Saharon Shelah- Special Subsets of ^CF(μ)-μ, Boolean Algebras and Mahram Measure Algebras

8/3/2019 Saharon Shelah- Special Subsets of ^CF()-, Boolean Algebras and Mahram Measure Algebras

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SPECIAL SUBSETS OF CF(), BOOLEAN ALGEBRAS ANDMAHARAM MEASURE ALGEBRAS

SH620

Saharon Shelah

Institute of MathematicsThe Hebrew University

Jerusalem, IsraelRutgers University

Mathematics DepartmentNew Brunswick, NJ USA

University of WisconsinMathematics Department

Madison, WI USA

Abstract. The original theme of the paper is the existence proof of there is =

: < which is a (, J)-sequence forI = Ii : i < , a sequence of ideals.This can be thought of as a generalization to Luzin sets and to Sierpinski sets, but

for the productQ

i


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2 SAHARON SHELAH

Annotated Content

0 Introduction

1. The framework and an illustration

We define when = : < is a ( , I , J )-sequence for I = Ii : i < ,which means (I = Jbd for simplicity) that each belongs to

i 20 is almost OK, but involves more pcf considerations),then is a free caliber of the Maharam measure algebra. We use: if > 2, f Ord for < , = f/J + f/J, then (almost always) for some ideal I on

kappa extending J and X [], f/I : < are pairwise =I.

3. Strong independence in Maharam measure algebrasWe define when is a super ( , I , J )-sequence for I. The strengthening is

that we now can deal with n-tuples (any n < ) and prove the easy existence(see 3.1, 3.2). We define for a set of intervals in a Boolean algebra variantsof independence and strong negation of it (3.4) and apply it to prove existenceof strongly -anti-independent set in Maharam Measure algebra (3.7), which (by3.8) suffices for having a subalgebra of dimension with no independent set ofcardinality . This completes the consistency part of the solution of a problem,which was to characterize all cardinals which can have this property.

We prove here, e.g., if = +1 = + , then there is a Hausdorff compactzero dimensional topological space with measure on the family of the Borel subsetssuch that it has dimension , so as a measure space is isomorphic to the Maharammeasure space B(), but there is no homomorphism from X onto 2 (see 3.9). Wefinish by some easy examples.

4. The interesting ideals and the direct pcf application


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SPECIAL SUBSETS 3

We return to our original aim: existence of -sequences for I. In 4.1 we considersome ideals (JbdA ,

2+1). We point out (4.9) that for I = Jbdi

: i < ,

if = tcf(i


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4 SAHARON SHELAH

results (so e.g. the first question is not answerable in ZFC) and phrasing the prin-ciples involved, so slightly sharpening the previous results. (See 6.14 - 6.17). Sotogether with the earlier part of the section we have answered [M2, Problems 35,36]

and [M2, Problems 32,33] in the case we are near a strongly limit singular cardinal.7. A nice subfamily of function exists

For completeness we deal with the following: f Ord for < are given,

2 < = cf() and we would like to get approximation to for some X ,|X| = , f : X is a -system, continuing [Sh 430, Claim 6.6D]. We phrasea special case (7.3) and deal with some variants.

8. Consistency of P(1) has a free caliber and discussion of pcfWe deal with another of Monks problems, [M2, Problem 37], proving the con-

sistency of there is no complete Boolean algebra B of cardinality 21 with empty

free caliber (in fact 1+1 = 21

is always a free caliber of B). The universe isobtained by adding 1+1 Cohens to a model of ZFC + GCH, and the proof uses7. We finish by discussing some pcf problems: pcf preserves being even; and westate a consequence of { : strong limit, cf() = 0, + = 2} being unbounded(here?).

9. Having a -sequence for a sequence of non-stationary idealsWe return to the original theme, for a more restricted case. We assume = cf(2)

where is strong limit singular, and in this section = 2 i.e. 2 is regular(for the singular case see 10). We get quite strong results: (fix n() < forsimplicity) for some ideal J on cf() (usually Jbdcf(), always close to it) we can find

i : i < cf(), i < j max(i) < min(j), i = i, : < n(), i,+1 > 2i,(i, regular of course, = supi


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SPECIAL SUBSETS 5

preparation for a proof of the consistency of a negative answer with strong violationof GCH. We use Boolean algebras generated by xs freely except for xxx = 0for {,, } W for some set W of triples with intersection having at most one

element. The point is that the properties of is a -sequence for I with suchideals I (unlike the ones associated with the Erdos-Rado theorem) are preservedby adding many Cohens to (where |Dom(Ii)| etc.).

12. Constructing ccc Boolean algebras with no large independent setsWe complete the consistency results for which the ground was prepared in 11.

We construct the relevant Boolean algebra using a (, J)-sequence for I, I as there,using, as building blocks, Boolean algebras generated e.g. from the triple system. Sowe will give sufficient conditions for the -c.c. and other properties of the Booleanalgebra.

13. The singular caseWe continue 11, 12 by dealing here with the case is singular but ( < )

(||


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6 SAHARON SHELAH

0 Introduction

Our original aim was to construct special subsets ofi


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SPECIAL SUBSETS 7

(a) I is an ideal on (if not mentioned, we assume I = Jbd ), Ii is an ideal onDom(Ii),

(b) J is an ideal on (if not mentioned, we assume J = Jbd ),

(c) i

j

j


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8 SAHARON SHELAH

m


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SPECIAL SUBSETS 9

Christian Carrieres, Donald Monk and the participants of the seminar in Madisonfor their influence, and mainly Mirna Dzamonja for god-mothering this paper inmany ways and to David Fremlin who lately informed me that 1.16 was well known

and 3.8, 3.15 have already appeared in Plebanek [Pl1], [Pl2].Concerning 3, the question was asked for = 1 by Haydon and appeared in

Fremlins book [Fre]. Haydon [Ha1], [Ha2] and Kunen [Ku81] independently provedit to be consistent for = 1 assuming CH. The question from [Ha1] and [Fre] waswhat happens with 1 under M A. Recently, Plebanek [Pl1], [Pl2] proved thatunder M A all regular cardinals 2 fail the property, and finally Fremlin [Fre]gave the negative answer to the original question of Haydon by showing that underM A the property fails for 1. Dzamonja and Kunen [DK1], [DK2] considered thegeneral case (any ) and topological variants.


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1 The framework and an illustration

We are considering a sequence Ii : i < of ideals, and we would like to finda sequence = : < of members ofi

j


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SPECIAL SUBSETS 11

2) We say I = Ii : i < is normal if

comp(Ii) > j


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Let Zi = { i : < }, so Zi j

(Dom(Ii)) to a set C of colours (or from [Dom(Ii)]() (or from [()]


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1.6

1.7 Comments: 1) Of course in 1.6 we can restrict ourselves to colouring of pairs.

Note that the conclusion works for all ds simultaneously. Also, additional proper-ties of the cis are automatically inherited by c, see 1.8 below.2) We can also be interested in colours of n-tuples, n > 3, where i < as in clause(e) of 1.6 does not exist.3) What is the gain in the conclusion?

A reasonable gain is catching more cardinals, i.e. if Ii = Jbdi

, I = Jbd , thenin addition to having an example for i we have one for . A better gain is whenI is simpler than the Iis. The best situation is when we essentially can get I =Jbd , J = J

bd for all normal I with |Dom(Ii)| : i < increasing with limit .

Assuming a case of G.C.H. this is trivially true.

* * *Normally we can find many tuples for which there is i < as in clause (e) of 1.6.

1.8 Fact. In 1.6 if = (2||)+, or at least = cf() & ( < )(|||| < ) then:

() for every X [], we can find Y [X] and i < and a 1-to-1 function hfrom Y into Dom(Ii) such that

c(0, . . . , n1) = ci(h(0), . . . , h(n1))

for 0, . . . , n1 Y (actually h() = (i), where for all we have i = for some

j


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14 SAHARON SHELAH

(c) = supi


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SPECIAL SUBSETS 15

In =: {X Dn :there is h : X 2n such that for no infinite subset

A of (n+3(n))+ is h (X [A]n) constant}.

1.12 Fact. In is an ideal.

1.13 Fact. The ideal In is not trivial (so Dn / In).[Why? By the Erdos-Rado Theorem, see 1.17 - 1.18 for a detailed explanation.]

1.14 Fact. In is +n -complete.

[Why? If hi : Dn 2n

(i < n), then there is h : Dn 2n

such that h(x) =h(y) i

hi(x) = hi(y)].

1.15 Conclusion. So, By Lemma 1.9, there is = i : i < + = which is a

-sequence for In : n < .

We apply Conclusion 1.15 to measure algebras getting a well known result: 1.16 ApplicationAssume = + and is a strong limit singular of cofinality 0 (i.e., as in 1.11).IfB is a measure algebra (Maharam) of dimension , we can find a B for < with Leb(a

) > 0 for each , such that for every X [] we can find

n < , 1, . . . , n X satisfying

B n=1

a = 0.

Proof. Let and In be as in conclusion 1.15 (all in the context of Example 1.11).Let xn, : n < , < n+3(n)+ be independent in the sense of measure, allelements ofB and of measure 1/2.

For any n 0 if n 2). Let


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16 SAHARON SHELAH

y =n5

y,n B.

So Leb(y) 1 2

n5 2n = 1 2 24 = 1 23 > 1/2. We let a = y for

< . We check that a : < is as required. Suppose X []. So, as I isnormal, for some n > 5 and

0 and such that z u

y(u) . Then

without loss of generality

< 2n 1 z xn, z 1 xn, .

Case 1: |{ : z xn,}| n. Let u [{ : < 2n 1}]n be such that

u(z xn,).

So zle

uxn, . But z y(u) 1B

u

xn,, a contradiction.

Case 2: Not Case 1. So necessarily |{ : z 1B xn,}| n and continue asabove using 1B xn, . 1.16

Let us elaborate on the ideals used above.


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1.17 Definition. For n,, let

ERJn,

= Jn,

= {A []n

: there is no w satisfying otp(w) = and [w]n

A}

ERIn,, = In,, = {A []

n : there are Ai Jn, for i < i() <

such that A =

i


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Ai =

Ai\

j


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SPECIAL SUBSETS 19

(So if X [], ( E) |X (, min(E\( + 1))]| 1 then f : X is


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2 There are large free subsets

The reader may wonder if really something like = cf() (, 2

] for stronglimit singular, is necessary for 1.16. As in [Sh 575], the answer is yes, though notfor the same reason.

Of course, in what follows, Maharam measure algebra can be replaced by anymeasure algebra. The interesting case is ()( < 0).

2.1 Fact. LetB be a Maharam measure algebra. If2 = 0 < cf() 2

and a B+, (so Leb(a) > 0) for < are pairwise distinct, then for some

X [] we have:

() any nontrivial Boolean combination of finitely many members of {a

: X} has positive measure.

Proof. Let {xi : i < i()} be a basis of the Maharam measure algebra (so eachxi has measure 1/2 and xis are measure-theoretically independent). So for each < we can find ordinals i(, n) < i() for n < , and a Boolean term suchthat a = (xi(,0), xi(,1), . . . ). Note that this equality is only modulo the idealof null sets. Remember

()0 we can replace x : < by x : X for any X [].

Without loss of generality, each is a countable intersection of a countable unionof finite Boolean combinations of the xis. Again without loss of generality, i(, n) :n < is with no repetition. Note that without loss of generality

i() = {i(, n) : < and n < }.

Hence without loss of generality i() , hence without loss of generality i() = .By Engelking Karlowicz Theorem [EK], clearly we can divide to sets X : < such that

()1 the sets A,n =: {i(, n) : X} for each satisfy: A,n : n < arepairwise disjoint.

As the number of possible terms is 20 by ()0 without loss of generality

()2 if , X then = , call it .


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SPECIAL SUBSETS 21

Note also

()3 if Y X then

ind(Y) =:

Y :for no m < and 0, . . . , m1 Y do we have :

a the complete subalgebra generated by

{xi(,n) : < m, n < }

satisfies |ind(Y)| + 20 |Y|.

[Why? We can prove by induction on ind(Y) that for some m < and0, 1, . . . , m1 ind(Y) we have a the complete subalgebra ofB generatedby {xi(,n) : < m,n < }, using the transitive character of this property. Now

for each m < and 0, . . . , m1 ind(Y), the number of a such that a (thesubalgebra generated by {xi(,n) : < m, n < }) is at most continuum.]

As cf() > , for at least one < , |X | = , hence by ()3 we have |ind(X)| =. So, without loss of generality

()4(a) the sets An = {i(, n) : < } are pairwise disjoint,

(b) = for < ,

(c) for no m < and 0 < < m < do we have am the completesubalgebra generated by {xi(,n) : < m, n < }.

Now for each < we define an ideal I on (thought apriori I = P() is

allowed): it is the ideal generated by the sets

Z, =: {n < : i(, n) = i(, n)} for < .

and (where chA(n) is 1 ifn A and 0 ifn / A for any A )

J =:

A :(x0, x2, . . . , x2n, . . . )

= (x0+chA(0), x2+chA(1), . . . , x2n+chA(n), . . . )}.

As {xi : i < i()} is free (in the measure theoretic sense),

()5 if

(a) for A J, and {m : n < } and {n : n < } such that n < i()are with no repetition and n < i() with no repetition, and

(b) (m, n < )[n = m n = m & n / A]then (x0 , . . . ) = (x0 , . . . ) in B, of course.


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22 SAHARON SHELAH

(Just apply the definition of J to x0 , x0 , x1 , . . . ). By transitivity of equality

(i.e., using n : n < such that n / A n = n = n and n A n / {m :m < } {m : m < } {m : m / n}) we get

()6 if (a) of ()5 then

(n < )[n A n = n] (x0 , . . . ) = (x0 , . . . ).

Hence J is closed under subsets and (finite) unions, that is J is an ideal on . LetI be the ideal on generated by I

J. By clause (c) of ()4 and ()6 we know

that / I; recall that I

is an ideal on though it is possible that singletons are

not in I (a violation of a convention in 0). [In fact we could have eliminated thisviolation, but there is no reason to put extra work for it.] Also J I.

Now, the number of possible ideals on is at most 2 < cf(), so it sufficesto prove

()7 if Y , Y I I, where I is an ideal on (so / I butsingletons may or may not belong to I) extending J, then any finite Booleancombination of {a : Y} has positive measure.

Proof of ()7. Let 0 < < m1 be from Y. Let

A = {n < : for some < k < m we have i(, n) = i(k, n)}.

By the definition of Z,, clearly A I. For Z i() let B[Z] be the complete

subalgebra ofB generated by {x : Z}. We let B =: B[Z] where Z =

{i(, n) : < m, n A}. Let B =: B

[{i(, n) : n A}].As B is complete, for each < m we can find b

, b

+ B

such that

(i) b a b+ ,

(ii) if c B then c a c b and c a c b

+ .

By the definition ofB and the assumptions on xi : i < i() and on a : < clearly

()8 if {i(, n) : n A} Z and {i(, n) : n \A} Z = and Z i()then

(ii)Z if c B[Z], then c a c b

and c a c b

+ .


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SPECIAL SUBSETS 23

Obviously, for some Boolean terms , + we have

b

=

(. . . , xi(,n), . . . )nAb+ = + (. . . , xi(,n), . . . )nA.

Now, as = for Y, clearly =

and + = + for some fixed and

+. Also b < b+ as otherwise \A J. Let b = b

+ b

so Leb(b) > 0, and for

some term , b = (. . . , xi(l,n), . . . )nA, and let b =

0 any Boolean combination of the a( < m) has positivemeasure.

[Why? prove it on {a : < m

} by induction on m

m using ()8.]For proving Leb

0, we define an equivalence relation E on :

n1En2 iff for every < k < m we have

i(, n1) = i(k, n1) i(, n2) = i(k, n2).

Clearly E has finitely many equivalence classes, say A0, A1, . . . , Ak()1. For k1 k() and = k : k1 k < k() satisfying k < m let

Zk1, =

(. . . , xi(n,n), . . . ) : for every k < k(), for some < m we have

n : n Ak = i(n, ) : n Ak,

but ifk k1 then = k

.

We prove by induction on k1 k() that for any appropriate

c =: Leb(

{b : b Zk1,}) > 0.

(In fact the measure does not depend on .)For k1 = k() we have {b : < m} Zk1, so this gives the desired conclusion.

The case k1 = 0:

It is trivial: Z0, is a singleton {(. . . , xi(n,n), . . . )}, where n An so obvi-

ously it has positive measure.

The case k1 + 1:


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24 SAHARON SHELAH

So let = k : k1 + 1 k < k(), and we know that for each n < the elementd = cn is > 0. For t < m let ft be a function from Y =

i(, ) : < such

that if Ak then k [k1 + 1, k()] n = k and k = k1 n = 0 and k < k1

n < m

into , ft is one to one, ft is the identity on Y

= {i(n, ) Y : / Ak1}and Rang(ft (Y\Y)) : t < m are pairwise disjoint and

Ak1 ft(i(0, )) = i(t, ).

Now we can imitate the beginning of the proof of ()5 and getn 0. Let

Yt = Rang(ft), and note that f0 is the identity and Y0 = Y. Clearly ft induces an

isomorphism from B[Y0] onto B[Yt]. Call it ft and easily dt =: ft(d0). So we canimitate the beginning of the proof of ()5 and get

n 0. But

ci =n scite{2.16} undefined2) We may wonder if 2 is necessary. Actually it almost is not (see 2.5below) but cf() > 20 is essential (see 3.11 below).

We shall see below (in 2.5) what we can get from the proof of 2.1.

2.3 Definition. For a Boolean algebra B we say a, b : < is an explicitly

independent sequence of intervals in B if:

(a) B a < b,

(b) if u0, u1 are finite and disjoint then

B u0

b u1

(a) > 0.

2.4 Claim. Assume

()Y[X](a) |X| = andB(X) is a Maharam measure algebra with free basis {xi : i X}. For Z X we letB(Z) be the complete subalgebra ofB(X) generatedby {xi : i Z}

(b)Y a B+ (i.e., Leb(a) > 0) for Y and < a = a, while

|Y| = , Y a set of ordinals for simplicity.


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SPECIAL SUBSETS 25

1) If = cf() > 1 then for some Y [Y], Z [X] |X| = and 1 < , = cov(, +1 , 1, 2) < then Y can be

represented as the union of subsets Y such that for each there is Z []1

satisfying {a : Y} B(Z).

4) If the clause () below holds then we can represent Y as the union of subsetsY each satisfying (c) below (and (b)Y),

(c)Y a = (. . . , xi(,n), . . . )n


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26 SAHARON SHELAH

2.5 Claim. In 2.1 we can weaken 2 to 20 or even cf() > 20except possibly when is singular but below fails:

for any countable seta of regulars, |pcf(a)| 0

or () from 2.6.

Proof. Without loss of generality we assume ()4 from the proof of 2.1 (as the proofof 2.1 up to that point works here too). Let J be as there, so J is an ideal on , so

(+) J is an ideal on and i(, n) : n < /J for < are pairwise distinct;

by the following observation 2.6 for some ideal I on extending J and X [],we have

X & X & = {n : i(, n) = i(, n)} I.

This is enough for continuing with the old proof of 2.1. 2.5

2.6 Fact. 1) If J is an ideal on , f/J : < are pairwise distinct functions inOrd and = cf() > 2 then for some ideal I on extending J and X [] wehave:

X & X & = f =I f

except possibly when

() is singular and , where for any set a of regular cardinals > we have |a| |pcf(a)| .

2) We can replace () by

() is singular and +, or ,, where

+, for no set a of regular cardinals > , do we have |a| and =sup( pcf(a))

, there are no , cf() = < < and increasing sequences = i :

i < of regular cardinals (2, ) such that max pcf{i : i < } : < is increasing with limit but for every ultrafilter D on wehave

sup

tcfi


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SPECIAL SUBSETS 27

Proof. 1) Follows by (2).2) The proof is split to cases.

Case 1: is regular. We apply 7.5 here which is [Sh 430, 6.6D] (or in more details[Sh 513, 6.2]).

Case 2: singular. First note Subfact: +, ,.

[Why? Let a exemplify +,, let pcf(a) \ {} be increasing for < withlimit . Let b a be such that = max pcf(b) and let : < list a and let be: if b and (2

)+ if b. Now = : < exemplifies ,.

First max pcf{ : < } = < and is increasing with limit sup( pcf(a)).

Secondly, for every ultrafilter D on for each we have tcf


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() if i : i A satisfies i A i <

i then for unboundedly many

ordinals < i B i < ,i <

i ,

i A\B ,i = i .

Can B = ? This would mean that for some unbounded X we have

X (i A)[,i = i ]

hence { : X} pcf{cf(i ) : i A

}, so {cf(i ) : i A} has pcf of

cardinality > 2 whereas |A| , contradiction, so really B = .As we are assuming ,, there is an ultrafilter D on A

such that

sup

tcf iA

,i/D

: <

.

Clearly

tcf iA

,i/D

<

(by the choice of A = A). Without loss of generality > for each < . So

we can choose, for each , a function h iA ,i such that() if < and = , while ,i : i A

D ,i : i A then

,i : i A


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SPECIAL SUBSETS 29

2.7 Claim. 1) If and = cf() < , and , (from 2.6) then for somef

(for < ) the conclusion of 2.6(1) fails.

Proof. 1) Let , i (i < , < ) be as in ,.

Let a =: {i : i < }, and = max pcf(a). Without loss of generality

: < is increasing with limit . By [\Sh:g }. Now

check. 2.7

2.8 Discussion. 1) So if 2 < , = cf() then 2.7 shows that 2.6 is the bestpossible. (Of course, we still do not know if , is possible). See more in 3.13.

2) Note: If cf() > 2, and

(a)(a Reg & |a| < min(a) |pcf(a)| |a|),

then , cannot occur as without loss of generality

J = {A : max pcf{i : i A} < max pcf{

i : i < }}

does not depend on .


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3 Strong independence in Maharam measure algebras

3.1 Claim. Assume

(a) Ii is a i-complete ideal on i for i < ,

(b) i >

j


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SPECIAL SUBSETS 31

Then

()1 Claim 3.1 applies,

()2 for every m < and X


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32 SAHARON SHELAH

3.5 Definition. We say that B = B(B, (a, b) : < ), or B is the free

extension of B for (a, b) : < , if:

()B

is the algebra freely generated byB

{x : <

} except for theequations:

(a) the equations which B satisfies,

(b) a x b, for < .

3.6 Observation. 1) In 3.4(3), ifB B(0), 0 + + 1 then we can embedB into B(1) over B.

2) There are obvious implications among the notion from Definition 3.4 and some

equivalences: independent (3.4(3)) with explicitly independent; and stronger inde-pendent with (a) of 3.4(1) and if 1, . . . , 2, 1, . . . , m < with no repetition,

B

n=1

a m=1

(b) > 0 .

3.7 Lemma. Assume is strong limit singular of countable cofinality and =+ = 2.Then inB(), (the Maharam measure algebra of dimension ) we can find a se-quence (a, b) : < such that:

(a) B() a < b,

(b) (a, b) : < is strongly -anti independent.

Remark. What is the difference with 1.16 Note that 3.4(ii)(b) speaks on no freesubset of the Boolean algebra, not just of the set.

Proof. 1) Let =

n


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SPECIAL SUBSETS 33

b,n =

{(1 xm : appears in (m)}.

We define by induction on n, the elements a,n, b,n as follows: for n < 5 let a,n =0, b,n = 1. For n 5 we let a,n = a,n1 (a

,nb,n) and b,n = b,n+1 (b

,n

a,n). We can prove by induction on n < that a,n1 a,n b,n b,n1.We can compute the measure, e.g., let (b,n a,n) =

{1 2(1) : 5 n}.

Let a =n


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34 SAHARON SHELAH

(We just do it by induction on using the definition ofk()1=k

In and the definition

of I.) So it suffices to show that dh(t) : t Sk() is not independent. For this

just note:

for every R>0 if n is large enough compared to k(), 1/, and m() islarge enough compared to n then for every ultrafilter D on B() we can by

downward induction on = k, . . . , k() 1 find us [us]m()/2k() and

s {k,... ,k()1}2 for s S such that: s t S1 and 1 < k() and

ut [xn D s(1) = 1].

Now let = (i.e., s for the unique s S0) and for m < k() letting Sm =

s Sm : if < m then s() [us]

n

, we have s Sk() dh(s) D or

s Sk() dh(s) / D.

So to prove that d : < is not independent it suffices to find S Sk()such that

SS

d

Sk()\S

d = 0,

or equivalently

S for no ultrafilter D on B() do we have

Sk() [d D S].

By the argument above it will suffice to have

S if us : s {S

: < k() satisfies: S

0 = S0, S

S,

s S us [us]

m()/22k()

andS+1 = {s

w : s S and w [us ]

n} then S Sk() / {, S}.

Now, not only that this is trivial by the probabilistic existence proof a la Erdos butthe proof gives much more than enough. 3.7

3.8 Claim. : Assume

() is regular > 0 and (a, b) : < is a strongly (or just mediumly)-anti-independent sequence of pairs fromB() satisfying a < b.


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SPECIAL SUBSETS 35

Then:

(a) There is B, such that:

() B

is a subalgebra ofB(),() B has cardinality and even dimension ,

() there is no subset of B of cardinality which is independent.

(b) Let B, x( < ) be as in 3.5, then the Boolean algebra in clause (a) canbe chosen isomorphic to x : < B.

Proof. Straight. Clause (a) follows from clause (b). For clause (b) apply Definition3.4(5) and 3.6. (Note: we can use B B( + )). It has already been done by

Plebanek [Pl1]. 3.8

3.9 Conclusion. For as in 3.7 (i.e., = + = 2, strong limit > cf() = 0)or just as in () of 3.8, we have

() there is a topological space X which is Hausdorff, compact zero dimensional,with a measure Leb on the Borel sets such that it has dimension , so as ameasure space is isomorphic to B() but there is no homomorphism fromX onto 2.

Proof. By 3.7(1) () of 3.8 holds so we can restrict ourselves to this case. So by 3.8we know that clause (a) of 3.8 holds. Now it follows that () holds, more specifically,that the CechStone compactification of B (i.e., the set of ultrafilters of B withthe natural topology) and the measure of B (which is just the restriction of theone on B()) satisfies () of 3.9. 3.9

3.10 Example: Assume B is a Maharam measure algebra of dimension and freebasis x : < , > cf() = 0. Then ()2, below holds, where

()2, there are positive pairwise distinct members a ofB() for < , suchthat for every X [] for some = from X, a a = 0.

Proof. Trivial: let =

n


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36 SAHARON SHELAH

have Leb m

b

= 0 and . Then we can find pairwise distinct a B()

for < such that for every X [] for some m < , 0, . . . , m X we have

Leb

m a

= 0, i.e., B() m

a = 0.

Proof. Like the proof of 3.10 replacing xn m


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SPECIAL SUBSETS 37

f() if

< & A ,

if


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38 SAHARON SHELAH

4 The interesting ideals and the direct pcf application

Our problem, the existence of ( , I , J )sequences for I, depends much on the

ideals Ii we use. Under strong set theoretic assumptions, there are -sequences by 1.9 (and 3.1); but we would like to prove their existence (i.e., in ZFC). Forsome ideals, by [Sh:g] we will have many cases of existence, e.g., when Ii is J

bdi

, iregular. But we are more interested in the existence for more complicated ideals.The first step up are Jbd

with a (finite) strictly increasing sequence of cardinals.

The proof for them is not much harder than with the Jbd s. We then consider thecentral ideal here: Jbd

for a (strictly) decreasing sequence of regular cardinals,

and explain why the existence of for these ideals is more useful. We also considertheir strong relative which comes from the nonstationary ideal. We would of courselove to have even stronger ideals but there are indications that for those which

we considered and failed, the failure is not completely due to incompetence, i.e.,there are related independence results (see later). We commence this section byreviewing some general definitions, some of them used earlier in the paper.

4.1 Definition. 1) For a set A of ordinals with no last element (mainly A = =cf())

JbdA = {B : B A is bounded}.

2) IfA Ord is such that cf otp(A)) > 0 and A stationary in sup(A), we let

J

nst

A = {B A : B is not a stationary subset of sup(A)}.3) IfA Ord, = cf() < cf(otp(A)) and

{ < sup(A) : A, cf() = }

is a stationary subset of sup(A), then let

Jnst,A = {B A : { B : cf() = } is a nonstationary subset of sup(A)}.

4.2 Definition. 1) For an ideal J let (J+

x)(x) mean that

{x Dom(J) : (x)} J+.

2) For an ideal J let (Jx)(x) mean

{x Dom(J) : (x)} J.


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SPECIAL SUBSETS 39

4.3 Definition. 1) J =


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40 SAHARON SHELAH

(d) Like (c), adding

(iv) = = .

Proof. Straight. For (b) (c) use induction on n = g(), see the proof at the endof the proof of 4.11, of () there. 4.4

4.5 Discussion. From 4.4, we see that for for X (Jbd

)+ there are patternswhich necessarily occur as subsets of X. These are essentially like the branches (=maximal nodes) of a tree with n levels, with a linear order on each level and withno dependencies between the different levels. These patterns were explored in [Sh

462], [RoSh 534], [Sh 575]. The patterns considered there can be represented as aset

k. Let J =


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SPECIAL SUBSETS 41

2) If [A & |A| < ] A J then we can add

(d) like (c), but adding

(iii) i < j if i < j < lg()+1.

Proof. Similar to 4.4.

4.7 Claim. Let = : < n be a decreasing sequence of regular cardinals.

(1) If > 2+1 for < n, then:

() for every A (Jbd

)+, there are A (Jbd

)+ such that


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42 SAHARON SHELAH

4.8 Definition. For a partial order P let tcf(P) = iff there is an increasingcofinal sequence of length in P (tcf stands for true cofinality); so e.g.,(,


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SPECIAL SUBSETS 43

and for some ideal J on , we have J Jbd and J is generated by

{(i, n) : n < ni, i A} : A J

.Then there is a (, J)-sequence for Jbd

i: i < .

2) Jbdi

: i < is normal (hence above is normal) if

()1 = and i < j < i,ni1 < j,0 or

()2

{i, : i < j, < nj} < j,0.

3) If we ask just to be normal it suffices to demand

()3 max pcf{i, : i < j, < nj} < j,0.

Proof. Again, let f = f : < be


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44 SAHARON SHELAH

[0, i,0] Xi, JbdQn1=1 i,

.

So for each such we have n=1

i, as given by the induction hypothesis. Let

=

0 + 1 if = 1,

{ : [0, i,0)} otherwise.

Why is the latter < i,? As i,0 < cf(i,).] 4.11

4.12 Question: Are there many cases fitting the framework of 4.11?

4.13 Answer: Not so few. E.g., for any , for many = cf() we have that

= tcf i || for i < and < ni and 0 < ni < . Then the following are equivalent

(a) for every ki : i < i


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SPECIAL SUBSETS 45

(exists by the pcf theorem). Now:

(a) (b)If

i,n

i,n/I does not have true cofinality , then for some A (I)+ we have

that

(i,n)A

i,n/I has true cofinality = (here we use the pcf theorem) and

without loss of generality A A A B, hence / pcf{i,n : (i, n) A}.Let B = {i < : (n < ni)[(i, n) A]}, so by the definition of I

we knowB I+. So, for i B we can choose ki {0, . . . , ni 1} such that (i, ki) A. So{(i, ki) : i B} A hence pcf{i,ki : i B} pcf{i,k : (i, k) A}, but doesnot belong to the later, hence not to the former, contradicting (a).

(a) (b)

So there is ki : i < i


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46 SAHARON SHELAH

5 -sequences for decreasing i by pcf

5.1 Discussion. Our aim here is to get decreasing from increasing (for Jbd ),

in some sense, to make gold from lead. We do this by using pcf assumptions,then proving that these assumptions are very reasonable.

(Note: when we cannot materialize the pcf assumptions the situation is close toSCH, and then we have other avenues for construction of -sequences for some I,e.g., (1.9, 3.1).)

* * *

In the following claim the interesting case is when are increasing, i = ,i :

< n decreasing sequence of regular cardinals, ,i > j


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SPECIAL SUBSETS 47

gi() = min m,i : if [, m,i) then Xi JbdQn1=m+1 ,i.This definition just unravels the definition of Jbd

i; note

() Xi = Xi Jbdi

,

() if Xi JbdQ lg()

,i then gi() < lg().

Now we choose by induction on m < n ordinals m < m such that for m n wehave

()m Bm =:

i B : Xif, (i):


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48 SAHARON SHELAH

()m

n1

=m

:

: < m X (Jbd[m,n)

)+.

So from ()n we get : < n X hence for all i we have Xif, (i):


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J =

A :A {(,i,) : < n, i < , < i} and;

(Ji)(Ji)[(,i,) / A]

,

and let

Ii, =


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5.7 Claim. Assume

(a) is strong limit,

(b) 0 = cf() < ,(c) 2 ++1 = [also +4++1 = is OK, or just = ++1 < pp+()

and cf(+) < ].

Then:1) We can find ,i, k such that ( i < ):

(A) i, < =

m,j m,j,

(B) 2+1,i < ,i and 20,i < 2i+1,i+1,

(C) tcf i


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SPECIAL SUBSETS 51

Proof of 5.7. 1) We will just give a series of quotations.First cf(+) = 0, so by [Sh:g, II,1.6], there is an increasing sequence i : i <

of regulars with limit + such that

= ++1 = tcfi . So without loss of generalityi

i > .

So let i = +ki , ki (0, ) strictly increasing. By [Sh:g, 5.9,p.408], we have

pp() > +ki . (We would like to have pp() = 2, but only almost proved.)This means by the no hole theorem [Sh:g, 2.3] that for some countable set a ofregulars < , = sup(a) and

+k pcf(a). So by the pcf theorem, without lossof generality +k = max pcf(a) and

+, . . . , +(k1) / pcf(a) (alternatively use

[Sh:g, VIII,1]).So necessarily

+k = tcf

a/Jbda

.

Let =

n i, ,i ai,

and (B) holds. So, as ,i ai and ,i is increasing with i, with limit , we have

tcf

i

,i/Jbd

= +k .

2) Let h : be such that (m)(0 i) (h(i) = m). Choose by induction on i,,i {h(i),m : m < } such that (b) + (c) of (2) hold. For each i we do this bydownward induction on . Then apply the last theorem. 5.8

We may deal with all ns at once, at some price. The simplest case is:

5.9 Claim. . Assume

(a) A : < is a sequence of pairwise disjoint sets,(b) = tcf

n


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52 SAHARON SHELAH

Then there is a (, J)-sequence for Jbdn, : (n, ) .

Proof. Straight.

5.10 Remark. 1) We can replace n : n < by i : i < .2) Another way to get an example for 5.4 is to have i : i < increasing contin-uous, = cf() > 0, < 0, = =

i .

3) In all the cases here we can get normality as in 4.4) See 1.19, 1.20.


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SPECIAL SUBSETS 53

6 Products of Boolean Algebras

Monk asks [M2, Problem 35,p.15]:

6.1 Monks Problem. Doesn B this holds trivially.2) FreeCal(B) = { |B| : is a free caliber of B}.We show that, e.g., if+ = 2 then the answer is NO.

6.4 Claim. Assume:

(a) there is a normal1 super (, J)-sequence for I = Ii : i < ,

(b) Ii = ERI2i,i

=:

X [i]2 : for some h : X i, |Rang(h) < i, and for

no u [i]0 do we have (h [u]2 constant) & [u]2 X

,

(c) < 1.

Then is not a free caliber of i 0 = cf(), then we canfind such i, i < and for = .

Proof. By renaming without loss of generality

()1 (i)

j


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54 SAHARON SHELAH

g() =k


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SPECIAL SUBSETS 55

Theni 2, cf() > 0 and ( < )(||0 < ) and each

i is a cardinal. Theni 0 and ( < )(|| < ),

or alternatively

() = cf() and > 2 are as in 7.5 below and we assume i < i .

(This was our first proof. It possibly covers all cases under some reasonable pcfhypothesis, and illuminates the method).

Let g i


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56 SAHARON SHELAH

(f) m\v f[] u = ,i : i u

,

(g) for every

viu,i for ordinals X we have,

i u & v ,i < f[] (i) <

,i,

(h) if v, X, i u then f[] (i) > sup

1,i1 :

1,i1

< ,i where

1 < m and i1 <

and < X implies: for every i u large

enough we have f[] (i) > max

f

[1] (i1) :

1,i1

= ,i and 1 < m

and i1 <

(the interesting case is i1 = i).

Now for any n < , and 0 < < n1 from X, we have

for every i u large enough

f0(i),f1(i), . . . , f n1(i) =0,i, : < m

, 1,i, : < m, . . . , n1,i, : < m

is as in a system, in fact

k(1),i,(1) = k(2),i,(2) (k(1), (1)) = (k(2), (2)) ((1) = (2) v).

As v = {0, 1, . . . , m 1} and in no variable is redundant clearly

for every i u large enough, (x0,i,0 , . . . ), (x1,i,0 , . . . ), . . . is inde-pendent.

This implies that g : < n is independent (ini


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SPECIAL SUBSETS 57

a Reg \, |a| max pcf(a) < .

6.9 Problem. 1) Which of the following statements is consistent with ZFC:

(a) is strong limit, cf() = 0, and for every Reg (, 2] and cardinalsn such that =

n


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58 SAHARON SHELAH

i,ni . As > 0 is regular and as we can replace ai : i < by ai : i Xwhenever X [], without loss of generality for some m,


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SPECIAL SUBSETS 59

Proof. 1) By 6.4, 6.5s proofs.2) If () use 6.7, if () the proof is similar. 6.11

6.12 Remark. We can deal with singular cardinals similarly as in the earlier proofs.

6.13 Claim. In the earlier claims if

()1 = ++, or at least if

()2 < , and [ < cf([], ) < ], = supi


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6.15 Definition. Pr,(, ) means:

iff is a partial function from to such that |Dom(f)| < for < ,

then some f

extends of the functions f.

If = we may omit it.

6.16 Claim. In Claim 6.14 the assumption on the density of box products can bereplaced by cases of Definition 6.15:

(a) 2.1 Assume B = B() is a Maharam measure algebra of dimension, cf() >20 and cf() > 2 = cf()

,0

. If Pr0(, ) then B has as afree caliber

(b) 6.6 Assume 2 < = cf(), = supi


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SPECIAL SUBSETS 61

1 in 3.7, we can have x = x : < satisfies

()B,x for every X [], m < , and (, k) < for < , k < 2m

pairwise distinct, for every n large enough there are pairwise distinct

0, . . . , 2n1 X such that

0 =


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7 A nice subfamily of functions exists

We expand and continue on [Sh 430, 6.6D], [Sh 513, 6.1].

7.1 Claim. Assume

(A) = cf() > 2,

(B) D is a complete2 filter on containing

(C) f : Ord for < ,

(D) D contains the co-bounded subsets of .

Then0) We can find w and = i : i < such that: i \ w cf(

i ) > 2

and for every i\w

i

for ordinals < (even a set inD+) we have and < 1 D is [||]+complete.

4) If in part (1) in addition (G) below holds, then we can add

()3 pcf-complete{i : i w

1 ; and < ni} if w

1 = , moreover

()4 if i < ni for i w1 then pcf-complete{cf(

ii

) : i w1 }.

(G)(i) ( < )(|| 0,

(ii) D is -complete

(iii) f = f for = (or just = X for some X D+).

5) If in part (1) in addition (H) below holds then we can add

()5 ifm < m, A Jm andi < ni fori \A (so w

0 A) then pcf{

ii

:i \ A}.

(H)(i) m < and Jm an 1-complete ideal on for m < m,(ii) D is -complete.

Remark. 1) Ifi, is singular we can replace it with a sequence i,1 : < cf(i,),and the index set : < i, by (, ) : < cf(i,) and < i,2, and i,1 ,are replaced by sequences of regular cardinals. Not clear if all this helps.2) The reader may concentrate on the case (F) + (G)(ii) holds.

Proof. 0) By part (1).

1) Let be regular large enough. Choose N such that

(i) N (H(), ),

(ii) 2 + 1 N and N = 2,

(iii) ,,,D and f : < belong to N,

(iv) N N.


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64 SAHARON SHELAH

Next choose () < which belongs to B =

{B D : B N}, which is theintersection of 2 < members ofD. Necessarily B D so () exists. Foreach i < let

Yi =: {A N : A a set of ordinals and f()(i) A},

clearly Yi = as


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SPECIAL SUBSETS 65

(1, 1)


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66 SAHARON SHELAH

[Why? The second inequality by ()4 and the choice of (i, i), the first inequalityas otherwise by renaming we can omit i+1 and contradict the otp{ Zi : k Zi}}. So

i Z

i N,

by renaming

Zi ik > sup{(i)i : k Zi},

and we get a contradiction to ()8 as in the proof of ()9 if ik1 = ik and as in

()12 if ik1 < ik.]

Hence

()15 without loss of generality i :


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[Why? Otherwise by ()12 we get a contradiction.]The conclusion can now be checked easily.2) Let a = {cf(i) :

i is singular and

i < } and use (E).

3) Easy.[Clearly D is +2 -complete where 2 =

2 = {||

: < 1}, so choose N asabove of cardinality 2.]4) Without loss of generality in clause (iii) of (G) we have < < f = f(otherwise replace D by D+ X and change f for \X in quite an arbitraryway).Assume that the desired conclusion fails. For this we choose not just one model Nbut an ( + 1)-tree of models. More precisely, we choose by induction on i asequence N : Ti such that

(a) Ti i,

(b) j < i & Ti j Tj ,

(c) |Ti| < ,

(d) N (H(), ) satisfies (i)(iv) from the proof of part (1),

(e) for Ti we have N and N : j


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SPECIAL SUBSETS 69

wk0 wk+10 , k k+1 and (i w

k1 )[(

i,k+1, i,k+1)


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()2() if > 0 then i is a limit ordinal (for every i u

), and cf(i ) : i

u is strictly increasing with limit and = tcf iu

cf(i )/Jbdu

and for every iu

i for ordinals < we have

(i u)(i < f(i) < i ),

()2() if = 0 then for some strictly increasing sequence = i : i u

of regular cardinals with limit , = tcf(iu

i/Jbdu) and for some

g = gi : i < , gi : Ord i, we have: for every iu

i for

ordinals < we have

i u i < fg(i) < i,

()3 i is a limit ordinal of cofinality for i u

and for some X []

we have: i u f(i) : X is strictly increasing with limiti and for X, the interval [f(i),

i ) is disjoint to

{f(j) : X; and j u\{i} & j = i or < and j u

}.

2) Assume > 0. For every sequence f = f : < of pairwise distinctfunctions from to >Ord such that |{f(i) : < }| < for i < , we can findu [] and n() [1, ) and v n nonempty and = ,i : < n

, i usuch that for each i

(a) for v we have that ,i is a limit ordinal, cf(,i) : i u

is strictly

increasing with limit and = tcf(iu

cf(,i)/Jbdu), and also for i < j in

u, and , k v we have cf(,i) < cf(k,j),

(b) for every ,i

,i

for ordinals < we have

(i < u)( v)[,i < (f(i))() < ,i] and

(i u)( n\v)[f(i))() = ,i].


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SPECIAL SUBSETS 71

3) In part (2) if = cf() > 0, we can replace u [] by u J+ for any normal

ideal J on . Moreover if { < : ( < cf())(||Ord is fine. If we omit the assumption |{f(i) : < }| < , instead

of v we have a partition (v1, v2, v3) of { : < n

} such that clause (a) holds for 2, clause (b) holds with v2 v3, v1 instead of v, n

\ v, and theparallel of ()3 holds for v3.

Proof. 1) By 7.1(0),(1),(2) we know that

there is i : i < and w such that letting u = \w we have:

(a) for every iu

i for ordinals < we have

i w f(i) = i ,

i u i < f(i) < i ,

and moreover (w0, w1 ),

i : i w

1 , < ni, X, f

: < , h = hi : i i +

i +

0, replacing ui : i < , without loss of generality ji = i and byFodor lemma, replacing ji : i < by a subsequence, without loss of generality

=:


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72 SAHARON SHELAH

sup{ji,m : i < , m < mji} < , and without loss of generality h1ji

(f(ji)) mji :

i < = x is the same for all X.Choose u = {ji : i < }, i = ji,mi , which is regular by 7.1(2) as then

assumption (E) is trivial. Now, j : j u is a strictly increasing sequence ofregular cardinals with limit , and hence

ju

j/Jbdu is directed and hence

directed. But, by 7.1, {h1ji (f(i))(mi) : i u : X} is unbounded in it (or

use max pcf{i, : i < , < ni} ). So = tcf ju

i/Jbdu

. Let gj be

defined by gj() = (h1j ())(mj), and we are done. We leave the case = 0 to

the reader.2) First without loss of generality g(f(i)) = n

, i.e., does not depend on ,secondly, e.g., by successive applications of part (1).3) Similar. 7.3

7.4 Conclusion. For1) In 7.3(1), ()2 and ()3 implies

()2 there are u, = i : i u

and X such that

(a) u [],

(b) X [],

(c) f u : X is 0, and fg u

: X

is 0 then for every iu

i there are ordinals X such

that

i u i < f(i) < i ,

(d)() if = 0, i = Rang(gi) then for every iu

i there are ordinals

X such that

i u i < fg(i) < i,

(e) if ()3 then:

(i) < from X f u < f u

,

(ii) if i = j are in u and i < j then X f(j) >

i ,

(iii) if i, j u, i = j and < are from X then f(i) < f(j).


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SPECIAL SUBSETS 73

2) Similarly for 7.3(2), getting from the proof of 6.7. [Saharon copied!]

there are u, m, v, , X such that

(a) u

[]

and X []

,(b) i u m(i) = m,

(c) v m but v = m,

(d) = ,i : < m, i u,

(e) v f[] u : X is sup

1,i1 :

1,i1

< ,i where

1 < m and i1 <

and < X implies: for every i u large

enough we have f[] (i) > max

f

[1] (i1) :

1,i1

= ,i and 1 < m

and i1 <

(the interesting case is i1 = i).

Proof. Straight. Choose the th member of X for < , by induction on .7.4

Similarly we can prove

7.5 Claim. Assume

(A) = cf() > 2,

(B) = min{ : }, cf() = > 0,

(C) if a Reg \ 2, |a| , pcf-complete(a), then for some b a,

= tcf( b/[b]


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74 SAHARON SHELAH

Why? Now b[a] : pcf(a) is well defined and pcf-complete(a) so lettingpcf(a) be : < , choose b[a] \

2 and n < ,

(B) f Ord for < n, < ,

(C) = f : < n = f : < n,

(D) ( < )(||0 < ).

Then we can find an ultrafilter D on (possibly a principal one) and X [],

v n and f Ord for < n such that

(a) for n\v and X we have f/D= f/D,

(b) for < from X and , m v such that f/D= fm/D (e.g., = m) wehave f/D< f

m /D,

(c) if ,m < n and f/D< fm/D and , are from X then f/D< f

/D.

2) Assume

(a) = cf() > 2 and ( < )(||


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SPECIAL SUBSETS 75

3) We can combine 7.1(1) with part (2) (having i, : < i ).

Remark. We can prove also the parallel of 7.1(5).

Proof. 1) Like the proof of 7.1(4) or by part (2) for = 1.2) We repeat the proof of 7.1(4) except that T

i 2.

Let J is the ideal on generated by

I {w :for some < () we have w \ w, and

> max pcf{cf(,i,) : i < w}}.

If for some , / J then we are done (choosing the -th member ofX by inductionon ). So assume that T J. We now choose by induction on < , asequence Ti such that < = j and

= + 1 {i < : ( < ())(,i, > ,i,)} = mod I.

For some < () and infinite Y we have:

Y Z = {i < : ,i, > ,i,+1} = mod I.

But for < we have ,i,+1 ,i, by ()1. Without loss of generality otp(Y) =

. As I is complete and 0, there is an i

{Z : Y}, and ,i, : Zis strictly decreasing, a contradiction.Now for = 0, limit there are no serious demands and for successor ordinal

we use J.3) Left to the reader (and not used). 7.7

7.4A Fact. Assume


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76 SAHARON SHELAH

(A) = +, > 2, = cf() > 0,

(B) |()|+ + 0 < ,

(C) f

Ord for < (), < ,

(D) ( < )(|| < ).

Then we can find a stationary S { < : cf() |()|+ + 0} and unboundedsubset X of and S S and f SOrd for < ()

(a) for < () we have X f S = f S,

(b) for 1 < () and < from X if S, = {i S : f(i) f(i)} \ S \ S is

unbounded in then f S, < f mod J

bdS,

,

(c) if , < (), f(i) < f(i), and X then f(i) < f(i),

(d) if 2|()| < then < () S {, S}.

Proof. Let f = f : < , let be large enough and : < be increasingcontinuous with limit , and choose by induction on < , an elementary submodelN of (H(), , scite{7.4A} undefined

Remark. We can demand S S in 7.7 if S { < : cf() |()|+ + 0} isstationary.

* * * * * * *


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SPECIAL SUBSETS 77

7.8 Discussion. We may wonder what occurs for ultraproducts of free Booleanalgebras

i 0 case for products;

()2 if

(A0, A1, . . . ) n 0 & i,0 = ) (ni > 0 & ()(i, = )),

and ifu0 = , then = max pcf{i, : i < , < ni} and+ / max pcf{i, :

i < , < ni}.


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78 SAHARON SHELAH

Proof. Let C = C : < + be such that otp(C)

+, [ C C =C ], C a set of successor ordinals and the set

S

= { < +

: cf() = +

and = sup(C)}

is stationary (exists by [Sh 420, 1]).Let f = f : < be given. Let be strong limit such that f H(). We

choose M by induction on < + such that

() M (H(), ,


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SPECIAL SUBSETS 79

Similarly choose A,i : i < , h,i : i < ,

,i, : i < , < n

,i, f

,, :

X as in 7.1(1). Let U = {(i, ) : i < , < ni}; this is Dom(f,,) for X

. Let

J = J, = {u U : + > max pcf{cf(,i,) : (i, ) u}}.

By the pcf theorem ([Sh:g, VIII,2.6]) there is W U such that:

+ / pcf{cf(,i,) : (i, ) U \ W

},

+ max pcf{cf(,i,) : (i, ) W}.

If W / J let h, = h,, : < + M be cf([], )] we can get7.9.


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80 SAHARON SHELAH

8 Consistency of P(1) has a free caliber and discussion of pcf

This solves [M2, Problem 37].

8.1 Claim. Assume for simplicity GCH andP is adding 1 Cohen reals. In VPwe have 20 = 1 , 2

1 = 1+1 and

() there is no complete Boolean algebraB of cardinality21 such that FreeCal(B) =. In fact for any complete Boolean algebra B of cardinality 21 we have1+1 FreeCal(B).

Proof. Clearly (as if the Boolean algebra B has cardinality 21 = 1+1 and satisfiesthe ccc then () holds, i.e., 1+1 FreeCal(B), because V

P (1)0 = 1 , and

otherwise we can reduce to the case B = P(1)) it is enough to show

()1 VP 1+1 FreeCal(P(1)).

So let p P

p P a : < 1+1 is a sequence of distinct elements ofP(1).

Note: P = {f : f is finite function from 1 to {0, 1}}. So PA = {f P :Dom(f) A} P for any A 1 .

For each < 1+1 and i < 1 there is a maximal antichain f,i,n : n < ofP and sequence of truth values t,i,n : n < such that

f,i,n P i a iff t,i,n.

Let A =

i


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SPECIAL SUBSETS 81

where f is defined by f(j) = ,j and get w : < 2, X [1+1]

1+1

and j, : j < j, < nj, and h = hj : j < j

as there. For i < 1 letwi = {j < j

: j w0 or j,nj1 i}.

We call (g0i , g1i , i) : i < 1 a witness above f if:

1(i) f, g0i , g

1i P and p

f,

(ii) f g0i ,

(iii) f g1i ,

(iv) Dom(gi ) j,

(v) Dom(g0i ) Dom(g1i )\Dom(f

) : i < 1 are pairwise disjoint,

(vi) g0i i a

,

(vii) g1i i / a

,

(viii) i < 1 and i = j for i = j.

Shrinking X (still unbounded in 1+1) we get:

2 if < are from X then there is i < 1 such that

j j \ wi j,nj1 > i (h1j (,j))(m) < (h

1j (,j))(m),

andj j \ wi j1 < j

,j1 = ,j .

Fact. There are f and a witness (g0i , g1i , i) : i < 1 above f

and X 1+1unbounded and an ideal J Jbd1 on 1 such that:

if = are in X then

{i : the functions g0,i, g1,i are compatible} J.

letting u,i = OPA,jDom(g0i

) Dom(g1

i

)\Dom(f),g,i =: g

i OPj,A .

We show how to finish the proof assuming the fact, and then we shall prove thefact. For some unbounded X 1+1 we have X f

= OPA,j(f) i.e.,

does not depend on X. (As there are |P| = 1 < 1+1 possibilities.)


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82 SAHARON SHELAH

We shall prove

f

P a : X is independent (as a family of subsets of

1), even modulo Jbd1

.

This is more than enough.If not then for some n < and pairwise distinct 1, . . . , 2n X, we have:

(f P n=1

a

2n=n+1

(1\a) is unbounded in 1)

So for some f1, f f1 P, and < 1 we have

3 f1 P

n=1

ai

n=n+1

(1\ai) .

Now lettingg0,i = g

0i OPj,A and

g1,i = g1i OPj,A

we have

Dom(g0,i) Dom(g1,i) {,j : j < j

}.

Let

B =: {i < 1 : i < },

B,m =: {i : u,i um,i = } for = m,

B =: {i : Dom(f1) (Dom(g0,i) Dom(g

1,i

)) = Dom(f)}.

Now

1 B J.[Why? By clause (viii) of 1 above and the choice of J.]


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SPECIAL SUBSETS 83

2 B,m J for = m {1, . . . , 2n}.[Why? By the Fact ? which we are assuming.]

> scite{8.1A} undefined

3 B J for {1, . . . , 2n}[by clause (v) of 1 above (in fact is finite).]

So we can find i 1\=m

B,m\

B\B (because the set of inappropriate is is in

J).

So f2 = f1 n=1

g0,i 2n

=n+1

g1,i P forces that the intersection from is not

, contradicting the choice of f1.

Proof of the Fact ?. We divide the proof into two cases, depending on the answer

> scite{8.1A} undefinedto:

Question: Is there < 1 such that: for no g0, g1 Pj above p and [, 1)

do we have

g0 w = g1 w , g

0 a

, g1 / a

?

Case A: The answer is YES.Let < 1 exemplify the yes. As GCH holds in V, clearly for some unbounded

X 1+1 and j : j w we have

j w & X ,j = j .

So a

is actually a P{j :jw}-name. So for X, adepends only on {f G

P :

Dom(f) {,i : i w}}. So f P a

= a

for , X, a contradiction.

Case B: The answer is NO.So for every < 1, we have , g

0 , g

1 giving the counterexample for ,

without loss of generality Dom(g0) = Dom(g1). As w : < 1 is increasing

continuous, by Fodors lemma we can find S 1 stationary and < 1 and n

such that

S (Dom(g0) w) (Dom(g1) w) w ,

and Dom(g0) Dom(g1 ) : S forms a system with heart v, and g

0

v = g1 v (being included in w) does not depend on , and we call it f. Also

Dom(g \ w) has n elements and 1 < S 1 < .


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84 SAHARON SHELAH

Let (i) : i < 1 be a (strictly) increasing sequence listing S, and i = (i).

For < n, X and i < 1 we let f(i) be the -th member of {,j : j Dom(g0(i)) \ w} . Shrinking X without loss of generality ,j : j w does

not depend on X (by 2); J = Jbd1 and X are as required. ?,8.1

8.2 Discussion. 1) Clearly we can replace 1, 1+1 by any , as in 7.7.2) Normally if is strong limit singular of cofinality , (at least large enough), wecan find long intervals ai of the Reg for i < , i < j sup(ai) < min(aj) suchthat (

i ai) [max pcf (Rang()) =

] for some [, 2], usually cf(2).This is a strong indication that Insup(ai),min(ai) : i < will have a -sequence,

so for example there is a (2)+-c.c. Boolean algebra of cardinality having noindependent subset of cardinality , for which even -Knaster property fails.

To make this happen for no , we need a very special pcf structure in the universe.But we do not know even if the following simple case is consistent.

8.3 Question: Is it consistent that

() for every set a of odd (or even) regular cardinals with |a| < Min(a) wehave max pcf(a) is odd (or even respectively) (we may moreover ask ()2 = +2)?

Essentially by [Sh 430, 5]:

8.4 Lemma. Assume > = cf(), strong limit, =

i


85/143


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86 SAHARON SHELAH

9 Having a -sequence for a sequence of non-stationary ideals

9.1 Lemma. Assume

(a) is a strong limit singular of cofinality ,

(b) = 2 = cf(),

(c) i regular increasing for i < with limit , < (usually = ),

(d) J is an ideal on extending Jbd ,

(e) = tcf(i


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SPECIAL SUBSETS 87

is not stationary. About ppJbd e.g. if |a| 0 |pcf(a)| 0 we then can get it,see [Sh:g, XI,5].

9.3 Remark. 1) This can be rephrased as having a (, J)-sequence for

Jnst,i,n :i < with i,n decreasing.

So compared to earlier theorems, the , i for which the Lemma applies arefewer, but the result is stronger: nonstationary ideal and we get also the superversion see ().2) Of course another variant is to start with Ii = J

nst,i

and get J = Jnst, .

3) Considering functions with finitely many s, < as parameters (i.e., ()+);

thinning f : < the conclusion follows.4) In ()+ instead n < we can ask n < if ( < )(||


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88 SAHARON SHELAH

(i) =


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SPECIAL SUBSETS 89

So {h, : < } is a subset ofiA

i of cardinality < , buti


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90 SAHARON SHELAH

= 2 = cf(2) = tcf n cf() = 0 strong limit singular,

= 2 = cf(2) pcf{n : n < },

n = cf(n) < ,

and kn : n < is as in 9.6. Then we can find i(n, ) < , < kn with norepetitions,

i(n, 0) > i(n 1, kn1 1) > > i(n 1, 0),

and letting


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SPECIAL SUBSETS 91

In =


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92 SAHARON SHELAH

Let

A = {i(n, ) : n < , kn > }.

So

A : < is a sequence of pairwise disjoint subsets of such that|A B| = 0.

We apply 9.1 for

An : n < , n : n < ,,.

9.7

9.9 Remark. If > cf() > 0, 2 regular, the parallel to 9.6 always occurs.9.7

If we use A = A0, A0 = in 9.1:

9.10 Conclusion. In 9.1 we get:

there is a (, J)-sequence for Ii : i < , even a super one.

9.11 Remark. By the proofs in [Sh 420, 1] we can replace Si : i < , S = { and c({, }) = .

2) In 9.13 we can add:

if e.g. = (2)+, for every X 2 for every < large enough, for < (), there is a sequence N : < as in the proof of 9.13 suchthat

() {i A : (i) = sup(N 1)} J.

9.15 Remark. In 9.14(1) we get even P r1(,,,).

Proof. 1) We relay on part 2).2) For < let c({, }) = if

{i A0 : f(i) f(i) or f(i) < f(i) & = otp(eif(i)

)} J,

and zero if there is no such .Let X, Y []. take X large enough, so that we can find N : asthere, with () for part (2). We can find N+1 Y such that sup(N i) :i < are as required.9.14

9.16 Claim. In 9.1

(1) Instead of > = cf() > || we can assume only

1 > = cf() and if u : < is a sequence of members of J theni < : = sup{ : i / u}

= mod J.

(2) Weakening the conclusion of 9.1 to weak (J, )-sequence, we can replace = cf() > || by

2 = cf() and if u : < is a sequence of members of J theni < : = sup{ : i / u}

J+.

(3) In part (1) and (2), if > 0, then we can find Ci = Ci : Si with

Ci a club of such that: we can replace Inst,i

by idai(Ci), see 9.13. above.


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10 The power of a strong limitsingular is itself singular: existence

10.1 Lemma. Assume

(a) strong limit singular,

(b) 2 is singular, = cf(2) (so 2 > > ),

(c) > = cf() > cf(),

(d) 2 = pp() (see discussion in 9).

Then

() we can find J, J, i = i : < cf() for i < and such that

(i)i

is an increasing sequence of regular cardinals < with limit fori < ,

(ii) = : < is an increasing sequence of regulars ( + , 2)with limit 2,

(iii) J J are ideals on cf(), cf()-complete,

(iv) = tcf

/J

,

(v) : < is


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SPECIAL SUBSETS 97

() Assume A : < is a partition of cf() to sets not in J. Then we canadd

(ii)+ For any sequence of functions

F = F : < cf(),

for some = F

, for every [, ) we have

() if < , n < , < for < n then

< cf() :F(. . . , , , . . . , 2cf(), we can replace by a subsequence, so without loss

of generality J J, so J is cf()-complete and is


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98 SAHARON SHELAH

Fi = Fi : < .

We choose < cf()

by induction on .

For a given < we choose A by induction on < . We will choose

A such that

() ifn < , 0, 1, . . . , n1 < and i < sup{ : < } (necessarily < ),

A :Fi(. . . , , . . . ,


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SPECIAL SUBSETS 99

11 Preliminaries to the construction of cccBoolean algebras with no large independent sets

Monk [M2] asks:

Problem 33. Assume cf() < < cf(). Is it possible in ZFC thatthere is a Boolean algebra of cardinality , satisfying the -cc with no independentsubset of cardinality ?

This is closely related to the problem of is a free caliber of such Boolean algebra(see also in Monk [M2]).

Why in ZFC? Because of earlier results under strong limit, 2 = +, I think.

The real problem seems to me is for regular, and we shall prove that almostalways there is such a Boolean algebra, so we prove the consistency of failure.

We shall use Jbdi,0,i,1 : i < with regular i,0 > i,1, but we use Booleanalgebras whose existence is only consistent.

So we shall use a (, J)-sequence for Jbdi,0,i,1 : i < , if = the

Boolean algebra B will have a dense subalgebra B which will be the free productof {Bn : n < }, x

t , x

+t Bn for t Dom(In) and B = B

, y : < wherey completion of B

is defined from x(n), x+(n)

: n < . We need special

properties of Bn, xt , x

+t (t Dom(In)). The construction continues [RoSh 534,

3]. Concerning the parallel to 6.16 see later.

For the case strong limit we can use instead subalgebras of the measure algebra.See 2. Now we have consistency (and independence) for , < 2, stronglimit singular, hence we concentrate on the other case where the behavior is differenti.e. when for some we have cf() < =


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then for some (x1, . . . , xn) T and pairwise distinct t1, . . . , tn X wehave

B (yt1 , yt2 , . . . , ytn) = 0.

11.2 Explanation. We think of having a (, J)-sequence for Ii : i < , andhaving (Bi, x

+i , x

i ) witnessing (Ii,T) for i < and using the sequence of intervals

(xi,(i), x+i,(i)

) : i < as a sequence of approximations for an element x of the

desired Boolean algebra B of cardinality .

But we may think not only of {x : < } has no independent subset of cardi-nality but of other subsets of B. So sometimes we use

11.3 Definition. 1) We say that (B, x, x+) strongly witnesses (I,T) if: (a)-(f)

as before, and

(g)+ If B B,B xt yt x

+t for t Dom(I),

b : m is a sequence of pairwise disjoint non-zero members of B,

m < and

X m=1

I+

,

and u [1, m], then we can find n, (x1, . . . , xn) T and distinct

t1

, . . . , tn

X, so tr

= tr : = 1, . . . , m, such that (ct1 , . . . , ctn) = 0where

ct = b0

[1,m]u

(b yt)

[1,m]/u

(b yt .)

2) We say that (B, x+, x) witness (I,T) m-strongly if we restrict ourselves tothis m. Similarly [m1, m2]-strongly.

Next we need our specific (B, x, x+, I). The following is essentially from [Sh 126,p.244-246].

11.4 Claim. 1) If = 2 = +, (or just []2) and 2 = +, then we can

find F = F : < + such that:

()F

(a) F : []2 is one to one

(b) IfA (Jbd+,)+, then for some (, i0), (, i1), (, i2) A we have F({i0, i1}) =

(, i2).


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SPECIAL SUBSETS 101

We write this also as

F({, i0}, {, i1}) = (, i2).

We can add that for every we have Rang(F) ({} )| 1 for > . We donot strictly distinguish F from F.2) The property ()

Fis preserved by forcing notions which have the (3, Jbd+,)

+-

c.c. (see 11.6 below).3) Let B = BF be the Boolean algebra freely generated by

x+,i = x+(,i); x

,i = x

(,i)( for (, i)

+ )

except that x,i x+,i and

x+(,i) x+(,j) x

+F(i,j)

= 0.

Then

(i) (B, x+, x) witness (Jbd+,, {x0 x1 x2 = 0})

(ii) B satisfies the c.c.c.

11.5 Remark. On more general Boolean algebras generated by such equations seeHajnal, Juhasz, Szemintklossy [HaJuSz].

11.6 Definition. For an ideal J and forcing notion P, we say that P satisfies the(n, J)-c.c. if for pt : t A, A J

+, there is B A, B J+ such that any nconditions in {pt : t B} have a common upper bound.

11.7 Fact. IfP is the forcing notion P, of adding Cohens for and


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102 SAHARON SHELAH

and if possible

F(i, j) AH({i,j}),

which occurs if AH({i,j}) .

2) Trivial. Let P be the forcing notion. Let p A

(Jbd+,)+ and it exemplifies

a contradiction to ()F

. Let A =: {(, i) : p (, i) / A

}. So A + and,

p A A

, A

(Jbd+,)+,

hence

A (Jbd+,)+.

For (, i) A there is p(,i) p such that

p(,i) (, i) A

.

Apply the (3, Jbd+,)-cc to p(,i) : (, i) A, and obtain B as in Definition 11.6.

As B (Jbd+,)+, by ()

Fwe can find (, i0), (, i1), (, i2) B such that

F({i0, i1}) = (, i2).

But by the choice of B there is q P such that

q p(,i0), p(,i1), p(,i2)

(hence q p). So

q (, i0), (, i1), (, i2) A

and F({i0, i1}) = (, i2).

But this contradicts the assumption on p, A

.

3) For clause (i), read the definition. For clause (ii): Call Z + closed ifF(t1, t2) = t3 & |{t1, t2, t2} Z| > 1 {t1, t2, t3} Z. Now

() if F(ti, si) = ri for i = 0, 1 then {t0, s0, r0} {t1, s1, r1} has 1 or 3elements.


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SPECIAL SUBSETS 103

[Why? As each F is one to one and

F = 1}.

Without loss of generality Z+ : < 1 is a -system with heart Z+.

Without loss of generality f {x+t : t Z

+} is constant.Without loss of generality Z Z

+ is constant.

So()4 If = < 1

F(t1, t2) = t3 and {t1, t2, t3} Z Z ,

then{t1, t2, t3} Z or {t1, t2, t3} Z.


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104 SAHARON SHELAH

[Why? Without loss of generality

|{t1, t2, t3} Z | 2.

So

{t1, t2, t3} Z+ .

Now if ti Z+ \Z , then ti Z (otherwise ti Z

+ Z

+ , hence ti Z

+, but

Z Z+ is constant). So {t1, t2, t3} Z .]

Now f f preserves the equations on Z Z and by the homomorphism itinduces, a a is mapped to 1, so BZZ

Saharon Shelah- Special Subsets of ^CF(μ)-μ, Boolean Algebras and Mahram Measure Algebras

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Transcript of Saharon Shelah- Special Subsets of ^CF(μ)-μ, Boolean Algebras and Mahram Measure Algebras