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Is a finite Hausdorff space necessarily discrete?

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closed as off-topic by Qiaochu Yuan, Andreas Blass, Emil Jeřábek, Dima Pasechnik, Alex Degtyarev Mar 29 at 22:16

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Seems a little harsh to retroactively pile on downvotes on a question that was welcomed in the first days of MO's existence. It was a more innocent time. :-) – Todd Trimble Mar 30 at 1:34

3 Answers 3

Yes. Let $X$ be finite and Hausdorff. It is enough to show that every point $x$ in $X$ is open. For every point $y$ different from $x$, there is an open neighborhood $U_{y}$ of $x$ not meeting $y$. The intersection of the $U_{y}$'s is open and equals $\left\{ x\right\}$.

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Just thought of another answer to this. Any topology on a finite set is compact. Any map from a discrete topology is continuous. Hence by the famous theorem on maps from compact spaces into Hausdorff spaces, the identity map on a finite space is a homeomorphism from the discrete topology to the given Hausdoff topology.

A certain phrase involving sledgehammers and walnuts springs to mind, though.

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LOL, that was funny (I mean about the walnuts...) – Carlo Von Schnitzel May 11 '10 at 7:16

Yes. Better, it works for T1, too: T1 is the axiom that one-point sets are closed.

Then since the set is finite, the complement of any point is also closed; the point is open. That's the discrete topology.

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protected by Qiaochu Yuan Mar 29 at 21:44

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