In: Advanced Math
( X, τ ) is normal if and only if for each closed
subset C of X and each open
set U such that C ⊆ U, there exists an open set V satisfies C ⊆ V
⊆clu( V) ⊆ U
Let be a
normal topological space. Using the same notation as that in
question note that
and
(the
complement of
in
) are disjoint closed
subsets of
(since
) and hence there exist open sets
and
of
so that
and
Now, we claim that
For, let if possible
Since
is
open and
it
follows that
which is a contradiction to
above and hence
and this implies that
Conversely, let be disjoint
closed subsets of
. Now, as
by the hypothesis there exists open set
such that
and if we take
then
is open and
and
and this shows that
is a
normal topological space.