saharon shelah- further cardinal arithmetic

8/3/2019 Saharon Shelah- Further Cardinal Arithmetic

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ISRAEL JOURNAL OF MATHEMATICS 95 (1996), 61114

FURTHER CARDINAL ARITHMETIC

BY

Saharon Shelah

Institute of Mathematics

The Hebrew University of Jerusalem, Jerusalem, Israel

and

Department of Mathematics

Rutgers University, New Brunswick, NJ, USA

ABSTRACT

We continue the investigations in the authors book on cardinal arithmetic,

assuming some knowledge of it. We deal with the cofinality of (S0(),)

for real valued measurable (Section 3), densities of box products (Section

5,3), prove the equality cov(,,+, 2) = pp() in more cases even when

cf() = 0 (Section 1), deal with bounds of pp() for limit of inaccessible

(Section 4) and give proofs to various claims I was sure I had already

written but did not find (Section 6).

Annotated Contents

1. Equivalence of two covering properties . . . . . . . . . . . . . . . . . 63

[We try to characterize when, say, has few countable subsets; for a given

(0, ), we try to translate to expressions with pcfs the cardinal

Min

|P|: P S


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62 S. SHELAH Isr. J. Math.

[We show that if > , = cov(, +, +, ) and cov(,,, 2) (or

), then cov(, +, +, 2) = cov(,,, 2). This is used in [Sh-f, Appendix,1]

to clarify the conditions for the holding of versions of the weak diamond.]

3. Cofinality ofS0() for real valued measurable and trees . . . . . . . 72

[Dealing with partition theorems on trees, RubinShelah [RuSh117] arrive at thestatement: > > 0 are regular, a S


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Vol. 95, 1996 FURTHER CARDINAL ARITHMETIC 63

1. Equivalence of Two Covering Properties

1.1 Claim: If pp = +, > cf() = > 0 then cov(,,+, 2) = +.

Proof: Let = 3()+; choose B : <

+ increasing continuous, such

that B (H(), ,


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a, be

Ch,() =

sup( N1) if N

1,

0 otherwise,

we have: Ch < f a, mod Jbda,

.

[Why? Clearly ChB(N0 ) B+1, so a, B+1, hence there are in B+1

elements b[a,]: pcf(a,) and fa,, : < : pcf(a,) as in [Sh

371, 2.6, 1]. So for some , (, +) we have Ch b+ [a,] < f , so it is

enough to prove a, b+ [a,] is bounded below but otherwise pp() = +

will be contradicted. Let = sup{,: N0}.]

(e) E =: { < +: < & |C| < and > } is a club of .

Now as S is stationary, there is () S E. Remember otp C() = +.

Let C() = {(), : < +} (in increasing order).

Let (for any < +) M0 be the Skolem Hull of

f(), : <

{i : i }, andlet M1 be the Skolem Hull ofa

f(), : <

{i : i }. Note: for < +

non-limit

f(), : <

=

f: C(),

. Clearly M0 : < +, M1 : 0, we have (0) (1) (2) = (3)

and ifcov(, 1, 1, 2) < they are all equal, where:

(0) =: is the minimal such that: ifa Reg +\, |a| then we

can find a: < such that a =


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1.2A Remark: (1) We can get similar results with more parameters: replacing

0 and/or 1 by higher cardinals.

(2) Of course, by assumptions as in [Sh410, 6] (e.g. | pcfa| |a|) we get

(0) = (3). This (i.e. Claim 1.2) will be continued in [Sh513].

Proof:

(1) (2): Trivial.

(2) (3): Let = 3((3))+ and for + we choose B (H(), ,


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For n = 0 we can define Na0 , Nb0 , An, trivially. Suppose N

am, N

bm, Am,,

Pm, are defined for m n, < and fm (m < n) are defined. Now an is well

defined and Reg +\ B and |an| . So an = an, and an, an,+1

where an, =: an An, and, of course, an, Reg +\ has cardinality .

Note that an, is not necessarily in B but

()1 every countable subset ofan, is included in some subset ofB which belongs

to Pn, and is Reg +\.

By the definition of (3) (see equivalently there), for each n, we can

find an increase sequence an,,k: k < of subsets of an, with union an, and

dn,,k [, (3)] pcf(an,,k), |dn,,k| < such that:

()2 ifb an,,k is countable then b is included in a finite union of some members

of {b[an,,k]: dn,,k} (hence max pcf(b) (3)).

By the properties of pcf:()3 for each , k < and c Reg +\ such that c Pn, we can find

e = e,kc (3)+ pcfc, |e| |dn,,k| < such that for every dn,,k we

have: cb[an,,k] is included in a finite union of members of{b[c]: ec}.

By [Sh371, 1.4] we can find fn an

such that:

()4 () sup(Nbn ) < fn();

() if c Pn,, , k < , c Reg +\ and e,kc pcf(c) [, (3)]

(where e,kc is from ()3) then for some m < , p + pcf(c)

and p

< p

, (for p

m) the function fn

(b

[c]) is included in

Maxpm fc,p bp [c] (the Max taken pointwise).

Note

()5 if b an,,k is countable (where , k < ) then there is c Pn,, |c| < ,

c Reg +\ such that b c.

By ()4 :

()6 if ,k < , c Pn,, c Reg +\, and dn,,k (3)

+ pcfc\ then

fn b[c] B.

You can check that (by ()2 ()6) :

()7 if b an,,k is countable then there is fn,,kb B, | Dom fn,,kb

| < such

that fn b fn,,kb

.

Let i(i < ) list the Skolem function of (H(), ,


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and Pn+1, = S


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1.3 Claim: Assume 0 < cf < < , pp() and

cov(, +, +, 2) < .

Then cov(,,+, 2) < .

Proof: Easy.

1.3A Definition: Assume = cf > = cf > 0.

(1) (C, P) T[, ] if (C, P) T[, ] (see [Sh420, Def 2.1(1)]), and

S(C) = sup(acc C) (note: acc C C), and we do not allow (viii)

(in [Sh420, Definition 2.1(1)]), or replace it by:

(viii) for some list ai: i < ofS(C) P, we have: S(C), acc C

implies {a : a P, a } {ai: i < }.

(2) For (C, P) T[, ] we define a filter D(C,P)

() on [S


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1.3C Claim: Let > = cf > 0, = +, (C, P) T[, ] then the

following cardinals are equal:

(0) = cf (S = cf > 0, = + and (C, P) T[, ]. Let

B1 be a rich enough model with universe 1 and countable vocabulary which is

rich enough (e.g. all functions (from 1 to 1) definable in (H((1)+), , 2


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of of cardinality is included in the union of < of them (exists by the

definition of = cov(, +, +, )). Let P0 = {Ai: i < }. Let P1 be a family of

cov(,,, 2) subsets of , each of cardinality < such that any subset of of

cardinality < is included in one of them.

Let P =:iaAi: a P1

; clearly P is a family of subsets of each

of cardinality , |P| |P1| = cov(,,, 2), and every A , |A| is

included in some union of < members ofP0 (by the choice ofP0), sayibAi,

b , |b| < ; by the choice of P1, for some a P1 we have b a, hence

A ibAi

iaAi P. So P exemplify cov(,

+, +, 2) cov(,,, 2).

Second we prove the inequality . If 0 then cov(, +, +, 2) =

and cov(,,, 2) = so trivially holds; so assume > 0. Obviously

cov(, +, +, 2) . Note, if is singular then, as cf+ > for some

1 < , we have = cov(, +

, +

, ) = cov(, +

, +

,

) whenever

[1, ]is a successor (by [Sh355, 5.2(8)]); also cov(,,, 2) sup{cov( ,,, 2):

[1, ] is a successor cardinal} and cov( ,,, 2) cov(, , , 2) when < ,

so without loss of generality is regular uncountable. Hence for any 1 < we

have

()1 we can find a family P = {Ai: i < 1}, Ai , |Ai| , such that any

subfamily of cardinality + has a transversal. [Why? By [Sh355, 5.4],

(=+) and [Sh355,1.5A] even for .]

Hence if 1 , cf1 < + (or even cf1 ) then ()1 . Now we shallprove below

(1) ()1 cov(1,,, 2) cov(, +, +, 2)

and obviously

(2) if cf then cov(,,, 2) =


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the definitions ofb[A] and Aa; note a S

and)

cov(, +, +, 2) cov(, +, +, ) + cov(,,, 2)

= + cov(,,, 2) = cov(,,, 2)

and

cov(,,, 2) cov(, +, +, 2) + cov(,,, 2),

hence, cov(,,, 2) = cov(, +, +, 2) + cov(,,, 2).

3. Cofinality of S0() for Real Valued Measurable and Trees

In RubinShelah [RuSh117] two covering properties were discussed concerning

partition theorems on trees, the stronger one was sufficient, the weaker one nec-

essary so it was asked whether they are equivalent. [Sh371, 6.1, 6.2] gave a partial

positive answer (for successor of regular, but then it gives a stronger theorem);

here we prove the equivalence.

In GitikShelah [GiSh412] cardinal arithmetic, e.g. near a real valued mea-

surable cardinal , was investigated, e.g. {2: < } is finite (and more); this

section continues it. In particular we answer a problem of Fremlin: for real

valued measurable, do we have cf(S = > 0. Then

the following conditions are equivalent:

(A) for every < we have cov(,,, 2) < ,

(B) if < and a S


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3.1A Remark: (1) Note that (B) is equivalent to: if a S


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in I[] (see [Sh420, 1.5]) and let S S+, C = C: S+ be such that: C

closed, otp C , [ nacc C C = C ], [otp C = S] and

for S+ limit, C is unbounded in (see [Sh420, 1.2]).

Without loss of generality C is definable in (B, , , ). Let 0 [, )

be minimal such that cov(0, ,, 2) , so 0 > , > cf0. We choose by

induction on < , A, a such that:

() A (H(), ,


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There is no problem to carry the definition: for n = 0 define N0 by (i)

[trivially (b) holds and also (c), as for (d), note that C A0 A() and

{i: i < } A() as C is definable in B hence {,, : S+, < , and

is the -th member of C} is a relation ofB hence each C+1( < ) is in A()

hence each {i: i < } is and we can compute the Skolem Hull in Aj for j <

large enough].

Next, choose Mn by (k), it satisfies (e) + (a). If Nn : < , Mn are

defined, we can find fn satisfying (f) + (g) + (h) by [Sh371,1.4] (remember ()).

For n + 1 define Nn by (j) and then Mn+1 by (k).

Next by [Sh400, 3.3A or 5.1A(1)] we have

()

n


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3.2 Conclusion: (1) If is real valued measurable then = cf [S = cf > 0, I is a -complete ideal on extending

Jbd and is -saturated (i.e. we cannot partition to sets not in I). Then for

< , cf(S


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() for every successor 2 there is a tree from [Sh355, 3.5]: cf

levels, every level of cardinality < and (cf)-branches,

() for every (, ), there is a tree T of cardinality with

branches of the same height,

() cf cf and even cf > 0 pp(cf)() =+ 2.

(C) Like (B) but we omit () and retain ().

Proof:

First Case: = 0. Trivially (A) holds.

Second Case: is regular uncountable. So and 2 = 2 and [ <

2 < 2] hence 2 ). Try to apply [Sh410, 4.3], its

assumptions (i) + (ii) hold (with here standing for there) and if possibility

(A) here fails then the assumption (iii) there holds, too; so there is as there; so(), () of (B) of 3.3 holds and let us prove (), so assume (, ), without

loss of generality, is regular, and we shall prove the statement in () of 3.3(B).

Without loss of generality is regular and (, )&cf pp() < ;

i.e. is (, +, 2)-inaccessible. [ Why? If is not as required, we shall show how

to replace by an appropriate regular [, ).]

Let (, ) be minimal such that pp() , (so cf ) now

pp() < (by the choice of) and =: pp()+, by [Sh355, 2.3] is as required] .

Let be minimal such that 2

. So trivially < and(2


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(, 2


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()n n < < n&cf

n ppn() < n hence pp

+n

(n) = pp+(n)

(n)

=

2n

.

Note that 2 m, so Max{cfn, cfm} = Max{n, m} m. As the class of cardinals

is well ordered we get S1 =: {n < : n S0, n n+1} is co-infinite and

S =: {n: n } is finite (so () of 3.4(2)(b) holds).

So for some n() < , S n() hence for every n [n(), ) for some

m (n, ), n < m. Note: n = m n = m (as their cofinalities are distinct)

and [n / S0 n / {m: m < }]. Assume n n(), if n > n+1, letm = mn = Min{m: m+1 > n and m n} (it is well defined as

k n < k

and k < k < = m+1 and ifm+1 / S0

trivially and if m + 1 S0 by one of the demands on m+1 (in its choice) and

[Sh355, 2.3] we have m+1 m; but m < n, so m+1 < n contradicting the

choice of m. So by the last sentence, n n() mn < mn+1. By [Sh355,

5.11] we get the desired conclusion (i.e. also part () of 3.4(2)). 3.4

Remark: It seemed that we cannot get more as we can get an appropriate prod-

uct of a forcing notion as in Gitik and Shelah [GiSh344].

4. Bounds for pp(1) for Limits of Inaccessibles

4.1 Convention: For any cardinal , > cf = 1 we let Y, Eq be as in

[Sh420, 3.1], is a strictly increasing continuous sequence of singular cardinals

of cofinality 0 of length 1, =i


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4.2 Theorem (Hypothesis [Sh420, 6.1C] ):

(1) Assume

(a) > cf = 1, Y = Y, Eq Eq,

(b) every D FIL(Y) is nice (see [Sh420, 3.5]), E = FIL(Y) (or at least

there is a niceE (see [Sh420, 5.25], E =

E = Min E, E is-divisible

having weak -sums, but we concentrate on the first case),

(c) < < pp+E(), inaccessible.

Then there are e Eq and x: x Y/e, a sequence of inaccessibles < and

a D FIL(e, Y) E nice to , D FIL(e, Y) such that:

()xY/e

x/D has true cofinality ,

() = tlimDx: x Y.

(2) We can weaken (b) to E FIL(Eq, Y) and for D E, in the game

wG(,D,e, Y) the second player wins choosing filters only from E.(3) Moreover, for given e0, D0, 0x: x Y/e0, if

xY/e0

0x/De0 is-directed,

then without loss of generality e0 e, D0 D and x x[e0] .

4.2A Remark: (1) We could have separated the two roles of (in the definition

of Y, etc. and in (, pp+E())) but the result is less useful; except for the

unique possible cardinal appearing later.

(2) Compare with a conclusion of [Sh386] (see in particular 5.8 there):

Theorem: Suppose > 2

1

, (weakly) inaccessible.(1) If 1 < i = cfi < for i < 1, D is a normal filter on 1,

i (x),(B)f,D = tcf

xY/e f(x)/D

.

Let K0 =:

(f, D): D E, f Y/e and conditions (A)f and (B)f,D hold }, so

K0 = . Now if (f, D) K0, for some

(C)f,D, in G(D,f,e, Y) the second player wins (see [Sh420, 3.4(2)])

* I.e.: ifa Reg, |a| < min(a), inaccessible then > sup( pcfa).


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hence K1 = where K1 =: {(f , D , ) K0 condition (C)f,D, holds}.

Choose (f1, D1, ) K1 with minimal. By the definition of the game

() for every A = mod D1 we have (f1, D1 + A, ) K1.

Let e1 = e(D1).

Case A: {x: f1(x) inaccessible} = mod D1. We can get the desired conclu-

sion (by increasing D1).

Case B: {x: f1(x) successor cardinal} = mod D1. By (), without loss of

generality f1(x) = g(x)+, g(x) a cardinal (so (x)) for every x Y/e. By

[Sh355, 1.3] for every regular (, ) there is f (Y/e) Ord satisfying:

(a) f < f1, each f(x) regular,

(b) tlimD1f = ,

(c)x f(x)/D1 has true cofinality .

By (a) we get

(d) f g.

By (b) we get, by the normality of D1, that for the D1-majority of x Y/e,

f(x) (x); as f(x) is regular (by (a)) and (x) singular (see 4.1) we get

(e) for the D1-majority of x Y/e, we have f(x) > (x).

Let be large enough, let N be an elementary submodel of (H(), ,

sup( N), so for some () < :

h N Y/e1 Ord

x Y/e1: f,()(x) h(x) < f(x)

= mod D1.

[Why? For any such h define h Y/e1Ord by: h(x) is h(x) if h(x) < f(x)

and zero otherwise, so for some h < , h < f,h mod D1. Let () =

sup

h: h N

Y/e1 N

; it is < as N < , and it is as required.]

Let f = f,()

. The continuation imitates [Sh371, 4], [Sh410, 5].

Let

K2 =

(D, B, jx: x Y/e1): D1 D E, player II wins GE (f

1, D),

e1 = e(D), B = < Bx,j : j < j0x (x) > : x Y/e1 N,

|Bx,jx| g(x) and jx < j0x (x),

{x Y/e1: f(x) is in Bx,jx} D

.


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Clearly K2 = . For each (D, B, jx: x Y/e1) K2 :

()1 letting h Y/e1 Ord, h(x) = |Bx,jx|, for some h =

, f1, 0, h

, for

some < and D player II wins in G,E (D, h, e1, Y).

So choose (D, B, jx: x Y/e1, 0) such that:

()2 (D, B, jx: x Y/e1) K2, ()1 for 0 holds and (under those restric-

tions) 0 is minimal.

So (as player I can move twice), for every A D+, if we replace D by D + A,

then ()2 still holds.

So without loss of generality (for the first and third members use normality):

()3 one of the following sets belongs to D:

A0, =

x Y/e1: cf|Bx,jx| > (x) and j

0x < }

(for some < 1 such that |Y/e1| < ),A1 =

x Y/e1: cf|Bx,jx| < (x) |Bx,jx|

,

A2, = {x Y/e1: |Bx,jx| and jx < } (for some < 1).

If A2, D then (for x Y/e1)

Bx =:

Bx,j: x Y/e1, j < j0x and |Bx,jx| < and j <

is a set of ordinals and

{x Y/e1: f(x) Bx} D

and Bx: x Y/e1 belongs to N (as (D, B, jx: x Y/e1) K2 and the

definition ofK2), contradiction to the choice of f (see , remember D1 D by

the definition of K2).

If A1 D, we can find B1 N, B1 = B1x,j: j < j

1x (x): x Y/e1,

|B1x,j| g(x) andj (x). Let

a =

cf|Bx,j |: cf|Bx,j| > (x), x Y/e1, j < j0x, j < and (x) >

,


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so a is a set of regular cardinals, and (remember |Y/e1| < ) we have |a| < Min a,

so let b = b[a]: pcfa be as in [Sh371, 2.6]. So as (by the Definition ofK2),

Bx,j: j < j0x: x Y/e1 N, clearly a N hence without loss of generality

b N. Let = sup[ pcfa], so by Hypothesis [420, 6.1(C)], < , but

N, so + 1 N.

By the minimality of the rank we have for every pcfa,

{x y/e1: cf|Bx,jx| b} = mod D hencex cf|Bx,jx|/D is -directed, hence

we get contradiction to the minimality of the rank of f1.

(2), (3) Proof left to the reader. 4.2

4.2B Remark:

(1) The proof of 4.3 below shows that in [Sh386] the assumption of the existence

of nice filters is very weak, removing it will cost a little for at most one place.

(2) We could have used the framework of [Sh386] but not for 4.3 (or use forcing).

4.3 Claim (Hypothesis 6.1(C) of [Sh420] even in any K[A]): Assume > cf =

1, > > 1, pp(,1)() > , inaccessible. Then for some e Eq,

D FIL(e, Y) and sequence of inaccessiblesx: x Y/e, we havetlimD x =

and = tcf(

x/D) except perhaps for a unique in V (not depending on

) and then pp+(,1)() +.

Proof: By the Hyp. (see [Sh513, 6.12]) for some a Reg , |a| < Min(a),

= max pcf(a), and

( < )(b)[b a & |b| < >]& > sup pcf 1complete

(b) > ],

J = J when

A . Choose A such that L[A] and for every < , there is a one

to one function f from || (i.e. ||V) onto , f L[A], so Card

L[A]

+ 1

=

CardV, and apply 4.2 to the universe K[A] (its assumption holds by [Sh420, 5.6]).

Second assume () in K[A] there is a Ramsey cardinal > when A +

and assume our desired conclusion fails. Let S be stationary [ S cf =

+], a: < , exemplify S I[] (exist by [Sh420, 1]). We can find a, J as

described above. Let f: < exemplify = tcf(a/J), now by [Sh355, 1.3]

without loss of generality = max pcfa. Let A0 be such that a, f: < ,

b[a]: pcfa are in L[A0]. Hence in L[A0] for suitable J, f/J: < is

increasing, and without loss of generality for some c: a: S L[A0],


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we have: for S, cf = |a|+, a a club of and f (a\c): a is


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4.5 Theorem: If is regular (> 1) A , Z K[A] a bounded subset of

then for some < , Z


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for some = they are equal and we get contradiction by g, g() = 0, g() = 1,

Dom g = {, }].

Also trivial is: for limit, d


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(D) for everyx


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(E) (G) when = : where { <

: } = mod J: Easy too.

Next assume every is regular, J an ideal on .

(G)(F): (F) is a particular case of (G), because (S


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Ci i, j Ci Cj = j Ci, otp(Ci) ||+ and S =: {i < : cf(i) = ||+, =

sup(Ci)} stationary: so wlog j Ci


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5.9 Observation: In several of the models of set theory in which we know

strong, singular, limit, 2 > + our sufficient conditions for dcf(, 2) = 2

usually hold by the sufficient condition 5.4(a) (simplest: if GCH holds below ,

cf = 0).

Remark: We could prove this consistency by looking more at the consistency

proofs, adding many Cohen subsets to in preliminary forcing; but the present

way looks more informative.

6. Odds and Ends

6.1 Lemma: Suppose cf() > +, I an ideal on , f Ord for < is

I-increasing. Then there are J, s, f( < ) such that:

(A) s = si: i < , each si a set of ordinals,(B)i


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For a set a sup


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()4 J is an ideal on extending I, in fact is the ideal generated by I{a,:

(, )}.

As f: < is I-increasing (i.e. ()1):

()5 J decreases with , in fact a,/I increases with , decreases with ,

()6 if D is an ultrafilter on disjoint to J, then f/D is a


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b a I. [Does this occur? E.g. for I = S fori < ,

J a -complete ideal on and = tcfi


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club E E of , for stationarily many S, g(C , E) E) (remember

g(C , E) = {sup( E): C ; > Min(E)}). Let C

, = { C

: =

Min(C\ sup(E)}, they have all the properties of the Cs and guess clubs in

a weak sense: for every club E of for some S E, if1 < 2 are successive

members of E then |(1, 2] C, | 1; moreover, the function sup(E )

is one to one on C, .

Now we define by induction on < , an ordinal and functions g

i


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(3) In () of 6.2, < can be replaced by < + (prove by

induction on ).

6.3 Observation: Assume < }. Then there are ,

andT

, satisfying the condition (

) below for = 2 or at least arbitrarily large

regular 2.

() T a tree with levels, (where ) with a set X of -branches, and

for < , .

Case 1:2, T =2 are O.K. (the set of branches

2 has cardinality 2).

Case 2: Not Case 1. So for some < , 2

, but by the choice of , 2

,so 2 = , < and so < 2|| = 2. Note |>2| = as .

Subcase 2A: cf() = cf(). Let >2 =j , for some ordinal j < we have

{ 2: j j} has cardinality .

As cf() = cf() and (by its definition) clearly < , hence |Bj | < .

Let

T = { : < g() and Bj} .

It is as required.

Subcase 2B: Not 2A so cf() = cf(). As ()[ < = 2

cf() = cf(2

) > ], clearly cf() so is regular. If = we get = , so singular. So if < , < i = cf(i) < for i < then (see

[Sh-g, 345a, 1.3(10)]) max pcf{i: i < } i


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is T i . In fact we can get any regular cardinal in (, pp+()) in the same

way. Let = min{: < , cf() = and pp() > }, so (by [Sh355,

2.3]), also has those properties and pp() pp(). So if pp+() = (2)+

or pp() = 2 is singular, we are done. So assume this fails.

If > 0, then (as in 3.4) < 2 cov(, +, +, ) < 2 and we can

finish as in subcase 2A (as in 3.4; actually cov(2


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Now choose by induction on < |a0|+, c, satisfying > max pcf{: < }.

If we are stuck in , maxpcf{: < } is the desired maximum by ()1. If we

succeed = maxpcf{: < |a0|+} is in pcf{: < } for some < |a0|

+ by

()2; easy contradiction. 6.4A

6.4

6.5 Conclusion: Assume 0 = cf() 0 < , [ (0, ) & cf()

pp() < ] and pp+ () > = cf() > . Then we can find n for n < ,

0 < n < n+1 < , =n


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Let i = maxpcf(b {j : j < i}). For each i let bi = b {j : j < i} and

fb,: < : pcfb be as in [Sh371, 1]. Let

T0i = Max=1,n fb

, bi: pcf(bi), < , n < .

Let Ti = {f T0i : for every j < i, f bj T0j moreover for some f

j


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6.6B Observation: Let < be regular uncountable, 2 < i < (for i < ),

i increasing in i. The following are equivalent:

(A) there is F such that:

(i) |F| = ,

(ii) |{f i: f F}| i,

(iii)

f = g F f =Jbd g

;

(B) there be a sequence i: i < such that:

(i) 2 < i = cf(i) i,

(ii) max pcf{i: i < } = ,

(iii) for j < , j maxpcf{i: i < j};

(C) there is an increasing sequence ai: i < such that pcfi


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()3 if {j: j < i} / Ji, then for every f F, f i || for < ,

and for each < , = : < is a sequence of ordinals. Then for every

X , X = mod D there is : < (a sequence of ordinals) and w

such that:

(a) \w cf( ) ,

(b) if and [ w

=

], then { X: for every < we have

and [ w

=

] } = mod D.

Proof: Essentially by the same proof as 6.6C (replacing i by Min{ X: for

every Y Ni D we have Y}). See more [Sh513, 6]. 6.6D

6.6E Remark: We can rephrase the conclusion as:

(a) B =: { X: if w then = , and: if w then

is <

but > sup{ : < , <

}} is = mod D.

(b) If < for w then { B: if w then >

} =

mod D.


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(c) w cf() is but .

6.6F Remark: (1) If |a| < min(a), F a, |F| = = cf pcf(a) and even

> = sup(+ pcf(a)) then for some g a, the set {f F: f < g}

is unbounded in (or use a -complete D as in 6.6E). (This is as a/J


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[exists by [Sh345a, Def. 3.3(2)b + Fact 3.4(1)]].

Let = (sup a)+, |a| < = cf < Min a (without loss of generality

there is such ) and N = Ni: i < be an increasing continuous sequence of

elementary submodels of (H(), ,


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()4 bi,j,(i,j) J[a].

For this end we define by induction on < |a|+ functions fa,, with domain

bi,j, for every < a, such that < f

a,, f

a,, , so the domain

increases with .

We let fa,,0 = fa, bi,j , fa,, =


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Now if < sup(N ) then for some (1), < (1) N , so letting

b =: bi,j, b[a] bi,j, , it belongs to J[a]J


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(i) b[a] increases with .

This will be proved below.

6.7B Claim: In 6.7A we can also have:

(1) if we let b[a] = b[a] = |a|, c Ni , c a (hence |c| < Min(c) and c a), then

for some finite d (pcfc) a we have c d b[a]. Similarly for

-complete, < cf() (i.e. we have clauses (h), (h)+ for = ).

(4) We can have continuity in when cf() > |a|, i.e. b = +3, let = +2,

let N, N, a, b (as a function), i: =: |a|+ be as in 6.7A but also

i: i < N0. So for j < , cj =: {i: i < j} N0 (and cj a0) hence

(by clause (h) of 6.7A), for some finite dj a1 pcfcj = Ni1 pcfa pcfcj we

have cj

djb1[a]. Assume j(1) < j(2) < . Now if a dj(1)

b1[a]

then for some 0 dj(1) we have

b1

0 [a

]; now 0 dj(1) pcf(

cj(1))

pcf(cj(2)) pcfdj(2)

b1[a]

=dj(2)

pcf(b1[a]) hence (by clause (g) of

6.7A as 0 dj(0) N1) for some 1 dj(2), 0 b11 [a]. So by clause (f)

of 6.7A we have b10 [a] b11 [a] so remembering b

10 [a], we have b

11 [a].

Remembering was any member ofadj(1)

b1[a], we have a

dj(1) b

1[a]

adj(2) b

1[a] (holds without a but not used). So a

dj

b1[a]: j <

is a non-decreasing sequence of subsets of a, but cf() > |a|, so the sequence is


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eventually constant, say for j j(). But

maxpcf

a

djb1[a]

maxpcf

dj

b1[a]

= maxdj

maxpcf(b1[a])

= maxdj

max pcf{i: i < j} < j

= max pcf

a

dj+1

b1[a]

(last equality as bj [a] b1[a] mod J maxpcf{j: j < i},

by 6.7C(3A), we will be stuck at some i, and by the previous sentence (and

choice of (a, b, ), i is limit, so pcf({j : j < i}) but it is pcf(b) +,

so = maxpcf{j : j < i}. For each j, by the minimality condition for some

bj b, we have |bj| |a|, j pcf(bj). So pcf{j: j < i} pcf(j


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(instead of strict inequality) and


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Proof of (h): So let , , c be given; let b[a]: pcfc( Ni+1) be a

generating sequence. We define by induction on n < , An, c, : An such

that:

(a) A0 = {}, c = c, = maxpcfc,

(b) An n, |An| < ,

(c) if An+1 then n An, c cn, < n and = max pcf(c),

(d) An, c, : An belongs to Ni+1+n hence Ni+1+n ,

(e) if An and pcfcomplete(c) and cb+1+n

[a] then

()[ An+1 & = 0] and c0 = c\b+1+n

[a] (so 0 =

maxpcfc0 < = max pcfc),

(f) if An and / pcfcomplete(c) then

c =

bi [

c]: i < in < , i An+1

,

and if = i An+1 then c = b [c],

(g) if An, and pcfcomplete(c) but c b+1nn

[a], then ()[

An+1].

There is no problem to carry the definition (we use 6.7F(1) below, the point is

that c Ni+1+n implies b[c]: pcf[c] Ni+1+n and as there is d as in

6.7F(1), there is one in Ni+1+n+1 so d a+1+n+1). Now let

dn =:

: An and pcfcomplete

(c) and c b+1+n

[a]

and d =:n


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If = 0 (i.e. clause (h)) we should havenAn finite; the proof is as above

noting the clause (f) is vacuous now. Son |An| = 1 and

nAn = , so

nAn

is finite. Another minor point is d Ni++1 ; this holds as the construction is

unique from Nj : j < i+, ij : j + , (ai(), b: ai()): + ;

no outside information is used so (An, (c, ): An): n < Ni++1 ,

so (using a choice function) really d Ni++1 . 6.7A

6.7E Proof of 6.7B: Let b[a] = b =


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|d| < such that {i: i < } d

b1[a]; hence by clause (g) of 6.7A

and 6.7F(0) we have a1 pcfcomplete({i: i < }) d

b1[a]. So for

< < |a|+, d a1 pcfcomplete{i: i < } a1 pcfcomplete{i: i sup pcfcomplete({i: i < }), whereas d

pcfcomplete{i: i < }, hence > sup d hence

()2 > supd max pcfb1[a] sup pcfcomplete d b

1[a] a .

On the other hand,()3 pcfcomplete{i: i < + 1} pcfcomplete

d+1

b1[a] a

.

For = () we get contradiction by ()1 + ()2 + ()3.

(5) Assume a, c, form a counterexample with minimal. Without loss of

generality |a|+3 < Min(a) and = maxpcfa and = maxpcfc (just let a =:

b[a], c =: c pcf[a

]; if / pcfcomplete(c) then necessarily pcf(c\c)

(by 6.7F(0)) and similarly c\c pcfcomplete(a\a) hence by 6.7F(2),(3)

pcfcomplete(a\a),

contradiction).Also without loss of generality / c. Let , , N, i = i(): ,

a = ai: i be as in 6.7A with a N0, c N0, N0, = |a|+, = |a|+3 > 0 using ranks and normalideals, in Cardinal Arithmetic, Chapter V, Oxford University Press, 1994.

[Sh400] S. Shelah, Cardinal arithmetic, in Cardinal Arithmetic, Chapter IX, Oxford

University Press, 1994.

[Sh410] S. Shelah, More on cardinal arithmetic, Archive for Mathematical Logic

32 (1993), 399428.

[GiSh412] M. Gitik and S. Shelah, More on ideals with simple forcing notions,

Annals of Pure and Applied Logic 59 (1993), 219238.

[Sh420] S. Shelah, Advances in cardinal arithmetic, Proceedings of the Banff Con-

ference in Alberta; 4/91, Finite and Infinite Combinatorics in Sets and

Logic (N. W. Saure et al., eds.), Kluwer Academic Publishers, Dordrecht

1993, pp. 355383.

[Sh460] S. Shelah, The Generalized Continuum Hypothesis revisited, submitted to

Israel Journal of Mathematics.

[Sh513] S. Shelah, PCF and infinite free subsets, submitted to Archive for Mathe-

matical Logic.

[Sh589] S. Shelah, PCF theory: applications, in preparation.

saharon shelah- further cardinal arithmetic

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