saharon shelah- pseudo pcf

8/3/2019 Saharon Shelah- Pseudo PCF

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PSEUDO PCF

SH955

SAHARON SHELAH

Abstract. We continue our investigation on pcf with weak form of choice.Characteristically we assume DC + P(Y) when looking and

Q

sY

s. We get

more parallel of theorems on pcf.

Anotated Content

0 Introduction, pg.2

1 On pseudo true cofinality, pg.4

[We continue [Sh:938, 5] to try to generalize the pcf theory for 1-completefilters D on Y assuming only DC + ACP(Y). So is similar to [Sh:b, ChXII].We suggest to replace cofinality by pseudo cofinality. In particular we getthe existence of a sequence of generators, get a bound to Reg pp()\0using the size of Reg \0 using a no-hole claim and existence of lub(unlike [Sh:835]).

2 Composition and generating sequences for pseudo pcf, pg.14

[We deal with pseudo true cofinality ofiZ

jYi

i,j and below also in the

degenerated case each i,j : j Yi is constant. We then use it to clarifythe state of generating sequences; see 2.1, 2.2, 2.4, 2.6, 2.12, 2.13.

3 Measuring Reduced products, pg.23

(3A) On ps-TD(g)

[We get that several measures of /D are essentially equal.]

(3B) Depth of reduced power of ordinals, pg.26

[Using the independence property for a sequence of filters we can bound

the relevant depth. This generalizes [Sh:460] or really [Sh:513, 3].]

(3C) On the Depth

Date: October 21, 2010.The author thanks Alice Leonhardt for the beautiful typing. First Typed - 08/July/28; revised

version of F728. Partially supported by the United States-Israel Binational Science Foundation(Grant No. 2006108). Part of this work was done during the authors visit to Mittag-LefflerInstitute, Djursholm, Sweden. He thanks the Institute for hospitality and support, Publication955.

1


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[We start by basic properties by No-Hole Claim (1.11(1)) and dependenceof /D only (3.20). We give a bound for +(1)/D (in Theorem 3.21, 3.23).]


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0. Introduction

In the first section we deal with generalizing the pcf theory in the direction asstarted in [Sh:938, 5] trying to understand pseudo true cofinality of small productsof regular cardinals. The difference with earlier works is that here we assumeACU for any set U of power |P(P(Y))| or actually working harder, just |P(Y)| when analyzing

tY

t, whereas in [Sh:497] we assumed ACsup{t:tY} and

in [Sh:835] we have (in addition to ACP(P(Y)) assumptions like (sup{t : t Y})

is well ordered). In [Sh:938, 1-4] we assume only AC


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Assume

(a) Y Ord

(b) cf(t) hrtg(P(Y) for every t Y

(c) t pcf1comp() for t Z, in fact, t = ps-tcf(,


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1. On pseudo true cofinality

We continue [Sh:938, 5].Below we give an improvement of [Sh:938, 5.16], omitting DC from the assump-

tions but first we observe{r15}

Claim 1.1. Assume ACZ .1) We have hrtg(Z) when = t : t Y and ps-pcf() and t Y cf(t) hrtg(Z).2) We have cf(rkD())) hrtg(Z) when = t : t Y, t Y cf(t) hrtg(Z).

Proof. Clearly is a regular cardinal.1) If we have AC for every < hrtg(Z) then we can use [Sh:938, 5.7(4)]. In generallet D be an 1-complete filter on Y such that = ps-tcf(,


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3 (a) D is non-empty

(b) D is an 1-complete filter on Y.

For A Y let DA = {D D : A / D} and let P = {A Y : DA = },equivalently P = {A Y : A = mod D}. As ACP(Y) holds we can findDA : A P such that DA DA for A P. Let D = {DA : A P}, clearly

4 (a) D = {D : D D}

(b) D D is non-empty.

As ACP holds clearly

()1 we can choose FA : A P such that FA exemplifies DA D as in

[Sh:938, 5.17,(1),(2)], so in particular is 0-continuous and without loss ofgenerality FA = ,F

A for every < .

For each < let

()2 F1 = {f = fA : A P : f satisfying A P fA FA }

()3 for f F1 let sup{fA : A P} be the function f YOrd defined by

f(y) = sup{fA(y) : A P}

()4 F1 = {sup{fA : A P} : f = fA : A P belongs to F1}.

Now

()5 (a) F1 : < is well defined, i.e. exist

(b) F1 .

[Why? As t Y cf(t) hrtg(P(Y)).]

()6 F1 = for < .

[Why? As for < , the sequence FA : A P is well defined (as FA : A

P is) and A P FA = , so we can use ACP(Y) to deduce F1 = .]

Define

()7 (a) for f and A P letA(f) = min{ < : f < g mod DA for every g F

A }

(b) for f let (f) = min{A(f) : A P}.

Now

()8 for A P and f , the ordinal A(f) < is well defined.

[Why? As FA : < is cofinal in (, ,


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(b) if < and f F1 then A(f) > .

[Why? Clause (a) holds because < g FA

g < f mod DA and = f FA f f mod DA. Clause (b) holds because for some fB : B P {FB : B P} we have f = sup{fB : B P} hence B P fB f hencein particular fA f, recalling (fA) > by clause (a) it follows that (f) > .]

()11 (a) for < let = min{(f) : f F1 }

(b) for < let F2 = {f F1 : (f) = }

()12 (a) F2 : < is well defined, i.e. exists

(b) if < then < < .

[Why? is the minimum of a set of ordinals which is non-empty by ()6 and ,by ()8, and all members are > by ()10(a).]

()13 for < we have F2 and F2 = .

[Why? By ()11 as F1 = and F

1 .]

()14 we try to define < by induction on the ordinal < = 0: = 0 limit: = { : < } = + 1: =

()15 (a) if < then < is well defined

(b) if < is well defined then < .

[Why? Clause (a) holds as is a regular cardinal so the case limit is O.K., the

= +1 holds by ()12. As for clause (b) recall A(f) > for f F1

by ()10(b).]

()16 if f , then for some g {F2

: < } we have f < g mod D.

[Why? Recall that A(f) for A P and (f) are well defined ordinals < andlet < be such that (f) < , exists as . As F

A is


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{r18}Claim 1.5. The Generator Existence ClaimLet Y(Ord\{0}).

1) J


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AC0 there is an -sequence Xn : n < with Xn witnessing An D1. Then

X = {Xn : n < } belongs to D and witness that A := {An : n < } D1

because D is 1-complete. Fourth, if A B Y and A D1 , then some X1

witness A D1, i.e. X D X X1 A DX ; but then X1 witness alsoB D1.]

(g) assume F : < is


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1 the domain of h is P(Y)

2 the range ofh is ps-pcf1comp(){0}{ : = sup( ps-pcf1comp())

and has cofinality 0 or just for some A P(Y)\J1comp


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\{0} hrtg(P(Y)), hence \{0} > sup( ) so the range ofh is as required in 2.

Second, if +

and cf() = 0 then clearly +

\ and we can find anincreasing sequence n : n < of members of ps-pcf1-comp() with limit . For

each n there is Xn J1compn

[]\J1-comp 0, then


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Proof. Obvious by Definition [Sh:938, 5.6] noting Conclusion 1.6 above and 1.11below. That is, letting = Reg \0, for every Y by 1.6 the set ps-

pcf1comp() is a subset of := Reg pp+Y()\0, of cardinality < hrtg(P(Y)).

By Claim 1.11 below we have = { ps-pcf1comp() : Y}. So there is a

function h with domain hrtg(P(Y)) Y such that < hrtg(P(Y)) Y (h(, ) is the -th member of ps-pcf1comp() if there is one, min() otherwise).So h is a function from hrtg(P(Y)) Y onto the set of cardinality , so we aredone. 1.10

{r22}Claim 1.11. The No Hole Claim[DC]1) If YOrd and 2 ps-pcf1comp(), for transparency t Y t > 0and hrtg(P(Y)) 1 = cf(1) < 2, then for some

we have 1 =ps-pcf1comp(

).2) In part (1), if in addition ACY then without loss of generality YReg.3) If in addition ACP(Y) + AC


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Third, we prove D1 is closed under intersection of < members, so assume < and A = A : < is a sequence of members of D1. Let A := {A : 0.1) For D a filter on Y let ps-TD() = sup{hrtg(F) : F is a family of non-emptysubsets of such that for every F1 = F2 from F we have f1 F1 f2 F2 f1 =D f2}, recalling f1 =D f2 means {y Y : f1(y) = f2(y)} D.2) Let ps-Tcomp() = sup{hrtg(F): for some -complete filter D on Y, F is asabove for D}.3) If we allow t = 0 just replace by

:= {f : f t

(t + 1) and {t : f(t) =

t} = mod D}.{r29}

Theorem 3.2. [DC + ACP

(Y) ] Assume that D is a -complete filter on Y and > 0 and g Y(Ord \{0}), if g is constantly we can write . The following

cardinals are equal or at least 1, 2, 3 are Fil1(D)-almost equal which means: for

1, 2 {1, 2, 3} we have 1 salS 2 which means if < 1 then is included in

the union of S sets each of order type < 2 :

(a) 1 = sup{|rkD1(g)|+ : D1 Fil1(D)}

(b) 2 = sup{+: there are D1 Fil1(D) and a


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Stage C: 3 salFil1(D)

1, 2.

Why? Let < 3. Let F : < exemplify < 3. For each <

let D = {dual(J[f, D]) : f F} so a non-empty subset of Fil1(Y). Now forevery D1 {D : < } let XD1 = { < : D1 D} and for XD1 letD1, = min{rkD(f) : f F and D1 = dual(J[f, D])} and let FD1, = {f F : D1 = J[f, D] and rkD(f) = D1,} so a non-empty subset ofF and clearly(D1,,FD1,) : XD1 exists.

Now

(a) D1, is a one-to-one function with domain XD1

(b) = {XD1 : D1 Fil1(Y)}

(c) if < are from XD1 and D1, < D1, , f FD1,, g FD2, then f < gmod D1.

[Why? For clause (a), if = X1

, f FD1,, g FD

1, then f = g mod D

hence by [Sh:938, 1.11] we have D1, = D1,. For clause (b), it follows by thechoices of D, XD1 . Lastly, clause (c) follows by [Sh:938, 1.11(2)].]

Hence (by clause (c))

(d) otp(XD1) is < 2 and is rkD1(g) for D1 {D : < } Fil1(D).

Together clauses (b),(d) show that < 1, 2 so we are done. 3.2

Concerning Theorem 3.2 we may wonder when does 1, 2 being S-almost equalimplies they are equal.

{r32}Definition 3.4. 1) We say the power ofU1 is S-almost smaller than the power ofU2, or write |U1| |U2| mod S or |U1| almS |U2| when: we can find a sequenceus : s S such that U1 = {us : s S} and s S |Us| |U2|.2) We say the power |U1|, |U2| are S-almost equal (or |U1| = |U2| mod S or|U1| =almS |U2|) when |U1|

almS |U2|

almS |U2|.

3) Let |U1| alm


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2) Otherwise without loss of generality 2 < 1 and by part (3) we have 1 alm


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Note that some of the main results of [Sh:835] can be expressed this way.

(D) rk-supY,() = rk-sup {rkD() : D is -complete filters on Y}

(E) for each non-empty X Y let

sp1(X) = {(D, J) : D an 1-complete filter on Y, J = J[f, D], = rkD(f) and f X}

sp1(X) = {sp1(X) : }

(F) question: If {sp(Xs) : s S} is constant, can we bound J?

(G) X, Y are called connected when sp(X1), sp(X2)) are non-disjoint or equal.

4) We hope to prove, at least sometimes := (Y) pp() that is we like toimmitate [Sh:835] without the choice axioms on . So there is f = f : <

witnessing < (Y). We define u = uf = {: there is no such that(it Y)(f(t) {fn(t) : n < }). You may say that uf is the set of < suchthat f is really novel.

By DC this is O.K., i.e.

1 for every < there is (uf ) such that (t Y)(f(t)) = {fn(t) :

n < }.

Next for uf we can define Df, , the 1-complete filter on Y generated by{t Y : f(t) = f(t)} : <

. So clearly = uf Df , = Df , f =D

f. Now for each pair D = (D1, D2) Fil4Y (i.e. for the 1-complete case) let

f,D = { uf : Df , = D1 and J[f, D1]} = dual(D2). So is the union of

P

(P

(Y))-sets (as |Y| = |Y| |Y|, well ordered.So

()1 hrtg((Y ())

()2 u is the union ofP(P())-sets each of cardinality < pp+Y,1

()

(I) what about hrtg() < ps-ppY,1()?

We are given F : < = F = ,F , = F F = .Easier: looking modulo a fix filter D.

()2 for D FilY,1 , let F,D = {f F : (g F)(g


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(3B) Depth of reduced power of ordinalsOur intention has been to generalize a relative of [Sh:460], but actually we are

closed to [Sh:513, 3] using IND but unlike [Sh:938] rather than with rank we dealwith depth.

{k1}Definition 3.9. 1) Let sucX() be the first ordinal such that we cannot find asequence Ux : x X of subsets of, each of order type < such that = {Ux :x X}.

2) We define suc[]X () by induction on naturally: if = 0 it is , if = + 1 it is

sucX(suc[]X ()) and if is a limit ordinal then it is {suc

[]X () : < }.

3) For a quasi-order P let the pseudo ordinal depth ofP, denoted by ps-o-Depth(P)be sup{: there is a


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3) If we assume IND(Y(n), D(n) : n < for every increasing , which is

quite reasonable then in Theorem 3.12 we can strengthen the conclusion, replacing

for infinitely many ns by for every n < large enough.4) Note that 3.12(2) is complimentary to [Sh:835].

{k8}Observation 3.14. If x is a filter -sequence and n < and IND(x[n, ) thenIND(x).

Proof. Let F = Fn,m : n < m < alg(x), so Fn,m : n [n, ) andm (n, ) belongs to alg(x[n, ) hence by the assumption IND(x[n, ))there is t = tn : n [n, )

nn

Yn such that tn / Fn,m(t(n, m)) when

n n < . Now by downward induction on n < n we choose tn Yn such thattn / Fn,m(tk : k [n + 1, m]) for m [n + 1, ). This is possible as the countableunion of members of dual(Dn) is not equal to Yn. We can carry the induction andtn : n < is as required to verify IND(x). 3.14

Proof. Proof of Theorem 3.12We concentrate on proving part (1), part (2) is easier.Assume this fails. So for some n < for every n [n, ) there is a counter-

example. As AC0 holds we can find a sequence Fn : n [n, ) such that

for n [n, )(a) Fn = Fn, :

(b) Fn, Yn is non-empty

(c) Fn is a


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[Why? We prove it by induction on n; first if n = m this is trivial. The uniquefunction g with domain {} and value . Next, if m < n we choose f Fn1,

hence the sequence G1m,n1,f(s) : s Yn1 is well defined and by the induction

hypothesis each set in the sequence is non-empty. As ACYn1 holds there is asequence gs : s Yn1 such that s Yn1 gs G1m,n1,f(s). Now define g as

the function with domain Y1m,n:

g() =

g(s) = gs() for Y1m,n1.

It is easy to check that g G1m,n, indeed so 4 holds.]

5 if g, h G1m,n and g() < h() then there is an (m, n)-witness Z for (h, g)which means (just being an (m, n)-witness means we omit (d)):

(a) Z Y1m,n is closed under initial segments, i.e. if Y0k,n Z andm k < n then [, n) Y0,n Z

(b) Z

(c) if Z Y0k+1,n, m k < n then {y Yk : y Z} Dk

(d) if Z then g() < h().

[Why? By induction on n, similarly to the proof of4.]

6 (a) we can find g = gn : n < such that gn G10,n, for n <

(b) for g as above for n < , s Yn let gn+1,s G1m,n be defined by

gn+1,s() = gn+1(s).

[Why? Clause (a) by 4 as AC0 holds, clause (b) is obvious by the definitions in2 +3.]We fix g for the rest of the proof

7 there is Zn,s : s Yn : n < such that Zn,s witness (gn, gn+1,s) for n 1 then for every > for infinitelymany n < we have (1 < )(cf() = n pp() ).

{k13}Claim 3.16. [DC] For x = Yn, Dn : n < with each Dn being an 1-completefilter on Yn, each of the following is a sufficient condition for IND(x), letting Y( and = sup( Reg).Toward contradiction assume this fails. We first choose a1, D1 such that

()1 (a) a1 Reg of cardinality <

(b) D1 as 1-complete filter on a1

such that(c) 1 = ps-tcf(a1, .

Without loss of generality

()2 (a) a1 ++ = and sup(a1) < and n3 = n() max{n() : a1} hrtg(Fil4(a1)).]

By the assumption toward contradiction, there is a pair (a2, D2) such that

1 (a) a2 Reg \+

(b) |a2| <

(c) D2 is a +n()-complete filter on a

(d) ps-tcf(a2, .

As hrtg(Fil4(a1)) < n3 and min(a1) > min(Reg\) by 1.11, the no-hole claim, weknow

2 for every a2 there is a sequence = , : a1 such that(a) , Reg \

(b) = ps-tcf(, (P(n2)), so we are done.]

7 (a) without loss of generality A S a1 A {a2, }.

[Why? By xxx.]

8 sup(ps-pcf1com(a3,)) for a1.

[Why? By 7 and the assumption on S.]

9 let D3 be the following filter on Y = a2 a1D2 D1 := {A a2 a1 : { a2 : { a1 : (, ) A} D1} D2}


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10 (a) ps-tcf(,


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6. Absoluteness for non-well founded ultra-powers

Question 6.1. (to [Sh:F1039])

This may be used in 5 to immitate [Sh:460]. Here we try to avoid using thesmooth closed generating sequences.

Check: What does this give directly?{m0}

Hypothesis 6.2.

(a) AC(), = () > 0(b) D is a uniform 1-complete ultrafilter on = ()

(c) P a forcing notion, D

a P-name of an ultrafilter on P()V extendingD, G P generic over V, in V[G]; D

[D] = D

(d) W = W()/D, so in general not well founded, computed in V[G]

(e) j = jG is the canonical embedding of V into W. {m1}Remark 6.3. 1) We may demand P(P) well ordered and ACP(P) holds.2) Natural to choose P = ({D : D and 1-complete filter on extending D}, ).

{m2}Claim 6.4. If1 +2 then when

1 (a) 1 < 2 < 3

2 = |(2)(1)/D2) can we use less? AC j() of cardinality j(3)

(f) W |= A : c where c = ps-pcf2com(a) is a generating force c(not just (a) as in gxxx)

(g) Y = { : W |= a}

(h) for Y let I be { : W |= < } linearly ordered by > are cardinals

(c) Dx,1 Dx,2 are filters on .{g4}

Definition 7.6. 1) We say x is large when the chooser has a winning strategy inthe game x defined below.2) The game x between the player cutter and chooser last moves in the n-thmove a set An+1 D

+x,2 is chosen, letting A0 = . In the n-th move the cutter

chooses n < and fn : An n, and the chooser chooses wn [n]


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rks(f) = B iff < = (rks(f) = ) for some t Ps for every g Ordsatisfying g


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8. Various

(7C){c3.2}

Definition 8.1. Assume D is a filter on Y.1) Let oq(Y) = oq(Y, D) = {f : f a function from Y onto some ordinal}.2) For f oq(Y) let ef = {(y1, y2) : y1 Y, y2 Y and f(y1) = f(y2)}.3) Let oeq(Y) = {ef : f oq(Y, d)}.4) For h oq(Y, D) let D/h be {x Rang(h) : h1(X) D}, a filter on Rang(f)which necessarily is an ordinal < hrtg(Y).5) For f YOrd let gf be the following function:

(a) Dom(gf) = otp(Rang(f))

(b) gf(i) = iff (y)(y Y f(y) = i = otp(f(y) Rang(f)).

6) For f YOrd let hf be the following function:

(a) Dom(hf) = Y

(b) hf(y) = otp(f(y) Rang(f)) oq(Y, d).

7) Assume D Fil1(Y) and f = f : < () is a


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of Ord, and |Rang(pr| = Dom(pr) as it is one-to-one and obviously |Dom(pr)| ;together pr is onto hrtg(Y). We define y : < hrtg(Y) by ypr(,) = f()

for < < hrtg(Y); let u = { < hrtg(Y) : ( < )(y = y)}, so easily < () +1 = |u [ , +1)|, hence |u| = hrtg(Y), hence y : hrtg(Y)exemplify (Y) > hrtg(Y), contradiction. 8.3

{c3.12y}Claim 8.4. Assume [?]

(a) FD : D ps-tcf-fil() is as in ?

(b) D = Di : i < i() < >ps-tcf-fil()

(c) for D as above and i

tcf(, hrtg(Fil

1(Y)) necessarily for some

E, |uE| = but uE = || hence otp(uE) = , so dual(E) is as required.2) By (3). ?

3) ? So J is from (1) and toward contradiction assume dual(J) D1 Fil1(Y)

and 1 , but rkD1(1) ; without loss of generality y Y 1,y > 0 andrkD1(1) = 1. Now we choose F

1 ,F

1,E , E2 as in the proof of part (1) starting

with 1, 1. ??{c14y}Claim 8.6. [DC] 1) If 0 < = cf() < then rkJbd () >

+.

Proof. 1) Clearly Jbd is a uniform -complete filter on . Let i : i < beincreasing continuous with limit , < 0. For each < + let

F = {f : f a one-to-tone function from some subset of onto }

G = {g : g for some f F} where for f F for some < + we let gf be

defined by


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48 SAHARON SHELAH

()0 Dom(gf) = and for every i < ,g(i) = otp({g() : < i Dom(f)})}

()1 F = for < +

()2 G = for < +

()3 G i j}. Let U = { Dom(f2) : f2() < 1} and f1 = f2U and letg1 = gf1 , so clearly g1 G1 . Now if i i1, 0 then {f1() : 1 Dom(gf1)} B1 {f2() : i Dom(f2)} and 1 {g2() : i Dom(f2)}, so clearlygf1(i) < gf2(i).

So g1 < g2 mod Jbd is as required.]For [+, ++) and we shall prove that rkD(g) for some g , this

suffices.As () there is w such that

() (a) u = wi : i <

(b) i < |wi| =

(c) = {wi : i < }.

As cf(+) = we can choose such that

() (a) = j : j <

(b) is increasing, j > ,

(c) is with limit +.

Now y Y let

() wy = {wi : i y}

() for y Y(a) |wg|

(b) ??


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9. Private Appendix

We can add to [Sh:938, 2.6,2.7]{k17}

Claim 9.1. The filter D2 4-commutes with the filter D1 (see [Sh:938, 3.1]) when :

(a) D Filcc(Y) for = 1, 2

(b) D1 is -complete

(c) if J1 {J[f, D1] : f Y1Ord} or just J1 is a -complete ideal extending

dual(D1) then A Y1 but dual(J1) {D1 + A : A D+1 }; this follows from

clause (b) + DC VACP(Y1) when D1 is -c.c., i.e. there is no sequence

Ai : i < of a pairwise disjoint sets from D+1

(d) DC and ACY1 , ACY2(e) () D1 is P(Y2)-complete or just

() if Bs : s A1 A(J+2 ) and A J

+1 , J {J[f, D] : f} for

= 1, 2 then for some B J+2 and we have A A, A J1we have s A Bs Bs.

Proof. Stage A:

Let A D2 and B = Bs : s A A(D2) and J

1 = J1t : t Y2 whereJ1t {J[f, D1) : f

Y2Ord} and J2 {J[f, D2) : f I2Ord}, i.e. as in the

assumption of4 of Definition [Sh:938, 2.1]. We should find A, B as there.

Stage B:

For each t I2 there is At D+1 such that J

1t = dual(D1 + At), hence as ACY2

holds such that At : t Y2 exist. Why? By clauses (b),(c) of the assumption.

Stage C:

Choice of B, A. Apply clause (d) of the assumption applied to (J2, At : t I2). 10.15{k19}

Remark 9.2. 1) We can weaken D1 is -complete, -c.c. to D2 is -complete,+-c.c. when we have some normality conditions.2) We can replace this by any J[f, D1] is of the form D1 + A for some A D

+1 .

We can add in [Sh:938, 4]{k23}

Conclusion 9.3. [AC


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10. Private Appendix

Remark 10.1. pcf inventory (August 2009)1) See [Sh:F663] lecture - [Sh:430, 6] is locality proved for pcf-com(), > |a|.2) See Rinot question [Sh:F893].3) See the notes for Larson [Sh:F814] - on HOD.4) Continue [?], see [Sh:F878].5) Failed try to continue [Sh:460, 5B], [Sh:F563].6) [Sh:F355] - on consistency - answer Gitik?7) [Sh:F354] = sup( pcf(a)) is weakly inaccessible.8) Densities of basic product [Sh:F132], covered by paper with Moti?9) [Sh:F50] to Shimoni.10) Hopes rank for precipiousness?11) Sort out? Yn is well ordered, need IND1(D)?12) (09.10.19) A related question: let x = (Yn, Dn, hn) : n < is here hn : Yn

Y and D a filter on Y and we try to prove

() for every f YOrd, for every large enough n we have rkDn(fhn) rkD(f)or similarly for Depth.

13) (09.10.26, old thought) As we pass from cofinality to pseudo-cofinality, iteratethis notion and then have strong dichotomies.14) (09.11.15) Think of a problem where:

(a) Depth((n),Dn) large given an answer.

15) Tasks (2010.1.08)

(a) if Y = , then we can replace ACP(Y) by DC+

(b) replace Y by all < (Y), just split to some ?(c) Definition dp-pcf(Y) = {x : regular and there is a filter D such that

= dp-tcf(,


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(d) no f/D < f/D satisfies (c), or do we use less?

(e) < , 1, , probably assuming 2 < maybe it is much less interesting

though we may get more than in [Sh:460], then D is in VP, |P| = 2(f) j Reg 2\1(j < )

(g) + = tcf(j

1,


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52 SAHARON SHELAH

Alternative: (2010.3.08)1) Assume DC


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(b) there are no tn = tn, : < n I10,n for n < , stipulating tn,n = x wehave m < n tn,mtn1,m / Fm(tn,m+1, tn+1,m).

2) Let IND((Yn, Dn) : n < ) means that there is no Fm,n : m,n < a witnessagainst it which means:

(a) Fm,n is a two-place function from In {} into dual(Dn)

(b) un,,,t {(1, 1 : 1 < xi1 } coming from (Fn,,Fn,).{k14}

Question 10.10. 1) If we try to prove 3.12 with choosing n

(I/n)?

2) Try n = oDepth(In,


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54 SAHARON SHELAH

Question 10.17. Interesting? Natural for a sequence ( Z)-complete filter, as inwe can use

a/By

: y Y.

Proof. We choose gn, Zn as in the proof of 3.12 using the definition.

Remark 10.18. 1) In (5B), ??(2) silly? We can find disjoint Y1Y2 with id(Y1) =id(Y2).

2) Definition ??(2) line 2: I J.

Discussion 10.19. Seemingly [Sh:835] connect well to [Sh:F955].So ssume i : i < is increasing with limit and that is we should deal with

a game, where..?


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11. Private AppendixUsing pure : July 2009

{m6}

Definition 11.1. We say s is a frame when s consists of the following objectssatisfying the following conditions:

(a) i : i < cf() is increasing with limit

(b) set D

(c) Dd a filter on Id = I[d] for d D

(d) for d D() (d) {(e, h) : e D and h a function from Ie onto Ie such taht

Dd = {h(A) : A De}

() pr(d) (d), a set of so called pure extensions

() ap(d) (d), a set of so called a-pure extensions such that (e, h)

ap(d) Ie = Id h = idId

() d pr(d) ap(d)

() transitivity of ? pr? ap?(e) j is a function from D to cf() and Dd is j(d)-complete and c par(d)

|S| < j(d)(?)

(k) par(d) and for p part(d), Xp = Xp,s : s Sp is a sequence of pairwisedisjoint subsets of Id with union Dd and ep,s : s S is such thatep,s D, Iep,s = Id, Dep,s = Dd + Xp,s so ep,s = d + Xp,s

(l) () if d1 pr(d0) and d2 ap(d0) then d1 +d0 d2 = d1 +sd0

d2is a well defined member ofDs and d3 pr(d2) ap(d1)

() above

() above if e (d1) 1(d2) then e (d).

Question 11.2. Maybe cf() replaced by a linear order (which can have a pseudocofinality)?

We now give examples{m8}

Definition/Claim 11.3. 1) Assume = n : n < , J = Jn : n < , whenJn is a n-complete ideal on In, and n < n+1 (or just n n+1)? We defines = s,J and prove that s is a pre-system as follows (so = s, etc.)

(a) = n and is given

(b) D is the st ofd : d = (, A) = (d, Ad) and for some m = md n = nd < we have() Fd = {F : F = Fm1,n1 : md < m1 n1 nd = F

dm1,n1 : md

m1 < n1 nd and Fm1,1 :n1

=m1+1

I() J(m1)

() = n, n 1, . . . , m

() Id =n

=m

I

() Dd = {X Id: there are X J for some F Fd for [m, n]such that A X { Id : (m1) / Fm1,n1([m1, n1]) whenevermd m1 < n1 nd}}


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() / Dd and nec Dd is m-complete(c) for d D

() let (d) be the set of pairs (e, h) such that e D, me = md nd ne, Fem1,n1 = F

dm1,n1 when nd m1 < n1 nd and h is h() =

[md, nd]

() pr(d) = {(d, h) (d) : Fem1,n1 is constantly when nd < n1 (and

md m1 < n1 ne)

() ap(d) = {(e, h) (d) : h = idId so ne = nd(d) for d Ds and A D+d let e = d + A D be defined naturally, it is

(d, Ad A, F)

(e) part (d) is the set of p = ((Xp,s, ep,s) : s S such that: for some so calledwitness G = langleGm1,n1 : md m1 < n1 nd, Gm1,n1 : I[m1+1,n1] m1 with bounded range letting S

= {m1,n1 : md m1 < n1 nd :m1,n1 < m1} and Aa = { Id : Gm1,n1([m1 + 1, n1]) = m1,n1

for m1 < n1 from [md, nd] we have Sp = {a A : Dd + Aa} anded,p,s = d + Aa

(question): should we allow |Rang(Gm1,n1)| be large, etc.?

(f) part(d) = {p par(d) : |Sp| < md}(question: should we have par(d) {(e, d , p) : (e, h) (d) and as

above}?

Discussion 11.4. (09.8.17) 1) Discuss (here?) to achieve our hope (dichotomyusing [Sh:835]). We would like for every Dx = dec


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(d) for non--maxiam Tt, (d+t,, h

+t,) pr(dt,) and p par(d

+t,) sat-

isfied sucTt() = {s : S Sp} and dt, = edt, and h,rho =

h+t for Tt, g() = m,g( ) = n then h, : Idt, Idt, is

ht, hm+1 . . .ht,n where ht,+1 := h+t,

and = for = m , . . . , n 1

(e) if Tt we have: h,

maps Ad into Adquestion: put this in Definition 11.1?

{m16}Definition 11.8. Given a candidate s we try to define a rank; (we may omit thesubscript s as its value is fixed).

If d Ds and f I[d]Ord we define rktrd (f) = rktrd (f, s) Ord {}; or we

may replace tr by 1 or omit it; by defining by induction on the ordinal whenrktrd (f) : it holds iff for every 1 < there is a pair (t, g) such that

(a) t tree(s) where Tt is well founded, i.e. with no -branch

(b) dt= = d(c) g = g : max(Tt)

(d) g : Id[dt,] Ord

(e) g < f ht,, mod Ddt,(f) rktrdt,(g) 1.

The choice in ?? though more transparent than the following relative, need moreuse of choice.

{m18}

Definition 11.9. Like 12.9 - FILL - rk2d(f), but maybe rk1 is enough.

Check.{m21}

Claim 11.10. Lets be a candidate and k = 0, 1.

1) The rank rkkd(f) for f I[d]Ord is well defined ( Ord {}).2) If (d2, h) pr(d1) and f1

I[d1]Ord then rkkd1(f) = rkkd2

(f h).

3) If d Ds and f I[d]Ord and p par(d) then rkd(f) = min{rked,s(f) : s Sp}.

Proof. 1) Easy.2) Use + on D - FILL.3) By induction - FILL.

{m25}Claim 11.11. For a free? s the following condition (a),(b) are equivalent: and ifs = s,J from 11.3 we can add (c), and if s = s,J is from 11.5 we can add clause(c)+:

(a) rkd(f) = for some d Ds and f I[d]

Ord(b) there t tree(s) andY Tt such that ( lim(Tt))(n)[n Y)

and f I[dt,]Ord for Y such that for any < from Y we have

f < f ht,

mod Ddt,

(c) IND(, J) when...?{m26}

Definition 11.12. For (, J) as in 11.3 or 11.5 let IND(, J) mean that:

Case 1: Definition 11.3 for every Fm,n : I[m+1,n] Jm for m < n < there is


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58 SAHARON SHELAH

Case 2: Definition 11.5[copied] 1) Above p1

Jis not well 0-founded iff: there are , f such that

,f (a) = i : i < is increasing

(b) f = fi,j : i < j <

(c) fi,j is a function from Ij ,j1,...,i+1 into Ji(d) for every

i then for some pair (e, g) we have: e (d) andg


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{m32}Definition/Claim 11.15. For a frame s let p be the following quasi rank system:

,D, ,j are as in Definition 11.1

rkd(f) is as in Definition ?{m34}

Claim 11.16. 1) If(, J) is as in Definition 11.3 or 11.5 and IND(, J) holds, seeDefinition 11.12 then ps(,J) is a rak system.2) Moreover it is a strict one.

Saharon copied. 1) As in the proof of e5.g of 4 of [Sh:938, 4,e5.g] or better seethe proof of 12.17 except that we use 12.9 instead of 12.8 which simplify clause (f),but is cumbersome in other places.2) We check Definition m4.3 of 3 of [Sh:938, 3,m4.3].

Clause (a): is singular.As =

n n and n < n+1 this is obvious.

Clause (b): Let d D, = d, J = Jn now clause () says (I) = (|I|) =(0), (0)+1 < so as for clause (), Dp is a filter on I, it holds by the choiceof p.

Clause (c): rkpd(f) = rkd(f, p) is an ordinal as defined in 12.9.

Clause (d):

Clearly (d) is of the right form.

Clause (e):

On j - see 12.13(2)(c).

Clause (f):

We prove by induction on the ordinal that:

() if d D and j(d) > and A = {A : < } Dd and f I[d]Ord andA D

+d rkd+A(f) then rkd(f) .

Now Definition 12.9 is tailored made for this.

Older version using 12.8 recheck:For = 0 and a limit ordinal this is obvious. For = + 1 let Y = { and A = {A : < } Dd and f I[d]Ord andA D

+d rkd+A(f) then rkd(f) .

Now Definition 12.9 is tailored made for this.

Older version using 12.8 recheck:For = 0 and a limit ordinal this is obvious. For = + 1 let Y = { d(0) and assume e Dj .

(2B) Revisiting

The simplest case below is: x consist In = n, n < n+1, J1,n = [n]


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Remark 12.21. 1) Definition 12.20? seemingly does not behave transitively.2) We may allow nd = md.

{k7}Definition 12.22. For x an -candidate, we define a p.c.s. p = p0x as follows:

(a) Dp = {d : d is an x-object}

(b) (d) = {d: for some d2 the triple (d, di1 , di2) is as in Definition 12.20

(c) j(d) is md

(d) = {n : n < }.{k9}

Claim 12.23. If x is an -candidate then p0x is a quasi rank sytem.

Proof. FILL. {k11}

Definition 12.24. 1) For an -candidate x we say it is well founded when thep.r.s. p0x is well founded, e.g. px is a weak rank system.2) For a well founded.

{k13}Claim 12.25. If x is a well founded -candidate then px is a strict rank system.

Proof. Stage A: We have to check clause (1) from Definition m4.6 of3 of [Sh:938].So assume d, , , f are as in there. Choose j < such that j > nd and

toward contradiction assume e, g are as in there.

Stage B: We find (e1, g1) satisfying of clause (j) of m4.6 of 3 of [Sh:938] andme1 = nd; note if we define as in [?](2) rather than as in 12.13(3), we would notneed this step, but then we may have to reconsider the proof of (f) of Definitionm4.3 of 3 of [Sh:938].

Stage C: We use ACI[e] we continue as in 12.18 and in 4. But see footnote to 3in in clause (j) of m4.6 of 3 of [?]. 12.25

13. Appendix: psuedo true cofinality

We repeat here [Sh:938, 5].Pseudo PCF

We try to develop pcf theory with little choice. We deal only with 1-completefilters, and replace cofinality and other basic notions by pseudo ones, see below.This is quite reasonable as with choice there is no difference.

This section main result are ??, existence of filters with pseudo-true-cofinality;

13.19, giving a parallel of J


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(c) if 1 < 2, p1 P1 and p2 P2 then p1


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F (a) F {f YOrd : f


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The rest should be clear.2) Follows.

3),4) Easy. 13.8

Concerning [Sh:835]{r9.yajan}

Claim 13.9. The Existence of true cofinality filter [ > 0 + DC + AC 0 but

f1, f2 Y Ord (f1 = f2 mod D) rkD(f1) = rkD(f2) hence without loss of

generality y Y y > 0.Let D = {D : D is a filter on Y extending D which is -complete}. So (D)

(Fil11(Y)) cf(). For any < rkD() and D D let

()2 (a) F,D = {f : rkD(f) = and D is dual(J[f, D])}

(b) FD = {F,D : < rkD()}

(c) ,D = { < rkD() : F,D = }

(d) F = {F,D : D D}.

Now

()3 if < rkD() then F = .

[Why? By [Sh:938, 1.8(2)=z0.23(2)] there is g YOrd such that g < f mod D andrkD(g) = and without loss of generality g . Now let D = dual(J[g, D]),so (D, D) Fil4(Y), D

D and g F,D , see [Sh:938, 1.7(2)=z0.23(2)], Claim[Sh:835, 0.10(2)], here we use AC


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[Why? The function D sup(,D) witness this.]

()5 the set in ()4 is bounded below rkD() so let () < rkD() be its supre-mum.

[Why? By ()4.]

()6 there is D D such that ,D is unbounded in (, (). By ()3 there for some f F()and D D we have f F(),D so by the choice of () the set ,D cannot bebounded in rkD().]

()7 if 1 < 2 are from ,D and f1 F1,D , f2 F2,D then f1


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(b) = y : y Y YOrd(c) D is an 1-complete filter on Y

(d) y Y cf(y) .Proof. Note that y Y y > 0. Toward contradiction assume cf() < so has a cofinal subset C of cardinality < . For each < for some f YOrd wehave rkD(f) = and f


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{r14}Theorem 13.19. The Canonical Filter Theorem Assume DC and ACP(Y).

Assume = t : t Y Y

Ord and t Y cf(t) (P(Y)) and ps-pcf1-comp() hence is a regular cardinal. Then there is D = D, an 1-

complete filter on Y such that = ps-tcf(/D) and D D for any other suchD Fil11(D).

{r14b}Remark 13.20. 1) By ?? there are some such .2) We work to use just ACP(Y) and not more.

Proof. Let

1 (a) D = {D : D is an 1-complete filters on Y such that (/D) haspseudo true cofinality },

(b) D = {D : D D}.

Now obviously

(c) D is an 1-complete filter on Y.

For A Y let DA = {D D : A / D} and let P = {A Y : DA = }.As ACP(Y) we can find DA : A P such that DA DA for A P. LetD = {DA : A P}, clearly

2 D = {D : D D} and D D is non-empty.

As ACP holds clearly

()0 we can choose FA : A P such that FA exemplifies DA D as in

13.17(1),(2), so in particular is 0-continuous.

For each < let F = {FA : A P}, now

()1 F .

[Why? As by 13.17(1)(c) we have FA for each A P.]

()2 if 1 < 2 < , f1 F1

and f2 F2

then f1 < f2 mod D.

[Why? As A P f1


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PSEUDO PCF SH955 73

[Why? We choose by induction on n, a sequence n = n,A : A P and asequence fn = fn,A : A P and a function fn such that

() n < and m < n m < n() 0 = and for n > 0 we let n = sup{m,A : m < n, A P}

() n,A (n, ) is minimal such that there is fn,A FAn,A

satisfying n =m + 1 fm < fn,A mod DA

() fn,A : A P is a sequence such that each fn,A are as in clause ()

() fn is defined by fn(t) = sup{fm,A(t) + 1 : A P and m < n}.

[Why can we carry the induction? Arriving to n first, fn is well defined by clause () as cf(t) (P) for t Y. Second by clause (), n,A : A P iswell defined. Third by clause () we can choose fm,A : A P as ACP .

Lastly, the inductive construction is possibly by DC.]

Let = {n : n < } and f = supfn : n < . Easily f {FA : A P}as each FA : < is 0-continuous.]

()6 if f then for some < and f F we have f < f

mod D.

[Why? By ()3 + ()4.]

So we are done. 13.19{r16.yajan}

Definition 13.21. For YOrd let J1-comp


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74 SAHARON SHELAH

14. Appendix: Definition of Rank-System

Moved from pg.3:

We define a function H from into {X : X D} by:

() (H(f))(X) = Min{ < X : if f FX then f f mod DX}.

We let

() D be the following filter on the set Y := D:Z D iff Z D and (X D)[Z {X D : X X}].

Now

() D is an 1-complete filter on Y

() if f1, f2 and f1 f2 mod D1 then H(f1) H(f2) mod D

() (tY

t, > 0 using ranks and normal ideals, Cardinal

Arithmetic, Oxford Logic Guides, vol. 29, Oxford University Press, 1994.[Sh:430] , Further cardinal arithmetic, Israel Journal of Mathematics 95 (1996), 61114,

math.LO/9610226.[Sh:460] , The Generalized Continuum Hypothesis revisited, Israel Journal of Mathematics116 (2000), 285321, math.LO/9809200.

[Sh:497] , Set Theory without choice: not everything on cofinality is possible, Archivefor Mathematical Logic 36 (1997), 81125, A special volume dedicated to Prof. Azriel Levy.math.LO/9512227.


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[Sh:513] , PCF and infinite free subsets in an algebra, Archive for Mathematical Logic 41(2002), 321359, math.LO/9807177.

[Sh:F563] , More on the revised GCH.[Sh:F663] , Pcf: the advanced pcf theorems.[Sh:F814] , On Choice and HODX.[Sh:829] , More on the Revised GCH and the Black Box, Annals of Pure and Applied Logic140 (2006), 133160, math.LO/0406482.

[Sh:835] , PCF without choice, Archive for Mathematical Logic submitted,math.LO/0510229.

[Sh:F878] , Questions of Eric hall, Sept 2007.[Sh:F893] , Rinot question.[Sh:938] , PCF arithmetic without and with choice, Israel Journal of Mathematics ac-cepted, 0905.3021.

[Sh:F955] , PCF with little choice.[Sh:F1039] , Bounds on pcf with weak choice using ranks and normal filters.

Einstein Institute of Mathematics, Edmond J. Safra Campus, Givat Ram, The He-

brew University of Jerusalem, Jerusalem, 91904, Israel, and, Department of Mathe-

matics, Hill Center - Busch Campus, Rutgers, The State University of New Jersey, 110

Frelinghuysen Road, Piscataway, NJ 08854-8019 USA

E-mail address: [email protected]: http://shelah.logic.at

saharon shelah- pseudo pcf

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