saharon shelah- the character spectrum of beta(n)

8/3/2019 Saharon Shelah- The Character Spectrum of beta(N)

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THE CHARACTER SPECTRUM OF (N)

SH915

SAHARON SHELAH

Dedicated to Kenneth Kunen

Abstract. We show the consistency of: the set of regular cardinals which arethe character of some ultrafilter on N can be quite chaotic, in particular canhave many gaps.

0. Introduction

The set of characters of non-principal ultrafilters on N, that we call the characterspectrum and denote by Sp, is naturlly of interest to topologists and set theoristsalike, see Definition 0.1 below. A natural question is what can this set of cardinalsbe? The first result on Sp is Pospsil proof that c Sp.

It is consistent that Sp = {20}, since Martins Axiom implies Sp = {2

0}.

Nevertheless, Sp = {20} is not a theorem of ZFC. Juhas (see [Juh80]) proved

the consistency of the existence of a non-principal ultrafilter D so that (D) < 20 .Kunen (in [Kun]) mentions that 1 Sp in the side-by-side Sacks model.

Those initial resulsts show that (D) is not a trivial cardinal invariant. But we

may wonder whether Sp is an interesting set. For instance, can Sp include morethan two members? Does it have to be a convex set? It is proved in [BnSh:642,6] that |Sp| large is consistent, e.g. 2

0 is large and all regular uncountable

20 (or just of uncountable cofinality) belong to it. It was asked there: amongregular cardinals is it convex? Now (proved in [Sh:846]) Sp does not have to beconvex. In the model of [Sh:846], there is a triple of cardinals (,,) such that < < ,, Sp but / Sp. In the present paper we show that Sp mayexhibit much more chaotic behavior.

To be specific, starting from two disjoint sets 1 and 2 of regular uncountablecardinals we produce a forcing notion P which forces the following properties:

(a) no cardinal (of V) is collapsed in VP

(b) 20 is an upper bound for the union of 1 and 2

(c) 1 Sp whereas 2 Sp = .

The proof requires that each element of 2 be measurable and that 1 satisfy


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2 SAHARON SHELAH

and hence that we also do not know for certain that there are successor cardinalsoutside Sp.

In the last section we show that we can, e.g. specify Sp , basically at will:if we have infinitely many measurable cardinals then we can make the intersectionbe {n : n u} for any subset of [1, ) that has no large gaps, i.e. for every nat least one of n and n + 1 belongs to u. If we assume infinitely many compactcardinals then we can realize any ground model subset of [1, ), e.g. Sp canbe even {p : p prime}.

Let us try to explain how do we do this. A purpose of [BnSh:642] is to createa large Sp. It provides a way to ensure many cardinals are in Sp. On the otherhand, [Sh:846] provides a way for guaranteeing a cardinal is not in Sp. Here wetry to combine the methods, hence creating a large set with many prescribed gapswhich establishes Sp in V

P.For adding cardinals we use systems of filters, so we deal with them and with

the one step forcing in 1; we use such systems indexed, e.g. by -trees, andin the end force by a suitable product of those trees, not adding reals. In thisdirection we do not need large cardinal assumptions. For eliminating cardinals weneed, essentially, measurables in the ground model. After the forcing with P, ourmeasurable cardinals become weakly inaccessible, and we show that they do notbelong to Sp.

We emphasize that for adding a cardinal to 1 Sp, we have to assume

=


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THE CHARACTER SPECTRUM OF (N) SH915 3

1. Preliminaries

This section is devoted to definitions and facts, needed for proving the mainresults of the paper. We present filter systems D = Dt : t I and we deal withthe one step forcing QD where D = D :

>, D a filter on N containing theco-finite subsets of N; when P1 QD

1

P2 QD2

, and with frames d = (Dd, Fd) foranalyzing Qd-name ofA

of subsets ofN modulo the filter on N which Fd generated,

in particular, a derived QD-name of an ideal idd.

{z4}Definition 1.1. For forcing notion P1, P2 (i.e. quasi orders).1) P1 P2 iff p P1 p P2 and for every p, q P1 we have P1 |= p q iffP2 |= p q.2) P1 ic P2 iff P1 P2 and for every p, q P1 we have p, q are compatible in P1iff p, q are compatible in P2.3) P1 P2 iff

1 P1 P2 and every maximal antichain of P1 is a maixmal antichain of P2,

equivalently

2 P1 ic P2 and for every p2 P2 for some p1 P1 we have p1 P1 p (p2, pare compatible in P2).

{cn.1}Definition/Observation 1.2. 1) For A P(N) let fil(A) = {B :


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{cn.14}Observation 1.4. Let I be a partial order.

0) ,

and

quasi order the set of I-filter systems and fil(Dt) : t I isan I-filter system for any I-filter system D and D1 D2 D1 D2 and

D1 = D2 D1 D2 D1 D2 and D1 D2 D1 D1 = D2 andD1

D2 D1 D1 =

D2.1) If As [N]0 for each s I and At As for s I t then there is an I-filtersystem D such that s I Ds = {As}.2) IfD is an I-filter system then for some ultra I-filter system D we have D D.3) If D is an I-filter system, s I and A and (t)[s I t A = mod fil(Dt)], then for some I-filter system D

we have D D and A Ds.4) If D : < is an -increasing sequence of I-filter systems then some I-filtersystem D is an upper bound of the sequence; in fact, one can use the limit, i.e.D,s = {D,s : < }; similarly for -increasing.5) If D is an I-filter system and D = fil(Dt) : t I then D D.

6) If D is an I-filter system and each Dt is an ultrafilter on then D is an ultraI-filter system and necessarily s I t Ds = Dt.7) If D1 is an ultra I-filter system and D2 is an I-filter system such that D1

D2then D1 D2.8) Assume P1 P2 and P1 D

is an I-filter system for = 1, 2. IfP1 D

1 D

2

then P2 D

1 D

2; also if P1 D

1 D

2 then P2 D

1 D

2.9) If P1 P2 and P1 D

is an I

-filter system for = 1, 2 and P1 D

1 is ultra

and D

1 D

2 then P2 D

1,t D

2,t and (fil(D

1,t)

+)V[P1] fil(D

2,t)+.

Proof. 0) Easy.1) Check.2) Use parts (3),(4), easy, but we elaborate. We try to choose D by induction on

< (2

0

+ |I|)+

such that

D is an I-filter system, <

D

D and foreach = + 1 for some t, D,t = D,t. For = 0 let D = D, for limit use part(4) and for = + 1 if D is not ultra, use part (3). By cardinality considerationfor some , D is defined but we cannot define D+1 so necessarily D is ultra asrequired.3)-9) Easy, too. 1.4

{cn.17}Claim 1.5. 1) Assume the quasi-order I as a forcing notion adds no new reals.An I-filter system D is ultra iff I {fil(Dt) : t GI} is an ultrafilter on .2) Assume the quasi-order I as a forcing notion adds no new 1-sequences of or-dinals and P is a c.c.c. forcing notion, (or just I is 1-complete or just P forc-ing with I add no new real). If P D

t : t I is an I-filter system then

PI

{fil(D t

) : t G I

} is an ultrafilter onN

iff P D t

: t I is an ultraI-filter system.

Proof. Easy. 1.5

{cn.19}Discussion 1.6. An I-filter system D may be degenerated, i.e. Dt = D is anultrafilter, the same for every t I. But in this case adding a generic set to I willnot add naturally a new ultrafilter, which is our aim here.

{c20}Definition 1.7. 1) For D = D :

>, each D a filter on N let QD be


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T : T > is closed under initial segments, and for some

tr(T) >

, the trunk of T, we have :(i) g(tr(T)) T = {tr(T) }(ii) tr(T) > {n : n T} D

ordered by inverse inclusion.2) For p QD let wfst(p, D) be the set of pairs (S, ) such that:

(a) () S { p : tr(p) p}

() tr(p) S

() tr(p) S S

(b) () is a function from S into 1

() if are from S then () > ()

() if S and () > 0 then {k : k S} = mod D.

3) If p QD and p then we let p[] = { p : or }.

4) IfD = D : >, D = D for > then let QD = QD and wfst(p,D) =wfst(p, D); we may write instead of p when this holds for some p QD withtr(p) = ; wfst stands for well founded sub-tree.

{c30}Claim 1.8. Assume >, D is a filter on N for > andY is a subsetof = = { : >}. Then exactly one of the following clauses holds:

(a) there is q QD such that() = tr(q)

() Yq = , equivalently q+ = q\{tr(q(t)) : < g(tr(q))}) is disjoint

to Y(b) there is a function such that(Dom(), ) wfst(, D) andmax(Dom())

Y; that is:

() Dom() is a set satisfying

(i) { : >}

(ii)

(iii) if and then

() (i) Rang() 1(ii) () < ()

() for every at least one of the following holds:

(i) Y

(ii) the set {n : n } belongs to D+ .

Proof. Similar to [Sh:700, 4.7] or better [Sh:707, 5.4].In full, recall = { : >}. We define when dp() for by

induction on the ordinal :

= 0: always

a limit ordinal: dp() iff rk() for every <

= + 1: dp() iff both of the following occurs:


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6 SAHARON SHELAH

(i) / Y

(ii) the following set belongs to D+ : {n : dp(n) }.

We define dp() Ord {} such that = dp() iff ( Ord)[dp() iff ].

Easily

for every , dp() 1 {}.

Case 1: dp() = .For each such that dp() = clearly there is A D such that

n A dp(n) = . Let q be

{ p : or and if then (g()) A}.

Clearly q is as required in clause (a) of 1.8.

Case 2: dp() < .We define

= { : and if k [g(), g())then k / Y and dp(k) > dp((k + 1))}.

We define : 1 by () = dp().Now check. 1.8

{cn.28}Claim 1.9. P1 QD

1 P2 QD

2 when :

(a) P1 P2 and D

= D

, : > for = 1, 2

(b) D

1, is a P1-name of a filter on N

(c) D

2, is a P2-name of a filter on N

(d) P2 D

1, D

2, and moreover (fil(D

1,)+)V[P1] fil(D

2,)

+, i.e. for

every A P(N)V[P1] we have A fil(D

1,) A fil(D

2,).

Proof. Like [Sh:700, 4] more [Sh:707, 5] but we elaborate.Without loss of generality P1 and P2 p for everyp P2. Clearly P1QD

1

P2 QD

2

by clause (d) of the assumption and moreover P1 P2 P2 QD2

recalling

Definition 1.1(1),(2). Now we can force by P1 so without loss of generality it istrivial, hence we have to prove that QD1 P2 QD

2

identifying q QD1 with(, q) P2 Q

D2

. By clause (d) of the assumption, this identification is welldefined and QD1 ic P2 QD

2

because for p1, p2 QD1 , p1, p2 are compatible iff(tr(p1) p2) (tr(p2) p1). It suffices to verify 1.1(3), requirement 2. Solet (p2, q

2) P2 QD

2 ; without loss of generality for some

from V we havep2

= tr(q

2), so > and of course:

()1 P2 q

2 QD2.

By 1.1(3), it suffices to find q QD1 such that

()2 q q QD1 (p2, q

2), q

are compatible; that is, (p2, q

2), (, q) are

compatible in P2 QD2

.


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Now we shall apply Claim 1.8 in V with , D1 here standing for , D there. Still

Y is missing, so let

Y = { : >and there is r QD1 such that = tr(r) and

(, r), (p2, q

2) are incompatible in P2 QD2

equivalently p2 P2 q

2, r are incompatible in QD

2}.

By Claim 1.8 below we get clause (a) or clause (b) there.

Case 1: Clause (a) holds, say as witnessed by q QD1 .We shall prove that in this case q is as required, i.e. q QD1 and [q QD1 r

QD1 (p2, q

2) P2 QD2

and r are compatible (in P2 QD2

)].

Why? Let = tr(r). Clearly ( q) hence by the choice ofq, i.e. 1.8(a)()we have / Y so r cannot witness Y hence r, (p2, q

2) are compatible in

P2 QD2 as required.

Case 2: Clause (b) holds as witnessed by the function .By the definition ofY, in V, we can choose q such that:

(a) q = q : Y

(b) q QD1 and tr(q) =

(c) q witness Y, i.e. p2 q, q

2 are incompatible in QD2

.

We define a P2-name q as follows:

q = { : or q

2 and if g(

) k < g()and k Y then qk, hence

k g() qk}.

Clearly P2 q QD

2

and tr(q) =

and QD2

|= q

2 q.

()3 if Y then and p2 P2 ( q

).

[Why? Otherwise there is p3 P2 such that p2 p3 and p3 P2 q

, as

tr(q) is forced to be

and tr(q) = , necessarily p3 P2 q , q are compatible.

But p2 P2 q

2 q, we get a contradiction to the choice of q .]

Now we know that Dom() and q hence S := { : Dom()

hence and p2 / q} is not empty. So as S Dom() the set

U = {() : S} is not empty, and by the choice of the function we haveU 1, hence there is a minimal U and let Dom() be such that () = .By the definition, if = 0 then by clauses () and () of 1.8(b), i.e. the choice of() we have Y and, of course, S. By ()3, p2 P2 ( q

) we get easy

contradiction to S, hence we can assume > 0. By the definition of S there isp P2 such that P2 |= p2 p and p P2 q

hence q

2 and, of course,

S. By the choice of the function , in V we have A := {n : n Dom()} = mod D1,, hence by clause (d) of the assumption of the claim P2 A = mod D

2,

and, of course, p P2 {n : n q} D

2,. Together p P2 there is n such

that n q Dom(), so let n and p P2 be such that P2 |= p p

and p P2 n q Dom().


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So ( n) is well defined, i.e. n belongs to Dom() hence ( n) and D [N]0 , / fil(D) for >

(b) F [N]0 and / fil(F).

1A) Above let Dd = Dd, : >, Dd, = fil(D), Fd = fil(F), Qd = QDd andif D = D for > we may write D, Dd instead of D, Dd, respectively.2) We say A

is a d-candidate when (d is a frame and):

(c) A

is a Qd-name of a subset of N.

3) We say A

is d-null when it is a d-candidate and is not d-positive, see below.4) We say A

is d-positive when for some p Qd, for a dense set of p p somequadruple (p,A, S, ) is a local witness1 for (A

, d) or for (, A

, d) when = tr(p)

or for (p,A

, d) or for A

being d-positive, which means:

(a) p Qd(b) A F+d(c) S = Sn : n A and = n : n A

(d) (Sn, n) wfst(p, D) for n A recalling Definition 1.7(2)

(e) if Sn and n() = 0 then p[] n A

.{c33}

Definition 1.11. 1) For a frame d = (D, F) let idd = id

(d) = {A

N : A

is a

Qd-name which is d-null}.2) If P d

is a frame then id

d

[P] is the P Qd

-name of idd

.{c35}

Claim 1.12. For a frame d, QD idd is an ideal on N containing the finite setsand N / id

d; moreover, for every A P(N) from V, we have A = mod Fd iff

Qd A idd.

Proof. It suffices to prove the following 1 4.

1 If QD if A1 A2 and A2 idd then A1 id

d.

[Why? If (p, A, S, ) is a local witness for (A

1, d) then obviously it is a lcoal witnessfor (A

2, d).]

2 if Qd if A1, A2 idd then A1 A2 id

d.

Why? It suffices to prove: if Qd A

1 A

2 = A

N and A

is d-positive then A

is d-positive for some {1, 2}. Let (p, A, S, ) be a local witness for (A

, d) and

we shall prove that there are {1, 2} and a local witness for (tr(p), A, d); by the

dense in Definition 1.10(4) this suffices.For any n A and Sn such that n() = 0 we choose (n,, n, , Sn,) such

that:

()2.1 (a) ,n {1, 2}

1An equivalent version is when we weaken clause (e) to: if Sn and n() = 0 then thereis q Qd such that tr(q) = , p q and q n A

, see ()2.2 in the proof. Moreover, we can

omit p q; hence actually only tr(p) is important so we may write tr(p) instead of p.


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(b) (Sn,, n,) wfst(p[], Dd)

(c) if n,() = 0 so Sn, then there is q Qd such that

p q, tr(q) = and q n An,; let qn, be such q.

[Why (n,, n,, Sn,) exists? We shall use 1.8; that is for {1, 2} let Yn,, = { : p and there is r QD such that tr(r) = and p r and r n A

}.

We apply for = 1, 2 Claim 1.8 with Dd, ,Yn,, here standing for D, ,Ythere. If for some {1, 2} clause (b) there holds as witness by the function ,easily the desired ()2.1 holds. If for both = 1, 2 clause (a) there holds then for = 1, 2 there is q Qd such that tr(q) = and q Yn,, = .

Necessarily q := q1 q2 p belongs to Qd and has trunk and is disjoint toYn,,1 Yn,,2. But Qd |= p[] q and q[] Qd n A

= A

1 A

2, hence

there are {1, 2} and r Qd such that q r and r Qd n A, but then

tr(r) Yn,, and tr(r) q q, contradicting the choice of q. So ()2.1 holdsindeed.]

()2.2 without loss of generality n,() = 0 g() > n.

[Why? Obvious.]

()2.3 for n A there are n, Sn, n such that

(a) (Sn, n) wfst(p, Dd)

(b) Sn Sn and max(Sn) = S

n max(Sn)

(c) n {1, 2} and max(Sn) n, = n.

[Why? Easy.]

()2.4 for n A letting Sn = {Sn, : max(S

n)} S

n, for some

n and qn we

have:

(Sn, n) wfst(p, D) { : n() = 0} = {: for some we have S

n, n() = 0, Sn,

and n,() = 0}

q = qn, : n() = 0

n() = 0 p qn, tr(qn,) =

qn, n An

[Why? Think.]

()2.5 there is {1, 2} such that A := {n A : n = } = mod Fd.

[Why? Obvious as A F+d .]

We now consider the quadruple (p, A, S, ) defined by:

p = { p: if tr(p) , n g() and max(Sn) then qn,}where Sn, qn, are from ()2.4.

[Why p Qd with tr(p) = tr(p)? Recall()2.2.]So together we have:

A is from ()2.5, so A F+d

S = Sn : n A where Sn is from ()2.4


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10 SAHARON SHELAH

= n : n A where n is from ()2.4.

Now check that (p, A, S, ) is a witness for (tr(p), A, D) hence 2 holds as said

in the beginning of its proof.

3 Qd idd; moreover if A = mod Fd is from V then A id

d.

Why? Because of clause (b) in Definition 1.10(4).

4 QD[d] N / idd, moreover if B F

+d and B V then B / id

d.

Why? This means that B is d-positive which is obvious: use the witness (p, A, S, )where p is any member ofQd, A = B, Sn = {tr(p)}, n(tr(p)) = 0. 1.12

{c36}Observation 1.13. Assume d1, d2 are frames and Dd1 = D = Dd2 and Fd1 Fd2then QD id

d1 idd2.

Proof. Should be clear. 1.13

{c37} Claim 1.14. We have P2 idd1 id

d2

and (idd1

)+[P1] (idd2

)+[P2] when :

(a) P1 P2(b) P d

is a frame for = 1, 2

(c) P2 Dd1, D

d2, for

>

(d) if A (D

+d1,

)V[P1] then A (D

+d2

)V[P2]

(e) P2 Fd1 Fd

2

(f) if A (F+d1

)V[P1] then A (F+d2

)V[P2].

Proof. Should be clear by 1.15 below recalling 1.9. 1.14{c39}

Claim 1.15. Letd be a frame and A

a QDd-name of a subset of N. We have A

is

d-null iff for a pre-dense set of p Qd we have tr(p) p there is no localwitness for (p[], A

, d) equivalently, for (, A

, d).

Proof. Straight. 1.15{c40}

Remark 1.16. The point of 1.15 is that the second condition is clearly absolute inthe relevant cases by 1.9, i.e. in 1.14.

{c41}Definition 1.17. 1) fin(I) is the set of finite functions from I to H(0).2) Let K be the set of forcing notions Q such that some pair (I, f) witness it, i.e.(I , f , Q) K+ which means:

(a) f is a function from Q to fin(I)

(b) if p1, p2 Q and the functions g(p1), g(p2) are compatible then p1, p2 have

a common upper bound p with g(p) = g(p1) g(p2).2) We define K=wkK by: (I1, f1, Q1)

wkK (I2, f2, Q2) means that:

(a) (I, f) witness Q K for = 1, 2

(b) I1 I2(c) f1 f2(d) Q1 ic Q2.

3) We define stK similarly adding:


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(d)+ Q1 Q2.

4) If q K+ let q = (Iq, fq, Qq).

Remark 1.18. We can use much less in Definition 1.17.


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2. Consistency of many gaps

We prove the first result promised in the introduction. Assume = 1and we like to build a c.c.c. forcing notion P of cardinality , such that VP is asrequired: Sp includes 1 and is disjoint to 2; really we force by P

T, the T

quite complete and translate P-names of ultra systems of filters to ultra-filters. Inorder to have 1 Sp, we shall represent P as an FS iteration P, Q

: , 2 and for each 2 we have a D = D,s : s T

a P-name of a ultra system of filters for unboundedly many < , increasing with; in the end we force by P {T : 1}. Toward this for each s T, 1we many times force by Q+D,s from 1.

But in order to have 2 Sp = , we intend to represent P as the union ofa -increasing sequence P : < and for each 2 for stationarily many < , cf() = and P+1 is essentially the ultrapower (P

)/E, E a -complete

ultra-filter on , so is a measurable cardinal.To accomplish both we define a set Q, each x Q consist of a FS-iteration ofP, Q : +, < + with D

s, : s {T : 1} for many < +,

increasing with and Q = QDt(), .In the end for suitable x, we shall use P for some < + of cofinality


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(h) if 1 then |T| is (1\) except when sup(1) is strongly inacces-sible and then the value is sup(1)

{c45}Choice 2.2. 1) Without loss of generality T : 1 is a sequence of pairwisedisjoint trees.2) Let T be the disjoint sum of {T : }, so it is a forest.3) Let t = ti : i S be a sequence of members ofT where S = { < + : cf() =cf()} such that if t T then { S : t = t} is a stationary subset of +; lett(i) = ti.4) Furthermore choose

() S0 { < + : cf() = 0} is stationary

() = ,t,n : S0, t T, n N; let (,t,n) = ,t,n() ,t,n : n < is an increasing -sequence of ordinals with limit

() ,t,n { S : t = t}

() guess clubs, i.e. ifE is a club of + then the set { S0 : C := {,t,n :t, n} E} is stationary.

Remark 2.3. If |T| < we can find such , but in general it is easy to force such.

{cn.47}

Claim 2.4. Assuming 2.1(1) only, a sequence T as in 2.1, clauses (f),(g),(h) (andalso t ,s,S, as in 2.2) exists, provided that 1 { : =

2, ). 2.4

{c49}Definition 2.5. Let Q be the set of objects x consisting of (below , +):

(a) P H

(++

) and I


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14 SAHARON SHELAH

(h) () if S0 E then Q is Qfil() with the generic

() if S0 E and C Ex, see 2.2(4)() then

u,t,n D,t, see below, whenever t T, n N and S Ex\( + 1)

() in clause () we let u,t,m = {

(,t,n)(k) : n N, n m

and k (n)}.

Discussion 2.6. 1) Later we shall use an increasing continuous sequence x : . Where and how will cofinality reappear? Well, we shall use P()[x] for some() Ex of cofinality . So why not replace

+ by above? We have a problemin proving the existence of a (canonical) upper bound to x : < , specifically infinding the D

i in the proof of Claim 2.11, i.e. completing an appropriate T-filter

system to an ultra one, e.g. in Case 3 in the proof of 2.11. To help we carry astrong induction hypothesis, see clause (i)()2 in there and then first find anR

j,+ [Pj x]-name, then reflect it to a i.

2) Note that it helps to have not only Q = QD, but possibly some related forcing

notions. First in proving there is a limit, see 2.11, in proving the reflectiondiscussed above lead us to use some unions. Second, using ultrapower by E,see 2.13, for limit of cofinality , the ultrapower naturally leads us to use someiterations.3) We may in 2.1 demand / 1, equivalently < min(), but let T be asingleton {t} and T is Tmin(1) T. In this case in 2.17 we get PT {} 1 Sp.

{c50}Definition 2.7. 1) For x Q, of course we let Qx = Qx = Q[x] = Q, Px =P[x] = P, Px = Px = P = Px+ , etc.2) We define a two-place relation Q on Q : x Q y iff:

(a) (Ix


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is a FS-iteration but dom(p), p P implies that P forces a value totr(p()) and also P D

is a T-filter system such that D

D

for <

and D is ultra when / S0; recall that in Definition 2.5 D is defined only for E S, but no harm in defining D

in more cases. For = 0, 1 this was done

above.For limit let P = {P : < } and D

= D

,t : t T where D

,t(i) =

{D,t(i) : < }. It is easy to see that P : is a -increasing continuous

sequence of c.c.c. forcing notions and P D is an T-filter system. If S0

let D = D

, otherwise by 1.4(2) we can find D

such that P D

is an ultra

T-filter system and D D

.

For = + 1 such that / S S0 let Q be trivial. Now let P = P Q

and

let D,t be D

,t. Easily P D

,t : t T is an T-filter system and choose D

as above, i.e. (a P-name of an) ultra T-filter system above D.

Next, assume = + 1, S; we let Q = QD,t()

and P = P Q

. Now

for s T, let D,s = D,s {N\A : A idd t(),s} where t() = t is from 2.2(3).

Note that P fil(D,s) fil(D

,t) iff s T t by the choice of the D

1,ss and the

D,ss, so the definition of idd

t(),s

depend on the truth value of t() I s.

Now (pedantically working in VP):

D,s [N]

0 by its definition

D,s D,s, by 1.12

/ fil(D,s) by 1.12

if A (D+,s)V[P ] then A ((D,s)

+)V[P+1] by 1.12

s I t D,s D

,t by 1.13 and the choice of the D

,ts.

We continue as in the previous case.

Lastly, assume = + 1, S0 and we shall define for . We let Q = Qfil()in VP and so

is defined as the generic and P+1 = P Q

. Note that u

,t,n is

well defined, (see clause (h) of Definition 2.5). By Claim 2.9 below letting D,t =

D,t {u

,s,n : n N and s T satisfies s T t} we have D

= D

,t : t T is

a P-name of a T-filter system above D and let D

be (a P-name of) an ultra

T-filter system above D.

Let I = {} for < +, I


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16 SAHARON SHELAH

(b) PQfil()

fil(D,t) = fil({u

,s,n : s T t and n N})

(c) PQ

fil()if t T and A {D,t : S} then u

,t,n A

for every large enoughn.

Proof. Straight; the point is PQfil()

fil(D,t) for t T, which holds as

()1 if A D

(,t,n) then for every large enough k,

(,t,n)(k) A

()2 if A D

+(,t,n)

in VP then for infinitely many k,

(,t,n)(k) A

()3 is a dominating real.

2.9{cn.53}

Observation 2.10. 1) Q partially orders Q.2) Px satisfies the c.c.c. and even is locally 1-centered

2 when x Q and +.

Proof. Easy. 2.10{cn.56}

Claim 2.11. The upper bound existence claim

If x : < is Q-increasing and is a limit ordinal < + then there is xwhich is a canonical limit of x : < , see below.

{cn.57}Definition 2.12. We say x = x is a canonical limit of x = x : < when xis Q-increasing, is a limit ordinal < + and (for every < +):

(a) x Q

(b) x Q x for < and Ex {Ex : < }

(c) I[x] = {I[x] : < }

(d) if has uncountable cofinality then() Px = {P

x : < }

() Px Dx,t = {D

x,t : < } for t T if Ex S

() Q

x = {Q

x : < }

() g

x = {g

x : < }.

(e) if has cofinality 0, then() if +\(S Ex) or S0 C

Ex then P[x] Q

[x] =

{Q[x] : < } and similarly g

[x] = {g

[x] : < }

() if S Ex then P[x] D,t[x] {D

,t[x] : < }

(f) in fact |Px | ({|Px | : < })

0 .

Proof. Let

0 (a) I = {I[x] : < } for < +

(b) I


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1 (a) p R iff p = (p1, p2) and some pair (, p0) witenss it which means < and p0 P[x], p1 P [x], p2 Q and one of the

following occurs() p1 = or p2 = recalling clause (c) of 2.5

() p0 P[x] p1 P[x]/P[x] and p2 Q/P[x](b) for p R let (p) be the minimal < such that (, p0)

witness p R for some p0(c) R |= p q iff letting = max{(p), (q)} we have

P[x] |= p1 q1 and Q |= p2 q2.

We note that:

2 (a) R, [Q, x] is a partial order

(b) above R, [Q, x] is a dense subset of R, [Q, x] where R, [Q, x] is

defined like R, [Q, x] when in 1(a) we omit subclause ().

[Why? Clause (a) by 3 below and clause (b) is easy.]So below we may ignore the difference between R, [Q, x] and R, [Q, x]

3 for (,, Q) as above; if (, p0) is a witness for p = (p1, p2) R, [Q, x]and (, ) then for some q0 P[x ] the pair (, q0) is a witness for(p1, p2) R, [Q, x].

[Why? As we can increase p0 in P[x], without loss of generality (p1) p0,where on recall Definition 2.5, clause (c). As (, p0) is a witness for (p1, p2) R, [Q, x] necessarily p0, p2 are compatible in Q hence they have a common upperbound q2 Q. As P[x ]Q, there is q0 P[x ] such that q0 q P[x ] q, q2are compatible in Q. As we can increase q0 in P[x ] and p0 q2 without loss

of generality p0 q0 but (p1) p0 hence (p1) q0. As x x andP[x ], Q

[x ] : < + is FS iteration and p1 P [x] P [x ], clearly q0 q

P[x ] q, p1 are compatible. So clearly (, q0) is a witness for p R, [Q, x] asrequired in 3.]

4 if ,, Q are as above and (1) + then R, [Q, x] R,(1)[Q, x].

[Why? We check the conditions from Definition 1.1(3), the second alternative.First, ifp = (p1, p2) R, [Q, x] we shall prove p R,(1)[Q, x]; as p R, [Q, x],some (, p0) witness it, easily it witnesses p R,(1)[Q, x] as P[x] P(1)[x].

Second, assume R, [Q, x] |= p q and we should prove R,(1)[Q, x] |= p q, this is obvious by the definition of the orders for those forcing notions. TogetherR, [Q, x] R,(1)[Q, x].

Third, we should prove R, [Q, x] ic R,(1)[Q, x] so assume p, q R, [Q, x]has a common upper bound r = (r1, r2) in R,(1)[Q, x]. Now easily (r1, r2) is acommon upper bound of p, q in R, [Q, x] as required.

Fourth, for p R,(1)[Q, x] we should find q R, [Q, x] such that ifR, [Q, x] |=q q then q, p are compatible in R,(1)[Q, x].

Now let p = (p1, p2) R,(1)[Q, x] and let (, p0) witness it; without loss ofgenerality P[x] |= (p1) p0.

Let q1 = p1 P[x], now q := (q1, p2) satisfies

q R, [Q, x].


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18 SAHARON SHELAH

Why? The pair (, p0) witness it because if p0 q P[x] then first p1, q has a

common upper bound r P(1)[x] hence r P [x] is a common upper bound

of q

, q1; second q

, p2 has a common upper bound in Q as (, p0) witness (p1, p2).So indeed (, p0) witness q = (q1, p2) R, [Q, x].

If q q R, [Q, x] then q, p are compatible in P(1)[x].

Why? Let q = (q1 , q2) and let r1 = (p1[, (1)))q

1 , easily (r1, q

2) R,(1)[Q, x]

is a common upper bound of q, p.This finishes checking the last demand for R, [Q, x]R,(1)[Q, x] so 4 holds.]

5 if Q satisfies the c.c.c. then R, [Q, x] satisfies the c.c.c..

[Why? Let pi = (p1,i, p2,i) R, [Q, x] for i < 1. Let (i, p0,i) be a witness for

(p1,i, p2,i). As before let qi Q be such that p0, p1,i, p2,i are below it.We can find an uncountable S such that f[xi ](p1,i) : i S are pairwise

compatible functions and i : i S is non-decreasing. As Q satisfies the c.c.c., forsome i < j from S there is a common upper bound q Q of qi, qj ; let { : < n}list in increasing order {} dom(p1,i) dom(p1,j)\ and let n = .

By induction on n we choose r P [xj ] such that:

if = 0 so = then r0 r P[xj ] r, q are compatible in Q

if = m + 1 then rm r

P [xj ] |= (p1,i) r and (p1,j) r.

For = 0 use q Q and P[xj ]Q. For = m+1, we shall choose r Pm+1[xi ]

as follows: if / dom(p1,i) then r = rm {(, p1,j())}; if / dom(p1,j)similarly; otherwise, i.e. if dom(p1,i) dom(p1,j) use the demands on g

recalling () of clause (c) and end of clause (d) of Definition 2.5.Having carried the induction, (rm, q) is well defined. Now let r P[xj ]

be above r0 such that r r P[xj ] rm, r are compatible. Also r r P[xj ] r0 r P[xj ] r, q are compatible in Q. So (j , r) witness (rm, q) R, [Q, x] and easily (rm, q) is above pi = (p1,i, p2,i) and above pj = (p1,j , p2,j), so5 holds indeed.]

6 for ,, Q as above, Q R, [Q, x] when we identify p2 Q with (, p2).

[Why? Again, first p Q p R, [Q, x] by the identification, and for p, q Qwe have Q |= p q R, [Q, x] |= p q by the definition of the order ofR, [Q, x]. So Q R, [Q, x] holds, moreover Q ic R, [Q, x] by the definition ofthe order.

Lastly, let q R, [Q, x], so by 2 without loss of generality q = (q1, q2) R, [Q, x] and we shall find p Q such that p p

Q p, (q1, q2) are compati-ble.

Let p = q2, i.e. (, q2), and the rest should be clear.]

7 for ,, Q as above we have P[x]R, [Q, x] when we identify p1 P[x]with (p1, ).


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[Why? Similarly.]

Now by induction on i + we choose i and P, f (when i and j < i

j < ), Q, g

(when < i and j < i j ) and

3 also Di (when i S)

such that

the relevant parts of clauses (a)-(e) of Definition 2.12 and of the definitionof x Q holds, in particular (all when defined):(a) P H(+) is a c.c.c. forcing notion

(b) () Px P and Px P = Px for <

() (I


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20 SAHARON SHELAH

[Why? This follows from clause (i) of .]Let us carry the induction, this clearly suffices.

Case 1: i = 0.Trivial.

Case 2: i is a limit ordinal.Let = i be {j : j < i}, clearly j : j i is increasing continuous and

i E. Below vary on .Let P = {P : < } and f = {f : < } and from 0 recall I


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Proof of ()3:

Let (0, p0) be a witness for (p1, p2) R, [P, x]; as we can increase 0, by 3,

and we can increase , without loss of generality 0 = .Without loss of generality p0, p2 P, as we can increase , moreover as sup( ) andT is -complete for 1, = cf() for simplicity; let f be a function with domain2 such that > f() > sup( ), f() > 1 is regular (sof()


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28 SAHARON SHELAH

cardinality , or better a set of representations of such names, (see Definition 2.14),which generates D

.

Now D has the same properties in VTQ

(see first and second above) sowe have SpV[P] so V

P |= 1 Sp.

Fourth, the main point, we would like to prove that 2 Sp = in VP.

So toward contradiction assume

1 2 and (p, r, q) P forces D

is an ultrafilter on N with (D

) = .

Let Q so essentially Q =Q Q> and Q = Q


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(c) (p, r, q) Px()

T


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30 SAHARON SHELAH

Let p2 : < (P()[x()]) be such that j()(p2 : < ) = p

2.Hence

()12 U1 = { U : p2 (C0 , . . . , C

n1) is (A

)

[t]} belongs to E.

Without loss of generality p2 : U1 are pairwise compatible hence by Lostheorem for some

()13 U1 so < and p2, p2 has a common upper bound p3 Px()+1, hence

p1, p3 has a common upper bound p4 Px().

So recalling q is from ()4,

()14 (p4, r, q) forces

(a) A D

(b) (C0 , . . . , C

n1) D

(c) (C 0 , . . . , C

n1)

(A)[

t], (A

)[

t].

Contradiction. 3.2{e.8}

Claim 3.4. In 3.2 (and 1.6) instead of E is -complete (so is measurable) wemay require that there is 2 2 such that:

(a) (2, f) are as in 3.2

(b) defining Q we use 2 if 2 then E is -complete

(c) if 2\2 then = max(

2 ) is well defined, [, ] 1 = and E

is a uniform -complete ultrafilter on so is a -compact cardinal.

Proof. Similar to 3.2. 3.4{e.11}

Remark 3.5. The situation is similar for any set { : u} of successor of regularcardinals.

{e.30}

Claim 3.6. In 3.1 above the sufficient conditions for / Sp inVP are sufficient

also for ()(cf()) = / Sp).

Proof. The same. 3.6

So we can resolve Problem (6) from Brendle-Shelah [BnSh:642, 8].{e.34}

Conclusion 3.7. If GCH and 1 < = cf() < = , is measurable,

then there is a forcing notion P of cardinality collapsing the cardinals in (, )but no others such that in VP, for every cardinal (, ) of cofinality , we have / Sp = sup(Sp ).


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4. Private Appendix

Moved 2011.7.04 from the old introduction:We investigate the Sp = character spectrum of non-principal ultrafilters on ,

see Definition 0.1 below. On background and early history, see [vD84], [Juh80]. Thefirst consistency result was of Juhasz who proves the consistency of the charactermay be < 20. Here we continue Brendle-Shelah [BnSh:642] by which Sp canbe very large and [Sh:846] by which it can be a non-convex set. We prove that ifthere are enough measurables then we can force that the character spectrum set,{(D) : D an ultrafilter on N} is quite chaotic.

Concerning the proof, on the one hand, as in [BnSh:642] we use a product ofa c.c.c. forcing and an Easton support product of a sequence T : ,Ta tree used to index a name in the c.c.c. forcing notion of a system of filters toget a witness for Sp, and on the other hand, as in [Sh:846] that is as in

[Sh:700] we use ultrapower of a c.c.c. forcing notion by a -complete ultrafilter toget the non-existence of a witness for Sp. Concerning the latter, the readermay be helped by the articles of Brendle, [?], [Bre07] which include exposition of[Sh:700] and probably also by [Bre03]. However, we do not rely on those works,and generally within reason we try to make this work self contained.

Moved 2010.11.30 from the proof of 1.8,pg.5:

Case 1: There is S with dp(S) = .Let (sn, n) : n < list {(s, ) : s I, s

>} each appearing infinitelymany times. We choose Sn by induction on n such that:

(a) Sn moreover dp(Sn) =

(b) if n = m + 1 then Sm Sn

(c) if (sn, n) Sn and k is minimal such that (s, nk) / Sn butdp(Sn {(sn, nk)) = , (sn, nk) Sn+1.

For n = 0 let S0 = S by the case assumption the demands hold.For n = m + 1, if (sn, n) / Sn let Sn = Sm; so assume (sn, n) Sn; if there is

k > n such that dp(S {(sn, nk)) 1, then such k is as required, so assumenote. Hence...SAHARON SORT OUT!

Moved 2010.11.30,pg.14:{cn.84}

Claim 4.1. In 2.13, recallT =

Tmin(1) is the product {

T : 1} withEaston support and assume 2 and > sup(1 ). Then there is no pair

(A

, z) such that

(a) A

is a Px T-name of a family of subsets of [N]0 of cardinality

(b) y z

(c) Pz fil(A) is an ultrafilter on satisfying (fil(A

)) = .

Proof. Let = sup( 1) and = (2), so < . Now we repeat the proof of

[Sh:846, 1.1] and anyhow we do a more complicated proof in 3.2 below. 4.1


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32 SAHARON SHELAH

Moved 2010.11.26 from the proof of 1.9,pg.4:Let S = Dom().

First, = because {(s, s) : s I0} belongs to Dom() by (b)() of 1.8 and

as witnessed by p2 as if q QD1I0 , tr(q(s)) : s I0 = then p2 P2 q2, q are

compatible in QD2I0 by the definition of QD

2I0 .

Hence there is S with the ordinal (S) minimal; so S satisfies at least oneof the clauses in (b)() of 1.8.

Subcase 2A: S satisfies (b)()(i).This means there is Y such that {(t, t) : t I0} S and let p

+2 witness S

QD1I0 witness Y. So p+2 forces that q

2, q3 QD

2I0

, s I0 s q

2(s)

and s I0 s = tr(q3, (s)) hence p+2 forces that q

2, q3 are compatible in QD

2I0

.

This implies that q3(p+2 , q

2) are compatible in P2 QD2I hence also q3, (p2, q

2) are.

But this contradicts the choice of q3 (witnessing Y).

Subcase 2B: S satisfies (b)()(ii) of 1.8.Let (s, ) S be as there. Let U1 = {n: there is p

2 such that P2 |= p

+2 p

2

and p2 n q2(s)}. By the definition of and as p+2 witness S ,

clearly n U1 S {(s, n)} . By the Definition of QD2I0

we have

p2 U fil(D

2,s). But U1 V hence by clause (f) of the assumption ofthe claim 1.9 we have proved really U1 fil(D1,s). But by the present subcaseassumption, U2 := {n : S { n} Dom() and (S { n}) < (S)} is = mod fil(D1,s). Together necessarily there is n U1 U2.

Let S1 = S {(s, )}, as U2, clearly S1 Dom() and (S1) < (S) and asn U1, S1 .

Together we get contradiction to the choice of S, i.e. S has minimal (S).

Together all cases lead either to the desired conclusion or to a contradiction, so weare done.

Moved 2010.11.24 from 0:{z8}

Definition 4.2. We say D is double-Ramsey when :

(a) D is a filter on N

(b) every co-finite subset of N belongs to D

(c) in the games (D), (D+) the PO player does not lose where for X P(N)the game (X) is defined as follows:() a play last -moves

() in the n-th move

the NU player chooses An X

the PO player choose kn An

() in the end, the PO player wins the play if {kn : n < } X.{z10}

Definition 4.3. 1) We say D is a double-Ramsey when :

(a) D = Ds : s I

(b) each Ds is a filter on N containing the co-bounded subset of N


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(c) in the games (D) and (D+) the PO player does not lose where D+ =D+s : s I and for X = Xs : s I,Xs P(N) the game X (x) is

defined as follows:() a player last moves

() in the n-th move

NU chooses a finite In I and An,s Xs for s In

PO chooses kn,s An,s for s In

() in the end PO wins when {kn,s : n satisfies s In} Xs for everys {Im : m < }.

2) We say D is strongly Ramsey when (a), (b) of part (1) holds and

(c)+ in the game drn(D) the player PO does not lose where:() a play last moves

() in the n-th move NU chooses finite In I, An,s D+s (for s Is)

PO chooses kn An,s

() in the end PO wins when g

n

In such that

if t In An,t Dt for infinitely many ns then{kn : t In, An,t Dt} Dt

if t In An,t D+t for infinitely many ns then

{kn : t In, An,t Dt} D+t .

{z12}Remark 4.4. With enough Cohens or just cov(meagre) = 20 , it is easy to constructsuch Ds

Moved 2010.11.22, from pgs.4,5:

Remark 4.5. (Was in 1.6, pg. 4):2) In 4.8(2) below, note that I2 may be the equality.

{cn.21}Definition 4.6. For an I-filter system D let QD be defined by

(A) p QD iff(a) p is a function with domain a finite subset of I

(b) for t Dom(p), p(t) is a perfect subtree of> with trunk tr(p(t)) p(t) such that tr(p(t)) p(t) sucp(t)() := {n : n p(t)} fil(Dt) (and, of course, p(t) g() g(tr(p(t)) tr(p(t))).

(B) p Q q iff(a) Dom(p) Dom(q)

(b) p(t) q(t), i.e. p(t) q(t) for every t Dom(p).(C) for p Q let p+ be the function with domain Dom(p) such that p+(t) =

{ : tr(p(t)) p(t)} for every t Dom(p).{cn.23}

Definition 4.7. 1) For a partial order (or a quasi order) I let EI be the finestequivalence relation on I such that s I t sEIt.2) For a quasi order I let E+I = {(s, t) : s I t and t I s}.


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34 SAHARON SHELAH

{cn.24}Observation 4.8. 1) IfI is a quasi order and D = Dt : t I is an I-filter system

then sE

+

I t fil(Ds) = fil(Dt).2) QD does not depend on the order of I, that is, if I1, I2 are quasi-orders with thesame set of elements, D an I-filter system, for = 1, 2 and D1,s = D2,s for s I1then QD1 = QD2 .

{cn.28ya}Claim 4.9. 1) The property D is an I-filter system is absolute (in order thatthis will hold we did not demand Ds is a filter on but just Ds P()).2) We have P1 QD

1 P2 QD

2 when :

Moved 2010.11.22, from pg.6:{cn.31}

Definition 4.10. We say (P1, D

1) (P2, D

2) when the assumptions (a)-(f) ofClaim 1.9(2) holds (hence P1 QD

1

P2 QD2

).

Moved 2010.11.22, from pg.7:{c36.yajan}

Claim 4.11. If P1 P2 and D is a P-name of a nonprincipal ultrafilter on for

= 1, 2 and P2 D

1 D

2, then P1 Q(D

1) P2 Q(D

2).

Moved 2010.11.22 from the end of the proof of 2.11, pg.11:Q is well defined by (f), (g).

We let P = P Q so by Claim 1.13 the forcing notion P satisfies the relevant

parts of clauses of . But we have also to define D. For t Ix let D

,t =

{Dx,t : < satisfies t Ix} D

,t.Now clearly

()1 D,t is a P-name of a family of subsets of ()2 if s


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(A) (a) R0 R1(b) E

is an R-name of a filter on N containing the co-finite sets

for = 0, 1(c) R1 E

0 E

1 moreover (E

+0 )

V[R0] (E

+1 )

V[R1]

(d) R+ = R QE

for = 0, 1

(e) R+2 = R QE

for = 0, 1

(f) E2 is a R2-name of a filter extending E

0 moreover(E

+0 )

V[R] (E

+2 )

V[R+2]

(B) (a) R3 / fil(E

E

).

Proof. It is enough to prove (check both are stated!) that 2 holds assuming 1when

1 (a) p3 R3

(b) A is an R-name of a member of E so subset of N for = 1, 22 p3 R3 A

1 A

2 = .

So we shall assume

()1 (a) p3 = (p1, q

3) hence R1 q

3 QE1

and without loss of generality

(b) p1 forces tr(q

3) = .

As R0 R1, there is p0 such that

()2 (a) p0 R0(b) if R0 |= p0 p then p, p1 are compatible in R1.

We can work in V[G0] where

()3 (a) G0 R0 is generic over V

(b) p0 G0

Let

()4 Y1 = {(n, ): for some q R1/G0 we have R1 |= p1 q and q n A

1

and q

3}.

4.12

Moved from pg.19:Alternative earlier assumption:

Note that Px() Px()+1 Px. As (p, r, q) forces that {A

: < } generates

D it also forces that D {A

,i : < ,i < } generates D. But Px {A

,i :


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36 SAHARON SHELAH

By the choice of j(), (q, A) it follows that there is a pair (B

, p)

() p4 Px()

p4, p3 are compatible in Px() B

is a P()-name of a subset of

(p4, q2, r2) A,i D

and [t = 1 B

= A

A

1,i1

] and

[t = 0 B

= \A A

1,i1

].

NowU E where U := { U such that (r2, q2), (r2 , q) T


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5. Private AppendixFurther results

We may look at further properties in such forcing extensions.{g.4}

Claim 5.1. In all the results above (2.17, 3.2, 3.4) we can add: in VP we have:

(a) u = Min(1)

(b) there is no MAD family of cardinality [, cf()]

(c) b = d.

Proof. Same proof. 5.1

More seriously in [Sh:700, 4] we have a more versatile framework.{g.42}

Hypothesis 5.2. As in 2.1, but only Min(1).{g.49}

Definition 5.3. 1) We define Q = Q

2

is as the class objects of x consisting ofS, P, Q, for < , D

,t,

,t (for < , t I) such that:

(a) Q = P, Q : < H(+) is a FS iteration and P = {P : < }

(b) D = D

,t : t I is a P-name of an I-filter system for

(c) moreover it is an ultra I-filter system

(d) P D D

if

(e) S { < : cf() / 2} is unbounded below

(f) P Q = QD

(g) P+1 ,t := {tr(p(t)) : p G

Q} satisfies Rang(

,t) D

+1,t.

1A) For x Q as above we let Q = Qx, Px = P, Ix = I, D,t = Dx,t,

x,t =

,t, Ix = I. In 3 we use only Q2 so we write Q and

x = .2) Q is the following two-place relation on Q : x Q y iff

(a) Px Py for

(b) Py Dx,t D

y,t for , t I

(c) Py x

y.

(note: clause (b) restrict ourselves concerning the T).3) (070729) Use several S : each optional:

(a) one for b = = d

(b) Sp

(c) A = {|A| : A [w]0 is MAD}.

{g.56}Claim 5.4. The upper bound existence claim

If x : < is Q-increasing continuous and is a limit ordinal < + thenthere is x which is a canonical limit of x : < which means

(a) x Q

(b) x x for <

(c) Ix = {Ix : < }

(d) if has uncountable cofinality then


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38 SAHARON SHELAH

() Px = {Px : < } and

() Px D

x,t = {D

x,t : < and satisfies t I}

(e) in fact |Px ({|Px| : < })0 .

Proof. Combine the proof of 2.11 and ??. {g.77}

Claim 5.5. If x Q and 2 then we can find a pair(y,j) such that

(a) x Q y Q

(b) j is an isomorphism from (Px)/E onto Py extending the canonical em-

bedding of Px into (Px)

/E

(c) j maps (Px)/E onto Py for < is a -embedding

(d) |Pg| |Px|/E.

Remark 5.6. Saharon: 1) Deal with Sp? Sp?2) Other invariants: b, d which now need:

(A) in QD, many the t dominating by a suitable side condition.


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6. Assignments

Moved 070726, pg.1:Saharon: (Oct. 2006)1) Fill a proof of 1.8 by [Sh:707], pg.7 here.2) Fill, like [Sh:509], proof of ()3, case 2, proof of 2.11.3) Check end of 4.1.4) (July 2007) S.5) S.6) More invariants?

Moved from end of old proof of 3.2,pg.10:

()6 for some representation q of a Px-name A of a subset of , we have: for

every < the set { < :Px A

A

is infinite iff A

A

is infinite}

belongs to E.

[Why? By the construction of x = x, the uses of ultra-power by E.]Letting G Px T


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40 SAHARON SHELAH

7. Remarks on -bases{p.2}

Definition 7.1. 1) A is a -base iff:

(a) A []0

(b) for some ultrafilter D on ,A is a -base of D, see below.

1A) We say A is a -base of D if (B D)(A A)(A B).1B) (D) = Min{|A| : A is a -base of D}.2) A is a strict -base iff :

(a) A is a -base of some D

(b) no subset ofA of cardinality < |A| is a -base.

3) D has a strict -base when D has a -base A which is a strict -base4) Sp = {|A| : D a non-principal ultrafilter on and A of D}.5) We say A is a [strict] -base if it is a [strict] -base of D for some ultrafilterD on .{p.4}Definition 7.2. For A []0 let IdA = {B : for some n < and partitionB : < n of B for no A A and < n do we have A

B}.{p.5}

Observation 7.3. For A []0 we have:

(a) idA is an ideal on P() including the finite sets, though may be equal toP()

(b) if B then: B []0\ IdA iff there is a (non-principal) ultrafilter Don to which B belongs and A is a -base of D.

Proof. Clause (a): Obvious.

Clause (b): The if direction: Let D be a non-principal ultrafilter on such that

B D and A is a -base of D. Now for any n < and partition B : < n ofB as B D as D is an ultrafilter clearly there is < n such that B < n hence byDefinition 7.1(1A) there is A A such that A B. By the definition of IdA itfollows that B / IdA but []


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{642.2a}Claim 7.5. In VP as in [?, 1.1], we have {, } Sp and 2 / Sp

.

Proof. Similar to the proof of [?, 1.1].


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42 SAHARON SHELAH

References

[Bre03] Jorg Brendle, The almost-disjointness number may have countable cofinality, Transac-

tions of the American Mathematical Society 355 (2003), no. 7, 26332649 (electronic).[Bre07] , Mad families and ultrafilters, Acta Universitatis Carolinae. Mathematica et Phys-

ica 48 (2007), no. 2, 1935.[Juh80] Istvan Juhasz, Cardinal functions in topologyten years later. Second edition, Mathe-

matical Centre Tracts, vol. 123, Mathematisch Centrum, Amsterdam, 1980.[Kun] Kenneth Kunen, Ultrafilters and independent sets, Trans. Amer. Math. Soc. 172, 299306.[vD84] Eric K. van Douwen, The integers and topology, Handbook of Set-Theoretic Topology

(K. Kunen and J. E. Vaughan, eds.), Elsevier Science Publishers, 1984, pp. 111167.[Sh:509] Saharon Shelah, Vive la difference III, Israel Journal of Mathematics 166 (2008), 6196,

math.LO/0112237.[BnSh:642] Joerg Brendle and Saharon Shelah, Ultrafilters on their ideals and their cardinal

characteristics, Transactions of the American Mathematical Society 351 (1999), 26432674,math.LO/9710217.

[Sh:700] Saharon Shelah, Two cardinal invariants of the continuum (d < a) and FS linearlyordered iterated forcing, Acta Mathematica 192 (2004), 187223, Also known under the title

Are a and d your cup of tea?. math.LO/0012170.[Sh:707] , Long iterations for the continuum, Archive for Mathematical Logic submitted,

math.LO/0112238.[Sh:846] , The spectrum of characters of ultrafilters on, Colloquium Mathematicum 111,No.2 (2008), 213220, math.LO/0612240.

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- the character spectrum of beta(n)

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