# Separation Axioms on Double Pointed Topology/T4 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 $T_4$ space if and only if $T$ is also a $T_4$ space.

## Proof

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

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

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

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

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

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 $A \times \left\{{a, b}\right\} \subseteq U \times \left\{{a, b}\right\}$ and $B \times \left\{{a, b}\right\} \subseteq V \times \left\{{a, b}\right\}$ and:

$U \times \left\{{a, b}\right\} \cap V \times \left\{{a, b}\right\} = \varnothing$

demonstrating that $T \times D$ is a $T_4$ 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 $U' \cap V' = \varnothing$.

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 $U, V \subseteq T$.

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

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

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

Hence the result.

$\blacksquare$