Saharon Shelah and Pauli Vaisanen- On the number of L-infinity-omega-1 equivalent non-isomorphic...

8/3/2019 Saharon Shelah and Pauli Vaisanen- On the number of L-infinity-omega-1 equivalent non-isomorphic models

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On the number of L 1-equivalent non-isomorphicmodels

Saharon Shelah Pauli V ais anen

October 6, 2003

Abstract

We prove that if ZF is consistent then ZFC + GCH is consistent withthe following statement: There is for every k < a model of cardinality

1 which is L 1 -equivalent to exactly k non-isomorphic models of car-dinality 1 . In order to get this result we introduce ladder systems andcolourings different from the standard counterparts, and prove the fol-lowing purely combinatorial result: For each prime number p and positiveinteger m it is consistent with ZFC + GCH that there is a good laddersystem having exactly pm pairwise nonequivalent colourings. 1

1 Introduction

If M is a model, card( M ) denotes the cardinality of the universe of M . SupposeM and N are two models of the same vocabulary and is a cardinal. We writeM N if M and N satisfy the same sentences of the innitary languageL . For a denition of L , the reader is referred to [Dic85]. For any modelM of cardinality , dene

No(M ) = card N / = | card( N ) = and N M ,

where N / = is the equivalence class of N under the isomorphism relation. We

study the possible values of No( M ) for models M of cardinality 1. In partic-ular, we prove the following theorem:

Theorem 1 Assuming ZF is consistent, it is consistent with ZFC + GCH that there is for every k < a model M (of a vocabulary of cardinality 1) such that card( M ) = 1 and No(M ) = k.

Thanks to GIF for its support of this research and also to University of Helsinki for fundinga visit of the rst author to Helsinki in August 1996. Pub. No. 646.

This paper is the second authors Licentiates thesis. The second author did his share of the paper under the supervision of Tapani Hyttinen.

1 1991 Mathematics Sub ject Classication. Primary 03C55; secondary 03C75, 03E05. Keywords and phrases. Number of models, ladder system, uniformization, innitary logic, iterated

forcing.

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When M is countable, No( M ) = 1 by [Sco65]. This result extends to structuresof cardinality when is a singular cardinal of countable conality [Cha68].

So the study of possible values of No( M ) is divided into the following casesaccording to the cardinality of M :

1) card( M ) is weakly compact;

2) card( M ) is singular of uncountable conality;

3) card( M ) is uncountable, regular, and non-weakly compact.

In [She82a] Shelah was able to show that when is a weakly compact cardinalthere is for every non-zero cardinal , a model M such that card( M ) = and No( M ) = . In a paper which is in preparation by the authors, theproblem of the possible value of No( M ) between and 2 for a model M of weakly compact cardinality is completely solved.

Shelah has considered the singular case in two of his papers [She85, She86].Let be a singular cardinal of uncountable conality. In the former paper itis shown that if one allows relation symbols of arbitrary large arity < and is a non-zero cardinal with cf( ) < , then there exists a model M of singularcardinality with No( M ) = . In the latter paper Shelah gives a general wayto build models M with relations of nite arity only and for which the valueof No(M ) is quite arbitrary: for every non-zero cardinal { cf( )}, thereexists a model M of cardinality such that No( M ) = and its vocabulary

consists of one binary relation symbol, provided that cf( )

< for all < .The paper [She86] together with a recent paper [SV] offer a complete answerto the singular case provided that the singular cardinal hypothesis holds. Forexample it follows that No( M ) = is possible, even in L.

If V = L and 1 is a regular cardinal which is not weakly compact, No( M )has either the value 1 or 2 for all models M having cardinality . For = 1this result was rst proved in [Pal77a]. Later Shelah extended the result to allother regular non-weakly compact cardinals in [She81b].

It seems that there are no published independence results about the case thatcard( M ) is a regular but not weakly compact cardinal. But it is known that

the independence result given in [She81a] implies the consistency of there isa model M of cardinality 1 such that No( M ) = 0 with ZFC + GCH.Namely, in [She81a] Shelah proves: it is consistent with ZFC + GCH thatthere is a group G for which the group of extensions of Z by G, in symbolsExt( G, Z ), is the additive group of rationals. Here Z is the additive group of integers. Then one extension of Z by G can be directly coded to a model Msuch that No( M ) = card(Ext( G, Z )) = 0. The L 1 -equivalence between twocoded models follows from the group theoretic properties of G (G is strongly

1-free). But Ext( G, Z ) is a divisible group and hence this coding mechanism isnot applicable to the case 1 < No(M ) < 0. So there was the problem left if isit consistent to have a model M of cardinality 1 for which 1 < No(M ) < 0.

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As Shelah did with the Whitehead problem, we transform Theorem 1 into aquestion of the nature of pure combinatorial set theory. The combinatorial

problem will be a variant of the uniformization principles and ladder systemsgiven for example in [She82b] or [EM90]. As a matter of fact the more compli-cated ladder systems used here retrace back to the papers [She80] and [She81a].

For the benet of the reader we sketch the standard notion of ( , 2)-uniformization.For a limit ordinal < 1, a ladder on is a strictly increasing -sequence of or-dinals with limit . Let S be a set of limit ordinals below 1. A ladder system onS is a function : S 1 such that each () is a ladder on . A 2-colouringon S is a function c : S {0, 1}. For all S and n < , a 2-colouring con S associates the element c,n (the ( n + 1)th element of the sequence c())for each step ,n of a ladder system on S , hence the name 2-colouring. A2-colouring c on S can be uniformized if there is a function f : 1 {0, 1}

satisfying that for all S there is m < such that for all n < , n > mimplies f (,n ) = c,n . Such a function f is called a uniformizing function andwe say that c is uniform with respect to . The ( , 2)-uniformization holds if every 2-colouring on S is uniform w.r.t. .For our purpose we need a different kind of ladder system. The main difference isthat instead of the principle all colourings are uniform we want to know whatthe number of nonuniform colourings can be. We consider colourings whichtake values in a eld, and hence we can dene a natural equivalence relationfor colourings. (The following denition is from [She80], see also [ES96] wherecolourings which take values in a group are considered.) For 2-colourings c andd on S let c d be the 2-colouring e on S dened for all S and n < by e,n { 0, 1} and ( e,n + d,n ) c,n (mod 2). Then 2-colourings c and don S are equivalent w.r.t. a ladder system on S if c d is uniform w.r.t. .The number of pairwise nonequivalent colourings is the number of equivalenceclasses of 2-colourings on S under the given equivalence relation. But as itis pointed out in [She80, Theorem 6.2], for all set S 1 of limit ordinalsand ladder systems on S , the number of pairwise nonequivalent colourings iseither 1 or 2 0 . In our transformation of Theorem 1 the value of No( M )will correspond to the number of pairwise nonequivalent colourings. So, all thecases 1 < No(M ) 0 are ruled out when only standard ladder systems areconsidered.

The main result concerning the combinatorial problem is that for all nite eldsF ,

it is consistent with ZFC + GCH that there are good ladder systemand good equivalence for colourings (which take values in F ) such thatthe number of pairwise nonequivalent colourings is card( F ).

Recall that all nite elds are of the size pm with p a prime number and m apositive integer.

In standard ladders each step is one ordinal. The principal idea of the goodladders will be answering to the following simple question: what happens if each step could be a nite set of ordinals, or even a linear combination of

standard steps?

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In order to make our presentation self contained we give proofs of some factswhich are essentially proved elsewhere (mainly in [She77] and [She81a]). In

Subsection 2.1 we give the exact denitions for the good ladder systems,colourings, and equivalence. In Subsection 2.2 we introduce some basic factsabout iterated forcing.

In Section 3 the combinatorial problem is reformulated in a precise form and asolution of the problem is presented. Some remarks concerning generalizationsare given in Subsection 3.3. Since ladder systems and uniformization principlesare also used in abelian group theory and general topology this section may beof independent interest.

Section 4 is devoted to the proof of Theorem 1. We take a good ladder sys-tem and code each colouring a to a model M a . Then all of the coded modelswill be L

1-equivalent, and moreover, they are isomorphic if and only if the

corresponding colourings are equivalent. So the main result really is a straight-forward consequence of the independence result concerning the combinatorialproblem. The coding technique we have used in the proof of Theorem 1 is anice trick, and may also be of independent interest. Hence Section 4 is writtenin a way that if the reader accepts Theorem 2 on faith, she or he can read onlySubsection 2.1 and then directly proceed to reading Section 4.

2 Preliminaries

For all sets X,Y,Z , ordinals and functions f : X Y :

the restriction f Z has the meaning f (Z dom( f )),

X Y is the set of all functions from X into Y ,

Y is the set of all -sequences of elements in Y , and < Y is


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that supp( y) 1, f (y) is a shorthand for the following element of F ,


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c) An F -colouring a on S is uniform w.r.t. x and D if there is f : 1 F satisfying f (x ()) D a () for all S . The set of all uniform F -

colourings on S w.r.t.x

and D is Unif x ,D .d) The set ColS,F forms a vector space over the eld F , when addition in

ColS,F and operation of F on ColS,F are dened componentwise, and theunit element for addition is the function which is constantly 0. Using theaddition of this space we dene a and b in ColS,F to be equivalent w.r.t.x and D , written a x ,D b , if a b is a uniform colouring w.r.t. x and D . We denote by a F the subspace of ColS,F generated by a ColS,F .

It is easy to see that the set Unif x ,D forms a subspace of Col S,F . So thefactor space Col S,F / Unif x ,D also forms a vector space over F , and consequently,for all a , b ColS,F , a x ,D b if and only if a and b belong to the samecoset of ColS,F / Unif x ,D . If A and C are subsets of Col S,F then A + C is{a + c | a A and c C }. Hence b F + Unif x ,D denotes the set

a + c | a b F and c Unif x ,D= (e b ) + c | e F and c Unif x ,D= d ColS,F | there is e F such that e b x ,D d .

Lemma 2.3 Suppose D is a lter over including all conite sets of , S 1is a set of limit ordinals, F is a eld, and x is a VecF -ladder system on S .

a) If a is an F -colouring on S , 0 < 1, and f 0 : 0 F uniformizesa 0 + 1 w.r.t. x and D , then for all 1 < 1 (0 + 1) , there is an extension f 1 : 1 F of f 0 which uniformizes a (1 + 1) w.r.t. x and D .

b) If S is nonstationary in 1, then all F -colourings on S are uniform w.r.t.x and D .

c) Let a be an F -colouring on S and g a function from 1 into F . If thereexists < 1 such that g(x ()) D a () for all S , then a isuniform w.r.t. x and D .

Proof. a) Suppose S is enumerated by { | < 1}, where < for all < < 1, and e,n F for , < 1 and n < , are coefficients such that

x ,n =


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By Denition 2.1(a.iii) the set supp( x ,n ) m


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b) Suppose a is an F -colouring on S , and C = { | < 1} is a closed andunbounded subset of 1 disjoint from S . We dene by induction on < 1

functions f : F such that 0 and for all < < , functions f , f , satisfying f f and f uniformizing a + 1, are dened.

If is a successor of the form + 1, let f : F be some extension of f which uniformizes a + 1. This is possible by (a). If is a limit ordinalthen f =


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The subscript P in the notation P is not written when P is obvious from thecontext.

Let G be a P -generic set over V . When is a P -name, the interpretation of in the generic extension V [G] is denoted by int G (). For an object o in V [G], aP -name for o is written o, i.e., int G (o) = o. The canonical name for the genericset G itself is G. If an object o is in V , we identify the name o with the objecto itself instead of using standard names . The only exceptions for these rulesare that the standard names for uncountable cardinals and collections Y X arewritten and ( Y X ) respectively, to distinguish them from the cardinals , > 0, and corresponding collections in the generic extension. If f is a P -namefor a function from X V into Y V and x X , a condition p P decidesthe value of f (x) when there is y Y satisfying p P f (x) = y.

If P is a forcing notion having 2-c.c. then P preserves all conalities 2,i.e., for all limit ordinals , if cf() = 2 in V then P cf() = .Hence P preserves all cardinals too, i.e., if 2 is a cardinal in V then

P is a cardinal.

Suppose that P, P , 1 P is a forcing notion in V and Q, Q , and 1 Q are P -names satisfying P Q, Q , 1 Q is a forcing notion. The two stage iteration P Q, P Q , 1 P Q is dened by

P Q = ( p, q) | p P and p P q Q ,

and for the elements in P Q, ( p, q) P Q ( p , q ) if both p P p and p P (q Q

q ) hold. So1

P Q is the pair (1

P ,1

Q ). We identify elements ( p, q), ( p , q ) P Qif both ( p, q) P Q ( p , q ) and ( p , q ) P Q ( p, q) hold. This iteration amountsto the same generic extension as does the composition where one rst forceswith P and then with Q.An iterated forcing of length 2 with countable support ,

P 2 , P 2 , 1 P 2 = CountLim P , Q | < 2is inductively dened for all 2 as follows.

a) The forcing notion P 0, P 0 , 1 P 0 is dened by 1 P 0 = , P 0 = {1 P 0 }, and P 0 = P 0 P 0.

b) Suppose for all < , Q , Q , 1 Q are given P -names and they satisfy

P Q , Q , 1 Q is a forcing notion .

Moreover, assume that for all < ,

P , P , 1 P = CountLim P , Q | <

are already dened. It follows from (a) that V = V [H ] for all P 0-genericsets H over V . Hence we assume that Q0, Q 0 , 1 Q 0 are standard namesand Q0, Q 0 , 1 Q 0 is a forcing notion in V .The set P is the collection of all functions p satisfying the following

requirements:

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i) The domain of p is , and for each < the value of p( ) is aP -name such that p P p( ) Q .

ii) The set { < | p P p( ) = 1 Q } is countable.c) For all 2 and p, q P , the order of these conditions is q P p if

either is a limit ordinal, and

for all < , q P p ,

or otherwise, is a successor ordinal of the form + 1, and

q P p ,q P q( ) Q p( ).

d) 1 P is the function which maps each < into 1 Q .

Remark. For all 2 and p P , we let dom( p) denote the set of ordinalsgiven in (b.ii) above. This set is usually called the support of p. So, one can aswell think that the domain of a condition p P really is the set dom( p). Wemay write f P , 2, when f is only a function satisfying dom( f ) and f { (, 1 Q ) | dom( f )} is a condition in P . We abbreviate P by and P by , or even more compactly by when the subscript isobvious.

For each < 2, P Q is isomorphic to P +1 via the mapping ( p, q) p q .

If G is a P -generic set over V then for each < , G denotes the P -genericset { p | p G }.

Fact 2.4 Suppose 2 and P = CountLim P , Q | < .

a) If P has 2-c.c. for all < , then P has 2-c.c.

b) If = 2, P 2 has 2-c.c., X is a set in V , and Y is P 2 -name satisfying 2 (Y X and card( Y ) < 2), then for all P 2 -generic sets G over V ,

there is < 2 such that the subset Y = int G (Y ) is already in V [G ].

c) Let S be a set of limit ordinals < 1 and F a eld of cardinality 1.

If 21

= 2 and card( Q ) = card(1) for all < , then there is a collection {c , | < 2} of P -names satisfying {c , | < 2} =ColS,F . Such a collection is called (P , 2)-enumeration for ColS,F .

For < 2, p P and q P such that p q the composition of these conditions, in symbols p q, is the function having domain and denedfor all < by

( p q)( ) = p( ) if < ;q( ) if < .

Then, as in [She77, Denition 1.1 and Fact 1.3] or [Gol93, Denition 1.12 and

Fact 1.13], p q is a condition in P and ( p q) q.

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We shall also need the quotient forcing notion P , , , , 1 , of an iteratedforcing P = CountLim P , Q | < , where < 2. The following

denition is from [Gol93]. The P -name P , is such that

P , = { p P | p G },

, is a P -name for which

, = P , ,

and 1 , is the standard name for 1 P . So, for all P generic sets H over V and p, q P , = int H (P , ), we have p , q in V [H ] iff p q in V , where , = int H ( , ). We abbreviate P , by , .

Fact 2.5 Suppose < 2, H is a P -generic set over V , o is a P -name,and is a formula. Then there is a P , -name o in V [H ] such that the following hold.

a) If p P , p H , and p (o) then in V [H ], there is q P , such that q , p and q , (o).

b) If in V [H ], r P , and r , (o) then in V , there is s P satisfying s r , s H , and s (o).

Fact 2.6 Suppose 2, p, q P , and H is a P -generic set over V .If both p H and q H hold, then there are p , q P such that p p,q q, and p = q H .

3 The Combinatorial Problem

This section is devoted to the proof of the following theorem which is a preciseform of the theorem described in the introduction.

Theorem 2 Assume the following properties hold in V :

the generalized continuum hypothesis, GCH ;

S is a set of limit ordinals below 1 and bistationary in 1;

F is a nite eld;

Vec is the vector space over F freely generated by x | < 1 ;

D is a lter over including all conite sets of .

Then there is a forcing notion P, , 1 of cardinality 2 such that P satises2-c.c., P does not add new countable sequences, and for every P -generic set G

over V , there is in V [G] a Vec-ladder system x on S such that card(Col S,F / Unif x ,D ) =

card( F ).

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colourings will be killed, a bookkeeping function will be needed. Fix to be afunction from 2 onto 2 2 such that whenever () = ( , ) then .

The bookkeeping function is useful only if we can ensure that the colouringscan be enumerated by 2. Since we assume GCH the cardinality of Col iscard( S ( F )) = (2 0 ) 1 = 2 1 = 2. Hence there is an enumeration {c 0, | 0 and p Q ={1 }. Observe that if dom( p) then p p( ) = 1 , and is not p-trivial.Note also that Uf(a ) = {1 } by Lemma 2.3(a). In fact, if p P and pforces a b + Unif then p forces Q to be a nontrivial forcing notion (seeLemma 3.4(d) below).

We have to check that the property (1) for = holds. We shall prove thatP does not add new countable sequences. Hence (< 1 F ) = < 1 F . Thisimplies that

card( Q ) card < 1 F = card (< 1 F ) = card(1),

since card( < 1 F ) = 2 0 = 1.

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Before proving that P does not add new countable sequences, we introduceuseful notations and lemmas. Let H , for , denote the model

H (), ,,S,F,D, P , , 1 | ,

where is some large enough cardinal, for example ( 2 )+ , and H () is the

set of all sets hereditary of cardinality < . The expansion of the model H with new constant symbols X 1, X 2, . . . is denoted by H (X 1, X 2 , . . . ).

A condition p in P has height , where and < 1, if for every dom( p), p dom( p( )) = . We say that p is of height < when p dom( p( )) < . The notion p is of height is dened analogously. Thesenotions are from [She81a].

If X is a set of pairwise compatible conditions in P , the composition of theseconditions, in symbols ( p X ) p, is the function f with dom( f ) = p X dom( p)and for each dom( f ), f ( ) is a P -name such that

f ( ) ={ p(0) | p X } if = 0;{ p( ) | p X } otherwise .

Observe that f is not necessarily a condition in P (as we pointed out earlier, bythis we mean that not even the extended function f { (, 1 ) | dom( f )}is a condition in P ).

Lemma 3.2

a) Suppose , pn | n < is a descending chain of conditions in P , < 1 is a limit ordinal not in S , and n | n < is an increasing sequence of ordinals with limit . Suppose also that for all < ,

i) there are innitely many m < for which pm dom( pm ( )) m , and

ii) there are innitely many n < such that pn dom( pn ( )) .

Then q = n 1 and for every < ,q P , q pn for all n < , and q has height . If is a limit ordinalthen the claim holds directly by the denition of P and height. Note thatdom( q) is countable even if has conality > since dom( q) is a countable

union of countable sets.

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Suppose = + 1 and dom( q) (if dom(q) then the claim follows fromthe induction hypothesis). By the denition of q, q n


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Lemma 3.3 If E n , n < , are dense and open subsets of P , then n


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c) For every P -generic set G over V , V [G] satises GCH and ( )V = for all ordinals .

d) For all nonzero < 2 and P +1 -generic sets G +1 over V , V [G +1 ] |=a b + Unif .

e) For every P -generic set G over V , V [G] |= card(Col / Unif) card( F ).

Proof. Even though all the properties are standard we sketch proofs for them.

a) The claim follows directly by the property (1) on page 13 and Fact 2.4(a).

b) If we assume that there is a new subset of in V [G], where G is a P -generic set over V , then by the 2-c.c. property of P and Fact 2.4(b) we canchoose < 2 such that the new subset is already in V [G ]. This contradicts

Lemma 3.3.c) The generalized continuum hypothesis is preserved by (a), (b), and by thefollowing well-known fact :

if card(P ) 2, P has 2-c.c., 2 1 = 2, is an uncountable cardinal,and = ( 2 )V , then P 2 .

By (a) the ordinals V , 2, are cardinals in the generic extension. Sinceby (b), 1V is not collapsed, the claim follows.

d) Let G +1 be a P +1 -generic set over V . If (Q = {1 }) holds in V [G ] thenby Denition 3.1 V [G ] |= a b + Unif. Since V [G +1 ] V [G ], the latter

formula is also satised in V [G +1 ].Suppose Q = Uf( a ) holds in V [G ]. By Lemma 2.3(a) for each < 1 thegeneric set G +1 contains a condition p for which p dom( p()). Thusdom( g ) = 1 in V [G +1 ]. Let f p be a shorthand for int G +1 ( p()). Then f puniformizes a (dom( f p) + 1) in V [G +1 ]. Consequently, g = {f p | p G }uniformizes a in V [G +1 ]. So V [G +1 ] |= a b + Unif.

e) Assume the claim fails. Since card( b F ) card( F ), let G be a P -genericset over V and d a colouring in V [G] for which d b + Unif. Since P has

2-c.c. and P card( d ) < 2 there must be, by Fact 2.4(b), < 2 suchthat d V [G ]. By the denition of the forcing P and Fact 2.4(c), {c , |

< 2} = Col holds in V [G ]. So there is < 2 with V [G ] |= d = c,

.By Denition 3.1 and since the bookkeeping function is surjective, there is < 2 such that ( a = c , ) holds in V [G ]. Then by (d), V [G +1 ] satisesa b + Unif. Since V [G + 1] V [G], c , = a = d b + Unif holdsin V [G] contrary to our initial assumption.

Remark. It can be seen from the constructions in Subsection 3.2 below that P is a proper forcing notion [She82b, Theorem 2.8(1) on page 86]. But this factdoes not, however, help with the main problem of Subsection 3.2.

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3.2 The Generic Colouring is Nonuniform

The main problem left after Lemma 3.4 is that maybe the size of Col / Unif issmaller than the size of F in the generic extension. Since card(Col / Unif) 2, (A) holds, and k = 0 for all < < 1. Analogously tothe case (1), to prove that there are c = e, d F for which (c,d,e,d ) Pos ( p),it suffices to show for arbitrary < 1 the existence of y Vec, and conditionsq , r in P satisfying

min(supp( y)) > ,q , r p,q = r ,q( )(y ) = c ,r ( )(y ) = e ,q h(y ) = d ,r h(y ) = d .

Let < 1 be given. Fix > max{,, dom( p())} and > such thatk = k . As in (1) x q , r p such that

q = r H,q( )() = 1 and q( )( ) = 1 ,r ( )() = 2 and r ( )( ) = 2 .

Dene e = k and e = k . Then ek + e k = 0, and e + e = 0 sincek = k . If we let y be (e x + e x ), then

q( )(y ) = e q ()() + e q( )( ) = e + e ,

andq h(y) = e (k + m ) + e (k + m )

= ( e k + e k ) + ( em + e m )= ( e m + e m ).

By a similar reasoning r satises

r ( )(y ) = 2( e + e ),r h(y ) = 2( e k + e k ) + ( e m + e m ) = ( e m + e m ).

Hence c = e + e (= 0), e = 2 c(= c), and d = em + e m are the desiredelements of F . 3.15

Now we can proceed with analyzing properties of condition trees. Recall that Ais a ltration of A . Suppose m < , T is an A m -condition tree, Ind( A m ),and p is a condition in P such that p T () for = max Am . We denea function T [/p ] by setting for all Ind( A m ) that

T [/p ]( ) = p if = ;

p T ( ) otherwise;

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where = max { Am | = }. Observe that for each Ind( A m ),T [/p ]( ) is a condition in P since p T = T . Hence, T [/p ] is

an A m -condition tree and T [/p ] T .Lemma 3.16 Suppose < 1 and T is an A m -condition tree. Then there isan A m -condition tree R T of height .

Proof. Suppose {i | i < k }, k < , is an enumeration of Ind( A m ). Wedene by induction on j k, A m -condition trees R j as follows. Let R0 be T .Suppose j < k , R i for all i j are dened, and the conditions R j (i ), i < j ,are of height . By Lemma 3.2(b) there is p R j ( j ) in P having heightgreater than . We dene R j +1 to be R j [ j /p ]. It follows that R k T is anA m -condition tree of height .

Denition 3.17 We x the following notation for each m < :Val( A m ) = { | is a function from Ind( A m ) into F },IInd( A m ) = { + 1 | Am and Ind( A m )},IVal( A m ) = { | is a function from IInd( A m ) into F }.

Let m < and T be an A m -condition tree. For all y Vec and (, )IVal( A m ) Val( A m ) we write

T [y] (, )

if for each Ind( A m ), both of the requirements

for each Am , either is T -trivial or T ( )(y) = ( + 1) ,

and T h(y) = (),

are satised. We dene TPos( A m ) to be the set of all (, ) IVal( A m ) Val( A m ) with the following property. For all A m -condition trees T thereexist an unbounded subset Y of Vec and for each y Y an A m -condition treeT y T satisfying T y [y] (, ).

Suppose m < and T is an A m -condition tree. We set

Dec(T ) = y Vec | for all Ind( A m ) and Am , is T -trivial or supp( y) dom( T ( )) ,

Dec(T ) = y Vec | for all Ind( A m ) and Am ,supp( y) dom( T ( )) ,

Dech (T ) = y Vec | for each Ind( A m ),T decides the value of h(y) .

For i = 0 , 1, (i , i ) IVal( A m ) Val( A m ) and ei F we dene the sum

e0 (0 , 0) + e1 (1, 1)

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to be the pair ( , ) IVal( A m ) Val( A m ), where for all IInd( A m ) and Ind( A m )

() = e0 0() + e1 1(), () = e0 0() + e1 1().

Lemma 3.18 Suppose m < and T is an A m -condition tree.

a) For every y Vec there is an A m -condition tree R T for which yDec(R) Dech (R).

b) For all y Dec(T ) and IVal( A m ) there are an A m -condition treeR T and Val( A m ) such that R[y] (, ).

c) For every IVal( A m ), there is Val( A m ) such that (, )TPos( A m ).

d) If (i , i ) TPos( A m ) and ei F , for i = 0 , 1, then i=0 ,1 ei (i , i ) isin TPos( A m ).

Proof. a) Suppose Ind( A m ) = {i | i < k }. Let R0 be T . Assume A m -condition trees R i , i j < k , are already dened.

(A) By Lemma 2.3(a) there is p R j ( j ) in P for which supp( y) dom( p())for all Am .

Assume q p in P decides the value of h(y), and dene R j +1 to be R j [ j /q ].Then y Dec(R k ) Dech (R

k ).

b) This is proved as (a). The only difference is that instead of (A) the followingis used:

by Lemma 3.5(b) there is p R j ( j ) in P satisfying for each Amthat either is p-trivial or otherwise p()(y) = ( j + 1).

Then the function Val( A m ) satisfying R k [y] (, ) is uniquely deter-mined by R k .

c) Since T and the domains of the conditions in T are countable there mustbe a limit T < 1 such that for every y Vec, min(supp( y)) > T impliesy Dec(T ). Hence, directly by (b), for every y Dec(T ) there are T y T

and y

Val( A m ) satisfying T y

[y] (, y

). Since Val( A m ) is countableand Dec( T ) uncountable, there must be an unbounded subset Y of Dec(T )and Val( A m ) such that = y for each y Y . Thus Y and the treesT y | y Y stronger than the arbitrary A m -condition tree T exemplify

(, ) TPos( A m ).

d) Since (0, 0) TPos( A m ) there are an unbounded subset Y of Vec andfor each y Y , an A m -condition tree T y0 T satisfying T

y0 [y] (0, 0).

Because ( 1, 1) TPos( A m ), there exist for each y Y an A m -conditiontree T y1 T

y0 and an element zy Vec such that max(supp( y)) < min(supp( zy ))

and T y1 [zy ] (1 , 1). Consequently, for all y Y ,

T y1 [e0y + e1zy ] e0 (0, 0) + e1 (1, 1).

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So the unbounded subset {(e0y + e1zy ) | y Y } of Vec and the trees T y1 |y Y stronger than an arbitrary T exemplify that i=0 ,1 ei (i , i ) is in

TPos( A m ). 3.18

We let 0 IValm be the 0-function of IVal( A m ) and 0Valm be the 0-function of Val( A m ).

For all Val( A m ), Ind( A m ), and d F , [ d] denotes the functionin Val( A m ) which is the same as except it maps into d.

Lemma 3.19 For every IVal( A m ) the pair ( , 0Valm ) is in TPos( A m ).

Proof. We shall prove the following claim.

For every 0 Ind( A m ) there are ( , ) TPos( A m ) and d1 F suchthat d1 = (0) and ( , [0 d1]) is in TPos( A m ).

This suffices, because if the claim holds then by Lemma 3.18(d)1

(0 ) d1 ((, ) (, [0 d1]))= 1 (0 ) d1 (0

IValm , 0

Valm [0 (0) d1])

= (0 IValm , 0Valm [0 1]) TPos( A m ),

for all 0 Ind( A m ). Furthermore, by Lemma 3.18(c), there is Val( A m )for which ( , ) TPos( A m ), and hence by Lemma 3.18(d),

( , ) 0 Ind( A m ) (0) (0IValm , 0

Valm [0 1])

= ( , ) (0IValm , )

= ( , 0Val

m ) TPos( A m ).

For the rest of the proof of the lemma let be the maximal element of Am ,T be an A m -condition tree, and 0 be an arbitrary element of Ind( A m ).By Lemma 3.15(c) there are c, d0 = d1 F , an unbounded subset Z of Vec,and conditions py0 , p

y1 T (0), for each y Z , exemplifying (c, d0 , c ,d1)

Pos (T (0)). This means that for all y Z , i = 0 , 1, and Am ,

py0 = py1 ,

pyi h(y) = di , py0 ( )(y) = p

y1( )(y) or is p

yi -trivial for both i = 0 and 1 .

By Lemma 3.18(a) there is an A m -condition tree T y T [0/p y0] for everyy Z such that y Dec(T y ) Dech (T

y ). Since Z is uncountable there mustbe an unbounded subset Y of Z and ( , ) IVal( A m ) Val( A m ) such thatT y [y] (, ) for all y Y . So Y and the trees T y | y Y stronger than anarbitrary tree T exemplify (, ) is in TPos( A m ). Observe that T y (0) py0implies T y (0) h(y) = (0) = d0.

Now, the functionR y = T y [0/ (T y (0) ) p

y1 ]

is a A m -condition tree for each y Y , since T y (0) py0 = py1 . Hence

Y and Ry

| y Y exemplify (, [0 d1]) is in TPos( A m ).

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3.3 Remarks

There is a forcing notion which gives the conclusion of Theorem 2 for all niteelds simultaneously. Namely, we dened an iterated forcing P k = CountLim P , Qk | < 2 for xed k. The extended result would follow if each Qk was replacedby Q2 Q3 . . . where Q i takes care of the case (i) = ( p,m) and is acoding for the pairs of primes and positive integers. So F i would be the eld of size pm where (i) = ( p,m). For example, to prove that for each coordinate ithe cardinality of Col S,F i / Unif x ,D is as wanted, it would suffice to concentrateon one coordinate i, and dene the condition trees and systems, Pos ( p), etc.,only for xed i. Hence an assumption that the size is wrong for some i wouldlead to a contradiction in the same way as in Subsection 3.2.

It is possible to have a Vec F -ladder system on S such that card(Col S,F / Unif x ,D ) =0. A proof of this fact would be a forcing argument just like the one we

have given. The only difference is that instead of one generic colouring b , oneshould add generic colourings b m | m < by dening Q0 = ILad ICol.Then by replacing b F + Unif with ( b 0, b 1, . . . F + Unif) the desired resultwould follow. The conclusion of such a generalized theorem would be P card ColS,F / Unif x ,D = card b 0, b 1, . . . F = 0. Other changes would be, forexample, that Lemma 3.6 would have the form 1 if b 0, b 1, . . . F then Unif,and analogous changes would be needed in Lemma 3.15.

We may also continue the iteration longer than 2 and get the consistency of our main result with CH + any reasonable value for 2 1 . The 2-c.c. for

such a forcing follows from the use of pic [She82b] or better [She98, Section 2of Chapter 8].

During the given proof, for example in Lemma 3.3, it is possible to use thegeneral claim on preservation of ( 1 S )-complete forcing notions and thepreservation of properness for the preservation of stationarity [She82b, Chapter5] or [She98, Chapter 5]. But this does not, however, help with the mainproblem.

4 The Models

As in the preceding sections, we assume that S 1 is a set of limit ordinals,F is a eld, D is a lter over including all conite sets of , Vec is the vectorspace over F freely generated by x | < 1 , x is a Vec-ladder system on S ,Col denotes the set of all F -colourings on S , and Unif is the set of all uniformcolourings.

Let M be a model of vocabulary , 0 < n < , and R a relation symbolwith n + 1 many places. We say that R is a partial function in M if thereare X M n and Y M such that the interpretation R M of the symbol Rin M is a function from X into Y . For all relations R , which are partialfunctions in M , R M (x) = y means x y R M , and atomic formulas R(x, y )

are written in the form R(x) = y.

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Denition 4.1 We dene a vocabulary and for all a Col models M a of vocabulary by the following stipulations:

a) Each model M a has the same domain (S F


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Lemma 4.2 Suppose a , b Col and 1.

a) If f : F uniformizes (b a ) + 1 , then M a + 1 = M b + 1 .

b) If M a + 1 = M b + 1 , then there is f : F which uniformizes(b a ) + 1 .

c) M a 1 M b .

Proof. a) Suppose f : F uniformizes ( b a ) + 1. We dene : M a +1 = M b + 1 by the following equations.

For all < ,(x , 0) = x , f () ,

and for all (y, c) Vec

F , we set

(y, c) = + c


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b) Suppose then : M a + 1 = M b + 1. We let f : F be the uniquefunction satisfying for all < and c F , f () = c iff (x , 0) = ( x , c).

Assuming that x ,n is of the form


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c) For each a Col, No(M a ) = card(Col / Unif) .

Proof.a) The claim holds by (a) and (b) of Lemma 4.2.

b) We let , for all S , be the following L 1 ()-sentence,

r ,n | n < s R I D n I

Pr n (s) = r ,nn I

Pr n (s) = r ,n .

For all S , holds in N since the interpretation r ,n = ( x ,n , a ,n ), for all S and n < , satises the formula in M a . We let r ,n | n < , S ,be a sequence of elements in N satisfying , and s be the unique element inR N which satises Pr n N (s) = r ,n for all n < .

We dene : (S F


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Thus for all n < and ( , u) S F


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[EM90] Paul C. Eklof and Alan H. Mekler, Almost free modules , North-Holland Math. Library, 46, North-Holland, Amsterdam, 1990.

[ES96] Paul C. Eklof and Saharon Shelah, New nonfree Whitehead groups by coloring , Abelian groups and modules (Colorado Springs, CO, 1995)(New York), Lecture Notes in Pure and Appl. Math., 182, Dekker,New York, 1996, pp. 1522.

[Gol93] Martin Goldstern, Tools for your forcing construction , Set theory of the reals (Ramat Gan, 1991) (Ramat Gan), Israel Math. Conf. Proc.,6, Bar-Ilan Univ., Ramat Gan, 1993, pp. 305360.

[Pal77a] E. A. Palyutin, Number of models in L , 1 theories, II , Algebra iLogika 16 (1977), no. 4, 443456, English translation in [Pal77b].

[Pal77b] E. A. Palyutin, Number of models in L , 1 theories, II , Algebra andLogic 16 (1977), no. 4, 299309.

[Sco65] Dana Scott, Logic with denumerably long formulas and nite strings of quantiers , Theory of Models (Proc. 1963 Internat. Sympos. Berkeley)(Amsterdam) (J. W. Addison, Leon Henkin, and Alfred Tarski, eds.),North-Holland, Amsterdam, 1965, pp. 32934.

[She77] Saharon Shelah, Whitehead groups may be not free, even assuming CH, I , Israel J. Math. 28 (1977), no. 3, 193204.

[She80] Saharon Shelah, Whitehead groups may not be free even assuming CH,II , Israel J. Math. 35 (1980), no. 4, 257285.

[She81a] Saharon Shelah, The consistency of Ext (G, Z ) = Q, Israel J. Math.39 (1981), no. 12, 7482.

[She81b] Saharon Shelah, On the number of nonisomorphic models of cardinal-ity L , -equivalent to a xed model , Notre Dame J. Formal Logic22 (1981), no. 1, 510.

[She82a] Saharon Shelah, On the number of nonisomorphic models in L ,when is weakly compact , Notre Dame J. Formal Logic 23 (1982),no. 1, 2126.

[She82b] Saharon Shelah, Proper forcing , Lecture Notes in Math., 940,Springer-Verlag, Berlin-New York, 1982.

[She85] Saharon Shelah, On the possible number no(M ) = the number of non-isomorphic models L , -equivalent to M of power , for singular ,Notre Dame J. Formal Logic 26 (1985), no. 1, 3650.

[She86] Saharon Shelah, On the no(M ) for M of singular power , Around clas-sication theory of models (Berlin), Lecture Notes in Math., 1182,Springer-Verlag, Berlin, 1986, pp. 120134.

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[She98] Saharon Shelah, Proper and improper forcing , second ed., Springer-Verlag, Berlin, 1998.

[SV] Saharon Shelah and Pauli Vaisanen, On inverse -systems and thenumber of L , -equivalent, non-isomorphic models for singular , Toappear in J. Symbolic Logic. No. 644 in the list of Shelahs publica-tions.

Saharon Shelah:Institute of MathematicsThe Hebrew UniversityJerusalem. Israel

Rutgers UniversityHill Ctr-BuschNew Brunswick. New Jersey [email protected]

Pauli V aisanen:Department of MathematicsP.O. Box 400014 University of [email protected]

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