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Definition: For a set $x$ define its cardinality by $|x|=\min\{\alpha\in On\mid\alpha \approx x\}$.

where $On$ is the calss of all ordinals, $\alpha\approx x$ means there is a bijection $f:\alpha\rightarrow x$.

In ZFC we can say that every set $x$ has a cardinality, because $x$ can be well ordered and thus is bijective to atleast one ordinal. Can we turn this around? That is, in ZF, can we construct (find) a set, that has no cardinality? That means $\{\alpha \in On\mid \alpha \approx x \}$ is empty, quivallently there is no well order on $x$.

Michal Dvořák
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    "Not assuming AC" and "assuming not-AC" are very different things. – Eric Wofsey Jul 21 '19 at 19:27
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    No, otherwise ZFC would be known to be inconsistent because ZF would prove the negation of the axiom of choice. – Matt Samuel Jul 21 '19 at 19:28
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    Assuming that you have the axiom of regularity, you can use Scott's trick to explicitly (but nonconstructively) define in ZF a set that will be well-orderable iff AC is true. – hmakholm left over Monica Jul 21 '19 at 19:34
  • Also relevant, https://math.stackexchange.com/questions/53770/defining-cardinality-in-the-absence-of-choice and https://math.stackexchange.com/questions/53752/theres-non-aleph-transfinite-cardinals-without-the-axiom-of-choice and https://math.stackexchange.com/questions/172316/non-aleph-infinite-cardinals, and as always, search before posting. – Asaf Karagila Jul 21 '19 at 21:50

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It depends what you mean.

First, note that (in ZF) choice holds iff every level of the cumulative hierarchy is well-orderable. So if choice fails there is a canonical witness: namely, the first level of the cumulative hierarchy which is non-well-orderable.

But that probably doesn't seem to satisfying. The problem is that we can't do much better: while it's consistent that very natural sets (like $\mathbb{R}$) are non-well-orderable, it's also consistent that choice holds for a very long time - that is, that the least such $\alpha$ with $V_\alpha$ being non-well-orderable is much bigger than any ordinal we actually care about in general mathematics. So you're not going to be able to do better than the above.

Noah Schweber
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  • Since the second claim of the second paragraph seems interesting to me, the "choice holds for a long time..." part, can you please elaborate more on that? Or better, can you please provide a paper or some material on it? Many thanks. – Ldddd Jul 22 '19 at 07:21