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Math Help - Are the set of integers not complete?

  1. #1
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    Are the set of integers not complete?

    Define the metric spaces with  ( \mathbb {Z} , \rho ) , the set of integers with the metric  \rho (x,y) = \mid x - y \mid

    Find sequence  \{ x_n \} in this metric space such that it is cauchy but do not converge.

    I understand I must find a sequence that takes only integer values but converge to non-integer. But how should I get it to be cauchy?

    Thank you.
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  2. #2
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    By setting Cauchy constant \varepsilon = 1/2 you easily see how all Cauchy sequences in this space must look like. And to what values they converge.
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    So suppose that  \{ x_n \} is a cauchy sequence in  \mathbb {Z} , then  \forall \epsilon > 0 , there exists  N \in \mathbb {N} \ s.t. \ \forall n,m \geq N , we have  \mid x_n - x_m \mid < \epsilon

    Set  \epsilon = \frac {1}{2} , we then have  \mid x_n - x_m \mid < \frac {1}{2}

    Am I on the right track here?
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  4. #4
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    Quote Originally Posted by tttcomrader View Post
    So suppose that  \{ x_n \} is a cauchy sequence in  \mathbb {Z} , then  \forall \epsilon > 0 , there exists  N \in \mathbb {N} \ s.t. \ \forall n,m \geq N , we have  \mid x_n - x_m \mid < \epsilon

    Set  \epsilon = \frac {1}{2} , we then have  \mid x_n - x_m \mid < \frac {1}{2}

    Am I on the right track here?
    It is not possible for |x_n-x_m| < \tfrac{1}{2} unless x_n = x_m because x_n,x_m are integers and so their difference is always an integer. Thus, the only way you can have an integer less than 1/2 is when that integer is zero.
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