saharon shelah- advances in cardinal arithmetic

8/3/2019 Saharon Shelah- Advances in Cardinal Arithmetic

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ADVANCES IN CARDINAL ARITHMETIC

SH420

Saharon Shelah

Institute of MathematicsThe Hebrew University

Jerusalem, Israel

Rutgers University

Mathematics DepartmentNew Brunswick, NJ USA

I thank Alice Leonhardt for typing (and retyping) the manuscript so nicely and accurately.First Typed - 9/90; revised 11/92; appeared in IJMLatest Revision - 09/Jan/17Partially supported by the BSF, Publ. 420

Typeset by AMS-TEX

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Annotated Content

1 I[] is quite large

[If cf = , + < cf = then there is a stationary subset S of { < : cf() = } in I[]. Moreover, we can find C = C : S, Ca club of , otp(C) = , guessing clubs and for each < we have:{C : nacc C} has cardinality < .]

2 Measuring S


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ADVANCES IN CARDINAL ARITHMETIC SH420 3

1 I[] is Quite Large and Guessing Clubs

On I[] see [Sh 108], [Sh 88a], [Sh 351, 4] (but this section is self-contained; seeDefinition 1.1 and Claim 1.3 below). We shall prove that for regular , , such that+ < , there is a stationary S { < : cf() = } in I[]. We then investigateguessing clubs in (ZFC).

1.1 Definition. For a regular uncountable cardinal , I[] is the family of A such that { A : = cf()} is not stationary and for some P : < we have:

(a) P is a family of < subsets of

(b) for every limit A of cofinality < there is x , otp(x) < = sup(x)such that < x {P : < }.

1.2 Observation. In Definition 1.1 we can weaken (b) to:

for some club E of x for every limit A E of cofinality < . . . .

Proof. Just replace P by {x : x {P : Min(E\ + 1)}}.

We know (see [Sh 108], [Sh 88a] or below)

1.3 Claim. Let > 0 be regular.1) A I[] iff (note: by (c) below the set of inaccessibles in A is not stationaryand) there is C : < such that:

(a) C is a closed subset of

(b) if nacc(C) then C = C (nacc stands for non-accumulation)

(c) for some club E of , for every A E, we have: cf() < and =sup(C), and cf() = otp(C)

(d) nacc(C) is a set of successor ordinals.

2) I[] is a normal ideal.

Proof. 1) The if part:Assume C : < satisfy (a), (b), (c) with a club E for (c). For each limit

< choose a club e of order type cf(). We define, for < :


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4 SAHARON SHELAH

P =: {C : } {e : } {e : Min(E\( + 1)}.

It is easy to check that P : < exemplify A I[].

The only if part:

Let P = P : < exemplify A I[] (by Definition 1.1). Without lossof generality

() if C P, and C then C\ P and C P

For each limit < let e be a club of satisfying otp(e) = cf() andcf() < cf() < min(e). Let i : i < be strictly increasing continuous,

each i a non-successor ordinal < , 0 = 0, and i+1 i 0 + |i

P| + |i|

and i A cf(i) < i.(Why? Let E be a club of such that E A cf() < , and then choosei E by induction on i < .)

Let Fi be a one to one function from (i

P)i into {+1 : i < +1 < i+1}.

Now we choose C as follows. First, for = 0 let C = . Second, assume isa successor ordinal, let i() be such that i() < < i()+1. If / Rang(Fi()),

let C = . If = Fi()

(x, ) hence necessarily x i()

P, < i()

) and x,

are unique. Let C be the closure (in the order topology) of C , which is defined

as:

Fj(x , ) : the sequence (j,,) satisfies ()

x,j, below

where

x,j, (i) x

(ii) otp(x ) e ,

(iii) j < i() is minimal such that x j

P

(iv) if x , otp(x ) e then

(j(1) < j)[x

j(1)

P]

(v) < Min(x).


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Third, for < limit, choose C: if possible, nacc(C) is a set of successor ordinals,C is a club of , [ nacc(C) C = C]; if this is impossible, let C = .Lastly, let C0 = and let E =: {i : i is a limit ordinal < }.

Now we can check the condition in 1.3(1).Note that for successor C = nacc(C).

Clause (a): C a closed subset of .If = 0 trivial as C = and if is a limit ordinal, this is immediate by the

definition. So let be a successor ordinal, hence, by the choice of i : i < as anincreasing continuous sequence of nonsuccessor ordinals with 0 = 0, clearly i() iswell defined, i() < < i()+1. Now if / Rang(Fi()) then C = and we are

done so for some x, we have = Fi()(x, ) hence necessarily x

i()

P and

< i()

. By the definition of C

(the closure in the order topology on , of the

set of C i.e. the set of Fj(x , ) for the pair (j,) satisfying x,j, it suffices to

show C , i.e.

() if the pair (j,) satisfies x,betaj, then Fj(x , ) < .

So assume (j,) satisfies x,j, but by clause (iii) we know that j < i() and so

Rang(Fj) j+1 i() < as required.

Clause (b): If nacc(C) then C = C .

If it is enough to show C = C and as C

= nacc(C), we have

C .

As C necessarily for some , j satisfying x,j, we have = Fj(x , ). Bythe choice of Fj necessarily

is a successor ordinal and j < < j+1.

Now any member (1) of C has the form Fj(1)(x (1), ) with j(1), (1)

satisfying x,j, ; clearly j(1) < (1) = Fj()(x (1), ) < j(1)+1 and j < =

Fj(x , ) < j+1. But (1) < (being in C ) so necessarily j(1)+ 1 j.

So j(1), (1) satisfy (i) (v) with x replaced by x , i.e., satisfy x,j, ; recall

by = Fj(x , ), so Fj(x)(x (1), ) C . So

C C ; similarly

C C , so we get the desired equality.

Clause (c): We shall show that E = {i : i is a limit ordinal < } is as required in

closed (c).Clearly E is a club of . So assume that A E we should prove: cf() cf() holds as we assume i A cf(i) < i. As E,by Es definition for some limit ordinal i() we have = i(). By the choice ofC it is enough to find a set C closed unbounded in of order type cf() such that nacc(C) successor & C = C .


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By the choice ofP, for some x , otp(x) < = sup(x) and


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(iv) C P & nacc(C) C P

(v) for any club E of for some S E and C P we have C E &

otp(C) = ().

Proof. Let C : < witness S I[] be as in 1.3(1); without loss of generalityotp(C) (). For any club E, consisting of limit ordinals for simplicity, let usdefine PE by induction on < :

PE =:{ g(C, E) : E and < Min[E\( + 1)]}

{C {} : E , C PE and otp(C) < ()}

where

g(C, E) =: {sup(E (+ 1)) : C and > Min(E)}.

Note that |PE| |Min(E\( + 1)| < .We can prove that for some club E of the sequence PE : < is as requiredexcept possibly clause (v) which can be corrected gotten by a right of E (just bytrying successively + clubs E (for <

+) decreasing with , see [Sh 365]). Notethat clause (iv) guaranteed by demanding E to consist of limit ordinals only and

the second set in the union defining P

E. 1.4

The following lemma gives sufficient condition for the existence of quite largestationary sets in I[] of almost any fixed cofinality.

1.5 Lemma. Suppose

(i) > > 0, and are regular

(ii) P = P : < , P a family of < closed subsets of

(iii) IP =: {S : for some club E of for no S E is there a club C of

, such that C E and [ nacc(C) C


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1.6 Remark. 1) The for stationarily many in the conclusion can be strengthenedto: a set whose complement is in the ideal defined in [Sh 371, 2].2) So if < then we can have {i < : cf(i) = } IP .

Proof. Let be regular large enough, N be an elementary submodel of (H(), ,


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Suppose T, so there is C, a club of such that C E and


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10 SAHARON SHELAH

Now the conclusion of 1.5 (see the moreover and choice of P i.e. ()) gives thedesired conclusion. 1.7

1.8 Conclusion. If > are uncountable regular, + < , then for some stationary

S { < : cf() = } and some P = P : < we have: ,P,S from the

conclusion of 1.4 holds.

Proof. As is regular apply 1.7 and then 1.4. 1.8

Now 1.8 was a statement I have long wanted to know, still sometimes we want tohave C E, otp(C) = (), () not a regular cardinal. We shall deal with such

problems.1.9 Claim. Suppose

(i) > > 0, and are regular cardinals

(ii) P = P, : < for = 1, 2, where P1, is a family of < closedsubsets of , P2, is a family of clubs of and [C P2, &

C C


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Proof. Same proof as 1.5. (Note that without loss of generality [C P1, & < < C P1,]).

1.10 Conclusion. If() is a limit ordinal and = cf() > |()|+ then we can findP = P

, : < for = 1, 2 and stationary S { < : cf() = cf(())}

such that:

,()

P1 ,P

2(A) P1, is a family of< closed subsets of each of

order type < ()(B) nacc(C) & C P1, C P

1,

(C) P2, is a family of clubs of

(yes, maybe = ) of order type

(), and [ nacc(C) & C P

2, C P

1,](D) for every club E of for some E S,

cf() = cf(()) and there is C P2, such that C E.

Proof. If = |()|++ (or any successor of regulars) use [Sh:e, ChIII,6.4](2) or [Sh365, 2.14](2)((c)+(d)).

If > |()|++ let = |()|++ and let S1 = { < ++ : cf() = cf(())}; ap-

plying the previous sentence we get P1 , P2 satisfying

++,()

P1 ,P

2 ,S1, hence satisfying

the assumption of 1.9 so we can apply 1.9. 1.10

1.11 Definition. +,()

P1,P2,Sis defined as in 1.10 except that we replace (C) by

(C)+ P2, is a family of < clubs of of order type ().

1.12 Remark. Note that ifP = P1, P2,, |P2,| 1, P1, = {C P :

otp(C) < ()},P2, = {C P : otp(C) = ()} then+

,()

P1,P2,S

,()

PS

mod.

1.13 Claim. Suppose = cf() > |()|+, () a limit ordinal, additively inde-

composable (i.e. < () + < ()), ,()

P1,P2,Sfrom 1.10 and

() S |P2,| ||.


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(Note: a non-stationary subset of S does not count; e.g. for successor cardinal

the with ||+ < . Note: +,()

P1,P2,Sholds by () and if is successor then

+,()

P1,P2,Ssuffice).

Then for some stationary S1 S and P = P : < we have: P P1, P2, and:

,()

P,S1(i) P is a family of closed subsets of , |P| <

(ii)otpC < () if C P, / S1(iii) if S1 then: P = {C}, otp(C) = (),

C a club of disjoint to S1(iv) C P & nacc(C) C P(v) for any club E of for some S1 we have C E.

1.14 Remark. Note there are two points we gain: for S1, P is a singleton(similarly to 1.4 where we have (1C P)[otp(C) = ()]), and an ordinal cannot have a double role C a guess (i.e. S1) and C is a proper initialsegment of such C. When () is a regular cardinal this is easier.

Proof. Let P2, = {C,i : i < } (such a list exists as we have assumed |P2,| ||, we ignore the case P2, = ). Now

()0 for some i < for every club E of for some S E we have C,i\E isbounded in [Why? If not, for every i < there is a club Ei of such that for no S E is C,i\E bounded in . Let E = { j < : j a limit ordinal,

j i


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()3 for some h : |()|+, for every S we have h() / {h() : C}

[Why? Choose h() by induction on .]

()4 for some < |()|+ for every club E of, for some S h1({}), C

E[Why? If for each there is a counterexample E then {E : < |()|

+}is a counterexample for ()2.]

Now we have gotten the desired conclusion. 1.13

1.15 Claim. If S { < : cf() = }, S I[], + < = cf(), then for some

stationary S1 S and P1 we have ,()P1,S1

.

Proof. Same proof as 1.4 (plus ()3, ()4 in the proof of 1.10). 1.15

1.16 Claim. Assume = +, |()| < and cf(()) = cf().Then we can find stationary S { < : cf() = cf()()} and P such that

,()

P,S.

Remark. This strengthens 1.10.

Proof. Case (). regular.By [Sh:e, Ch.III,6.4](2), [Sh 365, 2.14](2)((c)+(d)).

Case . singular.

Let =: cf(), =: |()|+ + + and = and for each < let =


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So as we can replace every C by { C : otp(C )} is even, without loss ofgenerality [because we can replace every C by { C : otp( C) is even},without loss of generality (check)]

()+ S1 & nacc C C {C : < }.

Without loss of generality [ A, C A,] (just note |C| < ) and A, A, A, . For S1 let C = {, : < }(, increasing in )and let , [,, ,+1) be mimimal such that C ,+1 = C, (exists as

S1 C E1). Without loss of generality every C is an initial segment ofsome C , S1 (if not, we redefine it as ).

()1 there are = () < and stationary S2 S1 such that for every club Eof , for some S2 we have: C E, and for arbitrarily large < ,

, A,+1, .

[Why? If not, for every < (by trying () = ) there is a club E of

exemplifying the failure of ()1 for . Let E =

> sup( C) & C = C].[Why? Define C for < :C0 = { : nacc(C) and ( A,())[ > sup( C) & C = C]}.C is: if S2, > sup(C

0)

closure of C0 otherwise.] Now C : < can be replaced by C :

< .]

()3 For some 1 = 1() < for every club E of for some E : cf() =

cf(()), and there is a club e of satisfying: e E, otp(e) is (), and forarbitrarily large nacc(e) we have e {C : A,1}.[Why? If not, for each 1 < there is a club E1 of for which there is

no as required. Let E =:1


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< ()} satisfies the requirements except the last. As cf(()) = cf(),for some 1() < , 1() () and { < () : , A,(),1()} is un-

bounded in (). Clearly =: ,(), e =: C satisfies the requirement.

Now this contradicts the choice of E1().]()4 For some club Ea of , for every club Eb Ea of , for some Eb we

have:

(a) cf() = cf(())

(b) for some club e of : e Eb, otp(e) = (), and for arbitrarily large nacc(e) we have e {C : A,1()}

(c) for every A,1() we have: C Ea C Eb (we could have

demanded C Ea = C Eb).[Why? If not we choose Ei for i <

+1()

by induction on i, [j sup{|P| : < }.3) If 1.17(2) if is strong limit we can have |P| 1 for each .

Remark. Compare with [Sh 186, 3].

Proof. Left to the reader (reread the proof of 1.16 and [Sh 186, 3].

1.18 Claim. 1) Let be regular uncountable and we have global choice (or restrictourselves to < ). We can choose for each regular > +, P = P : < (assuming global choice) such that:

(a) for each , P is a family of of closed subsets of of order type < .

(b) if is regular, F is the function P (for regular < ), 0 < =cf(), ++ < , x H() then we can find N = Ni : i , an increasingcontinuous chain of elementary submodels of (H(), , +

and , let e be a club of of order type cf(). For as above and for limit let P = {{i e : i < , otp(e i) C} : < has cofinality

++,and S}. Given x H(), we choose by induction on i < ++, Mi, Ni suchthat:


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Ni Mi (H(), ,


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2 Measuring []


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1A) If each = cf() and S I[] is stationary then there isC T0[, ] such that S(C) = S.

Proof. 1) Let S0 { < : cf() = } be stationary, C0 a club of of or-

der type for every S0. By [Sh 365, 2], for some club E of lettingS = S0 acc(E) and letting, for S, C = g(C0 , E) = {sup( E) : C

0 }

we have S / ida(C : S0), now use part (2).2) Check.3) Check.

4) By [Sh 351, 4], [Sh:e, Ch.IV,3.4](2) or [Sh 365, 2.14](2)((c)+(d)) but see [Sh:E12].

5) By 1.7 and 1.15 (so we use the non-accumulation points).6) Similarly. 2.2

Remember (see [Sh 52, 3]).


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2.3 Definition. 1) D is the filter generated by the family of clubs of .2) D


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2.5 Claim. 1) If > 0 and = +. Then the following four cardinals are equal for any (C, P) T[, ], recalling there aresuch (C, P) by 2.2:

(0) = cf([]


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In this case 2.6, 2.7(4) (and 2.5) remain true and we can strengthen 2.2.2) We can even use P with another order (not ).

Proof. Clearly (0) = (1) (2) (3) (the last by 2.5(2)). So we shallfinish by proving (3) (1), and let Q exemplify (1) = cov(,,, 2). LetS = S(C), etc.

Let be e.g. 3()+ and let M be the model with universe + 1 and allfunctions definable in (H(), ,


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2 if i S then h ai has the form Max{fai

,: < n} for some n < ,

pcf(a) and < for < nhence

3 if i S then h ai belongs to Mand obviously (as ai i < j1 < j2 sup( Nj1) < sup( Nj2))

4 Dom(h) cf(h()) = .

Let e be a definable function in (H(), ,


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(ii) if y {bi : i < }, z P and y


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(A) (a) for S, y P and n < we have My,,n N =

zP

Nz .

[Why? We prove this by induction on n. First assume n = 0, My,,n is the Skolemhull of y (N ) in the model M

, well defined as y hence y M

and

N . As y Ny N and M

N

y N

clearly My,,n N

.

Second, assume n = m + 1 and My,,m N. Now My,,n in the Skolem hull ofMy,,m {eh()() : My,,m Reg (

+\+) and y}, so it is enough toshow that: if My,,m (hence N

) and Reg

+\+ and y theneh()() N

. But by ()2(iii) this holds.

(b) for z y in P we have Mz,,n My,,n.

[Why? Just by their choice, i.e. we prove this by induction on n < .]

(c) for y P and m n we have My,,m My,,n.

[Why? Just by their choice, i.e. we prove this by induction on n.]

(d) M := {My,,n : y P and n < } is N.

[Why? By the above.]

(e) if z (hence C E), {y, z} P, sup(y) < , y


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[Why? By (A)(a) we have one inclusion, the . By the choice ofM and clause(D) the result follows by [Sh 400, 3.3A,5.1A] recalling N .] 2.6

But to get normality of the filter we better define

2.9 Definition. Assume = cf() > = cf() > 0, (C, P) T[, ] and X isa set, of cardinality for simplicity and let be large enough. We define a filterD(C,P)[X] on [X]


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Proof. 1) Trivial.2) Repeat the proof, the change is minor.3) We can find Q = {ui : i < } []

= cf() and < ||tr < and S(C) cf() = .

Proof. Fill.

2.13 Conclusion. Suppose > > 0 are regular cardinals and ( < )[cov(,,, 2) |b|. So forsome < the set T

2 := { T1 : h() = } is stationary and by () for some c

the set T2 := { T2 : b = c} is stationary. 2.13

2.14 Conclusion. If > > 0, and are regular cardinals and [ < < cov(,,, 2) < ] then { < : cf() < } I[].

Proof. Use (3) of 2.6.

2.15 Claim. Let(),, mean: ifai []


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3 Nice Filters Revisited

This generalizes [Sh 386] (and see there).See [Sh 410, 5] on this generalization of normal filters.

3.1 Convention. 1) n is a niceness context; we use , FILL, etc., for n, Filn =FIL(n) when dealing from the content.

3.2 Definition. We say the n is a niceness context or a -niceness context or a(, )-niceness context if it consists of the following objects satisfying the followingconditions:

(a) is a regular uncountable cardinal(b) I > is non-empty -downward closed with no -maximal member2

default value is {0n : n < }

(c) let be > and Y : i < is a sequence of pairwise disjoint sets andY {Yi : i < 1} so i < 1 |Y|, |Yi|

(d) the function with domain Y is defined by (y) = i when y Yi

(e) e is a set of equivalence relations e on Y refiningi and T = {} and >{0}. The defaultvalue will be >.


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3.3 Definition. Let n be a -niceness context.1) We say e1 e2 if e2 refines e1. If not said otherwise, every e is from e. Lete be the set of all such equivalence relations with < equivalence classes. Let

(x/e) = (x).2) FIL = FIL(n) is

ee

FIL(e, n). For D FIL, let e = e[D] be the unique e e

such that D FIL(e, n).3) For D FIL(e) let D[] = {X Y : X[] D}; see (5) below.4) For D FIL(n) and e(1) e(D), let D[e(1)] = {X Y/e(1) : X[] D[]}, see(5) below.5) For A Y/e,A[] = {(x/e) : (x/e) A}, and for e(1) e let A[e(1)] = {y/e(1) :y/e A}.

3.4 Definition. 1) For D FIL(e, n), let D+ be {Y Y/e : Y = mod D}.2) n is 1-closed if D FIL(n), A D+ D + A FIL(n).3) n is 0-closed if for every D1 FILn and A D

+1 there is D2 FIL2 such that

(D1 + A) (D2) D2.4) A niceness context n is full if

(a) for every e en, every filter on Yn/e which is normal (with respect to thefunction n) belong to FILn(e).

4A) A niceness content n is semi-full when: for every e1 en and D1 FILn(e1)and e2, e1 e2 en and A P(Yn/e2) lift(W) FIL(e2) whenever

()e1,e2,D1,W (a) e1 e2 in en

(b) D1 FILn(e2)

(c) 2(Y/e2) (or more ???)

(d) W []0 is stationary

(e) D2 = lift(W, D[e2]1 ) is normal (i.e. lift(W, D1)).

5) A niceness context n is thin when

Sucn = {(D1, D2) :D1 = D2 FILn and

D2 = D[e1]1 + A for some A (D

[e1]1 )

+}.

6) A niceness context n is thick if: Sucn = {(D1, D2) : D1, D2 FILn, e(D1)

e(D2) and D[e2]1 D2 and if = 2

|Yn/e2), W1 []0 is stationary and lift(W, D1) =D1 then for some stationary W2 W1 we have lift(W2, D2) = D2}.


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Remark. 1) On lift see Definition 3.17, HERE??2) We can use more freedom in the higher objects.

3.5 Claim. Assume(a) the -niceness context is thick

(b) D1 FILn(e1)

(c) e1 e2 ed(d) for each y Yn/e1, zy, : < y list {z/e2 : z y1}, dy, is a -complete

filter ony(e) D2 FILn(e2)

(f) if A D2 then {y Yn/e1 : { < y : zy, A} dy,} belongs to D1.

Then D2 Sucn(D1).

Discussion: We may consider allowing player I, in the beginning of each move tochoose Wn as above.

3.6 Definition. (0) For f : Y/e X let f[] : Y X be f[](x) = f(x/e). Wesay f : Y X is supported by e if it has the form g[] for some g : Y/e X.

If e1, e2 e and f : Y/e X for = 1, 2 then: we say f1 = f[e1]2 if f

[]1 = f

[]2 .

Writing f[] for f 1X we identify {i}, i < 1 with Yi.(1) Let Fc(T, e) = Fc(T, e,Y) be the family of g, a sequence of the form g : u, u fc(T) = the family of non-empty finite subsets of

> closed under taking

initial segments, and for each u we have g YOrd is supported by e. LetDom(g) = u, Range(g) = {g : u}. We let e = e(g), for the minimal possiblee assuming it exists and we shall say g


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32 SAHARON SHELAH

In the general case:

Player I chooses en and Dn,1 Ducn(Dn1) and let en = e(Dn1) and he choosesgn F (T, e(Dn1) which is extending gn1 then Dom(gn) (i.e. gn1 =

gn Dom(gn1), gn supported by e(Dn,1) and gn is Dn,1-decreasing.Player II chooses Dn = Dn,2 FIL(en) extending Dn,1.

In the end, the second player wins ifn


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3.9 Definition. 1) We say D FIL is nice to g Fc(T, e,Y), e = e(D), if playerII wins the game (D, g, e) (so in particular g is D-decreasing, g supported by e).2) We say D FIL is nice if it is nice to g for every g Fc(T, e).

3) We say D is nice to if it is nice to the constant function . We say D is niceto g Ord if it is nice to g[e(D)].4) Weakly nice is defined similarly but e is not changed.5) Above replacing D by n means: for every D FILn.

3.10 Remark. Nice in [Sh 386] is the weakly nice here, but

(a) we can use n with en = {e}

(b) formally they act on different objects; but if xey (x) = (y) we get a

situation isomorphic to the old one.

3.11 Claim. Let D FIL and e = e(D).1) If D is nice to f, f Fc(T, e), g Fc(T, e) and g f then D is nice to f.2) If D is nice to f, e = e(D) e(1) e then D[e(1)] is nice to f[e(1)].3) The games from 3.7(2) are determined and winning strategies do not needmemory.4) D is nice to g iff D is nice to g (when g Fc(T, e) is D-decreasing).

5) If e e and for simplicityi 1), |e| = cf() if is a limit cardinal.

Proof. Left to the reader. (For part (4) use 3.12(2) below).

3.12 Claim. 1) Second player wins (D, g, e) iff for some second player wins

(D, g, e).2) If second player wins (D , f , e) then for any D-decreasing g Fc(T, e), g

supported by e and,y

g(y) f(y), the second player wins in (D, g, e), when we

let

= + [max{(g() g() + 1) : satisfies Dom(g)}].


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34 SAHARON SHELAH

3) If u1, u2 Fc(T), h : u1 u2 satisfies [ h()h()] and for = 1, 2 wehave g Fc(T, e2), g1 g

2h() (for u1),

= : u is a -decreasing

sequence of ordinals, 2 2h() and the second player wins in

2(D, g2, e) then

the second player wins in 1(D, g1, e).

Proof. 1) The if part is trivial, the only if part [FILL] is as in [Sh 386].2), 3) Left to the reader.

The following is a consequence of a theorem of Dodd and Jensen [DoJe81]:

3.13 Theorem. If is a cardinal, S then:(1) K[S], the core model, is a model of ZF C+ ( )2 = +.

(2) If inK[S] there is no Ramsey cardinal > (or much weaker condition holds)then (K[S], V) satisfies the -covering lemma for + 1; i.e. if B V is aset of ordinals of cardinality then there is B K[S] satisfying B B andV |= |B| .(3) If V |= ( )()[ > + > 2] then in K[S] there is a Ramsey cardinal > .

3.14 Lemma. Suppose

(a) n is a semi-full niceness content thin or medium = 1

(b) f Ord, > 0 =: sup{(2|Y/e|0 ) : e en}

(c) for every A 0, in K there is a Ramsey cardindal > 0, then for everyfilter D FILn(e) is nice to f.

Remark. 1) The point in the proof is that via forcing we translate the filters fromFIL(e,Y) to normal filters on [for higher s cardinal restrictions are better].2) At present we do not care too much what is the value of 0, i.e., equivalently,how much we like the set S to code.

Saharon: compare with [Sh:g, V], i.e., improve as there! But if we use e = {e},the proofs are more similar to [Sh:g, V] we can consider just Levy(1), |D|), now insome proofs we may consider filters generated by |pcf(a)| set |a| < aleph.

First Proof. Without loss of generality (i)f(i) 2. Let S 0 be such that

[ < & A 2||0

A L[S]], e L[S] (see 3.11(6)) and: if g Ord, (i


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1)g(i) f(i) then g L[S] (possible asi 0 (see 3.13(3)). Let in K[S].

Let

Y0 = {X :X , X a countable ordinal > 0, {, 0} X,

moreover X 0 is countable}.

Let

Y = Y1 = {X Y0 : X has order type f(X )}.

Now for g Ord such thati


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36 SAHARON SHELAH

Case 2: General.Let P(Y/e) = {Ae : < 2

|Y/e|}.Now we describe a winning strategy for the second player. In the side we choose

also (pn, n, fn), n

, W n such that3

(where en, An are chosen by the second player):

(A)(i) Pn =n

Q where Q is Levy(1,Y/en)

(we could use iterations, too, here it does not matter).

(ii) pn Pn

(iii) pn increasing in n

(iv) fn is a Pn-name of a function from 1 to Y/en

(v) pn Pn fn(i) Yi/en

(vi) pn+1 fn+1(i) f

n(i) for every i < 1,

(vii) fn is given naturally it can be interpreted as the generic object of Qn

except trivialities.

(B)(i) n, gn have the same domain, n <

(ii) pn Pn Wn YD, W

n+1 W

n

(iii) n = n+1 Dom(n), Dom(n) = Dom(gn) and n is -decreasing

(iv) pn Pn {X YD : for {0,...,n}, f(X1) A and

Dom(gn)

g(X) =

and for {1, 0,...,n 1}, X 2|Y/e| we have:

Ae D f(X 1) A

e } W n

= mod FPn

(v) gn = gn+1 Dom(gn) [difference]

(C)(i) Dn = {Z Y/en : pn Pn {X JD : fn(X 1) / Z} = mod

(DPnn + Wn)}

(ii) gn is Dn-decreasing. [Saharon: diff]

Note that Dn K[S], so every initial segment of the play (in which the secondplayer uses this strategy) belongs to K[S].By (B)(iii) this is a winning strategy. 3.14

3For the forcing notions actually used below by the homogeneity of the forcing notion the valueof pn is immaterial


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Recall all normal filters on Y/e belong to FIL(e).

Alternate: We split the proof to a series of claims and definitions.

3.16 Definition. 1) W = {u : otp(u) f(u w1) and u is countable}.2) Let J be the following ideal on Y0:

W J iff for some model M on with countable vocabulary (with Skolemfunction) we have

W W {w W : w = cM(w)}.

3) For g i


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38 SAHARON SHELAH

2) Forcing with Q preserves y J+ (i.e. if W J+ thenQ W J+.

Proof. Easy, fill.

3.19 Definition. Assume e en and D FILn(e).1) Q = Qe = {f : f is a function with domain a countable ordinal such thati Dom(f) f(i) Yni }.2) f

e is the Q-name {f : f G

Qe}.

3) Let D/fe be the Qe-name of {A 1: for every B D for stationarily many

i < 1, fe(i) B} and nor(D, f

e) the normal filter which D/f

e generates.

4) For W J+ let lift(W, D) = {A Y/e for some B D :Qe {w W :

fe(w 1) B\A J (note that we have enough homogeneity for Qe.

3.20 Claim. Assume e en and D FILn(e).1) Q D

/fe is a normal filter on 1, (i.e. w1 / D

).

2) |Qe| |Yn/e|0 so Z|Qe| 0 hence Qe has 0 maximal antichains; in fact,equality holds as we have demand |Y/e| = | {Yi : i [i0, 1)}/e| for every e e.3) Combine scite3.2A(4) + 3.19 - FILL.

3.21 Definition. 1) We say that x = (e,D, g, ,f ,W ) is a good position (in the

content of proving 3.14) if(a) e en(b) D FILn(e)

(c) g = g : u Fc(T, e), so u = ux

(d) = : u, <

(e) p Qe(f) W = {w W : g(w) = for u} J

+

(g) p Qe Wx WD,fe J

+ and proj(Wx WD,fe) = D nor(D, fe) [FILL].

3.22 Observation. 1) Ifx = (e,D, g, ,p,W

) is a good position then

(a) is decreasing

(b) DW

.


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3.23 Claim. Ife en, D FILn(e) and g = g : u Fc(T, e) andg f[e] for every Dom(g) then we can find a good position x with gx = ex = e, gx = gand D Dx.

Proof. Let G Qe be generic over V and fe = fe[G]. So in V[G] the set WD,f

e[G]

belongs to J+ (by 3.17(3)), i.e., let AD1 : < list D1 and W,D,fe = {w W:

if w then fe(i) = fe[G](i) A}.

Also g defined in 3.16(3) is a choice function on WD,fe (see 3.17(4)), so as Jis a normal ideal and u finite, we can find = : u such that W = {w WD,fe : g(w) = for u} belongs to J

+. As all this holds in V[G]. So thereis a condition p Qe which forces this, and we are done.

3.24 Claim. Assume that

(a) x1 = (e1, D1, g1, 1, p , W 1) is a good position

(b) g2 = g2 : u2 Fc(T, n) and g2 u1 = g2

(c) e1 e2 in en and D2 FILn(e2) or justA P(Yn/e2),A= {A : , e2) where

= , 2, second player wins.4.4

4.5 Claim. Let e e, D FIL(e,Y).

1) For e e(D), A (D[e]+

, f Y/eOrd, f


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5 More on Ranks and Higher Objects

5.1 Convention.

(a) is a cardinal > 1 (using 1 rather than an uncountable regular is tosave parameters)

(b) Y a set of cardinality


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44 SAHARON SHELAH

2) We say E has -sums when: for every D E E and sequence Z : < < of subsets ofY/e(D) there is Z Y/(e/(D), such that: Z Z = mod D

and: [if (e(D), D) (e, D), e = e(D), D E[D] and

Z[e] = mod D

then

Z D].3) We say E has weak -sum if for every D E( E) and sequence Z : <

< of subsets ofY/e(D) there is D, D E[D] such that:

() if (e(D), D) (e, D), D E[D] and Z = mod D for < and

e(D) e(D) then D D (more exactly D[]

D[] and)

() Z = mod D for < .

4) If = we omit it. We say E is -divisible if every E E has. We say Ehas weak -sums if: [rest diff] for every E E and sequence Z : <

< of

subsets ofY/e(E) there is E, E E[E] such that:

() if (e(E), E) (e, E), E E and Z = mod Min(E) for < ande(E) e(E) then E E

() Z = mod Min(E) for < .

We now define variants of the games from 3.

5.3 Definition. For a given E, for every E E:1) We define a game G2(E, g). In the n th move first player chooses Dn

En1 (stipulating E1 = E) and choose gn Fc(

, e(Dn),Y

) extending gn1(stipulating g1 = g) such that gn is Dn-decreasing. Then the second player choosesEn, (En1)[Dn] En E.

In the end the second player wins ifn ] similarly to G2(D, g) but the second player in addition choosesan indexed set n of ordinals, Dom(n) = Dom(gn), n Dom(n1) = n1 and[ n, > n,].

5.4 Definition.1) We say

Eis nice to g Fc(

T, e,Y

) if for every E E

withe e(E) the second player wins the game 2(E, g).2) We say E is nice if it is nice to g whenever E E, e e(E), g Fc(T, e), g is(Min E)-decreasing, we have: E[E] is nice to g.3) If Dom(g) = {} we write g instead g.4) We say E is nice to if it is nice to the constant function .


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5.5 Claim. 1) IfE is nice to f, f Fc(T, e,Y), g Fc(T, e,Y), g f then Eis nice to f.2) The games from 5.4 are determined, and the winning side has winning strategy

which does not need memory.3) The second player wins G2(E, g) iff for some second player wins G

2(E, g).

4) If the second player wins G2(E, f), g Fc(T, e(E))g f for Dom(g) thenthe second player wins in G2(E, g) when we let

= + max{(g() g() + 1) : satisfies Dom(g)}.

5.6 Lemma. Suppose f0 (Y/e)Ord, e Eq and 0 =: sup{

xY

Ye(f[e]0 (x) + 1 : e

satisfies e0

e e}.1) If there is a Ramsey cardinal {f(x) + 1 : x Dom(f0)} then there is a-divisible E nice to f0 having weak

-sums.2) If for every A 0 there is in K[A0] a Ramsey cardinal > 0, then there is a-divisible E which has weak -sums and is nice to f.3) In part 2 if0 = 2


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46 SAHARON SHELAH

En,e,hp = {Dn1,e1,h1

q : p q P, n1 = n(q) and (n1, e1, h1) (n,e,h)}

where (n1, e1, h1) (n,e,h) means: n n1 < , e e1 Eq, h1 Hn1,e1 and fors J(n1), h

1(s)[e] = h(s n) and we define (p1, n1, e1, h1) (p,n,e,h) similarly.

Let

En,e,hp = {E

n1,e1,h1

q : p q P, n1 = n(q), (n1, e1, h1) (n,e,h)}.

Note: (p1, n1, e1, h1) (p,e,n,h) implies Dn1,e1,h1

p1 Dn,e,hp , E

n1,e1,h1

p1 En,e,hp

and En1,e1,h1

p1 En,e,hp . Now any E = E

n,e,hp (p P) is as required.

A new point is E is -divisible. So suppose E E = En,e,hp so E = En1,e1,h1qfor some (q, n1, e1, h1) (p,n,e,h). Let Zbe a set of cardinality < , so (n1)

|Z| =n1 ; let {h : <

= |Y/e1||Z| 2 n1} list all function h from Y/e1 toZ. Let S : < |Y/e1||Z| list a sequence of pairwise disjoint stationary subsetsof { < n1+1 : cf() = 0}. Let e2 Eq be such that e1 e2 and for everyy Y, {z/e2 : ze1y} = {x(y/e,t) : t Z}, we let q2, q q2 P be: q2 = {s

Jn1+1 : s n1 q and sup s

S}, lastly we define h2 : Jn1+1 Y/e1 by:

h2(s) = x(h1(s n1), h(s n1)) if s q2, sup s S (for s Jn1+1\q2 it doesnot matter). The proof that q2, e2, h

2 are as required is as in [RuSh 117] and more

specifically [Sh 212]. As for proving En,e,hp has weak -sums the point is thatthe family of fine normal filters on has -sum.2) Similar to 3.14(and 3.11(5),(6)).3) Similar to [Sh 386, 1.7]. 5.6


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6 Hypotheses: Weakening of GCH

We define some hypotheses; except the first we do not know now whether theirnegations are consistent with ZFC.

6.1 Definition. We define a series of hypothesis:

(A) pp() = + for every singular .(B) Ifa is a set of regular cardinals, |a| < Min(a) then |pcf(a)| |a|.(C) Ifa is a set of regular cardinals, |a| < Min(a) then pcf(a) has no accumulationpoint which is inaccessible (i.e. inaccessible sup( pcf(a) < ).(D) For every , { < : singular and pp() } is countable.(E) For every , { < : singular and cf() = 0 and pp() } is countable.(F) For every , { < : singular of uncountable cofinality, pp(cf())() } is

finite.(D),, For every , { < : > cf() [, ) and pp(,)() } hascardinality < .(A) If > cf() then pp() =

+ (or in the definition of pp() the supremumis on the empty set).(B), (C) Similar versions (i.e. use pcf).

We concentrate on the parameter free case.

6.2 Claim. : In 6.1, we have:

(1) (A) (B) (C)

(2) (A) (D) (E), (A) (F)

(3) (E) + (F) (D) (B). [Last implication by the localization theorem[Sh 371, 2]]

(4) if ()( > cf() = 0 the hypothesis (A) of 6.1 holds.[Why? By [Sh:g, xx].]

6.3 Theorem. Assume Hypothesis 6.1(A).

1) For every > ,

cov(, +, +, 2) =

+ if cf()

if cf() > .

2) For every > = cf() > 0, there is a stationary S [], |S| = + if

cf() and |S| = if cf() > .


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48 SAHARON SHELAH

3) For singular, there is a tree with cf() levels each level of cardinality < , andwith +(cf())-branches.4) If cf() < 2 then there is an entangled linear order T of cardinality

+

.

Proof. 1) By [Sh 400, 1].2) By part (1) and 2.6.3, 4) By [Sh 355, 4].


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REFERENCES.

[DoJe81] A. Dodd and Ronald B. Jensen. The core model. Annals of Mathemat-ical Logic, 20:4375, 1981.

[RuSh 117] Matatyahu Rubin and Saharon Shelah. Combinatorial problems ontrees: partitions, -systems and large free subtrees. Annals of Pureand Applied Logic, 33:4381, 1987.

[Sh:E12] Saharon Shelah. Analytical Guide and Corrections to [Sh:g].math.LO/9906022.

[Sh:e] Saharon Shelah. Nonstructure theory, accepted. Oxford UniversityPress.

[Sh 52] Saharon Shelah. A compactness theorem for singular cardinals, freealgebras, Whitehead problem and transversals. Israel Journal of Math-ematics, 21:319349, 1975.

[Sh 108] Saharon Shelah. On successors of singular cardinals. In Logic Collo-quium 78 (Mons, 1978), volume 97 of Stud. Logic Foundations Math,pages 357380. North-Holland, Amsterdam-New York, 1979.

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

[Sh 212] Saharon Shelah. The existence of coding sets. In Around classification

theory of models, volume 1182 ofLecture Notes in Mathematics, pages188202. Springer, Berlin, 1986.

[Sh 88a] Saharon Shelah. Appendix: on stationary sets (in Classification ofnonelementary classes. II. Abstract elementary classes). In Classi- fication theory (Chicago, IL, 1985), volume 1292 of Lecture Notes inMathematics, pages 483495. Springer, Berlin, 1987. Proceedings ofthe USAIsrael Conference on Classification Theory, Chicago, Decem-ber 1985; ed. Baldwin, J.T.

[Sh 351] Saharon Shelah. Reflecting stationary sets and successors of singular

cardinals. Archive for Mathematical Logic, 31:2553, 1991.[Sh 410] Saharon Shelah. More on Cardinal Arithmetic. Archive for Mathemat-

ical Logic, 32:399428, 1993. math.LO/0406550.

[Sh 371] Saharon Shelah. Advanced: cofinalities of small reduced products. InCardinal Arithmetic, volume 29 of Oxford Logic Guides, chapter VIII.Oxford University Press, 1994.


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[Sh 355] Saharon Shelah. +1 has a Jonsson Algebra. In Cardinal Arithmetic,volume 29 ofOxford Logic Guides, chapter II. Oxford University Press,1994.

[Sh 386] Saharon Shelah. Bounding pp() when cf() > > 0 using ranksand normal ideals. In Cardinal Arithmetic, volume 29 of Oxford LogicGuides, chapter V. Oxford University Press, 1994.

[Sh:g] Saharon Shelah. Cardinal Arithmetic, volume 29 of Oxford LogicGuides. Oxford University Press, 1994.

[Sh 400] Saharon Shelah. Cardinal Arithmetic. In Cardinal Arithmetic, vol-ume 29 of Oxford Logic Guides, chapter IX. Oxford University Press,1994. Note: See also [Sh400a] below.

[Sh 365] Saharon Shelah. There are Jonsson algebras in many inaccessible car-dinals. In Cardinal Arithmetic, volume 29 of Oxford Logic Guides,chapter III. Oxford University Press, 1994.

[Sh 430] Saharon Shelah. Further cardinal arithmetic. Israel Journal of Mathe-matics, 95:61114, 1996. math.LO/9610226.

saharon shelah- advances in cardinal arithmetic

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