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.

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

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.