## “$\kappa$ strongly inaccessible” = “every function $f:V_\kappa\to V_\kappa$ can be self-applied”?

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Strongly inaccessible cardinals are usually introduced either as (a) cardinalities of models of ZFC or (b) cardinals which are not the power set of a smaller cardinal nor the supremum of a set with hereditarily lesser cardinality. These seem to represent the model-theoretic and set-theoretic perspectives on strong inaccessibility.

Recently I learned that if $\kappa$ is a strongly inaccessible cardinal, then $(V_\kappa)^2\subseteq V_\kappa$, so any function $f:V_\kappa\to V_\kappa$ is a set of pairs whose coordinates are members of $V_\kappa$, and so any such function can be "applied" to any other such function:

`$$f(g) = \{\ z\ |\ \langle \langle x,y\rangle, z \rangle\in f\ \&\ \langle x,y\rangle\in g\  \}$$`

Therefore one can say that if $\kappa$ is strongly inaccessible, then $V_\kappa$ is closed under self-application (as defined above) of functions $f:V_\kappa\to V_\kappa$.

This seems to be sort of a "recursion-theoretic" characterization of strong inaccessibility: it identifies a definable operation under which all strongly inaccessible cardinals are closed.

Question 0: does this make sense?

Question 1: is the converse true, making this a complete characterization? (if $V_\kappa$ closed under self-application of functions then $\kappa$ is strongly inaccessible)?

Question 2: if so, I'm sure this has come up before in the literature. In what sorts of directions does this investigation lead?

This is one of the more-vague questions I've asked so far. I guess I'm sort of fishing for enlightenment here; it took me a long time to understand the point of inaccessibility, and I suspect that I might have caught on more quickly if this motivation (closure under self-application) had been introduced early on.

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## 1 Answer

Unfortunately, your characterizations of the strongly inaccessible cardinals are not quite correct. The correct definition is that κ is strongly inaccessible (also known as just plain inaccessible), if κ is an uncountable regular strong limit cardinal. The cardinal κ is regular if it is not the union of fewer than κ many sets of size less than κ. And κ is a strong limit cardinal if whenver β < κ, then the power set of β also has size less than κ.

This is not equivalent to the assertion that Vκ is a model of ZFC. (Although, to be sure, this false assertion has appeared surprisingly often in print and I have even heard a famous proof theorist make this assertion to a very large audience of hundreds of logicians.) The reason is that if κ is strongly inaccessible, then a Lowenheim-Skolem argument shows that there will be many γ < κ for which Vγ is elementary in Vκ, and so these also will be models of ZFC. It is an exercise to show that the least γ for which Vγ is a model of ZFC has cofinality ω, and so is definitely not inaccessible.

Also, since ZFC is a first order theory in a countable language, if it has any models at all, then it has models in every infinite cardinality. So it is not correct to characterize inaccessible cardinals as the sizes of models of ZFC in that way either.

It is also not equivalent to asserting that κ is regular and not the size of a power set of a smaller set. The reason is that if, say, CH failed, then ω1 would be regular and also not be the size of the power set of any smaller set (since 2ω would be already too large). But ω1 is not an inaccessible cardinal.

Your remark that Vκ is closed under pairs when κ is inaccessible actually doesn't need any amount of inaccessibility. If x and y are sets in any Vα, then the pair (x,y) appears just a few steps later (and actually, one can use flat pairing function that do not increase rank at all, for infinite rank sets), and so every Vλ is closed under pairing for any limit ordinal λ. If one uses a flat pairing function (instead of the common Kuratowski pairing function), then every Vα for every infinite ordinal α will be closed under pairing.

Finally, yes, if Vλ is closed under pairing, then you can apply such functions to themselves, and this idea is used quite often when we have elementary embeddings defined on models of ZFC. For example, if j:V to M, then j(j) is a function defined on M, into some structure j(M), which will be the union of j(VαM). This operation is called application.

There is a famous result of Laver concerning the left distributive algebra of nontrivial elementary embeddings j:Vλ to Vλ. The first results characterizing normal forms in the free algebra with one generator used such embeddings, with the accompanying very large large cardinal hypothesis. For example, Laver produced a decision procedure, which was only known to work under these enormous large cardinal assumptions. Later, the large cardinal hypotheses were removed and the algebra became studied apart from the large cardinals, but the basic properties were definitely inspired and discovered by knowledge of what the large caridnals were like. The basic operation in this algebra is known as application, and is exactly the operation that you mention.

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1
Hear hear. It is much too infrequently mentioned that there are many non-inaccessible a for which Va is a model of ZFC. I think it is true, though, that k must be inaccessible as soon as Vk is a model for "second-order" ZFC, right? – Mike Shulman Feb 16 2010 at 3:11

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@Mike: There might be no such V_a's, though you are correct that there are plenty if there is also an inaccessible. What you say about ZFC2 is true, this is an old theorem of Shepherdson (from one of his trilogy of papers titled Inner models for set theory). – François G. Dorais♦ Feb 16 2010 at 3:18

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Well, if by second order you just mean GBC=Goedel-Bernays set theory (with choice), then that is not quite right either, since every ZFC model can be given a second order part satisfying GBC. If you mean KM = Kelly Morse, then you have to go a bit further, but you still don't need an inaccessible. But if by "second-order" you mean that you have V_kappa with the full V_{kappa+1} as the second order part, then this does indeed imply that kappa is inaccessible. – Joel David Hamkins Feb 16 2010 at 3:18
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