〉 . Example 3: A Free Group The general linear group of all invertible n × n matrices with integer entries is commonly denoted by GL(n, ). Consider the subgroup G of GL(2, ) generated by the   1 2   1 0   set   ,   , and its Cayley graph Γ on that set. We make Γ a   0 1   2 1   2  a a12  11 topograph by defining t   aij2 . A case-by-case careful (but  =   a21 a22  i , j =1   tedious) inspection shows that there are no nontrivial peaks in Γ. For example,

∑

  if the edges e1 and e2 in a path e1e2 are both labeled by  1 2 , if e1 starts at  0 1   a b  u= G t ( v ) > t ( u ), then the end vertex w of e2 is  ∈ , and if it ends at v with  c d  such that t (w ) > t (v ), and hence the edges e1 and e2 do not make a peak. Justifying this  a claim is equivalent to proving that if t   c

b   a 