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 -- http://mail.python.org/mailman/listinfo/python-list