Von Neumann’s Problem ∗

Boban VelickovicEquipe de Logique, Universite de Paris 7

2 Place Jussieu, 75251 Paris, France

Abstract

A well known problem of Von Neumann asks if every ccc weakly distributive completeBoolean algebra is a measure algebra. A closely related question of Maharam, knownas the Control Measure Problem, asks if every complete Boolean algebra carrying acontinuous submeasure is actually a measure algebra. We survey some recent work onthese two problems and their relationship and present some new results. In particular,by strengthening either one of Von Neumann’s conditions one obtains that the Booleanalgebra is a Maharam algebra, i.e. carries a continuous strictly positive submeasure.

Introduction

We say that a complete Boolean B algebra is a measure algebra if it admits a strictlypositive σ-additive probability measure. Let B be a measure algebra. Then it is easyto see that it satisfies the following.

1. B has the countable chain condition (ccc), i.e. if A ⊆ B is such that a ∧ b = 0,for every a, b ∈ A such that a 6= b, then A is at most countable.

2. B is weakly distributive, i.e. if {bn,k}n,k is a double sequence of elements of B thenthe following weak distributivity law holds:

∧

n

∨

k

bn,k =∨

f :N→N

∧

n

∨

i<f(n)

bn,i

In 1937 Von Neumann asked if these two conditions are sufficient to characterizemeasure algebras (see [Mau]). In her work on Von Neumann’s problem Maharam [Mah]formulated the notion of a continuous submeasure and found an algebraic character-ization for a complete Boolean algebra to carry one. Recall that a submeasure on acomplete Boolean algebra B is a function µ : B → [0, 1] such that

1. µ(a) = 0 if and only if a = 0

2. If a ≤ b then µ(a) ≤ µ(b)

∗2000 Mathematics Subject Classification Primary 03Exx, Secondary 28Axx

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3. µ(a ∨ b) ≤ µ(a) + µ(b)

4. µ(an)→ µ(supn an), whenever {an}n is an increasing sequence.

If 4. is strengthened to

4’. If {an}n is a decreasing sequence and∧n an = 0 then µ(an)→ µ(a).

then the submeasure µ is called continuous.

If a complete Boolean algebra B carries a strictly positive continuous submeasure wecall it a Maharam algebras. Any Maharam algebra is both ccc and weakly distributiveso Von Neumann’s question can be decomposed into two questions.

Question 1 Is every Maharam algebra a measure algebra?

Question 2 Is every ccc weakly distributive complete Boolean algebra a Maharam al-gebra?

A significant amount of work has been done on Question 1 which is known tobe equivalent to the famous Control Measure Problem, i.e. the statement that anycountably additive vector valued measure µ defined on a σ-algebra of sets and takingvalues in an F -space, i.e. a completely metrizable topological vector space, admitsa control measure, i.e. a countable additive scalar measure λ having the same nullsets as µ. The most significant result in this area is a theorem of Kalton and Roberts[KR] who showed that a submeasure µ defined on a (not necessarily complete) Booleanalgebra B is equivalent to a measure if and only if it is uniformly exhaustive. Recallthat a submeasure µ on a Boolean algebra B is called exhaustive if for every sequence{an}n of disjoint elements of B we have limn µ(an) = 0. µ is uniformly exhaustive if forevery ε > 0 there is an integer n such that there is no sequence of n disjoint elementsof B of µ-measure ≥ ε. Clearly, every continuous submeasure on a complete Booleanalgebra is exhaustive. Moreover, if µ is an exhaustive finitely subadditive submeasuredefinied on a not necessarily complete Boolean algebra B the metric completion B ofB is a Maharam algebra. Thus, Question 1 is equivalent to the statement that everyexhaustive submeasure on a Boolean algebra B is uniformly exhaustive. It is easilyseen that if this statement is true for all countable Boolean algebras then it is true forall Boolean algebras. Hence this problem is equivalent to a Π1

2-statement and therefore,by Shoenfield’a Theorem, it is absolute between different models of set theory havingthe same ordinals. For a good survey on the Control Measure Problem see [Ka]).

Concerning Question 2, already Maharam [Mah] observed that a Souslin tree pro-vides a counterexample. It is well known that Souslin trees may or may not existdepending on additional axioms of set theory (see for instance [Ku]). Various otherexamples relatively consistent MA + ¬CH were provided by Glowczynski, the authorand others.

In the positive direction, in late 2003 it was shown by Balcar, Jech and Pazak [BJP]and the author [Ve3] that a positive answer to Question 2 is relatively consistent with

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ZFC, assuming the consistency of a supercompact cardinal. This was achieved by de-riving it from the P -ideal dichotomy, a combinatorial statement which was introducedby Abraham and Todorcevic [AT] and later generalized by Todorcevic [T]. In fact,it was shown in [Ve3] that by slightly strenghtening the countable chain condition weobtain a ZFC result. Say that a complete Boolean algebra B is properly indestructiblyccc if it remains ccc in any generic extension of the universe by a proper forcing notion.Then by Corollary 2 from [Ve3] every properly indestructibly ccc weakly distributivecomplete Boolean algebra B is a Maharam algebra. In fact, all that is needed is thatB remains ccc in any generic extension by a σ-distributive proper forcing notions. Aversion of this result for Souslin posets was also obtained by Farah and Zapletal in[FZ].

Concerning weak distributivity there is a natural two player infinite game G(B) ona complete Boolean algebra B such that B is weakly distributive if and only if playerI does not have a winning strategy in G(B). If, in addition, player II has a winningstrategy in this game we call B strategically weakly distributive. We will show in thispaper that a complete Boolean algebra B is a Maharam algebra if and only if it isccc and strategically weakly distributive. Finally, we mention that it was shown byFarah and the author [FV1] that some large cardinal assumptions are necessary for theconsistency of the postive answer to Question 2.

Coming back to Question 1, in order to gain some insight it is natural to analyzeMaharam algebras by asking if they share some combinatorial properties of measurealgebras. In this direction, it was shown in [Ve3] that every nonatomic Maharamalgebra adds a splitting real and in [FV2] that the product of two nonatomic Maharamalgebras adds a Cohen real. Both of these properties are well known for measurealgebras. However, many natural questions remain open.

The paper is organized as follows. In §1 we present some counterexamples toQuestion 2 under various additional set theoretic assumptions. In §2 we develop somefacts about the sequential topology and present positive results related to Question 1.In §3 we show that Maharam algebras share some combinatorial properties of measurealgebras. Finally, §4 we discuss some open problems and directions for future research.

1 Examples

We start by presenting some examples of ccc weakly distributive complete Booleanalgebras which are not Maharam algebras under additional set theoretic assumptions.

Example 1 A Souslin tree is an ω1 tree with no uncountable chains or antichains. Itis well known that Souslin trees may or may not exist. For instance, the combinatorialprinciple ♦ which holds in the constructible universe L implies that there is a Souslintree. On the other hand, MA + ¬ CH implies that there are no Souslin trees (see[Ku]). The regular open algebra of a Souslin tree is a ccc weakly distributive completeBoolean algebra which is not a Maharam algebra (see [Mah]).

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Example 2 Sacks forcing S consists of all perfect subtrees of 2<ω ordered underinclusion. A key property of this forcing is called the Sacks property.

Definition 1 A forcing notion P has the Sacks property iff for every P-name τ foran element of ωω and for every condition p ∈ P there is a condition q ≤ p and asequence 〈In : n < ω〉 such that In ∈ [ω]2

n, for all n, and q τ ∈ ΠnIn.

Clearly the Sacks property implies weak distributivity. The forcing notion S itselfis not ccc, but it is possible, under suitable assumptions, to extract a ccc suborder of Swhich still has the Sacks property. This was first done by Jensen [Jen] who constructeda ccc suborder of S in order to ’construct’ a nonconstructible Π1

2-singleton. He used ♦and a fusion argument. This construction was later extended by the author [Ve2]. Fora partial ordering P let CCC(P) denotes the following statement.

For every family D of 2ℵ0 dense open subsets of P there is a ccc perfectsuborder Q of P such that D ∩ Q is dense in Q, for every D ∈ D.

Here perfect suborder mean that the incompatibility relation of Q is the restriction ofthe incompatibility relation of P . In [Ve2] we showed the following.

Theorem 1 Let κ > ℵ1 be a regular cardinal such that κ<κ = κ. Then there is ageneric extension in which Martin’s Axiom holds, 2ℵ0 = κ, and CCC(S) holds.

Proposition 1 Assume CCC(S). Then there is a ccc perfect suborder of S which hasthe Sacks property.

PROOF: In order to do this we have to define a family of 2ℵ0 dense open subsets ofS. Suppose A = 〈An : n < ω〉 is a sequence of countable antichains in S. Let DA bethe set of all T ∈ S such that either there is n such that T is incompatible with allmembers of An or for all n there is Xn ⊆ An with |Xn| ≤ 2n such that T ≤ ∨Xn. Bya standard fusion argument one shows that each DA is dense and open in S. Let

D = {DA : A a sequence of countable antichains }.

By applying CCC(S) to D we obtain a ccc perfect suborder Q of S in which the setsDA and are all dense. It follows that Q has the Sacks property.

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Example 3 We now present a construction of Glowczynski from [Gl]. In order toobtain the assumptions of the theorem one needs a measurable cardinal, but the ad-vantage is that the construction is very simple.

Theorem 2 ([Gl]) Assume Martin’s Axiom holds and there is an uncountable car-dinal κ < 2ℵ0 which carries an ℵ1-saturated σ-ideal I. Then P(κ)/I is a ccc weaklydistributive complete Boolean algebra which is not a Maharam algebra.

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Remark. This situation is easy to obtain: start with a measurable cardinal κ andforce, using the standard ccc poset, MA + κ < 2ℵ0 . The dual of the measure on κ inthe ground model generates an ℵ1-saturated σ-complete ideal I.

Proof: Let B = P(κ)/I. Clearly, B is ccc.

Claim 1 B is weakly distributive.

PROOF: Fix a double sequence {an,k}n,k of elements of B. We may assume withoutloss of generality that

∨k an,k = 1, for each n. We can find subsets An,k of κ, for each

n and k, such that an,k = [An,k]I and such that the family {An,k}k is a partition of κ,for each n. For each α < κ, define a function gα ∈ ωω by:

gα(n) = min{k : α ∈ An,k}.Since we have MAκ there is a function g ∈ ωω such that gα ≤∗ f , for each α. Here≤∗ de-notes domination modulo finite sets. Define functions fl ∈ ωω by fl(k) = max{f(k), l}and let

Bl = {α : gα(n) ≤ fl(n), for all n}.It follows that κ =

⋃lBl. Thus, letting bl = [Bl]I , for each l, we have that

∨l bl = 1,

and on the other hand bl ≤∧n

∨k<fl(n) an,k.

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Claim 2 B is not a Maharam algebra.

PROOF: Suppose towards contradiction that B is a Maharam algebra and fix a con-tinuous submeasure µ on B. By lifting µ to κ we obtain a continuous submeasure µon P(κ) which gives value 0 to singletons and value 1 to the whole space. Notice thatwe can replace κ by any set of reals D of size κ. Since we have MAκ, any subset of Dis a Gδ-set. Therefore if Y is a subset of D of µ-submeasure 0 then, for every ε > 0,there is an open set G ⊇ Y such that µ(G) < ε. But now we can repeat the classicalresult that under MAκ any set of size κ of reals is Lebesgue null (see [Ku]).

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Example 4 We now present an example from [FV1] showing that if there is a singularstrong limit cardinal κ such that 2κ = κ+ and �κ holds then there is a completeBoolean algebra B of size κ+ which is not a Maharam algebra, but every subalgebra ofB of size ≤ κ is a measure algebra. The point of this example is that if there is no suchcardinal κ then there is an inner model with a measurable cardinal λ of Mitchell ordero(λ) = λ++ (see [GM]). Therefore the consistency of the positive answer to Question2 would require at least these large cardinal assumptions.

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Theorem 3 Assume κ is a singular strong limit cardinal of uncountable cofinality suchthat 2κ = κ+ and �κ holds. Then there is a complete Boolean algebra B of size κ+

which is not a Maharam algebra, but any subalgebra of B of size at most κ is a measurealgebra.

Proof: The point of these assumptions is that we can obtain a nonreflecting stationaryset S ⊆ κ+ consisting of limit ordinals of cofinality ω such that ♦(S) holds and suchthat there is a �κ-sequence (Cα : α < κ+& lim(α)) such that Cα ∩ S = ∅, for all limitα (see [FV1] for details).

For a set I let AI be the Boolean algebra of clopen subsets of {0, 1}I . If I ⊆ J thereis a natural projection πJI : {0, 1}J → {0, 1}I and we can consider AI as a subalgebraof AJ by identifying a set A ∈ AI with (πJI )−1(A). We shall consider finitely additiveprobability measures on various AI . Note that if µ is such a measure then it naturallyextends to a σ-additive measure µ on the σ-algebra BI of Baire subsets of {0, 1}I (see[Fr1]). We let Iµ denote the σ-ideal of µ-null sets. Given two finitely additive measuresµ and ν on a Boolean algebra A we say µ and ν are equivalent and write µ ∼ ν iffor every sequence (an)n of elements of A we have limµ(an) = 0 iff lim ν(an) = 0. Incase A = AI this means that the measures µ and ν have the same null sets. Finally,if I ⊆ J and µ is a finitely additive measure on AJ let µ � AI be the induced measureon AI .

Fix a nonreflecting stationary set S ⊆ κ+ consisting of ordinals of cofinality ω, a♦(S)-sequence (Dα : α ∈ S) and a �κ-sequence (Cα : α < κ+& lim(α)) such thatCα ∩ S = ∅, for all limit α.

We construct by induction a sequence µα, for α < κ+, such that µα is a finiteprobability measure on Aα such that the following condtions hold:

(1) If α < β then µα ∼ µβ � Aα.

(2) If α = β+1 for some β, then µα is the product of µβ and the uniform probabilitymeasure on {0, 1} at the β-th coordinate.

(3) If α /∈ S is a limit ordinal or α ∈ S and Dα is bounded in α, then µα is theproduct measure of µα0 and the µαξ+1

� A[αξ,αξ+1), where {αξ : ξ < δ} is theincreasing enumeration of Cα.

(4) If α ∈ S and Dα is unbounded in α fix a sequence {αn}n cofinal in α consistingof elements of Dα. Now let νn be the measure on A[αn,αn+1) which is the productof µαn+1 � A(αn,αn+1) and, at coordinate αn, the probability measure ρn on {0, 1}which gives measure 1/2n+1 to {0} and 1− 1/2n+1 to {1}. Finally, let µα be theproduct measure of µα0 and the νn, for n < ω.

It is easy to show by induction that condition (1) holds. The main point is showingthat once the construction is completed we have the following.

Claim 3 There is no finitely additive probability measure µ on Aκ+ such that for allα < κ+, µ � Aα ∼ µα.

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PROOF: Assume otherwise and fix such a µ. For each α let Eα be the set of allf ∈ {0, 1}κ+

such that f(α) = 1. Then Eα ∈ Aκ+, but we also consider it as anelement of AI , for all I such that α ∈ I. Formally, we are identifying Eα and πκ

+

I (Eα).

Claim 4 For every ε > 0 there is δ > 0 such that for every A ∈ Aα if µ(A) > ε thenµ(A \ Eα) > δ.

PROOF: Recall that µα+1 is the product of µα and the uniform probability measureon {0, 1} at the α coordinate. Therefore if A ∈ Aα then

µα+1(A \ Eα) =1

2µα(A).

Since µ � Aα+1 ∼ µα+1 the claim follows.

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Now for a given α < κ+ define the function fα ∈ ωω by letting fα(n) be the leastm such that for every A ∈ Aα if µ(A) ≥ 1/n then µ(A \ Eα) ≥ 1/m. By Claim 4 thisfunction is well defined. Since κ+ > 2ℵ0 there is an unbounded subset X of κ+ and afunction f ∈ ωω such that fα = f , for all α ∈ X.

Claim 5 Suppose {αn}n is a strictly increasing sequence of elements of X. Then

limn→∞

µ(⋂

i<n

Eαi) = 0.

PROOF: Assume otherwise and fix an integer k such that for µ(⋂i<nEαi) > 1/k, for

all n. Let l = f(k). Since fαn = f , for every n, and the sequence {αn}n is strictlyincreasing, it follows that for every n

µ(⋂

i<n+1

Eαi) < µ(⋂

i<n

Eαi)− 1/l,

which is a contradiction.

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Now since (Dα : α ∈ S) is a ♦(S)-sequence there is α ∈ S such that Dα = X ∩ αand this set is unbounded in α. It follow that at stage α we were using case (4) of ourconstruction. Let {αn}n be the chosen sequence of elements of Dα. By the definitionof µα the sets Eα are stochastically independant and µα(Eαn) = 1− 1/2n+1. But thenlimn µα(

⋂i<nEαi) ≥

∏n(1− 1/2n+1) > 0. Therefore µ � Aα � µα, a contradiction.

2

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Now, extend each µα to a σ-additive measure µα on the σ-algebra Bα of Bairesubsets of {0, 1}α. Let Iµα be the σ-ideal of µα-null sets. It follows by our constructionthat if α < β then Iµα = Iµβ � Bα in the obvious sense. Now let I =

⋃α<κ+ Iµα and

C = Bκ+/I. It follow that C is a complete Boolean algebra of size κ+ which is nota measure algebra but any complete subalgebra of C of size at most κ is a measurealgebra. Finally to see that B is not a Maharam algebra note that any Maharamalgebra which is not a measure algebra contains a countably generated subalgebrawhich is not a measure algebra ([Fr1, p. 584]).

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2 Positive results

In her work on Von Neumann’s problem Maharam [Mah] defined the concept of con-vergence in an arbitrary σ-complete Boolean algebra.

Definition 2 Let B be a σ-complete Boolean algebra. We say that an infinite sequence{xn}n of elements of a B converges strongly to x (in symbols xn → x) iff

lim supnxn = lim inf

nxn = x

It is fairly easy to verify.

1. If xn = x, for all n, then xn → x

2. If xn → x, then any subsequence of {xn}n also converges strongly to x.

3. A monotone sequence is strongly convergent

4. xn → 0 iff lim supn xn = 0

5. If xn → x then −xn → −x6. infn xn ∨ infn yn = infn,m xn ∨ ym

We can now define the closure A of a set A of elements of B by: x ∈ A iff there isa sequence {xn}n of elements of A such that xn → x. We now have:

1. {0} = {0}2. A ⊆ A

3. A ∪B = A ∪B

Note that B is not necessarily a topological space since A may not be closed. Inorder to get a topological space we would have to iterate this operation ω1 times.However, Maharam ([Mah], Theorem 2) showed that this is not necessary if B satisfiesthe following condition.

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(I) Given any double sequence {xn,k}n,k which, for each fixed n decreasesmonotonically to 0 as k →∞, there exists a function f : N× N→ N suchthat: ∧

i

∨

n

xf(n,f(i,n)) = 0

Moreover, it is easy to see that if B is weakly distributive and has the countable chaincondition then it satisfies (I). In this case let τs be the induced topology. Note thatthe boolean operations are not in general continuous as binary functions, but they areseparately continuous, i.e. if we fix x then the function y 7→ x ∨ y is continuous in y.Similarly for ∧ and ∆.

Now, if B is a measure algebra with a measure µ this topology is in fact metrizableand the metric d is given by d(x, y) = µ(x∆y), where ∆ is the symmetric difference. Inthis case (B,∆) is a topological group with the sequential topology. Maharam showedthat a necessary and sufficient condition for the existence of a positive continuoussubmeasure on B is that (B, τs) be metrizable in the topology defined above. Moreover,she found an algebraic condition equivalent to the metrizability of (B, τs). We restateher result in modern terminology. Recall that an ideal I of subsets of a set X is calleda P -ideal if for every sequence {An:n < ω} of elements of I there is A ∈ I such thatAn ⊆∗ A, for all n. Here ⊆∗ denotes inclusion modulo finite sets. We say that a setY is orthogonal to a family A provided the intersection of Y with any element of A isfinite. In this case we write Y⊥A and let A⊥ denote the collection of all all subsetsY of X which are orthogonal to A Note that A⊥ is always an ideal. Finally we saythat an ideal I on a set X is countably generated if there is a countable family {An}nof elements of I such that for any A ∈ I there is n such that A ⊆∗ An. Let B bea complete ccc weakly distributive Boolean algebra. Following Quickert [Qu1] let usdefine the following ideal I of countable subsets of B:

X ∈ I if there is a maximal antichain A of elements of B such that everymember of A is compatible with at most finitely many elements of X.

Note that X ∈ I iff for every enumeration {xn}n of X limn xn = 0 in the sequentialtopology. Now we can restate Maharam’s result.

Theorem 4 ( [Mah]) A complete Boolean algebra B carries a continuous submeasureif and only if the dual I⊥ of the Quickert ideal of B is countably generated.

2

Fix a ccc weakly complete Boolean algebra and let I be its Quickert ideal. Note thata set A is in I⊥ iff the closure A of A in the sequential topology does not contain 0.The following two facts are from [Qu2].

Lemma 1 I is a P -ideal.

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Proof: Suppose Xn ∈ I, for all n. Fix for each n a maximal antichain An such thateach element of An is compatible with at most finitely many members of Xn. Now,using the weak distributivity of B we can find a maximal antichain A which weaklyrefines each An, i.e. such that for each a ∈ A and each n there are at most finitelymany members of An compatible with a. Note that each element of A is compatiblewith at most finitely many members of Xn, for each n. Now, fix an enumeration {an}nof A and let X be the union of the sets Xn \ {x ∈ Xn : x ∧∨n

i=0 ai 6= 0}. Then X ∈ Ias witnessed by the antichain A and Xn ⊆∗ X, for all n.

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Lemma 2 There is no uncountable X such that [X]≤ω ⊆ I.

Proof: Suppose X is an uncountable subset of B. Since B satisfies the ccc there isb ∈ B such that every nonzero a ≤ b is compatible with uncountably many members ofX. Then again, using the ccc of B there is a countable A ⊆ X such that every nonzeroa ≤ b is compatible with infinitely many elements of A.

2

The following strengthening of Maharam’s theorem was proved by the author in[Ve3]. A similar result was obtained by Balcar, Jech and Pazak in [BJP].

Theorem 5 Let B be a weakly distributive ccc complete Boolean algebra and let Ibe the Quickert ideal of B. Assume B \ {0} can be covered by countably many setsorthogonal to I. Then B is a Maharam algebra.

Proof: Now, by our assumption we can write B\{0} =⋃nXn, whereXn is orthogonal

to I, for all n. By replacing Xn by its upward closure we may assume that if a ∈ Xn

and a ≤ b then b ∈ Xn. Second, we can replace Xn by its topological closure. Notethat the topological closure of Xn is still upward closed. Finally, by replacing Xn by⋃i≤nXi, we may assume that X0 ⊆ X1 ⊆ X2 ⊆ . . .. Let Un = B \Xn. Let U be the

collection of all open sets which are downward closed and contain 0. Then, each Unbelongs to U , U0 ⊇ U1 ⊇ U2 ⊇ . . . and

⋂{Un:n < ω} = {0}.Our first goal is to improve this sequence in order to have the additional property

that Un+1 ∨ Un+1 ⊆ Un, for every n, where U ∨ V = {u ∨ v:u ∈ U and v ∈ V }. Forthis, it is clearly sufficient to show that for every U ∈ U there is V ∈ U such thatV ∨ V ⊆ U .

For a subset V of B let V (x) = {a ∈ V :x ∨ a ∈ V }. Using the separate continuityof ∨ we see that if V is open then so is V (x), for every x. If V is also downward closedthen so is V (x) and finally if x ∈ V then V (x) is nonempty since it contains 0. So, ifV ∈ U and x ∈ V then V (x) ∈ U .

Lemma 3 For every V ∈ U there is W ∈ U such that W ∨W ⊆ V .

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Proof: Assume otherwise and fix a V ∈ U for which this fails. Let V0 = V ∩ U0.Then, by our assumption, there are x0 and y0 in V0 such that x0 ∨ y0 /∈ V . LetV1 = V0(x0) ∩ V0(y0) ∩ U1. Then V1 ∈ U and therefore there are x1 and y1 in V1 suchthat x1 ∨ y1 /∈ V . Let V2 = V1(x1) ∩ V1(y1) ∩ U2. Then again, V2 ∈ U . We pick x2 andy2 in V2 such that x2 ∨ y2 /∈ V . We proceed by recursion. Given Vn ∈ U we can pickxn and yn in Vn such that xn ∨ yn /∈ V . We then let Vn+1 = Vn(xn) ∩ Vn(yn) ∩ Un+1.Notice that for every n and k we have xn ∨ xn+1 ∨ xn+2 . . . ∨ xn+k ∈ Un.

Claim 6 {xn:n < ω} and {yn:n < ω} belong to I.

Proof: Since the statement is symmetric let us assume, towards contradiction, that{xn:n < ω} is not in I. Fix a nonzero b in B such that every nonzero a ≤ b iscompatible with infinitely many xn. Since B is weakly distributive we can pick astrictly increasing function f in ωω such that

c =∧

n

f(n+1)∨

i=f(n)+1

xi 6= 0.

Let zn =∨f(n)i=f(n)+1 xi. Then, by our construction, we have that zn ∈ Uf(n) and on

the other hand c ≤ zn, for all n. Since the Un are downward closed it follows thatc ∈ ⋂{Ul: l < ω} = {0}, a contradiction.

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Now, since both {xn:n < ω} and {yn:n < ω} are in I it follows that {xn ∨ yn:n < ω}is in I, as well. This means that the sequence {xn ∨ yn}n strongly converges to 0 andV was supposed to be a neighborhood of 0. Therefore, for almost all n, xn ∨ yn ∈ V ,a contradiction.

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Now, using Lemma 3 we can improve the original decreasing sequence U0 ⊇ U1 ⊇ . . .to have in addition that Un+1 ∨ Un+1 ⊆ Un, for all n. At this point we could useMaharam’s theorem, but it is equally easy to define a continuous submeasure directly.First, let us define a function ϕ:B → [0, 1] by:

ϕ(a) = inf{2−n: a ∈ Un}Now we define a submeasure µ:B → [0, 1] as follow

µ(b) = inf{l∑

i=1

ϕ(ai): b ≤l∨

i=1

ai} ∪ {1}

Lemma 4 µ is a positive continuous submeasure on B.

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Proof: It is clear that if a ≤ b then µ(a) ≤ µ(b) and that µ(a ∨ b) ≤ µ(a) + µ(b), forevery a, b ∈ B. We need to show that µ is positive on every nonzero element of B andthat it is continuous. The following fact is immediate.

Fact 1 Suppose n1 < n2 < . . . < nk and ai ∈ Uni+1, for i = 1, . . . k. Then∨ki=1 ai ∈

Un1.

2

From this it follows that if a /∈ Un then µ(a) ≥ 2−n, therefore µ is positive. Finally,to see that µ is continuous notice that by Lemma 1 of [Mah] it suffices to provethat if {xn:n < ω} ∈ I then limn µ(an) = 0. To show this fix an integer k. Since{xn:n < ω} ∈ I and Uk is large, there is n such that xl ∈ Uk for all l ≥ n. This meansthat

µ(xl) ≤ ϕ(xl) ≤ 2−k

for all l ≥ n. Since k was abritrary it follows that limn(an) = 0, as desired. Thisfinishes the proof of Lemma 4 and Theorem 5.

2

For a given uncountable cardinal κ we consider the following statement.

(∗)κ Let X be a set of size at most κ and let I be a P -ideal of countable subsets ofX. Then one of the following two alternatives holds:

(a) there is an uncountable subset Y of X such that [Y ]≤ω ⊆ I(b) we can write X =

⋃nXn, where Xn is orthogonal to I, for each n.

The P -ideal dichotomy is the statement that (∗)κ holds, for all cardinals κ. Thisprinciple, which follows from the Proper Forcing Axiom, was first studied by Abrahamand Todorcevic [AT] who proved that (∗)2ℵ0 is relatively consistent with CH and used itto show that some consequences of Martin’s Axiom and the negation of the ContinuumHypothesis are relatively consistent with CH. Later, Todorcevic [T] extended this resultby showing that the full version of the P -ideal dichotomy is relatively consistent withGCH assuming the existence of a supercompact cardinal. In particular, what is shownin [T] is the following.

Lemma 5 Let I be a P -ideal of countable subsets of a set X. Assume that X is notcovered by countably many sets orthogonal to I but that every subset of X of smallersize is covered by such a collection. Then there is forcing notion P which is < ω1-proper, complete for some simple σ-complete completeness system and such that in V P

there is an uncountable subset Y of X such that [Y ]≤ω ⊆ I.

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Let Σ be a class of forcing notions. Say that a forcing notion P is Σ-indestructiblyccc if P is ccc in V Q, for every Q in Σ. For example, forcing notions which are σ-linked, satisfy the σ-finite chain condition and Souslin forcings are all indestructiblyccc. On the other hand, a Souslin tree T is not indestructibly ccc since forcing with Titself destroys the ccc-ness of T . Now, using Theorem 5, Lemma 2 and Lemma 5 weimmediately have the following.

Corollary 1 Let B be a weakly distributive complete Boolean algebra which is Σ-indestructibly ccc, where Σ is the class of < ω1-proper σ-distributive forcing notions.Then B is a Maharam algebra.

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Similarly, we have the following.

Corollary 2 Assume the P -ideal dichotomy. Then every weakly distributive ccc com-plete Boolean algebra is a Maharam algebra.

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In [Sh1] Shelah showed that every ccc Souslin ∗forcing which is nowhere weaklydistributive adds a Cohen real. On the other hand, it is well known (see [JuSh]) thatthe ccc of Souslin forcing notions is indestructible. Thus we have the following.

Corollary 3 Let S be a nonatomic Souslin ccc forcing notion. Then either there isp ∈ S such that forcing with S below p adds a Cohen real or else the regular openalgebra RO(S) is a Maharam algebra.

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The sequential topology τs on a complete Boolean algebra was studied in detailby B. Balcar, W. Glowczynski, and T. Jech in [BGJ]. In particular they show thata complete Boolean algebra B is a Maharam algebra iff the sequential topology on Bis Hausdorff. In order to prove this result we will need a lemma which is similar toLemma 3. Let as before U denote the family of open sets which are nonempty anddownward closed. The difference between the following Lemma 6 and Lemma 3 is thatnow we do not assume that there is a decreasing family {Un}n of sets which are inU and such that

⋂n Un = {0}. In fact, Lemma 3 is not true without this additional

assumption.

Lemma 6 ([BGJ]) Assume B is ccc and weakly distributive. Then for every U ∈ Uthere is a V ∈ U such that V ∨ V ∨ V ⊆ U ∨ U .

∗A forcing notion (P ≤) is called a Souslin forcing if both P and ≤ are analytic sets of reals.

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Proof: Assume otherwise and fix a U ∈ U for which this fails. For any V ⊆ B andx ∈ B let as before V (x) denote the set of all y ∈ B such that x ∨ y ∈ V . So we havethat if V ∈ U and x ∈ V then V (x) ∈ U .

We define inductively a sequence {Vn}n of sets in U and three sequences {xn}n,{yn}n and {zn}n as follows. Let V0 = U . Then, by our assumption, there are x0, y0

and z0 in V0 such that x0 ∨ y0 ∨ z0 /∈ U ∨ U . Let V1 = V0(x0) ∩ V0(y0) ∩ V0(z0). ThenV1 ∈ U and therefore there are x1, y1 and z1 in V1 such that x1 ∨ y1 ∨ z1 /∈ U ∨ U .Let V2 = V1(x1) ∩ V1(y1) ∩ V1(z1). We proceed by recursion. Given Vn ∈ U we canpick xn, yn and zn in Vn such that xn ∨ yn ∨ zn /∈ U ∨ U . We then let Vn+1 =Vn(xn) ∩ Vn(yn) ∩ Vn(zn). Once we complete the construction let x = lim supn xn,y = lim supn yn and z = lim supn zn. Notice that x∨ y∨ z ≥ lim supn(xn ∨ yn ∨ zn) andsince xn ∨ yn ∨ zn /∈ U ∨ U , for all n, and U ∨ U is downward closed and topologicallyopen it follows that x ∨ y ∨ z /∈ U ∨ U .

On the other hand, notice that by our construction for every l < m < n we have

l−1∨

i=0

xi ∨m−1∨

i=l

yi ∨n−1∨

i=m

zi ∈ U.

Now, first find l such that a = x − ∨l−1i=1 xi ∈ U , then let U1 = U(a) and find m > l

such that b = y −∨m−1i=l yi ∈ U1, and finally let U2 = U1(b) and find n > m such that

c = z −∨n−1m zi ∈ U2. So we have that a ∨ b ∨ c ∈ U . Therefore we have that

x ∨ y ∨ z ≤ (

l−1∨

i=0

xi ∨m−1∨

i=l

yi ∨n−1∨

i=m

zi) ∨ (a ∨ b ∨ c) ∈ U ∨ U.

Since U ∨ U is downward closed it follows that x ∨ y ∨ z ∈ U ∨ U , a contradiction.

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Theorem 6 ([BGJ]) Suppose B is a ccc complete Boolean algebra for which the se-quential topology is Hausdorff. Then B is a Maharam algebra.

Proof: We first prove that B is weakly distributive. So, fix a sequence {An}n ofmaximal antichains in B. Enumerate each An as {an,k}k. Now, given a nonzero x ∈ Bwe need to find y ≤ x which is compatible with at most finitely many elements of An,for each n. Fix disjoint τs-open sets U and V such that 0 ∈ U and c ∈ V . Now,define a decreasing sequence {xn}n of elements of V as follows. Let x0 = x. Given xnconsider the sequence {xn∧

∨ki=0 an,i}k. Since An is a maximal antichain this sequence

converges to xn and since V is open and xn ∈ V there is kn such that xn∧∨kni=0 an,i ∈ V .

Let xn+1 = xn ∧∨kni=0 an,i. Note that the sequence {xn}n is decreasing, so it converges

to y =∧n xn. Since, the xn do not belong to U and U is open it follows that y /∈ U .

So in particular y is not zero and it is clearly compatible with at most finitely manymembers of An, for each n.

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Let, as before, U denote the family of all topologically open downward closednonempty subsets of B. We claim that for every nonzero a ∈ B there is a se-quence {Un}n of members of U such that a − ∨⋂n Un 6= 0. To see this fix a 6= 0.Since τs is Hausdorff we can find U ∈ U such that a /∈ U ∨ U . Using Lemma 6we can pick a decreasing sequence {Un}n of members of U such that U0 = U , andUn+1 ∨ Un+1 ∨ Un+1 ⊆ Un ∨ Un, for every n. It follows that

Un+k ∨ Un+k ∨ . . . ∨ Un+k︸ ︷︷ ︸k+2−times

⊆ Un ∨ Un

We claim that the sequence {Un}n works. Suppose towards contradiction that a ≤∨⋂n Un. Since B is ccc we can find a sequence {bn}n of elements of

⋂n Un such that

a ≤ ∨n bn. Since U1 is a downward closed neighbourhood of 0 there is an n such thata−∨n

i=0 bi ∈ U1. Since bi ∈ Un, for each i, we have that∨ni=0 bi ∈ U1 ∨ U1. Therefore,

a ∈ U1 ∨ U1 ∨ U1 ⊆ U ∨ U , a contradiction.Now, we can choose a maximal antichain A and, for each a ∈ A, a decreasing se-

quence {Uan}n of members of U such that a is incompatible with every element of

⋂n U

an .

Since B is ccc the antichain A is countable and it follows that {U an : n ∈ ω and a ∈ A} is

a countable family of members of U whose intersection is {0}. Therefore, by Theorem5, B is a Maharam algebra.

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We saw in Corollary 1 that by strengthening the ccc to indestructibly ccc we ob-tain a ZFC result. We now show that we can do the same by strengthening weakdistributivity. For this we will need the concept of a weak distributivity game G(B)on a complete Boolean algebra B which is played as follows.

I a0, A0 A1 A2 ...II a1 a2 a3 ...

Player I starts by playing a nonzero a0 ∈ B and a maximal antichain A0 in B.Then Player II chooses a nonzero element a1 ≤ a0 which is contained in the union offinitely many elements of A0. Player I then plays another maximal antichain A1 andPlayer II plays a nonzero a2 ≤ a1 contained in the union of finitely many elements ofA1. The game proceeds in this fashion ω moves. In the end we say that Player I winsiff∧n an = 0; otherwise Player II wins. A winning strategy τ for one of the players is

a rule which tells him what to play at any given position such that if he follows τ hewins the game. The following fact is well known and easy to prove (see e.g. [Jech]).

Proposition 2 Suppose B is a complete ccc Boolean algebra. Then B is weakly dis-tributive iff Player I does not have a winning strategy in G(B).

2

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Clearly, at most one player can have a winning strategy in this game. Note thatif B is a Maharam algebra then Player II has an easy winning strategy. Namely, fixa strictly positive continuous submeasure µ on B. Suppose I plays a0, A0 in the firstmove. Let α = µ(a0). II just makes sure to play elements an such that µ(an) > α/2.At stage n suppose we have an such that µ(an) > α/2 and I plays a maximal antichainAn. Let {xk}k be an enumeration of An. Since µ is continuous there is k such thatµ(a0 ∧

∨ki=0 xi) > α/2. Player II then plays an+1 = an ∧

∨ki=0 xi. In the end, since µ

is continuous we have that µ(∧n an) ≥ α/2, so Player II wins this run of the game.

Now, let us say that a complete Boolean algebra B is strategically weakly distributiveif Player II has a winning strategy in G(B). We now show that by strengthening weakdistributivity to strategic weak distributivity in Question 2 we obtain a positive resultin ZFC.

Theorem 7 Suppose B is a ccc strategically weakly distributive complete Boolean al-gebra. Then B is a Maharam algebra.

Proof: By Theorem 6 it suffices to show that the sequential topology τs on B isHausdorff. We first show that there are two disjoint τs-open sets which separate 1and 0. Let us fix a winning strategy σ for Player II in G(B). We say that a positionp = (a0, A0, a1, A1, . . . , An) is consistent with σ if ak = σ(a0, A0, a1, A1, . . . , Ak−1), foreach k ≥ 1.

Claim 7 There is a position p in G(B) consistent with σ such that for every twopositions q and r of odd length consistent with σ and extending p we have σ(q)∧σ(r) 6=0.

Proof: Otherwise we could build a tree {ps : s ∈ 2<ω} of positions of odd lengthconsistent with σ and such that σ(psb0) ∧ σ(psb1) = 0, for every s ∈ {0, 1}<ω. Ifα ∈ {0, 1}ω then pα =

⋃n pα�n is an infinite run of the game in which Player II follows

σ and therefore wins. Let {aα,n}n be the sequence of moves played by Player II in pα.It follows that aα =

∧n aα,n 6= 0. Since σ(psb0)∧σ(psb1) = 0, for each s, it follows that

{aα : α ∈ {0, 1}ω} is a antichain of size 2ℵ0 in B which contradicts the ccc of B.

2

Now fix a position p = (a0, A0, a1, A1, . . . , An−1, an) as in Claim 7. We may assumeit is Player I’s turn to play. Let U be the set of all a ∈ B such that there exists amaximal antichain A in B such that σ(pA) ≤ a and let V be the set {−a : a ∈ U}.Clearly, 1 ∈ V and 0 ∈ U . Note that by our assumption on p U and V are disjoint.We claim that they are both open. To see this for U suppose {xn}n is a sequenceconverging to 0. We can fix a maximal antichain A such that every element of A iscompatible with at most finitely many of the xn. Let a = σ(pA). Then a ∈ V and−a ∈ U . Since U is closed downward it follows that all but finitely many of the xnbelong to U . Thus, a tail of every sequence converging to 0 is contained in U and so

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0 belongs to the interior of U . A similar argument shows that V contains a tail ofany sequence converging to 1, i.e. 1 is in the interior of V . Given any nonzero a ∈ BPlayer I can start the game by playing a0 = a, so a similar argument gives two disjointopen sets separating a and 0. This implies that τs is Hausdorff and thus by Theorem6 B is a Maharam algebra.

2

3 Properties of Maharam algebras

One natural approach to Question 1 is to isolate some combinatorial properties ofmeasure algebras and prove that they are shared by any Maharam algebra. Onesuch properties is independence. Every nonatomic measure algebra has an infinitestochastically independent family. In general Boolean algebras there is a weaker notionwhich we call a splitting family. Let B be a complete Boolean algebra and say that asubset X of B is splitting if for

∨Y = 1 and

∧Y = 0, for every infinite subset Y of

X. This means that forcing with B adds with Boolean value 1 a splitting real, i.e. aset of integers x which neither contains nor is disjoint from an infinite set of integersfrom the ground model. Our first result from this section is from [Ve3] and says thatevery nonatomic Maharam algebra adds a splitting real.

Lemma 7 Let B be a ccc complete Boolean algebra and let I be the Quickert ideal ofB. Suppose B does not add a splitting real below any condition. Then every infinitesubset X of B has an infinite subset Y such that either

∧Y 6= 0 or Y ∈ I.

Proof: Fix an enumeration X = {bn:n < ω} and let τ be the name for an element of2ω defined by ||τ(n) = 1|| = bn. Since τ is forced not to be a splitting real there is aninfinite I0 ⊆ N and a nonzero c0 such that c0 ”τ � I0 is constant.” We recursively buildan antichain {cξ: ξ < δ} and a decreasing mod finite sequence I0 ⊇∗ . . . ⊇∗ Iξ ⊇∗ . . .such that cξ B ”τ � Iξ is almost constant”, for all ξ. At a countable limit stage λ wefirst diagonalize to find an infinite J such that J ⊆∗ Iξ, for all ξ < λ. If {cξ: ξ < λ}is not already a maximal antichain, by using the fact that τ � J is forced not to be asplitting real, we find cλ incompatible with all the cξ, for ξ < λ, and an infinite Iλ ⊂ Jsuch that cλ B ”τ � J is constant”. Since B is ccc the construction must stop aftercountably many steps. At this stage we get an infinite I such that ”τ � I is almostconstant”. Let Y = {bn:n ∈ I}. If there is c ∈ B \ {0} and an integer n such thatc B τ � (I \ n) ≡ 1, then it follows that c ≤ ∧Y . Otherwise B ”Y ∩ G is finite ”.

2

Theorem 8 Let B be a nonatomic Maharam algebra. Then forcing with B adds asplitting real.

17

Proof: Let µ be a continuous submeasure on B. If µ is uniformly exhaustive, by atheorem of Kalton and Roberts [KR] B is a measure algebra and therefore it adjoinsa random real. Assume now B is not uniformly exhaustive and fix an ε > 0 whichwitnesses this. We can now fix, for each n, a family An = {an,1 . . . , an,n} of pairwisedisjoint sets of µ-submeasure ≥ ε. Note that by Lemma 7 and the continuity of µif X is an infinite set of members of B each of µ-submeasure ≥ ε then there is aninfinite subset Y of X such that

∧Y 6= 0. Fix a family {fξ: ξ < ω1} of functions in∏

n{1, . . . , n} such that for ξ 6= η there is l such that fξ(k) 6= fη(k), for all k ≥ l. Webuild a tower of infinite subsets of N, I0 ⊇∗ I1 ⊇∗ . . . ⊇∗ Iξ ⊇∗ . . ., for ξ < ω1, suchthat

bξ =∧{an,fξ(n):n ∈ Iξ} 6= 0,

for each ξ. At a stage α we do the following. If α is a limit ordinal we first find aninfinite set J such that J ⊆∗ Iξ, for all ξ < α; if α = β + 1 let J = Iβ. Now, look atthe family {an,fα(n):n ∈ J}. By Lemma 7 we can find an infinite Iα ⊆∗ J such thatbα =

∧{an,fα(n):n ∈ Iα} 6= 0. Notice that if ξ 6= η then bξ and bη are incompatible.Therefore {bξ: ξ < ω1} is an uncountable antichain in B, a contradiction.

2

For the following corollary we only need a version of the P -ideal dichotomy for idealson 2ℵ0 whose relative consistency with CH was proved by Abraham and Todorcevic[AT] without any large cardinal assumptions. This follows from the fact that everynonatomic complete ccc Boolean algebra contains a nonatomic complete subalgebra ofsize at most 2ℵ0 .

Corollary 4 Assume ZFC is consistent. Then so is ZFC + CH + ”every nonatomicweakly distributive ccc forcing adds a splitting real”.

2

4 Open Questions

In this section we list some open problems in this area. One natural approach toQuestion 1 is to investigate combinatorial properties of Maharam algebras and showthat they have to resemble measure algebras.

Question 3 Suppose B is a Maharam algebra of uniform density κ. Does there existX ⊆ B of size κ such that

∧Y = 0 and

∨Y = 1, for every infinite Y ⊆ X?

Question 4 Can forcing with a Maharam algebra add a minimal real, i.e. a real rsuch that for every real s ∈ V [r] either s ∈ V or V [s] = V [r]?

Question 5 Suppose a ccc forcing P does not add splitting reals. Is P necessarilyweakly distributive?

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Question 6 Is it relatively consistent with ZFC that every nonatomic ccc forcing addsa Cohen or a random real?

Recall that a Boolean algebra B has the σ-finite chain condition if it can be writtenas B \ {0} =

⋃nXn such that Xn does not contain infinite antichains, for each n. We

say that B has the σ-bounded chain condition if there is such a partition such thatfor each n there is an integer kn such that Xn does not contain any antichain of sizeat least kn. Clearly, every Maharam algebra has the σ-finite chain condition; we cansimply take Xn = {a ∈ B : µ(a) ≥ 1/n}, where µ is a continuous strictly positivesubmeasure on B.

Question 7 Is there a Boolean algebra which has the σ-finite chain condition, but notthe σ-bounded chain condition? In particular, does every Maharam algebra have theσ-bounded chain condition?

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