saharon shelah- strong dichotomy of cardinality

8/3/2019 Saharon Shelah- Strong Dichotomy of Cardinality

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STRONG DICHOTOMY

OF CARDINALITY

SH664

Saharon Shelah

Institute of MathematicsThe Hebrew University

Jerusalem, Israel

Rutgers UniversityMathematics DepartmentNew Brunswick, NJ USA

Abstract. We investigate strong dichotomical behaviour of the number of equiva-lence classes and related cardinal.

Saharon: compare with Journal proofs!

2000 Mathematics Subject Classification. 20K99Key words and phrases: Dichotomies in uncountable cardinals, abelian groups, Ext, p-rank

Research supported by the German-Israeli Foundation for Scientific ResearchI would like to thank Alice Leonhardt for the beautiful typing.Written Spring 97First Typed - 97/Sept/2Latest Revision - 02/Jan/30

Typeset by AMS-TEX

1


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2 SAHARON SHELAH

Annotated Content

0 Introduction

1 Countable Groups

[We present a result on a sequence of analytic equivalence relations on P()and apply it to 0-system of groups getting a strong dichotomy: beinginfinite implies cardinality continuum sharpening [GrSh 302a].]

2 On -analytic equivalence relations

[We generalize theorems on the number of equivalence classes for analyticequivalent relations replacing 0 by regular, unfortunately this is only

consistent. Noting that if we just add many Cohen subsets to we getsomething, but first the dichotomy is +, = 2 rather than , = 2,second we assume much less.]

3 On -systems of groups

[This relates to 2 as the application relates to the lemma in 1.]

4 Back to the p-rank of Ext

[We show that we can put the problem in the title to the previous context,and show that in Easton model, 2 and 3 apply to every regular .]

5 Strong limit of countable cofinality

[We generalize the theorem on 0 systems of groups from 1, replace 0 bya strong limit uncountable cardinal of countable cofinality; this continues[GrSh 302a].]


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STRONG DICHOTOMY OF CARDINALITY 3

0

A usual dichotomy is that in many cases, reasonably definable sets, satisfies the

continuum hypothesis, i.e. if they are uncountable they have cardinality continuum.A strong dichotomy is when: if the cardinality is infinite it is continuum, as in [Sh273]. We are interested in such phenomena when = 0 is replaced by regularuncountable and also by = or more generally by strong limit of cofinality 0.

Question: Does the parallel of 1.2 holds for e.g. ? portion?This continues Grossberg Shelah [GrSh 302], [GrSh 302a] and see history there.

We also generalize results on the number of analytic equivalence relations, contin-uing Harrington Shelah [HrSh 152] and [Sh 202] and see history there.On the connection to the rank of the p-torsion subgroup see [MRSh 314] and history

there. See more [ShVs 719].On Ext(G,Z), rkp(Ext(G,Z) see [EM].


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4 SAHARON SHELAH

1 Countable groups

Here we give a complete proof of a strengthening of the theorem of [GrSh 302a],

for the case = 0 using a variant of [Sh 273].

1.1 Theorem. 1) Suppose

(A) is 0. Let Gm, m,n : m n < be an inverse system whose inverselimit is G with n, such that |Gn| < . (So m,n is a homomorphismfrom Gn to Gm, , , = , and , is theidentity).

(B) Let I be an index set. For every t I, let Htm, tm,n : m n < be

an inverse system of groups and Ht with tn, be the corresponding inverse

limit and Htm of cardinality .

(C) Let for every t I, tn : Htn Gn be a homomorphism such that all dia-

grams commute (i.e. m,n tn =

tm

tm,n for m n < ), and let

t be

the induced homomorphism from Ht into G.

(D) I is countable1

(E) For every < and t I there is a sequence fi G : i < such thati < j fif

1j / Rang(

t).

Then there is fi G : i < 2 such that

i = j & t I fif1j / Rang(

t).

2) We can weaken in clause (A) to (A)

replacing |Gn| < by |Gn| , if wechange clause (E) to

(E) for everyt I, m < there are n, f such that f is a member of G, n 2 (seebelow) such that P = {{ < : () = 1} : lim(T)}.2) T is a -perfect tree if:

(a) T >2 is non-empty

(b) T & < g() T

(c) if < is a limit ordinal, 2 and ( < )( T), then T

(d) if T,g() < < then there is , T 2

(e) if T then there are -incomparable 1, 2 T such that 1 & 1.

3) Lim(T) = { : g() = and ( < )( T)}.

Proof of 2.3.Let T = >2.

Let N and r 2 for < be as in Definition 2.2. We identify r with { < :r() = 1}.By clause () of Definition 2.2(2) clearly

()0 if < (), and = < , then E define an equivalence relation inN[r, r] on P()

N[r,r].

It is enough to prove assuming = < and < () that,

()1 rEr.

By clause () of Definition 2.2(2) it is enough to prove

()2 N[r, r] |= rE.

Assume this fails, so N[r, r] |= rEr then for some i <

(r i, r i) (>2)(>2) r1Er

2

and without loss of generality i > . Define r 2 by

r(j) =

r(j) if j = i1 r(j) if j = i

So also (r, r) is a generic pair for>2 >2 over N and (r i, r i) =

(r i, r i) hence by the forcing theorem

()3 N[r, r] |= rEr.


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But r, r, r N[r, r] = N[r, r]. As we are assuming that ()2 fail (towardcontradiction) we have N[r, r] |= rEr and by ()3 and the previous sentencewe have N[r, r] |= rEr so together by ()0 we have N[r, r] |= rEr hence

V |= rEr, a contradiction to assumption (b). 2.3

2.6 Definition. We call Q a pseudo -Cohen forcing if:

(a) Q is a nonempty subset of {p : p a partial function from to {0, 1}}

(b) p Q q p q

(c) Ii = {p Q : i Dom(p)} is a dense subset for i <

(d) define Fi : Ii Ii by: Dom(Fi(p)) = Dom(Fi(p)) and

(Fi(p))(j) = p(j) if j = i

1 p(j) if j = i

then Fi is an automorphism of (Ii, 2, ) by Q.

2.8 Observation: So if V |= G.C.H., P is Easton forcing, then in VP for everyregular , for Q = ((>2)V , ) we have: Q is pseudo -Cohen and in VP we have

is (, 2)-w.c.a.

2.9 Discussion: But in fact being (, 2)- w.c.a. is a weak condition.

We can generalize further using the following definition

2.10 Definition. 1) For r0, r1 2 we say (r0, r1) or r0, r1 is an R-pseudo Cohenpair over N if (R is a definition (in (H(+), )) of a relation on P() (or 2), thedefinition belongs to N and) for some forcing notion Q N and Q-names r

0, r

1

and G Q (G V) generic over N we have:

(a) r0[G] = r0 and r

1[G] = r1

(b) for every p G, for every i < large enough and () < 2 there is G Qgeneric over N such that: p G and (r

[G

])(j) = (r[G])(j) (j,) =

(i, ())

(c) for < (), R is absolute from N[G] and from N[G] to V.


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10 SAHARON SHELAH

2) We say is -p.c.a for R if for every x H(+) there are N, ri : i < suchthat:

(a) N is a transitive model of ZF C

(b) for i = j < , (ri, rj) is an R-pseudo Cohen pair over N.

3) We omit R if this holds for any -sequence of1

1 formula in H(+).

Clearly

2.11 Claim. 1) If is -p.c.a for E, E an equivalence relation on P() andA B & |B\A| = 1 AEB, thenE has equivalence classes.

2) Similarly if E =


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3 On -systems of groups

3.1 Hypothesis. = cf().We may wonder does 2.3 have any cases it covers?

3.2 Definition. 1) We say Y = (A, K, G, D, g) is a -system if

(A) A = Ai : i is an increasing sequence of sets, A = A = {Ai : i < }

(B) K = Kt : t A is a sequence of finite groups

(C) G = Gi : i is a sequence of groups, Gi tAi

Kt, each Gi is closed

and i < j Gi = {g Ai : g Gj} and

G = {g

tA

Kt : (i < )(g Ai Gi)}

(D) D = D : (a limit ordinal) , D an ultrafilter on such that < [, ) D

(E) g = gi : i < , gi G and g

i Ai = eGi = eKt : t Ai.

Of course, formally we should write AYi , KYt , G

Yi , D

Y , g

Yi , etc., if clear from the

context we shall not write this.2) Let Y be the same omitting D and we call it a lean -system.

3.3 Definition. For a -system Y and j + 1 we say f cont(j,Y) if:

(a) f = fi : i < j

(b) fi G

(c) if < j is a limit ordinal then f = LimD (f ) which means:

for every t A, f(t) = LimD fi(t) : i <

which means

{i < : f(t) = fi(t)} D.

3.4 Fact: 1) If f cont(j,Y), i < j then f i cont(i,Y).2) If f cont(j,Y) and j < is non-limit, and fj G then

f fj cont(j + 1,Y).


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12 SAHARON SHELAH

3) If f cont(j,Y) and j is a limit ordinal , then for some unique fj G wehave f fj cont(j + 1,Y).4) Ifj + 1, f G then f = f : i < j cont(j,Y).

5) If f , g cont(j,Y), then figi : i < j and f1i : i < j belongs to cont(j,Y).

Proof. Straight (for part (3) we use each Kt is finite).

3.5 Definition. Let Y be a -system.1) For g j(G) and j we define fg G by induction on j for all such g asfollows:

j = 0: fg = eG = eKt : t A

j = i + 1: fg = fgigi

j limit: fg = LimD fgi : i < j

2) We say g is trivial on X if i X g(g) gi = eG .3) For 2 let g = gi : i < g(), where

gi =

gi if (i) = 1

eG if (i) = 0

recall gi is part ofY (see Definition 3.2).

3.6 Claim. 1) If i j and g, g, g j(G), g i = g i, g is trivial on [i, j),

g [i, j) = g [i, j) and g is trivial on i, then:

fg = fgfg and fg = fgi.

2) For

2, f(g

) = Limf(gi

) : i < (i.e. any ultrafilter D

on containingthe co-bounded sets will do), so Y, a lean -system, is enough.

Proof. Straight.


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3.7 Claim. LetY be a -system (or just a lean one), H a subgroup of G for < () andE the equivalence relation [f

(f)1 H] and assume: > i gi / H.

(1) The assumption (B) of 2.3 holds with fA = f(g) when A , 2, A =

{i : (i) = 1}

(2) if in addition A, K, G K, D, g H(+) and H : < () is (, )-w.c.a., then also assumption (A) of 3.3 holds (hence its conclusion).

Proof. Straight.

3.8 Claim. Assume

(A) Y a -system (or just a lean one), Ai +, |Ai| , Gi H(+)

(i) () ,

(ii) H = Hi : i , < (),

(iii) i,j : Hj H

i a homomorphism,

(iv) for i0 i1 i2 we have i0,i1 i1,i2

= i0,i2 ,

(v) i : Ht Gi,

(vi) i i,j(f) = (

j (f)) Ai,

(vii) H, is the inverse limit (with

i,) of H

i ,

i,j,

i : i j < and

(viii) i < Hi H(+)

(B) H = Rang().

Then

() the assumptions of 3.7 holds

() if is (, )-w.c.a. then also the conclusion of 3.7, 2.3 holds so there areh G for M such that = < & < () ff

1 / H.

Proof. Straight.

We can go one more step in concretization.


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14 SAHARON SHELAH

3.9 Claim. 1) Assume

(a) L is an abelian group of cardinality

(b) p a prime number(c) if L L, |L| < , then Extp(L,Z) = 0

(d) is -w.c.a. (in V).

Then rp(Ext(L,Z)), see definition below.2) If (a), (b), (d) above, > , strongly inaccessible then rp(Ext(L,Z)) / [, ).

3.10 Remark. 1) For an abelian group M let prime p and rp(M) be the dimensionof the subgroup of {x M : px = 0} as a vector space over Z/pZ.

2) For an abelian group M let r0(M) be max{|X| : X M\Tor(M) and is inde-pendent in M/Tor(M).

Proof. Without loss of generality L is 1-free (so torsion free).Without loss of generality the set of elements of G is . Let A = A = , L = L,for j < , Aj a proper initial segment of such that Lj = L Aj is a pure subgroupof L, increasing continuously with j.Let Kt = Z/pZ, Gi = HOM(Li,Z/pZ).Let () = 1, so = 0; let Hi = HOM(Li,Z) and (

i (f))(x) = f(x) + pZ, M =

Rang() for i j let i,j : Gj Gi is i,j(f) = f Gi. We know that

rp(Ext(G,Z)) is (G : M0). By assumption (d) for each i < we can choosegi G\M such that g

i Li is zero. The rest is left to the reader (using 3.8 using

any lean -system Y with Gi, Kt, (), i,j, as above (and D for limit ordinal< , any ultrafilter as in 3.2). 3.9


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4 Back to the p-rank of Ext

For consistency of no examples see [MRSh 314].

4.1 Definition. 1) Let

Z =

: = p : p < prime or zero and for some

abelian (1-free) group L, p = rp(Ext(G,Z))

.

2) For an abelian group G let rk(G) = Min{rk(G) : G/G is free}.Clearly Z is closed under products. Let P be the set of primes.

Remember that (see [Sh:f, AP], 2.7, 2.7A, 2.13(1),(2)).4.2 Fact: In the Easton model if G is 1-free not free, G G, |G| < |G| G/G

not free then r0(Ext(G,Z)) = 2|G|.

4.3 Fact: 1) Assume is a strong limit cardinal > 0, cf() = 0, = +, 2 = +

and some Y []+

is -free, (equivalently +-free, see in proof).Let P0, P1 be a partition of the set of primes.Then for some 1-free abelian group L, |L| = +, 2 = r0(Ext(G,Z)) and p P1 rp(Ext(G,Z)) = 2

and p P0 rp(Ext(G,Z)) = 0.

Remark. On other cardinals see [MRSh 314], close to [MkSh 418, Th.12].

Proof. For notational simplicity assume P0 = . Let Y = {i : i < }. Letpr:2 be a pairing function, so pr(pr1(), pr2()) = . Without loss ofgenerality i(n) = j(m) n = m & i m = j m. Let L be


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16 SAHARON SHELAH

Arriving to i first choose a function let hi : i be such that j < i hi(j) > pjand {j : [hi(j), )} : j < i is a sequence of pairwise disjoint sets(possible as Y is +-free). Second choose gi such that clauses () + () holds

with n = hi(j), this is possible as the choice of h splits the problem, that is,the various cases of () (one for each j) does not conflict. More specifically, firstchoose g {x : ( j < i)()(hi(j) < = j()) as required in (),possible as L is free. Second by induction on m hi(j) we choose bm+1 such that0/pZ = bm+1/piZ + fi(xj(m)) fi(xj(m)) and then choose gi(xj(m)), g(xj(m))

such that the m-th equation in clause () for j holds. Let i =n 1 extensions to h

Hom(G+1,Z/pZ) hence Hom(G,Z/pZ) has cardinality 2+ > 2, whereas

every h Hom(L,Z) has at most one extension to h+ Hom(G,Z), sothe result follows.]

4 r0(Ext(G,Z)) = 2+

[Why? Similar to 3, i.e. using cardinality considerations.]


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4.3

4.4 Question: Do we have compactness for singular for Extp(G,Z) = 0?

4.5 Claim. [Omitted, see [Sh 724] and x.x.]

4.6 Question: If Z can we derive Z by increasing some ps?

4.7 Fact: If i = ip : p P {0} Z for i < and p =i


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5 Strong limit of countable cofinality

We continue [GrSh 302] and [GrSh 302a].

5.1 Definition. 1) We say A is a (, I)-system ifA = (, I, G, H, , ) whereG = G : , H = Ht : t I, Ht = Ht : , = ,,

t, :

, t I, = t : t I, ) satisfies (we may write A , t,A, , etc.)

(A) is 0 or generally a cardinal of cofinality 0

(B) Gm, m,n : m n < is an inverse system of groups whose inverse limitis G with n, such that |Gn| . (So m,n is a homomorphism from Gnto Gm, , , = , and , is the identity).

(C) I is an index set of cardinality . For every t I we haveHtm, tm,n : m n < is an inverse system of groups and Ht with tn,being the corresponding inverse limit Ht with

tm, and H

tm has cardinality

.

(D) for every t I, tn : Htn Gn is a homomorphism such that all diagrams

commute (i.e. m,n tn = tm

tm,n for m n < ), and let

t be the

induced homomorphism from Ht into G

(E) G0 = {eG0}, Ht0 = {eHt0} (just for simplicity).

2) We say A is strict if |Gn| < , |Htn| < , |I| < . Let Et be the following

equivalence relation on G : fEtg iff f g

1

Rang(t

).3) Let nu(A) = sup{ : for each n < , there is a sequence fi : i < such thatfi G and n,(fi) = n,(f0) for i < and i < j < & t I fiEtfj}.We write nu(A) =+ to mean that moreover the supremum is obtained. Letnu+(A) be the first such that for n = 0, there is no fi : i < as above(so nu(A) nu+(A) and if nu(A) > then nu(A) nu+(A) nu(A)+

and nu(A) < nu+(A) implies nu(A) is a limit cardinal and the supremum notobtained).4) We say A is an explicit (, J)-system if: A= (, J, G, H, , ) and

() = n : n < , J = Jn : n < () n < n+1, Jn Jn+1,

() letting A =n


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5) We add in (4), full if

() |Htn| n.

6) For an explicit (, J)-system A let nu+ (A) = sup{+:for every n < there is asequence fi : i < such that fi G, and n,(pi) = n,(f0) for i < and i < j < & t Jn fiEtfj}.7) For a -system A, we define nu+ (A) similarly, except we say: for some J =

Jn : n < such that I =n


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Clearly

()1() G is a subgroup of G

() {G :

increasing} is directed, i.e. if (n < )(n) (n)) where, then G G

() G = {G : (increasing)}.

First assume cf() = 0 so as cf() > 20 for some , strictly increasing, we

have

()2 sup{|X|+ : X G, and t I & f = g X f g

1 / t(Ht)}.

However, as , cf() = 0, cf() > 20 clearly > ; also if X1, X2 are as

in ()2 then for some X X2 we have |X| |X1| + |I| and X1 (X2\X2) is asrequired there. So we can choose , increasing such that

()3 there is X G of cardinality such that t I & f = g X f g1 /

t(Ht).

Second assume cf() = 0, so let =n n > . If > , for eachn there is a witness fn : < n to nu

+(A) > n, so fn G

A and as

n > |GAn |, without loss of generality n,(fn ) = n,(f

0) so as we can re-

place f

n

by f

n

+1(f

n

0 )

+1

, without loss of generality m n m,(f

n

) = eG.For each let n be increasing be such that n,(f

n ) Gn,(n). As

20 < cf(n) = n, for some increasing n we have (n < n),

n = n.

So, hence without loss of generality < n = n. Let be (n) =

Max{n(n) : m n}. So we have n < & < n n,(fn ) Gn. So

()4 for every n < and 0 < (in fact even i = n) there are f G for

< such that n,(f) = eGn and < < & t I

f g1 / t(Ht).

Lastly, if = , so cf() = 0 the proof is as in the case > & cf() = 0,

except that n,(fn) = n,o(f

n0 ) holds by the choice of f

n : < n instead of by

without loss of generality.

For each t Jn and strictly increasing let H

(t,) be the subgroup {g Ht :

for every n < we have n,(g) Htn,(n)}. So let J

n = {(t, ) : t J and

increasing}.

We define Gn,, a subgroup of Gn,(n), decreasing with by induction on :


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= 0: Gn, = Gn,(n)

= + 1: Gn, = {x : x Gn, and x Rang(n,n+1 G

n+1,)

and n > 0 n1,n(x) Gn1,(n1),}

limit: Gn, =


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5.6 Claim. Let (g, f) be a nice t-pair.1) The following statements are equivalent:

(a) rkt(g, f) = (b) there exists g Ht extending g such that

t(g

) = f.

2) If rkt(g, f) < , then rkt(g, f) < + where =

n


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24 SAHARON SHELAH

For limit , by the induction hypothesis for every < and every k we haverkt(gk, f) , hence by Definition 5.5(2)(b), rkt(gk, f) .

For = +1, by the induction hypothesis for every k n we have rkt(gk, f) .

Let k0 n be given. Since gk0 is below gk0+1 and rkt(gk0+1, f) , Definition5.5(2)(c) implies that rkt(gk0 , f) +1; i.e. for every k n we have rkt(gk, f) .So we are done.2) Let g Htn and f G be given. It is enough to prove that if rkt(g, f)

+

then rkt(g, f) = . Using part (1) it is enough to find g Ht such that g is below

g and f = t(g).

We choose by induction on k < , gk Htn+k such that gk is below gk+1 andrkt(gk, f)

+. For k = 0 let gk = g. For k +1, for every < +, as rkt(gk, f) >

by 5.5(2)(c) there is gk, Gn+k+1 extending gk such that rkt(gk,, f) . But thenumber of possible gk, is |H

tn+k+1| 2

n+k+1 < + hence there are a function

g and a set S +

of cardinality

+

such that S gk, = g. Then takegk+1 = g.3) Immediate from the definition.4) Check the definitions. 5.6

5.7 Lemma. 1) Let (g, f) be a nice t-pair. Then we have rk(g, f) rk(g1, f1).2) For every nice t-pair (g, f) we have rk(g, f) = rk(g1, f1).

Proof. 1) By induction on prove that rk(g, f) rk(g1, f1) (see moredetails in the proof of Lemma 5.8).

2) Apply part (1) twice. 5.7

5.8 Lemma. 1) Letn < be fixed, and let(g1, f1), (g2, f2) be nice t-pairs withg Htn( = 1, 2). Then(g1g2, f1f2) is a nice pair and rkt(g1g2, f1f2) Min{rkt(g, f) : = 1, 2}.2) Let n, (f1, g1) and (f2, g2) be as above. If rkt(g1, f1) = rkt(g2, f2), thenrkt(g1g2, f1f2) = Min{rkt(g, f) : = 1, 2}.

Proof. 1) It is easy to show that the pair (g1f2, f1, f2) is t-nice. We show by induc-

tion on simultaneously for all n < and every g1, g2 Htn that Min{rk(g, f) : = 1, 2} implies that rk(g1g2, f1f2) .

When = 0 or is a limit ordinal this should be clear. Suppose = + 1 andthat rk(g, f) + 1 for = 1, 2; by the definition of rank for = 1, 2 there existsg H

tn+1 extending g such that (g

, f) is a nice pair and rkt(g

, f) . By the

induction assumption rkt(g1g2, f1f2) and clearly (g

1g2) n = g1g2. Hence g

1g2

is as required in the definition of rkt(g1g2, f1f2) + 1.


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2) Suppose without loss of generality that rk(g1, f1) < rk(g2, f2), let 1 = rk(g1, f1)and let 2 = rkt(g2, f2). By part (1), rkt(g1g2, f1f2) 1, by Proposition 5.7,rkt(g

12 , f

12 ) = 2 > 1. So we have

1 = rkt(g1, f1) = rkt(g1g2g12 , f1f2f

12 )

Min{rkt(g1g2, f1f2), rkt(g12 , f

12 )}

= Min{rkt(g1g2, f1f2), 2} Min{1, 2} = 1.

Hence the conclusion follows. 5.8

5.9 Theorem. Assume (A is an explicit -system and)

(a) is strong limit > cf() = 0(b) nu(A) or just nu+ (A) .

Then nu(A) =+ 2.

The proof is broken into parts.5.10 Fact: We can choose by induction on n, fn,i : i < n such that

() fn,i G and fn,i Gn+1 = eGn+1() i < j < n & t Jn fn,iEtfn,j

() rkt(g, fn,if1

n,j) < for any t J

n, k n, g Ht

k and i = j < n() iff belongs to the subgroup Kn ofG generated by the {fm,j : m < n, j rkt(g, f)

(v) rkt(g, f) < rkt(g, fn,i0f

1n,i1

ffn,i2f1n,i3

)

(vi) rkt(g, f) < rkt(g, fn,i2f

1n,i3

ffn,i0f1n,i1

)

(vii) rkt(g, fi0f1i1

) <

(viii) rkt(g, fi1f1i0

) <


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26 SAHARON SHELAH

() for each t Jn one of the following occurs:

(a) for i0 < i1 i2 < i3 < n we have

rkt(eHtkt(n) , fn,i0f1n,i1) < rk(eHtkt(n)

, fn,i2f1n,i3)

(b) for some nt for every i < j < n we havent = rkt(eHtkt(n)

, fn,if1n,j).

Proof. We can satisfy clauses (), () by the definitions and clause () follows. Nowclause () is straight by Erdos Rado Theorem applied to a higher n.For clause () notice the transitivity of the order and of equality and there is nodecreasing sequence of ordinals of length . 5.10

5.11 Notation. For let T = k


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5.14 Definition. For T we have f G is well defined as the inverse limitof fn Gn : n < , so n < f Gn = fn. This being well defined followsby 5.13 and G being an inverse limit.

5.15 Proposition. Let , T be such that (n)((n) = (n)), (n) > 0, (n) >0. Ift I, then ff

1 /

t(H

t).

Proof. Suppose toward contradiction that for some g Ht we have t(g) = ff

1 .

Let k < be large enough such that t Jk, ()[k < () = ()]. Let = rkt(g H

tkt()

, f(+1)f1(+1)) and

= rkt(g Htkt(+1)

, f(+1)f1(+1))

(the difference between the two is the use of kt() vis kt( + 1)). Clearly

()1 f(+1)f1(+1) = (f,3()+1f

1,3())(ff

1)f,3()f

1,3()+1

[Why? Algebraic computations and Definition 5.12.] Next we claim that

()2 < for k ( < ).

Why?Case 1: () < ().

Assume toward contradiction = , but by clause () of 5.10 aboverkt(eHt

kt(), f,3()+2f

1,3()+1

) < = , hence by 5.8(2).

rkt(g Htkt()

, f,3()+2f1,3()+1f(+1)f

1(+1)

) = Min{rkt(eHtkt()

, f,2(()+2f1,2()+1),

rkt(g Htkt()

, f(+1)f1(+1)

)} =

rkt(eHtkt()

, f,2()+2f1,2()+1) < .

Now (by the choice off(+1), f(+1) that is Definition 5.12 that is ()1, algebraiccomputation and the previous inequality) we have

> rkt(g Htkt(),f,3()+2f

1,3()+1f(+1)f

1(+1)) =

rkt(g Htkt()

, (f,3()+2f1,3())(ff

1)(f,3()f

1,3()+1)).

This and the assumption = gives a contradiction to ()(i) of 5.10 (forn = and f = f,f

1 K (see 5.13(1)) and (i0, i1, i2, i3) being (3(), 3() +

2, 3(), 3() + 1) and being (3(), 3() + 1, 3(), 3() + 1); the contradiction is


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28 SAHARON SHELAH

that for the first quadruple we get rank < by the previous inequality by the lastinequality, for the second quadruple we get equality as we are temporarily assuming = , the definition of and ()1).

Case 2: () > ().Similar using ()(ii) of 5.10 instead of ()(i) of 5.10 (using () > 0).

So we have proved ()2.

()3 +1 for > k.

Why? Assume toward contradiction that +1 > .Let f = f(+1)f

1(+1), so

= rkt(g Htkt(+1)

, f) and using the choice of+1

and ()1 we have +1 = rkt(g Htkt(+1)

, f(+1),3(+1)+1f1+1,3(+1)

ff+1,3(+1)

f1

+1,3(+1)+1).

If < rkt(eHtkt(+1)

, f+1,3(+1)+1f1+1,3(+1)

) then by 5.10()(iii) also

< rkt(eHtkt(+1)

, f+1,3(+1)+1f1+1,3(+1)) hence using twice 5.8(2) we have first

= rkt(g Htkt(+1)

, f+1,3(+1)+1f1+1,3(+1)

f) and second (using also 5.7(2))

we have = rkt(g Htkt(+1)

, f+1,3(+1)+1f1+1,3(+1)f

f+1,3(+1)f1+1,3(+1)+1),

so by the second statement in the previous paragraph (on +1) we get = +1

contradicting our temporary assumption toward contradiction ()3; so we have rkt(eHt

kt(+1), f+1,3(+1)+1f

1+1,3(+1).

Also if rkt(eHtkt(+1) , f+1,3(+1)+1f

1

+1,3(+1)) = rkt(eHtkt(+1), f+1,3(+1)+1f

1

+1,3(+1)

then is not equal to at least one of them hence by 5.10()(iii) + (iv) also isnot equal to those two ordinals so similarly to the previous sentence, 5.8(2) gives 3

+1 = Min{rkt(eHtkt(+1)

, f+1,3(+1)+1f1+1,3(+1)

),

rkt(g Htkt(+1)

, f), rkt(eHtkt(+1)

, f+1,3(+1)+1f1+1,3(+1)

)} which is so +1

, contradicting our assumption toward contradiction, ()3.

Together the case left (inside the proof of ()3, remember 5.7) is:

= rkt(g Htkt(+1)

, f) rkt(eHtkt(+1)

, f+1,3(+1)+1f1+1,3(+1)

) =

rkt(eHtkt(+1) , f+1,3(+1)+1f1+1,3(+1)).

So in clause 5.10(), for n = + 1, case (b) holds, call this constant value .As, toward contradiction we are assuming +1 > during the proof of ()3; so by, +1 > hence we get, by computation and by 5.8 that if( + 1) > ( +

3as the three are pairwise non equal


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1) then rkt(g Htkt(+1)

, f+1,3(+1)+2f1+1,3(+1)

ff+1,3(+1)f1+1,3(+1)+1

) =

rkt(eHtkt(+1)

(g Htkt(+1)), (f+1,3()+2f1+1,3(+1)+1))(f+1,3(+1)+1f

1+1,3(+1)f

f+1,3()+1f1+1,3

rkt(eHtkt(+1)

, f+1,3(+1)+2f1

+1,3(+1)

) but by (b) of 5.10() proved above the later

is < +1 = rkt(g Htkt(+1), f+1,3(+1)+1f1+1,3()

ff+1,3(+1)f1+1,3(+1)+1

)

contradiction to 5.10()(v) for the two quadruples (3( + 1), 3( + 1) + 1, 3( +1), 3( + 1)+ 2) and (3( + 1), 3( + 1)+ 1, 3( + 1), 3( + 1)+ 1) and n = + 1.If ( + 1) < ( + 1) we use similarly f+1,3(+1)+2f

1+1,3(+1). So ()3 holds.

()4

[Why? Look at their definitions, as g Htkt(+1)

is above g Htkt()

. Now if

kt(), kt( + 1) are equal trivial otherwise use 5.6(3).]

()5 if kt( + 1) > kt() then < (so > 0)

[Why? Like ()4.]()6 +1 and if kt( + 1) > kt() then >

+1

[Why? By ()3+()4 the first phrase, and ()3+()5 for the second phrase.]

So : [k, ) is non-increasing, and not eventually constant sequence ofordinals, contradiction.

5.15

Proof of 5.9. Obvious as we can find T T, a subtree with 0 -branches suchthat = lim(T) ()() = () and lim(T) & n < (n) > 0.

Now f : lim(T) is as required by 5.15.

5.16 Conclusion: IfA is a (, I)-system, and is an uncountable strong limit ofcofinality 0 and nu(A) (or just nu

+ (A) ), then nu(A) =

+ 2.

Proof. So we assume > 0 hence > 20 and trivially nu+(A) nu(A) .

We apply 5.2(2) to A and = (so cf() = 0) and get an explicit (, J)-systemB such that nu+(B) nu(A) hence by 5.9 we have nu(B) =+ 2 hence bythe choice ofB also nu(A) =+ 2. The proof for nu+ (A) is similar. 5.16

5.17 Concluding Remarks. Can we weaken condition (E)+ in Theorem 1.1(2)? Canwe use rank?


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REFERENCES.

[EM] Paul C. Eklof and Alan Mekler. Almost free modules: Set theoreticmethods, volume 46 of NorthHolland Mathematical Library. NorthHolland Publishing Co., Amsterdam, 1990.

[EkSh 505] Paul C. Eklof and Saharon Shelah. A Combinatorial Principle Equiva-lent to the Existence of Non-free Whitehead Groups. In Abelian grouptheory and related topics, volume 171 of Contemporary Mathematics,pages 7998. American Mathematical Society, Providence, RI, 1994.edited by R. Goebel, P. Hill and W. Liebert, Oberwolfach proceedings.math.LO/9403220.

[GrSh 302] Rami Grossberg and Saharon Shelah. On the structure of Extp(G,Z).Journal of Algebra, 121:117128, 1989. See also [GrSh:302a] below.

[GrSh 302a] Rami Grossberg and Saharon Shelah. On cardinalities in quotients ofinverse limits of groups. Mathematica Japonica, 47(2):189197, 1998.math.LO/9911225.

[HrSh 152] Leo Harrington and Saharon Shelah. Counting equivalence classes forco--Souslin equivalence relations. In Logic Colloquium 80 (Prague,1980), volume 108 of Stud. Logic Foundations Math, pages 147152.North-Holland, Amsterdam-New York, 1982. eds. van Dalen, D., Las-car, D. and Smiley, T.J.

[MRSh 314] Alan H. Mekler, Andrzej Roslanowski, and Saharon Shelah. On thep-rank of Ext. Israel Journal of Mathematics, 112:327356, 1999.math.LO/9806165.

[MkSh 418] Alan H. Mekler and Saharon Shelah. Every coseparable groupmay be free. Israel Journal of Mathematics, 81:161178, 1993.math.LO/9305205.

[Sh 202] Saharon Shelah. On co--Souslin relations. Israel Journal of Mathe-matics, 47:139153, 1984.

[Sh 273] Saharon Shelah. Can the fundamental (homotopy) group of a space

be the rationals? Proceedings of the American Mathematical Society,103:627632, 1988.

[Sh:f] Saharon Shelah. Proper and improper forcing. Perspectives in Math-ematical Logic. Springer, 1998.

[Sh 724] Saharon Shelah. On nice equivalence relations on 2. Archive forMathematical Logic, 43:3164, 2004. math.LO/0009064.


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[ShVs 719] Saharon Shelah and Pauli Vaisanen. On equivalence relations sec-ond order definable over H(). Fundamenta Mathematicae, 174:121,2002. math.LO/9911231.

saharon shelah- strong dichotomy of cardinality

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