### Building models of ZFC with exactly 2 GBC-realizations

#### by kamerynwilliams

In a previous blog post I gave a construction of -like and rather classless models of . As an application of this, I showed that there are models of with exactly zero -realizations and models with exactly one -realization. (If is a model of and then is a *-realization for * if ). The original construction is due to Keisler. At the end of that blog post I asked whether this result could be generalized to : for which are there models of which have exactly many -realizations?

The answer is yes for . Before stating the theorem to this effect, observe that a model of has a definable (possibly with parameters) global well-order if and only if it satisfies . Because my interest here in this axiom is only that it gives a definable global well-order, I will call this axiom by the easier to write .

TheoremLet be a countable model of . Then has an elementary top extension to -like so that has exactly -realizations.

The construction of will follow roughly the same path as Keisler’s construction. However, we do not want to end with a rather classless model, as such a model will in this case have only one -realization. Instead, we will select a class and get so that is *-rather classless*, meaning that the amenable classes of are the classes definable from . If is carefully built, then will be as desired.

In the earlier blog post, I gave the construction for rather classless models where the base model did not have any additional predicates. However, it is not difficult to check that the same construction works if the model carries up to countably many strongly amenable classes , i.e. classes so that the model satisfies , meaning where the Separation and Collection schemata allow predicates for the classes. The important detail is that we need a global well-order for the construction of Skolem ultrapowers, but this is given by our base model satisfying .

This will be a certain perfect generic. In arithmetic, perfect generics are an adaptation of Sacks forcing to a forcing over models of arithmetic. A similar construction can be done for models of set theory, where the perfect generic is a certain function .

The following definitions take place over a fixed model of , possibly carrying up to countably many strongly amenable classes.

DefinitionLet denote the full binary tree on . For a perfect tree say that is-deciding forif for any formulae in the forcing language, there is an ordinal so that if has length at least , then decides in .

Note that if is definable then we can find -deciding subtrees of in a definable manner.

Definitionis aperfect genericif there is a sequence of definable perfect trees so that and is -deciding for .

An important point is that perfect generics really are generic. Because the sequence of trees decide more and more formulae in the forcing language, will decide all formulae in the forcing language. Also, note that if the height of our model of set theory has countable cofinality then we can construct perfect generics.

This lemma is a little more general than what is needed to get the theorem, but I don’t think that’s a bad thing.

Key LemmaLet be countable. Then there is a perfect generic so that

- Every is in ; and
- is minimal over , in the sense that if then either or .

*Proof:* Start with countable . Apply the key lemma to get . Now apply the rather classless construction to get where the top extension is -like and -rather classless. It is clear that has at least two -realizations, namely and . Let us see that these are the only two. Suppose that is a -realization. Necessarily, . By elementarity and the key lemma, either or else . In the first case, . In the second case, .

The key to proving the key lemma is to choose the correct sequence of perfect trees. Three constructions will be interleaved. The first construction is of an -deciding subtree. This will get that the sequence is of the right form, due to the easy fact that if and is -deciding for then is -deciding for . The second construction will ensure the minimality of the extension. The third construction will ensure that every is definable from the perfect generic. The second and third constructions are encapsulated by the following lemmata. I present them in reverse order, because the principality sublemma is easier than the minimality sublemma.

Principality sublemmaLet be countable and be a perfect tree. There is a perfect tree in so that if is any branch through then .

*Proof:* By , we may assume without loss that . We build by pruning . We will prune at every other splitting node, cutting off one branch depending on information about . Say that a splitting node in is an -node if the node has many splitting nodes below it. If is even, then keep both branches from . If is odd, then remove one of the branches. If , then remove the branch. If , then remove the branch. Do this for all splitting nodes to get .

Suppose that is a branch through . Then, we can recover from and . From we can find the odd splitting nodes. From we know whether the branch or branch was removed. That is, if and only if there is a -node in so that is an initial segment of .

Minimality sublemmaLet be a formula in the forcing language and be a definable perfect tree. Then there is definable so that either

- All sufficiently long sequences decide the same for all ordinals ; or
- For any ordinal there is so that if are distinct sequences of length which agree up to then there is some ordinal so that and decide the same.

It is probably not clear at the moment how this sublemma will ensure that is minimal over . To give a sneak preview of the proof of the main lemma, the first condition will correspond to the set defined by being definable over from and the second condition will correspond to the set defined by being able to define .

*Proof:* Pick large enough so that is . We start with which is definable and -deciding. We will further prune this tree to produce . We can pick so that there is an embedding which is onto the splitting nodes and decides . We find ourselves in one of two cases.

The first is if there is some sequence so that if are extensions of of the same length then and decide the same for all ordinals . In this case, set to be , the subtree of formed from nodes which extend or are extended by . This is easily seen to yield the former condition in the consequent of the statement of the sublemma.

If this does not happen, then we will inductively define an embedding .

- Set ;
- If is already defined then pick and to be nodes of the same length so that and decide differently for some . Note that this picking relies upon having a uniform way to pick elements from for arbitrarily long , and thus appeals to Global Choice.
- For limit stages simply take unions.

Now define to be the closure in of . Ponder for a moment and see that this yields the second condition in the statement of the sublemma.

With the sublemmata out of the way we are now ready to move to the proof of the key lemma.

*Proof:* Fix an enumeration of formulae in the forcing language so that the predicate for does not appear in for . As we will continually apply the principality sublemma to code the , this will ensure that in the end they are definable just from (and possibly set parameters). Further fix a sequence of ordinals cofinal in . We will inductively build a descending sequence of definable perfect trees

Pick to be -deciding and to not split below . This step of the construction will ensure that we get a perfect generic. Pick according to the minimality sublemma applied to . Finally, pick according to the principality sublemma applied to . Consider . I claim that is as desired.

As already mentioned, is a perfect generic. An easy inductive argument, using the principality sublemma, gives that every is in . Also, is a -realization because has a definable global well-order. It remains only to check that is minimal over . Take defined by . Without loss of generality assume that is a class of ordinals. Then in stage of the construction we ensured the minimality sublemma held for . Thus we are in one of the two situations. If the first condition holds then, for some ,

Manifestly, . Actually, we get the slightly stronger fact that because the definition of and only use up to .

If we are in the second situation, then we can define from and . Namely, is an initial segment of if and only if for every ordinal there is a condition extending of length so that for every , we have that implies and implies . Again, we get a slightly stronger fact, namely that .

Given this argument, the natural thing to try is to generalize it to . The strategy to try is to start with countable . Then apply the key lemma to get . Repeatedly apply the key lemma to to get . One would like to get that applying Keisler’s construction to would yield a model with exactly many -realizations.

What goes wrong?

The construction works and can be applied multiple times, so that is not the issue. Where we run into difficulties is getting that there are exactly many -realizations for which are contained in . Let’s illustrate the difficulty by looking at the case. We started with , found minimal over and then found minimal over . This gives us at least 3 -realizations, namely , , and . But could there be more? Suppose that and that is a -realization for . By the key lemma, we are in one of two cases. One case is that . This case is not problematic; just apply the key lemma again to get that either , whence , or else , whence . The other case is that . This is the problematic case. We know that we can define given both and , but how can we know that can be defined just from ? It is conceivable that there could be many new -realizations contained in which are -incomparable with . As such, this argument alone cannot give a model of with exactly 3 -realizations. Damn.

A similar argument can be done in the setting of arithmetic to produce models of arithmetic with precisely 2 -realizations.

** Some References **

Details about perfect generics in arithmetic can be found in Kossak and Schmerl’s book. They haven’t been used much in set theory, but Hamkins, Linetsky, and Reitz make use of them in their paper.

- J. D. Hamkins, D. Linetsky, and J. Reitz, “Pointwise definable models of set theory,”
*J. Symbolic Logic*, vol. 78, iss. 1, pp. 139-156, 2013. [link] - H. Jerome Keisler, “Models with tree structures,”
*Proceedings of the Tarski Symposium*(Proc. Sympos. Pure Math., Vol. XXV, Univ. California, Berkeley, Calif., 1971) Amer. Math. Soc., Providence, R.I., 1974, pp. 331–348. - R. Kossak and J. Schmerl,
*The Structure of Models of Peano Arithmetic*(Oxford Logic Guides, Vol. 50, Oxford, 2006).