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Math Help - intersections and compact sets

  1. #1
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    intersections and compact sets

    Show that if K is compact and F is closed, then K intersect F is compact.

    Alright this is what I have so far.

    Suppose K is compact, which implies that K is closed and bounded. Since F is closed and K is closed, by theorem that means that K intersect F is closed as well. Since K intersect F is closed, that means that every cauchy sequence in K int. F has a limit that is an element of K int. F.

    This is where I get stuck, I know that the definition of a compact set is:
    A set K in R is compact if every sequence in K has a sub sequence that converges to a limit that is also in K.

    I feel like that is what I concluded in my proof, so I could just say that means that K int. F is compact, however, it really isn't saying anything about sub sequences. Any help would be great!!!
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  2. #2
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    Quote Originally Posted by mathgirlie View Post
    Show that if K is compact and F is closed, then K intersect F is compact.

    Alright this is what I have so far.

    Suppose K is compact, which implies that K is closed and bounded. Since F is closed and K is closed, by theorem that means that K intersect F is closed as well. Since K intersect F is closed, that means that every cauchy sequence in K int. F has a limit that is an element of K int. F.

    This is where I get stuck, I know that the definition of a compact set is:
    A set K in R is compact if every sequence in K has a sub sequence that converges to a limit that is also in K.

    I feel like that is what I concluded in my proof, so I could just say that means that K int. F is compact, however, it really isn't saying anything about sub sequences. Any help would be great!!!
    If you're working in \mathbb{R}^n, then we can use the Bolzano-Weierstrass theorem which states that S\subset\mathbb{R}^n is sequentially compact if and only if it is closed and bounded.

    A proof for this theorem may be found here: Bolzano?Weierstrass theorem - Wikipedia, the free encyclopedia

    Since F,K are closed with K bounded, then F\cap K is closed and bounded, and thus by Bolzano-Weierstrass is sequentially compact. \blacksquare
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  3. #3
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    The result is true in general. This question appeared two days ago in other thread, Here is the argument.

    If  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
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  4. #4
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    Any compact set is closed so the "intersection of K and F" is closed.

    That is, F\cap K is a closed subset of a compact set and so is compact. "Bounded" or, in fact, even being in a metric space, is not necessary.

    Added: As Tonio pointed out to me in another thread, my cavalier statements about compact sets being closed and a closed subset of a compact set being compact are only true in a Hausdorff space.
    Last edited by HallsofIvy; October 27th 2009 at 05:43 AM.
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  5. #5
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    thanks for the help everyone!
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