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I want to prove the following proposition:

Let $X$ be a Hausdorff topological space such that $X$ is a normal (i.e., Hausdorff and such that disjoint closed subsets can be separated by open sets) $\Leftrightarrow$ for every closed subset $A$ of $X$, for every (open) neighbourhood $U$ of $A$ there exist an open neighbourhood $V$ of $A$ such that $\bar{V} \subseteq U$.

My attempt

I'm trying to do the right implication. Let $A$ be closed, $U$ be an open neighbourhood of $A$. Then $B=U^c$ is closed. Both $A$ and $B$ as closed subset of a Hausdorff space are therefore compact and disjoint, therefore by normality they can be separated by two disjoint open sets $U_A,U_B$. My intuition would be to try to prove that the required $V$ is $U_A$. I would only need to prove $\bar{V} \subseteq U$. I can't really go on from here, can anybody give me a hint on how to proceed?

Lilla
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1 Answers1

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The definition of $X$ being normal is that given disjoint closed sets $A$ and $B$ of $X$ there exists disjoint open sets containing $A$ and $B$ respectively.

Given $A$ closed and $U$ open containing $A,$ consider $B=X-U$ and apply the hypothesis of normality with $A$ and $B.$

Edit. Your $U_A$ works as $V$ since given $b\in B,$ $U_B$ is a neighborhood of $b$ disjoint from $U_A,$ so $\overline{U_A}\subset U.$

positron0802
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  • $\forall b\in B=X\setminus U \implies \exists U_b$ neighbourhood of $b$ that is disjoint from $U_A \implies \exists U_b$ neighbourhood of $b$ s.t. $U_b \subseteq X \setminus U_A \implies b \in \text{Ext}A=X\setminus\bar{A}$. Therefore $\bar{A} \subseteq U$? – Lilla Dec 17 '18 at 21:59
  • You're correct, just a typo on the end: $b\in X\setminus \overline{U_A}$ and therefore $\overline{U_A}\subset U.$ – positron0802 Dec 17 '18 at 22:11
  • Right, I wrote the wrong set, thank you – Lilla Dec 17 '18 at 22:14