GPT 6 Astra Ultra attempt on Math-6181

Erdős Problem 1127: distinct distances and the continuum hypothesis

Statement and answer.

For each fixed positive integer $n$, the assertion that $\mathbb R^n$ is a countable union of sets with all unordered pairwise distances distinct is equivalent, over ZFC, to the continuum hypothesis (CH). We prove the obstruction and the one-dimensional construction below, and explicitly use Kunen's theorem for the higher-dimensional positive direction. This reconstructs the known independence result; see https://www.erdosproblems.com/1127.

CH gives a partition of the line.

Assume CH and index a Hamel basis of $\mathbb R$ over $\mathbb Q$ as $\{b_\alpha:\alpha<\omega_1\}$. For each countable ordinal $\alpha$, fix an injection $e_\alpha:\alpha\to\mathbb N$. If $x\ne0$, express it uniquely as a finite rational linear combination of basis elements, and let $\alpha$ be the largest index appearing. Colour $x$ by its nonzero leading coefficient together with the finite list of pairs \[ \{(e_\alpha(\beta),q_\beta):\beta<\alpha,\ q_\beta\ne0\}. \] There are countably many possible colours. In a fixed colour, at most one real has any given largest index $\alpha$. Every finite set of distinct elements of that colour is therefore linearly independent: in a purported nontrivial rational relation, the uniquely largest basis index among the participating reals cannot cancel. Give $0$ a separate colour.

Rational linear independence implies distinct distances. An equality $|x-y|=|z-w|$ between two distinct unordered pairs, each consisting of distinct elements of one colour, yields $x-y=\pm(z-w)$, a nontrivial rational linear relation after collecting repeated elements.

Failure of CH forbids such a partition, already on a line.

If $2^{\aleph_0}>\aleph_1$, a Hamel basis has cardinality $2^{\aleph_0}$. Indeed the rational span of an infinite set has the same cardinality as that set. Choose disjoint subfamilies $\{u_i:i\in I\}$, $\{v_j:j\in J\}$ with $|I|=\aleph_1$, $|J|=\aleph_2$. Given any countable colouring of the reals, for each $j$ choose distinct $i_j,i'_j$ such that $u_{i_j}+v_j$ and $u_{i'_j}+v_j$ have a common colour $c_j$. Such a pair exists because $I$ is uncountable.

There are only $\aleph_1$ possible triples $(\{i_j,i'_j\},c_j)$. Two distinct $j,j'$ therefore give the same triple. The four reals \[ u_i+v_j,\quad u_{i'}+v_j,\quad u_i+v_{j'},\quad u_{i'}+v_{j'} \] are distinct by basis independence, are monochromatic, and have two distinct pairs at distance $|u_i-u_{i'}|$. This contradicts the required property. A partition in any higher dimension would restrict to such a partition of a line, so the same obstruction applies.

Explicit higher-dimensional input.

K. Kunen, Partitioning Euclidean space, Mathematical Proceedings of the Cambridge Philosophical Society 102 $1987$, 379--383, proves in Section 2: under CH, for every finite $n$ there is a countable colouring of $\mathbb R^n$ whose colour classes have distinct distances. The primary paper is https://doi.org/10.1017/S0305004100067426. That geometric transfinite-construction theorem is an external input here; the Hamel-basis argument above alone does not prove its higher-dimensional version. Together with the obstruction it proves the claimed equivalence for every positive $n$. The usual relative consistency and independence of CH are the additional foundational input for calling the question independent of ZFC.

Completion estimate.

The equivalence is established, with Kunen's higher-dimensional theorem and the standard independence of CH explicitly imported. COMPLETION: 100%