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Aug 28, 2023 at 12:11 comment added Joel David Hamkins @bof Yes, the assertion that all classes are equinumerous is true in some class theories and not others. Note that if the class V of all sets is equiunumerous with the class of ordinals Ord, then the universe is well-ordered, and this implies a strong form of the axiom of choice known as global choice. Meanwhile, it is consistent with ZFC including choice for sets, that V is not the same size as Ord. Indeed, it is consistent that the univese admits no global linear order (see mathoverflow.net/a/110823/1946).
Aug 28, 2023 at 11:08 comment added bof This stuff is over my head, but somehow I was under the inpression that all proper classes were the "same size". But in that case there'd be no point in asking about the Cantor-Bernstein theorem. Is "all proper classes are equivalent" something that only holds true is certain brands of set theory? Or is it all wrong?
Nov 24, 2010 at 21:27 history edited user5810 CC BY-SA 2.5
changed 3 instances of "\ran" to "\operatorname{ran}"
Sep 6, 2010 at 23:13 comment added Joel David Hamkins Yes, every $a\in A$ is the start of a unique back-and-forth sequence, by an instance of the Replacement axioms.
Sep 6, 2010 at 22:14 comment added Martin Brandenburg Ah ok. For me the point is the definabilty of the sequences. I thought it was difficult since the we cannot control the size of the range, which should be a subset (which set?) of the class $A \cup B$. But instead, we could just say, that $s$ is a set, which is a function with domain $\omega$, and such that for all $n$, $s(2n+1)=F(s(2n))$ and $s(2n+2=G(s(2n+1))$. I know this is trivial, but I didn't see it. ;)
Sep 6, 2010 at 21:32 comment added François G. Dorais +1: Thank you for clarifying, Joel!
Sep 6, 2010 at 20:35 history edited Joel David Hamkins CC BY-SA 2.5
added 112 characters in body; edited body
Sep 6, 2010 at 20:30 history answered Joel David Hamkins CC BY-SA 2.5