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

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.

Answer 4

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 $$3^{-m}<\beta-\alpha.$$

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Answer 5

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