On Jun 17, 4:18 pm, 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, 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
I was being incredibly lazy and using loads of handwaving, seeing as I posted that (and this!) while procrastinating at work. an even lazier argument: given the _/\_ construct, you prove that its vertical growth is bound: the height of / is less than 1/3 (given a length of 1 for ___), so, even if you were to build _-_ with the middle segment at height = 1/3, the maximum vertical growth would be sum 1/3^n from 1 to infinity, so 0.5. Sideways growth has a similar upper bound. 0.5 < 1, so the chebyshev distance between any two points on the curve is <= 1. Ergo, for any x,y, f(x) is at most at chebyshev distance 1 of (y). Induce the argument for "smaller values of one". -- http://mail.python.org/mailman/listinfo/python-list