In some set theories such as ZF+GAC, in which GAC is global axiom of choice, the Von Neumann universe $V$ bijects to $Ord$, the class of ordinals. It suggests us that proper classes may also have cardinality，in the example is $|V|=|Ord|$. In addition, if we are in ZF+GAC+ALS, it seems $|V|$ is the only cardinality which is not a cardinal number. Moreover, it seems some properties such as Cantor–Bernstein–Schroeder theorem also holds for cardinality of proper classes, but I'm not sure if it is well-defined and won't cause any paradox…
Set Theory – Can Proper Classes Have Cardinality?
cardinalsset-theory
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The idea behind Scott's trick of turning the equivalence classes into rather complicated sets is merely to allow working with the partial order of cardinalities within the theory with ease.
In the presence of AC, we can always pick a canonical example for each cardinality, namely the initial ordinal of the equivalence class.
It is consistent with ZF that no choice of canonical representatives exist. Namely, there is no definable class-function $C$ such that for all $X\in V$:
- $C(X)=C(Y)\iff |X|=|Y|$;
- $|C(X)|=|X|$
This sort of $C$ exists naturally with the axiom of choice, as I have mentioned above. It seems that we somewhat take for granted this existence.
With or without the axiom of choice we can consider $|X|$ as in Scott's trick, namely taking the equipollent sets of the least possible rank. However with the axiom of choice we can set $C(X)=\min\{a\in Ord\mid |X|=|a|\}$, and just assume $|X|=C(X)$.
The point is that without the axiom of choice we simply cannot have this luxury, and we are reduced to handling these complicated sets of cardinalities. This is just one more reason why the cardinal arithmetics become so heavy when leaving the axiom of choice behind.
When canonical representatives are not guaranteed, the use of Scott's trick become essential when writing theorems about cardinalities.
Suppose $A$ is amorphous (that is $B\subseteq A$ implies $B$ is finite or $A\setminus B$ is finite).
I want to describe $(\{|Y|\colon Y\subseteq A\},<)$. Using Scott's trick this is easily done, since $Y\mapsto |Y|$ is a definable function, the domain of this partially ordered set is definable nicely from $A$.
However using the first approach I am left to wonder what is the domain of the cardinalities of subsets of $A$? In this approach $|A|$ is a syntactic object, not semantic.
I can describe that this is a linearly ordered set (i.e. every two subsets of $A$ have comparable cardinalities) but can I prove that this set is exactly $\omega+\omega^*$? (that is to say, a linear order in which every point has either finitely many points above or finitely below; but not both) No, I cannot.
This is because $B_X=\{|B|\colon |B|<|X|\}$ cannot be described uniformly within the model, and so we cannot describe its size in a uniform way (that is as a function $X\mapsto |B_X|$).
As Andres commented on the main question, in many cases it is not a big issue. This is the main reason why this "example" seems a bit artificial. However it does help when you have a nice way to define cardinalities in the times you actually need it.
I should mention that ordinals are always well-ordered and therefore of an $\aleph$-number kind of cardinality, and such $C$ can be defined for the class of well-orderable sets. The thing is that without the axiom of choice we just tend to have sets which cannot be bijected with ordinals with the absence of choice.
For more information: T. Jech, The Axiom of Choice Ch. 11
Added note: Scott's trick makes a heavy use of the axiom of regularity (also: axiom of foundation), and I am not aware of a clean way for defining cardinalities with the lack of both regularity as well choice (or even with only the former absent).
Another important note is that Scott's trick is not only useful to define cardinalities when lacking choice, but also to define any other equivalent relation over classes. Things such as ultraproducts of the universe, for example, rely heavily on this construction.
What I mean by this is: how do we know it does not have unexpected consequences which may alter the rest of mathematics?
I will give a couple remarks, and also link to these MathOverflow discussions:
(1) In set theory, the study of large cardinals (much "larger" than just inaccessible) has been very fruitful. The existence of many of these large cardinals requires the existence of inaccessibles. So set theorists are interested in these large cardinals because of their useful (perhaps "startling") consequences. If there were no interesting consequences, set theorists would find other things to look at.
(2) From the skeptical POV, we don't know what the consequences might be. It could be that ZFC is consistent but ZFC plus the universe axiom is inconsistent. Many people come to feel that they have some intuition that the existence of universes is not inconsistent with ZFC. This belief often comes from thinking about the way that the cumulative hierarchy works. On the other hand, there is a manuscript by Kiselev (link) in which he claims to prove that the existence of even one inaccessible cardinal is inconsistent with ZFC.
We do know that ZFC cannot prove that there is even one inaccessible cardinal. And we know we cannot prove in ZFC that the existence of an inaccessible is consistent, because of limitations coming from the incompleteness theorems. So any argument that inaccessibles are consistent will have to use methods that cannot be formalized in ZFC.
(3) Temporarily adopt a Platonistic perspective, at least for the sake of argument. From this position, each "axiom" is either true or false, but it cannot alter the properties of mathematical objects, which exist separately from the axioms used to study them. Of course we can prove false statements from false axioms, but we can't actually change the objects themselves.
(4) Now temporarily reject Platonism, and think only about formal proofs. Then it will not make any difference to my conception of mathematics if someone else adopts an axiom that I don't accept. I will simply put a * beside all the theorems that use this axiom, and count them as dubious at best. I might even reprove some of the theorems without the new axiom just so I know they are OK. In this way, my personal conception of mathematics would also be unchanged by other people using different axioms.
I think that (3) and (4) start to indicate the way that philosophical issues will enter in when we ask about the effects of different axioms on "mathematics".
(This answer is marked as community wiki, as I already gave a different answer for this question. Please feel welcome to add more links to the list of links above.)
Best Answer
Two sets $A$ and $B$ have the same cardinality if and only if there is a bijective function $f : A \to B$. If we identify the function $f$ with its graph $F = \{ \langle x, y \rangle \in A \times B\, :\, f(x)=y \}$ then we can reformulate this to say that $|A|=|B|$ if and only if there is a set $F$ such that
The first two of these tell you that $f$ is a well-defined function $A \to B$ (or, rather, that $F$ is the graph of a well-defined function $A \to B$), the third gives you injectivity and the fourth gives you surjectivity.
If $A = \{ x:\phi \}$ and $B = \{ y:\psi \}$ are classes, where $\phi,\psi$ are unary predicates, then $x \in A$ really just means $\phi(x)$ and $y \in B$ really just means $\psi(y)$. So I guess you could translate the above definitions to refer to classes instead of sets. More precisely, say $|A|=|B|$ if and only if there exists a binary predicate $F$ such that
Notice that this notion of classes 'having the same cardinality' coincides with that of sets when we restrict to the case where $A$ and $B$ really are sets. However, unlike with sets, this is formulated by quantifying over formulae, so we have to work in the metatheory.
Also beware that this is a definition of 'having the same cardinality', not a definition of 'cardinality'; finding a good notion for the latter might be quite difficult.
Disclaimer: There's a chance that I'm going to be told that this is a load of rubbish. And indeed it might be, ZFC does weird things with classes. But it seems like one of the possible 'natural' extensions of the notion of bijection from sets to arbitrary classes.