Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $D \subset V$.

No idea how to show this. Can you please help?

share|cite|improve this question

3 Answers 3

up vote 6 down vote accepted

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.

share|cite|improve this answer
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) $C_{n+1} \cup \text{cl } V_n \subseteq V_{n+1} \subseteq \text{cl } V_{n+1} \subseteq X \setminus (\text{cl } D \cup \text{cl } V_n)$


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

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.

share|cite|improve this answer

Not a complete answer.

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 $\overline{T} \cap C,D=\emptyset$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.