Enumeration of rational numbers If $\Bbb Q=\{q_n:n\in \Bbb N\}$ be an enumeration of $\Bbb Q$, is it true that $|q_n|<1/n$ for infinitely many $n$?
I just come up with this question, it seemed simple but I can't solve it. Is there any idea?
 A: There are enumerations with and without this property. The first claim is easy (just let $q_n=\frac1{n^2}$ for odd $n$ and enumerate the rest among the even indices somehow).
To enumerate $\mathbb Q\setminus\{0\}$ such that $|q_n|>\frac 1n$ for all indices $n$, start with any enumeration $p_n$ of $\mathbb Q\setminus\{0\}$ and recursively define $q_n$ as $p_m$ where $m$ is minimal with $p_m\notin\{q_1,\ldots,q_{n-1}\}\cup[-\frac1n,\frac1n]$. Observe that this is indeed a complete enumeration because each $p_m$ will occur as some $q_n$ where $n\le m+\lceil\frac1{|p_m|}\rceil$.
A: Let $A_n$ be the rationals in $[-1/n,1/n]\setminus (-1/{n+1},1/(n+1))$ and $B$ the rest.
Enumerate the numbers in $A_n$ by $a_{n,k}$ and the numbers in $B$ by $b_n$.
To define $x_n$ we take the first two elements of $b_n$, then the first of $a_{1,k}$, then the next $3$ from $b_n$, then the next $2$ from $a_{1,k}$, and the first from $a_{2,k}$, then the next $4$ from $b_n$, then the next $3$ from $a_{1,k}$, then the next $2$ from $a_{2,k}$, then the first from $a_{3,k}$, ... and so on.
In each round we increase by one the number of elements that we took from the sequences we have already taken numbers from, and take the first one from the next sequence $a_{n,k}$ from which we haven't taken any number.
A: It is not necessarily true.  For each $n$ just take an enumeration of those rationals $q\in\mathbb{Q}$ which satisfy $|q|>1/n$.  Doing this for each $n$ we see that each rational lies in at least one enumeration.  Then enumerate $\mathbb{Q}$ from these enumerations by choosing from these enumerations in the following order: $1,1,2,1,2,3,1,2,3,4,\dots$ ($1$ means choose from the first enumeration, etc), and when doing so ignoring redundant choices.
