The Ehrenfeucht-Fra¨ıss´e-game of length ω1
논자는 이 게임이 결정될 수 있는지 여부를 연구하고 있습니다. 특히, 모델 A와 B가 카디날리티 ℵ2인 경우에 대해 논자는 Gω1(A,B)가 불결정적일 수 있는지에 대해 관심을 가지고 있습니다.
논자는 ✷ω1 이론을 사용하여 무한 개의 움직임 게임이 결정되지 않는다는 것을 증명합니다. 또한, 모델 A와 B가 카디날리티 ℵ2인 경우 Gω1(A,B)가 결정을 불가능하게 할 수 있는 토대는 있지만, 이는 measurable 카디날로 시작할 때만 가능하다고 말하고 있습니다.
논자는 또한 ✷ω1 이론을 사용하여 무한 개의 움직임 게임이 결정되지 않는다는 것을 증명합니다. 이를 통해, 모델 A와 B가 카디날리티 ℵ3인 경우 Gω1(A,B)가 결정을 불가능하게 할 수 있는 토대가 있음을 알 수 있습니다.
논자는 이 연구의 결과를 바탕으로, 무한 개의 움직임 게임이 결정되는지 여부에 대한 답을 찾기 위한 새로운 도구를 제시하고 있습니다. 또한, 카디날리티 ℵ2의 모델 A와 B에 대해 Gω1(A,B)가 결정을 불가능하게 할 수 있는 토대의 존재를 증명함으로써, 무한 개의 움직임 게임이 결정되는지 여부에 대한 이해를 더深각하고 있습니다.
결론적으로, 이 논문은 Ehrenfeucht-Fraïssé 게임의 길이가 ω1인 경우에 대해 연구하고 있으며, 무한 개의 움직임 게임이 결정을 불가능하게 할 수 있는 토대의 존재를 증명하고 있습니다. 이 연구는 무한 개의 움직임 게임을 다루는 데 있어 새로운 도구와 아이디어를 제공합니다.
The Ehrenfeucht-Fra¨ıss´e-game of length ω1
arXiv:math/9305204v1 [math.LO] 15 May 1993The Ehrenfeucht-Fra¨ıss´e-game of length ω1byAlan MeklerDepartment of MathematicsSimon Fraser UniversityBurnaby, B.C.CanadaSaharon ShelahInstitute of MathematicsHebrew UniversityJerusalem, IsraelJouko V¨a¨an¨anenDepartment of MathematicsUniversity of HelsinkiHelsinki, FinlandMay 8, 2019AbstractLet A and B be two first order structures of the same vocabu-lary. We shall consider the Ehrenfeucht-Fra¨ıss´e-game of length ω1 ofA and B which we denote by Gω1(A, B).
This game is like the ordinaryEhrenfeucht-Fra¨ıss´e-game of Lωω except that there are ω1 moves. It isclear that Gω1(A, B) is determined if A and B are of cardinality ≤ℵ1.We prove the following results:1
Theorem 1 If V=L, then there are models A and B of cardinality ℵ2such that the game Gω1(A, B) is non-determined.Theorem 2 If it is consistent that there is a measurable cardinal,then it is consistent that Gω1(A, B) is determined for all A and B ofcardinality ≤ℵ2.Theorem 3 For any κ ≥ℵ3 there are A and B of cardinality κ suchthat the game Gω1(A, B) is non-determined.1Introduction.Let A and B be two first order structures of the same vocabulary L. Wedenote the domains of A and B by A and B respectively. All vocabulariesare assumed to be relational.The Ehrenfeucht-Fra¨ıss´e-game of length γ of A and B denoted by Gγ(A, B)is defined as follows: There are two players called ∀and ∃.
First ∀plays x0and then ∃plays y0. After this ∀plays x1, and ∃plays y1, and so on.
If⟨(xβ, yβ) : β < α⟩has been played and α < γ, then ∀plays xα after which ∃plays yα. Eventually a sequence ⟨(xβ, yβ) : β < γ⟩has been played.
The rulesof the game say that both players have to play elements of A∪B. Moreover,if ∀plays his xβ in A (B), then ∃has to play his yβ in B (A).
Thus thesequence ⟨(xβ, yβ) : β < γ⟩determines a relation π ⊆A × B. Player ∃winsthis round of the game if π is a partial isomorphism.
Otherwise ∀wins. Thenotion of winning strategy is defined in the usual manner.
We say that aplayer wins Gγ(A, B) if he has a winning strategy in Gγ(A, B).Recall thatA ≡ωω B⇐⇒∀n < ω(∃wins Gn(A, B))A ≡∞ω B⇐⇒∃wins Gω(A, B).In particular, Gγ(A, B) is determined for γ ≤ω.The question, whetherGγ(A, B) is determined for γ > ω, is the subject of this paper. We shallconcentrate on the case γ = ω1.The notion∃wins Gγ(A, B)(1)2
can be viewed as a natural generalization of A ≡∞ω B. The latter impliesisomorphism for countable models.Likewise (1) implies isomorphism formodels of cardinality |γ|:Proposition 1 Suppose A and B have cardinality ≤κ.
Then Gκ(A, B) isdetermined: ∃wins if A ∼= B, and ∀wins if A ̸∼= B.Proof. If f : A ∼= B, then the winning strategy of ∃in Gκ(A, B) is to playin such a way that the resulting π satisfies π ⊆f.
On the other hand, ifA ̸∼= B, then the winning strategy of ∀is to systematically enumerate A ∪Bso that the final π will satisfy A = dom(π) and B = rng(π). ✷For models of arbitrary cardinality we have the following simple but usefulcriterion of (1), namely in the terminology of [15] that they are “potentiallyisomorphic”.
We use Col(λ, κ) to denote the notion of forcing which collapses|λ| to κ (with conditions of cardinality less than κ).Proposition 2 Suppose A and B have cardinality ≤λ and κ is regular.Player ∃wins Gκ(A, B) if and only if ⊩Col(λ,κ) A ∼= B.Proof. Suppose τ is a winning strategy of ∃in Gκ(A, B).
Since Col(λ, κ) is<κ–closed,⊩Col(λ,κ) “τ is a winning strategy of ∃in Gκ(A, B)”.Hence ⊩Col(λ,κ) A ∼= B by Proposition 1. Suppose then p ⊩˜f : A ∼= B forsome p ∈Col(λ, κ).
While the game Gκ(A, B) is played, ∃keeps extendingthe condition p further and further. Suppose he has extended p to q and ∀has played x ∈A.
Then ∃finds r ≤q and y ∈B with r ⊩˜f(x) = y. Usingthis simple strategy ∃wins. ✷Proposition 3 Suppose T is an ω-stable first order theory with NDOP.Then Gω1(A, B) is determined for all models A of T and all models B.Proof.
Suppose A is a model of T. If B is not L∞ω1-equivalent to A, then ∀wins Gω1(A, B) easily. So let us suppose A ≡∞ω1 B.
We may assume A andB are of cardinality ≥ℵ1. If we collapse |A| and |B| to ℵ1, T will remainω-stable with NDOP, and A and B will remain L∞ω1-equivalent.
So A and3
B become isomorphic by [18, Chapter XIII, Section 1]. Now Proposition 2implies that ∃wins Gω1(A, B).
✷Hyttinen [10] showed that Gγ(A, B) may be non-determined for all γwith ω < γ < ω1 and asked whether Gω1(A, B) may be non-determined.Our results show that Gω1(A, B) may be non-determined for A and B ofcardinality ℵ3 (Theorem 17), but for models of cardinality ℵ2 the answer ismore complicated.Let F(ω1) be the free group of cardinality ℵ1. Using the combinatorialprinciple ✷ω1 we construct an abelian group G of cardinality ℵ2 such thatGω1(F(ω1), G) is non-determined (Theorem 4).
On the other hand, we showthat starting with a model with a measurable cardinal one can build a forc-ing extension in which Gω1(A, B) is determined for all models A and B ofcardinality ≤ℵ2 (Theorem 14).Thus the free abelian group F(ω1) has the remarkable property that thequestionIs Gω1(F(ω1), G) determined for all G?cannot be answered in ZFC alone. Proposition 3 shows that no model of anℵ1-categorical first order theory can have this property.We follow Jech [11] in set theoretic notation.
We use Smn to denote theset {α < ωm : cf(α) = ωn}. Closed and unbounded sets are called cub sets.A set of ordinals is λ-closed if it is closed under supremums of ascending λ-sequences ⟨αi : i < λ⟩of its elements.
A subset of a cardinal is λ-stationaryif it meets every λ-closed unbounded subset of the cardinal. The closure of aset A of ordinals in the order topology of ordinals is denoted by A.
The freeabelian group of cardinality κ is denoted by F(κ).2A non-determined Gω1(F(ω1), G) with G agroup of cardinality ℵ2.In this section we use ✷ω1 to construct a group G of cardinality ℵ2 such thatthe game Gω1(F(ω1), G) is non-determined (Theorem 4). For background onalmost free groups the reader is referred to [4].
However, our presentationdoes not depend on special knowledge of almost free groups.All groupsbelow are assumed to be abelian.4
By ✷ω1 we mean the principle, which says that there is a sequence ⟨Cα :α < ω2, α = ∪α⟩such that1. Cα is a cub subset of α.2.
If cf(α) = ω, then |Cα| = ω.3. If γ is a limit point of Cα, then Cγ = Cα ∩γ.Recall that ✷ω1 follows from V = L by a result of R. Jensen.
For a sequenceof sets Cα as above we can let Eβ = {α ∈S20 : the order type of Cα is β}.For some β < ω1 the set Eβ has to be stationary. Let us use E to denotethis Eβ.
Then E is a so called non-reflecting stationary set, i.e., if γ = ∪γthen E ∩γ is non-stationary on γ. Indeed, then some final segment Dγ ofthe set of limit points of Cγ is a cub subset of γ disjoint from E. Moreover,cf(α) = ω for all α ∈E.Theorem 4 Assuming ✷ω1, there is a group G of cardinality ℵ2 such thatthe game Gω1(F(ω1), G) is non-determined.Proof.
Let Zω2 denote the direct product of ω2 copies of the additive groupZ of the integers. Let xα be the element of Zω2 which is 0 on coordinates ̸= αand 1 on the coordinate α.
Let us fix for each δ ∈S20 an ascending cofinalsequence ηδ : ω →δ. For such δ, letzδ =∞Xn=02nxηδ(n).Let ⟨Cα : α = ∪α < ω2⟩, ⟨Dα : α = ∪α < ω2⟩and E = Eβ be obtained from✷ω1 as above.
We are ready to define the groups we need for the proof: LetG be the smallest pure subgroup of Zω2 which contains xα for α < ω2 andzδ for δ ∈E, let Gα be the smallest pure subgroup of Zω2 which containsxγ for γ < α and zδ for δ ∈E ∩α, let F (= F(ω2)) be the subgroup of Zω2generated freely by xα for α < ω2, and finally, let Fα be the subgroup of Zω2generated freely by xγ for γ < α.The properties we shall want of Gα are standard but for the sake ofcompleteness we shall sketch proofs. We need that each Gα is free and forany β /∈E any free basis of Gβ can be extended to a free basis of Gα for allα > β.5
The proof is by induction on α. For limit ordinals we use the fact thatE is non-reflecting.
The case of successors of ordinals not in E is also easy.Assume now that δ ∈E and the induction hypothesis has been verified upto δ. By the induction hypothesis for any β < δ such that β /∈E, there is n0so thatGδ = Gβ ⊕H ⊕Kwhere K is the group freely generated by {xηδ(n): n0 ≤n} and xηδ(m) ∈Gβfor all m < n0.
ThenGδ+1 = Gβ ⊕H ⊕K′where K′ is freely generated by{∞Xm=n2m−nxηδ(m): n0 ≤n}.On the other hand, if δ ∈E and {xηδ(n) : n < ω} ⊆B, where B is asubgroup of G such that zδ /∈B, then G/B is non-free, as zδ + B is infinitelydivisible by 2 in G/B.Claim 1 ∃does not win Gω1(F, G).Suppose τ is a winning strategy of ∃.Let α ∈E such that the pair(Gα, Fα) is closed under the first ω moves of τ, that is, if ∀plays his firstω moves inside Gα ∪Fα, then τ orders ∃to do the same. We shall playGω1(F, A) pointing out the moves of ∀and letting τ determine the moves of∃.
On his move number 2n ∀plays the element xηα(n) of Gα. On his movenumber 2n + 1 ∀plays some element of Fα.
Player ∀plays his moves in Fαin such a way that during the first ω moves eventually some countable directsummand K of Fα as well as some countable B ⊆Gα are enumerated. LetJ be the smallest pure subgroup of G containing B ∪{zα}.
During the nextω moves of Gω1(F, A) player ∀enumerates J and ∃responds by enumeratingsome H ⊆F. Since τ is a winning strategy, H has to be a subgroup of F.But now H/K is free, whereas J/B is non-free, so ∀will win the game, acontradiction.Claim 2 ∀does not win Gω1(F, G).6
Suppose τ is a winning strategy of ∀. If we were willing to use CH, wecould just take α of cofinality ω1 such that (Fα, Gα) is closed under τ, andderive a contradiction from the fact that Fα ∼= Gα.
However, since we do notwant to assume CH, we have to appeal to a longer argument.Let κ = (2ω)++. Let M be the expansion of ⟨H(κ), ∈⟩obtained by addingthe following structure to it:(H1) The function δ 7→ηδ.
(H2) The function δ 7→zδ. (H3) The function α 7→Cα.
(H4) A well-ordering < of the universe. (H5) The winning strategy τ.Let N = ⟨N, ∈, .
. .⟩be an elementary submodel of M such that ω1 ⊆N andN ∩ω2 is an ordinal α of cofinality ω1.Let Dα = {βi : i < ω1} in ascending order.
Since Cβi = Cα ∩βi, everyinitial segment of Cα is in N. By elementaricity, Gβi ∈N for all i < ω1. Letφ be an isomorphism Gα →Fα obtained as follows: φ restricted to Gβ0 is the<-least isomorphisms between the free groups Gβ0 and F0.
If φ is defined onall Gβj, j < i, then φ is defined on Gβi as the <-least extension of Sj
To achieve this we have to show that,when ∃plays his canonical strategy based on φ the strategy τ of ∀directs ∀to go on playing elements which are in N, that is, elements of Gα ∪Fα.Suppose a sequence s = ⟨(xγ, yγ) : γ < µ⟩, µ < ω1, has been played. Itsuffices to show that s ∈N.
Choose βi so that the elements of s are inGβi ∪Fβi. Now s is uniquely determined by φ↾Gβi and τ.
Note that becauseCβi = Cα ∩βi, φ↾Gβi can be defined inside N similarly as φ was definedabove, using Cβi instead of Cα. Thus s ∈N and we are done.We have proved that Gω1(F, G) is nondetermined.
This clearly impliesGω1(F(ω1), G) is nondetermined. ✷Remark.
R. Jensen [14, p. 286] showed that if ✷ω1 fails, then ω2 isMahlo in L. Therefore, if Gω1(A, B) is determined for all almost free groups7
A and B of cardinality ℵ2, then ω2 is Mahlo in L.If we start with ✷κ,we get an almost free group A of cardinality κ+ such that Gω1(F(ω1), A) isnondetermined.3Gω1(F(ω1), G) can be determined for all G.In this section all groups are assumed to be abelian. It is easy to see that ∃wins Gω1(F(ω1), G) for any uncountable free group G, so in this expositionF(ω1) is a suitable representative of all free groups.
In the study of determi-nacy of Gω1(F(ω1), A) for various A it suffices to study groups A, since forother A player ∀easily wins the game.Starting from a model with a Mahlo cardinal we construct a forcing exten-sion in which Gω1(F(ω1), G) is determined, when G is any group of cardinalityℵ2. This can be extended to groups G of any cardinality, if we start with asupercompact cardinal.In the proof of the next results we shall make use of stationary logicL(aa ).
For the definition and basic facts about L(aa ) the reader is referredto [1]. This logic has a new quantifier aa s quantifying over variables s rangingover countable subsets of the universe.
A cub set of such s is any set whichcontains a superset of any countable subset of the universe and which isclosed under unions of countable chains. The semantics of aa s is defined asfollows:aa sφ(s, .
. .) ⇐⇒φ(s, .
. .) holds for a cub set of s.Note that a group of cardinality ℵ1 is free if and only if it satisfiesaa s aa s′(s ⊆s′ →s′/s is free).
(2)Proposition 5 Let G be a group. Then the following conditions are equiv-alent:(1) ∃wins Gω1(F(ω1), G).
(2) G satisfies (2). (3) G is the union of a continuous chain ⟨Gα : α < ω2⟩of free subgroupswith Gα+1/Gα ℵ1-free for all α < ω2.8
Proof. (1) implies (2): Suppose ∃wins Gω1(F(ω1), G).
By Proposition 2 wehave ⊩Col(|G|,ω1) “G is free.” Using the countable completeness of Col(|G|, ω1)it is now easy to construct a cub set S of countable subgroups of G such thatif A ∈S then for all B ∈S with A ⊆B we have B/A free. Thus G satisfies(2).
(2) implies (3) quite trivially. (3) implies (1): Suppose a continuouschain as in (3) exists.
If we collapse |G| to ℵ1, then in the extension thechain has length < ω2. Now we use Theorem 1 of [8]:If a group A is the union of a continuous chain of < ω2 freesubgroups {Aα : α < γ} of cardinality ≤ℵ1 such that eachAα+1/Aα is ℵ1-free, then A is free.Thus G is free in the extension and (1) follows from Proposition 2.
✷Let us consider the following principle:(∗) For all stationary E ⊆S20 and countable subsets aα of α ∈E such thataα is cofinal in α and of order type ω there is a closed C ⊆ω2 of ordertype ω1 such that {α ∈E : aα \ C is finite} is stationary in C.Lemma 6 The principle (∗) implies that Gω1(F(ω1), G) is determined for allgroups G of cardinality ℵ2.Proof. Suppose G is a group of cardinality ℵ2.
We may assume the domainof G is ω2. Let us assume G is ℵ2-free, as otherwise ∀easily wins.
If we provethat G satisfies (2), then Proposition 5 implies that ∃wins Gω1(F(ω1), G).To prove (2), assume the contrary.By Proposition 5 we may assumethat G can be expressed as the union of a continuous chain ⟨Gα : α < ω2⟩of free groups with Gα+1/Gα non-ℵ1-free for α ∈E, E ⊆ω2 stationary.By Fodor’s Lemma, we may assume E ⊆S20. Also we may assume thatfor all α, every ordinal in Gα+1 \ Gα is greater than every ordinal in Gα.Finally by intersecting with a closed unbounded set we may assume that forall α ∈E the set underlying Gα is α.
Choose for each α ∈E some countablesubgroup bα of Gα+1 with bα + Gα/Gα non-free. Let cα = bα ∩Gα.
Wewill choose aα so that any final segment generates a subgroup containingcα.Enumerate cα as {gn: n < ω} such that each element is enumeratedinfintely often. Choose an increasing sequence (αn: n < ω) cofinal in α sothat for all n, gn ∈Gαn.
Finally, for each n, choose hn ∈Gαn+1 \ Gαn. Let9
aα = {hn: n < ω} ∪{hn + gn: n < ω}. It is now easy to check that aα is asequence of order type ω which is cofinal in α and any subgroup of G whichcontains all but finitely many of the elements of aα contains cα.By (∗) there is a continuous C of order type ω1 such that {α ∈C :aα \ C is finite} is stationary in C. Let D = ⟨C S Pα∈C bα⟩.
Since |D| ≤ℵ1,D is free.For any α ∈C, letDα = ⟨(C ∩α)[(Xβ∈(C∩α)bβ)⟩.Note that D = Sα∈C Dα, each Dα is countable and for limit point δ of C,Dδ = Sα∈(C∩δ) Dα. Hence there is an α ∈C ∩E such that aα \ C is finiteand D/Dα is free.
Hence bα + Dα/Dα is free. Butbα + Dα/Dα ∼= bα/bα ∩Dα = bα/bα ∩Gα,which is not free, a contradiction.
✷For the next theorem we need a lemma from [6]. A proof is included forthe convenience of the reader.Lemma 7 [6] Suppose λ is a regular cardinal and Q is a notion of forcingwhich satisfies the λ-c.c.
Suppose I is a normal λ-complete ideal on λ andI+ = {S ⊆λ : S ̸∈I}. For all sets S ∈I+ and sequences of conditions⟨pα: α ∈S⟩, there is a set C with λ \ C ∈I so that for all α ∈C ∩S,pα ⊩Q “{β: pβ ∈˜G} ∈J +, where J is the ideal generated by I”.Proof.
Suppose the lemma is false. So there is an I-positive set S′ ⊆S suchthat for all α ∈S′ there is an extension rα of pα and a set Iα ∈I (note: Iαis in the ground model) so thatrα ⊩{β : pβ ∈˜G} ⊆Iα.Let I be the diagonal union of {Iα: α ∈S′}.Suppose now that α < β and α, β ∈(S′ \ I).
Since β /∈I, rα ⊩pβ ̸∈˜G.Hence rα ⊩rβ ̸∈˜G. So rα, rβ are incompatible.
Hence {rα: α ∈S′ \ I}is an antichain which, since S′ is I-positive, is of cardinality λ. This is acontradiction.
✷10
Theorem 8 Assuming the consistency of a Mahlo cardinal, it is consistentthat (∗) holds and hence that Gω1(F(ω1), G) is determined for all groups Gof cardinality ℵ2.Proof. By a result of Harrington and Shelah [7] we may start with a Mahlocardinal κ in which every stationary set of cofinality ω reflects, that is, ifS ⊆κ is stationary, and cf(α) = ω for α ∈S, then S ∩λ is stationary in λfor some inaccessible λ < κ.For any inaccessible λ let Pλ be the Levy-forcing for collapsing λ to ω2.The conditions of Pλ are countable functions f : λ × ω1 →λ such thatf(α, β) < α for all α and β and each f is increasing and continuous in thesecond coordinate.
It is well-known that Pλ is countably closed and satisfiesthe λ-chain condition [11, p. 191].Let P = Pκ. Suppose p ∈P andp ⊩“ ˜E ⊆S20 is stationary and ∀α ∈˜E(˜aα ⊆α is cofinal in α and of order type ω)”.LetS = {α < κ : ∃q ≤p(q ⊩α ∈˜E)}.For any α ∈S let pα ≤p such that pα ⊩α ∈˜E.
Since P is countablyclosed, we can additionally require that for some countable aα ⊆α we havepα ⊩˜aα = aα.The set S is stationary in κ, for if C ⊆κ is cub, then p ⊩C ∩˜E ̸= ∅,whence C ∩S ̸= ∅. Also cf(α) = ω for α ∈S.
Let λ be inaccessible suchthat S ∩λ is stationary in λ. We may choose λ in such a way that α ∈S ∩λimplies pα ∈Pλ.
By Lemma 7 there is a δ ∈S ∩λ such thatpδ ⊩Pλ “ ˜E1 = {α < λ : pα ∈˜G} is stationary.”Let Q be the set of conditions f ∈P with dom(f) ⊆(κ \ λ) × ω1. Notethat P ∼= Pλ ⊗Q.
Let G be P-generic containing pδ and Gλ = G ∩Pλ forany inaccessible λ ≤κ. Then Gλ is Pλ-generic and ω2 of V [Gλ] is λ. Letus work now in V [Gλ].
Thus λ is the current ω2, E1 = {α < λ : pα ∈Gλ}is stationary, and we have the countable sets aα ⊆α for α ∈E1. Since Qcollapses λ there is a name ˜f such that⊩Q “ ˜f : ω1 →λ is continuous and cofinal.”11
More precisely ˜f is the name for the function f defined by f(α) = β if andonly if there is some g ∈G so that g(λ, α) = β. Let ˜C denote the range of˜f.
We shall prove the following statement:Claim: ⊩Q {α < λ : aα \ ˜C is finite } is stationary in ˜C.Suppose q ∈Q so that q ⊩“ ˜D ⊆ω1 is a cub.” Let M be an appropriateexpansion of ⟨H(κ), ∈⟩and ⟨Ni : i < λ⟩, Ni = ⟨Ni, ∈, . .
.⟩, a sequence ofelementary submodels of M such that:(i) Everything relevant is in N0. (ii) If αi = Ni ∩λ, then αi < αj for i < j < λ.
(iii) Ni+1 is closed under countable sequences. (iv) |Ni| = ω1.
(v) Ni = Sj
Finally, let qω = S{qn : n < ω} and β = S{βn :n < ω}. Thenqω ⊩“β ∈˜D and aγ \ ˜f”β is finite.”The claim, and thereby the theorem, is proved.
✷Corollary 9 The statement that Gω1(A, B) is determined for every structureA of cardinality ℵ2 and every uncountable free group B, is equiconsistent withthe existence of a Mahlo cardinal.12
Remark. If Gω1(A, F(ω1)) is determined for all groups A of cardinality κ+,κ singular, then ✷κ fails.
This implies that the Covering Lemma fails for theCore Model, whence there is an inner model for a measurable cardinal. Thisshows that the conclusion of Theorem 8 cannot be strengthened to arbitraryG.
However, by starting with a larger cardinal we can make this extension:Theorem 10 Assuming the consistency of a supercompact cardinal, it isconsistent that Gω1(F(ω1), G) is determined for all groups G.Proof. Let us assume that the stationary logic Lω1ω(aa) has the L¨owenheim-Skolem property down to ℵ1.
This assumption is consistent relative to theconsistency of a supercompact cardinal [2]. Let G be an arbitrary ℵ2-freegroup.
Let H be an L(aa)-elementary submodel of G of cardinality ℵ1. ThusH is a free group.
The group H satisfies the sentence (2), whence so does G.Now the claim follows from Proposition 5. ✷Corollary 11 Assuming the consistency of a supercompact cardinal, it isconsistent that Gω1(A, B) is determined for every structure A and every un-countable free group B.4Gω1(A, B) can be determined for all A and Bof cardinality ℵ2.We prove the consistency of the statement that Gω1(A, B) is determined forall A and B of cardinality ≤ℵ2 assuming the consistency of a measurablecardinal.
Actually we make use of an assumption that we call I∗(ω) concern-ing stationary subsets of ω2. This assumption is known to imply that ω2 ismeasurable in an inner model.
It follows from the previous section that somelarge cardinal axioms are needed to prove the stated determinacy.Let I∗(ω) be the following assumption about ω1-stationary subsets of ω2:I∗(ω) Let I be the ω1-nonstationary ideal NSω1 on ω2.Then I+ has aσ-closed dense subset K.Hodges and Shelah [9] define a principle I(ω), which is like I∗(ω) except thatI is not assumed to be the ω1-nonstationary ideal. They use I(ω) to prove13
the determinacy of an Ehrenfeucht-Fra¨ıss´e-game played on several boardssimulataneously.Note that I∗(ω) implies I is precipitous, so the consistency of I∗(ω) im-plies the consistency of a measurable cardinal [12].Theorem 12 ([12])The assumption I∗(ω) is consistent relative to the con-sistency of a measurable cardinal.We shall consider models A, B of cardinality ℵ2, so we may as well assumethey have ω2 as universe. For such A and α < ω2 we let Aα denote thestructure A ∩α.
Similarly Bα.Lemma 13 Suppose A and B are structures of cardinality ℵ2. If ∀does nothave a winning strategy in Gω1(A, B), thenS = {α : Aα ∼= Bα}is ω1-stationary.Proof.
Let C ⊆ω2 be ω1-closed and unbounded. Suppose S ∩C = ∅.
Wederive a contradiction by describing a winning strategy of ∀: Let π : ω1 →ω1 × ω1 × 2 be onto with α, β, d ≤π(α, β, d) for all α, β < ω1 and d < 2.If α < ω2, let θα : ω1 →α be onto. Suppose the sequence ⟨(xi, yi) : i < α⟩has been played.
Here xi denotes a move of ∀and yi a move of ∃. Duringthe game ∀has built an ascending sequence {ci : i < α} of elements of C.Now he lets cα be the smallest element of C greater than all the elementsxi, yi, i < α.
Suppose π(α) = (i, γ, d). Now ∀will play θci(γ) as an elementof A, if d = 0, and as an element of B if d = 1.After all ω1 moves of Gω1(A, B) have been played, some Aα and Bα, whereα ∈C, have been enumerated.
Since α ̸∈S, ∀has won the game. ✷Theorem 14 Assume I∗(ω).
The game Gω1(A, B) is determined for all Aand B of cardinality ≤ℵ2.Proof. Suppose ∀does not have a winning strategy.
By Lemma 13 the setS = {α : Aα ∼= Bα} is ω1-stationary. Let I and K be as in I∗(ω).
If α ∈S,let hα : Aα ∼= Bα. We describe a winning strategy of ∃.
The idea of thisstrategy is that ∃lets the isomorphisms hα determine his moves. Of course,14
different hα may give different information to ∃, so he has to decide whichhα to follow. The key point is that ∃lets some hα determine his move onlyif there are stationarily many other hβ that agree with hα on this move.Suppose the sequence ⟨(xi, yi) : i < α⟩has been played.
Again xi denotesa move of ∀and yi a move of ∃. Suppose ∀plays next xα and this is (say)in A.
During the game ∃has built a descending sequence {Si : i < α} ofelements of K with S0 ⊆S. The point of the sets Si is that ∃has taken carethat for all i < α and β ∈Si we have yi = hβ(xi) or xi = hβ(yi) dependingon whether ∀played xi in A or B.
Now ∃lets S′α ⊆Ti<α Si so that S′α ∈Kand ∀i ∈S′α(xα < i). For each i ∈S′α we have hi(xα) < i.
By normality,there are an Sα ⊆S′α in K and a yα such that hi(xα) = yα for all i ∈Sα.This element yα is the next move of ∃. Using this strategy ∃wins.
✷5A non-determined Gω1(A, B) with A and Bof cardinality ℵ3.We construct directly in ZFC two models A and B of cardinality ℵ3 withGω1(A, B) non-determined. It readily follows that such models exist in allcardinalities ≥ℵ3.
The construction uses a square-like principle (Lemma 16),which is provable in ZFC.Lemma 15 [17, 19] There is a stationary X ⊆S31 and a sequence ⟨Dα : α ∈X⟩such that1. Dα is a cub subset of α for all α ∈X.2.
The order type of Dα is ω1.3. If α, β ∈X and γ < min{α, β} is a limit of both Dα and Dβ, thenDα ∩γ = Dβ ∩γ.4.
If γ ∈Dα, then γ is a limit point of Dα if and only if γ is a limitordinal.Proof. We shall sketch, for completeness, a proof of this given by Burke andMagidor [3, Lemma 7.7].15
Let <∗be a well-ordering of H(ω3). For each α ∈S31, let ⟨Nαδ : δ < ω2⟩be a continuously increasing chain of elementary submodels of ⟨H(ω3), ∈, <∗⟩such that(N1) (ω1 + 1) ∪{ω2, α} ⊆Nα0 .
(N2) |Nαδ | ≤ω1. (N3) Nαδ ∩ω2 ∈ω2.
(N4) Nαδ ∩ω3 ∈Nαδ+1.Let Aαδ = Nαδ ∩α for each α ∈S31. Since, α ∈Nαδ , Aαδ is cofinal in α. LetX ⊆S31 be stationary such that for some δ, ρ < ω2 and for all α ∈X we have1.
δ= least ordinal of cofinality ω1 with Nαδ ∩ω2 = δ.2. The order type of Aαδ is ρ + 1.Let f : ω1 →ρ be cofinal and continuous.
Let g : ρ + 1 ∼= Aαδ such that gfmaps successors to successors. Let Dα be the image of ω1 under gf.
✷Lemma 16 There are sets S, T and Cα for α ∈S such that the followinghold:1. S ⊆S30 ∪S31 and S ∩S31 is stationary.2.
T ⊆S30 is stationary and S ∩T = ∅.3. If α ∈S, then Cα ⊆α ∩S is closed and of order-type ≤ω1.4.
If α ∈S and β ∈Cα, then Cβ = Cα ∩β.5. If α ∈S ∩S31, then Cα is cub on α.Proof.
Let S and ⟨Dα : α ∈S⟩be as in Lemma 15. Let S′ = X ∪Y , whereY consists of ordinals which are limit points < α of some Dα, α ∈X.
Ifα ∈X, we let Cα be the set of limit points < α of Dα. If α ∈Y , we let Cαbe the set of limit points < α of Dβ ∩α, where β > α is chosen arbitrarilyfrom X.Now claims 1,3,4 and 6 are clearly satisfied.16
Let S30 = Si<ω2 Ti where the Ti are disjoint stationary sets. Since |Cα| ≤ω1, there is iα < ω2 such that i ≥iα implies Ci ∩Ti = ∅.
Let S′′ ⊆S′ bestationary such that α ∈S′′ implies iα is constant i. Let T = Ti.
Finally, letS = S′′ ∪S{Cα : α ∈S′′}. Claim 2 is satisfied, and the Lemma is proved.
✷Theorem 17 There are structures A and B of cardinality ℵ3 with one binarypredicate such that the game Gω1(A, B) is non-determined.Proof. Let S, T and ⟨Cα : α ∈S⟩be as in Lemma 16.
We shall constructa sequence {Mα : α < ω3} of sets and a sequence {Gα : α ∈S} of functionssuch that the conditions (M1)–(M6) below hold. Let Wα be the set of allmappingsGd0γ0 .
. .
Gdnγn,where γ0, . .
. , γn ∈S ∩α, di ∈{−1, 1}, G1γ means Gγ and G−1γmeans theinverse of Gγ.
Let W = Wω3. (Note that W consists of a set of partialfunctions.
)The conditions on the Mα’s and the Gα’s are:(M1) Mα ⊆Mβ if α < β, and Mα ⊂Mα+1 if α ∈S. (M2) Mν = Sα<ν Mα for limit ν.
(M3) Gα is a bijection of Mα+1 for α ∈S. (M4) If β ∈S and α ∈Cβ, then Gα ⊆Gβ.
(M5) If for some β, Gβ(a) = b and for some w ∈W, w(a) = b, then thereis some γ so that w ⊆Gγ. Furthermore if β is the minimum ordinal sothat Gβ(a) = b then γ = β or β ∈Cγ.In order to construct the set M = Sα<ω3 Mα and the mappings Gα wedefine an oriented graph with M as the set of vertices.
We use the terminol-ogy of Serre [16] for graph-theoretic notions. If x is an edge, the origin of xis denoted by o(x) and the terminus by t(x).
Our graph has an inverse edgex for each edge x. Thus o(x) = t(x) and t(x) = o(x).
Some edges are calledpositive, the rest are called negative. An edge is positive if and only if itsinverse is negative.
For each edge x of M there is a set L(x) of labels. Theset of possible labels for positive edges is {gα : α < ω3}.
The negative edges17
can have elements of {g−1α: α < ω3} as labels. The labels are assumed to begiven in such a way that a positive edge gets gα as a label if and only if itsinverse gets the label g−1α .
During the construction the sets of labels will beextended step by step.The construction is analogous to building an acyclic graph on which agroup acts freely. The graph then turns out to be the Cayley graph of thegroup.
The labelled graph we will build will be the “Cayley graph” of Wwhich will be as free as possible given (M1)–(M4).Condition (M5) is aconsequence of the freeness of the construction.Let us suppose the sets Mβ, β < α, of vertices have been defined. LetM<α = Sβ<α Mα.
Some vertices in M<α have edges between them and a setL(x) of labels has been assigned to each such edge x.If α is a limit ordinal, we let Mα = M<α. So let us assume α = β + 1.If β ̸∈S, Mα = Mβ.
So let us assume β ∈S. Let γ = sup(Cβ).
Noticethat since S consists entirely of limit ordinals and Cβ ⊆S, either γ = β orγ + 1 < β.Case 1. γ = β: We extend Mβ to Mα by adding new vertices {Pz : z ∈Z}and for each z ∈Z a positive edge xPzα with o(xPzα ) = Pz and t(xPzα ) = Pz+1.We also let L(xPzα ) = {gβ} ∪{gδ: β ∈Cδ}.Case 2. γ + 1 < β: We extend Mβ to Mα by adding new vertices {P ′z : z ∈Z \ {0}} for each P ∈Mβ \ Mγ+1. For notational convenience let P ′0 = P.Now we add for each P ∈Mβ \ Mγ+1 new edges as follows.
For each z ∈Zwe add a positive edge xP ′zα witho(xP ′zα ) = P ′z, t(xP ′zα ) = P ′z+1, L(xP ′zα ) = {gβ} ∪{gδ: β ∈Cδ}This determines completely the inverse of xP ′zα .This ends the construction of the graph. In the construction each vertexP in Mα+1,α ∈S, is made the origin of a unique edge xPα with gα ∈L(xPα).We defineGα(P) = t(xPα).The construction of the sets Mα and the mappings Gα is now completed.It follows immediately from the construction that each Gα, α ∈S, is abijection of Mα+1.
So (M1)–(M3) hold. (M4) holds, because gα is added tothe labels of any edge with gβ, where β ∈Cα, as a label.
Finally, (M5) is aconsequence of the fact that the graph is circuit-free.18
Let us fix a0 ∈M1 and b0 = Gβ0(a0), where β0 ∈Cα for all α ∈S. Notethat we may assume, without loss of generality, the existence of such a β0.If a0, a1 ∈M, letR(a0,a1) = {(a′0, a′1) ∈M2 : ∃w ∈W(w(a0) = a′0 ∧w(a1) = a′1)}.We letM = ⟨M, (R(a0,a1))(a0,a1)∈M2⟩A = ⟨M, a0⟩B = ⟨M, b0⟩and show that Gω1(A, B) is non-determined.The reduction of the language of A and B to one binary predicate is easy.One just adds a copy of ω3, together with its ordering, and a copy of M ×Mto the structures with the projection maps.
Then fix a bijection φ from ω3to M2. Add a new binary predicate R to the language and interpret R tobe contained in ω3 × M2 such that R(β, (a, b)) holds if and only if Rφ(α)(a, b)holds.
We can now dispense with the old binary predicates. We have replacedour structure by one in a finite language without making any difference towho wins the game Gω1(A, B).
The extra step of reducing to a single binarypredicate is standard.An important property of these models is that if α ∈S ∩S31, then Gα↾Mαis an automorphism of the restriction of M to Mα and takes a0 to b0.Claim 3 ∀does not win Gω1(A, B).Suppose ∀has a winning strategy τ. Again, there is a quick argumentwhich uses CH: Find α ∈S such that Mα is closed under τ and cf(α) = ω1.Now Cα is cub on α, whence Gα maps Mα onto itself.
Using Gα player ∃caneasily beat τ, a contradiction.In the following longer argument we need not assume CH. Let κ be a largeregular cardinal.
Let H be the expansion of ⟨H(κ), ∈⟩obtained by addingthe following structure to it:(H1) The function α 7→Mα. (H2) The function α 7→Gα.
(H3) The function α 7→Cα.19
(H4) A well-ordering <∗of the universe. (H5) The winning strategy τ.
(H6) The sets S and T.Let N = ⟨N, ∈, . .
.⟩be an elementary submodel of H such that α = N ∩ω3 ∈S ∩S31.Now Cα is a cub of order-type ω1 on α and Gα maps Mα onto Mα.Moreover, Gα is a partial isomorphism from A into B. Provided that τ doesnot lead ∀to play his moves outside Mα, ∃has on obvious strategy: he letsGα determine his moves.
So let us assume a sequence ⟨(xξ, yξ) : ξ < γ⟩hasbeen played inside Mα and γ < ω1. Let β ∈Cα such that Mβ contains theelements xξ, yξ for ξ < γ.
The sequence ⟨yξ : ξ < γ⟩is totally determined byGβ and τ. Since Gβ ∈N, ⟨yξ : ξ < γ⟩∈N, and we are done.Claim 4 ∃does not win Gω1(A, B).Suppose ∃has a winning strategy τ.
Let H be as above and N = ⟨N, ∈, . .
.⟩be an elementary submodel of H such that α = N ∩ω3 ∈T.Welet ∀play during the first ω moves of Gω1(A, B) a sequence ⟨an : n < ω⟩in A such that if αn is the least αn with an ∈Mαn, then the sequence⟨αn : n < ω⟩is ascending and sup{αn : n < ω} = α. Let ∃respond followingτ with ⟨bn : n < ω⟩.
As his move number ω player ∀plays some elementaω ∈M \ Mα in A and ∃answers according to τ with bω.For all i ≤ω, R(a0,ai)(a0, ai) holds. Hence R(a0,ai)(b0, bi) holds.
So thereis wi such that wi(a0) = b0 and wi(ai) = bi. Since Gβ0(a0) = b0, by (M5),for each i there is βi so that Gβi(ai) = bi.
We can assume that βi is chosento be minimal. Notice that for all i, βi > αi and for i < ω, βi ∈N .
Sosup{βi: i < ω} = α.Also, by the same reasoning as above, for each i < ω, R(ai,aω)(bi, bω)holds. Applying (M5), we get that Gβω(ai) = bi.
Using (M5) again and theminmality of βi, for all i < ω, βi ∈Cβω. Thus α is a limit of elements of Cβω,contradicting α ∈T.
✷References20
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