We denote $D^{\star} = D \setminus \{0\}$ where $D = \bigcup_{n \in \mathbb{N}^{\star}} D_n$ and $D_n = \left\{ \sum_{j=1}^{n} \frac{x_j}{2^j},\, (x_j)_{j \in \llbracket 1,n \rrbracket} \in \{0,1\}^n \right\}$. We set for all $(x_n) \in \{0,1\}^{\mathbb{N}}$ $$\Lambda\left((x_n)\right) = \begin{cases} \Psi\left((x_n)\right) & \text{if } \Psi\left((x_n)\right) \in [0,1[ \setminus D^{\star} \\ \frac{\Psi\left((x_n)\right)}{2} & \text{if } \Psi\left((x_n)\right) \in D \cup \{1\} \text{ and } (x_n) \text{ is eventually constant at } 1 \\ \frac{1 + \Psi\left((x_n)\right)}{2} & \text{if } \Psi\left((x_n)\right) \in D^{\star} \text{ and } (x_n) \text{ is eventually constant at } 0 \end{cases}$$ where $\Psi : \{0,1\}^{\mathbb{N}} \rightarrow [0,1]$, $(x_n) \mapsto \sum_{n=0}^{+\infty} \frac{x_n}{2^{n+1}}$. Show that $\Lambda$ realizes a bijection from $\{0,1\}^{\mathbb{N}}$ to $[0,1[$.
We denote $D^{\star} = D \setminus \{0\}$ where $D = \bigcup_{n \in \mathbb{N}^{\star}} D_n$ and $D_n = \left\{ \sum_{j=1}^{n} \frac{x_j}{2^j},\, (x_j)_{j \in \llbracket 1,n \rrbracket} \in \{0,1\}^n \right\}$. We set for all $(x_n) \in \{0,1\}^{\mathbb{N}}$
$$\Lambda\left((x_n)\right) = \begin{cases} \Psi\left((x_n)\right) & \text{if } \Psi\left((x_n)\right) \in [0,1[ \setminus D^{\star} \\ \frac{\Psi\left((x_n)\right)}{2} & \text{if } \Psi\left((x_n)\right) \in D \cup \{1\} \text{ and } (x_n) \text{ is eventually constant at } 1 \\ \frac{1 + \Psi\left((x_n)\right)}{2} & \text{if } \Psi\left((x_n)\right) \in D^{\star} \text{ and } (x_n) \text{ is eventually constant at } 0 \end{cases}$$
where $\Psi : \{0,1\}^{\mathbb{N}} \rightarrow [0,1]$, $(x_n) \mapsto \sum_{n=0}^{+\infty} \frac{x_n}{2^{n+1}}$.
Show that $\Lambda$ realizes a bijection from $\{0,1\}^{\mathbb{N}}$ to $[0,1[$.