Open Extension Space is Compact

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 a compact space.

By definition of open extension space, the only open set of $T^*_{\bar p}$ containing $p$ is $S^*_p$.

So any open cover $\CC$ of $T^*_{\bar p}$ must have $S^*_p$ in it.

So $\set {S^*_p}$ will be a subcover of $\CC$, whatever $\CC$ may be.

And $\set {S^*_p}$ (having only one set in it) is trivially a finite cover of $T^*_{\bar p}$.

Hence the result, by definition of compact space.

$\blacksquare$

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