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...

  • $\begingroup$ I think math.stackexchange.com/questions/2156812/… is very related to this question but it doesn't show as related. That question is about defining order type of well-ordered proper classes and this one is about defining cardinality of proper classes. $\endgroup$
    – Timothy
    Commented Jan 8, 2018 at 4:22

2 Answers 2


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.

  • $\begingroup$ No, I don't think this is a load of rubbish. Classes can be seen as unary relations over the set model, so maybe we can talk about it with Higher-order language. In detail, seems we can define it by $$Ep(X,Y) \leftrightarrow \exists F(\forall x \forall y\forall y'( F(x,y) \land F(x,y') \to y=y') \land \forall x \forall y (F(x,y) \to X(x) \wedge Y(y)) \land \forall x (X(x) \to \exists! y (Y(y) \land F(x,y))) \land \forall y (Y(y) \to \exists! x(X(x) \land F(x,y))))$$ $\endgroup$
    – Popopo
    Commented Mar 22, 2013 at 8:22
  • $\begingroup$ In which $X,Y$ are unary predicate variable symbols of type (0), $F$ is a binary predicate variable symbol of type (0,0), and $Ep$ is a binary predicate symbol of type ((0),(0)). $\endgroup$
    – Popopo
    Commented Mar 22, 2013 at 8:23
  • $\begingroup$ Oh, this is only for classes of sets, so it needs to be reconsidered too. $\endgroup$
    – Popopo
    Commented Mar 22, 2013 at 9:08

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.

  • $\begingroup$ Well, seems you are right. Besides, can the universe about cardinals of classes be argued within NBG? It seems in that proper classes can be talked easier $\endgroup$
    – Popopo
    Commented Mar 22, 2013 at 7:55
  • 1
    $\begingroup$ Well, in NBG you can write something about classes, but not about collection of classes. So you can't talk about the structure of class cardinals the same we can about cardinals in ZF. But NBG does make it somewhat easier to write something like "Every two classes have such and such property". $\endgroup$
    – Asaf Karagila
    Commented Mar 22, 2013 at 8:20

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