Andrzej Roslanowski and Saharon Shelah- Historic Forcing for Depth

8/3/2019 Andrzej Roslanowski and Saharon Shelah- Historic Forcing for Depth

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HISTORIC FORCING FOR Depth

ANDRZEJ ROSLANOWSKI AND SAHARON SHELAH

Abstract. We show that, consistently, for some regular cardinals < , thereexists a Boolean algebra B such that |B| = + and for every subalgebra B Bof size + we have Depth(B) = .

0. Introduction

The present paper is concerned with forcing a Boolean algebra which has someprescribed properties of Depth. Let us recall that, for a Boolean algebra B, itsdepth is defined as follows:

Depth(B) = sup{|X| : X B is well-ordered by the Boolean ordering },Depth+(B) = sup{|X|+ : X B is well-ordered by the Boolean ordering }.

(Depth+(B) is used to deal with attainment properties in the definition of Depth(B),see e.g. [5, 1].) The depth (of Boolean algebras) is among cardinal functions thathave more algebraic origins, and their relations to topological fellows is oftenindirect, though sometimes very surprising. For example, if we define

DepthH+(B) = sup{Depth(B/I) : I is an ideal in B },

then for any (infinite) Boolean algebra B we will have that DepthH+(B) is the

tightness t(B) of the algebra B (or the tightness of the topological space Ult(B) ofultrafilters on B), see [3, Theorem 4.21]. A somewhat similar function to DepthH+ isobtained by taking sup{Depth(B) : B is a subalgebra ofB }, but clearly this bringsnothing new: it is the old Depth. But if one wants to understand the behaviour ofthe depth for subalgebras of the considered Boolean algebra, then looking at thefollowing subalgebra Depth relation may be very appropriate:

DepthSr(B) = {(, ) : there is an infinite subalgebra B ofB such that

|B| = and Depth(B) = }.

A number of results related to this relation is presented by Monk in [3, Chapter4]. There he asks if there are a Boolean algebra B and an infinite cardinal suchthat (, (2)+) DepthSr(B), while (, (2

)+) / DepthSr(B) (see Monk [3, Problem14]; we refer the reader to Chapter 4 of Monks book [3] for the motivation andbackground of this problem). Here we will partially answer this question, showingthat it is consistent that there is such B and . The question if that can be donein ZFC remains open.

1991 Mathematics Subject Classification. Primary 03E35, 03G05; Secondary 03E05, 06Exx.Key words and phrases. Boolean algebras, depth, historic forcing.The first author thanks the KBN (Polish Committee of Scientific Research) for partial support

through grant 2 P03 A 01109.The research of the second author was partially supported by the Israel Science Foundation.

Publication 733.

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2 ANDRZEJ ROSLANOWSKI AND SAHARON SHELAH

Our consistency result is obtained by forcing, and the construction of the requiredforcing notion is interesting per se. We use the method of historic forcing which

was first applied in Shelah and Stanley [9]. The reader familiar with [9] will noticeseveral correspondences between the construction here and the method used there.However, we do not relay on that paper and our presentation here is self-contained.

Let us describe how our historic forcing notion is built. So, we fix two (regular)

cardinals , and our aim is to force a Boolean algebra B such that |B| =

+

and for every subalgebra B B of size + we have Depth(B) = . The algebra

B will be generated by xi : i U for some set U +. A condition p will be

an approximation to the algebra B, it will carry the information on what is thesubalgebra Bp = xi : i u

pB

for some up +. A natural way to describe

algebras in this context is by listing ultrafilters (or: homomorphisms into {0, 1}):

Definition 1. For a set w and a family F 2w we definecl(F) = {g 2w : (u [w]< )(f F)(f u = g u)},B(w,F) is the Boolean algebra generated freely by {x : w} except that

if u0, u1 [w]< and there is no f F such that f u0 0, f u1 1

thenu1

x u0

(x) = 0.

This description of algebras is easy to handle, for example:

Proposition 2 (see [6, 2.6]). LetF 2w. Then:

(1) Each f F extends (uniquely) to a homomorphism fromB(w,F) to {0, 1}(i.e. it preserves the equalities from the definition ofB(w,F)). IfF is closed,then every homomorphism fromB(w,F) to {0, 1} extends exactly one elementof F.

(2) If (y0, . . . , y) is a Boolean term and 0, . . . , w are distinct then

B(w,F) |= (x0 , . . . , x) = 0 if and only if(f F)({0, 1} |= (f(0), . . . , f (k)) = 1).

(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).

So each condition p in our forcing notion P will have a set up [+]


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HISTORIC FORCING FOR Depth 3

amalgamation (and natural limits) will be the only way to build new conditionsfrom the old ones, but the description above still misses an important factor. So

far, a condition does not have to know what are the reasons for it to be called toP. This information is the history of the condition and it will be encoded by twofunctions hp, gp. (Actually, these functions will give histories of all elements of up

describing why and how those points were incorporated to up. Thus both functionswill be defined on up ht(p), were ht(p) is the height of the condition p, that isthe step in our construction at which the condition p is created.) We will also wantthat our forcing is suitably closed, and getting (


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(i) for each p P, :

up [+]< , ht(p) , Fp 2up

is a non-empty closed set, gp is a

function with domain dom(gp) = up ht(p) and values of the form (, ),where < 2 and is a Boolean term, and hp : up ht(p) + 2 is afunction,

(ii) , pr are transitive and reflexive relations on P, , and

extends pr,

(iii) if p, q P, , p q, then up uq, ht(p) ht(q), and Fp = {f up :

f Fq}, and if p pr q, then for every i up and < ht(p) we have

hp(i, ) = hq(i, ) and gp(i, ) = gq(i, ),

(iv) if < then P, P

, , and

pr extends

pr, and

extends .

For a condition p P, , we will also declare that Bp = B(up,Fp) (the Boolean

algebra defined in Definition 1).

We define P,0 = { : < +} and for p = we let Fp = 2{}, ht(p) = 0

and hp

= = gp

. The relations 0pr and

0

both are the equality. [Clearly theseobjects are as declared, i.e, clauses (i)0(iv)0 hold true.]

If < is a limit ordinal, then we put

P =

p : < : ( < < )(p P, & ht(p) = & p

pr p)

,

P, =


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() otp(up0 ) = otp(up1 ) and if H : up0 up1 is the order isomorphismthen H u is the identity on u, Fp0 = {f H : f Fp1 }, H[v0 ] = v1

and(j up0 )( < )(hp0 (j,) = hp1 (H(j), ) & gp0 (j,) = gp1 (H(j), )).

We put P,+1 = P, P

+1 and for p =

, , n, u, p, v : < P+1 we

let up =


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(4) If i, j up are distinct, then there is < ht(p) such that = hp(i, ) =hp(j,) = .

(5) For each finite set X ht(p) there is i up

such that{ < ht(p) : hp(i, ) < } = X.

(6) Ifp pr q then there is aprincreasing sequence p : ht(p) P such

that pht(p) = p, pht(q) = q and ht(p) = (for ht(p)). (In particular, ifp pr q and ht(p) = ht(q) then p = q.)

(7) If ht(p) = is a limit ordinal, p = p : < , then for each i up and < :

i up if and only if ( < )( hp(i, ) ).

Proof. 1) Should be clear (an easy induction).

2) Suppose that p P and j up are a counterexample with the minimal

possible value of ht(p). Necessarily ht(p) is a limit ordinal, p = p : < ht(p),ht(p) = and < < ht(p) p pr p. Let < ht(p) be the first ordinal suchthat j up . By the choice of p, the set { : hp(j,) < } is finite, but clearlyhp(j,) for all (, ht(p)).

3) An easy induction on ht(q) (with fixed p).

4) We show this by induction on ht(p). Suppose that ht(p) = + 1, so p =, , n, u, p, v : < , and i, j u

p are distinct. If i, j up for some < , then by the inductive hypothesis we find < such that

= hp(i, ) = hp(i, ) = hp(j,) = hp(j,) = .

If i up \ u, j up \ u and , < are distinct, then look at the definitionof hp(i, ), hp(j,) these two values cannot be equal (and both are distinct from

). Finally suppose that ht(p) is limit, so p = p : < ht(p). Take < ht(p) suchthat i, j up and apply the inductive hypothesis to p getting < such thathp(i, ) = hp(j,) (and both are not ).

5) Again, it goes by induction on ht(p). First consider a limit stage, and supposethat ht(p) = is a limit ordinal, X []


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(1) If ht(q0) = ht(q1) = is a limit ordinal, q = q : < (for < 2), then

H uq0

is an isomorphism from q0 to q1 .

(2) If ht(q0) = ht(q1) = + 1, < , and q = , , n , u , q, v : <

(for < 2), then 0 = 1 ,

0 =

1 , n

0 = n

1, H u

q0 is an isomorphismfrom q0 to q

1 and H[v

0 ] = v

1 (for < ).

(3) Fq0 = {f H : f Fq1}.(4) Assume p0 q0. Then there is a unique condition p1 q1 such thatH u

p0

is the isomorphism from p0 to p1.[The condition p1 will be called H(p0).]

Proof. 1), 2) Straightforward (for (1) use Lemma 3(7)).3), 4) Easy inductions on ht(q0) using (1), (2) above.

Definition 6. By induction on < , for conditions p, q P, such that p q,

we define the ptransformation Tp(q) of q. If = 0 (so necessarily p = q) then Tp(q) = p. Assume that ht(q) = + 1, q = , , n, u, q, v : < .

If p q for some < , then let be such that p q . Next for < let q = TH,(p)(q), where H, is the isomorphism from q to q.

Define Tp(q) = , , n, u, q, v : < .

Suppose now that p = , , n, u, p, v : < and up = uq ,

p q (for < ). Let q = Tp(q) and put Tp(q) =

, , n, u, q, v : < .

Assume now that ht(q) is a limit ordinal and q = q : < ht(q).If ht(p) < ht(q) then p q for some < ht(q), and we may choose q

(for < ht(q)) such that ht(q) = , < < ht(q) q pr q

, and

q

= Tp(q) for [, ht(q)). Next we let Tp(q) = q

: < .If ht(p) = ht(q), p = p : < ht(p) and p q for > (for some

< ht(p)) then we define Tp(q) = p.

To show that the definition of Tp(q) is correct one proves inductively (parallelyto the definition of the ptransformation of q) the following facts.

Lemma 7. Assume p, q P, p q. Then:

(1) Tp(q) P, uTp(q) = uq, ht(Tp(q)) = ht(q),

(2) p pr Tp(q) q Tp(q),(3) ht(p) = ht(q) Tp(q) = p,

(4) if q P is isomorphic to q and H : uq uq

is the isomorphism from qto q, then H is the isomorphism from Tp(q) to TH(p)(q

),

(5) if q pr q then Tp(q) pr Tp(q).

Proposition 8. Every princreasing chain inP of length < has a prupper

bound, that is the partial order (P, pr) is (< )closed.

Let us recall that a forcing notion (Q, ) is (


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pi Di and pi pj for all j < i. The second player looses the play if for somei < he has no legal move.

It should be clear that (


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If ht(p) = 0 then every Z is pclosed; if ht(p) is limit, p = p : < ht(p), then Z is pclosed provided it is

pclosed for each < ht(p); if ht(p) = + 1, p = , , n, u, p, v : < and / Z, then Z is

pclosed whenever it is pclosed; if ht(p) = + 1, p = , , n, u, p, v : < and Z, then Z is

pclosed provided it is pclosed and

{ < : (j v {min(up \ u)})(hp (j,) < )} Z.

Lemma 13. (1) If p P and w [ht(p)]< , then there is a finite pclosed

set Z ht(p) such that w Z.(2) If p, q P are isomorphic and Z is pclosed, then Z is qclosed. If Z is

pclosed, < ht(p) andp is ancomponent of p, thenZ is pclosed.

Proof. Easy inductions on ht(p) (remember Lemma 3(2)).

Definition 14. Suppose that p P and Z ht(p) is a finite pclosed set. LetZ = {0, . . . , k1} be the increasing enumeration.

(1) We define

U[p,Z]def= {j up : ( < ht(p))(hp(j,) < Z)}.

(2) We let

p(Z) = , , n, g, h0, . . . , h

n1 : < k,

where, for < k, is an ordinal below , is a Boolean term, n < andg, h0, . . . , h

n1 : 2, and they all are such that for every (equivalently:

some) + 1component q = , , n, u, q, v : < of p we have: = , = , n = n and if v = {j0, . . . , jn1} (the increasingenumeration) then

(m < n)( < )(hm(

) = hq(jm, )),

and if i0 = min(uq \ u) then ( < )(g() = hq(i0, )). (Note that

, , n, g, h0, . . . , hn1 are well-defined by Lemma 11. Necessarily, for

all m < n and \ Z we have hq(i0, ), hq(jm, ) ; remember that

Z is pclosed.)

Note that if Z ht(p) is a finite pclosed set, = max(Z) and p is the + 1component of p satisfying p pr p (see 11(3)), then U[p,Z] u

p .

Lemma 15. Suppose that p P and Z0, Z1 ht(p) are finite pclosed setssuch that p(Z0) = p(Z1). Then otp(U[p,Z0]) = otp(U[p,Z1]), and the order

preserving isomorphism : U[p,Z0] U[p,Z1] satisfies() ( < k)(hp(i, 0) = h

p((i), 1)),where {x0 , . . . ,

xk1} is the increasing enumeration of Zx (for x = 0, 1).

Proof. We prove this by induction on |Z0| = |Z1| (for all p, Z0, Z1 satisfying theassumptions).

Step |Z0| = |Z1| = 1; Z0 = {00}, Z1 = {

10}.

Take the x0 + 1component qx of p such that qx pr p. Then, for x = 0, 1,

qx = , , n , ux, qx , vx : < , and for each i v

x , <

x0 we have h

qx (i, ) .


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Also, if ix0 = min(uqx \ ux) and < x0 , then h

qx (ix0 , ) . Consequently,n = |vx | 1, and if n = 1 then {i

x0} = v

x (remember Lemma 3(4)). Moreover,

U[p,Zx] = U[qx, Zx] = {Hx,(ix0 ) : < },

where Hx, is the isomorphism from qx to q

x . Now it should be clear that the

mapping : H0,(i00) H

1,(i

10) : U[p,Z0] U[p,Z1] is the order preserving

isomorphism (remember clause () of the definition ofP+1), and it has the propertydescribed in ().

Step |Z0| = |Z1| = k + 1; Z0 = {00, . . . , 0k}, Z1 = {

10, . . . ,

1k}.

Letp(Z0) = p(Z1) = , , n, g, h

0, . . . , h

n1 : k.

For x = 0, 1, let qx = , , n , ux, qx , vx : < be the

xk + 1component of

p such that qx pr p. The sets Zx xk (for x = 0, 1) are q

x closed for every

< , and clearly p(Z0 0k) = p(Z1 1k). Hence, by the inductive hypothesis,

otp(U[q0 , Z0 \ {0k}]) = otp(U[q1 , Z1 \ {1k}]) (for each < ), and the orderpreserving mappings : U[q0 , Z0 \{

0k}] U[q

1 , Z1 \{

1k}] satisfy the demand in

(). Let ix = min(uqx \ux). Then, as qx and q

x are isomorphic and the isomorphism

is the identity on ux, we have ( < k)(hp(ix , x ) = gk()). Hence (i

0) = i

1, and

therefore [u0 U[q0 , Z0 \ {0k}]] = u

1 U[q1 , Z1 \ {1k}]. But since the mappings

are order preserving, the last equality implies that (u0 U[q0 , Z0 \ {0k}]) =

(u0 U[q0 , Z0 \ {0k}]), and hence =


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contradiction (applying the inductive hypothesis to p). So let < be such thatj up \ u. We know that xj / xi : i u

(v up)Bp (remember clause

() of the definition of P+1), so we may take functions f0, f1 F

p

such thatf0 (u

(v up )) = f1 (u(v up )), f0(j) = 0, f1(j) = 1. Let g0, g1 : u

p 2be such that g u

p = f, g up = f0 H, for = (where H, is the order

isomorphism from up to up). Now one easily checks that g0, g1 Fp (remember

the definition of the term maj). By our choices, g0(i) = g1(i) for all i v, andg0(j) = g1(j), and this is a clear contradiction with the choice of i and v.

3) Suppose that a : < + is a Pname for a

+sequence of distinct members

of B and let p P. Applying standard cleaning procedures we find a set A

+

of the order type , an ordinal < and , n, u and p, v : A such thatp p, ht(p) = , p a = (xi : i v) and

qdef= 0, , n, u, p, v : A P

+1,

where A is identified with by the increasing enumeration (so we will think A = ).For < let =

(xi : i v) Bp . Since a were (forced to be) distinct weknow that Bq |= = for distinct , . Hence / xi : i u

Bp (for each )and therefore we may find functions f0 , f

1 Fp such that f0 u

= f1 u, and

f0 () = 0, f1 () = 1, and if < < , and H, is the isomorphism from p to

p , then f = f H, . Now fix < < and let

gdef=

3+2

f0

3+2


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Look at p(Z) (see Definition 14). There are only possibilities for the values ofp(Z), so we find 0 < 1 < + such that

(i) |Z0 | = |Z1 |, p(Z0) = p(Z1) = , , n, g, h0, . . . , hn1 : < k,(ii) if : Z0 Z1 is the order isomorphism then

Z0 Z1 is theidentity on Z0 Z1 ,

(iii) if : U[p,Z0 ] U[p,Z1 ] is the order isomorphism, then [w0 ] = w1 .

Note that, by Lemma 15, otp(U[p,Z0 ]) = otp(U[p,Z1 ]) and the order isomor-phism satisfies

(j U[p,Z0 ])( Z0)(hp(j,) = hp((j), ())),

and hence is the identity on U[p,Z0] U[p,Z1 ] (remember Lemma 3).For a function f Fp let G01(f) : u

p 2 be defined by

G01 (f)(j) =

f((j)) if j U[p,Z0 ],f(1(j)) if j U[p,Z1 ] \ U[p,0],

0 otherwise.

Claim 17.1. For each f Fp, G01 (f) Fp.

Proof of the claim. By induction on ht(p) we show that for each componentq of p, the restriction G01(f) u

q is in Fq.If is limit, we may easily use the inductive hypothesis to show that, for any

component q of p, G01(f) uq Fq.

Assume = +1 and let q = , , n, u, q, v : < be an componentof p. We will consider four cases.

Case 1: / Z0 Z1 .Then (U[p,Z0 ] U[p,Z1 ]) u

q uq and G01 (f) (uq \ u) 0 for each = .

Since, by the inductive hypothesis, G01(f) uq Fq for each < , we may use

the definition ofP+1 and conclude that G0

1(f) uq Fq (remember the definitionof the term maj).

Case 2: Z0 \ Z1 .Let Z0 = {0, . . . , k1} be the increasing enumeration. Then = for some

< k and = , = , n

= n. Moreover, if v = {j0 , . . . , j

n1

} (theincreasing enumeration), < , then for m < n:

( < )(hm() = hq(jm, )) and ( \ Z0)(h

q(jm, ) ).

Note that U[p,Z1 ] uq uq , so ifU[p,Z0 ] u

q = , then we may proceed as inthe previous case. Therefore we may assume that U[p,Z0 ] u

q = . So, for each Z0 \ we may choose i U[p,Z0 ] u

q such that

(i U[p,Z0 ] u

q

)(h

p

(i, ) = h

p

(i, ) = h

p

(i , ))(remember Lemma 11(1)). Let i = max{i : Z0 \ } (if = max(Z0), thenlet i be any element of U[p,Z0] u

q). Note that then

(i U[p,Z0 ] uq)( Z0 \ )(h

p(i, ) = hp(i, ) = hp(i, ))

[Why? Remember Lemma 11(1) and the clause () of the definition of P+1.] ByLemma 11, we find a (() + 1)component q = , , n, u, q, v

: < of p

such that (i) uq

and

(j uq

)( ((), ht(p)))(hp((i), ) hp(j,) ).


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We claim that then

() (j U[p,Z0 ] uq)((j) uq

U[p,Z1]).

Why? Fix j U[p,Z0 ] uq. Let r, r be components of p such that r pr p,r pr p, ht(r) = + 1, ht(r

) = () + 1 (so r and q, and r, q, are isomorphic).The sets Z0 ( + 1) and Z1 (

() + 1) are pclosed, and they have thesame values of , and therefore U[p,Z0 (+ 1)] and U[p,Z1 (

() + 1)] are

(order) isomorphic. Also, these two sets are included in ur and ur

, respectively.So looking back at our j, we may successively choose j0 u

r U[p,Z0 (+ 1)],

j1 ur U[p,Z1 (

() + 1)], and j uq such that

( )(hq(j,) = hr(j0, )),

( )(hr(j0, ) = hr(j1,

())), and

( ())(hr

(j,) = hq

(j, )).

Then we have

( )(hq(j,) = hq

(j, ()) and ( () \ Z1)(hq

(j, ) ).To conclude () it is enough to show that (j) = j. If this equality fails, thenthere is < ht(p) such that = hp((j), ) = hp(j, ) = . If (), thennecessarily Z1 , and this is impossible (remember h

p(j,) = hp((j), ())

for ). So > (). If hp((j), ) = + 1, then hp(j, ) < and (by thechoice of q) hp((i), ) < . Then Z1 and h

p(i, ()1()) < , and alsohp(i, ()1()) = hp(j, ()1()) = + 1 (by the choice of i), a contradiction.Thus necessarily hp((j), ) < (so Z1) and therefore

> hp(j, ()1()) = hp(i, ()1()) = hp((i), ) = hp(j, )

(as the last is not ), again a contradiction. Thus the statement in () is proven.Now we may finish considering the current case. By the definition of the function

(and by the choice of 0, 1) we have = ,

= , n = n, and [v] = v

for <

(and v is orderpreserving). Therefore

G01(f)((xi : i v)) = f(

(xi : i v)) (for every < ).

By the inductive hypothesis, G01(f) uq Fq (for < ), so as f Fp (and

hence f uq

Fq

) we may conclude now that G01(f) uq Fq.

Case 3: Z1 \ Z0Similar.

Case 3: Z0 Z1If U[p,Z0 ] u

q = = U[p,Z1 ] uq, then G01 (f) u

q 0 and we are easily

done. If one of the intersections is non-empty, then we may follow exactly as in therespective case (2 or 3).

Now we may conclude the proof of the theorem. Since

Bp |= (xi : i w0) < (xi : i w1),

we find f Fp such that f((xi : i w0)) = 0 and f((xi : i w1 )) = 1. Itshould be clear from the definition of the function G01(f) (and the choice of 0, 1)that

G01 (f)((xi : i w0)) = 1 and G01

(f)((xi : i w1)) = 0.


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But it follows from Claim 17.1 that G01(f) Fp, a contradiction.

Conclusion 18. It is consistent that for some uncountable cardinal there is aBoolean algebra B of size (2)+ such that

Depth(B) = but (, (2)+) / DepthSr(B).

Problem 19. Assume < =

Andrzej Roslanowski and Saharon Shelah- Historic Forcing for Depth

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