tapani hyttinen, saharon shelah and jouko vaananen- more on the ehrenfencht-fraisse game of length...

8/3/2019 Tapani Hyttinen, Saharon Shelah and Jouko Vaananen- More on the Ehrenfencht-Fraisse game of length omega-1

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wins. The notion of winning strategy is defined in the usual manner. The

game EFG(A,B) is defined like EFG(A,B) except that the players playsequences of length < at a time. Thus EFG(A,B) is the same game asEFG2(A,B).

It was proved in [8] that, assuming 1, there are models A and B ofcardinality 2 such that the game G1(A,B) is non-determined. In thispaper we weaken the assumption 1, to 2 is not weakly compact in L(Corollary 8), but we can do this only if we assume CH. We do not know ifthis is possible without CH. In the other direction, it was proved in [8] thatif the 1-nonstationary ideal on 2 has a -closed dense subset, then thegame EFG1(A,B) is determined for all A and B of cardinality 2. The

assumption is equivconsistent with the existence of a measurable cardinal.In this paper we weaken the assumption to a condition which is consistentrelative to the existence of a weakly compact cardinal (Corollary 13). Thuswe establish:

Theorem 1 The following statements are equiconsistent relative to ZFC:

1. There is a weakly compact cardinal.

2. CH andEF1(A,B) is determined for all models A andB of cardinality2.

In [8] we proved in ZFC that there are structures A and B of cardinal-ity 3 with one binary predicate such that the game EFG1(A,B) is non-determined. We now improve this result under some cardinal arithmeticassumptions. We prove:

Theorem 2 Assume that 2 < 23 and T is a countable complete first ordertheory. Suppose that one of (i)-(iii) below holds. Then there are A, B |= Tof power 3 such that for all cardinals 1 < 3, EF1(A, B) is non-determined.

(i) T is unstable.

(ii) T is superstable with DOP or OTOP.

(iii) T is stable and unsuperstable and 2 3.

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This result complements the result in [8] that if T is an -stable first

order theory with NDOP, then EFG1(A,B) is determined for all models Aof T and all models B. This is actually true under the weaker assumptionthat T is superstable with NDOP and NOTOP.

Notation: We follow Jech [5] in set theoretic notation. We use Smn todenote the set { < m : cof() = n}. Closed and unbounded sets arecalled cub sets. A set of ordinals is -closed if it is closed under supremumsof ascending -sequences i : i < of its elements. A subset of a cardinalis -stationary if it meets every -closed unbounded subset of the cardinal.

1 Getting a weakly compact cardinal

In this section we show that if CH holds and EFG1(A,B) is determinedfor all models A and B of cardinality 2, then 2 is weakly compact in L(Corollary 8). We use the results from [7] that if2 is not weakly compact inL, then there is a bistationary S S20 such that for all < 2 either Sor \ S is non-stationary.

IfI is a linear order, we use (I) to denote the reverse order ofI. We calla sequence s = (s)


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and f(2i + 1) J2 for all i < nf. We can make H(J1, J2) a linear order by

ordering the functions lexicographically, i.e.

f g m nf(i < m(f(i) = g(i))&(m < nf f(m) < g(m))).

Let I0 = H(Q, +(1)) and I1 = H(I0, 1). Thus I0 = (1+I0)(+(1)

)Qand I1 = (1 + I1) 1 I0. By using Q = Q+ 1 +Q, = 1 + and 1 = 1 + 1,one gets easily the following, first for I0, and then for I1:

I0 = I0 + 1 + I0 , I1 = I1 + 1 + I1. (1)

Let I be the set of f : I1 , where f(2i) I1 and f(2i + 1) for all i < ordered lexicographically. Thus I = I I1. In fact, I is ofthe form J Q, so (ii) is true. By (1) and = 1 + one gets immediately(v). As I = I (1 + I1) 1 I0, we get from (v) easily (iii) for = 1.From this and + 1 = 1 we get immediately (iii) for < 1. Note that = + (1)

+ . If we combine this with I = I I1 and (1) = (1)

+ 1,we get (iv).

As to (i), we only have |I| = 2. We use this lemma in a context whereCH is assumed, so we could simply assume it here. But actually the lemmais true without CH, as we can construct I in L. Then |I| = 1. Note thatour I0 and I1 are in L, and the only property of 1 that we used was that itis a limit ordinal. 2

Definition 4 Suppose S S20 . We define

(S) =i


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Proof. This is like Lemma 4.7.19 in [9]. We use Lemma 3 and induction on

.Let us first assume / S. If is a successor ordinal, then ,+1(S) =

I+ I = I by (iii). If has cofinality , then ,+1(S) = I + I = I. If has cofinality 1 and S is non-stationary, then I = I 1 + I = I. Finally,if has cofinality 1 and \ S is non-stationary, then I = I + I = I, by(v).

Let us then assume S. Thus has cofinality . Therefore ,+1(S) =I + I (1)

= I, by (iv). 2

Lemma 6 Assume S S20 is such that there is no S21 with both S

and (S S20) \ S stationary. Then 0,(S) = 0,() whenever S

21 and

S is not stationary.

Proof. Let ()


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plays only elements of rank < , then directs to play also elements of

rank < only. Let D S21 such that S is non-stationary. By Lemma 6there is an isomorphism f : 0,(S) 0,(). Now can beat by usingf. 2

Corollary 8 If CH holds and EFG1(A,B) is determined for all models AandB of cardinality 2, then 2 is weakly compact in L.

2 Getting determinacy from a weakly com-

pact cardinal

In this section we show that if is weakly compact, then there is a forcingextension in which the game EFG1(A,B) is determined for all A and B ofcardinality 2.

We shall consider models A,B of cardinality 2, so we may as well assumethey have 2 as universe. For such a model A and any ordinal < 2 we letA denote the structure A . Similarly B. Let us first recall the followingbasic fact from [8]:

Lemma 9 [8] Suppose A andB are structures of cardinality 2. If does

not have a winning strategy in EFG1(A,B), then

S = { : A = B}

is 1-stationary.

This shows that to get determinacy of EFG1(A,B) it suffices to givea winning strategy of under the assumption that the above set S is 1-stationary. In [8] an assumption I() was used. This assumption says thatthe non-1-stationary ideal on 2 has a -closed dense set. The rough ideawas that uses the Pressing Down Lemma on S to normalize his moves

so that he always has an 1-stationary sets of possible continuations of thegame. We use now the same idea. The hypothesis I() is equiconsistentwith a measurable cardinal. Since we assume only the consistency of a weaklycompact cardinal, we have to work more.

Suppose is a weakly compact cardinal. Let I denote the 11-ideal on ,i.e.. the ideal of subsets of generated by the sets { : (H(), , AH()) |=

6


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}, where A H() and is a 11-sentence such that (H(), , A) |= . We

collapse to 2 and then force a cub to the complement of every set S S21in I. In the resulting model the above normalization strategy of workseven though the non-1-stationary ideal on 2 may not have a -closed denseset.

Definition 10 LetF be a set of cardinality of regressive functions and S . The game PDG1(S, F) has two players called and . Theyalternately play 1 rounds. During each round first chooses fi F. Then chooses a subset Si of

j , and to 0 otherwise, andg is the regressive function mapping (= 0)to (h)

1() if > , and to 0 otherwise. Suppose has a winning strategyin PDG1(S, F). Then has a winning strategy in the game EFG

21

(A,B).

Proof. We present the proof for EFG21(A,B). The case of EFG21

(A,B)

is similar. H = {h : S}, where h : A = B for S. Let bea winning strategy of in the game PDG1(S, F). Suppose the sequence(xi, yi) : i < has been played, where < 1, xi denotes a move of and yi a move of . Suppose plays next x. During the game also playsPDG1(S, F). Let us denote his moves in PDG1(S, F) by Si. Thus Sj Sifor i < j < . The point of the sets Si is that has taken care that for alli < and j Si we have yi = hj(xi) or xi = hj(yi) depending on whetherxi A or xi B. Let S

=

i


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Proof. We may assume GCH. Suppose is weakly compact. Let Q be

the Levy-collapse of to 2. In VQ we define by induction a sequence P, < +, of forcing notions. Let (A), <

+, be a complete list of all sets inthe 11-idealI on such that every element of A has uncountable cofinality.If is limit of cofinality 1, then P is the inverse limit of all P, < .For other limit , P is the direct limit ofP, < . At successor stages welet P+1 = P R, where R is defined as follows: q R iffq is a boundedclosed sequence of elements of such that q A = . R is ordered bythe end extension relation. Thus each P is countably closed. Let P = P+.Now Q P satisfies the +-chain condition. Note also that for all < +,Q P has power . We prove that it is true in V

Q that P does not add

new subsets of of cardinality 1, hence remains 2 also after forcingwith P. It follows also that Q P and each Q P are countably closed.We show now that in VQP the claim is true. Suppose S and a set F =

{f : < }, of regressive functions are given in VQP such that (inVQP) S S21 is 1-stationary. Suppose <

+ is such that S, F and fi areQ P-names for S, F and fi, correspondingly. Since S is 1-stationary inVQP, S is not in the ideal generated by I in VQP. Suppose (p,q) S / I.For a contradiction, suppose also that (p,q) forces that does not have awinning strategy in the game BM1(S, F).

Let (B0, ) be a sufficiently elementary substructure of (V, ) such that|B

0

| = , B


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have the same interpretation in B1[G][K]. We do not distinguish j(X) and

Y. With this notation, there is r R which forces in B1[G], that j(S).Then there is some (p, q) (p,q) in G that in B1 forces the existence ofsuch R and r. So we may assume that G is generic over V and our VQP isthe same as V[G].

We describe in B1[G] a winning strategy of in the game BM1(S, F).This is a contradiction since all possible winning plays of are in B1[G] andbeing unbounded is absolute in transitive models. The strategy of is toplay on the side conditions qi in B1[G] and sets Si B0[G] with QP-namesSi in B0 such that

1. qi R.

2. q0 r.

3. i < k < 1 implies qk qi.

4. i < k < 1 implies Sk Si S.

5. qi R j(Si) in B1[G].

Suppose has followed this strategy, forming conditions qi and sets Sifor i < k. Let p = inf({qi : i < k}). If we let S to be

i


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3 Non-determinacy and structure theory

In this section we prove Theorem 2, which essentially establishes, undercardinality assumptions concerning the continuum, the existence of non-determined Ehrenfencht-Frasse games of length 1 for models ofnon-classifiabletheories. This complements the observation, made in [8], that the Ehrenfencht-Frasse game of length 1 is determined for models of classifiable theories.

We start be proving Theorem 2 under assumption (iii), which we considerthe most intereting case. That is, we start with a countable complete stableand unsuperstable first order theory and show that, assuming 2 3, ithas two models A and B of cardinality 3 for which EFG1(A,B) is non-determined. Actually, we construct A and B so that

does not have a

winning strategy even in EFG2+(A,B) and does not have a winningstrategy even in EFG31(A,B).

We then prove Theorem 2 under assumption (i), that is, we now start witha countable complete unstable first order theory and show that, assuming2 < 23, it has two models A and B of cardinality 3 for which EFG1(A,B)is non-determined.

Theorem 2 under assumption (ii) can be dealt with in the same wayas under assumption (i). The section ends with some remarks on possibleimprovements.

3.1 The stable unsuperstable case

We will prove Theorem 2, case (iii), in a series of lemmas. We assume 3 = 3all the time. Let T be a countable complete stable and unsuperstable firstorder theory. As usual, we work inside a large saturated model M of T.We start by fixing some notation. By a tree I we mean a lexicographicallyordered downwards closed subtree of


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We write u v if r(u, v) = r(u, u). For more on these definitions, see [3]. In

[3], it is shown that there are models A and Au, u P(I), and sequencesa from A{}, I, such that

(i) A =

uP(I)Au |= T,

(ii) if u v, then Au Av,

(iii) for all u, v P(I), Au Ar(u,v) Av,

(iv) for all u P(I), |Au| 3,

(v) if P() holds and is an immediate successor of , then

a A{} a.

These models are exactly what we want except that they are too large, wewant the models Au, u P(I), to be countable. In order to get this, we usethe Ehrenfeucht-Mostowski construction.

We extend the signature L ofT to L by adding 3 new function symbols,some of which will be interpreted in M so that they provide Skolem-functionsfor the L-formulas. In addition we interpret the functions so that if we writeSH(u) for the L-Skolem-hull of {a| {} u} then

(vi) for all u P(I), SH(u) = Au.

By the usual argument (using [11, Appendix Theorem 2.6] and compactness)we can interpret the new function symbols so that M remains sufficientlysaturated and the following holds

(vii) if U is a downwards closed subtree of I and f is an automorphism ofU, then there is an L-automorphism g of uP(U)Au such that for all U, g(a) = af().

Finally, it is easy to see that we can choose countable L1 L so that L L1,

L1 contains the Skolem-functions for the L-formulas and if we write SH1(u)for the L1-Skolem-hull of {a| {} u} then

(viii) for all u, v P(I), SH1(u) SH1(v) SH(v).

So we have proved the following lemma (for the notion proper for Ktr andthe Ehrenfeucht-Mostowski models EM1(J, ), see [4] Definition 8.1 or [11]).

11


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Lemma 14 There are countable L1 L and proper for Ktr such that the

following holds:

(a) For all J Ktr there are an L1-model EM1(J, ) |= T and sequences

a EM1(J, ), J, such that EM1(J, ) is the L1-Skolem-hullof {a| J} (i.e. {a| J} is the skeleton of EM

1(J, ) and asbefore foru J, SH1(u) denotes the L1-Skolem hull of{a| {} u}).

(b) If U is a downwards closed subtree of J and f is an automorphism of U,then there is anL1-automorphismg ofSH1(U) such that for all U,g(a) = af().

(c) Assume (i)i


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(4) for all and I, if P() does not hold, then there is an

immediate successor of such that I+1 I

(5) if (i)i


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(vi) player I does not have a winning strategy for the following game: The

length of the game is . At each move i < , first I chooses Mi andthen II chooses i < 3. I must play so that in the end (i), (ii) and(iv) above are satisfied. I wins if he has played according to the rulesand there is no < 3 such that ((Mi)i


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Lemma 17 does not have a winning strategy for EF31 (A, B).

Proof. For this it is enough to show that A does not have a winningstrategy for EF31 (I, J), which is clear by (2) and (3) above and Theorem 15.2

Lemma 18 does not have a winning strategy for EF2+(A, B).

Proof. For a contradiction, assume is a winning strategy of for thegame EF2+(A, B). We play a round of the game defined in Theorem 16(vi). We let player I play so that he follows the rules and

(i) for all i < , Mi (H(3), A, B, ),

(ii) for all , Mi, if , I Mi and P() does not hold, thenthere is (I+1 I) Mi+1 such that is an immediate successorof ,

(iii) the Skolem-hulls of {a| I Mi} and {a| J Mi} are subsetsof Mi+1,

(iv) A Mi is a subset of the Skolem hull of {a| I Mi+1} and B Miis a subset of the Skolem hull of {a| J Mi+1},

(v)

{rk()| I Mi}

{rk()| J Mi} Mi+1.

By Theorem 16 (vi), the round can be played so that loses. Let i, i < ,be the choices made and such that ((Mi)i


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3.2 The unstable case

We will prove Theorem 2, case (i), again in a series of lemmas. We assume3 < 2

3 . Let T be a countable complete unstable first order theory. Let Lbe the signature of T.

Theorem 19 ([11]) Assume T is a countable unstable theory in the sig-narute L. There are a countable signature L1 L, a complete Skolem theoryT1 T in the signature L1, a first-order L-formula (x, y) and properfor (, T1) (see [Sh1] Definition VII 2.6) such that for every linear order Ithere is an Ehrenfeucht-Mostowski model EM1(I, ) of T1 with a skeleton{a| I} such that

EM1(I, ) |= (a, a) iff I |= < .

We write EM(I, ) for EM1(I, ) L. Notice that by using the termi-nology from [12, Definition III 3.1], {a| I} is weakly (, )-skeleton-likein EM(I, ).

In order to use Theorem 19, linear orders are needed. If A is a linearordering, x A and B A, then by x < B we mean that for every y B,x < y, x > B and C > B, C A are defined similarly. By A we meanthe inverse of A. Again let S, U and C, S, be as in [8, Lemma 16], i.e.

Theorem 15 above, with the exception that 0 S and for all S {0},0 C. By induction on i < 3, we will define linear orders Ai and B

i,

< 3, and for i S, isomorphisms

Gi :


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We will do this by induction on i. However, in order to be able to show that

(3) holds in each step, we need additional machinery.Let C {A, B}. We say that (I, J) is a (C,i,)-cut if I is an initial

segment of Ci and J = Ci I. We say that the cut is basic if I = . We

define a notion of forbidden cut by induction on i as follows (we should talkabout i-forbidden cuts, but i is always clear from the context):

(a) for all limit , the basic (C, 0, )-cut is forbidden,

(b) if (I, J) is a (C,i,)-cut, j < i and (Cj I, Cj J) is forbidden, then

(I, J) is forbidden,

(c) if (I, J) is a forbidden (A,i,)-cut, I = I


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Proof. Immediate. 2

Now we are ready to do the construction: For i = 0, the linear ordersare defined by (1) and we let G0 be the only possible one. Clearly (1)-(6)hold. If i S or sup Ci = i, then we let Ai = j


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Now (1), (2) and (6) hold trivially, (4) is already shown to hold and by

Lemma 2.9 (iii), (4) implies (3). So we are left to show that

Lemma 21 (5) holds.

Proof. Assume (I, J) is a forbidden (E , i, )-cut, E {A, B}, and Sis such that < i and E J is coinitial in J, the other case is similar. If

j + 1 and both J (k


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Proof. For this, it is enough to show that A does not have a winning

strategy for EF31 (() 1, () 1), which follows easily from (2) in theconstruction of A and B and Theorem 15 (see e.g. [8, Claim 3 in the proofof Theorem 17]). 2

Lemma 23 There are < < 23 such that E does not have a winningstrategy for EF21(A, A).

Proof. By using the usual forcing notion, we collapse 3 to an ordinal ofpower 1. Since this forcing notion does not kill those stationary subsets of3 which consist of ordinals of cofinality and cofinalities 1 are preserved,in the extension, inv2

(

) = inv2

(

) for all = . Clearly, the skeletons

of the models A, remain weakly (, )-skeleton-like in A. So by (the proofof) [12, Lemma III 3.15 (1)], inv2() INV

2 (A, ) in the extension (for

the definition of INVn , see [12, Definition III 3.11] and notice that A= B

implies INV2 (A, ) = INV2 (B, )). Also by [Sh2] Lemma III 3.13 (1),

|INV2 (A, )| = 1. Since 3 < 2

3 in the ground model, in the genericextension, (23)V is a cardinal > 1. So there are < < (2

3)V such thatA = A in the extension. Since countable subsets are not added, E doesnot have a winning strategy for EF21(A, A) (in the ground model). 2

Now Lemmas 2.11 and 2.12 imply Theorem 2 (i). 2Before proving the theorem, we make some remarks which follow from

the proof.

Remark 24 In many cases in Theorem 2, the assumption on 2 can be re-moved. For example, this is true of linear orders. An easy proof for this isgiven in [2], alternatively this follows immediately from the proof of Theo-rem 2 (i) by checking where the assumption 2 < 23 was needed. Anothercase where the assumption on 2 can be removed is the case that = 3 inthe stable unsuperstable case. This follows from the proof of Theorem 2 (iii)by noticing that the black box can now be replaced by an argument from [3].Another remark is that in Theorem 2 (i) and (ii), 3 can be replaced by any

cardinal 3 such that is a successor of a regular cardinal and 2

>

.Finally, in Theorem 2 (iii), 3 can be replaced by any cardinal 3 suchthat is a successor of a regular cardinal and = .

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References

[1] Matthew Foreman. Games played on Boolean algebras. J. SymbolicLogic 48 (1983), no. 3, 714723.

[2] Taneli Huuskonen, Comparing notions of similarity for uncountablemodels, Journal of Symbolic Logic, 60, 1995, 11531167,

[3] Tapani Hyttinen and Saharon Shelah, On the number of elementarysubmodels of an unsuperstable homogeneous structure, MathematicalLogic Quarterly (44) 1998, 354358.

[4] Tapani Hyttinen and Heikki Tuuri, Constructing strongly equivalentnonisomorphic models for unstable theories, Annals of Pure and AppliedLogic (52) 1991.

[5] T. Jech, Set Theory, Academic Press, 1978.

[6] T. Jech, M. Magidor, W. Mitchell and K. Prikry, Precipitous ideals,Journal of Symbolic Logic 45 (1980), 18.

[7] Menachem Magidor, Reflecting stationary sets, Journal of SymbolicLogic, 47, 1982, 755771

[8] Alan H. Mekler, Saharon Shelah, and J. Vaananen. The Ehrenfeucht-Frasse-game of length 1. Transactions of the American MathematicalSociety, 339:567580, 1993.

[9] Taneli Huuskonen, Tapani Hyttinen and Mika Rautila. On potentialisomorphism and non-structure, to appear.

[10] Saharon Shelah. Reflecting stationary sets and successors of singularcardinals. Archive for Mathematical Logic, 31:2553, 1991.

[11] Shelah, Saharon, Classification theory and the number of nonisomor-

phic models, Second edition, North-Holland Publishing Co., Amster-dam, xxxiv+705 pp, Studies in Logic and the Foundations of Mathe-matics, Vol. 92, 1990

[12] Shelah, Non-structure Theory, to appear.

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[13] Shelah, Saharon and Stanley, Lee, A theorem and some consistency

results in partition calculus, Annals of Pure and Applied Logic, 36,1987, 119152.

[14] Shelah, Saharon and Stanley, Lee, More consistency results in partitioncalculus, Israel Journal of Mathematics, 81, 1993, 97110.

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tapani hyttinen, saharon shelah and jouko vaananen- more on the ehrenfencht-fraisse game of length...

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