Uri Abraham and Saharon Shelah- Coding with ladders a well ordering of the reals

8/3/2019 Uri Abraham and Saharon Shelah- Coding with ladders a well ordering of the reals

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Coding with ladders a well ordering of the reals

Uri AbrahamDepartment of Mathematics and Computer Science,

Ben-Gurion University, Beer-Sheva, Israeland

Saharon Shelah

Institute of MathematicsThe Hebrew University, Jerusalem, Israel

October 6, 2003

Abstract

Any model of ZFC + GCH has a generic extension (made with aposet of size 2) in which the following hold: MA+ 20 = 2+ there

exists a 21-well ordering of the reals. The proof consists in iteratingposets designed to change at will the guessing properties of laddersystems on 1. Therefore, the study of such ladders is a main concernof this article.

1 Preface

The character of possible well-orderings of the reals is a main theme in set

theory, and the work on long projective well-orderings by L. Harrington [4]can be cited as an example. There, the relative consistency of ZFC + MA+20 > 1 with the existence of a 13 well-ordering of the reals is shown. Adifferent type of question is to ask about the impact of large cardinals on

This research was supported by The Israel Science Foundation founded by the Israel

Academy of Sciences and Humanities. Publication # 485.

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definable well-orderings. Work of Shelah and Woodin [7], and Woodin [9] isrelevant to this type of question. Assuming in V a cardinal which is bothmeasurable and Woodin, Woodin [9] proved that if CH holds, then there isno 21 well-ordering of the reals. This result raises two questions:

1. If large cardinals and CH are assumed in V, can the 21 result bestrengthen to 22? That is, is there a proof that large cardinals and CHimply no 22 well-orderings of the reals?

2. What happens if CH is not assumed?

Regarding the first question, Abraham and Shelah [2] describes a poset

of size 2 (assuming GCH) which generically adds no reals and provides a22 well-ordering of the reals. Thus, if one starts with any universe with alarge cardinal , one can extend this universe with a small size forcing andobtain a 22 well-ordering of the reals. Since small forcings will not alter theassumed largeness of a cardinal in V, the answer to question 1 is negative.

Regarding the second question, Woodin (unpublished) uses an inaccessi-ble cardinal to obtain a generic extension in which

1. MA for -centered posets + 20 = , and

2. there is a 21 well-ordering of the reals.

Solovay [8] shows that the inaccessible cardinal is dispensable: any model ofZFC has a small size forcing extension in which the following holds:

1. MA for -centered posets + 20 = 2, and


In [3] we show how Woodins result can be strengthened to obtain thefull Martins axiom. We prove there that if V satisfies the GCH and containsan inaccessible cardinal , then there is a poset of cardinality that givesgeneric extensions in which

1. MA + 20 = , and


Our aim in this paper is to show that the inaccessible cardinal is notreally necessary, even to get the full Martins Axiom.

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Theorem 1.1 Assume 20

= 1 and 21

= 2. There is a forcing poset ofsize 2 that provides a cardinal preserving extension in which Martins Axiom+20 = 2 holds, and there is a 21 well-ordering of the reals. In fact, thereis even a 2[1] well-ordering of the reals there.

The concepts 21 and 2[1] will soon be defined, but first we shall point to

what we consider to be the main novelty of this paper, the use of laddersystems as coding devices. A ladder over S 1 is a sequence = | S where : is increasing and cofinal in . Two ladders over S, a subladder of , may encode a real (a subset of ). Namely the codingof a real r is expressed by the relationship between and (for every ).

Splitting 1 into 2 pairwise almost disjoint stationary sets, it is possible toencode 2 many reals (and hence a well-ordering) using 2 pairs of laddersequences. Of course, we need some property that ensures uniqueness ofthese ladders, in order to make this well-ordering definable. Such a propertywill be obtained in relation with the guessing power of the ladders. A laddersystem | S is said to be club (closed unbounded set) guessing if forevery closed unbounded C 1, []

C for some S. It turns out thatthere is much freedom to manipulate the guessing properties of ladders, and,technically speaking, this shall be a main concern of the paper.

We now define the 21 and 2[1] relations. The structure with the mem-

bership relation on the collection of all hereditarily countable sets is denotedH(1). Second-order formulas over H(1) that contain n alternations ofquantifiers are denoted 2n when the external quantifier is an existential classquantifier. Thus a 2n formula has the form

X1X2 . . . X n(X1, . . . , X n)

where may only contain first-order quantifiers over H(1) and predicatesX1, . . . , X n are interpreted as subsets of H(1). (One can either write Xi(s)treating Xi as a predicate, or s Xi treating Xi as a class.)

2 denotes theunion of all 2n formulas.

If the second-order quantifiers only quantify classes (subsets of H(1))of cardinality 1, then the resulting set of formulas is denoted 2[

1]n . So

2[1]1 for example denotes second order formulas of the form there exists

a subset X of H(1) of size 1 such that (X) where is a first orderformula. We write 2[1], without a subscript, for

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In Theorem 1.1 above, we get a well-ordering which is 2[1]

, and we willexplain now why MA + 20 > 1 implies that such a relation is necessarily21. This transformation which replaces any number of quantifiers over setsof size 1 with a single existential quantifier over arbitrary subsets of H(1)is a trick of Solovays that was used by him in [8]. The basic idea is to usethe almost-disjoint-sets coding (Jensen and Solovay [5]) in a way which willbe sketched here.

Theorem 1.2 (Solovay) Assume MA+20 > 1. Any 2[1] formula (x)

overH(1), with free variables x1, . . . , xn, is equivalent to a21 formula(x).

Proof. It seems easier to prove first that every 2[


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2 Ladder systemsThe notation A B is used for almost inclusion on subsets of1, meaningthat A \ B is finite. Similarly A = B is defined if A B and B A.A = B is the negation of A = B.

Definition 2.1 1. A ladder system over S 1 (consisting of limit or-dinals) is a sequence = | S, where is an increasing-sequence converging to . S is called the domain of , and is de-noted dom(). is called trivial if dom() is non-stationary. Therange of is denoted [] (so [] = {(i) | i }), andS [] isthe range of . So, [] C means that, except for finitely manyks, (k) C always holds.

2. Let and be two ladder systems. We say that and are almost dis-joint iff, for some club C 1, for any Cdom()dom(), [][] =

.

3. We say that is a subladder of iff the following holds for some clubC 1:

C dom() dom() and for C dom(), [] [].

In such a case we write . Also, = iff both and .That is, = iff there is a club set C 1 such that dom() C =dom() C, and [] = [] for dom() C.

4. The difference ladder = \ is defined by

[] =

[] if dom() \ dom()[] \ [] if this set is infiniteundefined otherwise

It is the -maximal ladder included in and (almost) disjoint from .

5. Given any A 1, the restriction ladder A is naturally defined,and its domain is A dom(). If x is infinite, then x meanssomething else: it is obtained by enumerating x = {xk | k } inincreasing order, and setting ( x) = where (k) = (xk) forevery dom().

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We shall define some properties of ladders (in fact, of =

equivalenceclasses).

Definition 2.2 Let be a ladder over S.

1. We say that is club-guessing iff for every club C 1 there is Ssuch that []

C. (So, in this case, C, and hence dom() isstationary if is club guessing.) For brevity, we may use the termguessing instead of club-guessing.

2. We say that is strongly club guessing (or just strongly guessing) iff for any club C 1 for some club D, if D S then []

C. If is strongly guessing and , then clearly is also stronglyguessing. (Be careful: if and is guessing, you cannot infer that is guessing, unless is non-trivial.) The trivial ladder is (trivially)strongly guessing, and hence we cannot say that a strongly guessingladder is always guessing. A strongly guessing non-trivial ladder is, ofcourse, guessing.

3. We say that a club set C 1 avoids iff for every S (except anon-stationary set), [] C =

.

4. We say that is avoidable iff some club set avoids . If every ladderover S is avoidable, then we say that S itself is avoidable. Hence, inparticular, if S is non-stationary, then S is avoidable. Remark that if is avoidable, then is non-guessing. So is strongly guessing andavoidable iff is trivial. The collection of all avoidable sets forms anideal which will be shown to be normal in the following subsection.

5. Maximal ladders. Suppose that is some strongly guessing ladderover S, and X S is a subset of 1. If every ladder over X and(almost) disjoint from is avoidable, then we say that is maximalfor X. In such a case, for every X X, X is maximal for X.

The trivial ladder is trivially maximal for any avoidable set. Ourterminology may be misleading because a maximal ladder for X is notnecessarily defined over X, it is rather the maximality which is for X.Thus, if is maximal for X, then for every strongly guessingladder over a subset of X. (Because \ is, in that case, stronglyguessing and disjoint from , and is hence avoidable. Thus dom( \ )

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is not stationary, and hence

.) Hence if both and are maximalfor X, then = . We denote this unique ladder, maximal for X, by(X).

It is easy to see that if is maximal for X and X0 X then X0 ismaximal for X0.

2.1 Ideals connected with ladders

We are going to define four ideals on 1: the ideal of non-guessing restrictions,denoted I, the ideal of avoidable sets, denoted I0, the ideal of maximal

guesses, denoted I1, and the ideal of bounded intersections, I(S). Then wewill prove that all are normal ideals.

Definition 2.3 The ideal of non-guessing restrictions. Let be a guess-ing ladder over X. The collection of all subsets S 1 for which Sis not guessing is a proper, normal ideal, denoted I.

The ideal of avoidable sets. S I0 iff every ladder system over S isavoidable.

The ideal of maximal guesses. The ideal I1 is the collection of all sets

X 1 such that there is a maximal ladder for X.So S I1 iff there is a strongly guessing ladder system such thatdom() S and any ladder over S and disjoint from is avoidable.As said above, this unique ladder is denoted (S). (Uniqueness is upto =, where non-stationary sets and finite differences do not count.)

In case S I0, then S I1, and (S) is the trivial (empty) ladder .So

I0 I1. (1)

The ideal of bounded intersections. Let S = Si | i 2 be a collec-

tion of 2 stationary subsets of 1 such that the intersection of anytwo is non-stationary (we say that S is a sequence of pairwise almostdisjoint stationary sets). The idealI(S) consists of those sets H 1for which

|{i 2 | H Si is stationary }| 1.

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Sets in I(S) will also be called S-small sets. It may seem that I(S)is not connected to ladders, but we will later show the consistency ofI(S) = I1.

Lemma 2.4 All four ideals are normal.

Proof. An ideal on 1 is said to be normal if it is closed under diagonalunions.

The ideal of non-guessing restrictions. Let be a guessing ladder overX. To prove normality of I, suppose A I, for < 1. Thus, for

every there is a club set C such that A X []

C. Let

A = 1Adef= { 1 | < ( A)}

be the diagonal union, and C = 1C be the diagonal intersectionof the club sets. Then A I because for A X, []

C.

The ideal of avoidable sets. We check that I0 is normal. Suppose S I0for 1, and let S = 1S be the diagonal union. Let = | S be any ladder over S, and we will show that is avoidable andhence that S I0. Indeed, a slightly more general fact will be used

later:

If S 1 are arbitrary sets, S = 1S, and is a ladderover S such that S is avoidable for every 1, then is avoidable.

To see this, let C for 1 be a club set that avoids S, and letC = 1C be their diagonal intersection. Then C avoids , as caneasily be checked.

The ideal of maximal guesses. We prove that I1 is normal. So suppose

that S I1 for 1 are given, and S = 1S is their diagonalunion. We must prove that S I1. First we claim that the sets{S | 1} may be assumed to be pairwise disjoint. Indeed, defineS = S \

{S |

< }. Then S = S , and the sets S are in I1 and

are pairwise disjoint. So we do assume now that the Ss are pairwisedisjoint. For every 1, (S) is a strongly guessing ladder over its

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domain S0

S (and S0

= when S I0). Define S0

= 1S0

.Clearly S0 S. For S0 define to be (S) for the (unique) < such that S0 .

Claim: = | S0 is maximal for S, and hence S I1.

Proof. We first prove that is strongly guessing. Well, if C 1is club, find for each 1 a club set D such that for D S0 , ((S))

C. Now define D = 1D to be the diagonalintersection. It follows that for every S0 D, []

C.

To prove maximality, assume is defined on S and is disjoint from. Then S is disjoint from (S) S \ ( + 1). Hence S isavoidable for every < 1, and by the proof of normality of I0, isavoidable.

The ideal of bounded intersections. Let S = Si | i 2 be a collec-tion of pairwise almost disjoint stationary subsets of1 defining I(S). IfH for 1 are in I(S), then there is a bound j0 < 2 such that for ev-ery j0 j < 2 Sj H is non-stationary. Hence Sj H = Sj His non-stationary, and thus H I(S).

2.2 A(S, )

In this subsection we formulate a statement, A(S, ), and show that it impliesI(S) = I1. The consistency of A(S, ) will be proved in the subsequentsections.

Definition 2.5 A(S, ) is the conjunction of the following six statements:

A1 S = Si | i 2 is a sequence of pairwise almost disjoint stationarysubsets of 1. is a ladder system, and

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A4 If X 1 is such that X Si is non-stationary for every i < 2, thenX is avoidable (equivalently, in view of (A2), X is avoidable).

A5 If X 1 is not S-small, is a ladder over X and , then thereexists i < 2 such that Si X and (Si) .

A6 For every i 2 either ((Si), Si) is clearly not encoding, or elser = d((Si), Si) is defined, and in this case r = d((Sj), Sj) forunboundedly many js. The meaning of this statement is clarified laterin this subsection.

A

(T , ) is the following statement: T = Ti | i < 2 is a sequence of pairwisealmost disjoint stationary subsets of 1. For every i < 2, Ti I1, and if Sidenotes dom((Ti)), then A(S, ) holds for S = Si | i < 2.

We first collect some simple consequences of the first five statements ofA(S, ).

Lemma 2.6 The first five statements of A(S, ) imply that:

1. If is avoidable, then dom() is S-small.

2. I0 I(S).

3. If is strongly guessing, then dom() is S-small.

4. Actually: If is strongly guessing, then dom() is S-small.

5. I1 = I(S).

Proof. To prove 1, assume but X = dom() is not S-small. Then (A5)implies that, for some i < 2, (Si) . Hence is not avoidable (by (A3)which says that (Si) is (strongly) guessing).

We prove 2. If X I0 (X is avoidable) then any ladder system overX, and in particular X, is avoidable. Hence (by item 1) dom( X) is

S-small. Thus X is S-small (because X = X0 X1 where X0 = X i Si

and X1 = X\ X0. X1 is clearly S-small, and X0 = dom( X)).To prove 3, assume that dom() is not S-small. Split into 1 and 2,

two halves defined by taking (1) to be an infinite co-infinite subset of

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(for every dom()), and letting 2

= \ 1

. If X = dom() is notS-small, then, by (A5) applied to 1, there is i such that Si X and

(Si) 1. (2)

Since is strongly guessing, 2 Si is strongly guessing (and non-trivial asits domain is the stationary set Si), but formula (2) shows that

2 Si isdisjoint from (Si), and this contradicts the maximality of (Si) for Si.

To prove 4, suppose that is a strongly guessing ladder over X. To showthat X I(S), we reduce this claim to the case that . Look at \ and its domain

X1 = { X | [] \ [] is infinite}.By (A2), \ is avoidable. But, as is strongly guessing, any subladder ofis also strongly guessing, and hence \ is strongly guessing and avoidable,which could only be if X1 is non-stationary.

Now set X2 = X\ X1, and = X2. Then is strongly guessing,and hence by the previous item dom() is S-small.

Finally we prove 5. If X I1 then X = X0 X1, where X0 I0 and X1is the domain of a strongly guessing laddernamely (X). Hence X I(S)by items 2 and 4.

Suppose now that X I(S). By definition, there is < 2 such that, for

i , X Xi is non-stationary. Let Tj | j 1 be an 1-enumeration ofthe collection {Si | i < }. Then each Tj I1 by (A3). Let T = j1Tj bethe diagonal union. By normality ofI1, T I1. Hence XT I1. But X\ Thas only countable intersections with each Tj (for in fact (X\T)Tj j +1),and hence, certainly, has non-stationary intersections with every Si, and isthus in I0 (by (A4)). As I0 I1 (by formula (1) in Definition 2.3), X I1.

We will prove next that if A(S, ) holds, then is determined, up to anI1 set, as that ladder for which (S)A(S, ).

Lemma 2.7 If the first five statements hold for A(S, 1) and A(T , 2), then

I(S) = I(T) = I1, and { 1 | [1 ] = [2 ]} I1.

Proof. Define S1 = dom(1 \ 2), and S2 = dom(2 \ 1). We claim thatS1, S2 I1. This implies the lemma because S1 S2 I1 follows. Bysymmetry, it suffices to deal with only one of these sets, for example withS1.

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that [1

] =

[2

] except for an I1 set, and I1 = I(S) = I(T) (Lemma 2.7).That is, if H = { 2 | [1] = [2 ]}, then H I1, and hence H I(S).

Thus there is an index i U such that

H Si is non-stationary. (3)

That is,

1. 1 Si = 2 Si (that is, [

1 ] =

[2 ] for all Si, except for anon-stationary set),

2. d((Si), 1 Si) = r.

Now (Si) is maximal for Si (a stationary set) and hence its domain Si is notavoidable. So by (A4) ofA(T , 2), for some j 2, X = Si Tj is stationary.Hence (Si) X is maximally guessing (and non-trivial). Similarly (Tj) Xis maximally guessing, and thus

(Si) X = (Tj) X

by the uniqueness of the maximal ladder over X (namely (X)). Since XHis non-stationary (by (3) above),

1 X = 2 X

and thus ((Tj), 2 Tj) encodes a real, and d((Tj), 2 Tj) = r.

3 The consistency ofA(S, )

Our aim in this section is to prove the followingTheorem. Assume that 20 = 1 and 2

1 = 2. Suppose that T = Ti |i < 2 is a collection of 2 pairwise almost disjoint stationary subsets of1,and is a ladder system such that

(1) Ti is guessing (but not necessarily strongly guessing) for every i < 2.

(2) range() Ti is empty for every i.

Then there is a generic extension in which A(T , ) and Martins Axiom hold.The extension is an iteration of the posets R(), and P(, C) described

below. Before proving this theorem, however, we review some notions fromproper forcing theory.

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3.1 Some proper forcing theoryThis short subsection assembles some known definitions and results on properforcing, such as -properness and S-properness for a stationary set S. Ournotations and terms are taken (with some minor changes) from Shelahs book[6] (see also [1]).

Recall that if P is a forcing poset and N H a countable elementarysubstructure, then a condition q P is N generic iff for every D N, densein P, every extension of q is compatible with some condition in D N. Aforcing poset P is proper is for some cardinal , for every countable N Hsuch that P N, every p P N has an extension that is N generic.

Definition 3.1 (of -properness.) Let be a countable ordinal. A posetP is said to be -proper iff for every large enough cardinal , if Ni | i is an increasing, continuous sequence of countable elementary submodels ofH such that P N0 and Nj | j i Ni+1 for every i < , then anyp0 P N0 can be extended to q P that is Ni-generic for every i .

Definition 3.2 Let S 1 be stationary. A forcing poset P is S-proper ifit is proper for structures M such thatM1 S. That is, P is S-proper ifffor sufficiently large , if M H is countable, S, P M, and M 1 Sthen any p P M can be extended to an M-generic condition.

A stronger property is that of a poset being S-complete. It means thatwhenever M H is countable, with P, S M, and M 1 S, thenevery increasing and generic -sequence of conditions in P M has an upperbound in P. (A sequence of conditions is generic if it intersects every denseset of P in M.)

The notion (E, )-properness is defined in Shelah ([6] (Chapter V). Justas properness is equivalent to the preservation of stationarity of S0(), sois (E, )-properness equivalent to the preservation of an appropriate notionof stationarity defined there. However, for our article, a notion of somewhatless generality suffices.

Let I be the collection of all increasing sequences of countable ordinals.We write = i | i < for I

. The club guessing property can beregarded as a notion of non-triviality of subsets of I.

Definition 3.3 1. A familyE I is stationary if for every club C 1there is E such that [] = {i | i } C.

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2. Let E I

be stationary. We say that the poset P is E-proper (or(E, )-proper, to emphasize that this notion is related to -properness)iff for every sufficiently large cardinal, wheneverMi H, fori < ,are countable with E, P M0 and are such that Mi Mi+1 for alli < , if

Mi 1 | i < E

then anyp PM0 can be extended to a condition which is Mi-genericfor everyi < .

In Shelah [6] it is proved that the countable support iteration of posets thatare S-proper (-proper or E-proper) is again S-proper ( -proper or E-proper, respectively). Also, if P is S-proper (E-proper), then, in VP, S(respectively E) remains stationary.

Lemma 3.4 If E I is stationary and P is an E-proper poset, then Eremains stationary in VP.

Proof. Let D be a name in VP forced by some p P to be a club subset of1. Define an 1 sequence Mi | i 1 where Mi H are countable withMi | i j Mj+1, and such that p,P,D M0. The set C = {Mi 1 | i 1 is closed unbounded in 1. Since E is stationary, there is Esuch that{n | n } C. Then n = Mi(n) 1 and Nn = Mi(n) is an increasingsequence of structures with Nn Nn+1 and such that Ni 1 | i < E.So there is an extension q P that is Ni-generic for every i < . So forevery i q i D. (Because q forces that D is unbounded below 1 =

Ni1 .)

Thus q [] D, as required.We shall define now two subsets of I, E and D, which will be used

later.

Definition 3.5 1. Let be a ladder system and S = dom(). Define

E I

by = i | i < E

iff

I and, for = sup{i | i < }, S and is an endsegment of (i.e., for some k, (k + i) = i for allis).

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It is obvious that E is stationary iff is club guessing. Thus, if isclub guessing and P is E-proper, then remains a guessing ladder inVP.

2. The set D I (D is for disjoint) is defined for any ladder asfollows: D iff for = sup{i | i < }, either dom() or[] [] =

. If is disjoint from , then E D. Thus, in thiscase, if P is (D, )-proper, then P is (E, )-proper as well.

3.2 The building blocks

Two families of posets are described in this subsection: R() and P(, C).The poset R(). Let be a ladder over a set S 1. The poset R()

introduces a generic club to 1 that avoids . So, naturally,

c R()

iff

c 1 is countable, closed (in particular max(c) c), and forevery S, [] c is finite.

The ordering on R() is end-extension.The cardinality of R() is the continuum. It is clear that R() is 1 \ Scomplete. A short argument is needed in order to prove that it is proper.

Observe first that for any condition q R() and dense set D R(), if0 = max(q) then there is a closed unbounded set of ordinals < 1, 0 < ,such that for every 1 with 0 < 1 < there is an extension q

D suchthat q and 1 q

is the successor of 0 in q. For example, the club

set can be obtained by defining a continuous, increasing chain N | 1of countable elementary substructures of some H with and the dense setD in N0. Then 1 N | 1 is as required.

Suppose that a countable M H and a condition p0

R() M aregiven. We want to define an increasing, generic sequence of conditions piextending p0 so that for = M 1, p =

i pi {} is a condition. The

case M 1 S is trivial and so assume that = M 1 S. The problemis that we may decide infinitely often to put (n) in

ipi, and then p is

not a condition. The preliminary observation enables the construction of the

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sequence pi in such a way that p [] p0 is finite. The point is thatwhen we need to extend a condition pi into a dense set D, we first considerthe club set formulated above (do it in the substructure M) and find a limitordinal in the club that is in M. Now 1 < is chosen so that the interval[1, ] is disjoint to []. (The fact that [] is only an sequence impliesthe existence of such an ordinal.

R() is not -proper. For suppose Mi, i < is an increasing sequenceof elementary submodels such that i = Mi 1 [] for infinitely manyis, where = sup{i | i < }. Then no condition can be generic for allof the Mis. However, if S or Mi 1 | i < is disjoint from [](or has only a finite intersection) then there is no problem in finding sucha generic condition. That is, R() is (D, )-proper. In fact, if pi is anysequence of increasing conditions where pi Mi is Mi1 generic, then

ipi

gives a condition. This property is stronger than (D, )-properness, but inapplication we shall mix proper forcings with R() forcings and hence theiteration itself is (D, )- proper.

Hence we have the following which will be used in Lemma 3.9.

Lemma 3.6 Suppose that is a ladder system and A, B dom() are suchthat A B is not guessing. Then R( A) is EB-proper.

Proof. Suppose that A, B dom() are such that A B is not guessing.Let C 1 be a club set such that, for every A B, []

C. Supposethat Mi H for i < are as in the definition of EB properness and = sup(Mi 1 | i < ). So, EB , R( A) M0. Hence A, B M0and thus C M0 can be assumed. Then Mi 1 C for every i. SinceMi 1 | i < EB, B and [] =

{Mi 1 | i < }. Thus[]

C and hence A B. So A and as R( A) is 1 \ A complete,there is no problem in finding a condition that is Mi-generic for every i.The poset P(, c). Let be a guessing ladder over a stationary co-stationaryset S, such that

S range() is non-stationary.

(See Definition 2.1 for range(). Then, for any club set c 1, the posetP(, c) introduces a generic club set D 1, such that for every D S, []

c. This may be viewed as forcing a club subset to the stationaryset { S | []

c} (1 \ S).

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Accordingly, we define d P(, c) iff d 1 is countable, closed (withmax(d) d), and for every d S, [] c.

The order is end-extension.It is easy to check that any condition has extensions to arbitrary heights

(as there are no restrictions on 1 \ S). The cardinality of P(, c) is thecontinuum.

P(, c) is not necessarily proper, because if, for = M 1, [] c,

then no M-generic condition can be found. Still, P(, c) possesses two goodproperties which allows its usage:

1. P(, c) is (1 \ S)-complete (the proof of this is obvious).

2. P(, c) is (E, )-proper. (E is stationary since is guessing.)

We check the second property it is for its sake that the requirement thatdom() range() is non-stationary was made. So let Mi | i < be anincreasing sequence of countable elementary submodels of H, with Mi Mi+1, and such that P(, c), , c M0. Denote i = Mi 1, and =sup{i | i < }.

The assumption is that i | i < E, and the desired conclusionis that any p0 P M0 can be extended to a condition that is generic forevery Mi. So the assumption is that S and i | i < is an end seg-ment of . Since dom() range() is non-stationary, i S (becauseM0contains a club that is disjoint from this intersection), and it is easy to find(in Mi+1) an Mi-generic condition extending any given condition (using the(1 \ S)-completeness). Thus, given p0 P(, c) M0, we may constructan increasing sequence of conditions pi Mi, such that pi+1 is Mi-generic.Then p = {}

i


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is used in this iteration of proper forcing posets, and hence the finalposet is proper. The final poset satisfies the 2-chain condition (see[6], Chapter VIII, or [1]). The length of the iteration is 2 so that eachpossible c.c.c poset of size 1 and each ladder are taken care of atsome stage.

2. Given a guessing ladder such that dom() range() = a model ofMA + 20 = 2 can be obtained in which is strongly guessing. Thistime posets of type P(, c) are iterated (varying club sets c 1) aswell as c.c.c. posets. The iteration is with countable support and oflength 2 as before. Put S = dom(). Then S is stationary (as is

guessing) and co-stationary (as S range() = ). Since each posetP(, c) is 1 \ S complete, and each c.c.c. poset is obviously 1 \ Sproper, we have here an iteration of 1 \ S proper posets. Thus thefinal poset itself is 1 \ S proper and 1 is not collapsed. Moreover,since the iterands (both P(, c) and the c.c.c. posets) are E proper,the final iteration is E proper. Hence remains guessing at each stageand in the final extension. It is strongly guessing since we took explicitsteps to ensure this.

3. Now we want to combine 1 and 2. We are given a guessing ladder system defined over a stationary co-stationary set T, such that Trange() =, and we want a generic extension in which is maximal for 1. Forthe iteration, decompose 2 into three sets 2 = JKI of cardinality2 each. At stage < 2 of the iteration, supposing that P has beendefined, define the poset Q in V

P as follows:

(a) If J, then Q is a c.c.c poset, and the iteration of all posetsalong J guarantees Martins Axiom.

(b) For K, Q will be of type R() where VP is a ladder

system disjoint from . R() is proper and it is (D, ) proper.Hence as E D, R() is E proper.

(c) For I, Q will be of type P(, c) where c is a club set in VP.These posets are 2 \ T complete, and E proper.

Any of the posets along the iteration is either proper or 2 \ T proper(namely, the P(, c) posets which are 2\T complete). So the iteration

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itself is 1 \ T proper, and thus 1 is not collapsed. Moreover, theposets are E proper, and hence retains its guessing property in theextension.

3.3 The iteration scheme

Recall that our aim is to prove the following theorem.

Theorem 3.7 Assume 20 = 1 and 21 = 2. Suppose

(1) A sequence T = Ti | i 2 of pairwise almost disjoint stationary sub-

sets of1. (Almost disjoint in the sense that Ti Tj is non-stationary.)

(2) A ladder system such that

1.

{Ti | i 2} dom(), and 1 \ dom() is stationary.

2. For every i, Ti is club guessing.

3. dom() range() = .

Then there is cofinality preserving generic extension in which A(T , ) andMA + 20 = 2 hold. (The definition of A

(T , ) is immediately after Defi-

nition 2.5.)

Proof. It is not difficult to get T and as in the theorem, and the followingsection contains a generic construction of such objects. Here we just assumetheir existence and prove the theorem. The generic extension is made viaP = P2 , obtained as an iteration, P | 2, with countable supportof posets of cardinality 1. At successor stages, P+1 = P Q, whereQ V

P is one of the following three types.

(1) A c.c.c. poset. (To finally obtain Martins Axiom.)

(2) A P(, c) poset, where VP

is a guessing ladder such that ,and c VP is a club set. Recall that P(, c) introduces a genericclub subset D such that D dom() implies

c. We havechecked that this poset is (1\ dom())-complete, and (E, )-proper(as dom() range() = ).

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(3) The third type of iterated posets is R() where VP

is a laddersystem. This forcing makes avoidable. We have seen that R() isproper, (D, )-proper, and (1 \ dom())-complete.

Each iterated poset Q is (1\dom()) proper in VP. Hence the iteration

itself is (1 \ dom()) proper, and it satisfies the 2-c.c.We must specify how to choose the posets Q for the iteration. Every P

will have cardinality 2 and will satisfy the 2-c.c. When we say that aname in VP satisfies property , we mean that it is forced by every conditionin P to satisfy . We say that a V

P name of a subset of 1 is standardiff it associates with every 1 a maximal antichain of conditions that

decide whether is in this subset or not. Every subset of 1 in VP has(an equivalent) standard name. For every poset P of size 2 that satisfiesthe 2-c.c., Fix an enumeration {E(P, ) | < 2} of all standard namesin VP of subsets of 1 and of ladder systems. Thus any ladder or subset of1 in V

P has a name of the form E(P, ) for some < 2. Fix a naturalwell-ordering of the pairs {, | , < 2} that has order-type 2. Soeach , has its place in 2. This will serve in the choice of Q.

To define the iteration, we partition 2 (in V):

2 = J K L

{Ii | i < 2},

where each set in this partition has cardinality 2. The type ofQ dependson the set in this partition that contains .

For J, Q is a c.c.c. poset of cardinality 1, and the iteration ofthese posets in J shall provide Martins Axiom. By now this is so standardthat no further details will be given.

For K, Q will be of type R(), where VP is a ladder systemdisjoint from (namely, is a name forced by every condition to be a ladder-system disjoint from ). The final result of iterating these posets along K isthat, every VP disjoint from is avoidable in VP. Thus property (A2)of A(T , ) can be assured.

Before going on, lets discuss the problem involved in the direct approachto obtain (A5) and why we do not get A(T , ) but rather A(T , ) (namelyA(S, ) where Si Ti). A possible approach to (A5) is to consider eachpossible ladder such that X = dom() is not T-small, and to findfor this some i 2 such that X Ti is stationary. Then, if possible, totransform X Ti into a maximally guessing ladder. For this to have any

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chance, it must be the case that Ti is guessing. Yet it is possible that Ti is non-guessing for every i. In this case we must shrink the Tis so asto make X S-small. This shows the need for defining subsets Si Ti. Butnow (A4) causes a problem because, if X 1 is such that in V

P X Si isnon-stationary for every i < 2, then we must be able to identify this X atsome intermediary stage of the iteration so as to make X avoidable. Yet,as the Si are not yet all defined in any intermediate stage, it is not clear howto identify these Xs.

We describe now in general terms how the sets Ii from the partition willbe used in the iteration. For every i < 2 let (i) be the first ordinal in Ii.A stationary subset Si Ti and a guessing ladder i over Si will be definedin VP(i). The iteration of the posets Q for Ii will make i maximal forTi, and will achieve (A3) by establishing (Ti) =

i. Finally, in VP, A(S, )will hold for S = Si | i 2.

To define Si and i we assume a function, , which assigns to any of

the form = (i) a name () VP that is one of the following.

1. If i is an even ordinal then ((i)) is a name of a real in VP. Thecomplete definition of is given in the following section where it isused to define the encoding of the well-ordering of reals. Here we onlyassume that () is defined.

2. If i is an odd ordinal, then ((i)) is determined as a name of theladder system (), defined as follows in VP. With respect to thewell-ordering of names in VP, () is the least ladder that is notof the form () for < , and is such that for D = dom()

D Ti is guessing.

Suppose that = (i). Instead of defining the names Si and Q directlyin VP, we let G P be V-generic and we shall describe the interpretationsof Si and Q. We will later see (Lemma 3.9) that Ti remains guessing in

V[G]. In V[G], collect all sets X 1 such that

1. a standard name of X appeared before in the well-ordering of thenames (i.e., for some and < 2,

, is placed before inthe well-ordering of2 2, and E(P, ), the th name in VP , givesX), and

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2. X is such that (Ti X) is not guessing (i.e., Ti X I).Let X | < 1 be an enumeration of these sets. Take their diagonal union

A = 1(X Ti). (4)

Then A I. Since Ti is guessing in V[G], Ti = Ti \ A I (that is,

Ti is guessing). (The reason for this specific definition of A and Ti will

only be apparent in the proof of item (A5) in VP.)Now () is either a real or a ladder system in V[G]. Accordingly the

definition of Si and i is split in two. Suppose that () is a real r in

V[G]. We want to encode r. Define Si = T

i , and let i

Si be a laddersystem over Si such thatd(i, Si) = r.

Since Si is guessing and i Si has domain Si,

i is also guessing.Suppose next that = (i) for i an odd ordinal and = () is (in V[G])

a ladder over X = dom() (such that , and XTi is guessing). Then X Ti is guessing, because Ti \ T

i I. It follows that X T

i is

guessing as well, because and X Ti dom(). In this case define

Si = X Ti ,

and define

i Si so that (i, Si) is clearly not encoding.

The iteration along Ii builds up the properties of i and establishes

(Ti) = i in VP. For this, the posets Q, for Ii, are of two types:

(1) P(i, c), where c runs over all possible clubs. This ensures that i

becomes strongly club guessing in VP. To enable the use of P(i, c)we rely on the assumption, proved later to hold, that i remains clubguessing at each stage.

(2) R(), where runs over all possible ladders over Si that are disjointfrom i. This ensures the maximality of i.

To satisfy item (A4) (in the definition of A(S, )), every X must be madeavoidable whenever all the intersections X Si are non-stationary. It suffices

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to show in such a case that the ladder X is avoidable to conclude thatX is avoidable, because any ladder disjoint from is necessarily avoidable inVP. It is the iteration along L that achieves this, by forcing with posets oftype R( X) as follows.

Given L and a generic filter G P, we will define Q in V[G]. Fori < 2 such that (i) < , the sets Si have been defined. For every i < 2define

Si =

Si if(i) < Ti otherwise

Using the well-ordering of standard names, take the least set X 1 (ifthere is one) that was not taken before at a stage in L, such that

i < 2 X Si is not guessing. (5)

Then define Q to be R( X) (or a trivial poset if no such X exists).This ends the definition of the iteration, but it is not yet clear why items

(A4) and (A5) hold in VP. To prove (A4) we shall first prove that if

(i < 2)X Si is non-stationary in VP,

then (5) holds at some stage VP , L, and hence X is avoidable in the

next step of the iteration. To see that this is indeed the case, we need thefollowing pivotal observation.

Lemma 3.8 Suppose G P2 is V-generic. If < 2 and V[G ] isa ladder over X such that, in V[G ] and

|{i 2 | X Ti is guessing}| = 2. (6)

Then there is i such that Si X Ti and i .

Proof. The proof of this lemma depends on the fact that for any i (with

(i) ) such that X Ti is guessing in V[G ], X Ti remainsguessing in V[G (i)] as well. Thus, as the turn of cannot be delayed 2many times, at some stage = (i), = () holds, and then Si = X T

i

and i Si were defined in V[G ].Now we can prove item (A4) in V[G]. For this, let X 1 be such that

(i < 2) X Si is non-stationary. We will show that, at some stage L,

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the poset R( X) was taken as Q. If, for some < 2, (6) of Lemma 3.8holds in V[G ], then Si X contradicts the fact that Si is stationary.Hence formula (6) never holds, and for such that X V[G ] there areonly boundedly many js for which X Tj is guessing. So let < j0 < 2be such that if X Tj is guessing, then j < j0. Since, in V[G], X Sjis non-stationary for every j < 2, there is a stage j1 such that for everyj < j0, X Sj is non-stationary in V[G j1]. Thus, for j1, in V[G ],for every i < 2, if i < j0 then Si is defined (that is, (i) ) and X Siis non-stationary, and hence X Si is non-guessing, and if i j0, then XTi is non-guessing. But this is exactly the condition required at stages L to force with R( X).

Finally, we turn to prove item (A5). So let with X = dom() begiven in the generic extension V[G]. Then for some < 2, V[G ],and a name of X appeared before in the well-ordering of names.

Case 1: In V[G ]: There is i0 < 2 such that, for every i i0,(i) > and

X Ti is not guessing.

Then, in defining Si for i i0, (i) > , and the set X appears as some X(in equation (4)), and hence X Si is at most countable (it is included in

+ 1). Thus, in Case 1, X Si is non-stationary (and even countable) fora co-bounded set of indices. That is, X is S-small. (It is for this argumentthat, in defining Si, we asked Si A = )

Case 2: Not Case 1. Hence (6) holds in V[G ]. So, by Lemma 3.8there is i such that Si X Ti and i , which establishes (A5).

Our proof relied on preservation claims that some ladders retain theirguessing property, and we intend now to prove these claims. First, set T =Ti. Then 1 \ T is stationary by assumption, and all the posets used are(1 \ T)-proper. (The c.c.c. posets are certainly proper. The P(, c) posets(defined for ) are (1 \ dom())-complete, and hence (1 \ T)-proper.

The R() posets are proper.) This secures the preservation of 1.

Lemma 3.9 Ti remains guessing in VP for = (i).

Proof. This follows from the fact that the posets iterated at stages < are all (ETi, )-proper:

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1. The c.c.c. posets are always -proper.2. The R() posets iterated at stages in K are defined for s that are

disjoint from . In such a case E D. But we remarked that R()is (D, )-proper.

3. The P(, c) posets introduced for < are defined along Ij onlyfor js such that (j) < , and thence for s such that Tj,implying the (ETi, )-properness. (Since Tj Ti is non-stationary,and range() Tj = .)

4. The R() posets defined along Ij for (j) < are defined for ladders

over Tj. As Tj is almost disjoint from Ti, these R()s are (ETi, )-proper.

5. The R( X) posets defined for L, < , are such that X Tiis non-guessing and the poset is thence (ETi , )-proper (by Lemma3.6).

Then, we must also show that the guessing ladder i Si defined inVP(i) remains guessing at every stage in Ii (and thus the posets P(

i, c) canbe applied). This is basically the same proof, done in VP(i) for the quotientposet P/P(i) which is again a countable support iteration of posets as above

that are Ei proper.

4 The 2[1] well-ordering

The main theorem, Theorem 1.1, is proved in this section. So 20 = 1 and21 = 2 are assumed in the ground model V. We need a sequence T ofpairwise almost disjoint stationary sets, and a guessing ladder system ; andwe are going to define them first.

Since we want to describe the 2 stationary sets in the language 2[1],

we need a compact form of generation for such sets. This is provided by the

following definition.

Definition 4.1 Let{0, 1} denote the set of all functions f : {0, 1}for . Ordered by function extension, this forms a tree. Define


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1. If f, g T and are such that f = g but f() = g(), thenf1{1} g1{1} .

2. T has 2 branches of length 1 and each gives a stationary set (thatis, the union of the nodes along any 1-branch forms a function f andf1{1} is a stationary subset of 1). It follows from item 1 that theintersection of any pair of these stationary sets is countable. Thus thebranches of T give 2 pairwise disjoint stationary sets enumerated asTi for i < 2.

3. We also require that U = i


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and for 1 = 2 f(1) = f(2). Thus T has 2 many 1-branches. The factthat every f() gives a stationary set requires a simple density argument.We will check now the following:

Claim 4.2 Any 1-branch of T in V[G] is some f().

Proof: Suppose, toward a contradiction, that p forces that is a branch ofT which is not f() for any 2. Observe first that since S is -closed, anycondition p can be extended to a condition q that describes up to height(p).Then, every condition p and dom(p) have an extension q such that fq()

diverges from the value of determined by q. Repeating this procedure 2

times, we finally get an extension q of p with = height(q) limit, and suchthat q determines as a branch ofTq of height which is different from eachof the branches fq(). Since Tq

{0, 1} consists only of the branches of theform fq(), the branch of is not in Tq. Then q forces T .

We denote with T = Ti | i < 2 the collection of stationary sets thusobtained from the branches f() of T. Let U = i


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branches is almost disjoint to any original branch and hence by (A4) itsstationary set is avoidable).We describe the 2[1] formula (x) that decodes this well-ordering ((x)

iff x E). First consider the 2[1] formula (T0, 0)(with class variables T0and 0) which says that T0 is a stationarity tree, and 0 is a ladder system suchthat A(T0, 0) holds (where T0 is the collection of non-avoidable stationarysets derived from the branches of T0). (The statement there are 2 indicessuch that. . . can be expressed by saying there is no 1-class containing allthe indices such that. . . ).

This enables a 2[1] rendering of the formula x E:

(x) there exists T0 and 0 such that (T0, 0) andx code(S0, 0)

Clearly, (x) holds for every x E (by virtue of the real T and ), andwe must also prove that if (x) then x E. But this follow from Lemma2.8.

References

[1] U. Abraham, Proper Forcing, in Foreman, Kanamori, and Magidor, eds,

Handbook of Set Theory.

[2] U. Abraham and S. Shelah, A 22 well-order of the reals and incompact-ness of L(QMM). Annals of Pure and Applied Logic 59 (1993) 1-32.

[3] U. Abraham and S. Shelah, Martins Axiom and 21 well-ordering of thereals. Archive for Mathematical Logic (1996) 35, 287298.

[4] L. Harrington, Long projective well-orderings. Annals of MathematicalLogic, 12 (1977), 124.

[5] R. B. Jensen and R. M. Solovay, Some applications of almost disjointsets, in: Y. Bar-Hillel, ed., Mathematical Logic and Foundation of SetTheory (North-Holland, Amsterdam, 1970) 84-104.

[6] S. Shelah, Proper and improper forcing, 2nd edition, Springer 1998.

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[7] S. Shelah and H. Woodin, Large cardinal imply every reasonably defin-able set is measurable, Israel J. Math. 70 (1990) 381-394.

[8] R. M. Solovay, a paper to be published in the Archive.

[9] H. Woodin, Large Cardinals and Determinacy, in preparation.

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Uri Abraham and Saharon Shelah- Coding with ladders a well ordering of the reals

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Transcript of Uri Abraham and Saharon Shelah- Coding with ladders a well ordering of the reals