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Is a finite Hausdorff space necessarily discrete?

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3 Answers 3

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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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. 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 {x}.

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