Saharon Shelah- Historic Iteration with ℵε-Support

8/3/2019 Saharon Shelah- Historic Iteration with -Support

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arXiv:math/9

607227v1

[math.L

O]15Jul1996

HISTORIC ITERATION

WITH -SUPPORTSH538

Saharon Shelah

Institute of MathematicsThe Hebrew University

Jerusalem, Israel

Rutgers UniversityDepartment of Mathematics

New Brunswick, NJ USA

Abstract. One aim of this work is to get a universe in which weak versions of Martin

axioms holds for some forcing notions of cardinality 0, 1 and 2 while on 2 club,the small brother of diamond, holds. As a consequence we get the consistency ofthere is no Gross space. Another aim is to present a case of Historic iteration.

I thank Otmar Spinas, Juris Steprans and Boban Velickovic for the helpful discussion.I thank Alice Leonhardt for the beautiful typing.Latest Revision 96/Aug/2

Originally typed 94/Jan/23 elsewhere

Typeset by AMS-TEX

1
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2 SAHARON SHELAH

1

When we make the continuum > 2, it in general takes effort not to add 2Cohen reals. We do it here, making a weak version of MA true without adding 2

Cohen reals (i.e. a function f : 2 {0, 1} which is quite generic for the forcingnotion {g : g a finite function from 2 to {0, 1}}). Moreover S20 is preserved.This was motivated by the following application to Gross spaces: the consistencyof there is no Gross space with ZFC, prove in 1.24 below; on Gross spaces seee.g. Shelah Spinas [ShSi 468].

We prove that for a restricted enough family of -centered forcing, (essentiallygiving almost intersection to definable filters) we have:

(a) for any 1 condtions, 1 of them belongs to a directed subset(b) for any 2 conditions, some 2 of them belongs to a directed subset(c) 2(d) 20 = 3.

This is enough; we can by the proof strengthen (a) (even to MA1(-centered)),it is less clear about (b). This was our first approach, but it seems considerablyclearer to prove just (b).

Definition 1.1 defines the family of c.c.c. forcing notions for which we will haveour approximation to (weak) MA. A main Definition is 1.11, where we define thefamily of iterations we shall use. For preserving S20 we want, among any given2 conditions pi : i < 2 to find (quite many) countable subsets w 2 suchthat {pi : i w} has an upper bound. So we have to marry the c.c.c. withsomewhat countable support hence the name -support. The support is fashionedafter historic forcing (see Shelah Stanley [ShSt 258]). We may continue thisgeneralization to a larger family of forcing, replacing 0 by a larger cardinal (see1.3(5)).

Note: S20 for some A 2 there is no L[A]-generic 2-sequence of Cohenreals p 2 (where p 2 means for some Ai D (for i < 2), D is a filter on containing all co-finite sets, there is no A []0 such that (i < 2)[A Ai]).

In 1.1-1.2 we deal with simple forcing, a very restricted class but sufficient forour need. In 1.3 we comment on generalizations, in 1.4 we fix some cardinal pa-rameters. In 1.5 - 1.10 we deal with creatures, they represent possible supportof a condition in the iteration to be defined later. In 1.11 we define our iterationand prove basic facts by simultaneous induction of length. In 1.12 - 1.18 we furtherinvestigate the iteration. In 1.19 - 1.23 we get the actual consistency results.

Notation: Let u, v denote creatures (see Definition 1.5 below), p, q, r denote condi-tions, (in the iteration), , , , denote countable ordinals (for depth of creatures

or conditions) and , , , denote ordinals. We use ,, , only as in Definition1.5.

The following looks extremely special but includes the forcing notion needed forthe application to Gross spaces. See Example 1.3(3); we may consider some gener-alizations.


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HISTORIC ITERATION WITH -SUPPORT SH538 3

1.1 Definition. 1) A forcing notion Q is a simple forcing notion if:

(a) for some S n


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4 SAHARON SHELAH

(a) S = {s : for some n, s is a function1 from {0, . . . , n 1} to U( ) suchthat s() : Dom(s) and s() = 0U} is linearly independent

(b) (s1, t) (s2, t2) iff s1 s2, t1 t2 and[n Dom(s2)\Dom(s1) & t t & Rang(t) K tHOM(s2(n)) = 0].

Here the following f, g are O.K.: f(n, m) = n + m + 1, g(s, {t k : < m}) = first

x U under the order of the natural numbers such that: x has the form and to allow todivide the set to finitely many convex parts, and this is done here.3) Can we get MA1 [-centered]? seems yes.4) We may use more strongly the 20 = 1 in the end: k = 2 the u are sub-model of some B.5) We can generalize replacing 0 by = , {1, 2} +, we use a strong version of +-c.c.(e.g. as in [Sh 80]).

1we could demand: if n1 < n2 < n,s(n) =m

amxm U\{0U}; then max(support(s(n1)) 1]

an ordinal (c) = : < , where <

(d) w = w : where:

=:

{ : }

=:

: = : < , <

and dif(, ) =: { : < and = }

is finite(we can think of as an alternate history of the condition from time on)

(e) F = F1,2 : (1, 2) where =

where

= {(1, 2) : 1 ,

2 and 1 [ + 1, ) = 2 [ + 1, )}.

(note: if k = 1 or = , has only pairs of the form (, ) and if 1 2 , 2 2 then for some

1 1 we have 2 = 1 1) such that:

(f) i f

1

,

2

and

1

=

2 [, ) then w

2

w1

and each w is a subsetof of cardinality <

(g) i f 0 and is a limit ordinal then w =


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6 SAHARON SHELAH

(i) for (1, 2) let (1, 2) be that (unique) ; for let () =

(j) F1,2 is a function, Dom(F1,2) is an initial segment ofw2 , Rang(F1,2)is an initial segment of w1

(k) F1,2 is a one to one and order preserving function and F2,1 = F11,2

(l) if (1, 2) , (2, 3) then F1,3 F1,2 F2,3 , F, = idw for

(m) for (1, 2) let 1 < 2 means 1 <

2

(n) if (1, 2) and 1 < 2 then Dom(F1,2) F12()

(o) i f +1, k = 2, < and for at least two 1 we have

1 [ + 1, ) = & w1 then:

() for every 1 satisfying 1 [ + 1, ) = we have w1

() if 1, 2 and 1 [ + 1, ) = 2 [ + 1, ) = thenF1,2() = (so is well defined and otp(w1 ) = otp(w2 ))

(p) i f 0 then w = (q) assume +1, k = 2; there is e, a convex2 equivalence relation on

with finitely many equivalence classes such that:

() if 1, 2 , 1 [+1, ] = 2 [+1, ) and 1, 2 are e-equivalent

then F1,2 is from w1 onto w2() if (1, 2) , (1, 3) (hence 1, 3) ) and 1e2 then

F1,3 = F1,2 F2,3(note: if ( )( ), which is the main case here, e can bereplaced by n < , with 1e

2 being replaced by

(n 1 & n 2) (

1 =

2).)

1.6 Explanation. The creature u can be thought of as a part of the creation ofa condition in Definition 1.11. For = 0 we have an atomic condition, wecould make it e.g. speaking on one iterand but we have made them empty. If = + 1, k = 2 we have conditions forming a -system of conditions ofdepth, length of history which we put together. If = + 1, k = 1 weextend a condition of depth to satisfy a density requirement on the domain. If is limit, we take a limit of a nicely increasing sequence of conditions. Now is the history: if we have just put together condition of depth from a-system, it is a free choice which one lies on the main history line and whichare just alternate histories. So is the set of possible alternate histories fromstage on w, and the domain of the condition which was the beginnings of thishistory and F1,2 is an isomorphism witnessing our having used a -system. So

for > 1, we shall get even for e.g.


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1.7 Definition. 1) Let an -creature mean an (, )-creature for some < .

2) Let CR, be the set of (, )-creatures, CR =


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8 SAHARON SHELAH

(a) if = u then 1 = 2 = and we let Fu1,2 = idDom(u)

(b) if < u = + 1, ku = 1 we let Fu1,2 = F

u[]

1[,),2[,)

(c) if < u = + 1, ku = 2 we let Fu

1

,

2 = Fu[1]

1[,),

2[,)

Fu

1

,

2

(d) if < u, u a limit ordinal let Fu1,2 = Fu[]

1[,),2[z,) for every < u

that is large enough such that 1 [, u] = 2 [, u] = u [, u).

1.8 Definition. 1) An -sequence is a = a : < such that a . We saya is a (, )-sequence if in addition | | < for < .2) For an -sequence a let CRa, be the set of u CR, such that:

() if < u and k = 2 and u and (1, 2) and 1 <

2

then w2 {a : w1 w2} w1 .

In this case, we say u is an (a, )-creature and we say u is an a-creature.

1.9 Claim. 1) If u is an (, )-creature, then u (see Definition 1.7(5))is a (, )-creature, (and [u ] = [u] of course). If a is an -sequence, u is an(, )-creature, then u is an (a , )-creature. If and u isan (, )-creature then u = (u ) . If u is a (, )-creature and then(u is an (, )-creature and) u = u.2) If u is an -creature then for a unique = u < ,u is an (, )-creature.3) e is a partial order on CR, [u1 e u2 (u1) (u2)] and [u1 e u2 &

[u1] = [u2] u1 u2] and [u1 e u2 ( u2)(u1 u[]2 )] and for an

-sequence a we have: u1 e u2 & u2 CRa u1 CRa.4) de is a partial order on CR, [u1 de u2 u1 e u2] and [u1 de u2 &

[u1] = [u2] u1 = u2] and [u1 de u2 ( u2)[u1 = u[]2 ].

5) u[] is the unique v such that v = & v de u (defined iff u), andfor u, (u[])[] = u[]. Also for u CR,, if and u then

u[] CR, and u[] e u.6) On CR, u u is an equivalence relation (see Definition 1.7(6)).7) If u1 e u2 then for some u2 u2 we have u1 de u

2.

8) If u1 e u2 and u1 e u

1 then for some u2 e u2 we have u1 = u2 .

9) If < then CR, CR, and for any -sequence a, CRa, CRa,.10) If u CR,, = + 1, ku = 1 then u

() = u[] = u[,0].

11) If u CR,, = + 1, ku = 2 then {v[]/ : v u} = {u(i)/ : i < u }.

12) If < is a limit ordinal, ui : < is a de-increasing sequence in CRthen for one and only one u CR we have: [i < ui de u] and u =

i


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(iv) Fu1,3 Fu1,2 F

u2,3

(v) Dom(Fu1,2) = wu2 Rang(F

u1,2) = w

u1

(vi) if in (v) the equalities hold then

Fu1,2(u[2]) = u[

1]

(vii) if in (v) the equalities fail and2 = Min(wu2\ Dom(F

u1,2)) and

1 = Min(wu1\ Dom(F

u2,1)) then F

u1,2(u

[2] 2) = u[1] 1.

14) Assume u CR,, v e u, u1 = u[], u+1, ku = 2, =

u ,

vi = u(i)1 (= u

[i], see Definition 1.7(10)).Then there is a convex equivalence relation e on with finitely many equivalenceclasses such that:

() iej implies:

(a) Fu1(i,j) is an order preserving map from Dom(vi) onto Dom(vj)

(b) Fu1(i,j) maps Dom(v) Dom(vi) onto Dom(v) Dom(vj).

15) If Dom(u), u CR, then for a unique < u we have: k = 1, and

u, =: { u+1 : wu} is non-empty but [

/ wu] and

[ >

u (1)(1 u, & = 1 ) and |a| < &

1 u, &

2 u, wu1 a = wu2 a.

16) Assume u+1, k = 2 and 1, 2 u , then

w1 w2 = { : Fu1,2() = } = Dom(F

u1,2) Dom(F

u2,1).

Proof. Straightforward, but we elaborate 1), 8) and 14).

1) Reading Definition 1.5, the least trivial clause to check is (q), so assume

< u , u+1 and ku = 2. But

u = u, u+1 = u+1 and

ku = ku and

u =

u , so as u CR, there is a convex equivalence relation e

1

on u as required on e in clause (q) of Definition 1.5 for u. We now define another

equivalence relation e2 on u ; letting

ie2j iff i,j < u,ie1j and

( w{,i})[ < Fu{,j},{,i}() < ].

Clearly e2

is an equivalence relation on u .

By clause (n) of Definition 1.5 clearly e2 is convex.Now if e2 has infinitely many equivalence classes, let j be minimal such that

{i/e2 : ie1j} is infinite. Let in be the first element in the n-th e2-equivalence

class j/e1 (or just in < in+1, ine2in+1, ine1j) and let n w{,in} witnessine2in+1; i.e. n =: F

u{,in+1},{,in}

(n) < n < .


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10 SAHARON SHELAH

Without loss of generality n is minimal under this condition, so by clause (n)of Definition 1.5 as in < in+1 we have n < n, hence necessarily n+1 <

n

hence Fu{,in},{,in+1}(n+1) < n (remember that by clause (q)() of Def-

inition 1.5, Fu{,in},{,in+1} is from wu{,in+1}

onto wu{,in}). HenceF

u

{,i0},{,in}(n) : n <

is a strictly decreasing sequence of ordinals;contradiction.8) Let u

2

be such that u1 = (u2)[] and let 1 u1 be such that

u1 = (u1 )[

1] (exists by the definition of e and the assumptions).

Clearly Dom() = [u1

, u2

) and Dom(1) = [u1 , u1). However, u

1 =

u1

, u1 = u

1

hence 1 u1

, so easily 1 u2

and Dom(1 ) =

[u1 , u2

), lastly let u1 = (u2)[

1]; check.

14) We prove it by induction of u. Let e1 be a convex equivalence relation on with finitely many equivalence classes which satisfies clause (a) of () (exists byclause (q) of Definition 1.5). We will refine it to satisfy clause (b) too.If v = u the conclusion is immediate: use e = e1, so we shall assume v = u.

Case 1: u = 0.Nothing to prove.

Case 2: u successor.

Subcase 2A: [u1] = u, so = u 1.

Let 1 be such that v = u[1], 1 u, note (as v = u) that necessarily Dom(1) =

, so Dom(1) and let i = 1 (< ) so v e u(i). Define e2, an equivalence

relation on with the classes {j : j < i}, {i}, {j : i < j < } (omitting anyoccurence of the empty set). Let e = e1 e2. By clauses (n), (o), of Definition 1.5we are done.

Subcase 2B: (u1) < u.

If ku[u]1 = 1, necessarily v e u[] (see Definition 1.7(9)), and u1 u[] and

[u[]] < u so we can use the induction hypothesis. So assume k[u]1 = 2, hence for

some i0, i1 < u[u]1 we have v e u(i0), u1 e u(i). Ifi0 = i1 as [u(i

0)] = u 1 we

can use the induction hypothesis; so assume i0 = i1. Let v = u[1] so 1[u]1 = i0,

let 2 = 2 : v < u be: 2 is

1 if [

v, u 1), and i1 if = u 1.

Now apply the induction hypothesis with u(i1), u1, u[2] here standing for u, u1, v

there and get an equivalence relation e. Clearly Fu(i1,i0) maps v to u[2]. Is e

as required? So assume j1, j2 < , j1ej2 and Dom(v) Dom(u(j1)1 ); then

necessarily Dom(u(i0)) Dom(u(i

1)) hence Fu(i1,i0)() = (see Definition 1.5

clause (o)) so Dom(u[2]) hence Dom(u[

2]) Dom(u(j1)1 ) hence by the

choice of e we have Dom(u[2

]) Dom(u(j2)1 ) hence Dom(u

(j2)1 ) hence

Dom(v) Dom(u(j2)1 ). By the symmetry between j

1 and j2 this suffices.

Case C: u is a limit ordinal.We are assuming v < u and as u1 is a successor ordinal u it is < u

so for every < u be large enough, v, u1 e u[], and we can use the induction


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hypothesis.1.9

1.10 Definition/Claim. 1) If , u1 CR, u0 =: u1 de u2 CR wedefined u = u1 u2 CR as follows. First

(a) u = u2 , ku = ku2 , u = u2 and u = u2 .

Second

(b) wu is defined as follows:

(i) if () u1 and [u1 , u2) = u2 [u1 , u2) thenwu = w

u1((Dom )[0,u1))

(ii) if (u1 , u2 ] and = u2 [, u2) thenwu = w

u2 w

u1

(iii) if both cases do not apply then w

u

= w

u2

.

Lastly

(c) Fu1,2 are defined naturally.

2) In part (1), u actually belongs to CR and u CR, u = u2 and u1 de u.3) If u1 = u2 and = { + 1 : < and Dom(u2) (equivalently Dom(u1)} then u1 = u2 .4) Let u1 rd u2 means: for some and u we have u1 = u , u e u2. We callsuch (, u) a witnessing pair to u1 rd u2, and if is minimal (as in (3)) we callsuch (, u) a good witnessing pair in u1 rd u2.5) rd is partial order (on creatures). In fact, if u0 rd u1 rd u2 and (, u) is

the witnessing pair for u rd u+1 and u

0 e u

1 is such that u0 = u

0

0 (existby 1.9(8)) then (0, u0 ) is the witnessing pair for u0 rd u2.

Proof. Straightforward.

We turn to the iteration

1.11 Definition/Lemma. We define and prove the following by induction on for any-sequence a:

(A) We define when

Q = P , Q : < is (an iteration) from

3

K

a

(B) for Q Ka we define the set of elements ofP = Lim(Q) = Lim(Q) = Lim,a(Q)

3Remark: Note that e.g. P Q P+1 is not included, in fact, this does not really interest

us. What we want is condition (b) of Conclusion 1.19(1) + 1.20(3).


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12 SAHARON SHELAH

(C) for p P and we define p P , u[p] and also p[u] P foru e u[p] (and show p is well defined and in P , u[p] is well defined andin CR andp

[u] is well defined and in P). We also define when F(p) = q,(D) we define the partial order P=


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() each pu (i.e. for u e u[p]) has the form(u, s, t) = u, su, tu = (u, su,p, tu,p) such that:

(a) su = su : Dom(u), su is a function with domain

n = Dom(su) and Rang(s

u) H(0)

(b)tu

= tu

: Dom(u), where tu

= { tu,

y : y Yu

}; (we shall saytu finite instead of Y

u is finite, similarly for etc., and demand

Yu { , , n : n < , < u}; (clearly tu = Y

u = )

(c) each t

u,y is a P-name of a member of

(but su , Yu are not names!).

Moreover, there is a v = v( t

u,y , p) e u (not depending on u) and for

each k < there is a setI = Iu,y,k I[u] =: {r : r P , u[r] e v }, each element of I

forcing a value to t

u,y (k) such that every q P with u e u[q] is

compatible with some r I and we even demand strongly compatible(see part (H)), note we use it to P ; i.e. to Q )

(d) Moreover, ifa has cardinality , then tu is finite andr Iu,y u[r] a (we could be more specific here).

We further require:

()(a) if u1 e u2 e u[p] thensu1 = su2 (Dom u1) (not required from t!)4

(so su,p can be written sp

)(b) if u1 u2 e u[p] then

tu1 = tu1

(c) if u1, u2 e u[p], < , u1 = u2 (hence [u1] = [u2], etc.) thentu1 = tu2 (for s this follows from the stronger conditions ()(a)).

()(a) If u e u[p] and u = + 1 and ku = 1, v = u[] then:

(i) sv = su (Dom v), (follows from ()(a))(ii) for Dom(v) we have tv t

u and equality holds for all but finitely

many s[we need the non-equal case e.g. for t tu whose name is not based on v,see conditions ()(b), (c) above]

(iii) for Dom(v) we have tu\tv is finite

(iv) if Dom(v) and a has cardinality < then tu = t

v

(v) { : Dom(u)\Dom(u[]) and su = or tu = } is finite and for

Dom(u)\Dom(u[]), Yu is finite.

(b) If u = + 1 and u e u[p], ku = 2,

u[p][u] and u = u[p]

[] and for

i < =: u[p] we let vi = u

(i) (see Definition 1.7(10)) then:

(i) for i < and Dom(vi)\

j


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14 SAHARON SHELAH

(ii) for i


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() p0 vpr p1 if in (), u de u[p0] and

Dom(u[p1])\ Dom(u[p0]) su[p1],p1

= & tu[p1],p1

= (of course by clause () we have

u[p0] (sp1

, tu[p1

],p1

) = (sp1

, tu[p0

],p0

))

() p0 apr p1 where p = pu : u e u[p]

so pu = (u, su,p, tup

)) holds iff:

(a) u[p0] u[p1]

(b) for every u e u[p1] and Dom(u), we have su,p

0

su,p1

and tu,p0

tu,p1

(c) for every Dom(u[p0]) we have (see Definition 1.1(2)):

p1 P Q |= (su[p0],p0

, tu[p0],p0

) (su[p1],p1

, tu[p1],p1

)

(Note: this trivially holds if su[p0],p0

= su[p1],p1

or if

su[p0],p0 s

u[p1],p1 & t

u[p0],p0 =

(d) if in clause (c) the set a is of cardinality < then we moreover have:

if t

tu[p0],p0

then (p1)[ t,p0]

force a value to t

(n) whenever

n Dom(su[p1],p1

)\ Dom(su[p0],p0

).

Moreover for any such n, if su[p1],p1

(n) = 0 then it is computed

by gQ

( V) from su[p1],p1

n and the values above.

() Now let P |= p r iff for some q P we haveP |= p pr q, and P |= q apr r

() p0 p1 means u[p0] u[p1]

and u e u[p

1

] p

0

u = p

1

u.

Part E:

() p x p if x {nr, vpr, pr, apr}[why? check]

() if p pr q and (, u) is as in clause () of Part (D) then u[p] rd u[q]hence Dom(u[p]) Dom(u[q]), moreover (, u) is a witnessing pair foru[p] rd u[q] and if is as in 1.10(3), even a good witnessing pair and wesay the pair witnesses p pr q.[why? check].

() if p pr q is witnessed by (, u) for = 0, 1 then p0 = p1[why? read the definitions]

() if p0 pr p1 pr p2 then p0 pr p1

[why? let (, u) witness p pr p+1 for = 0, 1, so by 1.10 we know some(0, u

0) witness u[p0] rd u[p2] and check that it witnesses p0 pr p1]

() if p0 vpr q then p0 pr q[why? demanded in the definition]

() if p0 vpr p1 vpr p2 then p0 vpr p2[why? as Dom(u[p0]) Dom(u[p1]) Dom(u[p2]), and if


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16 SAHARON SHELAH

Dom(u[p2])\ Dom(u[p0]) then for some {0, 1} we have

Dom(u[p+1])\ Dom(u[p]) hence su[p2],p2

= su[p+1],p

=

(as p vpr p+1) and similarly tu[p2],p2

= tu[p+1],p+1

= ]() if < and p q then in P, p x q

[why? check]() if p0 apr p1 apr p2 then p0 apr p2

[why? straightforward; for clause (c) note by clause () thatP |= p0 p1 p2 ]

() if u u[p] and p P and thenp[u] pr p as witnessed by (, u)

() if p0 apr p1 pr p2 then for some q we have p0 pr q apr p2.Why? Let (, u) be a good witness to p1 pr p.We define q as follows:

(a) u[q] = u[p2]

(b) su[q],q is: su[p2],p2

if Dom(u(p22))\Dom(u[p1])

su[p0],p0 if Dom(u[p1])

(c) su,q for u e u[q] is su[q],q

(d) tu,q for Dom(u), u e u[q] is:

tu,p0

if Dom(u[p0]) & u e u[p0]

tu,p if otherwise.

We now have to prove that q is as required.

q P: That is we have to check the definition in Part B.Almost all clauses of Part B are obvious, but we have to say something on

()(b)(iii) hence the definition of su,q . So let u e u[q], u = + 1, ku = 2, u =

u[q]

[]

where

u[q]

[u] and for i < =:

u[q]

we let vi = u

(i)

. So as u[q] = u[p

2

]we can replace q by p2 and so there is a convex equivalence relation e on suchthat iej Fuj,i maps (p

2)[vi] to (p2)[vj ]. Does e work for the q[vi]? The

problem is with the svi,q s and it works by 1.9(13).

p0 pr q Straightforward.

q apr p2 Straightforward.

(Compare with 1.14(1))

() P is a partial order.[why? follows by clause () as Ppr ,

Papr are transitive]

() if 1, 2 , p P then (p 1) 2 = p (Min{1, 2})[check]

() u[p ] = u[p] [why? check]

() for , x {nr, pr, apr,vpr}, we have Px P =Px

[why? read the definitions]


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() if and p Ppr q then (p ) Ppr (q

)

[why? let and u be as in Part (D) clause (), witnessing p Ppr q,let = Min{, }, so it suffices to prove that , u witness(p ) Ppr (q

), and we have to check (a),(b),(c) of () of Part (D).

Clause (a): u[p ] = u[p] = (u[q ]) = u[(q ) ],u[q Min{, }] = u[(q ) ] = u[q ] .

Clause (b),(c): Check.

() if and p Pvpr q then (p ) Pvpr (q

)[why? easy].

() if and p Papr q then (p ) Papr (q

)[why? think].

() if then p pr p[why? check].

Part (F): If p P, , p q P then we shall define r = p q andprove r P, p r, q r. The definition of r is as follows:

() u[p] de u[r] = u[p] u[q](see 1.10)

() su[r],r (Dom(u[p])\Dom(u[q])) su[p],p

() su[r],r Dom(u[q]) = su[q],q

() u e u[r] tu,r = tu,q

() u e up & Dom(u)\ tu,r = tu,p

() up e u e u[r] & Dom(u)\Dom(u[q]) tu,r = t[u[p],p .

Let u u[q] be such that u[p ] de u (exists by 1.9(7)). Clearly

[u[p]] [u]; for [u] let u be (u)[]. Let 0 = [u[p]] and 1 = [u]. Wenow define by induction on [0, 1], a condition r with u[r ] = u such thatr = qu : u e u, and r0 = p . There is no problem, (remembering ()(b)of part (B).)

Part (G): For < , P P P. Follows by parts (E),(F).

Part (H): We define: p1, p2 P are strongly compatible (inside u) if

(a) for < 2 we have u e u CR, p P and u[p

] = u

(b) if Dom(u0) Dom(u1) and su1,p0

= su2,p1

then for some {1, 2} we

have tu,p

= = su,p

or at least tu,p

= & su,p

su2,p2

.

Define q and their canonical common upper bound by:

u[q] = u, q = qv : v e u

sv,q = su0,p0

su1,p1


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18 SAHARON SHELAH

tv,q = tu0,p0

tu1,p1

.

Fact: q actually is a common upper bound of p0, p1. 1.11

Some properties of those partial orders:

1.12 Claim. 1) If p P then [u u[p] p[u] = p] and [u e u[p] p[u] p]moreover [u e u[p] p

[u] pr p].2) If p1, p2 P, u[p1] u[p2] and [u e u[p1] p1u = p

2u] then p

1 apr p2 aprp1; i.e. p1 p2.3) (P, vpr) is -complete, in fact any vpr- increasing sequences of length < has a lub.4) If P |= p apr q, < and r is defined by: u[r] = u[q],ru = u, su,q su,p [, ), tu,q tu,p [, ) then p apr r apr q.5) If P |= p q then for some p

0, p1, p2 P we have:p = p0 vpr p1 p2 apr q.

6) If p P,

u[p]

is a limit ordinal, < large enough as in 1.11(B)()(c),p[u(p)[]] apr q then we can find q P, p apr q such that q[u[q]

[]] = q, and1.11(B)()(c) holds for q, .7) If p P, u e u[p], of course p[u] pr p and p[u] apr q P then there is aunique r P such that p apr r, r[u] = q and for any v:[ Dom(v)\Dom(u[q]) & v e u[p] s

v,p = s

v,r & t

v,p = t

v,r ].

8) For any p P, u[p] and u = u[p][] and < u[p] there are E = Ep,u

and a set such that:

(a) E is an equivalence relation on

u[p],u = {

u[p] : [

u , u[p]) = }

(b) E has finitely many equivalence classes

(c) if 1E2 then Fu[p]

1

,2 (see 1.7(12) is an order preserving function from

u[p][2] onto u[p][

1]

(d) u[p],u is a finite set of representatives

(e) if1, 2 u[p], 1E3, 2E4 and Dom(u[p]3

) Dom(u[p]4

) then

Fu[p]1,3() = F

u[p]2,4()

(f) if 1E2 then Fu[p]1,2

maps p[u[1]] to p[u

[2]].

(g) If Dom(u[p]), p P and a is countable and t

= tu[p],py t

u[p0],p

then for a unique v = v[ t

, p] e u[p] and = : < u we have

ku[p] = 1, Dom(u

[+1,u[p])\ Dom(u), so necessarily Iu,y,k I[u[]].

Moreover is the same for all t

tu[p],p (and it is from 1.9(15)).

Proof. E.g.8) This is proved for each , by induction on u[p]. Now the case u[p] = is trivial

and the cases u[p] is limit or u[p] = + 1 & ku[p] = 1 follows easily by the

induciton hypothesis. So we can assume u[p] = + 1 > , ku[p] = 2; let e be an


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equivalence relation on u[p] as in clause ()(b)(iii) of Definition 1.11; Part (B). Let

i0 < .. . < in()1 be representatives of the e-equivalence classes. For n < n() let

pn = p[u[p](in)], so u[p

n] = < u[p], so by the induction hypothesis there is an

equivalence relation En on u[pn] as required. Now define the E we desire:

1E2 iff 1e2 and 1 [, )En2 [, ) for each n < n().The checking is easy. 1.12

The following claim helps as the Qs were required to be simple.

1.13 Claim. If a is an -sequence and < , Q Ka and n < then

In,0 =: {p P : Dom(u[p]) and su[p],p has domain including {0, . . . , n 1}}

is dense (subset of P) and even

In,1 = {p P : Dom(u[p]) and for some m [n, ) we have su[p],p (m) = 0}

is dense (subset of P).

Proof. Let p P; by clauses ()(a) + ()(a) of the part B of Definition 1.11, thereis q such that p q and Dom(u[q]). So by a variant of 1.11 Part (F) it isenough to prove (we shall use the simplicity of Q

; i.e. Definition 1.1(1)):

() ifp P, Dom(up) and p q P then for some m < for everyk < there are r and {s1, . . . , s

m} such that:

q apr r P and r P { t k : t t

u[p],p } = {s

1, . . . , s

m}.

If a has cardinality , this is easy (as by 1.11 of part (B), clause ( )(c) the set

tu[p],p is finite) so assume a has cardinality < .

Our problem is that we do not just have to force a value to the set

{t(n) : t

t

u[p],p } such that it is finite, but we need an apriory bound, one not

depending on the process. So the natural (and first) approach induction on p

seems problematic. Here our restriction in the case a is of cardinality < (in the

definition of the iteration, see clause ()(a)(iv) of Definition 1.11) help.Let be as in 1.9(15) for , p (and also as in 1.12(9)).

Let = { : u[p] and Dom(u

(+1,u[p]))\ Dom(u)} and assume

without loss of generality (u[p] ) de u[q] and let = u[q] [u[p], u[q]).

Let E = Eq,

and be as in 1.12(8). Let

= { q : there is such that belongs to /E}, for

= let t = tu[p]

1],p. Now let 1 = 1 and 0, . . . , n1 list and |t| n for .

Let u = u[r][].

Now choose m = nm; so to show () let k be given, choose q such that

q q P and q force value to t

(k) for k < k, t


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20 SAHARON SHELAH

(i) u[r] = u[q](ii) if Dom(u[q][]) for some then

su[q],r = s

u[r],r

(well defined by 1.12(8) clause (c)

(iii) if Dom(u[q]) and clause (ii) does not apply then su[r],r

is chosen as su[q],q

(iv) if Dom(u), u e u[q][], then tu,r = tu,r

(again by 1.12(8) it is well defined)(v) if Dom(u), u u[q] and clause (iv) does not apply then

tu,r = tu,q

{tv,q : v


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First Case: u

= 0.Trivial.

Second Case: u

= + 1, ku

= 1. Let v = (u)[], p = p

[v] . By the induction

hypothesis without loss of generality the conclusion of (3) holds forp : < 1, A = 1. By Definition 1.11 part(B),clause()(e) if = 1 the set{su

,p (Dom(u)\Dom(v) : < 1} is countable so without loss of generalitysu

,p (Dom(u)\Dom(v)) for A is constant. But even not assuming = 1,letting a = { : Dom(u) but / Dom(v) and su

,p = & t

u,p = }

is finite hence without loss of generality a : < 1 form a -system with heartsay a and su

,p a = s; i.e. is constant. Now clearly p : < 1 are pairwisestrongly compatible as required.

Third Case: u

= + 1, ku

= 2. For each < 1 we have an equivalence relation

e on u

, as in Definition 1.11 part (B), clause ()(f)(iii), so it is convex with

finitely many equivalence classes, so the number of possible es is |

nu

|n; as

u

< 1 without loss of generality e = e. Let a u

be a set of representatives.Let vi = (u)(i) for i < u

. By successive application of the induction hypothesis

to p[vi] : < 1 without loss of generality for i a we have p

(vi) : < 1

satisfies the conclusion for A = 1.

By use of the Fu

(i,j) this holds for any i < u

. Now it suffices to prove that

for any 1, 2 < 1, the conditions p1 , p2 are strongly compatible; so let

u[p1 ] = u[p2 ] = u. If /

i


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22 SAHARON SHELAH

Now for each limit < 1, there is < such that Dom((u)[ ]) & / Dom((u)[]) su

,p = . By Fodors lemma for some stationary s 1,

we have S = , and without loss of generality 1 < 2 in s 1 < 2 .

We now apply the induction hypothesis to p p[(u)[

]] : s.

4)? Like the proof of the last possibility in the proof of part (3) (cf(u

) = 1).5) If u CR and u then there is an equivalence relation E on u with

finitely many equivalence classes.

For each < 1 choose < )1 such that up e u so without loss ofgenerality > choose p P such that u[p

] u and p vpr p

and

let p p, u[p

] = u , and for limit < 1 choose < [u] such that

Dom

(u)[]

& / Dom

(u)[]

su ,p

= without loss of

generality = [u ], < and choose a stationary set of limit ordinals < 1

such that S = and 1 < 2 in S 1 < 2. Let v =?

Apply part (3) on (p)[u ] : S get an uncountable S2 S; now

p : S1 is as required. 1.15

1.16 Claim. Let p P and let

be a P-name of an ordinal. Then there is

q, p vpr q P such that:

() if q r P, r

= then for some r, letting be such that

u[q] = u[r][], q = (r)[] we have q

= .

Moreover

() there is a countable familyI {r : for some u, up de r de q, and q[u] apr r and r force a valueto } such thatq q & (q force a value to

) q is strongly compatible

with some r I.

Proof. Assume that there is no q satisfying (). We choose by induction on n we have (s

n, t) (s

m, t).

1.20 Claim. Assume is regular uncountable and a is a (, )-sequence. Assumefurther Q Ka and P = Lim

,aQ.1) P satisfies the (2 || ) and < |a| < .If S { < : cf() = 0} is stationary and S holds in V say exemplifies byc = c : S (and ), then forcing with P preserves it, even for the samec.3) Assume cf() > and = sup{ : a has cardinality < } and p P andp f

i

for i < . Then for some q and we have: p q P and q P

for unboundedly many i < we have fi t

u[r],r for some r GP

; hence for any

finite setw of such is, some n for every m > n we have (s n, {fi : i w}) (s n, {f

i : i w}).

1.21 Remark. 1) We are most interested in the case otp(c) = , but any countableordinal is O.K.


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24 SAHARON SHELAH

Proof. 1) Let pi P for i < (2


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Proof. We know cf([3]1 , ) = 3 hence we can find a = a : < 3 such thata []1 and (a)(a 3 & |a| 1 (3 < 3)(a a)) and (a)(a & |a| = 0 (3 < 3)(a = a)). Now choose Q Ka such thatfor every < 3 and P-name of a simple forcing notion Q with g, f V theset { : Q = Q[VP ]} is unbounded in 3. Alternatively deal only with Q of1.3(4) and derive the results for the others. Now use 1.20(1) for 2-c.c.c., 1.17 forproperness, 1.20(2) for part (a) and 1.19 for part (b) and 1.20(3) for part (c).

1.22

1.23 Conclusion. Let P3 be as in 1.22. Assuming for simplicity V |= GCH inVP3 there is no Gross space; i.e. for every countable field F and vector space Uover F with inner products (i) with dimension 0 for some U1,U2 U, we havedim U1 = dim U, dim U2 = 0 and U1U2.

Proof. Let = dim(U) and xi : i < be a basis of U.

Case 1: = 1.Use clause (b) of the conclusion of 1.22.

Case 2: = 2.Use clause (c) of the conclusion of 1.22.

Case 3: = 3.Use clause (a) of the conclusion of 1.22 (i.e. let c : S be a S ,

S { < 2 : cf() = 0} stationary.Define f : 2 F be f(i) = (x , xi). So for each i for some (i) S, f c

is constant. So for some , B = { < : (i) = } has cardinality . SoU1 = Span x : B,Ui = Span{xi xj : i, j c} are as required).

Case 4: = cf() > 3 or just cf() > 2.The proof is as in the case = 3.

Case 5: > cf() {0, 1, 2}.Cases of singular we reduce it to the problem of cf() as for (3, ) regular

for some B [], f : 2 F we have ( B)(j < 2)[(x, xj) = f(j)].1.23

1.24 Remark. 1) In 1.22 - 1.23 we can replace the choice ( ,,) = (1, 2, 3) byany other satisfying the requirements in 1.21 (e.g. 2


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26 SAHARON SHELAH

REFERENCES.

[B] James E. Baumgartner. Decomposition of embedding of trees. NoticesAmer. Math. Soc., 17:967, 1970.

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

[ShSi 468] Saharon Shelah and Otmar Spinas. On Gross Spaces. Transactions ofthe American Mathematical Society, 348:42574277, 1996.

[ShSt 258] Saharon Shelah and Lee Stanley. A theorem and some consistency re-sults in partition calculus. Annals of Pure and Applied Logic, 36:119152, 1987.

Saharon Shelah- Historic Iteration with ℵε-Support

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Transcript of Saharon Shelah- Historic Iteration with ℵε-Support