Let $M$ be a finite-volume hyperbolic manifold. If we chop off the cusps (are there only finitely many?) to get a compact surface with (finitely many?) boundary components, and then lift to the universal cover, we get hyperbolic space after deleting a family of horoballs with disjoint closures (known as neutered or truncated hyperbolic space). Apparently, it is true that:
- The set of basepoints of those horoballs is dense in the boundary sphere.
- There is an R > 0 such that no two horoballs in this family are at distance < R from each other.
Any idea why these statements are true? My rough thoughts are as follows (but even if I'm on the right track I don't know how to make the arguments precise). The first statement I think should have to do with $M$ being finite-volume. Maybe the idea is that a fundamental domain for the lattice action must have finite volume, but it won't if there's an open set of points on the boundary sphere that don't correspond to deleted horoballs. The second statement maybe has to do with how there are finitely(?) many boundary components that lift to the horospheres, and since the covering map is a local isometry, you can't find pairs of horospheres that are arbitrarily close.
UPDATE:
I have verified that indeed there are only finitely many cusps, and I have also verified statement 2.
As for statement 1, I tried looking through this book by Kapovich.
He defines a discrete subgroup $\Gamma$ of Isom($\mathbb{H}^n$) to be geometrically finite if every point in its limit set $\Lambda(\Gamma)$ is either a conical limit point or a cusped parabolic point. He defines a point in the limit set to be a cusped parabolic point if its stabilizer in $\Gamma$ is an almost abelian group $A$ containing parabolic elements such that $\Lambda(\Gamma)/A$ is compact. I believe that these cusped parabolic points are my horoball basepoints. (He even comments later that "parabolic fixed points are never conical limit points.")
Later he proves a theorem by Ahlfors which says that if $G \subset \text{Isom}(\mathbb{H}^3)$ is a geometrically finite discrete group, then its limit set either is all of $\hat{\mathbb{C}}$ or has Lebesgue measure zero. This seemed useful so I took a closer look, but in the proof, he says that a.e. limit point of $G$ is a conical limit point. This is troubling to me because this appears to be contrary to statement 1. Surely I am misunderstanding something, but I can't figure out what it is.
It sounds like you are getting off track by focussing on geometrically finite groups.
What you want to look for is a theorem which says that if $M = \mathbb H^n / \Gamma$ has finite volume, equivalently if $\Gamma$ has a finite volume fundamental domain, then the limit set of $\Gamma$ is the entire sphere $S^{n-1}_\infty$ (equivalently, if the limit set is not the entire sphere, then $M$ has infinite volume).
It follows, from the very definition of the limit set, that every orbit of the $\Gamma$ action on $S^{n-1}_\infty$ is dense.
In particular, for each horoball base point $x \in S^{n-1}_\infty$, its entire orbit $\Gamma \cdot x$ consists of horoball base points, and that orbit is dense.