# Equivalent condition for normal topological space

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?

• How do you know that $B$ is compact? – John Douma Dec 17 '18 at 17:16
• You're right, I don't know this, do you have a hint to prove the implication anyways? – Leonardo Dec 17 '18 at 17:24
• Your definition of normal is wrong: in any Hausdorff space we can separate disjoint compact subsets by open sets. So your "normal" is just Hausdorff. Normal is usually defined as being able to separate two disjoint closed subsets of $X$, maybe in combination with Hausdorff, that depends. – Henno Brandsma Dec 17 '18 at 17:49

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 $$\overline{U_A}\subset U.$$
• $\forall b\in B=X\setminus U \implies \exists U_b$ 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 $\bar{A} \subseteq U$? – Leonardo Dec 17 '18 at 21:59
• You're correct, just a typo on the end: $b\in X\setminus \overline{U_A}$ and therefore $\overline{U_A}\subset U.$ – positrón0802 Dec 17 '18 at 22:11