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Let us consider the set $\ell_\neq$ of bounded sequences of unequal real terms. We use the following descriptions. A sequence $x=(x_0,x_1,...)\in\ell_\neq$ is dense if, for all $\varepsilon>0$, whenever $\inf x<\xi<\sup x$, there is $i\in\Bbb N$ such that $|x_i-\xi|<\varepsilon$. A sequence is rational if all its terms are rational. The separation of a sequence $x=(x_0,x_1,...)\in\ell_\neq$ is defined as $$\operatorname{sep}x:=\dfrac{\inf\{|i-j|\,|x_i-x_j|:i,\!j\in\Bbb N;\,i\neq j\}}{\sup\{|x_i-x_j|:i,\!j\in\Bbb N\}} ,$$ and we write $\ell_+:=\{x\in\ell_\neq:\operatorname{sep}x>0\}$.

A dense sequence in $\ell_+$ is known, but it is not rational. Also known is a sequence of maximal separation, which is rational but not dense. The following questions arise.

(Q1) Is there a dense rational sequence in $\ell_+$?

(Q2---more demanding) Can a sequence in $\ell_+$ enumerate the rationals in an interval?

Ultimately, I am looking for the supremal separation (and, if possible, a sequence attaining that supremum) of sequences in $\ell_+$ of the following types: (A) dense; (B) dense and rational (if such exists), as in Q1; and (C) enumerating the rationals in an interval (if such exists), as in Q2.

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This is really only a half-answer since I haven't yet proved that it really works, but regarding (Q1), how about the following?

$1/2,1/4,3/4,1/8,5/8,3/8,7/8,1/16,9/16,5/16,13/16,3/16,11/16,7/16,15/16,\dots$

This sequence consists of all dyadic rationals inside the unit interval (and is therefore dense), in order of increasing denominator. Within each block of numbers with the same denominator, the fractions $p/2^k$ are ordered according to the reverse-lexicographic order on the binary expansion of $p$. For example, for $k=4$, the numerators are $0001,1001,0101,1101,0011,1011,0111,1111$.

Unless I'm missing something, this should have separation $1/4$.

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