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?
Best Answer
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 $\overline{U_A}\subset U.$