Cauchy Sequence

73

Patterns of Proof in Analysis

CHAPTER 6: CAUCHY SEQUENCES

Study Guide The problem in this chapter is to relate convergence of sequences with completeness of the real numbers by showing that there are enough numbers that sequences which "ought to" converge have a number to converge to. As a first step, we must find a way to characterize sequences that "ought to" converge in terms of their internal structure without having to establish their limit. Then we must prove that convergent sequences have the properties that characterize sequences which "ought to" converge. And, finally, we must prove that if a sequence "ought to" converge, then there is a number to serve as its limit.

After participating in class discussions, completing the readings, and doing the assigned and recommended homework, you should be able to do the following:

I. Disprove the following CONJECTURE: Suppose {an}n is a sequence of real numbers. If for every > 0 there exists a positive

integer N such that for every n > N, | an+1 – an | < , then {an}n converges.

(Are unbounded sequences the only counterexamples?)

II. State and prove the converse of the Conjecture in I. III. A. Define the phrase "{an}n is Cauchy sequence."

B. Prove from the definition that a given sequence of the form

{

An BCnD

}n (C > 0, D ≥ O) is a Cauchy sequence.

C. Prove from the (negation of) the definition that a given (divergent) sequence is not a Cauchy sequence.

IV. Prove that if a sequence {an}n converges, then {an}n is a Cauchy sequence.

V. Prove that if {an}n is a Cauchy sequence and its set of terms is finite, then {an}n is convergent.

VI. State and prove a theorem relating Cauchy sequences and bounded sequences. State the converse of this theorem and prove or disprove it.

VII. Define the phrase "x is an accumulation point for a set S of real numbers " in two different, but logically equivalent, ways.

VIII. A. Give examples of sets of real numbers which have given numbers as accumulation points or have a specified number of accumulation points.

B. Does an accumulation point for a set always belong to the set? If your answer is "yes," give a proof. If your answer is "no," give examples.

IX. Formulate and investigate conjectures relating convergence of a sequence with existence of accumulation points for its set of terms.

X. State and prove a theorem that gives a sequential characterization of accumulation points. Use this theorem to prove other results.

Chapter 6 Cauchy Sequences


74

XI. Prove the following result relating lub(E) with accumulation points. Proposition: Suppose E is a non-empty set of real numbers and is bounded above.

If x = lub(E), then either x is in E or x is an accumulation point of E (or both). Be able to state and prove this result and to use it to prove other results.

XII. A. Write a complete statement of the Bolzano-Weierstrass Theorem. B. Give examples to show that both hypotheses must be satisfied to assure the existence of an

accumulation point. C. State the converse of the Bolzano-Weierstrass Theorem and prove or disprove it.

XIII. Prove that if {an}n is a Cauchy sequence and its set of terms is infinite, then {an}n is convergent.

XIV. A. Write a complete statement of a theorem relating convergence of sequences with the Cauchy property.

B. Explain the role of each of the preliminary results included above in the proof. C. Explain how the definition of Cauchy sequence and this theorem are a solution to the (two part)

problem posed in the introduction to this study guide. D. I'll not ask you to reproduce the entire proof of this major result. However, you may be asked to

state and prove one or more of the preliminary results that are part of its proof (other than the Bolzano-Weierstrass Theorem).

XV. Be able to use the concepts and proof techniques developed to this point in the course to investigate conjectures and prove results you have not seen before analogous to those in the exercise sets.



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THE PROBLEM

The problem is to relate convergence of sequences with completeness of the real numbers by showing that there are enough numbers that sequences which "ought to" converge have a number available to serve as their limit. As a first step, we must find a way to characterize sequences that "ought to" converge in terms of their internal structure without having to establish their limit. Then we must prove that convergent sequences have the properties that characterize sequences which "ought to" converge and that sequences which "ought to" converge actually do converge. In other words, our problem is to discover some internal property of sequences that allows us to prove a theorem like this:

THEOREM: A sequence {an}n of real numbers [has some nice property that we have not discovered yet] if and only if {an}n converges.

From visualizing convergent sequences on the number line, we see that terms in a tail of a convergent sequence must be close to each other. This leads us to make the following

CONJECTURE: Suppose {an}n is a sequence of real numbers. If for every > 0 there exists a positive integer N such that for every n > N, | an+1 – an | < , then {an}n converges.

EXERCISES

1. Show that the sequence {

n }n is a counterexample for this conjecture.

2. Show that the sequence { ln(n) }n is a counterexample for this conjecture.

3. Are all the counterexamples to this conjecture unbounded sequences?

4. Prove the converse of the conjecture.

The conjecture seems very reasonable, but "for every > 0 there exists a positive integer N such that for every n > N, | an+1 – an | < " is not the nice property we are looking for. The problem is that this property

only requires successive terms in a tail of the sequence to be close to each other. We need all the terms in a tail to be close to each other. With this in mind, we make the following definition.

DEFINITION: We say that a sequence {an}n of real numbers is a Cauchy sequence to mean that for every

> 0 there exists a positive integer N such that for every pair of positive integers m and n greater than N, | an – am | < .

CAUCHY SEQUENCES: APPLYING THE DEFINITION

Example 1: Prove from the definition that {

3n 79n 2

}n is a Cauchy sequence.

SOLUTION:

We must prove that for every > 0 there exists a positive integer N such that for every positive integers

n, m > N we have |

3n 79n 2

–

3m 79m 2

| < .

Consider > 0, arbitrary.

Choose N to be an integer greater than 46

27. (This expression for N was found from scratch work.)



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Consider positive integers n, m > N, arbitrary. For such an n and m,

|

3n 79n 2

–

3m 79m 2

| = |

(27nm 6n 63m14) (27mn 6m 63n 14)(9n 2)(9m 2)

|

= |

69m 69n(9n 2)(9m 2)

| <

6981

|m n |mn

<

6981m nmn

=

2327

(1n

1m

) <

2327

( 1N

1N

) =

4627N

< QED

Example 2: Prove from the (negation of the) definition that {an}n = { (-1)n

n 1n

}n is not a Cauchy

sequence. Include a drawing which shows geometrically that the you chose will work.

SOLUTION:

We must prove that there exists > 0 such that for every positive integer N there exists m, n > N such that

| an – am | = | (-1)n

n 1n

– (-1)m

m 1m | ≥ .

Set = 2. [Note from the drawing that there are terms with large index that are more than 2 units apart.]

Consider positive integer N, arbitrary.

Set n = 2N and m = 2N + 1.

Then n > N and m > N as required. Furthermore,

| an – am | = | (-1)n

n 1n

– (-1)m

m 1m

|

= | (-1)2N

2N 12N

– (-1)2N+1

(2N 1)1(2N 1)

| = |

2N 12N

+

(2N 1)1(2N 1)

| = | 1 +

12N

+ 1 +

12N 1

| = 2 +

12N

+

12N 1

> 2 = . QED

Example 3: Prove from the (negation of the) definition that { an }n = { ln(n)}n is not a Cauchy sequence.

Solution: We must prove that there exists > 0 such that for every positive integer N there exists m, n > N such that

| an – am | = | ln(n) – ln(m) | ≥ .

Set = 1. (Since the sequence is unbounded, there are terms with large index arbitrarily far apart. As a result, any number could be used as .)

Consider positive integer N, arbitrary. Set m = N + 1. Choose n to be a positive integer greater than e(1 + ln(m)). Clearly, m > N. Also, since the function f(x) = ex is increasing,

n > e(1 + ln(m)) ≥ e ln(m) = m > N. Since the natural logarithm is an increasing function and n > m , ln(n) > ln(m) so ln(n) – ln(m) > 0. Also, since n > e(1 + ln(m)) we have ln(n) > ln e(1 + ln(m)) = 1 + ln(m). It follows that



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| ln(n) – ln(m) | = ln(n) – ln(m) > ln e(1 + ln(m)) – ln(m)

= (1 + ln(m)) – ln(m) = 1 = . QED

These examples suggest (correctly) being a Cauchy sequence is the internal property of sequences that characterizes convergence of a sequence. The following result lends further credence to this suggestion.

THEOREM: If { an }n is a convergent sequence of real numbers, then { an }n is a Cauchy sequence. PROOF: Suppose { an }n is a convergent sequence of real numbers.

We will prove that { an }n is a Cauchy sequence by showing that for every > 0 there exists a positive integer N such that for every positive integers n, m > N we have | an – am | < .

Consider > 0, arbitrary. Since { an }n converges, there exists A such that { an }n converges to A. Choose such an A. Now,

there exists a positive integer N such that for every n > N we have | an – A | < Choose such an N.

Consider n, m > N, arbitrary.

Since n > N, | an – A | < 2 and since m > N, | am – A | <

2 . From the triangle inequality,

| an – am | = | ( an – A ) + ( A – am ) | ≤ | an – A | + | A – am | < 2 +

2 ≤ .

Therefore, | an – am | < .

It follows that { an }n is a Cauchy sequence. QED

EXERCISES

1. Make up several sequences of the form {

An BCnD

}n (C > 0, D ≥ O) and prove from the definition

that each one is a Cauchy sequence.

2. Make up several sequences of the form { (-1)n

An BCnD

}n (A ≠ 0, C > 0, D ≥ O) and prove from the

negation of the definition that each one is not a Cauchy sequence.

3. Prove that {

n }n is not a Cauchy sequence.

4. Make up several unbounded sequences and prove from the negation of the definition that each one is not a Cauchy sequence.

5. Prove the following assertion directly from the definition of Cauchy sequence. CONJECTURE: If { xn }n and { yn }n are Cauchy sequences, then { xn + yn }n is a Cauchy sequence.

6. Prove or disprove the following assertion. CONJECTURE: If { xn }n and { yn }n are Cauchy sequences, then their merge { zn }n is a Cauchy

sequence. (The merge of two sequences is defined on page 53.)

7. Prove the following assertion. PROPOSITION: Suppose { xn }n and { yn }n are sequences of real numbers. If { xn }n is a Cauchy sequence and for every > 0 there exists positive integer N such that for every n > N | xn – yn | < , then { yn }n is a Cauchy sequence



78

CONVERGENCE OF CAUCHY SEQUENCES I

Our task now is to prove the converse of the theorem that convergent sequences are Cauchy sequences. The converse says that sequences which "ought to" converge -- that is, Cauchy sequences -- have a number to converge to. The converse is another way of saying that the real numbers are complete. Completeness is a fundamental and profound property of the real numbers. Proving another characterization of a fundamental property of the real numbers sounds difficult. It is. We must prove several preliminary results and develop a new idea to prove the converse. We'll construct the argument block by block as illustrated in the following figure.

We first prove convergence of Cauchy sequences of the simplest kind -- those with a finite number of distinct terms.

THEOREM: If { an }n is a Cauchy sequence and there exists a number that appears an infinite number of times as a term of the sequence, then { an }n converges.

PROOF: Suppose { an }n is a Cauchy sequence and there exists a number that appears an infinite number of times as a term of { an }n .

We'll prove that { an }n converges by showing that there exists a real number A such that for every > 0 there exists a positive integer N such that for every n > N we have | an – A | < .

By hypothesis, there exists a number that appears an infinite number of times as a term of { an }n .

Choose such a number as A.

Consider > 0, arbitrary. Since {an}n is a Cauchy sequence, there exists a positive integer N such that for every pair of positive integers n, m > N we have | an – am | < . Choose such a positive integer N.

Consider an arbitrary positive integer n > N.

Because appears an infinite number of times as a term of the sequence, there exists m > N such that am = . Choose such an m. Now m > N and n > N, so it follows that

| an – A | = | an – | = | an – am | < .

The conclusion follows.



79

COROLLARY: If { an }n is a Cauchy sequence and the set S of terms of the sequence is finite, then { an }n converges.

PROOF: Suppose { an }n is a Cauchy sequence and the set S = {an | n a positive integer} is finite.

Since there is a term of { an }n corresponding to each positive integer and since there are only a finite

number of numbers that are terms of the sequence, there must be a number from S that corresponds to an infinite number of positive integers. This number appears an infinite number of times as a term of the Cauchy sequence { an }n . It follows from the immediately preceding theorem that { an }n converges. QED

EXERCISES

1. Construct several different examples of sequences which have S = { 3, 52 ,

154 ,

103 } as their set of

terms. Do any of these sequences converge? Do any diverge? Are any of them Cauchy sequences?

2. In the proof that every Cauchy sequence that has a finite number of distinct terms converges (the Corollary to the Theorem), it is argued that there must be a member of the set S of terms of the sequence that appears an infinite number of times as a term of {an}n .

Could several members of the set of terms of a Cauchy sequence appear an infinite number of times as a term of {an}n?

Could several members of the set of terms of a sequence that is not a Cauchy sequence appear an infinite number of times as a term of {an}n?

3. Saying that there exists a number that appears an infinite number of times as a term of the sequence {an}n is the same as saying that there exists a real number such that for every positive integer K there exists k > K such that ak = . Explain why.

4. Prove or disprove the following assertion. CONJECTURE: If {an}n is a Cauchy sequence and there exists a real number such that for every positive integer K there exists k > K such that ak = , then the set S of terms of {an}n is a finite set.

5. Suppose S = { 3, 52 ,

154 ,

103 } is the set of terms of the sequence {an}n. Construct the set

D = { | ai – aj | : ai ≠ aj } = {d : d is the distance between distinct terms of the sequence {an}n }.

(D should consist of 6 different positive numbers.)

(a) What is the minimum of D ? Does a set D constructed in this way from a sequence with a finite number of distinct terms always have a minimum? Convince a classmate.

(b) Does a set D constructed in this way from an arbitrary sequence (which may have an infinite number of distinct terms) always have a minimum? Does it ever have a minimum? What if the sequence converges? Convince a classmate.

6. Prove the following results.

(a) PROPOSITION: If {an}n is a Cauchy sequence and S = { 3 , 52 ,

154 ,

103 } is the set of terms of

{an}n , then {an}n is eventually constant. (b) If {an}n is a Cauchy sequence and the set S of terms of {an}n is finite, then {an}n is eventually

constant.



80

ACCUMULATION POINTS

Think about what a proof of convergence of Cauchy sequences with an infinite number of distinct terms would require. The proof might begin like this:

Suppose { an }n is a Cauchy sequence and the set S = {an | n a positive integer} is infinite. We'll prove

that { an }n converges by showing there exists a real number A such that for every > 0 there exists a

positive integer N such that for every n > N we have | an – A | < .

The first (and hardest) problem is establishing the existence of a number A which is a candidate to be the limit. We can't expect to "serve up" a specific A. We must give a blueprint for building A from the materials at hand -- from the set S of terms of the sequence {an}n. Intuition suggests that the terms of {an}n will "pile

up" around any number which might be the limit of the sequence, so we might hope to build A as the "pile up" point of the set of terms of the sequence. We need to make this intuitive idea of the elements of a set "piling up" or "accumulating" around some number more precise.

DEFINITION Suppose that S is a set of real numbers. We say that a number is a strong accumulation point of S to mean that for every > 0 there exists an infinite number of members s of S such that | s – | < .

A less demanding mathematician might ask less of an accumulation point and formulate the following

DEFINITION Suppose that S is a set of real numbers. We say that a number is a weak accumulation point of S to mean that for every > 0 there exists a member s of S such that 0 < | s – | < .

PROPOSITION Suppose that S is a set of real numbers and is a number. The number is a strong accumulation point of S if and only if is a weak accumulation point of S.

Proof: (Remember, an "if and only if" assertion is a combination of two assertions that usually must be proven separately.)

First note that if is a strong accumulation point of S then is obviously a weak accumulation point of S. [If this is not obvious to you, write a complete proof that begins with "set the stage" and "plan" paragraphs.]

Next prove that if is a weak accumulation point of S, then is a strong accumulation point of S. We will argue contrapositively and prove that if is not a strong accumulation point of S then is not a weak accumulation point of S.

Suppose is not a strong accumulation point of S. We will prove that is not a weak accumulation point of S by showing that there exists > 0 such that for every member s of S\{} | s – | ≥ . Since is not a strong accumulation point of S, there exists > 0 such that there are only a finite number of members of S such that | s – | < . Choose such an . [We use instead of to avoid confusion with in the next sentence.] If there are no members of S other than such that | s – | < , set = . If there are members of S other than such that | s – | < , denote these members of S as s1, s2, … , sn. (Keep in mind that there is only a finite number of such members of S and none of them are

equal to ) Now, set = min[ , | s1 – | , | s2 – | , … , | sn – | ].

Then, > 0 and for every member s of S different from , | s – | ≥ . QED



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Because there is no logical difference between the two kinds of accumulation points we defined, we will drop the adjectives weak and strong and simply refer to accumulation points. Even though there is no logical difference between weak and strong characterizations of accumulation points, there is a psychological difference. If you were faced with the challenge of proving that a number is an accumulation point of a set of real numbers, which characterization would you prefer to use? Probably weak accumulation point. If you were going to use the fact that a certain number is an accumulation point of a set of real numbers to prove something else, which characterization would you prefer? Probably strong accumulation point.

EXERCISES

1. Prove that if a set S of real numbers has an accumulation point, then S is not the empty set.

2. Prove that if a set S of real numbers has an accumulation point, then S is an infinite set (i.e. has an infinite number of members).

3. Prove the following result. Proposition: Suppose S is a set of real numbers, is a real number, A = { x S | x < } and B = { x S | x > }. If is an accumulation point of S, then is either an accumulation point of A or of B (or both).

The conjectures in problems 4 - 8 involve limits of sequences, accumulation points, and bounds of sets of real numbers. Remember, sets have accumulation points but sequences do not. However, the set of terms of a sequence can have accumulation points. (a) Prove or disprove each conjecture. (b) Formulate and investigate similar conjectures that you form by making simple, reasonable changes in the

hypotheses and conclusions of these conjectures.

4. If a sequence {an}n converges to a number A, then A is an accumulation point of the set S of terms of

the sequence.

5. If a number A is an accumulation point of the set S of terms of the sequence {an}n , then {an}n

converges to A .

6. If a number A is the only accumulation point of the set S of terms of the sequence {an}n , then {an}n

converges to A .

7. If a sequence {an}n is bounded, then the set S of terms of the sequence has an accumulation point.

8. If a set S of real numbers is bounded above, then S has an accumulation point.

9. Prove the following result which relates the limits of certain sequences with accumulation points. PROPOSITION: Suppose {an}n converges to A. If for every N there exists n > N such that an ≠ A,

then A is an accumulation point of the set S of terms of the sequence.

10. This proposition relating the least upper bound of a set of numbers with accumulation points of the set will be used to prove the Extreme Value Theorem. Prove this result.

PROPOSITION: Suppose S is a nonempty set of real numbers and is bounded above. If x = lub(S), then either x is in S or x is an accumulation point of S. (My favorite approach is to exploit the language and restate the proposition so the conclusion does not involve "or". I can think of several ways to do this. The different restatements lead to different proofs.)



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To avoid an empty set fallacy and successfully carry out the proof that we planned in the paragraphs that began this discussion of accumulation points, we must know that the set of terms of a Cauchy sequence which has an infinite number of distinct terms has an accumulation point. (Remember, we want to use this accumulation point as the limit of the sequence.) The following theorem, which is equivalent to completeness of the real numbers, provides the tool we need to be sure that the accumulation point we believe should exist really does.

THE BOLZANO-WEIERSTRASS THEOREM: Every bounded infinite set of real numbers has at least one accumulation point.

PROOF Suppose S is a bounded infinite set of real numbers.

We will prove that there exists a number which is an accumulation point of S by constructing the number as the common limit of two monotone sequences and then showing that the number constructed is an accumulation point of the set S.

Since S is bounded, S is bounded below and bounded above. Choose a1 to be a lower bound for S and

b1 to be an upper bound for S. Now set c1 =

a1 b12

. Since S is infinite, either the interval [ a1 , c1 ] or the

interval [ c1 , b1 ] (or both) contains an infinite number of elements of S. If [ a1 , c1 ] contains an infinite number of elements of S, set a2 = a1 and b2 = c1. If not, set a2 = c1 and b2 = b1. Then [ a2 , b2 ] contains

an infinite number of elements of S, a1 ≤ a2 < b2 ≤ b1, and b2 – a2 =

b1 a1

2. Now, set c2 =

a2 b2

2. Since

[ a2 , b2 ] contains an infinite number of elements of S, either the interval [ a2 , c2 ] or the interval [ c2 , b2 ] (or both) contains an infinite number of elements of S. If [ a2 , c2 ] contains an infinite number of elements of S, set a3 = a2 and b3 = c2. If not, set a3 = c2 and b3 = b2. Then [ a3 , b3 ] contains an infinite number of

elements of S, a2 ≤ a3 < b3 ≤ b2 , and b3 – a3 =

b2 a2

2 =

b1 a1

4. Continue in this way. Having

constructed a1, a2, … , ak and b1, b2, … , bk construct ak+1 and bk+1 like this. Set ck =

ak bk2

. Since

[ ak , bk ] contains an infinite number of elements of S, either the interval [ ak , ck ] or the interval [ ck , bk ] (or both) contains an infinite number of elements of S. If [ ak , ck ] contains an infinite number of elements of S, set ak+1 = ak and bk+1 = ck. If not, set ak+1 = ck and bk+1 = bk.

In this way, we define two sequences { an }n and { bn }n which have the following properties: (i) for every n, an < bn, (ii) { an }n is increasing and { bn }n is decreasing, and

(iii) for every n, bn – an =

b1 a1

2n1 .

The sequences { an }n and { bn }n are bounded and monotone so, by the Monotone Convergence Theorem, they converge to numbers and . Choose such numbers and . By the Sum Theorem, { bn – an }n

converges to – . Since we also have that { bn – an }n = {

b1 a1

2n1 }n, { bn – an }n also converges to 0.

By the Uniqueness Theorem, – = 0 so = .

We will prove that is an accumulation point of S by showing that for every > 0 there are an infinite number of elements s of S such that | s – | < .



83

Consider > 0, arbitrary. Since { an }n converge to , there exists a positive integer N1 such that for every n > N1, | an – | = – an < . Choose such an N1. Since { bn }n converge to , there exists a positive integer N2 such that for every n > N2, | bn – | = bn – < . Choose such an N2. Set N = max[N1, N2]. Since N + 1 > N,

– < aN+1 < bN+1 < +

so the interval [ aN+1 , bN+1 ] is contained in the interval ( – , + ). The interval [ aN+1 , bN+1 ] was

constructed so that it contains an infinite number of elements of the set S. Consequently, there is an infinite number of elements s of the set S in the interval ( – , + ) or, to say it another way, there is an infinite number of elements s of the set S such that | s – | < . It follows that is an accumulation point of S. QED

EXERCISES

1. Prove or disprove the following conjecture. If a set S of real numbers does not have any accumulation points, then S is either finite or unbounded.

2. Prove or disprove the following conjecture. If a sequence {an}n is bounded and the set S of terms of {an}n has the number L as its only accumulation point, then {an}n converges to L.

3. The following conjecture is false. (a) Disprove the conjecture. (b) Find the error in the purported proof.

Conjecture If {an}n is bounded and its set of terms is infinite, then {an}n converges.

Purported Proof: The set S of terms of {an}n is bounded and infinite. Hence, by the Bolzano-

Weierstrass Theorem, S has an accumulation point. Choose an accumulation point of S as A. Consider > 0 arbitrary. Since A is an accumulation point of the set S of terms of {an}n , there is an infinite number of members of S, and hence an infinite number of terms ak of {an}n, such that |ak – A| < . Hence, there are an infinite number of positive integers k such that | ak – A | < . Choose a positive integer k as N. Then | aN – A | < .

Consider n > N, arbitrary. Since there is an infinite number of terms ak of {an}n such that | ak – A | < and aN is one of those terms, there is an infinite number of positive integers n greater than N such that |an – A| < . Since n > N, it follows that | an – A | < . QED

BOUNDEDNESS OF CAUCHY SEQUENCES In order to apply the Bolzano-Weierstrass Theorem and infer the existence of an accumulation point that can serve as the limit of a Cauchy sequence, we must know that the set of terms of a Cauchy sequence is bounded.

THEOREM: If {an}n is a Cauchy sequence of real numbers, then {an}n is bounded.

PROOF: Suppose {an}n is a Cauchy sequence.

We will prove that the sequence {an}n is bounded by showing that there exists a number B such that for every positive integer k, | ak | < B.

From the definition of Cauchy sequence (used with = 1), there exists a positive integer N such that for every n, m > N we have | an – am | < 1. Choose such an N. Set



84

B = max {1 + | aN+1|, | a1 |, | a2 |, | a3 |,… , | aN | }.

Consider positive integer k arbitrary.

Either k ≤ N or k > N. In the case k ≤ N, | ak | is one of the numbers used to specify B, so

| ak | ≤ max { 1 + | aN+1|, | a1 |, | a2 |, | a3 |,… , | ak |,… , | aN | } = B.

In the case k > N, both k > N and N + 1 > N, so, from the selection of N, we have that | ak – aN+1 | < 1.

Consequently, from the triangle inequality for absolute values,

| ak | ≤ | ak – aN+1 + aN+1 | ≤ | ak – aN+1 | + | aN+1 | < 1 + | aN+1 | .

Now 1 + | aN+1 | is one of the numbers used to specify B, so again

| ak | < 1 + | aN+1 | ≤ max { 1 + | aN+1|, | a1 |, | a2 |, | a3 |,… , | ak |,… , | aN | } = B.

Thus, regardless of the size of k, | ak | ≤ B.


EXERCISES

1. The Boundedness Theorem for Cauchy Sequences is similar to the Boundedness Theorem for Convergent Sequences. It could be proved as a consequence of preliminary results (lemmas) just as the Boundedness Theorem for Convergent Sequences was. (a) Formulate and prove those lemmas. (b) Then use the lemmas from (a) to give a proof of the Boundedness Theorem for Cauchy Sequences that is less technically complicated than the one above.

2. Prove the Boundedness Theorem for Convergent Sequences by an argument similar to the proof of the Boundedness Theorem for Cauchy Sequences given above (without relying on a preliminary lemmas).

3. Prove the Boundedness Theorem for Cauchy Sequences by a contrapositive argument.

CONVERGENCE OF CAUCHY SEQUENCES WHICH HAVE AN INFINITE NUMBER OF DISTINCT TERMS

THEOREM: If {an}n is a Cauchy sequence and the set S of terms of the sequence is infinite, then {an}n converges.

PROOF: Suppose that {an}n is a Cauchy sequence and the set S of terms of the sequence is infinite.

We will prove that the sequence {an}n converges by showing that there exists a number A such that for every > 0 there exists a positive integer N such that for every n > N we have | an – A | < .

Since, by a previously established theorem, every Cauchy sequence is bounded, the set S of terms of the sequence is bounded. By hypothesis, S is infinite. By the Bolzano-Weierstrass Theorem, the bounded, infinite set S has at least one accumulation point. Choose an accumulation point of S as A.


Since {an}n is a Cauchy sequence, there exists a positive integer N such that for all n, m > N we have

| an – am | < 2 . Choose such a positive integer N.



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Consider an arbitrary positive integer n > N.

Since A is an accumulation point of S, there is an infinite number of members of S, and hence an infinite

number of terms am of the sequence {an}n , such that | am – A | < 2 . Thus, there exists m > N such that

| am – A | < 2 . Choose such an m. Since n > N and m > N, we have | an – am | <

2 .

It follows that

| an – A | = | an – am + am – A | ≤ | an – am | + | am – A | < 2 +

2 ≤ .


EXERCISES

1. (a) Where does the reasoning involved in this proof that a Cauchy sequence with an infinite number of distinct terms converges fail when the sequence has only a finite number of distinct terms? (b) Conversely, where does the reasoning involved in the proof that Cauchy sequences that have a finite set of terms converges fail when the sequence has an infinite number of distinct terms?

2. (a) The Bolzano-Weierstrass Theorem guarantees the existence of at least one accumulation point. Many infinite sets have more than one accumulation point. Give examples of sets of real numbers that have more than one accumulation point.

(b) In the proof that a Cauchy sequence with an infinite number of distinct terms converges, what if the set of terms of the sequence has more than one accumulation point? (Formulate and prove a proposition that addresses this situation.)

SEQUENTIAL CHARACTERIZATION OF ACCUMULATION POINTS Sequences are our best tool for studying limiting processes. To make effective use of this tool, we need connections between sequences and other ideas related to limiting processes. The following result characterizes accumulation points in terms of sequences. Inductive construction, illustrated for the first time in the proof of the Bolzano-Weierstrass Theorem and used again here, is an important and much-used technique for establishing the existence of a sequence having specific properties. This result is important, but the technique of inductive construction illustrated in its proof may be even more important.

DEFINITION: Let A and B denote two sets of real numbers. By the difference of A and B, denoted A \B, we mean the set of all numbers that are members of A but not members of B. Thus, when E is a set of real numbers and xo is a number (that might or might not be a member of E), we write E \ { xo } to mean all members of E different from xo. When xo is not a member of E, the sets E and E \ { xo } are identical.

THEOREM: Suppose E is a set of real numbers. A number xo is an accumulation point of E if and only if there exists a sequence { xn }n of members of E, each distinct from xo such that { xn }n converges to xo.

PROOF: Suppose E is a set of real numbers.

First, we will prove that if there exists a sequence { xn }n from the set E \ { xo } that converges to xo, then xo is an accumulation point of E.



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Suppose there exists a sequence { xn }n from E \ { xo } that converges to xo. Choose such a sequence { xn }n.

We will prove that xo is an accumulation point of E by showing that for every > 0 there exists a member x of E, distinct from xo, such that | x – xo | < .


Since { xn }n converges to xo, there exists a positive integer N such that for every n > N, | xn – xo | < . Choose such an N. Set x = x

N+1. Since the terms of { xn }n are members of E distinct

from xo and xN+1 is a term of the sequence, x is distinct from xo. Now, N + 1 > N, so, from the selection

of N, 0 < | x – xo | = | x

N+1 – xo | < .

It follows that xo is an accumulation point of E.

Second, we prove that if xo is an accumulation point of E then there exists a sequence { xn }n of members of E, each distinct from xo, that converges to xo.

Suppose xo is an accumulation point of E.

We will prove there exists a sequence { xn }n of members of E, each distinct from xo, that converges to xo by giving an inductive construction of such a sequence.

Since xo is an accumulation point of E, there exists a member x of E, distinct from xo, such that | x – xo | < 1. Choose such a member of E as x1. Since xo is an accumulation point of E, there exists a

member x of E, distinct from xo, such that | x – xo | < 12 . Choose such a member of E as x2. Continue in

this way. After x1, x2 , … , xn have been chosen, choose xn+1 from the members of E different from xo

so that | xn+1 – xo | <

1n 1

. In this way we obtain a sequence { xn }n of members of E distinct from xo.

We'll prove that { xn }n converges to xo by showing that for every > 0 there exists a positive integer N such that for every n > N we have | xn – xo | < .


Choose positive integer N > 1 .

Consider n > N, arbitrary.

From the construction of the sequence { xn }n it follows that

| xn – xo | <

1n

<

1N

< .

Thus { xn }n converges to xo .

QED



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EXERCISES 1. Prove the following stronger version of the second part of the theorem by modifying the inductive

construction of the sequence { xn }n in the proof above. THEOREM: Suppose E is a set of real numbers. If xo is an accumulation point of E then there exists a

sequence {xn}n of members of E \ {xo}, each distinct from every other, such that { xn }n converges to xo.

2. Prove the following result. THEOREM: Suppose E is a set of real numbers. If xo is an accumulation point of a set E then there exists a sequence {xn}n of members of E \ {xo} such that { xn }n converges to xo and for every n | xn+1 – xo | < | xn – xo | .

3. Prove or disprove the following conjecture. CONJECTURE: Suppose xo is an accumulation point and also the greatest lower bound of a set E of real numbers. If { xn }n is a strictly decreasing sequence of members of E, then { xn }n converges to xo.

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