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Math Help - subsequence

  1. #1
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    subsequence

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

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

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

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

    This is Mine 11th Post!!!
    Last edited by ThePerfectHacker; July 7th 2008 at 06:04 PM.
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