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Dec 26, 2021 at 12:26 comment added Zuhair Al-Johar Yes, that was in my mind as a second result, I wanted to know the proof of the more general result, then this other question would follow. Thanks
Dec 26, 2021 at 12:25 vote accept Zuhair Al-Johar
Dec 26, 2021 at 10:17 comment added Farmer S But what about the following variant (or was this maybe what you were really thinking about?): Is ZF + $\neg$AC + "every set is cardinal-definable" consistent, where of course now the cardinals need not be wellorderable?
Dec 26, 2021 at 10:14 history answered Farmer S CC BY-SA 4.0