# Rudin Principles Page 42: Cantor set.

No segment of the form $\left(\dfrac{3k+1}{3^{m}},\dfrac{3k+2}{3^{m}}\right)$ where $k,m\in\mathbb{Z}^{+}$ has a point in common with the Cantor set. Is there a simple proof of this statement?

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Doesn't this follow directly from the construction of the Cantor set? All such segments are removed at step $m$. – Nate Eldredge Jan 16 '12 at 4:01

Consider $$P_m=\left\{0,\frac{1}{3^m},\ldots,\frac{3^m-1}{3^m},1\right\}$$ the uniform partition of $[0,1]$ in $3^m$ intervals of length $3^{-m}$, and $$I_m^j=\left(\frac{j-1}{3^m},\frac{j}{3^m}\right),$$ $j=1,2,\ldots, 3^m$, the intervals determined for that partition. If I'm not wrong with the indexation, letting you $k$ vary over $\{0,1,\ldots,3^{m-1}-1\}$ you get the $I_m^j$ with $j$ even, and that intervals are precisely the intervals removed in the $m$-th stage of the construction of the Cantor set. See this, for an explanation of the last sentence.

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This is a clear explanation. – MathMajor Jan 16 '12 at 4:38

I think it should be pointed out that some authors define the Cantor set to be the set of all $x$ in $[0,1]$ whose base three expansion can be written using only the digits $0$ and $2$. In this case, the problem is not so trivial.

But the interval $({1\over3}, {2\over3})$ is the set of numbers in $[0,1]$ whose base three expansion must have a $1$ in the first digit (the endpoints $1/3$ and $2/3$ can be written in base three using only the digits $0$ and $2$: ${1\over3}=0.02\overline2_3$ and ${2\over3}=0.2_3$). So, this interval is not in the Cantor set.

The intervals $({1\over9}, {2\over9})$ and $({7\over9}, {8\over9})$ give the numbers not already removed whose base three expansion must have a $1$ in the second digit (the endpoints can be written in base three using only the digits $0$ and $2$). So, these intervals are not in the Cantor set.

Continuing in this manner, one can show that $x\in[0,1]$ has a base three expansion using only the digits $0$ and $2$ if and only if it belongs to no segment of the form $\bigl( {3k+1\over 3^m}, {3k+2\over 3^m}\bigr)$.

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You can argue this by induction, using the definitions of $E_n$ Rudin uses to define the Cantor set. The inductive hypothesis would be $$E_n \text{ contains no segment of the form } \left(\frac{3k+1}{3^m}, \frac{3k+2}{3^m}\right) \text{ for } m = 1, \ldots , n.$$

Note that if $x \in E_n$, $1-x \in E_n$ so you can show that if $E_n$ contains a segment it contains one in the lower third. Note also that if $x \in E_{n+1}$ is in the lower third then $3x \in E_n$. You can now prove the inductive hypothesis by contradiction, and as the cantor set is the intersection of all $E_n$ the Cantor set contains no such segments.

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