Open Extension Topology is not Perfectly T4

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

This article is complete as far as it goes, but it could do with expansion. In particular: There are a number of missing results from this section of S&S. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by adding this information. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Expand}} from the code. If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.

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