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Construct a subset of [0,1] in the same manner as the Cantor set by removing from each remaining interval a subinterval of relative length $\theta$, $0<\theta <1$.

This is the first statement in a homework exercise. I don't know if my English parsing skills are lacking today or what, but it's unclear to me what this means. I just need help understanding what this construction "looks like."

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  • $\begingroup$ You may want to take a look at the construction in the wikipedia article: en.wikipedia.org/wiki/… If I read the question correctly, they want you to follow this same procedure but with an arbitrary length and presumably obtain a similar expression for C like at the end of the section of the article. $\endgroup$
    – WWright
    Oct 13, 2010 at 1:48

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The construction of the Cantor set begins with the removal of the interval $[1/3,2/3]$, an interval of length $1/3$. The remaining intervals have length $1/3$; you remove intervals of length $1/9$ from each of them. At stage $k$, you're removing $2^{k-1}$ intervals of length $1/3^k$. The total length removed is then $$\sum_{k=1}^\infty\frac{2^{k-1}}{3^k}=\frac{1}{3}\frac{1}{1-\frac{2}{3}}=1$$ so the Cantor set has measure zero.

Now let $\theta\ne 1/3$. You can repeat the same construction with $\theta$ instead of $1/3$: at step $k$, you remove $2^{k-1}$ intervals of length $\theta^k$. This is going to look very much like the Cantor set -- in particular, it has empty interior, which you can prove. If your homework problem is going where I'm guessing it's going, you should calculate the measure of the set you get for different $\theta$. Actually, you should do that no matter what the homework says. The results are quite surprising.

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  • $\begingroup$ At stage 1 you remove 1=2^0 intervals! Then your sum is really equal to 1 which is false right now. $\endgroup$
    – Rasmus
    Oct 13, 2010 at 16:55
  • $\begingroup$ I meant $2^{k-1}$ instead of $2^k$. Edited to fix. The sum I had is actually the right one, though. $\endgroup$ Oct 13, 2010 at 16:57

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