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A Ternary Cantor Set proof

  1. Feb 22, 2007 #1
    Hey all,

    I would really like help on this probably simple proof:

    That the map x |--> f(x) = (x+2)/3 on [0,1] is a contraction,
    and maps the ternary Cantor set into itself. Also, find it's fixed point.

    (1) I can easily show the fixed point (where f(x) = x) is 1.
    (2) I can also pretty easily show it is a contraction:
    where |f(x) - xo| <= q*|x - xo|, where q < 1, and xo is the fixed point.

    (3) However, I can't seem to find a way to tell whether it maps the ternary Cantor set into itself. I kno the definition of the ternary Cantor set is taking the interval [0,1] and deleting the middle-third of the interval, and then repeating the process on each remaining interval, infinitely.

    What does it mean by mapping the ternary Cantor set into itself? Does x have to start out being in the ternary Cantor set? If so, how is it possible if x can vary from [0,1]? What am I interpreting wrong?


  2. jcsd
  3. Feb 22, 2007 #2
    Look at how the function maps certain intervals. What sort of intervals must x be contained in for it to be in the Cantor set? What then happens to the intervals under function f(x)?
  4. Feb 22, 2007 #3
    ah thx. just the boost i needed =)
  5. Feb 22, 2007 #4


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    It means that if x is in the subset of [0,1] called the Cantor set, then f(x) is also in the Cantor set. You might use the fact that x is in the Cantor set if and only if its base 3 expansion contains no 1s.

    (I've edited to change "contains no 0s" to "contains no 1s"!)

    I've also given a little more thought to this. Since dividing by 3 base 3 just shifts the "decimal" point, this is trivial!
    Last edited: Feb 23, 2007
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