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Generalized Cantor Set

  1. Mar 15, 2010 #1
    "Given (rn), rn E (0,1), define a generalized Cantor set E by removing the middle r1 fraction of an interval, then remove the middle r2 fraction of the remaining 2 intervals, etc.

    Start with [0,1]. Take rn=1/5n. Then the material removed at the n-th stage has length < 1/5n, so the total length removed is < 1/5 + 1/52 + 1/53 +... = 1/4
    Thus the length of E is >3/4. "
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    I don't understand why the material removed at the n-th stage has length < 1/5n. How can we derive this? At the n-th stage, we are removing 2n-1 pieces, so don't we have to multiply that by 2n-1?
    I sat down and thought about this for half an hour, but I still can't figure it out.

    I hope someone can explain this! Thank you!
     
  2. jcsd
  3. Mar 15, 2010 #2

    jbunniii

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    It's true that you are removing [itex]2^{n-1}[/itex] intervals at stage [itex]n[/itex], but what is the LENGTH of each interval? It's not simply [itex]1/5^n[/itex], right? That is the FRACTION that you are removing from the preceding stage's intervals, so you have to multiply by the length of the preceding stage's intervals.

    Since there are [itex]2^{n}[/itex] intervals remaining at the end of stage [itex]n[/itex], each one has length at MOST equal to [itex]1/2^{n}[/itex]. In reality the lengths are smaller because this assumes no material has been removed.

    Therefore at stage [itex]n+1[/itex], you remove at most a length of [itex]2^n * (1/5^{n+1}) * (1/2^n) = 1/5^{n+1}[/itex]. (In reality you remove less.)
     
  4. Mar 16, 2010 #3
    I see what you're saying now: it is the FRACTION that you are removing from the preceding stage's intervals.

    But I don't follow the reasoning here. Why are there 2n intervals remaining at the end of stage n?

    And why each one has length at MOST equal to 1/2n? Can't it be the case that some of them has length less than 1/2n and some GREATER than 1/2n?

    Thanks for explaining!
     
  5. Mar 16, 2010 #4

    jbunniii

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    You start with one interval. In stage 1, you remove the middle part of that interval so at the end of stage 1 there are [itex]2 = 2^1[/itex] intervals left.

    In stage 2, you start with 2 intervals, and you remove the middle parts of each, so you are left with [itex]4 = 2^2[/itex] intervals at the end of stage 2. And so on.

    No, because the way the process works is symmetric - in a given stage, you are removing the MIDDLE part of each interval, and you are removing the same FRACTION of each interval. So all of the intervals at the end of a given stage have the same length. And that length cannot be more than [itex]1/2^n[/itex] because there are [itex]2^n[/itex] intervals and the original interval had length 1.
     
    Last edited: Mar 16, 2010
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