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Math Help - All Cauchy Sequences are bounded in a metric space

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    All Cauchy Sequences are bounded in a metric space

    Prove that all cauchy sequences are bounded in a metric space.

    Suppose that A is a metric space, with d the metric. And suppose that the sequence {Xn} is a cauchy sequence on A.

    I need to prove that it is bounded, meaning that I have to an element  a \in A and K such that  d(x_n,a) < K \ \ \ \ \ \forall n

    Now, since the sequence is cauchy I can set  \epsilon = 1 There exist N such that  \forall n,m > N , we have  d(x_n,x_m)<1

    Consider the following:

     | d(x_n,x_{N+1})-d(x_n,x_{N+2}) | \leq d(x_{N+1},x_{N+2}) (true by the reserve triangular inequality)

    implies that  d(x_n,x_{N+1}) \leq d(x_{N+1},x_{N+2}) + d(x_n,x_{N+2})

    Now, since  d(x_{N+1},x_{N+2}) < 1

    we would have
     d(x_n,x_{N+1}) < 1 + d(x_n,x_{N+2})

    Now, let K = max \{ d(x_1, x_{N+2}), d(x_2,x_{N+2}), . . . , d(x_{N+2},x_{N+2}) \} + 1 , therefore we have  d(x_n,x_{N+1}) < K

    But I know something is wrong with this proof because I'm confusing myself with the n, are there anything I can do to clear that up?

    Thank you very much!
    Last edited by tttcomrader; February 1st 2011 at 07:20 PM.
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    Quote Originally Posted by tttcomrader View Post
    Prove that all cauchy sequences are bounded in a metric space.
    Suppose that A is a metric space, with d the metric. And suppose that the sequence {Xn} is a cauchy sequence on A.
    I need to prove that it is bounded, meaning that I have to an element  a \in A and K such that  d(x_n,a) < K \ \ \ \ \ \forall n
    Now, since the sequence is Cauchy I can set  \epsilon = 1 \exist N such that  \for n,m > N , we have  d(x_n,x_m)<1
    This is just a suggestion to shorten things.
    Now you have that N. BTW it is m,n \ge N.
    Let M = \sum\limits_{1 \leqslant j < k \leqslant N} {d(x_j ,x_k )} . Your bound is now M+2.
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