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, where p(0)eQo. And ?(0)> ^ ^ ( O ) ) ittp(0)^q(0) in Q. (ii) If a = P + 1 then Pp = {p \fS\pePa} is an iteration of length /?, and there is some forcing notion QpsVPp such that pePa p^aq 50 can be extended to an (injective) homomorphism cp of A into B such that n((p(a)) = a for all aeA. Now let A be an arbitrary group. 5.1 , where p(0)eQo. And ^ , < # ) ) > iff p(0) ^ q(0) in QO. (ii) If a = /? + 1 then Pp = Pa\p = {p \p:pePa} is an iteration of length j3, and there is some forcing notion QPEVPP such that pePa p^aq as follows, by induction on /} < a. Having constructed p\PePp, we let p(/?) be (in FP0) a lower bound of {p^(jS):£< A}; if all the p^(jS), £ < / , are = 1, then we let p(p)= 1. This guarantees that supp(p) £ ( J ^ s u p p ^ ) and so p is a condition. • The next theorem deals with the chain condition. 7.9 such that for each n,p\n\\- p(n)eQn. For each n ^ 1, let P n = {p t^-pe^}- Pn is dense in Bn. The assumption is that for each n, \\-BnQn is a proper complete Boolean algebra. Let ^(Qn) denote the proper game on Qn (in VBn); by the assumption there are <jneVBn such that for all n \\-Bn6n is a winning strategy for II in $(&„) ,, such that for every n, <$„ satisfies <«0"4,-i>lhp H MJh Q l I Vm>n3fc a ^ a j " 1 ) . We claim that q = <„>„ satisfies ^lhVn3/c
iff iff
p\PePp p\P^pq\P
and
j and
p \ fi\\j p(P)
2 Finite support iteration
51
(iii) If a is a limit ordinal, then VjS^ a 4
iff Vj8 < a p\f}ePp iff
is
and for all but finitely many (I < a,
Vp<
The finite set {/?
Lemma If Pa is a finite support iteration and Pp = Pa\fl, yPp c J/V
then
Proof This can be proved either directly, by showing that if G is generic on Pa then Gp = {p \/3:peG} is generic on P, or by invoking Lemma 2.7 in Part I. The canonical projection h = haP:Pa^Pp and the canonical embedding i = ipa\Pp^»P0L are defined by
p\P, i(q) = frr..
•
(2.5)
Note that if a is a limit ordinal, then
which justifies the notation (2.2). The most important feature of finite support iteration is preservation of chain conditions. 2.7
Theorem Let K be a regular uncountable cardinal. Let Pa be a finite support iteration of (Qp'-P < a>, and let us assume that for each fi < a, \\jQp has the /c-c.c. Then Pa has the /c-chain condition. Proof By induction on a. If a = /? + 1 then Pa = Pp*Qp and the theorem follows from Theorem 1.10. Thus, let a be a limit ordinal. For each peP a , let supp(p) denote the support of p.
52
/ / Iterated forcing
Case I cf a # K. Let W £ Pa be a set of size *c. Since cf a # *c, there exists a jS < a and some ZczW of size /c such that VpeZ supp(p) c: /?. Then {p IP'.peZ} is a set of size K in P^, and since Pp has the K-C.C, there are p and g in Z such that p t/J and q \fi are compatible (in P^). But then p|g. Case II cf a = KJ. Let {a^:<J < /c} be a normal sequence with limit a, and let W= {p,*:£ < K) be a subset of P a of size zc. For each limit £, < K there is y(t;)<£ such that supp(p)na)«<=ay(^). By Fodor's theorem there is a stationary S a K and some y < K such that supp(p^)na^ c ay for all £eS. Furthermore, we easily construct a set Z a S of size /c so that for any I
if ; < a y if ay ^ i [pn(i) if a ^ i < a The condition r is (in Pa) stronger than both p^ and p^, and so p^ and p^ are compatible. • References Solovay, R. and Tennenbaum, S. (1971). Ann. Math. 94, 201-45. Baumgartner, J. (1983). In Surveys in Set Theory (Mathias, A.R.D., ed.), pp. 1-59, Cambridge University Press
Martins Axiom
The solution of Suslin's problem (see Chapter 4) in the late 1960s, by iterated forcing, led to a formulation of an 'internal forcing axiom' called Martin's Axiom. The use of Martin's Axiom often eliminates a separate forcing construction when one tries to establish the consistency of some conjecture. Martin's Axiom has proved to be very useful in combinatorial set theory, and has become particularly popular among point set topologists. 3.1
Martin's Axiom (MA) For every c.c.c. partial order (P, <), if X < 2*° and {Da:a < X) is a collection of dense sets in P, then there exists a filter G on P such that GnDa is nonempty, for every a < L We first remark that MA is a consequence of the continuum hypothesis: if P is any partial order (not necessarily c.c.c.) and {Dn:neco} are dense sets, then it is easy to find a filter that meets all the Dn\ let p0 ^ pl ^ •• • ^ pn^ -" be such that pneDn9 and let G be generated by {pn}n. We shall show that MA is consistent with the negation of the continuum hypothesis. The often used consequence of MA + 2X° > Xx is Martin's Axiom for collections of Xx dense sets: 3.2 If P is c.c.c. and if {Da:oc
3.4
Lemma MA is equivalent to the following variant: MA*If P is c.c.c. and \P\ < 2*°, and if {Da:a < X},X < 2K° are dense, then 53
54
/ / Iterated forcing
there is a filter G that meets all the Da. Proof Let us assume MA* and prove MA. Let P be a c.c.c. partial order, and let D a , a < X (X < 2X°) be dense sets in P. It is not difficult to find Q a P of size X such that (i) any p.qeQ that are compatible in P are compatible in Q, and (ii) for all <x,Dar\Q is dense in Q. Q is c.c.c. and so we apply MA* to Q and the sets Da n g. The filter we obtain generates a filter on P that verifies
MA.
•
3.5
Theorem Assume GCH and let K be a regular cardinal greater than Kx. There is a cardinal preserving notion of forcing P such that the generic extension by P satisfies Martin's Axiom and 2X° = K. Proof We construct P as a finite support iteration of length K of certain {<2a:a < K). At each stage, |— p Qa is c.c.c, so P is c.c.c, and so all cardinals are preserved. The construction is such that \Pa\ ^ K for each OL^K. This property is certainly true for a limit a if it is true for all /? < a, because P a is the direct limit of the P/s. At a successor stage, P a + x = P a * Qa, and the Qa will be such that lbrl6J
f6 {\}
if Q satisfies c.c.c otherwise
The 2 a , a < /c, determine the iteration P = PK. Now we have to prove that Vp satisfies MA as well as 2N° = K. Let G be a generic filter on P, and let Ga = GfP a for all OL
3 Martins Axiom 3.6
Lemma If X < K and X c X is in K[G], then XeVlGJ
55
for some
OL
Proof By the c.c.c, each Boolean value b^= \\£eX\\ is determined by a countable set of conditions in P (a maximal partition of b^ in P); hence, X is determined by at most X conditions in P. Each condition has a finite support and so in fact belongs to some P a , a < K. It follows that there is p OL
Theorem Assume MA. Then (a) the union of less than 2*° meagre sets is meagre;
(b) the union of less than 2X° Lebesgue null sets in null.
•
(Consequently, 2X° is a regular cardinal.) References Martin, D.A. and Solovay, R. (1970). Ann. Math. Logic 2, 143-78. Shoenfield, J. (1975). Am. Math. Monthly 82, 610-17. Rudin, M.E. (1977). In Handbook of Mathematical Logic (Barwise, J., ed.), pp. 491-501, NorthHolland, Amsterdam. Fremlin, D.H. (1984). Consequences of Martin's Axiom, Cambridge Tracts in Mathematics 84, Cambridge University Press.
4
Suslin's problem
The problem dates back to 1920. Let (L, < ) be a dense linear ordering without endpoints, Dedekind complete, and such that it has no uncountable family of disjoint open intervals. Is L isomorphic to the real line? The eventual solution of the problem led, among other things, to the introduction of iterated forcing and Martin's Axiom. Suslin's problem is independent of the axioms of set theory. A Suslin line is a linear ordering that satisfies the assumptions in the problem but is not isomorphic to the real line. There are models of set theory in which a Suslin line exists; in this chapter we show that the existence of a Suslin line is not provable in ZFC. We prove below that MAKX implies that Suslin lines do not exist. 4.1
Definition A Suslin tree is an uncountable partially ordered set (T, < ) such
that (i) for every teT, the set of all predecessors {ssT:s < t} is well ordered (and we call its order type the height of t)\ (ii) T has no uncountable chains (linearly ordered subsets) and no uncountable antichains (pairwise incomparable). Note that the height of each teT is a countable ordinal. 4.2
Lemma A Suslin tree exists if and only if a Suslin line exists.
Proof (a) Let (L, < ) be a Suslin line. By induction on a ea^, we construct a sequence {/a}a of open intervals of L as follows: at stage a, choose / a so that for every j? < a, either / a is disjoint from Ip, or / a c Ip. Since L is not isomorphic to the reals, the endpoints of the intervals {/^/J < a} are not a dense set in L, and so Ia exists. The set {/a:a
4 Suslin's problem
57
points of the / a 's in C form an increasing sequence {an\n KCD^ in L of order type col9 or the right endpoints form a decreasing sequence of length a)v But that is impossible because {(avan+l):rj
(4.3)
(To prove (4.3), use the fact that for every a
Theorem If MA holds and 2X° > Kl9 then there are no Suslin trees.
Proof Assume MAX l5 and let (T, < ) be a Suslin tree. Without loss of generality we may assume that T also satisfies (4.3). Let (P, < ) be the partially ordered set obtained by considering T upside down; i.e. T with the reverse order > . We apply MAKX to the notion of forcing P. If t and 5 are comparable in T then 11 s as forcing conditions; if t and s are incomparable in T, then t and s are incompatible as forcing conditions. As T is a Suslin tree, P has the countable chain condition. For each a
By (4.3), each Da is dense in P. Using MAKl9 there is a filter G on P that
58
/ / Iterated forcing
meets every Da. It follows that G is an uncountable chain in T, which is a contradiction. • References Souslin, M. (1920). Fund. Math. 1, 223. Solovay, R. and Tennenbaum, S. (1971). Ann. Math. 94, 201-45. Rudin, M.E. (1969), Am. Math. Monthly 76, 1113-19. Jech, T. (1971). J. Symb. Logic 36, 1-14. Devlin, K. and Johnsbraten, H. (1974). The Souslin Problem. Lecture Notes in Mathematics 405, Springer-Verlag, NY.
Whitehead's problem
In this chapter we describe an application of Martin's Axiom in the theory of infinite abelian groups. Throughout this chapter, a group means an infinite abelian group (with operation + ) . A group is free if it has a basis, i.e. a set of generators that are linearly independent. Free groups are characterized by the following property: let A be a free group, and let n:B-^A be a homomorphism of some group B onto A. Let X be a basis of A, and for each xeX, pick q>(x)eB so that n(
Definition A is a W-group if for any homomorphism TZ:B-+A onto A with kernel Z, there exists a homomorphism (p:A-+B such thatrc((p(a))= 0 for all aeA.
By the remark preceding Definition 5.1, every free group is a W-group. 5.2
Whitehead's problem Is every W-group free?
It had been known that every countable W-group is free. Shelah proved that Whitehead's problem is undecidable in set theory. Here we are concerned with the half of his solution that shows that it is consistent that an uncountable W-group exists that is not free: 5.3
Theorem (Shelah) MA^! implies that there is a W-group of cardinality Xx that is not
free. We shall outline the idea of Shelah's proof, but first we need some more definitions and facts on W-groups. 5.4
Definition A group A is torsion free if every nonzero aeA has order oo, i.e. 59
60
/ / Iterated forcing
n-a # 0 for all n = 0,1,2, A is Xrfree if every countable subgroup of A is free. We note that K r free implies torsion free because A is torsion free iff every finitely generated subgroup of A is free (this is a theorem in abelian group theory). It is known that every W-group is K r free. It had also been known that there exists a group A such that (i) (ii) (iii) (iv)
1^1 = Ki A is not free (5 5) ' A is K r free every countable subgroup of A is included in a countable subgroup B of A such that A/B is K r free
The proof of Theorem 5.3 consists of showing that under MAK ls the group A in (5.5) is a W-group, and so is a counterexample to Whitehead's problem. Let A be an K^free group of size Kx with property (5.5)(iv). A subgroup S a A is a pure subgroup if A/S is torsion free. Let n be a homomorphism of some B onto A. We use MAKj to construct a homomorphism cp:A-+B such that n(cp{a)) = a for all aeA. Let P be the set of all homomorphisms q>:S->B such that S is a finitely generated pure subgroup of A, and n{cp{a)) = a for all aeS. We consider P as a notion of forcing, partially ordered by 3 . For each aeA, let Da = {(peP.aedom (cp)} We need two things to apply MAXX: Every Da is dense in P P satisfies the countable chain condition
(5.6) (5.7)
If (5.6) and (5.7) hold then by M A ^ there is a filter G on P that meets every Da. Clearly, G yields a homomorphism of A into B verifying that A is a VK-group. It is relatively easy to prove that (5.6) holds. The proof requires a little bit of group theory and only uses the fact that A is K r free. The heart of the proof lies in showing that P has the countable chain condition. One employs a A-system argument, but the proof uses substantially the property (iv) from (5.5). References Shelah, S. (1974), Israel J. Math. 18, 243-56. Eklof, P. (1976). Am. Math. Monthly 83, 775-88.
6
Kaplansky's conjecture
This chapter deals with an application of forcing in the theory of Banach algebras. C[0,1] is the algebra of continuous functions on [0,1]. Kaplansky's problem Is every homomorphism on C[0,1] (into any commutative Banach algebra) continuous? Under the assumption of CH there exist homomorphisms on C[0,1] that are discontinuous. We devote this chapter to the following independence result: 6.1
Theorem (Solovay, Woodin) It is consistent that every homomorphism on C[0,1] is continuous.
Kaplansky's conjecture had been successively reduced to a purely set theoretic problem. For functions / , geof3, we define (a partial ordering) f
iff
3k
Mn>k
f(n)
(6.2)
Similarly, if U is an ultrafilter on co, we let f
iff
{n:f(n)
(6.3)
For geco™, we denote by
the initial segment {f/U:f
Theorem (Woodin) If there is a discontinuous homomorphism then there exists a nonprincipal ultrafilter U on co, an unbounded monotone function g and an order-preserving function 7i:Ult^)^(a; w ,<)
•
We shall construct a model in which Woodin's necessary condition fails, and consequently there is no discontinuous homomorphism in that model. 61
62
/ / Iterated forcing
The proof is in two steps, one using MA and the other constructing a model by iterated forcing. Consider the set {0,1}W1 ordered lexicographically. 6.5
Lemma Assume MA + not CH. For every nonprincipal U and every monotone unbounded g, {0,1}W1 can be embedded in
6.6
Lemma There is a model of MA + not CH which satisfies that {0,1}W1 cannot be embedded in (co03, <). These two lemmas and Theorem 6.4 establish the consistency result 6.1. We prove Lemma 6.5 first. 6.7
Lemma Let {/„}„ and {gn}n be two sequences of functions in co03 such that / o > / i > " • > / « > • • • > 0» > • • • > 0 i > 0 o
Then there are / and g in co03 such that for all n,
fn>f>Q>9n and, moreover, limfc (/(/c) — g(k)) = oo. Proof We find k0 < kx < ••• < kn < ••• such that for each n, kn is so that
fo(k)>-">fJik)>gH(k)>"'>go(k) Let / and g be so that for all n, if kH^k
+ l,
f(k)=fn(k) and
6.8
Lemma Assume MAX^ Let {/a } a and {ga}a be two c^-sequences of functions in co03 such that For every nonprincipal ultrafilter U on co there exists heco03 such that for all a <<*>!,
Proof Consider the following notion of forcing P: a condition peP is a triple (E,huh2\ where
6 Kaplansky's
conjecture
63
(i) E = {OL19..., am} is a finite subset of col9 ax < • • • < am (ii) hx = <MO), • • - M*,)> and fc2 = <MO), • - • > M M > are finite functions (into co), and (iii) kp is such that for every k> kp
(6.9)
U(k) >
>LJk)
> gam(k) >
> gai(k)
(*)
A condition q = (£, hx, /i2) is stronger than p = (£, /i t , /i2) if: (iv) £=>£ _ (v) /ix extends hl9 h2 extends h2, and (vi) VaeE V^> kp(k^ kq) either ga(k)< Mfc)
C/aim P satisfies the c.c.c. Proof As there are only countably many pairs (hl9h2), it suffices to show that any two conditions with the same hl and h2 are compatible. Let p = (E,huh2) and q = (F9hl9h2). For all k> kp = kp =fc^= length(/it), the functions {/ a ,# a :ae£} satisfy (6.9)(*), and so do the functions {/ a ,# a :aeF}. Let kr be such that all the {/ a ,# a :ae£uF} have the property (*) for all k>kr. We extend hx and h2 io hx and /i2 (of length kr) so that for every k between kp and /cr, if OLGE then /a(/c) > h^k) > ga(k% and if OLEF then /a(/c) > is stronger h2(k)>ga(k). It follows that the condition r = (EvF,h1,h2) than both p and g. • Now let Dn = {peP:Kp^n}
{ned)
and Aa = {pGP:aGjB
where
p = (E9hl9h2)}
(a
Each Dn and each Aa is dense. By MAKX there is a filter G on P that meets all the Dn and all the Aa. G yields a pair of functions huh2ecoto. If a < a^ then there is a peG such that aeE, p = (E,hl \kp,h2 \kp). For every k > kp9 either fa(k) > hx(k) > ga(k\ or fjk) > h2(k) > gjk). It follows that either (1) fa > {Jhl > vga, or (2) fa > vh2 > vga. Either (1) or (2) holds for uncountably many a's, let us say (1), but then (1) holds for all a, proving the lemma. • Proof of Lemma 6.5 Let U be a nonprincipal ultrafilter on co, and let geco03 be unbounded and monotone. To each £e{0,1} < W1, we assign a pair of functions
64
/ / Iterated forcing
ft and gt so that \imk(ft(k) — gt(k)) = oo, as follows (by induction on the length of t):firstlet g0
If r has limit length, we apply Lemma 6.7 to find / , and gt so that for every sc=t, gs
Definition (F, G) is a gap if there is no hecoco such that
F>h>G.
(F, G) is a strong gap if (i) for every a, /a(/c) > #a(/c) for all ICECO (ii) for all a ^ /?, either /a(/c) ^ ^(/c) for some /c, or fp(k) ^ ga(/c) for some k. 6.11
Lemma
A strong gap is a gap. Proof Assume that F>h> G. There is an uncountable We:co1 and some fixed k0 such that V/c^/c0, VaeW; fa(k)> h(k)> g^k). Furthermore, we may assume that for all ae W, the/ a 's have the same initial segment fa \k0, and similarly the ga \k0 are all the same. Now (i) and (ii) lead to a contradiction: let a ^ /? be in W. By (i) we have /a(/c) > #a(/c) = gp(k) for all /c < /c0, and for /c ^ k0 we have /a(/c) > h(k) > gp(k). Similarly, fp(k) > ga(k) for all /c; a contradiction. • Being a gap is of course not an absolute property for models of set theory, since (F, G) may be a gap in a smaller model, but not be a gap in a larger model. Being a strong gap, however, is an absolute property, as long as Kx remains Kx. This fact is used in the proof of Lemma 6.6 below.
6 Kaplansky's
conjecture
65
6.12
Lemma Let (F, G) be a gap. There is a c.c.c. notion of forcing such that in the generic extension there is a strong gap (F, G) such that for every a < co1 there is k0 so that L(k)=fM
and
ga(k) = gM
for all
/c^/c 0
(6.13)
Proo/ A condition is a finite set Eczco1 and functions J a , #a, ae£, that satisfy (6.13), and also satisfy (i) and (ii) from the definition of strong gap. As long as P satisfies the c.c.c, Kx is preserved and so a generic filter on P clearly yields a strong gap that satisfies (6.13). In order to show that P satisfies the c.c.c, let W a P be uncountable. W contains an uncountable A-system, with root A c c o t ; i.e. conditions p^ £
g* = min{#a: c
Claim There are £ ^ 77 so that /*(/c) ^ g*(k) for some /c. Proof Otherwise V£, 77 Vfc f*(k)>g*(k), and then it is easy to see that the function h defined by h(k) = sup {g%(k):£,
66
/ / Iterated forcing
dense in {0,1}" 1 and has size Kx. There are X2 names FeVp for an order-preserving map of D into CD™ in Vp; let Fa, a
Claim There is a b such that (F, G) is a gap. Proof Otherwise, for every b there is hb = hecoO} such that F>h> G. If £?! < b 2 then bl
Countable support iteration
Iteration of forcing with finite support was introduced in Chapter 2. We shall now give a general definition of iteration. We closely follow the notation of Chapter 2. 7.1
Definition Let a ^ 1. A forcing notion Pa is an iteration (of length a) if it is a set of a-sequences with the following properties: (i) If a = 1 then for some forcing notion Qo, Pl is the set of all 1-sequences
iff iff
p\pePp and \[Jp{p)eQfi p\P^fiq\P and p \ p \[jp(p) ^ q(p)
(iii) If a is a limit ordinal, then Vj8 < a, Pp = Pa \fi = {p \p:pePa} is an iteration (of length j8), and {a) \ePa (where l(^) = 1 for all £ < a) {b) if peP a , P < a and qePp is such that q^pp\P, then reP a , where for
Ip(^)
for
^<^ i3^
And p^a^ 7.2
iff
Vj?
Lemma If Pa is an iteration and Pp = Pa\P, then F p ^c vp*.
Proof As in Lemma 2.4 (using (iii)(«) and (fe)).
•
A finite support iteration is a special case of iteration of forcing. An iteration depends in general not only on the forcing notions Qp, but also on what happens at the limit stages of the iteration. Let a be a limit ordinal, and let Pa be an iteration of length a. Pa is a 67
68
/ / Iterated forcing
direct limit if, for every a-sequence p, pePa
->
3jS
(p \0ePp and V ^ p p(i) = 1)
(7.3)
P a is an inverse limit if, for every a-sequence p, <-> VjS
pf/JeP^
(7.4)
Most common iterations of forcing combine direct are inverse limits. Finite support iterations are exactly those that use only direct limits at all limit ordinals. If peP a , let supp(p) = {j8
(7.5)
be the support of p. If / is an ideal on a, then an iteration with I-support is an iteration P a that for each limit y ^ a satisfies that for every ysequence p, <-• V/?
p\PePp
and
supp(p)e/
Of particular interest are countable support iterations. 7.6
Definition Pa is a countable support iteration if for all limit ordinals y ^ a, /?ePy
iff
V jS < a
p\(iePp and for all but countably many
Equivalently, it is an iteration such that for all limit y ^ a, (a) if cf y = co then Py is an inverse limit, and (b) if cf y > co then Py is a direct limit. A countable support iteration is determined by the sequence {Qp}p
Lemma Let P and P' be countable support iterations of length a, of {Qp}p and {Q'p}p, respectively. Assume that for every /? < a, if B(Pp) = B(P'P) then \hpB(Qp) = B(Q'p). Then B(P) = B(F).
Proo/ By induction on a; assume that B(Pp) = B{P'p) for all /? < a. If a is a successor a = 0 + 1 then B(Pa) = B(Pp*Qp) = B(P'p*Qfp) = B(Pi). So let a be a limit ordinal. Without loss of generality we assume that Q'p = Bp = {B{Qp))yP\ for all p < a. We show that P a is dense in P'a. Let preP'a. We construct peP a so that
7 Countable support iteration
69
for all p < a, p \P ^ppr \p. Let p < a and assume that we have constructed
p\P^p'\p.
If p'\P\=pW=l
let p(P)=l. Otherwise, p\P\=fi{3qe&fi)
q ^ p\p\ hence 3 q such that p \ p \=fiq ^ p'(jB); let p(/?) = 4. Clearly, peP a is as desired. • We now turn our attention to preservation of cardinals under iterations. The following easy lemma shows that a countable support iteration of countably closed notions of forcing is countably closed: 7.8
Lemma Let K be a regular cardinal, and let / be an ideal on a limit ordinal a, closed under unions of K sets. Let P a be an iteration with /-support. If for each p < a, \\jQp is fc-closed, then P a is /c-closed. Proof Let {p^ < X] be a decreasing sequence in P a of length X ^ K. We find a lower bound p =
Theorem Let K be a regular uncountable cardinal. Let Pa be an iteration such that each Pp = P(X\P, /? < a, has the /c-chain condition, and that Pa is a direct limit. Furthermore, assume that if cf a = K then for a stationary set of P < a, P^ is a direct limit. Then P a has the /c-chain condition. Proof The proof is exactly like the proof of Theorem 2.7. Since P a is a direct limit, supp(p) is not cofinal in a, for any pePa. The proof of Case I (cf a ^ K) holds verbatim; in Case II (cf a = K), we restrict ourselves to a stationary set SOCZK such that for every £ < /c, P a is a direct limit (and so is bounded below a^). • 7.10
Corollary Let K be a regular uncountable cardinal, K ^ K2. Let P be a countable support iteration of length K such that for all P
70
/ / Iterated forcing
In practice, a countable support iteration is used either for K = K2, or so that K becomes K2 (and cardinals between X2 and K are collapsed). The main problem is then how to preserve K^ We shall now prove the Factor Lemma for countable support iteration. The lemma states that under certain reasonable assumptions, a countable support iteration Pa+n (of {Qp}p) is equivalent to Pp*P(*\ where Pj° is (in Fp«) the countable support iteration of {Qp}p><xThis formulation is not quite accurate. The name Qa+P is an element of Vp*+f>, while P(*} is claimed to be inside Fp«, an iteration of some Q{p] that are, technically, Boolean-valued names considered in VP<1. However, the factor lemma establishes (for every rj) an isomorphism between Vp*+r> and the Boolean-valued model Vpi^ defined in Vp% and so we can identify Q*+PeVP*+p and the Kp«-name for QfeVpT. We remark that a similar factor lemma holds for many other kinds of iterated forcing. For a < jS, we let [a, P) be the interval {£: a ^ £ < p). Let P*fi = Pfi\l*9P)={p\l*,P):pePfi}
(7.11)
In Vp\ let <: be the partial ordering of PaP defined as follows: let G be the generic ultrafilter on P a . p*kq
iff for some re G, fp^fq
(7.12)
It is easy to see that Pp is a dense subset of P a *(Pa/S , <:); thus (Pa/5, <:) is essentially Pp:Pa. The Factor Lemma Let Pa+n be a countable support iteration of {Q^}^<(X+r Assume that the model Vp
7.13
Corollary B(Pa+ri) = B(Pa*P(^). In other words, a countable support iteration Pa+rj of length a 4- rj is equivalent to a two-step iteration, of a countable support iteration P a , followed by a countable support iteration in
Py
\
Proo/ We work in K[G], where G is a generic filter on P a . We show, by induction on >/, that P^0L+n is isomorphic to Pj,°°. If ^7 = 1, then P a , a + 1 is isomorphic to Qa, which is isomorphic to P{'\ In general, we assume that for all P
7 Countable support iteration
71
and so P j 0 is defined. If rj is a successor ordinal, rj = /? + 1, then it is easy to show (in K[G]) that Pa,a + p+i is isomorphic to Pa.a + p*Qp- Thus, let rj be a limit ordinal. We find an isomorphism between P™ and Pa,a+ri by assigning to each qeP^ the appropriate corresponding element of Paa+r This correspondence is order preserving, and is onto Pa,a+r Let qeP{f\ and let qeVp* be a name for g. Now we work in V. For each P
Theorem (Shelah) There is a generic model in which 2xo = K2 and there are no p-points.
Shelah's proof uses a countable support iteration of length co2. The basic idea is that if U is p-point, if P(U) is the corresponding GrigoriefPs forcing, and if Q = Qv = P(L/)t0 is the (full) product of co copies of P{V\ then in VQ9 U cannot be extended to a p-point. The iteration uses {Qp}a<(O2 so that \\-pQp = Qu f° r s o m e p-point U, and so that all p-points in all Vp%(x
72
/ / Iterated forcing
7.15
Iteration with Easton support This is a type of iteration useful when dealing with large cardinals. Every condition has support S — supp(p) with the property | S n y | < y for every regular cardinal y
A typical application is Silver's consistency proof of 2K > K+ for a measurable cardinal K (using a supercompact cardinal in the ground model). References For more on Easton support iteration, and iterations in general see Menas, T. (1976). Trans. Am. Math. Soc. 223, 61-91. Baumgartner, J. (1983). In Surveys in Set Theory (Mathias, A.R.D., ed.), pp. 1-59. Cambridge University Press.
8
Borel's conjecture
One application of iteration of forcing with countable support is Laver's proof of consistency of Borel's conjecture. Let X be a subset of [0,1]. X has strong measure zero if for every sequence {zn}n<(O of positive reals there exists a sequence {/„}„ of intervals with length (/„) ^ sn such that X c {J?=0In. Clearly, every countable X has strong measure zero, and every strong measure zero has Lebesque measure 0. 8.1
BorePs conjecture All strong measure zero sets are countable.
Borel's conjecture is unprovable in ZFC, since counterexamples exist under the assumption of 2N° = K1. Laver proved its consistency by constructing a generic model in which Borel's conjecture holds and 2X° = K2. The model is obtained by the countable support iteration of length a>2 of {Qp}p, where each Qp is the Laver forcing (Part I, Section 3.16). Below we give a sketch of the proof. The central idea of the proof is the following. 8.2
Lemma Let P be the Laver forcing and let G be generic on P. Every set X of reals in the ground model that has strong zero in V[G] is countable.
In fact, we will show this: Let/eF[G] be the Laver real, and let en = 1//(«), for all n. If XeV is uncountable then, for some n 0 , the sequence {en}n>n is a counterexample to X having strong measure zero (i.e. X cannot be covered by {IH}H>nQ of lengths ej. Toward the proof of Lemma 8.2, we start with a property of Laver forcing: 8.3
Lemma Let p\\-
3i^k
such that
q\\-(Pi
(*)
Proof We use a fusion argument. Let us recall that for tep, S(t) = {a:taep} 73
74
/ / Iterated forcing
Let s = stem(p). Assume that the lemma fails. Then there are only finitely many aeS(s) such that 3q ^ o p \(sa) with property (*). By removing those finitely many nodes (and the nodes above them), we get px ^ op. For every saepl there are only finitely many beS(sa) such that 3q ^ opx \(sab) with property (*). By removing all such b we obtain p 2 ^ iP- Continuing in this way we get a fusion sequence p ^ 0 P i ^ 1P2 ^2'*•» a n d let r = P| n < w p n . By the construction of the pn's, if ter then there is no q ^ o r f t with property (*). But then no q^r forces(3i^k) cph a contradiction. • We need two more lemmas before we prove Lemma 8.2: 8.4
Lemma Let p be a condition with stem s and let x be a name for a real (in [0,1])- Then there is a condition q^op and a real u such that for every e > 0, for all but finitely many aeS(s), q\(sa)\H*-u\<s Proof Let {£„}„ be an enumeration of {sa':aeS(s)}. For each n we pick, by Lemma 8.3, a condition qn ^ o p f£n and an interval J n = [m/n, (m 4- l)/w] so that qn\\-xeJn. There is a sequence {kn}n<0> so that the Jkn form a decreasing sequence converging to a unique real u. Let q = []™=oqknl it is easy to see that q works. • 8.5
Lemma Let p be a condition with stem s and let {*„}„ be a sequence of names for reals. Then there exists a condition q ^ o p, and a set of reals such that for every s > 0 and every teq, Q s , for all but {ut:teq,t^s} finitely many aeS(t\ q\(ta)\h\xk~ut\<e where k = length (t) — length(s). Proof Using a fusion argument, by a repeated application of Lemma 8.4. First we get p x ^ o p and us. Then for every immediate successor t of s in Pi we get qt ^ o p x ft, and ut, and let p 2 = (J r q r . And so on; we get a fusion • sequence p ^ o P i ^ i P 2 ^ 2 > - - - , and let q = f]n<(Opn. Proof of Lemma 8.2 Let X e F be a subset of [0,1], and let peP be such that p\\-X has strong measure zero. We show that X is countable. Let s be the stem of p,\s\ = n. Let G be generic on P, and let / be the Laver real given by G.
8 Borel's conjecture
75
Consider the sequence {ek}k>n defined by % ~
(*>»)
There exists a sequence of intervals / k ,fc^n, of length ek, so that X c (J^°=w/fc. For each fc ^ w, let xk be the center of Ik. Let q ^ o p be a condition obtained by Lemma 8.5 (applied to p and {x k }^ n ); let {ut:teq} be the reals from Lemma 8.5. We shall show that
{} Let v${ut:teq} and show that v$X. Since p||— X c (J^ n / f c , it suffices to find some r ^ g such that r||—t;£/k, for all k^n. We construct r ^ g by induction on the levels of q\ at stage k^n we guarantee that r||— t?£/k. For example, the first step is as follows: let s = (l/2)-\v — us\. For all but finitely many aeS(s\ q\(sa)\\-\xn — us\<e. Also, for each a, q \(sa) \\-f(n) = a, and so q \(sa) |— en = I/a; thus, for all but finitely many a, q\(sa)\\-\xn — v\>en, in other words v$In. Thus, by removing finitely many immediate successors of s, we ensure that r\\-v$In. We continue in this way to get r^q such that r\\-v$\Jk^Jk. • 8.6
Theorem (Laver) Assume the GCH. Let P be the countable support iteration of {Qp} such that for each /? < co2, \\- pQp is the Laver forcing. In the generic extension K[G], all strong measure zero sets are countable. First we wish to show that the iteration preserves cardinals. Preservation of Kx is crucial: the forcing is neither co-closed, nor does it have the c.c.c. We state (without proof) the key lemma: 8.7
Lemma For every countable set XeV[G] of ordinals there is Ye V, countable in F, such that X c Y.
We remark that Lemma 8.7 follows from the more general property of countable support iteration of proper forcing, which will be proved in Part III, Chapter 5. Preservation of cardinals X2 and up follows from Corollary 7.10; but for what we need: 8.8
Lemma Assume the GCH. For every a < co2,p \<x has a dense subset of size Kx, and Again, we omit the proof and remark that the lemma holds for countable
76
/ / Iterated forcing
support iteration of proper forcings, provided that ||- p\Qp\ = 2*° for every
8.9
Corollary P has the K2-chain condition.
Proof Corollary 7.10.
•
8.10
Corollary F[G] satisfies 2*° = N 2 . Proof Because of the chain condition, every real in F[G] is in F [ G J for some a < co2- Hence, 2X° ^ N2 by Lemma 8.8. On the other hand, each Ga produces a Laver real, hence 2K° ^ K2. • 8.11
Lemma Every set of reals of size Xj in F[G] is in F [ G J for some a < co2.
Proof Follows from the chain condition.
•
We shall now sketch a proof of Theorem 8.6. We have to show that no uncountable set of reals in K[G] has strong measure zero. It suffices to show that if | X\ = X t then X does not have strong measure zero. Thus, let X e F [ G ] be a set of reals of size Kx . By Lemma 8.11, Xe K[GJ for some a < co2. Now it follows from Lemma 8.2 that in K[G a + 1 ] there is a sequence {en}n witnessing that X does not have strong measure zero. But that only means that X cannot be covered by intervals of length en in K[G a + 1 ]; we need, however, that X does not have strong measure zero in F[G]. The Factor Lemma 7.13 reduces the problem. By the lemma, the iteration P is equivalent to P a *(2, where Q is, in Vp*. the countable support iteration of Laver forcings of length co2. Thus, the following lemma completes the proof: 8.12
Lemma If X is an uncountable set of reals in V, then X does not have strong measure zero in K[G]. In fact, if/ is a Laver real over F, and en = l//(n) (n < co), then for some n0, X cannot be covered in F[G] by intervals {In}n^n of length en. We omit
8 Borel's conjecture
11
the proof of Lemma 8.12; it is not unlike the proof of Lemma 8.2, but its fusion arguments are more delicate, as it involves conditions peP rather than just Laver trees. References Laver, R. (1976). Acta. Math. 137, 151-69. Borel, E. (1919). Bull Soc. Math. France 47, 97-125.
PART III
Proper forcing
1
Stationary sets
We consider closed unbounded and stationary sets of countable ordinals. We assume that the reader is familiar with closed unbounded and stationary subsets of a regular uncountable cardinal K. With the exception of Chapter 7, we only consider stationary subsets of K for K = X x; later in this chapter we generalize these notions. A closed unbounded subset of a>x (a club) is an unbounded set C^col such that sup(Cna)eC for every a < co1. A set S c co1 is stationary if SnC ^ 0 for every club C. For any set A,A<(O denotes the set of all finite subsets of A. If F is a function on A<(O, we say that a nonempty X^A is c/osed under F if
1.1
Proposition (a) if F i c D i ^ - x o ! , then C F = {A < cox :k is closed under F}
is a club. (b) For every club C there is a function F:CO^0)-^(JJ1 such that C F ^ C . In fact, CF = C, the set of all limit points of C. Proo/ (a) Easy. (b) Let F(e) = least oceC greater than max(e). Note that every XeCF is a limit ordinal. •
Preservation of stationary sets If F[G] is a generic extension, then every club CeV remains a club in F[G], provided that Kx is preserved. But a stationary set S in F may no longer be stationary in F[G], as there may be a club CeV[G] disjoint from S. 1.2
Example Shooting a club through a given stationary set. For any stationary set S there is a forcing notion Ps that generically
80
1 Stationary sets
81
adds a club C so that C c S , and preserves X x. If S is costationary, i.e. if — S is also stationary, then — 5 is destroyed by Ps as it is disjoint from C in K[G]. Let S be a stationary set. Ps = the set of all (bounded) closed sets of ordinals p so that p c S p ^ g if /? is an end-extension of g (if q = p n a for some a) (1.3) Note that since p is closed, max (p) exists (a countable ordinal). If G is a generic filter, let
Clearly, C is a subset of S, and because, for every a
Lemma Ps is co-distributive.
Proof We prove that every countable set of ordinals in V[G~\ is in the ground model. Thus, let
we shall find a q ^ p and some / so that q\\-f = f. By induction on a we construct a chain {/la}a of countable subsets of Ps: Let ^ 0 = {p}, and ^ a = (J/?y a . Then we let The sequence {ya}a is increasing and continuous. Let C = {A: if a < X then ya < A} As C is a club and S is stationary, there is a limit ordinal X so that AeCnS. Let {%„}„ be an increasing sequence with limit X; then limnyan = X as well. There is a sequence {/?„}„ of conditions so that p0 = p, and for every n, pH+leAan+r pn+l^pn, and pn+l decides f(n). Since yan < max(p n + 1 )^y a , we have limn max (/?„) = A, and because the closed set
82
/ / / Proper forcing
is a condition. Since q ^ pn for each n, q decides each/(n), and so there is some / such that q\\-/ = / . • The following theorem gives sufficient conditions for a notion of forcing to preserve stationary sets: 1.5
Theorem (a) If a notion of forcing P satisfies the countable chain condition, then every club Ce K[G] has a club subset D such that De V. Consequently, every stationary set SeV remains stationary in F[G]. (b) If P is co-closed then every stationary set SeV remains stationary in F[G]. Proof (a) Let p\\-C is a club and let D = {<x:p\\-(xeC}. Clearly, p||—Z> c C; we prove that D is a club. It is not difficult to see that D is closed; we shall use the c.c.c. to show that D is unbounded. Let a 0 < co! and let us find a > a 0 so that p\\-aeC. There is a name y for an ordinal so that p\\-oc0
Similarly, we find a sequence OL1 < a 2 < ••• so that, for every n, p|H3yeCX
(b) Let S be stationary, and let p\\-C is a club We shall find a q^p so that <j||- S n C / 0 . We construct an increasing continuous ordinal sequence {ya}a and a decreasing sequence {pa}a of conditions as follows. Let Po^P and y0 be so that po\\-yoeC. Given pa and ya, we let
1 Stationary sets
83
p a +! ^ pa and ya +i > ya be so that pa+1 \\-ya+ xeC. When a is a limit, let pa be a lower bound of {pp}p
Closed unbounded and stationary subsets of \_AJ° Let A be an uncountable set. By
\_AT we denote the set all (at most) countable subsets of A. 1.6
Definition A set C <= [/4]40 is unbounded if for every xe[,4] w there is yeC such that x^y. C is closed if for every chain
in C, the union U^°=0^n is i n C. C is closed unbounded (club) in [/4]w if it is closed and unbounded. A set S c [yl]60 is stationary (in [/I]"0) if S n C ^ 0 for every club C in [,4] w. If ^4 is a set of cardinality Xx then as the following easy fact shows, the concept of club and stationary coincides essentially with the usual concept of club and stationary (note that cox a [co1']
Proposition The set co1 is a club in [cOi]". Moreover, a set C c OJX is a club in [cOi]*0 iff it is a club subset of co1. A set S ^ [ a ^ ] " is stationary iff S n co j is a stationary subset of a^; in particular, every stationary subset of co1 is stationary in [c^] 6 0 . D The closed unbounded subsets of [/I] w generate a normal countably closed filter: 1.8
Theorem (a) If {Cn}n are clubs then f)?=0Cn is a club. (b) The diagonal intersection
xef) flex
of clubs {Ca}aeA is a club.
84
/ / / Proper forcing
(c) Every choice function
f(x)ex on a stationary set S is constant on a stationary subset T^S, is aeA such that
i.e. there
f(x) = a for all xeT. Proof (a) First we show that the intersection of two clubs is a club. Let C = 0^(^02, where Cx and C2 are clubs. C is clearly closed. In order to show that C is unbounded, let xetyl] 00. We consider a chain x ^ x 0 ^ yo^xl^y1^ •••, where each xn is in C{ and each yn is in C2. Then y = fU%*i. = H ^ o ^ is in d n C 2 . In order to prove (a) it suffices now to consider a descending sequence C Q ^ C J ^ - ^ C ^ - of clubs. Their intersection C is certainly closed. xo^xl^ To show that C is unbounded, let xe[/4] w . We consider a chain x2 ^ •••, where each xn is a member of Cn. Then, y= \J™=oxn belongs to each Ck,k
iff
(Vaex)xeCa
First we show that C is closed. Let x 0 c xx c ... c xn c ... be a chain in C, and let x = (J^°=0^n- To show that xeC, let aex and let us show that xeCa. There is /c such that aexn for all n^k; hence, xneCa for all n ^ /c, and so xeC a . Now we show that C is unbounded. Let x o e[/l] a \ By induction, let x 0 c x x c ... ^ x n c ... be as follows: given xn, for every aexn choose and let xn + 1 = \J{z(n9a):aexn}. Let z(n,a)eCa such that xn^z{n,a\ y = lj*=o^n , and let us show that yeC. Thus, let aey and we show that yeCa. The sequence {z(n,a):n < co} is a chain in Ca, because
and so y = [j^=oz(n,a) belongs to Ca. (c) Let 5 be stationary, and let / be a choice function on S. In order to show that / is constant on a stationary subset of 5, let us assume that, on the contrary, for every aeA there is a club Ca disjoint from S such that /(x) ^ a on Ca. Let C be the diagonal intersection of {Ca:aeA}. Since S is stationary, and C is a club, there is some xeS such
1 Stationary sets
85
that xeCa for all aex. Hence,/(x) ^ a for all aex, contrary to the assumption that/(x)ex. • By the following lemma, the condition of closure under union of chains in Definition 1.6 can be replaced by unions of directed sets. A set D c \_Ay° is directed if for every x,yeD there is a zeD with 1.9
Lemma Let C^[y4] w . If C is closed then for every countable directed
D<=C, [jDeC. Proof Let D = {*„}„; we construct a chain {yn}n in C with the property that \J™=oyn = (J/X Let y0 = x 0 , and for each n let y n + 1 = xk, where /c • is the least k such that xk^ynvxn+l. We now characterize closed unbounded and stationary sets in terms of functions F:A
Proposition For every F:/l < t 0 ->[y4] w , the set CF = {xe[AY'{Veex<(O)F{e)
c x}
is a club. Proof Easy.
•
1.11
Theorem For every club C in [ 4 ] w there is a function F: / I < w -> ^1 such that C contains the club CF. Corollary A set S is stationary in [/I]™ if and only if for every function there is xeS closed under F. F:A
86
/ / / Proper forcing
As each G(e) is countable, we enumerate it by co, and obtain countably many functions Gk: A <}. Let us fix a pairing function n-+(kn,mn) and define F:A<
(1.12)
For every y<= [5] w , let e7}
(1.13)
1.14
Theorem Let i c f i . (a) If C is a club in [^] w then C* is a club in [B]". If 5 is stationary in [ £ ] " then 5 \A is stationary in •[i4]a>. (b) If C is a club in [ 5 ] " then C \A contains a club in [A]". If 5 is stationary in \_A~]
D = {xe[AY:(Veex<0>)G(e)nA c x } The set D is clearly a club. For every xeD, let y be the closure of x under • F. We have yeCG and }/n^ = x. Hence, xeCG \A and s o D ^ C f A The following is a useful proposition. We recall that if M is a model (for a given first order language) then the set of all elementary submodels of Jt is closed under unions of chains. Thus:
1 Stationary sets
87
1.15
Proposition Let Ji = < M,... > be an uncountable model. The set of all countable elementary submodels of Jt is a club in [M]*0. •
Preservation of stationary sets We now re-examine Theorem 1.5. 1.16
Theorem (a) If a notion of forcing P satisfies the countable chain condition, then for every uncountable A, every club CeV[G~] in \_Ay° has a club subset D such that DeV. Consequently, every stationary set S e F in [/4]w is a stationary set in \_A~y° in V[G~\. (b) If P is co-closed then every stationary Se V in \_AY remains stationary in F[G]. (Note that \_AY is possibly a larger set in F[G] then [A] w in V) Proof (a) Let p||—C is a club There is a name F for a function: A<
such that
Let G\A<
(1.17)
We have p\\-F(e)eG(e). Thus, if x is such that Veex<(OG(e)^x, then p\\-P(e)ex; therefore p\\-CG £ Q , and (1.17) follows. (b) Let S be a stationary set in [A]00. In order to show that 5 remains stationary in F[G], let
It suffices to find a condition q^p and some xeS such that q\hF(x<(a)^x Consider the model
(1.18)
88
/ / / Proper forcing
where K is a sufficiently large cardinal and A^VK. Let C be the set of all countable elementary submodels in Jt\ C is a club in [FJ W . By Theorem 1.14, there is N<J{ such that N n ^ e S . We claim that x = Nc\A satisfies (1.18). Enumerate x
/>„ Ih / K ) = an Such #„ exists in N because N is an elementary submodel of Jt. Then let q be a lower bound for [pn}n. Now (1.18) follows. • References Jech, T. (1973). Ann. Math. Logic 5, 165-98. Kueker, D. (1977). Ann. Math. Logic 11, 57-103. Shelah, S. (1982). Proper Forcing. Lecture Notes in Mathematics 940, Springer-Verlag, NY.
2
Infinite games on complete Boolean algebras
In this chapter we consider several properties of complete Boolean algebras, and of the corresponding forcing extensions, formulated in terms of infinite games. Throughout the chapter, let B be a complete Boolean algebra, let P be a (separative) partial ordering, dense in B, and Vp = VB (or K[G]) be the corresponding Boolean-valued (or generic) model. First we consider the descending chain game g0. Two players, I and II, choose successively the members of a descending chain a^b^a^b^-'-^a^b^'"
(2.1)
of nonzero elements of B. I chooses the an's and II chooses the fcn's. I wins the game (2.1) if and only if the sequence converges to zero (and II wins iff the sequence (2.1) has a nonzero lower bound). We shall concern ourselves with the existence of winning strategies in ^ 0 and in the other games defined below. It is easy to see that if the game ^ 0 is played on P instead of on B (i.e. the chain (2.1) is in F, and I wins if the chain has no lower bound) then I (or II) has a winning strategy in the game played on P if and only if he has a winning strategy in the game played on B. The same is true for the other games to be considered. 2.2
Proposition I has a winning strategy in ^ 0 if and only if B is not co-distributive.
Proof Assume that I has a winning strategy a. Let a0 =
90
/ / / Proper forcing
It is clear that if P is co-closed then II has a winning strategy in the game ^ 0 . By a theorem of Foreman, the converse is true if | P | < Kx. (Precisely: if II has a winning strategy and \P\ ^ K x then P has a enclosed dense subset.) It is an open problem whether the converse is true in general. One consequence of Foreman's result is the following equivalent of the existence of a winning strategy for II ([Jech, 1984], Addenda): 2.3
Proposition II has a winning strategy in &0 on P if and only if there is a Q such that P x Q has a dense co-closed subset. •
There is a B such that neither player has a winning strategy in # 0 . An example is the forcing notion P s from Example 1.2 (if S is costationary). Since Ps is co-distributive, I does not win. We omit the proof that II does not win, since we show in Chapter 3 that II does not win in the game ^ on Ps which is easier for II than # 0 . (All the games considered here are undetermined in general.) Next we consider the cut and choose game <&v Player I begins by selecting some peB and a partition Wo of p. II responds by choosing some uoeWo. At the nth move, I plays a partition Wn of p and II chooses uneWn: Wo,uO9W1,ul9...,WmuH9...
(2.4)
II wins the game ^ if and only if the sequence {un}n has a lower bound, i.e. iff Iq^p
Vn q^un
(2.5)
2.6
Proposition (Jech and Velickovic) The game ^ 0 and <^1 are equivalent, i.e. I (II) has a winning strategy in one iff I (II) has a winning strategy in the other.
Proof It is not difficult to prove that I has a winning strategy in <& x if and only if B is not co-distributive. Thus it suffices to show that II has a winning strategy in ^ 0 iff II has a winning strategy in ^v Assume that II has a winning strategy o in ^ 0 . We shall describe a strategy T for II in ^ Playing ^ l 9 player I chooses P and plays Wo. Let ao = p and apply a as if I plays a0 in ^ 0 ; i.e. let b0 = a(a0). There is uoeWo such that uo-bo^0. Let T(WO) = UO. When I plays # l 9 let a 1 =w 0 -b 0 , let b x = (7(00,60,0!), and let T(W^ O>M O»^I) b e s o m e
2 Infinite games on complete Boolean algebras
91
so that u1-bl^0. In general, let x(WO9uO9...9Wn) = uneWn be so that The strategy T is a un-bn^0, where bn = a(aO9bO9...9an)9 a^u^^b^^ winning strategy for II in &x. Conversely, assume that II has a winning strategy T in ^ . We shall describe a strategy a for II in ^ 0 . Playing ^ 0 , I chooses some a0. We observe that there is b0 ^ a0 with the property that for all u^b0 3 W partition of a0 such that u = r(a0; W)
(2.7)
(Otherwise, the set of all u ^ p for which (2.7) fails is dense below /?, and so there is a partition Z of p such that (2.7) fails for each weZ; a contradiction.) Thus, we (player II) choose such b0, and let a(ao) = bo. When I plays ax ^ b 0 , let Wo be a partition of a0 such that ax = x{aQ\ Wo). By a similar argument, there is bx ^ a1 with the property that for all u ^ b l 5 3 W partition of ax such that u = r(a 0 ; Wo, ^ ) We let cr(a0, ax) = b1. In general, at move n, we let a(a0,..., an) be some bn so that V u ^ f r ^ W partition of an with u = T(W0,...9Wn-l9W)9 and when I plays an+19 we find Wn so that an + x = r(W0,..., Wn). The strategy cr is a winning strategy for II in ^ 0 . D There is another version of the game &l9 involving the Boolean-valued model (the forcing version of ^ J as follows. Player I starts the game by selecting a condition /?, and chooses a name a 0 for an ordinal. Player II responds by choosing an ordinal j80. At the nth move, I plays a name for an ordinal, an, and then II plays an ordinal Pl&O, Po,&l, Pi,-.-,&n>Pn>-II wins the game if and only if
3q^p
Vn q\\-&n = Pn
Since ordinal names correspond to partitions, namely W = {/?• [a = 7]:y an ordinal} this game is just a reformulation of ^ . Next we consider the countable choice game ^ 0) . Player I begins by selecting a condition P, and chooses an ordinal name a 0 . Player II chooses a countable set of ordinals Bo. At the nth move, I plays an ordinal name an, and II plays a countable set Bn: p;&O9BO9&l9Bl9...9&n9BH>...
(2.10)
92
/ / / Proper forcing
II wins the game ^
Iq^p
if and only if
Vn qh&neBH
(2.11)
Clearly, ^ is an easier game to play for player II than yv Thus, if II has a winning strategy in <S1 then II has a winning strategy in ^ w , and, similarly, if I does not have a winning strategy in C§1 (if B is co-distributive) then I does not have a winning strategy in ^co . There are many examples of forcing notions, for which II wins the game ^ w ; this will be discussed in Chapter 4. A sufficient condition for I to win ^ w is when P collapses col9 or more generally when P adds a countable set A of ordinals that is not included in any countable set in the ground model. If that is the case, let p be a condition that forces it, and let {d0, a l 5 ..., a n ,...} be a sequence of names such that p\\-A
= {dn}n cannot be covered by a set countable in V (2.12)
Then I has a winning strategy in &„, playing the dn. The game ^ has a version formulated in terms of partitions: I selects p and at move n plays a partition Wn of p; II plays a countable subset Bn
o(Wn: p;W0,B0,...,Wn,Bn,...
(2.13)
II wins if and only if
ni(«:"4}^0
(2.14)
n= 0
(or, equivalently, 3q^p
Vrc Bn is predense below q)
Finally, we consider the proper game <& (due to Charles Gray). Player I begins by selecting a condition p, and chooses an ordinal name a0. Player II chooses an ordinal /?0- At the nth move, I plays an ordinal name dn, and II plays an ordinal /?„: j r , * o , / * o , <*i, 0 i , " • , < * „ , & , • • •
(2-15)
II wins the game <§ if and only if
3q^p
Vn q\{-3k
zn = pk
(2.16)
An equivalent version of the proper game is when I plays ordinal names aB and II plays countable sets Bn, and II wins just in case 3
Vn <jf|h3fc
aneBk
(2.17)
2 Infinite games on complete Boolean algebras
93
To see that the two versions are equivalent, note this: if II has a winning strategy in (2.17), then because B= {J?=0Bk *s countable, it is not hard to construct the set B as {Pn'.n < o} in co moves. Thus, II has a winning strategy in (2.16). Similarly, if II can win against any strategy for I in (2.17), II can accomplish the same in (2.16). Considering the version (2.17) it is easy to see that the game ^ is easier for II than the game ^ £0. Thus we have: II has a winning strategy in <^1 =>II has a winning strategy (2.18) in ^ W =>II has a winning strategy in ^ and similar implications for 'I does not have a winning strategy'. Note that if a forcing notion P has property (2.12) then I has a winning strategy in ^ as well. Consequently, if II has a winning strategy in the proper game then every countable set of ordinals in K[G] is included in a set in V that is countable in V, and hence Xx is preserved. In Chapter 3 we show that if P destroys a stationary set then II does not have a winning strategy in the proper game, and so Example 1.2 gives an example of an undetermined proper game. The proper game ^w also has a version formulated in terms of partitions: I selects p and at move n plays a partition Wn of p; II responds by choosing countable sets Bn0 c w0, B\ c wl9..., Bnn c Wn. II wins if and only if 00
3q^p
Vn
U flj is predense below q
(2.19)
k = n
We leave the proof of equivalence to the reader. We conclude this chapter by giving a (typical) example of a forcing notion for which II has a winning strategy in ^co . 2.20
Example II wins ^ w for the Sacks forcing. Let (P, < ) be the Sacks forcing, (Part I, Section 3.4). The proof that II has a winning strategy in ^ w is exactly like the proof of Lemma 3.5, Part I: let I play p and then ordinal names 6in. As in Lemma 3.5, Part I, we find successively conditions pn and countable (in fact finite) sets An so that ^lPl^2'"^nPn>n+l-'-
(2.21)
and that for all n, Pn\h&neAn
(2.22)
94
/ / / Proper forcing
Since (2.21) is a fusion sequence, it has a lower bound q, and
This gives a winning strategy for player II. References Jech, T. (1984). Ann. Pure and Appi Logic 26, 11-29. Shelah, S. (1982). Proper Forcing. Lecture Notes in Mathematics 940, Springer-Verlag, NY.
3
Proper forcing
Proper forcing was introduced by S. Shelah, to describe a wide class of forcing notions that can be iterated (with countable support) without collapsing Nx. The class of proper forcing includes all co-closed forcings, all c.c.c. forcings, as well as many standard notions of forcing that adjoin generic reals. We give four equivalent definitions of proper forcing. 3.1
A forcing theoretic definition A notion of forcing (P, < ) is proper if, for every uncountable set A, every stationary subset of \_A~Y° remains stationary in the generic extension
3.2
A game theoretic definition A notion of forcing (P, < ) is proper if player II has a winning strategy in the proper game (2.15) for P. 3.3
A model theoretic definition In order to state the definition, let (P, < ) be a notion of forcing. We say that X is sufficiently large if k is a cardinal, and the power set of P is in Vx. For any sufficiently large A, we consider the model KA = (F A ,e,P,<)
(3.4)
We recall that the set of all countable elementary submodels M •< Vk is a club subset of [F A]W. (As P and < are assumed to be in the language of M the restriction of |— to any such M is definable in M.) 3.5
Lemma Let X be sufficiently large, let M-<(K A ,e,P, <) and let q be a condition in P. The following are equivalent: (i) For every predense Wa P if WeM then Wr\M is predense below q. (ii) For every ordinal name a, if aeM, then q\\-deM 95
96
/ / / Proper forcing
i.e.
(or, in detail, V r ^ g 3 s ^ r 3 o r d i n a l /JeM s||— a = /?). (iii) g |— Gr\M is an M-generic filter o n P n M We say that q is (P9M)-generic
(3.6)
if it satisfies any of the equivalent conditions in Lemma 3.5. Proof We sketch proofs of some of the implications and leave the rest as an exercise. (i)->(ii): Let a be an ordinal name, aeM. The set
W={p:3p
p\h& = P}
is dense, and WeM (because M< Vx). By (i), WnM is predense below q. If pe W n M, then 3 /? p \\- a = /?, and because M is an elementary submodel, we have 3£eM p\\-i = p. Thus
is predense below q, and (ii) follows. (i)-»(iii): Let G be a K-generic filter on P such that qeG;we show that GnM meets every WeM that is predense in PnM. The statement ¥ e M is predense i n P n M ' is easily seen to be equivalent to M \\- (Wis predense in P). Since M is an elementary submodel, W is predense in P, and by (i) WnM is predense below q. Thus, W n M n G is nonempty. • 3.7
Definition A notion of forcing (P, < ) is proper if for some sufficiently large I there is a club set C^IV^ of countable elementary submodels M<(K A ,e,P, < ) with the following property: VpeM
3g ^ p
q is (P, M)-generic
(3.8)
(Below we prove that 'for some sufficiently large X can be replaced by 'for all sufficiently large K in Definition 3.7.) 3.9
A Boolean-algebraic definition This definition of properness is a variation on distributive laws. W is a matrix of partitions of a condition p, if W = {oLap:oiJ
(3.10)
3 Proper forcing
97
such that for every a, {aocP}p<x is a partition of p (with some aaP possibly equal to 0). 3.11
Definition A complete Boolean algebra B is proper if for every p > 0, every uncountable X and every matrix of partition (3.10) there is a club C c [^] w such that for every xeC, a/?#0
3.13
(3.12)
Theorem The four definitions of properness given above are all equivalent.
Proof We start by showing that the model theoretic Definition 3.7 implies the game theoretic property 3.2. Let us consider the version (2.19) of the proper game, i.e. I selects some p and plays partitions Wn while II responds by playing countable sets B"o, B\,...,Bnn. To find a winning strategy for II, let X be sufficiently large and let C <= [J/^]60 of countable models M -< Vx that all satisfy (3.8). When I selects p and plays WOi let us choose MoeC so that peM0 and WoeMo. Let Bo = WonMo be the first move of player II. In general, when I plays Wn, we choose MneC so that Mn^.Mn_l
and WneMn. We let Bno=WonMn,
Bn1 =
W1nMn,...,Bnn=WnnMn.
Since M o c M x c ... are all in C, their union M is also in C. There is g ^ p that is (P, M)-generic. For every n, WneM, and so W^nM is predense below q. But W n n M = U » = n ( ^ n M k ) = U r = n ^ . Hence, (2.19) holds, and the strategy described above is a winning strategy for player II in the proper game. Next we show that the game theoretic property of a complete Boolean algebra B implies the distributive law (Definition 3.11). Let p > 0, and let W= {aap} be a partition matrix of p. For each a, let Wa be the partition {aap'.p < A}. Let a be a winning strategy for II in the proper game (version (2.19)). We define a function F:/l < £ 0 ->[/]". If e is a finite subset of A, let F(e) be a countable subset of k such that if <x0,..., cck is any enumeration of e, and if {^OK; a r e t n e m o v e s of player II using the strategy a against Wao,..., Wak9 then for every a^eBl 0 is in F(e). By Proposition 1.10, the set C of all X G W 0 0 such that F(e) c x for all eex
98
/ / / Proper forcing
selects p and plays {W^}. Let {££} be IPs moves using the winning strategy o. It follows from the way we defined F that, for every n,
And because a is a winning strategy, there is some q ^ p such that (2.19) holds. Such a g witnesses (3.12). Next we show that if B satisfies Definition 3.11 then every stationary set remains stationary in the generic extension by B. Let X be uncountable, and let S c [A]" be stationary. To show that S is stationary in VB, let FeVB be such that
for some condition p. It suffices to find q^p
and some x e 5 such that
Let
{^}a
The collection {aap} is a partition matrix for p, and so there is a club and let C c [X]<° such that every xeC satisfies (3.12). Let xeSnCnD, aex /tex
Now it follows that q\\-
Veex<(OF(e)ex.
Finally we show that if P preserves stationary sets, then P satisfies the model theoretic definition of properness; that for all sufficiently large X there is a club C with property (3.8). Toward a contradiction, assume that for some sufficiently large k there is a stationary set S
if
WeM
then
%eM}
Since S remains stationary in K[G], there is some MeSnC.
For every
3 Proper forcing partition W of p, if WeM then qweWnM
99 and so
By the genericity of G, we have WeM
In other words, there is qeG such that each WnM is predense below q9 that is g is (P, M)-generic. This is a contradiction since Me5. • Reference Shelah, S. (1982). Proper Forcing. Lecture Notes in Mathematics 940, Springer-Verlag, NY.
4
Examples of proper forcing
One of the definitions of properness is that the forcing preserves stationary sets. It follows therefore from Theorem 1.16 that every c.c.c. notion of forcing is proper, and so is every co-closed forcing. In fact, this is easy to see when one uses the game definition: if P is co-closed, then II has a winning strategy in the proper game (even in the game ^ 0 ) . If P is c.c.c. then every partition is at most countable, and again II has a winning strategy in the proper game (say in the version (2.19)). 4.1
Proposition If P is proper then every countable set of ordinals X in V[G] is included in a set AeV such that A is a countable set in V.
Proof We have shown this in Chapter 2. For another proof, let X a X. The set S = [A] W (defined in V) remains stationary in K[G], and therefore meets the club set (in K[G]) {Ae[XY°'.A^X}. The proposition follows. • The class of proper forcings contains many other partial orderings in addition to the co-closed and the c.c.c. ones. Of particular interest are the examples of generic reals considered in Chapter 3 of Part I. 4.2
Theorem Sacks forcing, Prikry-Silver forcing, Mathias forcing, Laver forcing and Grigorieff forcing are all proper. We omit the proof for Grigorieff reals. The other four examples can be all shown proper in a uniform way. The following property of forcing was introduced by Baumgartner: 4.3
Axiom A A notion of forcing P satisfies Axiom A if there is a collection { <„}„ of partial orderings of P such that (i) p^oq implies p^q; (ii) P^n + ii implies p^nq;
100
4 Examples of proper forcing
101
(iii) if {pn}n is a sequence such that
then there is a q such that q ^npn for all n; (iv) for every peP and for every n < <x>, if W is predense below p then there is a q^np and a countable B^W that is predense below g. Note that (iv) is equivalent to: (iv') VpeP Vn Vd 3q ^np 3B countable such that
It is easy to see that every co-closed P satisfies Axiom A (with ^ w being ^ ) and so does every c.c.c. P (with ^ n being =). As for Sacks, PrikrySilver, Mathias and Laver forcings, the respective partial orders ^ n defined in Chapter 3, Part I satisfy (i)—(iv) in the definition of Axiom A. (See Lemmas 3.5, 3.12 and 3.20 in Part I for the property (iv'). Even though we have not explicitly found a q ^np for each n, the proofs of these lemmas can be easily so modified.) Thus, it remains to prove the following: 4.4
Lemma If P satisfies Axiom A then player II has a winning strategy in the game ^ w on P and so P is proper.
Proof Let I select poeP and an ordinal name d0. By (iv') there exists a condition p1 ^opo and a countable set Bo such that po\\-oioeBo. At the nth move, when I plays dn, there exists pn + i^npn and a countable set Bn with pn \\— dneBn. By (iii) there exists a condition q stronger than all the pn. This condition q verifies that II wins. • We briefly mention an example of a notion of forcing that is proper but does not satisfy Axiom A. 4.5
Example (Baumgartner) Adding a club with finite conditions A condition peP is a finite function with dom(p) c col9 ran(p) c co1, with the property that there exists a normal (i.e. increasing and continuous) function f:o1 -^co1 such that /?<=/. A condition q is stronger than p if q 3 p. A generic filter G yields a normal function fG:o>1-^co1. The forcing notion P has the property that player II wins in the proper game 0, but I wins in the game ^ w . We sketch the proof of this. For the proper game, we first observe that the countable ordinals of
102
/ / / Proper forcing
the form a = a / (indecomposable ordinals, which form a club set Co) have the property that for every condition p a a x a, the function pu{(a,a)} is also a condition. If W is an antichain, let C(W) = [XeC0\ Va < A3jS < AVp c a x a3 c j? x j % e W and g|p)} C(W) is a club. Player II wins the proper game as follows: when I selects p0 and plays a partition Wo of p 0 , II chooses XoeC(Wo) so that poa Ao x Ao and plays 5Q = { p e W V P ^ o x Ao}. At the nth move, when I plays Wm II chooses ln>K-i in C t ^ n - n C T O and plays Bnk = {peWk: p cz Xn x Xn} for fc = 0,..., n. We leave to the reader the proof that II wins. In the game <SW player I has a winning strategy: let / be the canonical name for the generic normal function fG. The first move of player I is the ordinal name /(0). When II responds by playing a countable set Bo a co1? I chooses an indecomposable ordinal oc1 greater than B o and plays / ( a ^ . The play continues in the obvious way, and we leave it again to the reader to verify that I wins. Reference For more examples and counterexamples, see Baumgartner, J. (1984). In Handbook of Set-theoretical Topology (Kunen, K. and Vaughan, J.E., eds.), pp. 913-59, North-Holland, Amsterdam.
5
Iteration of proper forcing
In this chapter we prove that properness is preserved by countable support iteration. It follows that when adding iteratively generic reals by proper forcing (such as Sacks reals, Laver reals, etc.), Xx is preserved. 5.1
Theorem Let Py be a countable support iteration of length y, of {Qp}p, such that for every j? < 7, \\j&p is proper. Then Py is proper. We know already that the theorem is true for y = 2, because when P preserves stationary sets and \\jQ preserves stationary sets, then P*Q preserves stationary sets. We shall nevertheless give another proof of this fact, since the following proof will be generalized to iterations of arbitrary length. The proof uses the game theoretic definition of properness. 5.2
Two step iteration We assume that P is proper, and that ||- Q is proper. We shall show that player II has a winning strategy in the proper game on P * Q. Let a be a winning strategy for II on P, and let xeVp be such that |fp f is a winning strategy for II on Q. We shall describe a strategy for II on P*Q'. Player I starts the proper game by selecting a condition (p9q)eP*Q, and an ordinal name &oeVp*Q. We describe IPs response, an ordinal y0. We intend to step inside the model Vp and apply f. The P*Qname a can be thought of as (or identified with) a P-name for a g-name for an ordinal. Thus, let us argue in Vp and apply IPs strategy f when I plays, in the proper game on Q, the condition q and the ordinal g-name a 0 . Let j50 be IPs move given by f. Back in the ground model, we consider the proper game on P, and use a to respond to Ps move p, /Jo. Let y0 = cr(p, j§0). At the nth move, I plays a P*Q-name dn. In Kp, dn is a Q-name, and we apply the strategy f to find an ordinal £ln. In F, we consider $„ (a P-name) as the nth move of I in the proper game on P, and use a to return an ordinal yn. Since f is a winning strategy for II (in Vp\ we have
dtn =
fa
(5.3) 103
104
/ / / Proper forcing
Hence, there is a qr such that p \\- q' ^ q and dn = j}m
(5.4)
Since a is a winning strategy for II, there is p' ^ p such that p'|hVm3fc
/5m = yk
(5 5)
Putting (5.4) and (5.5) together, we get
thus showing that the strategy in the proper game on P*Q that we described is a winning strategy. • Our next step in the proof is to show that iterations of length co preserve properness. We shall repeatedly use the following lemma on two step iterations: 5.6
Lemma Let a be a P*<2-name for an ordinal, and let (p,f)eP*Q. There exists a P-name /5 and some s such that p \\- s ^ r, and = /5
(5.7)
Proof There exists a partition W of p such that for every we W there exists j8(w) and s(w) with the property that w |— s(w) ^ f and (w, s(w)) |— dc = /?(w). Let s and j5 be the P-names such that, for every weW, ||S = 5(W)|| B = W,
Now (5.7) follows.
||j8 = j8(w)||=W
Q
5.8
Iteration of length co We consider an iteration of length co of proper notions of forcing {(2»K°=i- W e s h a l l assume that the g n 's are complete Boolean algebras; we are justified to assume that by Lemma 7.7 in Part II. Let Bo be the trivial algebra {0,1}, and for each n let Bn +1 = Bn * Qn. Let P be the iteration of length co, of the {Qn}n. Thus, conditions in P are sequences P =
We wish to show that II wins in the game
(5.9)
5 Iteration of proper forcing
105
Player I starts the game $(P) by selecting a condition p and chooses a P-name a 0 . Let us apply Lemma 5.6 to P = P1*Pl(O (where PUto = P\(l,co); see (7.11), Part II). Let po = p(0)eQo. There exist Pinames a® and 50 such that
and
and (Po>So)\\-<Xo
= (
so^pt[l^)
Xo
We apply the strategy o0 in the game ^(Qo) against Fs move Po>^o> and obtain an ordinal j?0. Player I then plays a P-name oi1. Let px e VPl be a name for a condition in g x such that p o lr-Pi = s o (l). We apply Lemma 5.6 to P2*f>2,co- There exist P2 -names d\ and s\ such that and The P 2 -name a} can be identified with a P^name for a 2 i - n a m e - Inside the model Vp\ we start the game ^(2i)- Player I begins by selecting px and by playing a}. Applying the strategy al9 II responds by playing an ordinal a? (so a? is a P r n a m e for an ordinal). Then we continue the game ^(Q0X whose first moves were p o ^ o a n d j?o- The second move of player I is a j , to which II responds by playing an ordinal Px. The game continues as in Table 5.10. At the nth move, I chooses a P-name an. Let pneVPn be a name for a condition in Qn such that ?0/>i •• .£„-!> \h Pn = sn-1{n). We apply Lemma 5.6 to Pn+i*Pn+i,a). There exist P w+1 -names a" and sn such that
•
»0
o
(x0
(X0
p0
A\
*?
pt
a
a
P2
At GC2
0C2
2
2
106
/ / / Proper forcing
In VPn, we start the game &{Qn) by letting I play pn and a". Applying <jn, II responds by playing a Pn-name ij"" 1 . Then we continue g(Qn-i) in VPn~l by letting I play aJ!"1, to which II responds (by <xn-i) by playing a"" 2 , and so on; eventually we obtain (by a0 in &(Q0)) an ordinal /?„. It remains to show that the strategy described above is a winning strategy for II in ^(P). Since dn is a winning strategy, we have 3«oeQo^o < Po such that q0 |h Vm3/c <x£ =
ft
(5.13)
and also, for every n,
3 ge
(5.14)
Consequently, there is a sequence <„>„ ^ p
an =
ft
(5.15)
To verify (5.15), let n
The iteration in general The generalization of the above proof from co to arbitrary limit ordinals is not trivial. To illustrate the difficulties, let y be an ordinal of cofinality co, say y = limnyn. By induction, we assume that each iteration Rn = PJn i7n is proper. Then the proof given in 5.8 shows that the ^-iteration R of {Rn}n is proper. However, the iteration R need not be in general equivalent to the y-iteration P. Given a sequence {fn)neR, each fn is forced to be a function in Py _ y with countable support; but there is in general no reason why there should exist a countable set S c y such that Vn the support of fn is included in S (so that (fn}n can be represented as a function in P, with countable support). We prove Theorem 5.1 by induction on the length of iteration. Simultaneously, we prove the following generalization of the Factor Lemma 7.13, Part II. 5.17
The 6>-Factor Lemma Let {yn}n be an increasing sequence of ordinals with limny,, = )>. Let P be an iteration of length y of proper {Qa}. Let Ro = P yo , and for
5 Iteration of proper forcing each n let Rn + 1=Pyy
107
, and let R be the (full) iteration of
{AH}m
a = j3k
(5.18)
It is not difficult to see that this version (seemingly more difficult for II) is equivalent to the proper game. To formulate the right induction hypothesis, let Py be an iteration of proper forcings, and consider the proper game ^ on P y . Let o be a strategy for II. We say that a is a good winning strategy if for every sequence of moves p, Ao,... ,An,... of player I, o produces a sequence {/?„}„ such that there exists a q ^ p that satisfies (5.18), and, moreover. support(q)^{Pn}n<(O
(5.19)
We shall prove Theorem 5.1 by induction on y. The inductive hypothesis (somewhat stronger than 'P is proper') reads as follows: VT/
(5.20)
We prove (5.20) by induction on y. If y is a successor ordinal then (5.20) follows easily from the earlier proofs. Thus, we assume that y is a limit ordinal. If cf y = a>, then we first prove Lemma 5.17. The proof of (5.20) incorporates the procedure used in the proof of Lemma 5.17. 5.21
Proof of the 6>-Factor Lemma Let y = limMyn, and let Py be an iteration of length y, of proper {<2Js< r For each n, let An = Pjm_l7u (and R0 = Pyo); l e t R b e the co-iteration of {Rn}. We wish to show that B(R) = B(Py). For any pePv let r = (rn}n<(O, where rn = p \[yn_ uyn). Thus, P y embeds in R; it suffices to show that Py embeds in R densely. Thus, let r = *„>„ be a condition in R. We wish to find pePy so that p^r. By the inductive hypothesis (5.20), each Rn has a good winning strategy on. We use these good strategies to produce p. We play the proper games &(Rn), simultaneously for all n. The game „) is initiated with the condition fn. The moves of player I are names
108
/ / / Proper forcing
for countable sets of ordinals; the moves of player II are according to the strategy 6n. At step 0, we start &(R0). Player I plays r0 and a name A% for the support of fl. Player II responds po. At step 1, we start ^(Rx) in VRo. Player I plays rx and a name A\ for the support of r 2 . Player II responds a?. Then we continue the game ^(R0Y player I plays a?, and II responds /?!. Generally at step n, we start %(Rn) in yR°*-*Rn-K Player I plays fn and a name i4J for the support of fn+1. Player II responds a"" 1 . Then, playing &(Rn- J , player I plays a"~x and II responds a"" 2 . And so on, until II plays pn in the game 9(R0). Since the 6n are good winning strategies, there exists a condition q = (qn}nER, stronger than
(5.22)
and the support of qn is included in the set of all ordinals played by II (5.23) Let S = {pn}n<(O. It follows from (5.22) and (5.23) that, for every n,
q\n\\- support(qn)^S We shall conclude the proof by constructing a condition pePy so that p = q (under the embedding of P in R). This we do by induction on £ < y: if ^ S we let p(£) = 1, and if £eS, then we let p(^) be the condition ieQ^ so that p\£\[-i = qn(£\ where n is the unique n for which yn_x ^ £
Proof of the inductive condition (5.20) Let 7 be a limit ordinal, and let Py be an iteration of proper
We want to show that (5.20) is true for y. It suffices to prove it for rj = 0: when Y\ < y is arbitrary, we do the same proof inside Vp", because by the Factor Lemma, P^ is in Fp* a countable support iteration (of length y - rj) of proper forcings, and by the induction hypothesis, |— P Vfy Vy such that rj ^ fj < y < y( \\- p_ II has a g.w.s. on P--) where g.w.s. = good winning strategy. Thus, we want to show that II has a good winning strategy in the proper game ^(P y ). We shall follow closely the proof of 5.8 (iteration of length a>). If cf y = o>, we choose beforehand a cofinal sequence {yn}n with limit y; if cf y > a) then we define a certain increasing sequence {yn}n below y in the course of the proof.
5 Iteration of proper forcing
109
Player I starts the game ^(Py) by selecting pePy and by choosing a Pyname a 0. If cf y > co, we choose some y0 < y (if cf y — a>, y0 is already given), and consider Ro = Pyo. Let po = p \y0', by Lemma 5.6 there are jR0-names &% and s0 such that p0 \\- RJ0 ^ p \ [y0, y) and (p0, s0) |h ag = a 0 . We start the game ^(Ro) by letting I play p0 and an # 0 -name for the countable set jag} usupport (s0). Player II uses a good winning strategy a0 to return /?0. We let this j80 be player IPs first move in &{Py). At the nth move, I chooses a Py-name an. If cf 7 > co, we choose some 7n
and
Sn^§n-X\lyn9y)
and We start the game ${&„) by letting I play pn and X", where A"n = {dnn}v support (sn).
II uses a good winning strategy 6n to play d^" 1. Then we continue &(Rn-1) by letting I play ann~ \ to which II responds d^"2. And so on, until II plays (by o0 in ^(Ro)) an ordinal fin. It remains to show that the strategy described above is a good winning strategy for II in &(Py). Let y^ = limnyn and S = {/?„}„. As in the proof of 5.8 we obtain a sequence q = <„>„ in the co-iteration R of {K,,},, such that
Since the an are good winning strategies, it follows, as in the proof of the co-Factor Lemma, that B{R) = B(Py ) and that q is a condition in P and support(^f)cS. Let us identify the condition geP ]oo with (jl l...ePy (if 7^ <7). Since S^y^ and for every n, g fn||- support (sn)^5, we have
It follows that ^f^p, q\\- Vn3/c otn = pk (and support(^f) ^ 5). Hence, the strategy is a good winning strategy. • 5.25
Iteration of length co2
By Theorem 5.1, iteration of proper forcing preserves K^ Assuming the CH, iteration of length at most co2 preserves other cardinals as well, provided the forcing notions iterated are of size at most Kx. This has applications when one adds iteratively generic reals, such as K2 Laver reals. We conclude this chapter by stating (without proof) the following lemma, the consequence of which is (see Corollary 7.10, Part II) that an iteration of
110
/ / / Proper forcing
length co2 of proper forcings of size Kx has the X2-chain condition and therefore preserves all cardinals: 5.26
Lemma Assume the CH. Let OL
6
The Proper Forcing Axiom
We consider a powerful internal forcing axiom that generalizes the version of Martin's Axiom (Part II, Section 3.2). 6.1
Proper Forcing Axiom (PFA) If P is a proper notion of forcing and if {Da:a < a^} are dense subsets of P, then there exists a filter G on P that meets all the Da. The aim of this chapter is to prove the following consistency result, due to Baumgartner and Shelah:
6.2
Theorem If there exists a supercompact cardinal then there is a generic model that satisfies PFA and 2N° = K2. 6.3 A few remarks before we start the proof of Theorem 6.2. Every c.c.c. notion of forcing is proper, and so PFA implies MAXX. Consequently, PFA implies 2 K °>K 1 . At present it is unknown whether PFA is consistent with 2S° > K2, but a stronger axiom than PFA, the so-called Martin's Maximum, implies that 2X° = K2. Unlike (the general version of) Martin's Axiom, PFA only deals with Xx dense sets at a time. It is impossible to generalize PFA to deal with X2 dense sets, as is easily seen when one considers the co-closed notion of forcing P that adjoins a collapsing map of co2 onto cot (with countable conditions): for each a < a>2, let Da = {pePiaerange (p)}; no filter on P can meet all the dense sets D a ,a < co2. The consistency proof given below uses a supercompact cardinal. There are consequences of PFA showing that some large cardinal assumption is necessary for the consistency of PFA. At present it is not clear exactly how strong large cardinal assumption is needed. We recall the definition of a supercompact cardinal: 6.4
Definition An uncountable cardinal K is supercompact if for every X ^ K there 111
112
/ / / Proper forcing
exists an elementary embedding with critical point K (i.e. K is the least ordinal such that ;(/<;) > K) such that J(K) > X and Mk a M, i.e. every A-sequence {aa:a < X) c= M is in M. A major tool in the proof of Theorem 6.2 is a Laver function: 6.5
Lemma Let /c be supercompact. There exists a function/: /c -• VK such that for every set x, for every sufficiently large X there exists an elementary embedding;: K-»M with critical point K such that j(k) > X, MA c M and (7(/))M = x
(6.6)
'Sufficiently large X" means X^K and X^ |TC(x)|, where TC(x) is the transitive closure of x. The proof of Lemma 6.5 can be found in Laver [1978]. The proof of Theorem 6.2 follows loosely the proof of the consistency of Martin's Axiom (see Theorem 3.5, Part II). We construct a notion of forcing by countable support iteration of length K. Each notion of forcing used in the iteration is proper, and so Kx is preserved. The iteration uses only forcings of size < /c, and so K and cardinals above K are preserved. Cardinals between K t and K are collapsed, and K becomes K2; and the model satisfies 2N° = K2. The crucial property of the model is of course that it satisfies PFA. This cannot be arranged as easily as in the case of MA, because we do not have an analog of Lemma 3.4, Part II (equivalence of MA with MA*). Instead, we use a Laver function, which makes it possible to handle all potential proper forcings in only K steps. 6.7
Proof of Theorem 6.2 Let K be supercompact, and l e t / b e a Laver function. We construct a countable support iteration PK of {(5a:a < K} as follows: At stage a, consider the set/(a). In Vp* we define Qa as follows: if/(a) is a pair (A Qj) of Pa -names such that P is a proper notion of forcing and @ is a ysequence of dense subsets of P, for some y
6 The Proper Forcing Axiom
113
2*° ^ K2, we shall be done as soon as we prove that K[G] satisfies PFA and ic = K2. We shall prove that K[G] satisfies If P is proper and 2 = {Da:a < y}, y < /c, is a sequence of dense sets then there is a filter F on P that meets each Da (6.8) The condition (6.8) clearly implies PFA in K[G]; let me show that it also implies K = K2: if y < K, let P be the notion of forcing that collapses y onto co1 with countable conditions. P is proper. For each a < y let Da = {/?eP:aerange (p)}. Applying (6.8), we obtain collapsing map of y onto a*!. Thus, K = X 2 . In order to prove (6.8) in K[G], let P be a proper forcing in F[G] and let 2 = {Da:oc < y} be a collection of y < /c dense subsets of P. Let P and ^ be PK-names for P and 3). Let A be a sufficiently large cardinal (say X > 2 ); we may also assume that P e l Since / is a Laver function, there exists an elementary embedding j : V-+M with critical point K such that j(k) > A, Mx a M, and (6.9) 6.10
Lemma P is a proper notion of forcing in M[G~\.
Proof P is proper in V[G]\ using Definition 3.7 (and the remark following it), this is witnessed by some club set C aiy^]™ of countable models, where t] > 2 |P| and r\ < k. Since Mk a M and because G is generic on PK that has the fc-chain condition, every >l-sequence of ordinals in V[G] belongs to M[G~\. It follows that the club C is in M[G]. Hence, C witnesses that P is proper in the model M[G\ • Now consider the notion of forcing j(PK) in M. It is an iteration (with countable support) of length J(K), using the Laver function^/). Since; \ VK is the identity, the first K stages are the same as in PK, i.e. PK = (j(PK)) \K. At stage K:, we havey(/)(/c) = (P, &) and P is proper in M[G\ and therefore P is the forcing used at stage K of j{PK). Thus, for some A, j(PK) = PK*P*R
(6.11)
Let H*K be a K[G]-generic ultrafilter on P*R9 and let us work in K[G*//*K]. We extend the elementary embedding j.V^M to an elementary embedding 7*:K[G]->M[G*iJ*K]
(6.12)
114
/ / / Proper forcing
This is done as follows: for every PK-name x, let j*(i/G)=j{i)/(G*H*K)
(6.13)
The definition of j * does not depend on the choice of the name x, for \\x = y\\eG implies \\j(x)=j{y)\\eG*H*K. (Because j(p) = p for every pePK.) Similarly, j * is elementary, since ||(p(x)||eG implies \\(p(j(x))\\e G*H*K. Also,;* =3;. The filter H on P is K[G]-generic and thus it meets every D a, a < y. Let
F = {j*(p):peH}
(6.14)
As we assume that P c A, we have F = {j(p):peH}, and because; f AGM, the set F is in M[G*H*K~]. Moreover, F generates a filter on;*(P) that meets every j*(Dx\ a < 7. Thus, M[G*/f *K] |h 3F on j*(P) that meets every Dej*{@) and, since ;* is elementary, ^[G] Ih 3F on P that meets every Z)e^ which proves (6.8).
•
References Baumgartner [1983] gives a number of applications of the Proper Forcing Axiom. Laver, R. (1978). Israel J. Math. 29, 385-8. Baumgartner, J. (1983). In Surveys in Set Theory (Mathias, A.R.D., ed.), Cambridge University Press, 1-59. Shelah, S. (1982). Proper Forcing. Lecture Notes in Mathematics 940, Springer-Verlag, NY.
7
Martins Maximum
The present chapter deals with a generalization of the Proper Forcing Axiom that, in a sense explained below, is a maximal consistent generalization of Martin's Axiom MAXX. The axiom PFA of 6.1 is an internal forcing axiom that generalizes MANj. The generalization is obtained by considering a more extensive class of forcings to which the axiom applies. To explore further generalizations, let ^ be an arbitrary class of forcing notions, and let us consider the principle 7.1
MA(^): If P is a notion of forcing in # and if {Da:a < a*!} are dense subsets of P, then there exists a filter G on P that meets all the Da: Thus MAX is MA(c.c.c) and PFA is MA (proper). An important generalization of PFA is Martin's Maximum. We say that P is stationary preserving if every stationary subset S of co1 remains stationary in K[G], for any generic filter on P. Martin's Maximum is the principle MA (stationary preserving): 7.2
Martin's Maximum (MM) If P is a stationary preserving notion of forcing and if {Da:a < a^} are dense subsets of P, then there exists a filter G on P that meets all the Da. Since every proper notion of forcing preserves stationary sets, the class of all stationary preserving P is at least as large as the class of all proper P, and so MM implies PFA. The maximality of the principle MM follows from this observation: 7.3
Lemma Let P be a notion of forcing with the property that there is a stationary set S e a>1 such that || S is not stationary || = 1
Then there are Kx dense sets such that no filter G on P meets them all. Proof Let C be a name such that || C is a club and CnS = 01| = 1
(7.4) 115
116
/ / / Proper forcing
We define dense sets Da and £ a , a < a^, such that no filter G can meet all the Da and all the Ea. For each a, let
= {p: eitherp |— aeC, or 3y < a such that for all ^ between 7 and a, p | | The sets Da and £ a are dense, because C is, respectively, unbounded and closed. Let G be a filter on P that meets all the Da and Ea and let C = {(x:3peG
p\\-oceC}
The set C is unbounded because G n Da ^ 0 for all a. Also, C is closed: Let a be a limit point of C and let peGc\Ea. It must be the case that p \\- aeC, because otherwise there is y < a such that p \\- £$C for all ^ between y and a, while some other qeG forces £eC for some £ between 7 and a. But p and g are compatible. Now S is stationary, and so there is some ccGSnC. Hence, some condition forces cceC for an oteS which contradicts (7.4). • The following theorem of Foreman, Magidor and Shelah establishes the consistency of Martin's Maximum (relative to a supercompact cardinal): 7.5
Theorem If there exists a supercompact cardinal then there is a generic model that satisfies MM. I will not give a proof of the consistency of MM here. Instead, an outline of the proof, as well as some discussion of the technique, is given below. In the second half of this chapter, we give three applications of MM. One consequence of MM is that 2X° = K2. This underscores the still open problem whether PFA is consistent with 2N° > K2. Another consequence of MM is that the club filter o n K i is K2-saturated. This is known to imply strong large cardinal properties, thus justifying the use of a supercompact cardinal in the consistency proof of MM (and possibly of PFA as well). 7.6 An obvious attempt at a generalization of the consistency proof of PFA given in Chapter 6 would be to iterate notions of forcing that are stationary preserving. Such an approach does not succeed, for the following reason. There is a stationary preserving notion of forcing Pf, that for a given sufficiently fast growing f u n c t i o n / : ^ -^cou adjoints generically a fast growing function g which is eventually below / . It is clear that such
7 Martin's Maximum
117
forcing cannot be iterated o times without collapsing Kx. (For this example, consult [Shelah, 1982], p. 255.)
Semiproper forcing We consider a property of forcing intermediate between proper and stationary preserving. For motivation, we refer the reader to Chapter 3, particularly to Theorem 3.13 showing that Definitions 3.1, 3.2, 3.7 and 3.11 are equivalent. 7.7
A game theoretic definition Consider the semiproper game on P. Player I selects a condition p, and chooses a name d0 for a countable ordinal. Player II chooses an ordinal /?0. At the nth move, I plays a name an for a countable ordinal, and II plays an ordinal /?„: p; <*o,/Mi> 0i, •.•>*,., ft,. •»
(7-8)
II wins the game if and only if Iq^p
Vn 4lh3/c
an =
fik
(7 9)
7.10
Definition A notion of forcing (P, < ) is semiproper if player II has a winning strategy in the semiproper game for P. 7.11
A model theoretic definition Let X be sufficiently large and let M be an elementary submodel of (VX9 e, P, <). A condition q is (P, M)-semigeneric if for every name a e M for a countable ordinal, q\\-3PeM
a=£
(7.12)
7.13
Definition A notion of forcing (P, < ) is semiproper if for some (for all) sufficiently large X there is a club set C ^ [VxY" °f countable elementary submodels M<(K A ,e,P, < ) with the following property: VpeM3q ^ p
q is (P, M)-semigeneric
(7.14)
Definitions 7.10 and 7.13 are weaker versions of the corresponding definitions of properness. Instead of arbitrary ordinal names, the definitions refer only to names for countable ordinals.
118
/ / / Proper forcing
A careful perusal of the equivalence proof in Chapter 3 will lead the reader to the following conclusion: 7.15
Theorem (a) The two definitions of semiproperness given above are equivalent. (b) If P is semiproper then P is stationary preserving. • The last part of the proof of Theorem 3.13 does not go through in the semiproper case, as the argument uses preservation of stationary subsets of [X]03 for some large A, not just X = X x . 7.16 It is consistent, relative to a large cardinal, that every stationary preserving P is semiproper (the converse of 7.15 (ft)); we discuss this later in some detail. As shown by Shelah, a large cardinal assumption is necessary for this implication. There is a notion of forcing (the Namba forcing) which is stationary preserving, but not semiproper unless 0 # exists.
Semiproper Forcing Axiom and revised countable support iteration As 'semiproper' is a property intermediate between 'stationary preserving' and 'proper', the forcing axiom MA (semiproper) is stronger than PFA and a consequence of MM: 7.17
Semiproper Forcing Axiom (SPFA) If P is semiproper and if {Da:a < a>1} are dense in P, then there is a filter G on P that meets all the Da. For the record*: MM-•SPFA-•PFA
(7.18)
The axiom SPFA is consistent relative to a supercompact cardinal. Its consistency (due to Shelah) is established in a manner similar to the proof of Theorem 6.2, namely by iteration of semiproper forcings. The iteration used in the consistency proof of SPFA is called revised countable support iteration. If one iterates proper forcing (with countable support), every countable set of ordinals at any stage a of the iteration is included in a countable set in the ground model. This fact is essential in the proof of the Factor Lemma, which enables us to replace a condition with countable support in K[G a ] by a condition with countable support in V. * S. Shelah proved recently that SPFA and MM are equivalent.
7 Martin s Maximum
119
When forcing with a semiproper partial order, however, the cofmality of an ordinal may change to co. With this in mind, the method of revised countable support uses a different kind of limit at the limit stages of the iteration, in a way anticipating when an ordinal changes its cofmality to co. Roughly speaking, the support of a condition is not a countable set of ordinals, but rather a countable set of potential names for ordinals. The execution of this idea is technically quite involved, and can be found in [Shelah, 1982]. An analog of Theorem 5.1 states that semiproperness is preserved under revised countable support iteration. This fact, along with the technique used in the proof of Theorem 6.2, establishes the consistency of SPFA relative to a supercompact cardinal. Forcing principles MA + ( We shall now consider a stronger version of the internal forcing axioms (7.1):
7.19
If P is a notion of forcing in V9 if {D^.OL < w1} are dense subsets of P and if S is a name such that || S is a stationary subset of col || = 1 then there exists a filter G on P that meets all the D a , and such that the set is stationary.
(7.20) +
For the particular classes # we use the notation MA^, PFA , SPFA + and M M + . Baumgartner has shown that MA^ = M A X ; the plussed version of the other forcing axioms is strictly stronger than the axioms themselves. The consistency proof of PFA in Chapter 6 can be modified in the obvious way to yield the consistency of PFA + . Similarly, the consistency proof of SPFA can be modified to give the consistency of SPFA + . Thus: 7.21
Theorem SPFA + is consistent relative to a supercompact cardinal.
We shall conclude our discussion on the consistency of MM by proving the following: 7.22
Theorem MA + (co-closed) implies that every stationary preserving forcing P is semiproper.
120 7.23
/ / / Proper forcing Corollary
And consequently, Theorem 7.21 yields Theorem 7.5, the consistency of Martin's Maximum. Note that by 7.16, the axiom MA + (co-closed) implies (at least) 0#. Toward the proof of Theorem 7.22, let P be a notion of forcing, let X be a sufficiently large cardinal, and consider the club set F of all countable elementary submodels of the model (F A ,e,P, < ) . An elementary chain is a sequence <Ma:a<0>
(O^coJ
(7.24)
of elements of F such that M a < Mp whenever a < /?, and My = \Ja< yMa for every limit y. 7.25
Lemma MA + (co-closed) implies that for every stationary set X ^ F there exists an elementary chain <M a :a < cox > such that M a => a for all a, and that the set {a:MaeX} is stationary. Proof Let Q be the following co-closed notion of forcing: a condition q is an elementary chain <M a :a < 6} for some countable 0, such that M a 2 a for all a; the ordering is by extension. A generic filter G on Q produces an elementary co^chain CG = <MG(a):a
7 Martins Maximum Xp = {MeT: peM
121 and Vg ^ p q is not (P, M)-semigeneric}
is stationary.
(7.26)
By Lemma 7.25 there exists an elementary chain <M a :a
(frM is an element of B(P% and is possibly 0). Let G be a generic filter on P with peG; we shall show that the set {oc:bM eG} contains a club. Let $£9 £
and
is a club (in V). Let, in K[G],
C is a club, and if a e C then for every SeMa there is geG such that q |h (3PeMa)S = p. Hence, bMaeG. Consequently, C c {a:fcM G G}. D
Consequences of Martin's Maximum Of the numerous applications of Martin's Maximum, we present three here: the continuum is equal to K2 , the Singular Cardinals Hypothesis holds, and the club filter on Kx is K2-saturated. 7.27
Theorem Assume MM. If K ^ K2 is a regular cardinal then K*° = K.
7.28
Corollary MM implies 2*° = X2.
7.29
Corollary MM implies the Singular Cardinals Hypothesis: for every singular
122
/ / / Proper forcing
(By Silver's theorem, it suffices to prove Corollary 7.29 for the case cf X = a>; this follows from Theorem 7.27 by letting K = A+.) Theorem 7.27 uses the following lemma: 7.30
Lemma Assume MM, and let K ^ X2 be regular. Let E = {
First we prove that Lemma 7.30 implies the theorem: let {Sf:i < K} be a pairwise disjoint collection of stationary subsets of K. Given a countable set X CZK, Lemma 7.30 provides an ordinal 9X
veAn
then f(v)eSn
(7.31)
The ordering of P is by extension. Once we prove that P is stationary preserving, and that for every a < co1 the set Da = {/eP:dom / ^ a + 1} is dense in P, Martin's Maximum yields a continuous increasing function f:co1^S with property (7.31), and 9 = sup v /(v) is as desired. In both proofs we shall make use of the following: 7.32
Lemma Let A c a>1 be stationary, and let T be a stationary subset of K such that cf £ = co for all £eT. Every model (KA ,e,...), where A^/c, has a countable elementary submodel M such that Mnco1eA and sup(Mn/c)GT.
Proo/ Since T is stationary in fc, there exists an elementary submodel N (of size K J such that co1czN and sup (AT nk)eT. N is the union of an elementary a>x -chain of countable models M a such that sup(M a n/c) = sup(iVn/c) for each a. Since ,4 is stationary in X l9 there is M = M a such
7 Martin's Maximum
123
We prove first that each Da is dense. We do it by induction on a. Assume that it is true for all ordinals less than a. Let peP; we find q 3 p of length at least a + 1. Let n be such that cceAn, and let T=Sn. By Lemma 7.32 there is a countable elementary submodel M of (KA,P,p, a) for sufficiently large X such that y = sup(Mnfc)eT. Let {an}n be a sequence of countable ordinals with limit a, and let {yn}n be a sequence converging to y. We construct a sequence of conditions p — p0 c px cz • • • c pn a - • as follows, such that pneM for each n: given pn, let prt +! be, inside M, an extension of pn such that a n edom(p n+1 ) and that At + i(°O ^ 7nJ s u c h Pn+i exists because M, an elementary submodel of KA, satisfies the induction hypothesis that Dan is dense. The function q = (J^ = o p n u{(a,y)} is a condition of length a + 1. Finally, we prove that P is stationary preserving. Let A be a stationary subset of col9 let peP, and let C be a name such that p \\- C is a club subset of a)x
(7.33)
We want a stronger g and some <xeA such that q\\-aeC. Let n be such that AnAn is stationary, and let T = Sn. By Lemma 7.32 there is a countable elementary submodel M of (KA,P,p, C) such that oc = Mr\co1eAnAn and y = sup(Mn/c)eT. Let {an}M be a sequence with limit a, and let {yn}n be a sequence with limit y. We construct conditions p = p0 <= ••• c P n <= •••, each in M, as follows: given pn, let pn+1eM be such that anedom(pn+1\ that pn + ^ a j ^ yn9 and that for some /^M ^ an in M (therefore
Definition The club filter on Kx is K2-saturated if there is no collection of stationary sets {At:i < K2} such that Ain Aj is nonstationary for all i ^j.
By the work of Solovay, Kunen and Mitchell, K 2-saturation of the club filter has very strong large cardinal consequences. By the following theorem, so does Martin's Maximum: 7.35
Theorem MM implies that the club filter on K t is K 2-saturated.
We start the proof by reviewing some facts on stationary sets of countable ordinals.
124
/ / / Proper forcing
7.36
Lemma For any collection {Ai'.ieZ} of stationary sets, with |Z| = X1? there is a stationary set A such that (i) At — A is nonstationary, for all ieZ, and (ii) every stationary subset of Z has stationary intersection with some A>v The set A is unique modulo the club filter, and is denoted A = £ i e Z At. Proof Assuming Z = cou we let A be the diagonal union:
(xeA^xxe ( J ^
D
(7.37)
Let S be a stationary set and let Ps be the notion of forcing that adds a club subset of S (Example 1.2). 7.38
Lemma If A c 5 is stationary, then A remains stationary in VPs.
Proof We follow closely the proof of co-distributivity (Lemma 1.4). Let p \\- C is a club we shall find q < p and xeA such that q \\- XeC. As in Lemma 1.4, we construct a chain {Aa}a of countable subsets of Ps and an increasing sequence ya. Given >4a, for each qeA^ we find P = f$(q) > ya and some r = r(g) ^ q so that r |— jSeC, and max (r) > ya. Then we let Aa+1 = Aav{r(q):q
Lemma Assume MM, and let {At:ieW} be a collection of stationary sets with the property that every stationary set has stationary intersection with some Av Then there is Z c w of size < Xx such that T.iez^i ^s i n the club filter; hence, every stationary set has stationary intersection with some Ah ieZ. The lemma clearly implies the theorem.
7 Martin's Maximum
125
Proof We apply MM to the following forcing notion P: first let Q be the forcing that collapses \W\ onto Xx with countable conditions. In F Q , consider S = Y.iew^i a n d l e t P = Q*P§Equivalently, let P (a dense set in Q*P$ be the set of all pairs (q,p) such that q: y + \->W for some y < CD1 , and p is a closed countable subset of col such that ocep implies ae (J A ^ }
(7.40)
A condition (q\ p') is stronger than (q, p) \iq' ^.q and /?' is an end-extension of p. Note that P is stationary preserving. If A is stationary, then for some ie W, A n Ai\s stationary. A n Ai remains stationary in VQ, and so An A(nS is stationary in VQ. By Lemma 7.38, AnA(nS remains stationary in Q P p V * *. Hence, A is stationary in V . For each a < col, let Da = {(q,p)eP:(x ^ max(p)}. Each Da is dense in P. By Martin's maximum, there is a filter G on P that meets all the Da. Let F = [J{q:(q,p)eG for some p] Z = range (F) and C = {J{p:(q,p)eG for some q] The set C is a club, and by (7.40)
In other words, C = Y^iez^i-
•
References Foreman, M , Magidor, M. and Shelah, S. (1986). Ann. Math, (in press). Shelah, S. (1982). Proper Forcing. Lecture Notes in Mathematics 940, Springer-Verlag, NY.
8
Well-founded iteration
In this final chapter, we discuss some of the ramifications of the iteration methods described in earlier chapters, as well as questions raised by the consistency proof of the Proper Forcing Axiom. One still unresolved question is whether the proper forcing axiom is consistent with 2*°>X 2 . ^ n v * e w °f Martin's Maximum, which implies 2K° = K2, this question may not be just a technical problem resulting from the particular method used in the consistency proof. Nevertheless, the inability to construct a model with 2 X °>X 2 by the method of Chapter 6 is caused by the failure of the countable support iteration to preserve cardinals above Xx. Indeed, it has been observed that countable support iterations of length > co2 do not necessarily preserve K2. One effect of this phenomenon is that in all applications of the countable support iteration one ends up with a model in which the continuum is exactly X2. This is in contrast with the finite support iteration which, as in the case of Martin's Axiom, enables us to construct models with the continuum arbitrarily large. We shall describe a more general method of iteration that makes it possible to preserve cardinals while adding a large number of generic reals. The method is a generalization of both iterated forcing and product forcing. In particular, it is then possible to give a partial answer to the question of consistency of PFA with 2N° > K 2 . Without going into details, let me state the result as follows. There is a class $ of proper forcing notions that includes, among others, Cohen forcing, Sacks forcing, Prikry-Silver forcing, Mathias forcing and Laver forcing, and for which the analog of PFA is consistent with 2N° > K2. 8.1
Theorem The axiom MA(^) is consistent with 2N<> > X 2 .
The class ^ is considerably smaller than the class of all proper forcings. Note that there is no mention of large cardinals in Theorem 8.1, as the consistency is relative just to ZFC. This is unlike the consistency of PFA for which large cardinals are necessary. 126
8 Well-founded iteration
127
We omit the proof of Theorem 8.1 but outline the method of wellfounded iteration which the proof uses. Well-founded iteration The following definition is a generalization of iteration of forcing described in Definition 7.1, Part II. Let (W, <) be a well-founded partial order. For xeW, let W[x] denote the initial segment {yeW:y < x}. In general, an initial segment of W is a set X c W with the property that xeX and y < x then yeX. Definition 8.2 is by induction on the height of W\ 8.2
Definition Let (W9 <) be a well-founded partial order. A forcing notion Pw is a (well-founded) iteration along W if it is a set of functions on W with the following properties: (i) For every xeW, PW[X] = Pw \ W[x] = {p \ W[x]\p£Pw} is an iteration along W[x] (ii) For every x e W there is a forcing notion &x e VPw^x\ and for every p e Pw \hPw[x]P(x)eQx (iii) For all
p,qePw iff for all xeW
and
(iv) lePW9 where l(x) = 1 for all xeW. If X is an initial segment of W, let Px = PW\X = {p\X\pePw). (v) If X is an initial segment of W and if pePw and qePx are such that q^p p\X then the following function r is a condition in Pw\ r{x)
\q{x) \p(x)
if if
xeX xeW-X
If the well-founded partial ordering Win Definition 8.2 is a well ordering (of length a) then Definition 8.2 is just Definition 7.1, Part II, of iteration of length a. The analog of Lemma 7.2, Part II, holds for well-founded iterations as well: 8.3
Lemma If X is an initial segment of W and Px = Pw \ X, then Vp* c Vpw.
•
128
/ / / Proper forcing
Note that well-founded iteration is also a generalization of product forcing: a product of forcing notions {P f :/e/} is in fact an iteration along /, where / is given the discrete partial order. As in the case of well-ordered iteration, we can define the forcing notion Pw by giving the names Qx9 xeW, and specifying the support of each condition. In particular, we may consider well-founded iterations with finite support, with countable support, etc. To illustrate the use of well-founded iterations, let us consider the problem of preserving X2. Suppose that Pw is such that the support of every pePw is countable, and is included in an initial segment X of w of size ^ K x . (Note that there are arbitrarily large partial orders with this property; only the height of W is limited to ^ X2.) Assuming that each PW[X] has the K2-chain condition, a standard A-system argument shows that Pw has the K2-chain condition as well. The preservation of Xx by well-founded iteration is an entirely different problem. In some cases, it suffices to use countable supports. Such is the case when iterating Sacks forcing, for example; a somewhat complicated fusion argument shows that Pw is proper (in fact, satisfies Axiom A). However, when applying well-founded iteration to other forcings, such as in the proof of Theorem 8.1, one needs a more general iteration. Note that in view of Propositions 5.16 and 5.17, Part I, and because iteration along W subsumes product forcing, neither finite support iteration nor countable support iteration is suitable when iterating Laver or Mathias forcing. The iteration used in the proof of Theorem 8.1 uses mixed support. We omit the (rather technical) definition of mixed support iteration, and instead give an example, a special case of mixed support iteration that is a prototype of the method. 8.4
Example (Baumgartner) Mixed support product of Mathias forcings Let us consider the following product P of a collection {P^.iel}, where each Pt is Mathias forcing (Section 3.11, Part I). Condition in P are certain functions p = (p(i):iel} such that each p(i) is a Mathias condition (sf, Sf). The support of p is the set
(Here 1 = ( 0 , co).) The root of p is the set root«p i :i6/» = {i:stem(pI.) * 0 }
8 Well-founded iteration
129
8.5
Definition The mixed support product of Mathias forcings {Pt:iGl} is the set of all functions p = (pt:ieI} whose support is countable and whose root is finite. The mixed support product of Mathias forcing preserves K^ 8.6
Proposition If p \\- X\CD -• V then there exists q ^ p and a countable A such that q \\- X c A. Moreover, we can find the q ^ p with the same root as /?, and, in fact, stem(^) = stem(pt) for every is I. Proof The proof is a variation on the proof of Lemma 3.12, Part I, or Theorem 5.12, Part I. To simplify matters, assume that p has empty root; we shall find q as desired, with empty root as well. Let {un}n be a sequence of natural numbers such that each u appears infinitely often. We construct a sequence p = p0 ^px ^p2 ^ •••; let Wn denote the support of pn. Let W= \J?=0Wn; by a suitable enumeration we also construct a sequence of finite sets F Q C F J C ^ C . . . with {J?=0Fn = W. For every n, we make sure that
This guarantees that for each ie W, p^i) = \imnpn(i) is a Mathias condition (with empty stem), and so p^ = (Poo(i)}i is a condition with support W and empty root. We also produce a sequence of finite sets {An}n with the iteration that p^ \\- X c A, where A = {J?=0An. At stage n, we already have pn = <(0,S,$)):ie/>. For each ieFm let Kn(i) be the set of first n elements of Sn(i). Let r l 9 ..., TZ be all the functions T on Fn such that T =
fl;
(8.7)
for some akn then we put akn into Xw, and let qk +1 be the condition (with empty root) <(0,S(fc+1)(O):*'<EO, where
[ 7(0
otherwise
130
/ / / Proper forcing
The sequence {pn}n converges to a condition p^. The rest of the proof is a verification of the fact that p^W-X ^ A. The argument follows closely the conclusion of the proof of Lemma 3.12, Part I. I omit it, as I believe that the readers who have been able to bear with me up to this point will be able to complete the proof on their own. • Reference Groszek, M. and Jech, T. Generalized Iteration of Forcing (1986) (in press).
Bibliography
Baumgartner, J. (1983). Iterated forcing. In Surveys in Set Theory (Mathias, A.R.D., ed.), pp. 1-59, Cambridge University Press. Baumgartner, J. (1984). Applications of the proper forcing axiom, in Handbook of Settheoretical Topology (Kunen, K. and Vaughan, J.E., eds.), pp. 913-59, North-Holland, Amsterdam. Borel, E. (1919). Sur la classification des ensembles de mesure nulle, Bull. Soc. Math. France 47, 97-125. Cohen, P. (1966). Set Theory and the Continuum Hypothesis, Benjamin, NY. Dales, H.G. (1979). A discontinuous homomorphism from C(X\ Am. J. Math. 101,647-734. Dales, H.G. and Woodin, W.H. (1986). An Introduction to Independence for Analysts, London Mathematical Society Lecture Notes 115, Cambridge University Press. Devlin, K. and Johnsbraten, H. (1974). The Souslin Problem. Lecture Notes in Mathematics 405, Springer-Verlag, NY. Easton, W. (1970). Powers of regular cardinals, Ann. Math. Logic 1, 139-78. Eklof, P. (1976). Whitehead's problem is undecidable, Am. Math. Monthly 83, 775-88. Ellentuck, E. (1974). A new proof that analytic sets are Ramsey, J. Symb. Logic 39, 163-5. Esterle, J. (1978). Sur l'existence d'un homomorphisme discontinu de C(K), Proc. London Math. Soc. 36, 46-58. Fleissner, W. (1984). The normal Moore space conjecture and large cardinals, in Handbook of Set-theoretical Topology (Kunen, K. and Vaughan, J.E. eds.), pp. 733-60, North-Holland, Amsterdam. Foreman, M., Magidor, M. and Shelah, S. (1986). Martin's maximum, saturated ideals and non-regular ultrafilters, Ann. Math, (in press). Fremlin, D.H. (1984). Consequences of Martins Axiom, Cambridge Tracts in Mathematics 84, Cambridge University Press. Gray, C.W. (1982). Iterated forcing from the strategic point of view, Thesis, Berkeley, CA. Grigorieff, S. (1971). Combinatorics on ideals and forcing, Ann. Math. Logic 3, 363-94. Groszek, M. and Jech, T. (1986). Generalized iteration of forcing (in press). Jech, T. (1971). Trees, J. Symb. Logic 36, 1-14. Jech, T. (1973). Some combinatorial problems concerning uncountable cardinals, Ann. Math. Logic 5, 165-98. Jech, T. (1978). Set Theory, Academic Press, NY. Jech, T. (1984). More game theoretic properties of Boolean algebras, Ann. Pure and Appl. Logic 26, 11-29. Kueker, D. (1977). Countable approximations and Lowenheim-Skolem theorems, Ann. Math. Logic 11, 57-103. Kunen, K. (1980). Set Theory, North-Holland, Amsterdam. Laver, R. (1976). On the consistency of Borel's conjecture, Acta Math. 137, 151-69. Laver, R. (1978). Making the supercompactness of K indestructible under /c-directed closed forcing, Israel J. Math. 29, 385-8. Martin, D.A. and Solovay, R. (1970). Internal Cohen extensions, Ann. Math. Logic 2, 143-78. Mathias, A.R.D. (1977). Happy families, Ann. Math. Logic 12, 59-111. Menas, T. (1976). Consistency results concerning supercompactness, Trans. Am. Math. Soc. 223,61-91.
131
132
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Rudin, M.E. (1969). Souslin's conjecture, Am. Math. Monthly 76, 1113-19. Rudin, M.E. (1977). Martin's axiom, in Handbook of Mathematical Logic (Barwise, J., ed.), pp. 491-501, North-Holland, Amsterdam. Sacks, G. (1971). Forcing with perfect closed sets, in Axiomatic Set Theory (Scott, D., ed.), pp. 331-55, American Math. Society, Providence, RI. Shelah, S. (1974). Infinite abelian groups, Whitehead problem and some constructions, Israel J. Math. 18, 243-56. Shelah, S. (1982). Proper Forcing. Lecture Notes in Mathematics 940, Springer-Verlag, NY. Shoenfield, J. (1971). Unramified forcing, in Axiomatic Set Theory (Scott, D., ed.), pp. 357-82, American Math. Society, Providence, RI. Shoenfield, J. (1975). Martin's axiom, Am. Math. Monthly 82, 610-17. Solovay, R. (1970). A model of set theory in which every set of reals is Lebesgue measurable, Ann. Math. 92, 1-56. Solovay, R. (1971). Real-valued measurable cardinals, in Axiomatic Set Theory (Scott, D., ed.), pp. 397-428, American Math. Society, Providence, RI. Solovay, R. and Tennenbaum, S. (1971). Iterated Cohen extensions and Souslin's problem, Ann. Math. 94, 201-45. Souslin, M. (1920). Probleme 3, Fund. Math. 1, 223. Woodin, W.H. (1987). Set theory and discontinuous homomorphisms of Banach algebras, Memoirs Am. Math. Soc. (in press).
Index of symbols
MA*
(P, <) 2 G 2 F 2 F[G] 2 a 2 Ih 2 fl(P) 3
pk 4 P-L 4 II9II 5 x/G 5
vp^yQ
53
f<9 61 f
8
sat(P) 11 ^ n 15, 17, 18, 20, 100 Sit) 19 «n 20 PxQ 23 Uipi 24 sip) 24 PK(/1) 40 [ A ] < K 40
P*Q 44 G*// 44 D\B 48 MA 53 MAX, 53
Ultygf 61 supp (p) 68 Pj,a) 70 ^a/, 70 A<(O 80 C F 80 Ps 81 [/!]" 83 ACa 83 a
&0 89 ^! 90 ^» 91 ^ 92 PFA 111 MM 115 SPFA 118 MA+(
MA (a) 126
133
Subject index
K2-saturated filter, 123 amalgamation, 16, 18, 20 antichain, 4 Axiom A, 100 Boolean-valued model, 3, 5 Boolean values, 5 branch, 57 branching point, 15 canonical embedding, 51 canonical names, 5 canonical projection, 51 cardinal, measurable, 39 real-valued measurable, 39 strongly compact, 40 supercompact, 111 cc.c, 11 chain condition, /c-, 11 closed, H>, 11 closed unbounded, 80, 83 Cohen real, 13 collapse, 35 compatible, 4 complete Boolean algebra, proper, 97 conditions, 2 costationary, 81 countable chain condition, 11 countable support iteration, 68 decides, 6 A-system, 24 dense, 2 dense set of conditions, 2 directed, 85 direct limit, 68 distributive, K>, 10 distributive laws, 10 dominates, 14 Easton product, 26 Easton support, 26 extends, 13 extension, generic, 2
134
Factor Lemma, 70 co-, 106 filter, K2-saturated, 123 fine ultrafilter, 40 finite support iteration, 50 forcing, 2 forcing conditions, 2 forcing relation, 2, 6 Forcing Theorem, 2 free abelian group, 59 fusion, 15, 17, 18,20 fusion sequence, 15 game, countable choice, 91 cut and choose, 90 descending chain, 89 proper, 92 semiproper, 117 gap, 64 strong, 64 generic, 2
(P,M)-, 96 generic extension, 2 Grigorieff real, 21 ground model, 2 group, abelian, 59 Krfree, 60 free, 59 torsion free, 59
W-,59 height, 56 homogeneous, weakly, 7 incompatible, 4 indecomposable, 102 initial segment, 127 inverse limit, 68 iteration, 50, 67 countable support, 68 Easton support, 72 finite support, 50 /-support, 68
Subject index revised countable support, 118 two step, 44 well-founded, 127 /c-chain condition, 11 K-distributive, 10 /c-product, 25 /c-saturated, 11 Laver function, 112 Laver real, 19 Laver tree, 19 uniform, 21 Levy collapse, 35 limit, direct, 68 inverse, 68 Martin's Axiom, 53 Martin's Maximum, 115 Mathias real, 17 matrix of partitions, 96 maximal antichain, 4 measurable cardinal, 39 measure algebra, 38 measure space, 38 measure, product, 39 mixed support, 129 model, Boolean-valued, 3 ground, 2 monotone function, 61 names, 5 normal Moore space conjecture, 40 notion of forcing, 2 co-Factor Lemma, 106 open dense, 3 p-point, 21 partition, 4 partitions, matrix, 96 perfect tree, 15 uniform, 17 predense, 3 Prikry-Silver real, 17 product, 23, 24 Easton, 26 infinite, 24
135 product forcing, 23 product measure, 39 proper forcing, 95-7 Proper Forcing Axiom, 111 proper game, 92 property (K), 23 pure subgroup, 60 random real, 13 real, 13 Cohen, 13 Grigorieff, 21 Laver, 19 Mathias, 17 Prikry-Silver, 17 random, 13 Sacks, 15 real-valued measurable cardinal, 39 refinement, 10 revised countable support iteration, 118 root, 24, 128 Sacks real, 15 saturated, K2-, 123 K-, 11 semigeneric, (P,M)-, 111 semiproper forcing, 117 Semiproper Forcing Axiom, 118 semiproper game, 117 separable, 13 separative, 4 separative quotient, 4 stationary, 80, 83 stationary preserving, 115 stem, 18 strong measure zero, 73 stronger than, 2 strongly compact cardinal, 40 sufficiently large, 95 supercompact cardinal, 111 support, 24, 68, 128 Suslin line, 56 Suslin tree, 56 tree, 15 Laver, 19 perfect, 15 Suslin, 56 two step iteration, 44
JC-,25
mixed support, 129 a-, 25
W-group, 59 well-founded iteration, 127
Author index
Baumgartner, J., 33, 100, 101, 111, 114, 128 Martin, D.A., vii, 53, 115, 121 Mathias, A.R.D., 17, 32, 126, 128 Borel, E., 71, 73 Mitchell, W., 123 Cohen, P., vii, 6, 13, 27, 126 Namba, K., 118 Nyikos, P., 41 Easton, W., 26, 28, 72 Foreman, M., 90, 116
Prikry, K., 17, 30, 126
Gray, C , vii, 92, 110 Grigorieff, S., 21, 34 Groszek, M., 130
Sacks, G., 15, 17, 30, 126 Scott, D., 38 Shelah, S., vii, 59, 71, 95, 110, 111, 116-19 Silver, J., 17, 30, 72, 121, 126 Solovay, R., vii, 37, 39, 48, 61, 123
Jech, T., 90, 130 Kaplansky, I., 61 Kunen, K., 41, 123
Suslin, M , vii, 53, 56 Tennenbaum, S., vii
Laver, R., 19,32,73,75, 112, 126 Levy, A., 35, 49 Magidor, M , 116
136
Velickovic, B., 90 Whitehead, J.H.C., 59 Woodin, W.H., 61