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I want to come up with an example which shows that there exists a sequence which is Cauchy under both metrics, say, $d$ and $e$, in the same space $X$, yet these metrics are not equivalent.

Since we know that if two metrics have the same convergent sequences then they are equivalent, I need to come up with an incomplete metric space. I've been thinking for quite a while but coulnd't come up with such $X$ and $d$ and $e$. One idea is to think of such a metric which will give $0$ for some two points, whereas the other metric will give some number greater than zero for the same two points in $X$. But what could that metric be?

Would appreciate a hint.

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    $\begingroup$ A constant sequence is always Cauchy $\endgroup$ Oct 15, 2017 at 21:18

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A less trivial example than taking constant sequences, is to take a sequence that converges in two non-equivalent metrics on a set. (Just having one such sequence is easy enough), e.g. $\frac{1}{n}e_n$ in $\ell_2$ in its $\ell_2$ metric and in the metric it inherits from $\mathbb{R}^\mathbb{N}$ in the product metric. In both metrics it converges to the $0$ sequence, and so is Cauchy in both of them.

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  • $\begingroup$ Can you please clarify what is the metric inherited on $\mathbb R ^ \mathbb{N}$ in the product metric? I've looked up the "Product metric" in Wikipedia, but I didn't quite understand the definition. How are the $n$-vectors formed there? Also, what is an $\ell_2$ metric, is it the square root of the infinite sum of the squares of all terms of the sequence? $\endgroup$
    – sequence
    Oct 15, 2017 at 22:41

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