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+2d]and[c+2d,d]. SinceAis infinite, at least one ofA∩[c,c+2d]
or A∩[c+2d,d] is infinite. LetI2 = [c,c+2d]if A∩[c,c+2d]is infinite and I2= [c+2d,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)|a2−a1| ≤ 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 IDetermine 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,∞)→Rbe 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∈Rbe 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.