Excluded Point Space is Compact/Proof 2
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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
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.
$\blacksquare$
Sources
- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $13 \text { - } 15$. Excluded Point Topology: $3$