#### 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 â€‰ 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, .

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 .

Proposition 9. *If , then .*

*Proof. *It is clear that . Conversely, let be the greatest element of and . Thus for each . By Lemma 8, . By (8) of Proposition 2, . Now, notice that and thus . We have
This implies . Therefore .

Lemma 10. *Let be the filter topological space and . If and , then .*

*Proof. *Let . Then . By (7) and (8) of Proposition 2,
Thus .

Theorem 11. *Let be the filter topological space and . Then
**
is a base of . Moreover,
**
is a base of for each , where is the set of all -neighborhoods of in . Clearly, is the smallest -neighborhoods of , where is the greatest element of satisfying .*

*Proof. *By Lemma 10, . Let . Then is -closed . Thus there exists such that . Let . Then and
Therefore is a base of .

Clearly, . Let and . Since is a base of , there are and such that and . We can assume and (otherwise, and can be replaced by and , resp.,). Then
Clearly, and
Therefore is a base of .

Clearly, if is the greatest element of satisfying , then is the smallest -neighborhoods of .

Let be a topological space and . The topology that was generated by is denoted by [5]. Clearly, .

Lemma 12. *Let be the indiscrete topology on and . Then .*

*Proof. *By and Proposition 7, the proof is obvious.

Theorem 13. *Let be a topological space and . Then , where is the topology generated by the base .*

*Proof. *Clearly, is a base of . Since , is also a base of . Therefore .

Corollary 14. *Let be the filter topological space and . Then .*

Corollary 15. *Let be the filter topological space and . Then*(1)*,*(2)*,*(3)*,*(4)*.*

*Proof. *(1) By (2) of Proposition 6, . Conversely, let . By Theorem 13, there exist and such that
Thus .

(2) By (1), Theorem 13, and Corollary 14,

Similarly, .

(3) By (1) and Theorem 13,
Therefore .

(4) Let . Then the proof follows from (2).

Theorem 16. *Let be the filter topological space, and . Then*(1)*,*(2)*.*

*Proof. *(1) By (3) of Proposition 2, . Conversely, . Then and . Let and . Then and . Thus
Thus which implies . Therefore .

(3) Let . Then . Then there exist and such that . We can assume , (otherwise, can be replaced by ). Thus or , (otherwise, which is a contradiction). Now, we take for example. Then
Since and , . Thus and so . Hence

Conversely, let . Then there exists such that . Let . Then which implies . Similarly, if , then . Therefore .

Corollary 17. *Consider .*

*Proof. *Let . The proof follows from (2) of Theorem 16.

Corollary 18. *Consider .*

*Proof. *By (2) of Proposition 6, . Conversely, if , then
Thus or . Thus or . Therefore .

#### 3. Power Idealization Filter Topological Quotient Spaces

Let and be two lattice implication algebras. A mapping from to is called lattice implication homomorphism, if for any . The set of all lattice implication homomorphisms from to is denoted by .

Let . Then, clearly, are topologies [4].

Lemma 19. *Let and be two implication algebras and let and , be power ideals.*(1)*If is injective, then .*(2)*If is surjective, then .*

*Proof. *(1) Since , then .

If and , then there exist such that . Thus and . Since is injective, .

If , then there exist such that and . One has . Since , . Therefore is a power ideal.

(2) By , .

If and , then there exists such that . Let . Then and . Since is surjective, .

If , then there exist such that and . Thus . Hence . Therefore is a power ideal.

Lemma 20. *Let and be two implication algebras and . Then*(1)*if is injective, then for each , is the smallest -neighborhood of ;*(2)*if is bijective, then for each , is the smallest -neighborhood of .*

*Proof. *(1) Clearly, is the smallest -neighborhood of . Then . Let . Then and . Thus . Therefore is the smallest one.

(2) Clearly, is the smallest -neighborhood of . Thus . Now, let . Then . Thus and so . Therefore (2) holds.

Lemma 21. *Let and be two implication algebras and .*(1)*If is injective and , then for each , is the smallest neighborhood of , where is the greatest element of satisfying .*(2)*If is bijective and , then for each , is the smallest neighborhood of , where is the greatest element of satisfying .*

Theorem 22. *Let and be two implication algebras and .*(1)*If is injective and , then
*(2)*If is bijective and , then
*

*Proof. *(1) Let and be the closure operators of the left side and the right side of the equation. We only need to prove .

Let and . Then and . By (1) of Lemma 21, . Thus there exists such that . Since and is injective, . Let be the greatest one satisfying . Then . Thus
By Lemma 21 and Proposition 1, . Therefore .

Conversely, let . By Proposition 1,
Thus , and . Then . Since is injective, . By (1) of Lemma 20, . Therefore and .

(2) Let and be the closure operators of the left side and the right side of the equation. We only need to prove .

Let . Then and . By (2) of Lemma 20, . Thus there exists such that . Since is bijective, . By , and so . Since is the greatest element of satisfying , and . By being bijective again, we have
By
. Hence .

Conversely, let . Since is the greatest element of satisfying , by (2) of Lemma 21, is the smallest neighborhood of . Thus . Since is bijective,
This implies and . Thus which implies . Since , . Therefore and so .

Generally, if is surjective but not bijective, then (2) of Theorem 22 fails.

*Example 23. *Let be the Hasse lattice implication algebra of Example 5. Let and , , , and . The Hasse diagram and the implication operator of are shown by Figure 2 and Table 2. Then it is clear that

A mapping from to is defined as It easy to check and is surjective. Let . Then and .

Since , by Theorem 4, Observe that . We have Moreover, by , .

In fact, we have Therefore