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

A $$T_1-$$space $$(X,\mathscr T)$$ is normal iff for each closed subset $$C$$ of $$X$$ and each open set $$U$$ such that $$C\subseteq U$$, there is an open set $$V$$ such that $$C\subseteq V$$ and $$\overline{V}\subseteq U.$$

My attempt:-

Proof. Let $$T_1-$$space $$(X,\mathscr T)$$ is normal. Let $$C$$ be a closed subset of $$X$$. Let $$U\in \mathscr T$$ such that $$C\subseteq U$$. We know that $$X\setminus U$$ is a closed subset of $$X$$ disjoint from $$C$$. By the normality, there is disjoint nonempty open sets $$V$$ and $$W$$ such that $$X\setminus U \subseteq V$$ and $$C\subseteq W$$. $$X\setminus U \subseteq V \implies X\setminus V \subseteq U.$$ $$W\subseteq X\setminus V \subseteq U.$$ Hence $$\overline{W}\subseteq X\setminus V \subseteq U$$.

Conversely, Suppose for each closed subset $$C$$ of $$T_1$$ space $$(X,\mathscr T)$$ and each open set $$U$$ such that $$C\subseteq U$$, there is an open set $$V$$ such that $$C\subseteq V$$ and $$\overline{V}\subseteq U.$$ Consider two closed disjoint subsets of $$X$$. We know that $$X\setminus C$$ is an open set. Additionally, we know that $$D\subseteq X\setminus C$$. Applying the assumption, there is an open set $$V$$ such that $$D\subseteq V$$ and $$\overline{V}\subseteq X\setminus C$$. So we found the disjoint open sets $$V$$ and $$X\setminus \overline{V}$$ such that $$D\subseteq V$$ and $$C\subseteq X\setminus \overline{V}.$$

Is my attempt true?

• Yes, your proof is fine, and essentially the same as mine here – Henno Brandsma Feb 12 at 15:26
• Thank you very much :) – Unknown x Feb 12 at 15:28