The PCF Conjecture and Large Cardinals

The PCF Conjecture and Large CardinalsAuthor(s): Luís PereiraSource: The Journal of Symbolic Logic, Vol. 73, No. 2 (Jun., 2008), pp. 674-688Published by: Association for Symbolic LogicStable URL: http://www.jstor.org/stable/27588654 .

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The Journal of Symbolic Logic Volume 73. Number 2. June 2008

THE PCF CONJECTURE AND LARGE CARDINALS

LU?S PEREIRA

Abstract. We prove that a combinatorial consequence of the negation of the PCF conjecture for

intervals, involving free subsets relative to set mappings, is not implied by even the strongest known large cardinal axiom.

?1. Introduction. In 1973 M. Magidor proved [21] the consistency of the follow

ing statement using large cardinals: K is a strong limit cardinal and 2^ = Hw+2. Moreover, he proved the consistency of H being a strong limit cardinal and 2Hw = Na+i> f?r anv co < a < co + co. In 1978 S. Shelah improved these re

sults [28]: for any countable infinite ordinal a it is consistent relative to large cardinals that ?( is a strong limit cardinal and 2^w ? V(a+\. This raised the ques tion: can one improve this? That is, is it possible to have H? strong limit and z ?

^COi + 1

In 1980 Shelah proved that if K is strong limit then 2H- < N(2?0)+ and in 1989

he proved his celebrated result that if ?( is strong limit then 2H < HW4. These results were obtained through the use of Shelah's PCF theory, the theory of reduced

products of small sets of regular cardinals. The result that if H is strong limit then 2Hw < NWl, would follow if one proved Shelah's PCF conjecture: if a is a set of

regular cardinals such that \a\ < min(<2) then |pcf(a)| =

\a\ (where pcf(a) is the result of the application of Shelah's topological closure operator pcf to the set a ). In fact, to obtain the bound on 2H?J above, it is only necessary to prove the PCF

conjecture for the interval of regular cardinals a = {Hn : 0 < n < co}. To talk about PCF theory we need the following definitions: a model TV -<

(H(6); e, <) is said to be internally approachable of type p, where p is a regular cardinal above the continuum, if |JV|

= p and N is an union of a continuous

elementary chain (N? : ?, < p) such that for every ? < p, N? U {(A^ : n < ?)} ?

N?+\. Its characteristic function xn is the function that to every regular cardinal k e N associates XnM

= sup(7V Pi k). These models and their characteristic

functions are the basic tool of PCF theory and of its applications to Cardinal Arithmetic.

Received April 24, 2007.

2000 Mathematics Subject Classification. 03E04, 03E05, 03E10, 03E35, 03E55.

Key words and phrases. PCF theory, PCF conjecture, free subsets, continuous tree-like scales, large cardinals.

Supported by the Portuguese Foundation for Science and Technology, ref. BD\16650\2004

? 2008, Association for Symbolic Logic 0022-4812/08/7302-0018/S2.50

674



THE PCF CONJECTURE AND LARGE CARDINALS 675

Another important tool are Skolem Hulls. The possibility of taking Skolem Hulls is given by the fact that we have a predicate < for a well ordering of a model

(H(6); G, <). We will use the notation Sk(X) to denote the Skolem Hull of a set X contained in a model H(9). Given a structure of the form (k\ <, Fn)neoj, where the

Fn are finitary functions, a subset X of the universe k of the structure is called free if for every x G X, x g Sk(Z \ {x}).

In [31] Shelah proved that the negation of the PCF conjecture for intervals implies that for a large class of set mappings there are infinite free sets of a specific form: they are subsequences of characteristic functions of internally approachable submodels.

More specifically and considering an illuminating case, he proved that if 2^? < H and K^? > H(D{ then for every internally approachable N of type smaller than V( there is an infinite sequence of cardinals (?in\ n G co) below tt0J such that for every distinct n\,ri2,... ,nk,

ZtfOO i Sk(N U {xN(jUn2), . . . , XN(Mnk)})

We call this the Approachable Free Subset Property. This property is not known to be consistent and if one could prove its negation in ZFC then it would follow that if ?( is a strong limit then 2Hw < H ].

Our main result says basically that not even the largest large cardinal axioms

imply the existence of an TV with an infinite approachable free subset. The biggest large cardinal axiom which does not involves questions about definability of well

orderings is II:

there is an embedding j : Vx+\ ?> Km

In this case, letting k be the critical point of the embedding, X is the supremum

ofthe^Ws.

Theorem 1. If\\ is consistent then II is consistent with the property that there is no internally approachable N such that there is an infinite sequence (?in : n G co) contained in (jn(n) : n G co) with the property that for every distinct n\, n^,..., n^,

XN&nx) i Sk{N U {xn?Mk), - - , XN(/?nk)})

Let (k? : n G co) be an increasing sequence of cardinals and let X be its supremum. Consider the hypothesis that each Kn is < A-supercompact, equivalently, Vx N "?? is supercompact", for each n.

Theorem 2. If it is consistent that for all n, Vx N "Kn is supercompact" then this

hypothesis is also consistent with the property that there is no internally approachable N such that there is an infinite sequence (jun : n G co) contained in (nn

' n G co) with the property that for every distinct n\, n^,..., n^,

XN?Vm) i Sk(N U {XN(Mn2)> - > XN(Mnk)}).

The structure of this paper is as follows. In the second section we recall some facts about PCF theory and free subsets. In the third section we prove our main results. In the final section we discuss the relation with Prikry sequences and some

approaches to proving the PCF conjecture for intervals. We follow the notation of [29] where we also refer the reader for background that

might be needed (easier-to-follow expositions are [13, 2, 15]).



676 LUIS PEREIRA

?2. Approachable free subsets.

2.1. Free subsets and large ideals. A subset X of the universe k of a structure

(?; <, Fn)neco, where the Fn are unitary functions, is called free if for every x e X, x ^Sk(X\{x}). We say that a cardinal k, has the Free Subset Property if for every structure

(K',<,Fn)ne , there exists a cofinal free subset.

The existence of free subsets is linked with partition properties (see for example [6,

chapters 44-46]) and it is obvious that if k is regular then the Free Subset Property

implies that k is Jonsson. The Free Subset Property for a singular cardinal also has large cardinal strength,

for example the Free Subset Property for H was proven by Koepke in [17] to be

equiconsistent with the existence of a measurable cardinal. The simplest way to guarantee this property or even stronger partition relations

for H^ is through the existence of large ideals over an infinite number of K?'s. This was the approach implicitly followed in [26, 27] and in [23]. We say that a

non-principal tt?-complete ideal / on 3P (N?) for n > 2 has the p-Laver Property if the Boolean algebra 3P(V<n)/I has a dense subset which is //-closed and p is

uncountable. The Laver Property allows us to assign a winning strategy to the good

player in Ulam's cut-and-choose game. For a very general version of this game see

Definition 1 of [30]. The intuition behind the use of this winning strategy to obtain

free subsets is described, for example, in section 6 of [4] (though one has to work

harder to get the Free Subset Property). For a given n > 2 the simplest way to obtain such an ideal is to consider a k

complete non-principal ultrafilter U over a cardinal k and then use the Levy collapse

Coll(tt?_i, < k) which collapses k to N?. In the Cohen extension we just consider

the ideal I dual to the filter generated by U. The Boolean algebra ?P(Hn)/I has a

dense subset which is < H?_i-closed. For proofs see Theorem 4 of [8] (who use a

normal ultrafilter) for the case n = 2 and the much more general Lemma 3 of [30]. To obtain a sequence of such ideals one uses an co-sequence of measurable cardinals

(nn : n < co) and collapse each nn to k,++x using a product with full supports of

Levy collapses. Here we are interested in a very strong version of the Free Subset Property. A

set X is said to be approachable if X ? range(/7v) for some approachable TV -<

(H(?)\ e,<) with 0 sufficiently big. The Approachable Free Subset Property for a

singular cardinal k is the modification of the Free Subset Property that demands

that for every internally approachable N of type smaller than k, there is a subset X

of range(xvO such that X is cofinal in k with order type cf (?) and for every x e X,

x?Sk(NUX\{x}). As a consequence of our main results we have that the existence of large ideals does not imply the Approachable Free Subset Property. We will come back to this at the

end of this section.

2.2. Some PCF theory. Fix a a set of regular cardinals such that |pcf(?)| <

min(a). Recall that there is a family fc:lG pcf(a)) of subsets of pcf(a) which are open for the topology defined by letting c C pcf(a) be closed iffpcf(c)

= c (we can even have clopen b?s, but this will not be necessary). These bx also have the

property that X = max(?^) and if p e bx then b^ ? bx. With this we can define a




well-founded partial order by // ̂ X iff ?i G bx. The rank of an element X for the relation r< is equal to the rank of bx in ({bM : ?i G pcf(a)}, C). We call the height of -< a Bonnet rank of the space.

The PCF conjecture says that |pcf(a)| =

\a\ for all a as above. The property of PCF spaces that the closure of an interval is also an interval has as a consequence that any Bonnet rank of a space that contradicts the PCF conjecture for intervals is large (see the proof in [25, Proposition 2.7.(4)(5)], where a stronger version with the Cantor-Bendixson rank of the following result is attributed to Foreman and

Magidor; in [31, Claim 2.7] this is sketched also):

Lemma 1. If a is an interval and |pcf(a)| > \a\ then any Bonnet rank of pe? (a) is> \a\

+ .

With this, Shelah proved in [31, Claim 2.3.1] (in fact Shelah only needs a very

general notion of rank) that:

Lemma 2. If a Bonnet rank of pe? (a) is> \a\+ then there is an increasing sequence

(jun : n < co) of elements of a with the property that for every I < k < co, there is a bx such that{jui,jui+i,...,juk} nb?

= {jui+u

... ,juk}.

One way to define the bxs is the following: we take an internally approachable TV that contains pe? (a) of type //, with |pcf(a)| < ju < min(a), and consider

bx:={^:xN(^)eSk(NU{xN(X)})}. Notice that this definition depends on the internally approachable TV. If we had

demanded that TV contains [TV]lpcf(^l then the ?/s would also be closed.

Applying the above Lemmas we have that:

Corollary 1. If a is an interval of regular cardinals with |pcf(a)| > \a\, then, for every internally approachable TV of sufficiently big type that contains pcf (a), there is a sequence (xn(Mh) : n < co) with the property that for every I and I < n\ < < nm

Xn(jUi) $ Sk(N U {xNiVm), ,Ztf CO})} Proof. Choose a k such that n\,..., nm are among / + 1,..., k. By the previous

Lemma there is a X such that Xn(mi) *s not *n Sk(TV U {xn(X)})} and yet this model contains Sk(TV U {xn(ju>i+i), ^XN^k)})}- This last model itself contains

Sk(TV u tevCO,..., xAvnJ})}. H

Remark. One should point out that if ju is regular and v < ju then xn'(h) =

Xn(ju) where Nf = Sk(TV U {xn(v)})- The proof is simple but instructive, see for

example [13, Lemma 4.4.3]: Sk(TV U v) = {/[v]: / G TV} and since for every / G TV, TV sees sup(/[v] n ?j), we have that XN'(ft)

= Xn(ju).

Consequently, if F : H< ?? H is an algebra and if F G TV, we have that the set

(XN(jUn) : n < co) is free with respect to F. In particular, if ?( is a strong limit and 2Hw > fyyj then ?( has the Free Subset Property. This is not so surprising if we

consider the fact that the negation of the Singular Cardinal Hypothesis implies that there are many measurable cardinals in some inner models and that in turn some of the //-complete ultrafilter s in the inner models should generate ideals with the Laver

Property. What is indeed remarkable is that here the free subsets are subsequences of characteristic functions, which a priori seems much more difficult to achieve and we even conjecture that it is impossible.



678 LUIS PEREIRA

To finish this subsection we mention that for a finite number of regular cardinals

k\ < < Km it is always possible to build an internally approachable TV of

type p for which xn(k\)i , XN^m) are free, assuming minor cardinal arithmetic

hypotheses. Proposition 1. Let k\ < < Km be regular cardinals such that for every 1 <

i < m and every cardinal v < ?;, we have v<? < Kj, then there is an internally

approachable N of type pfor which xn(k\), , XN^m) are free. Proof. (A similar argument can be found in [7].) Consider a model H(&) such

thati/(0) e H(e). We now consider a model Mm -< (H(?); e, <', {(H (9); e, <)}) built by recursion on the following way: Mm$ is any elementary submodel of

(H(&)\ e, <', {(H(6); e, <)}) such that Km-\ ? Mm0 and \Mm$\ < Km\ for ? < p limit we let Mm? be the union of the Mm^'s with n < ?. At successor steps we

demand that Mm?+l is the Skolem Hull in (if(0); e, <', {(H(6); e, <)}) of

Mm? U {(Mm.n : n < ?)} U [Mm?]<fl U (M^ H Km).

Finally, Mw = Mm^. Notice that [Mw]</? ? Mm.

We build Mm_i in the same way but we demand instead that Mm-\$ contains

Km-2 ? {Mm, (Mm? : t < p)} and |MOT_i5o| < Km-\\ at successor steps we demand that Mm-\?+\ is the Skolem Hull of

Mm_u U {(Mw_i,, : >7 < 0} U [Mm_up U (Mw_u n ?w_i).

We set, Mm_i =

Mm_M. We proceed in this manner until we get to M\ which contains p U {M2, (M2? :

? < //),..., Mw, (Mm^ : c < //)}. We construct TV in the following way: Nq =

Sk^(?) (0), at limit steps we take the union of the previous submodels and at successor

steps t + 1,

7V?-+1 =

SkH{e)(N? U {(Nr, : n < ?)} U {sup(Muf n /q),..., svp(Mm? n Km)}).

He have TV := N? ? Mi n n Mm by definability and the fact that all the

M/'s are closed under < //-sequences. Also, for all Mz's, xn(k>?) = M- n /^. It

is now evident that the ordinals xn(k\), , XN^m) are free because Sk^^TV U

tevfo+i)> - - -,Xn(^?i)}) Q Mi for every i. H

2.3. The structure of the proof. To explain the argument of the proof of our main results we need to recall a bit more of PCF theory.

Let FIN be the ideal of finite subsets of co. Given a sequence of regular cardinals

(pn : n e co), the product Y[n ?n nas true cofinality X modulo FIN for X a regular cardinal, if there is a sequence /

= (fa : a < X) contained in Yln jun which is

increasing modulo FIN, that is, if a < ? < X then fa <FIN f?, i.e, the set

{n : fa(n) > f?(n)} is finite; and for every f e]\npn there is a < X such that

/ <fin fa- The sequence / is called a scale. A function / is called an exact upper bound of a <jp/v-increasing sequence of

ordinal functions (fa : a < ?) if fa <FIN f for all a < S and for every function

g <fin f, there is an a < ? such that g <FIN fa. If ? > 0 and if an exact

upper bound exists then it is not difficult to see that it is unique modulo FIN. An exact upper bound always exists if cf(<5) > 2N? (see [15, Lemma 24.10] for a quick proof) and a sufficient condition for a function / to be an exact upper bound when ? has uncountable cofinality is the existence of a strictly increasing sequence




(a? : c? < ei(?)) o? supremum ? and such that there is a no such that for all n bigger than fl0 the sequence (fa?M

' ? < cf(<5)) is strictly increasing and has supremum

f(n). This fact will be used later. A scale /

= (fa : a < X) is continuous at a limit ordinal S < X if whenever there

exists an exact upper bound of (fa : a < ?) then f? is such an exact upper bound. A scale is continuous at cofinality ?i (continuous) if it is continuous at every limit ordinal of cofinality ?u (all ordinals of uncountable cofinality).

If / is a continuous scale at cofinality ?i and TV is an internally approachable submodel of type ju with / G TV and ?u < juq then we have the following crucial

property (see [3, Theorem 5.2], which even uses a weaker hypothesis)

(xn(Mh) : n<co) =FIN fXN(x)

Now we come to the most important concept of this paper:

Definition 1. A scale / =

(fa : a < X) is tree-like if for every n < co and a < ? < X, if fa(n) =

f?(n) then fa \ n = f? \ n.

If there is a scale then it is an easy observation that there is also a tree-like scale

(see also II.3.5 of [29] for a more general statement). Here we are interested in tree likeness in conjunction with continuity. Let v be the supremum of the /z?'s. If there exists a continuous tree-like scale / then we can define a function F : v ?> v by: if

Mn-i <?<jun and there is an a < X such that fa(n) = ? then F(S)

= fa(n

? 1);

F(?) = anything, otherwise.

Crucial Remark. If / and F belong to an internally approachable TV, we will have that there is an no such that for every n > no,

F(xn(Mh)) =

Z;v(/^-i).

In particular, the set {xn(^h) ' n G co} is not free. In the next section we will

show how to force the existence of continuous tree-like scales without changing the universe too much. We construct in particular a universe with (kh: n G co) an infinite increasing sequence of cardinals such that Vx \= uKn is supercompact", where X is the supremum of the Kn's, and there is a continuous tree-like scale in

n? &n, where (an : n G co) is an enumeration of the set {nn, n^ : n G co}.

Discussion. We now come back to the subject of large ideals and the Approach able Free Subset Property. We collapse the < X-supercompact k/s in the model referred in the last paragraph to the ̂ 2?'s, with n > no for some even ?0 ? &>> using the product with full supports P =

Yln< ?? ?ftne Posets Qo = Coll(H2?0+i > < Ko)> where ^2?0+i ^s anY cardinal that we wish to be above the continuum; Qn

=

Co\\(n^+n_x, < Kn[)+n) for n > 0. Then, remembering what we said in subsec

tion 2.1., there will be large ideals in the H2w's. Since P is cr-closed the Cohen extension has the same co-sequences than the ground model, and so the continuous tree-like scale remains a continuous tree-like scale. In particular, the Approachable Free Subset Property is false.

Notice that since Vx N "Kn is supercompact", we have a great freedom in choosing the Kn -complete ultrafilters that we want to use in the Cohen extension, in particular, the arguments of [23] are still valid.



680 LUIS PEREIRA

?3. The main results. Kunen in [19] proved that there is no embedding j : Vx+2 ?>

Vx+2 if we assume the existence of a well-ordering of Vx+\. The large cardinal axiom II (see also [16]) says that there is an embedding j: Vx+\ ?> Vx+\ and it is the strongest axiom that doesn't involve questions about the definability of a

well-ordering of Vx+\. It is natural to look at the sequence of cardinals (jn(K): n e co) where k is

the critical point of j. Let Kn = ju(k), then X = supw Kn. We will consider two

large cardinal hypotheses: II and the (weaker hypothesis of the) existence of an co

sequence of cardinals, which we will also denote by (?? : n e co) and its supremum by X, such that Vx 1= "?? is supercompact", for each n . The proof will be the same in both cases, we will first force an unary algebra F on X and then force a continuous tree-like scale / whose associated algebra is F.

The basic idea to build F is to add functions F : Kn+\ \ nn ? Kn with Cohen conditions and then to put them all together. Since we want to preserve large cardinals, we will build our model by first using a reverse Easton Iteration and the large cardinals will be preserved by standard arguments involving this kind of iteration (we give the references below).

Our basic tool will be the posets defined as follows:

Definition 2. g (a) is the set of g's such that q is a function from an ordinal

? + 1, with/? < a, into

Y\{y: 7 < a and 7 is regular uncountable}.

We suppose also that the empty set belongs to Q(a). The ordering is by reverse inclusion.

So, we are just adding Cohen functions with domain a and the ranges are varying along the regular uncountable cardinals below a. For simplicity we also assume that if ? < y < a then q(?)(y)

? 0, in particular, q(0)(y) is always 0. The length of a condition q is the ordinal ?. In case q is the empty set, we say that its length is ?1. Notice that Q(a) is < cf(a)-directed closed and also notice that the sets

Dy?? =

{q I there is n > ? in the domain of q such that q(n)(y) =

?},

are dense, for y < a, ? < y and ? < a. This will be used below.

Definition 3. The poset P is defined by recursion:

Qa is a full Pa-name for the Q(a) of VPa when a is a regular cardinal, otherwise it is the trivial poset. At limit stages a we take a direct limit if a is inaccessible and an inverse limit otherwise.

WesetP^ZV

Px has obvious closure properties and this together with the Factor Lemma (see for example [15]) and the uniformity of its definition imply that it preserves II

by [12, Lemma 5.2 and Theorem 5.3], because we will have j(Py) =

Pj(y) for all y < X. Px will preserve the < A-supercompactness of the ?w's by the proof of [22, Theorem 18]. We would like that in V[G] there were no infinite approachable free subsets,

but we do not know how to do this. Instead, we will now describe how to force over V[G] a continuous tree-like scale on the product of an arbitrary sequence of




regular cardinals with supremum X, with a ?-distributive poset. In section 3 of [4] the authors use a similar argument to force a squared scale (they called it Coherent

squares). We will identify G with the function \J{p'. p G G}. Fix a strictly increasing

sequence of regular cardinals ? ? (an : n < co) with supremum X. For an < ? <

an+\ let

Fa(?):=G(an+i)(?)(an).

In the following we omit the superscript ? and just write F for Fa.

Definition 4. R(an : n < co) is the set of q's such that:

(1) q is a function from an ordinal ? + 1, with ? < X+, into Yln< an;

(2) q is ̂/Tv-increasing and is such that for every y < ? + 1 of uncountable

cofinality, q(y) is an exact upper bound of the the set {q(y') : y' < y}', (3) finally, for every y < ? 4- 1 and n G co,

F(q(y)(n + l)) = q(y)(n). We suppose also that the empty set belongs to R(an : n < co). The ordering is by

reverse inclusion.

The length of a condition q is the ordinal ?. In case q is the empty set, we say that its length is -1. R =

R(?) is a cr-closed poset and it forces a continuous tree-like scale in Yln< a" of length X+. To see that the sequence of functions is cofinal, notice that by the density of the sets Dy?? referred above, the set F~l(?) is unbounded in an+2, for every an < ? < an+\. At successor steps we don't have to fulfill any continuity requirements, so as long has q(c\ + 1) is modulo FIN bigger than q(?) we

have freedom in choosing q(a, + 1 )(0) and then inductively choosing q(c;-\-\)(n + l) in F~l (q(tl + 1 )(?)). This means that given a condition q and a function / : co ?> X in V[G] we can construct a q' < q such that ?/'(length^) + 1) >fin f As R is

cr-closed, V[G] and F[G][//] have the same co-sequences. We will now describe how to prove that R is ?-distributive. Since X is singular

we only have to prove that it is < ?-distributive. The structure of the proof is the

following: We will use the Factor Lemma (see for example [15]) which says that, for every

y < X, we can factor the poset PA and the poset Px * R = Px+\, where R is a

iYname for the poset R(?) in Vp\ into Px = Py *

PJy) and Px+\ = Py *

P?\ respectively, where P/ , P/^

are Py -names. Moreover, consider isomorphisms

iy : Px ?> Py * i?

. With these we can define functions iy : VP/ ?> Vp*p> such that for all p e Px, for all formulas (??(xi,..., xn), and iVnames fi,..., i?, we have:

p Ihp. <p(ii,...,f?)iff/y(/?) \\-p^p<?) ip(iy(ii),...,iy(in)).

Similarly, there are functions jy : Vp*p? ?> (F^)^ such that for all (/?,^) G

Py *

PJ , for all formulas <^(xi,..., xn), and P7

* P[y

-names fi,..., zn, we have:

(P?) lhp.,*>(;) v(Ti,...,Tw)iff/7lhp., (^ Ih^,, <p(j(i\),...,j(in))).

Recall that a poset P is < v-strategically closed if for every a < v the Good

player has a winning strategy in the following game: Bad starts the game by playing



682 LU?S PEREIRA

a condition po in P, Good responds by playing a condition p\ below po and henceforth both players must play conditions below the ones that have appeared in the previous moves. Good plays at limit steps and he wins the game if he can

play until the a-th move. If v is a cardinal and P is < v-strategically closed then P is < v-distributive.

In the following, we will prove that for every n < co,

1 lhpa "PJ+i is < an -strategically closed".

In particular,

1 \\-Pan "P^l is < an-distributive".

Therefore, since

ll^?"JPi:"1,=â")*ia?(^(JR))", we have

1 \\-Pan "P^ * jan(Ian(R)) is < ^-distributive".

Then,

1 lhpQ/7 "? \\-p{a?) jan(ian(R)) is < a?-distributive".

Hence,

1 ll~p ?(a?) uian(R) is < a?-distributive".

Finally,

1 \\-p. "R is < an-distributive".

Since the supremum of the a?'s is X we will have that R is < /-distributive, and therefore /-distributive, over V[G].

Let Gm be an arbitrary generic object for Pa and work in V[Gno]. Consider

(^;+'? )G,? an<i iet us name it Px+\- There is no confusion as long as we remember

this poset starts with Q(ano). For shortness of proofs we identify this Px+i with the

poset given by the iteration of trivial posets up to a?0 followed by (P??^? )Gn?. The condition on a trivial poset is said to have length 0. Before proving that Px+\ is < ano-strategically closed in V[GnQ] and following the terminology of [4] we define:

Definition 5. A condition p is said to be flat if there are ordinals ?m, /??0+i and an ordinal ?A such that for every n > no?.

(1) p \ an I h length(/?(aj) - ?n, (2) p \ X Ih length(^?)) - h (3) p\Xhp(X)(??)(n)=?n.

Since Px+i is o -closed we have that:

Lemma 3. The set of flat conditions in Px+\ is dense.

Proof. Let p be an arbitrary element of PA+\. Let us see how to build a flat condition below p.




(1) Let r"o G P;l5 r"o < p \ X, be such that r0 decides the length of p(X), that is, there is ?x such that

r0 lh length^)) = ?x and r"o decides the values p(X)(?x)(n), that is, there are ?x,n such that

rohp(X)(??)(n) = ?x.n

Weputro = r0 ̂ {p(X)).

(2) Now, set riso = ro \ ̂o- By recursion on 0 < n < co, take F\n G Pa?, such that r\_n < ro \ an, in case n > no we demand r\_n < r"L?_i

^ (?,... ); we also

demand that F\M decides the length of ro(an), that is, there is ?\>n such that

rLn lhlength(r0(a?)) =

?\,n.

Let r\ G PA be such that for every n > no, h \ an is smaller than all the ru \ an with / > n and most importantly,

n f a? lh length(ri(a?)) > /?;.,?.

We can do this since every Pan is cr-closed and we take an inverse limit at X. We put r\=h~ (pti)).

(3) Now, by recursion on m < co, we set, again, rm+\$ = rm \ c?o and by recursion in n < co, we take rm+\ M G Pa?, such that rm+i? < rm \ an, in case n > no we also

demand rm+i,? < rm+i,?_i ̂ (h ) and fw+i,? decides the length of rm(an), that

is, there is ?m+\.n such that

rm+in lh length(rm(a?)) = ?m+hn.

Let rm+i be such that for every n > no, rm+\ \ an is smaller than all the rm+\j \ an with I > n and

rm+\ \ an lh length(rm+i(a?)) > ?m+i,n.

We put rm+i = fm+\ - (/??)). Let r be constructed by recursion: if a < X and a ^ cxn for all ? > no we take

any name r (a) such that

r T ck lh r(a) contains all the rm(a)'s.

In case that for all m, rm(a) = 1, we let r(a) ? 1. If a = an and n > no we also

demand that the name r (a?) is such that

r \ a?\\- length(r(a?)) =

sup{/?mj7 : m < co}, and

r \ an lh r(a?)(sup{jffw,? : m < co})(?n-\) =

sup{)8w,?_i : m < co},

it exists by fullness. Now, just take a name for a condition r(X) such that

r lh length(r?)) = ?k + 1 and F lh r(?)(/?? + \)(n)

= sup{/?m.? : m < co}. H

We are now in position to prove that PA+\ is < am-strategically closed. With what we said above and remembering that G?0 was arbitrary, this will finish our

proof of the main results.



684 LU?S PEREIRA

Proof of strategic closure: We describe the Good player's strategy: If Bad played a condition p^ let p'^+l be any flat condition such that

p^+l < p?.

Consider the ordinals ?an?+\ and ?x,c+\ such that

p'?+l \ an Ih length(/??(aw))

= fian?+\ and

p't+l r A Ih length(Û)) =

?.i+1

To define Good's play p^+\ < p^+l

we just take by recursion names p^+\ (an) and

Pz+i (X) for conditions such that for every n > no:

/?<*+! \ an Ih pz+\(oLn) 2 P'^n) andlength(^+i(a?)) =

j?an?+\ + co,

Pt+i \ X Ih pt+\(X) D p'?X) andlength(^+iU)) = ?x^+\ + co

and for every m > 1 :

p?+i \ X Ih Pz+\(X)(?x?+\ + rn)(n) = /L??+i + m,

P?+\ \ ?/i-hi II" ^+i(^+i)(â/7+1,e+i + m)(an) =

f}an?+\ + m,

if n < no we set:

Pt+l r^lKÛ)(?,i+i+u>)(w) = 0.

In particular, we demand for 1 < j < co:

Pc+i(ano)(?ano4+i +7')(ô-i) = 0

With ? limit we define Good's play p% by recursion, so that for every y < X,

Pi X y < Prj ? 7, for every n < ?, and we take names p?(an) for n > no and p?(X) which satisfy:

Pc \ an Ih length(/7^(a?)) =

sup{/?a,7^ : n < ?} =: /?a??,

p? \ X I h length^ U)) - sup{?^ : // < ?} =: &?

and

/?? f a?+i Ih ^(aw+i)(jffan+1^)(?w) =

?a??

finally, we demand for n>no'.

Pt \x^p?(x)(hdM = hu and for n < no'.

Pi \X\^Pi(X)(?xt)(n) = 0.

In particular,

Pc(ano)(?ano4)(ano-i) =0. H

To obtain Theorems 1 and 2 we just mention that if, in V[G], we force with

R(K,n: n e co) then in the extension K[G][#] our large cardinal hypothesis are

obviously preserved because Vx+\ is unchanged. Moreover, there is a continuous tree-like scale in Y[n &n which implies the non-existence of infinite free subsets which are subsets of (xn(^h) : n e co) for any internally approachable submodel N.

To conclude this section we mention that if we work with (?? : n e co) an co

sequence of < A-supercompact cardinals and in V[G] we force with R(an : n e co)




where (an: n G co) is an enumeration of the set {??,?+: n G co} then in the extension F[G][/T], Vx N uKn is supercompact", and there is a continuous tree like scale in Yln an. This gives the model to which we alluded at the end of sec tion 2.

?4. Remarks and open problems. We start by noticing that the iteration that leads to P need not have stopped at X. If we were working with an co-sequence of fully supercompact cardinals instead of just < ?-supercompact then we would even need that the iteration ran over all the ordinals, that is, P would have to be a class forcing. There is no problem in doing that as shown in [22] and in fact, it is that proof that

gives that < ?-supercompact cardinals are preserved in Vx. Now suppose there was a Mahlo cardinal 9 above X and we work on a model

(H(6);e,<,Pe), where we added Pq as a predicate. Notice that H (9) = Ve and

Vq^ =

Vo[Gq], for any generic G. Moreover, Pq is 9-ee and so we have fullness in H(9), hence, if TV -< (H(9)',e,<,Pe), then N[Ge] -< H(9)[G9]

= H(9)V^G\

Building on the proof of the existence of flat conditions we prove that:

Lemma 4. For every condition p G Pq and every regular fi such that ju^? = ju, there

are an extension q of p and a name TV for an internally approachable submodel of H (9) such that q forces that the functions Fa for ? G TV, capture the characteristic

function of? below X.

Proof. The proof is very similar to the proof of distributivity of the R(?)'s and the result may be seen as generalization of the result on the existence of flat conditions. We will avoid unnecessary repetition of details from those proofs.

The name TV will just be a name for TV[G#] where TV is an internally approachable submodel o? H(9) of type ju. We construct TV and q by recursion.

Letpo = p and TVo any submodel of size < ?i with ?i+ G TVo. TV^+i will always be an

arbitrary submodel that contains p%, (Nn : n < ?) and is closed under co-sequences. All the work will be done in the construction of p?+\.

First construct by recursion a decreasing sequence p?a where a runs through the regular cardinals bigger than ju and smaller than X in TV<* with the property that

for every such a, pâ \ a decides the length of p?(c?), that is, there is ?â such that

pLa \ a lh length^(a)) =

?â.

Moreover, we suppose that p^?+ decides the values of all names x in TV<* for sets in V. Since Pq is //-closed there is no problem doing this. Now, we build p%+\ by recursion: p%+\ is stronger than all the pâ and moreover for every a, p^+\(a) is a name for a condition in Qa of length bigger than Xn?(&) and ?â. Again, there is no problem doing this because we are using Easton supports and we started the iteration above//. At limit steps c; < ju we take TV^ to be the union of the previous ones and for

?? < ju, p? is any condition that is stronger than the previous ones. Notice that

Xn? (a) is equal to the sup of the ?nâ where n < c; and a G N?. When t; = ju we set TV = TV^ and we take a special p? which will also be our q:



686 LUIS pereira

We build Pu by recursion taking for every a e N, a name p?(a) for a condition

in Qa that extends the previous pv(a) and for every regular y in TV above p

Pfi \ ex Ih Pv(a)(xN(<*))($) = XN(y).

Since q Ih N[Ge] n K = TV, we have the result. H

As an immediate Corollary we have:

Corollary 2. The Approachable Free Subset Property for X is false in V[G].

As we saw in section 2, continuous tree-like scales guarantee us the existence of set

mappings without infinite approachable free subsets. The question of the existence, or not, of such scales in /-distributive extensions seems to us to be an important

question in current PCF theory, because the methods used to study this question might give us as a corollary a proof of the impossibility of the Approachable Free

Subset Property and consequently a proof of the PCF conjecture for intervals. In L there exist tree-like continuous scales, even tree-like squared scales (see [5],

the tree-like property comes just from condensation). In [9] Moti Gitik proved that

it is possible to build a model without tree-like continuous scales using the Extender based forcing of [10, 11], but it is not known if there exist infinite approachable sets in these models. So the question of the consistency of the Approachable Free

Subset Property still persists. To obtain continuous tree-like scales in a scenario without SCH, there is another

approach more in the spirit of the classical Prikry-Silver model. Suppose n is a

measurable cardinal with 2K = X = k,+?+1 where ? < k and there is a normal

ultrafilter U over k such that the ultrapower by U correctly identifies all cardinals

up to X. We force a Prikry sequence (nn : n e co) for U. Jech in [14] proved that the functions g : k ?? k in the ground model that represent ordinals in the

ultrapower below juM (the canonical functions) can be used to define the scales in the Cohen extension by defining functions f e]\n nn+\ by f(n) := g(nn) (we have g(nn) < Kn+\ eventually, before this point we just set f(n)

? 0). These scales

are continuous at points of cofinality smaller than k. For every a < ?, the scale in

Yin Kna+l is given by the canonical functions for ordinals less than n+a+l. So the

question about the existence of tree-like continuous scales in these models has an

easier answer if we prove that there exist tree-like canonical functions. If k is supercompact indestructible under < ^-directed closed forcing (see [20])

and instead of forcing X Cohen subsets we force a (n, X)-semimorass (see [18]) then

in the Cohen extension k is still supercompact because the forcing is < k -directed closed and we have tree-like canonical functions for ordinals below X [24]. This

canonical functions are even ordered by eventual dominance. So we reduced the

question to the canonical functions for ordinals between X and ju (k). Now, the ordinal juM is an ordinal between 2K ? X and X+, c?(ju(k)) is the true

cofinality of ]Jn Kn+\ and the canonical functions between X and juM are used to define the scale in Y\n nn+\. If there is a tree-like continuous scale in (kk, <ns) then there is one in Y[n ?w+i- Hans-Dieter Donder told us in an e-mail message that if there is a universal morass over k (an hypothesis that entails 2K = k+) then such a scale exists. A universal morass exists for example in the inner model

L[U] and Andrew Brooke-Taylor proved that the existence of an universal morass




is compatible with a supercompact cardinal [1]. The question of the existence of a tree-like continuous scale in (kk, <ns) when 2K > k++ is still open.

Even though for the moment we do not know how to capture characteristic functions when evaluated on the sequence (nn : n G co) we do know how to capture some of them when we collapse the cardinal k to H using Levy collapses. The basic idea already present in this section is that if we a force a regressive function with a

sufficiently closed forcing then we will always capture the characteristic functions of some approachable submodels. Here the regressive functions are given by the Levy collapses.

There are two ways to collapse a measurable cardinal k to Hffl: in the first one we start by forcing a Prikry sequence (nn : n < co) and then collapse each nn to a successor cardinal, say ?^, using a product with full supports. This kind of

collapse is simpler, but has the inconvenience that it forces SCH at H . In the second way of collapsing k to H we incorporate collapses in the Prikry conditions and make sure that the Prikry condition still holds.

In the author's Ph.D. thesis [24] it is proved that both these ways of collapsing add functions that capture characteristic functions.

The existence of scales similar to tree-like continuous scales is probably stronger than what we need for the negation of the Approachable Free Subset Property, it is not out of the question that we can prove this negation, and so the PCF conjecture for intervals, directly.

Acknowledgement. I wish to thank Stevo Todorcevic and the anonymous referee for their very useful remarks which improved the quality of this paper.

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The PCF Conjecture and Large Cardinals