1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Sequential Compactness

  1. Nov 19, 2006 #1
    How do I prove that every metric space that is sequentially compact and separable is compact? I can't seem to use either hypotheses.
     
    Last edited: Nov 19, 2006
  2. jcsd
  3. Nov 19, 2006 #2

    StatusX

    User Avatar
    Homework Helper

    A metric space is compact iff it is complete and totally bounded. Also, I don't think the separable condition is needed.
     
  4. Nov 19, 2006 #3
    That's if you use the sequential definition of compactness. Here 'compactness' alone is defined a la topology; i.e., open covers, etc. Basically, I want to prove that if X is sequentially compact and separable (sequential => separable, but I can assume it without proof, since I proved in some earlier place), then for all open cover there exists a finite subcover for X.
     
  5. Nov 19, 2006 #4

    StatusX

    User Avatar
    Homework Helper

    I thought if you couldn't just use that fact, it might at least be a good stepping stone. But actually it's probably easier to use sequential compactness as a stepping stone to get that fact.

    If so, I would try assuming there is some open cover that has no finite subcover and use it to construct a sequence with no convergent subsequence. Specifically, construct a sequence of infinitely many disjoint open balls.
     
  6. Nov 19, 2006 #5
    Even if you construct such a sequence of open balls, how would you construct a sequence of points in X that has no convergent subsequence? X is sequentially compact.

    The sequence of points in X must converge to something outside of X, but I can't think of what, since X could be complete.
     
  7. Nov 19, 2006 #6

    StatusX

    User Avatar
    Homework Helper

    If X is complete and non-compact, it is not totally bounded. Take the real numbers for example. One open cover of them are the sets (n-1,n+1) as n ranges over the integers. One infinite sequence with no convergent subsequence is n, as n ranges over the integers. The connection won't be as straightforward in general, but you should be able to find one.
     
  8. Nov 20, 2006 #7
    X is sequentially compact and separable, and hence if an open cover A does not have a finite subcover, it has a countable subcover since we can map a dense subset into a countable subcover of A. Take an element that is in X\Ai for i=1 to infty and form a sequence, that sequence has a convergent subsequence that cannot converge to anything in the subcover, and hence outside X. Contradiction.

    I don't know where you're getting at with completeness though.
     
  9. Nov 21, 2006 #8

    StatusX

    User Avatar
    Homework Helper

    I'll agree with your conclusion, but I don't know what "since we can map a dense subset into a countable subcover of A" means.

    Why not? For example, if X is the set of real numbers and your cover is (n-1,n+1), take x_i=1 for the index corresponding to (-1,1) and x_i=0 for all other terms and you have a convergent sequence. I think you'll need your sets to be disjoint, and then pick an element in each set. I'm not 100% sure though. Sorry, I'm really busy now, but I'll look over this more carefully later tonight. If you need help sooner, maybe someone else has some ideas...
     
    Last edited: Nov 21, 2006
  10. Nov 21, 2006 #9

    StatusX

    User Avatar
    Homework Helper

    Ok, I've thought about it a bit more. Your idea in post 7 was on the right track, but you need the open sets to be nested. Pick x_i outside of A_i, then you can show that each A_i can only contain finitely many of the x_i, and use this to show there can be no convergent subsequence.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Sequential Compactness
  1. Sequential compactness (Replies: 4)

  2. Sequential Compactness (Replies: 1)

Loading...