uri abraham and saharon shelah- ladder gaps over stationary sets

8/3/2019 Uri Abraham and Saharon Shelah- Ladder gaps over stationary sets

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Ladder gaps over stationary sets

Uri Abraham

Departments of Mathematics and Computer Science,

Ben-Gurion University, Beer-Sheva, Israel

Saharon Shelah

Institute of Mathematics

The Hebrew University, Jerusalem, Israel

October 6, 2003

Abstract

For a stationary set S 1 and a ladder system C over S, a newtype of gaps called C-Hausdorff is introduced and investigated. We

describe a forcing model of ZFC in which, for some stationary setS

,for every ladder C over S, every gap contains a subgap that is C-

Hausdorff. But for every ladder E over 1 \ S there exists a gap withno subgap that is E-Hausdorff.

A new type of chain condition, called polarized chain condition, is

introduced. We prove that the iteration with finite support of polar-

ized c.c.c posets is again a polarized c.c.c poset.

1 Introduction

We first review some notations and definitions related to Hausdorff gaps. Infact we follow here the terminology given by M. Scheepers in his monograph[3] on Hausdorff gaps, but since we restrict ourselves to (1,

1) gaps our

nomenclature is somewhat simpler. The collection of all infinite subsets of

The author would like to thank the Israel Science Foundation, founded by the Israel

Academy of Science and Humanities. Publication # 598.

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is denoted [], and for a, b [], a b means that a \ b is finite. In

this case X(a, b) (the excess number) is defined to be the least k such thata \ b k. Thus a \ X(a, b) b, but if X(a, b) > 0 then X(a, b) 1 a \ b.

A pre-gap is a pair of sequences g = {(ai | i I), (bj | j J)} whereI, J 1 are uncountable and ai, bj [] are such that

ai0 ai1

bj1 bj0

whenever i0 < i1 are in I and j0 < j1 in J. In most cases I = J = 1.Given a pre-gap as above, and uncountable subsets I I and J J, therestriction g (I, J) of g is the pre-gap {(ai | i I

), (bj | j J)}. We

write g I for g (I, I).

An interpolation for a pre-gap g is a set x such that

ai x aj

for every i and j. A pre-gap with no interpolation is called a gap. A famousconstruction of Hausdorff produces gaps in ZFC (which are now called Haus-dorff gaps). Specifically, a Hausdorff gap is a pre-gap g = {(ai | i 1), (bj |j 1)} such that for every 1 and n the set

{ | a \ n b}

is finite.A special (or Kunen) gap is a pre-gap g = {(ai | i 1), (bj | j 1)}

such that for some n0 :

1. a \ n0 b for every 1, and

2. for all < < 1,(a a) \ n0 b b

(equivalently, a \ n0 b or a \ n0 b).

The interest in these definitions arises from the fact (not too difficult to

prove) that these Hausdorff and Kunen pre-gaps are gaps and remain gapsas long as 1 is not collapsed: they have no interpolation in any extension inwhich 1 remains uncountable.

In this paper we define two additional types of special gaps: S-Hausdorffgaps where S 1 is a stationary set, and C-Hausdorff gaps, where C is aladder system over S.

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The motivation for this work is the desire to find an example with gaps of

the phenomenon in which 1 is split in a certain behavior on a stationaryset S 1 and an opposite behavior on its complement 1 \ S.

Definition 1.1 Let S 1 be a stationary set. A pre-gap g = {(ai | i 1), (bj | j 1)} is S-Hausdorff iff for some closed unbounded (club) setD 1, for every S D and for every sequence of ordinals (in | n ) increasing and cofinal in

limn

X(ain , b) = . (1)

That is, for everyk, there is only a finite number of n for whichain \k b. Since, for <

, b b, it follows that eventually X(ain , b)

X(ain, b), and hence (1) holds for every in 1. If the pre-gap g =

{(ai | i I), (bj | j J)} is defined only on uncountable sets I, J 1, wecan still define it to be S-Hausdorff if for some closed unbounded setD 1, for every S D, for every j J \ , and for every increasing sequenceof ordinals (in I | n ) cofinal in ,

limn

X(ain , bj) = (2)

Clearly, every Hausdorff gap is an 1-Hausdorff gap, and the closed un-

bounded set D can be taken to be 1. The converse of this also holds, in thesense that every 1-Hausdorff gap contains a Hausdorff gap. For suppose thatg = {(ai | i I), (bj | j J)} is some 1-Hausdorff gap, and let D 1 bethe closed unbounded set given by the definition ofg as an 1-Hausdorff gap.Define I I such that every two members of I contain a point from D inbetween. We claim that g = {(ai | i I), (bj | j J)}, is a Hausdorff gap.Indeed, if J then for every n the set E = { I | a \ n b}is necessarily finite. For if not, then let be an accumulation point ofE, andlet i E, for i , be increasing and converging to . Necessarily D,and ai \ n b shows that (2) does not hold.

Proposition 1.2 If S 1 is stationary, then any S-Hausdorff pre-gap isa gap. (So that any pre-gap containing an S-Hausdorff pre-gap is a gap.)

Proof. Assume that this not so and let x be an interpolation of an S-Hausdorff pre-gap g = {(ai | i I), (bj | j J)}. Then there is a fixed

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n0 such that for unbounded sets of indices I I J J, for every

I

, J

a \ n0 x \ n0 b.

And as a consequencea \ n0 b (3)

holds. Since g is assumed to be S-Hausdorff there exists a closed unboundedset D 1 as in the definition. We may assume that every D is anaccumulation point of I. Take now a limit S D and any sequencein I

increasing to . Take any j J \ . Then equation (3) impliesthat X(ain, b) n0, which is a contradiction.

A stronger notion than that of being S-Hausdorff can be defined if therate at which the sequences in (1) tend to infinity is uniform. For this wemust recall the definition of a scale or ladder system on a stationary set.

If S 1 is a stationary set, then a ladder (system) over S is a sequenceC = (c | S is a limit ordinal) such that every c = (c(n) | n ) isan increasing, cofinal in -sequence.

Definition 1.3 For a ladder system C over S, we say that a pre-gap g ={(ai | i I), (bj | j J)} is C-Hausdorff iff for some closed-unbounded setD 1 for all S D and j J \ there is k such that for everyn k in , if i I ( \ c(n)), then X(ai, bj) > n.

Every C-Hausdorff gap (where C is a ladder system over a stationary setS) is S-Hausdorff. Our aim in this paper is to prove the following consistencyresult.

Theorem 1.4 Assume G.C.H for simplicity. Suppose that is a cardinalsuch that cf() > 1. Let S be a stationary co-stationary subset of 1. Thenthere is a c.c.c poset of size such that in every generic extension made viaP 20 = and the following hold.

1. For every ladder system C over S, every gap contains a subgap that is

C-Hausdorff.

2. For every ladder system E over 1 \ S there is a gap g with no subgapthat is E-Hausdorff.

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2 Gaps introduced by forcing

Gaps can be created by forcing with finite conditions (a method due toHechler [1]). These gaps are not S-Hausdorff for any stationary set, as weare going to see.

Iff 2n (f is a function defined on n with range included in {0, 1}) thenf is a characteristic function and we let [f] = {k | f(k) = 1} be the subsetof n represented by f.

Let (I,


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Suppose that A I is such that , 0 A iff , 1 A. Let PA be the

subposet of P consisting of all conditions p such that dom(p) A. If p Pthen p A PA and p A p. We prove some additional properties of thisrestriction map taking p to p A.

In the definition of p q what really counts is the restriction of q to thedomain of p. That is, p q iff p q dom(p). It follows that p q impliesthat p A q A. It also follows that if p and q are conditions such thatfor C = dom(p) dom(q), p C = q C, then p and q are compatible. Infact, in this case, p q is the minimal extension of p and q.

Suppose that dom(p) = dom(q). Then p and q are compatible in P iffp q or q p.

For compatible conditions p and q, we define a canonical extension p qof both p and q. However, P is not a lattice and p q is not the minimum ofall extensions ofp and q. To define it, we first make an observation. ConsiderC = dom(p) dom(q). Then p C and q C are comparable in PC (sincethey are compatible and have the same domain), and hence we can assumewithout loss of generality that q p A and n = height(q) m = height(p)where A = dom(q) (the restriction on the heights is needed only in casep A = since it follows from q p A otherwise). Then r = p q isdefined as follows on dom(p) dom(q), and it will be evident that p q is anextension of p and q.

For i dom(q) define r(i) = q(i). For i dom(p) \ A define r(i) 2n bythe following two conditions:

p(i) r(i). (4)

[r(i)] \ [p(i)] =

{[q(k)] \ m | k


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by the definition of r. Since i A dom(p), i


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To define r, define first an extension p p by requiring that p(i)(k) =

1 and [p(, 0)] = [p(, 0)] for every , 0 C. This is possible since i

is never


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and the same restriction to the core. We also assume that the functions

p(j, 0) : n {0, 1} do not depend on , and that f() and m = mare fixed on T (m n). Now by Talayacos chain condition for P, thereis a stationary T T such that for every , T, p p is a commonextension. Pick some T that is an accumulation point of T (and suchthat for every < , j < ). Then find < , T

such that f() < .Then (as we shall see)

p p aj aj, j J, and > j > f().

Yetp aj \ m bk ,

and this is a contradiction to (8). Why does pp force aj aj? Becausethe functions p(j, 0) and p(j, 0) are the same, they describe aj nand aj n, so p p forces aj aj.

3 Specializing pre-gaps on a ladder system

Theorem 3.1 For every ladder system C over a stationary set S 1, andgap g, there is a c.c.c forcing notion Q = Qg,C such that in V

Q a restrictionof g to some uncountable set is C-Hausdorff (and hence S-Hausdorff). In

fact Q satisfies a stronger property than c.c.c, the polarized chain condition,which we shall define later.

Proof. Fix for the proof a ladder system C = c | S over a stationaryset S 1 consisting of limit ordinals, and a pre-gap g = {(ai | i 1), (bj |j 1)}. The forcing poset Q = Qg,C defined below is designed to make anuncountable restriction of g into a C-Hausdorff gap.

Define p Q iff p = (w, s) where

1. w [1] |c j|. It is easy to check that this is indeedan ordering defined on Q.

If G is generic over Q, define W =

{w | s(w, s) G}. We will provethat if g is a gap then Q satisfy the c.c.c. So 1 is preserved. Clearly,if p = (w, s) is a condition, then for any S, (w, s {}) extends p,and if j 1 and j > max(w), then (w {j}, s) extends p. (If, however,j < max(w), then (w {j}, s) may be incompatible with (w, s).) It follows

that W is unbounded in 1 and {(ai | i W), (bj | j W)} is C-Hausdorff.So the generic filter over Q selects an unbounded in 1 restriction of g

that is C-Hausdorff.If p = (w, s) is a condition then for every 1 the restriction p =

(w , s ) is defined. Clearly p p.If p = (w, s) and q = (v, r) are conditions in Q then define p q =

(w v, s r). If p and q are compatible in Q, then p q Q is the leastupper bound ofp and q.

The following lemma describes a situation in which the compatibility ofp1 and p2 can be deduced. This is the situation resulting when p1 and p2

come from a -system, with core fixed below , and such that p1 is boundedby some such that the domain ofp2 has empty intersection with the ordinalinterval [, ]. The proof is straightforwards.

Lemma 3.2 Suppose that

1. p1 = (w1, s1) and p2 = (w2, s2) are in P.

2. < < 1 are such that

(a) w1 , and p1 is compatible with p2.

(b) w2 , and s2 ( + 1) . p2 = p2 is compatiblewith p1.

3. DefineA =

{ai | i w1 \ }

B =

{bj | j w2 \ }

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and suppose that there is n A \ B such that, for every s2 \ ,

n > |c |.

Then p1 and p2 are compatible.

Proof. Form p = p1 p2 and prove that p1, p2 p. p1 p is immediate. Asfor p2 p, observe that X(ai, bj) > |c | for every i w1 \ , j w2 \ ,and s2 \ .

The following simple lemma is used in proving that Q is a c.c.c poset.

Lemma 3.3 Suppose g = {(Ai | i I), (Bj | j J)} is a pre-gap such thatfor everyi I and j J, i < j implies that Ai Bj. Theng is not a gap.

Proof. By throwing away a countable set of indices from J we can assumefor every n that ifn Bj for some j, then n Bj for uncountably manyjs. Define then x =

iIAi. Then x Bj for every j, because otherwise

there are some i I, j J, and n such that n Ai \ Bj . But then wemay find uncountably many indices j such that n Bj and in particularthere is such j > i. Thence Ai Bj, contradicting our assumption.

Theorem 3.4 Suppose that the domain of our ladder system C, namely S,is co-stationary.

1. If g is a gap then Q = Qg,C satisfies the c.c.c.

2. Suppose that T1, T2 1 \ S are stationary sets and p = (p | T),

for = 1, 2 , are two sequences of conditions in Q such that, for somefixed p Q, p p1 , p

2 , for every T1 and T2, is such

that p is compatible with every p1 and with every p2 . Then there

are stationary subsets T1 T1 andT2 T2 such that, for every1 T

1

and 2 T2, if 1 < 2 then p

11

and p22 are compatible in Q.

Proof. We prove 2 since the proof of 1 is similar. For any condition p =(w, s) define dom(p) = ws. Suppose that dom(p) . Then dom(p1)

, and dom(p2) , for every T1 and T2. We may assume thatif i < j then dom(pi) (dom(p

mj ) \ ) for , m {1, 2}.

Since T implies that S, it follows for p = (w, s) that the ladder

sequence ci for any i s is bounded below . So the finite union

{ci | i sp \ }

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is bounded below . Using Fodors lemma we may even assume that this

intersection is bounded below (extend if necessary) and has a fixed finitecardinality.

For every T1 define

A =

{ai | i wp1 \ }

Similarly, for T2 define

B =

{bi | i wp2 \ }.

Clearly, any interpolation for G = {(A | T1}, {B | T2}) is also aninterpolation for g, and hence G is a gap.

Let k be such that for every T, ifp = (w, s) and s \ , then| c |< k.

Now we find a stationary set T1 T1 such that for every n if n Afor some T1 then n A for a stationary set of s in T

1. Simply

throw away countably many non-stationary sets from T1. Similarly, find astationary T2 T2 such that if n B for some T

2 then n B for a

stationary set of T2.Now lemma 3.3 gives 1 T

1 and 2 T

2 with 1 < 2 such that

A1 \ k B2 . If we pick n A1 \ B2 such that n k then there arestationary sets T

1 T1 and T

2 T2 such that n A1 \ B2 for every

1 T

1 and 2 T

2 . Hence if1 T

1 , 2 T

2 , and 1 < 2, then p11and p22 are compatible in Q by lemma 3.2.

3.1 Polarized chain condition

Theorem 3.4 shows that the poset Qg,C for a gap g and ladder C over a sta-tionary co-stationary set S satisfies some kind of a chain condition, suitablefor two sequences indexed by stationary subsets of1 \ S. We formulate thiscondition in general and later prove that the iteration with finite supportpreserves this condition.

Definition 3.5 LetT 1 be a stationary set. A c.c.c poset P satisfies thepolarized chain condition (p.c.c) forT if it satisfies the following requirement.Suppose that

1.p = (p | T) for = 1, 2

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are two sequences of conditions in P, where T T are stationary for

= 1, 2.

2. p P is such that for each = 1, 2

p P { T | p GP} is stationary in 1,

where GP is the name of the generic filter over P.

Then there are stationary sets T T for = 1, 2 such that p11

and p22 arecompatible in P whenever 1 < 2 are in T

1 and T

2 respectively.

We want to prove that if g is a gap and C a ladder over a stationary set

S such that T = 1 \ S is also stationary, then Q = Qg,C satisfies the p.c.c.for T. The problem is that if p is as in the p.c.c. definition then it is notnecessarily of the form to which theorem 3.4 is applicable, and so we needsome argument to deduce that Q is p.c.c.

Recall that every club (closed unbounded) subset of1 in a generic exten-sion of V made via a c.c.c poset contains a club subset in V. The followingproperty of c.c.c posets is also needed.

Lemma 3.6 LetP be a c.c.c poset. Suppose that T 1 is stationary, andp | T is a sequence of conditions in P indexed along T. Then thereexists some p such that

p { T | p G} is stationary.

In fact, the set of these s is stationary in 1.

Prof. Assume that this is not the case and, for some club D 1, for every T D there is a club set C (necessarily in V) and an extension p

of

p such thatp C T p G. (9)

Let C = { 1 | ( < ) C} be the diagonal intersection of theseclub sets. Then C is closed unbounded in 1. Take a maximal antichain

(surely countable) from the set of extensions {p | T D}, and let 0be an index in T C D higher than all indexes of this countable antichain.Then p0 is compatible with some p

with < 0. But 0 C which leads

to a contradiction since p0 forces that p0 G, and p forces that p0 G

(by 9).Now we prove that Q is p.c.c. for T = 1 \ S.

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Lemma 3.7 If C is a ladder system over S, and T = 1 \ S is stationary,

then, for any gap g, Qg,c is p.c.c. over T.

Proof. Suppose that T1, T2 T are stationary, and p1, p2 are two sequencesof conditions indexed along T1 and T2. Let q

Q be such that for = 1, 2

q Q { T | p G} is stationary in 1. (10)

We claim first that we may assume that q p1 for every T1. Thiscan be achieved as follows. First, observe that the set of T1 for whichp1 and q

are compatible is stationary. Then use Fodors theorem on thisstationary set to fix p1 . Rename T1 to the the resulting stationary set.

Redefine p1 as p1 q, and finally apply lemma 3.6 to obtain 0 such that

p10 Q { T1 | p1 G} is stationary in 1.

Now q is as required. Since q extends the original q, it still satisfies (10)with respect to T2. Repeat this procedure for T2, and obtain two sequencesto which Theorem 3.4 is applicable.

We note here for a possible future use a stronger form of polarized chaincondition (strong-p.c.c) which is not used in this paper.

Definition 3.8 Let T 1 be stationary. A poset P is said to satisfy the

strong-p.c.c over T if whenever two sequences are given

p = (p | T) for = 1, 2

of conditions in P, where T T are stationary for = 1, 2, and for somep P, for every = 1, 2,

p P { T | p GP} is stationary in 1,

then there are stationary subsets T T for = 1, 2, and conditions q p2

for T2 such that:

For every T2 and q P such that q q there exists < such that for every that satisfies < T1

q and p1 are compatible in P.

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4 Iteration of p.c.c posets

Our aim in this section is to prove that the iteration with finite support ofp.c.c posets is again p.c.c. It is well known (by Martin and Solovay [2]) thatsince each of the iterands satisfies the countable chain condition the iterationis again c.c.c, but we have to prove the preservation of the polarized property.

Our posets are separative (and if not, they can be made separative). Aposet is separative iff p q implies that some extension of q is incompatiblewith p.

Lemma 4.1 Suppose thatP is a p.c.c. poset, and that Q VP is (forced byevery condition in P to be) a p.c.c. poset. Then the iteration P Q satisfiesthe polarized chain condition too.

Proof. Suppose that (p, q) P Q are given for T T and for

= 1, 2, such that for some condition (p,q) P Q

(p,q) { T | (p, q

) GPQ} is stationary (11)

for = 1, 2. Then

p P { T | p GP} is stationary.

Let G P be V-generic, with p G. In V[G] form the interpretationsq[G] (interpretation ofq) and Q[G] (interpretation ofQ). Then q[G] Q[G].Define the sets

T = { T | p G}, = 1, 2

(which are stationary) and define the sequences

q[G] | T

, for = 1, 2.

Then in V[G]

q[G] Q[G] { T

| q[G] H} is stationary

where H is the name for the V[G] generic filter over Q[G]. (This follows from(11) and since forcing with P Q is equivalent to the iteration of forcing withP and then with Q[G].)

Since Q[G] satisfies the polarized chain condition for T, there are station-ary sets T

T such that:

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if 1 T

1 , 2 T

2 , and 1 < 2, then q11

[G] and q22[G] are

compatible in Q[G].Back in V, let S1 and S2 be V

P names of T

1 and T

2 respectively, forcedto have these properties. The following short lemma will be applied to S1and to S2.

Lemma 4.2 Suppose that p | T is a sequence in P, S is a name of asubset of 1 and p P a condition such that

p P S { T | p G} and S is stationary in 1.

Then there is a stationary subset T T, and conditions p extending bothp

and p for each T such that p S.

Proof. Define T by the condition that T iff T and there is acommon extension of p and p that forces S. We must prove that T

isstationary. If C 1 is any closed unbounded set, find p

p and Csuch that p S. Then T and p p G. Hence p p

(becauseP is separative). So T.

Apply the lemma to S1 and find a stationary set T1 T1 and conditions

p1 p1 , p, for T

1 such that

p1 S1.

Then (lemma 3.6) find an extension of p, denoted p

, such thatp { T1 | p

1 G} is stationary.

Apply the same argument to S2, and find a stationary set T2 T2 and

conditions p2 p2 , p

for T2 such that p2 S2. Now p

p canbe found such that

p { T2 | p2 G} is stationary.

Since P satisfies the p.c.c., there are stationary sets T1 T1 and T

2 T

2

such that for every 1 < 2 in T1 and T

2 (respectively) p

11

and p22 arecompatible in P, say by some condition p extending both. But then p 1

S1 and 2 S2. It follows that (p11, q11) and (p22, q22) are compatible in P Qshowing that P Q satisfies the p.c.c. The point is that

p P q11

and q22 are compatible in Q

and hence for some q VP, p P q q11 , q

22

. That is, (pq) (p11 , q11

), (p22, q22

).

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Theorem 4.3 Let T be a stationary subset of 1. An iteration with finite

support of p.c.c. for T posets is again p.c.c. for T.

Proof. The theorem is proved by induction on the length, , of the iteration.For a successor ordinal, this is essentially Lemma 4.1. So we assume that is a limit ordinal, and P | is a finite support iteration, whereP+1 = P Q is obtained with Q V

P a p.c.c poset for T. Thusconditions in P are finite functions p defined on a finite subset dom(p) ,and are such that for every dom(p), p P p() Q. It is well-known that P satisfies the c.c.c, and we must prove the polarized property.

Suppose that p = pi | i T for = 1, 2 are two sequences of conditionsin P, where T T are stationary, and suppose also that p P is suchthat

p P {i T | pi G} is stationary for = 1, 2. (12)

We may assume that p is compatible with every pi (just throw away thoseconditions that are not). The case cf() > 1 is trivial, because the supportof all conditions is bounded by some < to which induction is applied).So there are two cases to consider.Case 1: cf() = . Let n | n be an increasing -sequence convergingto . For every pi there is n such that dom(p

i) n. It follows from

(12) that for some specific n , for some extension p p

p {i T | pi Pn G} is stationary for = 1, 2.

Now we can apply the inductive assumption to Pn.Case 2: cf() = 1. Let | 1 be an increasing, continuous, andcofinal in sequence. Intersecting T1 and T2 with a suitable closed unboundedset, we may assume that for every < T1 and T2, dom(p1) .

We claim that we may without loss of generality assume that, for some < , dom(p) for all T. We get this in two steps.

In the first step, find a stationary T1 T1 such that the sets dom(p1),

for T1, are bounded by some < . For each T1 let p

1 be a common

extension of p1 and p

. Then (use lemma 3.6) find an extension p

p

such thatp { T1 | p

1 G} is stationary.

Since p extends p, p {i T2 | p2i G} is stationary. We can again

assume that each p2i is compatible with p and get T2 T2 stationary such

that dom(p2) is bounded by some < (we rename to be the

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maximum of and ). Rename the stationary sets as T1, T2 and we have

our assumption.Apply induction to P and to the conditions p

. This yields two

stationary subsets which are as required.

5 The model

Theorem 5.1 Assuming the consistency of ZFC, the following property isconsistent with ZFC. There is a stationary co-stationary set S 1 suchthat

1. For every ladder system C over S, every gap contains a C-Hausdorffsubgap.

2. For every ladder system H over T = 1 \ S there is a gap g with nosubgap that is H-Hausdorff.

To obtain the required generic extension we assume that is a cardinalin V (the ground model) such that cf() > 1 and even

1 = . We shallobtain a generic extension V[G] in which 20 = and the two requiredproperties of the theorem hold. For this we define a finite support iterationof length , iterating posets P as in section 2, which introduce generic gaps,

and posets of the form Qg,C, as in section 3.1, which are designed to introducea C-Hausdorff subgap to g.

We denote this iteration P | < . So P+1 P R(), where the-th iterand R() is either some P or some Qg,C. The rules to determineR() are specified below. For any limit ordinal , P is the finite supportiteration of the posets P | < . We define P as our final poset, and weshall prove that in VP the two properties of the theorem hold.

Recall that P satisfies Talayacos condition and is hence a p.c.c. poset,and each Qg,C is p.c.c. over T = 1 \ S (by lemma 3.7).

Since the iterand posets satisfy the p.c.c over T, each P is a p.c.c. poset

over T (and in particular a c.c.c poset). It follows that every ladder systemand every gap in VP are already in some VP for < . It is obvious thatif g VP is forced by p P to be a gap in VP, then p forces it to bea gap already in VP.

To determine the iterands, we assume a standard bookkeeping schemewhich ensures two things:

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1. For every ladder system C over S and gap g in VP, there exists a stage

< so that C, g VP, and R() is Qg,C.

2. For some unbounded set of ordinals the iterand R() is P,producing a generic gap g, and the subsequent iterand R( + 1) is Qg,Cfor some ladder sequence C over S.

The first item ensures that, in VP, for every ladder system C over S,every gap contains a C-Hausdorff subgap. (A C-Hausdorff subgap in V+1remains C-Hausdorff at every later stage and in the final model).

Suppose now that H is a ladder over T = 1\S. Then H appears in someVP such that R() is the poset P, and R( +1) is the poset Qg,C where g is

the generic gap introduced by R(), and C is some ladder sequence over S.We want to prove that g is a gap that has no H-Hausdorff subgap in VP .We first prove that g remains a gap in VP. It is clearly a gap in VP+1 byLemma 2.3. Since g is C-Hausdorff in VP+2, it remains a gap in VP (byLemma 1.2).

This generic gap g satisfies the conclusion of lemma 2.4 in VP+1:

If J, K 1 are unbounded, then there is a club set D0 1such that for every D0 and k K \ there are m anda sequence j(n) J increasing and cofinal in such thataj(n)

\ m bk for all n .

Since P P+1 R, where the remainder R P/P+1 is interpreted inVP+1 as a finite support iteration of p.c.c. posets over T, we can view Pas a two-stage iteration in which the second stage is a p.c.c. poset over T.Thus, for simplicity of expression, we can assume that VP+1 is the groundmodel. The following lemma then ends the proof.

Lemma 5.2 Suppose in the ground model V a ladder system H over a sta-tionary set T 1, and a gap g that has the property quoted above (theconclusion of lemma 2.4). Suppose also a poset R that is p.c.c. over T.

Then in VR

the gap g contains no H-Hausdorff subgap.

Proof. Let g = {(ai | i 1), (bj | j 1)} and assume (for the sake of acontradiction) some condition q in R forces that g = {(a | A), (b | B)} is a H-Hausdorff subgap, where A and B are names forced by q to beunbounded in 1. Since every club subset of 1 in a c.c.c. generic extension

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contains a club subset in the ground model, we may assume that the club,

D, which appears in definition 1.3 (of g

being H-Hausdorff) is in V.For every T D define two conditions in R (extending the given

condition q):

1. p R is such that for some () 1 \ , p R () A. (This ispossible since A is forced to be unbounded.)

2. q R extending p is such that, for some () 1\, q R () B.Moreover, as g is forced to be H-Hausdorff, we can assume that forsome m ,

q R for every n m, if i A ( \ c(n)), then X(ai, b()) > n.

By Lemma 3.6 some condition forces that q G (and hence p G) fora stationary set of indices T D. Since R is p.c.c. for T, there arestationary subsets T1, T2 T such that any p1 is compatible with q2 if1 T1, 2 T2 and 1 < 2.

Consider now the two unbounded sets J = {() | T1}, and K ={() | T2}. Apply the conclusion of lemma 2.4 quoted above to J andK, and let D0 be the club set that appears there. Pick any D T2 D0.Consider k = (). Then k K\ , and so there are m and a sequencej(n) J cofinal in such that

aj(n) \ m bk for all n . (13)

Yet every j(n) is of the form (n) for some n T1 , and the ns tendto . So (13) can be written as

X(a(n), bk) m. (14)

It follows from the definition of T1 and T2 that pn and q are compatible inR. If q is a common extension, then q forces that for every n m if i =(n) c(n) then X(ai, bk) > n. It suffices now to take n max{m, m}to get the contradiction to (14).

References

[1] S. H. Hechler, Short nested sequences in N \ N and small maximalalmost disjoint families, General Topology and its Applications, 2, pp.139149, 1972.

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[2] D. A. Martin and R. Solovay, Internal Cohen extensions, Ann. Math.

Logic 2, 143,178, (1970).

[3] M. Scheepers, Gaps in , in Set Theory of the Reals (Ramat Gan, 1991),Israel Mathematical Conference Proceedings, Vol. 6, Bar-Ilan Univ., Ra-mat Gan (1993), 439561.

[4] D. Talayco, Applications of cohomology to set theory. I . Hausdorff gaps,Ann. Pure Appl. Logic, vol. 71, 1995, no. 1, 69106.

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uri abraham and saharon shelah- ladder gaps over stationary sets

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