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I have been reading through John Hutchinson's paper "Fractals and Self-Similarity" and some other things, and I haven't really found a definition of strict self-similarity to work with that makes much sense to me. Heuristically we want (any?) part of a strictly self-similar object to look like a smaller copy of the whole object, but how to we formalise this so we can test (ask) whether some set is strictly self-similar?

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  • $\begingroup$ A set is self-similar if it is the invariant set of an iterated function system, as described on the Wiki page for self-similarity. $\endgroup$ – Mark McClure Oct 22 '14 at 2:06
  • $\begingroup$ I understand that this is what we want, but is there a direct definition of self-similar coming from the idea of what we (heuristically) mean by self similar, so we could prove that the invariant set is self similar? I am wondering because it is not entirely clear that the invariant set satisfies my intuition of what a self similar set should be. $\endgroup$ – Dom Oct 22 '14 at 3:25
  • $\begingroup$ Well, it seems to me that this definition does exactly capture what we intuitively mean by self-similar - at least, in the case where the functions in the list are similarity transformations! Perhaps, though, I've used it too long. :) $\endgroup$ – Mark McClure Oct 22 '14 at 10:00
  • $\begingroup$ I kind of thought so too, but then I thought my intuition is that self-similar should mean that (I guess) every point should have a neighbourhood that looks like the whole object, but then you could have multiple similarities of the IFS mapping into this neighbourhood with their images overlapping, which may not look like the whole image anymore. I can't quite see how to resolve this. $\endgroup$ – Dom Oct 22 '14 at 11:05
  • $\begingroup$ You've heard of the Open Set Condition? It's a technical definition that captures the intuition behind "non-overlapping". The fact that Hutchinson had to define it means that he, too, found overlap to be problematic in the analysis of self-similar sets. So I'd say your intuition is spot-on. The resolution is to consider the sub-class of self-similar sets that satisfy the open set condition. Of course, this begs the question - what can we do lacking that condition? $\endgroup$ – Mark McClure Oct 22 '14 at 11:51

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