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?
On
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 $C \displaystyle=\lim_{n\to\infty}C_n=\cap_{n\in\mathbb N}C_n$.
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.