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Math Help - Show there exists a SUBSEQUENCE converging to L

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    Smile Show there exists a SUBSEQUENCE converging to L





    [note: also under discussion in Math Links forum]
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  2. #2
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    Quote Originally Posted by kingwinner View Post




    [note: also under discussion in Math Links forum]

    ** There is a subs. \{x_{n_k}^{(1)}\} converging to L_1\Longrightarrow\,\,\,for\,\,\,\epsilon=1\,\,\ex  ists\,K_1\in\mathbb{N}\,\,\,s.t.\,\,\,|x_{n_k}^{(1  )}-L_1|<1\,\,\,\forall\,\,n_k^{(1)}>K_1 . Choose now n_1:=n^{(1)}_{K_1+1} .

    ** There is a subs. \{x_{n_k}^{(2)}\} converging to L_2\Longrightarrow\,\,\,for\,\,\,\epsilon=1\slash 2\,\,\exists\,K_2\in\mathbb{N}\,\,\,s.t.\,\,\,|x_{  n_k}^{(2)}-L_2|<1\slash 2\,\,\,\forall\,\,n_k^{(2)}>K_2 . Choose now n_2:=n^{(2)}_{K_2+h_2} in such a way that n_2>n_1 (we can do this because we've an infinite subsequence here and thus there exists h_2 \,\,s.t.\,\,\,K_2+h_2>K_1+1 !)

    ** There is a subs. \{x_{n_k}^{(3)}\} converging to L_3\Longrightarrow\,\,\,for\,\,\,\epsilon=1\slash 3\,\,\exists\,K_3\in\mathbb{N}\,\,\,s.t.\,\,\,|x_{  n_k}^{(3)}-L_3|<1\slash 3\,\,\,\forall\,\,n_k^{(3)}>K_3 . Choose now n_3:=n^{(3)}_{K_3+h_3} in such a way that n_3>n_2 (we can do this because we've an infinite subsequence here and thus there exists h_3 \,\,s.t.\,\,\,K_3+h_3>K_2+h_2 !)

    Continue inductively in the above way, and in the m-th step:

    ** There is a subs. \{x_{n_k}^{(m)}\} converging to L_m\Longrightarrow\,\,\,for\,\,\,\epsilon=1\slash m\,\,\exists\,K_m\in\mathbb{N}\,\,\,s.t.\,\,\,|x_{  n_k}^{(m)}-L_m|<1\slash m\,\,\,\forall\,\,n_k^{(m)}>K_m . Choose now n_m:=n^{(m)}_{K_m+h_m} in such a way that n_m>n_{m-1} (we can do this because we've an infinite subsequence here and thus there exists h_m \,\,s.t.\,\,\,K_m+h_m>K_{m-1}+h_{m-1} !)

    Take it from here...

    Tonio
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    Smile

    Quote Originally Posted by tonio View Post
    ** There is a subs. \{x_{n_k}^{(1)}\} converging to L_1\Longrightarrow\,\,\,for\,\,\,\epsilon=1\,\,\ex  ists\,K_1\in\mathbb{N}\,\,\,s.t.\,\,\,|x_{n_k}^{(1  )}-L_1|<1\,\,\,\forall\,\,n_k^{(1)}>K_1 . Choose now n_1:=n^{(1)}_{K_1+1} .

    ** There is a subs. \{x_{n_k}^{(2)}\} converging to L_2\Longrightarrow\,\,\,for\,\,\,\epsilon=1\slash 2\,\,\exists\,K_2\in\mathbb{N}\,\,\,s.t.\,\,\,|x_{  n_k}^{(2)}-L_2|<1\slash 2\,\,\,\forall\,\,n_k^{(2)}>K_2 . Choose now n_2:=n^{(2)}_{K_2+h_2} in such a way that n_2>n_1 (we can do this because we've an infinite subsequence here and thus there exists h_2 \,\,s.t.\,\,\,K_2+h_2>K_1+1 !)

    ** There is a subs. \{x_{n_k}^{(3)}\} converging to L_3\Longrightarrow\,\,\,for\,\,\,\epsilon=1\slash 3\,\,\exists\,K_3\in\mathbb{N}\,\,\,s.t.\,\,\,|x_{  n_k}^{(3)}-L_3|<1\slash 3\,\,\,\forall\,\,n_k^{(3)}>K_3 . Choose now n_3:=n^{(3)}_{K_3+h_3} in such a way that n_3>n_2 (we can do this because we've an infinite subsequence here and thus there exists h_3 \,\,s.t.\,\,\,K_3+h_3>K_2+h_2 !)

    Continue inductively in the above way, and in the m-th step:

    ** There is a subs. \{x_{n_k}^{(m)}\} converging to L_m\Longrightarrow\,\,\,for\,\,\,\epsilon=1\slash m\,\,\exists\,K_m\in\mathbb{N}\,\,\,s.t.\,\,\,|x_{  n_k}^{(m)}-L_m|<1\slash m\,\,\,\forall\,\,n_k^{(m)}>K_m . Choose now n_m:=n^{(m)}_{K_m+h_m} in such a way that n_m>n_{m-1} (we can do this because we've an infinite subsequence here and thus there exists h_m \,\,s.t.\,\,\,K_m+h_m>K_{m-1}+h_{m-1} !)

    Take it from here...

    Tonio
    Thanks for your great help! But I still have some concerns about extracting the information given in the problem and the notations for different subsequences.

    As stated in the problem, we are using the notation (x_n_k) for the target subsequence (the one that we need to construct at the end), so I'm going to stick with that.

    "For each k≥1, there is a subsequence of (x_n) converging to L_k...".
    So if you're using (x_n_k (1) ), (x_n_k (2) ), ..., (x_n_k (m) ),..., wouldn't this notation mean that it is a (further) subsequence of our target subsequence (x_n_k)???
    Also the question says "For each k≥1...", so I think k should occur in the bracket (x_n_k (k) ), right? But then, we'll have two k's in it. Is that OK?

    So now I am confused with all those subscripts, indices and the relation between them. k seems to have a special meaning in this question that is tied to the subscript of our target subsequence. So we are going to fix k=1,k=2,k=3,... one at a time to construct our target subsequence.

    What is the usual (and correct) way to denote the DIFFERENT subsequences of (x_n)?

    Thanks for clarifying this! It has been a nightmare for me...
    Last edited by kingwinner; January 23rd 2010 at 10:56 AM.
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