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Let be $X$ a Hausdorff space where every open subspace of $X$ is compact. I need to prove that $X$ is finite.

As $X$ is Hausdorff, I have that for every distinct elements of $X$, exist disjoint neighborhoods in $X$ and I try to use the set of all these neighborhoods as an open cover of $X$. I think I need to use that $X$ is compact, and the fact that some pair of sets of this open cover are disjoint to prove the finiteness of $X$ but I could finish the proof.

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This seems to rely on the following two facts about Hausdorff spaces.

  • All singleton sets are closed. (This holds for the wider class of T1-spaces.)
  • All compact subsets are closed.

So given $x \in X$, $X \setminus \{ x \}$ is open, hence compact by assumption, hence closed by the above. Therefore the singleton $\{ x \}$ is open. As all singletons are open, $X$ is discrete. As $X$ is an open subset of itself, it is compact, and now note that the only compact discrete spaces are the finite ones.

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