Is a finite Hausdorff space necessarily discrete?
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8$\begingroup$ 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. :-) $\endgroup$– Todd TrimbleCommented Mar 30, 2015 at 1:34
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3 Answers
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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.
answered Oct 9, 2009 at 7:20
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$\begingroup$ LOL, that was funny (I mean about the walnuts...) $\endgroup$ Commented May 11, 2010 at 7:16
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Yes. Better, it works for $T_1$, too: $T_1$ 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.
answered Oct 12, 2009 at 7:39
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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 $\left\{ x\right\}$.
