Separation Axioms on Double Pointed Topology/T3 Axiom

Theorem
Let $T = \left({S, \tau}\right)$ be a topological space.

Let $D = \left({\left\{{a, b}\right\}, \vartheta}\right)$ be the indiscrete topology on two points.

Let $T \times D$ be the double pointed topology on $T$.

Then $T \times D$ is a $T_3$ space iff $T$ is also a $T_3$ space.

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

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

Then $F' = F \times \left\{{a, b}\right\}$ or $F' = F \times \varnothing$ by definition of the double pointed topology.

If $F' = F \times \varnothing$ then $F' = \varnothing$ from Cartesian Product Null, and the result is trivial.

So suppose $F' = F \times \left\{{a, b}\right\}$.

By the definition of the product topology it follows that $F$ is closed in $T$.

Let $y' = \left({y, q}\right) \in \complement_{S'} \left({F'}\right)$.

Then $y \notin F$.

Suppose that $T$ is a $T_3$ space.

Then by definition:
 * For any closed set $F$ of $T$ and any point $y \in S$ such that $y \notin F$ there exist disjoint open sets $U, V \in \tau$ such that $F \subseteq U$, $y \in V$.

Then $y' \in V \times \left\{{a, b}\right\}$ and $F' \subseteq U \times \left\{{a, b}\right\}$ and:
 * $U \times \left\{{a, b}\right\} \cap V \times \left\{{a, b}\right\} = \varnothing$

demonstrating that $T \times D$ is a $T_3$ space.

Now suppose that $T \times D$ is a $T_3$ space.

Then $\exists U', V' \in S': y' \in V'$ and $F' \subseteq U'$ such that $U' \cap V' = \varnothing$.

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

for some $U, V$ which are open in $T$.

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

It follows that $F$ and $y$ fulfil the conditions that make $T$ a $T_3$ space.

Hence the result.