Timeline for Does every ordinal appearing in a model of ZF appear in a model of ZFC?
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 24, 2023 at 10:27 | comment | added | Asaf Karagila♦ | No, because if $M$ is the FL model, and $N$ is a model of $\sf ZFC$ with the same ordinals, then either $\omega_1^M$ is countable, or $\omega_1^M=\omega_\omega^N$. | |
| Mar 24, 2023 at 10:08 | comment | added | Calliope Ryan-Smith | @AsafKaragila does FefLev necessarily not admit a model of ZFC agreeing on cardinalities of ordinals? We'd disagree on cofinality, but nothing immediately strikes me as contradictory. | |
| Mar 24, 2023 at 10:06 | vote | accept | Calliope Ryan-Smith | ||
| Mar 23, 2023 at 23:09 | comment | added | Asaf Karagila♦ | I mean, we covered the Feferman–Levy model in class... | |
| Mar 23, 2023 at 2:23 | answer | added | Farmer S | timeline score: 8 | |
| Mar 22, 2023 at 17:53 | comment | added | Noah Schweber | Note that we can always get choice by shrinking: if $M\models\mathsf{ZF}$ then $L^M\models\mathsf{ZFC}$ and $M$ and $L^M$ "agree on the ordinals" (although of course they may disagree about cardinalities). In the other direction, many but not all models of $\mathsf{ZF}$ have tame class forcing extensions with the same ordinals to models of choice. Gitik's model, in which all uncountable ordinals are singular, is an example where this can't be done. | |
| Mar 22, 2023 at 16:00 | history | asked | Calliope Ryan-Smith | CC BY-SA 4.0 |