In the integers:1,2,3... we can say that there isn't a largest number; but it turns out with the advent of set theory we can posit a largest number omega that doesn't lie within the integers and yet bounds them. In mathematical terms we can say that it is a limit to them.

Of course, with the advent of Cantorian set theory, it was also realised that one can define and investigate larger infinite numbers, which are ordinals and cardinals (cardinals are certain distinguished points in the ordinals and are very sparse). Then one can ask the same question at his new level: Is there a largest Cardinal?

Now, it turns out that Reinhardt cardinals are the largest cardinals yet defined in ZFC; To refine the question - are they provably the largest possible? Is there in fact provably, or conjecturally a largest cardinal wrt ZFC? (My intuition would say that there isn't).

(Choice is generally useful when dealing with properly infinitary objects, which is why I've asked with reference to ZFC, rather than ZF);

NBG, and NFU are alternative set theories that make formal use of a concept that ZFC informally uses, that is classes; as classes are not sets then formally the notion of cardinality or ordinality is applicable to them - but one could argue with judicious changes in the axioms that one could extend the notions to here; it would then seem a possibility of defining a class-cardinal that is the limit of all set-cardinals.

Is this this actually possible?

But then, one could argue that the same process could be iterated; in a much more profound sense the notion of mathematical infinity is always in potentia, though there are points at which we can say a certain actuality has been achieved (the limit points).

Is this a good argument?

  • -1 Sorry, but this is really a mathematical question and not related to phil of math.
    – DBK
    Jun 7, 2014 at 19:24
  • @DBK: Admittedly it is phrased mathematically; but questions on infinity usually are seen as philosophical - are you saying that a question of that type isn't part of philosophy of mathematics as usually/professionally understood? Jun 7, 2014 at 19:50
  • 1
    As you can see my observation referred to the first edit, which asked "Is there in fact provably, or conjecturally a largest cardinal wrt ZF?" The latter is simply a question about a mathematical fact/hypothesis. While we welcome questions about philosophical implications of mathematical facts (if there are any), questions about mathematical facts alone are off-topic and best asked on Math.SE (quite independently from the circumstance that there might be users who can answers math questions here as well).
    – DBK
    Jun 8, 2014 at 1:22
  • @DBK: I altered the question given your critique as perhaps what i was asking wasn't easily visible. Jun 8, 2014 at 1:29
  • Great, thanks! BTW, the cardinal hierarchy, i.e. large cardinals ordered by their consistency strength, clearly has an upper limit: 0=1 ;) (h/t to Kanamori)
    – DBK
    Jun 8, 2014 at 1:32

2 Answers 2


Note that Reinhardt cardinals do not provably exist in ZFC. If they were, it would contradict Gödel's theorem. It is the largest cardinal currently defined which is believed to be consistent with ZFC.

As Mozibur notes, you can't have a largest such cardinal, since given a consistent extension of ZFC, you can always (in theory) find a stronger theory which proves the existence of larger cardinals. However, it may be found next week that Reinhard cardinals are actually not consistent with ZFC. That's the tragedy of the incompleteness theorem.

It turns out, however, that if ZFC is consistent, there is a smallest cardinal which is not provably such in ZFC. To show this, you can simply consider the set of all uniquely defined syntactic objects which ZFC proves to be cardinals, and take the smallest cardinal not in that set.

Edit: I missed the fact that Reinhardt cardinals are inconsistent with Choice. You can replace ZFC by ZF everywhere in my comment though, or Reinhardt cardinals by some smaller cardinal numbers (superhuge for instance).

  • It’s open if it’s consistent with ZF but Kunen showed that Reinhardt cardinals contradict ZFC soon after they were proposed. Edit: I didn’t read the answer right below this which said the exact same thing
    – Lave Cave
    Aug 3, 2023 at 5:32

There is no largest large cardinal axiom as there is no maximal consistent and recursive extension of ZFC.

Furthermore though there is a definition of a Reinhardt Cardinal, However that they don't exist:

Shortly after this idea was introduced, however, Kunen famously proved that there are no such embeddings, and hence no Reinhardt cardinals in ZFC.

(In fact because of technical problems it was formulated in a class theory NBG and not ZFC, though there are work-arounds that allow it to work there).

This is formalised as the Kunen incocnsistency theorem, and shows that Reinhardt cardinals are inconsistent with Choice. In ZF or NBG it remains an open question.

Hamkins et al show that the Kunen inconsistency is a natural limit in other ways, as many natural definitions of larger cardinals than the Reinhardt one also do not exist.

  • Just to add to this, for any large cardinal axiom $\phi$, you could always lump an extra inaccessible on top (so $\phi$ + there is an inaccessible above $\phi$) to obtain a strictly stronger axiom (in fact, by Godel $\phi + Con(\phi)$ will do). Aug 30, 2016 at 19:42

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