Measuring Fractal Dimension ?

[Steven D'Aprano]

On Sun, 14 Jun 2009 14:29:04 -0700, Kay Schluehr wrote:

> On 14 Jun., 16:00, Steven D'Aprano
> <st... at 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).


> 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.



-- 
Steven