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Introductory Real Analysis Frank Dangello Michael Seyfried Shippensburg University

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Introductory Real Analysis

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Introductory Real Analysis

2000 Brooks/Cole, Cengage Learning

Frank Dangello and Michael Seyfried

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Printed in the United States of America 3 4 5 6 14 13 12 11

Contents Chapter 1

Preface

ix

Proofs, Sets, and Functions

1

1.1 1.2 1.3 1.4

Chapter 2

The Structure of 2.1 2.2 2.3 2.4

Chapter 3

Convergence Limit Theorems Subsequences Monotone Sequences Bolzano-Weierstrass Theorems Cauchy Sequences Limits at Infinity Limit Superior and Limit Inferior

Continuity 4.1 4.2 4.3 4.4 4.5 4.6

Chapter 5

Algebraic and Order Properties of The Completeness Axiom The Rational Numbers Are Dense in Cardinality

Sequences 3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8

Chapter 4

Proofs Sets Functions Mathematical Induction

Continuous Functions Continuity and Sequences Limits of Functions Consequences of Continuity Uniform Continuity Discontinuities and Monotone Functions

1 5 10 15

19 19 23 27 31

39 39 44 48 51 55 58 61 65

69 69 73 77 83 86 89

Differentiation

95

5.1 5.2

95 101

The Derivative Mean Value Theorems

v

vi

Contents 5.3 Taylor’s Theorem 5.4 L’Hôpital’s Rule

Chapter 6 Riemann Integration 6.1 6.2 6.3 6.4 6.5 6.6

Existence of the Riemann Integral Riemann Sums Properties of the Riemann Integral Families of Riemann Integrable Functions Fundamental Theorem of Calculus Improper Integrals

Chapter 7 Infinite Series 7.1 7.2 7.3 7.4

Convergence and Divergence Absolute and Conditional Convergence Regrouping and Rearranging Series Multiplication of Series

Chapter 8 Sequences and Series of Functions 8.1 8.2 8.3 8.4

Function Sequences Preservation Theorems Series of Functions Weierstrass Approximation Theorem

Chapter 9 Power Series 9.1 Convergence 9.2 Taylor Series

Chapter 10 The Riemann-Stieltjes Integral 10.1 10.2 10.3 10.4 10.5

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

Chapter 11 The Topology of 11.1 Open and Closed Sets 11.2 Neighborhoods and Accumulation Points 11.3 Compact Sets

104 107

111 111 117 120 126 131 135

143 143 148 156 161

167 167 174 180 187

191 191 198

207 207 215 220 226 233

239 239 245 249

Contents 11.4 11.5

Connected Sets Continuous Functions

vii 254 258

Bibliography

265

Hints and Answers

267

Index

285

Preface To the Student Is the prize worth the struggle? This book provides not only a complete foundation for the basic topics covered in the usual calculus sequence but also a stepping stone to higher-level mathematics. This book contains some beautiful mathematical results and opens the door for the student to see other beautiful mathematical results. This is the prize. The struggle is that the ability to see these beautiful mathematical results does not come easily. Put more succinctly, Introductory Real Analysis is a hard course. With the possible exception of mathematical geniuses, such as Euler, Gauss, and Riemann, people are not born knowing how to do proofs. The skill to prove a mathematical result is learned just as the skill to solve problems is learned, usually with lots of practice. We suggest covering up our proof of a statement. After understanding what the statement means, attempt your own proof. If you are stuck, look at the first line of our proof and see if you can proceed. If necessary, look at another line, and so on. The examples and exercises are essential components of this book. They will help clarify and strengthen your insight into the text material; and they will increase your mathematical maturity, which, like the ability to do proofs, is not an innate human trait. As much as possible, we have arranged the exercises in each section to correspond to the development of the text material, rather than ordering them from easy to hard. Learning to use previous results to obtain new results is part of mathematical maturity, and we freely draw on the exercises in this manner. Throughout the book we give overviews and commentaries in order to provide insights into what we are doing and where we are going. These features take the form of either short paragraphs or more formal remarks. Our purpose for dividing some of the commentaries into remarks is to separate different concepts for the student. In answer to our original question, those who have seen the beauty in mathematics definitely think the prize is worth the struggle.

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

x

Preface amount of material for most two-semester courses, while Chapters 10 and 11 offer additional topics depending on the instructor’s personal taste. As the table of contents indicates, our approach is sequential rather than topological. Because the table of contents lists the topics covered in the book, which for the most part should be done in order, we mention some aspects of the book that are not apparent from the table of contents. Chapter 1 is an introductory chapter whose purposes are to provide some necessary topics for proceeding in the book and to allow the students to develop some easy proofs. The only topic in Chapter 1 that most students have not seen before is the arbitrary collection of sets, which helps the students develop proofs with quantifiers. In our course, we cover Chapter 1 quickly. Because we do not fix notation until Chapter 2, the instructor can begin with Chapter 2, referring back to Chapter 1 as needed. We have meticulously provided appropriate references throughout the book. In Section 2.4 our purpose is to show that the rational numbers are countable and that the real numbers are uncountable. Our experience is that one could get bogged down in this section since cardinality is so fascinating to students; for example, how can there be just as many real numbers in the open interval from 0 to 1 as there are on the real line? Starting in Chapter 3, we use the terminology of neighborhoods, which provides a unifying aspect for both sequential and function limits. In our classes we usually cover about half of Section 3.8, our purpose being to understand what the limit superior and limit inferior are and what they mean in terms of sequential limits. Exercise 5 in this section is very valuable in this regard. We point out that Exercise 11 in Section 3.7 is needed only for the proof of Proposition 3.9 in Section 3.8. We use Section 3.8 in the proof of Mertens’s Theorem in Chapter 7 and in Chapter 9. In Chapter 7, our Ratio and Root tests use only limits, whereas in Chapter 9 we extend these tests using limit superior and limit inferior. Of course, Chapter 9 relies heavily on Chapters 7 and 8. Chapter 6 contains an extensive section on improper integrals with tests for convergence similar to those of infinite series. This section may be covered in depth or lightly depending on the instructor’s preference. Although we do not recommend this, Chapter 7 can be covered after Chapter 3 if one is willing to allow the Integral test without formally doing improper integrals or monotone functions. Our development of the Riemann-Stieltjes integral in Chapter 10 parallels our development of the Riemann integral in Chapter 6. Many of the proofs in Chapter 10 for monotone integrators are only slight modifications of the corresponding proofs in Chapter 6. We think Chapter 6 should be covered before Chapter 10, although both chapters can be taught simultaneously. Also, Section 10.4 on functions of bounded variation can be covered without doing anything else in Chapter 10; it depends only on Chapter 4. Thus, an instructor who does not reach Chapter 10 and who has only a couple of class periods left in the semester can cover Section 10.4. Chapter 4, of course, depends on Chapters 2 and 3. Chapter 11 gives a topological view of our previous concepts, and many of the results in Chapter 4 are now special cases of the action of a continuous function on a compact or connected set.

Preface

xi

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

To All A book such as this naturally contains many standard (to a mathematician) proofs that can be found almost anywhere. Some arguments are original with us in the sense that we developed them, and we have never seen them anywhere else. When we could not improve on another person’s proof or development, we have used it with an appropriate reference. However, all mistakes are ours, and we would appreciate being informed of any errors detected in the book so that we can correct them. We would like to acknowledge many people for their help in the preparation of this book. For their insightful criticisms and suggestions we thank the reviewers: Michael Berry, West Virginia Wesleyan College; David Gurney, Southeastern Louisiana University; Nathaniel F. Martin, University of Virginia; John W. Neuberger, University of North Texas; Alec Norton, University of Texas at Austin; Richard B. Thompson, University of Arizona; Guoliang Yu, University of Colorado; and Marvin Zeman, Southern Illinois University at Carbondale. We also thank our colleagues Douglas Ensley, Frederick Nordai, and William Weller of Shippensburg University for their input; our secretary Pamela McLaughlin for her assistance; and, most of all, our students for their invaluable comments when we did classroom testing of earlier versions of this book.

1

Proofs, Sets, and Functions O ne purpose of this chapter is to help improve the reader’s ability to understand and create proofs. We attempt to do this in Section 1.1. In Section 1.2, we consider sets, including arbitrary collections of sets. Sections 1.3 and 1.4 deal with functions and mathematical induction, respectively.

1.1 Proofs Many of the statements we prove in this section are known to the reader. It is the technique of proof we wish to emphasize. The typical mathematical assertion that 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 Proofs The 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 for some integer k. Then n2 = 4k 2 = 2(2k 2 ) is an even integer, because 2k 2 is an integer. 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, we need 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 the existential quantifier “for some” (equivalent to “there exists” and symbolically denoted by ∃). That is, given an even integer n, there is only one integer k such that n = 2k. The universal quantifier “for all” (equivalently, “for every,” “for each,” “for any,” and denoted symbolically by ∀) would have been incorrect in the proof above. Either at the very start of the proof or at the end of the first line 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 statement in 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 1

2

Chapter 1

Proofs, Sets, and Functions is two times an integer. One detail omitted in the proof above is why 2k 2 is an integer. The reader should answer this.

Converse The 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 its converse may or may not have the same truth value. As examples, the converse of the conditional statement in Proposition 1.1 is true, whereas the converse of the calculus theorem If a function f is differentiable at a point a, then f is continuous at a is false.

Biconditional Given statements p and q, the biconditional statement “p if and only if q” (also denoted 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 conditional statements. 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 for each 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 that x < a. Note that a depends on x. Definitions are to be interpreted in the “if and only if” sense, even though it is 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 exists a positive integer n such that 1/n < x. (Symbolically, ∀ x > 0, ∃ n in N such that 1/n < x.) Note that n depends on x. Proposition 1.2 N is unbounded above if and only if the Archimedean Principle holds. Proof Assume that N is unbounded above. We need to show that the Archimedean Principle is true. Let x > 0. Since N is unbounded above, there exists an n in N such that 1/x < n. Therefore, 1/n < x. Next, assume that the Archimedean Principle holds. We need to show that N is unbounded above. Let x > 0. (The reader should now ask: What do we have to do to finish the proof?) Then 1/x > 0 (see Exercise 6). By the Archimedean Principle, there exists an n in N such that 1/n < 1/x, and thus x < 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 have we 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 Archimedean Principle is a true statement, or that the statement “N is unbounded above” is logically equivalent to the Archimedean Principle.

Indirect Proofs There are two types of indirect proofs: the contrapositive argument and the contradiction 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 its contrapositive have the same truth value. Thus, to prove a conditional statement one can prove its contrapositive. Since we will have to negate many statements throughout the text, we state a basic rule for negation: Change all universal quantifiers to existential quantifiers; change all existential quantifiers to universal quantifiers; and negate the main clause. Proposition 1.3 Let n be an integer. If n2 is an odd integer, then n is an odd integer. Proof

This is the contrapositive of Proposition 1.1.

To prove the conditional statement “if p, then q” by contradiction, one assumes that p is true and that q is false and “hunts” for a contradiction. Once a 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 (for example, 0 = 1), and sometimes it is very unclear. To illustrate proof by contradiction in the next two propositions, we assume the usual order properties on R and that if a is a real number, then −a is the additive 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 hypothesis. Assume that a < ε for all ε > 0 and that a > 0. Then, by the Archimedean Principle, 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. Alternatively, instead of 1/n above, we could use ε = a/2. For the remainder of this section we need the following terminology. A rational number is a real number that can be expressed in the form m/n, where m and n are integers and n  = 0. An irrational number is a real number that is

4

Chapter 1

Proofs, Sets, and Functions not a rational number. A prime number (or simply a prime) is a positive integer greater than 1 whose only positive divisors are itself and 1. By divisors, we mean 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 3 is a divisor of 9. We will need the following theorem: If a prime divides a product of two integers, then the prime must divide at 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 exist √ positive 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 . Since 2 divides 2n2 , 2 divides m2 . Since 2 is a prime, 2 divides m. Thus there is a positive integer k such that m = 2k. Then 2n2 = 4k 2 and n2 = 2k 2 . As above, 2 divides n2 and hence 2 divides √ n. Thus m/n is not in lowest terms, which is a contradiction. Therefore, 2 is not a rational number. For the next theorem we need the result that each positive integer greater than 1 is divisible by a prime. Theorem 1.2 There are infinitely many primes. (Note: This result appears in Euclid’s Elements, Book IX, Proposition 20.) Proof Suppose there are only finitely many distinct primes, say p1 , p2 , . . . , pn . Let M = p1 p2 · · · pn + 1. Then M is an integer, and so there exists a prime that divides M. Thus some pi divides M. But pi divides p1 p2 · · · pn . Therefore, pi divides 1, which is a contradiction. From the proofs in this section, certain things should be clear. First, one must know what the terms in the theorem mean. Second, one usually needs to know 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 to prevent the line of reasoning from straying off course (McArthur).

Exercises √ 1. 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 a 2 + b2 = 0 if and only if a = 0 and b = 0. 4. Let a and b be real numbers. Prove that ab = 0 if and only if a = 0 or b = 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 a 2 = a, prove that either a = 0 or a = 1. 8. (Pigeonhole Principle) Suppose we place m pigeons in n pigeonholes, where m and n are positive integers. If m > n, show that at least two pigeons must be placed in the same pigeonhole. [Hint (from Robert Lindahl of Morehead State University): For i = 1, 2, . . . , n, let xi denote the number of pigeons that are placed in the ith pigeonhole; let xk denote the largest of n

the xi ’s; and let x = (  xi )/n denote the average of the xi ’s. Show that xk ≥ x = m/n > 1.]

i=1

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

Basic Results and Set Operations A 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 in the set. Each object of a set is called an element of the set, a point of the set, or a member of the set. If A is a set and x is a point, then x ∈ 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 stating a property of its elements. For example, A = {−1, 4} = {x ∈ R : x 2 − 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 : x 2 < 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.

6

Chapter 1

Proofs, Sets, and Functions Thus, to prove that two sets are equal, one must show that each is a subset of the other. Also note that since ∅ contains no elements, it follows from the definition 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 belong to 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. 3. 4. 5.

A ∪ ∅ = A and A ∩ ∅ = ∅; A ⊂ A ∪ B and A ∩ B ⊂ A; A ∪ B = B ∪ A and A ∩ B = B ∩ A (commutative property); A ∪ (B ∪ C) = (A ∪ B) ∪ C and A ∩ (B ∩ C) = (A ∩ B) ∩ C (associative property);

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 of sets, we need to show that (2) A ∪ (B ∩ C) ⊂ (A ∪ B) ∩ (A ∪ C) 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, by the 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

We now prove the first equality in part 7. We need to show that if A ⊂ B, then A ∪ B = B and that if A ∪ B = B, then A ⊂ B.

7

(4) (5)

To show (4), we assume A ⊂ B and note that we have to show A ∪ B = B. So we have to show that A ∪ B ⊂ B and that B ⊂ A ∪ B. The latter containment follows 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 to A, 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 of rational numbers—that is, if m  Q= : m and n are integers and n  = 0 , n then R \ Q is the set of irrational numbers. Also note that Q \ R = ∅. Proposition 1.7 Let A, B, and C be sets. Then 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 complement of an intersection is the union of the complements and the complement of a union is the intersection of the complements. Proof We prove the first equality in part 2, leaving the rest as an exercise. We need to show that A \ (B ∩ C) ⊂ (A \ B) ∪ (A \ C) and that (A \ B) ∪ (A \ C) ⊂ A \ (B ∩ C).

(6) (7)

To show (6), let x ∈ A \ (B ∩ C). By the definition of complement, x ∈ A and x∈ / 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, then x ∈ A \ C. So, in either case, x ∈ (A \ B) ∪ (A \ C).

8

Chapter 1

Proofs, Sets, and Functions 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 and x∈ / B ∩ C. Therefore, x ∈ A \ (B ∩ C).

Arbitrary Unions and Intersections We 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 = {x : x ∈ A for all A ∈ A}.



A, is defined as

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 do not define A. An equivalent formulation of Definition 1.5 can be given in terms of index sets. 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 } = Aα = Aα α∈I

α∈I

= {x : x ∈ Aα for some α ∈ I } and



A=



{Aα : α ∈ I } =

 α∈I

Aα =

 α∈I

Aα

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

 n n2, 3, . . . , n} for some n in N, we write i∈I Ai = i=1 Ai and  If I = {1, i∈I Ai = i=1 Ai .    ∞ If I = N, we write n∈N An = ∞ n∈N An = n=1 An and 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, there is 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 nonempty index set and let Aα bea 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 need to show that   X\ Aα ⊂ (X \ Aα ) (8) α∈I

α∈I

and that   (X \ Aα ) ⊂ X \ Aα . α∈I

(9)

α∈I

  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 \ A α by  the definition of complement. Thus, x ∈ (X \ A ). α α∈I  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 and B,denoted by A × B, is the set of all ordered pairs (a, b), where a is in A 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 pair or it may mean the open interval {x ∈ R : a < x < b}. The context should determine which meaning is appropriate. Proposition 1.9 Let A, B, and C be sets. Then 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.

10

Chapter 1

Proofs, Sets, and Functions

Exercises 1. 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 not imply B = C.  ∞ 8. Let An = (n, ∞) for each n ∈ N. Find ∞ An . n=1 An and ∞ n=1 ∞ 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∩





=

Aα

α∈I

and X∪







(X ∩ Aα )

α∈I

Aα

=

α∈I



(X ∪ Aα ).

α∈I

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 consider basic 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 into Y is a rule f that assigns to each element x in the set X a unique element f (x) in the set Y . The set X is called the domain of the function. The set {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 never defined. Because of this, an alternative definition is desirable. This form of the definition 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, denoted by 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 from X 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 they have 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. The direct 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 help verify the claims made about the direct and inverse images. Example 1.5 Let f : R → R be defined by f (x) = x 2 . 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 mapped to 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 ≤ 21 } = [0, 21 ].

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 . Then f −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 of f −1 (S ∪ T ), f (x) ∈ S ∪ T . So f (x) ∈ S or f (x) ∈ T . If f (x) ∈ S, then x ∈ 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 either x ∈ f −1 (S) or x ∈ f −1 (T ). If x ∈ f −1 (S), then f (x) ∈ S. If x ∈ f −1 (T ),

12

Chapter 1

Proofs, Sets, and Functions 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 (or a 1-1 function) if for any pair of distinct points x1 and x2 in X, f (x1 ) and f (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 only if 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 at how 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 , then f is onto Y . Equivalently, a function f from X into Y is onto Y if for each 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 both one-to-one and onto Y . Example 1.7 The function f in Example 1.5 is not one-to-one since f (−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

x − 1 if x ≥ 0 g(x) = x + 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 reader to 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 onto are possible. In a sense these ideas are independent of one another. Some consequences of these properties appear in the next proposition and in the exercises. 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

Proof 1. 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. So f (x) ∈ f (A) and f (x) ∈ f (B). Thus y = f (x) ∈ f (A) ∩ f (B). Hence f (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 the desired result. Let y ∈ f (A) ∩ f (B). Then y ∈ f (A) and y ∈ f (B). Since y ∈ 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) = x 2 . If we restrict our attention to the subset of R consisting of the nonnegative real numbers, we obtain a function that is one-to-one. This motivates the following definition. Definition 1.12 Let f be a function from X into Y . Let A ⊂ X. The restriction of f to A, denoted by f |A , is defined by f |A (x) = f (x) for all x in A. In a similar vein, let g be a function from A into Y . A function f : X → Y that satisfies f |A = g is called an extension of g to X.

Operations We assume that the reader is familiar with the basic algebraic operations on functions (the sum, difference, product, and quotient of two functions). An important operation on functions is that of composition. Definition 1.13 Let f be a function from X into Y. Let g be a function from Y into Z. The composition of f and g, denoted by g◦f , is a function from 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 function from 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.

14

Chapter 1

Proofs, Sets, and Functions 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 + 1 h : (1, ∞) → (0, 1) by h(x) = 1/x. The reader should draw the graphs of these functions to verify that each function is one-to-one and onto the appropriate set. By Proposition 1.12, the composition h ◦ 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 function associated with it and that this function in some sense reverses what the original function does. Proposition 1.13 Let f be a function from X into Y. Then f is a bijection from 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 to construct 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 in Y there is an element x in X with f (x) = y. This shows that g is defined for all elements 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 welldefined 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 a bijection. Suppose that f (x1 ) = f (x2 ). Since both f (x1 ) and f (x2 ) are in 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 so f is one-to-one. To show that f is onto Y, let y 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 inverse of f , and we write f −1 = g. Restating part of Proposition 1.13 (if f is a bijection), (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) = x 2 has no inverse. (Why?) However, if we restrict the domain of f to the set of nonnegative real numbers [0, ∞), then f |[0,∞) is a bijection onto [0, ∞). By Proposition 1.13, f |[0,∞) has an inverse. It should come as no surprise that the inverse of f |[0,∞) is the square root function.

Exercises In 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. Prove that 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 x

if if

x 1. Therefore n0 − 1 is a positive integer and p(n0 −1) is true (since n0 is the least positive integer for which the corresponding statement is false). By assumption 2, p(n0 ) = p[(n0 − 1) + 1] is true, which is a contradiction. Therefore, p(n) is true for every positive integer n. Next, suppose mathematical induction holds. We wish to show that N is well-ordered. Let A be a nonempty subset of N. Suppose A has no least element. 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) is true. 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 has a least element—namely, k + 1—which again is a contradiction to A having no least 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 element. Example 1.13 For each n in N, 12 + 22 + · · · + n2 = 16 n(n + 1)(2n + 1). For each n in N, let p(n) be the statement 1 12 + 22 + · · · + n2 = n(n + 1)(2n + 1). 6 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 +· · ·+k 2 = 16 k(k+1)(2k+1). We need to show that p(k+1) is true. We need to show that 12 +22 +· · ·+k 2 +(k+1)2 = 16 (k+1)(k+2)(2k+3). By the induction hypothesis, 1 12 + 22 + · · · + k 2 + (k + 1)2 = k(k + 1)(2k + 1) + (k + 1)2 6 1 = (k + 1)[k(2k + 1) + 6(k + 1)] 6 1 = (k + 1)(2k 2 + 7k + 6) 6 1 = (k + 1)(k + 2)(2k + 3). 6

Section 1.4

Mathematical Induction

17

Thus, p(k + 1) is true. By mathematical induction, p(n) is true for all n 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 1 2 ≥ 2 = 1 + 1. Let k ≥ 1 and assume that p(k) is true. That is, assume 2k ≥ k + 1. We want to show that p(k + 1) is true. We need to show that 2k+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 k 3 + (k + 1)3 + (k + 2)3 . We must show that 9 divides (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 + k 3 + 9k 2 + 27k + 27 = k 3 + (k + 1)3 + (k + 2)3 + 9(k 2 + 3k + 3). By the induction hypothesis, 9 divides k 3 + (k + 1)3 + (k + 2)3 . Since 9(k 2 + 3k + 3) is a multiple of 9, 9 divides it. Thus p(k + 1) is true and so p(n) is true for all n in N. Remark Sometimes mathematical induction is stated as follows. Let A be a subset 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 mathematical induction 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 an integer 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 the statement: p(n0 + n − 1). Note that q(1) = p(n0 ). A little thought shows that assumptions 1 and 2 imply by mathematical induction that q(n) is true for all n in N. Equivalently, p(n) is true for all n ≥ n0 . Remark The following represents a misuse of mathematical induction. Find the 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 the same color. p(1) is true since we have only one horse. Let k ≥ 1 and assume that p(k) is true. That is, assume that any set of k horses are all of the same color. We want to show that p(k + 1) is true. Let X = {x1 , x2 , . . . , xk+1 } be

18

Chapter 1

Proofs, Sets, and Functions a set of k + 1 horses. We want to show that all k + 1 horses are of the same color. 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, by the induction hypothesis, these are all of the same color. Thus, all k + 1 horses in X are of the same color. Therefore, p(n) is true for all n in N.

Exercises 1. 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. n

n

j =1

j =1

7. Prove that for each n in N,  (−1)j +1 j 2 = (−1)n+1  j. 8. Prove that if x1 , x2 , . . . , xn are n real numbers in the closed interval [a, b], x +···+x then 1 n 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 each n 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.

2

The Structure of B y the end of this chapter, the reader should understand the difference between the real numbers and the rational numbers. In Section 2.1, we note that both are fields with an order relation. However, in Sections 2.2 and 2.3, we show that the real numbers are “complete” whereas the rational numbers are not complete. In Section 2.4, we show that the rational 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 of rational 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 real numbers discussed in this section.

Field Axioms for R To each pair of real numbers a and b there correspond unique real numbers a + b and a · b that satisfy: Axiom 2.1 a+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) 19

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 1 b · = 1. (inverse elements) b Note that (a, b) → a + b and (a, b) → a · b actually define functions from R × R → R. We usually write ab instead of a · b. From these axioms we can derive the usual laws of arithmetic; for example, a · 0 = 0, −(−a) = a, and so on. The axioms above make (R, +, ·) into a field. However, (Q, +, ·) is also a field. This amounts to showing that the sum and the product of two fractions are again fractions.

Order Axioms for R On R there is an order relation, denoted by a if a < b and we define a ≤ b (or b ≥ a) if either a < b or a = b. From these axioms we can derive the usual properties of inequalities; for example, if a < b and c < 0, then ac > bc. The reader will find it instructive to 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 second one will be introduced in the next section. The first one, which was defined in Section 1.4, we now state: Axiom 2.10 The positive integers are well-ordered.

Decimal Representations The following is essentially given in Apostol, pp. 2–3. Our purpose in discussing decimals is in anticipation of Theorem 2.5. Every real number x has a decimal representation of the form x = n.a1 a2 a3 · · · where n is an integer and each ai is one of the digits from 0 to 9. The notation .a1 a2 a3 · · · is an abbreviation for the infinite series a1 a2 a3 + + + ···. 10 100 1000 For example 4 9 9 0.499999 . . . = + + + ··· 10 100 1000

Section 2.1

Algebraic and Order Properties of R

21

  2 4 9 1 1 = + ··· + 1+ + 10 100 10 10  4 9 1 = (the sum of a geometric series) + 1 10 100 1 − 10 4 9 10 = + 10 100 9 5 = 10 = 0.5. Thus, 21 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. For example, 41 = 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, 13 = 0.33333 · · · , 18 = 0.125000 · · · , and 1 = 0.9999999 · · · . Given the positive rational number m/n, in lowest terms with m < n, the process of long division produces the decimal expansion of m/n. Since at each step 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 with 0.49999 · · · above, a decimal that repeats forms a geometric series of the form

1 a + b(1 + r + r 2 + r 3 + · · ·) = a + b 1−r where a, b, and r are rational, with r being a power of a + b[1/(1 − r)] is rational.

1 . 10

Since Q is a field,

Absolute Value Definition 2.1 For x in R, the absolute value of x, denoted by |x|, is defined by

x if x ≥ 0 |x| = −x if x < 0. Note that |x| ≥ 0 for all x in R. Geometrically, |x| is the distance from x 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.

22

Chapter 2

The Structure of R 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, then ab > 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. If a < 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. However, we prefer the following: 0 ≤ (a + b)2 = a 2 + 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 any finite 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 part 2, 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

Exercises 1. 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 be rational. √ √ √ 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 axioms, 0  = 1. By the trichotomy property, either 0 < 1 or 1 < 0. Assuming 1 < 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| = a 2 . (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 a 2 ≤ 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, both of radius ε = 21 |a − b| , with U ∩ V = ∅.

2.2 The Completeness Axiom Boundedness Definition 2.2 The extended real number system, denoted by R# , consists 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# .

24

Chapter 2

The Structure of 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 bound of 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 subset of R# . However, the positive real numbers {x ∈ R : x > 0} are bounded below but not bounded above; the negative real numbers {x ∈ R : x < 0} are bounded above 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 elements of 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 in that set.

Section 2.2

The Completeness Axiom

25

Definition 2.5 Let S be a subset of R# and let s0 and s1 be elements of 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 Axiom The next axiom is our final assumption about the set of real numbers. We will see in the next section that this axiom distinguishes the real numbers from the rational numbers. Recall that bounded above (or below) means bounded above (or below) by a real number. Completeness Axiom for R Every nonempty subset of R that is bounded above has a supremum in R. Proposition 2.4 Every nonempty subset of R that is bounded below has an inf 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 the Completeness Axiom for R, α = sup A is a real number. We will show that α = inf S. Let s be in S. Then s is an upper bound of A. Since α is the least upper 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 bound of 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 real number 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# has a supremum and an infimum in R# , whereas a subset of R need not have a supremum or an infimum in R. This is why we chose to work in R# .

26

Chapter 2

The Structure of R We end this section with an example showing how to work with the supremum and the infimum. For a subset S of R and a point a in R, we define aS = {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 a real 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 bound of 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 in S, s ≤ γ /a and, hence, γ /a is an upper bound of S. Since the least upper bound 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. We wish to show that aβ = inf(aS). For each s in S, β ≤ s since β is a lower bound of 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 lower bound of aS. We need to show that γ ≤ aβ. For each s in S, since γ is a lower bound of aS, γ ≤ as. So, for each s in S, γ /a ≤ s and, hence, γ /a is a lower bound of S. Since β is the greatest lower bound of S, γ /a ≤ β. Thus, γ ≤ aβ.

Exercises 1. Find the supremum and the infimum of each of the following sets. (a) {x ∈ R : 0 < x 2 < 2} (b) {x ∈ R : x 2 < 2} 2. Prove part 2 of Proposition 2.3.

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

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 that inf(bS) = b sup S and sup(bS) = b inf S. 5. Let S 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 that sup(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 let β = sup B. Let C = {ab : a ∈ A and b ∈ B}. Show, by example, that αβ = sup C in general.

2.3 The Rational Numbers Are Dense in R In this section we establish three main results: first, that the Archimedean Principle 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 infimum. The next proposition states that we can approximate a real supremum or infimum by an element of the set. That is, we can find an element of the set that is as close as we would like to a real supremum or 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. S a−e

Figure 2.1

a

Proof We prove part 1, leaving part 2 as an exercise. Figure 2.1 illustrates the fact that α is an upper bound of S and α is in R. Suppose s ≤ α − ε for all s in S. Then α − ε is an upper bound of S, contradicting the fact that α is the least upper bound of S. 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 that α − 1 < n0 ≤ α. Since n0 + 1 is in N and α < n0 + 1, this contradicts the fact that α is an upper bound of N. We now state the Archimedean Principle, whose truth follows from Proposition 1.2.

28

Chapter 2

The Structure of R 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 between every 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 that A∩(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 such that 0 < 1/n < x, and 1/n is a rational number. 1 n

0

y−x

Figure 2.2

x

y

Case 3 0 < x < y. Refer to Figure 2.2. Since y − x is positive, by the 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 that m − 1 ≤ nx < m. So, m/n − 1/n ≤ x < m/n. Manipulating this expression and recalling how 1/n was chosen, we get m 1 x< ≤ x + < x + (y − x) = y, n n and m/n is a rational number. 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 bounded subset 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 is complete 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 by Theorem 1.1.) √ Since 2 is an upper bound of the set A, sup A ≤ 2. Suppose sup A < 2. √ Since Q is dense in R, there is a rational number q such that sup A < q sup A, which is a contradiction. So, sup A = 2. Thus, A is a nonempty bounded subset of Q that does not have a supremum in Q. Therefore, Q is not complete.

More on Suprema and Infima We end this section by expanding our techniques for working with the supremum and the infimum. We give two different arguments for the more difficult part of 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 Let A and B be nonempty subsets of R that are both bounded above. 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 the Completeness Axiom for R. We need to show that α + β = sup(A + B). For each a in A and b in B, a + b ≤ α + β, and so α + β is an upper bound of A + 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 upper bound of A + B, a + b0 ≤ γ . So, for each a in A, a ≤ γ − b0 , and thus γ − 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 element of B, b ≤ γ − α for all b in B. Hence, γ − α is an upper bound of B, and since 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 in B such that α−ε/2 < a0 and β−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 an arbitrary 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 is bounded 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 bounded range in R. Thus, for f : X → R and D a subset of X, f is bounded on D if and only if there exists an M > 0 such that |f (x)| ≤ M for all x in D. For example, f (x) = x 2 is bounded on (0, 1) while g(x) = 1/x is not bounded 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)} = ∞.

30

Chapter 2

The Structure of R 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}. By Proposition 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 two variables, a in A and b in B. To make Proposition 2.7 analogous to Proposition 2.6, 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 equality in Proposition 2.7. 2. Define f : [0, 2] → R by

1 if 0 ≤ x ≤ 1 f (x) = −1 if 1 < x ≤ 2 and define g : [0, 2] → R by

−1 if 0 ≤ x ≤ 1 g(x) = 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.

Exercises 1. 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 : x 2 < 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. Prove that 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) and sup 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 imply the conclusion of part (b).

2.4 Cardinality In this section we distinguish between “finite” and “infinite” sets and then we classify infinite sets as to whether they are “countable” or “uncountable.” The main results of this section are Theorems 2.3, 2.4, and 2.5 and Corollaries 2.4 and 2.5. Our goal is to prove that the set of rational numbers is countable and that 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, denoted by A ∼ B, if there exists a one-to-one function from A onto B (in other 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 “equivalence relation.” Such a relation has the properties of being reflexive, symmetric, and transitive. 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 bijections from A onto B and from B onto C, respectively, then g ◦ f is a bijection from A onto C, and so A ∼ C. As examples, note that N is equivalent to the set of even positive integers by the function n → 2n and that N is equivalent to the set of odd positive integers by the function n → 2n − 1. Also, R ∼ (0, 1) by Example 1.11.

32

Chapter 2

The Structure of R Definition 2.10 A set A is finite if either A is the empty set (which has cardinal number 0) or there is an n in N such that A ∼ {1, 2, . . . , n} (in which 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 }. From the Pigeonhole Principle (Section 1.1, Exercise 8), which states that if there are m pigeons and n pigeonholes with m > n, then at least two pigeons must get in the same hole, it is clear that there cannot be a bijection from a finite set onto a proper subset of itself. From the paragraph preceding Definition 2.10, this is not 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 proof is left as an exercise. Proposition 2.9 Let A and B be sets. 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 subset of A, which is finite by Proposition 2.8. To prove part 2, note that A is equivalent to a subset of B. Again, by Proposition 2.8, this subset of B, and hence B, is infinite. In reference to Proposition 2.9, since a function is single-valued, the range of a function cannot be larger than the domain with respect to cardinality. In the proof of part 1, the reader who wonders, or even questions, why we can choose an element from each member of a collection of sets indexed by the set B (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, N is bounded, which contradicts Theorem 2.1.

Section 2.4

Cardinality

33

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 and to denote the value x(n) as xn for each n in N. Thus, we think of the sequence as (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, . . .)  1 ∞ 2. n n=1 = (1, 21 , 13 , . . .)

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

The first two examples are sequences of distinct points, whereas the third is not. Theorem 2.3 A set X is infinite if and only if X contains a sequence of distinct 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 is infinite. Let X be an infinite set. We want to construct a sequence of distinct points in X. Since X is nonempty, there is some element x1 in X. Since X \ {x1 } is nonempty, there is an element x2 in X \ {x1 }. Suppose that distinct points 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 generate the first two or three terms and then say “continuing by induction, we obtain the rest.” Corollary 2.3 A set X is infinite if and only if there exists a one-to-one function from X onto a proper subset of X. Proof If X is finite, then no such bijection can exist by the Pigeonhole Principle. Let X be infinite. By Theorem 2.3, let (xn )∞ n=1 be a sequence of distinct points in X. Define 1−1

f : X → X \ {x1 } onto

by f (xn ) = xn+1 for all n in N f (x) = x otherwise.

34

Chapter 2

The Structure of R The reader should verify that this function is one-to-one and onto the desired set. The proof above provides a standard way to construct one-to-one functions on an infinite set. The function on the sequence of distinct points is really a function on N since you manipulate the subscripts, and the function on the rest of the set is the identity. There are obviously many more such functions. For example, we could have eliminated all the xn ’s with n odd by using f (xn ) = x2n .

Countable and Uncountable Sets So far, we have split sets into finite and infinite. Next, we divide the infinite sets 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 positive integers are all countably infinite. Informally, Theorem 2.3 states that the countably infinite sets are the “smallest” infinite sets. Also note that if A is countably infinite, we can write A as a sequence of distinct points. Proposition 2.11 Let A and B be sets. 1. 2. 3. 4.

If A is countable and A ∼ B, then B is countable. If B is uncountable and A ∼ B, then A is uncountable. If A is countable and B ⊂ A, then B is countable. 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, leaving parts 1 and 2 as exercises. We have that A is countable with B ⊂ A and we want to show that B is countable. If B is finite, then B is countable. So we can assume that B is infinite. Hence, A is countably infinite. So we enumerate A 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 smallest positive 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 . Proposition 2.12 Let A and B be sets. 1. If A is countable and there exists a function f from A onto B, then B is countable. 2. If A is uncountable and there exists a one-to-one function from A into B, then B is uncountable.

Section 2.4

Cardinality

35

Proof We first prove part 1. If B is finite, then B is countable; so we can assume that B is infinite. For each b in B, choose one ab in f −1 ({b}). Then the function b → ab is a bijection from B onto a subset of A. Thus, B is equivalent to 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, by Proposition 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 also applies here. Lemma 2.1 N × N is countable. Proof Define f : N × N → N by f (n, m) = 2n 3m . (Any two distinct primes will work.) We claim that f is one-to-one and hence N × N ∼ a subset of N and therefore N × N is countable. To see that f is one-to-one, suppose that f (n, m) = f (r, s). Then 2n 3m = 2r 3s . If n > r, then 2n−r = 3s−m . Since 2 divides 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 N onto 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

x if k < n f (k) = k 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 assume that each Aα is nonempty since empty Aα ’s add nothing to the union. Since I is countable, 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→ Aα α∈I

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

 We claim that h maps  N × N onto α∈I Aα and hence,  by Lemma 2.1 and Proposition 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 exists an m in N such that gα0 (m) = x. Since f maps N onto I , there is an n in N with 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}.

36

Chapter 2

The Structure of R If we define Qn = {(q1 , q2 , . . . , qn ) : qi ∈ Q for i = 1,2, . . . , n}, we can show 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 infinite matrix: f (1) = .a11 a12 a13 · · · f (2) = .a21 a22 a23 · · · f (3) = .a31 a32 a33 · · · .. . f (n) = .an1 an2 an3 · · · .. . 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

3 if ann  = 3 bn = 4 if ann = 3. Then x = .b1 b2 b3 · · · 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. Since R = Q ∪ (R \ Q), R \ Q is uncountable by Theorem 2.4. We make one final comment. The following axiom is independent of the other axioms of set theory. Axiom of Choice If I is a nonempty set and if Aα is a nonempty set for each α 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 reason we mention the Axiom of Choice is because we used the Axiom of Choice in the proofs of Propositions 2.9 and 2.12. We could have given alternative proofs using the fact that N is well-ordered, but we felt that this was a needless complication.

Section 2.4

Cardinality

37

Exercises 1. 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 the finite 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. Prove that 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. Show that A \ B is uncountable. 11. Let A be a collection of pairwise disjoint open intervals. That is, members of 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. Show that A is uncountable. 14. The purpose of this exercise is to show that given any set, there is always a larger set with respect to cardinality. For a set X, the power set of X, denoted by P(X), is the collection of all subsets of X. Recall from Exercise 9 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 that there does not exist a function from X onto P(X). [Hint: Suppose f is a function from X onto P(X). Let A = {x ∈ X : x ∈ / f (x)}. Then A is in P(X) and so A = f (x0 ) for some x0 in X. Ask yourself “where is x0 ?” to obtain a contradiction.] 15. Let A be the set of all real-valued functions on [0, 1]. Show that there does not exist a function from [0, 1] onto A.

3

Sequences T

his is an important chapter because we will use sequences throughout the text. The main theorem is the Bolzano-Weierstrass Theorem for sequences in Section 3.5. Other very important theorems are the Monotone Convergence Theorem (Section 3.4), Theorem 3.5 (Section 3.2), and Theorem 3.12 (Section 3.6).

3.1 Convergence Limit of a Sequence Recall that N = {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 a function 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 mean a sequence in R. For a sequence it is customary to use a letter such as x for the function and to denote the value x(n) as xn for each n in N. Thus, we think of a sequence 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 distinguish between the sequence (xn )n∈N , whose terms have an order induced by N, and the range {x  n : n ∈ N} of the sequence, which is not ordered. For example, the sequence (−1)n+1 n∈N = (1, −1, 1, −1, . . .) has range {−1, 1}. Some books use the notations (xn )n∈N and {xn : n ∈ N} interchangeably, whereas we will reserve the latter notation for the range of the sequence. 1 1 Sequences may be given explicitly, as in (1/n)∞ n=1 = (1, 2 , 3 , . . .), or recursively, 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 if there exists an n0 in N such that (xn )n≥n0 = (xn0 , xn0 +1 , xn0 +2 , . . .) has this property.

39

40

Chapter 3

Sequences For example, a constant sequence is a sequence whose range consists of a single number, whereas the sequence (1, 2, 3, 4, 4, 4, . . .) is eventually constant. Terminology In Definition 3.2, (xn )n≥n0 is 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 is a 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. The sequence (xn )n∈N converges to x (or has limit x) if for every neighborhood U of x the sequence (xn )n∈N is eventually in U . Notation and Terminology If the sequence (xn )n∈N converges to x, we write lim xn = x or lim xn = x or xn → x or simply xn → x, and we call n→∞

n

n

(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 proposition. 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 such that 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 is eventually 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 3.4 and Proposition 3.1. Usually, the smaller ε becomes, the larger n0 must be in order 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 all n in N. Then, xn → c. To see this, let ε > 0. Let n0 = 17. (Here n0 does not depend on ε. Any other n0 would also work.) Let n ≥ n0 . Then |xn − c| = |c − c| = 0 < ε. Example 3.2 lim (1/n) = 0. Let ε > 0. We want to determine a value of n→∞ 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 U

x x−e

123

n0

Figure 3.1 will indicate how to choose n0 . Let n ≥ n0 . Then   1   − 0 = 1 ≤ 1 < ε n  n n0 if we choose n0 in N such that 1/n0 < ε (by the Archimedean Principle), or equivalently, choose n0 in N with n0 > 1/ε since the positive integers are unbounded above (Theorem 2.1). Example 3.3 lim (2n + 3)/(3n + 5) = 2/3. Let ε > 0. Let n0 = ? Pretend n→∞ that you know n0 . For n ≥ n0 ,      2n + 3 2   3(2n + 3) − 2(3n + 5)       3n + 5 − 3  =   3(3n + 5) 1 9n + 15 1 < 9n 1 ≤ . 9n0 Now we know how to choose n0 . By the Archimedean Principle, choose n0 in N such that 1/n0 < 9ε. Then, for n ≥ n0 ,    2n + 3 2  1 1    3n + 5 − 3  < 9n < 9 (9ε) = ε. 0 =

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 = 21 |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. Therefore, U and V are disjoint neighborhoods of x and y. (The reader should compare this with Exercise 11 in Section 2.1.)

42

Chapter 3

Sequences 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 with xn → x and xn → y. We wish to show that x = y. Suppose that x  = y. By Lemma 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 . Since xn → 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 sequences. Namely, in order to have two conditions occurring simultaneously, we need to go out far enough in the sequence to guarantee that each condition holds. As in the above proof we accomplish this by taking a maximum.

Limits Do Not Always Exist The 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 xn  = x for all n→∞

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

exists a neighborhood U of x such that the sequence (xn )n∈N is not eventually in U ⇐⇒ there exists a neighborhood U of x such that for all n0 in N, there exists an 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 frequently outside the neighborhood − 21 , 21 of 0 and is frequently outside the   neighborhood 21 , 23 of 1. Let (xn )n∈N be a sequence in R and let x be in R. Negating the statements in Proposition 3.1, we have that lim xn  = x ⇐⇒ there exists an ε > 0 such n→∞ that for all n0 in N, there exists an n ≥ n0 with |xn − x| ≥ ε ⇐⇒ ∃ ε > 0 such that ∀n0 ∈ N, ∃ n ≥ n0 with |xn − x| ≥ ε. The reader should apply this to the sequences in Example 3.4.

Other Techniques for Convergence We end this section with two examples that utilize different techniques. In the first example we use Bernoulli’s inequality, given in Exercise 10 in Section 1.4, to derive an important result. Example 3.5 Let 0 < r < 1. Then lim r n = 0. To see this, let x = (1/r) − 1. n→∞

Then x > 0 and r = 1/(1 + x). By Bernoulli’s inequality, (1 + x)n ≥ 1 + nx for all n in N. Therefore, 1 1 1 0 < rn = ≤ < (1 + x)n 1 + nx 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 1 1 1 0 < rn < ≤ < (xε) = ε, nx n0 x x and so r n → 0. In the next example, we use results concerning the exponential function ex and the natural logarithm function ln x from calculus. This example also requires L’Hôpital’s rule. Example 3.6 lim [1 + (1/n)]n = e. n→∞ Write

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

Then lim [1 + (1/n)]n = lim en ln[1+(1/n)] = en→∞ n→∞

n→∞

(∗)

.

Since



ln[1 + (1/n)] 1 = lim lim n ln 1 + n→∞ n→∞ n 1/n (d/dn)[1 + (1/n)] 1 + (1/n) = lim n→∞ d 1 dn n 1 = lim n→∞ 1 + (1/n) = 1,

(by L’Hôpital)

lim [1 + (1/n)]n = e1 = e.

n→∞

The reason for the equality at (∗) is because ex is continuous on R (see Chapter 4). Remark A good exercise for the reader is to do Example 3.5 by the technique used in Example 3.6. This technique involves a sequence whose limit is negative infinity, which we consider in Section 3.7.

Exercises 1. Use Proposition 3.1 to establish the following limits. 1 n2 + 1 1 (a) lim 2 = 0 (d) lim = 2 n→∞ n n→∞ 2n + 5 2 3n sin n (b) lim =3 (e) lim =0 n→∞ n + 2 n→∞ n 3 3n + 7 −n (c) lim = (f) lim 2 =0 n→∞ 5n + 2 n→∞ n + 1 5

1/n if n is odd Does lim xn exist? 2. Let xn = 0 if n is even. n→∞

44

Chapter 3

Sequences

1/n if n is odd Does lim xn exist? 1 if n is even. n→∞ 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 not converge. 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.]

3. Let xn =

6. Show that a convergent sequence in R is bounded. That is, if (xn )n∈N converges, 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 lim r n = 0. n→∞

9. Establish the following limits. en (a) lim n = 0 n→∞ π (b) lim c1/n = 1 n→∞

(c > 0)

(c) lim n1/n = 1 n→∞

(d) lim (1 + n→∞

1 2n ) = e2 n

n2 =0 n→∞ n! n! (f) lim n = 0 n→∞ n

(e) lim

3.2 Limit Theorems In this section, we obtain many of the standard results concerning convergent sequences. Since a sequence is a function, the following definition is a special case of Definition 2.8. Definition 3.5 A sequence (xn )n∈N of real numbers is bounded if there exists 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. By Proposition 3.1 with ε = 1, there exists an n0 in N such that |xn − x| < 1 for all 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 converge. Hence, the converse of Proposition 3.2 is false. In Theorem 3.2 below, we obtain results for the usual combinations of convergent sequences. For the quotient of convergent sequences, we need the following lemma. Lemma 3.2 Let (xn )n∈N be a sequence in R. Let x be in R with x  = 0 and xn → x. Then there exist an ε > 0 and an n0 in N such that |xn | ≥ ε for all n ≥ 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 the sequence (|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 |x| |x| |xn | > |x| − ε = |x| − = = ε. 2 2 Theorem 3.2 Let (xn )n∈N and (yn )n∈N be sequences in R; let x and y be in R with xn → x and yn → y. Then 1. lim (xn + yn ) = x + y = lim xn + lim yn ; n→∞

n→∞

n→∞

2. lim xn yn = xy = lim xn · lim yn ; n→∞

n→∞

n→∞

3. lim cxn = cx = c lim xn for all real numbers c; n→∞

n→∞

lim xn

4. lim

n→∞

xn x n→∞ = = , provided yn  = 0 for all n in N and y  = 0. yn y lim yn n→∞

Proof For part 1, let ε > 0. By Proposition 3.1, there exist n1 and n2 in N such that ε |xn − x| < for all n ≥ n1 2 and ε |yn − y| < for all n ≥ n2 . 2 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 |xn yn − xy| = |(xn yn − xn y) + (xn y − xy)| ≤ |xn | |yn − y| + |y| |xn − x|. Let ε > 0. [Note: We wish to make each part less than ε/2. |y| , being a single number, is not a problem. However, |xn |, not being a constant, is a slight 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

46

Chapter 3

Sequences n0 = max{n1 , n2 }. Then n ≥ n0 implies that |xn yn − xy| ≤ B |yn − y| + |y| |xn − x|    ε  ε 0. Choose n2 in N such that if n ≥ n2 , then |yn − y| < ε |y|2 /2. If n0 = max{n1 , n2 }, then n ≥ n0 implies that   2 1   − 1  < 2 ε |y| = ε. y y  |y|2 2 n By mathematical induction, parts 1 and 2 of Theorem 3.2 can be extended to a finite number of convergent sequences. In particular, if (xn )n∈N converges to x and if k is in N, then lim (xnk ) = lim (xn · xn · · · · · xn ) = ( lim xn )k = x k .   n→∞ n→∞  n→∞ k times

Example 3.7 We redo Example 3.3 using Theorem 3.2: 2+0 2 2n + 3 2 + (3/n) lim = lim = = . n→∞ 3n + 5 n→∞ 3 + (5/n) 3+0 3 Example 3.8 Let p(t) = ak t k + ak−1 t k−1 + · · · + a1 t + a0 be a polynomial with real coefficients (the ai ’s are in R and k is in N ∪ {0}). Let xn → x, where x 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 of the 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 (xn +yn ) = 0, but it makes no sense to write lim xn + n→∞ n→∞ lim yn .

n→∞

We end this section with three important theorems whose proofs are straightforward. 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 U x

Figure 3.2

a

b

Limit Theorems

47

Proof Suppose that x < a. Let U be a neighborhood of x lying entirely to the left 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 lying entirely 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. Note that the argument above implies that if (xn )n∈N is a convergent sequence with xn ≥ 0 for all n in N, then lim xn ≥ 0. This is the reason why n→∞ x ≥ 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 sequences in R with xn ≤ yn ≤ zn for all n in N. Suppose that lim xn = n→∞ lim zn = x, where x is in R. Then lim yn = x.

n→∞

n→∞

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 from the Squeeze Theorem that lim |(sin n)/n| = 0. By Exercise 4(b) in Section n→∞ 3.1, lim (sin n)/n = 0. n→∞

Theorem 3.5 Let x be in R. Then there exist (1) a sequence of rational numbers that converges to x and (2) a sequence of irrational numbers that converges to x. Proof The proof of part 1 is based on the fact that the rational numbers are dense in R (Theorem 2.2), and the proof of part 2 is based on the fact that the irrational 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 a rational x2 in the open interval (x − 21 , x + 21 ). In general, choose a rational xn in the open interval (x − (1/n), x + (1/n)). Such rational numbers exist because Q 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 to construct a sequence converging to a given real number. By a slight modification, we could choose all the xn ’s smaller than x or all the xn ’s larger than x. Thus, we could construct a sequence of rational (or irrational) numbers converging to x from the left or from the right. For example, the sequence

14 141 1414 14,142 141,421 1, , , , , ,··· 10 100 1000 10,000 100,000 √ converges to 2 from the left.

48

Chapter 3

Sequences

Exercises 1. Find the limits of the following sequences. 

n 1 2 (a) (d) 3+ n + 2 n∈N n n∈N n √  √ (−1) n (b) (e) n − n + 1 n∈N n + 2 n∈N 2

  √ n + 4n 2+n (c) (f) n − n n∈N 2n2 + 5 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 lim (cos n)/n = 0. n→∞

6. Let (xn )n∈N be a bounded sequence in R (not necessarily convergent), and let (yn )n∈N be a sequence in R with yn → 0. Show that (xn yn )n∈N converges to 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 (xn yn )n∈N does not converge. (b) yn → 0 but (xn yn )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 use subsequences. 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 function 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 h

N ↑ N

x

→ 

R

y

commutes. Notation Given the sequence (xn )n∈N = (x1 , x2 , x3 , . . .), think of the subsequence (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 not just a subset of the original sequence. We usually write h(k) as nk so that the subsequence (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 a sequence is a subsequence of the original sequence. Example 3.11 A sequence is a subsequence of itself (let h be the identity map on N). Example 3.12 The sequence (0, 1, 0, 1, 0, 1, . . .) has as subsequences the constant sequence (0, 0, 0, . . .) and the constant sequence (1, 1, 1, . . .). It also has 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. Then every 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 → x or, equivalently, that lim xnk = x. We use Definition 3.4. Let U be a k

k→∞

neighborhood of x. Since xn → x, there is an n0 in N such that if n ≥ n0 , then n xn is in U. Let k ≥ n0 . Since nk ≥ k ≥ n0 , xnk is 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 subsequence that is converging, we suggest subscripting the arrow with the appropriate letter, as we did in the proof of Theorem 3.6. Example 3.14 From Theorem 3.6, it follows that if a sequence has two subsequences 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 sequence (0, 1, 0, 1, 0, 1, . . .) has a subsequence with limit 0 and a subsequence with limit 1. Thus, (0, 1, 0, 1, 0, 1, . . .) cannot converge. Also, the sequence (1, 2, 1, 3, 1, 4, 1, 5, . . .) has an unbounded subsequence and so this sequence cannot converge.

50

Chapter 3

Sequences The following characterization of nonconvergence in terms of subsequences will be used throughout the text. For now, the reader should apply this to the sequences in Example 3.14. Proposition 3.3 Let (xn )n∈N be a sequence in R and let x be in R. Then lim xn  = x if and only if there exist an ε > 0 and a subsequence (xnk )∞ k=1 of n→∞   (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 lim xn  = x. From the discussion following Example n→∞

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 cannot converge to x. (Why?) So, by Theorem 3.6, (xn )n∈N cannot converge to x.

Exercises 1. (a) Give an example of an unbounded sequence with a convergent subsequence. (b) Give an example of an unbounded sequence without a convergent subsequence. (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

2 1 n (a) xn = 1 + 2 n

1 n (b) xn = 1 + . [Hint: Recall Example 3.6.] 2n 3. Show that (sin n)n∈N does not converge. [Hint: Find a subsequence each of whose terms is in [ 21 , 1] and another subsequence each of whose terms is in [−1, − 21 ].] 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 if both (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 Convergence Theorem, the Nested Intervals Theorem, and the Monotone Subsequence Theorem. The first of these theorems allows us to show that a certain type of sequence converges without knowing (or guessing) the limit.

The Monotone Convergence Theorem Definition 3.8 A sequence (xn )n∈N in R is monotone increasing (respectively, strictly increasing) if xn ≤ xn+1 (respectively, xn < xn+1 ) for all n in N. A sequence (xn )n∈N in R is monotone decreasing (respectively, strictly decreasing) if xn ≥ xn+1 (respectively, xn > xn+1 ) for all n in N. A sequence is monotone (or monotonic) if it is either monotone increasing 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, 41 , 15 , 16 , . . .) is eventually monotone decreasing. Note that a monotone increasing sequence is bounded below by x1 , and a monotone decreasing sequence is bounded above by x1 . Recall that a convergent sequence must be bounded (Proposition 3.2) but that a bounded sequence need not converge. Theorem 3.7 (Monotone Convergence Theorem) A bounded monotone sequence 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 in R. We claim that (xn )n∈N converges to α. Let ε > 0. By Proposition 2.5, there is an n0 in N such that α − ε < xn0 . Let n ≥ n0 . Since (xn )n∈N is monotone increasing, α − ε < xn0 ≤ xn ≤ α. Hence, |xn − α| < ε for all n ≥ n0 , and so xn → α. Similarly, if (xn )n∈N is a bounded monotone decreasing sequence, then xn → inf{xn : n ∈ N}, which is a real number by Proposition 2.4. The details of this proof are asked for in Exercise 5. Prior to this section, to show that a sequence converges, our strategy has been basically to “guess” the limit and then prove that it is the limit, which is what we did in the proof of Theorem 3.7. Now, if we can show that a sequence is bounded and monotone, we know that the sequence converges, without knowing the limit. Of course, we know that the limit is a sup or an inf, but calculating this sup or inf may or may not be easy. √ Example 3.15 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 is strictly increasing. We show this by induction. Clearly, x1 < x2 . Assume that

52

Chapter 3

Sequences 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 = lim xn+1 = lim 6 + xn = 6 + x. n→∞

n→∞

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

2

Example 3.16 For each n in N, let n  1 1 1 1 xn = = 2 + 2 + ··· + 2. 2 k 1 2 n k=1 Since xn+1 − xn = 1/(n + 1)2 > 0, (xn )n∈N is strictly increasing. We next show that (xn )n∈N is bounded above by 2. For any n, 1 1 1 xn = 1 + + + ··· + 2·2 3·3 n·n 1 1 1 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 all n in N and such that an → π and bn → π. (Such sequences ∞ exist by Theorem 3.5 and the paragraph following Theorem 3.5.) Then n=1 [an , bn ] = {π}. Thus, ∞ , b ] ∩ Q) = ∅. This is another way to say that Q is not complete. ([a n n n=1 (We are using the fact that π is irrational; any other irrational number may be substituted for π .)

The Monotone Subsequence Theorem The following proof is taken from Bartle and Sherbert because the authors could not improve on their elegant argument. Theorem 3.9 (Monotone Subsequence Theorem) Every sequence in R has a monotone subsequence. Proof Let (xn )n∈N be a sequence in R. For the purpose of this proof, we call the mth term xm a peak if xm ≥ xn for all n ≥ m. That is, xm is a peak if xm is never exceeded by any term that follows it. Case 1 (xn )n∈N has infinitely many peaks. We pick off the peaks in order. Let m1 be the smallest positive integer such that xm1 is a peak. Let m2 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 each xmk is a peak, we have xm1 ≥ xm2 ≥ · · · and hence (xmk )k∈N is monotone decreasing.

54

Chapter 3

Sequences Case 2 (xn )n∈N has only a finite number of peaks. Let the peaks be (in order) xm1 , xm2 , xm3 , . . . , xmr . We go out beyond the last peak. Let n1 = mr + 1 (if the number of peaks is 0, let n1 = 1.) Since xn1 is not a 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 strictly increasing subsequence (xnk )k∈N of (xn )n∈N .

Exercises 1. Let x1 = 1 and xn+1 = 41 (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 converges, and find the limit. √ √ 3. Let x1 = 2 and xn+1 = 2 + xn for all n in N. Show that (xn )n∈N converges, and find the limit. 4. For each n in N, let 1 1 1 1 + + ··· + = . 2 3 n k n

xn = 1 +

k=1

5. 6.

7.

8.

Show that (xn )n∈N is monotone but does not converge. Complete the proof of Theorem 3.7. Find a nested downward sequence of half-open intervals whose intersection is 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.] 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.] In the notation of Theorem 3.8 and its proof, show that  (a) β = lim bn is in ∞ n=1 In ; n→∞ ∞ (b) [α, β] = n=1 In ;  (c) if lim (bn − an ) = 0, then ∞ n=1 In is a single point. n→∞

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 and that A contains a monotone decreasing sequence with limit β. [Hint: By Theorem 3.9 it suffices to find a sequence in A with limit α. Consider two cases: 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 and the Bolzano-Weierstrass Theorem for sets. The Bolzano-Weierstrass Theorem for sequences will be used in Sections 3.6 and 3.8 and also in Sections 4.4 and 4.5. The terminology set forth in our discussion of the Bolzano-Weierstrass Theorem for sets will be used in Section 4.3.

Bolzano-Weierstrass Theorem for Sequences For Theorem 3.10 below, we give two proofs. The first proof is an easy application of the results obtained in the previous section, but this proof cannot be generalized to Rn because it depends on the order axioms for R. The second proof, although more complicated, is extendable to Rn . Theorem 3.10 (Bolzano-Weierstrass Theorem for sequences) A bounded sequence in R has a convergent subsequence (that is, a subsequence that converges to a real number). Proof 1 Let (xn )n∈N be a bounded sequence in R. By the Monotone Subsequence Theorem, (xn )n∈N has a monotone subsequence. Since (xn )n∈N is bounded, this monotone subsequence is bounded. By the Monotone Convergence 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 an infinite number of times. Then (xn )n∈N has a constant subsequence (each term 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 least positive 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 numbers 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 be in at least one of these two subintervals, because otherwise {xn }n∈N would be finite. Let I1 be one of these two subintervals containing an infinite number of the xn ’s and choose an n1 in N with xn1 in I1 . Also note that the length of I1 is (b − a)/2. See Figure 3.4. I1 I3 a

Figure 3.4

I2

b

56

Chapter 3

Sequences We next bisect I1 into two closed subintervals of length (b −a)/22 . By an argument analogous to obtaining I1 , let I2 be one of these subintervals containing 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 each n and a subsequence (xnk )∞ k=1 of (x n∞)n∈N with xnk in Ik for each k in N. By the Nested Intervals Theorem, n=1 In  = ∅. (By Exercise 8 in Section  3.4, ∞ a single point.) n=1 In is  ∞ Let x be in ∞ n=1 In . We claim that (xnk )k=1 converges to x. Let ε > 0. n0 Choose n0 in N with (b − a)/2 < ε. For k ≥ n0 , nk ≥ k ≥ n0 , and so xnk ∈ Ik ⊂ In0 . Thus,   xn − x  ≤ length of In = (b − a)/2n0 < ε for all k ≥ n0 k

0

and, therefore, xnk → x. k

Remark Let (xn )n∈N be a sequence in Q with xn → π. Then (xn )n∈N is a n bounded sequence in Q. By Theorem 3.6, no subsequence of (xn )n∈N can converge to a point of Q. This is another way to say that Q is not complete.

Bolzano-Weierstrass Theorem for Sets Recall 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 an accumulation point of A if every neighborhood of x contains a point of A 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 neighborhood U of x, we have (U \ {x}) ∩ A = ∅. Some books use the terms “limit point” and “cluster point” for accumulation point. Informally, accumulation points of a set are “close” to the set. Also note that x is an isolated point of A if and only if there exists a neighborhood V of x such that V ∩ A = {x}. Example 3.17 The set of accumulation points of [0, 1] = the set of accumulation points of (0, 1) = [0, 1]. Example 3.18 The set of accumulation points of Q = R since Q is dense in R. Example 3.19 A finite set has no accumulation points. Example 3.20 N has no accumulation points. Thus, every point of N is isolated. 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} and each 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 is an accumulation point of A if and only if every neighborhood of x contains infinitely many points of A. Proof Let x be an accumulation point of A and let U be a neighborhood of x. We want to show that U ∩ A is infinite. Suppose U contains only a finite number of points of A, say (U \ {x}) ∩ A = {x1 , x2 , . . . , xn }. Let ε = 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 a contradiction. Thus, U ∩ A is infinite. The other implication is obvious. The following theorem is another way to say that R is complete. In Exercise 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 infinite subset of R has an accumulation point in R. Proof Let A be a bounded infinite subset of R. Since A is infinite, A contains a 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 sequences, 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 → x, the sequence (xnk )∞ k=1 is eventually in U and thus U k

contains an infinite number of points of A.

Exercises 1. 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 a subsequence that eventually is constant. 3. Show that a bounded sequence in R that does not converge has more than one subsequential limit. That is, show that a nonconvergent bounded sequence has 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 accumulation points. 6. Let A ⊂ R and let x be in R. Show that x is an accumulation point of A if and only if there exists a sequence of distinct points in A that converges to x.

58

Chapter 3

Sequences 7. Let A ⊂ R and let x be an isolated point of A. What are the only types of sequences in A that can converge to x? 8. Give an example of a bounded infinite subset of Q that has no accumulation point 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 in R but not in Q. This is another way to state that R is complete and Q is not complete. Definition 3.11 A sequence (xn )n∈N in R is a Cauchy sequence (or simply Cauchy) if for every ε > 0 there exists an n0 in N such that if n ≥ n0 and m ≥ n0 , then |xn − xm | < ε. Although the Cauchy sequences in R are precisely the convergent sequences in R (Theorem 3.12 below), the difference between Definition 3.11 and Proposition 3.1 is that you do not need to know the limit point of the sequence. 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 is an 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 ε < + = ε. 2 2 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 are   using 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 converges. 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 → x. We show that k xn → x. n

Let ε > 0. Since (xn )n∈N is Cauchy, there exists an N1 in N such that if n ≥ N1 and m ≥ N1 , then |xn − xm | < ε/2. Since xnk → x, there exists an N2 k   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 ε < + = ε. 2 2 Therefore, xn → x. n

Remark If each xn is in Q and (xn )n∈N is Cauchy, then (xn )n∈N need not converge 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 a Cauchy sequence in Q that has no limit in Q. Remark In Definition 3.11, we consider the nth and mth terms of the sequence. It is not sufficient to consider only the nth and (n + 1)st terms. For √ example, let xn = n for each n in N. Then (xn )n∈N is not bounded and hence not Cauchy. However, 1 1 |xn+1 − xn | = √ √ < √ −→ 0. 2 n n n+1+ n 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 provides us with another way of showing that a sequence converges without knowing the limit. Given (xn )n∈N , if we can show that (xn )n∈N is Cauchy, then we know that (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 suppose that |xn+1 − xn | < r n 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 | < r m−1 + r m−2 + · · · + r n = r n (1 + r + r 2 + · · · + r m−1−n ) 1 − r m−n = rn · 1−r rn (by Example 3.5). < →0 1−r n Let ε > 0. Then there is an n0 in N such that r n /(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: r n (1 + r + r 2 + · · · + r m−1−n ) < r n (1 + r + r 2 + r 3 + · · ·) 1 = rn · (sum of a geometric series). 1−r

60

Chapter 3

Sequences

Exercises 1. (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 1 1 1 1 + + ··· + = . 2 3 n k n

xn = 1 +

k=1

(a) Show that (xn )n∈N is not Cauchy. (b) Show that lim |xn+1 − xn | = 0. n→∞

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 converges if and only if it is eventually constant. 4. Let a < b. Let x1 = a, x2 = b, and xn+2 =

xn+1 + xn for n ≥ 1. 2

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 |
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 xn . n→∞

Section 3.7

Limits at Infinity

61

3.7 Limits at Infinity Recall Definition 2.2 for the ordering on R# = R ∪ {±∞}. In this section we extend the notion of the limit of a sequence to points in R# , thus allowing a sequence to have limit ±∞. However, the term “convergent” is still reserved for those sequences whose limits are real numbers. In this section we examine which theorems of Sections 3.1 through 3.4 have analogous results in 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 xn = x or lim xn = x or xn → x n→∞

n

as n → ∞, if for every neighborhood U of x, the sequence (xn )n∈N is eventually 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 if n ≥ n0 , then xn > α; and xn → −∞ if and only if for all β < 0, there exists an n0 in N such that if n ≥ n0 , then xn < β. Example 3.24 lim (n2 − 3n + 2) = ∞. Let α > 0. Choose n0 = ? Let n→∞ 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 lim (−n) = −∞. Let β < 0. Choose n0 in N such that n→∞ n0 > −β. Let n ≥ n0 . Then −n ≤ −n0 < β.

62

Chapter 3

Sequences Theorem 3.13 Limits of sequences are unique. Proof If x is in R, then U = (x − 1, x + 1) and V = (x + 1, ∞) are disjoint neighborhoods of x and ∞ while U and W = (−∞, x − 1) are disjoint neighborhoods of x and −∞. Clearly, (−∞, 0) and (0, ∞) are disjoint neighborhoods of −∞ and ∞. Thus, distinct points in R# can be separated by disjoint neighborhoods. The remainder of the proof is completely similar to the proof of Theorem 3.1. When the limits are ±∞, an exact analogue of Theorem 3.2 cannot be obtained 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 xn yn → ∞ and xn zn → −∞. 4. If −∞ ≤ x < 0, then xn yn → −∞ and xn zn → ∞. xn xn 5. If x is in R, then → 0 and → 0. yn zn Proof Note that the conditions in the different parts of the theorem do not allow 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 is an n1 in N such that xn > x/2 for all n ≥ n1 . Let α > 0. Since yn → ∞, there exists an n2 in N with yn > 2α/x for all n ≥ n2 . Let n0 = max{n1 , n2 }. Then n ≥ n0 implies that xn yn > (x/2)yn > (x/2)(2α/x) = α. Thus, xn yn → ∞. Let β < 0. Since zn → −∞, there exists an n3 in N with zn < 2β/x for all n ≥ n3 . Then, since zn < 0, n ≥ max{n1 , n3 } implies that xn zn < (x/2)zn < (x/2)(2β/x) = β. Thus, xn zn → −∞. 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 xn yn > yn and xn zn < zn . It follows that xn yn → ∞ and xn zn → −∞. 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 . Then n ≥ n1 implies that |xn /zn | ≤ B/|zn | < B(ε/B) = ε. The rest of the proof is left as an exercise.

Subsequences We 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 Definition 3.13.

Section 3.7

Limits at Infinity

63

To obtain the analogue of Proposition 3.3, we first negate the statements preceding Example 3.24. For (xn )n∈N a sequence in R, lim xn  = ∞ ⇐⇒ there exists an α > 0 such that for all n0 in N, there

n→∞

exists an n ≥ n0 with xn ≤ α ⇐⇒ ∃ α > 0 such that ∀ n0 ∈ N, ∃ n ≥ n0 with xn ≤ α

and lim xn  = −∞ ⇐⇒ there exists a β < 0 such that for all n0 in N, there

n→∞

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. lim xn  = ∞ if and only if (xn )n∈N has a subsequence that is bounded above; n→∞

2. lim xn  = −∞ if and only if (xn )n∈N has a subsequence that is bounded n→∞ below. Proof We prove part 1, leaving the proof of part 2 to the reader. Suppose that lim xn  = ∞. Then there exists an α > 0 such that for all n→∞

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 )∞ is never in (α, ∞) and hence (xnk )∞ k=1 k=1 cannot 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, the sequence (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 Sequences The 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 a limit in R# . Proof If (xn )n∈N is bounded, then (xn )n∈N has a limit in R by Theorem 3.7. Suppose (xn )n∈N is monotone increasing but unbounded. Then (xn )n∈N is unbounded 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 N such that xn0 > α. Since (xn )n∈N is monotone increasing, α < xn0 ≤ xn for all n ≥ n0 . Hence, xn → ∞. Similarly, if (xn )n∈N is monotone decreasing but unbounded, then xn → inf{xn : n ∈ N} = −∞.

64

Chapter 3

Sequences The reader should observe that the sequence in Exercise 4 in Section 3.4 has limit ∞. Remark If (xn )n∈N is a sequence in R, then (xn )n∈N has a monotone subsequence by Theorem 3.9. By Theorem 3.16, this monotone subsequence has a limit in R# ; and by Theorem 3.7, this limit is in R if (xn )n∈N is bounded. These observations are deduced at the beginning of the next section from a different point of view.

Exercises 1. Use the method of Examples 3.24 and 3.25 to establish the following limits. √ √ (a) lim n = ∞ (d) lim (n − 6 n) = ∞ n→∞ n→∞ √ √ n2 + 1 (b) lim =∞ (e) lim − n − 7 = −∞ √ n→∞ n→∞ n (c) lim (n2 − 6n + 1) = ∞ (f) lim (−n + sin n) = −∞ n→∞

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 lim xn = 0 if and only if lim

n→∞

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 a limit 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 (xn yn )n∈N converges. Show that yn → 0. 6. Let (xn )n∈N and (yn )n∈N be sequences of positive real numbers and suppose that xn /yn → L where 0 < L < ∞. Show that xn → ∞ if and only if yn → ∞. [Hint: Note that, eventually, 21 L < xn /yn < 23 L.] 7. Let (xn )n∈N and (yn )n∈N be sequences of positive real numbers and suppose that 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 with limit ∞; (b) if (xn )n∈N is unbounded below, then (xn )n∈N has a subsequence with limit −∞. [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 that A contains a monotone increasing sequence with limit α and a monotone decreasing sequence with limit β. [Hint: By Exercise 9 in Section 3.4, you need 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 and that A contains a monotone decreasing sequence with limit β. [Note: If α = −∞, 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 in R# = R ∪ {±∞}. If (xn )n∈N is bounded, then (xn )n∈N has a subsequence that converges 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 → x for some subsequence (xnk )∞ k=1 of (xn )n∈N }. k

Thus, E consists of all subsequential limits of (xn )n∈N in the extended reals and E is never empty. The limit superior of (xn )n∈N , denoted by lim sup xn , and the limit inferior of (xn )n∈N , denoted by lim inf xn , are n→∞

n→∞

given by lim sup xn = sup E n→∞

and lim inf xn = inf E. n→∞

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 in Definition 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

66

Chapter 3

Sequences E = {∞}, then inf E = ∞. If E  = {∞}, then inf E is a real number if E is bounded below and −∞ if E is not bounded below. Example 3.26 Let (xn )n∈N = (1, 21 , 2, 13 , 3, 41 , 4, 15 , . . .). 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# . Then xn → 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 Theorem 3.15. 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   ≥ ε for all k in N. By the first paragraph (xnk )∞ of (x ) such that x − x n n∈N nk k=1 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 is bounded above. This subsequence is either unbounded below or bounded below. Hence this subsequence has a subsequence with limit −∞ (Exercise 8 in Section 3.7) 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 . For the sake of clarity, we will postpone justifying this statement until Proposition 3.9 at the end of this section; for now we will use this as a given fact. Remark The advantage of the limit superior and limit inferior is that, given a sequence (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 no sense to write lim xn . A new way to show that (xn )n∈N has a limit is to show that lim sup xn = lim inf xn . This is illustrated in Exercise 5. For the purpose of the next proposition and some of the exercises, we extend 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 ). By the 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 → γ . The sequence (ank )k=1 has k # a subsequence (ank )∞ j =1 with limit u in R where u < ∞ since α  = ∞, and j

∞ # the sequence (bnk )∞ k=1 has a subsequence (bnk )j =1 with limit v in R where j

v < ∞ since β  = ∞. Then (ank + bnk ) → γ and (ank + bnk ) → u + v. By j j j j j j the uniqueness of limits, γ = u + v. Since α and β are the largest subsequential limits 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 and lim 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 → −∞ by Proposition 3.7 and so n 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 limit yi . To construct a subsequence of (xn )n∈N with limit α, we basically use a diagonalization 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 go directly down the diagonal. If each yi is real, we can choose x1,nk from row 1, x2,nk from row 2, 1

2

x3,nk from row 3, etc., with nki < nkj for i < j (we move to the right as we 3     go down the rows) and xi ,nk − yi  < 1/ i for each i in N. Then (xi ,nk )∞ is i i i=1 a subsequence of (xn )n∈N with limit α. Suppose ym = ∞ for some m in N. Then yi = ∞ for all i ≥ m. Choose xi ,nk from each row i such that nki < nkj for i < j with xi ,nk > i for all i i i ≥ m. Then (xi,nk )∞ i=1 is a subsequence of (xn )n∈N with limit α = ∞. i

68

Chapter 3

Sequences

Exercises 1. Find lim sup xn and lim inf xn if xn is given by (a) (−1)n , (b) (−1)n n,

1 n , (c) (−1) 1 + n

(d) cos(nπ ),

1 nπ (e) 1 + sin , n 2 (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 and lim 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 x1 + x2 + · · · + xn nx − n n (x1 − x) + · · · + (xn0 − x) (xn0 +1 − x) + · · · + (xn − x) = + n n

yn − x =

and, given ε > 0 and suitably choosing n0 ,

 

|x1 − x| + · · · + xn0 − x  n − n0 |yn − x| ≤ + ε. n n

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, . . .).]