I have already proved the following results:
Proposition 1. Let $p\in\mathbb{N}$, $p\ge 2$.
Let $x\in [0,1).$ Then exists a sequenece $\{c_k\}\subseteq \mathbb{N}$ such that $0\le c_k\le p-1$ for all $k\in\mathbb{N}$ and $$x=\sum_{k=1}^{\infty}\frac{c_k}{p^k}.$$
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Proposition 2. Let $p\in\mathbb{N}$, $p\ge2$. Let $x\in[0,1)$ such that \begin{equation} x=\sum_{k=1}^{\infty}\frac{a_k}{p^k}=\sum_{k=1}^{\infty}\frac{b_k}{p^k} \end{equation} with two distinct sequences $\{a_k\}$, $\{b_k\}\subseteq\mathbb{N}$, $0\le a_k\le p-1$, $0\le b_k\le p-1$ for all $k\in \mathbb{N}$.
Then exists $n\in\mathbb{N}$ such that $a_k=0$, $b_k=p-1$ for all $k\ge n+1$.
Let $x\in (0,1)$, then for Proposition 1, in particular, $$x=\sum_{k=1}^{\infty} \frac{c_k}{2^k}$$ with $\{c_k\}\in\{0,1\}^{\mathbb{N}}$. The map $$\varphi_2\colon (0,1)\to \{0,1\}^{\mathbb{N}},\quad\varphi(x):=\{c_k\}$$ defined by choosing binary development with a finite number $c_k\ne 0$ of in ambiguous cases (see Proposition 2.), is injective. In fact let $x,y\in (0,1)$, then exists $\{c_k\},\{d_k\}\in\{0,1\}^{\mathbb{N}}$ such that $$x=\sum_{k=1}^{\infty}\frac{c_k}{2^k}\quad\text{and}\quad y=\sum_{k=1}^{\infty}\frac{d_k}{2^k}.$$ We suppose that $\varphi_2(x)=\varphi_2(y)$, then $\{c_k\}=\{d_k\}$ this means that $c_k=d_k$ for all $k\in\mathbb{N}$. Therefore $$x=\sum_{k=1}^\infty\frac{c_k}{2^k}=\sum_{k=1}^\infty\frac{d_k}{2^k}=y,$$ then $\varphi_2$ is injective.
Question 1. It's correct?
Now, we cosider tha map $$\varphi_3\colon \{0,1\}^{\mathbb{N}}\to [0,1],\quad\varphi_3(\{c_k\}):=\sum_{k=1}^\infty\frac{c_k}{3^k}$$
Question 2 How can I show that $\varphi_3$ to be injective using the above proposition?
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Question 3. Is $\varphi_3$ onto?
Thanks!