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An iterated function system (IFS) is a finite set of contraction mappings on a complete metric space. Symbolically,

$$\{f_i :X \to X \mid i=1,2,..n \},\quad n \in \mathbb{N}$$

is an IFS if each $f_i$ is a contraction on the complete metric space $X$.

I think union, intersection or difference of different IFSs generate new IFSs. This is simply a fact. For example, the union of two IFSs also has an attractor. But, is there any relation between the attractor of union IFS and the attractors of the two IFSs? So, I wonder if it will make sense to use union, intersection or difference of two different IFSs. If it will, how is it interpreted for the attractors or other properties of IFSs?

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    $\begingroup$ Hmm, the standard Cantor set is the attractor of an IFS of two functions. ($x \mapsto 1/3 x$ and $x \mapsto 1/3x + 2/3$). This is the "union" of the two IFS consisting of just one function. Each of these sub-IFS have a single point as the attractor. So you may state trivialities like that the attractor of a subset is a subset of the attractor of the union. Beyond that some quite specific assumptions would seem necessary. $\endgroup$
    – Fabian Wirth
    Commented Oct 24, 2019 at 0:44
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    $\begingroup$ The question about "does it make sense" makes little sense (or rather is quite tautological, so is not really a question). The set of IFS on $X$ is the set of finite subset of a certain set (the set of contractions of $X$). So it inherits all operations which can be defined on the set of finite subsets of an arbitrary set. So it should be enough to focus the question on how the attractor operation behaves with respect to these operations. $\endgroup$
    – YCor
    Commented Oct 24, 2019 at 6:10

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If $F = f_1, \ldots, f_n$ is an IFS, then the attractor of any subset of F is a subset of the attractor of F. This is readily seen through coding points in the attractor (see Barnsley's book for the definition of "code space"), through which it is straightforward that codes with indices in the subset of $F$ are exactly the codes for the attractor of the subset.

What is more interesting to me is that if one topologizes the set of IFS -- for example, if one is using affine transformations on Euclidean space these have the standard topology on matrices -- then the process of taking an attractor is a continuous map from the space of IFS to the space of compact subsets of your metric space.

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  • $\begingroup$ Do you know of a good reference for your second statment? I'm curious as to what happens if you go from a totally disconnected attractor to a connected attractor by continuously varying the IFS. $\endgroup$
    – Zorngo
    Commented Nov 29, 2019 at 13:55
  • $\begingroup$ I recall this being proved in Barnsley's book. Examples of what you are looking for are plentiful. Consider the IFS on the real line with two affine maps which shrink to 0 and 1 with stretch factor s, so $s=1/3$ gives the Cantor set. As $s$ passes from $s < 1/2$ to $s = 1/2$ the attractor goes from being totally disconnected to connected. $\endgroup$
    – Dev Sinha
    Commented Nov 29, 2019 at 18:48
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    $\begingroup$ The underlying fact which makes this not so surprising is that totally disconnected spaces are dense in the space of all metric spaces. (Pointillism at work!) $\endgroup$
    – Dev Sinha
    Commented Nov 29, 2019 at 18:49

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