IMC 1998 · Day 2 · P11

For every $x \in M$ choose spheres $S, T \in S$ such that $S \ne T$ and $x \in S \cap T$; denote by $U$, $V$, $W$ the three components of $\mathbb{R}^n \setminus (S \cup T)$, where the notation is such that $\partial U = S$, $\partial V = T$ and $x$ is the only point of $\overline{U} \cap \overline{V}$, and choose points with rational coordinates $u \in U$, $v \in V$, and $w \in W$. We claim that $x$ is uniquely determined by the triple $\langle u, v, w \rangle$; since the set of such triples is countable, this will finish the proof.

To prove the claim, suppose, that from some $x' \in M$ we arrived to the same $\langle u, v, w \rangle$ using spheres $S', T' \in S$ and components $U'$, $V'$, $W'$ of $\mathbb{R}^n \setminus (S' \cup T')$. Since $S \cap S'$ contains at most one point and since $U \cap U' \ne \emptyset$, we have that $U \subset U'$ or $U' \subset U$; similarly for $V$'s and $W$'s. Exchanging the role of $x$ and $x'$ and/or of $U$'s and $V$'s if necessary, there are only two cases to consider: (a) $U \supset U'$ and $V \supset V'$ and (b) $U \subset U'$, $V \supset V'$ and $W \subset W'$. In case (a) we recall that $\overline{U} \cap \overline{V}$ contains only $x$ and that $x' \in \overline{U'} \cap \overline{V'}$, so $x = x'$. In case (b) we get from $W \subset W'$ that $U' \subset \overline{U} \cup \overline{V}$; so since $U'$ is open and connected, and $\overline{U} \cap \overline{V}$ is just one point, we infer that $U' = U$ and we are back in the already proved case (a).