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Math Help - Metric Spaces

  1. #1
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    Metric Spaces

    Let " alt="(X,d\" /> be a metric space. For arbitrary A\subset X and B\subset X put
    \displaystyle<br />
d\;(A,B)=\inf\limits_{x\in A, y\in B}d\;(x,y)<br />

    It can be proved, if (X,d) is compact space, then for every tho closed and disjoint subsets A and B is d\;(A,B) > 0

    Prove that in conected metric spaces holds converse, naimly, if for every two closed and disjoint subsets... this distance is positive, than " alt="(X,d\" /> is compact.
    Condition conected on metric space can't be omitted.

    Hint: My proff is relativly long and is based on contrary asumptation that " alt="(X,d\" /> isn't compact, and then using knowing equivalent of compacity in terms of sequences, conscruct efectivly tho closed and disjoint subsets A and B for which is d\;(A,B)=0. But there are some other equivalents of compacity, maybe someone find another solution. I have the appropriate counterexemple of totaly disconected metric space in which is d\;(A,B) >0 for suitable subsets, but space isn't compact.
    Last edited by ns1954; January 21st 2010 at 05:46 PM.
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  2. #2
    Super Member
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    Quote Originally Posted by ns1954 View Post
    Let " alt="(X,d\" /> be a metric space. For arbitrary A\subset X and B\subset X put
    \displaystyle<br />
d\;(A,B)=\inf\limits_{x\in A, y\in B}d\;(x,y)<br />

    It can be proved, if (X,d) is compact space, then for every tho closed and disjoint subsets A and B is d\;(A,B) > 0

    Prove that in conected metric spaces holds converse, naimly, if for every two closed and disjoint subsets... this distance is positive, than " alt="(X,d\" /> is compact.
    Condition conected on metric space can't be omitted.

    Hint: My proff is relativly long and is based on contrary asumptation that " alt="(X,d\" /> isn't compact, and then using knowing equivalent of compacity in terms of sequences, conscruct efectivly tho closed and disjoint subsets A and B for which is d\;(A,B)=0. But there are some other equivalents of compacity, maybe someone find another solution. I have the appropriate counterexemple of totaly disconected metric space in which is d\;(A,B) >0 for suitable subsets, but space isn't compact.
    Well, here's my attempt:

    Assume X is not compact then there exists a sequence (x_n) with no accumulation points. Construct another sequence as follows (y_n) such that d(y_n,x_n)<\min \{ \frac{a_n}{2} , \frac{1}{n} \} where a_n= \sup \{ r>0 : B_r(x_n) \cap (x_n)= \emptyset \} (we can guarantee that y_n\neq x_n since X is connected and so the singletons are not open). (x_n) and (y_n) have the same accum. points and so both are closed and disjoint but their distance is 0 by construction.

    For a counter-example take any infinite set with the discrete metric.
    Last edited by Jose27; January 23rd 2010 at 09:08 AM.
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