Independent families in Boolean algebras with some separation properties

We prove that any Boolean algebra with the subsequential completeness property contains an independent family of size continuum. This improves a result of Argyros from the 80ties which asserted the existence of an uncountable independent family. In fact we prove it for a bigger class of Boolean algebras satisfying much weaker properties. It follows that the Stone spaces of all such Boolean algebras contains a copy of the Cech-Stone compactification of the integers and the Banach space of contnuous functions on them has $l_\infty$ as a quotient. Connections with the Grothendieck property in Banach spaces are discussed.


Independent families
By an antichain in a Boolean algebra A, we will mean a pairwise disjoint subset of A, i.e., a B ⊆ A such that A ∧ B = 0 whenever A and B are two distinct elements of B. An independent family in a Boolean algebra A is a family for any two disjoint finite subsets F, G ⊆ I. For more information on Boolean algebras, see [8]. K A will stand for the Stone space of A and C(K) for the Banach space of real valued continuous functions on K.
[A] will stand for the clopen subset {x ∈ K A : A ∈ x} for any A in a Boolean algebra A, and 1 X for the characteristic function of X ⊆ K A . Unexplained notions concerning Banach spaces can be found in [2]. We will consider several separation properties in Boolean algebras. This is a natural generalization of the following well-known concept ( [5]).

Definition 1.2. A Boolean algebra
A is said to have the subsequential completeness property if given any countably infinite antichain {A n : n ∈ N} ⊆ A, there exists an infinite M ⊆ N such that in A, the supremum of {A n : n ∈ M } exists.
One more separation property introduced in [6] is the following. Definition 1.3. A Boolean algebra A is said to have the subsequential separation property if given any countably infinite antichain {A n : n ∈ N} ⊆ A, there is an A ∈ A such that the set The subsequential completeness property was introduced in [5]. That paper included applications of the subsequential completeness property in the theory of Banach spaces as well as a result of Argyros (Proposition 1G) that every Boolean algebra with the subsequential completeness property has an uncountable independent family. Here we strengthen this result.

Theorem 1.4. Suppose that A is an infinite
Boolean algebra that has a weak subsequential separation property. Then A contains an independent family of cardinality c.
Proof. Clearly, the Stone space of A cannot have a nontrivial convergent sequence and so cannot be a dispersed space; consequently, A is not superatomic and so contains a countable free infinite Boolean algebra B ⊆ A.
Let {0, 1} <N stand for finite sequences of 0s and 1s. A countable free Boolean algebra contains all countable Boolean algebras as subalgebras. We will identify a particular subalgebra C ⊆ A with generators {A s : s ∈ {0, 1} <N } such that whenever D is a countable free Boolean algebra generated by free generators {D s : s ∈ {0, 1} <N }, there is a surjective homomorphism h : D → C such that h(D s ) = A s and the kernel of h is generated by the elements of the form A s ∧ A t such that s t.
In particular, it follows that whenever s t, then Now we note that if n is an integer and u i ∈ {0, 1} <N for 1 ≤ i ≤ n, then unless u i ⊆ u j for some distinct 1 ≤ i, j ≤ n. Indeed, suppose that u i ⊆ u j for all distinct 1 ≤ i, j ≤ n. Consider an arbitrary element of the kernel of h; it may be assumed to be of the form as required for ( * * ). Now by ( * ) for every x ∈ 2 N , let us consider an antichain {A s : s ⊆ x}. By the subsequential separation property for each x ∈ 2 N , there are A x ∈ A as in Definition 1.1. We will show that {A x : x ∈ 2 N } is the required independent family. Let x 1 , . . . , x n be distinct elements of 2 N and let The sequences x 1 |m 1 , . . . , x n |m n satisfy the hypothesis of ( * * ) for u 1 , . . . , u n , so In the literature, there are several more weakenings of the subsequential completeness property that are stronger than our weak subsequential separation property, most notably the subsequential interpolation property introduced in [3] and applied in several other papers and the subsequential separation property of [6]. Hence, the above theorem applies to the algebras with these properties as well.

Corollary 1.5. If A is a Boolean algebra having a subsequential separation property and K
Proof. It is well known that a free Boolean algebra with c generators maps homeomorphically onto any Boolean algebra of cardinality c, and so, for example, onto ℘(N). Use the Sikorski extension theorem to obtain a homeomorphism of A onto ℘(N). By the Stone duality, it follows that the Stone space of ℘(N), which is homeomorphic to βN, is a subspace of K A . Restricting continuous functions on K A to a copy of βN gives, by the Tietze extension theorem, a norm one linear operator onto C(βN), which is known to be isometric to l ∞ .
It follows that many Banach spaces present in the literature have l ∞ as a quotient, in particular the spaces of [5] or [9]. In [9], besides Boolean separations a lattice version of the subsequential completeness property is considered for connected compact K. Namely (see [9,5.1]) we consider spaces K such that given any pairwise disjoint the supremum of (f n ) n∈M exists. It is not difficult to generalize the proof of Theorem 1.4 to conclude that such K always contain βN as well.

The Grothendieck property of Banach spaces
In this section, we would like to direct the attention of the reader to some links between the weak subsequential separation property and the Grothendieck property, which originated in the theory of Banach spaces. Definition 2.1. Let X be a Banach space. We say that X has the Grothendieck property if and only if weak * convergence of sequences in X * is equivalent to weak convergence. A Boolean algebra A has the Grothendieck property if and only if the Banach space C(K A ) has the Grothendieck property.
The Grothendieck property for Boolean algebras was introduced and first investigated by Schachermayer in [10]. It can be quite nicely characterized using finitely additive signed measures on Boolean algebras. Recall that the Riesz representation theorem says that the dual to a C(K) space is isometric to the space of Radon measures on K with the variation norm, i.e., all continuous functionals on C(K) are of the form fdμ for μ a Radon measure on K. A Radon measure means a signed, countably additive, Borel, regular measure. If K is totally disconnected (has a basis of clopen sets), then any finitely additive, bounded, signed measure on Clop(A) extends uniquely to a Radon measure on K. So there is a one to one correspondence between such measures and elements of the dual Banach space to C(K A ). For more information on this, see [11]. is not a convergent sequence of the reals.

Lemma 2.2. Suppose A is a Boolean algebra, K A its Stone space, and C(K
Proof. Suppose that C(K A ) has the Grothendieck property and A n , ε, and μ n are as above. Let x * n be elements of the dual to C(K A ) that are uniquely determined by the condition x * n (1 [A] ) = μ n (A) for A ∈ A. Using the Rosenthal lemma (see, for example, [2]) going to a subsequence, we may assume that if k is an even integer. In other words, the element of the bidual to C(K A ) corresponding to the Borel set n∈N [A 2n ] of K A witnesses the fact that (x * n ) n∈N is not weakly convergent. By the Grothendieck property, it is not weakly * convergent. As (x * n ) n∈N is bounded, it means that (μ n (A) : n ∈ N) is not a convergent sequence of the reals for some A ∈ A since the span of characteristic functions of clopen sets is dense in C(K A ) by the Stone-Weierstrass theorem.
For the converse implication, suppose that (x * n ) n∈N is a bounded sequence in the dual to C(K A ) that is not weakly convergent; assume the condition of the lemma, and conclude that (x * n ) n∈N is not weakly * convergent. (x * n ) n∈N , as a nonconvergent sequence in a compact dual ball, has at least two distinct accumulation points. If every sequence of (x * n ) n∈N contains a weakly convergent subsequence (and so weakly * convergent), then we conclude that (x * n ) n∈N is not weakly * convergent, as required. Otherwise, by choosing a subsequence of (x * n ) n∈N , we may assume that it does not have any weakly convergent subsequence. So by the Eberlein-Smulian theorem, {x * n : n ∈ N} is not relatively weakly compact, and hence by the Grothendieck-Dieudonne characterization of weakly compact subsets of the duals to Banach spaces C(K), we obtain an antichain {A n : n ∈ N} of A, an ε > 0, and an infinite M ⊆ N such that such that |ν n ([A n ])| > ε where ν n is the Radon measure on K A corresponding to x * n . Note that the restriction μ n of ν n to the family of all characteristic functions of clopen subsets of K A satisfies the condition of the lemma; hence, there is an A ∈ A as stated there. The function 1 [A] witnesses the fact that (x * n ) n∈N is not weak * convergent, as required.
It is proved by Haydon in [6, 6.3] that Boolean algebras with the subsequential separation property have the Grothendieck property.  Proof. Let {A n : n ∈ N}, ε > 0 and μ n , n ∈ N, be as in Definition 2.3. Applying [12, Lemmas 1 and 2] (see also [6, 6.2 and 6.3]), we may assume that there is an antichain {B n : n ∈ N} and finitely additive bounded measures λ n and ν n such that μ n = λ n + ν n , where ν n weakly converges, λ n (K) = η, and λ n (B n ) > 3η/4. By the Dieudonne-Grothendieck theorem applied to the μ n , we conclude that they do not form a relatively weakly compact set, and hence are not a weakly convergent sequence; hence, η = 0. Now use the weak subsequential separation property to obtain A ∈ A such that both of the sets M 1 = {n ∈ N : B n ≤ A} and M 0 = {n ∈ N : B n ∧ A = 0} are infinite. For each n ∈ M 1 , we have μ n (A) = λ n (A∩B n )+λ n (A\B n )+ν n (A) > 3η/4−η/4+ν n (A) = η/2+ν n (A), and for each n ∈ M 0 , we have As ν n (A) converges, η = 0; since weakly convergent sequences are weakly * convergent, we conclude that μ n (A) does not converge, as required.

Proposition 2.5. There is a Boolean algebra with the weak subsequential separation property that does not have the Grothendieck property and so does not have the subsequential separation property.
Proof. This is a classical example A (see, for example, [10]) of the Boolean algebra of all subsets M of N such that 2k ∈ M if and only if 2k + 1 ∈ M for all but finitely many k ∈ N. It is well known that μ n = δ 2n − δ 2n+1 form a weakly * convergent sequence in C(K A ) that is not weakly convergent and so A does not have the Grothendieck property. On the other hand, given an antichain {A n : n ∈ N} in A, there is an infinite M ⊆ N such that there are pairwise disjoint B n ∈ A with A n ⊆ B n for n ∈ M such that 2k ∈ B n if and only if 2k + 1 ∈ B n for all k ∈ N and al n ∈ M . Infinite unions of such B n provide elements witnessing the separation.
The Grothendieck property of C(K A ) does not imply in ZFC the existence of an independent family of cardinality c in A. Namely, assuming the continuum hypothesis, Talagrand proved in [12] that there is a Boolean algebra A such that C(K A ) has the Grothendieck property but l ∞ is not a quotient of C(K A ). In particular, βN is not a subset of K A and so A has no uncountable independent family. Moreover it is proved in [1] that it is consistent that there is a Boolean algebra A that has the Grothendieck property but has cardinality strictly smaller than 2 ω (ground model after adding Sacks reals). On the other hand, assuming p = 2 ω Haydon, Levy, and Odell proved in [7] that each nonreflexive Banach space with the Grothendieck property (in particular, each of the form C(K A ) for A infinite) has l ∞ as a quotient. However, we do not know the answer to the following. Question 2.6. Is it consistent that each Boolean algebra with the Grothendieck property has an independent family of cardinality c?

Efimov's problem
The affirmative answer to the question above could be considered a weak solution to the Efimov problem (see [4]) whether it is consistent that any compact K without a nontrivial convergent sequence has a subspace homeomorphic to βN or there is in ZFC a compact space without a nontrivial convergent sequence and without a copy of βN. Indeed, subsets of a compact space K can be considered as subsets of the dual ball to the Banach space C(K) or Radon measures (points of K correspond to pointwise measures) with the weak * topology. And so, instead of nontrivial convergent sequences or copies of βN in K, one can ask for the same subspaces in the dual ball or the Radon measures. The Grothendieck property of C(K), in a sense, asserts that there are no nontrivial convergent sequences among Radon measures (not just pointwise measures, here nontrivial means those which are not convergent in the weak topology, or those which can be separated by a Borel subset of the compact space, the notion changes as the dual ball always contains copies of intervals) and it easily implies the nonexistence of nontrivial (in the sense of having distinct terms) sequences of points of K. Of course, the negative answer to the above question would solve the original Efimov problem.
Note that Efimov's problem is equivalent to asking if an analogous property to our weak subsequential separation property for points of the Stone space instead of elements of the Boolean algebra implies the existence of an independent family of cardinality c.
Also, Talagrand proved (see [12]) that the dual ball to C(K) contains a copy of βN if and only if ∞ is a quotient of C(K). So another weak version of the Efimov problem would be to ask if it is consistent that whenever K has no convergent sequence, ∞ is a quotient of C(K). One should note here that consistently it is not the case (again the example of [12]) and that the result of [7] gives the consistency of the statement that for every compact K, the dual ball to C(K) with the weak * topology either contains a copy of βN or a convergent sequence that is not weakly convergent.