Abstract

The aim of this paper is to introduce power idealization filter topologies with respect to filter topologies and power ideals of lattice implication algebras, and to investigate some properties of power idealization filter topological spaces and their quotient spaces.

1. Introduction and Preliminaries

By generalizing Boolean algebras and Lukasiewicz implication algebras [1], Xu [2] defined the concept of lattice implication algebra which is regarded as an efficient approach to deal with lattice valued logical systems. Later, Xu and Qin [3] defined the concept of the filer topology of a lattice implication algebra which takes the set of all filters of the lattice implication algebra as a base. Based on these definitions and some results in [4], we introduce and study power idealization topologies with respect to filter topologies and power ideals of lattice implication algebras.

Now we recall some definitions and notions of lattice implication algebras and topological spaces.

Let be a bounded lattice with the greatest 1 and the smallest 0. A system is called a quasi-lattice implication algebra if is an order-reserving involution and is a map (called an implication operator) satisfying the following conditions for any :(1),(2),(3),(4) implies ,(5).

A quasi-lattice implication algebra is called a lattice implication algebra if the implication operator further fulfils the following conditions:(6),(7).

A lattice implication algebra will be simply denoted by .

Let be a lattice implication algebra and let be a subset of . We use to denote the complement , where . A subset is called a topology on , if satisfies the following:(1),(2) implies ,(3) implies .

Elements of are called -open sets and the complements of them are called -closed. The pair is called a topological space. A subset of is called a base of , if for each and each , there exists such that .

Let be an implication algebra. A subset of is called a filer, if satisfies the following: (1) ; (2) implies . The collection of all filters in is denoted by , or briefly. Clearly, consists a base of some topology , briefly . Usually, is called the filter topology generated by . And the pair is called the filter topological space. A subset is called -neighborhood of , or neighborhood of in if . The set of all -neighborhoods of is denoted by . Since and , is the smallest element of .

The closure operator and interior operator of are denoted by and . Clearly, for every , and . The following proposition describes .

Proposition 1. Let be the filer topology generated by . Then for , .

Proof. The proof is trivial since is the smallest -neighborhood of .

Let be a lattice implication algebra and let be the power set of . A nonempty subset of is called a power ideal of if satisfies the following: (1) and imply ; (2) implies . The collection of all power ideals in is denoted by , or briefly . Note that is the smallest power ideal and is the greatest power ideal. Moreover, if , then (1) ; (2) .

2. Local Functions and Power Idealization Filter Topologies

Let be a lattice implication algebra, let be the filter topology, and let be a power ideal. An operator   is defined as follows: for every .

The operator called the local function with respect to and . is called local function of . We usually write or instead of .

Clearly, if and only if . Thus . The following proposition gives some further details of .

Proposition 2. Let be the filter topological space and . Then(1) and ;(2)if , then ;(3)if , then ;(4);(5);(6)if , then ;(7)if , then ;(8)if , then ;(9);(10);(11);(12)if , then for each ;(13)if and , then .

Proof. (1) By Proposition 1, if and only if if and only if . Thus . Since for each , .
(2) Let and . Then . Since is a power ideal and , and so . Thus .
(3) Let and . Then . It follows that and so . Thus .
(4) If , then and so . Thus . This implies and so . Then .
It is clear that . Next, we prove .
Let . By Proposition 1, . Then there exists . By , . By , . Thus and so . Therefore .
(5) By (4), .
(6) Since for each , .
(7) Suppose that . Then . Thus and so . Hence which is a contradiction. Therefore .
(8) By (2), . Next, we prove the inverse inclusions.
If , then . Thus which follows from is a power ideal. This implies . Thus and so .
If , then . Since , Thus . This implies and so .
(9) Clearly, . Conversely, if , then . Let . Then . By (2), (7), (8), and , Thus and so This implies and so .
(10) is clear. Conversely, if , then . Thus . This implies . Therefore .
(11) We firstly prove . Assume that . Then and . Thus This implies which is a contradiction. Thus and so . Finally, follows from (2). Therefore .
(12) Since , for each . Assume that there exists such that . Let . Thus . Since , which is a contradiction. Therefore for each .
(13) Assume that there exists . Then which is a contradiction. Thus and . Since , follows from (12). Therefore .

Proposition 3. Let be the filter topological space and . The operator (briefly, ) on , defined by for , satisfies the following statements:(1); ,(2),(3),(4),(5) or implies .

Proof. (1) follows from . is clear.
(2) By (4) and (9) of Proposition 2,
(3) By (5) and (9) of Proposition 2, and .
(4) By (10) of Proposition 2,
(5) The result follows from (6) and (7) of Proposition 2.

Theorem 4. Let be the filter topological space and . The operator stated in Proposition 3 is the closure operator of a new topology which is finer than and the topology generated by (note that is not a topology since in general case). Such a topology is called a power idealization filter topology and often denoted by , , or .

Proof. Let . We prove that is a topology on .
(1) follows from (1) of Proposition 3.
(2) If , then and . Thus This implies .
(3) Let for , where is an index set. Then and Therefore .
Finally, by (6) and (7) of Proposition 3, .

Example 5. Let , , , , , , , and the implication operator be defined by for . Then is the Hasse lattice implication algebra (Figure 1 and Table 1). Then Let . Then is a power ideal. It is easy to check that Clearly, .

Figure 1 

Hasse diagram of .

Proposition 6. Let be the filter topological space and . Then(1) and ,(2)if , then .

Proof. (1) By (1) of Proposition 2, and . Thus if and only if . Similarly, for each . Therefore (1) holds.
(2) Let . By (3) of Proposition 2, for . Thus if , then . Therefore .

Clearly, if satisfies , then and so . Thus by (1) of Proposition 3, . If , we have the following proposition.

Proposition 7. Let be the filter topological space. If satisfies , then .

Proof. follows from Theorem 4. Conversely, suppose that . Then there exists such that . Let . Then . Thus . Put . We have . Since , . Thus and so . Hence . It is a contradiction. Therefore .

Lemma 8. Let be the filter topological space and . If satisfies for each , then .

Proof. is clear. Conversely, if , then and . Thus and . Since , . Observe that and is -closed. We have . Thus . Therefore .