saharon shelah- categoricity and solvability of a.e.c., quite highly

8/3/2019 Saharon Shelah- Categoricity and Solvability of A.E.C., Quite Highly

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arXiv:0808

.3023v1

[math.LO

]22Aug2008

CATEGORICITY AND SOLVABILITYOF A.E.C., QUITE HIGHLY

Saharon Shelah

The Hebrew University of JerusalemEinstein Institute of Mathematics

Edmond J. Safra Campus, Givat RamJerusalem 91904, Israel

Department of MathematicsHill Center-Busch CampusRutgers, The State University of New Jersey

110 Frelinghuysen RoadPiscataway, NJ 08854-8019 USA

0 Introduction

The hope which motivates this work is

0.1 Conjecture: IfK

is an a.e.c. then either for every large enough cardinal ,K

iscategorical in or for every large enough cardinal ,K is not categorical in .Why do we consider this a good dream? See Chapter N.

Our main result is 4.10, it says that if K is categorical in (ignoring few ex-ceptional s) and [LS(K), ) has countable cofinality and is a fix point of thesequence of the s, (moreover a limit of such cardinals) then there is a superlimitM K for which K[M] = K {M

: M = M} has the amalgamation property(and a good -frame s with Ks = K[M]). Note that Chapter IV seems to give astrong indication that finding good -frames is a significant advance. This may beconsidered an unsatisfactory evidence of an advance, being too much phrased inthe works own terms. So we prove in 5 - 7 that for a restrictive context we make

The author would like to thank the Israel Science Foundation for partial support of this research(Grant No. 242/03). Publication 734.I would like to thank Alice Leonhardt for the beautiful typing.

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2 SAHARON SHELAH

a clear cut advance: assuming amalgamation and enough instances of 2 < 2+

occurs, much more than the conjecture holds, see Chapter N on background.Note that as we try to get results on = > LS(K), clearly it does not

particularly matter if for (LS(K

), ) we use, e.g. 1 =

+

or 1 = (2

)+

(=1,1()) or even 1,7().After 4.10 the next natural step is to show that s has the better properties dealt

with in Chapter III, Chapter IV, see [Sh:F782]. Note that if we strengthen the assumption on in 4 (to =


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CATEGORICITY AND SOLVABILITY OF A.E.C., QUITE HIGHLY 3

2) We do not use but it is natural to define +1,0() = ,+1,+1() = ,()

with = (2+1,())+,+1,() =


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4 SAHARON SHELAH

superlimit.3) If every/some M K is locally superlimit then every/some M K is pseudosuperlimit.4) If some M K

is superlimit then every locally superlimit M K

is isomor-

phic to M.5) If M is superlimit in K then M is locally superlimit in K. If M is locally super-limit in K, then M is pseudo superlimit in K. If M is locally superlimit in K thenK has the joint embedding property iff M is superlimit.6) In Definition 0.5(1), clause (c) follows from

(c) LS(K) and K+ = .

7) M K is pseudo-superlimit iffK[M] is a -a.e.c. and K[M] is not the equality.

Also Definition 0.5(3A) is compatible with III. 600-0.33 .

0.7 Definition. For an a.e.c. K, let Ksl ,Kls ,Kpl be the class of M K which aresuperlimit, locally superlimit, pseudo superlimit respectively with the partial orderKsl , Kls , Kpl being K K

sl , K K

pl respectively.

0.8 Definition. 1) is proper for linear orders when:

(a) for some vocabulary = = (), is an -sequence, the n-th element acomplete quantifier free n-type in the vocabulary

(b) for every linear order I there is a -model M denoted by EM(I, ), gen-erated by {at : t I} such that s = t as = at for s, t I and

at0 , . . . , atn1 realizes the quantifier free n-type from clause (a) whenevern < and t0


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0.9 Claim. 1) Let K be an a.e.c. and M K be of cardinality 1,1(LS(K))recalling we naturally assume |K| LS(K) as usual.

Then there is a such that is proper for linear orders and:

(a) () K ,

() || = LS(K) + |K|

(b) for any linear order I the model EM(I, ) has cardinality |()| + |I| andwe have EM(K)(I, ) K

(c) for any linear orders I J we have EM(K)(I, ) K EM(K)(J, )

(d) for every finite linear order I, the model EM(K)(I, ) can be K-embeddedinto M.

2) If we allow LS(K) < |K| and there is M Kof cardinality 1,1(LS(K) + |K|),then there is or

LS(K)+|()|

[K] such thatEM(I, ) has cardinality LS(K) for I

finite. Hence E has 2LS(K) equivalence classes where E = {(P1, P2) : P1, P2 and P

EM(I,)1 = P

EM(I,)2 for every linear order I}.

3) Actually having a model of cardinality for every < (2LS(K)+|(K)|)+ suffice(in part (2)).

Proof. Follows from the existence of a representation ofK as a PC,2-class when

= LS(K) + |(K)| in I. 88r-1.4 (3),(4),(5) and I. 88r-1.8(or see [Sh 394, 0.6]). 0.9

0.10 Remark. Note that some of the definitions and claims below will be usedonly in remarks: K

sc() from 0.14(8), in 1.7; and some only in 6,7 (and part

of 5 needed for it): lin [2] from 0.11(5) (and even less lin [()] from Definition

0.14(9)). Also the use of , ie ,

is marginal.

0.11 Definition. We define partial orders , ie and

on

or [K] (for

LS(K)) as follows:1) 1 2 if(1) (2) and EM(K)(I, 1) K EM(K)(I, 2) and EM(I, 1) =EM(1)(I, 1) EM(1)(I, 2) for any linear order I.

Again for = LS(K) we may drop the .2) For 1, 2 or [K], we say 2 is an inessential extension of 1 and write1

ie 2 if 1

2 and for every linear order I, we have (note: there may be

more function symbols in (2)!)

EM(K)(I, 1) = EM(K)(I, 2).


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6 SAHARON SHELAH

3) Let lin be the class of proper for linear order and (producing a linear orderextending the original one, i.e.) such that:

(a) () has cardinality and the two-place predicate < belongs to ()

(b) EM{


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2) If is an automorphism of the linear order I then it induces a unique automor-phism of EM(I, ) (as above with I1 = I = I2).

0.13 Remark. 1) So in 0.11(2) we allow further expansion by functions definablefrom earlier ones (composition or even definition by cases), as long as the numberis .2) Of course, in 0.12 is true for trivial K.

So we may be interested in some classes of linear orders; below 0.14(1) is used muchmore than the others and also 0.14(5),(6) are used not so few times, in particularparts (8),(9) are not used till 5.

0.14 Definition. 1) A linear order I is -wide when for every < there is a

monotonic sequence of lenth +

in I.2) A linear order I is -wider if |I| 1,1().3) I2 is -wide over I1 if I1 I2 and for every < there is a convex subset of I2disjoint to I1 which is

+-wide. We say I2 is wide over I1 if I2 is |I1|-wide overI2.4) Klin[Klin ] is the class of linear orders [of cardinality ].5) Let Kflin be the class of infinite linear order I such that every interval hascardinality |I| and is with neither first nor last elements.6) Let the two-place relation Kflin on K

flin be defined by: I Kflin J iffI, J Kflin

and I J and either I = J or J\I is a dense subset of J and for every t J\I, Ican be embedded into J {s J\I : (r I)(s


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4) If I1 Klin then for some I2 Kflin we have: |I2| = |I1| + 0 and I1 Kflin I2;

and (I0)[I0 I1 I0 Kflin I0 Kflin I2].5) If I1 is -wide and I1


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2A) If M N are -models, then M L, N iff for some F clauses (a), (b)hold together with

(c) if A [M]M, f is a function

from Rang(a) to N such that (M, a) L, (N, f(a))} then M L, N iffF = iffF satisfies clauses (a),(b) of.3) If M is a -model, = cf() and = MM we have (M, a) L, (M, b) iff tp(a, , M) = tp(a, , M)

(b) for any -model N we have N L, M iff {tp(a, , N) : a >N} =

{tp(a, , M) : a >M}.

4) Assume > =


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(g) if I Kflin then I1 B Kflin and if I1 LS(K) (e.g. the first beth fix point > LS(K), see 3.5; in themain case has cofinality 0)

(b) K categorical in and orK

or just

(b) K is weakly/strongly/pseudo solvable in and orK

witnesses it; seebelow.

1.4 Definition. 1) We say K is weakly (, )-solvable when LS(K) andthere is or [K] witnessing it, which means that

or [K] and EM(K)(I, )

is a locally superlimit member ofK for every linear order I of cardinality . We

may say (K, ) is weakly (, )-solvable and we may say witness that K is weakly(, )-solvable.If = LS(K) we may omit it, saying Kor (K, ) is weakly -solvable in .

2) K is strongly (, )-solvable when LS(K) and some or [K] witnessit which means that if I Klin then EM[K](I, ) is superlimit (for K). We usethe conventions from part (1).3) We say K is pseudo (, )-solvable when LS(K) and there is or [K]witnessing it which means that for some -a.e.c. K with no K -maximal member,we have M K iff M = EM(K)(I, ) for some I K

lin iff M

= EM(K)(I, )

for every I Klin . We use the conventions from part (1).4) Let (, )-solvable mean weakly (, )-solvable, etc., (including 1.3)

1.5 Claim. 1) In Definition 1.3, clause (b) implies clause (b). Also in Definition1.4 Kis strongly(, )-solvable implies Kis weakly(, )-solvable which impliesK is pseudo (, )-solvable. Similarly for (K, ).2) Assume or [K]; if clause (b)

of 1.3 or just I(,K) < 2, or just 2 >

I(, {EM(K)(I, ) : I Klin }) for some satisfying LS(K) <

+ < then we candeduce that

() , really (K, ) has the -non-order property, where the -non-order prop-erty means that:

if I is a linear order of cardinality , t1

, t2

I form a -system pair (seebelow) and i(x) : i < lists the ()-terms (with the sequence x of vari-ables being xi : i < ) and at : t I is the indiscernible sequence gen-erating EM(I, ) (i.e. as usual at : t I is the skeleton of EM(I, ),so generating it, see Definition 0.8) then for some J I there is an auto-morphism of EM(K)(J, ) which exchanges i(at1

i: i < ) : i < and


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i(at2i : i < ) : i < .where

t1, t2 I is a -system pair when for some J I there are t J

for \{1, 2} such that t

: < is an indiscernible sequence forquantifier free formulas in the linear order J.

Proof. 1) The first sentence holds by Claim 0.9(1) and Definition 0.8 (and Claim0.6). The second and third sentences follows by 0.6.2) Otherwise we get a contradiction by [Sh 300, Ch.III] or better [Sh:e, III]. 1.4

1.6 Definition. 1) IfM is a class of linear orders and or [K] then we let

K[M

, ] = {EM(K)(I, ) : I M

}.2) Let K

u()-lin be the class of linear orders I of cardinality such that for some

scattered1 linear order J and proper for Klin such that < belongs to , || we have I is embeddable into EM{ I(, K[M, ]),

for M = Ksc()-lin or M

= Ku()-lin and restrict the conclusion () to I

Ksc()-lin. A gain is that, if > , every I Ksc()-lin is -wide so later K = K,

and being solvable is a weaker demand. But it is less natural. Anyhow we presentlydo not deal with this.1A) Note that K

sc()lin K

u()lin .

2) An aim of 1.8 below is to show that: by changing instead of assumingI1 I2 (I2 is -wide over I1) it suffices to assume I1 I2 (I2 is -wide).

1.8 Claim. For every 1 or [K] there is 2 such that

(a) 2 or [K] and if 1 witnesses K is weakly/strongly/pseudo (, )-solvablethen so does 2

(b) 1 2 and |2 | = |1 | + 0

(c) for any I2 Klin there are I1 and h such that:

() I1 Klin and even I1 Kflin, see 0.14(5)

1i.e. one into which the rational order cannot be embedded


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() h is an embedding of I2 into I1

() there is an isomorphismf fromEM(1)(I2, 2) onto EM(I1, 1) suchthat f(at) = ah(t) for t I2

() if J1 = I1 Rang(h) and we let E = {(t1, t2) : t1, t2 I1\J1 and(s J1)(s < t1 s < t2)} then: E is an equivalence relation andeach equivalence class has |I2| members andJ1 Kflin I1, see 0.14(6)

() [not used] if = J2 I2, J1 = {t I1: for some (2)-term(x0, . . . , xn1) and some t0, . . . , tn1 J2 we have f1(at) =EM(I2,2)(at0 , . . . , atn1)} and J

1 Rang(h)\J1 and t J

1 then

{s t/E : f1(as) belongs to the Skolem hull of {f1(ar) : r J1}in EM(I2, )} has cardinality |J1| and J

1 and its inverse can be

embedded into it; in fact, I1 and its inverse are embeddable into anyinterval of I2.

Remark. 1) We can express it by , see 0.11(4). So for some proper for linearorders such that is countable, the two-place predicate < belongs to and aboveEM{ 0 andt1,n n,g(2) > n, 1,n = 2,n < 0 andt2,n n and 2,n > 0(e) 1 n = 2 n,g(1) > n,g(2) = n and 1,n < 0.

We identify t I1 with the pair (1, t). Now check. 1.8


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1.9 Definition. 1) Let the language L,[K] or L,,K where 0 and ispossibly , be defined like the infinitary logic L,(K), except that we deal onlywith models from K and we add for i < the atomic formula {xi : i < i} is theuniverse of a K-submodel, with obvious syntax and semantics. Of course, it isinteresting normally only for > LS(K) and recall that any formula has < freevariables.2) For M a K-model and N K let M L, [K] N means that M N and if

(x, y) is a formula from L,[K] and N |= (x)(x, b) where b g(y)M, then forsome a g(x)M we have N |= [a, b].

1.10 Fact: 1) If > LS(K) and M, N are K-models and N K and M L, [K]N, then M K N and M K.2) The relation L, [K] can also be defined as usual: M L, [K] N iff M, N

K, M N and for every (x) L,[K] and a g(x)

M we have M |= [a] iffN |= [a].3) If N K and M is a K -model satisfying M L, N and > LS(K) thenM K, M K N and M L,[K] N.4) If N K, M a K-model and M L, N where > LS(K) then M K andM L,[K] N.5) The parallel of 0.17(2) holds for L,[K], i.e. there is F satisfying clauses (a),(b)there and

(d) if f F then

() M Dom(f) K M

() N Rang(f) K M.

6) Also the parallel of 0.17(2A) holds for L,[K].7) The parallel of 0.17(4) holds for L,[K].

Proof. Part (1) is straight (knowing I1 or [Sh 88, 1]). Part (2) is proved as in theTarski-Vaught criterion and parts (5),(6),(7) are proved as in 0.17.

Toward proving parts (3),(4) we first assume just

1 M, N are K-models, N K and M L, N and > LS(K) and [LS(K), )

and we define:

(a) I = I = {(f, M, N) : M M and N N and f is an isomorphism


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from M onto N and M and letting a list M we have(M, a) L, (N, f(a))}

(b) for t I let t = (ft, Mt, Nt)

(c) for = 0, 1, 2 we define the two-place relation I on I:let s I t hold iff

() = 0 and Ms Mt Ns Nt

() = 1 and Ms K Mt Ns K Nt

() = 2 and fs ft

(d) I1 = I1 := {t I0 : Nt K N} and let

I1

=I I1 for = 0, 1, 2.

Now easily

()0 () I = is partially ordered by I for = 0, 1, 2

() s 1I t s 0I t

() s 2I t s 0I t.

[Why? Straight, e.g. I = by 0.17(1).]

()1 if t I1 then Mt K and Nt K.

[Why? As t I1 by the definition of I we have Nt K (because Nt K N) and

Mt K as ft is an isomorphism from Mt onto Nt.]

()2 if s I, A [M] and B [N] , M L, and thedefinition of I.]

()3 if s 2I1 t then s 1I t, i.e. Ms K Mt and Ns K Nt.

[Why? As s, t I1 we know that Ns K N and Nt K N and as s 2I t we have

fs ft hence Ns Nt. By axiom V of a.e.c. it follows that Ns K Nt. NowMs K Mt as ft is an isomorphism from Mt onto Nt mapping Ms onto Ns (as itextends fs by the definition of 2I) and K is preserved by any isomorphism. Soby the definition of 1I we are done.]

()4 if s I then for some t I1 we have s 2I t (hence I1 = ).


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[Why? First choose N K N of cardinality such that Ns N, (possibly bythe basic properties of a.e.c. (see I1 or Chapter II.B)). Second we can find t Isuch that Nt = N fs ft by the characterization of L, as in ()2. So s

2I t

by the definition of

2

I and Nt = N

K

N hence t I1 as required. Lastly, I1 = as by ()0() we know that I = and apply what we prove.]

()5 if s 0I1 t then Ns K Nt.

[Why? As in the proof of ()3 by AxV of a.e.c. we have Ns K Nt (not the parton the Ms!)]

()6 if s I1, A [M] and B [M] then for some t we have s 2I1 t andA Mt, B Nt.

[Why? By ()2 there is t1 such that s 2I t1, A Mt1 and B Nt1 . By ()4 there

is t I1 such that t1 2I t hence by ()0() we have s 2I t. As s, t I1 this impliess 2I1 t.]

Note that it is unreasonable to have (I1, 2I1)-directed but

()7 (I1, 1I1

) is directed.

[Why? Let s1, s2 I1. We now choose tn by induction on n < such that

(a) tn I1

(b) Mtn includes {Mtk : k < n} Ms1 Ms2 if n 2

(c) Ntn

includes {Ntk

: k < n} Ns1

Ns2

if n 2

(d) t0 = s1

(e) t1 = s2

(f) if n = m + 1 2 then tm 0I1 tn

(g) if n = m + 2 then tm 2I tn hence tm 2I1

tn.

For n = 0, 1 this is trivial. For n = m + 2 2, apply ()6 with tm, {Mtk : k m + 1}, {Ntk : k m + 1} here standing for s,A,B there getting tn, so we gettn I1 in particular tm

2I1

tn, so clause (a) is satisfied by tn. By the choice of tnand as s1 = t0, s2 = t1, clauses (b) + (c) hold for tn. By the choice oftn, obviously

also clause (g). Now why does clause (f) hold (i.e. tm+1 0I tn)? It follows from

clauses (a),(b),(c), so tn is as required. Hence we have carried the induction. LetN = {Ntn : 2 n < }, so clearly by ()5 and clause (f) we have Ntn K Ntn+1for n 1, and clearly Mtn Mtn+1 for n 1. Let M

= {Mtn : 2 n < }.Note that by ()3 and clause (g) we have Mtn K Mtn+2 , so Mtn+2 : n < is -increasing, and for = 0, 1 the sequence Mt2n+ : n < is K-increasing


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with union M, hence by the basic properties of a.e.c. we have M2n+ K M.So Ms1 = Mt0 K M

, Ms2 = Mt1 K M. Now Ms1 , Ms2 Mt2 K M

hence Ms1 , Ms2 K Mt2 , Recall that Ns1 = Nt0 K Nt2 was proved above andNs2

= Nt1

K Nt2

was also proved above so t2

is a common 1I

-upper bound ofs1, s2 as required.]

()8 if s 0I1 t then s 1I1

t.

[Why? By ()7 there is t1 I1 which is a common 1I1-upper bound of s, t. So

Ms Mt (as s 0I1 t) and Ms K Mt1 (as s 1I1

t1) and Mt K Mt1 (as t 1I1

t1).Together by axiom V of a.e.c. we get Ms K Mt and by ()5 we have Ns K Nt.Together s 1I1 t as required.]

()9 Ms : s (I1, 1I1) is K-increasing, (I1, 1I1

) is directed and {Ms : s I1} = M.

[Why? The first phrase by the definition of 1I1 in clause (c)() of, the secondby ()7 and the third by ()6 + ()4.]

By the basic properties of a.e.c. (see I. 88r-1.6 ) we deduce

(a) M K

(b) t I1 Mt K M.

Now we strengthen the assumption 1 to

2 the demands in 1 and M L,[K] N.

We note

1 (a) if a M, || + LS(K) < then for some t I, ft(a) = a

(b) if M M and M then (idM , M, M) I

(c) if M1 N1 N and M1 M and N1 then for some t Iwe have Nt = N1 and idM1 ft.

[Why? Clause (a) is a special case of clause (b) and clause (b) is a special caseof clause (c). Lastly, clause (c) follows from the assumption M L,[K] N and0.17(2A),(2B).]

We next shall prove2 M K N.

By I. 88r-1.6 and ()9 above for proving 2 it suffices to prove:

3 if s I1 then Ms K N.


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[Why 3 holds? As M N there is N K N of cardinality such thatMs Ns N. By 1(c) there is t I such that Nt = N and idMs ft. AsN K N it follows that t I1. So by 1 (b) applied to s and to t we candeduce M

sK M and M

tK M. But as id

Ms f

tit follows that M

s M

thence by AxV of a.e.c. we know that Ms K Mt. But as t I clearly ft is anisomorphism from Nt onto Mt hence f

1t (Ms) K Nt, and as idMs ft this means

that Ms = f1t (Ms) K Nt. Recalling Nt K N and K is transitive it follows that

Ms K N as required.]Let us check parts (3) and (4) of the Fact. Having proved 1 (a), clearly in

part (4) of the fact the first conclusion there, M K, holds. The second conclusion,M L,[K] N holds by

4 if(x) L,[K] and |g(x)| + LS(K) < and t I and a g(x)(Mt)then M |= [a] N |= [ft(a)].

[Why? Prove by induction on the depth of for all simultaneously. For = 0,first for the usual atomic formulas this should be clear. Second, by ()4 there ist1 such that t

2I t1 I1 hence by 3+ clause (d) of+ clause (b) of we

have Mt1 K N Nt1 K N Mt1 K M respectively. So if u g(x) thenM Rang(a u) K M M Rang(a u) K Mt1 N Rang(f(a) u) KNt1 N Rang(f(a) u) K N. So we have finished the case of atomic formulas,i.e. = 0. For (x) = (y)(x, y) use ()2, the other cases are obvious.]

So part (4) holds. As for part (3), the first statement, M K holds by part (4),the second statement, M K N, holds by 2 and the third statement, M L,[K]N follows by 1(b) + 4. As we have already noted parts (1),(2),(5),(6) and part

(7) is proved as 4 is proved, we are done. 1.10

1.11 Claim. For a limit cardinal > LS(K):1) M L,[K] N provided that

(a) if < and (LS(K), ) then M L,[K] N

(b) for every < for some (, ) we have: ifa, b M and(M, a) L,[K](M, b) then (M, a) L,1 [K] (M, b) for every 1 [, ).

1A) M L,[K] N provided that

(a) if LS(K) < < then M L,[K] N

(b) as in part (1).

2) In parts (1) and (1A) we can conclude

(b)+ for every < for some (, ) we have: ifa, b M and(M, a) L,[K](M, b) then (M, a) L,[K] (M, b).


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3) If cf() = 0 then M = N when

(a) if < and (LS(K), ) then M L,[K] N

(b)as in part (1), i.e., for every

(LS(K), )for some

(, )we have: ifa M and b N and (M, a) L,[K] (N, b) then (M, a) L,1 [K] (N, b)

for every1 (, )

(c) M, N have cardinality .

Proof. 1) By 1.10(3) it suffices to prove M L, N, for this it suffices to applythe criterion from 0.17(2A).

Let F be the set of functions f such that:

() Dom(f) M has cardinality <

() Rang(f) N

() if a lists Dom(f) then for every (g(a), ) we have tpL,[K](a, , M) =tpL,[K](f(a), , N).

1A) Similarly.2) Similarly to part (1) using 1.10(4) and 0.17(2) instead 1.10(3),0.17(2A).3) Recall 0.17(1). 1.11

1.12 Claim. 1) Assume 1.3(a) + (b), i.e. K is categorical in > LS(K). If = LS(K) then for every M K N from K we have M L,[K] N

(and there are such M


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20 SAHARON SHELAH

that the other fails, say N |= [b, a] and M |= (y)(y, a). We choose by inductionon i < + a model Mi K, K-increasing continuous, and for each i in additionwe choose an isomorphism fi from M onto Mi and if i = j + 1 we shall choose anisomorphism g

jfrom N onto M

j+1extending f

j. For i = 0, let M

0= M, for i limit

let Mi =

j


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Clause (a): We prove that for (x) L,[K] we have

()(x) ifI1 I2 are -wide linear orders of cardinality and a g(x)(EM(K)(I, ))

then EM(K)(I1, ) |= [a] EM(K)(I2, ) |= [a].

This easily suffices as for any I Klin, the model EM(K)(I, ) is the direct limitofEM(I, ) : I I has cardinality , which is K-increasing and

+-directedand as we have:

M1 L,[K] M2 when:

(a) I is a -directed partial order

(b) M = Mt : t I

(c) s


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22 SAHARON SHELAH

i < +}}. So J1 J2, both linear orders have cardinality and are -wide aswitnessed by d : < + for both hence the conclusion of 1.12 holds, i.e.EM(J2, ) L,[K] EM(J1, ). Also I

I1 J2 and recall that a g(x)(EM(I

I1, )) hence a g(x)

(EM(J2, )). However, EM(K)(J1, ) |= [a], see above,hence by the last two sentences EM(K)(J2, ) |= [a].

So there is b g(y)(EM(K)(J2, )) such that EM(K)(J2, ) |= [b, a]. Let

J J2 be of cardinality such that b g(y)(EM(K)(J

, )) and I I1 J

recalling I [c0, ci)I2 = for i < +. Now let u + be such that J\I1 =

{d : u} so |u| < +. Let J3 = J2 {t : t J2 I1 or t = d > sup(u)or t = d u}; as cf( +) = + > |u|, clearly sup(u) < + hence|J3| = and J3 is -wide. So by the conclusion of 1.12 (or by the inductionhypothesis) also EM(K)(J3, ) |= [b

, a]. Let w = { < + : u or > sup(u) ( sup(u) < +)}, so otp(w) = +.

Let J4

= (J3

I1

) {d

: w}, so J4

is -wide as witnessed by I1

\ {[c0

, ci) :

i < +} or by {d : w} recalling () above and J4 J3 and J J4 hencea, b >(EM(J4, )) hence by the induction hypothesis EM(K)(J4, ) |= [b

, a].

Let J5 = J4 {ci : i < +}\{d : w} equivalently J5 = (J3 I1) {c :

< +} = (I1\ {[c0, ci)I1 : i < +}) {ci : i < +} so J5 I1 and let

h : J4 J5 be such that h(d) = cotp(w) for w and h(t) = t for others,i.e. for t J3 I1. So h is an isomorphism from J4 onto J5. Recalling 0.12 leth be the isomorphism from EM(J4, ) onto EM(J5, ) which h induces, so clearly

h(a) = a. Hence for some b we have b = h(b) g(y)(EM(K)(J5, )) andEM(K)(J5, ) |= [b

, a]. Note that by the choice of ci : i < +, (see ()above), we know that J5 is -wide. Also J5 I1 so by the induction hypothesis

applied to (y, x), J5, I1 we have EM(K)(I1, ) |= [b, a] hence by the definitionof satisfaction EM(K)(I1, ) |= [a], so we have finished proving the implication hence clause (a).

Clause (b): Without loss of generality for some linear order I we have I1 I, I2 Iand EM(I, ) EM(I, ) for = 1, 2 and use clause (a) twice.

Clause (c): Easy by now, e.g. using a linear order I extending I1, I2 which has anautomorphism h such that h(t1) = t

2 for < (). 1.14

1.15 Definition. Fixing or

K .1) For LS(K) let K , [let K ] [let K

, ] be the family ofM K isomorphic to

some EM(K)(I, ) where I is a linear order of cardinality [which is -wide][which

Kflin ]. More accurately we should write K,, K

,, K

,,; similarly below.

2) Let K is the class {K : a cardinal LS(K)}, similarly K,, K, K

,

etc.


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3) Let K = K = (K, K K).

4) Let K = K, be (K

,, K K

,).

1.16 Claim. 1) K is categorical in if LS(K) < , cf() = 0 and theconclusion of 1.12(2) hence of 1.14 holds for = (and ), e.g. K is pseudosolvable in as witnessed by and = LS(K) then K = K .

Proof. 1) By 1.14 and 0.17(1).2) Read the definitions.3) Recall 0.15(2). 1.16

1.17 Remark. 1) We will be specially interested in 1.16 in the case (, ) is a K-candidate (see Definition V.1.3) and = .2) Note that K in general is not a -a.e.c.3) If we strengthen 1.18(2) below, replacing (, ) by (, +) then categoricity ofK and in fact Claim 1.19(4) follows immediately from (or as in) Claim 1.16(1).

For the rest of this section we assume that the triple (,, ) is a pseudo K-candidate (see Definition 1.3) and rather than = we assume just the conclusionof 1.12, that is:

1.18 Hypothesis. 1) The pair (, ) is a pseudo K-candidate and witnesses this,so || LS(K) < = < and orK is as in Definition 1.4 so I K

lin

EM(K)(I, ) Kpl .

2) For every (LS(K), ) the conclusion of 1.12(2) holds hence also of 1.14 (if =


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24 SAHARON SHELAH

1.20 Remark. 1) What is the difference between say 1.19(3) and clause (a) of1.14? Here there is no connection between the additional ()-structures expandingM1, M2.2) Note that has the -non-order property (see 1.5(2)(*)) when LS(K), + < using 1.19(4).3) Concerning 1.19(2), note that ifM1 it is easy to deduce this from 1.18(2),i.e, 1.12(2). But the whole point in this stage is to deduce something on cardinals< .4) Note that the proof of 1.19(2) gives:

assume LS(K) and () = Min{(2)+, (2LS(K) + )} where on thefunction (), see II.A. 300a-1.2.3 ,II.A. 300a-1.2 ,if() then for some () < () we have:

if M1 K M2 are from K and M1 () then M1 L,+ [K]

M2.

5) Similarly for 1.19(3) so we can weaken the demand M K16) We use has countable cofinality, i.e. cf() = 0 in the proof of part (4) of1.19, but not in the proof of the other parts.7) Recall that for notational simplicity we assume LS(K) |K| hence ||.8) Note that for 1.19(2),(3) we can omit from Hypothesis 1.18.9) Note that we shall use not only 1.19 but also its proof.

Proof of 1.19. 1) The first phrase holds by part (2) noting that < if < as < =. The second phrase holds by 1.11 as its assumption holds by parts (1)and (3).2) We prove by induction on the ordinal that:

() ifM1 K M2 are from K and the formula (x) L,+ [K] has depth

(so necessarily g(x) < +) and a g(x)(M1) then M1 |= [a] M2 |=[a].

As in 1.12, the non-trivial case is to assume (x) = (y)(y, x) where a g(x)(M1)and M2 |= [a] and we shall prove M1 |= [a], so necessarily g(x) + g(y) <

+

and we can chooseb

g(y)

(M2) such that M2 |= [b, a]. For = 1, 2 as M K

there is an isomorphism f from EM(K)(I, ) onto M for some linear order I ofcardinality .

So we can find J I of cardinality for = 1, 2 such that a M1 where

M1 = f1(EM(K)(J1, )), and ab M2 where M

2 = f2(EM(K)(J2, )) and

without loss of generality M1 = M2 M1. By 1.18(1), i.e. 0.9(1), clause (c)


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clearly M K M and so by AxV of a.e.c. (see Definition III. 600-0.2

), we have M1 K M2 . First assume 2

LS(K); in fact it is not a real lossto assume this. By renaming without loss of generality there is a transitive set B

(in the set theoretic sense) of cardinality such that the following objects belongto it:

(a) J1, J2

(b) (i.e. and (EM(n, ), a)


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26 SAHARON SHELAH

By easy absoluteness (for clauses (a)1, (a)2 we use I. 88r-1.6 , I. 88r-1.7 and 3):

(a)1 ifC |= M K then MC K

(a)2 ifC |= M K N then MC K NC

(b)1 ifC |= I is a linear order then IC = I[C] is a linear order

(b)2 ifC |= I J as linear orders then IC JC

(c) similarly for -models

(d)1 ifC |= M = EM(I, ) then there is a canonical isomorphism fCIfrom EM(IC, ) onto MC (hence it is also an isomorphism fromEM(K)(I

C, ) onto MC (K))

(d)2 ifC |= I J as linear orders then fCJ extends fCI .

Now clearly JC

= J and IC

is a linear order of cardinality extending J for = 1, 2. Let M = (M

)

C for = 1, 2.So recalling clause (c) of we have: MC1 , M

C2 K

, M

C1 K M

C2 , M

K

MC , M1 K M

2 and f

C0 , f

CI

are isomorphisms from EM(K)(IC

, ) onto MC

, in

fact, fCI is the identity on EM(K)(JC , ) = EM(K)(J, ) and f

C maps it onto

M for = 1, 2.Now M2 |= [a, b], (why? assumed above) hence MC2 |= [a, b]

(why? By 1.14, clause (b) or (c) and the situation recalling 1.18(2), of course notingthat I2, I

C2 are of cardinality = 1,1() hence are

+-wide), hence MC2 |= [a](by definition of satisfaction), hence MC1 |= [a] (why? as M

C1 , M

C2 K

hence

MC

1 L,+ [K] MC

2 by and 1.18(2) and recalling 1.12(2)) hence M1 |= [a] (why?by clause (b) of 1.14 recalling 1.18(2)) as required in 1.19(2).]So we are done except for a small debt: the case < 2LS(K) and fC is an isomorphismfrom EM(K)(I

C , ).

In this case choose two sets B1, B2 such that |B1| = , |B2| = 2LS(K), B1 B2

and concerning the demands in above the objects from (a),(b),(d) and K belongto B1, the objects from (c) belong to B2.

Again, without loss of generality B1, B2 are transitive sets and B1, B2 serve asindividual constants ofB+ as well as each member of B1. Now concerning C wedemand that it is elementarily equivalent to B+; omit {x B1x = b : b B1} andfor some B+

1 B+ of cardinality we have B+

1 C and {b : C |= b B

2} B+.

This influences just the proof of3.3) Without loss of generality M = EM(K)(I, ) and I K

lin1

. As < +

and a, b M there is I1 I of cardinality such that a, b (M1) whereM1 = EM(K)(I1, ). As (M, a) L

,+ [K](M, b) necessarily there is I2 I of

cardinality and automorphism f of M2 = EM(K)(I2, ) mapping a to b such


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that I1 I2. Why? Recalling 0.17(2), by the hence and forth argument as in thesecond part of the proof of 1.10(3).Now as in the proof of part (2) there is a linear order I3 extending I1 of cardinality1

and an automorphism g of M3

= EM(K)(I

3, ) mapping a to b. Without loss

of generality for some linear order I4 we have I I4 and I3 I4.Let M4 = EM(K)(I4, ), now M L

,+ [K]M4 by part (2), M3 L

,+ [K]M4

by part (3) and (M3, a) L,+

[K] (M3, b) by using the automorphism g of M3 so

together we are done.4) So let M, N K (in fact, hence K

recalling K

= K

by 1.16(3) but not

used). By parts (1),(3) the assumptions of 1.11(3) holds with here standing for there, hence its conclusion, i.e. M = N.

1.19

Note: here the types below are sets of formulas.1.21 Definition. Assume M K, I M and L,L1,L2 are languages in thevocabulary K.1) We say that I is (L, , < )-convergent in M, if: |I| and for every b >M,for some J I of cardinality < for some2 p we have:

() for every c I\J, the L-type of cb in M is p.

2) Let AvL,,


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28 SAHARON SHELAH

1.22 Remark. 1) See definition of Sav(M) in 1.34(2) below.2) An alternative for clause (c) of 1.21(3) is:

(c) the set {AvL,, + LS(K) and I is (L, , )-convergent. Then the type p = AvL,,(I, M) belongsto Sfr

L(M), i.e., it is complete, recalling Definition 0.4 (no demand that it is

realized in some N, M K N!).2) Also I is (L, , )-based on some set of cardinality , even on J, for anyJ I of cardinality .

Proof. 1) By the definition.2) By the definitions: if b

+>M, = (x, y) L and g(b) = g(y), g(x) = ,then by the convergence

(x, b) p for all but < members a of I, M |= [a, b]

for all but < members of J, M |= [a, b].

So only tpL(b, J, M) matters hence the non-splitting required in clause (c) ofDefinition 1.21(3). 1.23

As in II.A. 300a-1.7 , we deduce non-splitting over a small set fromnon-order.

1.24 Claim. Assume M = EM(K)(I, ), + LS(K) < and 1,1() |I|where = (22

)+ or I is well ordered and = (2)+. If M L, [K] N then for

every a N there is B M of cardinality < such that tpL,+ [K]

(a ,M,N)

does not (L,+ [K],L,+[K])-split over B.

Proof. Let x = xi : i < g(a).We try to choose B, , a, b, c, (x, y) L,+ [K] by induction on <

such that

(a) B = {a : < }

(b) b, c M and < +


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(c) (x, y) L,+[K] such that g(y) =

(d) N |= [a, b] [a, c]

(e) a g(a)M realizes {(x, b) (x, c) : < } in M

(f) M |= [a, b] [a, c] for .

If we are stuck at () < then we cannot choose , b, c, (x, y) clauses(b),(c),(d), because then a as required in clauses (e),(f) exists because M L, [K]N. Hence B := {a : < ()} is as required. So assume that we have carriedthe induction. As < + < = cf() without loss of generality = < + forevery < .

Let 1 = (2)+.

Now by 1.25(5) below when I is not well ordered and by 1.25(4) below whenI is well ordered (and part (1) of 1.25(1), recalling I is +-wide as < and

1,1() |I|) clearly for some S of order type 1, the sequence abc : S is (L,+ [K], +, )-convergent and (L,+[K], < )-indiscernible in M

hence without loss of generality S = . But as 1 > + this contradicts

(e) + (f) of (if we use 1 = +, we can use a further conclusion of 1.25(1) stated

in 1.25(2), i.e., abc : S is a (L,[K], < )-indiscernible set not just asequence, contradiction to (e) + (f) of). 1.24

1.25 Claim. Assume M = EM(K)(I, ), I is +-wide, < and LS(K) +

< .1) Assume thatL = L,+ [K] and a = i(. . . , at(,i,), . . . ) . Assume further that letting t =t(,i,) : i < , < ni, the sequence t : < is indiscernible in I forquantifier free formulas (i.e. the truth values of t(1, i1, 1) < t(2, i2, 2) dependsonly on i1, 1, i2, 2 and the truth value of 1 < 2, 1 = 2, 1 > 2). Thena : < is (L, , )-convergent in the model M.2) In part (1), even dropping the assumption cf() > , moreover, the sequencea : < is (L,

+, )-convergent and (L, < )-indiscernible in M.3) In part (1) and in part (2), letting J0 = {t(0, i , ) : t(0, i , ) = t(1, i , ) andi < , < ni} assume J0 J I, J is

+-wide (e.g. J = {t(,i,) : < +, i 2.

Note

4 t realizes Avqf({t

: S}, I2) in the linear order I+2 .

Without loss of generality I+2 I3 = I2, so we can find a linear order I4 of cardinality such that I+2 I4 I3 I4. As I3 is

+-wide over I2 (see the assumptionand Definition 0.14(6)+(3)), there is a convex subset I3 of I3 disjoint to I2 whichcontains a monotonic sequence s : < +. Without loss of generality thereare elements s ( [

+, +) in I4 such that s : < + is monotonic

(in I4), and its convex hull is disjoint to I2. Let I3 = I2 {s : < +} andI3 = I2 {s : <

+}.Now we use 1.14 several times. First, EM(K)(I2, ) L

,+ [K]EM(K)(I

+2 , ) L

,+ [K]

EM(K)(I4, ) as I2 I+2 I4 are

+-wide, hence by 4 the sequence d realizesq := Av({(t

) : < : < +}, M2) = Av({a : < +}, M2) =Av(I, M2) in M

+2 and also in EM(K)(I4, ). Second, as |I2| < , I2 I

3 I4

and |I3 | = |I4| = , by 1.19(1) we have EM(K)(I3 , ) L,[K] EM(K)(I4, ) so

some d (EM(K)(I3 , )) realizes the type q in EM(K)(I

3 , ). Let w1

+

be of cardinality such that d belongs to EM(K)(I2 {s : w1}, ).Choose w2 + of order type + including w1, so EM((K)(I2 {s :

w2}, ) L,+ [K] EM(K)(I3 , ) and d belongs to the former hence realizes q in

it. But there is an isomorphism h from I2 {s : w2} onto I3 over I2, hence it

induces an isomorphism h from EM(K)(I2 {s : w2}, ) onto EM(K)(I3 , )

so h (d) realizes q in the latter. But I3 I3 are both +-wide hence by 1.14 the

sequence h(d) realizes q in M3 = EM(K)(I3, ) as required.


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36 SAHARON SHELAH

Clause (b):By part (2) we can find appropriate I and then apply clause (a). 1.26

1.27 Remark. 1) In fact in 1.24, we can choose B of cardinality , hence similarlyin the proof of 1.26(1).2) Also using solvability to get well ordered I we can prove : if A M =EM(K)(, ) and |A| < then the set of L,+ [K]-types realized in M over

A is (|A| + 2).

1.28 Claim. 1) If M K and LS(K) and = 1,1() , then fora, b M the following are equivalent: (the difference is using or )

(a) a, b realize the same L,[K]-type in M

(b) a, b realize the same L,[K]-type in M.

2) For M, , , as above, the number of L,[K]-types of a M where M =EM(K)(I, ), |I| is 2

.

Remark. Part (1) improves 1.19(3).

Proof. 1) Clearly (b) (a), so assume clause (a) holds. As M K withoutloss of generality there is a -wide linear order I such that M = EM(K)(I, );

hence for some J I, |J| = we have a, b (EM(K)(J, )). So for every (M1) the sequences ab, ac realize the sameL,[K]-type inM2. Then there are I3, M3 andf such thatI2 Kflin I3 K

flin , M3 =

EM(K)(I3, ) and f an automorphism of M3 over M1 mapping b to c.

2) Assume that for every a

>

(M1) the sequences ab, a

b realize the sameL,[K]-type in M2 (as in part (a)) and 1,1() |I1| and < . Then for

every a >(M1), the sequences ab, ac realize the same L,[K]-type in M2.3) Assume that cf() = 0 and |I1| = and recall = > LS(K). If M1 KM2 K

then for some I3, a linear order Kflin -extending I2 the model M

2 can

be K-embedded into M3 := EM(K)(I3, ) over M1.

Remark. 1) Under mild assumptions with somewhat more work in 1.29(1),(3) wecan choose I3 = I2 (but for this has to be more careful with the linear orders).Recall that for I Klin like I2 in 1.8(c) we have <

+ I can be embedded

into I and 1.4(1)(d).

Proof. 1) There is J2 I2 of cardinality such that b, c (EM(K)(J2, )); letJ1 = I1 J2.

We define a two-place relation E on I2\J2 : sEt iff (x J2)(x


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38 SAHARON SHELAH

2 (a) U i = Ui if i u0 u2

(b) U i = {t Ui : t < ti } if i u1 and

(c) U i = {t U : t


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As in the proof of 1.19 there is a (B+)-model C, such that

(a) for some unbounded S 2

() C is a first order elementarily equivalent to B+ for every S

() C omits every type omitted by B for every S.In particular this gives

() C omits the type {x = b x B : b B} so

() without loss of generality b B bC = b

(b) C is the Skolem hull of some infinite indiscernible sequenceyr : r I, where I an infinite linear order and yr QC for r I.

Without loss of generality I Kflin and I2 can be Kflin-embedded into I say bythe function g such that (t I2)(s1, s2 I)[s1


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40 SAHARON SHELAH

hence by the omitting type demand in (a)():

3 for t IC2 \J2 for some i < i() we have (s J2)[s


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Why its assumptions holds? The main point is to check that for every a m>(M1)the sequences a(bn+1 m), ab

m realize the same L,m [K]-type in M3,m. Now

a(bm+1 m), abm realize the same L,n[K]-type in M3,m by the induction

hypothesis. Also the sequencesbn+1

m,

bm+1

m satisfy for any a

m

(M1)the sequences a(bn+1 m), a(bm+1 m) realize the same L,m[K]-type inM3,m because the L,m [K]-type which a(bn+1 m) realizes in M3,m is the sameas the L,m[K]-type it realizes in M2 = M3,0 which (by the choice of bn+1) isequal to the L,m [K]-type which a(cn+1 m) realizes in M

2 which is the same

as the L,m [K]-type which a(cm+1 m) realizes in M2 which is equal to the

L,m [K]-type which a(bm+1 m) realizes in M3,m.By the last two sentences for every a m>(M1) the sequences a(bn+1

m), abm realizes the same L,m[K]-type in M3,m, so indeed the assumptionsof part (1) holds for the case we are trying to apply it, see above.

So we get the conclusion of part (1), i.e. we get I3,n, fn here standing for I3, f

there so I3,m


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42 SAHARON SHELAH

there. So there is a pair (I3, f2) such that I2


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44 SAHARON SHELAH

Let M = EM(K)(I, ) for = 0, 1, 2, 3 and let c2 = c and c1 = EM(I1,)(at10 , . . . , at1n1).

Let < be large enough such that tpL,+ [K]

(c, M0, M) for = 1, 2 be distinct

(exists by 1.29(1) because its conclusion fails by the toward contradiction). We

easily get contradiction to the non-order property (see () of 1.5(2)).Note that if in addition I1, : is


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identity on Rang(an) EM(K)(I0, ) hence f is the identity on

n

(Rang(an)

EM(K)(I0, ) = EM(K)(I0, ) so f is as required. 1.32

1.33 Exercise: 1) Assume K = (K, K) satisfies axioms I,II (and 0, presentedbelow) and amalgmation. Then tp(a,M,N) for M K N and a N and SK(M)are well defined and has the basic properties of types from III1.2) If in addition K satisfies AxIII

below and K is stable (i.e. |SK(M)| for M K) then every M K has a K-universal extension N which meansM K N and (N

)(M K N (f)[f is a K -embedding of N

into N overM]).3) AxIII (see III. 600-0.2 ) implies AxIII

where:

Ax0: K is a class of K

-models, K

a two place relation of K, both preservedunder isomorphisms

AxI: if M K N then M N (are (K)-models of cardinality

AxII: K is a partial order (so M K M for M K)

AxIII: In following game the COM player has a winning strategy. A play last moves, they construct a K -increasing continuous sequence M : . In the-th move M is chosen, by INC if is even by COM is is odd. Now Com winsas long as INC has legal moves.

AxIV: For each M K, in the following game, INC has no winning strategy:

a play lasts + 1 moves, in the -th move f, M, N are chosen such that fis a K-embedding of M into N, both are K -increasing continuous, f is -increasing continuous, M0 = M and in the -th move, M is chosen by INC, andthe pair is chosen by the player INC if is even and by the player COM if is odd.The player COM wins if INC has always a legal move (the player COM always has:he can choose N = M)

1.34 Definition. 1) Let


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46 SAHARON SHELAH

naturally.3) Let K = (K

, K K

,

K

), see 1.35 below but if (K, K K) is a -a.e.c.

then we omit K

.

1.35 Remark. 1) Note that the relation


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[Why? We can find M3i : i which is K

-increasing continuous and M30 = M20

and M3i < M

3i+1. Now apply the restricted version (i.e., with the assumption )

twice.]By induction on i we choose (fi, N1i , N

2i ) such that

(a) N1i , N2i belongs to K

(b) fi is an isomorphism from N1i onto N

2i

(c) N1i , N2i , fi are increasing continuous with i

(d) for i = 0, N1i = M1i , fi = f and N

2i is f(M

1i ) = M

2i

(e) if i > 0 is a limit ordinal then N1i = M1i and N

2i = M

2i

(f) when i = + 2n < we have

() N1+2n+1 = M1+2n+1

() N2+2n+1 K M2+2n+1

() N1+2n+2 K M1+2n+2

() N2+2n+2 = M2+2n+2.

Case 1: For i = 0 this is trivial by clause (d) and the assumption of the claim on f.

Case 2: i = + 2n + 1.Note that N2+2n = M

2+2n. (Why? If i = 0 (i.e. = 0 = n) by (d) and if

i is a limit ordinal (i.e. > 0 n = 0) by clause (e) of and if n > 0 by clause((f)() of).

Now we let N1i = N1+2n+1 := M

1+2n+1 and hence satisfying clause (f)()

of . So N1i1 = N1+2n K M

1+2n K M

1+2n+1 = N

1+2n+1 = N

1i ; and

note that N2i1 = N2+2n I there is an automorphism f of I over J which maps t to somemember of g(t)(J+).

Lastly, let N = EM(K)(J+, ), it is easy to check (see 1.4) that () holds.

If is a limit ordinal it is enough to prove for each < , a version of () with

< ; and this gives N. Now we choose N

such that < N K N

and(M, N) K,.3),4),5),6) We prove by induction on that if we let be {(x) : (x) hasquantifier depth < 1 + } then parts (3),(4),(5),(6) holds for . For all four parts,|| hence | | and it suffices to consider < +. For = 0 they aretrivial and for limit also easy (let r be the first regular and extend ||+ rtimes taking care of in stage + for each < ). So let = + 1.

We first prove (3), but we have two cases (see clause (c)) of the assumption. IfN and bi >(Ni+1) are such


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60 SAHARON SHELAH

that g(ai) = g(xi), g(bi) = g(yi) and

() b g(yi)(Mi) Mi i[bi, a] but

() Mi+1 i[bi, ai]

() for every b >(Mi+1) there is an automorphism of Mi+1 overNi mapping b into Ni+1.

If we succeed, by part (2) applied to the pair of models (

i(M). We know (by clause (d)() above)

that M+1 [b , a ] but M+1 K M hence M [a ,

b ]. Recall that fis an automorphism of M over N hence M [f(b), f(a)], but a >N so

f(a) = a hence M [b , f(a)] but M K M and a, f(b) are from M henceM [f(b), (a)]. However by clause (d)() of1 we have M [f(b), a].But as i is an odd ordinal the last two sentences contradicts clause (c) of1 appliedto i + 1.]Hence we are stuck for some i < +. Now for i = 0 clause (a) gives a permissiblevalue and for i limit take unions noting that clauses (c),(d) required nothing. Soi = j + 1; if j is even we apply the induction hypothesis to part (6) for the pair(Mi, Ni). Hence j is odd so we cannot choose j(y, x), aj, bj, recalling part (2) sothe pair (Mj, Nj) is as required thus proving (3) (for ).

Second, we prove part (4). We can now again try to choose by induction oni < + a pair (Mi, Ni) satisfying

2 (a) (M0, N0) = (M, N)

(b) (Mi, Ni) K, is K-increasing continuous

(c) if i = 2j + 1, then (Mi+1, Ni+1) is as in part (3) for with(Mi, Ni), (Mi+1, Ni+1) here standing for (M, N), (M1, N1) there

(d) if i = 2j then for some i(yi, xi) and ai (g(xi))(Ni) andbi (

g(yi))(Ni+1) we have Mi+1 i(bi, ai) butb g(yi)(Mi) Mi i[b, ai].

If we succeed, let S0 = { < + : cf() }, so by an assumption S is a stationarysubset of+, i.e. as by clause (b) we have = cf() < ; also for S0, asNi : i < is increasing with union N, and = 2 clearly a is well defined, so forsome i() < we have a >(Ni()) and without loss of generality i() = 2j()+1for some j() hence by clause (c) of2 the pair (Mi()+1, Ni()+1) is as required


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there contradiction as in the proof for part (3). Hence for some i we cannot choose(Mi, Ni).

For i = 0 let (Mi, Ni) = (M, N) so only clauses (a) + (b) of2 apply and aresatisfied. For i limit take unions. So i = j + 1. Ifj = 1 mod 2, clause (d) of

2is

relevant and we use part (3) for which holds as we have just proved it.Lastly, if j = 2 mod 2 and we are stuck then the pair (Mj, Nj) is as required.

Third, Part (5) should be clear but we elaborate.We prove by induction on that if (x) has quantifier depth < 1 +

then for every a g(x)(N1) we have M1 |= [a] N1 |= [a]. For atomic this is

obvious and for =

i LS(K)) instead =


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62 SAHARON SHELAH

By the categoricity ofK in or clause (b) of Hypothesis 2.1, K is categorical

in hence M1, M2 K are -generic. Suppose a (g(x))(M1), (x) , so

by M1 being -generic (or from the end of the proof of 2.11 applied to M2) we

have

()1 M1 |= [a] M1 [a] M1 |= [a]

and by M2 being -generic (or from the end of the proof of 2.11 applied to M2)

we have

()2 M2 |= [a] M2 [a] M2 |= [a]

and by the definition of M [a] recalling M1 K M2,

()3 if M1 [a] then M2 [a] for (x) {(x), (x)}.

So both M1 and M2 satisfy [a] if M1 satisfy it, but this applies to [a] too; sowe are done. 2.12

2.13 Claim. If K is categorical also in or just Hypothesis 2.7 apply also to ,too, (with the same ) and > > LS(K) and () below, then everyM K is L,[K]-generic and M1 K

M2 K

M1 K M2 M1 L,[K]

M2, i.e. the conclusions of 1.12, 2.12 hold where

() ifM K andA [M] then we can findN K M such thatA N K

and for every (x) L,[K

] and a

g(x)

N we have M

[a] N

[a].

Proof. We shall choose (Mi, Ni) K, by induction on i + such that not onlyMi K (see the definition of K,) but also Ni K

and this sequence of pairs

is K-increasing continuous. For i = 0 use any pair, e.g. M0 = EM(K)(, ) and

N0 = EM(K)(, ).For i limit take unions, recalling Mj , Nj are pseudo superlimit for j < i.For i = j + 1, let N+j K Mj be such that Nj N

+j K and (Mj, N

+j )

satisfies () of the claim (standing for (M, N)). Let

be as in 2.8 for . Thenby 2.11(5) with (, , ) here standing for (,,) there noting that in (c) therewe use the case


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So gtp (a, , N+j+1) is complete for every a

>(Nj), so also gtp(a, , N+) is

complete by monotonicity.

Now if a >(N+) then for some j < + we have a >(Nj), so by the above

pa := gpt

(a, , Mj+1) = gtp

(a, , N+

j+1) = gtp

(a, , N+) is complete and doesnot depend on j as long as j is large enough.

Now we prove that if a >(N+) then (x) pa N+ |= [a]; and weprove this by induction on the quantifier depth of (x); as usual the real case is

(x) = (y)(y, x). Let j < + be such that a g(x)(Nj), so pa = gtp

(a, Mj+1)so Mj+1 [a] and by the choice of (Mj+1, Nj+1) it follows that Nj+1 |= [a]hence for some b g(y)(Nj+1) we have Nj+1 |= [b, a] hence Mj+1 (b, a), hence(y, x) pba hence by the induction hypothesis N+ |= [b, a] hence N+ |= [a].

2.13

2.14 Conclusion. 1) For each LS(K) the family of > 2


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64 SAHARON SHELAH

2.15 Definition. 1) We define a two-place relation E = Eor [K] on

or [K], so

LS(K) : 1E2 iff for every linear orders I1, I2 there are linear orders J1, J2 ex-tending I1, I2 respectively such that EM(K)(J1, ), EM(K)(J2, ) are isomorphic.2) We define or

=or

[K], a two-place relation on or

[K] as in part (1) only in the

end EM(K)(J1, 1) can be K-embedded into EM(K)(J2, 2).

2.16 Claim. 1) The following conditions on 1, 2 or [K] are equivalent

(a) 1E2

(b) there are I1, I2 Klin of cardinality 1,1() such that EM(K)(I1, 1),

EM(K)(I2, ) are isomorphic

(c) there are 1, 2 satisfying

or [K] for = 1, 2 such that

1,

2

are essentially equal (see Definition 2.17 below).

2) The following conditions are equivalent

(a) 1 or 2 recall =or [K]

(b) there are I1, I2 Klin of cardinality1,1() such thatEM(K)(I1, 1) canbe K-embedded into EM(K)(I2, 2)

(c) for every I1 Klin there is I2 Klin such that EM(K)(I1, 1) can beK-embedded into EM(K)(I2, 2).

2.17 Definition. 1, 2 or [K] are essentially equal when for every linear order

I there is an isomorphism f from EM(K)(I, 1) onto EM(K)(I, 2) such thatfor any 1-term 1(x0, . . . , xn1) there is a 2-term 2(x0, . . . , xn1) such that:t0


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4) If 1 or [K] is strongly (, )-solvable and 2 exemplifies K is (, )-solvablethen 1E2.5) IfK is categorical in and > LS(K) then every or [K] is strongly(, )

-solvable.6) Assume (K, ) is pseudo (, )-solvable and 1,1() for = 1, 2. Then1E2 iff 1 or [K]2 2

or [K]1.

7) If1 or 2 and1 is strongly (, )-solvable or just pseudo (, )-solvable then1, 2 are E

or [K]-equivalent.

Proof. Easy, use 1.29(1) and its proof. 2.18

3 Categoricity for cardinals on a club

We draw here an easy conclusion from 2, getting that on a closed unboundedclass of cardinals which is 0-closed we get a constant answer to being categorical.This is, of course, considerably weaker than conjecture 0.1 but still is a progress,e.g. it shows that the categoricity spectrum is not totally chaotic.

We concentrate on the case the results of 1 holds (e.g. = ) for the s withwhich we deal. To eliminate this extra assumption we need 2. This section is notused later. Note that 3.4 is continued (and improved) in [Sh:F820] and Exercise3.8, [Sh:F782] improve 3.6; similarly 3.7.

In the claims below we concentrate on fix points of the sequence ofs.

3.1 Hypothesis. As in Hypothesis 1.2, (i.e. K is an a.e.c. with models of arbitrarilylarge cardinality).

3.2 Definition. 1) Let CatK be the class of cardinals in which K is categorical.1A) Let Sol = SolK, = Sol

1K, be the class of > LS[K] such that (K, ) is pseudo

-solvable. Let Sol2K,[Sol

3K,] be the class of > LS(K) such that (K, ) is weakly

[strongly] -solvable.2) Let mod-comK, be the class of pairs (, ) such that: > LS(K) andL,+ [K] is -model complete (on K

,, see Definition 2.3(3)(b), 2.3(5)).

3) Let CatK

be the class of CatK such that: 1,1(LS(K)) and if LS(K) and 1,1() then L,+ [K] is -model complete.3A) For or

Klet Solk,

K, be the class of SolkK, such that 1,1(LS(K))

and: if LS(K) and1,1() then the pair (L,+ [K], ) is -model complete.

Let Sol,


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66 SAHARON SHELAH

Let SolK, = Sol

1,K,. Instead k, we may write 3 + k.

4) Let C = { : = and cf() = 0}.

3.3 Exercise: 1) The conclusion of 1.12(1) equivalently 1.12(2) means that (, ) mod-comK,.2) Write down the obvious implications.

3.4 Claim. If > = > LS(K) and or [K], cf() = 0 then = LS(K) and 0 = cf().

Proof. Fix orK

, now clearly SolK, Cat

K

by their definitions.By the assumptions we can find n : n < such that = {n : n < },

LS(K) < n CatK and 1,1(n) < n+1 where

n = 1,1(n). As every M

Kn+1 is L,n [K]-generic (as Kn+1 K,n+1 and n+1 CatK

) easily

()0 if M K N are from K,n+1 then M L,n [K]N.

Let M K, for {1, 2}; so we can find a K-increasing sequence Mn : n <

such that Mn Kn , Mn K M

n+1 K M

and M = {Mn : n < }. Now

()1 Mn K

,n

.

[Why? As K is categorical in n = Mn.]

()2 if n, n < m < k and a, b (Mm) then:

(a) tpL,n [

K](a, , M

m) = t pL,n (

b, , M

m) iff tpL,n [

K](a, , M

k) =tpL

,n(b, , Mk).

(b) if tpL,n

[K](a, , Mk) = tpL,n [K]

(b, , Mk) then tpL,m [K](a, , Mk) =

tpL,m

[K](b, , Mk).


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[Why? Clause (a) by ()0, clause (b) by 1.19(3).]

()3 M1n

= M2n.

[Why? As K is categorical in n.]

We now proceed as in the proof of 1.38. Let Fn = {f: for some a1, a2 and < nwe have a (Mn+2) for = 1, 2, tpL,n+1[K](a1, , M

1n+2) = tpL,n+1[K](a2, , M

2n+1)

and f is the function which maps a1 into a2}, (actually can use = n).

By the hence and forth argument we can find fn Fn by induction on n < suchthat M1n Dom(f2n+2), M

2n Rang(f2n+2) and fn fn+1; hence {fn : n < }

is an isomorphism from M1 onto M1. 3.4

3.6 Claim. Kis categorical in when:

+ = > LS(K) and = otp(CatK C) and cf() = 0.

Proof. Fix as in the proof of 3.4. Let n : n < be increasing such that ={n : n < } and LS(K) < 0. For each n, by 2.14 we know { CatK : > nand the M K is not L,+n -generic} is not too large, i.e. is included in the

union of at most 2(n) intervals of the form [, n ]. Now we choose (n(), ) byinduction on < such that

(a) n() < and CatK

(b) if = k + 1 then n() > n(k), n() > k, CatK \+n() and the

M K is L,n() [K]-generic (hence L,+k

[K]-generic).

This is easy and then continue as in 3.5. 3.6

We have essentially proved

3.7 Theorem. In 3.5, 3.6 we can use SolK,, SolK, instead of CatK, Cat

K

.

3.8 Exercise: For Claim 1.38(2), Hypothesis 1.18 suffice.[Hint: The proof is similar to the existing one using 1.19.]


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68 SAHARON SHELAH

4 Good Frames

Here comes the main result of Chapter V: from categoricity (or solvability) as-

sumptions we derive the existence of good -frames.Our assumption is such that we can apply 1.

4.1 Hypothesis. 1)

(a) K is an a.e.c.

(b) > = > LS(K) and cf() = 0;

(c) orK

(d) K is categorical in or just

(d) (K, ) is pseudo superlimit in (this means Sol1K,; so 1.18(1) holds)

(e) also 1.18(2)(a) holds, i.e. the conclusion of 1.12(2) holds.

2) In addition we may use some of the following but then we mention them and(we add superscript when used; note that (g) (f) by 1.39)

(f) K is closed under K-increasing unions (justified by 1.38)

(g) n : n < is increasing, 0 > LS(K), = {n : n < } and theassumptions of 1.38 holds.

4.2 Observation. 1) K is categorical.2) K has amalgamation.3) (We assume (f) of 4.1(2)). K is a -a.e.c.

Proof. 1) By 1.16(1) or 1.19(4) as cf() = 0.2) By 1.30(1).3) As in 1.39, (i.e. as K

=K K, closure under unions ofK-increasing chains is

the only problematic point and it holds by (f) of 4.1(2)). 4.2

4.3 Remark. 1) Why do we not assume 4.1(1),(2) all the time? The main reason isthat for proving some of the results assuming 4.1(1),(2) we use some such resultson smaller cardinals on which we use 4.1(1) only.2) Note that it is not clear whether improvement by using 4.1(1) only will have anyaffect when (or should we say if) we succeed to have the parallel of IV 12.


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4.4 Claim. 1) Assume M0 K

M, < and a (M) for = 1, 2 and

:= 1,1(2()+) where := || + LS(K) so < . If tpL,[K](a1, M0, M1) =tpL,[K](a2, M0, M2) then tpK(a1, M0, M1) = tpK(a2, M0, M2).

2) IfM1 K

M2 then M1 L,[K] M2 for every < , and moreover M1 L,[K]M2.2A) If M0 K

M for = 1, 2 and tpK

(a1, M0, M1) = tpK

(a2, M0, M2) and

a (M0), < then tpL,[K](a1), M0, M1) = tpL,[K](a2, M0, M2).

2B) In part (1), ifM K

M for = 1, 2 thentpL,[K](a1, M , M 1) = tpL,[K](a2, M , M

2).

3) Assume that M0 K

M1 K

M2 K

M3, a (M2), < and =

1,1(|| + LS(K)) < < . Then

(a) fromtpL,[K](a, M1, M2) we can compute tpL,[K](a, M1, M2) andtpL,[K](a, M0, M3)

(b) fromtpL,[K](a, , M2) we can compute tpL,[K](a, , M2) and eventpL,[K](a, , M2)

(c) fromtpK

(a, M1, M2) we can compute tpL,[K](a, M1, M2) andtpK(a, M0, M3).

4) If M1 K

M2 and < < , I (M1), |I| > , I is (L,[K], )- conver-

gent inM1 for = 1, 2 and Av


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70 SAHARON SHELAH

2B) Should be clear by part (2).3) Clause (a):

By parts (1) + (2).

Clause (b): By 1.28(1).Clause (c): By part (2A) and the definition of tp.4) Easy, too. 4.4

4.5 Definition. Assume M0 K

M1 K

M2, < and a (M2) and p =

tpK

(a, M1, M2). We say that p does not fork over M0 (for K) when, letting

0 = || + LS(K), 1 =1,1(2(0)+), 2 = 21 , 2 =2(1) we have:

() for some N K M0 satisfying N 2 we have tpL,1 [K](a, M1, M2) doesnot split over N.

We now would like to show that there is s which fits Chapter III and Chapter IVand Ks = K

.

4.6 Observation. Assume that M0 K

M1 K

M2, a (M2), < , > 0

|| + LS(K), 1 = 1,1(2(0)+) and 2 = 2(1). Then the following conditionsare equivalent

(a) tpK

(a, M1, M2) does not fork over M0

(b) for some (+1 , 1)-convergent I (M0) of cardinality > 2 we have

tpL,1 [K](a, M1, M2) = Av 1(c) for every N K M0 of cardinality 2, if tpL,1 [K](a, M0, M2) does not

split over N then the type tpL,1 [K](a, M1, M2) does not split over N.

4.7 Remark. 1) See verification of axiom (E)(c) in the proof of Theorem 4.10.2) Note that have we used 7(1)+ instead of 1 in 4.5, 4.6, the difference wouldbe small.3) We could in clause (c) of 4.6 use for some N K M0 of cardinality < 1, tpL,1 [K]... The proof is the same.4) We can allow below M0 K M1 if M0 K2 .

Proof. (a) (b)Let 0, 1, 2 be as in Definition 4.5. By Definition 4.5 there is N K M0 of

cardinality 2 such that

()1 the type tpL,1 [K](a, M1, M2) does not split over N.


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By Claim 1.26(1) there is a (+1 , 1)-convergent set I (M0) of cardinality

+2

(convergence in M0, of course) such that tpL,1 [K](a, M0, M2) = Av


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72 SAHARON SHELAH

As above we have M0 L,1 [K] M1 by 1.19(1) hence Av


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Clause (A):By observation 4.2(3), [in the weak version using (f) from 4.9(1)].

Clause (B):

Categoricity holds by 1.16 (or 4.2(1)) and this implies there is a superlimitmodel, the non-maximality by K

holds by the choice of .

Clause (C):Observation 4.2(2) guarantee amalgamation, categoricity (ofK by 4.2(1)) im-

plies the JEP and no-maximal model holds by clause (B).

Clause (D)(a), (b):Obvious by the definition.

(D) (c) (density).

Assume M 2, so withoutloss of generality I Mi(), so by 4.6 we are done.

(E)(d) Transitivity of non-forkingWe are given M0 s M1 s M2 Ks M3 and a M3 such that tps(a, M+1, M3)does not fork over M for = 0, 1. So for = 0, 1 there is I M whichis (+1 , 1)-convergent in M+1 of cardinality

+2 such that Av


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74 SAHARON SHELAH

Av


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5 Homogeneous enough linear orders

5.1 Claim. Assume + = 1 = cf(1) < 2 = cf(2) < .1) Then there is a linear order I of cardinality such that: the following equivalencerelation E = EautI, on

I has 2 equivalence classes, where

1E2 iff there is an automorphism of I mapping 1 to 2.2) Moreover if I I has cardinality< 2 and n < then the following equivalencerelation E on nI has + |I| equivalence classes:sEt iff there is an automorphism h of I over I mapping s to t.3) Moreover, there is proper forKlin2 (i.e.

lin0

[2], see Definition 0.11(5),0.14(9))

with() countable such thatI = EM{


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76 SAHARON SHELAH

(M)/EautI, for any model M of cardinality 2 but 2, for any vocabulary M.

6) Claim 5.1(1),(2) holds also if we replace by [, 2).

Proof. 1) Fix an ordinal , < + such that cf() = 2, e.g., = 2 (almostalways cf() 2 suffice).

Let I1 be the following linear order, its set of elements is {(, ) : {2, 1, 1, 2}, < +} ordered by (1, 1) 2.

For t I1 let t = (t, t).

Let I2 be the set { : is a finite sequence of members of I1} ordered by1


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[Why? Let g : I1 I1 be g(t) = (t, t), clearly it is an anti-isomorphism ofI1. Let g : I

2 I

2 be defined by g() = g((m)) : m < g(), it is an anti-

isomorphism of I2 . Lastly g maps I2 onto itself, in particular by the character ofclause (d) of, i.e. the two cases are interchanged by g]

2 (a) I1, I2 , I2 have cofinality 0.

(b) if t I2 then I2,


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78 SAHARON SHELAH

Case 1: = 0 and there is (I1) such that (n < )(i < )(i n ).Let ni = g(i ), it is impossible that {i : ni = k} is infinite for some k, so

without loss of generality ni : i < is an increasing sequence and n0 > 0.For every i < we have (n

i+1)

i+1and

i+1 0 making =

impossible; now g() > 0 (as we have discarded the case J = , i.e. Case 2); and

let k = g() 1. Now we prove case 4 by splitting to several subcases.

Subcase 4A: (k) {2, 1}.Let 1 = ( k)(

(k), (k) + 1), note that 1 I2 as I2 ((k) <

(k)+1 < ) and (as (k) {2, 1}) clearly { : 1 I2} is a cofinal subset ofJ even an end segment. Now for n < we have 1(2, + n) I2 and it satisfies. (Why? As 1 I2, only n = k may be problematic, but (k) + 1 = 1(k) herestands for (n) there hence clause (b) of does not apply), so by the definitionof I2, clearly {1(2, + n) : n < } is I2 and is a cofinal subset of J so = 0 = cf(J) and clause (c) of3 holds.

Subcase 4B: (k) {1, 2} and (k) is a successor ordinal.Let 1 = ( k)(

(k), (k) 1), of course 1 I2 and as I2 clearly1 I2 so the set { : 1 I2} is an end segment of J and has cofinality 0because n < 1(2, + n) I2. (Why? It I2 and as 1 I2 checking only n = k may be problematic, but ((k), 2) here stand for ((n), (n+1)) therebut presently (k) {1, 2} contradicting clause (d) of). So clause (c) of3.

Subcase 4C: (k) {1, 2} and (k) = 0.Then let 1 = ( k)(

(k) 1, 0). Now 1 I2 as k I2 and for n = k 1clause (c) of fails and 1(2, + n) I2 because of 1 I2 and for n = k the

failure of clause (b) of so continue as in Subcase 4B above.Lastly,

Subcase 4D: (k) {1, 2} and (k) is a limit ordinal.Then {( k)((k), ) : < (k)} is I2 and is an unbounded subset of

J hence cf(J) = cf((k)). If cf((k)) = 0, then clause (c) in 3 holds, and

if cf((k)) [1, 1) then necessarily (k) = so being a limit ordinal < +

clearly (k) < so clause (d) from 3 holds. To finish this subcase note thatcf((k)) 1 is impossible.[Why impossible? Clearly for large enough i < we have i(m) (because = + as said in the beginning of the case) and recall i I2. We now show

that clauses (a)-(d) of hold with i, k here standing for , n there. For clause (a)recall g(i) g() + 1 and m = g() = k + 1. Now i(k+1) = i(m) = = 2as = 2 is part of the case, i(k) = (k) {1, 2} in this subcase, so clause (d)of holds. Also i(k+1) = i(m) as said above so clause (c) of holds andcf(i(k)) = cf((k)) 1 (as we are trying to prove impossible), so clause (b)of holds. Together we have proved (a)-(d) of. But i I2, contradiction.]


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80 SAHARON SHELAH

Now subcases 4A,4B,4C,4D cover all the possibilities hence we are done withcase 4.

Case 5: = 2, > 0 and (, ) / I2.Recalling is the limit of the increasing sequence i(m) : i < hence cf() =

> 0 and (2, ) / I2, necessarily = so = 2. As (2, ) / I2necessarily clauses (a) - (d) of hold for some n and as I2, clearly n = g()1(see clause (a) of) so we have g() > 0, and letting k = g() 1, by clause (d)of the (n+1) there stands for = 2 here so we have (k) {1, 2} and byclause (b) of we have cf((k)) 1. Hence {( k)(

(k), ) : < (k)} iscofinal in J and its cofinality is cf(

(k)) as ( k)((k), ) increase (by I2)with as (k) {1, 2}. But cf((k)) 1 and = 2 (see first sentence of thepresent case), so clause (e) of3 holds.

Case 6: (

, ) I2.Subcase 6A: (, ), (2, ) I2.

Note that for m = g() and the pair ( (, ), (2, ), m) standing for (, n)in , clauses (a),(c),(d) of hold (recall {2, 1}, see the discussion after case1) so necessarily clause (b) of fails hence cf() < 1 but = cf() so < 1.Now as (, ), (2, ) I2 clearly if < , then (, ), (2, + ) belongs toI2 hence { (, ), (2, + ) : < } is a cofinal subset of J by the choice of I2hence cf(J) = 0 so clause (c) of3 applies.

Subcase 6B: (, ), (2, ) / I2.

As (

, ) I2, necessarily clauses (a)-(d) of hold with ( (

, ), (2, ), m)here standing for (, n) there, recalling m = g() so by clause (b) of we knowthat cf() 1 but = cf() hence 1. Also { (, ), (2, ) : < } is asubset of I2 and cofinal in J and is increasing with so cf(J) = 2 so clause (e)of3 applies.

As the two subcases 6A,6B are complimentary case 6 is done.

Finishing the proof of3:It is easy to check that our cases cover all the possibilities (as after discarding

cases 0,1, if not case (6) then (, ) / I2, as not case (3), = 1 but (see clause(c) before case 2), {2, 1} so necessarily = 2, so case (4),(5) cover the

rest). Together we have proved 3.]

4 recall 0 < 1 < 2; if X I2, |X| < 2 then we can find Y such thatX Y I2, |Y| = + |X|, Y is unbounded in I2 from below and fromabove and for every I2\Y the following linear orders have cofinality 0:

(a) J2Y, = I2 { I2\Y : ( Y)(


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(b) the inverse ofJ2Y,

(c) JY, = I2 { I2 : ( J2Y, )(


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82 SAHARON SHELAH

is a limit ordinal (n) and 0 / W so cf(0) |W| < 2 but by the assumptionon W, (see clause (f) of2) we have cf(0) = 0. So ( n)(

(n), 0) J2Y, ;moreover

3 J2Y, iff I2 satisfies one of the following:

(a) (i) n = n, and (n) = (n),

(ii) (n) [0, 1)

(b) (i) n = n, and (n) = (n),

(ii) (n) = 1 and (n+1) [sup(W ), )

(iii) ((n+1), (n)) = ((n1), (n)) {(2, 2), (2, 1), (2, 1), (2, 2)}

(c) (i) 1 = and n > and ( n)((n), 1) / I2

(ii) ((n), (n1)) {(2, 2), (2, 1), (2, 2), (2, 1)}

(iii) cf((n)) 1 and (n) > sup(W (n))(iv) (n 1) = (n 1), (n1) = (n1)

(v) (n1) [sup((n 1) W), (n 1)).

[Why? First note that if J2Y, and k = k, (k) = (k), and k n

then necessarily k = n (k) = (k). We now proceed to check if. Let f :{2, 1, 1, 2} {2, 2} so that f1({2}) = {2, 1} and f1({2}) = {1, 2}.Case (a) is obvious. In case (b) in order for Y to separate between and it is necessary that (n + 1) = (n + 1), (n+1) = (n+1) = f((n)) and

that (n+1) , but then / I2. In case (c) in order to separate between and by Y there are two possibilities. Either n = n and then (n) =(n) = f((n1)) (recall that n((n), 1) / I2), and

(n) , but thenalso / I2. The other possibility is that (n 1) = (n 1),

(n1) = (n1)

and = (n1) is such that W and (n1) < < (n1) which is alsoimpossible by the choice of (n1). Showing that these are the only cases (theonly if direction) is similar and is actually done below.]

Now we proceed to check that clauses of4 hold.

Clause (a):

First assume (n)

{2, 1}, and let J = { n((n)

, 0), (2, +m) : m < }.Now J I2 [why? clearly if J then (n + 1) I2 so we only need tocheck for n, recall that cf(0) = 0 < 1, hence clause (b) of fails]. Now byclause (a) of3 we have that J J2Y, , and we claim that it is also cofinal in it.

[Why? Note that as (n) {2, 1} then n((n), 0)


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n((n), 0), (2, + m). If J2Y, is as in clause (c) of3 then (n) {2, 2}

by (ii) there, and as in this case (n) {2, 1}, necessarily (n) = 2 and so by(ii) of (c) of3 we have

(n1) {1, 2}, but then


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84 SAHARON SHELAH

Clause (c): As Y is unbounded from below in I2 (containing {(2, +n) : n < })it follows that JY, is non-empty, hence cf(J

Y, ) = 0, but what is cf(J

Y, )?

First, if (n) {1, 2} then {( n)((n), ) : < 0} is an unbounded

subset of J

Y, of order type 0 hence cf(J

Y, ) = cf(0) = 0 (see the assumptionon W and the choice of 0).Second, if (n) = {2, 1} and ( n)((n), 1) I2 and cf(1) 1 then as

in the proof of clause (a) we have {( n)((n), 1), (2, + m) / I2 for m < and again letting = sup(W) we have {( n)((n), 1), (2, ) : W}is included in I2 and in J

Y, and even is an unbounded subset ofJ

Y, of order type

otp(W ) which has the same cofinality as which is 0.Third, if (n) {2, 1} and ( n)((n), 1) I2 and cf(1) < 1, equiva-

lently cf(1) = 0, then {( n)((n), 1), (2, + m) : m < } is a subset of I2

(as cf(1) = 0) is included in JY, , unbounded in it and has cofinality 0, so we

are done.Fourth and lastly, if (n) {2, 1} and ( n)((n), 1) / I2 then as in

the proof of clause (a) we have 1 = and again letting = sup(W ) we have

cf() = 0 and ( n)((n), ) I2 and {( n)((n), ), (2, + m) :m < } is a subset of I2, moreover a subset of J

Y, unbounded in it and (

n)((n), ), (2, + m) is


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stationary) set S , m n and 0 = k0 < k1 < .. . < km = n stipulatingt,km = and letting () = Min(S) we have:

(a) for each i < m:

() if < are from S and 1, 2 [ki, ki+1) then t,1


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86 SAHARON SHELAH

one of the following holds:

() x, y Iu and x


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Why? The proof takes awhile. Toward contradiction assume = cf(Iu,x) is > 0and let b : < be an increasing sequence of members of Iu,x unbounded in it.So for each < there is a definable function f(x0, . . . , xn()1) where definable ofcourse means in the theory of real closed fields and t

,0 min(/e)Let h() = {h() : < } + +.

So clearly we can show by induction on < () that:

h() < + + () (2 + 2).

Now check.Also recalling + < clearly for < , < () we have h() < + .Now check.

Second (Ilin,(), Ilin,(),w) is a reasonable (, ())-base

Combine the proof of first with the parallel proof in part (1). 6.12

6.13 Definition. 1) Let I, J Klin()

. We say that E is an invariant (I, J)-

equivalence relation when:

(a) E is an equivalence relation on incJ(I), so E determines I and J

(b) ifh1, h2, h3, h4 incJ(I) and tpqf(h1, h2; I) = tpqf(h3, h4; I) then h1Eh2 h3Eh4.

2) We add non-trivial when:

(c) if eq(h1, h2) = {t J : h1(t) = h2(t)} is co-finite then h1Eh2

(d) there are h1, h2 incJ(I) such that (h1Eh2).

3) Let J, I1, I2 Klin()

. Then I1 1J I2 means that:

(a) I1 I2

(b) for every h1, h2, h3 incJ(I2) we can find h1, h2, h

3 incJ(I1) such that

tpqf(h1, h

2, h

3; I1) = tpqf(h1, h2, h3; I2).


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6.14 Claim. Assume J, I1, I2 Klin()

.

1) If I1 I2,E is an invariant (I2, J)-equivalence relation then E incJ(I1) is aninvariant (I1, J)-equivalence relation.

2) If I1 we define , e as follows:

2 (a) = ,t : t J


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(b) ,t = min((() + 1) N\h(t))

(c) e := {(s, t) : s, t J and ,s = ,t and ,s > h(s) and,t > h(t)}

(d) for N let X, := {t J : ,t = > h(t)}.

Note

()1 N.

[Why? As [N] N and |J| = and ,t N for every t J.]

()2 (a) e e(J), i.e. e is an equivalence relation on some subset of J witheach equivalence class a convex subset of J, see Definition 6.6(1)

(b) X, : {,t : t Dom(e)} hence X, = list the

e-equivalence classes.[Why? Think.]

()3 h := h (J\ Dom(e)) N.

[Why? By the definition of e we have t J t / Dom(e) h(t) N andrecall [N]

N.]

()4 if t Dom(e) then cf(,t) > .

[Why? As ,t N (H(), ) if cf(,t) = then there is a cofinal set B

of ,t of cardinality in N but + 1 N therefore B N . In particularas h(t) < ,t there is B so that h(t) < , but this contradicts the choice of,t.]

()5 e e(J, I).

[Why? Choose h incJ(I)N similar enough to h, specifically: t J\ Dom(e) h(t) = h(t) and t Dom(e) sup{,s : s J, s


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112 SAHARON SHELAH

Now we choose (n, hn) by induction on n < such that

(a) n n

(b) if n = m + 1 then m = n m(c) hn incJ(I)

(d) h0 = h

(e) if n = m + 1 then:

() hnEhm hence hnEh and Dom(en) Dom(em)

() hm (J\ Dom(em)) hn

() (hm {Xm,m, : < m}) hn

() hn (Dom(em)\ {Xm,m, : < m}) belongs to Nm

() moreover t Dom(em)\ {Xm,m,

: < m} implieshn(t) < hm(t)

() m > 0

(f) Ym+1 Ym where Ym := {Xm,m, : < }.

Why can we carry out the construction? For n = 0 we obviously can (chooseh0 = h). For n = m + 1 first choose h

m Nm as we choose in the proof of

()5. Now recalling Xm, m, : < m was chosen in ()6, and define hn byhn (Dom(em)\ {Xm,m, : < m}) = h

m (Dom(em)\ {Xm,m, :

(d) every type from S(M) is realized in N

(e) if n < , a n(N) then for some a n(N) and automorphism of N, (a) = a and maps M onto itself.

[Why? Clauses (a),(b) holds by clause (c) of Claim 0.9(1) recalling Definition0.8(2).Clause (c) holds by Clause (c) of the assumption of 7.2; you may note [Sh 394,6.7=6.4tex](2).Clause (d) holds as EM(K)(+ + J, ) K+ is saturated, and use the definitionof a type (or like the proof of claue (e) below using appropriate I + I instead+ + J); you may note [Sh 394, 6.8=6.5tex].Clause (e) holds as for every finite sequence t from J there is an automorphism of J such that: is the identity on Q, it maps I onto itself and it maps t toa sequence from J = Q + I, such exists as I is strongly 0-homogeneous andI I is infinite.]For any a = b from N let

(a, b) = Min{ : and if < then tpK(a, M, N) = tpK(b, M, N)}.

So (a, b) . Let

= sup{(a, b) : a, b N and (a, b) < }.

So is defined as the supremum on a set of cardinals < , which isa cardinal of cofinality cf() > , hence clearly < . Also as thereare a = b from M hence (a, b) = . Now suppose that [, ), M K is

saturated and p1 = p2 S(M) and we shall find M

K M, M

K such thatp1 M

= p2 M, this suffice.Clearly M K is saturated (by clause (c) of) hence the models M, M are

isomorphic so without loss of generality M = M. But by clause (d) of everytype from S(M) is realized in N, so l

saharon shelah- categoricity and solvability of a.e.c., quite highly

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