M.C. Laskowski and S. Shelah- The Karp Complexity of Unstable Classes

8/3/2019 M.C. Laskowski and S. Shelah- The Karp Complexity of Unstable Classes

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THE KARP COMPLEXITY OF UNSTABLE CLASSES

M. C. LASKOWSKI AND S. SHELAH

Abstract. A class K of structures is controlled if, for all cardi-nals , the relation ofL,-equivalence partitions K into a set ofequivalence classes (as opposed to a proper class). We prove thatthe class of doubly transitive linear orders is controlled, while anypseudo-elementary class with the -independence property is notcontrolled.

1. Introduction

One of the major accomplishments of model theory has been the dis-covery of a dividing line between those theories in a countable languagewhose models can be described up to isomorphism by a reasonable setof invariants and those whose models cannot be so described. Modelsofclassifiable theories are described up to isomorphism by an indepen-dent tree of countable elementary submodels, while the isomorphismtype of any unclassifiable theory cannot be described by any reason-

able set of invariants (see [9]). Unfortunately, the great majority ofclasses of structures studied in mathematics are unstable, and thus fallon the non-structure side of this divide. Thus, it is desirable to searchfor dividing lines between unstable classes of structures. Our thesis isthat while an unstable (pseudo-elementary) class necessarily has themaximal number of non-isomorphic models in every uncountable car-dinality, it is still possible to assign a set of invariants to some unstableclasses of structures. In some cases (see e.g., Example 3.6) the largenumber of non-isomorphic models is due simply to our ability to codearbitrary stationary sets into the skeletons of Ehrenfeucht-Mostowskimodels. In other words, for some classes of structures the reason for

the non-isomorphism of two structures in the class need not be veryrobust. Indeed, in such cases the structures can be forced to be iso-morphic by a forcing that merely adds a new closed, unbounded subset

Date: October 6, 2003.1991 Mathematics Subject Classification. 03C.Partially supported by NSF Research Grants DMS 9403701 and DMS 9704364.The authors thank the U.S.-Israel Binational Science Foundation for its support

of this project. This is item 560 in Shelahs bibliography.1


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2 M. C. LASKOWSKI AND S. SHELAH

of some cardinal to the universe. That is, although they are noniso-morphic, the structures are not very different from each other. On the

other hand, for other classes of structures (see Theorem 2.5) there aremore serious obstructions to a structure theorem.

Our ultimate goal is to determine to which unstable classes of struc-tures one can associate a reasonable set of structural invariants. Theseinvariants need not (and typically will not) determine the structuresup to isomorphism. Instead, we ask that any two structures with thesame invariants be very much the same. In this paper we focus onL,-equivalence for various cardinals and ask which unstable classesare partitioned into only a set of equivalence classes (as opposed to aproper class). We call a class K controlled if K has only a set ofL,-equivalence classes for all cardinals . Typically, L

,-equivalence does

not characterize models up to isomorphism even when we fix the cardi-nalities of the models. (In [8] the second author shows that for any un-stable pseudo-elementary class and any uncountable regular cardinal ,there are 2 non-isomorphic models of size that are L,-equivalent.)However, in some sense two L,-equivalent structures of the samecardinality are very much the same. For instance, if one uses the back-and-forth system witnessing their equivalence as a notion of forcing,then the two structures will become isomorphic in the correspondingforcing extension.

In this paper we obtain two complementary results. On one hand,

in Section 3 we analyze the pseudo-elementary class K2tr of doublytransitive linear orders. This class is unstable, hence the stigma ofnon-structure applies. Despite this, we prove that KC+(K2tr) (see Definition 2.3) for all uncountable cardinals , hence K2tr iscontrolled. This is one of very few theorems in which an unstablepseudo-elementary class shows any sign of structure. On the otherhand, in Section 4 we prove that any pseudo-elementary class with the-independence property (see Definition 4.4) is not controlled. In fact,if the language used in describing K is countable then KC(K) = for all cardinals 3.

There is still much that we do not know about the notion of control.

A fundamental question that remains open is whether there is an un-stable elementary class that is controlled. We conjecture, and hope toprove, that any pseudo-elementary class with the independence prop-erty is not controlled; this would substantially strengthen our secondresult.


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THE KARP COMPLEXITY OF UNSTABLE CLASSES 3

2. Controlled classes

In this section we state a series of definitions that lead to the conceptof a class of structures being controlled (see Definition 2.5). We applythese definitions to the theory of dense linear orders to illustrate why itis desirable to consider the -Karp complexity of a class for uncountablecardinals . We first reintroduce the notion of a partial isomorphism,but with a slight variation. As we are only concerned with the definablesubsets of structures (and not their quantifier complexity) we insist thatall partial isomorphisms are elementary maps.

Definition 2.1 Given two elementarily equivalent structures M and Nin the same language and an infinite cardinal , a -partial isomorphism

is a partial elementary map with domain of cardinality less than ,that is: a function f from a subset D of M into N of size less than satisfying

M |= (d1 . . . dn) if and only if N |= (f(d1), . . . , f (dn))

for all formulas (x1, . . . , xn) of the language and all d1, . . . , dn fromD. We denote the family of -partial isomorphisms by F(M, N). IfM = N we simply write F(M).

The complexity of F(M) is a measure of how deeply one needs tolook to understand the relationship of a small subset (i.e., of size lessthan ) with the rest of the model. In order to measure this depth weendow the family with the following rank.

Definition 2.2 For f F(M, N),

(1) Rank(f) 0 always;(2) For limit, Rank(f) if and only if Rank(f) for all

< ;(3) Rank(f) + 1 if and only if

(a) for all C M of size less than , there is g F(M, N)extending f with C dom(g) and Rank(g) ; and

(b) dually, for all C N of size less than , there is g F(M, N) extending f with C range(g) and Rank(g) .

The -Karp complexity KC(M, N) of the pair of structures M, N isthe least ordinal such that Rank(f) implies Rank(f) + 1for all f F(M, N). Again, if M = N we simply write KC(M).


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Note that if a class K is controlled, then it follows from Proposi-tion 2.4(2) that for every cardinal , the relation of L,-equivalence

partitions K into only a set of equivalence classes (as opposed to aproper class). Continuing our example, KC0(DLO) = 0, as DLO,the theory of dense linear orders with no endpoints is 0-categorical.However, this observation hides the fact that one can code arbitraryordinals into dense linear orders. This ability to code ordinals impliesthat the class DLO is not controlled. In fact, KC(DLO) = for alluncountable cardinals . To see this, fix an uncountable cardinal and,for each non-zero ordinal , let J be the linear order with universe( ) , where denotes the order type of the rationals. In light ofProposition 2.4(2) it suffices to show that J is not L,-equivalent toJ whenever = . So choose non-zero ordinals and such that Jis L,-equivalent to J. Let E be the equivalence relation such thatE(x, y) if and only if there are fewer than elements between x andy. Since E is expressible in the logic L,, this implies that the con-densation J/E is L,-equivalent to J/E. But (J/E, ) (, ),(J/E, ) (, ), and it is readily checked that distinct ordinals arenot even L,-equivalent. Hence must equal .

3. Doubly transitive linear orders

In this section we investigate the class K2tr of infinite doubly transi-tive linear orders. That is, (I, ) K

2tr

if and only if the linear order

I is dense with no endpoints and for all pairs a < b, c < d from I, theinterval [a, b] is isomorphic to the interval [c, d]. Such orders arise nat-urally: The underlying linear order of any ordered field is necessarilydoubly transitive. Clearly, there is only one countable structure in K2trup to isomorphism. The class K2tr is a pseudo-elementary (PC) classthat is visibly unstable, so by [9] there are 2 non-isomorphic structuresin K2tr of size for all uncountable cardinals . Further, by [8], forall uncountable regular cardinals there is a family of 2 structures inK2tr of size that are L,-equivalent, yet pairwise non-embeddable.

Nonetheless, the class of doubly transitive linear orders is not entirely

without structure. There are natural invariants one can associatewith such orders. These invariants will not determine the orders upto isomorphism, but they will be sufficient to demonstrate that the-Karp complexity of K2tr is bounded for all cardinals .

The most natural invariant of a doubly transitive linear order is theisomorphism type of its closed intervals. Accordingly, we call I0, I1 K2tr locally isomorphic and write

I0 I1


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if [a, b] [c, d] for a < b from I0 and c < d from I1. Evidently localisomorphism is an equivalence relation on K2tr and I J for any

infinite convex subset J I, if I K2tr.The second invariant was developed by Droste and Shelah in [4].

The definitions that follow are slight adaptations of similar notionsused there. The most notable variation is that in [4] there is no boundon the number of levels of the decomposition tree and the cardinals can be any uncountable regular cardinal.

For the whole of this section, fix an uncountable cardinal .

Definition 3.1 A -decomposition tree is a subtree T of

{ : ;

(2) If is not maximal in T, then SuccT() = { : C}for some club subset of cof(g()), and {g() : C}is continuous, strictly increasing, and has supremum g().

3: If dir() =RIGHT then

(1) is maximal in T if and only if one of the three conditionshold:

(a) is degenerate;(b) coi(g()) = 0;(c) coi(g()) > ;

(2) If is not maximal in T, then SuccT() = { : C}for some club subset of cof(g()), and {g() : C}is continuous, strictly decreasing, and has infimum g().


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A -representation (T, g) partitions I into a set of intervals {I : T} where I0 = I1 = (g(0), g(1)) for all T of limitlength, and if SuccT() = { : C} for a club C then

I =

(g(), g(+)) if dir() =LEFT;(g(+), g()) if dir() =RIGHT

where + is the least element of C larger than . It is easily shown byinduction that the intervals {I : T

} are pairwise disjoint forany fixed successor ordinal .

For any dense linear order I, one can build a -representation (T, g)of I level by level by successively choosing a continuous, strictly in-creasing [or decreasing] sequence g(

) : from the interval

I. At first blush, it appears that one has considerable freedom in sucha construction. However, our freedom is considerably limited by thefollowing observation.

Observation Let J be any linear order of cofinality > 0. For anyclub subsets C1, C2 of and any two continuous, strictly increasing,cofinal sequences ai : i C1 and bi : i C2 in J, the set D = {i C1 C2 : ai = bi} is a club subset of .

By repeatedly applying this observation to a pair of-representationsof a linear order, we see that they must agree on a club. More pre-cisely, call a subtree T of a -decomposition tree T a club subtree ifT

itself is a -decomposition tree and, for each T

that is not maximalin T, SuccT() and SuccT() are both indexed by club subsets of thesame regular cardinal. If (T1, g1) and (T2, g2) are two -representationsof I, then by using the observation above at each node there is a -representation (T, g) of I such that T is a club subtree of both T1 andT2 with g() = g1() = g2() for all T. More generally we havethe following definition and lemma.

Definition 3.4 A subset A of a -decomposition tree T is closed if Ais downward closed, (i.e., if A then | A for all < lg()) andA is closed under successor, (i.e., if A then SuccT() A).

Note that for any subset A T of size at most , there is a closedsubset B A of size at most .

Lemma 3.5. Suppose (T, g) is a -representation of I0, S T isclosed, and f0, f1 : I0 I1 are order-preserving, continuous partial

functions whose domains contain {g() : S T} that satisfyf0() = f1() and f0(+) = f1(+). Then there is a clubsubtree Y T such that

f0(g()) = f1(g())


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for all S Y.

Proof. We construct Y by induction on the levels of T. Assumethat we have found Y, a club subtree of T { : < } such that

f0(g()) = f1(g()) for all S Y . If is a limit ordinal or 0 then

put Y+1 = Y { } and there is nothing to check. If = + 1

where is a limit ordinal or 0, let Y+1 = Y

{SuccT() : Y}.Now if S Y+1 for some

, then | S Y for all < ,so f0(g(|)) = f1(g(|)) for all < . As both f0 and f1 are order-preserving and continuous, it follows that f0(g(i)) = f1(g(i))for i = 0, 1 so our inductive hypothesis is maintained.

Finally, assume = + n, where is a limit ordinal or 0 and n > 1.Fix Y +n1 and we specify its successors in Y+1:

If S or if SuccT() = , then let SuccY+1() = SuccT()and there is no problem.

If S and SuccT() = { : C} for some club sub-set of an uncountable regular cardinal , then our hypothe-ses imply that f0(g()) = f1(g()) and {g() : C} isa continuous, strictly increasing (or decreasing) sequence con-verging to g(). Thus, as both f0 and f1 are order-preservingand continuous, there is a club C C such that f0(g()) =f1(g()) for all C. So put SuccY+1() = { : C}.

As noted above, these invariants are not sufficient to determine theisomorphism type of an element of K2tr. In particular, the secondinvariant does not specify which elements of the representation are inI (as opposed to I). This affords considerable freedom in choosing theisomorphism type of the order. The family of structures in the examplebelow was first studied by Conway [3] and was later used as an exampleby Nadel and Stavi [6].

Example 3.6 There is a family of 21 locally isomorphic, L,1-equivalentdoubly transitive linear orders of size 1, all of whom have isomorphic1-representations; yet the orders are pairwise non-embeddable.

Let S be a collection of 21 stationary subsets of1 \ {0} with X\ Ystationary for all distinct X, Y S (see [13] for a construction of sucha family). As notation, let Q0 be the set Q [0, ). For X S, let

IX =i1

JXi where JXi =

Q if i X;Q0 if i X.

Clearly (a, b) = Q for all a < b from IX , so IX IY for all X, Y S.It was first noted by Silver that for any sets X, Y S, the set B(X, Y)


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of all order-preserving partial functions f : IX IY, whose domain Dis a proper initial segment ofIX such that IX \ D has no least element,

and whose range R is a proper initial segment ofIY such that IY \R hasno least element, is an 1-back and forth system; hence the orders IXand IY are L,1-equivalent. As the Dedekind completions of the IX sare isomorphic we can identify them. After this identification, each ofthe orders IX share the same 1-representation, namely (T, g), whereT = {0, 1} {1, : 1} and g(1, ) is the element of theDedekind completion realizing the cut preceding J for all > 0.

It remains to show that IX is not embeddable in IY whenever X = Y.(This was proved in [3] but is repeated here for convenience.) So fixX = Y and assume by way of contradiction that there is an embeddingf : IX IY . It is readily verified that the set

C = { 1 : f(i

JXi ) =i

JYi }

is a club subset of 1. Thus, since X \ Y is stationary, there is an C X\ Y. But IX \

i

JXi has a least element, whereas IY \i

JYi

does not, which is a contradiction.

Despite the limitations demonstrated by the example above, the in-variants described in this section do allow us to obtain an upper bound

on the Karp complexity of K2tr. The following definitions establishour notation.

Definition 3.7 For D I, a D-cut is a partition of D into twosets, D and D

+ (either may be empty) such that D

D

+ = D,

D D+ = , and D

is downward closed. We write = (D

, D

+ )

and let I() = {x I : D < x < D+ }.

Definition 3.8 Suppose I and J are two linear orders. IfD I andf : D J is any order-preserving function then f() is the f(D)-cut(f(D ), f(D

+ )). A function f : D J is proper if {, +} D

and f is order-preserving, continuous, f() = , f(+) = +,

and satisfies d I f(d) J for all d D.IfD I and f : D J is a proper function, then I\D and J\f(D)

are partitioned into corresponding families of D-cuts and f(D)-cuts.The following definitions measure the similarity of these cuts.

Definition 3.9 Two (possibly empty) linear orders Iand J are (+, )-equivalent, written I +, J, if I and J are elementarily equivalentand the empty function in F+(I, J) has Rank at least (see Defini-tion 2.2).


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THE KARP COMPLEXITY OF UNSTABLE CLASS ES 11

By allowing linear orders to be empty and by insisting on elementaryequivalence we intend that I = if and only if J = and |I| = 1 if

and only if |J| = 1 whenever I +, J for some ordinal .Definition 3.10 IfD I and f : D J is proper, then f is -strongif I() +, J(f()) for all D-cuts .

If f F+(I, J) has domain A and has Rank at least 2, then itis easily seen that f is continuous and extends uniquely to a properfunction

g : A lim(A) {, +} J,

where lim(A) denotes the set of limit points of A in I. Also, it iseasily established by induction on 1 that if g : D J is a properfunction with domain D I and the restriction f = g|(D I) is inF+(I, J), then g is -strong if and only if Rank(f) .

For 1 the class of-strong proper functions has desirable closureproperties. It is routine to show that the restriction of any -strongproper function to any set that contains {, +} is also proper and-strong. As well, we have the following lemma, which is proved by astraightforward induction on .

Lemma 3.11. Let 1. Suppose that D I, f : D J is an -strong proper function, and for each D-cut there is a set E I()and an -strong proper function g : E J(f()). Then f

g is

proper and -strong.

Lemma 3.12. Let I0, I1 K2tr satisfy I0 I1 and I0 +, I1 forsome ordinal 2. Assume that A I0 is of size at most andsatisfies

(1) A is bounded below or coi(A) = 0; and(2) A is bounded above or cof(A) = 0.

Then there is an f F+(I0, I1) with domain A of Rank at least .

Proof. We show that in fact A is contained in an interval of I0which is isomorphic to an interval of I1. This interval will be of theform (a, b), where a is a lower bound for A if one exists, or the symbol

, and b is defined similarly. Take as a typical case that in whicha I0 and b = . Then we claim that the interval (a, ) is isomorphicto (a, ) for any a I1. The point is that (a, ) has cofinality 0,hence (a, ) does by (+, )-equivalence. So we can build the desiredisomorphism in a countable sequence of steps, using double transitivityand the local isomorphism of I0 and I1.

As well, it follows from the relations I0 I1 and I0 +, I1 andanother instance of double transitivity that the intervals (, a) and


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(, a) are (+, )-equivalent. Thus, the the restriction of the iso-morphism to A has Rank at least .

The following Proposition is the key to the proof of Theorem 3.14.Before embarking on it, we introduce some more notation. For C ,let

C0 = { C : is a limit point of C }

and for T a -decomposition tree, let T0 be the club subtree of Tsatisfying SuccT0() = { : C0}, where SuccT() = { : C} for all non-maximal nodes T0 . Note that if (T, g) is a -representation of I, then (T0, g|T

0 ) is also a -representation of I with

the additional property that g() either has cofinality or coinitiality at

most for all T

0 \ {0, 1}.Proposition 3.13. Assume I0, I1 K2tr, I0 I1 and I0 +, I1. IfA I0 and |A| , then there is a function f : A I1 of Rank atleast .

Proof. Pick A I0 of size at most . In order to produce ah : A I1 of Rank at least , we first construct a desirable properfunction j : D I1. Choose a -representation (T, g) ofI0. By passingto the subtree T0 in the notation preceding this proposition, we mayassume that g() either has cofinality or coinitiality at most for all T \ {0, 1}. Let B = BL BR, where

BL = { T : dir() = LEFT and A is cofinal in I} and

BR = { T : dir() = RIGHT and A is coinitial in I}.

We claim that B has size at most . To see this, it suffices by symmetryto show that |BL| . Recall that for every successor ordinal , theintervals {I : T

} are disjoint. Since BL implies AI = ,this implies |BL | for all successor ordinals . Further, since|A| , we can choose a successor ordinal so that for every paira, a A, there is T of length less than satisfying a < g() < a

whenever there is any T with a < g() < a. But now, by ourchoice of , if , B

Lhave length > and have | = |, then

g() = sup(A I|) = g(),

so = and I| A = . Thus,

|{ BL : lg() > }| |{ T : lg() = and I A = }|

so |BL| .Let B B be a closed subset of T of size at most . As g() has

cofinality or coinitiality at most in I0 for each B\{0, 1}, there


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is a set X I0 of size at most such that g(B) lim(X){, +}.

Since I0 +, I1, for each n 2 we can choose an order-preserving

jn : X I1 of Rank at least n. As g(B ) lim(X), each jn extendsuniquely to a proper function (also called jn) from X g(B) to I1.As B T is closed, by Lemma 3.5 there is a club subtree Tn foreach n 2 such that jn(g()) = jn+1(g()) for all B Tn. LetY =

n2 T

n and let D = {g() : B

Y0}, where Y0 is the clubsubtree of Y described in the notation preceding this proposition. Asthe functions jn agree on D for all n 2, we let j : D I1 denotethis common (proper) function. As each jn was n-strong, the functionj is -strong.

By Lemma 3.11, in order to ascertain the existence of an -strongh : A I

1, it suffices to construct an order-preserving function f :

A I0() I1(j()) of Rank at least for every D-cut of I0. Sofix a D-cut = (D , D

+ ). We finish the proof by showing that the

hypotheses of Lemma 3.12 are satisfied for I0() and I1(j()). AsI0() and I1(j()) are convex subsets ofI0 and I1 respectively, I0() I1(j()). Since j is -strong, I0() +, I1(j()). Finally, assume byway of contradiction that A I0() is unbounded above in I0() andhas uncountable cofinality. (The case of A I0() unbounded below inI0() of uncountable cardinality is symmetric.) Let b = sup(A I0())and let = cof(A I0()). We will obtain a contradiction by showingthat b = sup(D ), which would make I0() empty. First, since Y0 is a

club subtree of T and b = inf(D+ ), b = g() for some Y0. As weassumed A cofinal below b, b B as well. There are now four cases to

consider, all of which imply b = sup(D ) or contradict our hypotheses.

Case 1. dir() =RIGHT and lg() = + 1 where is a limit ordinalor 0.

Say = 0. Since cof(b) = > 0 there is a strictly increasingsequence of limit ordinals i : i < such that b = sup{g(|(i + 1)) :i < }. Since B is closed, | B for all < lg(), so g(|) Dand b = sup(D ).

Case 2. dir() =RIGHT and lg() = + n for some n > 1.

Say = for some C0, where C is such that SuccY() ={ : C}. As cof(b) = there is a continuous, strictly increasingsequence of ordinals i : i < from C with limit . Again, as B isclosed, i B for all i . It follows that i B Y0 for alllimit ordinals i , so again b = sup(D ).

Case 3. dir() =LEFT and is not maximal in Y0.


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Say SuccY0() = {

: C0}. As A is unbounded below b

and > 0, there is a club C C0 such that A is unbounded below

g() for all C. Thus, B Y0 for all C, so againb = sup(D ).

Case 4. dir() =LEFT and is maximal in Y0.

As maximal in Y0 implies maximal in T, it follows from thedefinition of a -representation that cof(g()) = 0 or cof(g()) >. However, we assumed that cof(g()) > 0 and A witnesses thatcof(g()) , so both are impossible.

Our theorem now follows easily.

Theorem 3.14. KC+(K2tr) for all uncountable cardinals .

Proof. Fix I K2tr and an uncountable cardinal . Let f F+(I) have Rank at least . We claim that Rank(f) + 1. Tosee this, it suffices by symmetry to show that if A I, |A| thenthere is a function g F+(I) extending f of Rank at least with

A dom(g). So fix such a set A and let f denote the proper functionextending f with domain dom(f) {, +}. Since Rank(f) ,f is -strong. Now fix a dom(f)-cut . Clearly, I() +, I(f())and I() I(f()), so it follows from Proposition 3.13 that there isa function g : A I() I(f()) in F+(I(), I(f()) of Rank atleast . Thus, it follows from Lemma 3.11 that the proper functiong = f {g : a dom(f)-cut} is -strong, hence the restriction of gto A dom(f) has Rank at least .

4. The -independence property

This section is devoted to proving that any pseudo-elementary classwith the -independence property (see Definition 4.4) is not controlled.We begin the section by proving Proposition 4.3, which will provide uswith a method for concluding that KC(K) = by looking at thefamily of -partial isomorphisms from one element of K into another.

Definition 4.1 An -tree T is a downward closed subset of


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Clearly, dp(T) < if and only if T is well-founded. The mostinsightful example is that for any ordinal , the tree des() consisting

of all descending sequences of ordinals < ordered by initial segmenthas depth . The proof of the following lemma is reminiscent of theproof of Morleys Omitting Types Theorem.

Lemma 4.2. If T


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dp(T) +. By Lemma 4.2, there is a subtree S of T of depth suchthat the L,-types of the elements of S depend only on their level in

S. In particular, for each n there is an element Bn S at level n thathas a successor in S. Consequently, for each n the L,-formula

(Xn) = Yntp,(Xn, Yn) = tp,(Bn+1)

is implied by tp,(Bn). Applying this iteratively produces an elemen-tary partial function f : N M with domain

{An : n }, so T

has an infinite branch.

Definition 4.4 A class K ofL-structures has the -independence prop-erty if there is a set {n(x0, . . . , xn1, yn) : n } of L-formulas such

that for all M K there is a sequence ai : i < from M suchthat for all n and all functions f : n {0, 1} there is a sequencebi : i < n from M such that for all i < n,

M |= i(b0, . . . , bi1, ai) if and only if f(i) = 1.

As an example, the model completion of the empty theory in thelanguage L = {Rn : n } consisting of one n-ary relation for every nis a complete, simple theory with the -independence property. (In thisexample, the yns do not appear.) Clearly, ifK has the -independenceproperty, then K has the independence property. However, the theory

of the random graph has the independence property, but fails to havethe -independence property. We remark that despite this failure, thetheory of the random graph is not controlled. We do not attempt toprove this assertion here.

Our interest in the notion of -independence is largely captured bythe proposition given below.

Definition 4.5 An ordered multigraph is a structure (G,


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(3) MG |= n(bg1, . . . , bgn, an) if and only if G |= Rn(g1, . . . , gn) forall n and all g1, . . . , gn from G.

The proof of Proposition 4.6 is word for word like the proof ofthe existence of Ehrenfeucht-Mostowski models for unstable pseudo-elementary classes (see e.g., Section 11.3 of [5]) but with the Nesetril-Rodl theorem (see [7] or [1]) in place of Ramseys theorem.

The following lemma tells us that we need not explicitly consider theconstants {an : n } in the proof of Theorem 4.9.

Lemma 4.7. LetK be a class ofL-structures and letC be a set of fewerthan new constant symbols. Let K be the class of all expansions ofelements of K to L C-structures. Then KC(K

) KC(K).

Proof. For any M K, let M be its reduct to the language of L.For every partial function f F(M), let f F(M) be the extensionof f that is the identity on every element ofCM. It is easy to show byinduction that RankF(f) = RankF(f). Hence, KC(M

) KC(M),so KC(K

) KC(K).

The other theorem we will need is that there exist very complicatedcolorings of a number of cardinals. As notation, for x a finite subset of, let xm denote the mth element ofx in increasing order. Following thenotation in [11], let P r0(,, 0, 0) denote the following statement:

There is a symmetric two-place function c : suchthat for every n , every collection of disjoint, n-elementsubsets {x : } of , and every function f : n n ,there are < < such that

c(xm , xm

) = f(m, m)

for all m, m < n.

It is shown in [10] that P r0(,, 0, 0) holds for an uncountable car-dinal whenever there exists a nonreflecting stationary subset of. (Astationary subset S is nonreflecting ifS is not stationary in for

all limit ordinals < .) In particular, P r0(3, 3, 0, 0) holds. Morerecently, in [12] the second author has shown that P r0(2, 2, 0, 0)holds as well. This suffices for our purpose. See [11] for more of thehistory of P r0 and its cousins.

The following Lemma recasts P r0(,, 0, 0) into the form we willuse in the proof of Theorem 4.9.

Lemma 4.8. Let c : []2 witness P r0(,, 0, 0). For everyk, n , every collection {x : } of disjoint, n-element subsets


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of , and every family of colorings {fi,j : n2 : i < j < k}, thereare 0 < 1 < < k1 such that

c(xmi , xm

j) = fi,j(m, m

)

for all i < j < k and all m, m < n.

Proof. Fix k, n, {x : }, and {fi,j : i < j < k} satisfyingthe hypotheses. Without loss, we may assume that xn1 < x

0+1 for

all . For limit, let y =

{x+i : i < k} and let W0 = { : limit}. By induction on k k we will build a sequence 0 k and c(ynk+mk

, yjn+m

) = fk,j(m, m) for j > k}

has size , hence is a suitable choice for Wk+1. (If there were no suchk then one could successively build a subset Z of Wk of size onwhich there would be no < from Z satisfying the coloring.)

Theorem 4.9. Let L1 L0 be first order languages, let T1 be an L1-

theory and let K denote the class of reducts of models of T1 to L0.If K has the -independence property then K is not controlled. Moreprecisely, if a cardinal > |T1| is regular and there is a coloring of []

2

satisfyingP r0(,, 0, 0), then KC(K) = for all cardinals > .

Proof. First, by adding countably many constants to the languageL0 and invoking Lemma 4.7, we may assume that the -independenceofK is witnessed by formulas n(x0, . . . , xn1) with no additional con-stants. Second, by considering Meq in place of M for each M K,we may assume that each x is a singleton. Third, by expanding T1 ifnecessary, we may assume that it has built-in Skolem functions. Fix a

coloring c : []2

that witnesses P r0(,, 0, 0) and fix an ordinal. We will use the coloring to define two rather complicated orderedmultigraphs I and J and then use Proposition 4.6 to get Ehrenfeucht-Mostowski models M, N K that are built from I and J respectively.We will find a tree of -partial isomorphisms from N into M that iswell-founded, yet has depth at least . Since was arbitrary, itfollows immediately from Proposition 4.3 that KC(K) = . So, let

des() = {strictly decreasing sequences of ordinals < }


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and let (I,


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three stages. First, since > |T1| is regular, for every l there isan integer n(l), an L1-term l(x1, . . . , xn(l)), a subset Xl of of size ,

and functions tl,m : Xl I such that for each Xl

g((, l)) = l(dl()),

where dl() = tl,1(), . . . , tl,n(l)(). As notation, let W = {(l, m) :l and m [1, . . . , n(l)]} and for each (l, m) W, let l,m and l,mbe the functions with domain Xl satisfying

tl,m() = (l,m(), l,m()).

Next, we state two claims, whose proofs we defer until the end of theargument.

Claim 1. There is a sequence Yl : l such that each Yl Xl hassize and for each k

qftp(d0(0), . . . , dk1(k1)) = qftp(d0(0), . . . , dk1(

k1))

in the structure (I, 1 there is a sequence ml : l < k and apermutation of k such that

t(0),m(0)((0)) t(1),m(1)((1))

t(k1),m(k1)((k1))

for every sequence 0 < < k1 with l Yl for each l < k.

Given these two claims, it follows from Konigs Lemma (and the factthat the permutation is uniquely determined by the lengths of thets) that there is an infinite sequence ml : l and a permutation Sym() such that, letting l =

t(l),m(l) for each l ,

0((0)) 1((1)) . . .

for all sequences 0 < 1 < . . . satisfying l Yl for each l . Butthe existence of such a sequence is clearly impossible as each l() des().

Thus, to complete the proof of the theorem it suffices to prove the

claims. The proof of Claim 1 is tedious, but straightforward. First, bytrimming each of the sets Xl we may assume that for each (l, m) W,

(1) l,m is constant on Xl;(2) l,m() = for all Xl; or(3) {l,m() : Xl} is strictly increasing and disjoint from Xl.

We call (l, m) -constant if (1) holds and call (l, m) -trivial if (2)holds. Similarly, we may assume that for each (l, m) W,

lg(l,m()) is constant for all Xl and


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l,m is constant on Xl or else {l,m() : Xl} is strictlyincreasing (in lexicographic order).

Additionally, we may assume that for each pair (l, m), (l, m) Wwith the same l, the truth values of

l,m() < l,m(); l,m() l,m(); l,m() 1. Inlight of Claim 1, it suffices to find a sequence ml : l < k and a permu-tation ofk such that t(0),m(0)((0))

t(k1),m(k1)((k1))for some sequence 0 < < k1 with l Yl for each l < k. Con-sequently, we can trim the sets Yl still further. As notation, let Wk


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denote the finite set of all pairs (l, m) W with l < k. For each l < k,let hl enumerate Yl, i.e., hl() = the

th element of Yl.

By trimming each Yl for l < k, we may additionally assume that: The sets Yl are disjoint and l,m()

l |Wk|. As notation, for each

ordinal , let

Bl() = {l,m(hl()) : (l, m) W, (l, m) not -constant} {hl()}

For = 0 < 1 < < k1, let B() =l


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from the sets B() and B(). Since c can only attain the values ofp and q on pairs from B() and B(), our choice of p and q implies

that Rk is the only relation that can differ between D() and D().Thus, there are sequences s0, . . . , sk1 D() and s0, . . . , s

k1 D()

of corresponding elements such that

I |= Rk(s0, . . . , sk1) Rk(s0, . . . , s

k1).

In particular, s0 sk1 and for all i < j < k we have

si = sj

and c(si, sj) = p. As each si D(), there are functions l, m withdomain k such that

si = tl(i),m(i)(l(i)).

Now fix i < k. Since k > 1 and c(si , sj) = p C for any j = i, the

pair (l(i), m(i)) is not -constant. As well, the choice of the coloringof B() ensures that si = hr(

1r) Yr for some r < k. Thus, the

disjointness of the Yls imply that r = l(i) and the pair (l(i), m(i))is -trivial. That is, si = l(i). Further, since

si = sj wheneveri = j, the function l must be a permutation of the set k. So, lettingmi = m(l

1(i)) and = l, the sequence mi : i < k and permutation are as desired.

References

[1] Fred Abramson and Leo Harrington. Models without indiscernibles. Journalof Symbolic Logic, 43:572600, 1978.

[2] Jon Barwise. Syntax and semantics of infinitary languages, volume 72 of Lec-ture Notes in Mathematics. Springer-Verlag, Berlin, New York, 268 pp, 1968.

[3] John Horton Conway. PhD. thesis, Cambridge, England, 196? (cited in [6]).[4] Manfred Droste and Saharon Shelah. A construction of all normal subgroup

lattices of 2-transitive automorphism groups of linearly ordered sets. IsraelJournal of Mathematics, 51:223261, 1985.

[5] Wilfrid Hodges. Model theory, volume 42 of Encyclopedia of mathematics andits applications. Cambridge University press, 1993.

[6] Mark Nadel and Jonathan Stavi. L,-equivalence, potential isomorphism,and isomorphism. Trans. Amer. Math. Soc., 236:5174, 1978.

[7] J. Nesetril and V. Rodl. Partitions of finite relational and set systems. Journalof Combinatorial Theory, Series A, 22:289312, 1977.

[8] Saharon Shelah. Existence of many L,-equivalent, nonisomorphic models ofT of power . Annals of Pure and Applied Logic, 34:291310, 1987. Proceedingsof the Model Theory Conference, Trento, June 1986.

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

[10] Saharon Shelah. Strong negative partition relations below the continuum. ActaMathematica Hungarica, 58:95100, 1991.


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[11] Saharon Shelah. Cardinal Arithmetic, volume 29 of Oxford Logic Guides. Ox-ford University Press, 1994.

[12] Saharon Shelah. Coloring and non-productivity of2-c.c.Annals of Pure andApplied Logic, 84:153174, 1997.

[13] Robert Solovay. Real-valued measurable cardinals. Proc. Symp in Pure MathXIII, Part 1, pages 397428, 1971.

Department of Mathematics, University of Maryland, College Park,

MD 20742

E-mail address: [email protected]

Department of Mathematics, Hebrew University of Jerusalem and

Department of Mathematics, Rutgers University

M.C. Laskowski and S. Shelah- The Karp Complexity of Unstable Classes

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