On Wed, Jun 17, 2009 at 4:50 AM, Lawrence D'Oliveiro<l...@geek-central.gen.new_zealand> wrote: > In message <7x63ew3uo9....@ruckus.brouhaha.com>, wrote: > >> Lawrence D'Oliveiro <l...@geek-central.gen.new_zealand> writes: >> >>> I don't think any countable set, even a countably-infinite set, can have >>> a fractal dimension. It's got to be uncountably infinite, and therefore >>> uncomputable. >> >> I think the idea is you assume uniform continuity of the set (as >> expressed by a parametrized curve). That should let you approximate >> the fractal dimension. > > Fractals are, by definition, not uniform in that sense.
I had my doubts on this statement being true, so I've gone to my copy of Gerald Edgar's "Measure, Topology and Fractal Geometry" and Proposition 2.4.10 on page 69 states: "The sequence (gk), in the dragon construction of the Koch curve converges uniformly." And uniform continuity is a very well defined concept, so there really shouldn't be an interpretation issue here either. Would not stick my head out for it, but I am pretty sure that a continuous sequence of curves that converges to a continuous curve, will do so uniformly. Jaime -- (\__/) ( O.o) ( > <) Este es Conejo. Copia a Conejo en tu firma y ayúdale en sus planes de dominación mundial. -- http://mail.python.org/mailman/listinfo/python-list