# Show that every regular Lindelof space is normal

It would be appreciated if you could review my proof for correctness. Thanks!

Problem: Show that every regular Lindelöf space is normal

Proof:

Let X be a regular Lindelöf space. Let $$A$$, $$B$$ be closed sets in $$X$$ that are disjoint. Note that since $$X$$ is regular that implies that one point sets are closed.

I assert the following lemma:

If $$A$$ is a closed set in a Lindelöf space, then $$A$$ is Lindelöf.

Proof of the above:

Let $$A$$ be a closed set in $$X$$, where $$X$$ is Lindelöf.

Then let $$\mathcal{C}$$ be some open covering of $$A$$ by open sets in $$X$$. Then $$\mathcal{C} \cup \{X - A\}$$ is an open covering of $$X$$ since $$X - A$$ is open by that fact that $$A$$ is closed.

Since $$X$$ is Lindelöf there is a countable subcover which may or may not include $$(X - A)$$. If it does, simply remove the set $$X - A$$ and we are left with a countable open cover of $$A$$ as a countable subset of our original set $$\mathcal{C}$$. Hence $$A$$ is Lindelöf.

Now to the proof for the main statement:

Since $$X$$ is regular and $$B$$ is closed and $$A$$ is disjoint from $$B$$, there is then for each $$x \in A$$ an open set $$U_x$$ containing $$x$$ such that $$U$$ is disjoint from $$B$$. By regularity we can find an open set $$V_x$$ such that $$\overline{V_x} \subseteq U_x$$. We can do this same operation for any point $$x \in A$$. Do the similar thing for $$B$$ w.r.t. $$A$$.

After doing so, we arrive at two open coverings for $$A$$ and $$B$$. Since $$A$$ and $$B$$ are closed sets in a Lindelöf space we know that $$A$$ and $$B$$ are themselves Lindelöf. Then there are open subcovering of those covers of $$A$$ and $$B$$, call these open covers $$\mathcal{W}=\{W_n: n \in \Bbb N\}$$ and $$\mathcal{Z}=\{Z_n: n \in \Bbb N\}$$ for $$A$$ and $$B$$ respectively. We can now define the open sets $$W_n$$ and $$Z_n$$, $$n \in \Bbb N$$ as:

$$W'_n = W_n - \bigcup_{i=1}^n \overline{Z}_i$$

and

$$Z'_n = Z_n - \bigcup_{i=1}^n \overline{W}_i$$

Then consider the open sets $$W' = \bigcup_n W'_n,\, Z' = \bigcup_n Z'_n$$ Then $$W'$$ and $$Z'$$ are disjoint since suppose some $$y\in W'$$ and $$y\in Z'$$. Then we have for $$i \ge j$$ that $$y \in W_i$$ but $$y \notin Z_j$$ for any $$j \le i$$. But by the definition of $$Z'_i$$ we have also that $$y \in Z'_j$$ for some $$j \le i$$, hence we have a contradiction. The case where $$i \le j$$ works similarly.

Hence we have obtained two disjoint open sets $$W'$$ and $$Z'$$ that contain the closed sets $$A$$ and $$B$$. Hence $$X$$ is normal.

• It's a fine proof. Apr 10, 2019 at 10:44
• Many (most?) topologists do not require singletons to be closed in the definition of regular. You don't need it for your proof, so I would leave that bit out. The use of i and j in the proof that $W'$ and $Z'$ are disjoint is incoherent. I would rather something like: ... then there exist i and j such that $y \in W'_i$ and $y \in Z'_j$. But if $i \geq j$ then, by definition of $W'_i$, $y \notin Z'_j$ and if $j \geq i$ then $y \notin W'_i$. Contradiction. Apr 10, 2019 at 11:31
• You are explicit in showing openness of $X-A$ but spend no words on why the $Z'_n$ are open etc. If you're going to be detailed, be it everywhere! Apr 10, 2019 at 20:17
• I took the liberty to improve the mathjax, and some notational points for clarity. IMHO it looks a lot better now. Apr 10, 2019 at 20:20

The proof is in essence correct, but the disjointness argument could be slightly more to the point: suppose we had $$x \in W' \cap Z'$$, so $$x \in W'_i$$ for some $$i$$ and $$x \in Z'_j$$ for some $$j$$. Fix these $$i$$ and $$j$$, then we have two cases:

1. If $$i \ge j$$ then we know $$x \in W'_i$$, so $$x \in W_i$$ and $$x \notin \overline{Z_j}$$, so $$x \notin Z_j$$, so $$x \notin Z'_j$$ contradiction.

2. If $$j \ge i$$ we know that $$x \in Z'_j$$ so $$x \notin \overline{W_i}$$, so $$x \notin W_i$$, so $$x \notin W'_i$$, contradiction.

So in either case we have a contradiction and so $$W'$$ and $$Z'$$ must be disjoint.

You also do not give an argument why $$A \subseteq \bigcup_n W'_n$$ and likewise $$B \subseteq \bigcup_n Z'_n$$, but this is rather easy: $$A$$ is covered by the $$W_n$$ to begin with and all $$Z_n$$ were chosen so that $$A \cap \overline{Z_n} = \emptyset$$ which implies $$A \subseteq X - \bigcup_n \overline{Z_n}$$ (but the latter set is not necessarily closed so we cannot use it and we have to resort to finite unions to subtract, which we can, as we saw). The sets $$W'_n$$ and $$Z'_n$$ are also open as the difference between an open and a closed set, so indeed $$W'$$ and $$Z'$$ are open.

• Thanks! I’m curious why exactly the argument for defining W’ and Z’ works with a countable collection? Does countability give me the ability to define the sets in a way an uncountable collection wouldnt? More specifically, why would an uncountable collection breakdown in making those Z’ and W’ definitions...I put the proof together in reference to the chapter proof munkres gives so i wasnt totally clear on that aspect when writing it...thanks Apr 10, 2019 at 20:04
• @H_1317 Now you can make the sets disjoint by substracting finite unions of closed sets. If you had $\aleph_1$-size covers (so the first uncountable cardinal, for definiteness) you'd have to substract countable unions of closed sets which are not always closed so you have no guarantee the difference sets are open any more. Apr 10, 2019 at 20:15
• Ah shouldve gotten that one myself, thanks! Thought it was going to be something like an axciom or something, clearly not. Apr 10, 2019 at 20:17
• @HennoBrandsma Hi professor Brandsma, here I posted the same question and reading your answer I understand why $W'$ and $Z'$ are disjoint; nevertheless I don't understeand why $W'\subseteq A$ and $Z'\subseteq B$. Moreover in the above proof for $i=1$ it seems that $W'$ and $Z'$ are badly defined: so I think that they must be defined as $W'_1=W_1,Z'_1=W_1\setminus\overline{W'_1}, W'_2=W_2\setminus\overline{Z'_1},Z'_2=Z_2\setminus\overline{W'_1\cup W'_2}$, etc...Could you help me, please? Mar 20, 2020 at 12:30
• @AntonioMariaDiMauro Your question is about Munkres' proof, so that causes confusion. If $x \in A$, there is a minimal $n$ such that $x \in W_n$, show that $x \in W'_n$ too. For yet another approach (from Engelking) see thm 1 and 2 in my note here. Mar 20, 2020 at 13:44