I'm self teaching analysis and the second chapter is about some basic topology.

According to the book "Principles of Mathematical Analysis (3rd)" from Walter Rudin, the Cantor Set is constructed as follows, from what I know.

Let $E_0$ be the interval $[0,1] \subset \mathbb {R}$. Dividing this set in three equal parts and removing the inner segment $({1\over{3}},{2\over3})$, let $E_1$ be the union of intervals $[0,{1\over3}]\text{ and }[{2\over3},1]$. Next we will take the inner segment of each intervals in $E_1$ and so on...

The infinite set $P = \cap_{i=1}^{\infty} E_i $ is called the Cantor Set.

I understand that $E_1 \supset E_2 \supset E_3 ...$ and $E_i$ is the union of $2^i$ intervals each with length $3^{-i}$.

And I also understood why it will be a perfect set.

However, this is the part that I couldn't quite understand from the book. intuitively thinking I can see that P doesn't contain any segment $(a,b)$.

The book says, "No segment of the form" $$({3k+1\over 3^m},{3k+2\over 3^m})$$ "where k and m are positive integers, has a point in common with P. Since every segment $(a,b)$ contains a segment of this form, if $$3^{-m} < {b-a \over 6}$$ P contains no segment."

How did it come up with the number 6 ? Is it not enough to find a "large enough value of m" such that $3^{-m} < b-a$ ? It would be most helpful if someone could explain it with some diagram. Or if you are confident enough that you can help me out with just words, I am open to that, too.


One important idea here is the proof looses nothing by making $m$ bigger than it needs to be, so even if the statement holds for every $m$ with $3^{-m} < (b-a)$ I can still choose $3^{-m} < \frac{b-a}6$. It might just save the reader a couple of minutes thinking about the borderline case.

As for your suggestion suppose I set $a = \frac {3k + 1+\epsilon}{3^m}$, $b = \frac{3k + 5 -\varepsilon}6$ then $b-a = \frac{4-2\varepsilon}{3^m} > 3^{-m}$ but $(a,b)$ doesn't contain any interval in the form $\left(\frac {3k + 1}{3^m},\frac {3k + 2}{3^m}\right)$.

Mark Bennet's excellent answer has just popped up suggesting that $3^m < \frac{b-a}4$ is good enough. Which is true, and as this case shows is the smallest $m$ for which the statement holds.

The point about choosing $6$ is that an interval of length $ 6\cdot 3^{-m}$ defiantly contains at least four consecutive intervals in the form $\left(\frac{i}{3^m}, \frac{i+1}{3^m}\right)$, possibly always five but I don't want to have to think too hard about the end points. Then because I can do this for four consecutive $i$ I can choose $i = 3k+1$ for at least one of them. Maybe I only need three to do this, but I don't care, it's really obvious for four and I've saved myself a scintilla of mental exertion. On with the proof: with the minimum of intellectual effort expended on my choice of $m$.

  • $\begingroup$ From what you are saying, I picked up the fact that if I try to claim that there should be a large enough m such that 3^-m < b-a, then I have to prove when b-a contains the borderlines. Am I right ? $\endgroup$ – hyg17 May 5 '13 at 7:01
  • $\begingroup$ @hyg17: If you claimed you could choose $m$ such that $3^{-m} < b-a$, you'd be wrong. I could choose $a = 0.4$, $b = 0.8$. Then by your claim I could choose $m=1$ because $3^{-1}< 0.4$ but the interval $(0.4,0.8)$ doesn't contain an interval in the form $\left(\frac{3k+1}3, \frac{3k+2}3\right)$. So I need to make $m$ bigger. $\endgroup$ – Tim May 5 '13 at 8:02

This is definitely much simpler than what is in Rudin's text:

The Cantor set $P$ contains no segment $(\alpha,\beta)$ with $\alpha < \beta$.

Proof: By definition, $E_n$ does not contain any segments of length greater than $3^{-n}$. For some $m > 0$, the segment $(\alpha,\beta)$ satisfies $\beta - \alpha > 3^{-m}$ and so $(\alpha,\beta) \not\subset E_m$ and therefore, $(\alpha,\beta) \not\subset \bigcap_{n=1}^{\infty} E_n = P$. $_\Box$

  • $\begingroup$ How is this much simpler? $\endgroup$ – Andrés E. Caicedo Jun 1 '13 at 0:02
  • $\begingroup$ @AndresCaicedo Because his proof doesn't worry about the left-right positioning of the intervals, only the length, so it is sufficient to just invoke the Archimedean property on the reals. I'll agree that this is a much cleaner proof than Rudin's. $\endgroup$ – Mario Carneiro Jun 1 '13 at 0:21
  • $\begingroup$ Rudin's proof shows something stronger with half a line of effort. Bryan's proof is fine, but does not prove the stronger result. I agree, if all we care about is the result Bryan states, then the argument here is the way to go. $\endgroup$ – Andrés E. Caicedo Jun 1 '13 at 0:45
  • $\begingroup$ @AndresCaicedo What more than the statement highlighted in this answer does Rudin show? $\endgroup$ – knsam Sep 22 '14 at 12:13

I think you could get away with 4 rather than 6. The interval is being split into segments of length $3^{-m}$ (a standard segment, say), but you want to pick out a standard segment where the numerator at each endpoint is not a multiple of three. That is every third standard segment. Any segment of length greater than four standard segments contains three consecutive standard segments, and must therefore contain one of the form required.

Because you are picking out every third standard segment as special, you have to make sure you have included one in your interval, so you need the interval to be long enough (relative to $m$) to make sure you have covered it.

We could use 2 instead if the proof did not specify including the whole standard segment. Any segment of length $(2+\epsilon)3^{-m}$ $(\epsilon \lt 1)$ contains an overlap (or perhaps two) with one of the forbidden segments and one of those overlaps has length at least $\cfrac {\epsilon}{2\times3^m}$. And that is enough to show a contradiction.

So the proof is not quite tight as it could be, but it does work.

  • $\begingroup$ So, are you saying that the segment (a,b) must at least be covered by one of the subsets of P and that's the reason why 3^-m < b-a is not good enough ? $\endgroup$ – hyg17 May 5 '13 at 7:03
  • $\begingroup$ @hyg17 the text says - "every segment of length [ ] contains a segment of this form", and you need to use 4 to achieve that. But this is stronger than necessary to for the overall proof - "every segment of length [ ] has non-trivial intersection with a segment of this form" would be enough, and any constant greater than 2 will do for this. You have to prove that the segment includes a forbidden point. $\endgroup$ – Mark Bennet May 5 '13 at 8:22
  • $\begingroup$ @MarkBennet Thank you very much for this post. Your first paragraph really explains what is going on here and I finally understand the statement in Rudin. $\endgroup$ – user70962 May 31 '13 at 15:40

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