On Wed, 17 Jun 2009 07:35:35 +0200, Jaime Fernandez del Rio <jaime.f...@gmail.com> wrote:
>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. Nope. Not that I see the relvance here - the g_k _do_ converge uniformly. >Jaime -- http://mail.python.org/mailman/listinfo/python-list