saharon shelah- dependent t and existence of limit models

8/3/2019 Saharon Shelah- Dependent T and Existence of Limit Models

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DEPENDENT T AND EXISTENCE OF LIMIT MODELS

SH877

SAHARON SHELAH

Abstract. Does the class of linear orders have (one of the variants of) theso called (, )-limit model? It is necessarily unique, and naturally assumingsome instances of G.C.H. we get some positive results. More generally, lettingT be a complete first order theory and for simplicity assume G.C.H., for regular > > |T| does T have (variants of) a (, )-limit models, except for stableT? For some, yes, the theory of dense linear order, for some, no. Moreover,for independent T we get negative results. We deal more with linear orders.

0. Introduction

The first part of the introduction is intended for a general mathematical reader.Cantor proved that the structure the rationals as a linear order is characterizedup to isomorphism by being a dense linear order with neither first nor last elementwhich is countable. Hausdorff generalizes this as follows. For transparency assumethe G.C.H., the generalized continuum hypothesis then for every cardinal thereis a unique linear order I of cardinality + which is +-dense (i.e. if A < C aresubsets of cardinality then for some b I we have A < b < C) with neither firstnor last elements). This canonical linear order is, in later model theoretic notions,the unique saturated model of the theory Tord = Th(Q,


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What are our main results? First, a result meaningful also to one with very littleset theoretic background. If = |T|.

Note that if T has (any version of) a limit model of cardinality then there is

a universal M Mod(T). Now we know that if = 2 |T| then there is auniversal M Mod(T) (see e.g. [2]). But for other cardinals it is hard to havea universal model, see history [5] and [6]. E.g. if T has the strict order property,then, by Kojman-Shelah [5] there are ZFC non-existence results (a major case, forregular is when ()(+ < 2 > ). In at least one case, = 1 < 2

0

consistently we do not have a universal model, see [14].Stable theories have limit models (in many cases); hence it is natural to ask:

Question 1: Assume = > |T|. Does the existence of a (, )-md.-limitmodel of T imply T is stable?


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DEPENDENT T AND EXISTENCE OF LIMIT MODELS SH877 3

This is quite reasonable but in Theorem 1.1 we find a counterexample, in fact,one everyone knows about: the theory Tord of dense linear orders (see 0.12). This

per se is a continuation of Hausdorff result, revealing some canonical linear ordres.Returning to the family of elementary classes, i.e. first order theories, it is naturalto ask:

Question 2: Does T have a (, )-i.md.-limit model whenever = + |T|for every unstable T?

For non-existence results it is natural to look at T dissimilar to Tord.As Tord is prototypical of dependent theories, it is natural to look for independenttheories. A strong, explicit version of T being independent is having the strongindependence property (see Definition 2.4), e.g. Peano arithmetic has. We provethat for such T there are no limit models (2.3). But the strong independenceproperty does not seem a good dividing line. The independence property is a goodcandidate for being a meaningful dividing line.

Question 3: If T is independent, does T have a (, )-i.md.-limit model (with = > |T|)?

We work harder (than in 2.3) to prove (in 2.9) the negative answer for everyindependent T (for many cardinals), i.e. with the independence property though aweaker version meaning we prove non-existence of a stronger version of (, )-limitmodel.

This makes us {0.x.7}Conjecture 0.1. Any dependent T has (, )-i.md.-limit model.

Toward this end we intend to continue the investigation of types for dependentT.

We shall also consider a property Pr,(T) (and the stronger Pr2,(T)), see

Definition 2.5, which are relatives of there is no (, )-x-limit model; i.e. non-

existence results for independent T holds for = |T|. For > this strengthens there is no (, )-i.md.-limit model. But = is a newnon-trivial case and it is also a candidate to be an outside equivalent condition forT being dependent.

The most promising among the relatives (for having a dichotomy) is the followingconjecture (the assumption 2 = + is just for simplicity).

{0.x.16}Conjecture 0.2. The generic pair conjecture

Assume = |T| and 2 = + (for transparency) and M EC(T) is-increasing continuous for < + with

{M : < +} EC+(T) saturated.

Then T is dependent iff for some club E of+ for all pairs < < + from E bothof cofinality , (M , M) has the same isomorphism type (we denote this property

of T by Pr

2

(T)), see Definition 2.5).Here we prove that for independent T, a strong version of the conjecture holds.In 2, we also prove the parallel of what we say above. In 3 we prove that

(, )-superlimit models does not exist even for T = Tord. This work is continuedin [12], [10], [8], [11].

Now we define some versions of M is a (, S)-x-limit model and for them M

obeys a (, T)-x-function.


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4 SAHARON SHELAH

{y.1}Notation 0.3. 1) Let T denote a complete first order theory.

2) Let T = (T), M = (M) be the vocabulary of T, M respectively.{y.2}Definition 0.4. 1) For any T let EC(T) = {M : M a T-model of T}.2) EC(T) = {M EC(T) : M is of cardinality } and EC,(T) = {M EC(T) :M is -saturated}.3) We say M EC(T) is -universal when every N EC(T) can be elementarilyembedded into M.4) We say M EC(T) is universal when it is -universal for = M.5) For T T let

PC(T, T) = {M T : M is model of T}

PC(T, T) = {M PC(T, T) : M is of cardinality }.

{y.3}Definition 0.5. Given T and M EC(T) we say that M is a (, )-superlimitmodel when: M is a -universal model of cardinality and if < + is a limitordinal such that cf() = , M : is -increasing continuous, and M+1 isisomorphic to M for every < then M is isomorphic to M.

{y.4}Remark 0.6. We shall use:

(a) (, )-i.md.-limit in 1.1, (existence for Tord)

(b) (, )-wk-limit in 2.3, (non-existence from T is strongly independent)

(c) (, )-md.-limit in 2.9, (non-existence for independent T)

(d) (, )-i.st.-limit for Tord: 3.12 and 3.5(3), 3.7(3), (on characterization) forTord)

(e) (, )-superlimit in 3.10 (non-existence).Recall the definition of some versions of (, )-limit model.

{y.4d}Convention 0.7. In this owrk let M is (, S)-limit mean M is (, S)md-limit,see Definition below; similarly for (, ).

{y.5}Definition 0.8. Let be a cardinal |T|. For parts 3) - 5) but not 6), forsimplifying the presentation we assume the axiom of global choice; alternativelyrestrict yourself to models with universe an ordinal [, +). Below if S = { (M1)} onto tp(a, M, M2) : a >(M2)}

(actually even demanding just E is O.K., i.e. we can prove it); note that acts of M hence on S


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() a realizes the type p(y, c) = {(c , y) : < } in M iff a M.

The number of isomorphism types of T

-models M of cardinality is 2 whereasthe number of ci : i < for a given M

is < 2.For a given F the construction above works for every M EC(T), but

I(, T) = 2, see 0.14 as |T| + 1 so we can finish easily, or see more inpart (2).2) We can make the counterexample more explicit. For a model M and c g(x)Mfor < we define N = N[M, c : < ] as the following submodel of M (if welldefined): it is the submodel with universe the set A = {d M : M |= [c, d] forevery < }; (note that N is not necessarily an elementary submodel of M or evenwell defined, e.g. A = or A not closed under functions of M). For M EC(T)let M[M] = {N M : N is N[M, c : < ] for some c g(x)M for < }.Fixing F we can choose M EC(T) with universe (1 + ) such that

()1 if = 4+ 3 and 4 then M is not isomorphic to N M wheneverthere are c g(x)(M) for < such that N = N[M, c : < ]

()2 for < < + there is c

(g(x))(M+1) such that for every a M wehave M+1 |= [c

, a] a M.

As I(, T) = 2 and moreover for any theory T1 T of cardinality we haveI(, T1, T) = 2

and for every M EC(T), the number ofN M[M] is < 2

we get

for every appropriate F there is a -increasing continuous sequence M : < + of models of T obeying F such that if 1 = 2 < + has cofinality then M1, M2 are not isomorphic.

[Why? Without loss of generality 1 < 2, let 2 : < be increasing with limit2, all > 1 + 4. Now by ()2 we know that c

1+3

: < exemplified that in M2there is a sequence c : < which define M1+3, i.e. M1+3 = N[M2 , c

: = cf() then T has no (, ) md-limit models.2) Moreover, there is F such that


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(a) F is a function with domain

{K : < +odd} where K = {M : M amodel of T with universe (1 + )}

(b) if < + is odd and M K then M F(M) K+1(c) ifM K for < + is -increasing continuous andM2+2 = F(M2+1)

for odd < + then for no < + is the set{ : M = M and cf() = }stationary.

3) We can strengthen part (2) by adding in clause (c):

() there are c (M2+2) for < + such that: if , : < is anincreasing continuous sequence of ordinals < + with limit for = 1, 2and 1 = 2 then there is no isomorphism f from M1 onto M2 mappingc1, to c2, for < .

4) In part (2) we can replace K (for < +) by K


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(c) T,,2M,a = {(x)[P(x) tJ

(x, at)if((t))] for every finite J I and J2}

(so the vocabulary is +M

).

3) In (2) let M,a = T,M,a be the family of consistent sets of sentences in L(

+M)

such that is of the form M,a union with a subset of M,a = {(x)[P(x) (x, c)

tJ

(x, at)(t)] : J I is finite, J2, c g(z)M and (x, z) L(T)}.

4) For M,a let

(a) S = {p : p S(M) and {(x)(P(x) (x, c)) : (x, c) p(x)} isconsistent}

(b) for J I and J2 let S, = {p S : p include qM,a}

where

(c) q

M,a := q

T,,

M,a = {(x, at)

if((t))

: t Dom()}.5) For M,a, (x, z) L(T) and c

g(z)M let

M,a,((x, c)) = { fin(I) : is consistent with(x)[P(x) (x, c)

tDom()

(x, at)if((t))]}.

{2p.6g}Remark 2.12. 1) fin(I) = { : is a function from some finite J I to {0, 1}}.2) In parts (3) and (4) we could have used only (x, z) {(x, y), (x, y)}.

{2p.5}Observation 2.13. Let (M, a) be a (T, )-candidate.1) M,a M,a, i.e., M,a is consistent so M,a is non-empty.2) M,a is closed under increasing (by ) unions.

3) Any member of M,a can be extended to a maximal member of M,a.4) If M N then (N, a) is a (T, )-candidate and for every M,a the set N,a belongs to N,a.5) If I : is an increasing continuous sequence of linear orders and N : is -increasing continuous sequence of models of T, a = at : t I and(N, a I) is a (T, )-candidate for < then (N, a) is a (T, )-candidate.6) In part (5), if N,a for < is increasing continuous with then :=

{ : < } belongs to N,a.

7) In part (6) if is maximal in N,a for each < then is maximal inN,a.

8) If M,a, (x, z) L(T), c g(z)M and M |= (x)(x, c) then

(a) the empty function belongs to M,a,((x, c))

(b) if I1 I2 are finite subsets of I and M,a,((x, c)) (I1)2 then thereis M,a,((x, c)) (

I2)2 extending .

Proof. Straightforward. 2.13{2p.6}

Claim 2.14. Assume that (M, a) is a (T, )-candidate and T,M,a is maximal.

1) If (x, z) L(T) and c g(z)M and M,a,((x, c)) fin(I) then for

some we have fin(I) and / M,a,((x, c)).2) For every I2 there are N, b such that:


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(a) M N (and N M + |T|)

(b) b N

(c) if t I then N |= [b, at]if((t))

(d) if a g(z)M, = (x, z) L(T) and (x, a) tp(b,M,N) then isdisjoint to {(x)[P(x) (x, a)

tJ

(x, at)if((t))] : J I finite}.

Proof. 1) Assume that the conclusion fails. Consider the formula (x, c) :=tDom()

(x, at)if((t)) (x, c).

By the assumption of the claim + the assumption toward contradiction it followsthat fin(I) {(x)[P(x)

tDom()

(x, at)if((t)) (x, c)]} is consistent).

[Why? Just note that it is enough to consider fin(I) such that Dom() and we split to two cases: first when Dom() = then (x, c) adds nothing in

the conjunction (and use 2.11(2)(c)); second when Dom() = and we use theassumption toward the contradiction.]

So if N is a model of and we define N as N by replacing PN

by PN

={b PN

: N |= [b, c]} we see that {(x)[P(x) (x, c)]} M,a. By themaximality of it follows that (x)[P(x) (x, c)] . But this contradictsthe assumption M,a,((x, c)).2) Easy. 2.14

{2p.7}Claim 2.15. Assume that

(a) (M, a) is an (I , T , )-candidate

(b) = i : i < i() and i I2 for i < i()

(c) j() i()

(d) {i : i < j ()} is a dense subset of I2.

Then we can find N, c such that

() M N and N M + |T| + |i()|

() c = ci : i < i() and ci N

() if i < i() and t I then N |= [ci, at]if(i(t))

() for every a g(y)M at least one of the following holds:(i) [the perfect fakers] for some t I for every 0 fin(I\{t}) we can

find 1 fin(I\{t}) extending 0 such that: 1 i i < i() N |=[ci, a] [ci, at], i.e. for most i < i(), a, at are similar

(ii) [the rejected as] for no t I do we have i < j() N |= [ci, a] [c

i, a

t]

.Proof. By 2.9(1), M,a M,a hence by 2.9(3) there is a maximal M,a.Let N, ci : i < i() be such that

(a) M N and N = M + |T| + |i()|

(b) for i < i(), ci N realizes some pi S,i (see Definition 2.11(4)(b)).

Clearly clauses (), (), () of the desired conclusion hold, and let us check clause(). So assume that a g(y)M and clause (ii) there fails so we can choose t Isuch that i < j() N |= [ci, a] [ci, at].


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So it is enough to prove clause (i) for t; toward this assume 0 fin(I) satisfiest / Dom(0), i.e. 0 fin(I\{t}). Let 1 fin(I) extend 0 be such that 1(t) = 0.

By clause (d) of the assumption we know that for some i < j() we have 1 ibut (see above) N |= [ci, a] [ci, at] hence 1 M,a,((x, a)if(

i(t))) which

means that 1 M,a,((x, a)if(1(t))). Now apply claim 2.14(1) to (x, c) :=

(x, a)if(1(t)), so we know that for some we have 1 fin(I) and /M,a,((x, a)if (

1(t))) hence

()1 if i < i() satisfies i then N |= [ci, a]if(1(t)) which means N |=

[ci, a] [ci, at].

Let 2 Dom()2 be such that 2(t) = 1 and s Dom()\{t} 2(s) = (s). We

repeat the use of 2.14(1) for 2 instead of 1 and get such that 2

fin(I)and

()2 if i < i() satisfies i then N |= [ci, a]if(2(t)) which means thatN |= [ci, a] [ci, at].

Let 3 = (Dom()\{t}) and by ()1+()2 the function 3 fin(I) is as required

in subclause (i) (for our a ,t,0) in clause () of the claim, so we are done. 2.15{2p.11}

Definition 2.16. 1) For a model M of T, formula = (x, y) L(M), c =ci : i < i(), < such that c

i M let P(c, M) = {U i(): for some

a g(y)M for every < large enough U = u(a, ci : i < i(), M)} whereu(a, ci : i < i(), M) = {i < i() : M |= [c

i , a]}.

2) For a model M and = (x, y) L(M) let Pi(),(M) = {P(c, M) : c has

the form ci : i < i(), < with ci M}.

3) For M1

M2

and = (x, y) L(M

) and c = ci

: i < i() i()(M2

) letP(c, M1, M2) = {u(a, c, M2) : a

g(y)M1}.{2p.12}

Observation 2.17. For M, (x, y), i(), as in Definition 2.16, Pi(),(M) has

cardinality M|i()|+.{2p.13}

Claim 2.18. If I is a linear order of cardinality 2 then we can find a uniformfilter D on and a sequence U = Ut, : t I, {0, 1, 2} of members of []

such that:

1 (a) for each t I, Ut, : = 0, 1, 2 is a partition of

(b) Ut,2 D for t I

(c) P()/D has cardinality 2, moreover extend some free Boolean

Algebra of cardinality 2

2 if (M, a) is a (I , T , )-candidate and U = Ut : t I satisfies Ut,1

Ut Ut,1 U

t,2 then for some (N, c) we have

(a) M N and N M + |T| +

(b) c N

(c) ift I, a g(y)M andUt,1 u(a, c, N) Ut,1U

t,2 thenu(a, c, N) =

Ut mod D

(d) if t I thenUt P(c, M , N ).


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Proof. We replace by + .Let i : i < be a sequence of members of

I which is dense possible by [1].

For = 0, 1, 2 let Ut, = {i < : i(t) = or i(t) 3 = 2}. Notice that it isimportant that D is defined independently ofUt and we should therefore define ithere. But for clarity of exposition we will only define it later.

Let (where +U = { + : U})

()1 Ut,0 = Ut,0 ( +Ut,0)

()2 Ut,1 = Ut,1 Ut,2 ( +Ut,1)

()3 Ut,2 = +Ut,2.

Assume U = Ut : t I is such that

()4 Ut,1 Ut U

t,1 U

t,2 + .

Define i = Ui +2 for i < + by:

()5 i(t) =

0 i / Ut1 i Ut

Let = i : i < + .Notice that i : i < is dense in I2 by the choice of i in ()5 because Ut =Ut,1 U

t,2 and

i : i < was dense in

{0,1,2}I. By 2.15 applied to (M, a, ), i() = + , j() = we can find N, c as there and we should check that they are asrequired. Clauses (), (), () of the conclusion of 2.15 give the soft demand.

More specifically clause (a) of1 holds by the choice of the Ut,s; clauses (b),(c)

of2 holds by the conclusion of 2.15.

Clearly

()6 u(at, c, N) = {i < + : N |= [ci, at]} = {i < + : i(t) = 1} = Ut

hence

()7 t I Ut P(c, M , N ).

So we see that demand (d) of2 is satisfied - all the Ut are included. We still needto prove clause (c) of2, that is to show that there are no fakers. So assume

1 Ut1 u(a, c, N) U

t1 U

t1

for some t1 I and a g(y)M.

Denote U = u(a, c, N). We need to show U = Ut1 mod D.

By clause () of the conclusion of 2.15 for a one of the two clauses there (i),(ii)occurs.

Recall that

2 Ut1,1 U U

t1,1 U

t1,2.

So U = Ut1,1 = Ut1,1 Ut1,2.Now

3 for a clause (ii) of 2.15() fails.


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[Why? Because t1 witnesses this by the above equality and for each i <

i U i Ut1,1 Ut1,2 i[t1] = 1 N |= [ci, at1 ].]By 2.15() and 3 we can deduce:

4 for a, clause (i) of 2.15() holds so there is t2 witnessing it.

Next

5 t1 = t2.

Why? Toward contradiction assume t1 = t2 hence we can find 1 fin(I\{t2})such that

5.1 1 i i < + N |= [ci, a] [ci, at2 ].

Without loss of generality t1 Dom(1) and define 2 = {(t2, 1 1(t1))}, so1 2 fin(I). As {i : i < } was chosen as a dense subset of {0,1,2}I, there isi < such that 2

i , hence by 5.1

5.2 N |= [ci, a] [ci, at2 ]

but by the choice of i we have:

5.3 N |= [ci, at2 ]if(i(t2))

but i(t2) = 1 1(t1) hence together

5.4 N |= [ci, a]if(11(t1))

but by the choice of ci we have:

5.5 N |= [ci, at1 ]if(i(t1))

hence by 5.1

5.6 N |= [ci, at1 ]if(1(t1)).

But 5.4 + 5.6 contradict the choice of t1 as i < using 2 so 5 holds, i.e.t1 = t2.]

Now subclause (i) of 2.15() tells us

6 for every 0 fin(I\{t1}) there is 1 fin(I\{t1}) extending 0 such that(a) 1 i i < + N |= [ci, a] [ci, at1 ] hence

(b) if 1 i i < + i u(a, c, N) i Ut1 .

So let

D = {U + : for every 0 fin(I) there is 1,0 1 fin(I) such that i < + 1 i i U}.

Clearly the filter D satisfies clause 1(c) so we are done. 2.18

Proof. Proof of the Theorem 2.9(3) Like the proof 2.3 of the case T has the strongindependence property. 2.9


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{2p.8.3}Remark 2.19. 1) The F we construct works for all = cf() < for which =

simultaneously. {2p.9}Discussion 2.20. Can we prove 2.9 also for strongly inaccessible? Toward this

(a) we have to use c = c,i : i < , instead c,i : i <

(b) each M has a presentation M, : <

(c) for a club E of < , we use c,i : i < c to code U

(d) instead i, + i we use 2i, 2i + 1.

So the problem is: arriving to , we have already committed ourselves for thecoding ofU for E , what freedom do we have in ?

Essentially we have a set 2

2 quite independent, and for 1 < 2, there isa natural reflection, the set of possibilities in 2 is decreasing. But the amount offreedom left should be enough to code. We shall deal in [12] with the inaccessiblecase.

Question 2.21. Can we improve 2.9(3) in the case of T not strongly dependent?{2.2.14}

Claim 2.22. 1) Assume T has the strong independence property. If =cf(), 2min{2

,} > and > |T|, then Pr,(T).2) Assume T is independent. If , are as above, then Pr(, ).

Proof. 1) Let (x, y) exemplify T has the strong independent property, see Def-inition 2.4.

We choose F such that:

() if F(Mi : i + 1) M+2 then for every i for some c = c,i g(x)(M+2) the set {a Mi : M+1 |= [c, a]} does not belong to {{a

Mi : M+1 |= [d, a]} : d g(x)(M+1)}.

We continue as in the proof of 2.3.2) Similarly (recalling the proof of 2.9). 2.22


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3. More on (, )-limit for Tord

It is natural to hope that a (, )-i.md.-limit model is (, )-superlimit but inTheorem 3.10. we prove that there is no (, )-superlimit model for Trd, see Defi-nition 0.12(2).

We conclude by showing that the (, )-i.md.-limit model has properties in thedirection of superlimit. By 3.12 it is (, S)-limit+, that is if M : < + is a-increasing sequence of (, )-i.md-limit models for a club of < + of cofinality the model {Mi : i < } is a (, )-i.md.-limit model. Also in 1 the function Fdoes not need memory.

{nl.0}Hypothesis 3.1. 1) = = cf().2) We deal with ECTrd(), ordered by , so M, N denotes members of EC(Trd).

Recall Trd is from Definition 0.12(2) and recalling Definition 0.13.{nl.0.1}

Definition 3.2. 1) If M N and (C1, C2) is a cut of M let N[(C1,C2)] = N

{a N : a realizes the cut (C1, C2) of M which means c1 C1 c1


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(a) if = M =


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() M = M : is such that M from the proof of 1.1 holds andM = M.

{nl.11}Claim 3.10. For = = cf() then there is no (, )-superlimit model ofTord.

Remark 3.11. It is trivial to show that there is no superlimit M EC(T), but wedeal with (, )-superlimit.

Proof. Assume there is one, then by 1 it is a (, )-i.md.-limit model so there isM = Mi : i which witnesses this (i.e. such that M from the proof of 1.1).

As M0 is universal for EC(Trd), we can find c M0 for ( + 1) such that


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(d) if is a successor then is a successor

(e) if +1+1 ( + 1) then c (C+1(+11) ,

C+1(+11))M is disjoint to N .

There is no problem to carry the induction:

Case 1: = 0.Choose = 0, 0.

Case 2: = 1 + 1.Choose = h1 (1) + 6.Choose such that

(h1(1 )+5)1 .

Case 3: limit. = { : < }.Choose +1 such that < .Let =

{ : < }. So

( + 1) and c / N for every < but{N : < } = M

= M

, contradiction, so 1 holds indeed.

M is a (, )-.i.md.-limit model for < .

Why? We define M,i M for i < by: c M,i iff one of the following occurs:

(a) c Mi but for no do we have c B :=

{[c(+i), c)Mi : ( + i, )}

(b) c f(Mi) for some .

Let J, =

{(C , C)M : (, )}

J,, = (C , C)

J, : are pairwise disjoint

J,, is an initial segment of J, J, =

{J,, : (, )}.

We will make M,i J, bounded in J, for each i < .Now M,i : i < is a (, )-sequence, see Definition 3.3 hence by 3.5(1) the

model M is a (, )-i.md.-limit model. 3.10{sl.21}

Claim 3.12. If = = cf() then Trd has a (, )-limit+model, i.e.: if

M : < +

is -increasing continuous sequence of models EC(Trd) andM+1 is (, )-i.md.-limit model for every < + then : for a club of < + ifcf() = then M is a (, )-i.md.-limit.

Proof. Let M : < + be as in the theorem and M =


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26 SAHARON SHELAH

Let : be an increasing continuous sequence of ordinals from E and weshall prove that M is (, )-i.md.-limit; this suffices (really just E suffice).

Now M+1 is (, )-i.md.-limit hence there is an -increasing sequence M+1,i :i < witnessing M+1 is (, )-i.md.-limit model. Now M+1,i M =

{M+1,i M : < } hence without loss of generality M,0 M0 hascardinality hence Ni := M+1,i Mi EC(Trd).

Clearly

()1 Ni : i < is a -increasing sequence of members of EC(Trd) with unionM .

So it is enough to show that Ni : i < is a (, )-sequence by 3.5(1). By ()1,clause (a) from Definition 3.3 holds.

2 Ni : i < satisfies clause (b) of 3.3.

[Why? Let i < and (C1, C2) be a cut of Ni of cofinality = (, ). As C1, C2 Ni M+1,i by the properties of M+1,i : i < there is a M+1 such thatC1 < a < C2. If b Mi , C1 < b < C2 then a M and we are done. If not,a induces on Mi a cut (C

1, C

2), C1 C

1, C2 C

2, cf(C

1, C

2) = cf(C1, C2) = .

As i < E, by there is a M such that C1 < a < C2. So clause (b) ofDefinition 3.3 really holds.]

3 if a


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[15] Saharon Shelah. Classification of nonelementary classes. II. Abstract elementary classes. InClassification theory (Chicago, IL, 1985), volume 1292 of Lecture Notes in Mathematics,pages 419497. Springer, Berlin, 1987. Proceedings of the USAIsrael Conference on Classi-fication Theory, Chicago, December 1985; ed. Baldwin, J.T.

[16] Saharon Shelah. Classification theory and the number of nonisomorphic models, volume 92of Studies in Logic and the Foundations of Mathematics. North-Holland Publishing Co.,Amsterdam, xxxiv+705 pp, 1990.

[17] Saharon Shelah. Classification Theory for Abstract Elementary Classes, volume 18 ofStudiesin Logic: Mathematical logic and foundations. College Publications, 2009.

[18] Saharon Shelah. Dependent first order theories, continued. Israel Journal of Mathematic,173:160, 2009. math.LO/0406440.

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]: http://shelah.logic.at

saharon shelah- dependent t and existence of limit models

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