Convergent sequences in a metric space Consider the set N of natural numbers with the metric
$$d(m,n) = \left\lvert{\frac{1}{m}−\frac{1}{n}}\right\rvert \; \mathit{n, m ∈ \mathbb{N}} $$


*

*Describe all convergent sequences in this metric space and prove that the sequence $[{x_{n}]}^{∞}_{n=1}$ defined by $x_{n} = n$ is a Cauchy
sequence in this metric space.


So I can prove that its a cauchy sequence, im just not sure on what the convergent sequences are.
 A: First of all notice that (according to your example) this metric space is not complete. So there are non-convergent Cauchy sequences. If $x_{n_k}\to\infty,$ then it is not convergent in $\Bbb{N}$
(It may convergent in $\Bbb{N}\cup\{\infty\}$). So Any convergent sequence $(x_n)$ must be bounded. 
Let $N\in\Bbb{N}$ be fixed. Then for any sequence $(x_n)$converge to $N,$ we have $$d(x_n,N)=\left\lvert{\frac{1}{x_n}−\frac{1}{N}}\right\rvert\lt\epsilon$$ for sufficiently large $n\in\Bbb{N},$ where $\epsilon\gt 0$ is arbitrary.
In other words $$x_n\gt\dfrac{N}{\epsilon N+1}.$$ By letting $\epsilon\to 0$ we can conclude that $x_n\ge N$ for large $n.$
Again choosing $\epsilon =\dfrac{1}{N(N+1)}$ one can prove that $x_n\lt N+1$  for large $n$ values.

Hence any convergent sequence must be ultimately constant. 

A: If $m=n+k$ then $d(m,n)=|\frac{k}{n(n+k)}|\le\frac 1n$ since $k\le(n+k)$.
If $n\to\infty$ then $d(m,n)\to 0$.
We have also $d(n,n)=0$ and $d(0,n)=d(n,0)=+\infty$
$d(0,0)$ undefined but I think extending it to $d(0,0)=+\infty$ seems logical, making null sequences also divergent.
In all other cases, $d(m,n)$ is just $>0$
So the convergent sequences with this distance are :


*

*the ones which are not bounded 

*the ones which are eventually stationnary 

*plus for a sequence to be convergent it should not have infinitely many zeros


For last condition, if this was not the case then for any given $n_0$ there would exist $n>n_0$ such that $u_n=0$ and so $d(u_n,u_{n_0})=+\infty$.
Note that it excludes automatically sequences which are eventually zero with the extended distance.
Edit: seeing Nil's post, it's true that considering the particularity of this distance, I thought of convergence in $\overline{\mathbb N}$, but yes in $\mathbb N$ only it reduces to just stationary sequences, less interesting or maybe all the purpose of the exercise was to exhibit a distance that makes the space not complete.
A: Let $g(1)=1/2$ and $g(m)=1/(m^2-m)$ for $m>1.$ Then $d(m,n)\ge g(m)$ for all $m,n.$ So the open ball $B_d(m,r)=\{m\}$ for every $r\in (0,g(m)].$
Hence $(F(n))_n$ converges to $m$
iff $\{n: F(n)\not \in B_d(m,r)\}$ is finite for every $r>0$
iff $\{n:F(n)\not \in B_d(m,r)\}$ is finite for every $r\in (0,g(m)]$
iff $\{n: F(n)\not \in \{m\}\}$ is finite for every $r\in (0,g(m)]$
iff $\{n: F(n)\not \in \{m\}\}$ is finite
iff $\{n:F(n)\ne m\}$ is finite.
