A real analysis problem on series Associate to each sequence $a=\{\alpha_n\}$ in which $a=\{\alpha_n\}$ is 0 or 2, the real number $$x(a)=\sum_{n=1}^{\infty}\frac{\alpha_n}{3^n}$$
How to formally prove that the set of all $x(a)$ is precisely the set of all fractional decimal digits containing 0 or 2 represented in ternary notation.
 A: The cantor set is constructed by taking a segment of unit length $[0,1]$ and removing the middle third, $(1/3,2/3)$. That is, we are left with $[0,1/3]\cup [2/3,1]$ and then doing this ad infinitum. Let's consider $1/3$ in base three (ternary).
$$
\frac{1}{3} = 0\cdot 3^1 + 1\cdot\frac{1}{3} + 0\cdot\sum_{n = 2}^{\infty}\frac{1}{3^n} = 0.1
$$
Well this poses a problem since $\{\alpha_n\}$ is a sequence of $0$ or $2$. Suppose we can write $0.1$ as $0.0\bar{2}$ instead. 
$$
0.0\bar{2} = 2\Bigl(\frac{1}{3^2} + \frac{1}{3^3} + \cdots\Bigr) = 2\sum_{n=2}^{\infty}\frac{1}{3^n}=\frac{2}{9}\sum_{n=0}^{\infty}\frac{1}{3^n} =\frac{2}{9}\frac{1}{1-1/3} = \frac{1}{3} = 0.1\tag{1}
$$
Now, let's go back to the Cantor set. After the first iteration, in ternary, we removed $(0.1,0.2)$. That is, we removed all the terms with $0.1\ldots$ as the first digit but it appears that we kept $0.1$. From equation $(1)$, we see that we actually keep $[0,0.0\bar{2}]\cup[0.2,0.\bar{2}]$ where $1 = 0.\bar{2}$ by the same argument. With the second step, we remove all the digits with $0.01$ as the first digit and keep $0.01=0.00\bar{2}$. Since this continues ad infinitum, we are left with a set that is represented by the ternary expansion
$$
\sum_{n=1}^{\infty}\frac{\alpha_n}{3^n}
$$
where $\{\alpha_n\}$ contains only $0$ and $2$.
