TITLE: A minor question about the Cantor Set
QUESTION [6 upvotes]: 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.

REPLY [4 votes]: 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, $(\alpha,\beta) \not\subset \bigcap_{n=1}^{\infty} E_n = P$. $_\Box$