Condition for Open Extension Space to be T0 Space

Theorem

Let $T = \struct {S, \tau}$ be a topological space.

Let $T^*_{\bar p} = \struct {S^*_p, \tau^*_{\bar p} }$ be the open extension space of $T$.

By definition:

$\tau^*_{\bar p} = \set {U: U \in \tau} \cup \set {S^*_p}$

Let $T = \struct {S, \tau}$ be a $T_0$ space.

Let $\forall x, y \in S$ such that $x \ne y$.

First, suppose $x \ne p \ne y$.

Then by definition of $T$ as a $T_0$ space:

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

or:

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

Then as $U \in \tau^*_{\bar p}$, it follows that:

$\exists U \in \tau^*_{\bar p}: x \in U, y \notin U$

or:

$\exists U \in \tau^*_{\bar p}: y \in U, x \notin U$

Now suppose that $y = p$.

Then by definition:

$\forall U \in \tau^*_{\bar p}: x \in U, p \notin U$

unless $U = S^*_p$.

So we have shown that:

$\forall x, y \in S^*_p$ such that $x \ne y$, either:

$\exists U \in \tau^*_{\bar p}: x \in U, y \notin U$

or:

$\exists U \in \tau^*_{\bar p}: y \in U, x \notin U$

and so $T^*_{\bar p}$ is a $T_0$ space.

Now suppose that $T = \struct {S, \tau}$ is not a $T_0$ space.

Let $x, y \in S$ such that:

$\forall U \in \tau: x \in U \implies y \in U$

$\forall U \in \tau: y \in U \implies x \in U$

As $U \in \tau^*_{\bar p}$ it follows that:

$\forall U \in \tau^*_{\bar p}: x \in U \implies y \in U$

$\forall U \in \tau^*_{\bar p}: y \in U \implies x \in U$

and if $U = S^*_p$ the same applies.

Hence $T^*_{\bar p}$ is not a $T_0$ space.

$\blacksquare$

1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $16$. Open Extension Topology: $9$