Sahron Shelah- PCF and Abelian Groups

8/3/2019 Sahron Shelah- PCF and Abelian Groups

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PCF AND ABELIAN GROUPS

SAHARON SHELAH

Abstract. We deal with some pcf, (possible cofinality theory), investigationsmostly motivated by questions in abelian group theory. We concentrate onapplications to test problems but we expect the combinatorics will have rea-sonably wide applications. We almost always answer the original test problemwhich is proving the existence of -free abelian groups with trivial dual, i.e.,with no non-trivial homomorphisms to the integers. Combinatorially, we provethat almost always there are F which are quite free and have a relevantblack box. The qualification almost always means that except when we have

strong restrictions on cardinal arithmetic, in fact those restrictions are every-where. The nicest combinatorial result is probably the so called Black BoxTrichotomy Theorem proved in ZFC. Also we may replace abelian groups byR-modules, part of our motivation is that in some sense our advantage overearlier results becomes clearer in such context.

Date: August 5, 2010.Key words and phrases. cardinal arithmetic, pcf, black box, negative partition relations, trivial

dual conjecture, trivial endomorphism conjecture.First typed: 06.Dec.06 - Research supported by German-Israeli Foundation for Scientific Re-

search and Development. Publication 898.

1


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Annotated Content

0 Introduction, pp. 3-8

[We formulate the trivial dual conjecture for , TDU, and relate it topcf statements and black box principles. Similarly we state the trivialendomorphism conjecture for , TED, but postpone its treatment.]

1 Preliminaries, pp. 9-18

[We quote some definitions and results we shall use and state a majorconclusion of this work: the Black Box Trichotomy Theorem.]

2 Cases of weak G.C.H., pp. 19-27

[Assume C, < < 2 < 2 with minimal under those conditions.

Then for any < , a black box called BB(, +, , ) holds which for ourpurpose is very satisfactory.]

3 Getting large +

-free F

, pp. 28-40[The point is that to give the sufficient conditions for BB: see 2.10(2). Let C and = 2. We give sufficient conditions for the existence of+-free F of cardinality , which is quite helpful for our purposesimplying the existence of suitable black boxes. One such condition is (see3.6): the existence of < and < such that = . Recall that by 2assuming <


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PCF AND ABELIAN GROUPS 3

0. Introduction

We prove some black boxes, most notably the Black Box Trichotomy Theorem.Our original question is whether provably in ZFC the conjecture TDU holds andeven TED where:

{0p.7}Definition 0.1. 1) Let TDU, the trivial dual conjecture for mean:there is a -free abelian group G such that G has a trivial dual (i.e., Hom(G, Z) ={0}).2) Let TED, the trivial endomorphism conjecture for mean: there is a -freeabelian group with no non-trivial endomorphism, i.e., End(G) is trivial (i.e., End(G)= Z).

Much is known for = 1 (see , e.g., [GT06]). Note that each of the cases of0.1 implies that G is -free, not free, and much is known on the existence of -free,non-free abelian groups of cardinality (see , e.g., [EM02]). Also positive answers

are known if , e.g., V = L, see pg.461 of [GT06].Note that by singular compactness, for singular there are no counterexamples

of cardinality .By [Sh:883], if = n, then the answer to TDU is yes, for the cardinality =

n. It was hoped that the method would apply to many other related problems andto some extent this has been vindicated by Gobel-Shelah [GbSh:920]; on TED, =n see Gobel-Herden-Shelah [GbHeSh:970]. But we do not know the answer for = . Note that even if we succeed this will not cover the results of [Sh:883],[GbSh:920], [GbHeSh:970]; e.g. hence the cardinal is < and probably when wedeal with larger cardinals.

A natural approach is to prove in ZFC appropriate set-theoretic principles, andthis is the method we try here. This raised combinatorial questions which seem

interesting in their own right; our main result in this direction is the Black BoxTrichotomy Theorem 1.22. But the original question has bothered me and theresults are irritating: it is very hard not to answer yes in the following sense(later we say more on the set theory):

(a) failure implies strong demands on cardinal arithmetic in many , (e.g. ifcf() = 1 then +1 = cf(+1) = (+1)


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Obviously, e.g. clause (c) seems helpful for abelian groups; now at first sight isseen helpful that for every n there is an n-free non-free abelian group of cardinality

n, but this is not enough. More specifically presently this method does not resolvethe problem because for R = Z we know only that sp(R) includes {0, 1} (butunder MA there is no other one < 20).

Still we get some information: a reasonably striking set-theoretic result is theBlack Box Trichotomy Theorem 1.22 below; some abelian group theory conse-quences are given in 5.

A sufficient condition for a positive answer to TDU is (see 5.11):

0 TDU if BB(,,,J) when J is Jbd or J

bd1 and = 4.

Recall that{0p.9}

Definition 0.2. 1) A sequence of sets C = C : S is -free if for everyu [S]


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Variants are{0p.15}Definition 0.5. In Definition 0.3.

1) We may replace by (, ) which means there are S, C satisfying clause (A) ofDefinition 0.3 and

(B) if F = F : S and F is a function from (C) to , then for some cwe have:(d) c = c : S(e) c < (f) if c : , then c = F(c C) for some S [or if we replace

by I the set { S : c = F(c C)} does not belong to the idealI].

2) Replacing (, ) by, abusing notation, (, 1/), means that in clause (f) we re-place c = F(c C) by c = F(c C).3) We may replace by C so waiving the freeness demand, i.e. C is not necessarily

-free. Alternatively, we may replace by a set F of one-to-one functions from to when C lists {Rang(f) : f F }.4) Replacing by means that in (A)(b) we require just C (and not neces-sarily otp(C) = ).5) We may replace by < 1 meaning for every < 1.

Remark 0.6. Note the BB(,,,) is somewhat related to NPT(, ) from [Sh:g,Ch.II], i.e. BB(,,,) NPT(, ), but NPT has no predictive part.

In this work we shall show that it is hard for V not to give a positive answer(i.e. existence) for 0.1 via a case of 0.3 or variants; we review below the evidence

for this claim. By 5.11(1) we know that (really 2 21

can be weakened):

0 a sufficient condition for TDU is, e.g., BB(,, 2(21)+ , J), where J is Jbd

or Jbd1 (so is 0 or 1).Recall that C is the class of strong limit singular cardinals of cofinality when > 0 and most of them when = 0 (see Definition 1.1 and Claim 1.2).

Now the first piece of the evidence given here is that a failure of G.C.H. near C helps is the following fact:

1 BB(, +, , ) if < C and < < 2 < 2.

[Why? By Conclusion 2.7(1); it is a consequence of the Black Box TrichotomyTheorem 1.22.]

Note: another formulation is

1 if < C but BB(, +, , ) fails and := 2 then = 2}, so necessarily < 1; if 1 < 2

then

BB(1, +

, , ) holds by 1, so by our assumption 1 = 2

, so < 2

2 = 2 (2) = 2 2 = 2, but this means (2) is a treewith nodes and levels}.


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We have:

2

BB(2, ++1, , +) if < C

and ()( < 2 tr < 2).

[Why? See 3.1 or 3.2 recalling 2.10, remembering that tr is the -tree powerof , i.e., the supremum of the number of -branches of a tree with nodes and levels.]

So

2 if TDU fails, then for every C0 there is such that < + for < n and 0 is like above, then the conclusion appliesfor n.2) If + = 2 and {0, 1}, then there is an +1-free abelian group of cardi-nality + such that Hom(G, Z) = 0.

Note that we can prove TDU+1 if we answer positively to{0p.26}

Conjecture 0.8. If = + and = cf() and = 1 (or at least replacing the assumption = 1) then (D)S .

Related works are [Sh:922] and Gobel-Herden-Shelah ([GbHeSh:970]).We thank the referee for doing much more than duty dictates for pointing out

much needed corrections and clarifications.


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{0p.31}

Notation 0.9. 1) Usually C = C : S with S = S(C).

2) C is a -ladder system when S is a stationary subset of and C = sup(C)for S.3) C is a strict -ladder system when in addition otp(C) = cf().4) C is a strict (, )-ladder system when in addition S S where5) S := { < : cf() = }.


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1. Preliminaries

Most of our results involve C where{1.3.1}

Definition 1.1. Let C = { : is a strong limit singular cardinal and pp() =+

2}.2) C = { C : cf() = }.

Recall{1.3.3}

Claim 1.2.

(a) C and moreover, ppJbdcf()() =+ 2 when is a strong limit singular

cardinal of uncountable cofinality(b) if = > cf() and = 1 or just cf() > 0, then Ccf() and for a

club of < we have C

(c) if C and (, 2) or just = cf() < and (, pp+Jbd ()) thenthere is a +-free F of cardinality , even


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7) Let isspJ(F) means (< , ) isspJ(F) where (< 1, 2) isspJ(F) means

that (1, 2) isspJ(F) for some

1 < 1.

8) If we write isspJ(s : s I) we mean isspJ({s : s I}) but demand s1 = s2 I s1 = s2 .

{1.3.15}Observation 1.4. 1) If J is a -complete ideal on and F and 0 < 1 and J is an ideal on .Then f is not (, )-free as a set iff there is an increasing sequence : < of ordinals < such that the set S = { < : cf() and {i < : ( 2 1 = .1) F Ord is (2, 1)-free iff F is (, )-free for every regular [1, 2].

2) There is a (++1, )-free set F of cardinality iff for every n < thereis a (+n, +n)-free set F of cardinality .3) Assume > +, > = cf() and ( < )(|| < ). If F hascardinality for < , then we can findF of cardinality such that:

if 1 2 and some F is (2, 1)-free, then F is (2, 1)-free.

Remark 1.7. See 1.5, 3.5.

Proof. 1) By 1.5(2).2) By 3.10(1A).

3) Let i : i < be increasing with limit , i = i and let cdi : H(i) i

be one-to-one onto; F = {f : < }. Lastly, f

is defined by f(i) =cdi(f

(i i) : < ). 1.6

{1.3.32}Definition 1.8. 1) For let Jbd be the following ideal on : forU ,

we have U Jbd iff for every large enough < for every large enough i < thepair (, i) / U.2) For > we may identify Jbd with J

bd := {U : (

< )(i > = cf() and there is a -free F of cardinality andS S is stationary then there is a-free strictS-ladder systemC : = S.2A) In part (2) also for every = cf() (, ) and stationary S S there is(J, )-free strict S-ladder system C : S.

Proof. 1) If + = we shall construct F1 +(+ + ), but this is equivalent

so we shall ignore this point. For f F let gf : + be defined by:

() for < +, i < we letgf(, i) =

+ f(i) + + i.


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Let G = {gf : f F }, now

() if f1 = f2 F then gf1 = gf2 and moreover {(, i) + : gf1(, i) =

gf2(, i)} Jbd+.

[Why? By Definition 1.3(1) we know i() := sup{i < : f1(i) = f2(i)} < and{(, i) : < + and i < i()} Jbd+, so we are done.]

()2 assume G G is of cardinality < and we shall find u1g : g F

1 as

required.

Why? We can choose F F of cardinality < such that G = {gf : f F}.We can apply the assumption F is (, ++)-free and let uf : f F be as inDefinition 1.3(1); moreover let F : < () be as guaranteed in 1.4(3), so inparticular |F| +.

For each < () let f, : < |F| list F with no repetitions and let g, =gf,. First assume |F| , we let u

0,j = {i < : f,(i) : 1 is with some

repetition or i {uf,1 : 1 }}. As Jbd is -complete, clearly u0,j Jbd andlet u1g,j :=

+ u0,j.

Second, assume |F| = + and for each [, +) let (, j) : j < list without repetition and for j < , < + let

u0,,j = {i < : f,(,j1)(i) : j1 j is with some repetitions or i {uf,(,j1) :j1 j}}

and for < |F| letu1g, = {(, i) : (,

+), i < and i u0,,j where j is the unique j < such

that = (, j)}.Now check that u1g, : < () and < |F| is as required, i.e. witnessing the

freeness of F.2) Let f : = S be a sequence of pairwise distinct members of F and for

S let ,i : i < be an increasing sequence of ordinals with limit .Lastly, let C = {,i + f(i) : i < } for = S.

2A) Similarly. 1.9{1f.15}

Definition 1.10. 1) We say that G is an abelian group derived from F when G is generated by {x : < } {y,n : F and n < } freely excepta set of equations = { : F} where each has the form {a,ny,n+1 =y,n + x(n),n : n < } where a,n Z\{1, 0, 1}.2) We say that G is an abelian group derived from F 1 when G is generatedby {x,,n : < and < 1, n < } {y,,n : F, < 1, n < } {z,n : F and n < } freely except a set of equations = { : F} where each has the form

{a,,ny,,n+1 = y,,n + b,,n, z,,(n) + c,,n, x(,n),,n : < 1, n < }wherea,,n Z\{1, 0, 1}, b,,n Z\{0}, c,,n Z, , and 1 < 2 < 1

Rang(,1) Rang(,1) is finite.

Easily (see [EM02] on the subject).{1f.14}

Claim 1.11. If F or F 1 is (, Jbd1)-free, then any abelian groupderived from it is -free.

{a24}Remark 1.12. Here choosing ,

( + ) is O.K. but not for 5.


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Similarly to 1.9{1f.16}

Claim 1.13. 1) If F Dom(J) is (, +

2

, J)-free and J is a (2, +

1

)-regular1 and1-complete ideal then F is (, J)-free.2) Assume I, J is an ideal on S, T respectively. If F S is (,,I)-free, is a function fromT onto S and (J) I then F = {f : f F} T is(,,J)-free.

{a26}Definition 1.14. 1) Let (D)S mean that:

(a) = sup(S) is a regular uncountable cardinal(b) S is a stationary subset of (c) there is a witness P which means:

() P = P : S() P P() has cardinality < () for every subset U of , the set SU := { S : U P} is a

stationary subset of .2) Let (D)S be defined similarly but in (c)() we demand S\SU is not stationary.3) We write (D)D,S , (D)

D,S when D is a normal filter on and replace station-

ary by D+.{1f.18}

Discussion 1.15. 1) Of course, (D)S is a relative of the diamond, see [Sh:107].2) (D)S is equivalent to

S when is a successor cardinal but is not when is a

limit (regular) cardinal.3) Trivially (D)S (D)S .

{h.12}Definition 1.16. 1) The tree-power, tr is sup{|lim(T)| : T > is a treewith nodes and levels} where lim(T) = { : ( < )( T)}.2) Let tr + mean that there is a tree with levels, nodes and distinct

-branches.

{1.3.23}

Claim 1.17. 1) If = , then > sup{ < : = cf()and ( < )(||tr )}.3) If = , then { : = cf() and () < and (D)S or just

(D)S for some stationary S I[]} is finite.4) If = + and S is stationary, then (D)S is equivalent to S.

Proof. 1), 2), 3): See [Sh:829].

4) A result of Kunen; for a proof of a somewhat more general result see [Sh:247]1.17

Now by [Sh:775, 1.10] used in 1.22, 1.25.{1.3.25}

Theorem 1.18. Assuming C and = cf(2) we have BB(, C,< , J) when :

1that is, there are A J for < 2 such that u 2 |u| +1 {A : u} =

Dom(J)}.


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14 SAHARON SHELAH

(a) C, = cf(2)

(b) S S is stationary

(c) C = C : S, C , |C | < (d) < 2 tr < 2

(e) |{C : C}| < for < .{a38}

Remark 1.19. 1) Of course, if S I[] is stationary then there is C as in clauses(c) + (e) (and, of course, (b)).2) There are such stationary S as + < < .

{1.3.27}Definition 1.20. We say a filter D on a set X is weakly -saturated when there isno partition X : < of X such that < X D+ := {Y X : X\Y /D}.

A notable consequence of the analysis in this work is the BB (Black Box) Tri-chotomy Theorem 1.22.

{h.15}Remark 1.21. Using below C = C : S or using f = f : S does notmake a real difference.

{h.7}

The BB Trichotomy Theorem 1.22. If C and > = cf(), then atleast one of the following holds:

(A), there is a +-free F of cardinality 2

(B) (a) := 2 = |{C : C and S}|(so , e.g., the choice C = for S is alright forclause (ii)).

(C), (a) 2 = 2 is regular, < 2

< 2 and1 = min{ : 2

> 2} is (regular and) < 2

(b) like (b)(i)+(ii) of clause (B) for 2 but |C| < 1 for S(so C = is not alright).

(c) BB(JnstS , +, , )

for any < and any stationary subset S of 1.

{h.7d}Remark 1.23. 1) If = 0 above, then there is no infinite cardinal < asrequired, the proof still gives something say for = 1, and in this case we cannotget for every stationary S S, still by [Sh:829, 3.1] one has for all but finitelymany regular < for almost every stationary S S; maybe see 2.13.2) Assume C, = 2 = +. If is regular then (A) of 1.22 holds. By 3.12,there is C = C : S , divides , C = = sup(C), otp(C) = and C is+-free. If in addition =


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3) For = +, when singular, see below. What happens if := 2 is weaklyinaccessible? Now it is plausible to satisfy, for some 0

() (a) 0 < (b) < > cov(||, +0 , , 2)(b)+ < > cov(||, +0 ,

+0 , 2).

Now (b)+ implies (by [DjSh:562])

(c) there is P such that() P = P : < such that() |P| < () P {u : |u| 0, u is a closed subset of }() if u P, then u P.

This is enough for the argument above.4) Does clause (b) suffice?

Proof. Proof of 1.22:Recall that for every (, 2) there is a +-free F of cardinality (see

1.2(c)).If for some < 2 we have = 2 then by 3.6, clause (A) holds (when there

stands for here), so we can assume there is no such . If 2 is a singular cardinalthen by 3.10(3), clause (A) holds, so assume := 2 is regular. If =


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(B) BB(, +, < , ) where = min{ : 2 < 2}

(C) := 2 satisfies =


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2. Cases of weak G.C.H.

Note that if C and < 2 < 2, then we can find a +-free F of

cardinality (by the No hole Conclusion, 2.3 page 53, [Sh:g, II]) so by the SectionMain Claim 2.2 we can deduce BB(, +, (2, ), ) for < - see conclusion 2.7.

Observe below that if = 2, C = C : < , C (and 2 < 2), then easilyclause () of the conclusion of the Section Main Claim 2.2 below holds by counting -see 2.3(5). The point is to prove it for more colors, this is a relative of [Sh:775, 1.10]but this section is self contained. Also Definition 2.1 repeats Definition [Sh:775, 1.9].

This section is close to [Sh:775, 1] hence we try to keep similar notation.

{d.8}Definition 2.1. 1) Sep(,,,, ) means that for some f:

(a) f = f : < (b) f is a function from to

(c) for every

the set { : for every < we have f() =

()}has cardinality < .

2) We may omit if = . We write Sep(,, ) for Sep(,,,, ) and Sep(, )if for some = cf() 2 we have Sep(,,,, ) and Sep(< , ) if for some = cf() 2 and < we have Sep(,,,, ). Let Sep+(, ) meanSep(,,,,).

{d.6}The Section Main Claim 2.2. Assume

(a) 2 < 2

(b) D is a +-complete filter on extending the co-bounded filter(c) C = C : < , C ,(d) 2 and (or just D is +-complete)

(e) Sep(,, )(f) = Min{ : 2 > 2} or at least

(f) we have h (2) for < (2)+ such that = h =D h.

Then

() if satisfies < |C| , then we can find f = f : < satisfying f (C) such that (see 2.3(1)):for every f : , for some < , f f (and even for D+-many s)

() if F : (C)(2) for < , then we can find c = c : < suchthat() for any f : 2, for some < , F(f C) = c (even for

D+-many)() if = : < and < C , then we can find

f = f : < satisfyingf C

such that for everyf


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that clause (f) implies is regular (but not (f)) and clause (b) implies cf() > .5) Concerning clause () in 2.2, when = 2, this is easy: let D be the filter of

co-bounded subsets of , and let f : < 2

list

(2

), each appearing times.Now F := {1 F(f C)) : < : < 2} is a subset of 2 of cardinality2 < 2 = |2|. So every c 2\F is as required. We can use any filter D on such that |2/D| > 2,6) In the Main Claim 2.2 we can replace by any set of cardinality . E.g., >.Hence in replacing C by C = C : < , C

=

>(C) in () of 2.2 we canassume Dom(F) = {f : f a function from >(C) to 2}.7) We may wonder if clause (e) of the assumption of the Main Claim 2.2 is reason-able; the following Claim 2.5 gives some sufficient conditions for clause (e) of 2.2to hold.8) In 2.2 we implicitly assert that (f) (f); for completeness we recall thejustification (as there (2)+ 2).

{d.6.5}

Observation 2.4. We have (f) (f) in 2.2, i.e. if = min{ : 2 > 2} thenthere are h : 2 for < 2 such that < < 2 h = h mod Jbd .

Proof. As < |2| = 2|| 2 and 2 clearly >2 = {2 : < } hascardinality 2, so there is a one-to-one function g from >2 onto 2.

Let : < 2 list 2 and let h : 2 be defined by h() = g( ) for < .

Clearly h : < 2 is as required. 2.4{d.7}

Claim 2.5. Clause (e) of 2.2 holds, i.e., Sep(,, ) holds, when at least one ofthe following holds:

(a) = and = (b) U() = and 2 < and = (2)+

(c) UJ() = where for some we have J = []


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where Sol stands for solutions, so by clause (c) of the Definition 2.1(1) of Sep itfollows that

()1 |Sol| < .Let cd be a one-to-one function from (2) onto 2 such that

= cd( : < ) sup{ : < }

Let cd : 2 2 for < be such that < 2 = cd(cd() : < ).

Let H be a one-to-one function from 2 onto , such H exists as 2 byclause (d) of the assumption. For let Sol := { < 2

: H() Sol }, so

()2 |Sol | < .

Clearly in the assumption, if clause (f) holds, then clause (f) holds (see 2.4), so wecan assume that h : < (2

)+ are as in clause (f) so in particular h (2).

Fix < (2)+ for a while.For < let

()3, := H(h())

.

Let < . Recall that , for < and f is a function from

to so

f(, ) < . Hence we can consider the sequence c

= f(

, ) : <

as a

candidate for being as required in the desired conclusion () from clause () of theMain CLaim 2.2. If one of them is, we are done. So assume towards a contradictionthat for each < (recall we are fixing < (2)+) there is a sequence

(2)that exemplifies the failure of c to satisfy (), hence there is a set E

D, so

necessarily a subset of , such that

()4 E F(

C) = f(

, ).

Define (2) by

1 () = cd(() : < ) for < ; so

(2) for our < (2)+.

By clause (b) in the assumption of our Main Claim 2.2, the filter D is +-completehence

()5 E := {E

: < } belongs to D.

Now we vary < (2)+. For each such we have chosen (2), and clearly the

number of such s is |(2)| = (2) = 2 hence for some and unbounded

U (2)+ we have U = .

For < we define (2) by () = cd(

()). So by the choice of

2 if U, then < =

.

So by ()4 + ()5

3 if E where U then < F( C) = f( , ).So noting F(

C) : <

, clearly by ()0 we have:

4 if E where U, then

, SolF(C):


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As D is +-complete and also E := {Ei : i < } belongs to D. By theabove,

E i < hi() SolF(C): 2}, so if equality holds, by 2.4 there areh

(2) for < 2 such that = h =Jbd

h. So we can apply the Section

Main Claim 2.2 with D taken to be the club filter and with C, : < herestanding for C there; we get ci with domain let ci have domain Si, ci(i,) = c

i()

it is as required. If otherwise, i.e., > 0, the result follows by monotonicity of BBin .2) The proof is similar. 2.7d.11

{d.11p}Conclusion 2.8. Assume we add clause (g) and replace clause (b) by (b) + in theSection Main Claim 2.2 where

(g) = cf() and d > 2, recalling d = cf(,


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d = cf(,


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22 SAHARON SHELAH

[Why? Obvious, as in the proof of Erdos-Rado theorem; let i : i < be asequence with no repetition of members of R. For each j < , we try to choose by

induction on < ordinal i(j,), j such that(a) i(j,) < j is increasing with (b) j, = min{ : j() = i(j,)()}(c) i(j,) = min{i : i(j,) < i < j and i(j,) = i(j,) for < }.

If we succeed for some j we are done. Otherwise for each j < there is (j) < such that (i(j,), j,) is well defined iff < (j).

Let T = {(i(j,), j,) : < : j < and (j)} it is, under , a tree with levels, is normal, has a root and each node has at most immediate successors,hence |T |

i


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Proof. Easy, but we shall give details.1) Let S = [, ) and let S : < be a partition of S to sets each of cardinality

. Recalling Definition 0.3 it suffices to prove BB(, CS, , ) for each < , sofix now. Clause (A) in Definition 0.3 is obvious, so let us prove clause (B), so letus F : S and F : (C) be given and we should choose c (S).

Let f = f : S list, each appearing unboundedly often (and even

stationarily often), and choose c := F(f C). Now check.2) Look at the definitions. 2.10

{2b.99}Discussion 2.11. 1) We use 2.10, e.g. in 1.25.2) We may try to strengthen the results on Sep(,,) assuming = , a casewhich is unnatural for [Sh:771] but may be helpful.

{2b.107}Claim 2.12. Assume < + < = cf() and < cf([], ) < .1) If 2 < , = cf() and = cf() = , (, ) is regular

(b) = i : i < is a sequence of regular cardinals < with limJ() =

(c) J = J = {A : for every large enough < for every large enoughi < we have + i / A} Jbd is an ideal

(d) = tcf(i


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24 SAHARON SHELAH

{2b.112}

Claim 2.15. Assume = cf() < , = cf() =+ ppJbd ().

Then there is a (++1, J+)-free ladder system : S, S S+.

Remark 2.16. Used in 1.25.

Recall (see [Sh:g, ChII], more in [MgSh:204]).{2b.113}

Definition 2.17. Assume J is an ideal of and f = f : < () is a and we can find sequence

A = A : u witnessing is a good point of f which means:

u = sup(u) A J for u if < are from u and i \A\A then f(i) < f(i).

2) Let Sbdf

be the set of { < : cf() > and f has a \Sgdf

\Sbdf

.{2b.16}

Remark 2.18. The problem in proving TDU is to have (D)S assuming =


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3. Getting large +-free subsets of

Recall that = C pp() =+ 2 and easily (see 2.10(2))

if F is 1-free and = |F| = 2, then BB(, 1, , ) and hence

TDU1 holds.

This is a motivation of the investigation here, i.e., trying to get more cases of +-free subsets of of cardinality pp(). In 3.1 the case before our eyes is = , < < = +1(= 2), cf() = (, ).

{1f.7}Claim 3.1. There is a set F of cardinality satisfying if holds where

() the set F is -free() F is (+, (2)+)-free - see Definition 1.3

(a) < <

(b) = cf() < (c) is regular (naturally but not necessarily = cf())(d) < < or just = are < (e) < || < (f) J = J1 is a -complete ideal on (g) tr + as witnessed by T; i.e., the tree T has levels,

nodes and distinct -branches(h) pp+J1() > .

{1f.8}Claim 3.2. In Claim 3.1 we can replace by and() by() below, i.e. if holds then there is F of cardinality such that holds where:

() the set F is 1-free() F is (+, )-free

(a) < < (b) = cf() < (c) J2 is an ideal on (d) J = J1 is an ideal on (e) < || < hence < (f) J2 is an ideal on such that

() either 1 and J2 is 1-complete, or() J1 is cf()+-complete and J2 = Jbd and < of course

(g) there are for < such that < < { < :() = ()} J2

(h) there is a +-free F of cardinality

(i) () P()/J2 satisfies the -c.c. or just

() for some +-complete ideal J2 J2 of , sup min{|A| : A P()\J2 andA = B A A B J2}.

{1f.9}Remark 3.3. 1) Recall Definition 1.3 where we define notions of freeness for setsand for sequences.2) If in 3.1, F is not necessarily (, +)-free, > and = min{ : F is not(+, )-free}, then [+, ) (< , ) .3) In of Claim 3.1, we can even get a -Kurepa tree with distinct -branches.4) The proof of 3.1 is written so it can be adapted to become a proof of 3.2.


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26 SAHARON SHELAH

Proof of Claim 3.1: As cf() = < by clauses (b) + (d) of and < || < by clause (e), we can let i : i < be increasing with limit such that

(i)

= i > 2

. Let i =

j


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() = (). As this holds for every < , we conclude that = but : < is without repetitions hence = , so we are done.]

Now the main point is proving clauses () and () of.

Step 1: To prove clause () of, i.e., F is -free.Assume w and |w| < . Recalling ()1(b) and < , clearly the set {() :

w , < } being of cardinality < + is free, hence there is a sequences() : w, < of members of J1 such that: if (, ) w , for = 1, 2,and 1(1) = 2(2) and i \s1(1)\s2(2), then 1 (1)(i) = 2(2)(i).

Now as : w is a sequence of < distinct -branches ofT and 1(1) =2(2) 1 = 2 (as () [

, ) and those intervals are disjoint) and1() = 2() 1 = 2 by ()2, i.e., the choice of T and the regularityof we can find < such that ()) : w is with no repetitions, anddefine s = s() for w; now s

: w is as required. [Why? First

s J1 by the choice of s

of the s()s. Second, assume = are from

and i \s\s and we should prove (i) = (i). Now (()) = (()) by

the choice of () and s = s(()), s = s(()) hence i \s(()\s(())

so by the choice of () : , < we have (()) (i) = (())(i)

hence c()((i)) = (())(i) = (())(i) = cd()((i)) which implies that

(i) = (i).

Step 2: To prove clause () of.So let F { : < } have cardinality . Choose w such that F

= { : w}, so w [] and let u := {Rang() : w}, clearly u []. By thechoice of : < we can find a sequence s : u such that s J1 andi \(s1 s2) 1 = 2 {1, 2} u 1(i) = 2(i).

For w let t := {i < : the set of < such that i / s() belongs toJ2 = Jbd }.We shall now show that t := t : w is as required in Definition 1.3(1),(2);

that is, we have to prove that t J1 and that for any < and i < the set of such that i / t (i) = is small, i.e. of cardinality 2

; those demandsare proved below in ()4 and ()3 respectively. So let < and i < and letv = v,i = { w : i / t and (i) = }.

First we shall prove below that

()3 |v| 2.

This will do one half of proving t is as required in Definition 1.3(1),(2).Why does ()3 hold? Now if v, then i \t, hence (by the definitionof t) we have U,i := { < : i / s()} J

+2 . So if = are from v

and U,i U,i and () = (), then we have i / s() (as U,i)and i / s() (as U,i), and hence by the choice of s : u, we have()(i) = ()(i), so

cd((i)) = ()(i) = ()(i) = cd((i)).

Recall that (i) = = (i); as U,i U,i , this is a contradiction. Itfollows that v v = U,i U,i () = (); but = { < : () = ()} J2, hence this implies v v =


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28 SAHARON SHELAH

U,i U,i J2. As we have noted earlier that v U,i J+2 , it follows

that P()/J2 fails the |v|-c.c. But for the present proof, P() has cardinality 2,

hence P()/J2 satisfies the (2

)+

-c.c., and hence |v| 2

, as required in ()3. Forproving t is as required in Definition 1.3, we need also the second half:

()4 t J1 for v.

Why does ()4 hold? Firstly, assume < ; towards a contradiction assume thatt (J1)

+. By the choice of t, for each i t, the set { < : i / s()} belongs

to J2, but J2, being Jbd (and recalling is regular), is +-complete and |t| ,

hence the set

r :=it

{ < : i / s()} J2

hence we can choose < such that [, ) iti s(), so t s(),

but s() J1, and hence t J1 as required.Secondly, assume > ; towards a contradiction, assume t J

+1 . Again

i t { < : i / s()} J2, but J2 = Jbd , hence we can find = ,i :

i t (t) such that ,i = sup{ < : i / s()} < . However, J1 is

+-complete, so for some < , we have t := {i t : ,i <

} J

+1 . So

i t ,i < sup{ < : i / s()} <

i s() so t

s().

But s() J1, while t / J1, contradiction. 3.11f.7

Proof of 3.2: We note the points of the proof of 3.1 which have to be changed. Thechoice of = : < , i.e. ()1 is now done by using

(h). After ()2, insteadof defining J2 recall that it is given (see

(f)) and ifJ2 is not given (see (i)())

let J2 = J2. After ()2, instead of choosing : < it is given in (g) and the

tree T disappears.Now step 1 says that F is 1-free. For proving Step 1 we have to choose .

Of course, now |w| < 1 as we are proving F is 1-free.First, if clause () of(f) holds, as { < : () = ()} J2 for =

from w, but J2 is 1-complete, so U1, := { < : () = () for some =

from w} belongs to J2, hence there is < not in {U1, : = are from }.

Second, if clause () of(f) clearly < , so as J1 is -complete it sufficesto prove < < s, = {i < : (i) = (i)} J1 but for = wehave = hence for some < we have () = () hence s, {i < :()(i) = ()(i)} J1 so we are done.

We turn to Step 2, now for defining t we use belongs to J2; then ()3 shouldsay |v| < and in the proof instead of P()/J2 satisfies the (2

)+-c.c. we useclause (i)() if it holds and (i)(), as still = U,i U,i J2.

Lastly, in proving ()4 we use clause (f). 3.21f.8

{1f.10}Claim 3.4. In 3.1 assuming ( < )(|| < ), we can add F is -free when (a)or (b) or (c) hold where

Case (a) = ++1 and we can choose for < with no repetitions such

that + / isspJ({ : < }).

Case (b) + < and we can choose for < with no repetitions

such that < = cf() (< , ) / isspJ({ : < })


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Case (c) there are pairwise distinct for < and for < such

that for every regular ( + +, ) we have / issp({ : < }) or

/ issp({ : < }) but in1 of the proof of 3.1, in (i) code also (i).

Proof. By 1.4, Case (a) is a special instance of Case (b) and by Definition 1.3(7),Case (b) is a special case of clause (c), so we shall deal with clause (c) only.

We shall repeat the proof of 3.1 but we use : < , : < from theassumption (c).

Consider the statement

S is not a stationary subset of when:,S = cf() ( + +, ), < for < with no repetitions and S is

{ < : for some [, ), the set {i < : (i) {(i) : < }}

belongs to J+1 }.

It suffices to prove :

Why? We prove that { : < } is +-free by induction on < so letw , |w| . If + just note that = w {i < : (i) = (i)} J1,if recall () of 3.1. If + + is singular use compactness for singulars.So assume = cf() + + so by the induction hypothesis without loss ofgenerality |w| = and let : < list w and define S as in S above from : < . So as we are assuming , necessarily S is not a stationary subset ofso let E be a club of disjoint to S. Let () : < list in increasing order E{0}.For each < we apply the induction hypothesis to w := { : [(), ( +1))}and get the sequence s, J1 : w.

Lastly, for < let be such that [(), ( +1)) and s = s,{i < : (i)

belong to {(i) : < ()}}.Why holds?

Towards a contradiction, suppose that the set S is a stationary subset of =cf() ( + +, ) without loss of generality:

()5 (a) for some stationary S0 S, for every limit S0, itself can serveas the witness (in fact we can have S\S0 not stationary)

(b) for some club E of, if < and E (, ] = then {i < : (i) {(1)(i) : (1) < }} J1.

[Why? For clause (a) by renaming. For clause (b), it suffices to show that ( J2 is+-complete, clearly for each S1, for some j < , the set { < : ()

{j() : j < j}} belongs to J+2 . By Fodors lemma, for some j(), the set

S2 = { S1 : j j()} is a stationary subset of . Now { : S2}witnesses (< , ) usspJ2(lim(T)); but this contradicts a demand in case (c) ofthe assumption of 3.5.

Secondly, if < but recalling ()6 (see above) S0 cf() = and nowthe proof is similar.

Case B: < and S1 is not stationary.So necessarily S0\S1 is a stationary subset of . By the definition of S1 (and

()5) we can find s = s : (S0\S1) such that

()7 (a) s J2

(b) if 1 = 2 are from (S0\S1) and \s1

\s2 , then

1 () = 2 ().Let () = min(\s) for (S0\S1).

So for some stationary S2 (S0\S1), we have S2 () = () and so

()8 (()) : S2 is without repetitions.

Now ()2(b) in the proof of 3.1 says that : < is (+, J+1 )-free, apply this

to the subset { (()) : S2} which has cadinality < + hence (recall ()8)

()9 some s[ (())] : S2 witnesses that : S2 is free, i.e.

s (()) J1 for S1 and = S2 i \s[ (())]\s[(())]

(())(i) = (())(i).

As < , for some i() < , the set S3 := { S2 : i() s[ (())]} is a

stationary subset of . By ()6 we know that (i) = (i) i i for, < ,i < ; (i()) : S2 is a sequence without repetitions, but by thechoice of S we have S3 (i()) { (i()) : < }. However, thiscontradicts S3 S2 S0 S is a stationary subset of . 3.5

{1f.10}Claim 3.5. In 3.1, F satisfies: for+ < = cf() < , we have F is (, +, J1)-free iff(< , ) issp(F).

Proof. By the proof of the previous claim. 3.5

In 3.6, the case before our eyes is = 1 , = 1, = 0.

{1f.21}Claim 3.6. There is F of cardinality which is (+, J)-free when

(a) = cf() < = cf() < (b) =

(c) < < = (d) < || < (e) J is a +-complete ideal on (f) ppJ() =

+ .

Remark 3.7. Used in 1.22.


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Proof: Let i : i < be increasing such that (i) = i and let cd : and cd (for < ) be as in the proof of 3.1, noting that by clause ( a) of the claim

assumption,

= {

(i) : i < } = and let i = {j : j < i}.

As < ppJ(), by 1.2(c), i.e. [Sh:g, Ch.II] there is a sequence = : < of member of which is +-free. Let = : < with be pairwisedistinct.

Without loss of generality, i


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32 SAHARON SHELAH

()Pi for every u [i], we can find u and u Pi for < u such that

u {u : < u};

Pi exists by clause (d) of the assumption. Let P = {Pi : i < }, so |P| , P [].

By clause (b)(), let = : < with be such that < < implies =J2 , i.e. { < : () = ()} J2.

Lastly, for each < , for each i < , we know that {()(i) : < } [i],

hence we can find a sequence ui, : < of members of Pi such that {()(i) :

< } {ui, : < }.

For each < and i < , as J2 is a +-complete ideal on for some ,i > = cf(), < and < || < ,

and = sup{i : i < } and for each i < , there is a i-free F

of cardinality. Then there is a -free F of cardinality .1A) If = then < ||


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34 SAHARON SHELAH

sup{i : i < i()} = sup{i : i u}, so without loss of generality i : i < i() iswith no repetitions, and so i() + 1, and ifi() , we can find u as above of

cardinality < .]If for every i < i() clause () of (b) of the assumption holds then by 3.1 there

is i-free Fi of cardinality for each i < i() and by 3.10(1) if () of () the

conclusion holds. It holds by 3.6 if () of (b) apply for some i < i(). 3.111f.31{2b.91}

Claim 3.12. If C and = 2 = + and cf([], ) then

(a) there is a +-free F of cardinality 2 =

hence

(b) BB(, +, , ) for every < .

Remark 3.13. Actually as in [Sh:g, Ch.VII,6.1] and the no-hole claim.

Proof. By Definition 1.1 there is an ideal J on , a sequence i : i < of regular

cardinals < such that = tcf(i


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(b) = cf()(c) is regular

(d) < < < (e) J1 and J2 are +-complete ideals on , respectively(f) if < , then cov(||, +, +, +) or just(f) if < , then UJ2(||) (g) there is a set of pairwise J2-distinct members of .

Proof: Combine the proofs of 3.1 and 3.9. 3.15 1f.49{1f.51}

Claim 3.16. In 3.15:1) If in, clause (g) is exemplified by F2

which is (, J2)-free, > , thenF is (, +, J2)-free.2) If F F has cardinality > , then {Rang() : F } has cardinality .

Proof: We leave the proof to the reader. 3.16 1f.51{1f.53}

Claim 3.17. Assume C, J is a-complete ideal on and there is no (+, J)-

free F of cardinality := 2. Then the set = { : = cf() < , = and for some witness (, ) of cofinality we have ppJ() =+ for some-complete ideal J on } is a singleton or of the form {, +}.

Remark 3.18. This is intended to help in 5 to deal with R-moduels when sp(R)has at least 3 (or four!) members.

Proof: Let (1, 1, J1) be such that 1 and (1, J) a witness for 1 and 1

is minimal under those conditions.If 1 < , note that 1 < 1 hence we get a contradiction

by 3.6 to an assumption; so we can assume 1 > . Also by the choice of 1 asminimal, we have:

() < 1 cov(||, +, +, +) < .

If = {1} or = {1, +1 }, we are done; otherwise let (2, 2, J2) be such that

2 \{1, +1 } and (2, J2) witness that 2 , and 2 is minimal under those

requirements.Now

() there is a ++1 -free set F 1(1) of pairwise J1-distinct elements.

[Why? As 2 > 1, then necessarily 2 > +1 , hence such an F exists by 3.15 with

1, 2, , 1, 2, J1, J2 here standing for ,,,,,J1, J2 there. So assume 2 < 1;hence by 3.9 with 1, 2, , 1, 2, J1, J2 here standing for ,,,,,J1, J2 there,we are done.]

Now by () we can apply 3.5, case(c) and we are done. 3.17 1f.53


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36 SAHARON SHELAH

4. Propagating OBB(C) down by pcf

We deal here with the ordered black box, OBB and prove in ZFC that manycases occur.

{j.35}

Definition 4.1. 1) For a partial order J, a sequence C = C : S such thatC for every S and an ideal I on S, let OBBJ(C, I) mean that thereexists a sequence t : S with t J such that: if f :

S

C J, then

{ S : ( C)(f() J t)} = mod I.2) If sup(S) = sup(

S

C) is a regular uncountable cardinal and I is the non-

stationary ideal restricted to S, then we may omit I.3) If J = (,


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3) Easy. 4.3j.44

The main case in Lemma 4.4 below is: each Ji is a regular cardinal > i

. {j.45}Lemma 4.4. If clauses (a)-(f) below hold, then for some U I+ , for every i U,we have OBBJi(C, I), where:

(a) OBBJ(C, I) such that S = S(C), = (C) or just S otp(C) (b) I is an |i|+-complete ideal on S(c) J = Ji : i < i(d) Ji is a partial order such that Ji |= (s)(t)(s < t)(e) I is an ideal on i

(f) () J is a partial order() g = gt : t J() gt

i


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38 SAHARON SHELAH

()1 U0 := {i < i : si is a witness for OBBJi(C, I)}.

It suffices to prove

()2 U0 / I

[Why? Obviously.]Now for each i U1 := i

\U0 let fi : B Ji exemplify that si is not a witness

for OBBJi(C, I), i.e.,

()3 if i i\U0 then Wi = mod I where Wi := { S : ( C)(fi() Jisi)}.

If i U0, choose any fi : B Ji.Now

()4 W := {Wi : i U1} I.

[Why? By clause (b) of the assumption, the ideal I is |i

|+

-complete.]Now we choose for each B a function h as follows:

()5 (a) h i


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[Why? Because for each C() in the partial order J := (i cf() = , > = cf() = , and J is an idealon which is -complete (or just > J is +-complete).1) If < = cf() < pp+J(), S S

is stationary, and C = C : S is

a strict (, )-ladder system, tree-like when > , then for unboundedly manyregular < , we have OBB(C).2) Assume that for each regular (, pp+J()), S S

is stationary and C

=

C : S is a strict (, )-ladder system which is tree-like when > . Then


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40 SAHARON SHELAH

for some 0 < , for every regular (0, ), for some we have < = cf() } and replace i : i < byi()+i : i < .]

()2 let fi : < 2

i list (i)

()3 choose a sequence gi = gi : < 2

i of members of i() such that: if

< i and j < for j < then for some < i we have g() = fij

() :

j < .

[Why? As < 2i || < 2i and = by renaming it suffices to prove: ifF (i) has cardinality < 2i then for some g i we have (f F)( 2} as in 2.3) We may phrase the condition on F : not only when = cf().4) Like (3) but for our specific problem: Hom(G, Z) = {0}.

We look at another way to get cases of OBB. Recall the definition of d . If C isa (, )-ladder system then OBB(C) (and > ).2) Assume > are regular and = cf(dseq,) > . If C is a tree-like (, )-ladder

system then OBB(C).3) Assume > , are regular cardinals and D,. If OBB(C, I), I is+-complete, (C) = and (C) = then OBB(C).4) If > are regular cardinals then cf(d,) > cf(d,) D, andcf(d,


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44 SAHARON SHELAH

()2 (a) F = F : is -increasing continuous(b) g = g : < , g

(c) g is a (< )-bound of F for < (d) F = .

As we are assuing OBB(C) let t be such that:

()3 t = t : < is a witness for OBB(C).

For each i < let:

()4 ti = ti : S

be ti = gt (i).

Let

()5 U1 := {i < : ti is not a witness for OBB(C)}.

For each i < let (fi, Ei) be such that

()6 (a) fi : (b) Ei = S mod I(c) if i U1 and Ei then ( C)(fi() ti) that is

( C)(fi() gt (i)).

Let E = {Ei : i < } hence S\E I. For each < let h : beh(i) = fi() and s = min{ < : h F} so s is a function from to. By the choice of t we have

W := { S : ( C)(s t)} = mod I.

So we can choose () W E. Now {h : C()} F() hence by the choiceof gt() we have

()7 U2 = {i < : ( C)(h(i) gt(i)) equivalently ( C)(fi() ti)} = mod J

bd .

Now()8 U1 U2 =

hence

()9 U1 = mod Jbdhence

()10 \U1 = , i.e. for some i < , ti witness OBB(C).

This is enough.4) First, we have to show that := cf(d,) D, if cf(d,) > .

So let F be of cardinality d, such that no g is a ( )-bound ofF. LetF : < be -increasing with union F such that < |F| < |F| = d,.Let F+ = {g

: for some F F of cardinality < and i < we have

i < i < g(i) = sup{f(i) : f F

}}.1 F+ : < is -increasing.

Now we shall prove

2 {F+ : < } =.

Let h , by the choice of F the function h is not a (< )-bound of F hencethere is F F of cardinality < witnessing it which means that for every largeenough i < , h(i) sup{f(i) : f F}. Let F = {fj : j < j < }, leth(j) = min{ : fj F} and let h := sup{h(j) : j < j} < so clearly h F+hhence 2 is proved.


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Lastly, obviously

3 g is a (< )-bound of F+ .

Together we have shown that D, . Second, why = cf(d,) / [, ]?Let F : < , g : < be as above. Let g be a common


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46 SAHARON SHELAH

2) For some U I+ we have i U OBB0Ji,J

(C, I) when (a)-(f ) below holdswhere

(a) OBB0J,J(C, I), S = S(C), otp(C) = for S

(b) (e) as in 4.4

(f) () Ji is -directed

() U = U : Dom(C),U , |U| <

() if S, U = U : C,U I+ for C and C C U U U then {U : C} I+ .

3) Like part (2) but

(a)+ OBB1J.J(C, I), S = S(C), otp(C) = for S

(f) like (f) but in () we hvae U = i mod I.

Proof. Like 4.4. {k44}

Claim 4.23. 1) If J is linearly ordered, then OBB2J(C, I) OBB1J(C, I).

2) OBB1J(C, I) OBB0J(C, I) andOBB

1J(C, I) OBB

2J(C, I) andOBB

3J(C, I)

OBB2J(C, I).{k47}Claim 4.24. Assume that J is a -directed partial order.1) If J = Jbd , = cf() > 2

, =


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5. Conclusions for the -free trivial dual conjecture

We shall look at the following definition.{0f.07}

Definition 5.1. 1) For a ring R and a cardinal , let sp(R) be the class of regularcardinals such that there is a witness (G, h) which means:

(a) G = Gi : i + 1(b) G is an increasing continuous sequence of free left R-modules(c) if i < j + 1 and (i, j) = (, + 1), then Gj/Gi is free(d) h is a homomorphism from G to R as left R-modules(e) h cannot be extended to a homomorphism from G+1 to R(f) |G+1| .

2) For a ring R and cardinals , we define sp,(R) = sp1,(R) similarlyreplacing free by -free in clause (c). Writing sp


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48 SAHARON SHELAH

(A) Use BB(, C, , , J ) where = (sup{|R|i : i < }), where i = |Gi|+sup{|Hom(Gj ,RR)| : j < }

(B) We define SP,,,(R) as in 5.3 but(e) h = hi : i < , hi Hom(G,RR) and if i < j < , then hj hj

cannot be extended to any h Hom(G+1,RR)(f) |G+1| and |Hom(Gi, R)| i .

(C) In 5.6, we change(b) sp,,, or (C is tree-like, sp,,,) and J SP,,,

witnesses it)(e) BB(, C Si, (, ), J).

2) BB(, C Si, (, 1/), J) is sufficient for the right version of 5.3, see Definition0.3(2); really we need there to use = 2 and the guessing in an initial segmentof the possibilities, i.e. in 5.3 we need: without loss of generality |Gi| forevery i, given f Hom({Gi, i < }, Z) for < () < 2 we can find, e.g. a

permutation of , inducing G {Gi : i < } such that none of them can

be extended to f Hom(G, Z) or use: there is g Hom(

i (e) BB(, C,2(|R| + ), J) and , so I = JbdS recalling

JbdS = {U : U S and sup(U) < sup(S))(f) C is (, J)-free; recalling 1.3(1A); and .

Remark 5.7. 1) See Remark 5.10 and 0 in 0, needing fewer colours, even as in0.5(1A).2) The beginning of the proof can be stated separately.

Proof. By the definition of BB(, C,2(|R|+ ), J), there is a sequence S : < of pairwise disjoint subsets of S = S(C) such that BB(, C S,2(|R|+ )), J)for each < .

Without loss of generality S C S = and without loss of generalityS is a set of limit ordinals and each C is a set of successor ordinals and let C =


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{C : S}. We say that a D is C-closed when D C S and D S C D. So for every B C S there is a C-closed B C S such that

B

B

|B

| |B

| + . We can put of the sets Sis together, i.e.1 we can replace S : < by {S : U : < } provided that

U : < is a partition of ).

Also

2 we can replace C : S by C\h() : S when S h() C ,

hence without loss of generality

3 (a) < S S |S| < BB

(, C (S\S),2(|R| + ), J)

(b) if u has cardinality < then for ordinals < we have S sup(u) < min(C).

Without loss of generality0 |R|.

[Why? Replace by + |R| just check the demand on .]

1 there is a -free R-module G of cardinality := (2)+ such that

(a) G = {G, : < }(b) if G is a -free R-module of cardinality , then G is isomorphic to

G, for ordinals < (actually we need just Gr , + 1 from

Definition 5.3)(c) G, has cardinality for each < .

[Why? Because the number of such G up to isomorphism is 2|R|+ = 2.]

Let E = {(, ) : , < and G, = G,}, so E is an equivalence relation on and /E := { < : E} is the equivalence class of < under E. For < , let f1 be an isomorphism from G,min(/E) onto G,.

2 for any r R\{0} let xr = {(G, h) : (G, h) witness J SP,(R) for r, seeDefinition 5.3}

3 H := {G : C}

{K : S}, where each G is isomorphic to G under g

1

K isomorphic to G for S under g2 and

for < let G, = g1(G,), K, = g

2(G,)

4 let K


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50 SAHARON SHELAH

(b) by 6 +3 without loss of generality Si sup(Dxi

) < min(C)

8 H


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Let , < be such that G,,= G+1, and let f

0, be an isomorphism from

G+1 onto G,,, so g1 f

0, embeds G

+1 into H hence g

1 f

0, g

4 is a

homomorphism from G+1 into H (actually an embedding). Let

L = {g1 f

0,(x) g

4(x) : x G

}.

Clearly L is an R-submodule ofH. Now by the choice of (G , r )

()5 there is no homomorphism h from H into RR such that h1 h andh(x) = r

and h L = OL .

Lastly,

()6 (a) L := {L : S}, a sub-module of H

(b) H = H/L.

So()7 Hom(H, RR) = 0.

[Why? Assume h Hom(H, RR) = 0, so we can define h+ Hom(H,RR) byh+(x) = h(x + L). Let x H be such that h+(x) = 0, so for some i < we have(xi, ri) = (x, h

+(x)). By the choice ofh1 : Si the set { S : h K = h

1}

is unbounded in , so for some Si we have h K = h1, and by the choices of,,1, ,,2, , we are done as h+ L is zero.]

()8 H is -free.

[Why? As the case is easier we shall ignore it. Let H1 H be of cardinality< . So for some H2 H of cardinality < , we have H1 = {x + L : x H2}.So for some u of cardinality < , we have H2 {g1(G) : u S}

{g2(G) : S u}H. Without loss of generality, u C u, as|C| = < . Let H3 := {f1(G) : u} + {f

2 (G) : u S}.

So H (H3 + L)/L, and hence it is enough to prove that H3 + L/L is free.The rest should be clear.] 5.6

{5e.21}Claim 5.8. 1) In 5.6 if = and = , (i.e., for C the cardinality and amountof freeness coincide) we can deduce also sp(R).2) In 5.6, it suffices to assume

+ as in of 5.6 omitting (d) and strengthening clause (b) to

(b)+ sp,(R), moreover J SP,(R) is an ideal on .

Proof: This should be clear. 5.8 5e.21{5e.28}

Claim 5.9. 1) For R = Z, we have(a) Jbd0 belongs to SP0(R)

(b) Jbd1 belongs to SP1(R)

(c) Jbd10 belongs to SP1(R)

(d) if 20 = 1 or just 20 2 then Jbd2 belongs to SP2(R)

(e) if 20 = 1 or just 20 2 then J

bd21

belongs to SP2(R).

2) Similarly for R a proper subring of (Q, +).


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52 SAHARON SHELAH

{5e.30}Remark 5.10. 1) If we like the proof of TDU to be more direct, we have to add

Hom(G+1/G) = 0, otherwise we have to iterate.2) Claim 5.9 not novel but have found no direct quote. Clauses (b),(c) is essentiallyby [Sh:98] and clauses (d),(e) are the paralle for 1; we can continue (e.g. for higheris inductively.3) This is close to G is derived from F, see 1.10.4) We can phrase this more generally.5) Can we use this to prove TDU,+1 (Z) for some ? assuming CH. Or at leastis there k < k such that if 2 = +1 for < k then TDU,+1(Z) for some ?

Proof of 5.9: For part (1) let a Z be a prime, an = a (or can use, e.g. an = n!),for part (2) let a Z be a prime such that 1a / R and an = a; but could use anyan : n < such that anR R. We have to check Definition 5.3.

Clause (a):Let G+1 be the abelian group generated by {xn, yn : n < } freely except that

anyn+1 = yn + xn.

Let Gn = ZxnG = {Zxk : k < }.

Letting a


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For < 2 let G = {Rt : [1, 1 + 1)} and G2 = {G : < 2}.

2 G2+1/G1 is 2-free.

Why? Let H = {Ry : < 1} and for < 2, H be the subgroup of G2+1generated by G2 H {z,,n : < 1, n < 0} {x,

: >2}.

For 2 let H


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54 SAHARON SHELAH

{u : < 2} = 1.

We note that {F(f

u) : < 2 : f

(1)

R} is a subset of2 R

(b) h(t1++n) : n < ) = h(( + ))

(c) if < 1 is a limit ordinal, 2, +2 for = 1, 2and 1 = 2 then h(1) = h(2).

Now suppose h2 Hom(G2+1, R) extend h0, and define the two-place relation Eon 2 : E iff (

>2)(h2(x,

) = h2(x,

)). As 20 = 1 there are < < 2such that E. So there is a limit ordinal < 1 such that

=

, but

( + ) =( + ).

The rest should be clear.

Clause (e):

As in clause (d) but for A Jbd21 we let GA = {Rt1+ : (, ) A}.

5.95e.28

{5e.32}Conclusion5.11. 1) TDU holds, when BB(,, 2

(21)+ , J), where J {Jbd0 , Jbd10

}and has uncountable cofinality.2) Similarly BB(,, 2Dom(J), 2Dom(J), J).3) Similarly for TDU,.

Proof of 5.11: 1) By 5.6.2) By 5.14.3) Similarly. 5.115e.32

Remark 5.12. 1) The number, 2(21 )+ , of colours is an artifact of the proof. Actually

2 and even fewer colours (as in [Sh:f, Ap,1], 0.5(1A)) should suffice.2) See 1.9. But we can quote in 0 BB with 2 instead of 2(|R| + ).

{5e.35}Proof. Proof of 0.71) By 3 there is a +-free F of cardinality +, hence by BB(, C,, ) by2.10. By 5.6, 5.9 there is G as required.

Similarly for iterations.2) Similarly.

We can get more than in 5.6.


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{5e.38}Definition 5.13. 1) For cardinals ,,, let SP3,,(R) be the set of ideals J on

some such that for every r R\{0} some pair (G, h) witness it for r which means: (a) G = Gi : i < + 1 + is a sequence of R-modules each of

cardinality (b) G = {Gi : i < } and < G R R G+1+(c) if u J and < , then G+1+/ {Gi : i u} is a -free

left R-module(d) each Gi is a -free left R-module(e) h = h : < and h is a homomorphism from G to RR

for < (f) for each < or at most one < , there is a homomorphism

h+ from G,1+ to the left R-ideal RR extending hand mapping 1R to r.

{5e.43}Claim 5.14. A sufficient condition for TDU,(R) (i.e., there is a -free left R-module G of cardinality withHomR(G, R) = {0} with no non-zero homomorphismfom G to R as left R-modules) is:

(a) R is a ring with unit (1 = 1R)(b) J SP3,,(R) is an ideal on

(c) C = C : S is such that otp(C) = and C (d) is regular or at least cf() > and > (e) BB(, C, 2|R|+, , )(f) C is (, J)-free (but see 1.9).

Proof: Similar like the proof of 5.6 with some changes. First of all, instead of

1we use

0 let (Gr , hr) witness Definition 5.13 for r R\{0}

1 G is a -free R-module G of cardinality and for some ordinal () |R| + (a) G =

{G, : < ()}

(b) if r R\{0}, then for some 2|R|+ ordinals < () and sequence Gr

as in 5.13 we have j < Grj=fr,j

G,hence

(c) |G| + + |R|.

Secondly, after 8 we choose : Si such that

() and j < G,(j)

= Grij

.

Thirdly, we choose , Si such that

9.1 (a) if S, then 1 <

(b) if h Hom(H,RR), then for unboundedly many Si wehave: 1 = c

r(h

C

G) - see below

9.2 for Si and h Hom(K


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56 SAHARON SHELAH

() if j < , then x Grij f(x) = h(g(fri,j

(x)))).

The rest is similar. 5.145e.43{5e.47}

Conclusion 5.15. Assume that Jbdn SPn,n(R) and n < n+1 for n < .Then for some , for every large enough n, TDU,++1n holds.

Proof: We shall use 5.6 freely.Let C0 be greater than n for each n, and let n < be large enough.

Case 1: There is < 2 < 2

.Then we can apply 2.7.

Case 2: 2 is singular or just there is a +-free F of cardinality 2.By 2.10(2).

Case 3: Neither Case 1 nor Case 2.By Theorem 1.22 =


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6. Private Appendix{m1}

Discussion 6.1. We give details on the proof of 4.13.

Proof. Proof of 4.13 continued:We elaborate. Let F = F : S where F is a function from C(2) to .By assumption (e) we have Sep(, ) which means (see Definition 2.1(2)) that

for some = cf() 2 we have Sep(,,,, ).Let F = {f : < } where F exemplify Sep(,,,, ), see Definition 2.1(1) andfor let Sol =: { : for every < we have () = f()} where Solstands for solutions, so by clause (c) of the Definition 2.1(1) of Sep it follows that

()1 |Sol| < .

Let h be an increasing continuous function from into 2 with unbounded range;if 2 is regular, h = id is O.K. Recall that = cf(2). Let cd be a one-to-one

function from

(2

) onto 2

such that

= cd( : < ) sup{ : < }

and let the function cdi : 2 2 for i < be such that cdi(cd( : < )) = i.

In part (2) for < let F = { : h()}. For part (3) let F be (2) andlet F = F : < be -increasing with union F such that < |F | < .

Without loss of generality

() (a) F i < g() i F

(b) if < 2 || < 2 then F (i < g())((i) {(j) : F ,j < }) F

(c) if < 2 ||tr < 2 then F (i < )(i F) F.

For S let P =: { : is a function from C into h(t) such that for some Ft of length we have C () = (otp(C) ).

Clearly

() |P| |Ft | < 2.

For each P and < we define , C(h()) by ,() = cd(()) for

C . Now for P, clearly =: F(,) : < belongs to . Clearly{ : P} is a subset of of cardinality |P| which as said above is < 2.Recall the definition of Sol from the beginning of the proof and remember ()above, so P Sol has cardinality < = cf() 2

. Hence we canfind

\ {Sol : P}. [Why? if < 2 then | {Sol : P}|

{|Sol | : P} |P| < 2

=

= |

| and if = 2

then 2

is regular,|P| < 2, P |Sol | < 2

and again | {Sol : P}| < 2 = ||.]

Let < . Recall that for S and f from the list of members of

F is a function from to so f() < . Hence we can consider the sequence

c = f( ) : S S as a candidate for witnessing OBB(, C, (2, ), ) for F,

see Definition. If one of them is, we are done. So assume towards a contradictionthat for each < there is a sequence

(2) that exemplifies the failure of c,so there is a E , \E I such that

1 S E F( C) = f().


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58 SAHARON SHELAH

Define (2) by () = cd(() : < ).Next we define g : such that

2 (a) for part (2), let g() = Min{ < : () < h()}

(b) for part (3) so C is tree-like, let g() be the minimal < such thatfor some F we have:

if S C then g() = otp(C ) + 1 and i < g() C i = otp( C) (i) =

().

Let E := { S: if C then g() t}, so clearly

3 E I+.

By clause (a) in the assumption of our Theorem, the ideal I is +-complete henceE =: {E : < } E belongs to I+. Now choose E S, fixed for the restof the proof; clearly C P; just check the definitions of P and E, E. Let

< . Now recall that

C, is the function from C to h() defined by

C,() = cd(()).

But by our choice of clearly C cd(()) = (), so

C C,() = () so C, = C .

Hence F(C,) = F( C). As E E clearly F( C) = f( )

and as C P clearly C is well defined. Now easily C() =

F(C,) by the definition of C , so we have C() = f().

As this holds for every < it follows by the definition of SolC that

SolC . But C P was proved above so C { : P} hence

Sol

C

{Sol

: P

} whereas

has been chosen outside this set,

contradiction. 4.13


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7. private appendix

Moved 2010/7/26 from pg.41:

(h) in the following game the non-empty player has a winning strategy; aplay last or just sup{otp(C) : S}. In the -th move the empty playerchooses a function fi

i

Ji and the non-empty player chooses ti J. In

the end, the non-empty palyer wins the play when for every S there isi < i such that < otp(C) f(i) Ji gt(i).

Moved from pg.41:

Remark 7.1. (2009.06.09) 1) Revise the proof using clause (h) instead of clause (b).2) In 4.9 use a class of cardinals + instead some Old 09.5.16

[SAHARON: (09/4/02) - Rethink, maybe can prove upgrading OBB1 to OBB4

which is better.]

2) If in addition I is -complete, = cf() and ( < )(|||i

| < ), then {i < i :OBB5Ji(C, I

) for some -complete I I} = i mod I.

Moved 2010.7.26 from 4, i.e. ??, pg.54:{e59}

Conclusion 7.2. 1) If CH then TDU+1 . [Recheck last case.]2) If > > 0 are regular and , SP(R) then

++1 sp(R). [FILL!]3) If > 1 belongs to SP(Z) then ++1 sp(R). [FILL!]

Moved from pgs.54,55:

Proof. 1) The proof is divided to cases, clearly together they suffice.

Case 1: For some C2 and such that < 2 we have 1 = 2.

Let J = Jbd2 hence pp() =

+

2

.By conclusion 3.8 we have BB(2, +, , J ). By clause (d) of claim 5.9 we have

J SP1(Z) hence by claim 5.6 we have TDU2 ,+(Z), more than required.

Case 2: For some C0 letting = cf(2) we have <


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60 SAHARON SHELAH

Add to 5.10 (2010.7.27):

6) Assume n = cf(2n), n < n+1, ( < 2n(||) for n < , = tcf(

n

n,


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For each A I+ , we know that WA := { S: for every C, fA() Jt} = S mod I.

But the ideal I is (2|i|

)+

-complete by clause (b) of the assumption, hence W :={WA : A I+ } {W

i : i U1} = S mod I. Now for any W, considerfi() : i < i : C; it is a sequence of < members of i

J; hence by clause(f) for some A I+ , each fi() : i < i

A has a bound belonging to the set{gt A : t J} in (

iA

Ji,


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62 SAHARON SHELAH

()6 if = 2 then we get the conclusion.

[Why? Now by ??, ??(1) we can find stationary S S+ and a (+(+1), J+)-

free S-ladder system. We finish similarly to ()5 but here we use < 2i ||+ 0, > +, 0 then =

n


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64 SAHARON SHELAH

Proof. 1) Let h : < witness Sep(,, ), let for let Sol := { : ( < )(h() = ()} and let Sol = {Sol : F}. If < 2 then

|Sol| |F| + < 2

and if = 2

then 2

is regular hence Sol being the union ofa < 2 set each of cardinality < 2 we have |Sol| < 2. Choose \ Sol and letg be defined by g() = f(). So F / Sol


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For F , we can find G {Gf : f F } such that

(a) |G| = |F| + +

(b) if F is -free , then G is -free(c) I (guessing) !!!WHAT IS MEANT HERE?!!!

!!!THE FOLLWING REMARK IS TOO FRAGMENTARY TO EVALUATE.PLEASE REMEDY. !!!

Remark 7.20. 1) Do 1.3 and/or 1.9 really help? It helps us move from Jbd0 to

Jbd10 .2) But to use it, we need another version of 5.6, which we only remark above.3) Another approach is to try to develop a dual to [EkSh:505], even to R-modules,i.e., the existence of -free non-Whitehead !!! R-modules or groups. ?!!!.


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{k6}Conclusion 7.21. If sp(Z) has at least three members then TDU .

Remark 7.22. Use 3.17 (090810) - think.

Proof. Let sp(Z)\{0, 1}.

Case 1: If is large, let = { : C, as in the previous claim}. I.e.n < n(min()) we are done.

Case 2: C not as in the previous claim.By 3, we have BB(2, +, , ), so by 5 we can finish.

{k8}Claim 7.23. If 0 < 1 < 2 are members of sp(R) then - FILL, see 3.17.

Claim 7.24. Assume C0 , = 2 = |I2|(c) J2 is -complete, (|I2|, )-regular(d) J1 is (|I2|, |I1|)-regular [-complete].

Write

(e) if we have F is (, 2, J1)-free, J2 is regular on 2, J1 J2 then we get fr??

2) Let C0 , = 2 =


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A try to use cf(2n) : n < when n < n+1 < =n

n such that (

Documents

Sahron Shelah- PCF and Abelian Groups