Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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."

share|cite|improve this question
You may want to take a look at the construction in the wikipedia article:… 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. – WWright Oct 13 '10 at 1:48
up vote 3 down vote accepted

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.

share|cite|improve this answer
Fantastic, thank you – Bey Oct 13 '10 at 2:24
At stage 1 you remove 1=2^0 intervals! Then your sum is really equal to 1 which is false right now. – Rasmus Oct 13 '10 at 16:55
I meant $2^{k-1}$ instead of $2^k$. Edited to fix. The sum I had is actually the right one, though. – Paul VanKoughnett Oct 13 '10 at 16:57

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.