Saharon Shelah- Proper and Improper Forcing Second Edition: Chapter III: Proper Forcing

8/3/2019 Saharon Shelah- Proper and Improper Forcing Second Edition: Chapter III: Proper Forcing

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III. Proper Forcing

0. IntroductionIn Sect. 1 we introduce the property "proper" of forcing notions: preservingstationarity not only of subsets of \ but even of any S C < S < 0(). W e thenprove its equivalence to another formulation.In Sect. 2 we give more equivalent formulations of properness, and showthat c.c.c. forcing notions and Ni-complete ones are proper.

In Sect. 3 we prove that countable support iteration preserves properness(another proof, for a related iteration, found about the same time is given inIX 2.1; others are given in X 2 (with revised support) and XII 1 (by games)).Also we give a proof by Martin Goldstern (in 3).

In Sect. 4 it is proved that starting with V with one inaccessible ft, forsome forcing notion P: P is proper of cardinality ft, do satisfy the ft-c.c. andl^ p "if Q is a forcing notion o f cardinality N I , not destroying stationarity ofsubsets of \ and Ti C Q is dense for i < \, then for some directed G C ),/\i


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90 III. Proper Forcing

In sections 5, 6 we present known theorems on speciality of Aronszajntrees.

In Sect. 7 we prove: for V satisfying CH there is an M 2-c.c. proper P suchthat Ihp " ifo i s u p ( C Ai) > N I , Con(X) implies Con()(see 1.13,note that by 2.1 if > , Con2() implies Cn,2()) For this purposewe shall prove the fol lowing lemmas (1.11, 1.12, 1.13 and then return to theproof of 1.10).

1.11 Lemma. For any sets >, E we denote by DUE the set {x(Jy : x GD &yGE}. For all disjoint uncountable sets A,B we have: W is a closed unbounded(or stationary) subset of < > N 0 ( ^ 4 ) if f WUS#0(B) is a closed unbounded (orstationary) subset of SNO(A U B ).Proof. We can deal separately with the case any of the sets is empty, easily, sow.l.o.g. they are not empty. The proof that if W is closed unbounded in S# 0 (A)then WUS0(B) is closed unbounded in


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then, as we have shown, for some t S#0(B) wehave sUt e Sm(M). Howevers U t G WUS

Q(B) contradicting (WQS

Q(B)) Sm(M) = 0.

Thus we have proved the "only if" part. The "if"part can be provedsimilarly or by applying the "only if" part to W \ = S#Q(A) \ W. i.n

1.12 Claim. (1) If / is a one-to-one function f rom A into B, then for X C


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1 . Introducing Properness 97

complete Boolean algebra without 1. Let T be a stationary subset of < S 0() inV. To prove that T is also a stationary subset of S# 0 () in V[G] we have to provethat for every P-name M = (,F?)tn


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2. More on Properness 99

whose operations are not necessarily defined for all arguments, then Sm(M) isalso a closed unbounded subset of S N O (A).2.1 Claim. For < , Cn2(,P) = Con2(,P).

Proof. To see that let {pij : j < O L i , i < a} be as required by Con2(), i.e.,a < and c^ < for i < . Since < we can apply C n 2( ) and obtainD d= {s G < S* 0() : 3q[p < q G P& (V i 5 ) [{p : j G s Q() which establishes Con2() since D S 0() isexactly the set required fo r Con2(). U2.

2.2 Lemma. Con2(2'p') = ^ > (V > H o )Cn2() , and hence, since < \ andC n 2( ) = > Con2(), therefore (3 > (2\p\}(Con2(}) &( V ) C ro n 2( ) .Proof. It clearly suffices (see 2.1) to prove that for > 2' pl we have C n 2 ( 2 l p l )= > Con2(). Let p, (p^j : j < 0.1,i < a) be as in Con2(). Let 2^ denote thesubset {pij : j < Oi} of P. Let (Ji : i < 2 ' p l ) , be a listing possibly withrepetitions of all pre-dense subsets of P. Let (q^j : j < / % } be a listing ofthe members of Ji, then we can have i < \P\. We define a partial functionF : -> 2p| by F(i) = the first 7 such that J = Ii, for i < . Wedefine also two partial functions G and H on x , into by G(i,j) =the 7 such that p = < ? F ( i ) , 7 fr * < ? j < ? an p suchthat fo r alH G s , {pij : j G s ^} is pre-dense above q. This will establish


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100 III.Proper Forcing

Cori2(X) since the set which is required by Con 0().Let s G 5r a (7 V) ; since T V contains all the partial operations o f M w e have5 2'pl G Sra(M). Since Sra(M) C A we have s 2'pl G ,4; th erefo re there is6 q>p such that

(V i G 5 2p|)({^j : j G 5 i} is pre-dense above < ? ) .W e shall show that fo r this q we have (V i G s a)({ j : j G 5 ^} is

pre-dense above # ), which is all what is left to prove. Let i G s , since 5is closed under F also F ( i ) G 5 2p| (since Rang(F) C 2p|) hence, by ,{9F(i),j : j G s ft} is pre-dense above q. We shall see that {qp(i)j '- j s \ i} = {pij : j G 5 O L i } and this will establish that {pij : j G s ^} ispre-dense above q. For j G 5 ft we know qp(i)j Pi,H(ij) by the definition of#. Since i, j G 5 also f f ( i , j) G 5 and H(i,j) < ^ by the definition of #. Thus


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2. Mo re on Properness 101

2.5 Definition. Let N be an elementary substructure of (#(),G) and letP G N be a forc i ng notion. For q G P we say that q is ( T V , P)-generic, (orTV-generic if it is clear which P we are dealing with), if for every subset X of Pwhich is pre-dense and is in T V the set J TV is pre-dense above q.

2.6 Lemma. A condition q is ( T V , P)-generic z/f for every r which is a nameof an ordinal in the forc i ng notion P, if r G TV then q \ \ - " G T V " (i.e., if thename is in TV then q forces th e value to be in TV) iff fo r every P-name G T V ,q\\- " i f r V thenr G T V " .Proof . We prove only the first " i f f " , the second has the same proof. Assumethat q is TV-genericand let r G T V be a name of an ordinal. Let I = {r G P : r I h"r = ", for some ordinal a}. J is obviously pre-dense. J is def inable from Pan d r in (7J(), G ), hence T G TV .Since < ? is TV-generic, ZTV is pre-dense above< / . Let / be the funct ion on I defined by /(r) = that for which r Ih "r = ",then / is definable in (f(), G) f rom r, hence / GTV . Since J TV is pre-denseabove q, q Ih "GP (n A T ) /0", i.e., r Ih (3r G J T V ) r G G P". Therefore ifG is a subset of P generic over V and q G G then [G] = /(r) holds in V[G],where r G (J TV ) G . Since r G TV, also /(r) G TV (as / G TV ) thus r[G] G TVin F[G]. Therefore r Ih "r G N".

Now assume that for every P-name r of an ordinal, if r G TV then I h"r G T V " . Let J GTV be pre-dense in P. There is an / GH(\) which maps |J|onto J, hence there is such an / in TV .We take Min{i : f ( i ) GGp}. Since/, P G TVand r is definable f rom / and P in (f(), G ), also r G TV, and ris obviously a P-name of an ordinal. By our assumption q Ih "r G T V " , hence< ? Ih "(3i G N ) ( f ( i ) G GP)". Since / maps the members of TV \I\ to membersof TV J, being in N, we have q Ih "(3r G I T V ) ( r G G P)". Therefore ITVis pre-dense above q, which is what we had to prove. U2.62.7 Remark. For > |P| it does not matter in Gon2() whether we requirethat for each i < a the set {pij : j < oti} is pre-dense or whether this set ispre-dense above p. Why? on the face of it the version where we require the set


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{Pij '. j < & i} to be pre-dense is weaker since it makes a stronger assumption,but we now prove from it the stronger version. Suppose each {pij : j < ^}is pre-dense above p. Blow each such set by adding to it all members of Pincompatible with p, to get the set {pij : j < i}. Since \P\ < we know\i\ < so we can apply the weak version of Con2() (i may be > but sinceonly the cardinality figures here it is O.K. as long as i < +). We obtain a setA in T>0(X ) such that fo r s G A we have a q > p fo r which fo r each i G si th ese t {pij : j G s i} is pre-dense above q. For * < j < i, pij is incompatiblewith p an d hence also with , therefore also the set {pij : j G 5 ^} is pre-dense above q which establishes the stronger version of Con^X). For P beinga complete Boolean algebra with 1 omitted it suffice to add -p fo r p G P, p notminimal, so > N O suffice.2.8 Theorem. (1) Let > 2p|, regular and assume P G H( X) (this addslittle since H(X ) contains an isomorphic copy of P). P is a proper forcing notioniff fo r every countable elementary substructure N o f ( f f ( ) , ) satisfying P,p G N there is a condition q, p < q G P such that q is (N, P)-generic.(2) For > 2 l p l , P G H(), P is proper i f f t {N : T V - (#(), G) is countablean d there is an ( A T , P)-generic < ? > p } G T>0(H(X)) fo r every p G P.Proof. (1) We first prove that "if" part, i.e. suppose the condition of thetheorem holds, and we shall prove Con2(2'p') (suffice by 1.10 and 2.2). Let(pij : i < ,j < O i ) be as in Cn2(2 |p |). Let N - < (H(X),e) be such thatP > P > (Pj '. i < ,j < Q i ) e N. For i e N,Ii = {pij : j < 4} G N sinceit is definable in ((),G) from (p > p be( A T , P)-generic; since {PJ : j < ^} G N and it is pre-dense we have thatJi N = {pij : j G N c^} is pre-dense above < / . Therefore to establishCon2(2l pl) it suffices to prove that the set A = {N 2'pl : N - < (H(X) G ),and p,P,(pij : i < ,j < ) G J V , | 7 V | < H0} is in P K o(2 |P |). The setA * = {N e S* 0(H(X)) : N - < (ff{),) and p,P, (p^ : i < , j < < > G TV }

t We do not strictly distinguish between the set of Ps and the set of theiruniverses.


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2. More on Properness 103

is in >0(#()) by Theorem 2.3.This implies that also A G>Ho(2 |P |), by thetechnique of using a model with operations closed under composition which wehave already used several times, or more exactly by 1.12(1).

N ow we prove the other direction of the theorem, so let p G P, {p, P} CT V - < (H(\),e), N countable and we shall find q as required. Assume thatP is a proper forcing, i.e., Con2(2 |p |). Since > 2p| in H(\) there is asequence (pij : i < a,j < ^) where < |P|, < 2p| such that fori < a the set Xi = {pij : j < & i} varies over all the pre-dense subsetof P. By Con2(2'pl) there is a partial algebra with universe 2'p' such thatSm(M) C {s G S0(2 |p |) : (3 q > p) (V i G s a)[{pij : j G s j is pre-denseover q]}', necessarily M G H(\). Since there are such (ptj : i < a,j < ^}and M in H(\) there are such also in N. N obviously contains all the naturalnumbers. Since M is given as a mapping of on all partial operations of M, allthese operations belong to T V and hence N is closed under them (if you preferto see M as (|M|, F), F a function with domain the vocabulary of M, which iscountable, it works as well). Therefore T V 2'p' e Sm(M) and therefore thereis a q > p such that (V i G TV )({pij ' j G TV c^} is pre-dense over q) . Let Ibe any subset of P in TV which is pre-dense in P. Since in (H(), G) it is truethat "for every pre-dense subset of P there is an i < a such that X = Ti " (sincethis is the way we get (pij : i < a,j < ^} in ( f ( ) , G ) ) and this sequencebelongs to TV; clearly this is true in TV. Therefore X = X i fo r some i G TV .Fo r this i, if j G TV ^ then also pij G TV (since (p ) p to be ( T V , P)-generic.

(2) Left to the reader. D2.s2.9 Discussion. We shall now present another proof of the fact that if Psatisfies c.c.c. then it is proper. We shall prove that if r is a name of an ordinal,T V -< (#(), G) and r G TV then 0 I h "r G T V " . There is a maximal antichain Xof P such that for each p G J, p \ \ ~ "r = " for a unique . Because of the c.c.c.|I| < N O so we can take I = {p< : i < }, < ;. The sequence (pi : i < a) is


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in H(X ) and its properties can be formulated in ( f ( ) , G ). Therefore there issuch a sequence in N. Since C N we have pi G N fo r every < , and if ^ isthe ordinal such that p$ Ih "r = * " , then is defined in (if(), G) from P, rand pi hence $ G A/ " . As {p; : z < } is a maximal antichain in P we conclude0 Ih " G {* : i G }", but {* : i < a} C N (seeabove) so 0 Ih "r G A T " .

So for P satisfying th e c.c.c., for , p, A T as in 2.8(1), we have: p is ( T V , P)-generic.

2.10 Theorem. If the forcing notion P is Ni-complete then it is proper.Proof. Let be large enough, i.e., regular and > 2'p', let p, P G N -pand r >q and r \i = q (w e could add Dom(r) [i, j) TV [i, j)).


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For j = 0 the statement (*) is vacuously true. Now we assume (*) for jand prove it for j + 1. Since j + 1 G N also j G N. Therefore, since (*) holdsfo r j we may assume, without loss of generality that i = j. Let Gj be a genericsubset of Pj which contains q. By Theorem 2.11 we have

since Pj G N (because j, (Pi,Qi : i < a) G N and Pj is definable in(if(),e) from j and (Qi : i < }) and G N and hence Qj[Gj] G J V f G ^ ].Remember that Qj[Gj] is a proper forcing in V[Gj]. Since p,j e N alsoP(j) = P J N and pj [G^] G Qj[Gj] W [ G j ] ; since is still sufficientlylarge and Qj[Gj] is proper there is an T J G Qj[Gj] such that T J > pj[Gj]and T J is ( T V [Gj],Qj[Gj]) -generic. Since th e only requirement we had aboutthe generic subset Gj of Pj was that is contains q, q forces the existence of anT J as above. By the existential completeness lemma I 3.1 there is a name T Jsuch that q \ \ - j " T J G Q J & T J >PJ&J is (N[Gj], Qj )-generic" and w.l.o.g.\\-P. " T J GQj\ We set now r = q U { ( j , r j ) } Obviously r GP J+ I and r f j = g.Also since < ? > p f j and q I h "r^ >PJ" we have r >p. We still have to provethat r is ( A T , Pj+)-generic.

By the corollary in 2.13 in order to prove that r is ( A T , PJ+1)-genericit suffices to prove that for all generic subsets G of Pj+i which contain r, [ G ] n O r d = n O r d . Let Gj be the part of G up to j i.e. GnPj. Since r G Gclearly q GGj. Since q is (N , Pj)-generic we have N[Gj] Ord = N Ord. LetG * C Q j [ G j ] , be the "j-th component" of G. Since r G G clearly r,[G] G G * .Since r^Gj] is (^[G^^^G^D-generic wehave7V[Gj][G*]nOrd = T V f G ^ O r d ,and using the equality above we get N[Gj][G*] Ord = T V Ord. We haveto observe that N[G\ C N[Gj][G*], then we have N[G\ Ord = N Ord.For every Pj+i-name r in N there is a name r* G T V as in Lemma II 1.5(r* is definable f rom r and Pj+i, and is hence in N). By Lemma II 1.5,lG\= [Gj][G*}N[Gi]lG }.

Now wecome to deal with the case where j is a limit ordinal. Let (rn : n < ) be a sequence of all Pj-names of ordinals which are in N. Note that T V j


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3. Preservation of Properness Under Countable Support Iteration 111

is a countable set with no last element, so let i = I Q < i\ < in < , (n < ),be a sequence cofinal in it.

F or p 1 G Pj an d q f GPin such that q1 > p'\n le t q'pf denote q f U (p'\(j \in)). Since q' > p'\in, q'Qp' G Pj and q'Up' > pi. For p,p2 Pj we writePi w p2 for pi < p2 & P2 < P i-

We define now two sequences (qn : n < ) and (pn : n < ) such thatqQ q, P Q p and for all n < :

(1) fn GPin and #n is (N , Pirl ) -generic(2) qn+ \in =qn(3) pn G Pj an d Dom(pn) C T V j(4) qn>pn \in(5) Pn+i tin-pninand


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f t , s * are as required by (a) and (b). Therefore V[G] N (35*)(3^ G G)(s* and< ? are as in (a) and (b), and 5* is the first such element in some fixed wellordering of P j ) " . By the existential completeness lemma there is a P^-names* such that qn I h "(3q* G G) [s*,


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we have proved above qk\ \ \ ~ p ^ "pk() < ()"5 hence qk\j \\-pk "P() r7. Then

r / /f2 Ih r" satisfies (l)-(3),again a contradiction, because r / ;fa 2 > r.


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3.3G "Composition Fact", q e P+ is (P+, A^-generic iff:q\a is (P, 7V)-generic, and q\a Ih uq() is (Q, A[G AT])-generic."

Proof. See 2.3.3H Induction Lemma. For all N , for all N , all p T Vassume p is a P-name for a condition in P, and(a) q P(b) < ? is (P,7V)-generic.( c ) < ? l h p Q " p e G U V ' .

Then there is a condition q+ :(a)+ g + G P / 3 , g + t = g(b)"1"


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First we will define a sequence (pn : n < ), pn G N such that in N thefol lowing will hold:(0) pn is a P-name for a condition in PWlh+?n+l >* Pn(2) lhP+?n+1In(3) ll~ +1 "If P n n+ G Gan+l then pn+ n+ G G t tn+1".

For each n we thus get a name pn that is in N. For each n we can use the"preliminary lemma" (and Remark 3.3E before its proof) in N to obtain pn+-N ow we define a sequence (qn : n < ), qn GPttr,and gn satisfies (a), (b), (c)(if we write qn for g, pn for p, and n for ).


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4. Martin's Axiom Revisited 117

3.4 The General Associativity Theorem. Suppose (Pi,Qi : i < a) is a CSiterated forcing system each Qi proper then the parallel to II 2.4 holds.Proof. Left to the reader. Ds.4

3.5 Theorem. Suppose (Qj : j < a) is a (< /)-support iterated forcing,

If Ihp. "Q j < e Q],Qj a dense subset of QJ" and PJ = L im { Q t : i < j),t/ien Pj


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So why not include such forcing in MA? Because we can find pairwisedisjoint stationary sets S

nC \,\ Un }. Thus if G J 0 for n < ;, < i, then inV, each 5n is not stationary.

Y ou can still argue that CH is the cause of the problem but we shall provein Theorem 4.4 that even 2H > H,S C \ stationary co-stationary there isa forcing notion P of power H I, not changing cardinalities and cofinalities butstill collapsing S.

So it is natural to change the question to "P of power H I, not collapsingstationary subsets of H I " , and we shall answer it positively, assuming there isa model V of ZFC with a strongly inaccessible cardinal.

The natural scheme is to iterate (by CS iteration) proper forcing of powerH I, in an iteration of length % . However to prove th e consistency of almostanything by iterating proper forcing we usually have to prove the ft-chaincondition is satisfied, where K will be the new H 2 and the length of the iteration.We have a problem even if \P\ = H I, Ihp "|Q| = H I " , P * Q have a large powerbecause of the many names. We can overcome this either by using K stronglyinaccessible, or showing that the set of names which are essentially hereditarilycountable is dense.

Another problem is that "not destroying stationary subsets of \n is notthe same as "proper". However we shall prove that if P is not proper, then'^Levy(H 1 ,2i p i ) " destroys a stationary subset of \". So instead of "honestly"dealing with a candidate P i.e., a forcing notion which does not destroy sta-tionary subsets of i, but is not proper we cheat and make it to destroy astationary subset of \.

4.1 Theorem. Suppose Q = (Pi,Qi : i < K) is a CS iteration Ihp. "Qi is aproper forcing notion which has power < ", is regular and (V < )* < K.


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4. Martin's Axiom Revisited 1 1 9

Then PK = limQ satisfies the ft-c.c., and each Pi(i< ft) even has a dense subsetof power < ft. Hence for i < ft, Ihp. "2* < A C " .Proof. Easily (twice use 3.4),w.l.o.g. the set of elements of Qi is a cardinali < ft; (i.e., * is a P^-name of a cardinal < ft).

4.1A Definition. For a forcing notion P and P-name Q of a forcing notionwith set of elements (a P-name of cardinal) with minimal element 0g (candemand it to be 0) we define a hereditary countable P-name of a member of Q:it is the closure of the set of ordinals < (see(*) below) by the two operations(a) and (b) (seebelow):

(*) the names for a < or more exactly for an ordinal the P-namea is such that a[G] = a if < [Gp] and a[G] 0g[Gp] if >[Gp]. Of course we can restrict to a such that l/p >. Alsoif = we can use just , <

(a) if rn(n < ) are such names, and pn G P (for n < ;) ften let r bethe rnfor the least n satisfying pn G Gp, and 0g if there is no suchn.

(b) if rn>m(n < ;, m < ) are such names, let r be the least ordinal a < such that for every n, {n>m : m < } is pre-dense over r (in Q); and0g if there is no such . (Remember: the members of Q are ordinals


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subset of , f ( ) is a hereditarily countable P/-name of an ordinal < ;}. LetP C P inherit its order. We now can prove by induction on < , that Pis a dense subset of P, using the proof that properness is preserved by CSiteration. It is clear that |P^| < \\*, so for < / , P has a dense subset ofpower < ft. So we finish.

For = ft, if pi e P K for i < , clearly S = {i < K : c(i) = N I} isstationary, f ( ) = Sup[i Dom(pi)] < i is a pressing down fu nct io n, hence byFodor Lemma on some stationary 5 C 5, h has a constant value 7. There isa closed unbounded C C K such that: if G C, < , then Dom(p) C /?. So5 C is still stationary, hence has power K and for , / 3 e C S\,pa,p arecompatible iff P f 7 , > / 3 t 7 are compatible (in P7 or P, does not matter). Butwe have proved that P7 satisfies the K-chain condition, so we finish. U4.

We have proved

4.IB Claim. For Q = (Pi,Qi : i < ), a CS iteration of proper forcing, suchthat for each i < a it is forced that Qi is with set of elements C Ord, we have,fo r i < a:

1) PI = {/ G Pi', for j e Dom(/), f ( j ) is a hereditarily countable Pj-name}is a dense subset of P^, and i < j < a and / G P f\i G P/

2) If / is a Pa-name of a function f rom to Ord and cf () > N O en for adense open set of q G P, for some < a and P^-name # of a funct ionf rom to Ord, q \\- Pa "f = g".

4.2 Theorem. Suppose P is not proper, then there is an Ni-complete for c ingnotion Q, in fact Q-Levy(K!,2lpl) will do, such that |Q|


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4. Martin's Axiom Revisited 121

Fact A. \ \ ~ Q "S is a stationary subset of SQ()n .This is because Q is H I -complete hence, by 2.10, proper.

Fact B. The statement Ih p "5 C S0() is not stationary " is absolute, i.e., ifit holds in V it holds in VQ.

We just have to check that the P-names F LU continue to satisfy the suitablerequirement (and Q adds no new member to P and no new member to 5).

Fact C. \ \ - Q "the ordinal has power H I " .This is trivial.

Fact D. If forcing by P destroys a stationary subset of S0(A)(A = in ourcase ), A of power H 1 ? then forcing by P destroys some stationary subset of \.(follows f rom Lemma 1.5 and 1.12(3)). U4.24.3 Theorem. Suppose ZFC has a model with a strongly inaccessible cardinalK . Then ZFC has a model in which 2* = H 2 and(*) If P is a forcing notion of power H I not destroying stationary subsets of i, and Ti C P is dense for i < \ then there is a directed G C P satisfying

Q toi >,and |5| = K . Define by induction on i < K a CS iterated forcing system(Pi, Qi : i < K ). Let (


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Assuming (Pj, Q j : j < i) is already defined, le t_ ( i , if I l - P i "(i,


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5. On Aronszajn Trees 123

Q = {G : C a closed bounded subset of 5 which belongs to L[S, Gp}}GI < G2 if f GI = C2 ( M a x ( C ) + 1)Clearly Q is a forcing notion of power N I , so it cannot collapse cardinals or

regularity of cardinals except possibly K I (all finite subsets of S belong to Q).So we shall prove that P* Q does not collapse N I . So let (in V ) N - < (H(X), G ),T V countable, (p, q) G P * Q G N , (p, q) G T V , 5 = N \ e S an d supposeGp C P is generic over V and p G Gp. Note that as S is a stationary subsetof , there is such N (in V). So it is enough to find (qr,p') > (q,p) which is(N , P * Q)-generic. As P satisfies the countable chain condition, p is ( T V , P)-generic (by 2.9),hence N[GP] V = N. Clearly Q[GP] GL[S,GP] C V[GP],now N[Gp] does not necessarily belongs to L[S, Gp], but N[Gp] L[S, G p] isN[G] L5[5, Gp] = L[S, Gp] GL[5, GP] and is a countable set in L[5, GP].In LfS, Gp] we have an enumeration of Q[Gp] 7V[ Gp] (of length ), say(n : n < c;} (but not of the set of dense subsets ); in fact we have it evenin L[S, Gp \( + 1)] (and as we use L ev y( H o , < N I ) not Levy(^o? < ^) even inL[5, GP \]). Now in L[5, GP] there is a Cohen generic real over V[GP \( + 1)]say r* G and we use it to construct a sequence G (Cn : n < ) suchthat Cn GQ[Gp] in L[5, Gp] i.e. C L[S, Gp], e.g. we chooseCn by inductionon n; we let GO = q[GP] and we let Gn+ be m(n) where m(n) is the firstnatural number m such that: m > r*(n) and Q[Gp] |= "Gn < qmn.Let G ={q : f G Q[GP], an d q G A[G P] and for some n, Q[GP] |= q < Cn}. ClearlyG' C 7 V [ G p ] n Q [ G p ] is generic over V[GP\(+l)}. So qi = {JnCnU{} G Q[GP]is (A[Gp],(3[Gp])-generic. Going to names we finish. D 4 >4Remark. Also N in V [ G p ] is O.K. as it still belongs to some V[Gp \a] fo r somea


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tree which is a counterexample to the tree property of K is called a K - Aronszajntree. By the Knig infinity lemma H

0has the tree property.

2) A ft- Aronszajn tree in which every antichain is of cardinality < K iscalled a -Souslin tree. An H i - A ro n s za jn tree, and an Ni-Souslin tree are calledan Aronszajn and a Souslin tree, respectively. A + - A r o n s z a j n tree is saidto be special if it is the union of antichains, if = N O wemay omit it. Aspecial Aronszajn tree cannot be Souslin, since in a Souslin tree every antichainis countable, hence the tree, being uncountable, cannot be the union of N 0antichains. A -wide Aronszajn tree is a tree with \ levels, nodes and no

5.1A Remark. It is easy to show that an Aronszajn tree T is special if f thereis a function / : T > Q which is order preserving. A A ~ * ~ - t r e e which is specialis a +-Aronszajn tree. The following was proved by Aronszajn, [Ku35] (and5.3 is a well known generalization).

5.2 Theorem. There is a special Aronszajn tree T.Proof. The members of the -th level Ta of T will be increasing boundedsequences of rational numbers (of length ) with < as the tree relation. Whenwe come to define Ta we assume that for all < 7 < and for all x G Tand every rational q > Sup(Rang(x)) there is a y G T such that x < y andSup(Rang( i/) ) = q, and \T \ = N 0 If = 0 take Ta = {}. If a = 7 + 1take Ta = {z~ < q >: z G T&q G Q&g > Sup(Rang(z))}, where Q is theset of all rational numbers. Obviously \T a\ = \ T \ N 0 = N O - The inductionhypothesis holds also for < , as easily seen. If is a limit ordinal thenfor every x G U Sup(Rang(x)) we shall construct asequence y of length which extends x such that Sup(Rang(y)) = q. Let(n : n < ) be a (strictly) increasing sequence such that x G T0 andSup{/3n : n < } = a. Let (qn : n < ) be an increasing sequence of rationalesuch that < ? o > Sup(Rang(x)) and Supn


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5. On Aronszajn Trees 125

defined; Sup(Rang(zn)) = qn < qn+ then by the induction hypothesis there isan z

n+ GT

0n+lsuch that x

n< x

n+land Sup(Rang(z

n+)) = g

n+ Take

y =Un


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as strong as the consistency of the existence of a weakly compact cardinal; weshall see that these two consistency assumptions are equivalent. M itchell hadproved this theorem, and Baumgartner gave a simpler proof by proper forcing.

The following theorem is due to B aum gartner, M alitz and Reinhart [BMR].5.4 Theorem. For every tree T of height \ with no branch of length \ (norestrictions on its cardinality) there is a c.c.c. forcing notion P such that in thegeneric extension of V by P the tree T is special. If \T \ < 2H then by Martin 'saxiom it fol lows that T is special.

As a consequence, if we assum e M artin's axiom and 2 > N I , then allAron s za j n trees are special and hence there are no Souslin trees.Proof. Let P be the set of all finite functions p from T into such that ifp(x) p(y) then x and y are incomparable. For every x T the set X x of allmembers of P whose domain contains x is obviously dense in P, hence if G isa generic subset of P, F = UG is defined on all of T and if F(x) F(y) then xand y are incomparable. If we have Martin's axiom and \T\ < 2^ then there are< 2 N dense sets Ix and the directed set G can be taken to intersect all of them,and F = UG is as above, i.e., it specializes T since T Un{x : F(x) = n}.

We still have to prove that P satisfies the c.c.c. Suppose there is an un-countable subset W of P whose members are pairwise incompatible. Withoutloss of generality we can assume that all members of W have the same car-dinality, that their domains form a -system with the heart s and that forall p G W,p\s is the same function. Denote W = {pa : a < \} and letDom(p) \ s = {x t t f i , . . . , , n } Let a, < i, pa and p are incompatible,hence pa Up P. Since pa and p coincide on s and the rest of their domainsare disjoint we must have for some 1 < k,l < n, pa(xajk) = Pfaj) while xaand X tt are comparable. Let Ya,k, = { < ' " >P( ,/c) = p(x)and x >fc and X j are comparable }. As we saw U


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6. Ma yb e There Is No N 2-A ro n s z a j n Tree 127

y / 3 , f c , G E hence Y,;^ 10,*,* G an d therefore | y , / f e ^ y / j . f c , * ! = N I . Let7 y iy Mth

en x,/c and X tk are comparable with XJ. Now 7/swithdifferent 7 G Y a,k,t 1 0 , / c , . are different, an d since there ar e only countably

many members of T below xa,k or below X ^ (in T's sense) there must besome 7 G Y a,k,e n /3,M sucnthat x,^ is greater than both xa and x/^ (inT's sense) and since is a tree, xa is comparable with X ^ This holds for all, ^ 4 hence has a l inearly ordered subset of cardinality K I: {x ,f c ' & G -A},and therefore a branch o f length cj i , contradicting o ur assumption. D 5.4

6. Maybe There Is No N2-Aronszajn TreeToward this we mention (see for history, 6.2 below):6.1 Lemma. 1) As s um e V \ = "2H > tti&T is an N 2-Aronsza jn tree." Let Pbe an H i-complete forcing notion. Then V [ P ] N = " has no cofinal branches".(#2 m ay become of cardinality N I in V [ P ] so it does not have to stay an ^2-Aron s za j n tree.)2) As s um e :

(a) T is a tree with 5* levels such that c f ( < 5 * ) > N O(b) for no l imit < 5* of cofinality N 0 can we find pairwise distinct x G T

for G2 such that: [ < = > {x \a : G

2} is finite] ( \a is theunique y el > f for some < 2 and 5 :>2 > P such that:

(i) F() = the root of , 5() = p0(ii) for all x G2


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F() an d 5(77) ar e defined by induction on the length of 77 . Assume 5(77),F() are defined we shall define S(f (}), F(f ()) for I = 0,1. Since 5(77)>Ppo, S() has, for every < ^ , an extension which forces some member of T (i.e., the set of vertices of height in the tree) to be in B. If {t : F() p S() such that p Ih " G B"} was a set of pairwise comparablemembers of T then they would be a branch of T in V, contradicting ourhypothesis. Therefore there are two incomparable 's in this set, take one tobe F(~ < 0 >) and the other to be F(~ < 1 >) and choose S(~ < 0 >)and 5 (77 < 1 >) as conditions > S() such that S(~ < >) Ih "F(rf < i>) G B" fo r I G {0,1}. Since th e range of F is countable it is included in someT \ O L fo r some a < < 2 . Since P is H I -complete, fo r every 7 7 G 2 ,P contains acondition p which is an upper bound of {5(77 fn) : n < }. Since p > P Q thereis a > p and a t G T such that ^ Ih %G B". Let ^ 7 7 , and i / , 77 G 2.Let n be the least such that v\n 7 7 f n , then by requirement (iv)above we havethat F(v\n) and F(\n) are incomparable in . Now P |= " > p > S(fn)",hence also Ih "F(\n) G B". Since forces that is a branch of and that^,^(77fn) G B clearly t and F ( n f n ) are comparable in T. Since the heightof F(\ri) is < a and the height of t is we have F(\ri)


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6. Maybe There Is No N 2-Aronsza jn Tree 129

Proof. Let K be a weakly compact cardinal. W e shall use a system (P^, Q i : i q we know Qj is dense. Let G be thegeneric subset of Q then there is a p G G Qj such that p* G J, and for somen < , p * G n2 (as p* G Q = >2). Since p C F[G] we have t[G\ \n = p* , hencet[G\ \n G J, which establishes that t is a Cohen real. D6.36.4 Lemma. For every inaccessible < , V [ P ] ^ "Proof. Let G be a generic subset of P. We saw already that V [ G ] \ = " > N 2 " .Suppose now that V[G\ \ = "N 2 = ", where < . Let F G V[G\ be a functionon x i such that for all 0 < < we have {F(,/J) : / ? < i} = , i.e.,F(, ) is a m a p p i ng of i on . Let F be a name of F and let po G forcethat F is as we described. For each < and / ? < \ let J be a m a xi m a lantichain of members of P\ which are > po and which give definite values toF( , /3 ) . Since as we saw, P satisfies th e -c.c. condition, clearly \Ia\ < .Let J = U < >/3 < I >/3 . Since is regular \I\ < . Since each member ofJU {po} is a countable function on there is a 7 < such that JU {po} PLet G = G P7, then clearly F GV[G7] (since we define F(, ) in V[G7] tobe that 7 for which there is a f G I0i G such that g I h "F(, ) = 7"). ThenV[G7] N < 2, but since V [ G ] N " = 2" we have V [G ] N "-N 2 " . Butsince we force above V [G ] with Q$[G} which collapses th e H 2 of V[G7], wehave V [G ][G fl] N " < H 2" hence V[G\ N " < K 2 " , which is a contradiction.


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6. Ma yb e There Is No K 2-Aronsza jn Tree 131

Continuation of the Proof of Theorem 6.2. Now let us go on with th e proofof th e theorem. Assume that there is a P Q G P such that P Q lhp " there isan K2-Aronszajn tree," i.e., po "~p " there is a ft-Aronszajn tree T on ft anda function F on ft such that for a < ft, F() is the rank of a in T, (since bythe l e m m a V [PK ] \ = "K 2 ft"), i.e. po "~PK " there is a transitive relation Ton ft such that for all a,, 7 G ft we have: [T7&/3T7 = > T7] and there is afunction F from ft into ft such that for all ,/3 < ft: if T/3 then F() < F(/3),for all a G ft and 7 < F(a) there is a Ta such that F(/3) = 7, and for all7 ft there is a /? G ft such that for all G ft if F() = 7 then < / ? , and forall C ft there are a, G 5such that a -T--/?T, or else there isa < ft such that C /J". This implies, by the existential completeness thatthere are canonical names of T and F of relations on ft such that:

(*) Po "~ "T is a transitive relation on ft and ( V , / 3 , 7 < ^(aTj&Tj T7) and (V,^,7 < / c ) [ = / ? V /3 V /3T] and (3F : ft - )(Va, < ft)[(/3 -> F() < F(^)) & ( / 3 < F() - (37 < ft)(7T F(-) = ))} and(V 7


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indescribable there is an inaccessible cardinal < ft, from E such that (H(),G, T f x , Fx) is an elementary substructure of (H(), G, T, F) and satisfiesthe I!* statement mentioned above. P and lh p are definable in (H(), G, T, F)and the same definitions give P and lhP in (f(), G , T n x , F ( x ) .Therefore (T x ) [ G ] is a -Aronszajn tree in V [ G \ ] (where G is the genericsubset of P over V and G = G P). We claim that (T x )[G] is thepart of the tree T[G] up to level . Let ( , ) G (T x )[G] then , < Xand for some p G G , p lt-p {,/?} G ( x ), hence p \\-PK "(,/3) G "(by the relation between the two above mentioned structures), hence, sincep G G, (a,) G T[G], shows that ( x )[G] is included in the part of [G]up to level . The proof of the equality of these two trees will be completedonce we show that in T[G] all the ordinals in the levels below are < . Let < X then, since (*) holds for the structure (if(),...) there is a p GG\ andan ordinal < such that p lh P (V G ) (F x )() = - < /?),therefore p lh P (V G ft) (F() = > < /?), and since p G G w e have inV [ G ] that all the ordinals in the level are < < X.

Thus in V[P\] we know T ( x ) is a -Aronszajn tree, i.e., an ^2-A r o n s z a j n tree. Now we saw that in V[P\] there are at least real numbers,i.e., 2^ > ^2 m "[-P] an


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7. Closed Unbounded Subsets of \ Can Run Away from Many Sets 133

which there are no special ^2-Aronszajn trees ( i.e., trees which are the unionof N I antichains ) then it suffices to use a Mahlo cardinal K (see [B3] on this).

7. Closed Unbounded Subsets of Can RunAway from Many SetsBaumgartner [B3] has proved the consistency of the following with ZFC+2^ =N 2: if Ai C , for i < i, is infinite countable then there is a closed unboundedC C \ such that Ai C for every i < \. We prove a somewhat strongerassertion.

7.1 Theorem. ZFC + 2H = N 2 is consistent with:(*) i f A i C i has no last element and is nonempty and has order type

< SupAi, for i < , then there is a closed unbounded subset C of i, suchthat C ^4i is bounded in for every i

Remark. Abraham improved Baumgartner's result to:ZFC +2K = anything +

(**) there are N 2 closed unbounded subsets of K I, th e intersection of any K I ofthem is finite.Galvin proved previously that CH implies (*) fail. Our proof is similar to

[Sh80 4].Proof. We start with a model V satisfying CH, and use CS iterated forcing oflength u;2, such that in the intermediate stages CH still holds. So by 4.1, itsuffices to prove.7.2 Lemma. Suppose V satisfies CH. There is a proper forcing notion of powerN I which adds a closed unbounded subset of C of and Sup[C A] < Sup Afor any infinite A C \ with no last element and order type < Sup(^4), whichbelongs to V.


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Discussion. A plausible forcing to exemplify 7.2 is:Q = {C :C a closed countable subset of \ so that if A GV is a set

of ordinals < \, = Sup(A) is a limit ordinal, A has ordertype < < J , then S up [A C] < }

(so if > sup(C) this holds trivially), with the orderC < C2 if f C2 [ (M a xC i ) + 1] = Ci.

S o the elements of Q are approximations to the required C. It is clear that ageneric subset of Q gives a C as required, provided that N ! is not collapsed;hence the main point is to prove the properness of Q. Unfortunately it seemsQ is not proper, in fact has no infinite members. However if we want to adda C G Q which is ( A T , Q)-generic fo r some T V -< ( f ( ) , G ) a c.c.c., forcing isenough. So we could first force with some P, |P| = ^2?^ satisfies the c.c.c.such that

Ihp "2* = N 2 a n d M A holds".Then we define Q in V[G p] for Gp C P generic over V; similar to the definitionabove but members of the forcing notion are from V [ G p ] whereas th e A fo rwhich we demand sup(C) A < sup(A) are from V. So

Q = {C V [G]: C a closed countable subset of \, and if A G V isinfinite with no last element and (order type A) < SupA < MaxC, thenSup(CnA) < SupA}.

N ow Q is proper, and adds a C as required, so P*) adds a C as required.Unfortunately P* Q also collapses ^2, so if we are willing to use some stronglyinaccessible K , > N O > there is no problem. Otherwise, we use a restricted versionof M A, which is consistent with 2K = N I , so 7.3, 7.4, 7 .5 below prove Lemma7 .2 hence Theorem 7.1. Throughout we use the order < on Q defined above.7.3 Claim. Suppose

(a) V satisfies C H , P i s a forcing notion o f power H I satisfying th e c.c.c.and G C P is generic over V.


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7 . Closed Unbounded Subsets of Can Run Away from Many Sets 13 5

(b) < is a limit ordinal, ( V , / 3 < )[a + < ], R e V[G\ is acountable family of closed bounded subsets of , ordered by C\


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7.4 Claim. Suppose V satisfies CH. There is a forcing notion P of power H Isatisfying the c.c.c., such that the following statement is forced:

(*) Suppose < \ is limit, (V,/3 < ) [a + < ] (equivalently < 5 is anordinal power of ) an d R is a countable family of closed bounded subsets of such that (VG G)(V/3) (MaxC < < -> C U{/?} ) and for n < wehave: Zn is a dense open subset of R such that X\ = {(C, 0) : C GTn} C Qis pre-dense in Q (where QR is from 7.3 clause (c)). Then there are Cn G ,such that Cn < Cn+, SupnMaxGn = 5, Cn+ GInand for every A G V , A C 5of order type < 5, Sup[A (UnCn)] < 5; moreover we can choose CQ e Rarbitrarily.Proof. We can use an FS iteration (Pi, Qi : i < \) of forcing notions satisfyingth e c.c.c., such that for every possible R and (which are in V or appear inVP for some i < ) for uncountably many j < \ in Vp* we have Qj = QR.

7.5 Claim. Suppose V satisfies CH, P is as in 7.4, and Q = {C : C G Vpa closed bounded subset of \, such that for every infinite countable A G V ,A C cji with no last element, and (order type of A) < s u p , if S u p A < MaxCthen S u p ( A C) < Sup A} ordered by: C\ < C2 iff C\ = C2 ( maxC + 1).T/ien Q is proper (i.e., Ihp "Q is proper ") and has cardinality < H I.Proof. Let G C P be generic over V , be regular big enough and let in V[G]

S d= {N : N x (ff(),),P G T V , #(N2)V G T V , G G A T , an d there is asequence (N i : i < ) such that A^ X A T , (A^ : j < i) G A ^ + i , A ^ = Ui 0 ((/ f()) (in V [ G ] , remember we do not distinguish strictlybetween A / ' and its set of elements).


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8. The Consistency of SH + CH + There Are No Kurepa Trees 137

Fact B. In V [G\ : if N G 5, = N ] ^ R d= Q[G] T V , and J G T V is a denseopen subset of Q[G],(a) {(C, {(Ai,a.i) : i < n}) e QR : C eI\R} is a , dense subset of QR and

{(C, 0) : C G I\ R} is a pre-dense subset ofQR.(b) For every C G Q[G] A T there is C*, C < C* G Q[G] and C* is (Q[G\,N)-

generic.Proof. Let (A ^ : i < 5} be as in the definition of 5. Let pQ = (CO,

< < n}) GQ and let J GA f be a dense open subset of Q[G], so for some i < we have J, p GA^. Let < ^ be A^ . Let = {j < : A{ [j, j+) 0},so otp( ) < otp(A) < . hence as in the proof of 7.3 there is j G (i, ) whichdoes not belong to A' for t < . Now pi = (CO U {(^ },{( , ) : < n})belongs to R and is > p. There is C2 GQ[G] [G] such that C U{ } < C2(in Q[G]), hence there is such C % in A^-+ (as relevant parameters belong to it).N ow p2 = ( C - 2 , {( ,t}\i< n}) belongs to Q[G] N and is > pi > pQ. Thisproves clause (a).

Let (In : n < ) list the dense open subsets of Q[G] which belong to A fand C G Q[G] A T , and let Jn = ^ A T , and let = Q[G\ A T with th einherited order. Trivially C G R & max(C) < / ? < A T i = > C


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instead of i - K H . (B e careful: K H says "there are K u r e p a trees", but SH says"there are no Souslin trees"!)

Solovay proved that Kur e pa trees exist if V = L, m ore generally Jensen[Jn] proved th e existence of ft-Kurepa's trees follows from Jensen's < 0 +, whichholds in L fo r every regular uncountable ft which is not "too large" . B ut -KH) from th e consistency of "(3ft) ftinaccessible". This serve as a prelude and motivation to Chapter V (and evenmore Chapter V I ) , which deals with preservation of such properties. In chapterV we will show that moreover the iteration we construct here does not addreals, so (since we start from a m odel of CH ) we wil l get a mo del of "C H + S H+ -KH".


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8.3 Definition. A forcing notion Q is good for an i-tree T (so the -th levelof T is T O , etc.), if for any countable elementary submodel N -< H(, G) , fo rX large enough, with T, Q G T V , an d every condition p G N \ Q, there existsan ( A T , Q)generic condition q > p such that if G A T is a name, < / I h "either[Gp] is an old branch of T or [Gp] T


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to follow. In the second part we show how to extend the first part in order toget a full proof of the theorem.Let Q = (Pi,Qj : i < a,j < a) be a countable support iteration such that\\ -P. "Q j is proper" for all j < a. We fix some regular which is large enoughfo r what we need. As in III 3.2 we prove by induction on j < a the condition(*)j, which is stronger than the proper ness of PJ :(*)j Pj is good for T, and

(a) forcing with Pj add no i-branches to T and:(b) for all i < j and countable N - < (#(), G) such that i, j, Q e N andp G PJ T V and an ( T V , P^)- generic q G Pi which satisfies q >p\i thereis r such that:

(i) r e Pj(ii) r\i = q

(ii) r is ( T V , Pj)-generic(iv) p < r(v) om(q)n[iJ) = N[iJ).

The proof is split to cases. Note that though in the statement (*)j(b) wesay "for i < j" it holds for i = j too.Case 1. j = 0Trivial.Case 2. j a successor ordinalLet j j + 1, now (a) of (*)^ holds as (a) of (*)J1 holds and Qi is good for T,so we shall deal with (b) of (*)^. So by the induction hypothesis applied to j\and i (see remark above) w.l.o.g. i = j\ and continue as in the proof of III 3.2.Case 3. j a limit ordinalWe first look at a case of clause (b) of (*)j and/or of clause (b) of (*) j (by 8.4this suffice) so i, N, p, q are given as there.

As T V is countable, we can pick a sequence of ordinals of order type ,(in : n < ) which is cofinal in TV j and such that io = i and inG TV for all n.Let (Tn : n < ) , enumerate all the dense sets of Pj in TV . Let ((n, xn) : n < )


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enumerate the pairs (, x) where r is a P^-name f rom N and x G TN wherec def Ar _^y Y V u ; .We define by induction on n a condition gn and a P^-name of conditions pnsuch that:(a) gn G Pn is (W,P


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In V[Gin], for any p e Pj/Gin = {p Pj : p\in e Gin} let B % =5[GJ ={t G T : p \Yp. jGi "^ ^ In"}> equivalently { G T : for some p satisfyingp


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(*) if GJ C PJ is generic over V an d q G G j then pn[Gj P$J G Gj.So g is ( T V , Pj)-generic and as

clearly q is above p. This show that q is as required in clause (b) of (*)J . Butby the choice of p^+ (and the list {(n,xn) : n < )) necessarily q I h "forevery r G-/V[Gj], if r is not an old i-branch of T then r N[Gj] is not of theform {t :t

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