Timeline for Is it still impossible to partition the plane into Jordan curves without choice?
Current License: CC BY-SA 2.5
8 events
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| Apr 15, 2010 at 0:36 | comment | added | t3suji | Is the question about the Y-shaped subsets also without the AC? (The usual proof, which I can't remember the reference for, seems to use the AC.) | |
| Apr 14, 2010 at 16:51 | vote | accept | Benoît Kloeckner | ||
| Apr 14, 2010 at 16:51 | comment | added | Benoît Kloeckner | I accept this answer, and want also to thank damiano. Finally, I take the opportunity to give a question that is not a true one, in the sense that I saw a proof some years ago. Can you prove that one cannot partition the plane by Y shaped subsets? | |
| Apr 14, 2010 at 15:21 | comment | added | damiano | @t3suji: this is a very reasonable curiosity, given what the question asks! You can prove the Jordan Curve Theorem easily using homology. I am not sure that this does not use the axiom of choice, but it seems that you can do everything quite explicitly in the various arguments. As for your question about compact sets: if you already know that the sets are compact and it is a descending chain, they are required to have a common point by compactness! | |
| Apr 14, 2010 at 14:38 | comment | added | t3suji | Thanks for the comment, I completely agree it is better to have a determined construction. It is somehow my sloppy attitude that countable choice is acceptable even if continuum choice is not, because so many construction depend on it... Out of curiosity: we are using the Jordan Curve Theorem; does its proof rely on choice? In fact, don't we have to use that a descending chain of compact sets $K_n$ has non-empty intersection? How would one prove this without countable choice? | |
| Apr 14, 2010 at 12:30 | comment | added | Joel David Hamkins | I think it is not quite good enough, but Damiano's idea works here. The issue is that there could be many possible choices for C_i+1 satisfying your condition of being contained in the interior of C_i and not having U_i in its interior. But if you enumerate a countable dense subset of the plane, then each point is contained on exactly one curve, so the choice is completely determined by the well-ordering of the countable dense set: you pick the first point lying on a curve in the interior of the previous curve. | |
| Apr 14, 2010 at 12:15 | comment | added | t3suji | damiano typed his comment as I was typing this... | |
| Apr 14, 2010 at 12:13 | history | answered | t3suji | CC BY-SA 2.5 |