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Consider the topological space $(X,\tau)$ with $X=\{a,b,c\}$ and $\tau=\{\emptyset ,X,\{b\},\{a,b\},\{b,c\}\}$. This space is compact if every open cover contains a finite subcover. But there are a lot of open covers to consider. Is there a way to prove this space is compact without writing out every cover and showing each has a finite subcover?

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    $\begingroup$ Since $X$ if finite, every cover is finite... $\endgroup$ – J.-E. Pin Aug 9 '15 at 12:07
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All the open covers are finite, so they are themselves finite subcovers.

Note, you can see every open cover is finite without writing them all out. If $\mathcal{U}$ is an open cover of $X$, $\mathcal{U} \subseteq \tau$; as $\tau$ is finite, $\mathcal{U}$ is finite.

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Since $\tau$ has finitely many elements, every open cover of $X$ is also finite. Therefore $X$ is compact.

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