Proposition 3.1.4. Let $x \in \mathbb{D}\cap [0, 1)$, then there exists a unique $M(x) \subset \natp \cap [1, \text{rk}(x)]$ such that $x = \sum_{n \in M(x)}2^{-n}$.
Proof. First suppose that $\text{rk}(x) = 1$. In which case, $x \in \bracs{0, 1/2}$, and either $M(x) = \emptyset$ or $M(y) = \bracs{1}$.
Assume inductively that the proposition holds for all dyadic rationals of rank $n$. Let $x \in \mathbb{D}_{n+1}\setminus \mathbb{D}_{n}$. By Lemma 3.1.3, there exists a unique $y \in \mathbb{D}_{n}$ such that $x = y + 2^{-n-1}$. Since $x > 0$, $x \ge 1/2^{-n-1}$, so $y \in [0, 1)$. By the inductive assumption, there exists a unique $M(y) \subset \natp \cap [1, n]$ such that $y = \sum_{k \in M(y)}2^{-k}$. In which case, $M(x) = M(y) \cup \bracs{n}$ is the desired set.$\square$