We denote $D^{\star} = D \backslash \{0\}$. We set for all $(x_n) \in \{0,1\}^{\mathbb{N}}$ $$\Lambda\left((x_n)\right) = \begin{cases} \Psi\left((x_n)\right) & \text{if } \Psi\left((x_n)\right) \in [0,1[ \backslash D^{\star} \\ \frac{\Psi\left((x_n)\right)}{2} & \text{if } \Psi\left((x_n)\right) \in D \cup \{1\} \text{ and } (x_n) \text{ is eventually constant at } 1 \\ \frac{1 + \Psi\left((x_n)\right)}{2} & \text{if } \Psi\left((x_n)\right) \in D^{\star} \text{ and } (x_n) \text{ is eventually constant at } 0 \end{cases}$$ Show that $\Lambda$ realizes a bijection from $\{0,1\}^{\mathbb{N}}$ to $[0,1[$.
We denote $D^{\star} = D \backslash \{0\}$. We set for all $(x_n) \in \{0,1\}^{\mathbb{N}}$
$$\Lambda\left((x_n)\right) = \begin{cases} \Psi\left((x_n)\right) & \text{if } \Psi\left((x_n)\right) \in [0,1[ \backslash D^{\star} \\ \frac{\Psi\left((x_n)\right)}{2} & \text{if } \Psi\left((x_n)\right) \in D \cup \{1\} \text{ and } (x_n) \text{ is eventually constant at } 1 \\ \frac{1 + \Psi\left((x_n)\right)}{2} & \text{if } \Psi\left((x_n)\right) \in D^{\star} \text{ and } (x_n) \text{ is eventually constant at } 0 \end{cases}$$
Show that $\Lambda$ realizes a bijection from $\{0,1\}^{\mathbb{N}}$ to $[0,1[$.