CONSTRUCTING STRONGLY EQUIVALENT NONISOMORPHIC
우리는 unsuperstable 이론 T가 주어진 경우에 대해 다음 정리를 증명한다.
A와 B는 Ehrenfeucht-Fraisse 게임에서 ∃의 승리 전략을 가지고 있으며 모델이 매우 비슷해 보인다고 가정해 볼 때, 하지만 A ≠ B 인 경우에 대하여
1. unsuperstable 이론 T가 Skolem 함수를 가질 때,
2. unsuperstable 이론 T가 superstable 이하일 때,
두 경우 모두에 대해, unsuperstable 이론 T의 특성이 반영되는 E-M 모델 EM(I, Φ)와 index 모델 I ∈ Kκtr(θ)가 존재한다고 말할 수 있다.
이 정리는 unsuperstable 이론 T가 Skolem 함수를 가질 때, unsuperstable 이론 T는 E-M 모델 EM(I, Φ)를 통해 매우 비슷한 모델 A와 B를 만들수 있다는 것을示している. 또한 unsuperstable 이론 T가 superstable 이하일 때, unsuperstable 이론 T는 index 모델 I ∈ Kκtr(θ) 를 통해 매우 비슷한 모델 A와 B를 만들 수 있다는 것을 말하고 있다.
이 정리는 unsuperstable 이론의 특성에 따라서 많은 종류의 model과 Ehrenfeucht-Fraisse 게임에서 ∃의 승리 전략을 가지고 있는 모델들 간의 관계를 연구할 때 유용하게 사용될 수 있을 것이다.
CONSTRUCTING STRONGLY EQUIVALENT NONISOMORPHIC
arXiv:math/9202205v1 [math.LO] 15 Feb 1992CONSTRUCTING STRONGLY EQUIVALENT NONISOMORPHICMODELS FOR UNSUPERSTABLE THEORIES, PART BTapani Hyttinen and Saharon Shelah∗AbstractWe study how equivalent nonisomorphic models of unsuperstable theories can be. We measurethe equivalence by Ehrenfeucht-Fraisse games.
This paper continues [HS].1. IntroductionIn [HT] we started the studies of so called strong nonstructure theorems.
By strong nonstructuretheorem we mean a theorem which says that if a theory belongs to some class of theories then it hasvery equivalent nonisomorphic models. Usually the equivalence is measured by the length of theEhrenfeucht-Fraisse games (see Definition 2.2) in which ∃has a winning strategy.
These theoremsare called nonstructure theorems because intuitively the models must be complicated if they are veryequivalent but still nonisomorphic. Also structure theorems usually imply that a certain degree ofequivalence gives isomorphism (see f.ex.
[Sh1] (Chapter XIII)).In [HT] we studied mainly unstable theories. We also looked unsuperstable theories but we werenot able to say much if the equivalence is measured by the length of the Ehrenfeucht-Fraisse gamesin which ∃has a winning strategy.
In this paper we make a new attempt to study the unsuperstablecase.The main result of this paper is the following: if λ = µ+ , cf(µ) = µ, κ = cf(κ) < µ, λ<κ = λ,µκ = µ and T is an unsuperstable theory, |T | ≤λ and κ(T ) > κ, then there are models A, B |= Tof cardinality λ such thatA ≡λµ×κ B and A ̸∼= B.In [HS] we proved this theorem in a special case.∗Partially supported by the United States Israel Binational Science Foundation. Publ.
529.1
FromTheorem 4.4 in [HS] we get the following theorem easily: Let Tc be the canonical exampleof unsuperstable theories i.e. Tc = T h((ωω, Ei)i<ω) where ηEiξ ifffor all j ≤i, η(j) = ξ(j).1.1 Theorem.
([HS]) Let λ = µ+ and I0 and I1 be models of Tc of cardinality λ. Assumeλ ∈I[λ]. ThenI0 ≡λµ×ω+2 I1⇔I0 ∼= I1.So the main result of Chapter 3 is essentially the best possible.In the introduction of [HT] there is more background for strong nonstructure theorems.2.
Basic definitionsIn this chapter we define the basic concepts we shall use and construct two linear orders neededin Chapter 3.2.1 Definition.Let λ be a cardinal and α an ordinal. Let t be a tree (i.e.
for all x ∈t, the set{y ∈t| y < x} is well-ordered by the ordering of t). If x, y ∈t and {z ∈t | z < x} = {z ∈t | z < y},then we denote x ∼y, and the equivalence class of x for ∼we denote [x].
By a λ, α-tree t we meana tree which satisfies:(i) |[x]| < λ for every x ∈t;(ii) there are no branches of length ≥α in t;(iii) t has a unique root;(iv) if x, y ∈t, x and y have no immediate predecessors and x ∼y, then x = y.Note that in a λ, α-tree each ascending sequence of a limit length has at most one supremum.2.2 Definition.Let t be a tree and κ a cardinal. The Ehrenfeucht-Fraisse game of length tbetween models A and B, Gκt (A, B), is the following.
At each move α:(i) player ∀chooses xα ∈t, κα < κ and either aβα ∈A, β < κα or bβα ∈B, β < κα , we willdenote this sequence by Xα;(ii) if ∀chose from A then ∃chooses bβα ∈B, β < κα , else ∃chooses aβα ∈A, β < κα, we willdenote this sequence by Yα .∀must move so that (xβ)β≤α form a strictly increasing sequence in t.∃must move so that{(aβγ, bβγ)|γ ≤α, β < κγ} is a partial isomorphism from A to B. The player who first has to breakthe rules loses.We write A ≡κt B if ∃has a winning strategy for Gκt (A, B).2.3 Definition.Let t and t′ be trees.
(i) If x ∈t, then pred(x) denotes the sequence (xα)α<β of the predecessors of x, excluding xitself, ordered by <. Alternatively, we consider pred(x) as a set.
The notation succ(x) denotes theset of immediate successors of x. If x, y ∈t and there is z, such that x, y ∈succ(z), then we saythat x and y are brothers.
(ii) By t<α we mean the set{x ∈t| the order type of pred(x) is < α}.Similarly we define t≤α . (iii) The sum t⊕t′ is defined as the disjoint union of t and t′, except that the roots are identified.2.4 Definition.Let ρi, i < α, ρ and θ be linear orders.
(i) We define the ordering ρ × θ as follows: the domain of ρ × θ is {(x, y)| x ∈ρ, y ∈θ}, andthe ordering in ρ × θ is defined by last differences, i.e., each point in θ is replaced by a copy of ρ;(ii) We define the ordering ρ + θ as follows: The domain of ρ + θ is ({0} × ρ) ∪({1} × θ) andthe ordering in ρ + θ is defined by the first difference i.e. (i, x) < (j, y) iffi < j or i = j and x < y.
(iii) We define the ordering Pi<α ρi as follows: The domain of Pi<α ρi is {(i, x)| i ∈α, x ∈ρi}and the ordering in Pi<α ρi is defined by the first difference i.e. (i, x) < (j, y) iffi < j or i = j andx < y.2
2.5 Definition.We define generalized Ehrenfeucht-Mostowski models (E-M-models for short).Let K be a class of models we call index models. In this definition the notation tpat(x, A, A) meansthe atomic type of x over A in the model A.Let Φ be a function.
We say that Φ is proper for K , if there is a vocabulary τ1 and for eachI ∈K a model M1 and tuples as, s ∈I , of elements of M1, such that:(i) each element in M1 is an interpretation of some µ(as), where µ is a τ1 -term;(ii) tpat(as, ∅, M1) = Φ(tpat(s, ∅, I)).Here s = (s0, ..., sn) denotes a tuple of elements of I and as denotes as0 ⌢· · · ⌢asn .Note that if M1 , as , s ∈I , and M′1, a′s , s ∈I , satisfy the conditions above, then thereis a canonical isomorphism M1 ∼= M′1 which takes µ(as) in M1 to µ(a′s) in M′1. Therefore wemay assume below that M1 and as , s ∈I , are unique for each I .
We denote this unique M1 byEM 1(I, Φ) and call it an Ehrenfeucht-Mostowski model. The tuples as , s ∈I , are the generatingelements of EM 1(I, Φ), and the indexed set (as)s∈I is the skeleton of EM 1(I, Φ).Note that iftpat(s1, ∅, I) = tpat(s2, ∅, J),thentpat(as1, ∅, EM 1(I, Φ)) = tpat(as2, ∅, EM 1(J, Φ)).2.6 Definition.Let θ be a linear order and κ infinite regular cardinal.
Let Kκtr(θ) be theclass of models of the formI = (M, <, ≪, H, Pα)α≤κ,where M ⊆θ≤κ and:(i) M is closed under initial segments;(ii) < denotes the initial segment relation;(iii) H(η, ν) is the maximal common initial segment of η and ν ;(iv) Pα = {η ∈M | length(η) = α};(v) η ≪ν iffeither η < ν or there is n < κ such that η(n) < ν(n) and η ↾n = ν ↾n.Let Kκtr = S{Kκtr(θ) | θ a linear order }.If I ∈Kκtr(θ) and η, ν ∈I , we define η
Furthermore the properties of the models of Kκtr are reflected to theseE-M-models.2.7 Theorem.([Sh1]). Suppose τ ⊆τ1 , T is a complete τ -theory, T1 is a complete τ1 -theory with Skolem functions and T ⊆T1.
Suppose further that T is unsuperstable, κ(T ) > κ andφn(x, yn), n < κ, witness this. (The definition of witnessing is not needed in this paper.
See [Sh1]. )Then there is a function Φ, which is proper for Kκtr, such that for every I ∈Kκtr, EM 1(I, Φ) isa τ1 -model of T1, for all η ∈I , aη is finite and for η, ξ ∈P In, ν ∈P Iκ ,(i) if I |= η < ν , then EM 1(I, Φ) |= φn(aν, aη);(ii) if η and ξ are brothers and η < ν then ξ = η iffEM 1(I, Φ) |= φn(aξ, aν).Above φn(x, yn) is a first-order τ -formula.
We denote the reductEM 1(I, Φ) ↾τby EM(I, Φ). In order to simplify the notation, instead of aη , we just write η.
It will be clear fromthe context, whether η means aη or η.Next we construct two linear orders needed in the next chapter. The first of these constructionsis a modification of a linear order construction in [Hu] (Chapter 9).3
2.8 Definition.Let γ be an ordinal closed under ordinal addition and let θγ = ( <ωγ, <),where < is defined by x < y iff(i) y is an initial segment of xor(ii) there is n < min{length(x), length(y)} such that x ↾n = y ↾n and x(n) < y(n).2.9 Lemma.Assume γ in an ordinal closed under ordinal addition. Let x ∈θγ , length(x) =n < ω and α < γ .
Let Aαx be the set of all elements y of θγ which satisfy:(i) x is an initial segment of y (not necessarily proper);(ii) if length(y) > n then y(n) ≥α.Then (Aαx, <↾Aαx) ∼= θγ .Proof. Follows immediately from the definition of θγ .If α ≤β are ordinals then by (α, β] we mean the unique ordinal order isomorphic to{δ| α < δ ≤β} ∪{δ| δ = α and limit}together with the natural ordering.
Notice that if (αi)i<δ is strictly increasing continuous sequenceof ordinals, α0 = 0, β = supi<δαi and for all successor i < δ, αi is successor, then Pi<δ(θ ×(αi, αi+1]) ∼= θ × β , for all linear-orderings θ.2.10 Lemma.Let γ be an ordinal closed under ordinal addition and not a cardinal. (i) Let α < γ be an ordinal.
Thenθγ ∼= θγ × (α + 1). (ii) Let α < β < |γ|+.
Thenθγ ∼= θγ × (α, β].Proof. (i) For all i < α we let xi = (i).
Then by the definition of θγ ,θγ ∼= (Xi<αA0xi) + Aα(),where by () we mean the empty sequence. By Lemma 2.9(Xi<αA0xi) + Aα() ∼= θγ × (α + 1).
(ii) We prove this by induction on β . For β = 1 the claim follows from (i).
Assume we haveproved the claim for β < β′ and we prove it for β′ . If β′ = δ + 1, then by induction assumptionθγ ∼= θγ × (α, δ]and soθγ × (α, δ + 1] ∼= θγ + θγ ∼= θγby (i).If β′ is limit, then we choose a strictly increasing continuous sequence of ordinals (βi)i Thenθγ × (α, β′] ∼=Xi 2.11 Corollary.Let γ be an ordinal closed under ordinal addition and not a cardinal. Ifα < |γ|+ is a successor ordinal then θγ ∼= θγ × α.Proof. Follows immediately from Lemma 2.10 (ii).2.12 Lemma.Assume µ is a regular cardinal and λ = µ+ . Then there are linear order θ ofpower λ, one-one and onto function h : θ →λ × θ and order isomorphisms gα : θ →θ for α < λsuch that the following holds:(i) if gα(x) = y then x ̸= y and either(a) h(x) = (α, y)or(b) h(y) = (α, x)but not both,(ii) if for some x ∈θ, gα(x) = gα′(x) then α = α′ ,(iii) if h(x) = (α, y) then gα(x) = y or gα(y) = x.Proof. Let the universe of θ be µ × λ. The ordering will be defined by induction. Letf : λ →λ × λbe one-one, onto and if α < α′ , f(α) = (β, γ) and f(α′) = (β′, γ′) then γ < γ′ . This f is used onlyto guarantee that in the induction we pay attention to every β < λ cofinally often.By induction on α < λ we do the following: Let f(α) = (β, γ). We define θα = (µ×(α+1), <α),hα : θα →λ × θα and order isomorphisms (in the ordering <α)gαβ : θα →θαso that(i) if α < α′ then hα ⊆hα′ and <α⊆<α′ ,(ii) if α < α′ , f(α) = (β, γ) and f(α′) = (β, γ′) then gαβ ⊆gα′β ,(iii) if gαβ(x) = y then x ̸= y and either(a) hα(x) = (β, y)or(b) hα(y) = (β, x)but not both.The induction is easy since at each stage we have µ ”new” elements to use: Let B ⊆µ × α bethe set of those element from µ × α which are not in the domain of any gα′βsuch that α′ < α andf(α′) = (β, γ′) for some γ′. (Notice that B is also the set of those element from µ × α which arenot in the range of any gα′βsuch that α′ < α and f(α′) = (β, γ′) for some γ′.) Clearly if B ̸= ∅then |B| = µ.Let Ai , i ∈Z, be a partition of µ × {α} into sets of power µ. We first define gαβ so that thefollowing is true:(a) gαβ is one-one,(b) if B ̸= ∅then gαβ ↾A0 is onto B otherwise gαβ ↾A0 is onto A−1 ,(c) if B ̸= ∅then gαβ ↾B is onto A−1,(d) for all i ̸= 0, gαβ ↾Ai is onto Ai−1 .By an easy induction on |i| < ω we can define <α so that <α′⊆<α for all α′ < α and gαβ is anorder isomorphism. We define the function hα ↾(µ × {α}) as follows:(a) if B = ∅then hα(x) = (β, gαβ(x)),(b) if B ̸= ∅and i ≥0 and x ∈Ai then hα(x) = (β, gαβ (x)),(c) if B ̸= ∅and i < 0 and x ∈Ai then hα(x) = (β, y) where y ∈Ai+1 or B is the uniqueelement such that gαβ(y) = x.It is easy to see that (iii) above is satisfied.We define θ = (µ × λ, <), where <= Sα<λ <α, h = Sα<λ hα and for all β < λ we letgβ = S{gαβ| α < λ, f(α) = (β, γ) for some γ}. Clearly these satisfy (i). (ii) follows from the factthat if gαβ(x) = y then either x ∈µ × {α} and y ∈µ × (α + 1) or y ∈µ × {α} and x ∈µ × (α + 1). (iii) follows immediately from the definition of h.5 3. On nonstructure of unsuperstable theoriesIn this chapter we will prove the main theorem of this paper i.e. Conclusion 3.19. The idea of theproof continues III Claim 7.8 in [Sh2]. Throughout this chapter we assume that T is an unsuperstabletheory, |T | < λ and κ(T ) > κ. The cardinal assumptions are: λ = µ+ , cf(µ) = µ, κ = cf(κ) < µ,λ<κ = λ, µκ = µ.If i < κ we say that i is of type n, n = 0, 1, 2, if there are a limit ordinal α < κ and k < ωsuch that i = α + 3k + n.We define linear orderings θn , n < 3, as follows. Let θ0 = λ and θ1, h′ and gα , α < λ, as θ,h and gα in Lemma 2.12. Let θ2 = θµ×ω × λ, where θµ×ω is as in Definition 2.8.For n < 2, let J−n be the set of sequences η of length < κ such that(i) η ̸= ();(ii) η(0) = n;(iii) if 0 < i < length(η) is of type m < 3 then η(i) ∈θm.Letf : (λ −{0}) →{(η, ξ) ∈J−0 × J−1 | length(η) = length(ξ) is of type 1}be one-one and onto. Then we defineh : θ1 →J−0 ∪J−1and order isomorphismsgη,ξ : succ(η) →succ(ξ),for (η, ξ) ∈rng(f), as follows:(i) gη,ξ(η ⌢(x)) = ξ ⌢(gα(x)), where α is the unique ordinal such that f(α) = (η, ξ);(ii) Assume h′(x) = (α, y), α ̸= 0, and f(α) = (η, ξ). Then h(x) = ξ ⌢(y) if gα(x) = yotherwise h(x) = η ⌢(y). If h′(x) = (0, y) then h(x) = (0) (here the idea is to define h(x) so thatlength(h(x)) is not of type 2).3.1 Lemma.Assume η ∈J−0 and ξ ∈J−1 are such that m = length(η) = length(ξ) is of type2. Let m = n + 1. If gη,ξ(η′) = ξ′ then either(a) h(η′(n)) = ξ′or(b) h(ξ′(n)) = η′but not both.Proof. We show first that either (a) or (b) holds. So we assume that (a) is not true and provethat (b) holds. Let η′(n) = x, ξ′(n) = y and f(α) = (η, ξ). Now gα(x) = y, x ̸= y and eitherh′(x) = (α, y) or h′(y) = (α, x). Because (a) is not true h′(x) ̸= (α, y) and so h′(y) = (α, x). Wehave two cases:(i) Case y > x: Because gα is order-precerving, gα(y) > y > x. So gα(y) ̸= x and by thedefinition of h, h(y) = η ⌢(x) = η′ . (ii) Case y < x: As the case y > x.Next we show that it is impossible that both (a) and (b) holds. For a contradiction assume thatthis is not the case. Then (a) implies that there is β such that h′(x) = (β, y) and gβ(x) = y. Onthe other hand (b) implies that there is γ such that h′(y) = (γ, x) and gγ(y) ̸= x. By Lemma 2.12(iii), gγ(x) = y. By Lemma 2.12 (ii) β = γ . So h′(y) = (β, x) and h′(x) = (β, y), which contradictsLemma 2.12 (i).For n < 2, let J+n be the set of sequences η of length ≤κ such that(i) η ̸= ();(ii) η(0) = n;(iii) if 0 < i < length(η) is of type m < 3 then η(i) ∈θm.Let e : θ1 →λ be one-one and onto. We define functions s and d as follows: if i < length(η) isof type 0 then d(η, i) = η(i) and s(η, i) = η(i), if i < length(η) is of type 1 then d(η, i) = η(i) ands(η, i) = e(η(i)) and if i < length(η) is of type 2 and η(i) = (d, s) then d(η, i) = d and s(η, i) = s.For n < 2 and γ < λ, we defineJ+n (γ) = {η ∈J+n | for all i < length(η), s(η, i) < γ},J−n (γ) = J+n (γ) ∩J−n .Let us fix d ∈θ1 so that h(d) = (0).6 3.2 Definition.For all η ∈J−0and ξ ∈J−1such that n = length(η) = length(ξ) is oftype 1, let α(η, ξ) be the set of ordinals α < λ such that for all η′ ∈succ(η), s(η′, n) < α iffs(gη,ξ(η′), n) < α and e(d) < α. Notice that α(η, ξ) is a closed and unbounded subset of λ. Byα(β), β < λ, we meanMin\{α(η, ξ)| η ∈J−0 (β), ξ ∈J−1 (β), length(η) = length(ξ) is of type 1}.3.3 Definition.For all η ∈J+0 and ξ ∈J+1 , we write ηR−ξ and ξR−η iff(i) η(j) = ξ(j) for all 0 < j < min{length(η), length(ξ)} of type 0;(ii) for all j < min{length(η), length(ξ)} of type 1 ξ ↾(j + 1) = gη↾j,ξ↾j(η ↾(j + 1)).Let length(η) = length(ξ) = j + 1, j of type 1, and ηR−ξ. We write η →ξ if h(η(j)) = ξ. Wewrite ξ →η if h(ξ(j)) = η.3.4 Remark.If ξ →η and ξ →η′ then η = η′ and if ηR−ξ then η →ξ or ξ →η but notboth.3.5 Definition.Let η ∈J+0 −J−0and ξ ∈J+1 −J−1 . We write ηRξ and ξRη iff(i) ηR−ξ;(ii) for every j < κ of type 2, η and ξ satisfy the following: if η ↾j →ξ ↾j then s(η, j) ≤s(ξ, j)and if ξ ↾j →η ↾j then s(ξ, j) ≤s(η, j);(iii) the set W κη,ξ is bounded in κ, where W κη,ξ is defined in the following way: Let η ∈J+0 −J<δ0(see Definition 2.3 (ii)) and ξ ∈J+1 −J<δ1thenW δη,ξ = W δξ,η = V δη,ξ ∪U δη,ξ,whereV δη,ξ = {j < δ| j is of type 2 and ξ ↾j →η ↾j andcf(s(η, j)) = µ and s(ξ, j) = s(η, j)}andU δη,ξ = {j < δ| j is of type 2 and η ↾j →ξ ↾j andcf(s(ξ, j)) = µ and s(η, j) = s(ξ, j)}.Our next goal is to prove that if J0 and J1 are such that(i) J−n ⊆Jn ⊆J+n , n = 0, 1 and(ii) if η ∈J+0 , ξ ∈J+1 and ηRξ then η ∈J0 iffξ ∈J1,then (J0, <, Fromnow on in this chapter we assumethat J0 and J1 satisfy (i) and (ii) above.The relation R designed not only to guarantee the equivalence but also to make it possible toprove that the final models are not isomorphic. Here (iii) in the definition of R plays a vital role.The pressing down elements η such that cf(s(η, i)) = µ, i of type 2, in (iii) prevents us from addingtoo many elements to Jn −J−n , n < 2.For n < 2, we write Jn(γ) = J+n (γ) ∩Jn .3.6 Definition.Let α < κ. Gα is the family of all partial functions f satisfying:(a) f is a partial isomorphism from J0 to J1 ;(b) dom(f) and rng(f) are closed under initial segments and for some β < λ they are includedin J0(β) and J1(β), respectively;(c) if f(η) = ξ then ηR−ξ;(d) if η ∈J+0 , ξ ∈J+1 , f(η) = ξ and j < length(η) of type 2, then η and ξ satisfy the following:if η ↾j →ξ ↾j then s(η, j) ≤s(ξ, j) and if ξ ↾j →η ↾j then s(ξ, j) ≤s(η, j);7 (e) assume η ∈J+0 −J<δ0and {η ↾γ| γ < δ} ⊆dom(f) and letξ =[γ<δf(η ↾γ),then W δη,ξ has order type ≤α;(f) if η ∈dom(f) and length(η) is of type 2 then{i < λ| for all d ∈θ2, η ⌢((d, i)) ∈dom(f)} ={i < λ| for some d ∈θ2, η ⌢((d, i)) ∈dom(f)} ={i < λ| for all d ∈θ2, f(η) ⌢((d, i)) ∈rng(f)} ={i < λ| for some d ∈θ2, f(η) ⌢((d, i)) ∈rng(f)}is an ordinal.We define Fα ⊆Gα by replacing (f) above by(f’) if η ∈dom(f) and length(η) is of type 2 then{i < λ| for all d ∈θ2, η ⌢((d, i)) ∈dom(f)} ={i < λ| for some d ∈θ2, η ⌢((d, i)) ∈dom(f)} ={i < λ| for all d ∈θ2, f(η) ⌢((d, i)) ∈rng(f)} ={i < λ| for some d ∈θ2, f(η) ⌢((d, i)) ∈rng(f)}is an ordinal and of cofinality < µ.The idea in the definition above is roughly the following: If f ∈Gα and f(η) = ξ then ηRξand the order type of W δη,ξ is ≤α. If f ∈Fα then not only f ∈Gα but f is such that for all smallA ⊂J0 ∪J1 we can find g ⊃f such that A ⊂dom(g) ∪rng(g) and g ∈Fα .3.7 Definition.For f, g ∈Gα we write f ≤g if f ⊆g and if γ < δ ≤κ, η ∈J+0 −J<δ0,η ↾γ ∈dom(f), η ↾(γ + 1) ̸∈dom(f), η ↾j ∈dom(g) for all j < δ and ξ = Sj<δ g(η ↾j), thenW γη,ξ = W δη,ξ .Notice that f ≤g is a transitive relation.3.8 Remark.Let f ∈Gα . We define f ⊇f bydom(f) = dom(f) ∪{η ∈J0| η ↾γ ∈dom(f) for all γ < length(η)and length(η) is limit}and if η ∈dom(f) −dom(f) thenf(η) =[γ (ii) If δ < µ then Si<δ fi ∈Fα and fj ≤Si<δ fi for all j ≤δ.8 Proof. (i) We have to check that f = Si<δ fi satisfies (a)-(f) in Definition 3.6.Excludingpurhapse (e), all of these are trivial.Without loss of generality we may assume δ is a limit ordinal. So assume η ∈J+0 −J<β0and{η ↾γ| γ < β} ⊆dom(f) and letξ =[γ<βf(η ↾γ).We need to show that W βη,ξ ≤α.If there is i < δ such that η ↾γ ∈dom(fi) for all γ < β then the claim follows immediatelyfrom the assumption fi ∈Fα . Otherwise for all γ < β we let iγ < δ be the least ordinal suchthat η ↾γ ∈dom(fiγ). Let γ∗< β be the least ordinal such that iγ∗+1 > iγ∗. Because for allγ < β , fiγ ∈Fα , we get W γη↾γ,ξ↾γ has order type ≤α. If γ∗< γ′ < β then fiγ∗≤fiγ′ and soW γ∗η↾γ∗,ξ↾γ∗= W γ′η↾γ′,ξ↾γ′ . Because W βη,ξ = Sγ<β W γη↾γ,ξ↾γ , we get W βη,ξ ≤α. (ii) As (i), just check the definitions.3.10 Lemma.If δ < κ, fi ∈Gi for all i < δ and fi ⊆fj for all i < j < δ then[i<δfi ∈Gδ.Proof. Follows immediately from the definitions.3.11 Lemma.If f ∈Fα and A ⊆J0 ∪J1 , |A| < λ, then there is g ∈Fα such that f ≤gand A ⊆dom(g) ∪rng(g).Proof. We may assume that A is closed under initial segments. Let A′ = A ∩(J−0 ∪J−1 ). Weenumerate A′ = {ai| 0 < i < µ} so that if ai is an initial segment of aj then i < j . Let γ < λ besuch that A ∪dom(f) ∪rng(f) ⊆J0(γ) ∪J1(γ). By induction on i < µ we define functions gi.If i = 0 we define gi = f ∪{((0), (1))}.If i < µ is limit then we definegi =[j (i) n = length(ai) is of type 0 or 1: Then we choose gi to be such that(a) gj ≤gi ;(b) gi ∈Fα ;(c) if ξ ∈dom(gi) −dom(gj) then ξ ∈succ(ai);(d) if ξ ∈succ(ai) and s(ξ, n) < γ then ξ ∈dom(gi);(e) if ξ ∈succ(gj(ai)) and s(ξ, n) < γ then ξ ∈rng(gi).Trivially such gi exists. (ii) n = length(aj) is of type 2: Then we choose gi to be such that (a)-(c) above and (d’)-(f’)below are satisfied.Letβ = sup{i + 1 < λ| for all d ∈θ2, ai ⌢((d, i)) ∈dom(gj)}. (d’) if ξ ∈succ(ai) then s(ξ, n) < γ + 2 iffξ ∈dom(gi);(e’) if ξ ∈succ(gj(ai)) then s(ξ, n) < γ + 2 iffξ ∈rng(gi);(f’) gi ↾{η ∈succ(ai)| β ≤s(η, n) < γ + 1} is an order isomorphism to {η ∈succ(gj(ai))| β ≤s(η, n) < β + 1} and gi ↾{η ∈succ(ai)| γ + 1 ≤s(η, n) < γ + 2} is an order isomorphism to{η ∈succ(gj(ai))| β + 1 ≤s(η, n) < γ + 2}.By Corollary 2.11 it is easy to satisfy (d’)-(f’). Because gj ∈Fα , cf(β) < µ and we do not haveproblems with (a) and (b). So there is gi satisfying (a)-(c) and (d’)-(f’).Finally we defineg =[i<µgi.It is easy to see that g is as wanted (notice that f ≤g follows from the construction, not fromLemma 3.9).9 3.12 Lemma.If f ∈Gα and A ⊆J0 ∪J1 , |A| < λ, then there is g ∈Fα+1 such that f ⊆gand A ⊆dom(g) ∪rng(g).Proof. Essentially as the proof of Lemma 3.11.3.13 Theorem.If J0 and J1 are such that(i) J−n ⊆Jn ⊆J+n , n = 0, 1 and(ii) if ηRξ, η ∈J+0 and ξ ∈J+1 then η ∈J0 iffξ ∈J1,then (J0, <, Because ∅∈F0 , the theorem follows from the previous lemmas.3.14 Corollary.If J0 and J1 are as above and Φ is proper for T , thenEM(J0, Φ) ≡λµ×κ EM(J1, Φ).Proof. Follows immediately from the definition of E-M-models and Theorem 3.13.In the rest of this chapter we show that there are trees J0 and J1 which satisfy the assumptionsof Corollary 3.14 andEM(J0, Φ) ̸∼= EM(J1, Φ).3.15 Lemma. (Claim 7.8B [Sh2]) There are closed increasing cofinal sequences (αi)i<κ in α,α < λ and cf(α) = κ, such that if i is successor then cf(αi) = µ and for all cub A ⊆λ the set{α < λ| cf(α) = κ and {αi| i < κ} ⊆A ∩α }is stationary.We define J0 −J−0 and J1 −J−1 by using Lemma 3.15. For all α < λ we define Iα0 and Iα1 . LetI00 = J−0and I01 = J−1 . If 0 < α < λ, cf(α) = κ, and there are sequence (βi)i<κ and η ∈J+0 −J−0such that(i) (βi)i<κ is properly increasing and cofinal in α;(ii) for all i < κ, cf(βi+1) = µ, βi+1 > α(βi) and βi ∈{αi| i < κ};(iii) for all 0 < i < κ of type 0 or 2, s(η, i) = βi ;(iv) for all i < κ of type 1, η(i) = d;then we choose some such η, let it be ηα , and define Iα0 and Iα1 to be the least sets such that(i) {ηα} ∪Sβ<α Iβ0 ⊆Iα0 and Sβ<α Iβ1 ⊆Iα1(ii) Iα0 ∪Iα1 is closed under R.Otherwise we let Iα0 = Sβ<α Iβ0 and Iα1 = Sβ<α Iβ1 . Finally we define J0 = Sα<λ Iα0 and J1 =Sα<λ Iα1 .3.16 Lemma.For all α < λ and η ∈(J0 ∪J1) −(J−0 ∪J−1 ), the following are equivalent:(i) η ∈(Iα0 ∪Iα1 ) −(Sβ<α Iβ0 ∪Sβ<α Iβ1 ). (ii) sup{s(η, i)| i < κ} = α.Proof. By the construction it is enough to show that (i) implies (ii). So assume (i). Because oflevels of type 0, it is enough to show that for all i < κ, s(η, i) < βi+1 . We prove this by inductionon i < κ. If i is of type 0, the claim is clear. If i is of type 1 this follows from βi+1 > α(βi) ande(d) < α(βi) together with the induction assumption. For i is of type 2, i = j + 1, it is enough toshow that s(ηα, i) ≥s(η, i). This follows easily from the fact that ηα(j) = d and length(h(d)) ̸= i.3.17 Definition.Let g : EM(J0, Φ) →EM(J1, Φ) be an isomorphism. We say that α < λ isg-saturated ifffor all η ∈J0 and ξ0, ..., ξn ∈J1 the following holds: if(i) length(η) = l + 1 and for all i < l, s(η, i) < α;(ii) for all k ≤n and i < length(ξk), s(ξk, i) < α;(iii) g(η) = t(δ0, ..., δm), for some term t and δ0, ..., δm ∈J1;then there are η′ ∈J0 and δ′0, ..., δ′n ∈J1 such that(a) g(η′) = t(δ′0, ..., δ′m);(b) length(η′) = l + 1 and η′ ↾l = η ↾l;(c) s(η′, l) < α;(d) the basic type of (ξ0, ..., ξn, δ0, ..., δm) in (J1, <, ≪, H, Pj) is the same as the basic type of(ξ0, ..., ξn, δ′0, ..., δ′m).10 Notice that for all isomorphisms g : EM(J0, Φ) →EM(J1, Φ) the set of g-saturated ordinals isunbounded in λ and closed under increasing sequences of length α < λ if cf(α) > κ.3.18 Lemma.Let Φ be proper for T . ThenEM(J0, Φ) ̸∼= EM(J1, Φ).Proof. We write Aγ for the submodel of EM(J0, Φ) generated (in the extended language) byJ0(γ). Similarly, we write Bγ for the submodel of EM(J1, Φ) generated by J1(γ). Let g be anone-one function from EM(J0, Φ) onto EM(J1, Φ). We say that g is closed in γ , if Aγ ∪Bγ isclosed under g and g−1.For a contradiction we assume that g is an isomorphism from EM(J0, Φ) to EM(J1, Φ). ByLemma 3.15 we choose α < λ to be such that(i) cf(α) = κ, for all i < κ, g is closed in αi and for all i < κ, cf(αi+1) = µ and αi+1 isg-saturated;(ii) there are sequence (βi)i<κ and η = ηα ∈J0 −J−0satisfying (i)-(iv) in the definition of(J0 −J−0 ) ∪(J1 −J−1 ).Let g(η) = t(ξ0, ..., ξn), ξ0, ..., ξn ∈J1 . Now for all k ≤n, either ξk ∈J1(βi) for some i < κ orthere is j < κ such that s(ξk, j) ≥α or length(ξk) = κ, sup{s(ξk, j)| j < κ} = α and for all j < κ,s(ξk, j) < α. By Lemma 3.16, in the last case ξk has been put to J1 at stage α.We choose i < κ so that(a) i is of type 2 and > 2;(b) for all k < l ≤n, ξk ↾i ̸= ξl ↾i;(c) for all k ≤n, if length(ξk) = κ, sup{s(ξk, j)| j < κ} = α and for all j < κ, s(ξk, j) < αthen there are ρ0, ..., ρr ∈J0 ∪J1 such that(i) ρo = η and ρr = ξk ;(ii) if p < r then ρpRρp+1;(iii) if p < r then W κρp,ρp+1 ⊆i;(iv) for all p < q ≤r, ρp ↾i ̸= ρq ↾i;(d) for all k ≤n, if ξk ∈J1(βj) for some j < κ then ξk ∈J1(βi);(e) for all k ≤n, if s(ξk, j) ≥α for some j < κ then ξk ↾jk ∈J1(βi) and jk < i, wherejk = min{j < i| s(ξk, j) ≥α}.Let l ≤l′ ≤n + 1 be such that ξk ∈J1(βi) iffk < l, length(ξk) = κ, sup{s(ξk, j)| j < κ} = α andfor all j < κ, s(ξk, j) < α iffl ≤k < l′ and ξk ↾i ̸∈J1(α) iffl′ ≤k ≤n. (Of course we may assumethat we have ordered ξ0, ..., ξm so that l and l′ exist.) If l ≤k < l′ then there are ρ0, ..., ρr ∈J1 ∪J0satisfying (c)(i)-(c)(iv) above. By the choice of η(i −1), ρp ↾i ←ρp+1 ↾i, for all p < r, and soξk ↾(i + 1) ∈J1(βi). For all k ≤n we define ξ′k as follows:(α) if k < l then ξ′k = ξk ;(β ) if l ≤k < l′ then ξ′k = ξk ↾(i + 1);(γ ) if l′ ≤k ≤n then ξ′k = ξk ↾jk .Let g(η ↾(i + 1)) = u(δ0, ..., δm), u a term and δ0, ..., δm ∈J1(βi+1). Because βi is g-saturatedthere is η′ ∈J0(βi) and δ′0, ..., δ′m ∈J1(βi) such that(a) g(η′) = u(δ′0, ..., δ′m);(b) length(η′) = i + 1 and η′ ↾i = η ↾i;(c) the basic type of (ξ′0, ..., ξ′n, δ0, ..., δm) in (J1, <, ≪, H, Pj) is the same as the basic type of(ξ′0, ..., ξ′n, δ′0, ..., δ′m).Because for all l ≤k < l′, s(ξk, i + 1) ≥βi+1 and for all l′ ≤k ≤n, s(ξk, jk) > βi+1 , it is easyto see that the basic type of (ξ0, ..., ξn, δ0, ..., δm) in (J1, <, ≪, H, Pj) is the same as the basic typeof (ξ0, ..., ξn, δ′0, ..., δ′m).Let φn , n < κ, be as in Theorem 2.7. ThenEM 1(J1, Φ) |= φi+1(u(δ′0, ..., δ′m), t(ξ0, ..., ξn)).So η′ ̸= η ↾(i + 1), η′ ↾i = η ↾i andEM 1(J0, Φ) |= φi+1(η′, η).This is impossible by Theorem 2.7 (ii).11 3.19 Conclusion.Let λ = µ+, cf(µ) = µ, κ = cf(κ) < µ, λ<κ = λ and µκ = µ. Assume Tis an unsuperstable theory, |T | ≤λ and κ(T ) > κ. Then there are models A, B |= T of cardinalityλ such thatA ≡λµ×κ B and A ̸∼= B.References[HS] T. Hyttinen and S. Shelah, Constructing strongly equivalent nonisomorphic models for unsuper-stable theories, part A, to appear. [HT] T.Hyttinen and H.Tuuri, Constructing strongly equivalent nonisomorphic models for unstabletheories, APAL 52, 1991, 203–248. [Hu] T. Huuskonen, Comparing notions of similarity for uncountable models, Dissertation, Universityof Helsinki, 1991. [Sh1] S.Shelah, Classification Theory, Stud. Logic Found. Math. 92 (North-Holland, Amsterdam, 2ndrev. ed., 1990). [Sh2] S.Shelah, Non-structure Theory, to appear.Tapani HyttinenDepartment of MathematicsP. O. Box 400014 University of HelsinkiFinlandSaharon ShelahInstitute of MathematicsThe Hebrew UniversityJerusalemIsraelRutgers UniversityHill Ctr-BuschNew BrunswickNJ 08903USA12 출처: arXiv:9202.205 • 원문 보기