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Topic: Onto [0,1]
Replies: 40   Last Post: Apr 29, 2013 10:16 PM

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David C. Ullrich

Posts: 3,238
Registered: 12/13/04
Re: Onto [0,1]
Posted: Apr 25, 2013 10:42 AM
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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 |x-y| < eps.

(*) Now there exists delta > 0 such that if s, t are in the
Cantor set and |s-t| < 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, |s-t| < delta implies
that s and t lie in the same component of C_n,
hence |s-t| <= 1/3^n.)







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