TITLE: Defining cardinality in the absence of choice
QUESTION [28 upvotes]: Under ZFC we can define cardinality $|A|$ for any set $A$ as
$$
|A|=\min\{\alpha\in \operatorname{Ord}: \exists\text{ bijection } A \to \alpha\}.
$$
This is because the axiom of choice allows any set to be well-ordered, so that the set after $\min$ is nonempty.
If we don't assume the axiom of choice (i.e. work in ZF), then there's (at least) two approaches to cardinality. The first is that we use the same definition as above. However, the definition makes sense only for those sets that can be well-ordered. Hence the price we pay for the absence of choice is that cardinality of some sets (the nonwellorderable) is left undefined. But note that even though $|A|$ does not necessarily make sense for all sets $A$ in this approach, the various equalities and inequalities of the form $|A|=|B|$, $|A|\leq|B|$ etc. do, since we can always interpret them as shorthands for "there is a bijection/injection $A\to B$".
The second approach is that we define cardinality for all the sets in the universe using Scott's trick (hopefully I'll get it right):
$$
\gamma(A)=\min\{\alpha\in\operatorname{Ord}:\exists x\in V_\alpha\ \exists\text{ bijection }A\to x\}
$$
and
$$
|A|=\{x\in V_{\gamma(A)}:\exists\text{ bijection }A\to x\}.
$$
We manage to define cardinality for all the sets in such a way that $|A|=|B|$ iff there is a bijection $A \to B$. However, to me it seems that this time the price we have to pay is that the we get very unnatural cardinals compared to the first. For example

cardinals seem to be quite complicated sets compared to the plain and simple initial ordinals of the first approach,
$|\alpha|=\alpha$ does not hold for most (any?) of the initial ordinals $\alpha$ anymore,
$|0|=1$, $|1|=\{1\}$, etc.

My question is that what do we gain, if anything, using the second approach (besides managing to define cardinality for all the sets)? Is it an unnecessary complication having no actual advantage over the first approach (i.e. just a trick) or does it have a real use in ZF set theory?

REPLY [2 votes]: As pointed out in Andreas Blass's answer, Scott's cardinals are not so unnatural (and a large ordinal number is also a complicated object). Even with the Axiom of Choice, the initial ordinal definition makes cardinal & ordinal arithmetic confusing. For example, $2^\omega = \omega < \omega^2$, but $2^{\aleph_0} > \aleph_0 = \aleph_0^2$. Thus there is an advantage in ensuring that transfinite cardinals are not ordinal numbers.
The question does point out one serious problem with Scott's cardinals: his $0$ is the ordinal number $1$, which is confusing. Since finite ordinals and finite cardinals have the same arithmetic, I propose that (with or without AC) we apply Scott's trick to transfinite sets, and define the cardinal number of a finite (i.e. Dedekind finite & well orderable) set to be the appropriate ordinal number.