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.