At Finite topological space the user asked "Let $X$ be a Hausdorff topological space such that every closed subset has finitely many connected component. How can I verify that $X$ is finite?" I want to know is this true if every proper closed subset has finitely many connected component?
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The same answer works, essentially. Suppose $X$ is infinite and Hausdorff. Then $X$ has a discrete (in itself) infinite subspace $S$. Pick $p \in S$. Then $\overline{S\setminus\{p\}}$ is proper (as $p$ is not in the closure of $S\setminus \{p\}$), closed, and has all points of $S \setminus \{p\}$ as components so has infinitely many components (they're still isolated points, so cannot be in a strictly larger connected subset), contrary to assumption. So $X$ is finite.
Henno Brandsma
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How do you find $S$ to begin with? – Jason DeVito - on hiatus Nov 16 '15 at 18:58
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Same reference as my original answer: http://math.stackexchange.com/q/533051/4280 – Henno Brandsma Nov 16 '15 at 18:59
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Thanks! It makes perfect sense now. – Jason DeVito - on hiatus Nov 16 '15 at 19:02