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Math Help - Cauchy sequence in Q not converging to zero.

  1. #1
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    Cauchy sequence in Q not converging to zero.

    I have the following exercise:

    Let s_n be a cauchy sequence in Q(rationals) not converging to 0. Show that there exists an e(epsilon) >0 and a natural number N such that either for all n>N, s_n > e or for all n>N, -s_n >e.

    I know that since Q is not complete, we cannot assume that there exists a point (say s) such that s_n coverges to it since this s could well be in the real numbers.

    I am hessitant with the the answer I came up with since i didnt use the fact that s_n is cauchy. What i did is the following:

    From the fact that s_n does not converge to zero, then we can deduce that: there's an e>0 and N such that for all n>N, d(s_n, 0) is not less than e. so this implies that d(s_n, 0) >= e. So absolute value of (s_n - 0) > e in implies that s_n>e or -s_n>e. Now i need to show that s_n=e is not possible, but as of right now i havent been able to do so.

    Any advice and guidence will be greatly appreciated.
    Thank you very much.
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    Re: Cauchy sequence in Q not converging to zero.

    Quote Originally Posted by Arturo_026 View Post

    From the fact that s_n does not converge to zero, then we can deduce that: there's an e>0 and N such that for all n>N, d(s_n, 0) is not less than e. so this implies that d(s_n, 0) >= e. So absolute value of (s_n - 0) > e in implies that s_n>e or -s_n>e..

    The above is wrong
    Thanks from Arturo_026
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  3. #3
    MHF Contributor Siron's Avatar
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    Re: Cauchy sequence in Q not converging to zero.

    Out of the definition of a zero sequence it follows there exists a number e>0 wherefore:
    \forall n \in \mathbb{N}, \exists p>n: |s_p|>e
    (this is the negation of the definition of a zero sequence because it's supposed to be no zero sequence)
    Let \epsilon=\frac{e}{2} and because s_n is a cauchy sequence there exists a natural number N wherefore:
    \forall p,q \geq N: |s_p-q_q|<\epsilon
    either
    \forall p,q \geq N: s_p-\epsilon<s_q<s_p+\epsilon
    Now choose a term s_m wherefore at the same time m\geq N and |s_m|>e=2\epsilon.
    You can consider two cases:
    (1) s_m>0 so ... ?
    (2) s_m<0 so ... ?

    Finish it ...
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    Re: Cauchy sequence in Q not converging to zero.

    Quote Originally Posted by Arturo_026 View Post
    Let s_n be a cauchy sequence in Q(rationals) not converging to 0. Show that there exists an e(epsilon) >0 and a natural number N such that either for all n>N, s_n > e or for all n>N, -s_n >e.
    I know that since Q is not complete, we cannot assume that there exists a point (say s) such that s_n coverges to it since this s could well be in the real numbers.
    The real numbers are still complete.
    So your sequence still converges to some \sigma\in\mathbb{R} such that \sigma\ne 0.
    Therefore there is a neighborhood of \sigma such that \exists N\in\mathbb{Z}^+ such that all s_n has
    the same sign for n\ge N.
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