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 $\CC$ of $T$ must have $S$ in it.
So $\set S$ will be a subcover of $\CC$, whatever $\CC$ may be.
And $\set S$ (having only one set in it) is trivially a finite cover of $T$.
Hence the result, by definition of compact topological space.
$\blacksquare$