Rational distance implies countable set

Question

I am working on this problem for weeks without a good solution.

Let $S\subset\Bbb R^d$ be a set in which $\rho(s_1,s_2)\in\Bbb Q$ for any $s_1,s_2\in S$, where $\rho$ is the Euclidean distance in $\Bbb R^d$. Show that $S$ is countable.

If possible, can I request a proof not only some hints, since I've been working on it for weeks. I have some vague idea to attack it, but I can't make it rigorous.

Thank you!

Answer 1

This is based on Yuval Filmus' idea.

Let $n+1$ the maximum number of affinely independent points among the given ones. We have $n\le d$. Let $p_1$,$\ldots$, $p_{n+1}$, $n+1$ affinely independent and $A$ their affine span. Our set will be contained in $A$. For any $n+1$ nonnegative numbers $d_1$, $\ldots$, $d_{n+1}$ there exists at most one point $p$ in $A$ so that $d(p, p_i) = d_i$. We have therefore an imbedding of $A$ into $\mathbb{R}^{d+1}$ ( into an algebraic subset of $\mathbb{R}^{d+1}$ -- we have an algebraic relation between the distances of $d+2$ points in an affine $d$-dimensional space). Our set gets mapped injectively into $\mathbb{Q}^{d+1}$ so it is countable.

Answer 2

Consider any fixed $d$ points $p_1,\ldots, p_d$ and distances $x_1,\ldots, x_d$. There is a finite number of points at distance $x_i$ from $p_i$ for each $i$. In your case there are countably many choices for the distances.