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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 $\alpha^\text{th}$ point. If it is not yet covered, we may find a unit circle through this point that avoids the fewer-than-$\mathfrak{c}$-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-$\mathfrak{c}$-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 $\alpha^\text{th}$ 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 Jonsson, M.; Wästlund, J., Partitions of $\mathbb{R}^3$ into curves, Math. Scand. 83, No. 2, 192–204 (1998). ZBL0951.52018. JStor, 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.

LSpice
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Joel David Hamkins
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    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? – Asaf Karagila Apr 09 '12 at 22:27
  • Do we know that these things can't happen constructively? – Andrej Bauer Apr 10 '12 at 04:54
  • By the way, theorem 1.1 in Jonsson & Wästlund article is not constructive as far as I can tell. There is a problem with "each sphere intersects one of the circles in exactly two points". – Andrej Bauer Apr 10 '12 at 05:13
  • Andrej, I was using the term "construction" loosely, in classical logic, but could you explain why the argument isn't constructive in your sense? You have a circle in the $xy$ plane with center at $(4k+1,0,0)$ and a sphere of radius $r$ at the origin. So you are intersecting two circles; you can solve for the intersection points, and they vary step-wise continuously with $r$. (I am guesing that this "step-wise" step is the issue for constructive logic...) – Joel David Hamkins Apr 10 '12 at 06:38
  • And no, I don't think we know that we can't do all these things constructively. Perhaps there are nice explicit decompositions of each type. When there is, then we can prove that there is just by exhibiting it. But when there isn't, I don't see how we could prove that there isn't, and this is what my question is about. – Joel David Hamkins Apr 10 '12 at 08:43
  • Asaf, since it seems difficult to make any proof at all, it would be interesting to see that ZF alone, without DC, cannot give a certain kind of construction. So your idea of considering a model where the reals are a countable union of countable sets is a good one, but I don't see how to make it work. In that model, the two questions become in effect the same, since every set is Borel there. – Joel David Hamkins Apr 10 '12 at 10:00
  • @Joel: Your idea to use hyperboloids to construct skew lines is nice! – Joseph O'Rourke Apr 10 '12 at 13:06
  • Joseph, thanks! These lines are skew in the sense that any two of them have different directions, but not skew in the sense of your question, which I know how to achieve only by using AC. – Joel David Hamkins Apr 10 '12 at 13:15
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    @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. – Andrej Bauer Apr 11 '12 at 06:07
  • The constructive version is currently discussed at http://groups.google.com/group/constructivenews/browse_thread/thread/315f00722026af9e – Andrej Bauer Apr 11 '12 at 06:28
  • Thanks! I'm really looking forward to hearing the outcome for the constructive case---please post an answer when you arrive at something. I would be interested to know the connection between the constructive case and the case of continuous partitions in classical logic, which could of course be viewed as a stricter form of the Borel partitions I asked about. What I expect is that one could hope to rule out continuous partitions in many of the cases, but ruling out Borel partitions seems mysterious to me. – Joel David Hamkins Apr 11 '12 at 07:35
  • Hi Joel. The second question is whether some form (or consequence) of the Axiom of Choice not following from ZF (or ZF + DC$_{\mathbb{R}}$) follows from the existence of one of these decompositions? For instance, a selector for some Borel equivalence relation, or even a set of reals without the Baire property? It seems that you already know that full AC is not necessary, since the constructions all work from the existence of a wellordering of the reals. Is that right? Perhaps the existence of a selector for some equivalence relation is sufficient to carry out the constructions. – Paul Larson Jul 07 '14 at 10:56
  • Good morning, Paul! Yes, of course full AC is not necessary, since as you say a well-ordering of the reals is sufficient. My question was whether one can prove that ZF (or ZF+DC), if consistent, is not able to prove the existence of decompositions. I like your selector idea, and perhaps that is a good strategy. – Joel David Hamkins Jul 07 '14 at 11:11
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    Joel, you might be interested in this paper of Zoltan Vidnyansky: http://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! – Will Brian Sep 09 '16 at 16:00
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    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? – Will Sawin Dec 11 '16 at 18:58
  • Sounds like a good idea. Perhaps one could make such a kind of argument work with determinacy or $\text{AD}_{\mathbb{R}}$? – Joel David Hamkins Dec 11 '16 at 19:10
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    @LSpice Thanks! – Joel David Hamkins Aug 12 '22 at 20:33

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