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Consider the metric $$d(m,n) = \frac{|m-n|}{mn}$$

Is this metric on the natural numbers $(1,2,\ldots)$ complete?

I'm struggling but heres an idea I have from reading other similar questions.

The sequence given by $n^2 = \{1,4,9,16,\ldots\}$ is Cauchy in this metric space but the only way it converges is if there exists a natural number $m$ such that $d(n^2,m)=0$. But this implies $n=m=1$. This is a contradiction, right? So it is not a complete metric space?

Thanks for your time.

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It is true that $(\mathbb N,d)$ is not a complete metric space. Consider the sequence $1,2,3,4,\ldots$, which is a Cauchy sequence. However, it doesn't converge. Fix $N\in\mathbb N$. Then the distance from $N$ to any other $M\in\mathbb N$ (distinct from $N$) is at least $\frac1{N(N+1)}$. So, no injective sequence can possible converge to $N$.

Of course, the same argument shows that your sequence doesn't converge. However, your argument is not correct.

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No, not quite. It's perfectly possible for convergent (and hence Cauchy) sequences to never attain their limit. The fact that $d(m, n^2) > 0$ is not sufficient to show that $n^2$ fails to converge.

Here's another way to look at it. Note that the metric can be expressed like so: $$d(m, n) = \left|\frac{1}{n} - \frac{1}{m}\right|.$$ If we let $$f : \Bbb{N} \to [0, 1] : m \mapsto \frac{1}{m},$$ where the domain has the metric $d$ and the set of reciprocals has the usual absolute value metric, then $f$ is an isometry, that is, $|f(m) - f(n)| = d(m, n)$.

Note that the sequence $n^2$ maps to $\frac{1}{n^2}$, which converges in $[0, 1]$ to $0$. This makes $n^2$ Cauchy, since $f$ is an isometry. But $0$ is not in the range of $f$; if there were some $L \in \Bbb{N}$ such that $n^2 \to L$, then $f(n^2) \to f(L) \neq 0$. This would contradict the uniqueness of limits. Thus $n^2$ fails to converge.

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  • $\begingroup$ Thanks for the insight theo! I'm equipped to write (and understand) my proof now $\endgroup$ – Bruno Mar 8 at 19:55

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