MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Is a finite Hausdorff space necessarily discrete?

share|cite|improve this question

closed as off-topic by Qiaochu Yuan, Andreas Blass, Emil Jeřábek, Dima Pasechnik, Alex Degtyarev Mar 29 '15 at 22:16

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "MathOverflow is for mathematicians to ask each other questions about their research. See Math.StackExchange to ask general questions in mathematics." – Andreas Blass, Emil Jeřábek, Dima Pasechnik, Alex Degtyarev
If this question can be reworded to fit the rules in the help center, please edit the question.

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. :-) – Todd Trimble Mar 30 '15 at 1:34

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

share|cite|improve this answer

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.

share|cite|improve this answer

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.

share|cite|improve this answer
LOL, that was funny (I mean about the walnuts...) – Carlo Von Schnitzel May 11 '10 at 7:16

protected by Qiaochu Yuan Mar 29 '15 at 21:44

Thank you for your interest in this question. Because it has attracted low-quality 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?

Not the answer you're looking for? Browse other questions tagged or ask your own question.