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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.

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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.

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  • $\begingroup$ Is my proof correct? $\endgroup$ – Unknown x Feb 12 at 17:00
  • $\begingroup$ @Unknownx yes, in essence, but my version is simpler to read IMHO. $\endgroup$ – Henno Brandsma Feb 12 at 17:02
  • $\begingroup$ okay. Thank you. $\endgroup$ – Unknown x Feb 12 at 17:03
  • $\begingroup$ Does there exist a simple example for normal is not hereditary? When I google it I got Tychonoff plank space. which is beyond my level :( $\endgroup$ – Unknown x Feb 12 at 17:04
  • $\begingroup$ @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. $\endgroup$ – Henno Brandsma Feb 12 at 17:53

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