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As has been discussed here before, it is possible to define cardinality (a function $|\cdot|$ from the universal class to a class of "cardinals" such that $x\approx y\iff|x|=|y|$) using either AC or regularity. I saw somewhere else the statement that the following, strictly weaker than AC or regularity, is sufficient:

Every set is equinumerous to some well-founded set.

I don't know if that statement is provable in ZF without regularity. Is it known whether cardinality in the sense I have described must exist in ZF without regularity?

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dfeuer
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    You may want to take a look at this question, where I provide some references showing that the statement is not provable if we work in $\mathsf{ZFA}$, where atoms are allowed. As pointed out by Asaf Karagila in a comment there, David Pincus showed that if we further require that $x\approx |x|$, then $\mathsf{ZF}$ cannot prove the existence of such a function. The general case of your question (just as Mario Carneiro's) seems open. – Andrés E. Caicedo Jun 18 '13 at 00:28
  • @AndresCaicedo: if you can find a reference stating that it is an open problem, then I will accept that as the answer. – dfeuer Jun 18 '13 at 00:31

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First of all, much like Andres Caicedo writes in the comments, it is consistent with $\sf ZF$ that there is no choice function on the class of Scott cardinals (i.e. cardinals defined using Scott's trick). In fact it is consistent that there is no such function on a set of cardinals as well.

This is due to Jech and Sochor, although they claimed that the proof is trivially achieved from their embedding theorem. Pincus later showed they are wrong, and gave an improved embedding theorems which allowed this to be transfered from $\sf ZFA$. He later wrote a paper about cardinal representatives where he showed some conditions on the existence of a choice function on Scott cardinals. I am unaware of anyone else working on this topic since then.

As for the statement "Every set is equinumerous to a well-founded set", I don't think it is sufficient, and here is an example which I haven't sat down to write in details, but seems to work:

Consider the case where we have the axiom of choice and one Quine atom $x=\{x\}$ which generates the universe using iterated power set operations. Force (with symmetries of course) over the well-founded part in a way which violates the existence of cardinal representatives.

Note that the Quine atom $x$ still generates the universe. It is not hard to show that any set is equinumerous to a well-founded set (note that $V_1$ is a singleton, and therefore $\mathcal P^\alpha(x)\approx V_{1+\alpha}$).

Asaf Karagila
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  • Every well-founded set $S$ can be assigned a Scott cardinality, the set of all well-founded sets of minimal rank equinumerous to $S$. If every set is equinumerous to a well-founded set, then it can be assigned the Scott cardinality of that set. AoF implies this trivially. AC implies it via well-ordering. I'm not seeing how the choice function on cardinals (choice of representative) relates to the question, per se, although it is an interesting/important strengthening of the concept of cardinal. – dfeuer Jun 18 '13 at 06:15
  • @dfeuer: It's unclear from the formulation of your question, then, whether or not you want to know that the existence of cardinal representatives follow from the statement about equinumerosity, or something else. – Asaf Karagila Jun 18 '13 at 07:00
  • No, no, I'm looking for conditions under which it is possible to define a bare-bones (but total) cardinality function. – dfeuer Jun 18 '13 at 07:48
  • What do you mean cardinality function? Something like Scott cardinals, or something like cardinal representatives? – Asaf Karagila Jun 18 '13 at 07:52
  • Something like either. A function $|\cdot|$ with the property described in the question. – dfeuer Jun 18 '13 at 16:16
  • I'm still not sure what is it that you are asking for. What are the minimal axioms to prove the existence of such function, or does "every set is equinumerous to a well-founded set" is a sufficient condition, or what. – Asaf Karagila Jun 18 '13 at 17:07
  • What are the minimal axioms necessary to prove the existence of such a function? – dfeuer Jun 18 '13 at 17:23
  • I have no idea. It seems likely that the only "concrete" axiom would be the tautological "a cardinality function exists". – Asaf Karagila Jun 18 '13 at 19:55