# Is there some known way to create the Mandelbrot set (the boundary), with an iterated function system?

Is there some known way to create the Mandelbrot set (the boundary), with an iterated function system (IFS)?

Julia sets can be formed by iterating the two functions $z \mapsto \pm \sqrt{z-c},$ and this quickly converges to the Julia set. Formally, the two functions form a Hutchinson operator, and the Julia set is the unique compact fixed set. That is, $J = f_1(J) \cup f_2(J)$ where $f_1,f_2$ are the two functions, and $J$ is the Julia set.

However, Mandelbrot is the iteration of many functions, with start value 0, so there is no obvious way to define an IFS. Is there some non-obvious way?

That is, find maps $g_1,\dots,g_n$ s.t. the Mandelbrot $M$ can be expressed as $M = \bigcup_i g_i(M),$ where each $g_i$ maps $M$ to some subset of $M,$ but not the entire $M.$

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What do you mean by "create"? If $J_c$ is the Julia set corresponding to $c\in \mathbb C$, let $X_{c,n}$ be the set you create by taking the $n$-th iterate of your 2-valued function $z\longmapsto \sqrt{z−c}$, with the initial iterate being $z_0=0$. Do you want "create" to mean that $\cap_{n=1}^\infty \overline{\cup_{k=n}^\infty X_{c,k}} = J_c$ or something like that? –  Ryan Budney Sep 3 '11 at 21:26
Yes, something like that. I mean, the Julia set is the fixed set for a certain Hutchinson operator, en.wikipedia.org/wiki/Hutchinson_operator with the two functions given above. –  Per Alexandersson Sep 4 '11 at 7:27

Also it is not quite true that all Julia sets are fixed sets for Hutchinson operators, at least under the usual definition of Hutchinson operator which requires strict contractivity of the maps (the maps $z\to\pm\sqrt{z-c}$ are in general not contractions, even on the Julia set itself)