aa r X i v : . [ m a t h . L O ] J un GUIDE TO THE BRISTOL MODEL:GAZING INTO THE ABYSS
ASAF KARAGILA
Abstract.
The Bristol model is an inner model of L [ c ], where c is a Cohenreal, which is not constructible from a set. The idea was developed in 2011in a workshop taking place in Bristol, but was only written in detail by theauthor in [8]. This paper is meant as a guide for those who want to get abroader view of the construction. We try to provide more intuition that mightserve as a jumping board for those interested in this construction and in oddmodels of ZF . We also correct a few minor issues in the original paper, as wellas prove new results. For example, that the Boolean Prime Ideal theorem failsin the Bristol model, as some sets cannot be linearly ordered. In addition tothis we include a discussion on Kinna–Wagner Principles, which we think mayplay an important role in understanding the generic multiverse in ZF . Contents
1. Introduction 22. Symmetric extensions and other technical tools 33. The decoding apparatus 114. Cerberus, the three-headed guard of the underworld 205. Errata to “the Abyss” 236. Deconstructing constructibility 257. Gaps in the Multiverse 268. Kinna–Wagner Principles 279. Choice principles in Bristol 3010. Open questions related to the Bristol model 31References 33
Date : June 9, 2020.2010
Mathematics Subject Classification.
Primary 03E25; Secondary 03E35.
Key words and phrases. axiom of choice, symmetric extensions, iterations of symmetric exten-sions, Bristol model, generic multiverse, symmetric multiverse, Kinna–Wagner Principles.The author was supported by the Royal Society grant no. NF170989. Introduction
Mathematicians love classifications. We enjoy classifying objects into differentcategories, and for a good reason. Classifications teach us about abstract properties,and help us deepen our understanding of various objects and theories.Set theorists are generally interested in models of set theory. If V satisfies ZFC ,we want to classify models of set theory which lie between V and some genericextension, V [ G ]. In the case where “set theory” is understood as ZFC , Vopěnka’stheorem tells us exactly what the intermediate models are: they are generic exten-sions given by subforcings of the forcing which is used to introduce G over V .On the other hand, when we are interested in classifying arbitrary models of ZF , instead, even if we assume that V satisfied ZFC , the task becomes significantlyharder, and dare we say, nigh impossible. For a start, a generic extension of amodel of
ZFC cannot be a model of ZF + ¬ AC . One might be inclined to say thatsuch intermediate extension would still be a symmetric extension, which is a typeof inner model of a generic extension defined using automorphisms of the forcing.While this is true under some additional conditions on the intermediate model, itturns out that if M is an intermediate model between V and V [ G ], even if M isa symmetric extension of V , it might not be given by any forcing even remotelyrelated to the one for which G was generic.The reality is that intermediate models of ZF are far wilder than their ZFC -counterparts. The Bristol model is the first explicit example of such a model. Thisis a model intermediate to L [ c ], where c is an L -generic Cohen real, which is notconstructible from a set, let alone a symmetric extension of L (by any means, notjust the Cohen forcing). While there is a semi-canonical Bristol model, modulo aparticular choice of c , it is immediate from the construction that, in a very goodsense of the word, most models intermediate to L [ c ] are not even definable. We willclarify on this in section 7.The idea for this model came about in a small 2011 workshop in Bristol ontopics related to the HOD Conjecture. In attendance were Andrew Brooke-Taylor,James Cummings, Moti Gitik, Menachem Magidor, Ralf Schindler, Matteo Viale,Philip Welch, and W. Hugh Woodin, henceforth “the Bristol group”. The detailswere not written down in full, and the model remained as a folklore rumour untilthe author’s effort to formalise it. The details of the construction are given in [8],which was part of the author’s Ph.D. dissertation. This paper aims to give a bird’sview of the construction, from three different perspectives (for people coming fromdifferent walks of set theory). We will also correct a few minor mistakes in theoriginal paper, and prove a handful of new theorems about the Bristol model, andabout models of ZF in general.1.1. Structure of this Paper.
The Bristol model is presented in [8] as an itera-tion of symmetric extensions, starting from a Cohen real. The idea is to have, atsuccessor steps, a “decoding mechanism” which is a symmetric extension over anintermediate step such that two properties hold: (1) the decoding mechanism hasa generic (relative to the intermediate step) in the Cohen extension, and (2) thedecoding mechanism only adds subsets of sufficiently high rank.We will cover the basics of the technical tools in section 2. We will define sym-metric extensions, and briefly outline the main ideas related to iterating them (orrather, why it is hard to iterate symmetric extensions). We will also discuss thecombinatorial ideas needed for the decoding mechanism, both at successors of lim-its, as well as double successors.After covering the preliminary tools, we will present the decoding mechanism,and the generic argument needed for the proof to work. In section 4 we explain
UIDE TO THE BRISTOL MODEL 3 the three different approaches to constructing the Bristol model. All three areequivalent, but for different people some of these might be seen as “more natural”and can help understand the model better. We will not dive into the intimatedetails, though. The goal of this paper is to serve as a companion, and help providenot only the big picture of the construction, but also serve as a first step towardsreading and understanding the construction’s details presented in [8].Having discussed the construction of the model, we will then point towardssome minor gaps and typos in the original [8]. Then we will discuss Kinna–WagnerPrinciples which we expect to play a role in the study of choiceless models such asthe Bristol model. We will make some new observations, and suggest conjecturesfor future research. Finally, in section 9 we will prove that some sets in the Bristolmodel cannot be linearly ordered, and therefore the Boolean Prime Ideal theoremis false there. We finish the paper with a long list of open questions related to theBristol model.
Acknowledgements.
The author would like to express his deepest gratitude toDaisuke Ikegami for providing the opportunity to give a long tutorial on the Bristolmodel during the RIMS Set Theory Workshop in November 2019, “Set Theory andInfinity”, as well as to the audience, who sat and listened, asked questions, andpointed out difficulties, all of which contributed to this paper. We also want tothank David Asperó and Andrés E. Caicedo for providing thorough comments onearly versions of this manuscript.2.
Symmetric extensions and other technical tools
In this paper, the term forcing will denote a preordered set with a maximum,denoted by , unless explicitly mentioned otherwise. Of course, we will invariablythink about a forcing as a partially ordered set, a separative one, in fact, knowingfull well that this will not limit our generality. The elements of P are called conditions , and using them we define P -names . We refer the reader to any of[4, 6, 11] for the basic methodology of forcing.Let P be a notion of forcing, we follow the convention that if p, q ∈ P , then q ≤ p indicates that q is a stronger condition, and we will often say that q extends p . Two conditions are compatible if they have a common extension, and they are incompatible otherwise.Given a collection of P -names, { ˙ x i | i ∈ I } , that we want to transform into aname, we will denote by { ˙ x i | i ∈ I } • the name {h , ˙ x i i | i ∈ I } , and we say that aname is a • -name when it has this form. This extends naturally to ordered pairs,sequences, functions, etc. With this notation we can easily define the canonicalnames for ground model sets: ˇ x = { ˇ y | y ∈ x } • .Given two P -names, ˙ x and ˙ y , we say that ˙ x appears in ˙ y if there is some p ∈ P such that h p, ˙ x i ∈ ˙ y . We will use a similar terminology stating that p appears in ˙ y .2.1. Symmetric extensions.
As we remarked, a generic extension of a model of
ZFC is again a model of
ZFC . Symmetric extensions are intermediate models togeneric extensions where the axiom of choice may fail.Let P be a forcing, and let π be an automorphism of P . The action of π extendsto the P -names by recursion: π ˙ x = {h πp, π ˙ y i | h p, ˙ y i ∈ ˙ x } . We will, eventually, tickle class forcing. We still insist on the preorder definition, as it does make the definition of an iteration signif-icantly more manageable.
ASAF KARAGILA
Lemma (The Symmetry Lemma).
Let π be an automorphism of a forcing P ,and let ˙ x be a P -name. For every condition p , p (cid:13) ϕ ( ˙ x ) ⇐⇒ πp (cid:13) ϕ ( π ˙ x ) . Fix a group G ⊆ Aut( P ). We say that F is a filter of subgroups on G if it is afilter on the lattice of subgroups, namely, it is a non-empty collection of subgroupswhich is closed under finite intersections and supergroups. We will, unless statedotherwise, assume it is a proper filter, i.e. the trivial group is not in F . Finally, F is normal if whenever π ∈ G and H ∈ F , then πHπ − ∈ F as well. In mostcases we are interested not necessarily in a filter, but in a filter base, and we willignore the distinction between the two.Call h P , G , F i a symmetric system if P is a notion of forcing, G is a group ofautomorphisms of P , and F is a normal filter of subgroups on G . We shall fix asymmetric system h P , G , F i for the rest of this subsection.For a P -name, ˙ x , let sym G ( ˙ x ) denote the group { π ∈ G | π ˙ x = ˙ x } . If it is thecase that sym G ( ˙ x ) ∈ F , then we say that ˙ x is F -symmetric . And similarly, we saythat ˙ x is hereditarily F -symmetric if being F -symmetric is hereditarily true for allnames appearing in { ˙ x } • .We denote by HS F the class of hereditarily F -symmetric names. We denoteby (cid:13) HS the relativisation of the forcing relation to HS : we restrict the quantifiersand free variables to this class. It is not hard to check that the Symmetry Lemmaapplies for (cid:13) HS , provided that we use automorphisms from G . Theorem.
Let G ⊆ P be a V -generic filter, and let M denote the interpreted class HS G F = { ˙ x G | ˙ x ∈ HS F } . Then M is a transitive class model of ZF such that V ⊆ M ⊆ V [ G ] . Moreover, M | = ϕ ( ˙ x G ) if and only if there is some p ∈ G suchthat p (cid:13) HS ϕ ( ˙ x ) . The class M is also called a symmetric extension of V . It turns out that M isa symmetric extension of V if and only if M = V ( x ) for some x ∈ V [ G ]. We willdiscuss this in more detail in section 7.We will omit G and F from the notation and terminology when they are clearfrom context, which is usually what is going to happen.2.2. Example.
Let P be Add( ω, ω ). Namely, p ∈ P is a finite partial functionfrom ω × ω →
2. We let our G be the group of permutations of ω acting on P in the natural way: πp ( πα, n ) = p ( α, n ). Finally, for E ⊆ ω , let fix( E ) denote { π ∈ G | π ↾ E = id } , and set F = { fix( E ) | E ∈ [ ω ] <ω } .For every α < ω , let ˙ a α be the name of the α th Cohen real, {h p, ˇ n i | p ( α, n ) = 1 } ,and set ˙ A = { ˙ a α | α < ω } • . Claim 2.1.
For every π ∈ G and α < ω , π ˙ a α = ˙ a πα . Consequently, π ˙ A = ˙ A . (cid:3) As an immediate corollary, ˙ a α ∈ HS for each α < ω , as witnessed by fix( { α } ),and so ˙ A ∈ HS as well. Theorem 2.2. (cid:13) HS ˙ A cannot be well-ordered. Consequently, the real numberscannot be well-ordered, and therefore (cid:13) HS ¬ AC .Proof. Let ˙ f ∈ HS , and suppose that p is a condition such that p (cid:13) HS ˙ f : ˙ A → ˇ η forsome ordinal η . Let E be a countable set such that fix( E ) ⊆ sym( ˙ f ). We may alsoassume that π ∈ fix( E ) satisfies πp = p by adding a finite set to E , and replacingit with E ∪ { α | ∃ n h α, n i ∈ dom p } . There is no point in using the improper filter when taking a symmetric extension. However,for the sake of generality it should be noted that this can be useful when iterating. We promiseto never bring this up in the course of this paper again.
UIDE TO THE BRISTOL MODEL 5
Fix α / ∈ E , as P is a c.c.c. forcing, the set X = { ξ < η | ∃ q ≤ p : q (cid:13) HS ˙ f ( ˙ a α ) = ˇ ξ } is countable. Note that p (cid:13) HS ˙ f ( ˙ a α ) ∈ ˇ X .For any β < ω , let π β denote the 2-cycle ( α β ). Therefore, π β p (cid:13) HS π β ˙ f ( π β ˙ a α ) ∈ π β ˇ X. Easily, π β ∈ fix( E ) if and only if β / ∈ E . So for all β / ∈ E , p (cid:13) HS ˙ f ( ˙ a β ) ∈ ˇ X . Inparticular, p must force that ˙ f has a countable range. As p and ˙ f were arbitrary, wein fact have shown that every ordinal-valued function in the symmetric extensiondefined on A must have a countable range.To finish the proof we appeal to the c.c.c. of the Cohen forcing again, noting that ω is not collapsed, and therefore (cid:13) HS | ˙ A | 6 = ℵ . Therefore there is no injectionfrom A into the ordinals, as wanted. (cid:3) The standard arguments are usually presented in a slightly different way. Weusually extend p to a condition q , which decides the value of ˙ f ( ˙ a α ), and then showthat we find π ∈ fix( E ) such that πq is compatible with q , and πα = α . This thenshows that no extension of p can force ˙ f to be injective. Chain condition basedarguments are not very common in results of this type, and we hope that this paperwill help to popularise the idea. Iterations of symmetric extensions.
Iterating symmetric extensions is notan easy task. While some ad-hoc constructions can be found in the literature fromthe very early 1970s, the only systematic development of such framework was doneby the author in [9], and so far only for finite support iterations. The goal of thissection is not to fully develop and explain this technique, but instead provide anintuition as to how the technique works, and what are the difficulties that need tobe overcome in the case of the Bristol model.The intuition behind iterations of symmetric extensions is as naive and simpleas it can get. We want to have an iteration of forcing notions, in this case withfinite support, and we want to identify a class of names which correspond to theintermediate model that we would get if we were to iterate symmetric extensionsone step at a time.Looking at a two-step iteration, say P ∗ ˙ Q , we also need to associate G and F to P , and names for a group of automorphisms, ˙ H , and a filter of subgroups ˙ K . Now,a P ∗ ˙ Q -name is going to be in the iterated symmetric extension if its projection to P is in HS F , and it is guaranteed to be interpreted as a name in HS K , that is ahereditarily symmetric name in the second step.But we can weaken this slightly, and much like we only require that the P -nameis guaranteed to be a name in HS K , we can require that the P -name is guaranteedto be equal to some name in HS F . That we, we are allowed to “mix” names from HS F over an antichain. Consider, for example, Cohen’s first model, this is a model similar to the exampleabove, replacing ω by ω . Namely, we add an ω -sequence of Cohen reals, permutethem, and consider finite stabilisers. Let ˙ a n be the name for the n th real, then wecan define the name˙ a = {h p, ˇ m i | ∃ n ( p ( n,
0) = 1 ∧ ∀ k < n, p ( k,
0) = 0 ∧ p ( n, m ) = 1) } . See [10] for more examples of this sort. Examples include [12],[16],[17], and to some extent also [14]. In general a class of names X has the mixing property if when 1 (cid:13) ˙ x ∈ X , that is we can finda (pre-)dense set of conditions p and ˙ x p ∈ X such that p (cid:13) ˙ x = ˙ x p , then ˙ x ∈ X . ASAF KARAGILA
In clearer terms, this is the name for the first ˙ a n which contains 0. We are guar-anteed that ˙ a will be interpreted as one of the ˙ a n . But it is not hard to see that ˙ a ,as defined above, is not stabilised by fixing any finite subset of ω . Considering names like ˙ a seems like an unnecessary complication. But upona closer examination of the general construction of iterations, we see this idea issomehow necessary. Indeed, the conditions of the iteration are usually defined tobe h p, ˙ q i such that p ∈ P and P (cid:13) ˙ q ∈ ˙ Q . This definition generalises to non-finitesupports. We will call the forcing notions defined this way “Jech(-style) iterations”.Kunen, in his book, introduces iterations with a perhaps more naive approach:the conditions are pairs h p, ˙ q i such that p ∈ P and p (cid:13) ˙ q ∈ ˙ Q . This definitionfails to generalise nicely to infinite supports, but it is useful for understandingfinite support iterations. We will use “Kunen(-style) iterations” to refer to forcingiterations defined this way. Remark 2.3.
It is worth pointing out at this point that we assume
ZFC holds inour ground model. While the theory of symmetric extensions, as well as that ofiterated forcing, can be developed reasonably well in ZF , it is not clear to whatextent choice is truly necessary for developing the theory of iterated symmetricextensions. We utilise the mixing property quite significantly in the general theory,and while it is conceivable, and indeed it is our conjecture, that we can removechoice from the assumptions, there is a certain comfort and simplicity in assumingit. Moreover, as we want to start with L as our underlying model, ZFC is alreadya given. While V = L is far too strong of an assumption, we will see that at thevery least we will want GCH to hold, which implies choice anyway.When working with Jech iterations, we can quickly see that even for the con-ditions of P ∗ ˙ Q to be in the intermediate model, we need to allow this so-called“mixing property”. And so, if we require it to hold for the conditions, we will needto require it to also hold for the automorphisms of ˙ Q , etc., and so the idea itselfis important. We can now proceed towards finding a combinatorial definition thatwill allow us to directly define the class of names, much like we did with HS in thecase of a single symmetric extension. Definition 2.4.
Let P be a notion of forcing and let π be an automorphism of P .We say that π respects a name ˙ A if P (cid:13) π ˙ A = ˙ A . If h P , G , F i is a symmetricsystem we say that ˙ A is F -respected if there is some H ∈ F such that every π ∈ H respects ˙ A .Much like the definition of F -symmetric before it, this definition lends itself toa hereditary version. We will simply say “respected” when F is clear from context.If ˙ A carries an implicit structure (e.g., a forcing notion) then this structure is alsorequired to be respected.The idea is that being respected is “almost” being symmetric. We will soonweaken this property a bit further to accommodate the “mixing property” into therespected names.Respect is the foremost necessary condition for developing a combinatorial char-acterisation of iterated symmetric extensions. Given a symmetric system h P , G , F i and a P -name ˙ Q , in order to define an automorphism of P ∗ ˙ Q using some π ∈ G ,the first thing we need to ensure is that π respects ˙ Q . Otherwise, h p, ˙ q i 7→ h πp, π ˙ q i is not an automorphism of P ∗ ˙ Q . This shows that typically HS does not have the mixing property. This is not a bad thing,though, as we are often concerned with particulars when working with symmetric extensions, andhaving to specify witnesses is a good thing. UIDE TO THE BRISTOL MODEL 7
Definition 2.5.
Let h P , G , F i be a symmetric system, and let h ˙ Q , ˙ H , ˙ K i • be ahereditarily respected name for a symmetric system. Suppose that π ∈ G and ˙ σ isa name such that P (cid:13) ˙ σ ∈ ˙ H , then we denote by R h π, ˙ σ i the automorphism definedby h p, ˙ q i 7→ h πp, π ( ˙ σ ˙ q ) i .Here we utilise the mixing property quite significantly, by defining ˙ σ ˙ q to be thename guaranteed to be interpreted as the action of ˙ σ on the condition ˙ q . If we wereusing Kunen iterations, we would have to define R h π, ˙ σ i as a partial automorphismwhich is only defined when p (cid:13) ˙ σ ∈ G . This is not a formal problem, but it makesthe actual legwork a lot harder.We will denote by G ∗ ˙ H the group of all such automorphisms. In [9] we referto this group as the generic semidirect product . Turning our attention to the filters F and ˙ K , we define a support to be h ˙ H , ˙ H i where both of them are P ∗ ˙ Q -namessuch that (cid:13) ˙ H ∈ ˇ F and ˙ H ∈ ˙ K . Note that we can always extend whatevercondition to decide the actual group ˙ H , and likewise we can decide the actual P -name for ˙ H . However, using this approach allows us to take advantage of themixing property.We can now define the notion of respect relative to the supports. Namely, thereis a pair h ˙ H , ˙ H i such that whenever h p, ˙ q i (cid:13) h π, ˙ σ i ∈ h H , ˙ H i , which is to say h p, ˙ q i (cid:13) ˇ π ∈ ˙ H and ˙ σ ∈ ˙ H , we have that h p, ˙ q i (cid:13) R h π, ˙ σ i ˙ A = ˙ A . Whereas inDefinition 2.4 we required that ˙ H would actually be a concrete group, here weallow a bit more leeway. The result is that the respected names, in this sense, aretruly the closure of HS F under mixing.A symmetric iteration, or an iteration of symmetric extensions, is defined byspecifying a sequence of names for symmetric systems, h ˙ Q β , ˙ G β , ˙ F β | β < α i , anddefining P α as the finite support iteration of the forcings ˙ Q β ; G α as the finitesupport generic semidirect product of the groups ˙ G β ; and F α as the collection ofall supports.We define supports in the general case as P α -names for sequences h ˙ H β | β < α i such that (cid:13) α ˙ H β ∈ ˙ F β , and that (cid:13) α { ˇ β | ˙ H β = ˙ G β } • is finite. The last part iscrucial, as it allows us the flexibility in “knowing something has a finite definition”while not committing to its specifics just yet.Let IS α be the class of P α -names which are hereditarily respected. This class isnow the class of names which will be interpreted in the intermediate model of stage α . So to complete our definition of a symmetric iteration we need to require that h ˙ Q α , ˙ G α , ˙ F α i • is in IS α , and indeed that it is respected by all automorphisms in G α , as we pointed out before. We also have a forcing relation (cid:13) IS α which is definedby relativising the names and quantifiers to IS α .Using this definition we can show that when G is V -generic, then IS Gα is a modelof ZF , and if α = β + 1, then this model is a symmetric extension of IS G ↾ ββ usingthe β th symmetric system. On the other hand, if we want to continue a symmetriciteration, that is of course possible, but we need to make sure that the generic filterused is not only IS Gα -generic, but rather V [ G ]-generic, and we may need to shrinkthe groups ˙ G β in a few places. This may present an issue if we want to continue ouriteration by infinitely many steps, as we may need to shrink our groups infinitelymany times, which might not be possible.But here we arrive to our first obstacle when applying this to the Bristol model.We just required that the generic filter is not “pointwise”-generic, but rather V -generic for the entire iteration. While in the case of iterated forcing this is not anissue for the successor case, here we run into the first problem we had defining Although it can be an issue for the limit case.
ASAF KARAGILA the whole apparatus using mixing. The use of mixing allows us to use arbitraryantichains and predense sets to define the names in IS α , and so we end up withnames in IS α which encode some generic information in them.The “easy” way to leave this mess is to require that we relativise the definitionpointwise. That is, at each step we only take names which are in the intermediatemodel. But this adds a layer of complexity when defining the actual forcing P α ,or the automorphisms in G α , or using these in the same manner that we are usedto when working with symmetric extensions. Even worse, while for finite supportiterations all of these different constructions are equivalent, this is not the case ifwe want to extend our definition to other types of iterations.We take a different route instead. Looking at products as a type of degenerateiterations, we may want to mimic this definition here. But a copy-paste approachis bound to result in just a product of symmetric extensions. While this is fine, itis not what we are looking for. We want to force over the symmetric extensions,but with a “very canonically defined forcing”. The idea is that we want to iterate P ∗ ˙ Q , and in V [ G ], where G ⊆ P , the forcing ˙ Q G is isomorphic to a forcing in V ,and to some extent, this isomorphism does not even depend on G . This means thatwe are really taking a product. But in the symmetric extension given by P , theforcing ˙ Q G is not isomorphic to any partial order in V , maybe because it cannot bewell-ordered, or maybe due to a similar consideration. Nevertheless, it is distinctfrom forcings in V as far as HS G is concerned.We say that a symmetric iteration is productive if each ˙ Q α is a • -name whichallows us to use the Kunen-iterations for P α , which are simply the products of thesenames. We also require that the names of ˙ G α and ˙ F α have similar properties, inthat they are • -names and that α decides equality related propositions.Finally, in the definition of a support, needed for the definition of IS α , we requiremore. We require that h ˙ H β | β < α i is an excellent support , meaning that ˙ H β isa name appearing in ˙ F β , and in particular a P β -name, and the finite set of non-trivial coordinates is decided in advance. This is in line with the previous demands:everything is decided in advance, this is “almost a product”.Now that we have a condition on the iteration, we need a condition on the genericfilters as well. Definition 2.6.
Suppose that h P , G , Fi is an iteration of symmetric extensions. Wesay that D ⊆ P is symmetric if there is an excellent support ~H such that whenever p (cid:13) ~π ∈ ~H , then p (cid:13) D = { R ~π q | q ∈ D } . In other words, D is stable, as a set,under a large group of automorphisms.We say that G ⊆ P is symmetrically V -generic if it is a filter meeting everysymmetrically dense set in V .It turns out that symmetrically generic filters are exactly the filters needed tointerpret symmetric names. And we have the following theorem for productiveiterations: Theorem 2.7.
Let P be a productive iteration, and let p ∈ P and ˙ x ∈ IS be somesymmetric name. The following conditions are equivalent: (1) p (cid:13) IS ϕ ( ˙ x ) . (2) For every symmetrically V -generic filter G such that p ∈ G , IS G | = ϕ ( ˙ x G ) . (3) For every V -generic filter G such that p ∈ G , IS G | = ϕ ( ˙ x G ) . This is not to say that this is not doable, or maybe even a better way of iterating symmetricextensions. We hope that these remarks will inspire more people to work on these problems. These include, of course, actual products, as well as single-step symmetric extensions.
UIDE TO THE BRISTOL MODEL 9
It is not hard to check now that at least for the successor steps, the iteration of“pointwise” symmetrically generic filters is indeed a symmetrically generic filter.We finish this overview with a preservation theorem.
Theorem 2.8.
Suppose that h ˙ Q α , ˙ G α , ˙ F α | α < δ i defines a symmetric iteration,and G is V -generic for the iteration. Moreover, assume that for every α < δ , (cid:13) α ˙ Q α is weakly homogeneous and ˙ G α is rich enough to witness this. Let η be an ordinal such that there exists α < δ that for any α ∈ ( α , δ ) the followingequality holds: V IS G ↾ αα η = V IS G ↾ α +1 α +1 η . Then V IS Gδ η = V IS G ↾ α α η . In other words, if no sets of rank <η were added atsuccessor steps, none were added at limit steps either. This theorem is in stark contrast to the familiar case in the usual context ofiterated forcing: iterating, with finite support, forcings which are not c.c.c. willcollapse cardinals; and iterating non-trivial forcings, even if they are c.c.c., will addCohen reals at limit steps. But in the case of symmetric iterations, even if theforcings are non-trivial, as long as they are homogeneous and do not add reals, thelimit steps will not add reals either.As a consequence, we can extend our apparatus now to an Ord-length iterationwhile preserving ZF in the resulting model. Moreover, the result holds for produc-tive iterations with symmetrically generic filters, as one can state it in the languageof forcing, rather than talking about V η of various models. See also §9.2 of [9].2.4. Permutable families and scales.
The key mechanism in the constructionof the Bristol model is “decoding a long sequence from a short sequence”. This canmean a sequence of length ω from a Cohen real, or a sequence of length ω fromone of length ω . We use almost disjoint families in successor steps to repeatedlydecode these sequences, and we use a particular type of scale to succeed at thistask when we are at limit steps. This will be as good a place as any to remind thereader that we work in ZFC , especially when thinking about these combinatorialobjects that are used here.
Definition 2.9.
Let κ be a regular cardinal, and fix a family A = { A α | α < κ + } ofunbounded subsets of κ such that for α < β , sup( A α ∩ A β ) < κ . For permutations π : κ → κ and Π : κ + → κ + we say that that π implements Π if π “ A α = ∗ A Π ( α )for all α < κ + .Here we use = ∗ to mean equality up to a bounded subset of κ . Which κ , ofcourse, will be clear from the context, so we will spare the reader from using thesymbol = ∗ κ , or worse.We are looking for an abstract property of an almost disjoint family which willensure that it implements any bounded permutation of κ + , that is any permutationof κ + which is the identity on a tail can be implemented. Definition 2.10.
Let κ be a regular cardinal, { A α | α < κ + } ⊆ [ κ ] κ is called a permutable family if it is almost disjoint, that is for α = β , sup( A α ∩ A β ) < κ , andfor every I ∈ [ κ + ] <κ + there is a pairwise disjoint family { B ξ | ξ ∈ I } such that B ξ = ∗ A ξ and α ∈ I ⇐⇒ A α ∩ [ ξ ∈ I B ξ is unbounded in κ. We call { B ξ | ξ ∈ I } as in this definition a disjoint approximation , and in thecase where B ξ ⊆ A ξ , we say it is a disjoint refinement . We will say in this case that G α witnesses the homogeneity of Q α . Proposition 2.11. If A is a permutable family of subsets of a regular cardinal κ ,then it implements every bounded permutation of κ + .Proof. Let Π be a bounded permutation of κ + , fix η < κ + such that Π does notmove any ordinals above η . Next, set I = η and let { B ξ | ξ < η } be a disjointrefinement. Now let π be the function which is the order isomorphism from B α to B β when Π( α ) = β , and the identity elsewhere. Easily, π implements Π. (cid:3) Having fixed a permutable family, if π : κ → κ implements Π, we will denote thisby ι ( π ) = Π. Proposition 2.12.
Let κ be a regular cardinal, then a permutable family exists.Proof. Let h T α | α < κ + i be a ⊆ ∗ -increasing family of subsets of κ . Define A α as T α +1 \ T α , then { A α | α < κ + } is a permutable family. (cid:3) Remark 2.13.
It should be pointed out that one can construct an increasing familyof subsets from a permutable family. Recursively, set T = A , T α +1 = T α ∪ A α ,and for limit steps recursively construct T α as the union of a disjoint refinement of { A β | β < α } , which exists by definition of a permutable family. As long as α < κ + we can ensure that these disjoint refinements are also increasing in inclusion, whichguarantees that T α contains previous limit steps. Definition 2.14.
Given a permutable family on a regular cardinal κ , the derivedgroup is the group G of all permutations of κ which implement a bounded permu-tation of κ + . The derived filter is the normal filter of subgroups on G generated byfix( B ), for a disjoint approximation B , where fix( B ) = { π ∈ G | π ↾ S B = id } .While these definitions are given for regular cardinals, we will only use them inthe basis case and successor case. For the limit case, where κ is a limit cardinal, weneed to use a slightly different machinery, as the goal is to coalesce the informationfrom previous steps and use it as a kind of “short sequence”. In some way, inacces-sible cardinals are the “simpler case” compared to singular cardinals. Nevertheless,there is no need of separating the two. Definition 2.15.
Let λ be a limit cardinal, and let SC( λ ) denote { µ + | µ < λ } .Let { f α | α < λ + } be a scale in Q SC( λ ). Given a sequence of permutations ~π = h π θ | θ ∈ SC( λ ) i such that π θ is a permutation of θ , we say that ~π implements a function Π : λ + → λ + if for every α < λ + and for every large enough θ , f Π( α ) ( θ ) = π θ f α ( θ ) . We call a scale permutable if it implements every bounded permutation of λ + .As with the case of permutable families, we denote by ι ( ~π ) the permutation Πthat is implemented by ~π .The fact that inaccessible cardinals are the “simpler case” can be trivially seenas a consequence of the following proposition, while remembering that working in V = L we always have the wanted cardinal arithmetic. Proposition 2.16.
Suppose that λ is an inaccessible cardinal and λ = λ + , thenthere is a permutable scale on λ . This can be shown by a simple transfinite recursion, in a very similar fashion tothe permutable family case. We actually only need to have an increasing sequence, it is irrelevant that it is also bounding. The assumption on 2 λ can be completely removed by simply limiting ourselves to increasingsequences instead of scales. Nevertheless, as we are working under GCH anyway, this is justsimpler.
UIDE TO THE BRISTOL MODEL 11
Proposition 2.17.
Suppose that λ is a singular cardinal and (cid:3) ∗ λ , then there is apermutable scale on λ . The proof of this proposition can be found as Theorem 3.27 in [8]. The ideaof the proof goes back to the Bristol group, and utilises the work of Cummings,Foreman, and Magidor in [2] where it is shown that (cid:3) ∗ λ implies the existence of“better scales”. The aforementioned Theorem 3.27 show that a better scale is infact permutable.The key point in proving a better scale is a permutable scale is that given any I ∈ [ λ + ] <λ + , we can find a function d : I → SC( λ ) such that { f α “[ d ( α ) , λ ) | α ∈ I } is a family of pairwise disjoint sets. The proof of Theorem 3.27 also shows thateven if we are only allowing π θ to be a bounded permutation of θ , this is stillenough to ensure that we can implement every bounded permutation of λ + using apermutable scale. We say that a sequence of permutation groups of each θ ∈ SC( λ )is rich enough if we can require π θ to be in the relevant group when we find animplementing sequence. Definition 2.18.
Let λ be a limit cardinal, and for every θ ∈ SC( λ ), let G θ be arich enough group of permutations of θ . The derived group is the subgroup G of thefull support product Q θ ∈ SC( λ ) G θ consisting of all sequences ~π = h π θ | θ ∈ SC( λ ) i which implement a bounded permutation of λ + .For η < λ + and f ∈ Q SC( λ ) we let K η,f be the group { ~π ∈ G | ι ( ~π ) ↾ η = id and for all θ ∈ SC( λ ) , π θ ↾ f ( θ ) = id } . The derived filter is the filter generated by { K η,f | η < λ + , f ∈ Q SC( λ ) } .We can weaken the definition of K η,f and replace η by a bounded subset of λ + ,i.e., I ∈ [ λ + ] <λ + , and replace f by a sequence of bounded sets of each θ . But asthe definition is complicated enough as it is, it is easier to just use η and f as upperbounds. 3. The decoding apparatus
Assume V = L throughout this section. The Bristol model is constructed as asymmetric iteration, indeed a productive iteration. We will outline the constructionof the different intermediate steps in this iteration, and the arguments needed forutilising the iterations apparatus. Definition 3.1. A Bristol sequence is a sequence indexed by the ordinals such thatfor α = 0 or α successor we have A α = { A αξ | ξ < ω α +1 } which is a permutablefamily on ω α , and if α is a limit ordinal then we have F α = { f αξ | ξ < ω α +1 } whichis a permutable scale in the product Q SC( ω α ).Fix a Bristol sequence. By assuming V = L we not only have a Bristol sequence,indeed we have a canonical one, where we choose the < L -minimum permutablefamily or scale at each point. Our goal is to use these permutable objects and replaceat each stage, α , a sequence of length ω α by one of length ω α +1 . In particular thefirst step is to replace the Cohen real, which is a sequence of length ω , by a sequenceof length ω . But we want to be able and guarantee that the original sequence is not going to be definable from our longer sequence.In this sense, we want to decode from a Cohen real a sequence of length ω ,which captures “some crucial bits” of the Cohen real, but not really all of it. Thenwe want to decode from this ω sequence a new sequence of length ω , forget theone of length ω , and proceed. Example: first steps.
Let P be the Cohen forcing, and in this case we mean p ∈ P is a function from a finite subset of ω into 2. Let us omit the index from A ,as we are only concerned with the first step at the moment, so A α denotes the α thset in the first permutable family. We let G and F denote the derived group andfilter from A . The action of G on P is the natural one: πp ( πn ) = p ( n ) . We denote by ˙ c the canonical name for the Cohen real. For A ⊆ ω let P ↾ A denote the subforcing { p ∈ P | dom p ⊆ A } , and let ˙ c A denote the name {h p, ˇ n i | p ( n ) = 1 ∧ dom p ⊆ A } . Of course, ˙ c A is the canonical name of the generic real added by P ↾ A . We have nowthat π ˙ c A = ˙ c π “ A . In general we say that a name ˙ x for a set of ordinals is decent ifevery name appearing in it is of the form ˇ ξ for some ξ ∈ Ord. We say that a namefor a set of ordinals is an A -name if it is a P ↾ A -name. Proposition 3.2.
Suppose that ˙ x ∈ HS and (cid:13) ˙ x ⊆ ω , then there is some disjointapproximation B and a decent S B -name ˙ x ∗ such that (cid:13) ˙ x = ˙ x ∗ .Proof. Let B be a disjoint approximation such that fix( B ) ⊆ sym( ˙ x ) and define ˙ x ∗ as ˙ x ∗ = n h p, ˇ n i (cid:12)(cid:12)(cid:12) p (cid:13) ˇ n ∈ ˙ x ∧ dom p ⊆ [ B o . It is clear that (cid:13) ˙ x ∗ ⊆ ˙ x , to show equality it is enough to prove that if p (cid:13) ˇ n ∈ ˙ x ,then p ↾ S B (cid:13) ˇ n ∈ ˙ x . Let q ≤ p ↾ S B .By the very definition of a disjoint approximation S B is co-infinite, so we mayfind a finite set E such that E ∩ ( S B ∪ dom p ) = ∅ and | E | = | dom q \ S B| .Then let π be a permutation which maps dom q \ S B to E , and it is the identityelsewhere. Being a finitary permutation it implements the identity function, soindeed π ∈ G , and by its very definition π ∈ fix( B ), so π ˙ x = ˙ x . So if q had forcedˇ n / ∈ ˙ x , we would have πq (cid:13) π ˇ n / ∈ π ˙ x , which is the same as πq (cid:13) ˇ n / ∈ ˙ x . Alas, πq and p are clearly compatible, and so this is impossible. (cid:3) We let G be a V -generic filter and let M denote the symmetric extension HS G ,as is standard, we will “omit the dot” to indicate the interpretation of a name,so c is going to be ˙ c G , etc. The following is a very easy corollary from the aboveproposition. Corollary 3.3. c / ∈ M . (cid:3) This is where we start seeing the importance of the permutable family, as opposedto any almost disjoint family. If { X α | α < ω } is an almost disjoint family, then { c ∩ X α | α < ω } is a family of mutually generic Cohen reals, any finitely many ofthem are mutually generic over any other finite subfamily. But in our case, wherethe almost disjoint family is in fact a permutable one we get countable mutualgenericity. Any countable subset of { c ∩ A α | α < ω } is simultaneously mutuallygeneric over any other countable subset (provided they are pairwise disjoint).We can actually prove more. Nothing in the proof of Proposition 3.2 will changeif we assume that ˙ x is a name of an arbitrary set of ordinals. This shows that everyset of ordinals lies in an intermediate model given by c ∩ S B for some disjointapproximation. Another way to see this fact is to note that L [ c ] is, after all, onlya Cohen extension by a single real. So every set of ordinals is constructible from asingle real. Corollary 3.4. M | = ¬ AC . Or L -generic filter, to be explicit. UIDE TO THE BRISTOL MODEL 13
Proof.
Suppose that M | = AC , then there is a set of ordinals A which codes R M (e.g.by stacking the real numbers one after another, or by the usual coding of a set into aset of ordinals). Therefore there is r ∈ M such that R M = R L [ r ] . By Proposition 3.2we see this is impossible, indeed, if B is a disjoint approximation for which r has a S B -name, and A α is such that A α ∩ S B is finite, then c ∩ A α / ∈ L [ r ]. (cid:3) Of course, we can prove directly that R M cannot be well-ordered in M , and thisargument can be found in the proof of Theorem 2.7 in [8].We can see Proposition 3.2 as somehow indicating not only that the reals of M are generated by countable parts of A , but in fact if h T α | α < ω i is a towergenerated by A , then we actually have that R M is the increasing union of R L [ c ∩ T α ] .This was the original approach of the Bristol group.At this point, one might expect that the decoded sequence is h c ∩ A α | α < ω i or somehow h c ∩ T α | α < ω i . But of course, this is not the case. For starters,we want to somehow “fuzzy out” some of the information as to guarantee that c isnot constructible from the sequence. So instead of c ∩ A α , we will look at R L [ c ∩ A α ] .But more importantly, the decoded sequence is not even in M . Indeed, if we wantthis sequence to play the role of the Cohen real in the next step, that means thatit needs to be forced into M instead.Let us begin by understanding R L [ c ∩ A α ] . In what way does this set “fuzzy out”some information? Well, for one, c ∩ A α is not the obvious real from which weconstruct this model. Indeed, any finite modification would work, and many more.In fact, any π ∈ G for which ι ( π )( α ) = α will satisfy that π ˙ c A α is a name generatingthe same set of reals, as it is c ∩ A α up to a permutation of A α and a finite set.We say that a name ˙ x is an almost A -name if there exists B such that A = ∗ B and ˙ x is a B -name. We now define˙ R α = { ˙ x | ˙ x is a decent almost A α -name } • . Proposition 3.5.
For every π ∈ G , π ˙ R α = ˙ R ι ( π )( α ) . In particular, ˙ R α ∈ HS forall α , and { ˙ R α | α < ω } • ∈ HS as well.Proof. Observe that π “ P ↾ A = P ↾ π “ A . Therefore if ι ( π )( α ) = β we have that π “ A α = ∗ A β , and so an almost A α -name is moved to an almost A β -name, so π ˙ R α = ˙ R ι ( π )( α ) as wanted. We now have that { A α } is a disjoint approximationfor which fix( { A α } ) ⊆ sym( ˙ R α ), and thus witnessing that ˙ R α ∈ HS , and indeed G = sym( { ˙ R α | α < ω } • ) as wanted. (cid:3) Similar arguments as we have seen so far also prove the following statement.
Proposition 3.6. h ˙ R α | α < ω i • / ∈ HS , but for every countable I ∈ [ ω ] <ω ∩ L , h ˙ R α | α ∈ I i • ∈ HS . (cid:3) And here we arrive to the key point. Let ̺ denote the sequence h R α | α < ω i and let ˙ ̺ denote its name. We will also write ̺ I and ˙ ̺ I when I ⊆ ω for therestriction of the sequence to I , similar to ˙ c A . Indeed, ̺ is going to be the decodedsequence, and it is of course going to be M -generic, but we need to find a suitablepartial order. Proposition 3.6 provides us with a good clue: every initial segmentof this sequence is in fact in M , so we can “safely” approximate this sequence.It will be somewhat more convenient to use subsets of ω , as we did with theCohen forcing, rather than proper initial segments. This makes it easier to talkabout A -names and almost A -names. And again for convenience (and so we canclaim productivity, of course), we are also going to limit ourselves to subsets of ω which are already in L . Proposition 3.7.
Let ˙ Q denote { π ˙ ̺ I | π ∈ G , I ∈ [ ω ] <ω } • . Then ˙ Q ∈ HS , andindeed (cid:13) HS ˙ ̺ is HS -generic. In other words, ̺ is M -generic for Q . The idea behind Q is that we want thesmallest “reasonable” set which contains our generic filter (i.e. partial approxima-tions of ̺ ), and the easiest way to do that is to simply apply all permutations andobtain a set. But the true intuition behind Q , and really behind the whole decodingapparatus, comes from understanding π̺ I .The model M knows of the set R = { R α | α < ω } , it just does not know awell-ordering of this set. And we are trying to remedy that. As we know alreadyevery π ∈ G implements a permutation of ω , which induces a permutation of R moving countably many points. So π shuffles R and thus modifies the range of ̺ I . But R is an extremely impoverished set as far as M is concerned. This is notparticularly important for the construction, and can be skipped entirely, but it isan interesting fact. Proposition 3.8. M | = R is a strongly ℵ -amorphous set. That is, R cannot bewritten as a union of two uncountable sets, and every uncountable partition of R has countably many non-singleton cells.Proof. Suppose that ˙ X, ˙ Y ∈ HS and p (cid:13) HS “ ˙ X, ˙ Y ⊆ ˙ R and are uncountable”. Let B be a disjoint approximation such that fix( B ) ⊆ sym( ˙ X ) ∩ sym( ˙ Y ) and such thatdom p ⊆ S B . By uncountability, we can extend p to some q for which there are α, β such that:(1) q (cid:13) HS ˙ R α ∈ ˙ X and ˙ R β ∈ ˙ Y .(2) A α ∩ S B and A β ∩ S B are finite.By enlarging one of the sets in B , if necessary, we can also assume that dom q ⊆ S B as well. We can now find a permutation π ∈ fix( B ) such that ι ( π ) is the 2-cycleswitching α and β .Applying π to the first property of q we have that πq (cid:13) HS π ˙ R α ∈ π ˙ X, π ˙ R β ∈ π ˙ Y .But since π ∈ fix( B ) we have that πq = q and π ˙ X = ˙ X and π ˙ Y = ˙ Y . This meansthat q (cid:13) ˙ R β ∈ ˙ X and ˙ R α ∈ ˙ Y . In particular q forces that ˙ X and ˙ Y are not disjoint.Next we want to prove that every partition of R is almost entirely singletons.We will only sketch the idea behind the argument. If ˙ S ∈ HS and p (cid:13) HS “ ˙ S is apartition of ˙ R into uncountably many cells”, let B be an approximation such thatfix( B ) ⊆ sym( ˙ S ). Pick α, β as above, so that we may switch between them withoutinterfering with B , and we can implement the 2-cycle ( α β ) without changing anygiven condition. This means that any point that was in the same cell as α musthave moved to the cell containing β . But we only moved two points, so α and β must have been isolated as singletons. And therefore the only non-singleton cellscome from a partition of B itself. (cid:3) Remark 3.9.
The above implies that every permutation of R in M only movescountably many points, of course, as the orbits define a partition. Some of theseare new, as they can be encoded by some generic real, but this is irrelevant. We canprove that every permutation of R in M only moves countably many points with adirect argument in the style of Theorem 2.2: given a name for a permutation ˙ f and B a disjoint approximation such that fix( B ) ⊆ sym( ˙ f ), the c.c.c. condition ensuresthat there is δ such that for any α for which A α ∩ S B is infinite, ˙ R δ is not in thesame orbit as ˙ R α . But now we can utilise the same strategy as we did before andmove α to some other ordinal with a similar property, and therefore showing thateither f is the identity on a cocountable set, or it is constant there.Getting back to the matter at hand, we want to prove that ̺ is M -generic.Namely, if D ⊆ Q is a dense subset and D ∈ M , we want to prove that there issome α < ω such that ̺ α ∈ D . In [8] we prove this by proving a technical lemma UIDE TO THE BRISTOL MODEL 15 about names of dense open sets, Lemma 2.12. In the paper we use this lemmaalso as a means for proving that Q is σ -distributive, and therefore does not addany new reals to the model (and as a corollary, it does not force AC back into theuniverse somehow), which also finishes the proof that ̺ is indeed the sequence weare looking for.We will prove the genericity of ̺ using a simplified version of Lemma 2.12, andprovide a separate argument for the distributivity. Proposition 3.10. ̺ is M -generic. In other words, if ˙ D ∈ HS and p (cid:13) “ ˙ D is adense open subset of ˙ Q ”, then there is some η such that p (cid:13) ˙ ̺ η ∈ ˙ D .Proof. Let ˙ D be a name as above, and let B be a disjoint approximation such thatfix( B ) ⊆ sym( ˙ D ). Let p (cid:13) “ ˙ D is a dense open subset of ˙ Q ”. Let α be large enoughsuch that if A ξ ∩ S B is infinite, then ξ < α . Our strategy is to find “enough”extensions of ˙ ̺ α so that one of them will be both ˙ ̺ η , and in ˙ D .First we prove the following claim: suppose that q ≤ p and q (cid:13) π ˙ ̺ A ∈ ˙ D , where α ⊆ A and ι ( π ) ↾ α = id. In other words, π ˙ ̺ A is an extension of ˙ ̺ α which lies in˙ D . Then q (cid:13) ˙ ̺ A ∈ ˙ D . Of course, this is true because there is some τ ∈ fix( B ) suchthat τ q = q and ι ( τ ) = ι ( π ) − .The first property, of course, is trivial. The second one is also easy to obtainbecause π implemented the identity up to α , and therefore ι ( π ) − will also be theidentity up to α , and thus we can implement it using an automorphism in fix( B ).This completes the proof of the claim since τ q = q (cid:13) τ π ˙ ̺ A = ˙ ̺ ι ( τ ) ◦ ι ( π )“ A = ˙ ̺ A ∈ τ ˙ D = ˙ D. In turn, this is enough to prove the genericity of ̺ : find a maximal antichain below p of conditions which decide some π ˙ ̺ A as above, and by openness we can assumeeach such A is in fact an ordinal. Let η be the supremum of this countable set ofordinals, and we have that p (cid:13) ˙ ̺ η ∈ ˙ D . (cid:3) The final claim is that Q does not add new reals to M . This, as we remarked,ensures that M [ ̺ ] = L [ c ]. It has an added effect that Q is not adding any new setsof ordinals, or any countable sequences of ground model objects. In [8] the proofutilised a stronger version of the above proof which lets us intersect a countablesequence of dense open sets. Here we take a slightly different approach. Proposition 3.11. M | = “ Q is σ -distributive”. Namely, given h D n | n ∈ ω i ∈ M such that each D n is a dense open subset of Q , then T n ∈ ω D n is dense.Proof. In L [ c ], Q is naturally isomorphic to Add( ω , L . In L this forcing is σ -closed, and so in L [ c ] it is still distributive. If h D n | n ∈ ω i is a sequence of denseopen sets, then its intersection is still dense in L [ c ], and therefore in M . (cid:3) Outline of successor steps.
The first two steps are fairly indicative of thestandard successor step. The main change, of course, is that we need to understandwhat replaces R , L , and c . We have a hint as to what replaces c , namely, thesequence ̺ that we ended up with after forcing with Q . We also have an idea onwhat to replace R with, that would be P ( R ), and L is to be replaced by M itself.We can make this much clearer if we recast the example above by replacing R with V ω +1 . We can also replace c with a sequence of elements of V ω , of course,but this seems to needlessly complicate things. After all, the case of ω is separateanyway. There is a minor mistake in the statement of the original lemma, see subsection 5.1 for detailsand corrections. The approach we take here is mentioned in a remark at the end of §2 of [8].
For a more uniform approach, we denote by M α the α th step in the construction,which is a model of ZF intermediate between L and L [ c ]. We will also write ̺ α todenote the generic sequence for Q α , the α th forcing. So Q is Cohen forcing and ̺ is c itself.At each successor step we have M α +1 defined from ̺ α , which was the M α -genericsequence for Q α . We will assume that while ̺ α / ∈ M α +1 , for every ξ < ω α +1 , ̺ α ↾ A αξ ∈ M α +1 , where A αξ is the ξ th member of the permutable family we fixed inadvance. Moreover, for every I ∈ [ ω α +1 ] <ω α +1 ∩ L , h ̺ α ↾ A αξ | ξ ∈ I i is in M α +1 .We now want to define Q α +1 and ̺ α +1 . For this we replace R L [ c ∩ A ξ ] that wehad in the first step with V M α +1 [ ̺ α ↾ A αξ ] ω + α +2 , let us denote this as R ξ for now. The restis more or less the same as above, relying, of course, on the recursive fact that anyprevious M β were defined much in the same way as we are defining Q α +1 , ̺ α +1 ,and M α +2 .Let Q α +1 denote the set of approximations of ̺ α +1 = h R ξ | ξ < ω α +1 i whosedomains are in L . We are being vague, of course, as to what counts as “approxima-tion” in this context. The idea is that we may permute the different R ξ amongstthemselves using a permutation of ω α +1 which is coming from L .The lemmas in the general case are exactly the same as we had before. Thegenericity of ̺ α +1 is proved by the same argument as Proposition 3.10, and whilethe distributivity argument is also similar, it is worth writing down. Proposition 3.12. M α +1 | = Q α +1 is ≤| V M α +1 α | -distributive. In particular, no newsets of rank α + 1 are added.Proof. As in Proposition 3.11, L [ c ] | = Q α +1 ∼ = Add( ω α +1 , L , the latter of whichis ≤ℵ α -distributive in L [ c ]. Suppose now that { D x | x ∈ V α } ∈ M α +1 is a familyof dense open subsets of Q α +1 . By the c.c.c. of the Cohen forcing, and the fact weonly add a single real, V M α +1 α has the same cardinality as V Lα , which by GCH is ℵ α .Therefore T { D x | x ∈ V M α +1 α } is dense in L [ c ] and thus in M α +1 .Suppose ˙ f is a Q α +1 -name in M α +1 and q (cid:13) Q α +1 ˙ f : V α → ˇ M α +1 . Set D x as { q ′ ≤ q | q ′ decides ˙ f (ˇ x ) } , then every D x is dense and open below q . By the densityof their intersection, q has an extension q ′ ∈ T { D x | x ∈ V α } , which has to decidethe entire function ˙ f . In particular, no new subsets of V α are added. (cid:3) Finally, we define M α +2 as the symmetric extension obtained by applying thederived group and filter using the permutable family A α +1 . If we now consider V M α +1 [ ̺ α +1 ↾ A α +1 ξ ] ω + α +2 , this set is M α +2 . Indeed, every proper initial segment of ̺ α +2 isin M α +2 , where ̺ α +2 is the sequence of these V ω + α +2 , and it has length ω α +2 . Wecan now show that it is M α +2 -generic, etc., and thus all shall prosper.The successor case can be found in [8] as §4.4, as well as §4.7 and §4.10 forsuccessor of limit iterands. We are not separating the successor of limit steps inthis text as the idea is the same with very minor variations, so as far as outlinesgo, there is essentially no difference.3.3. Outline of limit steps.
The limit step is divided into two parts. We haveto contend with the iteration at a limit step, i.e. the finite support limit of theprevious steps, and we have to deal with the limit step itself. An observant readerwill notice that we did not utilise the full power of the framework of iteratingsymmetric extensions until now. Indeed, as far as successor steps are concerned anyautomorphism coming from coordinates before α itself will implement the identityfunction on the ( α + 1)th iterand, rendering it moot.It is here, at the limit, where we need to utilise the machinery as a whole. In fact,this machinery will do most of the heavy lifting at this stage. By Proposition 3.12 UIDE TO THE BRISTOL MODEL 17 we have that the rank initial segments of the universe are stabilising, indeed for any β > α we have V M α +1 ω + α +1 = V M β ω + α +1 . The work left at this stage is making sure thatthe generic sequences we collected thus far are symmetrically generic, and settingup the stage for the limit step iterand. So it is a good idea to understand the limitstep as a whole before proceeding to the details.The main idea is that limit steps coalesce the information we have up to thatpoint. Arriving to the limit is easy, as we said, the machinery of productive itera-tions is working for us there. But how do we proceed now? We are limited by twofactors that we need to ensure continue to hold when we deal with the α th step:(1) V ω + α is stable. That is, no new sets of low rank are added, and(2) whatever we do is coherent with the other limit steps.One simple way of ensuring this is by taking products of previous successor steps.This way, if α < β are two limit ordinals, then Q α is going to be, in some sense, arank initial segment of Q β . But we can think about this from a different angle.At each step, we gathered V ω + ξ of various intermediate models, for ξ < α ,our limit ordinal. But these are smoothed out, in a sense, as we progress up thehierarchy, as each V M ξ +2 ω + ξ +1 contains each V M ξ +1 ω + ξ . But what if we could pick just onesequence, and remember it? In that case we are not going to add bounded sets to V ω + α , at least not if we are being careful, and instead we only add this sequence.This idea should seem somewhat familiar to readers of all walks of set theory. Afterall, if we want to add a new subset to ℵ ω , it is easy to add Cohen subsets to each ℵ n first, and then choose a point from each one, creating a new cofinal sequence. Similar ideas, in one way or another, show up through Prikry-style forcings as well.The coherence of limit steps has another very important use for the limit iterand.One of the subtle, but important properties we used in the successor steps is upwardhomogeneity.
Definition 3.13.
We say that a two-step iteration P ∗ ˙ Q is upwards homogeneous if whenever h p, ˙ q i and h p, ˙ q ′ i are two conditions, there is an automorphism π of P that respects ˙ Q and such that πp = p and p (cid:13) π ˙ q = ˙ q ′ . In other words, we canmove conditions in ˙ Q by automorphisms of P . In the context of iterating symmetricextensions we require that π comes from the relevant automorphism group.So we want for the limit step that the iteration P α can move about conditions in Q α . If P α is a finite support iteration, then either Q α needs to have some finitaryflavour to its conditions, or somewhere along the iteration we had to condense theconditions from finitary to infinitary. Indeed, this is the very meaning of “coalesc-ing” the successor steps.To sum up, at the limit step of the iteration we use the properties of the iterationso far to ensure that the rank initial segments of the models stabilise, and indeedthat the sequence of generic sequences is symmetrically generic for the iteration.We then want to have a forcing such that the sequences which lie in the productof the successor-step generics combine to form a generic for it, here the permutablescales will come in naturally, as we are concerned with products of increasinglylonger sequences modulo the bounded ideal.We return to the context of the Bristol model’s construction. We denote by P α the iteration up to α and by Q α the α th iterand, as we did before, and for now wewill assume that those iterands were defined also for limit steps. If this proves tobe somewhat confusing, the section can be read twice, first assuming α = ω . Fact 3.14.
For every α , P α ∗ ˙ Q α is upward homogeneous. When forcing like this in the context of
ZFC these cofinal sequences are added automatically,of course, but if one does a symmetric iteration, the cofinal sequences are not added. Then onecan consider such a forcing in a more material sense.
We have seen this for the case of α being 0 or a successor. We will see the restof the cases in this section as we progress through it. But for now it is easier totake this as a working assumption.
Proposition 3.15.
Let α be a limit ordinal, then h ̺ β | β < α i is symmetrically L -generic for P α . This is essentially Proposition 4.4 ( α = ω ) and Proposition 4.15 ( α an arbitrarylimit ordinal) in [8]. We will prove this statement in subsection 5.2, as the originalproof had a minor gap that needs to be corrected anyway.But this means that we can understand the limit iterand fairly well now. Weknow that for β < α , V M β ω + β +1 = V M α ω + β +1 , and that not only we have a model of ZF which lies within L [ c ], but that it is in fact an iteration of symmetric extensions,which means that we understand exactly the objects which lie within it and thetruth value of statements about these objects from a forcing-theoretic point of view.We are now free to examine the iterand Q α .As we reiterate time and time again, we want to ensure that no sets of rank ω + α are added. In the successor steps we did that by making sure that Q α is sufficientlydistributive. For the limit step this will pose a problem. If we are to continue withour successor steps, then the ( α +1)th step needs to have a sequence of length ω α +1 .But without adding any sequences of length ω α , this would mean that the sequencemust have “mostly existed” already. So the forcing cannot be <ω α -distributive, letalone ≤| V ω + α | -distributive. In fact, if our plan is to add sequences of length α ofsets of the form V ω + β of some inner model, then at stages where cf( α ) = ω we musthave added an ω -sequence.The solution, as it turns out, is to not be distributive at all, but ensure that afterapplying the symmetric part of the step (rather than just the generic extension) wemanaged to remove any new set in V ω + α . So even if we do not have a distributiveforcing, we at least preserve the rank initial segments of the universe.We define ̺ α , where α is a limit, as a “copy of Q SC( ω α ) L ”. This means that forevery f ∈ Q SC( ω α ) L we define ̺ α,f to be the sequence h ̺ β +1 ( f ( β + 1)) | β < α i ,and ̺ α is defined as h ̺ α,f | f ∈ Q SC( ω α ) L i .How should we define Q α , then? The idea is always “bounded approximations”,but having the virtue of being a “two-dimensional object” this means that bound-edness has two sides to it. Let us first deal with the one-dimensional counterparts: ̺ α,f .If A is a subset of α , we write ̺ α,f ↾ A = h ̺ β +1 ( f ( β + 1) | β ∈ A i , and so ourrecursive hypothesis tell us that ̺ α,f ∈ M α for every f , and this lets us define Q α,f as the approximations of ̺ α,f . More rigorously, recall that we have defined the sym-metric iteration P α , along with the direct limit of the generic semidirect products, G α . Then ˙ Q α,f is the forcing ordered by reverse inclusion on the interpretation of (cid:8)R ~π ˙ ̺ α,f ↾ A (cid:12)(cid:12) sup A < α, R ~π ∈ G α (cid:9) • . For E ⊆ Q SC( ω α ) we may now define ̺ α ↾ E = h ̺ α,f | f ∈ E i , and so if E isbounded in the product, i.e. there is f ∈ Q SC( ω α ) such that for all g ∈ E and forall β < α , g ( β + 1) < f ( β + 1), and A is a bounded subset of α , our conditionsare going to be approximations of ̺ α , up to permutations of course, of the form h ̺ α,f ↾ A | f ∈ E i , which we denote by ̺ α ↾ ( E, A ). And so Q α is given by the name (cid:8)R ~π ˙ ̺ α ↾ ( E, A ) (cid:12)(cid:12) E, A are bounded and R ~π ∈ G α (cid:9) • . Proposition 3.16. P α ∗ ˙ Q α is upwards homogeneous. To be absolutely correct, we only talked about successors of 0 or other successor ordinals,but the successor of a limit will be just the same as before.
UIDE TO THE BRISTOL MODEL 19
We do not prove this statement here, but the idea is to simply utilise the upwardshomogeneity of the previous steps and “correct” the coordinates one by one. Onemight ask how do we deal with the case where α > ω , as a condition has seeminglyinfinitely many non-trivial coordinates. And the answer, as we repeatedly mentionhere, is utilising the previous limit cases where we condense this infinite amount ofinformation, also in the form of G α being a subgroup of the full support product Q β<α G β +1 . The complete proof can be found as Propositions 4.5 (for α = ω ) and4.15 in [8]. The action of G α is coordinatewise, and it is important to stress at thispoint that we have this action where the initial segments of ̺ α are “actual objectsin M α ”, so this is not applying automorphisms of a forcing, but rather applyingpermutations of each ω β +1 . Proposition 3.17. ̺ α is M α -generic for Q α . This again follows the same pattern as the successor steps, although here thereis a notable complication in the case where α > ω . We will outline the proof.
Proof.
Let ˙ D ∈ IS α be a name for a dense open subset of ˙ Q α , and let ~H be anexcellent support witnessing that ˙ D ∈ IS α . For each β < α , H β is a group of eitherthe form fix( B β ) when β is 0 or successor, or K η β ,f β when β is itself a limit. Wecan use these to define bounded sets E and A such that whenever ̺ α ↾ ( E, A ) isextended to a condition in ˙ D , we may permute this extension using the upwardshomogeneity without changing ˙ D .In the case where α = ω , the set E is simple. For each n < ω we let B n bea disjoint approximation such that fix( B n ) = H n , then E = Q n<ω dom B n , wheredom B n = I such that B n = { B ξ | ξ ∈ I } . Let A be the set { n < ω | H n = G n } ,which is also bounded by the virtue of ~H being excellent, and we have ourselvesthe condition ̺ ω ↾ ( E, A ).In the case where α > ω we need to take into consideration some limit point δ < ω , such that either δ + ω = α , or δ is the smallest limit ordinal such that ~H istrivial above δ . We call such δ the condensation point of ~H . So above δ there can beat most finitely many non-trivial coordinates, and only in the case where δ = α + ω .We then let E β for β < δ be decided by H δ = K η δ ,f δ by taking E β = f δ ( β ), andabove δ we do as with the case of ω .Now we may proceed as before, using the fact that any extension of ̺ α ↾ ( E, A )would have the form R ~π ̺ α ↾ ( E ′ , A ′ ), where π β will necessarily have π β ↾ E β = id,so we may find the needed automorphisms in ~H to complete the proof as in thesuccessor case. (cid:3) Now that we only need to take care of the preservation of V M α ω + α . Indeed, M α [ ̺ α ]contains the sequences ̺ β +1 for β < α , all of which have rank smaller than ω + α .Luckily, the symmetries of Q α will help us get rid of these unwelcome sets. Definition 3.18.
Working in M α , if ˙ x is a Q α -name we say that it is bounded by f ∈ Q SC( ω α ) L if whenever R ~π ̺ α ↾ ( E, A ) (cid:13) ˙ y ∈ ˙ x , then we may replace E by E ∩ f ↓ , where f ↓ = { g ∈ Q SC( ω α ) L | ∀ β, g ( β ) < f ( β ) } . Similarly, if β < α we saythat ˙ x is bounded by β if we can replace A by A ∩ β . Theorem 3.19.
Suppose that ˙ x ∈ M α is a Q α -name such that every name appear-ing in ˙ x is ˇ y for some y ∈ M α . (1) If ˙ x ∈ HS F α and K η,f ⊆ sym( ˙ x ) , then ˙ x is bounded by f . In the original paper the proof of α = ω is Lemma 4.7, and there is a minor mistake in theproof: B n should be dom B n − . Lemma 4.20, which is the general claim, has a correct proof. This also highlights the importance of the assumptions in Theorem 2.8 being only about thesymmetric extensions having the same V α . (2) If (cid:13) rank( ˙ x ) < ˇ ω + ˇ α , then ˙ x is bounded by some β < α . In combination we have that if ˙ x is a symmetric name for a set of small rank,then all of its elements are decided by conditions with a uniform bound, meaning ˙ x is equal to an object in M α . The proof of (1) is the standard homogeneity argument,and we have used it before in Theorem 2.2, so we will only outline the proof of (2). Proof.
Suppose that ˙ x ∈ IS α +1 , we denote by [ ˙ x ] the projection of the name to IS α .That is, [ ˙ x ] is a name in IS α which is interpreted in M α as a name in HS M α F α . Bythe assumption that each name appearing in ˙ x is of the form ˇ y for some y ∈ M α ,we may assume that [ˇ y ], which appears in [ ˙ x ] (in the broad sense of the term) is aname in IS β for β such that ω + β is an upper bound on the forced rank of ˙ x .Let β < α be large enough such that ~H is trivial above β , where ~H witnessesthat [ ˙ x ] ∈ IS α . Now we can use automorphisms which only move coordinates above β to move any names of conditions, R ~π ̺ α ↾ ( E, A ), by changing their “content above β ” to any value. Thus, we may conclude that we may reduce A to A ∩ β . (cid:3) The complete proof of the theorem appears as Lemma 4.10 and Lemma 4.27 in[8]. This almost completes our decoding apparatus. We only need to worry aboutthe ̺ α +1 now.We define R ξ for ξ < ω α +1 to be V M α [ ̺ α,fαξ ] ω + α +1 . That is, we use ̺ α,f αξ as the “guide”for a new sequence in V ω + α +1 . Those who kept track can guess now that ̺ α +1 is h R ξ | ξ < ω α +1 i . We utilise the fact that the scale is permutable to ensure that anybounded part will be in M α +1 , as well as the rest of the permutability apparatusto ensure the upwards homogeneity.This is also the point where we see why the definition of Q α works in general,despite P α being a finite support iteration. The ̺ α,f are initial segments of those ̺ β,f ∗ that come in the future, and even if cf( α ) > ω , we still end up with what wewanted to have.With this we finish the discussion on the decoding apparatus. This is the maintechnical part of the construction. It is our sword and shield in our journey down-wards. Now that we have that, we may venture deeper into the Hadean adventurethat is the Bristol model.4. Cerberus, the three-headed guard of the underworld
Much like Cerberus, the construction of the Bristol model can be seen as athree-headed dog, guarding the realm of the underworld of models of ZF . Thethree heads of Cerberus represent the three causes of strife: nature, cause, andaccident. The three heads of the Bristol model can be seen as representing threetypical approaches to its construction: nature, cause, and accident. All lead us tothe same construction, and indeed when getting down to brass tacks, the detailsbecome suspiciously similar in each approach. But the presentation of one may bemore appealing to some readers over another.Nature is the way in which the original group in Bristol went about to define themodel: defining the von Neumann hierarchy by hand, and defining a model L ( X )where X is a class of sets in the Cohen extension. Cause is the way in which [8]presents the construction: defining an iteration of symmetric extensions, and findinga symmetrically generic filter at each step of the way, thus constructing an iterationand the von Neumann hierarchy of the Bristol model in tandem. Finally, Accident isthe way in which we define a productive iteration of symmetric extensions, we studythis iteration in an abstract manner, and then we find that by “complete accident”we can find all the symmetrically generic filters inside the Cohen extension. UIDE TO THE BRISTOL MODEL 21
Fix a Bristol sequence, a permutable family A α = { A αξ | ξ < ω α +1 } for α = 0or a successor, and a permutable scale F α = { f αξ | ξ < ω α +1 } for α limit. We willdefine M , the Bristol model, in three different, yet equivalent ways. We will arguethat it is a model of ZF + ∀ x ( V = L ( x )).4.1. Nature.
In here we will define the Bristol model one step at a time by definingits von Neumann hierarchy in L [ c ]. For this purpose it would be easier at timesto use an increasing sequence modulo bounded sets, rather than the permutablefamilies. Since the two are equivalent, we let { T αξ | ξ < ω α +1 } denote a sequenceobtained from A α .We define the ̺ α and V Mω + α in tandem, and we will omit the M from the su-perscript where possible, as the definitions will be complicated enough. Let ̺ = c and, as V Mω is just V Lω = L ω , it is defined. Suppose that for α , ̺ α and V ω + α weredefined.If α is 0 or a successor ordinal, define V ω + α +1 = [ ξ<ω α +1 V L ( V ω + α ,̺ α ↾ T αξ ) ω + α +1 , (1) ̺ α +1 = D V L ( V ω + α ,̺ α ↾ A αξ ) ω + α +1 (cid:12)(cid:12)(cid:12) ξ < ω α +1 E . (2)If α is a limit ordinal, we define V α = S β<α V ω + β +1 , and we define ̺ α = Q β<α ̺ β +1 .And as before we write ̺ α,f to indicate the “thread” of the function f in thisproduct. To define the next step we need an analogue of the T αξ , we write ̺ α,f todenote the sequence of V L ( V ω + β,̺ β ↾ T βf ( β ) ) ω + β +1 for β < α .And we now define V ω + α +1 = [ ξ<ω α +1 V L ( V ω + α ,̺ α,fαξ ) ω + α +1 , (3) ̺ α +1 = (cid:28) V L ( V ω + α +1 ,̺ α,fαξ ) ω + α +1 (cid:12)(cid:12)(cid:12)(cid:12) ξ < ω α +1 (cid:29) . (4)Finally, M = S α ∈ Ord V Mω + α +1 . The handwritten notes passed on to us by someof the members of the Bristol group indicate that the original line of thought wasabout R M and P ( R ) M , rather than V ω +1 and V ω +2 . The arguments, moreover, asto why V ω +1 = V L ( V ω +1 ,̺ ↾ T αξ ) ω +1 , i.e. why no reals are added when defining P ( R ) M ,was not written down in these notes. Instead it merely suggests “condensation”.While restoring the original arguments is certainly beyond us, these would beequivalent, more or less, to the arguments we presented at the first step of section 3.4.2. Cause.
In here we will define the Bristol model in tandem: define a symmetricextension, find it a generic in L [ c ], define a symmetric extension again, repeat adordinalum.This definition was used in [8], and you can very clearly see that this is thedefinition that we have in mind, as it very much dominates the approach given insection 3. As such, we really have done all the work ahead of time.We define M α , ̺ α as in the decoding apparatus, i.e. as the generic objects for thesymmetric iteration. We now define M simply as S α ∈ Ord M α = S α ∈ Ord V M α +1 ω + α .4.3. Accident.
In here we will first define a class-forcing, and then argue that wecan just happen to find symmetrically generic filters in L [ c ].Despite being the guiding view on the construction of the Bristol model, theactual argument for M | = ZF in [8] is the one rising from this approach, as it is less“ad-hoc” and more structural and general. This is also the reason why the proof of the distributivity of successor steps in the original paper was proved directly,rather than the approach used in this paper, which is more in line with “Cause”.We first define the iteration. Let P = { } and Q = Add( ω, G and F arethe derived group and filter. Finally, ˙ ̺ is the canonical name for the Cohen real.Suppose that P α is the finite support iteration of the symmetric extensions de-fined so far. In the case where α = β + 1, we define ˙ Q α as the name (cid:26)R ~π (cid:28) ˙ V V [ ̺ β ↾ A βξ ] ω + α (cid:12)(cid:12)(cid:12)(cid:12) ξ ∈ I (cid:29) • (cid:12)(cid:12)(cid:12)(cid:12) R ~π ∈ G α , I ∈ [ ω α ] <ω α (cid:27) • . Where ˙ V V [ x ] ω + α denotes the name of the rank initial segment of the extension of IS β by the set x . We then define ˙ ̺ α as the name for the generic of ˙ Q α .If α is a limit ordinal, we define P α as the finite support iteration. We nextdefine ˙ Q α in the same spirit. For f ∈ Q SC( ω α +1 ) we define˙ Q α,f = { R ~π h ˙ ̺ β +1 ( f ( β + 1)) | β ∈ A i | R ~π ∈ G α , sup A < α } • , and we then define ˙ Q α by adding the additional dimension of a bounded subset of Q SC( ω α ). As before, G α and F α are the derived group and filter. Finally, ˙ ̺ α isthe name of the generic filter. Moving on, the definition ˙ Q α +1 and ˙ ̺ α +1 is in linewith what we have done so far.The properties of the decoding apparatus imply the distributivity of each iterand.We need to be a bit more careful here, as we are not allowed to argue in L [ c ],instead we carry on a recursive hypothesis that for successor steps P α satisfies achain condition that allows us to prove that ˙ Q α will be isomorphic to Add( ω α , L .The conditions of Theorem 2.8 are therefore satisfied. This implies that if G isany symmetrically L -generic for P , the class-length iteration, then IS G | = ZF , atthe very least.By recursion we can now show that each ˙ ̺ α has a symmetric interpretation inside L [ c ], and therefore we may find an interpretation of the model which is intermediatebetween L and L [ c ].4.4. The basic properties of the Bristol model.Proposition 4.1. M | = ZF .Proof. We have two slightly different proofs here, the first we mentioned in the“Accident” approach, utilising Theorem 2.8. The second approach works well for“Nature” and “Cause”.Since M is very clearly a transitive subclass of L [ c ] that contains all the ordinals,it is enough to verify that it is closed under Gödel operations and that it is almostuniversal to conclude that it is a model of ZF . The first part is easy: if x, y ∈ M ,there is some M α , or some large enough V ω + α in the “Nature” approach, such that x, y ∈ M α and therefore { x, y } , x × y , etc. are all in M α and therefore in M .The second part is the fact mentioned at the end of “Nature” about condensation;or in the case of “Cause” this follows from the fact that for each α , V M α ω + α = V M β ω + α for all α ≤ β , and therefore the model is almost universal. (cid:3) Proposition 4.2. M | = ∀ x ( V = L ( x )) .Proof. If x ∈ M , then there is some α such that x ∈ M α , and therefore L ( x ) ⊆ M α .But since M α ( M α +1 ⊆ M , we have that L ( x ) = M . In the “Nature” approachwe need to be slightly more careful, as we have to verify that if x ∈ V α , then V α +1 / ∈ L ( x ). One way of doing this would be to argue that ̺ α ↾ A αξ are all genericover L ( V α ), and therefore cannot be elements of this model. (cid:3) This is mentioned after Theorem 2.8, and proved as Theorem 9.4 in [9].
UIDE TO THE BRISTOL MODEL 23
We will see other ways of deducing Proposition 4.2 later on, in ways that will besignificantly more informative and more general.5.
Errata to “the Abyss”
Here we correct some minor gaps and mistakes in the original paper. Thesemistakes repeat, once with the cases of α = 0 and α = β + 2, and once with thecases of α = ω and a general limit ordinal.We repeat the remark we made in footnote 19: in [8] the proof of Lemma 4.7,dealing with the genericity of ̺ ω , contains a typo, whereas B n is defined as S B n ,and it should have been defined as dom B n − . This is somewhat inconsequential,as the more general proof where α is a limit, in Lemma 4.20, is written correctly.We also point out that there is a minor mistake in the construction’s inductionhypothesis, where the requirement that P α satisfies the ℵ α -c.c. will not hold forlimit ordinals, only for 0 and successor ordinals. We can modify this by defining anotion of “symmetric chain condition”, but this is an unnecessary complication.We simply do not use the chain condition assumption for limit iterands.5.1. Corrections to Lemma 2.12/4.1.
Lemma 2.12 is read in the context of thefirst steps. Namely, P is Cohen forcing, ˙ Q is the name of the forcing adding ̺ ,denoted in that lemma by σ . As such HS is the class of hereditarily symmetricnames defined in the very first step. Lemma ( [8] , Lemma 2.12).
Suppose that ˙ D ∈ HS and p (cid:13) ˙ D ⊆ ˙ Q is a denseopen set. There is some η < ω such that for every π and A such that p (cid:13) π ˙ σ A ∈ ˙ D and η ⊆ A , if τ ˙ σ A is a condition such that p (cid:13) π ˙ σ η = τ ˙ σ η , then p (cid:13) τ ˙ σ A ∈ ˙ D aswell. This lemma was used twice. The first consequence is the genericity of ̺ , andthe second is to show that ˙ Q is σ -distributive. The proof is very similar to theproof of Proposition 3.10. Nevertheless, when we have two permutations π and τ ,we want to move τ so that it agrees with π .For this we need not only that ι ( τ ) ↾ A = ι ( π ) ↾ A , but also that their inverses agreeon A . In the case of the genericity of ̺ we take π = id, in which case this propertyholds trivially. Indeed, this is the very strategy of the proof of Proposition 3.10.To prove that ˙ Q is σ -distributive we interpret the lemma above as saying thatfor a condition p ∈ P , there is an ordinal η , such that deciding whether a conditionfrom ˙ Q , whose domain is at least η , lies in ˙ D depends only on the permutation π . We then utilise the fact P is c.c.c. to show that this η can be bound uniformlydepending on ˙ D . Now, if ˙ D n are names for a sequence of dense open sets, wecan uniformly bound all of them by some η . Now if we take any condition whosedomain is at least this η , we may extend it into each ˙ D n , but this means that suchextension lies, in fact, in all the ˙ D n simultaneously, and therefore the intersectionis also dense.Once we add the condition that ι ( π ) − ↾ A = ι ( τ ) − ↾ A , the proof as written in [8]follows through. Alternatively, and perhaps more wisely, for proving the distribu-tivity one should rely on the absoluteness proof that we used in this manuscript,which is also mentioned in the “Abyss”.Lemma 4.1 is precisely the same, in the context of successor iterations. There isno need to repeat that which has been said. For example, h P , G , F i has κ -s.c.c. if every symmetrically dense open set contains a predenseset of size <κ . Or σ , in that context. Corrections to Proposition 4.4/4.15.
These two propositions are dealingwith limit iterations. They show that h ̺ β | β < α i is symmetrically generic for P α ,with Proposition 4.4 dealing with the case α = ω , which was treated separately inthe paper. Proposition ( [8] , Proposition 4.4).
Suppose that D ⊆ P ω is a symmetricallydense open set, then there is a sequence h β n | n < ω i such that h ˙ ̺ n ↾ β n | n < ω i ∈ D . In the paper the proof takes a symmetrically dense open set D , and an excellentsupport ~H witnessing that. We then generate some sequence of domains such thatwhen we extend the condition h ̺ β ↾ X β | β < α i into D , then we can “correct” itusing automorphisms which lie in ~H . Namely, each coordinate of ~H is of the formfix( B β ), and we take X β to be sup dom B β + 1, and therefore our starting conditionis h ̺ β ↾ X β | β < α i .The idea is fine, and it is somehow intuitively clear what should happen. Howeverthe proof described above, taken from the “Abyss”, will not work. The first hint isobvious: ̺ β has domain ω β , and X β is a bounded subset of ω β +1 . We can resolvethis issue by considering ̺ β +1 ↾ X β , but this raises the obvious question, what shouldwe do with limit steps and with ̺ ?Let us prove this proposition in its general form. Proposition 5.1 ( [8] , Proposition 4.15). h ̺ β | β < α i is symmetrically L -generic for P α .Proof. Let D be a symmetrically dense open set. Our goal is to find a condition ofthe form h ̺ β ↾ X β | β < α i , for appropriate X β , in D .Let ~H denote the excellent support witnessing that D is a symmetrically denseopen set. For β < α let ξ β +1 the ordinal defined by either,(1) when β is 0 or a successor let B β be such that H β = fix( B β ), and define ξ β +1 = sup dom B β ;(2) when β is a limit let ξ β +1 be an ordinal ξ such that for some f ∈ Q SC( ω β ), H β = K ξ,f β ;(3) when H β = G β , set ξ β +1 = 0.For a limit ordinal β , let E β ⊆ Q SC( ω β ) be defined by Q β<α ( ξ β +1 + 1), and let A β denote sup { γ < β | H γ +1 = G γ +1 } ; if β is a limit ordinal such that H γ = G γ for all γ ≥ β , set E β = A β = ∅ .. Now define Y β as 0 for β = 0, ξ β for successors,and ( E β , A β ) for limit ordinals.The goal is to find the “domain” over which ~H must be the identity, and usethat to start our journey into D . Let p be the condition h ̺ β ↾ Y β | β < α i . By itsvery definition, p is not moved by any ~π ∈ ~H . Using the density of D we can nowextend p to a condition p ∈ D . Moreover, since D is in L , we may assume thatthe first coordinate of p agrees with ̺ , i.e. the Cohen real c , and by shrinking H if necessary, we may assume that p (0) is not moved by automorphisms in H .If p is not compatible with h ̺ β | β < α i , let β be the least coordinate witnessingthat. By the choice of Y β we can find ~π ∈ H ↾ β , and if β is not a limit ordinalthen we may assume ~π has a single non-trivial coordinate too, such that by taking p = R ~π p , the following hold:(1) supp( p ) = supp( p ),(2) p ↾ β = p ↾ β ,(3) p ( β ) is indeed compatible with ̺ β .The reason we can do that is exactly upwards homogeneity combined with ourchoice of p , which p extended. As we did not increase the non-trivial coordinateswhen moving from p to p , we may proceed by recursion and after finitely manysteps the process must halt with some p n ∈ D . (cid:3) UIDE TO THE BRISTOL MODEL 25
Indeed, the main point is the fact that we may assume that the first coordinate,for which true genericity is already assumed, is compatible with the Cohen real, andthen we can modify any further coordinates recursively using upward homogeneity,combined with the fact that we only need to change things outside of the domainswe found, which means that the needed automorphisms can be found in ~H .6. Deconstructing constructibility
Now that we have constructed the Bristol model, and we have a good idea abouthow the construction works, we can ask the obvious question: do we really need V = L in the ground model? The answer, of course, is not really.We have merely used three assumptions: GCH , (cid:3) ∗ λ for singular λ , and globalchoice for fixing the Bristol sequence. Of those three, the last one can be easilydispensed, and we will discuss this in section 7. So we are left with only twoassumptions which are compatible with a large range of models, including all knowninner models from large cardinals below a subcompact.This raises an interesting question, of course. What kind of large cardinals canthe Bristol model accommodate, assuming they existed in the ground model? Proposition 6.1. If A is a set of ordinals in the Bristol model, then there is a realnumber r such that A ∈ V [ r ] , and moreover r ∈ V or r is Cohen over V .Proof. Note that V [ A ] is a model of ZFC intermediate to V and V [ c ], and therefore V [ A ] is equivalent to some V [ r ]. (cid:3) This leads to an immediate corollary: if κ is a large cardinal defined by theexistence (or lack thereof) of sets of ordinals, and this largeness is preserved byadding a Cohen real, then κ remains large in the Bristol model.For example, if κ is Mahlo, then the stationary set of regular cardinals remainstationary in M , and thus κ remains Mahlo. If we define a weakly compact cardinalby stating that every colouring of [ κ ] in 2 colours has a homogeneous subset, thistoo is given by sets of ordinals, and so it continues to hold in M .We will see in section 9 that this can be extended from sets of ordinals to sets ofsets of ordinals, and so on, as long as we iterate power sets less than κ times, thelargeness remains. So for example, any measurable on a ground model measurablecardinal will have a unique extension in the Bristol model. Strong cardinals, definedby extenders, are also preserved.6.1. Limitations.
Despite this handsome accommodation of large cardinal as-sumptions in the ground model, as well as in the Bristol model itself, we canput a stop to this. Indeed, if κ is a supercompact cardinal, then there is a notableshortage of (cid:3) ∗ λ sequences for singular λ such that cf( λ ) < κ < λ . We may askourselves, perhaps we can salvage permutable scales without having (cid:3) ∗ λ ? Theorem 6.2 (The Bristol group).
Suppose that κ is a supercompact cardinaland λ > κ > cf( λ ) . Then there are no permutable scales on Q SC( λ ) . Recall that we are still working under the assumption of
GCH . Removing it willrequire adding 2 λ < λ + , which itself is a harmless assumption as it holds for allstrong limit cardinals above κ . Proof.
Let F = { f α | α < λ + } be a scale in Q SC( λ ), and let π be a permutationof λ + , not necessarily bounded, which is not implemented by any sequence ofpermutations, ~π . Pick j : V → W an elementary embedding with critical point κ witnessing that κ is at least λ + -supercompact, so in particular j ( κ ) > λ + . Wedenote by G = { g α | α < j ( λ + ) } the scale j ( F ), which is a scale in j ( Q SC( λ ))which is equal to Q SC( j ( λ )). Let ν = sup j “ λ + , then ν is closed under j ( π ), so we can let τ denote j ( π ) ↾ ν ,which is a bounded permutation of j ( λ + ).Suppose that there was a sequence of permutations ~τ that implemented τ , thenfor α < λ + and µ < λ , we have g j ( α ) ( j ( µ + )) = j ( f α ( µ + )) and τ ( j ( α )) = j ( π ( α )).Combining this with the assumption that ~τ implements τ , we get that τ j ( µ + ) ( j ( f α ( µ + ))) = j ( f π ( α ) ( µ + )) . We use this to define π µ + : µ + → µ + . If τ j ( µ + ) ( j ( ξ )) = j ( ζ ) for some ζ , define π µ + ( ξ ) = ζ ; otherwise π µ + ( ξ ) = ξ . This is well-defined, since τ j ( µ + ) is itself apermutation of j ( µ + ), and so j ( ζ ) < j ( µ + ). This is also a permutation by similarreasons.Finally, we claim that ~π implements π . Fix α < λ + , then for any large enough µ + < λ , τ j ( µ + ) ( j ( f α ( µ + ))) = j ( f π ( α ) ( µ + )), which means that we defined π µ + to beexactly f π ( α ) ( µ + ).We have shown that if F can be used to implement all bounded permutations,then it can be used to implement all permutations, but that is definitely impossibleon grounds of cardinal arithmetic. (cid:3) Gaps in the Multiverse
The Bristol model, as we said at first, is a particularly striking counterexample toour understanding of intermediate models when the axiom of choice is not assumed.Working in V ’s meta-theory, we can consider the collection of all intermediatemodels between V and V [ c ]. We remarked before, those models which satisfy ZFC are exactly V itself or Cohen extensions of the form V [ r ] for some r ∈ R V [ c ] .What about models of ZF ? We have, of course, symmetric extensions. Thesewere studied extensively by Grigorieff in [3], and later by Usuba in [19]. Grigorieffproved that if M is a model such that V ⊆ M ⊆ V [ G ], then M is a symmetricextension given by the same forcing used to add G if and only if M = (HOD V ∪ x ) V [ G ] for some set x ∈ V [ G ]. Usuba extended this and showed that in general, M is asymmetric extension of V if and only if M = V ( x ) for some set x .The construction of the Bristol model shows that there is indeed a differencebetween the two results. Indeed, by counting the number of possible automorphismgroups and filters of groups we can see that there cannot be more than ℵ distinctsymmetric extensions of Add( ω,
1) over L . Yet, the construction of the Bristolmodel goes through a proper class of steps, and even if we did not formally provethat each separate one is of the form L ( x ), we may take L ( V Mα ) as our models. ByUsuba’s result, these are all symmetric extensions of L , but of course most of themare not symmetric extensions where the forcing used is the Cohen forcing.So when we consider the multiverse of symmetric extensions of L , even thosethat are landlocked inside L [ c ], we seem to have two different options. But we wantto study all intermediate models, and these include the Bristol model, which is verymuch not a symmetric extension of L , by any means, as Usuba’s result indicate.So what can we say about the multiverse of ZF models?First, let us ask, how many Bristol models are there? Of course, there is thecanonical one, given by the < L -minimal Bristol sequence. But there are certainlymore. Simply by removing some of the sets or functions in any given point in theBristol sequence we invariably create a new Bristol model. Proposition 7.1.
Assuming
GCH and that (cid:3) ∗ λ holds for all singular λ , there is aclass forcing which does not add sets whose generic is a Bristol sequence. (cid:3) We can improve this counting argument to show no more than ℵ , actually. UIDE TO THE BRISTOL MODEL 27
This is done, of course, by approximating the Bristol sequence with set-lengthoptions. This is also the way we can remove global choice from the assumptions.Indeed, if V was a model of ZFC , add a generic Bristol sequence, use it to definea Bristol model between V and V [ c ], where c is a Cohen real, and promptly forgetabout this generic sequence.As very clearly the Bristol model is definable from its Bristol sequence in V [ c ],this means that there may be undefinable Bristol models. And indeed, this canvery much be the case. Theorem 7.2.
Suppose that V is a countable transitive model of ZFC + GCH + (cid:3) ∗ λ for all singular cardinals λ , and let c be a Cohen real over V . Then there areuncountably many Bristol models intermediate between V and V [ c ] .Proof. Given two Bristol sequences such that A and A ′ are the two permutablefamilies on ω and such that S A ∩ S A ′ is finite, the two Bristol models aredistinct. This can be easily arranged, for example taking A to only have subsetsof even integers and A ′ only has subsets of odd integers. This can be extended toany other step in the Bristol sequence. We can now easily construct uncountablymany distinct Bristol models by simply considering with each subset of Ord V howto modify a given, or indeed a generic, Bristol sequence. (cid:3) This is in stark contrast to the case of intermediate models of
ZFC of which thereare only not just countably many, but there is a set of all the necessary generatorsin V [ c ], and the same can be said about symmetric extensions given by the Cohenforcing itself, as Grigorieff’s theorem shows. And while there is a proper class ofsymmetric extensions of V , it is still enumerated by the sets in V [ c ], making itcountable.Therefore between V and V [ c ] most models are models of ZF , they are notsymmetric extensions of V , and in fact they are not definable in V [ c ]. To makematters worse, inside each Bristol model, we can find a different real, and use thatreal to interpret the Bristol sequence. Just as well, we may also use one of the ̺ α +1 ↾ A αξ to interpret the Bristol model construction above a certain stage.Which truly indicates that the Bristol models are intertwined through this mul-tiverse of models. But to truly appreciate the Bristol model(s), and to understanda bit more its internal structure, we need to have a refined sense of choicelessness.8. Kinna–Wagner Principles
The axiom of choice can be simply stated as “every set can be injected into anordinal”, or in other words, “every set is equipotent with a set ordinals”. This, inconjunction with the following theorem, makes a nice way of understanding modelsof
ZFC . Theorem (Balcar–Vopěnka).
Suppose that M and N are two models of ZF withthe same sets of ordinals. If M | = ZFC , then M = N . Commonly this is stated when M and N are both models of ZFC , but that is infact unnecessary. The proof relies on the fact that a relation on a set of ordinals canbe coded as a set of ordinals. But can we extend the idea of this characterisation?Yes, yes we can.
Definition 8.1.
We say that A is an α -set of ordinals , or simply “ α -set”, if thereis an ordinal η such that A ⊆ P α ( η ). Definition 8.2.
Kinna–Wagner Principle for α , denoted by KWP α , is the state-ment that every set is equipotent with an α -set. We write KWP to mean ∃ α KWP α . For α = 0 this is simply AC . The principle for α = 1 was defined by Kinna andWagner, although formulated differently using selection functions, and was studiedextensively. One of the immediate results is that KWP implies that every set canbe linearly ordered, but as the work of Pincus shows in [15], it is stronger and infact independent of the Boolean Prime Ideal theorem (which also implies every setcan be linearly ordered).These general principles were defined by Monro in [12] for α < ω , and laterextended by the author in [9]. Monro extended the result of Balcar and Vopěnka,and this result can be further extended as well. Theorem 8.3 (The Generalised Balcar–Vopěnka–Monro Theorem).
Suppose that M and N are models of ZF with the same α -sets. If M | = KWP α ,then M = N .Proof. Note that for every α , we can code relations on α -sets by using α -sets in arobust and definable way by extending Gödel’s pairing function. So we can simplyrepeat the proof of Balcar–Vopěnka. If x ∈ M , we can encode tcl( { x } ) and itsmembership relation as an α -set, X . So X ∈ N , and by decoding the membershiprelation and applying Mostowski’s collapse lemma, x ∈ N . Therefore M ⊆ N .In the other direction, suppose that V Mη = V Nη , then we can encode it as an α -set in M ∩ N . Now given any x ⊆ V η in N , i.e. x ∈ V Nη +1 , by looking at the α -set encoding V η , we can identify the subset corresponding to x , in N . But as M and N share the same α -sets, this implies that this subset is in M as well, and wecan therefore find x ∈ M . Now by transfinite induction, V Mη = V Nη for all η , andequality ensues. (cid:3) Monro proved in [12] that
KWP n +1 KWP n for all n . This result was extendedto show that KWP ω KWP n for all n by the author in [9], and was later extendedas well by Shani in [18] to show that for all α < ω , KWP α +1 KWP α . Definition 8.4.
Small Violation of Choice holds if there exists a set A such thatfor any set x there is an ordinal η and a surjection f : η × A → x . We write SVC ( A )to specify this set A , or SVC to mean ∃ A SVC ( A ).This axiom is due to Blass in [1], and it turns out to play an important role inthe study of symmetric extensions. Proposition 8.5.
Suppose that
SVC ( A ) holds, if A is equipotent with an α -set,then KWP α +1 holds.Proof. We may assume that A is an α -set itself, and so if f : η × A → x is asurjection, by coding we may replace η × A by an α -set as well, say A α . Now thefunction mapping y ∈ x to y f = { a ∈ A α | f ( a ) = y } is injective, and each y f is an α -set. Therefore x is equipotent to the ( α + 1)-set { y f | y ∈ x } . (cid:3) Blass proved in [1] that
SVC is equivalent to the statement “The axiom of choiceis forceable [by a set forcing]”, and that
SVC holds in every symmetric extension.Usuba showed in [19] that the latter implication can be reversed. Namely, V | = SVC if and only if V is a symmetric extension of a model of ZFC .Before we return to study the Bristol model, let us study Kinna–Wagner Prin-ciples a bit more in depth.
Theorem 8.6.
Suppose that V | = KWP α , if V [ G ] is a generic extension, then V [ G ] | = KWP α ∗ , where α ∗ = sup { β + 2 | β < α } . In other words, α ∗ is the successor of α when α itself is a successor, or α itself otherwise. UIDE TO THE BRISTOL MODEL 29
Proof.
For every x ∈ V [ G ] there is a name ˙ x in V , so ˙ x ↾ G = {h p, ˙ y i ∈ ˙ x | p ∈ G } in V [ G ], and the interpretation map is a surjection onto x , which we can extend toa surjection from ˙ x itself. Therefore every set in V [ G ] is the surjective image of aset in V .If α < α ∗ the above completes the proof, as we may assume that ˙ x was an α -set, and conclude that x is an ( α + 1)-set. For α = 0 or a limit ordinal, we useLemma 8.7. (cid:3) Lemma 8.7.
Let A be an α -set for α = 0 or a limit ordinal. If f : A → x is asurjection, then there exists an α -set B which is equipotent with x .Proof. For α = 0, A is a set of ordinals, so we may simple choose the least ordinalfrom the pre-image of each y ∈ x . Suppose that α is a limit ordinal, and define for β < α , A β = { a ∈ A | a is a β -set } . For every y ∈ x , let β y be the least β such thatfor some a ∈ A β , f ( a ) = y .Now define B y = { a ∈ A β y | f ( a ) = y } , this is a ( β y + 1)-set, and y = y ′ implies B y = B y ′ . Therefore B = { B y | y ∈ x } is an α -set equipotent with x . (cid:3) On the other hand, ground models need not satisfy the same
KWP α as theirgeneric extension: as we have seen in the construction of the Bristol model, L [ c ]has a proper class of ground models, L ( V Mα ), and as we will see in the next section,the various KWP α fails as we go up the hierarchy.The next obvious question is whether or not Theorem 8.6 can be improved.Unfortunately, it cannot. The Cohen model famously satisfies KWP , but asMonro demonstrated in [13], there is a generic extension of the Cohen model inwhich there exists an amorphous set, which cannot be linearly ordered, in particular, KWP must fail in that generic extension. Nevertheless, by the theorem above, KWP must hold. This leads us to the following conjecture. Conjecture 8.8 (The α ∗ Conjecture) . Suppose that
KWP α ∗ holds in every genericextension of V , then KWP α must hold in V .It is also easy to see that if M | = ¬ KWP , then also any generic extension of M must satisfy this. Which points out to a particularly poignant feature of ageneric multiverse, and indeed a symmetric multiverse , KWP hold or fails uni-formly throughout the entire multiverse.
Conjecture 8.9 (The Kinna–Wagner Conjecture) . Suppose that V | = KWP and G is a V -generic filter. If M is an intermediate model between V and V [ G ] and M | = KWP , then M = V ( x ) for some set x .The Bristol model was an exercise in finding an intermediate model which isnot constructible from a set. And we conjecture that having any such model,intermediate to a generic extension, will fail KWP , and vice versa: any intermediatemodel of
KWP is constructible from a set over the ground model.It may also be the case that
KWP implies ground model definability, which isa notoriously difficult problem in ZF . Usuba proved that under certain conditionsground models are definable in ZF , but currently the only known models to satisfythese conditions are models satisfying SVC . Nevertheless, as α -sets can be used tocharacterise a model of ZF in the presence of KWP α , it stands to reason that itmay play a role in ground model definability as well. An implicit proof can be found as Lemma 5.25 in [7]. These are not directly related to the Bristol model, so we include them here instead ofsection 10 Allowing symmetric extensions and grounds. See [19] for more information.
We can also define
SVC α to mean that we replace the ordinal, η , by an α -set.And in that case we can easily see that SVC α is equivalent to “ KWP α ∗ is forceable”.And one is now left wondering if SVC α is equivalent to being a symmetric extensionof a model satisfying KWP α .One can also take a different approach and define SVC M , for a class M , wherewe may replace the ordinal η by a set from M , so SVC = SVC
Ord and
SVC α is ashorthand for SVC P α (Ord) . For this to be truly useful, we need to modify KWP α sothat 0-sets are subsets of M . These ideas may play a role in the ultimate answersregarding ground model definability in ZF , and we hope this discussion will help toinspire some of the readers to think about that. Question 8.10.
Is ground model definability equivalent to
KWP ?Note that this question is meaningful since as we observed,
KWP is absolutethrough the generic, and indeed the symmetric, multiverse.9.
Choice principles in Bristol
We want to investigate the failure of the axiom of choice in the Bristol model, M , and provide alternative proofs to the key property of the Bristol model, namely ∀ x ( V = L ( x )). Proposition 9.1 ( [8] , Theorem 5.5).
Let M denote the Bristol model, and M α bethe α th model in the construction. Suppose that A ∈ M is an α -set, then A ∈ M α +1 . We will not prove this proposition here, but the idea extends the homogeneityargument used in Proposition 3.2. Indeed, L ( V Mα +1 ) contains all the α -sets of M . Corollary 9.2. If β > α , then M β | = ¬ KWP α . In particular, M | = ¬ KWP .Proof.
M, M β and M α +1 have the same α -sets. If one of them would satisfy KWP α ,by The Generalised Balcar–Vopěnka–Monro Theorem, M = M β = M α +1 . (cid:3) Corollary 9.3. M | = ¬ SVC , and consequently M | = ∀ x ( V = L ( x )) , as well as “theaxiom of choice is not forceable (by a set forcing)”. (cid:3) It follows from this that at least for a proper class A ⊆ Ord, if α < β are bothin A , then KWP β KWP α . But we want to understand the gradation, or ratherthe degradation, of KWP through the construction.
Proposition 9.4. M | = ¬ KWP ∧ ¬ BPI
Proof.
We proved that R , the set of R ξ = V L [ c ∩ A ξ ] ω +1 for ξ < ω , is ℵ -amorphousin M in Proposition 3.8, and the same argument shows that it cannot be linearlyordered. In particular, it cannot be equipotent to any 1-set and it also witnessesthat BPI fails.Briefly, the argument starts by taking a name ˙ ≺ ∈ IS and some p (cid:13) IS “ h ˙ R, ˙ ≺i • isa linearly ordered set”. Let B be a disjoint approximation such that fix( B ) ⊆ sym( ˙ R )and dom p ⊆ S B . We pick α, β / ∈ dom B and distinct, and we let q ≤ p decide,without loss of generality, ˙ R α ˙ ≺ ˙ R β . But now we can find π such that ι ( π ) = ( α β )and πq = q . (cid:3) This is a remarkable point, as the elements of R themselves can be well-orderedseparately. So you may think we can replace them by 0-sets, but the truth is thatwe cannot do that uniformly, and this forces us to treat them as 1-sets instead. Proposition 9.5. M | = ¬ BPI . UIDE TO THE BRISTOL MODEL 31
Proof.
We show that R still cannot be linearly ordered when passing to M . Ofcourse the forcing that led us there, Q , linearly orders (and in fact well-orders) R . But it is easy to see that this well-order is promptly discarded, and instead weonly remember bits and pieces of it in the form of ̺ ↾ A ξ . To show that there isno linear ordering in M we need to use the full power of the symmetric iteration.We are going to start with a rather naive attempt, which may not work, but wecan identify the problem and circumvent it.Suppose that ˙ ≺ ∈ IS is such that h p, ˙ q i (cid:13) IS “ ˙ ≺ linearly orders ˙ R ”. Let B and B be disjoint approximations such that h fix( B ) , fix( B ) i is a support of ˙ ≺ .Let α, β < ω be such that α, β / ∈ dom B ∪ S B and moreover ˙ R α , ˙ R β are notmentioned in ˙ q .Let h p ′ , ˙ q ′ i be a condition extending h p, ˙ q i such that h p ′ , ˙ q ′ i (cid:13) IS ˙ R α ˙ ≺ ˙ R β . Wewould like to apply upwards homogeneity and consider π ∈ fix( B ) which imple-ments ( α β ) while also not moving p ′ . But we have to contend with the fact that π ˙ q ′ may have moved. Luckily, we know exactly where it moved to: π ˙ q simple permutesthe range of ˙ q , so if ˙ R α and ˙ R β appear in ˙ q ′ , which is the likely case, then we simplyneed to switch them back using some σ ∈ fix( B ), that is an automorphism of ˙ Q ,and that will be enough, since automorphisms of ˙ Q do not change ˙ R α and ˙ R β .Alas, we have a problem. For σ to exist, we need to make sure that ˙ R α and ˙ R β appear in ˙ q ′ in coordinates which are not in S B , otherwise we cannot move thesecoordinates at all. So this naive approach cannot work.Luckily, we assumed that ˙ R α and ˙ R β are not mentioned in our original ˙ q . So tofind ˙ q ′ , first add both of these in coordinates that are not in S B , and if this wasnot enough to decide how ˙ ≺ will order them we can extend further to find ˙ q ′ . Inother words, we may assume without loss of generality that ˙ q ′ mentions ˙ R α and ˙ R β in coordinates which are eligible to be switched from within fix( B ).Therefore, if h p ′ , ˙ q ′ i (cid:13) IS ˙ R α ˙ ≺ ˙ R β , then also h p ′ , ˙ q ′ i (cid:13) IS ˙ R β ˙ ≺ ˙ R α . Therefore h p ′ , ˙ q ′ i cannot force ˙ ≺ to be a linear ordering, which is a contradiction, since itextends a condition which did force just that. (cid:3) One may think that this is enough to prove that there are sets which cannot belinearly ordered in the Bristol model, as we just exhibited that the second symmetricextension will not linearly order R either. Alas, we already concluded that R is a2-set, so a linear ordering of R will also be a 2-set. But we know that 2-sets areonly determined in M . But luckily, we are not very far behind completing thispart of the journey. Theorem 9.6. M | = ¬ BPI .Proof.
Suppose that h p, ˙ q, ˙ r i ∈ P is a condition that for some ˙ ≺ ∈ IS forces that ˙ ≺ is a linear order of ˙ R . We can actually run the proof of Proposition 9.5 again. Firstof all, π will not affect the condition extending ˙ r at all, but more importantly, σ can be chosen as a permutation moving only two points which means it implementsthe identity. So it also will not modify the condition extending ˙ r .And so as long as we were careful to choose the extension ˙ q ′ in a way that allows σ to be taken from fix( B ), the argument is not affected. Therefore we showed that M | = “ R cannot be linearly ordered”, and therefore M does as well. (cid:3) Open questions related to the Bristol model
There is still so much to learn about the Bristol model, both in the specificcontext of L , as well as many natural questions that come up from generalisationsand details in the proof. We cannot possibly include all of these, but we will give fourfamilies of questions which are interconnected, but also seem to have independentinterest. The Bristol models in the multiverse.Question 10.1.
Is there a condition characterising the equivalence classes on Bris-tol sequences (definable or otherwise) based on the Bristol model they generateusing a fixed Cohen real?
Question 10.2.
Is the theory of any two Bristol models the same? Does the theorydepend on the sequence or its properties?
Question 10.3.
Are there any non-trivial grounds of any Bristol model?
Question 10.4.
Is there a Bristol model which is definable in its generic extensions,or maybe is there one that is not definable in some of its generic extensions?
Question 10.5.
Is there a generic extension of a Bristol model which itself is aBristol model?
Question 10.6.
Is there a maximal Bristol model, namely, is there a Bristol model M ⊆ L [ c ] such that for any x ∈ L [ c ] \ M , M ( x ) = L [ c ]? Question 10.7.
Is it true in general that for x ∈ L [ c ], either L ( x ) = L [ c ] or thereis a Bristol model M such that x ∈ M ?10.2. Large cardinals in the Bristol model.Question 10.8.
We saw that measurable cardinals remain measurable. Do theyremain critical cardinals in the sense of [5]? What about weakly critical cardinals?
Question 10.9.
Principles like (cid:3) ∗ λ are considered to witness failure of compactness.Suppose that λ is singular and no permutable scale exists on Q SC( λ ). Can thiscompactness be harnessed to restart the Bristol model construction? (Note thata positive answer would indicate that Woodin’s Axiom of Choice Conjecture ispossibly false, which may imply also the eventual failure of the HOD Conjecture.As such the answer to this question is most likely negative, and a positive answerwould be extremely hard to prove.) Question 10.10.
Suppose that elementary embeddings can be lifted and Woodincardinals are preserved. Starting from strong enough hypotheses, can we constructBristol model-like objects that satisfy AD ?10.3. Other type of Bristol models.Question 10.11.
Can we start the construction of the Bristol model with a dif-ferent type of real? Clearly not every real is useful, minimal reals do not haveintermediate models, for example. But what about reals that admit sufficientlymany automorphisms and intermediate models such as random reals? What about“Cohen + condition” type of reals (Hechler, Mathias, etc.)? Will this also impactthe type of forcing we need to do in the following steps (namely, will that force usto use something which is not isomorphic to Add( ω α , L in successor steps)? Question 10.12.
Can we start the construction with a Prikry-like forcings insteadof a Cohen forcing?
Question 10.13.
While there is no good definition for iteration of symmetricextensions with countable support, it is imaginable that for productive iterationssuch as the one used in the Bristol model this is doable by hand. What wouldthis be? Can we have an ω -Bristol model starting with an L -generic sequence forAdd( ω ,
1) for example?
UIDE TO THE BRISTOL MODEL 33
Weak choice principles.Question 10.14.
Does DC hold in the Bristol model? Question 10.15.
Does ω have free ultrafilters in the Bristol model? Question 10.16.
Does M α | = KWP α ∗ ? Question 10.17.
Can any choice principles be forced over the Bristol model?
Question 10.18.
Is countable choice true in the Bristol model? If so, is AC WO ,the axiom of choice for families that can be well-ordered, true? (Note that this willprovide a positive answer about DC , as well as the lifting of elementary embeddingsfor measurable cardinals.) Question 10.19.
Say that A is x -amorphous if it cannot be well-ordered, and itcannot be written as the union of two sets that are not well-orderable. That is, forsome κ , A is κ -amorphous. Are there any x -amorphous sets in the Bristol model? References
1. Andreas Blass,
Injectivity, projectivity, and the axiom of choice , Transactions of the AmericanMathematical Society (1979), 31–31.2. James Cummings, Matthew Foreman, and Menachem Magidor,
Squares, scales and stationaryreflection , J. Math. Log. (2001), no. 1, 35–98. MR 18383553. Serge Grigorieff, Intermediate submodels and generic extensions in set theory , Ann. of Math.(2) (1975), 447–490. MR 3738894. Lorenz J. Halbeisen,
Combinatorial Set Theory , Springer Monographs in Mathematics,Springer, London, 2012, With a gentle introduction to forcing. MR 30254405. Yair Hayut and Asaf Karagila,
Critical cardinals , Israel Journal of Mathematics (2020),no. 1, 449–472.6. Thomas Jech,
Set Theory , Springer Monographs in Mathematics, Springer-Verlag, Berlin,2003, The third millennium edition, revised and expanded. MR 19405137. Thomas J. Jech,
The Axiom of Choice , North-Holland Publishing Co., Amsterdam-London;Amercan Elsevier Publishing Co., Inc., New York, 1973, Studies in Logic and the Foundationsof Mathematics, Vol. 75. MR 03962718. Asaf Karagila,
The Bristol model: an abyss called a Cohen real , J. Math. Log. (2018),no. 2, 1850008, 37. MR 38784709. , Iterating symmetric extensions , J. Symb. Log. (2019), no. 1, 123–159. MR 392278810. , Realizing realizability results with classical constructions , Bull. Symb. Log. (2019),no. 4, 429–445. MR 406411311. Kenneth Kunen, Set Theory , Studies in Logic (London), vol. 34, College Publications, London,2011. MR 290539412. G. P. Monro,
Models of ZF with the same sets of sets of ordinals , Fund. Math. (1973),no. 2, 105–110. MR 034760213. , On generic extensions without the axiom of choice , Journal of Symbolic Logic (1983), no. 1, 39–52.14. Douglass Bert Morris, Adding total indiscernibles to models of set theory , ProQuest LLC,Ann Arbor, MI, 1970, Thesis (Ph.D.)–The University of Wisconsin - Madison. MR 262029315. David Pincus,
On the independence of the Kinna-Wagner principle , Z. Math. Logik Grund-lagen Math. (1974), 503–516. MR 36906616. Gershon Sageev, An independence result concerning the axiom of choice , Annals of Mathe-matical Logic (1975), no. 1-2, 1–184.17. , A model of ZF + there exists an inaccessible, in which the Dedekind cardinals con-stitute a natural non-standard model of arithmetic , Annals of Mathematical Logic (1981),no. 2-3, 221–281.18. Assaf Shani, Borel reducibility and symmetric models , arXiv (2018).19. Toshimichi Usuba,
Geology of symmetric grounds , arXiv (2019).
E-mail address , Asaf Karagila: [email protected]
URL : http://karagila.orghttp://karagila.org