# Prove that a regular Lindelöf topological space is normal

That a closed subset of a Lindelöf space is Lindelöf, is already proven. I use this in my proof of the following:

Prove that a regular Lindelöf topological space is normal.

Here is my proof:

Let $X$ be a regular Lindelöf topological space and let two disjoint closed sets $A$ and $B$ be given. Since $X$ is regular, for each $x\in A$ there exists open disjoint sets $U_x$ and $V_x$ such that $x\in U_x$ and $B\subset V_x$. Let $\mathcal{U}$ and $\mathcal{V}$ be the set of all such sets $U_x$ and $V_x$, respectively, for every $x\in A$. The set of sets defined by $\mathcal{W}=\{U_x\cap A \}$ for all $x\in A$ is an open covering of $A$. Since $A$ is a closed set in a Lindelöf space, $A$ is Lindelöf as well by the previous problem and there exists a countable subcollection $\mathcal{A}$ of $\mathcal{W}$ that covers $A$. Let $U_0$ be the intersection of all sets of $\mathcal{A}$ and let $V_0$ be the intersection of all $V_x$ corresponding to the sets of $\mathcal{A}$. Since $U_x\cap V_x=\emptyset$ for all $x\in A$, it is clear that $U_0\cap V_0=\emptyset$. Also, since a countable intersection of open sets is open, $U_0$ and $V_0$ are both open. Since $V\subseteq V_x$ for each $x\in A$, $V\subseteq V_0$ as well. In summary, $U_0$ and $V_0$ are disjoint open sets containing $A$ and $B$, respectively, and hence $X$ is normal.

Question: Is my proof correct? I think it has a couple of mistakes. Can I be sure that every set in $\mathcal{W}$ is open in $A$? And it is simply wrong that a countable intersection of open sets is open, right? How can I correct my proof, and is there a different and better proof?

• If $x\in A\cap U_x$ and $A\subset V_x$, then $x\in V_x$, so $U_x$ and $V_x$ are not disjoint... – Math1000 Nov 23 '16 at 14:45
• @User1006: Yes, that is a typo. I should be $B\subset V_x$. Thank you! – Barbara Nov 24 '16 at 13:43
• For those who were directed here while reading Munkres (e.g. because it is exercise 32.4), note that the proof to theorem 32.1 (that every regular space with a countable basis is normal) can be used to prove this statement after minor modifications – suncup224 Sep 12 '17 at 5:13
• @User1006 Please do not add the set-theory tag to questions where it does not belong. – Andrés E. Caicedo Nov 27 '17 at 17:45

At 4th line of your proof, "$\mathcal{W}=\{U_x\cap A \}$ for all $x\in A$ is an open covering of $A$" is wrong because $U_x\cap A$ may not be open any more ($A$ is closed).
Since $X$ is regular, for any $x\in A$, there are open sets $U_x, G_x$ such that $x\in U_x$ and $B\subset G_x, \:U_x\cap G_x=\varnothing$. So $G_x^c\subset B^c$ and $U_x\subset G_x^c$. Since $G_x$ is open, $G_x^c$ is closed. Thus $$U_x\subset \overline{U_x}\subset \overline{G_x^c}\subset G_x^c\subset B^c$$ Clearly $\mathcal{W}=\{U_x|x\in A\}$ is an open cover of $A$. Since $X$ is Lindelöf topological space, there is an countable sub cover of $\mathcal{W}$ for $A$, i.e. $$A\subset \bigcup_{n=1}^{\infty} U_n, \quad U_n\cap B=\varnothing\tag1$$ Likewise, there is an countable open cover for $B$, i.e. $$B\subset \bigcup_{n=1}^{\infty} V_n, \quad V_n\cap A=\varnothing\tag2$$ where $V_n$ is open and $V_n\subset \overline{V_n}\subset A^c$. Now let $$O_n=U_n\cap \bigcap_{i=1}^{n} \overline{V_i}^c\quad\text{and}\quad W_n=V_n\cap \bigcap_{i=1}^{n} \overline{U_i}^c$$ Since $\overline{U_i}, \overline{V_i}$ are closed, $\overline{U_i}^c, \overline{V_i}^c$ are open. So $O_n, W_n$ are open by the fact that finite intersection of open sets is open. Since $\overline{V_i}\subset A^c$, $A\subset \overline{V_i}^c$. So $$A\subset \bigcap_{i=1}^{n} \overline{V_i}^c$$ Thus by $(1)$ $$\bigcup_{n=1}^{\infty}O_n=\bigcup_{n=1}^{\infty} \left(U_n\cap \bigcap_{i=1}^{n} \overline{V_i}^c\right)=\bigcup_{n=1}^{\infty} U_n\cap\bigcap_{i=1}^{n} \overline{V_i}^c\supset A$$ Likewise since $B\subset \overline{U_i}^c$, by $(2)$ there is $$\bigcup_{n=1}^{\infty}W_n=\bigcup_{n=1}^{\infty} V_n\cap\bigcap_{i=1}^{n} \overline{U_i}^c\supset B$$ So $\bigcup_{n=1}^{\infty}O_n$ and $\bigcup_{n=1}^{\infty}W_n$ are open cover of $A$ and $B$ by the fact that arbitrary union of open sets is open. Furthermore, WLOG suppose $n\geqslant m$ $$O_n\cap W_m=\left(U_n\cap \bigcap_{i=1}^{n} \overline{V_i}^c\right)\cap \left(V_m\cap \bigcap_{i=1}^{m} \overline{U_i}^c\right)\subset \overline{V_m}^c\cap V_m\subset {V_m}^c\cap V_m=\varnothing$$ So $O_n\cap W_m=\varnothing$. Thus $$\bigcup_{n=1}^{\infty}O_n\cap \bigcup_{n=1}^{\infty}W_n=\bigcup_{n, m=1}^{\infty}(O_n\cap W_m)=\varnothing$$ i.e. $\bigcup_{n=1}^{\infty}O_n$ and $\bigcup_{n=1}^{\infty}W_n$ are disjoint.
Hence we have proved that $X$ is normal space.
• It's open in the subspace topology, which the OP meant (they use that $A$ is a Lindelöf space in its own right). – Daniel Fischer Nov 23 '16 at 21:47
• @User1006: I have a question to equation (2) in your proof. How do you know that $V_n$ does not intersect $A$? – Barbara Nov 24 '16 at 15:06
• It is similar to prove $U_n\cap B$, which I did not give details. Since $V_n\subset \overline{V_n}\subset \overline{H_n^c}\subset H_n^c\subset A^c$ where $H_n$ is open set and $A\subset H_n, \:V_n\cap H_n=\varnothing$, there is $V_n\cap A=\varnothing$. – Math Wizard Nov 24 '16 at 17:41