**2. SEQUENCES**

**2.4 THE BOLZANO-WEIERSTRASS THEOREM**

The Bolzano-Weierstrass Theorem is at the foundation of many results in analysis. It is, in fact, equivalent to the completeness axiom of the real numbers.

Theorem 2.4.1 — Bolzano-Weierstrass. Every bounded sequence {an} of real numbers has a convergent subsequence.

Proof: Suppose{an}is a bounded sequence. DefineA={an:n∈N}(the set of values of the

sequence{an}). IfAis finite, then at least one of the elements ofA, sayx, must be equal toanfor

infinitely many choices ofn. More precisely,Bx={n∈N:an=x}is infinite. We can then define

a convergent subsequence as follows. Pickn1such thatan1 =x. Now, sinceBxis infinite, we can

choosen2>n1such thatan2 =x. Continuing in this way, we can define a subsequence{ank}which

is constant, equal toxand, thus, converges tox.

Suppose now that Ais infinite. First observe there exist c,d∈Rsuch thatc≤an≤d for all

n∈_{N}, that is,A⊂[c,d].

We define a sequence of nonempty nested closed bounded intervals as follows. SetI1= [c,d].

Next consider the two subintervals[c,c+_{2}d]and[c+_{2}d,d]. SinceAis infinite, at least one ofA∩[c,c+_{2}d]

or A∩[c+_{2}d,d] is infinite. LetI2 = [c,c+_{2}d]if A∩[c,c+_{2}d]is infinite and I2= [c+_{2}d,d] otherwise.

Continuing in this way, we construct a nested sequence of nonempty closed bounded intervals{In}

such thatIn∩Ais infinite and the length ofIntends to 0 asn→∞.

We now construct the desired subsequence of{an}as follows. Letn1=1. Choosen2>n1such

thatan2∈I2. This is possible sinceI2∩Ais infinite. Next choosen3>n2such thatan3 ∈I3. In this

way, we obtain a subsequence{ank}such thatank∈Ikfor allk∈N.

Set In= [cn,dn]. Then limn→∞(dn−cn) =0. We also know from the proof of the Monotone

Convergence Theorem (Theorem2.3.1), that{cn}converges. Say`=limn→∞cn. Thus, limn→∞dn=

limn→∞[(dn−cn) +cn] =`as well. Sinceck≤ank≤dkfor allk∈N, it follows from Theorem2.1.5

that limk→∞ank =`. This completes the proof.

Definition 2.4.1 (Cauchy sequence). A sequence{an}of real numbers is called aCauchy sequence if for anyε>0, there exists a positive integerNsuch that for anym,n≥N, one has

|am−an|<ε.

Theorem 2.4.2 A convergent sequence is a Cauchy sequence.

Proof:Let{an}be a convergent sequence and let

lim n→∞

an=a.

Then for anyε>0, there exists a positive integerNsuch that

|an−a|<ε/2 for alln≥N.

For anym,n≥N, one has

|am−an| ≤ |am−a|+|an−a|<ε/2+ε/2=ε.

48 2.4 THE BOLZANO-WEIERSTRASS THEOREM

Theorem 2.4.3 A Cauchy sequence is bounded.

Proof:Let{an}be a Cauchy sequence. Then forε=1, there exists a positive integerNsuch that

|am−an|<1 for allm,n≥N. In particular,

|an−aN|<1 for alln≥N.

Let M=max{|a1|, . . . ,|aN−1|,1+|aN|}. Then, forn=1, . . . ,N−1, we clearly have |an| ≤M.

Moreover, forn≥N,

|an|=|an−aN+aN| ≤ |an−aN|+|aN| ≤1+|aN| ≤M. Therefore,|an| ≤Mfor alln∈Nand, thus,{an}is bounded.

Lemma 2.4.4 A Cauchy sequence that has a convergent subsequence is convergent.

Proof: Let{an}be a Cauchy sequence that has a convergent subsequence. For anyε >0, there

exists a positive integerNsuch that

|am−an| ≤ε/2 for allm,n≥N.

Let{ank}be a subsequence of{an}that converges to some pointa. For the aboveε, there exists a

positive numberKsuch that

|ank−a|<ε/2 for allk≥K.

Thus, we can find a positive integern`>Nsuch that

|an`−a|<ε/2.

Then for anyn≥N, we have

|an−a| ≤ |an−an`|+|an`−a|<ε. Therefore,{an}converges toa.

Theorem 2.4.5 Any Cauchy sequence of real numbers is convergent.

Proof: Let{an}be a Cauchy sequence. Then it is bounded by Theorem2.4.3. By the Bolzano-

Weierstrass theorem,{an}has a convergent subsequence. Therefore, it is convergent by Lemma

2.4.4.

Remark 2.4.6 It follows from Definition2.4.1that{an}is a Cauchy sequence if and only if for everyε>0, there existsN∈Nsuch that

|an+p−an|<ε for alln≥Nand for allp∈N.

Definition 2.4.2 A sequence{an}is called contractive if there existsk∈[0,1)such that

Theorem 2.4.7 Every contractive sequence is convergent.

Proof:By induction, one has

|an+1−an| ≤kn−1|a2−a1|for alln∈N.
Thus,
|an+p−an| ≤ |an+1−an|+|an+2−an+1|+· · ·+|an+p−an+p−1|
≤(kn−1+kn+· · ·+kn+p−2)|a2−a1|
≤kn−1(1+k+k2+· · ·+kp−1)|a_{2}−a_{1}|
≤ k
n−1
1−k|a2−a1|.

for alln,p∈_{N}. Since kn−1→0 as n→∞(independently of p), this implies {an}is a Cauchy

sequence and, hence, it is convergent.

Example 2.4.1 The conditionk<1 in the previous theorem is crucial. Consider the following

example. Letan=lnnfor alln∈N. Since 1<nn+2+1 <

n+1

n for alln∈Nand the natural logarithm is

an increasing function, we have

|an+2−an+1|=|ln(n+2)−ln(n+1)|= ln n+2 n+1 =ln n+2 n+1 <ln n+1 n =|ln(n+1)−lnn|=|an+1−an|.

Therefore, the inequality in Definition2.4.2is satisfied withk=1, yet the sequence{lnn}does not converge.

Exercises

2.4.1 _{I}Determine which of the following are Cauchy sequences.
(a) an= (−1)n.

(b) an= (−1)n/n. (c) an=n/(n+1). (d) an= (cosn)/n.

2.4.2 Prove that the sequence

an=

ncos(3n2+2n+1)

n+1 has a convergent subsequence.

50 2.5 LIMIT SUPERIOR AND LIMIT INFERIOR

2.4.3 Let f:[0,∞)→_{R}be such that f(x)>0 for allx. Define

an=

f(n)

f(n) +1.

Prove that the sequenceanhas a convergent subsequence.

2.4.4 Define

an= 1+2n

2n forn∈N.

Prove that the sequenceanis contractive.

2.4.5 Letr∈_{R}be such that|r|<1. Definean=rnfor n∈N. Prove that the sequence{an}is
contractive.

2.4.6 Prove that the sequence 1 n ∞ n=1 is not contractive.