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Math Help - minimum distance

  1. #1
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    minimum distance

    Let (X,d) be a metric space and let A be a compact subset of X. Let x be a point in X.

    I want to show that there exists a point a* in A such that d(x,a*) is less than or equal to d(x,a) for all a in A.

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

    Attempt:

    Let D = inf\{d(x,a): a \in A\}

    Then there is a sequence (a_n) in A such that d(x,a_n)\rightarrow D as n \rightarrow \infty

    Because A is compact there is a subsequence (a_{n_j}) in A which converges to a point in A, call it a*

    Then d(x,a_{n_j}) \rightarrow d(x,a^*) as n \rightarrow \infty

    Then... something... to give d(x,a*) = D which is equivalent to what I was trying to show

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

    I'm not sure... it's all a little cloudy to me... I feel like I'm on the right track but I may not be appreciating some of the subtleties...

    Also, the question came with a hint to construct a sequence in A satisfying d(x, a_n) less than or equal to D + 1/n, then use the compactnes of A... That's even cloudier to me :S


    Thanks for any help
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  2. #2
    Senior Member TheAbstractionist's Avatar
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    Quote Originally Posted by RanDom View Post
    Let (X,d) be a metric space and let A be a compact subset of X. Let x be a point in X.

    I want to show that there exists a point a* in A such that d(x,a*) is less than or equal to d(x,a) for all a in A.
    Hi RanDom.

    If x\in A then we can take a^*=x and the result is proved. So assume x\notin A.

    Then for each a\in A let d(x,a)=\epsilon_a>0. Then \{B_{\epsilon_a}(a):a\in A\} is an open cover for A and by compactness there is a finite subcover \{B_{\epsilon_{a_i}}(a_i):1\le i\le n\}. Hence there exists 1\le j\le n such that d(x,a_j) is a minimum. Set a^*=a_j.
    Last edited by TheAbstractionist; April 24th 2009 at 01:36 PM.
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  3. #3
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    Thanks, that makes sense.

    Can you show me how to do it following the hint that was given as well?
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