Theorem
A spaceis regular if and only if for every
and every open neighbourhood
of
contains a closed neighbourhood
of with
Proof
Supposeis regular and let
be a point of
Let
be an open neighbourhood of
then
is closed and
Hence open setsand
exist such that
Sincewe have
Since
we have
Hence
Now supposeand
is an open subset of
with
An open neighbourhood
of
exists with
Letand let
be any closed subset of
with
is an open neighbourhood of
then an open set
exists with
and
is open and
so
Henceand
are the required open sets and
is regular.