Property of a sequence being an enumeration of the rationals.

Question

Let $(r_n)$ be an enumeration of the rationals and $x\in\mathbb{R}$. Is it possible to find out whether the set $\left\{n\in\mathbb{N}:\left|x-r_n\right|<\frac{1}{2^n}\right\}$ is finite or infinite?

Answer 1

Let $x$ be an arbitrary real. As $\mathbb Q$ is dense in $\mathbb R$, we can surely find a sequence $(q_n)_{n\in\mathbb N}$ of mutually distinct rationals with $|x-q_n| < \frac1{2^{2n}}$ for all $n$. Let $(p_n)_{n\in\mathbb N}$ be an enumeration of all rationals (there are certainly infinitely many) not occurring in this sequence. For $n\in\mathbb N$, define $r_{2n+1}$ to be $p_n$ and $r_{2n}$ to be $q_n$, then $|x-r_{2n}|=|x-q_n|< \frac1{2^{2n}}$. So, for all even $n$ (and thus for infinitely many) we have $|x-r_{n}| < \frac1{2^n}$.

For the opposite direction, let $x$ again be an arbitrary real and let $(q_n)_{n\in\mathbb N}$ be an arbitrary enumeration of the rationals. Define $(r_n)_{n\in\mathbb N}$ recursively like so: \begin{align*} k_n &= \min\{k \in \mathbb{N} \mid q_k \notin \{r_m \mid m < n \} \land |x - q_k| \geq \frac1{2^n}\} \\ r_n &= q_{k_n} \end{align*} It should be obvious that $(r_n)_{n\in\mathbb N}$ is by definition injective. Furthermore, for each $n$ there will always be infinitely many $q_k$ with $|x-q_k| \geq \frac1{2^n}$, so $(r_n)_{n\in\mathbb N}$ is well-defined. Because we always pick the minimal $k$ in each step, each rational will eventually occur in the $(r_n)_{n\in\mathbb N}$ sequence, i.e. we have an enumeration. For this enumeration, by construction there is no single $n$ with $|x-r_{n}| < \frac1{2^n}$.

These two paragraphs together show that if we don't know anything else about $x$ and the enumeration, then your set can be infinite or finite (even empty).