Timeline for Concerning proofs from the axiom of choice that ℝ³ admits surprising geometrical decompositions: Can we prove there is no Borel decomposition?
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24 events
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| Aug 12, 2022 at 20:33 | comment | added | Joel David Hamkins | @LSpice Thanks! | |
| Aug 12, 2022 at 20:29 | history | edited | LSpice | CC BY-SA 4.0 |
Minor tidying while this is on the front page
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| Aug 12, 2022 at 20:24 | history | edited | Joel David Hamkins | CC BY-SA 4.0 |
fixed broken link
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Dec 11, 2016 at 19:10 | comment | added | Joel David Hamkins | Sounds like a good idea. Perhaps one could make such a kind of argument work with determinacy or $\text{AD}_{\mathbb{R}}$? | |
| Dec 11, 2016 at 18:58 | comment | added | Will Sawin | 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? | |
| Sep 9, 2016 at 16:00 | comment | added | Will Brian | 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! | |
| Jul 7, 2014 at 11:11 | comment | added | Joel David Hamkins | 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. | |
| Jul 7, 2014 at 10:56 | comment | added | Paul Larson | 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. | |
| Apr 11, 2012 at 7:35 | comment | added | Joel David Hamkins | 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. | |
| Apr 11, 2012 at 6:28 | comment | added | Andrej Bauer | The constructive version is currently discussed at groups.google.com/group/constructivenews/browse_thread/thread/… | |
| Apr 11, 2012 at 6:07 | comment | added | Andrej Bauer | @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. | |
| Apr 10, 2012 at 13:15 | comment | added | Joel David Hamkins | 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. | |
| Apr 10, 2012 at 13:06 | comment | added | Joseph O'Rourke | @Joel: Your idea to use hyperboloids to construct skew lines is nice! | |
| Apr 10, 2012 at 10:00 | comment | added | Joel David Hamkins | 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. | |
| Apr 10, 2012 at 8:43 | comment | added | Joel David Hamkins | 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. | |
| Apr 10, 2012 at 6:38 | comment | added | Joel David Hamkins | 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...) | |
| Apr 10, 2012 at 5:13 | comment | added | Andrej Bauer | 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". | |
| Apr 10, 2012 at 4:54 | comment | added | Andrej Bauer | Do we know that these things can't happen constructively? | |
| Apr 9, 2012 at 22:48 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
Added constructive version of lines partition
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| Apr 9, 2012 at 22:27 | comment | added | Asaf Karagila♦ | 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? | |
| Apr 9, 2012 at 21:00 | history | edited | Joel David Hamkins |
Fixed tag
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| Apr 9, 2012 at 20:38 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
deleted 3 characters in body; added 1 characters in body; added 5 characters in body
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| Apr 9, 2012 at 20:32 | history | asked | Joel David Hamkins | CC BY-SA 3.0 |