Theorem
A closed subsetof a compact spaceis compact.
Proof
Letbe a closed subset of a compact spaceand letbe an open cover ofso that
We have
Sinceis closedis open andis an open cover of
X is compact so the open coveris reducible to a finite subcover, sa
Sinceandare disjoint we obtain
Hence the open coveris reducible to a finite subcoverandis compact.