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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?

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    $\begingroup$ 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... $\endgroup$
    – Math1000
    Nov 23, 2016 at 14:45
  • $\begingroup$ @User1006: Yes, that is a typo. I should be $B\subset V_x$. Thank you! $\endgroup$
    – Barbara
    Nov 24, 2016 at 13:43
  • $\begingroup$ 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 $\endgroup$
    – suncup224
    Sep 12, 2017 at 5:13
  • $\begingroup$ @User1006 Please do not add the set-theory tag to questions where it does not belong. $\endgroup$ Nov 27, 2017 at 17:45
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    $\begingroup$ Nice attempt, in a sense every beginner would write that(simple/straightforward) kind of proof. Your proof would be valid, if we were working in compact set. Working with lindelof space(countable set) we run into the problem, countable intersection of open sets is not open. If we try to do it by setting $V_0=\bigcup_n V_{x_n}$, now we run into the problem of disjoint, i.e. $U_0 \cap V_0 \neq \emptyset$ in general. So there is no trivial construction of disjoint open set containing $A$ and $B$. $\endgroup$
    – user264745
    May 12, 2022 at 9:08

2 Answers 2

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At line 4 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 anymore ($A$ is closed).

The actual proof is much harder and as follows.

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 a Lindelöf topological space, there is a countable subcover 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 a 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 a finite intersection of open sets is open. Since for any $n$, $\overline{V_n}\subset A^c$, $A\subset \overline{V_n}^c$. Likewise, $B\subset \overline{U_n}^c$. So $$ A\subset \bigcap_{n=1}^{\infty} \overline{V_n}^c\quad\text{and}\quad B\subset \bigcap_{n=1}^{\infty} \overline{U_n}^c $$ Thus by $(1)$ $$ \bigcup_{n=1}^{\infty}O_n=\bigcup_{n=1}^{\infty} \left(U_n\cap \bigcap_{i=1}^{n} \overline{V_i}^c\right)\supset\bigcup_{n=1}^{\infty} U_n\cap\bigcap_{n=1}^{\infty} \overline{V_n}^c\supset A $$ Likewise, by $(2)$ there is $$ \bigcup_{n=1}^{\infty}W_n\supset\bigcup_{n=1}^{\infty} V_n\cap\bigcap_{n=1}^{\infty} \overline{U_n}^c\supset B $$ So $\bigcup_{n=1}^{\infty}O_n$ and $\bigcup_{n=1}^{\infty}W_n$ are open covers of $A$ and $B$ because 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 $X$ is a normal space.

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    $\begingroup$ It's open in the subspace topology, which the OP meant (they use that $A$ is a Lindelöf space in its own right). $\endgroup$ Nov 23, 2016 at 21:47
  • $\begingroup$ @User1006: I have a question to equation (2) in your proof. How do you know that $V_n$ does not intersect $A$? $\endgroup$
    – Barbara
    Nov 24, 2016 at 15:06
  • $\begingroup$ 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 $V_n\cap A=\varnothing$. $\endgroup$ Nov 24, 2016 at 17:41
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By regularity we can construct $U_x$ for all $x$ in $A$ such that $\lbrace U_x \rbrace$ is an open cover of $A$ and $U_x$ closure is disjoint from $B$. $A$ is Lindelöf as it is closed subspace of Lindelöf space $X$. So, the cover $\lbrace U_x \rbrace$ has a countable subcover $\lbrace U_n \rbrace$. Now similarly we can construct a countable open cover $\lbrace V_n\rbrace$ Of $B$ such that $U_n$ closure disjoint from $A$. Now we can proceed as we have proof in Munkres 32.1(normal space) for "regular space with countable basis is normal".

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