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Claim. Theorem 1: assuming the continuum hypothesis, $\mathbb{R}^2$ is the union of countably many sets in each of which any four points determine six different distances, so that all pairwise distances within a set are distinct. Theorem 2: conversely, if the plane, or even the line, is the union of countably many sets in each of which any four points determine six different distances, then the continuum hypothesis holds; for a set of at least four points that condition says that all pairwise distances are distinct. For the line and the plane the question of Problem 1127 is therefore equivalent to CH, and so independent of ZFC. For every $n$, Theorem 2 gives the negative half of the independence: a decomposition of $\mathbb{R}^n$ of the kind asked for restricts to one of a line inside it, so in a model where CH fails no such decomposition exists in any dimension. The positive half for $n\ge3$ is Kunen's, on his page, which carries the full claim.

Covers. The dimensions $n=1$ and $n=2$ completely, each equivalent to CH; and for every $n$ the half stating that the decomposition does not exist when CH fails, so that ZFC does not prove its existence.

Context. For $n=1$ the positive half under CH is older: Erdős and Kakutani proved that CH is equivalent to writing $\mathbb{R}$ as a countable union of sets each linearly independent over $\mathbb{Q}$ (on the corpus's source card and their claim page), and two distinct pairs at the same distance in such a set would give a nontrivial rational relation among at most four of its points. Davies's abstract says that Theorem 2 strengthens the converse half of that theorem by essentially its method, in a simplified form, and that Erdős had asked the question for the plane in a private communication. The site's commentary attributes the necessity of CH to Erdős and Hajnal, without a reference: when CH fails, in every decomposition of $\mathbb{R}$ into finitely many sets some set contains four points determining only four distances. As printed, with finitely many sets, that statement holds in ZFC and needs no hypothesis on the continuum: by van der Waerden's theorem every finite coloring of $\mathbb{R}$ has a monochromatic five-term arithmetic progression $a+id$ ($0\le i\le4$), whose points $a$, $a+d$, $a+3d$, $a+4d$ determine exactly the four distances $d,2d,3d,4d$. So the remark cannot show that CH is necessary, and countably many sets was presumably meant. No source for the statement is cited on the site or held by this corpus; the necessity of CH recorded on this page is Davies's Theorem 2.

Source. Roy O. Davies, Partitioning the plane into denumerably many sets without repeated distances, Proc. Cambridge Philos. Soc. 72 (1972), no. 2, 179--183, doi:10.1017/S0305004100046983. The issue is dated September 1972 and carries no day, so this page's date is the first of that month. The paper is not held by this corpus; its abstract, in the publisher's record, states both theorems and is the basis of this page, and the proofs are not part of that basis. Nothing on this page is independently reviewed by this project.

Acceptance. Refereed: the result is a journal paper in the Proceedings of the Cambridge Philosophical Society. Reviewed: the curator of erdosproblems.com, T. F. Bloom, labels Problem 1127 independent and credits Davies with the case $n=2$ under the continuum hypothesis (problem page last edited 30 December 2025); the curator credits the necessity of CH to Erdős and Hajnal rather than to Theorem 2, which is recorded here from the paper's own abstract. The curator is independent of the author.