1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ $B$ and $C$ are already closed in $X$, no need for closures. $\endgroup$ – Henno Brandsma Jan 9 '18 at 15:03
1
$\begingroup$

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 $A$.

$\endgroup$
  • $\begingroup$ Thanks for answer. If I use B and C instead of their closures will proof be true? $\endgroup$ – esrabasar Jan 9 '18 at 15:35
  • $\begingroup$ @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 $X$. $\endgroup$ – Henno Brandsma Jan 9 '18 at 15:54
  • $\begingroup$ 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 $\endgroup$ – esrabasar Jan 9 '18 at 19:26
  • $\begingroup$ @esrabasar that’s obvious. As $A \subseteq O_1$ we only intersect with $A$ to get an open set in $A$. $\endgroup$ – Henno Brandsma Jan 9 '18 at 19:35
1
$\begingroup$

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.

$\endgroup$

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.