Separation Axioms on Double Pointed Topology/T4 Axiom

Theorem
Let $T_1 = \struct {S, \tau_S}$ be a topological space.

Let $D = \struct {A, \set {\O, A} }$ be the indiscrete topology on an arbitrary doubleton $A = \set {a, b}$.

Let $T = \struct {T_1 \times D, \tau}$ be the double pointed topological space on $T_1$.

Then $T \times D$ is a $T_4$ space $T$ is also a $T_4$ space.

Proof
Let $S' = S \times \set {a, b}$.

Let $H' \subseteq S'$ such that $H$ is closed in $T \times D$.

Then $H' = H \times \set {a, b}$ or $H' = H \times \O$ by definition of the double pointed topology.

If $H' = H \times \O$ then $H' = \O$ from Cartesian Product is Empty iff Factor is Empty, and the result is trivial.

So suppose $H' = H \times \set {a, b}$.

From Open and Closed Sets in Multiple Pointed Topology it follows that $H$ is closed in $T$.

Suppose that $T$ is a $T_4$ space.

Then by definition:
 * For any two disjoint closed sets $A, B \subseteq S$ there exist disjoint open sets $U, V \in \tau$ containing $A$ and $B$ respectively.

Then $A \times \set {a, b} \subseteq U \times \set {a, b}$ and $B \times \set {a, b} \subseteq V \times \set {a, b}$ and:
 * $U \times \set {a, b} \cap V \times \set {a, b} = \O$

demonstrating that $T \times D$ is a $T_4$ space.

Now suppose that $T \times D$ is a $T_4$ space.

Then $\exists U', V' \in S': A' \subseteq U'$ and $B' \subseteq V'$ such that $U' \cap V' = \O$.

As $D$ is the indiscrete topology it follows that:
 * $U' = U \times \set {a, b}$
 * $V' = V \times \set {a, b}$

for some $U, V \subseteq T$.

From Open and Closed Sets in Multiple Pointed Topology it follows that $U$ and $V$ are open in $T$.

As $U' \cap V' = \O$ it follows that $U \cap V = \O$.

It follows that $A$ and $B$ fulfil the conditions that make $T$ a $T_4$ space.

Hence the result.