Rudin's proof that the Cantor set has no segments

Question

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.

Answer 1

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 $\frac1{3^n}\le\frac{\beta-\alpha}6$.

Answer 2

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 $3^{-n}\le\frac{\beta-\alpha}{4}$.

Answer 3

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 $(\frac{3k+1}{3^{m}}, \frac{3k+2}{3^{m}})$.