Timeline for Can Z + Ranks + Successor cardinals + Ordinal inaccessibility be equal to ZF?
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| May 2, 2021 at 19:56 | comment | added | Farmer S | If every wellordered set is bijective with a von Neumann ordinal (hence in fact has the ordertype some von Neumann ordinal), then yes, I agree you get replacement. By "coincide" I just mean "are equivalent". If you use Scott's ordinals instead of von Neumann's, I agree you get replacement, but I think to first get well ordered replacement, it seems to require a small argument to observe that there is no Scott ordinal with ordertype that of the class of all von Neumann ordinals... | |
| Apr 30, 2021 at 7:23 | comment | added | Zuhair Al-Johar | I don't know what you mean by "coincide" in your last remark. But it appears to me that should I've used Scott's ordinals instead of von Neumanns in the formulation of the above axioms, then the resulting system would prove full replacement, since every well ordered set would have a Scott ordinal and even an initial segement of Scott ordinals that is order isomorphic to it, then well ordered replacement would follow and this would prove full Replacement over Z + Foundation. | |
| Apr 30, 2021 at 5:45 | comment | added | Zuhair Al-Johar | so what we need to get replacement is to add to the above the axiom that every well ordered set is bijective to an ordinal. I think we won't need sucessor cardinals by then. | |
| Apr 29, 2021 at 22:09 | vote | accept | Zuhair Al-Johar | ||
| Apr 29, 2021 at 22:02 | history | answered | Farmer S | CC BY-SA 4.0 |