Timeline for Does Well-Ordered Interval Power Set "WOIPS" principle , prove $\sf AC$ in $\sf ZFA$?
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 3 at 6:19 | comment | added | Elliot Glazer | $\mathcal{P}(X) \cong \mathcal{P}(X^2) \supseteq \{\text{wellorderings of } X \}\ge^* \kappa \Rightarrow \mathcal{P}^2(X) \ge^* \mathcal{P}(\kappa) \ge^* \kappa^+ \Rightarrow \kappa^+ \le \mathcal{P}^3(X).$ | |
| Jun 2 at 22:56 | comment | added | Zuhair Al-Johar | @ElliotGlazer How did you construct the successor cardinal I believe you mean $\kappa^+$, won't we need the fourth power for that, I mean $\aleph(X) \leq \mathcal P^2(X)$ but $(\aleph(X))^+ \leq \mathcal P^4(X)$. What are the details of your construction that enabled doing matters in $\mathcal P^3(X)$? | |
| Jun 1 at 18:03 | comment | added | Elliot Glazer | @AsafKaragila The condition $X=X^2$ is enough to get $\aleph(X) \le \mathcal{P}^2(X),$ though I did need the third power set for the successor cardinal. I still suspect the claim holds for just a single power set, perhaps by listing out as many comparisons as possible among the cardinalities between $\mathcal{P}^k(X)$ and $\mathcal{P}^{k+1}(X)$ for $k$ up to say 10, and eventually finding a contradiction. I don’t think it would be a particularly enlightening endeavor though. | |
| Jun 1 at 14:56 | comment | added | Asaf Karagila♦ | I'm not sure that 3 power sets is really unnecessary. mathoverflow.net/questions/98365/… | |
| Jun 1 at 14:55 | comment | added | Asaf Karagila♦ | Another way of assuming $X^2=X$ is by simply replacing $X$ with $X^{<\omega}$ or $X^\omega$. Well-ordering a larger set is "more difficult" anyway. | |
| Jun 1 at 8:14 | comment | added | Elliot Glazer | See mathoverflow.net/a/471784/109573 | |
| Jun 1 at 7:18 | comment | added | Zuhair Al-Johar | Why $|X|=|X|^2$? | |
| Jun 1 at 1:01 | history | answered | Elliot Glazer | CC BY-SA 4.0 |