Skip to main content
5 events
when toggle format what by license comment
Nov 12, 2019 at 23:30 vote accept Noah Schoem
Nov 11, 2019 at 15:50 answer added Yair Hayut timeline score: 3
Nov 8, 2019 at 19:01 comment added Noah Schoem @YairHayut Yes, I believe you're right. In summary, even just with two measures on $\kappa$, $U_0\prec U_1$, a homogeneous set for $U_1$ must consist of measurables almost everywhere, whereas a homogeneous set for $U_0$ consists of nonmeasurables almost everywhere.
Nov 7, 2019 at 20:16 comment added Yair Hayut Isn't the function that sends an ordinal to its Mitchell order a counterexample for the conjecture?
Nov 6, 2019 at 21:17 history asked Noah Schoem CC BY-SA 4.0