5
$\begingroup$

Is a finite Hausdorff space necessarily discrete?

$\endgroup$
1
  • 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 Trimble
    Mar 30, 2015 at 1:34

3 Answers 3

17
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ LOL, that was funny (I mean about the walnuts...) $\endgroup$ May 11, 2010 at 7:16
14
$\begingroup$

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.

$\endgroup$
10
$\begingroup$

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

$\endgroup$

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