Prove that $C+C=[0,2]$, where $C$ is the Cantor set. I'd like someone to verify my sketch proof of the below exercise 3.3.7 from Abbott's, Understanding Analysis. If it's incorrect, could you hint/point at the correct approach to the proof. Thanks!

Prove that the sum $C+C=\{x+y:x,y\in C\}$ is equal to the closed interval $0,2]$.


Because $C\subseteq [0,1], C+C \subseteq [0,2]$, so we only need to prove the reverse implication. Given $s \in [0,2]$, we must find two elements $x,y \in C$, satisfying $x + y = s$.


(a) Show that there exists $x_1,y_1 \in C_1$, for which $x_1 + y_1 = s$. Show in general that, for an arbitrary $n \in \mathbf{N}$, we can always find $x_n,y_n \in C_n$ for which $x_n + y_s = s$.

Proof.
(a) Let $s \in [0,2]$. Then, atleast one of (1) $s \in \left[0,\frac{1}{3}\right]$ (2) $s \in \left[\frac{1}{3},\frac{4}{3}\right]$ (3) $s \in \left[\frac{4}{3},2\right]$ holds. If (1) holds, then we pick $x_1,y_1 \in \left[0,\frac{1}{3}\right]$. If (2) holds, we pick $x_1 \in \left[0,\frac{1}{3}\right]$, $y_1 \in \left[\frac{2}{3},1\right]$. If (3) holds, we may pick $x_1,y_1 \in \left[\frac{2}{3},1\right]$.
Now, $C_n$ is given by
$$
\left[0,\frac{1}{3^n}\right]\cup\left[\frac{2}{3^n},\frac{3}{3^n}\right]\cup\left[\frac{4}{3^n},\frac{5}{3^n}\right]\cup \ldots \cup \left[\frac{3^n-1}{3^n},1\right]
$$
If $x_n$ belongs to atleast one of the above sets, then
\begin{align*}
x_n \in \bigcup_{k}\left[\frac{2k}{3^n},\frac{2k+1}{3^n}\right]
\end{align*}
where $k \in \{0,1,2\ldots,\frac{3^n-1}{2}\}$. Suppose, $y_n$ belongs to one of the above sets, then
\begin{align*}
y_n \in \bigcup_{l}\left[\frac{2l}{3^n},\frac{2l+1}{3^n}\right]
\end{align*}
where $l \in \{0,1,2\ldots,\frac{3^n-1}{2}\}$.
By definition, $x_n + y_n$ belongs to
\begin{align*}
\bigcup_{k,l}\left[\frac{2k+2l}{3^n},\frac{2k+2l+2}{3^n}\right]=\left[0,\frac{2}{3^n}\right]\cup \left[\frac{2}{3^n},\frac{4}{3^n}\right] \cup \ldots \cup \left[\frac{2\cdot(3^n - 1)}{3^n},2\right]=[0,2]
\end{align*}

(b) Keeping in mind that the sequences $(x_n)$ and $(y_n)$ do not necessarily converge, show how they can nevertheless be used to produce the desired $x$ and $y$ in $C$ satisfying $x + y = s$.

$\star$ Any hints on how to proceed here, would be super-helpful!
 A: Here is - in essence - the same argument, written differently. Let $C$ be the Cantor set. Each element $a$ in $C$ can be written in base $\color{blue}3$ in the form
$$
\begin{aligned}
a 
&
= 0,a_1a_2a_3\dots_{\color{blue}{(3)}}
=\frac 13a_1+\frac 1{3^2}a_2+\frac 1{3^2}a_3 + \dots \ ,\qquad a_1,a_2,a_3\dots\in\{0,2\}\ ,\text{ so}
\\
b:=\frac a2 
&= 0,b_1b_2b_3\dots_{\color{blue}{(3)}}
=\frac 13b_1+\frac 1{3^2}b_2+\frac 1{3^2}b_3 + \dots \ ,\qquad b_1,b_2,b_3\dots\in\{0,1\}\ .
\end{aligned}
$$
So $b\in\frac 12C=:D$.
Fix $x\in[0,2]$, associate $y=\frac 12x$. We need to show that it can be written as the sum of two elements in $D$. Consider the digits of $y$ in base $3$. For each digit $y_k\in\{0,1,2\}$ write it as $y_k=b_k+b'_k$ according to $0=0+0$, $1=1+0$ (or $1=0+1$, here we have a choice, but let it be always the one...), and $2=1+1$. Build $b,b'\in D$ with these digits. Then $y=b+b'$.
(Note: $y=1$ can be written as $0,22222\dots$ in base three, and the above procedure applies.)
A: Note that $\{x_n\}$ is bounded, so there is a subsequence $\{x_{n_k}\}$ converging to some $x \in [0,1]$. Use contradiction to show that $x \in C$: If not, then $x$ belongs to some $[0,1] \backslash C_k$ (this is an open set) but $\forall n \ge k+1, x_n \notin [0,1] \backslash C_k$, contradicting the convergence. The corresponding $\{y_{n_k}\}$ sequence also converges to some $y$ because $x_{n_k}+y_{n_k}=s$. Again $y$ can be shown to lie in $C$.
