Open Extension Topology is not Perfectly T4
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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 $T$.
Then $T^*_{\bar p}$ is not a perfectly $T_4$ space.
Proof
By definition:
- $\tau^*_{\bar p} = \set {U: U \in \tau} \cup \set {S^*_p}$
We have that $S$ is an open set in $T$ and so open set in $T^*_{\bar p}$.
So $\set p = S^*_p \setminus S$ is closed in $T^*_{\bar p}$.
The only open set in $T^*_{\bar p}$ which contains $p$ is $S^*_p$.
So $\set p$ 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.
$\blacksquare$
Sources
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- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $16$. Open Extension Topology: $9$