Introductory Real Analysis 0395959330

Table of contents :
Chapter 1: Proofs, Sets, and Functions
Chapter 2: The Structure of R
Chapter 3: Sequences
Chapter 4: Continuity
Chapter 5: Differentiation
Chapter 6: Riemann Integration
Chapter 7: Infinite Series
Chapter 8: Sequences and Series of Functions
Chapter 9: Power Series
Chapter 10: The Riemann-Stieltjes Integral
Chapter 11: The Topology of R

Citation preview

it,poaucioPL) topu

Peal A

PIG

L)SIS

Frank Dangeilo Michael Seyfried Shippensburg University

H o u g h t o n M i f fl i n C o m p a n y B o s t o n

N e w Yo r k

ntiroducTorL)

Peal A

T i a

L)SIS

Dedicated to Sue and Rosa

Sponsoring Kclitor; Jack Shira Assistant Editor: Carolyn Johnson Editorial Assistant; Christine E, Lee

An Supervisor/Interior Design: Gary Crespo Marketing Manager: Michael Busnaeh Senior Manufacturing Coordinator: Prieilla Bailey Cover Design: Henry Raehlin

Cover Image: O Photodisc

Copyright 00 by Houghton Mifllin Company. All rights reserved. No part of this work may be reprodueed or transmitted in any form or by any means, electronic or mechanical, including pliotttcopying and recording, or by any infoniiation

storage or retrieval system without the prior written permission of Houghton Mifllin Company unless such copying is expressly permitted by federal copyright law. Address inquiries to College Permissions, Houghton Mifflin Company, 222 Berkeley Street, Boston,MA 02116-3764. Printed in the U.S.A.

Library of Congress Catalog Card Number: 99-71719 ISBN: 0-395-95933-0

23456789-DC-03 02 01 00

c

o n Te n

Preface

Chapter

1

Ix

Proofs,

Sets,

1.1 1.2

Functions

Proofs Sets

1.3 1.4

Chapter

and

1 5

Functions Mathematical

2

The

1 10 15

Induction

Structure

of

E

19

2.1 Algebraic and Order Properties of K 19 2.2 2.3

The The

Completeness

Rational

Numbers

2.4

3 Limit

3.3

Cauchy Limits Limit

Chapter

44

4.1

and

Theorems

Limit

Inferior

Continuity Continuous

Limits

and of

Consequences Uniform Discontinuities

5

and

Sequences Functions

of Continuity Continuity Monotone

Functions

Differentiation The Mean

Va l u e

51 55

58 61 65

69

Functions

Continuity

4.3

48

Sequences I n fi n i t y

at

Superior

4

5.1 5.2

39

Sequences

Bolzano-Weierstrass

3.6 3.7

Chapter

39

Theorems

Monotone

3.5

27

31

Subsequences

3.4

4.6

23 R

Convergence

3.2

4.4 4.5

in

Sequences

3.1

4.2

Dense

Cardinality

Chapter

3.8

Axiom

Are

Derivative Theorems

69

73 77

83 86 89

95 95 101 V

vi

Contents

5.3

Ta y l o r ' s

5.4

Chapter

Theorem

L'Hopital's

6

RIemann

104

Rule

107

Integration

111

6 . 1 E x i s t e n c e o f t h e R I e m a n n I n t e g r a l 111 6.2

RIemann

Sums

11 7

6.3 Properties of the RIemann Integral 120 6.4 Families ofRiemann Integrable Functions 126 6.5

Fundamental

6.6

Theorem

Improper

Chapter

7

i n fi n i t e

of

Calculus

131

Integrals

135

Series

143

7.1 Convergence and Divergence 143 7.2 Absolute and Conditional Convergence 148

7.3 Regrouping and Rearranging Series 156 7.4 Multiplication of Series 161

Chapter 8 Sequences and Series of Functions 167 8.1

Function

8.2 8.3

Sequences

Preservation Series of

167

Theorems Functions

174 180

8.4 Welerstrass Approximation Theorem 187

Chapter

9

Power

9.1

Series

191

Convergence

9.2

Ta y l o r

191

Series

198

Chapter 10 The Riemann-Stieltjes Integral 207 10.1 10.2 70.3 70.4

Monotone Increasing Integrators Families of Integrable Functions Riemann-Stieltjes Sums Functions

of

Bounded

Va r i a t i o n

207 215 220 226

1 0 . 5 I n t e g r a t o r s o f B o u n d e d Va r i a t i o n 2 3 3

Chapter

11 7

7.7

The Open

To p o l o g y and

of

Closed

R Sets

239 239

11 . 2 N e i g h b o r h o o d s a n d A c c u m u l a t i o n P o i n t s 2 4 5 7 7.3 Compact Sets 249

Contents

11 . 4 71.5

Connected Continuous

Sets Functions

Bibliography Hints Index

and

vii

254 258

265 Answers

267 285

p PG

a c G

To t h e S t u d e n t

Is Ihc prize wimh the struggle? This book provides not only a complete foun

dation for the basic topics covered in the usual calculus sequence but also a stepping stone to higher-level niaiheinatics. This book contains some beautiful

mathematical results and opens the door for the student to see other beautiful mathematical results. This is the prize. The struggle is that the ability to see these beautiful mathematical results does not come easily. Put more succinctly, Introductory Real Analysis is a hard course.

With the possible exception of mathematical geniuses, such as Euler,

Gauss, and Rieinann, people are not born knowing how to do proofs. The skill to prove a mathematical result is learned just as the skill to solve problems is learned, usually with lots of practice. We suggest covering up our proof of a sialemcnt. After under.standing what the statement means, attempt your own proof. If you arc stuck, look at the first line of our proof and see if you can proceed. If necessary, look at another line, and so on.

The examples and exercises are essential components of this book. They

will help clarify and strengthen your insight into the text material; and they will increase your mathematical maturity, which, like the ability to do proofs, is not an innate human trait. As much as possible, we have arranged the exercises in each section to correspond to the development of the text material, rather than ordering them from easy to hard. Learning to use previous results to obtain new results is part of malliematical maturity, and we freely draw on the exercises in this manner.

Throughout the book we give overviews and commentaries in order to

provide insights into what we are doing and where we are going. These features take the form of either short paragraphs or tnore formal remarks. Our purpo.se for dividing some of the commentaries into remarks is to separate different concepts for ilie student.

In answer to our original question, those who have seen the beauty in

mathematics delinileiy iliink the prize is worth the struggle. To t h e I n s t r u c t o r

Although the only prerequisite for this book is the usual calculus sequence,

another mathematics course (such as an introduction to abstract algebra) would help develop the student's sophistication. In this book we do not do analysis

from .scratch, nor do we construct the real numbers. Rather, we build on what

the student has learned in calculus. This book is designed to be used effec

tively in one- or two-setncster courses. The first nine chapters contain an ample ix

amount of material for most two-scmestcr courses, while Chapters 10 antl 11

offer additional topics depending on the instructor's personal taste. As the table of contents indicates, our approach is sequential rather than

topological. Because the table of contents lists the topics covered in the book,

which for the most part should be done in order, we mention some aspects of the book that arc not apparent from the table of contents. Chapter I is an introductory chapter whose purpo.ses are to provide some necessary topics for proceeding in the book and to allow the students to develop some easy proofs. The only topic in Chapter 1 that mo.st students have not seen before is the arbitrary collection of sets, which helps the students develop proofs with quantifiers. In our course, we cover Chapter 1 quickly. Because we do not fix notation until Chapter 2. the instructor can begin with Chapter 2, referring back to Chapter I as needed. We have meticulously provided appropriate references throughout the book. In Section 2.4 our purpose is to show that the rational numbers are countable and that the real numbers are uncounlable. Our experience is that one could get

bogged down in this section sitice cardinality is so fascinating to .students; for example, how can there be just as many real numbers in the open interval from 0 to I as there are on the real line?

Starting in Chapter 3. we use the terminology of neighborhoods, which

provides a unifyitig aspect for both .sequential and function limits. In our classes

we usually cover about half of Section .^.8, our purpose being to understand what the limit superior and limit inferior are and what they mean in terms of sequential limits. Exercise 5 in this section is very valuable in this regard.

We point out that Exercise 11 in Section 3.7 is needed only for the proof of

Proposition 3.9 in Seelion 3.8. We use Section 3.8 in (he proof of Mortens's Theorem in Chapter 7 and in Chapter 9. In Chapter 7, our Ratio and Root tests use only limits, whereas in Chapter 9 we extend these tests using limit superior and limit inferior. Of course. Chapter 9 relics heavily on Chapters 7 and 8. Chapter 6 contains an extensive section on improper integrals with tests for convergence similar to those of inliniie series. This section may be covered in depth or lightly depending on the insirnelor's preference. Although we do not recommend this, Chapter 7 can he covered after Chapter 3 if one is willing to allow the Integral test without formally doing improper integrals or monotone functions. Our development of the Rieniaiin-Slieltjes integral in Chapter 10

parallels our development of the Riemann integral in Chapter 6. Many of the proofs in Chapter 10 for monotone integrators are only slight modifications of the corresponding proofs in Chapter 6. We think Chapter 6 should be covered before Chapter 10, although both chapters can be taught simultaneously. Also. Section 10.4 on functions of bounded variation can be covered without doing

anything else in Chapter 10; it depends only on Chapter 4. Thus, an instructor who does not reach Chapter 10 and who has only a couple of cla.ss periods left in the semester can cover Section 10.4.

Chapter 4, of course, depends on Chapters 2 and 3. Chapter 11 gives a topological view of our previous concepts, and many of the results in Chapter 4 are now special ca.ses of the action of a eontimioiis function on a compact or connected set.

Preface

xi

Accompanying the text is an Instructor's Manual, which contains com plete solutions to every exercise in the text and additional exercises (also with

complete solutions) suitable for student take-home problems or projects. The Instructor's Manual also contains some suggestions on the text material and on the level of difficulty of some of the exercises. To A l l

A book such as this naturally contains many standard {to a mathematician) proofs that can be found almost anywhere. Some arguments arc original with us in the sense that we developed them, and we have never seen them anywhere else. When we could not improve on another person's proof or development, we have used it with an appropriate reference. However, all mistakes are ours, and we would appreciate being informed of any errors detected in the book so that we can correct thcin.

We would like to acknowledge many people for their help in the prepa ration of this book. For their insightful criticisms and suggestions we thank the reviewers; Michael Berry. West Virginia Wesleyan College; David Gurney, Southeastern Louisiana University; Nathaniel F. Martin, University of Virginia; Jtthn W, Neubergcr. University of North Texas; Alec Norton, University of Texas at Austin; Richard B. Thomp.son, University of Arizona; Guoliang Yu, University of Colorado; and Marvin Zenian, Southern Illinois University at Carbondalc. We also thank our colleagues Douglas Ensley, Frederick Nordai, and William WellerofShippcnsburg University for their input; our secretary Pamela McLaughlin Ibrhcra-ssistance; and, most of all, our students for their invaluable

comments when wc did classroom testing of earlier versions of this book.

Proofsj Sets, r o o s

Fu n c T i o n s

a n c

O '.ne purpose of this chapter is to help improve the reader's ability to understarid and create proofs. We attempt to do this in Section i.l. In Sec tion 1.2, we consider sets, including arbitrary collections of sets. Sections 1.3 and 1.4 deal with functions and mathematical induction, respectively.

1.1 Proofs Vlany of the statements we prove in this section are known to the reader. It is the technique of proof we wish to emphasize. The lypicaJ mathematical assertion that requires proof is the conditional .suuement (or the implication) if

p,

then

X = lll/ll > l.|

1.2 Sets Although the first part of this section should be familiar to the reader, the

technique of proof may not be. We adopt the viewpoint of "naive" set theory considering the notion of a .set as already known.

Basic Results and Set Operations A sei is a well-defined collection of objects. By "well-defined" we mean that, given a set and an object, it is possible to tell whether the object is or is not in the set. Each object of a set is called an element of the set, a point of the .set, or a member of the set. If A is a set and x is a point, then X e A denotes that x is an element of A while

X i A denotes that a is not an element of A.

A set can be defined either by listing the elements of the set or by stating a property of its elements. For exantple.

= {x e K : -v^ - 3.V -4 = 0) where M denotes the set of real numbers.

The empty set (void set. null set), denoted by 14, is the set with no elements. Thus

0 = (A' e K : .r- < 0) = : X #a:) and so on.

Definition 1.2 Let A and B be sets.

1. A is a subset of B, denoted by A c B or S 3 A, if for each .r in A, a is in B.

2. A is equal to B. denoted by A = B, if A c B and B c A. 3- A is a proper subset of B if A c B and A ^ B.

6 Chapter 1 Proofs, Sets, and Functions Thus, to prove that two sets are equal, one must show that each is a subset of the other. Also note that since H contains no elements, it follows from the definition of a subset that 0 C A for all sets A.

D e fi n i t i o n 1 . 3 L e t A a n d B b e s e t s .

1. The union of A and B. denoted by A U B. is defined as A U B = {.r : .t € A or.r e B). The word "or" is used in the inclusive sense, so that points that belong to both A and B also belong to the union. 2. The imcrsection of A and B. denoted by A H B. is defined as A n B = (.r : J e A and .t € B |. Thus A n B c A U B.

.3. A and B are disjoini if A fl B = 0.

Proposition 1.6 Let A. B. and C be sets. Then 1. A U A = A and A n A = A;

2. A U H = A and A n 0 = .1. A C A U B and A n B C A;

4. A U B = B U A and AO li = li D A (commutative property); 5. A U (B U C) = (A U B) U C and A n (B n C) = (A n B) n C (associative pn)perty):

6. AU(BnC) = (A UB)n{A UC) and An (BUC) = (AnB)U(AnO (distributive properly);

7. A C B if and only if A U B = B and A C B if and only if A n B = A. Proof Wc prove the first equality in part 6. By the definition of equality of sets, we need to show that

A

U(BnC)

C

(AUB)n(AUC)

(2)

and that

(A

U

B)

n

(A

U

C)

C

A

U

(B

n

C).

(3)

To show (2).let.r € A U(BnC). Then either .r € A or.v e BHC. If.v e A,

then by part 3. .r € A U B and .v e A U C. By the definition of intersection, -v e (A U B)n(AUC). If.r e BOC. then.v e B and j: e C. Again, by part 3,

.r € B U A and .v € C U A. By pan 4, .v e A U B and .v e A U C. Therefore. .«• € (A U Bf-ifAUC).

To show (3). let .r e (A U S) n (A U C). Then .v e A U B and .v e A U C.

If.r € A. then by pan 3. t e A U(B nC). So we may assume .v ( A. Then, by the definition of union, .r € B and .r e C and so .t € B D C. Again, by pan 3. -r e (BnC)U A = AU(BnC).

Section

1.2

Sets

7

We now prove the first equality in part 7. Wc need to show thai

if

A

C

B.

then

/t

U

B

=

B

(4)

B,

(5)

and that

ir>t

U

B

=

B,

then

/t

c

To show (4), we assume A c B and note that we have to show 4 U B = 6. So we have to show that A U B c B and that B c A U B. The latter containment

follows from part 3. To show A U B c 6, let a: € A U 6. Then either x e A orx € B. If A- e A, since A c B, .v € B.

To show (5), assume that A U B = B. We need to show that A c B. Let A" s A. Then .r e A U 6 by part 3. Since A U B = B. x e B.

The remaining parts of the proposition are left as exercises. ■

Definition 1.4 Let A and B be sets. The complement of B ivknive to A, denoted by A \ B. is defined as

A \ B = {.r e A ; X ^ S),

For example, if IR denotes the set of real numbers and Q denotes the set of rational numbers—that is, if m

Q =

— : m and ii arc integers and /i ^ 0 ,

then R \ Q is the set of irrational numbers. Also note that Q \ R = 0, Proposition 1.7 Let A. B. and C bc.set.s. Then 1. A \ 0 = A and A \ A = 0

2. DeMorgan's Laws: A\(BnC) = (A\B)U(A\C) and

A\(BUC) = (A\B)n(A\C),

DeMorgan's Laws are generally remembered as stating thai the comple

ment of an intersection is the union of the complements and the complement of a union is the intersection of the complements. Proof We prove the first equality in part 2, leaving the rest as an exerci.se. We need to show that

A\(BnC)C(A\B)U = '2) € /, tben t'l = vi.

Notationally, we write f : X ^ Y la denote that / is a function from X into y. and we often denote the ordered pair (x. y) € / by v = fix).

Two functions / and g are equal, denoted by / = g, provided tbat tbey have the same domain and that for each domain element .v. f{x) = g(x). Definition 1.9 Let / be a function from X into T. Let S C X. The tlirect linage of S, denoted by /(5). is defined as fiS) = [fix): .1 e S\.

Let T C y. The inverse image of T. denoted by is defined as

.r'iT) = lx€X:f(x)sT]. For the following examples the reader should graph the functions to help verify the claims made about the direct and inverse images. Example 1.5 Let / : R -♦ R be defined by /(.v) = x-. 1. Let S be the set of integers. Then f(S) = (0. 1.4.9. 16....]. 2. Let T = (64). Then /"'(T) = (±8). .3. Let T = (y : -3 < y < 4). Tben = {.r ; -2 < .v < 2). Observe

that for an element in T less than 0. there are no real numbers that are mapped to that element. In other words. /"'((—3.0)) = H. Example 1.6 Let / : R ^ R be defined by f(x) = sin .v.

1. Let S = (a-: 0 < x < n/6}. Then fiS) = [y : 0 < y < ^| = |0, ij. 2. Let r = (1). Then /"'(T) = [x : /(.v) = 1) = |;r/2 + 2k7T : k is an integer).

3. Let 7- = [y : 0 < y < I). Tben /"'(T) = (2n + I).tJ. 4. Let r = (y : tt/I < y < ,t). Then f~'iT) = 0. Proposition 1.10 Let / : A- ^ T. Let S and T be subsets of Y. Then

/-'(5U7-> = /-'(S)U/-i(r).

Proof We will show that /"'(JUT) c /-'(5)U /-'(T) and/-'(5') U

f~\T) c /~'(SU T). To begin, let x e f~\SLI T). Tben, by definition of /-' (5- U T). /(.r) e 5 U T. So fix) e 5 or fi.x) € T. If f(x) e S. tben -t e /-'(S). If/(.v) e T, then.v e f'fT). Sax e /-'(5> U/"'(T). Thus, /-'(5U7-) c/-'(S)u/-i(r). To prove the reverse inclusion, let-r e /"'(5) U/"'(T). Then either

X € /"'(S) or A- e /~'(7"). If.v e f~HS). then fi.x) 6 S. If x e f~fT),

12

Chapter 1 Proofs, Sets, and Functions

ihcn /U) € T. Therefore, /(a) e S UT and so a e / '(5 U T). Hence. f-HS)U f'HT) C f-^(SuT). ■ Definition 1.10 Afunclion / from X into Y isaone-zo-wiefunclionfor

a I-I function) if for any pair of distinct points .xi and X2 in X, f(x\) and J (.V2) are distinct points in Y [that is, if.vi # A2 in X, then /(.V|) ^ /(xz) in K]. Equivalently, using the contrapositive. / is one-to-one if and only if for each A| andAi in X, if /(.V|) = /(-t:), then.ti = xz.

Another way to classify a function between two sets X and Y is to look at how much of Y is taken up by the image of X. Definition 1.11 Let / be a function from X into K. If/(X) = K. then

/ is onto Y. Equivalently. a function / from X into Y is onto Y if for each y 6 Y, there is an .r e X with /(.r) = y. A function from X into f is a bijeciioii of X onto Y if it is both one-to-one and onto Y.

Example 1.7 The function / in Example L.'i is not one-to-one since/(-8) = /(8). Since no negative number is in the range of /, / fails to be onto R. Example 1.8 Define g : R-▶ R by A -

I

if A'

>

0

~ .r-i-I if A-) = y. Thus f{a) = f{b). Since / is a one-to-one function, a = h. That is.

« + I. /)(l) is true since

2' > 2 = I + I. Let A- > I and assume that p(k) is true. That is. assume 2* > A' + I. We want to show that p(k + I) is true. We need to show that 2' *' > (A- + I) + 1 = A- + 2. Obserse that 2'*' = 2' -2 > (A + 1) • 2 (by the induction hypothesis) = 2A + 2

>A + 2 (since 2A > A).

Thus, p(k + 1) is true and so pin) is true for each ;; in N.

Example 1.15 For each ii in N. 9 divides n^ + (h + I)' + {n + 2)' (where "divide-s" means with 0 remainder).

Fttrcach/i inN. let /»(/i)bethcstatetncnt: 9dividesn^ + (n+1 )^+{/j+2)\

Since h' + 2' + .1' = .16. />(l) is true. Let k > I and assume that p{k) is trtic. That is. assume that 9 divides A ' + (A- + 1)^ + (A- + 2)^ We mu.st show that 9 divides (A 4- I)' + (A + 2)^ + (A + ,1)^. Ob.serve that

(A + 1 )■ + (A + 2)' + (A + .1)-^ = (A + 1)-'' + (A + 2)^ + A"' + 9A- + 27A + 27 = + (A + I )-^ + (A + 2)-'' + 9(A- + .1A + 3). By the intlticiion hypothesis. 9 divides A-' 4- (A + I )^ + (A + 2)'. Since 9(A- 4.3A 4- 3) is a multiple of 9.9 divides it. Thus p(k 4-1) is true and so pin] is true for all n in N.

Remark Sometimes mathematical induction is stated as follows. Let A be a

sub.set of N .satisfying 1. I € /1 and

2. if A > I and A € A. then A 4- I 6 4.

'Iljcn A = N. That this is equivalent to the previous version of mathematical induction can be seen be letting 4 = (/i e N : pOi) is truel. Remark In mathematical induction, one need not start with 1. Let iid be an integer and suppose that />0i) is a statement for each n > no. Assume that (1) p{n„) is true and (2) for each A > no. if p{k) is true, then p(k 4- 1) is true. Ttien pin) is true for all n > no- To see this, for each n in N. let q{n) be the statement: pC'ii 4- n — 1). Note that /lu. Remark The following repre.sents a misuse of mathematical induction. Find the error. 1-or which A does the argutnent fail? For each ii in N. let pin) be the statement: any set of n horses arc all of the .same color, pi I) is true since we have only one horse. Let A > 1 and assume that /»(A) is true. That is. a.ssume that any set of A horses are all of the same color. We want to show that p(A 4- 1) is true. Let X = {.vj. .vs -Vi+r) be

Chapter 1 Proofs, Sets, and Functions a sei of k + 1 horses. Wc want to show that all k + i horses arc of the same

color. Since {.i' = [»;(« + 1 )/21-. 3. Prove that for each h in N. I + .3 + 5 H 1- (2n - I) = ir. 4. Prove that for each « >4.2" l)^ Taking square roots (sec Exercises 8 and 9), !"-!-/>!


|.

■

By mathematical induction, the triangle inequality can he extended to any finite number of elements in R. That is, for any oi, «2,..., a„ in R, 1"! + "2 "I h «nl 5 l"ll + + 1- jflni'

The following simple result, especially the second pan. will be used often in Chapters 3 and 4.

Corollary 2.1 For a and h in R, 1. k-/>| < |«n-|fc| and

2, ilai-j/'ii !.

Proof For part I, ja — /j| = In -i- (—f»)| < [a| -f |— /;| = |fl| -|- |/?j. For part 2. we also use the triangle inequality: |n| = |(fl - h) + b\ | < jn — b\. Reversing the roles of a and b, we get !/>! - |a| < -«[ = la - b\. Thus.||«t-|fc|j|.

■

Section 2.2 The Completeness Axiom

23

Exercises ■■ 1. Let -v be irrational. Show that if r is rational, then -v + r is irrational. Also show that if r is a nonzero rational, then rx is irrational.

2. Show, by example, that if .v and y arc irrational, then x + y and xy may be rational.

3. Show that >/2 + n/3 is irrational. [Him: Use the fact that V6 is irrational.] 4. From the order axioms for R, show that the set of positive real numbers, {a' € R : .r > 0], is closed under addition atid multiplication. 5. From the order axioms for R. show that 0 < I. [Him: From the field ax ioms, 0 1. By the trichotomy property, either 0 < I or I < 0. Assuming 1 < 0, get 0 < — I. Now use Exercise 4.J 6. Write 0.33474747 • • • as a fraction.

7. Prove parts 1 and 2 of Proposition 2.1.

8. For a in R, show that |a| = i/^. (Note that for h >0, Vh dctioles the nonnegalive .square root of h.) 5). Lei > 0 and h > 0. Show that a < h if and only if < /r.

10. Show that equality holds in the triangle inequality if and only if > 0. 11, l-'or.r in R and e > 0. let (a - f. a -I- £) be the open interval centered at a of radius s. That is, (.r — .r -!- £) = ly € R ; A — e < V < -V + £} = |y eR: |y-.v| < e).

Use the triangle inequality to show that if « and h are in R with a ^ h, then there exist open intervals U centered at a and V centered at /;, both

of radius £ = ^ jfl - , with U r\V =,H.

2.2 The Completeness Axiom Boundedness

Definition 2.2 The extended real mimher .nxtem, denoted by R*. con sists of the real number system together with two distinct symbols, co and —CO, neither of which is a real number. That is,

R" =RU{oolU|-ool.

We extend the usual ordering on R to R* by —oo < oo, —CO < X < CO for all x in R. and H < i; if either « < i; or n = u for all ii and u in R*.

24 Chapter 2 The Structure of K Wc also use llie symbol +00 for 00 where convenient. Definition 2.3 Let S be a subset of R*. Let 11 and u be in R*.

1. /(is an upper ImiiiuI of S if s < u for all s in S. 2. I) is a iower houiitl of S if u < .5 for all s in S.

3. S is ixHuuleit above if there is a real number that is an upper bound of S. 4. S 'Khdiuuiedheiow if there is a real numberlhal is a lower bound of 5. 5. S is bounded if i" is both bounded above and bounded below.

Obviously, 00 is an upper bound and —00 is a lower bound of every subset

of R*. However, the positive real numbers {a: e R : J > 0} are bounded below but not bounded above; the negative real numbers |.r € R : .t < 0} are bounded above but not bounded below; and the closed interval (0, 1] = {.r € R : 0 < .V < It is bounded.

Definition 2.4 Let S be a subset of R* and let a and ^ be elements ofR*.

1. Of is the lecisi upper bound of S or the supreimitii of S (a) if o is an upper bound of S and

(b) whenever y is an upper bound of S,a < y. Notation; o = luh S = supS. 2. is the f-reaiexi lower bound of 5 or the infunum of S (a) if is a lower bound of S and

(b) whenever y is a lower bound of S, y < fi. Notation: fi = gib S = inf i". As a simple exercise, ihercadersboiildshow that sup i'and inf 5 arcuniqtie. Example 2.1 The following illustrate the basic ideas. 1. supR =+00 and inf R =-00. 2. supjO, 1] = suptO, I) =5 I and inf[(), 1] = itif(0. I) = 0, where (0, I) = {.V € R : 0 < .V < It. 3. sup{.v e R : .r < ;t] = and inl'la: 6 R ; .t < ;r) = —00. 4. Pathologically, supH = —00 and inf H = +00. Observe that the supremum and inlimum of a set may or may not be in that set.

Section 2.2 The Completeness Axiom

2 5

Definition 2.S Let 5 be a subset of R* and let .vo and .i\ be clenients of 5,

1. i'o is the smuUest {least, ininimiim) element of S if .V(i < x for all x in S.

2. .?( is the greatest {largest, maximum) element of S if x < Sj for all x in S.

Notation: min S = so and max S = .V|.

Proposition 2.3 Let S be a subset of R*. 1. If 5 has a smallest element, then this smallest element is the inliimim of S.

2. If S has a greatest element, then this greatest element is the suprcmiim of S. Proof We prove pan 1, leaving part 2 as an exercise. Let i'o be the smallest clement of S. Then Sd is a lower bound of 5 since

S(, < .r for all .v in S. Suppose y is a lower bound of 5. Then y < x for all x i n S . I n p a r t i c u l a r, y < - Vu . T h u s , . t o = i n f S . ■

The Completeness Axiom The next axiom is our final assumption about the set of real numbers. We will see in the next section that this axiom distinguishes the real numbers from the rational numbers. Recall that bounded above (or below) means bounded above

(or below) by a real number.

Completeness Axiom for R Every nonempty subset of R that is bounded above has a supremum in R.

Proposition 2.4 Every nonempty subset of R that is bounded below has an m/imum in R.

Proof Let 5 be a nonempty subset of R that is bounded below. Let A = ft € R : -t is a lower bound of 5].

Then A is nonempty and A is bounded above by each point in S. By the

Completeness Axiom for R. « = sup A is a real number. We wilt show that a = inf S. Let s be in S. Then s is an upper bound of A. Since or is the least upper bound of A. a < s. Thus, a is a tower bound of S. To show that o- is the greatest lower bound of S. let y be a real lower bound

of S. (If y = -cc. then clearly y < a.) We need to show that)' < u. Since y is a lower bound of S, y is an element of A. Since a is an upper bound of A, y 0, there is an i in S such that .r < s). then sup S = +oo. Also,

if S is not bounded below, then inf S = —oc. Thus, every sub,set of R* has a supremum and an infimum in S", whereas a subset of R need not have a

supremum or an infimum in R, This is why we chose to work in R".

26

Chapter 2 The Structure of R We end this section with an example showing how to work with the supremum and the infimum. For a subset 5 of R and a point « in R, we define aS = {fl.9 : J € 5).

Example 2.2 Let S be a nonempty bounded subset of R and let« > 0. Then 1. sup(flS) = «sup5 and

2. inf(fl.S) = a inf 5.

To show part I, let a = .sup S. By the Completeness Axiom for R. a is a real number. We wish to show that aa = sup(«.S). First note that for each .v in S. s < a since a is an upper hound of S. So. tis < cia for each j in 5 and. therefore, aa is an upper bound of aS.

To .show that aa is the least upper bound ol'wS, let)/ be a real upper bound of (If J' = 00, then clearly y > aa.) We need to show that < /. For each s in S, since y is an upper bound or(;.S". as < y. So, for each s in •S- -I 5 y/c and, hence, y/a is an upper hound of S. Since the lea.st upper bound of 5 is a, o < y/a. Thus, aa < y. To show part 2. let fi = in!" S. By Proposition 2.4, is a real number. We wish to show thatfl/( = inf(«.S). Ft5rcach.v in .5./J < .v since/(is a lower bound of S. So < as for each .v in S and, therefore, afi is a lower bound of aS. To show that afi is the greatest lower bound of aS. let y be a real lower

bound of aS. We need to show that y < afi. For each .v in S, since y is a lower bound of y < as. ,So, for each .v in .V, y/a < s and. hence, y/a is

a lower bound of .V. Since ^ is the greatest lower bound of S.y/a < fi. Thus, y < 2)

(b) (.t € R ; .x^ < 2) (d) |.r e R : .x- > 2j 2. Prove part 2 of Proposition 2.3. 3. Let A and B be nonempty subsets of R with c B. Show that inf ft < inf A < sup/t < sup fi.

4. Let 5 be a nonempty bounded subset of R and let b J) = ft sup 5 and sup{ft5) = ft inf 5. 5. Let She a nonempty .subset ofR that is bounded above. Prove that lnf(-5) = — sup5. where -S = (-i).9 = [-.v: s e .S|. 6. Let 5 be a subset of R and let a be an element of R. Define « + 5 = {a + .s : s 6 SI. Assume that S is a nonempty bounded subset of R. Show that sup(n + S) = a + sup S and inf (o + S) = a + inf S.

Section 2.3 The Rational Numbers Are Dense in K

27

7. Let A and H be .subsets of R. Show that

su|)(/t U H] = maxlsup/1. sup fi] and

iiit'tA U IS) = miiilinf A. inf W).

S. Show that a nonempty litiile subset of R contains both a maximum and a minimum clement, \lliiii: Use Induction.1

♦L [.c( A and IS be nonempty bounded subsets of R. let a = stipA. and let

ft = sup li. Let r = {(lb -.(isA and h € IS]. Show, by example, that aft ^ sup C in general.

2.3 The Rational Numbers Are Dense in R In this section we establish three main results; first, that the Archimedean

Prirtciple holds; second, that the rationals are "dense" In the reals; and third, that the rationals arc not "complete." We also expand our techniques for working with the suprcmum and irtlimum. The next proposition stales that we can approximate a real supremum or

intimum by an element of the set. That is, we can find an element of the set that is as close as we would like to a real supremum or infimtnn.

The Archimedean Principle Proposition 2.5 Let S be a nonempty hounded subset of R. Let a = sup S and let ft = Inf S. Let e > 0. Then 1. there exists an .vo In 5 such that a — e < .vu and

2. there exists an .vj In S such that .vi < ft + e.

Proof We prove part I, leaving pail 2 as an exercise. Figure 2.1 illu.strates the fuel that a is an upper bound of S and a is in R. Siippo.se s 0. then there exists u positive integer h such that 1 /« < A.

It follows froin the Archimedean Principle that inf{l/n ; h e Nj = 0.

The Density of Q In K

^ -^finitlon 2.6 Let A be a subset ofR. Then A is r/tvi.vc in R if between every two real numbers there exists an clement of A. Equivalenily, A isdenseinRirandonly if for each a and y inR with a < v, one has that /tn(A. y) ^ 0, Here (a, y) is the open interval (z e R : a < z < y) in R.

Theorem 2.2 Q is dense in R. Proof We consider various cases.

Case 1 Ha is negative and y is positive, then a < 0 < y and 0 is rational.

Case 2 0 < a. By the Archimedean Principle, there is an n in N such that 0 < \/ii < A, and I fit is a rational number. J. n

HH

1

0

y-A

Figure 2.2

1

1-

A

y

Case 3 0 < a < y. Refer to l-igiire 2.2. .Since y - a is positive, by the Archimedean Principle, there is an n in N with 0 < 1 /;i < y - a. llnformally, if we add i/n to itself a sufficient niimher i>f limes, .say m times, then a < infn < y, because we cannot "jump" over the interval (a. y) since 1/n < y - a.) By Exercise 2, there is an in in N such that — I < jiA < in. So, in/ii — \/n < a < in/n. Manipulating this

expression and recalling how I /ii was chosen, we get in

I

n

n

A < — < A + - < A + (V - A) = V. '

and in/n is a rational number.

Case 4 a < 0. By Ca,se 2. there is an n in N such that 0 < \/n < -a, and .so A < — I /n < 0 and — 1 /« is a rational number.

Case S X < y < 0. By Ca,se 3, there is a rational number q with —y < q < —A. Then —q is rational and a < —q < y. ■

Definition 2.7 A subset C ofRiscom/'/c/c ifovery nonempty bounded sub.set of C has both a supremum and an intimum in C. For instance, [0, I] is complete but (0. 1) is not complete. Also. R is complete by the Completeness Axiom for R.

Section 2.3 The Rational Numbers Are Dense in R

2 9

^C oroa l ry2.2Therao tinalnumbersarenotcompe l te. Proof Let A = (.v € Q : 0 < .v < -/l]. (Recall that Vl is irrational by Theorem 1.1.) Since \/2 is an upper bound of the set A. sup A < v/2. Suppose sup A < v/2. Since Q is dense in R, there is a rational number such that sup A < q < s/2. Thus ^ is in A and q > sup A, which is a contradiction. So. sup A = s/2. Thus. A is a nonempty bounded subset of Q that does not have a supremum in Q. Therefore. Q is not cotnplele. ■

More on Suprema and Infima We end this section by expanding our techniques for working witli lite supremum and the inlimum. We give two different arguments for the more diliicult part of the next proof because both are elegant. For subsets A and B ofR, we define

A + B = \inAanda/;i) in /f suchthata—£/2 < ao'dndp—s/2 < fco.Thus.a+/^—f < tio+bo < y since y is an upper bound of A + 6. Hence, a + p < y +£. Sitice e is an arbitrary positive number, a + P ■ N by fOi.m) — 2"3"'. (Any two distinct primes will work.) We claim that / is one-to-one and hence N x N ~ a .subset of N and therefore M x N is countable. To see that / is one-to-one. suppose that

/(,?.,„) = /(r..y).Then2''3"' = 2'"3'. Ifn > r, then l"" = 3'-"'. Since 2 divides the left side, 2 divides 3'""" atid so 2 divides 3, which is a contradiction.

The case » < r is handled similarly. Thus, n — r and hence in = .v. ■

Note that if A is a countably infinite set, then there is a bijection from N onto A. Also, if A is finite atid nonempty, there is a function from N onto A,

because if A = {.ri.Jz .r„}, we can define a function / from N onto A by

|.r,, if k>ii. So. if A is countable and nonempty, there is a function from N onto A. Theorem 2.4 The countable union of countable sets is countable. That is,

if / is a countable index set and A^ is a countable set for each a In /, then

Utt€/ i** countable. Proof We can assume that / 7^ H since A^ = 0. and we can assume that each A„ is nonempty since empty A„'s add nothing to the union. Since / is countable, there exists a function / from N onto /. Since each Aa is countable, for each a in I there exists a function from N onto Aq. Define

/I : N X N ^ y A„ a e l

by bin. Ill) =

We claim that b maps N x N onto Aa and hence, by Lemma 2.1 and

Proposition 2.12. Aa is countable. Let a: be in Aa. Then there

exists an ao f such that a' is in A„y. Since maps N onto Aa„, there exists an m in N such that goo("') = Since / maps N onto I. there is an n in N with

/(/i)

=

ao-

Then,

h(n.

m)

=

=

a.

■

Corollary 2.4 The sets Z, Q, and Q x Q are countable. Proof We write each .set as the countable union of countable .sets.

Z= {...,-3.-2,-1) UfOjU (1.2.3, ...|.

Q = U^=|{"'/" :€ Z).

Q

X

Q

=

U,eQl('7-

P)

-

P

^

Ql- ■

36 Chapter 2 The Structure of R ir we define Q" = {(q\,qz qn) '■ qi € Q for /' = 1,2..... we can

show that Qi" is countable by induclionandby writingQ" = U^tQW"' ^ I'/l'Theorem 2.5 The closed interval [0, I ] is uncountable.

Proof The technique used here is called the Cantor diagonalization argument. Suppose that |0, 1] is countable. Then there exists a bisection / from N onto |(), 1]. We list the elements of [0, 1) in their decimal expansion as an infinite matrix:

/(I) = .«ii«i2nu--/(2) = .«2ia22«:.r • • •

/(3) = .fl3lfl32fl33

f(n) = .a„ia„2a„i---

where each is a digit from 0 to 9.

We now construct an element of the interval [0, 1 ] that is not in the range of / by "going down" the diagonal of the matrix. For each n in N, let h„ =

3 if a„„ ^ 3 4 if a„„ = 3.

Then x = .616263 - • • is in (0. 1], but for all n in N, x /(h) since 6„ 56 a„„. (The choice of 3 and 4 for6„ is done simply to avoid duplicate representations, which occur only with tails of nines or zeros.) ■

Corollary 2.5 The real numbers and the irrational numbers are uncountable. Proof That R is uncountable follows from part 4 of Proposition 2.11. Since R = Q U (R \ Q). R \ Q is uncountable by Theorem 2.4. ■ We make one final comment. The following axiom is independent of the other axioms of set theory.

Axiom of Choice If / is a nonempty set and if /)„ is a nonempty set for each « in /, we may choose an.r„ in Aa for each cr in /.

If / is countable, then wccan use mathematical induction to choose each j:„. Tlte strength of the Axiom of Choice comes when I is uncountable. The reason we mention the Axiom of Choice is because wc lused the Axiom of Choice

in the proofs of Propositions 2.9 and 2.12. We could have given alternative proofs using the fact that N is well-ordered, but we felt that this was a needless complication.

\

Seclion 2.4 Cardinality 37 xepcises

1. Compleiethe proof of Proposition 2.8. 2. Let / be a one-to-one function from A into B with B finite. Show that /\ is finite. J. If A and B are finite sets, show that .4 U B is a finite set. Conclude that the finite union of finite sets is finite.

4. If X is an infinite set and.v is in X. show that X ~ X \ |.vl. 5. Define explicitly a bijeclion frotn 10, l]onto(0. 1). 6. Complete the proof of Proposition 2.11. 7. Let / be a one-to-one function from A into B with B countable. Prove that A is countable.

8. l-orni in N. show that N — N \ (1.2 m). y. If A and B arc countable sets, show that A x B is countable. II). Let A be an uncountable set and let B be a countable subset of A, Show

that /4 \ is uncountable.

11. Lel?l beacollection of pairwise disjoint open intervals. Thai is, members of21 are open intervals in Rand any two distinct members of 21 aredisjoint. Show that 21 is countable.

12. Let (5 bo the set of all open intervals with rational endpoinls. Show that 0 is countabic.

13. Let A be the set of all sequences whose terms are the digits 0 and I. Show that A is uncountable.

14. The purpose of this exercise is to show that given any set, there is always a larger set with respect to cardinality. For a set X, the power set of X. denoted by P(X). is the collection of all sub.sets of X. Recall from Exercise 9 in Section 1.4 that a set with ii elements has 2" subsets fortj in N U {O}. The map.r -♦ {.r} is a one-to-one function from X into V{X). Show that there does not exist a function from X onto ViX). [Hint: Suppose / is a function from X onto "PfX). Let 4 = {.r 6 X : .r /(.v)|. Then A is in ■PfX) and-SO A = /(.to) for some jto in X. Ask yourself "where is.vy?" to obtain a contradiction.]

15. Let A be the set of all real-valued functions on [0. 1 ]. Show that there does not exist a function from [0, I ] onto A.

Sequences Thsi si anm i pora tntchape trbecausewewtliusesequencesh trough out the text. The main theorem is the Bolzono-Weierstrass Theorem for

sequences in Section 3.5. Other very important theorems are the Mono tone Convergence Theorem (Section 3.4), Theorem 3.5 (Section 3.2), and Theorem 3.12 (Section 3.6),

3.1 Convergence Limit of a Sequence Recall that N = 11. 2. 3 1 and lhat K is the set of real numbers.

Definition 3.1 A seqticnce of real numbers (or a .sequence in R) is a fiinclion whose domain is N and whose range is a subset of R. Thus, a sequence in R is a I'unclioti from N into R. Notation When wc use llie word "sequence"" without qualification, we mean a sequence in R. l-or a sequence it is customarj' to use a letter such as a' for the function and to denote the value x(n) as.v„ for each ii in N. Thus, we think of a .sequence as.v ; N -♦ R or as (ai. .rs, .1-3 ) ora.s or as Each .v,i is a term of the sequence. Following Banle and Sherbert. we distinguish between the sequence (.v„),|gf), whose terms have an order induced by N. and the range {a„ : n 6 N) of the sequence, which is not ordered. For example, the

sequence ('~l)"' = (1. — I. 1. — 1 ) has range { — 1, 1). Some books use the notations »"d (.v„ : /i € N) interchangeably, whereas we will reserve the latter notation for the range of the sequence,

.Sequences tnay be given explicitly, as in (I/h)^, = (I, 3, ^,...). or recursively, as in the riboncwci .sequence: let .V| = .V: = I let .*•„ = .r„.| + .v„_2 for n > 3.

Definition 3.2 A sequence eveniuaiiy has a certain property if there exists an no N such that

-V«g + I. A/iQ+i. • - •) has this property.

39

Chapter 3 Sequences For example, a constant sequence is a sequence whose range consists of a single number, whereas the sequence (I. 2.3.4,4.4 ) is eventually con stant.

Terminology In Definition 3.2. (.v„)„>„,| is a fu//of the sequence This tail is also a sequence. To see this, define a bijeciion / from N onto (/lo. /!() -F 1. "u 4- 2... .| by fin) = iio + n - ] for each « in N. Then .r o / is a sequence and .v„ = (.v o /)(/i + 1 — /jo) for n > na. A useful and important concept is that of a neighhorliooci of a point. Definition 33 For .v in R and £ > 0. the open interval (.V - f, .V + £) = {y 6 R : jy - .vj < e] centered at .v oi' radius £ is a ncighhoiiiooil qf.x. Definition 3.4 Let be a .sequence in R and let x be in R. The sequence (.r„ )„6)sj converges to .v (or has limit .v) if for every neighborhood U of .V the sequence is eventually in U.

Notation and Terminology If the sequence (.Vn)„jN converges to .v, we write lim x„ = .v or lim x,, = .r or .v or simply .r„ x. and we call 11 ^ " ^

II

n

(.v,i a convergent sequence. A sequence that is not convergent is divergent.

Faraphrasing Definitions 3.2 through 3.4, we obtain the following propo sition.

Proposition 3.1 Let he a sequence in R and let x be in R. Then (.v„)„gN converges to x if and only if for all e > 0 there exists an /lo in N such that if M > 11(1. then (.v„ - .v| < e.

Proof Combining Definitions 3.2 through 3.4. we have that x„ —» .t 0, (.v„)„eN eventually in (a' — e, .v -I- s) for all e > 0. eventually, j.tn — .r] < e. ■ In general, ;i(] depends on e. Figure 3.1 graphically depicts Definition 3.4 and Proposition 3.1. Usually, the smaller s becomes, the larger /lu must he in order for the distance from .v„ to .v to be less than e for ail n > in,.

Example 3.1 Let he the constant sequence c—thai is, .v,, = c for all n ill N. Then. .v„ ^ c. To see this, let s > 0. Let hq =17. (Here n[) docs not

depend on £. Any other no would also work.) Let n > iio. Then |.v„ - c| = jc - (■[ = 0 < £.

Example 3.2 lim (l//i) = (). Let e > 0. We want to detertnine a value of n - » c c

no as specified in Proposition 3.1. Right now we do not know how tochoo.se nn, and so we proceed as if we know no with the hope that the calculations below

/

Section 3.1 Convergence

.r

U

+

41

€

X

, t - €

-i-H123

"(I

Figure 3.1 will indicate how to choose /i(|. Let n > /i|j. Tlien -

-0

I =

-

1
0. Let no = ? Pretend that you know ho- For n > /iq. 2m-I-3

3 ( 2 M - P. 3 ) - 2 ( 3 M - f 5 ) 3(3n-|-5)

3m-t-5 i

9m 4- 15 1

9/;

9mi)

Now we know how to choose mq. I3y the Archimedean Principle, choose mq in N such that I /mq < 9£. Then, for n > /to,

2m

+

.3m

4-5

3

2 3

I

I

9n„

9

^

< ;-(9e) = c.

An important property of the limit of a sequertce is uniqueness. Lemma 3.1 Distinct points in R can he separated by disjoint neighborhoods. That is. if x and y arc in R with .v y. then there exist neighborhoods U of x and V of y such that f/ O V = M,

Proof Let.randy beinR witiiA- 56 y. Lei£ = ^|.r-y|.Lctf/ = (.t-E. x4-£) and V = (y — s, y 4- e)- We claim that U HV = H. Suppo.se that z is in f/ n V. Then, by the triangle inequality, \x - y| < !a- - cI 4- U - yl < e 4- £ = 2e = j.r - yj. This is a contradiction, becau.se a real number cannot be less than itself. There

fore, U and V are disjoint neighborhoods of x and y. (The reader should c o m p a r e t h i s w i t h E x e r c i s e 11 i n S e c t i o n 2 . 1 . ) ■

42 Chapter 3 Sequences Theorem 3.1 Limits of sequences are unique. Proof Let (-tnlngfj be a sequence in R. Suppose thai .r and v are in R with x„ —• ,v and .r„ —» _v. We wish to show that x = y. Suppose that .t ^ y. By Lemma 3.1. let U and V be disjoint neighborhiMxis of x and y, respectively. Since .v,, ^ .v. there is an tii in N such that .v„ is in U for all ii > it]. Since .v,i -» y. there is an ^2 in N such that .r,, is in V for alt ii > hi. Let m > max{H|. H2). Then .r„i is in fy fl V = H, which is a contradiction. Therefore. .V = y. ■ Remark The proof above illustrates a technique commonly used with se quences. Namely, in order to have two conditions tx-'cuning simultaneously, we need to go out far enough in the sequence to guarantee that each condition holds. As in the above proof we accomplish this by taking a maximunt.

Limits Do Not Always Exist The fact that limits do not always exist should come as no surprise to the reader. The sequence (.Vn),,g[u in R does not converge if and only if lim .t„ ji.vforall n

—

?

C

.V in R.

Let (.v,i)nes be a sequence in R and let x be in R. Using the basic rule for negation as given on page 3, we negate both sides of the "if and only iF" in Definition 3.4. Using~Definition 3.2, we have that lim Xn # x there exists a neighborhood U of.r such that the sequence (XnlnsN is not eventually in U there exists a neighborhood U of.t such that for all hq in N, there exists an II > III) such that .v„ is not in U. The last phrase is sometimes expressed as "the sequence fn'iiiienlly outside of U."

Example 3.4 The sequences (0. 1.0. 1.0. 1....). (1. — I. 1. —1. 1. —1....). and (1.2, 1,3, 1, 4,.,.) do not converge. For instance, the first sequence is fre

quently outside the neighborhood (—5. 5) of 0 and is frequently outside the neighborhood (5. j) of 1. Let be a sequence in R and let .v be in R. Negating the statements in Proposition 3. t, we have that lim .v„ x there exists an r > 0 such that for all /to in N, there exists an 11 > iin with i.r„ — .tj > f 4=i> 3 e > 0 such

that Vh() € N. 3 h > III) with l.v,, - .ij > s. The reader should apply this to the .sequences in Example 3.4.

Other Techniques for Convergence We end this .section with two examples that utilize different techniques. In the first example we use Bernoulli's inequality, given in Exercise 10 in Section 1.4, to derive an important result.

Example 3.5 LctO < r < I. Then lim r" = O.Toseethis,let j = (1/r) - 1. Then .t > 0 and r = I/( 1 -)- .r). By Benutulli's inequality, (I + .v)" > I -i- iix /I

—

OC

for all II in N. Therefore,

0 0, choose iio in N such lhal I /iio < xs. Then n > iin implies that 0 < r" < — < — < -(.vf) = e. iix

/Jn-V

.V

and so r" -» 0.

In the next example, we use results concerning the exponential function

t'" and the natural logarithm function In.v from calculus. This example also requires L'Hopital's nile. Example 3.6 lim [ I + ( I/h)|" = c. Write

^1+1^_ Then

litn

ll+(l/ii))"=

lim n Inj (4>( j/n)\

lim

c'l-'*'

«-»2c

{•)

Since

l + Inn /I , In I , +. - ) l u = lhm \ u / "—X

(l/n)| \/„

ut/tli\)\\ +(1/»)|

= lim y ^^ ^ (by L'Hopital) (III \ n J —

lim

K—X I + (l/n)

= 1.

lim [I +(l/n)r =e' = e. n

—

C

K

The reason for the equality at (*) is because e' is continuous on K (see Chapter 4).

Remark A good exercise for the reader is to do Example 3.5 by the technique used in Example 3.6, Thisteclmiqneinvolvcsasequenccwho.se limit is negative infinity, which we consider in Seclion 3.7.

x c p t i s e s

1. Use Proposition 3.1 to establish the following limits,

( a )n l i—m x— rI I - = ' I0- . 9 (Cd ) 2 / l1i-m + " 5= -2 3// (hi

fc)

>1

lim -oo

•, = II

3/! hm

-I-

+ =

2

+

3

? -

5

2

sin/I le) "

3 . (t)

'I—»

•>;

lim

. Itm

/I-

-2 I. I . L e1 t1x/,", i=f "( - i sD ooeds d h rm. , ..t -, , 0

il

«

IS

even.

ii-«.

=0

n

-n +

—-=0 I

extsi?

44 Chapter 3 Sequences ^

3.

.

( 1 / / =1 { i ,t D n IS odd „ , ocs lim .v„ cxisl.' I 1 if /I IS even. n—5C

l.et.Vn

4. Let (-v„)„gf( be a sequence in R and Ici .r be in S . (a) Show (hai if .t,, —» x. ihen |.r„| —* |.r|. lW;n/; Use Corollary 2.1.) (b) Show ihat if [.v,, | -» 0, ihen —» 0.

(e) Show, by example, thai (la^nDneN '"''y converge and (.tn)n€N converge.

5. Let .i„ > 0 for each n in N; let x be in R wiih .t„ -»• x. Note that x > 0.

Show that y/^, -Jx. {Hint: Make two cu-ses: .t = 0 and j: > 0. In the latter case, rationalize.)

6. Show thai a convergent sequence in R is bounded. That is. if (.*«),,5^ con verges, show that there is a B > 0 such that |.i„| < B for all n in N. [Hini; Use Proposition 3.1 with 4-= l.j 7. Show that the sequences = (1,4,9,...) and (—= ( — I, — 2, — 3,...) do not converge. 8. Show that if jr) < I. then lim r" =0.

9. Establish the following limits. / (a)

e"

lim

—=0

DC

JT"

(d)

lim

n-»Dc

1

(l-f—)""=e" /I

2

lb) lim f"" = 1 (c > 0) (c) lim ^ = 0 «—3c

ji;

I

;

( e ); i l i m

/t""

—»

='I 1

( 0— al: i m

—/ i " = 0

3.2 Limit Theorems In this section, we obtain many of the standard results concerning convergent sequences. Since a sequence is a function, the following definition is a special case of Definition 2.8.

Definition 3.5 A sequence (.v,, )„5f< of real numbers is Iwimdetl if there exists a S > 0 such that l.v„| < B for all >1 in N.

Proposition 3.2 A convergent sequence is bounded. Proof Let (-r„)„sri be a sequence in R and let .r be in R with .x„ .r. By Proposition 3.1 with £ = 1. there exists an i)n in N such that |.v„ — .t| < 1 for all " > lit)- By Corollary 2.1, U',,) — j.ti < |.r„ —.vj < 1, which implies that

|.v„l < 1 -I- j.rj for ail 11 > ii„. Let « = max{|.ril, |.r>| I-X/iq-'I • • + Then

S

>

0

and

|.y„|


0 and an iio in N such that |.v„| > f for all

II > nil- In other words, the ser|uence is eventually bounded away from 0. Proof Let e = l-rl/2. By Erxercise 4 in Section 3.1. |.v„| —* |.v|. and .so the sequence (|.v,il)„eH is eventually in the neighborhood (|.r| — p. |.r| + p) of |.r|. Hence, there is an iia in N such that if ii > then

I I l -- * '£l = |.v| - Y = — = f. ■ |.v|

I|.r„| I .> I

Theorem 3.2 Let (-r,i)n5« (.^''I'nsrt sequences in R; let .v and y be in R with x„ —• X and >•„ —» y. Then 1. lini (.v,i + v„) =.v + V = lim .i„ + lim v„: 2. lim .v„v„ = .vv= lim .v,, • lim v,,; 3. lim c.r„ = c.v = o lim .v„ for all real numbers /to implies that K-V/i + .v„) - (.V + y)| = |(.v„ - .1) + (y„ - y)| < |.v„ - .vj + |y,i - yi
0. [Noic: We wish to make each pan less than £/2. |y|. being a single number, is not a problem. However. !.v„|. not being a constant, is a slight problem.) First, choose n\ in N such that if/i > /ij. then |.v„ —.rj < £/|2(|yl +1)]. (iyj could be 0, .st) we added the I.) By Proposition 3.2, is a bounded sequence, and .so there is a I) > 0 such that i.v„| < B for all ti in N. Now choose 112 in N such that if n > 112. then |y„ — yj < e/2B. Let

46 Chapter 3 Sequences «o = maxj/7i. Hi). Then ii > hq implies that - .rv| < B |v„ - v| + |v| |.\r„ - x\
•„ | > lyl /2 for all H > H|. Hence.

. 2|y„-y|

lyf

y -

for all H > Hi.

Let e > 0. Choo.se 112 in N such that if n > 112. then |y„ - y| < s |y|- /2. If Ho = maxlHi, Hi), then 11 > ho implies that

y'n

■|yl= 2

By mathematical induction, parts I and 2 of Theorem 3.2 can be extended to a finite number of convergent sequences. In particular, if converges to .r and if k is in N, then

lim (.v*) = lim (.v„ ■ .t,, .t„) = (lim a:,,)* = x'^. n—•

00

-

n—

k

Example 3.7 We redo Example 3.3 using Theorem 3.2; hm

2h + 3 2 + (3//i) 2 + 0 2 =

lim

—

_

-.

'1-50 3H +5 ''-OO3 + (5/H) 3 + 0 3

Example 3.8 Let />(/) = hd implies that |.r„

—

x|


72 from the left.

48 Chapter 3 Sequences

Exercises 1. Find the limits of the following sequences,

V"+2A,„

vv

\ " + -^ /„€N

V2/i-

+

5/„5fj

V

^-i6N

2. Give examples of two sequences that do not converge but whose (a) sum converges.

(b) product converges. (c) quotient converges.

i. Let (x-„)„jN and (v„)„gN ^ sequences in R such that {.t„ + >'„)„eN and (-rn - .Vn)««s both converge. Show that (.t„)„€N and (y„)„gN both converge. 4. Find a convergent sequence in (0, 1) that docs not converge to a point in (0. l|. Note Theorem 3.3.

5. Use the Squeeze Theorem to show that lim (cos/i)//i = 0.

6. Let (.v„)„gs be a bounded sequence in R (not necessarily convergent), and let (yn)„gfj be a sequence in R withy„ -▶ 0. Show that (x„>'„)„g[ij converges loO.

7. In reference to Exercise 6. give an example of sequences (.Vn);i€N and ()'n)„gN where

(a) is bounded and (y„)„gn converges, but (.t„y„)„gN docs not con verge.

(b) y„ —» 0 but (.v„\'„)„gpj does not converge. 8. Prove part 2 of Theorem 3.5.

3.3 Subsequences Often, (he easiest way to show that a sequence does not have a limit is to use subsequences. The main result of this section is Theorem 3.6. Definition 3.6 A function h from N into N is Mricily increasing if, whenever «i and ii arc in N with in < it, then li(i") < hOi).

Definition 3.7 Let (-v„)„gN and (>',i)„gN be sequences in R. Then is a of (x„)neN 'f 'bere is a strictly increasing func tion h from N into N such that y„ = .r(,(n) for all n in N.

Section 3.3 Subsequences 49

Eiiuivulciuly. y = .t o h. ur the diagram /'

T

/

N c o m m u t e s .

Notation Given the sci|uence = (,V|. .vj. .tj....). think of the sub

sequence (y,i)„€N follows: yi = .v/.(i, where/i(l) > I. vi = .v/,(2i where h{2) > li{ I). y, = .v„(3i where /i(3) > /i(2), etc. Thus, a subsequence has the same ordering as the original sequence. It is not Just a subset of the original sequence. We usually write h{k) as so that the subsequence (yn)„eN where ;i| < jm < /ij < • • •. Note that m > k foreaeh k in F!.

A little thought should indicate that a subsequence of a sub.sequencc of a .sequence is a subsequence of the original sequence. Example 3.11 A sequence is a subsequence of it.self (let h be the identity map on N).

Example 3.12 The sequence (0. 1.0, 1.0. I ) has as subsequences the constant sequence (0.0, 0....) and the constant sequence (I. 1. I....). It also has many other subsequences.

Example 3.13 The sequence (I. 3.2.4. 5.6....) is not a subse(|uence of (1, 2. 3.4.5.6....) because the ordering is not the same. Theorem 3.6 Let be a sequence in R that converges to .v. Then every subsequence of (.r„ )„gr< converges to .r.

Proof Let (.Vni)^i f sub.sequencc of (.v,i),iepj. We want to show that .r or.equivalenily. that lim x„, = .v. We use Delinition 3.4. Lett/bea t

l

—

oo

neighborhood of .v. Since .i„—» x. there is an no in N such that if n > iio. then ' I

.v,i is in U. Let k > iin, Since lu > k > ii„, x„^ is in U. Thus the sub.sequencc (-*■"1 li^i eventually in ly. and so converges lo.r. ■ Remark In order to keep straight whether it is the sequence or the .subsequence that is converging, we suggest subscripting the arrow with the appropriate letter, as we did in the proof of Theorem 3.6. Example 3.14 From Theorem 3.6. it follows that if a sequence has two sub sequences each with a dilTcrenl limit, then the sequence itself has no limit. This is an easy method for verifying Example 3.4. For instance, the se quence (0. 1.0. 1.0. I....) has a subsequence with limit 0 and a sub.sequencc with limit 1. Thus. (0. 1.0. 1.0. 1....) cannot converge. Also, the .sequence (1.2. 1.3, 1.4,1.5....) has an unbounded subsequence and .so this .sequence cannot convetge.

50 Chapter 3 Sequences The following chanicicrizatioii of nonconvergencc in (crms of subsequences will be used throughout the text. For now, the reader should apply this to the sequences in Example 3.14.

Proposition 3.3 Let a sequence in R and let x be in R. Then liin .v„ 5^ J if and only if there exist an £ > 0 and a subsequence (.t,„ )5^. of

n-»9C

'•

'

(.r„)„£ri such that .t„, —.r| >£ for all A'in N. [The last part can be restated as: there exists a subsequence of (.CnlngN 'hat is bounded away froin.t.]

Proof Suppose that lim .t„ ^ x. From the discussion following Example 3.4. there is an f > 0 such that for all wo in N, there exists an n > /to with |,f„ — .t| > £. Let/!() = l.Then there exists an ni > /lo = I with tr;,

Lelting ho = »ji + 1. there exists an «> > hi + I with rill = + I. there exists an > "2 + I with .r„3 - .V

-

X

• X

>

£.

> £, Letting

> £. Continuing, we

obtain a sub,sequence (.v„j )^, of (.Vn),,^^ with j.r„^ — .v > e for all A in M. To show the reverse implication, suppose that (.c„),i6ri hfs.subsequence that is bounded away from x. Then the subsetiuence cannot converge to x. (Why?) So. by Theorem 3.6. (.r«),igN cannot converge lo x. ■

LXGPCISGS

I. (a) Give an example of an unbounded .sequence with a convergent subsequenee.

(b) Give an example of an unbounded sequence without a convergent sub\equcnce.

(e) Can you give an example of a bounded sequence that does not have a convergent subsequence? ' 2. Find the limit of (.v„),^n where

(a) .r„ = 1 +

(b) j:,, = ! + :;—}• \Hini: Recall Example 3.6.) ' 3. Show that (sin H)„^«i doe.s not converge. [Hiiu: Find a subsequence each of

who.se terms is in [ ^. I j and another subsequence each of whose terms is in 4. Let and (.Vn)„6N''^ two sequences in R. Let (:n)n6N be the sequence (.V|. yi. j:2- .V2.-r.t. .Vj. • •■)- Show that (:«)n5N bas a limit in R if and only if both (-tnlneH (.V/i)n€N have the same limit in R. 5. Let (.t„)„gN be an unbounded sequence in R. (a) If (-Vn)neN i'* unbounded above, show that (JTnJnsN has a .subsequence

(.v„j )^| with x„^ > A for all A in N.

(b) If is unbounded below, show that {A„)„6fj has a subsequence

(.v„j)^i with .r,^ < —A for all A in N.

Section 3.4 Monotone Sequences 51

3.4 Monotone Sequences In this section we prove three imponani theorems: the Monotone Convergence Theorem, the Nested iniervals Theorem, and the Monotone Subsequence The orem. The first of these theorems allows us to show that a certain type of sequence converges without knowing (or guessing) the limit.

The Monotone Convergence Theorem Definition 3.8 A sequence in R is inonofone imreasi/i^ (re spectively. siriclly iiiciviixiiig) if.v,, < (respectively. .v„ < .t„+i) for

all n in N. A sequence in R is monotone decreasing (respec tively. siriclly decreasing) if a'„ > .v,i+i (respectively, > .v„.j) for all n in N. A sequence is nuinoione (or monoionic) if it is either tiionotone increasing or monotone decreasing. For example, (l/M)„6t-i is strictly decreasing, («)„gri is strictly increasing, and a constant sequence is both monotone increasing and tnonotone decreasing,

whereas (( - l)">„gN is not monotone and (0. I, 2,2. j, 5. g. - • •) is eventually monotone decreasing. Note that a monotone increasing sequence is bounded below by .V|. and a monotone decreasing sequence is bounded above by .V|. Recall that a convergent sequence must be bounded (Proposition 3.2) but that a bounded sequence need not converge. Theorem 3.7 (Monotone Convergence Theorem) A bounded monotone se quence converges.

» Proof First, suppose that (.t:„),i6ri is abounded monotone increasing sequence. By the Coniplctenc.ss Axiom for R (Section 2.2), a = sup{.t„ : n e N| is in

R. We claim that converges to a. Let e > 0. By Proposition 2..5, there is an »!) in N such that or — £ < x„^y Let n > iiq. Since (.tn)n€N is monotone increasing, a — r < .v,,,, < .v„ < a. Hence. (.v„ — oj < £ for all n > n,), and so . x „ — o r.

Simihirly. if (.v,i)„gN is a bounded monotone decreasing sequence, then

.r„ —• inf(.v,i ; n e N), which is a real number by Proposition 2.4. The details of

this

proof

are

asked

for

in

Exercise

5.

■

Prior to this .section, to show that a sequence converges, our strategy has been basically to "gue.s.s" the limit and then prove that it is the limit, which is

what wedid in the proof of Theorem 3.7. Now. ifwe can show that a sequence is bounded and monotone, we know that the .sequence converges, without knowing the limit. Of course, we know that the limit is a sup or an inf. but calculating this sup or inf may or may not be easy. Example 3.15 Lei.ri =2andx-„+i = + .!« for all n in N. The first three

terms are .ri = 2. .vt = s/S, and .tj = Vb -I- s/8. So it appears that (j:„)„eN is strictly increasing. We show this by induction. Clearly, .vi < .ti- Assutne that

52 Chapter 3 Sequences Xk


k> »io). Then for 11 > iiq.

U-. - .v| < l.r„ - x„^ I + e

£

^2 + 2=^Therefore.

. r „ — » . v.

■

n

Remark If each .r„ is in Q and (.v,i)„sn is Cauchy, then need not converge to a point of Q. This is another way to say that Q is not complete. For exatnple. let (.v„)„eN be a sequence in Q with .v,, ^ n. Then is a

Cauchy sequence in Q that has no limit in Q. Remark In Definition 3.11, we consider the nth and «iih temis of the se

quence. It is not sufficient to consider only the ;ith and (« + I )st terms. For

example, let x„ = ^ for each n in N. Then (-r„),ieN is not hounded and hence not Cauchy. Howex er,

U„+i -x„\ -y/ii-p=4 F "= tV— + I -i- yj? 2y/n " So, given e > 0, there exists an no in N with — .r„| < £ for all 11 > hqAs with monotone .sequences, showing that a sequence is Cauchy provides us with another way of showing that a sequence converges without knowing the limit. Given (.rn)n€N» 'f we can show that (.r„)neM is Cauchy. then we know that (.tolnsN converges. We may or may not be able to (ind the limit. Example 3,23 Let be a sequence in K, Let 0 < r < I and suppose that |.Vn+i — x„\ < r" for all n in M. Then (.Vn)„eN converges. We show that (.r,,),,, :, is Cauchy. First suppose that in > 11. Then

l-v,„ - .r,il < !.r„, - .v„,_i| -1- + ■■■ + |.r„+i - x„\'' = r"(l + r +/•- -I- ...-i-f"'-'-'') 1

-

r"-"

0 (by Example .t..S). \ — r n

Let e > 0. Then there is an iin in N such that r"/(l — r) < t for all ti > ii„. So, if II > 11(1 and 111 > /to. then |.v„ — x,„\ < £\ hence (Jr„),n,ri is Cauchy. In the example above, we could have proceeded as follows:

r"(l -Hr < r"(I r +/-" -I-r' -f • • •) — r" (sum of u geometric series). 1 — r

Chapter 3 Sequences

E

x e r c i s e s

1. (a) Find a Cauchy sequence in (0, I) Ihat does not converge to a point of (0, I).

(b> Sltow that a Cauchy sequence in 10, I] must converge ton point of fO, 1], (See Theorem 3.3.) 2. For each it in N. lei

.

'I

1^1 km\

la) Show that (.v„)„6n is not Cauchy. llOShowthat lim |.v„., i --r,il = 0. I I — * 0 0

j.ei x„ be in Z( the integers) for each u in N. Show thai ilCv,,),,,:^ is Cauchy, ilien (j:„),,5fj is eveniually constant. Conclude that a setiuenec inZcortverges if and only if it is eventually con.stant. 4. Let a < h. Let .vj = «, .vo = b. and •rii+2

-tn+l +x,i ^ = torn >

, I.

Follow these steps to .show that (.r„),igN is Cauchy. (a) Draw a picture and let L — b - a.

(b) Use induction to show that |.v„+i - j:„| = L/l"~^ for each n. (c) Proceed as in Example 3.23 to show for m > ii that

-->^',1 < ^"7 "■ (d) Conclude that (. 0 and let .t,1+1 = 1/(2+.t„) for n > i. (a) Use Exercise 6 to .show that (.r„)„gN is Cauchy. (b) Find lim .t„.

Section 3.7 Limits at infinity 61

3.7 Limits at Infinity Recall Dofiniiion 2.2 for the ordering on R* = R U (±og). In this section wc extend the notion of the limit of a sequence to points in R*, thus allow ing a sequence to have limit ±oo. However, the term "convergent" is still reserved for those sequences whose limits are real numbers. In this section we ex.iniine which theorems of Sections .3.1 through .3.4 have analogous results in R*.

Basic Results

Definition 3.12 Let or and p be in R. The open ray (a. oc) = {.V € R : X > ff) is a iieifthhorliood of oo. while the open ray (-00. /I) = (x e R : X < is a Iieifthhorliood of—oc.

The following definition extends Definition .3.4 to R*. Definition 3.13 Let be a sequence in R and letx be in R*. Then has liinii x. denoted by lint x„ = x or limx„ = x or x„ x n

as n -* 00. if for every neighborhood U of x, the sequence (xn)„gfj is eventually in U. Note that if (-v„)„gN has limit ±oo. then is a divergent sequence. Paraphrasing Definitions 3.12 and 3.13, for •' sequence in R, we have that

.T,i —» 00 if and only if for all a > 0, there exists an no in N such that if n > iio, then .r„ > or; and

.r„ -» -00 if and only if for all p 0. Choose ii„ = ? Let n

—

o

c

II > U)). Por n > 3. II' — 3ij + 2 > n' — .3/t = n{n — 3) > 'i — 3 > "o — 3.

So choose 111) in N such that no > or + 3. Then ir — 3n + 2 > no — 3 > a. Example 3.25 lim (—n) = —oo. Let p < 0. Choose uq in N such that r t — ' O C

no > —p. Let n > no- Then —n < —no < p.

62

Chapter 3 Sequences Theorem 3.13 Limits of sequences are unique. Proof If.r is in R, then U = (-v - l..r + I) and V = (.v + l.oc) are

disjoint neighborhoods of x and oo while U and U' = (—oo. x — !) are dis joint neighborhoods of .r and —cc. Clearly, (—oo.O) and (0, co) are disjoint

neighborhoixis of —oo and oo. Thus, distinct points in R" can be separated by disjoint neighborhoiKis. The remainder of the proof is completely similar to the proof of Theorem .1.1. ■ When the limits arc ±oo. an exact analogue of Theorem 3.2 cannot be obtained since indeterminate forms can arise: for exatnple, oo - oo, 0 • oo, or oc/oc. However, we have the following. Theorem 3.14 Let (.v„)n€N, (v'nlneN. nnd (c,i)„6n be sequences in R. Let j:

be in R* and supptjse that x„ -* x. y„ oo, and —* —oc. 1. If —oc < .V < 00. then .t„ -f y„ ^ co. 2. If -00 < -t < 00. then .t„ -I- c„ ^ -oc.

3. If 0 < -v < 00, then .t,y„ —» oo and x„Zn —* —oo. 4. If —00 < .r < 0. then .t„y„ —» —oc and .v,,;,, -» oc. 5. If a: is in R. then — 0 and — ^ 0. yn

Proof Note that the conditions in the different parts of the theorem do not allow the limits to be indeterminate forms such as co — oo. 0 - oo, etc.

For part 3. first suppose that 0 < x < co. Then, as in Lemma 3.2, there is an Hj in N such that .r„ > x/2 for all n > n\. Let a > 0. Since y,, —♦ oc, there exists an /12 in N with y„ > 2afx for all 11 >112. Let »Jo = max|/i|, /12I. Then n > no implies that .«:„y„ > (.T/2)y„ > (.t/2)(2a/.r) = a. Thus, .r„y„ -» 00. Let ^1 < 0. Since Zn —* -co. there exists an tij in N with Zn < 2fi/x for all n > tiy. Then, since c„ < 0./i > max(«i. »i3l implies that .v„Jrt < {x/2)Zn < lx/2)(2pi.x) = p. Thus..v„c„ -oc. Next, let .*• s= 00. Tlien there is an A'l in N such that .r„ > I for all« > A'l.

Since eventually y„ > 0 and z„ < 0. we have that eventually x„y„ > and x„z„ < Zn. It follows that .t„y,i —♦ 00 and x„Zn -* -CO. For part 5, because Vn —♦ oc and Zt, -* —00, we assume that no y„ or Zn is zero. Since .r is in R, by Proposition 3.2, there is a li > 0 such that |a'„| < H for all II in N. Let f > 0. Since y„ —♦ 00. choose no ia N such that y,, > B/e for all II > /!(). Tlien n > tio implies that |.v„/y,i| < B/y,, < ll{e/B) = s. Since c„ -▶ -00. choose /i| in N such that < -(B/t:) for all ti > 'i|. Then n > ji| implies that |.v„/r„| < fl/|o„| < Bie/B) =f. The rest of the proof is left as an exercise. ■

Subsequences We first obtain the analogue of Theorem 3.6.

Theorem 3.15 Let be a sequence in R; let.v be in R* withx,, x. Then every subsequence of (-Vn)n6N bas limit x. Proof In the proof of Theorem 3.6, replace Definition 3.4 with Defini tion

3.13.

■

Section 3.7 Limits at Infinity

63

To obtain the analogue of Proposition 3.3. we lirst negate the statements preceding Example 3.24. For a sequence in IR, lim .r„ cc there exisl.s an a > 0 such that for all ;io in N. there exists an n > hq with .v„ < a 3 a > 0 such that Ve N. 3 » > no

with

.r„


no with .r„ > fi 3 < 0 such that V no € N. 3 n > no with .v„ > fi.

Proposition 3.6 Let be a sequence in R. Then 1. lint x„ ^ k; Hand only if (.r„)„gK hasasubsequence that isboundedabove; 2. lim .v„ 5^ -cc if and only if (.v,i)„eN has a subsequence that is bounded n - * 3 C

b e l o w.

Proof We prove part 1. leaving the proof of part 2 to the reader. Suppose that lim ^ oo. Then there exists an or >0 suclt that for all n - ^ c c

n,i in N. there exists an n > no with x„ < a.

Lei no = I. Then there exists an n| > no = 1 with a',,, < «. Leiiingno = m + I, there exists an nj > iii + I withjr,,, < u. Continuing,

we obtain a subsequence (-t/nltii «)f (.tdlBsN w'itit .r,,^ < a for all k in N.

For the reverse implication, if (.VnInert '' subsequence that is

bounded above, say by a, then (.Vn^)^| is never in (a. co) and henee (-trultli cannot have limit oc. So. by Tlicorem 3.15. (.r„)„gN cannot have limit oo. ■

As an example, the sequenee (1.2. 1.3. 1,4, 1,5....) has no limit in R" but has the constant sequence (I. I. 1. 1....) as a subsequence. Also, the

sequenee (I. —2.3. —4.5. —6....) has no limit in R* but has the subsequence (—2. —4. —6....). which is bounded above, and the subsequence (1,3.5....), which is bounded below.

Monotone Sequences 'Die following theorem is the analogue of Theorem 3.7. Theorem 3.16 If (x»)„gN " monotone sequenee in R. then (-v„)„£fj has a limit in R*.

Proof [f (.Vn)„^.'4 is bounded, then (-v„)„gN has a limit in R by Theorem 3.7. Suppose (An)ne.'tmonotone increasing but unbounded. Then is unbounded above. We claim that has limit oo. which is also supi.r,, : II e N). Let or > 0. Since (-Vn)ngN unbounded above, there is an «y in N

such that x„Q > a. Since is monotone increasing, a < a,,,, < x„ for all II > iif,. Hence, .v„ —▶ oc.

Similarly, if is monotone decreasing but unbounded, then x„ —♦ infj.r„

:

H

e

N1

=

-oo.

■

64 Chapter 3 Sequences The reader should observe that the sequence in Exercise 4 in Section 3.4 has limit oo.

Remark If (-r„)„gN 'S a sequence in K, then (.tn)„eN a monotone subse quence by Theorem 3.9. By Theorem 3.16. this monotone subsequence has a

limit in R*; and by Theorem 3.7, this limit is in R if (-tn)neN's bounded. These observations arc deduced at the beginning of the next section from a different point of view.

. x e p c i s e s

1. Use the method ofExampies 3.24 and 3.25 to establish the following limits,

(a)

lim

V"

=

1''"

("

—

6Vn)

=

oo

%/»--!- 1

(b) lim —— = 00 (c) lim - ^/i — 7 = —oo n

—

o

c

(c) lim (/I-- 6h + 1) = 00 (0 lim (-rj H-sinn) =-oc n - * 3 C

2. Let (a:,,),igfj be a sequence in R. (a) Show that x„ co if and only if —.r„ —* —oo. (b) If .v„ > 0 for all /i in N, show that lim x„ = 0 if and only if lim — = oc.

fi—'ac

n—•OC

.Vfl

3. Let be a sequence in R. (a) If .t„ < 0 for all n in N, show that .v„ -* —oo if and only if |a„ | —* oo. (b) Show, by example, that if |a,i| —* oo, then (.•„ -» 0. 6. Let and (y,,),ieN be sequences of positive real numbers and suppose thai —* L where ()',i)h6N is bounded, then ^ 0. 8. Let (A„),|gN ^ sequence in R. Show that (a) if (.v,i),|gpj is unbounded above, then (-Vn)«€N has a subsequence with limit 00:

(b) if (.Vnlnjfj is unbounded below, then (-ti,)n€N has a subsequence with limit —oc. [Hini: See Exercise 5 in Section 3.3.]

Section 3.8 Limit Superior and Limit Inferior

65

9. Lcl .r„ > 0 lor each h in N. Show thai if does noi have limit oo,

then (.v,i)„g'j has a convergent .subsequence. 10. Let .4 be a nonempty subset of R with t/ = sup/A and ^ = inf /t. Show that A contains a monotone increasing sequence with limit a and a monotone decreasing sequence with limit p. [Hiiii: By Exerei.se 9 in Section 3.4. you need only consider the ca.ses« = oc and p — —oc.|

11. l"orthe purpose of this exercise, wcallowoiir.sequence to have values in R'; that is. ±00 are permissible terms of the sequence. The term "monotone"

is defined as in Delinition 3.8, using the ordering on R*. Let /\ be a nonempty subset of R* with a = sup A and p = inf A. Show that A contains a monotone increasing sequence with limit a and that A contains a monotone decreasing sequence with limit p. [Note: If a = —oo, then A = {—oo). so that the constant sequence with each term —oc has litnit ff.]

3.8 Limit Superior and Limit Inferior In this section we utilize re.sults from Sections 3..S and 3.7. First ob.scrve that if

.sequence in R, then (.Vnlnerv u subsequence that has a litnit in R* = R U [±ocl. If (.v„)„gs is bounded, then (-v,i)„eM has a subsequence that converges to a real mitnberby the Bolzano-Weiersirass Theorem for sequences, whereas if is unbounded, then has a sub.sequence with limit ±oo by Exercise 8 in Section 3.7.

Definition 3.74 Let (.x„)„grj be a sequence in R. Let

I: = |,Y € R* : ■' 0 and a subsequence

of (.v,i)„€Pj such that j.r„^ -.r| > £ forall A" in N. By the first paragraph of this .section, the sequence (-rnt)ii| will have a subsequence (y„)„gN that has a limit y in R*. Then y ^ x since |y„ - .r| > e for all n in N, and y is in £ since (y„),i€N is a subsequence of If X = 00, then, by Proposition 3.6, (.rn)„6N has a subsequence that is bounded above. This subsequence is either unbounded below or bounded below. Hence this subsequence has a subsequence with limit-co (Exercise 8 in Section 3.7) or with limit a real number (Theorem 3.10). In either case, £ ^ {00). T h e c a s e . v = — 0 0 i s h a n d l e d s i m i l a r l y. ■

It is intuitively clear that lim sup.v„ is the largest sub.sequcntial limit of (-r„)n€N- mid that lim inf .r„ is the smallest subscquential limit of (-t„)„6N- F"r the sake of clarity, wc will postpone justifying this statement until Proposition 3.9 at the end of this section; for now we will use this as a given fact.

Remark The advantage of the limit .superior and limit inferior is that, given a sequence (.v„)„jn in K. lim,sup.v,, and lim inf .r„ always exist in R', whereas

(-^n)«6N miiy or may not have a limit in R*. and in the latter ca.se it makes no sense to write lim .t„. A new way to .show that (-v„)„eN has a limit is to show that lim sup.v,, = lim inl'.v,,. This is illustrated in Exerci.se 5. For the purpose of the next proposition and some of the exercises, we

extend addition to R" by; .V + 00 = oe + -v = 00 for — 00 < -v < oc and

.V + (-00) = (-CC) + -v = -00 for - oc < X < 00

Section 3.8 Limit Superior and limit Inferior

6 7

and we do not define —c 0. Since / is continuous at c. corresponding to this value of e there is a 5 > 0 such that if x is in D and jx - cj < 3, then |/(x) - /(c )| < £, and so /(c) - e < /(x) < /(c) + e. If /(c) > 0. then

fix) > /(c) - £ = /(c)/2. Similarly, if /(c) < 0. then fix) < /(c)/2. Hence. |/(x)| > |/(c)|/2 = e for all x \r\ U r\ D where U = (c - 3. c + 3).

The hypothesis of Lemma 4.1 implies that / is bounded away from 0 on U C\ D for some neighborhood U of c, In particular, if /(c) > 0, then / is positive on U r\ D, and if /(c) < 0, then / is negative on d/ n D. Figure 4.2 piclorially describes the situation in Leimna 4.1 for /(c) < 0. Letting V = jf(c)) be our neighborhood of /(c), by Definition 4.1, there must be a neighborhood U of c such that /(x) is in V for all x in d/ n D. In

particular, fix) < 5/(c) for all x in (7 Pi D. Proposition 4,2 Let f : D —* R and; D —» R be functions. Let c be an element of D and as.sume that / and g are continuous at c. Then the functions

72

Chapter 4 Continuity f + f;. f - K- / • K- d'o'" iiny ^ i." E) are continuous at c. Also, if ^ 0, then I'/a is continuous at c. Proof Wc lirsl show that / + ^* is continuous at c. Let f > 0. Since / is continuous at c. there is a 5| >0 such that if .v is in D and |.t — rj 0. By Lemma 4.1, there exist a Sj > Oanda li > Osuch that if .t is in I) and |.v - oj < S|, then |«(.r)| > B. Also, by the continuity of at c, there is a Si > 0 such that if .t is in W and [.t — f| 0 such that if .v is in B and |.v - /( 0 such that if .r i.s in A and |.v — ci < S, then

|/(.v) - /(t-)| < Sj. Thus, if.v is in A and |.t — c) < 5. then

Ks o /)(-v) - o/)(f)| = Ia'(/(y)) - ^?(/(c))| < e. and so .i; o / is continuous

at

c.

■

/

Section 4.2 Continuity and Sequences 73

-XGPCises

1. Prove that /(j) = U| is continuous on R.

2. Show that fix) = I /x is continuous at any c 56 0. [Him: Choose your 3 so that you stay away from ().| 3. Show that

fi x )

X sin - if .t 56 0

=

0

X

if

.r

^

=

0

is continuous on R.

4. Let / : i) -• R be continuous at r in D. Let h be in R with /(c) < li [respccliveiy, fic) > h]. Show that there is a neighborhood U o( c such that if.t is in t/ n D, then fix) < h [respectively, fix) > /i],

5. Let / : Z) -» R be a function. Assume that / is continuous at a point c in D. Show that there is a neighborhood U of c such that / is bounded on U D D .

6. Let / : D —» R be a continuous function whose range is a subset of the nonnegative real numbers. Prove that g(.r) = y/fix) is continuous on D. 7. Show that any polynomial is continuous on R. 8. Recall that a rational function is the quotient of two polynomials. Show that any rational function is continuous except at a finite set of real numbers (those where the denominator is 0).

9. Finish the proof of Proposition 4.2.

10. Let f.g : D —* R be continuous functions. Define/i(.r) = max{/(j:), g(.v)l andifa:) = min{/(.v). /?(.v)} for.t in D. Prove that both h and A" are

continuous functions. Observe that for any real numbers a and /?.

max(«. b\ = i(« + /> + [« - h\) and min{«, /;) = i(« + b - \a — 6)).]

4.2 Continuity and Sequences We first establish a.sequential criterion for continuity in Theorem 4.1. Wc then use this criterion to examine continuity or discontinuity of various functions.

Theorem 4.1 Let / : D -» R and let c be in D. Then / is continuous at f if and only if whenever (.r„)„6N is a sequence in D that converges to c, then (/(a-„))„6n converges to fic).

Proof Suppose that / is continuous at c and lot (x-„),i6n be a sequence in D with x„ c. We want to show that /(.r„) ^ /(c). Argument 1 Let £ > 0. Since / is continuous at l\ there is a 0 such

that if J is in D and j.r — cj < 5. then \fix) - fic)\ < e. Since x„ -» c,

74 Chapter 4 Continuity ihcrc is iin /lo in N such that if n > iin, then |a:„ — ci < 5. Therefore,

n > II,) implies that |/(.v,i) - /(f)l < e and so f(x„) -* /(c). Argument 2 Let V be a neighborhood of /(c). Since / is continuous at c. there is a neighborhood U of c such that if .t is in (/ n D. then f(x) is in V. Since .Vn c, (.v„)„gN is eventually in U and hence (/(.x,i))neN is eventually in V. Thus. /(.i„) -> /(c).

To show the reverse implication, suppose that / is not continuous at c. We will construct a sec]uence (.v,in D with (.tn)n«N converging to c but (/(•r/.)) 0 such thai for all 5 > 0. there is an .v in D with |.v - ci < i but |/(.v) - /(c)| > f. So for each II in N. using 5 = l//i, there is an x„ in D with l.v„ - c| < i//i but l/(Vn) — /('')! > e. Therefore. (a-„)„6n is a sequence in D. .v„ -»■ c. but (/(•t' 0 and a sequence (.r„),ieN in D with a',, -> c but l/(-v„) — /((■)! > £ for all n in N. Each .r„ is either rational or irrational.

Therefore, cither 4 = {« e N : .v„ € Q) or « = e N : x„ e R \ Q| is inlinite.

If A is infinite, then there exists a subsequence (-tK^.)^, of (A„)„6r( with each .v,n in Q (see the Remark below), By Theorem 3.6. Since

/(-Vnj) - /(c)| ^ £ for all k in N. (/(-rni))^i does not convetge to /(c).

which is a contradiction to supposition i. If W is infinite, then there exists a

76 Chapter 4 Continuity

subsequence of (A„)„grj consisting entirely of irrational numbers which gives a contradiction

to

supposition

2.

■

Remark For A infinite, we constmct the subsequence as follows. Let ui be the smallest positive integer such that is in Q. Let 112 be the smallest

positive integer greater than M| such that .i„, is in Q. Continue by induc tion.

Example 4.8 Define / : (0, 00) —♦ R by 0 if .r is irrational I

/(-V) =

III

- if .t is rational and x = — n

II

in lowest terms with in and 11 in M.

We show that / is continuous at every irrational number and discontinuous at every rational number in (0. oc). Let (■ be a rational number in (0. co). Then /(c) > 0. Let (.v,i)„eN be a

sequence of irrational numbers in (0. 00) with .v„ —♦ c. Since f(x„) = 0 for each II in N, /(.r„) —» 0 ^ /(c). Therefore, / is not continuous at c. Let c be an irrational number in (0.00). We use Lemma 4.2 to show that

/ is continuous at c. If (.v„)„6n is a sequence of irrational numbers in (0, 00) with .v„ - 0

/(.V) =

-1

if

-v


;(a) does Example 4.12 Let : . \ (0) -» R with i'(.r) = I/a. Then lim i - * 0 not exist.

Example 4.13 Let fix) = sin(l/A) for a ^ 0. Then lim fix) does not i - » 0

exist. This follows from Proposition 4.6 as follows. If = I/nn for each ii in N, ihen.Vn 0 and f(x„) = 0 0; but if y,, = l/[(7r/2) + 2/jn-| for each II in N. then .v„ —• 0 and /(>'„) = I —• 1.

Example 4.14 Let f(x) = Asin(l/A) for a 56 0. Then liin fix) = 0. To x

^

O

see this, let e > 0. Let 5 = t. Let a be in R with 0 < j.t - Oj < S. Then l/(.t)-()| =

A

SIM

-

< |A| < 5 = £.

A

Remark Let /J C R and let / ; /.) ^ R be continuous on D. Suppose that c is an accumulation point of D and c is not in /). The question is whether or not we can continuously extend / to c. That is, does there exist a continuous function F : DU (c) -» R .such that Fix) = f(x) for all a in /J? Such a continuous extension F of /' exists if and only if lim Fix) = F(c) if and only X — * C

if lim fix) = F{c) (since a 5^ c). Thus, / can be continuously extended to I-.,' •

D U (c) if and only if lim /(.v) is a real number, X

^

V

rrom Examples 4.13 and 4,14 it follows that sin(l/.v) cannot be contin uously extended to 0, but Asin(l/.v) can be continuously extended to 0 by detining Fix) =

A sin- if A 7^ 0 A

0

if

A

=

U.2-0.15- -

0.

Section 4.3 Limits of Functions

81

Given D C a and f : D R. we now consider the limit of /(.t) as X approaches oo or —oc. The only difference between this and what we have done previously is that mw there must be points in D "close to" co or -oo. If D is unbounded above (respectively, below), then every neighborhood of

CO (respectively, -co) contains points of D and hence D contains a sequence of points with limit oc (respectively, -oo). Tliis follows becau.se if D is un bounded above (respectively, below), then there exists an .r,, in D with .v„ > ii (respectively. .r„ < -/i) for each n in N. The point we make is that if D is

bttunded above (respectively, below), then it makes no sense to approach co (respectively. —oo) through points of l). Definition 4.4 Let I) be an unbounded subset of R. let f : D -* R.

and let L be in R*. If I) is unbounded above (respectively, below), then the limit of f at oc (respectively, — oc) is L if for every neighborhood V of L there is a neighborluxid U of oc (respectively, -oo) such that if .r is in C n D. then /(.t) is in V.

Notation Thelimitof / at oo is L is denoted by Mm f(x) = Lor Mm f = .r—-CC '

L or /(.v) —• L as .r —• 00. The limit of / at —oo is L is denoted by Mm /(.t) = L or lim f = Lot /(.v) L as .r —» —oc. .1

—-ac

J—-5c

The reader should .see the similarity between Delinitions 4.3 and 4.4. For the sake of completeness, we rewrite parts of Definition 4.4 using the definition of a neighborhood. The reader should rewrite the other parts. For L in R. lim /(.r) = Z, if and onlv if for all e > 0 there is an a > 0 A

-•

OC

such that if.v is in Dand.r > a. then |/(.v) - Lj < e. Here, F is (Z. -e. L -fc) and U is (cr. oo). See Figure 4.4. where lim fix) = 0 is illustrated. «-•

OC

Also, Mm /(.r) = oc if and only if for all a > 0 there is a a. Here, V is (a, 00) and U is (-oo.p).

We point out that Proposition 4.6, Corollary 4.1. and Proposition 4.7 have corresponding analogues for limits at ±oo. For example, lim fix) — L if and only if lim fix„) = L for all sct|ucncc.s ix„)„eti in D with x„ -» cc. / ) - » %

Also, limits are unique, and the limit of a sum. product, and quotient is the sum, product, and quotient of the limits if everything is defined. .Since the proofs ofTer nothing new. we omit them. Example 4.15 The reader is probably familiar with the statement that a polynomial behaves a.s its leading temi behaves for |.v| large. To see this, let fix)

=

(i„x"

-I

h«i.r

-Fflu.

where » is in N, each o, is real, and o„ 0. E-or x 0,

Since

lim

t—±c 0 such that i/(A)| > £ for all a in (t/ \ {(•[) n D.

*>. Let /';/)—» R with f (a) > 0 for all a in D. Show that Mm fix) = oo if and only if lim 1//(a) = 0.

Section 4.4 Consequences of Continuity 83 10. Lot / : iR -» K such that f{x-+ y) = fix) + f(y) for all x aiKl y in R. Such a function is called udduive.

(a) Show thai /(«) — n/(l) for all integers n. (h) Show that /(a ) = a:/( I) for all .r in Q. (c) Show that if / is contintious at stmie r in R. then / is cotuiiuioiis on

R. U.se the second Remark on page 78.| (d) Show that if / is continuous on R. then /(.r) = xf{ I) for all x in R. \llini: Use scc|uencos to pass from the rationals to the irraiionals.l

4.4 Consequences of Continuity In this section we derive some important properties of functions that arc con tinuous on an interval, usually a closed interval. These properties will be used often throughout the text. Proposition 4.8 If/: \a.h] -» Risconlinuouson[«. f>J. then / isboundcd on \a.h\.

Proof Suppose that / is not hounded on |«,/»|. Then for eacli n in N there is an element .r„ in [«. b] with l/Lr,,)! > n. The sequence is bounded since each x„ is in [a.b\. By Theorem 3.10 (Bolzano-Weierslrass Theorem for sequences), htts a convergent subsequence, say

Lei ^lim x„^ = x. Since each is in |(i. /;|, .t is in [a.h] by Theorem 3.3 By the continuity of / on \a. ft], f{x„^) -» /(.v). Proposition 3.2 then im plies that the sequence (/))t€fr's bounded. This is a contradiction since

/f-Ln )j > ^ for each k in N. I lcnce, / must be bounded on \a, b\. ■ By constructing examples, the reader can show that Proposition 4.8 need not hold if \ci.h\ is replaced by an open or half-open interval, or if / is not continuous.

Definition 4.5 Let / : D ^ R. Then / has an absolute imiximiim (respectively, tihsoliiie iiiiitiimiiii) on I) if ihea* is an element .V| in D (respectively, .vj in D) such that fix) < /(xi) (respectively, fix) > /(.V2)| for all .V in D. In general, a function may not have an ab.soliitc maximum or an absohiie minimum on its domain. For extimplc. consider fix) = x detincdon (0. I). • Theorem 4.2 If / : [a.h] R is continuous on (a./'|. then / has an absolute maximum and an absolute minimum on |u. h].

Proof We must show that there is an element M in [a. such that f{x) < /(iW) forall -v in \a,b]. Similarly, we must show that there is an element m in

84 Chapter 4 Continuity \ii.h] such ihai /(.t) > f(m) for all .v in |«. />j. We prove the second pari and leave the first part as an exercise. By Proposition 4.8, / is bounded on Ui.h]. By Proposition 2.4, a = inf{/(.r) : .v € |ri. />)! exists in K. We want to find an m in [ 0. However, if / is also continuous on ||. then Theorem 4.2 implies that a is in the range of /, and so or > 0. Thus. / is bounded away from 0 on [«. />]. Theorem 4.3 (Inlerinediate Value Tlieoreni) Lei / ; [«. ft] R he contin uous on [(I, ft). Assume that /(a) ^ /(ft). Then, for any k between fUi) and /(ft), there is a c in |i. The following very important result is needed in Chapter 6. Theorem 4.4 A continuous function on a closed interval is uniformly con tinuous there.

Proof Assume that / : /;| ^ R is not uniformly continuous on [a, b\. Then there is an c > 0 such that for all 5 > 0 there exist .v and y in [«. h\ with |.r — yl For each ;j in N. using 3 = l//i. there exist .x„ and y„ in [«. h\ with |.t„ - y„| < l//i but |/(.r„) - /(y„)| > e. Since each .\n is in la, fj]. by the Bolzano-Weierstrass Theorem for sequences (Theorem 3.10) and Theorem 3.3, there exist a sub.sequence of (a:,i),,6N 'inJ tt" x

in [a. /;] with .r. Also, the subsequence (y„^,)^, of (y„),ispi converges to

.r for

lim v„ = lint [x„.+(y„, --W)

k-"x

*

il-»

'

*

*

= A- -I- 0 = .V.

Since l/C-Xn^) - / f for all k in N. the two sequences and (/(v„^))igN"nn«"hoth converge to /(.v). By Theorem 4.1,/is not continuous at

. V.

■

We now consider uniform continuity and sequences. A continuous function need not take a Cauchy sequence to a Cauchy .sequence. For example, let fix) = I/.r for .V > 0. Then (l/»i)n€N is Cauchy, but (/(l/n))neN = (")neN is unbounded and hence not Cauchy. Theorem 4.5 Let / ; /J —» R be uniformly contiittious on D. If (.tn)„gN is

a Cauchy sequence in D. then (/(.v„))„«h is a Cauchy sequence in R. Proof Let £ > 0. By Dclinilion 3.11, wc want an ii(, in N .such that if/i > «(> and III > lie. then |/(.r„) - /(.v„,)| < f. Since / is unifonnly continuous on D. there is a 3 > 0 such that if .v and y are in D with ft - yj < 5. then |/(-t) - /(y)| < e. Since (.t„)„gpi is Cauchy. there is an ;»o in N such that if" > "a and in > no. then |.v„ - .r,„| < 3. Thus, for II > "0 and in > no. l/(-t„) - /(x„)| < s. ■

Example 4.20 Let / : (l.oo) ^ R be defined by /(.r) = l/(.v - 1). If .t„ = 1 + l/« for each n in N, then -v„ -♦ I. but /(.t„) = n for each n. Hence. (/(xn))neN is not Cauchy. By Theorem 4.5, / is not uniformly continuous on (I. oo). IThis sequence approach always works when the function has a vertical asymptote. The reader should redo Example 4.18 using this approach.]

88 Chapter 4 Continuity Remark The result of Theorem 4.5 may hold even if / is not uniformly continuous. By Example 4.19, /(.v) = is not uniformly continuous on R. However, lei Ia„)„jn he a Cauchy sequence in R. By Theorem .112, there is an -t fn R such that a,, -» x. Since / is continuous. /(.v„) ^ /(-v). Thus, (/(-l,)),icN is Cauchy. Theorem 4.6 Let Z? be a bounded sub.set of R and Icl f : D ^ R. Assume thai whenever (.v,i)„gH is a Cauchy .sequence in D. (/(.v„))„5n is a Cauchy sequence. Then / is uniformly continuous on I).

Proof Suppose / is not uniformly continuous on D. From the proofof Theo rem 4,4 (with In, h] replaced by D), there exist .sequences and (y,, . both in D. and an e > 0 such that |j:„ — y„| < \/ti and |/(.rn) — > e for each n in 1^.

Since D is bounded, by Theorem 3.10, there is a subsequence (-V/n)^! of (-r/ilneN 'iiid an Z, in R such that f-- Note that L may or may not be in D.

Also, as in the proof of Theorem 4.4, y,n -* L. Consider the sequence = (a,,,. y„,.x„2, .v«,. .r^j. >•„, ). Then z„—* L (Exercise 4 in Section 3.3) and hence (Zn)n€?i is Cauchy. For each k n

in N.

l/(:2t-i) - fizn)\ = |/(A„^) - /(.v„^)l > £. and

so

(/(Cn))n6N

is

not

C a u c h y.

■

Remark We give another proof of Theorem 4.4. Let / : [n.Z»] -♦ R be continuous on [a, b]. Let (AjJ^eh be a Cauchy sequence in («. Zj]. Then there is an X in !«. b\ with .t„ ^ x. Since / is continuous. f{x„) /(a). Thus. (/(A. For example./(.r) = l/.ron (0, I) cannot be continuously extended to 0. In the following theorem, wc sec that uniform continuity is sufficient to guarantee a continuous extension. Theorem 4.7 If/is uniformly continuous on (a. b). then /has a continuous extension to [a. fcj.

Proof We show that / can be continuously extended to a, leaving the con tinuous extension to h as an exorcise. By the Remark on page 80, we must show thai lim /(.v) is a real number. Let (.v„)„en be u sequence in (a, b) wiili .v„ —» a. By Theorem 4.3, (/(-C;i))n6N is a Cauchy sequence and so there is an Z. in R such that lim /(a,,) = Z.. We claim that lim /fA) = Z.. Let (y,,),,^^ It—*00

*

—). Since Zn -* « by Exercise 4 in Section 3.3, (/(z„))n€N converges to some M in R. Since (/(A„))„gN und (/(y„))„6N are subsequences of (/(z„))/.ePi. /(a,i) —» M and /(y„) —»• M. Since limits are unique, L = M = L\. ■

Section 4.6 Discontinuities and Monotone Functions

8 9

. x e p c i s e s

1. Show (hat /(.v) = .r' is not unifonnly continuous on R. 2. Show that /(x)= l/(.r - 2) is not uniformly continuous on (2, cc). 3. Show that /"(.t) = sind/.r) is not uniformly continuous on (0, ;r/2]. 4. Show that /(.r) = in.x + 6 is uniformly continuous on R.

5. Show that /(x) = l/.t^ is uniformly continuous on ll.oo), but not on (0. I].

6. Show that if / is unifonnly continuous on [«, b\ and uniformly continuous on D (where D is either [/>. c] or [h. do)), then / is uniformly continuous on [«. b] U D.

7. Show that fix) = ■^xh uniformly continuous on [1. co). Use Exercise 6 to conclude that / is uniformly continuous on [0. oo). 8. Show that if D is bounded and / is uniformly continuous on D. then / is bounded on D.

9. Let / and be uniformly continuous on D. Show (hat / +}{is uniformly continuous on D. Show, by example, that /}• need not be uniformly con tinuous on D.

10. Complete the prtmf of Theorem 4.7.

11. Give an example of a continuous function on Q that cannot be continuously extended to R.

12. Let / : Q -» R be unifonnly continuous on Q. Show that / has a unique continuous extension to R. [Him: Define ^ : R -♦ R by fi x ) f:ix) =

if.rsQ

Urn fix„) ifx e R \ Qand isa sequence in v

First show that g is well-defined; that is, lim /(.r„) is a real number and

if (>'/i)n€N another sequence in Q with y„ -* x. then liin^ fix,,) = lim /(>•„). Then show directly (t - 5 argument) that g is uniformly conr t - » O C

tinuuus on!

4.6 Discontinuities and Monotone Functions In this section we first define one-sided limits, then types of discontinuities, then monotone functions. Our main results arc Corollary 4.4 and Theorem 4.9. Much of this material is taken from Rudin.

Our setting in this section is as follows. / is an interval or ray in R—that is, / = ia, h), |rt, />], («. b\, [a, b), («, oc), [fl, co), (-oo, /;), (-oo, h] where a and h tue real or / = (-oo, oo); / is a function from I into R; and c is a real number (because limits at ±oo are automatically one-sided). Recall that

90 Chapter 4 Continuity for D C I. J\d denotes the restriction of / to D; that is, /|;j (jr) = /(.v) for ail .t in D.

One-Sided Limits

Definition 4.7 If there is a fc in R such that (c. b) C /. the riglu-lumd limit of f at c. denoted by /(c-l-)orlim /(.r). x

>

c

islim /|/n(c,3s):andifthereisan«inRsuchlhat(«.f) c l.\.\\ckfi-lumcl limit of f at c, denoted by /(C-) or lim f(x). T

X

—f




0


.

From Section 4.3 or by proofs completely similar to those of Section 4.3. it follows that

/(c-l-) = Z.(in R') if and only if for all neighborhoods V of L. there is a 5 > 0 such that if.v is in I and c < x < c + S. then f(x) is in V if and only if f{x„) L for all sequences in I with x„ > c for all n and.Vn —» c; and

/((-) - f.(tnR*) if and only if for all neighborhoods V of L, there is a 0 such that if x is in I and c — 5 < .t < c, then /(.v) is in V if and only if f(x„) L for all sequences (.v,i),i6h in I with x„ < v for all n and.r„ c.

Remark If c is an interior point of / (that is, c is not an endpoint), then it tnakes sense to consider both right-hand and left-hand limitsof / at c. Ifc is an endpoint of /. then it makes sense to consider only one of the one-sided limits. Also note that if c is an endpoint of /, then the appropriate one-sided limit of / at c and the limit of / at c as defined in Section 4.3 are the same.

Section 4.6 Discontinuities and Monotone Functions

91

ll should be clear thai for c an interior point of /, liin /(.v) exists if and only if/(c+) = /(c-) = lim /(.v). J — • C

Types of Discontinuities Definition 4.8 Let c be in I and suppose that / is discontinuous at c. If c is an interior point of /, then / has a discontinuity of the first kiml. or a

simple eliscontiniiit}.'. at c if both f{c+) and /(c-) exist in R; otherwise. / has a discontinuity of the jeco/u/t'iW ale. Ifcisanendpoini of /.then / has a discontinuity of the j7rs7 kind at c if the appropriate one-sided limit of / at c exists in R; otherwise, / has a discontinuity of the secaiui kind at c.

Example 4.23 The function in Example 4.21 has a discontinuity of the first kind at U.

Example 4.24 Let I if .r is rational

/(.V) =

0 if.v is irrational.

Since neither f(c+) nor f(c-) exists at any point c. / has a discontinuity of the second kind at every real number.

Example 4.25 Let 1 sin — if

fi x ) =

X

0

if

.r

=

0.

Then / is continuous at every .t ^ 0 and / has a discontinuity of the .second kind at 0. That /(0+) does not exist follows from Example 4.1.3.

Remark When / has a discontinuity of the first kind at a point c in I. there is a jump in the graph of / at c. Consequently, some authors call this a jump disconliniiiry. Forcan interior point of/.thiscan occurif(l)/(c—) f{c+) or if (2) f(c—) = f{c+) 56 /(c). In (2). the discontinuity is removable by redetining /(c) = /(c-F). whereas in (I) the discontinuity is not removable. Monotone Functions Definition 4.9 The function / is monotone increasini' (respectively,

strictly increasing) on / if whenever .v and y are in / with ,v < y, then fix) < /(y) [respectively./(.t) < /(y)|. Aho, f ismonoionedecreasing (respectively, strictly decreasing) on / if whenever x iind y are in I wiih.v < y, then /(x) > f(y) [respectively, f(x) > /(y)). / iswwiotoneon I if f is either monotone increasing or monotone decreasing on /.
0. Since p = inf(/(.r) : .r e / n (c, oc)}. there is an .vo in I n («•, 00) such that p < fixo) < ^ -f £. Let 5 = .vo - c. Let .\ be in / with

r < a: < r + 5 = .ro. Since / is monotone increasing, p < /(.r) < /(.to) < P -i- E. and so |/(j) — ^| < s. Hence, /(t-j-) = p.

In summary, for / monotone increasing and c an interior point of /, we have shown that

sup/(a) = fic-) < fic) < /({•+) = inf fi.x). .xiirn~x.ct

J€/n(c,cci

The analogous results for / monotone decreasing are asked for in Exerci.se

6.

■

Our main results, given below, are now an easy consequence of Theorem 4.8.

Corollary 4.4 Monotone functions have no discontinuities of the second kind.

Proof This follows from Detinilion 4.8 and Theorem 4.8.

Section 4.6 Discontinuities and Monotone Functions

9 3

Lemma 4.3 Lcl c and cl be in / with c < d. 11" / is monotone increasing (respectively, decreasing) on /. then /(c+) < f{d—) [respectively, f(c+) > Proof Assume that / is inonolune increasing on/. From the proof of The orem 4.8.

/(C4-) = inf/(.v) =inf/(.r). AStntc.oo)

tc;n(r.'().

0

By Definition 5.1, lim /i(v) = 0 = /i(vo) and so h is continuous at yo. Since / is continuous at c (Theorem 5.1) and h is continuous at yo = /(c), /i o / is continuous at f. Thus.

lim (It o f)(x) = {li 0 /)(c) = /K.Vo) = f). Letting y = f(x) in (I), we have (/io/)(.v) =

i*(/(-v)) - g(yo) - ^'(.vo) if f{x) ^ yo. /(-*•)-yo

Therefore,

^;(/(.v)) - gfyo) = \h(f(x)) + /(y,))ll/(Ar) - .voJ, and this equation holds even if /(x) = yo. f-tir X in /. .V ^ c. we obtain

S ( / ( . r ) ) - . ? ( / ( ( • ) ) = Wf(x))+g'(yo)] X

—

c

Thus,

(); o /)'(

98 Chapter 5 Differentiation The graph of / follows.

/

0.01 •

/

\

0.005-

i-ll 1 \ / v

aA/

'* V \/ \ o.i/

/ V _().005 /

-0.01-

\/

Thus, / is iliffcrcmiable on R. However. /' is not continuous at 0 since cos( I fx) in /'(.v) above does not have a limit as .r 0.

Suppose that, in the definition of fix) in Example 5.2, we replace .v^ by

x" where n is in N. In an analogous manner, the reader can show that / is differcntiable everywhere and /'(()) = t) if ii > 2. For the case /i = I, see Fxercise 5.

Remark Given / ; / -♦ R. /' is a new function whose domain consi.sts of lho.sc points in / where / is diffcrenliable (points where /' is ±oo are not allowed in order to keep /' real valued), and /' is also called the firxi (Icrmiiive of/. If the domain of /' is an interval or ray iiiR, then we can form the derivative

of /', denoted by /" and called the xecoiul Jerivaiive of /. In Example 5.2, /'(()) exists in R, but /"(O) does not exist since /' is not continuous at 0 (Theorem 5.1).

In general, the iiili derivative of /, denoted by is the linst derivative of n = 2. .T 4 For /'"'(f) to exist in R, /'""" must exist in an interval or ray containing c and / must be differcntiable at c. Definition5.2 Let / : / ^ Rwithcin /. Then /hasnlocalinaxiniiitn (re.spectivcly. local miiiiniiim) at c if there is a neighborhood U of c such that fix) < /((.') [respectively, fix) > /(c)j for all x \r\U n I.

Note that an absolute maximum or an absolute minimum, as defined in Chapter 4, is a local maximum or a local minimum. Recall that an interior

point of / is a point in / that is not an endpoini of /. Proposition 5.2 Suppose that / has a local maximum or a local minimum

at an interior point c of /. If / is differcntiable at c, then /'(c) = f). Proof Let / have a local maximum at c. Since c is an interior point of /. by Definition 5.2. there is a 5 > 0 such that fix) < f(c) for all .v such that

Section 5.1 The Derivative 99

c — S < X < c + 5. Fore < .v < c- + 5. i/(.v) — /l'")|/(.v — f) < 0. Since / is

differentiable al t\ by Dcfiniiion .5.1 and Exercise 4 in .Section 4.3. f'{c) < 0. Fore -8 < X < c. |/(.v) - /(c-)|/(.v - c) > 0. and llie same reasoning yields /'(t)>O.Thus /'((•) =0. The proof for./' having a local minimum al c is similar. ■ Although a derivative need not be continuous, the following theorem .shows ihai a derivative satisfies the intermediate value property (Exercise 10 in Section 4.4). This will be useful in Section 6.5 in connection with the Fundamental Theorem of Calculus.

Theorem 5.3 if f is differentiable on /. then /' has the intermediate value

property on I. That is, given in I with /'(«) ^ /'(/>) and r between /'(«) and f'{h). there is a c between a and b such that /'(c) = r. Proof Assume that u < b in / and f'{a) < r < f'{h). Define F : [«./'] R by /"(.v) = f(x) - rr. By Proposition 5.1. F is dilYerentiable and lience continuous on (Theorem 5.1). Since F'(.\) =

/'(.V) - r, it suffices to find a c in («. b) such that F'(r) = 0. By Theorem 4.2, F has an absolute maximum and an absolute minimum on 1«, /'|. If either of these oeeurs at an interior point c of I«, /?|, then F'(c) = 0 hy Proposition 5.2. So wc must show that al least t>nc of these occurs at an itucrior point. If neither of these occurs al an interior point, then we have two cases. Case 1 F() is an ab,solule minimum. Case2 FUi) is an absolute minimum and F(b) is an absolute maximum. Ill Case 1,

f\b)

-r=

F'ih)

=

l xi — i i ib
r > f'(h). ■

Corollary 5.1 If / is dilToremiable on / .then all discontinuities of /' arc of the second kind.

Proof Referring to Section 4.6. if /' has a discontinuity of the first kind at a

point c in /. then /' has a jump discontinuity at c. Thus /' cannot satisfy the

intermediate value properly in an interval around c. ■ The reader should note that in Example 5.2 the discontinuity of /' at 0 is of the second kind.

TOO

Chapters Differentiation We end this section with a proposition concerning sequences and deriva tives, We will use this proposition in Chapter 8 to obtain a surprising result— namely, an example ofacontinuous function on R that is nowhere differentiuble.

Intuitively, this is a continuous function with a "sawtooth" at every point.

Proposition 5.3 Let / ; / -> R be differentiable at the interior point c of I. Let («,i)„eN and be two sequences in / both converging to c with a„ < c < h„ for all n in N. Then

/((•)=

hm

;

.

Proof This proof is taken essentially from Rudin. For each »i. let A„ = {h„ - c)/(h„ - a„). Then, for each ii, 0 < A„ < I and /U>„) - /(rtfl) ~ f'ic) =

(2)

bn rj/j

/(/'„) - /(C) - /'(C)

+

b „ - c

(3)

/ ( « . ) - / ( c ) - fi e )

(I -x„)

a

„

-

c

By Definition 5.1. the two expressions in brackets have limit 0 as n ^ oo.

Since (A„)„gN and (1 — a„)„jn arc bounded sequences, it follows that (3). and hence

(2).

approaches

0

as

h

^

oo.

■

l_xepciscs

1. Let f{x) = ^ for J > 0. Find /'(.v) for a: > 0 and note that the tangent line to / at 0 is vertical.

2. Complete the proof of Proposition 5.1. 3. Show that a polynomial is differentiable on R. and that a rational function (the ratio of two polynomials) is differentiable wherever the denominator is not zero.

4. Let /aiidg be differentiable functions on / and suppose that = I for all .V in I. Show that

fix) g'(x) _ /(-V) g(x) 5. Let

f(x) =

.t

sin

-

if

ar

0

O' if.r = 0.

Show that / is continuous on R. dilTcrentiable at a: 51^ 0, but not differen tiable at a: = 0.

Section

5.2

Mean

Va l u e

Theorems

101

6. Given the notation and hypothesis of Theorem 5.2. what is wrong with the following attempt to prove it? Inn

.'—c

X—C

_

hm

/(-v)

—

—

f(c)

x—c

= g'(f {€))■/'{€). 7. Complete the proof of Proposition 5.2 when / has a local minimum at c. 8. Prove Rolle's Theorem: If / is continuous on 1), and /(«) = /(/?) = 0. then there is a c in (n, h) such that /'(r) = 0. [H'mi: If / is nonzero, use Proposition 5.2.) 9. Let / be differentiable on /. Show that (a) if / is monotone increasing on /, then f'(x) > 0 for all .r in /; (b) if / is monotone decreasing on /, then f'(x) < 0 for all x in I. 10. Suppose that / is differentiable on I and /' is monotone on /. Show that f is continuous on /. [Him: See Corollary 5.1.] 11 . L e t I t > 0 . S h o w t h a t t h e r e d o e s n o t e x i s t a d i ff e r e n t i a b l e f u n c t i o n o n

[0, 00) with /'(O) = 0 and fix) > li for all .v > 0. 12. Use mathematical induction to derive Leibnitz's formula for the nth deriva

tive of a product of two differentiable functions / and g:

* = 0

where /'®' = / and

\k) kHn-k)>

5 . 2 M e a n Va l u e T h e o r e m s The Mean Value Theorem, Theorem 5.5 below, is one of the most useful and

important results in analysis as the corollaries, examples, and exercises will indicate. For example. Corollary 5.4 gives us an easy proof that sin.r and cos x are unifonnly continuous on R. This theorem follows easily from the more

general result. Theorem 5.4, which will also be used in the proof of L'Hopilal's rule in Section 5.4.

Theorem 5.4 (Generalized Mean Value Theorem) Let /andg be continuous functions on [n. h] that are differentiable on («, />). Then there is a c in (n, h) such that

[/(/') - fMln'U-) = (glf*) - g(fl)]/'(c). Proof Define F : |«. (»] -+ R by F i x ) = l f ( b ) - / ( a ) \ g i x ) - l g ( b ) - g { a ) \ fi x ) .

102

Chapter 5 Differentiation Tlien /■' is continuous on [a. b\ and difTereniiable on {a, h), and /■" ( ( ' ) = - g ( b ) f ( a ) = F i b ) . Wc need to stiow that there is a c in («. b) such that F'{c) = 0.

If /•■ is a constant function, any c in (o, /;) will work, so we can assume that F is not a constant function. By Theorem 4.2, /-' has an absolute maximiini and an absolute minimum on [u, />]. Since F is nonconstant and /•"(«) = F{.b).

at least one of these occurs at an interior point c of (n, b). By Proposition 5.2, ■

r(c)=0.

Theorem S.S (Mean Value Theorem) If / is continuous on In, h] and differentiable on (a, b). then there is a c in ()). The Generalized Mean Value Theorem has the same geometric interpreta

tion when weconsider the curve as given parametrically by X = g(t). y ~ f (t) with / in \a. h].

Corollary 5.2 Let / satisfy the hypothesis of Theorem 5.5. 1. If f'{x) = 0 for all .V in (a.b). then / is a constant function.

Figure 5.1

2. If f'(x) > 0 for all x in (a. b). then / is monotone increasing. 3. If f'(x) > 0 for all .r in (a. b). then / is strictly increasing. 4. If /'(.r) < 0 for all .r in (a, h). then / is monotone decreasing. 5. If/'(.r) < 0 forall .V in (a./;). then / is strictly decreasing. All results for / are on (a. b].

Proof For any .ri and .r> in [ 0 with |/'(.t')| < M for all .r in /. Let /r > 0 and let S = s/M. If .V| and .vi are in I with 0 < [.V| — .r:! < 5, then, by Theorem 5.5, there is a c between .ri and X2 sucli that

!/Ui) - /(-V2)l = |/'(c)| l-vi --v:! < M \x, - X2\


/*h

fb

/ (/±i')= / /± /

J(i

Jit

Ju

and

f.f-.f, Ja

Ja

Proof We use Theorem 6.3. Let £ > 0. By Definition 6.7 there exist 5] > 0

andS: > Osuch that for P inPfn./j], if ||/>|) < 5,. then S(P.f)-f^f < e/2 and if ||P|| < ^2. then S(P,g)~f^g < e/2. Let 5 = min(5i, S2I and

Section 6.3 Properties of the Riemann Integral 121 let P = be inP[a.61 with ||P|1 < 5. Since n

S{Psf+g) = Y^ I/O,) + g(fi)l A.V,1 = 1

= X^/Oi)Ajr( + ^^Oi)Aj:/ 1 = 1

= S(/',/) + S(P, g) for all choices of the f/s in [x,_i, x,-],

S(P./ + s)-(^jf f + j^ g) S(P.f) + S(P.g)-(^j^ f +1 sj SiP.f)- [ / + S(P.g)- [ Ja

£

J

a

€

^ 2 2 ~

Therefore, / + g is in TZUi.h] and /|'(/ + g) = + g) = I'f + lU-

If c = 0. then every Rioinann sum for cf is zero and the result for cf

follows. Assume that c 0 and let e > 0. By Definition 6.7 there exists a

i > Osuchthatifi.sinP[rt. 61 with ||P|| 0 for all x in [a, b]. Since

L{P.h) > 0 for every P in Via. 6], /j' li = sup{L(P. 6) ; P e Via, h\] >0. Since (g - /)(x) > 0 for all x in [n, 6], by Proposition 6.3. i-h

fb

fb

0< (g- /)= f 8 - f f Ju

Ja

J fi

andso/j"/ O.andlet/'beinPio. fc] witht/(P,/)UP, f) < £- By passing to a relinenient and using Proposition 6.1, we can

assume that c is in P. Then P, = F n In. c] is in /"(«, c] and P2 = P n

[c.h] is in /'[c, fc). Since U(P,f) = U{P^, f) + U{Pi. f) and UP.f) = UP\. f) + UPi, f).U{Pi. f)-L(Pi. f) 0

with |0(y>| < K Tor all v in 1/n. M|. Letf' = e/{h-(t + 2K). By Theorem 4.4,

0 is uniformly continuous on [/«, Ml: so there exists a i > 0 such that 5 < s'. and if yi and .vj are in [w, M\ with lyi - .v:! < 5, then |0(>'i) - 0(>'2)l < f'Since f is in TCln, /;), by Theorem 6.2, (here is a partition !' = [fl./jj with

U ( l ' . f ) - L { P. f )

| Axi + j^fc) - /(c)| Axt+, < \g(z)-f{z)\2 ||P||.

Let tt > 0. Since / is in 72[«, ft], by Theorem 6.3. choose 0 < 5
:,•)—say. /(.r) = for.v,_i < .v < .v,- for / = 1,2 II. Such a function / is a xiep fiinciion. Show that / is in

7?|a,/?|and ]]','/= I yi^Xi. 1=1

6. Let / be in '/?[«.&] with w < f{.\) < M for all .v in [a,/a]. Show that there is a k in |mi, M\ with / = k(h — a). 7. Give an example in which |/| is in h\ but / is not.

H. If / is ill 7?.|«,/?), show that / = iiin /*/• 9. Let / be in 'R\-u,a], where a > 0. Show that

(a) if / is an even function I/(-.v) = f{x) (brail a: |. thenf ~2 /; (b) if / is an odd function [/(-.t) =-/(a) for all A-). then = 10. For / and }• in b\. show that

This is the Cauchy-Bunyakovsky-Schwarz inequality for integrals. [Hint:

For any real number x. expand jf U/ + g)- into a quadratic in .r and use the discriminant from the quadratic formula.]

11. Assuming that ^ and Vcosj; are in 7J[0. ;r/2]. show that Ir s' / .-r c,o s— .trf.r 0 such that if .t' and x" are in [a, h] with . *■ - A '

< S, then f(x') - f(x") < s/(b - a).

Section 6.4 Families of Riemann integrabie Functions

127

Let /' = |x/)"^be in Via, b\ with 1|/'|| < 5. For each i = 1,2 /i. by Theorem 4.2, there are points /i- and n" in {xi-j, .v,] such that /(mD = H 'ind = iiij (where A/,- and m; arc as in Dehnilion 6,2). Since ~ :5

||F|| < 5. A// -nii = \f(n\) - /(/i")| < £/(b - a), and so n

U{P, f) -UP.f) = Y.

■It.. h

—a

^

i k I

=

e.

By Theorem 6.2. / is in 7S(«, b].

Example 6.7 Let I — cos X if

0


0. Since / > 0 on la. b], by Proposition 6.5, fl >

f

rv

fb

fv

//=//+//+//>/ />0.

J

a

J

a

Ju

Jv

Ju

which is a contradiction. Therefore. / is identically 0 on [a, b]. ■ Remark Note that for / continuous and nonnegative on fa, b], the contrapositive of Corollary 6.2 states that if / is positive anywhere on [«. 6], then

r/>o.if 0 otherwise.

then /p / = 0. which implies that continuity is necessary in Corollary 6.2. Monotone

Functions

Theorem 6.9 If / is monotone on [a. 6). then / is in 72[a, b\. Proof First let / be monotone increasing on [a, b], if /(a) = /(/'). then / is a constant function and hence in 72[fl. b]-. so we assume that /(a) < /{b). U P = {a,}-^o is any partition of [a, b], then = /(a,) and m,- = /(Ai_i)

Section 6.4 Families of Riemann Integrable Functions

129

for each/ = 1.2 h.Therefore. rt

U{P. f) - UP. f) = J2 i s l n

/ /

t = i

= II/'I1[/Un)-/U«)] = I1/'II1/(M-/(«)]-

Given e > 0. choose a P in V\a.h\ with ||Pj| < e/{f{b) — /(fl)]. Then

UiP.f)- UP. f) < e. and so / is in Tlia.b] by Theorem 6.2. If / is monotone decreasing on (w./j], then -/ is monotone increasing on [(/, b\ (Exercise 5 in Section 4.6), and so / = —(—/) is in 7S[a. 6] by Proposition 6.3. *

Example 6.9 Let / be bounded on [a.h], Delineg and/i from [a. b] into R by g(x) = sup(/(.v): « < y < and

/i(.r) = infl/(y) : « < y < .vj.

Then g is monotone increasing and li is monotone decreasing on («, b], and so both g and h are in VXa. b\.

Theorem 6.10 (Second Mean Value Theorem) If / is monotone on [a, b\. then there is a c in [o. b] such that

f f = f{(i){c-a) + f(b)(l>-c). Jll

Proof Assumethat/ismonotoneincreasingon[n./>]. Since/(fl) < /(.r)
")

,

,

\i/'i

[Hint: First show that (/g /") ^ ^ for each n. For e > 0, there is an interval fc. d \ C 10. 1) with /(.r) > M - s/2 on [c. (i\. Conclude that, for

large n. f")"" > M - e.\ 6.5 Fundamental Theorem of Calculus Tlieorems 6.11 and 6.12 below are sometimes both called Fundamental Theo

rems of Calculus, When the integrand is continuous, some books call Theorem

6-11 a corollary of Theorem 6.12. These are very important theorems relating ititegration and differentiation.

Definition 6.9 Let / and (p be two functions defined on any interval I. \( 0 such that |/(/)| < M for all / in l Osuchthatc/2 < f(x)lg(x) < \c forallj: > a.Thus.x > a implies (c/2)g(j:) < f{x) < (3c/2)g(jr). Applying Theorem 6.13 twice to this last inequality proves part 1. For part 2, note that there is an A/ > o such that f{x)/g(x) < I or, equivalently, f(x) < g(x) for all x > M. Since f = f + / and

f converges by Theorem 6.13,/^ f converges. For part 3. note that for large x, f(x)/g{x) > I ot f(x) > g(x). Diver gence of / now follows from the divergence of g and Theorem 6.13.1 Example 6.16 sin(l/jr) 1,/j'^jl/lx''In Ar)] a. If \ f\ converges, then / converges.

Proof For.r > a, - |/(j:)| < f(x) < |/(.t)| and so I) < f{x) -I- |/(.t)|
I and conditionally convergent if 0 < /> < 1. In particular. /, [(sinA)/a:lr/A: converges.

Since |x"''sinar| < x''' for.t > 1 and x~''dx converges for /; > 1

(Example 6.14). x~'' sinxtf.t converges absolutely for p > 1. r

Let /3 > 0 and let r > 1. Integration by parts yields x"''sin.rr/.v = - P /",'x"'''^'*cos.T£/x. Since lim [(cos/)/?''] =0, - P c o s x '

'

f-*CC

x~'' sinx dx will converge if cos.v t/.r converges. This last in

tegral is actually absolutely convergent since |x~'''"'"" cos.v| < .v"''''''" and p-l- 1 > I. Now let 0 < p < 1. We have left to show that x~'' jsinxj dx docs not converge. For/iinN,»i > I, " pk:i

/:

x'''\s\nx\dx = y] / x-^\s\nx]dx ^^2 •'(t-il" "

1

/ • *k7i"

•SiL

jsinxj dx Djt

\ 1

The inequality on the second line follows from the fact that x"'' > (l/kTT)'' > ]/kn. The third line follows since jsin.tjf/x = 2. From calculus.

the harmonic series Z Uk diverges and so lim Z"'x"** jsinxj r/x = 00. i=l

11 - 0 0

Therefore. /,* x~'' jsin xj dx diverges. Definition 6.12 For / defined on R and c in R, / comei-ges and

r

J^x

f=f

J—oc

Jc

/+r

f

w

provided both integrals on the right of (4) converge. If either of the integrals on the right of (4) diverges, then / diverges.

Section 6.6 Improper Integrals

139

Remark From Proposition 6.5 it follows that the choice of c is immaterial. The Cuuchy principal value of / is lim f. By writing the integrals on the right of (4) in tcmis of limits, it should be clear that it f converges, then its value is the same as the Cauchy principal value. However, / may diverge but its Cauchy principal value may exist. For example, the Cauchy principal value of * dx is 0 but x dx diverges.

Improper Integrals of the Second Kind Definition6.13 Let/bein 72[rt, f| forall; in fa,/^) with / unbounded

at/?. The improper Rieiiimn integral of f o«|a.fcJis llm /'/. Sim/-•b"

"

ilarly, if / is in Tif/.fo] for all t in ta.bi with / unbounded at a, the

improper Riemann integral of f on [a, b] i.s lim /'' /. 1—11+

The improper integral is convergent (or converges) if the correspond ing limit is a real number; otherwise it is divergent (or diverges). Notation Although some authors use the notation / to denote that this integral is improper at a, we will simply use /. For / in K\a, 6], this is ju.stified by Exercise 8 in Section 6.3. It is incumbent on the reader to decide first whether f is proper or improper.

Example 6.18 By the symmetry (draw both graphs on the same axes) be

tween e' and In.r, it follows from Example 6.13 that /J Inx = -1. Doing this directly involves integration by parts and L'Hopital's rule to evaluate the limit.

Example 6.19 Let 6 > 0. Then/i'x"'d.rf -=« 0lim lnx|f =/ - .lim (in/j+ 0+ In /) = CO. For p \,

/,I-P _ fl - / ' ^ -I' 1 f■f 'hox-''dx =C Olim —j = I _ ^ ^ if p>l. Hence, for h > 0, x'P converges if and only if p < 1.

Remark In Example 6.19changing variable y = l/.r gives that J^^x~''dx = which, by Example 6.!4,convcrgesifand only if2-p > I or, equivalently, p < 1.

Similarly, an improper integral of the second kind can be changed into an improper integral of the first kind by the appropriate substitution. Thus, one would expect theorems analogous to Theorems 6.13, 6.14, and 6.15 to hold

for improper integrals of the second kind. Since their proofs are basically the saine, we simply stale these iheorem.s for integrals that are improper at h. Comparison test If /isinTSfa, r] forali/ in (a./j), with 0 < f{x) < g(x)

for all x in [a, b), then the convergence of /*" g implies the convergence of

/:/-

140 Chapters Riemann Integration Limit Comparison test If / is nonncgalive on [«, h). g is positive on [«. h), with both f and in 7v|«, i \ for all i in [«. h). and liin f(x)/g(x) = c. then j - h -

1. for 0 < f < 00, fll f ind fl^' g either both convene or both diverge; 2. for c = 0, /j' g converges implies that /j' / converges; 3. for c = 00. if /j' g diverges, then f^' f diverges. Absoiute convergence implies convergence If f is in TC[a. /1 for all t in Ifl. b), then |/| converges implies that / converges.

Example 6.20 c'~'x'"' dx is proper if p > 1. For 0 < /) < 1,

/J dx converges since < .v'"'. and JJ, x'"'dx converges if I — /) < I or, cquivalcnlly, p > 0.

Example 6.21 Consider .r"'' sin x dx. This is a proper integral if p < I

for lim (sin.tl/.v'' is 1 if /> = I and is 0 if 0 < /j < 1 (use L'Httpilal's rule). .1—11+

[f/> > I. .1—0+ lint (.V"''sin.v)/.v"'"'''= 1 and since _/J, converges if and only if /' — I < 1 p Ji) .V'sin.r r/A" converges if and only if/) < 2. Definition 6.14 If / is unbounded at an interior point c of (a. b). then f converges itnd

f

Ja

f= Ju

Jc

f

f+

I"

/

(5)

provided both integrals on the right of (5) converge. Othenvise, f diverges. The expression in (.^) is valid even when / is not unbounded at c. For example.

sinx , /•' sinx . sinx ,

/ d xX Jo

= J n/

dX x

+ Jt

/

X

dx

converges because the first integral on the right side is proper, and

/i'^|(sinx)/xlf/x converges by Kxample 6.17.

Section 6.6 Improper Integrals

141

Lxepcises

1. Show thai f^[l/(xy/x + converges but f^[]/(x^/x + \)]dx divcrges.

2. Show that /~sin(j^)i = a:^.| 3. Show that e"-'sin(l/.t)«/.v converges absolutely.

4. Show that /J ((l/.r)sin(l/.t)] 0 (here exists an nn in N such that if ;i > m > /lo. then I

cit


/lo and m > /lo. then |s,i — i,„| < E (Delinition 3. II)

if and only if for all f > 0 there is an iiq in N such that if n > m > iia. then E

< c (since .v„ - ,V„| = E K satisfies /(.v) > Oforall.r in 11, oo) and that / is monotone decreasing on [ I, oo).Letfl„ = /(«) for each n in N. Then the series L converges if and only if the improper nm I

integral f converges.

Proof Consider Figure 7. i and note that / is inlegrable on [ 1, n] for each ii in N since / is monotone (Theorem (>,9).

For;i>2, S is an upper sum and I is a lower sum of / on [1, »|, k = 2

and .so

I

2

3

Figure 7.1

t=i

t=2

if f converges, then the second inequality in (I) shows that the partial suiiis of E a„ are bounded above by /. By Theorem 7.2, E a„ converges ? c

and so E a„ converges by Exercise 4. n = l

If f diverges, then the first inequality in (1) shows that the partial sums fl C

00

of E are unbounded above, and so E «„ diverges by Theorem 7.2. ■ nsl

n=\ o c

Example 7,5 (p-series) E I/«''converges if and only if p > I.Thisfoll i s ]

lows from the Integral te.st and Example 6.14, where we showed that i/x'' converges if and only if p > I. The reader will find that the p-series is very u.scful in determining whether

other series converge or diverge. This will be demon.sirated in the remainder of this section.

Theorem 7.4 (Comparison test) Suppose that 0 < ci„ < h,, for all n in N. Then CO

00

1. if Z converges, Z 2 andn =E2 \/n

diverges. Also, E l/(H-+3)convergessincc l/(«^ + 3) < l/»i^and E l/rr 1=1 n=l converges by Example 7.5.

Theorem 7.5 (Limit Comparison lest) Let n„ > 0 and h„ > 0 for each it in N and suppose that lim a„/b„ = L. n—•

X

X

X

1. irO I. n s 2

X

oc

J

7. Let «„ > 0 for all n in N. Show that if Z ti„ converges, then Z 'ht converges. a;

CO

8. Let a„ > 0 for all ii in N. Show that if Z (i„ converges, then Z nsl

nsl

converges. [Hhii: Expand —(1/«)1^.I 9. Use Theorem 7.5 to show that

(a) Z - 2)/(3;j-- « + 6) dive^es n=2l

and 3C

,

(b) Z (Inn)/{tr + 4) converges. 10. Prove Theorem 7.5.

7.2 Absolute and Conditional Convergence In this section we first consider alternating series, We next distinguish between absolute and conditional convergence, and then we develop tests For each type of convergence.

Alternating Series Definition 7.3 An alternating series is one whose successive terms are alternately positive and negative.

Section 7.2 Absolute and Conditional Convergence

149

If > 0 for each n in N, then C C

^(—= fii — fi2 + rt3 ~ rij H r t = l

and C C

= -«1 + ll2 - Hi + H4 R = 1

are alternating series. Since the second series is the constant —1 times the first series, in view of Exercise I in Section 7.1, we consider the first series.

Theorem 7.6 (Alternating Series test) If («,i),i6pj is a monotone decreasing e

o

sequence of positive real numhers with lim a,, = 0. then E(—!)" n„con/l-^OS ,1=1 verges. In this case. If A and are, respectively, the sum and partial sums of this series, then |A — .v„| < h,i+i for each ii In N. Proof For ii In N, •i'ii = («l ~ H2) + («3 — H4) + h (Hjn-I " 0. By the Cauchy con.

.

"

dition (Theorem 7.1), there is an no in N .such that Z | |£J„| for all n > iin. 0

0

Hence lini ti„ ^ 0, and Proposition 7, i implies that E «« diverges n—oc

n=l

oc

30

,

3- The divergent scries E l/n and the convergent series E l/«" both have 1=1

Z,

1=1

= c

_

■

I.

c

Example 7.10 E it'./n" converges by the Ratio test. First note that + ' i )"!! ~ _ («+" " l)"" ~ ~ 1)""^' By L'Hopilal's rule or Example 3.6. lim («/(" + 1)1" = 1/e < 1 (this is a good exercise for the reader). iX)

Theorem 7.8 (Root test) Given the series E",nletZ-=: lim Then „=|

1—C50

C V

I. E «;i converges absolutely if Z. < I; fl s l

2- E (In diverges if Z. > 1; 3. the test is inconclusive if Z. = 1. Proof 1. First choose a number r such that L < r < 1. Then there is an iio in N

with !!/\(i7\ < r for all n > iiq or, equivalently, |«„| < r" for all n > mo. 00

1, |fl„| > I for all large ii and so lim «„ ^ 0. I I - . O S CO

00

.

3. By an application of L'Hopital's rule. E I/" and E l/'i both have n s l

L=

1.

■

Example 7.11 Our purpose here is to give an example in which the Ratio test fails but the Root test works, thus giving the correct impression that the Root test is more powerful than the Ratio test. (More precisely, if the Ratio test shows convergence, then so does the Root test; and if the Root test is inconclusive, then so is the Ratio test. See Rudin, p. .59.)

Consider E a„ = 2r+r- + 2r^+r'^+ 2r^ , which clearly converges

ifr =0. Forr^O.

' 2 |r| if n is even

Cn + I

- |r| if n is odd.

Therefore, lim \o„^\la„\ does not exist for /■ 0 and so the Ratio test fails. n

—

C

C

152 Chapter? Infinite Series However,

^ j|r"|"" =lr| if« is even Ipf-'li'" =2'"'|r| if«isodd.

Since iim 2'^" = I. lim = |r|. By the Root test. E «„ converges absolutely if |r| < I atiii diverges if |r| > I. If/-= ±1, then lini a„ # 0 and so this series diverges if |r| > 1. Our next test is often used when the Ratio and Root tests fail, c

c

Theorem 7,9 (Raabe'stcst)If E is a series of non/ero real numbers with n = l

L = lim /i(l - |(/„+|/«,||), then / r - * £ C 5 C

1. Z a„ converges iibsolutely if L > I; n = \

2. E rt;i diverges or converges conditionally if L < 1; /IB I

3. the test is inconclusive if L = 1. Proof

1. First choose a number r such that 1 < r < L. Then there is an iio > 2 in

N with /i(l - jn,i+|/a„|) > r for all n > no or, equivalently, |cj„+i |/1(7„| < I - (/•/«) for all II > /ifl. By manipulation it follows that (/I

-

1)|«„|

-II

|«„+i|

>

(r

-

l)|ti„!

{2)

for all n > no- Let in > iiq and rewrite (2) for/i = iio. /to -f 1 nr.

UU) - I) |rt„„

"O Ono +|l - ("0 -+- I )

"O

tno+2

> >

(r-l) (r-1)

""0 "no-*-'

(//() + 1) |rt„„t2| - ("0 + 2) |«n„+3| > ('■ - 1) [«nn o + 2 ("'-Dk'ml - '"|am + ll

> (r - I) |rt„,I,

Adding these inetjualiiies, we get

("0 - I) I"," D

—

Ill

"m+l > ('" — l)( «n() + |"n() + l + ' ' ' + dm )•

Selting.Y„= E Ifil. we obtain ("0 - I)

—

•"0

m

dm + l > (r -

Hence,

(r - 1 hm < ("0 - 1)

•"0

+ (/--Uinn-I - in

dm+i

< (/!()- I) + (/--l)j„„.l. Therefore, is a bounded sequence and so is bounded. 0

0

By Theorem 7.2, E k',, | converges.

Section 7.2 Absolute and Conditional Convergence

153

2. Since L < 1. there is an mo > 2 in N such that »i(l — |rt,i+iA'nl) < I for all II > lit) or, eqiiivalenily,

l"n+il > " - ' k'rJ

"

for all II > iio. Hence, ii j > (n - I) |«nl for all « > /i». and so " I'Ci+il > ("0 -

for all II > III,. Letting c = (ho - I) > 0, we have iH„+i| > c/ii for all H >/111. Since E c/h diverges. E |«„| diverges by the Comparison ;i=n()

/i=n()+-l

test, and SO E jo,, | diverges. >1=1

3. The divergent harmonic .series and the conditionally convergent alternating harmonic series both have L = I. If «2 = 2 and /) In II

(/I + I) In 11 + 2 " for/I > 2. then E «« has L = I and converges absolutely. The easiest nss2

way to show the absolute convergence is to use Kummer's test (see Fridy,

p. 157), which we have not developed. ■ Example 7.12 Coasider

^ 2-5-8 (3/1- I) 2 2-5 2-5-8 ^9-12-15 (3/1+6) ~ 9 9-12 9-12-15 By calculation. = (3/i +2)/(3// + 9) -*• I. and so the Ratio test fails. (The Root test does not look good either.) Since

"O-lvD-

3/1

+

9

^3 1 .

this series converges by Raabe's test.

Tests for Conditional Convergence Although there are many more tests for absolute convergence, we now turn our attention to tests for determining convergence when the series may not converge absolutely. So far we have only one of these, the Alternating Series test. We develop one such test and leave a second one as Exercise 7. Lemma 7.1 (Abel's partial summation formula) Let («n)>i6N i6N two sequences in R. and let A„ = E ih for each // in M. Then n

*=i

"

^ Ai(//n.i - f'i)t-i

OC

^

CC

In particular. E Ukl'k converges if both the series E At{bk+\ — Ih) and the /i=i

t=i

sequence (A„/;,|.n)„5pj converge. Proof Letting Ao = 0. we have II

n

n

n

^2'ikl'k — ~ At_i)/>A = ^ Atht - ^ Ai,bt+i + A„b„+\. k=\

and

k=\

so

4=1

(3)

*=l

follows.

■

154 Chapter 7 Infinite Series

Theorem 7.10 (Dirichlet's test) Let E c,, be a series with bounded partial ns

I

sums, and let be a monotone decreasing sequence with l)„ -» 0. Then cc

u

S ci,ib„ converges. /i=i

Proof Let A„ = E eik, and let M > 0 with |4„| < M for ull ii in N. Since ^

(V

lim A„t)„+\ =0.ByLeiiima7,],ilsur(iceslo,showihal S 4i.(/?nihk) converges. Since (/.k -Z't+i) for each k in N. Since n

52 C't l'k+\) ~ U'\ — l>2) + (l>2 — h-i) + ■ ■ ■ + {b„ — b„, ] ) t = l

= b\ — b„4.\ -* bj, I I

S (»i-«i+i)converges. By the Comparison (est. E Ak(bk,i -/>t) converges *=l

i=l

■

a b s o l u t e l y.

Note that the Alternating Series lest follows directly from Dirichlet's test,

because given the hypothesis of Theorem 7.6. E (-I)""'' has bounded partial ; i s l

sums and thc«„'s (which are the/>„'.s in Theorem 7.10) decrease monotonically to 0.

Example 7.13 For x ^ 0. ±2n. ±4n',.... the trigonometric identity cos i.c — cos(;j + 5).v = E. i k m, k x "-OS 2siniv

implies that

E sinA-.v

< l/jsin j.v|. Of course. .sin(2Hi;r) = 0 for any

i s l

integer»i. Thus, the partial sums of E sin/i.v are bounded for all .t in R. So if /t=i

(f'n),ieri is a monotone decreasing sequence with limit 0. Dirichlet's test implies that E ^„sin«.vconvergesforalLvinR. In particular. E (-sin/i)//; converges. "=1

n=l

- x c p c i s e s

1. Show that the following seric,s converge absolutely. (a) E rVn! where r is in R (b) E(n+2")/3" "=1

n=l

2. Show that the following series converge. (a)

E

l/(lnn)"

(b)

E

D"

Section 7.2 Absolute and Conditional Convergence 3. Show that

~

,

1-3-5

'

(2a-l>

2-4-6

(2«)

I

2

1-3

1.3.5

2-4

2-4-6

155

is conditionally convergent- [/•/mr; Use Raabe's test (the Ratio test fails) to show nonabsolute convergence; then use the Alternating Series test. To show ii„ -» 0, show that < 1 /{In + i) by induction.] 90

00

4. Suppose that E converges with eucli a„ > 0. and let »•„ = E for lie

I

k=n

each n in N. Show that

(a) {/"iilrieM is strictly decreasing with limit 0; 0 0

(b) E a„r„ converges; n = l O C

(c) L diverges. [Mini: Use die Cauchy condition.) CC

CO

5. Suppose that L converges and x is in [0, 1]. Show that Y. anx" con;i=l

1=1

verges.

6. U.se the trigonometric identity

^ sin(»i + i)* — sin ix >

COSlt.!:

=

—;

for -T # 2m;r (where m is an integer) to show that E h„ cos nx converges « = I

for all A- # Imn whenever (-b„),i6fj is a monotone decreasing sequence with limit 0. 90

90

7. This is Abel's test. Show that S a„h„ converges if S a„ converges and 1=1

1=1

the sequence (('i)i€n is monotone and bounded. [Hint: Use Dirichlet's test.] CO

"

8. Suppose that E a,, converges with each «„ > 0.andlei.s„ = E «t.Show ;j = l CO

CO

that E 0, and let s,, = E Ok(a) Show that E «i/.Vi diverges. [Hint: U.se the Caiichy condition. | 1 = 1

(b) Show that E a„/sf, converges, [Hint: ci„/sf, = (.v„ < nssl

C O

I/.?„_! - l/.r„ forn > 2and1 =E2 (l/si-i - i/.fi) converges.] 10. Consider

•

E

3-5

2-4.6

(2/1-1)'

t

\

f

/l-3.5\''

(2/1)

1 = 1

where p is in R. (a) UseRaabe'stestloshowthatthissericsconvergesif// > 2anddiverges if/; < 2.

156 Chapter 7 Infinite Series (b) When p — 2. Raabe's test is inconclusive. To show thai this series

diverges wlien /> = 2, lirsl do long division to obtain

Ah+I _ I _ 4/1 + 3 _ I I ^ 5// + 4 I I a„ 4//2 + 8/i + 4 n /i(4/|2 + 8/;+4) ii'

It follows that «„+i > (l//i)rt2 for// >2.

7.3 Regrouping and Rearranging Series Regrouping The first question wcconsider is what happens to the convergence or divergence of a series if the terms of the series arc regrouped by in.serting parentheses. For example, what happens when wc regroup the alternating harmonic series

The next definition makes the insertion of parentheses precise. Definition 7.S Let ^ be a function from N into N such that < .

.

^

OC

n+l = "^(n)4.| + «*'(n)4-2 H b

a

s

Then £ !>„ hn rennnipiiif{ o( £ (i„formctl hv inseriine luirenihvse.s. If "=l

ri«l

"

OC

_

ao

we start with the series £ />„ above, then the series £ «„ is a/'t'i'/T//f/////g cc

of £ h„ formed hy remnving ptm'iuheses. n=

«•!

I

In Definition 7.5, note that O C

y"! 6,1 = (a| +«2 d ii) + («^|I1+| +«^(||+2 H 1" >'2. followed by just enough negative terms so that pi + 1- Pnj — r/i — ■ • • — i;:l

1=1

5. Can a conditionally convergent series be rearranged to form an absolutely convergent series? 0 0

6. In the notation of Lemma 7.2, show that if E a„ is absolutely convergent. n = l

-x

Ctt

OC

DO

OC

then E and E (3i both converge and E fln = E /'n - E Qi. Ilsl

1

=

1

1=1

1=1

1

=

1

7. If E „. and their, tCauchy product £1 =c„0 converge to A, B, =0 and C, respectively, then C = AB.

Example 7.18 Consider

^

/I

,1=0

2"

+

I

"

'

. "

,3 ^3 '

4

2^

5

2'

This scries converges by the Ratio test, but to what di>es it converge? One way to answer this question is to consider the Cauchy product of £ .r" with itselt. 0 0

,1=0

Since £ .i" converges absolutely for j.vj < 1, the Cauchy product 1 = 1 )

/

X

ic

/

rt

E-v" E-^" =E

\n= 0 there is an /lu (depending only on e) in N such that if »i > wo, then |/„(.r) - F(.t)| < e for all x in D.

If (/n) „gjj converges uniformly to F on D. we will sometimes write /„ -» F unifonnly on D. Clearly, uniform convergence implies pointwise convergence. As Figure 8.3 illustrates with D = [ 0, the entire graph of /„ must lie within the 2£ band centered on the graph of F for all .suflicienily large n.

-^txampl Let f„ andonF[0.beS).asLet in Exampl e 8.1. For 0< < 1,that we5""show ' that Jne-*8.5 F uniformly £ > 0. Choose /lo in N5 such < £. If M > «(). \fn(x) — F(.t)| = .x" < .t"" < 5"!' < £ for all .r in [0,5]. Next, we show that (/„)„6n does not converge to F uniformly on [0. 1 j.

Suppose that /„ -♦ F uniformly on (0. 1] and let £ = ^. By Definition 8.2. there is an no in N such that .v" < ^ for all .t in [0. I > and for all n > itn. But if

(i)i/"o < A- < 1. then .r"'i > 5. which is a contradiction. So does not converge uniformly to F on [0. 1 ].

Example 8.6 Let f„ and F be as in Example 8.2. Then J„-*F uniformly on 5 becau.se if £ > 0. we can choose iiq in N with I /»io < £"• Then for all x

in R and for all n > ;j(j. |/n(J:) — F(.v)| < 1 /^ii < 1 Thus, even uniform convergence does not guarantee that the limit of the derivatives will be the derivative of the limit.

N

Section 8.1 Function Sequences 1 71

F n

b

a

Figure 8.3 Example 8.7 Let /,i(a ) = nx/( \ + for;7 in N (Figure 8.4).

o.s

1.5

2.5

Figure 8.4 Then (/,i)„gN converges poiniwisc to the zero function on (0. oo). Let i5 > 0. Tlien .t > 5 implies ihni

Given f > 0, choose no in N witli I /no < Then .r > ^ and n > no imply /nU) < l/(n()3) < E. Therefore. (/«)„ert converges uniformly on |5, oo) for all 5 > 0.

We now show that does not converge uniformly on (0. oo). The

key is to notice that /,i(I/ji) = ? for each n. Thus, given f = j and no in N. /„( l/n) > E for all n > /iq and so (fn)„es cannot converge uniformly on (0. oo). The reader should compare the idea here with that of Exercise 5. Our next result will be useful in the following sections.

1 72 Chapter 8 Sequences and Series of Functions Proposition 8.1 (Cauchy condition) Let '' function sequence on D. Then (/„) con\'erges unifonnly on D if and only if for each £ > 0 there IS an iio in N such (hat if in > «o and/i > iin, then |/,„(j:) — /n(.r)l < £ for all -r in D.

Proof Suppose that /„ F uniformly on D and let £ > 0. By Definition 8.2, choose «o in N such that if n > iiq. then |/„(a:) — F(j:)1 < e/2 for all x in D. Then in > no. " > "o. und x in D imply that \fmix) - /rt(-V)l < l/«W - FM\ + IP'ix) - /,,(x)| < £. Conversely, assume the statement after the if and only if. Then for each

X in D, (/H(r))neN ^ Cauchy sequence in R and hence converges to a real number, which we call F{x). Thus, fn ^ F pointwise on D. To show that the convergence is uniform, let £ > 0 and choose no in N such that m > no, II > no. and .r in I) imply that \ f„(x) - f,n(x)\ < f/2. Since fm(x) F{x) as III —» 00, holding n > no and x in D fixed and letting m ^ co gives |/„(x) — F{x)| < e/2. Hence, n > no implies |/„(x) - F(x)| < £ for all x in D. and so /„ -♦ F uniformly on D. ■ Applying Proposition 8.1 to the function sequence in Example 8.4 is one way to tell that the convergence is not uniform, because if n < /n, then l/m(''m) ~ /ufr".!!)! — '•

["xePcisGS 1. Let /„(x) = x/n forx in [0, ex). Show that (/n)„gpj converges uniformly on |0, h\ for all Z? > 0 but not on (0, oo).

2. Lot /„(x) = .r"/(l +x") forx in [0. IJ. Show that (/„),iepj converges uniformly on [0, b] for all 0 < ft < I but not on [0, I ].

0.5-

0.4 • •

0.3 - -

0.2 - -

0.1 - -

0.2

l).4

0.6

0.8

.1. Let /„(x) — nx(l - x)" for x in [0, 1]. Show that (/„),|gN converges pointwise but not uniformly to the zero function on [0, I].

Section 8.1 Function Sequences

0.35-*-

173

0 = 8

0.3 0.25

0.2 0.15

0.1 0.05

4. A function sequence (/„)„eni is imij'onnly bounded on D if ihcre is an Af > 0 such that |/„(.r)| < M for all x in D and all n in N. Show that if

(f,i)neu umfonnly convergent on D and each /„ is bounded on D, then (/„)„eH is uniformly hounded on D. Use this to conclude that the function sequence in Example 8.3 is not uniformly convergent. 5. Suppose that/„-♦ Fpoinlwiscon Dand let A/,, = sup |/,(.v) - F(.r)| for each H in N. Show that/„ -» Funiformly on/3 ifandonly if lim M„ = 0. ft. Let /„(.v) = l/(/i.r+ 1) and g„(.v) = x/(iix + 1) Ibr.v in (0. 1). Show that (fn)nen not converge uniformly on (0. I) but does converge uniformly on (0. I).

7. Let /,i(.v) = .r/(I+H.t-) forx in R (see graph below), Show that (/,i)„gpj convergesuniformly to the zero function on R. \/liiii: By calculus, the M„ defined in Exercise 5 is 1 /{2./n).] Then show thai 0 if X ^ ()

0.5-

0.4 • •

0.3 • •

0-2--

0-1 ••

0.5

1.5

2.5

174

Chapter 8 Sequences and Series of Functions 8. Let /,i(.r) = .V + (l/'i) lor-v in R.

(u) Show thai (/.,)„gri converges uniformly to F(x) = a: on R.

(h) Show that (/,?)„£» does not converge uniformly on R. Let and converge uniformly on D. (a) Show that (/„ ± con%crges uniformly on D. (h) If each /„ and each g,, are bounded, show that (/ngfllngp,- converges uniformly on £>. [Him: First use Exercise 4.]

111. Use Proposition 8.1 to show that if converges uniformly on Di and on 1)2, then {fn)nen converges uniformly on D] U Di ll. Let /„. F ■. D (f. d| for each n in N. If ^ F uniformly on D and g is u continuous function on [c. no- Fix an n > no- Since /„ is continuous at c, there is a (*> > 0 such thai if .v is in D and l.r — f| < 5. then |/„(.r) - /n(c)| < e/.l. Let -v be in D with |a' — c| < S. Then

If (.V) - f (f)| < |f(.r) - /„(.T)| + |/„(.r) - /„(c)| + |/„(c) - f (c)| S

E

E

^3+5+3 —

and

so

f

e.

is

eonlinuous

at

c.

■

From the material in Section 8.1 following the paragraph entitled Main Problem, it follows that if each f„ is continuous at an accumulation point c in D, then uniform convergence implies that lim[]im /„(a)1 = lim [lim/„(.t)].

Also note that Theorem 8.1 is an easy way to show that the convergence in Example 8.1 is not onifomi.

Section 8.2 Preservation Theorems

175

Remark Theorem 8.1 shows Ihat uniform convergence is suflicienl lo guaranCee lhat the limit function is continuous when each member of the function

sequence is continuous. By Example 8.3 or Exercise 1 in Section 8.1. uniform convergence is not necessary. The next theorem gives a partial converse to Theorem 8.1.

Theorem 8.2 l.et /„ —• F pointwise on («./?]. and let each f„ and F be continuous on [«. h\. If f„{x) > /„4-|(.v) for ail ii in N and all .v in | 0 for all n in N and all .v in |«. A]. Suppose that (/n)„6N does not converge uniformly to F on | t)

such that for all iio in N there exist an n > no and an.t in [«, A] with /,i(.v) > e. Letting ii(, = 1. there exist an ui > jjq and an .vi in \a. A] with (.vi) > e. Letting iio = ui + I, there exist an 112 > no > iii and an .vi in jrf.A] with /«•) (.r:) > Continuing, we obtain a strictly increasing sequence of positive integers ("il/gN and a sequence in [«, A] with /„.{.v,) > e for all / in N. By the Bol/ano-Weierstrass Theorem for sequences (Tlieorern 3.10) and

Theorem 3.3, there exist a subsequence of (.r,)igH and an .v in [«, Aj with X,,' ■-* X. Fix an mk in N. For k >~ m,

Since is continuous on [a, A]. /«,„ (x^.) Hence. > e, and this holds for all/'I inN. However, since f„(x)—* 0. there is an A'l in N such n

that n > N\ implies that f„(x) < e. In particular, for 111 > /V,. /„, (.t) < s, f/tl which is a contradiction. Therefore. /„ -» 0 uniformly on («. A). Next let F be arbitrary. Setting g„U) = f„(r) - F(/). we have that each

is continuous on l ,?„+i on [«. A] for each n in R By the previous case. f;„ 0 uniformly on [«. A), and hence f„ —» F uniformly on [«. A|. ■ The function .sequence (/«)„gH on (0. )) given in Exercise 6 in Section 8.1 has the property that f„ > /„+, on (0. 1). but the convergence is not uniform. Hence, an arbitrary domain will not work in Theorem 8.2. However. Theorem

8.2 remains valid if we change the monotonicity condition to f„{x) < /„♦ i(.r)

for all n in N and all x in 1«, A], because then (1 - /„)„gn and 1 - F satisfy the hypothesis of Theorem 8.2, and .so I — » I — F uniformly on [«. Aj. Hence,/„ —» /■'uniformly on [u. A). Example 8.8 Let

0

if

.v€R\Q.

Then each /„ is di.scontinuous at every point of R. Since |/„(.t)l < 1 fii for all .r in R and all n in N, converges uniformly to the zero function on 1, which is of course continuous. To quote Gelbaum and Olmsted, p. 76, "uniform convergence preserves good behavior, not bad behavior."

176

Chapters

Sequences and Series of Functions

Uniform Convergence and Integration TheoremS.B Il'yii ^ F uniforiiily on («./j] andeach/„ is in TCfn, fc], then

F is in TC((;, b] and /j' F = fnProof To slu>w thai F is in TZ\ 0 and note that we want a partition F of (rt, /;] such that l/(F, F) — L{P< F) < e. For any n we have U { F. F ) - m \ F ) < \ U U \ F ) - U ( p . f „ ) \ + \ U ( P. f „ ) - L ( P. f „ ) \ - ^ \ L { P, f „ ) - L { P. F ) \ .

(1)

Since /„ -» F uniformly on [«,/>], (here is an n in N such that

I/h(J^) -/"'U)! < £/[.^(/'- «)] for all J-in [a, ft). Since /„ is in ft), ihcrc isapanilion P = of [o. ft] such that U{P. /„)- L{P, f„) < e/Z. Now wc heed to consider the first and third summands on the right of (I).

Letting Mi = supiF(A) : .rf_i < .r < aJ and Mf = sup(/„(a) : a,-i < A < -v/1 for / = 1 k, we have (

\ U { P. F ) - U ( P. J „ ) \ = t = ]

it

< - Af,"! AA/ < = i k

3(ft - a)

Axj (by our choice of/n) £

" 3' Similarly, \L(P, /„) - HP. F)| < e/3. Combining these into (I), we have that U(P. F) - L{P, F) < e. and so F is in 7^[a, ft).

To show that /'f F = lim f[' /„, let e > 0 and choose wo in N such that n > U() implies that |/,i(.v) - F(a)| < £/|2(ft - «)| for all a in [ Ho.

/ F- f„ = / (F-/„:

Ju

Jil

Jn

< f'\F-f„\

(Corollary 6.1) (Proposition 6.4)




n.

Then {.fn)neN converges uniformly to the zero function F on [0, oo), However.

/„ = I foreach H, and so/J* F ,6 lim /„■ Uniform Convergence and Differentiation In Example 8.2 and Exercise 7 in Section 8.1, we have /„ —» F unit'ormly but

does not converge to F' (even pointwise). Exercise 6 provides us with an example of a sequence of differentiable functions that converges uniformly to a function that is not differentiable at a certain point. Thus the analogue of Theorems 8.1 and 8.3 for differentiation requires stronger hypotheses. Theorem 8.4 Suppose that (sequence of differentiable functions defined on a bounded interval / and that (/,i(x-o)),ier< converges for some point

X(i in /, II (/,!),igfj converges uniformly on /. then (/,i),|£rt converges uniformly on / to a differentiable function F and F'{x) = lim f'(x) for all .v in /. U - ' O C

Proof Let 0. Choose no i" N such that if ii > no i "«• 'hen

l/n(vo) -/,„(.Vo)| < I [since (./"/if.tnlL^rt Cauchy] and

for all .V in / (by Proposition 8.1). Since /„ — /,„ is differentiable on /, by the Mean Value Theorem (Theorem 5.5). for x ^ r in I. we have

l/„(.r) - /,„(.v) - [J„U) - f„,{i)\\ = |/:(.v,) - y;;,(.vi)| j.v -/[ (for some ,t] hciween x and f) K

2{b - u)


n^. Note that (2) holds if .r = t. Hence,

I./;,(-«•) - fmM\ < l/„(-r) - fmU) - [/„(j:t)) - f,Ax(i)]\ + |/„(.Vo) - /„(.vo)l < £ for all X in / whenever n and ni > iiq. By the Catichy condition (Proposition (,//i)neN converges uniformly on / to a function F.

178

Chapter 8 Sequences and Series of Functions Fix c in /. We want to show that F is differentiable at c and F'{c) =

lint /^(c). Define a new function sequence on / by fnix) - f„(c) Snix) =

X

—

if

.V

c

c

, f-{c) if -V = C, S i n c e i i , ' „ ( c ) = c o n v e r g e s . F o r. v 5 6 c . f;„(x) - gm(-v) =

/,i(.v) - /,„(-»•) - |/,i(c) .V

—

c

and so converges uniformly on / \ {c). By Exercise 10 in Section 8.1,

(A'«) „gf^ converges uniformly on I to a function f;. Since /„ is differentiable at c. lini "nil so cach^,, is conlinnous at c. By the proof of Theorem 8.1, is continuous at c. Thus ,§(c) = lim^'f.v) = lim

=

lim g„(.t)

lim

l i m

/;i(-V) - f„{c)

r t ^ O C

= liin

X

—

c

F{x) - F(c) X

—

c

since f,t—* F on I. Therefore, F is differentiable at c and F'{c) = g(c) = lim g„(c) = lim

r f ^ C O

Since c is an arbitrary point of !, litis completes the proof. ■ If I = [

n

n

ir.t

=

n

linear otherwise.

Show that (/n),i£p; converges pointwise but not uniformly to the zero func tion on [0.2], [Hint: Use Theorem 8.3.[

4. Use Theorem 8.3 to show that f„ (a') = irxi 1 - .v^)" converges pointwise hut not uniformly to the zero fiirtclion on (0. 11. The graph i.s similar to Mgure 8.2.

5. ].ei f„{x) ~ x"/ii on |(). 11. Show that converges uniformly to a dilTercniiable function on |(), 11, but docs not converge uniformly on (0. I).

6. Show that f„(x) = /.v- -f (I//1-) converges uniformly to |.t| on R. [Him: Rationalize to show that

/„(.v) -

< l/u for each n.)

7. Let f„ F uniformly on D with each /„ continuous on/). If.t is in Z) and (-3'r),ieN 'S a .sequence in D with .r„ -» .v. show that lim /„(.v„) = F(a).

8. If /„ F pointwise on D and each /„ is monotone increasing on D. show that F is monotone increasing on D. A similar re.sult hold.s for monotone decreasing.

180 Chapters Sequences and Series of Functions 9. Suppose that /„ ->■ F pointwise on [o. b], each /„ is monotone on [«, b], and F is continuous on Show that f„ ^ F uniformly on [Hini: F is uniformly continuous on [a. i"].] ' 10. Use the Fundamental Theorem of Calculus to provide a proof of Theorem 8.4 under the additional assumption that each /„' is continuous on / =

For.vinla.f?],/,^/; = fAx)-f„(.xo)Afguniformly

on 1«. /']. then Theorem 8.3 implies that lim {Lix) — /n(-vo)J = /,'0 S- h n—»5C follows that /„ -»• F pointwise on [a. b]. where Ffj) = lim f„(xo) + 5

C

/,], S- By Theorem 6.12. F'{x) = g(.r) on [a, b\. Now show that f„-* F uniformly on!«./'].]

8.3 Series of Functions We define convergence of a .scries of functions in terms of convergence of the corresponding partial sums. This allows us to obtain easily the series analogues of earlier theorems in this chapter for function sequences. We then obtain the

Weicrslrass ,V/-icsi, which gives sufficient conditions for a series of functions to converge uniformly; and we use this test to construct a continuous function on R that is nowhere differcnliablc.

Definition 8.3 Let (./".tliigfj be a sequence of functions defined on />. 0

0

Then E yif series of functions (or a function series) on D. For eacli n in N iitul x in P, let II

.v«(v) = Y^ft(x) = /|(-v) + fltx) + ••• + f„(x). l.=\

Then (.v,i)„gN is the function sequence of partial sums of the series of CC

3C

functions E /«• The function seines E fn converges pointwise (or is ;i«l

/7=I

pointwise convergent) to a function F on D, denoted by 90

5C

F = Y^f, ox F{x) = Y^fAx) ;t=l

"=l

for all -V in D. if the sequence of partial sums (i'n)„gN converges pointC C

wise to F on D. The function series E fn converges tmifonnly (or r t = l

is uuiformiy convergent) on D if the sequence of partial sums (inlnsN converges uniformly on D. 9

C

Remark In the notation of Definition 8.3, when E fn converges pointwise to F. then F is the limit of the sequence of partial sums and F is the sum of

Section 8.3 Series of Functions

181

the series E /„. If E f„ converges uniformly to F on D, we will sometimes nsl

1=1 c

c

write this as E /, = F uniformly on D. Clearly, uniform convergence implies 1 = 1

point wise convergence.

Example 8.10 For each n in N, define /„ : R ^ R by 2

For X ,6 0, = n s ]

X'

I +

(I +x'-r

+

I + 1 —

X'

I "(1/(1 +.t-)]

= I +X-. Hence. 3 C

F(x) = ^/„(.r) = 1 = 1

0 \+x-

if

x=0 if

x^O.

Thus, £ f„ converges pointwise to F on 1=

That this convergence is not

I

uniform is most easily seen from Theorem 8.5 below. We next state the analogues of Proposition 8.1 and Theorems 8.1, 8.3, and 8.4 for function series. Their validity follows by applying the hypothesis to the sequence of partial sums 2nd using the corresponding results for function sequences. 0 0

Proposition 8.2 (Cauchy condition) The function scries rEl = fn converges l uniformly on D if and only if for each e > 0 there is an hq in N such that if n > m > rtO'

E fk(x) i=ii>+|

< e for all x in D.

182

Chapter 8 Sequences and Series of Functions Proof Note that .v„(a)= L /t(.v) for/i > m and apply Proposi

tions.!.

t-'n+l •

^

^

Theorem 8.5 II' E f„ = F uniformly on D and each /„ is continuous on 'Is I

D, then F is continuous on D.

Proof Each s„ is continuous on D and -» F uniTonnly on D. Now apply ■

TheorctnS.I. n c

[f each /„ IS continuous at an accumulation point c in D and E /„ = F unifonnly on D, then Theorem 8.5 implies that

lim V /n(-v) = lim F{x) t

—

c

;i>l

= F(C) C C

txm\ O C

= Vlim/„(.t). r

-

*

r

n m \

Thus, in this silualion we can interchange the two operations oClimit and sum. 0

0

Theorem 8.6 If E S,^ = F imiformiy on ]«. /^] and each j'„ is in 72|«, /j|,

then r is in/)] and F = f'' E /„ = S /I*/„. (Thus, the function ls|

scries may be integrated term by term.)

isl

*

Proof Apply Theorem 8,3 to the sequence of partial sums (inlngn. ■ Theorem 8.7 Suppo.se that (/„)„gN is a .sequence of dilTcrcntiable functions defined on a bounded interval I and that E /nf.to) converges for some point X

5C

.To in /. If E /„ converges uniformly on I. then E /„ converges uniformly ;i

=

1

n-\ 0 0

on I to a diffcrentiable function F and F'(.r) = E for all x in /.

Proof Note that (.v,i(-vi)))„g|>j converges and that converges uniformly on

I.

Now

apply

Theorem

8.4.

■

The following theorem gives us a useful but different typo of result.

Theorem 8.8 (Wcicrsirass /W-test) Let be a sequence of functions on D. and let (MnlwsN a sequence of nonnegaiive numbers satisfying |/n(.r)| < oc

ac

M„ for all H in N and all .v in D. !f E M„ converges, then E fn converges uniformly on D.

Proof We use the Cauchy condition. Let e > 0. Since E converges, by tt=i

„

Theorem 7.1. there is an no in N .such that if n > m > /lo. then E Mt. < e. ksm-¥\

Then n > n\ > no implies E fkix)
it] implies

i e S

/w-/(^

.r*(l-X)

n—it

*o-^r* We now consider E v"* Choose M > 0 such that |/(a )1 < M for all x in k i T

[0. Ij. Iffc is in T. then |(/(-/;i) - .t| > l/ii'''-' and so [(it/ii) - .tp > l/zi'/^. Hence.

-V^l -.t)"-*

k i s r n - k »6T

i = 0

= {2Mh'/^) A(1

-.V)

[by (8)]

I M

Choose «2 in N such that I/^/iiT < e/(4 /12 implies

E >ek(x) k ^ T

e/2.

Let/Jo = max(/i|./i2l.Then/i > ;ioimplies,by(10).that|/(A) - B„(x)\ < e forallA in 10, Ij. Since/iowaschosen independently of.t, ^ / uniformly onIO,

1],

■

Theorem 8.10 (Weierstrass Approximation Theorem) If / is continuous on [a. (>], then there exists a sequence of polynomials that converges uniformly to / on {a.b]. Proof Define g on |0. 1] by g(r) = / 0, |g(f) - B„(l)\ < efor

190 Chapters Sequences and Series of Functions all I in(0. llandallsafTidcnlly large h. If.v is in [a, fe], Ihenr = {x-a)/{h-a) is in 10, I ] and g(r) = f (x). Hence. \f{x)-B„((x-a)l(b-i,))\ 0, once 5 and M (as in the proof of Theorem 8.9) have been found, we choose n such that

< 5 and IZ-Jn < e/(4M). For example, if fix) = ,/x on [0. I], then S = £-/8 (this is a good exercise for the reader) and M = I. Hence, we must

choose/i > max{4096/£*, 16/e-}. Ife = -j^. Ihen/i > 4096(10"). As in Exercise 4. however, it is often necessary to know only that there is a polynomial that uniformly approximates / on [a. h].

■

xcpcises

===g^==

1. Find B|. Bj. and Sj for (a) fix) = I - .t;

tb) fix) = xh (c) fix) = e'. 1. By considering /(.v) = I /x on (0. I), show that Theorem 8.10 would not hold if [a. b\ were replaced by («. h).

X Let fix) = jxj on \a. />], where a i

n=> 0

X

X "

x '

x^ +

to (2")! +

x)

■» (. _,)«X''+1 =

-

t

=

A (-irx^"^! =

„ 2/1+1 ~ _,2n+.

X

■

n = 0

nsQ

^ ^.2/1 a

s

(8)^ cosh.t =( 2y^, 1 ) !

x^

~ XT"

X

x-"

-^T - T " * -

/I + 1

(7) sinh jr = ^ ( 2 / 1 +

...

6!

x'

X5

3 "^

T-

x^

3! x"*

5!


that (1 + .r)'" = E C^.v" for an=0 "

in (— 1, 1), Although the reader may wish to try this using Theorem 9.6, we proceed in a completely different manner. For .r in (-1, 1) and p in R, let

fp{x)= E (iV''^yP'i"2ofCorollary9.1,

11 = 1

^

'

n=0

'

for j-rj < [.Since (/i + l)(„i;,) = p{'"„^). f'pix)

=

=

P/z-iW

(5)

for |a| < 1. Also for |.r| < 1, CC , ^ 1 \

(l+.t)/,,_l(A) = (l+A)^l''^^

^n+\

0

0

:rT V'/ = fp{x). t = j

204 Chapter 9 Power Series Combining this with (5), we have

(l+.r)/;,U)

=

/>/,,(.t)

(6)

for all X in (-1,1). One could now solve the dilTerentla! equation (6) directly

using the initial condition /,r(0) = 1. or one could observe that

^ [/p{.t)(i +xr"] = (1 +A)--'" [d +A)/;{A) - p/p(A)] = 0 by(6). Hence,/j,(a)/(1+A)''isaconstani function on (-1, D-LettingA =0,

we see that the constant value of this function is 1; that is. f,,{x) = (1 + a)'' for all A in (—1, I).

Let/J > 0 and not an integer. We now show that«E= o1(^)1 converges. Let (i„ = |(J^')| for u = 0, 1, 2 Then = (n - p)/{n + 1) whenever " S I/'] + 1, where [/;] denotes the largest integer less than or equal to p. For " > IflH- 1. (ji + l)n,i+i = iui„ - pa„ and so

-

(;i

+

l)a„+i

=/Jt;„

>

0.

(7)

Therefore, («a;i)^(p)+i is a decreasing sequence, and so it has a limit. Let lim na„ = y > 0. Since

y [na„ - (n + l)o„+!l = + l)rtn,+i ^ - X, m

n = i ) 0 0

the series E [no,, - (/i + I )a„+i 1 converges. By (7), a„ = 1 = 0 OO

.

(Ul>)[nOn - »11 lor /? > | /j| + 1, and so Ihc scriesn = I,0 o„ =H =I,0 ( J converges.

Since E ('') terminates if/) is a nonncgative integer, we have that E K^,) converges for all p > 0. Since < |(J|')| for |a| < I. E converges f f s O

absolutely on (-1, 11 for all p > 0 by the Comparison te.st. By Theorem 9.4

and Exercise 9 in Section 9.1, E (''}a" = (I + a)'' for all /? > 0 and for all a n

in|-l.ll.

Section 9.2 Taylor Series 205

- x e p c i s e s

1. Let

jf

,.^0

fi r ) = 0

if

.r

=

0.

Show that /""(O) = 0 for n = 0. 1, 2.3 2. Derive (7) of Table I.

3. (a) FindlheMaclaurinscricsanditsinicrvaioftotivcrgcncclor/(.v) = 2', (b) Find the Taylor series for«?'centered al I. [Hint: Use (2) of Table I for both part (a) and part {b).| 4. Suppose that / hasderivativesof all orders on an interval /. and lei c be an

interiorpointof /. Ifthereisan M > Osuch lhat |/'"'(.v)| < M" for all/i = 0, I. 2.3 and for all a- in /, show that f(x) = S l/'"'(c)/»!|(.v - f)" for all X in /.

5. (a) Derive (3) and (4) of Table 1.

(b) Replace .t by x — 7t/2 in (4) of Table 1 to obtain the Taylor series for sinj centered at nil. What is its interval of convergence?

6. Assume that (2) of Table 1 holds when x is a complex number. Replace aby ix. where i = and a is real, to show that t-'' = cos a + i sin a. Note that this implies 1 + e^' = 0.

7. Find the Maclaurin series and its interval of convergence for

8. Use (9) of Table 1 to obtain a series expansion of (3 + a)-'-'. What is the interval of convergence? 9. Use Raabe's test (Theorem 7,9) to show that the binomial series is ab

solutely convergent at a = I when p > 0. Conclude from this that the binomial series is absolutely convergent on [— 1. 11 when /> > 0, 10. (a) If /» < -1. show that the interval of convergence of the binomial series is(-l,I).

(b) If -1 < p < 0. show that the binomial seiies is ctinditionally con vergent at 1 and divergent at -1, Hence, its interval (tf convergence is (-1.1).

11. Derive the Maclaurin series and its interval of convergence for arcsiti.v. Use this to obtain 71

I

1

I

1-3

I

1.3.3

I

6 ~ 2 2 ' 3 -23 2~4 ' 5~2^ 2 • 4 • 6 ' [Hint: Use (9) of Table 1 to obtain a .series expansion of (1 -Letting p — — integrate this from 0 to a.)

1G Piemann-StiG Ijes Gq pa T he Riemann-Stieltjes integral on on interval [], then —/is nionototie increasing on \a, b], and so / = —(—/) is in '/2(a) on [u, /)| by part I of Proposition 10,). ■ The analogue ofTheorem 6.10, the Second Mean Value Theorem for Riemann integrals, is given in Exercise 6. In Theorem 10.9 below, we establish a remarkable connection between f da and a df under the appropriate hypothesis. First, we need a lemma.

Lemma 10.1 if / and a arc both monotone increasing on («, h\ and P is in "PUi. /)], then U(P, f.a) = f{b)a(h) - fUi)a(a) - L(P,a. f) and

L(P, f.a) = f(h)a{h) - /{(DaUi) - U(P. a. /).

Proof Let P = be in 'P{a.h\. Since / is monotone increasing on [rt./j]. I t

i/(/',/,a) = ^/(.t,)Aa, i s l

= /(-Vi)[a(.V|) -o(.r,))l + /(.v:) [af.rj) -a(.v,)] + • •• +

/(.r„_,)[a(.r„_|) - a(.r„_2)] + f(x„) [a(.v„) - o(.r„-i) = /(.v„)a(.r„) + a(.r„_i) [/(x-„_i) - /(.\:„)] + ••• + a(.vi)[/(.>:i) - /(.r:)] - /(.ri)a(,ro) = f(x„)a(x„) - /(.ro)a(.ro) + a(.V(i)|/(-ri)) - /(-ti/l + n - 1

Y^a(Xi) [/(j:,) - /(.t,.^!)] 1 s t 1 - 1

= f{b)a(h) - fUDaUt) - ^a(.r,) [/(.v, + i) - /(.r^)] i s O

= f(b)a{b) - f{a)a{a) - L{ P. a. f) since a is monotone increasing. The second equation follows by interchanging / and a in the first ■

equation.

Theorem 10,9 Suppose that / and a are both monotone increasing on\ci.b\. Then / is in 72(a) on [ti. />] if and only if a is in 72(/) on l].

4. Complete the priKif of Tlieorem 10.7 when / and or are both discontinuous from the right at c. where a 0 such that if P is in V with ||P|| < 5. then U(P.f.a) - HP, f.a) < e. But this is precisely what is done in the proof of Theorem 10.4. 3. Assume that / is in TZ(a) and or is continuous on («. fc). Let S > 0 with 1/U)! £ ^ for all -r i" [«• ^'1. and let e > 0. Since f da = f da,

there is a P' = {.r/1"_(, in f with

ViP'.f.a) < J/ / Osuch that if a: and _v are in [a. b] with |.r — _v| < 5|, then [a(.t) — ff(_v)| < £/(4/iB). Hence, if ||P|| < 5|. then each Ao,- < £/(4;iS). With the usual modilications as in part 1 above, the remainder of the proof is identical to the first part of the proof of Theorem 6.3. ■

Example 10.6 The purpose of this example is to show that the continuity assumptions made in Theorem 10.10 are necessary. Let 0

/(.V) =

if

0

]. Since a'is uniformly continuous oti jri. /r|. there is a 62 > 0 such that if .v and y are in [a, b\ with |.r — y| < 32,

then |tf'(.r) - o'(y)| < k/[2M(/) - «)|. If P = is in ■p\(i'.b\ with It/'II < 31. then (6) implies that

!«''■ /■ «>-«''■ /"'I ■= E= J' j a l

Combining this la.st inequality with (5) gives us that r h

S{P.f.a)- Jof fa"


]. If or(.t) = [.vj on (0. I], then a'Cr) = 0 for 0 < x < 1

and a'(l) dws not exist. By Theorem 6.5. defining a'(I) arbitrarily, a' is in

7J[0. I ]. However, for any continuous function / on [0. 1|, /J f da = /(I) bul/o'/a'=0. Remark One way to generalize what we have been doing would be to allow

our integrator a to be an arbitrary function )• delined on (o, />], and to delinc

/j" / df{ as the limit of Riemann-Stieltjes sums, S(P. f. g), as the mesh of P

tends to zero, whenever this limit exists. Our proof of Theorem 10.11 still holds in this generalized setting.

Example 10.7 f-T/2

JI o.r" ( y ) = J/fUi)

/

f(x)dx.

(7)

Ja

(Formally, this is obtained by letting x = a(y).] Proof By Exerci.se 8, a is strictly increasing and continuous on \fS(a), p(h\. Hence, both integrals in (7) exist and (4) of Theoretn 10.10 holds. l.et P = be in 'P[u.h\ and set y/ = fiiXi) for i = 0, 1,2,.,., n.

Then Q = {yllU's in Vma). /!(/»]. Also,

^ /(a(yi)) [or(yi) -or(y/_i)] = ^ /(.V/) 1=1

Xi — Aj_|

(8)

1=1

Since /I is uniformly continuous on [«,/>], ||f

0 itnplies ||(7|| 0.

Section 10.3 Riemann-Stieltjes Sums

225

Therefore, as ||/'|| —» 0, ihc two members of (8) lend to the corresponding members

of

(7).

■

Example 10.9 We reconsider JJ,' v" f/(v-) of Example 10.8. Letting fix) = ,v. fiix) = y/x, aiy) = )'• on |l). 11 (or simply .v = y^). Proposition 10.2 implies that I

^0

^

- x e P c i s G S

1. Use Theorem 10.11 to calculate |.r| (/(.r-) where ii is in N. Compare with Exercise 8 in Section 10.2.

2. Evaluate the following.

(a) /^''xdisinx) (b) /y"*.sin.vJCsinx) 3. Evaluate the following.

(a) f^x^tlix-) (b) 4. Evaluate the following. (a) CO.S- .r |, then both v and v — f are monotone increasing on 1«. /)|. Proof For w < a < y < h. Proposition 10.4 implies that !;(>) = iif.v) +

F, (a, y) or. et|ulvalent]y. i;(y) — n(A) = Vy(.v. y) > 0. Hence. i'(a) < i;(y) and so n is monotone increasing on |». b\. Also. IV-f) (y) -(V- f) (A) = |v(y) - t;(A)| - |/(y) - /(a)|

= V; ( - v. y ) - l / ( y ) - / { A ) J . From the detinition of F/(.v. y), using P = (a. yl as a partition of [a. y],

|/(y) - /(a)| < VfU. y).andso(ii - /) (y)-(u - /)(a) > O.Thus, u-/ is

monotone

increasing

on

jo.

/'].

■

Corollary 10.2 A function / is of bounded variation on [«. /^I if and only if f can be expressed as the difference of two monotone increasing functions on \a.h\.

Proof If /' is of bounded variation on |«. /;). then / = v — (u — /). The converse follows from Example 10.10 and Theorem 10.13. ■ Remark The decomposition of a function of bounded variation as the dif ference of two monotone increasing functions is not unique. If g is monotone increasing on 1«, l>\. then v -i- g and v — f 4- g are monotone increasing on [«, />| and / = (u 4-i') - (v - f + a).

Section 10.4 Functions of Bounded Variation

231

Corollary 10.3 If / is ot'bounded variation on [a. fo). then /(.v+) exists for u < X < h. /(-t—) exists foru < -v < h. and the set of discontinuities of / is countable.

Proof This follows from Corollary 10.2 and the corresponding results for monotone

functions

(Theorems

4.8

and

4.9).

■

Theorem 10.15 Let / be of botinded variation on 1«./;| and let c be in l|. Then / is continuous at c if and only if v is continuous at c. Proof First suppose that « < c < h and that / is continuous at c. We will show thai u is continuous from the right at c. Let e > 0. Then there exists a 5i > 0 such that if.v is in |o. f>) with j.r - cj < 5i, then |/(.v) - /(c)| < e/2.

By definition of V/(c, h), there is a /' = in Vic. b\ such that r j

Vfic.h)- ^ < J2\f{x^) - /(J:,-I)|. (11) f s l

Let 5 = min(6i, JTi — c}. and let .v > c with.v — c < S. From (11) we have g

"

Vf(c. b)--< |/(.v) - /(c)| + |/(.r,) - /(.v)| + |/(.Vi) - /(.v,_, )| 1 = 2

< I + I;' (.r. b) since (j:) U is in Vlx. h\. By Proposition 10.4,

u(.v) - i»(c) = V/(«,.y) - V,(rt. c) = V/(C..Y) = Vf(c.h)- Vf(x.b)


if

I

3-.r

if

<
0 such that (.r — s, .r -l-c) C G. The difference now is that neighborhoods need not be open intervals or even

open sets. Some people call (.r - E. x 4- e.) an E-neighborluiod of.v or a basic neighborhood of A'.

Example 11.6 1. |0, 11 is a ncighboriiood of eaeli point in (0, 1). but |0, 1| is not a neighbor hood of 0 or 1.

2. (0. I)U|21 isaricighborhoodofeuehpoint in (0, I) but is not a neighborhood of 2.

3. Z is not a neighborhood tif any of its points. Neither is N or Q. This is because none of these sets contains an open interval.

246

Chapter 11 The Topology of K 4. The Ciintor set is not a neighborhood of any of its points. See Exercise 8. 5. If G is a tieighborhoodof.v. then any set containing G is also a neighborhood of a .

Proposition 11.3 A set is open in R if and only if it is a neighborhood of

each of its points.

Proof If G is open iriRand.r is in G, taking U = C in Definition 11.3 shows that G is a neighborhood of x. Now suppose that G is a neighborhood of each of its points. For each .r in

G, by Definition 11.3. there is an open set (/, in R such that x 6 G, C G. By Proposition

II.

.r€6'

I,

G

=

U

G,

is

open

in

R.

■

From Exercise 2 in Section Il.l, it now follows that a set in R is a neigh borhood of precisely those points that lie in its interior. The reader is probably wondering how this new definition of neighborhood. Definition 11.3, affects earlier results in the text that involved the old definition

of neighborhood. Definition 3.3. It does not; the earlier results are still valid under Definition 11.3. This follows since now every neighborhood of .v must contain an old type neighborhood of.r. (.r - c-.v + £•), for some s > 0. For example, we restate Deliniiion 3.4 in our new setting.

Definition 71.4 The sequence (.r,i)„gN in R converges to the real iiuinhor.r iffor every neighborhood G of .v. the sequence is eventually in G.

Since the sequence is eventually in every neighborhood G of x if and only if (.v„)„gN is eventually in every basic neighborhood (.t — e,x + e) of .t, Deliniiions 3.4 and 11.4 are equivalent. Accumulation Points Definition 3.10 of an accumulation point carries over verbatim to our new set ting, as do all of the examples and results on pages 56-57. This is because everything on pages 56-57 was phrased in terms of neighborhoods. The.se results and definitions, which we now restate, are also summarized at the be

ginning of Section 4.3. Accumulation Point Results For A c S and .r in R, .v is an accumulation point of A if and only if every neighborhood of.r contains a point of A different from.v (that is, (G \ (a)) r\ A for all neighborhoods G of.v] if and only if cveiy neighborhood of.v contains infinitely many points of A

if and only if there exists a sequence of distinct points in A converging to a.

Section 11.2 Neighborhoods and Accumulation Points 247 Allhough we cannot characterize closed sets as we did open sets in Theorem 11.1. we can characterize closed sets in terms of sequences and accumulation points. Theorem 11.2 Let F C 3. The following are equivalent. 1. F is closed in R.

2. If is a sequence in F and .r is in R with converging to x, then .V is in F. (The reader should compare this with Theorem 3.3.) 3. F contains all of its accumulation points.

Proof Part I implies part 2. Let F be closed in R. and let (Xn)„gr< he a sequence in F that converges to a point a: in R. vSuppose that .v is not in F. ThcnR\Fisanopcnncighborhoodof.t. Since.v,, .v, the .sequence (.t,, )„gfj

is eventually in R \ F (Definition 11.4), which contradicts the fact that each x„ is in F. Therefore, x is in F.

Part 2 implies part 3. If .t is an accumulation point of F, then by the accumulation point results above, there is a sequence (of distinct points) in F that converges to .v. By pad 2. .v is in F.

Pad 3 implies pad 1. To show that F is closed, we will show that R \ F is tipen. Let y be in R \ F. By part 3, y is not an accumulation point of F. By the accumulation point results above, there is a neighborhood C of y such iliai G n F = fl. Therefore, y 6 C C R \ F. and so R \ F is a ncighborliood of y. Since y is an arbitrary element of R \ F. R \ F is a neighborhood of each of its points. By Proposition II .3. R \ F is open in R. ■ Example 11.7

1. j I : /I € N} is not closed in R since (I /« : n € N1 does not contain its accumulation point 0.

2. Q is not closed in R since every real number is an accumulation point of Q. 3. Since N and Z have no accumulation points, both N and Z are closed in R. We next characterize, in Theorem 11.3 below, the smallest (by containment) closed set that contains a given set. First, we need the following. Notation For A C R. let A' denote the set of accumulation points of A.

Proposition 11.4 A' is closed in R for each A C R. Proof Let A C R and let x be an accumulation point of A'. To show that A' isclo.scd in R, it suffices by Theorem II.2 to show that x is in A'. Thai is, we must show that .v is an accumulation point of A. Let G be a neighborhood of .v. By Dctinition 11.3. there is an open set V

in R such thai x € U C C. Since .t is an accumulation point of A' and U is a neighborhood of .t.

(G\f.i))nA'?tH. Let V e (G \ {.r})n A'. Since y is in A', y is an accumulation point of A. Since y 6 U and U is an open neighborhood of y. (/ 0 A is infinite. Since U C G. G

n

A

is

infinite.

Therefore,

x

is

in

A'.

■

248

Chapter n The Topology of K

Theorem 11.3 Lci ^ c R. Then_/t = A U 4' is closed in R. Moreover, if F is closed in R ami /t C F. then A c F, {A is called the closure of A in R. A is the smallest closed set in R containing A.)

Proof Let x be an accumulation point of A. By Exercise 2. either x is an accumulation point of A or.r is an accumulation point of A'. If .v is an accu mulation point of A. then x is in A'. If.r is an accumulation point of A', then

X is in A' by Proposition 11.4. Therefore, .r is in /\, and so A is closed in R by T h e o r e m 11 . 2 .

Let F be closed in S with A c F. If.r is an accumulation point of A, then, since A c F. x is an accumulation point of F. That is. A' c F'. ,Since F is closed in R, F' C F by Theorem 11,2. Therefore. A' C F and so A

=

AU

A'

C

F.

m

Example 11.8

1. [Ol = (oTT) = [(). 1], 2. Q=RandRTQ = S. 3. {I/«:«€N} = |I/h:;I€N|U(0|. At this point the reader may be wondering when a set cc|tials its set of

accumulation points. That is. when does A = A"1 By Proposition 11.4. such ;i set must be closed.

Definition 11.5 Let A C R. Then A is iicrfeci if A = A'. Example 11.9 1. If —CO < u < b < CO, then the closed interval [«, /;| is perfect. Also, la, oo) and (-oo, h\ arc perfect. 2. N. Z. and Q are not perfect.

3. A = [0. 1] U {2| is not perfect because 2 is an isolated point of A and not an accumulation pttint of A. (An isolated point was defined in Definition 3.J0.)

4. The Cantor set is perfect. That is, every point in the Cantor .set 0,

recall the length ofthe closed intervals in F„ to choose »i appropriately. Then the endpoints of this closed interval in F„ will both be in {x — x' + e).|

11.3 Compact Sets The reader can recall that in calculus there were many more theorems about continuous functions on a closed interval than on an open interval. In this section we show that a closed interval has a much stronger property than an

open interval—namely, that of compactness. The main result of this section is the Meinc-Borel Theorem, Theorem 11.4.

Definition 11.6 Let A be a subset of R and let t? be a collection of

subsets of R. Then C) is a cover of A (or G covers A) if A C Up. If 0 is a cover of A and if each member of G is open, then G is an open cover of A. If a subcoliection of G covers A. then this subcollection is called a siibcover of A.

250

Chapter 11 The Topology of R We are primarily inierested in open covers.

Example 11.10 Lcl Q = {(—«. n) : n e N}. Then Q is an open cover of R

with no finite subcoverof R; that is. no finite number of members of^ has tinion R (since a finite number of members of Q will only cover out to the maximum n). Also. 0 is an open cover of [0. I], and (-2.2) is a finite subcoverof [0, 1],

Example 11.11 Let Q = {(0, \/n) : n e N}. Then G is an open cover of (0. I) with a finite siibcovcr—namely. (0. 1).

Example 11.12 Let G = {(l/«. I) ; « = 2.3.4....). Then G is an open cover of (0. 1) with no finite subcoverof (0. 1).

Example 11.13 Let a and h be real numbers with a < h. Choose «o in

such that/?-(!/«,)) > a.Lc\G = {( «o). Then G is an open cover of |(;, h) with no linile subcover of [«,/»),

In Proposition 11.5 below, we e.stablish the fact that every open cover has a countable subcovcr. For the purpo.se of this section. let B be the collection of all open intervals in R with rational endpoinls. By Exerci.se 12 in Section 2.4. B is countable.

Lemma 11.1 For each open set U in Rand for each x' inf/, there is a member li of B with .r € B CU.

Proof Given x in the open set U. let («, h) be an open interval in R such that X e («, h) c U. Since Q is dense in R. there exist qi and beinR wi(h« ). [«. b). [«, y?i. (—00. />). (—CO. fr|. (c. oc), [«. oo). R.

Proof This cliaraciciization of (he connected subsets of R follows directly front

Theorem

11 . 6 .

■

Allowing {«! tobc the closed iniorval | 0. Le( x and v be in X wiih |.v - v| < 8. Then x is in C/j. for some

; in (1,2 ;i). Since |v-.Vi|) or («, 6| onto (a, (« + h)/2) or (a, (a + h)/2], respectively, with no fixed point in the interval, and g(x) = (x + b)/2 takes [rt, h) onto [(fl + b)/2. b) with no fixed point in \ii.h). ■

Exercises

■

—

In Exercises 1 to 6. X c R.

1. Complete the proof of Theorem 11.7 by .showing that parts 2 and 3 are

equivalent. [Him: /-'(R\ F) = X\/-'(F).| 2. Let / : X —» R be continuous on X, and let a be in R. Show that {x € X : /(x) < rt) and {.r e X : f(x) > «) are open in X and that {x € X : /(x) < a], ix e X : /(x) > «|, and (x € X : /(x) = a] are closed in X.

3. Let / and g be continuous functions from X into R. Then show that {.v 6 X : /(x) = g(x)) is closed in X. Conclude that if X = R and f = g on a dense subset of R, then f = g on R. [Iliiii: See Exercise 6 in Section 11.2.J

4. Let / be a continuous function from R onto R, Show that if A is den.se in R, then/(A) is dense in R. [Wt'/ir; Use Proposition ll.ll.| 5. U.se Propositions 11.11 and II. 12 to show that if A is a connected subset of R, then A is also connected. Then compare with Exercise 3 in Section 11 . 4 .

6. Show that a continuous function from [«, h\ into 1«, />| has a fixed point. In Exerci-ses 7 to 13, X and Y are subsets of R: and vse consider X and Y as

topological spaces, each endowed with its own relative topology induced by R (see Definition 11.8 and Exercise 9 in Section 11.3).

7. Let / be a function from X into Y. Show that the following are equivalent. (a) / is continuous on X.

(b) /~'(5) is open in X for each open set 6' in f. (c) /~'(F) is closed in X for each closed .set 7" in T.

264

Chapter 11 The Topology of R

8. A function / ; X —▶ 1'iscalled ano/7^/im«/> if/ft/) isopen in K for each open set t/ in X and is called a closed map if /(F) is closed in Y for each closed set F in X. Suppose that / is a one-to-one function from X onto Y. Show that the following are equivalent, (a) : K —»• X is continuous on Y.

(b) / is an open map. (c) / is a closed map. 9. Let / be a continuous function from X into Y. If X is compact, show that / is a closed map. U.se Proposition II.7,Theorem 11.8,and then Proposition 11.6.] 10. A hoiiH'omorphism from X onto K is a continuous one-to-one function /

from X onto Y such that /"' is continuous on Y. Show that if / is a continuous one-to-one function from X onto Y and X is compact, then / is a homeomorphism from X onto Y. 11. Two topological spaces are called humeomorphic if there exists a home omorphism from one onto the other. In Example 1.11 we actually con structed a homeomorphism from R onto the open interval (0, 1). A topologist considers homeomorphic spaces to be identical because they have the same topological properties; that is, they are lopologieally indistinguish able. For example, a lopologist considers R and (0, I) to be the same topological space. Show that (0. 1] and (0,1) are not homeomorphic. 12. Let X be connected. A point c in X is called a cut point of X if X \ (c) is disconnected. Show that if / is a homeomorphism from X onto Y. then c in X is a cut point of X if and only if f(c) is a cut point of Y. 13. Show that neither [0. 1) nor (0, 11 is homeomorphic to (0. I). Are [0. 1) and (0. 11 homeomorphic?

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Aposlol. T. M.. Mtilhcmdlktil Analysis, Addison-Wesley, Reading, MA. 1964. Burtle, R. G. and Slierl>en, D. R.. Iiiimdiiciion to Real Analysis, Second Edition, John Wiley and Sons, New York. 1992. Tridy, J. A., IninHliiciory Analysis, Harcourt Brace Jovanovich, Sati Diego, 1987.

Gclbaum. B. R. and OInisted, J. M., Ctiunierexainples in Analysis. Holden-Day. San Francisco, 1964. Hewitt, E. and Stmmberg, K.. Real and Ahstraa Analysis, Springer-Verlag, N e w Yo r k , 1 9 6 3 .

Kirkwood, J. R., An Introduction to Analysis, Second Edition, PWS Publishing Company, Boston, 1995. McArthur, W. G., Am Introduction to the Art of Mathematical Proof, Shippensburg Collegiate Press, Shippensburg, PA, 1969. Rudin, W., Principles of Mathematical Analysis, Second Edition. M c G r a w - H i l l . N e w Yo r k . 1 9 6 4 .

Simmons, G. P., Calculus Gems, .McGraw-Hill. New York. 1992.

Spiegel. M. R.. Advanced Calculus, Schauiti's Outline Series, McGraw-Hill, N e w Yo r k . 1 9 6 3 .

265

Hints and A n s w G P S Section 1.1 3. II'rt 0, ihcn(/^ + ^- > 0. 6. ir« > Oand 1/h 1. Let 112 be the first positiveintegergreaterthan«i with.Xn, > 2.Continue.getiing.v,n, > /c at each stage.

Hints

and

Answers

271

Section 3.4

!• (-'•'n)n€N is bounded above by 2 and strictly increasing, .tn 2. .t„ ^ I 3. .v„ 2

6- l//i| -0andn,r=iU - d/'O. D = « 8. (c) Let £ > 0. Choose jiq e N with /j,,,, - < e. Then 0 < - a < ^'"0 ~ ""0 Section 3.5 1. [0,71

3. First use Theorem 3.10. Then use Proposition 3.3. 4.

R

5. ( I/m)„€nU( I +(l//t)},ieN'-l{2+(l/M)l„6N hasO. 1. and 2 as accumulation points. 7. Sequences that are eventually the con.slant x. Section 3.6 2. (a) Combine Exerci.sc 4 in Section 3.4 and Theorem 3.12.

3. Let £ = [in Delinition 3.11. Then eventually |*„ — Xml < I. 4. (e) For m > ii. j.Vni -A■„ is 00, the result is clear. So assume that p < 00. Let

be a subsequence of (y„),ieN with y„j, —> p. The corresponding subsequence of (x„)„eN has a subsequence with limit in R*. Since -r'H 5 y'tn this limit must be < p. Section 4.1 L Choose & = s.

2. Choose ^ = min{|ti/2, c^£/2|. 4. i-"or /(c) < /i, let e = h ~ /(c) > 0. Then 3 a neighborhood U of c such that X e L/ n /7 => /(x) e (/(f) - e, /(c) +f). 5 . L e t fi

= 1.

6. is the composition of two continuous functions. 9. To show that fg is continuous at c, note Chat |/(x)g(x) - /(c)#(c)| < |/(x)| - g(c)| + [g(c)| |/(x) - /{c)|

and / is bounded on a neighborhood of c by Exercise 5. Now make each suimnand less than e/2. Section 4.2

2. For c e Q, let (x„),i6n be a .sequence in R \ Q with .r„ —> c. Then /(.V,.) -»■ 0 5^ /(c). For c 6 R \ Q. let (yn)ii6N be a sequence in Q withvn -» c. Then /(y„) ^ 1 56 /(c).

4. To show that / is continuous at ^. let 3 = e. To show chat / is discontinuous at c ^ proceed as in Exercise 2. 6. Forx e R \ Q. let (x,i)„gN be a sequence in Q with .v„ -» x.

7. Let /(.v) = 1/x and consider (l/'i)^-..

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273

Section 4.3

2. Fore € R. lei (-v„)„6n be asequence in Q\ (c) and (>'„)ngN be a sequence in (R\Q)\{fl with.r„ ^ cand v„ -▶ c. Then 1 and/(.v„) -* - I. Use Proposition 4.6. 4. If lim f U) < ft, llien /(.t) < a for x ^ c in some neighborhood of c. 6. Lei V = (/, - 1, Z, + I), where L = lim /(.r). 7. Let V = (/./2. (i/2)L). where L = lim fix). .t

—r

1(1. (a) /(()) = /(O) + /(O) => /(O) = 0. f{2) = /(1) + /(I) = 2/(I). Vh € H. 0 = /(O) - fl-n + n) = fl-ii) + /(/i). so that fl-n) = - fi l l ) . (h) l-ei /) € Z and I). /(I) < 0. fl2) > 0. Use the Intermediate Value Theorem. 5. Recall Example 4.15 and u,se the Intermediate Value Theorem.

7. Suppose that / is nonconstant and use the Intermediate Value Theorem.

8. Define g : |0. 5] -♦ R by g(.r) = fix) - fix + i). If ^(0) ^ 0. apply the Intermediate Value Theorem to g. Section 4.5 I. Let E = I and 5 > 0. Let .v > 0 and let y = x + (5/2). Show that

|/(-v) - /(y)| > (5/2)(3.t^). Now choo.se .v. .3. {l/((;r/2) + H7r])„eN is Cauchy, but its image under / is not Cauchy. 5. On 11. CO). 5 = s/2 will work. On (0. 11, consider the Cauchy sequence (l/")n€N.

8. Suppose that / is not bounded on I). Then 3 a .sequence (.r„)ngf( in D with |/(.v,i)| > n Vh € N. Use the Bol/.ano-Weicrstrass Theorem to obtain a Cauchy subsequence of (.r„)„6f(. 9. For fg, let fix) = /;(.r) = .r on S.

Section 4.6

1. / has a discontinuity of the first kind at each integer. Also, / is monotone increasing.

2. / has a discontinuity of the first kind at each integer. / is not monotone.

274 Hints and Answers

4. / has a discontinuity of the second kind at every c ^0. 9. Let /(j:) = laiiA'on (—jr/2. jr/2). The difference between this and The orem 4.8 is that —7r/2 and 7t/2 arc not in the domain of /. 10. If / is discontinuous at r [f(x) - f(c)]/(x - c) > 0 and a < c ^ [/(a) - f(c)]/ (a - c) > 0. Hence, /'(c) > 0. 11. No such function exists by Tlieoretn 5.3. Section 5.2 1. (a) and (b) follow from the Mean Value Theorem, while (c) follows from

the intermediate value property of /'. 3. Use the Mean Value Theorem on Iaj, -VtJ. where A| and at are two consec

utive roots of /. Example: fix) = x^ + 1. 5. Use the Mean Value Theorem to show that 's Cauchy. 7. For a < x < b. the Mean Value Theorem implies that [/(a) — /(«)!/ (a — fl) = f'(Ci), where ti < r, < A. Lei x a.

«. fix) = ^ II. For A > I, apply the Mean Value Theorem to fir) s= In r on [1, .v|. Section 5.3

a' a-'' a'' „ (sinc)A*

1. Piix) = ~ ^ ~ between 0 and a,

2. P^^x) = ' ~ ^ ^ ~ c between 0 and a. A^ * A^ 3. PAX) ^A-+ Ay-- ^ - +A ^-- -A ^+ -.A *RAx) = 0 < C < A.

4. For sin A and cos a, ;; =9, For ln(l + a), n = 10''.

—A^

Hints

and

Answers

275

Section 5.4 1. —oo 2. 0 3. I 4. 5.

2

6. 0. NolclliatO < i'/7T < 1. 7. 0 8.

Section 6.1

2- /o / - I 3. U(P. f) - UP. f) < 2 \\P{\ ^P- / = 1

4. /) < 6 ||/'|| wlieiiever I e P'.ff~ f =0 5. J(P, f) = I Lei 0 < e < I and choose n such thai \/ii < e/2. Let I P

0.

=

II

I n

3

=

MU-

2

V; = I II - 1. add Vj < z, between .t, and .r,>i with V; - -^1 < £'/4n and .v,-i - c, < £/4ii. For this retincment. the upper sum minus the lower sum is less than e. / = I.

7. L(P. f) = 0 VP, Show that U(P. /) > ;; VP by passing to a relinement containing 5. Section 6.2

2. Jo .V d.v 3.

-2)| sin .vi H 1-> 1,

then Pi + • •• + Pn, - F(-v).

9. Let £• > 0 and choose 0 < 5 < - a) hy the unitbnn continuity of

/•' corresponding to tt/S. Consider the partition {.v, lf^(J of [ci, f>), where .V() = a, = /), and .V; = a + i{5/2) V/ = 0, I, 2 k. where k is lite smallest positive integer such that h. Then use the pointwise convergence of (/„(-ti))„6pj Vi = (), 1,2 t 4- !. Section 8.3

1. (d) |/„(.r)| I. show that f'(x) = 0 on [«. b\. For a = 1, V/(n. b) < M{h~a).

(b) Note that if / satisfies a uniform Lipschitz condition on [a, 6], then / is continuous on [a. b\.

Section 10.5 1. (a) - 2

(b) 5

2. /J* \x\ (Iv = 71 6

3.

Itt

—6+cos

i

1 = 1

5.

/■I '

r

,

rb

< M V^.^Aa.b)

So /''"-fa /''"'■

8. ForP = {xAU^'P\"-''U rb

f sf'la-S{P,8,ft) = '^f lgU)-g{ti)]f(i) 0. By (a) there is an a in A with .r—£a] = /-'({«, c»)).

Hints and Answers 283

3. U € A- : /(.V) =A'(.v)) = (/-g)-'({0|)

7. (a) =?• (b). If 5 is open in V. then 5 = V n V. where V is open in K. Uy Theorem 11.7, is open in X. .Show that /~'(5) = /"'(V), (c) => (a). Given F elosetJ in K. /-'(/■') = /"'(/•• n Y) is clo.sed in X by (e). 1(>. Use Exercises 8 and 9.

11. |(), I ] is compact but (0. 1) is not compact. A homeomorphism must pre serve compactness.

13. Every point of (0, I) is a cut point of (0. I). whereas 0 is not a cut point of |{), 1). Also, fix) = I - is a honieoinorpliism from 10, 1) onto (0, I j.

n d e x

Abel's Partial Suniniaiion Formula, 153

first kind. 91

jump discontinuity. 91

range, 10 represented by a power series.

absolute value. 21

removable. 91

aecuinulation poinl. 56, 77.

second kind. 91

rcslriciion. 13

simple. 91

Fibonacci sequence, 39

right-hand limit. 90 .Squeeze Theorem. 82 step function, 126 slriclly decreasing, 91 strictly increasing. 48, 91

field axioms for the reals. 19

total variation. 226, 229

first derivative, 98

uniformly continuous, 86-89.

246-249

antiderivative. 131

Archimedean Principle. 2, 28

extended real number system, 23

Axiom of Choice, 36

Bernoulli's inequality. 18. 42 Bernstein Approximation Theorem. 188

function. lO-l I

198

103, 260

function sequence. 167

hiconciilional statement. 2

absolute maximum. 83 absolute minimum. 83

Bolzano-Weierslrass Theorem for

additive. 83

l i m i t fi i r i c l i o i i . 1 6 7

analytic. 199 bijection. 12

poinlwisc convergent. 167 uniformly bounded. 173 uniformly convergent. 170

Bernstein polynomial. 187

sequotKCs. 55 Bolzano-Wcierstrass Theorem for sets. 57

Cantor diagonal iziilion argument. 36

Cantor set, 242-245, 249, 254 cardinal nutnber. 31

Cauchy-Bunyakovsky-Schwarz. inequality, 126

bounded.29 of hounded variation. 226

Ciiiichy condilioii, 172

furiclion series. 180

composition, 13

Caiichy condition. 181

contiiuuHis. 69, 70. 258

Dirichlet's test, 186

continuous extension. 80. 82.

ptirlial sums. 180 pointwise convergent. 180 uniformly convergent. 180

88

differentiable, 95

direct image. 11

Wcierstrass M-lesi. 182

complete. 28-29

discontinuous. 69 domain. 10 extension. 13

Fundamental Theorem of

Completeness Axiom for the

fixed point. 85. 104.262. 263

Gamma function. 141

reals, 25 conclusion, I

inverse, 14

G e n e r a l i z e d M e a n Va l u e

conditional statement. I

left-liaiid limit. 90

contradiction. 3

limit. 78. 81

Hciiie-Borel Theorem. 251

contrapositivc. 3. 12

local maximum. 98

hypothesis. I

converse, 2

local minimum. 98

chain rule, 97

decimal representation. 20 DcMoivre's Theorem, 18 DeMorgan's Laws, 7, 9 derivative, 95 discontinuity, 91

inverse image, 11

map. 10 monotone. 91-94

monotone decreasing. 91 monotone increasing. 91 one-to-one. 12 onto. 12

Calculus. 131

Theorem. 101

implication, I improper Riemami integral. 1-35-141

absoltitcly convergent. 137 Comparison test, 137 conditionally convergent, 137 285

286

Index

convergent. 136. 139 divergent. 136, 139 Limit Compiirison test. 137 indeterminate form, 107

induction hypothesis. 16-18 intennediale value property. 8.S. 94.99 I n t e r m e d i a t e Va l u e T h e o rcni. 84.

refinement. 112

Riemann integral. 114 First Mean Value Theorem. 127

integrand, 114 integration by parts, 133 interval of integration. 114 lower Riemann integral, 112

monotone decreasing. 51 numolone increasing. 51 Monotone Subsequence Theorem. 53

lower sum, 112

Squeeze Theorem. 47 strictly decreasing, 5) strictly increasing. 51 subsequence. 48

Riemann integrable, 114

tail. 40

irrational miniber. 3

R i e m a n n s u m . 11 7

term. 39

isolated point, 56

Second Mean Valtie Theorem.

85

129

l.'l-lopital's rule. 107 Leibnitz's formula. 101

upper Riemann integral. 112 upper sum. 112

Riemann-Stieltjes integral. 208, mathematical induction. 15-18

233

M e a n Va l u e T h e o r e m . 1 0 2

First Mean Value Theorem. 215

ntcsh (of a partition), 115 Minkowski's inequality. 126

improper. 214 integrand. 209 integration by pans. 218 integrator. 209 interval of integration. 209 lower Riemann-Stieltjes

/ith derivative, 98

neighborhood, 40. 61. 245 Nested Intervals Theorem. 53.252

nonii (of a partition). 115

integral, 208 lower sum. 208

order axioms for the reals, 20

ordered pair, 9. 11

Riemann-Stieltjes integrable, 2{}8. 233

Rientann-Stieitjes sum. 221 partition. 111 Pigeonhole Principle. 5. 32 polynomial. 46. 73. 81 power series. 191

S e c o n d M e a n Va l u e T h e o r e m . 218. 236

Abel's limit theorem. 195

upper Riemann-Stieltjes integral, 208 upper sum. 208

binomial series. 202-205

Riemann /eta function. 186

c o e f fi c i e n t s . 1 9 1

Rolle's Tlioorem. 101

converges uniformly. 193 expansion. 198

second derivative, 98

interval of convergence, 192

scqiience. 33. 39

Maclaurin ,Series. 199

bounded, 44

Cauchy. 58. 87. 88

Ratio test. 197

convergent. 40

Root test. 191

divergent. 40

rational function. 73 rational niimhcr, 3

Abel's Partial Summation

Formula. 153 Abel's test. 155

ab.soluiely convergent. 150 alternating. 148 alternating harmonic, 149 Alternating Series lest, 149 Cauchy condition. 145 Cauchy product. 16). 196 Comparison test. 146 conditionally convergent. 150 convergent, 144 Dirichlet's test. 1.54

divergent. 144 geometric. 144 harmonic, 145

inner product .series, 165 Integral test. 146 Kummer's test. L53

Limit Comparison lest. 147 Menens's Theorem. 162 /Ith term, 143

/'-series. 146 partial sum. 143 Raube's lest. 152 Ratio test. 150

radius of convergence. 192

Taylor .scries. 199 prime number. 4 primitive, 131

series. 143

limit. 40. 61

rearrangement, 157 regrouping. 156 Root lest. 151 sum. 144

telescoping. 144 set. 5

limit inferior. 65

arbitrary collcclioii. 8 associative properly. 6

limit superior. 65

bounded. 24

monotone. 51

bounded above. 24

Monotone Convergence

bounded below. 24

Theorem. 51

Cartesian product. 9

Index

commutative propeily, 6

Heine-Borel Theorem, 251

complement. 7

proper subset, 5 sequence of subsets. 52

countable. 34

smallest element. 25

homeomorphic. 264 hotncomorphism, 264

couniably infmile, 34

subset, 5

Interior. 244

dense, 28. 249

supremum, 24 symmetric difference. 10

Lindeldf property, 250

disjoint, 6 distributive property, 6, 10

unbounded above, 2

element. 5

uncountable, 34

empty .set, 5

union, 6, 8

equality, 5

upper bound. 24

equivalent sets, 31

void set, 5

finite. 32

greatest element, 25 greatest lower bound, 24 index set. 8

indexed collection. 8

Taylor polynomial, 105 Taylor's Theorem, 104 topological space, 239 basic neighborhood. 245

interior point, 90. 244

Cantor set, 242-245, 249, 254 closed in R. 241 closed in X, 254

intersection. 6, 8

closed map, 264

least upper bound, 24

closure. 248

lower bound, 24

compact. 250-252, 260 component, 257

infinnim, 24 i n fi n i t e . 3 2

tncmber. 5 nested downward. 52

connected. 254. 261

nested upward, 52

cover, 249

null set, 5

cut point, 264

point, 5 power set, 37

disconnected. 254

discrete spttce, 254

287

neighborhood. 245 nowhere dense. 249

open cover, 249 open in R, 239 open in X, 253 open map. 264 perfect, 248, 249 relative topology, 252 subcover, 249

topology. 239 totally disconnected, 258 total variation function, 229

transitivity, 20 triangle inequality. 22 trichotomy, 20 uniform Lip.schitz condition, 233 Weierstrass Approximation Theorem, 189

well-ordered, 16, 36