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?

The set $C+C$ is a non-empty compact subset of $\mathbb R$. Since $C=\frac13C+\big\{0,\frac23\big\}$, we have $(C+C)=\frac13(C+C)+ \big\{0,\frac23,\frac43\big\}$. So both $C+C$ and the interval $[0,2]$ are fixed points of the map $F\mapsto \frac13 F+ \big\{0,\frac23,\frac43\big\}$. The latter is a contraction on the set $\mathcal K$ of all non-empty compact subsets of $\mathbb R$ w.r.to the Hausdorff distance $d_H$, so they coincide. $\square$
rmk. Note that since $(\mathcal K,d_H)$ is a complete metric space, you can also use the contraction principle to define  $C$ as the unique non-empty compact $C\subset \mathbb R$ such that $C=\frac13C+\big\{0,\frac23\big\}$; the iteration $C_{n+1}:=\frac13C_n+\big\{0,\frac23\big\}$ starting from $C_0:=[0,1]$ produces the usual construction $C \displaystyle=\lim_{n\to\infty}C_n=\cap_{n\in\mathbb N}C_n$.
