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I need to give an example of two metrics on a set that induce the same topology, but where a sequence is Cauchy relative to one of the metrics and not the other.

Any help would be appreciated! Thanks!

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Please make the bodies of your posts self-contained, not relying on the title for content. – Arturo Magidin Apr 6 '11 at 0:23
Closely related:… – Arturo Magidin Apr 6 '11 at 0:25

Hint: What is the topology induced on $\left.\left\{\frac{1}{n}\;\right|\; n\in\mathbb{Z}, n\gt 0\right\}$ by the standard metric?

Alternative example. Take $\mathbb{R}$ with the usual metric, and the metric $$d(x,y) = \Bigl|\arctan(x) - \arctan(y)\Bigr|.$$ Then consider the sequence $a_n = n$.

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This would be incomplete since the limit of 1/n= zero which is not in the set. So the sequence (sn)=1/n would not be Cauchy since it doesn't converge with this metric. If you use the discrete metric d(x,y)=1 x doesn't equal y d(x,y)=0 x=y then (sn) would be Cauchy with this metric. Is this correct? Thanks! – hawaii99 Apr 6 '11 at 0:31
You have it the wrong way around. The sequence is Cauchy (but not convergent) in the standard metric, and not Cauchy (so definitely not convergent) in the discrete metric. – Nate Eldredge Apr 6 '11 at 2:31
@caligurl11: "Cauchy" is not a synonym for "convergent." In any metric space, convergent implies Cauchy, but the converse is only true when the space is complete. The sequence $(1/n)$ is Cauchy in that space under the usual metric, but because the space is not complete, being Cauchy does not necessarily imply convergent. – Arturo Magidin Apr 6 '11 at 3:34
Thank you I showed that (sn) is Cauchy with the regular metric by using the definition with N=1/epsilon. I was wondering how to show that it is not Cauchy with the discrete metric. – hawaii99 Apr 6 '11 at 15:34
@caligirl11: Correction: $d(a_n,a_m)=0$ if $n=m$; but for any $N\gt 0$, you can find $n,m\gt N$ with $n\neq m$, and in that case you will have $d(a_n,a_m)=1\gt 1/2$. Otherwise, fine. – Arturo Magidin Apr 6 '11 at 15:45

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