Property of a sequence being an enumeration of the rationals. 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?
 A: 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).
