1550251515-Classical_Complex_Analysis__Gonzalez_

Topology of Plane Sets of Points

then for any point ( x, y) E Q k we have

d[(x, y), (a*, b*)] :::; d[(x, y), ( ak, bk)]+ d[( ak, bk), (a*, b*)]

./2l 1

<-+-8<8 2k 2

109

Therefore, Qk C Ns((a*, b*)) C Gcx 0 , so just one set from {Gcx} suffices to

cover Qk(k > I<). Since this contradicts the definition of Qk, the square

Q is compact.

Definition 2.36 A family of closed sets {Fcx : a EI}, where the index

set I is infinite, is said .to have the finite inter.section property iff for every
finite subset Io of I the set hcxEio Fcx is not empty.
A different but equivalent characterization of compactness depends on
the finite intersection property, as the following theorem shows:

Theorem 2.17 Let (S, d) be a metric space. A subset A of Sis compact

iff for· every infinite family {Fcx : a EI} of closed sets of A with the finite

i.ntersection property, the set ncxEJ Fcx is not empty.

Proof 1. Suppose A is compact and that {Fcx} is an infinite family of closed
sets of A with the finite intersection property. Assume that

By taking complements with respect to A (considered as a subspace), we
obtain

LJF~=A

ex EI

It follows that {F~} is an infinite open covering of A. Since A is compact,

there exists a finite subcovering of A, i.e., there is Io finite, Io CI, such that

LJ F~=A

cxEio

Taking complements again, we get

which is a contradiction since the family {Fcx} of closed sets has the finite
intersection property. Hence ncxEI Fcx is not empty.

2. Suppose that A is not compact. Then there exists an infinite open
   covering { G ex} (relatively open) that contains no finite sub covering. Thus

for every finite Io c I, UaEio Ga is a proper subset of A, so naEio G~