This is just the IFS { x -> (1/2) x - 1/3, x -> (1/2) x + 5/6 }. It satisfies the open set condition, so yes, it is homeomorphic to a Cantor set and has Hausdorff and similarity dimension = 1. -Thomas C On Tue, Sep 8, 2015 at 9:42 AM, James Propp <jamespropp@gmail.com> wrote:
Starting from S_0 = [0,1], iteratively define S_n (for n = 1, 2, 3, ...) as the set obtained from S_{n-1} by replacing each of the 2^n closed intervals [a,b] by the two closed intervals [a-1/3^n,(a+b)/2-1/3^n] and [(a+b)/2+1/3^n, b+1/3^n] (that is, we break the interval in half, duplicate the midpoint, shift the left half to the left by 1/3^n, and shift the right half to the right by 1/3^n).
It seems to me that at each stage we get 2^n disjoint closed intervals of length 1/2^n separated from one another by 1/3^n, and that every point (including the duplicate points created along the way) moves a finite distance and therefore has a well-defined location "at time infinity". So we seem to get a well-defined set "at time infinity".
Am I right? And do we obtain a set of measure 1 that's homeomorphic to the Cantor set?
I've seen constructions of "fat Cantor sets" on the web, where one modifies the culling procedure, but I've never seen one that uses sliding and preserves the measure throughout the procedure. So I'm worried that I'm overlooking something.
Jim Propp _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun