The countable sup property for lattices of continuous functions
aa r X i v : . [ m a t h . F A ] J un THE COUNTABLE SUP PROPERTY FOR LATTICESOF CONTINUOUS FUNCTIONS
M. KANDI ´C AND A. VAVPETI ˇC
Abstract.
In this paper we find sufficient and necessary condi-tions under which vector lattice C ( X ) and its sublattices C b ( X ), C ( X ) and C c ( X ) have the countable sup property. It turns outthat the countable sup property is tightly connected to the count-able chain condition of the underlying topological space X . Wealso consider the countable sup property of C ( X × Y ). Even whenboth C ( X ) and C ( Y ) have the countable sup property it is pos-sible that C ( X × Y ) fails to have it. For this construction oneneeds to assume the continuum hypothesis. In general, we presenta positive result in this direction and also address the questionwhen C ( Q λ ∈ Λ X λ ) has the countable sup property. Our resultscan be understood as vector lattice theoretical versions of resultsregarding products of spaces satisfying the countable chain condi-tion. We also present new results for general vector lattices thatare of an independent interest. Introduction
In topology, when one deals with continuous functions, there aretwo possibilities. One can work either by open sets or by nets (general-ized sequences). Although dealing with nets is maybe computationallyeasier, one needs to be cautious since there is a variety of differenttypes of nets and one can often make errors. When the topology of agiven space is metrizable (more general first-countable), the sequential
Mathematics Subject Classification.
Key words and phrases. vector lattices, continuous functions, countable supproperty, chain conditions, strictly positive functionals.The first author acknowledges the financial support from the Slovenian ResearchAgency (research core funding No. P1-0222). The second author acknowledges thefinancial support from the Slovenian Research Agency (research core funding No.P1-0292 and J1-7025). nature of the space enables us to work by sequences instead of nets.Since order convergence in vector lattices is also defined through nets,one would also like to pass from nets to sequences, of course, if possible.In the setting of vector lattices this notion is called the countable supproperty . It plays an important role in the recent research in vectorand Banach lattices. For example, in [AT17] it was used to prove thatevery function in C ( R m ) is the order limit of an order convergent se-quence of piecewise affine functions. Next, in [GTX] authors used itto prove that in some Banach function spaces over σ -finite measurespaces convex Koml´os sets are norm bounded. Last but not least, in[LC] authors proved that the universal completion of a vector latticewith a weak unit and with the countable sup property also has thecountable sup property. With this result they proved uo-completenessof the universal completion of some vector lattices (see [LC, Theo-rem 2.10]). Recall that a vector lattice E is said to be uo-completewhenever every uo-Cauchy net in E uo-converges in E . These daysuo-convergence plays a very important role in the research of vectorlattices. Although uo-convergence is very exciting on its own, its valueshows through its applications in Mathematical finance. For generalresults on uo-convergence and its unbounded norm version we refer thereader to [GX14, Gao14, KMT17, GTX, LC]. For applications of uo-convergence and its techniques to Mathematical finance we refer thereader to [GLX, GLX2].In this paper our interest is the countable sup property itself. Al-though we present some new general results which also extend someresults from [AT17] and [LC], our main concern is the countable supproperty for vector lattices of continuous functions on a given topologi-cal space. The paper is structured as follows. In Section 2 we introducenotation and basic notions needed throughout the text. In Section 3 weintroduce different chain conditions on a topological space X and provethat the existence of a strictly positive functional on C b ( X ) implies theweakest of them. In Section 4 we connect the countable sup property of C ( X ) to chain conditions from Section 3. It turns out that the count-able chain condition of a topological space X implies that C ( X ) has thecountable sup property and that, in general, they are not equivalent. HE COUNTABLE SUP PROPERTY 3
However, they are equivalent when X ∈ T . Also, for a metric space X , the countable sup property of C ( X ) is equivalent to separability of X . Along the way we extend two results from [AT17] and [LC]. InSection 5 we prove that vector lattices C c ( X ) and C ( X ) simultane-ously have the countable sup property or simultaneously fail to haveit. In the last section we consider the vector lattice C ( X × Y ). It ispossible for both C ( X ) and C ( Y ) to have the countable sup propertywhile C ( X × Y ) lacks it. This follows under continuum hypothesis fromGalvin’s example [Gal80] and Proposition 4.7. This example also leadsto an example of an extremally disconnected compact Hausdorff space X such that C ( X ) has the countable sup property while C ( X × X )lacks it (see e.g. [Roy89]). We also prove that whenever C ( X ) hasthe countable sup property and Y is separable, then C ( X × Y ) hasthe countable sup property. This result can be considered as a vectorlattice version of [Wis69, Theorem 3.3]. Last but not least, we alsoprove that C ( Q λ ∈ Λ X λ ) has the countable sup property whenever foreach finite family Λ ⊆ Λ the space C ( Q λ ∈ Λ X λ ) has the countablesup property. Again, this can be considered as a vector lattice versionof [Roy89, Theorem 2.2]. 2. Preliminaries
Throughout the paper, if not otherwise stated, vector lattices areassumed to be Archimedean. A vector x of a vector lattice E is said tobe positive if x ≥ . The set of all positive vectors of E is denoted by E + . A vector e ∈ E + is said to be a unit if for every x ∈ E there is some λ ≥ | x | ≤ λe . A positive vector e ∈ E is said to be a weakunit if | x | ∧ e = 0 implies x = 0. A vector sublattice F is order dense in E if for each x ∈ E + there is y ∈ F + satisfying 0 < y ≤ x. If for each x ∈ E there is some y ∈ F with x ≤ y , then F is said to be majorizing in E . Order dense sublattices are always regular , i.e., if x α ↓ F ,then x α ↓ E . It is easy to see that properties of being an orderdense, majorizing or a regular sublattice are transitive relations. Vectorlattice E is said to satisfy the countable sup property wheneverevery nonempty subset possessing a supremum contains a countablesubset possessing the same supremum. The countable sup property is M. KANDI ´C AND A. VAVPETI ˇC equivalent to the following fact: for each net ( x α ) in E that satisfies0 ≤ x α ↑ x there is an increasing sequence α n such that 0 ≤ x α n ↑ x (see e.g. [LZ71, Theorem 23.2]). If F is a regular sublattice of a vectorlattice with the countable sup property, it is easy to see that F has thecountable sup property as well. A positive functional ϕ on E is saidto be strictly positive if ϕ ( x ) = 0 for some vector x ∈ E + implies x = 0. If there exists a strictly positive functional on E , then E hasthe countable sup property (see e.g. [AB03, Theorem 1.45]).Let E be a vector lattice and let E δ be its order (Dedekind) com-pletion. Then E is an order dense and majorizing sublattice in E δ ,moreover these properties characterize E δ . If F is a regular sublatticeof E , then F δ is a regular sublattice of E δ (see e.g. [GTX, Theorem2.10]). For the unexplained terminology about vector lattices we referthe reader to [AB03, AB06].Since in this paper we will be mainly concerned with the vectorlattice C ( X ), we recall the most needed topological facts. A space X is said to satisfy T -separation axiom if X is Hausdorff and for eachclosed set F in X and x / ∈ F there is a continuous function f on X with f ( x ) = 1 and f ≡ F . We write X ∈ T for short when X satisfies T -separation axiom. For a function f on X , the set Z ( f ) ofzeros of f is called the zero set of f . The closure of X \ Z ( f ) is calledthe support of f and is denoted by supp f. A subset U of X is said tobe a cozero set whenever there exists a function f ∈ C ( X ) such that Z ( f ) = X \ U . The constant 1 function is denoted by . The order on C ( X ) is defined pointwise. Clearly is a unit in C b ( X ) while it is onlya weak unit in C ( X ). Also, when X is not compact, then / ∈ C ( X ).3. Chain conditions in topological spaces
Countability in Topology has a special role. One of the first prop-erties one encounters is separability of a topological space X , i.e.,the existence of a countable dense set in X . Although the notion ofseparability is very natural and intuitive, in Functional analysis thereis an abundance of nonseparable Banach spaces which are of great im-portance. As an illustration, consider a compact Hausdorff space K .It is well-known that C ( K ) is separable iff K is metrizable (see e.g. HE COUNTABLE SUP PROPERTY 5 [MN91, Proposition 2.1.8]). For real or abstract analysis sometimesthe full force of separability is not needed. In those cases, one can de-velop richer theory under relaxed assumptions however through moreinvolved and deeper proofs. One can relax the notion of separabil-ity through the so-called “chain conditions” imposed on a topologicalspace X which are introduced below.A topological space X is said to satisfy • Shanin’s condition if every point-countable family of opensets is countable, i.e. every point is a member of only countablemany members of a family, • calibre ℵ if every uncountable family of nonempty open setscontains an uncountable subfamily with a nonempty intersec-tion, • Knaster’s condition or property K if every uncountableset of nonempty open sets contains an uncountable subset inwhich no two elements are disjoint, • countable chain condition if every family of pairwise dis-joint open subsets of X is countable.We will write for short that X is CCC whenever X satisfies thecountable chain condition. In general, we have the following chain ofimplications regarding chain conditions.separability ⇒ Shanin’s condition ⇒ calibre ℵ ⇒ property K ⇒ CCCand none of them can be reversed. When X is metrizable, CCC impliesseparability (see e.g. [VR59, Theorem 1]). Hence, in the metrizablecase, all conditions that were introduced above are equivalent.Countable chain condition is also called Suslin’s condition and itoriginates from the so-called
Suslin’s problem about totally orderedsets posed by Suslin [Sus20]. Suslin asked whether there exists a totallyordered set R which is not order isomorphic to the real line R andsatisfies the following properties:(1) R does not have a least nor a greatest element;(2) between any two elements there is another;(3) every nonempty bounded set has a supremum and an infimum; M. KANDI ´C AND A. VAVPETI ˇC (4) every collection of pairwise disjoint nonempty open intervals in R is countable.Totally ordered set which satisfies (1)-(4) and is not order isomorphicto R is called a Suslin line . Suslin’s hypothesis says that there areno Suslin lines. Solovay and Tennenbaum [ST71] proved that Suslin’shypothesis is undecidable in ZFC. However, Suslin’s hypothesis is true[ST71] if one assumes the negation of the continuum hypothesis and
Martin’s axiom [MS70] which says that no compact Hausdorff spaceis a union of fewer than 2 ℵ nowhere dense sets. Regarding Suslin’shypothesis we refer the reader also to [Jech67] and [Ten68].So far we have seen that separability condition and CCC are thestrongest and the weakest chain condition we introduced so far, re-spectively. The CCC condition can be relaxed as follows. The space X is said to satisfy cozero CCC or CCC for cozero sets if anycollection of pairwise disjoint cozero sets is countable. Obviously CCCimplies CCC for cozero sets. We will see later that a space X whichsatisfies CCC for cozero sets does not necessary satisfy CCC (see Propo-sition 3.1 and Example 3.2).In general, even compact Hausdorff spaces do not satisfy CCC. Theexample of such space can be found in [SS70, p. 116-117]. On theother hand compact groups satisfy Shanin’s condition which is strongerthan CCC. This follows from the fact that every Cantor cube satis-fies Shanin’s condition and from the remarkable Ivanovski˘ı–Vilenkin–Kuzminov theorem which says that compact groups are continuousimages of Cantor cubes. More generally, Tkaˇcenko proved that ev-ery σ -compact group satisfies CCC. For details see [Tod97]. Compactgroups share another feature which is of more use to us within theframework of this paper. If G is a compact Hausdorff group, then thenormalized Haar measure λ on G induces a strictly positive functional f R G f dλ on C ( G ). In general, a compact Hausdorff space K doesnot admit strictly positive finite Borel measures. When they do, thensuch measures induce strictly positive functionals on C ( K ). Proposition 3.1.
If there exists a strictly positive functional on C b ( X ) ,then X satisfies CCC for cozero sets. HE COUNTABLE SUP PROPERTY 7
Proof.
Let ϕ be a strictly positive functional on C b ( X ) and let ( U λ ) bea family of pairwise disjoint cozero sets. Then for each λ there exists afunction f λ : X → [0 ,
1] such that f λ is nonzero precisely on U λ . Hence,if f is a finite sum of some functions in ( f λ ), then 0 ≤ f ≤ . For each n ∈ N define A n = { f λ : ϕ ( f λ ) ≥ n } . Suppose A m is infinite for some m ∈ N . Pick k · m distinct functions in A m and denote their sum by f . Then 0 ≤ f ≤ implies ϕ ( ) ≥ ϕ ( f ) ≥ k. This contradicts thefact that ϕ is a functional. Hence, each set A n is finite from where itfollows that the family ( f λ ) is countable. (cid:3) By applying Proposition 4.7 we conclude that although CCC or CCCfor cozero sets are weaker notions than separability, yet they are stillstrong enough to provide the right tool for order analysis on the vectorlattice C b ( X ).The following example shows that there exists a Hausdorff space X such that C b ( X ) admits a strictly positive functional; yet X does notsatisfy CCC. This example shows that CCC for cozero sets is weakerthan CCC. Example 3.2.
Let us define the topology τ on R as follows: a point( x, y ) ∈ ( R × Q ) ∪ ( Q × R ) has Euclidean neighborhoods, but a point( x, y ) ∈ ( R \ Q ) has { (( x − r, x + r ) ∩ ( R \ Q )) × { y } : r > } for its fundamental system. The resulting space is denoted by X .Because τ is stronger than Euclidean topology, continuous functionson X separate the points. In particular, X is Hausdorff. Because { R \ Q × { y } : y ∈ R \ Q } is an uncountable collection of disjoint opensets, the space X is not CCC.We claim that if f : X → R is a nonzero τ -continuous function, then f is nonzero on some Euclidean ball. First, we may assume f ≥ f is nonzero at a point ( e x, e y ) ∈ X and denote c := f ( e x, e y ) > e x, e y ) ∈ ( R × Q ) ∪ ( Q × R ), then by continuity f is nonzero onsome Euclidean neighborhood of ( e x, e y ). Suppose now e x and e y are bothirrational. Again, by continuity of f we can find δ > x ∈ ( e x − δ, e x + δ ) ∩ ( R \ Q ) we have f ( x, e y ) ≥ c . Pick x ⋆ ∈ M. KANDI ´C AND A. VAVPETI ˇC ( e x − δ, e x + δ ) ∩ Q . Since ( x ⋆ , e y ) has Euclidean neighborhoods, continuityof f implies that there is δ ⋆ > x, y ) ∈ K (( x ⋆ , e y ) , δ ⋆ )we have f ( x ⋆ , e y ) ≥ c . This proves the claim.We claim that there exists a strictly positive functional on C b ( X ).Let ( q n ) be one of the enumerations of the countable set Q . Then thewell-defined mapping ϕ : C b ( X ) → R given by ϕ ( f ) = P ∞ n =1 12 n f ( q n )defines a positive functional on C b ( X ). If f ∈ C b ( X ) is nonzero, then f ( q n ) > q n ∈ Q . Hence, ϕ ( f ) ≥ n f ( q n ) >
0, so that ϕ isstrictly positive.Although CCC for cozero sets is weaker than CCC, by Theorem 4.8they coincide when X ∈ T . Similar argument as in Example 3.2shows that C b ( X ) admits strictly positive functionals when X is sepa-rable. 4. The countable sup property
As we already observed, if F is a regular sublattice of E and E has the countable sup property, then F itself has the countable supproperty. If F is not regular in E , then directly from the definitionit is not clear whether F still has the countable sup property since0 ≤ x α ↑ x in F does not necessary imply 0 ≤ x α ↑ x in E . However,there is a way around through the following theorem whose proof canbe found in [LZ71, Theorem 29.3]. Recall that in Section 2 we assumedthat all vector lattices are Archimedean. Theorem 4.1.
A vector lattice E has the countable sup property iffevery disjoint system of positive vectors which is bounded from aboveis countable. Now it easy to see that sublattices of vector lattices with the count-able sup property have the countable sup property. The following ques-tion appears naturally.
Question 4.2.
Suppose F is a vector sublattice of a vector lattice E and suppose F has the countable sup property. Does E have thecountable sup property? HE COUNTABLE SUP PROPERTY 9
The following example reveals that the answer can be negative evenif F is an order dense ideal in E . Example 4.3.
Let Ω be an uncountable set and ℓ ∞ (Ω) be the vectorlattice of all bounded functions on Ω. Let E be the vector lattice ofall functions in ℓ ∞ (Ω) with a countable support. Then E is an orderdense ideal in ℓ ∞ (Ω), E has the countable sup property while ℓ ∞ (Ω)does not have it.Note that in Example 4.3 the vector lattice E has the countablesup property and contains an uncountable disjoint system of positivevectors { χ ω : ω ∈ Ω } . This is impossible when a given vector latticepossesses weak units. Proposition 4.4.
A vector lattice E with a weak unit has the countablesup property iff every disjoint system of positive vectors is countable.Proof. Suppose E has the countable sup property. Let e be a weak unitin E . Choose a disjoint system ( x λ ) λ ∈ Λ of positive vectors in E . Then( x λ ∧ e ) λ ∈ Λ is an order bounded disjoint system of positive vectors in E as well. Since E has the countable sup property, the system ( x λ ∧ e ) λ ∈ Λ is countable. Hence, x λ ∧ e = 0 for all but countably many λ ∈ Λ . Since e is a weak unit, we have x λ = 0 for all but countably many λ ∈ Λ . The converse implication is clear by Theorem 4.1. (cid:3)
Now we are able to provide some positive answers to Question 4.2.Suppose E is a vector lattice which contains a sublattice F with thecountable sup property. Then E has the countable sup property in thefollowing cases: • E = F δ is the order completion of F ; • F has a weak unit and E = F u is the universal completion of F .For the proofs of these facts we refer the reader to [LZ71, Theorem32.9] and [LC, Lemma 2.9], respectively. In the following theorem weextend the results above to a more general setting. Theorem 4.5.
Let F be an order dense vector sublattice of a vectorlattice E . If F has the countable sup property, then E has the countablesup property in the following cases: (a) F is majorizing in E . (b) F has a weak unit.Proof. (a) Since F is regular in E , we conclude that F δ is regular in E δ . Also, order density of F in E implies that F δ is order dense in E δ . Since F δ is order complete, by [AB03, Theorem 1.40] F δ is anideal in E δ . Since F δ is also majorizing in E δ , we conclude E δ = F δ . Now we apply the fact that the countable sup property is preservedunder passing to order completions. Hence, E has the countable supproperty.(b) Let u be a weak unit in F . Then order density of F in E impliesthat u is also a weak unit in both E and E δ .Suppose first that F is an ideal in E . Let ( x λ ) λ ∈ Λ be a disjoint systemof positive vectors in E . Then ( x λ ∧ u ) λ ∈ Λ is a disjoint system of positivevectors in F . Since F has the countable sup property, x λ ∧ u = 0 forall but countably many λ ∈ Λ. Since ( x λ ∧ nu ) ↑ n x λ in E , we have x λ = 0 for all but countably many λ ∈ Λ. Proposition 4.4 implies that E has the countable sup property.For the general case, consider order completions F δ and E δ of F and E , respectively. As in (a) we argue that F δ is an order dense ideal in E δ . Since the countable sup property is preserved under order completions,the first part of the proof implies that E δ has the countable sup prop-erty. Now, E as a sublattice of E δ also has the countable sup property.Finally we conclude that E has the countable sup property. (cid:3) It should be noted here that the vector lattice E in Example 4.3 doesnot have weak units and is not majorizing in ℓ ∞ (Ω).Suppose E is a vector lattice with the property that each disjointsystem of positive vectors in E is countable. In view of Proposition 4.4it is reasonable to suspect that E has a weak unit. However, the normedvector lattice c of all eventually zero sequences does not have weakunits, yet each disjoint system of positive vectors in c is countable.In Banach lattices this property implies the existence of weak units.We even have a stronger result. HE COUNTABLE SUP PROPERTY 11
Theorem 4.6.
Let E be a completely metrizable locally solid vectorlattice. Every disjoint system of positive vectors of E is countable iff E has a weak unit and the countable sup property.Proof. Assume that every disjoint system of positive vectors of E iscountable. By Theorem 4.1 it follows that E has the countable supproperty. To prove that E has a weak unit, pick a maximal disjointsystem F of nonzero positive vectors. Such systems exist by Zorn’slemma. By the assumption F is countable.Suppose first that F = ( f n ) n ∈ N . Since the topology of E is metriz-able, there exists a local basis ( U n ) of zero consisting of solid sets suchthat U n +1 ⊆ U n for each n ∈ N . We claim that there exists a local basis ( V n ) of zero consisting of solidsets such that V n +1 + V n +1 ⊆ V n for each n ∈ N . We first define V = U .Suppose that V n is already defined for some n ∈ N . Since addition iscontinuous in E , there exist solid neighborhoods S and S ′ of zero with S + S ′ ⊆ V n . Then V n +1 := S ∩ S ′ ∩ U n +1 satisfies V n +1 + V n +1 ⊆ V n . Since for each n ∈ N we have V n ⊆ U n , the family ( V n ) is a local basisfor zero.For each n ∈ N find λ n > f n ∈ λ n V n and define e n := f n λ n ∈ V n . For each n ∈ N we define s n = e + · · · + e n . We claim that( s n ) is Cauchy in E . To see this, pick a neighborhood V of zero. Thenthere is n ∈ N such that V n ⊆ V. If n > m ≥ n , then s n − s m = e m +1 + · · · + e n ∈ V n − ( n − m ) = V m ⊆ V n ⊆ V. Since E is complete the sequence ( s n ) converges to some element e .Since the cone of E is closed by [AB03, Theorem 2.21], we have 0 ≤ s n ≤ e for each n ∈ N .Suppose x ∧ e = 0. Then for each n ∈ N we have0 ≤ x ∧ e n ≤ x ∧ s n ≤ x ∧ e = 0 . Maximality of ( e n ) implies x = 0; hence e is a weak unit in E .If F is finite, then we define the vector e = P f ∈F f. Again, maxi-mality of F implies that e is a weak unit in E .The converse statement follows from Proposition 4.4. (cid:3) Although the countable sup property of a vector lattice C b ( X ) isa purely order theoretical notion, it can be connected to a particularchain condition of X . Proposition 4.7.
For a topological space X the following statementsare equivalent: (a) C ( X ) has the countable sup property. (b) C b ( X ) has the countable sup property. (c) X satisfies CCC for cozero sets.Proof. Since C b ( X ) is an order dense ideal in C ( X ), the equivalencebetween (a) and (b) follows from Theorem 4.5.(a) ⇒ (c) Let ( U λ ) be a family of disjoint cozero sets in X . For each λ there exists a nonnegative function f λ : X → [0 ,
1] such that f λ isnonzero precisely on U λ . Hence, ( f λ ) is a family of pairwise disjointfunctions in C ( X ) which is bounded by above by . Countable supproperty of C ( X ) implies that ( f λ ) is countable.(c) ⇒ (a) Suppose ( f λ ) is a family of nonnegative pairwise disjointfunctions. For each λ define U λ = f − λ ((0 , ∞ )). Then ( U λ ) is a familyof pairwise disjoint cozero sets in X . By the assumption this family iscountable. (cid:3) If X is CCC, then X satisfies CCC for cozero sets, so that by Propo-sition 4.7 the lattice C ( X ) has the countable sup property. The space X from Example 3.2 is an example of a space which is not CCC, yet C b ( X ) has the countable sup property. If X ∈ T , then X is CCC iff X satisfies CCC for cozero sets. Theorem 4.8.
For a topological space X ∈ T the following assertionsare equivalent. (a) X is CCC. (b) X satisfies CCC for cozero sets. (c) C ( X ) has the countable sup property.If C b ( X ) admits a strictly positive functional, then (a), (b) and (c)hold.Proof. That (a) implies (b) is obvious and that (b) implies (c) followsfrom Proposition 4.7. To see that (c) implies (a), pick a family ( U λ ) HE COUNTABLE SUP PROPERTY 13 of nonempty disjoint open sets in X . For each λ we find a nonzerofunction f λ : X → [0 ,
1] such that f λ ≡ X \ U λ . Then ( f λ ) is afamily of pairwise disjoint positive functions which are bounded by .The countable sup property of C ( X ) implies that ( f λ ) is countable. (cid:3) Since metrizable spaces are T , the following result immediatelyfollows from Theorem 4.8. Corollary 4.9.
A metrizable space X is separable iff C ( X ) has thecountable sup property. When the underlying topological space in Corollary 4.9 is a topolog-ical group, the countable sup property is connected to ℵ -boundednessof the group. Recall that a topological group G is said to be ℵ -bounded whenever for every open set U there is a countable set S such that G = U · S = [ s ∈ S U · s. Theorem 4.10.
Let G be a metrizable topological group. Then thefollowing assertions are equivalent. (a) G is separable. (b) G is CCC. (c) G is ℵ -bounded. (d) C ( G ) has the countable sup property.Proof. Since G is metrizable, (a) and (b) are equivalent. Also, metriz-ability of G implies G ∈ T , so that (b) and (d) are equivalent byTheorem 4.8. That (b) implies (c) follows from [Tka98, Proposition3.3].(c) ⇒ (b) Since G is metrizable, the unit element has a countablelocal basis ( U n ) consisting of symmetric sets. For each n ∈ N there is acountable set S n such that G = U n · S n . We claim that the countable set S := S ∞ n =1 S n is dense in G . To see this, pick an open set U in G and x ∈ U . Then there is n ∈ N such that U n x ⊆ U. Also, there are u ∈ U n and s ∈ S n such that x = us. Since s = u − us = u − x ∈ U n x ⊆ U , weconclude that S is dense in G . (cid:3) As a special case of Theorem 4.10 we consider normed spaces. Itis well-known that a normed space X is separable whenever its normdual X ∗ is separable. Applying Theorem 4.10 one sees that X is CCCwhenever X ∗ is CCC. Therefore, maybe the easiest example of a Ba-nach space which is not CCC is ℓ ∞ . This example also shows that CCCproperty is not closed under taking norm duals.Corollary 4.9 can be used as an argument why C ( R n ) has the count-able sup property. We actually do not require the full force of separabil-ity of R n . In [AT17, Theorem 5.4] authors argued as follows: the space R n can be exhausted by sets { mB n ∞ : m ∈ N } where B n ∞ denotes theclosed unit ball in R n with respect to k · k ∞ -norm. Since the Riemannintegral is a strictly positive functional on C ( mB n ∞ ) for each m ∈ N the space C ( mB n ∞ ) has the countable sup property. By exhausting R n with closed balls mB n ∞ they proved that C ( R n ) itself has the countablesup property. It should be noted here that C ( R n ) does not admit astrictly positive functional.We proceed to the extension of [AT17, Theorem 5.4]. Proposition 4.11.
If there exist sets ( A n ) in X such that C ( A n ) hasthe countable sup property and S ∞ n =1 A n is dense in X , then C ( X ) hasthe countable sup property.Proof. Suppose ( f λ ) λ ∈ Λ is a positive disjoint system in C ( X ). Then( f λ | A n ) λ ∈ Λ is a positive disjoint system in C ( A n ) which is countable bythe assumption. Hence, the set { λ ∈ Λ : f λ | A n = 0 for some n ∈ N } is countable. This implies that f λ = 0 on S ∞ n =1 A n for all but countablymany λ ∈ Λ . Since S ∞ n =1 A n is dense in X , we conclude f λ = 0 for allbut countably many λ ∈ Λ . (cid:3) Corollary 4.12.
Let A be a dense subspace of a topological space X .If C ( A ) has the countable sup property, then C ( X ) has the countablesup property. Suppose U is a dense set in X . It is easy to see that X is CCC iff U is CCC. Hence, when X ∈ T the lattice C ( X ) has the countable HE COUNTABLE SUP PROPERTY 15 sup property iff C ( U ) has the countable sup property. The followingexample shows that this does not hold when X / ∈ T . Example 4.13.
Let X be the topological space from Example 3.2.Since C b ( X ) has the countable sup property and X is not CCC, byTheorem 4.8 we conclude X / ∈ T . It is easy to see that the openset U = ( R \ Q ) is dense in X . Since for each λ ∈ R \ Q the set R \ Q × { λ } is both open and closed in X , the characteristic function f λ of R \ Q × { λ } is continuous. Hence, { f λ : λ ∈ R \ Q } is anuncountable positive disjoint system in C ( U ). This shows that C ( U )does not have the countable sup property.The following example shows that even when X ∈ T and C ( X )has the countable sup property, one can find a subset A ⊆ X such that C ( A ) does not have the countable sup property. Example 4.14.
Let R S be the Sorgenfrey line, i.e. the set R withthe topology defined by a basis { [ a, b ) : a, b ∈ R } . Since R S ∈ T ,we conclude that the product R S × R S ∈ T . Since Q is dense in R S × R S , the space R S × R S is separable and therefore it is CCC.Hence, C ( R S × R S ) has the countable sup property. On the otherhand, the set A := { ( x, − x ) : x ∈ R } is uncountable and discrete.This implies A is not CCC. Since the property T is hereditary, C ( A )does not have the countable sup property by Theorem 4.8.Although in Proposition 4.11 we exhausted the space X , we didnot exhaust the vector lattice C ( X ) since C ( A ) is not even a subsetof C ( X ). However, if one considers the case when X ∈ T the re-striction operator Φ A : f f | A induces a lattice isomorphism between C ( X ) / ker Φ A and an order dense sublattice of C ( A ) (see e.g. [KV,Proposition 3.5]). This remark and Proposition 4.11 lead us to the fol-lowing general “exhaustion-type” result for the countable sup propertyof general vector lattices. Proposition 4.15.
Let E be a vector lattice and suppose there existcountably many uniformly closed ideals J n in E such that E/J n has thecountable sup property for each n ∈ N . If T ∞ n =1 J n = { } , then E hasthe countable sup property. Proof.
Pick an order bounded disjoint system ( x λ ) λ ∈ Λ of positive vec-tors in E . Then ( x λ + J n ) λ ∈ Λ is an order bounded disjoint systemof positive vectors in E/J n . Since J n is uniformly complete, E/J n isArchimedean, so that Theorem 4.1 implies that ( x λ + J n ) λ ∈ Λ is count-able. Let us defineΛ ′ = { λ ∈ Λ : x λ + J n = 0 for some n ∈ N } . Then ( x λ ) λ ∈ Λ ′ is a countable disjoint system of positive vectors. If x µ ∧ x λ = 0 for each λ ∈ Λ ′ , then x µ ∈ J n for each n ∈ N , so that x µ ∈ T ∞ n =1 J n = { } . This implies that E has the countable sup property. (cid:3) Vector lattices C c ( X ) and C ( X )As was already observed in Proposition 4.7, a vector lattice C ( X ) hasthe countable sup property iff C b ( X ) has the countable sup property.This followed from the fact that C b ( X ) is an order dense ideal in C ( X )and that is a (weak) unit in C b ( X ). In this section we are interestedin finding necessary and sufficient conditions for the lattices C c ( X ) and C ( X ) to have the countable sup property in terms of properties of X .In order to establish the connection between CCC property of X andthe countable sup property of C ( X ), we first characterize weak unitsin C ( X ). Example 5.1.
Let Ω be a set equipped with the discrete topology.Then the vector lattice C (Ω) has weak units iff Ω is countable. Proposition 5.2.
Let X be a locally compact Hausdorff space. (a) A nonnegative function f ∈ C ( X ) is a weak unit in C ( X ) iff Int Z ( f ) = ∅ . (b) C ( X ) has a weak unit iff X contains an open dense σ -compactsubspace.Proof. (a) Since f is continuous, Z ( f ) is a closed subset of X . SupposeInt Z ( f ) = ∅ . If g ∧ f = 0, then g ≡ X \ Z ( f ). Since X \ Z ( f ) isdense in X and g is continuous, we have g = 0.Assume now that Int Z ( f ) is nonempty and pick x ∈ Int Z ( f ). ByUrysohn’s lemma for C ( X )-spaces, there exists a nonnegative function HE COUNTABLE SUP PROPERTY 17 g ∈ C ( X ) with g ( x ) = 1 and g ≡ X \ Int Z ( f ). Then f ∧ g = 0and since g = 0 we conclude that f is not a weak unit in C ( X ).(b) If f ∈ C ( X ) is a weak unit, then Int Z ( f ) = ∅ . Since for each ǫ > { x ∈ X : f ( x ) ≥ ǫ } is compact in X , the dense set X \ Z ( f ) = ∞ [ n =1 { x ∈ X : f ( x ) ≥ n } is σ -compact.For the converse, assume U ⊆ X is a σ -compact open dense subspacein X and define F = X \ U .Since U is open in X , it is locally compact and since it is σ -compact,[Dug66, Theorem X.7.2] implies that there exists an increasing sequenceof relatively compact open sets ( U n ) in U such that U n ⊆ U n +1 . ByUrysohn’s lemma for C ( X )-spaces there exists a function f n : X → [0 ,
1] with compact support such that f n ≡ U n and f ≡ X \ U n +1 . The function f := P ∞ n =1 f n n is strictly positive on U and zeroon F . By (a) we conclude that f is a weak unit in C ( X ). (cid:3) Remark 5.3. If X is not compact, then C c ( X ) does not have a weakunit. Indeed, if f would be a weak unit in C c ( X ), then the set X \ Z ( f )is an open σ -compact dense set in X . Since supp f is compact, weobtain that X = X \ Z ( f ) = supp f is a compact set.In the following example we construct a locally compact Hausdorffspace X which is not σ -compact, yet C ( X ) has a weak unit. This ex-ample shows that, in general, the existence of an open dense σ -compactset of a locally compact space does not imply that the whole space is σ -compact. Example 5.4.
Let X = R and for every x Q we pick a sequenceof rational numbers ( q n ( x )) n ∈ N with lim n →∞ q n ( x ) = x . Let τ be thetopology on X such that { x } is open for every x ∈ Q and { U n ( x ) = { x, q n ( x ) , q n +1 ( x ) , . . . } | n ∈ N } is a fundamental system for x Q . The space X is Hausdorff andlocally compact. Let A ⊂ X be such that A ∩ ( R \ Q ) = { x λ : λ ∈ Λ } is infinite.Then { U ( x λ ) : λ ∈ Λ } ∪ {{ x } : x ∈ A ∩ Q } is an open cover of A .Because U ( x λ ) is the only element of the cover containing x λ , thereis no finite subcover of A . Hence every compact subset of X containsonly finitely many irrational numbers. Therefore X is not σ -compactwhile Q ⊂ X is σ -compact, open and dense in X .Let ( r n ) be an enumeration of the rational numbers. Since the func-tion f = P ∞ n =1 12 n χ { r n } is continuous on X , strictly positive on Q andzero on R \ Q , by Proposition 5.2 f is a weak unit in C ( X ).In the rest of this section we consider when C ( X ) has the countablesup property. If X is CCC, then C ( X ) has the countable sup property.Therefore, C ( X ) as a sublattice of C ( X ), itself has the countable supproperty. The following is an example of a space X ∈ T which is notCCC while C ( X ) has the countable sup property. Example 5.5.
Let Ω be an arbitrary uncountable set equipped by thediscrete topology. Then Ω is a locally compact Hausdorff space whichis not CCC. Since each f ∈ C (Ω) has a countable support, C (Ω) hasthe countable sup property.It should be noted that C (Ω) from the preceding example does nothave a weak unit. Proposition 5.6.
Let X be a locally compact Hausdorff space. If thereis an open dense σ -compact set U in X and C ( X ) has the countablesup property, then X is CCC.Proof. Pick a disjoint family ( U λ ) of open sets of X . By Urysohn’slemma for each λ there exists a nonzero nonnegative function f λ ∈ C ( X ) with supp f λ ⊆ U λ . Hence, ( f λ ) is a disjoint system of nonneg-ative functions in C ( X ). Since C ( X ) has a weak unit by Proposi-tion 5.2, Proposition 4.4 implies that ( f λ ) is countable. Hence ( U λ ) iscountable and X is CCC. (cid:3) In particular, when X is locally compact σ -compact Hausdorff space,then X is CCC iff C ( X ) has the countable sup property. If C ( X ) hasthe countable sup property, then C c ( X ) as a sublattice of C ( X ) hasthe property itself. It goes the other way around when X ∈ T : HE COUNTABLE SUP PROPERTY 19
Theorem 5.7.
For X ∈ T the following assertions are equivalent. (a) C c ( X ) has the countable sup property. (b) C ( X ) has the countable sup property. (c) For each compact set K ⊆ X there exists a relatively compactcozero set U such that K ⊆ U and C ( U ) has the countable supproperty. (d) X can be covered by relatively compact cozero sets which satisfyCCC.Proof. (a) ⇒ (b) Pick a disjoint system ( f λ ) of functions in C ( X ) suchthat 0 ≤ f λ ≤ f for some f ∈ C ( X ) and each λ. For every n ∈ N we define K n = f − ([ n , ∞ )), U n = f − (( n , ∞ )) and Λ n = { λ ∈ Λ : f − λ ((0 , ∞ )) ∩ K n = ∅} . Since f n ∈ C ( X ), sets K n and U n are compactand open in X , respectively. Let us fix n ∈ N . Since K n ⊂ U n +1 , forevery x ∈ K n there exists ϕ x : X → [0 ,
1] with compact support suchthat ϕ x ( x ) = 1 and ϕ x ( X \ U n +1 ) = 0. Hence, { ϕ − x ((0 , x ∈ K n } is an open cover of K n . By compactness of K n there exists a finite set F ⊂ K n such that K n ⊂ S x ∈ F ϕ − x ((0 , ϕ := P x ∈ F ϕ x satisfies ϕ ( x ) > x ∈ K n and ϕ ( X \ U n +1 ) = { } . Since U n +1 is relatively compact, we conclude ϕ ∈ C c ( X ). Then { f λ · ϕ : λ ∈ Λ n } is a disjoint family of nonnegative functions satisfying0 ≤ f λ · ϕ ≤ f · ϕ ∈ C c ( X ). By the assumption the set Λ n is countableand since Λ = S ∞ n =1 Λ n we finally conclude that Λ is countable. That C ( X ) has the countable sup property follows from Theorem 4.1.(a) ⇒ (c) Pick a compact set K and find a nonnegative function ϕ ∈ C c ( X ) with ϕ ≡ K . The set U = { x ∈ X : ϕ ( x ) > } is a cozeroset with U = supp ϕ compact.We claim that C ( U ) has the countable sup property. To prove this,pick a disjoint system ( f λ ) of nonnegative functions in C ( U ) which isbounded from above by f ∈ C ( U ). For each λ we define a function g λ : X → R by g λ ( x ) = ( ϕ ( x ) f λ ( x ) : x ∈ U x ∈ X \ U . If x ∈ U \ U , then ϕ ( x ) = 0. Hence, the restrictions of g λ to U and X \ U , respectively, agree. Since they are continuous and { U , X \ U } isa closed cover of X , we conclude g λ ∈ C c ( X ). By a similar argumentwe see that g : X → R defined as g ( x ) = ( ϕ ( x ) f ( x ) : x ∈ U x ∈ X \ U is also in C c ( X ). By the assumption the family ( g λ ) is countable. Fromcontinuity of functions f λ and from ϕ > U we conclude that ( f λ )is also countable.(c) ⇒ (d) Pick x ∈ X and find a relatively compact cozero set U suchthat x ∈ U and C ( U ) has the countable sup property. By Theorem 4.8the space U satisfies CCC and since U is dense in U , U itself satisfiesCCC.(d) ⇒ (a) Pick a positive disjoint system ( f λ ) in C c ( X ) such that0 ≤ f λ ≤ f for some f ∈ C c ( X ) and each λ. Since the support supp f of f is compact, there exist cozero sets U , . . . , U n which satisfy CCCand supp f ⊆ U ∪ · · · ∪ U n . Since the set U := U ∪ · · · ∪ U n also satisfiesCCC, the family ( f λ | U ) of restrictions to U is countable. To finish theproof note that each function f λ is zero on X \ U . (cid:3) Here we would like to point out two things. First, if X is an un-countable discrete space, then C ( X ) has the countable sup propertywhile C ( X ) does not. Second, when X ∈ T , then C ( X ) is the normcompletion with respect to the the uniform topology of its order ideal C c ( X ). This leads us to the following question. Question 5.8.
Suppose E is a locally solid vector lattice with thecountable sup property. Does the topological completion b E of E alsohave the countable sup property?The answer is yes when E is metrizable and sits in b E as an orderideal. Theorem 5.9.
Let E be a metrizable locally solid vector lattice whichis an ideal in its topological completion b E . Then E has the countablesup property iff b E has the countable sup property. HE COUNTABLE SUP PROPERTY 21
Proof.
Assume that E has the countable sup property, pick a nonzerovector 0 ≤ x ∈ b E and a positive disjoint system ( x λ ) λ ∈ Λ which satisfies0 ≤ x λ ≤ x for each λ. Find a sequence ( y n ) in E with y n → x in b E .Since the lattice operations are continuous, by successively replacing y n first with y + n and then with y n ∧ x we may assume that 0 ≤ y n ≤ x for each n ∈ N . Since 0 ≤ y n ∧ x λ ≤ y n ∧ x in E , the disjoint system( y n ∧ x λ ) λ ∈ Λ is countable in E for each n ∈ N . Hence, the set { ( n, λ ) ∈ N × Λ : y n ∧ x λ = 0 } is countable. Since y n → x , for each λ ∈ Λ there is at least one n ( λ ) ∈ N such that y n ( λ ) ∧ x λ = 0, so that Λ is countable. By Theorem 4.1 weconclude that b E has the countable sup property. (cid:3) Example 5.10.
Let Λ be an uncountable set equipped with the dis-crete topology. We equip the vector lattice C c (Λ) by the topology ofpointwise convergence which is not metrizable since Λ is uncountable.The topological completion of C c (Λ) is the vector lattice R Λ of all realvalued functions on Λ which does not have the countable sup propertywhile C c (Λ) itself has the property.6. The countable sup property of the lattice C ( X × Y )The product of two CCC spaces is not necessary a CCC space. Wewill see that the product of two vector lattices E and E with thecountable sup property has the same property with respect to bothnatural orders on E × E . More interesting question is if a vectorlattice C ( X × Y ) has the countable sup property, provided the lattices C ( X ) and C ( Y ) have the property.Let E and E be vector lattices. On the product E × E of the sets E and E we can naturally define two orders: • ( x , y ) ≤ ( x , y ) iff x ≤ x and y ≤ y , • ( x , y ) (cid:22) ( x , y ) iff x < x or ( x = x and y ≤ y ).The second one is lexicographical order . Proposition 6.1. If E and E have the countable sup property thenboth lattices ( E × E , ≤ ) and ( E × E , (cid:22) ) have the countable supproperty. Proof.
Let 0 ≤ ( x α , y α ) ↑ ( x, y ) in ( E × E , ≤ ). Then 0 ≤ x α ↑ x in E and 0 ≤ y α ↑ y in E . Since E and E satisfy the countablesup property there exist increasing sequences ( α n ) and ( β n ) such that0 ≤ x α n ↑ x in E and 0 ≤ y β n ↑ y in E . Choose arbitrary γ > α , β . Inductively, for each n ∈ N one can find γ n +1 > α n +1 , β n +1 , γ n . Then γ n is an increasing sequence, 0 ≤ x γ n ↑ x in E , and 0 ≤ y γ n ↑ y in E .This yields 0 ≤ ( x γ n , y γ n ) ↑ ( x, y ) in ( E × E , ≤ ).Let ( x α , y α ) α ∈ Λ be a net such that 0 ≤ ( x α , y α ) α ∈ Λ ↑ ( x, y ) in ( E × E , (cid:22) ). If there is no α ∈ Λ such that x α = x , then ( x α , y α ) < ( x, z )for all α and all z ∈ E . Therefore y is the first element in E whichis a contradiction. Hence, the set e Λ = { α ∈ Λ : x α = x } is nonempty.From 0 ≤ ( x, y α ) α ∈ e Λ ↑ ( x, y ) we conclude 0 ≤ y α ↑ α ∈ e Λ y . Since E hasthe countable sup property, there exists an increasing sequence ( α n ) in e Λ such that 0 ≤ y α n ↑ y in E . Finally we obtain 0 ≤ ( x α n , y α n ) ↑ ( x, y )in ( E × E , (cid:22) ). (cid:3) So the product of function spaces C ( X ) and C ( Y ) with the countablesup property has the same property in both mentioned cases. Thesituation with the product of the form C ( X × Y ) is more difficult.We do not know whether C ( X × Y ) has the countable sup propertywhenever C ( X ) and C ( Y ) have it. It seems one needs to assume somespecial axioms to answer the question. This is not so surprising sincethe countable sup property of C ( X ) is tightly connected to CCC ofthe space X . If we assume the Continuum Hypothesis, there exists acompact Hausdorff space X which is CCC but X × X is not (see e.g.[Roy89]). Since compact Hausdorff spaces are T , this space X byTheorem 4.8 serves us also as an example of a space such that C ( X )has the countable sup property while C ( X × X ) does not have it.If we sharpen an assumption on one of the factor X and Y , we havea positive answer. Theorem 6.2. If C ( X ) has the countable sup property and Y hascalibre ℵ , then C ( X × Y ) has the countable sup property.Proof. Suppose there exists an uncountable family { f λ : X × Y → [0 ,
1] : λ ∈ Λ } of pairwise disjoint functions in C ( X × Y ). For every λ thereexists an open set U λ × V λ ⊂ f − λ ((0 , Y has calibre ℵ there HE COUNTABLE SUP PROPERTY 23 exist an uncountable subset Λ ′ ⊂ Λ and y ∈ Y such that y ∈ T λ ∈ Λ ′ V λ .Let us define g λ ( x ) = f λ ( x, y ). Then { g λ : λ ∈ Λ ′ } is an uncount-able family of nonzero pairwise disjoint functions in C ( X ) which is acontradiction. Hence C ( X × Y ) has the countable sup property. (cid:3) Since every separable space has calibre ℵ , the following corollarycan be seen as a vector lattice version of [Wis69, Theorem 3.3]. Corollary 6.3. If C ( X ) has the countable sup property and Y is sep-arable, then C ( X × Y ) has the countable sup property. It is known that the product of a CCC space and a space satisfyingKnaster’s condition is a CCC space. In particular, if X is CCC and Y satisfies Knaster’s condition, then C ( X × Y ) has the countable supproperty. However, we do not know if C ( X × Y ) has the countable supproperty provided C ( X ) has the countable sup property and Y satisfiesKnaster’s condition.The following theorem is a vector lattice version of [Roy89, Theorem2.2]. Theorem 6.4.
Let ( X λ ) λ ∈ Λ be a family of topological spaces such thatfor each finite family Λ ⊆ Λ the space C ( Q λ ∈ Λ X λ ) has the countablesup property, then C ( Q λ ∈ Λ X λ ) has the countable sup property.Proof. Suppose there exists an uncountable family of pairwise disjointfunctions { f α : Q λ ∈ Λ X λ → [0 ,
1] : α ∈ A } in C ( Q λ ∈ Λ X λ ). Forevery α ∈ A there is a basis set U α ⊂ f − α ((0 , α ⊂ Λ and for every λ ∈ Λ α there exists an open set V αλ ⊂ X λ such that U α = T λ ∈ Λ α p − λ ( V αλ ), where p λ : Q µ ∈ Λ X µ → X λ isthe projection. By delta system lemma [Roy89, p. 174] there exists anuncountable subset e A ⊆ A and a set Λ ⊂ Λ such that Λ α ∩ Λ β = Λ for all α, β ∈ e A for which Λ α = Λ β . If Λ α ∩ Λ β = ∅ for α, β ∈ e A the functions f α and f β are not disjoint, hence Λ = ∅ . For every λ Λ there is at most one α ∈ e A such that λ ∈ Λ α . Pick anelement x λ ∈ V αλ if there is α ∈ e A such that λ ∈ Λ α otherwise pickany x λ ∈ X λ . Let i : Q λ ∈ Λ X λ → Q λ ∈ Λ X λ be the slice embeddingsuch that p λ i ( x ) = x λ for every λ Λ . Then f α ◦ i is nonzero for every α ∈ e A and { f α ◦ i : α ∈ e A } is an uncountable family of pairwisedisjoint functions in C ( Q λ ∈ Λ X λ ) which is a contradiction. (cid:3) Corollary 6.5.
Let ( X λ ) λ ∈ Λ be a family such that C ( X λ ) has thecountable sup property for some λ ∈ Λ and X λ has calibre ℵ for all λ ∈ Λ \ { λ } . Then C ( Q λ ∈ Λ X λ ) has the countable sup property.Proof. Pick a finite subset Λ ⊆ Λ. If λ Λ , then Q λ ∈ Λ X λ has calibre ℵ , hence C ( Q λ ∈ Λ X λ ) has the countable sup property. If λ ∈ Λ , byinduction and Theorem 6.2 the space C ( Q λ ∈ Λ X λ ) has the countable supproperty. To finish the proof we apply Theorem 6.4. (cid:3) Of course we can replace the property “calibre ℵ ” with strongerproperty “separability” in the above Corollary to get the same result. References [AB03] C.D. Aliprantis and O. Burkinshaw,
Locally solid Riesz spaces withapplications to economics , 2nd ed., AMS, Providence, RI, 2003.[AB06] C. D. Aliprantis and O. Burkinshaw,
Positive operators , Springer, Dor-drecht, 2006, Reprint of the 1985 original.[AT17] S. Adeeb, V. G. Troitsky,
Locally piecewise affine functions and theirorder structure , Positivity (2017), 213–221.[Dug66] J. Dugundji, Topology , Allyn and Bacon, Inc., Boston, Mass. 1966.[Gal80] F. Galvin,
Chain conditions and products , Fund. Math. (1980),33–48.[Gao14] N. Gao, Unbounded order convergence in dual spaces,
J. Math. Anal.Appl. , 419, 2014, 347–354.[GLX] N. Gao, D.H. Leung, and F. Xanthos, Dual representation problem ofrisk measures. arXiv:1610.08806 [q-fin.MF][GLX2] N. Gao, D.H. Leung, and F. Xanthos,
Duality for unbounded orderconvergence and applications , arXiv:1705.06143 [math.FA][GTX] N. Gao, V.G. Troitsky, and F. Xanthos,
Uo-convergence and its ap-plications to Ces`aro means in Banach lattices , Isr. J. Math. (2017).doi:10.1007/s11856-017-1530-y[GX14] N. Gao and F. Xanthos, Unbounded order convergence and applicationto martingales without probability,
J. Math. Anal. Appl. , 415 (2014),931–947.[Jech67] T. Jech,
Non-provability of Souslin’s hypothesis , Comment. Math. Univ.Carolinae (1967), 291–305. HE COUNTABLE SUP PROPERTY 25 [KMT17] M. Kandi´c, M. Marabeh, V.G. Troitsky, Unbounded norm topology inBanach lattices,
J. Math. Anal. Appl. , 451 (2017), no. 1, 259-279.[KV] M. Kandi´c, A. Vavpetiˇc,
Topological aspects of order in C ( X ),arXiv:1612.05410 [math.FA][LC] H. Li, Z. Chen, Some loose ends on unbounded order convergence , Pos-itivity (2017). doi:10.1007/s11117-017-0501-1[LZ71] W.A.J. Luxemburg, A.C. Zaanen,
Riesz spaces I , North-Holland Math-ematical Library, 1971.[MS70] D. A. Martin, R. M. Solovay,
Internal Cohen extensions , Ann. Math.Logic (1970), 143–178.[MN91] P. Meyer-Nieberg, Banach lattices , Springer-Verlag, Berlin, 1991.[Roy89] N. M. Roy,
Is the product of CCC spaces a CCC space? , Publ. Mat. (1989), no. 2, 173–183[ST71] R. M. Solovay, S. Tennenbaum, Iterated Cohen extensions and Souslin’sproblem , Ann. of Math. (2) (1971), 201–245.[SS70] L. A. Steen, J. A. Seebach, Counterexamples in Topology , Holt, Rine-hart and Winston, New York-Montreal, Que.-London, 1970.[Sus20] M. Souslin,
Probl`eme 3 , Fund. Math. 1 (1920), 223.[Ten68] S. Tennenbaum,
Souslin’s problem , Proc. Nat. Acad. Sci U.S.A. (1968), 60–63.[Tka98] M. Tkaˇcenko, Introduction to topological groups , Topology ad its Ap-plications (1998), 179–231.[Tod97] S. Todorcevic, Topics in Topology , Lecture Notes in Math. 1652(Springer, Berlin, 1997).[VR59] V.S. Varadarajan, R. Ranga Rao,
On a theorem in metric spaces , Ann.Math. Stat., (1959), 612–613.[Wis69] M. R. Wiscamb, The discrete countable chain condition , Proc. Amer.Math. Soc. Fakulteta za Matematiko in Fiziko, Univerza v Ljubljani, Jadranskaulica 19, SI-1000 Ljubljana, Slovenija
E-mail address : [email protected] Fakulteta za Matematiko in Fiziko, Univerza v Ljubljani, Jadranskaulica 19, SI-1000 Ljubljana, Slovenija
E-mail address ::