Kenneth Hoffman-Analysis in Euclidean Space -Prentic-Hall _ Dover Publications (1975)

ANALYSIS IN

Kenneth Hoffman

ANALYSISIN

EUCLIDEANSPACE

ANALYSISIN

EUCLIDEANSPA CE

Kenneth HoffmanMassachusetts Institute of Technology

PRENTICE-HALL, INC., ENGLEWOOD CLIFFS, NEW JERSEY

Library of Congress Cataloging in Publication DataHOFFMAN, KENNETH.

Analysis in Euclidean space.Bibliography: P.1. Mathematical analysis. I. Title.

QA300.H63 515 74-18263ISBN 0-13-032656-9

1975 by PRENTICE-HALL, INC.Englewood Cliffs, New Jersey

All rights reserved. No part of this bookmay be reproduced in any form or by anymeans without permission in writing fromthe publisher.

10 9 8 7 6 5 4 3 2 1

Printed in the United States of America

PRENTICE-HALL INTERNATIONAL, INC., LondonPRENTICE-HALL OF AUSTRALIA, PTY. LTD., SydneyPRENTICE-HALL OF CANADA, LTD., TorontoPRENTICE-HALL OF INDIA PRIVATE LIMITED, New DelhiPRENTICE-HALL OF JAPAN, INC., Tokyo

Contents

Preface

Preface to the Student

ix

xur

Chapter 1. Numbers and Geometry 1

1.1. The Real Number System 11.2. Consequences of Completeness 61.3. Intervals and Decimals 111.4. Euclidean Space 161.5. Complex Numbers 191.6. Linear Geometry 24

Chapter 2. Convergence and Compactness 31

2.1. Convergent Sequences 312.2. Convergence Criteria 372.3. Infinite Series 422.4. Sequential Compactness 512.5. Open and Closed Sets 552.6. Closure and Interior 632.7. Compact Sets 662.8. Relative Topology 71

V

vi Contents

Chapter 3. Continuity 77

3.1. Continuous Functions 773.2. Continuity and Closed Sets 853.3. The Limit of a Function 913.4. Infinite Limits and Limits at Infinity 973.5. Continuous Mappings 1003.6. Uniform Continuity 108

Chapter 4. Calculus Revisited 119

4.1. Differentiation on Intervals4.2. Integration on Intervals4.3. Fundamental Properties of the Integral4.4. Integrability and Real-Valued Functions4.5. Differentiation and Integration in Rn4.6. Riemann-Stieltjes Integration

119

128

136144

149

160

Chapter 5. Sequences of Functions 169

5.1. Convergence 1695.2. Calculus and Convergence 1785.3. Real Power Series 1845.4. Multiple Power Series 1925.5. Complex Power Series 1965.6. Fundamental Results on Complex-Analytic Functions 2045.7. Complex-Analytic Maps 2145.8. Fourier Series 2215.9. Abel-Poisson Summation 2255.10. The Dirichlet Problem 236

Chapter 6. Normed Linear Spaces 241

6.1. Linear Spaces and Norms 2416.2. Norms on Rn 2496.3. Convergence and Continuity 2536.4. Completeness 2636.5. Compactness 2746.6. Quotient Spaces 2856.7. The Completion of a Space 293

Contents vii

Chapter 7. The Lebesgue Integral 296

7.1.7.2.7.3.7.4.7.5.7.6.7.7.7.8.7.9.

MotivationThe SettingSets of Measure ZeroThe Principal PropositionsCompleteness and ContinuityThe Convergence TheoremsMeasurable Functions and Measurable SetsFubini's TheoremOrthogonal Expansions

296301

305314322327336350359

Chapter 8. Differentiable Mappings 370

8.1. Linear Transformations 3708.2. Partial Derivatives 3728.3. Differentiable Functions 3768.4. Differentiable Maps 3838.5. Inverse Mappings 3938.6. Change of Variable 406

Appendix Elementary Set Theory 414

A.I. Sets and Functions 414A.2. Cardinality 417

List of Symbols 422

Bibliography 423

Index 425

To my parents

Preface

This textbook has been developed for use in the two-semester intro-ductory course in mathematical analysis at the Massachusetts Institute ofTechnology. The aim of the course is to introduce the student to basic con-cepts, principles, and methods of mathematical analysis.

The presumed mathematical background of the students is a solidcalculus course covering one and (some of) several variables, plus (perhaps)elementary differential equations and linear algebra. The linear algebrabackground is not necessary until the second semester, since it enters theearly chapters only through certain examples and exercises which utilizematrices. At M.I.T. the introductory calculus course is condensed into oneyear, after which the student has available a one-semester course in dif-ferential equations and linear algebra. Thus, over half the students in thecourse are sophomores. Since many students enter M.I.T. having had aserious calculus course in high school, there are quite a few freshmen inthe course. The remainder of the students tend to be juniors, seniors, orgraduate students in fields such as physics or electrical engineering. Sincevery little prior experience with rigorous mathematical thought is as-sumed, it has been our custom to augment the lectures by structuredtutorial sessions designed to help the students in learning to deal withprecise mathematical definitions and proofs. It is to be expected that atmany institutions the text would be suitable for a j unior, senior, or graduatecourse in analysis, since it does assume a considerable technical facilitywith elementary mathematics as well as an affinity for mathematicalthought.

The presentation differs from that found in existing texts in two ways.First, a concerted effort is made to keep the introductions to real and com-

ix

x Preface

plex analysis close together. These subjects have been separated in thecurriculum for a number of years, thus tending to delay the introductionto complex function theory. Second, the generalizations beyond Rn arepresented for subsets of normed linear spaces rather than for metric spaces.The pedagogical advantage of this is that the original material can bedeveloped on the familiar terrain of Euclidean space and then simplyobserved to be largely valid for normed linear spaces, where the symbolismis just like that of Rn. The students are prepared for the generalization inmuch the same way that high-school algebra prepares one for manipula-tion in a commutative ring with identity.

The first semester covers the bulk of the first five chapters. It empha-sizes the Four C's: completeness, convergence, compactness, and con-tinuity. The basic results are presented for subsets of and functions onEuclidean space of n dimensions. This presentation includes (of course) arigorous review of the intellectual skeleton of calculus, placing greateremphasis on power series expansions than one normally can in a calculuscourse. The discussion proceeds (in Chapter 5) into complex power seriesand an introduction to the theory of complex-analytic functions. Thereview of linear geometry in Section 1.6 is usually omitted from the,formalstructure of the first semester. The instructor who is pressed for time orwho is predisposed to separate real and complex analysis may also omitall or part of Sections 5.5-5.10 on analytic functions and Fourier serieswithout interrupting the flow of the remainder of the text.

The second semester begins with Chapter 6. It reviews the main resultsof the first semester, the review being carried out in the context of (subsetsof and functions on) normed linear spaces. The author has found that thestudent is readily able to absorb the fact that many of the arguments heor she has been exposed to are formal and are therefore valid in the moregeneral context. It is then emphasized that two of the most crucial resultsfrom the first semester the completeness of Rn and the Heine-Boreltheorem-depend on finite-dimensionality. This leads naturally to a dis-cussion of (1) complete (Banach) spaces, the Baire category theorem andfixed points of contractions, and (ii) compact subsets of various normedlinear spaces, in particular, equicontinuity and Ascoli's theorem. Fromthere the course moves to the Lebesgue integral on Rn, which is developedby completing the space of continuous functions of compact support. Mostof the basic properties of integral and measure are discussed, and a shortpresentation of orthogonal expansions (especially Fourier series) is in-cluded. The final chapter of the notes deals with differentiable maps onRn, the implicit and inverse function theorems, and the change of variabletheorem. This chapter may be presented earlier if the instructor finds itdesirable, since the only dependence on Lebesgue integration is the proofof the change of variable theorem.

A few final remarks. Some mathematicians will look at these notes

Preface xi

and say, "How can you teach an introductory course in analysis whichnever mentions partial differential equations or calculus of variations?"Others will ask, "How can you teach a basic course in analysis whichdevotes so little attention to applications, either to mathematics or to otherfields of science?" The answer is that there is no such thing as the intro-ductory course in analysis. The subject is too large and too important toallow for that. The three most viable foci for organization of an intro-ductory course seem to be (i) emphasis on general concepts and principles,(ii) emphasis on hard mathematical analysis (the source of the generalideas), and (iii) emphasis on applications to science and engineering. Thistext was developed for the first type of course. It can be very valuable for acertain category of students, principally the students going on to graduateschool in mathematics, physics, or (abstract) electrical engineering, etc.It is not, and was not intended to be, right for all students who may needsome advanced calculus or analysis beyond the elementary level.

Thanks are due to many people who have contributed to the develop-ment of this text over the last eight years. Colleagues too numerous tomention used the classroom notes and pointed out errors or suggestedimprovements. Three must be singled out: Steven Minsker, DavidRagozin, and Donald Wilken. Each of them assisted the author in improv-ing the notes and managing the pedagogical affairs of the M.I.T. course. Iam especially grateful to David Ragozin, who wrote an intermediate ver-sion of the chapter on Lebesgue integration. I am indebted to Mrs. SophiaKoulouras, who typed the original notes, and to Miss Viola Wiley, whotyped the revision and the final manuscript. Finally, my thanks to ArtWester and the staff of Prentice-Hall, Inc.

KENNETH HOFFMAN

Prefaceto the Student

This textbook will introduce you to many of the general principlesof mathematical analysis. It assumes that you have a mathematical back-ground which includes a solid course (at least one year) in the calculus offunctions of one and several variables, as well as a short course in dif-ferential equations. It would be helpful if you have been exposed to intro-ductory linear algebra since many of the exercises and examples involvematrices. The material necessary for following these exercises and examplesis summarized in Section 1.6, but a linear algebra background is notessential for reading the book since it does not enter into the logical devel-opment in the text until Chapter 6.

You will meet a laige number of concepts which are new to you, andyou will be challenged to understand their precise definitions, some oftheir uses, and their general significance. In order to understand the mean-ing of this in quantitative terms, thumb through the Index and see howmany of the terms listed there you can describe precisely. But it is thequalitative impact of the definitions which will loom largest in your ex-perience with this book. You may find that you are having difficulty fol-lowing the "proofs" presented in the book or even in understanding whata "proof" is. When this happens, look to the definitions because the chancesare that your real difficulty lies in the fact that you have only a hazyunderstanding of the definitions of basic concepts or are suffering from alack of familiarity with definitions which mean exactly what they say,nothing less and nothing more.

You will also learn a lot of rich and beautiful mathematics. To makethe learning task more manageable, the notes have been provided withsupplementary material and mechanisms which you should utilize:

xiii

xi v Preface to the Student

1. Appendix: Note that the text proper is followed by an Appendixwhich discusses sets, functions, and a bit about cardinality (finite, infinite,countable, and uncountable sets). Read the first part on sets and functionsand then refer to the remainder when it comes up in the notes.

2. Bibliography: There is a short bibliography to which you mightturn if you're having trouble or want to go beyond the notes.

3. List of Symbols: If a symbol occurs in the notes which you don'trecognize, try this list.

4. Index: The Index is fairly extensive and can lead you to variousplaces where a given concept or result is discussed.

One last thing. Use the Exercises to test your understanding. Mostof them come with specific instructions or questions, "Find this", "Provethat", "True or false?". Occasionally an exercise will come without in-structions and will be a simple declarative sentence, "Every differentiablefunction is continuous". Such statements are to be proved. Their occur-rence reflects nothing more than the author's attempt to break the mo-notony of saying, "Prove that ..." over and over again. The exercisesmarked with an asterisk are (usually) extremely difficult. Don't be dis-couraged if some of the ones without asterisks stump you. A few cff themwere significant mathematical discoveries not so long ago.

KENNETH HOFFMAN

ANALYSISIN

EUCLIDEANSPACE

1. Numbersand Geometry

1.1. The Real Number SystemThe basic prerequisite for reading this book is a familiarity with the

real number system. That familiarity should include a facility both withthe elementary algebra of real numbers and with a few inequalities derivedfrom the natural ordering of those numbers. This section is designed toemphasize some properties of the number system which may be less fami-liar.

The first thing we shall do is to list a few fundamental properties ofalgebra and order from which all of the properties of the real number sys-tem can be deduced. Let R be the set of real numbers.

A. Field Axioms. On the set R there are two operations, as follows.The first operation, called addition, associates with each pair of elementsx, y in R an element (x + y) in R. The second operation, called multiplica-tion, associates with each pair of elements x, y in R an element xy in R.These two operations have the following properties.

1. Addition is commutative,x+y= y+x

for all x and y in R.2. Addition is associative,

(x+y)+Z==x+(y+Z)for all x, y, and z in R.

3. There is a unique element 0 (zero) in R such that x + 0 = x forall x in R.

1

2 Numbers and Geometry Chap. 1

4. To each x in R there corresponds a unique element -x in R suchthat x 4- (-x) -- 0.

5. Multiplication is commutative,xy - yx

that

for all x and y in R.6. Multiplication is associative,

(xy)z -- x(yz)for all x, y, and z in R.

7. There is a unique element I (one) in R such that xl = x for all xin R.

8. To each non-zero x in R there corresponds a unique element x-1(or 1 /x) in R such that xx -' -- 1.

9. 1 # 0.10. Multiplication distributes over addition,

x(y j - z) _= xy xzfor all x, y, and z in R.

B. Order Axioms. There is on R a relation

Sec. 1.1 The Real Number System 3

1 -xn+1 =(1 x)(l +x+x2+ ... +xn).We could define the set of positive integers

Z+=[1,2,3, ...}(from the axioms) as the set consisting of the numbers 1 , 1 + 1 , 1 + 1 +-1, ... ; and then we could prove the principle of mathematical induction :If S is a subset of R such that

(a) I E S;(b) if x E S then (x +- 1) E S,

then S contains every positive integer. Then we could define the set ofintegers

Z= {..., -2, -1, 0, 1, 2,...}and the set of rational numbers

Q= m;m E Z,n E Z+n

and diligently verify thatyyl + p mq+npn q nq

and so on. Perhaps (logically) we should carry out those deductions; how-ever, that would be time-consuming and it would be of virtually no helpin understanding analysis.

A similar comment is applicable to a few inequalities which can bededuced (easily) from the order axioms (B). If x < y, then x + z < y + z;if x < y and 0 < c, then cx < cy. We use x < y to mean x < y or x = y.It is understood that y > x means the same thing as x < y. The absolutevalue of a number x is defined by

if x>01x1=

x,

J-X if x


that is to state that the system is characterized by two theorems: (i) Thereexists a complete ordered field. (ii) Any two such fields are isomorphic;that is, there exists a 1: 1 correspondence between their members whichpreserves addition, multiplication, and order. The second reason for listing(A), (B), and (C) is that it will help us understand the completeness prop-erty (C). A fair fraction of introductory analysis consists of learning themeaning of the completeness of the real number system and learning touse various reformulations of it.

As we have suggested, we shall not prove here that the real numbersystem exists or that it is unique. What we assume is a familiarity withcalculations in an ordered field. The one aspect of the number system withwhich we do not assume much familiarity is the completeness. In the nexttwo sections we begin to look at some implications of completeness.

Right now, let us try to be clear about what it says.Intuitively, property (C) is intended to say that if one thinks of real

numbers as corresponding to points on a line, then the line has no holes init. How can one subdivide the "line" R into the union of two non-emptysets, S and T, such that every number in S is less than every number in T?The only way to do that is to cut the line at some point, to let S be every-thing on one side of the cut and to let T be everything on the other side ofthe cut. Of course, the point where we cut must be put either in S or in T,and it will accordingly be the largest number in S or the smallest number inT.

Precisely, suppose we choose any real number c. From c we obtaintwo slightly different subdivisions as described in (Q:

S={s E R;sc}

or

S={s E R;sc}.

The completeness property states that there are no other examples, thefirst type being the one in which S has a largest member, the second typebeing the one in which T has a smallest member.

EXAMPLE 1. Let us look at the rational number system, which consistsof Q (the set of rational numbers), together with the addition, multiplica-tion, and ordering inherited from R. Since sums, differences, products,and quotients of rational numbers are rational, we see that if we substituteQ for R in (A), the field axioms are satisfied. Similarly, Q satisfies theorder axioms (B). Thus, the rational number system is an ordered field;however, it is not complete. Long ago, the Greeks noted that (looselyspeaking) the set of rational numbers had holes in it-for instance, atthe place where /__2 ought to be. More precisely, they proved that there

Sec. 1.1 The Real Number System 5

is no rational number x such that x2 = 2. That means that we can cut theset of rationals at ,/ and show that Q is not complete, because the setS = {s c Q; s < /T} has no largest member and the set T = {t c Q;t > /_2} has no smallest member. Let us describe the situation withoutmentioning / -2.

Suppose we defineS = {s c Q ; either s < 0 or 0 < s and s2 < 21(1.1) T={t E Q;02}.

Clearly S and T are non-empty subsets of Q and each number in S is lessthan each number in T. Here is the important point. Since there does notexist any x in Q with x2 = 2, it follows that

Q=SuT.Does T have a smallest member? If t c T, then t2 > 2. If r is a very smallpositive rational number, then we shall have (t - r)2 > 2 (as well as t - r> 0); i.e., we shall have (t - r) E T. Hence T has no smallest member.By similar reasoning, S has no largest member. Thus, the rational numbersystem does not have the completeness property (C).

Why can't we give the same example in the real number system? Ofcourse, the completeness property says that we cannot. But, let's try it andsee exactly what goes wrong. We define sets S and T as in (1.1), but replaceQ by R. Again, we conclude that S and T are non-empty and that everynumber in S is less than every number in T. Again, we can show that S hasno largest member and that T has no smallest member. The completenessproperty (C) leaves us with only one possibility, namely, that R # S u T,i.e., that some real number belongs neither to S nor to T. It is very easy tosee that if x is a real number such that x 0 S and x 0 T, then x2 = 2.Thus, one of the things which completeness guarantees is that there existsin R a square root for the number 2.

Exercises

In Exercises 1-10, deduce the stated properties of real numbers from thebasic properties (A), (B), and (C).1. Ifx


7. (a) (-x)y = -(xy). Hint: [x + (-x)]y = ?(b) (-x)(-y) = xy.

8. For each x in R, x2>0.9. If x

Sec. 1.2 Consequences of Completeness 7

bounded below; in fact, any x < 0 is a lower bound for R. But it is notbounded above. If b were an upper bound, we could deduce in order:b>0, (b+ 1) E R+,b>b+ 1, ??

EXAMPLE 3. The set of positive integers Z+ is bounded below. It is notbounded above. One of the properties of the real number system with whichthe reader is supposed to be familiar is the Archimedean ordering property,which states that if b c R, then there is a positive integer greater than b.We shall call that a theorem (Theorem 2), and prove it as an exercise inthe use of completeness.

Theorem 1. Let A be a non-empty subset of R which is bounded above.Then A has a least (smallest) upper bound.

Proof. Let T be the set of all upper bounds for A :T={b c R; x< b for allx c Al.

Let S be the complement of TS={x c R; x 0 T}.

We can see easily that(i) R=SUT;

(ii) if s c S and t c T, then s < t.We defined S so that (i) would be true. What does (ii) say? It says, if s 0 Tand t c T, then s < t; or, if t e T and s > t, then s c T. The last state-ment is clearly true. Look at the definition of T.

The hypothesis that A is bounded above is precisely the statement thatT is non-empty. The hypothesis that A is non-empty tells us that S is non-empty, as follows. Choose any x c A. Then S contains every numbery < x, because, if y < x then y is not an upper bound for A.

The completeness condition now tells us that either S has a largestmember or T has a smallest member. But S does not have a largest mem-ber. Let s e S, that is, let s 0 T. Then s is not an upper bound for A. Con-sequently there exists a number a c A with a > s. The number d =__-(a + s) satisfies

s

8 Numbers and Geometry Chap. I

Evidently, there is a companion result which asserts that, if a subsetA of R is non-empty and bounded below, it has a greatest lower bound.That is a number c such that

(i) c is a lower bound for A ;(ii) if b is a lower bound for A, then c > b.

Notation and Terminology. Let A be a non-empty subset of R. If A isbounded above, the least upper bound for A is also called the supremum ofA and is denoted by

sup A.If A is bounded below, the greatest lower bound for A is also called theinfimum of A and is denoted by

inf A.

One might wonder why we introduce other names for least upperbound and greatest lower bound. One reason is that they occur so oftenthat they must be abbreviated, and "lub" and "glb" leave a little to bedesired.

Theorem 1 is a reformulation of the completeness of the real numbersystem. In Section 1.1, if one assumes Theorem 1 instead of property (C),then it is easy to prove (C) as a theorem. The two properties are only slight-ly different. Let's use Theorem I to prove that the set of positive integersis not bounded.

Theorem 2 (Archimedean Ordering Principle). If x is a real number,there exists a positive integer n such that x < n.

Proof. Suppose Z+ is bounded above. Let c == sup Z. Since c is theleast upper bound for Z+, c - I is not an upper bound for Z. Therefore,there exists a positive integer n such that c - 1 < n. So c < n + 1. Butthat says that c is not an upper bound for Z. (?)

Corollary. If x > 0, there exists a positive integer n such that 1 /n < x.

Corollary. If y --- x > 1, there is an integer n such that x < n < y.Proof. According to Theorem 2, there exists an integer m such that

x < m. There are at most finitely many integers k such that x < k < M.(That follows from the principle of mathematical induction.) Let n be theleast of those integers. It is a simple matter to verify that x < n < y.

Corollary. If A is a bounded set of integers, then sup A and inf A areintegers.

Corollary. If x < y, there exists a rational number r such that x < r< Y.

Sec. 1.2 Consequences of Completeness 9

Proof. Choose a positive integer n such that n(y - x) > 1. Then findan integer m such that nx < m < ny. Let r = m/n.

Theorem 3. Let x be a positive real number and let n be a positive inte-ger. There is precisely one positive real number y such that yn = X.

Proof. Let us make a simple basic observation. If s > 0 and t > 0,then t > s if and only if to > Sn. That follows from the fact that

to sn (t s) f (t, S)where f (t, s) - to - I + tn-2 s + + tsn-2 + s'. Since f (t, s) > 0 un-less s = t = 0, the numbers tn - Sn and t - s have the same sign.

Obviously (then) we cannot have two distinct positive nth roots. Theonly problem is to prove that there exists at least one.

LetA={y c R;y>0andyn>x}.

Then A is bounded below. Furthermore A is non-empty. In case x < 1,we have 1 n > x so that 1 E A ; and, in case x > 1 we have

xn-x=x(xn-1-1)>0

so that x c A. Let c = inf A. Certainly c > 0, and the claim is that cn = X.First, we show that cn < X. Suppose cn > x. Then we can find a

small positive number r such that (c - r)n > x. (See following lemma.) Bythe definition of A, (c - r) E A. But, c - r < c and c is a lower boundfor A, a contradiction. It must be that cn < X.

The fact that no lower bound for A is greater than c will imply thatcn > X. Suppose Cn < X. We can find a small positive number r such that(c + r)n < x. Thus (c -F r)n < x < yn for ally c A, which yields c + r c; hence,something is wrong. We conclude that Cn > X.

Lemma. Let n be a positive integer. Let a, b, and c be real numbers suchthat a < cn < b. There exists a number a > 0 such that a < (c + r)n < bfor every r which satisfies I r I < o.

Proof. We have

wheret n Cn (t C)f(t, C)

f(t, C) = to-1 + t n-2c + . . . + tCn-2 d- cn-1.If we apply this with t = c + r, we obtain

(c+r)n-cn=rf(cd-r,c)and hence

I(c+r)n--cnI C IrIIf(c+r,c)I.

10 Numbers and Geometry

Now

I.f(c + r, c) I < (I CI + IrI)n-1 (ICI -I IrI)n-2I c1 _ .. .+ (IcI + IrI)ICIn-2 + ICIn-1.

If I r I < ],thenI

-+- r, c)I < ( I--+-

1)n-1 +- ( I 14- 1 4 -

4 -

4- .. .

+_ (ICI + 016n-2 + IcIn-1f(1 + IcI, IcI).

Therefore,

Chap. 1

I (c -+- r)n cnI < IrIf(1 -+- IcI, ICI), if I r I < 1.Although it is not necessary, we shall rewrite this inequality in a more con-crete form : Since (1 + I c I) - I c I = 1, the definition off. f tells us that

f(1 + IcI, IcI) _ (1 + I cl)n - IcIn.So our inequality says(1.5) I(c+r)n-cIn_

Sec. 1.3 Intervals and Decimals 11

4. Is the empty set bounded above? Does it have a least upper bound?5. Every subset of a bounded set is bounded. Any set which contains an un-

bounded set is unbounded. (Unbounded means not bounded.)6. If A is bounded above and B is bounded below, then the intersection A n B

is bounded.7. Prove that, if x is any real number, then

x = sup [r c Q; r < x}.8. If x < y, there exists an irrational (not rational) number t such that x < t

< Y.9. Verify that every non-empty set of positive integers contains its infimum.

10. Let A be a subset of R which has uncountably many points in it. Prove thatthere exists a non-empty set B c A such that sup B is not in B. (Uncountable isdefined in the Appendix.)11. Prove the completeness property (C) from Theorem 1.12. Prove that, if a subset S of R (with the inherited addition, multiplication, andordering) is a complete ordered field, then S = R.

*13. Let R and S be complete ordered fields. Show that R and S are isomorphic,i.e., show that there is a 1:1 correspondence between the members of R and themembers of S which preserves addition, multiplication, and order.

1.3. Intervals and Decimals

This is a short section, in which we shall discuss the decimal represen-tations of real numbers. We shall not use these representations very much.The purpose of the section is twofold. It provides us with some concreteobjects to which we can point and say, "There, if you will, are the realnumbers." More important, it will make us think about the relation ofintervals to the completeness of the real number system.

Definition. An interval is a set I c R such that(i) I contains at least two points;(ii) ifx


There are five types of unbounded intervals, to which we shall referoccasionally: (a,oo){xE R;a


Next, we subdivide the interval Ia, into 10 intervals, each of length1 / 102. The intervals are

[a1] 0-1 + k10-2, a110-1 + (k + 1)10-2), k -- 0, ... , 9.The one of those intervals which contains x determines the digit a2. Inother words,

a2 = sup [k c Z; a110-1 + k10-2 < x}.We repeat the subdivision process and continue. What do we end up withas a description of the decimal representation? It is a recursive definitionof the digits a1, a2, a3, ....

(i) a 1 is the largest integer k such that k 10-1 C x.(ii) After a 1, . . . , an_ 1 have been determined, an is the largest integer

k such thata110-1 + a210-2 + ... + an-1 10-(n-n- 110-cn- 11 + k 10-n < X.

What we have done is to place x successively in the semi-closed inter-vals J1, J2, J3, ... defined by

in = [a110-1 + a210-2 + ... + an10-n, a110-1 + a210-2(1.7) + . . . +. (an + 1)10- n).The intervals in are "nested"

J1 :-) J2 :-) J3and x belongs to the intersection of all the Jn. In fact

(1.8) ')Jn = {x} ;that is, no other number belongs to every Jn. Why? If a, b E in, thena - b I < l0-n. If y (as well as x) belongs to everyJn, then

Iy - xI< 10-n, n= 1,2,3,....Hence y-x=0.

We know that, by the scheme just described, there is associated witheach x c [0, 1) a sequence of digits a1, a2, a3, ... ; and, we know (1.8)that different x's have different sequences of digits associated with them.So, it is legitimate to employ the shorthand

x = .a1a2a3 . .What is really being abbreviated is

x = E an10-n,n=1

but we'll worry about that later.What interests us now is this. Is every sequence of digits a1, a2, a3, .. .

the decimal representation of a number x c [0, 1) ? Obviously not, if we


follow the ground rules thus far. Take an = 9 for every n. The intersectionn [ 1 - 10-n, 1)

n

is empty so that we cannot have x .999... for any x in [0, 1]. We knowhow to fix that. Had we considered [0, 1] instead of [0, 1), then .999.. .would have arisen as the decimal representation of the number 1. Butthere are still other sequences of digits which do not occur in our process.Obviously, in the scheme we described, no decimal representation willarise which is ultimately all 9's:(1.10) .a1a2...am999... .But, we know how to fix that also. We allow a very slight ambiguity in thedecimal representation by agreeing that (1.10) represents the same num-ber as does(1.11) .a,a2...(arn + 1)000...provided am # 9.

The fuss about repeating 9's is not, however, at the heart of the ques-tion of whether a sequence of digits a1, a2, ... need represent any realnumber. The central problem is this. If someone hands us a sequence ofdigits .a,a2a3... , where will we find the real number which it represents?The digits give us a nested sequence of intervals

J1 DJ2 DJ3 :.defined by (1.7). The x we want is supposed to be (in) the intersection ofthat sequence of intervals. But the intersection may be empty, because ofthe repeated 9's business. So, we must replace the semi-closed interval

in = [bn, cn)by the closed interval

in [bn , Cn] .The intersection of the sets in will catch the right-hand end point if the 9'srepeat. What we want to assert is that

n in = {x}n

where x E [0, 1]. Since the length of Jn is 10-n, there cannot be more thanone point in the intersection. It is the completeness of the real number sys-tem which guarantees that there exists at least one x in the intersection.

Theorem 4 (Nested Intervals Theorem). Let11 DI2=) I3=)

-

be a nested sequence of (bounded) closed intervals in R. Then there exists areal number x which belongs to every I.


Proof: Let In == [bn, cn]. Thenb1


1.4. Euclidean Space

Chap. 1

We presume that the reader knows something about Euclidean spaceof n-dimensions. If n is a positive integer, then

Rn=R x x Ris the set of all n-tuples of real numbers. The points in Rn will sometimesbe called vectors, and our standard notation for those points will be

X (x1, . . . , xn)Y = (Y1, ... , Yn)

and so on. The number xl is the ith (standard) coordinate of X.There is a natural (vector) addition on Rn, defined by adding the coor-

dinates:X

There is a product, called scalar multiplication, defined for vectors X E Rnand numbers c c- R by

cX = (cx1, . . . , cxn).With this addition and scalar multiplication, Rn is a vector space. Thismeans that the vector addition satisfies conditions A(l)-A(4) of Section 1.1and that the scalar multiplication satisfies

c(X+ Y) = cX+ cY(b+c)X=bX+cX

ix=X.The zero vector for addition is the origin 0 = (0, . . . , 0).

If X and Y are vectors in Rn, the (standard) inner product of X and Yis the number(1.12) = x1Y1 + ... + xnYnIn many books, this is called the dot product and is denoted by X. Y.Evidently, the inner product has these properties:

(i)_;(1.13) (ii) = c + < Y, Z>;

(iii) > 0; if = 0 then X = 0.If X E Rn, the length (norm) of X is

XI=112.

The distance from X to Y is I X - YI. In order to see that length and dis-tance have their expected properties, it is most convenient to verifyCauchy's inequality : If x1, ... , xn and y 1 , ... , yn are real numbers, then(1.14) (x1Y1+...+xnYn)2

Sec. 1.4 Euclidean Space 17

Lemma (Cauchy's Inequality). If X and Y are vectors in R I, then(1.15) II < IXI IYI.Furthermore, equality holds if and only if one of the two vectors is a scalarmultiple of the other.

Proof. If Y = 0, the inequality is a trivial equality. If Y # 0, use thefact that

O0; ifIXI = 0 then X= 0;

(ii) I cX I= I C I I XI;(iii) IX+YI


X + Y

FIGURE 2

having k rows and k columns:A = [al;], 1 < i < k, 1

Sec. 1.5 Complex Numbers 19

Proof.

Now

JAB 12 E LE airbrj] 2i, j r

a r E bs j (Cauchy).i, j r s

1Al2 a 2i,r

1BI2 = 37, bsjs, j

and so it is apparent thatIAB12 C 1A121B12.

We might remark that matrix multiplication can be used to expressinner products on the space of matrices in the following way(1.17) = trace (ABt).The trace of a matrix is the sum of its diagonal entries. The matrix Bt is thetranspose of B. Its i, ,j entry is b ji. In the case B = A, (1.17) says

A 12 = trace (AAt).

1.5. Complex NumbersThe complex number system is (essentially) obtained by adjoining to

the real number system a square root for the number -1. The enlarged"system" has (in one sense) less structure, because the ordering of the realnumbers does not extend to an ordering of the complex numbers. But, thecomplex system is richer in ways which make it indispensable for under-standing parts of mathematics. For instance, we obtain complex numbersby introducing a zero for the polynomial x2 +I ; but it turns out that everynon-constant polynomial with real (or even complex) coefficients has azero in the set of complex numbers. That is the so-called fundamentaltheorem of algebra, which we shall prove later.

We mildly caution the reader on two points: (1) This section is brief;however, the importance of complex numbers in this book should not beunderestimated. (2) Mathematicians have retained the mystical termi-nology of "complex" and "real" and "imaginary" numbers; however, theterms are not now intended to suggest anything about reality or the ab-sence thereof.

We list some properties which characterize the complex number sys-tem. Let C be the set of complex numbers.

1. C is a field; i.e., there is an addition and a multiplication on Cwhich satisfy the field axioms (A) of Section 1.1.

2. C contains R as a subfield.


3. There is an element i E C such that i2 = -1.4. If a subfield of C contains R and i, then that subfield is (all of) C.The subset S is a subfield if S contains 0 and 1 and is closed under the

formation of sums, differences, products, and quotients. Thus, (2) says thatR sits in C and that when we restrict to R the addition and multiplicationof C, they become the addition and multiplication of R. If S is a subfieldwhich contains R and i, then S contains every element of C of the formx + iy, with x and y in R. On the other hand, one can show that the set ofthose numbers is a subfield; hence, by (4) it exhausts C.

Therefore C consists of the numbers(1.18) z=x+iy, x,y E Radded and multiplied according to the usual rules of algebra (the fieldaxioms), with i2 -= -1. The representation (1.18) of the number z isunique. We call x the real part of z :

x = Re (z)and we call y the imaginary part of z:

y = Im (z).Notice that the imaginary part is real (a real number). The (complex) con-jugate of x is the number(1.19) z* =x- iy.Note that (z + w)* = z* + w* and (zw)* = z*w*. Since(1.20) zz*=x2+y2>0it makes sense to define the absolute value of z(1.21) I Z I = (Z`*)1/2.In connection with (1.20), we might remark that if w c C and we writew > 0, that is understood to mean that w is real and w > 0.

It is a straightforward matter to verify thatIZ+III -KIWI(1.22)

IzWI = IZI I WISince absolute value preserves products, the number z/I z I has absolutevalue 1 (if z # 0). Hence each non-zero complex number z is uniquelyexpressible in the form(1.23) z=rw, r>0, Iwl= 1.

There is a 1: 1 correspondence between complex numbers and pointsin R2 which is so immediate that we often identify C and R2 as sets. Thenumber z = x + iy corresponds to the point (x, y) in R2. Addition of com-plex numbers corresponds to the vector addition in R2 ; briefly, we addcomplex numbers by adding their real and imaginary parts:


Re (z + w) = Re (z) + Re (w)Im (z + w) = Im (z) + Im (w).

Thus, we have the parallelogram picture of addition in C. Absolute valuecorresponds to length in R2:

I z I = (x2 + y2)1/2.Conjugation corresponds to reflection about the real line.

The geometric interpretation of multiplication involves angles. Letus look at complex numbers w of absolute value 1:

w = U + ivu2+v2= 1.

These points (u, v) comprise the circle of radius 1 centered at the originusually called the unit circle. Each w on the unit circle is uniquely locatedby the angle 0 from the vector 1 to the vector w. (See Figure 3.) Further-more,

u = cos 0v = sin 0

because that is the usual definition of cos 0 and sin 0. Thus(1.24) w =cos 0 + i sin 0.If angles are measured by numbers, i.e., if 0 "is" a number in our discus-sion, then (1.24) determines a unique 0, 0 < 0 < 2ir. Any number 0 +2k7r, k E Z, would then serve as well.

FIGURE 3


Now

Chap. I

cos 0 + i sin 0 - ei9.(We'll discuss that carefully later.) It is then clear that each non-zero z C-C can be expressed(1.25) z = reie, r > 0, 0 E R.Of course r z I. There are many such 8's, all differing by integer multi-ples of 27r. Each 0 is an argument of z. If we also have

w= pelt, p> 0, t c- Rthen

zw - (rp)ei(t+e)In other words, multiplication of complex numbers multiplies the absolutevalues and adds the arguments. If the same fact is verified without usingthe exponential function, it amounts to the verification of the trigonometricidentities

(cos 0 + i sin 0)(cos t + i sin t) =cos (0 + t) + i sin (0 + t)that is,

cos (0 + t) = cos 0 cos t - sin 0 sin tsin (0 + t) = sin 0 cos t + cos 0 sin t.

Each non-zero complex number z has n distinct nth roots. Writez=IzIeie, 0


In terms of coordinates, Cauchy's inequality states that1 Z 1 w 1 + ... + Znwn 1 2 < (I Z 1 1 2 + ... + I Zn I2)(I w 112 + ... + I Wn I2)

There is a (real vector space) isomorphism between Cn and R2n. If z; _x; + iy;, then

(Z 1 , . . , Zn) * (X 1 , Y I 5 X2, Y25 . . , xn, yn)maps Cn onto R2n. It preserves sums, multiplication by real scalars, andlength:

IZ112+ ... +x2+yn.EXAMPLE 5. The discussion of matrices in Example 4 can be extended

immediately to the complex case. The space of k x k matrices with com-plex entries behaves like complex k2-space. In this case, the inner productis described this way:

= trace (AB*)where B* is the conjugate transpose of B. Its i, j entry is R. The verification71that

IABI


7. Each complex k x k matrix A is uniquely expressible in the form A = A 1+ iA2 where A; = A!' (the conjugate transpose). Is it true that

IAI2 = IA1I2 + IA2I2?

1.6. Linear GeometryWe shall review a few basic facts from linear analytic geometry. We

hope that the reader is acquainted with this material. It is not essential forreading this book; however, we shall refer to this material often in theexamples and exercises.

Suppose that X and Y are distinct points in R. The (straight) linethrough X and Y is

[tX + (1 - t) Y; t c R}.A subset S of Rn is called flat if it has this property: If X and Y are in

S, then the line through X and Y is contained in S. A (linear) subspace ofRn is a flat subset which contains the origin.

A subspace is more commonly defined as a non-empty subset S suchthat

(i) if X and Y are in S, then (X Y) is in S;(ii) if X is in S and c is any real number, then cX is in S.

In this formulation, one may replace (i) and (ii) by the single condition:If X and Y are in S, then cX + Y is in S for all c c- R. This is equivalent tothe flat subset characterization which we used as the definition. The essen-tial point is that if X1, ... , X, are vectors in the subspace S, then everylinear combination

c1X1+...+ckXkof those vectors is in S. If we start with any vectors X1, ... , Xk in Rn, theset of all linear combinations of those vectors is a subspace, called thesubspace spanned by X1, ... , X,.

Note that a line is a flat subset and consequently is a subspace if andonly if it passes through the origin. Planes through the origin will be(defined as) the 2-dimensional subspaces. Let us say something about di-mension.

The vectors V1, ... , Vk are linearly dependent if it is possible to ex-press the 0 vector as a linear combination of them in some non-trivial way

C1 V1 + - +CkVk-0, c;# 0 for at least one j.Linearly independent means not linearly dependent. If V 1, ... , Vk arelinearly independent, then Vi # 0 and also Vi # V; when i # j.

The most basic fact about linear dependence is this. I f V1, ... , Vkare vectors in Rn and i f k < n, then V1, ... , Vk are linearly dependent. Inshort, any n + 1 vectors in Rn are linearly dependent. That is a reformula-

Sec. 1.6 Linear Geometry 25

tion of the basic theorem on systems of linear equations. For, suppose wewish to find numbers c1, ... , Ck such that cl Vl + - + CkVk = 0. If

Vi - (vi l , ... , v1n), I < i < kthen the condition on the numbers ci is

Clvi l + . . +- CkVkl = 0C l v 1 2 + ... + C kV k 2 = 0

C1v1n + . .. + CkVkn = 0.

If k > n, this homogeneous system of n linear equations in k unknownshas a solution for (c1, ... , Ck) which is non-trivial (not every ci = 0).

If S is a subspace of Rn, the dimension of S is the maximum number oflinearly independent vectors which can be found in S. More precisely,dim S is the largest non-negative integer k such that some k-tuple of vec-tors in S is linearly independent. Of course dim S < n. It is not difficult tosee that dim S < n, except for the one subspace S = Rn.

An (ordered) basis for a subspace S is a k-tuple of vectors V1, . . . , Vksuch that

(i) V1, . . . , Vk are linearly independent;(ii) S is the subspace spanned by V1, ... , Vk.The simplest example of a basis is the standard basis for Rn

E1=(1,0,0,...,0)E2 = (0, 1, 0, ... , 0)

(1.26)

En=(0,0,0,..., 1).The unique expression for X = (xl, ... , xn) as a linear combination ofthose vectors is

X = xlE1 + ... +- xnEn.

Theorem 6. If S is a subspace of Rn, then S has a basis, and every basisfor S consists of precisely dim S vectors.

Proof. If d = dim S, then S has a basis consisting of d vectors: Wecan find vectors Y1, ... , Yd in S which are independent. By the definitionof d, we know that, for any X EE S, the vectors Y1, ... , Yd, X are depen-dent. But that means that X is a linear combination of the vectors Y;.

Suppose V1, ... , Vk is any (ordered) basis for the subspace S. EachX in S can be expressed as a linear combination(1.27) X = Cl V1 + . . . + CkVk.


Furthermore, the numbers c1, ... , Ck are uniquely determined by X (andthe basis). If we had another expression for X as a linear combination, wecould subtract and contradict the independence of V1, ... , Vk. The k-tuple (c 1, ... , Ck) is the k-tuple of coordinates of X relative to the orderedbasis V1, ... , Vk. It is clear that (1.27) defines a 1: 1 correspondencebetween the set of all vectors X in S and the set of all k-tuples (c 1, ... , COin Rk. If we add two vectors X and Y in S, the corresponding coordinatesadd; and, multiplication of X by the number t multiplies each coordinatec; by t. Thus, as far as linear operations are concerned, S behaves Justlike Rk (S is isomorphic to Rk). In particular, any k + 1 vectors in S arelinearly dependent. If d = dim S, there exist d independent vectors in S.Thus d < k. But k < d by the definition of d.

Suppose V1, . . . , Vn is a basis for Rn. Then we can describe each X inRn by its coordinates relative to that basis, as well as by its standard coor-dinates. The standard coordinates are the coordinates relative to the basisE1, ... , En (1.26). How do we get from one set of coordinates to the other?If(1.28) X = (x1, ... , xn) = C1 V1 + + CnVnthen

x; = c1v1 j .+. ... .+. Cnvn;, 1


if and only if the matrix Q =_ [vij] is invertible. In turn, that happens if andonly if the determinant of Q is different from zero.

The simplest bases are orthonormal bases, and we should say some-thing about them. The vectors X, Y in Rn are called orthogonal (perpen-dicular) if = 0. The Pythagorean property is immediate from thedefinition: If X and Y are orthogonal, then

IX+ Y12 = IX12 + I Y12.The k-tuple of vectors V1, ... , Vk is called orthogonal if Vi is ortho-

gonal to Vj for i #j. The k-tuple is called orthonormal if it is orthogonaland each V1 has length 1. Thus, orthonormality means(1.32) -- Sij.

Suppose we have an orthonormal k-tuple V1, ... , Vk. If X is a linearcombination of those vectors

X = C 1 V1 + . . . + C k Vk

it is easy to compute from (1.32) that(1.33) ci =, 1


This is a great simplification over the situation for a general basis, whereone needs to invert the matrix Q = [vi;] before many coordinates can becomputed. In the orthonormal case, the inverse of Q is the transposematrix Qt = [v;i]. The conditions = Si; say precisely that(1.36) QQt=1Such a matrix Q is called an orthogonal matrix.

The vectors V1, ... , Vn :

Vi (vi 1 , , vin)form an orthonormal basis for Rn if and only if the matrix Q = [vi;] is anorthogonal matrix. If Q is orthogonal, then the coordinates C = (c1, ... , cn)of the vector X relative to the orthonormal basis V1, . . . , Vn are given by

C = XQtor

ci = .If one wishes to study a subspace S, it is often most convenient to use

the orthogonal complement of S :(1.37) SJ-={XERn;=0allYESJ.

Theorem 8. Let S be a subspace of R. Each vector X in Rn is uniquelyexpressible in the form(1.38) X=Y+Z, YES, ZES1-.

Proof. Suppose we have X decomposed as in (1.38). Let V1, ... , Vkbe any orthonormal basis for S. Since Y is in S

Y = V1 + . . . + VkSince Z is orthogonal to each Vi, we have < Y, V1> _ . Hence,

(1.39)Y V1 + . . . + VkZ=X-Y.

That determines Y and Z uniquely. On the other hand, given X, we candefine Y and Z by those formulas and clearly Y is in S and Z is in SJ-.

If X is in Rn, the vector Y in Theorem 8 is called the orthogonal pro-jection of X on S. Even in the proof of the theorem we used a particularorthonormal basis to define Y (1.39) ; however, as the proof shows, Y isindependent of the basis. Notice how norms and inner products behaverelative to orthogonal projections. Suppose Y1 is the orthogonal projec-tion of X1 on S and Y2 is the orthogonal projection of X2 :

X1=Y1+Z1X2=Y2+Z2.


Then _ -f

IXiI2=IYjI2-IZjI2.We have described subspaces by bases. There is a dual point of view in

which we describe a subspace by giving a set of equations for it. The equa-tions are linear; that is, they are of the form f(X) = 0 where f is a linearfunction

fRn>Rf(cX+ Y)=cf(X)+f(Y).

If f is such a (real-valued) linear function on Rn, there are numbers a1, ... ,an such that f has the form

f (X) = a1x1 + ... + anxn.Suppose V1, ... , Vn is an ordered basis for R. There is an associated

k-tuple of coordinate functions f 1, ... , fn. The function fi assigns to thevector X its ith coordinate relative to the basis V1, ... , Vn. In short

X = f1(X)V1 + . . . +Jn(X)VnCertainly f1, ... , fn are linear functions. Their specific form is

fj(X) = p1 jx1 + + pnjxnwhere P = [pij] is the matrix of (1.30) :

Ej= pj1 V1 + ... + pinVnThe functions f 1, . . . j n are uniquely determined by the fact that they arelinear and that

fi(Vj) = Sij.Theorem 9. If S is a k-dimensional subspace of Rn then S is the solution

space for a system of (n - k) homogeneous linear equations.P r o o f . Let V1, ... , Vk be a basis for S. We can find vectors Vk+1,

... ,

V n such that V1, ... , V,, is a basis for Rn. Consider the correspondingcoordinate functions f1, . . . , fn. Obviously S consists of all vectors X inRn such that

fi(X)=0, i==k+ 1,...,n.Notice that in the proof we could use orthonormal bases. The con-

clusion is that there are (n - k) vectors Vk+ 1 , Vn in Rn such thatS={X;=O,i=k+ 1,...,n}.

A hyperplane in Rn is a level set of a non-zero linear function f:H = {X; f(X) = c}


A hyperplane is a flat subset of dimension (n - 1). For, if we are givensuch an H and if we choose any Xo such that f(X0) = c, then f(X - X0)= 0 for every X in H. Thus X is in H if and only if

X=X0+ Y, f( Y) =0.The set of vectors which, f sends into 0 is an (n - 1)-dimensional sub-space. Thus, every hyperplane is a subspace of dimension (n -- 1), trans-lated by a fixed vector in Rn. Similarly, we see that each non-empty flatsubset of Rn is a subspace, translated by a fixed vector. Theorem 9 saysthat every k-dimensional subspace of Rn is the intersection of n - k hyper-planes (through the origin). Thus, we see that every non-empty flat subsetF is an intersection of hyperplanes; in fact, F can be described by specify-ing (not more than n) linear conditions on the coordinates of the vectorsin F.

2. Convergenceand Compactness

2.1. Convergent Sequences

Analysis is founded upon the concept of limit; indeed, the centralrole played by limiting operations is essentially what defines the branch ofmathematics known as analysis. We shall first take up the idea of thelimit of a sequence. It is the simplest type of limit, and yet a thoroughcomprehension of it enables one to understand other limits rather easily.

We shall be working in Rm, Euclidean space of dimension m. Thefirst thing we need is some language for describing when points are nearto one another.

Definition. If Xo E Rm and r > 0, the open ball of radius r about thepoint X0 is(2.1) B(X0 ; r) _ {X; I X - X0 I < rJ.

The closed ball of radius r about the point X0 is(2.2) B(X0 ; r) _ {X; I X - X0 I < rJ.

A neighborhood of X0 is a subset (of Rm) which contains an open ballabout X0.

A neighborhood of X contains every point which is (sufficiently)close to X. The most important neighborhoods of X are the open ballsabout X. Notice that the concept of open [closed] ball reverts to open[closed] disk in R2 and (symmetric) open [closed] interval in R1.

31

32 Convergence and Compactness Chap. 2

Definition. The sequence {Xn} converges to the point X if every neigh-borhood of X contains Xn, except for a finite number of values of n. If {Xn}converges to X, then we write(2.3) X = lim X.

n

Suppose that X = lim X. If r > 0, the open ball B(X; r) containsX, except for certain integers n1, ... , nk, which presumably depend on r.Let Nr be the maximum of those numbers, and then Nr is a positive integersuch that

IX - XnI < r, n>Nr.Since every neighborhood of X contains some B(X; r), we see that Xnconverges to X if and only if, for each positive number r, there exists a posi-tive integer Nr such that(2.4) I X -XnI < r, n > Nr.

The terminology concerning convergence is frequently bent invarious ways. Often, we omit the braces and say "Xn converges to X". Wesay that the sequence [X,J converges (or, is convergent; or, has a limit) ifthere exists an X such that [Xn} converges to X. If no such X exists, wesay that the sequence diverges. If [Xn} converges to X, we sometimes callX the limit of the sequence and we sometimes write

X=limXnor

X = lim Xnn oo

instead of (2.3).Be careful about trying to bend the wording of the definition of con-

vergent sequence. There is a distinct difference between "contains Xn,except for a finite number of n's" and "contains all except a finite numberof Xn's". In R1, the sequence Xn = (- 1)n would converge to every x e R,if we used the second wording. Every open interval contains all except afinite number of xn's, because there are only two xn's. On the otherhand, no interval of length less than 2 has the property that it contains xn,except for a finite number of values of n. Hence, this sequence [xn}does not converge.

Lemma. IfX=limXn and Y=limYn

n n

then(i) for every real number c,

cX+Y=lim(cXn+Yn)

Sec. 2.1 Convergent Sequences 33

(ll) = lim .n

Proof. (i) Part (i) is trivial when c = 0. Suppose c # 0, and let r be apositive number. Since

I(cX+Y)-(cX +- Yn)I M.2IcISimilarly, since [Yn} converges to Y, there is a positive integer N such that

I Y- YnI < r, n>N.2

Let K = max (M, N). Then we shall haveI(cX+ Y)-(cXX+ Yn)I K.

(ii) There is a standard sort of manipulation for this type of argument:-=+.

Thus by Cauchy's inequality

I -I


Hence, a sequence has at most one limit. In part (ii), if Xn = Yn, we haveIX12=Jim IXnl2.

n

It is then easy to show that we can remove the exponents 2; but, note thatit is trivial by another means to see that

IXl=IimlXnln

if Xn converges to X, because

I I X I I X n l l C IX Xnl.In R1, part (ii) states that xnyn converges to xy if xn converges to x and ynconverges to y.

Theorem 1. The sequence {X,,} in Rm:Xn = (x1, . . . , xnm)

converges if and only if each of the m coordinate sequences [xn j} converges.If

x; = limxn;, 1

Sec. 2.1 Convergent Sequences 35

the space RI-x,'- [CI-x,'-], Euclidean space with the k2 coordinates arrangedin k rows and k columns. By Theorem 1, that is equivalent to convergenceof the corresponding entries. Note that, if An converges to A and Bn con-verges to B, then the matrix product AnBn converges to AB. The proof isas in part (ii) of the last lemma:

I AB --- AnBnI < I A I I B -- BnI + I A -- AnI I BnI-We have used here the complex version of Theorem 5 of Chapter 1:

ISTI

36 Convergence and Compactness

verges to (1 - z)-1. Symbolically,00

=1 : Zn IzIN.

7. If xn > 0 and xn converges to x, then ,V/ xn converges to 1/ X.8. Let A be a square matrix. Look at the sequence of its powers. Show that if

An converges to B. then AB = B. Give an example where the sequence [An} doesnot converge, yet I An I remains bounded.9. Let z be a complex number. Prove that the sequence

Zn

n!

is bounded. From the fact that it is bounded, show that it converges to 0. Fromthis fact prove that, if E > 0, there is a constant K such that

KEnn!

for all except a finite number of n's.10. Let S be a (linear) subspace of Rm. If X is a vector in Rm, let P(X) be theorthogonal projection of X onto the subspace S. Show that if Xn converges toX, then P(Xn) converges to P(X).

Sec. 2.2 Convergence Criteria 37

11. Let [A,,} be a sequence of invertible k x k matrices. Suppose An convergesto A but A is not invertible. Show that

"1imI An 1 = co"n

12. Prove that if xn E R and xn converges to x, then the sequence of arithmeticmeans

also converges to x.Sn 1 (XI + + xn)n

2.2. Convergence Criteria

It is extremely important to develop criteria which will ensure that asequence converges, in spite of the fact that we do not know the limitexplicitly. How can we tell if a sequence converges? One crude test is theboundedness of the sequence

JXJ


of least upper bounds. Evidently, there is a companion result aboutmonotone decreasing sequences.

Let x1, x2, x3, ... be any bounded sequence of real numbers. Foreach n, define

(2.8) an- inf{xk;k>n}bn = sup [xk ; k > n}.

Then we have two monotone sequences, one increasing and the otherdecreasing:

a1


From the definitions of an and bn, we then haveX -- E MX n < X - - E, n > N.

Thus,Ix - xnI < E, n> max(M,N).

One frequently sees the notations Jim xn and lim xn employed forlim sup x and lim inf xn, respectively. In attempting to picture the geomet-rical relationship of the superior and inferior limits to the sequence [xn},the following observation (which we state only for lim sup) is sometimesuseful. If S - Jim sup xn and e > 0, then xn > s +E holds for only afinite number of values of n and xn > s - E holds for infinitely manyvalues of n.

The completeness of the real number system will (evidently) bereflected in some type of completeness of Euclidean space Rm. The se-quential form of that completeness is conveniently phrased in terms ofthe Cauchy convergence criterion.

Start with a sequence [Xn} in Rm. If it converges to some point X,then the various Xn's with large subscripts must be close to one another,because they are all close to the limit point X.

Definition. A Cauchy sequence is a sequence [Xn} with this property.For each E > 0, there exists a positive integer N, such that(2.10) IXk - XnINE, n>NE.

We have just commented that every convergent sequence is a Cauchysequence. In that case, a positive integer N, (2.10) can be determinedprecisely this way. If Xn converges to X, choose N, so that

IX - XnI < E, n>NE.2Then (2.10) is satisfied.

Theorem 4 (Completeness of Rm). Every Cauchy sequence in Rm con-verges.

Proof. SupposeYn = (Xn1, . . . , Xnm), n = 1,2,3,...

is a sequence of points in Rm. For each coordinate index JIXkj Xnj1


Let [xn} be a Cauchy sequence of real numbers. How will we find anumber x c- R to which it converges? First, let us note that the sequenceis bounded. By the Cauchy condition, there is a positive integer N suchthat

IXk - XnI < 5, k>N, n>N.In particular,

IXN - XnI < 5, n>N.Let M be the largest of the numbers

IX119...,IXN-1195 + IXNI

and plainly I x I < M for all n.Now, let

x = lim inf xn.We remind the reader what that means:

an = inf{xk; k > n}a1


(2.13) IXn-Xn+1I


6. True or false? If the infinite series Z Xn converges, then the set of norms[I XnI; n E Z+}

has a largest number in it.7. Let [xn} be a bounded sequence of real numbers. Let A be the set of real

numbers t such that xn < t for only a finite number of n's. Show thatlim inf xn = sup A.

8. True or false? If z is a complex number and I z I > 1, then z" diverges.9. True or false? If A is a k x k matrix and I A > 1, then An diverges.

10. Letxn = (1 - 2-1)(1 - 2-2) ... (l - 2-n).

Prove that the sequence converges and lim xn # 0-11. Let [xn} and [ yn} be (bounded) sequences of real numbers. Prove that

lim sup (xn + yn) < lim sup xn -{- Jim sup yn.12. Prove this generalization of Example 4. If [ Xn} is a sequence in Rm such that

00

I Xn - Xn+ 1 I < 00n=1

then [ Xn} is a Cauchy sequence.13. True or false? In R', every convergent sequence is the sum of an increasingsequence and a decreasing sequence.14. Assume the monotone convergence theorem. Prove that every non-emptyset of real numbers which is bounded above has a least upper bound.15. If you knew that every Cauchy sequence in R converged, how would youprove the monotone convergence theorem? (You will need to use the Archi-medean ordering property.)

2.3. Infinite Series

Now we shall see what the convergence criterion of the last sectiontells us about infinite series. We shall concentrate on the two most impor-tant classes of infinite series, namely, series with positive terms and ab-solutely convergent series.

We shall make frequent use of the series analogue of sums and scalarmultiples of convergent sequences: If Z Xn = S and Z Yn = T, thenZ (cXn + Yn) = cS +T-

Suppose that we have an infinite series in which the terms are non-negative real numbers:

E xnn

xn>0.

Sec. 2.3 Infinite Series 43

Then the partial sums constitute an increasing sequence:

Sn = Xkk = I

Si


another while before he thought of proving it by grouping the terms asin (2.14). We showed that

E < 00n n

by verifying the inequality (2.15), which is probably not the first thing onewould think of, upon being confronted with that infinite series.

As one builds a stockpile of specific series which are known to con-verge or diverge, it becomes possible to test new series for convergence bycomparing them with the known series. If

E yn < 00n

and 0 < xn < yn, obviouslyxn < 00.

n

In fact, it is enough to know that xn C yn for all sufficiently large n :xn < Yn, n > N.

Similarly, ifE Yn = 00

n

and xn > yn >0 for all sufficiently large n, then E xn diverges.

EXAMPLE 7. In Section 2.1, we verified that the series Z z" convergesfor all complex numbers of absolute value less than 1:

00

1 = E z I z I< 1.1 - Z n=0In particular,(2.16) x" < 00, 0 < x < 1.

n

A number of different series can be seen to converge by comparingtheir terms with the terms of a geometrical series (2.16). For instance,(2.17) Y11

n


follows. Since

lim n = 11n n+

and t < 1, there exists an N such that2t< n , n>N.

n + 1This inequality may be rewritten

(n + 1)2tn+ 1 < n2tn, n > N.Thus the sequence [n2t} is monotone decreasing for large n and, ac-cordingly, it converges. If the limit were not 0, we could divide and obtainthe contradiction

1 _ lim n2tnlim (n + 1)2tn+ 1

= lim n2t-1

n + I

The reader should be aware that the seriesxnE

n n!

converges for every real number x. For the moment, let's worry only aboutnon-negative numbers x. If 0 < x < 1, the series obviously converges.The interesting point is that it converges for large x. Fix such an x. Whathappens to

xn

n!

as n gets large? We pass from the nth term to the (n + 1)th term by mul-tiplying by xl(n + 1). Once n is large enough so that n + 1 > x, we aremultiplying by a number less than 1. It should be clear now that the seriesconverges, and converges faster than a geometric series. If we want to beprecise, we can phrase it this way. Given x, choose a positive integer Nsuch that

x t1 n!

N-x

n N< i + x E tk

1 n. N. k=1

< 00.


The series which we have just been discussing is the power series for theexponential function and we shall say more about it shortly. The readershould be familiar with the case x -_ I which defines the number e:

fxl

e =_- -- 1 1n o n.

I -i 12!' 3!'Now that we have indicated what the monotone convergence theorem

tells us about non-negative series, let us see what the Cauchy convergencecriterion tells us about infinite series. Suppose that we have a (now vector-valued) series with terms Xn and partial sums Sn :

n

Sn -- E x1 c.k I

By definition, the series converges if the sequence [Sn} converges. Ac-cording to Theorem 4, the series converges if and only if [Sn} is a Cauchysequence. The crudest possible estimate for the distance from Sk to Sn(with n > k) is

sic Sn I -- I Xk + 1 -f I Xn II X1c +1 I -f . . . -I- I Xn I.

How can we guarantee that the last sum is small, provided that k and nare both large? Well

IXk+II -I . -+- IXnI < E, N


It is this fact which makes all rearrangements possible. Let A 1, A2, .. .be a sequence of subsets of the positive integers such that each n is inprecisely one A,:

Z,=UAnn

AnnAk=0,For each n, the series

n # k.

E XkkEAn

converges, because it converges absolutely:E IXkI


EXAMPLE 9. Let A be a k x k matrix with complex entries. SupposeA I < 1. Then the series

Any (A = I)n=0

converges absolutely because I An I C I A In, n # 0. It should converge to

1-Abut that doesn't exactly make sense. What does make sense is the inverseof the matrix I - A. Let

(2.22) B= >An, IAA < 1.c*0

It is easy to verify that(2.23) B(I - A) _ (I - A)B = Ii.e., I - A is invertible and B = (I - A)-1.

Let A be any k x k matrix with complex entries. We define the ex-ponential of A to be

(2.24)exp(A)=eA=I+A+ 1 A2+

2!C110 IE nAn0

Again, the series converges absolutely because (see Example 7)1 Ann!

that

< I A In = eJ A I - 1.I n!

If B is a matrix which commutes with A, AB = BA, then we can show

e(A+B)= eAeB = eBeA.

For any fixed matrix Mro

A' M = eAM.0

Thus,

Since

1 1 I AnBk I < o0n k n! k!

we may regroup to obtain

eAeB = E 1 AneBo n

CIOE 1 An Bko n! (k=_o k!

(Y)

eAe B AnBkN=ok+n=N k. n!


Since AB = BA, that sayseAe'3 = E 1 (A + B)N = e(A-F8).

N Nt

Exercises

1. Let x and t be any positive real numbers. Show thatxn

n 1.n


9. Show thatZ n-(3!2) < 00

n

Chap. 2

by using induction to verify thatsn < 3 - 2n-(121.

10. Let [xn} be a monotone decreasing sequence of positive numbers. Prove thatE X. converges if and only if

n

E 2kX2kk

converges.11. Use the result of Exercise 10 to investigate the convergence of

E n-q q > 0.n

12. Let x1, x2, ... be any sequence of non-negative numbers such thatE Xn < 00.n

Prove that there exist positive integers N1, N29 ... with these properties:N 2 1 +x

for all real numbers x. One procedure might be to verify in order:ex> 1 fix, x>0ex< 1 0 1 fix, x> -1ex> 1 fix, x c R.

14. Use the result of Exercise 13 to show that

expI z I

IzI1

< I 1 + z exp (IzI)

for all complex numbers z with I z I < 1.15. If A is a V x k matrix, show that matrix eA is invertible.16. Let A and B be k x k matrices of norm less than 1. Discover a simple rela-tionship between (I - AB)- 1 and (I - BA)-1. In what sense does that relation-ship hold for all A and B?17. Let [wn} be a sequence of complex numbers. The infinite product

wnn

Sec. 2.4 Sequential Compactness 5?

is said to converge if the sequence of partial productsn

Pn I I Wk - W1 W2 . . . Wnk 1

converges to a non-zero number w. Prove the following, by using the results ofExercises 7 and 13: If [zn} is a sequence of complex numbers (zn #-I) such that

I I Zn I < Con

then the infinite product

converges.I I (1 + zn)n

'18. Let E xn be a series of real numbers which converges but does not convergeabsolutely. Show that, if t is a real number, the order of the terms in the seriescan be rearranged so that the rearranged series converges to t. Hint: Take enoughpositive terms so that their sum just exceeds t, then add enough negative termsso the sum is just under t, etc.

*19. Let E Xn be a series of vectors in Rm which converges but does not con-verge absolutely. Let S be the set of all vectors which are sums of rearrangementsof the series. What kind of a set can S be?

*20. Prove thatdet eA = etr (A)

for all k x k complex matrices A (det = determinant function; tr = trace func-tion).

2.4. Sequential Compactness

The sequence [X} converges to the point X if Xn is near X for allsufficiently large n. There are many situations in which we are given asequence and we don't need to know that it converges. What we need toknow is that there is a point X at which the sequence accumulates, i.e.,Xn is near X for infinitely many values of n.

Definition. The point X is a point of accumulation (accumulationpoint) of the sequence [Xn} if every neighborhood of X contains Xn forinfinitely many values of n.

We can say it another way: X is an accumulation point of [Xn} if,for each e > 0 and each positive integer n, there exists k > n such that

IX --- XkI < E.If [X} converges to X, then clearly X is the unique point of accumulationof the sequence.


EXAMPLE 10. Let r 1 , r2, r3, ... be the sequence which consists of thepositive rational numbers, enumerated according to the scheme

Then, every non-negative real number is an accumulation point of thesequence [rn}.

EXAMPLE 11. Beware of working with coordinates when discussingaccumulation points. Consider in R2

Xn = (0, 1), n oddXn = (1, 0), n even.

The sequence of first coordinates is 0, 1, 0, 1, . . . which has two accumula-tion points in R, 0 and 1. The sequence of second coordinates is 1, 0, 1, 0,... and it has the same accumulation points. In particular, 0 is an accumu-lation point for the first coordinates and for the second coordinates. Wecannot conclude that (0, 0) is a point of accumulation of the sequence inR2.

Definition. The sequence Y1, Y2,.Y31 ... is a subsequence of the se-quence X1, X2, X3, ... if there exist positive integers n,, n2, n3, ... suchthat

(i)n1

Sec. 2.4 Sequential Compactness 53

for infinitely many n2's. Choose one so that n2 > n 1. Continue in that wayand obtain

IX-Xnkl < 1knl t for infinitely many n}and verify that x is an accumulation point of the sequence.

Corollary. A bounded sequence in Rm converges if and only if it hasprecisely one point of accumulation.

Let us outline another proof of the Bolzano-Weierstrass theoremwhich has a slightly different intuitive basis. We can take the boundedsequence, multiply it by a non-zero scalar and then translate it so thatthe new sequence is in the box

B=[X;0


Therefore, we may as well assume that the given sequence is in B. Thisbox is the Cartesian product of the closed unit interval with itself m times:

B= I x .. x I= [X; x; E I, 1

Sec. 2.5 Open and Closed Sets 55

After we had chosen the boxes Bn, we did not need to use the Cauchycriterion. We might, for example, have applied the nested intervals theoremin each coordinate.

Exercises

1. Prove this. In R 1, the bounded sequence [xn} has a smallest and a largestaccumulation point. They are (respectively) lim inf xn and lira sup xn.2. Let z be a complex number of absolute value 1:

z = eie, 0 < 0 < 27rWhat are the accumulation points of the sequence [Zn} ? Distinguish between thecase where 9 is a rational multiple of 27r and the case where it is not.3. Let [Xn} be a sequence. Suppose that Y1, Y29 ... is a sequence of accumula-

tion points of [Xn}. Show that any accumulation point of the sequence { Yn} is anaccumulation point of the sequence [X}.4. If [Xn} is a Cauchy sequence and if some subsequence converges to X, then

X=limXn.5. Let [xn} be a bounded sequence of real numbers such that I xn -- xn+ 1 I = 1 for

each n. Show that the sequence has only a finite number of accumulation points.6. If [Xn} is a bounded sequence of real numbers such that I xn - xn+ 1 I > I for

each n, can it have an infinite number of accumulation points?7. Suppose that [xn} is a bounded sequence of real numbers such that

lim sup xn < sup xn.n n

Show that, among the numbers xn, there is a largest one.8. Find a sequence [zn} of complex numbers with I zn I < 1 such that

(a) no point z with I z < I is a point of accumulation of the sequence;(b) every point z with z I = 1 is an accumulation point of the sequence.

*9. True or false? Let A be a k x k matrix with complex entries. The set of allaccumulation points of the sequence {An} is bounded.

2.5. Open and Closed SetsThe title of this section mentions two very special classes of sets. One

can do analysis in Euclidean space without either concept; however, quitea bit is lost if one does. The concepts of open set and closed set shed lighton questions of convergence and on the geometry of mappings.

Definition. The set U is open (in Rm) if it is a neighborhood of each ofits points.


Thus, U is open if and only if the following condition is satisfied: Foreach point X in U, there exists a positive number r such that the open ballB(X; r) is contained (entirely) in U.

Theorem 6. The union of any collection of open sets is open. The inter-section of any finite collection of open sets is open.

Proof. Suppose {Ua} is a family of open sets. LetU= U Um.

a

If X E U, then X E Ua, for some a. Therefore (that) Ua is a neighborhoodof X. Since U contains Ua, U is a neighborhood of X.

Suppose U1, . . . , Un are open sets. LetU=nUk.

k

Let X E U. For each k, 1 < k < n, there is a number rk > 0 such thatB(X; rk) C Uk.

Let r be the least of the numbers r 1, ... , r, ThenB(X;r)=U.

Theorem 6 is virtually trivial. It is called a theorem, because it statesproperties of open sets which are used so often.

EXAMPLE 12. Every open ball B(X; r) is an open set. If Y E B(X; r)then B(Y; t) c B(X; r) where

t=r - IX- Yl.Thus, the union of any collection of open balls is open:

U B(XX; ra)a

Furthermore, every open set is of the last type. It is the union of thoseopen balls which it contains.

EXAMPLE 13. Let us look at open sets in R1. Each open interval (a, b)is an open set in R I. On the other hand, an interval (a, b] is not open in R1,because b c (a, b] but no open interval about b is contained in (a, b]. Theunbounded interval (a, oo) is open in R1.

Every open set in RI is a union of open intervals (a, b). In this 1-di-mensional case, the open set U can be expressed as a union of intervalsin a very special way. That is because open intervals in R 1 have this specialproperty: If several open intervals have a point in common, their unionis an open interval. (See Exercise 11.) Take any x in the open set U. LetIx be the union of all those open intervals which contain the point x and arecontained in the set U. Then Ix is an open interval (possibly unbounded).


Therefore, for each x c U there is a largest open interval Ix which containsx and is contained in U. If y c I, then plainly I,, = Ix. In other words, allthe points in Ix belong to the same largest interval in U, namely Ix. Con-sequently, if x, y c U then either Ix = I,, or the intersection of Ix and I,,is empty. How many different intervals Ix are there? Only a countablenumber. (See Exercise 12 and Appendix.) Thus every open set in RI isuniquely expressible as the union of a countable collection of open intervalswhich are pairwise disjoint.

EXAMPLE 14. Let's look at the space of k x k matrices (real or com-plex entries). Let U be the set of invertible matrices, i.e., matrices A suchthat there exists A- 1 with AA-1 = A-1 A = I. Is U an open set ? What wouldthat say? It would say that, if the matrix A is invertible then every matrix(sufficiently) near A is also invertible. We showed earlier (Example 9) thatevery matrix near the identity matrix is invertible: If I T I < 1, then

orifl1-SI< 1, then(I - T)-I = E Tn;co

n=0

S-1 = (I - S) n.n=0

So maybe U is open. If A is invertible, how close must B stay to A toguarantee that B is invertible? Now

A-B=A(I-A-1B),or

Thus,

Suppose

Then

I-A-1B=A-1(A- B).I I- A- IB I< I A-' I I A- B1.

IA-RI < A-1

II - A-IBI < 1;hence, A B is invertible, and since A -1 is invertible, that makes B invert-ible. To summarize, the set U of invertible matrices is open because, ifA E U, then U contains the open ball of radius I A- 1 I- 1 about the pointA.

Definition. The point X is a cluster point of the set S if every neighbor-hood of X contains a point of S which is different from X.

Lemma. Let S be a subset of Rm and let X E Rm. The following areequivalent (all true or all false).


(i) X is a cluster point of the set S.(ii) Every neighborhood of X contains infinitely many points of S.

(iii) There exists a sequence {X n} in S such that X,, #- X and Xlimn X.

Proof Exercise.

The reader may have noticed the similarity of the concepts of "clusterpoint of a set" and "accumulation point of a sequence". It is importantto be clear about the relationship between the two ideas. If {Xn} is asequence in Rm, then its image

S=-= {Xn;n e Z_,}is a subset of Rm. It is the set of all distinct vectors which occur as one(or more) of the terms of the sequence. A point X is a cluster point of Sif each neighborhood of X contains infinitely many points of S. Such anX is surely an accumulation point of the sequence, because each neighbor-hood, containing infinitely many points of S, must contain Xn for infinitelymany values of n. On the other hand, if Y is a point of accumulation ofthe sequence it need not be a cluster point of the set S. A simple exampleshould make this clear. The sequence of real numbers

0,1,0,1,0,1,...has two points of accumulation, 0 and 1. The image of the sequence isS -- {0, 11, and it has no cluster points at all. If all of the terms of thesequence {Xn} are distinct, then every accumulation point of the sequenceis a cluster point of S.

The terms "cluster point", "limit point", and "accumulation point"are used interchangeably in most parts of mathematics. We have electedto apply "accumulation" to sequences and "cluster" to sets, as a reminderof the distinction discussed in the last paragraph.

Definition. The set K is closed if every cluster point of K is in K.

A closed set is one which is closed under (the process of taking) limits.From the last lemma, it should be clear that these conditions on a set Kare equivalent:

(i) K is closed.(ii) If {Xn} is a sequence of points in K and if the sequence converges,

then the limit of the sequence is in K.

Theorem 7. A set S is open if and only if its complement (complementaryset) is closed.

Proof. Let T be the complement of S:T = [XERm;X0S}.


To say that T is closed is to say that, if X 0 T, then X is not a clusterpoint of T. Think about that.

Corollary. The intersection of any collection of closed sets is closed.The union of any finite collection of closed sets is closed.

Proof. This follows from Theorems 6 and 7. We describe the proofof the first statement. Let {Ka} be a collection of closed sets. For each a,let Ua be the complement of K. Then each Ua is an open set and, there-fore, the union

U=UUaM

is an open set. This union is the complement ofK= nKa,

a

the intersection of the complements of the sets U. Since U is open, K isclosed.

We should remark that closed balls and closed intervals are closedsets. In view of Theorem 7, there is no need for a separate list of examplesof closed sets. Every example of an open set provides an example of aclosed set (and vice versa). But, the human mind being what it is, it doesn'tfollow that just because we know about open sets we'll recognize a closedset when we bump into it.

EXAMPLE 15. Let's look at a famous closed set the Cantor set. Weshall refer to it often. Start with the closed interval [0, 1] in R1. Removethe open middle one-third, i.e., the open interval (-, 2). What remains isthe union of two disjoint closed intervals. Remove the open middle one-third of each of those intervals. (See Figure 5.) Now, remove the openmiddle one-third of each of the four remaining closed intervals. Continuead infinitum. What remains is the Cantor set K.

out out out0

FIGURE 5

We obtained K by removing from [0, 1] an open set U. That set isthe union of the sequence of open intervals: (1, 2), (i-, 2), (7, 8), .... SinceU is open and [0, 1] is closed,

7

K= {x c [0, 1]; x 0 U}is a closed set. The set K is very thin. The lengths of the open intervals inU add up to


1 +2 +4 1 +3 2700 2n-1

=n=1 3n

CX)

(2)nn

1[(1-)-1-1]1.

Chap. 2

But, there are a lot of points in K-uncountably many. In particular, Kcontains many more points than the end points of the deleted intervals.(Those are the obvious points in K.)

We can describe the Cantor set very nicely, if we use the ternaryrather than the decimal expansion of points in the interval [0, 1]. Theternary expansion represents each x c [0, 1] as the sum of a series

00x=Ean3-n

n=1

where the "digits" an are 0, 1, or 2. The digits are defined by locating thepoint x in a sequence of intervals, the lengths of which go down by afactor of 3 each time:

a1=sup kEZ; k


such a sequence, "contains infinitely many X,'s" is the same as "containsX, for infinitely many n's".

Theorem 9. LetK1DI{2DI{3 ..

be a nested sequence of bounded closed sets in Rm. If each Kn is non-empty,then the intersection

flKnn

is non-empty.

Proof. For each n, there exists a point Xn in K. The sequence [Xn}is bounded and, accordingly, it has a point of accumulation X. Since Xk EK, for k > n and since K, is closed, X must be in K,

Here is an amusing application of the weaker result.

EXAMPLE 16. An analyst (of the mathematical variety) might proceedthis way to show that the medians of a triangle meet in a point. Let thetriangle be ABC and let X, Y, Z be the midpoints of the sides. By com-parison of similar triangles, it is clear that (1) the median lines of ABCare also the median lines of XYZ, (2) diam XYZ = -1 diam ABC. (SeeFigure 6.) Replace ABC by XYZ and repeat. Then repeat again, etc. Doyou see where Theorem 9 comes in?

z

FIGURE 6

One might think of Theorem 9 as a slightly more geometrical way ofstating the Bolzano-Weierstrass theorem. If (in Theorem 9) one knowsthat diam (Kn) converges to 0, then the intersection of all the Ku's will con-sist of precisely one point. That result is weaker than Theorem 9. It is(essentially) a reformulation of the fact that each Cauchy sequence in Rmconverges.


Exercises

Chap. 2

1. Show that there are (at least) two subsets of Rm which are both open andclosed-Rm and the empty set.

2. If S is a subset of Rm and if S is both open and closed, then either S = Rmor S is empty. (You might prove it first in R1.)3. If S is a subset of Rm, what is the union of all closed subsets of S?4. Which of the following sets of complex numbers are closed ? Which are

open ?(a) all z such that z = z* ;(b) all z such that zz* > 2;(c) all z such that I z < 1 and z #O;(d) all z such that I z is rational.

5. True or false? If S is closed, then S contains a cluster point of S.6. True or false? If S is a bounded infinite subset of R, then among the cluster

points of S there is a largest one.7. True or false? If every subset of S is closed, then S contains only a finite

number of points.8. True or false? If every subset of S is open, then S contains only a finite num-

ber of points.9. If S is a set, the X0-translate of S is X0 -F S = [X0 + Y; Y E S}. Show that

each translate of an open set is open and each translate of a closed set is closed.10. What can you say about scalar multiples of open [closed] sets?11. If {la; a E Al is a family of open intervals on the real line and if the inter-section

n Ia

is non-empty, then the union is an open interval. (Remember the definition ofinterval.)12. Let {lx; a c Al be a family of open intervals on the real line which arepairwise disjoint:

IanIQ=0,Prove that A is a countable set.13. Let A be a subset of Rm and let B be a subset of R. The Cartesian productA X B is a subset of Rm+n. Prove that

(a) if A and B are open, then A x B is open;(b) if A and B are closed, then A X B is closed.

14. Is the set of orthogonal k x k matrices open?15. Let M be a linear subspace of Rm. Let K be a closed subset of Rm. Project theset K orthogonally onto M. Do you end up with a closed set?

Sec. 2.6 Closure and Interior 63

16. Every (linear) subspace of Rm is closed.17. A set K is perfect if K is closed and every point of K is a cluster point of K.Show that the Cantor set is a perfect set.

*18. Every non-empty perfect set is uncountable.*19. Every closed set is the union of a perfect set and a countable set.

2.6. Closure and Interior

Let S be a subset of Rm. The closure of S is the intersection of allclosed sets which contain S. The interior of S is the union of all opensets which are contained in S. The boundary of S is the intersection ofthe closure of S with the closure of the complement of S.

The closure of S will be denoted S. Evidently S is a closed set. It isthe smallest closed set which contains S. If X is a point in Rm, these con-ditions are equivalent:

(i) X E S.(ii) Either X E S or X is a cluster point of S.

(iii) There exists a sequence of points in S which converges to X.The interior of S will be denoted S (although we sh

Kenneth Hoffman-Analysis in Euclidean Space -Prentic-Hall _ Dover Publications (1975)

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