Condition for Closed Extension Space to be T0 Space

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

Let $T^*_p = \left({S^*_p, \tau^*_p}\right)$ be the closed extension space of $T$.

Then $T^*_p$ is a $T_0$ (Kolmogorov) space $T$ is.

Proof
By definition:


 * $\tau^*_p = \left\{{U \cup \left\{{p}\right\}: U \in \tau}\right\} \cup \left\{{\varnothing}\right\}$

Let $T = \left({S, \tau}\right)$ be a $T_0$ space.

Then:
 * $\forall x, y \in S$ such that $x \ne y$, either:
 * $\exists U \in \tau: x \in U, y \notin U$
 * or:
 * $\exists U \in \tau: y \in U, x \notin U$

Let $x, y \in S^*_p$ such that $x \ne y, x \ne p, y \ne p$.

Then:
 * $\exists U \in \tau: x \in U, y \notin U$

or:
 * $\exists U \in \tau: y \in U, x \notin U$

Now let $U^* = U \cup \left\{{p}\right\}$.

It follows that:
 * $x \in U^*, y \notin U^*$

or:
 * $y \in U^*, x \notin U^*$

Now consider $x, p \in S^*_p$ such that $x \ne p$.

We have that $\left\{{p}\right\} \in \tau^*_p: p \in \left\{{p}\right\}, x \notin \left\{{p}\right\}$

So we have shown that:
 * $\forall x, y \in S^*_p$ such that $x \ne y$, either:
 * $\exists U \in \tau^*_p: x \in U, y \notin U$
 * or:
 * $\exists U \in \tau^*_p: y \in U, x \notin U$

and so $T^*_p$ is a $T_0$ space.

Now suppose that $T = \left({S, \tau}\right)$ is not a $T_0$ space.

Let $x, y \in S$ such that:
 * $\forall U \in \tau: x \in U \implies y \in U$
 * $\forall U \in \tau: y \in U \implies x \in U$

Now consider $U^* = U \cup \left\{{p}\right\}$ for any $U \in \tau$.

Then:
 * As $x \in U \implies y \in U$ it follows that $x \in U^* \implies y \in U^*$.


 * As $y \in U \implies x \in U$ it follows that $y \in U^* \implies x \in U^*$.

Hence $T^*_p$ is not a $T_0$ space.