If $X$ is normal and $A$ is a $F_{\sigma}$-set in $X$, then $A$ is normal. How could I prove this theorem?

Question

A topological space $X$ is a normal space if, given any disjoint closed sets $E$ and $F$, there are open neighbourhoods $U$ of $E$ and $V$ of $F$ that are also disjoint. (Or more intuitively, this condition says that $E$ and $F$ can be separated by neighbourhoods.) And an $F_{\sigma}$-set is a countable union of closed sets.

So I should be able to show that the $F_{\sigma}$-set has the necessary conditions for a $T_4$ space? But how could I for instance select two disjoint closed sets from $F_{\sigma}$?

Answer 1

$\newcommand{\cl}{\operatorname{cl}}$We can actually prove more. Let $X$ be a $T_4$-space, and let $A$ be an $F_\sigma$-set in $X$; if $H$ and $K$ are disjoint, relatively closed subsets of $A$, then there are disjoint open sets $U$ and $V$ in $X$ (not just in $A$) such that $H\subseteq U$ and $K\subseteq V$.

One proof of this is very similar to the usual ‘climbing a chimney’ proof that a regular, Lindelöf space is normal.

Proof: There are closed sets $F_n\subseteq X$ for $n\in\Bbb N$ such that $A=\bigcup_{n\in\Bbb N}F_n$, and $F_n\subseteq F_{n+1}$ for each $n\in\Bbb N$. Note that $H\cap\cl_XK=\varnothing=K\cap\cl_XH$.

For $n\in\Bbb N$ let $H_n=H\cap F_n$ and $K_n=K\cap F_n$; the sets $H_n$ and $K_n$ are closed in $X$. (To see this, note that $H_n=H\cap F_n=(\cl_XH\cap A)\cap F_n=\cl_XH\cap(A\cap F_n)=\cl_XH\cap F_n$, which is clearly closed in $X$, and similarly for $K_n$.)

Now use the normality of $X$ to carry out the recursive construction of open sets $U_n$ and $V_n$ in $X$ for $n\in\Bbb N$ such that:

1. $H_0\subseteq U_0\subseteq\cl_XU_0\subseteq X\setminus\cl_XK$;
2. $K_0\subseteq V_0\subseteq\cl_XV_0\subseteq X\setminus(\cl_XH\cup\cl_XU_0)$;
3. $H_{n+1}\cup\cl_XU_n\subseteq U_{n+1}\subseteq\cl_XU_{n+1}\subseteq X\setminus(\cl_XK\cup\cl_XV_n)$ for each $n\in\Bbb N$; and
4. $K_{n+1}\cup\cl_XV_n\subseteq V_{n+1}\subseteq\cl_XV_{n+1}\subseteq X\setminus(\cl_XH\cup\cl_XU_{n+1})$ for each $n\in\Bbb N$.

(1) is possible because $H_0$ and $\cl_XK$ are disjoint closed sets in $X$. Then (2) is possible because $K_0$ and $\cl_XH\cup\cl_XU_0$ are disjoint closed sets in $X$: we already knew that $K_0\cap\cl_XH=\varnothing$, and by construction $K_0\cap\cl_XU_0\subseteq\cl_XK\cap\cl_XU_0=\varnothing$. The argument that (3) and (4) can be carried out is a straightforward induction.

Now let $U=\bigcup_{n\in\Bbb N}U_n$ and $V=\bigcup_{n\in\Bbb N}V_n$; clearly $U$ and $V$ are open in $X$, $H\subseteq U$, and $K\subseteq V$; I’ll leave to you the easy verification that $U\cap V=\varnothing$. $\dashv$

Answer 2

Let us begin with a lemma ( see it in the Engelking's "General Topology" book, lemma 1.5.15):

If $X$ is a $T_1$ space and for every closed $F$ and every open $W$ that contains $F$ there exists a sequence $W_1$, $W_2$, ... of open subsets of $X$ such that $F\subset \cup_{i}W_i$ and $cl(W_i)\subset W$ for $i=$ 1, 2, ..., then the space $X$ is normal.

Suppose $X$ is normal and $A = \cup_n F_n \subset X$ is an $F_\sigma$ in $X$, where all the $F_n$ are closed subsets of $X$. Then $A$ is normal (in the subspace topology). To apply the lemma, let $F$ be closed in $A$ and $W$ be an open superset of it (open in $A$). Let $O$ be open in $X$ such that $O \cap A = W$, and note that each $F \cap F_n$ is closed in $X$ and by normality of $X$ there are open subsets $O_n$ in $X$, for $n \in \mathbb{N}$ such that $$ F \cap F_n \subset O_n \subset \overline{O_n} \subset O $$ and define $W_n = O_n \cap A$, which are open in $A$ and satisfy that the $W_n$ cover $F$ (as each $W_n$ covers $F_n$ and $F = F \cap A = \cup_n (F \cap F_n)$) and the closure of $W_n$ in $A$ equals $$\overline{W_n} \cap A = \overline{O_n \cap A} \cap A \subset O \cap A = W$$ which is what is needed for the lemma.