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Thread: Metric Spaces

  1. #1
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    Metric Spaces

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

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

    Prove that in conected metric spaces holds converse, naimly, if for every two closed and disjoint subsets... this distance is positive, than $\displaystyle (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 $\displaystyle (X,d\$ isn't compact, and then using knowing equivalent of compacity in terms of sequences, conscruct efectivly tho closed and disjoint subsets $\displaystyle A$ and $\displaystyle B$ for which is $\displaystyle d\;(A,B)=0$. 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 $\displaystyle d\;(A,B) >0$ for suitable subsets, but space isn't compact.
    Last edited by ns1954; Jan 21st 2010 at 04:46 PM.
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  2. #2
    Super Member
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    Quote Originally Posted by ns1954 View Post
    Let $\displaystyle (X,d\$ be a metric space. For arbitrary $\displaystyle A\subset X$ and $\displaystyle B\subset X$ put
    $\displaystyle \displaystyle
    d\;(A,B)=\inf\limits_{x\in A, y\in B}d\;(x,y)
    $

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

    Prove that in conected metric spaces holds converse, naimly, if for every two closed and disjoint subsets... this distance is positive, than $\displaystyle (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 $\displaystyle (X,d\$ isn't compact, and then using knowing equivalent of compacity in terms of sequences, conscruct efectivly tho closed and disjoint subsets $\displaystyle A$ and $\displaystyle B$ for which is $\displaystyle d\;(A,B)=0$. 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 $\displaystyle d\;(A,B) >0$ for suitable subsets, but space isn't compact.
    Well, here's my attempt:

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

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