Compact spaces associated to separable Banach lattices

We study the class of compact spaces that appear as structure spaces of separable Banach lattices. In other words, we analyze what C(K) spaces appear as principal ideals of separable Banach lattices. Among other things, it is shown that every such compactum K admits a strictly positive regular Borel measure of countable type that is analytic, and in the nonmetrizable case these compacta are saturated with copies of βN.\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\beta {{\mathbb{N}}}.$$\end{document} Some natural questions about this class are left open.


Introduction
Given a vector lattice E and a nonzero positive vector u ∈ E + , there is a natural way to associate a compact space K u (E) : Following [12], K u (E) can be described as the set of all functions ∶ E ⟶ [0, +∞] such that (u) = 1, (x + y) = (x) + (y), (x ∧ y) = (x) ∧ (y) and (rz) = |r| (|z|) for all x, y ∈ E + , z ∈ E and r ∈ ℝ. Here, [0, +∞] are the nonnegative reals extended by a point +∞ (satisfying usual conventions plus +∞ ⋅ 0 = 0 ), and the topology is the pointwise topology. When E is a Banach lattice, then the principal ideal becomes, by the Kakutani representation theorem, canonically isomorphic, as a vector lattice, to the space of continuous functions C(K u (E)). We have a one-to-one Banach lattice homomorphism of norm 1, whose range is E u , and that sends the constant one function to u. A vector lattice isomorphism between C(K) and C(L) that preserves the constant functions induces a homeomorphism between K and L, so K u (E) is uniquely determined by the existence of such T. For details, consult [12] or [13,Theorem III.4.5]. The purpose of this paper is to investigate the class of compact spaces that arise in this way from separable Banach lattices. Definition 1. 1 We say that a compact space K is sick 1 if it is homeomorphic to K u (E) for some separable Banach lattice and some u ∈ E + .
In the sequel, E will always be a Banach lattice. It is shown in [12,Proposition 2] that K u (E) is homeomorphic to K v (E) whenever E u = E v . The vector u is called a quasi-interior point of E if E = E u . Whenever such points exist (and this is the case for all separable Banach lattices), we can talk about K(E), the structure space of E, as the compact space homeomorphic to K u (E) for any quasi-interior point u ∈ E. An important feature is that we can represent E as an ideal of the vector lattice of continuous functions f ∶ K(E) ⟶ ℝ ∪ {−∞, +∞} with f −1 (ℝ) dense, (cf. [17,Theorem 3.5]). The class of sick compacta coincides with the class of structure spaces of separable Banach lattices.
Let us now discuss what we have found out about the class of sick compacta. In Sect. 2, we present the first obvious examples. The structure space of C(K) is the compact space K, and therefore all metrizable compact spaces are sick. More generally, for f ∈ C(K) + , K f (C(K)) is the Čech-Stone compactification of {x ∶ f (x) ≠ 0}, and therefore W is sick for any open subset W of a compact metric space. A particular case of this is ℕ, which also appears as the structure space of any Banach lattice with the lattice structure given by an unconditional basis, like c 0 or p . Finally the structure space of L 1 [0, 1] is the Stone space of the measure algebra of the Lebesgue measure. Indeed, the principal ideal generated by the constant one function is, as a vector lattice, identified with L ∞ [0, 1] = C( ). Thus, gives a nonseparable sick compactum. We also show that the class of sick compacta is not stable under subspaces or quotients, but it contains subspace-universal and quotient-universal spaces.
By the way that sick compacta are linked to separable Banach lattices, that are in particular Polish spaces, they should have low complexity in the sense of descriptive set theory. This idea is exploited in Sect. 3. We prove that every sick compactum K admits a strictly positive regular Borel measure of countable type that is moreover analytic. This means that, through the natural embedding, C(K) is an analytic subset of L 1 ( ). We also show that for any countable set D of G points of K, the set of restrictions {f | D ∶ f ∈ C(K)} form an analytic subset of ℝ D , and some more general results. These facts are reminiscent of Godefroy's results on Rosenthal compact spaces [7]. In a little disgression from the main topic, in Sect. 4 we observe that all measures on Rosenthal compacta are analytic. However, the two classes are very different. Both Rosenthal and sick compacta are of analytic complexity, but they are orthogonal from the Efimov's problem perspective: while Rosenthal compacta are full of convergent sequences, nonmetrizable sick compacta are full of copies of ℕ.
This last statement is made precise in the main result of Sect. 5: Every sick compactum contains a (possibly trivial) sequence of metrizable closed subsets K n , such that the closure of every discrete sequence x n ∉ K n is homeomorphic to ℕ. In particular, nonmetrizable sick compacta always contain copies of ℕ.
In Sect. 6, we connect sick compacta to a different problem. Given, a sequence {e n } of positive pairwise disjoint vectors in a Banach lattice, we can consider the family S E ({e n }) of all subsets A ⊆ ℕ for which the supremum sup n∈A e n exists in E. What are the families A of subsets of ℕ that can be represented as S E ({e n }) with E separable? The main result of this section is that, if ℕ ∈ A, this happens if and only if A is an algebra of sets containing the finite sets that is isomorphic, as a Boolean algebra, to the clopen subsets of a sick compactum.
Borodulin-Nadzieja et al. [2] have characterized the families that can be represented as S E ({e n }) with E a Banach lattice obtained from an unconditional basis. These are exactly the so called non-pathological analytic P-ideals. This does not give direct information on sick compacta because these families do not contain ℕ except in one trivial case, and the structure space of an unconditional Banach lattice is always ℕ. However, a variation of their construction obtained by adding an upper bound to the basis (similarly to how c is obtained from c 0 ) provides a new family of examples, described in Sect. 7. If I is a non-pathological analytic P-ideal, then the Stone space of the algebra generated by I is a sick compactum.
In spite of the number of properties and examples of sick compacta that we have exposed, a satisfactory understanding of this class-which means understanding the vector lattice structure of separable Banach lattices-is far from being reached. The paper is scattered with open questions that arise naturally. Perhaps the most fundamental is to have an intrinsic handable topological characterization. We collect some of these questions at the end of the paper.  K⧵W) ). Let C b (W) be the space of all bounded and continuous functions ∶ W ⟶ ℝ and let T ∶ C b (W) ⟶ C(K) be given by This function T is a vector lattice isomorphism onto its range. We claim that the range of T is exactly the ideal C(K) f generated by f in C(K). Once we have the claim proved, this ideal would be vector lattice isomorphic to C b (W) = C( W), which proves that K f (C(K)) is homeomorphic to W and the proof would be finished.

First examples
So let us prove the claim. To do so, if | | is bounded by ∈ (0, +∞), then |T | ≤ ⋅ f , and this proves that the range of T is contained in An elementary remark is that every separable Banach lattice is ccc, i.e. every pairwise disjoint family of positive elements is countable. This property is inherited by ideals, so C(K) is ccc for a sick compactum K, and so K itself is ccc (via Urysohn Lemma), in the sense that every pairwise disjoint family of nonempty open sets is countable. We shall see later that something stronger holds: every sick compactum supports a strictly positive measure. Anyway, we can already notice that the class of sick compacta is not stable under taking subspaces, since ℕ is sick but ℕ⧵ℕ is not because it fails ccc [16,Theorem 3.22]. The class of sick spaces is not stable under continuous images either. We shall see concrete examples later, but we can use a counting argument. There are only many different separable Banach lattices, but there are 2 non-homeomorphic separable compact spaces (cf. for instance [6]), all of them continuous images of ℕ.
Despite this unstability, we do have universal objects. The assignment E ↦ K(E) takes quotients to inclusions, cf. [12,Propositions 3]. Every separable Banach lattice is a quotient of the free Banach lattice FBL(ℕ), which is separable [10]. Thus, the structure space K(FBL(ℕ)) is a sick compactum that contains any other sick compactum as a subspace. The topological nature of this universal sick compactum looks mysterious to us. On the other hand, by Theorem 3.2, every sick compactum will be a continuous image of , cf. [5, 415Q-415R].

Analytic complexity
A principal ideal of a Banach lattice E is an F subset of E. So when K is a sick compactum, C(K) appears bijected with a nice subset of a Polish space E. This situation suggests that C(K) should be viewed as an object of low descriptive complexity, but it is not clear at first glance in what sense. In this section, we will present several results that make precise the idea that C(K) is of analytic complexity when K is sick. The first such result also encircles our class within the realm of compact spaces that admit strictly positive measures. By a measure on a compact space K we mean a regular Borel measure on K. Such a measure is of countable type if the corresponding space L 1 ( ) is separable, and it is strictly positive if all nonempty open sets have nonzero measure.
Definition 3.1 A measure on a compact space K will be called analytic if it is of countable type and the range of the formal identity i ∶ C(K) ⟶ L 1 ( ) is an analytic set.

Theorem 3.2 If K is sick, then K admits a strictly positive analytic measure.
Proof The composition x * •T gives a strictly positive functional on C(K), that is represented by a strictly positive measure on K. Let i ∶ C(K) ⟶ L 1 ( ) be the formal inclusion. Note that the natural identity T(C(K)) ⟶ i(C(K)) is a continuous operator between normed spaces, because for every f ∈ C(K). Since is a countable union of closed subsets of a separable Banach lattice E, we conclude that i(C(K)) is analytic. ◻

Corollary 3.3 If K is sick and {x
Note that j is a bounded operator and is a Borel function, so the composition is Borel. Since i(C(K)) is analytic, so is ( •j)(i(C(K))), which is the set in the statement. ◻

Corollary 3.4 Let B be a subalgebra of P(ℕ) that contains all finite sets. If the
Stone space of B is sick, then B must be an analytic subalgebra of P(ℕ).
Proof In such a Stone space there is a dense set of isolated points {x n ∶ n ∈ ℕ} and, under the natural bijection of P(ℕ) and {0, 1} ℕ , the algebra B is sent exactly to the set These results are saying that it is only among compact spaces that have an analytic definition that sick compacta are to be found, while other compacta constructed by set-theoretic techniques should be ruled out. Along the rest of the section, we prove some generalizations of Corollary 3.3.

Definition 3.5
Let F (K) be the least family of closed subsets of K that contains the closed G sets, and is stable under countable intersections and closures of countable unions. Sets in this family will be called -generated closed sets.

Lemma 3.6 Let K be sick and
Proof The first observation is that it suffices to prove that the set Let us prove the case when L is a closed G set. In that case, there exists g ∈ C(K) such that g ≥ 0 and L = g −1 (0). Then {f ∶ f | L = 0} is the closed ideal generated by g in C(K). So, Note that ‖ ⋅ ‖ ∞ is a Borel function on the closed set E x ⊂ E, because its graph is a Borel set. So we conclude that the set {Tf ∶ f | L = 0} is analytic as desired. It is enough to prove now that the family of all closed L such that {Tf ∶ f | L = 0} (and hence also {Tf ∶ ‖f � L ‖ ∞ ≤ r} ) is analytic is stable under countable intersections and closures of countable unions. The latter is obvious because Concerning intersections, let us start with an intersection of just two L 1 ∩ L 2 . In this case, the observation is that The image under T of that set of sums would be analytic, and by the same description as before, the image under T of the closed ideal it generates would also be analytic. From this we deduce that our family is closed under finite intersections. Finally, it is enough to check that the family is stable under decreasing intersections. But if L 1 ⊃ L 2 ⊃ ⋯ , then we have that and since we assume that each set A n,m = {f ∶ ‖f � L n ‖ ∞ ≤ 1∕m} is analytic, we conclude that is an analytic set. ◻ Theorem 3.7 Let {D n } n be a family of -generated closed subsets of a sick compactum. Then the following is an analytic set: Proof Let us start with the following Claim: Proof of the claim. The inclusion [⊆] is obvious. For the converse, suppose that (t n ) satisfies the property above to be in the right-hand side set S ′ . For each x ∈ ⋃ n D n , the set is a singleton. Indeed, for an intersection of closed intervals to be empty we would need to have among them at least two disjoint intervals [a, p] and [q, b] with p < q, and we would get that in such a way that {f (x)} = R x . It is clear that f | D n is constant equal to t n . By Tietze's extension theorem, it is enough to check that f is continuous. The argument follows a similar line as before, if f (x) = p < q and p < p ′ < q ′ < q then so K⧵ ⋃ t n ≥p ′ D n is a neighborhood of x where f (x) < q. An analogous argument works for the opposite inequality.
In the description of S in the above Claim one may demand that p and q are rational. So it is enough to show that for all p < q the set is analytic. For this, it is enough to show that is analytic, because S p,q is the preimage of S ′ under the measurable assignment But S ′ can be written as This is analytic by Lemma 3.6. ◻ We call a point x ∈ K -generated if {x} is a -generated closed set. Every G point is -generated. Notice that Corollary 3.8 does not hold if the assumption on the points of D is removed. The compact space ℕ is sick. If we fix a nonprincipal ultrafilter U, then it is a non G -point and taking is not analytic by the classical result of Sierpiński [14] that nonprincipal ultrafilters are not measurable. We remind an argument: if U is measurable in the product probability space {0, 1} ℕ , then it should have probability 1/2 since its complement coincides with its 0-1 switch, but by Kolmogorov's zero-one law its probability must be either 0 or 1.

A disgression: analytic measures on Rosenthal compacta
The results of the previous section remind us of Godefroy's characterization of Rosenthal compacta [7]: A separable compact space K is a Rosenthal compactum if and only if {f | D ∶ f ∈ C(K)} is an analytic subset of ℝ D for all countable dense subsets D of K. The word dense may be removed since the property of being Rosenthal is inherited by closed subspaces. Another fact first claimed by Bourgain, with proofs given by Todorcevic [15], Marciszewski and Plebanek [9] and Plebanek and Sobota [11] is that all measures on a Rosenthal compactum are of countable type. The following puts together both results.

Theorem 4.1 Every measure on a Rosenthal compactum K is analytic. A separable compactum K is Rosenthal if and only if all measures on K are analytic.
Proof We start proving the first statement. We take a measure on K that we know that is of countable type as have been mentioned above. By [7, Proposition 8] one can assume that the compact space K is separable. By [7, Theorem 9], we have that (C(K), cyl C(K) ) is Suslinean. Here, the cylindrical -algebra cyl X on a Banach space X is the least -algebra that makes all bounded linear functionals measurable. A measurable space is Suslinean if it is isomorphic to an analytic space with its Borel -algebra. Look at the formal identity i ∶ C(K) ⟶ L 1 ( ). Since it is a bounded operator, it is cyl C(K) -to-cyl L 1 ( ) -measurable. But in every separable Banach space, the cylindrical and the Borel -algebra coincide. So, i(C(K)), endowed with the Borel structure inherited from L 1 ( ) is the image of a Suslinean space under a measurable function. We conclude that i(C(K)) is analytic as desired. Finally, we look at the second statement. Suppose that K is a separable compact space where all measures are analytic. Let D = {x n ∶ n = 1, 2, 3, …} be any countable subset of K. Consider a discrete probability measure on K with {x n } = 2 −n , that must be analytic. Reproducing the proof of Corollary 3.3, this implies that {f | D ∶ f ∈ C(K)} is an analytic subset of ℝ D . By Godefroy's characterization, we just proved that K is Rosenthal. ◻

Omnipresence of ˇℕ
Theorem 5.1 Let K be a sick compactum. There exists an increasing sequence K n ⊆ K of metrizable closed (possibly empty) subspaces of K such that if {x n } n is a discrete sequence of points in K with x n ∉ K n for every n ∈ ℕ, then {x n } n is a copy of ℕ.
Proof Let us fix a sick compactum K and let T ∶ C(K) ⟶ E be a vector lattice isomorphism of C(K) onto a principal ideal E x of a separable Banach lattice E. We can assume that ‖x‖ = 1. We define the function N ∶ K ⟶ [0, 1] by the formula We check now that K is metrizable. Set Note that C(K ) is isomorphic to C(K)∕I. Furthermore, if we take J = T(I) ⊆ E, then J is a closed ideal of E and T induces an operator T ∶ C(K)∕I ⟶ E∕J through the formula T (h + I) = Th + J. We claim that T is an isomorphism, so the metrizability of K will follow from the separability of E∕J. Take any h ∈ C(K) with ‖h + I‖ = 1. We are going to show that ‖Th + J‖ ≥ and therefore T is an isomorphism. Suppose by contradiction that ‖Th + J‖ < . By definition of J, this is equivalent to the existence of a function g ∈ I such that ‖Th − Tg‖ = ‖T(h − g)‖ < . On the one hand, since ‖h + I‖ = 1, we have that |h(t)| ≤ 1 for every t ∈K and there exists s ∈K such that |h(s)| = 1. On the other hand, it follows from the definition But then s ∉K , which yields a contradiction and finishes the proof of Claim 1.
Set now K n ∶=K 1∕2 n for every n ∈ ℕ. It remains to show that if {x n } n is a discrete sequence of points in K with x n ∉ K n for every n ∈ ℕ then {x n } n is a copy of ℕ. This is an immediate consequence of the following.
Claim 2. If {x n } n is a discrete sequence of points in K with ∑ n N(x n ) < ∞, then {x n ∶ n ∈ ℕ} is a copy of ℕ.
Proof of the claim: By definition of the function N, there exists a sequence of norm-one functions {f n } n in B C(K) such that f n (x n ) = 1 and ‖Tf n ‖ < N(x n ) + 1 2 n . Furthermore, since {x n } n is discrete, we can take pairwise disjoint functions g n ∈ B C(K) such that g n (x n ) = 1. Set h n = |g n f n | for every n ∈ ℕ. Then {h n } n is a sequence of pairwise positive disjoint functions such that h n (x n ) = 1 and Since ∑ n (N(x n ) + 1 2 n ) < ∞, all subsequences of {Th n } n have a supremum, which indeed belongs to E x = T(C(K)) because Th n ≤ x for every n ∈ ℕ, so all subsequences of {h n } have a supremum in B C (K) . But this implies that any bounded function on {x n ∶ n ∈ ℕ} extends to a continuous function on {x n ∶ n ∈ ℕ}, so {x n ∶ n ∈ ℕ} becomes a copy of ℕ. ◻

Corollary 5.2 If K is a sick compactum, then every nonmetrizable closed subspace contains a copy of ℕ.
Proof By Theorem 5.1, we can write K = L ∪ ⋃ n K n , where K n is an increasing sequence of metrizable closed subspaces of K and such that if {x n } n is a discrete sequence of points with x n ∉ K n for every n ∈ ℕ then {x n ∶ n ∈ ℕ} is homeomorphic to ℕ. Thus, if S ⊆ K is an infinite closed subspace of K, then either S⧵K n is finite for some n ∈ ℕ and therefore S is metrizable, or else S contains a discrete sequence of points with x n ∉ K n for every n ∈ ℕ as desired. ◻ As another application, we show that products of sick compacta are not sick in general.

Corollary 5.3 If K × L is sick then either K or L is a metrizable compactum.
Proof Consider the metrizable closed subspaces Z n that Theorem 5.1 gives for K × L. Let K n and L n be the projections of Z n onto the first and second coordinates respectively. If neither K nor L are metrizable, then K ≠ K n and L ≠ L n for all n, so we can find discrete sequences {x n } and {y n } with x n ∈ K⧵K n and y n ∈ L⧵L n . Consider any function ∶ ℕ ⟶ ℕ with (n) > n 2 for all n, and look at the following sequence of points in K × L : For n ∈ ℕ and 1 ≤ k ≤ 2n − 1, let This is a discrete sequence of points with z n ∉ Z n , so its closure should be homeomorphic to ℕ. But this closure equals {x (n) } × {y (n) } while ℕ is not the product of two infinite compact spaces. ◻ We conjecture that if the product of two infinite compacta K × L is sick, then both K and L must be metrizable. We shall give some partial results supporting this in Sect. 6.

Algebras of sets associated to sick compacta
So far, we know that sick compacta support strictly positive measures, are of analytic complexity in the appropriate sense and they are full of copies of ℕ wherever they are not metrizable. But, as we shall see, even having all those features is not enough to be sick. To get a deeper insight into the class, we are going to focus now on the following class. Definition 6.1 An algebra A ⊂ P(ℕ) will be called an s-sick algebra if it contains all finite sets and is isomorphic to the algebra of clopen sets of a sick compactum.
One motivation to look at this class is that it characterizes the families of subsequences of a disjoint sequence with a supremum, whose suprema exist in a separable Banach lattice.  C(K, {0, 1}), and let T ∶ C(K) ⟶ Y be a lattice homomorphism onto Y e that takes the constant one function to e. The elements e n = T {n} satisfy the required conditions.
[2 ⇒ 3] Consider Y ′ the closed sublattice of Y generated by {sup i∈A e i ∶ A ∈ A}. We claim that (e n ) are atoms in Y ′ . Indeed, suppose y ∈ Y � satisfies 0 ≤ y ≤ e n for some n ∈ ℕ. Note that y is a limit of a sequence (y k ) ⊂ Y � such that each y k belongs to the sublattice generated by finitely many elements of the form (sup i∈A k Without loss of generality we can assume 0 ≤ y k ≤ e n for every k ∈ ℕ. For each k ∈ ℕ we can find a finite collection of pairwise disjoint B k 1 , … , B k q k ∈ A so that the subalgebra of A generated by {A k 1 , … , A k p k } coincides with the subalgebra generated by {B k 1 , … , B k q k }. Since the sequence (sup i∈B k j e i ) q k j=1 are pairwise disjoint it follows that y k belongs to the linear span of (sup i∈B k j e i ) q k j=1 . In this situation, the inequality 0 ≤ y k ≤ e n implies that y k = k e n for some 0 ≤ k ≤ 1. Taking the limit in k it follows that y = e n for some 0 ≤ ≤ 1. Hence, (e n ) n∈ℕ are atoms in Y ′ as claimed.
[3 ⇒ 1] Let e = sup n e n and let K be the sick compactum for which there is a vector lattice isomorphism T ∶ C(K) ⟶ Y e that takes the constant one function to e. Since e n is an atom of Y, it follows that T −1 e n is an atom of C(K). Atoms of a space of continuous functions are positive multiples of characteristic functions of isolated points, so T −1 e n = n 1 {p n } . Since sup n e n = e, we have that sup n T −1 e n is the constant one function. This implies that T −1 e n = 1 {p n } for all n. Now consider the function Ψ ∶ clopen(K) ⟶ P(ℕ) given by Ψ(B) = {n ∶ p n ∈ B}. We claim that Ψ is a Boolean isomorphism onto A. It is clear that Ψ is a Boolean homomorphism. It is one-to-one because the points {p n } are dense in K, as sup n 1 {p n } is the constant one function. The range of Ψ is A, because f = sup n∈A 1 {p n } in C(K) if and only if f is the characteristic function of {p n ∶ n ∈ A} and this set is clopen. ◻ It should be noted that we do not require in the definition of s-sick algebra that the compact space K is totally disconnected. The space K that appears in the proof of [3 ⇒ 1] above may not be totally disconnected. It seems like an elementary question, but we do not know if the Stone space of an s-sick algebra is a sick compactum.
Our next observation is, roughly speaking, that the elements of an s-sick algebra B that are hereditarily in B form a substantial part of B. We first introduce the following terminology. Given B, let us define which is an ideal of P(ℕ). For J ⊆ P(ℕ), set Proof Let {x n } be the associated sequence of atoms in a separable Banach lattice given by Theorem 6.2. We claim that which is countably generated by the sets A m = {n ∈ ℕ ∶ ‖x n ‖ > 1∕m}.
For the inclusion [⊆] suppose that inf n∈A ‖x n ‖ = 0. Then there exists an infinite set A ′ ⊂ A such that ∑ n∈A � ‖x n ‖ < +∞. But this means that A � ∈ J(B), and therefore A ∉ J(B) ⟂ .

For the inclusion [⊇]
, we must prove that if inf n∈A ‖x n ‖ = > 0 and B ∈ J(B) then C = A ∩ B is finite. If not, P(C) ⊆ B and for all D ⊆ C there exists x D = sup n∈D x n . If n ∈ D⧵D � , then x D − x D � ≥ x n . Therefore for any different D, D � ∈ P(C) we have ‖x D − x D � ‖ ≥ . Since there are uncountably many subsets of C, this contradicts that all those vectors belong to a separable E.
The same argument proves the last statement. ◻ As an application, we provide some partial results, that add to Corollary 5.3, supporting the conjecture that a product of two infinite compact spaces is sick only when both factors are metrizable. Proposition 6.4 If K × L is sick and has a dense countable set of isolated points, then both K and L have countably many clopen sets.
Proof Let fin ⊂ , ⊆ P(ℕ) be the algebras of clopens of K and L respectively, and let ⊗ ⊆ P(ℕ × ℕ) the product algebra, that represents the algebra of clopens of the product. Here recall that ⊗ is the algebra generated by the sets of the form A × B with A ∈ and B ∈ . If ⊗ is s-sick, then by Proposition 6.3, J( ⊗ ) ⟂ is countably generated. Note that We claim that the only way that the latter ideal can be countably generated is that J( ) ⟂ = J( ) ⟂ = P(ℕ). Indeed, assume for instance that there exists a set D ∉ J( ) ⟂ and X 1 , X 2 , … are generators of J( ⊗ ) ⟂ . For every n pick a different d n ∈ D such that (d n , n) ∉ X n . The set {(d n , n) ∶ n ∈ ℕ} ∈ J( ⊗ ) ⟂ but it is not contained in any X n , and this is a contradiction. By the last statement of Proposition 6.3 we conclude that and are countable. Every s-sick algebra B is analytic, and J(B) is a 1 2 set. We do not know if J(B) must actually be an analytic ideal. What we know is that not all analytic ideals may appear as J(B). One reason is that, if we think of the sequence {e n } in Theorem 6.2, all subsequences whose norms are summable belong to J(B) so, in the nontrivial cases, J(B) must contain summable ideals. Here is a concrete example: Example Let I be the ideal of subsets of ℚ 1 = ℚ ∩ [0, 1] with a finite number of accumulation points. Then I ≠ J(B) for any s-sick algebra B of subsets of ℚ 1 . In particular, the algebra made of I and the complements of I is not s-sick, and its Stone space is not a sick compactum.
Assume that I = J(B) and that B represents the subsequences of {x q ∶ q ∈ ℚ 1 } with a supremum as in Theorem 6.2. Since I ⟂ are just the finite sets, if follows from the proof of Proposition 6.3 that lim q ‖x q ‖ = 0. Enumerate all rational open intervals of (0, 1) as J 1 , J 2 , … , and choose by induction q n ∈ J n ∩ ℚ 1 , all different such that ‖x q n ‖ < 2 −n . Then ∑ n ‖x q n ‖ < +∞ but {q n ∶ n ∈ ℕ} ∉ J(B), a contradiction.

Non-pathological analytic P-ideals
Non-pathological analytic P-ideals were introduced by Farah [4] as those ideals that are of the form Exh( ) for some lower semicontinuous submeasure that is the supremum of those measures that it dominates. After decoding what all these words mean, one can rephrase the concept into a working definition like this: ∑ n∈A e n converges unconditionally}. When we look at a space with unconditional basis as a Banach lattice, the unconditional convergence of the series is the same as the existence of the supremum. This reminds us of what we studied in Sect. 6, with the difference that there we assumed that the global supremum sup n e n exists, which does not occur inside the span of the basis out of trivial cases. What we are going to do is just to add this supremum to the space from [2] to get the main result of the section: Theorem 7.2 If I is a non-pathological analytic P-ideal that contains the finite sets, then the Stone space of the algebra generated by I is a sick compactum.
Proof Observe that the algebra generated by I is of the form I ∪ F where F is the filter made of the complements of sets in I. Let C ⊆ c 00 ∩ B + 1 be the set such that I = I(C). Note that, since the finite sets belong to I, the ideal I(C) does not change if we add to C the sequences (1, 0, 0, 0, …), (0, 1∕2, 0, 0, 0, …), (0, 0, 1∕3, 0, 0, …) and so on. So we can suppose that n = sup{c n ∶ c ∈ C} > 0 for every n. We shall define now three Banach lattices X 0 , X 1 and X ∞ . In the terminology of [3], X 0 = EXH(Φ) and X ∞ = FIN(Φ), while X 1 will be the Banach lattice that we are looking for.
For a sequence x = (x n ) of real numbers, define the norm Proof Follows from Claims 1 and 2. ◻ It remains to show that K e (X 1 ) is homeomorphic to the Stone space of the algebra B = I ∪ F. This Stone space L is obtained by gluing together all ultrafilters of ℕ that contain F, and so C(L) ≃ {(x n ) ∈ ∞ ∶ lim F x n exists}. Note also that the ideal generated by e inside X 1 equals X 1 ∩ ∞ . So we just apply Claim 3.
◻ We refer to [2][3][4] for further information on examples and non-examples of nonpathological analytic P-ideals. We may just highlight two basic ones: If we fix a sequence ( n ) of positive numbers with lim n n = 0, and we take then we obtain the summable ideal I(C) = {A ⊆ ℕ ∶ ∑ n∈A n < ∞}. This is because On the other hand, if we take then I(C) is the ideal of statistically zero sets, because

Open questions
We have shown (see Sect. 3) that sick compacta always admit strictly positive regular measures of countable type which are analytic, and satisfy several other properties. However, the following is still open: Find an intrinsic characterization of sick compacta.
In connection with this we have a seemingly simpler question: If is a strongly countably determined measure on a sick compactum (that is, a G -point in P(K)), is an analytic measure?
We mentioned at the end of Sect. 2 that the sick compactum K(FBL(ℕ)) contains every sick compactum, which coincides with the structure space K(FBL(ℕ)) of the free Banach lattice FBL[ 1 ] generated by 1 (see Corollary 2.9 in [1]). With a bit more generality one can ask the following: