Saharon Shelah and Pauli Vaisanen- Almost free groups and Ehrenfeucht-Fraisse games for successors...

8/3/2019 Saharon Shelah and Pauli Vaisanen- Almost free groups and Ehrenfeucht-Fraisse games for successors of singular c

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Almost free groups and Ehrenfeucht-Frasse gamesfor successors of singular cardinals

Saharon Shelah Pauli V ais anen

October 6, 2003

Abstract

We strengthen non-structure theorems for almost free Abelian groupsby studying long Ehrenfeucht-Frasse games between a xed group of cardinality and a free Abelian group. A group is called -game-freeif the isomorphism player has a winning strategy in the game (of thedescribed form) of length . We prove for a large set of successorcardinals = + the existence of nonfree ( 1 )-game-free groups of cardinality . We concentrate on successors of singular cardinals. 1

1 Introduction

The problem of possible cardinals carrying a nonfree almost free Abeliangroup has already a long history, see e.g. [EM90]. Recall that if G is agroup of cardinality , G is L -equivalent to a free Abelian group if andonly if G is strongly -free [Ekl74]. Recall also that the L -equivalenceis characterized by an Ehrenfeucht-Frasse game of length (< elementsat the time). For an ordinal , a group G is called -game-free, if the iso-morphism player has a winning strategy in the Ehrenfeucht-Frasse gameof length between G and a free Abelian group (countable many elements

Research supported by the United States-Israel Binational Science Foundation. Pub-lication 787.

Research supported by the Academy of Finland grant 40734 and the Royal SwedishAcademy of Sciences.

1 2000 Mathematics Subject Classication: primary 20K20, 03C05, 03C75; secondary03E55.Key words: almost free groups, Ehrenfeucht-Frasse games.

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at the time). Let be a cardinal. We study existence of nonfree groups of cardinality + which are -game-free with < + . We concentrate ourattention on the case that is singular. This work continues [V ai01], wherethe case that is a successors of a regular cardinal is studied. The followingresult is the main theorem of the paper (presented in a more general formin Section 5).

Theorem 1 Let C0 denote the smallest set of cardinals such that 0 is in C0 and C0 is closed under the operations + and , + +1 . For every cardinal of the form + in C0 with 2, there exists a nonfree( 1)-game-free group of cardinality .

For an ordinal and of the form + , ( )-game-freeness (the ordinalmultiplication) of a group G implies that G is equivalent to a free Abeliangroup with respect to a deep innitary language L , even a strongerlanguage than L [Hyt90, Kar84, Oik97]. So our results can be interpretedas strengthened non-structure theorems for almost free Abelian groups.

Shelah proved in [She85] that the question of existence of nonfree almost freeAbelian groups is equivalent to a purely set theoretical question concerningexistence of nonfree almost free families of countable sets. A family is almostfree if all the subfamilies of cardinality strictly less than the whole family hasa transversal. A transversal for a subfamily is an injective choice function

whose domain is the subfamily. It is nowadays a standard custom to writethat NPT( , ) holds if there exists a family S of sets such that the elementsof S have cardinality , S is almost free, S is not free, and S has cardinality (in some papers NPT( , + ) is used instead of NPT( , )). For somehistory of NPT( , ) see, e.g., [She94, II.0] and [MS94, 0]. Recall thatNPT( , ) fail for all > if is a singular cardinal [She75]. Recall alsothat by [She96, 1], for every cardinal , NPT( , 0) holds iff there exists anonfree, -separable group of cardinality .

A transversal game on a family S of sets is a two players game, where on eachround i the rst player chooses a set s i from S and the second player, alsocalled transversal player, must answer with an element x i from s i so thatx i is distinct from the elements x j , j < i , the second player has chosen onthe earlier rounds (Denition 2.2). That means that the transversal playermust be able to choose an extendable transversal whose domain containsat least the sets chosen by the rst player. A family S is called -game-freeif the transversal player has a winning strategy in the transversal game of length on S . As can be expected, the question does there exist a nonfree

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disjoint sets r s | s R such that s r s has cardinality < for everys R;

then S and any family exemplifying NPT( , 0) can be amalgamated to geta family exemplifying NPT( , 0). The point of introducing R ()-familiesand R I ()-freeness in Section 4 is that the existence of such a family isalso necessary to build a canonical example of a family of countable sets. Inother words, the largest nicely incompact subset of (which denotes the rstcardinal xed point), dened in [She96, 2], coincides with the smallest setof cardinals below which contains 0 and is closed under amalgamationof R ()-families (Denition 5.1).

The main theorem of the paper is presented in Section 5. Namely we linktogether the pieces proved in [V ai01, Section 4], Section 3, and Section 4.For all nonzero n < , it is possible to build nonfree ( n n 1)-game-free families of cardinality n +1 level by level using the old methods,see [Vai01, Section 2]. To get an example of nonfree -game-free group of cardinality + with = , one needs Proposition 3.1 and [MS94, Theorem4 of 1]. To get examples for < < + , one may use an example for and apply [V ai01, Lemma 4.29], which just says that a certain type of anNPT( 2+ ,0)-skeleton is ( 2 1)-game-free, if 2 = 1+ and the previouslevel is 1-game-free (a group whose -invariant consists of ordinals of conality 2 is always 2-game-free, but not necessarily 2 + 2-game-free [Vai01, Section 2]). By [MS94, Theorem 4 of 1] and Proposition 3.1,one get examples for even larger s, e.g., = n for any n < . In factit follows that such examples exist for unbounded many s below the rstcardinal xed point.

Suppose that G is an Abelian group of cardinality + whose -invariant(G) contains stationary many conal ordinals, i.e., for some ltrationG | < + of G, the set cof( ) { < + | G +1 /G is not free} is

stationary in + . Assume, for a while, that ( G) cof() is in the idealI [+ ] of all good subsets of + (see, e.g., [She94, Analytical Guide 0]).Then by [V ai01, Lemma 2.11], ( ( +1))-game-freeness implies freeness forgroups G of cardinality + . Hence Theorem 2 is very close to the optimal

result provable from ZFC alone.What happens if is singular and ( G) is not in I [+ ]? (For the regularcase see [Vai01, Proposition 3.4].) In Section 6 we collect together somefacts about good points with respect to a scale for and we prove that,relative to the existence of a supercompact cardinal, it is possible to have anonfree group G of cardinality + (e.g. = n for any n < ), so that G

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closed under initial segments. Let S f denote the nal nodes of S , i.e., S f is the smallest subset of S with S = m | S f and m lh() . Aset S as above is called a -set if there exist uncountable regular cardinals | S S f such that

= ;

for every S S f , E S = { | S } is a stationary subset of (so S implies that < );

for every S S f and E S , card( ) (so > ).

The sequence | S S f is called the type of S . A -set S is said tohave height n if n is a nite ordinal such that lh() = n for all S f .

Denition 2.4 Suppose S is a -set of type | S S f . A subset S of S is called a sub- -set of S if S is a -set such that S f S f and thetype of S is the restriction | S S f of the type of S . A subset I of S f is small in S f if the set { m | I and m lh()} is not a sub--set of S .

Denition 2.5 [She85, Claim 3.2] We say that a -set S of type has a canonical form when the following demands are fullled. S has height n

and there exist sequences of cardinals n

| n < n and n

| n < n such that 0 = and for every S S f of length n ,

(i) either both of the following two properties are satised:

the set E S consist of regular limit cardinals and n = 0 (in thiscase n is called undened);

if n + 1 < n then for every E S , = and n +1 = 0(in this case n +1 is called undened);

(ii) or otherwise both of the following two properties are satised:

E S is a subset of cof(n ); if n + 1 < n then for every E S , = n +1 .

Fact 2.6 [She85, Claim 3.2] Every -set contains a sub- -set which is in a canonical form.

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Denition 2.7 [She85, Denition 3.4] Suppose is an uncountable regular cardinal, S is a -set of type = | S S f . An indexed family

= | S S f and <

is called a -system of type , if is the empty set (only for technical reasons) and for every S S f the sequence = | < satises that

is a strictly increasing continuous chain of sets;

the union of has cardinality ;

each has cardinality < .

A -system is called disjoint when the sets are innite regular cardinals. A tupleS, , S is called an NPT( , )-skeleton (of type and of height n) when

S is a -set of type ;

S is in a canonical form and its height is n;

the cardinals in are greater than ;

= | S S f and < is a disjoint -system of type ;

S is a family of sets of cardinality enumerated by {s | S f };

S is based on , which means that for every S f ,

s m n

m .

Additionally, the demands in [She85, Claim 3.6 and 3.7] are fullled (in case = 0 they are essentially equivalent to the beautifulness propertiespresented in [EM90, Denition VII.3A.2(16)]). Dene for every S f and m < n , that

sm +1 = s m +1 .Remark. sm +1 is denoted by sm in [She85].

For the readers convenience we repeat three of the additional demands (used in Section 3 ):

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(A) For all = S f , if s s = then there is an unique m < n such that s s = sm +1 sm +1 and for every l < n with l = m , (l) = (l).

(B) For every S f and m < n , if (m) has conality > , then thereis n < n with n = ( n is from the type of S ).

(C) In case = 0: For all S f and m < n , the sets sm +1 haveenumerations x,ml | l < with the following property: for every S f and n < , if x

,mn sm +1 then the initial segments x

,ml | l n

and x,ml | l n are equal.

Denition 2.9 Suppose = | < is an increasing sequence of regular cardinals approaching such that 0 > + . For functions f, g


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the cardinal for which E S cof() holds (and which exists by De-

nition 2.8) is such that cf() = .

Then there exists NPT( , 0)-skeleton S, , S such that S is -game-free for every < . Moreover, S is a sub--set of S and hence E S

cof()

holds ( S = S if does not contain limit cardinals).

The rest of the section is devoted to the proof of this proposition.

Suppose that S has height n(by the canonical form height is well-dened).Now note that nmust be at least 2: If has uncountable conality, thisfollows directly from our second demand. On the other hand, if has

countable conality, the claim follows from the demand that E S cof(0)together with Denition 2.8(B).

Remark. If n= 1 and E S cof(0), then S would not be 1-game-free

as explained in [V ai01, Example 4.4].

We let 1 denote the regular cardinal given by Fact 2.6, i.e., for every E S , = 1. By taking a suitable sub- -set S of S if necessary, we may assumethat

if cf() is uncountable, then there is a xed m < n such that for every S f , m = cf( ).

If there is no limit cardinals in , then Fact 2.6 guarantees that we canchoose S = S .

By Fact 2.10 choose a scale , f for (Denition 2.9), where = | < cf() and f = f | < .

The only modication of the family S = {s | S f } needed is thatthe rst coordinate ( s1) = s 1 of each s is slightly changed. Themodication depends on the conality of as follows: Fix from S f . If has countable conality, then dene

s1 = {F l | l < },

where F is a xed function from onto ran( f (0) ) (s1) . This deni-tion ensures that the property Denition 2.8(C) becomes fullled. In orderto choose the corresponding new -system, let , for every < , be

< 0 , and additionally, for all nonempty S S f and < ,

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dene to be the old . Otherwise, for the xed , cf() = m > 0 holds and we dene

s1 = f (0) (m) (s1) ,

and for every < , = (Denition 2.8(C) is satised since(s1) satises it). Note that now S, , S is an NPT( , 0)-skeleton, i.e., itfullls all the demands mentioned in Denition 2.8, because S , , S is anNPT( , 0)-skeleton.

In order to show that player II has a winning strategy in the game GT (S )for every < , it suffices to describe a winning strategy for player II in a

modied game of length for every regular cardinal with > > 1 ,where the rules of the modied game are exactly as in GT (S ), except thatplayer II is demanded to choose a transversal only if the index of the roundis a limit ordinal.

On every round i < player II chooses two elementary submodels M i+1and N i+1 of H ( ),, f , ,S, , S such that

is some large enough regular cardinal;

card( M i+1 ) = ;

card( N i+1 ) = ; N i+1 M i+1 ;

N i+1 contains all the elements chosen by player I;

+ 1 N i+1 ;

M i+1 is an ordinal denoted by i+1 ;

M i , N i N i+1 , where M 0 = N 0 = and for limit i, M i = j


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because then after i + < rounds, player II is able to continue (or start)with some transversal T i+ whose domain consist of the elements in S N i+ (contains the elements chosen by player I so far) and satises that T i T i+ (where T 0 = and for i which is a limit of limit ordinals, T i = j


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From this the claim (3.3) follows in the following way: By the denitionof the new rst coordinate there are two different cases according to theconality of . If has countable conality then choose a transversal T for s | S f i (s is the old rst coordinate), and dene the de-sired transversal T for s | S f i by setting for every S f i ,that T (s) is the nite restriction of F whose greatest element is the pairf (0) (), T (s) . Suppose then that cf( ) is uncountable. By the deni-

tion of s1 , when S f i , and the assumption (3.5), the following set issmall in S f i :

I 0 = S f i | s1 (R) = .

Choose any transversal U 1 for I 1 = S f i I 0 (I 1 S f has cardinality < ).By (3.2), I 1 = { I 1 | U 1() s1} must be small in S f i . Now the setI 0 I 1 is small in S f

i , and by (3.2) again, there is a transversal U 0 forI 0 I 1 with the property that U 0() s1 for all I 0 I 1. Because of Denition 2.8(A), the union T = U 0U 1 I 1 I 1 forms the desired transversalwitnessing that (3.3) holds. So (3.5) implies (3.3).

Note that { j | j < i } is a subset of N i . Dene to be the rst index incf() with > . For every j i dene a function h j < cf( ) bysetting h j () = 0 if < , and h j () = sup( N j ) otherwise. Clearlyhk () < h j () when k < j i and . For every j < i there is < iwith h j < f , since h j N i and N i i . Thus h j < f i for every j < i .To prove (3.5) it suffices to show that h i f i . For every j < i , there is thesmallest j < cf() satisfying that < cf() | h j () f i () j .By the denition of h j s, k < j for every k < j < i . By the assumptionE S

cof(cf()) = , cf(i) = cf( i ) = cf( ). Hence there exists suchthat j for every j < i . By the denition of j s, for every ,f i () sup{h j () | j < i } = h i (). Therefore (3.5) holds, and we haveproved (3.3).

Next we prove that there is some transversal for S i / R , and nally we explainhow to nd a transversal witnessing (3.4). In order to show that S i / R isfree, it suffices to conclude that for every j < i ,

(3.6) S ( j +1) / (S j R)

is free. Note that in this case i might be in E S or in its complement, but inboth cases S f j is a subset of N i for every j < i (because of the inequality > 1 > > , lh() > 1). Since ( j ) + 1 is not in E S , S ( j +1) / S (j )+1 is

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free. Since S ( j +1) / S (j )+1 is in N i , also

S ( j +1) / S (j )+1 / S ( j +1) / S (j )+1 N i = S ( j +1) / S (j )+1 (R S (j )+1 )

is free. Because both (S (j )+1 R) = ( S (j )+1 ) ( R ) and S j R hold, we have that S ( j +1) / (S j R) = S ( j +1) / S (j )+1 (R S (j )+1 ) .Hence (3.6) holds, and we have proved that S i / R is free.

Let X denote the set E S

i N i . Suppose, for the moment, that for every X , the set

(3.7) J = { S f | s s = for some S f i } is small in S f .

By (3.2) choose a transversal U for J , X , so that U () s1 hold forall J . By Denition 2.8(A) together with the fact that X N i is empty,ran( U ) R is empty. Consequently, the desired extension T in (3.4) canbe T W X U , where W is a restriction of some transversal witnessingthat S i / R is free into the set S i {s | J for some X }.

So it remains to prove (3.7). Fix some from X . Since f < f i , thereis with f () < f i () for each > . Assume that X S f

and S f i are such that s s is nonempty. From Denition 2.8(A) itfollows that s1 s1 = and there is with (m) = (m) = . From the newdenition of the rst coordinate it follows that f () = f i (). Therefore . This means that for every S of length m, the set { S f |s s = for some S f i } has cardinality < m = m = cf( ).Thus (3.7) holds and we have proved Proposition 3.1.

4 Building blocks of incompactness skeletons

In this section > are regular cardinals and is a decreasing sequence of regular cardinals. In the denitions below we present a variant NRT( , )of the notion NPT( , ). A simplest form of this variant is mentioned in[She94, Sh355: Fact 6.2(9)]. So we dene analogical notions of a transversalfor a subfamily and a free subfamily over another subfamily. There aretwo main motivations. Firstly, even if both NPT( 1 , 2) and NPT( 2 , 3)hold, for regular cardinals 1 > 2 > 3, NPT( 1, 3) does not need to hold:

Fact 4.1 [She97, Lemma 3.1] It is consistent with ZFC relative to existenceof two weakly compact cardinals, that for some regular cardinals > > 0,both NPT( , ) and NPT( , 0) hold, but NPT( , 0) does not hold.

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However, we show that an analogical transitivity property for NRT( , )holds, Corollary 4.19. Secondly, the introduced families and their canon-ical forms, called NRT( , 0)-skeletons, provide a unied picture of thoseconstruction methods of NPT( , 0)-skeletons presented in [MS94, 1] and[She97, 2] (see Fact 4.6). We shall deal with the ideal J nst of the productof nonstationary ideals of xed cardinals (the next denition) instead of considering just the ideal of the bounded subsets of . However, the readermay think in the beginning, if she or he wants, that = and I consistof the ideal of all bounded subsets of .

Remark. From Fact 4.5 below it follows that NRT( , )-skeletons arenecessary to pump up more complicated NPT( , 0)-skeletons. In fact

we shall see in Denition 5.1 and Conclusion 5.3, that the denition of anicely incompact set of regular cardinals from [She96, Denition 2.1] can beexpressed using NRT( , )-skeletons as building blocks. All together (whichmeans Fact 4.16, Denition 4.17, Lemma 4.18, and Lemma 5.2), NRT( , )-skeletons are handy tools to divide and amalgamate NPT( , 0)-skeletons.The reader is advised to look at Section 5 to understand our goal, if she orhe feel lost with the technicalities in this section.

Denition 4.2 Suppose = k | k < k is a nonempty decreasing sequence of innite cardinals. A family A is called an R ()-family ( R ()comes from the notion NRT( , )), when A fullls the following demands:

the elements of A are nite sequences of functions;

A is enumerated by {a i | i < card( A)};

each a i is of the form a i,l | l < lh( a i ) ;

a i,l s have domain .

Denition 4.3 Suppose is as in Denition 4.2. A sequence I = I k |k < k is called a -sequence of ideals, when each I k is a proper ideal on k containing all the bounded subsets of k . For two ideals I 0 and I 1 on 0 and 1 respectively, the product I 0 I 1 is the ideal of all subsets X of 0 1 such that

0 < 0 | 1 < 1 | 0, 1 X I 1 I 0.

For a regular cardinal , let J bd denote the ideal of all the bounded subsetsof , and if is uncountable, let J nst denote the ideal of all nonstationary

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subsets of . For a -sequence of ideals, let J nst denote the product of theideals in I k | k < k , where I k = J nst k if k is uncountable, and otherwise, I k = J bd0 .

Denition 4.4 Assume that I is a -sequence of ideals, A is an R ()- family, and A = {a i | i I } is a subfamily of A. A sequence b = bi |i I of functions is called an R I ()-transversal for A if the following properties are satised for every i I :

for some l < lh( a i ), bi a i,l and dom( bi ) is in I ;

the ranges of bi , i I , are pairwise disjoint.

A is called R I ()-free if there exists an R I ()-transversal for A (the empty sequence is transversal for the empty family). A family A is almost R I ()- free, when every subfamily of cardinality < card( A) is R I ()-free.

For a cardinal > max , we say that NRT I (, ) holds, if there exists an al-most R I ()-free R ()-family of cardinality which is not R I ()-free. In thecase = and I = J bd , NRT( , ) is an abbreviation for NRT I (, ).

First we see that existence of an NPT( , 0)-skeleton necessarily gives manyalmost free nonfree R ()-families (depending on the type of the skeleton).Later in Lemma 4.18 we shall see that existence of certain type of R ()-families is also sufficient condition for building NPT( , 0)-skeletons. Moregenerally, neat R ()-families can be transformed into a form of an NRT( , )-skeleton, which is an analog of NPT( , 0)-skeletons for R ()-families.

Fact 4.5 Let S, , S be an NPT( , 0)-skeleton of type and of height n. Suppose that there is a sequence n | n < n such that for every n < n and S of length n , n = holds ( does not contain regu-lar limit cardinals). Then for every n < n , NRT J nst n (n ,

n ) holds, wheren = m | n < m < n 0 and J nst n is the ideal given in Deni-tion 4.3 above. In fact, there exists an NRT( n , )-skeleton of type n ,see Denition 4.10.

Proof. Fix n < n and S S f of length n. Let kdenote the length of , i.e., k= n n + 1. Dene a family A to be a | E S , whereeach a is a function having domain and for every , the valuea ( ) is chosen in the following way. For xed and let , S f be suchthat , n = , , (n) = , and for every k < k 1, , (n + k) is the

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(k)s member of E S , n + k in -order. Dene a ( ) to be the (k 1)s

element of the countable set s n +1, = s, in the xed enumerationof it, recall Denition 2.8(C).

By Fodors lemma A cannot be R J nst ()-free. On the other hand A isalmost R J nst ()-free: Fix < n and let I denote the set { S f

| (n) n

must be small in S f

. It follows that the corresponding set Y

= { | , I } must be in J nst . So we may dene a transversal b | < for a | < by setting b = a Y . 4.5

There are many examples of nonfree almost free R I ()-families. Note thatthe examples below are NRT( , )-skeletons of height 1 for some of lengthat most 2.

Fact 4.6 Suppose is an uncountable regular cardinal.

(a) [She79, Lemma 23] If there exist a regular cardinal < and a nonre-

ecting stationary subset E of cof() (which means for all < ,E is nonstationary in ), then NRT( , ) holds.

(b) If = + and is a regular cardinal, then NRT( , ) holds (since cof() is a nonreecting stationary set, or for the other conalitiesuse (, ) given in the references of the next item).

(c) ([MS87] or [EM90, Theorem VI.3.9 and VI.3.10]) If is a singular cardinal of conality and V = L , then NRT( + , ) holds (becauseof (, )).

(d) [She79, Lemma 24] Suppose is a singular strong limit cardinal of co-

nality such that I [+

] = P (+

) (for a denition of I [+

], see e.g.[She91, Denition 2.1]). Then for every regular < with = ,NRT I (, , ) holds, where I = J bd , J bd .

(e) [MS94, Theorem 4] Suppose < are regular cardinals and = + +1 .Then NRT I (, , ) holds, where I = J nst , J bd .

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(f) [She97, Lemma 1.16] Suppose is a regular cardinal such that for =+ there is a scale , f whose good points S gd [f ] contains a closed and unbounded subset of + (Denition 2.9 and Denition 6.2). Then for every regular with < < , NRT I (+ , , ) holds, where I = J nst , J bd .

(g) [She94, Sh355:Claim II.1.5A] If is a singular cardinal of conality and pp I () >

+ , then NRT I (+ , ) holds. Moreover, when is un-countable, already pp( ) > + together with some weak assumptions im-ply NRT J bd (

+ , ) [She94, Analytical Guide 5.7(B) and Sh371: 0 and 1].

(h) [She94, Sh355:Theorem II.6.3] If is a singular cardinal of countableconality and cov(,, 1, 2) > + (e.g., strong limit and 0 > + ),then NRT( + ,0) holds.

The rest of this section is essentially based on a similar analysis and aconstruction of a -system as in [She85, 3]. Our presentation resembles[She85, Appendix] (or [EM90, VII.3A]). The next denition yields an analogof the notion a subfamily is free over an other subfamily.

Denition 4.7 Suppose A is an R ()-family. A sequence A | < iscalled a ltration of A if it is a continuous increasing chain of subfamilies,

such that the members have cardinality < and the union of the membersequals A.

For every A A, A denotes the set aA l< lh( a ) ran( a l).

Suppose I is a xed -sequence of ideals and A1, A2 are subfamilies of A.We denote by A2/ A1 the family {a L a | a A2 A1 and L a = }( I should be clear from the context), where L a denotes those indices l < lh( a) for for which ran( a l ) ( A1) is small, i.e.,

L a = l < lh( a ) | | a l ( ) A1 I .

We say that A2 is free over A1, if the family A2/ A1 is free (we omit theprex R I ()).

The next fact offers some basic facts needed to understand Lemma 4.9.

Fact 4.8 Suppose A is an R ()-family and I is a -sequence of ideals.

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(a) If A | < is a continuous chain of subfamilies of A such that A0is free and A +1 / A is free for every < , then the family max , every almost R I ()-free family of cardinality is free,i.e., NRT I (, ) does not hold.

Proof. The claim follows from [She75, Theorem 2.1] or [BD78, Theorem in0], because for a xed family A of cardinality , the relation A2 is freeover A1 for subfamilies of A satises the demanded axioms. However, webriey sketch why axioms IV of [Hod81, Theorem 5 in 4] hold. Denethe set S (A) of the subalgebras of A to be the set of all subfamilies of A. A subalgebra A1 S (A) is free when A1 has an R I ()-transversal, sayb , and a basis F of A1 is the set of all subalgebras A2 of A1 such thatthe corresponding restriction of b is a transversal for A1/ A2 (the proof of Fact 4.8(b)). Note that a basis F is fully closed unbounded above 0 asdemanded in axiom II (use h s as in the proof of Fact 4.8(b)). Axioms I,III, and IV hold by the denition. Axiom V can be proved by a similarconstruction as in the proof of Fact 4.8(a).

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Our next task is to prove that R ()-families can be transformed into acanonical form in the same way as families of countable sets are transformedinto incompactness skeletons (Lemma 4.13). In order to succeed in the proof we have to assume the R ()-family to be neat (Denition 4.12).

Denition 4.10 Suppose and I are as in Denition 4.4 and is a reg-ular cardinal greater than max . A tuple S, , A is called an NRT( , )-skeleton of type , when the following conditions hold (recall the denitionsof a canonical -set and a disjoint -system from Section 2):

S is a -set of type = | S S f ;

S has a canonical form and its height is n< ;

= | S S f and < is a disjoint -system;

A is an R ()-family of cardinality ;

A is enumerated by {a | S f };

every sequence in A has a xed length l< , where l n;

A is based on , i.e., for all S f and l < l , ran( a,l ) m lh( ) m ;

A is almost R J nst ()-free ( J nst

given in Denition 4.3).

Moreover, analogously to the denition of an NPT( , )-skeleton, we de-mand that for every n < n and S of length n the following conditionshold:

When the cardinal n , given by the canonical form of S (Denition 2.5), iswell-dened:

n > max implies that there is m < n such that the cardinal m ,given by the canonical form of S , is well-dened and m = n ;

if 0 < n max then n ran( ).

The index set l can be partitioned into n blocks Ln +1 | n < n (analo-

gously to the partition sn +1 | n < n of a set in an NPT( , 0)-skeleton)such that for every n < n , l Ln +1 , and S f , ran( a,l ) n +1 (by thedisjointness of , for all l Ln +1 and l Ln +1 , ran( a,l ) ran( a ,l ) = whenever n = n ).

For all in S S f , S , , A denotes the NRT( , )-skeleton of type | S S f , where

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S denotes the set { S | };

is the restriction of according to the new index set;

A denotes the family a L> lh( ) | S f , where

L> lh( ) =lh( ) n 0 and < (n 1), there are strictly less than indices S f such that and for some l Ln +1 , the set

| a,l ( ) is not in J nst .

Fact 4.11 If S, , A is an NRT( , )-skeleton, then there is no injectivechoice function for the family {s | S f }, where each s is the set

l< lh( a ) ran( a,l ). In particular, there is no R J nst ()-transversal for A.

Proof. The given family is based on the -system . Hence the claim followsfrom [She85, Claim 3.5] (or [EM90, VII.3.6]).

Denition 4.12 For technical reasons (needed in the proof of Lemma 4.13),we say that an R ()-family A is neat, when for all a, a A, l < lh( a), and l < lh( a ): a l = a l implies that the set

(0) | and a l( ) = a l ( )

is a nonstationary subset of 0, if 0 is uncountable, and otherwise, the set is nite.

Remark. We do not demand neatness in the denition of an NRT( , )-skeleton. Therefore Fact 4.6(d), Fact 4.6(e), and Fact 4.6(f) are examples of NRT( , )-skeletons.

In the denition of the neatness the most natural demand would be thatintersection of two different coordinates is small in the J nst sense, butsuch a denition would cause problems in the next lemma, since J nst is not0-complete when has length greater than 1.

Lemma 4.13 If A is a neat family witnessing that NRT J nst (, ) holds,then there is an NRT( , )-skeleton.

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Proof. We dene an NRT( , )-skeleton S , , A following the same ideasas in [She85, Claim 3.3] or [She85, Appendix: Proposition 4] (or [EM90,Proposition VII.3.7]), except that at the end we need a different trick.

Let be itself. By Fact 4.8(a) there is a ltration A | < of A and a stationary subset E of such that for every E, A +1 / A is not free. For every E, choose B A +1 A so that =card( B ) < is the smallest cardinal such that B is not free over A .Suppose > 0 holds. By the choice of B , the family B / A wit-nesses that NRT J nst ( , ) holds. By Lemma 4.9 is a regular cardinal.By Fact 4.8(a) there is a ltration A , | < of B and a station-ary subset E so that A , +1 is not free over A , A . Hence we

may continue choosing a subset B , of A , +1 (A A , ) having thesmallest possible cardinality , < for which B , is not free overA , A .Assume is a sequence of ordinals such that its length is n > 0 and n 1 >0 . Such sequences form the nal nodes S f of the desired -set S .Suppose also that all the sets A m , for nonzero m n, together with B arealready chosen so that B is not free over A , where A is an abbreviation for

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,i,l () = iff for each k < lh(), (k) is the (k)s ordinal (in -order) of the set { (k) | X ,i,l and k = k}.

The property of these maps used below is that if X ,i,l is not in J nst , thenfor every B :

(A) B J nst iff ,i,l 1[X ,i,l B ] J nst .

A new candidate for a could be a i,l X ,i,l ,i,l | a i,l a i A for some isuch that a i B . But then we face the real problem, why should such anA be almost free? A nal denition of a will be a concatenation of nitenumber of sequences of the lastly given form. How to choose a suitable niteset?We claim that the neatness of the original A guarantees that there is nochoice function picking different functions from the sequences in B / A ,i.e., there is no injective function f with domain B / A such that f (a) afor each a in the domain. Namely, assume that such a f exists and a | < enumerates B . By induction on < dene b to be the restrictionf (a ) ( A ), where

A = | f (a )( )


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of the coordinate a i,l i in a i / A . But then for all i = j I , a i,l i = a j,l jcontrary to the choice of F . Therefore, there is i I such that for somecoordinate a i,l from the nonempty sequence a i A , bi is a restriction of a i,l .Thus X ,i,l is not in J nst . By (A), dom( bi ) J nst implies that theset Y ,i is in J nst , where Y ,i = ,i,l 1[dom(bi ) X ,i,l ]. Hence wemay dene b = bi ,i,l Y ,i .

It remains to show that all the other demands listed in Denition 4.10 canbe fullled. Exactly as for the NPT( , 0)-skeletons, S , , and A can bemodied so that S is in the canonical form of type and height n, andmoreover, is disjoint.

By choosing a suitable sub- -set of S if necessary [She85, Claim 3.2(1)],we may assume that for all n < n and , S f , the sets l < lh( a) |ran( a,l ) n +1 and l < lh( a ) | ran( a,l ) n +1 are equal. Hencethere exists lsuch that lh(a) = l for every S f , and also, the requiredpartition Ln +1 | n < n of lexists.

When 0 < n < min is well-dened one can add n into (see Fact 4.16below), if not yet appeared there. So assume that n has the index k in and dene a,l ( ) = a,l ( ), (k ) for every l < l

and , where | < k is an increasing sequence of ordinals conal in (n) (change

accordingly).

For the rest of the properties (covering the case n > max ), a suitable mod-ication procedure is described in a very detailed way in [EM90, TheoremVII.3A.5] (e.g., the case that n > min and n ). 4.13

Corollary 4.14

(a) Suppose S, , A is an NRT( , 0)-skeleton of type . Then there is a family S of countable sets such that the triple

S, [ ]< 0 | S S f , S

forms an NPT( , 0)-skeleton.

(b) For every uncountable cardinal , NPT( , 0) holds if and only if thereis a neat family exemplifying that NRT( , 0) holds.

Proof. The reader may wonder why a family witnessing that NRT( , 0)holds is not straightforwardly an example of a family witnessing that

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NPT( , 0) holds. The answer is that, because R (0)-transversal is a strongernotion than the standard transversal, the non-freeness in the R (0)-sensedoes not guarantee the non-freeness in the standard sense.

(a) The existence of an NPT( , 0)-skeleton follows from Fact 4.11 togetherwith [She85, Claim 3.8] or [EM90, VII.3A.6]. To see that the structure of the -system can be almost preserved, consider a family {s | S f }such that each coordinate (w.r.t. to the slightly modied -system) s n +1corresponds to a well-chosen subset of all nite sequences of elements in

lL n +1 ran( a,l ) (the reason to use nite sequences is the same as in theproof of Proposition 3.1).

(b) From left to right the claim follows from [She85, Claim 3.8] or [EM90,VII.3A.6]. The other direction follows from Lemma 4.13 and (a). 4.14

The following two small facts are needed in the proof of Lemma 4.18.

Fact 4.15 For all NRT( , )-skeletons S, , A and I S f , I is small in S f if and only if there exists an R J nst ()-transversal for {a | I }.

Proof. A proof of this fact is similar to a proof of the property (3.2) inSection 3.

Fact 4.16 If there is an NRT( , )-skeleton of type , then for every regular cardinal < min , there exists an NRT( , )-skeleton of type ,where ran( ) = ran( ) { }.

Proof. Suppose = k | k < k is a decreasing enumeration of ran( ) { } and K denotes the index set {k < k | k = }. For every S f and l < lh( a ), replace the old coordinate a,l with a new one a,l , wherea,l has domain and for every , a,l ( ) = a,l ( K ). Dene anew family A to be {a | S f }, where a is a,l | l < lh( a ) . Theonly problem is that A should be almost R J nst ( )-free. However, for anyX , X J nst implies that { | K X } is in J nst aswell. Hence every R J nst

()-transversal b | I can be straightforwardlytransformed into the form of an R J nst ( )-transversal b | I , whereeach dom( b) equals | K dom( b) .

In the light of the previous fact and for purposes of the forthcoming tran-sitivity lemma, it makes sense to dene compatible skeletons.

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Denition 4.17 Suppose S , , A is an NRT( , )-skeleton of type and S , , A is an NRT( , )-skeleton of type . We say that S , , Ais compatible with S , , A if the following conditions are satised:

(A) min > max ;

(B) for all cardinals ran( ), if min then there is n below theheight of S such that for every S of length n , = .

Lemma 4.18 Suppose that an NRT( , )-skeleton S , , A of type iscompatible with an NRT( , )-skeleton S , , A of type . Then thereexists an NRT( , )-skeleton S, , A of type .

Proof. By Fact 4.16 we may assume that is an end segment of (doesnot have any effect on compatibility and if is 0 , just make sure that 0is the last element of ). By the assumption on , let N n be such thatthe initial segment of of cardinals not in is equal to n | n N . Thusfor every S f and , the concatenation N is a sequencein .

We dene S, , A in a straightforward manner by concatenating S , , Aand several copies of S , , A . So S f is dened to be { | S f and S f }, and S is { n | n < n }, where n= n + n , n is theheight of S , and n is the height of S .

To dene the other components, x an arbitrary = from S f (wherethis notation means S f and S f ). Let l denote l + l , where land l are the lengths of the sequences in A and A respectively.

Dene a = a,l | l < l , by setting for each l < l and ,

a,l ( ) = l, , a ,l N ,

and for every l with l l < l and ,

a,l ( ) = l, , a ,l l ( ) .

Dene a -system by setting for every S S f and < ,

= l S f ,

and for every S f , S S f , and < , set

= l { } .

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Clearly S and have the desired form and A = {a | S f } is based on. So to prove that S, , A is an NRT( , )-skeleton, it remains to showthat A is almost R J nst ( )-free.

Fix I S f of cardinality < . We dene an R J nst ( )-transversal b | I for {a | I }. The set I = { n | I } has cardinality < .Therefore, there is an R J nst ()-transversal b | I for {a | I }.Fix now I . Dene J to be the set of all S f such that

| N dom( b ) J nst .

Since dom( b ) is in J nst and is an end segment of , the complement

K = S f J must be small in S f (remember, X S f is small in S f if { N | X } is in J nst k , where k is the largest index below lh() withk > max ). By Fact 4.15 there exists an R J nst ( )-transversal b , | K for {a | K }.

Consider S f . If is in J , then dene

b = a,l | N dom( b ) ,

where l < l is the index with b a ,l . Otherwise is in K , and we candene

b = a,l dom( b , ).where l = l + m and m < l is the index with b , a ,l . 4.18

Corollary 4.19 For all regular cardinals > > , if NRT( , ) and NRT( , ) hold, then NRT( , ) holds.

5 Conclusion

Now we may put the pieces together. By Denition 2.5, if is a regularcardinal below the possible rst regular limit cardinal and S is a -set inthe canonical form of height nand type | S S f , then there existsequences = n | n < n and = n | n < n such that for every S S f of length n, both = n and E S cof(n ) hold. For the restof this section we assume that all the types of skeletons are given in thissimplied form.

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Remark. For simplicity we have chosen to be the domain of thefunctions appearing in the elements of R ()-families. Hence we must restrictourselves below the possible rst regular limit cardinal. It is possible toreplace with a full 0-set and J nst with small sets w.r.t. the xed0-set, but that makes the notation unnecessarily complicated.

Denition 5.1 Let CNRT denote the smallest set of cardinals such that

0 is in CNRT ;

if there exists an NRT( , )-skeleton of type such that there is noregular limit cardinal below + and ran( ) CNRT , then ran( ) CNRT .

Remark. One could expect that, analogously to Fact 4.5 and because of thetransitivity property Lemma 4.18, it would suffice to consider NRT( , )-skeletons of height 1 only (i.e., that in the denition above we could assume has length 1). However, an analogous proof does not work for NRT( , )-skeletons because, even in a xed level n, a transversal may choose fromseveral possible coordinates a,l , l Ln +1 , contrary to the NPT( , 0)-case,where on each level n there is only one coordinate, namely s n +1 .

The benet of Fact 4.5 is that we can separate levels of NPT( , 0)-skeletonsto independent building blocks and combine those blocks in various ways.

Of course in this countable case, we apply Corollary 4.14 (i.e., the coordi-nates a,l , l Ln +1 , of level n in the denition of an R (0)-family can beamalgamated to a single coordinate).

Lemma 5.2 Suppose < are cardinals in CNRT . There exists an NPT( , 0)-skeleton of type such that if is uncountable, ran( ).

Proof. If is the rst uncountable cardinal in CNRT , i.e., = 1, then theclaim holds by the denition. Suppose S 0, 0, A0 is an NRT( , )-skeletonof type , ran( )ran( ) CNRT , and the claim holds for all cardinals below. Note that the problematic case is max > > min .

By the induction hypothesis there is an NPT(max , 0)-skeleton S 1 , 1, S 1of type such that ran() ran( ) and if max > > 0 then ran( )holds too. Since S 0, 0 , A0 and S 1, 1, S 1 are compatible, it followsfrom Lemma 4.18 and Corollary 4.14, that there is an NPT( , 0)-skeletonS 2, 2 , S 2 of type .

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Suppose > min (hence ran( )) and n is the largest index with n > > n +1 . By the induction hypothesis there is an NPT( ,0)-skeleton S 3, 3, S 3 of such a type that it contains the sequence = m | n < m < n . By applying Fact 4.5 and Fact 4.16 to S 2 , 2 , S 2 ,

there is an NRT( , )-skeleton S 4 , 4 , A4 of type m | m n . NowS 3, 3 , S 3 and S 4 , 4 , A4 are compatible, and hence the claim follows

from Lemma 4.18 together with Corollary 4.14. 5.2

Conclusion 5.3

(a) For all cardinals in CNRT , NPT( , 0) hold. In particular, CNRT is a nicely incompact set of regular cardinals in the sense of [She97, Deni-tion 2.1].

(b) If K is a nicely incompact set of regular cardinals below the possible rst inaccessible cardinal, then K CNRT .

(c) All the cardinals from the set C0 dened in [MS94, Theorem 1 of 1] belong to CNRT (or look at Theorem 1). Particularly, all the regular cardinals below +1 are in CNRT , and CNRT is conal below the rst cardinal xed point.

Proof. (a) By Lemma 5.2.

(b) Suppose is a type of some NPT( , 0)-skeleton. By induction onincreasing order of the cardinals in , apply Fact 4.5 to show that ran( ) CNRT .

(c) By Fact 4.6(b) and Fact 4.6(e).

Our nal conclusion concerns game-free groups. As we have seen in Propo-sition 3.1, game-freeness of an NPT( , 0)-skeleton S, , S is closely con-nected to the possible value of E S

. Hence we have to look at a little bit

more restricted set of nicely incompact cardinals.

Denition 5.4 Let CGT denote the set of all cardinals in CNRT such that appears in a type of some NRT( , )-skeleton S, , A satisfying that CGT , and moreover, if is a successor of a singular cardinal , then E S

cof(cf()) = .

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Fact 5.5 All the cardinals from the set C0 (Conclusion 5.3 (c)) belong toCGT . In fact, all the examples in Fact 4.6, except the rst two of them, yield skeletons fullling the conality demand in the denition of CGT .

Theorem 2 Suppose is a cardinal such that both cf() and = + arein CGT . Then there exists a nonfree ( )-game-free group of cardinality ,where < is a regular cardinal such that if is a successor of a regular cardinal then = + , and otherwise = cf( ).

Proof. The proof proceeds by induction on increasing order of the cardinalsin CGT . By [Vai01, Lemma 4.17], [Vai01, Lemma 4.23], and [Vai01, Lemma4.29] it suffices to show existence of NPT( , 0)-skeleton such that if is

regular then S is -game-free, and if is singular, then S is -game-free forevery < .

The case successor of a regular cardinal follows from [V ai01, Fact 4.6 andLemma 4.29], since as an induction hypothesis, we may assume that thereexists a -game-free NPT( ,0)-skeleton S, , S , where is a cardinal suchthat = + and if is a regular cardinal, then = + and E S

cof().

The case successor of a singular cardinal follows from Conclusion 5.3 andProposition 3.1, since CGT CNRT and a modication of Lemma 5.2 for CGTholds (thus the demands of Proposition 3.1 can be fullled).

6 On cub-game and game-free groups

In this section is a singular cardinal, denotes the conality of , and is the successor cardinal of . We study existence of nonfree -game-freefamilies of cardinality , when > ( ) + . Our tools for that are someknown and modied results about cub-game for successors of singularcardinals. An analog study for successors of regular cardinals, which is aneasier case, is carried out in [Vai01, Section 2].

Denition 6.1 Suppose is an uncountable regular cardinal A , and

< is an ordinal. For notational purposes let x denote a set of regular cardinals below ( x tells in which conalities the limits are checked).We denote by GC x (A, ) the following two players cub-game. The players,player I (also called outward player) and player II (also called inwardplayer), choose in turns a sequence i | i < of ordinals such that

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player I chooses ordinals 0 < and i+2 < , for i < , with i+2 > i+1 ;

if i is a limit ordinal, player I must choose the ordinal i = sup j i .

Player II wins a play if for all limit ordinals i for which cf(i) x, i belongsto A.

We say that player II wins GC x (A, ) if there exists a winning strategy for player II in GC x (A, ). For a regular cardinal below , GC (A, ) is a shorthand for GC {}(A, ).

Denition 6.2 ([She97, Def. 1.9], [MS94, Def. 2 of 1], [She91, 1]) Sup-pose , f is a scale for a singular cardinal of conality (Denition 2.9),where = | < and f = f | < . We denote by S gd [f ] theset of all good points w.r.t. f , i.e., s below such that for some un-bounded set A and < , f () < f () holds whenever < A and < . Let S ngd [f ] denote the set of all limit ordinals < such that S gd [f ] and cf() > = cf( ).

Proposition 6.3 Assume that the following conditions are fullled:

is a singular cardinal, = + , and cf() = ;

there is a scale , f for such that S ngd [f ] is stationary in ;

there exists an NPT( , 0)-skeleton S, , S of type such that E S S ngd [f ];

if is uncountable then there is n below the height of S such that for every S of length n , = (where is from the type ).

Then there exists a nonfree group of cardinality , which is -game-free for every < (and even in a longer game).

Proof. By Proposition 3.1 we may assume that S, , S is chosen so thatS is -game-free for every < . By [Vai01, Lemma 4.23] we know that S is -game-free. Fix a below . As in [Vai01, Lemma 4.29], we can x altration S | < of S , and ensure using a suitable bookkeeping, that

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after any block of moves by the players of the transversal game GT (S ),there is some < such that the elements chosen by the players are fromS , and moreover, all the elements of S has been chosen.

Suppose is a regular cardinal below such that E S cof(). Note that

E S S ngd [f ] implies > . By Lemma 6.7 and Lemma 6.8, player II has

a winning strategy in the cub-game GC (S gd [f ], ).

During the transversal game GT (S ), player II can additionally use his win-ning strategy in the cub-game GC (S gd [f ], ) to ensure that after arbitrarymany rounds of the blocks of moves in the transversal game, the elementschosen by the players are exactly the elements in S for some which is notin E S

= { < | S / S is not -free} S ngd [f ] cof().

By [Vai01, Lemma 4.25], the family S / S = {s S | s S S } is-game-free. Therefore player II can continue the transversal game GT (S )one more round of the block of moves. During these new moves playerII uses his bookkeeping and winning strategy in the cub-game again, andso on, up to all the required moves. Now the claim follows from [Vai01,Lemma 4.17]. 6.3

Before changing the subject to the winning strategies in the cub-game, weask: for which singular cardinals can the demands of the last propositionbe fullled? We need few lemmas before the conclusion.

Lemma 6.4 [She79, Claim 27] Suppose is a supercompact cardinal, is a singular cardinal, = cf( ), < < , = + , and , f is a scale for . There exists a singular cardinal < such that cf() = and S ngd [f ] cof(+ ) is stationary in .

Proof. Let us rst show that S ngd [f ] is stationary in . Suppose, contraryto this claim, that C is a cub of with C S ngd [f ] = and j is an embeddingfrom V onto an inner model M satisfying that j () = for every < , j ( ) , and [M ] M . Let be sup j []. The set j [] is in M and is in j (C), because C is a cub of , < sup j (C) = j (), and j [] j (C).So is good w.r.t. j (f ) for j () in M and j [] is a conal subset of . By[MS94, Lemma 6] there exists a conal subset A of j [] and witnessing thegoodness of . Dene A to be the set j 1[A]. Then for every < A , theinequality f ( ) < f ( ) holds, since j is an elementary embedding, j ( ) = , j (), j ( ) A, and ( j (f )) j ( )( ) < ( j (f )) j ( ) ( ). This is a contradiction,since A has cardinality , and the set f ( ) | < has cardinality < < (where is from ).

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Now the desired conclusion follows from the fact that is a successor of thesingular cardinal of conality and has conality in M : If Cwhenever supC = and has conality , where is a successor of a singular cardinal having conality , then j [C] has this property too. Sothe assumption C S ngd [f ] = leads to a contradiction as above. 6.4

Lemma 6.5 Suppose is a regular cardinal, = > , = + , , f is a scale for , and moreover, 2 = + . Then S ngd [f ] cof(+ ) and S ngd [f ] is nonstationary in , for every < .

Proof. Suppose below has conality > + . We show that is good (of course, all the ordinals of conality < are good). By [She94, II.1.2A(3)],

2 = + implies that there exists an exact upper bound g of f | < (i.e., g is an < -upper bound for f | < such that for every h ,h < g implies that h < f for some < ).

Now argue as in [MS94, Case 1 of the proof of Lemma 5 in 1]: By [MS94,Lemma 7] the set < | cf g() > is in J bd . Assume, toward acontradiction, that is not good. By [MS94, Lemma 6] cf g() | < is not eventually constant, even though, its range is a subset of ( + 1)modulo J bd . Since there are < cardinals in the range of this sequence,there must exists different cardinals 1 and 2 such that both < |cf g() = 1 J bd and < | cf g() = 2 J bd hold. However,by [MS94, Lemma 8], 1 = = 2 must hold, a contradiction.

The claim on nonreection follows from the denition of goodness: if is inS gd [f ], then there is a closed unbounded subset of consisting of ordinalsin S gd [f ] only.

Using [She91, Fact 4.2] (similarly to [She79, Conclusion 29]) we get thefollowing conclusion.

Lemma 6.6 Suppose is a supercompact cardinal, GCH holds, < isa regular cardinal below , is + , and is + . Then there is a forcing extension, where ZFC + GCH holds, no bounded subset of is added, is

the singular cardinal , = +

, and all the assumptions of Proposition 6.3 hold too.

Proof. If NPT( , 0) does not hold, shoot a nonreecting stationary subsetF of cof(0) by a forcing notion described, e.g., in [She96, Proof of Lemma3.1]. This is for building the desired NPT( , 0)-skeleton at the end of this

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proof. Then no bounded subset of is added, remains supercompact, andGCH still holds.By Lemma 6.4, there is a cardinal which is a successor of a singular cardinal < so that cf( ) = and S ngd [f ]cof() is stationary in . Let E 1 denotethis stationary set. Note that by the form of and , < + = < holds.

Using Levy collapse Col( , < ), collapse all the cardinals below to .Because of GCH and cf( ) = , Col(, < ) has cardinality in V. Recallthat Col( , < ) is -complete. So in V Col( ,


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except . For every < , player II has a winning strategy in the gameGC= (S gd [f ], ).

Proof. First of all let be the least index below with < . Supposethat the players has already chosen the ordinals j | j i and playerII should choose i+1 . Assume that player II has picked during the earlierrounds also functions h j | j < i is odd satisfying for each odd j that

h j


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Lemma 6.8 Assume are cardinals, is + , A , and x a set of regular cardinals below . Suppose that player II wins GC x (A, ) for every

< , and moreover, if is a regular cardinal, player II wins GC x+1 (A, ).Then player II wins GC x (A, ) for every < .

Proof. This is a known fact, presented e.g. in [V ai01, Section 2].

It is again a known fact that the inward player has a winning strategy even inmuch longer games of the form GC xT (A, ), where the length of the gameis measured by the tree T of closed subsets of A of order type + 1, < (ordered by the end extension). Details are presented in [Vai01, Section 2].

References

[BD78] Shai Ben-David. On Shelahs compactness of cardinals. Israel J.Math. , 31(1):3456, 1978.

[Ekl74] Paul C. Eklof. Innitary equivalence of abelian groups. Fund.Math. , 81:305314, 1974. Collection of articles dedicated to An-drzej Mostowski on the occasion of his sixtieth birthday, IV.

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

[FM97] Matthew Foreman and Menachem Magidor. A very weak squareprinciple. J. Symbolic Logic, 62(1):175196, 1997.

[HHR00] Taneli Huuskonen, Tapani Hyttinen, and Mika Rautila. On po-tential isomorphism and non-structure. Preprint , 2000.

[Hod81] Wilfrid Hodges. In singular cardinality, locally free algebras arefree. Algebra Universalis , 12(2):205220, 1981.

[Hyt90] Tapani Hyttinen. Model theory for innite quantier languages.Fund. Math. , 134(2):125142, 1990.

[Kar84] Maaret Karttunen. Model theory for innitely deep languages.Ann. Acad. Sci. Fenn. Ser. A I Math. Dissertationes , (50):96,1984.

[MS87] Alan H. Mekler and Saharon Shelah. When -free implies strongly-free. In Abelian group theory (Oberwolfach, 1985) , pages 137148. Gordon and Breach, New York, 1987.

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[MS94] Menachem Magidor and Saharon Shelah. When does almost freeimply free? (For groups, transversals, etc.). J. Amer. Math. Soc. ,7(4):769830, 1994.

[Oik97] Juha Oikkonen. Undenability of -well-orderings in L . J.Symbolic Logic, 62(3):9991020, 1997.

[She75] Saharon Shelah. A compactness theorem for singular cardinals,free algebras, Whitehead problem and transversals. Israel J.Math. , 21(4):319349, 1975.

[She79] Saharon Shelah. On successors of singular cardinals. In Logic Col-loquium 78 (Mons, 1978) , pages 357380. North-Holland, Ams-terdam, 1979.

[She85] Saharon Shelah. Incompactness in regular cardinals. Notre DameJ. Formal Logic , 26(3):195228, 1985.

[She91] Saharon Shelah. Reecting stationary sets and successors of sin-gular cardinals. Arch. Math. Logic , 31(1):2553, 1991.

[She94] Saharon Shelah. Cardinal arithmetic , volume 29 of Oxford LogicGuides . The Clarendon Press Oxford University Press, New York,1994. Oxford Science Publications.

[She96] Saharon Shelah. If there is an exactly -free abelian group thenthere is an exactly -separable one. Journal of Symbolic Logic ,61:12611278, 1996.

[She97] Saharon Shelah. Existence of almost free abelian groups and re-ection of stationary set. Mathematica Japonica , 45:114, 1997.

[Vai01] Pauli Vaisanen. Almost free groups and longehrenfeucht-frasse games. 2001. Preprint,http://www.math.helsinki.fi/ pavaisan/ .

Saharon Shelah:Institute of MathematicsThe Hebrew UniversityJerusalem, Israel

Department of MathematicsRutgers University

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Pauli V aisanen:Department of MathematicsUniversity of [email protected]

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Saharon Shelah and Pauli Vaisanen- Almost free groups and Ehrenfeucht-Fraisse games for successors...

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