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Math Help - Metric space question...

  1. #1
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    Post Metric space question...

    I am having difficulty approaching this question... any help would be greatly appreciated! Thank you!

    Show that in a metric space the intersection of a compact set and a closed set is compact.
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  2. #2
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    Quote Originally Posted by Majialin View Post
    I am having difficulty approaching this question... any help would be greatly appreciated! Thank you!

    Show that in a metric space the intersection of a compact set and a closed set is compact.
    This should help you get started.

    In a metric space a set is compact iff it is closed and bounded.

    The finite intersection of closed sets is closed.

    Note that if A is your compact set and B is your closed set then

    A \cap B \subset A
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  3. #3
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    Quote Originally Posted by TheEmptySet View Post
    In a metric space a set is compact iff it is closed and bounded.
    [/tex]
    This is only true in finite dimensional normed spaces. For metric spaces compactness is equivalent to [completeness and totally-boundedness] and sequential compactness which I believe would be more useful in this case
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  4. #4
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    In addition, the set of rational numbers, with the metric d(x,y)= |x- y| has closed and bounded sets that are not compact.
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    Quote Originally Posted by HallsofIvy View Post
    In addition, the set of rational numbers, with the metric d(x,y)= |x- y| has closed and bounded sets that are not compact.

    Perhaps a slightly simpler and sharper example: the reals with the discrete topology (from the discrete metric), and X=\{1,2,3,4,...\}=\mathbb{N} with the inherited topology . X is closed and totally bounded, but not compact.

    Tonio
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  6. #6
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    The answer to the original question of Majialin doesn't need so subtle concepts as totally bounded. In a Hausdorff space is clear that a closed subset of a compact is again compact.

    I am not sure in the non Hausdorff case, I am not used to think in this spaces, it is not the point but, is it also true?
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  7. #7
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    It is not serious, but I can answer my question given the argument for the question. It follows from the very definition.

    If K is compact and F is a closed subset of K, then any open covering of F can be "extended" to an open covering of K by adding K\setminus F. Hence it follows compactness of F, even not assuming ithe space to be Hausdorff.
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  8. #8
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    Quote Originally Posted by Enrique2 View Post
    The answer to the original question of Majialin doesn't need so subtle concepts as totally bounded. In a Hausdorff space is clear that a closed subset of a compact is again compact.

    I am not sure in the non Hausdorff case, I am not used to think in this spaces, it is not the point but, is it also true?
    Well, every metric space is a paracompact Hausdorff space. So are you considering other spaces?
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  9. #9
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    Quote Originally Posted by Enrique2 View Post
    The answer to the original question of Majialin doesn't need so subtle concepts as totally bounded. In a Hausdorff space is clear that a closed subset of a compact is again compact.

    I am not sure in the non Hausdorff case, I am not used to think in this spaces, it is not the point but, is it also true?
    Yes, that's a very good suggestion. Any compact set, in a Hausdorff space, is closed and the intersection of two closed sets is closed so the intersection of a compact set and a closed set is compact. Then use the fact that a closed subset of a compact set is closed.

    And, of course, any metric space is Hausdorff.
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  10. #10
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    Quote Originally Posted by HallsofIvy View Post
    Yes, that's a very good suggestion. Any compact set, in a Hausdorff space, is closed and the intersection of two closed sets is closed so the intersection of a compact set and a closed set is compact. Then use the fact that a closed subset of a compact set is closed.

    And, of course, any metric space is Hausdorff.
    Yes, of course. I only introduced the non Hausdorff case because of my curiosity. And I believe the argument works, more detailedly

    if f<br />
F\subset K, F closed and F=\cup_{i\in I}(V_i\cap F),
    V_i open in K, then K=\cup_{i\in I}V_i\cup K\setminus F. If K is compact then there exists J\subset I finite such that K=\cup_{i\in J}V_i\cup K\setminus F . Hence F=\cup_{i\in J}V_i\cap F, i. e. F is also compact. We do not need HAusdorff as I believed in my first post, then the original problem in a metric space is a particular case, in any topological space the propertry is true, always closed intersected compact is compact (if hopefully I have not missed anything!). Sorry because my previous posts were somehow confusing.
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