Topology [12th printing ed.] 9780697068897

Table of contents :
Contents
I. Elementary set theory
Sets
Boolean Algebra
Cartesian Product
Families of Sets
Power set
Functions, or Maps
Binary relations; equivalence relations
Axiomatics
General Cartesian Products
Problems
II. Ordinals and Cardinals
Orderings
Zorn´s Lemma; Zermelo´s Theorem
Ordinals
Comparability of Ordinals
Transfinite induction and Construction
Ordinal numbers
Cardinals
Cardinal Arithmetic
The ordinal number omega
Problems
III. Topological spaces
Topological spaces
Bassis for a given topology
Topologizing of sets
Elementary concepts
Topologizing with preassigned elementary operations
Gfi, Fsigma and Borel sets
Relativization
Continuous maps
Piecewise definition of maps
Continuous maps into E1
Open maps and colse maps
Homeomorphism
Problems
IV. Cartesian products
Cartesian product topology
Continuity of maps
Slices in Cartesian Products
Peano curves
Problems
V. Connectedness
Conectedness
Applications
Components
Local Connectedness
Path-Conectedness
Problems
VI. Identification Topology; weak topology
Identification topology
Subspaces
General theorems
Spaces with equivalence relations
Cones and suspensions
Attaching of spaces
The relation K(f) for continuous maps
Weak topologies
Problems
VII. Separation axioms
Hausdorff spaces
Regular spaces
Normal spaces
Urysohn´s characterization of normality
Tietze´s characterization of normality
Covering characterization fo normality
Completely regular spaces
Problems
VIII. Covering axioms
Coverings of spaces
Paracomplact spaces
Types of refinements
Partitions of unity
Complexes; Nerves of Coverings
Second-countable spaces; Lindelöf spaces
Separability
Problems
IX. Metric spaces
Metrics on sets
Topoloty induced by a metric
Equivalent metrics
Continuity of the distance
Properties of metirc topologies
Maps of metric spaces into affine spaces
Cartesian products of metric spaces
The space l2(A); Hilbert cube
Metrization of topological spaces
Gauge spaces
Uniform spaces
Problems
X. Convergence
Sequences and nets
Filterbases in spaces
Convergence properties of filterbases
Closure in terms of filterbases
Continuity; convergence in cartesian products
Adequacy of sequences
Maximal filterbases
Problems
XI. Compactness
Compact spaces
Special properties of compact spaces
Countable compactness
Compactness in metric spaces
Perfect maps
Local compactness
sigma-compact spaces
Compactification
k-spaces
Baire spaces; category
Problems
XII. Function spaces
The compact-open topology
Continuity of composition; the evaluation map
Cartesian products
Application to identification topologies
Basis for Zy
Compact subsets of Zy
Sequential convergence in the c-Topology
Metric topologies; relation to the c-topology
Pointwise convergence
Comparison of topologies in Zy
Problems
XIII. The spaces C(Y)
Continuity of the algebraic operations
Algebras in C(Y;c)
Stone-Weierstrass theorem
The metric space C(y)
Embedding of Y in C(Y)
The ring C(Y)
Problems
XIV. Complete spaces
Cauchy sequences
Complete metrics and complete spaces
Cauchy filterbases; total boundedness
Baire´s Theorem for complete metric spaces
Extension of uniformly continuous maps
Completion of a metric space
Fixed-point theorem for complete spaces
Complete subspaces of complete spaces
Complete gauge structures
Problems
XV. Homotopy
Homotopy
Homotopy classes
Homotopy and function spaces
Relative homotopy
Retracts and extendability
Deformation retraction and homotopy
Homotopy and extendability
Applications
Problems
XVI. Maps into spheres
Degree of a map Sn a Sn
Brouwer´s theorem
Further applications of the degree of a map
Maps of spheres into Sn
Maps of spaces into Sn
Borsuk´s antipodal theorem
Degree and homotopy
Problems
XVII. Topology of En
Components of compact sets in En+1
Borsuk´s separation theorem
Domain invarience
Deformations of subsets of En+1
The jordan curve theorem
Problems
XVIII. Homotopy type
Homotopy type
Homotopy type invariants
Homotopy of pairs
Mapping cylinder
Properties of X in C(f)
Change of bases in C(f)
Problems
XIX. Path spaces; H-Spaces
Path spaces
H-structures
H-Homomorphisms
H-Spaces
Units
Inversion
Associativity
Path spaces on H-Spaces
Problems
XX. Fiber spaces
Fiber spaces
Fiber spaces for the class of all spaces
The uniformization theorem of Hurewicz
Locally trivial fiber structures
Problems
Appendix one: Vector spaces; polytopes
Appendix two: Direct and inverse limits
Index

Citation preview

TOPOLOGY

Other books in this series are

The Theory of Groups: An Introduction

Joseph J. Rotman

A Survey of Matrix Theory and Matrix Inequalities

Marvin Marcus and Henryk Minc

THis Book Is PArRT OF THE ALLYN AND BACON SERIES IN ADVANCED MATHEMATICS CoNSULTING EDITOR: IRVING KAPLANSKY UNIVERSITY OF CHICAGO

Other books in this series are

The Theory of Groups : An Introduction

Joseph J. Rotman

A Survey of Matrix Theory and Matrix Inequalities Marvin Marcus and Henryk Minc

TOPOLOGY

James

Dugundji Professor of Mathematics University of Southern California Los Angeles

ALLYN AND BACON, BOSTON « LONDON

INC. . SYDNEY

« TORONTO

To Merope

© Copyright 1966 by Allyn and Bacon, Inc. 470 Atlantic Avenue, Boston.

All rights reserved. No part of the material protected by this copyright notice may be reproduced or utilized in any form or by any means, electronic

or mechanical, including photocopying, recording, or by any informational storage and retrieval system, without written

permission from the copyright owner. Library of Congress Catalog Card Number: 66-10940 Printed in the United States of America ISBN " 0-205-00271-4

Twelfth printing . . . June, 1978

viii

Preface

concepts discussed in the book. Two appendices, one on linear topological spaces and the other on limit spaces, are included. Nearly every definition is followed by examples illustrating the use of the abstract concept in some fairly concrete situations. This device makes the book suitable for self-study. It also enables the instructor who uses the book as a text to proceed rapidly to those parts of the subject that he deems of greater importance. Remarks, in small type, call attention either to further developments, or to direct applications in other branches of mathematics. The problems, which are given at the end of each chapter, can all be solved by the methods developed in the book. Moreover, no proof in the text relies on the solution of any problem. Some of the problems are routine. Others are important theorems that complement the material in the text; these are accompanied by hints for their solution. The notation of symbolic logic used throughout the book is given immediately after the table of contents. I wish to thank E. A. Michael and Ky Fan, who read the original manuscript, for their valuable suggestions; and H.-J. Groh and P. A. White, for their help with the proofreading. I am particularly indebted to H. Salzmann for his constant willingness to discuss points of detail and content: his imaginative and penetrating criticisms and suggestions have led to many improvements. I also wish to thank Mrs. L. Syfritt, for her tidy and meticulous typing work; and the members of Allyn and Bacon who were involved with this book, for their patience and co6peration. Finally, I gratefully acknowl- edge the support given me by the National Science Foundation during the period that this book was being written.

James Dugundji The University of Southern California

Contents

I.

ll.

Elementary Set Theory

l

1

Sets

1

2

Boolean Algebra

3

3 4 5

Cartesian Product Families of Sets Power Set

7 8 10

6 7

Functions, or Maps ~ Binary Relations; Equivalence Relations

10 14

8 9

Axiomatics General Cartesian Products Problems

17 21 25

Ordinals and Cardinals 1

Orderings

2 3

Zorn’s Lemma; Ordinals

4

Comparability of Ordinals

5 6 7 8 9

'Transfinite Induction and Construction Ordinal Numbers Cardinals Cardinal Arithmetic 'The Ordinal Number £

Problems

»

29 29

Zermelo’s Theorem

31 36

'

38 40 41 45 49 54

57

Contents

X

lll.

Topological Spaces 1 2 3 4 5 6 7 8 9 10 11 12

IV.

Cartesian Products 1 2 3 4

V.

Cartesian Product Topology Continuity of Maps Slices in Cartesian Products Peano Curves Problems

Connectedness 1 2 3 4 5

VI.

Topological Spaces Basis for a Given Topology Topologizing of Sets Elementary Concepts 'Topologizing with Preasmgned Elementary Operations G, F, and Borel Sets Relativization ' Continuous Maps ‘ Piecewise Definition of Maps Continuous Maps into E? Open Maps and Closed Maps Homeomorphism Problems

Connectedness Applications Components Local ConnectednessPath-Connectedness Problems

Identification Topology; Weak Topology 1 2 3 4 5

Identification Topology Subspaces General Theorems Spaces with Equivalence Relations Cones and Suspensions

62 62 64 65 68 72 74 77 78 81 83 86 87 90

98 98 101 103 104 105

107 107 110 111 113 114 116

120 120 122 123 125 126

o0 NN

Contents

Separation Axioms

N

e

Vil.

Attaching of Spaces ‘ The Relation K(f) for Continuous Maps Weak Topologies Problems

Hausdorff Spaces Regular Spaces Normal Spaces

N

ON U

Urysohn’s Characterization of Normality

Covering Axioms

NN

VLR

WD

-

Vill.

Coverings of Spaces Paracompact Spaces Types of Refinements Partitions of Unity Complexes; Nerves of Coverings Second-countable Spaces; Lindelof Spaces Separability Problems

0NNV Wi

-

Metric Spaces

O

IX.

Tietze’s Characterization of Normality Covering Characterization of Normality Completely Regular Spaces Problems

Metrics on Sets Topology Induced by a Metric Equivalent Metrics Continuity of the Distance Properties of Metric Topologies Maps of Metric Spaces into Affine Spaces Cartesian Products of Metric Spaces The Space I%( 4. (2). Foreach4 C X and B C Y, f[f~Y(B) n 4] = B n f(4);

in particular, f[f~}(B)] = B N f(X). Given f: X— Y and g: Y— defined as the map x — g(f(x)).

maps f~1, g~ 6.6

Theorem

Z, their composition gof: X — Z is We can clearly compose the induced

and we have Letf: X—

Yandg:

Y—% Z.

Then(gof)™?

=f“iog‘1.

Proof:

xe(gof) 1 Cegof(x)e Cwflx)eg Y(C) s xef g (O = xeftogmH(C), Given an f: X — Y’ and a subset 4 C X, the map f considered only on A is called the restriction of f to A, is written f | A4, and can alternatively be defined as f| 4 = fN (A x Y). In the reverse direction, if 4 C X

13

Functions, or Maps

Sec. 6

and g: A— Y is a given map, a map G: X — Y coinciding with g on 4 (that is, satisfying G | A = g) is called an extension of g over X relative to Y. The following result is very useful:

Theorem Let X be any set, and {4, | « € &/} any family of subsets with {J 4, = X (a “covering” of X). For each o€/, let an

6.7

f.: Ay — Y be given, and assume thatf, | 4, N Ay = fz | A, N A for each

(¢, B) €/

x .

Then

there

exists one, and only one,

f: X— Y which is an extension of each f,; thatis, Vae o/ f| 4, = f,. Proof: which

For each x € X, define f(x) = f,(x), where o is any index for

x € 4,;

this

definition

is unique

because,

if also x € 4;,

the

= fo(x).of = ful®) f(x) gives A, N Ae | f = Ay O A, | f, hesis hypot x—> f(x) is therefore a map f: X — Y; it is evidently an extension each f, over X; and it is unique because each x belongs to some A4,, so any map that satisfies the requirements will have to assume the value f.(x) at x (that is, the same value as f).

If {A,| c €} is a covering of X, and if 4, " A; = & whenever « # B, then the family {4, | « € &/} is called a partition of X. We obtain at once the 6.8

Corollary If {4, |« e &} is a partition of X and if for each « € & there is given an f,: A,— Y, then there exists a unique f: X — Y which is an extension of each f,,.

Proof:

The

requirement

f,| 4, N Az =fz| Ae N Ay

of

6.7

is

evidently satisfied.

If f: X — Y takes on every value in its range, f is called surjective (or a surjection; or “onto”). Observe that for surjective f, 6.5(2) takes the simpler form: V B: B C Y = f[f ~Y(B)] = B. If f sends distinct elements of X to distinct elements of Y, f is called injective (or an injection; or one-to-one). Evidently f is injective if and only if [x # x'] = [f(x) # f(x')], or equivalently, [f(x)= f(x")] = [x = x']. The restriction of an injection to any subset is also an injection. If f is both injective and surjective, f is called bijective (or a bijection; or a one-to-one onto map). Ex. 9 The map F: 4 x B— B x A given by (a, b)— (b, a) is bijective. The map p,: X x Y— X is surjective. The map f of Ex. 5 is neither surjective nor

injective; however, f| [0, 1] is injective.

Chap. 1

14

Elementary Set Theory

Clearly, f: X — Y is bijective if and only if Vy e Y: f~{y} is a single point. Thus, with each bijection f: X— Y, we have also a map f~1: Y — X determined byy — f ~1({y}) [this differs from f~1: Z(Y)— W(X ), since the domains are not the same]; it is evident that f ~': ¥ - X is also bijective and that (f~!)~! = f. The followmg proposmon indicates a simple method for establishing that a given map f(resp. g) is injective (resp. surjective). 69

Letf: X— Yandg: and g is surjective.

Proof:

Y—

Xsatisfygo f = 14. Thenfis injective

'The map f is injective, since [f(x) = f(¥')] = x = go f(x) =

gof(x')= &'. g is surjective, since for any x, € X, x, = g[f(x,)]. Ex. 10 As map h: X — Let p,: X x that p; o f: X

7.

a trivial example illustrating the use of 6.9, we show that for any Y, the map f: X — X x Y defined by x — (x, 2(x)) is injective. Y — X be projection on the first coordinate; since it is immediate — X is 1y, it follows from 6.9 that f is injective.

Binary Relations; Equivalence Relations

A binary relation R in a set 4 is, intuitively, a proposition such that for each ordered couple (a, b) of elements of A, one can determine whether a R b (“aisin relation R to b”) is or is not true. We state this formally in terms of the set concept. 7.1

Definition A binary relation R in aset Aisasubset R C 4 x A. (a, b) € R is writtén a R b.

Ex.1 For any set 4, the diagonald = {(a,a) | ae A} C A x A is the relation of equality. Note that in binary relations, each pair of elements need not be re-

lated: If @ # b, neither (a, b) nor (b, @) is in 4. R = (4 x A) — 4 is the relation of inequality. Ex.2 Ex. 3

The relation

< between real numbers is the set

{(x,y) | x < y} C E' x E'. Ex.4

The relation of inclusion in Z(A4) is {(4, B) | A C B} C #(A4)

x P(A).

Ex. 5 An evident generalization of 7.1 is to define any subset of X x Y to be a binary relation between the elements of X and those of Y; thusamap f: X - Y would be a special type of binary relation.

Let A be a set with a binary relation R. IfB C A,then(3.4)B x B C A x A, and therefore R N (B x B) is a binary relation on B [corresponding to the statement “(xe€ B) A (yeB) A (xRy)’]; we call

.

Binary Relations; Equivalence Relations

Sec. 7

15

R N (B x B) the relation induced by R in B. In particular, a binary relation in 4 induces a definite binary relation in every subset of 4. Equivalence relations are very frequently used in all mathematics; they arise whenever one desires to regard all those members of a set that have some preassigned characteristic as a single entity. 7.2

Definition A relation if: (1)) (2). (3).

binary

relation

R

in A4 is called

an

equivalence

VaceA:aRa (reflexive). (@aRb)= (bRa) (symmetric). (@aRb) A (bRc)= (aRc) (transitive).

If a R b, we say that a and b are equivalent. Ex. 6

The relation of Ex. 1 is an equivalence relation.

Ex.7

In N, {(x,y) | x = y mod 2} is an equivalence relation.

In the set of vectors C E3, let a R b denote that a and b have the same Ex. 8 direction. Then R is an equivalence relation. Ex. 9 Letf: X— Y be a map. The relation {(x, x") | f(x) f(x)} is an equivalence relation in X. Each of the following relations satisfy exactly two of the requirements Ex. 10 of 7.2. (a) In E1, the relation [0, 1] x [0, 1] is not reflexive. (b) The relation of Ex. 3 is not symmetric. (c) In E!, the relation {(x, y) | |[* — »| < 1}is not transitive. Ex. 11 Any symmetric and transitive relation R that satisfies the condition Va3 b:a Rbis in fact an equivalence relation, as the reader will immediately verify,

Let R be an equivalence relation in A. For each a € 4, the subset = {be A | b R a} is called the equivalence class of a. 'The fundamental theorem on equivalence relations is a consequence of

7.3

lemma

(1). (2). (3). Proof:

Let R be an equivalence relation in 4. Then:

U{Ra|aecd}= A. IfaRb,then Ra = Rb. If w(aRb),then RaNnRb

=

Ad (1). Because R is reflexive, we have a € Ra, and therefore

U{Ra|ae A4} = Ad (2). Let x € R a; using transitivity of R gives xeRa; this is the “slab” in [ [ ¥, where each factor is Y, except the ath, which is C,.

Similarly, for finitely many indices o, - - -, «, and sets

C, the subset

9.6

Corollary

Ca,,>

InJ[{Y,| e}, ]___[ Ca

CY,,

NG, > = pall(Cal) N -

N

noted by X be the isomorphism onto an ideal of X. By 3.2(a) and the lemma 4.1, 0u | Sa N Sg =5 | Sy NSy for all (e, B)e x o, so (¢f. 1, 6.7), there is a unique map @: | S,— | @(S,), which is easily verified to be an isomorphism.

Since by 3.2(a) both | S, and |J ¢,(S,) are ideals,

we find that (J S, 2. If W e 2, we have an isomorphism as required (that is, alternative 1 or 2 of the theorem). If W € 2 then by 3.4 there must be some S, = W(w,) with W(w,) U {w,} € 2. In this case, we must have ¢,(S,) = X (that is, alternative (3) of the theorem), for otherwise, ¢,(S,)= X (%,) and extending ¢, by p(w,) = x, would give the contradiction W(w,) U{w,} e 2. 'The uniqueness of the isomorphisms follows from the lemma.

For subsets of a well-ordered set, the following consequence is of importance: 4.3

Corollary (a). Any subset A of a well-ordered set W is isomorphic either to W or to an initial interval of W. (b). No initial interval of W can be isomorphic to W.

Proof: (a). Regarding A4 as a well-ordered set, we show the possibility that W is isomorphic to an initial interval of A4 cannot occur. If g: W— A(a,) were such an isomorphism, then, using the identity map i: A— W, we would have a monomorphism goi: A > 4 satisfying g oi(ay) < ayand contradicting 4.1 on comparison with 1: 4— 4. (b) is a direct consequence of 4.2. Ex. 1 It is important to note that A C W may be isomorphic to W: Taking W = w and A as the subset of even integers, 2n — n is such an isomorphism. However, no initial interval of W can be isomorphic to W.

4.4

Corollary The class of all well-ordered sets is well ordered, if we define W < X to mean that W is isomorphic to an ideal of X (and W = X means that W is isomorphic to X).

40

Chap. 11

Proof:

Clearly, W < X is a preordering.

ordering, observe that (X < W)

Ordinals and Cardinals

To see that it is a partial

A (W < X) =

W = X, since by 4.2,

one set isomorphic to an initial interval of another excludes the possibility

of the second set being isomorphic to an ideal of the first. To show that this is a well-ordering, let # be any nonempty class and W e &; because each X € 4 that precedes W is isomorphic to an ideal of W, and because

I(W) is well ordered, there is a first element in £.

5.

Transfinite Induction and Construction

There are two induction principles associated with ordinals. 5.1

Theorem (Transfinite Induction) Let W be a well-ordered set, and let Q C W. If [W(x) C Q] = [xe€ Q] for each x € W, then 0=W.

Proof: 'The first element, 0, of W is in Q, since & = W(0) C Q, and evidently there can be no first among the elements not in Q. Remark 1: 'Theorem 5.1 is (a) Let {P(x) | x € W} be and (ii) for each x, the P(x) is also true. Then

frequently stated in the following form: a set of propositions. Assume both (i) P(0) is true, hypothesis that every P(«), « € W(x), is true implies every P(x) is true.

This is evidently 5.1, with Q = {x | P(x) is true}.

Remark 2: On w, the (b) Let {P(1) | { € w} and (ii) for each also true. Then

induction principle is equivalent to: be a set of propositions. Assume both (i) P(0) is true, n, the hypothesis that P(n — 1) is true implies P(n) is every P(n) is true.

The equivalence of (a) and (b), on w, follows & has an immediate predecessor; and since, as not always have this property, the analogue example, on w U {g}, where ¢ is adjoined as the establish that P(q) is true.

precisely because each element of we have seen, well-ordered sets do of (b) is not generally true. For last element, the form (b) does not

"The second induction principle is the important 5.2

Theorem

(Transfinite Construction)

and E an arbitrary class. Assume:

Let

W be a well-ordered set,

For each x € W, there is given a rule R, that associates with each

®: W(x) — E, a unique R (¢) € E. Then there is one, and only one, F: W—

R,[F | W(x)] for each x € W¥.

E such that F(x) =

Sec. 6

Ordinal Numbers

41

Proof: There is at most one such F. For, if F, G were two distinct such maps, there would be a first ¥ with F(x) # G(x); then, since

F| W(x) = G| W(x), we would have F(x) = G(x), contradicting the choice of x. We now use 3.4 to show that a map F, as required, exists. Let 2’ be the set of those ideals S of W for which there is a ¢,: S— E satisfying the stated condition. By what we have already proved, ¢, is unique on each S, and therefore, on (the ideal) SN .S’ we have ps| SNS"

=@,

| SN

S’; as in 4.2, we conclude that any union of

members of 2 belongs to Z. Let S = W(x) €Z; then W(x) U {x} € X, since we can extend ¢ by setting @(x) = R, (¢, | W(x)). from 3.4 that W e 2, proving the theorem.

It now follows

Remark 3: Writing 0 for the first element of W, note that W(0) = & since there is only one map, & -> E, the hypothesis of 5.2 tacitly assumes that @(0) 1s uniquely defined. Thus, in greater detail, 5.2 can be stated as follows: Let W be well-ordered, E an arbitrary class, and e € E a given element. Assume that for each x # 0, there is given a rule R, that assigns to each ¢: W(x) — E an element R,(p) € E. Then there is exactly one F: W — E such that F(0) = e and

F(x) = R,[F | W(x)] for each x # 0. Remark 4: In w, 5.2 can be given a simpler form because each element has an immediate predecessor: Let E be an arbitrary class and e € E a given element. Assume that for each 7 there is given a map R,: E — E. Then there exists one, and only one, F: w -> E such that F(0) = e and F(n + 1) = R, ., ,[F(n)] for each neEw.

Remark 5: The reader should observe that 5.2 cannot be proved by simply defining F: W — E by the formula F(x) = R,[F | W(x)]: such a definition would be circular in that it defines F in terms of itself. The whole purpose of 5.2 is to show that there exists a map F: W — E having this property.

6.

Ordinal Numbers

In the class of all ordinals, define W = X if W is isomorphic to X, This 1s evidently an equivalence relation, so it divides the class of all ordinals into mutually exclusive subclasses. We wish to attach to each ordinal an object, called its ordinal number (underscored in the subsequent text), so that two ordinals have the same ordinal number if and only if they are isomorphic. Following Frege, we could define the ordinal number of an ordinal to be the equivalence class of that ordinal. Though this definition is adequate for most mathematical purposes, it has the disadvantage that ordinal numbers are not sets: Without separate axiomatics for them, we could not, for example, legitimately consider any collection of ordinal numbers. In this section, we will define “ordinal number” within the framework of set theory. It will be shown

42

Chap. 11

Ordinals and Cardinals

that there exists a uniquely defined well-ordered class 2 such that each well-ordered set is i1somorphic to some initial interval of 2. The desired objective is attained by calling the members of & ordinal numbers and assigning to each ordinal W the « € & for which W = Z(«). The basic idea is to use the sets postulated by the axiom of infinity (I, 8, Axiom X): each ¢ € & will be a set whose elements are all the sets in & that precede it; in other words, each ¢ € 2 will be simply the initial interval Z(¢) in Z. To illustrate the mechanics, we write down the first few members of Z':

gi{o}{o,{{2} o}{ {2,e, {2}}} The

ordinal

number

{o,{0}, {@,{2}} 6.1

Definition

(1). (2).

of

{1, 2, 3},

in its natural

order,

is

then

An ordinal number is a set ¢ with the properties

(xee) A(yeg)=(xey) Vv (yex) Vv (y = x). (xey)A(yeo)= (xeq)

We say that “x precedes ¥” in ¢ if “x € y”; observe that € is not an ordering,

since by I, 8.11, x € x is not true.

The

e-relation in ordinal

numbers has the properties 6.2

(a). In each nonempty set A C ¢, there is a unique a € 4 (called the first element of A) such that (aex) v (¢ = x) for each xed. (b). The first member in ¢ is . (c). If 2 € g, then 2 1s also an ordinal number.

Proof: (a). By the axiom of foundation (I, 8, Axiom IX), there is an acAwithan A = g [thatis, (x € A) = —(x € a)]; in view of 6.1(1), the element a has the required property. If there were another element b e A with this property, we would have a € b and b € a, which contradicts I, 8.11(2). (b). If a is the first member of ¢, there can be no x € a because of 6.1(2). (c). Let x,yex; since (x,y€2) A (R€¢)= (x,y€¢), we find that 6.1(1) is valid for the members of 2. To verify 6.1(2), assume that (xey) A (¥ €3); by what we have just proved, we must have one of (xe2), (2€x), (* = x), and we need show only that the last two possibilities cannot be true. (1). If 2 €x, then the subset 4 = {x, y, 2} C « does not have a first

element, as required by (a). (ii). If = = x, we would have (xey) axiom of foundation [¢f. I, 8.11(2).]

A (y €x) which

violates the

43

Ordinal Numbers

Sec. 6

According to (c), ordinal numbers can be treated as sets or as elements of sets. We next establish the essential uniqueness of ordinal numbers.

6.3

If ¢, B are ordinal numbers, and ¢ # B, then ¢ C g if and only if ¢ € B (that is, an ordinal number consists of all those ordinal numbers that are proper subsets of it). (b). If ¢ and B are any two ordinal numbers, then either ¢ C 8 or B C a. (a).

Proof: (a). If ¢ €8, then by 6.1(2), xea => x€p, so that ¢ C 8. Conversely, assume ¢ C 8 and let x, € 8 be the first element of 8 — ¢. The definition of x, shows that y € x, = y € ¢, so that x, C . Now assume that' y € ¢; since the possibilities x, €y, x, = y are excluded and so because each implies x,€¢, we find that yeag=yex, e C x5. Thus¢ = x5 €.

(b). It is immediate that ¢ N B (that is, {x |x€a A x€f8}) is an ordinal number; we need show only that it is one of g, 8. If it were neither, then ¢ N B would be a proper subset of both ¢ and B, so, by (a), we would have (¢ N Beg) A (¢ N B€pB), which would then imply ¢ N B ea N B, in contradiction to I, 8.11.

We now have the comprehensive 6.4

Theorem

¢ - - -, then there is an z# such that ¢, = ¢, for all7 > n).

(6). Each well-ordered set W is isomorphic to a suitable 2(g). ¢ is called the ordinal number of W and is denoted by “ord

W.”

Proof: (1). < is clearly a partial ordering. To see that it is a wellordering, let £ C & be a nonempty set; we are to show that E has a first element. Choose ¢, € E and define A=¢,NE={x]|(xeg)

A (xeE)}.

44

Chap. 11

If A =

Ordinals and Cardinals

@, then by 6.3(a) and (b), we find that

xeE

=> —(x Cgy) >,

Cx

so that g, is first in E.

If 4 # for each

o, then by 6.2(a), there is an a € A with (ae x) v (a = x) x€ A.

We

now

have

a€ E and

a C x for

eachx e,

N E;

but since a C gy and gy C y for each ye E — (gq N E), we also have a C y for eachy € E — (go N E), so that a is first in E. (2). If Z were a set, we verify immediately that 2 would then be an ordinal number, and this would imply that & € &, which is impossible I, 8.11). Z(z) ={8|BeZ A Bega}; by 6.2(c), the (3). Given g%, (

condition 8 € & is redundant, so Z(¢) = {8 |Bea} = «. (4). Let E be any set of ordinal numbers. Then 8 = | is a set, easily verified to be an ordinal number.

{¢ | ¢ € E}

Since we have ¢ C 8

for each ¢ € E, it follows that ¢ < Bin Z. Noting that 8 U {8} is also an ordinal number and that 8 is a proper subset of 8 U {8}, we find 8 U {8} larger than each ¢ € E. The second part follows from (1). (5). Since Z is well-ordered, the set {g; | 7 € N} has a first element, which is some one of its members, say, «,. Then, for all i > n, we have o

=

Q.

:

(6). We will use transfinite construction. For each x € W and each ¢: W(x) - Z define R,(p) to be the smallest ordinal number greater than all the ordinal numbers in @(W(x)); this definition is legitimate, since by the axiom I, 8, VI of replacement, p(W(x)) C & is a set and (4) applies. By 5.2, there exists a map F: W— % such that F(x) = R,[F | W(x)] for each x € W, and it is easy to see that F is a monomorphism. Since F(IW) C Z is a set, we have by (4) that F(IW) C Z(p) for some B, and from 4.3 that W is isomorphic to some Z(). [The reader may show that, in fact, F(W) = Z(¢).]

According to 6.2(b), the smallest ordinal number is @, the null set; it is denoted by “0.”

The next one, as indicated before, is { @}, whose

only member is the null set, and is denoted by “1.” The ordinal number {0, 1} is denoted by 2, and in general, the ordinal number with the # elements {0, 1,---,n — 1} is denoted by n. If ¢ is any ordinal number, o U {¢} is also an ordinal number, written ¢ + 1. Note, however, that

the axiom of infinity (I, 8, X) states 'the existence of ordinal numbers not of the form ¢ U {¢}; these are called Lmit ordinal numbers and have no immediate predecessor in . A well-ordered set W is called finite if ord W = n for some n. The

ordinal number of the well-ordered set {1, 2, - - -} in its natural order is denoted by w. Taking the notation for intervals in E?, we denote Z(a) by [0, o[ ; obviously, ord[0, ¢[ = «.

6.5

45

Cardinals

Sec. 7

We have used the ordinal numbers as “counters”’—attention has Remark been directed to the.initial intervals of Z. However, if in any ordinal number ¢ we define x

2 Any b—a

7

El,as x —

open interval ]a, [ C E! is equipotent to ¥ = ]—1, +1[, since b+ a ¥ is equipotent with is a bijection ¥ — ]a, b[. Furthermore,

¥t 3 x

1 + |x]

shows.

Thus, by transitivity, card ]a, 8[ = card E!: each open

interval in E! has “just as many points” as E? itself. Ex. 3 Distinct ordinals may be equipotent. The ordinals [0, w[ and [0, w] in % are not isomorphic [¢f. 4.3(b)], but p(w) = 0, @(n) = n + 1 is a bijection of [0, w] onto [0, w[. This indicates an important distinction between finite and infinite ordinals: For infinite sets, the ordinal number depends both on the size of the set and the manner in which the set is counted. Thus the decomposition

of the class of all ordinals according to equipotence does not coincide with that according to isomorphism: each isomorphism class lies in one equipotence class, “ut an equipotence class generally contains many isomorphism classes.

Chap. 11

46

Ordinals and Cardinals

Let X be any set. The set of all maps X — {0, 1} ({0, 1} is the set consisting of the two integers 0, 1) is denoted by 2*. For any set 4 C X,

the map ¢, € 2% defined by c4(x) = A(%)

0

xeA

1

xe A

is called the characteristic function of A. The following simple result has important consequences.

7.2

Card 2* = card Z(X).

Proof: Define f: 2X — P(X) by ¢ — ¢ }(1); f is clearly surjective, since f(c,) = A. But f is also injective: ¢ # d =3 x: c(x) # d(x) = x belongs to exactly one of ¢1(1), d (1) = f(c) # f(d). To compare the size of sets, we make the 7.3

Definition For two sets, X, Y, we write card X < card Y if an injection X — Y exists.

Note that we use “ X,.

8.4

Definition Let X, X be two cardinal numbers. The exponentiation R® is the cardinal number of the set of all maps X — K.

This notion of exponentiation reduces to the usual concept for finite cardinals: #™ = n™. Note that 2¥ is simply the cardinal number of the set of all characteristic functions on X so, by 7.8(2), it is always true that X < 2% and, by 7, Ex. 7, that 2%

= ¢

If X, X are any two sets, let X¥ denote the set ofall maps X — X. It is trivial to verify that, if X(X) = X and X(X) = K, then N(XX) The usual rules for exponentiation apply: (R)X+¥ = R“-N“ , (R-R)¥ = R¥.R¥ and R&¥ — (R®)¥, We indicate a proof of the last equality; the others are proved similarly. Choosing X, ¥V, Z, such that R(X) = X, X(Y) = R, and R(Z) = X', associate with each fe X¥*Z% the map ¢;: Z— XY given by 2y — f(¥, 2); then f — ¢, is easily verified to be bijective. Ex. 3 X.X = N2 since Xl = X and N+l = X2; also X2 =1 from I, 6, Ex. 3; thus, with finite cardinals as exponents, exponentiation has the usual interpretation. Also from I, 6, Ex. 3, 0% =

Ex. 4

As with sums,

Ex. 3 extends to infinitely many

factors:

If X, = X

for each p € A, and if R(A) = X, then [ | R, = XY, For [| R, is the cardinal "

u

number of |[ X, where each factor X, is the same set X having ®(X) = R and, n

by I, 9, Ex. 2, [ [ X, is the set of all maps .# — X. n

has cardinal number

280 =

¢,

In particular, the Cantor set

Ordinals and Cardinals

Chap. Il

52

We have seen that X,- X, = X,; the fundamental theorem asserts that this is true for all transfinite cardinal numbers:

Theorem

8.5

Lemma

X-X = X for each X > X,

If X-X

Proof of Lemma:

= X for any R, thenalso X = 2-X = 3.X = X. K,

We have

< 0 KR=R, + IRNR 21R< R+ X= R=R Proof of Theorem: It suffices to prove that for some set M with X(M) = X, there is a bijection M — M x M. We use Zorn’s lemma, 2.1(2). Let X be any set with X(X) = X, and let o/ be the set of all couples x A is a bijection. Then (4, ¢,), where A C X and ¢,: 4—A4 o/ # @: since X is infinite, 7.8(3) shows that X contains a subset 4 with X(4) = X,, and we know that X,-X, = X,. Partially order&7 by (4, ¢,) < (B, pg)if both A C Band gy | 4 = ¢,. Each chain {(4;, ¢;) | B€ #} has an upper bound (| 4,4, ¢) because B of I, 6.7, and consequently there is a maximal (M, ¢,); since — M x M is bijective, we have R(M)-RX(M) = X(M), and oy: M the argument reduces to showing that (M)

= X.

Since M C X, we have X(M) < R(X) = X; we shall prove that N(M) < X(X) is impossible because it would contradict that (M, @) is maximal in /. Thus, assume that R(M) < R(X). Then we must have

R(M) < R(X — M).

(1). For, if R(X — M) < R(M),

then because X = MU

disjoint union, using the lemma would yield R(X) = X(M) + R(X - M) < X(M) + X(M)

(X — M)

is a

= 2X(M) = X(M),

which contradicts the assumed R(M) < R(X). Because of (1), there is a bijective map of M onto asubset ¥ C X — M,

and we have X(Y) = X(M), Y " M =

@. To extend ¢y to a bijection

p:(MUY)>(MUY)x (MuUY) M)U (Y x x (Y x UM x Y)U M) =M

Y),

we need show (I, 6.8) only that a bijection

Y->(Mx

YYU(Y x M)U(Y

x Y)

exists. This is, however, apparent, since on the right side the sets are

Sec. 8

Cardinal Arithmetic

pairwise

disjoint,

so from

53

R(Y)-RX(Y) = R(Y)

and the lemma,

the

Since

right side has cardinal number 3-X(Y) = X(Y).

(M, ou) < (M VY, o), this contradicts

the

maximality

of (M, ¢y). Thus

R(M) < R(X)

is

impossible, and as remarked, this proves the theorem.

The basic facts of cardinal arithmetic follow from 8.5. They show that the only way to obtain larger cardinal numbers is through exponentiation. 8.6

Corollary Let X, X be two cardinal numbers, at least one of which is transfinite. Then:

(a). X + R = R.R = max(R, N). (b). If X < X, then X — X = X; that is, removal of a set having a

smaller

cardinal

number

does

not

reduce

the

cardinal

number of the given set. (c). If2 < R < RN, then XX = 2¥, (d). X% = R for each finite z and transfinite X.

Proof:

(a). Assume that X < R. Then:

%l

X=1RgXR


some «,. Then there exists a B, < 2 with the following

property: As o increases, its image f(«) repeatedly returns value below B;,. In symbols: 38,V B3I « > B: f(«) < B, Proof:

to

Assume that the conclusion is false. Then

VBoABYa > B:f(e) > B and so we could define a map R: Z(2)— Z(L2) by sending each Bo€ Z(£2) to the smallest B satisfying this condition. By inductive construction, 5, Remark 4, there would exist a map ¢: [0, w[— [0, £[ with @(0) = «y and @(n + 1) = R[p(n)] for each n. Thus, writing ¢, = ¢(n), we would obtain a countable subset {a, | ne N} C 2(Q) such that for each n, (@ > «,,) = (f(e) > «,). By 9.1, {a, | ne N} would have a least upper bound @ < 2, and we would have the contradictory statements: (1). f(a) < (2). f(a) > showing that f(&) ‘This proves the

&, since @ = ¢(0) = «. abecause @ > «,,, for each n, so that also f(&@) > «,, is an upper bound for {«, | n € N}. theorem.

9.3

Remark The ordinal £ intervenes in any construction utilizing operations that involve at most countably many objects at a time. The general pattern and result is as follows:

9.4

Let X be a set, and %7 any family of maps [| X — X (that is, all factors in the 1

countable cartesian product are X).

For any 4

C X, write

) = Ui([14)1 re) Then, given any M, such that M, that

D).

L (My)

= Mo,

C .&/(M,), there exists an M,

O M, such

Ordinals and Cardinals

Chap. 11

56

(2). Mg, is the smallest set O M, and satisfying (1): that is, if M O M, and satisfies (1), then Mo C M. In particular, Mg is uniquely determined by M,.

(3). If max [R(&), R(Mo)] 2 2, then R(Mg)< [R()- R(Mo))]¥o. be the map 4 — 2/(A4).

Ad (1). Let R: Z(X) — P(X)

Proof:

By transfinite

construction (5.2), there exists a map ¢: Z(2) — #(X) with ¢(0) = M, and () = R[U {@(B) | B < a}] for each « < 2. Writingp(B) = M; C X, let Mg = U {M; | B < 3. Then: (a). Mg C 4 (My), since for each x € M, there is a first 8 in [1, .Q[ with x € M, so that x € (VU {M, | ¢ < B}) C A (My). (b). L (Mg) C Mg:

If {x}€ HMQ,

each

then

coordinate

x; € ¢(e;)

for

1

some o < £2; letting B < £ be an upper bound of U {x) C U {M.| e« < B}, so that & ({x}) € My,; C M.

{a;|i€ N},

we

have

Given M, let P(«) be the pro-

Ad (2). We proceed by transfinite induction.

position “|J {M; | B < a} C M.” P(0) is clearly true by hypothesis, and assuming that P(B) is true forall B < ¢, it follows that M, = &/ (U (Mg | B < o) C A(M) = M, so that P(«) will be true also. Ad (3). Again we use transfinite

induction.

Let

P(c)

be the proposition

“RU {M; | B < o)) < R()Ro- R(Mp)R0.” P(0) is obviously true. Assuming P(B) to be true for all B < o, we have [using 8.2, 8.3(a), and 8, Ex. 4] that

R(M,) < R()-[R(A)FoR(Mo)¥o]Ro = R(2Z)¥o- R(M)o, and consequently, by; 8.2 again,

| B < of) < Ry R()Fo- R(M)¥o A

R(U M

R()RoR(Mo)¥o,

the latter equality holding because one of R(&7), R(M,)> 2 and X; < 2%o, Thus P(«) is true, proving (3). Ex.1 A famxly of sets # C .@(X ) is called a Borel family if, whenever X, € #

for all7 € N, then U X, € # and fl X, € # also. Letting .# be any family of sets,

and

letting &/

consist

of the two

maps

N,

U: HQ’(X) — P(X)

defined by

1

{Xi} -

N X,

{Xi}—> U X,

1

respectively,

9.4

asserts

that

there

is

a

unique

1

smallest Borel family #(#) O #, and that in fact, R(B(A)) < R(A)¥. The existence of () can also be established more directly by observing (i) that P(X) is a Borel family, and (ii) that any intersection [ %, of Borel families is

also a Borel family. Then () is simply the intersection of all Borel families containing .#. However, this approach gives no information about the cardinality of the family #(A). Ex. 2 The existence of Mg in 9.4 can also be established directly, as follows: LetB=U{M|MCZM) A (My C M}. ThenB C &/(B),since [x € B] = [x

belongs to some M] = [x € /(M) C &(B)]. On the other hand, B C &/(B) implies that &/(B) C 2/(2/(B)); that is, &/ (B) is one of the sets M. Thus &/ (B) C B, showing that /(B) obtained.

= B. Again no information about the cardinality of B is

57

Problems

Problems Section |

1. Let 4 be the diagonal in A x A. Show that R onlyif4 CRand R-R = R. 2. In Z*, define m < n if n divides m. Show that every chain has an upper bound, and determine Let Z be the set of all real-valued functions of

C A x A is a preorder if and

this is a partial ordering, that the set of maximal elements. a real variable. Show that by defining f < g to mean “V x: f(x) < g(x),” (¥, 0: B(x;r,)

C G)

= [B(x;min (ry,- -, r,)) C r:]G’,].

(3) is trivial, This topology, 7, is called the Euclidean topology of E*; the topological space (E1, .7) is called the Euclidean 1-space. Note that.7 is not the indiscrete topology in the set E1, that is, 7 does not consist only of @ and E*: indeed, each B(y; r) belongs to 7, since, given any x € B(y; r), we haved = |y — x| < 7, and therefore x€ B(x;r — d) C B(y;r). Itis simple to see that J can also be described more directly as the family of all unions of open intervals.

64

Chap. 111

Topological Spaces

Ex. 5 Let E" be the set of all ordered n-uples of real numbers. Using vector and radius 7’ the set B(x; ) = {y ||y — x| < 7} notation (cf. I, 1) call “ball of centxer The Euclidean topology in E™ is determined by calling G C E™ “open” if for-each x € G there is some r > 0 such that B(x;r) C G. The verification that this

criterion describes a topology .7, and that in fact each B(x;r) belongs to 7, is contained in Ex. 4, since only reinterpretation of B(x;r) is involved. With this topology, E™ is called Euclidean n-space. As before, the open sets in E™ can be described equally as the arbitrary unions of balls.

As the examples show, a set X may have many topologies; with each, it is a distinct topological space. By regarding each topology as a subset of #(X), the topologies in X are partially ordered by inclusion; clearly, J C J C 2 for each topology J. We call 7, larger (or “with more open sets,” or “finer”) than J , whenever 7, C 7. It is trivial to verify

1.4

Let{7, | « € &} be any family of topologies in X. Then (N 7, =

{U|Vaesd:

UeT,}

is also a topology in X; however, | 7, ’

need not be a topology.

2.

Basis for a Given Topology

The task of specifying a topology is simplified by giving only enough open sets to “generate” all the open sets. 2.1

Definition Let (X, J) be a topological space. A family # C J is called a basis for J if each open set (that is, member of J) is ' the union of members of #.

A is also called a “basis for the space X,” and its members the “basic open sets of the topology 7.” Not only is each member of 7 the union of members

of %, but also, because of &

C J

and

I.1, each union of

members of # belongs to.7; thus a basis for J completely determines 7. Ex.1 J is a basis for 7 . Ex. 2 Let & be the discrete topology on basis for 2.

X. Then

# = {{x} | x€ X}

is a

:

In view of Exs. 1 and 2, a given J~ may have many bases. The families | % C T that can serve as a basis are characterized by

2.2

Theorem Let # C 7. The following two properties of % are equivalent: (1). 4 is a basis for 7. (2). For each GeJ and each x€ G there is a Ue # with xeU CG.

65

Topologizing of Sets

Sec. 3

Proof: (1) = (2). Let x€ G; since GeJ and # 1s a basis, G = € # with \U U,, where each U, e %. Thus there is at least one U,

e U, CG.

hx find U, e Zwithxe U, C G; (2) = (1). LetG e J;foreaecG, ,G}. | xe | {U then G = J

IneachE"n =1, % ={B(x;r)|x€E" r > 0} is a basis for the EucliEx.3 dean topology, as the descriptions in I, Exs. 4 and 5 show.

Ex. 4

E™ has a countable basis: the family & = {B(&, r) | £ has all coordinates

rational, and » > 0 is rational}. Exs. 4 and 5, there is a B(x, ) C The reader will easily verify that within a distance 7/3 of x; then

For, let G, and there is x € B(§,

Immediate consequences are: (a).

G be any open set, and x € G. By |, we can clearly assume that r is rational. a point £, with all coordinates rational, r/2) C B(x, r) C G, as required by 2.2,

'

Each set open in E™ is the union of at most countably many balls.

(b). The cardinal number of the Euclidean topology of E™ is 2%°,

By specifying a basis for 7, all the open sets are generated as unions. However, there is a more convenient way to describe the open sets:

2.3

. Then 4 is open (that is, Let 4 C J be a basis for Theorem isin .7) if and only if for each x€ A thereisa Ue & withxe U C A. Proof:

1f A is open, the condition follows from 2.2. Conversely, if

the condition holds, then (as in 2.2) we find 4 = |J {U, | ae€ 4}, where each U,e # C J; from l.1 it follows that 4 is open. In particular, since J is a basis for .7, 2.3 provides a very useful method for showing that a given set A4 is open.

3.

Topologizing of Sets

In this section, two general methods for introducing topologies in sets will be given. The first, and most popular, starts from any given family 2 C Z(X) and leads to a unique topology containing X'; thus we can to an extent preassign the notion of nearness desired.

3.1 '

Theorem

Given

there always

any family 2 = {4, | « € &/} of subsets of X,

exists a unique,

smallest topology .7 (2) O 2. The

family Z(Z) can be described as follows: It consists of &, X, all finite intersections of the A,, and all arbitrary unions of these finite intersections. X' is called a subbasis for 7 (2), and J (X)) is said to | be generated by 2.

Chap. 111

66

Topological Spaces

Proof: Let J(2) be the intersection of all topologies containing 2'; such topologies exist, since (X ) is one such. By 1.4, 7 (X) is a topology; it evidently satisfies the requirements “unique” and “smallest.” To verify that the members of J(2) are as described, note that since 2 C J(2), it follows from L.1 that 7 (X) must contain all the sets listed. Conversely, because [ distributes over N, the sets listed actually do

form a topology containing X, and which therefore contains J (2.

In 3.1, we started from X and obtained a topology 7 (2) O 2. I, conversely, we are given a topology .7, a family 2’ C J subbasis for J whenever J = J(2). Ex. 1 Ex. 2 Ex. 3

is called a

For any topology 7, is a subbasis for J . ‘The finite intersections of members of 2 are a basis for 7 (2). In the set E* of all real numbers, let X be all sets of form {x | x > a}

and {x | x < b). Then .7 (&) is precisely the Euclidean topology: Each finite open interval, being an intersection of two subbasic open sets, belongs to .7 (X)), and it is evident from the description of 7 (2) in 3.1 that the family & of all these finite

intervals forms a basis for 7 (2). But, by 2, Ex. 3, # is a basis for the Euclidean topology. Ex. 4 In the set E! of all real numbers, let 2 be all sets of the form {x | x > a} and {x | x < b}. The topology 7 (2) is called the upper limit topology and has the sets ]a, b] as basis. .7 (X) is not the Euclidean topology of E2, since the sets ]a, 5] do not belong to the Euclidean topology. The reader can verify that the Euclidean topology is smaller than 7 (2). The topological space (E*, 7 (X)) is denoted by E;. Ex. 5 Let I’ be any ordinal number and in [0, I'] use the topology generated by all sets of form {x | x > «} and {x | x £ B}. We call this topological space the

ordinal space [0, I']. Observe that the sets Jo, Bl = {x|x > e} N{x|x

< 8 + 1}

are a basis for the topology. Note also that [«, B[ is open if and only if @ = 0 or if a has an immediate predecessor.

Ex. 6 in X;

Let (X;,7,), x ---x

X,

-, (X, 7 ,) be spaces. The cartesian product topology

is- that having

as subbasis

all sets (¢f. I, 9.6) of form

{4,

where 4,67 ,i=1,---,n In view of I 9.6(1), a basis for this topology simply the family of cartesian products of open sets of the spaces Xj.

is

The construction of a topology from a subbasis loses some control over the open sets; they build up from the finite intersections of the 4, rather than topologizing constructed family to be

from the A4, themselves. In the second general method for a set, which we will now describe, the open sets are only by union from the given family; that is, by specifying a used as a basis for constructing the topology. Since inter-

sections are involved in topologies but no intersections are involved in forming open sets from a basis, it is to be expected that not every family can serve as the basis for some topology.

Sec. 3

3.2

67

Topologizing of Sets

Theorem

Let # = {U,|p€.#}

be any family of subsets of X

that satisfies the following condition: For each (u, A)e # x A and each xe U, N U,, there exists some

Then

U, withxe

U, C

the family J (%)

U, N U,.

consisting

of

@, X,

and

all unions

of

members of %, is a topology for X; that is, Z U {@} U {X} is a basis for some topology. .7 (%) is unique and is the smallest topology containing #. Proof: Using 3.1, we obtain a topology J (%) having # as subbasis. To sée that () actually has # as a basis, we need show only that each finite intersection of members of # is in fact a union of members of #; and, as in 2.2 (2 = 1), it suffices to show that foreachxe U; N --- N U, thereisa Ue # withxe U C U, N --- N U,. We proceed by induction, the assertion being true (by the hypothesis) for n = 2. If it is true for (n — 1), then writing xe Uy n---NnU, =(U;Nn---nU,_,)n U, the inductive hypothesis gives x € U N U, for some UC

Ulm“‘mUn_l

in # so, by the case n = 2, we find

xeU'CcCUnU,CUn---NnU, for some U’ € #, completing the induction and the proof of the theorem. The specification of a topology by giving a basis is generally accomplished by specifying for each x € X a family of nbds {U,(x) | « € &(x)}

(called a “basis at x” or a “complete system of nbds at x”') and verifying that the family # = {U,(x) | « € #(x), x € X} satisfies the requirement of 3.2. Ex. 7 Let C be the set of all continuous real-valued functions on [0, 1]. For each f € C and ¢ > 0, define

M(f,e) =g [ 1f — ol 0} is a basis for some topology #

in C. For, if

he M(f, &) » M(g, 1), let

= [ 1f=hl and r=[ lg-nl 1

1

and let { = min [e — r;, — 7,]; then { > 0 and M(h, {) C M(f, &) " M(g, ), since

peMm D= [ If—gl< [ 1f -+ 1

1

1

ol PY), f~1: PY) > P(X). Of these, f~! should be

want Jy), maps used

Sec. 8

Continuous Maps

79

to relate the topologies, since it is the only one that preserves the Boolean operations involved in the definition of a topology. Thus the suitable maps f: X — Y are those for which simultaneously f~*: 7y - Jx. Formally stated, 8.1

Let (X, 7 %) and (Y, Jy) be spaces. Amap f: X — Y Definition is called continuous if the inverse image of each set open in Y is open

in X [that is, if f ~! maps J y into J x]. A constant map f: X — Y is always continuous: The inverse image Ex.1 of any set U open in Y is either & or X, which are open. Ex. 2 Let X be any set, 7, J 3, two topologies on X. The bijective map, 1:(X,J 1) — (X, T 2) is continuous if and only if 7 5 C 7 ;. Note that a continuous map need not send open sets to open sets, and also that increasing the topology T, preserves continuity. Ex: 3 A map sending open sets to open sets is called an open map. An open map need not be continuous. 1:(X,J ;) — (X, J3) is open if and only if T, C ., but it is not continuous whenever 7 ; # J 5. Let Y C X. The relative topology 7 y can be characterized as the Ex. 4 if For, . nuous conti is X — Y ¢: map sion inclu the h whic for Y on logy topo est small U e, the continuity of ¢ requires i “}(U) = U N Y to be open in Y, so that any topology for which 7 is continuous must contain I y.

The elementary properties are

8.2

(1). (Composition.) Iff: X — Y and g: Y — Z are continuous, so — Z. X f: alsoisgo (2). (Restriction of domain.) Iff: X — Yis continuousand 4 C X is taken with the subspace topology, then f| 4: 4 — Y is | continuous. (3) (Restriction of range.) If f: X = Y is continuous and f(X) is taken with the subspace topology, then f: X — f(X) is continuous.

g': y and 7 ;T Proof: (a). We have (gof)™' =f~log~1;since f~1: Ty — T, the continuity of g o f follows. (b). Note f| A = fo i, where i: A — X; thus, apply Ex. 4 and (a). ). f~YUNf(X)) =fHU)Nnff(X) = f~1(U), and the statement is proved.

The basic theorem on continuity is: 8.3

Let X, Y be topological spaces, and f: X — Y a map. Theorem The following statements are equivalent:

(1). fis continuous. (2). The inverse image of each closed set in Y is closed in X.

Chap. 111

80

Topological Spaces

(3). The inverse image of each member of a subbasis (basis) for ¥ is open in X (not necessarily a member of a subbasis, or basis for X1). (4). For each x € X and each nbd W(f(x)) in Y, there exists a nbd V(x) in X such that f(V (x)) C W(f(x)). (5). f(A) C f(A) for every A C X.

(6). f~I(B) C f~Y(B) for every B C Y. < (2), since f~(Y Proof: (1)

— E) = X — f~}E)

for any E C X.

(1) < (3). Let {U, | « € &} be a subbasis for Y. If fis continuous, each f ~1(U,) is open. Conversely, if each f~1(U,) is open, then because any open U C Y can be written U=

L){U,,,1 Nn---nU,

I{al,---,an}

C o},

we have that

FHU) = U{F~ 1 (Us,) N -0 f7H(U,)} is a.union of open sets and so is open. (1) = (4). Since f~(W(x)) is open, we can use it for V (x).

(4) = (5). LetA C X and b € 4; we show f(b) € f(4) by proving each W(f(b)) intersects f(A).

For, finding V' (b) with f(V (b)) C W(f(d)),

bed= o £ VB NA

= o # f(V(b) 0 A) Cf(V (b)) nf(4) C W(f(b)) N f(A).

(5) = (6). Let

A =f~YB);

then f(4) C f(4) = f[f"(B)] =

B N f(X) C B, so that A C f~1(B), as required. (6) = (2). Let B C Y be closed; then f~1(B) C f~(B), and since always f~1(B) C f~(B), this shows [4.4(b)] that f~}(B) is closed.

The formulation (4) of 8.3 shows that continuity is a “local” matter, a fact having many applications. Precisely, 8.4

Definition

An f: X —

Y is continuous at x, € X if 8.3(4) is satisfied

at x,.

From this viewpoint, the equivalence of (1) and (4) in 8.3 asserts: f is continuous according to 8.1, if and only if it is continuous at each point of X. Ex. 5 Let f: E® — E*; that is, a real-valued function of 7 real variables. The continuity of f at xo € E" 1s simply the usual notion encountered in analysis. For,

Sec. 9

Piecewise Definition of Maps

.81

by 8.4, f is continuous at x, if for each open interval ]Jf(xo) — &, f(xo) + e[, there is a ball B(xo; 8) mapped into it by f; that is,

— flxo)] < &. :|x < 8= |f(x) — x0]8>0 Ve>033 Ex. 6 Let X be any space, and for each pair of maps f, g: X — Y, define p:X—Y x Y by x —( f(x), g(x)). Then @ is continuous if and only if both f and g are continuous. For, note that ¢~ 1(U x V) = f~YU) ng~YV); since a subbasis for ¥ x Y is all sets of form U x Y, Y x V, where U, V belong to a subbasis for Y, 8.3(2) gives the result. Thus, for example, the map E!' — E? given by x — (cos ¥, sin x) is continuous. Let £ be the first uncountable ordinal number, and let (Vickery). Ex. 7 [0, £[ be the subspace of the ordinal space [0, £] (cf. 3, Ex. 5). Then each continuous @: [0, [ — E* must be constant on a tail [B, £[.

We first assert that Vne Z* Ja, < @V € > o, | p(§) — p(e) | < 1/n. For, if this were not true, then Ang Vo < 23 ¢ > a: | ¢(€) — p(e) | = 1/n,. We could then use induction to construct a sequence {£; | i € Z*} such that both & < &1 and | (&) — @(€i+1) | = 1/n, for eachi: choosing £&; = 0and assuming &, - - -, & defined, let £, .1 be the first element in [&;, [ satisfying the hypothesis. The {£,} thus found would then have a least upper bound y < £ and then ¢ would not be continuous at y: any basic nbd ], ¥] contains some §¢; and therefore all ¢, & > 1, so cannot have image contained in the nbd Je(y) — 1/3n,, @(y) + 1/3n[ of @(y). Our assertion is therefore established. Now let 8 be an upper bound of the {c,};

then B < £ and ¢ is constant on [B, Q[: if {€[B, 2[, then we have both @) — plen) | < 1/nand | @(B) — ¢(a,) | < 1/n for every n,so | p(0) — @(B)| < 2/n

for all n, and therefore ¢({) = ¢(B).

9.

Piecewise Definition of Maps

In analysis, continuous functions are frequently defined piecewise: For example, a continuous function on [0, #] C E! may be constructed by using different formulas on each [, i + 1], adjacent functions agreeing on the common end point. We give here the general formulation of this process.

9.1

Definition A family {4, | « € .2} of sets in a space X is called nbd-finite if each point of X has a nbd V such that VN 4, ;é & for at most finitely many indices «.

This concept is related only to the “position” of the A4, in X, and is completely unrelated to the intersections that occur among the A4, themselves. Ex.1 {A,| o« €.} may be nbd-finite, even though each 4, intersects infinitely many other Ag:in E!, take the family {4, | ne N} with 4, = {x | x > n}. The family {4, | « € &/} may not be nbd-finite, even though no A, other A;: in E!, take the family {4, | x € E'}, where4, = {x}.

intersects

any ‘

|

|

82

Chap. III

Topological Spaces

The main property of nbd-finite families is 9.2

Let {4, | « € .2/} be a nbd-finite family in X. Then:

(1). {4, | @ € &} is also nbd-finite.

(2). For each & C o, | {4; | Be B} is closed in X. Proof: Ad(1). Given x, thereis anbd U(x)suchthat 4, " U(x) = & for all but at most finitely many «; since

A,

NU@K) = o > A, CEU@K) > A, CECU(x) > A,N U(x) = o,

this proves (1).

_

Ad (2). Let B = |J A4. For each x € B, there is, by (1), a nbd U meeting at most finitely many A, say, 4,, - -+, 4, ; then U N fi €A, is a nbd of x not meeting B so, by 2.3, ¥B is open.

1

The following simple result on coverings is very important; we shall derive several of its consequences as we proceed.

9.3

Theorem

Let {4, | ac .2/} be a family of sets that cover the space

X; that is, X = [

or

4,. Assume that either:

(1). All the A, are open, (2). All the A4, are closed, and form a nbd-finite family.

Then B C X is open (resp. closed) if and only if each BN 4, is open (resp. closed) in the subspace 4,,.

Proof: 'The necessity of the condition is clear, from definition of the relative topology. For the sufficiency: Case (1): Assume that each BN A4, is open in the open A4,; 7.4 shows BN A,openin X,sothat B=BNnX=Bn{J4d,=UBnNA,is also open in X.

If each B N 4, is closed in the open 4,, then

A, — (BN 4)=A,N¥B is open in 4,4, and we Case (2): Assume shows BN A4, closed so by 9.2, B = U {B

find that ¥B is open in X. that each BN 4, is closed in the closed A4,; 7.4 in X. Since {4,} is nbd-finite, {B N 4,} is also, N A4,}is closed in X. The remaining case, where

each B N A, is open in the closed 4,, is treated as above.

83

Continuous Maps into E*

Sec. 10

The general formulation of piecewise definition of maps is

9.4

Theorem Let X be a space, and {4, |« such that either: (1). The sets 4, are all open,

- or

€ &/} a covering of X

(2). The sets 4, are all closed, and form a nbd-finite family. For each ¢ € &7, let f,: A, —

Y be continuous and assume

fol AeN A = f3| A, N Ag for each

that

(«,f)eZ x of. Then there

exists a unique confinuous map f: X — each f,; that is, Va: f | 4, = f,.

Y, which is an extension of

Proof: 'The existence of f comes from I, 6.7. To show continuity, le¢ U C Y be open; then f~Y(U)N A, = f;*(U) and so is open in A, for each a. According to 9.3, this means that f~1(U) is open in X, as required. 9.5

9.6

10.

Remark In the case that the 4, are closed, it is evident that some restriction on their position is required: In E!, [a, 8] = v {x | x € [a, b]}, and if the closed sets {x} are used, it is clear that 9.4 cannot be true. Remark 'Theorems 9.3 and 9.4 have wide applicability. In many cases, it is possible to find a nbd-finite covering of a space X satisfying 9.3, by sets {4,} that have “nice properties”; in such cases, the topology of X can be expressed by using the sets {4,} alone, and the continuity of an f: X — Y follows from its continuity on each A4, (see VI 8).

Continuous Maps into E’

Let X be an arbitrary space, and f, g: X — E!. Because E! has an algebraic structure, maps can be combined by performing algebraic operations on their values at each point. Thus we can form

x — f(%) + g(x), x — f(x)-g(x), x — c-f(x) (¢ a real constant); these

are

denoted

by f + g, f-g,

operations performed. these maps. 10.1

c-f,

respectively,

In this section, we

to indicate

the

consider the continuity of

f: X — E' is continuous if and only if for each real b, both the sets {x | f(x) > b} and {x | f(x) < b} are open.

- Proof:

The sets involved are the inverse images of, respectively, the

subbasic open sets {y | ¥ > b} and {y | y < b} for the Euclidean topology of E1, so 8.3(3) applies. Ex. 1 The requirement that both types of sets be open cannot be relaxed: If A =10,1[ C E'andc,: E! — E'is its characteristic function, c, is not continuous [{x | ca(x) < 1} is not open], yet all sets of type {x | c4(x) > b} are open.

84

Chap. 111

Topological Spaces

Proposition 10.1 indicates that a splitting of the notion “continuity” for maps X — E! is appropriate. 10.2

Definition An f: X - Elis upper semicontinuous if for each real b, {x | f(x) < b}is open; it is lower semicontinuous if for each real

b, {x | f(x) > b} is open. Clearly, f is continuous if and only if it is both upper and lower

semicontinuous;

equivalently,

by

{x| —f(x) < =¥}, f is continuous upper (or lower) semicontinuous.

observing if and

that

{x|f(x) > 8} =

only if both f and

—f are

The proof of the following theorem is not the simplest, but it does generalize to other situations in analysis (for example, measure theory). 10.3

Theorem Continuity is preserved under the usual operations of analysis. Precisely, let f, g: X — E! be continuous. Then:

(1). (2). (3). (4).

| f|® is continuous for af + bg is continuous f-g is continuous. If f(x) # 0 on X, then tinuous wherever it is

each o > 0. for each pair of real constants, a, b. 1/f is continuous (that is, 1/f is condefined).

Proof:‘ Ad (1). For b < 0, we have {x| | f(x)|* < b} =

11 f @I > B = X.

& and

For b > 0, we find

o] [f)]* > b} = {x]f(x) > bV}U {x]|f(x) < —bY, /@ < b = {x[f(x) < 8V} {x|flx) > —bY}, which, by continuity of f and 10.1, are therefore open sets. Ad (2). We first show that for any real a, af is continuous:

Note

’

that

{x|af(x) > b} = {«| f(x) > bla}

if a >0

= {x|f(x) < bja} if a < 0;

this shows that (f continuous) = (af lower semicontinuous). In particular, —(a-f) = (—a)-f is lower semicontinuous also, so that af is continuous. Now we need show only that f + g is continuous; but

(] f(®) + g(x) > By = U x| f(#) > b — Y0 {x|g(x) > A} | Areal) is a union of open

since —f and

sets, so f + g is lower semicontinuous;

—g are continuous,

semicontinuous also.

similarly,

—(f + 2) = (—f) + (—g) is lower

35

Continuous Maps into E*

Sec. 10

Ad (3). Wehavef-g = [| f + £]2 — | f — £]?], so use (2) and (1). Ad (4). Since f(x) # 0 on X, we have {x | 1/f(x) > b} equal to

[{x /(%) > 03N {x]bf (%) < BJU x| f(x) < 0} {x | bf(x) > 1}]; since f and bf are continuous,

this shows

1/f is lower semicontinuous.

Similarly, —(1/f) is lower semicontinuous.

For maps into the extended real line £ (7, Ex. 2), the criterion for continuity is evidently formally the same as 10.1, so we can speak of upper

and

of lower

semicontinuous

maps.

Furthermore,

E!

has

the

feature that for any A C E, sup 4 and inf 4 always exist.

10.4

Corollary Let {f, | €2/} fo: X — E*. Then:

be any family of continuous

maps

(a). M(x) = sup{f,(x) | « € &} is lower semicontinuous. (b). m(x) = inf{f,(x) | « € &} is upper semicontinuous. Furthermore, if & is finite, both M and m are continuous. Proof: Since M(x) > b if and only if at least one f,(x) > b, we have the identity {x | M(x) > b} = |J {x | fu(x) > b}, which proves (a). Noting that m(x) =

—sup{—f,(x)} proves (b). Furthermore, whenever

R(7) is finite, then M(x) < b if and only if all f(x) < b, so that {x| M(x) < b} = (N {x ]| fu(x) < b}; being a finite intersection, this set is certainly open and proves continuity of M (x) in this case; that of m(x) follows as before.

We will need later the analogue of the Weierstrass M-test, 10.5

Letf,:X—E'i=1,2,...beasequence of continuous maps such @

that | fi(x)| < M, for each i, where Z M, is a convergent series i=1

of reals. Then f(x) = Z fi(x) exists and

is a continuous

map

1

f.: X —

EL. n

@

Proof:

Let s,(x) = Z fi(x); from domination by ZMi we obtain 1 1 Ve>03An,Vn >mn,Vx:|s,(x) — s, (x)] < e so that s, converges to funiformly on X. The continuity of f at each x, (¢f. 8.4) follows from | f(x) — f(xo)|


7}, | x < s}, where 7, s are rational. Show that 7 (2) is the Euclidean topology R. Is this still true if 7, s are restricted to be numbers of the form k/2" and n arbitrary)?

Section 4

SN

1. Determine the closure, derived set, interior, and boundary, of the following sets: (a) The rationals in E'; (b) the Cantor set in E'; (c) the set

{(r1, 72) | 71, 72 rational} C E2; (d) {(x,0) |0 < x < 1} C E2, Show that E®* x 0 C E® x E' = E**t s closed. Let A

C E!

be a bounded

set.

sup 4 € 4.

Show

Under

what conditions is

S & A

sup Ae A’? Prove: G is open in X if and only if GN A = G N A for every A C X. Show that (4" = @) = (A4 is closed). LetA = {1/m+ 1/n|mne Z*} C E*. Show that A’ ={1/n|ne Z*} v {0}, A" = {0}. 7. Let {A, | « €57} be any family of sets in X. Assume that |J A, is closed.

o >

Prove J 4, = U A.. Prove: Fr(4) = @ if and only if A4 is both open and closed. Prove the following formulas: (@). Ft[Fr{Fr(4)}] = Fr[Fr(4)]; (b). Fr[Int(4)] C Fr(4); (c). Int(4 — B) C Int(4) — Int(B).

10. Assume

that

Fr(A4) N Fr(B) =

@.

Prove:

Int(4 v B) = Int(4)

vV Int(B)

and Fr(4 N B) = [4 N Fr(B)] U [Fr(4) N B]. 11. For what spaces X is the only dense set X itself? 12. Let E and G be dense in X. Prove: If E and G are open, then E N G is also dense in X. 13. Let D be dense in X. Prove: D N G = G for every open G C X. 14. Leat % be a subbasis for X, and D C X suchthat U " D # & foreach Ue %. Does this imply that D is dense in X? 15. 1n 3, Problem 6, what is the closure of {x} in.7 ,? For what points is {x} a closed set? An open set? 16. Assume (X, J) a space with the properties: (a) The intersection of any family of open sets is open; (b) if ¥ # y, then there is at least one open set containing some one of x, ¥, and not the other. Definex < yif x€4§. Show that this

is a partial ordering and that the topology 7 j (3, Problem 6) is precisely .

17. In X x Y, show that 4 x B = 4 x B; Fr(4 x B) = (Fr(4) x BYu (4 x Fr(B)).

Int(4 x B) = Int(4) x Int(B);

92

Chap. 111 exterior,

18. The

19.

Ext(4),

of a set A C X

is defined

Topological Spaces by

Ext(4) = Int(€4).

Prove: (a). Ext(4 U B) = Ext(4) N Ext(B); (b). A N Ext(4) = &;(c). X = Ext (2); (d). Ext [€ Ext(4)] = Ext(A4). If every countable subset of a space is closed, is the topology necessarily

discrete? 20. A point a € A is called isolated whenever ae 4 — A4’. A set is called perfect if it is closed and has no isolated points. Prove: (1) If 4 has no isolated points, then A is perfect. (2) If a space X has no isolated points, then every open set and every dense set in X also have no isolated points. 21. Call a set residual if its complement is dense, and call it nowhere dense if its closure has empty interior. Prove: (1) A nowhere dense set is a residual set. (2) A is nowhere dense if and only if 4 C ¢A. (3) The union of a residual and a nowhere dense set is a residual set. (4) The boundary of a closed (or

open) set is nowhere dense. (5) For any set 4, both AN%A4 and AN €A are residual. (6) The boundary of any set is the union of two residual _sets. _ 22. An open set U is called regular if U = Int{(U); a closed set 4 is called regular if A

a. b. c. d. ~e. f. g.

= Int(A4).

Prove:

If A is closed, then Int(A) is a regular open set. If U is open, then U is a regular closed set. The complement of a regular open (closed) set is a regular closed (open) set. If U, V are regular open sets, then U C V if and only if U C V. If A, B are regular closed sets, then 4 C B if and only if Int(4) C Int(B). If A4, B are regular closed sets, so also is 4 U B. If U, V are regular open sets, so alsois U N V.

Section 5

1. Let A —» u(A) and 4 — v(A) be two closure operations. Assume that vou(A) is u-closed. Prove that A —» vou(A) is a closure operation and that vou(A) is in fact the intersection of all sets containing A that are closed in both # and v. Finally, show that uocv(A) C vou(A). 2. Let p: X — ZP(Y) be amap ofa set X into the set (Y). For 4 C X, letep(A4) = U {p(x) | xe A}; and for B C Y, letp~1(B) = {x | (p(x) C B) A (p(x) # 2)}Show that u(4) = ¢~ *op(A) satisfies (1)~(3) and (i) of 5.1. 3. Let X be a space, and let 7 be an operation associating with each pair of sub-

sets 4 and B a set 7(4, B) C X subject to the conditions:

a. b.

74, BuC)v (B, Cu A (2, X)= 2.

c. d.

(4, ¢4) C 4. 7(4,B) C A v B.

=14 v B,C)v14,B). |

Show that, necessarily, 7(4, B) = (4 nB) U (4 n B). Section 6 1. A set is bivalent if it is both an F, and a G;. Show that the complements, finite unions, and finite intersections of bivalent sets are bivalent.

For the remaining problems, we assume that all closed sets are G,-sets.

93

Problems

2. Prove: Every F, is the disjoint union of bivalent sets. (Write F = |J F; = FLulF,

3. Prove:

- Fjyv..-[F,

For

sequence

every

{H}

— (Fp-1V:--UF;)]

sequence

of F,-sets

{F;} of F,-sets,

with

U--.

there

H; C F; for every

and use Problem 1.)

exists ¢ and

a pairwise U H; =

U F,.

disjoint [Write

F, = L;J F;, where the Fy; are bivalent and disjoint; arrange the pairs (7, j) in a linear order, and repeat Problem 2.] Prove:

If {G;}

is any

sequence

of G;-sets

with

(| G; =

&,

there

exists

a

1

sequence of bivalent sets {B} with G, C B, and

(1] B, =

@.

(Set F; = €G; in

Problem 3 and take B, = ¥ H,.) Prove: (a) If G, H are disjoint G,-sets, there exists a bivalent set B with H C B and BN G = g. (b) If Gis a G, and F an F, such that G C F, there exists a bivalent B with GCBCF.

=

Section 7 In E1, show that for any set 4, sup 4 € 4. Describe the relative topology of {z | |2| = 1} as a subspace of E?2. Show that the rationals, as a subspace of E!, do not have the discrete topology. Let K C E! be the set of irrationals in ]0, 1[. Expressing each x, y € K as decimals, define d(x, y) = 1/nif x and y have their first (n — 1) digits identical, and their nth digits different. Let B(x; 1/n) = {y € K| d(x, y) < 1/n}. Show that the {B(x, 1/n) | x € K, n€ Z *} is a basis for a topology 7 on K, and prove that 7 coincides with the induced topology of K as a subspace of E*. Let A C B be open in B. Show: For any set S, 4 NS is open in B N S. Let Y,, Y, be (not necessarily disjoint) subspaces of X, and 4 C Y; n Y,. Assume that 4 is open (closed) in Y, and open (closed) in Y,. Prove: 4 is open (closed) in Y; U Y,. Let 4 C X be closed and U C 4 open in A. Let I be any set open in X with UCV. Prove: U v (V — A)is open in X. a. Let D be dense in X. Give an example to show D N A need not be dense in A.

b. Prove: If 4 is dense in B C X, then 4 is dense in B. Let X be the set of 3, Problem 6, with the topology 7 z. Let A C X, and using the the 10. For sets

Let

induced ordering on A, obtain 7 z(A). Show that 7 z(A) coincides with induced topology on A as a subspace of X. any linearly ordered set X, let J o(X) be the topology with subbasis all of form {x | x > a}, {x | x < b}. In E', .7 o(E!) is the Euclidean topology.

4 = {0} v {x| [x|] > 1}. Show

that

its topology 7 o(4)

as a linearly

ordered set does not coincide with its topology as a subspace of E*. . 11. Let X be any space, and let Y be a closed subspace. Let 4 C X be any set, and let Hbeanbdof AN Yin Y. Prove: AN (Y — H) = &. 12. Let X be any space, and assume that X = E, U E,, where E, and E, are closed in X. Let B C E; be such that BN E; C Q where Q is open in E,. Prove: B C Int(E, U Q).

94

Chap. 111

Topological Spaces

Section 8 1. Let & be Sierpinski space and let 2 be the discrete space {0, 1} (¢f. I, Ex. 3). Let f: & — 2 be the identity map. Show that f is not continuous, but that

f1:2>%is. 2. Let ¢ be the characteristic function of [0, 1[ C E*. Is ¢|[0, 1] continuous at ' = 1? On [0, 1]? x = 0? Atx 3. Let X be any space and ¢, the characteristic function of 4 C X. Show that c4: X — E* is continuous if and only if 4 is both open and closed in X. 4. Let C be the set of 3, Ex. 7. (a) Define ¢: C — E* by ¢( f) = f(1). Show that @ is continuous in the % topology, but is not continuous in the A topology. (b) Define : C — E* by

W = [ feax. 5. 6.

7.

8. 9.

10.

Show that i is continuous in both the 4 and % topologies. (c) Are either of these maps continuous in the .Z topology (3, Problem 5)? Under what conditions is the bijective map 1: (X, 7 ;) — (X, 2) not continuous, and with inverse not continuous. Let X be the space in |, Problem 1(a). State a necessary and sufficient condition that f: X — X be continuous. Let X be the space in 1, Problem 4. Show that f: X — X is continuous if and only if it is order-preserving. Let Z* be taken with the topology of I, Problem 5. Show that f: Z*+ — Z+ is continuous if and only if (m divides n) = (f(m) divides f(n)). Let X be the set of 3, Problem 6, with topology 7 . Derive a necessary and sufficient condition for continuity of f: X — X. Prove that the following three statements are equivalent: a. f: X — Y is continuous.

b. f(4") C f(A) for each A C X. c. Fr[f-1(B)] C f ! [Fr(B)] for each B C 11.

Y.

f | X f | X; and Let X = X, U Xy and f: X — Y. Assume at x € X; N X,. Show that f is continuous at x.

to be continuous

Let f: X — Y be a map and 4 C X. Give an example showing f| 4 continuous, although f is not continuous at any point of A4. t is f ~1(B) 13. Let f: X — Y be continuous. If B C Y is a Gy(resp. F,), show tha 12.

also a Gs(resp. Fy).

14. Construct an example of a map f: X x X — E* continuous in each variable separately but not continuous on X x X (use X = E?! for simplicity).

Section 9 1. Let {4,} be a nbd-finite

Ay, (¢ = 1,

closed covering

of X.

Let xo € X,

.-, n) containing x,, let V; be a nbd of x, in 4,;.

and

for each set

Show that 0 1

contains a nbd of x; in X, 2. Let X be any space, and {4, | n € Z*} a family of sets in X such that 4,

for each n € Z*. nbd-finite.

Prove that, if F]Hn 1

=

Vi

> 4.1

¢, then the family {4, |ne Z*}

is

Problems

95

3. Prove: a. Any open set in E' is the union of a pairwise disjoint family of open intervals. Give an example to show that this is not true for E*, n > 1, if “open intervals” is replaced by “open cubes.” b. Any open set in E" is the union of at most countably many nonoverlapping closed cubes [Hint: Generalize the following procedure for E2: For each neZ*, let H, be the network of closed squares formed by all the lines x = p/2", y = q/2", p,q€ Z. Let S; be the set of all the closed squares of H; contained in the open-set G, and proceeding inductively, let S,,; be the set of all the closed squares of H, ,; contained in G — (S; U

++-- U S,). Then G = G S,]. 1

c. Show that a non-empty open set in E™ cannot be the union of a nbd-finite family of nonoverlapping closed cubes.

Section 10 1. Letf: X — E* be such that for each rational 7, {x | f(x) > 7}is open.

f is lower semicontinuous. 2. Let .

sections.

Show that

'

be a family of sets in X closed under countable unions and finite inter-

Call f: X — E' an . -function if f~1(U) € .

for each subbasic U

of E'. Show that 10.3 is true, with “continuous” replaced by “.#-function.” 3. Let E be the space of real numbers with subbasis all sets of form {x | x > b}. Prove f: X — E is continuous if and only if f is lower semicontinuous, as a map into E*, v 4. Show that the sum, inf of finitely many, and sup of an arbitrary family, of lower semicontinuous functions is lower semicontinuous. Show that if f, g are >0 and lower semicontinuous, so also is f-g. 5. Prove: f is lower semicontinuous if and only if: Ve>0Vaxod

Ulxg): xe Uxo)

= f(x)

> f(xo) — &

= f(x)

< f(xo) + &

f is upper semicontinuous if and only if: Ve>0Vxod

U(xo):xe

U(xo)

Equivalently, f is lower (upper) semicontinuous if and only if for each real b, {x | f(x) < b} ({x | f(x) = b}) is closed. 6. Let C? be the set of all continuous real-valued functions on [0, 1] that have a continuous derivative on [0, 1]. Let J be the topology having

U f) = {g|Vx:|gx) — f®)| < &} as a basis. Define the length

If) = f: VT + fi(x)? dx. Show that [: C! — E! is lower semicontinuous. 7. Let { f»} be a sequence of continuous real-valued functions on a space X. Show that the set of points x € X for which the sequence { f,(x)} converges is an F,; (that is, a countable intersection of F,-sets). Section

|1

1. Let p: E' x E' — E* be the projection (x,y) — x. Show: (a) that p is an open mapping; (b) that p is not a closed map. (Consider p{(x, y) | xy = 1}.) Thus “bijective” in Ex. 2 cannot be replaced by “surjective.”

9

Chap. 111

Topological Spaces

2. Let p(x) be a polynomial on E!. Show that x — p(x) is a closed map of E1, 3. Show that a continuous bijective map f: E! — E! is an open map. 4. Give an example to show that a continuous open map need not map the interior of a set onto the interior of the image. 5. Let X be any space, and f: X — E™ an open mapping. Denote the distance of f(x) € E™ to the origin by | f(x) |. Let A C X be any set. Prove that | f(a) | < sup {| f(x) | | x € A} for every a € Int(4). [Hint: Given a € Int(4), there is, by 11.3(4), a ball B(f(a); €) C f(A)]. Since a nonconstant analytic function of a complex variable is an open map on its domain of definition, this result implies “maximum modulus theorem” of complex analysis. 6. Show that f: X — Y is open if and only if f~[Fr(B)] C Fr[f~1(B)]for each BC Y. 7. Letf: X — Yand g: Y — Z. Prove:

a. If f, g are open (closed), so also is g o f. b. If g o f is open (closed) and f is continuous, surjective, then g is open (closed). c. If gof is open (closed) and g is continuous, injective, then f is open (closed). 8. Let f: X — Y be open (closed). Show that if 4 = f~*(B) for some B C Y, then f| A is also an open (closed) map into B.

9. Let{B, | « € &} be an open, or nbd-finite closed, coveringof Y. Letf: X — Y be such that for each a € 27, f | f ~*(B,) is an open (closed) map of f ~1(B,) into B,. Show that f is an open (closed) map. 10. Prove: A surjective map p: X — Y is a closed map if and only if it satisfies the condition of 11.2(1). 11. Let p: X— Y be a closed surjection. Prove: If U C X is open, then

Fr[p(U)] C p(U) N p(X — V). 12.

Let p: X —

Y be closed.

Let U C X be open, and p~1(y) C

"y € Int[p(U)]. 13. Let p: X — Y be closed. Prove:

U.

Show that

If A C X is closed, then p| A4 is a closed

mapping of 4 into p(A4). 14. Prove that the following three statements are equivalent:

a. p: X — Y is a closed map. b. If U C X is open, then {y | p~(y) C U} is open in Y. c. If AC Xis closed, then {y | p~2(y) " 4 # @] is closed in Y. 15. Let c¢: I — 2 be the characteristic function of [0, 1] C I. Show that ¢ is surjective, open, closed, but not continuous. 16. Let p: X — Y be an open and closed map. Let ¢: X — I be continuous, and for eachy € Y, let ¢(y) = sup [p | p~2(3)]. Prove: ¢: Y — I is continuous. Section

12

1. In E? let V2 = B(0,1). Show V2 >~ I x I. 2. Show that the space of # X n matrices in 3, Problem 4, is homeomorphic E™, 3. Show: a. (Xx Y)x Z 2 X x (Y x 2Z), b. XxY2>Yx

X.

to

Problems

4. Letf: X —

97

Y be a map, and G(f) C X x Y the subspace {(x, f(x)) | x € X}. Show that x — (x, f(x)) is a homeomorphism if and only if f is continuous. 5. Let 4, B be countable dense subsets of E!. Show A ~ B.[There is even a homeomorphism of E! carrying A onto B.] Letf: X — Y and g: Y — X be embeddings. Show that (see 11, 7, Problem 7) X and Y can be written as disjoint unions: X = X; U X,, Y = Y; U Y, such thatf| X:: X; & Yiandg| Ya: Vo & Xo. Let {B, | « € &/} be an open, or nbd-finite closed, covering of Y. Letf: X - Y be continuous, and assume that for each « € &7, f | f ~1(B,) is a homeomorphism of f~Y(B,) and B,. Prove that f: X ~ Y. Show that the map x — e** of [0, 27[ onto S is not a homeomorphism.

Cartestan Products

IV

In this chapter, we define the cartesian product topology for an arbitrary family of spaces and derive some of its basic properties.

1.

Cartesian Product Topology

I.I

Definition

Let {Y, |« ./} be any family

of topological spaces.

For each e« e/, let J, be the topology for Y,. The cartesian product topology in [| Y, is that having for subbasis all sets

Uy = ps '(Up), where Uy ranges over all members of I, and B over all elements of /. If any one factor Y, = @, then the cartesian product is also empty and the topologies of the non-empty factors then play no role in determining that of the product; we wish to exclude such behavior, so we shall assume throughout this book that each factor of any cartesian product of topological spaces is non-empty. Ex. 1

n

If o7 is a finite set, the cartesian product topology of [|

Y, coincides

a=1

with that previously defined in III, 3, Ex. 6. Ex. 2 Because of I, 9.6, it follows at once that the basic open sets are all those sets of form Ual

X oo

X

Uy,

X

H{Yfilfi¢

1y * 0y O}

=

Op;il(Ua,),

where 7 is finite and each U,, is an open set in Y,,; we will denote this set by {Ugy, * + + Ug,>.

Thus,

R(&/)

whenever

= n


K, this is no longer true: 98

In fact, a set of form [|

U,, where

99

Cartesian Product Topology

Sec. 1

each U, # Y, and is open in Y,, is never an open set in the cartesian product: because each basic open set restricts only finitely many coordinates, no basic open set can be contained in [| Uy, so (111, 2.3) [| U, is not open. a

a

If R(&)

Ex. 3

>

R,, and each

Y, is a discrete space having more than one

point, [| Y, is never a discrete space, for since each a, € Y, is an open set, it a

follows from Ex. 2 that no point {a,} € [ [ Y, is an open set. The cartesian product a

topology is frequently used in this way, to create a nontrivial topology in infinite families of objects, starting from the discrete topology in each. Ex. 4 Itisuseful to observe thatif ¥ is any set open in [| Yy, then p (V) = for all but at most finitely many «: any such

-,

I/ contains a basic U = {(Uy,,

U,,> and po(U) C po(V) for all .

As Ex. 2 indicates, results for finite cartesian products may not carry over formally to infinite cartesian products, so these two cases will be treated separately.

1.2

Let ()

be arbitrary. Then, in the space [ {Y, | a € &7}:

(1). For each « € &, let 2, be a subbasis for the topology 7, of Y,.

Then the family {(V,) |all VyeZXy; all Beof} is also a subbasis for the cartesian product topology in 1_[ Y,.

(2). If A, C Y, for each @ € &, then 1_[ A, I_I A (that is, in contrast to Ex. 2, the cartesian product of closed sets is always closed). 3). If4,C Y, for each o € &, then I—I A, as a cartesian product

of spaces (each having the induced topology) has the same topology that it receives as a subspace of | [ Y.

(4). Let y° = {y;} be a fixed element in | [ Y,, and let = {{x.} | {x,} and {y;} differ in at most finitely many coordinates}. Then D is dense in [| Y. Proof:

(1). This is immediate from I, 9.5, and is left for the reader.

(2). Assume

{y,} €[] A,;

we

show

that

Va:y,€d,,

that

is,

{va€el ] A,. Let y, € U,, where U, is open in Y,; since {y,} € CU,>,. o

we must have (111, 4.3)

g # Uy N1 Ay = (U N Ay) x TT{4g|B # o} and so by I, 9, Ex. 1, we find U, N 4, # @. This proves y,€A4,.

The

100

Chap. IV

converse inclusion is established by ][] 4,, then for any nbd (Vo) €Uy, so that CU,, -+, Uy> N [T4,

#

Cartesian Products

reversing these steps: If {y,}e -+, Uy>, each U, N4, # o

.

(3). This is left for the reader. (4).

Let be any basic open

set; since all but the

@y, - - -, &, coordinates are unrestricted, this set contains a point having the coordinates y; for each 8 # «;, - - -, «,. By III, 4.13(3), D is dense.

For the case N(&)

1.3

< X, we have

n

Inthespace ] Y;: i=1

(1).

(HAi)

= (4} x Ay x x 4,

1

U(leAéxzsx---xZ,)

U--U (A

x Ay x---x

A).

2). Int (H A,) — [11 Int (4,). n

1 n

(3). Fr (1:[ Ai) = [Fr(4)) x 4y x --- 4,]

U [4; x Fr(d,) x -+ x 4,] V.- U[4; x 4y x -+ x Fr(4,)]. Proof:

(1) is proved by induction, from the result for two factors:

(a,)e(4 x BY < (a,b) (@ x B) = (@, b)

=[(x4 BJU[4a x (B)- })] =[(4 —a) x BJU[4 x (B - b)]

zZ,

and from this it follows that

if {al,'”’an}n'%u:{ajy"';ak}?é

if {a,

e iNA,

1%}

= o

(U,, ---, U, > isopenin |[ Z,. Finally,

q is bijective because {&/, | u € A} is a partition of /.

H

By III, 12.2, ¢ is

therefore a homeomorphism, and the proof is complete.

2.5

Theorem

Let X(&/) be arbitrary. Foreach ¢ € ., let f,: X, — Y,

be a map. Define [Tf,: [ [ X, =]

] Y, by {x.} = {fu(x.)}-

Then:

’

:

(1). If each f, is continuous, so also is [ ] f,. (2). If each f, is an open map, and all but at most finitely many are surjective, then [ f, is also an open map. Proof:

Ad (1). Let (V,>

be a subbasic

(IT1f) Vo = {f;(V,)>, which By II1, 8.3, [ ] f, is continuous.

is open

open set in [[

Y,; then

because f, is continuous.

103

Slices in Cartesian products

Sec. 3

Ad (2). Each basic NR,. Let Y, = Yforeacha,andlet Z = [ [ Y,. Then each cartesian product Z x Z x ---x to Z.

Zconsisting of < NR() factors Z is homeomorphic

Proof: If there are X < (&) factors Z, the cardinal number of the factors Y present in Z x --- x Z is X-R() = R(F), so 2.6 applies.

3.

Slices in Cartesian Products

In E? = E' x E1, we can identify a line parallel to the x-axis with E?; this idea extends to arbitrary cartesian products.

3.1

Definition

Let

[[

¥,

be

an

arbitrary

cartesian

product,

and

a

¥° = {y3} a given point. For each index B, the set

S8

=Yy x [T{yale#BCl]Y, a

is called the slice in [ [ Y, through y° parallel to Y. o

Ex.1 In E® = E! x E' x E, with y° = (xi, x3, x5), the slice S(y°;1) = {(x, x3, x3) | x € E'}, and so is a line parallel to the x-axis going through »°.

3.2

Theorem

The map s;: Y; — S(»°; B) given by

ys—>ys x [T{yz|a # B} is a homeomorphism of Y, with the subspace S(y°; B) = s.

104

Chap. 1V

Proof:

Cartesian Products

We first note that, because of I, 9.6, the open sets

N

S

of the subspace S are &, S, or S N (U,), where U, is open in Y. Thus S ({Uyyy -+, Ug,> NS) = @, Yy, or Uy, hence it is always open, and

therefore s; is continuous. Since the projection pg: [ [ Y,—

¥, is

a

continuous, so also is p = pg | S; and because pos; = 1, s;0p = 1, an application of III, 12.3 shows that both sz and p are homeomorphisms. 4.

Peano Curves

As one simple application of the results in this chapter, we prove the existence of space-filling curves. For

each

n = 1,2,..-,let

A, be the discrete space {0, 2}; in I, 9,

Ex. 3, we have seen that if C is the Cantor set, the map ¢: [[ 4, > C given by

" - a 1

is a bijection of the set [ 4, onto the set C. 4.1

Taking [] A, with the cartesian product topology, and C C E*! with the subspace topology, ¢: [ 4, ~ C.

| Proof:

¢@iscontinuous: Givenanyc = Za,- 3-1eC,and nbd B(c; )N C,

choose N so large that Z 2:-37t < e;theng({ay, - - -, ayy) C B(c; &) N C. N

¢ is open: Given any x = {4;} and nbd (g,, - - -; @;) containing x, let N = max {iy, - -, i} and W = B(e(x); 1/3¥+1) N C; then wC

so III, 11.3, applies.

42

(P()

Being bijective, ¢ is a homeomorphism.

Let] = [0,1] C E?, and let

: [] A, — I be the map

= a {a} — Z 2{+‘1' 1

Then ¢ 1s a continuous surjection.

Continuity follows as in the preceding proof and, by using Proof: dyadic expansions, the surjectivity is obvious.

The n-cube I C E™is | [ I; the space I*® = [| I is called the Hilbert 1 1 cube.

We

now have

105

Problems

4.3

For each k < o0, there exists a continuous surjection of the Cantor set on I*.

Let k < oo be given. We have ¢~1: C =~ ] 4,, and by 2.7,

Proof:

an h:[[ A, 2[4,

[1A4, (k < o

x---x

factors). Using

2.5

and

4.2, we obtain a continuous surjection: k

H¢HAn

Xooox

[[TAp—=I

I=1IF

x---x

1

so that

(fi¢)oho¢-1:c—>1k 1

is the desired map.

By a “curve” in a space Y is meant the image f(I) of a continuous map f: I— Y. From 4.3, the existence of space-filling curves (that is, Peano curves) follows easily: 4.4

Theorem (Generalized Peano) For each k < oo, there exists a continuous surjection I — I* (that is, a curve going through each

point of I*¥). Proof: Letf:C —1I¥ =1 x---x I be the continuous surjection of 4.3, and let p, o f: C — I be the coordinate functions. Recall that C C [ is obtained by successively dropping out middle thirds. Extend each pyo f: C — I to a continuous f;: I — I by defining f, to be linear on each omitted interval. Then, by 2.3, F(#) = {f(#)} is a continuous map

I — I¥, and since F | C = f, it is surjective.

Problems Section

|

1. Let {Y, | « € &/} be a family of spaces. Assume that each Y, has a basis of cardinal number < X. What is the cardinal of a basis for [| Y,?

2. Let X(%) be arbitrary and [| 4, C [] Y.. If all but at most finitely many factors A, = Y, prove Int ([| 4.) = [] [Int(A4,)]. 3.

Let R be the real numbers with the upper-limit topology (111, 3, Ex. 4). Show that R x R is not a discrete space, but that 4 = {(x, y) | x + y = 1}, as a subspace of R x R, has the discrete topology.

4.

Prove:

I—[ A, is dense in l—I Y. if and only if each 4,

C

Y, is dense.

106

Chap. IV

Cartesian Products

Section 2 1. Prove:

The

cartesian product

topology in' [|

Y, is the smallest topology for

which all projections pg: [ | Y — Y} are continuous. a

2. Let {Y, | « € &} be a family of spaces. For each # C &7, let pa:][{Ya

be the projection.

Let 4 C [

4=

l aefl}aH{Yfi

| B € %}

Y, be closed. Prove: &

Nfinite pg'lra(D]

Section 3

1. Let & be the Sierpinski space, and & x {0} the slice in & first factor. Is & X {0} closed in & x &?

x & parallel to the

Connectedness

V

Intuitively, a space is connected if it does not consist of two separate pieces. This simple idea has had important consequences in topology and has led to highly sophisticated algebraic techniques for distinguishing between spaces. In this chapter, we obtain the basic properties of this concept and those of some of its modifications.

I.

1.1

Connectedness

Definition A space Y is connected if it is not the union of two nonempty disjoint open sets. A subset B C Y is connected if it is connected as a subspace of Y.

Ex. 1 Sierpinski space % is connected: The only possible decomposition is 0, 1, and 1 is not open. The discrete space 2 is not connected. Ex. 2 The real number system with the upper-limit topology (III, 3, Ex. 4) is not a connected space, since {x | x > a} and {x | ¥ < a} are both open sets. Ex. 3 The rationals Q C E! are not connected, since {x | x > V2 n 0,

{x | x < V2} n Q is a decomposition as required.

1.2

The only connected subsets of E! having more than Theorem one point are E! and the intervals (open, closed, or half-open). 107

108

Chap.

V

Connectedness

Y connected implies that Y is an interval: For if Y is not an Proof: interval, then by definition there mustbea, b€ Y,c€ Y witha < ¢ < b,

YN{x|x

so YN{x|x>c¢c,

< ¢} is a decomposition of Y.

Y is an interval implies that Y is connected:

If Y were not connected,

then Y = A U B, where A, B are disjoint nonempty open sets, and there would be an ae€ 4, be B that (relabeling if necessary) satisfy

a < b. Define « = sup {x | [a, x[ C A4}; then « < b, and because Y is an interval, « € Y. Clearly, « € 4y; noting that 4 = % B is closed in Y, we must have « € 4.

However, A4 is also open in Y, and since Y is an

interval, an application of III, 2.3 shows there must be a nondegenerate Je — A, « + A C A, which contradicts the definition of «. The definition

1.3

1.1 can be formulated in handier fashion:

The following three properties are equivalent: (1). Y is connected. (2). The only subsets of Y both open and closed are @ and Y. (3). No continuous f: Y — 2 is surjective.

Proof: (1) = (2). If G C Yisbothopenandclosed,and G # &, 7, then ¥ = G U ¥G shows that Y is not connected. (2) = (3). If f: Y — 2 were a continuous surjection, then f~1(0) # @, Y, and because 0 is open and closed in 2, f ~(0) is open and closed in Y. 3)=(). If Y =Av B, 4, B disjoint nonempty open sets, then A, B

are also closed,

and

the characteristic

function

¢,: Y —2

is a

continuous surjection.

Connectedness is clearly a topological invariant; even more,

1.4

Theorem The continuous image of a connected set is connected. That is, if X is connected and f: X — Y is continuous, then f(X) is connected.

Proof:

The

map f: X — f(X)

is continuous;

if f(X)

were

not

connected, there would be, by 1.3, a continuous surjection g: f(X) — 2, and then go f: X — 2 would also be a continuous surjection, contradicting the connectedness of X.

I.5

Theorem Let Y be any space. The union of any family of connected subsets having at least one point in common is also connected.

Proof:

Let C = |J A,, y,€ () 4,, and f: C — 2 continuous.

each 4, is connected,

Since

no f| 4, is surjective, and because y, € 4, for

each «, f(y) = f(yo) for all y € 4, and all a. Thus f cannot be surjective.

109

Connectedness

Sec. 1

Ex. 4 In contrast, the intersection of even two connected sets need not be connected. Furthermore, if all the A,,7e€ Z* are connected, and 4, D A4, -- -,

still

)

A,

need

be

connected:

let

Y =12

— {(x,0) |4 1, are not homeomorphic.

Proof : Assume that h: E™ ~ E!; removing one point a € E", we must have h: E" — a > E' — h(a), by III, 12.4. However, this is impossible since, by 2.2, E* — a is connected whereas E! — h(a) is not. The theorem that E” is not homeomorphic to E™ for n # m is much deeper, involving more delicate topological invariants (although the

Components

Sec. 3

111

technique is the same). Concerning I" (n > 1) and I', a proof similar to 2.3 shows that they are not homeomorphic. Thus, though there is a bijective map of the set I' onto the set I", there is no bicontinuous bijection and, as we shall see later, not even a continuous bijection (X1, 2, Ex. 4). 2.4

In E! each closed interval is homeomorphicto [—~1, +1], Theorem each open interval to |—1, +1[, and each half-open interval to Furthermore, no two of these three intervals are homeo-

]—1, +1].

morphic. Proof:

Given an interval with end points @, b, a suitable one of the

b+a b-a x exhibits a homeomorphism. To see that maps x — :2 * 2 none of the three standard intervals are homeomorphic, note that we can remove 2, 0, 1 (respectively) points without destroying the connectedness.

3.

Components

A disconnected space can be decomposed uniquely into connected “components”; the number of components provides a rough indication of how “disconnected” a space is. 3.1

Definition Let Y be a space, and y € Y. The component C(y) of y in Y is the union of all connected subsets of Y containing y.

It is evident from 1.5 that C(y) is connected. Ex. 1 Let Q C E! be the subspace of rationals. The component of each y € Qis the point y itself. Thus, even though Y does not have the discrete topology, the components may reduce to points. Y is called totally disconnected if C(y) = y for each ye Y. Ex. 2 Let Y C E? be the subspace consisting of the segments joining the origin 0 to the points {(1, 1/n) | n € Z *}, together with the segment ]4, 1] on the x-axis. Asin I, Ex. 5, Y is connected, but Y — {0} is not: in Y — {0} the component of each point is the ray containing it.

Ex. 3

In ]

Y., the component of y = {y,} is C(y) = [| C(¥.), where C(y,)

is the component of the coordinate y, in Y,. Indeed, by 1.7, C(y) is connected, so it is contained in the component K of y. If K — C(y) # @, selecty°e K — C(y); then some coordinate y; € C(3,), so that p,(K) C Y, is connected and contains ye € C(34), in contradiction to C(y,) being the component of y,. In particular, though the infinite cartesian product of discrete spaces is never discrete (IV, 1, Ex. 3) it is always totally disconnected.

112

3.2

Chap.

V

Connectedness

Theorem (1). Each component C(y)is a maximal connected set in Y: there is no connected subset of Y that properly contains

C().

(2). The set of all distinct components in Y forms a partition of Y. (3). Each C(y) is closed in Y.

Proof: (1) follows from the definition. (2). fC(y) " C(y') # o,thenby L5, C(y) U C(y’)is connected, contradicting the maximality of C(y). (3). Since C(y) is connected, so also (1.6) is C(y); by the maximality of C(y), we must have C(y) C C(y), so that C(y) is closed. Ex. 4 Components need not be sets open in Y, as Ex. 1 and Ex. 3 show. In particular, a splitting of a disconnected space into disjoint nonempty open sets is not generally accomplished simply by taking any one component as one of the “pieces.”

The number and structure of each component topological invariant:

3.3

Theorem Let f: component of X h: X ~ Y, then components of homeomorphic.

of a space

Y is a

X — Y be continuous. Then the image of each must lie in a component of Y. Furthermore, if 4 induces a 1-to-1 correspondence between the X and those of Y, corresponding ones being

Proof: If fis continuous, then f(C(x)) C C( f(x)) follows from 1.4, since f(C(x)) is a connected set in Y containing f(x). If h: X ~ Y, then because / is bicontinuous and bijective, we have both #(C(x)) C C(k(x)) and 2~} C(h(x))) C C(x), which shows that A(C(x)) = C(h(x)). The rest of the proof is trivial. Ex. 5 Ex. 6

The Cantor set is totally disconnected, by Ex. 3 and 3.3. Components are also used to decide questions of nonhomeomorphism.

As an application, we show that a strong version of the Bernstein-Schréder theorem is not valid for topological spaces: There may be a continuous bijection f: X — Y and a continuous bijection g: Y — X; yet, X and Y need not be homeomorphic (C. Kuratowski). In E!, let

X=10,1{uv{2lul,4{u{5iu---Vdn,3n+1[V{3n + 2L Y=10,1]0l34uv{5lu.-- U]3n,3n + 1[U{B3n + 2}U ...

.-,

These spaces cannot be homeomorphic, since the component 10, 1] is not homeomorphic to any component of X. However, f(x) = x (x # 2), f(2) = 1 is a continuous bijection X — Y, and g: Y — X, where

glx) =< is also a continuous bijection.

x/[2

if x €10, 1]

(x — 2)2

ifxe]3, 4

x—3

otherwise

113

Local Connectedness

Sec. 4

4.

Local Connectedness

4.1

Definition A space Y is locally connected if it has a basis consisting of connected (open) sets.

Ex. 1 E™" is locally connected, since each ball B(x; r) is connected. Furthermore, each interval in E! is locally connected. For each n = 0, S" is locally connected. Ex. 2 A space may be locally connected, but not connected, as the discrete space 2 shows. A space may be connected, but not locally connected. Let Y be the Ex. 3 space of 3, Ex. 2, y =(3,0) and U = B(y;4) n Y. Then U, and any nbd V(y) C U, is not connected: For I must intersect a ray joining 0 to some (1, 1/n) and it is trivial to verify that this intersection is both open and closed in V. Thus no basis for Y can consist only of connected sets.

Y is locally connected if and only if the components of Theorem each open set are open sets.

4.2

Proof: Let G C Y be open, C a component of G, and {U} a basis consisting of connected open sets. Given y € C, then because y € G, there is a U with y € U C G'; but since C is the component ofy and U is connected, y € U C C, showing that C is open (I1I, 2.3). For the converse, note that the family of all components of all open setsin Yisa basis as required. Ex.

4.

Observe that 4.2 need not be true for nonopen sets, as {0} U {1} C E?

shows. Ex. 5. Let Y be the space of I, Ex. 5,and Z = Y U {0}. Then Z is not locally connected, since the components of Z N {(x,y) |y < 4} are not open in Z.

4.3

A cartesian product [|

Y, is locally connected if and only if all

the Y, are locally connected, and all but at most finitely many are also connected. Proof:

Assume that [|

most finitely many

I

Y.

Y, islocally connected.

Then: (1) All but at

Y, are connected: if V' is any connected open set in

we have (IV,

1, Ex. 4) that p, (V) = Y, for all but at most

finitely many «, and projections are continuous. (2) Each Y connected: given y; € Y, and an open U C Y; containing yg, is a nbd of some y having Bth coordinate y,, so there is an open V with ye V C {(U); since py(V) is an open connected ys € ps(V) C U, this shows that Y; has a basis consisting of sets.

is locally then {(U) connected set, and connected

114

Chap.

V

Connectedness

To prove the converse, let % C o/ be the finite set of indices for which the space

Y, is not connected.

Now let x€ [ ] Y, and let V be

any open set containing x. Then xe C V for suitable

ay, - -+, a,. For each «; let V, be a connected nbd of p,(x) such that V, C U, and for each By, -+, B€ B — {ay, - -+, «,}, let V; be a connected nbd of ps(x). Then (V,, -, V,, Vs, -+, V;> is a con-

nected (¢f. 1.7) nbd of x and is contained in . Thus, ][] Y, has a basis consisting of connected open sets, so it is locally o

connected. -

Ex. 6 The hypothesis that all but at most finitely many Y, be connected is essential: If 4, = {0, 2}, we have seen || 4, is totally disconnected. But though n

each A, is locally connected, || 4, is not: Its components are its points, and since n

|] 4x is not discrete, none is an open set. n

Ex. 7 Local connectedness is evidently a topological invariant, and therefore it can be used in questions of nonhomeomorphism. Thus the space Z of Ex. 5 cannot be homeomorphic to any interval in E1. Ex. 8 Local connectedness is not invariant under continuous maps. Let X be the discrete space {0, 1, 2, - - -}, Y the subspace 0 U{l/n|n =1,2,.-.} of E! and f: X — Y the map f(0) = 0, f(n) = 1/n. Then X is locally connected and f is a continuous bijection; but Y is not locally connected.

5.

Path-Connectedness

For most purposes of analysis, the natural notion of connectedness is joining by paths. In IV(4), we have defined a curve in a space Y to be a continuous image of the unit interval I. A path in Y is a continuous mapping f:I— Y, rather than the image f(/) in Y. Thus, a path is a continuous function, whereas a curve is a subset of Y; we shall see later (Chapter

XIX) one reason for this distinction f:1— Y is a path in Y, we call f(0) € and f(1) € Y the terminal (or end) point, from f(0) to f(1), or joins f(0) to f(1).

between paths and curves. If Y the initial (or starting) point, of the path f, and say that f runs If f runs from f(0) to f(1), it is

clear that the mapping ¢ — f(1 — ¢), t € I, is a path in Y running from

f(1) to £(0). 5.1

Definition A space Y is path-connected (or: pathwise connected) if each pair of its points can be joined by a path.

Ex. 1 E™ and S™n > 1) are path-connected. E™ — B is also path-connected.

For

any

countable

B C E*,

115

Path-Connectedness

Sec. 5

Ex. 2 Sierpinski space % is path-connected: The characteristic function of 1 € I, regarded as a map I — &, is a path joining 0 to 1. Ex. 3 A discrete space having more than one point is never path-connected. Every indiscrete space is path-connected.

A trivial, but useful, reformulation of 5.1 is given in

5.2

Let Y be a space, and y, € Y any element. Y is path-connected if and only if each y € Y can be joined to y, by a path.

Proof: 1If Y is path-connected, the condition is trivially true. Conversely, assume that the condition is satisfied and that y, ¥y’ € Y have been given. Let f: I — Y run from y to y, and g: I — Y from y, to y';

0 p~1(B)/R, be the projection. Then sq 1:p~YB)/Ry, — B is single-valued, and by 3.2, is. continuous. Similarly, gs~': B — p~!(B)/R, is continuous, and since gs~1, sqg~! are inverses of one another, each is a homeomorphism (I11, 12.3). (2) follows from 1.6, since R(U) is the p-load of U. (3) is simply 3.2 in this special case. Ex. 3 Let I be the unit interval and R the equivalence relation 0 ~ 1, x ~ x (x # 0,1), which “identifies” 0 and 1. Then I/R ~ S*. For, let p: I — I/R be the identification and h: I — S be the map x — 2™, Since kp ~! is single-valued, it is continuous and also bijective. T'o see that it is an open map, note that no open subset of I having form [0, b[, or]a, 1] satisfies U = R(U); therefore such intervals must always occur in pairs for sets that satisfy this condition. It follows that each p-saturated open set is mapped by A onto an open set in E*, so hp ~! is open. Ex. 4 If p: X — X/R is an open (closed) map, R is called an open (closed) relation. This is not related to the behavior of R as a subset of X x X: In I, define x ~ y by “|x — y| is rational”’; then p(]a, b[) = I/R for each open interval so that 7 (p) is indiscrete. Thus p is an open map, but R is not open in I x I.

4.3

Theorem Let X, Y be spaces with equivalence relations R, S, respectively, and let f:X — Y be a relation-preserving, continuous map. Then, passing to the quotient (I, 7.7), the map f«: X/R — Y|S is also continuous. Furthermore, f, is an identification whenever f is an identification. Proof:

We have the commutative diagram

|

X

X/R

f

— Y

—/—> *

YIS

Since ¢ o f is continuous, and f, o p = g o f, 3.2 shows that f, is continuous. Now assume that fis an identification; then 3.3 shows first that fe o P (= gof)is an identification and then that f, is also.

5.

Cones and Suspensions

5.1

Definition Foranyspace X, the cone T.X over X is the quotient space (X x I)/R, where R is the equivalence relation (x, 1) ~ (x’, 1) for all x, x" € X.

Equivalently, TX = (X x I)/(X x 1); intuitively, TX is obtained from X x Iby pinching X x 1 to a single point. The elements of TX are denoted by {x, t>. It is trivial to verify that the map x — {x, 0> is a homeomorphism, so we can identify X with the subspace { | x€ X} C TX.

127

Attaching of Spaces

Sec. 6

Ex. 1 Though TX is, geometrically, a cone, it may have more open sets than a cone formed by possibly more elementary methods. In E?, let X ={(n,0) |ne Z *}, and let CX be the subspace of E2 obtained by joining each x € X to w, = (0, 1) by a segment. It is easy to verify that there is a continuous bijection T.X — CX; but the spaces are not homeomorphic. Indeed, let V,, = all points on the segment from wq to (n, 0) within 1/n of wo, and V' = t;.J V,.; then V is evidently not open in

CX, but it is open in T.X since p (V') is open in X 52

A continuous f: X — {x, t) — { f(x), t.

Y

induces

a

x I.

continuous

:

Define f: X x I—-Y x I by (x,t) — (f(x), t). Noting Proof: relation-preserving, we pass to the quotient and use 4.3. 5.3

by

Tf: TX — TY

map

that

f

is

Let ¥ be the interval [—1, + 1]. For‘ any space X, the suspension Definition SX of X is the quotient space (X x ¥)/R, where R is the equivalence relation (x, 1) ~ (x, 1), (x, =1) ~ (x/, —1) for all x, x" € X.

Intuitively, SX is obtained from X x ¥ by pinching to points each of the sets X x 1, X x (—=1). The elements of SX are denoted by {x, t> also. Ex.2 It is easy to verify S S° >~ S!; we will see later, once compactness has been discussed, that S S" =~ S"*** for all n (¢f. XI, 2, Ex. 5).

5.4 .

(1). SX =~ TX/X. (2). TX is homeomorphic to the subspace {{x, £> € SX |t = 0}. (3). A continuous f: X— Y induces a continuous map Sf: SX—SY

by

fx), .

(1) This follows easily by using the map u: 7X — SX sending Proof: {x, ) — {x, 2t — 1) and passing to the quotient. (2) is straightforward, and (3) follows as in 5.2,

6.

Attaching of Spaces

The process of attaching a space X to a space Y by a map f has great importance in modern topology; it contains as special cases the cone and suspension constructions and also the identification of closed sets to oints.

P The free union X + Y of disjoint spaces X, Y is the set X U Y topology: U C X + Y is open if and only if U N X is open in X UNYisopenin Y. Since XNY = g, X and Y keep their topologies and are disjoint open sets in X + Y. Clearly, BC X is closed if and only if both BN X and B N Y are closed. 6.1

Definition Let X, Y be two disjoint spaces, 4 C X and f: A— Y continuous. In X + Y, generate relation R by a ~ f(a) for each ae A. The (X + Y)/R is said to be “X attached to Y by f,” X Ur Y; fis called the attaching map.

with and own + Y

a closed subset, an equivalence quotient space and is written

Identification Topology; Weak Topology

Chap. VI

128

In

intuitive

terminology,

fla)e Y. Ex.1

Let 4

we

“identify

each

ae A

with

its

C X be closed, and attach X to a point yo by f(4)

image

= yo; then

X Usyo =~ X/A. For, definingp: X — X + yoby p(x) = x,and¢: X + yo > X by ¥(x) = x, ¥(yo) € A, both @ and ¥ are continuous and relation-preserving; passing to the quotient gives continuous maps @4: X/A— X/A, which are inverses. In particular, attaching I to gives S, Ex.2 TXis obtained by attaching X x Itoapointp° TX = (X x I)/(X x 1); the suspension SX is obtained

X Ur o, Yg: X Ur yo— y, by f(0) = f(1) = yo by f(X x 1) = p°, since by attaching X x ¥ to

PP VP T by f(X x 1) =p% f(X x (=1)) =p~. Ex. 3 (x, 1, y) — spaces X, no two of X*xyy~

Attaching X x I x Y to the free y gives a space called the join X * Y Y together with line segments joining the segments have interior points in TX,and X * S° >~ SX.

union X + Y by (x,0, v) — «x, of X and Y. This consists of the each x € X to each y € Y, where common. The reader can verify

The construction of closed sets in X Uy Y is generally based on

6.2

Let p: X + Y— X UsY be the projection, andlet C C X + Y be such that C N X is closed in X. Then p(C) is closed in X Us ¥ if and only if (C N Y)J f(C N A)is closed in Y.

Proof: It is elementary to verify that forany C C X + Y, p~1p(C)= CUCAAUSHACN AU SfYC N Y); consequently,

pHAONNY = (CA Y)Y f(Cn A), PAONN X = (CAX)U S p7'p(C)N Y], Now

assume

that

C N X

is closed.

Then,

and f is continuous, p~'p(C) is closed (CNY)YUSF(Cn A)is closed in Y.

6.3

Theorem

because

4

isclosed

in X + Y if and

in X

only

if

Letp: X + Y — X Us Y be the projection. Then:

(1).

Y is embedded as a closed set, homeomorphicto Y,andp | ¥ i1s a homeomorphism. (2). X — A is embedded homeomorphically as an open set, and

? | X — A is a homeomorphism. Proof: (1) p| Y is continuous, and evidently bijective; by 6.2 it is also a closed map and hence is a homeomorphism. (2) is proved similarly. For subspaces,

6.4 Let Xbeattachedto Ybyf:4-—-Y. Let X; C Xand Y, C Y be closed subsets such that f(4 N X,;) C Y, and attach X, to Y, by

fi=f|ANnX,. Then subset of X Us Y.

X, s

Y, is homeomorphic

to a closed

129

The Relation K(f) for Continuous Maps

Sec. 7

Proof: Let i:X; + Y,—+>X + Y be the identity map; since potop ti X, Uy, Y1 — X Up Y is single-valued, it is continuous. It is clearly also injective. We now show that it is a closed map by proving (3.2) that p o i(C,) is closed for each closed C; C X, + Y, satisfying pripi(C1) = Cq. Since C; N X = C; N X, is closed in the closed X, it is closed in X. Furthermore, by 6.2, (C; N Y;)Jfi(C; N A) is closed in the closed Y, and therefore also in Y; since f; = f| 4 N X, and C; C X, + Y, thissetis (C, N Y)J f(C; N A), and so, by 6.2, we have p(C;) = p o i(C,) closed in X U, Y, completing the proof.

X Uy Y — Z is of frequent

The construction of continuous maps occurrence and is based on the important 6.5

and let Let X be attached to Y by f:4—Y, Theorem p: X + Y— X U Y be the identification map. Let ¢: X — Z and Y1 Y— Z be a pair of continuous maps, and let (p,): X + Y- Z be their unique common extension. If ¢ and ¢ are “consistent” (that p Y >Z 1 XU is, if p(a) = Y[f(a)] for each ae 4) then (p,¢) is continuous.

Proof: applies.

The

(@, $)p~!

map

transgressive

is single-valued,

so 3.2

7. The Relation K(f) for Continuous Maps If f: X— Y is a homomorphism of groups, we can construct f(X) from X and the kernel of f. We will see that a similar process for continuous maps of spaces is possible if and only if f is an identification. Let f: X — Y be any continuous map. We define the relation K( f) in X by x ~ &’ if f(x) = f(x"). This is clearly an equivalence relation in

X, and therefore we have the identification map p: X — X/K(f); X/K(f) is called the decomposition space of f. Since fp~?! is singlevalued, it is a continuous (in fact, injective) map X/K(f) — Y. Using 3.2, we have thus proved the first part of 7.1

(1). Let f: X—

Y be continuous.

Then f can always be factored as

XiX/K(f)i Y (that is,f = gop), wherep andg (= fp~ 1) are continuous, p is surjective, and g is injective. (2.) If f X —>Y

qol

P

zZ ——> h

W

Identification Topology; Weak Topology

Chap. VI

130

is a commutative diagram of continuous maps, there eontinuous A: X/K( f) — Z/K(h) such that the diagram ?

X —>

g

XK(f) l)\

Z ——

(gop=fi8°p

ltfi

ZIKhy

——

p’ g commutes in each square.

is a

=h)

W

Proof: (2) Because the given diagram commutes, it is trivial to verify that ¢: X — Z is relation-preserving; A is the map ¢,, obtained from ¢ by passing to the quotient (4.3). The verification of commutativity in the resulting diagram is trivial. If f: X — Y is surjective, then fp~1: X/K(f) — Y is bijective and continuous. It need not be a homeomorphism: 7.2

Theorem

Let f: X —

Y be a continuous surjection.

o

Then

XIK(f) 2 Y

if and only if f is an identification. Proof: Assume that f is an identification. We need fp~t is an open map. Using 3.2, if U = p~1p(U) for U C X, then since p~1p(U) = f~1f(U), we find that as required. Conversely, if fp~! is a homeomorphism, it tion, and by 3.2 and 3.3, so alsois fp~top = f. Let A, — X by h | 4, = h,. Then h is continuous and 4/K (k) =~ X.

Proof: 'The continuity of % is evident from 8.3, and its surjectivity is obvious. To prove the theorem, we need show (¢f. 7.2) only that 4 is an identification. For this, let U C X be such that 27}(U) is open in

> Ay; then, A~} (U) N A, = h;}(UN 4,) is open in A, for each q, and, since A, is a homeomorphism,

U N A4, is open in A,.

Thus

U is

open in X, completing the proof. Ex. 5 If X determined by weak topology is due to C. H.

has the weak topology determined by {4, |« € .9/} and Y that {B; | B € &}, the cartesian product topology X x Y may not be the induced by {A, x B;| (e, B) € o/ x H}. The following example Dowker:

133

Problems

Let be the set of all maps of Z* into itself. Let X be a real vector space having a basis {, | p € M} in 1-to-1 correspondence with the set ., and let Y be a real vector space with a basis {v, | n€ Z*}. We take both X and Y with the finite topology of Ex. 3. The induced weak topology in X x Y is simply the finite topology of the vector space X x Y; we will show that this is not the cartesian product topology of the spaces X and Y. LetP C X x Y be the set

1

1

{((';("n—) Ups 9’;‘(;) Un )

qae.//l,nEZ"'}-

We observe that P is closed in the finite topology of X x Y, since its intersection with any finite-dimensional flat is a finite set and so is closed in the Euclidean topology of that flat. However, P is not closed in the cartesian product topology of the spaces X and Y. For, if it were, ¥P would be open, and since the origin

0 € €P, there would be a basic nbd U x V with 0e U x V C €P.

Since U, V

are open in X, Y, respectively, then for each ¢ and each »n, there would be an a, and a, such that

Mg |0< A XxY

is continuous.

3. In Problem 2, let both {4, |« €%} and {Bs | 8 € #} be open coverings or nbd-finite closed coverings. Show that [ is then always a homeomorphism.

136 4.

Chap. VI

Identification Topology; Weak Topology

In Problem 2, assume that each x € X is in the interior of some A4, and also that each y € Yisin the interior of some Bs. Show thatthemapl: (X x YV;7)—>Xx Y is a homeomorphism.

5. Let {Y,|a €2/} be any family of spaces, and {b3} a fixed point in [ | Y,. Let o

PY, be the subset of all points in the set |[

Y, having at most finitely many

coordinates different from {b2}. For each finite # C &7, let s(%) be the slice through {62} parallel to [ [{Y,|ac€F}. Take each s(¥) with the cartesian product topology, and let PY, be given the weak topology determined by the

family

{s(¥#)}. Prove:

tinuous open map.

(1)

Each

projection

p; | PY,: PY,—

Y,;

is a con-

(2) If s(b; &) is the slice in [ | Y, through {b,} parallel to Y,

the map s,: Y, —>s(b; «) is continuous.

(3) This type of cartesian product is in

general not associative (hint: use Dowker’s example).

Separation Axioms

VII

So far, our only requirement for a topology has been that it satisfy the axioms. From now on, we will impose increasingly more severe additional conditions on it. With each new condition, we will determine

the invariance properties of the resulting topology: by this we mean (1) whether the topology is invariant under open or closed maps rather than only homeomorphisms; (2) whether the additional properties are inherited by each subspace topology; and (3) whether the additional properties are transmitted to cartesian products. We will also give various desirable features that each such topology has, and more important, we will determine to an extent its behavior under quotient-space formation. In this chapter, we will require of a topology that it “separate” varying types of subsets.

1.

Hausdorff Spaces

Hausdorff topologies have the weakest kind of separation that we will consider in this book; after this section, all spaces will be assumed to be Hausdorff, unless specifically stated otherwise.

I.I

Definition A space Y is Hausdorff (or separated) if each two distinct points have nonintersecting nbds, that is, whenever p # ¢ there are nbds U(p), V(g) suchthat UNn V = g. 137

Separation Axioms

Chap. VII

138

Ex. 1 E™ and discrete spaces are Hausdorfl. In any set Y, a topology larger than a Hausdorff topology is also Hausdorfl. Ex. 2 Sierpinski space and indiscrete spaces having more than one point are not Hausdorff. Ex. 3 'Hausdorff spaces are also called T-spaces in the literature. There are two weaker separation axioms. To—for each pair of distinct points, at least one has a nbd not containing the other, and T3—for each pair of distinct points, each one has a nbd not contamlng the other. Sierpinski space is Ty but not T;; an infinite set

7 = {@}U{U| €U is finite} is a T;-space that is not Hausdorf. with topology

The definition has several equivalent formulations: 1.2

The following four properties are equivalent:

(1). Y is Hausdorff. (2). Let pe Y. For each ¢ # p, there is a nbd U(p) such that

g€ U(p).

(3). ForeachpeYV, N {U| Ulsanbdofp}—(4). The diagonal 4 = {(y,y) | ye Y} is closed inY

x Y.

Proof: (1) = (2). Given q # lp, there are disjoint U(p), U(gq), which says that ¢ € U(p). (2) = (3). If g # p, thereis a nbd U(p) with g€ U(p), so that

geN{U| Ulsanbdofp} To prove the remaining 1mp11cat10ns observe that the statement “UNV =

g”isequivalent to “(U x V)YNn4d =

(Ux

V)Ynd

#

’ since

@¢3p:(p,p).eU>< vV = (pelU) A (pelV)
R, then there exists a discrete closed subspace D such that X(D) = X(&Z) and D C U 4,. Proof: (1). A = {y,, -, ¥y} is the uniom of its finitely many points, and by 1.2(3), each point (being the intersection of closed sets) is closed. (2). If the condition holds, y € 4’ is clear. If the condition does not hold, there is a nbd U(y) with UN (4 — {y}) = {91, -, ¥a}; by

(1), UNE{yy, -+, ¥a} is a nbd ofy not intersecting 4 — {y}, so y€A'. (3). To see that €S(»°, B) is open, observe that any y € S(y°, B) has some coordinate y, # 33, (y # B); because Y, is Hausdorfl, nbd U(¥,) not containing y;, and then y € C €S(»°; (4). Let I' be the initial ordinal of cardinal RX(7); we will injection ¢: [0, I'[ — |J 4, by transfinite construction. Let 8

there is a B). define an < I" and

assume @(y) = y, € Y defined for all y < B. Let B(y,)

= eed

|y, €4,

and let &, = U {%4(y,) |y < B}. Each R[#(y,)] < R, because the family is nbd-finite, so since R([0, B[) < R(LZ), we find (II, 8.3) that

140

‘

Chap.

VII

Separation Axioms

R(Z5) < R(H). Thus, & — HB; # @, and we define ¢(B) to be the first element in U {4, |pesl — %5} in some fixed well-ordering of Y. This completes the inductive step. Now let D = ¢([0, I); because ¢ is injective, we have (D) = R(). Furthermore, D is clearly a nbd-finite system of points in Y; since Y is Hausdorff, so that points are closed sets, 111, 9.2 shows that D and each

subset of D is closed in Y, consequently D is closed and discrete. For continuous maps into Hausdorff spaces,

I.5

Let X be arbitrary, ¥ be Hausdorff, and f, g: X — Y be continuous. Then: (1). {x | f(x) = g(x)} is closed in X. (2). IfD C Xisdense,and f| D = g | D, thenf = gon X. (3). The graph of the continuous f: X— Y is closed in X x Y. (4). If f is injective and continuous, then X is Hausdorff.

Proof: (1). {x|f(x) = g(x)} is the inverse image of the closed 4 C Y x Y under the continuous map x — ( f(x), g(x)) of X - YV x Y. (2). The set on which f and g agree is a closed set containing the dense set D; by III, 4.13, this must be X.

" (3). The graph of f is under the continuous map (4). The inverse map Hausdorff space f(X) onto

the inverse image of the closed 4 C Y x Y (¥, y) = (f(x),y) of X x Y- Y x Y. f~!: f(X) — X is a closed bijection of the X.

We now consider conditions under which an identification topology is Hausdorff. Given p: X — Y observe that 7 (p) is Hausdorff if and only if distinct fibers are contained in disjoint p-saturated open sets. Since this is a condition on both p and the topology in X, it cannot be expected that a simple requirement only on either p or the topology of X will suffice to assure J(p) is Hausdorff. One simple sufficient condition that frequently appears is 1.6

Let X be arbitrary, let R be an equivalence relation in X, and let p: X — X/R be the identification map. If both (1) RC X x Xisclosed in X x X, (2). p 1s an open map, then X/R is Hausdorff.

Proof: Let p(x), p(y) be distinct members of X/R; since x and y are not related, and R C X x X is closed, there is a nbd U x V of (x, y) such that U x V' C €R. Thus, p(U), p(V') are disjoint, and since p is an open map, are open in X/R.

141

Regular Spaces

Sec. 2

Ex. 5 The hypothesis (1) is essential: If X/R is Hausdorff, for any space X, then R must be closed in X x X, since R is the inverse image of 4 C X/R x X/R under the continuous map (x, x") — (p(x), p(x)) of X

x X — X/R

x X/R.

One other particularly simple condition that assures a quotient space is Hausdorff is 1.7

Let Y be Hausdorff, X be arbitrary, and f: X — Y any continuous map. Then X/K( f) is Hausdorff.

Proof:

Since the map

g=r"hXIK(f)=Y is continuous and injective, this follows from 1.5(4).

2.

Regular Spaces

A separation condition stronger than Hausdorff is obtained by replacing one of the points in 1.1 by a closed set: 2.1

Definition A Hausdorff space is regular (or: T3) if each y € Y and closed set A not containing y have disjoint nbds; that is, if 4 is closed and y € 4, then there is a nbd U of y and an open V' O 4 suchthat

UNnV

=

g.

Ex. 1 Discrete spaces, and E", are regular (the regularity of E" is easily seen from 2.2 below). Ex. 2 Every regular space is a Hausdorff space, but not conversely. Let E be the set of all real numbers, and J the topology having the open intervals and

the set Q of rationals as subbasis. Since J is larger than the Euclidean topology of E, (E, J) is a Hausdorff space. However, J is not regular: €Q is a closed set, but 1 and €Q do not have disjoint nbds. Observe also that, since J is larger than the Euclidean topology, this example shows that a topology larger than a regular topology need not be regular.

The definition has several equivalent formulations: 2.2

The following three properties are equivalent:

(1). Y is regular. (2). For each ye Y and nbd U of y, there exists anbd V of y with

yeV CV CU. (3). For each y € Y and closed 4 not containing y, there is a nbd VofywithVNnA = 2.

142

Chap. VI1I

Proof:

Separation Axioms

(1) == (2). Given U, theny and the closed #U have nbds V' D y,

W D %U with VN W = g; thus ¥V C €W, so that V C €W also, and from VNGU CVNW = o, wefindV C U. | (2) = (3). Usingy anditsnbd %4, find V satisfyingye V C V C ¥ 4; then VN4 = o. (3) = (1). Let A be closed, y€A. Choose a nbd V' D y such that

VNA=

o;thnACEV,and VNEV= 2.

Ex. 3 If the condition 2.2(2) is known to hold only for all subbasic open U, the space is still regular: for if G is any given nbd of y, there is a finite intersection

fi

U, of subbasic open

U; with y € O

U

C G; fromyeV,

CV,;

CU,

for each

1

7, we find

For invariance properties we have

2.3

Theorem (1). Every subspace of a regular space is regular. (2). [1{Y,|«€ o/} is regular if and only if each Y, is regular. Proof: (1). Given X C Y, let B C X be closed in X and x, € X — B.

Then B = X N A, where A4 is closed in Y, and since 4 does not contain

%o, there are disjoint open U D %y, V D A. Then UNn X and VN X are the required disjoint nbds of x;, and B in X. (2). Asin L3, it follows immediately from (1) thatif | | Y, is regular, then each Y, is regular. For the converse, let {y,} be given and (U,> be any sub-basic nbd; choosing ¥V, so that y, e V,C {yo:} €

C

=

V,C

U, gives

C

(¢f. IV, L2), and therefore, by Ex. 3, the regularity of [ [ Y, follows. The following special property of regular spaces is frequently useful:

2.4

Let Y be regular and 4 C Y any infinite subset. Then there exists a family {U, | » > 0} of open sets whose closures are pairwise disjoint and such that 4 N U, # @ for eachn > 1.

Proof: We proceed by induction, taking U, = @. Assume that Uy, -+, U, have been defined so that U, - - -, U, are pairwise disjoint, AnNnU,#

@ for 1 < k < n, and An=_A‘—Ln)l_]i 0

is an infinite set.

Choose a, b € 4,; since Y is regular, we find first an open V such that

143

Regular Spaces

Sec. 2

acVcCVc Y—[fiou---u » U {b}] and then an open W such CcWCY—[Uu---UU,U V] Now define U,,; =V W be that if VN A is finite, and U,,, = W otherwise. Then U,, ---, U, ,, are pairwise disjoint, A —

n+1

)

__

Uj is infinite, and AN

0 n + 1, so the inductive step is complete. We

now

consider

U, #

@ for 1 < k
4} satisfiesspe VC V C U 2.2(2) gives the result. Ex. 3 A regular space may not be completely regular. Let (¢f. 3, Ex. 4) T = [0, 2] x [0, w] — (2, w) and for each ne Z let T, = T x {n}; denote the elements of T, by (¢, x; n).

fications

In the

free union (VI, 8.4)

(£, x;2k + 1) ~ (£, x;2k + 2)

and

Z

T,, make the identi-

(o, w; 2k + 1) ~ (o, w; 2k)

for

each k€ Z, «, and «x, to get a “spiral staircase” space S. Let ¥V = SUa U b, where a, b, are two objects not in S, and take the nbds of a (resp. b) to be all sets of form {a} V€ \U {T, C S| n < N} (resp.{b} W€ U {T, C S|n = N3}). Clearly, Y is a regular space; we will show that Y is not completely regular. First note that for a continuous f: T — E?, (1) If f(a, w) = r > s for all large «, then f(£2, x) > s for all large x. (i) If (2, x) = r > s for all large x, then f(c, w) > s for all large «.

Indeed, one easily verifies (¢f. XI, 3, Ex. 2) that f has a continuous extension F: [0, 2] x [0, w] — E*; the hypothesis of either (i) or (ii) gives F(2, w) = »

since F takes values

>7 in each nbd of (£2, w); and because (£2, w) has a nbd

mapped by F into ]s, o[, the conclusions follow. Now let f: Y— E! be any continuous function with f(a) = 1; we will prove that f(b) = 0. For, fis >3 on some nbd of 4, and therefore also on some Tq ;.

Thus,

f(x, w;2k

+ 1) = fla, w; 2k) = %;

F(Q,x;2k —1) > 1

for

all

large x;

wusing

then

(ii)

(1)

gives

[f(2, x; 2k) =

gives flo, w;2k — 1) =

fle, w; 2k — 2) > 1 for all large «. By induction, f(o, w; 2k — N) > 1/2N for each N € Z* and all sufficiently large o; thus, f is positive at points in each nbd of b, so f(b) = 0. This result implies that Y is not completely regular: for, if it were, a continuous F: Y — E! with F(a) = 1, F(b) = 0 would exist, and then f = 2F — 1 would be a continuous function on Y with f(a) = 1, f(b) = —1. [More generally, f(a) = f(b) for each continuous f: Y — E.]

For invariance properties, we have 7.2

Theorem

(1). Every subspace completely regular.

of a completely

regular

space

is

(2). TI{Y, | @€/} is completely regular if and only if each Y, is completely regular.

Proof: Ad (1). Let B C Y be a subspace and pe V, where V is open in B. Since V = BN U, where U is open in Y, then letting @: Y — I satisfy 7.1 for p and €U, it is clear that ¢ | B satisfies the requirement for p and €,V

Ad (2). That the complete regularity of [ [ Y, implies that of each Y, follows in the usual manner from (1), since complete regularity is a topological invariant. For the converse, lety = {y,} and (U, ,---, U,>

Sec. 7

155

Completely Regular Spaces

g(y) = min{gop,(¥)|i=1,---,n} - which, according to III, 10.4, is continuous. Then g(y) = 1 and it is evident that g vanishes outside (U,, - -+, U, . Ex. 4 Any cartesian product of unit intervals is called a parallelotope. By 3, Ex. 1 and 7.2(1), (2), every parallelotope and all subsets of parallelotopes are completely regular (we will prove later that all parallelotopes are in fact normal).

For any space Y, let I¥ denote the set of all continuous maps f: ¥ — I.

Let {I; | f € I*} be a family of unit intervals indexed by IY, and let P¥ be ~ the parallelotope [] {I, | f € I'}; the points of P¥ are denoted by {z,}. The following theorem is a converse of Ex. 4.

7.3

If Y is completely regular, then it can be embedded in a Theorem parallelotope. Precisely, the map p: ¥ — P¥ defined by p(y) =

{f(»),} is a homeomorphism of Y and p(Y) C PY.

Proof: p is injective, for if x # y, there is a nbd containing x but not y, so by complete regularity, there is an fe IY with f(x) = 1, f(y) = 0; then p(x) # p(¥), since they differ at the fth coordinate. p is continuous: for the projection of p on the fth axis is p; o p(y) = f(¥), which is continuous. p: Y— p(Y) is an open map. First observe that the family of open sets {V, | V; = f~1]0, 1]} in Y form a basis: For any open Uand pe U

there is, by regularity, a V with pe ¥V C ¥V C U and therefore an f: Y— I with f(p) = 1, f(¥V) = 0.so that pe V, C U (¢f. 111, 2.2). Since for any basic set V;,, we have p(V,) = {t;} | t;, > 0}N p(Y); this shows that p(V;) is openin p(Y), and, by III, 11.3, proves that p is | an open map.

Since Y and p(Y) are homeomorphic, it is evident that for any space Z, each continuous k: Y— Z is uniquely of the form gop, where g: p(Y) — Z is continuous. If Z is also completely regular, we have 7.4

Corollary Let Y, Z be completely regular spaées, andleth: Y - Z be continuous. Then there exists a continuous H: P¥ — PZ (H is defined on the entire parallelotope P¥) such that the diagram h Y

—>Z

"

Y —— > pz PHP

is éommutati_ve.

In particular, H | p(Y): p(Y) — pi(Z).

Separation Axioms

Chap. VII

156

Since for each g € IZ we have go heI¥, define h;: PY — I, Proof: by h{t;} = t,., (that is, project P¥ onto the g o h-axis and identify I,, with I;). Since each A, is continuous, we obtain a continuous map

H: PY — PZ by H{t,} = {h,{t;}}. From the definition of H we have that H o p(y)

= H{f()s})

= {hlf(D)s}}

= {(g o H(¥))o}>

and from 7.3 that

p1° h(y) = pi(K()) = {&(h(y))s} so that the diagram is indeed commutative. Since the commutativity implies that H(p(Y)) C p,(Z), the continuity of H on PY shows that

H(p(Y)) C H(p(Y)) C py(Z), and the proof is complete.

Problems Section

|

1. Let

Y be Hausdorff.

Prove:

a. N {F|(peF) A (Fis closed)} = p. b. N{U|(peU) A (Uis open)} = p. Give examples to show that neither of these two properties is equivalent to “Hausdorff.” 2. Let {x,,-* -, x,} be a finite subset of a Hausdorfl space. Show that there exist pairwise disjoint nbds U(x,), « - -, U(xy). 3. Let X be a finite set. Prove that the only Hausdorff topology 7 in X is the discrete topology. 4. Prove that in Hausdorff spaces: (a) A’ is always closed; (b) (4) C A4’; and

(c) 4y = 4.

o %

5. Let f: X—> Y, g: Y—> X be continuous, with gof = 1x. Prove: If Y is Hausdorff so also is X, and f(X) is closed in Y. 6. Let Y = I U{§}, where £EI, with the identification topology determined by p:[—1, +11 - Y, where x — |x|, x # —1, and p(—1) = £. Show that (a)p is an open map; (b) Y is not Hausdorff; (c) the relation K(p) is not closed in X x X. 7. Prove: A necessary and sufficient condition that points be closed sets is that the topology be T}. . Prove: X/R is T, if and only if each equivalence class is closed in X. A point y, of a connected Hausdorff Y is called a dispersion point if Y — {y,} is

totally disconnected. 10.

Let X be Hausdorff,

Prove:

Y can have at most one dispersion point.

let f: X —

Y be continuous,

and let D

C X be dense.

Assume f | D is a homeomorphism into Y. Prove: f(X — D) C Y — f(D). 11. Let X be Hausdorff and f: X — X be continuous. Prove: {x | f(x) = x} is closed in X. 12. Prove: Every infinite Hausdorff space contains a countably infinite discrete subspace.

Problems

157

13. Let D be a dense subset in each of the two Hausdorff spaces X and Y. Let 1: D — D be extendable to a continuous f: X — Y and also to a continuous g: Y — X. Prove that f and g are homeomorphisms and that f = g~1, 14. Let Y be Hausdorff. Prove that the cone TY is Hausdorff. 15. Show that the space (Z*,7) of V, Problem 1. 10 is a Hausdorff space. [This example of a countable connected Hausdorff space is due to Morton Brown.] Section 2 1. Let Y be Hausdorff, and assume that each y € Y has a nbd ¥ such that V is regular. Prove: Y is regular. 2. Prove: If Y is regular, each pair of distinct points have nbds whose closures do not intersect. 3. Retopologize the real line by taking as complete system of nbds at each x the

sets

Up(x) = {x} U{y|(y

is

rational) A (|y — x| < 1/n)},

that this space is not regular, but that each pair of whose closures do not intersect. Thus, the converse 4. Show that the space (Z*,7) of Problem 1. 15 is closed sets {2n | n € Z*} and {1} do not have disjoint 5. Let X be regular, and 4 C X closed. Show that

A = N{U]|[Uisopen]

ne Z*.

Show

distinct points have nbds of Problem 2 is false. not regular. [Hint: The nbds.]

A [U D A].

Section 3 1. Show that if “Hausdorff” is omitted in the Definition 3.1, then the indiscrete spaces and Sierpinski space are normal. 2. Let X be normal and p: X — X/R be a closed and open map. Show that X/R _ is normal. 3 (a). Let X be the set of irrationals in E! and let u: X — E! be a map that is always positive. For each ne Z* let H, = {x | u(x) = 1/n}. Prove: There

exists an m and an open interval £, such that ,# N II,, # interval

_# C #,.

[Hint:

Assume

this

assertion

@ for every open

were

false;

letting

{ro. | n€ Z*} be an enumeration of the rationals, we could then find a sequence {J,} of intervals, each having rational end points such that for each

n, () r,€Jn; Q) JunNHy= &;3)Jns1 CJn; 4 0 < length J, < 1/n. By Cantor’s definition of real numbers () J, would be some real number ¢. But ¢ cannot be rational because of (1), and it cannot be

irrational

either

since vn: ¢ € H,, so u(§) = 0, an impossibility.] (b). Use part (a) to show that in E; x Ej the disjoint closed sets 4 = {(r, —7) | » rational} and B = {(x, —x) | x irrational} cannot be separated. [Hint: Let U > B be any open set. For each irrational x, let

u(x) =sup{A|Jx

— A x] x ]—x — A, —x] C U};

u(x) is never zero. Letting r € #,, every nbd of (r, —r) intersects U.] 4. Let T be the space of Ex. 4 and A4, B be the subsets indicated there. a. Attach T to a single point p, by the map f(4) = p,. Show that T \Usp, is Hausdorff, but not regular. b. Attach T to two points, p, U pg by the map f(4) = p,4, f(B) = pg. Show that T Us(p4 Y pp) is not Hausdorff.

158

Chap. VII

Separation Axioms

5. Let X be the upper half of the Euclidean plane E2, bounded by the Use the Euclidean topology on {(x, y) | ¥ > 0}, but define nbds of the (x, 0) to be [(x,0)] v [open disc in {(x,y) | y > 0} tangent to the x-axis at Prove that this space is not normal. 6. Show that the unit interval I is completely normal. 7. 'Let I be the unit interval, andJ its Euclidean topology. Let Q C I be

x-axis, points (x, 0)].

the set

of irrationals in I. Define a topology 7 ; in the set I by taking 7 U Z(Q) as’ ;) is normal. subbasis. Prove that (I,

Section 4 1. Let Y be a connected normal space having more than one point. lower bound for R(Y). 2.

Let X be normal.

Prove:

Determine a

Every F,-set in X is also normal.

3. Let X be perfectly normal. Prove that every subspace of X is also perfectly normal. 4. Let X be normal, 4 C X closed, and U an open set containing 4. Prove: There exists an open F,-set VV suchthat 4 C V C U.

Section 5 1. Show [a, b[ C E! is an AR (normal). 2. Prove: If Y is an ANR (normal), then every open subspace of Y is also an ANR (normal). Give an example to show that this need not be true for closed subsets. 3. A normal space Y is said to have property Q if whenever Z is normal and Y C Z is closed, there is a continuous r: Z — Y such that 7(y) = y for each y € Y. Prove: Y has property Q if and only if it is an AR (normal). Formulate a similar criterion for ANR (normal).

Section 6 1. Let Y be normal, and F,, - - -, F, closed subsets such that fn] F, = 1

There exist open sets V; D Fysuchthat 2. Prove:

V', n Vo n-.-

n ¥V, =

@. Prove:

& also.

Y is normal if and only if for each finite covering U,, ---, U, of ¥ by n

open sets, there exist n continuous maps f;: Y — each fi(y) = 3. Let Y be a covering {4, 4. Let {U, | « €

I such that Z fi{y) =1

and

1

O0Oforye Y — U, space having weak topology with respect to a countable closed ] n€ Z}. Prove: If each A4, is a normal space, then Y is normal. &} be a nbd-finite covering of the normal space Y by open sets.

For each (a, B) € &/ x .7, and each y € U, N Uy, let Vo 4(y) C U, n Uy be a given nbd of y. Prove that one can assign a nbd V' (y) to each y € Y such that:

a. ye Uy N U

= V(y) C Vag(y).

b. If V(y) N V(x)

#

&, there exists an « such that V(y) v V(x) C U,.

Problems

159

Section 7 1. Let Y be a T, space satisfying the condition: each y € Y and closed 4 not containing y have disjoint nbds. Prove: Y is a regular (Hausdorff) space.

2. Let Y be completely regular, 17 open, and p € V. Prove: A necessary and sufficient condition that there be a continuous ¢: ¥ — I such that ¢~1(1) = p,

e(@&V) = 0, is that p be a G,. 3. Prove: If Y is a connected and completely regular space with more than one point, then X(Y) > c. 4. Let X be completely regular. Show that the topology of X is precisely the projective limit topology determined by the family of all continuous maps f: X — I(cf. VI, I, Problem 4). 5. Let X be completely regular, and C(X) the set of all bounded continuous realvalued functions on X. Foreachx, f,elet U(x, f,€) = {y | | f(x) — fF(») ] < €. Prove: {U(x, f, €) | all f, x, and € > 0} is a subbasis for the topology of X. 6. Let Y be completely regular, and f: Y — E! lower semicontinuous. Show that f = sup f, for a suitable family {f, | @ € %} of continuous functions. Conversely, show that if Y is any space such that each lower-semicontinuous function is the sup of continuous functions, then Y is completely regular.

Covering Axioms

VIII

In VII, 6.1, a normal space was characterized by a property of certain types of coverings by open sets. In this chapter, we study some other important classes of spaces that are determined by some requirement on the behavior of their coverings.

I.

Coverings of Spaces

In this section, we shall be concerned primarily with finite, nbd-finite, and point-finite coverings of a space Y by arbitrary sets. It is clear that

any finite covering is nbd-finite and any nbd-finite covering is pointfinite; however, as 111, 9, Ex. 1, shows, a point-finite covering need not

be nbd-finite. By a subcovering of the covering {4, | c €%/} of Y is meant any subfamily {4, | « € &}, # C &, that is also a covering of Y. I.1

Theorem

Let {4, |«€ 2} be a point-finite covering of Y. Then

there exists an irreducible subcovering, that is, a subcovering that,

when any single set is removed, is no longer a covering of Y.

Proof: Call a family R C {4,} removable if {4,} — R still covers Y. Partially order the set Z of all removable families by inclusion. If {R,} is any chain in %, it has an upper bound, R = | J R, in Z: for, if

R€ Z,

n

there would be some y e Y such that the finitely many 160

sets

161

Coverings of Spaces

Sec. 1

> Aa, containing y are in R, and since {R,} is a chain, these 4,, Ay, would all lie in some one R,, which would contradict R, € #Z. By Zorn’s lemma, there is a maximal removable family R,, and so {Aa

I o€

‘%}

-

RO

is irreducible. The hypothesis of point-finiteness is essential: The covering of E! by

Ex. 1 the sets A,

= ]—n,n[,

n = 1, 2, - - -, has no irreducible subcovering.

The family of all coverings of a given space has a natural preorder:

[.2

Definition Let {4, |« € &/} and {B; | B € #} be two coverings of a space Y. {4,} is said to refine (or be a refinement of) {B,} if for each A, there is some By with 4, C Bs. We write {4,} < {Bs}.

It is clear that any subcovering of a given covering is a refinement of that covering. If {C,} < {4,} and {C,} < {Bg}, then {C,} is called a

common refinement of {4,} and {By}. Ex. 2 The relation < of refinement is easily seen to be a preordering in the set of all coverings of Y. It is not a partial ordering: In E?', each of the two A, = {x|x e, then by (2), we find that y € p(V

of

«,

shows

1., ,); therefore

we conclude that y e H, ;. ;. To complete the proof, we

need only modify the H,, slightly to each n. Choose a precise open nbd-finite

assure nbd-finiteness for refinement of {p~1(H,,) | (¢,n) e & x Z*}, and shrink it to get an open nbd-finite covering {K, ,} satisfying p(K,,) C H,,. For each n,

167

Types of Refinements

Sec. 3

let S, = {¥ | some nbd of y intersects at most one H, ,}; S, is open and contains the closed | p( K,,) = p( U K,.,), so by normality of ¥ we o

o

find an open G, with {J p(K,,)

C G, C G, C S,. The open covering

(G,NH,,|(e,n)esf x Z*}, with the decomposition {G,NH, ,|xcxl} forn = 1, 2, . - - satisfies the conditions of 2.3(2) for the given {U,}.

Types of Refinements

3.

In this section, we obtain characterizations of paracompact spaces by means of refinements that are not necessarily nbd-finite. Let I = {U,| « € .27} be a covering of a space Y. For any B C Y,

the set U {U, | BN U, # @} is called the star of B with respect to 1, and is denoted by St(B, 1). 3.1

Definition A covering B is called a barycentric refinement of a covering I whenever the covering {St(y, 8) | y € Y} refines 1.

32

Let

Y

be

covering.

Proof:

V,C

normal,

and

1 = {U, | « € &/}

an

nbd-finite

open

Then Ul has an open barycentric refinement.

Shrink U to an open covering 8 = {V, | «€ &/} such that

U, for each «; clearly, % is also nbd-finite. For each y € Y, define

W) = N{U | yeV3 O N{EV, | yEV,. We show that @ = {W(y) | y € Y} is the required open covering. Note first that each W(y)is open: the nbd-finiteness of B assures that the first term is a finite intersection, and that the last term, ¥ ) I_/B, 1S an open

set (111, 9.2). Next, ¥ is a covering, since y € W(y) for each ye Y. Finally, fix any y, € Y and choose a I/, containing y,. Now, for each y such that y, € W(y), we must have y € V,, also, otherwise W(y) C €V ,; and because y € V,, we conclude that W(y) C U,. Thus, St(y,, B) C U,, and the proof is complete.

3.3

Definition A covering 8 = {V,;|Be #} is called a star refinement of the covering U whenever the covering {St(V,, %8 |8 e %} refines U.

3.4

A barycentric refinement % of a barycentric refinement 8 of Uis a star refinement of U.

Proof: Given such that Wn St(z, ) C some that St(IW,, 8) C

W, e W, # V € B. St(y,,

%, choose a fixed y, e W,. For each We® @, choose a s3e WnN W,; then WU W, C Because each such V contains y,, we conclude B) C some U e ll.

Covering Axioms

Chap. VIII

168

Since it is clear that a barycentric refinement of any refinement of U is also a barycentric refinement of 1, it follows from 3.2 and 3.4 that each open covering of a paracompact space has an open barycentric, and an open star, refinement. Much more important, however, is that this property characterizes the paracompact spaces, not only among the Hausdorff spaces, but in fact also among the T, spaces:

3.5

Theorem (A. H. Stone) A T, space Y is paracompact if and only if each open covering has an open barycentric refinement.

Proof: Only the sufficiency requires proof. We first show that any open covering I = {U, | « € &/} has a refinement as required in 2.3(2). Let U* be an open star refinement of U (¢f. 3.4) and let {11, | n > 0} be a sequence of open coverings, where each 11, star refines 11, and U, star refines U*. Define a sequence of coverings inductively by

-, , } ) , B ® e V | ) , 0 , V ( S { = B, = U, B, =8V,

1} 0) | Ve®,

Each %, is an open refinement of 1,; in fact, each covering {St(V, 1,) | V € B,} refines U,: this is true for » = 1 and, proceeding by induction, if it is true for n = k — 1, its truth for n = k follows by noting that whenever V = St(V,, 1,) for some Ve B,_,, then St(V, W)= St[St(V,, 1), U] C St(V,, U, _,) because U, is a star refinement of 1, _;. Now well-order Y and for each (n,y)e Z+* x Y define

E.(y) = St(y, 8B,) — U {St(z, B,, | » precedes 1) y} Then

set A

€ = {E, () ] (n,y)e Z*

={z|pe

U

x

Y} is a covering:

St(z, B;)} is not empty,

given p€ Y, the

since p e 4; if y is the

i=1

first member of 4, thenp € St(y, B,,) forsomen e Z* and p € St(z, B, ;1) for all preceding ¥, so p € E (y). Moreover, since 8, refines U, we find that € refines 11*. Each U e U,,; can meet at most one E,(y): for, if UN E,(y) # &, then there is a Ve ®, withyeVand VN U # @,soyeVulUC Voe®B,,,and

U C St(y, B, .,).

Thus, if E (y) is the first set U meets,

it cannot meet any E, (p) for p following y.

Now let W,(y) = St(E,(y), 1,.5).

W)

Then

| (my)eZ* x Y}

is clearly an open covering of Y. Furthermore, ¥ refines U because & refines U*. Finally, for each fixed n € Z*, the family {W,(y) | ye Y} is nbd-finite:indeed, each U € 1, ,, can meet at most one W,(y), because

Sec. 4

169

Partitions of Unity

UNnWy(y) # o if, and 1s contained St(U, U,,,) meet at most one E, (y).

only if, E(y)n Sy(U, U,,,) # ¢ in some U,e U,,; which we know

and can

The theorem will follow from 2.3(2), once we show that Y is regular. To this end, let B C Y be closed and y € B. Since in a T; space each point is a closed set, U = {Y — y, €B} is an open covering. Let B be an open star refinement. Then St(y, 8) and St(B, B) are the required disjoint nbds of y and B: for if there were a V containing y and a V"’ meeting B such that VN V' # o, then St(V, 8) would contain y and points of B, which is impossible. The theorem is proved. This leads to still another characterization of paracompactness, this time based on a sequence of open coverings.

3.6

Definition Let I = {U, |« € .9} be an open covering of Y. A sequence {ll, | n € Z*} of open coverings is called locally starring for 1 if for each y € Y there exists an nbd V(y) and ann € Z* such that St(V, 1,) C some U,.

3.7

Theorem (A. Arhangel’skii) A T, space is paracompact if and only if for each open covering U there exists a sequence {ll, | n € Z*} of open coverings that is locally starring for 1.

Proof: “Only if” 1s trivial. “If”: We can assume that 11, ,; < U, for eachne Z*. Let

B ={Vopenin

Y|3In:[VC

Ueln,] A [S{V,11,) C some U,]}.

For each V € B, let n(V') be the smallest integer satisfying the condition. Because {11, | n € Z*} is locally starring for U, it follows that 8 is an open covering; we will show that 8 is in fact a barycentric refinement of 1.

Let ye Y be fixed, let n(y) = min{n(V)|(ye V) A (VeB)},

and

let V, € B be a set containing y such that n(V,) = n(y). For any 'e 8 containing y, we have n(}’) > n(y), and consequently

St(y, 8) C U {St(y, W) | 7 > n(y)}Since U,,; < U, for each ¢, this shows St(y, B) C St(y, Uyy,y) C St(Vo, Uyy,,) C some U,.

By Stone’s theorem,

4.

Y is therefore paracompact.

Partitions of Unity

Partitions of unity play an important role in modern topology; one of the reasons that paracompact spaces are useful is that “arbitrarily fine” such partitions exist on them. For any space Y, the support of a map f: Y — E?! is the closed set

170

Chap. VIII

Covering Axioms'

{y | f(y) # 0}; observe that y is not in the support of f if and only if y has a nbd on which f vanishes identically.

4.1

Definition Let Y be a Hausdorff space. A family {«, | « € &/} of continuous maps «,: ¥ — I is called a partition of unity on Y if: (1). The supports of the «, form a nbd-finite closed covering of Y. (2). Z ke(y) = 1 for each yeY (this sum is well-defined because each Kg)If {Uy;|Be %} is partition {k; | 8 € #} of each «; lies in the

y lies in the support of at most finitely many

a given open covering of Y, we say that a of unity is subordinated to {{/} if the support corresponding Uj,.

Clearly, every space has a partition of unity subordinated to the covering by the single set itself. 4.2

Theorem

Let

{U, | « € &} {U,}.

Y be paracompact.

Then

for each open

covering

of Y there is a partition of unity subordinated to

Proof: Shrink a precise nbd-finite refinement of {U,} to get a nbdfinite open covering {V,} with V, C U, for each a. Now shrink {V,} to get a nbd-finite open covering {W,} satisfying W, C V,. For each ae s/,

VII,

4.1, gives a continuous g,: ¥ — I, which is identically 1 on W, and vanishes on ¥V, (we take g, = 0 if V, = @); each g, has

its support in U,. Since {W,} is a nbd-finite covering, it follows that for each y € Y at least one, and at most finitely many, g, are not zero, consequently Zga is a well-defined real-valued function on Y and is never zero. z g, 1s continuous on Y': every point has a nbd on which «

all but at most finitely many g, vanish identically, so the continuity of 2. 8, on this nbd follows from that of each g,, and by 111, 9.4, 5 g, is therefore continuous on Y. The required partition of unity is given by the family of functions {«, | « € &7}, where

k(y) =

gY)

> 29)

We remark that in a normal space Y, the proof shows that a partition of unity subordinated to a given nbd-finite open cover exists; C. H. Dowker has shown that their existence for each open cover is equivalent to paracompactness of Y [cf. 5.5(2)].

Sec. 5

Complexes; Nerves of Coverings

171

To give an application of 4.2, note that if {x, | « € &} is a partition of unity on

¢,:

Y, and

if {p, | « €/}

Y — E', then the map

is any family of continuous maps

Y — E! given by y — Z Po(¥)(y) is also

continuous.

4.3

(C. H. Dowker) Let Y be paracompact. Assume that g is a lower, and G an upper, semicontinuous real-valued function on Y such that G(y) < g(y) for each y € Y. Then there exists a continuous ¢: Y — E' such that G(y) < ¢(y) < g(») foreachye Y.

Proof:

For each rational 7, let U, = {y | G(y) < r} n{y | g(y) > r};

due to the semicontinuities, this is open; and because for each y there is

some rational 7 with G(y) < 7 < g(¥), the family {U,} is in fact an open

covering of Y. Let {«,} be a partition of unity subordinated to {U,}; the required continuous

function

is ¢(y) = Z r-k(y).

For, let ye Y be

given, and let «,, - - -, x, be all those functions whose support contains y; then ye U, N ---N U, so that G(y) < r; < g(y) for each i = 1, ..., n, and therefore

G(y) = G(3) 2 1.,(9) < 21 (¥) 5.

= o(3) < &)

2 x,(¥) = &)-

Complexes; Nerves of Coverings

The concept of a partition of unity subordinated to a given open covering has an alternative, more geometrical, interpretation. To develop this, we need two preliminary notions.

(1). Let & be any set. By an n-simplex o” in &7 is meant a set (a,- - -, 0,,) of n + 1 distinct elements of &7; «y,- - -, a, are called the vertices of o”, and any 0? C o" is termed a g-face of o™.

5.1

Definition

An

abstract simplicial complex £

over 27 is a set of simplexes

in &7 with the property that each face of a 0 € )~ also belongs to %£". With each abstract simplicial space. For this we need

complex

we

will associate a standard topological

(2). Given (n + 1) independent points po,- - -, p, in an affine space, the open geometric n-simplex o™ spanned by po,- - -, pn is

{ i Apy

Sh=1,

0 Ro] = [[] ¥, is not 1° countable]. Indeed, let {y2} e[] Y, be given, and let {V,, | » € Z} be any countable family of its nbds.

With

cach V,, associate a finite set R(n) = {a;, -, ¢} C & such that {yayeU,,, -+, U,> C V,. Then |J R(n) is at most countable and, since

R(s7) > R,, there is a B € .o/ not in |) R(n), that is, no V,, restricts the B-coordinate. Choosing a nbd U, D yj; with U, # Y, which is possible

because Y; has more than one point, there is no V, contained in the nbd {Up) of {y;}.

190

Chap. I1X

Metric Spaces

The converse of 7.1 will follow from

7.2

Theorem Let {Y,|ne Z*} be acountably infinite family of metric spaces. In each Y, choose a metric d, giving its topology, and let 8,(Y,) be the diameter of Y, according to d,. For x = {x,}, y = {y,},

define p(x, ¥) = sup {d,(x,, ¥,) | ne Z*+}. Then:

(1). p is a metric in the set [|

Y, whenever the 8,(Y,) are uni-

formly bounded for all large n. (2). p metrizes the cartesian product topology of the space [ [ Y, if and only if §,(Y,) — 0. Proof: Ad (1). Since §,(Y,)< M < o for some fixed constant M andall# > some n,, it follows that pisin factdefinedon[] Y, x [ Y,; it clearly has the properties (1) through (3) of I.I, and the triangle inequality results from P(x’ 2)

=

sup dn(xn’ 2n)


0 x € B,(x, p) C U, since Y € By(x, p) = p(x,y)

and

we

have

< p = sup dy(x,, y,) < p

=> ‘v’nn: do(%,, ¥,) < 1y =>Vmn:y,e B(x,r,

(b). Let xe B,(x,7). Since §8,(Y,)—0, 8,(Y,) < n/2 for all n > ny; then

there

is

=>ye U.

an

n,

with

¥ e U = CBlssy2/2) - Bty 112> € Bl )

For, if y e U then d(x;, y;) < »/2 for 1.< 7 < ny, and since this holds also for¢ > n,, we have

ye=U p(x,y)< n/2=>y€B (, 7). (a) and (b) show that the metric and cartesian product topologies are the same, Now assume that §,(Y,) + 0; then we can find an e > 0 and infinitely many indices 7, for which there are points %7, y; € Y, that satisfy

191

The Space I>(«7); Hilbert Cube

Sec. 8

d, (%, ¥i.) = e If we select any x° €[] Y, having all the x] among its coordinates, there can be no cartesian product nbd U satisfying x°e U C B,(x° ¢): since U restricts only finitely many coordinates, at least one 7, is not restricted, so U contains the point 2 having coordinates 2 = x; (I # 1), %, = ¥;,, and therefore p(x°, 2) > =. 7.3

Corollary

Let {Y, |« €2/}

be a family of metrizable

spaces.

If

R(«) < R, then [ [ Y, is metrizable. Proof:

Case

N¥Z) = n < X;:

The

proof

of 7.2(2) given above

applies to show that if we choose any (not necessarily bounded) metric d, for each Y;, then p(x,y) = max {d(x;, ¥,) |1 < ¢ < n} metrizes the

cartesian product topology of [|

V..

1

Case R(«7) = X;: By 3.3, we can choose a metric d; for each that the requirement of 7.2(2) is satisfied.

Y, so

It should be noted that although in each case the metric p depends on the choices of the d;, the topology 7 (p) does not. Remark. We have seen that the cartesian product of paracompact spaces need not be normal. In view of 7.3 it is natural to ask about the nature of a cartesian product of a paracompact and a metric space. This question has recently been settled by E. A. Michael, who has shown that if X is paracompact and Y is metric, then again X X Y need not be normal, even though Y is separable metric. His example is the following: Let Y be the metric space of all irrationals, and let X be the paracompact space of VII, 3, Problem 7. Let R be the subset of rationals in X, and T the irrationals. Then the closed sets 4 = R x Y and B = {(2,2) | z€ T} in X x Y cannot be separated. The reader may construct a proof by making suitable modifications of that given in VII, 3, problem 3.

8. 8.1

The Space /*(«7); Hilbert Cube Definition

Let .2/ be any set, of arbitrary cardinal, and

let E¥)

be the cartesian product of R(&7) factors E'. The metric space

I1?(7) is that having for: Elements:

every x = {x,} € EX%

such that x, = 0 for all but

at most countably many « € o7, and > x2 converges.

Topology:

thatinduced by the metric d(x, y) =

/Z (%, — V)2

It must be verified that d(x, y) is indeed a metric. Its finiteness and the triangle inequality follow from Minkowski’s inequality: Given

x, ¥ € [2(2f), then summing over the at most countably many elements of

-

192

Chap. IX

Metric Spaces

&/ for which at least one of x, y has a nonzero coordinate, we find for each finite N that A/i(xa; 1

"ym)zsA/fix1 fi,

+

A/iyfi‘ -


V,: Then, in any countable collection of elements, we find nonzero entries

for at most countably many coordinates; if 8 is an index for which all members of the countable set have coordinate 0, the point x with x; = 1 and x, = 0 (« # B) has nbd B(x, }) not containing any member of the countable set.

8.3

Definition

'The subspace {{x,} € I%(X,) | |x,] < 1/n} is called the

Hilbert cube and is denoted by I=.

8.4

193

Metrization of Topological Spaces

Sec. 9

'The Hilbert cube I

to the cartesian product

is homeomorphic

o0

[ T 1 of countably many unit intervals. 1

Proof:

:

Sincel @ J=[—1,

+1],wehave]| [I 1

~ ]] J, so we prove 1

I* ~ []J. Define : I* —] [ J by @(x, %, - - ) = (%1, 25, 33~ - -); 1 1 clearly, this is bijective; and ¢ is continuous, since the projection on each o0

factor is continwous.

To see that ¢!

1

7.2, using

is continuous, metrize [ [ J as in 1

X, — dn(‘xm yn)

=

|_7174y—n|;

if p(x,y) < & in |[ J, we then have 1

do @, 7o) = 3 (= - 2)

2

n

xn


0, the function min(M, d) is also a gauge.

A family & - {d, | @ € o} of gauges on Y is called separating if for each pair of points x 7 y there is a d, € & such that d,(x, y) # 0.

10.2

Definition

Let

Y be a set and

9 = {d, | « € &/} a separating

family of gauges on Y. The topology 7 () having for a subbasis the family 8(2) = {B(y; dy, &) |y€ Y,d,€ D, e > 0} of balls is called the topology in Y induced by the family 2.

199

Gauge Spaces

-

Sec. 10

Because we require & to be separating, it follows that (a): the topology T (2) 1s always Hausdorff, and (b): if & consists of one gauge alone, then that gauge must be a metric and J(2) is the topology induced by that metric. As metric spaces show, distinct families of gauges in Y may lead to the same topology. 10.3

Definition A gauge structure for a topological space (Y, 7 ) is a separating family & of gauges such that J = J(2). A topological space that admits a gauge structure is called a gauge space.

If Y has the gauge structure &, a basis for its topology can be obtained by enlarging Z:

104

Let & = {d, | « € o/} be a separating family of gauges in Y. Let Z* be the family of gauges

{max(d,, ---,d, )| all finite subsets {«;, - - -, a,} C 2} Then the family B8(Z*) of all balls is a basis for 7(2). Proof:

1t is simple to verify that if d*

= max(d,,, -, d, ), then

B(y;d*,¢e)

= ('n] B(y;d,, ¢) for each ¢ > 0; thus, each member of 1 B(2*) is open in F(2). The proof that B(D+) is a basis for T (2) is - now entirely similar to that given in 2.1. Ex. 5 1In a gauge space (Y, 7 (2)), the expression of any given topological concept in terms of the gauges uses the basis B(Z*) [rather than the subbasis B(ZL)] and is similar to that given in terms of the metric in metric spaces: calling the ball B(y;d*,¢€) a (d*, €)-nbd of y, the idea of an e-nbd in metric spaces is replaced by that of a (d ¥, €)-nbd in gauge spaces. Thus, y € 4 if and only if each (d*, €)-nbd of ¥ meets A. Similarly, an f: Y — Z is continuous if for eachy e Y and open W D f(y)thereisad* € Z* and ane > Osuchthat f[B(y;d*, €] C W.

Ex. 6 The formula | d(x,y) — d(x’, ¥") | < d(x, x") + d(y, y’) is valid for any gauge; it shows that in any gauge space (Y, .7 (2)) each d* € 2+ is a continuous map of ¥ x Y into E1.

In addition to showing that subspaces and cartesian products of gauge spaces are gauge spaces, the following theorem also gives the construction of a gauge structure for such spaces starting ffom one in each factor: 10.5

Theorem (1). Let the space Y have the gauge structure & and let 4 be a subspace of Y. Let 2, be the family of gauges in &, each restricted to A x A. Then Z, is a gauge structure for the subspace 4.

Chap. IX

200

Metric Spaces

(2). Let {(Y,, 7(25)) | Be %} be any family of gauge spaces. For each B € 4, let 93 be the family of gauges induced on [

¥,

by the members of &,. Then the family {Dy | B e B} of gauges is a gauge structure for the cartesian product topology of the gauge spaces. Proof: (1) is left for the reader. (2). We need only observe that the sets (B(y; d,, €)> C [] Y, are a subbasis for both topologies. The position of gauge space topologies 1s 10.6

Theorem regular.

A space Y is a gauge space if and only if it is completely

Proof: Assume that Y has the topology 7 (Z) for some separating family 2 of gauges. Let y, € U, where U is an open set. By 10.4, there isad* e 2" suchthat B(y,;d*,e) C U;thenf(y) = min[1, e~ 1d* (¥, yo)] is a continuous map of Y into E! such that f(y,) = 0, f(¥U) = 1. Thus, Y is completely regular. Conversely, assume that Y is completely regular. Then Y can be embedded (VII, 7.3) in a suitable cartesian product P¥ of unit intervals I. Since I is a gauge space, so also (10.5) is PY and therefore also Y. 10.7

Corollary A completely regular space is metrizable if and only if it admits a countable gauge structure.

Proof:

Only the sufficiency requires proof. Letting @ = {d, |ne Z*}

be a countable gauge structure for the space, the reader can verify that

d(x, y) = § min[d,(x, ), 2-"] is a metric that induces the topologyJ (Z).

1. Uniform Spaces A metric d in a space Y can be régarded as providing a measure of nearness that is applicable throughout the space: for each ¢ > 0, we may consider all the sets B(y;¢) as being equally small. This notion of uniform smallness is not a topological concept: equivalent metrics specify different sets as being equally small. In this section, we consider the idea of uniformity in general topology due to A. Weil; it has many important applications, particularly in the study of topological groups.

201

Uniform Spaces

Sec. 11

notation of binary relations in a set ¥ will be used. Given CY x Y, we define U™! = {(y,%)|(x,9)e U} and UoV =

The UV

{(x, %) | 3y: [(%, ¥) € V] A [(3, 2) € U]}. we let Uyl diagonal.

={2|(»,2)eU};

and

Furthermore, 4 C Y x Y

for each yeY,

will

denote

the

Definitiorr A uniform structure in a set Y is a family & of subsets of Y X Y such that (). If Veg, thend C V.

1.1

(2). If V4, V,€, thenthereisa We Fsuchthat WC VNV, (3). If V € &, then there is a W e § such that Wo W1 C V.

4). If Vegand VC W, then We . The uniform structure is called separating if N {V | Ve g} = 4. A family satisfying only (1), (2), and (3) is called a base for a uniform structure or, more simply, a uniformity. Since{4d e #(Y x Y) |4 C 4} is a uniform structure, and since the intersection of uniform structures

is also a uniform structure, each uniformity generates a unique smallest uniform structure containing it. Two uniformities §;, s, if they generate the same uniform structure. It is simple is equivalent to §, if and only if for each V,.€ §, there is that V, C V;, and also for each W, € &, there is a W,

are equivalent to see that g, a I, € F, such € &, such that

W, C W,. For an interpretation of a uniformity g, call two points V-close whenever (x, v) € V or, equivalently, ¥ € V'[x]; in this terminology, (3) says that for each V € §, there is a W e & such that whenever x 1s Wclose to ¥ and y is W-close to 2, then x is V-close to 2. Thus, each member of the family can be regarded as a relation of uniform nearness defined over the entire set Y. Ex.

1

Let

Y be a metric space,

Vi) = {(x,y) e Y x

Y

|d(x,y) < €;

and d a metric for

then

the family

Y.

For each € > 0, let-

{I(¢) | e > 0} is a uni-

formity in Y, called the uniformity determined by d. Observe that equivalent metrics in a space Y need not determine equivalent uniformities: for example, in the space {x | x = 1} C E?, the uniformities determined by the equivalent metrics

d(x,y) = |x — y| and d’(x,¥) = |]x~! — y~| are not equivalent. Ex. 2

If(Y, 7 (2)) is any gauge space, then the family of sets

{(,9) C Y x Y|d(xy) 0is a uniformity in Y, called the uniformity determined by 2. Ex. 3 'The idea of uniform continuity has its natural expression in terms of uniformities: if X, Y are sets with uniformities ¥, &, respectively,amap f: X — Y is called uniform with respect to the given uniformities whenever for each 'e & there is a U € ¥ such that if (x, y)€ U, then (f(x), f(3)) € V. This leads to the idea of a uniformly continuous map of a gauge space (X, .7 (¥)) into another, (Y, 7 (2): it is a continuous map of X into Y that is uniform with respect to

202

Chap. 1X

Metric Spaces’

the uniformities determined by &, £2’. Stated directly in terms of the gauges: f: X — Y is uniformly continuous if for each d’ € &’ and € > 0, there exists a de 2 and a 8 > 0 such that if d(x, y) < §, then d’(f(x), f(»)) < e. Notice that even in metric spaces, a continuous map may be uniformly continuous if one pair of metrics is used, but not uniformly continuous when another pair of equivalent metrics is used; uniform continuity is therefore not a topological concept.

The concept of a uniformity can be expressed equivalently by coverings of the set Y:

1.2

Definition

A family 6 = {11, | « € &/} of coverings of a set

is called a uniformizing family (1). Each pair of coverings ment 11, € G. (2). For each 11, € G there The family is separating if for is some 11, such that y € St(x,

if 11,, Uz € G have a common

Y

refine-

is a barycentric refinement 11, € G. each pair x # y of elements, there U,).

The intuitive idea is that sets belonging to the same covering have the same “size”. The precise relation between Il.l and 11.2 is

1.3

(a). Let 6 = {ll,| « € &/} be a uniformizing family. For each ae,let V, = J{U x U| Uel,}. Then w(6) = {V, | x e} is a uniformity.

(b). Let & be a uniformity. For each Ve, let U(V) be the covering {V[y] |y € Y}. Then the family (%) = {II(V) | Ve g} i1s a uniformizing family. Furthermore, u[¢(F)] 1s a uniformity equivalent to . Proof: 'The proof of (a) is straightforward, and is omitted. The proof for the first part of (b) depends on the observation that, whenever WoW-=1 CV, then St(y, W(W)) C V[y] for each ye Y. For the second part, one first establishes that

that VC

Vyy, =

Vo V' =1, and then notes

Vo V1,

We now use a uniformity to derive a topology. Any uniformity & in Y gives a topology 7 (%) in Y by taking the family {V[y] | Ve, ye Y} as basis. With this topology in Y, it is easy to verify that in the space Y x Y, the family {Int ' | V € &} is a uniformity equivalent to & and inducing the same topology; in particular, when considering the space (Y, 7(g)) there is no loss in generality to assume that each Ve is open in Y x Y. If one starts with a uniformizing family G = {1, | @ € o/} rather than with , the induced topology in Y is that having the family {St(y, U,) | ¥ € Y, « € &7} as basis; then for each « € &7, each Uel, is open and, furthermore, the uniformity u(G) gives the same topology.

203

Uniform Spaces

Sec. 11

A space with a topology derived from a uniformity is called a uniform space; it is Hausdorff if and only if the uniform structure (or uniformizing family) is separating. in Y ) be an arbitrary topological space. A uniformity Now let (Y, is called compatible with the given topology J whenever 7 (§) = 7. Assume,

for example,

that

Y is a metric space, and § is a uniformity

compatible with the topology. We have seen that equivalent metrics do not necessarily give the same uniformity, so the question arises: is there a metric that gives the topology of the space and also determines the given uniformity? The general answer to this question, which accounts for the importance of gauge spaces, is

11.4

Let (Y, 7) be a topological space, and § a separating Theorem uniformity compatible with the topology. Then Y must be a gauge space, and there exists a family & of gauges such that (1). The topology 7 is precisely the topology 7 (2). (2). The uniformity determined by £ is equivalent to the given uniformity .

Proof: We may assume that the uniformity is given by a uniformizing family 6 = {Ul, | « € &} of open coverings. For each open covering 1, € G, we first construct a gauge d, as follows:

Let B, = {Y}, 8; = U,, and define a sequence {B, | B, € G} of open coverings inductively so that each %,,, is a star refinement of B, (cf. VIII, 3.4). For each pair of points x,y € Y, let

Mx, y) = inf {27 | x € St(y, B,)} (X,

- vy )

= A%, %1) + -

4+ AXp1, Xi),

d (%, y) = inf {u(xy, - - -, x,) | all finite sets {xy, ---, %} C Y such that x, = x, x, = y}.

Then d, is a gauge: d,(x, y) = 0 whenever x = y and d, (%, y) = d,(y, x) for each pair of points ¥, y. The triangle inequality is immediate from the definition of d,(x, y) as an infimum. We now determine the size of the d,-balls. Itis evident that d,(x, y) < A(x, y) so that St(y, 8,) C B(y; d,, 1/2"1). We will show that also B(y; d,, 1/2%) C St(y, B,_,); this will follow once we prove that Ao, %)

< 2p(xo, - - -5 Xy)

because then we would have A(x, y) < 2d,(x, y) so that d,(x,y) < 2-" gives A(x, ¥) < 27"*! and therefore that x € St(y, B,_,). The proposed inequality is proved by induction on &, the assertion being true for £ = 1. Assume it is true for all 2 < 7, and let u(x,, - - -, X, %, ,1) = a. Let s be the largest index for which u(x,, - -+, x;) < a/2;

then (%,1, -+, ¥, 41) < a/2 also, and by the induction hypothesis we +1) < @, and certainly A(x,, x;,,) < a. find AM(x,, x;) < 2a/2, M(%g4%1,

204

'

Now choose the largest 2™

Chap. IX

Metric Spaces

< a. By the definition of A we must have

AMxgy %) < 27™ Mg, X541) < 27™, and A% 41, %,41) < 27™. It follows from this that there is a V € 8,, containing {x, %,,} and that x, and

%, ., belong to St(V, 8B,,), which is contained in some U € %B,,_,. Thus, Mg, %, ,1) < 27™*1 < 2a and the inductive step is complete. We now complete the proof of the theorem rapidly. Let 2 =

{d, | « € o/} be the family of gauges obtained in the manner indicated above, one for each U,. From what we have proved, the family 8(2) of balls is a basis for the topology .7 (J) so that 7 (2) = J . Verification that the uniformity determined by & is equivalent to & being routine, the proof is complete. It is not hard to verify that the family & of gauges determining the topology J (&) and giving the uniformity ¢ can be described alternatively as follows: & consists of all gauges uniformly continuous when the uniform structure & is used in Y and that determined by the metric d(x,y) = |x — y| is used in E1.

Problems Section

|

1. Let d:' Y x Y —> E?! satisfy: (1) d(x,y) = 0 if and only if x = 3, and d(x,y) < d(2,x) + d(z, y) for all x, y, 2. Prove: d is a metric. 2. Letd;,i

=1,2,---,n

be n metrics in a set

Y.

(2)

For any constants a; = 0, not

n

all zero, show that Z a; di(x, y) is a metric. 1

Section 2 1. Give an example to show that in a metric space (X, 7 (d)), it is not necessarily

true that (a) Bu(a, r) = {x| d(x,a) < r}, and (b) Fr[B(a,r)] = {x|d(x, a) = r}. 2. Prove that a pseudometric d in Y induces a topology that has {Ba(y, r) | y € Y, r > 0} as basis.

Section 3

1. Let (Y, 7 (d)) be a metric space.

Show that

_

_dxy)

is a metric in Y and that p ~ d. 2.

Let Yy,--+, Show that d[(xl)

Y, be metric spaces, and d, - - -, d, be metrics for these spaces. ) xn))

(yl,”

',yn)]

n

is a metric for the space [| Y. 1

=

max{di(xbyi)

l 1=

1) T

n}

205

Problems 3. Let a. b. c.

X = ]J0, 1[ C E'. Show: d(x,y) = [x~* — y~1] is a metric on X. d is equivalent to the usual metric do(x, y) = |x — y| on X. There exists no metric on E1 that coincides with metric d on X.

Section 4 1. Let {4, | « € 27} be a nbd-finite family of sets in a metric space Y. Letd be a metric for Y. Prove that the map y — sup {d(y, €4,) | a € 2/} of Y into E' is continuous. 2. Prove:

@) VvV (A is a single point).

a. 8(A) = 0« (A4 = b. 8(A4) = 8(A4).

c. AC B = 84) < &B). d. f AN B # @, then 84 U B) < 84) + (B).

3. Prove:

e

o

O

. d(4, B) = d(4, B). d(4,C) < d(4, B) + d(B, C) + 8(B). . d(4,C) < d(4,B) + d(4 v B, C) + &B). . d(4, B U C) = min[d(4, B), d(4, C)]. e. (A C BC A) = (d(p, A) = d(p, B) = d(p, A)).

4. Let S be the set of all closed non-empty subsets of a metric space be a fixed point of Y. In the set S, define

Y, and let P

dy(4, B) = sup{ld(y, A) — d(y, B)] e P |y e Y}. Show that d, is a metric in the set S and that d, ~ d; 5. In the complex plane, let Y = {z| |2| < 1}, and define follows: Extend the line joining x, y so that it intersects points v, u, and take the order of the points to be v, x, ¥, distance measured along this line and let

plan9) = low [

for any two p, g€ Y. p(x,y) in Y x Y as the circumference at u. Let d be ordinary

d(x,u) d(y,v)]

T8

)

Prove that p is a metric in Y. 6. Prove: A connected metric space with more than one point must contain at least 2¥o points. 7. Let d be a pseudometric in Y, and let R be the equivalence relation xRy < d(x,y) = 0. Then Y/R is a metric space (see l). Prove that, if the pseudometric topology of 2, Problem 2, is used in Y, then the projection p: ¥ — Y/R is a continuous open and closed surjection. 8. Let (X, d) be a metric space, and# (X) the set of all nonempty bounded closed sets in X. For (4, B) e #(X) x #(X) define h(A4, B) = sup {d(x, B) | x € 4} and let p(4, B) = max [k(A, B), h(B, A)]. Prove: a. p is a metric (called the Hausdorff metric) in #(X).

b. p({a}, {6}) = d(a, b).

c. p(4, B) < eif and only if [A C U(B, €)] A [B C U(4, €)]. d. p(4, B) < eifand only if BC U(A,€) and Va € A: B(a,e) "B

#

&.

206

Chap. IX

Metric Spaces

Section 5 1. Let (Y, .7 (d)) be a path-connected metric space, and f: I — Y be a pathin For each finite subdivision ¢, = 0 < t; a VU(y,) Vadb>a

:¢(b)e U. :¢b)e U.

Ex. 3 With the natural ordering ¥, in 1.2 coincide with those in I.1. Ex. 4 The set of all open coverings of a space Y, with {U} < {V} meaning {V'} refines {U}, is a directed set, since any two coverings have a common refinement. Similarly, the set of all partitions of Y, preordered in the same way, is a directed set.

Definitions 1.2 (1), (2) can be stated more directly by first introducing 1.3

Definition

Let D

be a directed set. The

mined by anaeDis {be D

terminal

set

| a < b}.

Then, for a net : D — Y, we have:

=Y

if YU(y)3IT,

>y

f

:(T,) CU,

YU VT, :p (T)NU # y

3.

T, deter-

211

Filterbases in Spaces

Sec. 2

This formulation of the definitions shows clearly that the convergence of ¢ is not determined by the set cp(D) C Y, but rather by the behavior of the famlly of sets {p(T,) | @ € D} in Y. Now, if we regard the map ¢ as serving simply to define the family {p(T,)} of sets, then we can think of convergence as a notion that involves only the space Y and pertains to any family of subsets of Y having properties similar to those of the family {@(7,)}. Such families of sets are called filterbases.

2.

Filterbases in Spaces

2.1

Definition Let Y be a space. A filterbase % in Y is a family A = {4, | « € &} of subsets of Y having the two properties:

(). Vees 1 4,# 2, N A4, A, C 2. Va VB 3y :4, Ex. 1 Let Y« consist of one nonempty set; then U is a filterbase. infinite set, {4 C Y | R(FA) < R} is a filterbase in Y.

Ex. 2

If Y is an

Let ¢: D — Y be a net. Then the family W(p) = {p(T,) | a€ D}

is a

filterbase in Y. For, given ¢(T,) and ¢(T}), first find a ¢ € D such that a < c, b < ¢, and then observe that T, C T, N T, because < is transitive. 2(gp) is

called the filterbase determined by the net ¢. Let Y be a space and y, € Y. The family U(y,) of all nbds Ex. 3 clearly a filterbase, called the nbd filterbase of y,.

of y, is

The Boolean algebra of filterbases is giVen in

22

Let A={4,]|« e&i} in Y. Then:

and

B = {B; | B€ %} be two filterbases

(1. AU B ={4,U B;| (¢, B) e/ (2). If each 4, " B; # @, then

x A} is a filterbase.

AN DB ={A, "B, | (0, B)eL x B} is a filterbase. (3). For each finite family {4,,--:,4,}C ¥, there is an A,e % such that 4, C 4, N --- N A, . In particular, each

finite intersection of members of a filterbase is not empty. Proof: (1) and (2) are trivial, since U distributes over N, and conversely. We prove (3) by induction. According to 2.1, the assertion is true for n = 2. Assuming its truth for n = k, we proveitforn = k + 1 by first findingan 4, C A4, Nn..-N A, andthenan4,C 4, N4 O +1°

Chap. X

212

Convergence

The convergence concepts for filterbases are as follows:

‘2.3

Let % = {4, | « € o/} be a filterbase in Y. Then:

Definition

(1). A converges to y, (written A—y,) if: V U(y,)I 4, : A, C U. (2). A accumulates at y, (written A > y,) if:

Y Ulyo) Y Ay A, " U # 2. It is evident from the definition (¢f. I1I, 4.3) that % > y, if and only if y, € () 4,; this alternative characterization of the accumulation points of a filterbase will be used very frequently. Ex. 4

Let U consist of a single set, A.

If A has only one point, then U — a;

if A has more than one point, 2 accumulates at each point of A. If Y is an infinite discrete space, the filterbase {4 | R(€A) < Ry} does not converge and has no accumulation point.

Ex.

5

Let

A(p)

be the filterbase

determined

by the net ¢: D —

Y. Then

@ — yo [p > ¥o] if and only if WA(p) — v, [WA(p) > yol. Ex. 6 Let U(y,) be converge to other points, Letg: [a, b] — Ex.7 of [a, b] ¢f. 1, Ex. 4. For points, and let Ry be'the

the nbd filterbase at y,. Then U(yo) as in Sierpinski space, where 1(0) — E* be given, and let {P} be the directed each partition P, leta = xo < x; < -++ set of all real numbers .ilg(gi)[xi

—

— ¥o; it may also 0 and also to 1. set of all partitions < X, = bbe its end

X;5-1),

for all choices & € [x;_1, x;]. Define Ap = |J {Rg | P < Q}; then A = {Ap} is a filterbase in E*, and g is Riemann integrable on [a, b] if and only if 2 converges.

The idea of ”subsequénce” is given by

2.4

Definition

Let

filterbases on Y.

%A = {4,|ae .o/}

and

B = {B; | Be %} be two

B is subordinate to A, written B — U, if

V 4, 1B,: B, C A,. Ex.8 Let(p) be the filterbase in Y determined by asequenceg. If: Z+ — Y is a subsequence of @, then A(}) — W) follows at once. However, it is not true that if B — A(p), then B is a filterbase determined by some subsequence of g, or even that 5 has at most countably many members. For example, if Y is an uncountable set, the sets of A(p) together with {4 | R(€A) < R} form a filterbase B — W(p). Ex. 9 The relation is a preordering in the family of all filterbases in Y, but I is not a partial ordering: In the space Y = Z*, the two distinct filterbases and B — A. A — B y B = {Ty, |neZ*}andA = {T,, | ne Z*} satisf

The subordination relation has the useful properties:

Sec. 3

25

Convergence Properties of Filterbases

213

(1). IfA C B, then B - A. (2). If 8 — ¥, then each member of B meets every member of . 3). A—y,if y and only if A — 11(y,). y

Proof: (1) is obvious. (2). Assume 3 B; 3 4,: A, N B; = &; since B — A, for this 4, we can find a B, C 4,, and then B,N B; = @ contradicts " that B is a filterbase. (3). Compare the definition of

9 - U(y,) with 2.3(1).

Remark I: In any space Y, the concepts of convergence based on filterbases and on nets are “equivalent” in the sense that first: (¢f. Ex. 5) each net ¢ determines a filterbase (p) such that ¢ — 3y, according to 1.2, if and only if A(p) — 3y, according to 2.3; and second: (G. Bruns and J. Schmidt) each filterbase ¥ determines a net ¢: D — Y such that A — y, if and only if ¢ — y,. To prove the second statement, it suffices to show that ¢ and D can be selected so that for each A €U there is a de€ D such that o(T;) = A. Let D be the set of -all ordered couples (a, A), where a € 4 € U, and preorder D by setting (a, A) < (b, B) whenever B C A;then D is directed because 2 is a filterbase. Defining the netp: D — Y to be the map (a, A) — a, it 1s trivial to verify that if T is the terminal segment determined by any given (b, B), then ¢(T") = B. It frequently happens, in any given instance, that one of these two methods for expressing convergence is more convenient than the other. Remark

2: In the Boolean algebra Z(Y), a family of sets # with the properties

1) [4,BeF)=>[AnBeF], and 2) (AeF) A (A CB)=(BeXF), is called a dual ideal; it is a proper dual ideal, or filter, if also (3) @ €. Itis easy to see that a family {A, | « € 9/} is contained in a filter if and only if {4,} is a filterbase; {4,} is then called a base for the smallest filter containing it.

3.

Convergence Properties of Filterbases

Though a filterbase may converge to more than one point (¢f. 2, Ex. 6), this can occur only in non-Hausdorfl spaces. 3.1

Theorem Y is Hausdorff if and only if each convergent filterbase in Y converges to exactly one point.

Proof: Assume that Y 1s Hausdorff and that %A — y,. For any Y1 # Yo, find disjoint nbds U(y,), U(y,); since by hypothesis there is some A4, C U(y,) and since any two A,, A, have nonempty intersection, there can be no 4; C U(y,); thus, % cannot converge to y; # ¥,. Conversely, assume that Y is not Hausdorff. Then there must exist

Yo, ¥1 such that V U(y,) ¥V U(y1) : U(yo) N U(y:) # @; by 2.2, B = U(yo) N U(y,;) is therefore a filterbase, and evidently 8 — y,, B — y,.

Chap. X

214

Convergence

The usual relations between the convergence properties of sequences and their subsequences extend to filterbases in the form 3.2

(1). If o converges to y,, then 9 accumulates at y, and,

Theorem

in Hausdorff spaces, at no point other than y,. (2). Let ¥ — A. Then;

(a). [A— o] = [B — y,]. (b). [B > yo] = [U > yo)Proof: (1). Given U(y,), there is some A, C U(y,); since each 4, must intersect A,, it follows that V Agz: A, N\ U(y,) # &, so A > y,. Now let Y be Hausdorff, and let y, # y,; choosing disjoint nbds U(y,), U(y,), there must be some A,e %A contained in U(y,); then A, N U(y,) = &, and so A cannot accumulate at y,. (2a). V U(y,) 3 A4,: A, C U(y,); since B +— A, there is a B; C A,, so B —y, also.

(2b). Given

U(y,) and A4, there is some B; C 4,, and since @, wefindV 4,: A, n U(y,) #

V Bg: B, N U(y,) #

:

A > y.

3.3

@, which proves

Corollary

(1). A —>y,ifandonlyif VB A IC B : € — y,. (2). A > y,ifandonlyif 3B — A : B — y,.

Proof: Ad (1). The “only if” is trivial. “If”: Assume that 2 does not converge to y,, so that 3 U(y,) V 4,: 4, ¢ U(y,). It then follows from 2.2 that 8 = A N €U is a filterbase, and clearly 8 ~ A. Since y, is not an accumulation point of 8 we find from 3.2(1), 3.2(2b) that no filterbase subordinated to % can converge to y,. Ad (2). If B+ % and B —y,, then by 3.2, B > y,, so that A > yo. Conversely, assume U > y, and let B = AN U(y,); B is a filterbase because all 4, N U(y,) # &, and we evidently have B ~ %, and B — y,. Restricting subsequence of sequences and sidering now

% and B to be sequences, so that ¥ — % means B is a A, 3.2 and 3.3 become the usual statements involving subsequences in the spaces of elementary analysis. Conthese sequential statements in general topological spaces,

it is clear that 3.2 is also true, and we will prove in 6.1, that 3.3(1) is valid too; however,

3.3(2) fails: that is, a sequence may accumulate to a

point y, and yet have no subsequence converging to y,. 'This pathological behavior is one reason that sequences alone are unsatisfactory for the purposes of general topology. Ex.1

(R. Arens).

Let

Y = [Z+

x Z*]

'

dorff) topology: (a). Discrete topology on Z*

x Z*.

v {(0, 0)} with the following (Haus-

215

Continuity; Convergence in Cartesian Products

Sec. 5

(b). Nbds of (0, 0) are all sets U containing (0, 0) and satisfying the condition: 3N vVn = N : U contains all but at most finitely many points of n x Z*, The sequence giving the diagonal enumeration of Z* x Z* accumulates at (0, 0); but it is easy to see that no subsequence can converge to (0, 0).

Closure in Terms of Filterbases

4.

In this section, we show that all the basic topological concepts can be for this, it suffices to characterize the closure

by filterbases;

expressed

operation. 4.1

Theorem Let Y be a space, and 4 C Y. Then y € 4 if and only if there is a filterbase on A converging to y.

Proof: Let ye A; then U(y) N A # o for each Ue11(y), so that B =ANu(y) is a filterbase on A4, and clearly 8 — y. Conversely, assume that B is a filterbase on 4 converging to y; then

Y U(y) 3B, : B, C U, so that every U(y) contains points of 4, that is, y € 4. Ex. 1

Theorem

ordinal space

4.1 is not true if we restrict attention only to sequences.

[0, £2] we

have

In the

2 e [0, Q[, but there can be no sequence in [0,

]

that converges to £2 (I1, 9.1). Thus, beside the failure of 3.3(2), sequences alone are generally incapable of expressing all topological concepts.

4.2

(a). A C Yisclosed if and only if the accumulation points of each filterbase on A4 all lie in 4. (b). ae A’ if and only if there is a filterbase on 4 — {a} that converges to a.

The simple proofs are omitted.

5.

Continuity; Convergence in Cartesian Products

By using filterbases, the continuity of a map can be expressed in a manner analogous to that in elementary analysis. We begin with the simple observation that if 2 is a filterbase in X andf: X — Y is any map, then (1): f(A) = {f(4) | A € ¥} is a filterbase in Y, and (2): if A ~ B,

then f() — £(B). 5.

Theorem Let f: X— Y. Then f is continuous at x, € X if and only if the filterbase f(11(x,)) converges to f(x,).

216

Chap. X

Proof:

Convergence

'The statement that f is continuous at x; is

Y W(f(x0)) 3 Ulxo) : /(U) C W. This is exactly the statement that the filterbase f(11(x,)) converges to S (o)

5.2

Corollary f: X— Y is continuous on X if and only if f(UA)—f(x) for each x € X and each filterbase % — x. Proof:

Assume f continuous

A — U(x) so that f(A) — f(U(x)),

on

X,

and

and

let A — x;

then

[2.5(c)]

since f(U(x)) — f(x),

so also

does f(). Conversely, assume the condition holds; we prove that f(A) C f(A) for all A C X. Let ae A; then there is a filterbase % on A with % — a, so that f(%A) — f(a); since f(A) is on f(A), this shows that f(a) € f(A).

For maps of arbitrary spaces into regular spaces, this formulation of continuity gives rise to a process called extension by continuity:

53

Theorem

Let D be a dense subset of X, let Y be a regular space,

and let f: D — Y be continuous. Then f has a continuous extension F: X — Y if and only if the filterbase f(D N u(x)) converges for each x € X. If F exists, then F is unique.

Sufficiency: Because D is dense, we have DN U # & for each Proof: fixed x,€ X and each Ueli(x,); consequently D N 1(x,) is indeed a filterbase in X. For each x € X, define F(x) to be the limit of f(D N 1u(x)); F(x) is uniquely defined because Y is Hausdorff. To prove that F is continuous at x, let W be any nbd of F(x); because Y is regular, there — F(x) is an open V with F(x) € V C V C W, and since f(DN U(x)) there is a U = U(x) with f(UN D) C V. Now, 8, = VN f(D N Uz)) e U: Indeed, each Un U(z) is open, so that is a filterbase for each DN UnN U(z) # @, and thus

(DN

Because B, V, that F(z) Necessity: each x, and from VII,

UN

UR)

C VN DN

UR).

— f(D N 1(2)), we find that B, — F(2), and since B, is in € V. Thus F(U) C ¥V C W, and continuity has been proved. This is immediate from 5.1, since D N U(x) — U(x), for F| D = f. That F is uniquely determined by f follows

1.5.

217

Adequacy of Sequences

Sec. 6

For filterbases in arbitrary cartesian products,

5.4

Theorem ()

A filterbase % on [|

— 35 for each fixed y.

Y, converges to {35} if and only if

7

Proof: Necessity is clear, since each projection map p, is continuous. Conversely, let be any basic nbd of {y}}. For each i = 1,---,n the convergence of p,(A) to y; yields an A;€ A such that p,(4;) C U,,; by 2.2(3), there is an 4 € A such that 4 C (n\ A;, and 1

clearly, A

C U,

,---, U, ).

The statement of 5.4 for sequences is the most commonly used version:

Ex. 1

Y, converges to {y5} if and only if each coordinate

explicitly, a sequence {y%} in [| 7

Y

o

Yy > Yy

6.

Adequacy of Sequences

We now show that, in 1° countable spaces, sequences and subsequences not only behave properly, but also are adequate to express all topological concepts; this accounts for their great utility in the metric spaces of elementary analysis. 6.1

Leto:Z*—>Xbea

sequence..

Then: (1). ¢ — 1y, if and only if each subsequence ¢’ of ¢ contains a subsequence ¢” such that ¢” — y,. (2). Let X be 1° countable. Then ¢ > x, if and only if there is some subsequence ¢’ > x,.

Proof: (1). T'he “only if” is trivial. “If”: Assume that ¢ + x,; then 3 U(xo) VY T,: o(T,) ¢ U. Proceeding by induction, let #; > 1 be the first integer in T such that o(n;) € U, and assuming thatn; < --- < n, have been obtained, let n,,; be the first integer in T, ,; such that o(n.,,) € U. Let ¢’ be the subsequence defined by ¢'(k) = ¢(n,). Since ¢'(T,) C €U for cvery n, no subsequence of ¢’ can converge to y,,. (2). Only the existence of ¢’ requires proof. Let U; D U, D -be a countable basis at x, and assume that @ > x,; then YU, V T,:

#(T,) NU; # . Proceeding by induction, let n; > 1 be the first integer in T, with ¢(n,) € U;, and assuming that #, < --- < n;, have been defined, let n, ,, be the first integer in T, , ; with ¢(n,;) € Uy.

The subsequence ¢'(k) = ¢(n,) evidently converges to x,.

218

Chap. X

Convergence

According to 6.1(2), sequences in 1° countable spaces cannot display the pathological behavior of 3, Ex. 1; that they are in fact capable of expressing all topological concepts in such spaces follows from

6.2

Theorem Let X be 1° countable and let 4 C X. Then x €4 if and only if there is a sequence on A converging to x. of 4.1, it is enough

Because

Proof:

to prove:

For

each x;, € X and

. filterbase B — x,, there is a sequence ¢ lying on (J{B | B € 8} each such that ¢ — x,. To this end, let U; O U, D - - - be a countable basis at x,. For

each

U,

find

sequence ¢ defined by ¢(7) is complete.

a B,

C

U; and choose a b, € B,,; then the

= b,, evidently converges to x,, and the proof

For maps of 1° countable spaces, we have 6.3

Let X be 1° countable. Then for Y arbitrary, and f: X —

Y:

(1). f(u(x)) converges to y, if and only if f(x,)—y, for each sequence x, — X. (2). f is continuous at x, if and only if f(x,)— f(x,) for each sequence x, —> Xo. (3). Let Y be regular, D C X dense, and g: D — Y continuous.

Then g is extendable to a continuous G: X — Y if and only "if for each x € X and all sequences {d,} C D converging to x, the sequences { f(d,)} all converge and to the same limit. (1). Assume f(U(x)) —y, and let the sequence ¢ — x; Proof: since A(p) — U(x), we find f(A(p)) — ¥, as required. For the converse, first note that if ¥: U; D U, D - - - is a countable basis at x, then B is a filterbase and B +~ U(x) — B, so that f(11{x)) converges if and only if f(3B) does. Now assume that f(B) does not converge to y,; this means that 3 W(y,) V U;: f(U,) ¢ W; choosing x;€ U; so that f(x)€ W, we find a sequence x, — x, such that f(x,) does not converge to yj, completing the proof. Proofs of (2) and (3) follow at once from (1) and from 5.1 and 5.3 respectively.

7.

7.1

Maximal

Filterbases

Definition A filterbase M in Y is called maximal (or an ultrafilter base) if it has no properly subordinated filterbase; that is, if VA

A~

M

=M

+—

A

Sec. 7

Maximal

7.2

219

Maximal Filterbases filterbases are characterized in

The filterbase M in Y is maximal if and only if for each 4 C Y, one of the two sets A, €A, contains a member of 9.

Proof: Assume I = {M, | B e %} is a maximal filterbase. Clearly, we cannot have an M; C A and an M, C ¥ A4, since then M;NM,= o. Assume now that V My;: M, ¢~A4; then all M; N4 # @, so A = M N FA is a filterbase. Since A — I, also M — A; consequently,

using any M; N €A, there is an M, C M; N €A C €A. Conversely, assume that the condition is satisfied and that % — 9%; given any A, € ¥, the condition assures that either there is an M; C 4, or an M, C ¥A,; the latter possibility is excluded, since the assumption %A — M implies [2.5(2)] that all MgzN A, # @. Thus M ~ A and 9 is maximal.

Maximal filterbases exist: In any space

filterbase I

€ M} = {M C Y |y,

Y, choose y, € Y; then the

is maximal,

since it satisfies 7.2.

More important, a maximal filterbase subordinate to any given filterbase can always be found: 7.3

Theorem Let B be any filterbase maximal filterbase M +— B.

in

Y. Then

Let of be the family of all filterbases % — Proof: empty, since B € &. Preorder .o/ by %' < % if A — A'. an application of Zorn’s lemma, consider any chain {2} A = | J A,. We first note that A is a filterbase: if 4, C

there

B; To in e A

exists

a

o/ is not prepare for ./ and let are given,

u

say Ae U, Ce ¥, and A, — A,, then I Be A,: B C C and so there is an E€ %, C Asuchthat EC AN B C AN C. Next, by using. 2.5(1), it is evident that % — A, A B for each %A,; this shows that % € &/ and that 9 is an upper bound for {,}. Thus, the use of Zorn’s lemma is legitimate, and we conclude that there exists a maximal M € &/; the filterbase M clearly satisfies the requirements of the theorem. For convergence properties,

7.4

Let 9 be a maximal filterbase in Y. Then M > y, if and only if M — Yo

Proof: Only the implication (M > y,) = (M — ¥,) need be proved. Given U(y,), there is an M, C U or an M, C €U, since M > y,, so that VM,: M, N U # @, the latter possibility is excluded, and therefore M — y,.

220

Chap. X

Convergence

Maximality is preserved under arbitrary maps: 7.5

Let

9

be a maximal

filterbase on X and p: X —

Y any map.

Then p(M) is a maximal filterbase in Y. Proof: We show that 7.2 holds for p(9). Given 4 C Y, we consider p~Y(A) and p~Y(FA) = €p~1(A); since M is maximal, there is eitheran M, C p~1(4)oran M, C €p~1(A4), and the conclusion follows.

Problems Section

|

1. Let (Y, d) be a metric space, and {y,} a sequence in Y. Prove: y, — y, if and only if d(yn, yo) — 0.

Section 2 1. Let A = {4, | « € &7} be a filterbase in X, and B = {B; | B € #} a filterbase in Y. Show A x B = {4, x By | (o, B) € x A} is a filterbase in X x Y. 2. let f: X — Y be a map, and *B a filterbase in Y. Prove that

fH®) = {f~UB) | Be b} is a filterbase in X if and only if f ~1(B)

#

& for each B e B.

3. Let f: X — Y be a map and U a filterbase in X. Prove: f(A) = {f(4) | Ae W} is a filterbase in Y. 4. Letf: X — Y be a map, U a filterbase in X and B a filterbase in Y. Prove: a. f=1of(A) is a filterbase in X and A +— f ~ £ (A). b. If f~(®B) is a filterbase in X, then fo f~1(B) — B. 5. Prove that the set of accumulation points of any filterbase is always a closed (possibly empty) set. Section 3

1. Let

Y be Hausdorff, let 2A — y, and let B — A. Prove:

If B accumulates at

any point y, then y = y,. 2. If the filterbase 2 does not converge to y,, show that there is a filterbase B — such that no filterbase € — B converges to .

Section 4 1. Prove the following analog of the usual diagonal process: Let U = {4, | x € 27} be a filterbase on Y, with 2 — y,. Assume that for each o € &7, there is a filterbase B, = {By; | Be %} with B, > a, € A,. Then there is a filterbase on

U B, converging to yo. o,f

2. Prove that A C Y is closed converges to a point of 4.

if and

only if each

convergent

filterbase

on 4

Problems

221

Section 5 1. Let f: X — Y be an open surjection. Prove: For each x € X and filt ‘rbase B - f(x), there is a filterbase A — x such that f(A) — B and B — f(N).

Section 6 1. Let Y be 1° countable. Prove: 4 C Y 1s closed if and only if the accumulation points of each sequence on 4 all lie in 4.

Section 7 1. Let I be a maximal filterbase in Y, and let 4, B C AV BeIN. Prove: Either 4 € M or B e IMN.

Y be disjoint and such that

Compactness

X1

In this chapter, we consider spaces having a strengthened version of the Lindel6f property; such spaces play an important role in all branches of mathematics.

I. I.I

Compact Spaces Definition A Hausdorff space Y is compact if each open covering has a finite subcovering.

Ex. 1 A discrete space is compact if and only if it is finite. The proof in VIII, 2, Ex. 2, shows the ordinal space [0, {2] is compact; note that a compact space need not be even 1° countable. Ex. 2 In any space X, all finite subsets, and &, are compact subsets. If @: Z* — X is a sequence convergent to x,, then 4 = xo U @(Z*) is a compact subset of X: any set of an open covering of 4 that contains x, contains all but at most finitely many clements of A. Observe that an infinite subset of A that does not contain x, is not a compact set. Ex. 3 FE!'is not compact, since the open covering | —n, n[,n = 1, 2, - - - has no finite subcovering. However, each closed finite interval [a, b] is compact: In fact,

given any open covering {U,} of [a, 6], let ¢ = sup {x finitely many U,}; if ¢ < b, we derive a contradiction choosing any U, D ¢, observing that there is a B(c, [a, ¢ — (r/2)] can be covered by finitely many sets U, with U, are a finite open covering of [a, ¢ + (r/2)]. 222

| [a, x] can be covered by to the definition of ¢ by r) C U, and that since - -, U,,, these sets together

223

Compact Spaces

Sec. 1

The position of compact spaces is 1.2

Theorem

Every compact space is paracompact (hence, also normal).

Proof: A finite subcovering of a covering is evidently a nbd-finite refinement of the given covering. Ex. 4 The converse of 1.2 is false, as an uncountable discrete space shows. Ex. 5 'Though metric spaces are also among the paracompact spaces, compactness and metrizability are not related: E! is metrizable, but not compact, and [0, £2] is compact, but not metrizable. Ex. 6 To emphasize the difference between paracompact and compact spaces, we have: Y is compact if and only if each open covering has a nbd-finite subcovering (rather than nbd-finite refinement). If Y is compact, the condition is evident; conversely, given any open covering {U,}, choose U,, # @ and consider the open covering

by

sets

the

VV, =

U,,

Y U,;

a nbd-finite

subcovering

of {V,}

must

evidently be finite.

The definition has several equivalent formulations:

1.3

Theorem

The following four properties are equivalent:

(1). Y is compact. (2). The finite intersection property: For each family {F, | o € 2/} of closed sets in Y satisfying (" F, = @, there is a finite

subfamily F, ,- -, F, with (\F, = o. 1

(3). Each filterbase in Y has at least one accumulation point.

(4). Each maximal filterbase in Y converges. Proof: (1)< (2). These two statements are de Morgan duals: The assertion that an intersection of closed sets, fl By is empty is equivalent

Y, that is, that the complementary open

to the assertion that U €B;= sets cover Y.

(2) = (3). Let A = {4, | a &M}

be a filterbase in Y; since all finite

intersections of the A4, are nonempty, so also are all finite intersections

of the A,

and therefore (2) assures

() A, # @. Thus,

the set of

accumulation points of % is not empty.

(3) = (4). Since M > y,, and M is a maximal filterbase, M — y,. (4) = (3). Given %, there is a maximal filterbase M — A; since M — Yo, we find A > y,.

:

(3) = (2). Let {F, | « € &/} be any family of closed sets, and assume that each finite intersection is nonempty. The sets F,, together with

224

Chap. X1

their finite intersections, then form a filterbase % in Y.

lates at some y, € Y, we

have y,€ (" F, = [ F,, o

intersection is not empty.

Compactness

Since 9 accumu-

and

therefore

the

@

Ex.7 E!isnot compact. Observe that the sets F,, = {x | x > n} are closed and each finite family has a nonempty intersection; yet, () F,, = . n

For invariance properties, we have

1.4

Theorem (1). The continuous image of a compact set is compact. (2). A compact subset 4 ofa Hausdorff space X is closed in X; indced, for each x € A, there arc nonintersecting nbds U(4), U(x). (3). A subspace of a compact space is compact if and only if it is closed. (4). (A. Tychonoff) Let {Y,| « €27} be any family of spaces. T’hen [ ] Y, is compact if and only if each Y, is compact. o

Proof: (1). Let Y be compact, and f: Y- Z continuous. Let {U,} be any open covering of f(Y); then { f ~(U,)} is an open covering f =Y (U,,), - - -, [ "YU,); of ¥ and so can be reduced to a finite covering, it is evident that U,, - - -, U, 1is a finite covering of f(Y). (2). To show that ¥4 open, we prove that cach fixed x, € ¥4 has a nbd lying in 4. For each « € 4, find disjoint nbds U(a), U,(x,). Since {U(a) N A | a € A} is an open covering of A, reduce it to a finite n

covering U(a;) N A4, - - -, U(a,) N 4; then U(4) = Ln) U(a;) and U(x,)= 1 n

() U, (x) are disjoint open sets. 1

(3). Because of (2), we need show only that if ¥ is compact and A C Y is closed, then A is compact. Let {B, | « € 27} be any family of sets closed in 4 with

() B, =

also closed in Y, so (n) B, =

@; since 4 1s closed

in

Y, the B,

are

w for some finitc subfamily, and by 1.3(2),

1

A is therefore compact.

(4). If |[

Y,iscompact, then because each projectionps: 1Y,

> Y,

is a continuous surjection, we find from (1) that each Y, is compact. Conversely, assume that each Y, is compact, and let 9 be a maximal filterbase in | [Y,; the image of a maximal filterbase being also a maximal

filterbase, each p,(9M) converges to some y,; by X, 5.4, we find that M- > {y,}; and so, by 1.3(4), [ [ Y, is compact.

225

Compact Spaces

Sec. 1

Ex. 8 The extended real line E! (I11, 7, Ex. 2) is compact, since it is homeomorphic to [—1, +1]. If X is not Hausdorff, 1.4(2) may be false: In Sierpinski space, the Ex. 9 subspace {0} is compact, but it is not closed. The Cantor set (I, 9, Ex. 3) and, more generally, any cartesian product Ex. 10 of finite discrete spaces is compact. Since the cartesian product of any family of closed unit intervals is Ex. 11 compact, we have that (a) the Hilbert cube I® is compact and (b) a space is completely regular if and only if it is a subset of a compact Hausdorft (hence, normal) space (c¢f. VII, 7.3).

In (Hausdorff!) spaces, the compact subsets behave as points do and have the same scparation properties: .5

(a).

A finite union

of compact

subsets

of a Hausdorff space is

compact.

(b). T'wo disjoint compact subspaces of a Hausdorfl space have disjoint nbds. (c). If A is a compact subset of a regular space, then for each open U D A, thereisan open VwithAC VC V¢ U.

(d). Let {U, | « € &/} be a nbd-finite open covering of a space X. If A C X is compact, then 4 has a nbd meeting at most finitcly many sets U, (in precise terms: there exists a nbd U of A such that U N U, #

Proof: (a) 1s (b). Let 4, 1.4(2), for each open covering

¢ for at most finitely many indices o).

trivial. B be compact subsets of X, and let A " B = @. From be B there are disjoint nbds U,(A4), U(b). From the {U(b) N B | beB} of B, extract a finite subcovering

Ub)N B,.-, Ub,)NDB; then U U(b,), fl U, (A) are the required nbds.

(c). For each ae A there is a nbd extracting a finite subcovering gives

V(a) such

that

V(a) C U;

AcOw@cQw@cu (d). Each a € A has a nbd U(a) meeting at most finitely many U,; extract a finite subcovering U(a,), - - -, U(a,); since each U(a;) meets

only finitely many

U, so also does Lnj U(a)). 1

Families of compact subsets of a space have the useful property:

1.6

Let X be Hausdorff, let w, be an initial ordinal, and let {F, | p < w,} be a descending family of nonempty p p,. Proof:

For the proofs of (a) and (b) we can assume that X is compact,

otherwise we work in F;.

(a). Since each finite intersection of the F, is nonempty, so also is () F, = C; since each F is closed in the closed F;, C is closed in X.

*

(b). From N F, C U, we find €U C \J F,; since €U is H

u

closed, and hence is compact,

n

extract a finite covering | J €F,; then, 1

for any p > py = max(u,, - - -, u,), we have

F,CF, =(F,CU. 1

2.

Special Properties of Compact Spaces

Some properties of compact spaces that are frequently used, and which contribute to their importance, are given in this section. We have seen that a continuous bijection need not be a homeomorphism; one of the important features for maps of compact spaces is

2.1

Theorem Let Y be continuous. Then:

compact,

Z

be

Hausdorff,

and f: Y - Z

(1). fis a closed map. (2). If f is a continuous bijection, then f is a homeomorphism. Proof: (1). Let A C Y be closed; it is compact and consequently so is f(A). Since Z is Hausdorff, f(A4) is closed in Z. (2) is an immediate consequence [cf. III, 12.2]. ' Ex. 1 A continuous map of a compact space into a Hausdorff space need not be open, even if it is injective. Let Y = [0,1] v {2}, Z = [0, 1], and f: Y — Z be the map y — 1y. Y is a closed subset of [0, 2] and thus is compact; f is not open, since the open set {2} does not have open image. Ex. 2 The hypothesis that Z be Hausdorff is essential, as 1: 2 — %, the Sierpinski space, shows. : Ex. 3 The delicate position of a compact Hausdorff topology .7 in a set Y is shown by its two properties:

(a). If 7 is a proper subset of

,,

then (Y, .7 ,) is not compact.

(b). If 7 _ is a proper subset of 7, then (Y, .7 _) is not Hausdorff. For, in case (a), the continuous

1: (Y,

,) — (Y, 7))

would

be

a homeomor-

phism if 7, were compact; in case (b), the continuous 1:(Y, 7 )—(Y,7 _) would be a homeomorphism

if

_ were Hausdorff.

227

Special Properties of Compact Spaces

Sec. 2

Ex. 4 We have seen (IV, 4.4) that there is a continuous surjection I -— I™, 2 < n < o; according to 2.1, there can be no continuous bijection I — I", since we know that I and I" are not homeomorphic.

Corollary Letp: X — Y be an identification, and let ~: X — Z be continuous. If X is compact, if Zis Hausdorft, andif hp~1: ¥ — Z is bijective, then Ap~! is a homeomorphism.

2.2

Proof: According to VI, 3.2, hp~! is a continuous bijection, so we need to show only that Y is compact. Now, because Z is Hausdorff, VII, 1.5 (4) shows that Y is also Hausdorff and therefore, since p is surjective, Y is indeed compact. Ex.

5

The

cone

T'S™ over S"

(VI,

5.1) is homeomorphic

to the ball "*+1,

For, let p: S™ x I— T'S™ be the identification map, and 4: S® x I — IV"*! the map h(x, £) = (1 — t)x; then kp~1 is bijective, and since S™ x I is compact, 2.2 ' applies. Similarly, the suspension of S" is homeomorphic to S™**1,

The next general feature is 2.3

Let Y be compact and f: Y — E! be continuous. Then Theorem f attains its supremum and its infimum, and both are finite. Precisely, there is at least one y, € ¥ with f(y,) = sup {f(») |y € Y}

and at least one y, € Y with f(y,) = inf { f(y) |y € Y}. This is a consequence of the more general 2.4

Let Y be compact, and f: Y— E* a map. If f is lower (upper) semicontinuous, then it attains its infimum (supremum).

Proof: Assume f to be lower semicontinuous, and let m =inf f(Y). For each ¢ > m, the set F, = {y | f(») < ¢} is closed because f is lower semicontinuous; and it is not empty, by the definition of m. Since the intersection of any finite family of the sets F, is not empty, the compactness of Y assures (| F, # @; and for y, € (" F,, we clearly q>m

a>m

have f(y,) = m. The result for upper semicontinuous f follows from this by noting that —f is then lower semicontinuous.

In cartesian products, we know that each projection is an open map and that it need not be a closed map. We show now that projections “parallel to compact factors” are also closed maps. 2.5

Let Y be compact, Z Hausdorff, and p: Y x Z— Z Theorem the projection “parallel to the compact factor Y.” Then p is a . closed map.

228

Chap. X1

Compactness

Proof: Let F C Y x Z be closed; we show that Z — p(F) is open. Let 2 € Z — p(F'); then (Y x 2) N F = g, so that each point (7, 2,) has a nbd U(y) x U,(z,) not intersecting F. From this open covering of the compact Y x 2, extract a finite covering U(y;) x U, (),

¢ =1,---, n; then (n] U,,(20) is a nbd of %, that does not meet p(F). 1

The most frequently used version of this result is 2.6

Corollary Let A C X be arbitrary and let Y be compact. Let U beanbdof A x Yin X x Y. Then there is a nbd VV O A4 such that the tube V' x Y C U.

Proof: Let p: X x Y — X be the projection; since p is closed and p~1(4) = A x Y C U, the result follows immediately from III, 11.2(1). We have seen that a continuous map into a Hausdorff space has a closed graph; for maps into compact spaces, 2.7

Theorem Let X be Hausdorff and Y be compact. Then f: X — Y is continuous if and only if its graph G( f) is closed in X x Y.

Proof: f: X — Y closed in parallel to

We need show only that if G(f) is closed in X x Y, then is continuous. Let B C Y be closed; then py(B) N G(f) is the closed G( f) and therefore in X x Y. The projection py the compact factor Y is a closed map, and since

px[py {(B) N G(N)] = f~X(B), this establishes the continuity of f.

3.

Countable Compactness

A generalization of the compactness concept will be considered in this section. 3.1

Definition A Hausdorff space is countably compact countable open covering has a finite subcovering.

if

every

Ex. 1 Every compact space is evidently countably compact, but the converse is not true. The ordinal space [0, £[ is not compact, since it is not a closed subset of [0, £2] (or, alternatively, since it is not paracompact). We now show that [0, [ is in fact countably compact. Let {U, | n € Z*} be any open covering of [0, £[; observe that for each B < £, the interval [0, B] is compact, since it is closed in [0, £2], and therefore can be covered by finitely many of the U;. Now, if there were

229

Countable Compactness

Sec. 3

no finite subcovering for [0, [, then for each n &€ Z* we could find an element an € Uy Y-V Uy; since the «, have an upper bound o < 0, we would have an interval [0, ag] that cannot be covered by finitely many of the U,.

Countable compactness is characterized by the behavior of sequences only, rather than arbitrary filterbases, as the following equivalent formulations show: 3.2

Theorem The following three properties are equivalent: (1) Y is countably compact. (2). (Bolzano-Weierstrass property.) Every countably infinite subset of Y has at least one cluster point. (3). Every sequence in Y has an accumulation point.

Proof: (1) = (2). Assume 4 C Y were a countably infinite subset with no cluster point. Then 4 would be a closed discrete set, and so each point a;€ 4 would have an nbd U; not containing any other

member of 4. Thus, {U, |ie Z+} U {€ A4} would be a countable open covering of Y that has no finite subcovering. (2) = (3). Let p: Z* — Y bea sequence in Y. If X[p(Z*)] = KX,, then the set ¢(Z*) has a cluster point y,; since each nbd U(y,) contains infinitely many points of ¢(Z*), we have U(y,) N ¢(T,) # < for each meZ*,s0 @ >y, If R[p(Z*)] < KX, then there is some y, such that @(n) = y, for infinitely many #, so again ¢ > y,. (3) = (1). Assume that Y is not countably compact. Then there is a countable open covering {U; | i € Z*} that cannot be reduced to a finite covering.

n

For each n € Z* we can therefore find some y, € ¥ —~ (J U; ;

1

then n — y, defines a sequence ¢ in Y that does not have an accumulation point: each y € Y belongs to some U, and ¢(T,,) N U,y = 2. Countable compactness reducesto compactness in a class of spaces that includes the paracompact spaces. Calling a space metacompact if each open covering has a point-finite open refinement, we have 3.3

Theorem (R. Arens; J. Dugundji) A space is compact if and only if it is both countably compact and metacompact.

Proof: We need prove only that if Y is countably compact and metacompact, then Y is compact. Let {U, | « € &7} be any open covering of Y. By the metacompactness, {U,} has a point-finite open refinement

{Vs| Be #} and, by VIII, I.1, {V,;} has an irreducible subcovering {V, | v € ¢}. This minimal covering must be finite: for we can find in each V, a point y, belonging to no set other than V', and, if {V, | y € ¥}

Compactness

Chap. X1

230

were not finite, then {y, | y € 4} would be an infinite set in Y with no accumulation point. Thus, by choosing for each V/, some set U,,, D V,, we reduce the covering {U, | « € &/} to a finite covering. 3.4

Corollary Countable compactness is equivalent to compactness in (a): paracompact spaces and in (b): arbitrary Lindelof spaces.

Proof: (a) is immediate from 3.3. (b) Given any open covering, the Lindelof property gives a countable subcovering, and then the countable compactness permits a further reduction to a finite subcovering. We now study the topological properties of countably compact spaces. Since countable compactness is based entirely on sequences, we can expect it to impose some topological restrictions in the 1° countable spaces. Thus, although countably compact nonregular (Hausdorff) spaces exist, we do have 3.5

A countably compact 1° countable space Y is regular.

Letye Yandanbd

Proof:

Vofy be given,andlet V; O V, D - ..

be a countable nbd basis at y. From VII, 1.2(3), it follows that fi vV, = 1

{y}, consequently {F} U {€V,|ne Z*} is a countable open covering of Y. Because this covering has a finite subcovering, there is some

integer n such that ¥ = V U Lnj €V, =Vu¢¥ 1

(n] V,; therefore V, = 1

(n] V, C V, and the proof is complete. 1 For the invariance properties we have

3.6

Theorem (1). The continuous image of a countably compact space is countably compact. (2). A closed subspace of a countably compact space is countably compact. (3). If X is 1° countable, and 4 C X is countably compact, then A is closed in X.

(4). The

cartesian

product

need not be countably compact.

of two

countably

compact

spaces

However, if each X;, 1€ Z7%,1s 1°

countable, then [ [ X; is countably compact if and only if each X; 1

is countably compact. Proof:

'The proofs of (1) and (2) are left for the reader.

Sec. 3

231

Countable Compactness

(3). Let x € A; because X is 1° countable, there is a sequence ¢ lying on A and converging to x. Since A4 is countably compact, ¢ has an accumulation point in 4, and since x is the only accumulation point of ¢, we have x € A.

Thus, A4 is closed.

(4). We will give an example in 8 of a separable countably compact completely regular space Y such that ¥ x Y is not countably compact. To prove the second part, note first that because of (1), the countable o0

compactness of [| X; implies that of each X,. For the converse, we will | 1 use the Cantor “diagonal process” to show that each sequence ¢ in o]

] ] Xi has an accumulation point. Let p, denote the projection onto the 1 nth factor. Now, p, o @ is a sequence in X, and since X is 1° countable and countably compact, we can (X, 6.1) extract a subsequence ¢, of ¢ such that p, o ¢, converges to some x; € X;. Now consider p, o ¢,; for the same reason as before, we can extract a subsequence g, of ¢, such that p, o p, converges to some x; € X,. Proceeding by induction, we obtain a family {p, | n € Z*} of subsequences of ¢ such that ¢, , is a subsequence of ¢, for each n € Z*, and each sequence p, o p, converges to some x, € X,. Now let ¢ be the subsequence n — ¢,(n) of ¢; then, for each fixed &, we have {¢(s) | s > &} C {@i(s) | s = &}, so ¢ ~ ¢, and [X, 3.2 (2a)] therefore p; o ¢ — x;. By X, 5.4, we find that ¢ — {x,} and, because ¢ ~ o, it follows that ¢ > {x,}. The proof is complete. It is clear that each continuous real-valued function f on a countably compact space Y is bounded, because the countable open covering {y | |f(¥)| < n},ne Z*, has a finite subcovering. This behavior motivates an extension of the notion of countable compactness.

3.7

-

Definition A Hausdorfl space Y is pseudocompact real-valued function on Y is bounded.

if every

continuous

Ex.'2 Every countably compact space is pseudocompact, but the converse is not true. Let Y = [0, £2] x [0, w] — (£2, w); since the set {(2,n) | ne Z*} has no cluster point in Y, this space is not countably compact. To prove that Y is pseudocompact, it suffices to show that each continuous f: Y-— E* can be extended over the compact [0, £2] x [0, w], since 2.3 then applies. Now, by III, 8, Ex. 7, for each n < w there exists an a, € £ and a constant ¢, such that f(a, n) = ¢, for all @ = o,; letting oy = sup a,, we find o, < £2 and that f is constant on each strip {(e, n) | @ = «o}. The reader can verify that by defining f(£, w) = ¢4, the extended function is continuous at (£2, w). Pseudocompactness may be expected to be significant in completely regular spaces where there are sufficiently many nonconstant maps f: Y -— E!. For such spaces, we have the equivalent formulations

Compactness

Chap. X1

232

Let Y be completely regular. The following three properties are Theorem equivalent: (1). Y is pseudocompact. (2. If V, D V5 D - - is any descending sequence of nonempty open sets,

3.8

then NV,

#

.

1

closures cover

(1) = (2).

Proof:

open

countable

(3). Each

covering

of

Y has

a finite

D

/3

D ---

and

subfamily

whose

@.

Then

Y.

V;

that

Assume

F) V., = 1

{V,|ne Z*} must be a nbd-finite family: for if each nbd of some given y€ Y meets infinitely many V;, then because the I/; descend, each nbd of ¥ would meet V; and

all the

belong

therefore y would

to

{0'0] V..

ne Z*,

for each

Now,

1

gn: Y-—>E' gAY

V,)

—

be

a continuous

= 0. Because

map

that g,(y,)

such

{V,, | n € Z*} is nbd-finite, the function

and

y, €V,

= n for some

let

2 gn is well1

defined and continuous; it clearly is not bounded, so Y is not pseudocompact. (2) = (3). Let {U,|neZ*} be a given countable open covering, and let — C) U, for each ne Z*; we note that V.C 1

V,=Y Now,

if no

V,

were

empty,

then

we

would

have

(n]%_(_]ic 1 &

#

=

fi%fi 1

fi V,C 1

{i](gUi. thus

fi %U,; 1

0

Y # U U, thatis, {U, | n€ Z*} would not be a covering. 1

(3) = (1). Let f: Y-— E* be continuous. Setting U, = {y| |f(3)| < n} for each n € Z*, we obtain a countable open covering {U, | n € Z*}. Since there is an integer m such that U, v---U

U,

=

Y, the function is bounded.

Pseudocompactness reduces to countable compactness in a class of spaces that includes the normal spaces. A completely regular space is called weakly normal if each two disjoint closed sets, one of which is countable, have disjoint nbds. This

is actually weaker than normality: it is weakly normal. weakly

In

3.9

normal

spaces,

[0, 2]

but

x [0, 2] — (£, £) is not normal,

pseudocompactness

is equivalent

to countable

compactness.

Let Y be weakly normal. If Y is not countably compact, then there is Proof: a countably infinite discrete closed set D C Y. Now, according to VII, 2.4, we can find a system {U, | n € Z*} of open sets whose closures are pairwise disjoint, and such that DN U, # @ for each ne Z*. Choose y, € D N U,; since E = {yn|n€ Z*} is a countable closed set, we can find disjoint nbds W > E and

VO&®

G

U,.

Then

the covering

{VV} U {U, |ne Z*}

has no

finite

subfamily

1

whose Thus,

closures cover Y: each y, belongs Y is not pseudocompact.

only to

U, and

no y, belongs

For the invariance properties of psuedocompactness, we have (a). The continuous image of a pseudocompact space is pseudocompact.

to

V.

233

Compactness in Metric Spaces

Sec. 4

(b). Although a closed subspace of a pseudocompact space need not be pseudocompact, the reader can verify that if the pseudocompact space Y is completely regular, then the closure of each open subset of Y is pseudocompact. The example to be given in 8 will show that the cartesian product of pseudocompact spaces need not be pseudocompact.

Compactness in Metric Spaces

4.

In metric spaces, there i1s no distinction between countable compactness and compactness, since metric spaces are paracompact and 3.4 applies; this serves to explain the importance of the Bolzano-Weierstrass property in metric spaces. Furthermore, compact metric spaces are always 2° countable; in fact,

4.1

Theorem A countably compact space Y is metrizable if and only if it is 2° countable.

Proof: Sufficiency: 2° countability assures first (3.5) that Y is regular and then, by Urysohn’s theorem (IX, 9.2), that Y is metrizable. Necessity: By 3.4, a countably compact metric space Y is compact, and since a compact space is Lindelof, IX, 5.6 shows that Y is 2° countable. Ex. 1

Since [0, £2] is compact, but has no countable dense set, this is another

way to see that [0, £2] is not metrizable.

For subsets of metric spaces, 4.2

Let (Y, d) be a metric space, and 4 closed and bounded. Proof:

Due

to

1.4,

only

the

C

Y compact.

boundedness

of A

Then

requires

4 is proof.

Choose a4 € 4, and define f: 4 — E* by a — d(a, a,); being continuous,

f attains a finite maximum m and clearly 8(4) < 2m. The converse of 4.2 is not true, as an infinite discrete space (metrized so that distinct points have distance 1 from each other) shows. However, in one very important case, the properties are characteristic: 4.3

Theorem In E", a set is compact if and only if it is closed and bounded.

Proof:

Since A is bounded, 4 C [ ] I, where each I, is some finite 1 n

closed interval [a;, b;]. By 1.4(4) and I, Ex. 3, [[ I, is compact; since 4 1 n

is closed in E™, 4 is closed in || I, and is therefore compact. 1

234

Chap. X1

Compactness

We have seen (IX, 4, Ex. 3) that in metric spaces, two disjoint closed sets may have distance zero; we now show

4.4

Let X be a metric space, 4 C X closed, and C C. X compact. IfCnNnA4d= g@,thend(4,C) > 0.

Proof: Since d(c, A) is a continuous real-valued function on C, it attains its minimum at some point ¢,, and d(c,, 4) = inf{d(c, 4) | ¢ € C}

= d(C, A); since ¢, € A = A, we have d(c,, A) > 0.

Ex. 2 It need not be true that there are points a € 4, ¢ € C with d(4, C) = d(a,c). InX = E* — {0},letA ={— 1/n|ne Z*}and C = [1, 2].

Open coverings of compact metric spaces have the important property

4.5

Theorem Let (Y, d)bea compact metric space and {U,} be an open covering of Y. Then there exists a positive number A({U,}), called a Lebesgue number of the covering, with the following property: Each ball B(y, A) is contained in at least one U,,.

Proof:

Foreachye Y, choose r(y) > 0 so that B(y, r(y)) C some U,;

and from the open covering {B( v, n(y ))}

({0 5

extract a finite subcovering

2

=

Let

1,. }

=min[

then A > 0 is a Lebesgue

r(y1) 2

number.

r(yn)]. T2 For, given any B(y, A), we have

yE B(yi, 7(3’1)) for some ¢, and so for any 2z € B(y, A),

Az 3) < d(9) + dy3) < A+ "2 < (g i

that is, B(y, A) C B(y;, 7(y;)). result follows.

Since the latter set lies in some

U,, the

Ex. 3 'Theorem 4.5 frequently occurs in the dual form: If Y is compact metric and {Fa} 1s a family of closed sets with fl F, = @, there is a A > 0 with the followmg property: If M least one F.

C

Yis any set w1th 0(M)

< A, then M does not meet at

A basic application of 4.5 generalizes the classical theorem uniformity of continuity: 4.6

on the

Theorem Let Y be a compact metric space, Z an arbitrary metric space, and f: Y — Z continuous. Then for each ¢ > 0, there is a 8(¢) > 0, depending only on ¢, such that f(B(y, 8)) C B(f(»), ¢) for every y € Y (that is, f is uniformly continuous).

Sec. 5

235

Perfect Maps

Cover Z by the balls {B(z, Proof: number of the open covering { f ~1(B(z, lies in one of these sets, f(B(y, 6)) C cause f(y) € B(z, ¢/2), we find for any

¢/2)} and let § be a Lebesgue ¢/2))} of Y. Since each B(y, ) B(z, ¢/2) for some 2, and be¢ € B(y, 6) that

d(f(€), f(3)) < d(f(€), 2) + d(=, f(y)) < &]2 + ¢[2 = ¢, that is, f(B(y, 8)) C B(f(»), ¢). Let Y be a compact metric space. Because of 3.2(3) and X, 6.1(2), Remark: from each sequence in Y we can extract a convergent subsequence. However, it -is possible to define a notion of convergence in compact metric spaces according to which every sequence converges to a unique limit; such a generalized convergence is used to define a Haar measure in locally compact topological groups. In Z*, introduce one of the finitely additive measures u: #(Z+) — 2 of 1I, 2.5(3); recall that u(4) = 0 if 4 is finite, and w(Z*) = 1. Given a sequence ¢: Z* — Y, Y Hausdorff, define Lim¢ = yo if V U(yo): u{n | p(n) e U} = 1; it is trivial to verify that (1) Lim ¢, if it exists, is unique (depending on the u used); (2) if ¢ — yo, then Lim @ = y,, independently of the p used. In addition, if Y is compact metric, every sequence converges: covering Y by finitely many closed balls B(y, 1), at least one satisfies u[¢p ~1(B(y, 1))] = 1; covering that one by closed balls of radius 4, and repeating the process, 1.6 shows Lim ¢ exists. In the particular case that Y = E', we find: (a) Every bounded sequence converges; (b) lim %, < Lim x, < limx,

converge); (c¢) Lim x,-Lim y,. same Lim.

5.

(its value

depending

on

the p used

if {x,} does

not

Lim(x, + y,) = Limx, + Limy,; and (d) Lim(x,-y,) = However, a subsequence of a given sequence need not have the

Perfect Maps

We have seen that normality and paracompactness are preserved under continuous closed maps. In this section, we consider a type of map under which the image generally inherits all the properties of the mapped space. 5.1

Definition A map p: X — Y is called perfect (or proper) if it is a continuous closed surjection and each fiber p~1(y), (ye YY) is compact.

5.2

Theorem (1). (2). (3). (4).

Letp: X — Y be a perfect map. If If (5. If

Then:

X is Hausdorff, so also is Y. X is regular, so also is Y. Hanai) If X is metrizable, so also is Y. X is 2° countable, so also is Y.

236

Chap. XI

Compactness

Proof: Ad (1). Let y;,y, be distinct points of Y; then p~1(y,), p~(y,) are disjoint compact sets in X and therefore have disjoint nbds U,, U,. Since p is closed, there are (1II, 11.2) open V; D y, with P y) Cp V) C U,i=1,2, and then V,, V, are disjoint nbds of Y15 Y.

Ad (2). Given y € U in Y, there is by 1.5(b) an open IV C X with p YY) CV CV Cp-YU). Since p is a closed map, we find a nbd

W Dy with p~(y) C p~Y(W) C V; then W C p(V) C U, and since p(V)is closed,ye W C W C p(V)C U. Ad (3). We will show that Y satisfies the metrizability condition in IX, 9.5(2). Letd be a metric for X, and for eachn € Z* let &, be a nbdfinite closed refinement of the covering {B(x, 1/n)| x € X}. Define

&, = {p(F) | F € F,}; each @, is a closed covering of Y, and we first prove that each &, is also nbd-finite: Let y € Y and n be fixed. Since p~Yy) is compact, there is a nbd U D p~1(y) that intersects only finitely many members of §,, and since p is closed, there is a nbd W(y) such that p~}(W) C U; then W clearly meets at most finitely many members of &,,. Now let U be any nbd of y € Y.

Because p~1(y) is compact, we have

d(p~y), €p~*(U)) = 1/n > 0 for a suitable n; then St(p~1(y), Fes) C p~YU) and therefore St(y, G2,) = p St(p~), F2n) C U. Since Y is obviously 7, the space Y is metrizable. Ad (4). Let {U;} be a.countable basis for X, and let {V;} be the family of all finite unions of the Uj; by 11, 8.8, the family {V/;} is countable, and we show that the opensets W, =

Y — p[X — V] are a basis for Y.

Let y e W; then p~(y) C p~}(W), and since p~!(y) is compact, there are finitely many sets Uy, - - -, U, such that p~1(y) C Ln) U, Cp 1

Y (W);

letting V,, = C) Ui, it follows that y € W,, C W, completing the proof. I Perfect maps also preserve certain properties under inverse images; in fact they certainly preserve those properties determined by the behavior of open coverings:

5.3

Theorem (1). (2). (3). (4).

Let p: X — If If If If

Y Y Y Y

is is is is

Y be a perfect map.

Then:

paracompact, so also is X. compact, so also is X. Lindelof, so also is X. countably compact, so also is X.

Proof: Ad (1). Let {W, | « € &/} be any open covering of X. For each y € Y, extract a finite covering W,,, 7 = 1,2,---, n(y) for the

Sec. 6

237

Local Compactness

and, since p is a closed map, find a nbd V(y) of y ny ) such that p~1(V(y)) C U W,y Now let {U(y); be a precise nbdcompact fiber p ()

1

finite open refinement of the open covering {V(y) | ¥ € Y}, and for each

ye¥, i=1--ny),

let W, a() = p U] N W, The

family {W(y, «;(y))} is an open covering of X that clearly refines {W,}; and it is nbd-finite: For, given any x, € X, there is a nbd W of p(x,) intersecting at most finitely many sets U(y) and the nbd p~}(W) of x, then intersects at most finitely many W(y, «;(y)). The proofs of (2) and (3) are entirely analogous and are left for the reader.

Ad (4). Let {W,|ne Z*} be any countable open covering of X. The open sets V, =

¥ — p[X — C) W,

Y: given any

ne Z*, cover

i y € Y, the compactness of p~(y) assures that p~1(y) C Lmj W, for some 1

m, and therefore that y€ p[X — (n) W]. Since there is a finite subi covering Vy, - -+, Vi for Y, and since p~}(V;) C W, for each i, the sets Wi,

-+ -, W

are a finite subcovering for X.

As an application of 5.3, 5.4

Let X be compact. If Y is paracompact (resp. Lindelsf, resp. countably compact), then X x Y is also paracompact (resp. Lindel6f, resp. countably compact).

Proof: Because X is compact, it follows at once from 2.5 that the projection map p: X x Y — Y is perfect.

This result should be contrasted with the known fact that (VIII, 2.4(3)) the cartesian product of paracompact spaces need not be paracompact.

6.

Local Compactness

Many of the important spaces occurring in analysis are not compact, but have instead a local version of compactness. Calling a subset 4 of a space relatively compact whenever its closure 4 is compact, this local property is formalized in 6.1

Definition A Hausdorff space is locally compact if each point has a relatively compact nbd.

238

Chap. XI

Compactness

Ex.1 A compact space is locally compact. E™, and any infinite discrete space is locally compact, but not compact. The set of rationals in E*! is not a locally compact space. A Hilbert space /2(X), with X > R, is not locally compact, as IX, 8, Ex. 1, shows.

Equivalent formulations are given in 6.2

Theorem

The following four properties are equivalent

(1). X is locally compact. (2). For each x e X and each nbd U(x), there is a relatively compact open ¥V with xe V C ¥V C U. (3). For each compact C and open U D C, there is a relatively compact open V with CC V' C V C U. (4). X has a basis consisting of relatively compact open sets.

Proof: (1) = (2). There is some open W with xe W C Wand W compact. Since W is therefore a regular space, and W N U is a nbd of

x in W, there is a set G open in W such that xe G C Gy C Wn

U.

Now G = E N W, where E is open in X, and the desired nbd of x in XisV=EnW. (2) = (3). For each ¢e C, find a relatively compact nbd V(c) with V(c) C U; since C is compact, finitely many of these nbds cover C, and by 1.5(a), this union has compact closure. (3) = (4). Let & be the family of all relatively compact open sets in X; since each x € X is compact, (3) asserts that # is a basis. (4) = (1) is trivial. The reader should observe that the relative compactness of V' distinguishes (3) from 1.5(c). Furthermore, the basis specified in (4) is very large; we obtain better information in the useful

6.3

Let X be 2° countable. If X is locally compact, it has a countable basis consisting of relatively compact open sets.

Proof: Let {U,|ne Z*} be a basis for X. For each fixed n, cover U, by relatively compact open sets {V(y) |y e U,} such that each V(y) C U,. Since a subspace of a 2° countable space is 2° countable, we can extract a countable subcovering {V,,| i€ Z *} of U,. Repeating for each n, the family of sets {V, | (n,i)e Z* x Z*} forms the required basis. The position of locally compact spaces is given in 6.4

Theorem

Every locally compact space is completely regular.

Sec. 6

239

Local Compactness

Proof: Let pe X and let 4 be a closed set not containing p. Repeated applications of 6.2(3) give relatively compact open Vi, Vs, such that pe V, C V, C Vo, C V, C ¥A. Since V, is compact and therefore normal, there is a continuous f: ¥, — I having value 1 at p and 0 on

Vo — V,. Let F: X — I be the map coinciding with f on 7, and identically zero on €V,. Since each of the two functions F | V,, F | €V, are continuous

and

both

coincide

on

V, N €V,

application

of III,

9.4

shows F to be continuous; thus X is completely regular. Ex. 2 For an example of a nonnormal locally compact space, note first that because of 6.2(2), an open subset of a compact space is always locally compact. Next note that [0, w] being closed in [0, £2] is compact, so that the space [0, w] x [0, £2] is also compact (and, in particular, normal). The open subspace. [0, w] x [0, £2] — {(w, £2)} is consequently locally compact, but we have seen in VII, 3, Ex. 4, that it is not normal.

For invariance properties, we have

6.5

Theorem (1) Local compactness is invariant under continuous open mappings. (2). A locally compact subset 4 of a Hausdorff space Y is of the form V N F, where V is open and F is closed in Y. (3). A subspace of a locally compact space is locally compact if and only if it is of the form V" N F, where V is open and F'is closed.

(4). TI{Y, | @ € &} is locally compact if and only if all the Y, are locally compact and at most finitely many are not compact. Proof: (1). Let f: X — Y be continuous and open, and let ye Y be given. Choose x € X so that f(x) = y and choose a relatively compact nbd U(x). Because f is an open map, f(U) is a nbd of y, and because

f(U)

is compact,

we

find from f(U) C f(U) = f(U)

that f(U) is

compact. (2). Assume that 4 is locally compact. Each a € 4 has a nbd V(a) in Y such that 7(a) N A4 is compact, and therefore closed, in Y. Define V = U{V(a) | a € A}; then V is open in Y and contains 4 ; furthermore, the formula V(a) N A = V(a) N [V(a) N A] shows that each V(a) N 4 is closed in V{(a), so by III, 9.3 we conclude that 4 is in fact closed in V. | Thus A = V N F for some closed Fin Y. (3). Because of (2) we need show only that, if 4 = V N F, then A is locally compact. Given a € 4, choose any relatively compact open U

in Y satisfying ae U C U C V and consider the nbd UNn 4 of a in A. The clos of thisnbdi ure nAisUN A =UnVnNnF)=UnNF, which is a set closed in U, and consequently is compact.

240

Chap. XI

Compactness

(4). Assume the condition holds. Given {y,} €[] Y,, for each of o

the at most finitely many indices for which Y, is not compact, choose a relatively compact nbd V,(y,,); then (V,,.--, ¥V,> is a nbd of {y,} and (V-

Vo>

Conversely, assume

= 0 such that

$(B(2, 8)) C Bi(¥(2), &) for each ze Z. Then F(B(fy 8)) C By(F(fo), d*(f,fo) < 6 implies that d(f(¥),fo(¥)) < § consequently that di(bofoo@ ™ (y1), dofeo @ y1 € Y;. To see that the metric topology need

€), as required: For, for every y€ Y, and (y1)) < e for every not coincide with the

271

Metric Topologies; Relation to the c-Topology

Sec. 8

¢-topology, let Z be the discrete space 2 and Y the discrete space Z *; the metric topology in ZY is evidently discrete and, having 2% elements, is not 2° countable. However, by 5.2, the ¢-topology in ZY is in fact 2° countable. (3). Since the set f(Y) is compact for each continuous f: ¥V — Z, it is (X1, 4.2) bounded in Z, and consequently the set C(Y, Z;d) = Z?. We now show that the metric topology in Z¥ coincides with the ctopology of Z¥. (a). Each ball B( f, ¢) contains a c-nbd of f. For, the compact ¥ can be covered by finitely many open sets Uy, - - -, U, such that §f(U,) < ¢/4 for each { = 1, - - -, n. Letting V, be an ¢/4-nbd of the compact f(U)),

we have 8V, < 3s, and so fe () (T V) C B(J, ¢). 1

(b). Each ¢-nbd (4, V) of f contains a ball B( f, £). For, since f(A) is compact, ¢ = d(f(A4), Y — V) is positive because

fA ALY - V] =2,

and thereforfe€ B(f, ¢/2) C (4, V).

Unlike the ¢-topology, in any metric space C(Y, Z; d) we have 8.3

The evaluation map w: C(Y, Z;d)

Proof: open set

x Y — Zisalways continuous.

Given U = B( fo(yo), €) C Z, let V = fg*(U), which is an containing y,, and let W = B(f,,¢) C C(Y, Z;d); then

o(W, V) C B(fo(y0), 2¢), since [ fe W] = [d(f(3), fo(»)) < ¢ for all

y€ Y]and [ye V] = [d(fo(y), fo(30)) < e-

In the c-topology, sequential convergence is equivalent to uniform convergence on every compact set; in any metric topology, we have

8.4

In any metric space C(Y, Z;d):

(1). Asequence{f,} C C(Y, Z;d)convergestoanfe C(Y, Z; d) in the metric topology if and only if f, — f uniformly on Y. (2). If a sequence {f,} C C(Y, Z; d) converges to some function f uniformly on Y, then f is continuous and belongs to

C(Y, Z; d).

Proof:

a*(f, fo)-

(1). We. need

observe

only

that

always

d(f(y), f.(y))
1 indeterminates with no constant term. A(D) is unitary if D contains a nonzero constant function.

The Spaces C(Y)

Chap. XIII1

282

Proof: If A is any algebra containing D, then by 2, Ex. 4, 4 must certainly contain all the functions described above. However, the described set is clearly itself an algebra, so it must be 4(D).

The Stone-Weierstrass theorem states that if D satisfies a fairly simple condition, then the generated algebra A(D) is dense in C(Y; ¢). 3.2

Definition A sct D C C(Y;¢) is called separating if for each pair of points y # y’ in Y there is an f € D such that f(y) # f(»').

3.3

Theorem (M. Stone)

Let Y be an arbitrary space, and D C C(Y;¢)

a family that contains a nonzero constant function and is separating. Then A(D) is dense in C(Y; ¢). We give some examples before entering the proof. Ex. 1 The Stone-Weierstrass theorem contains the classical Weierstrass theorem. Let Y = I, and D = {8(y) = 1, ¢(y) = »}; this family is separating, so that A(D), which (3.1) consists of all polynomials in x, is dense in C(I; ¢). Because I is compact, C(I; ¢) is (XII, 8.2) homeomorphic to C(I, E*; d,) with the metric topology induced by dg, therefore sequences can be used to describe closure (X, 6.2) and convergence means uniform convergence on I. Thus, for each

feC(I; ¢) and each & > 0, there is a polynomial p.(x) such that | f(x) — pe(x) | < ¢ for all x € I. It contains the zn-dimensional form of the Weierstrass theorem: For Ex. 2 each f € C(I™; ¢) and each & > 0, there is a polynomial py(xy, - - -, ¥») such that

| Foeyy - - -y %) — pelxs, -+ o, xq)| < & for all (xy, -+, x,) € I". We need note only that the functions f(x1, - -, %) = 1, x4, - - -, &, satisfy the Ex. 3 Let Y = [0, oo[; then D = {8(x) = 1, f(x) = since Y is 2° countable, C(Y; ¢) is metrizable (IX, 9.2, and can be used to describe the closure of A(D). Thus: for each function f on the positive x-axis, there exists a sequence

requirement of 3.3. e *} separates points; XI I, 1.3) so sequences continuous real-valued

Nk

Pk(x)

=

Z

aze” "

n=0

such that p, — f uniformly on every compact subset (XII, 7.2).

Proof of Theorem:

'The proof breaks into three parts:

(1). For any pair x # y of points in Y and any constants a, b, there is an f € A(D) such that f(x) = a and f(y) = b. Indeed, since D is separating, there is a g € D with g(x) = « # B = g(»); since D contains a constant function, A(D) contains all constant functions, and so

a) — (b f=a+

E=o¥

is the required function in A(D).

Sec. 3

Stone-Weierstrass

283

Theorem

To prove the theorem, it evidently suffices to show that 4(D) is dense in C(Y;c);

that is, given f and any nbd

(n] (4;, V;) of f, there is a

g€ A(D) in this nbd. Letting ¢ = min {de(f(Al) Y-V)li=1,---,n}, which is positive (XI, 4.4), this will follow if we can findag A(D) such that | f(r) — g(r)| < & for all 7 in the compact R = Lnj A;. We produce 1 this g in two stages: (2). For each fixed 7, € R there is a g € A(D) such that: (a). (b).

&(ro) = f(ro), gr) < f(r) + «

for all r € R.

In fact, by (1), for each r € R there is some A, € A(D) such that h(ry) = f(ro) and h(r) < f(r) + ¢/2. Since h, is continuous on Y, there is a nbd V' (r) such that y e V(r) = h(y) < f(¥) + ¢; covering the compact R by finitely many sets V(ry), -, V(r,) and defining g = minfk, ,---, h, ], we find g € A(D) by 2.4. Furthermore, since each r € R belongs to some V(r)), it follows that g(r) < A, (r) < f(r) + &, as required.

(3). There is a g € A(D) with | f(r) — g(r)] < ¢ for all 7 € R. For each rq, let g, be the function obtained in (2), and find a nbd V(r,) such that ye V(ro) = g, (v) > f(¥) — &; cover R with finitely many such nbds V' (r,), - - -, V(r,), and define g =

which belongs to A(D).

max[g,l,

e "grn],

Since each r € R belongs to some V (r;), we have

gr)y = g, (r) > f(r) — ¢; and, by (2), we have g, (r) < f(r) + ¢, for every

n,sothat g(r) < f(r) + ¢also; consequently, | f(r)— g(r)| < e i =1, on R, and the proof of 3.3 is complete.

The Stone-Weierstrass theorem need not be true for complex-valued

functions. For example, with

Y = {z| |2| < 1}, the unit disc in the

complex plane, and Z = the complex plane, the family D C C(Y, Z; d,) consisting of {8(2)= 1, f(2) = =z} satisfies the requirement of 3.3, but any sequence of polynomials converging uniformly on every compact subset of Y necessarily has an analytic function for limit. The theorem does, however, extend to complex-valued functions by placing on D the following additional requirement: If f belongs to D, so also does its

complex conjugate f*. For, with this additional requirement, it is easy to see that the family { f + f*,¢(f — f*) | f € D} of real-valued functions is separating; therefore it can be used to approximate separately the real and imaginary parts of any given complex-valued function.

284

'

Chap. X111

The Spaces C(Y)

Somewhat surprisingly, Theorem 3.3 (that is, with no additional requirement on D) also holds for quaternion-valued functions (J. C. Holladay). A proof, which will be left for the reader, can be based on the following simple identity: if — jqj — kgqk. g=a+ bi+ ¢ + dk, then 4a = ¢ — iqi

4.

The Metric Space C(Y)

Let d, be the Euclidean metric d,(x, ) = |x — y| in E!. In this section, we consider two topologies on the set C(Y, E';d,) of all bounded continuous real-valued functions on Y, and investigate their interrelation.

4.1

Definition The set C(Y, E';d,), as a subspace of C(Y;c), is denoted by C(Y;¢). The same set, with the metric topology induced by the supremum metric d;, is denoted by C(Y).

If Y is compact, then [XII, 8.2(3)] we know that C(Y) However, this is not true in general.

=~ C(Y;¢).

Ex. 1 Let Q C E! be the subspace of rationals. Using the c-topology, XII, 1.2, shows that I® can be regarded as a subspace of (E!)?, and it is evident that I° C C(Q;c) C (EY)Q. If the space C(Q; c) were metrizable, so also would be I9, and this is impossible because (XII,

5, Ex. 3) I

is not first countable.

For the metric space C(Y) and its relation to C(Y; ¢), we have

4.2

C(Y) is a locally convex linear topological space. the identity map 1: C(Y) — C(Y; ¢) is continuous.

Furthermore,

Since C(Y)is a metric space, sequences can be used. Because Proof: convergence means uniform convergence on Y, we find from f, — f, gn — & and real numbers A, — Athatf, + g, —f + g, fo-ga—f-£ and — Af, and therefore the algebraic operations are continuous. For Aufn local convexity, it is simple to verify that if f, g € B( fo, ), then also M+ (1 — NgeB(fy,e) for all 0 < A < 1. [We remark that these results follow much more simply from the observation that C(Y) is a normed space (¢f. Appendix I).] To see that 1: C(Y)— C(Y;¢) is continuous, note that (XII, 8.3) the metric topology is conjoining; since the c-topology is splitting, it is smaller than the metric topology (XII, 10.3). It is clear that the set C(Y) can be described in purely topological terms as { f: Y— E' | f(Y) is compact}, and because compact sets are bounded, it therefore has the minimal property: C(Y) C C(Y, E*; d) for every metric d equivalent to d,. This shows that for each d equivalent to d,, we can use d* to impose a topology on C(Y). We now inquire

285

Embedding of Y in C(Y)

Sec. 5

whether all these metrics will yield the same topology on the set C(Y) [or, what is the same, whether the topology of the space C(Y') depends only on the topology of E' and not on the metric d,]. Although we know [XII, 8.2(1)] that the metric spaces C(Y, E'; d) need not be homeomorphic for different equivalent d, the possibility remains that all the metrics d * induce the same topology on their common subset C(Y). 4.3

'The Y is of Y same

metric space C(Y) is always a topological invariant of Y. If completely regular, then C(Y) is a joint topological invariant and E*; in particular, the metric topology in C(Y) is then the as that imposed by any metric in E* equivalent to d,.

Proof: 'The first assertion is trivial: if H: Y ~ Y’, then the formula d(f,g) = df[fo H, go H], which is valid for all f, g€ C(Y"’), shows that the bijective map f— fo H is a homeomorphism. Now let Y be completely regular, and let p: Y — B(Y) be its Stone-Cech compactification. Since each fe C(Y) is bounded, there exists (XI, 8.2) for each f a unique F:BY — E! such that f = Fo p, and consequently the map C(BY) — C(Y) defined by F — F o p is bijective. Since Y is dense in B(Y), wehaved} (F, F') = d}(Fop, F'op)sothat F— Fopis a homeomorphism of the metric spaces C(B(Y)) and C(Y). Since B(Y) is compact, C(B(Y)) ~ C(B(Y);c) =~ (E')PT; and because (XI, 8.2) B(Y) is a topological invariant of Y, the theorem has been proved. As Ex. 1 shows, in completely regular spaces Y the spaces C(Y) and C(Y; ¢) are, in general, distinct joint topological invariants of ¥ and E*.

5. 5.1

Embedding of Y in C(Y) Definition

Let (Y, d) and (Z, d’) be metric spaces. A map p: Y > Z

is called an zsometry if d'(¢(x), (¥)) = d(x, y) for all

(x,y)e Y x

Y.

It is immediate that an isometry is always injective, is always uniformly continuous, and is a homeomorphism of Y and ¢(Y); a surjective isometry is called an isomorphism. Note that an isometry is always relative to specified metrics in the two spaces. Referring to the proof of 4.3, we find: If Y & -Y’, then C(Y) is isoEx. 1 morphic to C(Y’); if Y is completely regular, then C(Y) is isomorphic to C(8(Y)).

286

Chap. X111

5.2

The Spaces C(Y)

Theorem The metric space (Y, d) can always be isometrically embedded in the metric space (C(Y), d;"). Furthermore, its image Y’ C C(Y) is closed in its convex hull.

Proof: Choose a point p € Y, which will remain fixed throughout the discussion. Let ¢: Y — C(Y) be the map a — f,, where fa(y)

=

d(y’ a) -

d(y’ P)

We verify f, € C(Y) by noting that | f(y) | < d(a, p) and that d(a, p) is a constant independent of y. To prove that ¢ is an isometry (that is,

d; (p(a), ¢(b)) = d(a, b)), we must show sup | fo(¥) — fo(¥) | = d(a, b). Since

| fo(9) = fi(») | = | d(3, @) — d(3,b) | < d(a, b) for each y € Y, the sup cannot exceed d(a, b). Selecting y = b shows the value d(a, b) is in fact attained, so that dj (f,, f;) = d(a, b), as asserted. Let H(Y') be the convex hull of ¢(Y) = Y’ in C(Y). We are to show that Y’ is closed in H(Y'). Let fe H(Y') — Y’; then

f= Z Mo where

¢, € Y, n > 2, z A =1, 1

we find

10) - £

and all A; > 0. Now,

for each ax Y,

= (3 4 d, @) - dy, )

Since n > 2, this gives

f() = fa() > [min(Ay, A2)]-[d(y, a1) + d(y, a2)] — d(y, @) > [min(Ay, A2)]-[d(ay, a2)] — d(y, a). Using ¥y = a, we conclude that 4 ( f, f;) = [min(}, A))][d(a;, a5)] > 0 for all a € Y, consequently f is not in the closure of Y’. This proves the theorem. As an immediate application, we show that if X is attached to Y by f: 4 — then whenever X, Y are metric spaces, the resulting set X Us Y can be given metric topology (in general, distinct from its identification space topology).

5.3

Y, a

(C. Kuratowski, F. Hausdorff) Let X, Y be metric spaces, 4 C X closed, and f: A — Y continuous. Then there exists a metric space Z D Y such that: ' 1. Yis closed in Z. 2. f has a continuous extension F: X — Z.

3. F| X — A is a homeomorphism of X — 4 and Z — Y.

Sec. 6

The Ring C(Y)

287

Proof: Let H(Y) be the convex hull of Y in C(Y); since (X, d) is a metric space and C(Y) is a locally convex linear space, then by IX, 6.1, there is an extension f*: X — H(Y) off. Let ¢: X— C(X) be the embedding of 5.2, and define

F: X— H(Y)

x E'

x C(X)

by F(x) = [f*(x), d(x, 4), d(x, A)-p(x)]. F is evidently continuous (IV, 2.2). Furthermore, F| X — A is injective: if F(x) = F(x'), then d(x, A) = d(x’, A) and @(x)-d(x, A) = o(x")-d(x’, A); since d(x, A) # 0, we have p(x) = @(x’), and since @ is injective, x = x’. To prove that F| X — A is a homeomorphism, we show its inverse continuous by proving [F(x,) — F(x0)] = [, — %x0]. Since F(x,) — F(x,), we find d(x,, 4) — d(xo, A) and @(x,) - d(xn, A) — @(x0) - d(x0, A); because d(xo, A) # 0, we have d(x,, A) # 0 for all large n, so that ' (2n)! ¢, n

Let

s(x) = f(x)

+ eb~" cos b(x — a);

then s € C® and d;(f®, s®) < &b~ " < ¢-2%~" for each k < n, consequently s € B(f, 2¢). However, [s@™(a) — f@V(a)| = eb" > (2n)! ¢*", so that s € T(a; ¢). This proof is due to H. Salzmann and K. Zeller.

5.

Extension of Uniformly Continuous Maps

Let X, Y be metric spaces and f: X — Y continuous. Although the image f() of a convergent filterbase 2 is convergent (X, 5.2) the image of a Cauchy filterbase in X need not be Cauchy in Y, as I, Ex. 1, shows. However,

5.1

If f: (X, d")— (Y, d) is uniformly continuous, then the image of a d’-Cauchy filterbase is a d-Cauchy filterbase.

Proof: Let U be d’-Cauchy. Given any ¢ > 0, let 8(¢) > 0 be such that V x € X: f(By(x, 8)) C By(f(x), ¢); finding an 4, € A with 8(4,) < 3, we have 8[ f(4,)] < ¢, as required. The uniformly continuous image of a complete space need not, however, Ex.1 be complete. The bijection f: (E?, d.) — (1—1, +1[, d.) given by x — x/(1 + [x]) is uniformly continuous, and even a homeomorphism, yet the image is evidently not complete,

Because of 5.1, uniformly continuous maps into complete spaces can be expected to have special significance. The following fundamental extension theorem has many applications; a standard one is to define the function &%, x real, (@ > 0) from a knowledge of 4", r rational.

5.2

Let (X, d’) be an arbitrary metric space, and 4 C X a Theorem dense subset. Let Y be d-complete and f: 4 — Y uniformly continuous.

Then there exists one, and only one, continuous extension

F: X —

Y of f, and F is also uniformly continuous.

Proof: For each xe X, the nbd filterbase” 1(x) N A 1s certainly d’-Cauchy, so by 5.1, the filterbase f(11(x) N A) is d-Cauchy and consequently convergent. By X, 5.3, f can be extended by continuity to a unique continuous F: X — Y. We now prove that F is uniformly continuous. Given & > 0, let § > 0 be such that d(f(a;), f(as)) < e whenever d’(ay, a;) < §; we will show that d(F(x;), F(xp)) < 3

303

Extension of Uniformly Continuous Maps

Sec. 5

whenever d’(x;, x3) < 8/3. In fact, since F is continuous, find nbds U(x,), U(xs) such that F(U(x;)) C B(F(x;), ¢), ¢ = 1, 2, and let W(x;) =

U(x;) N B(x;, 6/3).

Now 4 N W(x;) # @, since A is dense, and we choose a; € 4 N W(x,). Then, because

d'(ay; a5) < d'(ay, %) + d'(xy, x3) + d'(x3, a3) < 38/3, we find that

d(F(xy), F(x3)) < d(F(x1), f(a1)) + d(f(a1), f(a2)) + d(f(az), F(x2)) < 3e ‘

and this completes the proof.

Ex. 2 The hypothesis that f be uniformly continuous is essential. Let 4 = E! — {0} C E! and let Y = E'. The continuous map f(x) = x/|x| of 4 — E* cannot be extended over E*!. The hypothesis that Y be d-complete is essential. Let X = E!, 4 C X Ex. 3 be the subspace of rationals and Y = 4. The identity map f: 4 — A is uniformly continuous, but cannot be extended to a continuous F: E' — A4, since E' is connected and F cannot be constant.

If the map f in 5.2 is 2 homeomorphism, it is not in general true that its extension F' is also a homeomorphism. For example, let X = I, let A = Int(l), and let Y be the unit circle S!. Using the Euclidean metrics, the map x — 2% of Int(J) onto S* — {(1, 0)} is a uniformly continuous homeomorphism, yet its extension over I is not bijective. However, there is one type of homeomorphism for which such behavior

cannot occutr. A homeomorphism h: (X,d) @ (Y,d’) is called a uniform isomorphism of X and Y whenever both % and 2~' are uniformly continuous. A surjective isometry is a particularly important example of a uniform isomorphism, but the notion is more general: the map 1: (EY, d,) — (E', d'),

where d’ = min (1, d,), is a uniform isomorphism, although it is not an isometry.

5.3

Corollary Let Y be d-complete, let Y’ be d’-complete, and let ACY, A CY' be dense. Then each uniform isomorphism h: A ~ A" has an extension H: YV ~ Y’ that is also a uniform isomorphism. Furthermore, if % is an isometry, then so also is H.

Since k is uniformly continuous, it is extendable to a uniProof: formly continuous H:Y — Y’'. Since g = A~! is also uniformly continuous, it also has a uniformly continuous extension G: Y’ — Y. Because Go H| A = 1, and 4 is dense in Y, we have (VII, 1.5) that Go H =

1,

and,

similarly,

that

H o G = 1y..

It

now

follows

from

304

III,

Chap. XIV

12.3, that H is a uniform isomorphism.

mediate from the manner

6.

Complete Spaces

The second part is im-

in which the extension H is defined.

Completion of a Metric Space

Let Y be a metrizable space and d a given metric for Y. If Y is not d-complete, then some d-Cauchy sequences in Y do not converge. By emulating the Cantor process for getting the reals from the rationals, we successively adjoin ideal elements to Y, which act as limits for the nonconvergent d-Cauchy sequences, to get ultimately an enlarged space Y in which Y is dense, and an extension of d to a complete metric d for Y. Observe that this process depends on the metric d with which we start: Different metrics for the same space Y generally result in distinct enlargements.

The Cantor process can be described in a simpler manner, as the proof of the following theorem shows. 6.1

Theorem Let Y be a metrizable space and d a given metric for Y. Then Y can be isometrically embedded as a dense subset of a complete space (Y, ci) Y and d are unique up to an isometry: if

(Y, d) is isometrically embedded as a dense subset of any dcomplete Y,, then (7, dA) and (Y,, d,) are isometric. Proof: 'The uniqueness (up to isometry) of ¥ results from 5.3, Existence: Given (Y, d), thereis by XIII, 5.2, an isometric embedding i of Yinto (C(Y), d;}). Define Y = i(Y); then Y is dense in #(Y), and since (C(Y), d;) is complete (2.6), the closed subspace Y is also.

The metric space (Y, d ) is called the completion of the metric space (Y, d). Relating the completion process with uniformly continuous maps, we have

6.2

Corollary Let (Y, d), (Y,, d;) be metric spaces, and f: ¥ — Y a uniformly continuous map. Then there exists a unique uniformly continuous f: ¥ -> ¥, such that the diagram

1s commutative.

Sec. 7

Fixed-Point Theorem for Complete Spaces

305

Proof: The map iyofoi~t:4(Y)— Y, is evidently uniformly continuous; since Y, is complete and i(Y) is dense in Y, 5.2 gives a uniformly continuous extension f: ¥ — Y|, and f satisfies the requireA

S

S

A

.

.

ments.

7. 7.1

Fixed-Point Theorem for Complete Spaces Definition Let 7: Y— Y be a map of a space Y into itself. A point y, € Y is called a fixed point for T if T(y,) = y,.

Not every map has a fixed point; for example the map 7T': E! - E* given by x — x + 1 has no fixed point. Theorems asserting the existence of a fixed point for certain types of maps usually have important applications, as we shall see.

A map T:(Y,d)— (Y,d) of a metric space into itself is called d-contractive if there exists an « < 1 such that d(Tx, Ty) < ad(x, y) for all (v, y)e Y x Y. The following theorem is a topological version of the Picard “successive approximation” process in analysis: 7.2

Theorem (S. Banach) Let Y be d-complete, and let T: Y - Y be d-contractive. Then T is continuous and has exactly one fixed point.

Proof:

'The continuity of T is obvious.

Furthermore, T cannot have

more than one fixed point: T'(x,) = x4, T(vo) = Yo, and d(xg, y4) > 0 yield the contradiction d(x, ¥o) = d(Txy, Tyo) < ad(xy, yo) < d(Xq, ¥o)- 'T'o prove that T does have a fixed point, we first choose any ye Y and show that the sequence y, Ty, T(Ty) = T?y,--- of iterates is a d-Cauchy sequence. In fact, note that

d(Ty, T%) < «d(y, Ty), d(T?%y, T?y)

< ad(Ty,

T?%)

< o?d(y, Ty), so that, by induction, d(T"y,

T"*'y)

< «" d(y, Ty).

It follows that for

any n and any m > n we have

d(Tty, T"y) < d(T"y, T"*'y) + -+ < (e

+ -+

d(T" "1y, T™y)

o™ ) d(y, Ty)

Now, because o < 1, this formula shows that an

d(T™y, T™y) < T 4 for all m > n; therefore, since o — 0 as n —

sequence {T"y} is in fact d-Cauchy.

1) 00, we conclude that the

306

Chap. XIV

Complete Spaces

Since Y is d-complete, the sequence {T™y} converges to some y, € Y. The point y, is the fixed point of T. For, on the one hand Ty — y, and the continuity of T imply that T(T"y)— Ty,; on the other hand, the sequence {T(T™y)} = {T"*1y} is a subsequence of the d-Cauchy sequence {T™y} and therefore must converge to y,. Thus, y, = Ty, and the theorem is proved. Ex. 1 In analysis, the elements T"y are called the successive approximations to yo; note that T™y —> y,, regardless of which y € Y is used, and that the “error” d(yo, T™y) of the nth approximation is an


u(Y)

and, as in VII, 1.2 (4), we find that u(Y) is closed in [] Y,. Now let A be the mapping of 4 into the subspace [] U, C [ Y, which agrees

with pu | 4; then Ais an embedding of 4 and, since \(4) = w(Y) N[ U, the image of 4 is closed in [] U,. Our observation established, 2.5 (2) and (4) show that we need prove only that an open set U in a topologically complete space Y is topologically complete. We may obviously assume that U # Y, and shall again construct a suitable embedding. Let d be a metric for Y, and define f: U — E*! by f(u) = 1/d(u, Y — U); since f is continuous, the graph G C U x E* off is clearly homeomorphic to U. Let j: U x E' — Y x E?! be the inclusion map; then j(G) = G. Moreoever, j(G) is closed in ¥V x E1: forif h:' Y x E* — E' is the continuous map (y, t) - t-d(y, Y — U),

then j(G) 1s the closed set £71(1). Thus, U is homeomorphic to a closed subset of the complete space Y x E?! so, by 2.5 (2), U is topologically complete. The theorem is proved.

9.

Complete Gauge Structures

In this section, we will extend the notion of completeness to arbitrary gauge structures; since gauge spaces need not be 1° countable, this is done by working with filterbases rather than with sequences. Let d be a gauge in a space Y. The d-diameter of a set 4 C Y is defined, as for metrics, to be sup {d(x, y) | x, y € 4}. We next define d-Cauchy filterbase in Y exactly as in 3.1, and finally make the

Sec. 9

9.1

Definition A filterbase A = {4,|«e/} in a gauge space (Y, 7(2)) is called a &-Cauchy filterbase if it is d-Cauchy for eachde 2.

The 2-Cauchy marized in 1.2: 9.2

309

Complete Gauge Structures

Theorem

filterbases have properties analogdus to those sum-

Let (Y, .7(2)) be a gauge space. Then

(1). Every convergent filterbase is Z-Cauchy. (2). If A is Z-Cauchy and if B — A, then B is Z-Cauchy. (3). If A is 2-Cauchy, and if A > y,, then A — y,.

The proofs are formally the same as those given in 1.2; in particular, it follows from (3) that a 2-Cauchy filterbase either converges or has no convergent subordinated filterbase. 9.3

Definition A gauge structure & for a space Y is called complete if every @-Cauchy filterbase in Y converges. A completely regular space having a complete gauge structure 2 is called Z-complete.

From 9.2(3) and XI, 1.3(3), we find that every gauge structure for a compact space Y is complete moreover, because of 3.2, every dcomplete metric space is a d-complete gauge space. The invariance properties for Z-complete gauge structures differ from those of their metric analog (2.5) in the behavior of cartesian products: 9.4

(1). If YV is Z-complete and 4 C Y is closed, then A4 is Theorem 2 4,-complete. (2). If (Y, T (2)) is any gauge space, and if A C Y is complete, then 4 is closed in Y. (3). Let {(Y;, 7(2;)) | Be #} be any family of gauge spaces,

and let & be the induced gauge structure of [T Y. complete if and only if each &, is complete.

Then 9 is

The proofs are entirely similar to those given in 2.5, with that for (3) being immediate from the observation that the projection of a é-Cauchy filterbase on each factor Y, is Z;-Cauchy. We next consider total boundedness. Again, the definition of a totally bounded gauge is formally the same as that for a totally bounded metric, and we make the

9.5

Definition A gauge structure & for a space bounded if each d € Z is totally bounded.

Y is called totally

310

Chap. XIV

Complete Spaces

There is an important difference between the notion of total boundedness using gauges and that of its purely metric analog: because a completely regular space can be embedded in a parallelotope, it follows easily from 9.4(3) that every completely regular space has a totally bounded gauge structure. In particular, any nonseparable metric space has a totally bounded gauge structure, although (3, Ex. 3) it does not have any totally bounded metric. However, in exactly the same way as in 3.5, it follows that

9.6

A completely regular space is compact if and only if it has a gauge structure that is both complete and totally bounded.

We now take up the completion of a given gauge structure. Since we have a meaning (IX, 11, Ex. 3) for “uniformly continuous map of one gauge space into another,” we can introduce the notion of uniform isomorphism as in Section 5; from this viewpoint, the notion has the following significance: the identity map 1: (Y, 7(2)) — (Y, 7(2")) is a uniform isomorphism if and only if the uniform structures determined by 2 and &’ are equivalent. The proofs of the evident analogs of 5.2 and 5.3 are practically the same as before; in particular, we obtain the following analog of 5.3:

9.7

Let Y be Z-complete, let Y’ be 2'-complete, and let 4 C Y, A’ C Y’ be dense. Then any uniform isomorphism h: 4 = 4’ can be extended to a uniform isomorphism H: ¥ ~ Y.

We now prove: 9.8

Theorem Each gauge space (Y, 7(2)) is uniformly isomorphic to a dense subspace of a complete gauge space (¥, 7(2)). Further-

more, (Y, 7(2)) is unique, up to a uniform isomorphism. Proof: 'The uniqueness comes from 9.7; we now prove the existence. Let {Cy(Y)|de P} be a family of copies of the metric space (C(Y), d}) indexed by the gauges d e 9. The gauge structure for the

space [ [ {Cy(Y) | d e &} derived by using the metric d; in each factor is, by 94 and 2.6, a complete gauge structure. For each de & and ye, let y,: Y— E! be the continuous function y,(2) = d(3,y) — d(3, p), where p is a fixed element of Y, and let j: Y —JT{Cy(Y) | de 2} be the map given by j(y) = {y,}, that is, the dth coordinate of j(y) is y,. Because 2 is separating, the map j is readily verified to be injective, and because (as in XIII, 5.2) we have dgf (v,, x;) = d(v, %), it follows that j is a uniform isomorphism of (Y, 7(2)) onto (j(Y), 7(Z,.)). Thus, J(Y) is, by (9.4), the required complete gauge space.

Sec. 9

Complete Gauge Structures

311

We remark that, in this broad extension of the concept of completeness from metric spaces to completely regular spaces, one of the most important consequences of metric completeness is significantly weakened: there seems to be no generalization of Baire’s theorem to gauge spaces that does not involve some fairly restrictive hypotheses on the gauge structure (cf., 4, Ex. 1, for example).

DiacraM oF THE MAIN CLASSES OF TopoLoGICAL Sraces Discussep IN THis Book

Hausdorff Regular

Completely regular




Con'lplete

metric

metric

Without additional hypotheses, none of the implications is reversible.

Problems Section

|

1. Let d be the Euclidean metric in E".

Show that a sequence in E™ is d-Cauchy

&

if and only if each coordinate is d.-Cauchy. 2. Prove: A Cauchy sequence can converge to at most one point. Using the fact that every bounded monotone sequence in E! converges, show

that a d.-Cauchy sequence {a,} in E* converges to sup{ inf a,}. m

izm

312

Chap. X1V

Complete Spaces

Section 2

1. Let (Y, d) be a metric

space. Prove: Each arbitrary intersection and each finite union of d-complete subspaces is also d-complete. 2. Let [2(R,) be Hilbert space and d its metric. Show that d is complete (use the diagonal process). 3. Let (Y, d), (Z, d’) be metric spaces and f: Y ~ Z a homeomorphism such that d(y,y) < ad'(f(y), f(3")) for all y,y" € Y and some fixed « > 0. Prove: If d is complete, so also is d’. 4. Prove that every discrete space is topologically complete.

Section 3 1. Prove:

a. Any subordinate of a d-Cauchy filterbase is also a d-Cauchy filterbase. b. If a d-Cauchy filterbase has a convergent subordinate, then it itself converges. 2. Prove that Y is d-complete if and only if: For every descending sequence — 0, the Ay D Az D --- of nonempty closed sets having d-diameter 6(A;) intersection () A4; is not empty. i

3. Let Y be d-complete.

Let {4, |{ = 1, 2, - - -} be a countable family of arbitrary

sets such that 4, N A; # @ for all (3, 7), and such that > 8(4;) < .

Show

1

that there exists a yo € Y such that each nbd

U(y,)

contains all but at most

finitely many 4;.

4. Let Y be d-complete, let Z be an arbitrary (Hausdorff!) space, and letf: ¥ — Z be continuous. Assume 4; D A, D --- is a descending sequence of closed

nonempty sets and 8(A4;) — 0. Prove:

(A 4) = 0 fc. 5. Let Y be totally d-bounded. Show that d is a bounded metric for Y. 6. Prove: Y is totally d-bounded if and only if each infinite sequence in ¥ contains a d-Cauchy subsequence. 7. Let Y be d-complete, and let Z be any metric space. Let f: Y— Z be a continuous map with the property:

vr>03p=p0)>0VyeY :JIBO,N > BUG), pl-

Prove that f is an open mapping. for each ¢ > 0.]

[Hint: Show that f[B(y, r + €)] > B[f(¥), p]

Section 4 Any open and any closed subset of a topologically complete space is a

1. Prove:

Baire space. 2. Let Y be complete, .

and let each G, be a G4-set in

Y.

Assume

each G; to be

o

dense in Y; show that [ G; is dense in Y. 1

3. Prove that the Cantor set is nowhere dense in E*. 4. Show that if Y is topologically complete and has X(Y) > N,

no

isolated

points,

then

Problems

313

Section 5

1. Show that a d-complete space may be isometric with a proper closed subset of itself.

[Hint: Let f: 10, co[ -> ]0, oc[ be the map f(x) = x + 1.]

2. Let (X, d) be a compact metric space, and let A C X be dense. Let Y be d’complete. Prove: A map f: A — Y has a continuous extension over X if and only if it is"uniformly continuous. Show that the homeomorphism x — x° of (E!, d,) on itself preserves d,-Cauchy sequences, but is not a uniform isomorphism.

Section 6

1. Let X, Y be complete, and f: X — Y uniformly continuous. Prove:

If 4 is

relatively compact, so also is f(A4).

2. Let (Y, d) be a metric space, and A C Y any subset. embedding of Y in its completion

”

Y.

Letj: 4 —

A the completion of 4 (with metric d | A).

Let i: Y —

¥ be the

Y be the inclusion map, and

Show: ]A 14—

Pisan isomorphism

) onto the closure of 1(4) in Y. Let {(X,, d) | i€ Z*} be a family of metric spaces, and for eachie Z * let X,

be the completion

of (X, d;). Prove:

The

completion

of | [ X; is uniformly 1

@

isomorphic to [ [ X,. 1

. Let (Y, d) be a metric space, and let C be the set of all d-Cauchy sequences in Y. Prove: (1) If {x,}, {zx} € C, then lim d(x,, 2,) exists. (2) The map d: C x C — E*! given by d({x,}, {z:}) = limd(x,, z,) is a gauge on C. (3) If R is the equivalence relation xRz < j(x, 2) = 0 in C, and if p is the induced metric on C/R, then (C/R, p) is isometric to the completion of (Y, d). [Hint: Observe that the map Y — C/R given by y — R{y,}, where y, = y for all n, is an isometry onto a dense subset.]

Section 7

1. Let

G(x,y) be a continuous real-valued function on I, x I,. Assume that G(%o, ¥0) = 0 and Gy(xo, ¥0) # 0. Prove: There exists a positive s and a unique continuous A: I; — I, such that A(x,) = yo and G(x, h(x)) = 0 on [I.. [Hint: Define

G(x, y)

F(x,y)=y—yo—m v

’

and use the theorem in the text.] Let f(x,y) be continuous on I, x I, and

satisfy some Lipschitz condition | f(x, %) — f(x,9)| < k|ly — 3| on I, x I,. Prove: If @ is chosen so that d-max| f| < b and @-k < 1, then the differential equation y’ = f(x, y) has a unique solution h: I; — I, satisfying h(xo) = yo. [Hint: Consider the map

T@@® = yo + [ 1€ @) dt]

314

Chap. XIV

Complete Spaces

3. Let K(x,v, 2) be defined in R = {(x,5,2) [0 g'o f'. (2). (Restriction.) If f, g: X — Y are homotopic, then

flAd=gl4

for any 4 C X. (3). (Cartesian product.) f, g: X — [|

Y, are homotopic if and

only if p, o f = p, o g for each . (4). (Attaching.) Let X be attached to Y by f: A — Y, where ACX is closed, and let p:X + V> X Ur Y be the identification map. Let gy, g,: X — Z and hy, byt YV — Z, If ®:p, ~ g, and ¥: hy ~ h, are “consistent” (that is, if D(a, t) = ¥(f(a), t) for all (a,t) e A x I), then (g, hy)p? and (g4, h;)p~! are homotopic maps of X Uy Y into Z.

Proof:

(1). Let @:f ~ f'and ¥:g ~ g’; then

4

2D(x, 2t)

(%”z{wwuxm—u is clearly a continuous map X x I —

0]]Y,

is (IV,

2.3)

for each the map

continuous

and

«

shows f ~ g. (4). According

XII, 4.2, the map 4 = (D, ¥)(p x 1)~! of (X Us Y) x Iinto Z is continuous, and 4 provides the required homoto

topy.

For behavior of [Y, Z] under continuous maps, 2.3

Let X, Y be two spaces and ¢: X —

(a).

For

each

Z, ¢ induces a map

e[f1 = [fo el

(b). For

Y continuous.

each

Z, ¢

induces

a map

Then:

¢#:[Y, Z]— [X, Z], by ¢4: [Z, X]—[Z,

Y]

by

pylg] = ¢ - gl.

(c).

If ¢ =~ i, then the maps induced by ¢, ¢ are the same.

then

: Y —S,

(d). If

# = p#oth# (poq@)

and

(poyh),=

Py Py Proof: These proofs involve straightforward applications of 2.2(1), and are left for the reader.

It follows invariant

of

directly Y

and

from Z;

2.3(d)

that

intuitively,

[Y, Z]

[Y, Z]

is a joint

measures

the

topological number

of

“essentially different” ways that ¥ can be mapped into Z, and so provides information on the “complexity” of the topological structure of Z relative to that of Y. For example, X([E?, Z]) is the number of path components of Z (as follows from Ex. 1 and I, Ex. 2). The explicit determination of [Y, Z] in terms of calculable algebraic invariants of Y and Z is one of the main problems in homotopy. Only particular cases have been settled, even for such simple spaces as spheres; for example, if

n > 4, it is known that [S™®, S?] is a finite set, but the exact number of elements is not yet known for any large value of 7 (say, n > 50).

3.

Homotopy and Function Spaces

A topology in the set Z¥ decomposes it into path components. On the other hand, the relation of homotopy decomposes the set ZY into homotopy classes. One advantage of the c-topology in ZY is that under mild restrictions on Y, the homotopy classes are exactly the path components. This follows from

320 3.1

Chap. XV Theorem

Let ZY be given the c-topology.

Homotopy

Then:

(1). If f ~ g, then f and g lie in a common path component. (2). Iff andg lie in a common path component of Z¥, thenf ~ g, provided Y is a k-space.

Proof:

(1). Let

®@:f ~ g;

because

@: Y x I—

Z is continuous,

XII, 3.1, assures that the associated map &: 1 — Z¥ is also continuous,

and & is evidently a path in Z¥ joining f to g.

(2). Let @: I

Z¥ be a path joining ftog. Since (XII, 4.4) Y x ]

is a k-space, XII, 3.2 states that the associated

@:

¥V x I — Z is con-

tinuous, and clearly, @: f >~ g.

A further advantage of the c-topology is that homotopic maps induce homotopic maps of function spaces, again under mild restrictions. 3.2

Theorem In all function spaces, use the ¢-topology and let Z be an arbitrary space.

(1). If £y, fi: X — Y are homotopic, and or if X is a k-space, then the induced are homotopic. (2). If g% g': Y — Z are homotopic, and or if X is a k-space, then the induced are homotopic.

if Y is locally compact maps f§, f{: Z¥ — Z% if Y is locally compact maps g%, g1 : Y¥ —» Z%

Probf: In case Y is locally compact, the proof of (1) and (2) depends on XII, 2.2, that the composition T': Y% x ZY — Z¥ is continuous. Ad (1). Let @: f, ~ f;; because (XII, 3.1) the associated &: I >Yx

is continuous, so also (IV, 2.5)is & x 1: I x Z¥— YX x Z¥; thus To(® x 1) is continuous and shows that f§ ~ f#. The proof of (2) is similar. In case X is a k-space, the proof depends on XII, 5.3 and 4.4, that ZXXT

~

(ZX)

~

(Z1)X,

Ad (1). Let @: f, ~ f,; by XII, 2.1, the induced

b+: Z7 > ZXX1 x (ZX) is continuous, and [X1I, 3.1(2)] the associated continuous Z¥ x | — ZX provides the required homotopy. Ad (2). Let ®:g° ~ g'; then the associated $: Y- Z' is continuous, so that the induced (@)*: Y ¥ —» (Z)¥ ~ (ZXY is also continuous; and the associated continuous

Y¥

x I —

Z¥ shows g%

=~ g,

Sec. 4

Relative Homotopy

321

Relative Homotopy

4.

In many cases, attention is restricted only to maps f: Y — Z having special properties, and two such maps are called homotopic only if one can be deformed to the other in a certain way. The usual pattern for this notion is 4.1

Definition

Let

o# C ZY

be

some

subset,

and

call its members

A -maps. Two H#’-maps are J#’-homotopic if there is a homotopy ®:Y x I+ Zsuchthat @Y x tisan 5 -map for each ¢t € I. Ex.1

H

Letfe ZY be a given continuous map, let 4 C Y be any subset, and define

= {geZ¥ | g|A = f|A}.

Then

two H#-maps g, h are S#-homotopic

there is a homotopy of g to & such that the image of each point a € fixed [at f(a)] throughout the entire deformation. Such homotopies are considered; to denote that g, f are homotopic in this way, we shall in write “g ~ frel A.” Ex. 2 Let 4 C Y, BC Z, and define # = {fe Z¥ | f(4) C B}; set H# occurs frequently, an S -map will be denoted briefly by f: (Y; 4)

only if

4 remains frequently the future since this

— (Z; B). An J-homotopy of # -maps is now one in which the image of 4 remains in B (not necessarily pointwise fixed!) during the entire deformation. Ex.3 It is evident that J#-homotopy of J# -maps is more restrictive than ordinary homotopy, and the following trivial example shows this. Let # C I’ be {ff: I; FrI) — (I; FrI)}; then, although the J#-maps f(t) = ¢, g({) =1 — ¢ are homotopic, it is obvious that they are not J#-homotopic.

Let # relation in Regarding a k-space of

C Z¥. As before, ##-homotopy of #°-maps is an equivalence #; the set of #-homotopy classes is denoted by [Y, Z; 5#]. 5# as a subspace of the c-topologized Z¥, then whenever Y is the J#-homotopy classes are precisely the path components

. As an illustration of these ideas, we prove

42

(). W. Alexander) Let A: V®— V" be a homeomorphism such that h|S" 1 =1, Then & ~ 1 rel S*~1, with each stage of the deformation being a homeomorphism. Alternatively stated: If J# is the set of all homeomorphisms IV'™* — /" coinciding with the identity map on S"~!, then J is path-connected. Proof:

For each t € ]0, 1] define

D(x, t) = {

x

x| = ¢

th(x/t)

|x] < ¢,

which keeps all points outside the ball V] of radius ¢ fixed and duplicates on V7 — V7 the given transformation /4 on a diminished scale. @ is not defined for t = 0; however, we set @(x, 0) = x. It is readily seen that @: V" x [ — " is continuous, that @:1 ~ hrel S"~1, and that @ | /" x t is a homeomorphism for each t.

322

Chap. XV

Homotopy |

Retracts and Extendability

5.

Let X be a given space. In this and the next section we will be concerned with characterizing those subsets 4 C X having the two properties: For all spaces Z, (a) each f: 4 — Z is extendable over X; and (b) two maps F, G: X — Z are homotopic wheneveronly F | 4 ~ G | A. To see the importance of such sets, note that since the map ##: [X, Z] — [4, Z] induced by the inclusion map i: 4 — X is exactly [F]— [F | 4], it follows from (a) that ¢# is surjective, and from (b) that i# is injective; thus, for any space Z, the calculation of [X, Z] can always be reduced to the (perhaps easier, or known) determination of [4, Z]. We begin with the extension problem. Conditions for extending a given f: A — Z over X generally involve Z also. For example, if X is a connected normal space and A is the union of two disjoint closed subsets, every continuous f: A — E! is extendable over X (cf. VII, 5.1), whereas this is not true for continuous maps f: 4 — 2. However, important case where such conditions do not involve Z. 5.1

Definition

Let X be aspace and 4 C X.

identity map

there is an

A4 is a retract of X if the

1: A — A is extendable to a continuous r: X — 4;

such an extension is called a retraction.

Equivalently, 4 is a retract of X if there exists an r: X — 4 such that r(a) = a for each a € 4, or alternatively,

if r: X — 4

is surjective

and

r o #(x) = r(x) for each x € X. The concept of retract is evidently topologically invariant: if A: X ~ X', then A(4) is a retract of X’ if and only if A is a retract of X. Ex.1

Let X be any space.

Then X and each xo € X are retracts of X.

Ex. 2 'The unit ball " is a retract of E", as the map r(x) = «x/|x| (if |x] > 1 and 7(x) = x otherwise) shows. Further, S"~! is a retract of E" — {0}. Ex. 3 Let I_I Y, be any cartesian product, and for each «, let B, C Y, be a retract.

Then

l—I B, is a retract of H

Y,; for, if each r,:

Y, — B, is a retraction,

so also is H % 1_[ Y, — l—I Bg. In particular, X x 0 is a retract of X x I. Ex. 4 The Hxlbert cube I* is a retract of the Hilbert space [2(X,). In fact, for each n=1,2,--:, let r,: E* —~[—1/n,1/n] be a retraction; then the map ri 12(Ry) — 1%, given by {x,} — {rn.(x,)}, s continuous and is a retraction. Ex. 5 If ZY be given the c¢-topology, then (XII, 1.2) there is an embedding h: Z — Z¥, where h(z) = constant map sending Y to the point 2. A(Z) is a retract of Z¥: Choosing any y, € Y, the evaluation map wy,: Z¥ — Z is continuous and & o w, is clearly a retraction.

5.2

If X is Hausdorff, and A C X a retract of X, then A4 is closed in X,

Sec. 6

Deformation Retraction and Homotopy

323

Proof: 1f there were some x € A — A, then because 7(x) # x and X is Hausdorfl, there would exist disjoint nbds U D {x}, and V D {r(x)} such that r(U) C

V; however, since x € A, there must be some a € 4 in

U, and since a = r(a) € V, this contradicts the disjointness of U and V. Ex. 6 If X is not Hausdorfl, (¢f. Ex. 1).

5.2 need not be true, as Sierpinski space shows

We now show that any continuous map always be extended over that space; in fact

5.3

of a retract of a space can

Theorem Let X be any space and 4 C X. A necessary and sufficient condition that 4 be a retract of X is that for every space Z, each continuous f: 4 — Z is extendable over X.

Proof: 1If A is a retract of X and r: X — A a retraction, then for is an extension of any given f over X. Conversely, if the condition is satisfied, A is a retract of X: Choose Z = A and f: A — A to be the identity map. Remark: Any given extension question can always be formulated as a retraction problem. Let 4, X, Z be fixed and f: A — Z; then f is extendable over X if and only if Zis aretract of X s Z, For,ifp: X + Z — X Uy Z is the identification map and 7 is a retraction, 7 o p | X is an extension of f. Conversely, if an extension F off is given, define p: X + Z— Z by p(x) = F(x) if x€ X and p(z) = zif ze Z; then pp ~': X Us Z — Zissingle-valued, so it is continuous and is a retraction of X Uy Z onto Z. In particular, 4 is a retract of X if and only if under any attachment to any space Z by any continuous f: A -— Z, the subspace Z is a retract of X Uy Z.

6.

Deformation Retraction and Homotopy

In this section, we find conditions on A which assure, for all spaces Z and any two continuous F, G: X — Z, that

' ~ G whenever

F|4~G|A4. Such conditions generally involve Z also. Ex.1 Let X f: A — Z extends for x € [0, 1] and they would be if

=2 =1[0,11v][2,3] and 4 = [0, 1]. Then the inclusion map to the identity map F, and also to the map G, where G(x) = x G(x) = x — 2 otherwise. Clearly F' and G are not homotopic; Z were taken as [0, 3].

Again there is an important instance where the homotopy question does not involve Z.

324

6.1

Chap. XV

Homotopy

Definition Let 4, B be any two subsets of X. B is said to be deformable into 4 over X if the identity map 1: B — B is homotopic in X to a map of B into 4.

That is, we require a @: B x I — X such that @(b, 0) = b for each beBand &(B x 1) C 4. If we have B = X in the definition, we omit “over X” and say simply that X is deformable into 4; it is important to observe that we do not require that the image of 4 remain in 4 during the deformation. This concept is clearly topologically invariant: if £: X ~ X', then h(B) is deformable into A(A4) over X"’ if and only if B is deformable into A over X. Ex. 2 E" — {0}is deformableinto S* 1, as @(x,t) = (1 — )x + t-x/|x| shows. Ex. 3 S"~1isdeformable over '™ to {0}, by @(x, t) = (1 — ¢)-x; similarly, I'" is deformable into {0}.

Ex. 4

If each Y, is deformable into B, then | [ Y, is deformable into [| B,

[see 2.2(3)].

‘

‘

Ex. 5 Aset A C X may bearetract of X, and yet X may not be deformable into A, as any point x¢ of a discrete space (having more than one point) shows. Ex. 6 X may be deformable into 4, yet 4 may not be a retract of X. For example, I is deformable into Fr (1), say by ¢(x, t) = t-x; but Fr (/) is evidently not a retract of 1.

That this concept allows reduction of the homotopy question is 6.2

Theorem Let B be deformable into A over X. Let Z be any space and fo, f1: X — Z any two continuous maps. Then f, | 4 ~ f, | 4

implies f, | B =~ f; | B. In particular, if X is deformable into 4, thenf, ~ fyifand only if f, | 4 ~ f, | 4. Proof: We prove only the first statement; because of 2.2(2), the second is an immediate consequence. Let @: B x I — X be the deformation, and define ¢: B -~ 4 by ¢(b) = @(b, 1). Since f, | 4 =~ f, | 4, it follows from 2.2(1) that fyo @ =~ f; o . Since ¥(b, t) = fy o D(b, 1) shows f, | B = f, o ¢ and, in a similar manner, also, f; | B =~ f; o ¢, the proof is complete. In view of 5.3 and 6.2, we obtain subsets for which all maps are extendable, with preservation of homotopy relations, by combining the concepts of deformation and retraction. 6.3

Definition Let 4 C B C X. We call 4 a deformation retract of B over X if the identity map 15: B — B is homotopic in X to a retraction 7: B— A. The set A is called a strong deformation retract of B over X if r ~ 15 rel A4, that is, keeping the points of 4 fixed throughout the entire deformation of B into A.

325

Deformation Retraction and Homotopy

Sec. 6

Ex. 7 In Ex. 2 and 3, all the deformations are strong deformation retractions. {l/nxI|n=1,2.--}. LetX C E%2bethesubspace(0 x NV (I x 0O Ex.8 Then 0 x I is a deformation retract of X; but it evidently is not a strong deformation retract of X. Ex. 9 Let I’"be the unit n-ball in E™; then (V'™ x 0) U (S*~! x I) is a strong deformation retract of V'™ x I. A deformation retraction is obtained by projecting V" x Tonto(V" x 0) U (S" ! x I)from(0,---,0,2)€ E" x E';informulas,

D(x, 1) = (—,2—2—t) || ||

| > 221

el

)

2x

2 -t

For the characterization question in 5, we now have Let A C X.

Theorem

6.4

Then

A

retract of X if

is a deformation

and only if it has the following two properties: (1). For every space Z, each continuous f: A — Z is extendable over X, and

(2). For every space Z and each F, G: X — Z, F is homotopic to G whenever F | 4 ~ G | 4. Proof: Necessity is clear, from 6.2 and 5.3. Sufficiency: from (1) and 5.3, there is a retraction r: X — 4. Taking Z = X and noting that

the maps 7, 1x: X — X satisfy r | 4 = 14 | 4, we find from (2) that T

As remarked, 6.4 implies: If A is a deformation retract of X, then #:[X, Z]—[4, Z] is bijective for every Z. Thus, for example, the homotopy classes of maps of the punctured ball V" — {0} into any space are the same as those of S™~! into the same space. Observe that 6.4(1) is much stronger than required to assure that ¢# is surjective: for this purpose, it would suffice to know for each f: 4 — Z simply that it is homotopic to a map that is extendable. We now examine the concept of a deformation retract more closely. Definition 6.3 combines both retraction and deformation into one condition. That this conceptis simply the conjunction of the two properties follows from

Theorem

6.5

A is a deformation retract of B over X if and only if 4

is a retract of B and B is deformable into A4 over X.

Proof: tion,

and

Necessity is clear. For sufficiency, let r: B— A be a retrac@: B

x I —- X

a deformation

of B

over X,

into 4

where

//\ //\

0 A(b, t) = D(b, 2t) 3 ro Db, 2 — 2t) is continuous because D(b, 1) = 7o D(b, 1), and homotopic to r: B — A.

//\ //\

®(b, 0) = b. Then 1

2 1

shows

1: B— B

326

Chap. XV

Homotopy

“Transitivity” takes the form 6.6

Let A, B, C be subsets of X such that A C B N C. Assume that

A is a (strong) deformation retract of B over X. If C is deformable over X into B (rel A), then A4 is a (strong) deformation retract of C over X. Proof:

Let @ be a deformation of C into B, and

¥ a deformation

retraction of B onto 4. The map 4: C x I — X given by d(c, t)

D(c, 2t) V[P, 1),2t —

is the required deformation retraction whenever both @ and ¥ do so.

7.

02

is the desired retraction. The proof of (2) is similar. In the remainder of this section, we are going to characterize those subsets 4 C X for which (X x 0)uU (4 x I)is aretractof X x I to do this,

we will need the idea of a zero-set, so well as that of a halo.

A set Cin aspace X is called a zero-set in X if there exists a continuous p: X — I such that p~1(0) = A, that is, such that p vanishes on, and only on, A. Clearly, such sets are closed G;-sets; and if X is normal, the

zero-sets in X are exactly the closed G;-sets. Note that in any space Y, not necessarily Hausdorff, ¥ x 0 is a zero-set in Y x I, as the pro‘ jection p: Y x I —I shows. The second of the required notions is given in

Definition Let X be a space.andlet 4 C X. An open set U D 4 is called a halo of A4 if there exists a continuous £: X — I such that A C h~}0)and {x | i(x) < 1} C U. Clearly, X is always a halo of any subset 4 C X. In normal spaces X,

7.3

328

Chap. XV

Homotopy

every nbd of a closed A C X is a halo of 4; but in non-normal spaces, a closed A

C

Y may have nbds that are not halos: in VII, 7, Ex. 3, each

nbd of the point a that excludes the point b is not a halo of a. The desired characterization is given in

7.4

Theorem (D. Puppe) Let X be a (Hausdorff) space, and 4 C X closed. The following four statements are equivalent:

(1). (X x 0)u (A4 x I)is aretract of X x L. (2). 4 is a zero-set in X, and also a strong deformation retract over X of a halo U D 4. (3). There exists a continuous A: X — E! with 4 C A~1(0) and a continuous 4: X x I— X with 4(x, 0) = x for each x € X, such that 4[x, A(x)] € A whenever A(x) < 1. 4. (X x0)u (4 x I)is a zero-set in X x I and a strong deformation retract of X x 1. Proof: (1) = (2). Let r: X x I — (X x 0) U (4 x I) be a retraction, and let J = 0, 1]. Since 4 x Jisopenin (X x 0) U (4 x I), we haver=*(4 x J)openin X x [Iand therefore (X x 1)Nnr~1(4 x J) = U x 1, where U C X is an open set that necessarily contains A. Let Px:X x I —X be the projection and define @:U x I - X by D(u, t) = pxor(u, t); then @ is a strong deformation retraction of U over X into A. We now show that U is a halo of 4. Let w(x) = p; o r(x, 1), where pr: X x I — Iis the projection. For u e U, we have r(u, 1) e A x J, so w(u) > 0; for x € U we must have r(x, 1) € X x 0, so w(x) = 0. Since w(a) = 1 for all a € 4, the function k(x) = 1 — w(x) shows that U is a halo of A. To

show

that

4

is

a zero-set,

F(x,t) =t — pyeor(x, t), and p: X — 1 is also continuous:

consider

the

continuous

function

define p(x) = sup{F(x,t)|¢eI}. Then its lower semi-continuity follows from

ITI, 10.4 (a); and to see that p is also upper semi-continuous, observe that because I is compact, p(x,) > b if and only if F(x,, f) > b for some

t; thus {x | p(x) = b} = px{(x, t) | F(x, t) = b} and, by XI, 2.5, this set is closed. We now prove that 4 = p~1(0). If a € 4, then 7(a, t) = ¢ for each ¢, so p(a) = 0. Conversely, if p(x) = 0, then p; o r(x, t) > ¢t for each ¢ > 0 so that, in particular, p; o 7(x, £) > O for each ¢ > 0; thus r(x,t)e A

x I for each ¢ > 0 and, because 4

x [ is closed, we must

have (x, 0) = r(x, 0) ¢ A x I also, therefore x € A. (2) = (3). Let A: X — I show that U is a halo of A4, and let O:x.U I+ X

be a strong deformation retraction of U onto A. Let p: X — I show that

Sec. 7

Homotopy and Extendability

329

A is a zero-set, and define A(x) = 3-max[p(x), A(x)]. Then A~%(0) = 4, and {x | A(x) < 2} C U. Define H: X x I — X by

I

H(x, 1) = ®(x, t-min[2 — Xx), 1]) if Ax) < 2 x if Ax) > 2 This of H @(a, Now

is continuous, since if A(x) = 2, then x € U and the two definitions at (x, t) coincide. Moreover, H(x, 0) = x for eachx € X, H(a, t) = t) = afor each (a,t) e A x I, and II(x, 1) € A whenever A(x) < 1. define 4: X x I — X by

d(x, t) = H(x, 0) = H(x, min[1, t/A(x)])

if if

Ax) =0 Ax) # 0

This map is actually continuous on X x [. Indeed, it is clearly continuous at each (x, t) with A(x) # 0, so we nced only prove its continuity at each (a,71)€4 x I. Let W be any nbd of 4(aq,t) = l(a,0) = a; since H(a, I) = a and I{ is continuous, there is (XI, 2.6) a nbd V of a such that II(V, I) C W; thus, 4(a, t) e W whenever simply x € V' and therefore 4 is continuous at (a, t). It is now immediate that 4(x, 0) = x for each xe€ X, and that A(x, M(x)) = II(x, 1) € A whenever A(x) < 1. (3) > (1). Definer: X x I - (X x 0)u (4 x I) by r(x, t) = (d(x, t), 0) = (d(x, A(x)),t — Ax))

if if

¢ < Ax) > Ax)

It is clear that » is continuous, and it is trivial to verify that r is a retraction. IN)=>@). Let r: X x - (X x 0) U (A4 x I) be a retraction, and define I': (X x I) x I - (X x I) by I'[(x,2), s] =

[px o r(x, s1), (1

— s)t + sp; o 7(x, 1)]

Then I': 1 ~ r is the desired deformation retraction.

Moreover, because

(1) < (2), there is a p: X —> I showing A4 is a zero-sct in X; then P(x, t) = t p(x) shows (X x 0) U (4 x I)is a zero-set in X x I. (4) = (1) is trivial. Ex. 3 The requirement in 7.4(2) that 4 be a zero-sct is essential: even in normal spaces (where every nbd of a closed set is a halo) a closed set may be a strong nbd deformation retract without being a zero-set. For example, let I¢ be the cartesian product of 2¥o unit intervals, and let 0° be the origin; then 0¢ is a strong deformation retract of I¢, but 0¢ is not a G, so it is not a zero-set in I¢. In particular, (I¢ x 0) U (0¢ x I) is not a retract of It x I. Ex. 4 We call 4 C X a strong halo deformation retract in X if therc is a halo U D A and a strong deformation retraction @: U x I -> X of U onto 4. With this terminology, it follows from 7.2 that A has the AHEP in X if and only if 4 is a zero set in X and a strong halo deformation retract in X.

330

Chap. XV

Homotopy

Ex. 5 Let X be any (Hausdorff) space and let 7X be the cone over X. Then (TX x 0) U (X x I) is a strong deformation retract of TX x I. For, letting V={xt>|t 0 such that whenever

¢': T — 2" satisfies |p(p) — ¢'(p)] < ¢

proper vertex map for

every

vertex

p,

then

D(¢

T, @) = D¢,

T, ¢).

From now on, the common value D(§, T, ¢) will be denoted simply by

D( Ta

‘P)'

a

339

Degree ofa Map S™ — S™

Sec. 1

C. Let f: S*— X" be a continuous map. Since S™ is compact, e T. we can find a triangulation 7' of S™ such that éf(c) < 1 for each Replacing f by the proper vertex map ¢,: T — 2", given by setting oA p) = f(p) for each vertex p of T, we have

Lemma 4

The number D(7, ¢,) is independent of the triangulation T of.S™ (for which the associated ¢, is a proper vertex map).

Proof: Let T, T’ be two triangulations, and let ¢, ¢} be the associated proper vertex maps. Since 7T, T’ have a common triangulation 77, it suffices to show that both D(p,, T) and D(p}, T') are equal to D(gy, T7). This will follow by repetition if it is shown that introducing one new vertex to (say) T does not alter the value, and this is trivially true, since we may count at a point of 2" lying in an unaltered ¢(o). The common value D(T, ¢,), which therefore depends only on f, is called the degree of f, and is written D( f). Ex. 2 Let1:S" — Ex. 3 Let f: S®— ¢ not in the image, we Leta:S™— Ex.4 that D(e) = (—1)"*+% For maps S° Ex.5 one given above cannot to this case and define degree O.

S be the identity map; it is clear that D(1) = 1. 8" be a constant map; computing D(§, T, ¢;) at a point find D(f) = 0. S"be the antipodal map a(x) = —x. It follows from Ex. 1

— S9, a separate definition of degree is required, since the be used. We extend the results in the above three examples D(1) = 1, D(«) = —1, and the other two maps to have

We now show that D( f) is a homotopy class invariant. I.1

Theorem

Let

7n > 0. If

f,g:S"—>2"

are

homotopic,

then

D(f) = D(g)Proof: For n = 0, the theorem is evident. We thus assume n > 1. Let @:S" x I— X" be a homotopy of f and write D(x,t) = fi(x). and us, inuo cont mly for uni is @ map the t, pac com is I x S™ e Sinc therefore there is a 8 > 0 such that d(x, ¥') < 8 = | f(x) — fux")| < 1 for every t € I. Thus there is a single triangulation 7" of S™ such that 8f,(c) < 1 for each o € T and ¢ € I; from now on we work with 7. Given t, and fixing any £€Z2", lemma 3 gives an ¢ > 0 for which any evariation of the vertices {p;, (p)} does not change D(¢, T, ¢;,). By uniform continuity there is a §(¢) > 0 such that

|t — to] < 8= | filx) — ()] < e for every x€.S" and consequently we have D(f;) = D(f,,) for all |t — to| < 8. This says that the integer-valued function D(f;) is

340

Chap. XVI1

continuous at each point ¢, € I, and The theorem has been proved. The converse of this theorem 1.2

Remark

A

map

Maps into Spheres

therefore D(f,) is constant on I.

will be considered later in 7.

f:(V"**1; S") — (V"+1; 8"

is

called

a regular

map

of

'+l Since V"*! has triangulations T into (n + 1)-simplexes such that each n-face not on S™ of an (n + 1)-simplex of T is the face of exactly two (n + 1)-simplexes, we can define “degree” for regular maps in a manner analogous to that for maps S" — S". Indeed, given T, call a vertex map @: T— V"*1 regular if (1) each vertex of S™ maps to a point on S™ and (2) @ | S™ is a proper vertex map. Then the degree D,( f) of a regular map f is determined by choosing an associated regular vertex map @;: T — I/"+1, a £el/**! — S™ not on the boundary of any ¢(g), and letting D,(f) = (number of positive ¢;[o] containing £) — (number of negative ¢[o] containing £) the sign of p[o] being determined by taking [o] positive. The arguments in lemmas 1-4 and in 1.1 apply verbatim to show that D,( f) depends only on f and that two regular maps have the same degree whenever they are homotopic in such a way that the image of S" remains on S™ during the entire deformation. A useful consequence is

1.3

Let g: (V™**'; 8™ - (V™*'; S") — S™ Then D(f) = D,(g). S

be

a regular

map,

and

letf =

g

| S™:

Proof: We first extend f to another regular map F of '"*! by sending each — —> radius Ox of V"% ! linearly onto the radius Of (x). It is obvious that D(f) = D,(F), by taking a triangulation of IV"*! using only simplexes of form (po, - -, pn, 0). Furthermore, F is homotopic to g in such a way that the image of S™ remains in fact fixed during the entire deformation, as the homotopy @: V"*1 x [ — [/n+1 given by (v, t) — tF(v) + (1 — t)g(v) shows. Thus D,(F) = D,(g), and the proof is complete. We

2.

note that

1.3 holds also for n =

0 if the definitions in Ex. 5 are used.

Brouwer’s Theorem

Brouwer’s theorem plays a fundamental as we shall see in the next chapter.

2.1

Theorem (L. E. ). Brouwer) nullhomotopic.

role in the topology

of E™

The identity map 1: S™ — S™" is not

Proof: We have seen in I, Exs. 2 and 3, that D(1) = 1,and D(0) = 0; by 1.1, the identity map cannot be nullhomotopic.

Sec. 3

2.2

341

Further Applications of the Degree of a Map

Corollary Brouwer’s theorem is equivalent to each of the following two statements:

(1). There exists no continuous map F: V"*!— S" keeping the boundary points fixed (that is, S is not a retract of yrt

1).

(2). (Brouwer’s

f: Vr+l

fixed-point

theorem)

Every

continuous

map

. Pn*l has a fixed point.

Proof: (2.1) = (1). Assume By XV, L2, this would imply is nullhomotopic, contradicting (1)= (2). Assume that there f(x) # x for each x € P+l

— S" did exist. a retraction F: V"*! that the identity map F|S": S" — S* 2.1. were some f: V**! — I/"**! such that

The map F: V"*! — S

defined by _—

F(x) = point of S™ that lies on the directed ray f(x)x,

would evidently be a (continuous) retraction V*+! — S" (2) = (2.1). Assume the identity map were nullhomotopic. By XV, 1.2, the map would extend to an F: V"*!— S* and then x— — F(x) would be a map V**! — V"*1 with no fixed point. Clearly, Brouwer’s fixed-point theorem is valid in any space homeomorphic to ¥**+1; furthermore, we can extend 2.2(1) by 2.3

If U C E"*! is any bounded open set, then Fr(U) is not a retract

of U. Proof: Assume that there were a retraction 7: U — Fr(U). We can assume that 0 € U, and we can find a ball B(0, N) O U. Define

#: B0, N) — B(0, N) by

‘

fl@z{ @]

e

Nr(x)/|r(x

xeU

Nx/|x|

xe B0, N) - U.

According to 111, 9.4, f is continuous: the intersection of the two closed sets on which f is defined is Fr(U), and the two definitions agree on Fr(U). But we now have a contradiction to 2.2(1), since f is a retraction

of B(0, N) onto Fr[B(0; N)]. 3.

Further Applications of the Degree of a Map

We give here only two of the most immediate applications.

(a) The following contains the fundamental theorem of algebra:

342 3.1

Chap. XVI

Maps into Spheres

Let f be a continuous complex-valued function defined on the finite complex plane.

Assume

that lim 20

fz I

=

n

¢ #

0 for some nonzero positive or negative

integer n. Then the equation f(2) = 0 has at least one solution. We can evidently assume that ¢ = 1; with this modification, the hyProof: pothesis is that f(2) = 2"(1 + 7(2)), where n(z) — 0 as 2 — . We argue by contradiction, so assume that f(z) # 0 for all finite z. Let ¢t be a real parameter,

t = 0, and let { vary only on the unit circle |2| = 1. Then, for each ¢z > 0, F

@D

9 = 1762y =

is a continuous map S!— S?, and since any two of these maps are obviously homotopic, they all have the same degree. Because Fj is a constant map, we must therefore have D(F;) = 0 for all ¢t = 0. We now show that this is impossible. For, since 7n({t) — 0 as t— oo, there is a ¢, such that

1—_— + 7(Lto)

11 + n(Cto)] thus |F;,(0) — (" map { — {".

=1

< 2

< 2 for all {; consequently (XV,

for all {;

¢

1.2) F;, is homotopic

to the

Since the latter has degree n # 0, l.1 gives the desired contradiction.

(b) Given a subset A C E"*! and (n + 1) continuous real-valued functions +, Pr+1 defined on A, this situation can be interpreted in two ways: @1,

(). As amap @: A— E"*Y, where p(a) = (¢1(a), - - -, Pn+1(a))-

(i). As a continuous vector field ¢ on A4, where at each a € A we have the vector ®(a) with components (p1(a), * « *, Pn+1(a)).

These two viewpoints are obviously equivalent; we call ¢ the map associated with the vector field ®. A continuous vector field is monvanishing if it has no zero vector. With each continuous nonvanishing vector field @ on A, we can construct a map ¢,: 4 — S*" by a — @(a)/|p(a)| which involves only the directions and not the lengths of the vectors in ®. In case 4 = S*, the degree of ¢, is called the characteristic of the vector field .

If ® is a continuous nonvanishing vector field on "'"*1, then ® | S" has Ex.1 characteristic 0. For, ¢,: S® — S™ is extendable to a ¢: '**! — S§™ Ex. 2 'The field of outward-drawn normals on S™ has characteristic 1; by I, Ex. 4, the field of inward-drawn normals on S™ has characteristic (—1)**1, since ¢, is the antipodal map. If ®, ¥ are two continuous nonvanishing vector fields on S™ such that Ex. 3 &(x) and W(x) are not opposed for each x € S", then ® and ¥ have the same characteristic. For, c,(x) and cy(x) are never antipodal (¢f. XV, 1.2). We now derive some of the easier facts about vector fields on balls and spheres.

3.2

Each nonvanishing continuous vector field on ¥"*! must at least one outward- and one inward-drawn normal.

contain,

on S*,

Sec. 4

Maps of Spheres into S™

343

By Ex. 1, ® | S™ has characteristic 0; by Ex. 2, ¢ | S* and the field Proof: of inward (outward) normals have differing characteristics; by Ex. 3, € must contain an outward and an inward normal.

3.3

Every continuous nonvanishing Theorem (H. Poincaré and L. E. J. Brouwer) vector field on an even-dimensional S*" must contain at least one normal vector. In particular, there can be no continuous nonvanishing vector field of tangential directions on any .S2".

If nis even, Ex. 2 shows that the inward and outward normal fields Proof: have different characteristics. Since any vector field must therefore have characteristics differing from at least one of these two fields, the result follows from Ex. 3.

3.4

Corollary antipode.

Each f: S?" — S2" either has a fixed point or sends a point to its

—> If f: S2" — S2" has no fixed point, the vectors xf(x) form a continuous

Proof: nonvanishing vector field and must therefore contain a normal vector.

Ex. 4 Every odd-dimensional sphere has a nonvanishing continuous tangent vector field: for each x = (xy, -« -, X2,) € 82"~1 let ®(x) be the vector with components (—Xp41, ***, —Xan, ¥1, * * *, X,). Observe that the map associated with this vector field is a mapping S2"~! — S27~1 that has no fixed point and does not send any point to its antipode.

4.

Maps of Spheres into S”

So far we have discussed maps of S™ into a sphere S” of the same dimension; for n > 1, it is easy to see (¢f. end of 7) that there are infinitely many homotopy classes. We now consider [S¥, S"] for & # n. In case 2 > n > 1, the number of homotopy classes of maps S* — S* is in general unknown except in relatively few cases; their determinations require methods beyond the scope of this book. To cite some examples that indicate the complexity of the problem even for n = 2: it is known that there are exactly two homotopy classes of maps S* — S§2, §5 - §2, and S7 — S2, but that there are twelve classes of maps S® — S? and infinitely many classes of maps S3— S?2; on the other hand, for each k > 1, all maps S¥ - S are nullhomotopic. The situation for 2 < 7 is simple: Each map S* — S" is nullhomotopic. To prove this, we will first establish the general fact that a continuous f: §* — S" (k, n arbitrary!) is always homotopic to a

“piecewise” linear map.

’

Let f: S* — S™ be given, and let T be a triangulation of S* such that 8f(0) < 1 for each k-simplex o € T. Let ¢,: T— S™ be the associated proper vertex map; we extend ¢; to a map A;: S — S™ by mapping each spherical k-simplex o€ T barycentrically (VIII, 5) onto ¢(o).

344

Chap. XVI

Maps into Spheres

A; is continuous, since it is continuous on each ¢ and its definitions agree on each (k — 1)-face common to two k-simplexes (cf. I11, 9.4); A, is called the linear T-approximation of f. More generally, any continuous map A: S¥— S" that maps each simplex of a triangulation barycentrically is called a piecewise linear map.

4.1

of S*

Letf: S*— S" be given. Thenf =~ A, for each linear T-approximation A, of f.

Proof: We need show only (XV, 1.2) that A/(x) and f(x) are never antipodal. Let x € o = (po, - - -, pr), Where o € T. Since f(p;) = A(p,) fori = 0,---, k and 8f(c) < 1, it follows that each A/(p;) is contained in the ball B(A,(p,), 1). Because the ball is convex, it must contain the convex hull of the A, (p;) and consequently also A/(c). This shows that d(A,(x), A{(po)) < 1, and since

d( f(x), A(x)) < d(f(%),f(po)) + d(A(Po)s As(%)) < 2, the points f(x), A;(x) are not antipodal.

4.2

Theorem Proof:

If k < n, then each f: S* — S

is nullhomotopic.

Let A;: S — S" be a linear T-approximation to f. Since A,

is piecewise linear and k < n, it follows that A,(S*) lies on finitely many great S"~1 C S™; thus A(S*) # 8% and so (XV, 1.2) A, ~ 0. Ex. 1 Observe that an arbitrary continuous f:S¥ — S™ may be surjective, even if £ < n, as the existence of Peano curves shows. The role of A; is to show that whenever 2 < n, the map f can always be deformed to uncover a point of S™, Ex. 2 Letn > 1 and «, B: I — S™ be two paths, each starting at po € S™ and ending at p; € S™ Then o ~ Brel Fr(Z). Indeed, define ¢: Fr(I?) — S™ by o(t, 0) = a(t), @, 1) = B(t) ¢(0,s) = po ¢(1,s) = p1; the map ¢ is evidently continuous. Since S! =~ Fr(/?) we can regard ¢ as a map S!— S" so that @ ~ 0; @ is therefore extendable to a @: I? — S", which is the required homotopy.

We

have seen (VII, 5.3) that if X is normal and 4 C X 1s closed, a

continuous f: 4 — S can always be extended over a nbd U D> A4, and 2.2(1) furnishes an instance in which f is not extendable over X itself. We have two important cases where more information about the nature of the extension is available.

4.3

(1). LetA C S"beclosed. Thenany continuousf: 4— S* can be extended to a continuous F: S" — S™ (2). Let A C S*"*! be closed and f: A — S™ be continuous. Removing exactly one arbitrarily chosen point p; from each component U; of ¥4, there is an extension

Lemma

F:

Sn+1

—

y{pi}ésn'

Sec. 4

Maps of Spheres into S™

345

Proof: (1). Let U O A4 be a nbd over which f: 4 — S™ can be extended; since A is compact, « = d(4, €U) is positive. Triangulate S™ so that the diameter of each m-simplex is 0, and let f: S® — S™ be an antiTheorem (K. Borsuk) podal map. Then D(f) is odd; in particular, f is not nullhomotopic.

We will use repeatedly the trivial observation that if f is a barycentric map of the zn-simplex o into E", and if Intf(c) # &, then for each y € f(o) there is exactly one point x € ¢ such that f(x) = y. To prove 6.1, we need the

Lemma

Letn > 1. Let T be a triangulation of " and letg: V" — V" be any linear T-map such that 0 does not lie on the boundary of any g(a). Let w: V'™ — {0} = S~ 1 be the radial projection,

and

¢ =mo(g|S*1):S* 1 — S""1

Then,

if

¢

is

the

number of points mapped by g to 0, we have ¢ = D(p) mod 2.

Chap. XV1

348

Proof of Lemma:

Maps into Spheres

Define a regular h: V™ — V™ as follows:

h(tx) = g(2tx) = (2 - 28)g(x) + (2t — Dmog(x)

(xeS™1, (xe€S™,

0 AU

an

extension

F: AU

U—

S",

we

draw

a ball

U and define g: B — Fr(B) as follows:

g(x)=p+R~x_§

xeB

- U

xe U.

= p + R-F(x)

Then g would be continuous, since the intersection of the two closed sets lies in Fr(U) C A and both definitions of g agree on 4; but, since g is a retraction B — Fr(B), this would contradict Brouwer’s theorem. Conversely,

assume

that 4

does

not

separate

E®*!;

then ¥4

has

exactly one component, which is necessarily unbounded. By XVI, 4.4, any f: A — S" is extendable to an F: E"*! — S" and so (XVI, 5.2) is nullhomotopic. Ex. 1

The compactness of 4 is essential: E* C E? separates E?; yet, every f: E* — S is nullhomotopic (XV, Ex. 1).

Using XV, 3.1, Borsuk’s theorem has the elegant formulation

2.2

Corollary Let A4 C E"*! be compact. Then A4 does not separate E™*1 if and only if the function space (S™)4 is path-connected.

Since (S™)* is a topological invariant of 4, 2.2 can also be stated as 2.3

Corollary Let X beacompactspace. Then the property “separates E™*1” is a positional invariant of X rel E*+1,

358

Chap. XVII

Topology of En

An immediate application of 2.3 yields 2.4

Theorem (Jordan Separation Theorem) Every homeomorphic image " of S™ in E"*! separates E"*1, and no proper closed subset of ¥" does so. In particular, the entire set " is the complete boundary of each component of €¥".

Proof: Since S™ is compact and separates E™*1 (I, Ex. 1), it follows from 2.3 that so also does each ¥" C E"*l. Any proper subset F of S™ C E™*1 does not separate E"*!, since E"*! — F is evidently pathconnected. Whenever F is closed, 2.3 shows that no homeomorph of F can separate E™*1, Remark 1: 'The general Jordan theorem is 2.4 together with the additional statement that E"*1 — ¥" has exactly two components. This additional conclusion is nontrivial, and does not follow simply from the fact that " is the boundary of each residual component. There exist compact sets B that separate E"*! into infinitely many parts, each component of ¥ B having the entire set B as its complete boundary. In this book, we will prove the general Jordan theorem only for embeddings of S in EZ2 (5). Remark 2: 'The general Jordan theorem follows in obvious fashion from the more specific version of 2.3 due to J. W. Alexander: If 4 C E™*1! is compact, then the number of components in €4 is a positional invariant of 4 rel E"*1, A proof

of this requires techniques different from those used in this book (either homology theory, or cohomotopy

groups).

Remark 3: 'The general Jordan theorem immediately leads to the Schoenflies problem: If U is the bounded component of ¥* C E™*1,is U U ¥" always homeomorphic to I"**1? The answer is “yes” for n = 1, and “no” for each n > 1 (for n = 2, Alexander’s “horned sphere” H is the usual counterexample, and the (n — 2)-fold suspension of H is a counterexample in E"*!). It has recently been proved (M. Brown, B. Mazur, M. Morse) that if the embedding A: S™ — En*1, regarded as an embedding S™ x 0 — E"*', can be extended to an embedding H:S" x [—e, 6] — E™*! for some & > 0, (thatis, if 2 can be extended to an embedding of a “shell” containing S™) then the Schoenflies problem for 4(.S™) has an affirmative answer.

3.

Domain

Invariance

3.1

Theorem

(Brouwer

X, the property X rel En*1,

Domain

“open

Invariance

in E"*!”

Theorem)

For

is a positional

any

space

invariant

of

Proof: Let U C E™*! be open and A: U— E™*! be a homeomorphism; we are to prove that U; = A(U) is open in E**1, It suffices to show that for each x, € Uy, there is an open (in E™**!) set W such that

Sec. 4

359

Deformation of Subsets of E**1

x, € WC U;. Let x = h~(x;) and let B = B(x, ¢) be a ball such that B C U. Then:

(1). E**' — R(B) is connected, because B >~ V"*! and V"+! | does not separate E™*1, (2). (B — Fr[B]) = A(B) — h(Fr[B]) is connected, since it is homeomorphic to B(x, &).

Because of (1) and (2) the formula

1 — W(Fr[B]) = {E"** — W(B)} U (B — Fr[B])}

expresses E**1 — h(Fr[B]) as the union of two nonempty disjoint connected sets; these must therefore be the components of E**1 — h(Fr[B]), and since A(Fr[B]) is compact, each must be an open set (in E"*1l),

Setting W = k(B — Fr[B]) gives x, € W C Uj; as required. 3.2

Corollary Let B C E™*! be any set and h: B~ E"*! be a homeomorphism. Then if x is an interior (boundary) point of B, h(x) is also an interior (boundary) point of (B).

For “interior”: By 3.1, A[Int(B)] = Int[A(B)]. For “boundProof: ary”: x € Fr(B) == x € B — Int(B) = h(x) € i(B) — Int[A(B)].

By using only a connectedness argument, we were able to show that E'!is not homeomorphic to E" for any n» > 1; with 3.1, we can now prove 3.3

(Invariance

Theorem

of

Dimension

Number)

E™ and

E™

are not

whenever m # n.

homeomorphic

Proof: Let m > n. If E® were homeomorphic to E™ then because of 3.1, the image of E™ under any embedding in E™ would be open in E™, However, the image is not open under the embedding (xla

"',xn)é(xl’

ct oty X

O>

Ty

0)

and this proves the theorem.

4,

Deformations of Subsets of £"**

In this section, we derive some properties of the Borsuk maps and use them to express some intuitively evident statements about subsets that separate E"*1, 4.1

Let A C E™*! be compact.

Then:

(1). p and ¢ belong to the same component of €4 if and only if BP

| A

=

Bq

I A'

(2). p belongs to the unbounded component of ¥4 if and only if

B, 4= 0.

360

Chap. XVII

Topology of E™

Proof: (1). Assume that p, g are in a common component U; then [1.2(1)] there is a continuous a: ] — U such that «(0) = p, «(1) = ¢, a — oft) and the map ®: 4 x I — S™ given by P(a, t) = shows that B, | A ~ B, | A.

Conversely, assume p, g are not

o(d) = @ in the same component;

at least one (say, p) lies in a bounded component U. If B, | A were homotopic to B, | 4, then because B, | 4 can be extended over E**1 — ¢ XVI, 5.1 would yield an extension of 8, | A over E**! — g also; in particular, 8, would be extendable over A U U, which, by statement (1) in the proof of 2.1, is impossible. (2). We need show only that 8, | 4 ~ 0 for some p belonging to the unbounded component U of #A4. Choose 7 so large that A C B(0;7),

and let pe UN EB(0;r); By,| A ~ 0.

then B,(4) C S™ — {p/|p|}, and therefore

Since E™*! is contractible (XV, contractible (over E"*1) to a point. 4.2

I, Ex.

1), any 4 C E™*!

is also

Let A C E"*! be compact, and let p be any point of a bounded component of ¥4. Then, in any contraction of 4 over E"*! to a point, 4 must cross p (that is, A must contain p at some stage of the deformation).

Proof: Let @: A x I — E™*! be any contraction of 4 over E**! to a point where @(q,0) = a, P(a, 1) = x, € E"*! for each aec 4. If _ E DA

x I), then

D(a, t) — p —————

would

show

. A ~ 0 which,

b

4.1(2), is impossible. 4.3

Let A C E"*! be compact, and let p, ¢ be in distinct components of #A. If A is deformed over E™*?! into a set A,, and if 4 never crosses either p or ¢ during this deformation, then p, ¢ are still in distinct components of €A4,;.

Proof: 4.4

'This is immediate from 4.1(1) and XV, 6.2.

Let A C E"*! be compact, and B C 4 be a retract of 4. Then % B cannot have more components than €4.

We will show that if Uy is any bounded component of €B, Proof: then there is some component U, of €A contained in Ug. Note first that Uy — A # @: For, if Uy C A4, then because

Fr(Uz)

C B, the re-

traction r: 4 — B would give a retraction r | Ug: Ug -~ Fr(Uy), which

Sec. 5

The fordan Curve Theorem

361

(XVI, 2.3) is impossible. Thus there is a pe Ug — A4, and we let U, be the component of €A containing p. Since it is clear that (BQIA:;BpIA)

=

(BQIBZIBPIB)’

it follows at once that U, C Uy, as required.

5.

The jJordan Curve Theorem

Two maps of an arbitrary space X into S! can be combined by multiplying their values (regarded as complex numbers of modulus 1). Taking advantage of this fact [which evidently makes (S*)* into a group], we prove that a special type of separation theorem (5.2) is valid for subsets of the plane, and this then yields the general Jordan theorem for E2, 5.1

Theorem (S. Eilenberg) (1). Let X be any compact metric space and f: X — S be continuous. Then f ~ 0 if and only if there exists a continuous ¢: X — E! such that f(x) = ¢“® for all xe X. (2). f ~ gif and only if flg ~ 0.

Proof: (1). If f is of the stated form, then P(x, t) = ¢**® shows that f >~ 0. For the converse, we first make the following observation: If f is of the stated form and if |g(x)— f(x)| < 2 forall x€ X, then g is also of the stated form.

Indee

podal, so by defining ¢(x) to =1

to

g(x) f(®)

which

g(x)

7 be the

—1, since they are never anti-

length of that oriented arc from

does not contain 2 = —1, it follows that ¢(x) is

continuous and that fig; = e'9@; thus g(x) = f(x)-e¥® and g has the stated form also. We now prove the theorem. Let @:0 =~ f, and write @(x, t) = f(x). By uniform continuity, there isa & > 0 such that | f(x) — fi.(x)| < 2 for all x whenever [t — | < 8. Choose a subdivision 0 =¢, (X, 4) such that fog ~ 1y, keeping the image of B in B during the entire homotopy, and go f ~ 1,, keeping the image of A in 4 during the entire homotopy. With these restrictions on the deformations, the notions of a pair homotopy equivalence f: (X, 4) ~ (Y, B), and of left and right pair homotopy-inverses of a given g: (X, 4) — (Y, B), are defined in obvious fashion, and the analogue of 1.4 is evidently valid. By taking A = B = @, the pair concepts reduce to those discussed in |; in this sense, the concept of homotopic pairs is more general than that of homotopic spaces. Observe that if f: (X, 4) ~ (Y, B), then both f: X ~ Y and f|A4:4~B; however, X~ Y and A~ B do not imply that (X,4)~(Y,B), that is, that there is a map accomplishing both these equivalences simultaneously. Ex. 1 Let X=Y = St'uU{0}, and let yo = (1,0) € S'. Then X ~ Y and {0} ~ {yo}; yet there is no pair homotopy-equivalence f: (X, {0}) — (Y, yo), since any g: (Y, yo) — (X, {0}) must send all S* to {0}.

4.

Mapping Cylinder

The notion of mapping cylinder, due to J. H. C. Whitehead, is particularly important in homotopy considerations because it provides a link between the concepts “homotopy of maps” and “homotopy of spaces.” 4.1

Definition Let f: X- Y be continuous. The space (X x I) Us Y obtained by attaching X x I to Y by the map (x, 0) — f(x) is called the mapping cylinder of f and is denoted by C( f).

Sec. 4

Mapping Cylinder

369

The space C( f) can be regarded as a “cylinder” with X(= X x 1) at the top and with its base lying in Y, the “generators” of the cylinder being the line segments joining each x € X to f(x)€ Y. Indeed, let p:(X xI)+ Y— C(f) be the identification map; from VI, 6.3, follows that p | Y embeds Y homeomorphically as a closed subset of C(f) and that p | X x ]0, 1] embeds X x ]0, 1] homeomorphically as an open subset of C(f); in particular the map :: X — C(f) given by #(x) = p(x, 1) is a homeomorphism of X onto the upper face X x 1 of C(f). We will use the notation {x, t) for p(x, ¢) and {y) for p(y); observe (VI, 6.4) that a pair of continuous maps ¢: X x I - Z, : Y > Z satisfying ¢(x, 0) = ¢( f(x)) for each x e X, determines a continuous

(® #): C(f)= Z by (p, K, £) = @(x, t), ((P) ¢) = ‘A(y)’

and similarly for homotopies (XV, 2.2(4)). The “collapsing map” p: C( f) —

Y is defined by

plx, t) = f(x), ey =", and 1s evidently continuous. Letting j: Y— C(f) be the homeomorphism p | Y, then jo p is precisely the collapsing of C(f) onto its base. Thus, with each mapping cylinder, we have the diagram

< wheref = p o 1.

4.2

The map p: C(f)— Y is a homotopy equivalence, and has j as homotopy inverse. In particular, C( f) belongs to the homotopy type of its base, Y. Proof:

1Tt is evident that poj = 1;; to showjo p ~ 1, let e: (X x I) x I— C(f)

be the continuous map ¢[(x,?), 7] = {x,#1 — 7)) and ¢: ¥ x I the continuous map ¥(y,?) = (y>. Since for each 7€l we el(x, 0), 7] = {x, 0> = {f(x)> = Y[ f(x), 7], it follows (XV, 2.2) (ps ¥): C(f) x I — C(f) is continuous and clearly (¢, ¢): 1 =~ jo

C(f) have that p.

The argument in 4.2 shows that in C( f) itself, the base j(V) is a strong deformation retract of C( f).

Homotopy Type

Chap. XVIII

370 Remark:

From

4.2 and the diagram

for mapping cylinders, we observe that 4 o any map f: X — Y can be factored as X — C(f) — Y, where { is an injection and p is a homotopy equivalence; identifying Y with j(Y), we havejof ~ jopoi =~ so we can conclude that any map f is homotopic to an embedding map into a suitable space belonging to the homotopy type of Y. Thus, for work with homotopy, there is no loss in generality to assume that all maps are injective.

The connection between homotopy of maps and of spaces is provided

by 4.3

the

pair

Let®:f, ~ f,;foreachxe X, themapt— D(x,£) (0 < an arc in Y joining fy(x) to fi(x). We define F: C( fo) — identity map on the base, and to map each line segment linearly on the arc in C( f;) consisting of the segment above arc in the base joining f,(x) to fo(x). Precisely, I) + Y= C(fy), 7= 0,1, be the identification maps,

t < 1) C( f1) 2t>1

Fpo(x, t) = pq(x, 2t — 1)

= p,[D(x, 2t)]

1>2t>20

This is continuous, since

Fpo(x, 0) = p1[D(x, 0)] = p1[fo(x)] = Fpol fo(*)]-

1>t >t

Gpy(, 1) = po(x, 2t — 1) = po[P(x,1 = 2t)]

>4

+>21t20. Vv

Gpi(y) = po(¥)

Vv

We define a continuous G: C( f;) = C( f,) similarly by

Thus G o F maps po(Y) identically, and sends each “generator” (x, £) semilinearly on the arc consisting of {(x, t) plus the arc from fy(x) to fi(x) plus the retracing of that arc from f;(x) to fy(x). We perform a deformation that uniformly shrinks the retraced arcs over themselves and simultaneously alters the parametrization to the identity, as follows:

A[y>, sl = < A[{x, £, 5] = = {D(x, 3s — 41))

— (O, 20))

1>t>3s s >

3s >

Sec. 5

371

Properties of X in C(f)

This is continuous, since for each s € I we have

A[{x, 05, 5] = (D(x, 0)) = {fol(x)> = A

f(%o)s 5]

Furthermore, 4: 1 ~ G o F, and indeed the points of X(= X x 1) are kept fixed throughout the entire homotopy. In similar fashion, we verify Fo G ~ 1, and the theorem is proved.

5.

Properties of X in C(f)

Throughout this section, f: X — Y will be a fixed given map. It was seen that j(Y) is a deformation retract of C( f); we now determine the properties of i(X) in C( f). To cut down on symbolism, 7(X) is identified with X. 5.1

[C(f)

x 0] U [X x I]is always a strong deformation retract of

C(f) x I

Note that the map ¢: C(f) — 1 given by ¢lx, ) =1 — ¢, Proof: @{y> = 1 is continuous and shows X is a zero-set in C(f). Since X is a strong deformation retract of the halo p(X x ]0, 1]) of x in C(f), the proposition follows from XV, 7.4

5.2

X is a retract of C(f) if and only if f has a left homotopy inverse.

Assume Proof: r:C(f)— X by

®:gof ~ 1y for some g: ¥— r

=

X.

Define

a map

C ®:1 ~ pod; segment send each (2> pointwise fixed

1

)

373

Change of Bases in C(f)

Sec. 6

Assume diagram

now that p is a homotopy

inverse of A. We then have the

C(poAeof) where F, G are the homotopy equivalences in 4.3 constructed by using the homotopy D( f(x), t) of f to po Ao f. It follows directly from the definitions that @ = F o u,; consequently, ~ 1.

(-/_\OG)O;Z=X0G0F0;L+:X°;L+

Since also zo A, ~ 1, Theorem 1.4 shows that ji is a homotopy equivalence, with homotopy inverse Ao GofoA, ~ A,, and the proof is complete. Replacing the top base of C( f), we find 6.2

Given Z—>

f

X ——

Y, there is an induced continuous map

p*: [C(fop), Z] = [C(S), X]. If u has a right homotopy inverse, then u* has a pair right homotopy inverse; and if u is a homotopy equivalence, then p* is a pair homotopy equivalence. 'The map u* is determined by sending each segment (z, ) Proof: linearly onto the segment {u(2), t> and each (y)> to {y>;if P:1 =~ po A, then A: [C(f), X]-> [C(fo n), Z] is defined by mapping each segment {x, t> linearly onto the arc [segment {A(x), t> plus arc fo @(x, t)]. The rest of the proof follows as in 6.1.

6.3

p

f

A

Q be given. Y — Let P—> X— Theorem (J. H. C. Whitehead) If u has a right homotopy inverse and A has a left homotopy inverse, then there is a map ¢: (C(f), X) — (C(A o fo u), P) that has a pair left homotopy inverse; and if A, u are homotopy equivalences, ¢ is a pair homotopy equivalence.

Chap. XVIII

374

Proof:

Homotopy Type

We have u*: (C(fopn), P)— (C(f), X) and also

Ay (C(fom) P)—>(C(Aofop) P) If 7 is a right homotopy inverse of u, then defining 7 as in the proof of 6.2, we find that ¢ = A, o ¥ has the required properties. Remark: A pair (Y, B) is said to dominate a pair (X, A) if there exists an f: (X, A) -> (Y, B) that has a left homotopy inverse. With this terminology, 6.3 becomes: If P dominates X and Q dominates Y, then the pair [C(A o f o u), P] dominates [C( f), X].

We use 6.1 in another way to derive 6.4

Theorem

Letf: X —

(S. Eilenberg, M. Shiffman)

Y be a homotopy

equivalence, and attach '" to X by a: S®~! — X. Then, by attaching V"to Ywithf =foa:S" '—>Y,wehave XU, V"> YU, V", Proof: By 6.1 we have that f,:[C(«), S* ] —[C(foa),S"] isa homotopy equivalence. Let g be a homotopy inverse of f; thenf, o § ~ 1 andgof, ~ 1, keeping S"~! x 1 pointwise fixed. Thus, attaching V'™ to each mapping cylinder by the identity map S®~1—S8""! x 1, westill have the resulting spaces homotopic. Since these new spaces are obtained by attaching (V™ x 1)U(S™®* xI) to X and Y by «|S" ! x 0 and foa|S"1 x 0, respectively, and since [(V"™ x 1) U (8"~ x I), S"" ! x 0] is homeomorphic to (V'*, S™~1), the proof is complete. Ex. 1 Theorem 6.4 has many applications. To indicate only one of them, we construct models for the homotopy types of certain spaces (the abstract situation we describe occurs in the Morse critical point theory). Assume that a space X is decomposedas Ao

C B;

C A;

C By

C---C

A,

= X, where (1) Ay is contractible,

and for i > 1, both (2)B, ~ A;_; and (3) (4, — B;, Fr(B))) = (V™, S™ 1) for eachi. Then, by (1), 4, is homotopic to a space M, consisting of a single point; by (2), we have B; ~ M,; by (3), we can attach '™ to M, and get a space M, ~ A;. By proceeding in this manner, the space X is seen to belong to the homotopy type of n balls of varying dimensions suitably attached to each other.

Problems Section

|

1. Let Y ~ Z and let X be locally compact. Prove Y* ~ Z¥X, 2. Let X ~ Y and let X be locally compact. Prove Z¥ ~ Z%, 3. For any space X, show that X @ X x [ ~ X x E" ~ X x I*.

375

Problems

4. Give an example to show that (4 C X) A (4 ~ X) does not imply that 4 is a deformation retract of X, even though A4 is closed. 5. Prove: Being a homotopy equivalence is a homotopy class invariant; that is, if fi X~ Yandg ~ f,theng:X ~ Y. 6. Let f: X~ Y and g: Y ~ Z have homotopy inverses f’, g’, respectively. Show that g o f has f’ o g’ as homotopy inverse. 7. A space Y is called weakly locally contractible if each y € Y has a nbd U deformable over Y to y. Show that if X ~ Y and if Y is weakly locally contractible, then so also is X.

- Section 5 1. Let f: X— Y, and h: X — Z. Prove: There exists a g: Y — Z such that g of ~ hif and only if & can be extended to a map C(f) — Z. 2. Let g: Y— Z and h: X — Z. Prove: There exists an f: X — Y such that g o f ~ hif and only if % is homotopic over C(g) to a map of X into Y. 3. Let f: X— Y be extendable to a map F: TX — TY. Let Tf: TX — TY. Prove that Tf ~ Frel X.

Section 6 1. Let the diagram

Z— W be commutative. that the diagram

Show

that there

X —>

exists a continuous

C(f)-~£>

A: C(f) — C(g)

such

Y

o Z—i;—>

C(g)

T

w

is commutative in each square. 2. Let A4 be a strong deformation retract of X, and let Y be an arbitrary space. Prove:

If X is attached to Y by a continuous f: A — Y, then deformation retract of X Us Y. Using this result, show that x [is attached to Y by (a, 0) — f(a), then b. If X x 1 vA4 a.

Y is a strong

Y >~ XUrY. [X x1udxIlur (One can use this result to obtain an evident extension of the model construction described in 6, Ex. 1.)

Path Spaces; H-Spaces

XIX

In this chapter, we consider structures consisting of a space Y together

with a “multiplication” I': ¥ x Y -» Y that has a simple homotopy property. Such structures groups and path spaces.

I.

include,

as special

cases,

both

topological

Path Spaces

Recall that [V, 5] a path in Y is a continuous mapping of the unit interval I into Y, rather than a continuous image of I in Y; that is, a path in ¥ is an element of ¥Y/. The path « € Y7 is said to start at the point «(0) e ¥ and to end at the point «(1) € Y; a closed path, or loop, at y,€ Y isa path starting and ending at y,. The constant path n(/) = y, is called the null path, or loop, at y,.

of2t)

weB(f) = { B2t — 1) 376

0

t

A

Definition The inverse of a path «e€ Y! is the path «~'e ¥! defined by the rule ™ 1(#) = «(1 — #), (0 < ¢ < 1). The product of two paths «, B, written « * 3, is defined only in case o(1) = B(0), ' and is the path N

1.1

La~lof (Y;a,b) —

Y; b, a)is ahomeomorphism.

Proof: 'The map is clearly bijective. Now let p: I — I be the homeomorphism p(f) =+— ¢; by XII, 2.1, the induced map p*: Y’ — Y7 is continuous and is precisely the map «— o~'. Similarly, the map o~! — « is continuous. The following proposition is very useful.

1.3

Let p: I —>1 be continuous and such that p(0) = 0, p(1) =1 (p is called a “parameter transformation”). Then the map ¢ — a0 p of (Y, a, b) in itself is homotopic to the identity map.

Note first that p ~ 1rel Fr(I), as py(t) = (1 — s)p(t) + st Proof: (0 < s < 1) shows. By XV, 3.2, it follows that the induced map p*: Y! — Y!is homotopic to the identity map, and since p;" | £(Y; a, b) maps 2(Y; a, b) in itself, the assertion follows.

For the product operation, 1.4

(1).

The mapping (a, B) — a * B of QAY;a

b) x Y;b,c)—>AY;

a,c)

1s continuous. (2). If ye £(Y; b) is the null path, the map a—a*% of £(Y;a,b) in itself is homotopic to the identity map, and so also is the map B—n % B of Y;b,c) in itself.

Proof: Ad (1). This will follow from XII, 3.1, by showing that the associated map £2(Y;a,b) x (Y;b,¢) x I - Y is continuous. This latter map decomposes into (o, B, t) = (o1, ) — ( a*(B*vy)

AY;a,b)

x AY;b,c)

x &AY;¢,d)—

of

Y} a,d)are homo-

topic.

(2). The map o — o * a1 of (Y;a,b)— £2(Y;a)is nullhomotopic, and so also is the map ¢« > a1 x a.

Ps(t)

=

{

20 - (1 —s)

A

N

Proof: Ad (1). Let R(e, B, ¥) = « * (B ) and observe that there is a parameter transformation p such that (a x 8) * y = [a* (8 * y)] o p for all a, B, y. From 1.3 we find that p* o R ~ R, and the proof is complete. Ad (2). Define a homotopy 1 261 — ) 0 t

1 T"'; in this case we say I', I'" are equivalent H-structures on Y. It is obvious that

3.2

Equivalent Comp

Y.

H-structures

on

Y induce the same H-structure in

Sec. 4

H-Spaces

383

Ex. 4 All H-structures on a contractible space are equivalent. Ex. 5 Distinct H-structures on a discrete space (with more than one point) are never equivalent. It is true, but much more difficult to prove, that there exist

path~connected spaces having at least two nonequivalent H-structures.

4.

H-Spaces

It is not true that every space can carry an H-structure; spaces that do admit an H-structure are called H-spaces. Ex.1 Every group space, and every loop space, is an H-space; each contractible space is an H-space. Ex.2 89 S, and S® are group spaces (integers mod 2, complex numbers of norm 1, and quaternions of norm 1, respectively). S7 is an H-space (Cayley numbers of norm 1), but not a group space. It has been proved by J. F. Adams that, among the spheres, S° S, S3 and S7 are the only ones that are H-spaces. Ex.3 Any space with a non-abelian fundamental group (a figure 8, for example)

cannot be an H-space, as we shall see in 8.4.

The notion of H-space is significant only for path-connected spaces, because 4.1

1In order that Y be an H-space, it is necessary and, if Y is locally path-connected, also sufficient, that one path component of Y be an H-space.

Proof: Necessity follows from 2.2. Sufficiency: If P is a path component carrying an H-structure, define a composition in Y by preserving that on P, setting y-y’ = y'.y = p, € P whenever y, y' € P,

and p-y = y-p =y

whenever y € P, p e P. This composition is con-

tinuous, since in locally path-connected spaces each path component is both open and closed, and defines an H-structure on Y.

The concept of H-space is also a homotopy type invariant:

42

Let (Y,I') be an H-structure, X any space, and /: X > Y a homotopy equivalence with homotopy inverse g. Define a composition law in X by I'y(x, ') = g o T'( f(x), f(x)). Then (X, T',) is an

H-structure and both f, g are H-isomorphisms. Proof: We first show that I, determines an H-structure on X. Note first that from fo g = 1y, it follows that f(X) contains points from each path component of Y, so we can choose an ¢ € X such that f(e) belongs to the principal component of Y. Define k(y) = y-f(e); then

384

Chap. XIX

Path Spaces; H-Spaces

(2.2) we have & >~ 1; and the map x — I'y(x, ) is precisely the map gohof~gof~ 1, Similarly, x — I'y(e, x) is homotopic to 15 and (X, I'y) is an H-structure. once because fog ~ 1.

That f, g are H-homomorphisms

follows at

Note that the H-structure imposed on X depends on the particular homotopy equivalences that are used: If f’, g’ is another pair, and if f =~ f' (so that g =~ g’ also), then the H-structures imposed on X by the pairs ( f, 2) and ( f', ') are easily seen to be equivalent. Observe that to show (X, I'y) 1s an H-structure, we require only the hypotheses (1) g o f ~ 1, and (2) f(X) contains points of the principal component of Y'; in this case, however, it may not be true that f and g are

H-homomorphisms. Since the hypothesis (2) is always satisfied whenever Y is path-connected, we have 4.3

Every space dominated by a path-connected H-space.

H-space

is also an

Ex. 4 A retract of a path-connected H-space Y is also an H-space, since Y dominates each of its retracts. Ex. 5 SXY;a,b)is an H-space: for, by using 1.7, we find that £(Y; a, b) ~

. XY; a). Ex. 6 A homotopy type containing a group (or H-) space is composed entirely of H-spaces. The homotopy type of S7 consists only of H-spaces and H. Samelson has shown that it contains no group space.

5.

Units

By a unit of an H-structure Y is meant an element ¢ such that e-y = y-e =y for all y € Y. There can be no more than one unit, since for any two units e, ¢/, we would have e = e¢-¢’ = ¢'. Not every H-structure has a unit: For example, the natural H-structure (Y; y,) has no unit,

However, it is trivial to verify 5.1

Any H-structure on a discrete space has a unit. In particular, the induced H-structure on Comp Y always has a unit. The requirement that an H-structure have a unit is not an essential

restriction.

5.2

'T'o see this, we need

Lemma Let Y be an H-structure, let e be a point in the principal component, and let R be a homotopy of the map y —e-y to the identity, R(y, 0) = e-y. Then there exists a homotopy L of the map y — ¥ -e to the identity such that the two paths L.(¢) = L(e, t) and R,(t) = Rfe, t) from e-e to e are equivalent.

Sec. 5

Units

385

Proof: Let Il be any homotopy of y—>y-e to 1y, where I(y, 0) = y-e. We first show that there is a loop « at e such that

R, ~ (e-a)

[,

Let @ = R, % I;1. Regarding & as a map a:(S?, p,) = (Y, e-€) and noting that (S x 0) U (py x I) is a retract of St x I, it follows that we can deform & to a map «: (S, py) — (Y, €) in such a way that the image of p, traces R, during this deformation. The homotopy R(«(?), 1 — s), 0 < s < 1, transforms « to the path e-a. During these two deformations, the image of p, traces the path R, * R; ', which is equivalent to a null path, so that (XV, 8.3) & can be deformed to e-« without ever changing the image of p,; that is, @ ~ e-a«. Thus

Now

define

R xl; xl, ~ axl, ~ (e-a)*l,.

- a(2t)

L(y’t)={l 2, the nth homotopy group of any space is abelian.

Path Spaces; H-Spaces

Chap. XIX

390

Problems Section

|

1. Let a, a’, b, b’ be any four (not necessarily distinct) points lying in a single

path component of Y. Show that £(Y; a, a’)~ (Y 2.

b, b).

Prove that any two path components of .Q( Y'; a, b) belong to the same homotopy

type. 3. Let Y, Z be path-connected spaces belonging to the same homotopy type. py oto hom e sam the to ngs belo b) a, ; £XY of ent pon com path any that e Prov

type of any path component of £(Z; ¢, d). 4. Let yo€ Y and let po € S? be fixed points, and let A € YS! be the subspace of all maps sending po to ¥o. Prove /A is homeomorphic to £(Y; yo). 5. For any two subsets of 4, B of Y, let 24, B) = {ae Y|

a(0) € 4, «(1) € B}.

Assume that 4 is contractible over Y to a point y,.

(A4, B) ~ A

Prove that

x £Xyo; B).

Section 2 yo)? (Y, re ctu tru H-s ral natu the of ent pon com l cipa prin the is t Wha 1. 2. Let Y be a topological group and X a locally compact space. Prove:

a. The induced H-structure Y¥ is also a topological group. b. The induced H-structure Comp Y ¥ is also a topological group. c. Show that the principal component P of Y ¥ is a normal subgroup and that Comp Y ¥ is isomorphic to Y */P.

Section 5 1. Let Y be a HausdorfT space, and yo€

(Y x ¥0) V(30 x Y) and let f: A — Prove:

Y. Let 4 C

Y x

Y be the closed subset

Y be the map f(¥,¥0) = f(¥0,¥) = ».

Y carries an H-structure with unit y, if and only if Y is a retract of

(Yx Y)us Y.

Section 6 1. In the extension of Y to Y’ described in 5.3, show that whenever inversion, the inversion can be chosen in Y’ so that 1-' = 1.

Y admits

Section 7 1. Let Y be any path-connected space. For each integern > 1, let I" be the unit n-cube and £2,(Y, y,) the subspace of Y'" consisting of all

f: " Fr(I™) — (Y, yo).

Problems

391

Let k,(I™) = yo be the constant map.

a. Prove that 2(2.(Y; ¥o); k.) is homeomorphic to £2,, 4+ 1(Y; ¥o). [Represent In*1 a5 I" x I and define h: 2, .1 —>.Q(.Qn) by A[f(x, )]= [f(x)]() (x,t)eI™ x L] b. Prove that any path component of £2,(Y, o) belongs to the homotopy type T.(Y). 2. Let Y be an arbitrary space, and let q, b, be two points of Y. Let X be a locally compact space, and denote the poles of its suspension SX by p, and p_. Let {SX, Y} be the space of all maps of SX into Y sending p, toaand p_ to b. Show that {SX, Y} =~ [£XY; a, b)]%, and that consequently {SX, Y} has an H-structure such that Comp {SX, Y} is a group. Describe a composition operation in {SX, Y}. 3. Show that the fundamental group of S?! is an infinite cyclic group. [Hint: see XVI, 7.4] 4. Prove that the fundamental group of S™, n > 2, is trivial. 5. Show that the fundamental group of any contractible space is trivial.

Fiber Spaces

XX

In this chapter, we will consider only the covering homotopy property in fiber structures, and some of its more immediate consequences. Our aim is to establish that, under fairly general conditions, the local validity of the covering homotopy property implies its validity in the large.

I.

Fiber Spaces

1.1

Definition A fiber structure is a triple (E, p, B) consisting of two spaces E, B and a continuous surjection p: E — B.

The space E is called the total (or fibered) space; p is termed the projection, and B is the base space. We refer to (E, p, B) as a fiber structure

over B, and for each b € B, the set p~1(b) is called the fiber over b. Let (E, p, B) be a fiber structure, let X be an arbitrary space, and let

g: X — B be continuous, so that we have the diagram

Sec. 1

Fiber Spaces

393

A continuous §: X (or, more simply, identity map 1: B every cross section

— a — is

E such that p o § = g is called a lifting of g into E covering of g). In particular, a lifting o of the B into E is called a cross section; since poo = 1, evidently injective.

Ex. 1 A map g: fiber structure with the identity map g: directly from XVI, cross section.

X — B need not be liftable into E. Let (E?, p, projection p(x) = ' and let X = S!. It is easy X — S! cannot be lifted into E! [for example, 6.2(2)]. In particular, this fiber structure does

S*) to this not

be the see that follows have a

'The general question that we will consider is the following: if g: X — B can in fact be lifted to a §: X — E, then can every homotopy of g be covered by a homotopy of g? Fiber structures in which this can always be done are very important in modern topology. Ex. 2 If g: X — B can be lifted to §: X — E, then a homotopy of g may not be liftable to a homotopy of g. Let E = (I x 0) U (0 x [I) and let (E, p, I) be the fiber structure with projection p(x, y) = x. Let X be any space, let g: X — I be the constant map g(X) = 0, and lift g to the constant map g, where (X) = (0, 1). Then the homotopy @(X, t) = t of g cannot be covered by a homotopy of g.

Let ¢: X x I — E be a homotopy covering ¢. We say that ¢ is stationary with ¢ if for each x, € X such that ¢(x,, f) is constant as a function of #, the function $(x,, ?) is also constant. 1.2

Definition (1). A fiber structure (E, p, B) is called a fiber space (or fibration) for a class &/ of spaces if the following condition (called the covering homotopy condition) is satisfied: For each each continuous f: X x 0— E and each homotopy X e, @: X x I — B of pof, there exists a homotopy ¢ of f covering ¢. The fibration is regular if ¢ can always be selected to be stationary with ¢. (2). A fiber space for the class of all spaces is called a Hurewicz fibration.

Ex. 3 Let p: Y x Z— Y be the natural projection p(y,2) = y. Then (Y x Z,p, Y) is a regular Hurewicz fibration. In fact, let pz: Y x Z— Z be the projection; then for any continuous f: X X 0— Y x Z and homotopy of pof, the map @(x,t) = (p(x, t), pz o f(x,0)) is a covering p: X x I—Y homotopy, and is stationary with ¢. Ex. 4 Let B be any space, let Z be a normal locally compact space, and let 20 € Z be a Gs and strong nbd deformation retract of Z. Let w: B%Z — B be the evaluation map w(g) = g(20). Then (BZ, w, B) is a Hurewicz fibration. Indeed,

394

Chap. XX

let f: X x 0— BZ?

be

continuous,

and

let ¢: X x I— B

be

Fiber Spaces a homotopy

of

wof. Since Z is locally compact, the associated map f: Xx0x Z—B continuous. Let L = X x [(0 x Z) U (I x {20})], and define u: L — B by

is

(. 0, 2) = f(x, 0, 2), l-l'(x9

t)

2’0)

=

(P(x)

t):

which is continuous. Because (0 x Z) U (I X {z¢}) is a retract of I x Z, we find that L is a retract of X x I x Z, so that p is extendable to a Y: X x I x Z—B; the associated map lfi: X x I-—> BZ is the required covering homotopy. Ex. 5 1If(E, p, B) is a fibration for the class .2/ of spaces, then for each C C B, the fiber structure (p~3C), p | p~*(C), C) is also a fibration for the class 27 of spaces.

We derive some immediate consequences of the covering homotopy condition.

1.3

Let (E, p, B) be a fibration for the class o/ of spaces, and let Xed. (1). If f: X — B is nullhomotopic, then f can be lifted into E, In particular, if B € o and if B is contractible, then a cross

section always exists. (2). A §: X — E can be deformed into a single fiber if and only if p o £ is nullhomotopic.

Proof:

(1).

Let

¢: X x I— B

be a

nullhomotopy

of f,

with

(X, 0) = b,. Since the map X — b, can be lifted, so can the homotopy

@, and the map x — @(x, 1) is a lifting of f into E. The proof of (2) is entirely analogous. For any two spaces, X, Y, a surjective f: X — Y is called irreducible whenever all maps homotopic to f are also surjective. Obviously, if ¥ has more than one point, an irreducible map is never nullhomotopic; in the particular case Y = S", the irreducible maps are exactly those maps that are not nullhomotopic.

1.4

Let (E, p, B) be a fibration for the class &/ of spaces, and let X e If f: X — Eis irreducible, then so also is p o f. In particular, if E € o and if 1: E — E'is irreducible, thenso alsois p: E— B ' and 1: B — B.

Proof: A homotopy of p o f that frees a point of B would be covered by a homotopy of f that frees an entire fiber, so the first assertion is

proved. To establish the remaining ones, we need observe only that 1z: E— E is a lifting of p: £— B, so that a homotopy of p (or of 1j) can be covered by a homotopy of 1;.

Sec. 2

1.5

Fiber Spaces for the Class of all Spaces

395

Let (E, p, B) be a regular fibration for the class &/ of spaces, and . If C C B is a strong deformation retract of B, then let E€ p~Y(C) is a strong deformation retract of E.

Let ¢: B x I — B be a strong deformation retraction; Proof: write @(b, t) = @b), and let ¢, = 1p. Since 1, covers @, © p, there homotopy @, of 15 covering ¢, ° p, and which is stationary with ¢; Thus, ¢, (E) C p~1(C) and ¢e) = 15(e) = e for each ee p~*(C) t € I, so ¢, is the required deformation retraction.

we is a o p. and

Let (E, p, B) be a fiber structure, let X be any space, and letg: X — B be any continuous map into the base B. Let E(g) C E x X be the subspace E(g) = {(e, x) | p(e) = g(x)} of the cartesian product, and let q: E(g) — X be the projection g(e, x) = x. The fiber structure (E(g), ¢, X) is called the fiber structure over X induced by the map d foun r fibe the x over ing plac by med for is E(g) ely, itiv intu B; — X g: over g(x). Letting =: E(g) — E be the projection 7(e, x) = e, it is immediate from the definitions that the diagram

Eg

Jo

—>

E

X

"

B

]

is commutative.

1.6

Let (E, p, B) be a fibration for the class o/ of spaces, let X be an arbitrary space, and let g: X — B be any continuous map. Then (E(g), ¢, X) is also a fibration for the class & of spaces.

Let YesZ, let f: Y x 0— E(g) be continuous, and let Proof: X be any homotopy of gof. Thengog: ¥ x I+ Bisa ¢: Y x I— homotopy ofg o go f = pomof, so there is a homotopy @: ¥ x [ - E of m o f covering g o p. The continuous map ¢: ¥ x [ — E x X given by (v, t) = (®(y, t), ¢(y, t)) actually maps Y x I into E(g), since po®D(y,t) =gop(y,t) forall (y,2)e Y x I, and is evidently a homotopy of f covering o.

2.

Fiber Spaces for the Class of All Spaces

In this section, we obtain an intrinsic characterization of the Hurewicz

fibrations, and use this to derive some of their special properties. Let (E,p, B) andf: X x 0 — E be given, and let 9: X x I — B bea h pat a nes defi t) ¢(x, — ¢ map the X, € x each For f. o p of py oto hom

396

Chap. XX

Fiber Spaces

@, in B. The problem of finding a covering homotopy is essentially that of lifting each path ¢, to a path in E starting at f(x, 0) in such a way that

the family {g, | x € X} is lifted “continuously” into E. Thus, we may expect to have an intrinsic characterization of the Hurewicz fibrations if we require that the family of all paths in B can be “continuously” lifted in a similar manner. To state this lifting process precisely, we use the c-topology in both B! and E!, and make the

2.1

Definition Let (E, p, B) be a fiber structure, and let 4 C E x B! be the subspace 4 = {(e, @) | p(e) = «(0)} of the cartesian product. A lifting function for (E, p, B) is a continuous map A: 4 — E’ such that A(e, «)[0] = e and p o A(e, @)[t] = of¢] for each (e, @) €4 and t € I. We say that A is regular if A(e, @) is a constant path whenever

a is a constant path. A lifting function therefore associates with each e € E and each path « in B starting at p(e), a path Afe, &) in E starting at e and covering «. Since the c-topology is used in E’; the continuity of A is equivalent to that of the associated A: 4 x I — E. Ex.1 In the fiber structure (Y x Z, p, Y) of |, Ex. 3, the map A[(y, 2), «](¢) = (a(2), 2) is a regular lifting function.

2.2

Theorem (M. L. Curtis; W. Hurewicz) The fiber structure (E, p, B) is a (regular) Hurewicz fibration if and only if a (regular) lifting function exists.

Proof: Sufficiency. Assume that (E, p, B) has a lifting function. Let f: X x 0— E be given, and let ¢: X x I-- B be a homotopy of p o f, For each xe X, let ¢, be the path t-— ¢(x,?). Then ¢(x,12) = ALf(x, 0), @,)(¢) is the required covering homotopy, and is stationary with ¢ whenever A is regular. Necessity. Let X = A,andletp: 4 x I— Bbethe map ¢[(e, @), {] = a(t). Let f:4 x 0— E be given by f[(e, «), 0] = e. Since p(e) = «(0) for each (e, ) €4, we have pof = ¢ |4 x 0. Letting $:4 x I — E be a homotopy covering ¢, the associated map A: 4 — E! is a lifting func“tion, which is regular whenever ¢ is stationary with g.

2.3

Corollary. Let (E, p, B) be a fibration for the class of metric spaces. If both E and B are metric spaces, then (E, p, B) is a Hurewicz fibration.

Proof: In this case, 4 C E x B! is a metric space, and the proof of the “necessity” in 2.2 shows that a lifting function exists.

2.4

397

Fiber Spaces for the Class of all Spaces

Sec. 2

Corollary. Let (E, p, B) be a Hurewicz fibration. If B is a metric space, then (E, p, B) is a regular Hurewicz fibration.

Proof: Let A:4 — E'" be a lifting function for (E, p, B). Choose a metric d for B such that §(B) < 1, and for each « € B! let d(e) = o[«([)]. Define t < d(«) a(t) = aft/d(«)]

~ of1]

t > do),

and let Ae, «)[t] = A(e, &@)[t - d(«)]. Then A is a regular lifting function. In the remainder of this section, we establish some special properties of Hurewicz fibrations that illustrate the use of lifting functions. A Hurewicz fibration always has an extended lifting function A which, with each e € E, each s € I and each path « in B going through p(e) at time ¢ = s, associates a path in E going through e and covering o 2.5

Let (E, p, B) be a Hurewicz fibration. Let D C E x B! x I be the subspace D = {(¢, o, 5) | p(¢) = «(s)}. Then there exists a continuous A: D — E’ such that A(e, «, s)[s] = e and p o Ale, o, 5)[t] = «oft]

for each (e, o, s) e D, tel.

If (E, p, B) is regular, then A can be

selected to send constant paths to constant paths. Proof:

o) =

Let A be a lifting function.

s —2) o(0)

02%, then with the finite topology it is neither a linear topological space nor a locally convex space.

Proof: The following proof is simply an immediate adaptation of C. H. Dowker’s example (VI, 8, Ex. 5). Let L be a vector space having a basis % such that X(#) > 2%. Let {u,} be a set of basis vectors in fixed 1-to-1 correspondence with the positive integers, and let

{u} C B — {u,|ne Z"*} be a subset in a fixed 1-to-1 correspondence with the set .# of all maps f of the positive integers into themselves. Since R({u;}) = 2% and X({x,}) = R,, two such subsets of # certainly exist.

For each n and f, let 1

fm) "

dn'f=——u

Clearly, each a,; # 0. Let

1

A={a,,|(n,

f(n)

+'_uf-

e Z*

x M};

since each a,, lies in the linear subspace spanned by the linearly independent u,, u,, a finite-dimensional linear subspace can contain at most finitely many members of A. Thus the intersection of 4 with each finite-dimensional linear subspace is closed in the Euclidean topology of the linear subspace, and therefore A4 is a closed set in the finite topology of L.

Sec. A

Vector Spaces

417

Consider now the nbd U = €4 of the origin. We will prove that addition is not continuous at 0 by showing that there is no nbd W of 0 such that W + W C U. Indeed, given any W, then for each u, € 4, we can determine a real number A, > 0 such that Au,e W for all '0< A< A, Definenowamape: Z* — Z* by p(n) = max[n, (1/A,)] + 1; then ¢ € #, so it corresponds to some u,, and we choose 7 such that (i) > 1/A,. It follows that both

!

u

and

o(7) "

1

o(7)

u

are in W; yet, their sum q, , is not in U. Neither is the space L locally convex: If F = .4, then V = €F is a nbd of 0 containing no convex nbd of 0. For, as we have seen, any nbd

'

.. : 1 1 W of the origin contains some —— u, and —

fm)

"

the segment containing them is not in V.

f(n)

. . u,, but the mid-point of

Though vector spaces with the finite topology need not be locally convex, they exhibit a behavior analogous to local convexity in certain instances: 4.4

Let X be either locally compact, or first countable, and let L be any vector space with the finite topology. If f: X — Liscontinuous, then each xe€ X has a nbd U(x) mapped into some finitedimensional flat.

Proof: Assume that X is first countable and that the assertion is false at xo. Let {V,, | ne Z*} be a nbd basis at x, (where V; D V, D ...); by induction we construct a sequence {x,}, with x, € V,, as follows: Choose x, € V, such that f(x;) # f(x,), and having selected x,, - - -, x,,_1, let x, € V,, be such that f(x,) is not in the linear subspace spanned by f(x1),++, f(%,-1).

Then

)

4= fs

is closed in L (in fact, it is discrete), since each finite-dimensional linear subspace contains at most finitely many points of 4. Since x, — x,, each nbd of x, has points with images not in the nbd L — A4 of f(x,), which contradicts the assumed continuity of f at x, and completes the proof. The demonstration is similar for locally compact X: Each nbd with compact closure maps into a finite dimensional linear subspace. This leads to the class of spaces described in the text (IX, 6):

4.5

Definition An affine space is of type m if for each first countable space X and every continuous f: X — L, the following is true: For

418

Appendix One

Vector Spaces; Polytopes

each x€ X and nbd W D f(x), there exists a nbd convex set C C L such that f(U) C C C W.

U D x and a

Thus all locally convex linear topological spaces (and also all vector spaces with the finite topology) are of type m.

B. 5.

Polytopes By a triangulation of a set X is meant a family 4 = {5, | « € &7} of closed geometric simplexes such that:

(1). U{o,| e} = X. (2). Each face of a ¢, € u also belongs to u. (3). For each («, B) e & x &, G, N 7,4 1s either empty or a face of both &, and ;. We do not require the dimensions of the simplexes to have a finite upper bound, and we allow the set of simplexes incident with any given one to have any finite or transfinite cardinal number. A set X together with a definite triangulation u is called a geometric complex, and is denoted by (u, X). If (u, X) is any geometric complex, the topological space consisting of the set X together with the weak topology in X, determined by the Euclidean topology on each closed 6 € u, is called a polytope and is denoted by X (u). The n-simplexes of u are called the n-cells of X (u). Asin VIII, 5.2 et seq., every polytope X (u) possesses a (homeomorphic) model that has its vertices at the unit points in a vector space with finite topology. X (u) and each of its models are homeomorphic in such a way that cells are mapped linearly onto cells. By a subdivision of a given geometric complex (#, X) is meant a geometric complex (#;, X) such that:

(a). Each closed simplex of #; is contained in a closed simplex of w. (b). Each closed simplex of u is the union of at most finitely many closed simplexes of u;. It is trivial to verify that the polytope X (u,) is then homeomorphic to the polytope X (u). The barycentric subdivision (', X) of (1, X) is defined as follows:

(i). Vertices: the barycenter p, of each open simplex o. (ii). Simplexes: (pg,, -, Ps,) is a simplex of «’ if and only if in the finite sequence oy, - - -, o,, each simplex is a proper face of its successor. A generalization is the barycentric subdivision mod Q, where Q is a subcomplex of (u, X); here, the vertices are those of the 5 € # and the

Sec. B

Polytopes

419

barycenters of all open simplexes not in Q (so that Q is not subdivided). Let X(u) be a polytope. With each subdivision (v, X) of (u, X) is associated a covering of the space X (u) by the (open) stars of the vertices of X(v). We now show that “arbitrarily fine” coverings can be obtained in this way. 5.1

Theorem (J. H. C. Whitehead) Let X (u) be a polytope. Given any open covering of X (u), there exists a subdivision (v, X) of (u, X) such that each closed vertex star of X (v) is contained in some set of the given covering.

Proof: Denote by X® the union of all closed k-cells of X () [that is, the k-skeleton of X (u)]. Assume that there is a subdivision X7?~?! of X(m~1 that satisfies the requirements of the theorem. With each vertex p’ of X711 associate a definite open set U(p’) of the given open covering such that the closed star of p’ in X! is contained in U(p’). Let o be a closed n-cell of X (u) and select an & > 0 such that (1) each subset of 6™ having diameter ~

YB

Dir.@

Lim

Dirfl

Ya,

for any direct spectrum {Y,; .} over &7. Proof: Let B® = LimDirg Y; and let gz Y;— B® be canonical map. Let ¢4: Y;— Y*® be the canonical map of the spectrum. It is evident that {p; | B€ %} is a continuous map of Z-spectrum into Y* so 1.5 gives a continuous 4: B® — Y*® such the diagram Y

)

the /the that

Ps

95 1 B®

._}_l__)

Y®

commutes for each 8 € %.

(1). & is injective. Let h(b) = h(b’); as previously remarked, we can always find representatives by, by, for b, b’, lying in some one Y, B € 4.

Since

@g(bs) = ho qubs) = h(b) = h(b') = @g(bp),

bg, b;, have a common successor in some

Y, « € &7

1.3(2)

shows

that

since & is cofinal,

bg, by, have a common successor in some Y,, y € &, so they represent the same element in B® and therefore b = b'. (2). h is surjective. We first note that

U {pu(Ye) | e et} = U {es(Yy) | B € 5}, since given any ¢,(Y,), the cofinality gives some B € #, « < B, and then Pl Y.) = 95 0 945(Yy) C ¢s(Y;). The required surjectivity now follows Y% = LaJ P Ye) = LBJ os(Yp) = %) hoqy(Yy) = h[g) 95(Yp)] = from

h(B*). (3). his an open ®z [M(U)] is openin since the bijectivity and ¢, is continuous.

map. Let U C B> be open; we are to show that Y, for each @ € /. First, this is true for each € %,

of & gives ¢; 1[A(U)] = ¢z A" [WU)] = ¢5 }(U), It now follows for each « € .7, since by choosing

Sec. 1

Direct Limits

425

B € & such that « < B, we have g 1[H(U)] = pg5 o o5 1[H(U)], and g, is continuous. The theorem therefore is proved.

This theorem frequently permits comparison of direct limit spaces when the direct spectra are over different directed sets. As an example: Let {Y,; @ashr {Zs; s} be direct spectra over &7, &, respectively. Assume there is a cofinal # C & and a relation-preserving map r of &4 into & such that r(%) is cofinal in . Then a continuous map {hB I Igeg}:

{YB; ‘Pafi}'»{zd; Y,

I g, TGT(Q)}

induces a continuous /&: Y° — Z*®, which is a homeomorphism whenever each £, is. The simple proof of this result is left for the reader. Direct limits behave poorly for subspaces and for cartesian products, - in a sense that will be made precise for each case.

1.8

(Subspaces) aeo,

Let{Y,; @,5} be a direct spectrum over .

let A, C

Y,

and

assume

that

¢,45(4,) C 4;

For each whenever

a < B. Then {A,; @, | 4,} is a direct spectrum over o, and there is a continuous injection hA: A® — Y, Proof:

Let 1,: A, —

Y, be the inclusion.

map {Ay; Pup | Ae} = {Yo; Pus}y

Since {i,} is obviously a

1.5 yields a continuous z: A — Y=,

which is injective because each i, is.

By the “poor behavior” of subspaces we mean that, in general, 4% is not homeomorphic to the subspace A(4*) C Y*. Ex.4 topology and take Ay C Y,

Let Y bethe set {0} U {x (c¢f. 111, 7, Problem 10). {Y,; @.x} to be the trivial be the subspace {0} W {x

bijective and

| [x|] > 1} of real numbers taken with the order Let 27 be the directed set of positive integers, spectrum over ./ with space Y. For each n, let | |x| > 1 + 1/n}; then h: A — Y® is in fact

satisfies #{0} = {0}; however,

{0} is open in A*®, whereas

it is not

open in Y%,

1.9

(Cartesian Products)

Let {Y,; ¢}

and {Z,; ¢,,} be direct spectra

over directed sets &7, &, respectively.

(1). Preorder o x % by(e, 0) < (B, 7)if both « < Bande < 7. With this preordering, &/ x & is a directed set. (2). For each («, 0) < (B, 7), let pp oy

be the {Y, x form a (3). There

Yo X Z;—

Y, x Z,

continuous map ¢,z x ¢,,. Then the family of spaces Z,| (o, 0) eZ x &}, together with the maps p 4 (5.0 direct spectrum. is a continuous bijection

h:LimDir , , »{Y,

x Z,} —LimDir,

Y, x Lim Diry, Z,.

426

Appendix

Two

Direct and Inverse Limits

Proof: Since (1) and (2) are trivial, we prove only (3). For (o, @), let by5y: Yy x Z;— Y™ x Z* be the continuous map ¢, The family {hq. | (¢, 0) € & x S} is readily verified to be a map spectrum into Y*® x Z%, so with the continuous 2 = Lim Dir we obtain for each («, o) the commutative diagram

each x . of the % (a,a)

Y, x Z, (Pax‘)l‘o'

Ma,0>

Lim Dir(Y,

x Z,)

—“—h"‘%’

Y®

x Z®

To show that % is bijective, it suffices (I, 6.9) to construct a map g:Y® x Z° - Lim Dir(Y, x Z,)) (g is not asserted to be continuous!) such that both Aog = 1 and goh =1. We define g as follows: Given (y, 2)e Y ® x Z %, choose any representative y,, of y, 2, of z, and set 2(y, 2) = .oV %) £ 1s uniquely defined: if the selected representatives were y;, %, then y,, y, (resp. 2, 2,;) have a common successor y, (resp. z,); since y, x 2, is clearly a successor of both Yo X Zzand y; x 2, we have [1.3(2)] that w0 Var 25) =

s 0(Ver 25)-

It is now evident that both 2o g = 1 and gok = 1, so the proof is complete. By the “poor behavior” of cartesian products we mean that in general h is not a homeomorphism. Indeed, it may fail to be 2 homeomorphism even though &/ = & = positive integers, each Y, is a topological group, and

each

Y; = Z;. This

is one

reason

that,

when

direct

spectra

of

topological groups are considered, each group is usually taken with the discrete topology. 1.10

Remark 'The notions of direct spectrum and of direct limit can be extended to include the case where more than one connecting map between pairs of spaces is allowed. This extension is as follows:

Let .o/ be a directed set, and [Y, | « € &7} a family of spaces indexed by

&Z. For each pair of elements «, 8 such that « < 83, let (¢hs | p € M(e, B)} be a given nonempty family of continuous maps ¢fs: Y, — Y;, which we call connecting maps [the cardinal of each M(«, 8) need not be finite, and may vary with (¢, 8)]. Assume that the family of all connecting maps satisfies the following two conditions:

(1). If ¢ < B