Open Extension Topology is not Perfectly T4

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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 not a perfectly $T_4$ space.


Proof

By definition:

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


We have that $S$ is an open set in $T$ and so open set in $T^*_{\bar p}$.

So $\left\{{p}\right\} = S^*_p \setminus S$ is closed in $T^*_{\bar p}$.

The only open set in $T^*_{\bar p}$ which contains $p$ is $S^*_p$.

So $\left\{{p}\right\}$ can not be the intersection of open sets of $T^*_{\bar p}$, whether that intersection be countable or not.

So by definition $T^*_{\bar p}$ is not a perfectly $T_4$ space.

$\blacksquare$


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