I have already proved that:
Proposition 1. Let $x\in[0,1)$, then exists $\{c_k\}_{k\in\mathbb{N}}\subseteq\mathbb{N}$, with $0\le c_k\le p-1$ such that $$x=\sum_{k=1}^{+\infty}\frac{c_k}{p^k},$$ where $p\in\mathbb{N},p\ge2.$
I proved that actually, $$c_k=\text{floor}\big(p\{p^{k-1}x\}\big),$$ where $\{x\}:=x-\text{floor(x)}.$
We know that the Cantor set is the intersection of the sets $K_n$, where $K_1=[0,1/3]\cup[2/3,1]$, $K_2=[0,1/9]\cup[1/9,1/3]\cup[2/3,7/9]\cup[8/9,1]$, and, in general $K_n$ is defined as the union of $2^n$ disjoint intervals of the type $[k/3^n,(k+1)/3^n]$, then $K_{n+1}$ is obtained by removing the open middle third of each of these intervals.
We consider the open interval $I:=\big(\frac{1}{3},\frac{2}{3}\big)$. We fix $p=3.$ I want to show that all $x\in I$ have the first coefficient $c_1=1$. We consider $$c_1=\text{floor}(3x)\le 3x<3\cdot\frac{2}{3}=2,$$ then $c_1<2$, since $c_1\in\mathbb{N}$ we have that $0\le c_1\le 1$. We suppose that $a_1=0$, then $$c_1=\text{floor}(3x)=0\Longrightarrow 0<3x<1\;\text{for all}\;x\in I.$$ This leads to a contradiction because $3x>3\cdot\frac{1}{3}=1.$ Therefore $c_1=1.$
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Moreover I have proved that
Proposition 2. Let $p\in\mathbb{N}$, $p\ge 2.$
$(i)\;$ Let $x=q/p^{n}$ with $q\in\mathbb{N}$ such that $1\le q\le p^{n}-1$. Then exists exactly two developments in base $p$ of $x$. The first is $$x=\sum_{k=1}^{+\infty}\frac{c_k}{p^k},$$ where \begin{equation} c_k= \begin{cases} \text{floor}(p\{p^{k-1}x\}) & \text{if$\;k=1,\dots, n$}\\ 0 & \text{if$\;k\ge n+1$} \end{cases} \end{equation} The second is $$x=\sum_{k=1}^{+\infty}\frac{d_k}{p^k},$$ where \begin{equation} d_k= \begin{cases} 0 & \text{if$\;k=1,\dots, n-1$}\\ q-1 & \text{if$\;k=n$}\\ p-1 &\text{if$\;k\ge n+1$}. \end{cases} \end{equation} $(ii)\;$ For all $x\in[0,1)$ that is not of the form above exists a unique development in base $p$.
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Now, I want to show that all $x\in [0,1/3)$ have the first coefficient $c_1\ne1$. We consider $$c_1=\text{floor}(3x)\le 3x<3\cdot\frac{1}{3}=1.$$ Therefore $c_1=0$ or $c_1=2$.
Obviously in $[0,1/3)$ there are such numbers of the form $q/3^n$ with $n\ge 2$. The first development in base $3$ we have just studied it as a function in $[0,1/3)$ the second follows from proposition $2$, which assures us that every number of $[0,1/3)$ of the form $q/3^n\;n\ge 2$ has the first coefficient $d_1=0$. Therefore every $x\in[0,1/3)$ has the first coefficient of ternary development $0$ or $2$.
Question. Is this procedure correct?
Thanks!