closed Hausdorff neighbourhoods, a theorem on

Theorem. 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
Canonical name	ClosedHausdorffNeighbourhoodsATheoremOn
Date of creation	2013-03-22 18:30:46
Last modified on	2013-03-22 18:30:46
Owner	yark (2760)
Last modified by	yark (2760)
Numerical id	7
Author	yark (2760)
Entry type	Theorem
Classification	msc 54D10