Skip to main content
11 events
when toggle format what by license comment
Nov 3, 2012 at 12:46 comment added Joel David Hamkins See related question on whether there can be a global linear order without a global well-order: mathoverflow.net/questions/111370/…
Nov 3, 2012 at 7:19 vote accept Asaf Karagila
Nov 2, 2012 at 21:58 comment added Asaf Karagila Oh. I see. You're pointing out that ZFC cannot prove that there is an ordinal definable linear ordering. How does that help to answer the question, though? After all, ZFC does not prove V=OD.
Nov 2, 2012 at 21:17 comment added Ashutosh By Solovay's model I meant the model obtained by Levy collapsing an inaccessible - AC holds here; the choice free model is obtained by passing to HOD(R).
Nov 2, 2012 at 20:59 comment added Asaf Karagila I should also point out that there are simpler ways to prove there is no definable linear ordering of the universe without the axiom of choice. Simply add a non-linearly ordered set (e.g. amorphous sets).
Nov 2, 2012 at 20:46 comment added Asaf Karagila Ashutosh, while you are correct, you should also note that I asked about ZFC, not about ZF.
Nov 2, 2012 at 19:46 comment added Ashutosh In Solovay's model, there is no ordinal definable non measurable set of reals. Hence there cannot be an ordinal definable linear ordering of power set of reals because one can pass, in a definable way, from such an ordering to a non measurable set as follows: Work in Cantor space. Write x~y iff they agree on all but finitely many bits. Restrict your ordering to the ~ equivalence classes. Let X be the set of those reals x such that the eq class containing x precedes the one containing x' where x' is obtained by changing all bits of x. Then X is not Lebesgue measurable.
Oct 28, 2012 at 17:19 vote accept Asaf Karagila
Nov 2, 2012 at 13:53
Oct 27, 2012 at 18:32 answer added Ali Enayat timeline score: 8
Oct 27, 2012 at 11:47 answer added Joel David Hamkins timeline score: 15
Oct 27, 2012 at 1:20 history asked Asaf Karagila CC BY-SA 3.0