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Dec 9, 2015 at 12:35 answer added Grigor timeline score: 4
Nov 17, 2015 at 21:00 vote accept Miha Habič
Nov 17, 2015 at 1:36 answer added Trevor Wilson timeline score: 18
Nov 11, 2015 at 15:23 comment added Asaf Karagila Yeah, I meant of course the latter case. How about trying to show that in that situation there is no uncountable model without an inaccessible?
Nov 11, 2015 at 14:57 comment added Miha Habič @AsafKaragila Well, that will give you an uncountable model with lots of inaccessibles. One would then have to argue that there are also models which are quite far from $L$, where all of the inaccessibles have been killed. This is simple enough to do if we additionally assume that in $L$ there is a transitive model containing $L_{\kappa+1}$ (where $\kappa$ is the Mahlo), but if there isn't I am not sure how to proceed.
Nov 11, 2015 at 9:02 comment added Asaf Karagila My usual solution when I'm stuck is to try and engineer a counterexample. Have you tried starting with $L$ with a single Mahlo and collapse it to be $\omega_1$?
Nov 11, 2015 at 5:46 history asked Miha Habič CC BY-SA 3.0