Adding members of the Cantor set to get $[0,2]$ 
Let $C$ be the Cantor tertiary set. Show $C+C \equiv \{x+y : x,y\in C\}=[0,2]$ 

My guess is that I should utilize the standard base-3 representation for $x,y \in C$.
It is immediate that $C+C \subset [0,2]$. I cannot show the other inclusion. Here's where I'm stuck: let $\alpha \in [0,2]$ (we assume $\alpha$ has a base-3 representation). We must find an $x,y \in C$ such that $x+y=\alpha$. My thought is to break up $\alpha$ as such: $$\alpha = \sum_{n=0}^{\infty}\frac{x_n}{3^n}+ \sum_{n=1}^{\infty}\frac{y_n}{3^n} \equiv \alpha_1 + \alpha_2$$ with $x_n \in \{0,1\}$ and $y_n \in \{0,2\}$. Certainly, $\alpha_2 \in C$. I also note that $2*\alpha_1 \in C$. But I can't get any further insight from this. 
My other attempt was to approach the problem from a base-2 perspective, but I was not able to advance at all here. Any hint or bump in the right direction would be greatly appreciated! 
 A: You showed that $\frac12\alpha=\sum\limits_{n\geqslant1}\frac{z_n}{3^n}$ where $z_n=\frac12(x_n+y_n)$, $x_n\in\{0,2\}$ and $y_n\in\{0,2\}$. 
Thus, $x_n+y_n\in$ $___$, hence $z_n\in$ $___$, which proves that $\frac12\alpha$ spans $[0,1]$. 

This shows how to find $x$ and $y$ in $C$ such that $\alpha=x+y$, for every $\alpha$ in $[0,2]$:


*

*Write down a ternary expansion of $\frac12\alpha$, that is, $\frac12\alpha=\sum\limits_{n\geqslant1}\frac{z_n}{3^n}$ where $z_n\in\{0,1,2\}$.

*For each $n$:

*

*if $z_n=0$, let $x_n=y_n=0$,

*if $z_n=1$, let $x_n=2$ and $y_n=0$,

*if $z_n=2$, let $x_n=y_n=2$.


*Consider $x=\sum\limits_{n\geqslant1}\frac{x_n}{3^n}$ and $y=\sum\limits_{n\geqslant1}\frac{y_n}{3^n}$. Then $x$ and $y$ belong to $C$ and $\alpha=x+y$.

A: HINT: Show that $\frac12 C+\frac12 C=[0,1]$.
A: $\frac{1}{2}C$ contains those numbers in $[0,\frac{1}{2}]$ which in their ternary expansion contain only ones and zeros. 
Try to write any $x\in[0,1]$, written in its ternary expansion, as a sum of two members of   $\frac{1}{2}C$.
