On $\mathbb{Z}\cup\left\{\pm\infty\right\}$ I would like to consider the metric $d$ which is defined by the arclength metric $d_{arc}$ on the inverse stereographic projection with projection Center $(0,1)$ of the semi-circle on the real-line (see the Picture).
That is, I would like to consider $$ d(p,q):=d_{arc}(p',q'):=\text{arclength}(p',q'). $$
My first problem is to find a formula for the inverse stereographic projection in order to identify the Points on the real line with their Points on the semi-circle.
Let $p\in\mathbb{R}$. Then the line through $(0,1)$ and $(p,0)$ should be $$ f(x)=-\frac{x}{p}+1. $$ The semi-circle should have parametrization $$ g(x)=-\sqrt{1-x^2}+1. $$ Hence, in order to determine the Point $p'$ on the semi-circle, I should solve $$ -\frac{x}{p}+1=-\sqrt{1-x^2}+1, $$ yielding $$ x_1=-\frac{p}{\sqrt{p^2+1}},~~~x_2=\frac{p}{\sqrt{p^2+1}}. $$ I think, only $x_2$ makes sense here. Putting $x_2$ in $f$ or $g$ yields $f(x_2)=g(x_2)=-\frac{1}{\sqrt{p^2+1}}+1$.
So the Point $p'$ on the semi-circle should be given by $$ p'=\left(\frac{p}{\sqrt{p^2+1}},-\frac{1}{\sqrt{p^2+1}}+1 \right) $$
Hence, the inverse stereographic projection $S^{-1}$ here should be given by $$ S^{-1}(x)=\begin{cases}\left(\frac{x}{\sqrt{x^2+1}},-\frac{1}{\sqrt{x^2+1}}+1 \right), & x\in\mathbb{R}\\(1,1), & x=+\infty\\(-1,1), & x=-\infty\end{cases} $$
Am I right?
My second Problem now is to define the arclength for Points on the semi-circle.
The length of the arc $p'q'$ is the angle $\theta = \angle PAQ$ (in radians) which can be found by computing the dot product. More explicitly, we have
$$ \cos \theta = \frac{ \left< \vec{AP}, \vec{AQ} \right>}{\| \vec{AP} \| | \vec{AQ} \|} = \frac{ \left< (p,-1), (q,-1) \right>}{ \| (p,-1) \| \| (q, -1) \|} = \frac{pq + 1}{\sqrt{(p^2 + 1)(q^2 + 1)}} $$
and so
$$ d(p,q) = \arccos \left( \frac{pq + 1}{\sqrt{(p^2 + 1)(q^2 + 1)}} \right). $$