Some compact subspaces Let’s say we have a compact space . A
subset may not be itself compact, but there’s one useful case in
which it will be. If is closed, then is compact. Let’s take an
open cover of . The sets are open subsets of , but they may not
be open as subsets of . But by the definition of the subspace
topology, each one must be the intersection of with an open
subset of . Let’s just say that each is an open subset of to
begin with. Now, we have one more open set floating around. The
complement of is open, since is closed! So between the
collection and the extra set we’ve got an open cover of . By
compactness of , this open cover has a finite subcover. We can
throw out from the subcover if it’s in there, and we’re left
with a finite open cover of , and so is compact. In fact, if we
restrict to Hausdorff spaces, must be closed to be compact.
Indeed, we proved that if is compact and is Hausdorff then any
point can be separated from by a neighborhood . Since there is
such an open neighborhood, must be