B7s is the math problem-solving club at MIT. We meet every Saturday 8-10PM at Random Hall to work on a hard problem, such as one from the Miklos Schweitzer competition.
Here’s the problem from Sat. 4/17.
[MS 1964/9] Let be the set of functions from to . Prove that there does not exist a topology on such that converges to on the topology if and only if converges to a.e. (almost everywhere).
Explanation: Recall that a sequence converges to in a topology if for every open set containing , all but finitely many points of are in . A sequence of functions converges to almost everywhere if it converges pointwise except on a set of measure 0. (Measure is like the size of a set, a generalization of the length of an interval. For this problem, we only need to know measure is countably subadditive- – and the measure of an interval is the length of the interval.)
Solution: It seems like the measure of the set where two functions differ could act like a “distance” between and … so let . Inspired by this idea, we formulate a condition on a set that we hope will be satisfied by open sets.
Condition: If then there exists so that contains all functions differing from on a set of measure , i.e. if then .
We show that if “ converges to almost everywhere implies converges to in the topology” is true, then every open set satisfies the condition. Then we show that if “ converges to in the topology implies converges to almost everywhere” is true, then some open set does not satisfy the condition. Thus both statements cannot simultaneously hold, and the problem will be solved.
For the first part, let be an open set. Suppose by way of contradiction the condition fails for . Then for all we can choose so that . By choosing , we ensure that does not converge to in the topology. We claim that converges to almost everywhere. Indeed, for every we can find so that . Then the measure of the set of such that does not converge a.e. to is at most the measure of the set of such that for some , which is
Hence converges to on a set of measure for all ; it must converge to on a set of measure 1. In summary, converges to almost everywhere, but does not converge to in the topology, violating the hypothesis.
For the second part, consider the sequence defined by
where denotes the characteristic function of the set – it equals 1 on and 0 elsewhere. If the condition holds for all open sets, then any open set containing must contain all functions such that , for some . Since converges to 0 (the intervals are getting shorter), converges to the zero function in the topology. But every point is in an infinite number of intervals of the form , as well as absent from an infinite number of those intervals, so does not converge pointwise to . Thus converges to nowhere! This contradicts the hypothesis.
Here’s the problem Sat. 4/24.
Let be a closed subset of . Prove that there exists a function that is zero on , positive outside , and whose partial derivatives all exist.
(I wasn’t able to attend the meeting, and I haven’t solved this yet.)