Open Extension Space is Compact

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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 a compact space.


Proof

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

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

So $\left\{{S^*_p}\right\}$ will be a subcover of $\mathcal C$, whatever $\mathcal C$ may be.

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

Hence the result, by definition of compact space.

$\blacksquare$


Sources