The Bristol model (5/5) Asaf Karagila University of East Anglia 22 November 2019 RIMS Set Theory Workshop 2019 Set Theory and Infinity Asaf Karagila (UEA) The Bristol model (5/5) 22 November 2019 1 / 23 Review Bird’s eye view of Bristol We started in L by choosing a Bristol sequence, that is a sequence of permutable families and permutable scales, and we constructed one step at a time models, Mα, such that: 1 M0 = L. Mα Mα+1 2 Mα+1 is a symmetric extension of Mα, and Vω+α = Vω+α . 3 For limit α, Mα is the finite support limit of the iteration, and Mα S Mβ Vω+α = β<α Vω+β. By choosing our Mα-generic filters correctly, we ensured that Mα ⊆ L[%0], where %0 was an L-generic Cohen real. We then defined M, the Bristol model, S S Mα as α∈Ord Mα = α∈Ord Vω+α. Asaf Karagila (UEA) The Bristol model (5/5) 22 November 2019 2 / 23 Properties of the Bristol model Small Violations of Choice Definition (Blass) We say that V |= SVC(X) if for every A there is an ordinal η and a surjection f : X × η → A. We write SVC to mean ∃X SVC(X). Theorem If M = V (x) where V |= ZFC, then M |= SVC. Theorem If M is a symmetric extension of V |= ZFC, Then M |= SVC. The following are equivalent: 1 SVC. 2 1 ∃X such that X<ω AC. 3 ∃P such that 1 P AC. 4 V is a symmetric extension of a model of ZFC (Usuba). Asaf Karagila (UEA) The Bristol model (5/5) 22 November 2019 3 / 23 Properties of the Bristol model Kinna–Wagner Principles Definition We say that A is an α-set of ordinals (or just α-set) if there is an ordinal η such that A ⊆ Pα(η). Definition The axiom KWPα states that every set is equipotent to an α-set. KWP is a shorthand for ∃α KWPα. So AC is KWP0, and KWP1 implies all sets can be linearly ordered, but KWP2 is already pretty loose and it is compatible with some pretty terrible counterexamples to choice. Proposition If M |= SVC, then M |= KWP. Asaf Karagila (UEA) The Bristol model (5/5) 22 November 2019 4 / 23 Properties of the Bristol model Kinna–Wagner Principles Theorem (Generalized Balcar–Vopěnka–Monro Theorem) Suppose that M and N are two models with the same α-sets of ordinals. If M |= KWPα, then M = N. Proof. Note that every relation on an α-set can be coded as an α-set, so if x ∈ M, then we can code htcl({x}), ∈i as an α-set, and therefore it is in N. Now apply the Mostowski collapse lemma in N to obtain that x ∈ N. In the other direction, let x ∈ N be such that x ⊆ M. And let η ∈ M be such M M that x ⊆ Vη . Let E be an α-set in M and f : E → Vη a bijection, then 0 0 E, f ∈ N, and therefore there is some E ⊆ E such that f E is a bijection with x. But E0 ∈ M, since it is an α-set. So x ∈ M as well. Now by induction M N on η we get Vη = Vη for all η, so M = N. Asaf Karagila (UEA) The Bristol model (5/5) 22 November 2019 5 / 23 Properties of the Bristol model Bristol is not small Theorem Let M be a Bristol model, then M |= ∀x(V 6= L(x)). Proof. M If x ∈ M, then x ∈ Vω+α. Therefore x ∈ Mα. Therefore L(x) ⊆ Mα. Since Mα ( Mα+1 ⊆ M, M 6= L(x). Proof. If M = L(x), then M |= SVC, then M |= KWP. It is therefore enough to show M |= ¬KWP. It is therefore enough to prove show any α-set in M, is already in Mα+1. Then by the GBVM theorem, M |= ¬KWPα for all α. ˙ IS ˙ Prove by induction on α. Suppose that A ∈ IS, and p A is an α-set. By the induction hypothesis we may assume that every a˙ appearing in A˙ is in ISα. ~ ˙ IS ˙ If H is a sequence of groups witnessing that A ∈ IS, and q a˙ ∈ A, where a˙ ∈ ISα, then we can find appropriate automorphisms in H~ which allow us to ˙ restrict q to q α + 1. So A is essentially in ISα+1. So A ∈ Mα+1. Asaf Karagila (UEA) The Bristol model (5/5) 22 November 2019 6 / 23 Improvements and limitations Better grounds We assumed our ground model was L. Indeed, this was necessary for obtaining M 6= L(x). For obvious reasons. But do we really need this assumption to construct such a model? Remember the question from the very beginning of this week: Question Suppose that G is a (set) generic filter over V . What kind of models lie between V and V [G]? This question is about arbitrary models. And by now you might think, did we even require V to satisfy ZFC or just ZF? Asaf Karagila (UEA) The Bristol model (5/5) 22 November 2019 7 / 23 Improvements and limitations Better grounds So how did L factor into the construction? We used it in three specific places: 1 ∗ λ + λ for singular cardinals (and 2 = λ for inaccessible cardinals) was used to obtain the permutable scale. 2 GCH in general, was used to ensure that successor steps, which were Add(κ+, 1)V , do not collapse cardinals. 3 Global Choice was used to choose a Bristol sequence to begin with. Of those three, Global Choice is indeed very disposable. If V satisfies the first two properties, let C be the class forcing adding a global choice function, F , without adding new sets. Consider the Bristol model over V [F ], then and promptly forget about F . Asaf Karagila (UEA) The Bristol model (5/5) 22 November 2019 8 / 23 Improvements and limitations Better grounds But what about the other two? We mentioned already that some combinatorial assumption on singular cardinals is in fact necessary. Whether or not it is ∗ exactly λ is kind of irrelevant: we need to have a certain combinatorial assumption. Finally, GCH. In principle, we can “space out” our coding so that even if cardinals are collapsed we can still continue. And that would be fine for constructing a model of ZF + ¬KWP. But if we want to ensure that the model is intermediate to a Cohen real, then we are not allowed to collapse cardinals. If we want it to be intermediate to some generic extension, then we are only allowed to collapse set-many cardinals. In particular that means that GCH needs to hold on a tail of cardinals. But in general GCH is a harmless axiom. Asaf Karagila (UEA) The Bristol model (5/5) 22 November 2019 9 / 23 Improvements and limitations Large cardinals ∗ If we only need to assume GCH and λ for singular cardinals, we can ask about Bristol models generated over many other type of models of ZFC. In particular inner models for “medium sized large cardinals” which satisfy the three assumptions. For example, if κ is a measurable cardinal, and D is a normal measure on κ, then L[D] is a model in which there is a measurable cardinal and the three assumptions hold. Suppose that we start with L[D], is κ a measurable in M? Since sets of ordinals are not added after the first step, it is immediately clear that any large cardinal property defined by sets of ordinals (e.g. Mahlo-ness) is preserved. This is also the case for a measurable cardinal. If κ is measurable and D is a measure, then it is a collection of sets of ordinals. It is a 1-set. Therefore we only need to know that there is a measure in M2 to know that there is one in M. But the first two steps of the iteration are small compared to a measurable. So any measure can be extended uniquely. Asaf Karagila (UEA) The Bristol model (5/5) 22 November 2019 10 / 23 Improvements and limitations Large cardinals The same argument shows that extenders, and all the other measure-based large cardinal properties are preserved. You could argue that measures are not the important part. It’s the embeddings that matter. Definition (Hayut–K.) We say that κ is a critical cardinal if it is the critical point of an elementary embedding j : Vκ+1 → X, where X is a transitive set. Easily, every critical cardinal is measurable. But the opposite is not necessarily true, if κ is critical then it is a strong limit cardinal. But it is consistent that ω1 is measurable. Question If κ is measurable in V , is it critical in M? We can also define weakly critical to mimic the embedding based definition of weakly compact cardinals. The same question applies here as well. While we do have a theorem for lifting elementary embedding to symmetric extensions, it is not suitable for the Bristol model. Asaf Karagila (UEA) The Bristol model (5/5) 22 November 2019 11 / 23 Improvements and limitations Large cardinal barriers While it seems plausible that critical cardinals are preserved, we know that supercompact cardinals are outright forbidden. In fact, a simple compactness argument shows that permutable scales fail to exist a long time before we even reach the first supercompact. Is there a way around this? In principle, the failure of square principles implies some compactness. And compactness is good at “coalescing small information”.