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, and since any curve can be parametrized in many different ways any proof of uniform continuity should specify exactly which parametrization is in use. Mark -- http://mail.python.org/mailman/listinfo/python-list