David C. Ullrich <ullr...@math.okstate.edu> writes:
> On Wed, 17 Jun 2009 08:18:52 -0700 (PDT), Mark Dickinson
> <dicki...@gmail.com> wrote:
>
>>On Jun 17, 3:46 pm, Paul Rubin <http://phr...@nospam.invalid> wrote:
>>> Mark Dickinson <dicki...@gmail.com> 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".
I think it is quite common to refer to call 'curve' the image of its
parametrization. Anyway there is a representation theorem somewhere
that I believe says for subsets of R^2 something like:
A subset of R^2 is the image of a continuous function [0,1] -> R^2
iff it is compact, connected and locally connected.
(I might be a bit -or a lot- wrong here, I'm not a practising
mathematician) Which means that there is no need to find a
parametrization of a plane curve to know that it is a curve.
To add to this, the usual definition of the Koch curve is not as a
function [0,1] -> R^2, and I wonder how hard it is to find such a
function for it. It doesn't seem that easy at all to me - but I've
never looked into fractals.
--
Arnaud
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