chanoch havlin and saharon shelah- existence of ef-equivalent non isomorphic models

8/3/2019 Chanoch Havlin and Saharon Shelah- Existence of EF-Equivalent Non Isomorphic Models

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EXISTENCE OF EF-EQUIVALENT NON ISOMORPHIC

MODELS

CHANOCH HAVLIN AND SAHARON SHELAH

Abstract. We prove the existence of pairs of models of the same car-dinality which are very equivalent according to EF games, but notisomorphic. We continue the paper [4], but we dont rely on it.

Contents

1. Introduction 22. Games with trees for regular = 0 33. Games with trees for singular = 0 94. regular > 135. > cf() > 18References 23References 23

1991 Mathematics Subject Classification. MSC (2000) 03250.Key words and phrases. model theory, infinary logics, EF equation, non-isomorphic.We would like to thank the Israel Science Foundation for partial support of this research

(Grant No. 242/03). Publication 866 of the second author.

1


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2 CHANOCH HAVLIN AND SAHARON SHELAH

1. Introduction

There had been much study of equivalence relations between models. Whenwe study such an equivalence relation, the basic question is whether thisrelation is actually trivial, i.e., if equivalent models are isomorphic. Forexample, countable models which are elementary equivalent in L1, areisomorphic. (Scott showed this in [10] for countable vocabulary, and Changgeneralized it in [1] for any vocabulary). For = cf() > 0, Morely gave(without publishing) a counter example - a pair of L, equivalent modelsof size which are not isomorphic. Shelah [9, Ch. II, 7] gave such anexample for almost every singular .

Those questions also relate to classification theory: The existence ofstrongly equivalent models which are not isomorphic is a non-structureproperty for a class of models. On the other side, if a not too strong

equivalence relation is actually the isomorphism relation, this is a structureproperty.(See [8] and [2]).

One of the equivalence relations studied in this context, is equivalenceunder EF(Ehernfeucht-Fraisse) games. A detailed discussion of EF gamesand their history can be found in [3] and in [Vaa95]. The general structureof an EF game on a pair of models is as follows: There are two players -isomorphism player, who we call ISO and anti-isomorphism player, who wecall AIS. During the game, AIS chooses members of the models, and ISOdefines interactively a partial isomorphism between the models - in everymove he has to extend that partial isomorphism so that the elements chosenby AIS will be contained in the domain or in the range. The isomorphism

player loses the game if at some point he cannot find a legal move. If hedoes not lose, he wins. We limit the length of the game and the numberof elements that AIS may choose at each move. (Because, if AIS can listall the members of one of the models, then the game is not interesting).In [4], the games were with fixed length. In this paper, we deal with EFgames approximated by trees - the length of the game is limited by addingthe demand that in each move, AIS has to choose a node in some fixed treeT (with certain properties) such that the sequence of nodes formed by hischoices is strictly increasing in the order


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EXISTENCE OF EF-EQUIVALENT NON ISOMORPHIC MODELS 3

in [2] who investigated such games in the context of classification theory).Second, we give results also for > without the assumption =

0,

where we use PCF theory to have some approximation instead of = 0.In Section 2 we prove that for regular = 0 for some class of reasonably

large trees (see detailed discussion justifying the choice, in the beginning ofSection 2) for every tree from that class there are non isomorphic models ofsize which are equivalent under EF games approximated by that tree suchthat in each move AIS is allowed to choose < members of the models (seeDefinition 2.1). Sec1-Def-Game-

With-TreeIn Section 3 we do the parallel for singular . But for singular , if weallow AIS to choose < elements in each move, and the tree has a branchof length cf(), then the game is not interesting, because AIS can chooseall the members of the models during the game. So we have to be more

careful - we allow AIS to choose only one element in each move. This isstill a generalization of the result for such in [4] - see the discussion at thebeginning of Section 3.

In Section 4 we prove that for regular > , for every tree of size without a branch of length there are non-isomorphic models of size whichare equivalent under the EF game approximated by that tree such that ineach move AIS is allowed to choose < members of the models.

In Section 5 we prove a similar result for > cf() > . As we explainedabove, because of the singularity of , we have to restrict the number ofelements that AIS is allowed to choose at each move - in stage , AIS has tochoose < 1 + members of the models. For a further work in preparationof the second author on this subject see [?] on his web. nisux

2. Games with trees for regular = 0

In [2] there is a construction of non-isomorphic models of size which areequivalent under EF games approximated by trees of size with no branch,when =


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We take the middle road: we do not limit explicitly the size of the tree,but we demand that the tree will be definable enough - the cause of not

having a branch of length is that the nodes of the tree are actually partialfunctions from to which satisfy a certain local condition. By local wemean that a function f satisfies the condition iff any restriction of f to acountable set satisfies it. The tree order is inclusion, and there is no functionfrom to which satisfies the condition. By Remark 2.4 this result is indeedSec1-Remark-

Strength-Of-Game

a generalization of for every tree of size and without a branch.

LABEL Sec1-Def-Game-With-Tree

Definition 2.1. For a tree T, a cardinal , and models with common vo-cabulary M1, M2, we define the game T,(M1, M2) between the playersISO and AIS as follows: After stage in the game we have the sequencef : , which is an increasing continuous sequence of partial isomor-phisms from M1 to M2, and the sequence z : which is an increasingcontinuous sequence in T.

Stage in the game is as follows : First, AIS chooses z of level of Tsuch that for every < z >

T z. Then:

1. If = 0, then f = .2. If is limit, then f =


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Then there are non-isomorphic models M1, M2 of size which are EFT,equivalent.

Proof. First, we shall define a tool for constructing models.LABEL Sec2-2.6

Definition 2.6. x is a structure parameter if it consists of the followingobjects:

a set I,a set Js for each s I, such that if s1 = s2, then Js1 Js2 = (denote J =

sIJs),

sets S, T such that S I I and T J J,LABEL Sec2-2.7

Definition 2.7. For a given structure parameter xwe define a model M = Mxin the following way: First for each s I let Gs be an abelian group gen-erated freely by {xt : t Js} except of the relation x(2x = 0). (We could

have also used a free group or a free abelian group, but this choice makes theproof a bit simpler). We demand also that if s1 = s2, then Gs1 Gs2 = .For (s1, s2) S, let Gs1,s2 be the subgroup of Gs1 Gs2 generated by{(xt1 , xt2) : (t1, t2) T (Js1 Js2)}. The universe of M is

sI Gs. The

vocabulary of M consists of

1. for each a M, a unary function symbol Fa;2. for each s I, a unary relation symbol Ps;3. for each (s1, s2) S, a binary relation symbol Qs1,s2.

The interpretation of the symbols in M is as follows:

1. for each b M, s I, a Gs, if b Gs, then FMa (b) = a + b, elseFMa (b) = b;

2. for each s I, PMs = Gs;3. for each (s1, s2) S, Q

Ms1,s2

= Gs1,s2.LABEL Sec1-Lemma-Cond-Of-BeingAuto

Lemma 2.8. Suppose I I and f is a function, f :sI Gs M. Then

f is a partial automorphism of M iff the following hold:

1. For each s I, f(0Gs) Gs.2. For each s I and a Gs we have f(a) = f(0Gs) + a.3. For each s1, s2 I

, if (s1, s2) S, then (f(0Gs1 ), f(0Gs2 )) Gs1,s2.

Proof. Suppose f is a partial automorphism. Then we have:

1. For each s I, 0Gs Gs = PMs which implies f(0Gs) P

Ms = Gs.

2. For each s I and a Gs, f(a) = f(FMa (0Gs)) = F

Ma (f(0Gs)) =

f(0Gs) + a.3. For each s1, s2 I

, if (s1, s2) S, then (0Gs1 , 0Gs2 ) Gs1,s2 (be-

cause it is a subgroup ofGs1 Gs2) but Gs1,s2 = QMs1,s2

, thereforewe have (f(0Gs1 ), f(0Gs2 )) Gs1,s2.

Similar arguments show the other direction. ?? Sec1-LemmaCondOfBeinNow we shall define a structure parameter x and put M = Mx. Then we will

choose elements a, b M, define M1 = (M, a), M2 = (M, b), and showthat M1, M2 are as required in Theorem 2.5. Sec1-Theorem-

Eq-T-lambda


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Let x = x,F be the following structure parameter:

1. I = []0 .

2. For u I, Ju consists of the quadruples t = (u,g,h,), where:(a) g, h are functions from u into ;(b) is a function from supRang(g) u into ;(c) F;(d) g, h are weakly increasing;(e) if g(x) = g(y) then h(x) = h(y);(f) h(x) > x.

For t = (u,g,h,) we will denote u = ut, g = gt, h = ht, = t.3. S = {(u1, u2) : u1, u2 I and u1 u2}.4. T = {(t1, t2) : t1, t2 J, u

t1 ut2, gt1 gt2 , ht1 ht2, t1 t2 }.

Let M = M,F = Mx be the corresponding model. Note that |I| = 0 =

and for each u I, |Ju| = 0 = , therefore ||M|| = . Define a = 0G , b = x(,,,),M1 = (M, a), M2 = (M, b).

LABEL Sec1-Claim-Equiv-Models

Claim 2.9. M1, M2 are EFT, equivalent.

Proof. We start with

LABEL Sec1-Def-G-of-lambda Definition 2.10. We define a set of functions G = G() with a partial order

G in the following way:

1. For an ordinal < , G is the set of functions g which satisfy(a) g : , < ;

(b) g is weakly increasing.2. G = g(x)} {}).

4. g1 G g2 if g1 g2 and hg1 hg2 .

LABEL Sec1-Claim-G-of-lambda

Claim 2.11. 1. If g(x) = g(y), hg(x) = hg(y).2. hg(x) > x.3. hg is weakly increasing.4. For every g1, g2 G, g1

G g2 iff(a) Dom(g1) = 1 2 = Dom(g2) and g1 g2;

(b) if 1 < 2, then g2(1) > g2(x) for every x < 1.5. If g1 G and Dom(g1) < < , then there is g2 G+1 such that

g1 G g2 and Dom(g2) = .

6. If < and we have g : < such that g G and < implies g

G g, then g =


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4. If there is x < 1 such that g2(1) = g2(x), then hg2(x) = hg2(1) > 1 hg1(x),so g1

G g2. On the other direction, if g1 g2 and g2(1) > g2(x)

for every x < 1, then for every such x: If there is y < 1 such thatg1(y) > g1(x), let y

be the minimal y which satisfies this. We gethg1(x) = hg2(x) = y

. If there is no such y, we get hg1(x) = hg2(x) = 1.Therefore we have hg1 hg2 .

5. Define g2 : + 1 by

g2(x) =

g1(x) if x Dom(g1), if x \ Dom(g1).

By 4. we get that g1 G g2.

6. Remember that is regular, therefore


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2. For each > 0,

f(a) = f(0G) = x(,,,) = b.

Therefore f is a partial isomorphism from M1 = (M, a) into M2 =(M, b).

This completes the proof of Claim 2.9. 2.9Sec1-Claim-Equiv-Models

Sec1-Claim-Equiv-Models

LABEL Sec1-ClaimNotIso

Claim 2.12. M1, M2 are not isomorphic.

Proof. It is enough to show that M is rigid (i.e. it does not have a non-trivialautomorphism).

Assume towards contradiction that f = id is an automorphism ofM. Foreach u I we define cu = f(0Gu). By Lemma 2.8, for each u w I we

Sec1-Lemma-Cond-Of-

BeingAuto

have (cu, cw) Gu,w.

For each u w I and t = (w,g,h,) Jw we define w,u(t) Ju byw,u(t) =: (u, gu, hu, supRang(gu) u).

By the definition of T we have that if t Jw and r Ju, then (r, t) T iffr = w,u(t). We define a homomorphism w,u : Gw Gu by w,u(xt) = xr,where r = w,u(t). We get that Gu,w is the subgroup ofGu Gw generatedby {(w,u(xt), xt) : t Jw}. Since {xt : t Jw} generate Gw, we get that

Gu,w = {(w,u(c), c) : c Gw}.

Now define n(u) to be the length of the reduced representation of cu as asum of the generators {xt : t Ju}. For u w I we get n(u) n(w),since cu = w,u(cw) and w,u sends one generator to one generator. If for

every u I there is w I such that n(w) > n(u), we can find a sequenceun : n < such that un I and n(un) < n(un+1). Define w =n 0.

Choose t Ju such that xt appears in the reduced representation ofcu.For each u w I there is a unique t(w) Jw such that w,u(t(w)) = tand xt(w) appears in the reduced representation of cw. Such t(w) existsbecause cu = w,u(cw). It is unique because if there were two such ts,t1, t2, then

w,u(xt1) = w,u(xt2) = xt.

Since in Gu x(2x = 0), it implies n(w) > n(u), which contradicts themaximality of n(u).

Note that ifu w z I, then z,u = w,u z,w . Therefore, by unique-ness of t(w), ifu w z I then t(w) = z,w(t(z)). For each u w I,

define gw = gt(w), hw = ht(w), w = t(w). If u w1, w2 I, then thefunctions gw1, hw1 , w1 and gw2 , hw2, w2 are respectively compatible, since

t(w1) = z,w1(t(z)) and t(w2) = z,w2(t(z)),

where z = w1 w2. Define

g = {gw : u w I}, h = {hw : u w I}, = {

w : u w I}.


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We get:

1. Dom(g) = Dom(h) = .

2. g, h are weakly increasing.3. h(x) > x.4. Ifg(x) = g(y), then h(x) = h(y).5. F (this is by Definition 2.2(2)). Sec1-Def-Square-

F-lambda6. supRang(g) Dom().

By Definition 2.2(3), Dom() = . Therefore by 6., supRang(g) < . Since Sec1-Def-Square-F-lambdag is weakly increasing and is regular, there is 0 < such that for every

0 < < , g() = g(0). By 4. we get that for every 0 < < , h() = h(0).Choose > h(0) > 0 and get that h() < , contradicting 3.

This completes the proof of Claim 2.12. 2.12 The proof of Theorem Sec1-ClaimNotIso

Sec1-ClaimNotIso

2.5 is now finished. 2.5

Sec1-Theorem-Eq-T-lambda

Sec1-Theorem-Eq-T-lambda

3. Games with trees for singular = 0

It is clear that for singular we cannot expect the same result as in theprevious section, since the AIS player would be able to list all the membersof M1, M2. Thus, we prove a weaker result - we allow AIS to choose onlyone element in each turn. We also remark in Remark 3.2 that this result

Section2-Generalize-866

generalizes the result in [4] for such .

LABEL Sec2-Theorem-Eq-T,1

Theorem 3.1. Suppose

cf() < = 0 ,F, holds,

T = TF.Then there are non-isomorphic models M1, M2 of size which are EFT,1equivalent.

LABEL Section2-Generalize-866Remark 3.2. We can show that Theorem 3.1 generalizes the result in [4] by

Sec2-Theorem-Eq-T,1

choosing appropriate F. The result there shows the existence of two non-isomorphic models of size which are equivalent under every EF game oflength < cf(), which consists of sub-games of length < , such that AISchooses the length of each sub-game before it starts, and in every sub-gamehe chooses one element in each move - see the definitions there. Now, anappropriate F can be chosen by looking at the proof there, but we will takea shortcut - we will use the result instead of the proof. Let us choose a pairof models M1, M2 as in the result in [4]. Without loss of generality assumethat the universe of M1 is {1} and the universe of M2 is {2}. We cantake F to be the set of functions f which satisfy the following conditions:

1. Dom(f) , Rang(f) .2. The partial function f from M1 to M2 defined by

Dom(f) = Dom(f){1}, where for every Dom(f), f((, 1)) = (f(), 2)

is a partial isomorphism.


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Now, it is not hard to see that EFTF,1 equivalence implies equivalence as inthe result of [4].

Proof of theorem 3.1. Denote = cf(). ( > 0 because = 0 .) LetSec2-Theorem-Eq-T,1 i : i < be an increasing and continuous sequence such that 0 = 0,

+i < i+1 = cf(i+1), gi > 0 for i > 0, andi x;(g) Dom() = u

{i(x) : x u and h(x) = i(x)+1}

For t = (u,g,h,) we denote u = ut, g = gt, h = ht, = t.3. S = {(u1, u2) : u1, u2 I, u1 u2}.4. T = {(t1, t2) J : u

t1 ut2, gt1 gt2 , ht1 ht2 , t1 t2}.

Let M = MF, = Mx be the corresponding model. Define a = 0G, b = x(,,,).Define M1 = (M, a) and M2 = (M, b).

LABEL Sec2-Claim-Equiv-

Models

Claim 3.3. M1, M2 are EFT,1 equivalent.

Proof. We start withLABEL Sec2-Def-W Definition 3.4. A partially ordered set of functions (W, W), which de-

pends on the sequence i : i < , is defined in the following way:

1. we define a set B such that B iff(a) = i : i < , i i i+1;(b) there is j = j() < such that i < j() iff i = i+1.

2. For B we define W to be the set of functions g which satisfy(a) Dom(g) = i g(x)} {i})

LABEL Sec2-Claim-W Claim 3.5. 1. If g(x) = g(y), then hg(x) = hg(y).

2. hg(x) > x.3. hg is weakly increasing.4. If x [i, i+1), then hg(x) [i, i+1].


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5. Suppose that g1 W1 , g2 W2. Then g1 W g2 iff

(a) g1 g2 (therefore for every i < 1i

2i );

(b) for every i < if 1i < 2i , then for every x [i, 1i ), g2(x) . We choose i < i+1 such that

u [i, i+1) [i, i).Now j() = i() + 1, so by Claim 3.5(6) we can find g Wi()+1Sec2-Claim-Wsuch that

g W g and

i x.6. Ifh(x) = i(x)+1, then i(x) Dom().7. F.


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By 7. we get that Dom() = , therefore by 6. there exists i < such thatif i(x) = i, then i(h(x)) = i. By 2., i(x) = i implies g(x) +i . By 3., g is

weakly increasing. Since i+1 = cf(i+1) > +i , we can find 0 such that if0 x < i+1, then g(x) = g(0). By 5., h(0) > 0. By the choice of i weget that h(0) < i+1. Choose h(0) < x < i+1. We get h(x) > x > h(0)but g(x) = g(0). This contradicts 4. Therefore we proved that M is rigid.

This completes the proof of Claim 3.6. 3.6 Sec2-ClaimNotIso

Sec2-ClaimNotIso

Ths proof of Theorem 3.1 is now also completed. 3.1

Sec2-Theorem-Eq-T,1

Sec2-Theorem-

Eq-T,1

4. regular >

In this section we show a result which holds for every regular > . Inthe previous sections we used the assumption = 0 . Here we use insteadof it the existence of a set P []0 of size which is dense. By densewe mean that for every A [] there is B A, B P.

LABEL Sec4-4.1Remark 4.1. 1. Looking at the proof, one can see that instead of > ,it is enough to assume the following:(a) > 20.(b) There is P []0 such that

i. |P| = ;ii. for every A [], there is B P, such that B A.

2. It is possible that it can be proved in ZFC that every > 20 satisfies1.(b) (it is a problem in cardinal arithmetic).

LABEL Sec3-Theorem-EquivTheorem 4.2. Suppose

= cf() > ,T is a tree of size with no branch of length .

Then there are models M1, M2 of size which are EFT, equivalent but notisomorphic.

Proof. Let be large enough cardinal (for example = 7()).

LABEL Sec3-Claim-exists-inner-model

Claim 4.3. We can findM such that the following hold:

1. M is elementary sub-model of H().2. + 1 M.3. ||M|| = .4. For every (xi, zi) : i < such that xi M and zi T for every

i < there exists an increasing sequence in : n < such that(a) (xin , zin) : n < M;(b) if in addition for i < j < the level of zi (in T) is strictly less

than the level of zj , then zin : n < is an antichain in theorder T.

In the proof Claim 4.3 we use a partial version of the RGCH Theorem Sec3-Claim-exists-inner-model

(see Shelah [5]).


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LABEL RGCH-Theorem

Theorem 4.4. (RGCH Theorem, partial version). If , then there is

regular < and P []


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By Claim 4.6 we can choose for every j < a triple i0(j), i1(j), i2(j) such Sec3-small-1that

1. i0(j) < i1(j) < i2(j) < ;2. j < j i2(j) < i0(j

);3. zi0(j) (since in such game AIS will be able to list all theelements of the two models). Therefore, we define another type of game.

LABEL Sec5-5.1

Definition 5.1. Let M1, M2 be models with common vocabulary. Let T bea tree. The game T(M1, M2) is defined in the same way as the definition


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of T, (see 2.1) except that in stage we demand that the sets A1, A2

Sec1-Def-Game-With-Tree

chosen by AIS will satisfy |A1 A2| < 1 + instead of |A1 A2| < 1 + .

We say that M1, M2 are EFT equivalent if ISO has a winning strategy forEFT(M1, M2).

LABEL Sec5-5.2Remark 5.2. Note that in Theorem 3.1, if we replace EFT,1 with EF

T Sec2-Theorem-

Eq-T,1we do not get a stronger result, because for every tree T which satisfiesthe conditions there, we can construct another tree T which satisfies theconditions, so that EFT,1 equivalence would imply EF

T equivalence.

LABEL Sec4-Theorem-EF*-T-Equiv

Theorem 5.3. Suppose that:

1. > cf() = > ;2. T is a tree of size without a branch.

Then there are non-isomorphic models M1, M2 of size which are EFT

equivalent.

Proof. Let be a large enough cardinal(for example = 7()).LABEL Sec4-Claim-exists-inner-model

Claim 5.4. We can findM such that the following hold:

1. M is elementary sub-model of H().2. + 1 M.3. For every (xi, zi) : i < such that xi M and zi T for every

i < there exists an increasing sequence in : n < such that :(a) (xin , zin) : n < M;(b) if in addition for every < there is i < such that the level

of zi is greater than , then we can also have that zin : n <

is an antichain in T.

Proof. The same proof as the proof of Claim 4.3 (we are using the fact that Sec3-Claim-exists-inner-model

is regular and > ). 5.4

Sec4-Claim-exists-inner-model

Let M be as in Claim 5.4. Let i : i < be an increasing and continuous

Sec4-Claim-exists-inner-model

sequence such that 0 = 0, +i + 0 < i+1 = cf(i+1), and

i


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(d) the level of z in the tree T. then is minimal with regardto the condition that i(x) for every x such that h(x) =

i(x)+1 or there are f1, f2 such that F(f1, f2) = 1 andf2(x) = i(x)+1;

(e) there is a witness (g, h) for t, which means thati. Dom(g) = Dom(h) , Rang(g) Rang(h) ,

ii. (s) Dom(g),iii. g g, h h,iv. for every (f1, f2)

2

F(f1, f2) = 1 iff f1 g f2 h,v. g, h are weakly increasing,

vi. h(x) > x,(i) if g(x) = g(y), then h(x) = h(y),

viii. g(x) [i(x), +

i(x)],ix. h(x) [i(x), i(x)+1].

3. S = I2.4. T consists of the pairs (t1, t2) J

2, where :(a) t1, t2 have a common witness;(b) zt1 , zt2 are comparable in the order T.

Fact 5.6. Suppose

s I, z T,g, h satisfy i-ii, v-ix from Definition 5.52.(e),Sec4-Def-

Parameter

{i(x) : h(x) = i(x)+1} , where is the level of z.

Then the following hold:1. There is a unique t Js such that (g, h) is a witness for t and

zt T z. We denote t = t(s, g, h, z).2. If

(a) g, h, z satisfy the conditions in 1.,(b) z, z are comparable in T,(a) g, h are compatible with g, h, respectively,

then t(s, g, h, z) = t(s, g, h, z).

Let M = Mx be the corresponding model. We can check that ||M|| = .Let a = 0G(,), b = x(,,,,,z), where z is the root of T (without loss

of generality there is a root). Define M1 = (M, a), M2 = (M, b).LABEL Sec4-Claim-Equiv Claim 5.7. M1, M2 are EF

T equivalent.

We describe a winning strategy for ISO - this is very similar to the proofof Claim 3.3, so we omit the details. We use the definitions in DefinitionSec2-Claim-

Equiv-Models 3.4. In every stage of the game ISO will choose a function g such that:

Sec2-Def-W 1. g0 = .2. g Wi()+1.

3. < gW g.


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Proof. See the proof of Claim 4.11. 5.9 Sec3-Claim-unique-projection

Sec4-Claim-unique-projection

LABEL Sec4-Claim-exists-projection

Claim 5.10. For every A there is r Ws+ such that (r, t

) T.

Proof. See the proof of Claim 4.12. 5.10Sec3-Claim-exists-projection

Sec4-Claim-exists-projection

Now, using Claim 5.10, we choose for each A an r Ws+ such that

Sec4-Claim-exists-projection

(t, r) T.

LABEL Sec4-claim-r-ies-distinct

Claim 5.11. If < , then r = r.

Proof. If we are in Case (*1): zt , zt are not comparable. But zr T zt

because they are comparable and zr is on greater level, since that levelis determined by Definition 5.5,2.(d). By the same argument, zr T zt .

Sec4-Def-Parameter

Therefore, zr , zr are not comparable, so r = r.If we are in Case (*2): Assume towards contradiction that r = r = r.

Let (g, h) be a witness for r. Then by Claim 5.9, gt , gt g and h

t , ht h.

Sec4-Claim-unique-projection

Since we are in Case (*2) we get that

g() = g() and h() < < h()

which contradicts the definition of a witness (see Definition 5.5,2.(e)). 5.11Sec4-Def-Parameter

Sec4-claim-r-ies-distinct

We got that Ws+ is infinite - contradiction. Therefore, M must be rigid,which proves Claim 5.8. 5.8

Sec4-Claim-Not-Isomorphic

Sec4-Claim-Not-

Isomorphic

The proof of Theorem 5.3 is now complete 5.8

Sec4-Theorem-EF*-T-Equiv

Sec4-Claim-Not-Isomorphic


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References

[1] C.C Chang, Some remarks on the model theory of infinitary languages, in The syntax

and semantics of infinitary languages, Lecture notes in Mathematics, 72, J. Barwiseed. (Springer, Berlin, 1968) pp. 36-63

[2] T.Hyttinen and H. Tuuri, Constructing strongly equivalent nonisomorphic models forunstable theories, Annals of pure And Applied Logic 52(1991) pp.203-248

[3] Hodges, Wilfrid Model theory. Encyclopedia of Mathematics and its Applications, 42.Cambridge University Press,Cambridge, 1993. ISBN: 0-521-30442-3 (Reviewer: J. M.Plotkin)

[4] S.Shelah A long EF equivalence non isomorphic models - to appear (SH 836 in Shelaharchive)

[5] S.Shelah, The generalized continum hypothesis revisited, Israel J. Math 116 (2000) pp.285-321

[6] S.Shelah, Existence of many L, non isomorphic models of power for singularwith = , Notre Dame J. Formal Logic 25(1984) pp. 97-104

[7] S.Shelah, Existence of manyL, non isomorphic models of

Tof power

, Annals ofPure and Applied Logic 34 (1987) pp. 291-310

[8] S.Shelah, Classification Theory, Stud. Logic Found. Math. 92 (North Holland, Ams-terdam, 2nd rev. ed. 1990)

[9] S.Shelah, Cardinal Arithmetic, Oxford Logic Guide 29, Oxford University Press, 1994.[10] D.S. Scott, Logic with denumerably long formulas and finite strings of quantifiers

in The theory of models , J.W.Adisson, L.Henkin and A.Tarsky , eds. (North holland,Amsterdam, 1965) pp. 329-341

[11] J. Vaananen, Games and trees in infinitary logic: A survay, in Quantifiers , M. Kryn-icki, M. Mostowski and L. Szczerba, eds. (Kluwer, 1995) pp. 105-138.

References

[Vaa95] Jouko Vaananen. Games and trees in infinitary logic: A survey. In M. Mostowski

M. Krynicki and L. Szczerba, editors, Quantifiers, pages 105138. Kluwer, 1995.

Institute of Mathematics The Hebrew University of Jerusalem Jerusalem

91904, Israel

E-mail address: [email protected]

Institute of Mathematics, The Hebrew University of Jerusalem, Jerusalem

91904, Israel, and, Department of Mathematics, Rutgers University, New

Brunswick, NJ 08854, USA

E-mail address: [email protected]: http://www.math.rutgers.edu/~shelah

chanoch havlin and saharon shelah- existence of ef-equivalent non isomorphic models

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