Excluded Point Space is Compact

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Theorem

Let $T = \left({S, \tau_{\bar p}}\right)$ be an excluded point space.


Then $T$ is a compact space.


Proof 1

We have:

Excluded Point Topology is Open Extension Topology of Discrete Topology
Open Extension Space is Compact

$\blacksquare$


Proof 2

By definition of excluded point space, the only open set of $T$ which contains $p$ is $S$.

So any open cover $\mathcal C$ of $T$ must have $S$ in it.

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

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

Hence the result, by definition of compact topological space.

$\blacksquare$