# 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 $U\cap {V}_{0}={U}_{0}\cap {V}_{0}=\mathrm{\varnothing}$. 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 |