Cantor set + Cantor set =$[0,2]$ I am trying to prove that

$C+C =[0,2]$ ,where $C$ is the Cantor set.

My attempt:
If $x\in C,$  then  $x= \sum_{n=1}^{\infty}\frac{a_n}{3^n}$ where $a_n=0,2$
so any element of $C+C  $ is of the form  $$\sum_{n=1}^{\infty}\frac{a_n}{3^n} +\sum_{n=1}^{\infty}\frac{b_n}{3^n}= \sum_{n=1}^{\infty}\frac{a_n+b_n}{3^n}=2\sum_{n=1}^{\infty}\frac{(a_n+b_n)/2}{3^n}=2\sum_{n=1}^{\infty}\frac{x_n}{3^n}$$
where $x_n=0,1,2, \ \forall n\geq 1$. 
Is this correct?

By $1-C=C$ we can just do it for all $x\in [0,1]$. Then denote $x=x_1x_2...x_m...$(where $x=\sum_n3^{-n}x_n$ with $x_n=0,1,2$) and we then construct as $a=\sum_n3^{-n}a_n$,$b=\sum_n3^{-n}b_n$ with $a_n,b_n=0,2$ and $a+b=x$.
We do it iteratively, suppose $x_1=0$, we set $a_1=b_1=0$; and suppose $x_1=1$, we set $a_1=b_1=0$; and suppose $x_1=2$, we set $a_1=2,b_1=0$. Then we see $x-a_1-b_1=0x_2x_3...$(first case) or $x-a_1-b_1=1x_2x_3...$(second case); and then in the first case, and if $x_2=2$, we set $a_2=2,b_2=0$; and if $x_2=1$, we set $a_2=0,b_2=0$; and if $x_2=0$, we set $a_2=0,b_2=0$. And in the second case, if $x_2=1$, we set $a_2=b_2=2$; if $x_2=2$, we set $a_2=b_2=2$; and if $x_2=0$, we set $a_2=2,b_2=0$. Then by this construction, we see $x-a_1a_2-b_1b_2=00x_3x_4...$ or $01x_3x_4...$.
And iterate this process again and again to construct a $a=a_1...a_n...$ and $b=b_1b_2...b_n...$ such that $x-a_1...a_n-b_1b_2...b_n=0...0x_{n+1}x_{n+2}....$ or $0...1x_{n+1}x_{n+2}...$. Then we can see $x=a+b$ indeed.
