saharon shelah- theories with ef-equivalent non-isomorphic models

8/3/2019 Saharon Shelah- Theories with EF-Equivalent Non-Isomorphic Models

1/32

(897)

revision:2010-10-

15

modified:2010-1

0-17

THEORIES WITH EF-EQUIVALENT NON-ISOMORPHIC

MODELS

SH897

SAHARON SHELAH

Abstract. Our long term and large scale aim is to characterize the firstorder theories T (at least the countable ones) such that: for every ordinal there are ,M1,M2 such that M1,M2 are non-isomorphic models of T ofcardinality which are EF+

,-equivalent. We expect that as in the main gap

([Sh:c, ?]) we get a strong dichotomy, so in the non-structure side we havestronger, better examples, and in the structure side we have a parallel of [Sh:c,Ch.XIII]. We presently prove the consistency of the non-structure side for T

which is 0-independent (= not strongly dependent), even for PC(T1, T).

Anotated Content

0 Introduction

1 Games, equivalences and the question

[We discuss what are the hopeful conjectures concerning versions of EF-equivalent non-isomorphic models for a given complete first order T, i.e.how it fits classification. In particular we define when M1, M2 are EF,-

equivalent and when they are EF+,,-equivalent and discuss those notions.]

2 The properties ofT and relevant indiscernibility

[We recall the definitions of T strongly dependent, T is strongly4 de-pendent, and prove the existence of models of such T suitable for provingnon-structure theory.]

3 Forcing EF+-equivalent non-isomorphic models

[We force such an example.]

4 Theories with order

[We prove in ZFC, that for regular there are quite equivalent non-isomorphicmodels of cardinality +.]

Date: October 15, 2010.Key words and phrases. Ehrenfuecht-Fraisse games, Isomorphism, Model theory, Classification

theory (as in the J.).The author would like to thank the Israel Science Foundation for partial support of this research

(Grant No.242/03). I would like to thank Alice Leonhardt for the beautiful typing.

1


2/32

(897)

revision:2010-10-

15

modified:2010-1

0-17

2 SAHARON SHELAH

1. Introduction

1(A). Motivation.We first give some an introduction for non-model theorists. A ma jor theme in

the authors work in model theory is to find main gap theorem. This means,considering the family of elementary classes (e.g. the classes of the form ModT =the class of models of a (complete) first order theory T, each such class is eithervery simple or is very complicated; expecting that we have much knowledge togain on the very simple ones and even on approximations to them.

Of course, this depends on the criterion for simple. Essentially the main theo-rem of [Sh:c] does this for countable T, with complicated interpreted as I(, T),the number of models in ModT of cardinality , is maximal, i.e. 2, for every .See more e.g. in [Sh:E53]. Here we are interested with interpreting complicatedas for arbitrarily large cardinals, there are models M1, M2 ModT of cardinality which are very similar but not isomorphic, where very similar is interpreted

as a relative of the game of the following form. The isomorphism player constructsduring the play, partial isomorphism of cardinality < , in each move the anti-isomorphism player demands some elements to be in the domain or the range, theisomorphism player has to extend the partial isomorphism accordingly; in the playthere are moves, < ; and the isomorphism player wins the play if he has alegal move in each stage (see Definition 2.5, 2.7).

In the present paper we try to deal with suggesting the right variant of thegame, (see Definition 2.6), and give quite weak sufficient conditions for ModT beingcomplicated.

Our aim is to prove (on PC(T1, T), see Definition 1.3(3))

ifT T1 are complete first order theories such that T is not strongly stable,

is an ordinal and > |T| (or at least for many such s) then() there are M1, M2 PC(T1, T) of cardinality which are EF

+,-

equivalent for every < but not isomorphic (for the definition ofEF+,, see Definition 2.7 below, it is a somewhat stronger relative of

EF,-equivalent).

1(B). Related Works.Concerning constructing non-isomorphic EF+,-equivalent models M1, M2 (with

no relation to T) we have intended to continue [Sh:836], or see more Havlin-Shelah[HvSh:866] and see history in Vaananen in [Vaa95]. Those works leave the case = 1 open; a recent construction [Sh:907] resolve this but whereas it applies toevery regular uncountable , it seems less amenable to generalizations.

By [Sh:c] we essentially know for T a countable complete first order when thereare L,(T)-equivalent non-isomorphic models of T of cardinality for some ,see 4; this is exactly when T is superstable with NDOP, NOTOP; (see [Sh:220]).

On restricting ourselves to models of T for EF,-equivalent non-isomorphic,Hyttinen-Tuuri [HT91] started, then Hyttinen-Shelah [HySh:474], [HySh:529], [HySh:602].The notion EF+,-equivalent is introduced here, in Definition 2.7.

By [HySh:474], if T is stable unsuperstable, complete first order theory, =+, = cf() |T|, then there are EF,-equivalent non-isomorphic models ofT (even in PC(T1, T)) of cardinality . But by the variant EF

+,-equivalent, such


3/32

(897)

revision:2010-10-

15

modified:2010-1

0-17

THEORIES WITH EF-EQUIVALENT NON-ISOMORPHIC MODELS SH897 3

results are excluded; by it we define our choice test problem the version of beingfat/lean, see Definition 1.1.

Why EF+

? See Discussion 2.6.Concerning variants of strongly dependent theories see [Sh:783, 3],[Sh:863] (and

maybe [Sh:F705]), most relevant is [Sh:863, 5], part (F). The best relative for us isstrongly4 dependent, a definition of it is given below but we delay the treatmentto a subsequent paper, [Sh:F918]. There we also deal with the relevant logics andmore.

We prove here that if T is not strongly stable then T is consistently fat. Morespecifically, for every = |T| there is a -complete class forcing notion Psuch that in VP the theory T is fat. The result holds even for PC(T1, T). Thisgives new cases even for PC(T) by 1.2.

Also if T is unstable or has the DOP or OTOP (see 1.7 below or [Sh:c]) then itis fat, i.e. already in V.

Of course, forcing the example is a drawback, but note that for proving there isno positive theory it is certainly enough. Hence it gives us an upper bound on therelevant dividing lines.

On Eherenfeucht-Mostowski models, see [Sh:e, Ch.III] or [Sh:c, Ch.VII] or [Sh:h],[Sh:e, Ch.III,1]. I thank a referee for pointing out on earlier version that Hyttinen-Shelah [HySh:474] was forgotten hence as Definition 2.7 was not yet written, themain result 4.1 had not said anything new.

I also thank referees for many helpful remarks.

1(C). Notations and Basic Definitions.{0z.3}

Definition 1.1. Let T be a complete first order theory.1) We say T is fat when for every ordinal , for some (regular) cardinality > there are non-isomorphic models M1, M2 of T of cardinality which are EF

+,,,-

equivalent for every < (see Definition 2.7 below).2) If T is not fat, we say it is lean.3) We say the pair (T, T1) is fat/lean when (T1 is first order T and) PC(T1, T) :={M T : M a model of T1} is as above.4) We say (T, ) is fat when for every first order T1 T the pair (T, T1) is fat. Wesay (T, ) is lean otherwise.

Our claims (mainly 4.1) seem to make it clear that some stable T has NDOTand NOTOP which falls under 4.1, but a referee asks for an example, see [Sh:863,5(F)] for details.

{0z.5}Example 1.2. 1) There is a stable NDOP,NOTOP countable complete theorywhich is not strongly dependent; (moreover not is not strongly4 stable), see [Sh:863,5(G)].2) T = Th(1(Z2), En)n


4/32

(897)

revision:2010-10-

15

modified:2010-1

0-17

4 SAHARON SHELAH

2) PC(T) = {M : M a model of T} where T is a theory or a sentence, inwhatever logic, in a vocabulary T ; if = T we may omit .

3) If T T1 are complete first order theories then PC(T1, T) = PC(T)(T1).{0z.11}Notation 1.4. 1) g(a) is the length of a sequence a.2) a b means that a is an initial segment of a.3) a is the unique initial segment of a of length for g(a).

{0z.13}Definition 1.5. 1) For a regular uncountable cardinal let I[] = {S : somepair (E, a) witnesses S I(), see below}.2) We say that (E, u) is a witness for S I[] when:

(a) E is a club of the regular cardinal

(b) u = u : < , a and a a = a(c) for every E S, u is an unbounded subset of of order-type < (and

is a limit ordinal).{0z.17}

Notation 1.6. 1) For a model M, a M, B M and a set of formulas, we areinterested in formulas of the form (x, y), x = xi : i < , so may be infinite,but the formulas here are normally first order, so all but finitely many of the xisare dummy variables.1A) tp(a ,B,M) = {(x, a) : (x, y) and b

g(y)A and M |= [a, b]}.2) If qf is the set of quantifier-free formulas in L(M), we may write tpqf insteadof tp.3) I(, T) is the number of isomorphic types of models of T of cardinality .

4) I(, T) is the number of isomorphic types ofM , M a model ofT of cardinality.5) IE(, T) is the supremum of {|K| : K PC(T) and M K M = no

M Khas an elementary embeding into any N K\{M}, writing

I

E(, T) =

+

we mean the supremum is obtained if not said otherwise.6) IE(, T) = IE(T)(, T).

{0z.23}Definition 1.7. Let T be a first order complete theory.1) T has OTOP when T is stable and for some n, m letting x = x : < n, y =y : < n, z = z : < m, there are complete types p(x, y, z) such that: forevery there is a model M of T and a nM for < such that:

(a) a : < is an indiscernible set

(b) for = B < the type (p(a, ab, z) is realized in M iff < .

1A) T has the NOTOP when it is stable but fail the OTOP.2) T has NDOP when T is stable and we can find |T|+-saturated models M of

T for 3 such that M0 M M3 for = 1, 2 and tp(M1, M2) does not forkover M0, M3 is |T|+-prime over M1 M2 but not |T|+-minimal over it; equivalentlyfor every c >(M3) the type tp(c, M1 M2, M3) is |T|+-isolated but there is noinfinite I M3 which is indiscernible over M1 M2.2A) T has DOP when T is stable and fail to have the NDOP.

{0z.25}Definition 1.8. 1) For a complete first order theory T, we can say that is a(,,T)-candidate when:

(a) L+,() for some vocabulary T of cardinality


5/32

(897)

revision:2010-10-

15

modified:2010-1

0-17


(b) PC(T)() EC(T)

(c) for some1 texttr satisfying and EM(, ) |= for every

(equivalent some) and witness T is not superstable.

Recall that by [Sh:c, Ch.VII]:{0z.29}

Claim 1.9. If a first order complete theory T is not superstable, then for some tr2 , see Definition 3.2, 2 () of cardinality , witness T is notsuperstable, i.e. for some formulas n(x, yn) L(T), if I = , M = EM(I, )then for ,n < and < we have M |= n[a, a(n)] iff = (n).

{0z.37}Definition 1.10. 1) For any structure I we say that at : t I is indiscernible(in the model C, over A, if A = we may omit it) when: (at g(at)C and) g(at),which is not necessarily finite depends only on the quantifier-free type of t in I and:

if n < and s = s0, s1, . . . , sn1, t = t0, . . . , tn1 realize the samequantifier-free type in I then at := at0 . . . atn1 and as = as0 . . . asn1realizes the same type (over A) in C.

2) We say that au : u [I]I)-incr


6/32

(897)

revision:2010-10-

15

modified:2010-1

0-17

6 SAHARON SHELAH

2. Games, equivalences and questions

What is the meaning in using EF+,? Consider for various s the game , (M1, M2)where M1, M2 ModT(), T complete first order L()-theory. During a play wecan consider dependence relations on short sequences from M (where 2||+0

is the default value), definable in a suitable sense. So ifT is a well understood unsu-perstable T like Th(, En)nM and M M M, M 2|T| and

T

tp(M , {M : T

, (g() + 1) = (g() + 1)) does notfork over M , i.e. M = M : T is a non-forking tree of models with r(T)many levels.

We think the right (variant of the) question is from 2.2. Probably a reasonableanalog is the situation in [Sh:c, Ch.XII,XIII]: the original question was on the

function I(, T), the number of non-isomorphic models; but the answer ismore transparent for IE(, T).

If = +, = |T| = cf(), T = Th(, En)n |T| and vocabulary 1 T and L,(1)such that PC() ECT has members of arbitrarily large cardinality we have(a) (b) where

(a) for every cardinal in PC() := {M : M a model of } there is a-saturated member

(b) for every for arbitrarily large there are M1, M2 PC() of cardinality with non-isomorphic -reducts which are EF+,-equivalent.

Version (B)0: Like (B)1 for EF,.

Version (C)1: Like (B)1 using = T1 where T1 is first order T.

Version (C)0: Like (B)0 using = T1 where T1 is a first order T.


7/32

(897)

revision:2010-10-

15

modified:2010-1

0-17


{1a.5}Discussion 2.3. 1) For reasons to prefer version (B) over (C) - see [Sh:E53].

2) Now by the works quoted above, (see [HySh:529, 3.19] quoted in 5.1 below): Tsatisfies (A)0 iff T is superstable NDOP, OTOP iff (B)0. Of course if we changethe order of the quantifier (to for aribitrarily large some for every < ,...)this is not so, but we believe solving (A)1 and/or (B)1 will eventually do much alsofor this.

So all this means{1a.6}

Conjecture 2.4. 1) For a complete (first order) T the following are equivalent:

(a) for every ordinal for some there are non-isomorphic, EF+,-equivalent

models M1, M2 ECT(N)

(b) for arbitrarily large for every < there are non-isomorphisms, EF+,-

equivalent models M1, M2 ECT()

(c) for every large enough regular there are non-isomorphisms M1, M2 ECT() which are E,-equivalent for every < .

2) Similarly for some T1 T, PC(T1, T) is lean.We conjecture that proving that if we prove that a (countable) fat T is close

enough to superstable, will enable us to generalize proofs in [Sh:c, Ch.XII] onlynow the tree has 1 levels rather than .

We can also return to the ordinals (,+).

Now we shall actually look at the games.

{1a.7}Definition 2.5. 1) We say that M1, M2 are EF-equivalent if M1, M2 are models(with same vocabulary) and is an ordinal such that the isomorphism player has

a winning strategy in the game G1 (M1, M2) defined below.1A) Replacing by < means: for every < ; similarly below.2) We say that M1, M2 are EF,-equivalent or G

-equivalent when M1, M2 are

models with the same vocabulary, an ordinal, M a cardinality such that theisomorphism player has a winning strategy in the game G (M1, M2) defined below.3) For M1, M2, , as above and partial isomorphism f from M1 into M2 we definethe game G (f, M1, M2) between the players ISO, the isomorphism player and AIS,the anti-isomorphism player as follows:

(a) a play lasts moves

(b) after moves a partial isomorphism f from M1 into M2 is chosen, increas-ing continuous with

(c) in the (+ 1)-th move, the player AIS chooses A,1 M1, A,2 M2 suchthat |A,1| + |A,2| < 1 + and then the player ISO chooses f+1 fsuch that A,1 Dom(f+1) and A,2 Rang(f+1)

(d) if = 0, ISO chooses f0 = f; if is a limit ordinal ISO chooses f = {f : < }.

The ISO player loses if he had no legal move for some < , otherwise he wins theplay.4) If f = we may write G (M1, M2). If is 1 we may omit it. We may write instead of +.


8/32

(897)

revision:2010-10-

15

modified:2010-1

0-17

8 SAHARON SHELAH

{1a.8}Discussion 2.6. 1) Why do we need EF+?

First, if we like a parallel of [Sh:c, Ch.XIII], i.e. a game in which set of smallcardinality are chosen, say |T| or 2|T| or whatever rather than just < = M,clearly EF, cannot help.2) Also, consider = +, = cf() > |T| and an ordinal < and ask for whichT: for any two models M1, M2 of T of cardinality , EF,-equivalence impliesisomorphisms? (The EF,-equivalence means that the isomorphism player wins inthe game of length , in each step adding elements to the domain and rangeof the partial isomorphism.)

Now we know (by earlier works, see 5.5) for countable T that if [, ]that the answer (for the pair (, )) is as in the main gap for IE (T superstablewith NDOP and NOTOP). But for larger < this is not so, as e.g. for theprototypical stable unsuperstable T for = ( + 2) we get yes, it is low.3) Looking at the reason for this, i.e. why we need ( + 2) moves, not ( + 2)moves we formulate EF+. We think that with EF+,,, for small ,, and just =

M we get the desired dichotomy. In general, we expect the results will be robustunder choosing such an exact game; and will resolve the case ( (, 2), )case above.4) More specifically, the reason EF,-equivalence does not imply isomorphisms forM1, M2 EC(T), even in the case T = Th(

, En)m


9/32

(897)

revision:2010-10-

15

modified:2010-1

0-17


Case 1: The AIS player chooses A = A M for = 1, 2 such that |A1| + |A2| (M) (see

Definition 2.8 below) and A (M) of cardinality for = 1, 2 such that:

(a) if k = 0 then R = [>(M)]

(M) which the dependence relation, i.e. ER , i.e. 2.8(6) and equality oflength refine for = 1, 2 and choose a function h from the family ofE1-equivalence

classes onto the family of E2-equivalence classes which preserve cardinality up to; that is, if h(a1/E1) = a2/E2 then g(a1) = g(a2)and min{dim(a1/E1), ) =min{dim(a2/E2), }.

Then the AIS player chooses a pair (a1, a2) such that a >(M) for = 1, 2

such that h(a1/E1) = (a2/E2) and ISO has to choose f+1 f such that f(a1) =a2.

Subcase 2B: The player ISO chooses f+1 f as required such that for somen < and a1

Dom(f) for n we have: {a10, . . . , a1n1} is R1-dependent iff

{f(a10), . . . , f (a1n1)} is not R2-dependent.

{1a.15}Definition 2.8. 1) We say R is a pre-dependence relation on X when R is asubset of [X]


10/32

(897)

revision:2010-10-

15

modified:2010-1

0-17

10 SAHARON SHELAH

(a) R is a subset of [X]


11/32

(897)

revision:2010-10-

15

modified:2010-1

0-17


(d) () xn is an initial segment of a play of the game +2,0,0,(M1, M2)

() in xn only finitely many moves have been played(can specify), the last one is m(xn )

() in xn , the player ISO uses his winning strategy st

(e) if 1 T1,n and 2 = hn(1), then for some b1 Dom(fxnm(xn )

)

we have

() b1 a1/EM1n

() fxnm(xn )

(b1) a2hn()/EM2n

(f) if T1,n then xg() is an initial segment of xn .

Why can we carry the induction?

For n = 0:Note that h0 is uniquely determined. As for x

0, any x as in (d) is O.K., as

long as at least one move was done (note that EM0 has one and only one equivalenceclass.

For n = m + 1: So hm, xm has been chosen.Let 1 T1,m and let 2 = hm(1) and

F1 := {(1, 2) : 1 sucT1(1), 2 sucT2(2)and there is x as in (d) such that

xm1 is an initial segment of x and for someb1 Dom(f

xm(x)) we have

b1 a

1

1/E

M1

n and f

x

(b1) a

2

2/E

M2

n }.Now

to do the induction step, it suffices to prove that: if 1 T1,m then thereis a one-to-one function hn,1 from sucT1(1) onto sucT2(2) such that sucT1(1) (, hn,1()) F1 .

However by case 2 in Definition 2.7 this holds.

Stage E:So we can find hn : n < , x : T1 as in . Let h := {hn : n < },

clearly it is an isomorphism from T1 onto T2 and h is well defined, see fromStage C.

So it is enough to check the sufficient condition for M1 = M2 then, i.e. T1, 1, = 2,h(). But if T1, then xn : n < is a sequence of initialsegments of a play ofG with ISO using his winning strategy st, increasing with n,each with finitely many moves. So x, defined as the limit xn : n < , is aninitial segment of the play G, with moves and f

xm(x)

= {fxnm(xn)

: n < }.

Clearly n < f(a1n)EM2n a

2hn(n)

. As we have one move left and can use

case 2 in Definition 2.7(2) we are done. 2.9

The following claim says that the game in 2.5, 2.7 when = +, < divisibleenough are equivalent, i.e. the ISO player wins one iff he wins the other.


12/32

(897)

revision:2010-10-

15

modified:2010-1

0-17

12 SAHARON SHELAH

{1a.23}Claim 2.10. 1) M1, M2 are EF

+,,,-equivalent when :

(a) M1, M2 are -models

(b) = +1 , 1 and and and cf() < 1 > and I[], see Definition 1.5

(c) M1, M2 are EF(),-equivalent where () = 1 (see Definition 2.5(2))

(d) M = = cf() cf())

and he actually plays fx A1, i.e. is an initial segment of a play ofG, of length

in which the ISO player uses the strategy st such that [1 < x1 is an initialsegment of x].

The only problem is when = + 1 and in Definition 2.7, Case 2 occurs, i.e.with the AIS player choosing R1, R

2. We may for notational simplicity choose

< and deal only with A (M) for = 1, 2.We can consider x extending x; if it is as required in subcase (2B) of Definition

2.7 we are done. Let

F1 = {(a1, a2) : for some < , a (M) for = 1, 2

and there is a candidate x for the-th move such that fx(a1) = a2}.

Let

F2 = {(a2, a1) : (a1, a2) F1}

A1

= A

A2

= {a A : the number ofb such that (a, b) F is }

A3

= A2

{a : for some b A2

3 we have (a, b) F}.

So |A3 | by clause (d) of the assumption and let a : < list A

3 possibly

with repetitions

() it is enough to take care ofA3 A for = 1, 2.


13/32

(897)

revision:2010-10-

15

modified:2010-1

0-17


[Why? By the basic properties of dependence relation.]So we can continue.

Let S be the set of limit ordinals < such that: for a club of [, ) ofcofinality 0 we can find b

: [, ) for = 1, 2 such that:

() b {a : [, )}

() (b1, b2) F

1

() b : [, ) is R-independent over {a : < }

() if < and a A then a does R-depend on {a

: < } {b

:

[, )}.

If S is not stationary we can easily finish (we start by playing moves in G ). Soassume S is stationary, hence for some regular 1 the set S

= { S : cf() =} is stationary. By playing + moves (recalling I[]) we get a contradiction

to the definition of S.2),3) Obvious. 2.12

Remark 2.11. In 2.10(1), to get the exact (), we combine partial isomorphisms.

So we simulate two plays and use the composition of the fxi s from two plays where

in each ISO use a winning strategy st.{1a.22}

Claim 2.12. We can use a variant of Definition 2.7(2) as follows: we can in case2 make a R dependence relation on >(M), but equivalently C >(M) fora set C of cardinality , but

(a) it seems to help presently relevant only for 2|(M1)|+0

(b) if 22. now we define F : >(M) (2(M) as follows: for a >(M) let

i(a) = min{i : 2i g(a) or 2i + 1 < g(a) a2i = a2i+1}

a = T.V.(a2i(a)+2+2j = a2i(a)+2+2j+1) : j 0 and 2i(a) + 2 + 2j + 1 < g(a)

where T.V. stands for truth value.

(a) = Min{ : if < then = a}.

Finally, F(a) is ((a), a2j : j < i(a)) if i(a) < g(a) (a) < , and is (0, a) ifotherwise.

Let

R := {A : A>(M) and {F(a) : a A} R

or for some a = a A we have F(a) = a}.


14/32

(897)

revision:2010-10-

15

modified:2010-1

0-17

14 SAHARON SHELAH

Now check. 2.12{1a.24}

Claim 2.14. M1, M2 are EF+,,,-equivalent when :

(a) K a class of 0-structures and [K], see 3.2(8), used here for K =Kor = the class of linear orders and Koi, see Definition 3.1

(b) the structures I1, I2 K are EF+,,,-equivalent

(c) M = EM(I, ) for = 1, 2 for some (d) 0 and || < .

Proof. Let St be a winning strategy of the ISO player in the game G+,,,(I1, I2).

We define a strategy st of the ISO player in the game G+

,,,(M1, M2) as follows.During a play of it after moves a partial isomorphism f from M1 to M2 has

been chosen, but the ISO player also simulates a play ofG+,,,(I1, I2) in which wecall the function h, and in which he uses the winning strategy st and

f h where h is defined byh(

M1 (at0, . . . , atn1)) = M(ah(t0), . . . , ah(tn1)) for n < , (x0, . . . , xn1)

a term of and t0, . . . , tn1 Dom(h).

Why can the player ISO carry this strategy st? Suppose we arrive to the -thmove. The point to check is Case 2 in Definition 2.7(2), so the AIS player haschosen R1, R2,A1,A2 as there.

Let { (x ) : < 2(M) = {M (t) : < 2


15/32

(897)

revision:2010-10-

15

modified:2010-1

0-17


3. The properties of T and relevant indiscernibility

In [Sh:c, Ch.VIII], [Sh:e, Ch.VI] we use as indiscernible sets trees with + 1levels, suitable for dealing with unsuperstable (complete first order) theories.

Here instead we use a linear order and family of -sequences from it, suitablefor our case. The letters oi stands for order + increasing (-sequences).

{2b.1}Definition 3.1. 1) Koi is the class of structures J of the form (J,Q,P


16/32

(897)

revision:2010-10-

15

modified:2010-1

0-17

16 SAHARON SHELAH

(b) at M for t IJ, a = a M1 for P

J

(c) for PJ, t QJ and n < we have M |= n[a, at] iff Fn() = t;

(pedantically we should write n(a, at g(yn)))(d) at : t J is indiscernible in M for the index model J

(e) M is a model of T1

(f) in fact (not actually used, see 3.2) there is oi|T1| depending on

T1, only such that M = EM(J, ), in fact if n < , t = t : < n J then tpqf(at0 . . . atn1, , M) = (tpqf(t, , J)).

Proof. Let I = (QJ,


17/32

(897)

revision:2010-10-

15

modified:2010-1

0-17


free type mean the truth value of the inequalities Fn1(r) = Fn2(r

)(including F) and the order between those terms).

Proof. Straight (or as in [Sh:e, Ch.III] = [Sh:E59]). 3.5

Remark 3.6. We could have replaced QJ by the disjoint union of QJn : n < , (I) and some automorphism of I over I, maps

s to t}.

()7 Let Y be the set of e-equivalence classes.

Note

1 for {1, 2}, n < and y0, . . . , yn Y the following are equivalent:(a) some a yn depend (by R1) on y0 . . . yn1(b) every a yn depend (by R1) on y0 . . . yn1.

So R1 induce a 1-dependence relation on Y1, so let yi : i < i() be a maximalindependent subset of Y1[i < i() a yi a does not depend on {yj : j |T| and M1, M2 ModT() are L,(T)-equivalent then

M1, M2 are isomorphic(B)2 like (B)1 for some = cf() > |T|

(C) if = cf() > |T| and M1, M2 ModT(+) are EF,-equivalent thenM1, M2 are isomorphic

(D) for some regular > |T|, if M1, M2 ModT(+) are EF,+ -equivalent

then they are isomorphic.

Proof. Clause (A), clause (B)1, clause (B)2 are equivalent because: as proved in[Sh:c, Ch.XIII,Th.1.11], we have (A) (B)1and(B)2 and the inverse implicationholds by [Sh:220]. Now by the definitions trivially (B)1 (C) (D).

Lastly, by [HySh:529], i.e. by 5.1 we have (A) (D), i.e. (D) (A) so wehave the circle. 5.5

So 5.5 tells us what we know about Qustion (A)0 of 2.2. Similarly concerning(B)0 of Question 2.2.

{4d.27}Conclusion 5.6. (ZFC) For first order countable complete first order theory T and 20 the following conditions are equivalent:

(A) T is unsuperstable

(B) for every > |T| and (, T)-candidate (see Definition 1.8), and or-dinal < satisfying ||+ = | || , there are EF,-equivalentnon-isomorphic models M1, M2 PC(T)() of cardinality


31/32

(897)

revision:2010-10-

15

modified:2010-1

0-17


(C) for some > |T|, for no (, T)-candidate is the class PC(T)()categorical in .

Proof. First, assume T is superstable, so clause (A) holds. By the proofs of [Sh:c,Ch.VI,4] there is a (, T)-candidate , PC(T)() is the class of saturated models of

T, (in details, ifn < , a nC, tp(b, a, C) is stationary, q = tp(b, ,C), p = p(x, y) =tp(ab, ,C) then let p,q be such that M |= p,q iff for every b

nM realizingthe type q(y), the function c FMp,q(c, b

) is one-to-one and if k < , c0, . . . , ck

M are pairwise distinct then tpL((T))(FM

p,q(ck), {FM

p,q(c0), . . . , F M

p,q(ck1)} b, M)

extends p(x, b) and does not fork over M.Lastly, = {p,q : p, q as above} so L+,. So in the present case also

(B), (C), (D) holds.Second, assume T is not superstable, so clause (A) holds and we shall prove the

rest. Let be a (, T)-candidate.By 1.9 and let there is 1-tr witnessing this hence witnessing unsupersta-

bility and now we can use Theorem 5.1 quoted above. 5.6


32/32

)

revision:2010-10-

15

modified:2010-1

0-17

32 SAHARON SHELAH

References

[HT91] Tapani Hyttinen and Heikki Tuuri, Constructing strongly equivalent nonisomorphic mod-

els for unstable theories, Annals Pure and Applied Logic 52 (1991), 203248.[Vaa95] Jouko Vaananen, Games and trees in infinitary logic: A survey, Quantifiers

(M. Mostowski M. Krynicki and L. Szczerba, eds.), Kluwer, 1995, pp. 105138.[Sh:c] Saharon Shelah, Classification theory and the number of nonisomorphic models, Studies in

Logic and the Foundations of Mathematics, vol. 92, North-Holland Publishing Co., Amsterdam,xxxiv+705 pp, 1990.

[Sh:e] , Nonstructure theory, vol. accepted, Oxford University Press.[Sh:h] , Classification Theory for Abstract Elementary Classes, Studies in Logic: Mathe-

matical logic and foundations, vol. 18, College Publications, 2009.[Sh:E53] , Introduction and Annotated Contents, 0903.3598.[Sh:E59] , General non-structure theory and constructing from linear orders .[Sh:220] , Existence of manyL

,-equivalent, nonisomorphic models ofT of power, An-nals of Pure and Applied Logic 34 (1987), 291310, Proceedings of the Model Theory Conference,Trento, June 1986.

[Sh:300] , Universal classes, Classification theory (Chicago, IL, 1985), Lecture Notes in

Mathematics, vol. 1292, Springer, Berlin, 1987, Proceedings of the USAIsrael Conference onClassification Theory, Chicago, December 1985; ed. Baldwin, J.T., pp. 264418.

[HySh:474] Tapani Hyttinen and Saharon Shelah, Constructing strongly equivalent nonisomor-phic models for unsuperstable theories, Part A, Journal of Symbolic Logic 59 (1994), 984996,math.LO/0406587.

[HySh:529] , Constructing strongly equivalent nonisomorphic models for unsuperstabletheories. Part B, Journal of Symbolic Logic 60 (1995), 12601272, math.LO/9202205.

[HySh:602] , Constructing strongly equivalent nonisomorphic models for unsuperstabletheories, Part C, Journal of Symbolic Logic 64 (1999), 634642, math.LO/9709229.

[Sh:F705] Saharon Shelah, Representation over orders of elementary classes.[Sh:783] , Dependent first order theories, continued, Israel Journal of Mathematic 173

(2009), 160, math.LO/0406440.[Sh:836] , On long EF-equivalence in non-isomorphic models, Proceedings of Logic Col-

loquium, Helsinki, August 2003, vol. Lecture Notes in Logic 24, ASL, 2006, math.LO/0404222,pp. 315325.

[Sh:863] , Strongly dependent theories, Israel Journal of Mathematics accepted,math.LO/0504197.

[HvSh:866] Chanoch Havlin and Saharon Shelah, Existence of EF-equivalent Non-IsomorphicModels, Mathematical Logic Quarterly 53 (2007), 111127, math.LO/0612245.

[Sh:907] Saharon Shelah, EF equivalent not isomorphic pair of models, Proceedings of the Amer-ican Mathematical Society 136 (2008), 44054412, 0705.4126.

[Sh:918] , Many partition relations below density, Israel Journal of Mathematics accepted,0902.0440.

[Sh:F918] , Theories with EF-Equivalent Non-Isomorphic Models II.

Einstein Institute of Mathematics, Edmond J. Safra Campus, Givat Ram, The He-

brew University of Jerusalem, Jerusalem, 91904, Israel, and, Department of Mathe-

matics, Hill Center - Busch Campus, Rutgers, The State University of New Jersey, 110

Frelinghuysen Road, Piscataway, NJ 08854-8019 USA

E-mail address: [email protected]

URL: http://shelah.logic.at

saharon shelah- theories with ef-equivalent non-isomorphic models

Documents