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Math Help - A Cauchy sequence of real numbers is bounded

  1. #1
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    Question A Cauchy sequence of real numbers is bounded

    Hello,

    There is one step that I did not get in the following proof I found.

    Let Xn be a sequence of real numbers. If Xn is Cauchy then the sequence is bounded.

    Proof:
    Given 1 there exist N such that for all n>=N, |xn-xN|<1.
    It follows that|xn|<=|xN|+1.
    Let b the maximum of |x1|,...,|xN|,|XN|+1. Then b is a bound for the sequence.

    I am facing troubles understanding how did you get from |xn-xN|<1 to |xn|<=|xN|+1
    Any hint will be very appreciated !!
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  2. #2
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    Re: A Cauchy sequence of real numbers is bounded

    Quote Originally Posted by matemauch View Post
    I am facing troubles understanding how did you get from |xn-xN|<1 to |xn|<=|xN|+1
    Any hint will be very appreciated !!
    It is very simple.
    We know that |A|-|B|\le |A-B|.
    So |x_n|\le |x_N|+1
    Thanks from matemauch
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