Open Extension Topology is not T3

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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 $T_3$ space.


Proof

As $S$ is (trivially) open in $T$, it is also open in $T^*_{\bar p}$.

As $S^*_p = S \cup \left\{{p}\right\}$, it follows that $\left\{{p}\right\}$ is closed in $T^*_{\bar p}$.

But by definition, the only open set of $T^*_{\bar p}$ that contains $\left\{{p}\right\}$ is $S^*_p$ itself.

So there can be no $x \in S^*_p: x \notin \left\{{p}\right\}$ contained in an open set of $T^*_{\bar p}$ which is disjoint from $S^*_p$.

Hence the result.

$\blacksquare$


Sources