Measuring Fractal Dimension ?

David C. Ullrich ullrich at
Tue Jun 16 14:57:57 EDT 2009

On 15 Jun 2009 04:55:03 GMT, Steven D'Aprano
<steven at> wrote:

>On Sun, 14 Jun 2009 14:29:04 -0700, Kay Schluehr wrote:
>> On 14 Jun., 16:00, Steven D'Aprano
>> <st... at> 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

Limits, as in calculus, as in that limit above, are not

>> 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.

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