Proving that a specific function $2^{\mathbb{N}}\to\mathbb{R}$ is injective.

Question

Consider $\mathbb{R}$ as a set of Dedekind cuts. $A \subseteq \mathbb{Q}$ is a Dedekind cut if

• $A \notin \{\varnothing,\mathbb{Q}\}$,

• $(\forall q \in A)(\forall p \in \mathbb{Q})(p < q \Longrightarrow p \in A)$,

• $(\forall q \in A)(\exists p \in A)(p < q)$.

Consider a function $\rho\colon 2^\mathbb{N} \to \mathbb{R}$ sending each function $f\colon\mathbb{N}\to\{0,1\}$ to $\bigcup_{n \in \mathbb{N}} \{q \in \mathbb{Q} \mid q < \sum_{k = 0}^{k = n} \frac{f(k)}{10^{k+1}} \}$.

It's pretty straightforward to check that a union of Dedekind cuts is a Dedekind cuts unless it is equal to $\mathbb{Q}$, but for any $n \in \mathbb{N}, 0 \leq \{q \in \mathbb{Q} \mid q < \sum_{k = 0}^{k = n} \frac{f(k)}{10^{k+1}} \} \leq 1$.

What I need to do is to prove that $\rho$ is injective. That is, that given two function $f,g\colon\mathbb{N}\to\{0,1\}$, $$\bigcup_{n \in \mathbb{N}} \{q \in \mathbb{Q} \mid q < \sum_{k = 0}^{k = n} \frac{f(k)}{10^{k+1}} \} = \bigcup_{n \in \mathbb{N}} \{q \in \mathbb{Q} \mid q < \sum_{k = 0}^{k = n} \frac{g(k)}{10^{k+1}} \}.$$

Answer 1

If $f,g\colon\mathbb{N}\longrightarrow\{0,1\}$ are distinct, consider that smallest $n\in\mathbb N$ such that $f(n)\neq g(n)$. You can assume without loss of generality that $f(n)=0$ and that $g(n)=1$. Then$$m\geqslant n\implies\sum_{k=0}^m\frac{f(k)}{10^{k+1}}<\sum_{k=0}^m\frac{g(k)}{10^{k+1}}.$$But then $\rho(f)\varsubsetneq\rho(g)$.