closed Hausdorff neighbourhoods, a theorem on

Theorem.If $X$ is a topological space in whichevery 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 whichevery 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 $a\otin N$ , 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.

单词	ClosedHausdorffNeighbourhoodsATheoremOn
释义	closed Hausdorff neighbourhoods, a theorem on Theorem.If $X$ is a topological space in whichevery 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 whichevery 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 $a\otin N$ , 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.
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