Is there a subset $S$ of the rationals between 0 and 1 $S \subset \mathbb{Q} \cap[0,1] \stackrel{\text{def}}= [0,1]_\mathbb{Q}$ such that
- is dense in $[0,1]_\mathbb{Q}$, in the sense that $\forall q_1, q_2 \in [0,1]_\mathbb{Q}, q_1<q_2,\;\exists s\in S, q_1<s<q_2$?
- has the property that given any two numbers in $S$, their midpoint (defined in the usual way on $\mathbb{Q}$) is not in $S$, i.e.: $\forall x, y \in S,\;\frac{x+y}{2} \notin S$?
I can easily define a non-dense set with the property (2): given $q\in\mathbb{Q}, 0\leq q<\frac12$, then this set has no midpoints: $\{q^n, n \in \mathbb{N}\}\cup\{0\}$.
I tried to get something that is dense (maybe starting subdividing [0,1] in some iterative way), but I got stuck pretty soon.
(This is not for any practical application. I was thinking about the different strategies required in software engineering to deal with rounding numbers to the nearest integer when their decimal part is 0.5, and the question about the existence of a set where there's never a tie to break came to mind.)
Notes:
- My requirement is weaker than asking for a "dense rational subset such that no two pairs of points are the same distance apart", so if a solution is known to the latter problem, it'd satisfy mine as well
- My requirement is stronger than asking for a similarly constrained subset of the real numbers. This seems intuitively to exist, while for the rational I'm genuinely unsure whether it does.
Best Answer
An explicit such set would be $A=\{q^3:q\in [0,1]_\mathbb{Q}\}$, the set of all cubes of rationals in that interval. This set has none of its midpoints: If there were three rational numbers $q,r,s$ such that $$\frac{1}{2}(q^3+r^3)=s^3,$$ we could rearrange to obtain $$\left(\frac{q}{s}\right)^3+\left(\frac{r}{s}\right)^3=2.$$ But this is impossible (unless $q=r$, which is irrelevant to the question).
Therefore, it just remains to see that $A$ is dense, which is more straightforward: Note that the map $x\mapsto x^3$ is a homeomorphism from $[0,1]$ to itself. As such, it maps dense sets to dense sets. As $A$ is just the image of $[0,1]_\mathbb{Q}$, we are done.
(Note that the "just-build-it"-approach in the comments can be made constructive by using your favourite well-order of $[0,1]_\mathbb{Q}$. This approach then just as easily yields a subset with the stronger condition of no repeated distances. I do not know a more "explicit" such set though.)