Saharon Shelah- Successor of Singulars: Combinatorics and Not Collapsing Cardinals ≤ κ in (<...

8/3/2019 Saharon Shelah- Successor of Singulars: Combinatorics and Not Collapsing Cardinals in (< )-Support Iterations

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SUCCESSOR OF SINGULARS: COMBINATORICS

AND NOT COLLAPSING CARDINALS

IN (< )-SUPPORT ITERATIONS

SH667

Saharon Shelah

The Hebrew University of Jerusalem

Einstein Institute of MathematicsEdmond J. Safra Campus, Givat RamJerusalem 9l904, Israel

Department of MathematicsHill Center - Busch Campus

Rutgers, The State University of New Jersey110 Frelinghuysen Road

Piscataway, NJ 08854-8019 USA

Department of MathematicsUniversity of Wisconsin

Madison, WI USA

Abstract. On the one hand we deal with (< )-supported iterated forcing no-

tions which are (E0, E1)-complete, have in mind problems on Whitehead groups,uniformizations and the general problem. We deal mainly with the caes of a succes-sor of the singular cardinal. This continues [Sh 587]. On the other hand we deal withcomplimentary ZFC combinatorial results.

I would like to thank Alice Leonhardt for the beautiful typing.This research was supported by The Israel Science Foundation founded by the Israel Academy of

Sciences and Humanities.Publ. 667; Notes - Spring 96Received November 16, 1998 and in revised form March 5, 2001.Corrected after Proofreading for the Journal.First Typed - 97/June/30Latest Revision - 03/Apr/30

Typeset by AMS-TEX

1


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2 SAHARON SHELAH

Annotated Content

1 GCH implies for successor of singular no stationary S has uniformization

[For strong limit singular, for stationary S S+

cf() we prove strong nega-

tion of uniformization for some S-ladder system and even weak versions ofdiamond. E.g. if is singular strong limit and 2 = +, then there arei < increasing in i < cf() with limit for each S such that forevery f : + < for stationarily many S, for every i we havef(2i) = f(

2i+1).]

2 Forcing for successor of singulars

[Let be strong limit singular = + = 2, S Scf() stationary not

reflecting. We present the consistency of a forcing axiom implying e.g.: ifh is a function from A to , A = sup(A), otp(A) = cf(), < then for some h : for every S we have h

h.]

3 +-c.c. and +-pic

[In the forcing axioms we would like to allow forcing notions of cardinality> ; for this we use a suitable chain condition (allowed here and in[Sh 587]). This sheds more light on the strongly inaccessible case and wecomment on this (and forcing against cases of diamonds).]

4 Existence of non-free Whitehead groups (and Ext(G,Z) = 0) abelian groupsin successor of singulars

[We use the information on the existence of weak version of the diamond

for S S+

cf(), strong limit singular with 2 = +, to prove that there are

some abelian groups with special properties (from reasonable assumptions).We also get more combinatorial principles on = +, > cf() (even ifjust = 2

).]


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SUCCESSOR OF SINGULARS 3

1 GCH implies for successor of singularno stationary S has unformization

We show that a major improvement in [Sh 587] over [Sh 186] for inaccessible (everyladder on S has uniformization rather than some ladder on S) cannot be done forsuccessor of singulars. This is continued in 4.1.1 Fact: Assume

(a) is strong limit singular with 2 = +, let cf() =

(b) S { < + : cf() = } is stationary.

Then we can find < i : i < >: S such that

()

i is increasing (with i) with limit () if < and f : + then the following set is stationary:

{ S : f(2i) = f(2i+1) for every i < }.

Moreover

()+ if fi : + i, i < for i < then the following set is stationary:

{ S : fi(2i) = fi(

2i+1) for every i < }.

Proof. This will prove 1.2, too. We first concentrate on () + () only.

Let = i 2i, 0 > 2

. For

< +, let =i 0. For S let b

i : i < be such

that: bi i , |b

i | i, b

i is increasing continuous with i and =

i


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4 SAHARON SHELAH

(A) i,j, : < j is a strictly increasing sequence of ordinals

(B) i < i,j, <

i+1, (can even demand

i,j, <

i + )

(C)

i,j, / {

i1,j1,1 : j1 < j, 1 < j1 (and i1 < , really only i1 = i matters)}(D) for every 1, 2 bj , the sequence Min{j , f

1

(2, i,j,)} : < j is

constant i.e.: one of the following occurs:

() < j (2, i,j,) / Dom(f1

)

() < j f1

(2, i,j,) = f

1

(2, i,j,0), well defined

() < j f1(2, i,j,) j , well defined.

For each i < j < we use is strong limit > j j1


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j we can define a function h : by h(i) = Min{j : j > i + 1 and < j and[i+1] b

j}. So h G hence for some () < 2

we have h = g(). Now looking

at the choice of i,h(i),0,

i,h(i),1 we know (remember 2

< 0 bj and < h(i))

( < 2)( bh(i))[Rang(f) & Dom(f

)

i+1 f

(,

i,h(i),0)

= f(, i,h(i),1)].

In particular this holds for = (), = [i+1], so we get

f[i+1]

((), i,h(i),0) = f[i+1]

((), i,h(i),1).

By the choice of f and of[i+1] this means

fg()(i,h(i),0) = fg()(

i,h(i),1))

but h = g() and the above equality means fg()

(g(),

2i ) = fg()

(g(),

2i+1 ), and

this holds for every i < , and E Eg() so we get a contradiction tothe choice of (fg() , E()).

So we have finished proving () + ().

How do we get ()+ of 1.1, too?The first difference is in phrasing the question, now it is, for g G:

Questiong : Does g, : S satisfy:

(f0

+0)(f1 +1) . . . (fi

+i) . . .

i


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6 SAHARON SHELAH

1.2 Fact: 1) Under the assumptions (a) + (b) of 1.1 letting = i : i < be increasingly continuous with limit such that 2 < 0, 2

i < i+1 we have()1 + ()2 where

()1 we can find < : < >: S such that

() is increasing in with limit

()+ if fi : + i+1, for i < , then the following set is stationary

{ S : fi( ) = fi( ) when , [i, i+1) for every i < }

()2 moreover if Fi : [+]


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2 Case C: Forcing for successor of singular

We continue [Sh 587].

2.1 Hypothesis. 1) strong limit singular = cf() < , = +, , 2 = +.

2.2 Definition. 1) Let C


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8 SAHARON SHELAH

() Nii Nj0 for i < j

() Mi Mi0, Mi M

i0.

2) We say above that (Mi : i < , Ni : i < ) is an (E0, E1)-approximation toM.3) Let C


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The player INC first decides what is i0 < and then it chooses a conditionp Q Mi0 stronger than r. Next, at the stage i [i0 1, ) of the game, COM

chooses pi Q Mi+1 such that:

(i) p Q pi

(ii) (j < i)( < j)(qj,Qpi),

(iii) if i is a non-limit ordinal, then pi Q is minimal satisfying (i) + (ii)

(iv) if i is a limit ordinal, then pi Q.

Now the player INC answers choosing an increasing sequence qi = qi, : < isuch that pi Q qi,0 and qi is (Ni [, i],Q)-generic for some < i (see [Sh 587,B.5.3.1]) and < i qi (+ 1) Mi,+1.The player COM wins if it has always legal moves and the sequence pi : i <

has an upper bound in Q.2) We say that the forcing notion Q is complete for (E0, E1) or (E0, E1)-complete if

(a) Q is strongly complete for E0 and

(b) for a large enough regular , for some x H(), for every sequence M

ruled by (E0, E1) with an E0-approximation (Mi : i < , Ni : i < ) andsuch that x M0 and for any condition r Q M0, the player INC doesnot have a winning strategy in the game G

M,Mi:i


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10 SAHARON SHELAH

(a) qt {q rk(t) : q I}, and

(b) for each Dom(qt), one of the following occurs:

(i) qt() = pt()

(ii) P qt() Q (not just in the completion Q

)

(iii) P there is r Q such that Q

|= pt() r qt() (not really

needed).

Proof. Just like the proof of [Sh 587, B.7.1].

Our next proposition corresponds to [Sh 587, B.7.2] which corresponds to [Sh 587,A.3.6]. The difference with [Sh 587, B.7.2] is the appearance of the M , Mi.

2.8 Proposition. Assume that E C


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(iv) rki(t) = min(w {}\ dom(t)) and


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12 SAHARON SHELAH

So in limit cases i < : the existence of limit is by the clause () of Definition 2.3.In the end we use the winning of the play and then need to find a branch in the

tree of conditions of level : like Case A using E0. 2.9

2.9 Theorem. Suppose that (E0, E1) C


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clauses (),(c)(i)-(iv) of 2.8 and such that T0 N10

w0 = M0w0 = M0

(remember cf(0) > 2M0). So, in particular, ift T0, dom(t) then t(0) M1

is either of a P-name for an element ofQ

.

Moreover, we additionally require that (T0, p0) is the


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14 SAHARON SHELAH

So now, if there is a P-name

for an element ofQ such that

pit P (j < i)(pj

tj()

Q

),

then we take the P-name of the lub of pj

tj() : j < i, pj

tj() = in Q,

and we continue. If there is no such

then we decide that t / T+i and we

stop the procedure2.

Now, let T+i consist of those t Ti for which the above procedure resulted in a

successful definition of pit Prki (t). It might not be clear at the moment if T+i

containss anything more than , but we will see that this is the case. Note that

T+i Ti

j


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q P (i < )((t+1)(i) = i < i

0 ) and the sequence

i0 , pi0

projTTi0

(t+1)

(), (t+1)(i), p

i

projTTi

(t+1)

() : i0 i <

is a result of a play of the game GM[G

],Ni[G

]:i


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16 SAHARON SHELAH

Then for some (E0, E1)-complete forcing notion P of cardinality we have

P forcing axiom for (E0, E1)-complete forcing notionof cardinality and < of open dense sets

and in VP the set S is still stationary (by preservation of (E0, E1)-nontrivial).

2) If clauses (a),(c) holds and S , then for some a, if we define E1 as in clause (d)then clause (b),(d),(e) holds.

Proof. 1) See more in the end of 3.2) Easy. 2.11

2.12 Application: In VP of 2.11:

(a) if

(i) < , A = sup(A) for S,

(ii) |A| <

(iii) h = h : S, h : A

(iv) A

{a,i+1\a,i : i < },

then for some h : and club E of we have ( S E)[h h]

where h h means that sup(Dom(h)) > sup{ : Dom(h) and / Dom(h) or Dom(h) & h() = h()}

(b) if we add: h constant, then we can omit the assumption (iii)

(c) we can weaken |A| < to |A a,i+1| |a,i|

(d) in (c) we can weaken |A| |A a,i+1| |a,i| to h a,i+1 belongsto Mi+1 Ni for some < i(remember cf(sup a,i+1) >

i ).

2.13 Remark. 1) Compared to [Sh 186] the new point in the application is (b).

2) You may complain why not having the best of (a) + (b), i.e. combine their goodpoints. The reason is that this is impossible by 1, 4; the situation is different inthe inaccessible case.

Proof. Should be clear. Still we say something in case h constant, that is (b).Let


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Q = {(h, C) : h is a function with domain an ordinal

< = +,C a closed subset of + 1, C

and ( C S ( + 1))(h h)}.

with the partial order being inclusion.For p Q let p = (hp, Cp).

So clearly if (h, C) Q and = Dom(h) < then for some h1 we haveh h1 Q1, Dom(h1) = ; moreover, if < & / S then (h, C)

(h [,], C {}) Q.The main point is proving Q is complete for (E0, E1). Now Q is strongly complete

for E0 is proved as in [Sh 587, B.6.5.1,B.6.5.2] (or 3.14 below which is somewhatless similar). The main point is clause (b) of 2.5(2); that is, let M, Mi : i scite{2.10} undefined

demand ( S)(h h) provided that we add in ?, recalling S does not

> scite{2.10} undefinedreflect is a set of limit ordinals and

A = A : S, A = sup(A)


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18 SAHARON SHELAH

satisfies

() 1 = 2 in S sup(A1 A2) < 1 2.

2) If ( S)(otp(A) = this always holds.

Proof. We define Q = {h : Dom(h) is an ordinal < and h() = 0 Dom(h) ( S)[h() = h()] and (Dom(h) + 1) S implies h

h}ordered by . Now we should prove the parallel of the fact:

if p Q, = Dom(p) < < then there is q such that p q Q andDom(q) = .

Why this holds? We can find A

: S(+1) such that A

A, sup(A\A

) < and A = A : S (+ 1) is pairwise disjoint.

Now choose q as follows

Dom(q) =

q(j) =

p(j) if j <

h(j) if j A\ and S (+ 1)\( + 1)

0 if otherwise.

Why does A exist? Prove by induction on that for any A1, A : S ( + 1)as above and satisfying < < , we can end extend A1 to A : S(+ 1)which is as above. 2.14

2.15 Remark. Note: concerning inaccessible we could immitate what is here:

having Mi+1 =

Nii ,i


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3 +-c.c. and +-pic

We intend to generalize pic of [Sh:f, Ch.VIII,1]. The intended use is for iterationwith each forcing > - see use in [Sh:f]. In [Sh 587, B.7.4] we assume each Qi ofcardinality . Usually = +.Note: E0 is as in the accessible case, in [Sh 587] but this part works in the other

cases. In particular, in Cases A,B (in [Sh 587]s context) if the length of a E0 is

< (remember = +), then we have (< )-completeness implies E0-completeness

AND in 3.7 even a E0 g(a) = is O.K.In Case A on the S0 S

if g(a) = , a S0 is O.K., too. STILL can start

with other variants of completeness which is preserved.

3.1Context: We continue [Sh 587, B.5.1-B.5.7(1)] (except the remark [Sh 587,B.5.2(3)]) under the weaker assumption = 0, so is not necessarily

strongly inaccessible; also in our Es we allow a such that |a | = || is stronglyinaccessible.

3.2 Definition. Assume:

(a) = cf() > || 0, , E0 {a : a anincreasing continuous sequence of members of []


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() N0 and

(i) if = 1, then for some club C of for every S we have(N, p) : S C belong to N0

(ii) if = 2, then for some club C of for every S C andi < we have (N

, (i + 1), p (i + 1)) : S Cbelongs to Ni+1

() we define a function g with domain S as follows: g() = (g0(), g1())

where g0() = N(


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We let Y = [0]


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22 SAHARON SHELAH

(e) (Mj ,Pj , Y

j , p

i ) : j i, S belongs to N

i+1 (hence P

j M

j+1,

etc.)

(f) pi,c Q Ni+1

(g) ifc Pi and pj,cj : j c has an upper bound then pi,c is such a bound

(h) pi,c {I : I Mi is a dense open subset ofQ}.

Can we carry the induction?For i limit let Mi = {M

j : j < i} and choose Y

i ,P

i by clause (d) i.e. by

the rules of the game ,(E0) and pi by clause (g) + (h) (possible as forcing by

Q adds no new sequences of length < of members of V). For i non-limit, let

xi = (Mj ,P

j , Y

j , p

j ) : j i, S let Y

i = {a : a [

]


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3.8 Remark. We can omit the assumption Lim(Q) add no bounded subsets of if

we add the assumption c(E0) C


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24 SAHARON SHELAH

disjoint to it. But as P satisfies the -c.c. without loss of generality C = C so

C = C

wy

C is a club of. Now choose 1 < 2 from Sy C and we choose

by induction on w = wy {0, g(Q)} a condition q P such that:

3(i) 1 < q1 = q 1(ii) q is a bound to {p

1u : i < 1} {p

2i : i < 2}.

For = 0 let q0 = . We have nothing to do really if is with no immediatepredecessor in w, we let q be {q1 : 1 < , 1 w

}. So let = 1 + 1, 1 w;now ifq G P1,2, G generic over V, then 1, 2 S

y,1 [G], hence S

y, [G] C1

is non-empty, hence is stationary, and we use Definition 3.2.

Case 2: p = 2.

Similar proof. 3.7

3.9 Claim. Assume = cf() > , ( < )(||


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Let Ax1,2(E0, E1, E), the forcing axiom for (E0, E1, E), and = (0, 1, 2) be thefollowing statement:

if

(i) Q is a focing notion of cardinality < 1

(ii) Q is complete for (E0, E1), see Definition [Sh 587, B.5.9(3)]

(iii) Q satisfies (0, S, E)-pic

(iv) Ii is a dense subset ofQ for i < i < 2,

then there is a directed H Q such that (i < i)(HIi = ).

3.12 Theorem. Assume of Definition 3.11 and =


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26 SAHARON SHELAH

3.14 Claim. Assume

(a) ,S, E0, E1 are as in 3.13

(b) if Sbd =: { < : strongly inaccessible, S is stationary in and S}is not a stationary subset of

(c) A = A : S, A

(d) Q = QA1 is as in Definition 3.15 below

(e) E E0 is nontrivial.Then

() Q is complete for (E0, E1)

() Q satisfies the (, +, E)-pic

() Q satisfies the +

-c.c.

3.15 Definition. For = cf(), S = sup(S), A = A : S, with A we define the forcing notions Q = Qad

Aas follows:

(a) p Q iff

(i) p = (c, A) = (cp, Ap)

(ii) c is or a closed bounded subset of hence has a last element

(iii) A sup(c) such that

(iv) if C S then A = A

(b) p q iff

(i) cp is an initial segment of cq

(ii) Ap = Aq sup(cp).

Proof of 3.14. We concentrate on part (1), part (2)s proof is similar. Now

()1 for every < ,I = {p Q : < sup(cp)} is dense open.

[Why? If p Q, let = sup(cp) + 1 + and q = (cp {}, Ap), sop q I.]

()2 If < is a limit ordinal, pi : i < is Q-increasing and sup(cpi) i+1 .

Then we can choose (g, : < ) : S such that

1(i)

: < is strictly increasing with limit

(ii) if < +1 then h0( ) = h0(

) = , and h1( ) = h1(

) =

(iii) h a partial function from to , sup(Dom(h)) <

for S

2 for every f : , B [] 2

. Let Mi

(H((2)+), ,


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let = i : i < be increasing continuous with limit , i divisible by

and > 0. For S let bi : i < be such that: bi

i , |b

i | i, b

i is

increasingly continuous in i and = i


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32 SAHARON SHELAH

(ii) in clause (A) we demand i,g, : g G, < i+1 belongs to Mi+1 (hence

also j,g, : g G, < j+1 : j i belongs to Mi+1)

(iii) clause (c) is replaced by:

i,g, Fi({

j,g(j+1), : < j+1 and j < i}).1.2

Proof of 4.1. 1) We apply 4.2 to the A : S from 4.1, and any h0, h1 as inclause (d) of 4.2.Let {ti,j + Gi : <

i,j} be a free basis of Gj/Gi for i < j . If i = 0, j =

we may omit the i, j, i.e. t = t0, and =

0,. Let + 0 = |G| < ; actually

,+1 < is enough; without loss of generality < 1 in 4.2. Let ,i =

(,i)

where (, i) =


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(, ] and for no S(, ] do we have {,i : i < , and < }};

moreover G+1 is freely generated (as an extension of G+1). So G

+1/G

+1

is free, as also G1 is free we have shown Fact A.

Fact B: G is not Whitehead.

Proof. We choose by induction on , an abelian group H and a homomorphismh : H G

= {x : < } {y

: < , S}G increasing continuous in

, with kernel Z, h0 = zero and k : G H is a not necessarily linear mapping

such that h k = idG . We identify the set of members of H, G,Z withsubsets of (1 + ) such that OH = OZ = 0.Usually we have no freedom or no interesting freedom. But we have for = + 1,

S. What we demand is (G - see before Fact A):

()2 letting H = {x H+1 : h+1(x) G}, if s = g(x()) Z\{0}(g from 4.2), then there is no homomorphism f : G

H suchthat

() f(x,i

) k(x,i

) Z is the same for all i (


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34 SAHARON SHELAH

f(x) = 0 and5 |f(x)| is minimal under this, so without loss of generality it is 1.Hence for some S we have:

()3 f(g(0)) = 1Z()4 f(x+1+1+) = f(x

) for +1\

that is f(x,

) = f(xzeta

)

(in fact this holds for stationarily many ordinals S).So we get an easy contradiction.

3) The proof is included in the proof of part (2). 4.1

We also note the following consequence of a conclusion of an instance of GCH.

4.3 Claim. Assume

(a) = + and > = cf()

(b) = where = 2

(equivalently = + > 2)

(c) S { < : cf() = } is stationary

(d) = : S with an increasing sequence of length with limit .

Then we can find A : S such that:

() A = Ai : i <

() Ai []: S with A

i .

2) We can add to the conclusion Ai (i + 1) if guess clubs.

Proof. Let i : i < be increasing continuous with limit . Let : < list , s o = , : < and without loss of generality , . Foreach S let bi : i < be an increasing continuous sequence of subsets of with union such that |bi | < and sup(b

i ) < ; for ()

+, moreover |bi | i;

5What does this mean? f(x) is an integer so its absolute value is well defined


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36 SAHARON SHELAH

REFERENCES.

[Fu] Laszlo Fuchs. Infinite Abelian Groups, volume I, II. Academic Press, NewYork, 1970, 1973.

[Sh 186] Saharon Shelah. Diamonds, uniformization. The Journal of SymbolicLogic, 49:10221033, 1984.

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

[Sh 576] Saharon Shelah. Categoricity of an abstract elementary class in twosuccessive cardinals. Israel Journal of Mathematics, 126:29128, 2001.math.LO/9805146.

[Sh 587] Saharon Shelah. Not collapsing cardinals in (< )support iterations.

Israel Journal of Mathematics, 136:29115, 2003. math.LO/9707225.

Saharon Shelah- Successor of Singulars: Combinatorics and Not Collapsing Cardinals ≤ κ in (<...

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