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Math Help - Compactness; show inf is achieved

  1. #1
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    Compactness; show inf is achieved


    ===================================


    I don't really understand the hint in part a (actually I'm confused). HOW can we find such sequences x_n and y_n? And how can we find a subsequence x_{n_k} that converges?

    Any help is appreciated!


    [note: also under discussion in math link forum]
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by kingwinner View Post

    ===================================


    I don't really understand the hint in part a (actually I'm confused). HOW can we find such sequences x_n and y_n? And how can we find a subsequence x_{n_k} that converges?

    Any help is appreciated!


    [note: also under discussion in math link forum]
    What have you tried mon capitain?
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  3. #3
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    I am not sure how to begin because I don't know how to FIND x_n and y_n? Do we need to produce an actual concrete example for which d(xn,yn)->d(X,Y)? Or can we just say such sequences exist?

    Also, since X is compact, and by definition we know there EXISTS a convergence subsequence x_nk whose limit lies in X., isn't it? But then the hint asks us to FIND such a subsequence, why?? What does that mean? Don't we know such a subsequence x_nk already exists by definition?

    Can someone help me, please?
    Last edited by kingwinner; February 22nd 2010 at 11:52 PM.
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  4. #4
    MHF Contributor Drexel28's Avatar
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    This is a lot harder than I initially thought. The idea behind the following is that this is obvious if Y is bounded, and if it isn't we can "erase" the parts that are far enough away from X to be ignored.

    So, we know that X is compact and thus totally bounded (specifically bounded). Thus, X\subseteq \left\{\bold{x}\in\mathbb{R}^n:\|\bold{x}\|\leqsla  nt M\right\} for some constant M. Let N=\left\{\bold{x}\in\mathbb{R}^n:\|\bold{x}\|\leqs  lant M+d(X,Y)+1\right\}. This is a compact subspace of \mathbb{R}^n and so G\cap N=K is compact (since the intersection of a compact set and a closed set is compact). Now, we next claim that d(E,G)=d(E,K), but I leave this for you to prove. Thus, we must merely show that d(E,G)=d(E,K)=d(e,k) for some e\in E,k\in K. To do this we note that since G,K are compact that G\times K is compact by Tychonoff's theorem. Also, it is clear that d:G\times K\mapsto\mathbb{R} is continuous and so it assumes it's minimum on G\times K and the conclusion follows.
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  5. #5
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    Quote Originally Posted by Drexel28 View Post
    This is a lot harder than I initially thought. The idea behind the following is that this is obvious if Y is bounded, and if it isn't we can "erase" the parts that are far enough away from X to be ignored.

    So, we know that X is compact and thus totally bounded (specifically bounded). Thus, X\subseteq \left\{\bold{x}\in\mathbb{R}^n:\|\bold{x}\|\leqsla  nt M\right\} for some constant M. Let N=\left\{\bold{x}\in\mathbb{R}^n:\|\bold{x}\|\leqs  lant M+d(X,Y)+1\right\}. This is a compact subspace of \mathbb{R}^n and so G\cap N=K is compact (since the intersection of a compact set and a closed set is compact). Now, we next claim that d(E,G)=d(E,K), but I leave this for you to prove. Thus, we must merely show that d(E,G)=d(E,K)=d(e,k) for some e\in E,k\in K. To do this we note that since G,K are compact that G\times K is compact by Tychonoff's theorem. Also, it is clear that d:G\times K\mapsto\mathbb{R} is continuous and so it assumes it's minimum on G\times K and the conclusion follows.
    hmm...I don't have enough background to understand your proof...there are concepts and theorems that I've never seen...Thanks though...

    But I have some ideas regarding the method suggested in the hint:
    d(xnk,ynk) is convergent, hence bounded
    Since X is bounded, d(xnk,0) is bounded
    With this and the triangle inequality, I think we can show that (ynk) is bounded, and so it has a convergent subsequence.
    But I can't figure out the other parts of this proof...


    Any ideas about part b of this problem? I don't know how to construct a counterexample...
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