On Jun 17, 2:18 pm, pdpi <pdpinhe...@gmail.com> wrote: > On Jun 17, 1:26 pm, Jaime Fernandez del Rio <jaime.f...@gmail.com> > wrote: > > > P.S. The snowflake curve, on the other hand, is uniformly continuous, right? > > The definition of uniform continuity is that, for any epsilon > 0, > there is a delta > 0 such that, for any x and y, if x-y < delta, f(x)-f > (y) < epsilon. Given that Koch's curve is shaped as recursion over the > transformation from ___ to _/\_, it's immediately obvious that, for a > delta of at most the length of ____, epsilon will be at most the > height of /. It follows that, inversely, for any arbitrary epsilon, > you find the smallest / that's still taller than epsilon, and delta is > bound by the respective ____. (hooray for ascii demonstrations)
I think I'm too stupid to follow this. 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', in the language of precalculus 101), so I'm confused. Here's an alternative proof: Let K_0, K_1, K_2, ... be the successive generations of the Koch curve, so that K_0 is the closed line segment from (0, 0) to (1, 0), K_1 looks like _/\_, etc. Parameterize each Kn by arc length, scaled so that the domain of the parametrization is always [0, 1] and oriented so that the parametrizing function fn has fn(0) = (0,0) and fn(1) = (1, 0). Let d = ||f1 - f0||, a positive real constant whose exact value I can't be bothered to calculate[*] (where ||f1 - f0|| means the maximum over all x in [0, 1] of the distance from f0(x) to f1(x)). Then from the self-similarity we get ||f2 - f1|| = d/3, ||f3 - f2|| = d/9, ||f4 - f3|| = d/27, etc. Hence, since sum_{i >= 0} d/(3^i) converges absolutely, the sequence f0, f1, f2, ... converges *uniformly* to a limiting function f : [0, 1] -> R^2 that parametrizes the Koch curve. And since a uniform limit of uniformly continuous function is uniformly continuous, it follows that f is uniformly continuous. Mark [*] I'm guessing 1/sqrt(12). -- http://mail.python.org/mailman/listinfo/python-list