j. baldwin, m. c. laskowski and s. shelah- forcing isomorphism

8/3/2019 J. Baldwin, M. C. Laskowski and S. Shelah- Forcing Isomorphism

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Forcing Isomorphism

J. Baldwin

Department of Mathematics

University of Illinois, Chicago

M. C. Laskowski


University of Illinois, Chicago

S. Shelah


Hebrew University of Jerusalem

October 6, 2003

If two models of a first order theory are isomorphic then they remainisomorphic in any forcing extension of the universe of sets. In general, how-ever, such a forcing extension may create new isomorphisms. For example,any forcing that collapses cardinals may easily make formerly non-isomorphicmodels isomorphic. Certain model theoretic constraints on the theory andother constraints on the forcing can prevent this pathology.

A countable first order theory is said to be classifiable if it is superstableand does not have either the dimensional order property (DOP) or the omit-ting types order property (OTOP). Shelah has shown [7] that if a theory Tis classifiable then each model of cardinality is described by a sentence of

Partially supported by N.S.F. grant 90000139.Visiting U.I.C. from the University of Maryland thanks to the NSF Postdoctoral

program.The authors thank the U.S. Israel Binational Science Foundation for its support of

this project. This is item 464 in Shelahs bibliography.

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L,. In fact this sentence can be chosen in the L

. (L

is the result ofenriching the language L,+ by adding for each < a quantifier sayingthe dimension of a dependence structure is greater than .) Further work([3], [2]) shows that + can be replaced by 1. The truth of such sentenceswill be preserved by any forcing that does not collapse cardinals andthat adds no new countable subsets of , e.g., a -complete forcing. Thatis, if two models of a classifiable theory of power are non-isomorphic, theyremain non-isomorphic after a -complete forcing.

In this paper we show that the hypothesis of the forcing adding no newcountable subsets of cannot be eliminated. In particular, we show thatnon-isomorphism of models of a classifiable theory need not be preserved by

ccc forcings. The following definition isolates the key issue of this paper.

0.1 Definition. Two structures M and N are potentially isomorphic if thereis a ccc-notion of forcing P such that ifG is P-generic then V[G] |= M N.

In the first section we will show that any theory that is not classifiablehas models that are not isomorphic but are potentially isomorphic. In thesecond, we show that this phenomenon can also occur for classifiable theo-ries. The reader may find it useful to examine first the example discussed inTheorem 2.3.

1 Non-classifiable Theories

We begin by describing a class (which we call amenable) of subtrees of Q

that are pairwise potentially isomorphic. Then we use this fact to show thatevery nonclassifiable theory has a pair of models that are not isomorphic butare potentially isomorphic.

1.1 Notation.

i) We adopt the following notation for relations on subsets of Q.

denotes the subsequence relation; < denotes lexicographic ordering; for , lev is a unary predicate that holds of sequences of length (level); is the operation on two sequences that produces their largestcommon initial segment. We denote the ordering of the rationals by


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ii) For Q

, let D = { Q

: (2n) = (n)} and S = { D :(2n + 1) is 0 for all but finitely many n}. Let C = QS.

iii) The language Lt (for tree) contains the symbols ,


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iv) For all n with 2n + 1 k, p(ai)(2n + 1) = 0.

Now define an equivalence relation on P by p q if n(p) = n(q),k(p) = k(q) and letting a1 . . . an enumerate (in lexicographic order) domp,a1 . . . a

n enumerate (in lexicographic order) dom q, for each i, ai|k(p) =

ai|k(p) and p(ai)|k(p) = q(ai)|k(p). Then since Q is countable and the do-

mains of elements of P are finite, has only countably many equivalenceclasses; we designate these classes as the Fn. We must show that if p qthen p q P. It suffices to show that for all i, j if C |= ai < a

j then

C |= p(ai) < q(aj).

Case 1: i = j. By the definitions of k = k(p) and , we must havej > i, ai|k < aj |k and p(ai)|k < q(a

j)|k. This suffices.

Case 2: i = j. Choose the least t such that ai(t) = ai(t). Then

ai(t) < ai(t). Note that t > k and t must be even since for any odd

t > k, ai(t) = ai(t) = 0.

Suppose ai S and ai S.

We now claim p(ai)|t = q(ai)|t. Fix any < t. If < k, p(ai)() =q(ai)() by the definition of . If k is odd then the fourth condi-tion in the definition of k(p) guarantees that p(ai)() = q(a

i)() = 0.

Finally, if k and = 2u, since p and q preserve the P, p(ai)() =(u) = ai() and q(a

i)() = (u) = a

i() but these are equal by the

minimality of t.

It remains to show p(ai)(t)


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Case 1: as|m = a|m = as+1|m = c. Suppose m is odd. Let bs = p(as)and bs+1 = p(as+1). Then bs|m = bs+1|m and bs(m) < bs+1(m). By thedefinition of amenability for any r with bs(m) < r < bs+1(m), there isan B S with |(m + 1) = bs

r so p {< a, >} P as required.

If m is even choose as the image of a, p(c)(m/2).

Case 2: as|m = a|m = c but as+1|m = c. Again, let bs = p(as)and bs+1 = p(as+1) and denote bs|m by b

. By amenability there is an B S with |m = b. Any such is less than bs+1. If m is even > bs is guaranteed by (m) > (m); ifm is odd by Remark 1.3 whichrecasts the definition of amenability we can choose (m) > bs(m). In

either case is required image of a.

We deduce three results from this lemma. First we note that there arenonisomorphic but potentially isomorphic suborderings of the reals. Thenwe will show in two stages that any countable theory that is not classifiablehas a pair of models of power 20 that are not isomorphic but are potentiallyisomorphic.

1.6 Theorem. Any two suborderings of C,


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Proof. We may assume that T1 is Skolemized. Note there is no assumptionthat T1 is stable. Let M be a reasonably saturated model ofT1. By VII.3.5(2)of [4] there are L-formulas i(x, y) for i and a tree of elements a M : Q such that for any n , Q, and Qn+1 if |n = |n thenn+1(a, a) if and only if . By VII.3.6(3) of [4] (applied in L1!) wemay assume that the index set is a collection of L1-tree indiscernibles.

Let Y = a M : Q


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that an isomorphism between linear orders I1, I2 induces an isomorphism ofthe corresponding models.Let us expand on why we quote [6] above. In [6], Theorem III 3.10 it is

proved that for all uncountable cardinals and all vocabularies , if there isa formula (x, y) such that for every linear order (J,


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Then define the quantifiers and connectives as usual. We will only be con-cerned with the satisfaction of sentences of this logic for models of superstabletheories.

1.13 Remarks. i) The property coded in Condition (1) of Lemma 1.11is expressible by a formula (x, y) in the logic L+,. Each formula

in L+, and in particular this formula is absolute relative to anyextension of the universe that preserves cardinals. More precisely isabsolute relative to any extension of the universe that preserves +.

ii) If T has OTOP the formula can be taken in the logic L+,. So in

this case is preserved in any forcing extension.

iii) Alternatively, the property coded in Condition (1) of Lemma 1.11 isalso expressible in L+,+. That is, there is a formula (x, y) L+,+(in the original vocabulary L) so that

MI |= (ai, aj) if and only if i


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{a : Q

} and anL

+

,-formula (x, y) so that (a, a) holds inM if and only if is lexicographically less than . Further, the state-ments M |= (a, a) and M |= (a, a) are preserved under anyccc forcing. Note that this L1-order indiscernibility certainly implies L1-tree-indiscernibility in the sense of Definition 1.7.

Thus, the construction of potentially isomorphic but not isomorphic mod-els proceeds as in the last few paragraphs of the proof of Theorem 1.8 oncewe establish the following claim.

Claim. For any Q there is a collection p(x) of boolean combina-tions of (x, a) as a ranges over Y such that for any Q and any L1-term

f, if f(a1 , . . . an) realizes p in M then some i = .Proof. The conjunction of the (x; a|n) and (x; a|n)that define the cut of a will constitute p. Now if is not among the ichoose any n such that 1|n, 2|n , . . . k|n, |n are distinct. Then the se-quences 1, . . . k, |n

(n) + 1 and 1, . . . k, |n(n) 1 have the

same type in the lexicographic order so

M |= (f(a1 , . . . ak); a|n) (f(a1, . . . ak); a|n).

Thus, f(a1 , . . . ak) cannot realize p.

1.15 Remarks. i) Note that in Theorem 1.8 we were able to use anyexpansion of T as T1 so the result is actually for P C-classes. InTheorem 1.14 our choice of T1 was constrained, so the result is truefor only elementary as opposed to pseudoelementary classes. The caseof unstable elementary classes could be handled by the second methodthus simplifying the combinatorics at the cost of weakening the result.

ii) While we have dealt only with models and theories of cardinality 2,the result extends immediately to models of any larger cardinality andstraightforwardly to theories of cardinality with = .

2 Classifiable examples

We begin by giving an example of a classifiable theory having a pair of non-isomorphic, potentially isomorphic models. We then extend this result to aclass of weakly minimal theories.

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Let the language L0 consist of a countable family Ei of binary relationsymbols and let the language L1 contain an additional uncountable set ofunary predicates P. We first construct an L0-structure that is rigid butcan be forced by a ccc-forcing to be nonrigid. Our example will be in thelanguage L0 but we will use expansions of the L0-structures to L1-structuresin the argument.

We now revise the definitions leading up to the notion of an amenablestructure in Section 1 by replacing the underlying structure on Q by onewith universe 2. In particular, D, S, and C are now being redefined.

2.1 Notation.

i) For 2, let D = { 2 : (2n) = (n)} and S = { D : (2n + 1) is 0 for all but finitely many n} {b}, where b is anyelement of D satisfying b(2n + 1) = 1 for infinitely many n. LetC = 2S.

ii) Let M be the L1-structure with universe 2 where Ei(, ) holds if

|i = |i, and the unary relation symbol P holds of the set S. LetM1 be the L1-substructure of M

with universe C.

iii) Any subset A of C inherits a natural L1 structure from M1 with P

interpreted as S A.

2.2 Definition. An L1-substructure M0 of M1 is amenable if for all 2,

all n and all s 2n, if there is a P(M1) with |n = s then there isa P(M0) with |n = s.

Note that any L1-elementary substructure ofM1 is amenable. Moreover,it easy to see that i) each D is a perfect tree, ii) 2

is a disjoint union ofthe D and iii) for each s 2


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2.4 Proposition.There is an L1-elementary substructure M0 of M1 suchthat

i) |P(M1) P(M0)| 1.

ii) M0|L0 is rigid.

Proof. Note that each automorphism ofM1|L is determined by its restric-tion to the eventually constant sequences so there are only 2 such. Thus wemay let fi : i < 2

enumerate the nontrivial automorphisms of M1|L. Wedefine by induction disjoint subsets Ai, Bi of M1 each with cardinality lessthan the continuum. We denote i


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Now define an equivalence relation on P by p q if n(p) = n(q), k(p) =k(q) and letting a1 . . . an enumerate domp, a1 . . . a

n enumerate dom q,

for each i, ai|k(p) = ai|k(p) and p(ai)|k(p) = q(ai)|k(p). Then has only

countably many equivalence classes; these classes are the Fn. We must showthat if p q then p q P. It suffices to show that for all i, j if M0 |=En(ai, a

j) then M1 |= En(p(ai), q(a

j)).

Case 1: i = j. By the definition of k = k(p), M0 |= Ek(ai, aj). Let be maximal so that M0 |= E(ai, aj). Then n since M0 |= En(ai, a

j).

Since aj|k = aj|k, is also maximal so that M0 |= E(aj, ai). As p isan isomorphism, M1 |= E(p(ai), p(aj)). But p(aj)|k = q(a

j)|k, so

is also maximal with M1 |= E(q(aj), p(ai)). Since n we concludeM1 |= En(q(a

j), p(ai)) as required.

Case 2: i = j. Suppose ai S and aj S. We have M0 |= Ek(ai, aj)

by the definition of k and similarly, M1 |= Ek(p(ai), q(ai)). Now weshow by induction that for each m k, M0 |= Em(ai, ai) if and onlyif M1 |= Em(p(ai), q(a

i)). Assuming this condition for m we show it

for m + 1. If m + 1 is odd, the result is immediate by parts iii) andiv) of the conditions defining . If m = 2u then ai(m) = (u) andai(m) = (u). Since p and q are L1-isomorphisms p(ai)(m) = (u) andq(ai)(m) = (u). But M0 |= Em(ai, a

i) implies (u) = (u) so we have

M1 |= Em(p(ai), q(ai)).

To show that the generic object is a map with domain M0, it suffices toshow that for any p P and any a M0 domp there is a q P with q pand dom q = domp {a}. Choose n so that the members of domp {a} arepairwise En-inequivalent. Fix any L1-automorphism g ofM

that extends p.Let s = g(a)|n and Ws = { 2

: s }. Since there is a C Ws andB is amenable, there is a B Ws. Choosing for b, p a, b is therequired extension of p.

Since M0 and M1 are isomorphic in a generic extension for this forcing,

we complete the proof.

2.6 Remark. The notion of a classifiable theory having two non-isomorphic,potentially isomorphic models is not very robust, and in particular can be lostby adding constants. As an example, let F ER be an expansion of F ER

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formed by adding constants for the elements of a given countable model ofF ER. Then every type in this expanded language is stationary and theisomorphism type of any model of F ER is determined by the number ofrealizations of each of the 2 non-algebraic 1-types. Thus, if two modelsof F ER are non-isomorphic then they remain non-isomorphic under anycardinal-preserving forcing.

Similarly, non-isomorphism of models of the theory CEF of countablymany crosscutting equivalence relations (i.e., Th(2, Ei)i, where Ei(, )iff (i) = (i)) is preserved under ccc forcings.

We next want to extend the result from Theorem 2.3 to a larger class

of theories. Suppose T is superstable and there is a type q, possibly over afinite set e of parameters, and an e-definable family {En : n } of properlyrefining equivalence relations, each with finitely many classes that determinethe strong types extending q. Let T be such a theory in a language L andlet M be a model ofT. Let L0 be a reduct of L containing the Ens.

We say a M : X 2 is a set of unordered tree L-indiscerniblesif the following holds for any two sequences , from X:

If and realize the same L0-type then a1 , . . . an and a1, . . . ansatisfy the same L-type.

We say that a superstable theory T with a type of infinite multiplicity

as above embeds an unordered tree if there is a model M of T containing aset of unordered tree L-indiscernibles indexed by 2. We deduce below theexistence of potentially isomorphic nonisomorphic models of weakly minimaltheories which embed an unordered tree. Every small superstable, non--stable theory has a type of infinite multiplicity with an associated family of{En : n < } of refining equivalence relations and a set of tree indiscerniblesin the sense of [1]. The existence of such a tree of indiscernibles sufficesfor the many model arguments but does not in itself suffice for this result.Marker has constructed an example of such a theory which does not embedan unordered tree. However, an apparently ad hoc argument shows thisexample does have potentially isomorphic but not isomorphic models.

2.7 Notation. Given A = {a : 2} a set of unordered tree indis-cernibles let D = {a A : (n) = 0 for all but finitely many n}. For 2

let p(x) S1(D) be q(x) {En(x, a) : a D and |n = |n}. Note that

D is a dense subset ofA, each a realizes p and each p is stationary.

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2.8 Lemma.Let T be a weakly minimal theory that embeds an unorderedtree. Fix A and D as described in Notation 2.7. There is a set X satisfying

the following conditions:

i) X A is independent over the empty set;

ii) for anyY with D Y A, and any 2, p is realized in acl(XY)if and only if p is realized in Y;

iii) for any Y with D Y A, acl(XY) is a model of T.

Proof. It is easy to see from the definition of unordered tree indiscernibility

that if X = , then conditions i) and ii) of the Lemma are satisfied for anyY A. We will show that for any X and Y with D Y A with XYsatisfying conditions i) and ii) and any consistent formula (v) over acl(XY)that is not satisfied in acl(XY) it is possible to adjoin a solution of toX while preserving the conditions. By iterating this procedure we obtain amodel ofT.

Now suppose there is a Y with D Y A, such that acl(XY) is not anelementary submodel of the monster. Choose a formula (x,c,a) with c Xand a Y such that (x,c,a) has a solution d in M but not in acl(XY). Ifwe adjoin d to X we must check that conditions i) and ii) are not violated.

Since T is weakly minimal and d acl(XY), XAd is independent. Supposefor contradiction that for some a Y, p is not realized in a

but p isrealized in acl(Xda) by say e. Since condition ii) holds for XY, e acl(Xa).Therefore by the exchange lemma d acl(Xea). Let (v, c, a, e) with c Xand a Y witness this algebraicity. Then

(c, c, a , a, z) = (x)[(x,c,a) (x, c, a, z)] (=mx)(x, c, a, z)

is a formula over Xaa satisfied by e. Moreover, e acl(Xaa). For, ifso, transitivity would give d acl(Xaa) acl(XY). Now tp(e/Xaa)and in particular (c, c, a , a, z) is implied by p and the assertion that

z acl(Xaa). Since XA is independent, it follows by compactness thatthere is b D such that (c, c, a , a, b) holds. So there is a solution of(x,c,a) in the algebraic closure ofXY. This contradicts the original choiceof so we conclude that condition ii) cannot be violated.

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2.9 Theorem.If T is a weakly minimal theory in a language of cardinalityat most 20 that embeds an unordered tree then T has two models that are

not isomorphic but are potentially isomorphic (by a ccc-forcing).

Proof. Let L be the language of T. Assume that the type q is based ona finite set e. Let T be the expansion of T formed by adding constants fore. Let M be a large saturated model of the theory T and let the sets A andD be chosen as in Lemma 2.8 applied in L to T.

Recall the definition ofC from Notation 2.1. For any W C, let MW bethe L-structure with universe acl(X {a : W}) and denote MW|L byMW. We will construct an amenable set W such that MW MC. Since both

are amenable, there is a forcing extension where W C as L1-structures.Since {a : C} is a set of unordered tree L-indiscernibles, the inducedmapping of {a : W} into {a : C} is L-elementary. Thus, MW LMC and a fortiori MW L MC.

To construct W, let {f : 2} enumerate all L-embeddings ofDe into

MC. Note that each p can be considered as a complete L-type over De.Let W = 2S where S

= S {b} if MC realizes f(pb) and S

=

S if MC omits f(pb). (see Notation 2.1.)Suppose for contradiction that g is an L-isomorphism between MW and

MC. Then for some , g|D = f . Now if c W, the definition of W

yields f(p) is not realized in MC. This contradicts the choice of g asan isomorphism. But if c is not in W then by the construction of W,f(c) = g(c) does not realize g(p). But this is impossible since g is ahomomorphism.

References

[1] J.T. Baldwin. Diverse classes. Journal of Symbolic Logic, 54:875893,1989.

[2] Steve Buechler and Saharon Shelah. On the existence of regular types.Annals of Pure and Applied Logic, 45:207308, 1989.

[3] Bradd Hart. Some results in classification theory. PhD thesis, McGillUniversity, 1986.

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[4] S. Shelah. Classification Theory and the Number of Nonisomorphic Mod-els. North-Holland, 1978.

[5] S. Shelah. Existence of many L,-equivalent non-isomorphic models ofT of power . Annals of Pure and Applied Logic, 34, 1987.

[6] S. Shelah. Universal classes: Part 1. In J. Baldwin, editor, ClassificationTheory, Chicago 1985, pages 264419. Springer-Verlag, 1987. SpringerLecture Notes 1292.

[7] S. Shelah. Classification Theory and the Number of Nonisomorphic Mod-els. North-Holland, 1991. second edition.

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j. baldwin, m. c. laskowski and s. shelah- forcing isomorphism

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