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This post is a follow up to my previous question enquiring whether it is always possible to extend a homeomorphism conformal on a region $R$ minus a Cantor set to the whole of $R$. From the answers I received it is clear that one can't say in general whether this is possible for some function $f$ without knowing further properties of either $f$ or the Cantor set. Thus I'd like to expand upon the particular cases I'm interested in, namely iterated function systems with Cantor set limit set, and ask whether the question has been investigated in these cases. I'm mainly interested in the cases where one has only two generators, so will describe these in more detail below, but am also interested in the cases where one has an arbitrary finite or infinite number of generators.

Let $R$ be a simply connected open region of $\mathbb{C}$ (if need be we can specify that the boundary $\partial R$ of $R$ is piecewise smooth). Let $R_0, R_1 \subset R$ and let $f_0$ and $f_1$ be homeomorphisms mapping $R$ to $R_0$ and $R_1$ respectively (again, if need be we can specify that the boundaries $\partial R_0$ and $\partial R_1$ are piecewise smooth). Letting $\overline{R}_0$ and $\overline{R}_1$ be the closures of $R_0$ and $R_1$ respectively we stipulate that $\partial R_0$ and $\partial R_1$ are disjoint, and thus $R \setminus (\overline{R}_0 \cup \overline{R}_1)$ is simply connected. Furthermore, we make the assumtion that each of $\partial R_0$ and $\partial R_1$ intersects $\partial R$ at at most one point. This gives us three possibilities, examples of which are shown below.

The three types of IFS

Let $f_0$ and $f_1$ be homeomorphisms, mapping $R$ onto $R_0$ and $R_1$ respectively. If $f_0$ and $f_1$ are holomorphic I think I'm correct in saying that the limit set $\Lambda$ (a.k.a. the accumulation set) will always be a Cantor set.

Now, assume that we only know that $f_0$ and $f_1$ are holomorphic on $R \setminus \Lambda$, but that we can show that $\Lambda$ is a Cantor set (we can always show this when the boundaries of $R$, $R_0$, and $R_1$ are all pairwise disjoint, and perhaps it is also true for the other two types?). If need be we can also assume that any point of the pairwise intersection of $\partial R$ and $\partial R_0$ or $\partial R_1$ is a fixed point of either $f_0$ or $f_1$ respectively. Can we conclude that they are holomorphic on the whole of $R$?

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This question was posed a long time ago, so you may already have found a solution for it yourself. In the case where the boundaries of your domains are disjoint, the resulting limit set will indeed be holomorphically removable. This is because it has absolute area zero. You may be interested in looking at Jeremy Kahn's PhD thesis (Holomorphic Removability of Julia Sets, available on the arXiv) for more background on holomorphic removability. –  Lasse Rempe-Gillen Nov 24 '12 at 13:25
    
Hi Lasse, my understanding is that absolute area zero is not sufficient in general, and the set needs to be of Hausdorff dimension less than 1 to guarantee holomorphic removability. Is this not the case? If absolute area zero is sufficient then this is good news as I'm pretty sure that the other two examples are also of absolute area zero, due to the 'parabolic' points of the limit set being countable. –  uncooltoby Nov 24 '12 at 18:16

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