## February 1998 #1

Problem Statement: Let $\{A_n\}$ be an uncountable collection of open sets in $\mathbb{R}$. Let $A=\bigcup\limits_{\alpha}A_{\alpha}$. Prove there exists a countable collection $\{I_n\}$ of open intervals satisfying (1) $\bigcup\limits_{n=1}I_n=A$ and (2) for every positive integer $n$ there exists an $\alpha$ such that $I_n\subseteq A_{\alpha}$.

Proof: Since the rationals are countable we may enumerate the rationals and any subset of the rationals. Let $\{q_n\}$ be an enumeration of the rationals which are contained in $A$. Then each $q_n\in A_{\alpha}$ for some collection of $A_{\alpha}'s$. Define $I_n$ to be the largest open interval around $q_n$ such that the interval is still completely contained within some $A_{\alpha}$.

First note that since each $I_n$ is determined by $q_n\in\mathbb{Q}$ it follows that $\{I_n\}$ is a countable collection of open intervals since we may simply use the enumeration on $\{q_n\}$ to enumerate $\{I_n\}$.

Furthermore, by construction it follows that for every positive integer $n$, there exists some $\alpha$ such that $I_n\subseteq A_{\alpha}$. Thus property (2) is satisfied.

Now we must show that $\bigcup\limits_{n=1}I_n=A_{\alpha}$. As stated above each $I_n\subseteq A_{\alpha}$ for some $\alpha$, thus $\bigcup\limits_{n=1}I_n\subseteq A$.

Let $p\in A$. We wish to show that $p\in I_n$ for some $n$. If $p\in\mathbb{Q}$ then $p=q_m$ for some positive integer $m$ and so $p\in I_m$.

If $p\notin \mathbb{Q}$ then $p\in A_{\alpha}$ for some $\alpha$. Create the open interval $J_p$ with radius $\delta>0$ such that $\delta$ is the largest radius possible for $J_p$ to still remain inside $A_{\alpha}$. Now consider the interval $(p-\frac{\delta}{2},p+\frac{\delta}{2})$. Since $\mathbb{Q}$ is dense in $\mathbb{R}$ we may find a $q_n\in (p-\frac{\delta}{2},p+\frac{\delta}{2})$. Now put an open interval around $q_n$ with radius $\varepsilon>0$ such that $\varepsilon$ is the largest possible radius such that $(q_n-\varepsilon,q_n+\varepsilon)\subset (p-\delta,p+\delta)$. Then this interval is contained in $I_n$ by construction. Since we chose $q_n\in (p-\frac{\delta}{2},p+\frac{\delta}{2})$ it follows that $p\in (q_n-\varepsilon,q_n+\varepsilon)$. Thus, $p\in I_n$ and so $A\subseteq \bigcup\limits_{n=1}I_n$. Since we have set containment both ways, condition (1) is satisfied.

Therefore, there exists a countable collection of open intervals satisfying the conditions.

$\Box$

Reflection: This was not a straightforward problem. The thing that makes it work though is how we construct the $I_n$. Originally we constructed them based on which $A_{\alpha}$ we were in, but then we came up the issue that any rational could be in several $A_{\alpha}$ since they can overlap and so this meant that we were creating uncoutably many $I_n$. No bueno.  So, we had to come up with a way to fix that. Dr. Retsek had the great idea to define our $I_n$ so that they were as large as possible. Also, the second condition is kind of hinting at that since your $I_n$ must be contained in $A_{\alpha}$ for some $\alpha$. Also, drawing a picture really helped us get an idea of what was going on here.

When trying to go from something that is uncountable to something that is countable you want to try and find a countable subset of your uncountable thing. The rationals worked really well since they are dense.

On another note, 10 days remaining.