Endpoints of Cantor Set

  1. How can I write down a closed form expression for the endpoints used in the construction of the Cantor set? i.e., 0, 1, 1/3, 2/3, 1/9, 2/9, 7/9, 8/9, etc.
  2. jcsd
  3. StatusX

    StatusX 2,565
    Homework Helper

    You don't see any patterns there? Try writing down more terms. Or is it that you're trying to prove some expression is correct?
  4. I have to show that these numbers are dense in the Cantor set. I found another way without using the closed form.
  5. AKG

    AKG 2,576
    Science Advisor
    Homework Helper

    The closed form is k/3n for all n, and for all 0 < k < 3n.
  6. HallsofIvy

    HallsofIvy 41,267
    Staff Emeritus
    Science Advisor

    No, it isn't. when n= 1, the first "cut", the enpoints are 1/3 and 2/3 so that the intervals left are [0, 1/3], [2/3, 1] but when you remove the middle third of those, the endpoint are 0, 1/9, 2/9, 1/3= 3/9, 2/3= 6/9, 7/9, 8/9, 1. In your notation, k/3n, with n= 2, k is not 4 or 5. It gets worse with higher values of n.

    Dragonfall, think in terms of base 3. Writing a number between 0 and 1 in base 3, the first "digit" (trigit?) may be 0, 1, or 2. If 0, the number is between 0 and 1/3; if 1, between 1/3 and 2/3; if 2, between 2/3 and 1. When you remove the middle third you remove all numbers have a "1" as first digit. Now, all numbers between 0 and 1/3 must have .00, .01, or .02 as first two digits, all numbers between 2/3 and 1 must have .20, .21, or .22 as first two digits. When you remove the middle third of each you remove all numbers that have 1 as the second digit. Do you see what happens in the limit?
  7. AKG

    AKG 2,576
    Science Advisor
    Homework Helper

    Sorry, just wasn't thinking.
  8. Yes, I was able to prove that the Cantor set contains those and only those numbers whose triadic expansion contains only 0 or 2. I did not use this fact for the denseness; however, but I did use it to prove that any number in [0,2] can be written as a sum of 2 'Cantor numbers'.

    For denseness, I simply showed that since anything between two endpoints is either entirely in or out of the Cantor set, and if there exists a non-endpoint such that some closed ball of radius e about it contains no endpoint, then the Cantor set has at least length 2e, which is impossible.
    Last edited: Sep 22, 2006
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