saharon shelah- nnr revisited

8/3/2019 Saharon Shelah- NNR Revisited

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NNR REVISITED

SH656

Saharon Shelah

Institute of MathematicsThe Hebrew University

Jerusalem, Israel

Rutgers UniversityMathematics DepartmentNew Brunswick, NJ USA

Abstract. We are interested in proving that if we use CS iterations of forcingnotions not adding reals that satisfies additional conditions then the limit forcingdoes not add reals. As a result we prove that we can amalgamate two earlier methodsand prove the consistency with ZFC + G.C.H. of two statements gotten separatelyearlier: SH and non-club guessing. We also prove the consistency of further cases ofstrong failure of club guessing.

I would like to thank Alice Leonhardt for the beautiful typing.First Typed - 97/June/17Written - 1 in 97/April/21-22Latest Revision - 2000/Mar/14

Typeset by AMS-TEX

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Anotated Content

0 Introduction

[We give lengthy explanations of the problems and proofs for No-New-Realsiterations (or proper forcing, CS iterations).]

1 Preservation of not adding reals

[We present sufficient conditions for CS iteration of proper forcing not toadd reals. For this we define reasonable parameters p and we have twomain demands. One (clause (c) of Definition 1.8) is a weakening of -proper for every < 1. This time it has the form (on Qi), p-properwhich informally says that: if p N, Y {M N : M appropriate} is

-large then for some (N, Qi)-generic condition q p, q forces that {M Y : M[GQ] V = M} is -large (the meaning of large depends on p),

hence without loss of generality, p has length 1. The other main demand(clause (d) of Definition 1.8) is a weak diamond preventive.We then show that -properness for < 1 is sufficient for the first maindemand (in 1.16(3)). The demand on the games for p helps to prove thepreservation ofp-properness.]

2 Delayed properness

[The preservation theorem in the first section does not, for standard p, cover

shooting a club C 1 running away for C = sup(C), C small (see3). For this we will use (p, , )-proper for enough pairs < g(p) (sostarting from -large we get -large; for many we can choose = butduring the inductive proof we pass through cases of < ).Here we introduce various definitions and basic facts needed. We discuss ax-ioms, version of the properties preserved by CS iterations and strengtheningof the iteration Lemmas of 1.]

3 Example: shooting a thin club

[We present the natural forcing showing = 2 is interesting (not only

= 0) (from [Sh:b, Ch.VIII,4]).We show that the natural forcing (see above) for running away from C ,of small order type (see [Sh:f, Ch.XVIII,2]) falls under our framework fordelayed properness. We give examples: running away from C,0, C,1 : then j = otp([i, j) M)

() M = N0

() M M0 M1() 1 n

m2 & 2 n

m2 G1m M = G2m M so we call it

Km() M0 = M1 when

nm2 call it N

() N Eg() for (mm>)2

() Ym = M[K0m , N] M[K

1m , N] Dj(N).


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1.14

1.15 Conclusion. Let Q be a CS iteration and p a non--loser reasonable parameter,2 n() 0, g(p) = 1.Then Q is a p N N R0n()-iteration iff for each i < g(Q)

()i Qi is a proper forcing and Pi, Q

i satisfies clauses (d) + (c) of the Definition

p NN R0n()-iteration with i, i + 1 here standing for i, j there.

Proof. By induction on j = g(Q). For j = 0 there is nothing, for j limit use 1.9(or 1.14) and if j successor just read the definitions. 1.15

We point out here that clause (c) of Definitions 1.8 and 1.13 really follows fromearlier properties which play parallel roles.

1.16 Claim. 1) Assume thatp is a standard reasonable parameter, and is standardand < g(p), N Ep , Y D(N) and 1 N a limit ordinal.Then we can find sequences N = Ni : i < , i : i < such that:

(a) Ni N (for i < ) is countable, N Ni and Ni Y

(b) Ni

N

(H(p), ), and Ni Ni H(p

) (H(p

), )

(c) i < j Ni Nj

(d) for i limit Ni =j


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NNR REVISITED SH656 21

2) Ifp is a standard, reasonable parameter, Q is a CS iteration, g(Q) = +1, Q is p N NR0

k()-iteration and

P Q

is proper and ( sup(C C,)}

(c) p obeys the P-ultrafilterD and is a reasonable parameter.

Then Q is a p N N R0 forcing notion.2) In part (1) if we add: N E & D

p(N) = {Y : {n : Mn Y} DN},DN a

P-point ultrafilter (see above) and < k {n < : Mn1 C,} = modDN,then we can allow C,0 = C,1.3) Assume


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(a) D is a family of subsets of Dom(D), the intersection Y of any < 1 + ofthem satisfies

() n(y1, . . . , yn Y)[ > sup(

n

=1

C,y)]

(b) C = C,x : x Dom(D) and is a limit ordinal < 1

(c) C,x : x Dom(D) is a sequence of pairwise disjoint subsets of

(d) X sup(C C,x))}ordered by: end extension

(f) D is a P-ultrafilter sequence.

Then Q is a p N N R0 forcing notion p which obeys D.

Proof. Straight.1) So let V = VP0 ,P Q = QC

is as above and in V0, N0 E0, N0 N1

E = {Mn : n < }, DN0 as in Definition 3.13(2), p P+1 N, Gm N1 Pis generic over N1 for m < k < 1 + (so k is now fixed). So for each we haveCN01,[Gm] is a closed subset of and for 1 < 2 < kN0 we have C N01,1

[Gm]

CN01,2 [Gm] = so for some (m) {0, 1, . . . , kN0 1} we have = (m)

CN01,[Gm] = mod DN0 . Now let B = {n :if < k, / {(m) : m < k} and

< n then Mn 1 / CN01,[Gm] for m < k and p Mn} belongs to DN0 .

Clearly, without loss of generality Mn : n < N1, DN0 N1, B N1. LetB = {ni : i < }, ni increasing with i and In : n < list the dense open subsetsof P+1 which belongs to N0. We can choose pi by induction on i < such that:

pi Nni P+1

pi m 0}

p pi


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4 Second preservation of not adding reals

We shall concentrate on the simple case.

4.1 Definition. Let p be a reasonable parameter. We say that Q is a p N NR1iteration (where 2 0 as we omit = 1 for convenience) if:

(a) cf(g(p)) > g(Q) (for simplicity)

(b) Q is a countable support iteration of proper forcing notions[in [Sh 311] we will relax the requirement that each iterand is proper]such that i < g(Q) Pi adds no reals(follows by other parts (close (d)) even for Pi+1, i < g(Q)

(c) [long properness] for each i, for some fi Fp

club we have Pi Qi is (p, fi)-

proper, fi increasing continuous, see Definition 2.8(2) + 2.8(4)(d) [-anti w.d.] if fi() < g(p), {Q,i,,} N0 E, N0 N1

E, k < 1 + , j = i +1 and for = 0, 1, . . . , k 1 we have q Pi is (N1, Pi)-generic and is (N0, Pi)-generic, q G

Pi N1 = G

, G N0 = G, p

Pj N0 satisfies p i G and lastly Y =:


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3) No real reason to use some , but also no real reason to use 2. For each to find G means essentially forcing with Qi adds no reals. The point is thecommon solution.

4) In clause (c) for a club C of g(p) we catch our tail, that is fi() C = C\ fora club of < g(p). We could use less, no real point now.5) In clause (d) much of the freedom/variation will be due to the decision howsimilar are G : < k such that G exists. Here we demand

() Y D(N1).

In [Sh:f, Ch.VIII,2] it is essentially required that

() G0 G1 Gk1 (Pi Pi) N1 (k times) isgeneric over N1.

In [Sh:f, Ch.V]

() the common Y is a predetermined increasing sequence of models.

We have a trade-off; clause () makes demand (d) in 4.1 easier, but the parallel of(c) harder compared to clause ().This explains why the present work doesnt supercede Ch.XVIII,2.6) It is harder to win with a slower function, so the assumption above is thestrongest possible - although practically makes no difference, probably.

4.3 Claim. Let Q be a p N NR1 iteration (see Definition 4.1), {2, 0} andp is a reasonable parameter and p is a f-winner for some f Fpclub or at leastf-non loser.

1) Forcing withPg(Q) = Lim(Q) does not add reals (so consequently no -sequences,as we are assuming properness).2) If i j g(Q), then

(b) Pj/Pi is proper

(c) Pj/Pi is (p, fi,j)-proper, wherefi,j F

p

club is increasing continuous and is computable from the f Fp

for [i, j),

(d) we have the parallel of clause (d), this time without assuming that j = i + 1.

Proof. By induction on g(Q). For notational simplicity we assume

all fi are even in Fp

end so consider them as functions from g(p) to g(p)(increasing continuous), and we demand also the fi,j are like that, increasing


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continuous and moreover fi,j(fi,j()) = fi,j() and they are f, f Fpend increasing continuous, p is f-winner or at least

f-non-loser.

Case 1: g(Q) = 0Trivial.

Case 2: g(Q) = i() + 1

Part (1): Pi() adds no reals by the induction hypothesis, Qi() adds no reals by

clause (d) in Definition 4.1, hence Pi()+1 = Pi() Qi() adds no reals.

Part (2): Clause (b) is known, (namely Pj/Pi is proper).

Clause (c):Given i j g(Q) if j < i() + 1 the conclusion follows by the induction

hypothesis. So assume j = i() + 1. Ifi = j, the required demand is trivial, soassume i < j. If i = i(), use clause (c) of Definition 4.1 for i. So assume i < i().Let fi,j,0 = fi(), fi,j,m+1 = fi() fi,j() fi,j,m and fi,j() = sup

m


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Recall that we have demanded fi,j(fi,j()) = fi,j() (see at the beginning ofthe proof).

Let N0, N1, G,,,i ,j,p ,k,q (for < k), G

(for < k) and G be as in the

assumption of clause (d) (from Definition 4.1) be given.Without loss of generality i < j, i < i(), j = i() + 1 (as in the proof of clause (c)the other cases are trivial).First, choose G for G, , , i , i(), G N1. For each < k, if for some s G

we have

s Pi in Pi()/GPi there is an upper bound to G

then (possibly increasing s, recalling G is generic over N) there are s G, r

Pi() N1 such that s forces that r is an upper bound to G and without loss

of generality r i s. Now without loss of generality [G1 = G2 s1 = s2 ]and [G1 = G2 s1 , s2 incompatible]. Now choose r Pi() N1 with domain

i()\i as follows: Dom(r) =


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q is (M+1, Pi())-generic

q forces a value to GPi() M+1

q is (N0, Pi())-generic

q Y+1 M[GPi() , M]

q i G.

(In the older version we have to increase times ifk not given, arriving to a fixedpoint.)

No problem, now apply clause (d) of the definition for i(), N0, Mk,q : < k, Yk, G

and get G as required.

Comment: The Y transfers information between the various generics. In[Sh:f, Ch.V], in the first proof after steps we are lost, but having the commontower of models help us.

Case 3: = g(Q) is limit.

Part (1): Follows from clause (d) in part (2) and part (2) is proved below.

Part (2): Let fi,j be fast enough (you can collect the demands used below).Clause (b) is obvious. We first prove clause (d) and later (c).

Clause (d): So again without loss of generality i < j = .Let N0, N1, p , G

, , , k < 1 + and G, q for < k be as in the assumption ofclause (d) in Definition 4.1 but for i < j.We for simplicity use = 0. We choose N0, < < such that is largerenough, in particular: i i < j < j fi,j() = .Choose im : m < N1 such that:

i0 = i

im < im+1

supm


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Choose y N H() coding enough. We choose M0, M1, M2, M3, M4 N1 E Y such that

N0 M0 M1 M2 M3 M4

Y Mm D(Mm) for m < 5.

Choose q G M4 above G M3 so

q is (M0, Pi)-generic,(M1, Pi)-generic,(M2, Pi)-generic, (M3, Pi)-generic.Let Im : m < M0 list the dense open subsets of Pj from N0.

Explanation. Now we shall use the diagonal argument, choose G

P N0, pm

Pim N0, rm Pin as usual, using in the m-th step fim,im+1+ relevant (d) things.

We fulfill the above in M4, so in the end can find a solution in N1, by using acanonical construction, e.g. each time the


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(e) if < k,J Pim is dense open and J M2, then for some finiteJ I M2,J is predense above q rm(like in the proof of-bounding)

(f) nm < , and < nm Gm is a subset ofPim M0 generic over M0,and Gm M1

(g) Gm+1 Pim {Gm : < nm}

(h) n0 = k, G0 = G M1

(i) q rm GPi,m M1 {G

m : < nm}

(j) Gm is a subset ofPim N0 generic over N0

(k) Gm Gm, so G

m G

m+1, G

0 = G

(l) pm Pj N(m) p0 = p

(n) pm im Gm(o) pm pm+1 Im

(p) Ym M[Gm, M0, y]

(q) Ym D(M0).

Why is this sufficient? If we can carry out the induction, then without loss ofgenerality the construction belongs to N1.

So G = {s Pj N0 : n


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nm.

We choose sm Pim+1 M1, above Gm+1, such that s

m im G

h()m . Now repeat

the argument of the successor stage of shrinking Y (but now we have a fixed !).

So we can find tm Pim+1 M1, above s

m, t

m im H

m, t

n GPim+1 M0 =:

Gm+1 such that

Ym+1 =:

i


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there is pm+1 pm, (Mm+1, Pim+1)-genericpm+1 im G

m

and pm+1 forces GPim+1

N and

forces Zn = Yn M[GPim+1 , Mm, y] Dn(Mm).As the challenger does not have a winning strategy, there is a play he wins, givingn


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Xx,m such that Zm Xm+1 and m+1 N and any m+1 N\pim+1,imr (m+1), y

m+1 N H(

pm+1

) and any m N\n\nand Mm+1 Xm+1 Em+1 , ym+1 Mm+1 H(

pm

) satisfying ym+1

Mm+1 and Ym+1 N DXm+1(Mm+1) and any Mm+1 Ym+1 En

(Mm+1 {Mm+1 H(pm

)} satisfying ym+1, ym+1 Mm+1.

Then

() there is y Tm+1 such that xy and (Zy,m, Xy,m+1, y,m+1, y,m+1, y

y,m+1,

y,m+1, yy,m+1, My,m+1, Yy,m+1, My,m+1) is equal to

(Zm, Xm+1, m+1, m+1, y

m+1, m+1, ym+1, Mm+1, Ym+1, M

m+1).

Why? Because fim,im+1(m) m, hence we know that Pim+1/Pim is(p, n, n)-proper and let Gim Pim be generic over V, rm Gim to themodel Mx,m and the set Yx,m. So now we can describe a strategy for the

challenger in the game ,(N, p) (or ,(N, N, p)) delaying his choiceofMm, Zm to the (m+1)-th move, he just chose on the side xm Tm whichcode they play so far, and preserve xm xm+1.

By 3 this is O.K. - all possible choices of the chooser are allowed, that is thisgives a well defined strategy for the challenger (will he have some free choice, thisdoes not hurt). But the chosen does not lose the game, so there is such a choice

xm : m < with {(Mxm+1,m

} Yxm+1,m : m < } D(N) andm


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() N Eg() for (mm>)2

() Y = M[K0, N] M[K1, N] Djg()(N).

Discussion: We may be interested in non-proper forcing, say semi-proper and UPones (see [Sh:f, Ch.X,XI,XV] and [Sh 311]). Here the change from (reasonable)

parameter p = pV to pV[G] is more serious as {N : N EpV[G]

} is in generalnot equal to {N : N E

p}. We intend to spell it out in [Sh 311].

We may be interested in combining this work with [Sh 587], [Sh:F259]. Intendto deal with it later (see also [RoSh 655]).


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5 Problematic forcing

5.1 Discussion: 1) In the examples the barely adding reals but not so may occuron some stationary E []0 and otherwise just properness is asked. We do notbother to do this in the examples.2) We may like to put the present lemmas and [Sh:f, Ch.XVIII,2] and more to-gether. The way is clear, we concentrate on 0.

So instead of a tower with six levels we have one with n() + 1 levels.

5.2 Definition. 1) We say Q is an p N N R00,Pr,

-iteration for p is:

(a) Q is a CS iteration of proper notions forcing, p-proper

(b) forcing with Lim(Q) = Pg(Q) does not add reals

(c) P r = P r : < n(), = : n(), g(p) g(p)

(d) P r(N, G, P) implies: P a forcing notion, N is a countable elementarysubmodel of (H(), ), G = Gm : m < k, k < , Gm N P is genericover N

(e) if = 0, P r(N, G, P), then

G = G0

(f) if = n() 1, then P r(N, G, P), iff the demand in clause (d) holds

(g) if Pr(N, G, P) and G G, g(G) finite and Rang(G) = Rang(G) then

P r(N, G, P)

(h) P r0(N, G, P) holds iff (the condition in clause (c) and)

G = G0

(i) if < n(), i < j g(Q), N0 N1 (H(), ), p N0, N0, N1 are count-able and for some (, ) N0 we have N0H(

p) E, N1H(

p

)

E, P r(N0, G, Pi), P r+1(N1, H, Pi), g(G) = g(H), Gk Hk, p = pk :k < g(G), p N0 Pj , pk i Gk, = 0 pk = p0, then we can findG+ = G+k : k < g(G) N1 such that P r(N0, G

+, Pj) and pk G+k

(j) assume P r(N, G, Pi), in < in+1, in N, i0 = i, supn


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5.3 Comment: 0) Of course, we can also use a reasonable parameter for induction.1) In [Sh:f, Ch.XVIII,2] for N N R2 we can use weaken clause (5): t a tree with 2branches if Q N (H(), ), if i < , Pi,i =: {(q, p0, p1) Pi : p0 i

= p1

i

}, then for i < j , i non-limit we have: Pi,i/Pi,j is E2-proper.In light of the previous proofs this is straight.

2) We can replace D(N) by D,i,j(N) for i < j , the supremum of thelength of the iterations we consider.


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saharon shelah- nnr revisited

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