Is a finite Hausdorff space necessarily discrete?
closed as offtopic by Qiaochu Yuan, Andreas Blass, Emil Jeřábek, Dima Pasechnik, Alex Degtyarev Mar 29 '15 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 $\left\{ x\right\}$. 


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. 


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. 


protected by Qiaochu Yuan Mar 29 '15 at 21:44
Thank you for your interest in this question.
Because it has attracted lowquality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site.
Would you like to answer one of these unanswered questions instead?