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In the hyperbolic-ball model, the limit set of a Kleinian group is the boundary accumulation set of an orbit $Go$ of an interior point: $\Lambda (G)=\overline{Go}\cap \partial {\mathbb{H}}^3$. The choice of $o$ does not matter, since any two such orbits remain a fixed hyperbolic distance apart. It is closed and $G$-invariant. The complicated/non-elementary hypothesis excludes relatively compact groups and elementary groups with at most two limit points; in particular the limit set is nonempty. Having no finite boundary orbit alone would not exclude a compact rotation group, whose interior orbits are bounded.

If $\Lambda (G)$ were finite and nonempty, the orbit of any of its points would be a finite boundary orbit, contrary to the hypothesis. Thus it is infinite. In fact it cannot be countable. We give the essential minimality/perfectness argument. A sequence of Möbius transformations sending an interior point toward $\xi \in \Lambda$ has a subsequence converging on the boundary away from one exceptional point to the constant $\xi$. This follows by normalizing the representing $2\times 2$ matrices: a diverging sequence has a rank-one projective limit, with one kernel line and one image line. For any boundary point $\eta$, the absence of a finite orbit allows choosing $h\in G$ with $h\eta$ different from that exceptional point; then $g_{j}h\eta \to \xi$. Hence the orbit of every $\eta \in \Lambda$ is dense in $\Lambda$.

If $\xi$ were an isolated limit point, density of every orbit would make every point belong to its orbit: each orbit must hit the open singleton $\{\xi \}$. Every point would then be isolated by equivariance, making the compact set $\Lambda$ discrete and therefore finite, a contradiction. Thus $\Lambda$ is a nonempty perfect compact metric space. Repeatedly choosing two disjoint smaller neighbourhoods meeting this perfect set, with diameters tending to zero, associates distinct nested limits to all infinite binary sequences. The limit set is therefore uncountable.

Jorgensen inequality states that for a non-elementary discrete group generated by lifts $A,B\in \mathrm{SL}(2,\mathbb{C})$,For a concrete failure of discreteness, take $A=\mathrm{diag}(e^{s/2},e^{-s/2})$ and $B=\left(\begin{array}{cc}1& 1\\ -1& 0\end{array}\right)$ with $s>0$ small. The fixed points of $A$ and of $BA{B}^{-1}$ collectively include $0,\infty ,-1$, so the group is non-elementary. Direct multiplication gives both summands equal to $4{\sinh }^2(s/2)$. Their sum is less than one when $s$ is sufficiently small, proving this two-generator group is not discrete.

For a connected-limit-set example, the modular group generated by $z\mapsto z+1$ and $z\mapsto -1/z$ is discrete because its matrices have integer entries. It preserves the upper half-plane; its limit set is the extended real line, a circle. Indeed the orbit of the parabolic fixed point infinity consists of the rational points and infinity, dense in that circle. For a disconnected example, choose four pairwise disjoint round disks and two loxodromic maps pairing the exterior of each negative disk with the interior of its positive disk. The Schottky group generated by these maps is free and discrete by the disjoint-domain ping-pong construction. Successive reduced words give nested disjoint disks with shrinking diameters; their limiting points form a Cantor set. Thus its limit set is disconnected, unlike that of the modular group.

Suppose a connected subset $C\subset F$ contains distinct points $a,b$. The distance function $x\mapsto |x-a|$ is $1$-Lipschitz, and its image on $C$ is connected and contains $0$ and $|b-a|$. It therefore contains the whole interval between those values. A Lipschitz map cannot increase Hausdorff dimension: taking images of arbitrarily fine covers multiplies their diameter powers by at most a fixed constant. Since a nondegenerate interval has Hausdorff dimension one, ${\dim }_{H}C\ge 1$, contradicting ${\dim }_{H}F<1$. Thus every connected subset is a singleton, so $F$ is totally disconnected.

A Möbius transformation pairs two disks when it maps their boundary circles to one another and takes the exterior of the first disk onto the interior of the second. Two pairs of disks with disjoint interiors give the ping-pong lemma construction of a free two-generator group. Tangencies are allowed in the convention suggested by the question; this is often called a Tangent-disk Schottky group, rather than a strictly separated Schottky group.

Here is a concrete configuration that also makes the dimension argument precise. Take $A(z)=z+2$ and $B(z)=z/(iz+1)$. The first pairs the half-plane disks $\Re z<-1$ and $\Re z>1$, which are disks on the sphere tangent at infinity. The second pairs $|z-i|<1$ and $|z+i|<1$, tangent at zero; their interiors are also disjoint from the half-plane disks. These pairings give a free discrete Kleinian group by ping-pong lemma, with a nonempty open fundamental region.

Set $C=AB{A}^{-1}{B}^{-1}$. Conjugating the subgroup generated by $B,C$ by $M(z)=iz/(2-z)$ gives matricesThe product of the second matrix with the first is $\left(\begin{array}{cc}1& 2\\ 0& 1\end{array}\right)$. These are the standard parabolic generators of the principal level-two modular group, whose ideal-quadrilateral fundamental region tiles the upper half-plane and whose Kleinian limit set is the entire extended real line. Consequently the original group's Kleinian limit set contains the circle $|z-1|=1$, as well as its image under $A$, the distinct tangent circle $|z-3|=1$.

It follows that the limit set has Hausdorff dimension at least one and is not homeomorphic to a circle. Indeed a proper subset of a circle cannot itself be a circle: after deleting a missing point it lies in a line, which cannot contain an embedded circle. A space containing two distinct complete circles thus cannot be homeomorphic to a single circle. This explains why dimension at least one does not force circle topology. The tangent-circle terminology is consistent with the primary discussion of T-Schottky configurations in arxiv.org/pdf/math/0701572 .

A Schottky group is a free, discrete group of Möbius transformations constructed by pairing disjoint Jordan curves on the Riemann sphere, taking the exterior of one bounded disk to the interior of its partner. For round circles it is classical. Strictly separated defining disks give a Cantor set as the limit set of a Kleinian group. The tangent-disk extension permits parabolic elements and is often called a noded or T-Schottky group; its Kleinian limit set need not be a Cantor set.

A Tangent-disk Schottky group uses side-pairing disks with disjoint interiors but permits boundaries to meet. This convention must be distinguished from the strictly separated, purely loxodromic Schottky group convention. For example, the maps $A(z)=z+2$ and $B(z)=z/(iz+1)$ pair the half-plane disks $\Re z<-1,\Re z>1$ and the disks $|z-i|<1,|z+i|<1$. The ping-pong lemma gives a free discrete group. With $C=AB{A}^{-1}{B}^{-1}$, the subgroup generated by $B,C$ is conjugate by $M(z)=iz/(2-z)$ to the principal level-two modular group, generated by $\left(\begin{array}{cc}1& 0\\ 2& 1\end{array}\right)$ and $\left(\begin{array}{cc}1& 2\\ 0& 1\end{array}\right)$. Thus its Kleinian limit set contains the circle $|z-1|=1$ and its translate $|z-3|=1$. Its Hausdorff dimension is at least one and it cannot be homeomorphic to a single circle.