Theorem
A space
is regular if and only if for every
and every open neighbourhood
of
contains a closed neighbourhood
of with![]()
Proof
Suppose
is regular and let
be a point of
Let
be an open neighbourhood of
then
is closed and![]()
Hence open sets
and
exist such that![]()
Since
we have
Since
we have![]()
Hence![]()
Now suppose
and
is an open subset of
with
An open neighbourhood
of
exists with![]()
Let
and let
be any closed subset of
with![]()
is an open neighbourhood of
then an open set
exists with
and![]()
is open and![]()
so![]()
Hence
and
are the required open sets and
is regular.