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Re: Onto [0,1]
Posted:
Apr 25, 2013 10:42 AM


On Thu, 25 Apr 2013 00:18:49 0700, William Elliot <marsh@panix.com> wrote:
>On Sun, 21 Apr 2013, baclesback@gmail.com wrote: > >> More specifically, use the representation of x in C Cantor set in base 3 >> with only 0's and 2's in the expansion of 3, and map >> >> f: x=0.a1a2..... > 0.b1b2....... >> >> Wheref(bi)= 0 , if ai=0 , f(bi)=1 , if ai=2 . >> >> This f is continuous, but not absolutely continuous ( which >> preserves sets of measure zero. ) > >Why is f continuous?
Let eps > 0. There exists N such that if x, y are in [0,1] and they have binary expansions that agree in the first N "digits" then xy < eps.
(*) Now there exists delta > 0 such that if s, t are in the Cantor set and st < delta then the ternary expansions of s and t agree in the first N "digits".
So s, t in C, s  t < delta implies f(s)  f(t) < eps.
(To see why (*) is true: Go back to the construction of the Cantor set: C is the intersection of the C_n, where C_n is the union of 2^n intervals each of length 1/3^n. Choose delta > 0 so the smallest gap between a pair of these 2^n intervals is larger than delta. Then s, t in C, st < delta implies that s and t lie in the same component of C_n, hence st <= 1/3^n.)



