Complete metric space

**jackie** · March 12th 2011, 05:17 PM

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.

**Tinyboss** · March 12th 2011, 05:41 PM

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'.

**jackie** · March 12th 2011, 08:09 PM

Originally Posted by Tinyboss

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?

**Tinyboss** · March 12th 2011, 08:46 PM

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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