1549901369-Elements_of_Real_Analysis__Denlinger_

162 Chapter 3 111 Topology of the Real Number System

Theorem 3.3.19 A set A of real numbers is compact if and only if, for every
collection C of closed sets with the finite intersection property relative to A,
n C contains at least one point of A.

Proof. Part 1 ( ==>): Suppose A is a compact set of real numbers, and let C
be a collection of closed sets with the finite intersection property relative to A.
For contradiction, suppose n C contains no points of A. Then

nc~N.

Therefore, A~ (n C)c, and by de Morgan's law,

A~ u {cc: c EC}.

But then {cc : C E C} is an open cover of A. Since A is compact, A can be

covered by finitely many of these sets:

A ~ Cf U Ci U · · · UC~.

Then, (Cf U q U · · · U C~)c ~Ac. By de Morgan's law, this says

C1 n C2 n · · · n Cn ~ N,

which implies that An ( C 1 n C2 n · · · n Cn) = 0. But this intersection must be
nonempty, since C has the finite intersection property relative to A. Contradic-
tion!
Therefore, n C contains at least one point of A.

Part 2 ( <:==:): Suppose that for every collection C of closed sets with the finite
intersection property relative to A , n C contains a point of A.

For contradiction, suppose that A is not compact. Then there is an open
cover U of A that has no finite subcover of A. Define

C = {Uc : U E U}.

Then C is a collection of closed sets. Suppose {Uf, Ui,, · · · , U~} is a finite
subcollection of C. By de Morgan's law,

(1)

n
Since U has no finite subcover of A, U Ui cannot contain all of A. That is,
i=l

some point of A is in (Q
1

Ui) c. In view of (1), this says that iol Vic has a

nonempty intersection with A. Thus, C is a collection of closed sets with the
finite intersection property relative to A.