I am stumped on a problem in a text book. This is not homework. I'm a physicist doing some self study on Lebesgue integrals and Fourier theory. I'm starting with the basics, and reading up on measure theory.

The problem is to show that $\frac{1}{4}$ is an element of the Cantor set. My first thought would be t0 find a ternary expansion consisting of only 0's and 2's.

However, what I'm having trouble with is imagining that anything remains following the infinite intersection creating the Cantor set other than the interval endpoints. I imagine that if I pick a real number not lying on some interval endpoint I could find an $N$ large enough that the portion of the real line the number belongs to would be deleted. I'd like to see why this argument breaks down.

  • 1
    $\begingroup$ Here's a random argument you might find interesting: For any element of the Cantor set that is the endpoint of an interval, you can choose a rational from that interval. Since every interval is counted at most twice this way (once for each of it's endpoints) then the uncountability of the Cantor set means that we have uncountably many disjoint intervals. Since you can choose one rational that is in each interval, this would mean we have uncountably many rational numbers, which is a contradiction. So, in fact, most elements of the Cantor set are not endpoints of intervals. $\endgroup$
    – Carl
    Oct 12 '12 at 5:35
  • $\begingroup$ Related: math.stackexchange.com/questions/210735/… $\endgroup$
    – Asaf Karagila
    Oct 12 '12 at 13:05
  • 1
    $\begingroup$ Thank you for this question. I was wondering exactly the same thing. This post and the subsequent answer helped me a lot. $\endgroup$ Jan 7 '13 at 1:53

Your first thought is a good one. $(\frac 14)_{10}=0.\overline{02}_3=\sum \frac 2{9^i}=\frac {\frac 29}{1-\frac 19}$. Since each stage deletes the numbers left that have a $1$ at that point in the expansion, it never gets deleted. It isn't the endpoint of an interval, but it is a limit of endpoints of intervals.

  • 1
    $\begingroup$ Thanks for the response. Slight notation change. I believe you want $\sum 2/9^i$ in your answer, not $\sum (\frac{2}{9})^i.$ Should I think of the remaining numbers that aren't endpoints as a consequence of completeness of the reals? $\endgroup$
    – ncRubert
    Oct 12 '12 at 3:20
  • 1
    $\begingroup$ @ncRubert: I would say it is a consequence of the Cantor set being closed. We know it is because we have removed a union of open intervals, which is therefore open. $\endgroup$ Oct 12 '12 at 12:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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