saharon shelah- on monk's questions

8/3/2019 Saharon Shelah- On Monk's Questions

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ON MONKS QUESTIONSSH479

Saharon Shelah

Institute of MathematicsThe Hebrew University

Jerusalem, Israel

Rutgers University

Department of MathematicsNew Brunswick, NJ USA

Partially supported by the Deutsche Forschungsgemeinschaft, Grant Ko 490/7-1.I would like to thank Alice Leonhardt for the beautiful typing.Publication 479

Typeset by AMS-TEX

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Anotated Content

1 Introduction

2 Existence of subalgebras with a preassigned algebraic density

[We first note (in 2.1) that if (B) = cf() then for some B B we have(B) = . Call this statement (). Then we give a criterion for(B) = > cf() (in 2.2) and conclude for singular that for a club of < the () above holds (2.2A), and investigate the criterion (in 2.3). Our main aimis, starting with = but for no B B do we have (B) = (in fact (B B)[(B) = = cf() cf() ] for every |B|. Toward this, wedefine the forcing (Definition 2.5: how x : WP generate a Boolean algebra,

BA[p], Wp [] has no non-zero member ofx : Wp BA[p]below it). We prove the expected properties of the generic (2.6), also the forcinghas the expected properties (-complete, +-c.c) (in 2.7). The main theorem (2.9)stated, the main point brings that if < cf() < for B BA[G], (B) = ;we use the criterion from above, a lemma related to -systems (see [Sh 430],6.6D,[Sh 513],6.1) quoted in 2.4; to reduce the problem to some special amalgamationof finitely many copies (the exact number is in relation to the arity of the termdefining the relevant elements from the xs). The existence of such amalgamationwas done separately earlier (2.8).

Lastly in 2.10 we show that the cf() > above was necessary by proving the

existence of a subalgebra with prescribed singular algebraic density : (B) > and ( < )[< cf() < ]].

3 On and of Products of Boolean Algebras

[If e.g. 0 < = cf() < < = cf() < , ( < )( < ) we show

that for some Boolean algebras Bi (for i < ) : i


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ON MONKS QUESTIONS SH479 3

1 Introduction

Monk [M] asks: (problems 13, 15 in his list; is the algebraic density, see 1.1 below)

For a (Boolean algebra) B, 0 (B), does B have a subalgebra B

with(B) = ?

If is regular the answer is easily positive (see 2.1), we show that in general itmay be negative (see 2.9(3)), but for quite many singular cardinals - it is positive(2.10); the theorems are quite complementary. This is dealt with in 2.

In 3 we mainly deal with (see Definition 3.2) show that the of an ul-traproduct of Boolean algebras is not necessarily the ultraproduct of the s.Note: in Koppelberg Shelah [KpSh 415] ,Theorem 1.1 we prove that if SCH holds,

(Bi) > 2

for i < then i


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2 Existence of subalgebras with a preassigned algebraic density

Note

2.1 Observation. If (B) > = cf() |Y| + 0 and Y B then for somesubalgebra A of B, Y A and (A) = .

Proof. Without loss of generality |Y| = . Let Y = {y : < }. Choose byinduction on subalgebras A of B, increasing continuous in , |A| < ,y A+1 such that: for each < , some x A

++1 is not above any y A

+ .

This is possible because for no < can A+ be dense in B.Now A = A is as required. 2.1

2.2 Claim. Assume B is a Boolean algebra, (B) = > cf() 0 (see Defini-tion 1.1). Then for arbitrarily large regular <

()B for some set Y we have:

()B [Y] Y B+, |Y| = , and there is no Z B+

of cardinality < , dense in Y

i.e.

yY

zZ

z y

.

2.2A Conclusion. If B is a Boolean algebra, (B) = > cf() > 0 and : < cf() is increasing continuously with limit (so < ) then for someclub C of cf() for every C for some B B we have (B) = .

Proof of 2.2. Let Z B+ be dense, |Z| = .If the conclusion fails, then for some < , for no regular (, ) does ()B

hold. We now assume we chose such , and show by induction on that:

: if Y B+, |Y| then for some Z B+, |Z| and Z is dense in Y.

First Case. .

Let Z = Y.

Second Case. < and cf() < .Let Y =

{Y : < cf()}, |Y | < . By the induction hypothesis for each

< cf() there is Z B+ of cardinality which is dense in Y .


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Now Z =:

< cf()

Z has cardinality + cf() < , hence by the induction

hypothesis there is Z B+ dense in Z with |Z| . Easily Z is dense in

Y, |Z|

and Z B+

so we finish the case.

Third Case. < , regular.If for this Y, ()B [Y] holds, we get the conclusion of the claim. We are assuming

not so; so there is Z B+, |Z| < , Z dense in Y. Apply the induction hypothesisto Z and get Z as required.So we have proved .

We apply to = , Y = Z and get a contradiction. 2.2

2.3 Claim. 1) If B, , , Y are as in 2.2 (so ()B

[Y] and is regular) then wecan find y = y : < contained in B+ and a proper -complete filter D on

containing all cobounded subsets of such that:

By,D for every z B+ , { < : z y} = mod D.

2) If in addition is a successor cardinal then we can demand that D is normal.

Remark. Part (2) is for curiousity only.

Proof. 1) Let Y = {y : < }. Define D:

for U ; U D iff for some 0 < < and z B+ for

< , we have U { < :


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6 SAHARON SHELAH

y Y


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4) IfG Q, is generic over V then BA[G] ispG BA[p] (see 2.6(2),(3) below).

Here BA[p] =: BA[wp, Fp].

2.6 Claim. 0) For p Q,, BA[p] is a Boolean algebra; also for f Fp andordinal wp (or ) we have f[] Fp.1) If f Fp, p Q, then f induces a homomorphism we call f

hom from BA[p]to the two members Boolean algebra {0, 1}. In fact for a term in {x : wp},BA[p] |= = 0 iff for some f Fp, fhom() = 1.2) If p q then BA[p] is a Boolean subalgebra of BA[q].3) Hence BA[G

] is well defined, p BA[p] is a Boolean subalgebra of BA[G

].

4) Forp Q,, for wp, x is a non-zero element which is not in the subalgebra

generated by {x : < } nor is there below it a non-zero member ofx : < BA[p].

Proof. Check. Part (0) should be clear, also part (1). Now part (2) follows by2.5(2)(b) and the definition ofBA[p]; so (3) should become clear. Lastly, concerningpart (4), x is a non-zero member of BA[p] by clause () of 2.5(2)(a). For w

p,by 2.5(2)(a)() there is f1 Fp, f1() = 1, and by 2.5(2)(a)() there is f0 Fp,f0() = 0, f0 (w ) = f1 (w ); together with part (1) this proves thesecond phrase of part (4). As for the third phrase, let be a non-zero element ofthe subalgebra generated by {x : < }, so for some f Fp, fhom() = 1. By2.5(2)(a)(), letting f1 = f

[], we have f1() = 0 and f1 Fp and f1 (w ) f.Hence fhom

1() = fhom() = 1 and f

1() = 0, hence fhom

1(x

) = 0. This proves

BA[p] |= x. 2.6

2.7 Claim. Assume =


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8 SAHARON SHELAH

F = {f w2 : for every < we have f w F}.

Clearly for every < and f F there is g = gf F extending F. Lastly, let

F = {gf : f n + 1.

Then there is q Q such that:

() p q for < m; and wq =


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f0 (01), . . . , f

0 (

0n)

= 0

f1 (01), . . . , f 1 (0n)

= 1.

Now there is wp0

{} such that (f0 )[] = f0 & (f

1 )

[] = f1 (e.g. = ).Choose such (, f0 , f

1 ) with minimal.

Let wq =m1=0

wp

. We define a function g (wq)2 as follows:

First Case. g wp0

= f1

Second Case. For odd [1, m), g wp = f1 H0, and

Third Case. For even [1, m), (but not = 0!) g wp

= f0 H0,.

Now g is well defined by clause (A) above. Let us define q:

Fq =

m1=0

f H0,

[]: wq {} and f Fp0

g[] : wq {}

.

q = (wq, Fq).

Let us check the requirements.

First Requirement: q Q.Clearly wq []


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Second Requirement: For < m, p q.

By the choice of q clearly wp

wq.

Also if f Fp

then (f H,0) Fp0 and

k


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First Case. For some wq {} and f0 Fp0

we have

f = n + 1, so for some k, 0 < k < k + 1 < m and jk = jk+1.

So (as i1 < <

in) n

j=1

f

xkj

= f

xk+1j

,

hencenj=1

fhom

xkj

= fhom

xk+1

j

hence

fhom xk1 , . . . , xkn = fhom xk+11 , . . . , xk+1n ,hence (see Definition of ) fhom() = 0 hence

fhom

x01 , . . . , x0n

= 0,

as required.

Second Case. For some wq {}, f = g[].

Let again = Minwp {}\, = H0,() (or = = ), and =

if <

if

j = Minj : j = n + 1 or j {1, . . . , n} and 0j

. So

: < m, : < m are non-increasing and so is j : < m. Here is theordinal we chose before defining q1 just after (B) in the proof.

If for some k, 0 < k < k + 1 < m, jk = jk+1 n,(hence k+1

k < ) then

fhom

x0

1, . . . , x

0

n = 0 as fhom() = 0 which holds because

fhom

xk1 , . . . , xkn

= fhom

xk+11, . . . , xk+1n

(the last equality holds

by the choice of ; i.e. if inequality holds then the triple (k, (f0 )

[k], (f1 )[k])

contradicts the choice of as minimal). But j ( = 1, m 1) is non-increasing


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12 SAHARON SHELAH

hence we can show inductively on = 1,...,n + 1 that jm . So necessarilyj1 = n + 1 but as j is non-increasing clearly j0 = n + 1 hence

fhom

x01 , . . . , x0n

= g[]

x01 , . . . , x0n

= g

x01 , . . . , x0n

= f1

x01 , . . . , x0n

= 1

hence

fhom

x01 , . . . , x0n

= 0,

as required. 2.8

2.9 Theorem. Suppose = then for no B BA[G] is (B) =

(c) BA[G] has non-zero pairwise disjoint elements but no +

such ele-ments (so its cellularity is )

(d) if a B+ then BA[G] a satisfies (a), (b), (c) above (also (e))(e) if and cf() then for some B BA[G] we have (B) = (f) in BA[G] for every < , {x : < +

+} B+ is dense in{x : < }BA[G].

2.9A Remark. 1) This shows the consistency of a negative answer to problems 13

+ 15 of Monk [M].2) We could of course make 2 bigger by adding the right number of Cohen subsetsof .

Proof. By Claim 2.7 clearly parts (1), (2) hold. We are left with part (3), by 2.6(3)BA[G] is a Boolean algebra, by 2.6(4) it has cardinality . As for clause (a), itis exemplified by x : < BA[G] (by 2.6(4)). Clause (c), the first statement iseasy by the genericity of G (i.e. as for p Q, \wp we can find q, p q Q,wq = wp {} and in BA[q], x is disjoint to all y J, for any ideal J of BA[p]).As for clause (c), second statement, it follows from the -system argument and theproof of 2.7(2).

Clause (d): concerning the generalization of clause (a), let a (BA[G])+, so

we can find finite disjoint u, v such that 0 ).

Hence w i w. For every < and i i\w, let ,i < ,i be such

that the interval [,i, ,i) is disjoint to {

,j : < , j < i

} {i : i < i},

and as we can omit an initial segment of i : i < without loss of generality[,i ,

i ) is disjoint to {

j : j < i

}. Choose for each < , A , such that

i i\w ,,i [,i + , ,i)

. Let

Y

= the Boolean subalgebra generated by

, : q Gand p, G

. This

set has cardinality , and we shall prove

() q0 Q Y

\{0} is dense in

0, : A0

.

This contradicts the choice q0 ()B

y0,D0, A

= mod D0.

To prove () assume q0 r0 Q, we can choose < and r+ , for [

, ) such

that r0 r+0 , q r+ , p, r+ and r : = 0 or [, ) is as in 2.8; apply2.8 and get a contradiction. 2.9

A theorem complementary to 2.9 is:


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2.10 Theorem. If (B) > and

(A) cf() = 0 or

(B)


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3 On and of Products of Boolean Algebras

3.1 Theorem. Suppose0 < = cf() < < = cf() < () and


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note also

(BA(I)) = Max{cf(Id

), cf((Iu

)

) : (Id

,Iu

) a Dedekind cut of I}.

Now by the assumption

(and [Sh:g],II,5.4 + VIII,1]), we can find a (strictly)

increasing sequence i : i < of regular cardinals, < i < , =i


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y Y+1

(y Y)(y > 0 y y).

There is no problem in doing this for i = 0, let Zi = {0}, for i limit let Zi =


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3.6 Claim. Assume that for i < , Ai is an infinite Boolean algebra, D is a non-

principal ultrafilter on and A =:i


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n0i elements and |Yi| = ni n0i + 1 and e.g. Yi is a set of natural numbers; note that

|Xi| > ni. Let X =: Xi : i < /D, Y =: Yi : i < /D, for y Y (in Cs sense)

let Sy =: {x X : y x}. Let n

1

i =: |Xi|, notei


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ADDED IN PROOF

We can add to claim 2.3:

CLAIM 2.3(3): Assume that B is a Boolean algebra and < = cf() and()B [Y] and for no (, ) and Y do we have ()B [Y

].Then in the conclusion of 2.3(1) we can add: D is normal (hence in 2.2 we get

that for arbitrarily large < , there are a normal filter D on and yi : i < asin 2.3(1) ).

Remark: If = + then 2.3(2) gives the conclusion.

Proof: Let Y = {yi : i < }, we choose by induction of n < a club En of andsequence yn = yni : i < of non zero members of B such that:

(a) letting Yn = {yni : i < }, we have Y = Y0, Yn Yn+1 and En+1 En

(b) if En+1 and if < < min(En\( + 1)) then for some < , yn+1 B y

n.

Let for n = 0, y0 list Y and E0 = { < : a limit ordinal divisible by }.For n = m +1, for each En, let = min(En\( + 1)) and let Zn be a subset

ofB+ of cardinality dense in {ymi : i < } which exist by the proof of 2.2.Let Zn = {z

n+i : i < }. (no double use of the same index).

Also for each < let Dn be the normal filter on generated by the subsets of

of the form {i < : z yni+} for z B+ ; by our assumption toward contradiction

there are zn B+ for < and club En of such that if E

n and < then

for some < we have z, yn+Let En+1 be a club of included in En and in each E

n and choose y

n+1 such that:

its range include the range of y

n

and for 1 < 2 En+1 we have {y

n+1

+ } : < }include {yn+ : < } Zn and {y

n+1i : i < } include each z

n, for < and

< . The rest as as in the proof of 2.3(2).


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REFERENCES

[KpSh 415] Sabine Koppelberg and Saharon Shelah. Densities of ultraproducts ofBoolean algebras. Canadian Journal Of Mathematics. Journal Cana-dien de Mathematiques, 47:132145, 1995. math.LO/9404226.

[M] J. Donald Monk. Cardinal functions of Boolean algebras. circulatednotes.

[Sh:92] Saharon Shelah. Remarks on Boolean algebras. Algebra Universalis,11:7789, 1980.

[Sh 420] Saharon Shelah. Advances in Cardinal Arithmetic. In Finite and Infi-nite Combinatorics in Sets and Logic, pages 355383. Kluwer Academic

Publishers, 1993. N.W. Sauer et al (eds.). 0708.1979.[Sh 355] Saharon Shelah. +1 has a Jonsson Algebra. In Cardinal Arithmetic,

volume 29 ofOxford Logic Guides, chapter II. Oxford University Press,1994.

[Sh:g] Saharon Shelah. Cardinal Arithmetic, volume 29 of Oxford LogicGuides. Oxford University Press, 1994.

[Sh 430] Saharon Shelah. Further cardinal arithmetic. Israel Journal of Math-ematics, 95:61114, 1996. math.LO/9610226.

[Sh 513] Saharon Shelah. PCF and infinite free subsets in an algebra. Archivefor Mathematical Logic, 41:321359, 2002. math.LO/9807177.

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