Abstract
The aim of this paper is to establish the prime filter theorem for multilattices.
1. Introduction
Benado’s pioneering work on posets [1] laid the foundation for a new theory called multilattice theory. This theory will be consolidated by several authors including Olga [2] and Hansen [3] who proposed many characterizations of multilattices. Subsequently, the work of Cabrera et al. [4, 5] led to an algebraization of multilattices. Multilattices appear under two inseparable aspects: both as an ordered structure generalizing ordered lattices and as an algebraic hyperstructure generalizing algebraic lattices.
If several authors lean towards this theory, it is precisely because of its numerous applications in decision making, in uncertainty modeling, in fuzzy logic, and so on [6]. To further broaden the scope of multilattices, mathematical concepts such as filter/ideal, congruence, and homomorphism [4, 5, 7] will be studied. However, in this new structure, the notion of associativity is replaced by a stronger notion of m-associativity. We thus lose the interest to define the concepts in a canonical way as well as the notion of distributivity is nontrivial [7], which requires more caution when extending some results such as the prime filter theorem.
I.P. Cabrera et al. proved in [5] that the set of all filter of a multilattice is a lattice with respect to inclusion. We observe that for some multilattices, the derived filter lattice is distributive. This allows us to use the notion of distributivity in multilattices.
The paper is divided into two sections. Section 1 concerns the preliminaries: we recall the definitions and results necessarily to understand the paper and we develop some examples to illustrate the proofs. Section 2 brings out the newly established results.
2. Preliminaries
Let be a poset, and let be a nonempty subset of . An element is said to be an upper bound of if it satisfies for all . Dually, is a lower bound of if it satisfies for all .
The set of upper bounds (resp., lower bounds) of is denoted by (resp. ). In particular, if , then we write (resp., ) instead of (resp., ). We also use the following notations and definitions: and .
Definition 1 (see [1]). A multisupremum (resp., multinfimum) of is a minimal (resp., maximal) element of (resp., ).
We will denote by (resp., ) the set of multisuprema (resp., multinfima) of . When has a unique multisupremum (resp., multinfimum), it is denoted by or (resp., or ). We will also use the following notations and definitions: for all nonempty subsets and of ,
Definition 2 (see [1]). Let be a poset. is said to be an ordered multilattice if the following conditions hold for all : M: and imply that there exists such that M: and imply that there exists such that The previous definition gives rise to two hyperoperations which allow us to look at multilattice from an algebraic perspective.These hyperoperations satisfy the following properties called axioms of multilattices. In [2, 3], many characterizations are proposed. AM1: For all , ; AM2: For all , ; AM3: For all , ; AM4: For all , ; AM5: For all , .
Definition 3 (see [5]). An algebraic multilattice is any triple which satisfies the properties AM1-AM5.
Ordered multilattices and algebraic multilattices are connected as follows.
Theorem 1. The following assertions are satisfied:I. Ifis an ordered multilattice, setThen, is an algebraic multilattice.
II. Conversely ifis a multilattice, setThen, is an ordered multilattice.
As the two aspects mentioned in Theorem 1 are inseparable, it is possible to manipulate multilattices using both the underlined order and the deduced hyperoperations. We can, therefore, use the following three notations to designate multilattice without fear of ambiguity:We simply write assuming that the underlined order and the corresponding hyperoperations are understood.
Definition 4. (1)is said to be a complete multilattice if every subset ofhas at least one multisupremum and at least one multinfimum.(2)Afull multilattice (resp.,-full multilattice ) is a multilattice in which(resp.,) for all. A multilattice is said to be full if it is bothfull andfull.There are several proposals of the definition of homomorphism between hyperstructures. The most used definition was introduced by Benado in [1].
Definition 5 (see [5]). A map between multilattices and is said to be a homomorphism if
Remark 1. If the initial multilattice is full, then the condition to be a multilattice homomorphism can be expressed in terms of equalities as follows:
Notation 1. Any lattice induces a multilattice given by and for all .
In the framework of multilattices, the notions of distributivity and associativity are not quite canonical as stated in [7].
Proposition 1 (see [7]). Let be a multilattice. Then, the following conditions hold:(1)If eitheroris canonically associative, thencollapses to a lattice(2)If(resp.) is canonically distributive to(resp.,), thencollapses to a distributive latticeThe notion of filter is studied in [5] in relation to homomorphism and congruences.
Definition 6 (see [5]). A nonempty subset of is said to be a filter if it satisfies the following conditions: F-1: For all,; F-2: For alland,; F-3: For allsuch that,. Dually, a nonempty subset of is said to be an ideal if it satisfies the following conditions: I-1: For all, I-2: For alland, I-3: For allsuch that,
Notation 2. The set of all filters of will be denoted by .
Proposition 2 (see [5]). Let be an -full multilattice. Then, is a lattice under the set inclusion.
The lattice is given by and is the smallest filter of containing both and . When is not -full, it is necessary to add the empty set to (lifting) in order to obtain a lattice. That is, is always a lattice ordered by set inclusion.
Example 1. The following diagrams give two multilattices and their filter lattices.
The smallest filter of containing is the intersection of all filters of containing . It is denoted by , and if , we write instead of .
3. Main Results
For every nonempty subset of , set
When , we denote instead of . We shall offer a description of the filter generated by . We first list some properties of the operator .
Proposition 3. For every nonempty subsets and of , the following conditions hold:(i)(ii)(iii)We have the following characterization of filters.
Lemma 1. Let be a nonempty subset of . Then, is a filter of if and only if and .
Proof. Suppose that is a filter of .
Let . If , then for some . Since is a filter of , and so . Thus, . If , then . Thus, , and it follows that .
Clearly, (see Proposition 3). If , then there is such that and , but (any filter is upward closed) and then , so and consequently . Therefore, .
Conversely, suppose that and .
Let . If , then , and if , then ; finally, if , then . Hence, is a filter of .
We define a sequence recursively as follows:The following summarizes the main properties of this sequence.
Proposition 4. Let be a nonempty subset of . Then, the sequence is increasing with respect to the inclusion.
Proof. Clearly, and also . Therefore, .
Theorem 2. Let be a nonempty subset of . Then, .
Proof. Let . If and for some positive integers and with , then , and so ; if , , then ; if , then . Hence, is a filter of . Clearly, (in fact ).
Let be a filter of containing . We will prove by induction that . We have . If , then and then (see Lemma 1). Hence, for all and it follows that .
Example 2. We consider the multilattices of Figure 1. Let , . We apply Theorem 2 to obtain the filters of and , respectively, generated by and .Therefore, and .
The next lemma follows directly from the definition of filter.
Lemma 2. Let . Then, the following assertions hold:(1);(2);(3).The inclusion of (2) will generally be strict. For instance, in the multilattice of Figure 1, we have but and with .
Figure 2 presents the filter lattice of the multilattices and of Figure 1. We can easily claim that is not distributive since it contains a copy of while is distributive. This allows us to use the following concept of distributivity.
Definition 7. is said to be a distributive multilattice if is a distributive lattice.
Lemma 3. Let . Then, the following conditions hold:(1);(2).
Proof. For (1), we claim that is a filter of . Let and , , and , then , where and . If , then , and if , then since is a filter of . Therefore, is a filter of . It is easy to see that and , and every filter of containing necessarily contains .
For (2), let . We claim that is a filter of . Let and , , and , then , where and. If , then , and if , then since is a filter of . Therefore, is a filter of . For all and for all , we have and ; thus, . Therefore, . It obvious that . In fact, implies and and then we write .
Definition 8. A proper filter of is said to be prime if for all , implies or .
From (2) of Lemma 3, we deduce the following result.
Corollary 1. Let be a filter of . Then, the following conditions are equivalent:(1)is a prime filter(2)For all,implies
Proof. The first implication follows directly from Definition 8. For the converse, suppose that but and . Let and . Then, neither nor . From (2), we have . However, from Lemma 3, which is a contradiction.
Let , .
Theorem 3. Let be a nonempty subset of . Then, is a prime filter of if and only if there exists a lattice homomorphism from to the lattice such that .
Proof. Suppose that is prime, then is also prime in and then there is a lattice homomorphism from to the lattice such that . However, we observe that . It follows that .
Conversely, if is a lattice homomorphism from to , then is a prime filter of . This implies is a prime filter of .
Corollary 2. Let a filter of . Then, is a prime filter if and only if there exists a multilattice homomorphism from to such that .
Proof. Let be an homomorphism form to . Let , then on the one hand, we have since . Hence, . On the other hand, we have since which implies . This proves that is a multilattice homomorphism. Then, we use Theorem 3 to obtain the existence of .
Lemma 4. Let and be two filters of , and let be an ideal of . Then,
Proof. Let and . Then, firstly, since is an ideal of ; secondly, and also . Thus, . Therefore, .
A. Rahanamai–Barghi established the prime ideal theorem for meet-hyperlattices [8] and for distributive hyperlattices [9]. In the following theorem, we state and prove the prime filter theorem for multilattices.
Theorem 4 (prime filter theorem). Let be a distributive multilattice. Let be a proper filter of and a proper ideal of such that . Then, there exists a prime filter of such that and .
Proof. Let be the set of all filters of containing and disjoint from .We claim that satisfies Zorn’s lemma. Since , the set is not empty. Let be a chain in , and let . Let . Firstly, if , then and for some , ; since is a chain, either or . If , then and so . Secondly, if and , then for some ; since is a filter, . Thirdly, if , then for some ; since is a filter, . Thus, is a filter of . It is obvious that and , verifying that . Therefore, Zorn’s lemma has a maximal element, say . We claim that is prime. Indeed suppose to the contrary that is not prime. Then, there exists such that , , and . and both are filters which strictly contain . Because of the maximality of , clearly and . Thus, from Lemma 4, . Since is distributive, we have . This implies , a contradiction.
Corollary 3. Let be a distributive multilattice with top element, . If is a proper ideal of , then there exists a prime filter such that .
Proof. Since is proper, we have . Hence, we apply Theorem 4 with to obtain the desired conclusion.
4. Conclusion
We have established the prime filter theorem and proposed some characterizations.
Data Availability
No data were used to support this study.
Conflicts of Interest
The authors declare that there are no conflicts of interest.