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?

Short answer for this question is "your argument is correct ". To justify the answer consider some particular $n_{0}\in \mathbb{N}$. Since $$\sum_{n=1}^{\infty}\dfrac{a_{n}}{3^{n}},\sum_{n=1}^{\infty}\dfrac{b_{n}}{3^{n}}\in C$$ we have that $ x_{n_{0}}=\dfrac{a_{n_{0}}+b_{n_{0}}}{2}\in \{0,1,2\} $. ($ x_{n_{0}}=0 $ if $ a_{n_{0}}=b_{n_{0}}=0 $. $ x_{n_{0}}=2 $ if $ a_{n_{0}}=b_{n_{0}}=2 $. Otherwise $ x_{n_{0}}=1 $. )

Then clearly $$\sum_{n=1}^{\infty}\dfrac{x_{n}}{3^{n}}\in [0,1].$$ So $$2\sum_{n=1}^{\infty}\dfrac{x_{n}}{3^{n}}=\sum_{n=1}^{\infty}\dfrac{a_{n}}{3^{n}}+\sum_{n=1}^{\infty}\dfrac{b_{n}}{3^{n}}\in [0,2].$$ Hence $ C+C\subseteq [0,2] $. To complete the proof you must show that the other direction as well. To show $ [0,2]\subseteq C+C $ it is enough to show $ [0,1]\subseteq \dfrac{1}{2}C+\dfrac{1}{2}C $.

Observe that $ b\in \dfrac{1}{2}C $ if and only if there exists $ t\in C $ such that $ b=\dfrac{1}{2}t $.

Hence $$ b\in \dfrac{1}{2}C\text{ if and only if }b=\sum\limits_{n = 1}^\infty \frac{b_n}{3^n}\text{ ; where }b_{n}=0\text{ or }1. $$ Now let $ x\in [0,1] $. Then $$ x= \sum\limits_{n = 1}^\infty \frac{x_n}{3^n}\text{ ; where }x_{n}=0,1\text{ or }2. $$ Here we need to find $ y,z\in \dfrac{1}{2}C $ such that $ x=y+z $. Let's define $ y=\sum\limits_{n = 1}^\infty \frac{y_n}{3^n} $ and $ z=\sum\limits_{n = 1}^\infty \frac{z_n}{3^n} $ as follows.

For each $n\in \mathbb{N}$, $ y_{n}=0 $ if $ x_{n}=0 $ and $ y_{n}=1 $ if $ x_{n}=1,2 $.

For each $n\in \mathbb{N}$, $ z_{n}=0 $ if $ x_{n}=0,1 $ and $ z_{n}=1 $ if $ x_{n}=2 $.

Thus $y,z\in \dfrac{1}{2}C $ and for each $n\in \mathbb{N}$, $ y_{n}+z_{n}=0 $ if $ x_{n}=0 $ , $ y_{n}+z_{n}=1 $ if $ x_{n}=1 $ and $ y_{n}+z_{n}=2 $ if $ x_{n}=2 $.

Therefore $x=y+z\in \dfrac{1}{2}C+\dfrac{1}{2}C$ and hence $[0,1] \subseteq \dfrac{1}{2}C+\dfrac{1}{2}C$. $\square $

We need to show both inclusions. $C+C\subseteq [0,2]$ is pretty obvious, however $[0,2]\subseteq C+C$ is no longer. For the proof of the second statement I recommend the book written by Steven G. Krantz - "A Guide to Topology", which has a really good analytic. You can find it online http://books.google.com/books?id=O3tyezxgv28C&printsec=frontcover&hl=pl&redir_esc=y#v=onepage&q=cantor&f=false look at page 35.

Since $C \subset [0,1]$ we have $1/2(C+C) \subset [0,1]$ by the convexity of $[0,1]$.

The inclusion $[0,1] \subset \frac{1}{2}C + \frac{1}{2}C$ was explained by @Tgymasb, anyways, the way I see it

$$\frac{1}{2}C = \sum_{n \ge 1} \frac{a_n}{3^n}$$ with $a_n \in \{0,1\}$ while $$[0,1] = \sum_{n \ge 1} \frac{c_n}{3^n}$$ with $c_n \in \{0,1,2\}$ and the point is that any function from some domain to $\{0,1,2\}$ is a sum of two functions mapping to $\{0,1\}$, and this is because we have the obvious decompositions: $$2 = 1 + 1\\ 1 = 1 + 0\\ 0 = 0 + 0$$

$\tiny{ \text{( pointwise addition so no carries)}}$