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A cardinal $\lambda$ is weakly inaccessible, iff a. it is regular (i.e. a set of cardinality $\lambda$ can't be represented as a union of sets of cardinality $<\lambda$ indexed by a set of cardinality $<\lambda$) and b. for all cardinals $\mu<\lambda$ we have $\mu^+<\lambda$ where $\mu^+$ is the successor of $\mu$. Strongly inaccessible cardinals are defined in the same way, with $\mu^+$ replaced by $2^\mu$. Usually one also adds the condition that $\lambda$ should be uncountable.

As far as I understand, a "large cardinal" is a weakly inaccessible cardinal with some extra properties. In set theory one considers various "large cardinal axioms", which assert the existence of large cardinals of various kinds. Notice that these axioms are quite different from, say the Continuum Hypothesis. In particular, one can't deduce the consistency of ZFC + there exists at least one (uncountable) weakly inaccessible cardinal from the consistency of ZFC, see e.g. Kanamori, The Higher Infinite, p.19. I.e., assuming ZFC is consistent, these axioms can not be shown independent of ZFC.

The "reasonable" large cardinal axioms seem to be ordered according to their consistency strength, as explained e.g. here This is not a theorem, just an observation. A list of axioms according to their consistency strength can be found e.g. on p. 472 of Kanamori's book mentioned above. (Noticeably, it starts with "0=1", which is a very strong axiom indeed.)

Large cardinals appear to occur seldom in "everyday" mathematics. One such instance when they occur is when one tries to construct the foundations of category theory. One of the ways to do that (and the one that seems (to me) to be the most attractive) is to start with the set theory and to add Grothendieck's Universe axiom, which states that every set is an element of a Grothendieck universe.

(As an aside remark, let me mention another application of large cardinal axioms: incredibly, the fastest known solution of the word problem in braid groups originated from research on large cardinal axioms; the proof is independent of the existence of large cardinals, although the first version of the proof did use them. See Dehornoy, From large cardinals to braids via distributive algebra, Journal of knot theory and ramifications, 4, 1, 33-79.)

Translated into the language of cardinals, the Universe axiom says that for any cardinal there is a strictly larger strongly inaccessible cardinal. I have heard several times that this is pretty low on the above consistency strength list, but was never able to understand exactly how low. So I would like to ask: does the existence of a (single) large cardinal of some kind imply (or is equivalent to) the Universe axiom?

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up vote 27 down vote accepted

A Grothendieck universe is known in set theory as the set Vκ for a (strongly) inaccessible cardinal κ. They are exactly the same thing. Thus, the existence of a Grothendieck universe is exactly equivalent to the existence of one inaccessible cardinal. These cardinals and the corresponding universes have been studied in set theory for over a century.

The Grothendieck Universe axiom (AU) is the assertion that every set is an element of a universe in this sense. Thus, it is equivalent to the assertion that the inaccessible cardinals are unbounded in the cardinals. In other words, that there is a proper class of inaccessible cardinals. This is the axiom you sought, which is exactly equivalent to AU. In this sense, the axiom AU is a statement in set theory, having nothing necessarily to do with category theory.

The large cardinal axioms are fruitfully measured in strength not just by direct implication, but also by their consistency strength. One large cardinal property LC1 is stronger than another LC2 in consistency strength if the consistency of ZFC with an LC1 large cardinal implies the consistency of ZFC with an LC2 large cardinal.

Measured in this way, the AU axiom has a stronger consistency strength than the existence of any finite or accessible number of inaccessible cardinals, and so one might think it rather strong. But actually, it is much weaker than the existence of a single Mahlo cardinal, the traditional next-step-up in the large cardinal hierarchy. The reason is that if κ is Mahlo, then κ is a limit of inaccessible cardinals, and so Vκ will satisfy ZFC plus the AU axiom. The difference between AU and Mahloness has to do with the thickness of the class of inaccessible cardinals. For example, strictly stronger than AU and weaker than a Mahlo cardinal is the assertion that the inaccessible cardinals form a stationary proper class, an assertion known as the Levy Scheme (which is provably equivconsistent with some other interesting axioms of set theory, such as the boldface Maximality Principle, which I have studied a lot). Even Mahlo cardinals are regarded as rather low in the large cardinal hierarchy, far below the weakly compact cardinals, Ramsey cardinals, measurable cardinals, strong cardinals and supercompact cardinals. In particular, if δ is any of these large cardinals, then δ is a limit of Mahlo cardinals, and certainly a limit of strongly inaccessible cardinals. So in particular, Vδ will be a model of the AU axiom.

Rather few of the large cardinal axioms imnply AU directly, since most of them remain true if one were to cut off the universe at a given inaccessible cardinal, a process that kills AU. Nevertheless, implicit beteween levels of the large caridnal hiearchy are the axioms of the same form as AU, which assert an unbounded class of the given cardinal. For example, one might want to have unboundedly many Mahlo cardinals, or unboundedly many measurable cardinals, and so on. And the consistency strength of these axioms is still below the consistency strength of a single supercompact cardinal. The hierarchy is extremely fine and intensely studied. For example, the assertion that there are unboundedly many strong cardinals is equiconsistent with the impossiblity to affect projective truth by forcing. The existence of a proper class of Woodin cardinals is particularly robust set-theoretically, and all of these axioms are far stronger than AU.

There are natural weakenings of AU that still allow for almost all if not all of what category theorists do with these universes. Namely, with the universes, it would seem to suffice for almost all category-theoretic purposes, if a given universe U were merely a model of ZFC, rather than Vκ for an inaccessible cardinal κ. The difference is that U is merely a model of the Power set axiom, rather than actually being closed under the true power sets (and similarly using Replacement in place of regularity). The weakening of AU I have in mind is the axiom that asserts that every set is an element of a transitive model of ZFC. This assertion is strictly weaker in consistency strength thatn even a single inaccessible cardinal. One can get much lower, if one weakens the concept of universe to just a fragment of ZFC. Then one could arrive at a version of AU that was actually provable in ZFC, but which could be used for most all of the applications in cateogory theory to my knowledge. In this sense, ZFC itself is a kind of large cardinal axiom relative to the weaker fragments of ZFC.

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Why AU? (French ordering?) – François G. Dorais Jan 24 '10 at 3:02
I thought it was usually called the Axiom of Universes. Is this wrong? – Joel David Hamkins Jan 24 '10 at 3:08
Joel, thanks! You say that "few of the large cardinal axioms imply AU directly". But are there any that do? If so, which one is the weakest one? – algori Jan 24 '10 at 3:13
I don't know. Wiki seems to agree with you. They also state all axioms of ZF as "axiom of blah" instead of the "blah axiom." – François G. Dorais Jan 24 '10 at 3:20
@Mike: Yes, I agree. (Although in set theory we work inside all different kinds of various universes, and have to pay attention to what is absolute and what is not, so I think it would be possible to do it and get used to it.) But for convenience, carry on with AU, in my opinion. I look forward to the future category theoretic arguments that (need to employ) more of the large cardinal hierarchy. For exmple, I could imagine constructions that ask for a stationary tower of universes, which would be something like a Mahlo cardinal. – Joel David Hamkins Jan 24 '10 at 4:22

It is very near the bottom of Kanamori's chart. The very bottom of the chart is the level of a (strongly) inaccessible cardinals, which is the smallest large cardinal axiom. Right above the inaccessibles in the chart are the α-inaccessible cardinals. It turns out that the Universe Axiom (UA) is strictly weaker than the existence of a 2-inaccessible cardinal. In fact, κ is 2-inaccessible if and only if κ is regular and Vκ ⊧ ZFC + UA.

Specifically, a cardinal κ is:

  • 0-inaccessible iff κ is regular,
  • 1-inaccessible iff κ is a regular strong limit of 0-inaccessibles,
  • 2-inaccessible iff κ is a regular strong limit of 1-inaccessibles,
  • etc.

So an inaccessible cardinal is exactly the same as a 1-inaccessible cardinal, which are also precisely the regular cardinals κ such that Vκ ⊧ ZFC. If κ is 2-inaccessible, then there are unboundedly many inaccessibles λ < κ. These are inaccessible in Vκ too, which is why Vκ satisfies UA.

Note that the existence of a 2-inaccessible cardinal κ does not directly imply the Universe Axiom. Indeed, κ may well be the last inaccessible cardinal, which means that there may be no universe that contains κ itself. However, if κ is 2-inaccessible then the universe Vκ does satisfy UA, which means that the existence of a 2-inaccessible proves the consistency of ZFC + UA.

Although UA is indeed a large cardinal axiom, there is no way to formulate UA as the existence of a single large cardinal. However, morally speaking, you can think of UA as saying "the class of all ordinals (viewed as a cardinal number) is 2-inaccessible." Of course, this doesn't make sense since the class of all ordinals is not a set, but this is exactly what κ looks like when viewed from inside Vκ.

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Thanks, F.G.! Sorry, I'm not sure I get it. Suppose $\kappa$ is 2-inaccessible and $\lambda$ is arbitrary. How does one construct a 1-inaccessible cardinal greater than $\lambda$? – algori Jan 24 '10 at 2:18
5… just for people who wanted to see the chart. – Harry Gindi Jan 24 '10 at 2:27
Francois, we're stepping on each other's toes this evening, aren't we? Sorry! – Joel David Hamkins Jan 24 '10 at 2:27
@Joel: I think it's pretty funny, I'm having fun! People must be thinking we're racing or something. (You write better than I do, so I always learn something by comparing our answers.) – François G. Dorais Jan 24 '10 at 2:37
@Harry: Every infinite kappa is like that, since kappa^2=kappa in infinite cardinal arithmetic. – Joel David Hamkins Jan 24 '10 at 3:11

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