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Math Help - Converging Subsequence

  1. #1
    MHF Contributor harish21's Avatar
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    Converging Subsequence

    Suppose that a sequence { x_n} is monotone and that it has a convergent subsequence. Show that { x_n} also converges.

    I think that "since the subsequence is convergent, the sequence { x_n} is bounded", and
    according to the monotone convergence theorem,a monotone sequence is convergent if and only if it is bounded.

    I am not sure about the first part of my reasoning, because there is a theorem that says "every bounded sequence has a convergent subsequence". Based on that, I am stating that "since the subsequence is convergent, the sequence is bounded".

    Is this the correct reasoning or are there any other ways this statement can be proved? Suggestions will be appreciated
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by harish21 View Post
    Suppose that a sequence { x_n} is monotone and that it has a convergent subsequence. Show that { x_n} also converges.

    I think that "since the subsequence is convergent, the sequence { x_n} is bounded", and
    according to the monotone convergence theorem,a monotone sequence is convergent if and only if it is bounded.

    I am not sure about the first part of my reasoning, because there is a theorem that says "every bounded sequence has a convergent subsequence". Based on that, I am stating that "since the subsequence is convergent, the sequence is bounded".

    Is this the correct reasoning or are there any other ways this statement can be proved? Suggestions will be appreciated
    Assume WLOG that x_n is increasing. We know that there exists some \{x_{\varphi(n)}\} such that x_{\varphi(n)}\to x. So, let \varepsilon>0 be given. Then, there exists some N\in\mathbb{N} such that N\leqslant n\implies \left|x-x_{\varphi(n)}\right|=x-x_{\varphi(n)}<\varepsilon. But, if \varphi\left(N\right)\leqslant n\implies x_{\varphi(N)}\leqslant x_n\implies x-x_n\leqslant x-x_{\varphi(N)}<\varepsilon
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