1549901369-Elements_of_Real_Analysis__Denlinger_

112 Chapter 2 • Sequences

Thus, we see that the cluster points are 0, 1, and +oo. D

BOLZANO-WEIERSTRASS THEOREM

The following theorem has profound consequences, which we shall see in

later sections and chapters. It should be regarded as of major importance.

Theorem 2.6.16 (Balzano-Weierstrass Theorem for Sequences) Every

bounded sequence has a convergent subsequence.

Proof. Suppose { Xn} is a bounded sequence. Then ::3 A , B E JR 3 t:/n E N,

A::::; Xn::::; B. We construct a subsequence {xnk} of {xn} as follows.

Let Xn 1 = X1.

Consider the closed interval Ii = [a 1 , bi] = [A, B]. One of the two halves

of this interval, [ a 1 , ai ; bi] or [ ai ; bi , b 1 ] must contain Xn for infinitely

many n, and hence must contain some Xn 2 where n2 > n 1. Pick one of the

half-intervals that does and call it 12 = [a2, b2]· Note: b2 - a2 = Hb1 - a1).

Consider the two halves of this interval, [a 2 , a^2 ; b^2 ] and [ a^2 ; b^2 , b 2 ].

One of them must contain Xn for infinitely many n, hence must contain some Xn 3

where n3 > n2. Pick one of the half-intervals that does and call it 13 = [a 3 , b 3 ].

Note: b3 -a3 = ~(b2 -a2) = ~(b1 - a1).

Continuing in this way, we develop a sequence of closed intervals {h} where

Ii 2 12 2 · · · 2 l k 2 · · · , and each interval h = [ak, bk] contains Xnk, with

n1 < n2 < · · · < nk < nk+1 <···.Thus, {xnk} is a subsequence of {xn}·

Moreover, each interval h has length

1

bk - ak = --(b1 2k-l - a1).

Thus, lim (bk - ak) = lim [ k

1

k->oo k->oo 2 _^1 (b1 - a1)] = 0.

Therefore, we can apply Cantor's Nested Intervals Theorem to conclude

that there is a unique real number L such that

00

k=l
Claim: lim Xnk = L.
k->oo
Proof: 'Ilk E N, L E h. But also, Xnk E h. Since Xnk and L both belong
to the interval h , lxnk - LI ::::; (bk - ak), the "length" of h. But, as observed
above, (bk - ak) -> 0. Therefore, by the squeeze principle,

lim Xnk = L.

k->oo

Thus, {xn} has a convergent subsequence, {xnk}. •