Rudin's proof that the Cantor set has no segments In Principles of Mathematical Analysis, on page 42, Rudin proves that the Cantor set has no segments by stating that every segment $(\alpha,\beta)$ contains a segment of the form $(\frac{3k+1}{3^{n}},\frac{3k+2}{3^{n}})$ if $3^{-n}\lt\frac{\beta-\alpha}{6}$. I don't know how $\frac{\beta-\alpha}{6}$ is obtained.
 A: Your denominators should be $3^n$, not $3^{-n}$.
Suppose that $\alpha$ is just a little bigger than $\frac{3k+1}{3^n}$ for some $k$, so that $(\alpha,\beta)$ is just barely too far to the right to contain $\left(\frac{3k+1}{3^n},\frac{3k+2}{3^n}\right)$ no matter how big $\beta-\alpha$ is. The next interval of that form to the right is
$$\left(\frac{3k+4}{3^n},\frac{3k+5}{3^n}\right)\;;\tag{1}$$
if we ensure that $\frac{3k+5}{3^n}\le\beta$, the interval $(\alpha,\beta)$ will contain the interval $(1)$. If $\alpha=\frac{3k+1}{3^n}+\epsilon$, then 
$$\frac{3k+5}{3^n}=\alpha-\epsilon+\frac4{3^n}\;,$$
so we need to make sure that $\beta\ge\alpha-\epsilon+\frac4{3^n}$ no matter how small $\epsilon$ is. This will certainly be the case if $\beta\ge\alpha+\frac4{3^n}$, i.e., if $\beta-\alpha\ge\frac4{3^n}$. In other words, if we choose $n$ big enough so that
$$\frac1{3^n}\le\frac{\beta-\alpha}4\;,$$
the interval $(\alpha,\beta)$ is certain to contain an interval of the form $\left(\frac{3k+1}{3^n},\frac{3k+2}{3^n}\right)$.
Clearly this will be the case if $\frac1{3^n}\le\frac{\beta-\alpha}6$.
A: We want to show that there is an integer $k$ such that $\alpha\le\frac{3k+1}{3^n}$ and $\frac{3k+2}{3^n}\le\beta$, 
so we want to have $\frac{\alpha(3^n)-1}{3}\le k\le\frac{\beta(3^n)-2}{3}$.  
Such an integer $k$ will exist if $\frac{\beta(3^n)-2}{3}-\frac{\alpha(3^n)-1}{3}\ge1$, so
$(\beta-\alpha)3^n\ge4$ gives $3^{-n}\le\frac{\beta-\alpha}{4}$.
A: 
If $\alpha$ and $\beta$ satisfy $3^{-m} < \frac{\beta-\alpha}{5}$, then every segment $(\alpha, \beta)$ contains a segment which was removed when the Cantor set was constructed.
See the above picture.
A: The condition $3^{-m} < \frac{\beta - \alpha}{6}$ is equivalent to  $3^{-(m-1)} < \frac{\beta - \alpha}{2}$.
Now if $3^{-(m-1)} < \frac{\beta - \alpha}{2}$, then $[\alpha, \alpha+\frac{\beta - \alpha}{2}]$ contains $\frac{k}{3^{m-1}}$ for some integer $m\geq 2$ and $k\geq 0$.
We can conclude that $[\alpha, \beta] \supset [\frac{k}{3^{m-1}}, \frac{k+1}{3^{m-1}}]$, which is equivalent to $[\frac{3k}{3^{m}}, \frac{3k+3}{3^{m}}]$. Then clearly $(\alpha, \beta) \supset (\frac{3k+1}{3^{m}}, \frac{3k+2}{3^{m}})$.
What I have shown is that the condition you asked, $3^{-m} < \frac{\beta - \alpha}{6}$, is sufficient to conclude that every segment $(\alpha, \beta)$ contains a segment of the form $(\frac{3k+1}{3^{m}}, \frac{3k+2}{3^{m}})$.
A: If Cantor set contains a segment $(\alpha,\beta)$, then it contains an interval $$\left(\frac {3k+1}{3^m}, \frac {3k+2}{3^m}\right)$$ for some $k$ and $m$. This surely contradicts the claim that the Cantor set has no common points with any such intervals. 
From the above, we can see that $$\beta-\alpha >3^{-m}$$ is needed. If we have $$\frac {\beta-\alpha}{6}>3^{-m},$$
then it is more than needed. 
-------Edited by the author: to see why we just need $3^{-m}<\beta-\alpha$. ----
Suppose that Cantor set $P$ contains an interval $(\alpha,\beta). $ We expand $\alpha$ in base-3. 
$$\alpha = \frac {a_1}{3}+\frac {a_2}{3^2}+\cdots+ \frac {a_n}{3^n}+\cdots,a_i=0,1,2,$$
We want to find two rational numbers, one is strictly larger than $\alpha,$ the other is strictly less than $\beta$. Let $m$ to be determined. We construct the first one from the $m$-th position of the decimal expansion of $\alpha$: $a_m$ is at most $2$; moreover the tail of the series expansion is at most $3^{-m}$; we sum the first $m-1$ terms using common denominators to see that there exists an integer $k$. To be strictly larger than $\alpha,$, adding one more $\frac {1}{3^m}$, we obtain the first rational is $$\frac {3k+2+1+1}{3^m}.$$
For the second rational, in order for the numerator of the above adding $1$ so that $(\frac {3k+2+1+1}{3^m}, \frac {3k+2+1+1+1}{3^m})$ is in $(\alpha,\beta)$, we need 
$$\frac {1}{3^m}<\beta-\alpha.$$
This alternatively gives $$3^{-m}<\beta-\alpha.$$ 

