Exercise: If we define $$d(m,n) = \left|\dfrac{1}{m} -\dfrac{1}{n}\right|$$ for $m,n\in\mathbb{N}$, show that $d$ is equivalent to the usual metric on $\mathbb{N}$ but that $(\mathbb{N},d)$ is not complete.
What I've tried: I know that $d(x,y)$ and the usual/euclidian metric $\hat{d}(x,y)$ are equivalent if $d(x_n,x)\to 0$ iff $\hat{d}(x_n,x)\to 0$. So what I need to show is that there exists a $N\in\mathbb{N}$ such that $\left|\dfrac{1}{x_n} - \dfrac{1}{x}\right| <\epsilon \Leftrightarrow \left|x_n - x\right|<\epsilon$ whenever $n\geq N$. I've tried rewriting the norms but I don't get very far. $\left|\dfrac{1}{x_n} - \dfrac{1}{x}\right|=\left|\dfrac{x - x_n}{x_nx}\right|<\dfrac{\epsilon}{\left|x_nx\right|}$ doesn't get me anywhere.
I know that $(\mathbb{N},d)$ is complete if every Cauchy sequence in $\mathbb{N}$ converges to a point in $\mathbb{N}$. So if I can show that there exists a sequence in $\mathbb{N}$ that converges to a point outside $\mathbb{N}$, I've shown that $(\mathbb{N},d)$ is not complete. The problem is that I don't know what this sequence would look like.
Question: How do I solve this exercise?
Thanks!
If $d (x_n,x)\to0$, then by definition $\frac1 {x_n}\to\frac1x $. As the function $t\longmapsto\frac1t $ is continuous, we get that $x_n\to x $. That is, $\hat d (x_n,x)\to0$.
Conversely, if $x_n\to x $, again we could use the continuity of $t\longmapsto1/t $ as above, or simply note that $x_n $ is eventually constant.
As for completeness, take $x_n=n $ to produce a sequence that is Cauchy but not convergent.