Timeline for Is discriminative choice provable in ZFC?
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 15 at 4:12 | comment | added | Monroe Eskew | @ZuhairAl-Johar Yes. Let $\phi$ be the equality relation. For all collections $F$ of pairwise disjoint sets, replace each member $f \in F$ with $f \times F$ to meet the cardinality requirement. A choice function for this family gets a choice function for the original one. | |
| Jun 14 at 17:00 | comment | added | Zuhair Al-Johar | Is it equivalent to $\sf AC$ over $\sf ZF$? | |
| Jun 14 at 13:05 | comment | added | Zuhair Al-Johar | Nice. The point is that $\kappa$ must actually be the cardinality of $F$. If $\kappa$ is not a cardinal, then this method fails. | |
| Jun 14 at 13:04 | vote | accept | Zuhair Al-Johar | ||
| Jun 14 at 12:13 | history | answered | Monroe Eskew | CC BY-SA 4.0 |