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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?

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

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