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For fractals defined iteratively (via subdivision) like the Koch curve or Sierpinsky triangle, what is the Hausdorf dimension of the intermediate iterates?

Specifically, for a fractal S defined as \Lim_n S_n, where each Sn is constructed by finite operations, is it correct to state that all S_n have integral dimension and only S has fractal dimension?

If not, how does the dimension converge?

firdaus
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    Your statement that "each $S_n$ is constructed by finite operations" is quite imprecise. If you're talking about the iterative procedure where we apply an IFS ${f_i}{i=1}^m$ recursively from an initial compact set $S_0$ to obtain a sequence via the formula $$S{n+1}=\bigcup_{i=1}^m f_i(S_n),$$ then the answer is - each $S_n$ has the same dimension as $S_0$ since it's just a countable collection of copies of $S_n$. Of course, what you've discovered is that the limit in the Hutchinson metric does not respect the Hausdorff dimension. – Mark McClure Apr 22 '15 at 01:33
  • Thanks for that. Does it also hold that for each point in $S_n$ it's dimension - locally - remains integral? Moreover is there a notion of local HD for a fractal? By local, I mean in an open neighborhood around each point. – firdaus Apr 22 '15 at 01:44
  • Specifically I'm trying to figure out if this argument is well formed: define $f:S_n \rightarrow R$ as the local dimension (topological or Hausdorf) then it is continous. And although $S_n \rightarrow S$, $f(S_n)$ does not converge to $f(S)$. – firdaus Apr 22 '15 at 01:46
  • Tip: you can use @username to tell something directly to that user, if you do that, the person gets a notification for this. – Pedro Apr 22 '15 at 01:48
  • Again, the dimension of $S_n$ need not be integral. If $S_0$ is the Cantor set, then each $S_n$ will have dimension $\log(2)/\log(3)$. You can certainly make reasonable notion of local dimension regardless - just look at $$\lim_{r\rightarrow0}\dim(B_r(x)\cap S).$$ You might prefer a $\limsup$ or $\liminf$ in some contexts. I don't believe there's any reason to expect the local dimension to be continuous, though. – Mark McClure Apr 22 '15 at 01:51
  • F or the specific case where $S_0$ is integral (like in the Koch curve) then it is the case that each $S_n $ is point wise of dimension 1, except for the limiting case which may, pointwise, have non constant dimension? Would you have any references that talk abFt pointwise topological properties of IFS constructions? – firdaus Apr 22 '15 at 02:01
  • If $S_0$ is point-wise of dimension one (which it might or might not be when constructing the Koch curve), then yes each $S_n$ will be pointwise of dimension 1. The limit (if we're talking about a strictly self-similar set) will certainly have constant dimension. Topological properties of self-similar sets is quite a tricky topic! – Mark McClure Apr 22 '15 at 02:13
  • @Mark McClure So given these assumptions my previous statement regarding the convergence of this local dimension function holds? Thanks again. Everything regarding self similar sets is a tricky topic :) – firdaus Apr 22 '15 at 02:24

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