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Math Help - Real Analysis - Sequences 3

  1. #1
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    Real Analysis - Sequences 3



    Please help!!

    Need counter-example I believe
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Phyxius117 View Post


    Please help!!

    Need counter-example I believe
    a) What about x_n=(-1)^n,y_n=(-1)^{n+1}+\frac{1}{n}
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  3. #3
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by Phyxius117 View Post


    Please help!!

    Need counter-example I believe
    d.) Assuming this sequence is in the metric space (M,d), if \{b_n\} converges, then it is bounded. ( \exists x_0\in M, R\in\mathbb{R} such that d(x_0,b_n)<R~\forall n)

    So d(a_n,x_0)-R<d(a_n,b_n)~\forall n. (Triangle Inequality)

    But since \{a_n\} is unbounded, so is d(a_n,x_0), because x_0 is a fixed point. Therefore d(a_n,b_n) is unbounded as well.

    @ Drexel: x_n needs to converge.
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Drexel28 View Post
    a) What about x_n=(-1)^n,y_n=(-1)^{n+1}+\frac{1}{n}
    Quote Originally Posted by redsoxfan325 View Post
    d.) Assuming this sequence is in the metric space (M,d), if \{b_n\} converges, then it is bounded. ( \exists x_0\in M, R\in\mathbb{R} such that d(x_0,b_n)<R~\forall n)

    So d(a_n,x_0)-R<d(a_n,b_n)~\forall n. (Triangle Inequality)

    But since \{a_n\} is unbounded, so is d(a_n,x_0), because x_0 is a fixed point. Therefore d(a_n,b_n) is unbounded as well.

    @ Drexel: x_n needs to converge.
    a) is redonkulous then. Let \lim\text{ }\{x_n+y_n\}=L and \lim\text{ }x_n=L'. Since they both converge we see that L-L'=\lim\text{ }\{x_n+y_n\}-\lim\text{ }x_n=\lim\text{ }\{x_n+y_n-x_n\}=\lim\text{ }y_n
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