Theorem
A space
is regular if and only if, given
and a neighbourhood
of
with
there is a neighbourhood
of
such that![]()

Proof
Suppose
is regular. Let
represent an open neighbourhood of
then
is closed and![]()
Hence open sets
and
exist such that
and![]()
Since
we have
and since
we have![]()
Hence![]()
Now suppose
and
is an open neighbourhood of
Then an open neighbourhood
of
exists such that![]()
Let
and let
be a closed subset of
with![]()
is a neighbourhood of
Then an open set
exists such that
and![]()
is open and
and
is an open subset of
containing![]()
Hence
and
and
are the required sets and
is regular.