For a xed value of, we will prove that there are some simple constraints on the triple of cardinals (b(); d(); 2 ). We will also prove that any triple

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1 Cardinal invariants above the continuum James Cummings Hebrew University of Jerusalem Saharon Shelah Hebrew University of Jerusalem y February 25, 1998 Abstract We prove some consistency results about b() and d(), which are natural generalisations of the cardinal invariants of the continuum b and d. We also dene invariants b cl() and d cl(), and prove that almost always b() = b cl() and d() = d cl() 1 Introduction The cardinal invariants of the continuum have been extensively studied. They are cardinals, typically between! 1 and 2!, whose values give structural information about!!. The survey paper [2] contains a wealth of information about these cardinals. In this paper we study some natural generalisations to higher cardinals. Specically, for regular, we dene cardinals b() and d() which generalise the well-known invariants of the continuum b and d. Supported by a Postdoctoral Fellowship at the Hebrew University. y Partially supported by the Basic Research Fund of the Israel Academy of Science. Paper number

2 For a xed value of, we will prove that there are some simple constraints on the triple of cardinals (b(); d(); 2 ). We will also prove that any triple of cardinals obeying these constraints can be realised. We will then prove that there is essentially no correlation between the values of the triple (b(); d(); 2 ) for dierent values of, except the obvious one that 7! 2 is non-decreasing. This generalises Easton's celebrated theorem (see [3]) on the possible behaviours of 7! 2 ; since his model was built using Cohen forcing, one can show that in that model b() = + and d() = 2 for every. b() and d() are dened using the co-bounded lter on, and for >! we can replace the co-bounded lter by the club lter to get invariants b cl() and d cl(). We nish the paper by proving that these invariants are essentially the same as those dened using the co-bounded lter. Some investigations have been made into generalising the other cardinal invariants of the continuum, for example in [7] Zapletal considers s() which is a generalised version of the splitting number s. His work has a dierent avour to ours, since getting s() > + needs large cardinals. We are indebted to the referee and Jindrich Zapletal for pointing out a serious problem with the rst version of this paper. 2 Denitions and elementary facts It will be convenient to dene the notions of \bounding number" and \dominating number" in quite a general setting. To avoid some trivialities, all partial orderings P mentioned in this paper (with the exception of the notions of forcing) will be assumed to have the property that 8p 2 P 9q 2 P p < P q. Denition 1: Let P be a partial ordering. Then U P is unbounded if and only if 8p 2 P 9q 2 U q P p. D P is dominating if and only if 8p 2 P 9q 2 D p P q. b(p) is the least cardinality of an unbounded subset of P. 2

3 d(p) is the least cardinality of a dominating subset of P. The next lemma collects a few elementary facts about the cardinals b(p) and d(p). Lemma 1: Let P be a partial ordering, and suppose that = b(p) and = d(p) are innite. Then = cf() cf() jpj: Proof: To show that is regular, suppose for a contradiction that cf() <. Let B be an unbounded family of cardinality, and write B = S <cf() B with jb j <. For each nd p such that 8p 2 B p p, then nd q such that 8 < cf() p q. Then 8p 2 B p q, contradicting the assumption that B was unbounded. Similarly, suppose that cf() <. Let D be dominating with cardinality and write D = S <cf() D where jd j <. For each nd p such that 8p 2 D p p, and then nd q such that 8 < cf() p q. Then 8p 2 D q p, contradicting the assumption that D was dominating. The next result shows that we cannot hope to say much more. Lemma 2: Let and be innite cardinals with = cf() and < =. Dene a partial ordering P = P(; ) in the following way; the underlying set is [] <, and (; x) (; y) if and only if and x y. Then b(p) = and d(p) =. Proof: Let B P be unbounded. If jbj < then we can dene = sup f j 9y (; y) 2 B g < ; x = [ f y j 9 (; y) 2 B g 2 [] < : But then (; x) is a bound for B, so jbj and hence b(p). On the other hand the set f (; ;) j < g is clearly unbounded, so that b(p) =. 3

4 Let D P be dominating. If jdj < then S f y j 9 (; y) 2 D g 6=, and this is impossible, so that d(p). On the other hand GCH holds and cf(), so that jpj = < =. Hence d(p) =. Denition 2: Let P, Q be posets and f : P! Q a function. f embeds P conally into Q if and only if 8p; p 0 2 P p P p 0 () f(p) Q f(p 0 ). 8q 2 Q 9p 2 P q Q f(p). That is, rge(f) is dominating. Lemma 3: If f : P! Q embeds P conally into Q then b(p) = b(q) and d(p) = d(q). Proof: Easy. Lemma 4: Let P be any partial ordering. Then there is P P such that P is a dominating subset of P and P is well-founded. Proof: We enumerate P recursively. Suppose that we have already enumerated elements hb : < i into P. If f b j < g is dominating then we stop, otherwise we choose b so that b b for all <. Clearly the construction stops and enumerates a dominating subset P of P. To see that P is well-founded observe that b < b =) <. Notice that the identity embeds P conally in P, so b(p ) = b(p) and d(p ) = d(p). We also need some information about the preservation of b(p) and d(p) by forcing. Lemma 5: Let P be a partial ordering with b(p) =, d(p) =. 4

5 Let V [G] be a generic extension of V such that every set of ordinals of size less than in V [G] is covered by a set of size less than in V. Then V [G] b(p) =. Let V [G] be a generic extension of V such that every set of ordinals of size less than in V [G] is covered by a set of size less than in V. Then V [G] d(p) =. Proof: We do the rst part, the second is very similar. The hypothesis implies that is a cardinal in V [G], and since \B is unbounded" is upwards absolute from V to V [G] it is clear that V [G] b(p). Suppose for a contradiction that we have C in V [G] unbounded with V [G] jcj <. By our hypothesis there is D 2 V such that C D and V jdj <, but now D is unbounded contradicting the denition of. With these preliminaries out of the way, we can dene the cardinals which will concern us in this paper. Denition 3: Let be a regular cardinal. 1. If f; g 2 then f < g i 9 < 8 > f() < g(). 2. b() = def b(( ; < )). 3. d() = def d(( ; < )). These are dened by analogy with some \cardinal invariants of the continuum" (for a reference on cardinal invariants see [2]) known as b and d. In our notation b = b(!) and d = d(!). Lemma 6: If is regular then + b(). b() = cf(b()). b() cf(d()). 5

6 d() 2. cf(2 ) >. Proof: The rst claim follows from the following trivial fact. Fact 1: Let f f j < g. Then there is a function f 2 such that 8 < f < f. Proof: Dene f() = sup f f () + 1 j < g. < <. Then f() > f () for The next three claims follow easily from our general results on b(p) and d(p), and the last is just Konig's well-known theorem on cardinal exponentiation. We will prove that these are essentially the only restrictions provable in ZFC. One could view this as a renement of Easton's classical result (see [3]) on 7! 2. 3 Hechler forcing In this section we show how to force that certain posets can be conally embedded in ( ; < ). This is a straightforward generalisation of Hechler's work in [4], where he treats the case =!. We start with a brief review of our forcing notation. p q means that p is stronger than q, a -closed forcing notion is one in which every decreasing chain of length less than has a lower bound, and a -dense forcing notion is one in which every sequence of dense open sets of length less than has non-empty intersection. If x is an ordered pair then x 0 will denote the rst component of x and x 1 the second component. 6

7 Denition 4: Let be regular. D () is the notion of forcing whose conditions are pairs (s; F ) with s 2 < and F 2, ordered as follows; (s; F ) (t; F 0 ) if and only if 1. dom(t) dom(s) and t = s dom(t). 2. s() F 0 () for dom(t) < dom(s). 3. F () F 0 () for all. We will think of a generic lter G as adding a function f G :! given by f G = S f s j 9F (s; F ) 2 G g. It is easy to see that G = f (t; F ) j t = f G dom(t); dom(t) =) F () f G () g; so that V [f G ] = V [G] and we can talk about functions from to being D ()-generic. Lemma 7: Let < =, and set P = D (). Then 1. P is -closed. 2. P is + -c.c. 3. If g :! is P-generic over V then 8f 2 \ V f < g. Proof: 1. Let < and suppose that h(t ; F ) : < i S is a descending - sequence of conditions from P. Dening t = f t j < g and F : 7! sup f F () j < g, it is easy to see that (t; F ) is a lower bound for the sequence. 2. Observe that if (s; F ) and (s; F 0 ) are two conditions with the same rst component then they are compatible, because if H : 7! F ()[F 0 () the condition (s; H) is a common lower bound. There are only < = possible rst components, so that P clearly has the + -c.c. 7

8 3. Let f 2 \ V, and let (t; F ) be an arbitrary condition. Let us dene F 0 : 7! (F () [ f()) + 1, then (t; F 0 ) renes (t; F ) and forces that f() < f G () for all dom(t). If = cf() > = < then it is straightforward to iterate D () with < -support for steps and get a model where b() = d() =. Getting a model where b() < d() is a little harder, but we can do it by a \nonlinear iteration" which will embed a well chosen poset conally into ( ; < ). Theorem 1: Let = <, and suppose that Q is any well-founded poset with b(q) +. Then there is a forcing D (; Q ) such that 1. D (; Q ) is -closed and + -c.c. 2. V D(;Q) Q can be conally embedded into ( ; < ) 3. If V b(q) = then V D(;Q) b() =. 4. If V d(q) = then V D(;Q) d() =. Proof: We will dene the conditions and ordering for D (; Q ) by induction on Q. The idea is to iterate D () \along Q" so as to get a conal embedding of Q into. It will be convenient to dene a new poset Q + which consists of Q together with a new element top which is greater than all the elements of Q. We will dene for each a 2 Q + a notion of forcing P a. If a 2 Q + then we will denote f c 2 Q j c < a g by Q=a. It will follow from the denition that if c < a then P c is a complete subordering of P a, and that the map p 2 P a 7! p Q=c is a projection from P a to P c. Suppose that for all b < Q a we have already dened P b. 1. p is a condition in P a if and only if (a) p is a function, dom(p) Q=a and jdom(p)j <. 8

9 (b) For all b 2 dom(p), p(b) = (t; F _ ) where t 2 < and F_ is a P b -name for a member of. 2. If p; q 2 P a then p q if and only if (a) dom(q) dom(p). (b) For all b 2 dom(q), if p(b) = (s; _H) and q(b) = (t; _ I) then i. t = s dom(t). ii. p (Q=b) Pb dom(t) < dom(s) =) s() > I(). _ iii. p (Q=b) Pb 8 _H() I(). _ We dene D (; Q ) = P top, and verify that this forcing does what we claimed. The verication is broken up into a series of claims. Claim 1: D (; Q ) is -closed. Proof: Let < and let hp : < i be a descending -sequence S of conditions. We will dene a new condition p with dom(p) = dom(p ). For each b 2 dom(p ) let p (b) = (t (b); F_ (b)). Let p(b) = (t; F _ S ) where t = f t (b) j b 2 dom(p ) g and F _ (b) is a P b - name for the pointwise supremum of f F_ (b) j b 2 dom(p ) g. Then it it is easy to check that p is a condition and is a lower bound for hp : < i Claim 2: D (; Q ) is + -c.c. Proof: Let hp : < + i be a family of conditions. Since = < we may assume that the domains form a -system with root r. We may also assume that for b 2 r, p (b) = (t b ; F (b)) where t b is independent of. It is now easy to see that any two conditions in the family are compatible. 9

10 Claim 3: If c < a then P c is a complete subordering of P a, and the map p 7! p Q=c is a projection from P a to P c. Proof: This is routine. If G is D (; Q )-generic, then for each a 2 Q we can dene f a G 2 \ V [G] by f a G = S f t(a) 0 j t 2 G g. It is these functions that will give us a conal embedding of Q into ( \ V [G]; < ), via the map a 7! f a G. Claim 4: If a < Q b then f a G < f b G. Proof: Let p be a condition and let F_ be the canonical P b -name for fg. a Rene p to q in the following way; q(c) = p(c) for c 6= b, and if p(b) = (t; _H) then q(b) = (t; I) _ where I _ names the pointwise maximum of F_ and _H. Then q forces that fg() b is greater than fg() a for dom(t). Notice that by the same proof f b G dominates every function in V Pa. Claim 5: If a Q b then f a G f b G. Proof: If b < Q a then we showed in the last claim that fg b < fg, a so we may assume without loss of generality that b Q a. Let p be a condition and let <. Choose large enough that fdom(p(a) 0 ); dom(p(b) 0 ); g. Let p(b) = (t; F _ ), and nd q 2 P b such that q p (Q=b) and q decides F _ ( + 1). Let p 1 be the condition such that p 1 (c) = p(c) if c Q b and p 1 (c) = q(c) if c < Q b. Then p 1 renes p and p 1 (a) = p(a), p 1 (b) = p(b). Let p(a) = (s; _H). Find r 2 P a such that r p 1 (Q=a) and r decides _H ( + 1). Let p 2 be the condition such that p 2 (c) = p 1 (c) if c Q a and p 2 (c) = r(c) if c < Q a. Then p 2 renes p 1 and p 2 (a) = p(a), p 2 (b) = p(b). 10

11 Now it is easy to extend p 2 to a condition which forces f a G() > f b G(). Claim 6: The map a 7! f a G embeds Q conally into ( ; < ) in the generic extension by D (; Q ). Proof: We have already checked that the map is order-preserving. It remains to be seen that its range is dominating. Let G be D (; Q )-generic and let f 2 \ V [G]. Then f = ( f) _ G for some canonical name f, _ and by the + -c.c. we may assume that there is X Q such that jxj = and f _ only involves conditions p with dom(p) X. Now b(q) + so that we can nd a 2 Q with X Q=a. This implies that f _ is a P a -name for a function in, so that f < f a G and we are done. Claim 7: If V b(q) = then V D(;Q) b() =. Proof: Let G be D (; Q )-generic. By lemma 3 it will suce to show that V [G] b(q) =. This follows from lemma 5, the fact that D (; Q ) is + - c.c. and the assumption that b(q) +. Claim 8: If V d(q) = then V D(;Q) d() =. Proof: Exactly like the last claim. This nishes the proof of Theorem 1. 11

12 4 Controlling the invariants at a xed cardinal In this section we show how to force that the triple (b(); d(); 2 ) can be anything \reasonable" for a xed value of. Theorem 2: Let = < and let GCH hold at all cardinals. Let ; ; be cardinals such that + = cf() cf(), and cf() >. Then there is a forcing M (; ; ; ) such that in the generic extension b() =, d() = and 2 =. Proof: In V dene Q = P(; ), as in lemma 2. We know that V b(q) = and V d(q) =. Fix Q a conal wellfounded subset of Q, and then dene a new well-founded poset R as follows. Denition 5: The elements of R are pairs (p; i) where either i = 0 and p 2 or i = 1 and p 2 Q. (p; i) (q; j) i i = j = 0 and p q in, i = j = 1 and p q in Q, or i = 0 and j = 1. Now we set M (; ; ; ) = D (; R). It is routine to use the closure and chain condition to argue that M makes 2 =. Since R contains a conal copy of Q, it is also easy to see that M forces b() = and d() =. 5 A rst attempt at the main theorem We now aim to put together the basic modules as described in the previous section, so as to control the function 7! (b(); d(); 2 ) for all regular. A naive rst attempt would be to imitate Easton's construction from [3]; this almost works, and will lead us towards the right construction. Let us briey recall the statement and proof of Easton's theorem on the behaviour of 7! 2. 12

13 Lemma 8 (Easton's lemma): If P is -c.c. and Q is -closed then P is - c.c. in V Q and Q is -dense in V P. In particular < ON \V PQ = < ON \V P. Theorem 3 (Easton's theorem): Let F : REG! CARD be a class function such that cf(f ()) > and < =) F () F (). Let GCH hold. Then there is a class forcing P which preserves cardinals and conalities, such that in the extension 2 = F () for all regular. Proof:[Sketch] The \basic module" is P() = Add(; F ()). P is the \Easton product" of the P(), to be more precise p 2 P i 1. p is a function with dom(p) REG and p() 2 P() for all 2 dom(p). 2. For all inaccessible, dom(p) \ is bounded in. P is ordered by pointwise renement. There are certain complications arising from the fact that we are doing class forcing; we ignore them in this sketch. If is regular then we may factor P as P < P() P > in the obvious way. P is always -closed. It follows from GCH and the -system lemma that if is Mahlo or the successor of a regular cardinal then P < is -c.c. On the other hand, if is a non-mahlo inaccessible or the successor of a singular cardinal, then P < is in general only + -c.c. In particular for regular P = P < + is always + -c.c. so that by Easton's lemma ON \ V P = ON \ V P. This implies that in the end we have only added F () many subsets of. It remains to be seen that cardinals and conalities are preserved. It will suce to show that regular cardinals remain regular. If is Mahlo or the successor of a regular cardinal, then Easton's lemma implies that < ON \ V P = < ON \ V P<, and since is regular in V P< (by -c.c.) is clearly regular in V P. Now suppose that = + for singular. If becomes singular in V P let its new conality be, where we see that < and is regular in V. ON \ V P = ON \ V P, so that will have conality in V P. This is 13

14 absurd as P is + -c.c. and + < <. A very similar argument will work in case is a non-mahlo inaccessible. Suppose that we replace Add(; F ()) by P() = M (; (); (); ()), where 7! ((); (); ()) is a function obeying the constraints given by Lemma 6. Let P be the Easton product of the P(). Then exactly as in the proof of Easton's theorem it will follow that P preserves cardinals and conalities, and that 2 = () in V P. Lemma 9: If is inaccessible or the successor of a regular cardinal then b() = (), d() = () and 2 = () in V P. Proof: For any, \ V P = \ V P. b() and d() have the right values in V P() by design, and these values are not changed by -c.c. forcing. So assuming P < is -c.c. those invariants have the right values in V P, and hence in V P. We need some way of coping with the successors of singular cardinals and the non-mahlo inaccessibles. Zapletal pointed out that at the rst inaccessible in an Easton iteration we are certain to add many Cohen subsets, so that there really is a need to modify the construction. 6 Tail forcing Easton's forcing to control 7! 2 can be seen as a kind of iterated forcing in which we choose each iterand from the ground model, or equivalently as a kind of product forcing. Silver's \Reverse Easton forcing" is an iteration in which the iterand at is dened in V P. The \tail forcing" which we describe here is a sort of hybrid. We follow the conventions of Baumgartner's paper [1] in our treatment of iterated forcing, except that when have _ Q 2 V P and form P _ Q we reserve the 14

15 right not to take all P-names for members of Q (as long as we take enough names that the set of their denotations is forced to be dense). For example if Q 2 V we will only take names ^q for q 2 Q, so P ^Q will just be P Q. We will describe a kind of iteration which we call \Easton tail iteration" in which at successor stages we choose iterands from V, but at limit stage we choose Q _ in a dierent way; possibly Q =2 V, but we will arrange things so that the generic G factors at many places below and any nal segment of G essentially determines Q. This idea comes from Magidor and Shelah's paper [5]. We assume for simplicity that in the ground model all limit cardinals are singular or inaccessible. In the application that we intend this is no restriction, as the ground model will obey GCH. Denition 6: A forcing iteration P Easton tail iteration i with iterands h _ Q : + 1 < i is an 1. The iteration has Easton support, that is to say a direct limit is taken at inaccessible limit stages and an inverse limit elsewhere. 2. _Q = 0 unless is a regular cardinal. 3. If is the successor of a regular cardinal then Q 2 V. 4. For all regular, P +1 is + -c.c. 5. For a limit cardinal P is ++ -c.c. if is singular, and + -c.c. if is inaccessible. 6. For all regular with + 1 < there exists an iteration P dense in P such that P +1 = P +1, and for with + 1 < (a) P factors as P +1 P ( + 1; ). (b) If is inaccessible or the successor of a singular, and p 2 P, then p() is a name depending only on P ( + 1; ). (c) P ( + 1; ) is + -closed. 15

16 Clause 6 is of course the interesting one. It holds in a trivial way if P is just a product with Easton supports. Clauses 6a and 6b should really be read together, as the factorisation in 6a only makes sense because 6b already applies to <, and conversely 6b only makes sense once we have the factorisation from 6a. The following result shows that Easton tail iterations do not disturb the universe too much. Lemma 10: Let P be an Easton tail iteration. Then 1. For all regular <, ON \ V P = ON \ V P P preserves all cardinals and conalities. Proof: Exactly like Theorem 3. In the next section we will see how to dene a non-trivial Easton tail iteration. If + is the successor of a regular then it will suce to choose Q + 2 V as any + -closed and ++ -c.c. forcing. The interesting (dicult) stages are the ones where we have to cope with the other sorts of regular cardinal, here we will have to maintain the hypotheses on the chain condition and factorisation properties of the iteration. It turns out that slightly dierent strategies are appropriate for inaccessibles and successors of singulars. 7 The main theorem Theorem 4: Let GCH hold. Let 7! ((); (); ()) be a class function from REG to CARD 3, with + () = cf(()) cf(()) () () and cf(()) > for all. Then there exists a class forcing P 1, preserving all cardinals and conalities, such that in the generic extension b() = (), d() = () and 2 = () for all. 16

17 Proof: We will dene by induction on a sequence of Easton tail iterations P, and then let take a direct limit to get a class forcing P 1. The proof that P 1 has the desired properties is exactly as in [3], so we will concentrate on dening the P. As we dene the P we will also dene dense subsets P intended to witness clause 6 in the denition of an Easton tail iteration. Much of the combinatorics in this section is very similar to that in Section 3. Accordingly we have only sketched the proofs of some of the technical assertions about closure and chain conditions. The easiest case to cope with is that where we are looking at the successor of a regular cardinal. So let be regular and assume that we have dened P + (which is equivalent to P +1 since we do trivial forcing at all points between and + ), and P (which is equivalent to P +1) for all. + Denition 7: Q + = M ( + ; ( + ); ( + ); ( + )) as dened in Section 4. P + +1 = P + Q +, and P + +1 = P + Q + for. It is now easy to check that this denition maintains the conditions for being an Easton tail iteration. Since P + is + -c.c. and we know that + + \ V P1 = + + \ V P + +1, we will get the desired behaviour at + in V P1. Next we consider the case of a cardinal +, where is singular. Suppose we have dened P + (that is P ) and P appropriately. Let R be a + well-founded poset of cardinality () with b(r) = () and d(r) = (), as dened in Section 4. Let R + be R with the addition of a maximal element top. We will dene P + Q _ a by induction on a 2 R +, and then set P + +1 = P + Q _ top. In the induction we will maintain the hypothesis that, for each <, P Q _ + a can be factored as P +1 (P ( + 1; + ) Q _ a ). Let us now x a, and suppose that we have dened P + Q _ b for all b below a in R +. Denition 8: Let b < a, and let _ be a for a function from + to +. Then _ is symmetric i for all <, whenever G 0 G 1 and G 0 0 G 1 are two generics for P +1 (P ( + 1; + ) Q b ), then _ G 0G 1 = _ G0 0 G 1. 17

18 Of course the (technically illegal) quantication over generic objects in this denition can be removed using the truth lemma, to see that the collection of symmetric names really is a set in V. Denition 9: (p; q) is a condition in P + _Q a i 1. p 2 P q is a function, dom(q) R + =a and jdom(q)j. 3. For each b 2 dom(q), q(b) is a pair (s; F _ ) where s 2 <+ + and F_ is a symmetric P + Q _ b -name for a function from + to +. Denition 10: Let (p; q) and (p 0 ; q 0 ) be conditions in P + _Q a. (p 0 ; q 0 ) renes (p; q) i 1. p 0 renes p in P dom(q) dom(q 0 ). 3. For each b 2 dom(q), if we let q(b) = (s; F ) and q(b 0 ) = (s 0 ; F 0 ), then (a) s 0 extends s. (b) If 2 lh(s 0 ) lh(s) then (p 0 ; q 0 (R + =b)) s 0 () F (). (c) For all, (p 0 ; q 0 (R + =b)) F 0 () F (). Denition 11: We dene P + +1 as P + _ Q top. If < then we dene P + +1 as P + _ Q top. It is now routine to check that this denition satises the chain condition and factorisation demands from Denition 6. The chain condition argument works because 2 = +, and any incompatibility in Q top is caused by a disagreement in the rst coordinate at some b 2 R. The factorisation condition (clause 6a) holds because symmetric names can be computed using any nal segment of the P -generic, and the closure condition (clause 6c) follows from the fact that the canonical name for the pointwise sup of a series of functions with symmetric names is itself a symmetric name. Now we check that we have achieved the desired eect on the values of b( + ) and d( + ). 18

19 Lemma 11: The function added by P + Q _ top at b 2 R + eventually dominates all functions in + + \ V P + Q _ b. Proof: It suces to show that if F_ is a P + Q _ b -name for a function from + to + then there is a symmetric name F_ 0 such that 8 F () F 0 (). For each regular < we factor P + Q _ b as P +1 (P (+1; + ) Q _ b ). In V P (+1; + ) Q _ b we may treat F_ as a P +1 -name and dene G () = sup(f j 9p 2 P +1 p _ F () = g): Notice that we may also treat G as a P + Q _ b -name, and that if < then G G. Now let F 0 be the canonical name for a function such that 8 F 0 () = sup < G (), then it is easy to see that F 0 is a symmetric name for a function from + to + and that 8 F () F 0 (). Lemma 12: In V P + +1 there is a copy of R embedded conally into + +. Proof: Let f _ name a function from + to + in V P As P + +1 has the ++ -c.c. we may assume that f _ only depends on + many coordinates in R, and hence (since b(r) = ( + ) ++ ) that f _ is a P + Q _ b -name for some b 2 R. By the preceding lemma the function which is added at coordinate b will dominate f. _ It remains to be seen what we should do for inaccessible. The construction is very similar to that for successors of singulars, with the important dierence that we need to work with a larger class of names for functions in order to guarantee that we dominate everything that we ought to. This in turn leads to a slight complication in the denition of P. Suppose that we have dened P and P appropriately. Let R be a poset with the appropriate properties (jrj = (), b(r) = (), d(r) = ()) and let R + be R with a maximal element called top adjoined. As before we dene P Q _ a by induction on a 2 R +. In the induction we will maintain 19

20 the hypothesis that for each < there is a dense subset of P _ Q a which factorises as P +1 (P ( + 1; ) _ Q a ). Let us x a, and suppose that we have dened everything for all b below a. Denition 12: (p; (; q)) is a condition in P _Q a i 1. p 2 P. 2. <, is regular. 3. q is a function, dom(q) R + =a and jdom(q)j <. 4. For each b 2 dom(q), q(b) is a pair (s; F _ ) where s 2 <+ + and F_ is a P ( + 1; ) Q _ b -name for a function from + to +. Denition 13: Let (p; (; q)) and (p 0 ; ( 0 ; q 0 )) be conditions in P + _Q a. (p 0 ; ( 0 ; q 0 )) renes (p; (; q)) i 1. p 0 renes p in P dom(q) dom(q 0 ) For each b 2 dom(q), if we let q(b) = (s; F ) and q(b 0 ) = (s 0 ; F 0 ), then (a) s 0 extends s. (b) If 2 lh(s 0 ) lh(s) then (p 0 ; q 0 (R + =b)) s 0 () F (). (c) For all, (p 0 ; q 0 (R + =b)) F 0 () F (). Notice that since any P _ Q b -generic induces a P (+1; ) _ Q b -generic, there is a natural interpretation of any P (+1; ) _ Q b -name as a P _ Q b - name. We are using this fact implicitly when we dene the ordering on the conditions. We need to maintain the hypothesis on factorising the forcing, so we make the following denition. 20

21 Denition 14: Let be regular with <. P +1 (P ( + 1; ) _ Q a ) is dened as the set of (p 0 ; (p 1 ; (; q))) such that p 0 2 P +1, p 1 2 P ( + 1; ) and (p 0 _ p 1 ; (; q)) 2 P _ Q a with. The key point here is that the factorisation makes sense, because for such a condition q(b) depends only on P ( + 1; ) _ Q b. Denition 15: We dene P +1 as P _ Q top, and P +1 as P _ Q top, As in the case of a successor of a singular, it is straightforward to see that we have satised the chain condition and factorisation conditions. To nish the proof we need to check that the forcing at has achieved the right eect, which will be clear exactly as in the singular case when we have proved the following lemma. Lemma 13: The function added by P _ Q top at b 2 R eventually dominates all functions in \ V P _ Q b. Proof: We do a density argument. Suppose that f_ names a function in \ V P Q _ b, and let (p; (; q)) be a condition in P _Q top. We factor P Q _ b as P +1 (P ( + 1; ) Q _ b ), and use the fact that P +1 has + -c.c. in V P (+1;) Q _ b to nd a P ( + 1; ) Q _ b -name _g such that 8 f() g(). Now we can rene (p; (; q)) in the natural way by strengthening the second component of q(b) to dominate _g. This gives a condition which forces that the function added at b will eventually dominate f. _ This concludes the proof of Theorem 4. 21

22 8 Variations In this appendix we discuss the invariants that arise if we work with the club lter in place of the co-bounded lter. It turns out that this does not make too much dierence. All the results here are due to Shelah. Denition 16: Let be regular. 1. Let f; g 2. f< cl g i there is C closed and unbounded in such that 2 C =) f() < g(). 2. b cl() = def b(( ; < cl )). 3. d cl() = def d(( ; < cl )). Theorem 5: d cl() d() d cl()!. Proof: If a family of functions is dominating with respect to < it is dominating with respect to < cl, so that d cl() d(). For the converse, let us x D such that D is dominating with respect to < cl and jdj = d cl(). We may assume that every function in D is increasing (replace each f 2 D by f : 7! S f()). Let g 0 2. Dene by induction f n, g n, and C n such that 1. f n 2 D. 2. C n is club in, and 2 C n =) g n () < f n (). 3. C n+1 C n. 4. g n is increasing. 5. g n+1 () > g n () for all. 6. g n+1 () > f n (min(c n ( + 1))) for all. 22

23 T S Now let = min( n<! C n ). We will prove that g 0 () n f n () for >. Fix some >. For each n we know that C n \ 6= ;, so that if we dene n = sup(c n \ ( + 1)) then n is the largest point of C n less than or equal to. Notice that min(c n ( + 1)) = min(c n ( n + 1)). Since C n+1 C n, n+1 n, so that for all suciently large n (say n N) we have n = for some xed. We claim that g 0 () f N +1 (), which we will prove by building a chain of inequalities. Let us dene Then = min(c N ( + 1)) = min(c N ( + 1)): g 0 () g N () g N () < f N () < g N +1 () < f N +1 () f N +1 (); where the key point is that 2 C N +1 and hence g N +1 () < f N +1 (). Now it is easy to manufacture a family of size d cl()! which is dominating with respect to <, so that d() d cl()!. Theorem 6: b cl() = b(). Proof: If a family of functions is unbounded with respect to < cl it is unbounded with respect to <, so that b() b cl(). Suppose for a contradiction that b() < b cl(), and x U 0 such that ju 0 j = b() and U 0 is unbounded with respect to <. We may assume without loss of generality that every function in U 0 is increasing. We perform an inductive construction in! steps, whose aim is to produce a bound for U 0 with respect to <. By assumption U 0 is bounded with respect to < cl, so choose g 0 which bounds it modulo the club lter. Choose also club sets f Cf 0 j f 2 U 0 g such that 2 Cf 0 =) f() < g 0 (). For each f 2 U 0 dene another function f [0] by f [0] : 7! f(min(cf 0 ( + 1))) 23

24 For n 1 dene U n = U n 1 [ f f [n 1] j f 2 U n 1 g. By induction it will follow that ju n j = b(), so that we may choose g n such that g n () > g n 1 () for all and g n bounds U n modulo clubs. We choose clubs f Cf n j f 2 U n g such that 1. 2 C n f =) f() < g n (). 2. If f 2 U n 1, then C n f C n 1 f. 3. If n 2 and f 2 U n 2 then C n f C n 1 f [n 2], where this makes sense because in this case f [n 2] 2 U n 1. For each f 2 U n we dene f [n] : 7! f(min(cf n ( + 1))) to nish round n of the inductive construction. Now we claim that the pointwise sup of the sequence hg n : n <!i is an upper bound T for U 0 with respect to <. Let us x f 2 U 0, and then let = min( Cf n ). We will now give a very similar argument to that of Theorem 5. Fix >. We dene n = sup(c n f \ ( + 1)) and observe that n+1 n, so we may nd N and such that n N =) n =. Let = min(cf N ( + 1)) = min(cf N ( + 1)). We now get a chain of inequalities f() f() = f [N ] () < g N +1 () g N +1 (): This time the key point is that 2 C N +2 f C N +1 f [N], so that f [N ] () < g N +1 (). We have proved that > =) f() S n g n (), so that every function f 2 U 0 is bounded on a nal segment of by 7! S n g n (). This contradicts the choice of U 0 as unbounded with respect to <, so we are done. It is natural to ask whether the rst result can be improved to show that b cl() = b(). This can be done for suciently large, at the cost of using a powerful result from Shelah's paper [6]. 24

25 Denition 17: Let = cf() <. 1. P () = f X j jxj = g. 2. [] is the least cardinality of a family P P () such [ that 8A 2 P () 9B P (jbj < ^ A B): One of the main results of [6] is that ZFC proves a weak form of the GCH. Theorem 7: Let > i!. Then [] = for all suciently large < i!. It is easy to see that if P P is such that 8A 2 P () 9B P (jbj < ^ A [ B); then 8A 2 P () 9C 2 P ja \ Cj =. This is all we use in what follows, and in fact we could get away with 8A 2 P () 9C 2 P ja \ Cj 0. Theorem 8: Let = cf() > i!. Then d() = d cl(). Proof: Let = d cl(). Then > i!, so that we may apply Theorem 7 to nd a regular < i! such that [] =. let us x P P () such that jp j = and 8A 2 P () 9C 2 P ja \ Cj =. Now let D be such that jdj = and D is dominating in ( ; < cl ). We may suppose that D consists of increasing functions. Enumerate D as hh : < i, and then dene h A : 7! S 2A h () for each A 2 P. Since < i!, h A 2. We will prove that f h A j A 2 P g is dominating in ( ; < ). We will do a version of the construction from Theorem 5. Let g 0 2. Dene by induction f, g, and C for <, with the following properties. 1. f 2 D. 2. C is club in, and 2 C =) g () < f (). 3. < =) C C. 25

26 4. g is increasing. 5. If <, then g () > g () for all. 6. If <, then g () > f (min(c ( + 1))) for all. This is easy, because <. By the choice of P we may nd a set A 2 P such that jf h j 2 A g \ f f j < gj =. Enumerate the rst! many such that f 2 f h j 2 A g as h n : n <!i. We may now repeat the proof of Theorem 5 with f n, g n and C n in place of f n, g n and C S n. We nd that for all suciently large we have g 0 () g 0 () n f n (). By S the denition of h A and the fact that f f n j n <! g f h j 2 A g, n f n () h A () for all, so that g 0 < h A. This shows f h A j A 2 P g to be dominating, so we are done. We do not know whether it can ever be the case that d cl() < d(). This is connected with some open and apparently dicult questions in pcf theory. References [1] J. E. Baumgartner, Iterated forcing, Surveys in set theory, (A. Mathias, editor), LMS Lecture Notes 87, Cambridge University Press, Cambridge, 1983, pp. 1{59. [2] E. K. van Douwen, The integers and topology, Handbook of Set- Theoretic Topology, (K. Kunen and J. E. Vaughan, editors), North- Holland, Amsterdam, 1984, pp. 111{167. [3] W. Easton, Powers of regular cardinals, Annals of Mathematical Logic, vol. 1 (1964), pp. 139{178. [4] S. Hechler, On the existence of certain conal subsets of!!, Axiomatic Set Theory, Proceedings of Symposia in Pure Mathematics, vol. 13 part 26

27 II, American Mathematical Society, Providence, Rhode Island, 1974, pp. 155{173. [5] M. Magidor and S. Shelah, When does almost free imply free?, Journal of the American Mathematical Society (to appear). [6] S. Shelah, The Generalised Continuum Hypothesis revisited, Israel Journal of Mathematics (to appear). [7] J. Zapletal, Splitting number and the core model, (to appear). 27