Can proper classes also have cardinality?

Solution 1:

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

• $\forall x \forall y \forall y' (\langle x,y \rangle \in F \wedge \langle x, y' \rangle \in F \to y=y')$
• $\forall x \forall y (\langle x,y \rangle \in F \to x \in A \wedge y \in B)$
• $\forall x (x \in A \to \exists! y(\langle x,y \rangle \in F))$
• $\forall y (y \in B \to \exists! x(\langle x,y \rangle \in F))$

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

• $\forall x \forall y \forall y' (F(x,y) \wedge F(x,y') \to y=y')$
• $\forall x \forall y (F(x,y) \to \phi(x) \wedge \psi(y))$
• $\forall x (\phi(x) \to \exists! y F(x,y))$
• $\forall y (\psi(y) \to \exists! x F(x,y))$

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.

Solution 2:

There is absolutely no problem with extending the definition of a cardinal to classes, except that we cannot argue within the universe about cardinals of classes as we do for sets. Every argument of the form "All classes such that ..." would be a meta-argument. Of course, one can use a stronger set theory which allows classes, but that's a slightly different story.

Besides the above point, it is not very difficult to prove that Cantor-Bernstein theorem for classes (i.e. the existence of two injections implies the existence of a bijection). And so we can really ask whether or not there is a class function with such and such properties (injective, bijective, etc.)

It is important to note that just as when removing the axiom of choice it is possible that there are surjections which cannot be reversed, without global choice it is possible to have class-surjections which do not have an inverse injection. So it is important to stick to the definition by injections, because that definition works without any use of choice.