Prove that the sequence in the metric space $(\Bbb N, d)$ with the property that as $j\rightarrow\infty$, $n_j\rightarrow\infty$ diverges? Given the metric space $(\Bbb N,d)$ where $d$ is defined by $\forall m,n\in\Bbb N, d(m,n)=|\frac{1}{m}-\frac{1}{n}|.$ 
I need to use epsilon-delta to show that any sequence $(n_j)_{j \in \Bbb N}$ in this metric space satisfying the property that as $j\rightarrow\infty$, $n_j\rightarrow\infty$ diverges. 
To formalize the property, we have the following:
$(\forall \epsilon \in \Bbb N)(\exists N \in \Bbb N)(\forall j>N)(n_j >\epsilon)$
That is I need to prove 
$(\forall n\in \Bbb N)(\exists \epsilon>0)(\forall N\in \Bbb N)(\exists j\ge N)
(|\frac{1}{n_j}-\frac{1}{n}|\ge \epsilon).$
What I need to do is to fix an arbitrary $x\in \Bbb N.$ Then take a suitable $\epsilon.$ Can someone help me find a suitable $\epsilon$? I understand there are other ways to prove this. But I just want to use the epsilon delta definition to prove it. Thanks so much. 
 A: Note that $f: (\mathbb{N},d) \rightarrow (\mathbb{R},|.|)$ given by $f(n)= \frac{1}{n}$ is an isometry (and so a homeomorphism between $\mathbb{N}$ and $Y = f[\mathbb{N}]$. So if $n_j\rightarrow x$, $j \rightarrow \infty$ converged to some $x$, the image sequence $f(n_j)$ would converge to $f(x) \in Y$. But it also converges to $0$ in the reals. Limits are unique in metric spaces.... 
(added on request) If you want to do it the harder way: note that 
$$\left|\frac{1}{n_j} - \frac{1}{n}\right| \ge \epsilon \text{ iff } \frac{\left|n_j - n\right|}{n_j n} \ge \epsilon \text{ iff} |n_j - n| \ge n_j n \epsilon$$
So fo some fixed $n$ take $\epsilon = \frac{1}{n}$ and take $N$ such that $j \ge N$ implies $n_j > 2n$, which can be done for all fixed $n$. Then the above inequalities say that $d(n_j,m) = \left|\frac{1}{n_j} - \frac{1}{n}\right| \ge \epsilon$ as $n_j - n > n$ (note $n\epsilon = 1$ by choice).
It's clear by the first paragraph that $(n_j)_j$ is Cauchy (as its image is (convergent hence) Cauchy under an isometry). But is has no limit in $\mathbb{N}$, as we saw, so $(\mathbb{N},d)$ is not complete.
