 Solution

Fix an interior point $o\in {\mathbb{H}}^3$. The limit set of a Kleinian group is the set of boundary accumulation points of its orbit, orThe boundary compactification may be realized as the closed unit ball. A discrete isometry group has only finitely many orbit points in each compact interior region: isometries moving $o$ into a compact set form a compact family after accounting for the compact rotation stabilizer, and infinitely many distinct group elements there would contradict discreteness. The stabilizer of $o$ is finite for the same reason. Thus an infinite group has an infinite orbit escaping every compact interior set and a nonempty boundary accumulation set. A finite group has empty limit set. This also shows the displayed intersection does not introduce ordinary interior accumulation points.

For another point $o^{\prime }$, the corresponding points satisfy $d(\gamma o,\gamma o^{\prime })=d(o,o^{\prime })=:C$. In the Poincaré ball model the distance identity isIf ${\gamma }_{j}o\to \xi \in S_{\infty }^2$, bounded distance impliesSo ${\gamma }_j{o}^{\prime }\to \xi$ as well. Exchanging $o,o^{\prime }$ proves equality of the two accumulation sets. Hence the definition is independent of the interior base point.

It is closed as an intersection of the compact orbit closure with the closed boundary, and it is $\Gamma$-invariant because multiplying an orbit by an element of $\Gamma$ permutes that orbit. Changing ball coordinates or conjugating the geometric realization transports the set by the induced boundary homeomorphism. Different nonconjugate representations of an abstract group are not asserted to have the same subset of one fixed sphere. None of these arguments requires finite generation.