andrzej roslanowski and saharon shelah- forcing for hl and hd

8/3/2019 Andrzej Roslanowski and Saharon Shelah- Forcing for hL and hd

1/32

651

re

vision:2001-11-12

modified:2001-11-12

FORCING FOR hL AND hd

ANDRZEJ ROSLANOWSKI AND SAHARON SHELAH

Abstract. The present paper addresses the problem of attainment of thesupremums in various equivalent definitions of hereditary density hd and hered-itary Lindelof degree hL of Boolean algebras. We partially answer two prob-lems of J. Donald Monk, [13, Problems 50, 54], showing consistency of differentattainment behaviour and proving that (for the considered variants) this is thebest result we can expect.

0. Introduction

We deal with the attainment problem in various definitions of two cardinal func-tions on Boolean algebras: the hereditary density hd and the hereditary Lindelofdegree hL. These two cardinal functions are closely related, as it is transparentwhen we pick the right variants of (equivalent) definitions. Also they both aresomewhat related to the spread s of Boolean algebras. So, for a Boolean algebraB, we define

s(B) = sup{ : there is an idealindependent sequence of length }, hd(B) = sup{ : there is a leftseparated sequence of length }, hL(B) = sup{ : there is a rightseparated sequence of length }.

Let us recall that a sequence a : < of elements of a Boolean algebra is idealindependent ifa

w

a for each < and a finite set w \ {},

leftseparated if a w

a for each < and a finite set w \ ( +1),

rightseparated if a w

a for each < and a finite set w .

The above definitions of the three cardinal functions are of special use, see e.g.[15, 1]. However, neither these definitions explain the names of the functions, northey are good enough justifications for the interest in them. But all three functionsoriginate in the cardinal functions of the topological space Ult(B) (of ultrafilters onB). And thus, for a Boolean algebra B, we may define (or prove that the followingequalities hold true):

s(B) = sup{|X| : X Ult(B) is discrete in the relative topology },

1991 Mathematics Subject Classification. Primary 03E35, 03G05, 54A25; Secondary 03E05,06Exx.

Key words and phrases. Boolean algebras, spread, hereditary density, hereditary Lindelof de-gree, attainment.

The first author thanks the KBN (Polish Committee of Scientific Research) for partial supportthrough grant 2 P03 A 01109.

The research of the second author was partially supported by the Israel Science Foundation.Publication 651.

1


2/32

651

re

vision:2001-11-12

modified:2001-11-12

2 ANDRZEJ ROSLANOWSKI AND SAHARON SHELAH

hd(B) = sup{d(X) : X Ult(B)}, whered(X) = min{|Y| : Y X is dense in X },

hd(B) = sup{L(X) : X Ult(B)}, whereL(X) = min{ : every open cover of X has a subcover of size }.

The respective pairs of cardinal numbers are defined using sup, so even if we knowthat they are equal we still may expect different attainment properties: one of thefamilies of cardinals may have the largest member while the other not. Also wemay ask if the sup has to be attained. Situation may seem even more complicated ifone notices that there are more than just two equivalent definitions of the cardinalfunctions s, hd, hL: Monk [13] lists six equivalent definitions for spread (see [13,Theorem 13.1]), nine definitions for hd, and nine for hL (see [13, Theorems 16.1,15.1]). Fortunately, there is a number of dependencies here.

First, all of the equivalents of spread have the same attainment properties. More-over, spread is always attained for singular strong limit cardinals and for singular

cardinals of countable cofinality (for these and related results see Hajnal and Juhasz[3], [4], [5], Juhasz [8], [9], Roitman [14], Kunen and Roitman [11], Juhasz and She-lah [10]). Then Shelah [20] proved that 2cf(s(B)) < s(B) implies that the spread isattained (see 1.3 here). Finally, it is shown in Shelah [18, 4] that, e.g., if is asingular strong limit cardinal such that < cf() < 2, then there is a Booleanalgebra B such that |B| = s(B) = and the spread is not obtained. Thus, to someextend, the problem of attainment for spread is settled.

Many of the results mentioned above can be carried out for (some) variants ofhd and hL. However, the difference between these two cases and the case of thespread is that the various equivalent definitions of the respective cardinal functionmight have different attainment properties.

Let us introduce some of the equivalents of hL, hd. They will be called hL(n),hd(n), with the integer n referring to the respective cardinal n as used in the

proofs of [13, 15.1 and 16.1], respectively. Also, we will have hd+(n) and hL+(n) to

have proper language to deal with the attainment questions. Let us start withthe hereditary Lindelof degree hL. First, for a topological space X we define theLindelof degree L(X) of the space X as

L(X) = min{ : every open cover of X has a subcover of size }.

Definition 0.1. Let B be an infinite Boolean algebra. For an ideal I in a Booleanalgebra B we let

cof(I) = min{|A| : A I and (b I)(a A)(b a)}.

Now we define

hL(+)

(0)(B) = sup{L(X)(+) : X is a subspace of Ult(B) },

hL(+)(1)

(B) = sup{cof(I)(+) : I is an ideal ofB },

hL(+)(7) (B) = sup{

(+) : there is a rightseparated sequence a : < in B }.

The superscript (+) in the above definitions means that each of the formulas hastwo versions: one with + and one without it.

The cardinals mentioned in 0.1 are among those listed in [13, Theorem 15.1], andso hL(0)(B) = hL(1)(B) = hL(8)(B). The attainment properties can be describedusing the versions with +: hL+(i)(B) = hL(i)(B) means that the supremum is not


3/32

651

re

vision:2001-11-12

modified:2001-11-12

FORCING FOR hL AND hd 3

obtained; hL+(i)(B) = hL+(j)(B) means that the respective two definitions of hL have

the same attainment for B. It is not difficult to note that

hL+(7)(B) = hL(7)(B) hL+(1)(B) = hL(1)(B)

and

hL(0)(B) = hL+(0)(B) is a regular cardinal hL

+(7)(B) = hL(7)(B)

(and the attainment of hL in senses not listed in 0.1 can be reduced to those three;see [13, p. 190, 191] for details). Also, if hL(B) is a strong limit cardinal or if ithas countable cofinality, then hL(7)(B) < hL

+(7)(B) (see Juhasz [9, 4.2, 4.3]).

In 1.4 we will show that if hL(B) is a singular cardinal such that 2cf(hL(B)) 0 is a regular cardinal. Then, using the

assumption that the claim fails, we may find a set Z I such that |Z| < |Y| and(b Y)(a Z)(b a). Now apply the induction hypothesis to Z and get a setZ I of size 0 such that (a Z)(c Z

)(a c) clearly the set Z worksfor Y too.So suppose now that Y I and |Y| is a singular cardinal > 0. Let Y =


11/32

651

re

vision:2001-11-12

modified:2001-11-12


conclude that D0 is a proper filter on . Since / Fa , we get that D0 extends thefilter of co-bounded subsets of .

Claim 2.2.3. The set A+ def= { < : a idB({a : < })} does not belong tothe filter D0.

Proof of the claim. Assume toward contradiction that A+ D0. Thus we havea sequence b : < I, < , such that


12/32

651

re

vision:2001-11-12

modified:2001-11-12


(3) if w w, F 2w

and

(f F)(g F)(f g) and (g F)(g w cl(F))

thenB(w,F) is a subalgebra ofB(w,F).

Remark 2.6. Let F 2w. We will use the same notation for a member f of F andthe homomorphism from B(w,F) determined by it. Hence, for a Boolean term , afinite set v w and f F, we may write f((x : v)) etc.

Proposition 2.7. LetB be a Boolean algebra.

(1) A sequence a = a : < of elements ofB is: ideal independent if and only if for each < there is a homomor-

phism f : B {0, 1} such that

f(a) = 1 and ( < )( = f(a) = 0);

leftseparated if and only if for each < there is a homomorphismf : B {0, 1} such that

f(a) = 1 and ( < )( < f(a) = 0);

rightseparated if and only if for each < there is a homomorphismf : B {0, 1} such that

f(a) = 1 and ( < )(f(a) = 0).

(2) If the algebra B is generated by a sequence x : < , and there is anideal independent (leftseparated, rightseparated, respectively) sequence ofelements ofB of length , then there is such a sequence with terms of theform

a =

k


13/32

651

re

vision:2001-11-12

modified:2001-11-12


Definition 3.2. Let S = (,, ) be a convenient parameter and let the set

XSdef

= {(i, ) : i < cf() & < i}be equipped with the lexicographic order S (i.e., (i, ) S (i

, ) if and only ifeither i < i , or i = i and < ).

(1) We define a forcing notion QS as follows.A condition is a tuple p = wp, up, fpi, : (i, ) u

p such that

(a) wp [cf()]< , up [XS ]< ,

(b) (i wp)((i, 0) up) and if (i, ) up then i wp,(c) for (i, ) up, fpi, : u

p 2 is a function such that

(j,) up & (j,) S (i, ) fpi,(j,) = 0,

and fpi,(i, ) = 1,the order is given by: p q if and only if() wp wq , up uq , and() ((i, ) up)(fpi, f

qi,), and

() for each (i, ) uq one of the following occurs:either fqi, u

p = 0up ,

or i wp and for some , i we have (i, ) up and fqi, u

p =

(fpi,), where (fpi,) : u

p 2 is defined by

(fpi,)(j,) =

0 if j = i, < ,fpi,(j,) otherwise,

or i / wp and fqi, up = (fpj,) for some (j,) u

p and , j ,

where (fpj,) is defined as above.

(2) We say that conditions p, q QS

are isomorphic if the linear orders

(up, S up) a n d (uq, S u

q)

are isomorphic, and if H : up uq is the Sisomorphism then:() H(i, ) = (j, 0) if and only if = 0,() fpi, = f

qH(i,) H (for (i, ) u

p).

In this situation we may call H an isomorphism from p to q.

Remark 3.3. (1) Of course, S is a well ordering of XS in the order type .(2) The forcing notion QS is a relative of the one used in [16, 7].(3) There are isomorphism types of conditions in QS (remember


14/32

651

re

vision:2001-11-12

modified:2001-11-12


(i, ) up and fqi, up = (fpi,) . Now look at the definition of the operation ( )

it should be clear that fri, up = (fpi,) for some

.

Case 2: i / wp.If i wq then for some , we have fri, u

q = (fqi,) and fqi, u

p = (fpj,) for

some j, , . Now, since i / wp we may write fri, up = (fqi,) u

p = (fpj,)

and we are done. Suppose now that i / wq . Then fri, uq = (fqj,) (for some

j,,) and we ask if j wp. If so, then for some , we have fqj, up = (fpj,)

and hence fri, up = (fpj,) (for some

). If not (i.e., if j / wp) then like before

we easily conclude that fri, up = (fqj,) u

p = fqj, up = (fpj,) (for some

j, , ).

Thus QS is a forcing notion. To check that it is (< )complete suppose that < and p : < QS is an increasing sequence of conditions. Putwp =


15/32

651

re

vision:2001-11-12

modified:2001-11-12


(1) fi, : XS 2 (for (i, ) XS) is such that fi,(i, ) = 1 and

((j,) XS)((j,) S (i, )

fi,(j,) = 0).(2) B0S is the Boolean algebraB(XS,F) (see 2.4), where

F = {(fi,) : (i, ) XS , i}

and (fi,) : XS 2 is such that

(fi,)(j,) =

0 if j = i, < ,

fi,(j,) otherwise,(for (j,) XS).

(3) The sequence xi, : (i, ) XS is rightseparated in B0S (when we considerXS with the well ordering S).

Proof. Should be clear (for the third clause remember that each fi, extends to ahomomorphism from B0S to {0, 1}, see 2.5; remember 2.7).

Theorem 3.6. Assume S = (,, ) is a convenient parameter. Then

QS there is no ideal I B0S such that cof(I) = .

Proof. Let I be a QSname for an ideal in B0S , p QS, and suppose that p QScof(I) = .

Fix i < cf() for a moment.

It follows 2.2 that we may choose pi, i, ni, Di, ei and ti such that

() pi QS is a condition stronger than p, i is a regular cardinal, +i < i <

and ni ,() Di is a QSname for a (< i)complete filter on i extending the filter ofco-bounded subsets of i,

() QS ei : i ni XS and ti : i ni 2 ;

for < i let ai be a QSname for an element of B

0S such that

QS ai =


16/32

651

re

vision:2001-11-12

modified:2001-11-12


and Gi < such that the set

XG

i

def

=

< i : G

i and pi

Gi ,Gi , p

i

,Gi G and w

piG

i

,G

i = w

pi,G

i ,the conditions pi

Gi ,Gi

, pi,Gi

are isomorphic, and

if H : upiGi,Gi u

pi,G

i is the isomorphism then

( < ni)(H(eGii (

Gi , )) = e

Gii (, ) & t

Gii (

Gi , ) = t

Gii (, ))

and if j i, (j,) XS then

(j,) upiGi,Gi (j,) u

pi,G

i

is not modulo DGi (remember that in V[G] we still have cf()


17/32

651

re

vision:2001-11-12

modified:2001-11-12


the conditions r1, r2 are isomorphic and ifH is the isomorphism from r1 tor2 then H(e

ii (i, )) = e

ii (, ) and t

ii (i, ) = t

ii (, ) (for < ni),

wr1 = w

r2 and the isomorphism H is the identity on u

r1 u

r2 ,

(j,) S H(j,) for (j,) ur1 \ ur2 , and if j i, (j,) XS then (j,) ur1 (j,) ur2.

Why is the choice possible? Let G QS be generic over V such that qi G.It follows from clauses (ii), (iii) that we may find BGi \ (i + 1) such that( i\vi)(si,i S s

,i ). Then the two ordinals i, have the required properties

in V[G], and hence clearly in V too.Next we let wr = wr1 = wr2 , ur = ur1 ur2 and for (j,) ur we define

frj, : ur 2 as follows.

If (j,) ur1 ur2 then frj, = fr1j, f

r2j,,

if (j,) ur1 \ ur2 then frj, = fr1j, f

r2H(j,),

if (j,) ur2 \ ur1 then frj, = (fr1H1(j,)

) fr2j,.

Check that the functions frj, are well defined and that

r = wr, ur, frj, : (j,) ur QS

is a condition stronger than r1, r2. Let 1 =


18/32

651

re

vision:2001-11-12

modified:2001-11-12


(d) if \ v then for all j C+ i and < we have

s,j

S s

, s

,j

S s

,i

and s

j ,j

S s

, s

j ,j

S s

,i

,

(e) some condition stronger than qi forces that i Bi (see clause (ii) earlier),(f) if \ v then for all j C+ i and < we have

s,j S s,i s

,j S s

i,i and s

j ,j

S s,i s

j ,j

S si,i ,

(g) if v, s, = (j,) then j < min(C+).

Note that then

i, j C+ & , < & sj ,j

= si,i = v v

.

So si,i : < : i C+ is a system of sequences with the heart s, :

v v. Let u = {s, : v v} and w = {j < cf() : (j, 0) u}.Pick i C+ such that |C+ i| = .

Claim 3.6.2.

qi QS ( Bi)(j1, j2 C+)(ai

aj1j1

aj2j2 & pj1 , pj2 QS ) .

Proof of the claim. We are going to show that for every condition q qi and anordinal < i such that q Bi , there are a condition r q and ordinalsj1, j2 C

+ such that

r ai

aj1j1

aj2j2 & pj1 , pj2 QS .

So suppose q qi and q Bi . We may assume that pi

,i q (see the

definition of Xi , Bi). Choose j1 C+ i and j2 C+ \ (i

+ 1) such that

uq upj1

j1 ,j1 = uq upj2

j2 ,j2 = u and sup(wq) < min(wpj2

j2 ,j2 \ w).

(Remember that {upj

j,j : j C+} forms a system with heart u and hence

{wpj

j,j : j C+} forms a system with heart w.)

To make the notation somewhat simpler let q0 = pi

,i, q1 = pj1j1 ,j1

and

q2 = pj2j2 ,j2. Note that the conditions q0, q1, q2 are pairwise isomorphic, and

the isomorphisms are the identity on the u (which is the common part of any two

uqk

s). Put

0 =


19/32

651

re

vision:2001-11-12

modified:2001-11-12


(3) If (i, ) uq \ u then we look at fqi, uq0 . If it is 0uq0 then we let

fri, = fqi, 0uq1 0uq2 . Otherwise we find (j,) u

q0 and j such

that fqi, uq0 = (fq

0

j,) and if i wq0 then i = j, and we define:

() ifj w, j < i sup(w) then fri, = fqi,(f

q1

H0,1(j,))j(f

q2

H0,2(j,))j ,

() if i = j w then fri, = fqi, (f

q1

H0,1(j,)) (fq2

H0,2(j,)) , where

= max{, },

() if j w, i < j then fri, = fqi, (f

q1

H0,1(j,)) (f

q2

H0,2(j,)),

() if either i > sup(w) or j / w then we first choose j wq2

and

j such that (j, ) uq2

and (fq2

j,)(j, ) = 0 whenever

(j, ) uq2

, j w, and (fq2

j,) (2) = 1 if possible (under our

conditions); next we let fri, = fqi, 0uq1 (f

q2

j,) .

(4) If (i, ) uq2

\ u

, i w

then we let fri, = (f

q

H2,0(i,)) (fq1

H2,1(i,)) fq2

i,.(5) If (i, ) uq

2

, i / w then we put fri, = 0uq 0uq1 fq2

i,.

It should be a routine to check that in all cases the function fri, is well defined

and that r = wr, ur, fri, : (i, ) ur QS is a condition stronger than q, q1, q2

(and thus stronger than pj1 , pj2). [Remember that w min(C+), so for j w

we have (j,) uq0

(j,) upi

i ,i and hence, when checking clause 3.2(1c)in Case 1, we may use clauses (d), (f) of the choice of the set C+. They imply

that if (i, ) uq1

, i w then (i, ) S H1,0(i, ) S H1,2(i, ). Considering Case

3() with j / w, use the fact that min(wq0

\ w) sup(w) (it follows from our

choices). Similarly in Case 2 remember min(wq1

\ w) sup(w).]We claim that Br |= 0 1 2 and for this we have to show that there is no

function f Fr with f(0) = 1 and f(1) = f(2) = 0 (see 2.5). So suppose towardcontradiction that f Fr is such a function. Note that f cannot be 0ur as then thevalues given to all the terms would be the same (remember they are isomorphic).So for some (i, ) ur and i we have f = (fri,). Let us look at all thecases appearing in the definition of the functions frj, s (we keep labeling as thereso we do not repeat the descriptions of the cases).

Case 1: Clearly fri,(0) = fri,(1). It follows from the demands (d), (f ) of the

choice of C+ that if i w, (i, ) uq0

, (i, ) = H0,1(i, ), then i = i and .Consequently, we may use 3.6.1 to conclude that (fri,)(0) (f

ri,)(1), what

contradicts the choice of f.

Case 2: Plainly (fri,)(0) = (fri,)(2).

Case 3: Note that fri,(0) = fri,(1) and, as j < i sup(w), necessarily

i / wq0

wq1

. Hence easily (fri,)(0) = (fri,)(1).

Cases 3, , 4: Like in cases 1, 3 we conclude (fri,)(0) (fri,)(1).

Case 3: It follows from the choice of , , j there that fri,(0) fri,(2). If

i / wq0

then (as also i / wq2) we have f(0) = fri,(0) and f(2) = f

ri,(2), so we

are done. If i wq0

then i = j and we easily finish by the choice of , , j.

Case 5: Clearly (fri,)(0) = (fri,)(1), a contradiction.


20/32

651

re

vision:2001-11-12

modified:2001-11-12


Thus we may conclude that r ai

aj1j1

aj2j2 , finishing the proof of theclaim.

Now we may easily finish the theorem: take a generic filter G QS over Vsuch that qi G and work in V[G]. Since the filter DGi is (< i)complete andcf() < i , we find j1, j2 C+ such that pj1 , pj2 G and

{ BGi : (ai

)G (aj1j1 )

G (aj2j2 )G} = mod DGi

(remember BGi = mod DGi by (ii)). But then also (a

j1j1

)G (aj2j2 )G IG, so we

get a contradiction to clause ().

Conclusion 3.7. It is consistent that there is a Boolean algebra B of size such thatthere is a rightseparated sequence of length in B, (so hL+(7)(B) =

+), but there is

no ideal I B with the generating number (and thus hL+(1)(B) = hL(1)(B) = ).

Problem 3.1. Does there exist a Boolean algebra B as in 3.7 in semi-ZFC? I.e.,can one construct such an algebra for from cardinal arithmetic assumptions?

4. Forcing for hd

Here we deal with a problem parallel to the one from the previous section andrelated to the attainment question for hd. We introduce a forcing notion PS comple-mentary to QS and we use it to show that, consistently, there is a Boolean algebraB of size in which there is a strictly decreasing sequence of ideals but everyhomomorphic image of B has algebraic density less than . This gives a partialanswer to [13, Problem 54]. Again, we do not know if an example like that can beconstructed from cardinal arithmetic assumptions.

Definition 4.1. Let S = (,, ) be a good parameter (see 3.1) and let XS, Sbe as defined in 3.2.

(1) We define a forcing notion PS as follows.A condition is a tuple p = wp, up, fpi, : (i, ) u

p such that

(a) wp [cf()]< , up [XS ]< ,(b) (i wp)((i, 0) up) and if (i, ) up then i wp,(c) for (i, ) up, fpi, : u

p 2 is a function such that

(j,) up & (i, ) S (j,) fpi,(j,) = 0,

and fpi,(i, ) = 1,the order is given by: p q if and only if() wp wq , up uq , and() ((i, ) up)(fpi, f

qi,), and

() for each (i, ) uq one of the following occurs:either fqi, u

p = 0up ,

or i wp and for some , < i we have (i, ) up and fqi, u

p =

(fpi,), where (fpi,)

: up 2 is defined by

(fpi,)(j,) =

0 if j = i, < i,fpi,(j,) otherwise,


21/32

651

re

vision:2001-11-12

modified:2001-11-12


or i / wp and either fqi, up = (fpj,)

(defined above) for some

(j,) up, < j or fqi, u

p = (fpj,)j for some (j,) up and

j j, where (fpj,)j : up 2 is defined by

(fpj,)j (j, ) =

0 if j j,fpj,(j

, ) otherwise.

(2) Conditions p, q PS are said to be isomorphic if the well orderings

(up, S up) and (uq, S u

q)

are isomorphic, and if H : up uq is the Sisomorphism then:() H(i, ) = (j, 0) if and only if = 0,() fpi, = f

qH(i,) H (for (i, ) u

p).

Proposition 4.2. Let S = (,, ) be a good parameter. ThenPS is a (< )complete +cc forcing notion.

Proof. Plainly PS is a (< )complete forcing notion (compare the proof of 3.4).To verify the chain condition suppose that A PS , |A| = +. Apply the lemma and standard cleaning to choose isomorphic conditions p, q A such thatif H : up uq is the isomorphism from p to q then H (up uq) is the identityon up uq. Put wr = wp wq , ur = up uq and for (i, ) ur define a functionfri, : u

r 2 as follows.

If (i, ) up, i wp wq then fri, = fpi, (f

q

H(i,))+1,

if (i, ) uq, i wp wq then fri, = (fp

H1(i,))+1 fqi,,

if (i, ) up, i wp \ wq then fri, = fpi, (f

qH(i,))i,

if (i, ) uq, i wq \ wp then fri, = (fp

H1(i,))i f

qi, .

It is a routine to check that the functions fr

i, are well defined and that they satisfythe demand 4.1(1c). Hence r = wr, ur, fri, : (i, ) ur PS and one easily

checks that it is an upper bound to both p and q.

For a condition p PS let

Fp = {(fpi,), (fpi,)j : (i, ) u

p, < i, j i},

where (fpi,), (fpi,)j : u

p 2 are defined like in 4.1(1):

(fpi,)(i, ) =

0 if i = i, ,fpi,(i

, ) otherwise,

(fpi,)j(i, ) =

0 if j i,fpi,(i

, ) otherwise.

Like in the previous section, Bp is the Boolean algebra B(up,Fp) (see 2.4) (note that

p q implies that Bp is a subalgebra ofBq). Let B1S be a PSname such that

PS B1S =

{Bp : p PS} ,

and for s XS let fs be a PSname such that

PS fs =

{fps : s up, p PS} .

Proposition 4.3. Assume that S = (,, ) is a good parameter. Then in VPS :


22/32

651

re

vision:2001-11-12

modified:2001-11-12


(1) For s XS , fs : XS 2 is such that fs(s) = 1 and

(s

XS)(s S s

fs(s

) = 0).

(2) B1S is the Boolean algebraB(XS,F) (see 2.4), where

F = {(fi,), (fi,)j : (i, ) XS , < i, j i},

and (fi,), (fi,)j : XS 2 are such that

(fi,)(i, ) =

0 if i = i, ,

fi,(i, ) otherwise,

(fi,)j(i

,

) = 0 if j i,

fi,(i, ) otherwise.

(3) The sequence xs : s XS is leftseparated in B1S (when we consider XSwith the well ordering S).

Theorem 4.4. Assume S = (,, ) is a good parameter. Then

PS there is no ideal I B1S such that (B

1S/I) = .

Proof. Not surprisingly, the proof is similar to the one of 3.6. Let I be a PSnamefor an ideal in B1S , p PS, and suppose that p PS (B

1S/I) = .

Fix i < cf(). Use 2.3 to choose pi, i, ni, Di, ei and ti such that

() pi PS is a condition stronger than p, i is a regular cardinal, +i < i < and ni ,

() Di is a PSname for a (< i)complete filter on i extending the filter ofco-bounded subsets of i,

() PS ei : i ni XS and ti : i ni 2 ;for < i let a

i be a PSname for an element of B

1S such that

PS ai =


23/32

651

re

vision:2001-11-12

modified:2001-11-12


and Gi < such that the set

XGi

def

=

< i : Gi and p

iGi ,

Gi , p

i,Gi G and w

piG

i,G

i = w

pi,G

i ,the conditions pi

Gi ,Gi

, pi,Gi

are isomorphic, and

if H : upiGi,Gi u

pi,G

i is the isomorphism then

( < ni)(H(eGii (

Gi , )) = e

Gii (, ) & t

Gii (

Gi , ) = t

Gii (, ))

and if j i, (j,) XS then

(j,) upiGi,Gi (j,) u

pi,G

i

is not modulo DGi . Let Gi = otp(u

piGi,Gi , S) and for XGi let s

,i : <

Gi

be the Sincreasing enumeration of upi,G

i . Apply Lemma 2.1 to find a sequences,i : <

Gi XS and a set v

Gi

Gi such that

(i) ( Gi \ vGi )(+i cf({s XS : s S s,i }, S) i),

(ii) the set

BGidef=

XGi : if vGi then s

,i = s

,i , and

if Gi \ vGi then

supS{s,i : <

Gi , s

,i S s

,i } S s

,i S s

,i

is not modulo the filter DGi ,

(iii) ifs S s,i for

Gi \ v

Gi then

{ BGi : ( Gi \ v

Gi )(s

S s

,i )} = mod D

Gi .

We may assume that Gi B

Gi .

Now, in V, we choose a condition qi PS, ordinals i, i, i, a set vi and asequence s,i : < i XS such that qi p

ii,i

, and qi forces that these objects

are as described above. If some condition stronger than qi forces that Bi, then

we will use s,i : < i to denote the Sincreasing enumeration of upi,i .

Next, like in the proof of 3.6, we pick an unbounded set Y cf() and n < , < , v such that for i, j Y:

ni = n, i = , vi = v, and the conditions pii,i , p

jj ,j

are isomorphic, and the isomorphism maps eii (i, )

and tii (i, ) onto ejj (j , ), t

jj (j, ), respectively.

Now use Lemma 2.1 to find a sequence s, : < XS {(cf(), 0)} and a set

v

such that(a) ( \ v)(cf({s XS : s S s,}, S) = cf()),(b) the set

Cdef=

i Y : if v then s,i = s,, and

if \ v then

supS{s, : < , s, S s,} S s,i S s,

is unbounded in cf(),


24/32

651

re

vision:2001-11-12

modified:2001-11-12


(c) ifs S s, for \ v, then the set

{i C : ( \ v)(s S s,i )}

is unbounded in cf().

Next choose a set C+ [C]cf() and ordinals i < i < i (for i C+) such thatfor every i C+:

(d) if \ v then for all j C+ i and < we have

s,j S s, s,j S s

,i , and

sj ,j

S s, sj ,j

S s,i , and

sj ,j

S s, sj,j

S s,i ,

(e) some condition stronger than qi forces that i, i Bi,(f) if \ v and x {i, i}, then for all j C+ i and < we have

s,j

S s,i

s,j

S sx,i

and s

j ,j

S s,i

s

j ,j

S sx,i

, andsj ,j

S s,i s

j ,j

S sx,i and s

i,i S s

,i s

i,i S s

i,i ,

(g) if v, s, = (j,) then j < min(C+).

Then si,i , si,i : < : i C

+ forms a system of sequences with hearts, : v v; but note that si,i = s

i,i for v. Let u

= {s, : v v}and w = {j < cf() : (j, 0) u}.

Claim 4.4.1. For each i0 C+:

qi0 PS ( Bi0)(i C+)(ai

i (ai

i) ai0 & pi PS)

(where Bi0 was defined in (ii)).

Proof of the claim. Let i0 C+. We will show that for every condition q qi0 and

an ordinal < i0 such that q Bi0 , there are i C+ and a condition rstronger than both q and pi , and such that r a

i

i (ai

i) ai0 .

So suppose q qi, q Bi0 . We may assume that pi0,i0

q. Choose

i C+ \ (i0 + 1) such that

uq upi

i ,i = uq upi

i ,i = u and wq i,

Let pi0,i0= q0, pi

i ,i= q1, pi

i ,i= q2, and

0 =


25/32

651

re

vision:2001-11-12

modified:2001-11-12


(2) If (i, ) uq1

uq2

, i / wq then we first choose such that, if possible,

(fq0

H1,0(i,)) (0) = 1, and then we let f

ri, = (f

q

H1,0(i,)) fq

1

i, fq2

i, .

[Note that H1,0(i, ) S (i, ), and thus if H1,0(i, ) = (j,) then j < i,j / wq

1

.]

(3) If (i, ) uq2

\ uq1

(so i > i sup(wq)) then we first choose such

that, if possible, (fq0

H2,0(i,))

(0) = 1, and then we let fri, = (f

q

H2,0(i,))

fq1

H2,1(i,) fq2

i,.

[Note that H2,0(i, ) S (i, 0) S H2,1(i, ) S (i, ); remember wq1

= wq2

.

Also, if H2,0(i, ) = (j,), then j / wq1

.]

(4) If (i, ) uq1

\ uq2

then, like above, we choose such that if possible

then (fq0

H1,0(i,))(0) = 1, and next we put f

ri, = (f

q

H1,0(i,)) fq

1

i,

(fq2

H1,2(i,))+1.

(5) If (i, ) uq \ uq1 then we look at fqi, uq0 . If it is 0uq0 then we let

fri, = fqi, 0uq1 0uq2 . Otherwise, we consider the following three cases.

() Suppose i wq0

. Then for some < i, + 1 we have

fqi, uq0 = (fq

0

i,) and we let:

if i wq1

then fri, = fqi, (f

q1

H0,1(i,)) (fq

2

H0,2(i,)),

if i / wq1

then fri, = fqi, (f

q1

H0,1(i,))i (fq2

H0,2(i,))i.

[Note that if i wq1

then (i, ) S H0,1(i, ) = H0,2(i, ) S (i + 1, 0),

and if i / wq1

then (j, 0) S H0,1(i, ) S H0

,2(i, ) S (j + 1, 0) for somej > i.]

() Suppose i / wq0

(so i / wq1

) and fqi, uq0 = (fq

0

i,) , (i, ) uq

0

,

< i . If i wq

1

and i < i, then put fri, = fqi, (f

q1

H0,1(i,))

(fq2

H0,2(i,)) .

If i wq1

and i < i, then we put fri, = fqi, (f

q1

H0,1(i,))i

(fq2

H0,2(i,))i.

If i / wq1

, then let fri, = fqi, (f

q1

H0,1(i,))i (fq2

H0,2(i,))i.

() Suppose i / wq0

and fqi, uq0 = (fq

0

i,)j , j min{i, i}, (i, ) uq

0

.

Let fri, = fqi, (f

q1

H0,1(i,))j (fq2

H0,2(i,))j .

Verifying that the functions fri, are well defined and that r = wr , ur, fri, : (i, )

ur PS is a condition stronger than q, q1, q2 is left to the reader. Let us arguethat Br |= 1 (2) 0. If not then we have a function f Fr such thatf(0) = f(2) = 0 and f(1) = 1. Clearly f cannot be 0ur , so it is either (f

ri,)

or (fri,)j . Let us look at the definition of the functions fri, and consider each case

there separately.

Cases 1, 5,,: Plainly fri,(1) = fri,(2) and also (f

ri,)j(1) = (f

ri,)j(2)

(remember wq1

= wq2

). As far as the operation () is concerned, note that ({i}


26/32

651

re

vision:2001-11-12

modified:2001-11-12


i) uq1 = ({i} i) u

q2 , so (in these cases) we easily get (fri,)(1) = (f

ri,)

(2),a contradiction.

Case 2: Again, fri,(1) = fri,(2) and (f

ri,)j(1) = (f

ri,)j(2) (for each j). So

suppose that f = (fri,) for some , and look at the choice of in the current

case. Since 1 = (fri,)(1) = (f

q1

i,)(1), we conclude that 1 = (f

q0

H1,0(i,))(0) =

fri,(0) = (fri,)

(0), a contradiction.

Case 3: Note that fri,(1) = fri,(2) (and also (f

ri,)j(1) = (f

ri,)j(2)). Now,

if for some we have (fri,)(1) = 1, then look at the choice of

necessarily

(fri,)(0) = f

ri,(0) = 1 (remember (i, 0) S H

2,1(i, ) S (i, )).

Case 4: Like above: if for some we have (fri,)(1) = 1, then necessarily

fri,(0) = (fri,)

(0) = 1. Moreover, (fri,)j(1) = (f

ri,)j(2) for all j i.

In all cases we get a contradiction showing that Br |= 1 (2) 0, and hencer ai

i (ai

i) ai0 , finishing the proof of the claim.

Finally we note that 4.4.1 and clauses (), () give an immediate contradiction,showing the theorem.

Conclusion 4.5. It is consistent that there is a Boolean algebra B of size suchthat there is a leftseparated sequence of length in B (and thus hd+(5)(B) =

+),

but there is no ideal I B with (B/I) = (so hd+(7)(B) = hd(7)(B) = ).

Problem 4.1. Can one construct a Boolean algebra B as in 4.5 for from anycardinal arithmetic assumptions?

5. More on the attainment problem

In this section we will assume the following:

Hypothesis 5.1. S = (,, ) is such that , are cardinals satisfying

=


27/32

651

re

vision:2001-11-12

modified:2001-11-12


For a topological space X, a (0, 1)Lusin set in X is a set L X such that|L| = 0 and for every meager subset Z of X the intersection Z L is of size less

than 1. (See, e.g., Cichon [1] for a discussion of sets of this type.) Below, thespace is equipped with the topology generated by sets of the form

[] = { : }

for


28/32

651

re

vision:2001-11-12

modified:2001-11-12


Necessarily, there are X [] and p P such that p = p for X. Then also0 < 1 for 0 < 1 from X. Shrinking X a little we may also demand that for

some sequences j lh(p

) + 2 (for j < cf()) we have

X & j() = j j .

Now pick 0 < 1 from X such that letting j0 = j(0) and j1 = j(1) we have

j0 < j1 and j0 = j1 and j0


29/32

651

re

vision:2001-11-12

modified:2001-11-12


Lett(k) {0, 1} for k < k. Then

Bb |= k


30/32

651

re

vision:2001-11-12

modified:2001-11-12


(x) ifi1 < i2, i1, i2 S, then ji1k < i2 (for k < k

),(xi) if i1, i2 S are distinct, 1 Si1 and 2 Si2 , then

{(1, k) : k < k} {(2, k) : k < k} = {k : k < k}.

Let S =iS

Si. For < and k+ k < k let

SL,k =

S : ( S)

> lh((,k) (,k)) or (,k) (,k) / A or(,k) lex (,k)

,

SR,k =

S : ( S)

> lh((,k) (,k)) or (,k) (,k) / A or(,k) lex (,k)

.

We claim that both |SL,k| < and |SR,k| < . Why? Assume, e.g., |S

L,k| = .

Note that, by (v)+(vi)+(x), (, k) < (, k) for < from S. Pick and

a set X [SL,k] such that ( X)( (,k)). By 5.2(1)(c), there are distinct

, X such that (,k) (,k) A. Clearly lh((,k) (,k)) and weeasily get a contradiction with , SL,k. Similarly for S

R,k.

For k+ k < k let

Sk =

S : for all < there exists S such that (,k)


31/32

651

re

vision:2001-11-12

modified:2001-11-12


procedure as in 5.6.3 getting S, Si, , k , tk, j

ik etc such that clauses (i)(xi) are

satisfied. We let S = iS

Si and for < and k+ k < k we define

S+,k =

S : (S )

> lh((,k) (,k)) or (,k) (,k) / Aor (,k) lex (,k)

,

S,k =

S : (S )

> lh((,k) (,k)) or (,k) (,k) / Aor (,k) lex (,k)

.

Then both |S+,k| < and |S,k| < . [It is like before: assume, e.g., |S

+,k| = .

Pick and a set X [S+,k] such that ( X)( (,k)). Note that

(, k) < (, k) for < from S. Use 5.2(2)(c+) to find < , both from X,such that (,k) (,k) A and (,k)


32/32

re

vision:2001-11-12

modified:2001-11-12


[17] Boris Sapirovski. Canonical sets and character. Density and weight in bicompacta. (Russian).Dokl. Akad. Nauk SSSR, 218:5861, 1974.

[18] Saharon Shelah. Constructing Boolean algebras for cardinal invariants. Algebra Universalis,accepted, math.LO/9712286.

[19] Saharon Shelah. PCF and infinite free subsets. Archive for Mathematical Logic, accepted,math.LO/9807177.

[20] Saharon Shelah. Remarks on the numbers of ideals of Boolean algebra and open sets of atopology. In Around classification theory of models, volume 1182 of Lecture Notes in Math-ematics, pages 151187. Springer, Berlin, 1986.

[21] Saharon Shelah. Further cardinal arithmetic. Israel Journal of Mathematics, 95:61114, 1996,math.LO/9610226.

[22] Saharon Shelah. On Monks questions. Fundamenta Mathematicae, 151:119, 1996,math.LO/9601218.

[23] Saharon Shelah. Special Subsets ofcf(), Boolean Algebras and Maharam measure Algebras.Topology and its Applications, 99:135235, 1999, math.LO/9804156. 8th Prague TopologicalSymposium on General Topology and its Relations to Modern Analysis and Algebra, Part II(1996).

Department of Mathematics, University of Nebraska at Omaha, Omaha, NE 68182-

0243, USA, and Mathematical Institute of Wroclaw University, 50384 Wroclaw, Poland

E-mail address: [email protected]: http://www.unomaha.edu/aroslano

Institute of Mathematics, The Hebrew University of Jerusalem, 91904 Jerusalem,

Israel, and Department of Mathematics, Rutgers University, New Brunswick, NJ 08854,

USA

E-mail address: [email protected]: http://www.math.rutgers.edu/shelah

andrzej roslanowski and saharon shelah- forcing for hl and hd

Documents