# Finite Hausdorff spaces [closed]

Is a finite Hausdorff space necessarily discrete?

• 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 $\left\{ x\right\}$.