Wikipedia describes the process of creating the Cantor set at https://en.wikipedia.org/wiki/Cantor_set

We begin with one segment $[0,1]$ and remove $ \left( \frac{1}{3},\frac{2}{3} \right)$

This leaves us with two segments $\left[ 0,\frac{1}{3}\right] \left[ \frac{2}{3},1\right]$ and remove $ \left( \frac{1}{9},\frac{2}{9} \right) \left( \frac{7}{9},\frac{8}{9} \right)$

This leaves us with four segments $\left[ 0,\frac{1}{9}\right] \left[ \frac{2}{9},\frac{1}{3}\right] \left[ \frac{2}{3},\frac{7}{9}\right] \left[ \frac{8}{9},1\right]$

Wiki says we do this "ad infinitum".

Will we then have any segments left as opposed to just an infinite number of points, and if we do have segments, will the number of segments be countable or uncountable?

  • 1
    $\begingroup$ I don't understand "compliment". Is not $\left( \frac{1}{3},\frac{2}{3} \right) \subset [0,1] \setminus C$ where C is the cantor set? Then is not $\left( \frac{1}{3},\frac{2}{3} \right)$ uncountable? $\endgroup$
    – Ivan Hieno
    Apr 6, 2017 at 16:30
  • $\begingroup$ @NoahSchweber Oh my god, you're totally correct. Total screw-up on my part. I don't know what I was thinking. IvanHieno, your concern is correct, and I was wrong, and Noah pointed out. I aplogize. What a ridiculous mistake! I deleted the comment so it doesn't do any harm to others. $\endgroup$
    – The Count
    Apr 6, 2017 at 17:41
  • 1
    $\begingroup$ @TheCount Not a problem, happens to the best of us. (I once spent a good minute or so explaining to a professor why my proof worked because "you just take an infinite set of odd numbers divisible by six." He was very nice about it.) $\endgroup$ Apr 6, 2017 at 17:51
  • 1
    $\begingroup$ @NoahSchweber I do that sort of stuff all the time. My most famous and common one is writing inequalities in the wrong direction at first, then doing everything else right, and being sure I've proved something amazing. Sigh. Well, thanks for being nice about it. Ha. $\endgroup$
    – The Count
    Apr 6, 2017 at 17:57

1 Answer 1


There are no segments of positive length left (this may be seen by, for instance, looking at the lengths of the segments at each iteration; after the $n$th iteration, the segments have length $3^{-n}$). There are, however, uncountably infinitely many points in the cantor set.


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.