1550251515-Classical_Complex_Analysis__Gonzalez_

112 Chapter 2

Notation If {xn}~ converges to x 0 , we write (as in r~al analysis)

n-+oo lim {xn} = Xo or, briefly, n-+oo lim Xn = Xo

Also, we write
{ Xn} --t xo or Xn --t Xo as n --t oo
Definition 2.40 A metric space is said to be sequentially compact if every
sequence in it has a convergent subsequence.
Theorem 2.20 A metric space has the Bolzano-Weierstrass property iff
it is sequentially compact.
Proof 1. Suppose that the metric space (S, d) is sequentially compact. Let
A be an infinite subset of S, and let {an} be any infinite sequence of distinct
points of A. Since S is sequentially compact, the sequence {an} contains a
subsequence { ank} that converges to a point x E S. This point x is clearly
-an accumulation point of the range of the subsequence, and since this set
is a subset of A, it follows that x is also an accumulation point of 4.

2. Now suppose that (S, d) has the Bolzano-Weierstrass property. Let

{ Xn} be any infinite sequence in S. If the range of { Xn} is finite, there is

an x E S such that Xn = x for infinitely many values of n. 'J:;he terms for

which this equality holds form a constant subsequence that is convergent.

If the range of the sequence, say B, is an infinite subset of S, it has at least

one point of accumulation b, and we can choose a subsequence of { x n} that

converges to b.

Theorem 2.21 Every compact metric space has the Bolzano-Weierstrass

property.

Proof Let (S, d) be a compact metric space and A any infinite subset of S.

Suppose that A has no point of accumulation. Then for every point x E S

there is a neighborhood N 0 ( x) that contains no point of A except possibly x

itself. The set UxES N 0 ( x) is an open covering of S, and since Sis compact

there is a finite sub covering. But each N 0 ( x) contains at most one point

of A. Hence A must be fini_te, which is a contradiction.

Theorem 2.22 If the metric space (S, d) is sequentially compact, then

it is totally bounded.

Proof Choose c: :> 0 arbitrarily, and let x 1 be some point ~f S, If

d(x, x1) < c: for all x E S, then S = Ne( xi) and S can be covered by

this c:-neighborhood of x1. If this is not so, then there exists a point x 2 E S

such that d(x 2 , xi) ~ c:. If for every point x E S either d(x, x 1 ) < e' or

d(x,x2) < c:, then S = Ne(x 1 ) UNe(x 2 ) and there are two c:-neighborhoods

covering S. However, if this is not the case, then there exists X3 E S such