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2013 by Institut Mittag-Leffler. All rights reserved

Forcing axioms and the continuum hypothesis.

Part II: transcending ω 1 -sequences of real numbers

by

Justin Tatch Moore

Cornell University Ithaca, NY, U.S.A.

Dedicated to Fennel, Laurel and Stephanie.

1. Introduction

In his seminal analysis of the cardinality of infinite sets, Cantor demonstrated a procedure which, given an ω-sequence of real numbers, produces a new real number which is not in the range of the sequence. He then posed the continuum problem, asking whether the cardinality of the real line wasℵ1or some larger cardinality. This problem was eventually resolved by the work of G¨odel and Cohen who proved that the equality |R|=ℵ1 was neither refutable nor provable, respectively, within the framework of ZFC. In particular, there is no procedure in ZFC which takes anω1-sequence of real numbers and produces a new real number not in the range of the original sequence.

The purpose of this article is to revisit this type of diagonalization and to demon- strate that models of the continuum hypothesis (CH) exhibit a strong form of incom- pactness. The primary motivation for the results in this paper is to answer a question of Shelah [9, Problem 2.18] concerning whichforcing axioms are consistent with the CH.

The forcing axiom for a classCof partial orders is the assertion that if Dis a collection of ℵ1-cofinal subsets of a partial order, then there is an upward directed set G which intersects each element ofD. The well-knownMartin’s axiom for ℵ1-dense sets (MA1) is the forcing axiom for the class of partial orders which satisfy the countable chain con- dition. Foreman, Magidor and Shelah have isolated the largest class of partial ordersC for which a forcing axiom is consistent with ZFC. Shelah has established certain suffi- cient conditions on a classCof partial orders in order for the forcing axiom forC to be consistent with CH. His question concerns to what extent his result was sharp. The main

The research of the author presented in this paper was supported by NSF grant DMS-0757507.

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result of the present article, especially when combined with those of [1], show that this is essentially the case.

The solution to Shelah’s problem has some features which are of independent in- terest. Suppose thatT is a collection of countable closed subsets ofω1 which is closed under taking closed initial segments and thatT has the following property:

(1) Wheneversand tare two elements ofT with the same supremumδand lim(s)∩lim(t)

is unbounded inδ, thens=t. (Here lim(s) is the set of limit points ofs.)

As usual, we regardT as a set-theoretic tree by declarings6t ifsis an initial part oft. Observe that this condition implies thatT contains at most one uncountable path:

any uncountable path would have a union which is a closed unbounded subset ofω1 and such subsets ofω1 must have an uncountable intersection. In fact it can be shown that this condition implies that

{(s, t)∈T2: (htT(s) = htT(t))∧(s6=t)}

can be decomposed into countably many antichains (T2 is special off the diagonal).

Jensen and Kunen, in unpublished work, have constructed examples of ω1-Baire trees which are special off the diagonal under the assumption of Jensen’s principle♦, but this article represents the first such construction from CH.

The main result of this article is to prove that if the continuum hypothesis is true, then there is a treeT which satisfies (1) together with the following additional properties:

(2) T has no uncountable path;

(3) for eacht∈T, there is a closed unbounded set ofδsuch thatt∪{δ} is inT; (4) T is proper as a forcing notion and remains so in any outer model with the same set of real numbers in which T has no uncountable path; moreover T is complete with respect to a simpleℵ1-completeness systemD, and in fact isℵ1-completely proper in the sense of [7];

(5) ifX is a countable subset ofT, then the collection of bounded chains which are contained inX is Borel as a subset of P(X).

This provides the first example of a consequence of the continuum hypothesis which justifies condition (b) in the next theorem.

Theorem1.1. ([8], [4]) Suppose that hPα;Qα:α∈θiis a countable support iteration of proper forcings which

(a) are complete with respect to a simple 2-completeness system D; (b) satisfy either of the following conditions:

(i) are weakly α-proper for every α∈ω1;

(ii) are proper in every proper forcing extension with the same set of real numbers.

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Then forcing with Pθ does not introduce new real numbers.

Results of Devlin and Shelah [2] have long been known to imply that condition (a) is necessary in this theorem. The degree of necessity of condition (b)—and (i) in particular—has long been a source of mystery, however. Shelah has already shown [8,

§XVIII.1] that there is an iteration of forcings in L which satisfies (a) and which in- troduces a new real at limit stage ω2. This construction, however, has two serious limitations: it is not possible in the presence of modest large cardinal hypotheses—such as the existence of a measurable cardinal—and it does not refute the consistency of the correspondingforcing axiom with the continuum hypothesis. Still, this construction is the starting point for the results in the present article.

Properties (3) and (4) of T imply that forcing with T adds an uncountable path throughT and does not introduce new real numbers. Thus T can have an uncountable branch in an outer model with the same real numbers. By condition (1), this branch is in a sense unique, in that once such an uncountable branch exists, there can never be another. (Of course different forcing extensions will have different cofinal branches throughT and in any outer model in which ωV1 is countable, there are a continuum of paths throughT whose union is cofinal inω1V.)

Property (5) of the tree T is notable because it causes a number of related but formally different formulations of completeness to be identified [7]. This is significant because it was demonstrated in [1] that in general different notions of completeness can give rise to incompatible iteration theorems and incompatible forcing axioms.

The existence of the tree T can be seen as the obstruction to a diagonalization procedure forω1-sequences of reals. For each ω1-sequence of realsr, there will be an associated sequenceT!rof trees of length at mostω2 which satisfy (1). These sequences are such that they have positive length exactly whenω1L[r]V1, and ifξis less than the length of the sequence, then

(6) Tξrhas an uncountable path if and only if it is not the last entry in the sequence;

(7) if ξ0∈ξ, then every element ofTξ is contained in the set of limit points of Eξ0, whereEξ0 is the union of the uncountable path throughTξr0;

(8) if Tξr is the last entry in the sequence, then eitherL[r] contains a real number not in the range ofr, or elseTξris completely proper in every outer model of L[r] with the same real numbers;

(9) if the length of the sequence isω2andδis in the intersection ofT

ξ∈ω2Eξ, then hEξ∩δ:ξ∈ω2iis not inL[r] (in particular,ris not an enumeration ofR).

Moreover the (partial) function r7!T!ξris Σ1-definable for eachξ∈ω2. Thus in any outer model, the sequenceT!r may increase in length, but it maintains the entries from the inner model. Observe that ifris an enumeration ofRin order-typeω1, thenT!rhas

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successor lengthηr+1.

Shelah has proved (unpublished) that an iteration of completely proper forcings of length less thanω2does not add new real numbers. Thus, ifris a well ordering ofRin typeω1 andξ∈ω2, then it is possible to go into a forcing extension with the same reals such thatξ6ηr. The above remarks show that this result is optimal. Also, for a fixed ξ∈ω2, the assertion that there is an enumeration rofRin typeω1 withηr at leastξ is expressible by a Σ21-formula. By Woodin’s Σ21-absoluteness theorem (see [6]), this means that, in the presence of a measurable Woodin cardinal, CH implies that for eachξ∈ω2 there is an enumerationrofRin typeω1 such thatηr is at leastξ.

While the construction of the sequence of trees is elementary, the reader is assumed to have a solid background in set theory at the level of [5]. Knowledge of proper forcing will be required at some points, although the necessary background will be reviewed for the reader’s convenience; further reading can be found in [3], [4] and [8]. Notation is standard and will generally follow that of [5].

2. Background on complete properness

In this section we will review some definitions associated with proper forcing, culminating in the definition of complete properness. This is not necessary to understand the main construction; it is only necessary in order to understand the verification of (4). Aforcing notion is a partial order with a least element. Elements of a forcing notion are often referred to asconditions. Ifpandqare conditions in a forcing, thenp6qis often read “q extends p” or “q is stronger thanp.” If two conditions have a common extension, they arecompatible; otherwise they areincompatible.

LetH(θ) denote the collection of all sets of hereditary cardinality at most θ. If Q is a forcing notion inH(θ), then a countable elementary submodelM ofH(θ) issuitable forQif it contains the power set ofQ. IfM is a suitable model forQand ¯qis inQ, then we say that ¯qis (M, Q)-generic if whenever ¯q6randD⊆Qis a dense subset ofQin M, r is compatible with some element ofD∩M. A forcing notionQis proper if whenever M is suitable forQandqis inQ∩M, there is an extension ofqwhich is (M, Q)-generic.

IfM andNare sets, then−−→

M N will denote a tuple (M, N, ε), whereεis an elementary embeddingεof (M,∈) into (N,∈) such that the range ofεis a countable element ofN. Such a tuple will be referred to as an arrow. We will write M!N to mean −−→

M N and also to indicate “−−→

M N is an arrow”. IfM!N andX denotes an element ofM, thenXN will be used to denoteε(X), whereεis the embedding corresponding toM!N.

IfQ is a forcing, M is suitable for Q and M!N, then a filter G⊆Q∩M is −−→

M N- prebounded if, wheneverN!P and the imageG0 ofGunder the composite embeddings

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is inP,P satisfies “G0 has a lower bound”. IfQis a forcing notion, then a collection of embeddingsM!Ni,i∈I, will be referred to as aQ-diagram. A forcingQisλ-completely proper if, wheneverM!Ni,i∈γ, is aQ-diagram forγ∈1+λ, there is an (M, Q)-generic filterG⊆Q∩M which is−−→

M Ni-prebounded for all i∈γ.

Observe that, asλincreases,λ-complete properness becomes a weaker condition. In [7] it was shown that λ-complete properness implies D-completeness with respect to a simpleλ-completeness system D in the sense of Shelah (see [7] for undefined notions).

Moreover, the converse is true for forcing notions Qwith the property that, whenever Q0⊆Qis countable,{G⊆Q0:Ghas a lower bound inQ} is Borel [7].

3. The construction

Assume CH and fix a bijection ind:H(ω1)!ω1such that, ifxis a countable subset ofω1, then sup(x)∈ind(x). Ifxandy are inH(ω1), we will abuse notation and write ind(x, y) for ind((x, y)). Fix a Σ1-definable map ξ7!ξ which maps the countable limit ordinals intoω1 such that, if %∈ω1, then {ξ∈lim(ω1):ξ=%} is uncountable (for instance define ξ to be the unique ordinal%such that, for someς,ξ=ως+ω·%andω·%∈ως).

If δ is a limit ordinal, letCδ denote the cofinal subset of δ of order-type ω which minimizes ind; setCα+1={α}. Leteβ:β!ω be defined byeβ(α)= ¯%1(α, β), where ¯%1 is defined fromhCα:α∈ω1ias in [10]. In what follows, we will only need that the sequence heβ:β∈ω1iis determined by ind,eβ:β!ω is an injection, and ifβ∈β0∈ω1, then

{α∈β:eβ(α)6=eβ0(α)}

is finite.

LetE⊆ω1be a club. DefineT=TE=TEindto be alltwhich are countable closed sets consisting of limit points ofE such that the following properties hold:

• if ν is a limit point of t, then t∩ν has finite intersection with every ladder inν which either has index less than ind(E∩ν) or else is Cν;

• ifν is a limit point of t, then min(E\ν)∈ind(t∩ν);

• for allα∈β, ind((t∩(β+1))\α)∈min(t\(β+1)).

Define Te=TeEind⊆T by recursion. Begin by declaring ∅∈T. Now suppose that wee have definedTe∩P(ξ+1) for all ξ∈δ. Before defining Te∩P(δ+1), we need to specify a family of logical formulas which will be needed in the definition. We will useTeν to denote{t∈Te:ind(t)∈ν}. Letβ be a fixed ordinal less than δand let t∈Te be such that t⊆β. Consider the following recursively defined formulas about a closed subsetxofβ:

θ0δ(x, t, β): max(t)∈min(x),t∪xis in Teβ, and

otp(E∩min(x))= ind(t, n) for some n∈ω;

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θδ1(x, t, β): ifDis a dense subset ofTeν for some limit ordinalν∈β, ind(D)∈β and otp(E∩min(x))= ind(t, eδ(ind(D))),

thent∪xis in D.

θδ2(x, t, β): ify⊆β,eδ(min(y))∈eδ(min(x)) and θδ0∧θ1δ∧θδ2∧θδ3(y, t, β), thenx∩y⊆{min(x)}.

θδ3(x, t, β): ifs, z⊆β, min(z)=min(x) and θδ0∧θ1δ∧θδ2(z, s, β), then ind(x)6ind(z).

The truth ofθδi(x, t, β) is defined by recursion on the tuple (eδ(min(x)),ind(x), i)

equipped with the lexicographical ordering (δis fixed). Observe that, sinceeδis injective, there is at most one setD which satisfies the hypotheses of θ1δ(x, t, β). In particular, if θδ1(x, t, β) holds, thenθ1δ(x, t, β0) holds for allβ∈β0∈δ.

Now, if tis an element of T with sup(t)=δand δis not a limit point of t, definet to be in Te if and only if t∩δ is in Te. Ift is an element of T with sup(t)=δ and δ is a limit point oft, then we definet to be inTeif t∩α+1 is in Te for allα∈δ and if, for all but finitely many consecutive pairsα∈β in Cδ for which (α, β]∩tis non-empty,

θ0δ∧θδ1∧θδ2∧θ3δ(t∩(α, β], t∩α+1, β).

Lemma3.1. If E and E0 are clubs such that E0∩δ=E∩δ for some δ∈ω1,then TeE∩P(δ+1) =TeE0∩P(δ+1).

Proof. This follows from the observation that, under the hypotheses of the lemma, TE∩P(δ+1)=TE0∩P(δ+1) and the fact thatTeE∩P(δ+1) is defined by recursion from TE∩P(δ+1) andTeEδ.

Lemma3.2. If tand t0 are in Teand δ∈ω1 is a limit point of lim(t)∩lim(t0),then t∩δ=t0∩δ.

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Proof. Let t, t0∈Te and δ be a limit point of lim(t)∩lim(t0). It is sufficient to show thatt∩α=t0∩αfor cofinally manyα∈δ. Suppose that this is not the case. Letδ0∈δbe arbitrary andα∈β be consecutive elements ofCδ such thatδ0∈α, (α, β]∩lim(t)∩lim(t0) is non-empty, and such that

θiδ(t∩(α, β], t∩α+1, β) and θiδ(t0∩(α, β], t0∩α+1, β) hold fori=0,1,2,3. Without loss of generality, we may assume that

eδ(ind(t∩(α, β]))∈eδ(ind(t0∩(α, β])).

Define x=t0∩(α, β] and y=t∩(α, β]. Observe that, since x and y have common limit points, x∩y is in particular not contained in {min(x),min(y)}. Since θδ2(x, t∩α, β) is true, it follows that min(x)=min(y) and hence, byθδ0(y, t∩α+1, β) andθ0δ(x, t0∩α+1, β), thatt∩α=t0∩α.

Lemma 3.3. The tree

{(s, t)∈T2: (htT(s) = htT(t))∧(s6=t)}

is a union of countably many antichains.

Proof. Let U denote the set of those (s, t)∈T2 such that htT(s)=htT(t), s6=t and max(s)6max(t). By symmetry it is sufficient to prove that U is a countable union of antichains. For each (s, t) in U, define Osc(s, t) to be the set of all ξ>min(s4t) such that either

• ξ is insand min(t\ξ+1)∈min(s\ξ+1), or

• ξ is intand min(s\ξ+1)∈min(t\ξ+1).

Let osc(s, t) denote the order-type of Osc(s, t), observing that Lemma 3.2 implies that osc(s, t)∈ω2 for all (s, t) inU. If max(s)∈max(t), define

β(s, t) = min{β∈t: max(s)∈β} and n(s, t) =eβ(s,t)(max(s)),

and setn(s, t)=ωif max(s)=max(t). Observe that if (s, t) and (s0, t0) are inU ands<s0 andt<t0, then Osc(s, t) is an initial part of Osc(s0, t0) and that no element of Osc(s, t) is greater than max(s). Consequently, either min(s0\s) is in Osc(s0, t0)\Osc(s, t), and hence osc(s, t)6=osc(s0, t0) or else β(s, t)=β(s0, t0) and n(s, t)6=n(s0, t0). It follows that, for eachξ∈ω2andk∈ω+1,

{(s, t)∈U: (osc(s, t) =ξ)∧(n(s, t) =k)}

is an antichain. Sinceω2×(ω+1) is countable, this finishes the proof.

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Lemma3.4. Suppose that M is a countable elementary submodel of H((2ω1)+)with Te in M, and suppose that ti, i∈n, is a sequence of elements of Te such that there is a club C⊆ω1 in M such that C∩M⊆S

i∈nti. Then Tehas an uncountable chain.

Proof. LetM, ti, i∈n, and C be as in the statement of the lemma. By replacing C with its limit points if necessary, we may assume thatC∩M⊆S

i∈nlim(ti). Without loss of generalityti,i∈n, are such that ifi6=j, thenti∩M6=tj∩M. By Lemma3.2, there is aζ∈M∩ω1 such that if ν is a limit point of lim(ti)∩lim(tj)∩M, thenν∈ζ. Leti∈n be such that lim(ti)∩C0∩M is non-empty for every clubC0inM. Such aniexists since otherwise there would exist clubsCi0 inM for eachi∈nsuch thatC∩S

i∈nCi0 is empty.

LetN be a countable elementary submodel ofH(ω2) inM such thatCandζare in N andδ=N∩ω1is in lim(ti). Sinceδis not in lim(tj) forj6=i, there is aζ0∈δsuch that tj∩δ⊆ζ0 wheneverj6=i. LetC0 be the set of elements ofC which are greater thanζ0. If ν is inC0∩N, thenν is in lim(ti)\lim(tj) for eachj∈nwhich is different fromi. Define bto be the set ofp∈Tesuch that C0∩sup(p) is an infinite cofinal subset of lim(p). Since it is definable from parameters inN,b is inN. Furthermore,bis uncountable since it is not contained in N—it has ti∩(δ+1) as an element. By Lemma 3.2, b is a chain inTe and hence satisfies the conclusion of the lemma.

Lemma3.5. Suppose that E⊆ω1 is a club and TeE does not contain an uncountable chain. Then TeE is completely proper.

Proof. LetTedenote TeE. Suppose that M!Ni, i∈ω, is aTe-diagram and thatt0 is in Te∩M. Defineδ=M∩ω1 and let η∈ω1 be an upper bound forωN1i for eachi∈ω. In particular, ifX⊆M∩ω1 is inNi, then indNi(X)∈η. LetDn,n∈ω, enumerate the dense subsets ofTein M and letXn,n∈ω, enumerate the collection of all cofinal subsetsX of δ of order-type ω which are in Ni for some i∈ω. We will construct tn, αn and βn for eachn∈ω by induction, so that the following conditions hold for alln∈ω:

(10) tn+1extends tn and is inDn∩M; (11) αn∈βn∈αn+1∈δ;

(12) ifi∈nthentn+1\tn is contained inβnn and is disjoint fromXi ifi∈n;

(13) ifi∈ntheneNδi(ξ)=eδ(ξ) wheneverξ∈βnn;

(14) ifi∈n, then there are consecutive elements ¯α∈β¯ofCδNisuch that ¯α∈αn∈βn∈β¯; (15) there is a limit ordinalν such that max(tn+1)∈ν∈βn+1 and

otp(E∩min(tn+1\tn))= ind(tn, eδ(ind(Dν)));

(16) ifi∈n, then

Ni|=θ0δ∧θδ1∧θ2δ∧θδ3(tn+1\tn, tn, βn+1).

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Assuming that this can be accomplished, then define ¯t=S

n∈ωtn∪{δ}. It follows that, ifi∈ω andNi!Ne with ¯t∈Ne, then ¯tis inTNe. If ¯tis inTNe, then moreover we have arranged that

Ni|=θδ0∧θ1δ∧θδ2∧θδ3(¯t∩(α, β],¯t∩α+1, β)

wheneverα∈β are consecutive elements ofCδNi such that max(ti)∈α. In particular ¯t is inTeNe. Thustn,n∈ω, is−−→

M Ni-prebounded inTefor eachi∈ω.

Now suppose that tn is given. Let M0 be a countable elementary submodel of H((2ω1)+) such that

• M0 is inM;

• Dn is inM0;

• M0 is an increasing union of an∈-chain of elementary submodels ofH((2ω1)+).

Setβn+1=M0∩ω1 and letF⊆M0 be a finite set such that, ifi∈n, then CδNi∩M0⊆F, Xi∩M0⊆F and {ξ∈M0∩ω1:eNδi(ξ)6=eδ(ξ)} ⊆F.

Fix an elementary submodelM00 ofH((2ω1)+) such that F⊆M00∈M0. Set ν=M00∩ω1

and ζ=ind(Dn∩M00), and let αn+1∈ν be such that max(F)∈αn+1. Observe that, by elementarity,ζis inM0since it is definable fromν,Dn and ind. Furthermore,ν6ζsince otherwiseDn∩M00 would be inM00. Observe thateNδi(ζ)=eδ(ζ) for alli∈n. Letξ be a limit point ofE such thatαn+1∈ξ∈ν and

otp(E∩ξ)= ind(tn, eδ(ζ)).

Letyi,i∈l, list the closed subsetsy ofδsuch that

• for somej∈n, we havey⊆( ¯α,β], where ¯¯ α∈β¯are the consecutive elements ofCδNj with ¯α6αn+1∈βn+16β;¯

• eNδj(min(y))∈eNδj(ξ);

• θ0δ∧θδ1∧θδ2∧θ3δ(y, tn,β).¯

Observe that θ3δ ensures that there are only finitely many such y’s. Furthermore, Lemma 3.4 and our hypothesis implies that S

i∈lyi does not contain a set of the form C∩M00, whereC is a club inM00. Thus there is a countable elementary submodelM000 ofH(ω2) inM00such thatE,T,e Dn,ξand ind are inM000 andM000∩ω1is not inS

i∈lyi. Letηbe a limit point ofE which is inM000such thatξand max(S

i∈lyi) are less thanη.

SinceDn is dense, elementarity of M000 ensures that there is an extension of tn∪{ξ, η}

which is inDn∩M000. Such an extension ˜tnecessarily satisfies that ˜t\tn is disjoint from yi\{ξ}for alli∈l. We have therefore arranged thatθδ0∧θδ1∧θ2δ(˜t\tn, tn, βn+1) holds. Let tn+1 the element ofTebe such that

min(tn+1\tn) =ξ, θ0δ∧θδ1∧θ2δ(tn+1\tn, tn, βn+1) holds,

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and ind(tn+1\tn) is minimized. It follows thatθ3δ(tn+1\tn, tn, βn+1) holds. This finishes the inductive construction and, therefore, the proof.

Theorem3.6. AssumeCH. Then there is a clubEsuch that TeE has no uncountable branch. In particular, CH implies the negation of the forcing axiom for the class of completely proper forcing notions.

Proof. Suppose for contradiction that CH holds and TeE contains an uncountable branch for every club E⊆ω1. Let A⊆ω1 be such that ind is in L[A]. In particular, R⊆L[A] and ω11L[A]. Inductively construct clubs Eξ, ξ∈ω2, as follows. Since L[A]

satisfies♦, there is a functionh:ω11such that, ifE is a club inL[A], then for some limit pointδofE, there is a ladderX⊆δ withX∩E infinite and ind(X)∈h(δ). LetE0

be the <L[A]-least club in L[A] such that h(δ)∈min(E0\(δ+1)) for allδ. GivenEξ, let Eξ+1be the union of the branch throughTeEξ. IfEξ,ξ∈η, has been defined for a limitη less thanω2, defineEη=T

ξ∈ηEξ. Observe thatE2is not inL[A] since, wheneverδis a limit point ofE1,

h(δ)∈min(E0\δ+1)∈ind(E1∩δ),

and henceE2 has the property that wheneverδ is a limit point ofE2,E2∩δis disjoint from X wheneverX is a ladder in δ with index less than h(δ). Observe that, if ξ∈η, thenEη is contained in the limit points ofEξ. Also, if δis a limit point ofEη, then

min(Eξ\δ+1)∈ind(Eη∩δ)∈min(Eη\δ+1).

In particular, ifδ∈T

ξ∈ω2Eξ, then, for allk∈ω, sup

i∈ω

ind(Eω·k+i∩δ)∈Eξ

wheneverξ∈ω·(k+1).

Letδbe the least element ofT

ξ∈ω2Eξ. Observe thathEξ∩δ:ξ∈ω2iis inL[A]. We will obtain a contradiction once we show that hEξ:ξ∈ω2i is in L[A]. Working in L[A], defineνα andtα(ξ) for eachα∈ω1 andξ∈ω2 by simultaneous recursion as follows. The setstα(ξ) will satisfy that they areEξ∩ναand the ordinals να will each be elements of T

ξ∈ω2Eξ. Setν0=δandt0(ξ)=Eξ∩δ. Given ναandtα(ξ),ξ∈ω2, define να+1,k= sup

ξ∈ω·k

ind(tα(ξ)) and να+1= sup

k∈ω

να+1,k.

Next we define tα+1(ξ), ξ∈ω2, by recursion on ξ. We set tα+1(0)=E0∩να+1. Given tα+1(ξ), define tα+1(ξ+1) to be the unique element t of TeEξ∩P(να+1+1) such that supt=να+1andνα+1,k∈tfor allk(here we are employing Lemmas3.1and3.2). Ifη∈ω2 is a limit ordinal, set

tα+1(η) =\

ξ∈η

tα+1(ξ).

(11)

Ifναhas been defined for allα∈β, setνβ=supα∈βνα andtβ(ξ)=S

α∈βtα(ξ).

It is now easily seen that tα(ξ)=Eξ∩να, and therefore that Eξ is in L[A] for all ξ∈ω2. This is a contradiction, however, sinceE2is not inL[A].

We will finish by remarking that if r is an ω1-sequence of reals, then the sequence T!r described in the introduction is defined as follows. Ifω1L[r]1, then defineT!rto be the empty sequence. Otherwise, let ind be the<L[r]-least bijection betweenH(ω1)∩L[r]

andω1. LetA⊆ω1be such thatL[A]=L[r] and defineTξr=TEind

ξ as detailed in the proof of Theorem3.6.

References

[1] Asper´o, D., Larson, P. & Moore, J. T., Forcing axioms and the continuum hypothesis.

Acta Math., 210 (2013), 1–29.

[2] Devlin, K. J. & Shelah, S., A weak version of♦which follows from 20<21.Israel J.

Math., 29 (1978), 239–247.

[3] Eisworth, T., Milovich, D. & Moore, J. T., Iterated forcing and the continuum hy- pothesis, in Appalachian Set Theory 2006–2012, London Math. Society Lecture Notes Series, 406, pp. 207–244. Cambridge Univ. Press, Cambridge, 2013.

[4] Eisworth, T. & Nyikos, P., First countable, countably compact spaces and the contin- uum hypothesis.Trans. Amer. Math. Soc., 357 (2005), 4269–4299.

[5] Kunen, K.,Set Theory. Studies in Logic and the Foundations of Mathematics, 102. North- Holland, Amsterdam, 1983.

[6] Larson, P.,The Stationary Tower. University Lecture Series, 32. Amer. Math. Soc., Prov- idence, RI, 2004.

[7] Moore, J. T.,ω1 and−ω1 may be the only minimal uncountable linear orders.Michigan Math. J., 55 (2007), 437–457.

[8] Shelah, S.,Proper and Improper Forcing. Perspectives in Mathematical Logic. Springer, Berlin–Heidelberg, 1998.

[9] — On what I do not understand (and have something to say). I.Fund. Math., 166 (2000), 1–82.

[10] Todorˇcevi´c, S.,Walks on Ordinals and their Characteristics. Progress in Mathematics, 263. Birkh¨auser, Basel, 2007.

Justin Tatch Moore Department of Mathematics Cornell University

Ithaca, NY 14853-4201 U.S.A.

justin@math.cornell.edu Received October 27, 2011

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