Mirna Dzamonja and Saharon Shelah- On ◅^*-Maximality

8/3/2019 Mirna Dzamonja and Saharon Shelah- On ^*-Maximality

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On -maximality

Mirna DzamonjaSchool of Mathematics

University of East AngliaNorwich, NR4 7TJ,UK

[email protected]

http://www.mth.uea.ac.uk/people/md.html

andSaharon Shelah

Mathematics DepartmentHebrew University of Jerusalem

91904 Givat Ram, [email protected]

http://shelah.logic.at/

May 23, 2011

Abstract

This paper investigates a connection between the semantic notionprovided by the ordering among theories in model theory and thesyntactic (N)SOPn hierarchy of Shelah. It introduces two proper-ties which are natural extensions of this hierarchy, called SOP2 andSOP1. It is shown here that SOP3 implies SOP2 implies SOP1. In

[Sh 500] it was shown that SOP3 implies -maximality and we provehere that -maximality in a model of GCH implies a property calledSOP2. It has been subsequently shown by Shelah and Usvyatsov thatSOP2 and SOP2 are equivalent, so obtaining an implication between

-maximality and SOP2. It is not known if SOP2 and SOP3 areequivalent.

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Together with the known results about the connection between

the (N)SOPn hierarchy and the existence of universal models in theabsence of GCH, the paper provides a step toward the classificationof unstable theories without the strict order property. 1

Changes from the published version:In the published version of this paper it is claimed that witnesses

to being SOP1 can be chosen to be highly indiscernible, and this is justified by a certain notion of 1-fbti. The definition of this notion(Definition 2.10) has a typo in a crucial place, and in addition Claim2.11 for t = 2 is incorrect and for t = 1 the proof is incomplete. Inthis version we clarify these statements and proofs by introducting anew notion of indescernibility 3-fbti. The corrected statement is that

witnesses to being SOP1 can be chosen to be 3-fbti.That there are inconsistencies in the notions we used in the original

paper was first observed by Lynn Scowl (September 2008), ByunghanKim (May 2009) and Enrique Casanovas and Martin Ziegler (July2010). Whilst a Ph.D. student at UEA in 2008, Mark Wong alsoobserved some incosistencies and made partial progress in rectifyingthem. There is a paper by Kim and Kim (to appear in APAL as ofMarch 2011) which gives a different notion of 1-fbti and shows thatwitnesses can be chosen with that kind of indiscernibility.

0 IntroductionThis paper investigates a connection between the ordering among theories

in model theory and the (N)SOPn hierarchy of Shelah and as such provides

a step toward the classification of unstable theories without the strict order

property. The thesis we pursue is that the syntactic property SOP2 is closely

1This publication is numbered 692 in the list of publications of Saharon Shelah. Theauthors thank the United States-Israel Binational Science Foundation for a grant support-ing this research and the NSF USA for their grant numbered NSF-DMS97-04477. MirnaDzamonja thanks EPSRC for their support through grant number GR/M71121, as well as

the Royal Society for their support through grant number SV/ISR/NVB. We would alsolike to acknowledge the support of the Erdos Research Center in Budapest, during theWorkshop on Set-theoretic Topology, July 1999, and support of the Hebrew University ofJerusalem and the Academic Study Group during July 1999. Finally, warm thanks aredue to Alex Usvyatsov for his comments and improvements to the manuscript.

Keywords: classification theory, unstable theories, SOP hierarchy, oak property.AMS 2000 Classification: 03C45, 03C55.

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related to the semantic property of being maximal in the -order. We

shall now give the relevant definitions and explain the motivation behind thepaper as well as noting our main results. For the purpose of this introductory

discussion we shall limit ourselves to countable (complete first order) theories.

The following order among theories was introduced and investigated by

Keisler in [Ke].

Definition 0.1 (1) For any cardinal , the Keisler order among theoriesis defined as follows: T0 T1 if whenever Ml(l < 2) is a model of T0, T1respectively and D is a regular ultrafilter over , then the +-compactness

of M1 /D implies the +-compactness of M0 /D.

(2) We say T0 T1 if for all we have T0 T1.

The relevance of this order to the project of classifying unstable theories

without strict order property lies in the two following theorems of Shelah

(note that the second one implies the first).

Theorem 0.2 (Shelah [Sh c], VI4.3) Any (countable) theory with the strict

order property is -maximal.

As stated in [Sh c], pg xiv, Ch VI of [Sh c] gives a rather complete picture

of Keislers order and to complete it we should know more about unstabletheories without the strict order property. Paper [Sh 500] started a classifi-

cation of such theories by introducing the hierarchy SOPn for n 3 and in

particular it is stated there that being maximal in the Keisler order is not a

characterisation of theories with the strict order property,

Theorem 0.3 (Shelah [Sh 500], see also [ShUs 844]) Any theory with SOP3is -maximal.

Details of the proof are given in [ShUs 844]. Precise definitions of prop-

erties SOPn for n 3 will be repeated below in 2 but for the moment we

note that it was proved in [Sh 500] that for n 3

strict order property = SOPn+1 = SOPn = not simple

and that all the implications are irreversible. One may now wonder if having

SOP3 is a characterisation of theories that are maximal in the Keisler order,

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giving us a semantic equivalent to the syntactic notion of SOP3. This would

be consistent with what is known about this order, see the Introduction toCh VI of [Sh c]. This question remains open but instead one may attempt

to give a characterisation of SOP3 or SOPn in terms of some other similarly

defined order. This is suggested by [Sh 500] which in fact gives a theorem

stronger than 0.3, namely

Theorem 0.4 (Shelah [Sh 500], see also [ShUs 844]) Any theory with SOP3is -maximal.

The definition of this order will be recalled in 1 where we shall also prove

that being -maximal implies being maximal in the Keisler order. Giventhis fact one may now ask if being -maximal characterises theories with

SOP3. To test this claim it is natural to investigate a prototypical example of

an NSOP3 theory that is still not simple, which is T

feq. In 1 we shall recall

the definition of this theory and show that in fact it is not -maximal, as it

is consistently strictly below the theory of a dense linear order with no first

or last element (all we need for the consistency is a partial GCH assumption).

This naturally leads to the question of the possibility of refining the dis-

tinction between simplicity and SOP3. Definition of the SOPn hierarchy

from [Sh 500] does not immediately give way to such a refinement as SOPnis roughly speaking, defined in terms of omitting loops of size n. However in

2 we introduce two properties SOP2 and SOP1 that in fact satisfy

SOP3 = SOP2 = SOP1 = not simple.

We then ask if these properties in any way characterise the maximality in .

To this end in 3 we prove that any theory that is-maximal in a model of a

sufficient amount of GCH must satisfy a syntactic property SOP2 . Together

with a subsequent result of Shelah and Usvyatsov in [ShUs 844] that proved

that SOP2 is equivalent to SOP2 we hence obtain that

-maximality in anymodel of a sufficiently rich fragment of GCH implies SOP2. (See 3 for the

definition of SOP2 and the exact reference from [ShUs 844]). To summarise,

our main result, appearing as Corollary 3.9(1) below is

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Theorem 0.5 Suppose that T is a theory that is -maximal in some uni-

verse of set theory in which 2 = + holds for all large enough regular .Then T has SOP2.

Several questions remain open. The main one of course is if SOP2 is

actually equivalent to -maximality. Recall from the discussion above that

we know that SOP3 implies-maximality. It is not known if SOP3 and SOP2

are actually equivalent. We also note that Shelah and Usvyatsov have proved

in [ShUs 844] a local version of the implication SOP2 = maximality,

see 3 for a more detailed discussion.

A burning question also is that we in fact do not know almost anything

about the reverse of other implications in the (consistent) diagram

SOP3 = maximality = SOP2 = SOP1 = not simple,

apart that not all of them may be equivalences, as Tfeq is not simple but is

NSOP3. In fact [ShUs 844] proves that T

feq is not even SOP1.

Before laying down the organisation of the paper let us also mention the

connection of the SOPn hierarchy with another semantic property, which is

the possibility of having a universal model at in some universe of set the-

ory where a sufficient amount of GCH fails (under GCH every countable first

order theory has a universal model in every uncountable cardinal). The con-

nection between this property and unstable theories without the strict order

property has been investigated in a series of papers, notably in [KjSh 409]

where it is proved that if GCH fails sufficiently then there are no universal

dense linear orders. It was proved in [Sh 500] that SOP4 is already sufficient

for such a negative universality result. The question of universality is inter-

esting also for classes that are not elementary classes of models of a first order

theory, for example for classes without amalgamation the most interesting

case is the strong limit singular of cofinality 0. In [GrSh 174] it is proved

that for such and < a strongly compact cardinal the class of modelsof any L,-theory of cardinality < admits a universal model of cardinal-

ity . A rather detailed description of what is known about the connection

of unstable theories without the strict order property and the universality

problem may be found in the introduction to [DjSh 710].

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The paper is organised as follows. In the first section we investigate the

theory Tfeq. This is simply the model completion of the theory of infinitelymany parametrised equivalence relations. We show that under a partial GCH

assumption, this theory is not maximal with respect to , as it is strictly

below the theory of a dense linear order. In the second section of the paper

we extend Shelahs NSOPn hierarchy by introducing two further properties

SOP1 and SOP2, and we show that their names are justified by their position

in the hierarchy. Namely SOP3 = SOP2 = SOP1. Furthermore, SOP1theories are not simple. The last section of the paper contains the main

result showing that -maximality in a model of a sufficiently rich fragment

of GCH implies SOP

2 , and hence SOP2 by Shelah-Usvyatsov.The following conventions will be used in the paper.

Convention 0.6 Unless specified otherwise, a theory stands for a first

order complete theory. An unattributed T stands for a theory. We use

(T) to denote the vocabulary of a theory T, and L(T) to denote the set of

formulae of T.

By C = CT we denote a -saturated model of T, for a large enough

regular cardinal and we assume that any models of T that we mention are

elementary submodels ofC.

,, stand for infinite cardinals.

1 On the order

Definition 1.1 (1) For (first order complete) theories T0 and T we say that

= R(xR) : R a predicate of (T0) or a function symbol of (T0) or =,

(where we have xR = (x0, . . . xn(R)1)), interprets T0 in T, or that is an

interpretation of T0 in T, or that

T is a model ofT0,

if each R(xR) L(T), and for any M |= T, the model M[] described below

is a model of T0. Here, N = M[] is a (T0) model, whose set of elements

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is {a : M |= =(a, a)} (so M[] M) and RN = {a : M |= R[a]} for a

predicate R of T0.For any function symbol f of (T0) we have that N |= f(a) = b iff

M |= f(a, b), while

M |= f(a, b) = f(a, c) = b = c

for all a ,b,c.

(2) We say that the interpretation is trivial if R(xR) = R(xR) for all

R (T0), so M[] = M (T0), for any model M of T.

(The last clause in Definition 1.1(1) shows that we can in fact restrictourselves to vocabularies without function or constant symbols.)

We use the notion of interpretations to define a certain relation among

theories. This relation was introduced by S. Shelah in [Sh 500], section 2

and one can see [ShUs 844] for a more detailed exposition. The reason we are

interested in this ordering is Shelahs Theorem 0.3 quoted in the Introduction

and we shall now start developing methods for the proof of our main result

3.9.

Definition 1.2 For (complete first order) theories T0, T1 we define:

(1) A triple (T, 0, 1) is called a (T0, T1)-superior iff T is a theory and lis an interpretation of Tl in T, for l < 2.

(2) For a cardinal , a (T0, T1)-superior (T, 0, 1) is called -relevant iff

|T| < .

(3) For regular cardinals , we say T0 , T1 if there is a min(, )-

relevant (T0, T1)-superior triple (T, 0, 1) such that in every model M

of T in which M[1] is -saturated, the model M[0] is -saturated. If

this happens, we call the triple a witness for T0

, T1.

(4) We say that T0 , T1 over if , and T0

, T1 as witnessed

by a (T, 0, 1) with |T| < .

(5) If = , we write in place of,.

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(6) We say that T1 T2 iff T1

T2 holds for all large enough regular .

(6) T is -maximal iff T T

holds for all T. The notion of-

maximality is defined analogously.

(7) We say T0 ,= T1 iff T0

T1 but (T1

T0).

Although in this paper we do not consider this in its own right, it is

natural to define the local versions of the -relation. This is used by Shelah

and Usvyatsov in [ShUs 844] to obtain their local converse to the implication

-maximality = SOP2, see 3 for more discussion on this.

Definition 1.3 Relations ,l, and ,l are the local versions of

, and

respectively, where by a local version we mean that in the definition of

the relations, only types of the form

p {(x, a) : a lg(y)M}

for some fixed (x, y) are considered.

Observation 1.4 (0) If T0 , T1 and l < 2, then there is a witness

(T, 0, 1) such that l is trivial, hence Tl T.

(1) is a partial order among theories (note that T T for every complete

T of size < , and that the strict inequality is written as T1 ,= T2).

(2) IfT0 , T1 over and T1

, T2 over , then T0

, T2 over .

[Why? (0) Trivial.

(1) Suppose that Tl Tl+1 for l < 2 over , as exemplified by (T

, 0, 1)

and (T, 1, 2) respectively. Without loss of generality, 1 is trivial (apply

part (0)), so as T is complete we have T1 T. Similarly, without loss of

generality, 1 is trivial and so, as T is complete, we have T1 T. As

T1 is complete, without loss of generality, T and T agree on the common

part of their vocabularies, and hence by Robinson Consistency Criterion,

Tdef= T T is consistent. Also |T| + |T| < , hence |T| < . Clearly

T interprets T0, T1, T2 by 0, 1 = 1 and 2 respectively and T is com-

plete. We now show that the triple (T, 0, 2) is a (T0, T2)-superior which

witnesses T0 T2 over . So suppose that M is a model of T in which

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M[2] is -saturated. As (T, 1, 2) witnesses T1 T2, we can conclude

that M[1] = M[1] is -saturated. We can argue similarly that M[0] is-saturated.

(2) is proved similarly to (1).]

In this section we consider an example of a theory which is a prototypical

example of an NSOP3 theory that is not simple (see [Sh 457]). It is the model

completion of the theory of infinitely many (independent) parametrised equiv-

alence relations, formally defined below. We shall prove that for such that

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Note 1.6 One may easily check that every model ofTfeq can be extended to

a model of T+feq and that T+feq has the amalgamation property and the jointembedding property. This theory also has a model completion, which can be

constructed directly, and which we denote by Tfeq. It follows that T

feq is a

complete theory with infinite models, in which F is a full function.

Remark 1.7 Notice that Tfeq has been defined somewhat differently than

in [Sh 457, 1], but the difference is non-essential, as the following Claim 1.8

shows that the two theories have the same model completion. This claim

also shows the origin of the name infinitely many independent equivalence

relations for Tfeq

.

Claim 1.8 Let T be the theory defined (in [Sh457, 1]) by

(a) T has unary predicates P and Q and a three place relation E writen as

y Exz,

(b) the universe of any model of T is a disjoint union of P and Q,

(c) y Exz = P(x) & Q(y), Q(z),

(d) for any fixed x P the relation Ex is an equivalence relation on Q.

Then Tfeq is the model completion of T.

Proof of the Claim. Let M be a model of Tfeq, we shall extend M to a

model of T as follows. Each E-equivalence class e = a/E gives rise to an

equivalence relation Ee on P given by:

z1Eez2 iff z1, z2 P and F(a, z1) = F(a, z2).

This definition does not depend on a, just on a/E. Let PN and QN be QM

and PM

respectively. Define y EN

x z iff y Eez where e = x/EM

. Clearly N isa model of T.

Now suppose that we have a model M of T and we shall extend it to

a model N of Tfeq. Let PN and QN be QM and PM respectively. Define

x ENx iff for every y, z we have y Exz iff y Exz. Choose a representative of

each E-equivalence class and for any z QN and such a representative x let

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F(x, z) = x. Then for x QN which has not been chosen as a representative

of any equivalence class, find x which has been chosen as its representativeand define F(x, z) = F(x, z) for all z PN.

This shows that Tfeq and T are cotheories ([ChKe], 3.5.6(2)). Being the

model completion of Tfeq, T

feq is its cotheory, and hence a cotheory of T.

Hence Tfeq is a model companion ofT. In order to prove that it is the model

completion of T it suffices to show that T has the amalgamation property

([ChKe], 3.5.18) which is easily seen directly. 1.8

Observation 1.9 Tfeq has elimination of quantifiers and for any n, any

model of Tfeq is a model of Tnfeq.

Notation 1.10 Tord stands for the theory of a dense linear order with no

first or last element.

The following convention will make the notation used in this section sim-

pler.

Convention 1.11 Whenever considering (Tord, T

feq)-superiors (T, , ) we

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N must omit most of the Dedekind cuts induced by the tree of embeddings,

and that most of these cuts are not definable over N. After an application ofan appropriate automorphism ofC this finishes the proof of the Main Claim.

The proof of the theorem then follows by induction. The cardinal arithmetic

assumptions are used in Stage D and in the inductive proof of the theorem.

Definition 1.12 For a -relevant (Tord, T

feq)- superior T, the statement

[M, a, b] = [M, a, b , T, ]

means:

(i) M is a model ofT of size ,

(ii) a = ai : i < , b = bi : i < , are sequences of elements of IM[]

such that

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Case 1. (This will be the main case) p(z) implies that z P and it

determines which elements of QM are R-connected to z. Hence for somefunction f : QM QM which respects E, i.e.

a E b = f(a) = f(b),

and

f(a) a/EM;

we have

p(z) = {P(z)} { b R z : b Rang(f)}

and no a PM

satisfies p.Case 1A. Like Case 1, but f is a partial function and

p(z) = {P(z)} { f(b) R z : b Dom(f)}

{ (b R z) : (b/EM Rang(f)) = }.

(This Case will be reduced to Cases 1-3 in Subclaim 1.15).

Case 2. p(z) determines that z Q and that it is E-equivalent to some

a QM, but not equal to any old element. Note that in this case if

b realises p(z), we cannot have bRc for any c PM, as this would imply

F(a

, c) = b

M[]

(and we have assumed that p is not realised in M[]

).Hence, the complete M-information is given by

p(z) = {Q(z)} {a E z} {a = z : a a/EM}.

Case 3. p(z) determines that z Q, but it has a different E-equivalence

class than any of the elements of QM. As p is complete, it must deter-

mine for which c PM we have z R c, and for which c, d PM we have

F(z, c) = F(z, d). Hence, for some f : PM {yes, no} and some E, an

equivalence relation on PM such that cEd = f(c) = f(d), we have

p = {Q(z)} { (a E z) : a QM} {(z R b)f(b) : b PM}

{(F(z, c) = F(z, d))ifcEd : c, d PM}.

In the above description, we have used

Notation 1.14 For a formula we let yes and no .

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Subclaim 1.15 It suffices to deal with Cases 1,2,3, ignoring the Case 1A.

Proof of the Subclaim. Suppose that we are in the Case 1A. Let

{di/EM : i < i }

list the d/EM for d QM such that d d/EM = (d R z) p(z). We

choose by induction on i i a pair (Mi, ci) such that

(a) M0 = M, ||Mi|| = ,

(b) Mi : i i is an increasing continuous elementary chain,

(c) [Mi, a, b]

(d) ci (di/EMi+1) \ Mi, for i < i

.

For i limit or i = 0, the choice is trivial. For the situation when i is a

successor, we use Case 2.

Let ci/EMi : i [i, i) list without repetitions the c/EMi which are

disjoint to M. Note that |i| . Let

p+(z)def= p(z) {ci R z : i < i

}.

Then p+(z) is a complete type (for M[]i ), and [Mi, a, b] holds by (c). Ifp+(z) is realised in Mi , we can let N = Mi and we are done. Otherwise,

p+(z) is not realised in Mi and is a type of the form required in Case 1, so

we can proceed to deal with it using the assumptions on Case 1. 1.15

Stage B. Let us assume that p is a type as in one of the Cases 1,2 or 3,

which we can do by Subclaim 1.15. We shall define M : < , an -

increasing continuous sequence of elementary submodels of M, each of size

< , and with union M, such that:

(a) In Case 1, each M is closed under f,(b) In Case 2, a M0,

(c) For every < ,

(M, {(aj, bj ) : j < } M) (M, {(aj , bj) : j < }),

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Hence, for some club C of consisting of limit ordinals , we have that

for all C,aj M bj M j < ,

(c IM)(j < ) [c 2 for l = 0, 1 and 0 1 = , then:

(i) N0 N1 = N.

(ii) In addition, if al QNl for l = 0, 1 and a0 E

CTa1, then for some

a QN we have alECTa for l = 0, 1. (Equivalently, if al Q

Nl

and (a N)(l


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(a)

0def=

(x0j0, . . . , x

0jn1

; h(c)) : n < & c Mi & j0, . . . jn1 < ji &

Mi+1 |= (dij0

, . . . , dijn1; c)

,

(taking care of one side (for 0 or 1) of the part () above)

(b) 1, defined analogously to 0, but with x0j0

, . . . , x0jn1 replaced every-

where by x1j0, . . . , x1jn1

,

(taking care of the remaining side of () above, interchanging 0 and 1)

(c) 2 = {(x0

0, x0

1)I (x1

0, x1

1)I = },(this says that the above intervals in


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First note that requiring 5 6 throughout the construction indeed guaran-

tees that ()(ii) can be satisfied. Namely, suppose that the realisations of x0j0and x1j1 are E-equivalent. Then by 5 we must have that for some l < 2 and

a Mi we have that dijl Ea. By transitivity then the realisation of x

1lj1l

would have to be E-equivalent to h(a), which might only be precluded by

dij1l being E equivalent to some c such that a and c are not E-equivalent.

This cannot happen by 6.

We conclude that, if is consistent, as C is -saturated, the functions hlcan be defined. Namely, for a realisation {clj : j < j

i , l < 2} of , we can

define gl(dij) = c

lj, and then we let h0 = g0 if c

01


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By extending h to an automorphism h ofC, and applying (h)1, we may

assume that h = idMi . We can also assume that no clk is an element ofMi ,as otherwise the relevant inequalities can be absorbed by .

We shall use the following general

Fact 1.16 Suppose that N C and e mC is disjoint from N, while N A.

Then

r(x)def=tp(e, N) {xk = d : d A \ N, k < m}

{(xkEd) : d A & (d/E) N = , k < m},

is consistent (and in fact, every finite subset of it is realised in N).

Proof of the Fact. Otherwise, there is a finite r(x) r(x) which is

inconsistent. Without loss of generality, r(x) is the union of sets of the

following form (we have a representative type of the sets for each clause)

{ (x, c)} for some c N and such that |= [e, c].

{xk = csk : k < m} for some cs0, . . . c

sm1 A \ N and s < s

< ,

{(xk Edtk) : k < m} for some d

t0, . . . d

tm1 A \ N and t < t

<

such that (dtk/E) N = .

By the elementarity of N, there is e N with N |= [e, c]. By the choice of

the rest of the formulae in r(x), we see that e satisfies them as well, which

is a contradiction. 1.16

Let xl = (xl0, . . . , xln1) for l < 2. Let

0def= {(x0) : (x0j0, . . . , x

0jn1

) (0 03

06)}.

Applying the last phrase in the above Fact to 0(x0

), the model Mi andd, we obtain a sequence e0 = (e00, . . . e

0n1) Mi which realises 0(x

0). For

k, m such that (x0jkEx1jm

) 5 we have (a Mi)(aEdijk

aEdijm). So

(xkEe0m) (e

0m E xm) tp(d/Mi).

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Let now

1(x1) = {(x1k E e0m) (e0k E x1m) : (x0jk E x1jm) 5}

{x1k = e0m : k, m < n} {(x

1) : (x1j0 , . . . , x1jn1

) (1 13

16)}.

1(x1) is a finite set to which we can apply the last phrase of Fact 1.16.

In this way we find e1 = (e10, . . . e1n1) Mi realising 1(x

1). Now we

show that e0 e1 realises \ 2. So suppose (x0jkEx1jm

) 5, then

(x1k E e0m) 1, hence (e

1kEe

0m). Also k,m


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(a) 0(z) defines an initial segment of I,

(b) 1(z) defines an end segment of I,

(c) the segments defined by 0(z) and 1(z) are disjoint,

(d) 0(Mi) 1(Mi) = I Mi .

[Why? Note that (e0, d) D. Hence (d) holds. As for every

a I Mi we have a I bi , the conclusions follows.]

(e) 0(ai) and 1(bi) hold.

[Why? Again because

(d) holds.]

The above arguments show that {x : (y)[((y) x 2

of models and embeddings can be defined as required, and we consider

pdef= >2h(p Mlg()).

We shall show that p is finitely satisfiable, hence satisfiable. Let p

be finite. Recalling the analysis of p from Stage A, we consider each of the

cases by which p could have been defined (ignoring Case 1A, as justified by

Subclaim 1.15.)

Case 1. In this case there is a function f : QM QM respecting E,

with aEf(a) for all a QM, and without loss of generality there are some

0, . . . m1 >2 and {bji : j < m,i < nj} Rang(f) such that

= {P(z)} j


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The only possibility for such a contradiction is that for some j0, j1 and

bj0i , bj1k we have

hj0 (bj0i ) = hj1 (b

j1k ) hj0 (b

j0i )E hj1 (b

j1k )

and hj0(bj0i )R z , hj1(b

j1k )R z

. In such a case, any c which would realise

would contradict part (c) of the definition of T+feq. Suppose that bj0i , b

j1k and

0, 1 are as above. Let ldef= jl for l < 2 and let = 0 1. By part ()(ii)

in the definition of h, there is b N such that h0(bj0i )Eb and h1(b

j1k )Eb.

For some b Mlg() we have

h0(b) = h1(b) = h(b) = b,

so applying the elementarity of the maps, we obtain

bj0i E bE bj1k .

On the other hand, by the definition of p we have bj0i R z p(z) and

bj1k R z p(z). By the the demands on p this implies that bj0i = b

j1k / Mlg()

and f(b) = bj0i , contradicting the fact that Mlg() is closed under f.

Case 2. For a fixed a M0 we have

p(z) = {Q(z)} {a

Ez} {z = c : c a

/EM

},

so without loss of generality

= {Q(z)} {a E z} {z = hj (cj) : j < m}

for some c0, . . . , cm1 a/EM and 0, . . . , m1

>2, as h = idM0. As

a/E is infinite in any model of Tfeq, the set is consistent.

Case 3. We may assume that for some equivalence relation E on PM,

a function f from PM into {yes, no}, sequences 0, . . . n1 >2, and

{aki : i < m, k < n} QM and {bki , c

ki , d

ki : i < m, k < n} P

M we

have e1 Ee2 = f(e1) = f(e2) and

(z) ={Q(z)}

k


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We could have a contradiction if for some k1, k2, i1, i2 we had f(bk1i1

) =yes,

f(bk2i2 ) =no, but hk1 (bk1i1 ) = hk2 (bk2i2 ), which cannot happen by (i) and thefact that each h is 1-1. Another possibility is that for some b

k1i1

, bk2i2 we have

f(bk1i1 ) = f(bk2i2

)=yes, but hk1 (bk1i1

) = hk2 (bk2i2

) while hk1 (bk1i1

) E hk2 (bk2i2

). To

see that this cannot happen, we distinguish various possibilities for bk1i1 , bk2i2

and use part ()(ii) in the choice of h.

Yet another possible source of contradiction could come from a similar

consideration involving the last clause in the definition of (z), which cannot

happen for similar reasons.

Stage D. Now we can conclude, using = 2 N N

, such that p

is realised in N

.For 2, let h

def= i


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Proof. We prove (1), and (2) is proved similarly. Using the Main Claim

1.13, we can construct M of size +, by an -increasing continuous sequenceMi : i

+, with ||Mi || = satisfying [Mi, a, b] for each i +, and

letting M = M+ . The Main Claim 1.13 is used in the successor steps. To

assure that M is +-saturated for Tfeq, we use the assumption 2 = +, to

do the bookkeeping of all Tfeq-types involved. 1.17

Conclusion 1.18 Under the assumptions of Theorem 1.17, the theory Tfeqis

+strictly below the theory of a dense linear order with no first or last

elements.

[Why? It is below by Shelahs Theorem 0.4 above.]

We recall that our motivation for studying is to try to characterise

SOP3 (or SOP2) theories by the -maximality. As we explained in the

Introduction this has origins in the connection between the maximality in

the Keisler order and having the strict order property, so we should show

here what is the connection between the maximality in Keislers order and

the maximality in the order . The following Claim 1.19 does that for

countable theories.

Claim 1.19 Suppose that T is a countable theory that is -maximal.

Then it is maximal in the Keisler order .

Proof of the Claim. Suppose otherwise and let T1 be a theory that is

-maximal but not maximal in the Keisler order . In particular wehave Tord

T1, so there is a -relevant (Tord, T1)-superior triple (T, 0, 1)-

exemplifying this. By Observation 1.4(0) we may assume that the interpre-

tation 1 is trivial, so T1 T- for simplicity.

Since T is not maximal in the Keisler order , by [Sh c] 4.2 (1) there

is a regular ultrafilter D which is not good and a model M of T such thatM/D is nevertheless +-compact. We can extend M to a model N of T

and consider N = N/D. This is a model of T and by the Extension

Theorem for ultrafilters we have that [N]1 = M/D, so it is +-compact

and hence it is +-saturated. Again by the Extension Theorem we have that

[N]1 = (N1)/D. Now on the one hand we have by the -maximality of

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T1 that (N1)/D must be +-saturated, hence +-compact. But on the other

hand (N1)/D cannot be +-compact because D is not a good ultrafilterand Tord is maximal in the Keisler order, contradicting [Sh c] 4.2 (1). 1.19

2 On the properties SOP2 and SOP1

In his paper [Sh 500], S. Shelah investigated a hierarchy of properties un-

stable theories without strong order property may have. This hierarchy is

named NSOPn for 3 n < , where the acronym NSOP stands for not

strong order property. The negation of NSOPn is denoted by SOPn. It wasshown in [Sh 500] that SOPn+1 = SOPn, that the implication is strict

and that SOP3 theories are not simple. In this section we investigate two

further notions, which with the intention of furthering the above hierarchy,

we name SOP2 and SOP1. The original definitions of SOPn for n 3 do not

immediately lend themselves to extending the hierarchy for n = 1, 2, but the

properties we define nevertheless fulfill that role. In section 3, a connection

between this hierarchy and -maximality will be established.

Recall from [Sh 500] one of the equivalent definitions of SOP3. (The

equivalence is established in Claim 2.19 of [Sh 500]).

Definition 2.1 (1) A (complete) theory T has SOP3 iff there is an indis-

cernible sequence ai : i < and formulae (x, y), (x, y) such that

(a) {(x, y), (x, y)} is contradictory,

(b) for some sequence bj : j < we have

i j = |= [bj, ai] and i > j = |= [bj , ai],

(c) for i < j, the set {(x, aj), (x, ai)} is contradictory.

(2) NSOP3 stands for the negation of SOP3.

Definition 2.2 (1) T has SOP2 if there is a formula (x, y) which exempli-

fies this property in C = CT, which means:

There are a C for >2 such that

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(a) for every 2, the set {(x, an) : n < } is consistent,

(b) if , >2 are incomparable, {(x, a), (x, a)} is inconsistent.

(2) T has SOP1 if there is a formula (x, y) which exemplifies this in C, which

means:

There are a C, for >2 such that:

(a) for 2 the set {(x, an) : n < } is consistent.

(b) if 0 >2, then {(x, a), (x, a1)} is inconsistent.

(3) NSOP2 and NSOP1 are the negations of SOP2 and SOP1 respectively.The following Claim establishes the relative position of the properties

introduced in Definition 2.2 within the (N)SOP hierarchy.

Claim 2.3 For any complete first order theory T, we have

SOP3 = SOP2 = SOP1.

Proof of the Claim. Suppose that T is SOP3, as exemplified by ai : i < ,

bj : j < and formulae (x, y) and (x, y) (see Definition 2.1), and we

shall show that T satisfies SOP2. We define

(x, y0 y1) (x, y0) (x, y1), where lg(y0) = lg(y1).

Let us first prove the consistency of

def=

T {(x)[(x, y) (x, y)] : in>2}

n


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(i) = < < + 1 < < ,

(ii) 0 < 1.

For n2 let adef= a a . We show that C and {a

:

n2} ex-

emplify that is consistent. So, if n2 then we have

kn [b+1, ak]

as for every k n we have k < + 1, so [b+1, ak ] holds, but

also for all k n, as k , we have k > + 1, so [b+1, ak ]

holds. Hence (x)[

kn (x, ak)]. On the other hand, if l l for

l < 2, then {(x, a0), (x, a1

)} is contradictory as the conjunction implies

(x, a0 ) (x, a1 ), which is contradictory by 0 < 1 and (c) of Defini-

tion 2.1. This shows that is consistent, hence we have also shown that is consistent.

Having shown that is consistent, we can find witnesses {a : >2}

in C realising . Now we just need to show that {(x, an) : n < } is

consistent for every 2. This follows by the compactness theorem and

the definition of . Hence we have shown that SOP3 = SOP2.

The second part of the claim is obvious (and the witnesses for SOP2 can

be used for SOP1 as well). 2.3

Question 2.4 Are the implications from Claim 2.3 reversible?

Claim 2.5 If T satisfies SOP1, then T is not simple. In fact, if (x, y)

exemplifies SOP1 of T, then the same formula exemplifies that T has the

tree property.

Proof of the Claim. Let (x, y) and {a : >2} exemplify SOP1.

Then

def= {(x, a0n1) : n < }

for >2 consists of pairwise contradictory formulae. (Here 0n denotes

a sequence consisting of n zeroes.) For n < and n let

def= 0(0)+1 1 0(1)+1 . . . 0(n1)+1 1,

so >2 and = . For n let b = a . We observe

first that {(x, b k) : k < } is a set of pairwise contradictory formulae,

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for n; namely, if k0 = k1, then (x, bkl) for l < 2 are two different

elements of . On the other hand, {(x, bn) : n < } is consistent forevery . Hence (x, y) and {b :

>} exemplify that T has the

tree property, and so T is not simple. 2.5

This ends the discussion of the properties of SOP1 and SOP2 that are

directly relevant to the main thesis of the paper-the reader only interested

in the connection with the order can now turn directly to 3. The rest of

this section however contains some further syntactic developments of these

properties which are of interest if one wishes to understand the type theory

induced by them. The indescernibility results we have here were recently

used by Shelah and Usvyatsov [ShUs 844] to define a rank function on NSOP1theories (see Theorem 2.22)).

The definition of when a theory has SOP1 can be made in another equiv-

alent fashion.

Definition 2.6 Let (x, y) be a formula ofL(T). We say (x, y) has SOP1iff there is a :

>2 in CT such that

(a) {(x, an)(n) : n < } is consistent for every 2, where we use the

notation

l =

if l = 1, if l = 0

for l < 2.

(b) If 0 >2, then {(x, a), (x, a)} is inconsistent.

We say that T has property SOP1 iff some formula of L(T) has it.

Claim 2.7 (1) If(x, y) exemplifies SOP1 of T then (x, y) (hence T) has

property SOP1.

(2) If T has property SOP1 then T has SOP1.

Proof of the Claim. (1) Let {a : >2} and (x, y) exemplify that T

has SOP1. For >2 we define b

def= a1. We shall show that (x, y)

and {b : >2} exemplify SOP1.

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Given 2. Let c exemplify that item (a) from Definition 2.2(2)

holds for . Given n < , we consider [c, bn](n). If (n) = 1, then,as bn = an1 = a(n+1), we have that [c, bn]

(n) = [c, a(n+1)] holds.

If (n) = 0, then

( n) 0 = (n + 1).

As [c, a(n+1)] holds, by (b) of Definition 2.2(2), we have that [c, an1]

cannot hold, showing again that, [c, bn](n) = [c, an1] holds. This

shows that {(x, bn)(n) : n < } is consistent, as exemplified by c.

Suppose 0 >2 and that [d, b] [d, b] holds. So both

[d, a1] and [d, a1] hold. On the other hand, as 0 ,

clearly 0 1, and so (b) of Definition 2.2(2) implies that{(x, a1), (x, a1)} is contradictory, a contradiction. Hence the set

{(x, b), (x, b)} is contradictory

(2) Define first for 2 an element 2 by letting

(3k) = (k),

(3k + 1) = 0,

(3k + 2) = 1,

and iflg() = m < , then lg() = 3m. Note that for 2 and k < wehave k = (3k).

Let (x, y) and {a : >2} exemplify property SOP1. We pick c0 = c1and define for >2

b1def= a a1 c0, c1,

b0def= a0,0 a c0, c1,

bdef= c02n+2,

where ck stands for the sequence of k entries, each of which is c, and

n = lg(y) in (x, y). We define

(x, z) (x, z0 z1 w0 w1)

[w0 = w1] [(x, z0) (x, z1)],

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where z = z0 z1 w0, w1 and lg(z0) = lg(z1) = lg(y). We now claim

that (x, z) and {b : >2} exemplify that SOP1 holds for T. Before westart checking this, note that for >2 we have:

1 (d, b) holds for any d,

2 (d, b0) holds iff(d, a0,0) (d, a) holds,

3 (d, b1) holds iff(d, a1) (d, a) holds.

Let us verify 2.2(2)(a), so let 2. Pick c such that [c, an](n) holds

for all n < . We claim that

[c, bn] holds for all n. ()

The proof is by a case analysis of n.

Ifn = 0, this is trivially true. If n = k + 1 and (k) = 0, then we need to

verify that [c, ak0,0] holds and [c, ak ] holds. We have

k 0, 0 = (3k + 2),

and (3k + 2) = 1. Hence [c, ak0,0] holds by the choice of c. On

the other hand, we have k =

(3k), and (3k) = (k) = 0, hence[c, ak ] holds.

If n = k + 1 and (k) = 1, then we need to verify that [c, ak ] holds

while [c, a(3k)1] does not. As k = (3k), and (3k) = (k) = 1,

we have that [c, ak] holds. Note that [c, a(3k+2)] holds as (3k+2) = 1.

We also have ( (3k + 1)) 0 (3k + 2). Hence [c, a(3k+1)]

by part (b) in Definition 2.6. But

[c, a(3k+1)] [c, a(3k)1] [c, ak1]

holds, so we are done proving ().Let us now verify 2.2(2)(b). So suppose 0 and consider

{(x, b1), (x, b)}. Let and l be such that = l.

Case 1. = .

Hence l = 0. So (x, b) = (x, a) and (x, b1) = (x, a ),

by 2 and 3 respectively, showing that {(x, b), (x, b1)} is inconsistent.

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Case 2. and l = 0.

Hence 0 . Clearly 0 0, 0, as

(lg()) = (lg()) = 0.

We have (x, b1) = (x, a ) by 3 and (x, b) = (x, b 0)

implies (x, a0,0) by 2, while the two formulae being implied are con-

tradictory, by (b) in the definition of SOP1.

Case 3. and l = 1.

Observe that (x, b) = (x, a) by 3 and (x, b1) = (x, a ).

As above, using 0 , we show that the set {(x, a ), (x, a)} is

inconsistent. 2.7

Conclusion 2.8 T has SOP1 iff T has property SOP1 from Claim 2.7.

Question 2.9 Is the conclusion of 2.8 true when the theory T is replaced

by a formula ?

Start changes

It turns out that witnesses to being SOP1 can be chosen to be highly

indiscernible.

Definition 2.10 (1) Given an ordinal and sequences l = l0,

l1, . . . ,

lnl

for l = 0, 1 of members of >2, we say that 0 1 1 iff

(a) n0 = n1,

(b) the truth values of

lk = ,

lk1 lk2 lk3

lk4

do not depend on l,

(c) lk1 lk2

= 0k1(lg(0k1

0k2)) = 1k1

(lg(1k1 1k2

))

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for k1, k2, k3, k4 n0.

(2) We say that the sequence a : >2 of elements ofC (for an ordinal )is 1-fully binary tree indiscernible (1-fbti) iff whenever 0 1 1 are sequencesof elements of >2, then

a0def= a00 . . . a0n0

and the similarly defined a1 , realise the same type in C.

(3) Suppose that is a limit ordinal > 0. Define h = h :>2 >2 by

letting for >2

lg(h

()) = 2lg() + 1,

i < lg(h()) = h()(2i) = 0, h()(2i + 1) = (i),

h()(2lg()) = 1.

For n < and n(>2) we define h() = h(l) : l < n.

We say 2 iff h() 1 h(). We define 2-fbti like 1-fbti but using2 in place of1.

Observation 2.10 A The following can be easily checked:

(1) Let , n(>2) and let and be the closures of , , respectively,under intersections. Then 1 iff 1 .(2) Ifa :

>2 is 1-fbti then ah() : >2 is 2-fbti.

(3) h() is never , and h(0) is never a strict initial segment of h(1).

Claim 2.11 If t {1, 2} and b : >2 are of given constant length,

and is a (limit for t = 2) ordinal, then we can find a : >2 such

that

(a) a : >2 is t-fbti,

(b) if = m : m < n, where each m >2, is given, then we can find

m >2 (m < n) such that with def= m : m < n, we have t

and sequences a and b realise the same type in C.

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Proof of the Claim.2 Let us first deal with t = 1. By Observation 2.10 A

(1) above, we may reduce to checking clause (b) only for tuples b where is closed under intersections. By Compactness Theorem it suffices to assume

= . The proof goes through a series of steps through which we obtain in-

creasing degrees of indiscernibility. We shall need some auxiliary definitions.

Let be an infinite ordinal.

Definition 2.12 (1) Given = 0, . . . , k1, a sequence of elements of>2,

and an ordinal . We define = cl() as follows:

= , 0, 0 , 1, 1 , 0 1, 2, 2 , 0 2, 1 2 . . ..

We also define u[] = {l : lg(l) > }.

(2) We say that ,n iff def= cl() and

def= cl() satisfy

(i) = l : l < m and = l : l < m are both in

m(>2) for some m,

(ii) for l < m we have l 2 l

2, and for such l we have

l = l ,

(iii) n |u[]|,

(iv) l, k u[] = [lg(l) < lg(k) lg(l ) < lg(k)],

(v) l1 l2

l1 l2

, and the same holds for the equality,

(vi) ifl1 is not an initial segment ofl2

, then l1(lg(l1

l2)) = l1

(lg(l1l2

)).

(3) a : >2 is (, n)-indiscernible iff for every k, for every , k(>2)

with ,n , the tuples a and a realise the same type.(4) ( , n)-indiscernibilityis the conjunction of (, n)-indiscernibility for all

.

(5) We say that a : >

2 is 0-fbti iff it is (, n)-indiscernible for all and n.

2Note that the definition of1,2 has changed from the one given in the publishedversion of this paper, but the following proof is basically the same as the one there.

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Note 2.12 A (1) cl0() is simply the closure of under intersections,

joined with in appropriate places.(2) ,n iff cl() ,n cl().

Subclaim 2.13 If a kC for >2 are tuples of constant length and

closed under intersections, then

for any we can find a = a : >2 such that

(x) a is 0-fbti,

(xx) for every m and a finite set of formulae, we can find h : m2 >2

such that

() a : m2 and ah() :

m2 realise the same -type,

() h satisfies h() l h( l) for m>2 and l < 2, and

lg(1) = lg(2) = lg(h(1)) = lg(h(2)).

Proof of the Subclaim. By Compactness Theorem it suffices to work with

= .

Let (),n be the conjunction of the statement (x),n given by

a is ( , n)-indiscernible,

and (xx) above. We prove by induction on n and then that for any

we can find a for which (),n holds.

n = 0. We use a = a.

n + 1. By induction on , we prove that there is a for which

(),n+1 + (),n + (xx) holds.

< .

Without loss of generality, the sequence a : >2 is ( , n)-indiscernible,

as (xx) as a relation between a : >2 and a : >2 is transitive.

Suppose we are given , satisfying ,n+1 . By Note 2.1.2 A, we mayassume , to be the same as their cl closures and the same will hold for

any , that we mention in this context.

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If |u[]| n, the conclusion follows by the assumptions. We shall as-

sume |u[]| > n, so |u[]| = n+1. Moreover, if min(u[]) = min(u[])and for any l with lg(l ) = min(u[

]) we have l = l , then using

(x)min(u []),n, we get that a and a realise the same type. By the same ar-

gument, fixing a finite set of formulae, for every , we get that the tp(a)

depends just on the

/ ,n+1def= and {l : l < lg()}

min(u [])2 = {l : l v}

for some v lg(). Let us define F0, by F0

,(l : l v) = tp(a).

By the closure properties of and the definition of,n+1, we get that forl1 = l2 v

the truth value of l1 (+ 1) = l2 (+ 1) depends only on

. We can hence replace v by a set v v such that l : l v are the

representatives under the equality of the restrictions to + 1.

As we have fixed , there is a finite set A of s that can be used as

representatives for the values of F0,. Let r be the size of the range ofF0

,.

Let k = 2+1 (so finite) and let {k : k < k} list +12. We define a function

F, onk(>2) by letting

F,(x0, . . . , xk, . . .)k


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all nodes on any common level of S are mapped by f to a common

level of T.

(2) A nonempty subtree of > is well-behavedif it is finitely branching

and has no maximal nodes (hence it has levels).

Halpern-Lauchli theorem Let r, d < . Suppose that Ti : i < d

are well-behaved trees and that c is a colouring of

n2 >2 such that

h 2 is the identity,

lg(h()) depends just on lg() (not on ),

h() l h( l) for l = 0, 1,

for some c we have that for all m <

{k : k < k} m2 = F((h(0), h(1), . . . , h(k), . . .)k


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Let a for >2 be defined to be a if

>2, and otherwise ah() for the

unique such that (+ 1) = k and = k .We have obtained the desired conclusion, but localized to . The induc-

tion step ends by an application of the compactness theorem.

= This is vacuously true.

Having carried the induction, the conclusion of the Subclaim follows fromn()0,n. 2.13

Now we go back to the proof of the Claim. Given b : >2 as in the

assumptions, by the Subclaim we can assume that they are 0-fbti. We chooseby induction on n a function hn :

n2 >2 as follows. Let h0() = . If

hn is defined, let

kndef= max{lg(hn()) + 1 :

n2}

and let

hn+1() = , hn+1(1 ) = 1 hn(), hn+1(0 ) = 0, . . . , 0hn(),

where the sequence of 0s in the last part of the definition has length kn. The

point of the definition of hn is that if l = l0, . . . ,

lnl

for l = 0, 1 are given

and n = lg(cl0(0)), then

0 1 1 = hn(

00), . . . , hn(

0n0

) 0,n hn(10), . . . , hn(

1n1

).

To check this, we verify the six relevant items of the definition of0,n .

(i) Follows because n0 = n1 by the definition of1.

(ii) Ifhn(0i )hn(

0j ) = then

0i

0j = so

1i

1j = by the definition

of1, and hence hn(1i )hn(1j ) = . The opposite implication holdsby symmetry.

(iii) Follows by the definition of n.

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(iv) Suppose

0 < lg(hn(0i ) hn(

0j )) < lg(hn(

0k) hn(

0s )).

Let m n be the first such that

0 < lg(hn(0i m) hn(

0j m)) < lg(hn(

0k m) hn(

0s m)).

Clearly, m > 0. To simplify the notation, let us assume that m = n.

Let 0t = lt 0t for t {i,j,k,s} and for some lt {0, 1} depend-

ing on t. The situation we describe can happen iff li = lj = 1 and

lk = ls = 0, by the definition of hn. By the definition of1 this can berecognised by the 1-type of 0.

(v), (vi) Follow because the corresponding properties are preserved by hn.

Fix an n < and define a = bhn() for n2. By the above argument

it follows that a : n2 are 1-fbti. As n was arbitrary, we can finish by

compactness.

For t = 2, we use exactly the same proof. 2.11

The following Theorem 2.15 will finally tell us that witnesses for SOP1can be chosen with a certain degree of indiscernability. We need to introduce

a new notion of indiscernability:

Definition 2.14 (1) Given an ordinal and sequences l = l0, l1, . . . ,

lnl

for l = 0, 1 of members of >2, we say that 0 3 1 iff

(a) n0 = n1,

(b) the truth values of

lk = , lk1

lk2 lk3

lk4

do not depend on l,

(c) lk1 lk2

= 0k1(lg(0k1

0k2)) = 1k1

(lg(1k1 1k2

)),

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(d) lk1 lk2

= 0k1 = (0k1

0k2) 1 iff 1k1

= (1k1 1k2

) 1.

for k1, k2, k3, k4 n0.

(2) We say that the sequence a : >2 of elements ofC (for an ordinal )

is 3-fully binary tree indiscernible (3-fbti) iff whenever 0 1 1 are sequencesof elements of >2, then

a0def= a00 . . . a0n0

and the similarly defined a1 , realise the same type in C.

Theorem 2.15 Suppose that T has SOP1 as witnessed by (x, y) and a

sequence a = a : >2. Then there is d = d : >2 exemplifying

that (x, y) has SOP1 and d : >2 \ {} is 3-fbti.

Proof. Let k = lg(y). First define b for >2 by b = a0 a1.

Let for any z0, z1 of length k and l {0, 1}, l(x, z0 z1) (x, zl).

Now we use Claim 2.11 applied to b : >2. Therefore we can find

c = c : >2 such that

(a) c is 1-fbti,

(b) for any finite n and n

(>

2) there is n

(>

2) such that 1 andb and c realise the same type in C.

Let d for k

(>2) be defined by induction on the length of so that

d0 d1 = c and d = c. This is possible by the choice of b and c.

Claim 2.16 If 0 then (x, d) and (x, d1) are incompatible.

Proof of the Claim. Let = l for some l {0, 1}. Consider

{l(x, c), 1(x, c)}, we claim that this set is inconsistent. We know that

l(x, c) (x, dl) (x, d), 1(x, c) (x, d1).

By the 1-fbti property of c and the choice of c with respect to b it suf-

fices to check that {l(x, b), 1(x, b)} is inconsistent. This means that

{(x, a), (x, a1} is inconsistent, which is true by the choice of a. 2.16

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Claim 2.17 For any 2, {(x, dn) : n < } is consistent.

Proof of the Claim. It suffices to show that for any

0 1 . . . k

the set {(x, dl+1lg(l)l+1(lg(l)) : l > k} {(x, d)} is consistent. This

means {l+1(lg(l))(x, cl+1lg(l)) : l < k} {(x, c)} is consistent. By the

choice of b and c this is to say {l+1(lg(l))(x, bl+1lg(l)) : l < k} {(x, a)}

or {(x, al+1lg(l)) : l < k} {(x, a)} is consistent, but this is true by the

choice of a. 2.17

Claim 2.18 d : 2 \ {0} is 3-fbti.

Proof of the Claim. Suppose that 0 3 1 and consider d0 and d1 . Foreach kl let

lk be such that

lk =

lk m

lk for some m

lk {0, 1} and let 0, 1

be defined from lk(l {0, 1}, k < lg(0)). Then 0 3 1 = 0 3 1,hence c0 and c1 realise the same type, which implies that d0 and d1 do.

2.18

2.15

End changes.

As we mentioned before, it would be really interesting to know if SOP2and SOP1 are equivalent. A step towards understanding this question is

provided by the next claim which shows that in the case of theories which

are SOP1 and NSOP2, the witnesses to being SOP1 can be chosen to be

particularly nice. note a change here to 3-fbti from the old version

Claim 2.19 Suppose that (x, y) satisfies SOP1, but for no n does the for-

mula n(x, y0, . . . , yn1) k2 for (x, y) satisfying SOP1 which in addition satisfy:

(c) if X >2, and there are no , X such that 0 , then

{(x, a) : X} is consistent.

(d) a : >2 is 3-fbti.

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(In particular, such a formula and witnesses can be found for any theory

satisfying SOP1 and NSOP2.)

Proof of the Claim. We shall be using the following colouring theorem,

for which we could not find a specific reference and so we include a proof of

it.

Lemma 2.20 Suppose cf() = and we colour >2 by < colours. Then

there is an embedding h : >2 >2 such that h()l h(l) and

Rang(h) is monochromatic.

Proof of the Lemma. Let c be a colouring as in the assumptions and let

{ai : i < } list Rang(c). We claim that there is >2 and j < such

that for every >2 satisfying there is >2 with and

c() = j. For otherwise, we can choose by induction on i a member

i >2 with i < j = i j such that for no

2 do we have i+1

and c() = i, using < cf(). As < , we obtain a contradiction.

Having found such , j we define h() for n2 by induction on n < .

For n = 0 we choose h() to satisfy h() and c(h() = j, which is

possible by the choice of and j. For n + 1, for any n+12 we choose for

l = 0, 1 a member h( l) of >2 which is above h() l and on which

c is j, which again is possible by the choice of and j. 2.20

Let (x, y) be a SOP1 formula which is not SOP2, and moreover assume

that for no n does the formula n defined as above satisfy SOP2. By Theorem

2.15, we can find witnesses a : >2 which are 3-fbti. By the compact-

ness theorem, we can assume that we have a 1-fbti sequence a : 1>2

with the properties corresponding to (a) and (b) of Definition 2.2(2), namely

(a) for every 12, the set {(x, a) : < 1} is consistent,

(b) if 0 1

>

2, then {(x, a1), (x, a)} is inconsistent.

We shall now attempt to choose and w for 1>2, by induction on

lg() = < 1 so that:

(i) 1>2,

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(ii) < = ,

(iii) < = (lg()) = (),

(iv) w 1>2 is finite and w = lg() < lg(),

(v) if lg() is a limit ordinal > 0, then w = ,

(vi) if 2 and l < 2, then wl { 1>2 : l } and

max{lg() : wl} < lg(l),

(vii) for each there is = such that

() 12,

() |{ < 1 : () = 1}| = 1,

() letting

p(x)def= {(x, a) : w for some lg()},

we have that for all large enough , the set

p(x) {(x, a) : > () = 1}

is consistent,

(viii) p(x) {(x, a) : w0 w1} is inconsistent.

Before proceeding, we make several remarks about this definition. Firstly,

requirements (vii) and (viii) taken together imply that for each 1>2 we

have that w0w1 = . Secondly, the definition ofwl for l {0, 1}

implies that

l=0,1l wl = 0 1.

Thirdly, in (vii), any which satisfies that and |{ : () = 1}| = 1can be chosen as , by indiscernibility.

Now let us assume that a choice as above is possible, and we have made

it. Hence for each 1>2 there is a finite q p such that

q(x) {(x, a) : w0 w1} ()

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is inconsistent. Notice that there are q and 12 such that

(1)[ 1

1>2 = (2 1>2)(1 2 q2 = q)].

Namely, otherwise, we would have the following: each p is countable, hence

for every there is g() with g() 1>2 and

g() 1 = q1 p.

Let 0def= , and for n < let n+1 = g(n). Let

def= n2} such that

(x, b)

q

{(x, a) : w}.

We claim that (x, y) and b : >2 exemplify SOP2 of T, which

is then a contradiction (noting that is a formula of the form n for some

n, where n was defined in the statement of the Claim). We check the two

properties from Definition 2.2(1).

To see (a), let 2 be given. We have that p is consistent, and q p.

For n < , we have

(x, bn) q {(x, a) : wn}.As this is a conjunction of a set of formulae each of which is from p, we have

that {(x, bn) : n < } is consistent. To check (b), suppose >2.

Let n be such that n = n but (n) = (n). Hence

(x, b)

q

{(x, a) : wn(n)}

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and

(x, b)

q

{(x, a) : wn(n)},

so taken together, the two are contradictory by the choice of q.

We conclude that the choice of and w cannot be carried throughout

1>2. So, there is < 1 and 2 such that , wl, l for l < 2

cannot be chosen, and is the first ordinal for which there is such . Let

0 1>2 2 : } is finite and

for some < 1 the set

p {(x, a) : () = 1 & [, 1)}

{(x, a) : w}

is consistent

.

We make several observations:

(0) If (, , w) and w w with w finite and w \ w is contained in

{ : () = 1}, then (, , w) .

[This is obvious.]

(1) If (l, , wl) and for some 1>2 with 0 we have l l

for l < 2, while 0 and 1 are eventually equal, then (l, lg(), w0 w1) .

[Why? We have wl { 1>2 : l } is finite, so clearly

w0 w1 {

1>

2 :

} is finite. By the assumption, we have thatfor some l < 1 for l < 2

p {(x, al) : > l l() = 1} {(x, a) : wl}

is consistent. Suppose that (1) is not true with l = 0 and let max{0, 1}

be such that < 1 and for > the equality 0() = 1() holds. Hence

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we have that

p {(x, a0) : > 0() = 1} {(x, a) : w0 w1}

is inconsistent. By increasing w0 if necessary, (0) implies that

p {(x, a) : w0 w1}

is inconsistent. Let def= , for l < 2 let wl = wl, and let l

def= l

l

for a large enough l so that < l and max({lg() : wl}) <

l .

This choice shows that we could have chosen , wl as required, contra-

dicting the choice of .](2) If 0

12 for some such that there are 1 many < 1 with

() = 1, and lg(0 ) < 1, then (,, ) .

[Why? By the choice of p and the remark about the freedom in the

choice of that we made earlier.]

Now we use the choice of to define witnesses to T being SOP1 which

also satisfy the requirements of the Claim. For >2, let bdef= a0. Let

us check the required properties. Properties (a),(b) and (d) follow from the

choice of{a : 1>2}. Let X >2 be such that there are no , X

with 0 , we need to show that {(x, b

) : X} is consistent.

It suffices to show the same holds when X replaced by an arbitrary finite

X X. Fix such an X. Clearly, it suffices to show that for some , ,

letting w = {0 : X}, we have (, , w) .

Let 12 be such that 0 and () = 1 for 1 many . By

induction on ndef= |X| we show:

there is 12 such that for some max{lg() : w}, we have

(, , w) and > = () = (), while () = 1.

n = 0. Follows by observation (2) above.

n = 1. Let X = {} and = lg() + lg(0). Let 12 be such that

= 0 , () = 1 and > = () = (). By observation

(2) above, we have that (,, ) . Then, by observation (0), we have

(, , w) .

n = k + 1 2. Case 1. w is linearly ordered by .

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Let w be of maximal length, so clearly w\{} = 1 .

Let 12 be such that 1 and > lg(), while () = ().Now continue as in the case n = 1.

Case 2. Not Case 1.

Let 1>2 be -maximal such that ()( w = ). This

is well defined, as w = is finite. Let wldef= { w : l }, so

w0 w1 = but neither of w0, w1 is empty. Now we have that / w, as

otherwise we could choose w0 such that 0 , obtaining an easy

contradiction with our assumptions on X. Hence w = w0 w1. We can now

use observation (1) and the inductive hypothesis. 2.19

To complete this discussion of the syntactic properties (N)SOP1, 2 we

shall quote a result from [ShUs 844] in which the understanding of SOP1and the witnesses for SOP1 developed here was used to show that NSOP1theories admit a rank function.

Definition 2.21 Given (partial) types p(x), q(y) and a formula (x, y). By

induction on n < we define when

rk1(x,y)(p(x), q(y)) n.

n = 0. This happens iff both p(x) and q(y) are consistent.

n + 1. The rank is n + 1 iff for some c realising q(y) both

rk1(x,y)(p(x) {(x, c)}, q(y)) n

and

rk1(x,y)(p(x), q(y) {(x)((x, y) (x, c))}) n.

If the rank is n for all n then we say it is inifinite, otherwise we say it is

finite.

Theorem 2.22 (Shelah-Usvyatsov [ShUs 844]) A theory T is NSOP1 iff

rk1(x,y)(x = x, y = y) <

for every formula (x, y).

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3 -maximality revisited

In this section we come back to our main thesis, which is that properties

SOP2 and the maximality in the -order are closely connected.

Our main proof will use two auxiliary notions. The first is the order ,

which is a version of the -order.

Definition 3.1 (1) For (complete first order theories) T1, T2 and a regular

cardinal > |T1|, |T2|, let T1 T2 mean:

There is a -relevant (T1, T2)-superior (T, , ) (see Definition 1.2) such

that T has Skolem functions and if T T is complete with |T| <

then() there is a model M of T of size and an M[]-type p omitted by

M such that for every elementary extension N of M of size which omits p

and a type q (in one variable) over N[], there is an elementary extension of

N of size which realises q and omits p.

(2) Let T1 T2 mean that T1

T2 holds for all large enough regular .

(3) T1 is said to be -maximal iff there is no T2 such that T1

T2.

Similarly for .

The connection between this notion and is given by the following

claim:

Claim 3.2 Suppose that T1, T2 are theories and > |T1|, |T2| satisfies 2 = +.

Then

T1 + T2 = (T2

T1).

Proof. This statement is just a reformulation of the beginning of the proof

of Theorem 1.17. In other words, let (T, 1, 2) show that T1 + T2. This

means that |T| < + but since |T1|, |T2| we may assume

that |T| < . Namely since there is a consistent theory T 1 2 in

which l interprets Tl, and each Tl has size < , there is a consistent theory

T of size < which does the same. Without loss of generality T T. In

particular |(T)| < so by extending T to a complete subtheory of T and

renaming we may assume T is complete. Any model M of T has a reduct

N that is a model of T and that satisfies M[] = N[] and similarly for .

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Hence (T, , ) is a -relevant (T1, T2)-superior that exemplifies T1 + T2,

so by renaming we may assume |T| < .Suppose for contradiction that T2

T1 and let (T

, , ) exemplify this.

Without loss of generality, 1 = and 1 = and the common vocabulary

of T and T is ( 1) ( 2). Hence T = T T is consistent by Robinson

Consistency Criterium. Without loss of generality T is complete. Hence

let M be a model of T of size and p be a M[] type omitted by M

exemplifying the definition of . Using the assumption 2 = + we can

build by induction an elementary extension N of M with |N| = +, with N

omitting p and being -saturated. This is a contradiction with the choice of

T. 3.2

Corollary 3.3 Suppose that for all large enough regular we have 2 = +.

Then any -maximal theory is also -maximal.

Proof. Suppose otherwise and let T exemplify this. Hence for every there

is regular > such that T is not -maximal and 2 = +. Hence T is

not +

-maximal by Claim 3.2, a contradiction. 3.3

The next notion we need is a syntactic property.

Definition 3.4 Let T be a theory.

(1) For a formula (x, y) we say that (x, y) has SOP2 iff for some [by

compactness equivalently all] regular > |T| there is a sequence

e : = 0, . . . n1, 0 1 . . . n1

> and lg(i) a successor

such that

() for each , the set(x, e) : = (0 + 1), (1 + 1), . . . (n1 + 1)

and 0 < 1 < . . . n1 <

is consistent

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() for every large enough m, if g : nm > satisfies

= g() g()

and

nm = lg(g()) is a successor,

while for l < n 1

(g()) l g( l),

then

{(x, eg(1),g(2),...g()) : n

m}

is inconsistent. Here n = lg(y) in (x, y).

(2) T is said to have SOP2 iff some (x, y) exemplifies it.

Our theorem 3.6 is phrased in terms of SOP 2. Answering a question from

an earlier version of this paper Shelah and Usvyatsov proved in [ShUs 844]

the following Theorem 3.5, which then can be used together with theorem

3.6 to prove Corollary 3.9 which states that -maximality implies SOP2.

Theorem 3.5 (Shelah-Usvyatsov [ShUs 844]) For any theory T, T has SOP2iff it has SOP2.

Main Theorem 3.6 For any theory T and regular cardinal > |T|, ifT is

-maximal then T has SOP2 .

Proof. Let T be a given theory and let = cf() > |T|. We shall assume

that T is -maximal and prove that T has SOP2. To make the reading of

the proof easier we shall break it into stages.

Stage A. Let Tntreedef= Th(n2,


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seen to exist), and let , be the interpretations ofTtree and T in T, respec-

tively. We can without loss of generality, by renaming if necessary, assumethat L(T) L(T), so the interpretation is trivial.

As |T|, |T| < , we can find A which codes T and T. Working in

L[A], we shall define a model M of T of size as follows. Let

def= T {=(x, x) :

>}

{x }

{(x )}.

By a compactness argument and the fact that interprets Ttree in T, we see

that is consistent. Let M be a model of of size = let a be the realisation of x in M. For , let

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Hence the interpretation of Ttree in T satisfies the same statement

claimed about Ttree. We conclude: if M N and p is not realised in N, then there is no (x, c) with

c N such that (a, c) for all < holds and every two elements of N

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extension ofM in which q is realised by a. As T has Skolem functions, we

have M N. Let N1 be the submodel of N with universe

Adef= M

iE

{(a, b) : b Mi and a term of T}.

Note that the size ofN1 is . Clearly, N1 is closed under the functions ofT,

so M N1 N. As T has Skolem functions, we get that M N1 N.

By the third part of the definition of E, p is omitted in N1. This is in

contradiction with our assumptions, as a N1 realises q(x).

Hence we can conclude

for every club E of , the set E is inconsistent.

Stage D. Now we start our search for a formula that exemplifies that T

has SOP2. In the following definitions, we shall use the expression an almost

branch or the abbreviation a.b. to stand for a set linearly ordered by


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choosing n1, while chooses n1 (n1, ) such that bn1 lg(zn1)Mn1 .

Player wins this game iff for some e lg(t)M we have

(x, e) q and M |= (x)[(x, e) = (x, a0 , . . . , an1,k


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there is no loss of generality in assuming that bj,l = bj for all l < lj for j < n.

We shall omit the subscript 0 from,, e. Let n = n2 and let us definej(x, yj, zj) for j < n by

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Lemma 3.8 There are sequences

N : >, h :

>

such that

(i) h is an elementary embedding of Mlg() into CT with range N,

(ii) = h h,

(iii) for = < and > we have

h(alg()) h(alg()),

(iv) N0 N1 = N01 for all 0, 1.

Proof of the Lemma. This Lemma has the same proof as that of Main

Claim 1.13 Stage B. In the notation of that proof, ignore bi . When defining

use

=

byusing {x : < } in place of {x0, x1}. Assumptions on 0 , +2 and

3 are

analogous to the ones we made in that proof. Fact 1.16 still holds, except

that we drop the last set from the definition of r(x). The rest of the proof

is the same, recalling that the branch induced by {ai : i < } is undefinable

in M. 3.8

Stage F. For , let hdef=

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be as in the statement of (). For nm let egdef= eg(1),...g() (note that

this is always defined). We shall now show that the set

{(x, eg) : nm}

is inconsistent. Suppose otherwise, so let d CT realise it. For each nm,

let g() and let

def= 0, . . . ,

n1 satisfy lg(g( k)) =

k + 1

for k n, so for each k < n we have g( (k + 1)) = (k + 1). Now we

have that for each nm

(i) (x, eg) (x, h(n1+1)(e)) q (x)

(ii) N |= (x)[(x, eg) = (x, h(a1 ), . . . h(a ),

j


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For each nm and k < n we have that b(k+1)

Mk+1 (by the

choice of E), so in particular bj

Mj1

+1, and hence hs (bj

) is

a fixed b. By the choice of d and definitions of j, l and T , there are

s = t < lj l such that hs (as(j

+1) ) and ht (at(j+1) ) are on the

same almost branch. Now note that for all we have

a

(j+1) M

(j+1)+1

\ M

(j+1)

and (j+1)

> j. Hence hs (as(j+1) ) and ht(at(j

+1) ) are incom-

parable, by property (iii) in Lemma 3.8, a contradiction. This shows ()

from the definition of SOP2, so finishing the proof. 3.6

Putting this together with Corollary 3.3 and Shelah-Usvyatsov theorem

3.5 above we get the following corollary 3.9.

Corollary 3.9 (1) Suppose that T is a theory that is -maximal in some

universe of set theory in which 2 = + holds for all large enough regular .

Then T has SOP2.

(2) Suppose that T is a theory that is +-maximal in some universe of set

theory in which is regular and 2 = +. Then T has SOP2.

Proof. (1) Let W be a universe of set theory in which 2 = + holds for all

large enough regular and in which T is -maximal. Hence by Corollary

3.3 T is -maximal in W and hence by Main Theorem 3.6 in W it satisfies

SOP2. By Shelah-Usvyatsov Theorem 3.5 above T satisfies SOP2 in W.

An application of the Compactness Theorem shows that satisfying SOP2 is

absolute, hence T satisfies SOP2 in V.

(2) This follows similarly, but more directly, from Main Theorem 3.6 and the

Shelah-Usvyatsov Theorem 3.5. 3.9

This section hence provides us with the proof of one side of our thesis thatSOP2 and

-maximality are closely connected. Recall that Shelah proved in

[Sh 500] that SOP3 implies -maximality. So an important open question

(provided that SOP3 are not actually equivalent, which we still do not know)

is

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Question 3.10 Does SOP2 imply -maximality?

In a partial answer to this question posed in an earlier version of the

paper Shelan and Usvyatsov in Theorem 3.12 of [ShUs 844] provided a local

positive answer to this question, where by local we mean that they proved

that any theory with SOP2 is above Ttree when only types localised by a

certain formula are considered (see Definition 1.3).

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References

[ChKe] C.C Chang and H. J. Keisler, Model Theory, 3rd ed., Studies in

Logic and Foundations of Mathematics, vol. 73, North-Holland,

1990.

[DjSh 614] M. Dzamonja and S. Shelah, On the existence of universals and

an application to Banach spaces and triangle free graphs, submit-

ted.

[DjSh 659] M. Dzamonja and S. Shelah, Universal graphs at the successor of

a singular cardinal, Journal of Symbolic Logic, 68 (2) (June 2003),pg. 366-387.

[DjSh 710] M. Dzamonja and S. Shelah, On properties of theories which pre-

clude the existence of universal models, submitted.

[GrSh 174] R. Grossberg and S. Shelah, On universal locally finite groups,

Israel Journal of Mathematics, 44 (1983), 289-302.

[HaLa] J.D. Halpern, H. Lauchli, A Partition Theorem, Transactions of

the American Mathematical Society, vol. 124, Issue 2 (Aug. 1966),

pg. 360-367.

[Ke] H. J. Keisler, Six classes of theories, Journal of Australian Math-

ematical Society 21 (1976), 257-275.

[Ki] B. Kim and A. Pillay, Simple theories, Annals of Pure and Applied

Logic, 88 (2-3) (1997), 149-164 .

[KjSh 409] M. Kojman and S. Shelah, Nonexistence of universal orders in

many cardinals, Journal of Symbolic Logic 57 (1992), 875-891.

[KjSh 447] M. Kojman and S. Shelah, The universality spectrum of stableunsuperstable theories, Annals of Pure and Applied Logic, 1992.

[Sh c] S. Shelah, Classification theory and the number of nonisomorphic

models, vol. 92 of Studies in Logic and the Foundation of Mathe-

matics, North-Holland, Amsterdam, 1990.

58


59/59

n:2011-03-22

modified:2011-05-23

[Sh 457] S. Shelah, The Universality Spectrum: Consistency for more

classes in Combinatorics, Paul Erdos is Eighty, Vol. 1, 403-420,Bolyai Society Mathematical Studies, 1993, Proceedings of the

Meeting in honor of P. Erdos, Keszthely, Hungary 7. 1993, an

improved version available at http://www.math.rutgers.edu/ she-

laharch

[Sh 500] S. Shelah, Toward classifying unstable theories, Annals of Pure

and Applied Logic 80 (1996) 229-255.

[ShUs 844] S. Shelah and A. Usvyatsov, More on SOP1 and SOP2, Annals

of Pure and Applied Logic 155 (2008) 16-31.

[PiHa] D. Pincus and J.D. Halpern, Partitions of Products, Transactions

of the American Mathematical Society 267 no. 2 (1981) 549-568.

Documents

Mirna Dzamonja and Saharon Shelah- On ◅^*-Maximality