I've run across the statement, "The existence of a strongly inaccessible cardinal implies the consistency of ZFC" in several places (Cohen's Set Theory and the Continuum Hypothesis p. 80, for one). His argument is that the set of all sets of rank less than that large cardinal form a model. It seems to me that since we have access to all of those sets without the large cardinal, we could make a model of ZFC without the large cardinal axiom, and thus prove ZFC consistent. This, of course, is false. So, how does the large cardinal provide a model of ZFC that doesn't exist without the large cardinal?
[Math] Large Cardinals Imply a Model of ZFC
large-cardinalsmodel-theoryset-theory
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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.
This may not be the sort of thing you had in mind, but here goes anyway: The easiest way to prove Borel determinacy (which is a theorem of ZFC) is to assume there's a measurable cardinal and prove analytic determinacy. (Both results are due to Tony Martin. The proof of analytic determinacy from a measurable cardinal came well before the proof of Borel determinacy in ZFC. The exact consistency strength of analytic determinacy is the existence of sharps of all reals.)
Best Answer
A model of ZFC is a set $E$ together with a binary relation $R$ satisfying the axioms of ZFC (and
we can alwaysif we can suppose that $E$ is transitive and that $R=\in|_{E\times E}$ then the model is said standard). So we not only need to find a class of sets satisfying the axioms of ZFC, but this class must be itself a set.If you have a strongly inaccessible cardinal $\kappa$, then there is a set $V_\kappa$ which is a (standard) model of ZFC, hence ZFC is consistent. But without the existence of $\kappa$, even if you could consider the class of all sets of rank “less than $\kappa$”, this would not be a set and the theorem ”there is a model implies the consistency of ZFC” would not be applicable (because there is no model).