Measuring Fractal Dimension ?

David C. Ullrich ullrich at
Thu Jun 18 14:26:56 EDT 2009

On Wed, 17 Jun 2009 08:18:52 -0700 (PDT), Mark Dickinson
<dickinsm at> wrote:

>On Jun 17, 3:46 pm, Paul Rubin <http://phr...@NOSPAM.invalid> wrote:
>> Mark Dickinson <dicki... at> writes:
>> > It looks as though you're treating (a portion of?) the Koch curve as
>> > the graph of a function f from R -> R and claiming that f is
>> > uniformly continuous.  But the Koch curve isn't such a graph (it
>> > fails the 'vertical line test',
>> I think you treat it as a function f: R -> R**2 with the usual
>> distance metric on R**2.
>Right.  Or rather, you treat it as the image of such a function,
>if you're being careful to distinguish the curve (a subset
>of R^2) from its parametrization (a continuous function
>R -> R**2).  It's the parametrization that's uniformly
>continuous, not the curve,

Again, it doesn't really matter, but since you use the phrase
"if you're being careful": In fact what you say is exactly
backwards - if you're being careful that subset of the plane
is _not_ a curve (it's sometimes called the "trace" of the curve".

>and since any curve can be
>parametrized in many different ways any proof of uniform
>continuity should specify exactly which parametrization is
>in use.

Any _closed_ curve must have [a,b] as its parameter 
interval, and hence is uniformly continuous since any
continuous function on [a,b] is uniformly continuous.


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