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.

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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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