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". >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. >Mark -- http://mail.python.org/mailman/listinfo/python-list