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In ZFC, a cardinal is an isomorphism class of sets. However ZFC doesn't explicitly have classes; NBG, which is a conservative extension of ZFC does.

There is no largest cardinal by Cantors Theorem

There is no set of all sets - it is in fact a class.

Classes do not have cardinalities, as these have been only defined for sets - but if one could define a cardinality for classes - wouldn't this, in some sense be a 'limit' of all cardinals in Set, including all large cardinal axioms?

Thus, is it possible to extend the notion of cardinality in any significant way to classes?

Apologies for the loose phrasing of this question. It was originally going to be a posting on Philosophy.SE, but I thought I would get better answers here.

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"Wouldn't this, in some sense be a 'limit' of all cardinals in Set, including all large cardinal axioms?" No. Unless you mean "limit" in a trivial way, it would be something else. Maybe it would be useful from a mathematical point of view, who knows? Work on it for a while, prove some theorems, show us that it is useful, or convince yourself that it isn't. It would certainly not be a "largest large cardinal". –  Andres Caicedo Jun 6 at 5:17
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This is a very mathematical question in my opinion... I haven't done much set theory, but the question seems legit. +1 –  Patrick Da Silva Jun 6 at 5:17
    
@Caicedo: Well, there isn't much point in replicating work thats already done; and theres no harm in asking if such work has been done, suggested or thought about. 'It would be something else' - sure thats why I said in some sense. –  Mozibur Ullah Jun 6 at 5:25
    
You need to define things first before we can make sense of them. As it stands, this is unanswerable. –  Andres Caicedo Jun 6 at 5:28
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You almost answer your own Question, to the extent it is well posed. Since cardinality is an equivalence relation on sets, it cannot be extended to a similar notion on classes in ZFC or NBG. In other set theoretic frameworks the cardinality of the universal set is a valid construction, and does give a largest cardinal, for example Quine's NF (or the second edition of Quine's Mathematical Logic for a version with proper classes, an inessential extension). –  hardmath Jun 6 at 6:43

3 Answers 3

Yes, you can extend the notion of cardinality to classes, and it is consistent that there is only one cardinal for classes (e.g. $V=L$ implies that all classes have the same size), or it is consistent that there are classes which are incomparable (e.g. if you add two Cohen subsets to a proper class of cardinals, the class of the pairs cannot be definably well-ordered and it is incomparable in size with the class of ordinals). If you agree to violate the axiom of choices then there are even more options here (classes incomparable with sets).

But I don't think there is a lot written on this topic, and it's scattered throughout many papers. Some obvious consequences here, some minor mentioning there.

As for being a large cardinal, there are two type of properties to consider here. Small properties, which only affect the sets below the cardinal (e.g. inaccessible cardinals) and large properties, which affect sets which are not smaller than the cardinals (e.g. measurable cardinals). There are also very large properties which affect pretty much everyone in the universe (e.g. supercompact cardinals).

Since $\sf ZFC$ has a very limited access to proper classes, having the class ordinals as a large cardinal of a large property is meaningless. But you can easily have small properties by taking $V_\kappa$ for some $\kappa$ satisfying the wanted property and considering it as a model on its own. If $\kappa$ is a Mahlo cardinal, then in $V_\kappa$ every closed and unbounded set of ordinals includes an inaccessible cardinal.

Note that requiring that every class of ordinals which is closed [and unbounded] has an inaccessible cardinal actually requires less than a Mahlo ordinals and cutting off the universe, because we have less classes to worry about.

Another peculiar example is a Woodin cardinal. Being a Woodin cardinal is having a small property, which itself is a very large property (if $\kappa$ is a Woodin cardinal, it might not even be weakly compact, but $V_\kappa$ is incredibly rich in internally large cardinals). So you can think of the similar case where the ordinals behave like a Woodin cardinal to some extent. Although, I'm not sure why you would do that.

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If one could define a cardinality for classes - wouldn't this, in some sense be a 'limit' of all cardinals in Set...

Sure, if you modify the definitions to allow this then the cardinalilty of any proper class would be considered the largest cardinal. (Cantor's proof would then only apply to cardinals that are sets.) However, this is not very interesting.

... including all large cardinal axioms?

No, large cardinal axioms are quite different objects from cardinals, and the relationship between largeness of cardinals and "largeness" (strength) of large cardinal axioms is not as straightforward as you might think. (And anyway, there is no largest large cardinal axiom because there is no maximal consistent recursive extension of $\mathsf{ZFC}$.)

Many examples of large cardinal axioms take properties of the class of cardinals, e.g. infinitude, regularity (closure under limits of short sequences,) and closure under power sets, and posit that some (set-sized) cardinal already has these property. Large cardinal axioms therefore provide a sense in which the set/class distinction is made relative, rather than absolute.

This is why I said that the possibility of considering the size of a proper class as a "largest cardinal" is not interesting; if your model of $\mathsf{ZFC}$ is realized as $V_\kappa$ where $\kappa$ is an inaccessible cardinal in my model of $\mathsf{ZFC}$, then a proper class in your model is a set in my model. Moreover, I can go even farther and define $V_{\kappa+1}$, $V_{\kappa+2}$, etc. So the large cardinal axiom "there is an inaccessible cardinal" transcends the notion of cardinals in $\mathsf{ZFC}$ to a much greater extent than the "one-step" extension of the notion of cardinals that you get by allowing proper classes.

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There's also no largest large cardinal axiom because I can always "one up" any given axiom. If it's an axiom asserting a proper class of something, then I can say that there is an inaccessible cardinal that in $V_\kappa$ there is a proper class of that something; and if not then I can always say that $\kappa$ is "one up for $\varphi$" if $\kappa$ is an inaccessible cardinal larger than a cardinal satisfying $\varphi$. :-) –  Asaf Karagila Jun 6 at 16:50
    
@Asaf Sure, although I'm not sure how to rule out the possibility that someday we may find a large cardinal axiom for which this natural strengthening results in inconsistency. (I admit that this seems very unlikely.) –  Trevor Wilson Jun 6 at 19:52
    
Well, strictly speaking, if you assume choice, Reinhardt cardinals are an example! :-) –  Asaf Karagila Jun 6 at 19:57

Being a large cardinal isn't a question of mere size, is a question of complexity.

Consider $\lambda<\kappa$ with $\lambda$ measurable and $\kappa$ simply strongly inaccesible. Then, $V_\kappa$ is a model of $\mathsf{ZFC}$, $\lambda$ is measurable in this model and as $|V_\kappa|=\kappa$, $\kappa$ is the largest "cardinal" in your sense, but from the outside $\kappa$ isn't very big (isn't measurable).

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