Functional analysis is one branch of mathematics that focuses on the study of vector spaces equipped with specific structures. One of the theories within it is the theory of Riesz spaces, first investigated independently by Frigyes Riesz, Leinid Kantorovich, and Hans Freudenthal around 1935. Riesz spaces consist of vector spaces equipped with a partially ordered relation (usually denoted by ) that forms a lattice and is compatible with the structure of the vector space. Furthermore, there is a structure derived from Riesz spaces known as Banach lattices. This structure comprises Riesz spaces equipped with a norm and satisfying certain axioms.

One of the important concepts in the theory of Riesz spaces is the order convergence. This concept is significant because its definition involves only the order relation inherent in the Riesz space and does not depend on a specific topology.

Recall that a net in a Riesz space is a function from a directed set to . A net in the Riesz space is said to converge in order to , denoted by , if there exists a net in with , and for each , there exists such that for every with . Throughout this paper, unless ambiguity arises, we write in place of . The concept of order convergence has many applications not only in functional analysis but also in other fields such as financial mathematics.

Some researchers then attempted to generalize the concept of order convergence. In 1948, Hidegoro Nakano [1] introduced a concept of convergence that he called individual convergence. Subsequently, in 1964, Ralph DeMarr [2] defined the unbounded order convergence, abbreviated as uo convergence, which is essentially equivalent to individual convergence. The term uo convergence later became more popular and widely used to this day. A net in is said to converge uo to , denoted as for every . Over time, many researchers have examined the properties and applications of uo convergence, including Kaplan [3] investigating the relationship between uo convergence and weak units, Gao and Xanthos [4] discussing applications of uo convergence in martingales without probability, and Niushan Gao [5] studying uo convergence in dual spaces.

In this paper, we introduce a novel form of convergence that generalizes the concept of unbounded order convergence. This new convergence reduces the condition from "For all " to “for all ", where is specified, and is referred to as A-unbounded order convergence, abbreviated as A-uo convergence. We will investigate some properties of this new concept, including the limit uniqueness and its relation to other convergences. Before that, we establish a new criterion for , denoted order density on positive cones, inspired by the notion of order dense Riesz subspaces. It will become apparent later in the paper that this criterion is intricately linked to the uniqueness of the limit of A-uo convergent nets.

Throughout this paper, denotes a Riesz space. The set of all positive elements of is denoted by . Analogously, for all is the set of all positive elements of . Moreover, we use the symbols and to represent the ideal and band generated by , respectively. Definitions and fundamental concepts of Riesz spaces are referenced from [6].

We start this section by recalling the definition of order dense Riesz subspaces given by Aliprantis and Burkinshaw [7]. A Riesz subspace of is said to be order dense in if for any , there exist such that and . For the rest of the paper, we write "" to denote with . Inspired by this definition, we propose the definition of order density on a positive cone as follows.

Definition 2.1.A set is said to be order dense in if for any u , there exist such that .

Clearly, this definition is related to the concept of order dense Riesz subspace as is order dense in if and only if is an order dense ideal. It is also important to notice that is order dense in if and only if is also order dense in .

We now present two lemmas of order dense sets on the positive cone that will be usefull in the next discussion.

Lemma 2.2.Let be an Archimedean Riesz space and . Then, is order dense in if and only if .

Proof. Consider that

Hence, it is true that is order dense set in if and only if .

Lemma 2.3.Let and be order dense ideals in . Then, is an order dense set in .

Proof. Let with 0 \end{document} ]]> be arbitrary. Since is order dense in , then for some . Next, since is also order dense in , we can find such that . The fact that is an ideal gives , so that , and we also have . Therefore, it can be concluded that is an order dense set in .

We end this section by giving some examples.

Example 2.4.

After performing the idea in defining order dense sets on the positive cone, we continue our discussion to the new concept of convergence we mentioned before. We propose the definition as follows.

Definition 3.1.Given a nonempty set . A net in is said to be A-unbounded order convergent to , denoted by for all . Subsequently, the net is said to be A-unbounded order convergence if there exists such that . In this case, is called the limit of .

As mentioned before, we write "A-uo" as an abbreviation of “A-unbounded order”. If , we call the element as A-uo limit of . For , the convergence provided in Definition 3.1 is equivalent to the definition of unbounded order presented in Section 1. So, we will use the term "uo" instead of "".

Before exploring some properties, it is important to notice the following equivalence:

Thus, when studying some properties of A-uo convergence, it is suficient to study the A-uo convergence of positive nets to zero.

We now discuss the first property. The set in the definition of A-uo convergence can be "extended” to in the following sense.

Theorem 3.2.Given and a nonempty set . For any net in , we have if and only if .

Proof. Since , the backward implication is trivial. To prove the forward one, we may assume that is in and according to equivalence (1).

Now, take any . Then, there are , , and nonnegative numbers such that

From the assumption that , we have for all , so

Hence, as desired.

Theorem 3.2 provides convenience in proving the A-uo convergence of nets. If is an ideal, instead of showing that for all , it is suficient to show the convergence for all u in a set that generates .

Example 3.3.

Now, consider the sequence in . For any with m \end{document} ]]>, we have , where is the zero sequence. This means the sequence converge to 0 in order for all . Thus, it gives and it implies by Theorem 3.2.

Example 3.3(2) shows that the A-uo limit of a net is not unique in general. We provide a suficient condition for the uniqueness in the theorem below together with some algebraic properties of A-uo limit.

Theorem 3.4.Let be a nonempty set, and be two nets in , and . If and , then

Proof. According to Theorem 3.2, it can be assumed that for some ideal in .

which implies , a contradiction. As conclusion, or equivalently, .

Therefore, , as desired.

so that . Hence, . Furthermore, combining with part (2), we obtain

The first result in Theorem 3.4 states that the order density of in guarantees the uniqueness of the A-uo limit of A-uo convergence net. We now show that this fact can be strengthened. Specifically, if and are order dense set in , the A-uo and B-uo limit of any net (if exists) must be same. The converse is also true.

Theorem 3.5.Let be two nonempty sets. These two statements are equivalent.

In particular, if , the order density of in is necessary and suficient to ensure the A-uo limit uniqueness.

Proof. First, we may assume that and for some ideals and . Now, it’s known that and are order dense in . Since , then . On the other hand, since we also have Notice that is order dense in according to Lemma 2.3. Using Theorem 3.4(1), we can conclude that .

For the reverse direction, it’s known that for every net in , and then . Suppose isn’t order dense in . It means that there is with 0 \end{document} ]]> and for any satisfying , we must have . Now, pick arbitrarily. Since with is an ideal, then and , so that must hold. It implies, . However, it’s also true that . Hence, we must have , which is a contradiction. Therefore, must be order dense in , and it can be shown analogously that so is .

In Theorem 3.2, it has been shown that the set in the definition of A-uo convergence can be expanded to . Now, it’s natural to ask whether we can extend to the larger set, such as . This theorem below states that such expansion is possible in the Archimedean Riesz spaces.

Theorem 4.1.Let be an Archimedean Riesz space, be a nonempty set, and . For every net in , we have if and only if .

Proof. We first assume that for some ideal in . Also, we may assume that is in and based on equivalence (1). The backward implication of this theorem is trivial since . So, it remains to show the forward one.

For the proof, we will use the notion of Dedekind completion of , denoted by . Readers who are not familiar with this concept may refer to [8] for details. We also use the fact provided in [9], which states in is equivalent to in for any net in and . The assumption of Archimedean space for is needed since only Archimedean spaces have Dedekind completion.

Now, take any . Consider as a net on . Since it is bounded above by and bounded below by , we have

exists in . Next, we are going to show that .

Suppose on the contrary that is true. This means that we can find a lower bound for the net with . Without loss of generality, we may assume that 0 \end{document} ]]> (otherwise, pick to replace . Using the definition of Dedekind completion, there exists such that . Notice that for every index ,

so, since and is an ideal in , we get .

Now, using Theorem 7.8 in [6], . Since 0 \end{document} ]]>, we can find satisfying . Futhermore, also satisfy , so

0.\end{aligned}\end{equation*} \end{document} ]]>

On the other hand, and on , so we have on which implies di . It contradicts with the fact that 0 \end{document} ]]> di . Hence, it must be true that in .

Last, notice also that in , so it can be conclude that di which gives in . As conclusion, .

An element 0 \end{document} ]]> in is called weak (order) unit if . If is Archimedean and has a weak unit Samuel Kaplan [3] and Gao and Xanthos [4] gave a fact that to verify the uo convergence of to , it is suficient to check whether is convergent in order to . In this paper, we have a stronger statement as a corollary of Lemma 2.2 and Theorem 4.1.

Corollary 4.2.Given an Archimedean Riesz space , a nonempty set , , and a net in .

In general, A-uo convergence is not equivalent to B-uo convergence for two diferent sets . It raises a question about when A-uo and B-uo convergence are actually equivalent. Theorem 4.1 gives a suficient condition for it in an Archimedean space, that is when . In fact, using this theorem, we can deduce a suficient and necessary condition for the equivalence.

Theorem 4.3.Let be an Archimedean Riesz space and be two nonempty sets. The following two statements are equivalent.

Proof. The implication is clear according to Theorem 4.1. It remains to show the reverse implication. To do that, we may assume at first that and for some ideals based on Theorem 3.2.

Now, take any and let . For each , and hold, so that . Consequently, is a directed set with the same order relation as . Next, define a net with for each . It is clear that .

Pick any . We obtain and is the lower bound of the net . Next, take any lower bound of the net, say . It is obtained that is also its lower bound or in other words, for every . This fact gives , so that for all since for some ideal . We also have . Therefore, , which implies for all . Using mathematical induction, it can be shown that for every and , we have

From the result above, we obtain for all . It implies that must be zero since is Archimedean. This results in in , so that . So, since is arbitrary, we obtain .

Finally, using the assumption (2), we get , so that or equivalently, . Consequently , so that

Based on Theorem 7.8 in [6], we obtain , which implies . Analogously, we can show . Thus, it is proven that .

As a remark, we note that Theorem 4.3 may not be hold in non-Archimedean Riesz spaces. For instance, on the Riesz space with lexicographic order relation, let . It can be shown that is a band in . Now, take any net in and with . We have , so there is a net with and for each index , there is index such that for every index with is hold.

Since is totally ordered, there must be an index satisfying . Consequently, for any index with , we have

which gives . So, for each index , there is an index which is an upper bound of such that for each index with , it is true that

which means .

Since an order convergent net is also uo convergent, we obtain . This shows that is equivalent to for each net in and . However, it is easy to see that . Thus, Theorem 4.3 does not hold on the Riesz spaces with lexicographic order relation.