Measuring Fractal Dimension ?

[Kay Schluehr]

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. 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.