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Distance between compact subsets

  1. Nov 7, 2012 #1
    1. The problem statement, all variables and given/known data
    Let A,B be two disjoint, non-empty, compact subsets of a metric space (X,d).

    Show that there exists some r>0 such that d(a,b) > r for all a in A, b in B.

    Hint provided was: Assume the opposite, consider a sequence argument.
    2. Relevant equations

    N/A

    3. The attempt at a solution

    I've tried a few different characterizations of compactness, but neither one has led me anywhere particularly useful. I'm missing something such that the teacher's hint isn't helping me much.

    If I assume the opposite,

    Assume that for all r>0, there exists some pair (a,b) such that d(a,b) <= r

    I sense that I am supposed to reach a contradiction regarding the property that any sequence in a compact A or B should have a convergent subsequence. But how do I show that I would reach such a contradiction. Can anyone give me a push in the right direction?
     
  2. jcsd
  3. Nov 7, 2012 #2

    Dick

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    Make some sequences. Assume d(a_n,b_n)<=1/n for all n. Now pick subsequences.
     
  4. Nov 7, 2012 #3
    Do you mean:

    Assume, by way of contradiction, that for all r>0, there exists some pair (a,b) such that d(a,b) <= r.

    Let {a_n}, {b_n} be sequences in A and B respectively.

    ---------------

    I'm unclear on your next assumption. Are you using n to index the sequences as well as to bound the distances with 1/n, or was that an unintentional typo? If not, why may I /should I make that assumption?
     
  5. Nov 7, 2012 #4
    If I assume that for all r>0, there exist some pair such that d(a,b) <= r , that allows me to at least construct two sequences in A and B respectively having the property:

    d( a_1, b_1 ) <= 1/1
    d( a_2, b_2 ) <= 1/2
    ...

    and so on, is that what you meant? I hope I'm understanding you correctly.

    If I am trying to show that this property implies there is no convergent subsequence, how does it help. It doesn't tell me about the distances between each of the elements of the {a_n} sequence and the {b_n} sequence, does it?

    -----------------------------

    Another intuition I had which I'm trying to formalize, but I thought I would run past the forum is this. If I assume that for all r>0, there exists an (a,b) such that d(a,b) <= r, would this not imply that the two sets share a limit point? Since they are compact they must each contain their limit points, and then they would not be disjoint? Can I reach my desired contradiction in that manner?

    --------
     
  6. Nov 7, 2012 #5

    Dick

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    The second route. If a_n has a limit point a and b_n has a limit point b, what's d(a,b)?
     
  7. Nov 7, 2012 #6
    Here goes my full attempt:

    ---------------------------
    Let (X,d) be a metric space. Let A and B be non-empty, disjoint, compact subsets of (X,d).
    Then there exists an r>0 such that d(a,b) > r for all a in A, b in B.

    Proof:

    Assume by way of contradiction that for all r>0, there exists a pair (a,b) such that d(a,b) <= r.

    Under this assumption, we may construct the two sequences in A and B respectively, having the property that:

    d(a_n, b_n) <= 1/n , for all n in the natural numbers.

    Claim: If we consider this sequence of distances, its limit is 0.

    Subproof: For any ε > 0, there is an n in the natural numbers such that 1/n < ε . It follows that for any ε > 0, we have some N(ε) such that if n >= N(ε), | d(a_n, b_n) - 0 | < ε .
    And so the sequence of distances has limit 0.

    From this it follows that the limit of {a_n} is equal to the limit of {b_n}. So A and B share a limit point, and since both contain their limit points, the sets A and B cannot be disjoint. This is a contradiction, and so it follows that there exists some r>0 such that d(a,b) > r for all a in A, b in B.

    ---------------------------------------------------------

    Is this okay?
     
  8. Nov 7, 2012 #7

    Dick

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    The idea is fine. I'm not sure the proof really measures up. Skip to picking convergent subsequences and suppose d(a_n,b_n)<1/n and a_n->a and b_n->b. Use d(a,b)<=d(a,a_n)+d(a_n,b_n)+d(b_n,b) using the triangle inequality. Can you show the right side converges to 0?
     
  9. Nov 8, 2012 #8
    The right hand side terms should all go to zero in the limit as

    d(a, a_n) --> d(a,a) = 0
    d(b, b_n) --> d(b,b) = 0

    and

    d(a_n, b_n) --> 0 using the definition of a limit of a sequence in metric spaces, and the fact that d(a_n, b_n) < 1/n for all n.

    I think I have it now, thank you for your help.
     
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