Building models of ZFC with exactly 2 GBC-realizations

by kamerynwilliams

In a previous blog post I gave a construction of {\omega_1}-like and rather classless models of {\mathsf{ZFC}}. As an application of this, I showed that there are models of {\mathsf{ZFC}} with exactly zero {\mathsf{GBC}}-realizations and models with exactly one {\mathsf{GBC}}-realization. (If {M} is a model of {\mathsf{ZFC}} and {\mathfrak{X} \subseteq \mathcal P(M)} then {\mathfrak X} is a {\mathsf{GBC}}-realization for {M} if {(M,\mathfrak X) \models \mathsf{GBC}}). The original construction is due to Keisler. At the end of that blog post I asked whether this result could be generalized to {n > 1}: for which {n} are there models of {\mathsf{ZFC}} which have exactly {n} many {\mathsf{GBC}}-realizations?

The answer is yes for {n = 2}. Before stating the theorem to this effect, observe that a model of {\mathsf{ZFC}} has a definable (possibly with parameters) global well-order if and only if it satisfies {\exists x\ V = \mathrm{HOD}(\{x\})}. 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 {\mathsf{DGWO}}.

Theorem Let {M} be a countable model of {\mathsf{ZFC} + \mathsf{DGWO}}. Then {M} has an elementary top extension to {\omega_1}-like {M^*} so that {M^*} has exactly {\mathsf{GBC}}-realizations.

The construction of {M^*} 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 {\mathsf{GBC}}-realization. Instead, we will select a class {G \subseteq M} and get {(M,G) \prec_{\mathsf{top}} (M^*,G^*)} so that {M^*} is {G^*}-rather classless, meaning that the amenable classes of {M^*} are the classes definable from {G^*}. If {G} is carefully built, then {M^*} 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 {\vec A}, i.e. classes so that the model satisfies {\mathsf{ZFC}(\vec A)}, meaning {\mathsf{ZFC}} 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 {\mathsf{DGWO}}.

This {G} 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 {\mathrm{Ord} \rightarrow 2}.

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

Definition Let {\mathbb{B}} denote the full binary tree on {\mathrm{Ord}}. For a perfect tree {\mathbb{P} \subseteq \mathbb{B}} say that {\mathbb{Q} \subseteq \mathbb{P}} is {k}-deciding for {\mathbb{P}} if for any {\Sigma_k} formulae {\varphi} in the forcing language, there is an ordinal {\alpha} so that if {p \in \mathbb{Q}} has length at least {\alpha}, then {p} decides {\varphi} in {\mathbb{P}}.

Note that if {\mathbb{P}} is definable then we can find {k}-deciding subtrees of {\mathbb{P}} in a definable manner.

Definition {G \in {}^\mathrm{Ord}2} is a perfect generic if there is a sequence {\mathbb{B} = \mathbb{P}_0 \supseteq \mathbb{P}_1 \supseteq \cdots} of definable perfect trees so that {G = \bigcup \bigcap_i \mathbb{P}_i} and {\mathbb{P}_{i+1}} is {i}-deciding for {\mathbb{P}_i}.

An important point is that perfect generics really are generic. Because the sequence of trees decide more and more formulae in the forcing language, {G} 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 Lemma Let {(M,\vec A) \models \mathsf{ZFC}(\vec A) + \mathsf{DGWO}} be countable. Then there is a perfect generic {G \subseteq M} so that

  • Every {A_i} is in {\mathrm{Def}(M,G) \models \mathsf{GBC}}; and
  • {(M,G)} is minimal over {(M,\vec A)}, in the sense that if {B \in \mathrm{Def}(M,G)} then either {B \in \mathrm{Def}(M,\vec A)} or {G \in \mathrm{Def}(M,B,\vec A)}.

Proof: Start with countable {M \models \mathsf{ZFC} + \mathsf{DGWO}}. Apply the key lemma to get {(M,G)}. Now apply the rather classless construction to get {(M,G) \prec_{\mathsf{top}} (M^*,G^*)} where the top extension is {\omega_1}-like and {G^*}-rather classless. It is clear that {M^*} has at least two {\mathsf{GBC}}-realizations, namely {\mathrm{Def}(M^*)} and {\mathrm{Def}(M^*,G^*)}. Let us see that these are the only two. Suppose that {\mathrm{Def}(M^*,B)} is a {\mathsf{GBC}}-realization. Necessarily, {B \in \mathrm{Def}(M^*,G^*)}. By elementarity and the key lemma, either {G^* \in \mathrm{Def}(M^*,B)} or else {B \in \mathrm{Def}(M^*)}. In the first case, {\mathrm{Def}(M^*,B) = \mathrm{Def}(M^*,G^*)}. In the second case, {\mathrm{Def}(M^*,B) = \mathrm{Def}(M^*)}. \Box

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 {i}-deciding subtree. This will get that the sequence is of the right form, due to the easy fact that if {\mathbb{R} \subseteq \mathbb{Q} \subseteq \mathbb{P}} and {\mathbb{Q}} is {i}-deciding for {\mathbb{P}} then {\mathbb{R}} is {i}-deciding for {\mathbb{P}}. The second construction will ensure the minimality of the extension. The third construction will ensure that every {A_i} 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 sublemma Let {(M,\vec A, B) \models \mathsf{ZFC}(\vec A,B) + \mathsf{DGWO}} be countable and {\mathbb{R} \in \mathrm{Def}(M,\vec A)} be a perfect tree. There is {\mathbb{P} \subseteq \mathbb{R}} a perfect tree in {\mathrm{Def}(M,\vec A)} so that if {G} is any branch through {\mathbb{P}} then {B \in \mathrm{Def}(M,\vec A, G)}.

Proof: By {\mathsf{DGWO}}, we may assume without loss that {B \subseteq \mathrm{Ord}}. We build {\mathbb{P}} by pruning {\mathbb{R}}. We will prune at every other splitting node, cutting off one branch depending on information about {B}. Say that a splitting node in {\mathbb{P}} is an {\alpha}-node if the node has {\alpha} many splitting nodes below it. If {\alpha} is even, then keep both branches from {\alpha}. If {\alpha = 2\beta + 1} is odd, then remove one of the branches. If {\beta \in B}, then remove the {0} branch. If {\beta \not \in B}, then remove the {1} branch. Do this for all splitting nodes to get {\mathbb{P}}.

Suppose that {G} is a branch through {\mathbb{P}}. Then, we can recover {B} from {G} and {\mathbb{R}}. From {\mathbb{R}} we can find the odd splitting nodes. From {G} we know whether the {0} branch or {1} branch was removed. That is, {\beta \in B} if and only if there is a {(2\beta +1)}-node {s} in {\mathbb{R}} so that {s {}^\smallfrown 1} is an initial segment of {G}. \Box

Minimality sublemma Let {\varphi(x)} be a formula in the forcing language and {\mathbb{Q}} be a definable perfect tree. Then there is definable {\mathbb{R} \subseteq \mathbb{Q}} so that either

  • All sufficiently long sequences {p \in \mathbb{R}} decide {\varphi(\xi)} the same for all ordinals {\xi}; or
  • For any ordinal {\alpha} there is {\beta > \alpha} so that if {p,q \in \mathbb{R}} are distinct sequences of length {\beta} which agree up to {\alpha} then there is some ordinal {\xi} so that {p} and {q} decide {\varphi(\xi)} the same.

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

Proof: Pick {i} large enough so that {\varphi} is {\Sigma_i}. We start with {\mathbb{Q}' \subseteq \mathbb{Q}} which is definable and {i}-deciding. We will further prune this tree to produce {\mathbb{R}}. We can pick {\mathbb{Q}'} so that there is an embedding {f : \mathbb{B} \rightarrow \mathbb{Q}'} which is onto the splitting nodes and {f(s)} decides {\varphi(\mathrm{len}(s))}. We find ourselves in one of two cases.

The first is if there is some sequence {s \in \mathbb{B}} so that if {t,t'} are extensions of {s} of the same length then {f(t)} and {f(t')} decide {\varphi(\xi)} the same for all ordinals {\xi}. In this case, set {\mathbb{R}} to be {\mathbb{Q}' \upharpoonright f(s)}, the subtree of {\mathbb{Q}'} formed from nodes which extend {f(s)} or are extended by {f(s)}. 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 {g : \mathbb{B} \rightarrow \mathbb{Q}'}.

  • Set {g(\emptyset) = f(\emptyset)};
  • If {g(s)} is already defined then pick {g(s {}^\smallfrown 0)} and {g(s {}^\smallfrown 1)} to be nodes {p_i} of the same length so that {p_0} and {p_1} decide {\varphi(\xi)} differently for some {\xi}. Note that this picking relies upon having a uniform way to pick elements from {{}^\alpha2} for arbitrarily long {\alpha}, and thus appeals to Global Choice.
  • For limit stages simply take unions.

Now define {\mathbb{R}} to be the closure in {\mathbb{Q}'} of {g''\mathbb{B}}. Ponder for a moment and see that this yields the second condition in the statement of the sublemma. \Box

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

Proof: Fix an enumeration {\varphi_i(x)} of formulae in the forcing language so that the predicate for {A_i} does not appear in {\varphi_j} for {j \le i}. As we will continually apply the principality sublemma to code the {A_i}, this will ensure that in the end they are definable just from {G} (and possibly set parameters). Further fix a sequence {\alpha_i} of ordinals cofinal in {\mathrm{Ord}}. We will inductively build a descending sequence of definable perfect trees

\displaystyle  \mathbb{B} = \mathbb{P}_0 \supseteq \mathbb{Q}_0 \supseteq \mathbb{R}_0 \supseteq \mathbb{P}_1 \supseteq \mathbb{Q}_1 \supseteq \mathbb{R}_1 \supseteq \cdots

Pick {\mathbb{Q}_i \subseteq \mathbb{P}_i} to be {i}-deciding and to not split below {\alpha_i}. This step of the construction will ensure that we get a perfect generic. Pick {\mathbb{R}_i \subseteq \mathbb{Q}_i} according to the minimality sublemma applied to {\varphi_i}. Finally, pick {\mathbb{P}_{i+1} \subseteq \mathbb{R}_i} according to the principality sublemma applied to {A_i}. Consider {G = \bigcup \bigcap_i \mathbb{P}_i}. I claim that {G} is as desired.

As already mentioned, {G} is a perfect generic. An easy inductive argument, using the principality sublemma, gives that every {A_i} is in {\mathrm{Def}(M,G)}. Also, {\mathrm{Def}(M,G)} is a {\mathsf{GBC}}-realization because {M} has a definable global well-order. It remains only to check that {(M,G)} is minimal over {(M,\vec A)}. Take {B \in \mathrm{Def}(M,G)} defined by {\varphi_i(x)}. Without loss of generality assume that {B} is a class of ordinals. Then in stage {i} of the construction we ensured the minimality sublemma held for {\varphi_i}. Thus we are in one of the two situations. If the first condition holds then, for some {\alpha},

\displaystyle  B = \{ \xi \in \mathrm{Ord} : \forall p \in \mathbb{R}_i\ \mathrm{len}(p) > \alpha \Rightarrow p \Vdash_{\mathbb{Q}_i} \varphi_i(\xi) \}.

Manifestly, {B \in \mathrm{Def}(M,\vec A)}. Actually, we get the slightly stronger fact that {B \in \mathrm{Def}(M, A_0, \ldots, A_{i-1})} because the definition of {\mathbb{R}_i} and {\mathbb{Q}_i} only use up to {A_{i-1}}.

If we are in the second situation, then we can define {G} from {\vec A} and {B}. Namely, {p} is an initial segment of {G} if and only if for every ordinal {\alpha} there is a condition {q \in \mathbb{R}_i} extending {p} of length {> \alpha} so that for every {\xi}, we have that {q \Vdash_{\mathbb{Q}_i} \varphi_i(\xi)} implies {\xi \in B} and {q \Vdash_{\mathbb{Q}_i} \neg \varphi_i(\xi)} implies {\xi \not \in B}. Again, we get a slightly stronger fact, namely that {G \in \mathrm{Def}(M, B, A_0, \ldots, A_{i-1})}. \Box

Given this argument, the natural thing to try is to generalize it to {n > 2}. The strategy to try is to start with countable {M \models \mathsf{ZFC} + \mathsf{DGWO}}. Then apply the key lemma to get {G_1}. Repeatedly apply the key lemma to {(M,G_1, \ldots, G_{k-1})} to get {G_k}. One would like to get that applying Keisler’s construction to {(M,G_n)} would yield a model with exactly {n} many {\mathsf{GBC}}-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 {n} many {\mathsf{GBC}}-realizations for {M} which are contained in {\mathrm{Def}(M,G_n)}. Let’s illustrate the difficulty by looking at the {n = 3} case. We started with {M}, found {G} minimal over {M} and then found {H} minimal over {(M,G)}. This gives us at least 3 {\mathsf{GBC}}-realizations, namely {\mathrm{Def}(M)}, {\mathrm{Def}(M,G)}, and {\mathrm{Def}(M,H)}. But could there be more? Suppose that {A \in \mathrm{Def}(M,H)} and that {\mathrm{Def}(M,A)} is a {\mathsf{GBC}}-realization for {M}. By the key lemma, we are in one of two cases. One case is that {A \in \mathrm{Def}(M,G)}. This case is not problematic; just apply the key lemma again to get that either {A \in \mathrm{Def}(M)}, whence {\mathrm{Def}(M,A) = \mathrm{Def}(M)}, or else {G \in \mathrm{Def}(M,A)}, whence {\mathrm{Def}(M,A) = \mathrm{Def}(M,G)}. The other case is that {H \in \mathrm{Def}(M,G,A)}. This is the problematic case. We know that we can define {H} given both {G} and {A}, but how can we know that {H} can be defined just from {A}? It is conceivable that there could be many new {\mathsf{GBC}}-realizations contained in {\mathrm{Def}(M,H)} which are {\subseteq}-incomparable with {\mathrm{Def}(M,G)}. As such, this argument alone cannot give a model of {\mathsf{ZFC}} with exactly 3 {\mathsf{GBC}}-realizations. Damn.

A similar argument can be done in the setting of arithmetic to produce models of arithmetic with precisely 2 {\mathsf{ACA}_0}-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).