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Math Help - Proof in a Metric Space

  1. #1
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    Proof in a Metric Space

    Let (X, d) be a metric space, and let V,W contained in X be disjoint (ie the intersection of V and W equals zero), nonempty and closed. Prove that there exist disjoint open V' contained in X and W' contained in X such that V is contained in V' and W is contained in W'.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by TexasGirl
    Let (X, d) be a metric space, and let V,W contained in X be disjoint (ie the intersection of V and W equals zero), nonempty and closed. Prove that there exist disjoint open V' contained in X and W' contained in X such that V is contained in V' and W is contained in W'.
    Sketch of a proof (you may have to fill in detail):

    1. There exist a \delta >0 such that for all v\epsilon V and w\epsilon W d(v,w)>\delta.
    For if this were not the case we could construct a sequence v_i \epsilon V, i=1,2,3.. with
    limit w \epsilon W, but this contradicts V closed; as closed means the limit must be in V.
    (A closed set contains all its limit points)

    2. Let W' be the union of all open balls centred on points in W of diameter <\delta/2,
    and similarly for V'. Then V' and W' have the required properties.

    RonL
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    thanks

    Many thanks!
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    Quote Originally Posted by CaptainBlack
    Sketch of a proof (you may have to fill in detail):

    1. There exist a \delta >0 such that for all v\epsilon V and w\epsilon W d(v,w)>\delta.
    For if this were not the case we could construct a sequence v_i \epsilon V, i=1,2,3.. with
    limit w \epsilon W, but this contradicts V closed; as closed means the limit must be in V.
    (A closed set contains all its limit points)
    I think there is a hidden assumption here that the space is complete.

    I will have to look at this again to see if I can get around the problem.

    A possible way around this problem is to work with the completion of X?

    Further thought shows that this cannot be adapted to non-complete
    metric spaces, sorry

    RonL
    Last edited by CaptainBlack; January 19th 2006 at 03:29 AM.
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    Super Member Rebesques's Avatar
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    Not really, the proof is fine.



    (we don't need completeness. but spoke too fast, rgep is right. )
    Last edited by Rebesques; January 19th 2006 at 11:46 PM.
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    Quote Originally Posted by CaptainBlack
    1. There exist a \delta >0 such that for all v\epsilon V and w\epsilon W d(v,w)>\delta.
    No: consider the disjoint closed sets in \mathbf{R}^2 given by V=\{(x,y) : x = 0\} and W=\{(x,y) : xy=1\}. You're implicitly assuming the space is compact.

    Define d(x,V) = \inf\{d(x,v) : v \in V\}, observe that this is non-zero for x \not\in V and then consider the set \{ x : d(x,V) > d(x,W)\}.
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  7. #7
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    Quote Originally Posted by rgep
    No: consider the disjoint closed sets in \mathbf{R}^2 given by V=\{(x,y) : x = 0\} and W=\{(x,y) : xy=1\}. You're implicitly assuming the space is compact.
    Compactness may suffice, but since I know what I was assuming
    I can assure you that my assumption was in fact completness.

    I had in fact constructed your counter example while investigating
    the problems with the proof.

    Define d(x,V) = \inf\{d(x,v) : v \in V\}, observe that this is non-zero for x \not\in V and then consider the set \{ x : d(x,V) > d(x,W)\}.
    Yes, I had got to this point, but had not yet had time to show that this
    last set is open. I'm probably missing something obvious here

    RonL
    Last edited by CaptainBlack; January 20th 2006 at 12:28 AM.
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