Baby Rudin - Cantor Set. A question.

  • Thread starter julypraise
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  • #1
julypraise
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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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  • #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.
 
  • #3
HallsofIvy
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I do not get the second sentence of the paragraph in the image. What segment does he refer to when he says "no segment"?
Not surprizingly, when he says "no segment", he is not talking about any segment!

And why is it 3^-m < (beta - alpha)/6? Why 6?
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.
 
  • #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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