In ZFC set theory, we know that the size of every set can be measured through its cardinality, but what about proper classes? We can view a proper class W which has a 'cardinality' larger than any set, the question is, can we construct another class V which has greater 'cardinality' than W?

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    $\begingroup$ I would ask more broadly than ZFC, so as to get a better handle on proper classes. In that case, I some theories, no. But see discussion at en.m.wikipedia.org/wiki/Axiom_of_limitation_of_size, where it mentions a model of NBG without Choice such that $Ord$ is smaller than $V$, both proper classes $\endgroup$
    – David Roberts
    Apr 4 at 21:42
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    $\begingroup$ It's consistent with $\mathsf{ZFC}$ that every proper is in bijection with $V$ (more precisely, the scheme saying that every $\Sigma_n$-definable proper class is in bijection with $V$ for each $n$ is consistent with $\mathsf{ZFC}$), but it's also consistent that this fails. $\mathsf{ZFC}$ does prove, appropriately phrased, that every proper class surjects onto $\mathsf{Ord}$, but again there can be proper classes into which $\mathsf{Ord}$ does not inject. $\endgroup$ Apr 4 at 23:10
  • $\begingroup$ Does this answer your question? Bijective-equivalent collections of proper classes in set theory $\endgroup$ Apr 4 at 23:10

1 Answer 1


First I want to address the main issue of "the cardinality of a proper class": they don't have one.

Proper classes are so large, that they don't really have a cardinality. For example, take the class $\text{Card}$. It obviously doesn't have a cardinality, as it is transitive and well-ordered by $\in$ and such a set cannot contain its cardinality (the existence of some $X$ so that $|X| \in X$, $X$ is transitive and $X$ is well-ordered by $\in$ violates the axiom of regularity).

Similar things trivially hold for $V$ and similar classes. For the rest of proper classes, we can use Limitation of Size to show that $|C| = |V|$ for any proper class, yet $|V|$ is undefined.

Next, regarding "given a proper class $C$, can we find some proper class $D$ larger than $C$?" Well, working in something such as $\textsf{NBG}$, the answer is a strong no due to the aforementioned limitation of size. In other systems which don't have limitation of size, I believe that the statement "given a proper class $C$, we can find some proper class $D$ larger than $C$" is independent.

  • $\begingroup$ >Proper classes are so large, that they don't really have a cardinality $$$$ Fun fact 1: In $ZF$, this statement (when state correctly) is equivalent to AC. $$$$ Fun fact 2: In $ZF$, even tho classes can be "not big", there exists a (class) function $f:V→Ord$ such that a class $C$ is proper if and only if $f''C$ is a proper class (An example of such $f$: the rank function) $$$$ Fun fact 3: It is consistent with $ZF$-regularity that there is no function like the above $f$ $\endgroup$
    – Holo
    May 6 at 20:12
  • $\begingroup$ Neat, I didn't realize this. By the way, you're in TAU, right? $\endgroup$
    – Binary198
    May 7 at 21:15

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