saharon shelah- non cohen oracle c.c.c

8/3/2019 Saharon Shelah- Non Cohen Oracle C.C.C

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NON COHEN ORACLE C.C.C

SH669

SAHARON SHELAH

Abstract. The oracle c.c.c. is closely related to Cohen forcing. During aniteration we can omit a type; i.e. preserve the intersection of a given familyof Borel sets of reals is empty provided that Cohen forcing satisfies it. Wegeneralize this to other cases. In 1 we replace Cohen by nicely definablec.c.c., do the parallel of the oracle c.c.c. and end with a criterion for extractinga subforcing (not a complete subforcing, !) of a given nicely defined one suchthat the subforcing satisfies the oracle.

0. Introduction

This answers a question from [Sh:b, Ch.IV] (the chapter dealing with the oraclec.c.c.) asking to replace Cohen by e.g. random. Later we will deal with the parallelfor oracle proper and for the case is a (definition of a) nep forcing from [Sh:630].An application will appear in a work with Tomek Bartoszynski [BrSh:926].

How do we use this framework? We start with a universe satisfying 1 andprobably 21 = 2 and choose S

i : i < 2, S

i S

1 such that Si /D1 is

strictly increasing and for every i < 2, Si+1\S

iholds and for simplicity Si S

i+1

where D1 is the club filter on 1. We choose by induction on i < 2, a c.c.c.forcing Pi of cardinality 1, a sequence Mi = Mi : Si of countable models

(H(1), ) of some version of ZFC; without loss of generality transitive and a1-commitment mainly connected to, for each Si, a Pi-name

i which is, e.g.

random over Mi (and the commitment is that if, j > i, Pj

= Pj/Pi is represented(i.e. replaced by an isomorphism copy) such that it has set of elements 1, G Pjis generic over VPi , then for a club of Si ,

i V

Pi is random also over Mi[G]which naturally is defined as Mi[G ]. They are increasing in the relevant senseand the work at limit stages is done by the general claims here. In stage i, bybookkeeping we are given a task connected with a Pi-name X

i, we have some

freedom in choosing Pi+1, usually Pi+1 = Pi Q

i. So, working in VPi , Qi has tosatisfy a 0-commitment on Si , and we like it to satisfy that task, usually connectedwith X

i

R

V[Pi], say Xi

= X i

[GPi

]. We essentially have to choose Mi+1 such thatMi+1 Si = M

i but we have freedom (in addition to choosing Q

i) in choosing

Mi+1 : Si+1\S

i and a 0-commitment on S

i+1\S

i . Also the reals generic for

the chosen forcing notion as well as Mi+1 for Si+1\S

i can be chosen considering

Date: September 24, 2010.Key words and phrases. set theory, forcing, set theory of the reals, preservation theorems.The author would like to thank the Israel Science Foundation for partial support of this re-

search, Grant No. 242/03. The author thanks Alice Leonhardt for the beautiful typing. Firstversion done 2010/Feb.

1


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2 SAHARON SHELAH

Xi. E.g. Mi+1 can be the Mostowski Collapse of some countable M (H(2), )

to which Pi, Mi and X

i belongs.

Really this corresponds to the omitting type as in [Sh:e, XI]. This was origi-nally part of [Sh:630], particularly close to the so called faking; on more generaltreatments, including replacing 0 by a larger cardinal, see [Sh:895], [Sh:F1000].We thank the referee and (in 2008-2010) Jakob Kellner, Martin Goldstern, HeikeMildenberger and Wolfgang Wohofsky for corrections and comments.


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NON COHEN ORACLE C.C.C SH669 3

1. Definitions and preliminaries{12.1}

Hypothesis 1.1.(a) we assume CH, moreover S where S

{ < 1 : limit} is stationary

(b) let ZFC be a large enough subset of ZFC satisfied by (H(2), ) (or just(H(), ) for some with minor changes in the proof.

{1.1a}Definition/Notation 1.2. 1) M denotes an oracle, i.e., a sequence of the formM : S, M a transitive countable model of ZF C satisfying + 1 M andS S is stationary satisfying: for every X 1, the set { S

: X M}is stationary. For such M let SM = S.2) D denotes a normal filter on 1 usually extending DM which is defined in 1.6(1)below (of course, the default value is DM, see 1.8(1)).3) For a countable forcing P, a wide P-name is a Borel function giving for everyG P an object, i.e. the input of such a Borel function are truth value(p G) :p P, the output is normally in {x H(1) : rk(x) < } for some < 1. So ifp P p P then any wide P-name is still a wide P-name hence a P-name butit is natural to restrict ourselves, e.g. to the case P ic P, see below.4) P ic P when: P, P are forcing notions, i.e. quasi-orders, P P and if p, q Pthen they are incompatible in P iff they are incompatible in P.5) P P iff P ic P and every predense subset of P is a predense subset of P

(equivalently demands this for maximal antichains).6) For a forcing notion P let

(a) L1,(P) be the set ofL1, sentences in propositional logic considering themembers ofP as propositional variables

(b) for a generic G P and L1,(P) let [G] be the truth value (so if

= p P we have [G] = true iffp G, etc. (see [Sh:f, Ch.IX])(c) let p P means p P [G

] = true for p P, L1,(P)

(d) let P be the following quasi orderset elements: { L1,(P): for some p P we have p [G

] = true}

quasi order: 1 2 iffP if 2[G

] = true then 1[G

] = true.{1.1C}

Observation 1.3. 1) So P P as quasi orders, in fact, P is a dense subset of P(so no real difference as forcing notion).

2) If P1 P2 then P1 P2.3) Assume M H(1) is a countable transitive model of ZFC, M |= P is aforcing notion, the pair (,

) is as in 1.11(1) below and is a (Q,

)-generic

over M. Then

(a) P = (P Q

)/(

= ) is a forcing; note notion in M[]

(b) if P ic P and every predense suset of P which belongs to M is a predensesubset of P then P G

P is a subset of P generic over M and some

G P Q

generic over M extends G

P and satisfies

[G]

(c) P ic P (L1,(P))M[]; note: ifM |= P uncountable use (L|P|+,(P))

M[]

or do it outside M, see explanation

(d) if in addition M |= P1 P2 then (P1 Q)M (P2 Q)M and (P1 Q)M/(

= ) (P2 Q))M/(

= ).


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4 SAHARON SHELAH

Discussion 1.4. How P is interpreted as L1,(P)M[].

We should assume M |= P is a c.c.c. or use (L|P|+,(P))M which is L2,(P)

V

() if G P is generic over P and is (Q,

)-generic for M[G] then there is

() there is a unique G+ (P Q)M/(

= ) such that it is generic for(P Q), M

() G G+

() G QM = G

() there is a sequence of Borel functions Bp : p (PQ)M with infiniteP, in fact, in M and output a truth value G+ = {p (P Q)M :Bp(. . . , truth value(r G

), . . .)rP = truth, (in fact uniformly in ,

but not used)() we can identify p = (p0, p1) (P Q)M with Bp(. . . , truth value(r

G), . . .)rP.

2) The meaning of

is generic for Q mean then for every r Q we cancompute in a Borel way truth value(r G

) from the value of

(in our case ).

3) For random reals: if we interpret as the family of closed sets then

( a closed subset r of2)is O.K.4) In the case of random (p,q

) (P Q)M/

= is interpreted as (now F

qn : I

qn =

{T n2 : T a perfect subtree of >2}

p

n

(

rIn

(r n Fqn(r))).

Remark 1.5. In 1.3(2)(c), note that

() P1 P2 by the assumption as this is upward absolute from M as M istransitive

() the conclusion holds in M[] which is a generic extension of M hence in Vas above.

We first give the old definitions from [Sh:f, IV]{12.2}

Definition 1.6. 1) DM is{X 1 : for some Y 1 we have : Y M SM X}.2) A forcing notion P of cardinality 1 satisfies the (M , D)-c.c. iff SM D

+ andfor some (equivalently any) one to one f : P 1 the set S defined below belongsto D and P has minimal element P where:

S :=

: if SM and X M and {y P : f(y) < and f(y) X}is predense in P {y P : f(y) < } then X is predense in P

3) If D = DM we may write M-c.c.. Recall that D+ = {A 1 : 1\A / D}.

4) Let M1 M2 if M = M : S and { : S1\S2 or S1 S2 andM1 = M

2 } is not stationary; let M

1 D M2 be defined similarly (i.e. the set is= mod D); we call E a witness when E D is disjoint to those sets.4A) Let M1 M2 when M = M : S and { : S1\S1 or S1 2and M1 M

2 } is not stationary; let M

1 D M2 be defined similarly.


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5) A forcing notion P satisfies the (M , D)-c.c. iff: |P| 0 or for every X P ofcardinality 1 there is P1 P of cardinality 1 which includes X and satisfies the

(M, D)-c.c.

Remark 1.7. For the M-c.c. the order from 1.6(4A) is natural, but here it iseasier to use from 1.6(4).

{12.3}Fact 1.8. 1) DM is a normal filter on 1.2) The M-c.c. implies the c.c.c., and if DM D (or just there is a normal filterD DM D) then the (M , D)-c.c.c. implies the c.c.c. and if D2 D1 DM arenormal filters, then the (M , D1)-c.c. implies the (M , D2)-c.c.3) We can find S : < 2 such that S

S

, < S S mod DM, S

S+1 and S+1\S

D

+M

, moreover < < 2 S \S

is countable.

4) If M1 M2 and the forcing notion P2 satisfies the (M2, D)-c.c. and P1 P2,then P1 satisfies the (M1, D)-c.c.

5) Like part (4) when M1

M2.

Proof. See [Sh:f, Ch.IV], but for the readers convenience we prove part (4).4),5) Without loss of generality P2 has cardinality 1 and even its set of elementsis 1. As P1 P2 there is a function f : P2 P1 such that

()1 q P2 f(q) P1 p P1 p, q are compatible in P2.

Let g : P2 P2 P2 be such that

()2 if p, q P2 are compatible then g(p,q) is a common upper bound andp,q P1 g(p,q) P1.

So there is a club E1 of 1 which is closed under f, g so

()3 if 1 E Dom(M1) and I P1 is predense in P1 then I ispredense in P2 .

[Why? Ifq P2 then f(q) P1 so by the assumption on I, f(q) is compatiblewith some r1 I P1 , so there is r2 P1 above f(q) and r1, e.g. g(f(q), r1).By the definition of f the conditions r2, q are compatible in P2 hence g(r2, q) is acommon upper bound of them in P2 ]

()4 there is E2 D such that if E2 and I P1 and I M2 is predensein P2 then it is predense in P2.

[Why? As we are assuming that P2 satisfies the (M2, D)-c.c.]

()5 if I is a predense subset of P2 and is P1 then I is a predense subset ofP1.

[Why? As P1 P2 is assumed.]

()6 there is E3 D such that E M1 M

2 .

[Why? As M1 M2 implies M1 M2 and see the definition.]Putting together ()3, ()4, ()5, ()6 we are done. 1.8


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{12.3d}Definition 1.9. 1) We define when is an absolute definition of a c.c.c. forcing

notion with generic (for ZFC) say with a parameter, a real and countable ordinalmeans that for any transitive model N of ZFC , to which the parameters belong Q =Qn is a forcing notion, the property p Q, p Q q,p,q Q are incompatible,I is a countable predense subset are preserved in forcing extensions1 of N.2) We add with the generic

when

is a Q-name, the parameter including the

relevant information of a member of 2 or of (0) < and the generic is recon-structible from

.

3) We say is a nep forcing problem with the generic

as in [Sh:630]. See below.{12.3A}

Remark 1.10. 1) Note that below when Q is the older case ([Sh:f, IV]) we justpreserve every predense set, so in M (in the cases the commitment, see below,is obeyed) the forcing is countable.2) We may forget to mention this case as it is by now easy.

{12.4}

Definition 1.11. 1) We say Y = (S, ,

, ) = (SY, Y, Y, Y) is a 0-commitment

for M iff for some E DM:

(a) S S, S D+M

(b)

= : S, = : S and if S E then

M andM |= is an absolute definition of a c.c.c. forcing notion called Q =

Q with generic real so M |= (Q M[G

Q] = M[

]); note,

absolute here means that forcing extensions of M, preserve predensity ofcountable sets (in the sense of M), preserve order and its negation andpreserves incompatibility

(c) = : S where and for every S E the real

is (Q, )-generic over M or is (M, Q

, )-generic sequence whichmeans that for some G Q generic over M we have =

[G].

(d) we can allow Q :

1

2 for some ordinals 1

, 2 from Mand so : 1

2

.

We shall ignore M if clear from the context. We can replace M by (M , D) if aboveE D, S D+; similarly below.1A) A forcing notion P of cardinality 1 satisfies the 0-commitment Y = (SY, Y,

Y, Y)

for an 1-oracle M (we may suppress this if clear from the context) when P is aforcing notion and for any one-to-one mapping h : P 1 for some E DM forevery E we have

(e) every predense subset I of {p P : h(p) < } for which {h(p) : p I}

M is a predense subset of P(f) if S E then P the real is a (Q,

)-generic real over M[

hGP]; moreover

(f)+ letting Ph = h(P) the forcing notion Ph = Ph1 belongs to M and (P

h

Q

)M/(

= ) is ic Ph and, moreover, any predense subset of it from

M[] is predense in P, (in fact this implies clause (d), i.e. any predensesubset ofPh from M is predense in P

h)

1may restrict ourselves to the relevant forcing


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(g) so we get for the old case: if S E and Q is a singleton (hence M, a degenerated case), this actually follows from (d)

(h) the old case, i.e. Q = singleton occurs for a set of s from D+M (nec?);we may add M |= || = 0 most naturally in (e), but not used so far.

2) Let P H(2) be an M-c.c. forcing notion. We say that Y = (S,

,

,

) =

(SY, Y,

Y, Y) is a 1-commitment on P for M if: for any N satisfying ()1 below,

the clauses (a)-(d) of ()2 below hold

()1 N = N : < 1 is increasing continuous, N (H(2), ) is countable,N ( + 1) N+1 and {M, P} N for some < 1 and h N0 is aone-to-one mapping from P into 1

()2 (a) S Dom(M) S, S D+

M

(b) = : S, =

: S so (

,

) is a P-name of a pair

as in 1.11(1)(b), both are wide names over P

(c)

= : S and

a wide P-name of a real

(d) the set of the S satisfying the following belongs to DM + S: M, MosColN(N) M, and letting

2 h = MosColNwe have P = h(P

N) M so e.g. the set of members of P is{h(x) : x P N}, we have M |=

is a wide P-name of an

absolute definition of a c.c.c. forcing with generic real and

P is (Q

, )-generic over M[ h(G

P)]

(d)+ moreover if I M is a predense subset of P ifM |= (p,

) L, propositional sentence in the propositional

variable p, listing P and the

(i), i < g() such thatPQ (p,

) = true, then

P (h1 (p), ) = true.

Hence in particular if I M is a predense subset of P then h1 (I) is a predense

subset of P.For transparency the reader may concentrate below on the case (,

) :

S V.3) Let IS = {(P, Y, M) : P H(2) is an M-c.c. forcing notion and Y

is a 1-commitment on P for M}.We shall omit M if clear from the context. We can replace M by (M , D) naturally

and write ISD, but the claims are the same.4) For (P, Y, M) IS( = 1, 2) let (P1, Y1, M1) (P2, Y2, M2) means M1 M2, P1 P2 and for some E DM1 we have

(a) SY1

E SY2

E,(b)

Y1 (SY1 E) =

Y2 (SY

1

E),

(c)

Y1 (SY1

E) =

Y2 (SY1

E) and

(d) Y1 (SY

1

E) = Y2 (SY

1

E).

We call E a witness to (P1, Y1, M1) (P2, Y2, M2). Note that it does not matterif we demand E DM2 or E DM1 .

2this holds for a set of s which belongs to DM


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8 SAHARON SHELAH

We point out the connection between 0-commitment and 1-commitments.{12.4A}

Fact 1.12. 1) If Y is a 1-commitment on P for M, so P is an M-c.c. forcingnotion of cardinality 1, then P Y[G

P] = (SY,

Y[G

P],

Y[G

P],

Y[G

P]) is a

0-commitment so we call it Y[GP]. Note that each

[G

P] is still a name.

2) If P = {} (the trivial forcing) then: Y is a 1-commitment on P iff Y is a0-commitment.3) If Mi : i < is -increasing, < 2 and Dom(M

i)\S is not stationary whileS S, then there is M satisfying Dom(M) = S such that i < Mi M.4) Assume M2 is oracle, S M belongs to D+

M2and M1 = M2S. If Y2 =

(S2, 2,

2, 2) is a 0-commitment for M2 and S S2 then Y1 := (S, 2S,

2S, 2S)

is a 0-commitment for M1 and any forcing notion satisfying Y2 satisfies Y1. IfY2 =(S2,

2,

2,

2) is a 1-commitment on P for M2 and S S2 then (S,

2S,

2S, S)

is a 1-commitment on P for M1.5) If a forcing notion P satisfies the 0-commitment Y for the 1-oracle M andS = { SY : Q is a singleton (i.e., = Y and is of the old case for Y)}and S = mod DM then

(a) P satisfies the (M S)-c.c. which is defined as in [Sh:f, IV]

(b) if S S and S = mod DM and M |= X 2 is not meagre for

every S then

SX is not meagre in V[G

P].

As a warm-up (see [Sh:630] for more){12.4.xad}

Claim 1.13. 1) Assume

(a) M is a countable transitive model of ZFC, M |= P1 is a forcing notion

(b) M |= is an absolute definition of a c.c.c. forcing notion Q with generic

: 1 2 so 1, 2 OrdM

1 M(c) is a (M, Q,

)-generic sequence, recalling it means that there is G

(Q)M generic over M such that =

[G].

Then we can find a countable P2 such that:

() P1 ic P2 and every J M which is predense in P1 is predense in P2() P2 is (M

, Q,

)-generic sequence where M = M[GP2 P1].

2) Similarly for defining a nep forcing.

Proof. 1) In M we can define P+ = P1 (Q

)M[G P1], now as Q is absolutely

c.c.c., we know that q (, q) is a complete embedding of (Q)M into P+. Soif G (Q)M is generic over M such that =

[G] then let P2 = {(p,q

)

P1 (Q

)[G P1 ] : (p,q

) is compatible with (, q) for every q G}. Now check.2) See [Sh:630]. 1.13


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2. The iteration theorem{12.5}

Crucial Claim 2.1. In IS, any -increasing -chain has an upper bound.

Remark 2.2. 1) The -limit is the crucial one not the 1-limit. Actually for the1-limit we take the union and we preserve what we need by using the square (andhaving done something toward it in earlier limits or stages of cofinality 0).2) When is the union not an upper bound? If, e.g., for some S SY, S D+

Mand

for every S the forcing notion Y is random real forcing, we have in particularto preserve { : S} is non-null, but the union normally adds a Cohen.

Proof. So assume (Pn, Yn, Mn) IS and (Pn, Yn, Mn) (Pn+1, Yn+1, Mn+1)for n < , let M be such that M Mn for each n; so let En DM witnessboth, see the last line of Definition 1.11(4) and Definition 1.6(4). For simplicityassume that above any p Pn there are two incompatible elements (in Pn), and0 P is minimal in all Pn, i.e. is Pn. Without loss of generality the set of

elements of Pn is Xn 1 and 1\

n


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10 SAHARON SHELAH

where n() = Min{n : SYn

}. Defining P, F, Fn,, is harder, so we first

define AP, a set of approximations to them. A member t of AP has the form

(t, Pt, t, Ft, Ftn,,)n


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(iv) Ftn,, Fsn,, for n, <

(v) t is an initial segment of s.

Fact A: AP = .

Proof. Easy: choose E, let t = , Pt =

n nsuch that p Pm.

Lastly, t = the empty sequence. It is easy to check that t AP, as required.

Fact B: If t AP and t < E and is an old case for some Yn, then there iss satisfying t s AP with s , s = t.

Proof. Without loss of generality t, Pn : n < , Xn : n belong to M.Note

() any J M which is a predense subset of Pn is a predense subset ofPnand n < m Pn Pm , and of course Pn M.

Let A = Pt X t, B =

n


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12 SAHARON SHELAH

[Why? Recall that Q1 Q2 iffQ1 Q2 and for every p2 Q there is p1 Q1 suchthat p1 p1 Q1 p

1, p2 have a common upper bound in Q2; we shall use this

criterion. Let (p0

, q0

) Q, of course, we can replace this pair by any larger one, soby clause () above without loss of generality some m [n, ) is a possible valuefor n(p0, q0) so we have q0 Pm , hence recalling that Pn Pm there isq1 Pn such that:

(r Pn)(Pn |= q1 r r, q0 compatible in Pm ).

Assume q1 r Pn . So r, q0 are compatible in Pm hence has a commonupper bound q2 Pm .In particular q0 q2 Pm so by clause () we have (p0, q2) Q and (p0, q0) Q

(p0, q2); also r = (0, r) (p0, q2) as r q2 together r, (p0, q0) are compatible in Q,so [q1 r Pn (p0, q0), r = (0, r) are compatible in Q]. As (p0, q0) Q wasarbitrary we are done.]

() if p1, p2 Pt are incompatible in Pt then they are incompatible in Q.

[Why? Look at the order of Q].

() if < t then pt is a maximal antichain in Q.

[Why? Let (p, q) Q and we shall prove that it is incompatible in Q with some(pt,k , 0) with k < . Let n < be a possible value of n(p

, q) so q Pn andthere is a witness r q, r Pn t for (p, q) Q.By clause () in the definition of t AP we know that for some r Pn t wehave:

(i) r It,n,p

(ii) q, r are compatible in Pn

As q, r are compatible and r q also r, r are compatible in Pn hence in Pn t,so by the demand on r, we have: r, p are compatible in Pt. So in clause () ofthe definition of AP, in the definition of It,n,p for our r subclause (i) fails hence

subclause (ii) holds so there are k, p as in subclause (ii) there. Also let q1 Pn be a common upper bound of q, r. So r witnesses that (p, q1) Q with n apossible value of n(p, q1). Clearly it is above (p, q) and above pt,k so we are

done.]Let s = . Clearly Q M and M |= |Q| = || so as X \

n n such that

q Pm except when q = 0 then Fsn,,((p, 0)) = Ftn,,(p). Now it is easy to

check clause () of the definition of s AP. Also clauses (), () hold because theconstruction is made in M recalling () above since the construction is made inM.

Lastly, let s = t.


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Fact C: Ifti AP and ti ti+1 for n < then there is t such that i < ti

t AP and t =

i


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14 SAHARON SHELAH

()4 (a) the set of elements ofPt, be A ((X t) (X \

n


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Lastly, to prove the ic-embedding, toward contradiction assume that p, q Pt

are compatible in Pt but some r Q is a common Q-upper bound of j(p),j(q),

let r = (r1, r2) = ((r1, r

1 ), r2). By the definition of Q we have P

t

|= p r1 and

Pt |= q r2, so p, q Pt are compatible in Pt so we are done proving ()7.]

()8 (a) j M

(b) Q\ Rang(j) is dense in Q.

[Why clause (a)? Think.Why clause (b)? Assume that p Pt so j(p) = ((p, ), Ps,). Now Pt,n() P

t

hence there is p0 Pt,n() such that [p0 p0 Pt,n() p0, p are compatible in

Pt] and without loss of generality p0 = Pn . Let and p2 be such that p2 Ps,nis above p0 and forces

(t) = . Consider the triple ((p, ), p2), it suffices to

prove that it belongs to Q, as p2 = Pn and the demand j(p) ((p, ), p2) holds

trivially.Now in clause (A) above, subclauses (a),(b),(c) are obvious and as for subclause(d) we prove more than needed. Let n [n(), ) and G Pn = Ps,n genericover M be given such that p2 G, so as p2 Gs,n() and p2

(t) = clearly

= t()[G] and let =

[G]. We should find G

as there.

Recall G Pt,n is a subset ofPt,n generic over M(t) and M(t) |= Pt,n Pt and

every member of Pt,n compatible with p0 is compatible with p.As p0 G and G Pt,n is directed and p0 belongs to it. Clearly Mt,[GPt,n] |=

p Pt/(G Pt,n). As is generic for ( , M(t)[G Pt,n]) so Mt()[G Pt,n]

[]

is a generic extension of Mt()[G Pt,n] and it belongs to M[G] which is a genericextension of M.

In M(t) let P = (Pt,n Q

)

M(t) and P = (Pt Q

)M(t) so we know

() in M(t)(a) P P

(b) (p0, Q) P

(c) (p, ) P

(d) if (p0, Q) r P then r, (p, Q) are compatible in P

.

Now G Pt,n is a subset of Pt,n generic over M(t) and is (, Q)-generic over

M()[G Pt,n]. Hence there is H QM(t)[GPt,n] generic over M(t)[G Pt,n] such

that [H] = .

So H = (GPt,n)H is a subset ofP generic over M(t). As p0 GPt,n H

by ()(a) + (d) there is G P extending H generic over M(t) such that p G

hence G Pt,n G and note that P = P+t, so we are done.]

()9 We identify p P+t with j(p).

Next

1 (a) let gen(Q) = {G : G is a directed subset of Q such that G Pt

is generic over M(t) and GPs,k is generic over M for each k < }


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16 SAHARON SHELAH

(b) gen+(Q) = {G Q : G is directed, G Ps,n generic over Mfor n < , G Pt is generic over M(t), and letting

= [G Ps,n()] for some we have[G Ps,n()] = and

[G P

+t,] = }

Now

1.1 note that gen+(Q) = {G gen(Q): letting =

[G Ps,n()] we have

[G Ps,n()] is (

, Q)-generic for M(t)[G Pt,]}

1.2 if G Gen+(Q) and =

[G Ps,n()] then G P

+t, is a subset of P

+t,

generic over M(t).

A major point is

2 if p Q then there is: G gen+(Q) to which p belongs.

[Why? So let p = (p1, p2), p1 = (p1, p2 ) and by ()8 without loss of generalityp / Rang(g) and so p2

= , so n = n(p1, p2) = min{n < : n n() and

p2 Ps,n} is well defined.Next, let I1,k : k [n, ) list the dense open subsets of Pt, from M(t) and let

(jk,I2,k) : k [n, ) list the pairs (j,I) such that j < and I M is a denseopen subset ofPs,k and without loss of generality jk k and letI

+2,k = {q Ps, : q

is above some q I2,k}. By induction on k [n, ) we choose qk such that

2.1 (a) qk = (qk,1, qk,2) = ((qk,1, q

k,2), qk,2) Q hence

qk,1 Pt, so qk,1 P

t, qk,2 is a Pt-name of a member of Q

qk,2 Ps,k(b) if k = n then qk = p

(c) if k = m + 1 > n, then Q |= qm qk

(d) if k = m + 1 > n, m odd, then qk,1 I1,m.

Clearly if we succeed then {q Q : q qk} for some k < is as required (notingthat for every n the set {p Pn() : p forces a value to

n} {I2,k : k < }) so

it suffices to carry the induction.For k = n there is nothing to do.Let k = m + 1 > n and assume qm has already been chosen. Let Gm be a subset

of Ps,m generic over M such that q2,m Gm

2.2 M[Gm] the forcing notion ((Pt,k/(Gm Pt,m)) Q

)Mt() [GmPt,m] is a

complete subforcing (Pt

/(Gm Pt,m)) Q)Mt(0)[GmPt,m]

.Let :=

[Gm] =

[Gm Ps,n()].

Recalling qm Q, there is Gm such that

2.3 Gm is a subset of P

+t, = (P

t Q

)M(t) generic over M(t) such that

[Gm] = , Gm Pt,m G

m and qm,s G

m.

So there is qk,1

2.4 qk,1 = (qk, q

k,1) G

m is above qm,1 and belongs to G

m.


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By 2.2 we have

2.5 (Pt,k Q

)Mt() Gm is a subset of (Pt,k Q

)

M(t) generic over Mt() andstill for it

is interpreted as .

Hence there is a condition (r1, r1 ) (Pt,k Q

)

Mt() Gm such that

2.6 (a) r1 Pt,k/(Gm Pt,m)

(b) if r1 r Pt,k then r, qk,1 are compatible in Pt,

moreover

2.7 (a) (r1, r

1 ) (Pt,k Q

)

Mt()

(b) if (r1, r1 ) (r

2, r

2 ) (Pt,k Q)

Mt() then (r2, r2 ), (q

k,1, q

k,1) are

compatible in ((Pt/(Gm Pt,m))) Q

)Mt()[Gm].

So noting Gk is directed and r

1, qm,

2 Gk clearly there is qk,

2such that

2.8 (a) qk,2 Gk Ps,k(b) r1 qk,2 in Ps,k(c) qm,2 q in Ps,k(d) qk,2 I

+2,m

(e) qk,2 P2,k satisfies r

1 .

We next show that

2.9 qk = (qk,1, qk,2) is as required.

Now why does clause 2.1(a), i.e. why qk Q? Now in the definition (A) of Qclauses (a),(b),(c) obviously holds, and if we succeed to prove (d) there then clauses

(c),(d) of 2.1 are obvious as 2.4 and 2.8(d) respectively and clause (b) of 2.1holds emptily, so it is enough to prove (d) of (A).

So why does (A)(d) hold? Let Gk Ps,k be generic over M such that qk,2 Gkand let =

[Gk]. Now we work in M[Gk], so M(t)[Gk Pt,k] is a generic exten-

sion ofM(t), both are countable and qk,1[Gk Pt,k] is a member ofQ

M(t)[GkPt,k]

which satisfies, i.e. there is a generic subset H of QM(t) [Gk Pt,k] to which itbelongs and

[H1] = .

So H+ = (Gk Pt,k) H is a subset of (Pt,k Q

)M(t) generic over M(t).

Recalling 2.2 + 2.7 there is H a subset of Pt, = (Pt Q

)

M(t) generic over

M(t) which extends H+. So clearly Gk = H

Pt, is as required in (A)(d), soindeed qk Q so indeed we have carried the induction, but as said after 2.1 itsuffices to carry the induction for proving 2 so we are done proving 2.]

3 We identify p Pt with j(p) and q Ps, with (, q) pedantically replace Qby the quasi order Q with set of elements {p : p Pt or p = (Pt, q), q Q}and Q |= r1 r2 iff Pt |= r1 r2 or r1 Pt, r2 = (Pt, q2), Q |=j(r1) q2 or r1 = (Pt, q1), r2 = (Pt, q2), Q |= q1 q2

4 in M let Q be the forcing generated freely in M for our requirements(see [Sh:f, Ch.IX]), i.e.(a) let be the set of M such that: is a propositional L1,-

sentence generated by the set Q of propositional variables


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18 SAHARON SHELAH

(b) for and G we define the truth value [G] by induction of such that: if Q then [G] = truth G the connecting act

as usual(c) let Q+ set of elements be the set of such that there is G

gen+(Q) such that the truth value [G] is true, i.e. G is a model of

(d) Qt |= 1 2 iff (G gen(Q)) [if 2[G] is true then also 1[G] istrue]

Martin and Wolfgang would like to know if Q is necessary. They think that everyfilter G Q which is Q-generic over M will satisfy the following (and thereforebelong to gen+(Q)):

G P+t, is generic over M(t) (the same holds for G Pt)

G Ps,n is generic over M (for each n < )

[G Px,n()] =

[G P+t,].

[Why? Because there is no condition ((p1, p1 ), p2) Q that could force the oppo-

site.] 2.1

Discussion 2.3. On the one hand M(t) know Pt,n = Pn(t) is countable buthave the commitment about

(t). On the other hand M knows Ps,n1 but have

no commitment (generic old).So M |= Q is countable and for Qt we do not have to think about a commit-

ment in .Now

()1 Q M.

[Why? Formally as M |= P() exist, the class of all relevant sentences exist and

is of cardinality continuum in the sense ofM exist. Think ofMLevy(0,2

0) and use

absoluteness (the danger is that the ordinal depth > M Ord, but by the abovethis does not occur). So definition of Q in M is {: as above + Levy(0,20)there is G}.]

Next

()2 M |= Ps,n Q.

[Why? Let I M be a predense subset of Ps,n and let Q. So there is

G gen+

(Q) such that [G] = true; of course G gen(Q). By the definition ofG gen(Q) necessarily G I = ; let p G I, and let 1 = p, so 1 Ghence 1 Q. Also obviously Q |= 1 and Q |= p 1 so , r arecompatible in Q, so we are done.]

()3 M |= if I M(t) is predense in Pt then I is predense in Q.

[Why? Similarly.]

()4 Q ic Q (ignoring separability) moreover Q+ G Q

+

Q is directed.


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[Why? Assume p, q Q, Q+, Q+ |= p q and we shall finda common upper bound r Q of p, q which is compatible with in Q+, this

clearly suffices. As Q+

there is G Gen+

(Q) such that [G] = true. AsQ+ |= p q clearly p, q G. Let n < be large enough such thatp2, q2 Pr,n and n() n. We continue as in the proof of 2.]

()5 (a) if 1, 2 Q then 1 2 is the Q-lub of 1, 2, 1 2 Q(b) 1, 2 Q are incompatible in Q if 1 2 / Q(c) if , Q then they are incompatible and every Q

is compatible with at leaset one of them

(d) if Q, / Q then every Q is compatible with

(e) , are equivalent (both belong or does not belong to Q, etc.)

(f) like (a),(b) for : i < for < 1.

[Why? Obvious.]

()6 ifG Q is generic over M then for Q we have G [GQ] =truth.

[Why? We prove this by induction on .

Case 1: QObvious.

Case 2: = By ()5(c), (d).

Case 3: =i


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20 SAHARON SHELAH

[Why? By ()7, ()8, ()9.]We define s = (s, Ps, s, Fs, F

sn,,)


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{12.6}

Definition 2.4. Let C = C : < 2 a limit ordinal (and C = otherwise)

be a square sequence and X

= Xi : i < 1 be an increasing sequence of subsets

of 1, |Xi \j


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22 SAHARON SHELAH

The answer is that we should look at the definition of satisfaction of-commitmentin 1.11, so for notational simplicity consider = 0, i.e. 1.11(1A) and assume P 1

and h is the identity and E as there. For E note that clause (e) of 1.11(1A)says that P G

P is a subset of P := P generic over M. Note that

P M because E and M |= || = 0 hence M |= P is countable.

So M[GP] is a Cohen extension of M, (or is equal to M in a degenerate case).

So how can we satisfy the demand P is (Q

,

)-generic over M[G

P]? This

is not problematic but is not you may say in the next stage; see more in 4 but weshall elaborate. So in clause (e) of 2.4 letting i = we have j,,, ,

,

, Q, tias there let P = fj(Pj) for j C

i when = otp(j C

j ) and = otp(j C

i )

and let = otp(Ci ). Similarly let

= fj(

) so

is a P-name, hence is acountable object using members of P only.

We choose > from S[Yj, ] such that M := M

Yj is well defined, (M, P

:

< ) M and P : < and P GP

is a subset of P generic

over M. Of course, we can demand = .Now in M we define P = {P : < }. So M |= P is a countable forcingnotion union of the -increasing sequence P : < , M a countable transitivemodel P : < is -increasing, if < then every predense subset I Mof P, < is predense in P

and

is (Q

,

)-generic over M[GP ] when

[, ].Now we choose P M such that M |= P is an -extension ofP Q

/(

= )

and of each P, < , as in the proof of Fact D.Lastly, ti0 = t

, ti1 = and continue the proof of Fact D to choose t

i.

Proof. If cf() = 0 we use 2.1 but taking care of clause (e) of Definition 2.4, thisjust dictates to us how to start the induction there (as is done by Main Fact Dfrom inside the proof of 2.1). Note that if > sup acc(C ) we still use 2.1, just

our work is easier as we do not have to take care of clause (e). If cf() = 1, thenby the square bookkeeping (see clause (e) in Definition 2.4) our work is done (usingf = {f : acc(C )}). 2.5

{12.7A}Claim 2.7. 1) Assume

(a) Y = (S, ,

,

) is a 1-commitment on the forcing notion P H(2) for M,an oracle with domain SM

(b) GP P is generic over V, 0 = 0 : S where 0 =

[GP],

M1 = M[GP] = M[f(GP)] : SM for some one to one function ffrom P into 1

(c) inV[GP], Y1 = (S1, 1,

1, 1) is a 0-commitment, S S1 mod DM[GP], 1

(S S1

) = (S S1),

1

(S S1) = (S S1), 1

(S S1) = 0

(S S1) and (S1, 1,

1) V

(d) in V[GP], Q is a forcing notion satisfying the 0-commitment Y1 for M1.

Then for some P-name Q

and 1-commitment Y2 we have:

() (P, Y, M) (P Q

, Y2, M)

() SY2

= S1, Y2

= 1,

Y2 =

1, Y2 [GP] = 1

() Q

[GP] = Q.


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2) If for everyGP P generic overV there are Q satisfying some 1 and(S1, 1,

1, 1) V[GP] as above satisfying some 2, then we can demand

() P Q

[GP], Y2 as above satisfy 1, 2 respectively.

3) We may allow (, ) : S1 be a sequence of P-names and even (P Q

)-

names.

Proof. Straight. {12.7F}

Definition 2.8. For a pair (,

) as in Definition 1.11, we say Z 2 is positivefor (,

) when : for large enough, the set {N : N (H(), ) is countable,

(,

) N and there is Zwhich is (N, Q,

)-generic} is stationary, equivalentlynot empty.

{12.8}Claim 2.9. (iteration in successor case: increasing the commitment).

Assume

(a) (Pi, Yi, fi) : i < is an M-iteration and = + 1, SY S S, S Dom(M)

(b) (

, ) : S\S

Y is as required in Definition 1.11.

Lastly

(c) Z

2 is a P-name of a positive set for (

, ) for every such .

Then we can find (P , Y , f) such that

(i) (Pi, Yi, fi) : i < + 1 is an M-iteration

(ii) P = P, SY = S,

(iii) (

Y ,

Y ) = (

,

) andP

Y Z

when S\SY .

Proof. Straight. {12.9}

Claim 2.10. (iteration at successor: increasing the forcing)Suppose

(a) assume in claim 2.9

(b) Q

is a P-name satisfying, for every G P generic over V, the following:(i) Q

[G] is a c.c.c. forcing notion of cardinality 1

(ii) { SM : P M and G M[] is a generic subset of RY

Q/( = ) over M[] whichDM-almost always occurs andQ

[G]

M[G ] and there is R M such that RY R and R

Q

M[R]

P Q

DM[G].Then we can find (P+, Y+) such that (P, Y) (P+, Y+) IS and the P-nameP+/G

P is equivalent to Q

.

Proof. Straight.Now we draw an easy conclusion: consistently 20 = 2 and for each ideal defined

naturally by (,

) where absolutely defines a c.c.c. forcing Q and

is a widename for a read then for some set of 1 reals is positive for this ideal and more.


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3. Conclusions{12.10}

Conclusion 3.1. Assume (C, X) is as in 2.4. Let be a set of definitions offorcing notions with some real parameters, and Si : i < 2 is as in 1.8 for DM forsome M.

We can find (Pi, Yi, fi, Mi) : i < 2 such that

(a) it is an (C, X)-iteration (for this we have to allow = 2 in Definition2.4)

(b) P = {Pi : i < 2} is a c.c.c. forcing notion of cardinality 2 (so in VP, 20 2) and except in degenerated cases equality holds)

(c) SYi = Si from 1.8(3)

(d) if in VPi we have (,

) is a case of as in 1.11, moreover Pi {

Si+1\Si : M

i+1 [f

i (G

Pi)] |= (,

) as required in 1.11} D+

Mi+1(even

less with more bookkeeping) and Z (2)VP

is positive for (,

), then

() { SYi+1\SYi : ( , )/GPi = (,

) and

[GPi ] Z} D

+M

: infact the set is forced to include such old set (from V) by this we canget

() for some j > i, SYj+1\SYj ( , ) = (,

),

[GPi ] Z

(e) ifH is a pregiven function such that for every i < 2 and (P, Y, M) satisfy-ing the demands on (Pi, Yi, M): we have (P, YM)

H(P, Y, M) IS suchthat H(P, Y, M) satisfies the demands from (a) + (c) on (Pi+1, Yi+1, Mi+1),then we can demand (2j)[(Pj+1, Yj+1, Mj+1) = H(Pj, Yj , Mj)]; more-over, if SH { < 2 : cf() = 1} is stationary we can demand{j SH : (Pj+1, Yj+1, Mj+1) = H(Pj , Yj , Mj)} is stationary.(Of course, we can promise this for 2 such functions)

(f) similarly for SH { < 2: cf() = 0, / {C

i : i}} but the domain is asequence (Pn, Yn, Mn) : n < , SYn = Sin , in : n < increases to i.

Proof. Put together the previous claims. (Concerning clause (f) without loss ofgenerality {i < 1 : otp(C

i ) = 0} is stationary) so in those stages we have no

influence of clause (f) of 2.4; anyhow the influence of 2.4(f) is minor. {12.11}

Discussion 3.2. We discuss here some possible extensions. We can add a versionof the conclusion without the oracles, etc.

{12.10a}Claim 3.3. Assume Si : i < 2 is a sequence of pairwise almost disjoint station-ary subsets of 1, each with diamond and i < j Si S

+j mod D1, so S

+i 1

and Si S+i = and S

+i /D1 : i < 2 is increasing

Then in the following game between the bookkeeper and the forcer, the bookkeeper

has a winning strategy.A play lasts 2 moves, before the -th move a sequence (Pi, Q

i, M

i, Yi) : i < isdefined such that

(a) Pi a c.c.c. forcing notion of cardinality 1, say H


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(f) Yi is a Pi-name of an Si-commitment.

In the i-th move:(a) the bookkeeper chooses Pi and a Pi-name (Ni, Y

+) of an S+i -oracle and

0-commitment

(b) the forcer chooses Q

i and(M

i, Y

i), Pi-names such thatQ

i satisfies (Ni, Y

+i )

and (M

i, Y

i).

In the end the bookkeeper wins if

i < j < 2 Pj/Pi satisfies (M

i, Y

i).

Proof. Similar to earlier proofs.

We give an easy criterion for existence of forcing notion satisfying a given 0-commitment and a (not complete) sub-forcing of given nicely definable one.The following uses more from [Sh:630].

{12.12}Claim 3.4. Assume

(a) (P, , n)n


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26 SAHARON SHELAH

Case 1: = 0.Trivial.

Case 2: = + 1, non-limit or (P, ) / M.Let (P, ) = (P , ).

Case 3: limit.Let (P, ) = (


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4. Relatives and closing remarks

The comments below try to connect [Sh:669], [Sh:895], [Sh:F699], [Sh:F1000]. Arelated work is [Sh:895], the main difference is that there the continuumis forced tobe + and normally > 1. Now if there is = 1, still the construction is moregeneral. Here, e.g.

() if the forcing notion P satisfies the o-obligation Y then P (2)V is non-meagre.

[Why? Without loss of generality |P| 1 hence without loss of generality P 1.So if B

is a P-name of countable union of no-where-dense subsets of 2, then for

some SY we have B

M and P := P M and P GP is a subset of

P which is generic over M.So P G

P is generic for (P, M) hence (

2)M is non-meagre in M[GP ]hence has member / B

.

By absoluteness P (2)V BV[P], so we are done.]Now [Sh:895], for the case = 1, this proof does not work. In fact we can avoid

this but it requires some care.We now describe one version of [Sh:895].

Definition 4.1. 1) A 2-commitment base is a sequence p = (P ,


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28 SAHARON SHELAH

D a normal filter on 1 such that Sp D

if u : < 1 is as above

then for some W D for every W Sp,u letting h be as there then for somefunction h h , () u bbPq,, Rang(h) h

1(Pq,,) P.

Remark 4.2. 1) This definition fits [Sh:F699].2) Replacing 0 by -see [Sh:F1000].3) We can translate to proof here is a winning strategy st is [Sh:895] frame.4) We may replace the sequence P by a tree, then we can demand success on a club(or member of the filter). Necessary if we like to demand no such that G

p is

generic for (P , M), M |= || = 0.5) A natural question, continuing 3 is: for transparency assume V |= GCH andfor = 1, 2 let Q be a set of pairs (,

) so from V, we ask: is there a generic

extension VP of V such that:

if (,

) Q1 then there is a (,

)-positive set of cardinal 0

if (,

) Q2 then there is no (,

)-positive set of cardinal 1.

We surely can phrase sufficient conditions, but can we phrase sufficient and neces-sary conditions?6) More complicated when we have

above, so, i.e. allow real parameter in

, so

we will have 20 such cases rather than 1.


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5. Private Appendix

Discussion 5.1. (based on 2010.3.18 remark) after coments of Hecke1) In Definition 1.11(2) in ()2(d) should we separate for obligations of kind 1.2) In the proof of ()8 in Main Fact D inside the proof of 2.1, the paragraph onwhy clause (b)? Hecke thought n not n().

References

[Sh:b] Saharon Shelah, Proper forcing, Lecture Notes in Mathematics, vol. 940, Springer-Verlag,Berlin-New York, xxix+496 pp, 1982.

[Sh:e] , Nonstructure theory, vol. accepted, Oxford University Press.[Sh:f] , Proper and improper forcing, Perspectives in Mathematical Logic, Springer, 1998.[Sh:630] , Properness Without Elementaricity, Journal of Applied Analysis 10 (2004),

168289, math.LO/9712283.[Sh:669] , Non-Cohen Oracle c.c.c., Journal of Applied Analysis 12 (2006), 117,

math.LO/0303294.

[Sh:F699] , Revisiting con(borel conjecture + dual borel conjecture).[Sh:895] , Large continuum, oracles, Central European Journal of Mathematics 8 (2010),213234, 0707.1818.

[BrSh:926] Tomek Bartoszynski and Saharon Shelah, Dual Borel Conjecture and Cohen reals,Journal of Symbolic Logic 75 (2010), 12931310.

[Sh:F1000] Saharon Shelah, Higher Oracle iterations.

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- non cohen oracle c.c.c

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