If $C$ is the Cantor set, then $C+C=[0,2]$. 
Question : Prove that $C+C=\{x+y\mid x,y\in C\}=[0,2]$, using the following steps:
We will show that $C\subseteq [0,2]$ and $[0,2]\subseteq C$.
  
  
*
  
*a) Show that for an arbitrary $n\in\mathbb{N}$ we can always find $x_n,y_n\in C_n$, where $$C_n=\left[0,\frac1{3^n}\right]\bigcup \dots \bigcup\left[\frac{3^{n}-1}{3^n},1\right]$$such that for a given $s\in[0,2]$ we have  $x_n+y_n=s$.
  
*b) Then we will set $x=\lim x_n$ and $y=\lim y_n$,  then $x+y=s$.
  

My progress : Showing that $C+C\subseteq [0,2]$ is obvious, and I did part a) by showing that if $x_n,y_n$ are in different subintervals then concluding that $x_n+y_n$ covers $[0,2]$ (can be done using induction on $n$). 
My difficulty is in the second part. The sequence $x_n$ is bounded so it must have a convergent subsequence $(x_{n_{k}})$. If we set $x=\lim x_{n_{k}}$, then we can conclude $\lim y_{n_{k}}=y=s-x$, thus $x+y=s$. First I thought that $x,y$ will be in $C$ as it is closed. However $(x_{n_{k}})$ may not necessarily be in $C$, so $x$ can't be in $C$ for sure. 
Is this last thought correct? How do I overcome this last gap in my solution? Thanks for your help.
 A: To carry out (b), note that if $\ell\le n_k$, then $x_{n_k}\in C_\ell$. Thus, for each $\ell$ there is an $m(\ell)$ such that $x_{n_k}\in C_\ell$ whenever $k\ge m(\ell)$. Let $\ell$ be arbitrary. Clearly 
$$x=\lim_{k\ge m(\ell)}x_{n_k}\;,$$
and $C_\ell$ is closed, so $x\in C_\ell$. Thus, $x\in\bigcap_\ell C_\ell=C$.
(You might be interested in comparing your argument with the first proof here; they’re basically the same idea, just expressed a bit differently.)
A: You can prove $C+C=[0,2]$ by using the ternary representation of the Cantor set. Let $x,y\in C$ where
$$
x=\sum_{n=1}^{\infty}\frac{a_n}{3^n},\:a_n=0,2\quad\text{and}\quad y=\sum_{n=1}^{\infty}\frac{b_n}{3^n},\:b_n=0,2
$$
Then it can be proved that $x+y$ can cover all $[0,2]$. Let $z\in [0,2]$. Then $$
z=\sum_{n=0}^{\infty}\frac{c_n}{3^n},\:c_0=0,1,\:c_n=0,1,2,\:n>0
$$
If $c_0=0$, then $z\in [0,1]$. If $c_0=1$, then $z\in [1,2]$.
Given $c_n$, we construct $a_n,b_n\:(n\geqslant1)$ as follows:

*

*If $c_n=0$ and no carry-over from $a_{n+1}+b_{n+1}$, then set $a_n=0, \:b_n=0$.
If $c_n=0$ and has carry-over from $a_{n+1}+b_{n+1}$, then set $a_n=2, \:b_n=0$.

*If $c_n=1$ and no carry-over from $a_{n+1}+b_{n+1}$, then set $a_n=2, \:b_n=2$.
If $c_n=1$ and has carry-over from $a_{n+1}+b_{n+1}$, then set $a_n=0, \:b_n=0$.

*If $c_n=2$ and no carry-over from $a_{n+1}+b_{n+1}$, then set $a_n=2, \:b_n=0$.
If $c_n=2$ and has carry-over from $a_{n+1}+b_{n+1}$, then set $a_n=2, \:b_n=2$.

*$c_0=0$ or $c_0=1$, depends on whether there is carry-over from $a_{1}+b_{1}$.

Hence we show that
$$
z=\sum_{n=0}^{\infty}\frac{c_n}{3^n}=\sum_{n=1}^{\infty}\frac{(a_n+b_n)}{3^n}=x+y
$$
Since $z$ cover entire $[0,2]$, we have $C+C=[0,2]$.
A: It is not clear from your description of $C_n$ but I suppose that it is the $n$'th level of the Cantor set (i.e. with omission of the middle $1/3^n$ sets). Now $C\subset C_n$ for all $n$ and if $x_n\in C_n$ there is $y\in C$ with $|x_n-y|< 3^{-n}$.
Given the subsequence   $x_{n_k}\in C_{n_k}$ that you have extracted, $\epsilon_k =|x_{n_k}-x|$ goes to zero as $k\rightarrow \infty$. But then for every $k$, the interval $(x-\epsilon_k-3^{-n_k},x+\epsilon_k+3^{-n_k})$ intersects $C$ non-trivially. As $C$ is closed we must have $x\in C$.
Doing the same for $y$ you may conclude in the way you have already sketched.
