Is a finite Hausdorff space necessarily discrete?
put on hold as offtopic by Qiaochu Yuan, Andreas Blass, Emil Jeřábek, Dima Pasechnik, Alex Degtyarev yesterdayThis question appears to be offtopic. The users who voted to close gave this specific reason:



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. 


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


protected by Qiaochu Yuan yesterday
Thank you for your interest in this question.
Because it has attracted lowquality answers, posting an answer now requires 10 reputation on this site.
Would you like to answer one of these unanswered questions instead?