Timeline for Does "$X \not\to (\omega)^\omega_2$ for every infinite $X$" imply ${\sf AC}$?
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 29, 2022 at 20:18 | comment | added | Asaf Karagila♦ | Do we even know that WOC is forceable over Solovay models (without forcing full AC, that is)? | |
| Mar 29, 2022 at 17:55 | comment | added | Andrés E. Caicedo | Yes, that's the situation at the moment. And sure, if we knew that the Mathias relation gives you an inaccessible, we would be done. The alternative is to force $\mathsf{WOC}$ over, say, a Solovay model, while preserving the relation. | |
| Mar 29, 2022 at 16:50 | comment | added | Asaf Karagila♦ | So, really it's just WOC + $\omega\nrightarrow(\omega)^\omega_2$, which is a consequence of WOC + well-orderable continuum. And the question remains whether or not the partition relation just follows from WOC, and can therefore be eliminated as an explicit assumption? If $\omega\to(\omega)^\omega_2$ has some LC strength to it, that means that we can arrange all kind of models where the continuum is not well-orderable, but the partition relation fails and WOC holds, just by taking symmetric extensions over some LC-challenged model. | |
| Mar 29, 2022 at 8:12 | vote | accept | Dominic van der Zypen | ||
| Mar 29, 2022 at 3:59 | history | answered | Andrés E. Caicedo | CC BY-SA 4.0 |