$X$ is normal if and only if $A\subseteq U$ implies there exists $V$ such that $A\subseteq V\subseteq\overline{V}\subseteq U$

Question

I want to prove this: $X$ is normal if and only if for every closed $A$ and open $U$ such that $A\subseteq U$, there exists $V$ such that $A\subseteq V\subseteq\overline{V}\subseteq U$.

Suppose $X$ is normal. Let $A\subseteq U$ where $A$ is closed and $U$ is open. Then $X\setminus U$ is closed and $(X\setminus U)\cap A=\emptyset$. But since $X$ is normal, there exist open sets $V$ and $V'$ such that $A\subseteq V$, $X\setminus U\subseteq V'$, and $V\cap V'=\emptyset$. Clearly $V\subseteq \overline{V}$, and also $V\subseteq U$ because $V\cap V'=\emptyset$.

But how does that guarantee the closure of $V$ is also contained in $U$?

Conversely, suppose for each $A\subseteq U$ with $A$ closed and $U$ open, there exists an open set $V$ such that $A\subseteq V\subseteq\overline{V}\subseteq U$. Let $C_1, C_2$ be closed and $C_1\cap C_2=\emptyset$. Then $X\setminus C_2$ is open and $C_1\subseteq X\setminus C_2$. By assumption, there exists an open $V$ such that $C_1\subseteq V\subseteq\overline{V}\subseteq U$, hence $V\cap C_2=\emptyset$. Similarly there exists open $V'$ such that $C_2\subseteq V'\subseteq\overline{V'}\subseteq U$ with $V'\cap C_1=\emptyset$.

But how do we get that $V\cap V'=\emptyset$?

Answer 1

For the forward direction, we let $A\subseteq U$ for some closed $A$ and open $U$. Then, $U^c$ is closed, and $A\cap U^c = \emptyset$, as you pointed out. Therefore, we have some disjoint open $V_1$ and $V_2$ such that $A\subseteq V_1$ and $U^c\subseteq V_2$. As orangeskid points out, $V_1\cap V_2 = \emptyset$ implies $V_1\subseteq V_2^c\subseteq U$, and $V_2^c$ is closed, so $\overline{V_1}\subseteq V_2^c\subseteq U$. Therefore, $A\subseteq V_1\subseteq \overline{V_1}\subseteq U$.

You might also consider an application of Urysohn's lemma: let $f : X\to [0, 1]$ be a continuous function such that $f(A) = \{1\}$ and $f(U^c) = \{0\}$. Then, consider that $S = f^{-1}((1/2, 1])$ (or $S = f^{-1}((a, 1])$ for any $0 < a < 1$) is open in $X$ and satisfies $A\subseteq S\subseteq U$. The fact that $\overline{S}\subseteq U$ follows from $f(\overline{S})\subseteq [1/2, 1]$ (which implies that $\overline{S}\cap U^c = \emptyset$).

For the backward direction, we let $C_1$ and $C_2$ be two disjoint closed subsets of $X$ as you did. Then, we consider some open $V$ such that $C_1\subseteq V\subseteq \overline{V}\subseteq C_2^c$. As $(\overline{V})^c$ is open, is disjoint from $V$, and contains $C_2$, we have disjoint open neighborhoods $V\supseteq C_1$ and $(\overline{V})^c\supseteq C_2$, and therefore $X$ is normal.

Answer 2

Since $V\cap V'= \emptyset$, we have $V \subset X \backslash V'$, a closed set, so $\bar V \subset X\backslash V'$.