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Math Help - Cauchy Sequences

  1. #1
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    Cauchy Sequences

    I am working on a problem and was just reading a thread where a user posted :

    (X,d) is metric space. for sequence (x_n)_{n\in \mathbb{N}} \subset  X we say that is Cauchy sequence if

    (\forall \varepsilon >0) (\exists n_0=n_0(\varepsilon) \in \mathbb{N}) (\forall n,m \in \mathbb{N}) (n,m\ge n_0 \Rightarrow d(x_n,x_m) < \varepsilon )

    meaning that, sequence is Cauchy sequence if

    \displaystyle \lim _{n,m \to \infty } d(x_n,x_m) = 0

    Is it sufficient to show just

    \displaystyle \lim _{n,m \to \infty } d(x_n,x_m) = 0

    for a sequence to be cauchy? Or should it be done using the epsilon definition (which i am struggling with).

    Basically i'm trying to construct a metric on the real numbers so that  (R,d) is not complete. I have created such a metric (i think) with
     d(x,y)= |exp(-x) - exp(-y)| but i am struggling to show with the epsilon definition that the sequence of natural numbers is cauchy on it.
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  2. #2
    Senior Member Tinyboss's Avatar
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    If n\le m (both positive integers) then d(n,m)\le e^{-n}.
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  3. #3
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    n\le m (both positive integers) then d(n,m)\le e^{-n} < e^{-N} < epsilon
    ?

    So i should take N > ln(1/epsilon)?
    Last edited by Scopur; October 10th 2010 at 08:48 PM.
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  4. #4
    Senior Member Tinyboss's Avatar
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    Quote Originally Posted by Scopur View Post
    n\le m (both positive integers) then d(n,m)\le e^{-n} < e^{-N} < epsilon
    ?

    So i should take N > ln(1/epsilon)?
    I would just note that e^{-n}\to 0, so an N can be found for any positive epsilon.
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