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Thread: subsequence

  1. #1
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    subsequence

    Prove that if $\displaystyle a_n $ is a sequence such that every subsequence of $\displaystyle a_n $ has a further subsequence converging to $\displaystyle 0 $, then $\displaystyle a_n \to 0 $.

    Let $\displaystyle \sigma: \mathbb{N} \to \mathbb{N} $ be a strictly increasing function. Suppose that $\displaystyle b_{n} = a_{\sigma(n)} $ is a subsequence of $\displaystyle a_n $. Then $\displaystyle c_{n} = b_{\sigma(n)} \to 0 $ is a subsequence of $\displaystyle b_n $. If $\displaystyle \varepsilon > 0 $, then there is an $\displaystyle N \in \mathbb{N} $ such that $\displaystyle n \geq N $ implies that $\displaystyle c_{n} \in (-\varepsilon, \varepsilon) $.

    From here, I have to show that $\displaystyle a_n \in (-\varepsilon, \varepsilon) $.

    Now what? $\displaystyle c_n \leq b_n \leq a_n $. Could I use the Squeeze theorem?
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  2. #2
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    actually use the following theorem twice: $\displaystyle a_n \to L \Longleftrightarrow $ every subsequence of $\displaystyle a_n $ converges to $\displaystyle L $?

    Is this correct?
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  3. #3
    MHF Contributor kalagota's Avatar
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    possibly yes..

    EDIT: but if i read and interpret it correct, it doesn't tell you that every subsequence of the subsequence of a_n also converges to 0.
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by particlejohn View Post
    "every subsequence of $\displaystyle a_n $ converges to $\displaystyle L $"
    you do not have this condition. we are told "every subsequence of has a further subsequence converging to " it is a different statement
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  5. #5
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    yes, but this is a general theorem about subsequences. A subsequence is also a sequence right? Regardless, is my approach correct?
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  6. #6
    MHF Contributor kalagota's Avatar
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    if you are going to use that, you have to show that every subsequence of the subsequence of a_n also converges to 0.
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  7. #7
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    Quote Originally Posted by kalagota View Post
    possibly yes..

    EDIT: but if i read and interpret it correct, it doesn't tell you that every subsequence of the subsequence of a_n also converges to 0.

    Oh okay. It just has $\displaystyle 1 $ subsequence converging to $\displaystyle 0 $?
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  8. #8
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    Let $\displaystyle \{a_n\}$ be a sequence such that for any subsequence of $\displaystyle \{a_n\}$ we can find a subsequence which is convergent to zero. Assume that $\displaystyle a_n$ does not converge to zero, then it means for some $\displaystyle \epsilon > 0$ there are some $\displaystyle a_{n_1},a_{n_2}, ... $ such that $\displaystyle |a_{n_k}|\geq \epsilon$. The sequence $\displaystyle \{ a_{n_k}\}$ is a subsequence of $\displaystyle \{a_n\}$ but it does not have any subsequence which is convergent to zero since all its terms are $\displaystyle \geq \epsilon$ in absolute value. This is a contradiction. Thus, $\displaystyle \{ a_n\}$ must be convergent to zero.

    This is Mine 11th Post!!!
    Last edited by ThePerfectHacker; Jul 7th 2008 at 05:04 PM.
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