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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

  1. is dense in $[0,1]_\mathbb{Q}$, in the sense that $\forall q_1, q_2 \in [0,1]_\mathbb{Q}, q_1<q_2,\;\exists s\in S, q_1<s<q_2$?
  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.
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    $\begingroup$ Notationally, speaking, $\Bbb Q^{[0,1]}$ typically refers to the set of all functions $[0,1]\to\Bbb Q.$ I have much more frequently seen $[0,1]_{\Bbb Q}$ to indicate the interval in question, but $\Bbb Q\cap[0,1]$ is perfectly unambigous. Intriguing question, by the way! $\endgroup$ Apr 22 at 10:34
  • $\begingroup$ @user14111's comment could (should?) be an answer. $\endgroup$ Apr 22 at 11:12

5 Answers 5

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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.)

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  • $\begingroup$ Nicely inspired. Why not squares instead of cubes? or another power? $\endgroup$
    – Piquito
    Apr 22 at 13:40
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    $\begingroup$ Well, squares don't work (for instance $\frac{1}{100}$ and $\frac{49}{100}$ have the midpoint $\frac{1}{4}$) and cubes already did :) So there was no need to look further. The question for general powers seems (to me) just as hard as FLT, on which I am no expert, so this is were I stopped. $\endgroup$ Apr 22 at 13:51
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    $\begingroup$ Perhaps this will interest you: the elliptic curve $x^3+y^3=2$ has the rational point $P= (1,1)$ so it could have an infinite number of rational points if it were not torsion, as in fact it is. If it were not torsion, then your conclusion would not be valid. ($P$ is of order $2$ or $3$ I don't remember well, I think its order is $2$) $\endgroup$
    – Piquito
    Apr 22 at 16:35
  • $\begingroup$ Interesting! Just from plotting it, you can see that the order is $2$. (The order $3$ points should be $(\sqrt[3]{2},0)$ and $(0,\sqrt[3]{2})$, if I am not mistaken.) $\endgroup$ Apr 22 at 16:53
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    $\begingroup$ And in fact there are no solutions besides $\pm(1,1,1)$ to that equation (Darmon and Merel). $\endgroup$
    – K B Dave
    Apr 22 at 21:22
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Before Tim Seifert's simple and explicit solution, another user had suggested in the comments that there must clearly be an iterative process that creates the desired set. By their reasoning: this is because, while constructing a solution, at any step we have an infinite amount of points among which to choose our next addition, and only a finite number of exclusions. How to achieve density was not discussed.

(The comment was later deleted).

For the records, I post a sketch here of what I imagine this could look like:

Iterative construction

  1. Step 1: start with the set $\{0, 1\}$, let's call it $S_{1}$

  2. Iteratively (at each step $k \geq 2$):

    1. identify a maximal gap between elements of $S_{k-1}$
    2. take an increasingly narrow interval $I_k$ around the midpoint of this gap (for instance take it centered in the gap's midpoint, with width $= 2^{-k-1}$, which, is easy to show, is entirely contained in the gap)
    3. take $I_{{\rm valid}, k}$ = $I_k \setminus (M(S_{k-1}) \cup P(S_{k-1}))$ where $M(S)$ is the set of all midpoints of $S$, and $P(S) = \{ (2s_1 - s_2)|s_1, s_2 \in S\}$ is the set of all "prolongations" of $S$ pairs.
    4. take the element $s_k$ in $I_{{\rm valid}, k}$ that is the "smallest" ("first") under a well-order of $\mathbb{Q}$
    5. Add $s_k$ to the set being iteratively constructed: $S_k = \{s_k\} \cup S_{k-1}$
  3. the required set is $S = \bigcup_{k=1\dots \infty} S_k$

This has no midpoints because at any step we avoid all midpoints, as well as all points that would make an existing point a midpoint.

It is also dense because one can easily see that the width of the largest gap at any step is bound by some function of $k$ that (slowly) goes to zero.

Tim Seifert has noted:

This approach then just as easily yields a subset with the stronger condition of no repeated distances

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  • $\begingroup$ (+1) It seems that the exclusion of $S_{k-1}$ from $I_k$ in step 3 is redundant, since steps 2.1 and 2.1 already ensure this exclusion. $\endgroup$ Apr 25 at 19:01
  • $\begingroup$ @JohnBentin, that's right. Removed $\endgroup$
    – Nicola Sap
    Apr 26 at 9:53
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Let $R:=\left\{3^i4^{-j}:i,j\in \Bbb N\right\}$ and $S:=R\cap[0\,\pmb,\,1]$. Two arbitrary distinct elements of $S$, say $3^i4^{-j}$ and $3^k4^{-l},$ where $j\leqslant l,$ have midpoint $$s:=\frac{3^i4^{l-j}+3^k}{2\cdot4^l}.$$ If $l\neq j$, then the numerator of $s$ is odd, so that the factor $2$ cannot cancel from the denominator, and hence $s\notin S$. In the case that $l=j,$ the distinctness of the original points requires that $i\neq k$, and so $s$ is of the form $$s=\frac{3^m(3^n+1)}{2\cdot4^l},$$ where $m=\min\{i,k\}$ and $n=|i-k|\neq0$. Now suppose $s\in S.$ Then the factor $3^n+1$ must equal $2\cdot4^q$, for some $q\in\Bbb N.$ Consequently, $$3^n-4^q=4^q-1=3\sum_{r=0}^{q-1}4^r.$$ But this is impossible, since $3$ cannot divide $3^n-4^q$ unless $n=0$. It follows that $s\notin S$.

It remains to show that $S$ is dense in the unit interval. The logarithmic map $x\mapsto\log_4 x\;(x\in\Bbb R_+)$ is continuous and invertible. So, for $R$ to be dense in $\Bbb R_+$, it is sufficient to show that the positive real numbers of the form $i\log_43-j\;(i,j\in \Bbb N)$ are dense in $\Bbb R_+$. Since $\log_43$ is irrational, this is guaranteed by Dirichlet's approximation theorem, which has an elementary proof here.

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    $\begingroup$ Not sure if I'm not parsing the notation correctly but, say $(m_1, n_1)$ and $(m_2, n_2)$ are $(1, 2)$ and $(3,2)$. This defines the elements $s_1 = \frac{3}{10}$ and $s_2 = \frac{7}{10}$. Their midpoint is $\frac{5}{10}$, achievable for any $(m, n), m=n$. Please point out if I'm reading this wrong $\endgroup$
    – Nicola Sap
    Apr 22 at 19:24
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    $\begingroup$ Thank you, @NicolaSap, for pointing out my blunder. I have tried to be more careful with this new answer. $\endgroup$ Apr 25 at 18:33
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Let $(r_n)$ be a labelling of the rationals. Now choose $k_n$ such that $\frac{k_n-1}{p_n} < r_n \le \frac{k_n}{p_n}$ ( $p_n$ is the $n$-th prime). We got ourselves a dense subset of rationals $s_n\colon = \frac{k_n}{p_n}$ (eliminate any $1$'s). The average of any distinct $s_n$, $s_m$ has denominator divisible by $p_n$ and $p_m$, so it is not an $s_k$.

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$S = \{\frac{2^k}{p} : 0 \le k < \log_2(p)\}$ for all primes $p.$

Given $q<1<c,$ let $n = 2^k/q$ and pick $k$ large enough such that there is a prime in $[n, cn],$ guaranteed by Bertrand's strengthened postulate. Thus, $S$ contains an element from $[q, cq],$ which implies density.

In general, it seems either you get a messy constructive example and simple density proof, or a simple example but more involved density proof.

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