How do we know ternary expansions with only $0$'s and $2$'s are unique? Let $c  \in  [0,1]$ and consider one of its ternary expansions $\sum_{n \ge 1} c_n / 3^n$ s.t. each $c_n = 0$, $1$, or $2$.  This ternary expansion needn't be unique.  For example:
$$
0.0222222\ldots = 0.10000\ldots
$$
But if we restrict our attention to ternary expansions which only contain $0$'s and $2$'s, it seems these expansions are unique.  But what is the rigorous way to show this?
Attempt: At first I thought we could carve out $0,2$-ternary expansions into cases where the expansion tails trail off in $0000\ldots$ and cases where the expansion tails trail off in $22222\ldots$.  But then I realized that cases like $020202\ldots$ are possible.  So how do we know that all of these expansions are unique?
 A: Suppose that $\langle c_n \rangle_{n=1}^\infty$ and $\langle d_n \rangle_{n=1}^\infty$ are two different $0$,$2$-sequences.  Then there is a least $N$ such that $c_N \neq d_N$, and without loss of generality we may assume that $c_N = 0$ and $d_N = 2$.  In order for $\sum_{n=1}^\infty c_n 3^{-n} = \sum_{n=1}^\infty d_n 3^{-n}$ we must at least have that $\sum_{n=N+1}^\infty c_n 3^{-n} \geq 2 \cdot 3^{-N}$ (in order to make up the difference after the $N$th "ternary place"), however $$\sum_{n=N+1}^\infty c_n 3^{-n} \leq \sum_{n=N+1}^\infty 2 \cdot 3^{-n} = 3^{-N}.$$  Therefore $\sum_{n=1}^\infty c_n 3^{-n} \neq \sum_{n=1}^\infty d_n 3^{-n}$.
A: Let $\overline{\alpha}=\sum\limits_{n=1}^{\infty}\dfrac{\alpha_n}{3^n}$ and $\overline{\beta}=\sum\limits_{n=1}^{\infty}\dfrac{\beta_n}{3^n}$ where $\alpha_n, \beta_n\in \{0,1\}$. If $\quad\overline{\alpha}=\overline{\beta}$ then $\alpha_n=\beta_n$ for $\forall n \in \mathbb{N}.$
Proof: We have $$0=\sum\limits_{n=1}^{\infty}\dfrac{(\alpha_n-\beta_n)}{3^n}=\sum\limits_{n=1}^{\infty}\dfrac{\gamma_n}{3^n} \qquad (*)$$ where $\gamma_n=\alpha_n-\beta_n$ and $\gamma_n \in \{0, \pm 2\}$. 
Multiplying both side of $(*)$ to $3$ we got: $$0=\sum\limits_{n=1}^{\infty}\dfrac{\gamma_n}{3^{n-1}}$$ $$0\leqslant|\gamma_1|\leqslant \left|\frac{\gamma_2}{3^1}\right|+\left|\frac{\gamma_3}{3^2}\right|+\left|\frac{\gamma_4}{3^3}\right|+\cdots\leqslant\frac{2}{3^1}+\frac{2}{3^2}+\cdots=1$$Hence $|\gamma_1|=0$ and $\gamma_1=0$ then $\alpha_1=\beta_1$.
And we can proceed this process by math induction.
