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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.

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  • $\begingroup$ If $x_n$ is eventually constant, then it will be convergent. What would a convergent $x_n$ look like if it were not eventually constant but converged to $\ell$? $\endgroup$ – BigMathTimes Mar 6 '17 at 22:44
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    $\begingroup$ What, you couldn't cover your tracks by deleting the question, so you thought defacing it would work to hide it? The revisions are still visible to the general public, and people on the site usually notice these things. $\endgroup$ – Asaf Karagila Mar 7 '17 at 13:04
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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.

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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.

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