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

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 $F_{\sigma}$?

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You don't select as much as say "Let $C$ and $D$ be closed subsets of $A$." Then your task is to find $U,V$ open, disjoint such that $C\subset U$ and $D\subset V$. – anon271828 Jan 13 '13 at 23:19
If you want to think of the closed sets as being "selected", then you should imagine that they are selected by your worst enemy. Then, no matter how devious the enemy was, you still have to find appropriate open neighborhoods. – Andreas Blass Jan 13 '13 at 23:50

Let us begin with a lemma ( see it in the Engelking's book):

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.

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I am a little bit confused.. Did we just prove that A is normal? – omar Jan 15 '13 at 9:14
Yes. Don't you want to prove $A$ is normal? – Paul Jan 15 '13 at 9:33

$\newcommand{\cl}{\operatorname{cl}}$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 $U\cap V=\varnothing$. $\dashv$

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