Does the sequence $\{a_n\}_n$ converge in a special metric of $(\mathbb R_+,d)$

Question

I am trying to prove or disprove convergence of the Cauchy sequence $\{a_n\}$ with $a_n = n$ for all $n\in \mathbb N$ in the metric: $$ d(x,y) = \left|\frac 1x - \frac 1y\right|. $$ I can see that as $n$ increases, the distance of $a_n$ to $0$ is decreasing, but under this metric, I can't have $0$ because $1/0$ is undefined. Does this mean it does not converge? And thus, the space $(\mathbb R_+,d)$ is not complete?

Answer 1

If $a_n \to a$ then, given $\epsilon >0$ there exists $n_0$ such that $|\frac 1n -\frac 1 a|<\epsilon$ for all $n \geq n_0$. This gives $\frac 1 a <2\epsilon$ if $n \geq n_0$ and $n \geq \frac 1 {\epsilon}$. Since $\epsilon$ is arbitrary this gives $\frac 1 a=0$ which is a contradiction. Hence, the sequence is not convergent.

Answer 2

$\Bbb R_+$ is ambiguous. Let me assume it means $(0,+\infty)$ (but the answer would essentially be the same with $[0,+\infty)$).

The map $x\mapsto\frac1x$ is an isometric bijection between your $\left((0,+\infty),d\right)$ and the subspace $(0,+\infty)\subset\Bbb R$ with the usual metric, which we shall denote by $\delta.$ It maps your $a_n=n$ to $b_n:=\frac1n.$

Since (for $\delta$) $b_n\to0\notin(0,+\infty),$ $(b_n)$ is Cauchy but has no limit in $\left((0,+\infty),\delta\right).$ Therefore, $(a_n)$ is Cauchy but has no limit in $\left((0,+\infty),d\right).$