Is the set of natural numbers with this metric complete? Let $\mathbb{N}$, the set of all natural numbers, be given the metric $d$ defined as follows: 
$$ d(m,n) \colon= | m^{-1} - n^{-1} |$$ for all $m$, $m$ in $\mathbb{N}$. 
Then how to determine if $\mathbb{N}$ is complete with this metric or not? 
Of course, the sequence $x_n \colon = n^2 $ for $n = 1, 2, 3, \ldots$, for example, is a Cauchy sequence in this metric space. But how to show if this sequence converges or not? 
 A: Let's use the example that you brought up.
Consider the sequence $\{n^2:n\in\mathbb N\}\subset\mathbb N$.
As you said,
this is clearly a Cauchy sequence.
Now,
you would like to settle if this sequence converges in $\mathbb N$ or not,
that is,
you want to know if you can find a fixed number $m\in\mathbb N$ such that
$$d(n^2,m)=\left|\frac1{n^2}-\frac 1m\right|\to0.$$
The answer is that no such number $m$ exists.
Hint: $\frac1m$ is fixed,
but is there a limit to how small $\frac1{n^2}$ can get as $n\to\infty$? Think of what the value of $|n^{-2}-m^{-1}|$ looks like when $n$ is much bigger than $m$. If a concrete example helps, try $m=5$ and $n=10,15,20...$, and compare
$$\left|\frac1{10^2}-\frac15\right|,\left|\frac1{15^2}-\frac15\right|\text{ and }\left|\frac1{20^2}-\frac15\right|$$
versus $|1/5|$.
A: Fix $m\in\Bbb Z^+$. Show that if $m\ne n\in\Bbb Z^+$, then $d(m,n)\ge\frac 1m-\frac1{m+1}$. Conclude that the space has the discrete topology, and hence that the only convergent sequences are those that are eventually constant. In particular, the Cauchy sequence that you mention in your question cannot converge.
