to learn more about brooks/cole, visit …mullai.uah.edu/~ravindra/chapters1-3.pdf · introductory...

To learn more about Brooks/Cole, visit www.cengage.com/brookscole

Purchase any of our products at your local college store or at our preferred online store www.cengagebrain.com 9 7 8 0 3 9 5 9 5 9 3 3 6

9 0 0 0 0

Introductory Real Analysis

Frank DangelloMichael Seyfried

Shippensburg University

Australia • Brazil • Japan • Korea • Mexico • Singapore • Spain • United Kingdom • United States

2000 Brooks/Cole, Cengage Learning

ALL RIGHTS RESERVED. No part of this work covered by the copyright

herein may be reproduced, transmitted, stored, or used in any form or by

any means graphic, electronic, or mechanical, including but not limited to

photocopying, recording, scanning, digitizing, taping, Web distribution,

information networks, or information storage and retrieval systems, except

as permitted under Section 107 or 108 of the 1976 United States Copyright

Act, without the prior written permission of the publisher.

Library of Congress Catalog Card Number: 99-71719

ISBN-13: 978-0-395-95933-6

ISBN-10: 0-395-95933-0

Brooks/Cole Cengage Learning

20 Davis Drive

Belmont, CA 94002-3098

USA

Cengage Learning is a leading provider of customized learning solutions with

office locations around the globe, including Singapore, the United Kingdom,

Australia, Mexico, Brazil, and Japan. Locate your local office at

www.cengage.com/global

Cengage Learning products are represented in Canada by

Nelson Education, Ltd.

To learn more about Brooks/Cole, visit www.cengage.com/brookscole

Purchase any of our products at your local college store or at our

preferred online store www.cengagebrain.com


Frank Dangello and Michael Seyfried

Sponsoring Editor: Jack Shira

Assistant Editor: Carolyn Johnson

Editorial Assistant: Christine E. Lee

Art Supervisor/Interior Design: Gary Crespo

Marketing Manager: Michael Busnach

Senior Manufacturing Coordinator:

Pricilla Bailey

Cover Design: Henry Rachlin

Cover Image: © Photodisc

Printed in the United States of America

3 4 5 6 14 13 12 11

For product information and technology assistance, contact us at

Cengage Learning Customer & Sales Support, 1-800-354-9706

For permission to use material from this text or product,

submit all requests online at www.cengage.com/permissions

Further permissions questions can be e-mailed to

[email protected]

Dedicated to Sue and Rosa

Contents

v

Preface ix

Chapter 1 Proofs, Sets, and Functions 11.1 Proofs 11.2 Sets 51.3 Functions 101.4 Mathematical Induction 15

Chapter 2 The Structure of 192.1 Algebraic and Order Properties of 192.2 The Completeness Axiom 232.3 The Rational Numbers Are Dense in 272.4 Cardinality 31

Chapter 3 Sequences 393.1 Convergence 393.2 Limit Theorems 443.3 Subsequences 483.4 Monotone Sequences 513.5 Bolzano-Weierstrass Theorems 553.6 Cauchy Sequences 583.7 Limits at Infinity 613.8 Limit Superior and Limit Inferior 65

Chapter 4 Continuity 694.1 Continuous Functions 694.2 Continuity and Sequences 734.3 Limits of Functions 774.4 Consequences of Continuity 834.5 Uniform Continuity 864.6 Discontinuities and Monotone Functions 89

Chapter 5 Differentiation 955.1 The Derivative 955.2 Mean Value Theorems 101

5.3 Taylor’s Theorem 1045.4 L’Hôpital’s Rule 107

Chapter 6 Riemann Integration 1116.1 Existence of the Riemann Integral 1116.2 Riemann Sums 1176.3 Properties of the Riemann Integral 1206.4 Families of Riemann Integrable Functions 1266.5 Fundamental Theorem of Calculus 1316.6 Improper Integrals 135

Chapter 7 Infinite Series 1437.1 Convergence and Divergence 1437.2 Absolute and Conditional Convergence 1487.3 Regrouping and Rearranging Series 1567.4 Multiplication of Series 161

Chapter 8 Sequences and Series of Functions 1678.1 Function Sequences 1678.2 Preservation Theorems 1748.3 Series of Functions 1808.4 Weierstrass Approximation Theorem 187

Chapter 9 Power Series 1919.1 Convergence 1919.2 Taylor Series 198

Chapter 10 The Riemann-Stieltjes Integral 20710.1 Monotone Increasing Integrators 20710.2 Families of Integrable Functions 21510.3 Riemann-Stieltjes Sums 22010.4 Functions of Bounded Variation 22610.5 Integrators of Bounded Variation 233

Chapter 11 The Topology of 23911.1 Open and Closed Sets 23911.2 Neighborhoods and Accumulation Points 24511.3 Compact Sets 249

vi Contents

Contents vii

11.4 Connected Sets 25411.5 Continuous Functions 258

Bibliography 265

Hints and Answers 267

Index 285

ix

Preface

To the StudentIs the prize worth the struggle? This book provides not only a complete foun-dation for the basic topics covered in the usual calculus sequence but also astepping stone to higher-level mathematics. This book contains some beautifulmathematical results and opens the door for the student to see other beautifulmathematical results. This is the prize. The struggle is that the ability to seethese beautiful mathematical results does not come easily. Put more succinctly,Introductory Real Analysis is a hard course.

With the possible exception of mathematical geniuses, such as Euler,Gauss, and Riemann, people are not born knowing how to do proofs. Theskill to prove a mathematical result is learned just as the skill to solve problemsis learned, usually with lots of practice. We suggest covering up our proof ofa statement. After understanding what the statement means, attempt your ownproof. If you are stuck, look at the first line of our proof and see if you canproceed. If necessary, look at another line, and so on.

The examples and exercises are essential components of this book. Theywill help clarify and strengthen your insight into the text material; and they willincrease your mathematical maturity, which, like the ability to do proofs, is notan innate human trait. As much as possible, we have arranged the exercises ineach section to correspond to the development of the text material, rather thanordering them from easy to hard. Learning to use previous results to obtain newresults is part of mathematical maturity, and we freely draw on the exercises inthis manner.

Throughout the book we give overviews and commentaries in order toprovide insights into what we are doing and where we are going. These featurestake the form of either short paragraphs or more formal remarks. Our purposefor dividing some of the commentaries into remarks is to separate differentconcepts for the student.

In answer to our original question, those who have seen the beauty inmathematics definitely think the prize is worth the struggle.

To the InstructorAlthough the only prerequisite for this book is the usual calculus sequence,another mathematics course (such as an introduction to abstract algebra) wouldhelp develop the student’s sophistication. In this book we do not do analysisfrom scratch, nor do we construct the real numbers. Rather, we build on whatthe student has learned in calculus. This book is designed to be used effec-tively in one- or two-semester courses. The first nine chapters contain an ample

x Preface

amount of material for most two-semester courses, while Chapters 10 and 11offer additional topics depending on the instructor’s personal taste.

As the table of contents indicates, our approach is sequential rather thantopological. Because the table of contents lists the topics covered in the book,which for the most part should be done in order, we mention some aspects ofthe book that are not apparent from the table of contents.

Chapter 1 is an introductory chapter whose purposes are to provide somenecessary topics for proceeding in the book and to allow the students to developsome easy proofs. The only topic in Chapter 1 that most students have notseen before is the arbitrary collection of sets, which helps the students developproofs with quantifiers. In our course, we cover Chapter 1 quickly. Becausewe do not fix notation until Chapter 2, the instructor can begin with Chapter2, referring back to Chapter 1 as needed. We have meticulously providedappropriate references throughout the book.

In Section 2.4 our purpose is to show that the rational numbers are countableand that the real numbers are uncountable. Our experience is that one could getbogged down in this section since cardinality is so fascinating to students; forexample, how can there be just as many real numbers in the open interval from0 to 1 as there are on the real line?

Starting in Chapter 3, we use the terminology of neighborhoods, whichprovides a unifying aspect for both sequential and function limits. In our classeswe usually cover about half of Section 3.8, our purpose being to understandwhat the limit superior and limit inferior are and what they mean in terms ofsequential limits. Exercise 5 in this section is very valuable in this regard.We point out that Exercise 11 in Section 3.7 is needed only for the proof ofProposition 3.9 in Section 3.8. We use Section 3.8 in the proof of Mertens’sTheorem in Chapter 7 and in Chapter 9. In Chapter 7, our Ratio and Root testsuse only limits, whereas in Chapter 9 we extend these tests using limit superiorand limit inferior. Of course, Chapter 9 relies heavily on Chapters 7 and 8.

Chapter 6 contains an extensive section on improper integrals with tests forconvergence similar to those of infinite series. This section may be covered indepth or lightly depending on the instructor’s preference. Although we do notrecommend this, Chapter 7 can be covered after Chapter 3 if one is willing toallow the Integral test without formally doing improper integrals or monotonefunctions. Our development of the Riemann-Stieltjes integral in Chapter 10parallels our development of the Riemann integral in Chapter 6. Many of theproofs in Chapter 10 for monotone integrators are only slight modifications ofthe corresponding proofs in Chapter 6. We think Chapter 6 should be coveredbefore Chapter 10, although both chapters can be taught simultaneously. Also,Section 10.4 on functions of bounded variation can be covered without doinganything else in Chapter 10; it depends only on Chapter 4. Thus, an instructorwho does not reach Chapter 10 and who has only a couple of class periods leftin the semester can cover Section 10.4.

Chapter 4, of course, depends on Chapters 2 and 3. Chapter 11 gives atopological view of our previous concepts, and many of the results in Chapter4 are now special cases of the action of a continuous function on a compact orconnected set.

Preface xi

Accompanying the text is an Instructor’s Manual, which contains com-plete solutions to every exercise in the text and additional exercises (also withcomplete solutions) suitable for student take-home problems or projects. TheInstructor’s Manual also contains some suggestions on the text material and onthe level of difficulty of some of the exercises.

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

We would like to acknowledge many people for their help in the prepa-ration of this book. For their insightful criticisms and suggestions we thankthe reviewers: Michael Berry, West Virginia Wesleyan College; David Gurney,Southeastern Louisiana University; Nathaniel F. Martin, University of Virginia;John W. Neuberger, University of North Texas; Alec Norton, University ofTexas at Austin; Richard B. Thompson, University of Arizona; Guoliang Yu,University of Colorado; and Marvin Zeman, Southern Illinois University at Car-bondale. We also thank our colleagues Douglas Ensley, Frederick Nordai, andWilliam Weller of Shippensburg University for their input; our secretary PamelaMcLaughlin for her assistance; and, most of all, our students for their invaluablecomments when we did classroom testing of earlier versions of this book.

1

1 Proofs, Sets,and Functions

One purpose of this chapter is to help improve the reader’s ability to un-derstand and create proofs. We attempt to do this in Section 1.1. In Sec-tion 1.2, we consider sets, including arbitrary collections of sets. Sections1.3 and 1.4 deal with functions and mathematical induction, respectively.

1.1 Proofs

Many of the statements we prove in this section are known to the reader. It is thetechnique of proof we wish to emphasize. The typical mathematical assertionthat requires proof is the conditional statement (or the implication)

if p, then q (or p implies q). (1)

Here, p and q are statements each of which is either true or false but not both.In (1), p is called the hypothesis and q is called the conclusion.

Direct ProofsThe direct method of proving conditional statement (1) assumes the truth of p

and deduces the truth of q. We illustrate the direct method below.

Proposition 1.1 If n is an even integer, then n2 is an even integer.

Proof Assuming the hypothesis that n is an even integer, we have n = 2k forsome integer k. Then n2 = 4k2 = 2(2k2) is an even integer, because 2k2 is aninteger.

In Proposition 1.1, the hypothesis p is the statement “n is an even integer”and the conclusion q is the statement “n2 is an even integer.” First of all, weneed to know the definitions of the terms in the proposition. In Proposition 1.1,we need to know what an even integer is. Also, in the proof, note the use of theexistential quantifier “for some” (equivalent to “there exists” and symbolicallydenoted by ∃). That is, given an even integer n, there is only one integer k suchthat n = 2k. The universal quantifier “for all” (equivalently, “for every,” “foreach,” “for any,” and denoted symbolically by ∀) would have been incorrect inthe proof above. Either at the very start of the proof or at the end of the firstline of the proof, we should ask ourselves

“What is it that I must do to show that n2 is an even integer?”

Basically, you are asking yourself “what is the next to the last statementin the proof?”—the last statement typically being “therefore q” or, in this case,“therefore n2 is an even integer.” In Proposition 1.1, we have to show that n2

2 Chapter 1 Proofs, Sets, and Functions

is two times an integer. One detail omitted in the proof above is why 2k2 is aninteger. The reader should answer this.

ConverseThe converse of the conditional statement “if p, then q” is the statement “if q,

then p.” A little thought should indicate that a conditional statement and itsconverse may or may not have the same truth value. As examples, the converseof the conditional statement in Proposition 1.1 is true, whereas the converse ofthe calculus theorem

If a function f is differentiable at a point a, then f is continuous at a

is false.

BiconditionalGiven statements p and q, the biconditional statement “p if and only if q” (alsodenoted p ⇔ q or p iff q) means

if p, then q and if q, then p.

To prove the biconditional statement, one must prove both conditionalstatements. To illustrate this, we need some terminology. Throughout the text,R denotes the real numbers, N denotes the positive integers {1, 2, 3, . . .}, and< denotes the usual ordering on R.

Definition 1.1 Let A be a subset of R. A is unbounded above if foreach positive real number x there exists an a in A such that x < a.Symbolically, A is unbounded above if ∀ x > 0, ∃ a in A such thatx < a. Note that a depends on x.

Definitions are to be interpreted in the “if and only if” sense, even though itis common practice not to state them this way. For example, in Definition 1.1,the “if” is actually “if and only if.”

Archimedean Principle If x is a positive real number, then there existsa positive integer n such that 1/n < x. (Symbolically, ∀ x > 0, ∃ n inN such that 1/n < x.) Note that n depends on x.

Proposition 1.2 N is unbounded above if and only if the Archimedean Prin-ciple holds.

Proof Assume that N is unbounded above. We need to show that the Archime-dean Principle is true. Let x > 0. Since N is unbounded above, there exists ann in N such that 1/x < n. Therefore, 1/n < x.

Next, assume that the Archimedean Principle holds. We need to show thatN is unbounded above. Let x > 0. (The reader should now ask: What dowe have to do to finish the proof?) Then 1/x > 0 (see Exercise 6). By theArchimedean Principle, there exists an n in N such that 1/n < 1/x, and thusx < n. Therefore, N is unbounded above.

Section 1.1 Proofs 3

We point out that we have not shown that N is unbounded above, nor havewe shown that the Archimedean Principle is true. What we have shown is that“N is unbounded above” is a true statement if and only if the ArchimedeanPrinciple is a true statement, or that the statement “N is unbounded above” islogically equivalent to the Archimedean Principle.

Indirect ProofsThere are two types of indirect proofs: the contrapositive argument and thecontradiction argument. The contrapositive of the conditional statement “if p,then q” is the statement “if not q, then not p.”

A little thought should indicate that both a conditional statement and itscontrapositive have the same truth value. Thus, to prove a conditional statementone can prove its contrapositive. Since we will have to negate many statementsthroughout the text, we state a basic rule for negation:

Change all universal quantifiers to existential quantifiers; change allexistential quantifiers to universal quantifiers; and negate the mainclause.

Proposition 1.3 Let n be an integer. If n2 is an odd integer, then n is an oddinteger.

Proof This is the contrapositive of Proposition 1.1.

To prove the conditional statement “if p, then q” by contradiction, oneassumes that p is true and that q is false and “hunts” for a contradiction. Oncea contradiction is reached, it follows that if p is true, then q must also be true.Where do you find the contradiction? Sometimes the contradiction is clear (forexample, 0 = 1), and sometimes it is very unclear.

To illustrate proof by contradiction in the next two propositions, we assumethe usual order properties on R and that if a is a real number, then −a is theadditive inverse of a (so −a + a = 0).

Proposition 1.4 Let a be in R. If a > 0, then −a < 0.

Proof Assume a > 0 and that the statement −a < 0 is false. Then −a ≥ 0.

Since a > 0, −a + a > 0 or 0 > 0, which is a contradiction. Therefore,−a < 0.

Proposition 1.5 Let a be in R. If a < ε for all ε > 0, then a ≤ 0.

Proof The key to this proof is the universal quantifier “for all” in the hypothe-sis. Assume that a < ε for all ε > 0 and that a > 0. Then, by the ArchimedeanPrinciple, there exists a positive integer n such that 0 < 1/n < a. With ε = 1/n

(that is, using 1/n as a particular value of ε), this is a contradiction. Therefore,a ≤ 0.

The Archimedean Principle will be established in Theorem 2.1. Alterna-tively, instead of 1/n above, we could use ε = a/2.

For the remainder of this section we need the following terminology. Arational number is a real number that can be expressed in the form m/n, wherem and n are integers and n �= 0. An irrational number is a real number that is


not a rational number. A prime number (or simply a prime) is a positive integergreater than 1 whose only positive divisors are itself and 1. By divisors, wemean integers that divide a given number exactly (that is, with zero remainder).For example, 2, 3, 5, 7, and 11 are primes, whereas 9 is not a prime because 3is a divisor of 9. We will need the following theorem:

If a prime divides a product of two integers, then the prime must divideat least one of the two integers.

Theorem 1.1√

2 is an irrational number.

Proof Suppose√

2 is a rational number. Since√

2 is positive, there existpositive integers m and n such that

√2 = m/n and m/n is in lowest terms.

(We can always reduce a fraction to its lowest terms.) Then 2n2 = m2. Since2 divides 2n2, 2 divides m2. Since 2 is a prime, 2 divides m. Thus there is apositive integer k such that m = 2k. Then 2n2 = 4k2 and n2 = 2k2. As above,2 divides n2 and hence 2 divides n. Thus m/n is not in lowest terms, which isa contradiction. Therefore,

√2 is not a rational number.

For the next theorem we need the result that each positive integer greaterthan 1 is divisible by a prime.

Theorem 1.2 There are infinitely many primes. (Note: This result appearsin Euclid’s Elements, Book IX, Proposition 20.)

Proof Suppose there are only finitely many distinct primes, sayp1, p2, . . . , pn.Let M = p1p2 · · · pn +1. Then M is an integer, and so there exists a prime thatdivides M . Thus some pi divides M . But pi divides p1p2 · · · pn. Therefore,pi divides 1, which is a contradiction.

From the proofs in this section, certain things should be clear. First, onemust know what the terms in the theorem mean. Second, one usually needs toknow facts (axioms, propositions, theorems, etc.) to use in the proof. Third,one must know the end of the proof; and keeping the end in mind helps toprevent the line of reasoning from straying off course (McArthur).

Exercises1. Prove that

√6 is irrational.

2. Prove that√

p is irrational, where p is a prime number.

3. Let a and b be real numbers. Prove that a2 + b2 = 0 if and only if a = 0and b = 0.

4. Let a and b be real numbers. Prove that ab = 0 if and only if a = 0 orb = 0.

5. Let a be a real number. Prove that if a < 0, then −a > 0.

6. Let a be a real number. Prove that if a > 0, then 1/a > 0.

Section 1.2 Sets 5

7. Let a be a real number. If a2 = a, prove that either a = 0 or a = 1.

8. (Pigeonhole Principle) Suppose we place m pigeons in n pigeonholes, wherem and n are positive integers. If m > n, show that at least two pigeonsmust be placed in the same pigeonhole. [Hint (from Robert Lindahl ofMorehead State University): For i = 1, 2, . . . , n, let xi denote the numberof pigeons that are placed in the ith pigeonhole; let xk denote the largest of

the xi’s; and let x = (n

�i=1

xi)/n denote the average of the xi’s. Show that

xk ≥ x = m/n > 1.]

1.2 Sets

Although the first part of this section should be familiar to the reader, thetechnique of proof may not be. We adopt the viewpoint of “naive” set theoryconsidering the notion of a set as already known.

Basic Results and Set OperationsA set is a well-defined collection of objects. By “well-defined” we mean that,given a set and an object, it is possible to tell whether the object is or is not inthe set. Each object of a set is called an element of the set, a point of the set, ora member of the set. If A is a set and x is a point, then

x ∈ A denotes that x is an element of A

while

x /∈ A denotes that x is not an element of A.

A set can be defined either by listing the elements of the set or by statinga property of its elements. For example,

A = {−1, 4}= {x ∈ R : x2 − 3x − 4 = 0}

where R denotes the set of real numbers.The empty set (void set, null set), denoted by ∅, is the set with no elements.

Thus

∅ = {x ∈ R : x2 < 0} = {x : x �= x}and so on.

Definition 1.2 Let A and B be sets.

1. A is a subset of B, denoted by A ⊂ B or B ⊃ A, if for each x in A, x

is in B.

2. A is equal to B, denoted by A = B, if A ⊂ B and B ⊂ A.

3. A is a proper subset of B if A ⊂ B and A �= B.


Thus, to prove that two sets are equal, one must show that each is a subsetof the other. Also note that since ∅ contains no elements, it follows from thedefinition of a subset that ∅ ⊂ A for all sets A.

Definition 1.3 Let A and B be sets.

1. The union of A and B, denoted by A ∪ B, is defined as

A ∪ B = {x : x ∈ A or x ∈ B}.The word “or” is used in the inclusive sense, so that points that belongto both A and B also belong to the union.

2. The intersection of A and B, denoted by A ∩ B, is defined as

A ∩ B = {x : x ∈ A and x ∈ B}.Thus A ∩ B ⊂ A ∪ B.

3. A and B are disjoint if A ∩ B = ∅.

Proposition 1.6 Let A, B, and C be sets. Then

1. A ∪ A = A and A ∩ A = A;2. A ∪ ∅ = A and A ∩ ∅ = ∅;

3. A ⊂ A ∪ B and A ∩ B ⊂ A;

4. A ∪ B = B ∪ A and A ∩ B = B ∩ A (commutative property);

5. A ∪ (B ∪ C) = (A ∪ B) ∪ C and A ∩ (B ∩ C) = (A ∩ B) ∩ C (associativeproperty);

6. A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) and A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)

(distributive property);

7. A ⊂ B if and only if A ∪ B = B and A ⊂ B if and only if A ∩ B = A.

Proof We prove the first equality in part 6. By the definition of equality ofsets, we need to show that

A ∪ (B ∩ C) ⊂ (A ∪ B) ∩ (A ∪ C) (2)

and that

(A ∪ B) ∩ (A ∪ C) ⊂ A ∪ (B ∩ C). (3)

To show (2), let x ∈ A ∪ (B ∩ C). Then either x ∈ A or x ∈ B ∩ C. If x ∈ A,

then by part 3, x ∈ A ∪ B and x ∈ A ∪ C. By the definition of intersection,x ∈ (A∪B)∩ (A∪C). If x ∈ B ∩C, then x ∈ B and x ∈ C. Again, by part 3,x ∈ B ∪ A and x ∈ C ∪ A. By part 4, x ∈ A ∪ B and x ∈ A ∪ C. Therefore,x ∈ (A ∪ B) ∩ (A ∪ C).

To show (3), let x ∈ (A ∪ B) ∩ (A ∪ C). Then x ∈ A ∪ B and x ∈ A ∪ C.

If x ∈ A, then by part 3, x ∈ A∪ (B ∩C). So we may assume x /∈ A. Then, bythe definition of union, x ∈ B and x ∈ C and so x ∈ B ∩ C. Again, by part 3,x ∈ (B ∩ C) ∪ A = A ∪ (B ∩ C).

Section 1.2 Sets 7

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

if A ⊂ B, then A ∪ B = B (4)

and that

if A ∪ B = B, then A ⊂ B. (5)

To show (4), we assume A ⊂ B and note that we have to show A ∪ B = B. Sowe have to show that A ∪ B ⊂ B and that B ⊂ A ∪ B. The latter containmentfollows from part 3. To show A ∪ B ⊂ B, let x ∈ A ∪ B. Then either x ∈ A

or x ∈ B. If x ∈ A, since A ⊂ B, x ∈ B.To show (5), assume that A ∪ B = B. We need to show that A ⊂ B. Let

x ∈ A. Then x ∈ A ∪ B by part 3. Since A ∪ B = B, x ∈ B.

The remaining parts of the proposition are left as exercises.

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

A \ B = {x ∈ A : x /∈ B}.

For example, if R denotes the set of real numbers and Q denotes the set ofrational numbers—that is, if

Q ={m

n: m and n are integers and n �= 0

},

then R \ Q is the set of irrational numbers. Also note that Q \ R = ∅.


1. A \ ∅ = A and A \ A = ∅2. DeMorgan’s Laws:

A \ (B ∩ C) = (A \ B) ∪ (A \ C)

and

A \ (B ∪ C) = (A \ B) ∩ (A \ C).

DeMorgan’s Laws are generally remembered as stating that the comple-ment of an intersection is the union of the complements and the complementof a union is the intersection of the complements.

Proof We prove the first equality in part 2, leaving the rest as an exercise.We need to show that

A \ (B ∩ C) ⊂ (A \ B) ∪ (A \ C) (6)

and that

(A \ B) ∪ (A \ C) ⊂ A \ (B ∩ C). (7)

To show (6), let x ∈ A \ (B ∩ C). By the definition of complement, x ∈ A andx /∈ B ∩ C. By the definition of intersection, either x /∈ B or x /∈ C (note that“and” would be incorrect here). If x /∈ B, then x ∈ A \ B, and if x /∈ C, thenx ∈ A \ C. So, in either case, x ∈ (A \ B) ∪ (A \ C).


To show (7), let x ∈ (A\B)∪(A\C). Then either x ∈ A\B or x ∈ A\C.If x ∈ A \B, then x ∈ A and x /∈ B. Hence, x ∈ A and x /∈ B ∩C. Therefore,x ∈ A \ (B ∩ C). If x ∈ A \ C, then x ∈ A and x /∈ C. Hence, x ∈ A andx /∈ B ∩ C. Therefore, x ∈ A \ (B ∩ C).

Arbitrary Unions and IntersectionsWe now generalize Definition 1.3 to an arbitrary collection of sets. For example,

{An : n ∈ N} where An = (0, n) for each n in N.

The reader should draw these sets on the real line.

Definition 1.5 Let A be a collection of sets.

1. The union of A, denoted by⋃

A, is defined as⋃A = {x : x ∈ A for at least one A ∈ A}.

2. If A is nonempty, the intersection of A, denoted by⋂

A, is defined as⋂A = {x : x ∈ A for all A ∈ A}.

This definition extends the notions of union and intersection given previ-ously, for if A = {A, B}, where A and B are sets, then

⋃A = A ∪ B and⋂

A = A ∩ B. If A is an empty collection of sets, then⋃

A = ∅ and we donot define

⋂A.

An equivalent formulation of Definition 1.5 can be given in terms of indexsets. Let I be a set, called the index set. Suppose Aα is a set for each α in I .Then A = {Aα : α ∈ I } is an indexed collection of sets. Notationally,⋃

A =⋃

{Aα : α ∈ I } =⋃α∈I

Aα =⋃

α∈IAα

= {x : x ∈ Aα for some α ∈ I }and ⋂

A =⋂

{Aα : α ∈ I } =⋂α∈I

Aα =⋂

α∈IAα

= {x : x ∈ Aα for all α ∈ I } (for I �= ∅).

If I = {1, 2, 3, . . . , n} for some n in N, we write⋃

i∈I Ai = ⋃ni=1 Ai and⋂

i∈I Ai = ⋂ni=1 Ai.

If I = N, we write⋃

n∈N An = ⋃∞n=1 An and

⋂n∈N An = ⋂∞

n=1 An.

Example 1.1⋃∞

n=1(0, n) = (0, ∞) and⋂∞

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

Example 1.2⋃∞

n=1{n} = N and⋂∞

n=1{n} = ∅.

Example 1.3⋃∞

n=1(−n, n) = R and⋂∞

n=1(−n, n) = (−1, 1).

Example 1.4⋃∞

n=1(0, 1/n) = (0, 1) and⋂∞

n=1(0, 1/n) = ∅. To show the

latter, suppose x ∈ ⋂∞n=1(0, 1/n). Then 0 < x < 1/n for all n in N. But, by

Section 1.2 Sets 9

the Archimedean Principle (which is proved in Section 2.3), since x > 0, thereis an n0 in N with 0 < 1/n0 < x. This contradicts x < 1/n for all n in N.Therefore,

⋂∞n=1(0, 1/n) = ∅.

Proposition 1.8 (DeMorgan’s Laws) Let X be a set. Let I be a nonemptyindex set and let Aα be a set for each α ∈ I . Then X\⋃α∈I Aα = ⋂

α∈I (X\Aα)

and X \⋂α∈I Aα = ⋃α∈I (X \ Aα).

Proof We prove the first equality, leaving the second as an exercise. We needto show that

X \⋃α∈I

Aα ⊂⋂α∈I

(X \ Aα) (8)

and that⋂α∈I

(X \ Aα) ⊂ X \⋃α∈I

Aα. (9)

To show (8), let x ∈ X \⋃α∈I Aα . Then x ∈ X and x /∈ ⋃α∈I Aα . So x /∈ Aα

for all α ∈ I by the definition of union. Hence, for each α ∈ I , x ∈ X \ Aα bythe definition of complement. Thus, x ∈ ⋂α∈I (X \ Aα).

To show (9), let x ∈ ⋂α∈I (X \ Aα). Then x ∈ X \ Aα for each α ∈ I .

So x ∈ X and x /∈ Aα for each α ∈ I. So x /∈ ⋃α∈I Aα . Therefore, x ∈

X \⋃α∈I Aα .

Cartesian Product

Definition 1.6 Let A and B be sets. The Cartesian product of A andB,denoted by A × B, is the set of all ordered pairs (a, b), where a is inA and b is in B. Thus

A × B = {(a, b) : a ∈ A and b ∈ B}.

For example, R × R is the Cartesian plane. For real numbers a and b,the notation (a, b) has two different meanings: it may mean the ordered pairor it may mean the open interval {x ∈ R : a < x < b}. The context shoulddetermine which meaning is appropriate.


A × (B ∩ C) = (A × B) ∩ (A × C).

Proof To show that A × (B ∩ C) ⊂ (A × B) ∩ (A × C), let (x, y) ∈A × (B ∩ C). Then x ∈ A and y ∈ B ∩ C. Thus, y ∈ B and y ∈ C. So, x ∈ A

and y ∈ B imply (x, y) ∈ A×B while x ∈ A and y ∈ C imply (x, y) ∈ A×C.Therefore, (x, y) ∈ (A × B) ∩ (A × C).

To show the other containment, reverse the above steps.


Exercises1. Finish the proof of Proposition 1.6.

2. Finish the proof of Proposition 1.7.

3. Finish the proof of Proposition 1.8.

4. Let A and B be sets. The symmetric difference of A and B is (A ∪ B) \(A ∩ B). Show that (A ∪ B) \ (A ∩ B) = (A \ B) ∪ (B \ A).

5. Show that if A ⊂ B, then A = B \ (B \ A).

6. Show that if A ⊂ B, then A ∪ (B \ A) = B.

7. Show, by example, that for sets A, B, and C, A ∩ B = A ∩ C does notimply B = C.

8. Let An = (n, ∞) for each n ∈ N. Find⋃∞

n=1 An and⋂∞

n=1 An.

9. Let An = [0, 1/n] for each n ∈ N. Find⋃∞

n=1 An and⋂∞

n=1 An.

10. Let X be a set and let Aα be a set for each α in a nonempty index set I .Prove the distributive properties:

X ∩(⋃

α∈I

Aα

)=⋃α∈I

(X ∩ Aα)

and

X ∪(⋂

α∈I

Aα

)=⋂α∈I

(X ∪ Aα).

11. Let A, B, and C be sets. Prove that

A × (B ∪ C) = (A × B) ∪ (A × C).

1.3 Functions

The concept of a function is central to mathematics. In this section we considerbasic results about functions that we will need throughout the text.

Basic Definitions

Definition 1.7 Let X and Y be sets. A function (or map) from X intoY is a rule f that assigns to each element x in the set X a unique elementf (x) in the set Y . The set X is called the domain of the function. Theset {f (x) : x ∈ X} is called the range of the function.

Definition 1.7 is somewhat vague in the sense that the term rule is neverdefined. Because of this, an alternative definition is desirable. This form of thedefinition identifies a function and its graph.

Section 1.3 Functions 11

Definition 1.8 A function from a set X into a set Y is a subset, denotedby f, of the Cartesian product X×Y such that if (x, y1) ∈ f and (x, y2) ∈f, then y1 = y2.

Notationally, we write f : X → Y to denote that f is a function fromX into Y , and we often denote the ordered pair (x, y) ∈ f by y = f (x).

Two functions f and g are equal, denoted by f = g, provided that theyhave the same domain and that for each domain element x, f (x) = g(x).

Definition 1.9 Let f be a function from X into Y . Let S ⊂ X. Thedirect image of S, denoted by f (S), is defined as

f (S) = {f (s) : s ∈ S}.Let T ⊂ Y . The inverse image of T , denoted by f −1(T ), is defined as

f −1(T ) = {x ∈ X : f (x) ∈ T }.

For the following examples the reader should graph the functions to helpverify the claims made about the direct and inverse images.

Example 1.5 Let f : R → R be defined by f (x) = x2.

1. Let S be the set of integers. Then f (S) = {0, 1, 4, 9, 16, . . .}.2. Let T = {64}. Then f −1(T ) = {±8}.3. Let T = {y : −3 < y < 4}. Then f −1(T ) = {x : −2 < x < 2}. Observe

that for an element in T less than 0, there are no real numbers that are mappedto that element. In other words, f −1((−3, 0)) = ∅.

Example 1.6 Let f : R → R be defined by f (x) = sin x.

1. Let S = {x : 0 ≤ x ≤ π/6}. Then f (S) = {y : 0 ≤ y ≤ 12 } = [0, 1

2 ].2. Let T = {1}. Then f −1(T ) = {x : f (x) = 1} = {π/2 + 2kπ : k is an

integer}.3. Let T = {y : 0 ≤ y ≤ 1}. Then f −1(T ) = ⋃∞

n=−∞[2nπ, (2n + 1)π ].4. Let T = {y : π/2 < y < π}. Then f −1(T ) = ∅.

Proposition 1.10 Let f : X → Y . Let S and T be subsets of Y . Thenf −1(S ∪ T ) = f −1(S) ∪ f −1(T ).

Proof We will show that f −1(S ∪ T ) ⊂ f −1(S) ∪ f −1(T ) and f −1(S) ∪f −1(T ) ⊂ f −1(S ∪ T ). To begin, let x ∈ f −1(S ∪ T ). Then, by definition off −1(S ∪ T ), f (x) ∈ S ∪ T . So f (x) ∈ S or f (x) ∈ T . If f (x) ∈ S, thenx ∈ f −1(S). If f (x) ∈ T , then x ∈ f −1(T ). So x ∈ f −1(S) ∪ f −1(T ). Thus,f −1(S ∪ T ) ⊂ f −1(S) ∪ f −1(T ).

To prove the reverse inclusion, let x ∈ f −1(S) ∪ f −1(T ). Then eitherx ∈ f −1(S) or x ∈ f −1(T ). If x ∈ f −1(S), then f (x) ∈ S. If x ∈ f −1(T ),


then f (x) ∈ T . Therefore, f (x) ∈ S ∪ T and so x ∈ f −1(S ∪ T ). Hence,f −1(S) ∪ f −1(T ) ⊂ f −1(S ∪ T ).

Definition 1.10 A function f from X into Y is a one-to-one function (ora 1-1 function) if for any pair of distinct points x1 and x2 in X, f (x1) andf (x2) are distinct points in Y [that is, if x1 �= x2 in X, then f (x1) �= f (x2)

in Y ]. Equivalently, using the contrapositive, f is one-to-one if and onlyif for each x1 and x2 in X, if f (x1) = f (x2), then x1 = x2.

Another way to classify a function between two sets X and Y is to look athow much of Y is taken up by the image of X.

Definition 1.11 Let f be a function from X into Y . If f (X) = Y , thenf is onto Y . Equivalently, a function f from X into Y is onto Y if foreach y ∈ Y, there is an x ∈ X with f (x) = y.

A function from X into Y is a bijection of X onto Y if it is bothone-to-one and onto Y .

Example 1.7 The functionf in Example 1.5 is not one-to-one sincef (−8) =f (8). Since no negative number is in the range of f , f fails to be onto R.

Example 1.8 Define g : R → R by

g(x) ={x − 1 if x ≥ 0x + 1 if x < 0.

This function is onto R but fails to be one-to-one since g(−1) = g(1).

Example 1.9 Define h : R → R by h(x) = 2x + 7. Then

1. h is a one-to-one function. Suppose h(x1) = h(x2). Then 2x1 +7 = 2x2 +7.This implies that x1 = x2.

2. h is onto R. Suppose y ∈ R. Then h((y − 7)/2) = 2[(y − 7)/2] + 7 = y.

Example 1.10 Define f : R → R by f (x) = ex. We leave it to the readerto verify that f is one-to-one but not onto R . [Hint: Graph the function.]

Examples 1.7 to 1.10 show that all combinations of one-to-one and ontoare possible. In a sense these ideas are independent of one another. Someconsequences of these properties appear in the next proposition and in theexercises.

Proposition 1.11 Let f be a function from X into Y .

1. For any subsets A and B of X, f (A ∩ B) ⊂ f (A) ∩ f (B).

2. If f is a one-to-one function, then f (A ∩ B) = f (A) ∩ f (B).

Section 1.3 Functions 13

Proof1. Recall that f (A∩B) = {f (x) : x ∈ A∩B}. Let y ∈ f (A∩B). Then there

is some x ∈ A ∩ B with f (x) = y. Since x ∈ A ∩ B, x ∈ A and x ∈ B. Sof (x) ∈ f (A) and f (x) ∈ f (B). Thus y = f (x) ∈ f (A) ∩ f (B). Hencef (A ∩ B) ⊂ f (A) ∩ f (B).

2. We must show that if f is a one-to-one function, then f (A) ∩ f (B) ⊂f (A ∩ B). If this holds, then combining this with the first part yields thedesired result.

Lety ∈ f (A) ∩ f (B). Theny ∈ f (A) and y ∈ f (B). Sincey ∈ f (A),there is an a ∈ A with f (a) = y. Similarly, there is a b ∈ B with f (b) = y.

Thus f (a) = f (b). Since f is a one-to-one function, a = b. That is,a ∈ A ∩ B. Therefore, y = f (a) ∈ f (A ∩ B) and so f (A) ∩ f (B) ⊂f (A ∩ B).

In Example 1.5 we defined a function f : R → R by f (x) = x2. Ifwe restrict our attention to the subset of R consisting of the nonnegative realnumbers, we obtain a function that is one-to-one. This motivates the followingdefinition.

Definition 1.12 Let f be a function from X into Y . Let A ⊂ X. Therestriction of f to A, denoted by f |A, is defined by f |A(x) = f (x) forall x in A. In a similar vein, let g be a function from A into Y . A functionf : X → Y that satisfies f |A = g is called an extension of g to X.

OperationsWe assume that the reader is familiar with the basic algebraic operations onfunctions (the sum, difference, product, and quotient of two functions). Animportant operation on functions is that of composition.

Definition 1.13 Let f be a function from X into Y. Let g be a functionfrom Y into Z. The composition of f and g, denoted by g◦f , is a functionfrom X into Z defined by (g ◦ f )(x) = g(f (x)) for all x in X.

This definition can be extended to any finite number of functions.

Proposition 1.12 Let f be a function from X into Y and let g be a functionfrom Y into Z.

1. If both f and g are one-to-one, then so is g ◦ f.

2. If both f and g are onto functions, then so is g ◦ f.

3. If both f and g are bijections, then so is g ◦ f.

Proof See Exercise 6.


Example 1.11 We construct a bijection from R onto the open interval (0, 1).Define

f : R → (0, ∞) by f (x) = ex

g : (0, ∞) → (1, ∞) by g(x) = x + 1h : (1, ∞) → (0, 1) by h(x) = 1/x.

The reader should draw the graphs of these functions to verify that each functionis one-to-one and onto the appropriate set. By Proposition 1.12, the compositionh ◦ g ◦ f is a one-to-one map of R onto (0, 1).

We conclude this section by showing that a bijection has a natural functionassociated with it and that this function in some sense reverses what the originalfunction does.

Proposition 1.13 Let f be a function from X into Y. Then f is a bijectionfrom X onto Y if and only if there is a function g from Y into X such that(g ◦ f )(x) = x for all x in X and (f ◦ g)(y) = y for all y in Y.

Proof First assume that f is a bijection from X onto Y . We show how toconstruct a function with the stated properties.

For y in Y we define g(y) = x if f (x) = y. Since f is onto Y , for any y inY there is an element x in X with f (x) = y. This shows that g is defined for allelements of the set Y . To show that g is well-defined, let y be an element of Y

and assume that g(y) = x1 and g(y) = x2. Then, by definition of g, f (x1) =y = f (x2). Since f is a one-to-one function, we have x1 = x2 and so g is a well-defined function from Y into X. The properties of the composites follow at once.

Next, assume that such a function g exists. We must show that f is abijection. Suppose that f (x1) = f (x2). Since both f (x1) and f (x2) arein Y, we can apply g to them and obtain g(f (x1)) = g(f (x2)). Thus, x1 =g(f (x1)) = g(f (x2)) = x2, and sof is one-to-one. To show thatf is ontoY, lety be in Y . Then g(y) is in X. So, applying f, we get f (g(y)) = (f ◦g)(y) = y

by assumption. Hence, we have found an element x in X—namely, x = g(y)—with f (x) = y. This shows that f is onto Y .

The function g that was constructed in Proposition 1.13 is called the inverseof f , and we write f −1 = g. Restating part of Proposition 1.13 (if f is abijection), (f −1 ◦f )(x) = x for all x in X and (f ◦f −1)(y) = y for all y in Y .

Example 1.12 The function f : R → R defined by f (x) = x2 has no in-verse. (Why?) However, if we restrict the domain of f to the set of nonnegativereal numbers [0, ∞), then f |[0,∞) is a bijection onto [0, ∞). By Proposition1.13, f |[0,∞) has an inverse. It should come as no surprise that the inverse off |[0,∞) is the square root function.

ExercisesIn the exercises below, X, Y , and Z are sets.

1. Let f be a function from X into Y . Let A and B be subsets of X. Provethat f (A ∪ B) = f (A) ∪ f (B).

Section 1.4 Mathematical Induction 15

2. Define a function f from R into R by

f (x) ={−1 if x < 0

x if x ≥ 0.

Find each of the following.

(a) f ([−1, 1]) (d) f −1({−2})(b) f −1([−1, 1]) (e) f −1(f ((−∞, 0)))

(c) f −1({−1}) (f) f (f −1((−∞, 0)))

3. Let f be a function from X into Y. Let S and T be subsets of Y . Show thatf −1(S ∩ T ) = f −1(S) ∩ f −1(T ).

4. Let f be a one-to-one function from X into Y and let A ⊂ X. Show thatf −1(f (A)) = A. Give an example showing that equality need not hold iff is not one-to-one.

5. Let f be a function from X onto Y and let B ⊂ Y . Show that f (f −1(B)) =B. Give an example showing that equality need not hold if f is not onto Y .

6. Prove Proposition 1.12.

7. Let f be a function from X into Y. Let g be a function from Y into Z.Assume that g ◦ f is one-to-one. Prove that f is one-to-one. Give anexample showing that g need not be one-to-one.

8. Let f be a function from X into Y. Let g be a function from Y into Z.Assume that g ◦ f is onto Z. Prove that g is onto Z. Give an exampleshowing that f need not be onto Y.

9. Find a bijection from (0, ∞) onto (0, 1).

10. Let a and b be real numbers with a < b. Find a bijection from (0, 1) onto(a, b). Combine this result with that of Exercise 9 to produce a bijectionfrom (0, ∞) onto any open interval (a, b).

11. Let f be a bijection from X onto Y . Prove that f −1 is a bijection from Y

onto X.

12. Let f be a function from X into Y. Let g be a function from Y into Z. LetB be a subset of Z. Show that (g ◦ f )−1(B) = f −1(g−1(B)).

1.4 Mathematical Induction

Mathematical induction will be used often throughout the text. For instance,in Chapter 3 we will use it to construct sequences inductively. The examplesfollowing the proof of Theorem 1.3 illustrate the method of mathematical induc-tion. Recall from Section 1.1 that statements are either true or false, but not both.

Mathematical Induction. Let p(n) be a statement for each n in N. Assumethat

1. p(1) is true

and

2. for each k ≥ 1, if p(k) is true, then p(k + 1) is true.

Then p(n) is true for all n in N.


In assumption 2, “ p(k) is true” is called the induction hypothesis. Intu-itively, p(1) true implies that p(2) is true by assumption 2; p(2) true impliesthat p(3) is true by assumption 2; and so on.

Before illustrating the method of mathematical induction, we examine therelationship between mathematical induction and the following concept, whichwill be assumed as an axiom in Section 2.1.

N is Well-Ordered. Every nonempty subset of N has a least element. Thatis, if A ⊂ N and A �= ∅, then there is an a0 in A such that a0 ≤ a for all a in A.

Theorem 1.3 N is well-ordered if and only if mathematical induction is true.

Proof First assume that N is well-ordered. For each positive integer n, letp(n) be a statement satisfying assumptions 1 and 2 of mathematical induction.We want to show that p(n) is true for all positive integers n. Suppose this isfalse. Since N is well-ordered, let n0 be the least positive integer such that p(n0)

is false. By assumption 1, n0 > 1. Therefore n0 − 1 is a positive integer andp(n0−1) is true (sincen0 is the least positive integer for which the correspondingstatement is false). By assumption 2, p(n0) = p[(n0 − 1) + 1] is true, whichis a contradiction. Therefore, p(n) is true for every positive integer n.

Next, suppose mathematical induction holds. We wish to show that Nis well-ordered. Let A be a nonempty subset of N. Suppose A has no leastelement. For each n in N, let p(n) be the statement

A ∩ {1, 2, . . . , n} = ∅.

Suppose p(1) is false. Then A ∩ {1} �= ∅ and A has a least element—namely,1—which is a contradiction to A having no least element. Therefore, p(1) istrue. Let k ≥ 1 and assume that p(k) is true. Thus, A ∩ {1, 2, . . . , k} = ∅.

Suppose that p(k + 1) is false. Then A∩ {1, 2, . . . , k, k + 1} �= ∅. Thus, A hasa least element—namely, k + 1—which again is a contradiction to A having noleast element. Therefore, p(k + 1) is true.

By mathematical induction, p(n) is true for all n in N. Therefore A = ∅,which is a contradiction to A being nonempty. Therefore, A has a least ele-ment.

Example 1.13 For each n in N, 12 + 22 + · · · + n2 = 16n(n + 1)(2n + 1).

For each n in N, let p(n) be the statement

12 + 22 + · · · + n2 = 1

6n(n + 1)(2n + 1).

p(1) is true since 12 = 16 (1)(2)(3). Let k ≥ 1 and assume that p(k) is true. That

is, assume 12+22+· · ·+k2 = 16k(k+1)(2k+1).We need to show thatp(k+1) is

true. We need to show that 12+22+· · ·+k2+(k+1)2 = 16 (k+1)(k+2)(2k+3).

By the induction hypothesis,

12 + 22 + · · · + k2 + (k + 1)2 = 1

6k(k + 1)(2k + 1) + (k + 1)2

= 1

6(k + 1)[k(2k + 1) + 6(k + 1)]

= 1

6(k + 1)(2k2 + 7k + 6)

= 1

6(k + 1)(k + 2)(2k + 3).

Section 1.4 Mathematical Induction 17

Thus, p(k + 1) is true. By mathematical induction, p(n) is true for alln in N.

Example 1.14 For each n in N, 2n ≥ n + 1.For each n in N, p(n) is the statement: 2n ≥ n + 1. p(1) is true since

21 ≥ 2 = 1 + 1. Let k ≥ 1 and assume that p(k) is true. That is, assume2k ≥ k + 1. We want to show that p(k + 1) is true. We need to show that2k+1 ≥ (k + 1) + 1 = k + 2. Observe that

2k+1 = 2k · 2

≥ (k + 1) · 2 (by the induction hypothesis)

= 2k + 2

≥ k + 2 (since 2k ≥ k).

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

Example 1.15 For each n in N, 9 divides n3 + (n + 1)3 + (n + 2)3 (where“divides” means with 0 remainder).

For each n in N, let p(n) be the statement: 9 divides n3+(n+1)3+(n+2)3.

Since 13 + 23 + 33 = 36, p(1) is true. Let k ≥ 1 and assume that p(k) is true.That is, assume that 9 divides k3 + (k + 1)3 + (k + 2)3. We must show that 9divides (k + 1)3 + (k + 2)3 + (k + 3)3. Observe that

(k + 1)3 + (k + 2)3 + (k + 3)3 = (k + 1)3 + (k + 2)3 + k3 + 9k2 + 27k + 27

= k3 + (k + 1)3 + (k + 2)3 + 9(k2 + 3k + 3).

By the induction hypothesis, 9 divides k3 + (k + 1)3 + (k + 2)3. Since 9(k2 +3k + 3) is a multiple of 9, 9 divides it. Thus p(k + 1) is true and so p(n) is truefor all n in N.

Remark Sometimes mathematical induction is stated as follows. Let A be asubset of N satisfying

1. 1 ∈ A

and

2. if k ≥ 1 and k ∈ A, then k + 1 ∈ A.

Then A = N. That this is equivalent to the previous version of mathematicalinduction can be seen be letting A = {n ∈ N : p(n) is true}.Remark In mathematical induction, one need not start with 1. Let n0 be aninteger and suppose that p(n) is a statement for each n ≥ n0. Assume that (1)p(n0) is true and (2) for each k ≥ n0, if p(k) is true, then p(k + 1) is true.Then p(n) is true for all n ≥ n0. To see this, for each n in N, let q(n) be thestatement: p(n0 + n − 1). Note that q(1) = p(n0). A little thought shows thatassumptions 1 and 2 imply by mathematical induction that q(n) is true for alln in N. Equivalently, p(n) is true for all n ≥ n0.

Remark The following represents a misuse of mathematical induction. Findthe error. For which k does the argument fail?

For each n in N, let p(n) be the statement: any set of n horses are all of thesame color. p(1) is true since we have only one horse. Let k ≥ 1 and assumethat p(k) is true. That is, assume that any set of k horses are all of the samecolor. We want to show that p(k + 1) is true. Let X = {x1, x2, . . . , xk+1} be


a set of k + 1 horses. We want to show that all k + 1 horses are of the samecolor. Since {x1, x2, . . . , xk} is a set of k horses, by the induction hypothesis,these are all of the same color. Since {x2, x3, . . . , xk+1} is a set of k horses, bythe induction hypothesis, these are all of the same color. Thus, all k + 1 horsesin X are of the same color. Therefore, p(n) is true for all n in N.

Exercises1. Prove that for each n in N, 1 + 2 + · · · + n = n(n + 1)/2.

2. Prove that for each n in N, 13 + 23 + · · · + n3 = [n(n + 1)/2]2.

3. Prove that for each n in N, 1 + 3 + 5 + · · · + (2n − 1) = n2.

4. Prove that for each n ≥ 4, 2n < n!.5. Prove that for each n in N, 4 divides 7n − 3n.

6. Prove that for each n in N, 5 divides n5 − n.

7. Prove that for each n in N,n

�j=1

(−1)j+1j 2 = (−1)n+1n

�j=1

j.

8. Prove that if x1, x2, . . . , xn are n real numbers in the closed interval [a, b],then x1+···+xn

n is in the closed interval [a, b].9. Prove that a set with n elements has 2n subsets for each n ∈ N ∪ {0}.

10. Prove Bernoulli’s inequality: If x > −1, then (1 + x)n ≥ 1 + nx for eachn in N.

11. Prove DeMoivre’s Theorem: for t a real number,

(cos t + i sin t)n = cos nt + i sin nt

for each n in N, where i = √−1.

19

2 The Structure of

By the end of this chapter, the reader should understand the differencebetween the real numbers and the rational numbers. In Section 2.1, wenote that both are fields with an order relation. However, in Sections 2.2and 2.3, we show that the real numbers are “complete” whereas therational numbers are not complete. In Section 2.4, we show that therational numbers are “countable” but that the real numbers are“uncountable.”

2.1 Algebraic and Order Properties of R

Throughout the text we use the following notation.

R is the set of real numbers.

N = {1, 2, 3, . . .} is the set of positive integers (or natural numbers).

Z = {. . . , −2, −1, 0, 1, 2, . . .} is the set of integers.

Q = {m/n : m, n ∈ Z, n �= 0} = {m/n : m ∈ Z , n ∈ N} is the set ofrational numbers.

R \ Q, the complement of Q in R, is the set of irrational numbers.

In this text we are not going to construct the set of real numbers. Rather,we assume that the reader is familiar with many of the properties of the realnumbers discussed in this section.

Field Axioms for R

To each pair of real numbers a and b there correspond unique real numbersa + b and a · b that satisfy:

Axiom 2.1a + b = b + a

a · b = b · a (commutative laws)

Axiom 2.2 For all c in R,a + (b + c) = (a + b) + c

a · (b · c) = (a · b) · c (associative laws)

Axiom 2.3 For all c in R,

a · (b + c) = a · b + a · c (distributive law)

Axiom 2.4 There exist distinct real numbers 0 and 1 such that for all a in R,a + 0 = a

a · 1 = a (identity elements)

20 Chapter 2 The Structure of R

Axiom 2.5 For each a in R, there is an element −a in R such that

a + (−a) = 0

and for each b in R, b �= 0, there is an element b−1 = 1/b in R such that

b · 1

b= 1. (inverse elements)

Note that (a, b) → a +b and (a, b) → a ·b actually define functions fromR × R → R. We usually write ab instead of a · b. From these axioms we canderive the usual laws of arithmetic; for example, a · 0 = 0, −(−a) = a, and soon.

The axioms above make (R, +, ·) into a field. However, (Q, +, ·) is alsoa field. This amounts to showing that the sum and the product of two fractionsare again fractions.

Order Axioms for R

On R there is an order relation, denoted by <, that satisfies:

Axiom 2.6 For all a and b in R, exactly one of the following holds:

a = b, a < b, b < a (trichotomy).

Axiom 2.7 For all a, b, and c in R, if a < b, then a + c < b + c.

Axiom 2.8 For all a, b, and c in R, if a < b and 0 < c, then ac < bc.

Axiom 2.9 For alla, b, and c in R, ifa < b andb < c, thena < c (transitivity).

We write b > a if a < b and we define a ≤ b (or b ≥ a) if either a < b ora = b. From these axioms we can derive the usual properties of inequalities; forexample, if a < b and c < 0, then ac > bc. The reader will find it instructiveto show from the axioms above that 0 < 1 (see Exercises 4 and 5).

We will assume two more axioms on the set of real numbers. The secondone will be introduced in the next section. The first one, which was defined inSection 1.4, we now state:

Axiom 2.10 The positive integers are well-ordered.

Decimal RepresentationsThe following is essentially given in Apostol, pp. 2–3. Our purpose in dis-cussing decimals is in anticipation of Theorem 2.5.

Every real number x has a decimal representation of the form

x = n.a1a2a3 · · ·where n is an integer and each ai is one of the digits from 0 to 9. The notation.a1a2a3 · · · is an abbreviation for the infinite series

a1

10+ a2

100+ a3

1000+ · · · .

For example

0.499999 . . . = 4

10+ 9

100+ 9

1000+ · · ·

Section 2.1 Algebraic and Order Properties of R 21

= 4

10+ 9

100

[1 + 1

10+(

1

10

)2

+ · · ·]

= 4

10+ 9

100

(1

1 − 110

)(the sum of a geometric series)

= 4

10+ 9

100

(10

9

)

= 5

10= 0.5.

Thus, 12 has two decimal expansions. More generally, if a number has a decimal

expansion ending in zeros, it also has a decimal expansion ending in nines. Forexample, 1

4 = 0.2499999 · · · = 0.250000 · · · . Except in situations such as this,decimal expansions are unique.

A real number is rational if and only if its decimal expansion is repeating.For example, 1

3 = 0.33333 · · · , 18 = 0.125000 · · · , and 1 = 0.9999999 · · · .

Given the positive rational number m/n, in lowest terms with m < n, theprocess of long division produces the decimal expansion of m/n. Since at eachstep of the division a remainder that is an integer in the set {0, 1, 2, . . . , n − 1}occurs, after at most n steps, the quotient will repeat. Conversely, as with0.49999 · · · above, a decimal that repeats forms a geometric series of the form

a + b(1 + r + r2 + r3 + · · ·) = a + b

(1

1 − r

)where a, b, and r are rational, with r being a power of 1

10 . Since Q is a field,a + b[1/(1 − r)] is rational.

Absolute Value

Definition 2.1 For x in R, the absolute value of x, denoted by |x|, isdefined by

|x| ={

x if x ≥ 0−x if x < 0.

Note that |x| ≥ 0 for all x in R. Geometrically, |x| is the distance fromx to 0.

Proposition 2.1 Let a, b, and c be in R. Then

1. |a| = 0 if and only if a = 0;

2. | − a| = |a|;3. |ab| = |a||b|;4. if c ≥ 0, then |a| ≤ c if and only if −c ≤ a ≤ c.


Proof The proofs of parts 1 and 2 are left as exercises.Proof of part 3. If either a or b is 0, then both |ab| and |a||b| are 0. If

a > 0 and b > 0, then ab > 0 and so |ab| = ab = |a||b|. If a > 0 and b < 0,then ab < 0 and so |ab| = −(ab) = a(−b) = |a||b|. If a < 0 and b < 0, thenab > 0 and so |ab| = ab = (−a)(−b) = |a| |b| .

Proof of part 4. Suppose that −c ≤ a ≤ c. If a ≥ 0, then |a| = a ≤ c. Ifa < 0, then |a| = −a ≤ c since −c ≤ a. In either case, |a| ≤ c. Conversely,assume that |a| ≤ c. If a ≥ 0, then −c ≤ 0 ≤ a = |a| ≤ c. If a < 0, then−a = |a| ≤ c implies that a ≥ −c and so −c ≤ a < 0 ≤ c.

Proposition 2.2 (triangle inequality) For a and b in R,

|a + b| ≤ |a| + |b|.

Proof One could make a case-by-case argument as in Proposition 2.1. How-ever, we prefer the following:

0 ≤ (a + b)2

= a2 + 2ab + b2

= |a|2 + 2ab + |b|2≤ |a|2 + 2|ab| + |b|2= |a|2 + 2|a||b| + |b|2= (|a| + |b|)2.

Taking square roots (see Exercises 8 and 9),

|a + b| ≤ |a| + |b|.

By mathematical induction, the triangle inequality can be extended to anyfinite number of elements in R. That is, for any a1, a2, . . . , an in R,

|a1 + a2 + · · · + an| ≤ |a1| + |a2| + · · · + |an|.The following simple result, especially the second part, will be used often

in Chapters 3 and 4.

Corollary 2.1 For a and b in R,

1. |a − b| ≤ |a| + |b|and

2. | |a| − |b| | ≤ | a − b| .

Proof For part 1, |a − b| = |a + (−b)| ≤ |a| + |− b| = |a| + |b|. For part2, we also use the triangle inequality:

|a| = |(a − b) + b| ≤ |a − b| + |b|.Therefore, |a| − |b| ≤ |a − b|. Reversing the roles of a and b, we get

|b| − |a| ≤ |b − a| = |a − b|.Thus, | |a| − |b| | ≤ |a − b|.

Section 2.2 The Completeness Axiom 23

Exercises1. Let x be irrational. Show that if r is rational, then x + r is irrational. Also

show that if r is a nonzero rational, then rx is irrational.

2. Show, by example, that if x and y are irrational, then x + y and xy may berational.

3. Show that√

2 +√3 is irrational. [Hint: Use the fact that

√6 is irrational.]

4. From the order axioms for R, show that the set of positive real numbers,{x ∈ R : x > 0}, is closed under addition and multiplication.

5. From the order axioms for R, show that 0 < 1. [Hint: From the field ax-ioms, 0 �= 1. By the trichotomy property, either 0 < 1 or 1 < 0. Assuming1 < 0, get 0 < −1. Now use Exercise 4.]

6. Write 0.33474747 · · · as a fraction.

7. Prove parts 1 and 2 of Proposition 2.1.

8. For a in R, show that |a| = √a2. (Note that for b ≥ 0,

√b denotes the

nonnegative square root of b.)

9. Let a ≥ 0 and b ≥ 0. Show that a ≤ b if and only if a2 ≤ b2.

10. Show that equality holds in the triangle inequality if and only if ab ≥ 0.

11. For x in R and ε > 0, let (x − ε, x + ε) be the open interval centered at x

of radius ε. That is,(x − ε, x + ε) = {y ∈ R : x − ε < y < x + ε}

= {y ∈ R : |y − x| < ε}.Use the triangle inequality to show that if a and b are in R with a �= b,

then there exist open intervals U centered at a and V centered at b, bothof radius ε = 1

2 |a − b| , with U ∩ V = ∅.

2.2 The Completeness Axiom

Boundedness

Definition 2.2 The extended real number system, denoted by R#, con-sists of the real number system together with two distinct symbols, ∞and −∞, neither of which is a real number. That is,

R# = R ∪ {∞} ∪ {−∞}.We extend the usual ordering on R to R# by

−∞ < ∞, −∞ < x < ∞ for all x in R,

and u ≤ v if either u < v or u = v for all u and v in R#.


We also use the symbol +∞ for ∞ where convenient.

Definition 2.3 Let S be a subset of R#. Let u and v be in R#.

1. u is an upper bound of S if s ≤ u for all s in S.

2. v is a lower bound of S if v ≤ s for all s in S.

3. S is bounded above if there is a real number that is an upper boundof S.

4. S is bounded below if there is a real number that is a lower bound of S.

5. S is bounded if S is both bounded above and bounded below.

Obviously, ∞ is an upper bound and −∞ is a lower bound of every subsetof R#. However, the positive real numbers {x ∈ R : x > 0} are bounded belowbut not bounded above; the negative real numbers {x ∈ R : x < 0} are boundedabove but not bounded below; and the closed interval [0, 1] = {x ∈ R : 0 ≤x ≤ 1} is bounded.

Definition 2.4 Let S be a subset of R# and let α and β be elementsof R#.

1. α is the least upper bound of S or the supremum of S

(a) if α is an upper bound of S

and

(b) whenever γ is an upper bound of S, α ≤ γ.

Notation: α = lub S = sup S.

2. β is the greatest lower bound of S or the infimum of S

(a) if β is a lower bound of S

and

(b) whenever γ is a lower bound of S, γ ≤ β.

Notation: β = glb S = inf S.

As a simple exercise, the reader should show that sup S and inf S are unique.

Example 2.1 The following illustrate the basic ideas.

1. sup R = +∞ and inf R = −∞.

2. sup[0, 1] = sup(0, 1) = 1 and inf[0, 1] = inf(0, 1) = 0, where (0, 1) ={x ∈ R : 0 < x < 1}.

3. sup{x ∈ R : x < π} = π and inf{x ∈ R : x < π} = −∞.

4. Pathologically, sup ∅ = −∞ and inf ∅ = +∞.

Observe that the supremum and infimum of a set may or may not be inthat set.

Section 2.2 The Completeness Axiom 25

Definition 2.5 Let S be a subset of R# and let s0 and s1 be elementsof S.

1. s0 is the smallest (least, minimum) element of S if s0 ≤ x for all x

in S.

2. s1 is the greatest (largest, maximum) element of S if x ≤ s1 for all x

in S.

Notation: min S = s0 and max S = s1.

Proposition 2.3 Let S be a subset of R#.

1. If S has a smallest element, then this smallest element is the infimum of S.

2. If S has a greatest element, then this greatest element is the supremum of S.

Proof We prove part 1, leaving part 2 as an exercise.Let s0 be the smallest element of S. Then s0 is a lower bound of S since

s0 ≤ x for all x in S. Suppose γ is a lower bound of S. Then γ ≤ x for all x

in S. In particular, γ ≤ s0. Thus, s0 = inf S.

The Completeness AxiomThe next axiom is our final assumption about the set of real numbers. We willsee in the next section that this axiom distinguishes the real numbers from therational numbers. Recall that bounded above (or below) means bounded above(or below) by a real number.

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

Proposition 2.4 Every nonempty subset of R that is bounded below has aninf imum in R.

Proof Let S be a nonempty subset of R that is bounded below. Let

A = {x ∈ R : x is a lower bound of S}.Then A is nonempty and A is bounded above by each point in S. By theCompleteness Axiom for R, α = sup A is a real number. We will show thatα = inf S. Let s be in S. Then s is an upper bound of A. Since α is the leastupper bound of A, α ≤ s. Thus, α is a lower bound of S.

To show that α is the greatest lower bound of S, let γ be a real lower boundof S. (If γ = −∞, then clearly γ ≤ α.) We need to show that γ ≤ α. Since γ

is a lower bound of S, γ is an element of A. Since α is an upper bound of A,γ ≤ α.

Let S be a subset of R#. If S is not bounded above (that is, for each realnumber x > 0, there is an s in S such that x < s), then sup S = +∞. Also,if S is not bounded below, then inf S = −∞. Thus, every subset of R# hasa supremum and an infimum in R#, whereas a subset of R need not have asupremum or an infimum in R. This is why we chose to work in R#.


We end this section with an example showing how to work with the supre-mum and the infimum. For a subset S of R and a point a in R, we defineaS = {as : s ∈ S}.Example 2.2 Let S be a nonempty bounded subset of R and let a > 0. Then

1. sup(aS) = a sup S

and

2. inf(aS) = a inf S.

To show part 1, let α = sup S. By the Completeness Axiom for R, α is areal number. We wish to show that aα = sup(aS). First note that for each s

in S, s ≤ α since α is an upper bound of S. So, as ≤ aα for each s in S and,therefore, aα is an upper bound of aS.

To show that aα is the least upper bound of aS, let γ be a real upper boundof aS. (If γ = ∞, then clearly γ ≥ aα.) We need to show that aα ≤ γ .For each s in S, since γ is an upper bound of aS, as ≤ γ . So, for each s inS, s ≤ γ /a and, hence, γ /a is an upper bound of S. Since the least upperbound of S is α, α ≤ γ /a. Thus, aα ≤ γ .

To show part 2, let β = inf S. By Proposition 2.4, β is a real number. Wewish to show that aβ = inf(aS). For each s in S, β ≤ s since β is a lower boundof S. So aβ ≤ as for each s in S and, therefore, aβ is a lower bound of aS.

To show that aβ is the greatest lower bound of aS, let γ be a real lowerbound of aS. We need to show that γ ≤ aβ. For each s in S, since γ is alower bound of aS, γ ≤ as. So, for each s in S, γ /a ≤ s and, hence, γ /a isa lower bound of S. Since β is the greatest lower bound of S, γ /a ≤ β. Thus,γ ≤ aβ.

Exercises1. Find the supremum and the infimum of each of the following sets.

(a) {x ∈ R : 0 < x2 < 2} (c) {x ∈ R : 0 < x and x2 > 2}(b) {x ∈ R : x2 < 2} (d) {x ∈ R : x2 > 2}

2. Prove part 2 of Proposition 2.3.

3. Let A and B be nonempty subsets of R with A ⊂ B. Show that

inf B ≤ inf A ≤ sup A ≤ sup B.

4. Let S be a nonempty bounded subset of R and let b < 0. Show thatinf(bS) = b sup S and sup(bS) = b inf S.

5. LetS be a nonempty subset of R that is bounded above. Prove that inf(−S) =− sup S, where −S = (−1)S = {−s : s ∈ S}.

6. Let S be a subset of R and let a be an element of R. Define a +S = {a + s :s ∈ S}. Assume that S is a nonempty bounded subset of R. Show thatsup(a + S) = a + sup S and inf(a + S) = a + inf S.

Section 2.3 The Rational Numbers Are Dense in R 27

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

sup(A ∪ B) = max{sup A, sup B}and

inf(A ∪ B) = min{inf A, inf B}.8. Show that a nonempty finite subset of R contains both a maximum and a

minimum element. [Hint: Use induction.]

9. Let A and B be nonempty bounded subsets of R, let α = sup A, and letβ = sup B. Let C = {ab : a ∈ A and b ∈ B}. Show, by example, thatαβ �= sup C in general.

2.3 The Rational Numbers Are Dense in R

In this section we establish three main results: first, that the ArchimedeanPrinciple holds; second, that the rationals are “dense” in the reals; and third,that the rationals are not “complete.”

We also expand our techniques for working with the supremum and infi-mum. The next proposition states that we can approximate a real supremum orinfimum by an element of the set. That is, we can find an element of the setthat is as close as we would like to a real supremum or infimum.

The Archimedean Principle

Proposition 2.5 Let S be a nonempty bounded subset of R. Let α = sup S

and let β = inf S. Let ε > 0. Then

1. there exists an s0 in S such that α − ε < s0

and

2. there exists an s1 in S such that s1 < β + ε.

Proof We prove part 1, leaving part 2 as an exercise. Figure 2.1 illustratesthe fact that α is an upper bound of S and α is in R. Suppose s ≤ α − ε for alls in S. Then α − ε is an upper bound of S, contradicting the fact that α is theleast upper bound of S.

S

a − e a

Figure 2.1

Theorem 2.1 The set N of positive integers is unbounded above.

Proof Suppose N is bounded above. By the Completeness Axiom for R,α = sup N is a real number. By Proposition 2.5, there is an n0 in N such thatα − 1 < n0 ≤ α. Since n0 + 1 is in N and α < n0 + 1, this contradicts the factthat α is an upper bound of N.

We now state the Archimedean Principle, whose truth follows from Propo-sition 1.2.


Archimedean Principle If x > 0, then there exists a positive integer n

such that 1/n < x.

It follows from the Archimedean Principle that inf{1/n : n ∈ N} = 0.

The Density of Q in R

Definition 2.6 Let A be a subset of R. Then A is dense in R if betweenevery two real numbers there exists an element of A.

Equivalently, A is dense in R if and only if for each x and y in R with x < y,one has thatA∩(x, y) �= ∅. Here (x, y) is the open interval {z ∈ R : x < z < y}in R.

Theorem 2.2 Q is dense in R.

Proof We consider various cases.

Case 1 If x is negative and y is positive, then x < 0 < y and 0 is rational.

Case 2 0 < x. By the Archimedean Principle, there is an n in N suchthat 0 < 1/n < x, and 1/n is a rational number.

Case 3 0 < x < y. Refer to Figure 2.2. Since y − x is positive, bythe Archimedean Principle, there is an n in N with 0 < 1/n < y − x.

[Informally, if we add 1/n to itself a sufficient number of times, say m

times, then x < m/n < y, because we cannot “jump” over the interval(x, y) since 1/n < y − x.] By Exercise 2, there is an m in N such thatm − 1 ≤ nx < m. So, m/n − 1/n ≤ x < m/n. Manipulating thisexpression and recalling how 1/n was chosen, we get

x <m

n≤ x + 1

n< x + (y − x) = y,

and m/n is a rational number.

0 y − x yx

1n

Figure 2.2

Case 4 x < 0. By Case 2, there is an n in N such that 0 < 1/n < −x,and so x < −1/n < 0 and −1/n is a rational number.

Case 5 x < y < 0. By Case 3, there is a rational number q with−y < q < −x. Then −q is rational and x < −q < y.

Definition 2.7 A subset C of R is complete if every nonempty boundedsubset of C has both a supremum and an infimum in C.

For instance, [0, 1] is complete but (0, 1) is not complete. Also, R iscomplete by the Completeness Axiom for R.

Section 2.3 The Rational Numbers Are Dense in R 29

Corollary 2.2 The rational numbers are not complete.

Proof Let A = {x ∈ Q : 0 < x <√

2}. (Recall that√

2 is irrational byTheorem 1.1.) Since

√2 is an upper bound of the set A, sup A ≤ √

2. Supposesup A <

√2. Since Q is dense in R, there is a rational number q such that

sup A < q <√

2. Thus q is in A and q > sup A, which is a contradiction.So, sup A = √

2. Thus, A is a nonempty bounded subset of Q that does nothave a supremum in Q. Therefore, Q is not complete.

More on Suprema and InfimaWe end this section by expanding our techniques for working with the supremumand the infimum. We give two different arguments for the more difficult partof the next proof because both are elegant.

For subsets A and B of R, we define

A + B = {a + b : a ∈ A and b ∈ B}.Proposition 2.6 LetA andB be nonempty subsets of R that are both boundedabove. Then sup(A + B) = sup A + sup B.

Proof Let α = sup A and β = sup B. Both α and β are real numbers by theCompleteness Axiom for R. We need to show that α + β = sup(A + B). Foreach a in A and b in B, a + b ≤ α + β, and so α + β is an upper bound ofA + B.

Let γ be a real upper bound of A + B. We must show that α + β ≤ γ.

Argument 1 Fix b0 in B. Then, for each a in A, since γ is an upperbound of A + B, a + b0 ≤ γ. So, for each a in A, a ≤ γ − b0, and thusγ − b0 is an upper bound of A. Since α is the least upper bound of A,α ≤ γ − b0 or, equivalently, b0 ≤ γ − α. Since b0 is an arbitrary elementof B, b ≤ γ − α for all b in B. Hence, γ − α is an upper bound of B, andsince β is the least upper bound of B, β ≤ γ − α. Therefore, α + β ≤ γ.

Argument 2 Let ε > 0. By Proposition 2.5, there exist an a0 in A and a b0

inB such thatα−ε/2 < a0 andβ−ε/2 < b0.Thus, α+β−ε < a0+b0 ≤ γ

since γ is an upper bound of A + B. Hence, α + β < γ + ε. Since ε is anarbitrary positive number, α + β ≤ γ (see Proposition 1.5).

Definition 2.8 Let f be a function from a set X into R. Then f isbounded on X if the range of f is a bounded subset of R. For a subset D

of X, f is bounded on D if the restriction of f to D, f |D, has boundedrange in R.

Thus, for f : X → R and D a subset of X, f is bounded on D if and onlyif there exists an M > 0 such that |f (x)| ≤ M for all x in D.

For example, f (x) = x2 is bounded on (0, 1) while g(x) = 1/x is notbounded on (0, 1). Also, inf{f (x) : x ∈ (0, 1)} = 0, sup{f (x) : x ∈ (0, 1)} =1, inf{g(x) : x ∈ (0, 1)} = 1, and sup{g(x) : x ∈ (0, 1)} = ∞.


Proposition 2.7 Let f and g be bounded functions from a nonempty set X

into R. Then

inf{f (x) : x ∈ X} + inf{g(x) : x ∈ X} ≤ inf{f (x) + g(x) : x ∈ X}.Proof Let α = inf{f (x) : x ∈ X} and let β = inf{g(x) : x ∈ X}. ByProposition 2.4, both α and β are real numbers. For each x in X,

α + β ≤ f (x) + g(x).

So, α + β is a lower bound of {f (x) + g(x) : x ∈ X}. Since

inf{f (x) + g(x) : x ∈ X}is the greatest lower bound of {f (x) + g(x) : x ∈ X},

α + β ≤ inf{f (x) + g(x) : x ∈ X}.The difference between the two previous propositions is that in Proposition

2.7 we have only one variable, x in X, while in Proposition 2.6 we have twovariables, a in A and b in B. To make Proposition 2.7 analogous to Proposition2.6, we would need to consider inf{f (x) + g(y) : x ∈ X and y ∈ X}.Example 2.3 1. If we let f (x) = g(x) = 1 for all x in X, we obtain equalityin Proposition 2.7.

2. Define f : [0, 2] → R by

f (x) ={

1 if 0 ≤ x ≤ 1−1 if 1 < x ≤ 2

and define g : [0, 2] → R by

g(x) ={−1 if 0 ≤ x ≤ 1

1 if 1 < x ≤ 2.

The reader should graph these functions. Then inf{f (x) : x ∈ [0, 2]} =inf{g(x) : x ∈ [0, 2]} = −1 and inf{f (x) + g(x) : x ∈ [0, 2]} = inf{0} = 0.Thus, we obtain strict inequality in Proposition 2.7.

Exercises1. Prove part 2 of Proposition 2.5.

2. Let 0 < x. Show that there is a unique m in N such that m − 1 ≤ x < m.

[Hint: Consider {n ∈ N : x < n} and use the well-ordering of N.]

3. Find the sup and inf of

(a) {1 − 1/n : n ∈ N} (c) {n − 1/n : n ∈ N}(b) Q (d) {x ∈ Q : x2 < 2}.

4. Show that the irrational numbers are dense in R. [Hint: Use the fact that√2 is irrational.]

5. Let A and B be nonempty subsets of R that are both bounded below. Provethat

inf(A + B) = inf A + inf B.

Section 2.4 Cardinality 31

6. Let f and g be bounded functions from a nonempty set X into R. Show that

sup{f (x) + g(x) : x ∈ X} ≤ sup{f (x) : x ∈ X} + sup{g(x) : x ∈ X}.Show by examples that both equality and strict inequality can occur.

7. Let f and g be bounded functions from a nonempty set X into R.

(a) Prove that if f (x) ≤ g(x) for all x in X, then inf f (X) ≤ inf g(X) andsup f (X) ≤ sup g(X).

(b) Prove that if f (x) ≤ g(y) for all x and y in X, then sup f (X) ≤inf g(X).

(c) Give an example showing that the hypothesis of part (a) does not implythe conclusion of part (b).

2.4 Cardinality

In this section we distinguish between “finite” and “infinite” sets and then weclassify infinite sets as to whether they are “countable” or “uncountable.” Themain results of this section are Theorems 2.3, 2.4, and 2.5 and Corollaries 2.4and 2.5. Our goal is to prove that the set of rational numbers is countable andthat the set of real numbers is uncountable.

The Cardinality of a Set

Definition 2.9 Two sets A and B have the same cardinal number, de-noted by A ∼ B, if there exists a one-to-one function from A onto B (inother words, if there exists a bijection from A onto B).

If A ∼ B, then A and B are said to be equivalent sets and, intuitively, A

and B have the same number of elements. Thus, ∼ is read as “is equivalent to.”From a previous course, the reader may be familiar with the term “equivalencerelation.” Such a relation has the properties of being reflexive, symmetric, andtransitive. Observe that ∼ is an equivalence relation:

1. ∼ is reflexive since A ∼ A by the identity function;

2. ∼ is symmetric, because if A ∼ B and if f is a bijection from A onto B,then f −1 is a bijection from B onto A and so B ∼ A;

3. ∼ is transitive, because if A ∼ B and B ∼ C, with f and g being bijectionsfrom A onto B and from B onto C, respectively, then g ◦ f is a bijectionfrom A onto C, and so A ∼ C.

As examples, note that N is equivalent to the set of even positive integersby the function n → 2n and that N is equivalent to the set of odd positiveintegers by the function n → 2n − 1. Also, R ∼ (0, 1) by Example 1.11.


Definition 2.10 A set A is finite if either A is the empty set (which hascardinal number 0) or there is an n in N such that A ∼ {1, 2, . . . , n} (inwhich case A has cardinal number n or, equivalently, A has n elements).A set is infinite if it is not finite.

Observe that if A has n elements, we can write A as {x1, x2, . . . , xn}. Fromthe Pigeonhole Principle (Section 1.1, Exercise 8), which states that if there arem pigeons and n pigeonholes with m > n, then at least two pigeons must get inthe same hole, it is clear that there cannot be a bijection from a finite set ontoa proper subset of itself. From the paragraph preceding Definition 2.10, this isnot the case with infinite sets.

Proposition 2.8 Let A and B be sets.

1. If A is finite and A ∼ B, then B is finite.

2. If B is infinite and A ∼ B, then A is infinite.

3. If A is finite and B ⊂ A, then B is finite.

4. If B is infinite and B ⊂ A, then A is infinite.

Proof For part 1, first note that if A is empty, then so is B. Otherwise, A ∼{1, 2, . . . , n} for some n in N. Since ∼ is transitive, B ∼ {1, 2, . . . , n}.

For part 4, note that it is the contrapositive of part 3. The rest of the proofis left as an exercise.


1. If A is finite and there exists a function f from A onto B, then B is finite.

2. If A is infinite and there exists a one-to-one function from A into B, then B

is infinite.

Proof We first prove part 1. We can assume that B is nonempty. For each b

in B, f −1({b}) is nonempty since f maps A onto B. Choose one ab in f −1({b})for each b in B. Then the function b → ab is a bijection from B onto a subsetof A, which is finite by Proposition 2.8.

To prove part 2, note that A is equivalent to a subset of B. Again, byProposition 2.8, this subset of B, and hence B, is infinite.

In reference to Proposition 2.9, since a function is single-valued, the rangeof a function cannot be larger than the domain with respect to cardinality. Inthe proof of part 1, the reader who wonders, or even questions, why we canchoose an element from each member of a collection of sets indexed by the setB (when, at the time of doing this, all we know about B is that B is nonempty)should see the end of this section.

Proposition 2.10 N is infinite.

Proof Suppose that N is not infinite. Then, by Exercise 8 in Section 2.2, Nis bounded, which contradicts Theorem 2.1.


Sequences and Infinite Sets

Definition 2.11 A sequence in a set X is a function from N into X.

For a sequence it is customary to use a letter such as x for the function andto denote the value x(n) as xn for each n in N. Thus, we think of the sequenceas (x1, x2, x3, . . .) or as (xn)n∈N or as (xn)

∞n=1.

Example 2.4 The following are examples of sequences.

1. (n)∞n=1 = (1, 2, 3, . . .)

2.(

1n

)∞n=1

= (1, 12 , 1

3 , . . .)

3. ((−1)n)∞n=1 = (−1, 1, −1, 1, . . .)

The first two examples are sequences of distinct points, whereas the thirdis not.

Theorem 2.3 A set X is infinite if and only if X contains a sequence ofdistinct points.

Proof If X contains the sequence (xn)∞n=1 of distinct points, then N ∼ {xn :

n ∈ N} by the function n → xn. Hence, X has an infinite subset and so X isinfinite.

Let X be an infinite set. We want to construct a sequence of distinctpoints in X. Since X is nonempty, there is some element x1 in X. SinceX \ {x1} is nonempty, there is an element x2 in X \ {x1}. Suppose that distinctpoints x1, x2, x3, . . . , xk have been chosen in X. Since X \ {x1, x2, x3, . . . , xk}is nonempty, there exists an element xk+1 in X \ {x1, x2, x3, . . . , xk}. Then(xn)

∞n=1 is a sequence of distinct points in X.

In reference to the proof above, we constructed the sequence inductively.Once we acquire facility with this type of construction, we need only generatethe first two or three terms and then say “continuing by induction, we obtainthe rest.”

Corollary 2.3 A set X is infinite if and only if there exists a one-to-onefunction from X onto a proper subset of X.

Proof If X is finite, then no such bijection can exist by the Pigeonhole Prin-ciple.

Let X be infinite. By Theorem 2.3, let (xn)∞n=1 be a sequence of distinct

points in X. Define

f : X1−1→onto

X \ {x1}by

f (xn) = xn+1 for all n in N

f (x) = x otherwise.


The reader should verify that this function is one-to-one and onto the de-sired set.

The proof above provides a standard way to construct one-to-one func-tions on an infinite set. The function on the sequence of distinct points isreally a function on N since you manipulate the subscripts, and the function onthe rest of the set is the identity. There are obviously many more such func-tions. For example, we could have eliminated all the xn’s with n odd by usingf (xn) = x2n.

Countable and Uncountable SetsSo far, we have split sets into finite and infinite. Next, we divide the infinitesets into two classifications.

Definition 2.12 Let A be a set.

1. A is countably infinite (or denumerable) if A ∼ N.

2. A is countable if either A is finite or A is countably infinite.

3. A is uncountable if A is not countable.

Note that N, the set of even positive integers, and the set of odd positiveintegers are all countably infinite. Informally, Theorem 2.3 states that thecountably infinite sets are the “smallest” infinite sets. Also note that if A iscountably infinite, we can write A as a sequence of distinct points.


1. If A is countable and A ∼ B, then B is countable.

2. If B is uncountable and A ∼ B, then A is uncountable.

3. If A is countable and B ⊂ A, then B is countable.

4. If B is uncountable and B ⊂ A, then A is uncountable.

Proof Note that part 4 is the contrapositive of part 3. We prove part 3, leavingparts 1 and 2 as exercises. We have that A is countable with B ⊂ A and wewant to show that B is countable. If B is finite, then B is countable. So wecan assume that B is infinite. Hence, A is countably infinite. So we enumerateA as a sequence of distinct points (xn)

∞n=1. Recall that N is well-ordered. Let

n1 be the smallest positive integer such that xn1 is in B. Let n2 be the smallestpositive integer greater than n1 such that xn2 is in B. Continuing by induction,we get B = {xn1 , xn2 , xn3 , . . .}. Hence, N ∼ B by the function k → xnk

.


1. If A is countable and there exists a function f from A onto B, then B iscountable.

2. If A is uncountable and there exists a one-to-one function from A into B,then B is uncountable.


Proof We first prove part 1. If B is finite, then B is countable; so we canassume that B is infinite. For each b in B, choose one ab in f −1({b}). Then thefunction b → ab is a bijection from B onto a subset of A. Thus, B is equivalentto a subset of A, which is countable by Proposition 2.11.

To prove part 2, note that A is equivalent to a subset of B. Again, byProposition 2.11, this subset of B, and hence B, is uncountable.

In reference to the proof of part 1, the comment after Proposition 2.9 alsoapplies here.

Lemma 2.1 N × N is countable.

Proof Define f : N×N → N by f (n, m) = 2n3m. (Any two distinct primeswill work.) We claim that f is one-to-one and hence N × N ∼ a subset of Nand therefore N × N is countable. To see that f is one-to-one, suppose thatf (n, m) = f (r, s). Then 2n3m = 2r3s . If n > r , then 2n−r = 3s−m. Since 2divides the left side, 2 divides 3s−m and so 2 divides 3, which is a contradiction.The case n < r is handled similarly. Thus, n = r and hence m = s.

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

because if A = {x1, x2, . . . , xn}, we can define a function f from N onto A by

f (k) ={xk if k < n

xn if k ≥ n.

So, if A is countable and nonempty, there is a function from N onto A.

Theorem 2.4 The countable union of countable sets is countable. That is,if I is a countable index set and Aα is a countable set for each α in I, then⋃

α∈I Aα is countable.

Proof We can assume that I �= ∅ since⋃

α∈∅ Aα = ∅, and we can assumethat each Aα is nonempty since empty Aα’s add nothing to the union. Since I iscountable, there exists a function f from N onto I . Since each Aα is countable,for each α in I there exists a function gα from N onto Aα. Define

h : N × N →⋃α∈I

Aα

by

h(n, m) = gf (n)(m).

We claim that h maps N × N onto⋃

α∈I Aα and hence, by Lemma 2.1 andProposition 2.12,

⋃α∈I Aα is countable. Let x be in

⋃α∈I Aα . Then there

exists an α0 in I such that x is in Aα0 . Since gα0 maps N onto Aα0 , there existsan m in N such that gα0(m) = x. Since f maps N onto I , there is an n in Nwith f (n) = α0. Then, h(n, m) = gf (n)(m) = gα0(m) = x.

Corollary 2.4 The sets Z, Q, and Q × Q are countable.

Proof We write each set as the countable union of countable sets.

Z = {. . . , −3, −2, −1} ∪ {0} ∪ {1, 2, 3, . . .}.Q = ⋃∞

n=1{m/n : m ∈ Z}.Q × Q = ⋃

q∈Q{(q, p) : p ∈ Q}.


If we define Qn = {(q1, q2, . . . , qn) : qi ∈ Q for i = 1, 2, . . . , n}, we canshow that Qn is countable by induction and by writing Qn = ⋃

q∈Q(Qn−1×{q}).Theorem 2.5 The closed interval [0, 1] is uncountable.

Proof The technique used here is called the Cantor diagonalization argument.Suppose that [0, 1] is countable. Then there exists a bijection f from N onto[0, 1]. We list the elements of [0, 1] in their decimal expansion as an infinitematrix:

f (1) = .a11a12a13 · · ·f (2) = .a21a22a23 · · ·f (3) = .a31a32a33 · · ·

...

f (n) = .an1an2an3 · · ·...

where each aij is a digit from 0 to 9.We now construct an element of the interval [0, 1] that is not in the range

of f by “going down” the diagonal of the matrix. For each n in N, let

bn ={

3 if ann �= 34 if ann = 3.

Then x = .b1b2b3 · · · is in [0, 1], but for all n in N, x �= f (n) since bn �= ann.

(The choice of 3 and 4 for bn is done simply to avoid duplicate representations,which occur only with tails of nines or zeros.)

Corollary 2.5 The real numbers and the irrational numbers are uncountable.

Proof That R is uncountable follows from part 4 of Proposition 2.11. SinceR = Q ∪ (R \ Q), R \ Q is uncountable by Theorem 2.4.

We make one final comment. The following axiom is independent of theother axioms of set theory.

Axiom of Choice If I is a nonempty set and if Aα is a nonempty set for eachα in I , we may choose an xα in Aα for each α in I.

If I is countable, then we can use mathematical induction to choose each xα .The strength of the Axiom of Choice comes when I is uncountable. The reasonwe mention the Axiom of Choice is because we used the Axiom of Choicein the proofs of Propositions 2.9 and 2.12. We could have given alternativeproofs using the fact that N is well-ordered, but we felt that this was a needlesscomplication.


Exercises1. Complete the proof of Proposition 2.8.

2. Let f be a one-to-one function from A into B with B finite. Show that A

is finite.

3. If A and B are finite sets, show that A∪B is a finite set. Conclude that thefinite union of finite sets is finite.

4. If X is an infinite set and x is in X, show that X ∼ X \ {x}.5. Define explicitly a bijection from [0, 1] onto (0, 1).

6. Complete the proof of Proposition 2.11.

7. Let f be a one-to-one function from A into B with B countable. Provethat A is countable.

8. For m in N, show that N ∼ N \ {1, 2, . . . , m}.9. If A and B are countable sets, show that A × B is countable.

10. Let A be an uncountable set and let B be a countable subset of A. Showthat A \ B is uncountable.

11. Let A be a collection of pairwise disjoint open intervals. That is, membersof A are open intervals in R and any two distinct members of A are disjoint.Show that A is countable.

12. Let S be the set of all open intervals with rational endpoints. Show that S

is countable.

13. Let A be the set of all sequences whose terms are the digits 0 and 1. Showthat A is uncountable.

14. The purpose of this exercise is to show that given any set, there is alwaysa larger set with respect to cardinality. For a set X, the power set of X,denoted by P(X), is the collection of all subsets of X. Recall from Exercise9 in Section 1.4 that a set with n elements has 2n subsets for n in N ∪ {0}.The map x → {x} is a one-to-one function from X into P(X). Show thatthere does not exist a function from X onto P(X). [Hint: Suppose f is afunction from X onto P(X). Let A = {x ∈ X : x /∈ f (x)}. Then A is inP(X) and so A = f (x0) for some x0 in X. Ask yourself “where is x0?” toobtain a contradiction.]

15. Let A be the set of all real-valued functions on [0, 1]. Show that there doesnot exist a function from [0, 1] onto A.

39

3 Sequences

This is an important chapter because we will use sequences through-out the text. The main theorem is the Bolzano-Weierstrass Theorem forsequences in Section 3.5. Other very important theorems are the Mono-tone Convergence Theorem (Section 3.4), Theorem 3.5 (Section 3.2), andTheorem 3.12 (Section 3.6).

3.1 Convergence

Limit of a SequenceRecall that N = {1, 2, 3, . . .} and that R is the set of real numbers.

Definition 3.1 A sequence of real numbers (or a sequence in R) is afunction whose domain is N and whose range is a subset of R.

Thus, a sequence in R is a function from N into R.

Notation When we use the word “sequence” without qualification, we meana sequence in R. For a sequence it is customary to use a letter such as x for thefunction and to denote the value x(n) as xn for each n in N. Thus, we think of asequence as x : N → R or as (x1, x2, x3, . . .) or as (xn)n∈N or as (xn)

∞n=1. Each

xn is a term of the sequence. Following Bartle and Sherbert, we distinguishbetween the sequence (xn)n∈N, whose terms have an order induced by N, andthe range {xn : n ∈ N} of the sequence, which is not ordered. For example, thesequence

((−1)n+1

)n∈N

= (1, −1, 1, −1, . . .) has range {−1, 1}. Some booksuse the notations (xn)n∈N and {xn : n ∈ N} interchangeably, whereas we willreserve the latter notation for the range of the sequence.

Sequences may be given explicitly, as in (1/n)∞n=1 = (1, 12 , 1

3 , . . .), orrecursively, as in the Fibonacci sequence:

let x1 = x2 = 1

let xn = xn−1 + xn−2 for n ≥ 3.

Definition 3.2 A sequence (xn)n∈N eventually has a certain property ifthere exists an n0 in N such that

(xn)n≥n0= (xn0 , xn0+1, xn0+2, . . .)

has this property.

40 Chapter 3 Sequences

For example, a constant sequence is a sequence whose range consists ofa single number, whereas the sequence (1, 2, 3, 4, 4, 4, . . .) is eventually con-stant.

Terminology In Definition 3.2, (xn)n≥n0is a tail of the sequence (xn)n∈N.

This tail is also a sequence. To see this, define a bijection f from N onto{n0, n0 + 1, n0 + 2, . . .} by f (n) = n0 + n − 1 for each n in N. Then x ◦ f isa sequence and xn = (x ◦ f )(n + 1 − n0) for n ≥ n0.

A useful and important concept is that of a neighborhood of a point.

Definition 3.3 For x in R and ε > 0, the open interval

(x − ε, x + ε) = {y ∈ R : |y − x| < ε}centered at x of radius ε is a neighborhood of x.

Definition 3.4 Let (xn)n∈N be a sequence in R and let x be in R. Thesequence (xn)n∈N converges to x (or has limit x) if for every neighborhoodU of x the sequence (xn)n∈N is eventually in U .

Notation and Terminology If the sequence (xn)n∈N converges to x, wewrite lim

n→∞ xn = x or limn

xn = x or xn→n

x or simply xn → x, and we call

(xn)n∈N a convergent sequence. A sequence that is not convergent is divergent.

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

Proposition 3.1 Let (xn)n∈N be a sequence in R and let x be in R. Then(xn)n∈N converges to x if and only if for all ε > 0 there exists an n0 in N suchthat if n ≥ n0, then |xn − x| < ε.

Proof Combining Definitions 3.2 through 3.4, we have that xn → x ⇐⇒(xn)n∈N is eventually in every neighborbood of x ⇐⇒ for all ε > 0, (xn)n∈N iseventually in (x − ε, x + ε) ⇐⇒ for all ε > 0, eventually, |xn − x| < ε.

In general, n0 depends on ε. Figure 3.1 graphically depicts Definition 3.4and Proposition 3.1. Usually, the smaller ε becomes, the larger n0 must be inorder for the distance from xn to x to be less than ε for all n ≥ n0.

Example 3.1 Let (xn)n∈N be the constant sequence c—that is, xn = c for alln in N. Then, xn → c. To see this, let ε > 0. Let n0 = 17. (Here n0 does notdepend on ε. Any other n0 would also work.) Let n ≥ n0. Then

|xn − c| = |c − c| = 0 < ε.

Example 3.2 limn→∞ (1/n) = 0. Let ε > 0. We want to determine a value of

n0 as specified in Proposition 3.1. Right now we do not know how to choose n0,and so we proceed as if we know n0 with the hope that the calculations below

Section 3.1 Convergence 41

x + e

xU

x − e

123 n0

Figure 3.1

will indicate how to choose n0. Let n ≥ n0. Then∣∣∣∣1n − 0

∣∣∣∣ = 1

n≤ 1

n0< ε

if we choose n0 in N such that 1/n0 < ε (by the Archimedean Principle), orequivalently, choose n0 in N with n0 > 1/ε since the positive integers are un-bounded above (Theorem 2.1).

Example 3.3 limn→∞ (2n+3)/(3n+5) = 2/3. Let ε > 0. Let n0 = ? Pretend

that you know n0. For n ≥ n0,∣∣∣∣2n + 3

3n + 5− 2

3

∣∣∣∣ =∣∣∣∣3(2n + 3) − 2(3n + 5)

3(3n + 5)

∣∣∣∣= 1

9n + 15

<1

9n

≤ 1

9n0.

Now we know how to choose n0. By the Archimedean Principle, choose n0 inN such that 1/n0 < 9ε. Then, for n ≥ n0,∣∣∣∣2n + 3

3n + 5− 2

3

∣∣∣∣ <1

9n0<

1

9(9ε) = ε.

An important property of the limit of a sequence is uniqueness.

Lemma 3.1 Distinct points in R can be separated by disjoint neighborhoods.That is, if x and y are in R with x �= y, then there exist neighborhoods U of x

and V of y such that U ∩ V = ∅.

Proof Let x and y be in R with x �= y. Let ε = 12 |x−y|. Let U = (x−ε, x+ε)

and V = (y − ε, y + ε). We claim that U ∩V = ∅. Suppose that z is in U ∩V.

Then, by the triangle inequality,

|x − y| ≤ |x − z| + |z − y| < ε + ε = 2ε = |x − y|.This is a contradiction, because a real number cannot be less than itself. There-fore, U and V are disjoint neighborhoods of x and y. (The reader shouldcompare this with Exercise 11 in Section 2.1.)


Theorem 3.1 Limits of sequences are unique.

Proof Let (xn)n∈N be a sequence in R. Suppose that x and y are in R withxn → x and xn → y. We wish to show that x = y. Suppose that x �= y. ByLemma 3.1, let U and V be disjoint neighborhoods of x and y, respectively.Since xn → x, there is an n1 in N such that xn is in U for all n ≥ n1. Sincexn → y, there is an n2 in N such that xn is in V for all n ≥ n2.

Let m ≥ max{n1, n2}. Then xm is in U ∩ V = ∅, which is a contradiction.Therefore, x = y.

Remark The proof above illustrates a technique commonly used with se-quences. Namely, in order to have two conditions occurring simultaneously,we need to go out far enough in the sequence to guarantee that each conditionholds. As in the above proof we accomplish this by taking a maximum.

Limits Do Not Always ExistThe fact that limits do not always exist should come as no surprise to the reader.The sequence (xn)n∈N in R does not converge if and only if lim

n→∞ xn �= x for all

x in R.

Let (xn)n∈N be a sequence in R and let x be in R. Using the basic rulefor negation as given on page 3, we negate both sides of the “if and only if”in Definition 3.4. Using Definition 3.2, we have that lim

n→∞ xn �= x ⇐⇒ there

exists a neighborhood U of x such that the sequence (xn)n∈N is not eventually inU ⇐⇒ there exists a neighborhood U of x such that for all n0 in N, there existsan n ≥ n0 such that xn is not in U . The last phrase is sometimes expressed as“the sequence (xn)n∈N is frequently outside of U .”

Example 3.4 The sequences (0, 1, 0, 1, 0, 1, . . .), (1, −1, 1, −1, 1, −1, . . .),

and (1, 2, 1, 3, 1, 4, . . .) do not converge. For instance, the first sequence is fre-quently outside the neighborhood

(− 12 , 1

2

)of 0 and is frequently outside the

neighborhood(

12 , 3

2

)of 1.

Let (xn)n∈N be a sequence in R and let x be in R. Negating the statementsin Proposition 3.1, we have that lim

n→∞ xn �= x ⇐⇒ there exists an ε > 0 such

that for all n0 in N, there exists an n ≥ n0 with |xn − x| ≥ ε ⇐⇒ ∃ ε > 0 suchthat ∀n0 ∈ N, ∃ n ≥ n0 with |xn − x| ≥ ε.

The reader should apply this to the sequences in Example 3.4.

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

Example 3.5 Let 0 < r < 1. Then limn→∞ rn = 0.To see this, letx = (1/r) − 1.

Then x > 0 and r = 1/(1 + x). By Bernoulli’s inequality, (1 + x)n ≥ 1 + nx

for all n in N. Therefore,

0 < rn = 1

(1 + x)n≤ 1

1 + nx<

1

nx

Section 3.1 Convergence 43

for all n in N. Given ε > 0, choose n0 in N such that 1/n0 < xε. Then n ≥ n0

implies that

0 < rn <1

nx≤ 1

n0x<

1

x(xε) = ε,

and so rn → 0.

In the next example, we use results concerning the exponential functionex and the natural logarithm function ln x from calculus. This example alsorequires L’Hôpital’s rule.

Example 3.6 limn→∞ [1 + (1/n)]n = e.

Write(1 + 1

n

)n

= eln[1+(1/n)]n = en ln[1+(1/n)].

Then limn→∞ [1 + (1/n)]n = lim

n→∞ en ln[1+(1/n)]=(∗)

elim

n→∞ n ln[1+(1/n)].

Since

limn→∞ n ln

(1 + 1

n

)= lim

n→∞ln[1 + (1/n)]

1/n

= limn→∞

(d/dn)[1 + (1/n)]1 + (1/n)

d

dn

(1

n

) (by L’Hôpital)

= limn→∞

1

1 + (1/n)

= 1,

limn→∞ [1 + (1/n)]n = e1 = e.

The reason for the equality at (∗) is because ex is continuous on R (seeChapter 4).

Remark A good exercise for the reader is to do Example 3.5 by the techniqueused in Example 3.6. This technique involves a sequence whose limit is negativeinfinity, which we consider in Section 3.7.

Exercises1. Use Proposition 3.1 to establish the following limits.

(a) limn→∞

1

n2= 0 (d) lim

n→∞n2 + 1

2n2 + 5= 1

2

(b) limn→∞

3n

n + 2= 3 (e) lim

n→∞sin n

n= 0

(c) limn→∞

3n + 7

5n + 2= 3

5(f) lim

n→∞−n

n2 + 1= 0

2. Let xn ={

1/n if n is odd0 if n is even.

Does limn→∞ xn exist?


3. Let xn ={

1/n if n is odd1 if n is even.

Does limn→∞ xn exist?

4. Let (xn)n∈N be a sequence in R and let x be in R .

(a) Show that if xn → x, then |xn| → |x| . [Hint: Use Corollary 2.1.](b) Show that if |xn| → 0, then xn → 0.

(c) Show, by example, that (|xn|)n∈N may converge and (xn)n∈N may notconverge.

5. Let xn ≥ 0 for each n in N; let x be in R with xn → x. Note that x ≥ 0.

Show that√

xn → √x. [Hint: Make two cases: x = 0 and x > 0. In the

latter case, rationalize.]

6. Show that a convergent sequence in R is bounded. That is, if (xn)n∈N con-verges, show that there is a B > 0 such that |xn| ≤ B for all n in N. [Hint:Use Proposition 3.1 with ε = 1.]

7. Show that the sequences(n2)n∈N

= (1, 4, 9, . . .) and (−n)n∈N =(−1, −2, −3, . . .) do not converge.

8. Show that if |r| < 1, then limn→∞ rn = 0.

9. Establish the following limits.

(a) limn→∞

en

πn= 0 (d) lim

n→∞ (1 + 1

n)2n = e2

(b) limn→∞ c1/n = 1 (c > 0) (e) lim

n→∞n2

n! = 0

(c) limn→∞ n1/n = 1 (f) lim

n→∞n!nn

= 0

3.2 Limit Theorems

In this section, we obtain many of the standard results concerning convergentsequences. Since a sequence is a function, the following definition is a specialcase of Definition 2.8.

Definition 3.5 A sequence (xn)n∈N of real numbers is bounded if thereexists a B > 0 such that |xn| ≤ B for all n in N.

Proposition 3.2 A convergent sequence is bounded.

Proof Let (xn)n∈N be a sequence in R and let x be in R with xn → x. ByProposition 3.1 with ε = 1, there exists an n0 in N such that |xn − x| < 1 forall n ≥ n0. By Corollary 2.1, |xn| − |x| ≤ |xn − x| < 1, which implies that|xn| < 1 + |x| for all n ≥ n0. Let B = max{|x1| , |x2| , . . . ,

∣∣xn0−1

∣∣ , 1 + |x|}.Then B > 0 and |xn| ≤ B for all n in N.

Section 3.2 Limit Theorems 45

Note that (0, 1, 0, 1, 0, 1, . . .) is a bounded sequence that does not con-verge. Hence, the converse of Proposition 3.2 is false.

In Theorem 3.2 below, we obtain results for the usual combinations ofconvergent sequences. For the quotient of convergent sequences, we need thefollowing lemma.

Lemma 3.2 Let (xn)n∈N be a sequence in R. Let x be in R with x �= 0 andxn → x. Then there exist an ε > 0 and an n0 in N such that |xn| ≥ ε for alln ≥ n0. In other words, the sequence is eventually bounded away from 0.

Proof Let ε = |x|/2. By Exercise 4 in Section 3.1, |xn| → |x|, and so thesequence (|xn|)n∈N is eventually in the neighborhood (|x| − ε, |x| + ε) of |x| .Hence, there is an n0 in N such that if n ≥ n0, then

|xn| > |x| − ε = |x| − |x|2

= |x|2

= ε.

Theorem 3.2 Let (xn)n∈N and (yn)n∈N be sequences in R; let x and y be inR with xn → x and yn → y. Then

1. limn→∞ (xn + yn) = x + y = lim

n→∞ xn + limn→∞ yn;

2. limn→∞ xnyn = xy = lim

n→∞ xn · limn→∞ yn;

3. limn→∞ cxn = cx = c lim

n→∞ xn for all real numbers c;

4. limn→∞

xn

yn

= x

y=

limn→∞ xn

limn→∞ yn

, provided yn �= 0 for all n in N and y �= 0.

Proof For part 1, let ε > 0. By Proposition 3.1, there exist n1 and n2 in Nsuch that

|xn − x| <ε

2for all n ≥ n1

and

|yn − y| <ε

2for all n ≥ n2.

Let n0 = max{n1, n2}. Then n ≥ n0 implies that

|(xn + yn) − (x + y)| = |(xn − x) + (yn − y)|≤ |xn − x| + |yn − y|< ε.

For part 2, first note that

|xnyn − xy| = |(xnyn − xny) + (xny − xy)|≤ |xn| |yn − y| + |y| |xn − x|.

Let ε > 0. [Note: We wish to make each part less than ε/2. |y| , being asingle number, is not a problem. However, |xn|, not being a constant, is aslight problem.] First, choose n1 in N such that if n ≥ n1, then |xn − x| <

ε/[2(|y|+1)]. (|y| could be 0, so we added the 1.) By Proposition 3.2, (xn)n∈N

is a bounded sequence, and so there is a B > 0 such that |xn| ≤ B for all n

in N. Now choose n2 in N such that if n ≥ n2, then |yn − y| < ε/2B. Let


n0 = max{n1, n2}. Then n ≥ n0 implies that

|xnyn − xy| ≤ B |yn − y| + |y| |xn − x|< B

( ε

2B

)+ |y|

[ε

2(|y| + 1)

]< ε.

The proof of part 3 follows from part 2 by letting (yn)n∈N be the constantsequence (c, c, c, . . .).

For part 4, because of part 2, it suffices to show that 1/yn → 1/y, whereyn �= 0 for all n in N and y �= 0. First note that∣∣∣∣ 1

yn

− 1

y

∣∣∣∣ = |yn − y||yn| |y| .

By the proof of Lemma 3.2, there is an n1 in N such that |yn| ≥ |y| /2 for alln ≥ n1. Hence,∣∣∣∣ 1

yn

− 1

y

∣∣∣∣ ≤ 2 |yn − y||y|2 for all n ≥ n1.

Let ε > 0. Choose n2 in N such that if n ≥ n2, then |yn − y| < ε |y|2 /2. Ifn0 = max{n1, n2}, then n ≥ n0 implies that∣∣∣∣ 1

yn

− 1

y

∣∣∣∣ <2

|y|2ε |y|2

2= ε.

By mathematical induction, parts 1 and 2 of Theorem 3.2 can be extendedto a finite number of convergent sequences. In particular, if (xn)n∈N convergesto x and if k is in N, then

limn→∞ (xk

n) = limn→∞ (xn · xn · · · · · xn)︸︷︷︸

k times

= ( limn→∞ xn)

k = xk.

Example 3.7 We redo Example 3.3 using Theorem 3.2:

limn→∞

2n + 3

3n + 5= lim

n→∞2 + (3/n)

3 + (5/n)= 2 + 0

3 + 0= 2

3.

Example 3.8 Let p(t) = aktk + ak−1t

k−1 + · · · + a1t + a0 be a polynomialwith real coefficients (the ai’s are in R and k is in N ∪ {0}). Let xn → x, wherex is in R. By Theorem 3.2, p(xn) → p(x).

Example 3.9 Part 1 of Theorem 3.2 states for sequences that the limit ofthe sum is the sum of the limits if the limits exist in R. If we let (xn)n∈N =(1, −1, 1, −1, . . .) and (yn)n∈N = (−1, 1, −1, 1, . . .), then (xn + yn)n∈N =(0, 0, 0, . . .). Thus, lim

n→∞ (xn+yn) = 0, but it makes no sense to write limn→∞ xn +

limn→∞ yn.

We end this section with three important theorems whose proofs are straight-forward. Their importance will be demonstrated in the following sections.By a closed interval [a, b] we mean that a and b are in R with a ≤ b and[a, b] = {y ∈ R : a ≤ y ≤ b}.Theorem 3.3 Let (xn)n∈N be a sequence in the closed interval [a, b]. Let x

be in R with xn → x. Then x is in [a, b].

Section 3.2 Limit Theorems 47

Proof Suppose that x < a. Let U be a neighborhood of x lying entirely to theleft of a (see Figure 3.2). Since (xn)n∈N is eventually in U , xn < a eventually,which is a contradiction. Similarly, if x > b, let V be a neighborhood of x lyingentirely to the right of b. Then (xn)n∈N is eventually in V and hence xn > b

eventually, which is a contradiction. Thus, a ≤ x ≤ b.

x

U

ba

Figure 3.2

Note that the argument above implies that if (xn)n∈N is a convergent se-quence with xn ≥ 0 for all n in N, then lim

n→∞ xn ≥ 0. This is the reason whyx ≥ 0 in Exercise 5 in Section 3.1.

Theorem 3.4 (Squeeze Theorem) Let (xn)n∈N , (yn)n∈N , and (zn)n∈N be se-quences in R with xn ≤ yn ≤ zn for all n in N. Suppose that lim

n→∞ xn =lim

n→∞ zn = x, where x is in R. Then limn→∞ yn = x.

Proof Let ε > 0. By Proposition 3.1, there is an n1 in N such that |xn − x| < ε

for all n ≥ n1 and there is an n2 in N such that |zn − x| < ε for all n ≥ n2.

Then n ≥ max{n1, n2} implies that

−ε < xn − x ≤ yn − x ≤ zn − x < ε,

and so

|yn − x| < ε

for all n ≥ max{n1, n2}. Thus, yn → x.

Example 3.10 Since 0 ≤ |(sin n)/n| ≤ 1/n and 1/n → 0, it follows fromthe Squeeze Theorem that lim

n→∞ |(sin n)/n| = 0. By Exercise 4(b) in Section3.1, lim

n→∞ (sin n)/n = 0.

Theorem 3.5 Let x be in R. Then there exist (1) a sequence of rationalnumbers that converges to x and (2) a sequence of irrational numbers thatconverges to x.

Proof The proof of part 1 is based on the fact that the rational numbers aredense in R (Theorem 2.2), and the proof of part 2 is based on the fact that theirrational numbers are dense in R (Exercise 4 in Section 2.3). We prove part 1,leaving part 2 as an exercise.

Choose a rational x1 in the open interval (x − 1, x + 1). Next, choose arational x2 in the open interval (x− 1

2 , x+ 12 ). In general, choose a rational xn in

the open interval (x − (1/n), x + (1/n)). Such rational numbers exist becauseQ is dense in R. Then (xn)n∈N is a sequence in Q. We claim that xn → x.Let ε > 0. Choose n0 in N such that 1/n0 < ε. Then n ≥ n0 implies that|xn − x| < 1/n ≤ 1/n0 < ε, and so xn → x.

The technique illustrated in the proof of Theorem 3.5 is a standard way toconstruct a sequence converging to a given real number. By a slight modifi-cation, we could choose all the xn’s smaller than x or all the xn’s larger thanx. Thus, we could construct a sequence of rational (or irrational) numbersconverging to x from the left or from the right. For example, the sequence(

1,14

10,

141

100,

1414

1000,

14,142

10,000,

141,421

100,000, · · ·

)converges to

√2 from the left.


Exercises1. Find the limits of the following sequences.

(a)

(n

n + 2

)n∈N

(d)

((3 + 1

n

)2)

n∈N

(b)

((−1)nn

n + 2

)n∈N

(e)(√

n − √n + 1

)n∈N

(c)

(n2 + 4n

2n2 + 5

)n∈N

(f)(n − √

n2 + n)

n∈N

2. Give examples of two sequences that do not converge but whose

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

3. Let (xn)n∈N and (yn)n∈N be two sequences in R such that (xn + yn)n∈N and(xn − yn)n∈N both converge. Show that (xn)n∈N and (yn)n∈N both converge.

4. Find a convergent sequence in (0, 1] that does not converge to a point in(0, 1]. Note Theorem 3.3.

5. Use the Squeeze Theorem to show that limn→∞ (cos n)/n = 0.

6. Let (xn)n∈N be a bounded sequence in R (not necessarily convergent), andlet (yn)n∈N be a sequence in R with yn → 0. Show that (xnyn)n∈N convergesto 0.

7. In reference to Exercise 6, give an example of sequences (xn)n∈N and (yn)n∈N

where

(a) (xn)n∈N is bounded and (yn)n∈N converges, but (xnyn)n∈N does not con-verge.

(b) yn → 0 but (xnyn)n∈N does not converge.

8. Prove part 2 of Theorem 3.5.

3.3 Subsequences

Often, the easiest way to show that a sequence does not have a limit is to usesubsequences. The main result of this section is Theorem 3.6.

Definition 3.6 A function h from N into N is strictly increasing if,whenever m and n are in N with m < n, then h(m) < h(n).

Definition 3.7 Let (xn)n∈N and (yn)n∈N be sequences in R. Then(yn)n∈N is a subsequence of (xn)n∈N if there is a strictly increasing func-tion h from N into N such that yn = xh(n) for all n in N.

Section 3.3 Subsequences 49

Equivalently, y = x ◦ h, or the diagram

Nx→ R

h ↑ ↗y

N

commutes.

Notation Given the sequence (xn)n∈N = (x1, x2, x3, . . .), think of the sub-sequence (yn)n∈N as follows:

y1 = xh(1) where h(1) ≥ 1,

y2 = xh(2) where h(2) > h(1),

y3 = xh(3) where h(3) > h(2), etc.

Thus, a subsequence has the same ordering as the original sequence. It is notjust a subset of the original sequence. We usually write h(k) as nk so that thesubsequence (yn)n∈N is (xnk

)∞k=1 where n1 < n2 < n3 < · · · . Note that nk ≥ k

for each k in N.

A little thought should indicate that a subsequence of a subsequence of asequence is a subsequence of the original sequence.

Example 3.11 A sequence is a subsequence of itself (let h be the identitymap on N).

Example 3.12 The sequence (0, 1, 0, 1, 0, 1, . . .) has as subsequences theconstant sequence (0, 0, 0, . . .) and the constant sequence (1, 1, 1, . . .). It alsohas many other subsequences.

Example 3.13 The sequence (1, 3, 2, 4, 5, 6, . . .) is not a subsequence of(1, 2, 3, 4, 5, 6, . . .) because the ordering is not the same.

Theorem 3.6 Let (xn)n∈N be a sequence in R that converges to x. Thenevery subsequence of (xn)n∈N converges to x.

Proof Let (xnk)∞k=1 be a subsequence of (xn)n∈N. We want to show that

xnk→k

x or, equivalently, that limk→∞ xnk

= x. We use Definition 3.4. Let U be a

neighborhood of x. Since xn→n

x, there is an n0 in N such that if n ≥ n0, then

xn is in U. Let k ≥ n0. Since nk ≥ k ≥ n0, xnkis in U . Thus the subsequence

(xnk)∞k=1 is eventually in U , and so

(xnk

)∞k=1

converges to x.

Remark In order to keep straight whether it is the sequence or the subsequencethat is converging, we suggest subscripting the arrow with the appropriate letter,as we did in the proof of Theorem 3.6.

Example 3.14 From Theorem 3.6, it follows that if a sequence has two sub-sequences each with a different limit, then the sequence itself has no limit.This is an easy method for verifying Example 3.4. For instance, the se-quence (0, 1, 0, 1, 0, 1, . . .) has a subsequence with limit 0 and a subsequencewith limit 1. Thus, (0, 1, 0, 1, 0, 1, . . .) cannot converge. Also, the sequence(1, 2, 1, 3, 1, 4, 1, 5, . . .) has an unbounded subsequence and so this sequencecannot converge.


The following characterization of nonconvergence in terms of subsequenceswill be used throughout the text. For now, the reader should apply this to thesequences in Example 3.14.

Proposition 3.3 Let (xn)n∈N be a sequence in R and let x be in R. Thenlim

n→∞ xn �= x if and only if there exist an ε > 0 and a subsequence (xnk)∞k=1 of

(xn)n∈N such that∣∣xnk

− x∣∣ ≥ ε for all k in N. [The last part can be restated as:

there exists a subsequence of (xn)n∈N that is bounded away from x.]

Proof Suppose that limn→∞ xn �= x. From the discussion following Example

3.4, there is an ε > 0 such that for all n0 in N, there exists an n ≥ n0 with|xn − x| ≥ ε. Let n0 = 1. Then there exists an n1 ≥ n0 = 1 with

∣∣xn1 − x∣∣ ≥ ε.

Letting n0 = n1 + 1, there exists an n2 ≥ n1 + 1 with∣∣xn2 − x

∣∣ ≥ ε. Letting

n0 = n2 + 1, there exists an n3 ≥ n2 + 1 with∣∣xn3 − x

∣∣ ≥ ε. Continuing, we

obtain a subsequence (xnk)∞k=1 of (xn)n∈N with

∣∣xnk− x

∣∣ ≥ ε for all k in N.To show the reverse implication, suppose that (xn)n∈N has a subsequence

(xnk)∞k=1 that is bounded away from x. Then the subsequence (xnk

)∞k=1 cannotconverge to x. (Why?) So, by Theorem 3.6, (xn)n∈N cannot converge to x.

Exercises1. (a) Give an example of an unbounded sequence with a convergent subse-

quence.(b) Give an example of an unbounded sequence without a convergent sub-

sequence.(c) Can you give an example of a bounded sequence that does not have a

convergent subsequence?

2. Find the limit of (xn)n∈N where

(a) xn =(

1 + 1

n2

)n2

(b) xn =(

1 + 1

2n

)n

. [Hint: Recall Example 3.6.]

3. Show that (sin n)n∈N does not converge. [Hint: Find a subsequence each ofwhose terms is in [ 1

2 , 1] and another subsequence each of whose terms is in[−1, − 1

2 ].]4. Let (xn)n∈N and (yn)n∈N be two sequences in R. Let (zn)n∈N be the sequence

(x1, y1, x2, y2, x3, y3, . . .). Show that (zn)n∈N has a limit in R if and only ifboth (xn)n∈N and (yn)n∈N have the same limit in R.

5. Let (xn)n∈N be an unbounded sequence in R.

(a) If (xn)n∈N is unbounded above, show that (xn)n∈N has a subsequence(xnk

)∞k=1 with xnk> k for all k in N.

(b) If (xn)n∈N is unbounded below, show that (xn)n∈N has a subsequence(xnk

)∞k=1 with xnk< −k for all k in N.

Section 3.4 Monotone Sequences 51

3.4 Monotone Sequences

In this section we prove three important theorems: the Monotone ConvergenceTheorem, the Nested Intervals Theorem, and the Monotone Subsequence The-orem. The first of these theorems allows us to show that a certain type ofsequence converges without knowing (or guessing) the limit.

The Monotone Convergence Theorem

Definition 3.8 A sequence (xn)n∈N in R is monotone increasing (re-spectively, strictly increasing) if xn ≤ xn+1 (respectively, xn < xn+1) forall n in N. A sequence (xn)n∈N in R is monotone decreasing (respec-tively, strictly decreasing) if xn ≥ xn+1 (respectively, xn > xn+1) for alln in N. A sequence is monotone (or monotonic) if it is either monotoneincreasing or monotone decreasing.

For example, (1/n)n∈N is strictly decreasing, (n)n∈N is strictly increasing,and a constant sequence is both monotone increasing and monotone decreasing,whereas ((−1)n)n∈N is not monotone and (0, 1, 2, 2, 1

4 , 15 , 1

6 , . . .) is eventuallymonotone decreasing. Note that a monotone increasing sequence is boundedbelow by x1, and a monotone decreasing sequence is bounded above by x1.

Recall that a convergent sequence must be bounded (Proposition 3.2) butthat a bounded sequence need not converge.

Theorem 3.7 (Monotone Convergence Theorem) A bounded monotone se-quence converges.

Proof First, suppose that (xn)n∈N is a bounded monotone increasing sequence.By the Completeness Axiom for R (Section 2.2), α = sup{xn : n ∈ N} is inR. We claim that (xn)n∈N converges to α. Let ε > 0. By Proposition 2.5, thereis an n0 in N such that α − ε < xn0 . Let n ≥ n0. Since (xn)n∈N is monotoneincreasing, α − ε < xn0 ≤ xn ≤ α. Hence, |xn − α| < ε for all n ≥ n0, and soxn → α.

Similarly, if (xn)n∈N is a bounded monotone decreasing sequence, thenxn → inf{xn : n ∈ N}, which is a real number by Proposition 2.4. The detailsof this proof are asked for in Exercise 5.

Prior to this section, to show that a sequence converges, our strategy hasbeen basically to “guess” the limit and then prove that it is the limit, which iswhat we did in the proof of Theorem 3.7. Now, if we can show that a sequence isbounded and monotone, we know that the sequence converges, without knowingthe limit. Of course, we know that the limit is a sup or an inf, but calculatingthis sup or inf may or may not be easy.

Example 3.15 Let x1 = 2 and xn+1 = √6 + xn for all n in N. The first three

terms are x1 = 2, x2 = √8, and x3 =

√6 + √

8. So it appears that (xn)n∈N isstrictly increasing. We show this by induction. Clearly, x1 < x2. Assume that


xk < xk+1. Then

xk+1 = √6 + xk <

√6 + xk+1 = xk+2,

and so (xn)n∈N is strictly increasing.We next show that (xn)n∈N is bounded above by 3, using induction. Clearly,

x1 < 3. Assume that xk < 3. Then

xk+1 = √6 + xk <

√6 + 3 = 3.

By Theorem 3.7, (xn)n∈N converges to a real number, say x.We next find x using a new technique. Since (xn)n∈N converges to x, by

the limit theorems in Section 3.2 and by Exercise 5 in Section 3.1,

x = limn→∞ xn+1 = lim

n→∞√

6 + xn = √6 + x.

So, x2 = 6 + x or x2 − x − 6 = 0 or (x − 3)(x + 2) = 0. Hence, x = 3 orx = −2. Since xn ≥ 2 for all n, x = 3.

Example 3.16 For each n in N, let

xn =n∑

k=1

1

k2= 1

12+ 1

22+ · · · + 1

n2.

Since xn+1 − xn = 1/(n + 1)2 > 0, (xn)n∈N is strictly increasing. We nextshow that (xn)n∈N is bounded above by 2. For any n,

xn = 1 + 1

2 · 2+ 1

3 · 3+ · · · + 1

n · n

< 1 + 1

1 · 2+ 1

2 · 3+ · · · + 1

(n − 1)n

= 1 +(

1

1− 1

2

)+(

1

2− 1

3

)+ · · · +

(1

n − 1− 1

n

)

= 1 + 1 − 1

n(by “telescoping”)

= 2 − 1

n< 2.

Thus, by Theorem 3.7, (xn)n∈N converges to a real number x with 1 ≤ x ≤ 2.

At this point we are not in a position to find x. However, it follows from infinite

series that the series∞�

k=1(1/k2) converges and that x is the sum of this series.

From other techniques, it is known that x = π2/6 (Simmons).

The Nested Intervals Theorem

Definition 3.9 Let An be a subset of R for each n in N. Then (An)n∈N

is a sequence of subsets of R, which is nested upward if An ⊂ An+1 foreach n in N and is nested downward if An ⊃ An+1 for each n in N.

For example, ((0, 1/n))n∈N is a nested downward sequence of open inter-vals whose intersection is empty.

Section 3.4 Monotone Sequences 53

Theorem 3.8 (Nested Intervals Theorem) If (In)n∈N is a nested downwardsequence of closed intervals, then

∞⋂n=1

In �= ∅.

Proof For each n in N, let In = [an, bn], where an and bn are real numberswith an ≤ bn and [a1, b1] ⊃ [a2, b2] ⊃ [a3, b3] ⊃ · · · . See Figure 3.3. Then(an)n∈N is a monotone increasing sequence that is bounded above by b1. ByTheorem 3.7, there is an α in R such that an → α. We claim that α is in⋂∞

n=1 In.

a1 a2 a4 b4

a3

b2 b1

b3

Figure 3.3

Since α = sup{an : n ∈ N}, an ≤ α for each n in N. To show that α is in⋂∞n=1 In, it suffices to show that α ≤ bn for all n in N. Fix an n in N. We show

that bn is an upper bound of {ak : k ∈ N}. If k ≤ n, then ak ≤ an ≤ bn, whileif k > n, ak ≤ bk ≤ bn since (bk)k∈N is monotone decreasing. Therefore, bn

is an upper bound of {ak : k ∈ N}; and since α is the least upper bound of{ak : k ∈ N}, α ≤ bn.

Remark Let (an)n∈N and (bn)n∈N be sequences in Q with an < π < bn for alln in N and such thatan → π andbn → π. (Such sequences exist by Theorem 3.5and the paragraph following Theorem 3.5.) Then

⋂∞n=1[an, bn] = {π}. Thus,⋂∞

n=1 ([an, bn] ∩ Q) = ∅. This is another way to say that Q is not complete.(We are using the fact that π is irrational; any other irrational number may besubstituted for π .)

The Monotone Subsequence TheoremThe following proof is taken from Bartle and Sherbert because the authors couldnot improve on their elegant argument.

Theorem 3.9 (Monotone Subsequence Theorem) Every sequence in R hasa monotone subsequence.

Proof Let (xn)n∈N be a sequence in R. For the purpose of this proof, we callthe mth term xm a peak if xm ≥ xn for all n ≥ m. That is, xm is a peak if xm isnever exceeded by any term that follows it.

Case 1 (xn)n∈N has infinitely many peaks. We pick off the peaks inorder. Let m1 be the smallest positive integer such that xm1 is a peak. Letm2 be the smallest positive integer larger than m1 such that xm2 is a peak.Continuing, we obtain the subsequence (xmk

)k∈N of (xn)n∈N. Since eachxmk

is a peak, we have xm1 ≥ xm2 ≥ · · · and hence (xmk)k∈N is monotone

decreasing.


Case 2 (xn)n∈N has only a finite number of peaks. Let the peaks be (inorder) xm1 , xm2 , xm3 , . . . , xmr . We go out beyond the last peak. Letn1 = mr + 1 (if the number of peaks is 0, let n1 = 1.) Since xn1 is nota peak, there is an n2 > n1 such that xn1 < xn2. Since xn2 is not a peak,there is an n3 > n2 such that xn2 < xn3. Continuing, we obtain a strictlyincreasing subsequence (xnk

)k∈N of (xn)n∈N.

Exercises1. Let x1 = 1 and xn+1 = 1

4 (2xn + 3) for all n in N. Show that (xn)n∈N

converges, and find the limit.

2. Let x1 = 3 and xn+1 = 2 − (1/xn) for all n in N. Show that (xn)n∈N


3. Let x1 = √2 and xn+1 = √

2 + xn for all n in N. Show that (xn)n∈N


4. For each n in N, let

xn = 1 + 1

2+ 1

3+ · · · + 1

n=

n∑k=1

1

k.

Show that (xn)n∈N is monotone but does not converge.

5. Complete the proof of Theorem 3.7.

6. Find a nested downward sequence of half-open intervals whose intersectionis empty. [Half-open intervals are of the form (a, b] = {x ∈ R : a < x ≤ b}or of the form [a, b) = {x ∈ R : a ≤ x < b}, where a and b are in R.]

7. Find a nested downward sequence of open rays (respectively, closed rays)whose intersection is empty. [Open rays are of the form (a, ∞) = {x ∈ R :x > a} or (−∞, a) = {x ∈ R : x < a}, and closed rays are of the form[a, ∞) = {x ∈ R : a ≤ x < ∞} or (−∞, a] = {x ∈ R : −∞ < x ≤ a},where a is in R. Also, (−∞, ∞) is an open ray.]

8. In the notation of Theorem 3.8 and its proof, show that

(a) β = limn→∞ bn is in

⋂∞n=1 In;

(b) [α, β] = ⋂∞n=1 In;

(c) if limn→∞ (bn − an) = 0, then

⋂∞n=1 In is a single point.

9. Let A be a nonempty bounded subset of R with α = sup A and β = inf A.Show that A contains a monotone increasing sequence with limit α andthat A contains a monotone decreasing sequence with limit β. [Hint: ByTheorem 3.9 it suffices to find a sequence in A with limit α. Consider twocases: α in A and α in R \ A.]

Section 3.5 Bolzano-Weierstrass Theorems 55

3.5 Bolzano-Weierstrass Theorems

In this section we prove the Bolzano-Weierstrass Theorem for sequences andthe Bolzano-Weierstrass Theorem for sets. The Bolzano-Weierstrass Theoremfor sequences will be used in Sections 3.6 and 3.8 and also in Sections 4.4 and4.5. The terminology set forth in our discussion of the Bolzano-WeierstrassTheorem for sets will be used in Section 4.3.

Bolzano-Weierstrass Theorem for SequencesFor Theorem 3.10 below, we give two proofs. The first proof is an easy appli-cation of the results obtained in the previous section, but this proof cannot begeneralized to Rn because it depends on the order axioms for R. The secondproof, although more complicated, is extendable to Rn.

Theorem 3.10 (Bolzano-Weierstrass Theorem for sequences) A boundedsequence in R has a convergent subsequence (that is, a subsequence that con-verges to a real number).

Proof 1 Let (xn)n∈N be a bounded sequence in R. By the Monotone Sub-sequence Theorem, (xn)n∈N has a monotone subsequence. Since (xn)n∈N isbounded, this monotone subsequence is bounded. By the Monotone Conver-gence Theorem, this subsequence converges to a real number.

Proof 2 Let (xn)n∈N be a bounded sequence in R.

Case 1 {xn}n∈N is finite. Then some value, say a, must be repeated aninfinite number of times. Then (xn)n∈N has a constant subsequence (eachterm is a) that, of course, converges to a. (To construct this subsequence,let n1 be the least positive integer such that xn1 = a. Let n2 be the leastpositive integer greater than n1 such that xn2 = a. Continue by induction.)

Case 2 {xn}n∈N is infinite. Since (xn)n∈N is bounded, there are real num-bers a and b such that a ≤ xn ≤ b for all n in N. Consider the intervals[a, (a + b)/2] and [(a + b)/2, b]. An infinite number of the xn’s must bein at least one of these two subintervals, because otherwise {xn}n∈N wouldbe finite. Let I1 be one of these two subintervals containing an infinitenumber of the xn’s and choose an n1 in N with xn1 in I1. Also note that thelength of I1 is (b − a)/2. See Figure 3.4.

ba

I3

I1

I2

Figure 3.4


We next bisect I1 into two closed subintervals of length (b−a)/22. Byan argument analogous to obtaining I1, let I2 be one of these subintervalscontaining an infinite number of the xn’s. Choose a positive integer n2 > n1

with xn2 in I2.Continuing by induction, we obtain a nested downward sequence

(In)n∈N of closed intervals with the length of In = (b − a)/2n for eachn and a subsequence (xnk

)∞k=1 of (xn)n∈N with xnkin Ik for each k in N.

By the Nested Intervals Theorem,⋂∞

n=1 In �= ∅. (By Exercise 8 in Section3.4,

⋂∞n=1 In is a single point.)

Let x be in⋂∞

n=1 In. We claim that (xnk)∞k=1 converges to x. Let ε > 0.

Choose n0 in N with (b − a)/2n0 < ε. For k ≥ n0, nk ≥ k ≥ n0, and soxnk

∈ Ik ⊂ In0 . Thus,∣∣xnk− x

∣∣ ≤ length of In0 = (b − a)/2n0 < ε for all k ≥ n0

and, therefore, xnk→k

x.

Remark Let (xn)n∈N be a sequence in Q with xn→n

π. Then (xn)n∈N is abounded sequence in Q. By Theorem 3.6, no subsequence of (xn)n∈N canconverge to a point of Q. This is another way to say that Q is not complete.

Bolzano-Weierstrass Theorem for SetsRecall that a neighborhood of a real number x is an open interval centered at x.

Definition 3.10 Let A be a subset of R and let x be in R. Then x is anaccumulation point of A if every neighborhood of x contains a point ofA different from x. If x is in A and x is not an accumulation point of A,then x is an isolated point of A.

Thus, x in R is an accumulation point of A if and only if for every neigh-borhood U of x, we have (U \ {x}) ∩ A �= ∅.

Some books use the terms “limit point” and “cluster point” for accumula-tion point. Informally, accumulation points of a set are “close” to the set. Alsonote that x is an isolated point of A if and only if there exists a neighborhoodV of x such that V ∩ A = {x}.Example 3.17 The set of accumulation points of [0, 1] = the set of accumu-lation points of (0, 1) = [0, 1].Example 3.18 The set of accumulation points of Q = R since Q is densein R.

Example 3.19 A finite set has no accumulation points.

Example 3.20 N has no accumulation points. Thus, every point of N isisolated.

Example 3.21 The set of accumulation points of A = [0, 1] ∪ {2} is [0, 1],while 2 is an isolated point of A.

Section 3.5 Bolzano-Weierstrass Theorems 57

Example 3.22 The set of accumulation points of {1/n : n ∈ N} = {0} andeach 1/n is an isolated point of {1/n : n ∈ N}.Proposition 3.4 Let A be a subset of R and let x be in R. Then x isan accumulation point of A if and only if every neighborhood of x containsinfinitely many points of A.

Proof Let x be an accumulation point of A and let U be a neighborhoodof x. We want to show that U ∩ A is infinite. Suppose U contains onlya finite number of points of A, say (U \ {x}) ∩ A = {x1, x2, . . . , xn}. Letε = min{|x − x1|, |x − x2|, . . . , |x − xn|}. Then ε > 0 and the neighborhood(x − (ε/2), x + (ε/2)) of x contains no points of A distinct from x, which is acontradiction. Thus, U ∩ A is infinite.

The other implication is obvious.

The following theorem is another way to say that R is complete. In Exer-cise 8, we ask the reader to show that the corresponding result in Q is false.

Theorem 3.11 (Bolzano-Weierstrass Theorem for sets) Every bounded in-finite subset of R has an accumulation point in R.

Proof Let A be a bounded infinite subset of R. Since A is infinite, A containsa sequence (xn)n∈N of distinct points by Theorem 2.3. Since A is bounded,(xn)n∈N is a bounded sequence. By the Bolzano-Weierstrass Theorem for se-quences, there exist a subsequence (xnk

)∞k=1 of (xn)n∈N and an x in R such that(xnk

)∞k=1 converges to x.We claim that x is an accumulation point of A. Let U be any neighborhood

of x. Since xnk→k

x, the sequence (xnk)∞k=1 is eventually in U and thus U

contains an infinite number of points of A.

Exercises1. Let 0 < xn < 7 for each n in N. By Theorem 3.10, (xn)n∈N has a convergent

subsequence. In what interval will the limit of this subsequence lie?

2. Let (xn)n∈N be a bounded sequence of integers. Show that (xn)n∈N has asubsequence that eventually is constant.

3. Show that a bounded sequence in R that does not converge has more than onesubsequential limit. That is, show that a nonconvergent bounded sequencehas two subsequences each with a different limit.

4. What is the set of accumulation points of the irrational numbers?

5. Give an example of a bounded set of real numbers with exactly three accu-mulation points.

6. Let A ⊂ R and let x be in R. Show that x is an accumulation point of A ifand only if there exists a sequence of distinct points in A that converges to x.


7. Let A ⊂ R and let x be an isolated point of A. What are the only types ofsequences in A that can converge to x?

8. Give an example of a bounded infinite subset of Q that has no accumulationpoint in Q. (This is another way to say that Q is not complete.)

3.6 Cauchy Sequences

We show that convergent sequences and Cauchy sequences are equivalent inR but not in Q. This is another way to state that R is complete and Q is notcomplete.

Definition 3.11 A sequence (xn)n∈N in R is a Cauchy sequence (orsimply Cauchy) if for every ε > 0 there exists an n0 in N such that ifn ≥ n0 and m ≥ n0, then |xn − xm| < ε.

Although the Cauchy sequences in R are precisely the convergent se-quences in R (Theorem 3.12 below), the difference between Definition 3.11and Proposition 3.1 is that you do not need to know the limit point of thesequence. This will be illustrated in Example 3.23 and the exercises.

Proposition 3.5 A convergent sequence is Cauchy.

Proof Let (xn)n∈N converge to x and let ε > 0. By Proposition 3.1, there isan n0 in N such that if n ≥ n0, then |xn − x| < ε/2. Let n ≥ n0 and m ≥ n0.Then

|xn − xm| = |(xn − x) + (x − xm)|≤ |xn − x| + |x − xm|<

ε

2+ ε

2= ε.

Therefore, (xn)n∈N is Cauchy.

Lemma 3.3 A Cauchy sequence is bounded.

Proof Let (xn)n∈N be a Cauchy sequence and let ε = 1. By Definition 3.11,there exists an n0 in N such that if n ≥ n0, then

∣∣xn − xn0

∣∣ < 1 (we areusing m = n0). From Corollary 2.1, |xn| < 1 + ∣∣xn0

∣∣ for all n ≥ n0. Let-ting B = max{|x1| , |x2| , . . . ,

∣∣xn0−1

∣∣ , 1 + ∣∣xn0

∣∣}, we have |xn| ≤ B for all n

in N.

Theorem 3.12 A sequence in R is Cauchy if and only if the sequence con-verges.

Proof That convergence implies Cauchy is Proposition 3.5.Let (xn)n∈N be a Cauchy sequence in R. By Lemma 3.3, (xn)n∈N is

bounded. By the Bolzano-Weierstrass Theorem for sequences, there exist a

Section 3.6 Cauchy Sequences 59

subsequence (xnk)∞k=1 of (xn)n∈N and an x in R with xnk

→k

x. We show thatxn→

nx.

Let ε > 0. Since (xn)n∈N is Cauchy, there exists an N1 in N such that ifn ≥ N1 and m ≥ N1, then |xn − xm| < ε/2. Since xnk

→k

x, there exists an N2

in N such that if k ≥ N2, then∣∣xnk

− x∣∣ < ε/2. Let n0 = max{N1, N2}. Fix a

k ≥ n0 (so nk ≥ k ≥ n0). Then for n ≥ n0,

|xn − x| ≤ ∣∣xn − xnk

∣∣+ ∣∣xnk− x

∣∣<

ε

2+ ε

2= ε.

Therefore, xn→n

x.

Remark If each xn is in Q and (xn)n∈N is Cauchy, then (xn)n∈N need notconverge to a point of Q. This is another way to say that Q is not complete.For example, let (xn)n∈N be a sequence in Q with xn → π. Then (xn)n∈N is aCauchy sequence in Q that has no limit in Q.

Remark In Definition 3.11, we consider the nth and mth terms of the se-quence. It is not sufficient to consider only the nth and (n + 1)st terms. Forexample, let xn = √

n for each n in N. Then (xn)n∈N is not bounded and hencenot Cauchy. However,

|xn+1 − xn| = 1√n + 1 + √

n<

1

2√

n−→

n0.

So, given ε > 0, there exists an n0 in N with |xn+1 − xn| < ε for all n ≥ n0.

As with monotone sequences, showing that a sequence is Cauchy providesus with another way of showing that a sequence converges without knowingthe limit. Given (xn)n∈N, if we can show that (xn)n∈N is Cauchy, then we knowthat (xn)n∈N converges. We may or may not be able to find the limit.

Example 3.23 Let (xn)n∈N be a sequence in R. Let 0 < r < 1 and supposethat |xn+1 − xn| < rn for all n in N. Then (xn)n∈N converges.

We show that (xn)n∈N is Cauchy. First suppose that m > n. Then

|xm − xn| ≤ |xm − xm−1| + |xm−1 − xm−2| + · · · + |xn+1 − xn|< rm−1 + rm−2 + · · · + rn

= rn(1 + r + r2 + · · · + rm−1−n)

= rn · 1 − rm−n

1 − r

<rn

1 − r→n

0 (by Example 3.5).

Let ε > 0. Then there is an n0 in N such that rn/(1 − r) < ε for all n ≥ n0.

So, if n ≥ n0 and m ≥ n0, then |xn − xm| < ε; hence (xn)n∈N is Cauchy.In the example above, we could have proceeded as follows:

rn(1 + r + r2 + · · · + rm−1−n) < rn(1 + r + r2 + r3 + · · ·)= rn · 1

1 − r(sum of a geometric series).


Exercises1. (a) Find a Cauchy sequence in (0, 1) that does not converge to a point of

(0, 1).(b) Show that a Cauchy sequence in [0, 1] must converge to a point of [0, 1].

(See Theorem 3.3.)2. For each n in N, let

xn = 1 + 1

2+ 1

3+ · · · + 1

n=

n∑k=1

1

k.

(a) Show that (xn)n∈N is not Cauchy.(b) Show that lim

n→∞ |xn+1 − xn| = 0.

3. Let xn be in Z (the integers) for each n in N. Show that if (xn)n∈N is Cauchy,then (xn)n∈N is eventually constant. Conclude that a sequence in Z convergesif and only if it is eventually constant.

4. Let a < b. Let x1 = a, x2 = b, and

xn+2 = xn+1 + xn

2for n ≥ 1.

Follow these steps to show that (xn)n∈N is Cauchy.

(a) Draw a picture and let L = b − a.

(b) Use induction to show that |xn+1 − xn| = L/2n−1 for each n.(c) Proceed as in Example 3.23 to show for m > n that

|xm − xn| <L

2n−2 −→n

0.

(d) Conclude that (xn)n∈N is Cauchy.

5. Refer to Exercise 4. By Theorem 3.12, we have that (xn)n∈N converges tosome x in R. Follow these steps to find x.

(a) From your picture (or by induction) and from Exercise 4(b), show that

x2k+2 − x2k+1 = L

22k.

(b) Use induction to show that

x2n+1 = a + L

2+ L

23 + · · · + L

22n−1 for n ≥ 1.

(c) x = limn→∞ x2n+1 and

x2n+1 = a + L

2

⎡⎢⎣1 −

(14

)n

1 − 14

⎤⎥⎦.

6. Let (xn)n∈N be a sequence in R. Let 0 < r < 1 and suppose that

|xn+2 − xn+1| ≤ r |xn+1 − xn| for n ≥ 1.

Show that (xn)n∈N is Cauchy. [Hint: First show that |xn+2 − xn+1| ≤rn |x2 − x1|; then proceed as in Example 3.23.]

7. Let x1 > 0 and let xn+1 = 1/(2 + xn) for n ≥ 1.

(a) Use Exercise 6 to show that (xn)n∈N is Cauchy.(b) Find lim

n→∞ xn.

Section 3.7 Limits at Infinity 61

3.7 Limits at Infinity

Recall Definition 2.2 for the ordering on R# = R ∪ {±∞}. In this sectionwe extend the notion of the limit of a sequence to points in R#, thus allow-ing a sequence to have limit ±∞. However, the term “convergent” is stillreserved for those sequences whose limits are real numbers. In this section weexamine which theorems of Sections 3.1 through 3.4 have analogous resultsin R#.

Basic Results

Definition 3.12 Let α and β be in R. The open ray

(α, ∞) = {x ∈ R : x > α}is a neighborhood of ∞, while the open ray

(−∞, β) = {x ∈ R : x < β}is a neighborhood of −∞.

The following definition extends Definition 3.4 to R#.

Definition 3.13 Let (xn)n∈N be a sequence in R and let x be in R#. Then(xn)n∈N has limit x, denoted by lim

n→∞ xn = x or limn

xn = x or xn → x

as n → ∞, if for every neighborhood U of x, the sequence (xn)n∈N iseventually in U .

Note that if (xn)n∈N has limit ±∞, then (xn)n∈N is a divergent sequence.Paraphrasing Definitions 3.12 and 3.13, for (xn)n∈N a sequence in R, we

have that

xn → ∞ if and only if for all α > 0, there exists an n0 in N such that ifn ≥ n0, then xn > α;

and

xn → −∞ if and only if for all β < 0, there exists an n0 in N such that ifn ≥ n0, then xn < β.

Example 3.24 limn→∞ (n2 − 3n + 2) = ∞. Let α > 0. Choose n0 = ? Let

n ≥ n0. For n ≥ 3,

n2 − 3n + 2 > n2 − 3n = n(n − 3) ≥ n − 3 ≥ n0 − 3.

So choose n0 in N such that n0 > α + 3. Then n2 − 3n + 2 > n0 − 3 > α.

Example 3.25 limn→∞ (−n) = −∞. Let β < 0. Choose n0 in N such that

n0 > −β. Let n ≥ n0. Then −n ≤ −n0 < β.


Theorem 3.13 Limits of sequences are unique.

Proof If x is in R, then U = (x − 1, x + 1) and V = (x + 1, ∞) aredisjoint neighborhoods of x and ∞ while U and W = (−∞, x − 1) are dis-joint neighborhoods of x and −∞. Clearly, (−∞, 0) and (0, ∞) are disjointneighborhoods of −∞ and ∞. Thus, distinct points in R# can be separated bydisjoint neighborhoods. The remainder of the proof is completely similar tothe proof of Theorem 3.1.

When the limits are ±∞, an exact analogue of Theorem 3.2 cannot beobtained since indeterminate forms can arise: for example, ∞ − ∞, 0 · ∞, or∞/∞. However, we have the following.

Theorem 3.14 Let (xn)n∈N, (yn)n∈N, and (zn)n∈N be sequences in R. Let x

be in R# and suppose that xn → x, yn → ∞, and zn → −∞.

1. If −∞ < x ≤ ∞, then xn + yn → ∞.

2. If −∞ ≤ x < ∞, then xn + zn → −∞.

3. If 0 < x ≤ ∞, then xnyn → ∞ and xnzn → −∞.

4. If −∞ ≤ x < 0, then xnyn → −∞ and xnzn → ∞.

5. If x is in R, thenxn

yn

→ 0 andxn

zn

→ 0.

Proof Note that the conditions in the different parts of the theorem do notallow the limits to be indeterminate forms such as ∞ − ∞, 0 · ∞, etc.

For part 3, first suppose that 0 < x < ∞. Then, as in Lemma 3.2, there isan n1 in N such that xn > x/2 for all n ≥ n1. Let α > 0. Since yn → ∞, thereexists an n2 in N with yn > 2α/x for all n ≥ n2. Let n0 = max{n1, n2}. Thenn ≥ n0 implies that xnyn > (x/2)yn > (x/2)(2α/x) = α. Thus, xnyn → ∞.

Let β < 0. Since zn → −∞, there exists an n3 in N with zn < 2β/x for alln ≥ n3. Then, since zn < 0, n ≥ max{n1, n3} implies that xnzn < (x/2)zn <

(x/2)(2β/x) = β. Thus, xnzn → −∞.

Next, let x = ∞. Then there is an N1 in N such that xn > 1 for all n ≥ N1.

Since eventually yn > 0 and zn < 0, we have that eventually xnyn > yn andxnzn < zn. It follows that xnyn → ∞ and xnzn → −∞.

For part 5, because yn → ∞ and zn → −∞, we assume that no yn or zn

is zero. Since x is in R, by Proposition 3.2, there is a B > 0 such that |xn| ≤ B

for all n in N. Let ε > 0. Since yn → ∞, choose n0 in N such that yn > B/ε

for all n ≥ n0. Then n ≥ n0 implies that |xn/yn| ≤ B/yn < B(ε/B) = ε.

Since zn → −∞, choose n1 in N such that zn < −(B/ε) for all n ≥ n1. Thenn ≥ n1 implies that |xn/zn| ≤ B/|zn| < B(ε/B) = ε.

The rest of the proof is left as an exercise.

SubsequencesWe first obtain the analogue of Theorem 3.6.

Theorem 3.15 Let (xn)n∈N be a sequence in R; let x be in R# with xn → x.

Then every subsequence of (xn)n∈N has limit x.

Proof In the proof of Theorem 3.6, replace Definition 3.4 with Defini-tion 3.13.

Section 3.7 Limits at Infinity 63

To obtain the analogue of Proposition 3.3, we first negate the statementspreceding Example 3.24. For (xn)n∈N a sequence in R,

limn→∞ xn �= ∞ ⇐⇒ there exists an α > 0 such that for all n0 in N, there

exists an n ≥ n0 with xn ≤ α ⇐⇒ ∃ α > 0 such that ∀ n0 ∈ N, ∃ n ≥n0 with xn ≤ α

and

limn→∞ xn �= −∞ ⇐⇒ there exists a β < 0 such that for all n0 in N, there

exists an n ≥ n0 with xn ≥ β ⇐⇒ ∃ β < 0 such that ∀ n0 ∈ N, ∃ n ≥n0 with xn ≥ β.

Proposition 3.6 Let (xn)n∈N be a sequence in R. Then

1. limn→∞ xn �= ∞ if and only if (xn)n∈N has a subsequence that is bounded above;

2. limn→∞ xn �= −∞ if and only if (xn)n∈N has a subsequence that is bounded

below.

Proof We prove part 1, leaving the proof of part 2 to the reader.Suppose that lim

n→∞ xn �= ∞. Then there exists an α > 0 such that for all

n0 in N, there exists an n ≥ n0 with xn ≤ α.

Let n0 = 1. Then there exists an n1 ≥ n0 = 1 with xn1 ≤ α.

Letting n0 = n1 +1, there exists an n2 ≥ n1 +1 with xn2 ≤ α. Continuing,we obtain a subsequence (xnk

)∞k=1 of (xn)n∈N with xnk≤ α for all k in N.

For the reverse implication, if (xn)n∈N has a subsequence (xnk)∞k=1 that is

bounded above, say by α, then (xnk)∞k=1 is never in (α, ∞) and hence (xnk

)∞k=1cannot have limit ∞. So, by Theorem 3.15, (xn)n∈N cannot have limit ∞.

As an example, the sequence (1, 2, 1, 3, 1, 4, 1, 5, . . .) has no limit in R#

but has the constant sequence (1, 1, 1, 1, . . .) as a subsequence. Also, thesequence (1, −2, 3, −4, 5, −6, . . .) has no limit in R# but has the subsequence(−2, −4, −6, . . .), which is bounded above, and the subsequence (1, 3, 5, . . .),which is bounded below.

Monotone SequencesThe following theorem is the analogue of Theorem 3.7.

Theorem 3.16 If (xn)n∈N is a monotone sequence in R, then (xn)n∈N has alimit in R#.

Proof If (xn)n∈N is bounded, then (xn)n∈N has a limit in R by Theorem3.7. Suppose (xn)n∈N is monotone increasing but unbounded. Then (xn)n∈N isunbounded above. We claim that (xn)n∈N has limit ∞, which is also sup{xn :n ∈ N}. Let α > 0. Since (xn)n∈N is unbounded above, there is an n0 in Nsuch that xn0 > α. Since (xn)n∈N is monotone increasing, α < xn0 ≤ xn for alln ≥ n0. Hence, xn → ∞.

Similarly, if (xn)n∈N is monotone decreasing but unbounded, then xn →inf{xn : n ∈ N} = −∞.


The reader should observe that the sequence in Exercise 4 in Section 3.4has limit ∞.

Remark If (xn)n∈N is a sequence in R, then (xn)n∈N has a monotone subse-quence by Theorem 3.9. By Theorem 3.16, this monotone subsequence has alimit in R#; and by Theorem 3.7, this limit is in R if (xn)n∈N is bounded. Theseobservations are deduced at the beginning of the next section from a differentpoint of view.

Exercises1. Use the method of Examples 3.24 and 3.25 to establish the following limits.

(a) limn→∞

√n = ∞ (d) lim

n→∞ (n − 6√

n) = ∞

(b) limn→∞

√n2 + 1√

n= ∞ (e) lim

n→∞ − √n − 7 = −∞

(c) limn→∞ (n2 − 6n + 1) = ∞ (f) lim

n→∞ (−n + sin n) = −∞2. Let (xn)n∈N be a sequence in R.

(a) Show that xn → ∞ if and only if −xn → −∞.

(b) If xn > 0 for all n in N, show that

limn→∞ xn = 0 if and only if lim

n→∞1

xn= ∞.

3. Let (xn)n∈N be a sequence in R.

(a) If xn < 0 for all n in N, show that xn → −∞ if and only if |xn| → ∞.

(b) Show, by example, that if |xn| → ∞, then (xn)n∈N need not have alimit in R#.

4. Finish the proof of Theorem 3.14.

5. Let (xn)n∈N and (yn)n∈N be sequences in R such that xn → ±∞ and(xnyn)n∈N converges. Show that yn → 0.

6. Let (xn)n∈N and (yn)n∈N be sequences of positive real numbers and supposethat xn/yn → L where 0 < L < ∞. Show that xn → ∞ if and only ifyn → ∞. [Hint: Note that, eventually, 1

2L < xn/yn < 32L.]

7. Let (xn)n∈N and (yn)n∈N be sequences of positive real numbers and supposethat xn/yn → 0. Show that

(a) if xn → ∞, then yn → ∞;(b) if (yn)n∈N is bounded, then xn → 0.

8. Let (xn)n∈N be a sequence in R. Show that

(a) if (xn)n∈N is unbounded above, then (xn)n∈N has a subsequence withlimit ∞;

(b) if (xn)n∈N is unbounded below, then (xn)n∈N has a subsequence withlimit −∞. [Hint: See Exercise 5 in Section 3.3.]

Section 3.8 Limit Superior and Limit Inferior 65

9. Let xn > 0 for each n in N. Show that if (xn)n∈N does not have limit ∞,

then (xn)n∈N has a convergent subsequence.

10. Let A be a nonempty subset of R with α = sup A and β = inf A. Show thatA contains a monotone increasing sequence with limit α and a monotonedecreasing sequence with limit β. [Hint: By Exercise 9 in Section 3.4, youneed only consider the cases α = ∞ and β = −∞.]

11. For the purpose of this exercise, we allow our sequence to have values in R#;that is, ±∞ are permissible terms of the sequence. The term “monotone”is defined as in Definition 3.8, using the ordering on R#.

Let A be a nonempty subset of R# with α = sup A and β = inf A.Show that A contains a monotone increasing sequence with limit α andthat A contains a monotone decreasing sequence with limit β. [Note: Ifα = −∞, then A = {−∞}, so that the constant sequence with each term−∞ has limit α.]

3.8 Limit Superior and Limit Inferior

In this section we utilize results from Sections 3.5 and 3.7. First observe that if(xn)n∈N is a sequence in R, then (xn)n∈N has a subsequence that has a limit inR# = R ∪ {±∞}. If (xn)n∈N is bounded, then (xn)n∈N has a subsequence thatconverges to a real number by the Bolzano-Weierstrass Theorem for sequences,whereas if (xn)n∈N is unbounded, then (xn)n∈N has a subsequence with limit±∞ by Exercise 8 in Section 3.7.

Definition 3.14 Let (xn)n∈N be a sequence in R. Let

E = {x ∈ R# : xnk→k

x for some subsequence (xnk)∞k=1 of (xn)n∈N}.

Thus, E consists of all subsequential limits of (xn)n∈N in the extendedreals and E is never empty. The limit superior of (xn)n∈N, denoted bylim sup

n→∞xn, and the limit inferior of (xn)n∈N, denoted by lim inf

n→∞ xn, are

given by

lim supn→∞

xn = sup E

and

lim infn→∞ xn = inf E.

We will usually omit n → ∞ in the notation above. Clearly, lim inf xn ≤lim sup xn. For the remainder of this section, E will denote the set given inDefinition 3.14. If E = {−∞}, then sup E = −∞. If E �= {−∞}, then sup E

is a real number if E is bounded above and +∞ if E is not bounded above. If


E = {∞}, then inf E = ∞. If E �= {∞}, then inf E is a real number if E isbounded below and −∞ if E is not bounded below.

Example 3.26 Let (xn)n∈N = (1, 12 , 2, 1

3 , 3, 14 , 4, 1

5 , . . .). Then lim inf xn =0 and lim sup xn = ∞.

Example 3.27 Let (xn)n∈N = (1, −2, 3, −4, 5, −6, . . .). Then lim inf xn =−∞ and lim sup xn = ∞.

Example 3.28 Let (an)n∈N = (0, 1, 0, 1, 0, 1, . . .) and (bn)n∈N = (1, 0, 1, 0,

1, 0, . . .). Then lim inf an = lim inf bn = 0 and lim sup an = lim sup bn = 1.

The sequence (an + bn)n∈N = (1, 1, 1, 1, . . .). Thus,

lim inf an + lim inf bn = 0 < 1 = lim inf(an + bn)

and

lim sup(an + bn) = 1 < 2 = lim sup an + lim sup bn.

Proposition 3.7 Let (xn)n∈N be a sequence in R and let x be in R#. Thenxn → x if and only if lim sup xn = lim inf xn = x.

Proof If xn → x, then all subsequences of (xn)n∈N have limit x by Theorem3.15. Thus, E = {x} and sup E = inf E = x.

Suppose that (xn)n∈N does not have limit x. We will show that E �= {x}and so sup E and inf E cannot both be x.

If x is in R, then, by Proposition 3.3, there exist an ε > 0 and a subsequence(xnk

)∞k=1 of (xn)n∈N such that∣∣xnk

− x∣∣ ≥ ε for all k in N. By the first paragraph

of this section, the sequence (xnk)∞k=1 will have a subsequence (yn)n∈N that has

a limit y in R#. Then y �= x since |yn − x| ≥ ε for all n in N, and y is in E

since (yn)n∈N is a subsequence of (xn)n∈N.

If x = ∞, then, by Proposition 3.6, (xn)n∈N has a subsequence that isbounded above. This subsequence is either unbounded below or bounded below.Hence this subsequence has a subsequence with limit −∞ (Exercise 8 in Section3.7) or with limit a real number (Theorem 3.10). In either case, E �= {∞}.

The case x = −∞ is handled similarly.

It is intuitively clear that lim sup xn is the largest subsequential limit of(xn)n∈N, and that lim inf xn is the smallest subsequential limit of (xn)n∈N. Forthe sake of clarity, we will postpone justifying this statement until Proposition3.9 at the end of this section; for now we will use this as a given fact.

Remark The advantage of the limit superior and limit inferior is that, givena sequence (xn)n∈N in R, lim sup xn and lim inf xn always exist in R#, whereas(xn)n∈N may or may not have a limit in R#, and in the latter case it makes nosense to write lim xn. A new way to show that (xn)n∈N has a limit is to showthat lim sup xn = lim inf xn. This is illustrated in Exercise 5.

For the purpose of the next proposition and some of the exercises, weextend addition to R# by:

x + ∞ = ∞ + x = ∞ for − ∞ < x ≤ ∞and

x + (−∞) = (−∞) + x = −∞ for − ∞ ≤ x < ∞

Section 3.8 Limit Superior and Limit Inferior 67

and we do not define

−∞ + ∞ or ∞ + (−∞).

Proposition 3.8 Let (an)n∈N and (bn)n∈N be sequences in R. Then

lim sup(an + bn) ≤ lim sup an + lim sup bn

whenever the right side is defined.

Proof Let α = lim sup an, β = lim sup bn, and γ = lim sup(an + bn). Bythe right side being defined, we mean that α + β is not of the form ∞ + (−∞)

or −∞ + ∞. We want to show that γ ≤ α + β. We may assume that γ �= −∞and α + β �= ∞.

Since γ is a subsequential limit of (an + bn)n∈N, there is a subsequence(ank

+ bnk)∞k=1 of (an + bn)n∈N with ank

+ bnk→k

γ . The sequence (ank)∞k=1 has

a subsequence (ankj)∞j=1 with limit u in R# where u < ∞ since α �= ∞, and

the sequence (bnk)∞k=1 has a subsequence (bnkj

)∞j=1 with limit v in R# where

v < ∞ since β �= ∞. Then (ankj+ bnkj

) →j

γ and (ankj+ bnkj

) →j

u+ v. By

the uniqueness of limits, γ = u+v. Since α and β are the largest subsequentiallimits of (an)n∈N and (bn)n∈N, respectively, u ≤ α and v ≤ β. Therefore,γ = u + v ≤ α + β.

Proposition 3.9 Let (xn)n∈N be a sequence in R. Then lim sup xn andlim inf xn are both in E.

Proof We show that lim sup xn is in E, the proof for lim inf xn being similar.Let α = lim sup xn. If α = −∞, then xn →

n−∞ by Proposition 3.7 and so

E = {−∞}. Assume that α > −∞.

By Exercise 11 in Section 3.7, there exists a monotone increasing sequence(yn)n∈N in E with yn → α. Thus, each yi is a limit of a subsequence of(xn)n∈N. For each i in N, let (xi,nk

)∞k=1 be a subsequence of (xn)n∈N with limityi . To construct a subsequence of (xn)n∈N with limit α, we basically use adiagonalization argument on the following array:

x1,n1 x1,n2 x1,n3 · · · → y1

x2,n1 x2,n2 x2,n3 · · · → y2...

...

xi,n1 xi,n2 xi,n3 · · · → yi

......

↓α

Note that the nk’s in each row are not necessarily the same. So we cannot godirectly down the diagonal.

If each yi is real, we can choose x1,nk1from row 1, x2,nk2

from row 2,

x3,nk3from row 3, etc., with nki

< nkjfor i < j (we move to the right as we

go down the rows) and∣∣∣xi,nki

− yi

∣∣∣ < 1/i for each i in N. Then (xi,nki)∞i=1 is

a subsequence of (xn)n∈N with limit α.

Suppose ym = ∞ for some m in N. Then yi = ∞ for all i ≥ m. Choosexi,nki

from each row i such that nki< nkj

for i < j with xi,nki> i for all

i ≥ m. Then (xi,nki)∞i=1 is a subsequence of (xn)n∈N with limit α = ∞.


Exercises1. Find lim sup xn and lim inf xn if xn is given by

(a) (−1)n, (d) cos(nπ),

(b) (−1)nn, (e)

(1 + 1

n

)sin

nπ

2,

(c) (−1)n(

1 + 1

n

), (f) e−n.

2. Give an example of a sequence (xn)n∈N in R with lim inf xn = ∞.

3. Let xn ≤ yn for each n in N. Show that lim inf xn ≤ lim inf yn andlim sup xn ≤ lim sup yn.

4. Let (an)n∈N and (bn)n∈N be sequences in R. Show that

lim inf an + lim inf bn ≤ lim inf(an + bn)

whenever the left side is defined.

5. Let (xn)n∈N be a sequence in R. For each n in N, let

yn = x1 + x2 + · · · + xn

n.

Show that if (xn)n∈N converges to x, then (yn)n∈N converges to x. [Hint:Write

yn − x = x1 + x2 + · · · + xn

n− nx

n

= (x1 − x) + · · · + (xn0 − x)

n+ (xn0+1 − x) + · · · + (xn − x)

n

and, given ε > 0 and suitably choosing n0,

|yn − x| ≤ |x1 − x| + · · · + ∣∣xn0 − x∣∣

n+(

n − n0

n

)ε.

Now take the limit superior of both sides of this inequality.]

6. Refer to Exercise 5. Show that there are nonconvergent sequences (xn)n∈N

for which (yn)n∈N converges. [Hint: Consider (0, 1, 0, 1, 0, 1, . . .).]

to learn more about brooks/cole, visit …mullai.uah.edu/~ravindra/chapters1-3.pdf · introductory...

Documents