Infinitely generated Kleinian groups Is it true that the Hausdorff dimension of the limit set of a Kleinian group is the supremum of Hausdorff dimensions of finitely generated subgroups [perhaps under addtional hypotheses?]I can't seem to find a reference which proves (or disproves) this fact. This is related to my previous question on commutator of $\Gamma(2)$...
EDIT: It is a Theorem of D. Sullivan (1979) that the critical exponent $\delta(\Gamma)$ is the $\sup$ of critical exponents for finitely generated subgroups, and it is another theorem of Sullivan (1984) that the Hausdorff dimension of the radial limit set is equal to the critical exponent.
 A: No.  Here's how you can construct, given $\epsilon > 0$, an infinitely generated Kleinian group whose limit set has Hausdorff
dimension 2, but every finitely generated subgroup has Hausdorff dimension $< \epsilon$.
The easiest examples are infinite free groups.  First choose a sequence of bounds for the target Hausdorff
dimensions of the subgroup generated by first $k$ generators,
$$0 = \epsilon_1 < \epsilon_2 < \dots < \epsilon .$$
Let $\{x_i\}$ be a countable dense sequences of elements of the Riemann sphere.
Start with a cyclic group, generated say be a hyperbolic element with translation length 1, so its limit set is
two points, with Hausdorff dimension 0, where one of the points is $x_1$. Call this group
$G_1$.
For the second generator, we will take the free product with another similar cyclic group with translation length 1. For its axis, choose a  line ending at $x_2$ (or $x_3$ if $x_2$ is in the limit set of the existing group $G_1$).
Claim:  if the hyperbolic distance between the two axes is sufficiently large, the  group is discrete and its limit set is a Cantor set of Hausdorff dimension $< \epsilon_2$.
  This is easy to prove: make a graph consisting of the axes, their common perpendicular,
and all the images of these under the group.  Construct the perpendicular bisecting plane of the common perpendicular, and all its images under the group.  All but a countable set
of infinite words in the group correspond to geodesics that cross these planes infinitely
often. The diameters of the circles at infinity decrease geometrically, by an arbitrarily
large constant.
Inductively, let $x_i$ be the first of the sequence of points that is not in the limit 
set of the group $G_k$ generated by the first $k$ elements, choose a line
ending at $x_i$ that is far from the convex hull of the limit set of $G_k$, and define
$G_{k+1}$ by adjoining a generator that translates by distance 1 along the new generator.
By reasoning similar to the above, the Hausdorff dimension of the limit set of $G_{k+1}$
can be made arbitrarily close to the Hausdorff dimension of the limit set for $G_k$:
construct (this means imagine) the perpendicular bisecting planes to the shortest
geodesic from the axis of the new generator to the convex hull of the limit set of
the old.  The limit set for the new group consists of a countable union of copies of
the limit sets for $G_k$ and the new cyclic free factor, together with points that are
accessible only by passing through an infinite number of copies of these planes.
If the planes are spaced far enough apart, then the sum of the visual radii to the $\epsilon_{k+1}$ power can be made to converge. (This uses the fact that the Hausdorff dimension of $G_k$ is less than $\epsilon_k < \epsilon_{k+1}$). By omitting a finite number
of them  the sum becomes arbitrarily small, hence the Hausdorff dimension is $< \epsilon_{k+1}$.
Every finitely generated subgroup is contained in some $G_k$, so its limit set has
Hausdorff dimension less than $\epsilon$.  But the limit set for $G = \cup_i G_i$
is the entire sphere, so it is $2$.
A: If in the original question we replace limit set by radial limit set than the answer is yes by the two Sullivan's theorems. So in your exmple there are planty of non-radial limit points. What types are they?  
