# closed Hausdorff neighbourhoods, a theorem on

If $X$ is a topological space in which every point has a closed Hausdorff neighbourhood, then $X$ is Hausdorff.

Note. In this theorem (and the proof that follows) neighbourhoods are not assumed to be open. That is, a neighbourhood of a point $x$ is a set $A$ such that $x$ lies in the interior of $A$.

Proof of theorem. Let $X$ be a topological space in which every point has a closed Hausdorff neighbourhood. Suppose $a,b\in X$ are distinct. It suffices to show that $a$ and $b$ have disjoint neighbourhoods. By assumption, there is a closed Hausdorff neighbourhood $N$ of $b$. If $a\notin N$, then $X\setminus N$ and $N$ are disjoint neighbourhoods of $a$ and $b$ (as $N$ is closed).

So suppose $a\in N$. As $N$ is Hausdorff, there are disjoint sets $U_{0},V_{0}\subseteq N$ that are open in $N$, such that $a\in U_{0}$ and $b\in V_{0}$. There are open sets $U$ and $V$ of $X$ such that $U_{0}=U\cap N$ and $V_{0}=V\cap N$. Note that $U$ is a neighbourhood of $a$, and $V$ is a neighbourhood of $b$. As $N$ is a neighbourhood of $b$, it follows that $V\cap N$ (that is, $V_{0}$) is a neighbourhood of $b$. We have $U\cap V_{0}=U_{0}\cap V_{0}=\varnothing$. So $U$ and $V_{0}$ are disjoint neighbourhoods of $a$ and $b$. QED.

Title closed Hausdorff neighbourhoods, a theorem on ClosedHausdorffNeighbourhoodsATheoremOn 2013-03-22 18:30:46 2013-03-22 18:30:46 yark (2760) yark (2760) 7 yark (2760) Theorem msc 54D10