saharon shelah- was sierpinski right iv?

8/3/2019 Saharon Shelah- Was Sierpinski Right IV?

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WAS SIERPINSKI RIGHT IV?

SH546

Saharon Shelah

Institute of MathematicsThe Hebrew University

Jerusalem, Israel

Rutgers UniversityDepartment of Mathematics

New Brunswick, NJ USA

Abstract. We prove for any = +.

Typed 5/92 - (20 , k22-Mahlo, [2]23; some on models)

I thank Alice Leonhardt for the beautiful typing.

Latest Revision 97/Dec/4 - Revised and Expanded 5/94 based on lectures,Summer 94, Jerusalem.Partially supported by the basic research fund, Israeli Academy of Sciences.

Typeset by AMS-TEX

1


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2 SAHARON SHELAH

0 Introduction

An important theme is modern set theory is to prove the consistency of small

cardinals having a large cardinal property. Probably the dominant interpre-tation concerns large ideals (with reflection properties or connected to genericembedding). But here we deal with another important interpretation: partitionproperties. We continue here [Sh 276, 2], [Sh 288], [Sh 289], [Sh 473], [Sh 481]but generally do not rely on them except in the end (of the proof of 1.19) when itbecomes like the proof of [Sh 276, 2]. This work is continued in Rabus and Shelah[RbSh 585].

We thank the participants in a logic seminar in The Hebrew University, Spring94, and Mariusz Rabus for their comments.

Preliminaries

0.1. Let


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WAS SIERPINSKI RIGHT IV? SH546 3

Definition 0.4. []n holds provided that whenever F is a function from []n

to , then there is A of order type and t < such that [w [A]n F(w) = t].

Definition 0.5. []n, if for every function F from []n to there is A of order type such that {F(w) : w [A]n} has power . If we write < instead of we mean that the set above has cardinality < .

Definition 0.6. A forcing notion P satisfies the Knaster condition (has propertyK) if for any {pi : i < 1} P there is an uncountable A 1 such that theconditions pi and pj are compatible whenever i, j A.

What problems do [Sh 276], [Sh 288], [Sh 289], [Sh 473] and [Sh 481] raise? The

most important minimal open, as suggested in [Sh 481] were:

A Question. (1) Can we get e.g. CON(20 [2]23) (generally raise + in part

(3) below to higher cardinals). We solve it here.(2) Can we get CON( > 2

0 [1]23) (the exact n seems to me less exciting).(3) Can we get e.g. CON(2 > [+]23)?

Also

B Question. (1) Can we get the continuity on a non-meagre set for functionsf : 2 2? (Solved in [Sh 473].)(2) What can we say on continuity of 2-place functions (dealt with in Rabus Shelah

[RbSh 585])?(3) What about n-place functions? (continuing in this respect [Sh 288] probablyjust combine [RbSh 585] with)

C Question. (1) [Sh 481] for > 0.(2) Can we get e.g. CON(20 2, and ifP is 20-c.c., Q is 2-c.c., then P Q is20-c.c.).(3) Can we get e.g. CON(20 > > 0, and ifP is -c.c., Q is 1-c.c. then P Qis -c.c.); more general is CON( = 0+ if P is 2-c.c. Q is

+-c.c. thenP Q is 2-c.c).So a large number are solved. But, of course, solving two of those problems does

not necessarily solve their natural combinations.


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4 SAHARON SHELAH

1

We return here to consistency of statements of the form []2,2 (i.e. for every

c : []2 there is A [] such that on [A], c has at most two values), (when2 > , of course). In [Sh 276, 2] this was done for = 0, = 2, =1, 2 < < and quite large (in the original universe is an Erdos cardinal).Originally, [Sh 276, 2] was written for any = 0 case (in [Sh 288,1.4]) and using only k22-Mahlo (for a specific natural number k

22)(an improvement

for = 0 too), explaining that it is like [Sh 289]. A side benefit of the present

paper is giving a full self-contained proof of this theorem even for 1-Mahlo. Themain point of this work is to increase , and this time write it for = 0,too.

The case = + is easier as it enables us to separate the forcing producing thesets admitting few colours: each appear for some < , cf() = +, is connectedto a closed subset a of unbounded in of order type

+, so that below < inP we get little information on the colouring on the relevant set. Here there is lessseparation, as names of such colouring can have long common initial segments, butthey behave like a tree and in each node we divide the set to sets, each admittingonly 2 colours.

As we would like to prove the theorem also for > 0, we repeat material on+-c.c., essentially from [Sh 80], [ShSt 154a], [Sh 288].


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1.1 Definition. : 1) Let D be a normal filter on + to which { < + : cf( = }belongs. A forcing notion Q satisfies D where is a limit ordinal < , if player Ihas a winning strategy in the following game D[Q] defined as follows:

Playing: the play finishes after moves.In the -th move:

Player I if = 0 he chooses qi : i < + such that qi Q

and ( < )(Di < +)pi qi and he chooses a

function f : + + such that for a club of i < +, f(i) < i;

if = 0 let qi = Q, f = is identically zero.

Player II he chooses pi : i < + such that (Di)qi p

i and p

i Q.

The Outcome: Player I wins provided that for some E D: if

< i < j < +

, i , j E, cf(i) = cf(j) = and


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6 SAHARON SHELAH

1.3 Definition. 1) Let F = F : < be a strategy for player I in the

game D[Q] for = 1, 2. We say F1 F2 equivalently, F2 is above F1 if any

play (q, f , p) : < in which player I uses the strategy F2 (that is letting

(qi : i < +, f) = F(p : < ) we have i < + qi qi and for someE D, i E & j E f(i) = f(j) f(i) = f(i)) is also a play in which playerI uses the strategy F1.

2) Let < < , St be a winning strategy for player I in the game Q. We say

F : < is an increasing sequence of strategies of player I in D[Q] obeyingSt if:

(a) F is a winning strategy of player I in D[Q]

(b) for < < , F is above F

(c) if(q , f , p) : < is a play ofD[Q], Player I uses his strategy F, then

for any i < +, letting F

(p

: < ) = (q,

, f,) we have:

Q |= St(q,i : < ) q,i .

3) Similarly to (1), (2) for the game Q (instead D[Q]).

1.4 Observation. 1) Assume Q is -complete. If < and F : < is anincreasing sequence of winning strategies of player I in D[Q], then some winningstrategy F of player I in D[Q] is above every F

( < ).2) Assume < and Q is -strategically complete with a winning strategy St.

If < and F : < is an increasing sequence of winning strategies of playerI in D[Q] obeying St, then for some F

, F : < + 1 is an increasing sequenceof winning strategies of player I in D[Q] obeying St.3) Similarly with Q instead of

Q[D].

Proof. Straight.

1.5 Definition. Assume P, R are forcing notions, P R, P R.1) We say is a restriction operation for the pair (P, R) (or (P,R, ) is a strongrestriction triple) if (P, Q are as above, of course, and) for every member r R, r

P P is defined such that:

(a) r P r,

(b) ifr P p P then r, p are compatible in R in fact have a lub

(c) r1 r2 r1 P r2 P

(d) ifp P then p P = p and Q P = P

(so this is a strong, explicit way to say P R).1A) We say weak restriction triple if we omit in clause (b) the have a lub.2) We say (P,R, ) is -strategically complete if

() is a restriction operation for the pair (P, R)() P is -strategically complete

() if St1 is a winning strategy for player I in the game,

p, then in the game = [P, R; St1] the first player has a winning strategy St2.


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Playing: Just like a play of [R], except that

in addition, for every limit ordinal < , in the -the move first the secondplayer is allowed to choose r

i : i < +

such that:


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A play last moves, in the -th move player I chooses x P such that +.

During the play, player I in the -th move also chooses an ordinal , increasescontinuously with , 0 = 0 as follows:

+1 = min{ < : (i < +)( )(pi , q

i P)}

and he will make qi P , and the rest is as in Case 3.

Case 5: cf() = +.

Let : < + be increasing continuously with limit , 0 = 0, cf() ,

and we imitate Case 4, separating to different plays according to the value of

ji = Min{j < i : for each < we have pi i Pj and q

i i Pj}. 1.16

1.17 Claim. Assume

(a) Q = P, Q, a, I : <

(b) a limit ordinal(c) for every < we have Q K,.

Then Q K,.

Proof. Check.

1.18 Claim. Assume

(a) Q K,

(b) a is Q-closed, |a| <

(c) I {b a : b is Q-closed }

(d) I is closed under finite unions, I is closed under increasing unions oflength < and I

(e) Q is a P

a

-name of a forcing notion of cardinality <

(f) if b I then (Pb, Pa

Q) satisfies

(g) P = LimQ.

Then QP, Q, a, I belongs to K,

+1.

Proof. Check.


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1.19 Theorem. Suppose = 2,) we know that in VR, is still measurable. LetQ = P, Q

, a : < be

GR and P be the limit so P

= P P is a

dense subset, those are R-names. Now R P

is the forcing P we have promised.

The non-obvious point is RP

[]2,2 (where < , < ). So suppose

(r, p

) R P

and (r

, p

) the colouring

: []2 is a counterex-

ample. Let 1 = (2)+. Let GR R be generic over V, r

GR. By [Sh289], but the meaning is explained below in VR we can find an end extensionstrong (1,,, 2

++2, ( + + 2)+, )-system M = Ms : s [B] +, then in VP , each subset of QQ

+1

, of cardinality + is

included in the union of sets, each directed and with any two elements

having a lub.

Note that by the definition of QQ

, we have

()6 a family of < members of QQ

, has a common upper bound iff any twoof them are compatible, and then the union is a lub of the family.

So if cf() > +, we are done as by ()5 + ()6 we have P Q satisfies and

can use 1.7(4).

So we can assume = () + 1 and cf() = cf() = +, and let

(i) : i < + be increasing continuous with limit and cf((i)) for i < +.Let b I , hence b is a bounded subset of a . So by the induction hypothesis and1.7(4) without loss of generality b =

{M{0 ,1} : 0 < 1 < (0)}.


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Define c0 = b0 = b and for [(0), ()) let

b1, = b0

M{}

1

pi b0 if (i)

q2,,i =

qi b1, if (i) >

qi b0 if (i)

we have:

p2,,i : i < +, q2,,i : i <

+, f2,, : <

is a play of[P

b0

, Pb1,] in which the first player uses the strategy St1,.

(c) For any pair = (1, 2) of ordinals in , let

(i, ) = (i) is the 1-th member of Dom(q2i )\(i)


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20 SAHARON SHELAH

p3,,i = OPb1,(0),b1,(i)

(pi b1,(i))

q3,,i = OPb1,(0),b1,(i)

(qi b1,(i)),

we demand that

p3,,i : i < +, q3,,i : i <

+, f3,, : <

is a play of[P

b0

, Pb1,(0)] in which the first player uses the strategy St1,(0).

So for each i < , for 1 < too large (i, ) is not well defined and we stipulatethe forcing conditions are .

(d) f(i) codes f1,(i), f2,,(i) : [(0), ()) and ( b1,\b0)[pi () =

0Q

] and f3,,(i) : , and (i) is well defined and the informa-

tion on pi (()) and it codes

j1, 1, 2 :, the 2-th member of Dom(p

i )

satisfies : j1 = Min{j : Dom(pj)},

is the 1-th member of Dom(pj1

)

and

j,1, 2 : for some , , the 1-th member of Dom(p

i ),

belongs to b1,\b0 and satisfies :

j = Min{j : (Dom(pj) (b1,\b0) = }

and the 2-th member of Dom(pj) belongs to b1,\b0

(note: for each 2 < , i < + we have:{1 < : (1,2)(i) is well defined} is a bounded subset of ).

Check that such St2 exists, (note that the number of times we have to increasepi b0 is < ).

Clearly c2 c3 are Q-closed, hence there is a winning strategy St3 of the firstplayer in [P

c2

, Pc3 ] above St2.

(e) For any = (1, 2) such that (0) 1 < 2 < (), and defining

p4,, = pi b2,,

q4,,i = qi b2,(can behave similarly in clause (b)),

we have:

p4,,i : i < +, q4,,i : i <

+, f4,, : <

is a play of [Pb2,

] in which the first player uses the strategy St2,.


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Lastly, let St4 be a strategy of the first player in [Pc3

, Pc4 ] which is above St3and it guarantees:

() if pi

: i < +, q

i

: i < +, f4

: < is a play of the game in whichthe first player uses his strategy St4 then:

() qi a forces a value to pi (())

() if 1 = 2 are from (the value forced on) qi (()) then q

i a is

above the relevant parts of witnesses to this.

Clearly St4 is (essentially) a strategy of the first player in [Pb0

, Pc4 ] (for thealmost case above St0). All we have to prove is that St4 is a winning strategy.

So let pi : i <

+, qi : i < +, f4 : < be a play of

[P

b0

, Pc4 ] in which

the first player uses the strategy St4.By the definition of the game [Pb0

, Pc4 ] without loss of generality for some club

E1 of + (see clause (a)):

()1 if {i, j} S+

E1 (see ?) and> scite{0.I} undefined


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22 SAHARON SHELAH

Last

()5 E is a club of+ included in E1 E2 E3 E4 such that:

i < j E Dom(p

i c3) Dom(q

i c3) (j).

The rest is as in [Sh 276, 2]. 1.12

1.21 Theorem. We can in 1.19 replace measurable, by (strongly) Mahlo.

1.22 Remark. It is not straightforward; e.g. we may use the version of squareddiamond given in Fact 1.24 below.

We first prove two claims.

1.23 Claim. Suppose is a strongly inaccessible Mahlo cardinal, > > = sup{N{j1,i1} : j1 < i1 < i}

(h) Ns, Ns have cardinality when defined.

Proof. Let 1 = 2, 2 = 2

1 . Let A and be such that:

strongly inaccessible

A L+2,2C

A . Next choose


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Bi L+2,+2

C, increasing continuous in i for i < , Bi : i j Bj+1, Bj sup(a )

(c)

C = C : a(d) C = cf() & otp(C)

(e) C a & C = C

(f) C = C closed

(g) A = A : a

(h) A is a P-name of a subset of

(i) C A = A

order p q iff ap aq, Cp = Cq ap, Ap = Aq ap.

2) [Claim]: < P P.3) [Claim]: If p P, < , then p = (ap , C (a ), A (a )) belongsto P and: ifp q P then p, q are compatible in a simple way: p & q is alub of{p,q}.

4) [Claim]: If is strongly inaccessible and > then P =


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(b) C S, [ C C = C ], otp(C) , C a closed subset of so [sup(C) = C has no last element)

(c) B , otp(B) = otp(C), C B = B

(d) each Ms : s [B]2 is as in 1.23 (and B B) and C & s [B]

2 Ms = Ms

(e) diamond property: ifB is an expansion of (H(), , >

(c) kn as in [Sh 228] (see below).

Then for some +-c.c. forcing notion P of cardinality we have:

P 2 = []n

+1kn+1

, moreover for < ,

[]n+1

,kn.

1.27 Remark. 1) What is kn?

Case 1: = 0; define on [2]n

an equivalence relation E: if w1 = { : 0.Choose


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REFERENCES.

[JMMP] Thomas Jech, Menachem Magidor, William Mitchell, and Karel Prikry.On precipitous ideals. J. of Symb. Logic, 45:18, 1980.

[L] Richard Laver. Making the supercompactness of indestructible under-directed closed forcing. Israel J. of Math., 29:385388, 1978.

[RbSh 585] Mariusz Rabus and Saharon Shelah. Covering a function on theplane by two continuous functions on an uncountable square - theconsistency. Annals of Pure and Applied Logic, 103:229240, 2000.math.LO/9706223.

[Sh 80] Saharon Shelah. A weak generalization of MA to higher cardinals.Israel Journal of Mathematics, 30:297306, 1978.

[Sh 228] Saharon Shelah. On the no(M) for M of singular power. In Aroundclassification theory of models, volume 1182 ofLecture Notes in Math-ematics, pages 120134. Springer, Berlin, 1986.

[Sh 276] Saharon Shelah. Was Sierpinski right? I. Israel Journal of Mathemat-ics, 62:355380, 1988.

[Sh 289] Saharon Shelah. Consistency of positive partition theorems for graphsand models. In Set theory and its applications (Toronto, ON, 1987),volume 1401 ofLecture Notes in Mathematics, pages 167193. Springer,Berlin-New York, 1989. ed. Steprans, J. and Watson, S.

[Sh 288] Saharon Shelah. Strong Partition Relations Below the Power Set:Consistency, Was Sierpinski Right, II? In Proceedings of the Con- ference on Set Theory and its Applications in honor of A.Hajnal andV.T.Sos, Budapest, 1/91, volume 60 of Colloquia Mathematica Soci-etatis Janos Bolyai. Sets, Graphs, and Numbers, pages 637668. 1991.math.LO/9201244.

[Sh 473] Saharon Shelah. Possibly every real function is continuous on a nonmeagre set. Publications de LInstitute Mathematique - Beograd, Nou-velle Serie, 57(71):4760, 1995. math.LO/9511220.

[Sh 481] Saharon Shelah. Was Sierpinski right? III Can continuumc.c. timesc.c.c. be continuumc.c.? Annals of Pure and Applied Logic, 78:259269, 1996. math.LO/9509226.

[ShSt 154a] Saharon Shelah and Lee Stanley. Corrigendum to: Generalized Mar-tins axiom and Souslins hypothesis for higher cardinals [Israel Jour-nal of Mathematics 43 (1982), no. 3, 225236; MR 84h:03120]. IsraelJournal of Mathematics, 53:304314, 1986.

saharon shelah- was sierpinski right iv?

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