I was reading that the ordinal numbers do not form a set because there are too many of them, instead they form a proper class. Is there a maximum cardinality for a set?

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    $\begingroup$ It is more that such a collection is not compatible with our definition of set - it is not a set that is "too big" $\endgroup$
    – ocg
    Dec 5, 2015 at 20:19
  • $\begingroup$ Expanding on Julien Godawatta's comment, In theories that allow classes of classes (most don't), if $A$ is a proper class, then so is $\{A\}$, even though it has only one element. A proper class cannot be the element of a set. $\endgroup$ Dec 5, 2015 at 20:53
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    $\begingroup$ @PaulSinclair That's usually not true - $\{A\}$ is not a set, but it's not a class, either. Classes can't be elements of classes in most class theories. (E.g., in NBG, $x$ is a set iff $\exists y(x\in y)$.) $\endgroup$ Dec 6, 2015 at 0:39
  • $\begingroup$ There is a largest cardinal in NF and like theories, but generally no. $\endgroup$ Dec 6, 2015 at 0:58
  • $\begingroup$ Is it wrong to say "sets are constructed and may be filtered" (we can construct $\mathbb{R}$ via completing $\mathbb{Q}$ which is bijective with $\mathbb{N}$ and so forth) and filter out things (eg level sets, functions...) and classes are implicitly defined rather than constructed. Eg "the class of smooth functions", we can pull a smooth function out of nowhere to show the class exists, but we don't construct it. $\endgroup$
    – Alec Teal
    Dec 6, 2015 at 9:52

5 Answers 5


No, there is no maximum cardinality for a set.

If there were, then there would (by definition) be some set $A$ of this cardinality, and then $\mathcal P(A)$ would be a set with an even larger cardinality, by Cantor's theorem, so the cardinality of $A$ was not maximal at all, a contradiction.

This is not really more mysterious than, say, the fact that there is no maximum finite integer.

It is not uncommon to think of all proper classes as being "of the same size", which would then in some arguable sense be a least upper bound for the possible cardinalities of sets. However, it is tricky to make this intuitive claim precise.

  • $\begingroup$ It's actually not too hard to come up with a precise notion of "size" fir proper classes: we just talk about (parameter-)definable injections, surjections, and bijections. Plenty of things break down - e.g., Cantor-Bernstein for definable proper classes is independent of ZFC - but some things are provable, such as that any proper class surjects onto ON. $\endgroup$ Dec 5, 2015 at 22:21
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    $\begingroup$ @NoahSchweber: Well, the tricky thing is to find something that is both precise and true. For example, if every proper class bijects definably with ON, that would give us universal choice, which I believe is too strong. $\endgroup$ Dec 5, 2015 at 22:52
  • $\begingroup$ My emphasis was on the definition - indeed, the statement "every proper class definably bijects with ON" is equivalent to $V=HOD$. I just wanted to point out that even in ZFC, we can say certain things. $\endgroup$ Dec 5, 2015 at 22:53
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    $\begingroup$ @Noah: Okay then. We can all relax now. :-) $\endgroup$
    – Asaf Karagila
    Dec 6, 2015 at 6:22
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    $\begingroup$ @Noah: I would, but I'm really tired... :-P $\endgroup$
    – Asaf Karagila
    Dec 6, 2015 at 6:31

As the other answers have pointed out, there is no largest set.

However, there is a sense in which the answer to your question is "yes":

Suppose $C$ is any class. Then $C$ is a proper class (i.e., not a set) if and only if there is a (class) surjection from $C$ onto the ordinals.

In a sense, this says that the size of $\{$ordinals$\}$ is the smallest proper class "cardinality", analogously to how - while there is no "largest" finite set - $\aleph_0$ is the smallest infinite cardinality. (My use of "the" here is of course not quite correct, but oh well.) On the other hand:

It is consistent that there is no (class) surjection from the ordinals onto $V$.

So there can be different "sizes" of proper classes, and in particular the first statement above is non-trivial.


No, there is not. It is a basic result of Cantor that if $A$ is any set, $|\wp(A)|>|A|$.

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    $\begingroup$ (serious = "off") Hmm... What is the Weierstrass $\wp$-function of an arbitrary set?... (serious = "on") $\endgroup$ Dec 5, 2015 at 22:10

No, there is not. Not in the standard set theory, which is the ZFC Set Theory. The reason is the power set (the set of all subsets of a given set), as explained by the other answers here, we have $|\mathcal{P}(A)|>|A|$, for any set $A$.

But there are other models of set theory, with different axioms that allow something like that:


In this link, this set is called universe $U$ and surprising $|\mathcal{P}(U)|<|U|$. But remember, this DOES NOT occur in ZFC set theory.


A maximum cardinality, of course not. In ZF(C) you can always take the power set of a set to obtain a larger one.

I would actually say that the point of set theory is to describe ways you can make very big sets without making sets that are too big in some intuitive sense, because that leads to inconsistencies or paradoxes. For instance, "the set of all sets" is "too big." It leads to Russell's paradox when you use it to form "the set of all sets that don't contain themselves," which neither contains itself nor doesn't contain itself.

ZF(C) gives you both an infinite set by way of the natural numbers (defined using the successor function) and the ability to take power sets. These are powerful operations and via repeated application of the power set in some creative ways, enormously huge sets may be created, but so far as we know, ZF(C) is still consistent.

Other areas of math follow a similar path: it can be said that calculus was formalized because people kept making ridiculous derivations with power series, like $1+2+3+4+\ldots = -1/2$ and what-have-you. Now calculus provides explicit conditions for when power series will have sensical results. And computer science went through a similar (and somewhat different) path in describing what a computable function is.

  • $\begingroup$ Some mathematicians have proposed extending V and ON to "hyper-sets" and "hyper-ordinals" like ON, ON+1,etc.I don't know if any attempt at this, by modifying the axioms, have succeeded in anything consistent that does not produce the same theorems about sets . $\endgroup$ Dec 6, 2015 at 6:03

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