1550251515-Classical_Complex_Analysis__Gonzalez_

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106 Chapter^2

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14. If Ai and A 2 are starlike regions with respect to the same point z 0 ,


prove that both Ai U A 2 and Ai n A 2 are starlike with respect to z 0 •


  1. Show that the set JR.^2 - Q^2 is connected.


2.11 COMPACT SETS

In this section we consider another class of subsets of a metric space called
compact sets. This class has special importance in complex analysis since
there are many properties that hold good on compact subsets only. Several
equivalent definitions of compactness in metric spaces have been proposed.
However, it has been found most convenient to base the definition on that
of open covering. This approach was first proposed (for general topological
spaces) by Alexandroff and Urysohn in 1929 [2].


Definitions 2.34 Let (S, d) be a metric space. A family of open sets
{Ga : a E I} is an open covering of a set A C S if A is contained in the
union of the sets Ga, i.e., if A C UaEI Ga.
A subcovering is a subfamily with the same property. A finite
subcovering is one that consists of a finite number of sets.


Definition 2.35 A set A C S is compact iff every open covering of A
contains a finite subcovering. In particular, this definition applies to the
space S itself.
We remark that in the definition of a compact subset the covering


UaEI Ga is by open sets of S. However, each set Gan A is open in -A

(relatively open) and each open set in A (A regarded as a subspace) can be
expressed in the form Ga n A with Ga open in S. Hence it is immaterial
in what sense the definition is taken. -
The property embodied in the definition -of compactness is known as
the Heine-Borel property. It is the generalization to metric spaces (or to
topological spaces; see Section 2.13) of the following theorem, first used,
implicitly by E. Heine (1872) and proved by E. Borel in 1895, for closed
finite intervals in JR.i (a particular kind of compact set): Any covering of
the closed interval [a, b] by open intervals contains a finite subcovering.
Clearly, every finite subset of a metric space is compact.


Theorem 2.14 A compact subset of a metric space is closed and bounded.


Proof Let A be a compact subset of a metric space (S, d), and let b E A'.
For each a E A let v =^1 / 2 d( a, b ). Then we have


Nv(a) n Nv(b) = 0
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