Let $C$ to be the Cantor middle third set in $[0,1]$. Clearly every element of Cantor set has ternary representation and vice versa.
Question: Show that $$C-C=[-1,1],$$ where $C-C = \{x-y:x,y\in C \}.$
My attempt: Since $C\subseteq [0,1],$ therefore $$C-C\subseteq [-1,1].$$ Now, let $z\in [-1,1].$ I can use the following fact $$C+C=[0,2].$$ Since $z+1\in [0,2],$ therefore, $$z+1 = \sum_{k=1}^\infty\frac{a_k}{3^k} + \sum_{k=1}^\infty\frac{b_k}{3^k}$$ where $a_k,b_k\in\{0,2\}$ for all natural number $k.$ Observe that $$z= \sum_{k=1}^\infty\frac{a_k}{3^k} - \left(1 - \sum_{k=1}^\infty\frac{b_k}{3^k}\right)$$ where $$1 - \sum_{k=1}^\infty\frac{b_k}{3^k} = \sum_{k=1}^\infty\frac{2-b_k}{3^k}$$ is an element of Cantor set, as $2-b_k\in\{0,2\}$ for all natural number $k.$ Therefore, $z\in C-C.$
Is my proof correct?
A proof not starting from $C+C$ uses the balanced ternary system, in which every number $x\in [-1/2, 1/2]$ is represented as $$ x = \sum_{n=1}^\infty \epsilon_n 3^{-n},\quad \epsilon_n\in \{-1, 0, 1\} $$ Write each such $\epsilon_n$ as $(1-0)$ or $(0-1)$ or $(0-0)$ as appropriate, and redistribute the terms: $$ x = \sum_{n=1}^\infty a_n 3^{-n} - \sum_{n=1}^\infty b_n 3^{-n} $$ with $a_n\in \{0, 1\}$ and $b_n\in \{0, 1\}$. It remains to multiply this by $2$, and the result follows: $2x = a-b$ where $a, b$ are in the Cantor set $C$.