Theorem
A compact set is countably compact.
Proof
Suppose
is compact. Let be a subset of
with no accumulation points in
Each point
belongs to some open set
which contains at most one point of
Consider the family of sets
We have
Hence
is an open cover ofSinceis compact, a finite subcover
exists with
Since each
contains at most one point of
is finite. Therefore every infinite subset ofcontains an accumulation point in
so thatis countably compact.