saharon shelah- on nicely definable forcing notions

8/3/2019 Saharon Shelah- On Nicely Definable Forcing Notions

1/22

(711)

revision:2004-03-

05

modified:2004-0

3-07

ON NICELY DEFINABLE

FORCING NOTIONS

SH711

Saharon Shelah

The Hebrew University of JerusalemEinstein Institute of Mathematics

Edmond J. Safra Campus, Givat Ram

Jerusalem 91904, IsraelDepartment of MathematicsHill Center - Busch Campus

Rutgers, The State University of New Jersey110 Frelinghuysen Road

Piscataway, NJ 08854-8019 USA

Abstract. We prove that ifQ is a nw-nep forcing then it cannot add a dominatingreal. We also show that amoeba forcing cannot be P(X)/I if I is an 1-completeideal. Furthermore, we generalize the results of [Sh 480].

I would like to thank Alice Leonhardt for the beautiful typing.This research was partially supported by the Israel Science Foundation founded by the IsraelAcademy of Sciences and Humanities.Publication 711First Typed at RU - 01/10/24Latest Revision - 04/Mar/5

Typeset by AMS-TEX

1


2/22

(711)

revision:2004-03-

05

modified:2004-0

3-07

2 SAHARON SHELAH

0 Introduction

Nicely definable forcing notions have been studied since the mid-eighties, espe-

cially for the case when nicely definable was interpreted as Souslin (see, e.g.,[Sh 480], Judah and Shelah [JdSh 292] or Goldstern and Judah [GoJu]). Recently,in [Sh 630], we have initialized investigations of a wide class of reasonably de-finable forcing notions which satisfy the properness demand for countable modelswhich are not necessarily elementary submodels of some H(): the nep forcings.The present paper continues that research in two directions.

In the first section we introduce a very strong variant of nep forcing, where thethe candidates (i.e., the models for which we postulate the existence of genericconditions) do not have to be well-founded (Definition 1.1). We show that thoseforcing notions, called nw-nep, cannot add dominating reals (Theorem 1.4). Then,

by a similar proof, we show that a proper forcing notion which adds a dominatingreal and has sufficient amount of absoluteness for being predense, must force thatb = 1 (see Theorem 1.6). This result applies to forcing notions like Amoeba formeasure, the Hechler forcing and the Universal Meagre forcing (see Corollary 1.7),so in some sense it continues older work of Brendle, Judah and Shelah [BJSh 477]and Brendle [Bn95]. As a conclusion to 1.7 we get that a Boolean Algebra P(x)/J,where J is a 1-complete ideal on x, cannot be isomorphic to the Boolean Algebraof the Amoeba forcing or the Universal Meagre forcing. This answers a questionof Kamburelis (though this solution was obtained already in 1977 and discussed in[Sh 666, 4]).

In the second section we try to extend the results of [Sh 480] to nep forcing.There we showed that if a Souslin c.c.c. forcing notion Q adds an unbounded real,then it adds a Cohen real. Here we weaken the demands on Q (it is just nep c.c.c.),but the Q-name for a Cohen real is constructed in VP, where P is a forcing notionadding a dominating real and preserving Q-candidates.

We refer the reader to [Sh 666, 4] and [Sh 630] for more background and refer-ences. This paper will be continued in [Sh:F642].

We would like to thank Jakob Kellner and Andrzej Roslanowski for reading thepaper extremely carefully resulting in considerable improvement of the presentationand readability.


3/22

(711)

revision:2004-03-

05

modified:2004-0

3-07

ON NICELY DEFINABLE FORCING NOTIONS SH711 3

1 Nwnep forcing notions

In [Sh 630] we introduced nep (non-elementary proper) forcing notions as the

ones with reasonable definitions and such that the generic conditions exist overmany countable models (not necessarily elementary submodels ofH()). Still,those models (called candidates) were well-founded, see [Sh 630, 1] for details.

Here we consider a related property, allowing the candidates to be non-wellfounded (so the new notion has a flavour of a stronger property). The defini-tion below is ad-hoc to simplify the presentation. The nw stands for non-wellfounded, of course.

1.1 Definition. 1) We say that N is x 1-nw-candidate if, fixing some stronglimit , (a) or (b) holds where:

(a) N (H(), ) is countable, x N(b) for some N1 as in (a), N is an elementary extension ofN1 not increasing

N;i.e., ifN |= g < then g N1, and ifN |= y is a subset ofH(0) theny = {n N : N |= n y} (so really it should have one two-place relationE, EN is the membership relation in N; but we shall write N |= x y).

2) We say that N is a standard x 2-nw-candidate if (for as above) (a) holds or

(b) for some N1 as in (a), N is a forcing extension of N1 (and the demand in(b) on subsets ofH(0) holds).

3) Let Q

2 or just Q P(H(0)). We say that Q is a 1-nw-nep forcing notionifQ is a pair of formulas = (0(x), 1(x, y)), in the language of set theory (withparameter r) such that (below we write Q-candidate instead ofr 1-nw-candidate):

(a) 0(x) defines a set of reals (= set of members ofQ)

(b) 1(x, y) defines a set of pairs of reals, a quasi order on {x : 0(x)}, this isQ

(c) 0, 1 are 11-formulas; equivalently

1, are upward absolute from Q-candidates,i.e., for Q-candidates N1 N2, x Q

N1 x QN2 x Q andx N1Q y x

N2Q y x Q y

(d) if N is a Q-candidate and p QN

(i.e. N |= p Q) then there is q suchthat: q Q, p Q q and q is N, Q-generic, i.e.

q QN G is a subset ofQN, directed by NQ and G IN =

whenever N |= I Q is predense.

1because the candidates here are not necessarily well founded


4/22

(711)

revision:2004-03-

05

modified:2004-0

3-07

4 SAHARON SHELAH

4) We say Q is 2-nw-nep-forcing if above we replace r 1-nw-candidate by r 2-candidates.5) Let nw-nep mean 1-nw-nep.

The examples of nw-nep forcing notions include all -bounding forcing notionsfrom [RoSh 470], that is the class Kdefined in [BaRo02, Def.0.4]. So, in particular,the Silver forcing notion and the Sacks forcing notion are nw-nep. In the realm ofc.c.c. forcings, the natural examples of nw-nep are the Cohen forcing and therandom real forcing. They both are nw-nep because they are very Souslin c.c.c.,where:

1.2 Definition. A forcing notion Q is very Souslin c.c.c. if it is Souslin c.c.c. andthe relation

rn : n < is a maximal antichain in Q

is 11.

1.3 Proposition. Very Souslin c.c.c. forcing notions are nw-nep.

Roslanowski and Shelah [RoSh 672, 1.3.4](3),1.5.15 give more examples of very

Souslin c.c.c. (and thus also nw-nep) forcing notions.1.4 Theorem. Assume Q is nw-nep. Then forcing withQ does not add a domi-nating real.

Proof. Toward contradiction assume p Q

is a dominating real.

Without loss of generality, p

is strictly increasing,

(n) > n. Let

0 = { > : strictly increasing and () > for < g()};

so p Q

lim(0). As Q is nw-nep there is p such that

1 p Q p and for each n there is a countable Jn Q which is an antichainpredense above p, such that each p Jn forces a value to

(n) and is

above p and above some p Jm for each m < n.


5/22

(711)

revision:2004-03-

05

modified:2004-0

3-07


[Why? Take a countable model N (H(), ) such that r,

, p N (so N is a

Q-candidate). Inside N by induction on n < we choose Jn N as above exceptcountability. Let p be above p and be N, Q-generic.]

Clearly above any p p there are two incompatible elements ofQ, so withoutloss of generality

2 ifm < n and p Jm then there are infinitely many members ofJn which

are above p.

Let denote a subset of 0 closed under initial segments such that and (n)(n ). We can find , k and choose p such that:

3() p = p : and p Q (in fact p

{p

}

n

Jn ),

() Q |= p p ,() for n [1, ) we have: p :

n is an antichain ofQ predense abovep,

() k = k : where k < ,

() if >, m, n , then pm,n forces a value to

(m), which we

call k and n > km,n > m,

() p = p (= no information) and k = 1.

So we choose n and k, p (for n) by induction on n with =

p n

Jn .

For n = 0 let p = p, k = 1.

For n = 1 we declare that 1 = {< m >: m > 0}, p : m > 0 is anenumeration ofJ0 and k = 1.

For n + 1, n 1, for each n let m = sup Rang() and let p,j : j < listthe members ofJm which are above p, so for some k,j we have p,j Q

(m) =

k,j . Let f : be strictly increasing such that k,j < f(j) and m < f(j)for j < , and lastly, let n+1 = {f(j) :

n and j < } andk = k,j and p

f(j) = p,j .

Let be the Q-name of the -branch of such that p Q p

n GQ for eachn < . We claim that:

if h : , then

p Q for every large enough n < we have

(n) > h(

n).


6/22

(711)

revision:2004-03-

05

modified:2004-0

3-07

6 SAHARON SHELAH

[Why? Let fh : be

fh(n) = sup{h() : and sup Range() n} + 1,

Note that the supremum is over a finite set as every is strictly increasing. Soassume p G Q, G is generic over V, =

[G], =

[G], and we shall find

n as required. Clearly for some n > 2 we have m [n, ) fh(m) < (m). Weshall prove that m [n, ) h( (m + 1)) < (m + 1).

So assume m [n, ), then:h( (m + 1)) fh((m)) by the definition of fh, being increasing,fh(

(m)) < ((m)) as (m) m n and the choice of n,((m)) = k(m+2) by clause () andk(m+2) <

(m + 1) by ().]

For a limit ordinal < 1, let = { : = : where is limit andeach is a (strictly) increasing -sequence converging to }. For , we definea function g from to + 1, defining g() by induction on lg() as follows:

(B)(a) g() = ,

(b) if g() = + 1 and , then g() = ,

(c) if g() = , a limit ordinal and , then g() = (),

(d) if g() = 0 and then g() = .

Let An, = { : g() = and |{ < lg() : g( ) = }| = n}, so An, is afront of , and it is above Am, for m < n. Hence for each m we have p

has

an initial segment in Am,. Let a Q-name h of a function from to be such

that v An, pv h(n) =

(sup Rang(v)). Clearly In, = {p : An,} is

predense above p, so2 p Q h .

Now there is a Q-candidate N, with (1)N not well ordered, and p, pp N

such that N |= all the above statements on p, .[Why? Let L be a countable fragment ofL1,({}) and let N1 L (H(), ) besuch that Q, p, p N1 and N1 L N, N as above;

N = N1 as in the Definition1.1. See Keisler [Ke71]; alternatively, for a model N1 with Skolem functions define

PN1 = {Y : for some n = n(Y), Y is a set of

decreasing sequences of countable cardinals such that if

n(Y) > 0 then ( < 1)( Y)( < min Rang())}

2the h-s are here for clarification only, but they will be necessary in the proof of 1.6 below.


7/22

(711)

revision:2004-03-

05

modified:2004-0

3-07


Y1 Y2 iff n(Y1) n(Y2) & ( Y2)( Y1)( ).

If GPN1 is generic over N

1, we define N such that N

1 N,

n N, N =

Sk(N1 {n : n < }) and Y G 0, . . . , n(Y)1 Y.]Choose n for n < such that N |=

n+1 <

n are countable ordinals.

Without loss of generality N |= n is a limit ordinal, and so for some we haveN |=

0. So clearly N |= In, is predense above p for each n. By

1.1(3)(d), there is r Q above p which is N,Q-generic. Then

1 INn, is predense above r

for each n < .

There is in V (not in N!) a function f0 : such that

n f0(

). So for some 1, q forces that

1 < <

n

n 1

2 +1k

for some k (0, )

() limsupam,k : m < 12

1k

for some k (0, ).

If () holds, let r exemplify it, so for some 1 > 0, Leb(lim(r)) >12 + 1, and

Leb(lim(p r)) 12 for < n. We can find, for < n, a clopen subset B of

2such that Leb(lim(p r) B) > 0, Leb(B) from V we have P (n)g

(n) > h(g

n).

[Why? As, e.g., we can replace g

by

, where

(0) = g

(0) and

(n+1) = g

(

(n)+1),

note that

is strictly increasing as g

is.]

Let the Q-name

1 2 be such that for every < we have

1() = 1

Rang(f

1) and let

2 =

.

()6 N |= PQ t

2 is new.

[Why? For t = 1 by ()3, for t = 2 even easier immitating the proofs of ()3 and()4.]

For q Q let

3can use the ord collapse ofM


16/22

(711)

revision:2004-03-

05

modified:2004-0

3-07

16 SAHARON SHELAH

Tq[

t] = { >2 : q Q

},

so()7 Tq[

t] is a non empty subtree of>2 with no maximal nodes and no isolated

-branches,

()8 if t = 1, then in V and also in N[G] for each generic G PN over N, forevery q Q

(a) for some n < :for every m (n, ) we have

q Q ()(n < m &

t() = 1);

therefore

(b) for every m > n there is an m2 Tq [

t] such that for every

[n, m), we have () = 0,

(c) for some lim(Tq[t]), the restriction [n, ) is constantly zero.

[Why? By ()3.] Consequently,

()9 t = 1: q1, q2 Q are incompatible if for no n < and lim(Tq1 [

t])

lim(Tq2 [

t]) do we have [n, ) () = 0 and ()( 1

Tq1 [

t] Tq2 [

t])

t = 2: q1, q2 Q are incompatible if lim(Tq1 [

t]) lim(Tq2 [

t]) has finitely

many -branches.

We say that u []0 is large for (q,

t) if for every r Q above q the following

holds:

r,u Case t = 1: For some n u, for every n, m u such that n < n < m,

for some m2 Tr[

t] we have [n, m) () = 0 but ()(n 2, J0 = {}, k Jk, Jk+1 =(Jn\{k}) {k < 0 >, k < 1 >} and n < & Jn (m n)(m = )and k < n < g(k) g(n). Moreover, we require that if < g(k) =


18/22

(711)

revision:2004-03-

05

modified:2004-0

3-07

18 SAHARON SHELAH

g(n), k = n and k() = 0, n() = 1, then k < n. We choose pk =

p, k, m, n : Jk by induction on k such that (after we choose pk we have

already chosen pk+1 (Jk+1\{k < 0 >, k

< 1 >})):

0 (a) p P p P

N

(b) m < n

(c) if m > n then for some q PN we have p P q and

q P (i)(ni n m

i > m)

(d) pk P pk Ig(k) for = 0, 1

(e) pk P (i)(ni nk & m

i > mk)

(f) mk > sup{n : Jk}.

Let us carry out the induction.In step k = 0 let p = p

, n is chosen as below, m is immaterial. If wehave defined pk, first choose mk to satisfy clause (f), then choose p

k

pk to satisfy clause (e) (possible by clause (c)) and choose pk p

k

to

satisfy clause (d). Lastly, choose nk to satisfy clause (c); this is possible bythe observation below.

For each 2 let G = {p PN : p P p for some < }, clearly G( V)is a subset ofPN generic over N (by clause (d)). Now q = q

[G] and T = Tq[

t],

are well defined in N[G] hence in V. It is easy to see the following.

1 Assume that 1 = 2 2 and 1 k = 2 k. Then() if n > n1k, n2k then for some i k we have

n [n1i, m1(i+1)) or n [n2i, m2(i+1))

() ift = 2 and T1 T2 and g() > n1k, n2k, then has at mostone successor in T1 T2 .

[Why? Clause () follows from clauses (b) + (f) of0. Clause () followsfrom clause () and ()10(b).]

Hence

2 () if t = 1 and 1 = 2 2 and lim(T1) lim(T2) and () = 0for every < large enough then the set

{ < : 1 lim(T1) lim(T2)}

is finite

() if t = 2 and 1 = 2 2 then lim(T1) lim(T2) is finite.


19/22

(711)

revision:2004-03-

05

modified:2004-0

3-07


Now for each 2, N[G ] KP and N[G ] |= q Q, hence there is q+ Q

which is N[G ],Q-generic and Q |= q q+ . Hence really q+ Q

t

lim(T), so by 2 and ()9 we have

3 if 1 = 2 2 then q+1 , q

+2

are incompatible in Q.

But this contradicts assumption (a) of 2.2, i.e., the c.c.c. Subclaim

Observation. Assume

() p P ni < n

i+1 < , n

i < m

i < for every i < and for every h

V

for infinitely many i we have h(ni) < m

i.

Then we can find n < such that for every m [n, ) there is q satisfyingp P q and q (i)(n

i n

& mi m).

Proof. We define a function h : ( + 1) by

h(n) = sup{m :n < m and for some q P we have

p P q and q P (i)(ni n m m

i)}.

If for some n, h(n) = we are done. Otherwise h and hence p P(i)(h(n

i) < m

i). So there are n,m,q,i such that p P q and

q h(ni) < m

i & n

i = n & m

i = m.

But by the definition ofq, q witnesses that h(n) m, a contradiction.observation

Continuation of the proof of 2.2:

The conclusion of the subclaim holds in V as N (H(), ), and this gives theconclusion of part (2) of 2.2 when t = 2, and the conclusion of part (1) of 2.2 whent = 1 is similar to [Sh 480, 1.12,p.168] but we give details. By the subclaim, asN (H(), ), clearly in VP we have: for every q Q some infinite u w islarge for (q,

t). Fix such q, u. We concentrate on t = 1 as the case t = 2 is

obvious by this point.


20/22

(711)

revision:2004-03-

05

modified:2004-0

3-07

20 SAHARON SHELAH

Let u\{0} = {ni : 1 i < } be such that n0 =: 0 < n1 < n2 < . . . , letk(i, ) : < be such that i = k(i, )2

where k(i, ) {0, 1}, so k(i, ) = 0when 2 > i. For m < let m = k(i, ) : [log2(i + 1)] where i = iu(m)

is the unique i such that ni m < ni+1. We define a Q-name (of a member

of (2)VQ

): let {ki : i < } list in the increasing order the elements of the set

{k < :

t(k) = 1} and

be k0

k1

k2

. . . .

Clearly for every p Q and n < we have p

t(k) = 0 for every k n.

Hence Q {k < :

t(k) = 1} is infinite, hence Q

2.

It is enough to prove that q Q

is a Cohen real over VP. So let T VP be

a given subtree of >2 which is nowhere dense, i.e., ( T)()[ >2\T],

and we should prove q

Q / lim(T). So assume q

q Q and we shall findq, q q Q such that q Q

/ lim(T), this suffices. We apply the choice

of u so for some n u, if n, m u, n < m < n then for some n2 Tq[

t]

we have (n, m) () = 0 b u t ()(m < n &

() = 1]. As

n u for some i() we have n = ni() and =: {m : m < n} is finite hence =: {k0

. . .k1 : k0 < . . . < k < n} is finite. As T is nowhere dense wecan find a sequence >2 such that: / T and choose i > i()such that ni . This is possible by the definition of

ni

, i.e., it is enough that

i > 2g() and i = {()2 : < g()} mod 2g(

).As said above we can find q q such that q if n < ni then

t() = 0

but for some [ni, ni+1) we have

t() = 1. So q forces that ni appears in the

choice of

and before it we have a concatenation of finite sequences which belong

to , so we are done. 2.2


21/22

(711)

revision:2004-03-

05

modified:2004-0

3-07


REFERENCES.

[BaRo02] Tomek Bartoszynski and Andrzej Roslanowski. Towards Mar-tins Minimum. Archive for Mathematical Logic, 41:6582, 2002.math.LO/99041634

[Bn95] Joerg Brendle. Combinatorial properties of classical forcing notions.Annals of Pure and Applied Logic, 73:143170, 1995.

[BJSh 477] Jorg Brendle, Haim Judah, and Saharon Shelah. Combinatorial proper-ties of Hechler forcing. Annals of Pure and Applied Logic, 58:185199,1992. math.LO/9211202.

[GiSh 357] Moti Gitik and Saharon Shelah. Forcings with ideals and simple forcingnotions. Israel Journal of Mathematics, 68:129160, 1989.

[GiSh 412] Moti Gitik and Saharon Shelah. More on simple forcing notions andforcings with ideals. Annals of Pure and Applied Logic, 59:219238,1993.

[GiSh 582] Moti Gitik and Saharon Shelah. More on real-valued measurable cardi-nals and forcing with ideals. Israel Journal of Mathematics, 124:221242, 2001. math.LO/9507208.

[GoJu] Martin Goldstern and Haim Judah. Iteration of Souslin Forcing, Pro-jective Measurability and the Borel Conjecture. Israel Journal of Math-ematics, 78:335362, 1992.

[JdSh 292] Jaime Ihoda (Haim Judah) and Saharon Shelah. Souslin forcing. TheJournal of Symbolic Logic, 53:11881207, 1988.

[Ke71] H. Jerome Keisler. Model theory for infinitary logic. Logic with count-able conjunctions and finite quantifiers, volume 62 of Studies in Logicand the Foundations of Mathematics. NorthHolland Publishing Co.,AmsterdamLondon, 1971.

[RoSh 470] Andrzej Roslanowski and Saharon Shelah. Norms on possibilities I:forcing with trees and creatures. Memoirs of the American Mathemat-ical Society, 141(671):xii + 167, 1999. math.LO/9807172.

[RoSh 672] Andrzej Roslanowski and Saharon Shelah. Sweet & Sour and otherflavours of ccc forcing notions. Archive for Mathematical Logic, 43:583663, 2004. math.LO/9909115.

[Sh:F642] Saharon Shelah. More on nw-nep forcing notions.

4References of the form math.XX/ refer to arXiv.org


22/22

711)

revision:2004-03-

05

modified:2004-0

3-07

22 SAHARON SHELAH

[Sh 480] Saharon Shelah. How special are Cohen and random forcings i.e.Boolean algebras of the family of subsets of reals modulo meagre or null.Israel Journal of Mathematics, 88:159174, 1994. math.LO/9303208.

[Sh 666] Saharon Shelah. On what I do not understand (and have some-thing to say:) Part I. Fundamenta Mathematicae, 166:182, 2000.math.LO/9906113.

[Sh 630] Saharon Shelah. Properness Without Elementaricity. Journal of Ap-plied Analysis, 10:168289, 2004. math.LO/9712283.

saharon shelah- on nicely definable forcing notions

Documents