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Mar 27 at 21:45 vote accept Toby Meadows
Mar 27 at 21:45
Mar 24, 2022 at 12:55 answer added Farmer S timeline score: 8
Mar 24, 2022 at 0:36 comment added Farmer S @JasonZeshengChen In fact the filter $U$ is uniquely determined by the critical point, assuming we take it to be normal in $L[U]$; anyway if we allow it to be non-normal, we still get $U'$ such that $L[U']=L[U]$.
Mar 23, 2022 at 21:11 comment added Jason Zesheng Chen (continued) From a cursory look at the proof in Jech. I think the proof of (2) carries over to the case of forcing with Silver collapse. But of course, I could be wrong.
Mar 23, 2022 at 21:08 comment added Jason Zesheng Chen Asaf, I think for some ground model and $U$, it's possible for there to be no generic filter for the Levy collapse in this situation. For example, start with $V_0=L[U]$ with a measurable $\kappa$ and normal measure $U$. Forcing to collapse it to $\omega_1$ with Silver collapse, then in $V=V_0[G]$, $\kappa$ is $\omega_1$ and has a precipitous ideal. If we happen to pick the $U$ we started with, then $L[U]$ here is just the ground model. But since $V$ is a Silver collapse extension, there is no $V_0$-generic filter for Levy collapse (see here).
Mar 23, 2022 at 10:34 comment added Asaf Karagila @Farmer: Oh, I should write a book titled "On the Dangers of Posting Comments Immediately After Waking Up". Regardless, since $\omega_1$ is the measurable of $L[U]$, there's a filter for the point wise collapse of all the ordinals below it. I guess the only missing ingredient is to combine them together?
Mar 23, 2022 at 10:26 comment added Farmer S @AsafKaragila Why is there a Levy collapse filter in $V$?
Mar 23, 2022 at 6:43 comment added Asaf Karagila Yes, just do the usual Levy collapse. The tricky part is if you want the original ideal to be generic over L[U], which I guess could be impossible.
Mar 22, 2022 at 23:20 history asked Toby Meadows CC BY-SA 4.0