saharon shelah- large continuum, oracles

8/3/2019 Saharon Shelah- Large Continuum, Oracles

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LARGE CONTINUUM, ORACLES

SH895

SAHARON SHELAH

Abstract. Our main theorem is about iterated forcing for making the con-tinuum larger than 2. We present a generalization of [Sh:669] which dealswith oracles for random, (also for other cases and generalities), by replacing1,2 by , + (starting with =

1). Well, we demand absolutec.c.c. So we get, e.g. the continuum is + but we can get cov(meagre) = and we give some applications. As in non-Cohen oracles, [Sh:669], it is apartial countable support iteration but it is c.c.c.

0. Introduction

Starting, e.g. with V |= G.C.H. and = 1, we construct a forcingnotion P of cardinality +, by a partial of CS iteration but the result is a c.c.c.forcing.

The general iteration theorems (treated in 1) seem generally suitable for con-structing universes with MA


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2 SAHARON SHELAH

(C) 1 < b = < 20 the dual -Borel conjecture (i.e. A 2 is strongly

meagre iff |A| < )

(D) 1 < b = < 20+ the dual 20-Borel conjecture(E) combine (A) and (C) and/or combine (A) and (D).

Parallely Steprans suggests:{h.14}

Problem 0.3. 1) Is there a set A 2 of cardinality 2 of p-Hausdorff measure> 0, but for every set of size 2 is null (for the Lebesgue measure)?2) The (basic product) I think b = d d = 20 gives an answer, what aboutcov(meagre) = < 20?

We shall deal with the iteration in 1, give an application to a problem from[Sh:885] in 2 (and 3,4).

Lastly, in 5 we deal with Bartoszynskis test problem (B), in fact, we get quitegeneral such results.

It is natural to ask{h.21}

Discussion 0.4. 1) In 1, we may wonder if we can give a reasonable sufficientcondition for b = 1 or b = < ? The answer is yes. It is natural to assume thatwe have in V a


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LARGE CONTINUUM, ORACLES SH895 3

The following tries to describe the iteration theorem, this may be more useful tothe reader after having a first reading of 1.

We treat as the vertical direction and +

as the horizontal direction, themeaning will be clarified in 2; our forcing is the increasing union of Pk : < +where k K2 (so k gives an iteration P[k] : < , i.e. a -increasingcontinuous sequence of c.c.c. forcing notions) and for each such k each iterandPp[k] is of cardinality < and for each <

+ the forcing notion Pk is the unionof the increasing union of continuous sequence Pp[k] : < . So we can saythat Pk is the limit of an FS iteration of length , each iterand of cardinality < and for (, +), k gives a fatter iteration, which for most S( ), is areasonable extension.

{h.42}

Question 0.6. Can we get something interesting for the continuum > + and/or getcov(meagre) < ? This certainly involves some losses! We intend to try elsewhere.

{h45}Definition 0.7. 1) For a set x let otrcl(x), the transitive closure over the ordinalsof x, be the minimal set y such that x y (t y)(t / Ord t y).2) For a set u of ordinals let H


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4 SAHARON SHELAH

1. The iteration theorem

If we use the construction for = 1, the version we get is closer to, but not thesame as [Sh:669]; in this case it may be more convenient to have the forcing locallyCohen.

Here there are atomic forcings used below coming from three sources:

(a) the forcing given by the winning strategies s(see below), i.e. the quotientPq/Pp, see Definition 1.11

(b) forcing notions intended to generate MA


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How do we amalgamate? There are two natural ways which say that we leavePp[k]/Pp[k] as it is.

First way: We decide that Pp [k+] is Pp[k] ((Pp[k+]/Pp[k]) (Pp [k]/Pp[k])).[This is the do nothing case, the lazy man strategy, which in glorified fashion

we may say: do nothing when in doubt. Note that Pp[k+]/Pp[k] and Pp[k]/Pp[k]are Pp[k]-names of forcing notions.]

Second way: Pp[k]/Pp[k] is defined in some way, e.g. is a random real forcing inthe universe V[Pp[k]] and we decide that Pp[k+]/Pp[k+] is defined in the sameway: the random real forcing in the universe V[Pp[k+]]; this is expressed by thestrategy s.

[That is: retain the same definition of the forcing in the -th place, so in somesense we again do nothing novel.]

{it.1}Context 1.1. Let = cf() > 1 or just

1 = cf() 1.

Below, +K1 is used in defining k K2f as consisting also of +K1-increasingcontinuous sequence p : E (so increasing vertically).

{it.7}Definition 1.2. 1) Let K1 be the class of p such that:

(a) p = (u,P, ) = (up, Pp, p) = (up,Pp)

(b) u Ord,

(c) P is a set H


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6 SAHARON SHELAH

4) If p = p : < is K1-increasing and cf() = 1 implies { < : p theexact limit ofp or just


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{it.5.21}Claim 1.8. Assume () < and

(a) p : () is +K1-increasing continuous for = 1, 2

(b) (p1 , p2)

K1

(p1+1, p2+1) for < ().

Then

() p1() K1 p2()

() for < () we have (p1, p2)

K1

(p1 , p2).

Proof. Easy. 1.8

For the successor case horizontally, limit case vertically when the relevant game,i.e. the relevant winning strategy is not active we shall use

{it.5}

Claim 1.9. Assume () < is a limit ordinal and(a) p : () is

+K1

-increasing, and q : < () is +K1

-increasing

(b) p K1 q for < ()

(c) if < < () then (p, q) K1

(p , q)

(d) if < () is a limit ordinal then P[p] Pq/GP[p] = {Pq/(G

P[p]

Pp) : < }.

Then we can choose q() such that

() p() K1 q()() (p, q) K1 (p(), q()) for every < ()

() clause (d) holds also for = ().

Remark 1.10. We can replace K1 by K1

in (c) and () of 1.9 and (b), () of 1.8.

Proof. Should be clear. 1.9

The game defined below is the non-FS ingredient; (in the main application below, = ), it is for the horizontal direction; it lasts steps but will be used inK2

f-increasing subsequences of ki : i < +.

{it.14}Definition 1.11. For < and let , be the following game between theplayer INC (incomplete) and COM (complete).

A play last moves. In the -th move a pair (p , q) is chosen such that p +K1

q and (1) < (p(1) K1 p) (q(1) K1 q) (up uq(1) = up(1)) andup

= and uq

= uq0

+ 1.In the -th move first INC chooses (p , u) such that p satisfies the require-

ments and u satisfies the requirements on uq (i.e. {uq : < } up u

[+]


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8 SAHARON SHELAH

{it.16}

Remark 1.12. It is not problematic for COM to have a winning strategy. But

having interesting winning strategies is the crux of the matter. More specifically,any application of this section is by choosing such strategies.

Such examples are the

(a) lazy man strategy: preserve Pq = Pq0 Pp0 Pp recalling Claim 1.7

(b) it is never too late to become lazy, i.e. arriving to (p(), q()) the COMplayer may decide that () Pq = Pq() Pp() Pp

(c) definable forcing strategy, i.e. preserve Pq/Pp is a definable c.c.c. forcing(in VP[p]).

{it.20}Definition 1.13. We say f is -appropriate if

(a) f ( + 1)

(b) < f() < ()[f() = + 1](c) if < +, u : < is an increasing continuous sequence of subsets

of of cardinality < with union then { < : otp(u) < f()} is astationary subset of .

{it.20b}Convention 1.14. Below f is -appropriate function.

We arrive to defining the set of approximations of size (in the main applicationf is constantly ); we shall later connect it to the oracle version (also see theintroduction).

{it.21}Definition 1.15. For f a -appropriate function let K2f be the family of k suchthat:

(a) k = E, p, S, s, g, f(b) E is a club of

(c) p = p : E

(d) p K1(e) p K1 p for < from E

(f) if acc(E) then p = {p : E }

(g) S is a stationary set of limit ordinals

(h) if S E (hence a limit ordinal) then + 1 E

(i) s = s : E S

(j) s is a winning strategy for the player COM in ,f(), see 1.16(1)

(k) g = g : S E

(l) g is an initial segment of a play of,f() in which the COM playeruses the strategy s

if its length is < f() then g has a last move

(p , p+1) is the pair chosen in the last move, call it mv(g)

let S0 = { S E : g has length < f()} and S1 = S E\S0(m) if < are from E then p

+K1

p , so in particular P/P is absolutelyc.c.c. that is ifPP and P is c.c.c. then P PP is c.c.c.; this strengthensclause (e)


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(n) f

(o) if S E then f() + 1 is the length of g

(p) for every E, if f() < then f() otp(up).{it.24}

Remark 1.16. 1) Concerning clause (j), recall (using the notation of Definition 1.11)that during a play the player INC chooses p and COM chooses q, f() andrecalling clause (o) we see that (pf(), qf()) there stands for (p , p+1) here. Youmay wonder from where does the (p, q) for < f() comes from; the answer isthat you should think of k as a stage in an increasing sequence of approximationsof length f() and (p, q) comes from the -place in the -approximation. This ischeating a bit - the sequence of approximations has length < +, but as on a clubof this reflects to length < , all is O.K.2) Below we define the partial order K2 (or K2

f) on the set K2f , recall our goal

is to choose an K2-increasing sequence k : < + and our final forcing will be

{Pk : < +

}.3) Why clause (d) in Definition 1.17(2) below? It is used in the proof of the limitexistence claim 1.23. This is because the club Ek may decrease (when increasingk).

Note that we use K1f

economically. We cannot in general demand (in 1.17(2)

below) that for < from Ek2\() we have (pk1 , p

k1 )

K1

(pk2 , pk2 ) as the

strategies s may defeat this. How will it still help? Assume k : < () isincreasing, () < for simplicity and {Ek : < ()}

{Sk : 1 and = cf() = then for

some c.c.c. forcing notion P of cardinality we have:P 2

0 = , p = and for no regular < is there a peculiar (, )-cut sot= .

Proof. We choose Q = P,Q : , < such that:

(a) Q is an FS-iteration

(b) Q is a -centered forcing notion of cardinality <

(c) if < ,Q

is a P-name of a -centered forcing notion of cardinality< then for some [, ) we have Q

= Q

(d) Q0 is adding Cohens, r : < say r

.

Clearly in VP we have 20 = , also every -centered forcing notion of cardinality< , is from VP for some < , so as is regular we have

() MA for -centered forcing notions of cardinality < and < dense sets.

Hence by 2.4 there is no peculiar (1, 2)-cut when 1 1 < 2 = (even1 < 2 < , 1 < < ).

Lastly,

for , in VP1+ for every for every < large enough we haver Jbd .

[Why? We prove this by induction on . For = 0 this holds by (d). For limit of uncountable cofinality recall ()V[P] = {()V[P] : < }. For limit of cofinality 0 use Q is a FS-iteration. Lastly, for = + 1 use the ofcardinality < of clause (c) of.] 2.7


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3. Some specific forcing{p.1}

Definition 3.1. Let =: : < be a sequence of members of which is


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(c) Q p G iff p

(n)(g(p) n < (n)

(n) gp(n))

(d) Q

F, i.e.

(n) FV[Q ]

(e) Q for every f ()V we have f F iff f FV iff

Jbd f.

Proof. 1) Trivial.2) Assume p Q for < 1. So {(p) : < 1} is a set of 1 ordinals < .But cf() > 1 hence there is () < such that < 1 (p) < (). For each let n = Min{n: for every k [n, ) we have (p)(k) ()(k) g

p(k)}.It is well defined because (p) 1). Hence QV,Fis c.c.c. even in V[G] as required. Turning to part (4), letting F = (F)

V[P],clearly P2 Q

1 = Q

,F1

for 1 < 2 < . Now about the c.c.c., as P is c.c.c.,it preserves cf() > 1, so the proof of part (1) works.6) Easy, too. 3.2

{p.14}Definition 3.3. Assume A = A : < is a -decreasing sequence of mem-bers of []0 . We define the forcing notion QA and the generic real w

by:

(A) p QA iff(a) p = (w,n,A) = (wp, np, A(p)),

(b) w is finite,

(c) < and n < ,(B) p QA q iff

(a) wp wq wp (A(p)\np)

(b) np nq(c) A(p)\np A(q)\nq

(C) w

= {wp : p GQA

}.


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{p.15}Claim 3.4. Let A be as in Definition 3.3.

1) QA is a c.c.c. and even a -centered forcing notion.2) QA w []0 is A for each <

and V[G

] = V[w

].3) Moreover, for every p QA we have p G

iff wp w

(A(p)\np) wp.

Proof. Easy. 3.4{p.21}

Claim 3.5. Assume () is Jbd -increasing.1) IfF F is downward cofinal in(F,


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(d) FV(e) p q

(f) q P

.

So we are done.2A) Similarly.3) If cf() < cf() let U be unbounded of order type cf() and Q

A= {p

QA : p U}, it is dense in QA and has cardinality 0 + cf(

) < cf(), so weare done.

If cf() > cf() and X [P]cf(), let () = sup{p : p X} and Y = {p QA :

p = ()}.The rest should be clear. 3.7


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4. Proof of Theorem 2.5{pt.3}

Choice 4.1. 1) S { < : cf() > 0} stationary.2) is as in 4.2 below, so possibly using a preliminary forcing of cardinality 2 wehave such .

{pt.7}Definition/Claim 4.2. 1) Assume = cf() [2, ) and = : < isan


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(e) for each Sk, sk = s, is from 4.2 or better 4.4

(f) gk is (pk , p

k+1), mv(g

k ) = 0, only one move was done.

2) If k K2 k then Pk the pair ( : < , i : i < ) is a (, )-peculiar

cut.

Proof. Clear (by 4.2). 4.6{pt.28}

Definition 4.7. Let P be the following forcing notion:

(A) the members are k such that(a) k K2 k K

2

(b) u[k] = {u[pk] : Ek} is an ordinal < + (but of course )

(c) Sk = Sk and sk = s

k for Sk

(B) the order: K2

.{pt.35}

Definition 4.8. We define the P

-name Q

as

{Pk : k GP} = {Pp[pk] : Ek and k G

P}.

{pt.42}Claim 4.9. 1) P has cardinality +.2) P is strategically ( + 1)-complete hence add no new member to V.3) P Q

is c.c.c. of cardinality +.4) P Q

is a forcing notion of cardinality + neither collapsing any cardinal norchanging cofinalities.5) If k P then k P Pk Q

hence P Pk Q

.

Proof. 1) Trivial.2) By claim 1.23.3) G

P is (< +)-directed.

4),5) Should be clear. 4.9{pt.45}

Claim 4.10. If k P and G Pk is generic over V then

(a) [G Pk] : < is


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Proof. Proof of Theorem 2.5 We force by P Q

where P is defined in 4.7 andthe P-name Q

is defined in 4.8. By Claim 4.9(4) we know that no cardinal is

collapsed and no cofinality is changed. We know that PQ 20 + because

|P| = + and P Q

has cardinality +, so P Q

has cardinality +, see4.9(3),(4).

Also PQ

20 + as by 4.9(2) it suffices to prove: for every k1 P thereis k2 P such that k1 K2 k2 and forcing by Pk2/Pk1 adds a real, which holdsby 4.11(2).

Lastly, we have to prove that (i : i < , : < ) is a peculiar cut. In

Definition 2.1 clauses (), (), () holds by the choice of k. As for clauses (), ()to check this it suffices to prove that for every f they hold, so it is suffice tocheck it in any sub-universe to which (, ), f belong. Hence by 4.9(1) it suffices tocheck it in VPk for any k P. But this holds by 4.6(2). 2.5


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5. Quite general applications{bt.7}

Theorem 5.1. Assume = cf() = 2 and 2 = + and ( < )(0 + instead of+, so Ek is always ,P[p(k)]is of cardinality < .5) Each game , has actually only countable many moves. Moreover, we mayhave by a p = p : < which is K1

-increasing u = {u[p] : < } and

N H

saharon shelah- large continuum, oracles

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