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INTERNATIONAL SERIES IN PURE AND APPLIED MATHEMATICS
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McGraw-Hill, Inc. New York St. Louis San Francisco Auckland Bogota Caracas Lisbon London Madrid Mexico City Milan Montreal New Delhi San Juan Singapore Sydney Tokyo Toronto
WALTER RUDIN
Professor of Mathematics University of Wisconsin-Madison
Principles of Mathematical Analysis
THIRD EDffiON

This book was set in Times New Roman. The editors were A. Anthony Arthur and Shelly Levine Langman; the production supervisor was Leroy A. Young. R.R. Donnelley & Sons Company was printer and binder.

This book is printed on acid-free paper.

Library of Congress Cataloging in Publication Data

Rudin, Walter, date Principles of mathematical analysis.

(International series in pure and applied mathematics)

Bibliography: p.

Includes index.

1. Mathematical analysis. I. Title.

QA300.R8 1976

515

75-17903

ISBN 0-07-054235-X

PRINCIPLES OF MATHEMATICAL ANALYSIS
Copyright © 1964, 1976 by McGraw-Hill, Inc. All rights reserved.
Copyright 1953 by McGraw-Hill, Inc. All rights reserved. Printed in the United States of America. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of the publisher.
28 29 30 DOC/DOC O9 8 7 6 5 4 3 2 1 0

CONTENTS

Preface

ix

Chapter 1 The Real and Complex Number Systems

1

Introduction

1

Ordered Sets

3

Fields

5

The Real Field

8

The Extended Real Number System

11

The Complex Field

12

Euclidean Spaces

16

Appendix

17

Exercises

21

Chapter 2 Basic Topology

24

Finite, Countable, and Uncountable Sets

24

Metric Spaces

30

Compact Sets

36

Perfect Sets

41

vi CONTENTS

Connected Sets

42

Exercises

43

Chapter 3 Numerical Sequences and Series

47

Convergent Sequences

47

Subsequences

51

Cauchy Sequences

52

Upper and Lower Limits

55

Some Special Sequences

57

Series

58

Series of Nonnegative Terms

61

The Number e

63

The Root and Ratio Tests

65

Power Series

69

Summation by Parts

70

Absolute Convergence

71

Addition and Multiplication of Series

72

Rearrangements

75

Exercises

78

Chapter 4 Continuity

83

Limits of Functions

83

Continuous Functions

85

Continuity and Compactness

89

Continuity and Connectedness

93

Discontinuities

94

Monotonic Functions

95

Infinite Limits and Limits at Infinity

97

Exercises

98

Chapter S Differentiation

103

The Derivative of a Real Function

103

Mean Value Theorems

107

The Continuity of Derivatives

108

L'Hospital's Rule

109

Derivatives of Higher Order

110

Taylor's Theorem

110

Differentiation of Vector-valued Functions

111

Exercises

114

CONTENTS vii

Chapter 6 The Riemann-Stieltjes Integral

120

Definition and Existence of the Integral

120

Properties of the Integral

128

Integration and Differentiation

133

Integration of Vector-valued Functions

135

Rectifiable Curves

136

Exercises

138

Chapter 7 Sequences and Series of Functions.

143

Discussion of Main Problem

143

Uniform Convergence

147

Uniform Convergence and Continuity

149

Uniform Convergence and Integration

151

Uniform Convergence and Differentiation

152

Equicontinuous Families of Functions

154

The Stone-Weierstrass Theorem

159

Exercises

165

Chapter 8 Some Special Functions

172

Power Series

172

The Exponential and Logarithmic Functions

178

The Trigonometric Functions

182

The Algebraic Completeness of the Complex Field

184

Fourier Series

185

The Gamma Function

192

Exercises

196

Chapter 9 Functions of Several Variables

204

Linear Transformations

204

Differentiation

211

The Contraction Principle

220

The Inverse Function Theorem

221

The Implicit Function Theorem

223

The Rank Theorem

228

Determinants

231

Derivatives of Higher Order

235

Differentiation of Integrals

236

Exercises

239

Chapter 10 Integration of Differential Forms

245

Integration

245

.uJ OONTBNTI

Primitive Mappings

248

Partitions of Unity

251

Change of Variables

252

Differential Forms

253

Simplexes and Chains

266

Stokes' Theorem

273

Closed Forms and Exact Forms

275

Vector Analysis

280

Exercises

288

Chapter 11 The Lebesgue Theory

300

Set Functions

300

Construction of the Lebesgue Measure

302

Measure Spaces

310

Measurable Functions

310

Simple Functions

313

Integration

314

Comparison with the Riemann Integral

322

Integration of Complex Functions

325

Functions of Class ft'2

325

Exercises

332

Bibliography

335

LJst of Special Sym~I•

337

Index

339

PREFACE
This book is intended to serve as a text for the course in analysis that is usually taken by advanced undergraduates or by first-year students who study mathematics.
The present edition covers essentially the same topics as the second one, with some additions, a few minor omissions, and considerable rearrangement. I hope that these changes will make the material more accessible amd more attractive to the students who take such a course.
Experience has convinced me that it is pedagogically unsound (though logically correct) to start off with the construction of the real numbers from the rational ones. At the beginning, most students simply fail to appreciate the need for doing this. Accordingly, the real number system is introduced as an ordered field with the least-upper-bound property, and a few interesting applications of this property are quickly made. However, Dedekind's construction is not omitted. It is now in an Appendix to Chapter l, where it may be studied and enjoyed whenever the time seems ripe.
The material on functions of several variables is almost completely rewritten, with many details filled in, and with more examples and more motivation. The proof of the inverse function theorem-the key item in Chapter 9-is

X PREFACE
simplified by means of the fixed point theorem about contraction mappings. Differential forms are discussed in much greater detail. Several applications of Stokes' theorem are included.
As regards other changes, the chapter on the Riemann-Stieltjes integral has been trimmed a bit, a short do-it-yourself section on the gamma function has been added to Chapter 8, and there is a large number of new exercises, most of them with fairly detailed hints.
I have also included several references to articles appearing in the American Mathematical Monthly and in Mathematics Magazine, in the hope that students will develop the habit of looking into the journal literature. Most of these references were kindly supplied by R. B. Burckel.
Over the years, many people, students as well as teachers, have sent me corrections, criticisms, and other comments concerning the previous editions of this book. I have appreciated these, and I take this opportunity to express my sincere thanks to all who have written me.
WALTER RUDIN

1
THE REAL AND COMPLEX NUMBER SYSTEMS
INTRODUCTION A satisfactory discussion of the main concepts of analysis (such as convergence, continuity, differentiation, and integration) must be based on an accurately defined number concept. We shall not, however, enter into any discussion of the axioms that govern the arithmetic of the integers, but assume familiarity with the rational numbers (i.e., the numbers of the form m/n, where m and n are integers and n =I- 0).
The rational number system is inadequate for many purposes, both as a field and as an ordered set. (These terms will be defined in Secs. 1.6 and 1.12.)
For instance, there is no rational p such that p2 = 2. (We shall prove this
presently.) This leads to the introduction of so-called "irrational numbers" which are often written as infinite decimal expansions and are considered to be "approximated" by the corresponding finite decimals. Thus the sequence
1, 1.4, 1.41, 1.414, 1.4142, ...
"tends to J2." But unless the irrational number j2 has been clearly defined,
the question must arise: Just what is it that this sequence "tends to"?

2 PRINCIPLES OF MATHEMATICAL ANALYSIIJ

This sort of question can be answered as soon as the so-called "real number system" is constructed.

1.1 Example We now show that the equation

(1)

p2 =2

is not satisfied by any rational p. If there were such a p, we could write p = m/n
where m and n are integers that are not both even. Let us assume this is done. Then (1) implies

(2)

= m2 2n 2 ,

This shows that m2 is even. Hence m is even (if m were odd, m2 would be odd), and so m2 is divisible by 4. It follows that the right side of (2) is divisible by 4, so that n2 is even, which implies that n is even.
The assumption that (1) holds thus leads to the conclusion that both m and n are even, contrary to our choice of m and n. Hence (1) is impossible for rational p.
We now examine this situation a little more closely. Let A be the set of all positive rationals p such that p2 < 2 and let B consist of all positive rationals
p such that p2 > 2. We shall show that A contains no largest number and B con-
tains no smallest. More explicitly, for every p in A we can find a rational q in A such that
p < q, and for every p in B we can find a rational q in B such that q < p. To do this, we associate with each rational p > 0 the number

p2 -2 2p + 2

(3)

q=p- p+2 = p+2.

Then (4)

2 2 - 2(p2 - 2)
q - - (p + 2)2 •

If p is in A then p 2 - 2 < 0, (3) shows that q > p, and (4) shows that q2 < 2. Thus q is in A.
If pis in B then p2 - 2 > 0, (3) shows that 0 < q < p, and (4) shows that q2 > 2. Thus q is in B.

1.2 Remark The purpose of the above discussion has been to show that the rational number system has certain gaps, in spite of the fact that between any two rationals there is another: If r < s then r < (r + s)/2 < s. The real number system fills these gaps. This is the principal reason for the fundamental role which it plays in analysis.

THB UAL AND COMPLBX NUMBER SYSTEMS 3

In order to elucidate its structure, as well as that of the complex numbers, we start with a brief discussion of the general concepts of ordered set and field.
Here is some of the standard set-theoretic terminology that will be used throughout this book.
1.3 Deftnltlou If A. is any set (whose elements may be numbers or any other objects), we write x e A. to indicate that x is a member (or an element) of A..
If x is not a member of A., we write: x ¢ A.. The set which contains no element will be called the empty set. If a set has at least one element, it is called nonempty. If A. and B are sets, and if every element of A. is an element of B, we say that A. is a subset of B, and write A. c B, or B :::, A.. If, in addition, there is an element of B which is not in A, then A is said to be a proper subset of B. Note that A. c A. for every set A.
If A. c B and B c A., we write A. =B. Otherwise A. ,,= B.
1.4 Deflnldon Throughout Chap. 1, the set of all rational numbers will be denoted by Q.

ORDERED SETS

1.5 Deflnltion Let S be a set. An order on Sis a relation, denoted by <, with the foilowing two properties:

(i) If x e S and ye S then one and only one of the statements

is true.

x<y, x=y, y<x

(ii) If x, y, z e S, if x < y and y < z, then .x < z.

The statement "x < y" may be read as "xis less than y" or "xis smaller than y" or "x precedes y".
It is often convenient to write y > x in place of x < y.
The notation x S y indicates that x < y or x =y, without specifying which
of these two is to hold. In other words, x S y is the negation of x > y.

1.6 Deflnidon An ordered set is a set S in which an order is defined. For example, Q is an ordered set if r <sis defined to mean thats - r is a
positive rational number.

1.7 Definition Suppose S is an ordered set, and E c S. If there exists a
Pe S such that x S Pfor every x e E, we say that Eis bounded above, and call Pan upper bound of E.
Lower bounds are defined in the same way (with ~ in place of S ).

4 PRINCIPLBS OF MATHBMATICAL ANALYSIS
1.8 Deflnldon Suppose S is an ordered set, E c: S, and E is bounded above. Suppose there exists an ex e S with the foilowing properties:
(i) ex is an upper bound of E.
(ii) If y < ex then y is not an upper bound of E.
Then ex is called the least upper bound of E [that there is at most one such ex is clear from (ii)] or the supremum of E, and we write
IX= supE.
The greatest lower bound, or i,ffimum, of a set E which is bounded be]ow is defined in the same manner: The statement
ex= inf E
means that ex is a lower bound of E and that no /J with /J > IX is a lower bound
of E.
t.9 Examples
(a) Consider the sets A. and B of Example 1.1 as subsets of the ordered set Q. The set A. is bounded above. In fact, the upper bounds of A. are exactly the members of B. Since B contains no smallest member, A. has no least upper bound in Q.
Similarly, Bis bounded below: The set of all lower bounds of B consists of A. and of all re Q with r S 0. Since A. has no lasgest member, B has no greatest lower bound in Q. (b) If IX= sup E exists, then ex may or may not be a member of E. For
instance, let E1 be the set of all r e Q with r < 0. Let E2 be the set of all
r e Q with r S 0. Then
sup E1 = sup E2 =0,
and 0 ¢ Ei, 0 e E2 •
(c) Let E consist of all numbers 1/n, where n = 1, 2, 3, .... Then sup E = 1, which is in E, and inf E = 0, which is not in E.
t.10 Deflnltlon An ordered set Sis said to have the least-upper-boundproperty if the following is true:
If E c: S, Eis not empty, and Eis bounded above, then sup E exists in S. Example 1.9(a) shows that Q does not have the least-upper-bound property. We shall now show that there is a close relation between greatest lower bounds and least upper bounds, and that every ordered set with the least-upperbound property also has the greatest-lower-bound property. '

THE REAL AND COMPLEX NUMBER SYSTEMS 5

1.11 Theorem Suppose Sis an ordered set with the least-upper-bound property,

B c S, B is not empty, and B is bounded below. Let L be the set of all lower

bounds of B. Then

exists in S, and oc = inf B.

IX= supL

In particular, inf B exists in S.

Proof Since B is bounded below, L is not empty. Since L consists of exactly those ye S which satisfy the inequality y ~ x for every x e B, we
see that every x E B is an upper bound of L. Thus L is bounded above.
Our hypothesis about S implies therefore that L has a supremum in S;
call it IX.
If y < IX then (see Definition 1.8) y is not an upper bound of L, hence y ¢ B. It follows that IX ~ x for every x E B. Thus oc EL.
If IX < f3 then /3 ¢ L, since IX is an upper bound of L. We have shown that IX EL but /3 ¢ L if f3 > IX. In other words, oc is a lower bound of B, but f3 is not if /J > IX. This means that IX= inf B.

FIELDS

1.12 Definition A field is a set F with two operations, called addition and multiplication, which satisfy the following so-called "field axioms" (A), (M), and (D):

(A) Axioms for addition
(Al) If x e F and ye F, then their sum x + y is in F. (A2) Addition is commutative: x + y = y + x for all x, y E F. (A3) Addition is associative: (x + y) + z = x + (y + z) for all x, y, z e F. (A4) F contains an element Osuch that O+ x = x for every x E F.
(AS) To every x e F corresponds an element - x E F such that
X +(-x) = 0.

(M) Axioms for multiplication

(Ml) (M2) (M3) (M4) (MS)

If x E F and y E F, then their product xy is in F.
Multiplication is commutative: xy = yx for all x, y E F.
Multiplication is associative: (xy)z = x(yz) for all x, y, z E F.
F contains an element 1 =/: 0 such that lx = x for every x e F.
If x E F and x =/: 0 then there exists an element 1/x E F such that

X' (1/x) = 1.

6 PRINCIPLES OP MATHBMAnCAL ANALYSIS

(D) 1be diltrlbutive law holds for all x, y, z e F.

x(y + z) =xy + xz

1.13 Remarks
(a} One usually writes (in any field)
x - y, y-X , x + y + z, xyz, x2, x3, 2x, 3x, ...
in place of
G), x + (-y}, x • (x + y) + z, (xy)z, xx, xxx, x + x, x + x + x, .. , .
(b) The field axioms clearly hold in Q, the set of all rational numbers, if addition and multiplication have their customary meaning. Thus Q is a field. (c) Although it is not our purpose to study fields (or any other algebraic structures) in detail, it is worthwhile to prove that some familiar properties of Q are consequences of the field axioms; once we do this, we will not need to do it again for the real numbers and for the complex numbers.

1.14 Proposidon The axioms for addition imply the following statements.
(a) If x + y == x + z then y = z. (b) If x + y = x then y = 0. (c) lfx + y =0 theny = -x.
(d) -(-x) = x.
Statement (a) is a cancellation law. Note that (b) asserts the uniqueness of the element whose existence is assumed in (A4), and that (c) does the same for (AS).
Proof If x + y == x + z, the axioms (A) give
y =0 + y =(-x + x) + y = -x + (x + y)
= -x + (x + z) == (-x + x) + z = 0 + z = z. This proves (a). Take z = 0 in (a) to obtain (b). Take z = -x in (a) to
obtain (c).
Since -x + x =0, (c) (with -x in pl~ of x) gives (d).

THE REAL AND COMPLEX NUMBER SYSTEMS 7
1.15 Proposition The axioms for multiplication imply the following statements.
(a) If x # 0 and xy = xz then y = z.
(b) If x # 0 and xy = x then y = 1. (c) If x # 0 and xy = 1 then y = 1/x. (d) If x # 0 then 1/(1/x) = x.
The proof is so similar to that of Proposition 1.14 that we omit it.
1.16 Proposition The field axioms imply the following statements.for any x, y, zeF.
(a) Ox= 0. (b) If x # 0 and y # 0 then xy # 0. (c) (-x)y = -(xy) = x(-y).
(d) (-x)(-y) = xy. Proof Ox+ Ox= (0 + O)x = Ox. Hence l.14(b) implies that Ox= 0, and
(a) holds.
Next, assume x # 0, y # 0, but xy = 0. Then (a) gives
1 = (;)G)xy=G)G)o=O,
a contradiction. Thus (b) holds. The first equality in (c) comes from
(-x)y + xy = (-x + x)y = Oy = 0,
combined with l.14(c); the other half of (c) is proved in the same way. Finally,
(-x)(-y)= -[x(-y)]= -[-(xy)]=xy
by (c) and l.14(d).
1.17 Definition An ordered field is a field F which is also an ordered set, such that
(i) x + y < x + z if x, y, z E F and y < z, (ii) xy > 0 if x E F, y E F, x > 0, and y > 0.
If x > 0, we call x positive; if x < 0, xis negative. For example, Q is an ordered field. All the familiar rules for working with inequalities apply in every ordered
field: Multiplication by positive [negative] quantities preserves [reverses] inequalities, no square is negative, etc. The following proposition lists some of these.

8 PRINCIPLES OF MATHEMATICAL ANALYSIS
1.18 Proposition The following statements are true in every ordered field.
(a) lf x > 0 then - x < 0, and vice versa. (b) lf x > 0 and y < z then xy <xz. (c) If x < 0 and y < z then xy > xz.
(d) If x :/= 0 then x2 > 0. In particular, 1 > 0. (e) /JO< x <Y then O< 1/y < 1/x. Proof
(a) If x > 0 then O= -x + x > -x + 0, so that -x < 0. If x < 0 then 0 = -x + x < -x + 0, so that -x > 0. This proves (a).
(b) Since z > y, we have z - y > y - y = 0, hence x(z - y) > 0, and
therefore
xz = x(z - y) + xy > 0 + xy = xy.
(c) By (a), (b), and Proposition 1.16(c),
-[x(z -y)] = (-x)(z -y) > 0,
so that x(z - y) < 0, hence xz < xy.
(d) If x > 0, part (ii) of Definition 1.17 gives x2 > 0. If x < 0, then
-x > 0, hence (-x)2 > 0. But x 2 = (-x)2, by Proposition l.16(d). Since 1 = l2, 1 > 0. (e) lfy > 0 and vs; 0, thenyv s; 0. Buty •(1/y) = 1 > 0. Hence 1/y > 0.
Likewise, 1/x > 0. If we muJtiply both sides of the inequality x < y by the positive quantity (1/xXl/y), we obtain 1/y < 1/x.
THE REAL FIELD
We now state the existence theorem which is the core of this chapter.
1.19 Theorem There exists an ordered.field R which has the least-upper-bound property.
Moreover, R contains Q as a subfield.
The second statement means that Q c R and that the operations of addition and multiplication in R, when applied to members of Q, coincide with the usual operations on rational numbers; also, the positive rational numbers are positive elements of R.
The members of Rare called real numbers. The proof of Theorem 1.19 is rather long and a bit tedious and is therefore presented in an Appendix to Chap. I. The proof actually constructs R from Q.

THE REAL AND COMPLEX NUMBER SYSTEMS 9
The next theorem could be extracted from this construction with very little extra effort. However, we prefer to derive it from Theorem 1.19 since this provides a good illustration of what one can do with the least-upper-bound property.
1.20 Theorem
(a) If x e R, ye R, and x > 0, then there is a positive integer n such that
nx>y.
(b) [fx e R,y e R, and x < y, then thereexistsap e Q such that x < p < y.
Part (a) is usually referred to as the archimedean property of R. Part (b) may be stated by saying that Q is dense in R: Between any two real numbers there is a rational one.
Proof (a) Let A be the set of all nx, where n runs through the positive integers. If (a) were false, then y would be an upper bound of A. But then A has a least upper bound in R. Put a= sup A. Since x > 0, a - x < a, and a - x is not an upper bound of A. Hence a - x < mx for some positive integer m. But then a< (m + l)x e A, which is impossible, since a is an upper bound of A. (b) Since x < y, we have y - x > 0, and (a) furnishes a positive integer n such that
n(y-x) > I. Apply (a) again, to obtain positive integers m1 and m2 such that m1 > nx, m1 > -nx. Then
-m1 <nx <m1. Hence there is an integer m (with -m1 s; m s; m1) such that
m - 1 s; nx < m.
If we combine these inequalities, we obtain
nx<mS: 1 +nx<ny.
Since n > 0, it follows that
m x<-<y.
n
This proves (h), with p = m/n.

10 PRINCIPLES OP MATHEMATICAL ANALYSIS
We shall now prove the existence of nth roots of positive reals. This proof will show how the difficulty pointed out in the Introduction (irrationality of ,/2) can be handled in R.
1.21 Theorem For every real x > 0 and every integer n > 0 there is one
and only one po~itive real y such that y' = x.
ix This number y is written Or xlln,
Proof That there is at most one such y is clear, since O< y1 < y2 implies
Y': <)1.
Let E be the set consisting of all positive real numbers t such that t"<x.
If t = x/(1 + x) then Os; t < l. Hence t• St < x. Thus t e E, and
Eis not empty.
If t > 1 + x then t" ~ t > x, so that t rt E. Thus 1 + x is an upper
bound of E. Hence Theorem 1.19 implies the existence of
y = supE.
To prove that y' = x we will show that each of the inequalities y' < x
and y' > x leads to a contradiction. The identity b" - a"= (b - a)(b"- 1 + b"-2a + • • • + a"- 1) yields
the inequality b" - a"< (b - a)nb"- 1
when O<a< b. Assume y' < x. Choose h so that O < h < 1 and
x-y'
h < n(y + l)n-1
Put a = y, b = y + h. Then
(y + h)" -y' < hn(y + h)"- 1 < hn(y + 1r-1 < X -y'.
Thus (y + h)" < x, and y + h e E. Since y + h > y, this contradicts the
fact that y is an upper bound of E. Assume y' > x. Put
k-1'-x. nyi-1
Then O< k < y. If t ~ y - k, we conclude that
y' - t" s y' - (y- k)" < knyi- 1 = y' - x.
Thus t• > x, and t ¢ E. It follows that y - k is an upper bound of E.

THE REAL AND COMPLEX NUMBER SYSTEMS 11

But y - k < y, which contradicts the fact that y is the least upper bound
of E.
Hence y" = x, and the proof is complete.

Corollary If a and b are positive real numbers and n is a positive integer, then
(ab)lfn = alfnbtfn.

Proof = Put IX = a11", /3 b11". Then
ab = a."/3" = (a./3)",
since multiplication is commutative. [Axiom (M2) in Definition 1.12.J The uniqueness assertion of Theorem 1.21 shows therefore that
(ab)tfn = a./3 = alfnblfn,

1.22 Decimals We conclude this section by pointing out the relation between real numbers and decimals.
Let x > 0 be real. Let n0 be the largest integer such that n0 ~ x. (Note that the existence of n0 depends on the archimedean property of R.) Having chosen n0 , n1, ... , nk- i, let nk be the largest integer such that

n1

nk

no + IO + ••• + 10k ~ x.

Let E be the set of these numbers

(5)

(k = 0, 1, 2, ...).

Then x = sup E. The decimal expansion of xis
(6)
Conversely, for any infinite decimal (6) the set E of numbers (5) is bounded above, and (6) is the decimal expansion of sup E.
Since we shall never use decimals, we do not enter into a detailed discussion.

THE EXTENDED REAL NUMBER SYSTEM

1.23 Definition The extended real number system consists of the real field R

and two symbols, + oo and - oo. We preserve the original order in R, and define

for every x e R.

-oo<x<+oo

12 PRINCIPLES OF MATHEMATICAL ANALYSIS

It is then clear that + oo is an upper bound of every subset of the extended
real number system, and that every nonempty subset has a least upper bound.
If, for example, E is a nonempty set of real numbers which is not bounded
above in R, then sup E = + oo in the extended real number system.
Exactly the same remarks apply to lower bounds.

The extended real number system does not form a field, but it is customary to make the following conventions:

(a) If x is real then

x+oo=+oo,

x-oo=-oo,

X

X

-=-=0.

+co -oo

(b) Ifx>0thenx·(+oo)= +oo,x·(-oo)= -oo.

= - (c) If X < 0 then X • ( + 00)

= 00, X • ( - 00) + 00.

When it is desired to make the distinction between real numbers on the
one hand and the symbols + oo and - oo on the other quite explicit, the former
are called.finite.

THE COMPLEX FIELD

1.24 Definition A complex number is an ordered pair (a, b) of real numbers.

"Ordered" means that (a, b) and (b, a) are regarded as distinct if a-:/= b.
Let x = (a, b), y = (c, d) be two complex numbers. We write x = y if and only if a= c and b = d. (Note that this definition is not entirely superfluous;

think of equality of rational numbers, represented as quotients of integers.) We

define

x + y = (a + c, b + d),

xy = (ac - bd, ad+ be).

1.25 Theorem These definitions of addition and multiplication turn the set of all complex numbers into a field, with (0, 0) and (1, 0) in the role ofO and 1.
Proof We simply verify the field axioms, as listed in Definition 1.12. (Of course, we use the field structure of R.)
Let x =(a, b), y = (c, d), z =(e,f).
(Al) is clear. ~~ x+y=~+~b+~=~+~d+~=y+~

THE REAL AND COMPLEX NUMBER SYSTEMS 13

(A3) (x + y) + z = (a + c, b + d) + (e,f)
= (a + c + e, b + d + f) = (a, b) + (c + e, d + f) = x + (y + z). (A4) x + 0 = (a, b) + (0, 0) = (a, b) = x. (AS) Put -x = (-a, -b). Then x + (-x) = (0, 0) = 0.
(MI) is clear.
(M2) xy = (ac - bd, ad+ be) = (ca - db, da + cb) = yx.
(M3) (xy)z = (ac - bd, ad+ bc)(e,f)
= (ace - bde - adf - bcf, acf - bdf + ade + bee)
= (a, b)(ce - df, cf+ de) = x(yz).
(M4) Ix= (1, 0)(a, b) = (a, b) = x.
(MS) If x '# 0 then (a, b) '# (0, 0), which means that at least one of the
real numbers a, b is different from 0. Hence a2 + b2 > 0, by Proposition
1.18(d), and we can define

Then

-1= ( - - a , --b-) .
x a2 + b2 a2 + b2

x

• -I x

=

(a,

b)

(

~ a

b+ a

2

'

a----Y+b-b2) = (I, 0) = 1.

(D) x(y + z) = (a, b)(c + e, d + f)

= (ac + ae - bd - bf, ad+ af +be+ be)

= (ac - bd, ad+ be)+ (ae - bf, af + be)

= xy + xz.

1.26 Theorem For any real numbers a and b we have
(a, 0) + (b, 0) = (a + b, 0), (a, 0)(b, 0) = (ab, 0).
The proof is trivial.

Theorem 1.26 shows that the complex numbers of the form (a, 0) have the same arithmetic properties as the corresponding real numbers a. We can therefore identify (a, 0) with a. This identification gives us the real field as a subfield of the complex field.
The reader may have noticed that we have defined the complex numbers without any reference to the mysterious square root of -1. We now show that the notation (a, b) is equivalent to the more customary a+ bi.
1.27 Definition i = (0, 1).

14 PRINCIPLES OF MATHEMATICAL ANALYSIS

1.28 Theorem i2 = - 1.
Proof i2 = (0, 1)(0, 1) = (-1, 0) = -1.

1.29 Theorem If a and b are real, then (a, b) = a + bi.

Proof

a+ bi= (a, 0) + (b, 0)(0, 1) = (a, 0) + (0, b) = (a, b).

1.30 Definition If a, b are real and z = a + bi, then the complex number z = a - bi is called the conjugate of z. The numbers a and b are the real part
and the imaginary part of z, respectively. We shall occasionally write
a= Re(z), b = Im(z).

1.31 Theorem If z and w are complex, then
(a) z + w = z + w,
(b) zw = z • w, (c) z + z = 2 Re(z), z - z = 2i Im(z), (d) zz is real and positive (except when z = 0).
Proof (a), (b), and (c) are quite trivial. To prove (d), write z = a + bi, and note that zz = a2 + b2 •

1.32 Definition If z is a complex number, its absolute value zI I is the non-
negative square root of zz; that is, lzl = (zz) 112• The existence (and uniqueness) of Izl follows from Theorem 1.21 and
part (d) of Theorem 1.31.
Note that when xis real, then x = x, hence lxl =Jx2 . Thus lxl = x ifx~0, lxl = -xifx<0.

1.33 Theorem Let z and w be complex numbers. Then
(a) lzl > 0 unless z = 0, IOI = 0, (b) lzl = lzl,
(c) lzwl = lzl lwl,
(d) IRezl S lzl,
(e) lz + wl s lzl + lwl,

THE REAL AND COMPLEX NUMBER SYSTEMS 15

Proof (a) and (b) are trivial. Putz= a+ bi, w = c + di, with a, b, c, d
real. Then

lzwl 2 = (ac -bd)2 +(ad+ bc)2 = (a2 + b2)(c2 + d 2) = lzl 2 1wl 2

or Izw I 2 = (Iz I I w1)2• Now (c) follows from the uniqueness assertion of
Theorem 1.21.

To prove (d), note that a2 s; a2 + b2, hence

P J IaJ =

:S: a2 + b2•

To prove (e), note that zw is the conjugate of zw, so that zw + zw =

2 Re (zw). Hence

Iz + w12 = (z + w)(z + w) = zz + zw + zw + ww
= JzJ 2 + 2 Re (zw) + JwJ 2
s: lzl2 +2lzwl + 1w12
= lzl 2 + 2Jzl Jwl + lwJ 2 = (lzl + Jwl)2.

Now (e) follows by taking square roots.

1.34 Notation If x1, ••• , x,. are complex numbers, we write
x1 + x2 + ••• + x,, = L" x1 • J• 1
We conclude this section with an important inequality, usually known as the Schwarz inequality.

1.35 Theorem If a 1, ... , a. and bi, ... , b,, are complex numbers, then

= o Proof Put A I: Ia1 I 2, B = I: Ib1J2, C = I:a1 1 (in all sums in this proof,
j runs over the values 1, ... , n). If B = 0, then b1 = ••• = b,, = 0, and the
conclusion is trivial. Assume therefore that B > 0. By Theorem 1.31 we
have
= L 1Ba1 - Cb112 L (Ba1 - Cb1)(Ba1 - Cb1) = B2 L la1l 2 - BCL a1°1 -BCLii1b1 + ICl 2 L lb1l 2 = B2A- BJCl 2
= B(AB - ICl 2).

16 PRINCIPLES OF MATHEMATICAL ANALYSIS
Since each term in the first sum is nonnegative, we see that
B(AB - ICl 2) :;;:; 0. Since B > 0, it follows that AB - ICl 2 :;;:; 0. This is the desired inequality.

EUCLIDEAN SPACES

1.36 Definitions For each positive integer k, let Rk be the set of all ordered
k-tuples
= X (x1, X 2 , , , . , Xk),

where x1, ... , xk are real numbers, called the coordinates of x. The elements of Rk are called points, or vectors, especially when k > 1. We shall denote vectors
by boldfaced letters. If y = (y1, ... , Yk) and if ix is a real number, put

x + Y = (x1 + Yi, ... , xk + Yk),
= IXX (1XX1, , , , , IXXk)

so that x + y E Rk and ixx E Rk. This defines addition of vectors, as well as

multiplication of a vector by a real number (a scalar). These two operations

satisfy the commutative, associative, and distributive laws (the proof is trivial,

in view of the analogous laws for the real numbers) and make Rk into a vector

space over the real field. The zero element of Rk (sometimes called the origin or

the null vector) is the point 0, all of whose coordinates are 0.

We also define the so-called "inner product" (or scalar product) of x and

y by

and the norm of x by

k
x· y = LX;Y;
i= 1
(t Jxl = (x. x)1;2 = xf) 112_

The structure now defined (the vector space Rk with the above inner product and norm) is called euclidean k-space.

1.37 Theorem Suppose x, y, z E Rk, and ix is real. Then
(a) lxl :;;:; O;
(b) \x\ = 0 if and only ifx = O;
(c) Jixxl = lixl Jxl;
(d) Ix· y\ ;s; Ix\ IYI; (e) \x + y\ ;s; lxl + \yJ;
(f) \x-zl ;s; \x-y\ + ly-z\.

THB RIIAL AND OOMPLBX NUMBl!ll SYSTEMS 17
Proof (a), (b), and (c) are obvious, and (d) is an immediate consequence of the Schwarz inequality. By (d) we have
Ix + y12 == (x + y) • (x + y)
==x·x+2x·y+y·y
~ lxl 2 +21xl IYI + IYl 2 == (lxl + IYl>2,
so that (e) is proved. Finally, (f) follows from (e) if we replace x by x-yandybyy-z.
1.38 Remarks Theorem 1.37 (a), (b), and (f) will allow us (see Chap. 2) to regard R" as a metric space.
R1 (the set of all real numbers) is usually called the line, or the real line. Likewise, R2 is called the plane, or the complex plane (compare Definitions 1.24 and 1.36). In these two cases the norm is just the absolute value of the corresponding real or complex number.
APPENDIX
Theorem 1.19 will be proved in this appendix by constructing R from Q. We shall divide the construction into several steps.
Step 1 The members of R will be certain subsets of Q, called cuts. A cut is, by definition, any set IX c Q with the following three properties.
(I) IX is not empty, and IX "F- Q. (II) If p e IX, q e Q, and q <p, then q e IX. (III) If p e IX, then p < r for some r e IX.
The letters p, q, r, ... will always denote rational numbers, and IX, p, y, ...
will denote cuts. Note that (III) simply says that IX has no largest member; (II) implies two
facts which will be used freely:
Ifp e IX and q ¢ IX then p < q. If r ¢ IX and r < s then s ¢ IX.
Step 2 Define "IX< P" to mean: IX is a proper subset of p.
Let us check that this meets the requirements of Definition 1.5.
If IX< Pand P< y it is clear that IX< y. (A proper subset ofa proper sub-
set is a proper subset.) It is also clear that at most one of the three relations
IX <P, IX== p, p < IX

18 PRINCIPLES OF MATHEMATICAL ANALYSIS
can hold for any pair ex, {J. To show that at least one holds, assume that the
first two fail. Then ex is not a subset of {J. Hence there is ape ex with p ¢ p. If q e /J, it follows that q < p (since p ¢ P), hence q e ex, by (II). Thus p c ex. Since p ,;: ex, we conclude: fJ < ex.
Thus R is now an ordered set.
Step 3 The ordered set R has the least-upper-bound property. To prove this, let A be a nonempty subset of R, and assume that fJ e R
is an upper bound of A. Define l' to be the union of all ex e A. In other words, p el' if and only if p e ex for some oc e A. We shall prove that l' e R and that
l' = sup A.
Since A is not empty, there exists an ex0 e A. This oi0 is not empty. Since
oc0 c l', l' is not empty. Next, l' c /J (since oc c Pfor every oc e A), and therefore
y,;, Q. Thus l' satisfies property (I). To prove (II) and (III), pick p e y. Then p e oi1 for some ex1 e A. If q < p, then q e iXi, hence q e l'; this proves (II). If r e ex1 is so chosen that r > p, we see that r e l' (since ex1 c y), and therefore y satisfies (III).
Thus l' e R. It is clear that oi s; l' for every oi e A. Suppose <> < y. Then there is an s e l' and that s ¢ <>. Since s e l', s e oi for some ex e A. Hence<>< ex, and<> is not an upper bound of A.
This gives the desired result: l' = sup A.
Step 4 If oc e Rand pe R we define oc + pto be the set of all sums r + s, where re ex and s e /J.
We define O* to be the set of all negative rational numbers. It is clear that O* is a cut. We verify that the axioms for addition (see Definition 1.12) hold in R, with O* playing the role of0.
(Al) We have to show that oi + p is a cut. It is clear that ex+ /J is a nonempty subset of Q. Take r' ¢ oc, s' ¢ p. Then r' + s' > r + s for all choices of re ex, s e p. Thus r' + s' ¢ oi + p. It follows that ex+ /J has
property (I). Pickpeex+/J. Thenp=r+s, with reoc, sep. lfq<p, then
q - s < r, so q - s e oc, and q = (q - s) + s e oi + /J. Thus (II) holds.
Choose t e oi so that t > r. Then p < t +sand t + s e oi + p. Thus (111)
holds.
(A2) oc + pis the set of all r + s, with r e oc, s e p. By the same definition,
p + oc is the set of all s + r. Since r + s =s + r for all re Q, s e Q, we
have oi + /J = {J + ex.
(A3) As above, this follows from the associative law in Q.
(A4) If re oc ands e O*, then r + s < r, hence r + s e oi. Thus ex+ O* c oc. To obtain the opposite inclusion, pick p e oi, and pick r e ex, r > p. Then

THE REAL AND COMPLEX NUMBER SYSTEMS 19
p - reO*, and p = r +(p-r)eoc + O*. Thus oc c::: oc + O*. We conclude
that oc + O* = oc.
(AS) Fix oc e R. Let Pbe the set of all p with the following property:
There exists r > 0 such that - p - r ¢ oc.
In other words, some rational number smaller than - p fails to be in oc.
We show that pe Rand that oc + P= O*. Ifs¢ IX and p = -s - 1, then -p - 1 ¢ IX, hence p e {J. So fJ is not empty. If q e cc, then -q ¢ {J. So fJ-,,. Q. Hence p satisfies (I).
Pick p e {J, and pick r > 0, so that -p - r ¢ cc. If q <p, then -q - r > -p - r, hence -q - r ¢ IX. Thus q e {J, and (II) holds. Put
t=p+(r/2). Then t>p, and -t-(r/2)= -p-r(JIX, so that te{J.
Hence p satisfies (III). We have proved that Pe R. If r e IX and s e p, then -s ¢ oc, hence r < -s, r + s < 0. Thus
cc+ /Jc::: O*.
To prove the opposite inclusion, pick v e O*, put w = -v/2. Then
w > 0, and there is an integer n such that nw e oc but (n + l)w f: IX, (Note
that this depends on the fact that Q has the archimedean property I) Put
= p -(n + 2)w. Then pep, since -p - w¢ IX, and
v = nw + p e cc+ p.
Thus O* c::: IX + p.
We conclude that IX+ P=O*.
This p will of course be denoted by - IX.
Step 5 Having proved that the addition defined in Step 4 satisfies Axioms (A) of Definition 1.12, it follows that Proposition 1.14 is valid in R, and we can prove one of the requirements of Definition 1.17:
/foe, P, y eR and P< y, then oc + P< oc + y. Indeed, it is obvious from the definition of + in R that IX + p c::: IX + y; if we had ix + /J = ix + y, the cancellation law (Proposition 1.14) would imply P= y.
It also follows that 01 > O* if and only if - IX < O*.
Step 6 Multiplication is a little more bothersome than addition in the present context, since products of negative rationals are positive. For this reason we confine ourselves first to R+, the set of all ix e R with IX> O*.
If IX e R + and Pe R +, we define 1XP to be the set of all p such that p ::;; rs for some choice of re IX, s e p, r > 0, s > 0.
We define 1* to be the set of all q < 1.

20 PRINCIPLES OF MATHBMATICAL ANALYSIS

Then the axioms (M) and (D) of Definition 1.12 hold, with R+ in place ofF, and with 1* in the role of 1.
The proofs are so similar to the ones given in detail in Step 4 that we omit them.
Note, in particular, that the second requirement of Definition 1.17 holds:
If IX> 0* and p > O* then 1X/3 > O*.

Step 7 We complete the definition of multiplication by setting 1XO* = O*IX = O*,
and by setting

( -1X)(-P) IX/3= { -[(-IX)/31
-[c,:. (-/3)]

if IX < O*, /J < O*,
ifo < O*, p > 0*,
if c,: > O*, /3 < O*.

The products on the right were defined in Step 6. Having proved (in Step 6) that the axioms (M) hold in R+, it is now
perfectly simple to prove them in R, by repeated application of the identity y = -( -y) which is part of Proposition 1.14. (See Step 5.)
The proof of the distributive law

IX{P + y) = IX/3 + IXY

breaks into cases. For instance, suppose IX> O*, /3 < O*, /3 + y > O*. Then y = (P + y) + ( -/3), and (since we already know that the distributive law holds
in R+)
IXY = IX{P + y) + IX ' ( - /3).

But IX· (-/3) = -(1X/3). Thus

IX/3 + IXY = 1X(/3 + y).

The other cases are handled in the same way. We have now completed the proof that R is an ordered field with the least-
upper-bound property.

Step 8 We associate with each re Q the set r* which consists of all p e Q such that p < r. It is clear that each r* is a cut; that is, r* e R. The,;e cuts satisfy the following relations:
(a) r* + s* =(r + s)*,
(b) r*s* = (rs)*,
(c) r* < s* if and only if r < s.
To prove (a), choose per* + s*. Then p = u + v, where u < r, v < s.
Hence p < r + s, which says that p e (r + s)*.

THE REAL AND COMPLEX NUMBER SYSTEMS 21
Conversely, suppose p e (r + s)*. Then p < r + s. Choose t so that
2t = r + s - p, put
r' = r - t, s' =s - t.
Then r' er*, s' es*, andp = r' + s', so thatp er*+ s*.
This proves (a). The proof of (b) is similar.
If r < s then res*, but r ¢r*; hence r* < s*. If r* < s*, then there is apes* such that p ¢ r*. Hence rs p < s, so that r < s.
This proves (c).
Step 9 We saw in Step 8 that the replacement of the rational numbers r by the corresponding "rational cuts" r* e R preserves sums, products, and order. This fact may be expressed by saying that the ordered field Q is isomorphic to the ordered field Q* whose elements are the rational cuts. Of course, r* is by no means the same as r, but the properties we are concerned with (arithmetic and order) are the same in the two fields.
It is this identification of Q with Q* which allows us to regard Q as a subfield of R.
The second part of Theorem 1.19 is to be understood in terms of this identification. Note that the same phenomenon occurs when the real numbers are regarded as a subfield of the complex field, and it also occurs at a much more elementary level, when the integers are identified with a certain subset of Q.
It is a fact, which we will not prove here, that any two orderedfields with the least-upper-bound property are isomorphic. The first part of Theorem 1.19 therefore characterizes the real field R completely.
The books by Landau and Thurston cited in the Bibliography are entirely devoted to number systems. Chapter 1 of Knopp's book contains a more leisurely description of how R can be obtained from Q. Another construction, in which each real number is defined to be an equivalence class of Cauchy sequences of rational numbers (see Chap. 3), is carried out in Sec. 5 of the book by Hewitt and Stromberg.
The cuts in Q which we used here were invented by Dedekind. The construction of R from Q by means of Cauchy sequences is due to Cantor. Both Cantor and Dedekind published their constructions in 1872.
EXERCISES
Unless the contrary is explicitly stated, all numbers that are mentioned in these exercises are understood to be real.
* 1. If r is rational (r 0) and x is irrational, prove that r +x and rx are irrational.

l2 PRINCIPLES OF MATHEMATICAL ANALYSIS

2. Prove that there is no rational number whose square is 12.
3. Prove Proposition 1.15.
4. Let Ebe a nonempty subset of an ordered set; suppose IX is a lower bound of E
and /3 is an upper bound of E. Prove that IX '5:. {3.
5. Let A be a nonempty set of real numbers which is bounded below. Let -A be the set of all numbers -x, where x EA. Prove that

inf A= -sup(-A).

6. Fix b > 1. (a) If m, n, p, q are integers, n > 0, q > 0, and r = m/n = p/q, prove that
(b"')ll• = (b')1'•.

Hence it makes sense to define b' = (b"')1'"·

(b) Prove that hr+•= b•b• if rands are rational.

(c) If x is real, define B(x) to be the set of all numbers b', where t is rational and

t '5:. x. Prove that

b' =supB(r)

when r is rational. Hence it makes sense to define

b" =supB(x)

for every real x. (d) Prove that bH' b"b' for all real x and y.
7. Fix b > 1, y > 0, and prove that there is a unique real x such that b" = y, by
completing the following outline. (This xis called the logarithm ofy to the base b.) (a) For any positive integer n, b• - 1 :2: n(b - 1).
(b) Hence b - 1 :2: n(b11" - 1).
(c) If t > 1 and n > (b - 1)/(t - 1), then b11• < t.
(d) If w is such that bw <y, then bw+ctt•> <y for sufficiently large n; to see this, apply part (c) with t = y • b-w.
(e) If bw > y, then bw-<1 1•> > y for sufficiently large n. (/) Let A be the set of all w such that bw < y, and show that x = sup A satisfies
b"=y. (g) Prove that this x is unique.

8. Prove that no order can be defined in the complex field that turns it into an ordered
field. Hint: -1 is a square.
9. Suppose z = a + bi, w = c + di. Define z < w if a < c, and also if a = c but b < d. Prove that this turns the set of all complex numbers into an ordered set.
(This type of order relation is called a dictionary order, or lexicographic order, for
obvious reasons.) Does this ordered set have the least-upper-bound property?
10. Suppose z = a + bi, w = u + iv, and

THE REAL AND COMPLEX NUMBER SYSTEMS 23
z Prove that z2 = w if v Oand that (z)2 = w if v ~ 0. Conclude that every complex
number (with one exception!) has two complex square roots.
z 11. If z is a complex number, prove that there exists an r Oand a complex number
w with Iwl = 1 such that z = rw. Are wand r always uniquely determined by z? 12. If zi, ... , z. are complex, prove that
\z,+z2+"·+z.l ~!zi! + \z2\ +"·+Jz.l.
13. If x, y are complex, prove that
!lxl - !YI!~ \x-y\. 14. If z is a complex number such that \zl = 1, that is, such that zz= 1, compute
\1 +z\2+ ll -z\2.
15. Under what conditions does equality hold in the Schwarz inequality? 16. Supposekz3,x,yERt, lx-yl =d>O,andr>O. Prove:
(a) If 2r > d, there are infinitely many z E Rt such that
\z-x\ = \z-y\ =r.
(b) If 2r = d, there is exactly one such z. (c) If 2r < d, there is no such z.
How must these statements be modified if k is 2 or 1? 17. Prove that
Ix+ Yl 2 + lx-y\ 2 = 2\x1 2 + 2IYl 2
if xERt and ye Rt. Interpret this geometrically, as a statement about parallelograms.
z 18. If k 2 and x E Rt, prove that there exists y E Rt such that y # 0 but x •y = O.
Is this also true if k = 1? 19. Suppose a E Rt, b E Rt. Find c E Rt and r > 0 such that
\x-a! =2\x-b\
if and only if Ix - c I = r. (Solution: 3c =4b-a, 3r = 2lb-al .)
20. With reference to the Appendix, suppose that property (III) were omitted from the definition of a cut. Keep the same definitions of order and addition. Show that the resulting ordered set has the least-upper-bound property, that addition satisfies axioms (Al) to (A4) (with a slightly different zero-element!) but that (A5) fails.

2
BASIC TOPOLOGY
FINITE, COUNTABLE, AND UNCOUNTABLE SETS We begin this section with a definition of the function concept. 2.1 Definition Consider two sets A and B, whose elements may be any objects whatsoever, and suppose that with each element x of A there is associated, in some manner, an element of B, which we denote by f(x). Then/ is said to be a function from A to B (or a mapping of A into B). The set A is called the domain off (we also say f is defined on A), and the elements f(x) are called the values off. The set of all values offis called the range off. 2.2 Definition Let A and B be two sets and let f be a mapping of A into B. If E c A,f(E) is defined to be the set of all elements f(x), for x e E. We call f(E) the image of E under/. In this notation, f(A) is the range of/. It is clear
that/(A) c B. Iff(A) = B, we say that/maps A onto B. (Note that, according
to this usage, onto is more specific than into.) If E c B,f- 1(E) denotes the set of all x EA such that f(x) e E. We call
1- 1 (E) the inverse image of E under f. If ye B,f- 1(Y) is the set of all x e A

BASIC TOPOLOOY 25
= such that /(x) y. If, for each ye B,/-1(y) consists of at most one element
of A, then f is said to be a 1-1 (one-to-one) mapping of A into B. This may also be expressed as follows: f is a 1-1 mapping of A into B provided that /(x1):;: /(x2) whenever x 1 :;, x 2 , Xie A, x 2 eA.
(The notation x1 :;, x2 means that x1 and x2 are distinct elements; other-
= wise we write Xi X2 .)
2.3 Definidon If there exists a 1-1 mapping of A onto B, we say that A and B can be put in 1-1 correspondence, or that A and B have the same cardinal number, or, briefly, that A and B are equivalent, and we write A~ B. This relation clearly has the following properties:
It is reflexive: A ~ A. It is symmetric: If A~ B, then B ~ A. It is transitive: If A ~Band B ~ C, then A ~ C.
Any relation with these three properties is called an equivalence relation.
2.4 Definidon For any positive integer n, let Jn be the set whose elements are the integers l, 2, ... , n; let J be the set consisting of all positive integers. For any set A, we say:
(a) A is.finite if A~ Jn for some n (the empty set is also considered to be finite).
~ (b) A is infinite if A is not finite.
(c) A is countable if A J. (d) A is uncountable if A is neither finite nor countable. (e) A is at most countable if A is finite or countable.
~ Countable sets are sometimes called enumerable, or denumerable.
For two finite sets A and B, we evidently have A B if and only ifA and B contain the same number of elements. For infinite sets, however, the idea of "having the same number of elements" becomes quite vague, whereas the notion of 1-1 correspondence retains its clarity.
2.5 Example Let A be the set of all integers. Then A is countable. For, consider the following arrangement of the sets A and J:
A: 0, 1, -1,2, -2,3, -3, ... J: 1, 2, 3, 4, 5, 6, 7, ...

26 PRINCIPLES OP MATHEMATICAL ANALYSJS

We can, in this example, even give an explicit formula for a function/ from J to A which sets up a 1-1 correspondence:

/(•)_Jl-i n---1 2

(n even), (n odd).

2.6 Remark A finite set cannot be equivalent to one of its proper subsets. That this is, however, possible for infinite sets, is shown by Example 2.5, in which J is a proper subset of A.
In fact, we could replace Definition 2.4(b) by the statement: A is infinite if A is equivalent to one of its proper subsets.

2.7 Definidon By a sequence, we mean a function/ defined on the set J of all
positive integers. If/(n) = Xn, for n e J, it is customary to denote the sequence
/by the symbol {xn}, or sometimes by x1, x2 , x 3 , .... The values off, that is, the elements Xn, are called the terms of the sequence. If A is a set and if Xn e A for all n eJ, then{xn} is said to be a sequence in A, or asequenceofelementsofA.
Note that the terms x1, x2 , x3 , ... of a sequence need not be distinct. Since every countable set is the range of a 1-1 function defined on J, we may regard every countable set as the range of a sequence of distinct terms. Speaking more loosely, we may say that the elements of any countable set can be "arranged in a sequence." Sometimes it is convenient to replace J in this definition by the set of all nonnegative integers, i.e., to start with Orather than with 1.

2.8 Theorem Every infinite subset ofa countable set A is countable.
Proof Suppose E c A, and E is infinite. Arrange the elements x of A in a sequence {xn} of distinct elements. Construct a sequence {n1} as follows:
Let n1 be the smallest positive integer such that Xn, e E. Having
chosen ni, •.• , n1 _ 1 (k = 2, 3, 4, ...), let n1c be the smallest integer greater
than n1c-i such that x,.,. e E.
Putting/(k) = Xn,. (k = 1, 2, 3, ...), we obtain a 1-1 correspondence
between E and J. The theorem shows that, roughly speaking, countable sets represent
the "smallest" infinity: No uncountable set can be a subset of a countable set.
2.9 Definition Let A and O be sets, and suppose that with each element « of
A there is associated a subset of O which we denote by E«.

BASIC TOPOLOGY 27

The set whose elements are the sets E,. will be denoted by {E,.}. Instead of speaking of sets of sets, we shall sometimes speak of a collection of sets, or a family of sets.
The union of the sets E,. is defined to be the set S such that x E S if and only if x e E,. for at least one a: e A. We use the notation

(I)

S= LJ E,..

«&A

If A consists of the integers I, 2, ... , n, one usually writes

n

(2)

S= LJ Em

m•l

or

(3)

If A is the set of all positive integers, the usual notation is

IX)

(4)

S= LJ Em.

m=- 1

The symbol oo in (4) merely indicates that the union of a countable collection of sets is taken, and should not be confused with the symbols + oo, - oo, introduced in Definition 1.23.
The intersection of the sets E,. is defined to be the set P such that x e P if and only if x e E,. for every a: e A. We use the notation

(5)

P= nE,.,

«eA

or

(6)

n n
P = Em = E1 n E2 n ••• n E~,

m=l

or

(7)

as for unions. If An Bis not empty, we say that A and B intersect; otherwise they are disjoint.

2.10 Examples
(a) Suppose E1 consists of 1, 2, 3 and E2 consists of 2, 3, 4. Then
E1 u E2 consists of 1, 2, 3, 4, whereas E1 n E2 consists of 2, 3.

28 PRINCIPLES OF MATHEMATICAL ANALYSIS

(b) Let A be the set of real numbers x such that 0 < x s; 1. For every x e A, let E:r be the set of real numbers y such that O< y < x. Then

(i)

E:r c:: Es if and only if 0 < x S z S 1;

(ii) (iii)

UE:r =E1 ;
:re.t
nEx is empty;

X&A

(i) and (ii) are clear. To prove (iii), we note that for every y > 0, y ti Ex if X < y. Hence y 'nx&A Ex.

2.11 Remarks Many properties of unions and intersections are quite similar
to those of sums and products; in fact, the words sum and product were some-
n. times used in this connection, and the symbols l: and II were written in place
of Uand
The commutative and· associative laws are trivial:

(8)

AuB=BuA;

A riB=BnA.

(9) (A u B) u C = A u (Bu C); (A n B) n C = A n (B ri C).

Thus the omission of parentheses in (3) and (6) is justified. The distributive law also holds:

(10)

A n (B u C) =(A ri B) u (A ri C).

To prove this, let the left and right members of (10) be denoted by E and F,
respectively. Suppose x e E. Then x e A and x e B u C, that is, x e B or x e C (pos-
sibly both). Hence x e A n B or x e A n C, so that x e F. Thus E c:: F. Next, suppose x e F. Then x e A n B or x e A ri C. That is, x e A, and
x e B u C. Hence x e A n (B u C), so that F c:: E.
It follows that E = F.
We list a few more relations which are easily verified:

(11)

Ac:: Au B,

(12)

An B c:: A.

If 0 denotes the empty set, then

(13) If Ac B, then (14)

A u0 =A, A n0 =0. AuB=B, A. n B=A.

BASIC TOPOLOGY 29

2.12 Theorem Let {E.}, n = 1, '),, 3, ... , be a sequence of countable sets, andput

(15) Then Sis countable.

uCl)
s = E•. n= 1

Proof Let every set En be arranged in a sequence {x.k}, k = 1, 2, 3, ... ,
and consider the infinite array

(16)

in which the elements of En form the nth row. The array contains all elements of S. As indicated by the arrows, these elements can be arranged in a sequence
( 17)
If any two of the sets E. have elements in common, these will appear more than once in (17). Hence there is a subset T of the set of all positive
integers such that S ~ T, which shows that S is at most countable
(Theorem 2.8). Since E1 c S, and E1 is infinite, S is infinite, and thus countable.

Corollary Suppose A is at most countable, and, for every !Y. EA, Ba is at most

countable. Put Then Tis at most countable.

T = LJ Ba, a EA.

For Tis equivalent to a subset of (15).

2.13 Theorem Let A be a countable set, and let Bn be the set of all n-tuples
(a1, ... , an), v,.•here ak E A (k = 1, ... , n), and the elements a1 , ..• , a. need not be
distinct. Then B. is countable.

Proof That B1 is countable is evident, since B1 = A. Suppose Bn-t is countable (n = 2, 3, 4, , ..). The elements of B. are of the form

(18)

(b, a) (b E Bn-1, a EA).

For every fixed b, the set of pairs (b, a) is equivalent to A, and hence countable. Thus Bn is the union of a countable set of countable sets. By Theorem 2.12, B" is countable.
The theorem follows by induction.

30 PlUNCJPLES OF MATHEMATICAL ANALYSIS
Corollary The set ofall rational numbers is countable.
Proof We apply Theorem 2.13, with n = 2, noting that every rational r
is of the form b/a, where a and b are integers. The set of pairs (a, b), and therefore the set of fractions b/a, is countable.
In fact, even the set of all algebraic numbers is countable (see Exercise 2).
That not all infinite sets are, however, countable, is shown by the next theorem.
2.14 Theorem Let A be the set ofall sequences whose elements are the digits 0 and 1. This set A is uncountable.
The elements of A are sequences like I, 0, 0, I, 0, I, I, I, ....
Proof Let E be a countable subset of A, and let E consist of the sequences s1, s2 , s3 , .... We construct a sequences as follows. If the nth digit in sn is 1, we let the nth digit of s be 0, and vice versa. Then the sequence s differs from every member of E in at least one place; hence s ¢ E. But clearly s e A, so that Eis a proper subset of A.
We have shown that every countable subset of A is a proper subset of A. It follows that A is uncountable (for otherwise A would be a proper subset of A, which is absurd).
The idea of the above proof was first used by Cantor, and is called Cantor's diagonal process; for, if the sequences s1, s2 , s3 , ... are placed in an array like (16), it is the elements on the diagonal which are involved in the construction of the new sequence.
Readers who are familiar with the binary representation of the real numbers (base 2 instead of 10) will notice that Theorem 2.14 implies that the set of all real numbers is uncountable. We shall give a second proof of this fact in Theorem 2.43.
METRIC SPACES
2.15 Definition A set X, whose elements we shall call points, is said to be a metric space if with any two points p and q of X there is associated a real number d(p, q), called the distance from p to q, such that
(a) d(p, q) > 0 if p,;, q; d(p, p) = O;
(b) d(p, q) = d(q,p);
(c) d(p, q) ~ d(p, r) + d(r, q), for any re X.
Any function with these three properties is called a distance function, or a metric.

BASIC TOPOLOGY 31

2.16 Examples The most important examples of metric spaces, from our standpoint, are the euclidean spaces R'k, especially R1 (the real line) and R2 (the complex plane); the distance in Rt is defined by

(19)

d(x, y) = Ix - y I (x, ye R").

By Theorem 1.37, the conditions of Definition 2.15 are satisfied by (19). It is important to observe that every subset Y of a metric space Xis a metric
space in its own right, with the same distance function. For it is clear that if conditions (a) to (c) of Definition 2.15 hold for p, q, re X, they also hold if we restrict p, q, r to lie in Y.
Thus every subset of a euclidean space is a metric space. Other examples are the spaces (l(K) and !i72(µ), which are discussed in Chaps. 7 and 11, respectively.

2.17 Definition By the segment (a, b) we mean the set of all real numbers x such that a < x < b.
By the interval [a, b] we mean the set of all real numbers x such that a~ x ~ b.
Occasionally we shall also encounter "half-open intervals" [a, b) and (a, b]; the first consists of all x such that a ~ x < b, the second of all x such that a <x ~ b.
If a;< b; for i = l, ... , k, the set of all points x = (x1, ... , x") in R" whose
coordinates satisfy the inequalities a1 ~ X; ~ b; (1 ~ i ~ k) is called a k-cell. Thus a I-cell is an interval, a 2-cell is a rectangle, etc.
If x e R" and r > 0, the open (or closed) ball B with center at x and radiu~ r is defined to be the set of all y e R" such that Iy - x I < , (or Iy - x I ~ r).
We call a set E c R" convex if
AX + (1 - A)Y e E
whenever x e E, y e E, and O< i < I.
For example, balls are convex. For if IY - xi < ,, lz - xi < ,, and
0 < i < 1, we have
IAY+(l-i)z-xl = IA(y-x)+(l-A)(z-x)I
s;ily-xj +(1-i)jz-xl <Ar+(l-i)r
=r.
The same proof applies to closed balls. It is also easy to see that k-cells are convex.

32 PRINCIPLES 011 MATHEMATICAL ANALYSIS
2.18 Definition Let X be a metric space. All points and sets mentioned below are understood to be elements and subsets of X.
(a) A neighborhood of p is a set N,(p) consisting of all q such that
d(p, q) < r, for some r > 0, The number r is called the radius of N,(p).
(b) A point p is a limit point of the set E if every neighborhood of p contains a point q :/: p such that q e E.
(c) If p e E and p is not a limit point of E, then p is called an isolated point of E.
(d) E is closed if every limit point of E is a point of E. (e) A point p is an interior point of E if there is a neighborhood N of p
such that N c: E. (f) E is open if every point of E is an interior point of E. (g) The complement of E (denoted by E") is the set of all points p e X
such that p ¢ E. (h) E is perfect if E is closed and if every point of E is a limit point
of E. (i) E is bounded if there is a real number M and a point q e X such that
d(p, q) < M for all p e E.
U) E is dense in X if every point of X is a limit point of E, or a point of E (or both).
Let us note that in R1 neighborhoods are segments, whereas in R2 neighborhoods are interiors of circles.
2.19 Theorem Every neighborhood is an open set.
= Proof Consider a neighborhood E N,(p), and let q be any point of E.
Then there is a positive real number h such that
= d(p, q) r - h.
For all points s such that d(q, s) < h, we have then
d(p, s) ~ d(p, q) + d(q, s) < r - h + h = r,
so that s e E. Thus q is an interior point of E.
2.20 Theorem If p is a limit point of a set E, then every neighborhood of p contains in.finitely many points ofE.
Proof Suppose there is a neighborhood N of p which contains only a finite number of points of E. Let qi, ... , q" be those points of N n E, which are distinct from p, and put
r = min d(p, q,,.) 1 smsn

BASIC TOPOLOGY 33

[we use this notation to denote the smallest of the numbers d(p, q1), ... , d(p, q,,)]. The minimum of a finite set of positive numbers is clearly posi-
tive, so that r > 0.
The neighborhood N,.(p) contains no point q of E such that q "F p,
so that p is not a limit point of E. This contradiction establishes the
theorem.

Corollary A.finite point set has no limit points.

2.21 Examples Let us consider the following subsets of R2 :

(a) The set of all complex z such that Iz I < 1. (b) The set of all complex z such that IzI :S 1.
(c) A nonempty finite set.
(d) The set of all integers.
(e) The set consisting of the numbers 1/n (n = 1, 2, 3, ...). Let us note that this set E has a limit point (namely, z =0) but that no point of E is
a limit point of E; we wish to stress the difference between having a lilnit
point and containing one. (f) The set of all complex numbers (that is, R2).
(g) The segment (a, b).

Let us note that (d), (e), (g) can be regarded also as subsets of R1• Some properties of these sets are tabulated below:

Closed Open Perfect Bounded

(a)

No

Yes

No

Yes

(b)

Yes

No

Yes

Yes

(c)

Yes

No

No

Yes

(d)

Yes

No

No

No

(e)

No

No

No

Yes

(f)

Yes Yes

Yes

No

(g)

No

No

Yes

In (g), we left the second entry blank. The reason is that the segment
(a, b) is not open if we regard it as a subset of R2, but it is an open subset of R1•

(V r 0 2.22 Theorem Let {E.} be a (finite or infinite) collection ofsets E•. Then

(20)

E. = (E:).

'u. Proof Let A. and B be the left and right members of (20). If x e A, then
X E.' hence X ' E. for any cc, hence Xe E: for every cc, so that X en E!.
Thus A. c: B.

34 PRINCIPLES OP MATHEMATICAL ANALYSIS

Conversely, if x e B, then x e E; for every 01:, hence x ¢ E,,, for any 01:,
hence x ¢ U. E,,,, so that x e ( U. E,,,)':. Thus B c: A.
It follows that A =B.
2.23 Theorem A set Eis open ifand only ifits complement is closed.
Proof First, suppose Ee is closed. Choose x e E. Then x ¢ C, and x is not a limit point of Ee. Hence there exists a neighborhood N of x such that Ee r. N is empty, that is, N c: E. Thus x is an interior point of E, and E is open.
Next, suppose E is open. Let x be a limit point of Ee. Then every neighborhood of x contains a point of Ee, so that xis not an interior point of E. Since E is open, this means that x e C. It follows that Ee is closed.
Corollary A set Fis closed ifand only ifits complement is open.

2.24 Theorem
(a) For any collection {G,,,} ofopen sets, U. G,,, is open.
n. (b) For any collection {F,,,} ofclosed sets, F,,, is closed.
m. (c) For anyfinite collection Gi, ... , Gn ofopen sets, 1 G1 is open.
(d) For any finite collection Fi, ... , Fn of closed sets, Ui. 1 F1 is closed.

Proof Put G =U. G,,,. If x e G, then x e G,,, for some 01:. Since x is an
interior point of G,,,, x is also an interior point of G, and G is open. This proves (a).
By Theorem 2.22,
(21)

and F; is open, by Theorem 2.23. Hence (a) implies that (21) is open so
n. that F,,, is closed.
m. Next, put H = 1 G1• For any x e H, there exist neighborhoods
N 1 of x, with radii r,. such that N 1 c: G1 (i = 1, ... , n). Put

r = min (ri, ... , rn),

and let N be the neighborhood of x of radius r. Then N c: G1 for i = l,
... , n, so that N c: H, and H is open. By taking complements, (d) follows from (c):

(lJ F,)c = (\(Ft).

l•l

1•1

BASIC TOPOLOGY 35

2.25 Examples In parts (c) and (d) of the preceding theorem, the finiteness of

1, 1) the collections is essential. For let Gn be the segment (-

(n = 1, 2, 3, ...).

Then Gn is an open subset of R1• Put G = n:°= 1 Gn. Then G consists of a single point (namely, x =0) and is therefore not an open subset of R1.
Thus the intersection of an infinite collection of open sets need not be open. Similarly, the union of an infinite collection of closed sets need not be closed.

2.26 Deftnidon If X is a metric space, if E c X, and if E' denotes the set of
all limit points of E in X, then the closure of E is the set E = E u E'.

2.27 Theorem If Xis a metric space and E c: X, then
(a) E is closed,
(b) E =E ifand only ifEis closed,
(c) E c Ffor every closed set F c X such that E c F.
By (a) and (c), E Is the smallest closed subset of X that contains E.
Proof (a) Ifp e X and p ¢Ethen pis neither a point of E nor a limit point of E. Hence p has a neighborhood which does not intersect E. The complement
of E is therefore open. Hence E is closed.
(b) If E =E, (a) implies that Eis closed. If Eis closed, then E' c:: E
= [by Definitions 2.18(d) and 2.26], hence E E.
(c) If Fis closed and F=> E, then F=> F', hence F=> E'. Thus F=> E.

2.28 Theorem Let Ebe a nonempty set ofreal numbers which is bounded above.
Let y =sup E. Then ye E. Hence ye E ifEis closed.

Compare this with the examples in Sec. 1.9.
Proof If ye Ethen ye E. Assume y ¢ E. For every h > 0 there exists then a point x e E such that y - h < x < y, for otherwise y - h would be an upper bound of E. Thus y is a limit point of E. Hence ye E.

2.29 Remark Suppose E c: Y c X, where Xis a metric space. To say that E is an open subset of X means that to each point p e E there is associated a positive number r such that the conditions d(p, q) < r, q e X imply that q e E. But we have already observed (Sec. 2.16) that Y is also a metric space, so that
our definitions may equally well be made within Y. To be quite explicit, let us
say that E is open relative to Y if to each p e E there is associated an r > 0 such that q e E whenever d(p, q) < r and q e Y. Example 2.2l(g) showed that a set

36 PRINCIPLES OP MATHEMATICAL ANALY81S
may be open relative to Y without being an open subset of X. However, there is a simple relation between these concepts, which we now state.
2.30 Theorem Suppose Y c X. A subset E of Y is open relative to Y if and
= only ifE Y n G for some open subset G of X.
Proof Suppose Eis open relative to Y. To each p e E there is a positive
number rP such that the conditions d(p, q) < rP, q e Y imply that q e E. Let VP be the set of all q e X such that d(p, q) < rp, and define
u G= VP. peE
Then G is an open subset of X, by Theorems 2.19 and 2.24. Since p e VP for all p e E, it is clear that E c G n Y. By our choice of VP, we have VP n Y c E for every p e E, so that
G n Y c E. Thus E = G n Y, and one half of the theorem is proved.
Conversely, if G is open in X and E = G rs Y, every p e E has a
neighborhood VP c G. Then VP rs Y c E, so that Eis open relative to Y.
COMPACT SETS
2.31 Definition By an open cover of a set E in a metric space X we mean a
collection {G11} of open subsets of X such that E c U G 11 11 •
2.32 Definition A subset K of a metric space X is said to be compact if every open cover of K contains a.finite subcover.
More explicitly, the requirement is that if {G11} is an open cover of K, then there are finitely many indices a:1, ... , «n such that
KC G111 U • • • U Glln •
The notion of compactness is of great importance in analysis, especially in connection with continuity (Chap. 4).
It is clear that every finite set is compact. The existence of a large class of infinite compact sets in ~ will follow from Theorem 2.41.
We observed earlier (in Sec. 2.29) that if E c Y c X, then E may be open relative to Y without being open relative to X. The property of being open thus depends on the space in which E is embedded. The same is true of the property of being closed.
Compactness, however, behaves better, as we shall now see. To formulate the next theorem, let us say, temporarily, that K is compact relative to X if the requirements of Definition 2.32 are met.

BASIC TOPOLOGY 37

2.33 Theorem Suppose K c:: Y c:: X. Then K is compact relative to X if and only ifK is compact relative to Y.

By virtue of this theorem we are able, in many situations, to regard compact sets as metric spaces in their own right, without paying any attention to any embedding space. In particular, although it makes little sense to talk of open spaces, or of closed spaces (every metric space Xis an open subset of itself, and is a closed subset of itself), it does make sense to talk of compact metric spaces.

Proof Suppose K is compact relative to X, and let {V11} be a collection
of sets, open relative to Y, such that Kc:: U« V11 • By theorem 2.30, there are sets Gu open relative to X, such that V11 = Y n G,., for all a; and since
K is compact relative to X, we have

(22)

Kc:: G,.1 u ••• u G,."

for some choice of finitely many indices ai, ... , °'n. Since Kc:: Y, (22) implies

(23)

K c:: V«. u ••• u VGI".

This proves that K is compact relative to Y. Conversely, suppose K is compact relative to Y, let {G,.} be a col-
lection of open subsets of X which covers K, and put V11 = Y n G,.. Then
(23) will hold for some choice of a1, ... , an; and since V,. c:: G11 , (23) implies (22).
This completes the proof.

2.34 Theorem Compact subsets ofmetric spaces are closed.
Proof Let K be a compact subset of a metric space X. We shall prove that the complement of K is an open subset of X.
Suppose p e X, p ¢ K. If q e K, let Vq and Wq be neighborhoods ofp and q, respectively, of radius less than ½d(p, q) [see Definition 2.18(a)J. Since K is compact, there are finitely many points q1, ... , qn in K such that
KC Wq, u ••• u Wq" = w.
If V = Vq, n • • • n Vq", then V is a neighborhood of p which does not
intersect W. Hence V c Kc, so that p is an interior point of Kc. The theorem follows.

2.35 Theorem Closed subsets ofcompact sets are compact. Proof Suppose F c:: Kc X, Fis closed (relative to X), and K is compact.
r Let {V,.} be an open cover of F. If is adjoined to {V11}, we obtain an

38 PRINCIPLES OF MATHEMATICAL ANALYSIS
n open cover of K. Since K is compact, there is a finite subcollection Cll
of n which covers K, and hence F. If P is a member of Cll, we may remove
it from Cl) and still retain an open cover of F. We have thus shown that a finite subcollection of {V11} covers F.
Corollary If Fis closed and K is compact, then F n K is compact.
Proof Theorems 2.24(b) and 2.34 show that F n K is closed; since F n K c: K, Theorem 2.35 shows that F n K is compact.
n 2.36 Theorem If{K11} is a collection ofcompact subsets ofa metric space X such
that the intersection of every finite subcollection of {K..} is nonempty, then K..
is nonempty.
Proof Fix a member K1 of {K..} and put G11 = K;. Assume that no point
of K1 belongs to every-K11 • Then the sets G11 form an open cover of K1 ;
°'" and since K1 is compact, there are finitely many indices a:1, ... , such
that K1 c: G111 u ••• u G11,.. But this means that
K1 n K111 n • • • n K11,. is empty, in contradiction to our hypothesis.
Corollary If {Kn} is a sequence of nonempty compact sets such that Kn=> Kn+t
(n = l, 2, 3, ...), then ni Kn is not empty.
2.37 neorem If E is an infinite subset of a compact set K, then E has a limit point inK.
Proof If no point of K were a limit point of E, then each q e K would have a neighborhood Vq which contains at most one point of E (namely, q, if q e E). It is clear that no finite subcollection of {Vq} can cover E; and the same is true of K, since E c K. This contradicts the compactness of K.
2.38 Theorem If {In} is a sequence of intervals in R1, such that In=> In+t
(n =1, 2, 3, ...), then ni In is not empty. Proof If In = [an, bn], let E be the set of all an. Then E is nonempty and
bounded above (by b1). Let x be the sup of E. If m and n are positive integers, then
0,. S am+n S bm+n Sbm,
so that x s b,,. for each m. Since it is obvious that a. :s; x, we see that x e I,,. form = l, 2, 3, ....

BASIC TOPOLOGY 39
2.39 Theorem Let k be a positive integer. If Vn} is a sequence of k-cells such
that In::::, In+ 1(n = 1, 2, 3, ...), then ni In is not empty. Proof Let In consist of all points x = (xi, ... , x,.) such that
= an,J ~ Xj ::S: bn,J (l ~j ~ k; n 1, 2, 3, ...),
and put In,J = [an,J, bn,Jl• For each j, the sequence {In,J} satisfies the
hypotheses of Theorem 2.38. Hence there are real numbers xj(l ~j ~ k) such that
(1 ~j ~ k; n = 1, 2, 3, ...).
Setting x• == (xT, ... , x:), we see that x• e In for n = 1, 2, 3, . . . . The
theorem follows.
2.40 Theorem Every k-cell is compact.
Proof Let I be a k-cell, consisting of all points x = (x1, .•• , x,.) such
that a1 ~x1 ~ b1 (1 ~j ~ k). Put
{t o= (b1 - a1)2r2.
o, Then Ix - y I ~ if x e I, y e I.
Suppose, to get a contradiction, that there exists an open cover {Gil}
of I which contains no finite subcover of I. Put c1 = (a1 + b1)/2. The
intervals [a1 , c1] and [c1 , b1] then determine 2" k-cells Q1 whose union is I. At least one of these sets Q;, call it I1, cannot be covered by any finite subcollection of {Gil} (otherwise I could be so covered). We next subdivide I1and continue the process. We obtain a sequence Vn} with the following properties:
(a) I ::::> I1 ::::, I2 ::::, I3 ::::, • • • ; (b) In is not covered by any finite subcollection of {Gil};
(c) ifx e In and ye In, then Ix - YI ~ r"o.
By (a) and Theorem 2.39, there is a point x• which lies in every In. For some ex, x• e Gil. Since Gil is open, there exists r > 0 such that
Iy - x• I < r implies that ye Gil. If n is so large that r"o < r (there is such an n, for otherwise 2" ::S: o/r for all positive integers n, which is
absurd since R is archimedean), then (c) implies that Inc Gil, which contradicts (b).
This completes the proof.
The equivalence of (a) and (b) in the next theorem is known as the HeineBorel theorem.

40 PRINCIPLES OP MATHEMATICAL ANALYSIS

2,41 neorem If a set E in R" has one of the following three properties, then it has the other two:

(a) Eis closed and bounded. (b) Eis compact. (c) Every infinite subset ofE has a limit point in E.

Proof If (a) holds, then E c: / for some k-cell /, and (b) follows from Theorems 2.40 and 2.35. Theorem 2.37 shows that (b) implies (c). It remains to be shown that (c) implies (a).
If E is not bounded, then E contains points x" with
(n = 1, 2, 3, ...).

The set S consisting of these points Xn is infinite and clearly has no limit point in R", hence has none in E. Thus (c) implies that Eis bounded.
If E is not closed, then there is a point x0 e R" which is a limit point
of E but not a point of E. For n = 1, 2, 3, ... , there are points Xn e E
such that Ix" - x0 I < 1/n. Let S be the set of these points x". Then S is infinite (otherwise Ix" - x0 I would have a constant positive value, for
infinitely many n), S has x0 as a limit point, and S has no other limit point in R". For if ye R", y ¢ x0 , then

lxn - YI ~ lxo -yl - lxn - Xol

~

1 lxo-Yl-,;~

l
2lxo-YI

for all but finitely many n; this shows that y is not a limit point of S (Theorem 2.20).
Thus S has no limit point in E; hence E must be closed if (c) holds.

We should remark, at this point, that (b) and (c) are equivalent in any metric space (Exercise 26) but that (a) does not, in general, imply (b) and (c). Examples are furnished by Exercise 16 and by the space !i'2, which is discussed in Chap. 11.

2,42 Theorem (Weierstrass) Every bounded infinite subset of R" has a limit
point in R".
Proof Being bounded, the set E in question is a subset of a k-cell / c: R". By Theorem 2.40, / is compact, and so E has a limit point in /, by Theorem 2.37.

BASIC TOPOLOGY 41
PERFECT SETS
2.43 Theorem Let P be a nonempty perfect set in Rt.. Then Pis uncountable. Proof Since P has limit points, P must be infinite. Suppose P is countable, and denote the points of P by xi, x2, x 3, . •. . We shall construct a sequence {Vn} of neighborhoods, as follows.
Let Vi be any neighborhood of x1• If V1 consists of all y e ~ such
that Iy - x1 I < r, the closure Vi of V1 is the set of all y e Rk such that IY -xii S: r.
Suppose Vn has been constructed, so that Vn n Pis not empty. Since every point of Pis a limit point of P, there is a neighborhood Vn+t such that (i) V..+ 1 c: Vn, (ii) xn¢ V..+i, (iii) Vn+i nP is not empty. By (iii), Vn+1 satisfies our induction hypothesis, and the construction can proceed.
Put Kn = Yn n P. Since Yn is closed and bounded, V,. is compact. Since xn ¢ Kn+ i, no point of P lies in nf Kn. Since Kn c:. P, this implies that nf Kn is empty. But each Kn is nonempty, by (iii), and Kn=> Kn+i,
by (i); this contradicts the Corollary to Theorem 2.36.
Corollary Every interval [a, b] (a< b) is uncountable. In particular, the set of all real numbers is uncountable.
2.44 The Cantor set The set which we are now going to construct shows that there exist perfect sets in R1 which contain no segment.
Let E0 be the interval [O, I]. Remove the segment (¼, ¼), and let E1 be the union of the intervals
[O, ¼] [¼, 1].
Remove the middle thirds of these intervals, and let E2 be the union of the intervals
[o, -H [¾, ¾1, [f, H [¾, 11.
Continuing in this way, we obtain a sequence of compact sets En, such that (a) E1 => E2 :::, E3 => • • • ; (b) En is the union of 2n intervals, each of length rn.
The set
is called the Cantor set. P is clearly compact, and Theorem 2.36 shows that P is not empty.

42 PRINCIPLES OF MATHEMATICAL ANALYSIS
No segment of the form
(24)
where k and m are positive integers, has a point in common with P. Since every segment (ci:, /3) contains a segment of the form (24), if
3-m <{3---C,(
6
P contains no segment. To show that Pis perfect, it is enough to show that P contains no isolated
point. Let x e P, and let S be any segment containing x. Let In be that interval of En which contains x. Choose n large enough, so that In c S. Let xn be an endpoint of In, such that Xn '# x.
It follows from the construction of P that xn e P. Hence x is a limit point of P, and P is perfect.
One of the most interesting properties of the Cantor set is that it provides us with an example of an uncountable set of measure zero (the concept of measure will be discussed in Chap. 11).
CONNECTED SETS
2.45 Definition Two subsets A and B of a metric space X are said to be separated if both A n B and An Bare empty, i.e., if no point of A lies in the closure of Band no point of B lies in the closure of A.
A set E c X is said to be connected if E is not a union of two nonempty separated sets.
2.46 Remark Separated sets are of course disjoint, but disjoint sets need not be separated. For example, the interval [0, 1] and the segment (1, 2) are not separated, since l is a limit point of (l, 2). However, the segments (0, 1) and (1, 2) are separated.
The connected subsets of the line have a particularly simple structure:
2.47 Theorem A subset E of the real line R1 is connected ifand only ifit has the
following property: If x e E, ye E, and x < z < y, then z e E.
I
Proof If there exist x e E, y e E, and some z e (x, y) such that z ¢ E, then
E = A,. u B,. where
A,. =En (-oo, z), B,. =En (z, oo).

BASIC TOPOLOGY 43
Since x e A. and ye B., A and Bare nonempty. Since A. c:: ( - oo, z) and B11 c:: (z, oo), they are separated. Hence E is not connected.
To prove the converse, suppose E is not connected. Then there are
nonempty separated sets A and B such that A u B = E. Pick x e A, ye B,
and assume (without loss of generality) that x < y. Define
z = sup (An [x, y]).
By Theorem 2.28, z e A; hence z ¢ B. In particular, x ~ z < y. If z ¢ A, it follows that x < z < y and z ¢ E. If z e A, then z ¢ B, hence there exists z1 such that z < z1 < y and z1 ¢ B. Then x < z1 < y and z1 ¢ E.
EXERCISES
1. Prove that the empty set is a subset of every set.
2. A complex number z is said to be algebraic if there are integers a0 , ••• , a., not all
zero, such that
aoz" + a1z•- 1 + ··· + an-1Z +a.= 0.
Prove that the set of all algebraic numbers is countable. Hint: For every positive
integer N there are only finitely many equations with
n+ laol +lad+···+ la.I =N.
3, Prove that there exist real numbers which are not algebraic. 4, Is the set of all irrational real numbers countable? 5. Construct a bounded set of real numbers with exactly three limit points. 6, Let E' be the set of all limit points of a set E. Prove that E' is closed. Prove that
E and £ have the same limit points. (Recall that£ =Eu E'.) Do E and E' always
have the same limit points? 7. Let Ai, A1, A3, ... be subsets of a metric space.
(a) If B. = Ur-1 A,, prove that B. = Ur-1 A,, for n = 1, 2, 3, ....
(b) If B = Ul°-1 A,, prove that B :::> Ui!.1 A,.
Show, by an example, that this inclusion can be proper.
8. Is every point of every open set E c R2 a limit point of E? Answer the same question for closed sets in R2•
9. Let E" denote the set of all interior points of a set E. [See Definition 2.18(e); E0 is called the interior of E.] {a) Prove that E" is always open.
(b) Prove that Eis open if and only if E0 = E.
(c) If G c E and G is open, prove that G c £ 0 • {d) Prove that the complement of E 0 is the closure of the complement of E. (e) Do E and £ always have the same interiors? (/) Do E and E0 always have the same closures?

44 PRINCIPLES OF MATHEMATICAL ANALYSIS

10. Let X be an infinite set. For p e X and q e X, define

d(p,q)= {~

(ifp ¢ q) (ifp =q).

Prove that this is a metric. Which subsets of the resulting metric space are open? Which are closed? Which are compact? 11, For x e R1 and ye R1, define

d1(x, y) = (x - y) 2,

dz(X, y) = VIX - YI,
da(x,y) = lxz -yzl, d4(X, y) = Ix - 2yj,

ds(x,y)

=

lx-yl 1 +lx-yl.

Determine, for each of these, whether it is a metric or not.
12. Let Kc: R1 consist of 0 and the numbers 1/n, for n = 1, 2, 3, . . . . Prove that K is
compact directly from the definition (without using the Heine-Borel theorem). 13. Construct a compact set of real numbers whose limit points form a countable set. 14. Give an example of an open cover of the segment (0, 1) which has no finite sub-
cover. 15. Show that Theorem 2.36 and its Corollary become false (in R1, for example) if the
word "compact" is replaced by "closed" or by "bounded."
16. Regard Q, the set of all rational numbers, as a metric space, with d(p, q) = IP - q I, Let Ebe the set of all p e Q such that 2 <p2 < 3. Show that Eis closed and
bounded in Q, but that Eis not compact. Is E open in Q? 17, Let Ebe the set of all x e (0, 1] whose decimal expansion contains only the digits
4 and 7. Is E countable? Is E dense in (0, 1]? Is E compact? Is E perfect? 18. Is there a nonempty perfect set in R1which contains no rational number? 19. (a) If A and B are disjoint closed sets in some metric space X, prove that they
are separated. (b) Prove the same for disjoint open sets.
(c) Fix p e X, 8 > 0, define A to be the set of all q e X for which d(p, q) < 8, define B similarly, with > in place of <. Prove that A and Bare separated.
(d) Prove that every connected metric space with at least two points is uncount• able. Hint: Use (c). 20. Are closures and interiors of connected sets always connected? (Look at subsets of R2.) 21. Let A and B be separated subsets of some Rt, suppose a e A, be B, and define
p(t) = (1 - t)a + tb
for teR1• Put Ao =p- 1(A), Bo =p- 1(B). [Thus teA 0 if and only ifp(t)eA.]

BASIC TOPOLOGY 45
(a) Prove that Ao and Bo are separated subsets of R1•
(b) Prove that there exists toe (0, 1) such that p(to) ¢ A u B.
(c) Prove that every convex subset of Rt is connected. 22. A metric space is called separable if it contains a countable dense subset. Show
that Rt is separable. Hint: Consider the set of points which have only rational coordinates.
23. A collection {V«} of open subsets of X is said to be a base for X if the following
is true: For every x e X and every open set G c X such that x e G, we have x e V« c G for some ex, In other words, every open set in X is the union of a subcollection of {V«l•
Prove that every separable metric space has a countable base. Hint: Take all neighborhoods with rational radius and center in some countable dense subset of X. 24. Let X be a metric space in which every infinite subset has a limit point. Prove that
Xis separable. Hint: Fix 8 > 0, and pick x1 e X. Having chosen Xi, .•. , x1 e X, choose x1 + 1 e X, if possible, so that d(x,, x1+ 1)~ 8 for i = 1, ... ,j. Show that
this process must stop after a finite number of steps, and that X can therefore be
covered by finitely many neighborhoods of radius 8. Take 8 = 1/n (n = 1, 2, 3, ...),
and consider the centers of the corresponding neighborhoods.
25. Prove that every compact metric space K has a countable base, and thclt K is therefore separable. Hint: For every positive integer n, there are finitely many neighborhoods of radius 1/n whose union covers K.
26. Let X be a metric space in which every infinite subset has a limit point. Prove that X is compact. Hint: By Exercises 23 and 24, X has a countable base. It
follows that every open cover of Xhas a countable subcover {G.}, n = 1, 2, 3, ....
Ifno finite subcollection of {G.} covers X, then the complement F. of G1 u ••• u G. is nonempty for each n, but () F. is empty. If Eis a set which contains a point from each F., consider a limit point of E, and obtain a contradiction.
27. Define a point p in a metric space X to be a condensation point of a set E c X if every neighborhood of p contains uncountably many points of E. Suppose E c Rt, Eis uncountable, and let P be the set of all condensation points of E. Prove that P is perfect and that at most countably many points of E are not in P. In other words, show that pc r. E is at most countable. Hint: Let
{V.} be a countable base of Rt, let W be the union of those v. for which E r. v.
is at most countable, and show that P = we.
28. Prove that every closed set in a separable metric space is the union of a (possibly empty) perfect set and a set which is at most countable. (Corollary: Every countable closed set in Rt has isolated points.) Hint: Use Exercise 27.
29. Prove that every open set in R1 is the union of an at most countable collection of disjoint segments. Hint: Use Exercise 22.

46 PRINCIPLES OF MATHEMATICAL ANALYSIS

30. Imitate the proof of Theorem 2.43 to obtain the following result:
u If Rk = rFn' where each Fn is a closed subset of Rt, then at least one Fn
has a nonempty interior.

Equivalent statement: If Gn is a dense open subset of Rk, for n = l, 2, 3, ... ,

nrc. then

is not empty (in fact, it is dense in Rk).

(This is a special case of Baire's theorem; see Exercise 22, Chap. 3, for the general case.)

3
NUMERICAL SEQUENCES AND SERIES
As the title indicates, this chapter will deal primarily with sequences and series of complex numbers. The basic facts about convergence, however, are just as easily explained in a more general setting. The first three sections will therefore be concerned with sequences in euclidean spaces, or even in metric spaces.
CONVERGENT SEQUENCES 3.1 Definition A sequence {Pn} in a metric space Xis said to converge if there
is a point p e X with the following property: For every s > 0 there is an integer
N such that n ~ N implies that d(pn, p) < e. (Here d denotes the distance in X.) In this case we also say that {Pn} converges to p, or that p is the limit of
{Pn} [see Theorem 3.2(b)], and we write Pn ➔ p, or lim Pn =p.
n ➔ oo
If {Pn} does not converge, it is said to diverge.

48 PRINCIPLES OF MATHEMATICAL ANALYSIS

It might be well to point out that our definition of "convergent sequence"
depends not only on {Pn} but also on X; for instance, the sequence {1/n} con-
verges in R1 (to 0), but fails to converge in the set of all positive real numbers
[with d(x, y) = Ix - y I]. In cases of possible ambiguity, we can be more
precise and specify "convergent in X" rather than "convergent."
We recall that the set of all points Pn (n = 1, 2, 3, ...) is the range of {Pn}.
The range of a sequence may be a finite set, or it may be infinite. The sequence
{Pn} is said to be bounded if its range is bounded.
As examples, consider the following sequences of complex numbers
(that is, X = R2):

(a) If sn = 1/n, then limn.. 00 sn = 0; the range is infinite, and the sequence

is bounded.
(b) lf sn = n2, the sequence {sn} is unbounded, is divergent, and has

infinite range.
(c) If Sn = 1 + [( - 1)"/n], the sequence {sn} converges to I, is bounded,

and has infinite range.

(d) If Sn= i", the sequence {sn} is divergent, is bounded, and has finite

range.
(e) Ifs"= 1 (n = 1, 2, 3, ...), then {sn} converges to 1, is bounded, and

has finite range.

•

We now summarize some important properties of convergent sequences in metric spaces.

3.2 Theorem Let {Pn} be a sequence in a metric space X.
(a) {Pn} converges top e X if and only if every neighborhood ofp contains Pn for all but finitely many n.
(b) If p e X, p' e X, and if{Pn} converges top and top', then p' = p.
(c) If {Pn} converges, then {Pn} is bounded. (d) If E c X and ifp is a limit point ofE, then there is a sequence {Pn} in E
such that p = lim Pn .
n ➔ co
Proof (a) Suppose Pn ➔ p and let V be a neighborhood of p. For
some e > 0, the conditions d(q, p) < e, q e X imply q e V. Corresponding to this e, there exists N such that n ~ N implies d(pn, p) < e. Thus
n ~ N implies Pn e V. Conversely, suppose every neighborhood of p contains all but
finitely many of the Pn . Fix e > 0, and let V be the set of all q e X such that d(p, q) < e. By assumption, there exists N (corresponding to this V) such that Pn e V if n ~ N. Thus d(Pn,P) < e if n ~ N; hence Pn ➔p.

NUMERICAL SEQUENCES AND SERIES 49

(b) Let e > 0 be given. There exist integers N, N' such that
n~N implies d(p., p) < 2e'

n~N'

implies

d(pn,

, p)

<

2e -

Hence if n ~ max (N, N'), we have

d(p, p') :::;; d(p, p.) + d(p., p') < e.

Since e was arbitrary, we conclude that d(p, p') = 0.
(c) Suppose P. - p. There is an integer N such that n > N
implies d(p., p) < 1. Put

r = max {l, d(Pi, p), ... , d(pN, p)}.
Then d(p.,p) ::S: r for n = 1, 2, 3, ....
(d) For each positive integer n, there is a point p. EE such that
d(p., p) < 1/n. Given e > 0, choose N so that Ne > 1. If n > N, it follows that d(p., p) < e. Hence Pn - p.
This completes the proof.

For sequences in Rk we can study the relation between convergence, on the one hand, and the algebraic operations on the other. We first consider sequences of complex numbers.

3.3 Theorem Suppose {s.}, {t.} are complex sequences, and limn ➔ oo s. = s,
limn- 00 tn = t. Then

(a) lim(s.+tn)=s+t;
•
(b) Jim cs.= cs, lim (c + sn) = c + s,for any number c;

n-+oo

n ➔ oo

(c) Jim s.t. = st;

n ➔ oo

(d) lim .!_=~,provided s. # 0 (n = 1, 2, 3, ...), ands# 0.
n ➔ oo Sn S
Proof
(a) Given e > 0, there exist integers N1, N 2 such that

n ~ N 1 implies

n ~ N 2 implies

50 PRINCIPLES OF MATHEMATICAL ANALYSIS

If N = max (Ni, N 2), then n :2:: N implies

l(s,.+tn)-(s+t)I S lsn-sl + ltn-tl <s.

This proves (a). The proof of (b) is trivial.

(c) We use the identity

(1)

Sntn - st = (sn - s)(tn - t) + s(tn - t) + t(sn - s).

Given s > 0, there are integers Ni, N 2 such that
J;, n :2:: N1 implies lsn - sl < J;. n :2:: N 2 implies Itn - ti <
If we take N = max (Ni, N 2 ), n :2:: N implies
l(sn-s)(t,.-t)I <s,

so that
lim (sn - s)(t,. - t) =0.
n ➔ oo
We now apply (a) and (b) to (1), and conclude that
lim (sntn - st) = 0.
n ➔ oo
(d) Choosing m such that ISn - s I < ½Is I if n :2:: m, we see that

(n :2:: m).

Given s > 0, there is an integer N > m such that n :2:: N implies

lsn - sl < ½lsl 2s.

Hence, for n :2:: N,

I 1-S1n

-

-11
s

=

- - Sn - SI
sn3

<I-s2\l 2sn-sl

<s.

3.4 Theorem
(a) Suppose x,. e Rk (n =1, 2, 3, ...) and

Then {xn} converges to x = (a1, ..• , a1c) if and only if

(2)

lim rx1,n = rx1 (lSjSk).

n ➔ oo

NUMERICAL SEQUENCES AND SERIES 51

(b) Suppose {x.}, {yJ are sequences in R', {/Jn} is a sequence ofreal numbers, and x,, ➔ x, y,. ➔ y, /J,. ➔ /J. Then

lim (X,, + y,.) = X + y,
11 ➔ 00

lim x,, •Yn == X. y,
n ➔ co

lim /J,. x,, = /Jx.
n ➔ co

Proof (a) If x,. ➔ x, the inequalities

101:1,11-•01:JI S lxn-xl,
which follow immediately from the definition of the norm in R', show that
(2) holds. Conversely, if (2) holds, then to each e > 0 there corresponds an
integer N such that n c?: N implies

Jk8
101:J,h - 01:JI <

(1 SjS k).

Hence n c?: N implies

={f lx,.-xl

l01:1,n-01:Jl 2}112 < 8 ,

J=1

so that Xn ➔ x. This proves (a). Part (b) follows from (a) and Theorem 3.3.

SUBSEQUENCES
3.5 Definition Given a sequence {pn}, consider a sequence {nk} of positive integers, such that n1 < n2 < n3 < •••. Then the sequence {p,.,} is called a subsequence of {p,.}. If {p,.1} converges, its limit is called a subsequential limit o f {p,.}.
It is clear that {p,.} converges to p if and only if every subsequence of {Pn} converges top. We leave the details of the proof to the reader.
3.6 Theorem
(a) If {p,.} is a sequence in a compact metric space X, then some subsequence of{Pn} converges to a point of X.
(b) Every bounded sequence in R' contains a convergent subsequence.

52 PRINCIPLES OP MATHEMAnCAL ANALYSIS

Proof
(a) Let Ebe the range of {Pn}. If E is finite then there is a p e E and a
sequence {n1} with n1 < n2 < n3 < ••·, such that
Pn, -Pna - ••• -p.
The subsequence {p,.1} so obtained converges evidently to p. If E is infinite, Theorem 2.37 shows that E has a limit point p e X.
Choose n1 so that d(p, Pn.) < 1. Having chosen ni, ... , n,-i, we see from
Theorem 2.20 that there is an integer n1 > n1_ 1 such that d(P,Pn,) < 1/i. Then {p,.,} converges to p.
(b) This follows from (a), since Theorem 2.41 implies that every bounded subset of ,RA Jies in a compact subset of ,RA,

3.7 Theorem The subsequential limits of a sequence {p,.} in a metric space X form a closed subset of X.

Proof Let E* be the set of all subsequential limits of {Pn} and let q be a
limit point of E*. We have to show that q e E*.

Choose n1 so that Pni r/t- q. (If no such n1 exists, then E* has only
one point, and there is nothing to prove.) Put 6 =d(q, p,..). Suppose

n1, ... , n1_ 1 are chosen. Since q is a limit point of E*, there is an x e E*

r'6. with d(x, q) < i-16. Since x e E*, there is an n1 > n1_ 1 such that

d(x,p,.1) <

Thus

d(q,pn,) s 21 - 16

= for i 1, 2, 3, .... This says that {p,.,} converges to q. Hence q e E*.

CAUCHY SEQUENCES
3.8 Definition A sequence {p,,} in a metric space X is said to be a Cauchy sequence if for every e > 0 there is an integer N such that d(p,., p,,,) < e if n ~ N andm~N.
In our discussion of Cauchy sequences, as well as in other situations which will arise later, the following geometric concept will be useful.
3,9 Definition Let E be a nonempty subset of a metric space X, and let S be the set of all real numbers of the form ld(p, q), with ,p e E and q e E. the sup of S is called the diameter of E.

NUMERICAL SEQUENCES AND SERIES 53

If(pn}is a sequence in X and if EN consists of the pointspN, PN+t,PN+ 2 , ••• ,

it is clear from the two preceding definitions that (Pn} is a Cauchy sequence

if and only if

lim diam EN = 0.

3.10 Theorem

(a) If E is the closure of a set E in a metric space X, then
diam E = diam E.

(b) If Kn is a sequence of compact sets in X such that Kn::::, Kn+l (n = 1, 2, 3, ...) and if
lim diam Kn = 0,
n n ➔ ao
then j Kn consists of exactly one point.

Proof

(a) Since E c E, it is clear that

diam E ::,;; diam E.

Fix e > 0, and choose p e E, q e E. By the definition of E, there are points p', q', in E such that d(p, p') < e, d(q, q') < e. Hence

d(p, q) :s; d(p, p') + d(p' q') + d(q', q)
< 2e + d(p', q') :s; 2e -+ diam E.

It follows that

diam E :s; 2e + diam E,

n and since e was arbitrary, (a) is proved.
(b) Put K = j Kn. By Theorem 2.36, K is not empty. If K contains more than one point, then diam K > 0. But for each n, Kn ::::, K, so that
diam Kn ~ diam K. This contradicts the assumption that diam Kn ➔ 0.

3.11 Theorem
(a) In any metric space X, every convergent sequence is a Cauchy sequence. (b) If Xis a compact metric space and if{Pn} is a Cauchy sequence in X,
then {Pn} converges to some point of X. (c) In Rk, every Cauchy sequence converges.
Note: The difference between the definition of convergence and the definition of a Cauchy sequence is that the limit is explicitly involved in the former, but not in the latter. Thus Theorem 3.ll(b) may enable us

54 PRINCIPLES OF MATHEMATICAL ANALYSIS

to decide whether or not a given sequence converges without knowledge of the limit to which it may converge.
The fact (contained in Theorem 3.11) that a sequence converges in Rk if and only if it is a Cauchy sequence is usually called the Cauchy criterion for convergence.

Proof
(a) If Pn -+ p and if e > 0, there is an integer N such that d(p, Pn) < e
for all n ~ N. Hence
d(pn, Pm) ~ d(Pn, p) + d(p, Pm) < 2e
as soon as n ~ N and m ~ N. Thus {pn} is a Cauchy sequence.

(b) Let {Pn} be a Cauchy sequence in the compact space X. For

N = l, 2, 3, ... , let E 11 be the set consisting of PN, PN+t, PN+2, ... .
Then

(3)

lim diam EN= 0,

N-+~

by Definition 3.9 and Theorem 3.lO(a). Being a closed subset of the
compact space X, each EN is compact (Theorem 2.35). Also EN::::> EN+ 1, so that EN::::> EN+1•
Theorem 3.l0(b) shows now that there is a unique p e X which lies in every EN.
Let e > 0 be given. By (3) there is an integer NO such that diam EN< e if N ~ N0 . Since p e EN, it follows that d(p, q) < e for every q e EN, hence for every q e EN. In other words, d(p, Pn) < e if n ~ NO • This says precisely that Pn ➔ p.

(c) Let {xn} be a Cauchy sequence in Rk. Define EN as in (b), with x1 in place of Pi. For some N, diam EN < 1. The range of {xn} is the union of EN and the finite set {x1, ... , xN_ 1}. Hence {x,,} is bounded. Since
every bounded subset of K has compact closure in K (Theorem 2.41),
(c) follows from (b).

3.12 Definition A metric space in which every Cauchy sequence converges is said to be complete.
Thus Theorem 3.11 says that all compact metric spaces and all Euclidean spaces are complete. Theorem 3.11 implies also that every closed subset E of a complete metric space X is complete. (Every Cauchy sequence in Eis a Cauchy sequence in X, hence it converges to some p e X, and actually p e E since Eis closed.) An example of a metric space which is not complete is the space of an
rational numbers, with d(x, y) = Ix - y I,

NUMERICAL SEQUENCES AND SERIES 55
Theorem 3.2(c) and example (d) of Definition 3.1 show that convergent sequences are bounded, but that bounded sequences in R" need not converge. However, there is one important case in which convergence is equivalent to boundedness; this happens for monotonic sequences in R1.
3.13 Definition A sequence {sn} of real numbers is said to be
(a) monotonically increasing if Sn s; Sn+i (n = 1, 2, 3, ...); (b) monotonically decreasing if Sn~ Sn+i (n = 1, 2, 3, ...).
The class of monotonic sequences consists of the increasing and the decreasing sequences.
3.14 Theorem Suppose {sn} is monotonic. Then {s.} converges if and only if it is bounded.
Proof Suppose Sn s; Sn+i (the proof is analogous in the other case). Let E be the range of {sn}. If {sn} is bounded, let s be the least upper bound of E. Then
s. s; s (n = 1, 2, 3, ...).
For every s > 0, there is an integer N such that
S- 8 <SN~ S,
for otherwise s - s would be an upper bound of E. Since {sn} increases, n ~ N therefore implies
s- 8 < s,, S s, which shows that {s,,} converges (to s).
The converse follows from Theorem 3.2(c).
UPPER AND LOWER LIMITS
3.15 Definition Let {s"} be a sequence of real numbers with the following property: For every real M there is an integer N such that n ~ N implies s,, ~ M. We then write
s,, ➔ +oo. Similarly, if for every real M there is an integer N such that n ~ N implies s,, s; M, we write

56 PRINCIPLES OF MATHEMATICAL ANALYSIS
It should be noted that we now use the symbol ➔ (introduced in Definition 3.1) for certain types of divergent sequences, as well as for convergent sequences, but that the definitions of convergence and of limit, given in Definition 3.1, are in no way changed.
3.16 Definition Let {sn} be a sequence of real numbers. Let Ebe the set of numbers x (in the extended real number system) such that snk ➔ x for some subsequence {snk}. This set E contains all subsequential limits as defined in
Definition 3.5, plus possibly the numbers + oo, - oo.
We now recall Definitions 1.8 and 1.23 and put
s• = supE, s. = inf E.
The numbers s*, s* are called the upper and lower limits of {sn}; we use the notation
Jim sup Sn= s*,
ff ➔ OO
3.17 Theorem Let {sn} be a sequence of real numbers. Let E ands* have the same meaning as in Definition 3.16. Thens• has the following two properties:
(a) s* e E. (b) If x > s*, there is an integer N such that n:?: N implies Sn< x.
Moreover, s* is the only number with the properties (a) and (b).
Of course, an analogous result is true for s•.
Proof
(a) Ifs* = +oo, then Eis not bounded above; hence {sn} is not bounded above, and there is a subsequence {snk} such that snk ➔ + oo.
Ifs* is real, then Eis bounded above, and at least one subsequential limit exists, so that (a) follows from Theorems 3.7 and 2.28.
If s• = - oo, then E contains only one element, namely - oo, and
there is no subsequential limit. Hence, for any real M, Sn > M for at most a finite number of values of n, so that sn--+ - oo.
This establishes (a) in all cases. (b) Suppose there is a number x > s* such that Sn ~ x for infinitely many values of n. In that case, there is a number y e E such that y ~ x > s*, contradicting the definition of s*.
Thuss* satisfies (a) and (b). To show the uniqueness, suppose there are two numbers, p and q, which satisfy (a) and (b), and supposep < q. Choose x such thatp < x < q. Since p satisfies (b), we have Sn < x for n ~ N. But then q cannot satisfy (a).

NUMERICAL SEQUENCES AND SBRIBS 57

3.18 •Examples

(a) Let {sn} be a sequence containing all rationals. Then every real number is a subsequential limit, and

lim sup S11 = + 00,
n..,.00

lim inf Sn= -oo.
n..,.00

(b) Let Sn = ( -1 ") /[1 + (1/n)]. Then

Jim sups"= 1,
11 ➔ 00

Jim inf Sn = - 1.
n-+00

(c) For a real-valued sequence {sn}, lim Sn= s if and only if
n-+00
Jim sup Sn = lim inf Sn = S.

We close this section with a theorem which is useful, and whose proof is quite trivial:

3.19 Theorem Ifs,. :s; t11 for n ~ N, where N is fixed, then

Jim inf s,. :s; Jim inf tn,

,.... 00

,. ... 00

Jim sup Sn =S; Jim sup tn.

n-+oo

n-+oo

SOME SPECIAL SEQUENCES
We shall now compute the limits of some sequences which occur frequently. The proofs will all be based on the following remark: If O:s; x11 :s; s11 for n ~ N, where N is some fixed number, and if Sn ➔ 0, then Xn ➔ 0.

3.20 Theorem

(a) If p > 0, then lim 1., = 0.
..... oon
(b) If p > 0, then lim ::fp = 1.
,. ... 00
(c) Jim :jn = 1.
n..,.00

(d)

If p

>

0

and oc

is

real,

then

lim (l
11-+00

,f" )
+ P

11

=

0.

(e) Iflxl <1,thenlimx"=0.
,. .... 00

58 PRINCIPLES OF MATHEMATICAL ANALYSIS

Proof
(a) Take n > (1/e)11". (Note that the archimedean property of the real number system is used here.)
(b) If p > 1, put Xn = ::/P - 1. Then Xn > 0, and, by the binomial
theorem,

so that

0<x II

s

:p--n -1 ·

Hence Xn ➔ 0. Ifp = 1, (b) is trivial, and if O< p < 1, the result is obtained

by taking reciprocals.

::/n - (c) Put xn =

1. Then x,, ~ 0, and, by the binomial theorem,

x;. n = (1

+

Xn)"

~

n(n 2

1)

Hence

0 ~ Xn s:Jn2-1

(n ~ 2).

(d) Let k be an integer such that k > «, k > 0. For n > 2k,

(l

)"
+ P

>

'") k = n(n
u P

-

1) .. • (n k!

k + 1) k nky.
p > 2kk!

Hence
0 n,. 2kk! m-k < (1 +pt< p" n
Since« - k < 0, n"'-" -+O, by (a).
(e) Take « = 0 in (d).

(n > 2k).

SERIES
In the remainder of this chapter, all sequences and series under consideration will be complex-valued, unless the contrary is explicitly stated. Extensions of some of the theorems which follow, to series with terms in R", are mentioned in Exercise 15.

NUMERICAL SEQUENCES AND SERIES 59
3.21 Definition Given a sequence {a.}, we use the notation
n=p
to denote the sum aP + ap+t + ••• + aq. With {a.} we associate a sequence
{sn}, where

For {sn} we also use the symbolic expression

or, more concisely,

a1 + a2 + a3 + •••

(4)

The symbol (4) we call an in.finite series, or just a seriPs. The numbers s. are called the partial sums of the series. If {s.} converges to s, we say that the series converges, and write

L00 a.= s.
n= l

The number s is called the sum of the series; but it should be clearly under-

stood that s is the limit of a sequence of sums, and is not obtained simply by

addition.

If {s.} diverges, the series is said to diverge.

Sometimes, for convenience of notation, we shall consider series of the

form

00

(5)

La ..

n=O

And frequently, when there is no possible ambiguity, or when the distinction is immaterial, we shall simply write I:an in place of (4) or (5).
It is clear that every theorem about sequences can be stated in terms of
series (putting a1 = s1, and a. = s. - s. _1 for n > 1), and vice versa. But it is
nevertheless useful to consider both concepts. The Cauchy criterion (Theorem 3.11) can be restated in the following
form:

3,22 Theorem I:a. converges if and only iffor every B > 0 there is an integer
N such that
(6)
ifm~n~N.

60 Pll.INCIPLES OF MATHEMATICAL ANALYSIS

In particular, by taking m =n, (6) becomes

In other words:

(n ~ N).

3.23 Theorem /JI.an converges, then limn-+ao an= 0.

The condition a,, ➔ 0 is not, however, sufficient to ensure convergence of I.a•. For instance, the series
r -1 00
n=l n
diverges; for the proof we refer to Theorem 3.28.

Theorem 3.14, concerning monotonic sequences, also has an immediate counterpart for series.

3.24 Theorem A series of nonnegative1 terms converges if and only if its partial sums form a bounded sequence.

We now tum to a convergence test of a different nature, the so-called "comparison test."

3.25 Theorem
(a) If Ia. I S Cn for n ~ N0 , where N0 is some fixed integer, and if "£c,,
converges, then Ian converges. (b) If a,,~ dn ~ 0 for n ~ N0 , and if "£d,, diverges, then I.a,, diverges.

Note that (b) applies only to series of nonnegative terms a,,. Proof Given e > 0, there exists N ~ N 0 such that m ~ n ~ N implies
L"' c" ~ e,
k=n
by the Cauchy criterion. Hence
IJn°"' s,,tla1:I ~J/"~e,
and (a) follows. Next, (b) follows from (a), for if Ian converges, so must Id,, [note
that (b) also follows from Theorem 3.24].
1 The expression " nonnegative" always refers to real numbers.

NUMERICAL SEQUENCES AND SERIES 61
The comparison test is a very useful one; to use it efficiently, we have to become familiar with a number of series of nonnegative terms whose convergence or divergence is known.

SERIES OF NONNEGATIVE TERMS The simplest of all is perhaps the geometric series.

3.26 Theorem //0 s; x < 1, then

I00

x

"

=

1
--

·

n=O

1- X

If x ~ 1, the series diverges.

Proof If x ~ 1,

L n

1-x"+l

s n

=

/c=O

J

!

=1

--

-
X

·

The result follows if we let n--+ oo. For x = 1, we get

1+1+1+"',

which evidently diverges.

In many cases which occur in applications, the terms of the series decrease monotonically. The following theorem of Cauchy is therefore of particular interest. The striking feature of the theorem is that a rather "thin" subsequence of {an} determines the convergence or divergence of I:an.

Li~= 3.27 Theorem Suppose a1 ~ a2 ~ a3 ~ • • • ~ 0. Then the series 1 an con-
verges if and only if the series

(7) converges.

00
L 2"a2k = a1 + 2a2 + 4a4 + 8a8 + •••
/c=O

Proof By Theorem 3.24, it suffices to consider boundedness of the partial sums. Let
Sn = a1 + a2 + ••• + an '
t1c = a1 + 2a2 + ••• + 21ca2k.

62 PRINCIPLES OF MATHEMATICAL ANALYSIS

For n < 2k,
Sn s a1 + (a2 + a 3) + ••• + (a2k + ••• + a2 k+i- 1) s a1 + 2a2 + ••• + 2ka2k

so that (8)

On the other hand, if n > 2\

so that

sn ~ a1 + a2 + (a3 + a4 ) + ••• + (a2k- 1 +1 + ••• + a2k) ~ ½a1 + a2 + 2a4 + ••• + 2k- 1a2k = ½tk'

(9)

By (8) and (9), the sequences {s"} and {tk} are either both bounded or both unbounded. This completes the proof.

L - 3.28 Theorem

I converges if p > I and diverges if p s 1.

nP

Proof If p s 0, divergence follows from Theorem 3.23. If p > 0,
Theorem 3.27 is applicable, and we are led to the series

I = I "' 2k . k1p

"' 2(1 - p)k_

k=O 2

k=O

Now, 21 -p < 1 if and only if 1 - p < 0, and the result follows by com-
parison with the geometric series (take x = 21 - P in Theorem 3.26).
As a further application of Theorem 3.27, we prove:

3.29 Theorem If p > I,

(10)

"' I

J2n(logn)P

converges,· ifp s I, the series diverges.

Remark "log n" denotes the logarithm of n to the base e (compare Exercise 7,
Chap. 1); the number e will be defined in a moment (see Definition 3.30). We
let the series start with n = 2, since log 1 = 0.

NUMERICAL SEQUENCES AND SERIES 63

Proof The monotonicity of the logarithmic function (which will be discussed in more detail in Chap. 8) implies that {log n} increases. Hence {l/n log n} decreases, and we can apply Theorem 3.27 to (10); this leads us to the series

00

1

00

1

1 1 00

(11)

!"" • 1"'f 21(log 21)P =1~1 (k log 2)P = (log 2)P1~1 kP'

and Theorem 3.29 follows from Theorem 3.28.

This procedure may evidently be continued. For instance,

00

1

(12)

n~3 n log n log log n

diverges, whereas

00

1

(13)

n"'f3 n log n(log log n)2

converges.

We may now observe that the terms of the series (12) differ very little from those of (13). Still, one diverges, the other converges. If we continue the process which led us from Theorem 3.28 to Theorem 3.29, and then to (12) and (13), we get pairs of convergent and divergent series whose terms differ even less than those of (12) and (13). One might thus be led to the conjecture that there is a limiting situation of some sort, a "boundary" with all convergent series on one side, all divergent series on the other side-at least as far as series with monotonic coefficients are concerned. This notion of "boundary" is of course quite vague. The point we wish to make is this: No matter how we make this notion precise, the conjecture is false. Exercises ll(b) and 12(b) may serve as illustrations.
We do not wish to go any deeper into this aspect of convergence theory, and refer the reader to Knapp's "Theory and Application of Infinite Series," Chap. IX, particularly Sec. 41.

THE NUMBER e
f 3.30 Definition e = ..!_• 11-on!
Here n! = 1 • 2 • 3 .. •n if n ~ l, and O! = 1.

64 PRINCIPLES OF MATHEMATICAL ANALYSIS

Since

1

1

1

s• =1+11+ ·2 -+ 1·2-·-3-+···1+·2-..-·-n

1 I

1

<

1

+

1

+2

-

2+2

-

+

"

·

+2·--

<
1

3 '

the series converges, and the definition makes sense. In fact, the series converges very rapidly and allows us to compute e with great accuracy.
It is of interest to note that e can also be defined by means of another limit process; the proof provides a good illustration of operations with limits:

3.31 Theorem lim (1 + !)" = e.

n ➔ oo

n

Proof Let

n 1
s. = L k'' k=O •

By the binomial theorem,

Hence t. :$; s., so that

(14)

lim sup tn :$; e,

n ➔ oo

by Theorem 3.19. Next, if n ;:=: m,

t. ;:=:

1 +

I

+ -21! ( 1 -

-I)
n

+ ••• + m-1!

( 1 -

-I) ••• ( 1 -
n

- m --1) ·
n

Let n ➔ ex:,, keeping m fixed. We get

so that

11. m m. f
n ➔ cfJ

t.

;:=:

1 +

1 +

1
21
•

+

..

• +

1
m1 .

,

n ➔ OO
Letting m ➔ oo, we finally get

(15)

e :$; lim inf t• .

n ➔ oo

The theorem follows from (14) and (15).

NUMERICAL SEQUENCES AND SERIES 65

The rapidity with which the series L _!_ converges can be estimated as nl
follows: If Sn has the same meaning as above, we have

1

1

1

e - Sn = (n + 1) ! + (n + 2) I + (n + 3) ! + •••

so that (16)

< -1- - {1 + -1- + - -1- + · · ·}=1-

(n+l)! n+l (n+1)2

n!n

O

<

e

-

s n

<n-!1n·

Thus s10 , for instance, approximates e with an error less than 10- 1• The inequality (16) is of theoretical interest as well, since it enables us to prove the irrationality of e very easily.

3.32 Theorem e is irrational.

Proof Suppose e is rational. Then e = p/q, where p and q are positive

integers. By (16),

1

(17)

0 < q!(e -

Sq)< - • q

By our assumption, q!e is an integer. Since

q1•sq =q 1•(1 + 1 + 2_!!_ + ••• + q_!!_)
is an integer, we see that q!(e - sq) is an integer. Since q ~ l, (17) implies the existence of an integer between O and 1.
We have thus reached a contradiction.

Actually, e is not even an algebraic number. For a simple proof of this, see page 25 of Niven's book, or page 176 of Herstein's, cited in the Bibliography.

THE ROOT AND RATIO TESTS
3.33 Theorem (Root Test) Given tan, put ix = lim sup::/ Ian 1-
n ➔ oo
Then
(a) if ix < 1, tan converges; (b) if ix > 1, tan diverges,·
(c) if ix = 1, the test gives no information.

66 PRINCIPLES OF MATHEMATICAL ANALYSIS
Proof If ex < 1, we can choose P so that ex < p < 1, and an integer N
such that
vTciJ <P
for n ~ N [by Theorem 3.17(b)]. That is, n ~ N implies
lanl < pn. Since O< p < 1, r,pn converges. Convergence of !:a" follows now from
the comparison test.
If ex > 1, then, again by Theorem 3.17, there is a sequence {nk} such
that
Hence Ia" I > 1 for infinitely many values of n, so that the condition an ➔ O, necessary for convergence of "£a", does not hold (Theorem 3.23).
To prove (c), we consider the series

For each of these series ex= 1, but the first diverges, the second converges.

3.34 Theorem (Ratio Test) The series "£a"

I (a) converges 1i/11•m sup l-an+l < 1,

n ➔ oo

an

I !~ ~ (b) diverges if 0:: 1 1for all n n0 , where n0 is some fixed integer.

Proof If condition (a) holds, we can find p < 1, and an integer N, such
that

for n ~ N. In particular,
laN+il < PlaNI,
IaN +2 I < PI aN +1 I < P2 IaN I,

NUMERICAL SEQUENCES AND SERIES 67
That is, lanl < laNIP-N' pn
for n ~ N, and (a) follows from the comparison test, since r.p• converges. If Ian+ 1 1 ~ Ia. I for n ~ n0 , it is easily seen that the condition a. ➔ 0
does not hold, and (b) follows.
Note: The knowledge that lim a.+ 1/a. = 1 implies nothing about the
convergence of r.a•. The series 'f.1/n and 'f.1/n2 demonstrate this.
3.35 Examples (a) Consider the series 11 1 1 1 1 1 1
2+ 3+ 22 + 32 + 23 + 33 + 24 + 34 +
for which
lim inf an+l = lim (~)" = 0,
n ➔ oo an n ➔ oo 3

h•m sup :n:/-a. -_ h•m 2~n --1;;_ - ✓_ 1-•

n ➔ oo

•""oo 2

2

I1•mn ➔s<X>u

pa.-+-1
a.

=

n1➔1. moo

-1 (3- )" 2 2

=

+ oo.

The root test indicates convergence; the ratio test does not apply. (b) The same is true for the series

I

11 1 1 1 1

2 + 1 + 8 + 4 + 32 + 16 + 128 + 64 + ••• '

where but

I1.m m. fa-.+-1 =-1, n ➔ oo a. 8
= lim sup an+l 2,
n➔ oo a.

68 PRINCIPLES OF MATHEMATICAL ANALYSIS

3.36 Remarks The ratio test is frequently easier to apply than the root test, since it is usually easier to compute ratios than nth roots. However, the root test has wider scope. More precisely: Whenever ihe ratio test shows convergence, the root test does too; whenever the root test is inconclusive, the ratio test is too. This is a consequence of Theorem 3.37, and is illustrated by the above examples.
Neither of the two tests is subtle with regard to divergence. Both deduce divergence from the fact that an does not tend to zero as n-+ oo.

3.37 Theorem For any sequence {en} ofpositive numbers,

It•m t'nf -Cn-+l :S: lim inf VnCen,

n-+oo Cn

n-ooo

v h'm sup nCen :S: h'm suCpn+-l·

n ➔ oo

n ➔ oo Cn

Proof We shall prove the second inequality; the ~roof of the first is quite similar. Put

= oi I1,m supC-n+-l• n-+oo Cn
If oi = + oo, there is nothing to prove. If oi is finite, choose P> oi. There
is an integer N such that

for n ~ N. In particular, for any p > 0,
(k = 0, 1, ... , p - 1).
Multiplying these inequalities, we obtain
CN+,, :S: P"cN, or
(n ~N). Hence

so that (18)

lim sup ::Jc,. :S: P,
11-000

NUMERICAL SEQUENCES AND SERIES 69
by Theorem 3.20(b). Since (18) is true for every p > ix, we have
lim sup :fc,. s; ix.
n ➔ oo

POWER SERIES

3.38 Definition Given a sequence {c,.} of complex numbers, the series

(19)

L00 c,.z"

11=0

is called a power series. The numbers c,. are called the coefficients of the series; z is a complex number.
In general, the series will converge or diverge, depending on the choice of z. More specifically, with every power series there is associated a circle, the circle of convergence, such that (19) converges if z is in the interior of the circle
and diverges if z is in the exterior (to cover all cases, we have to consider the
plane as the interior of a circle of infinite radius, and a point as a circle of radius zero). The behavior on the circle of convergence is much more varied and cannot be described so simply.

3.39 Theorem Given the power series !cnz", put

a =lim sup .Ylc:"I,

1 R=-·
(X

(//a = 0, R = + oo; if a = + oo, R = 0.) Then 2::c,. z" converges if Iz I < R, and diverges if izl > R.

Proof Put a,.= c,.z", and apply the root test:

limsupy,.l;a-,.!= izl limsup,v,.1-!c,.I =Rlzl·

n ➔ co

n-+oo

Note: R is called the radius of convergence of Ic,. z".

3.40 Examples

(a) The series In" z" has R = 0.

L (b) The series

~
n.

has R

=

+ oo.

(In this case the ratio test is

easier

to

apply than the root test.)

70 PRINCIPLES OF MATHEMATICAL .ANALYSIS

(c) The series :Ezn has R = 1. If lzl = 1, the series diverges, since {zN}
does not tend to Oas n ➔ co.

L (d) The series !'.'. has R = 1. It diverges if z = 1. It converges for all n
other z with lzl = 1. (The last assertion will be proved in Theorem 3.44.)

L (e) The series

~
n

has R = 1.

It converges for all z with

lzl = 1, by

the comparison test, since Izn/n2 j = 1/n2.

SUMMATION BY PARTS 3.41 Theorem Given two sequences {an}, {b,.}, put

ifn~0;putA_ 1 =0. Then, i/0-5.p-5.q, we have
(20) Proof

and the last expression on the right is clearly equal to the right side of (20).
Formula (20), the so-called "partial summation formula," is useful in the investigation of series of the form :Ean bn, particularly when {bn} is monotonic. We shall now give applications.
3.42 Theorem Suppose
(a) the partial sums An of:Eanform a bounded sequence,· (b) ho ~ bi ~ b2 ~ .. • ;
(c) lim bn = 0.
n ➔ oo
Then :Ean b,. converges.

NUMERICAL SEQUENCES AND SERIES 71

Proof Choose M such that IA. I ~ M for all n. Given e > 0, there is an integer N such that bN ~ (e/2M). For N ~ p ~ q, we have

I 1-1 I Zq:a.b. n=p

qI-A1.Cb. -
n=p

b.+1)+Aqbq -

Ap-lbp,I

I ~ M

Iq-1
J}b• -

b,,+ 1) + bq + bPj

= 2MbP ~ 2MbN ~ e.

Convergence now follows from the Cauchy criterion. We note that the first inequality in the above chain depends of course on the fact that b. - b.+t ~ 0.

3.43 Theorem Suppose
(a) led~ ic2I ~ ic3I ~ .. ·;
(b) C2m-l ~ 0, C2m ~ 0 (m = 1, 2, 3, ...); (c) lim. ➔ 00 c. = 0.
Then I:c. converges.

Series for which (b) holds are called "alternating series"; the theorem was known to Leibnitz.
Proof Apply Theorem 3.42, with a.= (-1)"+ 1, b. = Ic. I,
3.44 Theorem Suppose the radius of convergence of :Ee. z" is 1, and suppose
= c0 ~ c1 ~ c2 ~ • • ·, lim. ➔ 00 c. 0. Then :Ec.z" converges at every point on the
circle Iz I = 1, except possibly at z = I.
Proof Put a.= z", b. = c11 • The hypotheses of Theorem 3.42 are then
satisfied, since

if Iz I = 1, z =1= 1.
ABSOLUTE CONVERGENCE The series I:an is said to converge absolutely if the series I: Ian I converges. 3.45 Theorem lfI:a. converges absolutely, then I:a. converges.

72 PRINCIPLES OF MATHEMATICAL ANALYSIS
Proof The assertion follows from the inequality
~ IJnakl ktlakl,
plus the Cauchy criterion.
3.46 Remarks For series of positive terms, absolute convergence is the same as convergence.
If I.an converges, but I, Ian I diverges, we say that I.an converges non-
absolutely. For instance, the series
I(-- 1n )"
converges nonabsolutely (Theorem 3.43). The comparison test, as well as the root and ratio tests, is really a test for
absolute convergence, and therefore cannot give any information about nonabsolutely convergent series. Summation by parts can sometimes be used to handle the latter. In particular, power series converge absolutely in the interior of the circle of convergence.
We shall see that we may operate with absolutely convergent series very much as with finite sums. We may multiply them term by term and we may change the order in which the additions are carried out, without affecting the sum of the series. But for nonabsolutely convergent series this is no longer true, and more care has to be taken when dealing with them.
ADDITION AND MULTIPLICATION OF SERIES
3.47 Theorem If I.an = A, and r.b" = B, then r.(an + b") = A + B, and I.can = cA, for any fixed c.
Proof Let
Then
n
L An + Bn = (ak + bk). k=O
Since limn ➔ oo An= A and limn ➔ oo Bn = B, we see that
The proof of the second assertion is even simpler.

NUMERICAL SEQUENCES AND SERIES 73

Thus two convergent series may be added term by term, and the resulting series converges to the sum of the two series. The situation becomes more complicated when we consider multiplication of two series. To begin with, we have to define the product. This can be done in several ways; we shall consider the so-called "Cauchy product."

3.48 Definition Given ta. and r.b., we put

n
L c. = akbn-k k=O

(n=0,1,2, ...)

and call r.c. the product of the two given series. This definition may be motivated as follows. If we take two power
series ta. z" and th. z", multiply them term by term, and collect terms containing the same power of z, we get

ro

oo

'Lna z" • 'Lnb z" = (aO + aI z + a2 z2 + .. ·)(b0 + b1z + b2 z2 + ••·)

n=O

n=O

= a0 b0 + (a0 b1 + a1b0 )z + (a0 b2 + a1b1 + a2 b0 )z2 + •••

= Co + C1Z + Cz z2 + ''' ,

Setting z = 1, we arrive at the above definition.

3.49 Example If

and An ...... A, B• ...... B, then it is not at all clear that {C.} will converge to AB,
c. since we do not have = A. Bn . The dependence of {C.} on {A.} and {B.} is
quite a complicated one (see the proof of Theorem 3.50). We shall now show
that the product of two convergent series may actually diverge.

The series

I (-1)" = 1--1 +-1 _ _1 + ...
n=oJn + 1 J2 J3 J4

converges (Theorem 3.43). We form the product of this series with itself and obtain

74 PRINCIPLES OF MATHEMATICAL ANALYSIS

so that Since

n

1

Cn=(-ltL--:=====•
k=0 J(n - k + l)(k + 1)

(n-k+l)(k+l)=(i+tf-(i-kf :S;(i+1f.

we have

I C n I

f
:.?: k.=£..on-

2 _ 2(n + 1)
-+-2- - "n-+-2- ,

so that the condition en ➔ 0, which is necessary for the convergence of ten, is not satisfied.

In view of the next theorem, due to Mertens, we note that we have here considered the product of two nonabsolutely convergent series.

3.50 Theorem Suppose

(a)
(b) (c)
(d) Then

00
L an converges absolutely,
n•0
00
}:an =A,
n=0
00
Lb,. =B,
n=0 n
Cn = L akbn-k (n = 0, 1, 2, ...). k=0
00
L Cn =AB.
n=0

That is, the product of two convergent series converges, and to the right value, if at least one of the two series converges absolutely.

Proof Put

Then
en= aobo + (aob1 + a1bo) + ''' + (aobn + a1bn-l + ''' + a,.bo) = a0Bn + a1 Bn-i + ••• + an Bo = ao(B + Pn) + a1(B + Pn-1) + ••• + anCB + Po)
= AnB + aoPn + a1Pn-1 + •••+ anPo

NUMERICAL SEQUENCES AND SERIES 75
Put
Yn = aoP. + a1P.-1 +.••+a.Po •
c. We wish to show that ➔ AB. Since A. B ➔ AB, it suffices to
show that (21)
Put
[It is here that we use (a).] Let e > 0 be given. By (c), P. ➔ o. Hence we can choose N such that IP. I ~ e for n ~ N, in which case
IY.I ~ lfJoa.+ .. ·+fJNan-NI + IPN+1an-N-1+·"+fJ.aol ~ IPoa. + .. •+ fJNa.-NI + B(X.
Keeping N fixed, and letting n ➔ oo, we get
Jim sup IY. I ~ B(X,
since ak ➔ 0 ask ➔ oo. Since e is arbitrary, (21) follows.
Another question which may be asked is whether the series re., if convergent, must have the sum AB. Abel showed that the answer is in the affirmative.
3.51 Theorem If the series ra., l:b., re. converge to A, B, C, and c. = a0 b0 + ••• + a. b0 , then C = AB.
Here no assumption is made concerning absolute convergence. We shall give a simple proof (which depends on the continuity of power series) after Theorem 8.2.
REARRANGEMENTS
3.52 Definition Let {kn}, n = 1, 2, 3, ... , be a sequence in which every
positive integer appears once and only once (that is, {kn} is a 1-1 function from J onto J, in the notation of Definition 2.2). Putting
(n = 1, 2, 3, ...),
we say that I:a~ is a rearrangement of I:an .

76 PRINCIPLES OF MATHEMATICAL ANALYSIS

If {s.}, {s~} are the sequences of partial sums of La., La~, it is easily seen that, in general, these two sequences consist of entirely different numbers. We are thus led to the problem of determining under what conditions all rearrangements of a convergent series will converge and whether the sums are necessarily the same.

3.53 Example Consider the convergent series

(22)

and one of its rearrangements

(23)

1 +t-½+t+t-¼+½+fr-¾+ ...

in which two positive terms are always followed by one negative. If s is the sum of (22), then

Since

s<l-½+-½=i-

for k ;?:: 1, we see that s3< s6 < s~ < ••• , where s~ is nth partial sum of (23).
Hence
Jim sups~ > s3= i, n-->oo
so that (23) certainly does not converge to s [we leave it to the reader to verify that (23) does, however, converge].
This example illustrates the following theorem, due to Riemann.

3.54 Theorem Let La. be a series of real numbers which converges, but not absolutely. Suppose
P - 00 S Ct S S 00.

Then there exists a rearrangement La~ with partial sums s~ such that

(24)

lim inf s~ = et, lim sups~= p.

n ➔ oo

n ➔ oo

Proof Let

(n = 1, 2, 3, ...).

NUMERICAL SEQUENCES AND SERIES 77

Then Pn - q" = an, Pn + qn = Ian I, Pn ;;:: 0, qn ~ 0. The series r.pn, r,qn
must both diverge. For if both were convergent, then

!,(pn + qn} = !:Ian I

would converge, contrary to hypothesis. Since

L L L L N

N

N

N

an = (pn - qn) = Pn - qn •

n=l

n=1

n=1

n=l

divergence of r.pn and convergence of r.qn (or vice versa) implies divergence of !:an , again contrary to hypothesis.
Now let Pi, P 2 , P3 , •.. denote the nonnegative terms of r.an, in the order in which they occur, and let Q1, Q2 , Q3 , ••• be the absolute values of the negative terms of r.an, also in their original order.
The series r,pn, r.Qn differ from r.pn, r.qn only by zero terms, and are therefore divergent.
We shall construct sequences {mn}, {kn}, such that the series

(25} P1 +•••+Pm, - Q1 - ••• - Qk, +Pm,+1 + •·•
+ pm2 - Qk, +1 - •• • - Qk2 + •••,
which clearly is a rearrangement of r.an, satisfies (24).
Choose real-valued sequences {ot:n}, {Pn} such that ot:n ➔ ot:, /3n ➔ /3,
<Xn < Pn, Pt > 0.
Let mi, k1 be the smallest integers such that
Pi+•••+ Pm,> P1,
Pi + •· • + Pm1 - Ql - • • • - Qk, < <X1;
let m2 , k2 be the smallest integers such that

Pi+··· +Pm, - Q1 - •·• - Qk, +Pm,+1 + ••• +Pm2 > /32,
P1 +•••+Pm, - Q1 - ••• - Qk, +Pm,+1 + ··• +Pm2 - Qk,+1
- ••• - Qk2 < ot:2;

and continue in this way. This is possible since r,pn and r.Qn diverge. If Xn, Yn denote the partial sums of (25) whose last terms are Pmn,
-Q""' then

Since Pn ➔ O and Qn ➔ Oas n ➔ oo, we see that Xn ➔ P, Yn ➔ ot:.
Finally, it is clear that no number less than ot: or greater than /3 can
be a subsequential limit of the partial sums of (25).

78 PRINCIPLES OF MATHEMATICAL ANALYSIS

3.55 Theorem lf'E.an is a series ofcomplex numbers which converges absolutely, then every rearrangement of'I.an converges, and they all converge to the same sum.

Proof Let ta; be a rearrangement, with partial sums s; . Given e > 0,
there exists an integer N such that m ~ n ~ N implies

m

(26)

Ela,l:s;e.

l=n

Now choose p such that the integers 1, 2, ... , N are all contained in the set k1, k2 , •.• , kP (we use the notation of Definition 3.52). Then if n > p, the numbers a1, ... , aN will cancel in the difference Sn - s~, so that
s; Is" - I :s; B, by (26). Hence {s;} converges to the same sum as {s"}.

EXERCISES
1. Prove that convergence of {s.} implies convergence of {ls.I}. Is the converse true?
2. Calculate lim (vn2 + n - n).
ft ➔ OO
3. If S1 = vi, and
(n = 1, 2, 3, ...),
prove that {s.} converges, and thats.< 2 for n = 1, 2, 3, ... .
4. Find the upper and lower limits of the sequence {s.} defined by

5. For any two real sequences {a.}, {h.}, prove that

Jim sup (a. + h.) ~ Jim sup a. + Jim sup h.,

n ➔ OO

n ➔ OO

provided the sum on the right is not of the form co - co. 6. Investigate the behavior (convergence or divergence) of !:a. if

(a) a. =Vn + 1- v'n;

(b) a_.-v'-n-+-l--v-n,,
n

(c) a.= (~n - 1)•;

1
(d) a. = 1 + z•'

for complex values of z.

7. Prove that the convergence of !:a. implies the convergence of

if a.;,::o.

~v'a.
""' n '

NUMERICAL SEQUENCES AND SERIES 79

8. If l:a. converges, and if {b.} is monotonic and bounded, prove that l:a.b. converges.
9. Find the radius of convergence of each of the following power series:
2•
(b) I; ;;j z•,

(c)

I:

2• n2

z•,

10. Suppose that the coefficients of the power series I;a. z" are integers, infinitely many
of which are distinct from zero. Prove that the radius of convergence is at most 1.
11. Suppose a. > 0, s. = 01 + ···+ a., and l:a. diverges.

(a) Prove that I: 1 ~•a.diverges.

(b) Prove that

0N+1 +•••+QN+t ~l-.!..!!_

SN+1

SN+k

SN+k

and

deduce

that

I

:~
s.

diverges.

(c) Prove that

and deduce that "~t"'2sa.. converges.

(d) What can be said about

I: a.
1 +na.

and

I: 1 +0n"2a. ?

12. Suppose a. > 0 and l:a. converges. Put
.....,
r. = I: a., .

(a) Prove that

if m< n, and deduce that I:~ diverges. r.

80 PRINCIPLES OF MATHEMATICAL ANALYSIS
(b) Prove that

and deduce that I:,a;.-converges.
-v r.
13. Prove that the Cauchy product of two absolutely convergent series converges absolutely.
14. If {sn} is a complex sequence, define its arithmetic means a. by

(n =0, 1, 2, ...).

(a) If Jim s. = s, prove that lim a. = s.
(b) Construct a sequence {s.} which does not converge, although Jim an= 0. (c) Can it happen thats.> 0 for all n and that Jim sup Sn= oo, although Jim a.= 0? (d) Put a.= s. - Sn-1, for n ~ 1. Show that

Sn -

a.

=

1

n

-+ 1

L• kat.
k=l

Assume that Jim (na.) = 0 and that {a.} converges. Prove that {s.} converges.
[This gives a converse of (a), but under the additional assumption that na.- 0.)
(e) Derive the last conclusion from a weaker hypothesis: Assume M < oo, Ina. I-;;;, M for all n, and lim a. = a. Prove that Jim Sn = a, by completing the
following outline: If m <n, then

Sn -

a.= - nm-+-m1(a. -

Um)

+

-n-1-m

L• (sn
1~m+l

-

s,).

For these i,

(n - i)M (n - m - l)M
Is. - s,I -::;;, i + 1 -::;;, m + 2 •

Fix e > 0 and associate with each n the integer m that satisfies n-e
m-::;;,1+e<m+1.

Then (m + 1)/(n - m)-;;;, 1/e and Isn - s, I < Me. Hence lim sup ls.- al ~Me.
•~co
Since e was arbitrary, Jim s. = a.

NUMERICAL SEQUENCES AND SERIES 81
15, Definition 3.21 can be extended to the case in which the a. lie in some fixed Rt.
Absolute convergence is defined as convergence of I: Ia. I, Show that Theorems
3.22, 3.23, 3.25(a), 3.33, 3.34, 3.42, 3.45, 3.47, and 3.55 are true in this more general setting. (Only slight modifications are required in any of the proofs.)
16, Fix a positive number or:. Choose x1 > v;;, and define x 2, x3, x4, ••• , by the
recursion formula
~= Xn +t (Xn + :.) .
(a) Prove that {x.} decreases monotonically and that Jim x. = v;;, (b) Put e. = x. - v;, and show that

so that, setting f3 = 2v;;,
(ir· e.+1 </3

(n = 1, 2, 3, ...).

(c) This is a good algorithm for computing square roots, since the recursion
formula is simple and the convergence is extremely rapid. For example, if or: = 3
and Xi= 2, show that eiff3 < / 0 and that therefore
£5 <4. 10-i6,

17. Fix or:> 1. Take Xi > v;, and define

or:+ Xn

or:-x!

X n +1i.= ..'L.x-. = x . + 1 +x-..

(a) Prove that Xi > X3 > Xs > •••.
(b) Prove that X2 < X4 < x 6 < ••• .
(c) Prove that Jim x. = v;,
(d) Compare the rapidity of convergence of this process with the one described
in Exercise 16. 18. Replace the recursion formula of Exercise 16 by

p-1 or:

Xn+i =--x.+-x.-p+i

p

p

where p is a fixed positive integer, and describe the behavior of the resulting sequences {x.}. 19. Associate to each sequence a= {or:.}, in which or:. is O or 2, the real number

I: x(a) = "' 3tXn•. n=i
Prove that the set of all x(a) is precisely the Cantor set described in Sec. 2.44.

82 PR.INCIPLJ!S OP MAmBMATICAL ANALYSIS

20. Suppose {p.} is a Cauchy sequence in a metric space X, and some subsequence {p.,} converges to a point p e X. Prove that the full sequence {p.} converges top.
21. Prove the following analogue of Theorem 3.lO(b): If {E.} is a sequence of closed nonempty and bounded sets in a complete metric space X, if E. => Ea+i. and if
.....Jim diam E. =0,
n then f E. consists of exactly one point. n 22. Suppose Xis a nonempty complete metric space, and {G.} is a sequence of
dense open subsets of X. Prove Baire's theorem, namely, that fG. is not empty. (In fact, it 1s dense in X.) Hint: Find a shrinking sequence of neighbor-
hoods E. such that E. c: G., and apply Exercise 21.
23. Suppose {p.} and {q.} are Cauchy sequences in a metric space X. Show that the sequence {d(p.,q.)} converges. Hint: For any m, n,
d(p., q.) ~d(p.,p,.) + d(p,,.,q,.) + d(q,., q.);

it follows that

Id(p.' q.) - d(p,. ' q,,.) I

is small if m and n are large. 24. Let X be a metric space.
(a) Call two Cauchy sequences {p.}, {q.} in X equivalent if
.....Jim d(p. , q.) = O.

Prove that this is an equivalence relation. (b) Let X* be the set of all equivalence classes so obtained. If P e x•, Q e x•, {p,.} e P, {q.} e Q, define
..... t:i.(P, Q) = limd(p.,q.);

by Exercise 23, this limit exists. Show that the number t:i.(P, Q) is unchanged if {p.} and {q.} are replaced by equivalent sequences, and hence that t:i. is a distance function in X*. (c) Prove that the resulting metric space X* is complete. (d) For each p e X, there is a Cauchy sequence all of whose terms are p; let P, be the element of X* which contains this sequence. Prove that

t:i.(P,, P4) =d(p,q)
for all p, q e X. In other words, the mapping <p defined by <p(p) = P, is an isometry
(i.e., a distance-preserving mapping) of X into X*.
(e) Prove that <p(X) is dense in x•, and that <p(X) = x• if Xis complete. By (d),
we may identify X and <p(X) and thus regard X as embedded in the complete metric space x•. We call x• the completion of X. 25. Let X be the metric space whose points are the rational numbers, with the metric
d(x, y) =Ix - y I, What is the completion of this space? (Compare Exercise 24.)

4
CONTINUITY
The function concept and some of the related terminology were introduced in Definitions 2.1 and 2.2. Although we shall (in later chapters) be mainly interested in real and complex functions (i.e., in functions whose values are real or complex numbers) we shall also discuss vector-valued functions (i.e., functions with values in Rk) and functions with values in an arbitrary metric space. The theorems we shall discuss in this general setting would not become any easier if we restricted ourselves to real functions, for instance, and it actually simplifies and clarifies the picture to discard unnecessary hypotheses and to state and prove theorems in an appropriately general context.
The domains of definition of our functions will also be metric spaces, suitably specialized in various instances. LIMITS OF FUNCTIONS 4.1 Definition Let X and Y be metric spaces; suppose E c: X, f maps E into Y, and p is a limit point of E. We write f(x)-+ q as x-+ p, or
limf(x) =q
x ➔p

84 PRINCIPLES OP MATHEMATICAL ANALYSIS

if there is a point q e Y with the following property: For every s > 0 there exists a 6 > 0 such that

(2)

dr(/(x), q) < s

for all points x e E for which

(3)

0 < dx(x,p) < 6.

The symbols dx and dr refer to the distances in X and Y, respectively. If X and/or Y are replaced by the real line, the complex plane, or by some euclidean space~. the distances dx, dr are ofcourse replaced by absolute values, or by norms of differences (see Sec. 2.16). It should be noted that p e X, but that p need not be a point of E in the above definition. Moreover, even if p e E, we may very well have f(p) 'F lim,. ➔,f(x). We can recast this definition in terms of limits of sequences:

4.2 Theorem Let X, Y, E, f, and p be as in Definition 4.1. Then

(4)

,l.i..m., /(x) =q

if and only if

(5)

lim f(pn) =q

for every sequence {Pn} in E such that

(6)

Pn 'F P, lim Pn =p.

n ➔ ao

Proof Suppose (4) holds. Choose {Pn} in E satisfying (6). Let s > 0 be given. Then there exists /, > 0 such that dy(f(x), q) < s if x e E and O< dx(x, p) < 6. Also, there exists N such that n > N implies 0 < dx(pn ,p) < /J. Thus, for n > N, we have dr(/(pJ, q) < s, which
shows that (5) holds.
Conversely, suppose (4) is false. Then there exists some s > 0 such that for every /J > 0 there exists a point x e E (depending on /J), for which
drif(x), q) ~ s but O< dx(x, p) < 6. Taking/," = 1/n (n = 1, 2, 3, ...), we
thus find a sequence in E satisfying (6) for which (5) is false.

Corollary Iff has a limit at p, this limit is unique. This follows from Theorems 3.2(b) and 4.2.

CONTINUITY 85

4.3 Definition Suppose we have two complex functions, f and g, both defined
on E. By f + g we mean the function which assigns to each point x of E the number f(x) + g(x). Similarly we define the difference f - g, the product Jg,
and the quotient/jg of the two functions, with the understanding that the quotient is defined only at those points x of E at which g(x) ,;,. 0. Iff assigns to each point x of E the same number c, then f is said to be a constant function, or simply a constant, and we write f = c. If f and g are real functions, and if f(x) ~ g(x) for every x e E, we shall sometimes write/~ g, for brevity.
Similarly, if f and g map E into Ji', we define f + g and f • g by
(f + g)(x) = f(x) + g(x), (f • g)(x) = f(x) • g(x);
and if l is a real number, (M)(x) = lf(x).

4.4 Theorem Suppose E c: X, a metric space, p is a limit point of E, f and g

are complex functions on E, and

lim /(x) =A,
x-+p

lim g(x) = B.
x-+p

Then (a) lim (f +g)(x) = A + B;

x➔p
(b) lim (fg)(x) =AB;
x-+p

(c)

xli➔m,

(!.)(x) g

=

~,
B

if B #- 0.

Proof In view of Theorem 4.2, these assertions follow immediately from the analogous properties of sequences (Theorem 3.3).

Remark If f and g map E into Rk, then (a) remains true, and (b) becomes (b') lim (f • g)(x) = A • B.
x-+p
(Compare Theorem 3.4.)

CONTINUOUS FUNCTIONS
4.5 Definition Suppose X and Y are metric spaces, E c: X, p e E, and f maps E into Y. Then f is said to be continuous at p if for every e > 0 there exists a '5 > 0 such that
dr(/(x),f(p)) < e
for all points x e E for which dx(x, p) < '5. If/is continuous at every point of E, then/is said to be continuous on E. It should be noted that f has to be defined at the point p in order to be
continuous at p. (Compare this with the remark following Definition 4.1.)

86 PRINCIPLES OF MATHEMATICAL ANALYSIS
If p is an isolated point of E, then our definition implies that every function / which has E as its domain of definition is continuous at p. For, no matter which e > 0 we choose, we can pick J > 0 so that the only point x e E for which
dx(x, p) <bis x =p; then
d1(f(x),f(p)) = 0 < e.
4.6 Theorem In the situation given in Definition 4.5, assume also that p is a
limit point of E. Then f is continuous at p if and only if limx➔ p/(x) =f(p).
Proof This is clear if we compare Definitions 4.1 and 4.5.
We now turn to compositions of functions. A brief statement of the following theorem is that a continuous function of a continuous function is continuous.
4.7 Theorem Suppose X, Y, Z are metric spaces, E c X, f maps E into Y, g maps the range off,f(E), into Z, and his the mapping of E into Z defined by
h(x) = g(f(x)) (x e E).
Iff is continuous at a point p e E and ifg is continuous at the point f(p), then h is continuous at p.
This function h is called the composition or the composite of/ and g. The notation
h =go/
is frequently used in this context.
Proof Let e > 0 be given. Since g is continuous at f(p), there exists 1'/ > 0 such that
dz(g(y), g(f(p))) < e if d1(y,f(p)) < 11 and y ef(E).
Since/ is continuous at p, there exists b > 0 such that
dr(f(x),f(p)) < 1'/ if dx(x, p) <band x e E.
It follows that
d2(h(x), h(p)) = d2(g(f(x)), g(f(p))) < e
if dx(x, p) < b and x e E. Thus h is continuous at p.
4.8 Theorem A mapping f of a metric space X into a metric space Y is con-
tinuous on X if and only if1- 1( V) is open in X for every open set V in Y.
(Inverse images are defined in Definition 2.2.) This is a very useful characterization of continuity.

CONTINUITY 87

Proof Supposef is continuous on X and Vis an open set in Y. We have
to show that every point of f- 1(V) is an interior point of 1- 1(V). So,
suppose p EX andf(p) EV. Since Vis open, there exists e > 0 such that y E V if d y(f(p), y) < e; and since f is continuous at p, there exists o > 0
such that dy(f(x),f(p)) < e if dx(x, p) < o. Thus x Ef- 1(V) as soon as
dx(x, p) < o.
Conversely, suppose f- 1(V) is open in X for every open set Vin Y.
Fix p EX and e > 0, let V be the set of ally E Y such that dy(y,f(p)) < e.
o Then Vis open; hencef- 1(V) is open; hence there exists > 0 such that
x E/- 1(V)as soon as dx(P, x) < o. But if x E f- 1(V), then f(x) EV, so
that dy(f(x),f(p)) < e. This completes the proof.
Corollary A mapping f of a metric space X into a metric space Y is continuous if and only iff- 1( C) is closed in X for every closed set C in Y.
This follows from the theorem, since a set is closed if and only if its com-
u- plement is open, and since/- 1(Ec) = 1(E)Y for every EC Y.
We now turn to complex-valued and vector-valued functions, and to functions defined on subsets of Rk.
4.9 Theorem Let f and g be complex continuous functions on a metric space X.
Thenf + g,fg, andf/g are continuous on X.
In the last case, we must of course assume that g(x) #- 0, for all x E X.
Proof At isolated points of X there is nothing to prove. At limit points, the statement follows from Theorems 4.4 and 4.6.

4.10 Theorem

(a) Let / 1 , .h ... , be real functions on a metric space X, and let f be the
mapping of X into Rk defined by

(7)

f(x) = (fi(x), ... ,/',ix)) (XE X);

then f is continuous ifand only ifeach ofthe functionsJi., ... ,I,. is continuous. (b) If f and g are continuous mappings of X into Rk, then f + g and f • g
are continuous on X.

The functions Ji., ... , I,. are called the components off. Note that
f + g is a mapping into Rk, whereas r•g is a real function on X.

88 PRINCIPLES OF MATHEMATICAL ANALYSIS

Proof Part (a) follows from the inequalities

={ f 2}\ J.tj(x) -1.,(y)J :S: Jf(x) - f(y)J

1/,(x) - /i<.v)J

1=1

for j = l, ... , k. Part (b) follows from (a) and Theorem 4.9.

4.11 Examples If Xi, .•. , x,. are the coordinates of the point x e R", the functions 4>, defined by

(8)

are continuous on Rk, since the inequality
14>,(x) - </>i(y) I :S: Ix - YI shows that we may take D= e in Definition 4.5. The functions q,1 are sometimes
called the coordinate functions. Repeated application of Theorem 4.9 then shows that every monomial

(9)

where ni, ... , nk are nonnegative integers, is continuous on R". The same is true of constant multiples of (9), since constants are evidently continuous. It follows that every polynomial P, given by

(10)

(x e R"),

is continuous on R". Here the coefficients cni•"nk are complex numbers, n1, ... , n"
are nonnegative integers, and the sum in (10) has finitely many terms. Furthermore, every rational function in Xi, ... , xk, that is, every quotient
of two polynomials of the form (10), is continuous on Rk wherever the denominator is different from zero.
From the triangle inequality one sees easily that

(11)
Hence the mapping x-+ Ix I is a continuous real function on R".
If now f is a continuous mapping from a metric space X into Rk, and if 4>
is defined on X by setting q,(p) = If(p) J, it follows, by Theorem 4.7, that 4> is a
continuous real function on X.

4.12 Remark We defined the notion of continuity for functions defined on a subset E of a metric space X. However, the complement of E in X plays no role whatever in this definition (note that the situation was somewhat different for limits of functions). Accordingly, we lose nothing of interest by discarding the complement of the domain off This means that we may just as well talk only about continuous mappings of one metric space into another, rather than

CONTINUITY 89
of mappings of subsets. This simplifies statements and proofs of some theorems. We have already made use of this principle in Theorems 4.8 to 4.10, and will continue to do so in the following section on compactness.

CONTINUITY AND COMPACTNESS

4,13 Definition A mapping f of a set E into R" is said to be bounded if there is
a real number M such that If(x) I ~ M for all x e E.

4.14 Theorem Suppose f is a continuous mapping of a compact metric space X into a metric space Y. Thenf(X) is compact.

Proof Let {V,.} be an open cover of/(X). Since/is continuous, Theorem
4.8 shows that each of the sets /- 1(V,.) is open. Since Xis compact,
there are finitely many indices, say oci, ... , °'n, such that

(12)

Xc.f- 1(V,.) u ... u/- 1(V,.").

Since/(f- 1(£)) c. E for every E c. Y, (12) implies that

(13)

f(X) c. v ..l u .. • u v ..n.

This completes the proof.

Note: We have used the relation /(r- 1(£)) c. E, valid for E c. Y. If
E c. X, then/- 1(/(E)) => E; equality need not hold in either case. We shall now deduce some consequences of Theorem 4.14.

4.15 Theorem // f is a continuous mapping of a compact metric space X into R", then f(X) is closed and bounded. Thus, f is bounded.

This follows from Theorem 2.41. The result is particularly important when.f is real:

4.16 Theorem Suppose f is a continuous real function on a compact metric space X, and

(14)

M= sup/(p), m = inf /(p).

peX

peX

Then there exist points p, q e X such thatf(p) = M andf(q) = m.

The notation in (14) means that M is the least upper bound of the set of all numbersJ(p), where p ranges over X, and that m is the greatest lower bound of this set of numbers.

90 PRINCIPLES OF MATHEMATICAL ANALYSIS

The conclusion may also be stated as follows: There exist points p and q in X such that f(q) Sf(x) Sf(p) for all x e X; that is,/ attains its maximum (at p) and its minimum (at q).
Proof By Theorem 4.1 S, f ( X) is a closed and bounded set of real numbers; hence/(X) contains
M== sup/(X) and m =inff(X),
by Theorem 2.28.

4.17 Theorem Suppose f is a continuous 1-1 mapping of a compact metric
space X onto a metric space Y. Then the inverse mapping 1- 1 defined on Y by
(xe X)
is a continuous mapping of Y onto X.
Proof Applying Theorem 4.8 to 1- 1 in place of/, we see that it suffices
to prove that/(V) is an open set in Y for every open set Vin X. Fix such a set V.
The complement vc of Vis closed in X, hence compact (Theorem 2.35); hence/(V1 is a compact subset of Y (Theorem 4.14) and so is closed in Y (Theorem 2.34). Since/is one-to-one and onto, /{V) is the complement off(Vc). Hence/(V) is open.

4.18 Definition Let/be a mapping of a metric space X into a metric space Y.
We say that/is uniformly continuous on X if for every e > 0 there exists b > 0
such that

(15)

dy(f(p),f(q)) < e

for all p and q in X for which dx(p, q) < b.
Let us consider the differences between the concepts of continuity and of
uniform continuity. First, uniform continuity is a property of a function on a set, whereas continuity can be defined at a single point. To ask whether a given
function is uniformly continuous at a certain point is meaningless. Second, if
/ is continuous on X, then it is possible to find, for each e > 0 and for each pointp of X, a number 8 > 0 having the property specified in Definition 4.5. This 8 depends on e and on p. If/is, however, uniformly continuous on X, then it is possible, for each e > 0, to find one number b > 0 which will do for all points
pof X. Evidently, every uniformly continuous function is continuous. That the
two concepts are equivalent on compact sets follows from the next theorem.

CONTINUITY 91

4.19 Theorem Let f be a continuous mapping of a compact metric space X into a metric space Y. Then f is uniformly continuous on X.

Proof Let e > 0 be given. Since f is continuous, we can associate to each point p e X a positive number cp(p) such that

(16)

q e X, dx(p, q) < cp(p) implies dy(f(p), f(q)) < 26°

Let J(p) be the set of all q e X for which

(17)

dx(P, q) < ½cf>(p).

Since p e J(p), the collection of all sets J(p) is an open cover of X; and since Xis compact, there is a finite set of points Pi, ... , p. in X, such that

(18) We put
(I 9)

Then [; > 0. (This is one point where the finiteness of the covering, inherent in the definition of compactness, is essential. The minimum of a finite set of positive numbers is positive, whereas the inf of an infinite set of positive numbers may very well be 0.)
Now let q and p be points of X, such that dx(P, q) < <>. By (18), there is an integer m, 1 ~ m ~ n, such that p e J (Pm); hence

(20)

dx(P, Pm) < ½¢(Pm),

and we also have

dx(q, Pm) ~ dx(P, q) + dx(P, Pm) < <> + ½¢(Pm) ~ ¢(pm).

Finally, (16) shows that therefore

dy(f(p),f(q)) ~ dy(f(p),f(pm)) + dy(f(q),/(Pm)) < 6.

This completes the proof.

An alternative proof is sketched in Exercise 10. We now proceed to show that compactness is essential in the hypotheses of Theorems 4.14, 4.15, 4.16, and 4.19.

4.20 Theorem Let E be a noncompact set in R1. Then
(a) there exists a continuous function on E which is not bounded,· (b) there exists a continuous and bounded function on E which has no maximum.
If, in addition, E is bounded, then