Is every finitely generated classical Schottky group quasifuchsian? $\DeclareMathOperator\PSL{PSL}$(Classical, finitely generated) Schottky groups are groups generated by finitely many hyperbolic elements of $A_i\in \PSL(2,\mathbb{C}), $ $i<n$ such that the isometric circles of $\{A_i,A_i^{-1}\}_{i<n}$ are pairwise disjoint.
Quasi-Fuchsian groups are discrete subgroups $\Gamma$ of $\PSL(2, \mathbb{C})$ such that their limit set is contained in a Jordan curve, invariant under $\Gamma$.
Are there any examples of Schottky groups that are not quasi-Fuchsian?
It is a theorem of Denjoy-Riesz that any Cantor set on a plane is contained in a Jordan curve. Can the Jordan curve in this case be taken to be invariant under $\Gamma$?
 A: Yes - every Schottky group is quasi-fuchsian.  See Lemma 1 of Chuckrow's paper "On Schottky Groups with Applications to Kleinian Groups" published in Annals of Mathematics, 1968.
The argument there is nice.  Start with a different, classical, Schottky group $\Gamma'$ where all of the circles are perpendicular to one, given round circle $C'$.  So $\Gamma'$ is fuchsian (and thus quasi-fuchsian).  Now use the fact (!) that the space of Schottky groups is connected - so we are given a quasi-conformal map conjugating $\Gamma'$ to the desired group $\Gamma$ and taking $C'$ to the desired Jordan curve $C$.
This proof is very slick, but does use some machinery.  There is a more direct proof given by "connecting-the-dots".  (However, in some sense, this proof is not really different.)  For the sake of an easy life, let's restrict to the classical case. "Carefully" choose a pair of points in each isometric circle. (The points need to match up.)  Since there are $2n$ circles, this gives us $4n$ points.  Now "carefully" choose $4n$ (smooth) arcs meeting the isometric circles only in their endpoints, which are at the chosen points.  (The arcs need to connect the points in a good cyclic order.)  Now act on these arcs using the group and take the union.  The result is the desired Jordan curve.
