Define the limit set for a Kleinian group. If your definition of the limit set requires an arbitrary choice of a base point, you should prove that the limit set does not depend on this choice.

Let $\Delta_{1}, \Delta_{2}, \Delta_{3}, \Delta_{4}$ be the four discs $\{z \in \mathbb{C}:|z-c| \leqslant 1\}$ where $c$ is the point $1+i, 1-i,-1-i,-1+i$ respectively. Show that there is a parabolic Möbius transformation $A$ that maps the interior of $\Delta_{1}$ onto the exterior of $\Delta_{2}$ and fixes the point where $\Delta_{1}$ and $\Delta_{2}$ touch. Show further that we can choose $A$ so that it maps the unit disc onto itself.

Let $B$ be the similar parabolic transformation that maps the interior of $\Delta_{3}$ onto the exterior of $\Delta_{4}$, fixes the point where $\Delta_{3}$ and $\Delta_{4}$ touch, and maps the unit disc onto itself. Explain why the group generated by $A$ and $B$ is a Kleinian group $G$. Find the limit set for the group $G$ and justify your answer.