On 15 Jun 2009 04:55:03 GMT, Steven D'Aprano <ste...@remove.this.cybersource.com.au> wrote:
>On Sun, 14 Jun 2009 14:29:04 -0700, Kay Schluehr wrote: > >> On 14 Jun., 16:00, Steven D'Aprano >> <st...@removethis.cybersource.com.au> wrote: >> >>> Incorrect. Koch's snowflake, for example, has a fractal dimension of >>> log 4/log 3 ? 1.26, a finite area of 8/5 times that of the initial >>> triangle, and a perimeter given by lim n->inf (4/3)**n. Although the >>> perimeter is infinite, it is countably infinite and computable. >> >> No, the Koch curve is continuous in R^2 and uncountable. > >I think we're talking about different things. The *number of points* in >the Koch curve is uncountably infinite, but that's nothing surprising, >the number of points in the unit interval [0, 1] is uncountably infinite. >But the *length* of the Koch curve is not, it's given by the above limit, >which is countably infinite (it's a rational number for all n). No, the length of the perimeter is infinity, period. Calling it "countably infinite" makes no sense. You're confusing two different sorts of "infinity". A set has a cardinality - "countably infinite" is the smallest infinite cardinality. Limits, as in calculus, as in that limit above, are not cardinailities. > >> Lawrence is >> right and one can trivially cover a countable infinite set with disks of >> the diameter 0, namely by itself. The sum of those diameters to an >> arbitrary power is also 0 and this yields that the Hausdorff dimension >> of any countable set is 0. > >Nevertheless, the Hausdorff dimension (or a close approximation thereof) >can be calculated from the scaling properties of even *finite* objects. >To say that self-similar objects like broccoli or the inner surface of >the human lungs fails to nest at all scales is pedantically correct but >utterly pointless. If it's good enough for Benoît Mandelbrot, it's good >enough for me. -- http://mail.python.org/mailman/listinfo/python-list
