1550251515-Classical_Complex_Analysis__Gonzalez_

114 Chapter^2

Fig. 2.rn

Proof Suppose that n::'=l Fn = 0. This implies that U::'=i F~ = s. Be-
cause S :is compact, it is the union of a finite number of the open sets F~.

But F{ c F~ c F~ c · · ·, so that we have S = F~ for some n. Thus we

get Fn = 0, a contradiction.

Corollary 2.2 If 6.(Fn) , 0 as n , oo, then there is just one point that

belongs to all the Fn.

Example Figure 2.10 shows a nest of squares each of which (after the
first) has half the side of the preceding square .. The first square may be
thought as the compact space S. By the corollary there is just one point
common to all the squares.

EXERCISES 2.6

1. Show that ( C*, x) is compact.

2. If A and B are two compact subsets of a metric space, prove that AU B

and A n B are compact.

3. If A and B are disjoint compact subsets of a metric space, prove that

d(A, B) > 0. Also,' prove that there exists disjoint open subsets G 1 and

G2 such that A C G1 and B C G2.

4. Show that every bounded sequence of complex numbers has a convergent
   subsequence.


5. Let A be a nonempty open set and B a compact subset of A. Show that

there is a compact subset J( with the property BC J( CA, J( "# B.

6. Consider the metric space (I, d), where

I = { x : x E JR, 0 ::S: x ::S: 1}

and d denotes the usual metric in R Clearly, the subspace (K, d) where

J{ = {x: x E Q, 0:::;: x:::;: 1} is not compact_ (Theorem 2.14). Find an

open covering of K with no finite subcovering.