Theorem
A closed subset
of a compact space
is compact.
Proof
Let
be a closed subset of a compact space
and let
be an open cover of
so that![]()
We have![]()
Since
is closed
is open and
is an open cover of![]()
X is compact so the open cover
is reducible to a finite subcover, sa![]()
Since
and
are disjoint we obtain![]()
Hence the open cover
is reducible to a finite subcover
and
is compact.