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.

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This entry was posted in Analysis, Countable, Dense, Math. Bookmark the permalink.

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