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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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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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SUCCESSOR OF SINGULARS 5
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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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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SUCCESSOR OF SINGULARS 7
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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() 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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SUCCESSOR OF SINGULARS 9
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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SUCCESSOR OF SINGULARS 11
(iv) rki(t) = min(w {}\ dom(t)) and
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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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SUCCESSOR OF SINGULARS 13
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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SUCCESSOR OF SINGULARS 15
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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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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SUCCESSOR OF SINGULARS 17
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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SUCCESSOR OF SINGULARS 19
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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SUCCESSOR OF SINGULARS 21
We let Y = [0]
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(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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SUCCESSOR OF SINGULARS 23
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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SUCCESSOR OF SINGULARS 25
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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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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SUCCESSOR OF SINGULARS 31
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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SUCCESSOR OF SINGULARS 33
(, ] 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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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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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.