Compact subsets of a Hausdorff space are closed
Let
Proof
Let
. For each assign an open neighbourhood of and likewise an open neighbourhood of such that for all (the Hausdorff property). Then is an open cover of and thus has a finite subcover . Then is an open subset of . Since every has an open neighbourhood , is open, whence is closed.
Similarly, Closed subsets of a compact space are compact.