Compactification of the integers With the usual discrete topology, $\mathbb{Z}$ is non-compact.
I wonder if it is possible to make $\mathbb{Z}$ compact; I think there are many different ways to do so.
I've heard about the following idea:
Consider the stereographic projection of the closed lower semi-circle as illustrated in the Picture below, where $\mathbb{Z}$ is embedded in the $x$-axis and where the Points $(\pm 1,1)$ represent $-\infty$ and $+\infty$, respectively. I guess
1) Is this a well known idea? 
2) Why is this now compact?
3) As a metric one can use the arclength metric?
4) I guess this has nothing to do with one-point compactification?
5) I do not really see how the Points $(x,1)$ on the closed lower semi-circle which are different from $(\pm 1,1)$ are projected.

 A: Observe that this is similar to embedding $\mathbb R$ in the extended reals $\overline{\mathbb R}$. The extended reals are just $\mathbb R$ with $-\infty$ and $\infty$ joined to each end. The extended reals are of course compact (since the space turns out to be homeomorphic to the interval $[0,1]$). Basically, what your construction does is considers $\mathbb Z$ first as a subspace of $\overline{\mathbb R}$ and then takes the closure - which includes $-\infty$ and $\infty$. Then, a closed subspace of a compact space is compact.
Note that, you can show your construction gives a compact space rather directly, since it identifies $\mathbb Z$ with a closed and bounded subset of the plane, which must be compact. You can also directly show that, if we let $\overline{\mathbb Z}$ be the new topology, then a subset $U$ is open if and only if it satisfies two conditions:


*

*If $\infty\in U$, then there is some integer $N$ such that for all $n\geq N$ we have $n\in U$.

*If $-\infty\in U$, then there is some integer $N$ such that for all $n\leq N$ we have $n\in U$.
Then, note that, if you have an open cover of $\overline{\mathbb Z}$, then all the large enough elements are covered by any open set containing $\infty$ and all the small enough elements are covered by any open set containing $-\infty$, so only finitely many elements remain to be covered, implying compactness directly.
I don't know if I'd call this "well-known", but it is implicit in a lot of places - for instance, if you take a limit
$$\lim_{n\rightarrow\infty}(1+1/n)^n$$
then, you are implicitly using this topological space to define $\infty$.
The arc length metric would work fine. 
This is somewhat related to the one-point compactification. Actually, if you use a stereographic projection to wrap the integers around a circle and then add the top point, you get the one point compactification. You seem to have conflated this projection with the gnomonic projection in your question.
