# $F_\sigma$-subsets in a normal space can be separated

Let $X$ be a normal Hausdorff space and let $C,D$ be two $F_{\sigma}$ subsets of $X$ such that $\overline{C} \cap D = \emptyset$ and $C \cap \overline{D} = \emptyset$.

Prove there exists disjoint open subsets $U,V$ such that $C \subset U$ and $D \subset V$.

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Write $C = \bigcup_{n=1}^\infty C_n$ as a union of countably many closed sets $C_n$. Since $C_n \subset C$ does not intersect $\overline D$, we can find an open set $U_n\supset C_n$, such that $\overline U_n \cap \overline D = \emptyset$ (by normality) for every $n$. Theses sets $U_n$ form a covering of $C$.

Similarly we can construct an open covering $\{V_n\}_{n\in \mathbb N}$ of $D$ such that the closures of the $V_n$ do not intersect $\overline C$.

One is tempted to choose $U = \bigcup_n U_n$ and $V = \bigcup_n V_n$, but these sets need not be disjoint. However we can use the following trick:

Define $U'_n = U_n \setminus \bigcup_{i = 1}^n \overline V_n$ and $V_n' = V_n \setminus \bigcup_{i = 1}^n \overline U_n$. And now let $$U' = \bigcup_{n = 1}^\infty U'_n, \qquad V' = \bigcup_{n = 1}^\infty V_n'$$

I claim that these sets satisfy

• $C \subset U'$, $D \subset V'$
• $U'$ and $V'$ are open
• $U' \cap V' = \emptyset$

The claim, I think, I can leave to you to verify.

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How to verify $C \subset U', D \subset V'$? The two others are straight forward.. – omar Jan 15 '13 at 14:27

You want to use the same 'climbing a chimney' technique that's used to prove that every regular Lindelöf space is normal. (You might want to stop reading here and take a look at that proof to see whether you can adapt the idea on your own.)

Let $C = \bigcup \limits_{n \in \omega} C_n$ and $D = \bigcup \limits_{n \in \omega} D_n$, where the sets $C_n$ and $D_n$ are closed and for each $n \in \omega$ $C_n \subseteq C_{n+1}$ and $D_n \subseteq D_{n+1}$. By normality there are open sets $V_0$ and $W_0$ such that $C_0 \subseteq V_0 \subseteq \text{cl } V_0 \subseteq X \setminus D$ and $D_0 \subseteq W_0 \subseteq \text{cl } W_0 \subseteq X \setminus C$. Given $V_n$ and $W_n$ for some $n \in \omega$, use normality to get open sets $V_{n+1}$ and $W_{n+1}$ such that

(1) $C_{n+1} \cup \text{cl } V_n \subseteq V_{n+1} \subseteq \text{cl } V_{n+1} \subseteq X \setminus (\text{cl } D \cup \text{cl } W_n)$

and

(2) $D_{n+1} \cup \text{cl } W_n \subseteq W_{n+1} \subseteq \text{cl } W_{n+1} \subseteq X \setminus (\text{cl } C \cup \text{cl } V_n)$.

Now let $V = \bigcup \limits_{n \in \omega} V_n$ and $W = \bigcup \limits_{n \in \omega} W_n$; these are disjoint open sets containing $C$ and $D$, respectively.

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How to prove these sets are disjoint? I tried to assume they aren't, which would mean $V_i\cap W_i\neq \emptyset$ for some $i$, but I don't see how to prove this can't happen. Where to use the fact $C\cap \overline D=\overline C\cap D=\emptyset$? – Exterior Jan 28 at 14:24
@Exterior: There was a very bad typo in (1) and (2) that seems to have escaped notice for almost five years. I’ve fixed it, and you should now be able to show easily that $V$ and $W$ are disjoint. (The fact that $C\cap\operatorname{cl}D=D\cap\operatorname{cl}C=\varnothing$ is used when we observe that any $C_m$ is disjoint from any $D_n$.) – Brian M. Scott Jan 28 at 20:00

An $F_\sigma$ set is a countable union of closed sets. So $C=\cup F_n$; we can suppose that $F_n \subset F_{n+1}$, or elseway define $F_n:=\cup_{k=1}^n F_k$. Therefore we can write $C=\bigcup C_n$ where the sequence $C_n$ of closed sets is increasing. Also, we can do the same with $D$: $D=\bigcup D_n$, with $D_n$ increasing.
By the relations presented, we have that $\overline{D} \cap C_n=\emptyset$ for every $n$, and then by regularity we can find open sets $C_n \subset O_n,\ \overline{D} \subset K_n$ with $O_n \cap K_n=\emptyset$. Pick $O=\bigcup O_n$. Then $O \cap \overline{D}=\emptyset$ and $C \subset O$. In the same way, we find an open set $U$ with $D\subset U$ and $U \cap \overline{C}=\emptyset$.
I don't know how to make $U,O$ disjoint. I was thinking of defining $T=U\cap V$ and prove somehow that $\overline{T} \cap C,D=\emptyset$.