Uncountability of the Cantor Set

Theorem

The middle thirds Cantor Set $K_{\infty }$ consists of all numbers in the interval $[0,1]$ that can be expressed in base 3 (ternary) using only the digits $0$ and $2$

Consider in base 3:

$$r=0.{\overline{02}}_3\in K_{\infty }$$

how many copies of:

$$\underset{base10}{\underbrace{3^0\cdot 3^{-1}\cdot 3^{-2}\cdot 3^{-3}}}$$ $$\begin{array}{rl}r& =\frac{2}{9}+\frac{2}{81}+\frac{2}{729}+\dots \\ & =\frac{2}{9}\left(1+\frac{1}{9}+\frac{1}{81}+\dots \right)\\ & =\frac{2}{9}\cdot \frac{1}{1-\frac{1}{9}}=\frac{2}{9}\cdot \frac{9}{8}=\frac{1}{4}\end{array}$$

What about $x=\frac{1}{3}=0.1_3=0.0{\overline{2}}_3$

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Example

The set of left endpoints of middle thirds Cantor Set intervals is countable

Example

subset of $K_{\infty }$ with a finite number of repeartint ternary digits ( e.g., ending in $\overline{0}$)

$\mathbb{N}$	ternary # in $K_{\infty }$	$a_{ij}\in \{0,2\}$
1	$r_1=0.a_{11}{a}_{12}{a}_{13}\dots$
2	$r_2=0.a_{21}{a}_{22}{a}_{23}\dots$
3	$r_3=0.a_{31}{a}_{32}{a}_{33}\dots$
…	…
n	$r_n=0.a_{n_1}{a}_{n_2}{a}_{n_3}\dots a_{nn}$

what is wrong with this? doesnt this have all items?

thanks to Cantor’s diagonal argument, this will never work, as given that it is irrational, we can always construct another $r$ not in the list.