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Math Help - Complete metric space

  1. #1
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    Complete metric space

    I am working on this problem, but I am not sure about one of my arguments. Could someone help me?
    Let (X,d) be a metric space. Define d'(x,y)=min\{d(x,y),1\}. Prove that (X,d) is complete iff (X,d') is complete.
    Suppose (X,d) is complete. Let \{x_n\} be a Cauchy sequence in (X,d'). Let \epsilon >0. Then there exists N such that for all m,n \geq N then d'(x_m,x_n)< \epsilon

    If 0< \epsilon <1, then min\{d(x_m,x_n),1\} < \epsilon implies that d(x_m,x_n) < \epsilon.

    If \epsilon \geq 1, then I want to conclude that d(x_m,x_n) < \epsilon, but I can't justify this.

    So \{x_n\} is Cauchy in (X,d). Since (X,d) is complete, \{x_n\} converges to some x \in X. This implies there exists M such that if n \geq M, then  d(x_n,x) < \epsilon. Let \epsilon =1. Then d(x_n,x) <1. So, if n \geq M then d'(x_n,x)=min\{d(x_n,x), 1\}=d(x_n,x) < \epsilon. So, \{x_n\} converges to x \in X.

    I think the converse is analogous, if I can show that any Cauchy sequence in (X,d') is Cauchy in (X,d), then the result would follow.
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  2. #2
    Senior Member Tinyboss's Avatar
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    Right, the point is just that Cauchy sequences in one metric are Cauchy in the other, because d=d' when either one is small. Likewise, a sequence converges under d if and only if it converges under d'.
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  3. #3
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    Quote Originally Posted by Tinyboss View Post
    Right, the point is just that Cauchy sequences in one metric are Cauchy in the other, because d=d' when either one is small. Likewise, a sequence converges under d if and only if it converges under d'.
    Thanks for your help, Tinyboss. But could you help me answer the part I can't justify?
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  4. #4
    Senior Member Tinyboss's Avatar
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    You can always assume \varepsilon<1 and forget the other case. Because if you find an N that works for \varepsilon<1, it always works for any larger \varepsilon.
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