Condition for Open Extension Space to be T0 Space

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Theorem

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

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


Then $T^*_{\bar p}$ is a $T_0$ (Kolmogorov) space if and only if $T$ is.


Proof

By definition:

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


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

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

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

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.

$\blacksquare$


Sources