Saharon Shelah- Proper and Improper Forcing Second Edition: Chapter V: α-Properness and Not Adding Reals

8/3/2019 Saharon Shelah- Proper and Improper Forcing Second Edition: Chapter V: -Properness and Not Adding Reals

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V. -Properness andN o t Adding Reals

0. IntroductionNext to not collapsing N I , not adding reals seems the most natural requirementon a for cing notion. There are many works deducing various assertions fromC H an d many others which do it from diamond of H I . I f w e want to show thatthe use of diamond is necessary, we usually have to build a model of ZFC inwhich CH holds but the assertion fails, by iterating a suitable for cing. A crucialpart in such a proof is showing that the forcing notions do not add reals evenwhen we iterate them. So we want a reasonable condition on Qi (in VP) whichensures that for cing with Pa does not add reals when (Pi,Qi : i < ) is aC S iterated for cing system. Another representation of the problem is "find aparallel o f M A consistent with G . C . H . " .

The specific question which drew my attention to the above was whetherthere m a y b e a non-free Whitehead group of power K I (from [Sh:44] w e knowthat there is no such group if V = L or even if Os holds for every stationaryS C , and that there is such a group if MA +2^ > N ! holds). This isessentially equivalent to: "Is there a stationary 5 C i, and for each E 5an unbounded subset A S of order-type , such that A = (A$ : E S ) has theu n i f o r m i z a t i o n property" (see I I 4.1, i.e. if h = (h $ : E S), h g a functionf ro m A to 2 = {0,1} then fo r some h : (J A S > 2 for every 5, h g C * h i.e.S{a E AS ' hs(a) h(a)} is finite). It is easy to see that Os implies (Ai : i G 5}


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does no t have th e u niform iza t ion property (and in II 4.3,w e proved, from ZFC+ MA -f2^ > H I , that A has the u niform iza t ion property).The solution was surprising. By Devlin and Shelah [DvSh:65] (see AP 1here), there is a weak form ^ of 0 which follows from CH; in fact is equiv-alent to 2H < 2N l. This statement implies many consequences of the diamond(see on it in Appendix 1; see [Sh:87a], [Sh:87b], [Sh:88] and a systematic de-velopment in Abraham and Shelah [AbSh:114] and [Sh:192] and lately [Sh:576],[Sh:600]). I n particular (A$ : G S ) does not have th e uniformization propertywhen S G ) i.e. S contains a closed unbounded subset of \. This still leavesopen the question for 5 a stationary costationary subset of \. Now for suchsets it was proved in [Sh:64] that the u niform iza t ion property may hold for afixed stationary costationary subset S of i , i.e. for all (A : G 5), A S C u n b o u n d e d of order type . However O u > \ s (and even 0 \5) may still hold.On the situation for > H I see [Sh:186], Mekler and Shelah [MkSh:274] and[Sh:587]. More i n f o r m a t i o n on the connection between u nifom iza t ion and grouptheoretic questions see [Sh:98] or see the book [EM] and lately Eklof, Mekler,Shelah [EMSh:441], [EMSh:442], Eklof, Shelah [EkSh:505].

We also deal with "when does a CS iteration of proper for cing add no newreals?" For this we need two properties. One is D-completeness (see 5) whichis a way to exclude the impossible cases, and another is -proper for < i,where we replace a countable elementary submodel by tower of height a ofsu ch models (in 3, and in 2 for more general case). The iteration theoremis proved in 7, but to apply it to the classical problem of SH we need "goodforcing notion", this is done i n 6; Jensen's original proof use a different for cing.Lastly, in 8 we deal with KH giving a proof in our context to results of Silverand Devlin. We also start investigating preservation of additional property: in4 we deal with ^-bounding.

Notation. In this chapter , will stand for uncountable cardinals, if notexplicitly stated otherwise.


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1. ^-Completeness - a S u f f i c i e n t Condition forNot Adding Reals1.1 Definition. Let 8 be a fa m il y of subsets of < S 0( ) (we assume alwaysS o ( ) G , so is reconstru ctible from 8 bu t u s i n g specific 8 w e m a y forg et tow r i t e SN O (M))- I n a n abuse of notation, instead of a singleton {}, E C < S 0(),we w r i t e E. W h e n 8 = {a G S 0( ) : a \ G 5} we w r i t e {5} or 5 (hereSS^o^i) orjust 5 a subset of i ) , s imilarly if 5 C 50(), we may interpretit as { G < S 0( ) : i G 5}. R e m e m b e r > 0(A) is the filter {A C S 0( A ) : Ainclu de some clu b of < S 0(A)} (see Definition III 1.4), it is N i-com pl et e, fine (i.e.x G A = > { : z G } G > 0(A)) and normal (i .e . Ax G > Q(A ) fo r x G A i m p l i e s

(1) We say that 8 is n o n t r i v i a l if for every larg e enou g h, there is a cou ntableN - < (if (), G ) such that 8 G N an d A T G A fo r every A G T V . Wesay in such cases that A T is suitable fo r 8.(2) We say, for a nontrivial , that a for cing notion P is ^-complete if for every large enou g h, and N -< (if (), G) countable, suitable for , to which Pbelong s, the pair ( N , P ) is complete (see below).

(3) The pair ( T V , P) is complete if every generic sequence ( pn : n < ) fo r( T V , P) has an u p p e r b o u n d in P, where:

(4) (pn :n


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answer for all for which < S 0( ) Gif(), so we may replace the universalquantifier on by an existential.3) If 8 is nontrivial, it has the finite (in fact, countable) intersection property.

4) If S C is stationary, then 8 {S} is nontrivial (for = \).5) If P is ^-complete for some nontrivial , then P does not add reals.6) If N - < ( - f f ( ) , G) is suitable for 8, P GN a forcing notion, q GP is generic

fo r ( T V , P ) , then q \\-P UN[GP] is suitable for S" .7) If S C 7 > ( < S Ho ()), i = ||+, 5= {Xi : t < |1} and *-{ G S ():

if i G and i < || then G Xi], then: S is nontrivial if f {E*}is nontrivial; also for any forcing notion P, P is ^-complete if f P is E*-complete.

8) If P, Q are ^-complete, then P x Q is -complete.9) If N X (if (), G) and , E G A T and E C S we have (if(),G) |= "(3 #())[(*) an f() GH(\) of course, and i is as required in Definition 1.1(1). Let N I - < (/f(),G)be suitable for . Then for some G ^ f(), T V i |= "(*)". Now considerT V = N I - f f () and get a contradiction. The other two sentences are easy too;on the normal fine filter on < S 0( ) which S generates see 1.4.2) Easy, by an argument similar to III 2.2.3)-9): Easy, (for (6) see 1.3(1)). D . 21.2A Explanation of 1.1(4). is large enough to ensure everything about P,forcing, etc., is expressible in f(), now as TV is an elementary submodel, it islegitimate to ask what goes on when you f o r c e with P starting in T V , of coursea generic sequence is not far f r o m being a generic subset, so what (4) says isthat for any generic extension N[G\, there is p G P which knows everythingabout it (so G C P n TV is generic over T V ) .


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1.3 Theorem.( 1 ) I f P is -complete (so nontrivial in V ) , then I hp " is nontrivial".(2) If Q = (Pi,Qi : i < ) is a countable support iteration, Ihp. "Q ^ is

-complete", then Pa limQ is ^-complete.1.3A Remark. So in (2) it is enough to assume is not trivial in V and Ihp. "if is not trivial then Qi is ^-complete."Proof.(1) Note: S0()v = SQ()v . Let be large enough and p G P. Let N X(ff(),) be suitable for , P G N, p G W , hence (7V,P) is complete (seeDefinition 1.1(2)). Choose (pn : n < ), a generic sequence for ( A T , P), p0 = pand choose p* > pn for all n < . Since p* is ( T V , P)-generic by Corollary III2.13 (see clauses (a), (f)) we have, p* I h "(TV[G],G) X (#()vlGl, G); G N [ G \and T V [ G ] = N a nd H ( X)V nN = H ( X)V JV[G]" and as E V alsoN [ G \ = N , hence 5^0() A T = < SNo( ) N[ G\ (as forcing by P addsno new c ountable subsets of ) hence

A= AA A e N [ G )

So p* > p forces A [ G ] to exempli f y that is not tr ivial .(2) Let be large enough, N -< (H(X),e) suitable for . Let ( pn : n < )be a generic sequence fo r ( A T , P), (note pn G A T , Dom(pn) countable henceDom(pn) C N ). Define p * G P : its domain is T V , and for i G / , p*(i) isa m e m b e r of Qi w hic h is an u p p e r b o u n d for {pn(i) : n < and i GDom(pn)}if there is suc h upper bound ( in Qi, say first such upper bound in some wellordering pn \i for every n (note i G Dom(pn) for every n large enough as{p G P : ^ G Dom(p)} is a dense open subset of P w hic h belongs to T V ) . Thereare no spec ial problems.

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1.4 Claim.(1) The minimal normal fine filter on S0() which includes 8 is T > = T>(8)which is defined by:

A G V if and only if there is C G > N 0 ( / - 0 and G U{S0()} for i < ,such that { G C : (V i G ) G AJ C A

(2) 8 is nontrivial if and only if 0 >( ).(3) P is -complete if and only if P is D()-complete.

Proof. Easy. .41.5 Lemma. Assume 2K = N I . If 0 = (Pi, Qi : i < ) is a countable supportiteration, I hp . |Qi| = N I " , a f a m i l y of subsets of < S 0( ) which is nontrivial,and each Qi is 8-complete. Then P = LimQ satisfies the N2-chain condition.Proof. Let {p i GP : i < ^2} be given. We shall find two compatible conditionsamong them. Pick regular large enough, fo j every i < N 2 , le t N i - < (-ff(), G )be countable such that {Q,p;,z,} C N i and A ^ is suitable fo r .

1.5A Fact. We can find i < j < ^ and an isomorphism h : N i > J V j (ontoA ^ ) such that h(pi) = P J and h\(Ni A^ ) = iProo/ o/ e Fc. Denote 5 = {7 < ^2 : cf (7) = K I } , clearly it is a stationaryset; define 7(7) = Um{ : A^ 7 (U


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h : N i > N j be the isomorphism. A s each c G 5* is an individual constant, h isan isomorphism over J5* (i.e., it is the identity on J B * ) and similarly /i(p^) = PJ.This is the isomorphism we promised in the Fact so we have proved the fact.

Continuation of the proof of 1.5: Let i < j and h be as in Fact 1.5A. Nowchoose {pf e Pa Ni : n < } such that pi = p


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^-suitable and (pf() : n < ) is a generic sequence fo r (Ni[G],Q)" . Hencer\ \ \ - p "(p?() : n < ) has an u p p e r b o u n d in Q" , so we finish.3. G N J \ J V i , s y m m e t r i c p r o o f to 2 (u sing th e choice of h).4. e Ni\ Nj\ r e m e m b e r that w.l .o.g . Q (set of elements) is and as aboveby the indu ction hypothesis r\ is (^P^-generic and (^P^-generic. SinceNii an d N J \\ ar e initial segments of \ an d N i = N J (and \ G NiiNj)clearly T V u i = N J i. A l s o r\ determines Gp T V ^ an d Gp^ iNj hence fo revery ra there is an n G , and m GA^ i, such that pj1 f lhp " (0= m" .To this relation in Ni we can a p p l y h , which yields p! ? f lhP "p( ) = m"(since m G N i A^ ) Hence r\ lhp (m(0 = p (0" for all m < . Nowcontinue as in the previous case 2.

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1.6 Theorem. S u p p o s e that CH holds in F, and 8 is a nontrivial family ofsubsets o f S N O ( / / ) , **1 = c f ( ) . T h e n V has a generic extension V\ byp r o p e r for cing in w h i c h :

(*) (a) CH holds, 8 is not trivial, 2K l = , and

(b) If P is an ^-complete proper forcing notion, |P| = K I and Xi C P isdense for z < i$ < c f ( ) , then there is a directed G C P such that

^ 0 for i < i0.

1.6A Remark.(1) Properness is not essential in the proof of the theorem (except for having

it in the conclusion), its use will appear in 1.7.(2) A l s o the reader should be aware of the fact that ^-completeness andproperness is more than properness alone, otherwise 1.6 w o u l d say:

& C H ,


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which is of course impossible.Proof. We use countable support iterated f o rc i n g systems (Pi,Qi : i < ) suchthat(*) a < + and lhp i "\Qi\ #,Qi is proper and is ^-complete".

For any such system Q, LimQ is an -complete for cing notion which satisfiesthe ^2-chain condition (by 1.3(2) and 1.5 respectively). Also by III 3 LimQ isproper.

By usual bookkeeping it is enough to prove the subfact below (note thatfo r R{f: for some < i, / : a > {0,1}} ordered by inclusion, always (a)of (iii)below holds).1.6B Subfact. If Q 1 satisfies (*),Pl = LimQ1 and R is a P1-name of a f o rc i n gnotion, then there is a Q2 such that:

(i) Q2 satisfies (*).(ii) Q1 is an initial segment ofQ2.

(iii) for some maximal antichain J of P2, for every p G T (where P2 LimQ2):either (a) p lh P 2 " there is a directed subset of R, generic over VPl" (in fact, it

is the generic subset of some Q, G fe(Q1),^g(Q2)]),or (b) for no Q > and q do we have: Q satisfies (*) and Q2 is as initial segment

of Q and p < q G LimQ, and q H- LimQ "? *saproper ^-completef o rc i n g with universe \".

Proof. Immediate. . 6 B , i 6

1.6C Remark. 1) This is different from the situation of II 3.4,where we had"c.c.c." instead of "^'-complete for some suitable "'.

2) So the Schemma of the proof of 1.6 is more general than the one in II 3.4.3) Assume that P, R are f o rc i n g notions in V, 8 C < S < 0( ) is nontrivial:(a) if R is not proper in V, P is proper, then R is not proper inVp (use

the equivalent definition in III 1.10(1) and for simplicity the set ofmembers of Q is an ordinal): similarly for 8-proper (see Definition2.2(5) below). The proof is included in in the proof of III4.2.


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b) If R is not ^-complete in V, P is e.g. ^-complete, then R is not -complete inV

p .c) Without "P is proper", clause (a) is not necessarily true.

4) By (3)(a) (of the Remark 1.6C), as \ < a = P/P a is proper], we canuse in the proof of 1.6 the older Schemma.

5 ) We can omit = cf() in 1.6 and replace i < I Q < cf() by i < whereX i < X is regular. (And use iterated forcing of length < +, cf() \.Instead + we can use an inaccessible.)

1.7 Conclusion. In the model from 1.6, if S = { \ \ S}, S C i , stationarycostationary the following holds:

a) for any (A j : G S), such that A C unbounded of order type fo r G S, we have: ((A$ : G S ), N O ) has the uni f ormi z ati on property.

b) S is still stationary (after the forcing, by properness).Remark. Remember, we say a family P = {Aa : a G S} of sets has the K -u niformization property (or ( P, K ) has the uniformization property) i f for everyf a m i l y {/ : a G5}, / a function from A to , there is / : UesAa > K suchthat / =e /f^*,where/ =e 9 if |{ : /() 7 ^ ^(), 6 Dom(/)}| ;, and Q = = { / : / a functionfrom some < i to , such that for every limit < , G 5we havef\A =ae /s}, ordered by C .W e have to check the following four facts:Fact A. If p GQ , D o m ( p ) < a < \ then, there is q, p < q GQ , Dom(') = a.Fact B. Q is 5-complete.Fact C. I f p G Q , ^1 C \ \ D o m ( p ) is finite, / a function from A to , /ienthere is g, p < f G Q , / C g.

. Q is proper.


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Proof of Fact A. Let {i : i < j < } be a list of all limit ordinals G 5, suc hthat D o m ( p ) < < a. As Al has order type , and Sup(A5J = ^ clearlyA S Agm is finite for m < i, hence we can define by induction on < , < i suc h that > D o m ( p ) and > Max(A^ A ) for m < I. Nowdefine , a f u n c t i o n from a to :

p() i G Dom(p)fi(i) ie(A\t)0 otherwise (but i < a)

N ow q is well defined as the were defined such that the A \ are pairw isedisjoint and disjoint to D o m ( p ) . It is tr ivial to check p < q G Q .Proof of Fact B. Trivial. (Note that if ( pn : n < ) is an increasing sequenceof m e m b e r s of Q, then (J pn satisfies almost all the requirements, the prob-n


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It is also clear that U f c < J V f c = J V t , so every pre-dense I C Q , X G N belongsto N ^ hence to s o m e A f c . N o w , define by i n d u c t i o n on n,pn such that:

a) P =P>Pn pn, p^ G Q,p| 2 r(^n+ nA d\Dom(pn)) and by Fact A w . l . o . g . an C Dom(^) (if i5,w e use only fact A). As An+1 - ( f f ( ) , ,


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206 V . -Properness and Not A d d i ng Reals

increasing, countable, free, and Gi/Gj is free when j 5, j < i, then G is aWhitehead group. D I . S

Also note that if 8 is a normal filter on i, it is well know that by -complete forcings we can "shoot" through all A e 8 closed unbounded subsetsof \. We can in 1.7 (hence 1.8) replace { \ 5}, by a normal nontrivial idealon ] which we can assume is dense in the sense that every stationary S C \contains a subset in the ideal.The reason for 1.8 is1.8A Fact. For G = (Gi : i < i), S C \ as in 1.8, H an abelian group,h a homomorphism from H onto G with kernel Z, letting Hi = h~l(Gi),the following f o rc i n g notion is proper and ( \ 5)-complete: P {g : g ahomomorphism from Gi into Hi such that ( h \ H i ) o g = id }. I.SA

Note

1.9 Claim. If P is {Sj-complete, S a stationary subset of \ and in V we haveOs then in Vp we also have OsProof. Straightforward (as in IV, we use Os to given the isomorphism type ofa countable elementary submodel and a name of a subset of i). D 1-9

2. Generalizations of PropernessWe shall repeat most of this section in the next one, with more details and lessgenerality.

2.1 Definition. For an uncountable cardinal , countable ordinal and < :(1) Let SQS^(X) be the set of sequences (Ni : i < a) such that:

a) N i a countable submodel of (-ff(), G).


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b) i GNi and (Nj : j


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(5) A f o rc i n g notion which is (, 0, 0)-proper will be called ^-proper.(6) A f o rc i n g notion will be called (,)-proper if it is ( {S

0()}, , )-proper.

2.3 Theorem. Assume 8 is (, ^)-nontrivial. Countable support iteration pre-serves (,,)-properness, provided that 0 or a is a limit ordinal.Before proving we show 2.4, 2.5 below.2.3A Remark. There are examples that the notions are distinct, proper is({5Ho()},0,0)-proper.

2.4 Claim.1) I f P G N - < ( H ( X ) , e , < * ) , N countable, G C P generic over V, then

N [G] = [[G] : r G N a P-name}.2) If T V G SQS%(X),P G T V 0 , G C P is generic, then Vp 1 = N [ G\ G

3) If T V G SQS%(\), Q = (P,Qt : < n) N0 is iterated f o rc i n g, re aP^-name of a member of Q , is (N\G^,Qt, fc)-generic (G C P^ the genericset), thenr= (ro,r, . . . ,rn_) is ( T V , P n,nA:)-generic.

4) If P is (, I m p r o p e r and a = , then P is (/?, 0)-proper.5) If P is (,)-proper, Q I > 2, ^i < ^2, then P is (2, ^-proper. Also P

is (0 , 0)-proper iff P is proper.Proof. 1) Straightforward.

2) like 1.3(1).3) Left to the reader.

4), 5) Check. Q24

2.5 Lemma. Consider the following properties of a forcing notion P, p G P,countable limit ordinal , regular large enough and 3, are equivalent:


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(1) There is a function F, Rang(F) C S p, such that for every < and J G a maximal antichain of P,J F({Aj : j < i)) is pre-dense above q.

(2) For every N i C (tf (), G , p such that(*) for every i < an d P-name of an ordinal, / ? G 7V;, we have

where(a) (N j : j < i) G A^+i, A^ continuously increasing, N i is countable an d

P G A 0, G A 0.(b) For every (first-order) f o r m u l a < / ? ( x , ), G A^ , i < , (J^(), G ,


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take the


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Proof of 2.3. W e seperate th e proof to the two natural cases. Let (P, Q : G T/}is a finite subset of Ji 7V 0 pre-dense above (r, #); moreover clearly we canchoose # G 7V2, in fact < ? G 7 V .

Unfortunately, there is no reason to asssume q\ exists. Look at the extremecase T^ = {n} (e.g. when P is Ni-complete and r determines Gp N Q ) . Soin V[G ] we know #, ( < ;) and we know {q, q^t : I < 0} is compatible forevery Q < ; this is not a good reason to assume {q,q : < } is compatible,except when Q is Ni-complete and any two compatible members have a leastupper bound.Second Try:Let JtNo = {(z4,j : m < } (Ji f r o m (e) of 4.4) and as N0 e N^(Ji : < ) e NI, we can assume that (((p^q^) : m < ) : < ) G N.Let 5 = {m < : p G Gp}. This is a P-name, 5 G A^i and even(Si : I < ) e NI. If A 0[G ] V = N0 then in F[G] there is a function/ : > such that

=? [ G ] / \ V 1]


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N ow in V w e have a P-name of it , / G NI such that \\ -P UA 0 [GP] V = Oimplies that / is as above; also in any case / G ()v"; so in particular rforces / to be as above. As r is ( / , P ) - g e n e r i c we have just countably manycandidates for / G()v N I i.e.for /[G]; and clearly there is in V a f u n c ti o n/* G such that for every g GN I we have, g


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N o w return to the beginning of the argument in the second try. We knowthat fo r every g1" Q[G\ N Q[G ] there is a / < G ()v[G] such that

V & [ \is consistent (i.e., as said above, some member of Q fores all those pieces ofinformation). In particular for every ( m o , . . . , mn-)(ii < ) there isan / = / e (")vM such that: if {q[G\, &,(!},&M- .C JG]} iscompatible (in Q [ G ] ) then

(*) qWt q G]?, /\ V &[G}t


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/t f\j) < /*(j).Also there is a unique -sequence of ordinals such that \i GT^ , j/^G(2 for1 < < ; (from the first try; remember {p^ : GT/} is a maximal antichainof P and p < p^). So, (p^,r*)-(Pm > < ? m J for some* < . B Y thedefinition of fc(), and m ^ we have mi < k() < /*(-).Let = ( m o , . . . ,mi-) (where i was chosen above). Then by (*)r?,

V

is consistent, so th e result follows, since / > i = > f ( ) < f+() < f*() < f * ( ) .4.4

Continuation of the proof of the Theorem 4 -3:We were proving by induction on a that if I h p . "Q ^ is -proper and has the -bounding property", Q = (Pi, Qi : i < ) is a CS iteration then Pa = LimQ is-proper and has the ;-bounding property. The -properness follows by 3.5,and for the ;-bounding property only th e case of cf () = N 0 w as left. N o wby III 3.3 w.l.o.g. a = . Let / be a PQ-name, p GP, p I h "/ Gn, and wehave to find g ()v and q satisfying p < q G Pa such that q I h "/ < ^". Wecan assume w.l.o.g. that

(*) f(n) isaPn- name(this follows from the proof that P is proper, see III 3.2).

Let NI - < (H(\), G) ( large enough) be an increasing chain such that, p,(Pn,Qn n < ) G N Q ,N G NI+I each NI countable (Note that NI X NI+Ifollows from NI G ^ + i , NI countable, an d N^Nt+i - < (f(), G)).

We want to find q GP, g > p, f lhF "/ < #" for some ^ G ()v. Forthis we now define by induction on n a sequence (qn : n G ), where each qn isin Pn such that the following will hold:

1) qn+\n = qn,P\n


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2) qn is ( J V f c , Pn)-generic for k = 0, and n 4-1


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5. Which Forcings Can We Iterate WithoutAdding RealsIn Sect. 1 we have proved that we can iterate f o rc i n g notions of special kind(^-complete) without adding reals. As a result we get a parallel of MA for suchforcings and get the consistency of some u n i f o r m i z a t i o n property (see morein Chapters VII, VIII). However this axiom, quite strong in some respects,is consistent with diamond on H I : (see [Sh:64], [Sh:98] or 1.9 here). On somestationary subsets of \ it can say m u c h , but on others nothing.

So we shall try here to find another property of f o rc i n g notions, so thatfor cing with LimQ,0 ( P % , Q i : < ) a CS iteration o f such f o rc i n g, doesnot add reals.

5.1 Example. Assume 2H = N I (or even 2N < 2*1 suffices).Let AS C be unbounded of order type , for < \ limit, so by [DvSh:65](or see AP 1), (As : < ) does not have the u n i f o r m i z a t i o n property, hence

there ar e f '. A > {0,1} such that for no / : \ > {0,1}, is f\A& =ae ffo r every . Let / = ( f $ : < > ) , Pj {/ : Dom(/) is an ordinal a < i,


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5. Which Forcings Can We Iterate Without Adding Reals 225

Now, Pf is a very nice for cing - e.g. it is -proper for every a < , butour desired property should exclude it. The fol lowing is a try to exclude thiscase by a reasonable condition.

We shall return to this subject in VIII, 4 (going deeper but also havingpresentational variations of the definitions).

5.2 Definition.(1) We call D a completeness system if for some , B is a function defined

on the set of triples ( T V , P , p ) , p G TV P, P G TV, TV X (f(),e), T Vcountable such that ( P is meant here as a predicate on T V , i.e., P T V ) :B(JV,P,P) B ( T V , P,p) is a filter, or even a f a m i l y of nonempty subsets ofGen(TV, P) = {G : G C TV P, G directed and G I 0 for any densesubset X o f P which belongs to TV } such that if G G Gen(TV, P ) belongs toan y member of B(JV ,P, P)> then p G G .

(2) W e call D a -completeness ( m a y also be finite or NO or N I ) system if eachf a m i l y D(jv,p,p) has the property that the intersection of any i elements isnonempty for < 1 -f (so for > NO , B{JV,P,P} generates a filter). Now,such D can be naturally extended to include TV - < ( i f ( t ) , G ) , G T V , < f by D(TV, P,p) = D ( T V if () , P,p). We do not distinguish strictly.

(3) We say D is on . We not always distinguish strictly between D and itsdefinition.

5.3 Definition.(1) Suppose P is a f o rc i n g notion, 8 a nontrivial f a m i l y of subsets of < S^ 0( )

and B a completeness system on .We say P is (,B)-complete if for every large enough , if P, ,B G T V ,p e P TV , T V X (if (), G ) , T V countable, , 4 e n T V = > T V n e A , thenthe fol lowing set contains some member of B{;v,p,p) (ie> D { w n H ( ) , p , p ) ):

Gen+(TV, P ) = {G G Gen(TV, P ) : p G G an d there isan upper bound for G in P}

(2) If 8 {


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5.4 Remark.(1) We can think of D ( , P , P ) as a filter on the f a m i l y of directed subsets G of

P TV generic over TV , to which p belongs. T he demand "(,B)-complete"means that (for > ( )-m a j orit y of such TV's) th e "majority" of such G 'shave an upper bound in P hence the name (,D)-completeness.

(2) In some sense the definitions above are trivial: if P is -proper and does notadd reals, then there is a ^-completeness system D such that P is (,B)-complete for all K simultaneously. Because, given (TV, P, p), w e extend p toq G P which is ( T V , P)-generic. If {In : n < ] is a list of the dense subsetsof P which belong to TV ,Jn N {pnjk 0 < k < u;}, we can define aP-name x:

x {(n, k) : n < and k is m i n i m a l such that pn^ G Gp}

Clearly q I hp "x G", and since P does not add reals there is an x* G()v, and r, q < r G P, r lhP "x* = x". LetGr = {pt G P N : pt = {{Gr}}

So what is the point of such a definition? We shall use almost alwayscompleteness systems restricted in some sense: O ( J V , P ,P) is defined in areasonably simple way. The point is that usually when we want to decidewhether some G G Gen(TV, P) has an upper bound, we do not need toknow the whole P, but rather some subset of A T , e.g. a function / from f to itself. Check th e example w e discussed before: if = N i, thenwe just need to know f\. But two f\'s may give incompatible demands,so for it the system is only a 1-completeness system. So if we deal withNo-completeness system, we exclude it (in fact later we shall discuss even2-completeness system).

An explication of "defined in a reasonably simple way" is:


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5. Which Forcings Can We Iterate Without Adding Reals 227

5.5 Definition.(1) A completeness system D is called simple if there is a first order f o r m u l a

such that:B ( 7 V , P,p) {Ax :x a finitary relation on T V ,

i.e. x C Nk, fo r some k G ]where

Ax = {Ge G e n ( 7 V , P ) : ( 7 V U P(N), G , p , P , O 1 = [G,x}}( 2) A completeness system D is called almost simple over V b (V b a class, usually

a subuniverse) if there is a first order f o r m u l a such that:D(7V, P,p) = {AXtZ : x a relation on N, z G V b }

where

where e= {(x, y) : x G A, y G A , x G y}.(3) If in (2) we omit z w e call D simple over V Q .

5.6 Claim.(1) A -completeness system (see Definition 5. 2(2) ) is a *-completeness sys-

tem for every * < .(2) P is (,D)-complete for some D if and only if P is -proper and (forcing

with P) does not add new reals.Proof. (1) Trivial.

(2) The direction 4= (i.e. "if") was proved in Remark 5.4(2) above. So, letus prove the "only if" part. So P is (5, Incomplete.

Suppose T V - < (ff(),e), p G P and {p,P,,D,} e N, N countable andT V G A So = {G G Gen(, P) : G has an upper bound and


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p G G} G D ( r , p , p), hence B 0 (by Definition 5.2) and let G GB. So G hasan upper bound q (by the definition of (,O)-completeness), G G G e n ( 7 V , P),and by Definition 5.3, p G G. So < ? > p is ( A T , P)-generic. If p lhP "/ G "",/ GN, then for every n,Jn= {r G P : r I h "/() = fc" for some / c < } is adense subset of P which belongs to A T , hence Jn G 0. Hence g determinesthe value of /(n). So g determines /(n) for every n. Hence it determines /, i.e./ is not a new real. Now if there were a new real, some p would without loss ofgenerality f o rce / is such a real. Choosing N as above we get a contradiction.

5.6

5.7 Example. Forcing with a Souslin tree T is not D-complete for any simple2-completeness system B.

Let N -X (i(), G) be countable, T GN, = Ni. Note that Gen(N, P)consists of all branches of T N, and G e n + ( , P) consists of the branches ofT N which have an upper bound, i.e. Ax = {y GT : y < x], where x GT $ =the 5-th level of T. Now N "does not know" what is the set of such branchesof T T V , and two disjoint sets are possible.

The above is an argument, not a proof. To be exact, we can, assumingdiamond of N I , build a Souslin tree, such that no first order formula definesa simple 2-completeness system for which T is D-complete.

6. Specializing an Aronszajn Tree WithoutAdding RealsThe traditional test for generalizing MA has been Souslin Hypothesis. Jensenhas proved the consistency of the Souslin Hypothesis with G.C.H. (see Devlinand Johnsbraten [DeJo]). He iterates forcing notions of Souslin trees, in limitpoints of cofinality N O he uses diamond to refine the inverse limit of the trees,in limit points of cofinality N I he uses the square on ^2 (and preparatorymeasures in previous steps). In successor stages he specializes a specific tree,


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6. Specializing an A r o n s z a j n Tree Without Adding Reals 229

by first forcing a closed unbounded set and then building a Souslin tree usingO S (more precisely he adds N 2 closed unbounded subsets in the beginning).We shall prove that there is a D-complete forcing notion PT specializingan Aronszajn tree T, for D a simple HI-completeness system. The proof is closeto Jensen's successor stage. We feel that the ideas of the proof are applicableto related problems, see [AbSh:114], [AbSh:403], [DjSh:604].

Notation. For an Ni-tree T, T; is the i-th level, T\i = U T}, and for x G T/j,O L


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(*) for < n we have pn(x\ht ( pn)) + l/2n > pn+(xt\ht (pn+))This is not difficult by itself, but we have also to ensure the genericity of

(pn : n < ). So it clearly suffices to prove, for each n(**)n if p G P T V , n < ;,Jan open dense subset of P which belongs to T V ,#0, - , #n- T/vni and > 0 ( a rational or real), ften there is a # G P T V ,p < < / , G J and

p ( x ^ t ht (p)) + > q(x \ ht ()) for i < nU n f o r t u n a t e l y we see no reason fo r (**)n to hold.I n fact, it is false, and for every natural number n, Zn = {p G P^ : for everyx G ht (p), we have p(x) > n} is dense.Second Approximation:We can remedy this by using Pf = {/ : / G P, and: i / / 3 < ht(/),x G 2>and > 0 and T is a Souslin tree, then for some y we have x < y G T1^/) and

N ow for n = 1, (**)n is true; more generally for any n < , (**)n is true if:(* * *)n {(2/0, , 3 / n - ) : *< n 2 f e T & j f c < z*} is generic for G/V,T n)} .(Tn - the n-th power of T i.e. the set of elements is in [J and ^< V

Why? Though it is not used we shall explain. For a given p and rational letR = Rp,, = {(yo, - - , yn-i) :for some a < , a > ht (p) &y GT, and forsome GPj1,^ < q G J, ht () = , q(yt) < q(xi \ ht (p)) 4- }. Now#is a densesubset of {(2/0, 2/n-) : fo r some 7, for each ^ , z ^ f ht (p ) < y G T7}. [Why?as given (y0, , 2/n-) n(7) we can find r, p < r G P, ht (r) =7, r(yt) 0 there are infinitely many pairwise dis joint y G n ( T ) suchthat f ( x i ) < f ( y t ) < f ( x i ) + for t < n and x < y.

6.5 The Main Definition. P =P = {(/,C,) : (/,C) G PI, and is acou ntable set of promises which (/,C) fulfills }

(/,C,)p, and itq > / ? ,(2) moreover, if G f| C() and > tp, then we can have ^g = ,re^p(3) moreover, if m < u;, j / o 2 / m - ^^/35 > 0 we can in addition to (2 )

d e m a n d / p( y< f f t p ) < / g ( i / i ) < / p ( 2 / < p) 4- for i < m.Proof. (1) Clearly f] C() is a closed u nbou nded su bset of \ ( as i s

6Pcou ntable and each C() is a closed unbounded subset of i). Hence thereis an o r d i n a l / ? t , \ > ^ , / ? t > #p and / j te c(), and a p p l y (2).rep(2) Let a = tip. We define Cq = CP\J {}, = p, so we still have todefine /9, but as we want to have /p C /, we have to define just fq\ T . Wehave tw o d e m a n d s on it, in order that < j G P:

( i) monotonicity: / p ( x ) = / 9 ( x ) < fq( x) G Q f o r x G T( i i ) from Definition 6.4 for a < 2 in Cq \ M i n ( C ( ) ) (hence in C(),

Gp-9, x G n()(T ) w h e n 2 = / 3 (for 2 < / ? usep GP)).I f we succeed to define fq\T such that it satisfies (i) and (ii), then q is

well defined, and trivial ly belong s to P and is > p.


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6. Specializing an Aronszajn Tree Wi tho u t Addi n g Reals 23 3

N o w , (ii) consists of count ably many demands on the existence of infinitelymanyG nCZ.

Let {(m,7m,xm) : m < ] be a list of the triples (,7,z), G p,x G n()(T), 7 < / ? , 7 G Cp C(), each appearing infinitely often (if thisf a m i l y is empty, we have no work at all).

We now define by induction on m, a function /m such that:a) fm is a f u n c ti o n from a f inite subset of T to Q such that fp(x\a) 7m > MinC()). So there is a z G n n(T ) such that xm < z,and /p(^) < fp(xm) 4- l/3m. Now we apply the argument above, replacing xmby f .

(3) The same proof as that of (2), using our freedom to choose /0.So we finish the proof of Fact 6.6. De.e


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N ow w e shall prove th e crux of the matter: th e parallel o f (**).6.7 Fact.(1) If N - < ( f ( ) , e ) ( large enough) P,p G N, p G P, N countable,

N \ = 5, > 0 a n d X Q , . ?^n- T (are distinct) and X e N isan open dense subset of P, en there is a q GZ T V , q >p, tq = andfq(xi\ttq) ,^ n(T 7 ) ,


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6. Specializing an Aronszajn Tree W i t h o u t Addi n g Reals 23 5

Let Ci = l r e pC()\etP' soaSinC GN-ythesubfact7A>NN " forevery 7 G Ci there is a y G i n(T 7)" bu t A T X (-ff(), G ), hence also th euniverse V satisfies the statement.

Our plan is to get a promise C i in , and show that r = (/p, Cp, P U{}) G P, j p < r, and above r there is no member of J, thus getting acontradiction to "Jis an open dense subset of P" .

Let 2 = {y G i: there are uncountably many z ,y < z}. By theabove, x G2. We shall prove later:

6.7B. Subfact. There is a closed unbounded C* C i, = MinC*, C* Cdsuch that = {y G 2: for some i G C*, ? / G n(T)} is a promise.

Let us show that this will be enough to prove 6.7, hence Theorem 6.1except checking simplicity.As before, we can assume C* G N ; and as MinC() = MinC* = a = tp,clearly pt = (/p, Cp, p U {}) e P\N and p < pt. AsJis an open and densesubset of P there is a # > p^ in J. As q G P, (/g, C g) satisfies the promise ,- so as G C() Cq, also / ? = tq GC() C g. Hence by the definition of"fulfi l l ing a promise" and as x G (see above), there is a y Gn(T/3) suchthat x < y and fq(y) < fp(xt)+ for each < n. So by IYs definition, y i(as GJ) but y G C 2 C i. We arrive at a contradiction thus provingFact 6.7, except that we need:Proof of Subfact 6. 7B. Note that

a) if z G U n(Ti), z < y G i, then z G u so clearly 2 has thisi>aproperty too.Next note that

b) for any y G2 the set {z G2 : y < z} is uncountable.(Why? If not, for some 7 such that < 7 < \ there is no z G 2,y


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z Gn(Tj) the set {i : 7 < i < , 2 J f 7 = zt for I < n} is uncoun table, hencez G 2).c) for any y G 2 n(T;), i < j < \ there is a z G n(T7 ) 2, y < z.

This is just a combination of a) and b).d) for any y G 2 n(Ti) there is a j such that i < j < \ and disjoint

O t he rw i s e for i < j < , let zj G 2 n(7 ), z < zj (by c)). So fori < < C < i , for s o m e and fc,z| < z (otherwise use (c) on z^ with , here standing for z,j there to get a contradiction). This contradicts T beingA r o n s z a j n by the proof of Theorem I I I 5.4.

e) for any y G 2 n ( T i ) and m < there are j < i, j > i and pairwisedisjoint z1, . . . , zm G 2 n(T J ) j / < for f = 1, . . . , m.

Just by induction on n, using d) and c) .N ow w e prove the subfact itself. For any y G 2 there are (by (e)) jm(y)

(m < ) such that there are m pairwise disjoint members of 2 n(Tjm^)w hic h are > y. By c) this holds for any j >jm(y). N o w , if j > \J jm(y), thenm y. Let {z1 : i < i0} be a maximal subset of {z G2 n( T j ) : y < z, j G}w h o s e m e m b e r s are pairwise disjoint. I f I Q < , choose another such se t {y1 :I < niQ + 1} (exists as j > \Jjm(y))' N o w , at least one y* should be disjointmfrom all zz's, a contradiction to the m a x i m a l i t y of {z1 : i < i0}- Hence I Q isinfinite. Let C* = {i : i is or i > , i G C\ an d y G 2 U \J n( T j ] impliesj U jm(y)}m codes th e branches of T \ N (use x which is eventuallyconstant on each 5-branch of T N where = N i ), then G en +( A, P) canbe written as Ax with some suitable ^as in 5.5. The point is that f}i


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7. Iteration of (,D)-Complete Forcing Notions 237

7. Iteration of (, D)-CompleteForcing NotionsThe discussion in the two previous sections lacked the crucial point that we caniterate such forcing notions without adding reals. In order to get a reasonableform of MA we need to iterate up to some regular K > K I and have the -c.c.For K K2, Lemma 1.5 does not suffice as \P\ = # 2 (Pr from the proof ofTheorem 6.2;not to speak of the lack of 8-completeness) but meanwhile Kstrongly inaccessible will suffice (seeVII 1, or VIII 2 for eliminating this).An aesthetic drawback of the proof is that we do not prove that the forcing weget by the iteration enjoys the same property we require f ro m the individualforcing notions but see VI4, which contains more detailed proofs of strongertheorems.

7.1 Theorem. Let Q = (Pi,Qi : i < a) be a countable support iteration,Pa LimQ; 8 a nontrivial family of subsets of S#0().(1) If each Qi is /3-proper for every < \, and (,D^)-complete for some

simple NI-completeness system D^ (so D^ is a P^-name), then Pa does notadd reals.

(2) We can replace in (1) "simple" by "almost simple over V" (note: V andnot VP).Combining the ideas of the proofs of 7.1 and of 4.3 we can prove

7.2 Theorem. In Theorem 7.1 we can weaken "Ni-completeness system" to"Ho-completeness system".

However we shall not prove it now (see VIII 4 for more).Proof of Theorem 7.1. Note: D is a function with domain , B is a Pi-name (ofan HI-completeness system or more acurately a definition of such a system).For clarity of presentation we first deal with the case = (for < thereis nothing to prove).


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Let N i - < (H(X), G) be countable (for i < ) , D , Q G 7 V " o , N i G Ni+ (henceN i - < Ai+1) each N i is suitable fo r 8 (remember Definition 1.1(1), really just / o suitable suffice) an d p G P 7 V , and / G A 0 be a P^-name of a real.

N ow we shall define by induction on n < conditions rn,pn such that:( A ) (1) rn G Pn, rn =r n + 1 fn

(2) rn is (]Vi, Pn)-generic for i = 0 and n-}~\ pn, so r decides all

values o f f ( k ) (as for each k fo r some n w e have Zn = {q G P^ : q force a valueto /(f c ) }.

Let us carry out the induction.n = 0: Trivial (Note P0 = {0}).n + 1: We shall first define Pn+> then C*+1, and finally rn+.

5ep. We want to find pn+ G P A 0, pn+ > pn, pn--N ^ G* andPn+i G Jn. A s Jn C P is dense and Jn G N O ? above every q G P U > 7V 0there is r G P O , A 0 Jn, r > < ? . Let J7n = {r G Pn: there is an r* G P,r* > Pn, ^ * * ^ ^n, ^ * f ^ 7*}, clearly it is dense above pn\n (in Pn). Butpnn G G * G G e n ( A ^ 0, P n ), an d Jn G 7V 0, hence there i s r G G* Jn, and sothere is an r* GP, r* > pn, r* G Jn, r* f n = r. So, clearly, r* can be chosenas pn+ we need.

Second step. Let Gn C Pn be generic over V, rn GGn (hence G C Gn). Weshall try to see what are the demands on G+1. Really we want in V[Gn], tofind a member of G e n ( A / r0[G n], Q n[G n]) which has an upper bound in Q[Gn]and pn(n) belongs to it.


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7. Iteration of ( , D ) - Co m p l e te Forcing Notions 23 9

Note that Dn is a Pn-name which belongs to 7V 0. So there is also a Pn-name n for the formula appearing in the definition of simplicity (or almostsimplicity), and it belongs to N Q . As we "know" Gn NQ = G , we "know" n n[Gn], ie some member of G* force n = n. So we know that forsome AXjy ( x a relation on T V o , y G V, see Definition 5.5) every G G AXV hasan upper bound (in Qn[Gn]), whereAI>y = {Ge Gen(N0(Gn},Qn[Gn}) : (V U N0(Gn]

ey, N0[G,]^(N0[G])t pn+1(n)[Gn},Qn[Gn],V,N0[Gn}) N [G,x,y}}.So we have Pn-names x,y for x and ? / . Now x, 2 ; are quite unlikely to be

in N0 (as their definitions used 7V 0 as a parameter) but they can be chosen inNn+[Gn] (remember Nn+l[Gn] - < ( f ( ) [ G n] , G)as rn GGn, rn is (7Vn+1,Pn)-generic, and 7V 0[G n] G JVn+[Gn]).

Moreover, though we need to know Gn to be able to find their exactvalues, we know that they are in V (remember P(N[Gn}) GV and Pn does notadd reals and even -seqeunces of a member of V); well formally Nn+[Gn]cannot be in V as it has members like Gn, but the isomorphism type of(Nn+[Gn],Nn+,Gn,c)CNn+l does, and so does the isomorphism type of themodel appearing in the definition of AXty. If you are still confused see VIII 4,where essentially we make the set of members of Q

n[G

n] and Nn[Gn] to be a

set of ordinals.So as rn is (7Vn+,Pn)-generic rn I h "x, y G f n + " , so we have just

countably many possible pairs (those in Nn+). Now Nn+ "thinks" thereare uncountably many possibilities, but as -/V n+ G A f n + 2> ^n+2 ^ "Nn+i iscountable", in N n+ 2 we "know" that {^ z ,2 / : x > 2 / ^ ^n+i} is nonempty(remember G * G 7Vn+ hence A^ 0[G*] G Nn+). So, in / n + 2 there is a Gn CQ n [Gn] NQ [G^] which belongs to the intersection. So though we do not knowexactly what x and y will be, we know (as long as rn GGn)) that Gn G Ax^y.From Gn, G; we can easily compute G*+1 G Gen+^o, Pn+), G*+1Pn = G * ,Gn+ ={( 9 0 , 9 1 , . - , ? n > ( 9 o , 9 , - . . , 9 n - > ^G*, n N0 and^n[Gn] GGn}.


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24 0 V . -Properness and No t A d d i ng Reals

Third step. We now have to define rn+ GPn+, so as we require rn+ \n rn,we just have to define r

n_ ( _ ( n ) . What are the requirements on it? Looking at

the induction demand, just:fn \ \ ~ p n "rn+(ri) is above each q(n)[Gn] for members q of G*+1 and is

(Ni[GPn],Qn[Gpn])- generic forn + 2 < i < ",and there is no problem in this. We have finished the proof of 7.1 for = u;.Now we have to turn to the general case i.e. with no restriction on .Let p, Q G NQ - < ( # ( ) , G ) , 0 countable, NQ suitable fo r 8 and p G Pa.

Let NQ ( + 1) = {A : i < ], i < j = > ft < / f y (so A0 ( + 1) hasorder type 7 + 1). Now we can find N i - < (-ff(), G) for i < 7, Ni countable,(N i : j


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8. The Consistency of SH 4- C H 4- There Are No Kurepa Trees 24 1

8. The Consistency of SH + CH + There AreNo Kurepa Trees

W e wish now to give yet another applicationof the techniqueof not addingreals in iterations.8.1 Definition. For any regular ft, a ft-Kurepa tree is a ft-tree such that then u m b e r of its /^-branches is > ft. Let the ft-Kurepa Hypothesis (in short ftKH)be the statement "there exists ft-Kurepa tree". We may write "KH" instead of i - K H . (Be careful: KH says "there are Kurepa trees", but SH says "there areno Souslin trees"!)

Solovay proved that Kurepa trees exist if V = L, more generally Jensen[Jn] proved the existence of ft-Kurepa's trees follows from Jensen's 0+, whichholds in L for every regular uncountable ft which is not "too large". But -KHis consistent with of ZFC -f GCH, which was first shown by Silver in [Si67],starting from a strongly inaccessible ft. The method of his proof is as follows:collapse every , \ < X < ft using Levy's collapse L e vy ( K , < ft) = {p : \p\ ft. Suppose now thatT Vp is an i-tree. So it has appeared already at an earlier stage along theiteration, say T GVp , where Vp is obtained from Vp> by an Ni- completeforcing. In Vp' the tree T has at most 2**1 branches, and this is less than ft.Note that by 6.1(2) the tree T can have no new i-branches in Vp. So T isnot a Kurepa tree in Vp.

Devlin in [Del] and [De2]has shown, starting from a strongly inaccessible,the consistency of GCH - f- SH + - i K H . For a proof by iteration see Baumgartner[B3].8.1A Remark. In both proof s the inaccessible cardinal is necessary, for - K Himplies that ^2 is an inaccessible cardinal of L.


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8.2 Definition. Suppose T is an i-tree, and that Q is a forcing notion. Wesay that Q is good for T if for every p G Q and a countable elem entary subm odelN - < ( - f f ( ), G ) , for large enough such that T , Q , p e A T , there is an ( A T , Q )-generic condition q >p such that fo r every name r G N of a branch of T, either? lhQ "[Q\ N andisanold cofinal branch of T" or q \\ -Q "[G] T ( ( N ) )is not 6() = {x T ( ( N ) ) : s < a} for any G

8.2A Fact. A forcing notion Q is good for an i-tree T i f f Q is proper and inV there are no new cofinal branches of T.

Proof. = > : S u p p o s e Q is g o o d for T. The properness of Q follows trivially.L et p \ \ - Q " is a new branch of T", and we shall derive a contradiction; le t{T,p, Q} G N - < (H(), G ), x large enough an d A T countable. So let q > p beas in the definition of good.

I f [G] is an old branch w e a r e done. I f not, [G] T < < $ N Bx {y :y p which is (Af, Q)-generic,and a generic subset G of P over V with r G G . S o [G] G A [ G ] - H ( , G)[G],and r[G] is either an old i-branch, or is not an i-branc h at all. In the firstcase we are done. Now if [G] is not an cji-branch, then either (3a)[G]}T a = 0or 3x,y G [G] such that x,y are not c o m p a r a b l e in T. By elementarily ofN[G], such an a or such x, y exist also in N[G]. So g forces what is required bythe definition. D8.48.3 Lemma. If Q is an K i -c o m p l e te for cing notion and is an i-tree, thenQ adds no new cofinal branches to T.Proof. It is e n o u g h to show that Q is good for . S u p p o s e that N -< (f(), G )is a countable elementary submodel an d that T , Q G A T . Let (X n : n < ) bea list of all dense open subsets of Q which belong to T V . L e t p G Q A T . Let


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8. The Consistency of SH + CH + There A re N o K u r e p a Trees 24 3

{( m In) : n < ) be a list of all pairs (x n, n) such that xn G Tg(N) and n G TVis a Q-iiame of a branch in of T.

By induction on n we construct a sequence of conditions pn such that:(1) Po = P(2) Pn


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24 4 V . -Properness and N ot A d d ing Reals

Let (In : n < ) be an enumerations of all dense open subsets of P whichbelong to N. Let ((xn, n) : n < ) be a enumeration of T$(#) x {r GN : isa P-name of a branch of T}.

By induction on n we pick pn, qn as in the proof of preservation ofproperness under countable support iteration in III, 3. In addition to theconditions there we demand:(*) qn I I - "Pn+i I f - (3)rt {x 6 () x


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8. The Consistency of SH + CH 4- There Are No Kurepa Trees 245

i ) . Incase Qj is L e v y ( H , H 2 ), this is known from Lemma 8.3, because the collapseis Ni-complete. The remaining case is that Qi = Q ( T i ) is the forcing notiondefined in 6 for specializing a given Aronszajn tree 2V So our proof is finishedonce we know that this forcing notion does not add branches to an i-tree T.

8.7 Claim. Suppose T is an Ni -Aronsz ajn tree, and T* is any Ni-tree. ThenQ = Q(T) is good for T, where Q ( T ) is the forcing for specializing T definedin 6.Proof. Suppose that T V -< ( ( ), ) is a countable elementary submodel suchthat Q, T, T* G N and p e Q A T . We shall find a condition q >p such thatq is (Q, T V ) -generic an d such that fo r every Q-name r G T V of a branch of T*,q I h "r is an old branch of T* or r * T V b(a) for all e ?()">

I n the proof we shall follow the proof of Theorem 6.1, in which the proper-ness of Q was shown.

8.8 Definition. Suppose that p Q and > it(p). For x = (x0, ,#n-) n(T$) and > 0 we say that q > p with t(q) < respects x by iff

< fp(x \lt(p)) - h for all I < n.The main point is the following claim which is the parallel of 6.7.

8.9 Claim. Suppose that TV - < (H(), ) is as usual, and p 6 TV is a conditionand r T V is a Q-name of an i-branch of T*. Let = TV \, x n(T$ ) , G T^ and > 0. T e n there is a condition q N, q>p and < ? respects x by and such that:

(i) q I h "\ b(a) = {y : y


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246 V . -Properness and N ot Addi n g Reals

Proof. Suppose there is no q > p in N as required and which satisfies (i).We shall see that there is one which satisfies (ii). Define the set of "bad" y asfollows: i is the collection of all y G n(T) such that > a and(a) For every 7 < and z G rn(T), ra < u;, there are 7' G [7,/3) and an

extension of p of height 7' < which respects (zt \^ ' : I < gx) by andwhich determines the value of r f7,

(b) There are no two extensions, q $ and # , of p such that q & respects y by (for k = 1,2) and < ? and go force contradictory inform a t ion about r.Let -B() be the set of levels for which there is y G i of height .

Observe that x G IV Why? Clause (a) follows from 6.7; if (b) does not holdthen there is a condition q such that q !h "r is not 6( )" , contrary to ourassumptions, remembering is constant in 8.9. Observe also that by 6.7, thereis a club E of \ such that if z G i and y < z is of height G ", then y G i.W.l.o.g. E GN. As x Gi, GJ5() 7 . Therefore B() n is unbounded,and clearly it is closed.

Let 2 = {y Gi: there are uncountably many z Gi such that y < z}.For each y G 2 of height /?, define t(y) to be the set of nodes t GT such thatsome extension of p of height / 3 , w h i c h respects y by forces that G r. Then(y) is linearly ordered and contains in fact a branch of height . If y\ and 2 /2are any n-tuples in 2, then t(y\) and t(y2) are compatible. (Why? By clause(b).) Thus the function t defines a cofinal branch of T and the intersection ofthis branch with 5* is just 6(). Since 2 is d e finab le in N, this branch is anold branch in N. Now, as before, we get a promise out of 2 and we add thispromise to p in order to obtain the desired q. It follows now that q I h "r is thisold branch".

.9,8.6,8.5

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