Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

a. Show that the collection of "right hand" end point in $\mathbb{F}$ is denumerable ($\mathbb{F}$ is denoted as the Cantor set). Show that if all these end points are deleted from $\mathbb{F}$, then what remains can be put onto one-one correspondence with all of $[0,1)$. Conclude that the set $\mathbb{F}$ is not countable.

b. Show that $\mathbb{F}$ is not the union of a countable collection of closed intervals.

I do not know how to prove this. I know that the Cantor set is not countable, there are proofs in google that showed me that, but they used the diagonal method which isn't helpful for me because I haven't been taught that method yet. In b, are they referring to nested intervals?

share|cite|improve this question
Do you mean the right hand endpoints of the closed intervals formed by deleting the "middle thirds"? – David Mitra Feb 10 '13 at 2:45
If you're trying to do uncountability proofs, and you haven't learned about the diagonal method, you're doing things backwards. May I suggest you learn about the diagonal method, and quick. – Gerry Myerson Feb 10 '13 at 2:46
@Gerry: I never really learned the (classical) diagonal argument before arriving to MathOverflow on the last week of my undergrad studies. And that was well into my second advanced course in set theory. – Asaf Karagila Feb 10 '13 at 2:47
@DavidMitra not sure if it is closed but I am sure it is middle thirds. – Q.matin Feb 10 '13 at 2:48
For b., you could first show that the Cantor set contains no non-degenerate interval (see this post). So, if you write $\Bbb F$ as a countable union of closed intervals, then each of those intervals must be a singleton point. – David Mitra Feb 10 '13 at 2:56
up vote 2 down vote accepted

Do you know that the right endpoints are rational numbers?

Do you know that the rationals are denumerable?

Do you know that the remaining elements of the Cantor set, when written in base $3$, are precisely the numbers that have only zeros and twos?

Can you see how to map that to (the binary expansions of) the reals?

Do you know the measure of the Cantor set? And the measure of a closed interval?

share|cite|improve this answer
I did not know the right endpoints are rational but since they are defined to be $(0,1)$ which are obviously rational. I do know the rational are denumberable. For the third question no I did not, not sure what you mean by that only having zeros and twos. – Q.matin Feb 10 '13 at 3:01
The point is that each interval in the construction has rational endpoints, because if an interval has rational endpoints then so does its middle-third, and you argue by induction. Do you know what it means to write a real number in base $3$? – Gerry Myerson Feb 10 '13 at 3:07
No I do not know what it means to write a real number in base 3. Is it when you can only use 0,1,2 ? – Q.matin Feb 10 '13 at 3:12
Yes. Just as every real can be written as a decimal, where $.d_1\ d_2\ d_3\ \dots$, with $0\le d_i\le9$, means $\sum d_i10^{-i}$, so every real can be written in base 3, where $.e_1\ e_2\ e_3\ \dots$, with $0\le e_i\le2$, is now interpreted as $\sum e_i3^{-i}$. – Gerry Myerson Feb 10 '13 at 8:45
Thank you! I am going to google this some more to try and hopefully understand! – Q.matin Feb 10 '13 at 23:24

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.