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Analysis Compactness Question

  1. Oct 20, 2013 #1
    1. The problem statement, all variables and given/known data
    Let ##{X_n}## be a sequence of nonempty compact subsets of the metric space Ω such that ##{X_n+1}## ##\subseteq## ##{X_n}##, with ##n: 1→∞##. Prove that their intersection is nonempty.

    **By the way, I mean to subscript "n+1", not have ##{X_n}## + 1 as it seems.

    2. Relevant equations
    One can use sequential compactness to describe the compactness of each subset. Every sequence in ##{X_n}## has a subsequence ##{X_k}## where ##x_k##→##x_o## ##\in## ##{X_n}##.


    3. The attempt at a solution
    I tried using sequential compactness to find some limit point that's found in ##{X_n}## and all of its subsets, but I have no clue on how to do so. I've tried setting up singleton sequences around the points in each nonempty set to say they must have a subsequence that converges to said point, implying said point is in ##{X_n}##, but I don't know if this implies that these points lie in the other sets, as well.
    On another line of thought, since ##{X_n+1}## is a subset of ##{X_n}##, there's some sequence in ##{X_n}## that's also in ##{X_n+1}##, both of which should converge to a point in ##{X_1}##.
    Help would be appreciated.
     
  2. jcsd
  3. Oct 20, 2013 #2

    Dick

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    Start by picking any point ##x_n## in each ##X_n##. The sequence ##x_n## has a convergent subsequence, right? If L is the limit of the convergent subsequence, what can you say about what sets it must belong to?
     
  4. Oct 20, 2013 #3
    When you say ##x_n##, first as a point and then a sequence, do you mean the single point sequence ##{x_n}##?
    Shouldn't ##L## be in ##X_n##, implying ##L## is in every ##X_n## before it, up until ##X_1##?
     
  5. Oct 20, 2013 #4

    Dick

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    No, I mean a sequence of possibly different points. Each point ##x_n## is a element of the corresponding set ##X_n##. So the sequence is ##x_1,x_2,x_3,...## with ##x_1## in ##X_1##, ##x_2## in ##X_2##, etc etc.
     
  6. Oct 20, 2013 #5
    Oh, so what you're doing is creating a subsequence of points from each ##X_n##, and saying that this sequence has a limit ##L##?
    In this case, shouldn't the limit lie in ##X_1##, and since it's a limit point for a sequence containing points from each ##X_n+1## it should get arbitrarily close to ##x_n## in the sequence after a certain point? This would imply that some ##X_n+k## will eventually contain points infinitesimally close to ##L##, and these points would be in all sets before and after it, implying the intersection of these sets are nonempty.
     
  7. Oct 20, 2013 #6

    Dick

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    You might have the right idea, but it's hard to tell from your description of it. Could you try to write that down in a clearer way (i.e. more like a proof)?
     
  8. Oct 20, 2013 #7
    Sorry:
    Create a sequence ##{x_n}## where ##x_n## ##\in## ##X_n##.
    The sequence ##{x_n} \subseteq {X_1}##.
    Since the sequence of ##{X_n}## is compact, ##\forall ε >0##, ##\exists x_k##, ##k \in N## such that ##d(x_k, L) < ε ## for ##k \geq n##, and where ##L## is the limit point of the subsequence.
    ##\forall ε## create a ball ##B(L,ε)##. This ball contains ##x_k## for ##k \geq n##.
    This implies that the ball contains points from ##X_n## whose ##n \leq k##, and since ##X_n+1 \subseteq X_n##, ##B(L,ε) \subseteq X_n## whose ##n<k##.
    Thus any element from the ball is an intersection point of these ##X_n##.
    Is this right?
     
  9. Oct 20, 2013 #8

    Dick

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    You are saying both more and less than you have to. Here, I'll get you started. Start with the sequence ##x_1,x_2,x_3,...## that we chose before. Now you can't just assume that is convergent. But because ##X_1## is compact and that is a sequence in ##X_1## (since all of the other sets are subsets of ##X_1##), then it has a convergent subsequence, call that ##x_{k_1},x_{k_2},x_{k_3},...## that does converge to some limit L. Now, sure, L must be in ##X_1##. Now can you try and tell me clearly why L has to also be in, say for example, ##X_{100}##?
     
    Last edited: Oct 20, 2013
  10. Oct 20, 2013 #9
    Hmm... would it be because if ##L \not\in X_100##, then that would imply that the ball ##B(L,ε)## would not be a subset of ##X_100##, though we've established that when ##n \geq k##, ##\exists k \in N##, ##x_k \in B(L,ε) \subseteq X_(n \geq k)## for ##\forall ε > 0##, implying ##X_(n \geq k) \subseteq X_100##?
     
  11. Oct 20, 2013 #10

    Dick

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    Sort of. But no talk of balls is needed. Can you think of a way to define a sequence that converges to L in ##X_{100}##?? And don't start from the beginning. There's an easy way to find a subsequence of ##x_{k_1},x_{k_2},x_{k_3},...## that does.
     
  12. Oct 20, 2013 #11
    Hmm... I am really pulling a blank here.
    I know the general definition of a convergent sequence is that ## \exists k \in N##, such that ##d(x_n,L)< ε##, ##\forall ε > 0##, ##k \geq n##.
    Maybe construct a sequence ##x_kn## where ##d(x_kn,L)<ε##, for ##ε>0##? Would this just be the terms from ##x_n## whose index is ##k \geq n## in relation to each term being infinitesimally close to L? Such a sequence would obviously converge to L.
    I am sorry if this is not it.
     
    Last edited: Oct 20, 2013
  13. Oct 20, 2013 #12

    Dick

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    Don't be sorry. I know you've got exactly the right idea. You just aren't expressing it as clearly as you might. Talking about the balls is just making in confusing. If you know ##x_{k_1},x_{k_2},x_{k_3},...## converges to L then ##x_{k_{100}},x_{k_{101}},x_{k_{102}},...## also converges to L. The neat way to do this is to prove that ##x_{k_{100}}## must be in ##X_{100}## and so must ##x_{k_{101}}## etc. But you don't need to be that neat. If ##x_{k_j}##, for ANY j, is in ##X_{100}## then so are ##x_{k_{j+1}}##, ##x_{k_{j+2}}## etc. Remember that ##x_{k_j}## is in ##X_{k_j}## and ##k_j<k_{j+1}<k_{j+2}<...## since it is a subsequence.
     
  14. Oct 21, 2013 #13
    I see what you mean, actually, that's what I thought concerning ##x_kj+c \in X_j## for ##c \in N##, ie ##x_k100+c \in X_100##.
    I think I can write a coherent proof now that you've helped me resolve my questions. Thank you very much for having the patience to help me! I appreciate it.
     
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