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Continuity of a given map

  1. Apr 29, 2009 #1
    1. The problem statement, all variables and given/known data
    Consider the map phi : C -> I which maps each point of the middle third Cantor set C, considered as a subset of real numbers between 0 and 1 written in base 3 and containing only digits 0 and 2, to the set of real numbers I=[0,1] written in base 2, according to the rule: 0.a_1a_2a_3... -> 0.b_1b_2b_3... where b_i = a_i / 2

    (1) Prove that phi is a continuous map of C onto I.
    (2) Prove that phi is not bijective.


    2. Relevant equations



    3. The attempt at a solution
    I decided to use the definition of continuity that all open sets in I must have open pre-images in C. I tried saying pick (1/2,1) in I (that is, all elements of the form 0.1b_2b_3...). This open intervals pre-image in C would be the intersection of (2/3,1) with the cantor set, C.

    It's really sketchy in my head and I would love some help. Also proving that phi is not bijective. I feel it may have something to do with certain decimal expansions.

    Thanks.
     
  2. jcsd
  3. Apr 29, 2009 #2
    There are more elegant ways to do this, but I would just get my hands dirty and prove phi is continuous at each x in the domain, the boring epsilon-delta way.

    Here is a hint, by looking at how to prove phi is continuous at x=1/4. Note 1/4 has ternary expansion 0.020202.... , so phi(1/4)=0.010101... in binary, i.e. 1/3.

    Let epsilon > 0.

    Find N such that 1/2^N < epsilon. For this hint, suppose N=5 works.

    (Note: I'm not going to be careful and figure out whether I meant N or N-1 or N+1 or N+2 in this hint.)

    So go out to the fifth (or sixth) binary "decimal" place in f(1/4)=1/3, so you have 0.010101..., and furthermore L < 1/3 < U, where L=0.01010100000000... and U= 0.01010111111111..... (binary).

    Note U-L<1/2^5 (or 6 or whatever).

    Now L and U are the images of 0.020202000000... and 0.0202022222222.... (ternary) respectively, and these two numbers differ by 0.00000022222... (ternary) which is 1/3^5 (or 6 or 7), so now you have your delta (or divide it by 2).

    Just write this up in general.

    You may or may not have to make a special case if x ends with 0 repeating or 2 repeating (ternary). For example 1/3 = 0.100000... = 0.0222222... (ternary) so 1/3 is in C. What is phi(1/3)? Also 2/3 = 0.12222222... = 0.2000000.... (ternary) so 2/3 is in C. What is phi(2/3)?
     
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