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Feb 13, 2023 at 16:22 vote accept Zuhair Al-Johar
Feb 11, 2023 at 17:41 history edited LSpice CC BY-SA 4.0
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Feb 11, 2023 at 13:23 answer added Farmer S timeline score: 8
Feb 11, 2023 at 9:55 answer added bof timeline score: 2
Feb 11, 2023 at 9:21 comment added Zuhair Al-Johar @FarmerS, Yes! Thanks. I've corrected it.
Feb 11, 2023 at 9:21 history edited Zuhair Al-Johar CC BY-SA 4.0
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Feb 11, 2023 at 9:18 comment added Farmer S Your formal rendering of "$X$ is a family of pairwise disjoint nonempty sets" is not right; it is in fact equivalent to saying that $X=\emptyset$, because otherwise we can let $a\in X$ and apply it with $a=b$, from which we get $a\neq\emptyset$ and $a\cap a=\emptyset$, a contradiction.
Feb 11, 2023 at 7:42 review Close votes
Feb 16, 2023 at 3:01
Feb 11, 2023 at 7:41 history edited Zuhair Al-Johar CC BY-SA 4.0
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Feb 11, 2023 at 7:41 history undeleted Zuhair Al-Johar
Feb 11, 2023 at 7:23 history deleted Zuhair Al-Johar via Vote
Feb 11, 2023 at 7:22 history undeleted Zuhair Al-Johar
Feb 11, 2023 at 6:20 history deleted Zuhair Al-Johar via Vote
Feb 11, 2023 at 5:42 review Close votes
Feb 11, 2023 at 6:27
Feb 10, 2023 at 21:52 comment added Noah Schweber I think this is equivalent to asking whether $\mathsf{ZF}$ proves that every set is in bijection with a transitive set (certainly this principle implies $\mathsf{DpC}$, the latter of which is incidentally perhaps-more-clearly phrased as "choice for sets of transitive sets").
Feb 10, 2023 at 21:35 history asked Zuhair Al-Johar CC BY-SA 4.0