Theorem
A subspace of a completely regular space is completely regular.
Proof
Let
be a completely regular space and let
be a subspace.
Let
be closed in
and let![]()

Then
for some
closed in![]()
Since
and
we have![]()
Since
is completely regular a continuous function
exists such that
and
for
Then
is continuous with
and for![]()
Then
is completely regular.