Every continuous function bounded implies compact

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 U \subset \mathbf{R}^n be such that every real-valued continuous function on {}U is bounded. Prove that U is compact.

Solution: By the Heine-Borel Theorem, it suffices to show that U is closed and bounded. Since the norm function x \mapsto \Vert x \Vert is continuous, it is bounded on U by assumption; but this is the very definition of what it means for U to be a bounded set. To prove that U is closed, suppose y is adherent to U but not in U. Then, because y \notin U, the map x \mapsto \Vert x - y \Vert ^{-1} is a continuous function on U. However, as y is adherent to U, for every \epsilon > 0 there exists a point u \in U such that \Vert u-y \Vert < \epsilon (so that \Vert u-y\Vert^{-1} > 1/\epsilon). Thus x \mapsto \Vert x - y \Vert ^{-1} is unbounded, contrary to the hypothesis that every continuous function on U 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 U — so how can we possibly write down a suitable function, one that is unbounded specifically on U? 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.

This entry was posted on Wednesday, February 18th, 2009 at 3:50 pm and is filed under mathematics. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

4 Responses to “Every continuous function bounded implies compact”

  1. Tortoise Says:
    March 9, 2009 at 1:53 pm | Reply

    When I first read the front page of this post, my eyes glossed over the fact that U was in R^n, so I tried to prove it for all spaces in general. At first I thought I hit upon of way of turning an open cover of U into an open cover of f(U), taking a finite subcover of f(U) (since f(U) is bounded, and must be closed, or else f could be composed with a function g:R->R which sends an excluded boundary point of f(U) to infinity, creating an unbounded function), but this method has at least two flaws. Namely, the image of an open set in U need not be open in f(U), and f need not be 1-to-1, so the preimage of f(T) need not equal T, for some open subset T of U. Still, it intuitively feels to me that the statement ought to hold true in general; at least, I haven’t been able to figure out what a counterexample would look like. Any thoughts for attacking the general case?

  2. James Cook Says:
    March 11, 2009 at 11:48 am | Reply

    Tortoise,

    Perhaps surprisingly, the general case is actually false. For a counterexample, see here.

    The property of every continuous real-valued function being bounded is called “pseudocompactness”. Apparently it’s equivalent to countable compactness for normal Hausdorff spaces. (I don’t know off the top of my head how difficult a result this is; it might be worth trying to come up with a proof.)

  3. Jaime Says:
    March 13, 2009 at 6:10 am | Reply

    For proofs of the equivalence in metric spaces, see:
    http://www.math.niu.edu/~rusin/known-math/99/cpt_metric

  4. C.Yang Says:
    November 12, 2009 at 4:07 pm | Reply

    Great posts…thanks a lot.

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