3.3 * Compact Sets 161
In either case, { Xn} has a subsequence converging to a point of A. Therefore,
by Theorem 3.3.13, A is compact. •Compactness is also closely related to Cantor's Nested Intervals Theorem,
as we shall see from the following results.Definition 3.3.15 A collection C of sets is said to have the finite intersec-
tion property if every finite subcollection of sets of C has nonempty intersec-
tion.
Example 3.3.16 A nested sequence of nonempty intervals Ii 2 I 2 2 · · · 2
In 2 · · ·, such as occurs in Cantor's Nested Intervals Theorem, has the finite
n
intersection property. In fact, n h , =Im, where m = max{k 1 , kz, · · · , kn}·
i=l
Theorem 3.3.17 If C is a collection of compact sets with the finite intersection
property, then n C is nonempty.
Proof. Suppose C is a collection of compact sets with the finite intersection
property. For contradiction, suppose n C = 0. Consider the family
U ={Cc: CE C}.
Since each set C in C is compact, it is closed. Hence, U is a family of open sets.
Consider a fixed set K E C. Since n C = 0, no point of K belongs to every
C in C. Thus, every point of K belongs to one of the sets cc in U. Thus, U
is an open cover of K. But K is compact. Hence, U has a finite subcover Cf,
Ci, · · ·, C~ of K; i.e.,
K ~ Cf U Ci U · · · U C~.
By de Morgan's law, this says
That is, Kn ( C 1 n C 2 n · · · n Cn) = 0. But this intersection must be nonempty,
since C has the finite intersection property. Contradiction!
Therefore, n Ci= 0. •
Definition 3.3.18 A collection C of sets is said to have the finite intersec-
tion property relative to a set A if every finite subcollection of sets of C
has nonempty intersection with A.