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Math Help - Compact set in metric space (finite and infinite dimensional)

  1. #1
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    Question Compact set in metric space (finite and infinite dimensional)

    hello

    can anyone explain why for compactness of the set in finite and infinite dimensional metric spaces necessary and sufficient conditions are not the same ?

    thanks for any help I'm desperate right now
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  2. #2
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    Quote Originally Posted by sedam7 View Post
    can anyone explain why for compactness of the set in finite and infinite dimensional metric spaces necessary and sufficient conditions are not the same ?
    Please give some more details to clarify exactly what you mean.
    This question could mean several different things.
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  3. #3
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    Quote Originally Posted by Plato View Post
    Please give some more details to clarify exactly what you mean.
    This question could mean several different things.
    hmmm...
    It's like in finite dimensional metric space if set i bounded and closed then that set is compact .... but in infinite dimensional metric spaces some "another story" (sorry but I can't express myself totally in English so I'm trying with at least google translate to get some badly needed words )

    necessary condition for compactness of the set M subset metric space X is that for \forall \varepsilon >0 exists finite \varepsilon - net (or web... i don't know how to translate it) of the set M.
    If metric space X is complete then that is sufficient condition ....
    Last edited by sedam7; May 30th 2011 at 10:26 AM.
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  4. #4
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    Quote Originally Posted by sedam7 View Post
    It's like in finite dimensional metric space if set i bounded and closed then that set is compact .... but in infinite dimensional metric spaces some "another story"
    Thank you for the clarification.
    You see, the definition for compactness is the same in any metric space: Any open cover has a finite subcover.
    But you are quite right that conditions that imply compactness do differ depending on the space and completeness.
    In any finite Euclidean space \mathbb{R}^n a closed and bounded set is compact. That is necessary and sufficient.
    But seems that you are confused about the notion of totally bounded sets, sets that have an \epsilon\text{-net}.
    That is also known as precompact and in a metric space that implies compactness.
    That is a bit broad, but I hope it helps.
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  5. #5
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    thank you very very much
    I lost it in my head somewhere that every finite dimensional metric space is complete....
    and that implies compactness
    thank you very much again
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