# Closed Subspace of a Normal Topological Space is Normal

If $(X,\tau)$ is a normal space then any closed subspace of $(X,\tau)$ is normal.

Let $(X,\tau)$ is normal and $A\subseteq X$ is closed in $(X,\tau)$. We must show that $(A,\tau_A)$ is normal.

Let $B \subseteq A$ $\subseteq X$ and $C \subseteq A$ $\subseteq X$ .Take $\overline B$ (closure of B) and $\overline C$ from $(X,\tau)$ such that $\overline B$ $\cap$ $\overline C$=$\emptyset$ . (They are always closed)

Since A is closed $\overline B$ $\cap$ $A$ and $\overline C$ $\cap$ $A$ are closed in $(X,\tau)$. Since they are closed there exists disjoint subsets $O_1$ and $O_2$ such that $\overline B$ $\cap$ $A$ $\subseteq$$O_1$ and $\overline C$ $\cap$ $A$ $\subseteq$ $O_2$ because $(X,\tau)$ is normal space.

$\overline B^A$ = $\overline B$ $\cap$ $A$ , $\overline C^A$ = $\overline C$ $\cap$ $A$ we have $\overline B^A$ $\subseteq$ $O_1$ and $\overline C^A$ $\subseteq$ $O_2$

If we take intersection with $A$ from both sides of $\subseteq$ in $\overline B^A$ $\subseteq$ $O_1$ and $\overline C^A$ $\subseteq$ $O_2$ we have $\overline B^A$ $\subseteq$ $O_1 \cap A$ and $\overline C^A$ $\subseteq$ $O_2 \cap A$

Since $\overline B^A$ (closure of B in $(A,\tau_A)$) and $\overline C^A$ are closed disjoint sets in $(A,\tau_A)$ and $O_1 \cap A$ and $O_2 \cap A$ disjoint open sets in $(A,\tau_A)$ we have $(A,\tau_A)$ is normal too.

What is the mistake in this proof or what are the missings? Could someone correct me please? Thanks in advance.

• $B$ and $C$ are already closed in $X$, no need for closures. – Henno Brandsma Jan 9 '18 at 15:03

## 2 Answers

You overcomplicate things:

If $A \subseteq X$ is closed and $C,D \subseteq A$ are closed (and disjoint) in $A$, then they're also closed in $X$.

So we separate them by open sets $U$ of $V$ of $X$ as $X$ is normal, and then $U \cap A$ and $V \cap A$ are the separating open sets in $A$.

• Thanks for answer. If I use B and C instead of their closures will proof be true? – esrabasar Jan 9 '18 at 15:35
• @esrabasar yes closures can go; no need to intersect with $A$ too, as they’re already subsets of $A$. You can just separate $B$ and $C$ in $X$. – Henno Brandsma Jan 9 '18 at 15:54
• I know they are subsets of $A$ but how can I say they are subsets of $O_1 \cap A$ and $O_2 \cap A$. Thanks again – esrabasar Jan 9 '18 at 19:26
• @esrabasar that’s obvious. As $A \subseteq O_1$ we only intersect with $A$ to get an open set in $A$. – Henno Brandsma Jan 9 '18 at 19:35

You don't need to choose $\bar B$ and $\bar C$ $B$ and $C$ are already closed and $O_1\cap A$ and $O_2\cap A$ seperate closed sets in $(A,\tau_A)$

I think remainder parts are true for B and C instead of their closures.