Timeline for Regarding extenders
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 23, 2018 at 21:08 | comment | added | Jing Zhang | @Collapse: here is a counter example: suppose 0# exists, and $j: L\to L$ with $crit(j)=\aleph_\omega$. (since $V$-cardinals are indiscernibles so this is possible.) Then the $L$-ultrafiler derived from $j$ is not $\omega$-complete (you can translate this to the extender setting), but for each $n$, $\aleph_\omega \backslash \aleph_n+1$ is in the ultrafilter. | |
| Apr 23, 2018 at 20:59 | comment | added | Collapse | see also math.cmu.edu/users/jcumming/Appalachian/steel_cmu_2015_files/… page 19 def 4.5.. | |
| Apr 23, 2018 at 20:54 | comment | added | Stefan Mesken | @Collapse I agree that what is written is your interpretation but it seems to me that this wasn't intended by the author. As I've suspected, this property is meant to capture well-foundedness of the ultrapower and the version I've proved suffices to conclude that. Maybe the stronger version is true as well, but I remain doubtful about that. | |
| Apr 23, 2018 at 20:42 | comment | added | Collapse | See Kanamori pages 354-355. | |
| Apr 23, 2018 at 20:34 | comment | added | Stefan Mesken | @Collapse I'm fairly certain that this is the general case. If you don't have $((a_n, x_n) \mid n < \omega) \in M$, I suspect there will be counterexamples. Also note that you want $\omega$-completeness (that's the name for the property we consider here) to show that $\mathrm{Ult}(M;E)$ is well-founded. The prove above suffices to conclude that. | |
| Apr 23, 2018 at 20:32 | comment | added | Collapse | Thank you very much for the answer, but i'm more intrested in the general case. | |
| Apr 23, 2018 at 17:56 | comment | added | Stefan Mesken | In this answer I assume that $((a_n,x_n) \mid n < \omega) \in M$ but not that $E \in M$. I also get a slightly better result than required since $f \in M$. If we don't assume that the sequences $(a_n \mid n < \omega), (x_n \mid n < \omega)$ are in $M$, I don't even know where to begin... | |
| Apr 23, 2018 at 17:51 | history | answered | Stefan Mesken | CC BY-SA 3.0 |