Saharon Shelah- Proper and Improper Forcing Second Editon: Chapter I: Forcing, Basic Facts

8/3/2019 Saharon Shelah- Proper and Improper Forcing Second Editon: Chapter I: Forcing, Basic Facts

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I. Forcing, Basic Facts

0. IntroductionIn this chapter we start by introducing f orc ing and state the most importanttheorems on it (done in 1); we do not prove them as we want to put the stresso n applying them. Then we give two basic proofs:in 2, we show why CH (the continuum hypothesis) is consistent with ZFC,and in 3 why it is independent of ZFC. For this th e NI-completeness andc.c .c .(=countable chain conditions) are used, both implying th e forcing does no tcollapse K I th e later implying th e forcing collapse no cardinal. In 4 we computeexactly 2* in the f orc ing from 3 ( in 3 we prove just V[G\ \ = "2* > "; wealso explain what is a "Cohen real"). In 5 we explain canonical names.

Lastly in 6 we give more basic examples of forcing: random reals, forcingdiamonds. The content of this chapter is classical, see on history e.g. [J]. (Except7 , 7.3 is A. Ostaszewski [Os] and 7.4 is from [Sh:98, 5], note that laterBaumgartner has fou nd a proof without collapsing and further works are:

P. Komjath [Kol], continuing the proof in [Sh:98] proved it consistent tohave MA for countable partial orderings +-CH, and Then S. Fuchino, S.Shelah an d L . Soukup [FShS:544] proved th e same, without collapsing H I andM . D z a mo n ja and S.Shelah [DjSh:604] prove that A is consistent with SH (noSouslin tree, hence -iCH).)


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2 I. Forcing, Basic Facts

1. Introducing Forcing1.1 Discussion. Our basic assumption is that the set theory ZFC is consistent.By Godes completeness theorem it has a countable model. We make thefol lowing further assumptions about this model.(a) The membership relation of the model is the real membership relation;

and therefore th e m o de l is of the form (V, G).(b) The universe V of the m o de l is a transit ive set, i.e., x E y G V > x V .

A ssu m p t io ns (a) and (b) are not essential but it is customary to assumethem, and they simpli fy the presentation. So "V a model of Z F C " , wil l mean"a countable model of ZFC sat is fy ing (a) and (b)", and the letter V is usedexclusively for such models.

Cohen's f orc ing method is a method o f extending V to another model V ^of ZFC. It is obvious that whatever holds in the model V ^ cannot be refutedby a proof from the axioms of ZFC, and therefore it is compatible with ZFC.If we show that a statement and its negation are both compat ible with ZFCthen we know that the statement is undecidable in Z F C .

Why do we look at extensions of V and not at submodels of V? Af ter all,looking at subsets is easier since their members are already at hand: To answerthis question we have to mention Godes constructibil ity. The construct iblesets are the sets which must be in a universe of set theory once the ordinals ofthat universe are there. Godel showed that the class L of the constructible setsis a model of ZFC and that one cannot prove in ZFC that there are any setswhich are not constructible. Therefore, for al l we know, V may contain onlysets which are construct ible and in this case every transitive subclass V ^ of Vwhich contains all ordinals of V and wh ich is a model of ZFC must coincidewith V, and therefore it gives us nothing new.1.2 Discussion. Now we come to the concept of forcing. A f orc ing notionP G V is just a partial ly ordered set (notempty of course). Usually a partialorder is required to satisfy p < q&q < p = > p = q, but we shall not (this is


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1. Introducing Forcing 3

just a technicality), this is usually called pre-partial order or quasi order. It isalso called a for c ing notion. We normally assume that P as a minimal elementdenoted by 0p, i.e.

really f rom Chapter II on, we do not lose generality as by adding such a memberwe get an equivalent for c ing notion, see 5. We want to add to V a subset Gof P as follows.(1) G is directed (i.e., every two members of G have an upper bound in G)

and downward closed (i.e., if x < y G G then also x G G ).Trivial examples of a set G which satisfies (1) is the empty set 0 and {x : x < p}fo r p G P .

The fo l lowing should be taken as a declaration of intent rather than anexactly formulated requirement.(2 ) We want that G $ . V and moreover G is "general" or "random" or "without

any special property" .We aim at constructing a (transitive) set V[G] which is a model of ZFC withthe same ordinals as V, such that V C V[G] and G G V[G], and which isminimal among the sets which satisfy these requirements.

So we can look at P as a set of approximations to G, each p G P givingsome information on G, and p < q means q gives more information; this viewis he lp ful in constructing suitable for c ing notions.

Where does the main problem in constructing such a set V[G] lie? In theuniverse of set theory the ordinals of V are countable ordinals since V itselfis countable. But an ordinal of V may be uncountable f rom the point of viewo f V (since V is a model of ZFC and the existence of uncountable ordinalsis provable in ZFC). Since for each ordinal G V the information that iscountable is available outside V, G may contain i.e. code that information foreach GV. In this case every ordinal of V (and hence of V[G]) is countable inV[G] and thus V[G] cannot be a model of ZFC. How do we avoid this danger?


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By choosing G to be "random" we make sure that it does not contain all thatinformation.

While w e choose a "random" G we do not a im for a random V[G], butwe want to construct a V[G] with very definite properties. Therefore we canregard p as the assertion that p G G and as such p provides some informationabout G. All the members of G, taken together, give th e complete informationabout G.

N ow we come back to the second requirement on G and we want to replacethe nebulous requirement above by a strict mathematical requirement.

1.3 Definition. (1) A subset J of P is said to be a dense subset of P if itsatisfies

(2 ) Call I C P open (or upward closed) if for every p,q e P

1.4. Discussion. Since we want G to contain as many members of P as possiblewithout contradicting the requirement that it be directed, we require:(2)' G X 0 for every dense open subset X of P which is in V.1.4A Definition. A subset G of P which satisfies requirements (1) and (2)' iscalled generic over V ( w e usually omit V ), where this adjective means that Gsatisfies no special conditions in addition to those it has to satisfy.The forcing theorem will assert that for a generic G , V[G] is as we intended itto be.

Does (2)' imply that G V I N o t without a further assumption, since if Pconsists of a single member p then G = {p} satisfies (1) and (2)' and G G V.However if we assume that P has no trivial branch, in the sense that aboveevery member of P there are two incompatible members, then indeed G V(incompatible means having no common upper bound). To prove this notice


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that if G G V, then P \ G is a dense open subset of P in V, remember thatG is downward closed, and by (2)' we would have G (P \ G) 0, which is acontradiction.

1.5 The Forcing Theorem, Version A. (1) If G is a generic subset of Pover V, then there is a transitive set V[G] which is a model of ZFC, V C V [ G ] ,G G V [ G ] and V and V[G ] have the same ordinals and we can allow V as aclass of V[G] (i.e. in the axioms guaranting (first order) def iniable sets exists"a: GV" is allowed as a predicate).(2 ) P has a generic subset G,moreover for every p G P there is a G C P genericover V , p G G . .51.6 Discussion. We shall not prove 1.5(1), but we shall prove 1.5(2). SinceV is countable, P has at most N O dense subsets in V; let us denote themwith 2 2 ) 2 . . . we shall construct by induction a sequence pn. We take anarbitrary p0 We choose pn+ so that pn


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(note if = 0,rk() = 0), the union of a set of ordinals is an ordinal, hencerk() is an ordinal if denned, and by the axiom of regularly r k ( ) is defined forevery . So1.8 Definition. We define what is a P-name (or name for P or a name in P)r of rank


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(3) G[G] = G.(4) rk

r(r),rk

n(r) are well defined ordinals, for any P-name r.

Proof. Trivial. DI.IO

1.11 Discussion. Notice that while every name belongs to V, the values ofthe names are not necessarily in V since the definition of the interpretation ofa name cannot be carried out in V. It turns out that these names are sufficientin the sense that the set of their values is a set V[G] as required:1.12 The Forcing Theorem (strengthened), Version B. In version A, inaddition V [G\ = {r(G} \ r e V, and r is a P-name }. D.2

We want to know which properties hold in V [ G \ . The properties we areinterested in are the first order properties of V[G], i.e., the properties given byformulas of the predicate calculus. W e shall refer to the members o f V[G] bytheir names so we shall substitute the names in the formulas.

1.13 Definition. If r\...,n are names, for the f orc ing notion P, (x,..., xn)a first-order f o r m u l a of the language of set theory with an additional unarypredicate for V, then we write p Ihp "(\..., n)" (p forces (r\..., n) forthe f orc ing P) if for every generic subset G of P which contains p we have:

([G]... ,rn[G]) is satisfied (=is true) in V[G],in symbols V [G\ \ = V([G],... ,rn[G])".

1.14 The Forcing Theorem, Version C. If G is a generic subset of P then(inaddition to the demands in versions A and B we have:) for every (r\..., rn) asabove there is a p G G such that p Ihp "->(..., n)" or p Ihp "(..., n)".Therefore V [G\ 1 = u(n[G\... ,n[G])" if f fo r some p G G p lhP V(n . . . ,rn)".Moreover I h (as a relation) is definable in V. . i4

This is finally the version we shall actually use, but we shall not prove thistheorem either.


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8 I. Forcing , Basic Facts

The forcing relation \\-p clearly depends on P. If we deal with a fixed Pwe can drop the subscript P. We refer to P as the forcing notion.

The rest of the section is devoted to technical lemmas which will help touse the forcing theorem.

1.15 Definition. For p, q G P we say that p and q are compatible if theyhave an upper bound. J C P is an antichain if every two members of I areincompatible. X C P is a maximal antichain if T is an antichain and there isno antichain J C P which properly includes X. We say J C P is pre-dense(above p P) if for every q G P ( < ? > p ) some < / G I is compatible with g .We say Z C P is dense above p G P i f for every # G P such that q > p thereis r, ^


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lemma there is an antichain J C J which is maximal among the antichains inJ, i.e. the antichains ofP which are subsets ofJ. Weclaim that J is a maximalantichain. Let r P, we have to prove that r is compatible with some memberof I (and hence J cannot be properly extended to an antichain). Since J isdense there is a p G J such that p > r. Since p e J and I is an antichainmaximal in J necessarily p is compatible with some member q of J, hence r isalso compatible with g; so we have finished proving "J is a maximal antichainof P". So by our present assumption \G\I\ = 1 hence G J D G \I 0.

Secondly to see that G is directed let q, r e G and let J = {p e P : p > q, ro r p is incompatible with q or p is incompatible with r}. Clearly J e V, toprove that J' is dense let s G P. If s is incompatible with # then s J.O th erwi se there is a G P such that 5, # < t. If is incompatible with r then J, and we know that t > s. Otherwise there is a w G p such that w > , r.Since > 5,q we have w >q,r and hence w e J. Since w > > 5 we know ,7is dense. By what we have shown above, G J ^0. Let p G \ J'. We shallse e that p cannot be incompat ible with q o r with r, therefore, since p G v7,p > q,r. We still have to prove that no two members of G, such as p and q,are incompatible. Suppose p, q G and p and < ? are incompatible. W e extendthe antichain {p, < ? } , by Zorn's lemma to a maximal antichain X V. We haveJ G 2 {p,g}, contradicting 1 1 G| = 1.

As part of the assumption of 1.16 is "G C P is downward closed", and wehave proved G is directed, and [J G V is a dense subset of P = > G J ^0],we have proved that G is a generic subset of P over V (see Definition 1.4A).Hence we have finished proving also the if part of the lemma. i . i e1.17 Lemma. If J is a pre-dense subset of P in V and G is a generic subsetof P then G n J 0.Proo/. L et J^ = {p P : (3 q G v7)p > < ? } . Let us prove that J^ is a denseopen subset of P. Now J^ is obviously upward-closed. Let r G P. Since J' ispre-dense there is a < ? G J such that < ? is compatible with r. Therefore, there isa p P such that p > q,r. By the definition of J^ we have p e J . Thus we


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have proved that for every r e P there is a p GJ^ such that p >r, and so Jis dense. Since J e V and j" is constructed f rom J in F we have J V.Since G is generic over V we have Gf J 0. Let p G l J . By the definitionof J^ there is a < / G J such that q

r,q. Since J is pre-dense above q, necessarily 5 is compatible with somemember of J, and hence r is compatible with the same member of I whichneccessarily is also in X^ . Thus we have shown that 2 is pre-dense. Let G be ageneric subset of P such that q G. Since ft is pre-dense and ft GV we haveG Z t 7^ 0. Let t G G Hit. Since , < ? G G , t is compatible with < / , hence by thedefinit ion of ft we must have t G Z and thus G G Z 0. DI.IS

1.19 Lemma. Let Z = {p^ : i < IQ } be an antichain in P and {r^ : i < i$} acorresponding indexed family of P-names (in V). Then there is a name r suchthat: fo r every i < Q and for every generic G, if P i G G then [G] = ^[G] (and[G] = 0 if G {p i : i < i0} = 0). (We recall that a generic G contains at mostone member of I and if J is a maximal antichain of P then G contains exactlyone member of X) .

1.19A Remark. This means we can define a name by cases.Proof. Suppose 7 $ = {(Pij^j) '- J < Ji}, (o f course j 0 is possible) and letT = {{r, Tij) : j < ji, i < 0, r >pitj and r > pi}. D i . i gWe note also:


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2. The Consistency of CH (The Continuum Hypothesis) 11

1.20 Claim. Let G be a downward closed directed subset of P. The fo l lowingare equivalent:(a) G is generic.(b) G J 7^ 0 for every dense open subset I of P.(c) G I /0 for every dense subset I of P.(d) G I 0 for every pre-dense subset T of P.(e) G J 0 for every maximal antichain J of P.

1.20A Remark. Clearly for J C P,(1) J is dense open = > J is dense = > J is pre-dense,(2) J is a maximal antichain of P = > J is pre-dense.Proof. By Remark 1.20A(1) clearly (d)=(c)=^(b), by 1.20A(2) clearly (d)=(e)by 1.16 (e)=^(a), trivially (a)=(b); by the closing up of subsets of P clearly(b)=>(c)=>(d); together we have finished. .2

2. The Consistency of CH (The ContinuumHypothesis)Usually the consistency of CH, i.e.,of 2K = N 1 ? is proved by showing that itholds in L (the class of constructible sets) but we do not want to go in thisway. So

2.1 Theorem. Model A. There is a model of ZFC in which 2H = N I.Proof. Let us first review the main points of the construction of the model. Westart with a countable transitive model V of ZFC in which 2N is either HI orgreater. We shall extend it to a model V[G] in which G will essentially be acounting o f length HI of all sets o f natural numbers, i.e. a function g f rom \onto the family of sets of natural numbers. Each condition, i.e.,each memberof P, is an approximation of the generic object G and therefore will consisto f partial inf ormat ion about the counting. Since every two members of G are


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compatible, th e members of G , taken together, yield a counting o f subsets of .The three things about which we have to worry in the proof are the fol lowing:

a) How do we know that every subset of in V will occur in the countinggiven by G?

This will be answered by showing that the set of all partial countings in whicha occurs is a dense subset of P, and hence G contains such a partial counting.

b) How will the new subsets o f cj, i.e., th e subset o f which are in V[G] butnot in V, be counted when the members of P, being in V, can count onlysubsets o f which are in V I

Here we shall make sure that V[G] has no new subsets o f .c) Is N I of V, which we have mapped on the set of all subsets of in V[G]

also th e N I of V[G}7Here the answer is easily positive because V and V[G] have the same sets ofnatural numbers.

In V[G] we want to obtain a function g f rom N x (where N x denotesth e ordinal which is the N I o f V[G], i.e., th e least ordinal such that V[G] doesnot contain a mapping of onto ) onto P()v (where P() is the powerset of and the superscript V[G] means that P()v^ is the power set of in V [ G ] ) . It will turn o ut that N ^ K]7, and the only subsets o f availablein V[G] are the members of P()v. By easy considerations for every countableordinal , the f unct ion g\a has to belong to V (as it can be coded by a set ofnatural numbers). Therefore partial in format ion about g is given by funct ionsfrom countable ordinals into P() in V. Thus it is natural to define

P = {/ : / is a function f rom a countable ordinal into P()}where th e def init ion is inside V : with /


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2. Th e Consistency of CH (The Continuum Hypothesis) 13

Let G be a generic subset of P. By the definit ion of the concept of a genericobject every two members of G are compatible, hence |JG \J{f : f G G} isa function, we shall denote it with /


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We have now to prove the following facts.

2.4 Fact. Every A G P()v is in the range of /


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2. Th e Consistency of CH (The Continuum Hypothesis) 15

2.7 Lemma. Our present set P is countably complete in V.Proof. Let (pn : n < ) be a nondescending sequence of members of P in V.Since the pn's are pairwise compatible p = \Jnpn for every n < . U2.7

2.8 Theorem. 1) For every countably complete forc in g notion P in V andevery generic subset G of P, V[G] contains no new -sequence of members ofV, i.e., if (an : n < ) GV [ G ] and an G V for n < then also (an : n


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We define a sequence (pn : n < ) of conditions and a sequence (an :n < ) of members of V as follows. Let us mention now that the forthcomingdefinition is carried out entirely within V and therefore the obtained sequencesare members of V. We set P Q = q . For n > 0 we choose pn+ and an so thatPn+1 >P n and pn+ Ih "(n) n", where (n) = n is an abbreviation of "then-th term of the sequence r is n" and n is the P-name of an (see Definition1.9(2) and 1.9A). Do such pn+ and an exist? To prove their existence we goout of V, but this does not matter since once we know they exist the definitionproceeds entirely within V. Let G^ be any generic subset which contains pn.

In V[Gt] we have r[G^] is an -sequence of members of V, since q r,Pn then also pn+ Ih "(n) = n", andPn+ and n are as required. In order to choose a definite pn+ in P we assumethat we have some fixed well ordering of P in V and pn+ is chosen to be theleast member of P in that well-ordering for which there exists an n so thatPn+i Pn and pn+ Ih r(n) = n". Note that n is uniquely determined bypn__ since if also for some b an we have pn+ I h "r(n) = 6" then for everygeneric G^ which contains pn+ we have that the n-th term of [G^ ] is both anand 6, which is impossible.Since P is countably complete there is a p G P such that p > pn for alln < . We have, obviously p >po < 7 ^ and for every n < we know p >pn+and hence p Ih "r(n) = n". Thus for every generic subset G^ of P whichcontains p we have: [G^ ] is an -sequence and r[G^](n) = n for every n < ;,hence [G^ ] = (n : n < ) G V (note (n : n < ;} G V since this sequence wasdefined in V). By the definition of the forcing relation we have p Ih "r G V ",which is what we had to prove.

If G V[G] and C then let (an : n < ) be the characteristic functionof a. Since each n is 0 or 1 we have (an : n < } is a sequence of membersof V and hence, by the present theorem (an : n < ) G V and can be easilyobtained f r o m (n : n < } within V'.


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3. On the Consistency of the Failure of CH 17

2) Lef t to the reader. 02.8,2.5,2.1

3. On the Consistency of the Failure of CHWe first prove a technical lemma, and then prove that 2H = H is possible foralmost any .

3.1 The Existential Completeness Lemma. 1) If p0 Ihp "(3x)(x)n thenthere is a name such that po "~P () where (x) is a formula which maymention names.2) Moreover for every formula (x) as above for some P-name r

Ihp "(3x)[(x)] -> ()n and lhP u-*(3x)[(x)] -> r = 0".Proof. 1) The idea of the proof is as fol lows. The condition pQ, "knows" that(3x)(x) but this does not tell us directly that po knows a particular name ofa set x which satisfies (x). However with more information than that in po weknow names of sets which satisfy (x). What we have to do is to combine thevarious names to a single name which equals each of those names just whenthe name satisfies (x).

Let

J = {q q\\-p "~ (3z)y?(z)" or for some name r we have q Ihp V(r)"}

J is defined in V, hence J e V. We shall now see that J is dense. Let r G P,but r does not force -*(3x)(x). Then there is a generic G C P such that r G G,and V[G] N "(3x)(^(x)", by r's choice. Since every member of V[G] is the valueof some name we have V [ G \ N "([G])" for some P-name r. By the forc in gtheorem there is an r^ G G such that r^ Ihp V(l)" Since G is directed andr e G without loss of generality r t > r and by definition, r^ G J7. Now ifr Ihp "-(z)y?(z)", trivially r e J. Thus we have shown that J is dense. Let Ibe a maximal subset of J of pairwise incompatible members. We shall see thatalso T is pre-dense. Let q G P and suppose, in order to obtain a contradiction,


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that q is incompatible with every member of J. By what we proved about Jthere is a q^ e J such that q ^ >q. Also q ^ is, clearly, incompatible with everymember of J. NowJUJgt} is a subset of J of pairwise incompatible memberswhich properly includes J, which contradicts our choice of J.

Let J = {qi : i < }, since X C J there is for every i < a a name T; suchthatqi Ihp VdO" or9< Ihp "-.(3z)y?()". Letr be the name

f i t i f f c l l - p XrO"10 otherwise

which we have proved to exist (1.19) for pairwise incompatible q^s. We claimthat po I I - 'V(l)" T prove that let G be a generic subset of P and p0 G. SinceJ is pre-dense we have G l I 0 and hence for some i < a we have q ^ e G. If r*is not defined q ^ Ihp "->(b)(/?(x)", but then < f t , p o are incompatible, but both arein G, contradiction. So I h 'V(l)" hence we have V[G] 1 = (Ti[G]). Also sinceft G G we have, by the definit ion of r, [G]-r;[G], hence V [ G ] N "([G\)n ,which establishes po 1 ^ 'V(l)"

2) The second part in the lemma was really proved too. DS.I

3.2 Theorem. Model B. There is a model in which the continuum hypothesisfai ls . Moreover, for every cardinal there is a f o r c i n g notion P such thatIhp "2H > and every cardinal of V is a cardinalConvention. We use the word "real" as meaning a subset of or its character-istic f u n c t i o n .Proof. We want to add to V real numbers ( i.e.,fun ct ion s f r o m into 2).Each condition gives us some i n f o r m a t i o n about them, so we have to make surethat the information contained in a single condition will not suff ice to computeone of the reals since in this case this real will be already in V. Therefore weshall def ine the conditions so that each one will contain only a finite amount ofinformation, and therefore each condition is clearly insuff icient for computingany of the reals. We shall regard the reals, each of which is an -sequence of


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3. O n t h e Consistency of the Failure of CH 19

O 's and 1's, as written in a long sequence o ne after th e other to form a sequenceof length of O 's and 1's. The members o f P will be finite approximations tothis member of 2. Therefore we take P = {/ : / is a finite function from into{0,1}} where by " a finite function from " we mean a function from a finitesubset of . For the partial order on P we take proper inclusion, i.e.,/ < g if/ C g. This forcing is called "adding Cohen reals."

3.3 Lemma. Ihp "there are at least reals".Proof of the Lemma. W e shall prove th e existence of the reals by giving themnames. Since for every generic G there is a function g which satisfies g = (J/eG /we have Ihp "(3x)(x = U / e G P /)" and bytne first l e m m a in this section, 3.1there is a name g such that Ihp "g = jJ/eGP /" Using again th e same methodwe get, fo r every i < X a name ^ defined by a^(n) = g( i -h n). N ow a^ is forcedto be a name of a real number provided g is a name of a function on into{0,1}. We shall prove it in the next sublemma.

3.4 Sublemma. g is a function from into {0,1} (i.e., this is forced).Proof of the Sublemma. g is a function since G is a directed set of functions,hence its union g is a function. Also g is into {0,1} since each member of G isinto {0,1}. Next om(g) C since for every / GG we have / GP and henceDom(/) C ; we still have to prove that Dom(


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Let p P. Since Dom(p) is finite there is an no such that for every n > Qi +n Dom(p) and an n\ such that for every n > n\ we have j -f n Dom(p) .We set r = p\J{(i + & , 0 ) , (j -f fc, 1), where fc > n 0,n. Clearly r is a functionsince i + A : , j 4- k Dom(p) and i 4- k j -f k since z j. Obviously p < r and(r forces that ) ( f c ) = 0, aj(k) = 1 hence O i ( f c ) 7 ^ j ( j ) DS.SContinuation of the Proof of 3.2. We started with , which is a cardinal of Vand we proved that in V[G] there are at least real numbers, but is in V[G]the "same" cardinal as it was in V I As the matter stands now we do not evenknow whether th e continuum hypothesis fails in V[G] since even though maybe a large cardinal in V i t may be countable o r H I in V [G]. W e shall no w provethat all the cardinals o f V are still cardinals in V[G] so for example if = Nfthen is still the third inf inite cardinal in V[G\ and thus = N^''. We shallprove that the cardinals of V are not collapsed in V[G] for forcing notions Pwhich satisfy the countable chain condition and that P satisfies this condition.I.e. the cardinals of V are still cardinals in V[G], and as V, V[G] have thesame ordinals and no non-cardinals o f V are cardinals o f V[G] we have: V ,V[ G \ have the same cardinals; this is done in 3.6, 3.8 below. This is importantgeneral theorem which we shall use a lot. This will finish the proof of 3.2.3.5 Definition.(1) A forc ing notion Q satisfies th e countable chain condition (c.c.c.) if Q

has no uncountable subset of pairwise incompatible members, i.e., if everyuncountable subset of Q contains two compatible members.

(2 ) A forcing notion Q satisfies th e -chain condition (-c.c.) if there are no pairwise incompatible members of P.

3.6 Lemma. If a forcing notion Q satisfies the c.c.c. then(i ) forc ing with Q does no t collapse cardinals and cofinalities, (i.e., \ \ - Q "everycardinal o f V is a cardinal (o f V[6?])), and the cofinality is preserved".


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3. On the Consistency of the Failure of CH 21

(ii) For every ordinal a and every Q-name r there is, in V, a function F froma (to V) such that for every < a we have |F(/3)| < K

0and Ihg "if r is a

function from into V then ( V / 3 < )[r(/3) G F(/3)]".Proo/ of Lemma 3.6. Proof of (ii): We define the function F on by F() ={ G F : (3 q G Q)(r Ihg "r is a function from and () = ")}. W e haveto prove that the right hand side is a set, and not a proper class of V andmoreover is countable. We shall assume it now and prove it later. The righthand side is the set of all possible values of r() in all the V[G]'s in which r[G]is a function from into V. To see this suppose G is a generic set such that[G] is a function f rom a into V. Then fo r some G V V[G] N "r is a functionfrom a and (/3) = ". By the f orc ing theorem there is a q G Q such thatq I f - Q "r is a function from and () = ". Hence G F(/J) and thereforeV[ G \ N "(V/? < )[r(/J) G F()]n. Since this is the case fo r every generic G wehave what is claimed in (ii).

Now we shall prove not only that the class { G V : (3 q G Q)(q \ \ ~ Q Tis a function on a and () = ")} is a set but even that it is a countableset, and thus \F()\ < N 0 Suppose {^ : i < } is a subset of this class.For each such there is a qi G Q such that < & Ihg "r is a function on and(/?) = di". Since Q satisfies the c.c.c. there must be some i j such that q ^and are compatible. Let q > qi,qj and let G C Q be a generic subset ofQ which contains q. We have q^qj G G and hence, since Ihg "r(/3) = ^"an d Ihg "() = ^ ", we have V[G] 1 = "^ = r(/?) = /', hence ^ = ^ .Thus we have shown that at least two of the ^'s must be equal therefore theclass { G V : 3 q G Q (q \ \ - Q "r is a function from and () ")} must becountable. The proof that F() is a set is similar.Proof of (i). Let be an uncountable cardinal of V and suppose is not acardinal in V [ G ] . Then there is an ordinal a < X and a function / G V[G]w h i c h maps a onto . Let r be a name for / in V[ G ] . By part (ii) of our lemma(already proved) there is a function F f rom a in V such that f ( ) G F() an d\F()\ < ^o for every < a. We have therefore = Rang(/) C (j


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of V and a < , and therefore also \a < . Thus we have proved that theuncountable cardinals are not collapsed in V [ G } . Also N O is not collapsed sincethe finite ordinals and o f V are also finite ordinals and in V[G].The preservation of the cofinality is proved similarly. D3.6

Similarly one can prove fo r uncountable .3.7 Claim. If Q satisfies the -c.c. then

(i) forcing by Q preserve cardinals and cofinalities wh ich are > .(ii) for every Q-name and ordinal a there is a function F with domain ,

(F e V) such that \\ -Q " if r is a function from to V then () G F()fo r every < ", and \F()\ < X for < a.

3.8 Lemma. The forc ing notion P wh ich we use here for Model B satisfies thec.c.c.3.8A Remark. O n c e we prove this lemma we know that all the cardinals inV here are cardinals also in V [ G ] and therefore is a cardinal also in V[G],and if is the -th infinite cardinal tt in V it is also the -th inf inite cardinalin V [ G ] .Proof. Suppose {/$ : i < N I} C P, in order to prove Lemma 3.8, it is enoughto prove that two of the functions are compatible. By 3.10 below for someuncountable A C \ and finite w for every i j from A , we have D om ( / i ) D o m ( / j ) = w. The nu m be r of possible fi\w is finite (i.e. < 2 ' ' ) , so withoutloss of generality fi\w = /* for every i GA. Now for any i j A, we knowfi U f j G P is a co m m o n u p p e r bo u nd o f /i, f j . DS.SWe have promised the so called -system lemma.3.9 Definition. A family F of finite sets is called a -system if there is a setw such that for any A , B in F we have A B = w.In this form ulat ion our problem reduces to :


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4. More on the Cardinality 2^and Cohen Reals 23

3.10 Lemma. Given an indexed family F of finite sets, \F\ = N I there isF

1" C F; |F

f| = N I such that F f is a -system.

Proof. The cardinali ty of each A in F is a memb er of ; bu t there are N Ielements in F, so by the pigeonhole principle some n is obtained uncountablymany times. In similar cases in the future we wil l just say: w.l.o.g. \ A\ = n fo rany A in F (since all we need is a fami ly of H I finite sets.)

N ow we proceed by induction on n:For n = 1: this is the pigeonhole principle for HI.For n > 1 we distinguish two cases.

1) There is a (J{A : A F} such that there are uncountably many B in Fsuch that a , then you have an easy induction step ( you take a "out"and put it back after using induction hypothesis).

2 ) If there is no such a then we build a sequence (Aa : a < \) such thata =Aa\A = (Aa G F ) ; suppose A is defined fo r < . N o w\J


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a real such that Ihp " if r is a real then r'". Since a real is a function from into {0,1}, a real r is given by telling for each n < whether r(n) = 0 orr(n) 1. For a P-name r of a real the answer whether (n) 0 or (n) = 1depends o n which condition is in G therefore w e shall have a maximal antichain(Pn,i ' i


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4. More on the Cardinality 2K and Cohen Reals 25

G is a generic subset of P which contains q then in V[G] either [G] is not areal or [G] is a real in which case r(G\(n) = 0 or [G](n) = 1. By the f orc ingtheorem some member p of G forces the statements mentioned above whichhold in V[G], and, without loss of generality, we can assume p > q. This p isin Jn}. Let Tn be a maximal antichain contained in Jn. Since P satisfies thec.c.c. we have |I< N O and we take n = |Jn|,Jn = {pn^ : i l h p '


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have f ( i ) = [G](i) = *[G] = *[G]-[G\( j) = /(j), contradicting the factthat / is one-to-one. So we have proved 4.1. QI.I

4.3 Definition. Cohen Generic Reals. Let us take P = {/ : / is a f ini tefunction f r o m into {0,1}}, with p Pi for all i < n. Obviously p = { ( f c - j , $) : i < n}


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4. More on the Cardinality 2* an d Cohen Reals 27

C we take A = (2 n V \C and C* = cl \ A*, we can write it also asC* = {r G"2 : (V n < ) [S n C

y0]}. One can easily see that if A

visa clopen (i.e. closed and open) set then A* is the same set whether we regard

Av as an open set or as a closed set.We shall use Av only when Av G V; in fact for every Borel set A G V

there is a unique Borel set A* in the universe such that they have a commond e f in i t ion (in V) and then A* V = A.

The close connection between Cohen f orc i ng and Cohen (generic) real is aquite universal phenomena for the f orc i ng in use.

4.4 Theorem. A real number r is a Cohen real (or Cohen generic real) overV iff r belongs to no


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2 8 I. Forcing, Basic Facts

coincides with r on m which includes Dom(pi), hence r f D o m ( p i ) = pi. LastlyrfDom(^) G since it is a finite subset of r, hence pi G G. Since Pi GI wehave G Z 0 and G is generic. U4.4

4.5 Corollary. The set of all Cohen reals over a model V is a comeager subsetof "2.Proof. The reals which are not Cohen reals are exactly th e reals which do notbelong to some set A* where A

vis a dense open subset of

p"} 0; let q be such that p" < q G G Q ,so # > p" > p,pt. Thus we have seen that G Q is directed. Let I G Vbe dense in Q, then, as easily seen, J is also dense in P. Therefore we haveI (Q G)-(I Q) G-I G 0. We can conclude by 1.20 that G Qis a generic subset of Q (over V).

Obviously {p P : (3r G G Q)p < < ? } C G. In the other direction, ifr G G then, as we have seen for p" above, there is a q G G \Q such that g > p.Therefore r G {p G P : ( 3 < ? G G Q)p


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In fact in 5.1 we have proved that P,Q are equivalent where

5.2 Definition. The forcing notions P,Q are equivalent if there are ,, aP-name and a Q-name respectively such that:(1) I hp "r is a generic (over V) subset of Q".(2) I f - Q " is a generic subset of P"(3) for G C P generic, G = [[G}}(4) for G C Q generic, G = [[G\]

Clearly equivalence of f orc i ng notion is an equivalence relation.

5.3 Definition. (1) A fun ct ion / f rom P into Q is called a complete embeddingif: for any maximal antichain T C P , f(X) = {f(p) : p G X} is a maximalantichain of Q and / \ is one to one of course and P N "p p G Q, for p, 9 G P wehave P 1 = "p < < ? " Q N "p < < / " ) , and the identity mapping is a completeembedding of P into Q.

5.4 Lemma. 1) If / is a complete embedding of P into Q, /ien there is aQ-name , I h g " is a generic subset of P". If in addition the range of / is adense subset of Q, then P,Q are equivalent.2) If P < $ Q, and pi, ^2 P then: pi, ^2 arecompatible in P iff pi, P2 arecompatible in Q.3) Assume P C Q. Then P < Q iff every pre-dense J C P is pre-dense in Qtoo.Proof. Easy. D.55.5 Definition. (1) For a f orc i ng notion P, and p, g G P, we say p w q (p,are equivalent ) if any r G P is compatible with p iff it is compatible with q.Clearly is an equivalence relation on P.


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5. Equivalence of Forcings Notions, and Canonical Names 31

(2) W e define P/ as follows: th e members are p/ (for p e P) and we definea partial order: (p/) < ( < ? / w ) z j f f there ar e p1" G p/ an d r t G r/ such thatevery r G P compatible with q ^ is compatible with p t.5.6 Claim. (1) In Definition 5.5 we have:(a) (p/*)


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(2) RO(P) is the partial order whose set of elements is the nonempty openregular subsets of P, and A < B (for A,B G O(P)), if B C A, equiv-alently if every p G P incompatible with ^4 (i.e.incompatible with everyg GA) is incompatible with B.

(3) For p G P let ro(p) = {q G P : there is no r >q incompatible with p}.

5.8 Theorem. (1) The mapping p > ro(p) is a complete embedding of P into

(2) PO(P)/, if we add to it a maximal element 1,becomes a complete Booleanalgebra;(3) If B is a complete Boolean algebra, and P = B \ {!#}, (with the usualordering of a Boolean algebra ) then RO(P)/& is isomorphic to P, moreover,we can use the isomorphism p \ > ro(p).Proof. Well known. Ds.g

5.9 Theorem. The f orc i ng notions P, Q are equivalent iffO(P)/w and O(Q)/ are isomorphic.

Proof. The proof is easy, using:

5.10 Claim. Suppose A G RO(P) andJa maximal antichain such that foreverypeleither p G ^4 or p is incompatible with A. TTien A, UPeAn:rro(p)are equal (in PO(P), union as in a complete Boolean algebra). s. 10,5.9

5.11 Definition. For any G W and V C V^ we define rky(): rkv() 0if f a G V, and rky() = U{rky(6) + 1 : 6 G} otherwise.

5.12 Definition. We def ine when a P-name r is canonical by induction on itsrank : if r = {(pi,Ti) : i < Z Q } , then r is canonical if:

(1) If I h p rk(r) < /?", then > , and if lhP rk v(r) < then


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5. Equivalence of Forcings Notions, and Canonical Names 33

(2) if I h p " r has power < ", is a regular cardinal and P satisfies the-c.c. then IQ < "

(3 ) each T I is canonical, moreover: if pi \ \ - "rk(i) < " then rkn(r^) < / ? ,and if pi Ih ukv(li) < " then > 1 + rk r(r;) and if lhP uTi haspower < ", a regular cardinal and P\{p : p > P i] satisfies the -c.c. then |r^| < .

5.13 Theorem. For every P-name r there is a canonical P-name such thatIhp "r = ".Proof. We prove by induction on the rank of r, that

(*) if r P, r Ih "rkv([G]) < a and kv([G\) < , and [G] has power< ", and is a regular cardinal, and P\{rf : r < r' GP} satisfies the -c.c.,then we can find a canonical P-name such that: r Ih "r = " and rkn() ) : 7 < sup{p : p ] and g G J }


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As P\{rf : r < r' G P} satisfies the -c.c., \I\ < , and as r I h "[G] haspower < " , each \p(p G I) is < . As is regular, UP


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6. Random Reals, Collapsing Cardinals and Diamonds 35

5.18 Claim. 1) If P is a f orc ing notion satisfying the c.c.c. then the numberof canonical P-names of an ordinal < is < (|P| + + H

O) N O .

2) In (1) the number of canonical P-names of a function from to , is3) If P satisfies the ft-c.c., K regular, th e numbers in (1) and (2) should be(|P| + + K0) - . Since there are K I i's and H O positive integers thereis an uncountable subset 5 of H I and an n < such that U i n for all i G S.Since for i,j < HI, i j,Pi and p^ are incompatible we have Leb(p^ pj) =0.Let T C 5 contain 2n2 4- 1 members, then Leb(( n , n ) ) > Leb(UTpi


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(-n,n)) =;eLeb(P* (-n,n)) > =( + 1) =2n+iwhichis a contradiction.

For every positive real in V the set {[n, (n + l)]: n is an integer } isobviously a maximal antichain, therefore G contains a closed interval of length. For p PC let p denote the closed set with the same description in V[G]as p has in V. If Pi GG C Pc for i = 1,..., n we have Leb(p|=1 pY ) /0. Sincep [G1 2 pf we haveLi 1 p such that q G P C G , i.e., q C p. Since th e genericreal g belongs to qv G\ and since q C p implies q C p^Pl also < ; Gp^'G'.Thus for every Borel set pv in G we have g G p v^l.

To reconstruct G f rom # it suff ices to tell which closed sets belong to G,since if we deal with P B or PM , G will consist of all members,of P B or PM whichare < these closed sets. Now we shall see that in V[Gpc], G = {Av GPC g G^N}. We have already seen that if Av G G then # GA .Now we assumethat g G AVIG1 and prove that Av G G. Assume Av G and extend {Av}to an antichain S of PC which is maximal among the antichains which consistof pairwise disjoint sets. We shall see that 5 is a maximal antichain of PC- Ifthis is not the case there is an E GPC such that E is incompatible with everymember of 5, i.e., Leb(E\B) = 0 for every B G 5. By the c.c.c. S is countable,hence Leb(\US) = Leb(S)\Leb(E|j5 ), butLeb(EnljS') 0.N ow E \ U 5 is a Borel set of positive measure, hence it includes a closed setW of positive measure. For every B GS we know that W \B CW \\JS = 0,contradicting the maximality of S. Now that we have proved that S is amaximal antichain we know that G contains some member Bv of 5, which


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6. Random Reals, Collapsing Cardinals and Diamonds 37

is different from Av (since Av G) and therefore BV(AV = 0. Since Bv G Gwe have g GBVW, hence g GBVW AVW. Let [n,n + l]v^ be an intervalwhich contains g, then BVM A [n, n +1]^0.O n the other hand, since (Bv [n, n + l]y) [Ay [n, n 4- 1]v) = 0 thedistance between B^ [n,n 4- 1] and A^ [n,n 4- I]17 is some d > 0. Lete be a rat ional number < d. There is a finite number of intervals of length ewhich separate V [n, n + 1]vfrom A v [n, n + 1]v.The "same" intervalsseparate also BVW [n,n + 1}VW from AvtG] [n,n +1] contradicting

6.2 Theorem. A real number r is random over V if f for every Borel set Av G Vsuch that Leb(Av) = 0 we have r A* if f this holds for every G set A^ (whereA* is the Borel set in the universe with the same description as A.)Proof. A s s u m e that r is random over V , r Gn{Bv^ : Bv G G C P}. LetAv be a Borel set with Leb(Av) = 0. As easily seen the complement R \ Av ,where Ry is the set of all reals in V, is such that {R^ \ Av} is a maximalantichain in PB and hence R \ Av G G. Therefore r G IR l \AVW C R\ A * ,i.e., r ^A * .

N ow assume that r ^ A * for every G - set Av (i.e. countable intersectionof open sets) such that Leb(Av) = 0. Define G C Pc by Bv G G i f f r G S*.If A y D5yGG then also A * D5* and r GBv, hence r GA * and Av G G.If Av ,5VG G then r G A * 5*. If Leb(y n 5V)-0 then since Av Bvis closed and therefore is a G-set, we would get r A * B*. ThereforeLeb(Av Bv) > 0, and since r GA * n B*, we get A y G G. Finally, let Sbe a maximal antichain in PC, we have to prove G ( 5 0 in order to prove thatG is generic. Let Ev = R \ (J{A : A G 5}, so is obviously a G - set. SinceS is ma x i ma l Leb ) = 0, therefore r E*. Since Ev = R \ \J{A : A G 5},so we have E* = R \ |JsB*. Since r g E* we have r G B* for some B G 5,and therefore G G and G S 0. D6.2


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T he Levy Collapse.6.3 Definition. 1) Levy(H0,) = {/ : / is a finite f u nct i on f r o m into },where is an uncountable cardinal. For the partial order on Levy (Ho, ) wechoose inclusion.2) Levy(tt, ) = {/ : / is a partial f u nct i on f r o m K to such that |Dom(/)| < }ordered by inclusion.6.4 Discussion. In V[G], where G is a generic subset of L evy( H o , ) it is easilyseen that g = (J G is a f u nct i on on onto . Therefore ||^[Gl H Q .

Since in V [ G] there are no cardinals between and + all the ordinals< + are countable in V[G] so (+)y = H - ^ ^ provided + is a cardinal inV[G], thus we must check what is the fate of the chain condition for Levy(H0, ).Levy (Ho, ) has pairwise incompatible members, for example {{0, ) : < }.However, it is easy to see that |L evy( H o , )| = and hence L evy( H o , ) satisfiesthe + chain condition and the cardinal + is not collapsed by f o r c i n g withLevy (Ho, ).

We define6.5 Definition. 1) Levy (Ho, < ) {/ : / is a finite f u nct i on f r o m x into such that / ( O , n) =0 and for /0 we have /(, n) < }. The partial orderon Levy (Ho, < ) is inclusion.2) L evy( / , < ) = {/ : / is a partial f u nct i on f rom x into such that|Dom(/)| < K and / ( O , ) = 0 and > 0 = > /(, i) < a} ordered by inclusion.6.6 Discussion. Let G be a generic subset of Levy(H0, < ) over V, and letfo = \JG. Obviously for every 0 < a < X the f u nct i on /(, -) is a mapping of onto , and hence is countable in V [ G ] . What about , is it, too,countablein V [ G \ or else is it H^'G'? If is singular in V it stays singular also in V[G],hence it cannot be H j ^ ^ and it is countable.6.7 Theorem. If is regular then Levy(H0, < ) satisfies the -chain condition,and hence = H^[GI in V[G\.


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6. Random Reals, Collapsing Cardinalsand Diamonds 39

We shall prove the following more general version.

6.8 Theorem. Let be a regular uncountable cardinal, \W\ = , let (Ax : xGW ) be such that \A X\ < for x GW and let P = {/ : / is a finite funct ion onW such that f ( x ) G Ax fo r each x G Dom(/)} be a for c ing notion such that: if/, g GP agree on Dom(/) D o m ( ^ ) then /, g are compatible. Then P satisfiesthe -chain condition.

Proof. Without loss of generality we can assume that W = and Ax C foreach x G W. Let ( f a : a < ) be a sequence of members of P. For a < X wedefine h(a) Max[{0}(J(Dom(/) )]. Since fa is a finite funct ion we haveh() < a for a > 0. Thus h is a regressive funct ion on \ {0} and therefore ithas a fixed value, 70 on a stationary subset S of by Fo dor's Lemma. Let ptbe the set of all members of P whose domain is included in 70 -h 1. For everyfinite subset u of 70 4- 1

|{/ G Pf :Dom(/)-u] =|{/e P :Dom(/)-u} \ =

The number of finite subsets u of 70H - 1 is < |7o| + N O < hence, since isregular |pt |=uc7o+i,uis finite K/ P : Dom(/) = u}\ < . For each g G ptlet Sg = {a G S : fa f (70 4- 1) = g}. Clearly U p e p t Sg =Ssince for every G 5 we have G Sy^^o+i)- Since |pt| < one of the 5p's say Sgo mustbe stationary (since the union of < nonstationary sets is nonstationary) . LetC = { < : a limit ordinal satisfying (V < )[om(fa) C 5]}, clearly Cis a closed unbounded subset of , hence 5 = Sgo C is a stationary subsetof . Let a G Sgo and let G 5 0 be such that < G 5 hence / ? is a strictupper bound of Dom(/), we shall see that fa and /^ are compatible. Since Sgo C S, clearly Max(Dom(/^) ) = 70 and since Dom(/) C wehave Dom(/) om(f) C 70 -f 1. Since , /? G 55o we have /f(7o + 1) =/ / 3 (7o + 1) = # o > hence / an d //? ar e compatible. .8,6.7


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On Diamonds.6.9 Theorem. If is regular uncountable cardinal, 5 C stationary, then forsome forc ing notion P:

(1) ll-p "Os holds " (see below)(2 ) forcing with P, preserve the cardinals < X (in fact it is -complete

and hence add no new -sequences of ordinals for a < .(3) I - P | < soif p such that q (Ai : i G S ), < , A i C iand q Ih "i C&A i A? for some i G S ; this will prove thatthe set of such < ? ' s is dense above / hence that G contains one of them soV[G] t = "{ G 5 : A a = Aa and a GC} / 0" and as this holds for any clubC G V[ G \ clearly V [O\ t = "{ G S : Aa = Aa} is stationary" i.e., V [G\ N "O sholds as exemplified by A" where A = \J{f : f G G}.

So let us find < / , w e define by induction o n < , a < , p = (Ai : i GS ) , -B an d / ? such that

1 ) for < C, < c, p^


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7. Does NotImply 0 41

2 ) Pc+1 i h GC"3) p

c+ Ih "A

c= B" for some B

GV such that B

C

c.

4) for limit C, we have p = (JB^ = B a. In the end { : < X} is a club in V, but 5 is a stationarysubset of (in V) hence for some limit C , 5, but then p^ Ih "for < " , l = and e C hence A = B = \J^


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"(A : G 5} is as above, and for every unbounded X in there are stationarilymany points G 5 such that A

aC X".

Proof. Assume the first form holds, let X be unbounded and let C be anarbi t rary c lub in . We must show that there is a point a GC S such thatAa C X. Define by induct ion an increasing sequence (i : i < ) of ordinals inX as follows: 7^ is the first member of C greater than all (i : j


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7. * Does Not Imply 0 43

j : KI - > K I as follows: j(a) = minimal j greater than all the ordinals in{j() - a (the first counterexampleyields an immediate contradiction). So there is a club C whose members areclosed under j.Set X' as the range of j. This is clearly an unbounded set bythe monotonicity of j. So there are stationarily many GS for which A$ C X'.Hence there are stationary many G S C for which A$ C X'. Fo r each such, let us prove that D $ is exactly X : As ^5 contains only members of X1',which are in particular indices of bounded subsets of the form X , it is clearthat D $ - which is the union of these sets - is contained in X\. To see equality,fix an arbitrary < . It suffice to show that in the union D $ appears an initialsegment of the form X a with a > . As C, clearly is closed underthe f unct ion j so j ( ) < 5, and since A is cofinal in 5, there is an ordinal 7such that > 7 > j ( ) and 7 GA$. But A$ C X' hence 7 j() for some ,no w by the monotonicity of j, > /?. So by definition of j, Bj(a) is an initialsegment of X which is obtained by cutting X somewhere higher than j ( ) , butthe latter is greater or equal to . So D $ includs X for arbitrarily large < , hence D $ include X ; hence equality follows. D7.s

7.4 Theorem. A does not imply ODiscussion.

It is clear that the principle 0 implies A, and under CH, we have seen thatA implies O It was asked whether > O As we shall now see, the answer isnegative. W e shall build a model of ZFC in which A holds, but CH fails. Astrivially 0 > CH,this necessarily implies that 0 also fails.Proof. Out intention is to begin with a ground model satisfying GCH (or just2^ N 1 } 2*1 = K 2 ; for example a model of V = L) which has a 0-sequenceo n ^2 Using the 0 we will define a ^-sequence on K 2 which is immune to NI-complete forcing, i.e. is still a ^-sequence in the universe after f orc ing with anNi-complete f orc ing notion. The next step is adding N 3 subsets to KI using anNi-complete f orc ing notion. The last stage will be collapsing K I, thus making


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of the cardinal that previously was H 2 our present HI. Although this will notbe done by an K

2-complete f orc ing, o ur Jfc-sequence on N

2, which has has now

dropped to HI, will remain a ^-sequence, while there will be K 2 subsets o f HO(which were formerly H S subsets o f HI). So we change our universe three times.We will present each stage of our plan in detail:Stage A: defining a ^-sequence on H 2 using

We first note here that a 0-sequence can guess not only ordinary subsetsof H 2, but also elementary substructures of any structure over H 2 . Supposea structure M over H 2 (in a countable language) is given. For each relationsymbol or a function symbol R fix a subset of AR H 2 of size H 2 and a 1-1function FR : H 2 ' > AR such that two different ARS are disjoint where welet g(R) = n if R is an n-place relation symbol, and g(R) = n + 1 if R is ann-place function symbol. Thus the set FR(RM) codes the relation RM . Let Abe the union of all FR(RM). So A codes the structure M. If (Ai : i S) is a0-sequence, then A is guessed stationarily often, namely for stationarily manyi G S we have A i = Ai. Clearly, the set C = {i < H 2: for all R we haveFR(RM^) C i} is closed unbounded in H 2 . Moreover, the set of such thatM \a is an elementary substructure of M is a club by the Skolem-Lwenheintheorem and the continuity of elementary chains. So the intersection of thetwo clubs with our stationary set is the stationary set SM of all such thatM\a is an elementary substructure of M and for all R we have FR(RM^) C and Aa = \jFR(RM^a) (that is, M\a contains its own coding and Aa equalsRthis coding). In short we say that for every a. in 5M,Aa guess the elementarysubstructure M\OL.

L et S be the subset o f H 2 of all ordinals having cofinality H Q . W e wish tomake use of Os Why does this principle hold in our ground model? Because,e.g. we could have forced it easily by a preliminary forcing which appeares in6.9.

N o w we come to def ining the Jfr-sequence. Let us choose coding for alanguage with two relation signs,


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7.*Does Not Imply 0 45fo r every structure M with universe ^2 fo r this language, fo r stationarily many S the substructures M of it are guessed by our 0-sequence i.e. M\ = M S .Restrict attention now only to those places in which the guessed substructuresatisfies th e fol lowing sentences (the set of such G S will be called S'):(i) )Note that there are stationarily many such places (by guessing one such struc-ture on ^2 ? for example). In each such place we define now a subset of of order type which we call D $ : le t be fixed, and let (^ : n < ) be acofinal increasing sequence in < J ; we define by induction o n n < a sequencen : < ) such that

2 - >

So let 7# , $ < < J be such that ( 7 o > f o )&$) < oexistsbycause(ni) above,and the induction step i.e. choosing 7*+1, +1 such that 7


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what we are about to check now. We must check first, however, the weakercondition, that this forc in g preserves stationarity of stationary subsets of N

2namely that a subset 5 of N 2 which intersects every "old" club ( "old" meaning"in the universe befor e the f o r c i n g " ) intersects also every "new" club. Thisfollows f rom the fol lowing couple of claims:

Claim D. PI satisfies the N 2-.c.c.Claim E. Every P that satisfies the N 2-.c.c preserves the stationarity of subsetsfN2.

Proof of Claim D. let be any regular cardinal which is large enough to havethe power set of PI in H ( ) . Suppose that J is a maximal antichain of PI whosecardinality is greater than N I . Now define an increasing sequence of elementarysubmodels o f (/f(), G) o f length N I called (N i : i < N I ) , as follows:(i ) PI, X are members of 7V 0, and 7V0 is countable, has cardinality N I .(ii) every countable subset of Ni is an element of N+I.(in) j < i = > Nj ^Ni.

There is no problem to carry out the construction, because at each stage inthe construction the cardinality of the model at hand is at most N I, thereforeit may have at most N I countable subsets (we have CH in the ground model).So close these NI elements together with everything you already have in anelementary submodel of cardinality N I using the Skolem-Lwenheim method.At limits take unions, for example. Let us denote the union of the increasingchain as T V . So T V is an elementary submodel of (f(),E) of cardinality N I .Furthermore, every countable set of elements of A T is a subset of one of themodels along the construction (say Ni) by the regularity of N I, therefore it isan element of N+I, and therefore of N. In short we say that " A T is closed undercountable subsets".

We recall that PI, X N. As X is greater in cardinality than JV, theremust be an element p of J which is not a member of N. (Remember that p is a


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7. Does Not Imply 0 47

countable funct ion f rom K 3 to 2.) Look at pf, which w e define as p N . Fromthe countability of p,p' is clearly countable, and as N is closed under countablesets and p 1 is certainly a countable subset of TV, we have p' N. Now (H(), e)satisfies the sentence "there is a member of I which extends p 1" (it is p). AsN is an elementary submodel of it, and contains as members all the constantsmentioned in this sentence - therefore there must be an element q G I Nwhich according to T V extends p'. We lack only one more detail to derive acontradiction: the domain of q is countable, therefore there is an enumerationof it in N. This implies that D o m ( g ) is contained in T V . So inspect the unionq U p. Clearly q extends the part of p which is in N, and contradicts nothingthat is outside of T V , as we just saw. Therefore p and q are compatible membersof J.Proof of Claim E. Let S be a stationary subset of H 2 in V, and suppose C isa PI-name for a club o f N 2 with th e condition p in the generic set for c ing p I h"C is a club in N 2 " . Attach to each ordinal < N 2 a maximal antichain Ia ofextentions o f p such that fo r each q G Ia there is a (q) such that q I h "/?(#)is the minimal member of C above ". Define as Ba the set of all ordinals 7fo r which some member of 2a forces that 7 is the first member of C above a.Clearly Ba has cardinality < |J| which is < N 2 hence Ba is bounded in N 2Using our standard argument, we see that C* : { < N 2 : for all a < the setBa is included in } is a club of K 2 . But this club, being defined in V, is an oldc l u b ! Therefore it meets 5, let us say in . As for each Ia, a < the genericset must choose a condition f rom J, and5is contained in 5, is a limit ofthe realization of C, therefore in it.Stage F. So we know now that our S f f rom the definition of the it-sequence isstill a stationary subset of N 2 after for c ing with P I. But is the sequence still aJk-sequence? We verify this now.

Let X be a Pi-name, p e PI and p lhPl "X C N 2 is unbounded and Cis a club of N 2 " We show that the set of conditions which force "there exists G S C such that L^ C X" is dense above p. Fix an elementary submodel


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N of (if (), G ), \N\ = H 2, N closed under subsets of size < HI, with p, PI, Xand C members o f T V . Enumerate in the sequence (p i : i < H 2) th e memberof PI M which are > p and define a structure M with the same languageas above with universe H 2 with: i < * j if f PI \ = P i < PJ, and R(i,x) if f asPi Ih "x G X". From th e definition of our diamond sequence (Ma : a G 5'),stationarily many elementary substructures of this structure are guessed by iti.e. S" = {a e S f : Ma = N\a} is stationary. Let a be one such coordinateso Da = {: n < } where< +1 < < * = U , #N 7-


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7. * Does Not Imply 0 49

From the lemma we easily deduce both the preservation of stationary sets(in a new club there is an old unbounded set, th e closure o f which is an oldc lub; alternatively use Claim E) and the preservation of our Jk-sequence, forinstead of guessing a new set, it is enough to guess an old subset. D.4

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Saharon Shelah- Proper and Improper Forcing Second Editon: Chapter I: Forcing, Basic Facts

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