aa r X i v : . [ m a t h . L O ] A ug DESTRUCTIBILITY AND AXIOMATIZABILITY OF KAUFMANN MODELS
COREY BACAL SWITZER
Abstract.
A Kaufmann model is an ω -like, recursively saturated, rather classless model of PA . Such models were constructed by Kaufmann under ♦ and Shelah showed they exist in ZFC by an absoluteness argument. Kaufmann models are an important witness to the incompactnessof ω similar to Aronszajn trees. In this paper we look at some set theoretic issues related to thismotivated by the seemingly na¨ıve question of whether such a model can be “killed” by forcingwithout collapsing ω . We show that the answer to this question is independent of ZFC andclosely related to similar questions about Aronszajn trees. As an application of these methodswe also show that it is independent of
ZFC whether or not Kaufmann models can be axiomatizedin the logic L ω ,ω ( Q ) where Q is the quantifier “there exists uncountably many”. Introduction
A Kaufmann model is an ω -like, recursively saturated, rather classless model of PA (theseterms are defined below). Kaufmann first constructed such models in [4] under ♦ and in [10] She-lah showed that Kaufmann models exist in ZFC by an absoluteness argument. These structuresform an important class of models of arithmetic that have been extensively studied, see [7, Chapter10] and the references therein. There are several reasons for this. First of all Kaufmann modelsrepresent a counterexample to the analogue of several theorems about countable recursively sat-urated models of PA holding at the uncountable including most notably the fact that countablerecursively saturated models of PA have inductive partial satisfaction classes, see [7, Theorem 1.9.3,Proposition 1.9.4]. They also are a witness of set theoretic incompactness at ω . For instance, thefollowing is immediate from the fact that all countable, recursively saturated models of PA havesatisfaction classes and Tarski’s theorem on the undefinability of truth. Proposition 1.1.
Let M be a Kaufmann model. By rather classlessness M cannot have a partialinductive satisfaction class. However, there is a club of countable elementary submodels N ≺ M so that N carries a satisfaction class. Kaufmann models are also very closely related to trees. This was used in Shelah’s absolutenessproof and also features prominently in Schmerl’s work on generalizations of Kaufmann models tohigher cardinals [9]. The analogy with trees is the jumping off point for the current work. Ourna¨ıve question that started this work was whether there could be a Kaufmann model which couldbe killed by forcing without collapsing ω . Note that this is similar to asking whether there is anAronszajn tree to which an uncountable branch can be added by forcing with out collapsing ω .The answer in that case is independent: if there is a Souslin tree the answer is ”yes” while if allAronszajn trees are special the answer is “no”. In the case of Kaufmann models the answer turnsout to be the same. Specifically we prove the following theorem (proved as Theorems 2.1 and 3.1respectively). Main Theorem 1.2. (1) Assume MA ℵ holds. If M is a Kaufmann model and P is a forcingnotion so that (cid:13) P “ M is not Kaufmann” then P collapses ω . Mathematics Subject Classification.
Primary: 03C62 Secondary: 03C80, 03E50.
Key words and phrases.
Kaufmann Models, strong logics, destructibility, Martin’s Axiom. (2) Assume ♦ holds. There is a Kaufmann model M and a Souslin tree S so that forcing with S adds a satisfaction class to M . It remains unclear whether the property of “being destructible by ω -preserving forcing” has acompletely combinatorial or model theoretic characterization but the models used in the proof ofMain Theorem 1.2 can be used to show the following, which is the second main theorem of thispaper (See Theorem 4.3 below). Main Theorem 1.3.
Let Q be the quantifier “there exists uncountably many ...” and L ω ,ω ( Q ) be the infinitary logic L ω ,ω enriched by this quantifier. The following hold:(1) Under MA ℵ there is an L ω ,ω ( Q ) sentence ψ in the language of PA enriched with a singleunary function symbol f , L PA ( f ) , so that a model M | = PA is Kaufmann if and only ifthere is an expansion of M to an L PA ( f ) -structure satisfying ψ .(2) Under ♦ there is a Kaufmann model M so that given any expansion L ′ of the language of PA and any expansion of M to an L ′ -structure, M ′ , and any countable set L of L ω ,ω ( Q ) sentences in the signature L ′ there is a model N which agrees with M ′ about the truthof every L -sentence but carries a satisfaction class for its L -reduct. In particular, the L -reduct of N is not rather classless. Informally the Main Theorem 1.3 can be phrased as saying it’s independent of
ZFC if Kaufmannmodels can be axiomatized by an L ω ,ω ( Q ) sentence. This logic is a natural one to consider inthe context of such models since being ω -like and recursively saturated are expressible here hencethe question is really about inexpressibility of rather classlessness. Moreover this logic plays animportant role in Shelah’s aforementioned absoluteness result, [10, Theorem 6], and is used inseveral other applications of abstract model theory to ω -like structures, see [5]. In fact, part 1can be deduced as an immediate corollary of the proof of [10, Theorem 6]. I do not know if thiswas observed by Shelah at the time. Part 2, as far as I know, is completely new.The rest of this paper is organized as follows. In the remaining half of this introduction wegive some basic definitions and background that will be used throughout. In Section 2 Part 1 ofMain Theorem 1.2 is proved. In Section 3 Part 2 of Main Theorem 1.2 is proved. In Section 4Main Theorem 1.3 is proved. Section 5 concludes with some open questions and lines for furtherresearch. Acknowledgments.
I would like to thank Roman Kossak and Bartosz Wcis lo for several veryinformative and helpful conversations relating to the material in this paper and Ali Enayat forpointing out the papers [6] and [3] to me.1.1.
Basic Definitions.
Throughout we will be interested in the language L = L PA of PA , whichfor us includes a symbol ≤ for the natural ordering definable in PA . All of the results belowwork equally well for any countable extension of L PA and any theory PA ∗ , that is PA in thatlanguage with induction extended to formulas in that language. Given a first order structure suchas M , N , M α etc we always let the associated non-calligraphic letter, M , N , M α etc denote theuniverse of the model. When it won’t cause confusion this won’t be stated explicitly. For ease ofnotation by a model we will always mean an L structure M modeling PA unless otherwise stated.Also, throughout definable means definable with arbitrary parameters unless specified otherwise.Since we’re looking at applications of set theory to models of arithmetic, and hoping to appeal toresearchers in both these fields, we have included more definitions and proof sketches than usual inorder to make this paper more self contained for the reader who is an expert in only one of thesesubjects. For all undefined terms in the model theory of PA we suggest the reader consult [7]. Forset theory we recommend [8]. Definition 1.4.
Let κ be a cardinal.(1) A model M is κ - like if it has size κ but for all a ∈ M | [0 , a ] | < κ . ESTRUCTIBILITY AND AXIOMATIZABILITY OF KAUFMANN MODELS 3 (2) By arithmetizing the language of arithmetic we can think of L -formulas as coded com-putably by natural numbers. As such it makes sense to talk about a set of formulas asbeing e.g. computable, arithmetic etc. A model M is recursively saturated if it realizesevery computable type with parameters.(3) If M is a model, then a class is a subset A ⊆ M so that for all a ∈ M the set A ∩ a := { b ∈ A | M | = b < a } is definable M .(4) A model M is rather classless if every class is definable.(5) A κ -Kaufmann Model is a model M which is κ -like, recursively saturated and ratherclassless. If κ = ω then we simply say M is a Kaufmann model .We recall a brief sketch of the existence of Kaufmann models under ♦ as ideas from thesearguments will be used repeatedly throughout the paper. Theorem 1.5 (Kaufmann [4]) . If ♦ holds then every countable, recursively saturated model hasan elementary end extension which is Kaufmann. Before sketching the proof we need to note a few things. First, recall that given two models M and N we say that N is an elementary end extension of M , denoted M ≺ end N if M ≺ N andfor every y ∈ N \ M and x ∈ M we have N | = x ≤ y , i.e. ( M, ≤ ) is an initial segment of ( N, ≤ ).The foundational MacDowell-Specker theorem states that every model of PA has an elementaryend extension, see [7, Theorem 2.2.8]. The proof of Theorem 1.5 uses the following lemma, whichis also due to Kaufmann. Lemma 1.6 (Kaufmann [4]) . Let M be a countable recursively saturated model and A ⊆ M . If A is not definable, then there is a countable, recursively saturated model N so that M ≺ end N and A is not coded into N i.e. there is no a ∈ N so that a codes an N -finite sequence s a and M ∩ s a = A .Proof of Theorem 1.5. Fix a countable recursively saturated model M and a ♦ sequence ~A = h A α | α < ω i . We want to define a continuous chain hM α | α < ω i of countable, recursivelysaturated models so that M α ≺ end M α +1 for all α < ω and the union of all the M α ’s will bea Kaufmann model. This is done recursively. The universe of each model will be a countableordinal. Note that there will necessarily be a club of δ < ω so that M δ = δ. At limit stageswe have to take unions since the chain is continuous so it remains to say what to do at successorstages. Suppose M α has been defined. If A α ⊆ M α is undefinable let M α +1 be as in Lemma1.6, namely a countable, recursively saturated elementary end extension of M α in which A α is notcoded. If A α is not an undefinable subset of M α (either because it’s not a subset or because it’sdefinable) then let M α be any countable, recursively saturated elementary end extension of M α .This completes the construction.Let M = S α<ω M α . Clearly this model is an ω -like, recursively saturated elementary endextension of M . The hard part is to show that it is rather classless. This is shown as follows:suppose A ⊆ M is an undefinable class. It’s straightforward to show that the set of α so that A ∩ M α is undefinable in M α is club, thus by ♦ there is an α so that A ∩ M α = A α . But then A ∩ M α is not coded into M α +1 by our construction contradicting the assumption that A is aclass. (cid:3) The sequence above hM α | α < ω i is commonly called a continuous, end-extensional filtration .For short we will refer to such a sequence as simply a filtration . Definition 1.7. A filtration is an ω -length sequence h M α | α < ω i of countable models so thatfor α < β , M α ≺ end M β and for limit ordinals λ < ω M λ = S ξ<λ M ξ . The filtration is said to be recursively saturated if every M α is recursively saturated. Here M δ is the universe of M δ , conforming to the convention mentioned in the first paragraph of this subsection Without the extra qualifiers this is not entirely standard, but this is the only type of filtration we will consider inthis paper so no confusion will arise.
SWITZER
We will need the notion of a partial inductive satisfaction class. This idea has generated anenormous amount of research and is central in the study of models of PA . We will only need a fewfacts, which we cherry pick below, and refer the reader to the excellent monograph [7] for moredetails. The definition we give, which comes from [9], is not standard but it’s easily seen that amodel has a partial inductive satisfaction class in the sense below if and only if it has one in thesense defined e.g. in [7, Definition 1.9.1]. Recall that for each standard n < ω there is (provably,in PA ) a Σ n formula T r n ( x, y ) so that for all Σ n formulas ϕ ( z ) PA ⊢ ∀ y [ ϕ ( y ) ↔ T r n ( ϕ, y )]. Givena model M | = PA let W Mn denote the set of pairs ( ϕ, a ) so that ϕ ( x ) is a Σ n formula with one freevariable from the point of view of M and M | = T r n ( ϕ, a ) i.e. M thinks that a satisfies ϕ . Definition 1.8.
Let M be a model. A set S ⊆ M is called a partial inductive satisfaction class if (1) For all x ∈ M S x := { y | h y, x i ∈ S } is a set of pairs ( ϕ, a ) so that ϕ is a formula from thepoint of view of M and a ∈ M .(2) For all n < ω we have S n = W n .(3) ( M, S ) satisfies the induction scheme in the language expanded with a predicate for S .Partial inductive satisfaction classes are the only types of satisfaction classes that will be dis-cussed in this paper so we drop the qualifiers and refer to them simply as “satisfaction classes”.Note that the definition above is unchanged if we fix a nonstandard a ∈ M and insist that forevery b ≥ a the set S b = ∅ .As mentioned above κ -Kaufmann models can be seen as a witness to incompactness at a cardinal κ . Schmerl has formalized this in the following striking way. Theorem 1.9 (Schmerl, Theorem 3 of [9] ) . If there is a κ -Kaufmann model, then there is a κ -Aronszajn tree. Roughly speaking the tree is the “tree of attempts to build a satisfaction class”.
Proof.
Let M be a κ -Kaufmann model. We will define a subset T ⊆ M and a tree-like order on T so that the levels of T are indexed by the elements of M , T has sequences of every order typein M , the set of such sequences in a given order type has size less than κ , and there is no subset B ⊆ T in order type ≤ M . Clearly then any cofinal, well-founded subset of this “tree” will be a κ -Aronszajn tree.Fix a ∈ M non-standard. Let W n denote the complete Σ n -set (as defined in M ), which wethink of an an M -indexed list of 0’s and 1’s corresponding to its characteristic function on the setof pairs consisting of Σ n formulas and elements of M (using some standard pairing function). Thetree T is the set of b ∈ M so that there is a d ∈ M and b codes a d × a sized matrix whose entriesare 0 or 1 and for which for each natural number n < ω the n th -column of b is an M -finite initialsegment of W n . For elements b , b ∈ T coding matrices of size d × a and d × a respectively welet b ⊑ T b if d < d and b = b ↾ ( d × a ). In words, b is below b if and only if b codes alarger matrix whose restriction to the coordinates ( d × a ) is b (end extend each column). This isclearly a tree like order, it remains to see that it forms a tree as described in the first paragraph.First let’s see that the levels have size <κ . Let T d := { b ∈ T | b codes a binary matrix of size d × a } . Then since b ∈ T d implies b ∈ M and codes a sequence of size d × a there are at most 2 d × a elements of T d (as computed in M ) so by κ -likeness | T d | < κ .Now lets see that the tree has height κ . This follows immediately by recursive saturation. Foreach b ∈ T and i < a let b n denote the n th column of the matrix coded by b . For any d ∈ M consider the type p d ( x ) := {∃ e > d ( x codes a matrix of size e × a ) } ∪ { x n ⊆ W n | n < ω } . Clearlythis is a finitely consistent, recursive type so it has a realization in M . But any such realization isan element of height greater than d .Finally there is no cofinal branch. This follows by rather classlessness: from any cofinal branchwe could define a satisfaction class by the definition of the tree, but since any satisfaction class ESTRUCTIBILITY AND AXIOMATIZABILITY OF KAUFMANN MODELS 5 is undefinable this can’t exist. See [9, Lemma 4.1] for a more detailed discussion of this lastpoint. Note that if κ is an uncountable regular cardinal then any class is inductive, see [7, pp.258-259]. (cid:3) As Schmerl notes, what the proof above shows is that if κ has the tree property then every κ -likerecursively saturated model has a satisfaction class. Regardless of the properties of the order typeof M , the proof shows that given any recursively saturated model M , there is an associated tree T Msat whose levels are cofinal in the model. Moreover, if M is κ -like for some regular κ then T Msat hasa cofinal branch if and only if M has a satisfaction class. We will call such a tree the satisfactiontree for M (relative to a ).Fix a Kaufmann model M . Suppose P is a forcing notion, when does (cid:13) P “ ˇ M is not Kaufmann”? Obviously, if P collapses ℵ to be countable, then ω -likeness is killed. Moreover, any cofinalsequence in ordertype ω in a model is an undefinable class. What about if P does not collapse ω ?This motivates the following definition. Definition 1.10.
A Kaufmann model M is destructible if there is an ω -preserving forcing notion P so that (cid:13) P “ ˇ M is not Kaufmann”.In this language, an immediate corollary of Main Theorem 1.2 is the following. Corollary 1.11.
The existence of destructible Kaufmann models is independent of
ZFC . Before ending this section, let us make one observation about destructibility of Kaufmann modelsthat will guide the rest of the paper. Suppose M is a Kaufmann model and P is an ω -preservingforcing notion. Then in V P M is still ω -like, and by absoluteness, there cannot be any newrecursive types, so M is still recursively saturated. Therefore, if P kills the Kaufmann-ness of M it’s because it added an undefinable class. This is what we will use to kill Kaufmann models.Finally, let us note that some similar ideas to those presented here were previously explored byEnayat in [3]. In particular, in Theorem 4.2 of that paper Enayat observes that there are ratherclassless models of ZFC − + V = H ℵ which remain rather classless in any forcing extension pre-serving ω . Thus in the language of this paper Enayat shows that there is always an indestructiblemodel of this theory. 2. Killing Destructible Kaufmann Models
In this section we prove the first part of Main Theorem 1.2. Specifically we show the following.
Theorem 2.1.
Assume MA ℵ . Then there are no destructible Kaufmann models. The rest of this section is devoted to proving Theorem 2.1. Towards this, for the rest of thesection unless otherwise stated, assume MA ℵ and fix a Kaufmann model M . We will show thatin any forcing extension if M has a new class, then ω is collapsed.To prove Theorem 2.1 it will be convenient to relax the definition of a tree to allow the rankingfunction to map into a model of PA as opposed to ordinals. For a given model M I call this an M -tree . For instance, the tree T Msat of Theorem 1.9 is an M -tree. Most well known facts anddefinitions about trees hold without change for M -trees. With the exception of the first, thefollowing definitions when applied to ordinal-ranked trees are standard. Definition 2.2.
Let T = h T, ≤ T i be a tree-like order.(1) We say T is ω -like if there is a cofinal subset S ⊆ T which is well-founded and has height ω .(2) If T is ω -like we say that T is Aronszajn if it has no uncountable, linearly ordered subset.(3) We say that T is special if there is a function f : T → ω so that if x ≤ T y then f ( x ) = f ( y ).(4) We say that T is weakly special if there is a function f : T → ω so that if x ≤ T y, z and f ( x ) = f ( y ) = f ( z ) then y and z are comparable in the ≤ T ordering. SWITZER
Note if M is ω -like then any M -tree is ω -like. The application of MA ℵ needed to proveTheorem 2.1 is the following fact, due to Baumgartner, Malitz and Reinhardt. Fact 2.3 (Theorem 4 of [1]) . Assume MA ℵ . Let T be an ω -like tree-like partial order of size ℵ .If T is Aronszajn then T is special. We also need the following, well known fact.
Lemma 2.4.
Suppose T is an ω -like tree-like order. If T is weakly special then any forcing addinga cofinal branch collapses ω .Proof. Suppose f : T → ω witnesses that T is weakly special, P is a forcing notion and (cid:13) P “˙ b ⊆ ˇ T is a new, cofinal branch”. Let G ⊆ P be generic over V and let b = ˙ b G . We claim that (in theextension) for each n < ω the set f − ( { n } ) ∩ b is bounded. Note that this implies the lemma sincewe will have that b , which is a set of size ℵ V can be covered by countably many countable sets.To see the claim, suppose for some n < ω we have that p (cid:13) ˇ f − “( { ˇ n } ) ∩ ˙ b is unbounded”. Bystrengthening if necessary, we may assume that p decides some x ∈ T is in ˙ b and f ( x ) = n . Nowsince ˙ b is forced to be new there are incompatible extensions p and p of p and incompatibleelements x and x extending x so that for i < p i (cid:13) x i ∈ ˙ b and f ( x i ) = n . But this contradictsthe defining property of f . (cid:3) The above lemma applies in particular to M -trees. The M -tree that will be relevant here is thetree of M -finite sequences, T Mfin . This tree is defined as follows. Given an ω -like model of M let T Mfin be the set of all x ∈ M thought of as binary M -finite sequences with x ≤ fin y just in case M | =“The finite binary sequence coded by x is an initial segment of the finite sequence coded by y ”. Note the order is definable in M . An element x is on level a just in case the binary sequenceit codes has length a . Observe that this tree has uncountably many levels since M thinks thereare finite binary sequences of length a for every a ∈ M but each level is countable since there areonly 2 a many such sequences (as computed in M ) and, externally, by ω -likeness, 2 a is countable. Lemma 2.5.
Under MA ℵ T Mfin is weakly special for any Kaufmann model M . The proof of this lemma uses the fact that if the conclusion of Fact 2.3 holds then any tree ofcardinality ℵ with at most ℵ many uncountable branches is weakly special. For (well-founded)trees, this result is well known, see [2, Corollary 7.8]. The proof goes through verbatim for M -treeswhen M is ω -like, but we give the details below for the sake of completeness, as well as to presentthe proof to model theorists of arithmetic who may not be as familiar with these ideas as settheorists. Proof.
Since M is rather classless T Mfin has ℵ -many classes and hence ℵ many uncountablebranches. Enumerate all the uncountable branches by B = { b α | α < ω } . Fix an injection g : B → T Mfin so that for each α g ( b α ) ∈ b α . By [2, Lemma 7.6], one can choose g so that whenever g ( b α ) < fin g ( b β ) then g ( b β ) / ∈ b α . Now let S = { t ∈ T Mfin | ∀ b ∈ B if t ∈ b then t ≤ fin g ( b ) } . This isa tree-like order with the order inherited from T Mfin . Moreover, it’s uncountable since it containsthe range of g . It has no uncountable branches. To see this, towards a contradiction, suppose that b were an uncountable branch through S . Let ¯ b = { t ∈ T | ∃ s ∈ b t < fin s } i.e. the downwardclosure of b in T Mfin . This must be an uncountable branch through T . But then since g (¯ b ) ∈ ¯ b weget an s ∈ b with g (¯ b ) ≤ fin s contradicting the definition of S .Applying Fact 2.3, MA ℵ implies that S is special. Let f : S → ω be such a specializing function.Let t ∈ T Mfin \ S . We extend f to include t as follows. Since t / ∈ S there is a branch b so that t ∈ b but g ( b ) ≤ T Mfin t . This branch is unique: If g ( b α ) < fin g ( b β ) < fin t with t ∈ b α ∩ b β thenin particular g ( b β ) ∈ b α which contradicts the choice of g . Now let f ( t ) = f ( g ( b )) for this uniquebranch. ESTRUCTIBILITY AND AXIOMATIZABILITY OF KAUFMANN MODELS 7
Claim 2.6. f : T → ω has the property that if f ( s ) = f ( t ) = f ( u ) and s ≤ T t, u then t and u arecomparable, i.e. it witnesses that T Mfin is weakly special.Proof.
Let s ≤ T t, u be as in the claim. Since f ( t ) = f ( s ) at least one of t and s is not in S since f is injective on chains in S . In fact neither s nor t are in S unless s = g ( b ) for some b . To see this,first note that if s ∈ S then, since t / ∈ S we would have that there is some b so that b is the uniquebranch with t ∈ b and g ( b ) ≤ T Mfin t and, since s ∈ b as well and s ∈ S we have that s ≤ fin g ( b ) andso either s = g ( b ) or f ( s ) = f ( g ( b )) = f ( t ) which is a contradiction. Similarly if t ∈ S then since s / ∈ S there is some branch c so that s ∈ c but g ( c ) ≤ fin s and since g ( c ) , t ∈ S and g ( c ) < fin t wehave that f ( g ( c )) = t but this is a contradiction since f ( g ( c )) = f ( s ) = f ( t ).Now, let b be the unique branch so that t ∈ b and g ( b ) ≤ fin t . As noted before, s ∈ b as well.If s < fin g ( b ) then there is a branch c = b so that s ∈ c and g ( c ) ≤ T s (since either s = g ( c ) or isabove it, by the argument in the previous paragraph). But now g ( c ) , g ( b ) ∈ S and g ( c ) < fin g ( b )so f ( g ( c )) = f ( g ( b )) but this is a contradiction since f ( s ) = f ( g ( c )) and f ( t ) = f ( g ( b )). Therefore g ( b ) ≤ fin s , b = c and hence s ∈ b . A symmetric argument allows one to conclude the same for u so t, s, u ∈ b and hence are comparable. (cid:3) Since the claim is proved the lemma is as well. (cid:3)
Let’s now conclude the proof of Theorem 2.1.
Proof of Theorem 2.1.
Assume MA ℵ and fix a Kaufmann model M . Combining Lemmas 2.4 and2.5, it remains to show that if P is a forcing notion adding a class to M then P must add a branchto T Mfin . To see this, note that if A ⊆ M is a class, then for every x ∈ M the characteristic functionof A ∩ x is in M so it is an element of T Mfin . Hence the characteristic function of A is a branchthrough T Mfin . Since forcing cannot add new elements to M , or change definablility it must be thecase that T Mfin stays the same in any forcing extension and hence adding a class adds a branch tothis tree. This completes the proof though since adding the branch collapses ω by Lemma 2.4. (cid:3) Before moving on to the proof of the second part of Main Theorem 1.2, let’s observe some easyextensions of Theorem 2.1. These involve the following two observations from the proof: first wasthat we did not need MA ℵ only that every Aronszajn, ω -like tree-like partial order of cardinality ℵ which embeds into an ω -like tree-like partial order with countable levels is special and secondis that we didn’t use the fact that M was rather classless (or recursively saturated), only thatit had ℵ -many classes. Therefore we actually have the following result which gives a strongerconclusion from a weaker hypothesis. Theorem 2.7. If M is ω -like, has ≤ ℵ -many classes and every Aronszajn M -tree which embedsinto T Mfin is special then there is no ω -preserving forcing adding a class to M . Using the forcing of [11], the above hypothesis can be forced over a model of CH without addingreals so it’s consistent with CH that there are no destructible Kaufmann models. In fact thefollowing is consistent. Corollary 2.8.
It’s consistent that CH holds and for all ω -like models M , if M has ≤ ℵ -manyclasses then there is no ω -preserving forcing adding a class to M . Finally let us note that if there is an ω -like model with more than ℵ -many classes then T Mfin is a Kurepa tree. Since it’s consistent (relative to an inaccessible) that there are no Kurepa treesit’s consistent that there is no ω -preserving forcing notion adding a class to any ω -like model of PA . Corollary 2.9.
From an inaccessible it’s consistent, both with CH and without CH , that any forcingnotion adding a class to an ω -like model of PA collapses ω . SWITZER
Proof.
By what has been said it suffices to note that from an inaccessible, MA ℵ can be forcedalongside the failure of Kurepa’s hypothesis and (for the CH case) a countable support iteration ofthe main forcing from [11] of length κ for κ inaccessible plus some routine bookkeeping works. (cid:3) Note that it was observed by Keisler [6] that if there is a Kurepa tree, then there is a model M so that T Mfin is Kurepa, so the inaccessible is needed.3.
Building a Destructible Kaufmann Model
In this section we prove the second part of Main Theorem 1.2. Specifically we show the following.
Theorem 3.1.
Assume ♦ . Then there is a Kaufmann model M so that the satisfaction tree T Msat contains a Souslin subtree and hence is destructible.
The “hence” part follows by observing that forcing with the Souslin tree is ccc, and therefore ω -preserving, but the generic branch will define a satisfaction class for M as explained in theproof of Theorem 1.9. The idea behind the proof is to use the diamond sequence to weave to-gether Kaufmann’s original argument for the existence of a Kaufmann model with Jensen’s classicargument of the existence of a Souslin tree.Now we prove Theorem 3.1. Proof.
Fix a diamond sequence ~A = h A α | α < ω i , a countable, recursively saturated model M and a nonstandard element a ∈ M . As in the proof of Theorem 1.5, we will build a filtrationof countable, recursively saturated models hM α | α < ω i however this time we will also build a ⊆ -increasing continuous sequence of sets h S α | α < ω i so that for all α < ω we have S α ⊆ M α ,and M = S α<ω M α is a Kaufmann model and S := S α<ω S α is a Souslin subtree of T Msat relativeto a .We construct ( M α , S α ) recursively. The construction essentially mirrors Kaufmann’s originalconstruction of a Kaufmann model from ♦ done at the same time as Jensen’s original constructionof a Souslin tree from ♦ . Given any M α let T αsat be the satisfaction tree for M α relative to a . Wealready gave M , let S be T sat . Assume that we have constructed ( M ξ , S ξ ) for all ξ < α , andthat for each ξ < α M ξ is a countable recursively saturated end extension of is predecessors, S ξ is a a subset of T ξsat which intersects every level d ∈ M ξ and so that each t ∈ S ξ has extensions onall levels above it. Without loss, we can assume that each M ξ is a set of countable ordinals. Asbefore there will be a club of ξ so that M ξ = ξ .Case 1: α is a limit ordinal. By the requirements we have, M α = S ξ<α M ξ and S α = S ξ<α S ξ .Case 2: α = β + 1 for some β . If A β ⊆ M β is an undefinable class then extend M β as in Lemma1.6 so that A β is not coded into M α . Otherwise let M α be any countable, recursively saturatedend extension. Note the priority: we have M α now and will use it to define S α .If A β ⊆ S β is a maximal antichain, then do as follows. First choose a level b ∈ M α \ M β and,for each of the countably many t ∈ S β choose exactly one s t ∈ A β comparable with t and oneelement u s t ,t ∈ T αsat on the b th level that extends both s t and t . Note that by the maximalityof A β there is such an s for each t and by recursive saturation in M α there is such a u s t ,s . Theset of all such u s t ,t will be the b th level of the Souslin tree we’re constructing. Specifically, let S − α = S β ∪ { u s t ,t | t ∈ S β } and let S α be the downward closure of S − α in T αsat alongside everyextension of an element in S − α in T αsat to the levels b ′ > b in M α .If A β is not a maximal antichain of S β then let S α be simply the collection of all extensions in T αsat of every node in S β to every level in M α \ M β . This completes the construction.Let M = S α<ω M α and let S = S α<ω S α . The verification that M is Kaufmann is verbatimas in Theorem 1.5.To see that S is a Souslin tree, suppose that A ⊆ S is a maximal antichain. I claim that there isa club of ξ so that A ∩ S ξ is a maximal antichain in S ξ . Let C denote the set of all such ξ . Clearly C is closed since any increasing union of maximal antichains will again be a maximal antichain. ESTRUCTIBILITY AND AXIOMATIZABILITY OF KAUFMANN MODELS 9
To see that C is unbounded, fix an ordinal ξ := ξ . If A ∩ S ξ is not maximal then, for each of thecountably many elements t ∈ S ξ not comparable with anything in A ∩ S ξ find some element a t ∈ A which is comparable with them. Let ξ > ξ be such that A ∩ S ξ contains all of these a t ( ξ iscountable since there are only countably many things to add). Continuing in this way, recursivelydefine for each n < ω a countable ordinal ξ n +1 > ξ n so that every a ∈ A ∩ S ξ n is comparable withsomething in A ∩ S ξ n +1 . Finally let ξ ω := sup n ∈ ω ξ n . Clearly A ∩ S ξ ω is maximal by the continuityrequirement of the construction.It follows by ♦ that there is an ξ so that A ξ = A ∩ S ξ . But then there is a level d ∈ M ξ +1 sothat every element of A ξ is comparable with a node t of height d by our construction so if s ∈ T Msat is of height greater than d then s / ∈ A since it’s comparable with some node in A ξ ⊆ A . Thus A isbounded and therefore countable. (cid:3) As a remark, let us note that the above construction can also be done via forcing: let ( M , S , a )be as above and P be the set of pairs ( M , S M ) so that M ≺ M , M is recursively saturated,countable, S ⊆ S M and S M ⊆ T Msat , which has non-empty intersection with every level in M .The order is pairwise by elementary end extension and end extension as a partially ordered set.This forcing is countably closed and the verification that it adds a destructible Kaufmann modelgoes through exactly as in the proof of the theorem, replacing the ♦ construction by a collectionof density arguments. I do not know if the second coordinate is necessary or if forcing with themodels alone will make the resulting generic Kaufmann model destructible, though I suspect thatthis is the case. However, this forcing construction is weaker than the proof from diamond sincethe forcing, being countably closed and adding a subset to ω , adds a diamond sequence.Finally we note that even though ♦ implies CH , it’s consistent that there are destructibleKaufmann models and the continuum is arbitrarily large. Proposition 3.2.
Assume ♦ , then there is a destructible Kaufmann model in the extension byany number of Cohen reals.Proof. Suppose M and S are as in the proof of Theorem 3.1 (the ♦ hypothesis guarantees theirexistence). Let P be the forcing to add λ many Cohen reals for your favorite λ . Since P is ccc, itpreserves ω hence M remains an ω -like recursively saturated model. Moreover, Cohen forcingneither kills Souslin trees nor adds branches to ω -trees (like T Mfin ) so it cannot add a class to M nor kill the Souslin-ness of S . Hence M is still a Kaufmann model and S is still a ccc forcingadding a satisfaction class. (cid:3) Axiomatizability of Kaufmann Models
In this section we prove Main Theorem 1.3. The proof involves the logic, L ω ,ω ( Q ) the infinitarylogic L ω ,ω enriched with the quantifier Q where the interpretation of Qxϕ ( x ) is “there existuncountably many x so that ϕ ( x ) holds”. Recall from [5] that a standard model of L ω ,ω ( Q ) is astructure M = h M, [ M ] ≥ ω , ... i so that for any formula ϕ (¯ x, y ) and any ¯ a ∈ M ln (¯ x ) we have that M | = Qyϕ (¯ a, y ) if and only if the set { y ∈ M | M | = ϕ (¯ a, y ) } is uncountable. There is also arelatively straightforward, arithmetic Hilbert-style notion of proof for this logic, see [5, p. 69]. In[5, Theorem 4.10] Keisler proved the following completeness theorem. Theorem 4.1 (Keisler) . For any sentence of L ω ,ω ( Q ) ψ we have that ⊢ ψ if and only if for everystandard model M in the vocabulary of ψ we have M | = ψ . Note that this theorem implies that if an L ω ,ω ( Q ) sentence from V has a model in some forcingextension, then it has one in the ground model via generic absoluteness. This is the key step inShelah’s argument that there are Kaufmann models in ZFC . One thing to note is that formulas of L ω ,ω ( Q ) are coded by reals so in forcing extensions adding reals, one adds new formulas.I will need the following, elementary observation. Observation 4.2.
Suppose M is an L -structure for some L and P is a forcing notion that preserves ω . Then for any L ω ,ω ( Q ) formula ψ (¯ x ) and any tuple ¯ a in M we have that M | = ψ (¯ a ) if andonly if (cid:13) P “ ˇ M | = ψ (¯ a ) ”. Roughly this observation amounts to saying that L ω ,ω ( Q ) truth cannot be changed by ω -preserving forcing. Proof.
The proof is by induction on ψ . Since L ω ,ω satisfaction is absolute between forcing exten-sions and grounds the only non obvious case is when ψ is of the form Qxϕ ( x, ¯ y ). However, thisfollows immediately by the inductive hypothesis and the fact that P preserves ω . (cid:3) Using these results and the proofs of Theorems 2.1 and 3.1 we will show the following.
Theorem 4.3. (1) Under MA ℵ there is an L ω ,ω ( Q ) sentence ψ in the language of PA en-riched with a single unary function symbol f , L PA ( f ) , so that a model M | = PA is Kauf-mann if and only if there is an expansion of M to an L PA ( f ) -structure satisfying ψ .(2) Under ♦ there is a Kaufmann model M so that given any expansion L ′ of the language of PA and any expansion of M to an L ′ -structure, M ′ , and any countable set L of L ω ,ω ( Q ) sentences in the signature L ′ there is a model N which agrees with M ′ about the truthof every L -sentence but carries a satisfaction class for its L -reduct. In particular, the L -reduct of N is not rather classless.Remark . The wording of Part 2 is a little verbose. The point is that, even enriching M withany amount of extra structure, we can always find a model which agrees with M on any L ω ,ω ( Q )sentence and has a satisfaction class. Thus, in contrast to the case under MA ℵ , no amount ofextra structure suffices to axiomatize Kaufmann models in L ω ,ω ( Q ).As mentioned in the introduction Part 1 of the above theorem can be inferred easily from theproof of [10, Theorem 6]. I’m not sure if this was observed at the time. We give a complete, selfcontained proof here however for the convenience of the reader. Note that in the proof we willoften write sentences in the signature of L PA involving natural numbers, n < ω . By this we willalways mean the formal term n := S ( S ( ... ( S (0)) ... )) (with n iterations of the successor function S ). Since every model of PA contains a copy of the natural numbers there is no ambiguity in this. Proof of Part 1 of Theorem 4.3.
Assume MA ℵ holds. First observe that if M is a model then onecan easily write down being ω -like and recursively saturated in L ω ,ω ( Q ) as follows.(1) M is ω -like if and only if it satisfies Qx ( x = x ) ∧ ∀ y ¬ Qx ( x ≤ y )(2) M is recursively saturated if and only if it satisfies ∀ ¯ y V p ( x, ¯ y ) a computable type ( V Φ( x, ¯ y ) finite subset of p ( x, ¯ y ) ∃ x Φ( x, ¯ y ) → ∃ x V ϕ ( x, ¯ y ) ∈ p ( x, ¯ y ) ϕ ( x, ¯ y ))Therefore, what we need to show is that there is a sentence ψ in the language L PA ( f ) so that amodel M is rather classless if and only if there is a function f M : M → M so that h M, ..., f M i | = ψ .The idea is that f will be a weak specializing function for the tree T Mfin (which exists by MA ℵ )and using this function we will be able to say that all uncountable branches are definable. Shelah’ssentence from [10] says more or less the same, though because we’re not working in the generalsetup he works in there we can simplify things slightly. First note that f being essentially specialcan be expressed as follows: ES ( f ) := ∀ x [ _ n<ω f ( x ) = n ] ∧ ∀ s, t, u [ f ( s ) = f ( t ) = f ( u ) ∧ s ≤ fin t, u → ( t ≤ fin u ∨ u ≤ fin t )]So it remains to show that, for an essentially specializing function f , we can write down that f witnesses that M is rather classless. The sentence is as follows, below “ RC ” means “ratherclassless”: ESTRUCTIBILITY AND AXIOMATIZABILITY OF KAUFMANN MODELS 11 RC ( f ) := ∀ s W n<ω ( f ( s ) = n ∧ Qt ( f ( s ) = f ( t ) = n ∧ s (cid:12) fin t )) → ∃ ¯ a W ϕ ∈L PA [ ∀ yϕ ( y, ¯ a ) ↔∃ t ( s ≤ fin t ∧ t ( y ) = 1 ∧ f ( t ) = n )]Note that since elements of T Mfin are (coded) binary sequences the notation “ t ( y ) = 1” makes sense.The reader should convince themselves that in English the above says the following:“For all s , if for some n f ( s ) = n and there are uncountably many t so that s ≤ fin t and f ( t ) = n then there are an ¯ a and a formula ϕ ∈ L PA so that for all y ϕ ( y, ¯ a ) if and only if t ( y ) = 1 for some t with s ≤ fin t and f ( t ) = n .”Since, by the proof of Theorem 2.1, we know that every Kaufmann model’s T Mfin is essentiallyspecial, we need to show that M is Kaufmann if and only if its essentially specializing function f satisfies RC ( f ). Here are the details. First suppose that M is an ω -like, recursively saturatedmodel of PA which has an expansion to L PA ( f ) satisfying ES ( f ) ∧ RC ( f ). Fix such an f M : M → M . Let b be an uncountable branch through T Mfin . We need to show that there is a formula ϕ anda tuple ¯ a so that for all y ∈ M , ∪ b ( y ) = 1 if and only if M | = ϕ ( y, ¯ a ). By RC ( f M ) then there isan ¯ a and a formula ϕ ∈ L PA so that for all y ϕ ( y, ¯ a ) if and only if there is a t above s with t ( y ) = 1and f ( t ) = n . By the property of weak specializing functions, if s ≤ fin t and f ( t ) = n then t ∈ b .Therefore ∪ b ( y ) = 1 if and only if ϕ ( y, ¯ a ) as required.For the converse, suppose M is a Kaufmann model and let f M be a weak specializing for T Mfin (which exists by MA ℵ ). We claim that this f M satisfies RC ( f ). To see this, fix s ∈ T Mfin and n < ω and suppose that f M ( s ) = n there are uncountably many t above s in T Mfin with f M ( t ) = n . Thenthe set of these t must generate a cofinal branch b by weak specialness so we can define that branchas ∪ b ( y ) = 1 if and only if M | = ϕ ( y, ¯ a ) by rather classlessness, hence RC ( f ) is satisfied. (cid:3) Before continuing on to the proof of Part 2, let me comment on the relation between this proofand Shelah’s [10, Theorem 6]. This theorem, despite being foundational in the field seems tohave been very rarely written down aside from in the original article. Restricted to the case ofKaufmann models, Shelah’s proof shows much the same as what is shown above. The difference isthat he replaces the application of MA ℵ by a concrete use of a ccc forcing to specialize T Mfin . Asa result his proof shows (in our language) that every L PA reduct of a model of ES ( f ) ∧ RC ( f ) isKaufmann (this is identical to the backward direction above) and, for every Kaufmann model M there is a ccc forcing extension of V in which M has an expansion to a model of ES ( f ) ∧ RC ( f )(using MA ℵ instead of forcing this is the forward direction). By composing this result withTheorem 1.5 Shelah gets that every model of set theory has a forcing extension in which there is amodel of ES ( f ) ∧ RC ( f ). By Keisler’s completeness theorem it follows that in V this sentence isconsistent and hence has a model. But then that model’s reduct to L PA is Kaufmann thus provingthat ZFC suffices to prove the existence of Kaufmann models. A natural question is whether thedetour through forcing extensions was necessary in this argument. Part 2 will show that, at leastsometimes, the answer is “yes”.
Proof of Part 2 of Theorem 4.3.
Let M be the model constructed in the proof of Theorem 3.1 andlet S be the Souslin subtree of T Msat . The existence of this model is the only application of ♦ . Let L ′ ⊇ L PA be any language extending the language of PA and M ′ be any expansion of M to an L ′ structure. We need to show that there is an N which agrees with M ′ on any countably many L ω ,ω ( Q ) sentences but whose L -reduct has a satisfaction class (and hence is not rather classless).Since S is Souslin, the L ω ,ω ( Q ) theory of M is the same in V as in any generic extension of V by S by Observation 4.2 plus the fact that, since any Souslin tree is ω -distributive, S won’t addnew reals and hence it won’t add new L ω ,ω ( Q ) sentences either. Let G ⊆ S be generic and workin V [ G ]. In this model, the branch G codes a satisfaction class A G for M . Consider a new theory, T in the language L ′ enriched with a unary predicate A giving the L ω ,ω theory of M ′ in L ′ plus“ A is a satisfaction class”. This theory is consistent, since h M ′ , ..., A G i is a model and, moreover, it is V since it’s the union of a theory in V with a simple set of additional sentences, definable inany model of set theory. Since consistency is absolute between models of set theory with the samenatural numbers, V | =“ T is consistent”. Hence by Keisler’s completeness theorem, any countablesubtheory ¯ T ⊆ T has a model N with a satisfaction class. Consider the reduct of N to L ′ . Thismodel is exactly what we wanted so the proof is complete. (cid:3) It’s tempting to conclude in the above proof that N can be made to be fully L ω ,ω ( Q ) equivalentto M ′ but Keisler’s theorem is sentence by sentence and since L ω ,ω lacks a compactness theorem,it’s not clear that this conclusion can be made, hence the restriction to countable subtheories. I’mnot sure whether the stronger conclusion is consistent or not, though I suspect that it is.5. Conclusion and Open Questions
There remain many open questions in this area. I want to finish this paper by listing some. Themost interesting is the following.
Question . Is there a non-forcing-theoretic characterization of destructible Kaufmann models? Isthis related to some sort of resplendency or something truth theoretic?Regarding the construction of destructible Kaufmann models by forcing:
Question . Does forcing with countable, recursively saturated models ordered by end extensionadd a Kaufmann model whose satisfaction tree is Souslin (not just having a Souslin subtree)?Also, it’s worth asking:
Question . What tree types can a satisfaction tree take? In particular, can the satisfaction treefor a Kaufmann model be Souslin (and not just contain a Souslin subtree)? What about treestypes for trees of the form T Mfin ?This paper is not the first to consider strong logics in the context of Kaufmann models. Sur-prisingly though the following appears to be open.
Question . Which logics extending L ω,ω can axiomatize Kaufmann models provably in ZFC ?Consistently?Finally, while this entire discussion has concerned ℵ -Kaufmann models, there seems to be awealth of possible directions in studying general κ -Kaufmann models. Note that by Schmerl’sTheorem 1.9, if κ has the tree property then there are no κ -Kaufmann models. The converse ofthis appears to be open. Question . Does the statement “there are no ℵ -Kaufmann models” imply the tree property on ℵ ? What is the consistency strength of “there are no ℵ -Kaufmann models”? References [1] J. Baumgartner, J. Malitz, and W. Reinhardt. Embedding trees in the rationals.
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