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Thread: 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 $\displaystyle (X,d)$ be a metric space. Define $\displaystyle d'(x,y)=min\{d(x,y),1\}$. Prove that $\displaystyle (X,d)$ is complete iff $\displaystyle (X,d')$ is complete.
    Suppose $\displaystyle (X,d)$ is complete. Let $\displaystyle \{x_n\}$ be a Cauchy sequence in $\displaystyle (X,d')$. Let $\displaystyle \epsilon >0$. Then there exists N such that for all $\displaystyle m,n \geq N$ then $\displaystyle d'(x_m,x_n)< \epsilon$

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

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

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

    I think the converse is analogous, if I can show that any Cauchy sequence in $\displaystyle (X,d')$ is Cauchy in $\displaystyle (X,d)$, 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 $\displaystyle \varepsilon<1$ and forget the other case. Because if you find an N that works for $\displaystyle \varepsilon<1$, it always works for any larger $\displaystyle \varepsilon$.
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