### Beta-models of GBC can satisfy very little transfinite recursion

#### by kamerynwilliams

The reason that the fact mentioned in the title is interesting is because this gives us another place where the analogy between arithmetic and set theory fails. (Or, perhaps more accurately, this is another facet of an already-known disanalogy.)

Say that a two-sorted structure , either a model of second-order arithmetic or of a second order set theory, is a *beta-model* if it is correct about well-foundedness. Naturally, this implies that must be transitive, in the case of set theory, or, in the case of arithmetic, . For models of arithmetic, being a beta-model implies quite a bit more. Let’s say that is a model of arithmetical comprehension, for concreteness. (Weaker requirements are enough, but let’s not worry about how low we can go.) Then, if is a beta-model, it automatically satisfies transfinite-recursion for first-order properties. That is, if is a well-founded relation then recursion for first-order properties can be iterated along it to get a solution in . If you’re familiar with the big five theories, what I’ve just said is that every beta-model of is a model of the stronger theory .

This is actually pretty easy to see, taking as given the following fact: for models of arithmetical comprehension, the formula “ is a well-order” is -universal. Thus, being a beta-model is equivalent to having as your first-order part plus being correct about assertions. Note that a recursion for a first-order property having a solution is itself a assertion. Since a solution exists in the ambient universe, because really is well-founded, we thus get that a beta-model must also see the solution.

This argument does not generalize to models of set theory. In this context, “ is a well-order” does not have a hope of being -universal. Indeed, the obstacle here is exactly what will give us the fact from the title. The following observation encapsulates what’s happening here.

## Observation.

If is a beta-model of a second-order set theory and is so that is also a model of set theory, then is a beta-model.

*Proof:* Suppose that so that thinks is ill-founded. Then, there is a *set* which is an infinite descending sequence in . But then thinks is ill-founded, so by betaness really is ill-founded. Thus is correct about well-foundedness.

As an immediate consequence of this observation we get lots and lots of beta-models of set theory which don’t satisfy transfinite recursion for first-order properties. Indeed, we can get beta-models whose second-order part is as small as possible. For example, take a beta-model of KM (or some other theory which proves transfinite recursion for first-order properties). More, assume that has a definable global well-order. Then, we get that is a model of GBC. By the observation, it is a beta-model. Yet it will fail to have solutions to many transfinite recursions. This even happens for transfinite recursions whose rank is as low as , such as the transfinite recursion to define a truth predicate in the Tarskian manner!

Thanks for the interesting post, Kameryn! This is a great observation. Maybe one should make a post of all the disanalogies that have been discovered so far.

Thanks, Vika!

As I intimated in my parenthetical at the beginning, the underlying disanalogy here is that for arithmetic, being ill-founded is a second-order property—the infinite descending sequence is a proper class. But for set theory, being ill-founded is a first-order property. The things I mentioned in this post—whether being a well-order is -universal, whether submodels of a beta-model are beta, whether beta-models see solutions to transfinite recursions—all differ between arithmetic and set theory because of this one fact.

In fact, I think this fact can be used to answer a question you had a few months ago. ETR implies that (class) well-orders are comparable, but what about the converse? In arithmetic is equivalent to the comparability of well-orders (over ). I think the answer is no for set theory: it’s possible for a model of GBC to think well-orders are comparable without satisfying ETR.

To sketch the possible argument: start with Hachtman’s as from my talk on Friday, i.e. thinks there is a largest cardinal and that is inacessible and sees a ranking function for all trees on . Then we should get that is bigger than , the first admissible ordinal after (because something like Kleene’s O should be in but not have a ranking function in ).

Now consider . This is a beta-model of GBC (because it is a submodel of the beta-model ) but isn’t a model of ETR. Now take two well-orders . They really are well-orders, by betaness. The function witnessing their comparison should be in and hence in , as it is definable by a simple recursion, simple enough that KP should be able to see that it works.