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

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.

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.
• @Unknownx $D^2$ where $D$ is the so-called Double Arrow space is a compact Hausdorff space that contains the Sorgenfrey plane as a subspace. – Henno Brandsma Feb 12 at 17:53