aa r X i v : . [ m a t h . F A ] M a r Convergence via filter in locally solid Rieszspaces
March 8, 2018
Abdullah AYDIN
Department of Mathematics, Mu¸s Alparslan University, Mu¸s, Turkey.
Abstract
Let (
E, τ ) be a locally solid vector lattice. A filter F on the set E is said to be convergeto a vector e ∈ E if, each zero neighborhood set U containing e , U belongs to F . Westudy on the concept of this convergence and give some basic properties of it. Let give some basic notation and terminology that will be used in this paper. Let E bereal vector space. Then E is called ordered vector space if it has an order relation ≤ that means it is reflexive, antisymmetric and transitive, and also it satisfies the axioms;if y ≤ x then y + z ≤ x + z for all z ∈ E , and if y ≤ x then λy ≤ λx for all 0 ≤ λ . Anordered vector E is said to be vector lattice (or, Riesz space ) if, for each pair of vectors x, y ∈ E , the supremum x ∨ y = sup { x, y } and the infimum x ∧ y = inf { x, y } both exist inE. Also, x + := x ∨ x − := ( − x ) ∨
0, and | x | := x ∨ ( − x ) are called the positive part, the negative part, and the absolute value of x , respectively. In a vector lattice E , a subset A is called as solid if, for y ∈ A and x ∈ E with | x | ≤ | y | , we have x ∈ A . Also, two vector x , y in a vector lattice is said to be disjoint whenever | x | ∧ | y | = 0; for more details see[2, 3].Let E be vector lattice E and τ be a linear topology on it. Then ( E, τ ) is said a locallysolid vector lattice (or, locally solid Riesz space ) if τ has a base which takes place withsolid sets, for more details on these notions see [2, 3, 4]. It is known that every lineartopology τ on a vector space E has a base N for the zero neighborhoods satisfying thefollowings; for each V ∈ N , we have λV ⊆ V for all scalar | λ | ≤
1; for any V , V ∈ N there is another V ∈ N such that V ⊆ V ∩ V ; for each V ∈ N there exists another U ∈ N with U + U ⊆ V ; for any scalar λ and for each V ∈ N , the set λV is also in N ; see forexample [1, 7]. Hence, every locally solid vector lattice satisfies these properties. Also, itfollows from [2, Thm.2.28] that a linear topology τ on a vector lattice E is a locally solidiff it is generated by a family of Riesz pseudonorms { ρ j } j ∈ J , where Riesz pseudonorm is arealvalued map ρ on a vector space E if it satisfies the following conditions; ρ ( x ) ≥ x ∈ X ; if x = 0 then ρ ( x ) = 0; ρ ( x + y ) ≤ ρ ( x ) + ρ ( y ) for all x, y ∈ X ; if lim n →∞ λ n = 0 Keywords: filter, locally solid Riesz space, vector lattice2010 AMS Mathematics Subject Classification: 46A40, 46A55e-mail: [email protected] n R then ρ ( λ n x ) → R for all x ∈ X ; if | x | ≤ | y | then ρ ( x ) ≤ ρ ( y ). Moreover, ifa family of Riesz pseudonorms generates a locally solid topology τ on a vector lattice E then x α τ −→ x iff ρ j ( x α − x ) → R for each j ∈ J . In this article, unless otherwise, thepair ( E, τ ) refers to as a locally solid vector lattice, and the topologies in locally solidvector lattices are generated by families of Riesz pseudonorms { ρ j } j ∈ J . In this paper,unless otherwise, when we mention a zero neighborhood, it means that it always belongsto a base that holds the above properties. Recently, there are some studies on locallysolid Riesz spaces; see for example [1, 6, 8].Next, we give the concept of filters. Let X be a set. A subset F of the power set of X is said to be filter on X if the followings hold; ∅ / ∈ F ; if A ∈ F and A ⊆ B then B ∈ F ; F is closed under finite intersections; see [4]. The second condition says that the set X belongs to the filter on it. The filter can be defined tanks to its base. A nonempty subset B ⊆ F is called a filter base for a filter F , if F = { F ⊆ X : ∃ B ∈ B , B ⊆ F } . A base B satisfies the following properties; B is nonempty; each B ∈ B is nonempty; for each B , B ∈ B , there is another B ∈ B such that B ⊆ B ∩ B .Filter can be defined with nets. A given partially ordered set I is called directed if, foreach a , a ∈ I , there is another a ∈ I such that a ≥ a and a ≥ a . A function from adirected set I into a set E is called a net in E . Now, we give a relation between net andfilter convergence. Let ( x α ) ( α ∈ I ) be a net in the set E . The filter F which is associatedof ( x α ) ( α ∈ I ) is defined as follows; let ˆ x α = { x α : α ∈ I, α ≥ α } , and so the collection B = { ˆ x α : α ∈ I } is a filter base and the filter that is generated by B is the associatedfilter of ( x α ) ( α ∈ A ) . So, we can give the following natural example. Example 1.1.
Let E be a set and ( x α ) ( α ∈ I ) be a net in E . The elementary filter associatedto ( x α ) ( α ∈ I ) is F ( x α ) = { F ⊆ E : ∃ α , x α ∈ F for all α ≥ α } . The filter convergence is defined on topological spaces with respect to the neighborhoodof limit points; see [5, Def.3.7]. In here, we define this concept on locally solid vectorlattices.
Definition 1.1.
Let (
E, τ ) be a locally solid vector lattice and F be a filter on the set E . A vector e ∈ E is said to be a limit of F (or, F is said to converge to e ) if each zeroneighborhood set containing e belongs to F , abbreviated as F τ −→ e . Also, a vector x ∈ E is said to be a cluster point of F if each zero neighborhood set containing x intersectsevery member of F . Example 1.2.
Suppose ( E, τ ) is a locally solid vector lattice. For a vector e ∈ E , thefilter generated by e is F e = { U ⊆ E : U is zero neighborhood and contains e } . So, we can see F τ −→ e Consider a topological convergence net ( x α ) ( α ∈ I ) τ −→ e ∈ E . Then the elementary filterassociated to ( x α ) ( α ∈ I ) is also convergent to e since every zero neighborhood containing e belongs to that filter. Remark . Let (
E, τ ) be a locally solid vector lattice. Then( i ) If a filter F τ −→ e then e is a cluster point of F . Indeed, let U be zero neighborhoodset and it contains e , and F ∈ F . Since U and F in F , we have U ∩ F ∈ F . Thus, we get U ∩ F = ∅ since ∅ is not in the filter F .( ii ) Let F , F be filters on E with F ⊆ F . So, if F τ −→ e then F τ −→ e for e ∈ E .Indeed, every zero neighborhood containing e is in F , and so is in F . Hence, F τ −→ e .In the filter convergence, the limit point may not be unique. That means a filter F ona locally solid vector lattice ( E, τ ) can be convergence both e , e ∈ E . roposition 1.1. Let ( E, τ ) be a locally solid vector lattice, and F be a filter on E . Thenthe followings hold;(i) If F τ −→ e for some e ∈ E then F τ −→ e + x for all x ∈ E whenever e and x are positive,or disjoint;(ii) If F τ −→ e for some e ∈ E then F τ −→| e | (iii) If F τ −→ e for some e ∈ E then F τ −→ e + and F τ −→ e − Proof. ( i ) Let U be a zero neighborhood that contains e + x . So, we show U ∈ F .Under the condition of positivity of e and x , we have e, x ∈ U since U is a solid set and e, x ≤ e + x . Therefore, by using the convergence F to e , we get U ∈ F . On the otherhand, if e and x are disjoint then we have | e + x | = | e | + | x | ; see [3, Thm.1.7(2-7)]. So, bysolidness of U , we get e, x ∈ U . Therefore, U ∈ F since F τ −→ e .( ii ) Suppose F τ −→ e and U is a zero neighborhood containing | e | . By the formula | e | ≤ (cid:12)(cid:12) | e | (cid:12)(cid:12) = | e | and by the solidness of U , we have e ∈ U . Therefore, U ∈ F since F τ −→ e .( iii ) Assume F τ −→ e and U is a zero neighborhood which contains e + . By the formula | e | = e + + e − ; see [3, Thm.1.5(2)], and by using the solidness of U , we get e + ∈ U .Therefore, we have U ∈ F since F τ −→ e . The other part of the proof is analog.In the next three results, we give some basic properties of filter convergence on locallysolid vector lattices. Theorem 1.2.
Let ( E, τ ) be a locally solid vector lattice. If F and F are filters on E such that F τ −→ e and F τ −→ x for some e, x ∈ E then the set F = { F ∪ F : F ∈F and F ∈ F } is also a filter on the set E , and F τ −→ e + x whenever e, x are positive,or disjoint.Proof. Firstly, we show that F is a filter. ( i ) ∅ / ∈ F since ∅ / ∈ F and ∅ / ∈ F ; if F ∈ F and F ∈ F , and F ∪ F ⊆ A ⊆ E then A ∈ F and A ∈ F , so A ∈ F ; if F , G ∈ F and F , G ∈ F then F ∪ F , G ∪ G and ( F ∪ F ) ∩ ( G ∪ G ) in both F and F , andso ( F ∪ F ) ∩ ( G ∪ G ) ∈ F .Next, for the first case, we show F τ −→ e + x . Let U be a zero neighborhood and itcontains e + x . By the properties of zero neighborhoods in locally solid vector lattice, wehave e, x ∈ U since they are positive and so that e, x ≤ e + x . Therefore, since F and F are filters on the set E , and also F τ −→ e and F τ −→ x , we get U ∈ F and U ∈ F . So,we get U ∈ F .Now, suppose e and x are disjoint in E . Then we have | e + x | = | e | + | x | ; see [3,Thm.1.7(2-7)]. So, for given any zero neighborhood U containing e + x , we have e, x ∈ U .Then the proof follows like first case. Theorem 1.3.
Suppose ( E, τ ) is a locally solid vector lattice, and F and F are filters onthe set E such that F τ −→ e and F τ −→ x for some e, x ∈ E . Then the class F = { F ∩ F : F ∈ F , F ∈ F , and F ∩ F = ∅} is also a filter on E , and also F τ −→ e + x whenever e, x are positive, or disjoint.Proof. We show F is a filter. ( i ) ∅ / ∈ F ; if F ∈ F and F ∈ F , and F ∩ F ⊆ A ⊆ E then A ∈ F and A ∈ F , so we get A ∈ F ; if F , G ∈ F and F , G ∈ F such thatintersection of for each pair of them is non empty then F ∩ G ∈ F , F ∩ G ∈ F , so( F ∩ F ) ∩ ( G ∩ G ) = ( F ∩ G ) ∩ ( F ∩ G ) is non empty and it is also in F . Then F isa filter on E , and by using the similar way in the proof of Theorem 1.2 we get the desiredresult.Let ( x α ) α ∈ I be a net in any topological space. A point x is called cluster point of( x α ) α ∈ I if, for each neighborhood U of x , ( x α ) α ∈ I is frequently in U , or equivalently, foreach neighborhood U of x , we have ( U \ { x } ) ∩ { x α } 6 = ∅ ). On the other hand, let F be a filter on a locally solid vector lattice ( E, τ ). A vector e ∈ E is said to be cluster oint of F (with respect to F ) if every zero neighborhood which contains e intersects witheach member of F ; see [5, Def.3.7]. So, we give the following result which has a relationbetween filter convergence and cluster point. Theorem 1.4.
Let ( E, τ ) be a locally solid vector lattice and F be a filter on the set E .Then the vector e ∈ E is a cluster point of F iff there exists another filter F containing F such that F τ −→ e .Proof. Suppose e ∈ E is a cluster point of F . Then the set B = { U ∩ F : F ∈F , U is a zero neighborhood and contains e } is a filter base. Indeed, we show the prop-erties of filter base. ( i ) B is non empty since, for each zero neighborhood U containing e intersects with every member of F ; ( ii ) for U ∩ F ∈ B , we have U ∩ F = ∅ since U ∩ F ∈ F and ∅ / ∈ F ; ( iii ) for U ∩ F and U ∩ F in B , we can take B = ( U ∩ U ) ∩ ( F ∩ F ) ∈ B .So, we assume it generates the filter F . For each F ∈ F , we have F = E ∩ F ∈ B , andso we get F ⊆ F . Therefore, for given a zero neighborhood U containing e , we have U = U ∩ E ∈ B , and so we get F τ −→ e .Conversely, assume such filter F exists it means that F is a filter on the set E with F ⊆ F and F τ −→ e . So, all zero neighborhoods containing e is in F . So, by thedefinition of filter, each zero neighborhood containing e intersects with member of F ,otherwise ∅ ∈ F . Therefore, in particular, intersects each set in F , and so we get e is acluster point of F .Now, we use net to define filter on locally solid vector lattice. Let ( x α ) α ∈ I be a net inthe locally solid vector lattice ( E, τ ). Then we define its associated filter F on the set E as follow; consider the tail ˆ x β = { x α : α ∈ I, α ≥ β } and B = { ˆ x β : β ∈ I } . So, B is afilter base. Indeed, ( i ) B is not empty; ( ii ) every ˆ x β ∈ B is not empty since I is directedset; ( iii ) for any ˆ x β , ˆ x β ∈ B , consider the index β = max { β , β } so that ˆ x β ⊆ ˆ x β ∩ ˆ x β and ˆ x β ∈ B . Thus, the filter which is generated by B is the associated filter of ( x α ) α ∈ I . Theorem 1.5.
Let ( E, τ ) be a locally solid vector lattice and ( x α ) ( α ∈ I ) be a net in E .Assume F is the associated filter of ( x α ) ( α ∈ I ) and e ∈ E . Then x α τ −→ e as a net iff F τ −→ e as a filter. Moreover, e is a cluster point of ( x α ) ( α ∈ A ) iff e is a cluster point of F .Proof. We show only the convergence part of proof, the cluster point case is analogous.Suppose x α τ −→ e as a net. Since every locally solid vector lattice has a base of zero neigh-borhoods, we can consider any zero neighborhood U which contains e . So, by definitionof F , we get U ∈ F . Thus, we have F τ −→ e as a filter.Conversely, assume the filter F converges to e . Let V be a zero neighborhood in E andcontains e . Then V ∈ F . By definition of F , there exists a index α such that ˆ x α ⊆ V .Therefore, we get x α ∈ V for every α ≥ α , and so x α τ −→ e . Corollary 1.6.
Let ( E, τ ) be a locally solid vector lattice which is generated by a familyof Riesz pseudonorms { ρ j } j ∈ J , ( x α ) ( α ∈ A ) be a net in E and F be associated filter of it,and e ∈ E . Then F τ −→ e iff ρ j ( x α − e ) → for all j ∈ J .Proof. It follows from [2, Thm.2.28] and Theorem 1.5.We can give convergence of filters with respect to continuous function. Let (
E, τ ) and( F, ´ τ ) be locally solid vector lattices, and F be a filter on E . For a function f : E → F ,the set ´ f F = { B ⊂ F : f − ( B ) ∈ F } is the filter on E generated by { f ( A ) : A ⊆ F } . Proposition 1.7.
Suppose ( E, τ ) and ( F, ´ τ ) are locally solid vector lattices. Then afunction f : E → F is continuous iff ´ f F ´ τ −→ f ( e ) in F for each F τ −→ e in E . roof. Suppose f is continuous. For fixed e ∈ E , consider the following set N e = { A ⊆ E : ∃ U zero neighborhood s.t. , e ∈ U ⊆ A } . Thus, we get N e τ −→ e . Note also that a filter F converges to e iff N e ⊂ F . Take a filter F such that F τ −→ e . The continuity of f implies that N f ( e ) ⊆ ´ f N e . Therefore, if F τ −→ e then N e ⊆ F , and so N f ( e ) ⊆ ´ f N e ⊆ ´ f F so that ´ f F ´ τ −→ f ( e ).Assume ´ f N e is the set of all subsets of E whose preimage is a neighborhood of e . Since N τ −→ e , we conclude that the preimage of any neighborhood of f ( e ) is a neighborhood of e . Hence, f is continuous. References [1] A. Aydın,
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