Another equivalent condition for being $T_1$−space $(X,\mathscr T)$ is normal.

Question

A $T_1-$space $(X,\mathscr T)$ is normal iff for each pair of disjoint closed subsets $C$ and $D$ of $X$ there are open sets $U$ and $V$ such that $C\subseteq U$, $D\subseteq V,$ and $\overline{U}\cap \overline{V}=\emptyset.$

My attempt:-

Proof. Suppose $T_1-$space $(X,\mathscr T)$ is normal. for each pair of disjoint closed subsets $C$ and $D$ of $X.$ We know that $X\setminus C$ is an open set. Additionally, we know that $D\subseteq X\setminus C$.By the theorem, there is and open set $V$ such that $D\subseteq V$ and $\overline V \subseteq X\setminus C$. Applying the same theorem. We get another open set $W$ such that $D\subseteq W$ and $\overline W \subseteq X\setminus C.$ Hence, we get $C\subseteq X\setminus \overline V$. Let $U= X\setminus \overline V$. $C\subseteq X\setminus \overline W\implies D\subseteq \overline W \subseteq V. $

We have $\overline U\cap \overline W\subseteq \overline U\cap V= \overline{X\setminus \overline{V}}\cap V=\emptyset$.($\because$ $X\setminus \overline{V} \subseteq X\setminus V$, $X\setminus V$ is closed so $\overline {X\setminus V}= X\setminus V)$. Hence $U$ and $W$ are the desired sets.

Conversely, Let $(X,\mathscr T)$ be a $T_1-$space, for each pair of disjoint closed subsets $C$ and $D$ of $X$ there are open sets $U$ and $V$ such that $C\subseteq U$, $D\subseteq V,$ and $\overline{U}\cap \overline{V}=\emptyset.$ So,$U\cap V \subseteq \overline{U}\cap \overline{V}=\emptyset.$ By the definition of normal space, $T_1-$space $(X,\mathscr T)$ is normal.

Answer 1

Slightly simpler put: If we can separate disjoint closed $C$ and $D$ with open sets with disjoint closures, then the open sets are already disjoint and we have normality.

On the other hand if we have normality and disjoint closed $C,D$ we apply normality thrice to get neighbourhoods with disjoint closure: once to get disjoint open $U_1,U_2$ separating $C$ and $D$. Then another time to find an intermediate $U$ with $C \subseteq U \subseteq \overline{U} \subseteq U_1$ and once more for $D$ and $U_2$ giving us an open $V$ with $D \subseteq V \subseteq \overline{V} \subseteq U_2$. Then the closures of $U$ and $V$ are disjoint as $U_1$ and $U_2$ are.

Remark: You cannot do this with Hausdorffness: the condition that distinct points have open neighbourhoods with disjoint closures is strictly stronger than Hausdorff.