Alan Mekler, Saharon Shelah and Jouko Vaananen- The Ehrenfeucht-Fraisse-game of length omega-1

8/3/2019 Alan Mekler, Saharon Shelah and Jouko Vaananen- The Ehrenfeucht-Fraisse-game of length omega-1

1/22

416

re

vision:1993-08-29

modified:1993-08-29

The Ehrenfeucht-Frasse-game of length 1

by

Alan Mekler

Department of Mathematics

Simon Fraser UniversityBurnaby, B.C.

Canada

Saharon Shelah

Institute of Mathematics

Hebrew University

Jerusalem, Israel

Jouko Vaananen

Department of Mathematics

University of Helsinki

Helsinki, Finland

October 6, 2003

Abstract

Let A and B be two first order structures of the same vocabu-lary. We shall consider the Ehrenfeucht-Frasse-game of length 1 ofA andB which we denote by G1(A,B). This game is like the ordinaryEhrenfeucht-Frasse-game ofL except that there are 1 moves. It isclear that G1(A,B) is determined if A and B are of cardinality 1.We prove the following results:

1


2/22

416

re

vision:1993-08-29

modified:1993-08-29

Theorem 1If V=L, then there are models A and B of cardinality 2such that the game G1(A,B) is non-determined.

Theorem 2 If it is consistent that there is a measurable cardinal,

then it is consistent that G1(A,B) is determined for all A and B ofcardinality 2.

Theorem 3 For any 3 there are A and B of cardinality such

that the game G1(A,B) is non-determined.

1 Introduction.

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-Frasse-game of lengthofA andB denoted by G(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(x, y) : < has been played and < , then plays x after which plays y. Eventually a sequence (x, y) : < has been played. The rulesof the game say that both players have to play elements of A B. Moreover,

if plays his x in A (B), then has to play his y in B (A). Thus thesequence (x, y) : < determines a relation A B. Player winsthis round of the game if is a partial isomorphism. Otherwise wins. Thenotion of winning strategy is defined in the usual manner. We say that aplayer wins G(A, B) if he has a winning strategy in G(A, B).

Recall that

A B n < ( wins Gn(A, B))

A B wins G(A, B).

In particular, G(A, B) is determined for . The question, whetherG(A, B) is determined for > , is the subject of this paper. We shallconcentrate on the case = 1.

The notion wins G(A, B) (1)

2


3/22

416

re

vision:1993-08-29

modified:1993-08-29

can be viewed as a natural generalization of A B. The latter impliesisomorphism for countable models. Likewise (1) implies isomorphism formodels of cardinality ||:

Proposition 1 Suppose A and B have cardinality . Then G(A, B) isdetermined: wins if A = B, and wins if A = B.

Proof. If f : A = B, then the winning strategy of in G(A, B) is to playin such a way that the resulting satisfies f. On the other hand, ifA = B, then the winning strategy of is to systematically enumerate A Bso that the final will satisfy A = dom() and B = rng(). 2

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(, ) to denote the notion of forcing which collapses|| to (with conditions of cardinality less than ).

Proposition 2 Suppose A and B have cardinality and is regular.Player wins G(A, B) if and only ifCol(,) A = B.

Proof. Suppose is a winning strategy of in G(A, B). Since Col(, ) is< closed,

Col(,) is a winning strategy of in G(A, B).

Hence Col(,) A = B by Proposition 1. Suppose then p f : A = B forsome p Col(, ). While the game G(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. 2

Proposition 3 Suppose T is an -stable first order theory with NDOP.

Then G1(A, B) is determined for all models A of T and all models B.

Proof. Suppose A is a model ofT. IfB is not L1-equivalent to A, then wins G1(A, B) easily. So let us suppose A 1 B. We may assume A andB are of cardinality 1. If we collapse |A| and |B| to 1, T will remain-stable with NDOP, and A and B will remain L1-equivalent. So A and

3


4/22

416

re

vision:1993-08-29

modified:1993-08-29

B become isomorphic by [18, Chapter XIII, Section 1]. Now Proposition 2implies that wins G1(A, B). 2

Hyttinen [10] showed that G(A, B) may be non-determined for all with < < 1 and asked whether G1(A, B) may be non-determined.Our results show that G1(A, B) may be non-determined for A and B ofcardinality 3 (Theorem 17), but for models of cardinality 2 the answer ismore complicated.

Let F(1) be the free group of cardinality 1. Using the combinatorialprinciple 21 we construct an abelian group G of cardinality 2 such thatG1(F(1), G) is non-determined (Theorem 4). On the other hand, we show

that starting with a model with a measurable cardinal one can build a forc-ing extension in which G1(A, B) is determined for all models A and B ofcardinality 2 (Theorem 14).

Thus the free abelian group F(1) has the remarkable property that thequestion

Is G1(F(1), G) determined for all G?

cannot be answered in ZFC alone. Proposition 3 shows that no model of an1-categorical first order theory can have this property.

We follow Jech [11] in set theoretic notation. We use Smn to denote the

set { < m : cf() = n}. Closed and unbounded sets are called cub sets.A set of ordinals is -closed if it is closed under supremums of ascending -sequences i : i < of its elements. A subset of a cardinal is -stationaryif it meets every -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 is denoted by F().

2 A non-determined G1(F(1), G) with G a

group of cardinality 2.

In this section we use 21 to construct a group G of cardinality 2 such thatthe game G1(F(1), 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


5/22

416

re

vision:1993-08-29

modified:1993-08-29

By21 we mean the principle, which says that there is a sequence C : < 2, = such that

1. C is a cub subset of .

2. If cf() = , then |C| = .

3. If is a limit point of C, then C = C .

Recall that 21 follows from V = L by a result of R. Jensen. For a sequenceof sets C as above we can let E = { S20 : the order type ofC is }.For some < 1 the set E has to be stationary. Let us use E to denote

this E. Then E is a so called non-reflecting stationary set, i.e., if = then E is non-stationary on . Indeed, then some final segment D ofthe set of limit points of C is a cub subset of disjoint from E. Moreover,cf() = for all E.

Theorem 4 Assuming21, there is a group G of cardinality 2 such thatthe game G1(F(1), G) is non-determined.

Proof. Let Z2 denote the direct product of 2 copies of the additive groupZ of the integers. Let x be the element ofZ

2 which is 0 on coordinates = and 1 on the coordinate . Let us fix for each S20 an ascending cofinalsequence : . For such , let

z =

n=0

2nx(n).

Let C : = < 2, D : = < 2 and E = E be obtained from21 as above. We are ready to define the groups we need for the proof: LetG be the smallest pure subgroup ofZ2 which contains x for < 2 andz for E, let G be the smallest pure subgroup of Z2 which containsx for < and z for E , let F (= F(2)) be the subgroup ofZ2

generated freely by x for < 2, and finally, let F be the subgroup ofZ2

generated freely by x for < .The properties we shall want of G are standard but for the sake of

completeness we shall sketch proofs. We need that each G is free and forany / E any free basis of G can be extended to a free basis of G for all > .

5


6/22

416

re

vision:1993-08-29

modified:1993-08-29

The proof is by induction on . 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 E and the induction hypothesis has been verified upto . By the induction hypothesis for any < such that / E, there is n0so that

G = G H K

where K is the group freely generated by {x(n): n0 n} and x(m) Gfor all m < n0. Then

G+1 = G H K

where K is freely generated by

{

m=n

2mnx(m): n0 n}.

On the other hand, if E and {x(n) : n < } B, where B is asubgroup ofG such that z / B, then G/B is non-free, as z + B is infinitelydivisible by 2 in G/B.

Claim 1 does not win G1(F, G).

Suppose is a winning strategy of . Let E such that the pair(G, F) is closed under the first moves of , that is, if plays his first moves inside G F, then orders to do the same. We shall playG1(F, A) pointing out the moves of and letting determine the moves of. On his move number 2n plays the element x(n) of G. On his movenumber 2n + 1 plays some element of F. Player plays his moves in Fin such a way that during the first moves eventually some countable directsummand K of F as well as some countable B G are enumerated. LetJ be the smallest pure subgroup of G containing B {z}. During the next moves ofG1(F, A) player enumerates J and responds by enumeratingsome H F. Since 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 G1(F, G).

6


7/22

416

re

vision:1993-08-29

modified:1993-08-29

Suppose is a winning strategy of . If we were willing to use CH, wecould just take of cofinality 1 such that (F, G) is closed under , andderive a contradiction from the fact that F = G. However, since we do notwant to assume CH, we have to appeal to a longer argument.

Let = (2)++. Let M be the expansion ofH(), obtained by addingthe following structure to it:

(H1) The function .

(H2) The function z.

(H3) The function C.

(H4) A well-ordering < of the universe.

(H5) The winning strategy .

Let N = N, , . . . be an elementary submodel of M such that 1 N andN 2 is an ordinal of cofinality 1.

Let D = {i : i < 1} in ascending order. Since Ci = C i, everyinitial segment of C is in N. By elementaricity, Gi N for all i < 1. Let be an isomorphism G F obtained as follows: restricted to G0 is the


8/22

416

re

vision:1993-08-29

modified:1993-08-29

A and B of cardinality 2, then 2 is Mahlo in L. If we start with2,we get an almost free group A of cardinality + such that G1(F(1), A) is

nondetermined.

3 G1(F(1), G) can be determined for all G.

In this section all groups are assumed to be abelian. It is easy to see that wins G1(F(1), G) for any uncountable free group G, so in this expositionF(1) is a suitable representative of all free groups. In the study of determi-nacy of G1(F(1), A) for various A it suffices to study groups A, since for

other A player easily wins the game.Starting from a model with a Mahlo cardinal we construct a forcing exten-

sion in which G1(F(1), G) is determined, when G is any group of cardinality2. 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 aas 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 is

closed under unions of countable chains. The semantics of aa s is defined asfollows:aas(s , . . .) (s , . . .) holds for a cub set of s.

Note that a group of cardinality 1 is free if and only if it satisfies

aas aas(s s s/s is free). (2)

Proposition 5 Let G be a group. Then the fol lowing conditions are equiv-alent:

(1) wins G1(F(1), G).

(2) G satisfies (2).

(3) G is the union of a continuous chain G : < 2 of free subgroupswith G+1/G 1-free for all < 2.

8


9/22

416

re

vision:1993-08-29

modified:1993-08-29

Proof. (1) implies (2): Suppose wins G1(F(1), G). By Proposition 2 wehave Col(|G|,1) G is free. Using the countable completeness ofCol(|G|, 1)it is now easy to construct a cub set S of countable subgroups ofG 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 1, then in the extension thechain has length < 2. Now we use Theorem 1 of [8]:

If a group A is the union of a continuous chain of < 2 freesubgroups {A : < } of cardinality 1 such that eachA+1/A is 1-free, then A is free.

Thus G is free in the extension and (1) follows from Proposition 2. 2

Let us consider the following principle:

() For all stationary E S20 and countable subsets a of E such thata is cofinal in and of order type there is a closed C 2 of ordertype 1 such that { E : a \ C is finite} is stationary in C.

Lemma 6 The principle () implies thatG1(F(1), G) is determined for allgroups G of cardinality 2.

Proof. Suppose G is a group of cardinality 2. We may assume the domainofG is 2. Let us assume G is 2-free, as otherwise easily wins. If we provethat G satisfies (2), then Proposition 5 implies that wins G1(F(1), G).

To prove (2), assume the contrary. By Proposition 5 we may assumethat G can be expressed as the union of a continuous chain G : < 2of free groups with G+1/G non-1-free for E, E 2 stationary.By Fodors Lemma, we may assume E S20 . Also we may assume thatfor all , every ordinal in G+1 \ G is greater than every ordinal in G.Finally by intersecting with a closed unbounded set we may assume that forall E the set underlying G is . Choose for each E some countable

subgroup b of G+1 with b + G/G non-free. Let c = b G. Wewill choose a so that any final segment generates a subgroup containingc. Enumerate c as {gn: n < } such that each element is enumeratedinfintely often. Choose an increasing sequence (n: n < ) cofinal in sothat for all n, gn Gn . Finally, for each n, choose hn Gn+1 \ Gn . Leta = {hn: n < } {hn + gn: n < }. It is now easy to check that a is a

9


10/22

416

re

vision:1993-08-29

modified:1993-08-29

sequence of order type which is cofinal in and any subgroup of G whichcontains all but finitely many of the elements of a contains c.By () there is a continuous C of order type 1 such that { C :

a \ C is finite} is stationary in C. Let D = C

C b. Since |D| 1,D is free.

For any C, let

D = (C )

(

(C)

b).

Note that D =

CD, each D is countable and for limit point of C,

D =

(C) D. Hence there is an C E such that a \ C is finiteand D/D is free. Hence b + D/D is free. But

b + D/D = b/b D = b/b G,

which is not free, a contradiction. 2

For the next theorem we need a lemma from [6]. A proof is included forthe convenience of the reader.

Lemma 7 [6] Suppose is a regular cardinal andQ is a notion of forcing

which satisfies the -c.c. Suppose I is a normal -complete ideal on andI+ = {S : S I}. For all sets S I+ and sequences of conditionsp: S, there is a set C with \ C I so that for all C S,

p Q {:p 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 S there is an extension r of p and a set I I (note: Iis in the ground model) so that

r { : p G} I.

Let I be the diagonal union of {I: S}.Suppose now that < and , (S \ I). Since / I, r p G.

Hence r r G. So r, r are incompatible. Hence {r: S \ I}is an antichain which, since S is I-positive, is of cardinality . This is acontradiction. 2

10


11/22

416

re

vision:1993-08-29

modified:1993-08-29

Theorem 8 Assuming the consistency of a Mahlo cardinal, it is consistentthat () holds and hence that G1(F(1), G) is determined for all groups Gof cardinality 2.

Proof. By a result of Harrington and Shelah [7] we may start with a Mahlocardinal in which every stationary set of cofinality reflects, that is, ifS is stationary, and cf() = for S, then S is stationary in for some inaccessible < .

For any inaccessible let P be the Levy-forcing for collapsing to 2.The conditions of P are countable functions f : 1 such thatf(, ) < for all and and each f is increasing and continuous in the

second coordinate. It is well-known that P is countably closed and satisfiesthe -chain condition [11, p. 191].

Let P = P. Suppose p P and

p E S20 is stationary and E(a is cofinal in and of order type ).

LetS = { < : q p(q E)}.

For any S let p p such that p E. Since P is countablyclosed, we can additionally require that for some countable a we have

p a = a.The set S is stationary in , for if C is cub, then p C E = ,

whence C S = . Also cf() = for S. Let be inaccessible suchthat S is stationary in . We may choose in such a way that S implies p P. By Lemma 7 there is a S such that

p P E1 = { < : p G} is stationary.

Let Q be the set of conditions f P with dom(f) ( \ ) 1. Notethat P = P Q. Let G be P-generic containing p and G = G P forany inaccessible . Then G is P-generic and

2of V[G] is . Let

us work now in V[G]. Thus is the current 2, E1 = { < : p G}is stationary, and we have the countable sets a for E1. Since Qcollapses there is a name f such that

Q f : 1 is continuous and cofinal.

11


12/22

416

re

vision:1993-08-29

modified:1993-08-29

More precisely

f is the name for the function f defined by f() = if andonly if there is some g G so that g(, ) = . Let C denote the range off. We shall prove the following statement:

Claim: Q { < : a \ C is finite } is stationary in C.

Suppose q Q so that q D 1 is a cub. Let M be an appropriateexpansion of H(), and Ni : i < , Ni = Ni, , . . ., a sequence ofelementary submodels of M such that:

(i) Everything relevant is in N0.

(ii) If i = Ni , then i < j for i < j < .(iii) Ni+1 is closed under countable sequences.

(iv) |Ni| = 1.

(v) Ni =j


13/22

416

re

vision:1993-08-29

modified:1993-08-29

Remark. If G1(A, F(1)) is determined for all groups A of cardinality +

, singular, then 2 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 G1(F(1), G) is determined for all groups G.

Proof. Let us assume that the stationary logic L1(aa) has the Lowenheim-Skolem property down to 1. This assumption is consistent relative to the

consistency of a supercompact cardinal [2]. Let G be an arbitrary 2-freegroup. Let H be an L(aa)-elementary submodel ofG of cardinality 1. ThusH is a free group. The group H satisfies the sentence (2), whence so does G.Now the claim follows from Proposition 5. 2

Corollary 11 Assuming the consistency of a supercompact cardinal, it isconsistent that G1(A, B) is determined for every structure A and every un-countable free group B.

4 G1(A,B) can be determined for all A and Bof cardinality 2.

We prove the consistency of the statement that G1(A, B) is determined forall A and B of cardinality 2 assuming the consistency of a measurablecardinal. Actually we make use of an assumption that we call I() concern-ing stationary subsets of 2. This assumption is known to imply that 2 ismeasurable in an inner model. It follows from the previous section that somelarge cardinal axioms are needed to prove the stated determinacy.

Let I() be the following assumption about 1-stationary subsets of2:

I() Let I be the 1-nonstationary ideal NS1 on 2. Then I+ has a

-closed dense subset K.

Hodges and Shelah [9] define a principle I(), which is like I() except thatI is not assumed to be the 1-nonstationary ideal. They use I() to prove

13


14/22

416

re

vision:1993-08-29

modified:1993-08-29

the determinacy of an Ehrenfeucht-Frasse-game played on several boardssimulataneously.Note that I() implies I is precipitous, so the consistency of I() im-

plies the consistency of a measurable cardinal [12].

Theorem 12 ([12])The assumption I() is consistent relative to the con-sistency of a measurable cardinal.

We shall consider models A, B of cardinality 2, so we may as well assumethey have 2 as universe. For such A and < 2 we let A denote thestructure A . Similarly B.

Lemma 13 Suppose A and B are structures of cardinality 2. If does nothave a winning strategy in G1(A, B), then

S = { : A = B}

is 1-stationary.

Proof. Let C 2 be 1-closed and unbounded. Suppose S C = . Wederive a contradiction by describing a winning strategy of : Let : 1 1 1 2 be onto with , , d ( , , d) for all , < 1 and d < 2.

If < 2, let : 1 be onto. Suppose the sequence (xi, yi) : i < has been played. Here xi denotes a move of and yi a move of . Duringthe game has built an ascending sequence {ci : i < } of elements of C.Now he lets c be the smallest element of C greater than all the elementsxi, yi, i < . Suppose () = (i , , d). Now will play ci() as an elementof A, if d = 0, and as an element ofB if d = 1.

After all 1 moves ofG1(A, B) have been played, some A and B, where C, have been enumerated. Since S, has won the game. 2

Theorem 14 Assume I(). The game G1(A, B) is determined for all A

and B of cardinality 2.

Proof. Suppose does not have a winning strategy. By Lemma 13 the setS = { : A = B} is 1-stationary. Let I and K be as in I(). If S,let h : A = B. We describe a winning strategy of . The idea of thisstrategy is that lets the isomorphisms h determine his moves. Of course,

14


15/22

416

re

vision:1993-08-29

modified:1993-08-29

different h may give different information to , so he has to decide whichh to follow. The key point is that lets some h determine his move onlyif there are stationarily many other h that agree with h on this move.

Suppose the sequence (xi, yi) : i < has been played. Again xi denotesa move of and yi a move of . Suppose plays next x and this is (say)in A. During the game has built a descending sequence {Si : i < } ofelements ofK with S0 S. The point of the sets Si is that has taken carethat for all i < and Si we have yi = h(xi) or xi = h(yi) dependingon whether played xi in A or B. Now lets S

i


16/22

416

re

vision:1993-08-29

modified:1993-08-29

Let

Alan Mekler, Saharon Shelah and Jouko Vaananen- The Ehrenfeucht-Fraisse-game of length omega-1

Documents

Transcript of Alan Mekler, Saharon Shelah and Jouko Vaananen- The Ehrenfeucht-Fraisse-game of length omega-1