Pacific Journal of Mathematics

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1 Pacific Journal of Mathematics COMPLETE EMBEDDINGS OF THE COHEN ALGEBRA INTO THREE FAMILIES OF C.C.C. NON-MEASURABLE BOOLEAN ALGEBRAS Natasha Dobrinen Volume 214 No. 2 April 2004

2 PACIFIC JOURNAL OF MATHEMATICS Vol. 214 No COMPLETE EMBEDDINGS OF THE COHEN ALGEBRA INTO THREE FAMILIES OF C.C.C. NON-MEASURABLE BOOLEAN ALGEBRAS Natasha Dobrinen The Cohen algebra embeds as a complete subalgebra into three classic families of complete atomless c.c.c. non-measurable Boolean algebras; namely the families of Argyros algebras and Galvin-Hajnal algebras and the atomless part of each Gaifman algebra. It immediately follows that the weak (ω ω-distributive law fails everywhere in each of these Boolean algebras. 1. Introduction. Von Neumann conjectured that the countable chain condition and the weak (ω ω-distributive law characterize measurable algebras among Boolean σ- algebras [Mau]. Consistent counter-examples have been obtained by Maharam [Mah] Jensen [J] Glówczyński [Gl] and Veličković [V]. However whether von Neumann s proposed characterization of measurable algebras fails within ZFC remains an open problem. In searching for possible counter-examples to von Neumann s proposed characterization of measurable algebras within ZFC we investigated three families of complete c.c.c. non-measurable Boolean algebras namely the Argyros Galvin-Hajnal and Gaifman algebras to find out whether these Boolean algebras sustain any weak form of distributivity. By the Cohen algebra we mean the completion of any countable atomless Boolean algebra. By the κ-cohen algebra we mean the completion of the free Boolean algebra on κ-many generators. We found the following: Theorem It is possible to construct both atomless Gaifman algebras and Gaifman algebras with atoms. 2 The cf (2 ω -Cohen algebra embeds as a complete subalgebra into each Galvin-Hajnal algebra. 3 The Cohen algebra embeds as a complete subalgebra into each Argyros algebra and related variants and into the atomless part of each Gaifman algebra. We show 1 in 4 2 in 2 and 3 in 3 and

3 224 NATASHA DOBRINEN Throughout this paper let B denote a Boolean algebra and P denote a partial ordering. Let B + denote B\{0}. Galvin and Hajnal and Argyros constructed families of partial orderings and Gaifman constructed a family of Boolean algebras to establish strict implications between various chain conditions including the c.c.c. the σ-bounded c.c. and CUP(B +. Recall the following definitions: Definition 1.2 ([Ko]. P satisfies the countable chain condition (c.c.c. if for each pairwise incompatible subset X P X ω. Definition 1.3 ([F]. P satisfies the σ-bounded chain condition (σ-bounded c.c. if there exist subsets X n P n < ω such that P = n<ω X n where n < ω each pairwise incompatible subset of X n has cardinality n + 1. Definition 1.4 ([GP]. Let S P be a nonempty set. For s 0... s n S not necessarily distinct let (1.1 α (s 0... s n = 1 n + 1 max{ I : I n + 1 p P i I s i p}. The intersection number of S is (1.2 α(s = inf{α (s 0... s n : n ω s 0... s n S}. We say that CUP(P holds if P is a countable union of subsets of P each of which has positive intersection number; i.e. there exist subsets X n P n < ω such that P = n<ω X n and n < ω α(x n > 0. The above three notions are defined for a Boolean algebra B by replacing P with B + in Definitions Note that for a partial ordering P the above three notions are preserved under completions. That is P satisfies the c.c.c. (σ-bounded c.c. iff r.o.(p satisfies the c.c.c. (σ-bounded c.c. respectively and CUP(P holds iff CUP(r.o.(P + holds. It is easy to see that CUP(P implies the σ-bounded c.c. which in turn implies the c.c.c. Gaifman showed that CUP(P is strictly stronger than the σ-bounded c.c. [Ga]. Galvin and Hajnal showed that the σ-bounded c.c. is strictly stronger than the c.c.c. [CN]. Definition 1.5 ([Ko]. B satisfies the weak (ω ω-distributive law (weak (ω ω-d.l. if for each family (b ij i<ωj<ω of elements of B b ij = (1.3 b ij i<ω j<ω f:ω ω i<ω j f(i provided that j<ω b ij for each i < ω i<ω j<ω b ij and i<ω j f(i b ij for each f : ω ω exist in B. We say that the weak (ω ω-d.l. fails everywhere in B if there exist (b ij i<ωj<ω B such that i<ω j<ω b ij = 1 and f:ω ω i<ω j f(i b ij = 0.

4 COMPLETE EMBEDDINGS OF THE COHEN ALGEBRA 225 Definition 1.6 ([Ko]. A Boolean σ-algebra B is measurable if there exists a function µ : B [0 which is strictly positive (µ(b = 0 b = 0 and σ-additive (µ( i<ω b i = Σ i<ω µ(b i for all pairwise disjoint {b i : i < ω} B. Theorem 1.7 ([Ko]. Every measurable Boolean algebra B satisfies the weak (ω ω-d.l. and the c.c.c. Theorem 1.8 (Kelley [Ke]. CUP(B + holds iff B carries a strictly positive finitely-additive measure µ : B [0. The following Theorem 1.9 of Kelley completely characterizes measurable algebras among Boolean σ-algebras in ZFC by strengthening the c.c.c. to CUP(B +. Theorem 1.9 (Kelley [Ke]. If B is a σ-algebra then B is measurable iff B satisfies the weak (ω ω-d.l. and CUP(B + holds. Theorem 1.10 (Folklore. The weak (ω ω-d.l. fails everywhere in the Cohen algebra. Each of our complete embeddings of the Cohen algebra will involve the following notions and lemmas: Definition 1.11 ([Ko]. A subalgebra A of a Boolean algebra B is a regular subalgebra of B if for each M A such that A M exists in A B M exists in B and A M = B M. Lemma Let B be a Boolean algebra P a dense subset of B + and A a subalgebra of B. A is a regular subalgebra of B iff p P a p A + such that whenever a A and a p a 0 then p a 0. Definition 1.13 ([Ko]. A subalgebra A of a Boolean algebra B is a complete subalgebra of B if for each subset M A such that B M exists A M exists and B M = A M. Definition 1.14 ([Ko]. A monomorphism f : A B is complete if for each M A for which A M exists f( A M = B {f(b : b M}. The following lemma is a natural consequence of the Sikorski Extension Theorem: Lemma 1.15 ([Ko]. If B is a complete Boolean algebra and A is a regular subalgebra of B then there is a complete monomorphism from r.o.(a + into B. Throughout this paper let P GH P A and B G denote members of the families of Galvin-Hajnal partial orderings Argyros partial orderings and Gaifman algebras respectively. Galvin and Hajnal Argyros and Gaifman

5 226 NATASHA DOBRINEN showed that in r.o.(p GH r.o.(p A and B G respectively CUP(B + fails; thus by Kelley s Theorem 1.9 these three Boolean algebras are not measurable. However this is not the only reason measurability fails in these algebras. By Theorem 1.10 completely embedding the Cohen algebra into r.o.(p GH r.o.(p A and the atomless part of r.o.(b G + shows that no weak form of distributivity holds in these Boolean algebras. Remark We thank the referee for pointing out the following: By a result of Shelah in c.c.c. Suslin forcings adding a Cohen real is equivalent to the weak (ω ω-d.l. failing everywhere [S]. Hence to prove 3 of Theorem 1.1 it would suffice to show that the weak (ω ω-d.l. fails everywhere in the Argyros and Gaifman algebras. However Shelah s result does not apply to the Galvin-Hajnal algebra since P GH is not Suslin. For κ = cf (2 ω in the case of Galvin-Hajnal algebras and for κ = ω in the case of Argyros algebras and the atomless part of Gaifman algebras the method we employ for completely embedding the κ-cohen algebra is the following: Choose κ many independent elements {c i : i < κ} B in such a way that the subalgebra C κ generated by {c i : i < κ} satisfies the conditions of Lemma Then C κ is isomorphic to the free Boolean algebra on κ-many generators and is a regular subalgebra of B. Since B is complete Lemma 1.15 implies that the completion of C κ the κ-cohen algebra embeds into B as a complete subalgebra. 2. A complete embedding of the Cohen algebra into the Galvin-Hajnal algebra. Galvin and Hajnal constructed a separative atomless partial ordering P GH which satisfies the c.c.c. but not the σ-bounded c.c. [CN]. To do this they used the following family of sets: Recall that for a set S [S] 2 = {{β γ} : β γ S and β γ} the collection of all two-element subsets of S. In this section we fix a well-ordering on 2 ω. Lemma 2.1 (Galvin-Hajnal [CN]. There is a family of sets {S α : α < 2 ω } with the following four properties: (S1 α < 2 ω S α α; (S2 α < 2 ω [S α ] 2 γ<2 ω{{β γ} : β S γ}; (S3 α < 2 ω o.t.(s α ω; (S4 If S 2 ω [S] 2 γ<2 ω{{β γ} : β S γ} and o.t.(s ω then α < 2 ω such that S = S α. The following properties of the collection of sets {S α : α < 2 ω } will be used extensively:

6 COMPLETE EMBEDDINGS OF THE COHEN ALGEBRA 227 Lemma 2.2. Suppose η ζ are ordinals with η ζ < 2 ω and S η = ω. Then there exist ζ < α < β < 2 ω such that S α S β = S η. Proof. Let X = {S α : α ζ S α S η } {S η \S α : α ζ}. X < 2 ω so choose some S S η such that S X and S η \S X. [S] 2 [S η \S] 2 [S η ] 2 α<2 ω{{β α} : β S α} by (S2. (S4 implies α β < 2 ω such that S α = S and S β = S η \S. α β > ζ since S X. Lemma 2.3. Given α 0 < 2 ω there is a sequence α 0 < α 1 < α 2 < < λ < 2 ω of order type ω + 1 such that for each i < ω (2.1 S αi+1 = {α 0... α i }; S λ = {α i : 0 < i < ω}. Proof. Let α 0 < 2 ω. [{α 0 }] 2 = so α 1 < 2 ω for which S α1 = {α 0 } by (S4. (S1 implies α 1 > α 0. Given α 0 < < α n where for each 0 < j n S αj = {α 0... α j 1 } the set [{α 0... α n }] 2 α<2 ω{{β α} : β S α}. By (S1 and (S4 there is some α n+1 > α n such that S αn+1 = {α 0... α n }. By our choice of the α n [{α 1 α 2... }] 2 α<2 ω{{β α} : β S α}. By (S1 and (S4 there is a λ < 2 ω such that S λ = {α i : 0 < i < ω} and λ > α j for all j < ω. Galvin and Hajnal constructed the following partial ordering P GH : α < 2 ω let (2.2 V α = {f : 2 ω 2 : β S α (f(β = 0 and f(α = 1}. That is V α is the collection of all functions from 2 ω into 2 which send each element of S α to 0 and send α to 1. Let { P GH = V α : F [2 ω ] <ω and } (2.3 V α. α F P GH is the collection of nonempty intersections of finitely many of the V α s partially ordered by inclusion. Theorem 2.4 (Galvin-Hajnal [CN]. P GH is a separative atomless partial ordering which satisfies the c.c.c. but not the σ-bounded c.c. Let e : P GH r.o.(p GH denote the canonical embedding of P GH into its completion r.o.(p GH. We shall call r.o.(p GH the Galvin-Hajnal algebra. Since the σ-bounded c.c. fails in P GH CUP(r.o.(P GH + fails. By Kelley s Theorem 1.9 r.o.(p GH is not measurable. Theorem 2.5. The cf (2 ω -Cohen algebra embeds as a complete subalgebra into r.o.(p GH. α F

7 228 NATASHA DOBRINEN Proof. Let κ denote cf(2 ω. We construct a subalgebra C κ of r.o.(p GH and show that C κ is isomorphic to the free Boolean algebra on κ-many generators and is a regular subalgebra of r.o.(p GH. Construction of C κ : By Lemma 2.3 there exist κ-many sequences each of order type ω + 1 of the form α(i 0 < α(i 1 < α(i 2 < < α(i j < α(i j + 1 < < λ(i i < κ such that the following hold: i < κ j < j < ω α(i j < α(i j < λ(i; i < i < κ λ(i < α(i 0; and i < κ 0 < j < ω (2.4 S α(ij = {α(i k : k < j}; S λ(i = {α(i j : 0 < j < ω}. Note that the sets {α(i j : j < ω} {λ(i} i < κ are pairwise disjoint. For each i < κ let (2.5 c i = e(v α(ij in r.o.(p GH. Let (2.6 0<j<ω C κ = c i : i < κ the subalgebra of r.o.(p GH generated by {c i : i < κ}. The elements V λ(i i < κ will be used in Proposition 2.11 to ensure that the generators of C κ are independent. The following simple facts will be useful: Fact 2.6. p q P GH e(p e(q = e(p q. Fact 2.7. For each finite F 2 ω α F V α iff ( α F S α F =. Fact 2.8. If {p i : i < ω} P GH is infinite then i<ω e(p i =. If {p i : i < ω} is infinite and q i<ω e(p i then every f q must take infinitely many elements of 2 ω to 1. There are no such q P GH. Fact 2.9. Given i < κ if F is a finite subset of 2 ω and α F S α S λ(i then e( α F V α c i. In particular i < κ e(v λ(i c i. Suppose α F S α S λ(i. Then p e( α F V α f p 0 < j < ω f(α(i j = 0. However q c i f q and 0 < j < ω such that f(α(i j = 1. So e( α F V α c i =. Fact i < κ c i = 0<j<ω e(v α(ij. Proof. Clearly c i 0<j<ω e(v α(ij. Suppose p 0<j<ω e(v α(ij. Let F be the finite subset of 2 ω such that p = α F V α. 0 < j < ω p V α(ij so S λ(i F =. By Lemma 2.2 γ < β < 2 ω with γ > sup(f {λ(i} such that S γ S β = S λ(i. By Fact 2.7 V γ V β p. By Fact 2.9 e(v γ V β p c i so e(p c i. Since c i is an open subset of P GH e(p c i implies p c i. Thus c i 0<j<ω e(v α(ij.

8 COMPLETE EMBEDDINGS OF THE COHEN ALGEBRA 229 Next we show that the generators of C κ are independent. Proposition For finite sets I J κ i I c i j J c j I J. = 0 iff Proof. Let I J κ be disjoint finite sets and let r = i I V α(i1 j J V λ(j. Fact 2.7 implies r. By Fact 2.9 i I c i j J c j i I e(v α(i1 j J e(v λ(j = e(r > 0. Since C κ is generated by cf(2 ω -many independent elements C κ is isomorphic to the free Boolean algebra on cf(2 ω -many generators. The next proposition will aid us in constructing elements c p satisfying the conditions of Lemma Proposition Given p P GH there are at most finitely many i < κ for which either e(p c i = 0 or e(p c i = 0. Proof. Suppose p P GH and let J κ be defined by j J e(p c j = 0. j J (2.7 0 = e(p c j = e(p ( 0<k<ω e(v α(jk = 0<k<ω ( e(p e(v α(jk the second equality following from Fact Thus j J 0 < k < ω p V α(jk =. Let F be the finite subset of 2 ω such that p = α F V α. The following can be shown by an easy induction argument using Fact 2.7: If e(p c j = 0 then either α(j 0 F or S λ(j α F S α. α(j 0 F for at most finitely many j < κ since F is finite. By (S3 o.t.( α F S α < ω ω so S λ(j α F S α for at most finitely many j < κ. Hence e(p c j = 0 for at most finitely many j < κ. Thus J < ω. Let I κ be defined by i I e(p c i = 0. Then e(p c i = (2.8 e(v α(ik = e(v α(ig(i i I i I i I 0<k<ω g:i ω\{0} where the first equality follows from Fact If I is infinite then Fact 2.8 implies that g : I ω\{0} i I e(v α(ig(i = contradicting p. Thus I < ω. Now we use Proposition 2.12 to show that C κ satisfies the conditions of Lemma Proposition p P GH c p C κ + such that c C κ if c p c 0 then e(p c 0.

9 230 NATASHA DOBRINEN Proof. Fix p P GH and let F be the finite subset of 2 ω such that p = α F V α. Let I J κ be the finite disjoint sets of Proposition Let c p = ( i I c i ( j J c j. c p C + κ by Proposition (c p is actually the minimal cover for e(p in C. Suppose c = ( k K c k ( l L c l C κ and c p c 0. Then I L = J K =. J K = implies k K 0 < m k < ω such that p V α(kmk. By Fact 2.7 (2.9 ( ( F {α(k m k : k K} α F S α {α(k m : k K m < m k } =. I L = implies l L e(p c l. Thus l L 0 < n < ω e(p e(v α(ln which implies α(l n F. Therefore (2.10 F ( S λ(l =. l L Using Lemma 2.2 l L choose β l α l > sup(f {λ(n : n K L} such that S αl S βl = S λ(l. Let (2.11 r = (V αl V βl. k K V α(kmk l L r by Fact 2.7. e(r c by Fact 2.9. Furthermore r p by Fact 2.7 (2.9 (2.10 and the fact that l L α l β l are larger than sup(f {λ(n : n K L}. Hence 0 < e(r p c so e(p c 0. It follows from Lemma 1.12 and Proposition 2.13 that C κ is a regular subalgebra of r.o.(p GH. Thus by Lemma 1.15 r.o.(c κ + the κ-cohen algebra embeds into r.o.(p GH as a complete subalgebra. That is P GH adds cf(2 ω -many side-by-side Cohen reals. 3. A complete embedding of the Cohen algebra into the Argyros algebra. Agyros constructed a separative atomless partial ordering P A in which the σ-bounded c.c. holds but CUP(P A fails and assuming CH property K 3 also fails [A]. He constructed P A using three basic types of elements. In this section let 2 ω denote the set of functions from ω to 2. For X Y [ω] <ω let (3.1 B X = {f 2 ω : x X f(x = 1} B Y = {f 2 ω : y Y f(y = 0}. For the third type of element Argyros constructed a tree T [ω] 2 as follows: Let {S nm : n < ω 1 m 3 n } be a family of sets such that n m < ω

10 COMPLETE EMBEDDINGS OF THE COHEN ALGEBRA 231 S nm [ω] 3 and S nm S n m = whenever n m n m. For each n < ω let Lev(n = 1 m 3 n[s nm] 2. For each n < ω index the elements of Lev(n as s nj 1 j 3 n+1. The partial ordering on T is defined at level n + 1 as follows: For s = s nj Lev(n and t Lev(n + 1 s t t [S n+1j ] 2. Let T = n<ω Lev(n. T is the Argyros tree. For s = {k l} T let (3.2 K s = (B {k} B {l} (B {l} B {k}. K s is the set of all functions in 2 ω which are nonconstant on s. Let Br be the set of all branches (finite and infinite of T. For σ Br let (3.3 A σ = s σ K s. A σ is the set of all functions in 2 ω which are nonconstant on every node s σ. Let { (3.4 P A = B X B Y σ Σ A σ : X Y [ω] <ω and Σ [Br] <ω } { } the collection of all nonempty intersections of finitely many elements of the three forms B X B Y and A σ partially ordered by inclusion. We call P A the Argyros partial ordering. Theorem 3.1 (Argyros [A]. P A is a separative atomless partial ordering which satisfies the σ-bounded c.c. but not CUP (P A and assuming CH does not satisfy property K 3. Let e : P A r.o.(p A be the canonical embedding of P A into its completion. We shall call r.o.(p A the Argyros algebra. By Argyros Theorem 3.1 r.o.(p A is a complete atomless Boolean algebra which satisfies the σ-bounded c.c. but not CUP(r.o.(P A + and hence by Kelley s Theorem 1.9 is not measurable. Definition 3.2. We will say that s t T are siblings if s t and v T such that s and t are both immediate successors of v. s t u T are triplets if they are pairwise siblings. Note: If s and t are siblings then there exist unique m n < ω such that s t [S mn ] 2 and K s K t. If s t u are triplets then s = t u (the set-theoretic difference of t and u in ω and K s K t K u =. Remark 3.3. The elements of P A are not uniquely represented by the form B X B Y σ Σ A σ. For instance if s = {k l} σ Br then B {k} A σ = B {k} B {l} A σ = B {l} A σ. We shall hold to the following convention: Given S T and X Y [ω] <ω the representation B X B Y s S K s of a subset of 2 ω is said to be in the normal form if and only if X Y =

11 2 NATASHA DOBRINEN s S(s (X Y = and S contains no triplets. If B X B Y s S K s is the normal form representation of some B U B V σ Σ A σ P A then X U Y V X Y = and S {σ : σ Σ}. It is not hard to see that for each element p P A there is a unique normal form representation of p. The normal form is not necessary for the proceeding proofs but rather serves to simplify notation. Lemma 3.4. Let p = B X B Y s S K s be a subset of 2 ω (not necessarily in the normal form and not necessarily in P A. Then p iff the following four conditions hold: (L1 p X Y = ; (L2 p s S s X and s Y ; (L3 p S does not contain any triplets; (L4 p If s t S are siblings then either (s t X = or (s t Y =. In particular if p = B X B Y s S K s is in the normal form then p. Proof. The forward direction is trivial. Assume (L1 p-(l4 p hold. We show there is a partial function f p with domain X Y S such that every total extension of f p is in B X B Y s S K s. By (L1 p B X B Y. By (L3 p we can divide S into two disjoint sets: Γ = {s S : s has no sibling in S} and Θ = {s S : s has exactly one sibling in S}. s Γ by (L2 p s X and s Y. Hence there is a partial function f s such that dom(f s = X Y s f s X 1 f s Y 0 and f s s is nonconstant. Every extension of f s to a total function is in B X B Y K s. For each pair of siblings t u Θ there is a partial function f tu such that dom(f tu = X Y t u f tu X 1 f tu Y 0 and every extension of f tu to a total function is in B X B Y K t K u. To define such an f tu on t u while preserving f tu X 1 and f tu Y 0 we consider 3 cases. If (t u X then (t u X = by (L2 p and (t u Y = by (L4 p; so let f tu (t u 1 and f tu (t u 0. If (t u Y then by (L2 p (t u Y = ; so let f tu (t u 1 and f tu (t u 0. If (t u X and (t u Y = let f tu (t u 0 and f tu (t u 1. Since two elements of S have nonempty intersection only if they are siblings the partial function f p = {f s : s Γ} {f tu : t u are siblings in Θ} is well-defined. Every total extension of f p is in B X B Y s S K s. Theorem 3.5. The Cohen algebra embeds into r.o.(p A as a complete subalgebra. Proof. We construct a countable atomless regular subalgebra C of r.o.(p A.

12 COMPLETE EMBEDDINGS OF THE COHEN ALGEBRA 233 Construction of C. Choose an infinite branch of T and call it β 0. n < ω let t n be the unique element in β 0 Lev(n and choose one s n+1 T \β 0 such that s n+1 and t n+1 are siblings. 0 < n < ω let β n be an infinite branch in T which contains s n. For 0 < m < n β m β n = {t 0... t m 1 }. Define the following sets: { T C = t T : t β n or t is a sibling of some s } (3.5 β n (3.6 n<ω N C = T C = {k < ω : l < ω such that {k l} T C }. T C T and N C ω. Let Br C denote the set of all branches (finite and infinite of T C. Br C is countable since T C has only countably many infinite branches. Let (3.7 { C = B X B Y } A σ P A : X Y [N C ] <ω and Σ [Br C ] <ω. σ Σ Let (3.8 C = {e(p : p C} the subalgebra of r.o.(p A generated by the set {e(p : p C}. Note that C = ω since N C and Br C are countable. Remark 3.6. The idea behind the choice of T C and N C is as follows: T C was chosen so that for any finite set of branches Σ Br C σ Σ A σ. However the subalgebra generated by {e( σ Σ A σ : Σ [Br C ] <ω } is not a regular subalgebra of r.o.(p A. To enlarge it to a regular subalgebra we chose N C so that we can tell exactly how elements of the form σ Σ A σ Σ [Br C ] <ω interact with members of P A. This allows us to find for each p P A a minimal cover for e(p in C and thus ensure that C is a regular subalgebra of r.o.(p A. The following are two simple facts which we shall use without mention in subsequent proofs: Fact 3.7. x ω e(b {x} = e(b {x}. Fact 3.8. p q P A e(p e(q = e(p q. Proposition 3.9. {e(p : p C} is dense in C +. Proof. Every element of C is a finite disjunction of elements of the form ( e B X B Y (3.9 A σ e(a γ σ Σ γ Γ n<ω

13 234 NATASHA DOBRINEN where X Y [N C ] <ω and Σ Γ [Br C ] <ω. Let c C + be of the form (3.9 and let p P A be such that e(p c. For each γ Γ e(p e(a γ = 0 = p s γ (B s B s. Fix f p. For each γ Γ choose an s γ Γ for which f (B sγ B sγ. Define Γ = {γ Γ : f B sγ } and Γ = {γ Γ : f B sγ }. Let (3.10 q = B X ( S {sγ:γ Γ } B Y ( S {sγ:γ Γ } σ Σ A σ. f q so q. By its construction q C. Furthermore γ Γ e(b sγ e(a γ ; γ Γ e(b sγ e(a γ ; and q B X B Y σ Σ A σ. Hence e(q c. Proposition C is atomless. Proof. It suffices to show that {e(p : p C} is atomless. Suppose p C and B X B Y s S K s is the normal form of p. Choose some z N C \(X Y such that z is neither in any member of S nor in any sibling of any member of S. Then B X {z} B Y s S K s is the normal form of p B {z} ; so by Lemma 3.4 p B {z}. Hence p B {z} C. To see that e(p B {z} < e(p note that e(b {z} e(p = e(b X B Y {z} s S K s. This is strictly greater than 0 by Lemma 3.4 since B X B Y {z} s S K s is in the normal form. We now show that C satisfies the conditions of Lemma Proposition p P A c p C + such that c C if c p c 0 then e(p c 0. Proof. Let p = B X B Y s S K s P A. p so by Lemma 3.4 (L1 p-(l4 p hold. Let X = X N C Y = Y N C S = S T C and q = B X B Y s S K s. q p and q C since S is a finite union of branches in T C. Let c p = e(q. c p C + and c p e(p > 0. (c p is actually the minimal cover for e(p in C. It suffices to show that r C if c p e(r > 0 then e(p e(r > 0. Suppose r = B U B V w W K w C and c p e(r > 0. Then (3.11 q r = B (X U B (Y V K t. Thus Lemma 3.4 implies t S W (L1 q r (X U (Y V = ; (L2 q r s S W s X U and s Y V ; (L3 q r S W has no triplets; (L4 q r s t are siblings in S W = ((s t (X U = or (s t (Y V =.

14 COMPLETE EMBEDDINGS OF THE COHEN ALGEBRA 235 It suffices to show that p r. First note that (3.12 p r = B ((X\X X U B ((Y \Y Y V s (S\S S W where (X\X N C = (Y \Y N C = ; X U Y V N C ; (S\S T C = ; and S W T C. Claim. (L1 p r-(l4 p r hold. (L1 p r and (L2 p r follow naturally from (L1 p (L1 q r (L2 p and (L2 q r. S W and S\S each contain no triplets by (L3 q r and (L3 p respectively. S\S T \T C and S W T C imply S\S and S W have no common siblings. Hence S W contains no triplets so (L3 p r holds. Suppose s t are siblings in S W. Then either s t S\S or else s t S W. Suppose s t S\S. Then (s t N C = so (s t (X Y U V =. Further (L4 p implies either (s t X = or else (s t Y =. On the other hand if s t S W then s t N C so (s t ((X\X (Y \Y =. By (L4 q r either (s t (X U = or else (s t (Y V =. Thus in both cases either (s t (X U = or else (s t (Y V =. Hence (L4 p r holds. By Lemma 3.4 p r. Thus e(p e(r > 0. By Proposition 3.11 and Lemmas 1.12 and 1.15 r.o.(c + is a complete subalgebra of r.o.(p A. This completes our construction of a complete embedding of the Cohen algebra into the Argyros algebra. Remark We are very grateful to the referee for suggesting that we investigate Argyros other variants of this example in which stronger chain conditions hold to see whether the Cohen algebra completely embeds in these. We have found that it does. Theorem 3.13 (Argyros [CN] p For each 2 m < ω there is a family of atomless separative partial orderings P m such that each P m satisfies the σ-bounded c.c. and property K m CUP(P m fails and assuming CH property K m+1 fails. Our preceding construction can be easily modified to completely embed the Cohen algebra into each r.o.(p m. Theorem For each 2 m < ω the Cohen algebra embeds as a complete subalgebra into r.o.(p m. The modification is as follows: Let m be given. Argyros constructed a family of trees T such that each T is an (m + 1-branching tree of height ω with the following properties: For each node σ T there is a finite subset dom(σ ω such that σ is a set of functions from dom(σ to 2. All K s

15 236 NATASHA DOBRINEN siblings have the same domain and any two nodes which are not siblings have disjoint domains. Any m siblings have nonempty intersection but the intersection of (m + 1-many siblings is empty. (See [CN] p. 156 for the precise definition of T m. For σ T let A σ = {f 2 ω : f dom(σ σ}. P m = {B X B Y σ Σ A σ : X Y [ω] <ω and Σ is a finite set of branches in T }\{ } partially ordered by inclusion. To construct a countable atomless regular subalgebra of r.o.(p m take as before ω-many infinite branches β i in T such that for each node σ i<ω β i σ has at most one sibling in i<ω β i. Let T C be the subtree of T consisting of i<ω β i and all siblings of nodes in i<ω β i. Let N C = {dom(σ : σ TC }. Let C be the subalgebra of r.o.(p m generated by the elements of the form B X B Y σ Σ A σ such that X Y are finite subsets of N C and Σ is a finite set of branches in T C. The proof that this subalgebra is atomless and regular proceeds as before. 4. Atomless and non-atomless Gaifman algebras. Gaifman constructed a family of Boolean algebras B G as follows [Ga]: Let Clop(2 (01 denote the clopen subsets of 2 (01. For X Y [(0 1] <ω let { } (4.1 B X = f 2 (01 : x X f(x = 1 { } B Y = f 2 (01 : y Y f(y = 0. Let {T i : 2 i < ω} be an enumeration of the open subintervals of (0 1 with rational endpoints. For each 2 i < ω choose i 2 -many disjoint open subintervals of T i and label them T i1 T i2... T ii 2. We let T i0 = (0 1\ 1 j i 2 T ij so that {T ij : 0 j i 2 } is a partition of (0 1. Let I be the set of those elements of Clop(2 (01 of the form B X B Y such that for some 2 i < ω X intersects at least i-many of the open intervals T i1... T ii 2. Let I be the ideal generated by I in Clop(2 (01. The Gaifman algebra is the quotient algebra (4.2 B G = Clop(2 (01 /I. For c Clop(2 (01 we shall denote the equivalence class of c in B G by [c]. Notice that the set {[B X B Y ] : X Y [(0 1] <ω and B X B Y I} is dense in B G +. We will use this fact implicitly in this and the next section. Theorem 4.1 (Gaifman [Ga]. B G satisfies the σ-bounded c.c. but does not satisfy CUP(B G +. By Kelley s Theorem 1.9 B G is not measurable. Depending on how the T i s and T ij s are chosen B G may or may not have atoms. To show this we will use Lemmas 4.2 and 4.4 and Fact 4.3.

16 COMPLETE EMBEDDINGS OF THE COHEN ALGEBRA 237 Lemma 4.2 (Gaifman [Ga]. For each B X B Y Clop(2 (01 with X Y = B X B Y I iff B X I. The next fact follows easily from Lemma 4.2. Fact 4.3. Suppose [B U B V ] [B X B Y ] in B G +. Then X U and (V Y U =. If in addition U\X then [B U B V ] < [B X B Y ]. The following lemma gives necessary and sufficient conditions for an element of B G + to be an atom: Lemma 4.4. b B G + is an atom for some X Y [(0 1] <ω b = [B X B Y ] > [0] and for each x (0 1\(X Y B X B {x} I. Proof. Suppose b is an atom. Then there must exist X Y [(0 1] <ω such that b = [B X B Y ]. If x (0 1\(X Y such that B X B {x} I then by Lemma 4.2 and Fact 4.3 [0] < [B X B {x} B Y ] < [B X B Y ]. Contradiction. Conversely suppose b = [B X B Y ] > [0] and x (0 1\(X Y B X B {x} I. Suppose also that [B U B V ] is such that [B U B V ] [B X B Y ] > [0]. Then X V = so v V \Y B X B {v} I. Thus [B V \Y ] [B X ]. Furthermore B U B X I so U must be contained in X. Therefore [B U B V ] [B X B Y ]. Hence b is an atom. Depending on the intervals T i T ij used in the construction a Gaifman algebra may have atoms. The following is an atom in many Gaifman algebras: Example 4.5 (Some Gaifman algebras have atoms. Let T 2 = (0 1 T 3 = (0 31 and T 4 = ( and choose the following T ij s in these T i s: { (0 ( 3 {T T 24 } = ( ( } { (0 ( 1 {T T 39 } = ( ( ( ( } ( { (0 ( 1 {T T 416 } = 1 Let (4.3 ( ( X = ( ( ( ( ( 3 5 ( ( ( 5 7 ( ( { }. ( 7 9 ( ( ( }.

17 238 NATASHA DOBRINEN For many choices of T 5 T 6 T 7 s [B X ] is an atom in the corresponding Gaifman algebra. For instance if T 5 T 6 T 7 ( 31 1 then no matter how the T i for i 8 and T ij for i 5 are chosen [B X ] > [0]. Furthermore for any x (0 1\X B X {x} I since x must lie in at least one of the T 2j s T 3j s or T 4j s (j 1 which X does not intersect. Thus by Lemma 4.4 [B X ] is an atom. Remark 4.6. Every Gaifman algebra has at most countably many atoms since the c.c.c. holds. Every Gaifman algebra has a large atomless part. Let (4.4 E = (T i0 \int(t i0. 2 i<ω E is the set of all endpoints of the intervals T ij 2 i < ω 1 j i 2. Lemma 4.7. If z (0 1\E then there are no atoms in B G below [B {z} ]. Proof. Suppose z (0 1\E and [0] < [B U B V ] [B {z} ]. Then z U by Fact 4.3. For each i 2 let j(i i 2 be such that z int(t ij(i and let N = U + 2. Then int( 2 i N T ij(i so choose a z int( 2 i N T ij(i\(u V. 2 i N z z T ij(i so U {z } has nonempty intersection with at most (i 1-many T ij s. Thus [0] < [B U {z } B V ]. Fact 4.3 implies [B U {z } B V ] < [B U B V ] since z U. Hence [B U B V ] is not an atom. Remark By Lemma 4.7 [B X B Y ] is an atom only if X E. Moreover if [B X B Y ] is an atom then X 2 i X +1 (T i0\int(t i0. 2. Atomless Gaifman algebras do exist. For example if the T ij s are nested so that (i < k 0 j i 2 1 l k 2 and T ij T kl = T kl T ij then the resulting Gaifman algebra is atomless. 5. The Cohen algebra completely embeds into the atomless part of each Gaifman algebra. In this section we work in the atomless part of Gaifman algebras. Let B G be a Gaifman algebra. We identify B + G with its image under the canonical dense embedding of B + G into r.o.(b + G and work in r.o.(b + G. Let ( { ( a = [1] b r.o. + (5.1 BG : b is an atom } in r.o.(b G + and define (5.2 A G = r.o.(b G + a.

18 COMPLETE EMBEDDINGS OF THE COHEN ALGEBRA 239 A G is the atomless part of r.o.(b G +. Since r.o.(b G + satisfies the σ- bounded c.c. A G also satisfies the σ-bounded c.c. Since CUP(r.o.(B G + + fails and r.o.(b G + has at most countably many atoms CUP(A G + must fail. Thus by Kelley s Theorem 1.9 A G is non-measurable. To avoid confusing notation between elements of r.o.(b G + and A G we will hold to the convention that [B X B Y ] always refers to an element of r.o.(b G +. We will often use without mention the fact that {[B X B Y ] r.o.(b G + : [0] < [B X B Y ] a} is a dense subset of A G +. Lemma 5.1. If F (0 1 is finite and [B X B Y ] A G + then z (0 1\(F X Y such that [0] < [B X {z} B Y ]. Proof. Claim. [B X B Y ] A G + z (0 1\(X Y for which [0] < [B X {z} B Y ]. Suppose [B X B Y ] A G +. Since A G is atomless [B U B V ] A G + such that [B U B V ] < [B X B Y ]. By Fact 4.3 U X so [B U B V ] = [B U B V Y ]. Since A G is atomless Lemma 4.4 implies z (0 1\(U V Y such that B U {z} I. Lemma 4.2 implies [B U {z} B V Y ] > [0]. Hence the Claim holds. Let n = F. By the Claim we can inductively choose a sequence of distinct elements z 0... z n (0 1\(X Y such that [0] < [B X {z0...z n} B Y ]. Thus there is some 0 i n such that z i F X Y and [0] < [B X {zi } B Y ]. Theorem 5.2. The Cohen algebra embeds as a complete subalgebra into A G. Proof. We construct a countable atomless regular subalgebra C of A G. Recall that E denotes the set of all endpoints of the intervals T ij 2 i < ω 1 j i 2 (see (4.4. Our construction uses two types of sets: F i s which keep track of elements of E and X i s which keep track of elements of (0 1\E. We start by constructing the F i s recursively. Construction of C. Let E 2 = T 20 \int(t 20 the endpoints of the open intervals T 21 T 22 T 23 T 24. Let (5.3 F 2 = E 2. Let E 3 = T 30 \(int(t 30 F 2. Recall that a is the complement of the supremum of the atoms in r.o.(b G + (see (5.1. F F 2 for which [B F ] a > [0] choose one x F (0 1\F 2 such that [B F {xf }] a > [0]. This is possible by Lemma 5.1 and the fact that {[B X B Y ] r.o.(b G + : [B X B Y ] a} is dense in A G +. Let (5.4 F 3 = E 3 {x F : F F 2 [B F ] a > [0]}.

19 240 NATASHA DOBRINEN Given F 2... F n let E n+1 = T n+10 \(int(t n+10 2 i n F i. Again F 2 i n F i for which [B F ] a > [0] choose one x F (0 1\( 2 i n F i such that [B F {xf }] a > [0]. Let F n+1 = E n+1 x F : F (5.5 F i [B F ] a > [0]. 2 i n The sets F i are finite and have the following properties: (F1 E 2 i<ω F i; (F2 2 i < j < ω F i F j = ; (F3 2 j < ω F 2 k<j F k such that [B F ] a > [0] there is an x F F j for which [B F {xf }] a > [0]. Taking the F i s into consideration we construct finite sets X i recursively. For i 2 let (5.6 J i = { s = s(2... s(i + 1 i+1 k=2 ( k : int ( i+1 k=2 T ks(k J i since 2 k i {T kj : j k 2 } is a partition of (0 1 into finitely many open intervals and a finite union of closed intervals. For each s J i choose one x s int( 2 k i+1 T ks(k\( 2 k<ω F k 2 k<i X k. Let (5.7 X i = {x s : s J i }. The sets X i F i have the following properties: (XF1 ( 2 i<ω X i ( 2 j<ω F j = ; (XF2 2 i < j < ω (X i F i (X j F j =. For each 2 i < ω define (in r.o.(b + G c i = [ ] [ ] (5.8 B{x} B{f} a. x X i f F i Note that by Lemma 4.7 for each x 2 i<ω X i [B {x} ] A G since x E. Hence c i = x X i [B {x} ] ( f F i [B {fi }] a. Let (5.9 C = {c i : 2 i < ω} the subalgebra of A G generated by {c i : 2 i < ω}. By our notational convention the complement of c i in A G will be written as c i a where c i denotes the complement of c i in r.o.(b G +. Proposition 5.3. The generators of C are independent. }.

20 COMPLETE EMBEDDINGS OF THE COHEN ALGEBRA 241 Proof. Suppose K L are finite disjoint subsets of { }. (5.10 c k ( c l a = k K l L k K u X k F k [ ] [ ] B{u} B S l L (X l F l a. If K = the right-hand side of (5.10 becomes [B S l L (X l F l ] a which is greater than [0]. Suppose K. It suffices to find a sequence x k : k K k K X k for which [B {xk :k K} B S l L X l F l ] > [0]. Let n = K + 1. Order the elements of K as k 2 < k 3 < < k n. Fix an x 2 X k2. 2 i k n + 1 let s(i i 2 be such that x 2 T is(i. By (F1 and (XF1 x 2 E so x 2 int( 2 i k n+1 T is(i. Thus 3 m n s(2 s(3... s(k m + 1 J km so choose x m X km int( 2 i k m+1 T is(i. Let X = {x m : 2 m n}. For each 2 i n 1 k i i so {x 2 x i 1... x n } all lie in T is(i. Therefore X intersects at most (i 3-many T ij s. Since X K < n B X I. By (XF2 and Lemmas 4.2 and 4.7 a [B X B S l L X l F l ] > [0]. By Proposition 5.3 C is atomless. Proposition 5.4. For each d A G + there exists a c d C + such that whenever c C and c c d [0] then c d [0]. Proof. It suffices to show the proposition for all d A G + of the form [B X B Y ]. Let d = [B X B Y ] A G +. By (XF2 we can fix an N X Y + 1 such that (X Y ( N<i<ω (X i F i =. Define I = {2 i N : X (X i F i } and J = {2 i N : X (X i F i = }. Let (5.11 c d = i I c i j J( c j a. c d > [0] by Proposition 5.3. (Unlike in our constructions of complete embeddings of the Cohen algebra into the families of Galvin-Hajnal and Argyros algebras c d is not necessarily a minimal cover for d in C. Suppose c = k K c k l L ( c l a C + where K L are finite subsets of { } and c d c > [0]. Then by Proposition 5.3 (I K (J L =. Let K = K {2... N} and K = K\K. K J = implies K I. Since d a c d = [ ] [ ] (5.12 B{u} B X B Y Sl L (X l F l. k K u X k F k k K (X k F k X since K I. Thus by Lemma 4.2 to show that c d > [0] it suffices to find a sequence u k : k K k K (X k F k for which:

21 242 NATASHA DOBRINEN (a [B {uk :k K } X] > [0]; and (b ({u k : k K } X (Y l L (X l F l =. Let m = K 1 and list the elements of K as N + 1 k 0 < k 1 < < k m. There are two cases for (a. If X E then since E 2 i<ω F i and X ( N<i<ω F i = it follows that X k K F k. [B X ] a > [0] and (F3 imply f k0 F k0 such that [B X {fk0 }] a > [0]. By induction using (F3 we obtain a sequence f k0... f km 0 j m F k j for which [B X {fk0...f km }] a > [0]. Thus (a holds for the sequence f k : k K. If X\E then fix an element x X\E. Let s 2 i k m+1 (i2 + 1 be such that x int( 2 i k T m+1 is(i. For each 0 j m since int( 2 i k j +1 T i s(i our construction of X kj guarantees that x kj X kj int( 2 i k j +1 T is(i. Choose one such x kj. For 2 i N + 2 X {x k : k K } intersects exactly the same T ij s as X; hence X {x k : k K } intersects less than i-many T ij s. For N + 3 i k m {x k : k K k i 1} lies in T is(i as does x. Hence X {x k : k K } intersects at most (i 2-many T ij s. Since X {x k : k K } k m (a holds for the sequence x k : k K. Y ( N<i<ω X i N<i<ω F i = and K L =. Hence ( k K (X k F k (Y l L (X l F l =. X Y = since d > [0]. X ( 2 i<ω X i F i i I (X i F i and I L = = X ( l L X l F l =. Thus (b holds for the sequences f k : k K and x k : k K. Thus c d [0]. Proposition 5.3 shows that C satisfies the conditions of Lemma Hence C is a regular subalgebra of A G. By Lemma 1.15 the Cohen algebra r.o.(c + embeds as a complete subalgebra of A G. 6. Remarks and acknowledgments. Remark 6.1. After the completion of the work in this paper it was brought to our attention that A. Kamburelis has done some similar work on these three Boolean algebras in an unpublished paper [Ka]. Our work extends some and gives a new perspective on some of his results in that paper. Specifically Kamburelis used forcing methods (in contrast to our constructive purely Boolean-algebraic methods to show that the weak (ω ω-d.l. fails in the Galvin-Hajnal Agryros and Gaifman algebras. In this paper Kamburelis mentions K. Skandalis remark that his proof of the failure everywhere of the weak (ω ω-d.l. in the the Argyros algebra can be easily modified to produce a Cohen real. By the result of Shelah [S] mentioned in Remark 1.16 we now know that Kamburelis proof of the failure everywhere of the weak (ω ω-d.l. actually implies the Argyros algebra

22 COMPLETE EMBEDDINGS OF THE COHEN ALGEBRA 243 adds a Cohen real. In addition Kamburelis showed that the Gaifman algebra adds a Cohen real although to do this he assumed that the Gaifman algebra contains no atoms which as we showed is not always the case. Acknowledgment 6.2. We are enormously grateful to the referee for many detailed and helpful comments and suggestions. In particular we would like to thank the referee for pointing out that our original proof that the Cohen algebra completely embeds into the Galvin-Hajnal algebra in essence showed that it actually adds cf(2 ω -many Cohen reals and for suggesting that we investigate the variants of Argyros algebra. Acknowledgment 6.3. Infinite gratitude goes to my thesis advisor K. Prikry for his idea of trying to completely embed the Cohen algebra into the Argyros Galvin-Hajnal and Gaifman algebras. Thanks also goes to him for pointing out the connections between von Neumann s problem and weak distributivity in c.c.c. Boolean algebras and for many helpful discussions. [A] [BJ] [CN] [F] References S.A. Argyros On compact spaces without strictly positive measure Pacific J. Math. 105(2 ( MR Zbl S.T. Bartoszyński and H. Judah Set Theory on the Structure of the Real Line A.K. Peters Ltd MR Zbl W.W. Comfort and S. Negrepontis Chain Conditions in Topology Cambridge University Press MR Zbl D. Fremlin Measure Algebras Handbook of Boolean Algebra 3 North-Holland MR Zbl [Ga] H. Gaifman Concerning measures on Boolean algebras Pacific J. Math. 14 ( MR Zbl [GP] F. Galvin and K. Prikry On Kelley s intersection numbers Proc. Amer. Math. Soc. 129(2 ( MR Zbl [Gl] W. Glówczyński Measures on Boolean algebras Proc. Amer. Math. Soc. 111 ( MR Zbl [J] [Ka] R. Jensen Definable sets of minimal degree Mathematical Logic and Foundations of Set Theory Proc. Internat. Coll. Jerusalem 1968 North-Holland MR Zbl A. Kamburelis On weak distributivity of some topological spaces. Preprint. [Ke] J. Kelley Measures on Boolean algebras Pacific J. Math. 9 ( MR Zbl [Ko] S. Koppelberg Handbook of Boolean Algebra 1 North-Holland 1989 MR Zbl [Mah] D. Maharam An algebraic characterization of measure algebras Ann. Math. 48(1 ( MR Zbl [Mau] D. Mauldin The Scottish Book Birhäuser 1981 MR Zbl

23 244 NATASHA DOBRINEN [S] S. Shelah How special are Cohen and random real forcings i.e. Boolean algebras of the family of subsets of reals modulo meagre or null Israel J. Math. 88 ( MR Zbl [V] B. Veličković CCC posets of perfect trees Comp. Math. 79 ( MR Zbl Received December and revised July The Pennsylvania State University Department of Mathematics 218 McAllister Building University Park PA address: dobrinen@math.psu.edu