Is a finite Hausdorff space necessarily discrete?
closed as offtopic by Qiaochu Yuan, Andreas Blass, Emil Jeřábek, Dima Pasechnik, Alex Degtyarev Mar 29 at 22:16This question appears to be offtopic. The users who voted to close gave this specific reason:



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 $\left\{ x\right\}$. 


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. 


Yes. Better, it works for T1, too: T1 is the axiom that onepoint 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. 


protected by Qiaochu Yuan Mar 29 at 21:44
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