It occurs to me that it might be nice to post solutions to miscellaneous mathematical exercises at least once in a while.
I saw this one on a chalkboard earlier today; evidently the room was serving as the venue for an analysis class. It’s exactly the sort of elementary exercise that usually takes me a day to solve, if I’m lucky. But this time, I’m happy to report, I managed to figure it out in just a few minutes (while ostensibly listening to a lecture on something else).
Problem: Let be such that every real-valued continuous function on is bounded. Prove that is compact.
Solution: By the Heine-Borel Theorem, it suffices to show that is closed and bounded. Since the norm function is continuous, it is bounded on by assumption; but this is the very definition of what it means for to be a bounded set. To prove that is closed, suppose is adherent to but not in . Then, because , the map is a continuous function on . However, as is adherent to , for every there exists a point such that (so that ). Thus is unbounded, contrary to the hypothesis that every continuous function on is bounded.
Easy, right? Well, it’s certainly not hard to hit upon the idea of constructing an unbounded function on a non-compact set. The difficulty is that we have absolutely no information about the set — so how can we possibly write down a suitable function, one that is unbounded specifically on ? This is where the norm comes in — it’s the one explicit function that comes free with your space, and which you can always reach for when you have nothing else. Add in the Heine-Borel Theorem, which relates compactness to the behavior of the norm, together with the idea that largeness is the reciprocal of smallness (e.g. of distance), and you’re pretty much done.