# On Power Idealization Filter Topologies of Lattice Implication Algebras

Authors: Shi-Zhong Bai; Xiu-Yun Wu
Journal: The Scientific World Journal
 (2014)

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## 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.


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## Body

## 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(L,∧,∨,0,1) be a bounded lattice with the greatest 1 and the smallest 0. A system (L,∧,∨,′,→,0,1) is called a quasi-lattice implication algebra if ′:L→L is an order-reserving involution and →:L×L→L is a map (called an implication operator) satisfying the following conditions for any x,y,z∈L:(1)
x→(y→z)=y→(x→z),(2)
x→x=1,(3)
x→y=y′→x′,(4)
x→y=y→x=1 implies x=y,(5)
(x→y)→y=(y→x)→x.A quasi-lattice implication algebra(L,∧,∨,′,→,0,1) is called a lattice implication algebra if the implication operator → further fulfils the following conditions:(6)
(x∨y)→z=(x→z)∧(y→z),(7)
(x∧y)→z=(x→z)∨(y→z).A lattice implication algebra(L,∧,∨,′,→,0,1) will be simply denoted by L.LetL be a lattice implication algebra and let φ be a subset of 2L. We use φc to denote the complement {L∖A:A∈φ}, where L∖A={x∈L:x∉A}. A subset τ⊆2L is called a topology on L, if τ satisfies the following:(1)
∅,L∈τ,(2)
A,B∈τ implies A∩B∈τ,(3)
{At∈τ:t∈T}⊆τ implies ∪t∈TAt∈τ.Elements ofτ are called τ-open sets and the complements of them are called τ-closed. The pair (L,τ) is called a topological space. A subset B of τ is called a base of τ, if for each A∈τ and each x∈A, there exists B∈B such that x∈B⊆A.LetL be an implication algebra. A subset F of L is called a filer, if F satisfies the following: (1) 1∈F; (2) x,x→y∈F implies y∈F. The collection of all filters in L is denoted by F(L), or F briefly. Clearly, F consists a base of some topology TF(L), briefly TF. Usually, TF is called the filter topology generated by F. And the pair (L,TF) is called the filter topological space. A subset U⊆L is called TF-neighborhood of x∈L, or neighborhood of x in TF if x∈U∈TF. The set of all TF-neighborhoods of x is denoted by NTF(x). Since F⊆TF and [x)=∩{F:x∈F∈F}∈F, [x) is the smallest element of NTF(x).The closure operator and interior operator ofTF are denoted by c and i. Clearly, for every A⊆L, c(A)=∩{L∖[x):x∈L,[x)∩A=∅} and i(A)=∪{[x):x∈L,[x)⊆A}. The following proposition describes c(A).Proposition 1.
Let(L,TF) be the filer topology generated by F(L). Then for A⊆L, c(A)={x∈L:[x)∩A≠∅}.Proof.
The proof is trivial since[x) is the smallest TF-neighborhood of x.LetL be a lattice implication algebra and let 2L be the power set of L. A nonempty subset I of 2L is called a power ideal of L if I satisfies the following: (1) A,B∈2L and A⊆B∈I imply A∈I; (2) A,B∈I implies A∪B∈I. The collection of all power ideals in 2L is denoted by I(L), or briefly I. Note that I∅={∅} is the smallest power ideal and IL=2L is the greatest power ideal. Moreover, if I,J∈I, then (1) I∩J∈I; (2) I∨J={I∪J:I∈I,J∈J}∈I.


## 2. Local Functions and Power Idealization Filter Topologies

LetL be a lattice implication algebra, let TF be the filter topology, and let I be a power ideal. An operator on⁡*  2L is defined as follows:
(1)A*(I,TF)={x∈L:∀U∈NTF(x),A∩U∉I}
for every A⊆L.The operatoris* called the local function with respect to TF and I. A* is called local function of A. We usually write A*(I) or A* instead of A*(I,TF).Clearly,x∈A* if and only if [x)∩A∉I. Thus A*={x∈L:[x)∩A∉I}. The following proposition gives some further details of A*.Proposition 2.
Let(L,TF) be the filter topological space and I,J∈I. Then(1)
A*(I∅)=c(A) and A*(IL)=∅;(2)
ifA⊆B, then A*(I)⊆B*(I);(3)
ifI⊆J, then A*(J)⊆A*(I);(4)
A*(I)=c(A*(I))⊆c(A);(5)
(A*)*(I)⊆A*(I);(6)
ifA⊆I, then A*(I)=∅;(7)
ifA∈TFc, then A*(I)⊆A;(8)
ifB∈I, then (A∪B)*(I)=A*(I)=(A∖B)*(I);(9)
(A∪A*(I))*(I)=A*(I);(10)
(A∪B)*(I)=A*(I)∪B*(I);(11)
A*(I)∖B*(I)=(A∖B)*(I)∖B*(I)⊆(A∖B)*(I);(12)
if{1}∉I, then [x)*(I)=L for each x∈L;(13)
if{1}∉I and 1∈A⊆L, then [A)*(I)=[A*(I))=L.Proof.
(1) By Proposition1, x∈A*(I∅) if and only if [x)∩A≠∅ if and only if x∈c(A). Thus A*(I∅)=c(A). Since [x)∩A∈2L=IL for each x∈L, A*(L)=∅.
(2) LetA⊆B and x∈A*(I). Then [x)∩A∉I. Since I is a power ideal and [x)∩A⊆[x)∩B, [x)∩B∉I and so x∈B*(I). Thus A*(I)⊆B*(I).
(3) LetI⊆J and x∈A*(I). Then [x)∩A∉J. It follows that [x)∩A∉I and so x∈A*(I). Thus A*(J)⊆A*(I).
(4) Ifx∉c(A), then x∈L∖c(A)∈TF and so [x)⊆L∖c(A). Thus [x)∩A⊆(L∖c(A))∩A=∅∈I. This implies x∉A*(I) and so A*(I)⊆c(A). Then c(A*(I))⊆c(c(A))=c(A).
It is clear thatA*(I)⊆c(A*(I)). Next, we prove c(A*(I))⊆A*(I).
Letx∈c(A*(I)). By Proposition 1, [x)∩A*(I)≠∅. Then there exists y∈[x)∩A*(I). By y∈A*(I), [y)∩A∉I. By y∈[x), [y)⊆[x). Thus [x)∩A∉I and so x∈A*(I). Therefore c(A*(I))⊆A*(I).
(5) By (4),(A*(I))*(I)⊆c(A*(I))=A*(I).
(6) Since[x)∩A⊆A∈I for each x∈L, A*(I)=∅.
(7) Suppose thatx∈A*(I)∖A. Then x∈L∖A∈TF. Thus [x)⊆L∖A and so [x)∩A⊆(L∖A)=∅∈I. Hence x∉A*(I) which is a contradiction. Therefore A*(I)⊆A.
(8) By (2),(A∖B)*(I)⊆A*(I)⊆(A∪B)*(I). Next, we prove the inverse inclusions.
Ifx∉(A∖B)*(I), then ([x)∩A)∖B=[x)∩(A∖B)∈I. Thus [x)∩A⊆I∪B∈I which follows from I is a power ideal. This implies x∉A*(I). Thus A*(I)⊆(A∖B)*(I) and so A*(I)=(A∖B)*(I).
Ifx∉A*(I), then [x)∩A∈I. Since B∈I,
(2)[x)∩(A∪B)⊆([x)∩A)∪B∈I.
Thus x∉(A∪B)*(I). This implies (A∪B)*(I)⊆A*(I) and so (A∪B)*(I)=A*(I).
(9) Clearly,A*(I)⊆(A∪A*(I))*(I). Conversely, if x∉A*(I), then [x)∩A∈I. Let [x)∩A=I. Then A⊆I∪(L∖[x)). By (2), (7), (8), and [x)∈TF,
(3)A*(I)⊆(I∪(L∖[x)))*(I)=(L∖[x))*(I)⊆L∖[x).
Thus A∪A*(I)⊆(L∖[x))∪A and so
(4)[x)∩(A∪A*(I))⊆((L∖[x))∪A)∩[x)=A∩[x)=I∈I.
This implies x∉(A∪A*(I))*(I) and so (A∪A*(I))*(I)⊆A*(I).
(10)A*(I)∪B*(I)⊆(A∪B)*(I) is clear. Conversely, if x∉A*(I)∪B*(I), then [x)∩A,[x)∩B∈I. Thus [x)∩(A∪B)=([x)∩A)∪([x)∪B)∈I. This implies x∉(A∪B)*(I). Therefore (A∪B)*(I)⊆A*(I)∪B*(I).
(11) We firstly proveA*(I)∖B*(I)⊆(A∖B)*(I). Assume that x∈(A*(I)∖B*(I))∖(A∖B)*(I). Then [x)∩(A∖B)∈I and [x)∩B∈I. Thus
(5)[x)∩A⊆[x)∩((A∖B)∪B)=([x)∩(A∖B))∪([x)∩B)∈I.
This implies x∉A*(I) which is a contradiction. Thus A*(I)∖B*(I)⊆(A∖B)*(I) and so A*(I)∖B*(I)⊆(A∖B)*(I)∖B*(I). Finally, (A∖B)*(I)∖B*(I)⊆A*(I)∖B*(I) follows from (2). Therefore A*(I)∖B*(I)=(A∖B)*(I)∖B*(I).
(12) Since{1}∉I, 1∉I for each I∈I. Assume that there exists x∈L such that [x)≠L. Let y∉L∖[x)*(I). Thus [y)∩[x)∈I. Since [x∨y)=[y)∩[x), 1∉[x∨y) which is a contradiction. Therefore [x)*(I)=L for each x∈L.
(13) Assume that there existsy∈L∖A*(I). Then 1∈[y)∩A∈I which is a contradiction. Thus A*(I)=L and [A*(I))=L. Since 1∈A, L=[1)*(I)⊆[A)*(I) follows from (12). Therefore [A)*(I)=L.Proposition 3.
Let(L,TF) be the filter topological space and I∈I. The operator cI* (briefly, c*) on 2L, defined by c*(A)=A∪A* for A⊆L, satisfies the following statements:(1)
c*(∅)=∅; c*(L)=L,(2)
c*(c*(A))=c*(A)⊆c(A),(3)
(c*(A))*=c*(A*)=A*,(4)
c*(A∪B)=c*(A)∪c*(B),(5)
A∈TFc or A∈I implies c*(A)=A.Proof.
(1)c*(∅)=∅ follows from ∅*=∅. c*(L)=L is clear.
(2) By (4) and (9) of Proposition2,
(6)c*(c*(A))=(A∪A*)∪(A∪A*)*=A∪A*=c*(A)⊆c(A).
(3) By (5) and (9) of Proposition2,
(7)(c*(A))*=(c*(A)∪(c*(A))*)*=(A∪A*∪(A∪A*)*)*=(A∪A*)*=A*
and c*(A*)=A*∪(A*)*=A*.
(4) By (10) of Proposition2,
(8)c*(A∪B)=(A∪B)∪(A∪B)*=(A∪A*)∪(B∪B*)=c*(A)∪c*(B).
(5) The result follows from (6) and (7) of Proposition2.Theorem 4.
Let(L,TF) be the filter topological space and I∈I. The operator c* stated in Proposition 3 is the closure operator of a new topology which is finer than TF and the topology generated by Ic (note that Ic is not a topology since ∅∉Ic in general case). Such a topology is called a power idealization filter topology and often denoted by TF*(I,TF), TF*(I), or TF*.Proof.
LetTF*={A⊆L:c*(L∖A)=L∖A}. We prove that TF* is a topology on L.
(1)∅,L∈TF* follows from (1) of Proposition 3.
(2) IfA,B∈TF*, then c*(L∖A)=L∖A and c*(L∖B)=L∖B. Thus
(9)c*(L∖(A∩B))=c*((L∖A)∪(L∖B))=c*(L∖A)∪c*(L∖B)=L∖(A∩B).
This implies A∩B∈TF*.
(3) LetAt∈TF* for t∈T, where T is an index set. Then c*(L∖At)=L∖At and
(10)L∖(∪t∈T)=∩t∈T(L∖At)⊆c*(∩t∈T(L∖At))=∩t∈T(L∖At)∪(∩t∈T(L∖At))*⊆∩t∈T((L∖At)∪(L∖At)*)=∩t∈Tc*(L∖At)=∩t∈T(L∖At)=L∖(∪t∈TAt).
Therefore ∪t∈TAt∈TF*.
Finally, by (6) and (7) of Proposition3, TF,Ic⊆TF*.Example 5.
LetL={0,a,b,c,d,1}, 0′=1, a′=c, b′=d, c′=a, d′=b, 1′=0, and the implication operator → be defined by a→b=a′∨b for a,b∈L. Then (L,∧,∨,′,→) is the Hasse lattice implication algebra (Figure 1 and Table 1). Then
(11)F(L)={{1},{a,1},{b,c,1},{a,b,c,d,1},L},TF(L)={∅,{1},{a,1},{b,c,1},{a,b,c,1},{a,b,c,d,1},L}.
Let I={∅,{0},{a},{0,a}}. Then I is a power ideal. It is easy to check that
(12)TF*={∅,{1},{a,1},{b,c,1},{a,b,c,1},{b,c,d,1},{0,b,c,d,1},{a,b,c,d,1},L}.
Clearly, TF⊆TF*.Table 1
The implication operator ofL={0,a,b,c,d,1}.
→
0
a
b
c
d
1
0
1
1
1
1
1
1
a
c
1
b
c
b
1
b
d
a
1
b
a
1
c
a
a
1
1
a
1
d
b
1
1
b
1
1
1
0
a
b
c
d
1Figure 1
Hasse diagram ofL2={0,a,b,c,d,1}.Proposition 6.
Let(L,TF) be the filter topological space and I,J∈I. Then(1)
TF*(I∅)=TF and TF*(IL)=2L,(2)
ifI⊆J, then TF*(I)⊆TF*(J).Proof.
(1) By (1) of Proposition2, A*(I∅)=c(A) and A*(IL)=∅. Thus cI∅*(A)=A if and only if c(A)=A. Similarly, cIL*(A)=A for each A⊆L. Therefore (1) holds.
(2) LetI⊆J. By (3) of Proposition 2, cJ*(A)⊆cI*(A) for A⊆L. Thus if A∈TF*(I), then A∈TF*(J). Therefore TF*(I)⊆TF*(J).Clearly, ifI∈I satisfies TFc⊆I, then L∈I and so I=2L=IL. Thus by (1) of Proposition 3, TF*=Ic=2L. If TFc∖{L}⊆I, we have the following proposition.Proposition 7.
Let(L,TF) be the filter topological space. If I∈I satisfies TFc∖{L}⊆I, then TF*=Ic∪{∅}.Proof.
I
c
∪
{
∅
}
⊆
T
F
* follows from Theorem 4. Conversely, suppose that TF*⊈Ic∪{∅}. Then there exists B∈TF*∖(Ic∪{∅}) such that (L∖B)*⊆L∖B≠L. Let y∈(L∖B)∖(L∖B)*. Then y∉(L∖B)*. Thus [y)∩(L∖B)∈I. Put [y)∩(L∖B)=I. We have L∖B⊆I∪(L∖[y)). Since TFc∖{L}⊆I, L∖[y)∈I. Thus L∖B⊆I∪(L∖B)∈I and so L∖B∈I. Hence B∈Ic. It is a contradiction. Therefore TF*⊆Ic∪{∅}.Lemma 8.
Let(L,TF) be the filter topological space and I∈I. If A⊆L satisfies A∩I=∅ for each I∈I, then c*(A)=c(A).Proof.
c
*
(
A
)
⊆
c
(
A
) is clear. Conversely, if x∉c*(A), then x∉A and x∉A*. Thus I=[x)∩A∈I and A⊆I∪(L∖[x)). Since A∩I=∅, A⊆L∖[x). Observe that x∉L∖[x) and L∖[x) is TF-closed. We have x∉c(A). Thus c(A)⊆c*(A). Therefore c*(A)=c(A).Proposition 9.
IfI∈TFc, then TF*=TF.Proof.
It is clear thatTF⊆TF*. Conversely, let IM be the greatest element of I and A⊆L. Thus (A∖IM)∩I=∅ for each I∈I. By Lemma 8, c*(A∖IM)=c(A∖IM). By (8) of Proposition 2, (A∖IM)*=A*. Now, notice that A∩IM∈I⊆TFc and thus c(A∩IM)=A∩IM. We have
(13)c(A)=c((A∖IM)∪(A∩IM))=c(A∖IM)∪c(A∩IM)=c*(A∖IM)∪(A∩IM)=(A∖IM)∪(A∖IM)*∪(A∩IM)=(A∖IM)*∪A=A*∪A=c*(A).
This implies c*=c. Therefore TF*=TF.Lemma 10.
Let(L,TF) be the filter topological space and I∈I. If I∈I and U∈TF, then U∖I∈TF*.Proof.
LetP=L∖U. Then P∈TFc. By (7) and (8) of Proposition 2,
(14)c*(P∪I)=(P∪I)∪(P∪I)*=(P∪I)∪P*=P∪I.
Thus L∖(P∪I)=U∩(L∖I)=U∖I∈TF*.Theorem 11.
Let(L,TF) be the filter topological space and I∈I. Then
(15)BTF*={U∖I:U∈TF,I∈I}
is a base of TF*. Moreover,
(16)BTF*(x)={V∖I:V∈NTF(x),x∉I∈I}
is a base of NTF*(x) for each x∈L, where NTF*(x) is the set of all TF*-neighborhoods of x in (L,TF*). Clearly, [x)∖Ix is the smallest TF*-neighborhoods of x, where Ix is the greatest element of I satisfying x∉Ix.Proof.
By Lemma10, BTF*⊆TF*. Let B⊆L. Then B∈TF*⇔L∖B is TF*-closed ⇔(L∖B)*⊆(L∖B)⇔B⊆L∖(L∖B)*. Thus x∈B⇒x∉(L∖B)*⇒ there exists U∈NTF(x) such that (L∖B)∩U∈I. Let I=(L∖B)∩U. Then L∖B⊆I∪(L∖U) and
(17)x∈U∖I=U∩(L∖I)=L∖(I∪(L∖U))⊆B.
Therefore BTF* is a base of TF*.
Clearly,BTF*(x)⊆NTF*(x). Let A∈NTF*(x) and y∈A. Since BTF* is a base of TF*, there are U,V∈TF and I,J∈I such that x∈U∖I⊆A and y∈V∖J⊆A. We can assume y∉I and x∉J (otherwise, I and J can be replaced by I∖{y} and J∖{x}, resp.,). Then
(18)(U∖I)∪(V∖J)=(U∩(L∖I))∪(V∩(L∖J))=[(U∪V)∩((L∖I)∪(L∖J))]∩[(U∪(L∖J))∩(V∪(L∖I))]⊇((U∪V)∖(I∩J))∩(L∖(I∪J))=(U∪V)∖(I∪J).
Clearly, y∈(U∪V)∖(I∪J)∈BTF*(x) and
(19)(U∪V)∖(I∪J)⊆(U∖I)∪(V∖J)⊆A.
Therefore BTF*(x) is a base of NTF*(x).
Clearly, ifIx is the greatest element of I satisfying x∉Ix, then [x)∖Ix∈NTF*(x) is the smallest TF*-neighborhoods of x.Let(L,τ) be a topological space and I∈I. The topology that was generated by B={U∖I:U∈τ,I∈I} is denoted by T*(I,τ) [5]. Clearly, T*(I,TF)=TF*(I).Lemma 12.
Letψ={∅,L} be the indiscrete topology on L and I∈I. Then T*(I,ψ)={∅}∪Ic.Proof.
ByTc∖{L}={∅}∈I and Proposition 7, the proof is obvious.Theorem 13.
Let(L,τ) be a topological space and I∈I. Then T*(I,τ)=τ∨T*(I,ψ), where τ∨T*(I,ψ) is the topology generated by the base {U∩V:U∈τ,V∈T*(I,ψ)}.Proof.
Clearly,BT*={U∖I:U∈τ,I∈I} is a base of T*(I,τ). Since BT*={U∩V:U∈τ,V∈T*(I,ψ)}, BT* is also a base of τ∨T*(I,ψ). Therefore T*(I,τ)=τ∨T*(I,ψ).Corollary 14.
Let(L,TF) be the filter topological space and I∈I. Then TF*=TF∨T*(I,ψ).Corollary 15.
Let(L,TF) be the filter topological space and I,J∈I. Then(1)
T*(I∨J,ψ)=T*(I,ψ)∨T*(J,ψ),(2)
TF*(I∨J)=T*(I,TF*(J))=T*(J,TF*(I)),(3)
TF*(I∨J)=TF*(I)∨TF*(J),(4)
T*(I,TF*(I))=TF*(I).Proof.
(1) By (2) of Proposition6, T*(I∨J)⊇T*(I)∨T*(J). Conversely, let ∅≠A∈T*(I∨J). By Theorem 13, there exist I∈I and J∈J such that
(20)A=L∖(I∪I)=(L∖I)∩(L∖J)∈T*(I)∨T*(J).
Thus T*(I∨J)⊆T*(I)∨T*(J).
(2) By (1), Theorem13, and Corollary 14,
(21)TF*(I∨J)=TF∨T*(I∨J,ψ)=TF∨T*(I,ψ)∨T*(J,ψ)=TF*(I)∨T*(J,ψ)=T*(TF*(I),J).
Similarly,TF*(I∨J)=T*(TF*(J),I).
(3) By (1) and Theorem13,
(22)TF*(I∨J)=TF∨T*(I∨J,ψ)=TF∨T*(I,ψ)∨T*(J,ψ)=TF∨T*(I,ψ)∨TF∨T*(J,ψ)=TF*(I)∨TF*(J).
Therefore TF*(I∨J)=TF*(I)∨TF*(J).
(4) LetI=J. Then the proof follows from (2).Theorem 16.
Let(L,TF) be the filter topological space, I,J∈I and A⊆L. Then(1)
A*(I∩J,TF)=A*(I,TF)∪A*(J,TF),(2)
A*(I∨J,TF)=A*(I,TF*(J))∩A*(J,TF*(I)).Proof.
(1) By (3) of Proposition2, A*(I,TF)∪A*(J,TF)⊆A*(I∩J,TF). Conversely, x∉A*(I,TF)∪A*(J,TF). Then [x)∩A∈I and [x)∩A∈J. Let [x)∩A=I and [x)∩A=J. Then A⊆I∪(L∖[x)) and A⊆J∪(L∖[x)). Thus
(23)A⊆(I∪(L∖[x)))∩(J∪(L∖[x)))=(I∩J)∪(L∖[x)).
Thus [x)∩A⊆I∩J which implies x∉A*(I∩J,TF). Therefore A*(I∩J,TF)⊆A*(I,TF)∪A*(J,TF).
(3) Letx∉A*(I∨J,TF). Then [x)∩A∈I∨J. Then there exist I∈I and J∈J such that [x)∩A=I∪J. We can assume I∩J=∅, (otherwise, I can be replaced by I∖(I∩J)). Thus x∉I or x∉J, (otherwise, x∈I∩J which is a contradiction). Now, we take x∉I for example. Then
(24)([x)∖I)∩A=[x)∩A∩(L∖I)=J.
Since [x)∈TF and x∈[x)∖I, [x)∖I∈BTF*(I). Thus x∉A*(J,TF*(I)) and so x∉A*(J,TF*(I))∩A*(I,TF*(J)). Hence
(25)A*(J,TF*(I))∩A*(I,TF*(J))⊆A*(I∨J,TF).
Conversely, letx∉A*(TF*(I),J). Then there exists I∈I such that ([x)∖I)∩A∈J. Let ([x)∖I)∩A=J. Then [x)∩A=I∪J which implies x∉A*(I∨J,TF). Similarly, if x∉A*(TF*(J),I), then x∉A*(I∨J,TF). Therefore A*(I∨J,TF)⊆A*(J,TF*(I))∩A*(I,TF*(J)).Corollary 17.
ConsiderA*(I,TF)=A*(I,TF*(I)).Proof.
LetI=J. The proof follows from (2) of Theorem 16.Corollary 18.
ConsiderTF*(I∩J)=TF*(I)∩TF*(J).Proof.
By (2) of Proposition6, TF*(I∩J)⊆TF*(I)∩TF*(J). Conversely, if A∉TF*(I∩J), then
(26)(L∖A)*(I,TF)∪(L∖A)*(J,TF)=(L∖A)*(I∩J,TF)⊈(L∖A).
Thus (L∖A)*(I,TF)⊈(L∖A) or (L∖A)*(J,TF)⊈(L∖A). Thus A∉TF*(I) or A∉TF*(J). Therefore TF*(I)∩TF*(J)⊆TF*(I∩J).


## 3. Power Idealization Filter Topological Quotient Spaces

Let(L1,∧,∨,′,→1,01,11) and (L2,∧,∨,′,→2,02,12) be two lattice implication algebras. A mapping f from L1 to L2 is called lattice implication homomorphism, if f(x→1y)=f(x)→2f(y) for any x,y∈L1. The set of all lattice implication homomorphisms from L1 to L2 is denoted by hom(L1,L2).Letf∈hom(L1,L2). Then, clearly,
(27)Tl(TF(L2),f)={f-1(U):U∈TF(L2)},Tr(TF(L1),f)={f(U):U⊆L2,f-1(U)∈TF(L1)}
are topologies [4].Lemma 19.
LetL1 and L2 be two implication algebras and let f∈hom(F1,f2) and I∈2L1, J∈2L2 be power ideals.(1)
Iff is injective, then f-1(J)={f-1(J):J∈J}∈I(L1).(2)
Iff is surjective, then f(I)={f(I):I∈I}∈I(L2).Proof.
(1) Since∅∈J, then ∅=f-1(∅)∈f-1(J).
IfI2∈f-1(J) and I1⊆I2, then there exist J2∈J such that I2=f-1(J2). Thus f(I1)⊆f(I2)=J2 and f(I1)∈J. Since f is injective, I1=f-1(f(I1))∈f-1(J).
IfI1,I2∈f-1(J), then there exist J1,J2∈J such that I1=f-1(J1) and I2=f-1(J2). One has I1∪I2=f-1(J1)∪f-1(J2)=f-1(J1∪J2). Since J1∪J2∈J, I1∪I2∈f-1(J). Therefore f-1(J) is a power ideal.
(2) By∅∈I, ∅=f(∅)∈f(I).
IfJ2∈f(J) and J1⊆J2, then there exists I2∈I such that J2=f(I2). Let I1=f-1(J1). Then I1⊆I2 and I1∈I. Since f is surjective, J1=f(I1)∈f(I).
IfJ1,J2∈f(J), then there exist I1,I2∈I such that J1=f(I1) and J2=f(I2). Thus I1∪I2∈I. Hence J1∪J2=f(I1)∪f(I2)=f(I1∪I2)∈f(I). Therefore f(I) is a power ideal.Lemma 20.
LetL1 and L2 be two implication algebras and f∈hom(F1,F2). Then(1)
iff is injective, then for each x∈L1, f-1([f(x)))∈Tl(TF(L2),f) is the smallest Tl-neighborhood of x;(2)
iff is bijective, then for each y∈L2, f([f-1(y)))∈Tr(TF(L1),f) is the smallest Tr-neighborhood of y.Proof.
(1) Clearly,[f(x)) is the smallest TF(L2)-neighborhood of f(x). Then x∈f-1([f(x)))∈Tl(TF(L2),f). Let x∈V∈Tl(TF(L2),f). Then f(x)∈f(V)∈TF(L2) and [f(x))⊆f(V). Thus f-1([f(x)))⊆f-1(f(V))=V. Therefore f-1([f(x))) is the smallest one.
(2) Clearly,[f-1(y)) is the smallest TF(L1)-neighborhood of f-1(y). Thus y∈f([f-1(y)))∈Tr(TF(L1),f). Now, let y∈U∈Tr(TF(L1),f). Then f-1(y)∈f-1(U)∈TF(L1). Thus [f-1(y))⊆f-1(U) and so f([f-1(y)))⊆f(f-1(U))=U. Therefore (2) holds.Lemma 21.
LetL1 and L2 be two implication algebras and f∈hom(F1,F2).(1)
Iff is injective and J∈I(L2), then for each x∈L1, [f(x))∖JM∈TF*(J,TF(L2)) is the smallest neighborhood of f(x), where JM is the greatest element of J satisfying f(x)∉JM.(2)
Iff is bijective and I∈I(L1), then for each y∈L2, [f-1(y))∖IM∈TF*(I,TF(L1)) is the smallest neighborhood of f-1(y), where IM is the greatest element of I satisfying f-1(y)∉IM.Theorem 22.
LetL1 and L2 be two implication algebras and f∈hom(F1,F2).(1)
Iff is injective and J∈I(L2), then
(28)T*(f-1(J),Tl(TF(L2),f))=Tl(TF*(J,TF(L2)),f).(2)
Iff is bijective and I∈I(L1), then
(29)T*(f(I),Tr(TF(L1),f))=Tr(TF*(I,TF(L1)),f).Proof.
(1) Letc2* and cl be the closure operators of the left side and the right side of the equation. We only need to prove c1*=cl.
LetA⊆L1 and x∉c1*(A). Then x∉A and x∉A*(f-1(J),Tl(TF(L2),f)). By (1) of Lemma 21, f-1([f(x)))∩A∈f-1(J). Thus there exists J∈J such that f-1([f(x)))∩A=f-1(J). Since x∉A and f is injective, f(x)∉J. Let JM∈J be the greatest one satisfying f(x)∉JM. Then f-1([f(x)))∩A⊆f-1(JM). Thus
(30)∅=f-1([f(x)))∩A∩(L1∖f-1(JM))=f-1([f(x)))∩f-1(L2∖JM)∩A=f-1([f(x))∩(L2∖JM))∩A=f-1([f(x))∖JM)∩A.
By Lemma 21 and Proposition 1, x∉cl(A). Therefore cl(A)⊆c1*(A).
Conversely, lety∉cl(A). By Proposition 1,
(31)∅=f-1([f(x))∖JM)∩A=f-1([f(x)))∩A∩(L1∖f-1(JM)).
Thus x∉A, f-1([f(x)))∩A⊆f-1(JM) and [f(x))∩f(A)⊆JM. Then J1=[f(x))∩f(A)∈J. Since f is injective, f-1([f(x)))∩A=f-1(J1). By (1) of Lemma 20, x∉A*(f-1(J),Tl(TF(L2),f)). Therefore x∉c1*(A) and c1*(A)⊆cl(A).
(2) Letc2* and cr be the closure operators of the left side and the right side of the equation. We only need to prove c2*=cr.
Lety∉c2*(A). Then y∉A and y∉A*((f(I),Tr(TF(L1),f))). By (2) of Lemma 20, f([f-1(y)))∩A∈f(I). Thus there exists I∈I such that f([f-1(y)))∩A=f(I). Since f is bijective, [f-1(y))∩f-1(A)=I. By y∉A, f-1(y)∉f-1(A) and so f-1(y)∉I. Since IM∈I is the greatest element of I satisfying f-1(y)∉IM, [f-1(y))∩f-1(A)⊆IM and [f-1(y))∩(L1∖IM)∩f-1(A)=∅. By f being bijective again, we have
(32)∅=f(∅)=f([f-1(y)))∩(L2∖f(IM))∩A=f([f-1(y))∖IM)∩A.
By
(33)f-1(y)∈[f-1(y))∖IM=f-1(f([f-1(y))∖IM))∈TF*(I,TF(L1)),y∉cr(A). Hence c2*(A)⊆cr(A).
Conversely, letz∉cr(A). Since IM is the greatest element of I satisfying f-1(z)∉IM, by (2) of Lemma 21, [f-1(z))∖IM∈TF*(I,TF(L1)) is the smallest neighborhood of f-1(z). Thus f([f-1(z))∖IM)∩A=∅. Since f is bijective,
(34)∅=f-1(∅)=([f-1(z))∖IM)∩f-1(A)=[f-1(z))∩(L1∖IM)∩f-1(A).
This implies [f-1(z))∩f-1(A)⊆IM and [f-1(z))∩f-1(A)∈I. Thus f([f-1(y)))∩A∈f(I) which implies y∈A*((f(I),Tr(TF(L1),f))). Since z∉cr(A), z∉A. Therefore z∉c2*(A) and so c2*(A)⊆cr(A).Generally, iff∈hom(L1,L2) is surjective but not bijective, then (2) of Theorem 22 fails.Example 23.
LetL1={01,a,b,c,d,11} be the Hasse lattice implication algebra of Example 5. Let L2={02,e,h,12} and 02′=12, e′=h, h′=e, and 12′=02. The Hasse diagram and the implication operator of L2 are shown by Figure 2 and Table 2. Then it is clear that
(35)TF(L1)*={∅,{11},{a,11},{b,c,11},{a,b,c,11},{b,c,d,11},{0,b,c,d,11},{a,b,c,d,11},L},TF(L2)*={∅,{e,12},{h,12},{e,h,12},{12},L2}.Table 2
The implication operator ofL2={02,e,h,12}.
→
0
2
e
h
1
2
0
2
1
1
2
1
2
1
2
e
h
1
2
h
1
2
h
e
e
1
2
1
2
1
2
0
2
e
h
1
2Figure 2
Hasse diagram ofL2={02,e,h,12}.A mappingf from L1 to L2 is defined as
(36)f(01)=02,f(11)=12,f(a)=f(d)=e,f(b)=f(c)=h.
It easy to check f∈hom(L1,L2) and f is surjective. Let I={∅,{c}}. Then I∈I(L1) and f(I)={∅,{h}}∈I(L2).Since{h}∈f(I), by Theorem 4,
(37){02,e,12}=L2∖{h}∈T*(f(I),Tr(TF(L1),f)).
Observe that {b,c}*(I,TF(L1))={01,b,c,d}⊈{b,c}. We have
(38){0,a,d,1}=L1∖{b,c}∉TF*(I,TF(L1)).
Moreover, by f-1({02,e,12})={02,a,d,12}, {02,e,12}∉Tr(TF*(I,TF(L1)),f).In fact, we have(39)Tr(TF*(I,TF(L1)),f)={∅,{12},L2},T*(f(I),Tr(TF(L1),f))={∅,{12},{02,e,12},L2}.
Therefore Tr(TF*(I,TF(L1)),f)≠T*(f(I),Tr(TF(L1),f)).Corollary 24.
LetL1 and L2 be two implication algebras and let f∈hom(F1,f2) be bijective.(1)
IfJ∈I(L2), then
(40)Tr(T*(f-1(J),Tl(TF(L2),f)),f)=T*(J,TF(L2)).(2)
IfI∈I(L1), then
(41)Tl(T*(f(I),Tr(TF(L1),f)),f)=T*(I,TF(L1)).Proof.
The proof follows from Theorem22.Corollary 25.
LetL1 and L2 be two implication algebras and f∈hom(f1,f2).(1)
Iff is injective and J1,J2∈I(L2), then
(42)Tl(TF*(J1∨J2,TF(L2)),f)=Tl(TF*(J1,TF(L2)),f)∨Tl(TF*(J2,TF(L2)),f).(2)
Iff is bijective and I1,I2∈I(L1), then
(43)Tr(TF*(I1∨I2,TF(L1)),f)=Tr(TF*(I1,TF(L1)),f)∨Tr(TF*(I2,TF(L1)),f).Proof.
(1) Clearly,f-1(J1∨J2)=f-1(J1)∨f-1(J2). By (3) of Corollary 15 and Theorem 22,
(44)Tl(TF*(J1∨J2,TF(L2)),f)=T*(f-1(J1∨J2),Tl(TF(L2),f))=T*(f-1(J1),Tl(TF(L2),f))∨T*(f-1(J2),Tl(TF(L2),f))=Tl(TF*(J1,TF(L2)),f)∨Tl(TF*(J2,TF(L2)),f).
(2) Is similar to (1).Corollary 26.
LetL1 and L2 be two implication algebras and let f∈hom(F1,f2) be bijective. If J∈I(L2) and I∈I(L1), then
(45)T*(f-1(J),Tl(TF*(J,TF(L2)),f))=T*(f-1(J),Tl(TF(L2),f)),T*(f(I),Tr(TF*(I,TF(L1)),f))=T*(f(I),Tr(TF(L1),f)).Proof.
The proof follows from (4) of Corollary15 and Theorem 22.



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