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Proving an equivalence concerning sequences

  1. Dec 1, 2011 #1
    given closed subsets, A and B, of R^d with A bounded prove the equivalence of:
    1) There exists a pair of sequences x_n in A and y_n in B such that |x_n - y_n| -> 0 as n -> infinity
    2) A intersection B is non empty.


    I have attempted this question but am a bit stuck on proving 1 implies 2 and not sure about my attempt at 2 implies 1.

    For 2 implies 1:

    Let p be a point of intersection between A and B.
    Then let x_n be a sequence in A that tends to p as n -> infinity.
    Let y_n be a sequence in B that tends to p as n -> infinity.
    Then we have that ||x_n - y_n|| -> 0 as n ->infinity.
    Thus we have shown a pair of sequences exists.

    For 1 implies 2:
    I'm actually clueless as to where to start. I've been told the Bolzano-Weierstrass theorem may help. I know this says that each bounded sequence in R^d has a convergent subsequence. I don't know how to use this though or really where to begin at all.

    Any help would be appreciated.

    Thanks in advance
     
  2. jcsd
  3. Dec 1, 2011 #2

    Dick

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    x_n is a sequence in A. A is bounded. So x_n must be a bounded sequence in R^d. Now keep going.
     
  4. Dec 1, 2011 #3
    x_n is a sequence in A. A is bounded. So x_n must be a bounded sequence in R^d.
    Then using B-W we know a bounded sequence has a a convergent subsequence let us call it x_n_k and so we have x_n_k -> Q.
    Now ||x_n - y_n|| -> 0 and ||x_n_k - Q|| -> 0 then the subsequence of y_n, y_n_k -> Q since
    ||y_n_k - Q||-> 0 = ||y_n_k - x_n_k + x_n_k - Q||-> 0 = ||y_n_k - x_n_k|| + ||x_n_k - Q|| -> 0
    and we know ||x_n_k - Q|| tends to 0 and so ||y_n_k - x_n_k|| must tend to 0 ans this implies x_n_k = y_n_k and so y_n_k also tends to Q.
    This implies the sets E and F have a intersection point and thus E intersection F is non empty.

    Is this correct?
    Apologies for the fact the proof above is probably rather messy.

    Also was my proof for 2 implies 1 correct?
     
  5. Dec 1, 2011 #4

    Dick

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    Yes, the proof is written pretty sloppily. I'd try and clean it up. But you didn't say why Q is in either A or B! You've got the right idea, though. For 2->1 you can't just say that there is a sequence a_n->p. You'd better show that there is one.
     
  6. Dec 1, 2011 #5
    Would the explanation of why Q is in A be because A is closed and bounded and so the sequence must tend to a point in A? And then since Q is in A and we have a sequence in B that converges to Q, the point Q must be in both A and B and hence A intersection B is non-empty? Also where I bought in a subsequence and used y_n_k in the proof would it have worked if I had just used y_n and showed that converges to Q?

    How would I show there is a sequence a_n->p? Could you give me a hint please.

    Thanks for the help s far and any future help, it is much appreciated!
     
  7. Dec 1, 2011 #6

    Dick

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    Closed sets contain all of their limit points, don't they? And {p,p,p,p,p,...} is a in A. What does it converge to?
     
  8. Dec 2, 2011 #7
    Oh right, so Q must be in A since A is a closed set and so contains it's limit points?

    And {p,p,p,p,...} is a constant sequence and so converges to p?
     
  9. Dec 2, 2011 #8

    Dick

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    Why the question marks?
     
  10. Dec 2, 2011 #9
    Just wanted to make sure :)

    Thanks for the help
     
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