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Math Help - Variation on the Cantor Set

  1. #16
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    Opalg's Avatar
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    Quote Originally Posted by redsoxfan325 View Post
    So would 0.2020202020... = \frac{3}{4} be in this set?
    Yes. It is in the Cantor set because its ternary expansion contains no 1s. But it is not an endpoint of one of the complementary intervals, because it cannot be expressed with a ternary expansion that ends in an infinite sequence of all 0s or all 2s.
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  2. #17
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    Quote Originally Posted by Opalg View Post
    Yes. It is in the Cantor set because its ternary expansion contains no 1s. But it is not an endpoint of one of the complementary intervals, because it cannot be expressed with a ternary expansion that ends in an infinite sequence of all 0s or all 2s.
    This is a really nice problem.
    Now, I agree an Opalg's argument.
    3/4 still remains in the OP's cantor set. As n goest to infinity, the radius of an open interval at 3/4 goes to 0, but 3/4 is intact.

    As mentioned by Opalg, the below is a key idea that OP's cantor set is not empty,

    "It may appear that only the endpoints are left, but that is not the case either. The number 1/4, for example is in the bottom third, so it is not removed at the first step, and is in the top third of the bottom third, and is in the bottom third of that, and in the top third of that, and so on ad infinitum—alternating between top and bottom thirds. Since it is never in one of the middle thirds, it is never removed, and yet it is also not one of the endpoints of any middle third. (wiki) "
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  3. #18
    Super Member Gamma's Avatar
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    Yeah, that was why I picked that point. Wanted something that wasnt an endpoint of an interval but was in the regular cantor set because all we adjusted were the endpoints of the removed intervals to be in the complement instead.

    Good problem, and I think we nailed it. Thanks for the question, thoroughly enjoyed thinking about it.
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  4. #19
    Super Member redsoxfan325's Avatar
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    Yeah, I enjoyed talking about it too.

    To recap:

    1.) This set contains an uncountable number of points.
    2.) The points in this set are all numbers between 0 and 1 with a ternary representation that does not contain a 1 or end in an infinite string of 0's or 2's.

    Good work to everyone involved!
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