Theorem
A spaceis regular if and only if, given
and a neighbourhood
of
with
there is a neighbourhood
of
such that
Proof
Supposeis regular. Let
represent an open neighbourhood of
then
is closed and
Hence open setsand
exist such that
and
Sincewe have
and since
we have
Hence
Now supposeand
is an open neighbourhood of
Then an open neighbourhood
of
exists such that
Letand 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
Henceand
and
are the required sets and
is regular.