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Aug 4, 2017 at 11:42 history edited Joseph O'Rourke CC BY-SA 3.0
Image link broken; now fixed.
Apr 3, 2012 at 12:25 vote accept Joseph O'Rourke
Apr 2, 2012 at 20:55 comment added Joel David Hamkins In particular, for any Borel partition the functions that map each point to the basic information about the line it is on, such as the slopes and the intercepts and unit directional vector and so on, would all be Borel functions. My only idea at the moment for refuting such a Borel partition is this: if there were a Borel partition, then the same Borel code would give a partition in every forcing extension, since the assertion that a code works has complexity $\Sigma^1_2$, which is absolute. Thus, the ground model partition would extend as-is to forcing extensions with new reals.
Apr 2, 2012 at 20:32 comment added Joel David Hamkins One way to formalize what it means is that the relation of "being on the same line" for the partition, which is a binary relation on points in $\mathbb{R}^3$, would be a Borel subset of $\mathbb{R}^6$.
Apr 2, 2012 at 20:29 comment added Joseph O'Rourke I would appreciate a definition of a Borel partition (or, a pointer to where I can learn). Thanks!
Apr 2, 2012 at 20:25 comment added Joel David Hamkins Let's try to prove that there can be no Borel such partition of space into lines...
Apr 2, 2012 at 20:21 comment added Joseph O'Rourke And I even supplied one of the answers to that question... Thanks, Anton! This accords with Andreas's sketch.
Apr 2, 2012 at 20:05 answer added Andreas Blass timeline score: 10
Apr 2, 2012 at 20:01 comment added Anton Petrunin The same construction as in the following question should give a YES answer. mathoverflow.net/questions/28647/…
Apr 2, 2012 at 19:45 history asked Joseph O'Rourke CC BY-SA 3.0