This question follows up on a comment I made on Joseph O'Rourke's recent question, one of several questions here on mathoverflow concerning surprising geometric partitions of space using the axiom of choice.

  1. Joseph O'Rourke: Filling $\mathbb{R}^3$ with skew lines

  2. Zarathustra: Is it possible to partition $\mathbb{R}^3$ into unit circles?

  3. Benoît Kloeckner: Is it still impossible to partition the plane into Jordan curves without choice?

The answers to these questions show that the axiom of choice implies, by rather easy and flexible arguments, that space admits several surprising geometrical decompositions. To give a few examples:

  • $\mathbb{R}^3$ is the union of disjoint circles, all with the same radius. Proof: Well-order the points in type $\frak{c}$. At stage $\alpha$, consider the $\alpha^{\rm th}$ point. If it is not yet covered, we may find a unit circle through this point that avoids the fewer-than-$\frak{c}$-many previously chosen circles. QED

  • $\mathbb{R}^3$ is the union of disjoint circles, with all different radii. Proof: at stage $\alpha$, find a circle of a previously unused radius that avoids the fewer-than-$\frak{c}$-many previously chosen circles. QED

  • $\mathbb{R}^3$ is the union of disjoint skew lines, each with a different direction. Proof: at stage $\alpha$, put a line through the $\alpha^{\rm th}$ point with a different direction than any line used previously. QED (Edit: it appears that one might achieve this one constructively by using the $z$-axis and a nested collection of hyperboloids, which unless I am mistaken would give lines of different directions.)

  • $\mathbb{R}^3$ is the union of disjoint lines, each of which pierces the unit ball.

  • And so on.

The excellent article M. Jonsson, J. Wästlund, "Partitions of $\mathbb{R}^3$ into curves" Math. Scand. 83 (1998) 192-204 proves among other things that $\mathbb{R}^3$ can be partitioned into unlinked unit circles, either all with the same radius or all with different radii, and they have a very general theorem about partitioning $\mathbb{R}^3$ into isometric copies of any family of continuum many real algebraic curves.

So the general situation is that the axiom of choice constructions are both easy and flexible and not entirely ungeometrical.

Meanwhile, there are also a few concrete constructions, which do not use the axiom of choice. For example, Szulkin (see theorem 1.1 in the Jonsson, Wästlund article) shows that $\mathbb{R}^3$ is the union of disjoint circles by a completely explicit method. And there are others. But my question is not about these cases where there is a concrete construction. Rather, my question is:

Question 1. Can we prove, in any of the cases where there is no concrete construction, that there is none?

For example, taking Borel as the measure of "explicit", the question would be: can we prove, for any of the kinds of decompositions I mention above or similar ones, that there is no Borel such decomposition? This would mean a decomposition of $\mathbb{R}^3$ such that the relation of "being on the same circle" would be a Borel subset of $\mathbb{R}^6$, and in this case, the function that maps a point to the nature of the circle on which it was to be found (the center, radius, and so on) would also be a Borel function.

A closely related question would be:

Question 2. Can we prove for any of the decomposition types that it requires the axiom of choice?

For example, perhaps there is an argument that in one of the standard models of $\text{ZF}+\neg\text{AC}$ that there is no such kind of decomposition.

My only idea for an attack on question 1 is this: if there were a Borel decomposition of $\mathbb{R}^3$ into circles or lines of a certain kind, then the assertion that this particular Borel definition was such a partition would be $\Pi^1_2$ and hence absolute to all forcing extensions. So the same Borel definition would work to define such a partition even in forcing extensions of the universe having additional reals. Furthermore, the ground model partitions would be a part of that new partition. And this would seem to be a very delicate geometric matter to fit the new reals into the partition in between the ground model objects. But I don't know how to push an argument through.

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    $\begingroup$ Shooting from the hip, what happens when the real numbers are a countable union of countable sets? Should you require DC as well, as it ensures that most mathematics behaves normally, and at least ensures that there are more sets than Borel sets? $\endgroup$
    – Asaf Karagila
    Apr 9, 2012 at 22:27
  • $\begingroup$ Do we know that these things can't happen constructively? $\endgroup$ Apr 10, 2012 at 4:54
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    $\begingroup$ @Joel: The trouble, constructively, is in showing that each sphere intersects the circles in exactly two points. When the points of intersection pass from one circle to another, as we vary the radius of the spehere, there is a discontinuity (the points jump suddenly). This is a "model-theoretic" way of seeing that something isn't right constructively. $\endgroup$ Apr 11, 2012 at 6:07
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    $\begingroup$ Joel, you might be interested in this paper of Zoltan Vidnyansky: arxiv.org/pdf/1209.4267.pdf. In it, he provides a "black box" theorem that allows one to deduce, under the assumption $V=L$, that sets like these, which are typically produced by transfinite recursion in the way you describe, can be coanalytic. His construction is still by transfinite recursion, but he shows how to do it very carefully so that the end result is coanalytic. This does not really answer either of your main questions, but it seems relevant so I thought I should make you aware of it! $\endgroup$
    – Will Brian
    Sep 9, 2016 at 16:00
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    $\begingroup$ I really like these questions and every so often I go back and think about them. For me the obvious strategy to attempt is analytic - under some convenient assumption like "all sets of reals are measurable", show that such a decomposition cannot exist by some argument that involves, e.g. the probability distribution of the angle of the unique line through a random point. However I've been unable to make anything like this work. Is there a philosophical reason why this sort of arugment cannot be the right way to go? $\endgroup$
    – Will Sawin
    Dec 11, 2016 at 18:58


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