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Baby Rudin - Cantor Set. A question.

  1. Feb 22, 2012 #1
    I do not get the second sentence of the paragraph in the image. What segment does he refer to when he says "no segment"? And why is it 3^-m < (beta - alpha)/6? Why 6?
     

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  3. Feb 22, 2012 #2

    lavinia

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    He is just saying that the Cantor set can not contain any segments. The proof is that
    any segment must contain a middle third (or ninth or 27'th or ...) and thus can not be contained in the Cantor set.

    The proof seems a bit pedantic. Try doing it yourself without the book.
     
  4. Feb 22, 2012 #3

    HallsofIvy

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    Not surprizingly, when he says "no segment", he is not talking about any segment!

    Again, you are misunderstanding. He is not saying that 3^-m< (beta- alpha/6, he says "If" 3^-m< (beta- alpha)/6. That is an hypothesis.
     
  5. Feb 23, 2012 #4

    Bacle2

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    I think the more rigorous statement of whatthe author is claiming is that the Cantor set has an empty interior. In R--where the Cantor set sits -- it means that , for any c in the Cantor set, and any e>0 , the interval:

    (c-e,c+e)

    Is not contained in the Cantor set.

    You can use the characterization of the points of C in terms of their base-3 expansion to show this.
     
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