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An iterated function system is defined as a finite set of contraction mappings, defined over a complete metric space $X$, and iteration is defined as sequential composition of these contraction mappings, where each mapping has some nonzero probability of being used at every iteration. Hutchinson (1981) proved that such iterated function systems (IFS) have invariant sets that iteration converges to, call this set $S \subset X$, so that for $n$ different contraction mappings defining our IFS, we have $$ S = \bigcup_{i=1}^n f_i(S) $$ I am wondering if such invariant sets $S$ are necessarily fractal if they are generated by an IFS. It is well know that many fractals can be generated by an IFS, but it is not clear if IFS's necessarily generate fractal invariant sets.

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    $\begingroup$ What definition of "fractal" are you working with here? $\endgroup$ Jan 28, 2021 at 5:02
  • $\begingroup$ @NoahSchweber For definition of fractal I suppose I am most interested in proving something is a Cantor set or "Cantor set like", so that it has self similarity. I have an iterated function system that appears to have an attractor that is a Cantor set, but I do not know how to prove that it is one. $\endgroup$
    – user918212
    Jan 30, 2021 at 17:00

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It slightly depends on your definition of fractal, but typically no it is not true that IFS always generate fractals. Let $X = [0,1]$ and consider $f_1$ and $f_2$ defined on $X$ by $f_1(x) = \frac{x}{2}$ and $f_2(x) = \frac{1+x}{2}$. These are both contractions with a contractive factor of $\frac{1}{2}$. It's easy to check that $$ [0,1] = f_1([0,1]) \cup f_2([0,1]),$$ so the invariant set is $X$ itself.

There's also the dumb example where the IFS consists of a single contraction, then the attractor is just a point.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Xander Henderson
    Jan 30, 2021 at 17:38

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