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Thread: 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 $\displaystyle a \in A $ and K such that $\displaystyle d(x_n,a) < K \ \ \ \ \ \forall n$

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

    Consider the following:

    $\displaystyle | 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 $\displaystyle d(x_n,x_{N+1}) \leq d(x_{N+1},x_{N+2}) + d(x_n,x_{N+2}) $

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

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

    Now, let $\displaystyle 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 $\displaystyle 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; Feb 1st 2011 at 06:20 PM.
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  2. #2
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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 $\displaystyle a \in A $ and K such that $\displaystyle d(x_n,a) < K \ \ \ \ \ \forall n$
    Now, since the sequence is Cauchy I can set $\displaystyle \epsilon = 1 \exist N $ such that $\displaystyle \for n,m > N $, we have $\displaystyle d(x_n,x_m)<1 $
    This is just a suggestion to shorten things.
    Now you have that $\displaystyle N$. BTW it is $\displaystyle m,n \ge N$.
    Let $\displaystyle M = \sum\limits_{1 \leqslant j < k \leqslant N} {d(x_j ,x_k )} $. Your bound is now $\displaystyle M+2$.
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