$X$ is normal $\iff$ for all $B,C$ closed, $B\cap C=\varnothing,$ there exists $U$ open s.t. $B\subseteq U$ and $\overline U\cap C=\varnothing$

Question

Prove that $X$ is normal
$\iff$ for all $B,C$ closed, $B\cap C=\varnothing,$ there exists $U$ open s.t. $B\subseteq U$ and $\overline U\cap C=\varnothing.$

I've been asked to prove that a space $X$ is normal if and only if whenever $B$, $C$ are disjoint closed subspaces of $X$, there exists an open set $U$ containing $B$ and disjoint from $C$ such that $\bar{U} \cap C = \emptyset$.

I think if I can prove that $B$ and $C$ can be separated by neighbourhoods, that would help, but I'm struggling with that too.

I think it is axiom $T_4$.

Please help, thanks.

Answer 1

Suppose $X$ is normal. Let $U,V$ be disjoint open neighborhoods of $B$ and $C$. Suppose for sake of contradiction that $x \in \bar{U} \cap C$. Since $U \cap C \subseteq U \cap V = \varnothing$, we see that $x$ must be a limit point of $U$. But this is also impossible: since $x \in C \subseteq V$, one can find a neighborhood of $x$ that lies entirely in $V$.

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For the other direction, suppose $\bar{U} \cap C = \varnothing$. Consider $V := (\bar{U})^c$.

Answer 2

If $X$ is normal, and $B,C$ are closed and disjoint we can find $O_1, O_2$ open and disjoint so that $B \subseteq O_1, C \subseteq O_2$. But then define $U=O_1$ and note that $O_1 \subseteq (X\setminus O_2) \subseteq X\setminus C$ so as the middle set is closed we see that $\overline{U} = \overline{O_1} \subseteq X\setminus O_2 \subseteq X \setminus C$ and it follows that $\overline{U} \cap C = \emptyset$.

For the converse, if for $B$,$C$ disjoint we can find such a $U$ with $B \subseteq U$ with $\overline{U} \cap C = \emptyset$, we define $V = X\setminus \overline{U}$ which is open (as closures are closed) and also note that $V$ contains $C$ and (as $U \subseteq \overline{U}$) we have $U \cap V = \emptyset$. So $X$ is normal.