# Interesting Normality Condition (topology)

I just got through an inquiry-based course in introductory topology, and I enjoyed the experience a lot. If you're unfamiliar, this roughly means that we were given a text full of theorems and the class consisted of providing as many proofs as possible, with student presentations in lieu of lectures.

There was one theorem which I got a lot of use out of, but was never able to prove. It wasn't assigned, and I talked to the professor about it a couple times, but he didn't have any advice for me that I hadn't figured out myself.

The text calls it the "Normality Lemma" if that helps, but it doesn't appear to be a standard term.

Let $A$ and $B$ be subsets of a topological space $X$ and let $\{U_i\}_{i\in\mathbb N}$ and $\{V_i\}_{i\in\mathbb N}$ be two collections of open sets such that:

• $A\subseteq \bigcup_{i\in\mathbb{N}} U_i$ and $B\subseteq \bigcup_{i\in\mathbb{N}} V_i$
• $\overline{U_i}\cap B$ and $\overline{V_i}\cap A$ are empty for all $i\in\mathbb N$.

Then there are disjoint open sets $U$ and $V$ such that $A\subseteq U$ and $B\subseteq V$.

The thing that really puzzles me about this is the countability condition. Is this doing anything more than saying the collections need to be infinite?

Another interesting quirk about this theorem is that it doesn't show that $X$ is normal explicitly. Rather it guarantees disjoint open sets around a specific pair of (sufficiently ''cushioned'') arbitrary sets for any space.

Can anyone help me through a proof of this?

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Did you want $A\cap\operatorname{cl}V_i$ to be empty, rather than $A\cap V_i$? –  Brian M. Scott May 26 '13 at 22:49
Yes! Thank you :) –  Eric Stucky May 26 '13 at 22:50

$\newcommand{\cl}{\operatorname{cl}}$This lemma is actually the basis of the usual proof that every regular, Lindelöf space is normal. The countability condition isn’t to make the collections of open sets infinite: it’s to make sure that they aren’t uncountable.

Let $G_0=U_0$ and $H_0=V_0\setminus\cl U_0=V_0\setminus\cl G_0$. Suppose that for some $n\in\Bbb N$ you’ve defined $G_n$ and $H_n$ so that $G_n\subseteq\bigcup_{k\le n}U_k$, $H_n\subseteq\bigcup_{k\le n}V_k$, and $G_n\cap H_n=\varnothing$. Let

$$G_{n+1}=(G_n\cup U_{n+1})\setminus\cl H_n$$

and

$$H_{n+1}=(H_n\cup V_{n+1})\setminus\cl G_{n+1}\;;$$

clearly $G_{n+1}\subseteq\bigcup_{k\le n+1}U_k$, $H_{n+1}\subseteq\bigcup_{k\le n+1}V_k$, and $G_{n+1}\cap H_{n+1}=\varnothing$, so the recursive construction can continue.

Now let $G=\bigcup_{n\in\Bbb N}G_n$ and $H=\bigcup_{n\in\Bbb N}H_n$. Clearly $G$ and $H$ are open. I’ll let you see if you can show that $A\subseteq G$, $B\subseteq H$, and $G\cap H=\varnothing$; feel free to ask for help if you need it.

Added: You might be interested in an old paper of mine in The American Mathematical Monthly: A ‘More Topological’ Proof of the Tietze-Urysohn Theorem, Vol. $85$, No. $3$ (Mar., $1978$), pp. $192$-$193$, JSTOR; the proof in question uses a lemma that is proved by the same ‘climbing a chimney’ technique that I used here.

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Huh. That is a very interesting construction. I'm not entirely sure I understand the chimney analogy. It seems like these sets are, for the most interesting kinds of cases, getting 'close' to one another. By this I mean that when the $U_i$ and $V_i$ overlap significantly, then $\cup U_i \cup \cup V_i$ differs from $G\cup H$ by some closed set which is 'thin' in some sense. Is that intuition reasonable, and is that what you meant? –  Eric Stucky May 26 '13 at 23:35
@Eric: No, the chimney analogy is very physical: one way to climb a rock chimney (or a house chimney, if it’s big enough) is to brace your feet against one side and your shoulders against the other. Then you step up with your feet (building $G_{n+1}$ from $G_n$) and then use your hands behind you to slide your shoulders up the same amount (building $H_{n+1}$). –  Brian M. Scott May 26 '13 at 23:39