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Feb 5, 2018 at 20:39 history edited Fernando Muro
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Feb 5, 2018 at 19:38 vote accept N. de Rancourt
Feb 5, 2018 at 16:52 answer added Taras Banakh timeline score: 32
Feb 5, 2018 at 16:04 history edited N. de Rancourt CC BY-SA 3.0
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Feb 5, 2018 at 16:00 comment added N. de Rancourt Ok, sorry, I was unclear. My question about Borel colourings was in ZFC, I edit to make it clearer.
Feb 5, 2018 at 15:57 comment added Asaf Karagila Well, there is a whole question as to what does "Borel" even mean when you remove AC completely. If $\Bbb R^n$ is a countable union of countable sets, is every set Borel, or do you mean necessarily something with a Borel code (so you can make some appeal to absoluteness later)? If the appeal to a finite lower bound requires choice, then it might be relevant here; but then also the definition of Borel matters. So in my opinion, you should edit to reflect that you're interested in a ZFC result (which could be about Borel sets), and after that we'll see where to go next.
Feb 5, 2018 at 15:30 comment added HenrikRüping You might be interested in Engelking "Dimension Theory". For example Lemma 1.5.2 might give lower bounds on the number of subsets needed to decompose $\mathbb{R}^n$.
Feb 5, 2018 at 15:09 comment added N. de Rancourt @AsafKaragila That's what I tried to do in my formlation of the question, I never asked anything without choice, but simply suggested that choice could play a role. Do you think that there is something unclear that I should modify? Only the last paragraph mentions the idea of restricting to Borel partitions, do you think I should remove it?
Feb 5, 2018 at 14:46 comment added N. de Rancourt @JānisLazovskis Your partition doesn't work: the curve with polar equation $r = \theta$ is entierly contained in the first part of your partition.
Feb 5, 2018 at 14:44 comment added Asaf Karagila Noé, I would separate this, then. First ask what happens in ZFC. If it turns out that the answer appeals to choice in a strange way, ask again with focus on that.
Feb 5, 2018 at 14:42 comment added N. de Rancourt @AsafKaragila Sorry, for the moment I am mostly interested by an answer in ZFC ;) If it turns out that an upper bound is obtained by a construction using AC, then yes, I would be interested to know if this upper bound is still valid without AC, of for partition into Borel sets.
Feb 5, 2018 at 14:33 comment added Jānis Lazovskis I think $\chi(\mathbf{R}^2)= 2$ as well, by stereographically applying your splitting of $\mathbf R$ onto the circle, then taking all the rational points (from $\mathbf R$) in circles at rational radii, and all the irrational points (from $\mathbf R$) at irrational radii to be one set. The other set would then be all the rational points at irrational radii and all the irrational points at rational radii. I think the point at $\infty$ can be added to either set. If that works, it may extend to $\mathbf {R}^n$.
Feb 5, 2018 at 14:25 comment added Wojowu @AsafKaragila OP is interested in the answer both using and without using the axiom of choice, I believe.
Feb 5, 2018 at 14:24 comment added Asaf Karagila So, are you asking the question without choice?
Feb 5, 2018 at 14:08 history asked N. de Rancourt CC BY-SA 3.0