Research ArticleOn Power Idealization Filter Topologies ofLattice Implication Algebras

Shi-Zhong Bai12 and Xiu-Yun Wu13

1 School of Mathematics and Computational Science Xiangtan University Xiangtan 411105 China2 School of Mathematics and Computational Science Wuyi University Jiangmen 529020 China3Department of Mathematics and Computational Science Hunan University of Science and Engineering Yongzhou 425100 China

Correspondence should be addressed to Xiu-Yun Wu wuxiuyun2000126com

Received 24 June 2014 Accepted 2 August 2014 Published 28 August 2014

Academic Editor Jianming Zhan

Copyright copy 2014 S-Z Bai and X-Y Wu This is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work is properlycited

The aim of this paper is to introduce power idealization filter topologies with respect to filter topologies and power ideals of latticeimplication 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 implica-tion algebras [1] Xu [2] defined the concept of lattice impli-cation algebra which is regarded as an efficient approach todeal with lattice valued logical systems Later Xu and Qin [3]defined the concept of the filer topology of a lattice impli-cation algebra which takes the set of all filters of the latticeimplication algebra as a base Based on these definitions andsome results in [4] we introduce and study power idealiza-tion topologies with respect to filter topologies and powerideals of lattice implication algebras

Now we recall some definitions and notions of latticeimplication algebras and topological spaces

Let (119871 and or 0 1) be a bounded lattice with the greatest 1and the smallest 0 A system (119871 and or

1015840 rarr 0 1) is called aquasi-lattice implication algebra if 1015840 119871 rarr 119871 is an order-reserving involution and rarr 119871 times 119871 rarr 119871 is a map (called animplication operator) satisfying the following conditions forany 119909 119910 119911 isin 119871

(1) 119909 rarr (119910 rarr 119911) = 119910 rarr (119909 rarr 119911)(2) 119909 rarr 119909 = 1(3) 119909 rarr 119910 = 119910

1015840 rarr 1199091015840(4) 119909 rarr 119910 = 119910 rarr 119909 = 1 implies 119909 = 119910(5) (119909 rarr 119910) rarr 119910 = (119910 rarr 119909) rarr 119909

A quasi-lattice implication algebra (119871 and or1015840 rarr 0 1) is

called a lattice implication algebra if the implication operatorrarr further fulfils the following conditions

(6) (119909 or 119910) rarr 119911 = (119909 rarr 119911) and (119910 rarr 119911)(7) (119909 and 119910) rarr 119911 = (119909 rarr 119911) or (119910 rarr 119911)

A lattice implication algebra (119871 and or1015840 rarr 0 1) will be

simply denoted by 119871Let 119871 be a lattice implication algebra and let 120593 be a subset

of 2119871 We use 120593119888 to denote the complement 119871 119860 119860 isin 120593where 119871 119860 = 119909 isin 119871 119909 notin 119860 A subset 120591 sube 2119871 is called atopology on 119871 if 120591 satisfies the following

(1) 0 119871 isin 120591(2) 119860 119861 isin 120591 implies 119860 cap 119861 isin 120591(3) 119860

119905isin 120591 119905 isin 119879 sube 120591 implies cup

119905isin119879119860119905isin 120591

Elements of 120591 are called 120591-open sets and the complementsof them are called 120591-closed The pair (119871 120591) is called atopological space A subsetB of 120591 is called a base of 120591 if foreach 119860 isin 120591 and each 119909 isin 119860 there exists 119861 isin B such that119909 isin 119861 sube 119860

Let 119871 be an implication algebra A subset 119865 of 119871 is calleda filer if 119865 satisfies the following (1) 1 isin 119865 (2) 119909 119909 rarr 119910 isin 119865

implies 119910 isin 119865 The collection of all filters in 119871 is denotedby F(119871) or F briefly Clearly F consists a base of some

Hindawi Publishing Corporatione Scientific World JournalVolume 2014 Article ID 812145 8 pageshttpdxdoiorg1011552014812145

2 The Scientific World Journal

topology 119879F(119871) briefly 119879F Usually 119879F is called the filtertopology generated by F And the pair (119871 119879F) is called thefilter topological spaceA subset119880 sube 119871 is called119879F-neighbor-hood of 119909 isin 119871 or neighborhood of 119909 in 119879F if 119909 isin 119880 isin 119879FThe set of all 119879F-neighborhoods of 119909 is denoted by N

119879F(119909)

Since F sube 119879F and [119909) = cap119865 119909 isin 119865 isin F isin F [119909) is thesmallest element ofN

119879F(119909)

The closure operator and interior operator of 119879F aredenoted by 119888 and 119894 Clearly for every119860 sube 119871 119888(119860) = cap119871[119909)

119909 isin 119871 [119909) cap 119860 = 0 and 119894(119860) = cup[119909) 119909 isin 119871 [119909) sube 119860 Thefollowing proposition describes 119888(119860)

Proposition 1 Let (119871 119879F) be the filer topology generated byF(119871) Then for 119860 sube 119871 119888(119860) = 119909 isin 119871 [119909) cap 119860 = 0

Proof The proof is trivial since [119909) is the smallest 119879F-neighborhood of 119909

Let 119871 be a lattice implication algebra and let 2119871 be thepower set of 119871 A nonempty subsetI of 2119871 is called a powerideal of 119871 if I satisfies the following (1) 119860 119861 isin 2119871 and 119860 sube

119861 isin I imply 119860 isin I (2) 119860 119861 isin I implies 119860 cup 119861 isin I Thecollection of all power ideals in 2119871 is denoted by I(119871) orbrieflyI Note thatI

0= 0 is the smallest power ideal and

I119871= 2119871 is the greatest power ideal Moreover ifIJ isin I

then (1)IcapJ isin I (2)IorJ = 119868cup 119869 119868 isin I 119869 isin J isin I

2 Local Functions and Power IdealizationFilter Topologies

Let 119871 be a lattice implication algebra let 119879F be the filter topo-logy and let I be a power ideal An operator lowast on 2119871 isdefined as follows

119860lowast(I 119879F) = 119909 isin 119871 forall119880 isin N

119879F(119909) 119860 cap 119880 notin I (1)

for every 119860 sube 119871The operator lowast is called the local function with respect to

119879F andI 119860lowast is called local function of 119860 We usually write119860lowast(I) or 119860lowast instead of 119860lowast(I 119879F)Clearly 119909 isin 119860

lowast if and only if [119909) cap 119860 notin I Thus 119860lowast =

119909 isin 119871 [119909) cap 119860 notin I The following proposition gives somefurther details of 119860lowast

Proposition 2 Let (119871 119879F) be the filter topological space andIJ isin I Then

(1) 119860lowast(I0) = 119888(119860) and 119860lowast(I

119871) = 0

(2) if 119860 sube 119861 then 119860lowast(I) sube 119861lowast(I)(3) ifI sube J then 119860lowast(J) sube 119860lowast(I)(4) 119860lowast(I) = 119888(119860lowast(I)) sube 119888(119860)(5) (119860lowast)lowast(I) sube 119860lowast(I)(6) if 119860 sube I then 119860lowast(I) = 0(7) if 119860 isin 119879

119888

F then 119860lowast(I) sube 119860

(8) if 119861 isin I then (119860 cup 119861)lowast(I) = 119860lowast(I) = (119860 119861)

lowast(I)

(9) (119860 cup 119860lowast(I))lowast(I) = 119860lowast(I)

(10) (119860 cup 119861)lowast(I) = 119860lowast(I) cup 119861lowast(I)

(11) 119860lowast(I)119861lowast(I) = (119860119861)

lowast(I)119861

lowast(I) sube (119860119861)

lowast(I)

(12) if 1 notin I then [119909)lowast(I) = 119871 for each 119909 isin 119871

(13) if 1 notin I and 1 isin 119860 sube 119871 then [119860)lowast(I) = [119860lowast(I)) =

119871

Proof (1) By Proposition 1 119909 isin 119860lowast(I0) if and only if [119909) cap

119860 = 0 if and only if 119909 isin 119888(119860) Thus 119860lowast(I0) = 119888(119860) Since

[119909) cap 119860 isin 2119871 = I119871for each 119909 isin 119871 119860lowast(119871) = 0

(2) Let 119860 sube 119861 and 119909 isin 119860lowast(I) Then [119909) cap 119860 notin I SinceI is a power ideal and [119909) cap 119860 sube [119909) cap 119861 [119909) cap 119861 notin I and so119909 isin 119861

lowast(I) Thus 119860lowast(I) sube 119861lowast(I)(3) Let I sube J and 119909 isin 119860lowast(I) Then [119909) cap 119860 notin J It

follows that [119909) cap 119860 notin I and so 119909 isin 119860lowast(I) Thus 119860lowast(J) sube

119860lowast(I)(4) If 119909 notin 119888(119860) then 119909 isin 119871 119888(119860) isin 119879F and so [119909) sube

119871 119888(119860) Thus [119909) cap 119860 sube (119871 119888(119860)) cap 119860 = 0 isin I Thisimplies 119909 notin 119860lowast(I) and so 119860lowast(I) sube 119888(119860) Then 119888(119860lowast(I)) sube

119888(119888(119860)) = 119888(119860)It is clear that 119860

lowast(I) sube 119888(119860lowast(I)) Next we prove119888(119860lowast(I)) sube 119860lowast(I)

Let 119909 isin 119888(119860lowast(I)) By Proposition 1 [119909) cap 119860lowast(I) = 0Then there exists 119910 isin [119909) cap 119860lowast(I) By 119910 isin 119860lowast(I) [119910) cap 119860 notin

I By119910 isin [119909) [119910) sube [119909)Thus [119909)cap119860 notin I and so119909 isin 119860lowast(I)Therefore 119888(119860lowast(I)) sube 119860lowast(I)

(5) By (4) (119860lowast(I))lowast(I) sube 119888(119860lowast(I)) = 119860lowast(I)

(6) Since [119909) cap 119860 sube 119860 isin I for each 119909 isin 119871 119860lowast(I) = 0(7) Suppose that 119909 isin 119860lowast(I) 119860 Then 119909 isin 119871 119860 isin 119879F

Thus [119909) sube 119871 119860 and so [119909) cap 119860 sube (119871 119860) = 0 isin I Hence119909 notin 119860

lowast(I) which is a contradiction Therefore 119860lowast(I) sube 119860(8) By (2) (119860 119861)

lowast(I) sube 119860lowast(I) sube (119860 cup 119861)

lowast(I) Next

we prove the inverse inclusionsIf 119909 notin (119860119861)

lowast(I) then ([119909)cap119860) 119861 = [119909)cap (119860119861) isin I

Thus [119909) cap 119860 sube 119868 cup 119861 isin I which follows fromI is a powerideal This implies 119909 notin 119860

lowast(I) Thus 119860lowast(I) sube (119860 119861)lowast(I)

and so 119860lowast(I) = (119860 119861)lowast(I)

If 119909 notin 119860lowast(I) then [119909) cap 119860 isin I Since 119861 isin I

[119909) cap (119860 cup 119861) sube ([119909) cap 119860) cup 119861 isin I (2)

Thus 119909 notin (119860 cup 119861)lowast(I) This implies (119860 cup 119861)

lowast(I) sube 119860

lowast(I)

and so (119860 cup 119861)lowast(I) = 119860lowast(I)

(9) Clearly 119860lowast(I) sube (119860 cup 119860lowast(I))lowast(I) Conversely if

119909 notin 119860lowast(I) then [119909) cap 119860 isin I Let [119909) cap 119860 = 119868 Then 119860 sube

119868 cup (119871 [119909)) By (2) (7) (8) and [119909) isin 119879F

119860lowast

(I) sube (119868 cup (119871 [119909)))lowast

(I) = (119871 [119909))lowast

(I) sube 119871 [119909)

(3)

Thus 119860 cup 119860lowast(I) sube (119871 [119909)) cup 119860 and so

[119909) cap (119860 cup 119860lowast

(I)) sube ((119871 [119909)) cup 119860) cap [119909)

= 119860 cap [119909) = 119868 isin I(4)

This implies119909 notin (119860cup119860lowast(I))lowast(I) and so (119860cup119860lowast(I))

lowast(I) sube

119860lowast(I)(10) 119860lowast(I) cup 119861lowast(I) sube (119860 cup 119861)

lowast(I) is clear Conversely

if 119909 notin 119860lowast(I) cup 119861lowast(I) then [119909) cap 119860 [119909) cap 119861 isin I Thus

The Scientific World Journal 3

[119909) cap (119860 cup 119861) = ([119909) cap 119860) cup ([119909) cup 119861) isin I This implies119909 notin (119860 cup 119861)

lowast(I) Therefore (119860 cup 119861)

lowast(I) sube 119860lowast(I) cup 119861lowast(I)

(11) We firstly prove 119860lowast(I) 119861lowast(I) sube (119860 119861)lowast(I)

Assume that 119909 isin (119860lowast(I) 119861lowast(I)) (119860 119861)lowast(I) Then

[119909) cap (119860 119861) isin I and [119909) cap 119861 isin I Thus

[119909) cap 119860 sube [119909) cap ((119860 119861) cup 119861)

= ([119909) cap (119860 119861)) cup ([119909) cap 119861) isin I(5)

This implies 119909 notin 119860lowast(I) which is a contradiction Thus119860lowast(I) 119861lowast(I) sube (119860 119861)

lowast(I) and so119860lowast(I) 119861lowast(I) sube (119860

119861)lowast(I)119861lowast(I) Finally (119860119861)

lowast(I)119861lowast(I) sube 119860lowast(I)119861lowast(I)

follows from (2) Therefore 119860lowast(I) 119861lowast(I) = (119860 119861)lowast(I)

119861lowast(I)(12) Since 1 notin I 1 notin 119868 for each 119868 isin I Assume that

there exists 119909 isin 119871 such that [119909) = 119871 Let 119910 notin 119871[119909)lowast(I)Thus

[119910) cap [119909) isin I Since [119909 or 119910) = [119910) cap [119909) 1 notin [119909 or 119910) which isa contradiction Therefore [119909)lowast(I) = 119871 for each 119909 isin 119871

(13) Assume that there exists 119910 isin 119871 119860lowast(I) Then 1 isin

[119910) cap 119860 isin I which is a contradiction Thus 119860lowast(I) = 119871 and[119860lowast(I)) = 119871 Since 1 isin 119860 119871 = [1)

lowast(I) sube [119860)

lowast(I) follows

from (12) Therefore [119860)lowast(I) = 119871

Proposition 3 Let (119871 119879F) be the filter topological space andI isin I The operator 119888lowastI (briefly 119888lowast) on 2119871 defined by 119888lowast(119860) =

119860 cup 119860lowast for 119860 sube 119871 satisfies the following statements

(1) 119888lowast(0) = 0 119888lowast(119871) = 119871(2) 119888lowast(119888lowast(119860)) = 119888lowast(119860) sube 119888(119860)(3) (119888lowast(119860))

lowast= 119888lowast(119860lowast) = 119860lowast

(4) 119888lowast(119860 cup 119861) = 119888lowast(119860) cup 119888lowast(119861)(5) 119860 isin 119879119888F or 119860 isin I implies 119888lowast(119860) = 119860

Proof (1) 119888lowast(0) = 0 follows from 0lowast = 0 119888lowast(119871) = 119871 is clear(2) By (4) and (9) of Proposition 2

119888lowast(119888lowast

(119860)) = (119860 cup 119860lowast) cup (119860 cup 119860

lowast)lowast

= 119860 cup 119860lowast= 119888lowast

(119860) sube 119888 (119860) (6)

(3) By (5) and (9) of Proposition 2

(119888lowast

(119860))lowast

= (119888lowast

(119860) cup (119888lowast

(119860))lowast

)lowast

= (119860 cup 119860lowastcup (119860 cup 119860

lowast)lowast

)lowast

= (119860 cup 119860lowast)lowast

= 119860lowast

(7)

and 119888lowast(119860lowast) = 119860lowast cup (119860lowast)lowast= 119860lowast

(4) By (10) of Proposition 2

119888lowast

(119860 cup 119861) = (119860 cup 119861) cup (119860 cup 119861)lowast

= (119860 cup 119860lowast) cup (119861 cup 119861

lowast) = 119888lowast

(119860) cup 119888lowast

(119861) (8)

(5) The result follows from (6) and (7) of Proposition 2

Theorem 4 Let (119871 119879F) be the filter topological space andI isin I The operator 119888lowast stated in Proposition 3 is the closure

operator of a new topology which is finer than 119879F and thetopology generated byI119888 (note thatI119888 is not a topology since0 notin I119888 in general case) Such a topology is called a poweridealization filter topology and often denoted by 119879lowastF(I 119879F)119879lowastF(I) or 119879lowastF

Proof Let 119879lowastF = 119860 sube 119871 119888lowast(119871 119860) = 119871 119860 We prove that119879lowastF is a topology on 119871

(1) 0 119871 isin 119879lowast

F follows from (1) of Proposition 3(2) If119860 119861 isin 119879

lowast

F then 119888lowast(119871119860) = 119871119860 and 119888

lowast(119871119861) = 119871119861

Thus

119888lowast

(119871 (119860 cap 119861)) = 119888lowast

((119871 119860) cup (119871 119861))

= 119888lowast

(119871 119860) cup 119888lowast

(119871 119861) = 119871 (119860 cap 119861)

(9)

This implies 119860 cap 119861 isin 119879lowastF(3) Let 119860

119905isin 119879lowastF for 119905 isin 119879 where 119879 is an index set Then

119888lowast(119871 119860119905) = 119871 119860

119905and

119871 (cup119905isin119879

) = cap119905isin119879

(119871 119860119905) sube 119888lowast(cap119905isin119879

(119871 119860119905))

= cap119905isin119879

(119871 119860119905) cup (cap

119905isin119879(119871 119860

119905))lowast

sube cap119905isin119879

((119871 119860119905) cup (119871 119860

119905)lowast

)

= cap119905isin119879

119888lowast(119871 119860

119905) = cap119905isin119879

(119871 119860119905) = 119871 (cup

119905isin119879119860119905)

(10)

Therefore cup119905isin119879

119860119905isin 119879lowastF

Finally by (6) and (7) of Proposition 3119879FI119888 sube 119879lowastF

Example 5 Let 119871 = 0 119886 119887 119888 119889 1 01015840 = 1 1198861015840 = 119888 1198871015840 = 1198891198881015840 = 119886 1198891015840 = 119887 11015840 = 0 and the implication operator rarr bedefined by 119886 rarr 119887 = 1198861015840 or 119887 for 119886 119887 isin 119871 Then (119871 and or 1015840 rarr )

is the Hasse lattice implication algebra (Figure 1 and Table 1)Then

F (119871) = 1 119886 1 119887 119888 1 119886 119887 119888 119889 1 119871

119879F(119871) = 0 1 119886 1 119887 119888 1 119886 119887 119888 1 119886 119887 119888 119889 1 119871

(11)

LetI = 0 0 119886 0 119886 ThenI is a power ideal It is easyto check that

119879lowast

F = 0 1 119886 1 119887 119888 1 119886 119887 119888 1 119887 119888 119889 1

0 119887 119888 119889 1 119886 119887 119888 119889 1 119871 (12)

Clearly 119879F sube 119879lowastF

Proposition 6 Let (119871 119879F) be the filter topological space andIJ isin I Then

(1) 119879lowastF(I0) = 119879F and 119879lowastF(I

119871) = 2119871

(2) ifI sube J then 119879lowast

F(I) sube 119879lowast

F(J)

Proof (1) By (1) of Proposition 2 119860lowast(I0) = 119888(119860) and

119860lowast(I119871) = 0 Thus 119888lowastI0(119860) = 119860 if and only if 119888(119860) = 119860 Simi-

larly 119888lowastI119871(119860) = 119860 for each 119860 sube 119871 Therefore (1) holds

4 The Scientific World Journal

Table 1 The implication operator of 119871 = 0 119886 119887 119888 119889 1

rarr 0 119886 119887 119888 119889 1

0 1 1 1 1 1 1

119886 119888 1 119887 119888 119887 1

119887 119889 119886 1 119887 119886 1

119888 119886 119886 1 1 119886 1

119889 119887 1 1 119887 1 1

1 0 119886 119887 119888 119889 1

ab

c d

1

0

Figure 1 Hasse diagram of 1198712= 0 119886 119887 119888 119889 1

(2) Let I sube J By (3) of Proposition 2 119888lowastJ(119860) sube 119888lowastI(119860)

for 119860 sube 119871 Thus if 119860 isin 119879lowastF(I) then 119860 isin 119879lowastF(J) Therefore119879lowastF(I) sube 119879lowastF(J)

Clearly if I isin I satisfies 119879119888F sube I then 119871 isin I and soI = 2119871 = I

119871 Thus by (1) of Proposition 3 119879lowastF = I119888 = 2119871 If

119879119888F 119871 sube I we have the following proposition

Proposition 7 Let (119871 119879F) be the filter topological space IfI isin I satisfies 119879119888F 119871 sube I then 119879lowastF = I119888 cup 0

Proof I119888 cup 0 sube 119879lowastF follows from Theorem 4 Converselysuppose that 119879lowastF sube I119888 cup0 Then there exists 119861 isin 119879

lowast

F (I119888 cup0) such that (119871 119861)

lowastsube 119871 119861 = 119871 Let 119910 isin (119871 119861) (119871 119861)

lowastThen 119910 notin (119871119861)

lowastThus [119910)cap(119871119861) isin I Put [119910)cap(119871119861) = 119868We have 119871119861 sube 119868cup (119871 [119910)) Since 119879119888F 119871 sube I 119871 [119910) isin IThus 119871119861 sube 119868cup (119871 119861) isin I and so 119871119861 isin I Hence 119861 isin I119888It is a contradiction Therefore 119879lowastF sube I119888 cup 0

Lemma 8 Let (119871 119879F) be the filter topological space andI isin

I If 119860 sube 119871 satisfies 119860 cap 119868 = 0 for each 119868 isin I then 119888lowast(119860) =

119888(119860)

Proof 119888lowast(119860) sube 119888(119860) is clear Conversely if 119909 notin 119888lowast(119860) then119909 notin 119860 and 119909 notin 119860lowastThus 119868 = [119909)cap119860 isin I and119860 sube 119868cup(119871[119909))Since 119860 cap 119868 = 0 119860 sube 119871 [119909) Observe that 119909 notin 119871 [119909) and119871 [119909) is 119879F-closed We have 119909 notin 119888(119860) Thus 119888(119860) sube 119888lowast(119860)Therefore 119888lowast(119860) = 119888(119860)

Proposition 9 IfI isin 119879119888F then 119879lowastF = 119879F

Proof It is clear that 119879F sube 119879lowastF Conversely let 119868119872 be thegreatest element of I and 119860 sube 119871 Thus (119860 119868

119872) cap 119868 = 0

for each 119868 isin I By Lemma 8 119888lowast(119860 119868119872) = 119888(119860 119868

119872)

By (8) of Proposition 2 (119860 119868119872)lowast

= 119860lowast Now notice that119860 cap 119868119872

isin I sube 119879119888F and thus 119888(119860 cap 119868119872) = 119860 cap 119868

119872 We have

119888 (119860) = 119888 ((119860 119868119872) cup (119860 cap 119868

119872))

= 119888 (119860 119868119872) cup 119888 (119860 cap 119868

119872)

= 119888lowast(119860 119868

119872) cup (119860 cap 119868

119872)

= (119860 119868119872) cup (119860 119868

119872)lowast

cup (119860 cap 119868119872)

= (119860 119868119872)lowast

cup 119860 = 119860lowastcup 119860 = 119888

lowast

(119860)

(13)

This implies 119888lowast = 119888 Therefore 119879lowastF = 119879F

Lemma 10 Let (119871 119879F) be the filter topological space andI isin

I If 119868 isin I and 119880 isin 119879F then 119880 119868 isin 119879lowastF

Proof Let 119875 = 119871 119880 Then 119875 isin 119879119888F By (7) and (8) ofProposition 2

119888lowast

(119875 cup 119868) = (119875 cup 119868) cup (119875 cup 119868)lowast= (119875 cup 119868) cup 119875

lowast= 119875 cup 119868

(14)

Thus 119871 (119875 cup 119868) = 119880 cap (119871 119868) = 119880 119868 isin 119879lowastF

Theorem11 Let (119871 119879F) be the filter topological space andI isin

I Then

B119879lowast

F= 119880 119868 119880 isin 119879F 119868 isin I (15)

is a base of 119879lowastF Moreover

B119879lowast

F(119909) = 119881 119868 119881 isin N

119879F(119909) 119909 notin 119868 isin I (16)

is a base of N119879lowast

F(119909) for each 119909 isin 119871 where N

119879lowast

F(119909) is the set

of all 119879lowastF-neighborhoods of 119909 in (119871 119879lowastF) Clearly [119909) 119868119909is

the smallest 119879lowastF-neighborhoods of 119909 where 119868119909is the greatest

element ofI satisfying 119909 notin 119868119909

Proof By Lemma 10B119879lowast

Fsube 119879lowastF Let 119861 sube 119871 Then 119861 isin 119879lowastF hArr

119871 119861 is 119879lowastF-closed hArr (119871 119861)lowastsube (119871 119861) hArr 119861 sube 119871 (119871 119861)

lowastThus 119909 isin 119861 rArr 119909 notin (119871 119861)

lowastrArr there exists 119880 isin N

119879F(119909) such

that (119871119861)cap119880 isin I Let 119868 = (119871119861)cap119880Then 119871119861 sube 119868cup(119871119880)

and

119909 isin 119880 119868 = 119880 cap (119871 119868) = 119871 (119868 cup (119871 119880)) sube 119861 (17)

ThereforeB119879lowast

Fis a base of 119879lowastF

ClearlyB119879lowast

F(119909) sube N

119879lowast

F(119909) Let 119860 isin N

119879lowast

F(119909) and 119910 isin 119860

SinceB119879lowast

Fis a base of 119879lowastF there are 119880119881 isin 119879F and 119868 119869 isin I

such that 119909 isin 119880 119868 sube 119860 and 119910 isin 119881 119869 sube 119860 We can assume

The Scientific World Journal 5

119910 notin 119868 and 119909 notin 119869 (otherwise 119868 and 119869 can be replaced by 119868 119910

and 119869 119909 resp) Then

(119880 119868) cup (119881 119869) = (119880 cap (119871 119868)) cup (119881 cap (119871 119869))

= [(119880 cup 119881) cap ((119871 119868) cup (119871 119869))]

cap [(119880 cup (119871 119869)) cap (119881 cup (119871 119868))]

supe ((119880 cup 119881) (119868 cap 119869)) cap (119871 (119868 cup 119869))

= (119880 cup 119881) (119868 cup 119869)

(18)

Clearly 119910 isin (119880 cup 119881) (119868 cup 119869) isin B119879lowast

F(119909) and

(119880 cup 119881) (119868 cup 119869) sube (119880 119868) cup (119881 119869) sube 119860 (19)

ThereforeB119879lowast

F(119909) is a base ofN

119879lowast

F(119909)

Clearly if 119868119909is the greatest element ofI satisfying 119909 notin 119868

119909

then [119909) 119868119909isin N119879lowast

F(119909) is the smallest 119879lowastF-neighborhoods of

119909

Let (119871 120591) be a topological space andI isin IThe topologythat was generated byB = 119880 119868 119880 isin 120591 119868 isin I is denotedby 119879lowast(I 120591) [5] Clearly 119879lowast(I 119879F) = 119879lowastF(I)

Lemma 12 Let 120595 = 0 119871 be the indiscrete topology on 119871 andI isin I Then 119879lowast(I 120595) = 0 cup 119868119888

Proof By 119879119888 119871 = 0 isin I and Proposition 7 the proof isobvious

Theorem 13 Let (119871 120591) be a topological space and I isin IThen 119879lowast(I 120591) = 120591 or 119879lowast(I 120595) where 120591 or 119879lowast(I 120595) is thetopology generated by the base 119880cap119881 119880 isin 120591 119881 isin 119879lowast(I 120595)

Proof Clearly B119879lowast = 119880 119868 119880 isin 120591 119868 isin I is a base of

119879lowast(I 120591) Since B119879lowast = 119880 cap 119881 119880 isin 120591 119881 isin 119879lowast(I 120595)

B119879lowast is also a base of 120591 or 119879lowast(I 120595) Therefore 119879lowast(I 120591) = 120591 or

119879lowast(I 120595)

Corollary 14 Let (119871 119879F) be the filter topological space andI isin I Then 119879lowastF = 119879F or 119879lowast(I 120595)

Corollary 15 Let (119871 119879F) be the filter topological space andI

J isin I Then

(1) 119879lowast(I orJ 120595) = 119879lowast(I 120595) or 119879lowast(J 120595)(2) 119879lowastF(I orJ) = 119879lowast(I 119879lowastF(J)) = 119879lowast(J 119879lowastF(I))(3) 119879lowastF(I orJ) = 119879lowastF(I) or 119879lowastF(J)(4) 119879lowast(I 119879lowastF(I)) = 119879lowastF(I)

Proof (1) By (2) of Proposition 6 119879lowast(I or J) supe 119879lowast(I) or

119879lowast(J) Conversely let 0 = 119860 isin 119879lowast(I or J) By Theorem 13there exist 119868 isin I and 119869 isin J such that

119860 = 119871 (119868 cup 119868) = (119871 119868) cap (119871 119869) isin 119879lowast

(I) or 119879lowast

(J)

(20)

Thus 119879lowast(I orJ) sube 119879lowast(I) or 119879lowast(J)

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

119879lowast

F (I orJ) = 119879F or 119879lowast(I orJ 120595)

= 119879F or 119879lowast(I 120595) or 119879

lowast(J 120595)

= 119879lowast

F (I) or 119879lowast(J 120595) = 119879

lowast(119879lowast

F (I) J)

(21)

Similarly 119879lowastF(I orJ) = 119879lowast(119879lowastF(J)I)(3) By (1) andTheorem 13

119879lowast

F (I orJ) = 119879F or 119879lowast(I orJ 120595)

= 119879F or 119879lowast(I 120595) or 119879

lowast(J 120595)

= 119879F or 119879lowast(I 120595) or 119879F or 119879

lowast(J 120595)

= 119879lowast

F (I) or 119879lowast

F (J)

(22)

Therefore 119879lowastF(I orJ) = 119879lowastF(I) or 119879lowastF(J)(4) LetI = J Then the proof follows from (2)

Theorem16 Let (119871 119879F) be the filter topological spaceIJ isin

I and 119860 sube 119871 Then

(1) 119860lowast(I capJ 119879F) = 119860lowast(I 119879F) cup 119860lowast(J 119879F)(2) 119860lowast(I orJ 119879F) = 119860lowast(I 119879lowastF(J)) cap 119860lowast(J 119879lowastF(I))

Proof (1) By (3) of Proposition 2 119860lowast(I 119879F) cup 119860lowast(J 119879F) sube

119860lowast(I cap J 119879F) Conversely 119909 notin 119860

lowast(I 119879F) cup 119860

lowast(J 119879F)

Then [119909) cap 119860 isin I and [119909) cap 119860 isin J Let [119909) cap 119860 = 119868 and[119909)cap119860 = 119869 Then119860 sube 119868cup (119871 [119909)) and119860 sube 119869cup (119871 [119909)) Thus

119860 sube (119868 cup (119871 [119909))) cap (119869 cup (119871 [119909))) = (119868 cap 119869) cup (119871 [119909))

(23)

Thus [119909) cap 119860 sube 119868 cap 119869 which implies 119909 notin 119860lowast(I cap J 119879F)

Therefore 119860lowast(I capJ 119879F) sube 119860lowast(I 119879F) cup 119860lowast(J 119879F)(3) Let 119909 notin 119860lowast(I orJ 119879F) Then [119909) cap 119860 isin I orJ Then

there exist 119868 isin I and 119869 isin J such that [119909) cap 119860 = 119868 cup 119869We can assume 119868 cap 119869 = 0 (otherwise 119868 can be replaced by119868 (119868 cap 119869)) Thus 119909 notin 119868 or 119909 notin 119869 (otherwise 119909 isin 119868 cap 119869 whichis a contradiction) Now we take 119909 notin 119868 for example Then

([119909) 119868) cap 119860 = [119909) cap 119860 cap (119871 119868) = 119869 (24)

Since [119909) isin 119879F and 119909 isin [119909) 119868 [119909) 119868 isin B119879lowast

F(I) Thus 119909 notin

119860lowast(J 119879lowastF(I)) and so 119909 notin 119860lowast(J 119879lowastF(I)) cap 119860lowast(I 119879lowastF(J))Hence

119860lowast(J 119879lowast

F (I)) cap 119860lowast(I 119879

lowast

F (J)) sube 119860lowast(I orJ 119879F)

(25)

Conversely let 119909 notin 119860lowast(119879lowastF(I)J) Then there exists 119868 isin

I such that ([119909) 119868) cap 119860 isin J Let ([119909) 119868) cap 119860 = 119869 Then[119909) cap 119860 = 119868 cup 119869 which implies 119909 notin 119860

lowast(I orJ 119879F) Similarlyif 119909 notin 119860lowast(119879lowastF(J)I) then 119909 notin 119860lowast(I or J 119879F) Therefore119860lowast(I orJ 119879F) sube 119860lowast(J 119879lowastF(I)) cap 119860lowast(I 119879lowastF(J))

Corollary 17 Consider 119860lowast(I 119879F) = 119860lowast(I 119879lowastF(I))

Proof LetI = J The proof follows from (2) ofTheorem 16

6 The Scientific World Journal

Corollary 18 Consider 119879lowastF(I capJ) = 119879lowastF(I) cap 119879lowastF(J)

Proof By (2) of Proposition 6119879lowastF(IcapJ) sube 119879lowastF(I)cap119879lowastF(J)Conversely if 119860 notin 119879lowastF(I capJ) then

(119871 119860)lowast(I 119879F) cup (119871 119860)

lowast(J 119879F)

= (119871 119860)lowast(I capJ 119879F) sube (119871 119860)

(26)

Thus (119871 119860)lowast(I 119879F) sube (119871 119860) or (119871 119860)

lowast(J 119879F) sube (119871 119860)

Thus119860 notin 119879lowastF(I) or119860 notin 119879lowastF(J)Therefore119879lowastF(I)cap119879lowastF(J) sube

119879lowastF(I capJ)

3 Power Idealization Filter TopologicalQuotient Spaces

Let (1198711 and or 1015840 rarr

1 01 11) and (119871

2 and or 1015840 rarr

2 02 12) be

two lattice implication algebras A mapping 119891 from 1198711to 1198712

is called lattice implication homomorphism if 119891(119909rarr1119910) =

119891(119909)rarr2119891(119910) for any 119909 119910 isin 119871

1 The set of all lattice

implication homomorphisms from 1198711to 1198712is denoted by

hom(1198711 1198712)

Let 119891 isin hom(1198711 1198712) Then clearly

119879119897(119879F(1198712)

119891) = 119891minus1

(119880) 119880 isin 119879F(1198712)

119879119903(119879F(1198711)

119891) = 119891 (119880) 119880 sube 1198712 119891minus1

(119880) isin 119879F(1198711)

(27)

are topologies [4]

Lemma 19 Let 1198711and 119871

2be two implication algebras and let

119891 isin hom(F1 f2) andI isin 21198711 J isin 21198712 be power ideals

(1) If 119891 is injective then 119891minus1(J) = 119891minus1(119869) 119869 isin J isin

I(1198711)

(2) If 119891 is surjective then 119891(I) = 119891(119868) 119868 isin I isin

I(1198712)

Proof (1) Since 0 isin J then 0 = 119891minus1(0) isin 119891minus1(J)If 1198682isin 119891minus1(J) and 119868

1sube 1198682 then there exist 119869

2isin J such

that 1198682= 119891minus1(119869

2) Thus 119891(119868

1) sube 119891(119868

2) = 1198692and 119891(119868

1) isin J

Since 119891 is injective 1198681= 119891minus1(119891(119868

1)) isin 119891minus1(J)

If 1198681 1198682isin 119891minus1(J) then there exist 119869

1 1198692isin J such that

1198681= 119891minus1(119869

1) and 119868

2= 119891minus1(119869

2) One has 119868

1cup 1198682= 119891minus1(119869

1) cup

119891minus1(1198692) = 119891minus1(119869

1cup 1198692) Since 119869

1cup 1198692isin J 119868

1cup 1198682isin 119891minus1(J)

Therefore 119891minus1(J) is a power ideal(2) By 0 isin I 0 = 119891(0) isin 119891(I)If 1198692isin 119891(J) and 119869

1sube 1198692 then there exists 119868

2isin I such

that 1198692= 119891(119868

2) Let 119868

1= 119891minus1(119869

1) Then 119868

1sube 1198682and 1198681isin I

Since 119891 is surjective 1198691= 119891(1198681) isin 119891(I)

If 1198691 1198692isin 119891(J) then there exist 119868

1 1198682isin I such that 119869

1=

119891(1198681) and 119869

2= 119891(119868

2) Thus 119868

1cup 1198682

isin I Hence 1198691cup 1198692

=

119891(1198681) cup119891(119868

2) = 119891(119868

1cup1198682) isin 119891(I) Therefore 119891(I) is a power

ideal

Lemma 20 Let 1198711and 119871

2be two implication algebras and

119891 isin hom(F1 F2) Then

(1) if 119891 is injective then for each 119909 isin 1198711 119891minus1([119891(119909))) isin

119879119897(119879F(1198712)

119891) is the smallest 119879119897-neighborhood of 119909

(2) if 119891 is bijective then for each 119910 isin 1198712 119891([119891minus1(119910))) isin

119879119903(119879F(1198711)

119891) is the smallest 119879119903-neighborhood of 119910

Proof (1) Clearly [119891(119909)) is the smallest119879F(1198712)-neighborhood

of 119891(119909) Then 119909 isin 119891minus1([119891(119909))) isin 119879119897(119879F(1198712)

119891) Let 119909 isin 119881 isin

119879119897(119879F(1198712)

119891)Then119891(119909) isin 119891(119881) isin 119879F(1198712)and [119891(119909)) sube 119891(119881)

Thus 119891minus1([119891(119909))) sube 119891minus1(119891(119881)) = 119881 Therefore 119891minus1([119891(119909))) isthe smallest one

(2) Clearly [119891minus1(119910)) is the smallest 119879F(1198711)-neighborhood

of 119891minus1(119910) Thus 119910 isin 119891([119891minus1(119910))) isin 119879

119903(119879F(1198711)

119891) Now let119910 isin 119880 isin 119879

119903(119879F(1198711)

119891) Then 119891minus1(119910) isin 119891minus1(119880) isin 119879F(1198711) Thus

[119891minus1(119910)) sube 119891minus1(119880) and so 119891([119891minus1(119910))) sube 119891(119891minus1(119880)) = 119880Therefore (2) holds

Lemma21 Let 1198711and 119871

2be two implication algebras and119891 isin

hom(F1 F2)

(1) If 119891 is injective and J isin I(1198712) then for each 119909 isin

1198711 [119891(119909)) 119869

119872isin 119879lowastF(J 119879F(1198712)

) is the smallestneighborhood of 119891(119909) where 119869

119872is the greatest element

ofJ satisfying 119891(119909) notin 119869119872

(2) If 119891 is bijective and I isin I(1198711) then for each 119910 isin

1198712 [119891minus1(119910)) 119868

119872isin 119879lowastF(I 119879F(1198711)

) is the smallestneighborhood of 119891minus1(119910) where 119868

119872is the greatest

element ofI satisfying 119891minus1(119910) notin 119868119872

Theorem 22 Let 1198711and 119871

2be two implication algebras and

119891 isin hom(F1 F2)

(1) If 119891 is injective andJ isin I(1198712) then

119879lowast(119891minus1

(J) 119879119897(119879F(1198712)

119891)) = 119879119897(119879lowast

F (J 119879F(1198712)) 119891)

(28)

(2) If 119891 is bijective andI isin I(1198711) then

119879lowast(119891 (I) 119879

119903(119879F(1198711)

119891)) = 119879119903(119879lowast

F (I 119879F(1198711)) 119891)

(29)

Proof (1) Let 119888lowast2and 119888119897be the closure operators of the left side

and the right side of the equationWe only need to prove 119888lowast1=

119888119897Let 119860 sube 119871

1and 119909 notin 119888lowast

1(119860) Then 119909 notin 119860 and 119909 notin

119860lowast(119891minus1(J) 119879119897(119879F(1198712)

119891)) By (1) of Lemma 21 119891minus1([119891(119909)))cap119860 isin 119891minus1(J) Thus there exists 119869 isin J such that 119891minus1([119891(119909))) cap119860 = 119891minus1(119869) Since 119909 notin 119860 and 119891 is injective 119891(119909) notin 119869Let 119869119872

isin J be the greatest one satisfying 119891(119909) notin 119869119872 Then

119891minus1([119891(119909))) cap 119860 sube 119891minus1(119869119872) Thus

0 = 119891minus1

([119891 (119909))) cap 119860 cap (1198711 119891minus1

(119869119872))

= 119891minus1

([119891 (119909))) cap 119891minus1

(1198712 119869119872) cap 119860

= 119891minus1

([119891 (119909)) cap (1198712 119869119872)) cap 119860

= 119891minus1

([119891 (119909)) 119869119872) cap 119860

(30)

By Lemma 21 and Proposition 1 119909 notin 119888119897(119860) Therefore 119888

119897(119860) sube

119888lowast1(119860)

The Scientific World Journal 7

Conversely let 119910 notin 119888119897(119860) By Proposition 1

0 = 119891minus1

([119891 (119909)) 119869119872) cap 119860

= 119891minus1

([119891 (119909))) cap 119860 cap (1198711 119891minus1

(119869119872))

(31)

Thus 119909 notin 119860 119891minus1([119891(119909))) cap 119860 sube 119891minus1(119869119872) and [119891(119909)) cap

119891(119860) sube 119869119872 Then 119869

1= [119891(119909)) cap 119891(119860) isin J Since 119891 is

injective 119891minus1([119891(119909))) cap 119860 = 119891minus1(1198691) By (1) of Lemma 20

119909 notin 119860lowast(119891minus1(J) 119879119897(119879F(1198712)

119891)) Therefore 119909 notin 119888lowast1(119860) and

119888lowast1(119860) sube 119888

119897(119860)

(2) Let 119888lowast2and 119888119903be the closure operators of the left side

and the right side of the equationWe only need to prove 119888lowast2=

119888119903Let 119910 notin 119888lowast

2(119860) Then 119910 notin 119860 and 119910 notin 119860lowast((119891(I)

119879119903(119879F(1198711)

119891))) By (2) of Lemma 20119891([119891minus1(119910)))cap119860 isin 119891(I)Thus there exists 119868 isin I such that 119891([119891minus1(119910))) cap 119860 = 119891(119868)Since119891 is bijective [119891minus1(119910))cap119891minus1(119860) = 119868 By119910 notin 119860119891minus1(119910) notin119891minus1(119860) and so 119891minus1(119910) notin 119868 Since 119868

119872isin I is the greatest

element ofI satisfying119891minus1(119910) notin 119868119872 [119891minus1(119910))cap119891minus1(119860) sube 119868

119872

and [119891minus1(119910)) cap (1198711 119868119872) cap 119891minus1(119860) = 0 By 119891 being bijective

again we have

0 = 119891 (0) = 119891 ([119891minus1

(119910))) cap (1198712 119891 (119868119872)) cap 119860

= 119891 ([119891minus1

(119910)) 119868119872) cap 119860

(32)

By

119891minus1

(119910) isin [119891minus1

(119910)) 119868119872

= 119891minus1

(119891 ([119891minus1

(119910)) 119868119872)) isin 119879

lowast

F (I 119879F(1198711))

(33)

119910 notin 119888119903(119860) Hence 119888lowast

2(119860) sube 119888

119903(119860)

Conversely let 119911 notin 119888119903(119860) Since 119868

119872is the greatest element

of I satisfying 119891minus1(119911) notin 119868119872 by (2) of Lemma 21 [119891minus1(119911))

119868119872

isin 119879lowastF(I 119879F(1198711)) is the smallest neighborhood of 119891minus1(119911)

Thus 119891([119891minus1(119911)) 119868119872) cap 119860 = 0 Since 119891 is bijective

0 = 119891minus1

(0) = ([119891minus1

(119911)) 119868119872) cap 119891minus1

(119860)

= [119891minus1

(119911)) cap (1198711 119868119872) cap 119891minus1

(119860)

(34)

This implies [119891minus1(119911))cap119891minus1(119860) sube 119868119872and [119891minus1(119911))cap119891minus1(119860) isin

I Thus 119891([119891minus1(119910))) cap 119860 isin 119891(I) which implies 119910 isin

119860lowast((119891(I) 119879119903(119879F(1198711)

119891))) Since 119911 notin 119888119903(119860) 119911 notin 119860 Therefore

119911 notin 119888lowast2(119860) and so 119888lowast

2(119860) sube 119888

119903(119860)

Generally if 119891 isin hom(1198711 1198712) is surjective but not bijec-

tive then (2) of Theorem 22 fails

Example 23 Let 1198711

= 01 119886 119887 119888 119889 1

1 be the Hasse lattice

implication algebra of Example 5 Let 1198712= 02 119890 ℎ 1

2 and

010158402= 12 1198901015840 = ℎ ℎ1015840 = 119890 and 11015840

2= 02 The Hasse diagram and

Table 2 The implication operator of 1198712= 02 119890 ℎ 1

2

rarr 02

119890 ℎ 12

02

1 12

12

12

119890 ℎ 12

ℎ 12

ℎ 119890 119890 12

12

12

02

119890 ℎ 12

e

h

12

02

Figure 2 Hasse diagram of 1198712= 02 119890 ℎ 1

2

the implication operator of 1198712are shown by Figure 2 and

Table 2 Then it is clear that

119879lowast

F(1198711)= 0 1

1 119886 1

1 119887 119888 1

1 119886 119887 119888 1

1

119887 119888 119889 11 0 119887 119888 119889 1

1 119886 119887 119888 119889 1

1 119871

119879lowast

F(1198712)= 0 119890 1

2 ℎ 1

2 119890 ℎ 1

2 12 1198712

(35)

A mapping 119891 from 1198711to 1198712is defined as

119891 (01) = 02 119891 (1

1) = 12 119891 (119886) = 119891 (119889) = 119890

119891 (119887) = 119891 (119888) = ℎ(36)

It easy to check 119891 isin hom(1198711 1198712) and 119891 is surjective LetI =

0 119888 ThenI isin I(1198711) and 119891(I) = 0 ℎ isin I(119871

2)

Since ℎ isin 119891(I) by Theorem 4

02 119890 12 = 1198712 ℎ isin 119879

lowast(119891 (I) 119879

119903(119879F(1198711)

119891)) (37)

Observe that 119887 119888lowast(I 119879F(1198711)) = 0

1 119887 119888 119889 sube 119887 119888 We

have

0 119886 119889 1 = 1198711 119887 119888 notin 119879

lowast

F (I 119879F(1198711)) (38)

Moreover by 119891minus1(02 119890 12) = 0

2 119886 119889 1

2 02 119890 12 notin

119879119903(119879lowastF(I 119879F(1198711)

) 119891)In fact we have

119879119903(119879lowast

F (I 119879F(1198711)) 119891) = 0 1

2 1198712

119879lowast(119891 (I) 119879

119903(119879F(1198711)

119891)) = 0 12 02 119890 12 1198712

(39)

Therefore 119879119903(119879lowastF(I 119879F(1198711)

) 119891) = 119879lowast(119891(I) 119879119903(119879F(1198711)

119891))

8 The Scientific World Journal

Corollary 24 Let 1198711and 119871

2be two implication algebras and

let 119891 isin hom(F1 f2) be bijective(1) IfJ isin I(119871

2) then

119879119903(119879lowast(119891minus1

(J) 119879119897(119879F(1198712)

119891)) 119891) = 119879lowast(J 119879F(1198712)

)

(40)

(2) IfI isin I(1198711) then

119879119897(119879lowast(119891 (I) 119879

119903(119879F(1198711)

119891)) 119891) = 119879lowast(I 119879F(1198711)

) (41)

Proof The proof follows fromTheorem 22

Corollary 25 Let 1198711and 119871

2be two implication algebras and

119891 isin hom(f1 f2)(1) If 119891 is injective andJ

1J2isin I(119871

2) then

119879119897(119879lowast

F (J1orJ2 119879F(1198712)

) 119891)

= 119879119897(119879lowast

F (J1 119879F(1198712)

) 119891) or 119879119897(119879lowast

F (J2 119879F(1198712)

) 119891)

(42)

(2) If 119891 is bijective andI1I2isin I(119871

1) then

119879119903(119879lowast

F (I1orI2 119879F(1198711)

) 119891)

= 119879119903(119879lowast

F (I1 119879F(1198711)

) 119891) or 119879119903(119879lowast

F (I2 119879F(1198711)

) 119891)

(43)

Proof (1) Clearly 119891minus1(J1orJ2) = 119891minus1(J

1) or119891minus1(J

2) By (3)

of Corollary 15 andTheorem 22

119879119897(119879lowast

F (J1orJ2 119879F(1198712)

) 119891)

= 119879lowast(119891minus1

(J1orJ2) 119879119897(119879F(1198712)

119891))

= 119879lowast(119891minus1

(J1) 119879119897(119879F(1198712)

119891))

or 119879lowast(119891minus1

(J2) 119879119897(119879F(1198712)

119891))

= 119879119897(119879lowast

F (J1 119879F(1198712)

) 119891) or 119879119897(119879lowast

F (J2 119879F(1198712)

) 119891)

(44)

(2) Is similar to (1)

Corollary 26 Let 1198711and 119871

2be two implication algebras and

let 119891 isin hom(F1 f2) be bijective If J isin I(1198712) and I isin

I(1198711) then

119879lowast(119891minus1

(J) 119879119897(119879lowast

F (J 119879F(1198712)) 119891))

= 119879lowast(119891minus1

(J) 119879119897(119879F(1198712)

119891))

119879lowast(119891 (I) 119879

119903(119879lowast

F (I 119879F(1198711)) 119891))

= 119879lowast(119891 (I) 119879

119903(119879F(1198711)

119891))

(45)

Proof The proof follows from (4) of Corollary 15 andTheorem 22

Conflict of Interests

The authors declare that there is no conflict of interests regar-ding the publication of this paper

Acknowledgments

This work is supported by the National Natural ScienceFoundations of China (no 11471202) and the Natural ScienceFoundation of Guangdong Province (no S2012010008833)

References

[1] J Lukasiewicz ldquoOn 3-valued logicrdquo Ruch Filozoficzny vol 5 pp169ndash170 1920

[2] Y Xu ldquoLattice implication algebrardquo Journal of Southwest Jiao-tong University vol 89 pp 20ndash27 1993

[3] Y Xu and K Y Qin ldquoOn filters of lattice implication algebrasrdquoThe Journal of FuzzyMathematics vol 1 no 2 pp 251ndash260 1993

[4] Y Xu D Ruan K Qin and J Liu Lattice-Valued Logic Springer2003

[5] D Jankovic and T R Hamlett ldquoNew topologies from old viaidealsrdquo The American Mathematical Monthly vol 97 no 4 pp295ndash310 1990

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Stochastic AnalysisInternational Journal of

2 The Scientific World Journal

topology 119879F(119871) briefly 119879F Usually 119879F is called the filtertopology generated by F And the pair (119871 119879F) is called thefilter topological spaceA subset119880 sube 119871 is called119879F-neighbor-hood of 119909 isin 119871 or neighborhood of 119909 in 119879F if 119909 isin 119880 isin 119879FThe set of all 119879F-neighborhoods of 119909 is denoted by N

119879F(119909)

Since F sube 119879F and [119909) = cap119865 119909 isin 119865 isin F isin F [119909) is thesmallest element ofN

119879F(119909)

The closure operator and interior operator of 119879F aredenoted by 119888 and 119894 Clearly for every119860 sube 119871 119888(119860) = cap119871[119909)

119909 isin 119871 [119909) cap 119860 = 0 and 119894(119860) = cup[119909) 119909 isin 119871 [119909) sube 119860 Thefollowing proposition describes 119888(119860)

Proposition 1 Let (119871 119879F) be the filer topology generated byF(119871) Then for 119860 sube 119871 119888(119860) = 119909 isin 119871 [119909) cap 119860 = 0

Proof The proof is trivial since [119909) is the smallest 119879F-neighborhood of 119909

Let 119871 be a lattice implication algebra and let 2119871 be thepower set of 119871 A nonempty subsetI of 2119871 is called a powerideal of 119871 if I satisfies the following (1) 119860 119861 isin 2119871 and 119860 sube

119861 isin I imply 119860 isin I (2) 119860 119861 isin I implies 119860 cup 119861 isin I Thecollection of all power ideals in 2119871 is denoted by I(119871) orbrieflyI Note thatI

0= 0 is the smallest power ideal and

I119871= 2119871 is the greatest power ideal Moreover ifIJ isin I

then (1)IcapJ isin I (2)IorJ = 119868cup 119869 119868 isin I 119869 isin J isin I

2 Local Functions and Power IdealizationFilter Topologies

Let 119871 be a lattice implication algebra let 119879F be the filter topo-logy and let I be a power ideal An operator lowast on 2119871 isdefined as follows

119860lowast(I 119879F) = 119909 isin 119871 forall119880 isin N

119879F(119909) 119860 cap 119880 notin I (1)

for every 119860 sube 119871The operator lowast is called the local function with respect to

119879F andI 119860lowast is called local function of 119860 We usually write119860lowast(I) or 119860lowast instead of 119860lowast(I 119879F)Clearly 119909 isin 119860

lowast if and only if [119909) cap 119860 notin I Thus 119860lowast =

119909 isin 119871 [119909) cap 119860 notin I The following proposition gives somefurther details of 119860lowast

Proposition 2 Let (119871 119879F) be the filter topological space andIJ isin I Then

(1) 119860lowast(I0) = 119888(119860) and 119860lowast(I

119871) = 0

(2) if 119860 sube 119861 then 119860lowast(I) sube 119861lowast(I)(3) ifI sube J then 119860lowast(J) sube 119860lowast(I)(4) 119860lowast(I) = 119888(119860lowast(I)) sube 119888(119860)(5) (119860lowast)lowast(I) sube 119860lowast(I)(6) if 119860 sube I then 119860lowast(I) = 0(7) if 119860 isin 119879

119888

F then 119860lowast(I) sube 119860

(8) if 119861 isin I then (119860 cup 119861)lowast(I) = 119860lowast(I) = (119860 119861)

lowast(I)

(9) (119860 cup 119860lowast(I))lowast(I) = 119860lowast(I)

(10) (119860 cup 119861)lowast(I) = 119860lowast(I) cup 119861lowast(I)

(11) 119860lowast(I)119861lowast(I) = (119860119861)

lowast(I)119861

lowast(I) sube (119860119861)

lowast(I)

(12) if 1 notin I then [119909)lowast(I) = 119871 for each 119909 isin 119871

(13) if 1 notin I and 1 isin 119860 sube 119871 then [119860)lowast(I) = [119860lowast(I)) =

119871

Proof (1) By Proposition 1 119909 isin 119860lowast(I0) if and only if [119909) cap

119860 = 0 if and only if 119909 isin 119888(119860) Thus 119860lowast(I0) = 119888(119860) Since

[119909) cap 119860 isin 2119871 = I119871for each 119909 isin 119871 119860lowast(119871) = 0

(2) Let 119860 sube 119861 and 119909 isin 119860lowast(I) Then [119909) cap 119860 notin I SinceI is a power ideal and [119909) cap 119860 sube [119909) cap 119861 [119909) cap 119861 notin I and so119909 isin 119861

lowast(I) Thus 119860lowast(I) sube 119861lowast(I)(3) Let I sube J and 119909 isin 119860lowast(I) Then [119909) cap 119860 notin J It

follows that [119909) cap 119860 notin I and so 119909 isin 119860lowast(I) Thus 119860lowast(J) sube

119860lowast(I)(4) If 119909 notin 119888(119860) then 119909 isin 119871 119888(119860) isin 119879F and so [119909) sube

119871 119888(119860) Thus [119909) cap 119860 sube (119871 119888(119860)) cap 119860 = 0 isin I Thisimplies 119909 notin 119860lowast(I) and so 119860lowast(I) sube 119888(119860) Then 119888(119860lowast(I)) sube

119888(119888(119860)) = 119888(119860)It is clear that 119860

lowast(I) sube 119888(119860lowast(I)) Next we prove119888(119860lowast(I)) sube 119860lowast(I)

Let 119909 isin 119888(119860lowast(I)) By Proposition 1 [119909) cap 119860lowast(I) = 0Then there exists 119910 isin [119909) cap 119860lowast(I) By 119910 isin 119860lowast(I) [119910) cap 119860 notin

I By119910 isin [119909) [119910) sube [119909)Thus [119909)cap119860 notin I and so119909 isin 119860lowast(I)Therefore 119888(119860lowast(I)) sube 119860lowast(I)

(5) By (4) (119860lowast(I))lowast(I) sube 119888(119860lowast(I)) = 119860lowast(I)

(6) Since [119909) cap 119860 sube 119860 isin I for each 119909 isin 119871 119860lowast(I) = 0(7) Suppose that 119909 isin 119860lowast(I) 119860 Then 119909 isin 119871 119860 isin 119879F

Thus [119909) sube 119871 119860 and so [119909) cap 119860 sube (119871 119860) = 0 isin I Hence119909 notin 119860

lowast(I) which is a contradiction Therefore 119860lowast(I) sube 119860(8) By (2) (119860 119861)

lowast(I) sube 119860lowast(I) sube (119860 cup 119861)

lowast(I) Next

we prove the inverse inclusionsIf 119909 notin (119860119861)

lowast(I) then ([119909)cap119860) 119861 = [119909)cap (119860119861) isin I

Thus [119909) cap 119860 sube 119868 cup 119861 isin I which follows fromI is a powerideal This implies 119909 notin 119860

lowast(I) Thus 119860lowast(I) sube (119860 119861)lowast(I)

and so 119860lowast(I) = (119860 119861)lowast(I)

If 119909 notin 119860lowast(I) then [119909) cap 119860 isin I Since 119861 isin I

[119909) cap (119860 cup 119861) sube ([119909) cap 119860) cup 119861 isin I (2)

Thus 119909 notin (119860 cup 119861)lowast(I) This implies (119860 cup 119861)

lowast(I) sube 119860

lowast(I)

and so (119860 cup 119861)lowast(I) = 119860lowast(I)

(9) Clearly 119860lowast(I) sube (119860 cup 119860lowast(I))lowast(I) Conversely if

119909 notin 119860lowast(I) then [119909) cap 119860 isin I Let [119909) cap 119860 = 119868 Then 119860 sube

119868 cup (119871 [119909)) By (2) (7) (8) and [119909) isin 119879F

119860lowast

(I) sube (119868 cup (119871 [119909)))lowast

(I) = (119871 [119909))lowast

(I) sube 119871 [119909)

(3)

Thus 119860 cup 119860lowast(I) sube (119871 [119909)) cup 119860 and so

[119909) cap (119860 cup 119860lowast

(I)) sube ((119871 [119909)) cup 119860) cap [119909)

= 119860 cap [119909) = 119868 isin I(4)

This implies119909 notin (119860cup119860lowast(I))lowast(I) and so (119860cup119860lowast(I))

lowast(I) sube

119860lowast(I)(10) 119860lowast(I) cup 119861lowast(I) sube (119860 cup 119861)

lowast(I) is clear Conversely

if 119909 notin 119860lowast(I) cup 119861lowast(I) then [119909) cap 119860 [119909) cap 119861 isin I Thus

The Scientific World Journal 3

[119909) cap (119860 cup 119861) = ([119909) cap 119860) cup ([119909) cup 119861) isin I This implies119909 notin (119860 cup 119861)

lowast(I) Therefore (119860 cup 119861)

lowast(I) sube 119860lowast(I) cup 119861lowast(I)

(11) We firstly prove 119860lowast(I) 119861lowast(I) sube (119860 119861)lowast(I)

Assume that 119909 isin (119860lowast(I) 119861lowast(I)) (119860 119861)lowast(I) Then

[119909) cap (119860 119861) isin I and [119909) cap 119861 isin I Thus

[119909) cap 119860 sube [119909) cap ((119860 119861) cup 119861)

= ([119909) cap (119860 119861)) cup ([119909) cap 119861) isin I(5)

This implies 119909 notin 119860lowast(I) which is a contradiction Thus119860lowast(I) 119861lowast(I) sube (119860 119861)

lowast(I) and so119860lowast(I) 119861lowast(I) sube (119860

119861)lowast(I)119861lowast(I) Finally (119860119861)

lowast(I)119861lowast(I) sube 119860lowast(I)119861lowast(I)

follows from (2) Therefore 119860lowast(I) 119861lowast(I) = (119860 119861)lowast(I)

119861lowast(I)(12) Since 1 notin I 1 notin 119868 for each 119868 isin I Assume that

there exists 119909 isin 119871 such that [119909) = 119871 Let 119910 notin 119871[119909)lowast(I)Thus

[119910) cap [119909) isin I Since [119909 or 119910) = [119910) cap [119909) 1 notin [119909 or 119910) which isa contradiction Therefore [119909)lowast(I) = 119871 for each 119909 isin 119871

(13) Assume that there exists 119910 isin 119871 119860lowast(I) Then 1 isin

[119910) cap 119860 isin I which is a contradiction Thus 119860lowast(I) = 119871 and[119860lowast(I)) = 119871 Since 1 isin 119860 119871 = [1)

lowast(I) sube [119860)

lowast(I) follows

from (12) Therefore [119860)lowast(I) = 119871

Proposition 3 Let (119871 119879F) be the filter topological space andI isin I The operator 119888lowastI (briefly 119888lowast) on 2119871 defined by 119888lowast(119860) =

119860 cup 119860lowast for 119860 sube 119871 satisfies the following statements

(1) 119888lowast(0) = 0 119888lowast(119871) = 119871(2) 119888lowast(119888lowast(119860)) = 119888lowast(119860) sube 119888(119860)(3) (119888lowast(119860))

lowast= 119888lowast(119860lowast) = 119860lowast

(4) 119888lowast(119860 cup 119861) = 119888lowast(119860) cup 119888lowast(119861)(5) 119860 isin 119879119888F or 119860 isin I implies 119888lowast(119860) = 119860

Proof (1) 119888lowast(0) = 0 follows from 0lowast = 0 119888lowast(119871) = 119871 is clear(2) By (4) and (9) of Proposition 2

119888lowast(119888lowast

(119860)) = (119860 cup 119860lowast) cup (119860 cup 119860

lowast)lowast

= 119860 cup 119860lowast= 119888lowast

(119860) sube 119888 (119860) (6)

(3) By (5) and (9) of Proposition 2

(119888lowast

(119860))lowast

= (119888lowast

(119860) cup (119888lowast

(119860))lowast

)lowast

= (119860 cup 119860lowastcup (119860 cup 119860

lowast)lowast

)lowast

= (119860 cup 119860lowast)lowast

= 119860lowast

(7)

and 119888lowast(119860lowast) = 119860lowast cup (119860lowast)lowast= 119860lowast

(4) By (10) of Proposition 2

119888lowast

(119860 cup 119861) = (119860 cup 119861) cup (119860 cup 119861)lowast

= (119860 cup 119860lowast) cup (119861 cup 119861

lowast) = 119888lowast

(119860) cup 119888lowast

(119861) (8)

(5) The result follows from (6) and (7) of Proposition 2

Theorem 4 Let (119871 119879F) be the filter topological space andI isin I The operator 119888lowast stated in Proposition 3 is the closure

operator of a new topology which is finer than 119879F and thetopology generated byI119888 (note thatI119888 is not a topology since0 notin I119888 in general case) Such a topology is called a poweridealization filter topology and often denoted by 119879lowastF(I 119879F)119879lowastF(I) or 119879lowastF

Proof Let 119879lowastF = 119860 sube 119871 119888lowast(119871 119860) = 119871 119860 We prove that119879lowastF is a topology on 119871

(1) 0 119871 isin 119879lowast

F follows from (1) of Proposition 3(2) If119860 119861 isin 119879

lowast

F then 119888lowast(119871119860) = 119871119860 and 119888

lowast(119871119861) = 119871119861

Thus

119888lowast

(119871 (119860 cap 119861)) = 119888lowast

((119871 119860) cup (119871 119861))

= 119888lowast

(119871 119860) cup 119888lowast

(119871 119861) = 119871 (119860 cap 119861)

(9)

This implies 119860 cap 119861 isin 119879lowastF(3) Let 119860

119905isin 119879lowastF for 119905 isin 119879 where 119879 is an index set Then

119888lowast(119871 119860119905) = 119871 119860

119905and

119871 (cup119905isin119879

) = cap119905isin119879

(119871 119860119905) sube 119888lowast(cap119905isin119879

(119871 119860119905))

= cap119905isin119879

(119871 119860119905) cup (cap

119905isin119879(119871 119860

119905))lowast

sube cap119905isin119879

((119871 119860119905) cup (119871 119860

119905)lowast

)

= cap119905isin119879

119888lowast(119871 119860

119905) = cap119905isin119879

(119871 119860119905) = 119871 (cup

119905isin119879119860119905)

(10)

Therefore cup119905isin119879

119860119905isin 119879lowastF

Finally by (6) and (7) of Proposition 3119879FI119888 sube 119879lowastF

Example 5 Let 119871 = 0 119886 119887 119888 119889 1 01015840 = 1 1198861015840 = 119888 1198871015840 = 1198891198881015840 = 119886 1198891015840 = 119887 11015840 = 0 and the implication operator rarr bedefined by 119886 rarr 119887 = 1198861015840 or 119887 for 119886 119887 isin 119871 Then (119871 and or 1015840 rarr )

is the Hasse lattice implication algebra (Figure 1 and Table 1)Then

F (119871) = 1 119886 1 119887 119888 1 119886 119887 119888 119889 1 119871

119879F(119871) = 0 1 119886 1 119887 119888 1 119886 119887 119888 1 119886 119887 119888 119889 1 119871

(11)

LetI = 0 0 119886 0 119886 ThenI is a power ideal It is easyto check that

119879lowast

F = 0 1 119886 1 119887 119888 1 119886 119887 119888 1 119887 119888 119889 1

0 119887 119888 119889 1 119886 119887 119888 119889 1 119871 (12)

Clearly 119879F sube 119879lowastF

Proposition 6 Let (119871 119879F) be the filter topological space andIJ isin I Then

(1) 119879lowastF(I0) = 119879F and 119879lowastF(I

119871) = 2119871

(2) ifI sube J then 119879lowast

F(I) sube 119879lowast

F(J)

Proof (1) By (1) of Proposition 2 119860lowast(I0) = 119888(119860) and

119860lowast(I119871) = 0 Thus 119888lowastI0(119860) = 119860 if and only if 119888(119860) = 119860 Simi-

larly 119888lowastI119871(119860) = 119860 for each 119860 sube 119871 Therefore (1) holds

4 The Scientific World Journal

Table 1 The implication operator of 119871 = 0 119886 119887 119888 119889 1

rarr 0 119886 119887 119888 119889 1

0 1 1 1 1 1 1

119886 119888 1 119887 119888 119887 1

119887 119889 119886 1 119887 119886 1

119888 119886 119886 1 1 119886 1

119889 119887 1 1 119887 1 1

1 0 119886 119887 119888 119889 1

ab

c d

1

0

Figure 1 Hasse diagram of 1198712= 0 119886 119887 119888 119889 1

(2) Let I sube J By (3) of Proposition 2 119888lowastJ(119860) sube 119888lowastI(119860)

for 119860 sube 119871 Thus if 119860 isin 119879lowastF(I) then 119860 isin 119879lowastF(J) Therefore119879lowastF(I) sube 119879lowastF(J)

Clearly if I isin I satisfies 119879119888F sube I then 119871 isin I and soI = 2119871 = I

119871 Thus by (1) of Proposition 3 119879lowastF = I119888 = 2119871 If

119879119888F 119871 sube I we have the following proposition

Proposition 7 Let (119871 119879F) be the filter topological space IfI isin I satisfies 119879119888F 119871 sube I then 119879lowastF = I119888 cup 0

Proof I119888 cup 0 sube 119879lowastF follows from Theorem 4 Converselysuppose that 119879lowastF sube I119888 cup0 Then there exists 119861 isin 119879

lowast

F (I119888 cup0) such that (119871 119861)

lowastsube 119871 119861 = 119871 Let 119910 isin (119871 119861) (119871 119861)

lowastThen 119910 notin (119871119861)

lowastThus [119910)cap(119871119861) isin I Put [119910)cap(119871119861) = 119868We have 119871119861 sube 119868cup (119871 [119910)) Since 119879119888F 119871 sube I 119871 [119910) isin IThus 119871119861 sube 119868cup (119871 119861) isin I and so 119871119861 isin I Hence 119861 isin I119888It is a contradiction Therefore 119879lowastF sube I119888 cup 0

Lemma 8 Let (119871 119879F) be the filter topological space andI isin

I If 119860 sube 119871 satisfies 119860 cap 119868 = 0 for each 119868 isin I then 119888lowast(119860) =

119888(119860)

Proof 119888lowast(119860) sube 119888(119860) is clear Conversely if 119909 notin 119888lowast(119860) then119909 notin 119860 and 119909 notin 119860lowastThus 119868 = [119909)cap119860 isin I and119860 sube 119868cup(119871[119909))Since 119860 cap 119868 = 0 119860 sube 119871 [119909) Observe that 119909 notin 119871 [119909) and119871 [119909) is 119879F-closed We have 119909 notin 119888(119860) Thus 119888(119860) sube 119888lowast(119860)Therefore 119888lowast(119860) = 119888(119860)

Proposition 9 IfI isin 119879119888F then 119879lowastF = 119879F

Proof It is clear that 119879F sube 119879lowastF Conversely let 119868119872 be thegreatest element of I and 119860 sube 119871 Thus (119860 119868

119872) cap 119868 = 0

for each 119868 isin I By Lemma 8 119888lowast(119860 119868119872) = 119888(119860 119868

119872)

By (8) of Proposition 2 (119860 119868119872)lowast

= 119860lowast Now notice that119860 cap 119868119872

isin I sube 119879119888F and thus 119888(119860 cap 119868119872) = 119860 cap 119868

119872 We have

119888 (119860) = 119888 ((119860 119868119872) cup (119860 cap 119868

119872))

= 119888 (119860 119868119872) cup 119888 (119860 cap 119868

119872)

= 119888lowast(119860 119868

119872) cup (119860 cap 119868

119872)

= (119860 119868119872) cup (119860 119868

119872)lowast

cup (119860 cap 119868119872)

= (119860 119868119872)lowast

cup 119860 = 119860lowastcup 119860 = 119888

lowast

(119860)

(13)

This implies 119888lowast = 119888 Therefore 119879lowastF = 119879F

Lemma 10 Let (119871 119879F) be the filter topological space andI isin

I If 119868 isin I and 119880 isin 119879F then 119880 119868 isin 119879lowastF

Proof Let 119875 = 119871 119880 Then 119875 isin 119879119888F By (7) and (8) ofProposition 2

119888lowast

(119875 cup 119868) = (119875 cup 119868) cup (119875 cup 119868)lowast= (119875 cup 119868) cup 119875

lowast= 119875 cup 119868

(14)

Thus 119871 (119875 cup 119868) = 119880 cap (119871 119868) = 119880 119868 isin 119879lowastF

Theorem11 Let (119871 119879F) be the filter topological space andI isin

I Then

B119879lowast

F= 119880 119868 119880 isin 119879F 119868 isin I (15)

is a base of 119879lowastF Moreover

B119879lowast

F(119909) = 119881 119868 119881 isin N

119879F(119909) 119909 notin 119868 isin I (16)

is a base of N119879lowast

F(119909) for each 119909 isin 119871 where N

119879lowast

F(119909) is the set

of all 119879lowastF-neighborhoods of 119909 in (119871 119879lowastF) Clearly [119909) 119868119909is

the smallest 119879lowastF-neighborhoods of 119909 where 119868119909is the greatest

element ofI satisfying 119909 notin 119868119909

Proof By Lemma 10B119879lowast

Fsube 119879lowastF Let 119861 sube 119871 Then 119861 isin 119879lowastF hArr

119871 119861 is 119879lowastF-closed hArr (119871 119861)lowastsube (119871 119861) hArr 119861 sube 119871 (119871 119861)

lowastThus 119909 isin 119861 rArr 119909 notin (119871 119861)

lowastrArr there exists 119880 isin N

119879F(119909) such

that (119871119861)cap119880 isin I Let 119868 = (119871119861)cap119880Then 119871119861 sube 119868cup(119871119880)

and

119909 isin 119880 119868 = 119880 cap (119871 119868) = 119871 (119868 cup (119871 119880)) sube 119861 (17)

ThereforeB119879lowast

Fis a base of 119879lowastF

ClearlyB119879lowast

F(119909) sube N

119879lowast

F(119909) Let 119860 isin N

119879lowast

F(119909) and 119910 isin 119860

SinceB119879lowast

Fis a base of 119879lowastF there are 119880119881 isin 119879F and 119868 119869 isin I

such that 119909 isin 119880 119868 sube 119860 and 119910 isin 119881 119869 sube 119860 We can assume

The Scientific World Journal 5

119910 notin 119868 and 119909 notin 119869 (otherwise 119868 and 119869 can be replaced by 119868 119910

and 119869 119909 resp) Then

(119880 119868) cup (119881 119869) = (119880 cap (119871 119868)) cup (119881 cap (119871 119869))

= [(119880 cup 119881) cap ((119871 119868) cup (119871 119869))]

cap [(119880 cup (119871 119869)) cap (119881 cup (119871 119868))]

supe ((119880 cup 119881) (119868 cap 119869)) cap (119871 (119868 cup 119869))

= (119880 cup 119881) (119868 cup 119869)

(18)

Clearly 119910 isin (119880 cup 119881) (119868 cup 119869) isin B119879lowast

F(119909) and

(119880 cup 119881) (119868 cup 119869) sube (119880 119868) cup (119881 119869) sube 119860 (19)

ThereforeB119879lowast

F(119909) is a base ofN

119879lowast

F(119909)

Clearly if 119868119909is the greatest element ofI satisfying 119909 notin 119868

119909

then [119909) 119868119909isin N119879lowast

F(119909) is the smallest 119879lowastF-neighborhoods of

119909

Let (119871 120591) be a topological space andI isin IThe topologythat was generated byB = 119880 119868 119880 isin 120591 119868 isin I is denotedby 119879lowast(I 120591) [5] Clearly 119879lowast(I 119879F) = 119879lowastF(I)

Lemma 12 Let 120595 = 0 119871 be the indiscrete topology on 119871 andI isin I Then 119879lowast(I 120595) = 0 cup 119868119888

Proof By 119879119888 119871 = 0 isin I and Proposition 7 the proof isobvious

Theorem 13 Let (119871 120591) be a topological space and I isin IThen 119879lowast(I 120591) = 120591 or 119879lowast(I 120595) where 120591 or 119879lowast(I 120595) is thetopology generated by the base 119880cap119881 119880 isin 120591 119881 isin 119879lowast(I 120595)

Proof Clearly B119879lowast = 119880 119868 119880 isin 120591 119868 isin I is a base of

119879lowast(I 120591) Since B119879lowast = 119880 cap 119881 119880 isin 120591 119881 isin 119879lowast(I 120595)

B119879lowast is also a base of 120591 or 119879lowast(I 120595) Therefore 119879lowast(I 120591) = 120591 or

119879lowast(I 120595)

Corollary 14 Let (119871 119879F) be the filter topological space andI isin I Then 119879lowastF = 119879F or 119879lowast(I 120595)

Corollary 15 Let (119871 119879F) be the filter topological space andI

J isin I Then

(1) 119879lowast(I orJ 120595) = 119879lowast(I 120595) or 119879lowast(J 120595)(2) 119879lowastF(I orJ) = 119879lowast(I 119879lowastF(J)) = 119879lowast(J 119879lowastF(I))(3) 119879lowastF(I orJ) = 119879lowastF(I) or 119879lowastF(J)(4) 119879lowast(I 119879lowastF(I)) = 119879lowastF(I)

Proof (1) By (2) of Proposition 6 119879lowast(I or J) supe 119879lowast(I) or

119879lowast(J) Conversely let 0 = 119860 isin 119879lowast(I or J) By Theorem 13there exist 119868 isin I and 119869 isin J such that

119860 = 119871 (119868 cup 119868) = (119871 119868) cap (119871 119869) isin 119879lowast

(I) or 119879lowast

(J)

(20)

Thus 119879lowast(I orJ) sube 119879lowast(I) or 119879lowast(J)

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

119879lowast

F (I orJ) = 119879F or 119879lowast(I orJ 120595)

= 119879F or 119879lowast(I 120595) or 119879

lowast(J 120595)

= 119879lowast

F (I) or 119879lowast(J 120595) = 119879

lowast(119879lowast

F (I) J)

(21)

Similarly 119879lowastF(I orJ) = 119879lowast(119879lowastF(J)I)(3) By (1) andTheorem 13

119879lowast

F (I orJ) = 119879F or 119879lowast(I orJ 120595)

= 119879F or 119879lowast(I 120595) or 119879

lowast(J 120595)

= 119879F or 119879lowast(I 120595) or 119879F or 119879

lowast(J 120595)

= 119879lowast

F (I) or 119879lowast

F (J)

(22)

Therefore 119879lowastF(I orJ) = 119879lowastF(I) or 119879lowastF(J)(4) LetI = J Then the proof follows from (2)

Theorem16 Let (119871 119879F) be the filter topological spaceIJ isin

I and 119860 sube 119871 Then

(1) 119860lowast(I capJ 119879F) = 119860lowast(I 119879F) cup 119860lowast(J 119879F)(2) 119860lowast(I orJ 119879F) = 119860lowast(I 119879lowastF(J)) cap 119860lowast(J 119879lowastF(I))

Proof (1) By (3) of Proposition 2 119860lowast(I 119879F) cup 119860lowast(J 119879F) sube

119860lowast(I cap J 119879F) Conversely 119909 notin 119860

lowast(I 119879F) cup 119860

lowast(J 119879F)

Then [119909) cap 119860 isin I and [119909) cap 119860 isin J Let [119909) cap 119860 = 119868 and[119909)cap119860 = 119869 Then119860 sube 119868cup (119871 [119909)) and119860 sube 119869cup (119871 [119909)) Thus

119860 sube (119868 cup (119871 [119909))) cap (119869 cup (119871 [119909))) = (119868 cap 119869) cup (119871 [119909))

(23)

Thus [119909) cap 119860 sube 119868 cap 119869 which implies 119909 notin 119860lowast(I cap J 119879F)

Therefore 119860lowast(I capJ 119879F) sube 119860lowast(I 119879F) cup 119860lowast(J 119879F)(3) Let 119909 notin 119860lowast(I orJ 119879F) Then [119909) cap 119860 isin I orJ Then

there exist 119868 isin I and 119869 isin J such that [119909) cap 119860 = 119868 cup 119869We can assume 119868 cap 119869 = 0 (otherwise 119868 can be replaced by119868 (119868 cap 119869)) Thus 119909 notin 119868 or 119909 notin 119869 (otherwise 119909 isin 119868 cap 119869 whichis a contradiction) Now we take 119909 notin 119868 for example Then

([119909) 119868) cap 119860 = [119909) cap 119860 cap (119871 119868) = 119869 (24)

Since [119909) isin 119879F and 119909 isin [119909) 119868 [119909) 119868 isin B119879lowast

F(I) Thus 119909 notin

119860lowast(J 119879lowastF(I)) and so 119909 notin 119860lowast(J 119879lowastF(I)) cap 119860lowast(I 119879lowastF(J))Hence

119860lowast(J 119879lowast

F (I)) cap 119860lowast(I 119879

lowast

F (J)) sube 119860lowast(I orJ 119879F)

(25)

Conversely let 119909 notin 119860lowast(119879lowastF(I)J) Then there exists 119868 isin

I such that ([119909) 119868) cap 119860 isin J Let ([119909) 119868) cap 119860 = 119869 Then[119909) cap 119860 = 119868 cup 119869 which implies 119909 notin 119860

lowast(I orJ 119879F) Similarlyif 119909 notin 119860lowast(119879lowastF(J)I) then 119909 notin 119860lowast(I or J 119879F) Therefore119860lowast(I orJ 119879F) sube 119860lowast(J 119879lowastF(I)) cap 119860lowast(I 119879lowastF(J))

Corollary 17 Consider 119860lowast(I 119879F) = 119860lowast(I 119879lowastF(I))

Proof LetI = J The proof follows from (2) ofTheorem 16

6 The Scientific World Journal

Corollary 18 Consider 119879lowastF(I capJ) = 119879lowastF(I) cap 119879lowastF(J)

Proof By (2) of Proposition 6119879lowastF(IcapJ) sube 119879lowastF(I)cap119879lowastF(J)Conversely if 119860 notin 119879lowastF(I capJ) then

(119871 119860)lowast(I 119879F) cup (119871 119860)

lowast(J 119879F)

= (119871 119860)lowast(I capJ 119879F) sube (119871 119860)

(26)

Thus (119871 119860)lowast(I 119879F) sube (119871 119860) or (119871 119860)

lowast(J 119879F) sube (119871 119860)

Thus119860 notin 119879lowastF(I) or119860 notin 119879lowastF(J)Therefore119879lowastF(I)cap119879lowastF(J) sube

119879lowastF(I capJ)

3 Power Idealization Filter TopologicalQuotient Spaces

Let (1198711 and or 1015840 rarr

1 01 11) and (119871

2 and or 1015840 rarr

2 02 12) be

two lattice implication algebras A mapping 119891 from 1198711to 1198712

is called lattice implication homomorphism if 119891(119909rarr1119910) =

119891(119909)rarr2119891(119910) for any 119909 119910 isin 119871

1 The set of all lattice

implication homomorphisms from 1198711to 1198712is denoted by

hom(1198711 1198712)

Let 119891 isin hom(1198711 1198712) Then clearly

119879119897(119879F(1198712)

119891) = 119891minus1

(119880) 119880 isin 119879F(1198712)

119879119903(119879F(1198711)

119891) = 119891 (119880) 119880 sube 1198712 119891minus1

(119880) isin 119879F(1198711)

(27)

are topologies [4]

Lemma 19 Let 1198711and 119871

2be two implication algebras and let

119891 isin hom(F1 f2) andI isin 21198711 J isin 21198712 be power ideals

(1) If 119891 is injective then 119891minus1(J) = 119891minus1(119869) 119869 isin J isin

I(1198711)

(2) If 119891 is surjective then 119891(I) = 119891(119868) 119868 isin I isin

I(1198712)

Proof (1) Since 0 isin J then 0 = 119891minus1(0) isin 119891minus1(J)If 1198682isin 119891minus1(J) and 119868

1sube 1198682 then there exist 119869

2isin J such

that 1198682= 119891minus1(119869

2) Thus 119891(119868

1) sube 119891(119868

2) = 1198692and 119891(119868

1) isin J

Since 119891 is injective 1198681= 119891minus1(119891(119868

1)) isin 119891minus1(J)

If 1198681 1198682isin 119891minus1(J) then there exist 119869

1 1198692isin J such that

1198681= 119891minus1(119869

1) and 119868

2= 119891minus1(119869

2) One has 119868

1cup 1198682= 119891minus1(119869

1) cup

119891minus1(1198692) = 119891minus1(119869

1cup 1198692) Since 119869

1cup 1198692isin J 119868

1cup 1198682isin 119891minus1(J)

Therefore 119891minus1(J) is a power ideal(2) By 0 isin I 0 = 119891(0) isin 119891(I)If 1198692isin 119891(J) and 119869

1sube 1198692 then there exists 119868

2isin I such

that 1198692= 119891(119868

2) Let 119868

1= 119891minus1(119869

1) Then 119868

1sube 1198682and 1198681isin I

Since 119891 is surjective 1198691= 119891(1198681) isin 119891(I)

If 1198691 1198692isin 119891(J) then there exist 119868

1 1198682isin I such that 119869

1=

119891(1198681) and 119869

2= 119891(119868

2) Thus 119868

1cup 1198682

isin I Hence 1198691cup 1198692

=

119891(1198681) cup119891(119868

2) = 119891(119868

1cup1198682) isin 119891(I) Therefore 119891(I) is a power

ideal

Lemma 20 Let 1198711and 119871

2be two implication algebras and

119891 isin hom(F1 F2) Then

(1) if 119891 is injective then for each 119909 isin 1198711 119891minus1([119891(119909))) isin

119879119897(119879F(1198712)

119891) is the smallest 119879119897-neighborhood of 119909

(2) if 119891 is bijective then for each 119910 isin 1198712 119891([119891minus1(119910))) isin

119879119903(119879F(1198711)

119891) is the smallest 119879119903-neighborhood of 119910

Proof (1) Clearly [119891(119909)) is the smallest119879F(1198712)-neighborhood

of 119891(119909) Then 119909 isin 119891minus1([119891(119909))) isin 119879119897(119879F(1198712)

119891) Let 119909 isin 119881 isin

119879119897(119879F(1198712)

119891)Then119891(119909) isin 119891(119881) isin 119879F(1198712)and [119891(119909)) sube 119891(119881)

Thus 119891minus1([119891(119909))) sube 119891minus1(119891(119881)) = 119881 Therefore 119891minus1([119891(119909))) isthe smallest one

(2) Clearly [119891minus1(119910)) is the smallest 119879F(1198711)-neighborhood

of 119891minus1(119910) Thus 119910 isin 119891([119891minus1(119910))) isin 119879

119903(119879F(1198711)

119891) Now let119910 isin 119880 isin 119879

119903(119879F(1198711)

119891) Then 119891minus1(119910) isin 119891minus1(119880) isin 119879F(1198711) Thus

[119891minus1(119910)) sube 119891minus1(119880) and so 119891([119891minus1(119910))) sube 119891(119891minus1(119880)) = 119880Therefore (2) holds

Lemma21 Let 1198711and 119871

2be two implication algebras and119891 isin

hom(F1 F2)

(1) If 119891 is injective and J isin I(1198712) then for each 119909 isin

1198711 [119891(119909)) 119869

119872isin 119879lowastF(J 119879F(1198712)

) is the smallestneighborhood of 119891(119909) where 119869

119872is the greatest element

ofJ satisfying 119891(119909) notin 119869119872

(2) If 119891 is bijective and I isin I(1198711) then for each 119910 isin

1198712 [119891minus1(119910)) 119868

119872isin 119879lowastF(I 119879F(1198711)

) is the smallestneighborhood of 119891minus1(119910) where 119868

119872is the greatest

element ofI satisfying 119891minus1(119910) notin 119868119872

Theorem 22 Let 1198711and 119871

2be two implication algebras and

119891 isin hom(F1 F2)

(1) If 119891 is injective andJ isin I(1198712) then

119879lowast(119891minus1

(J) 119879119897(119879F(1198712)

119891)) = 119879119897(119879lowast

F (J 119879F(1198712)) 119891)

(28)

(2) If 119891 is bijective andI isin I(1198711) then

119879lowast(119891 (I) 119879

119903(119879F(1198711)

119891)) = 119879119903(119879lowast

F (I 119879F(1198711)) 119891)

(29)

Proof (1) Let 119888lowast2and 119888119897be the closure operators of the left side

and the right side of the equationWe only need to prove 119888lowast1=

119888119897Let 119860 sube 119871

1and 119909 notin 119888lowast

1(119860) Then 119909 notin 119860 and 119909 notin

119860lowast(119891minus1(J) 119879119897(119879F(1198712)

119891)) By (1) of Lemma 21 119891minus1([119891(119909)))cap119860 isin 119891minus1(J) Thus there exists 119869 isin J such that 119891minus1([119891(119909))) cap119860 = 119891minus1(119869) Since 119909 notin 119860 and 119891 is injective 119891(119909) notin 119869Let 119869119872

isin J be the greatest one satisfying 119891(119909) notin 119869119872 Then

119891minus1([119891(119909))) cap 119860 sube 119891minus1(119869119872) Thus

0 = 119891minus1

([119891 (119909))) cap 119860 cap (1198711 119891minus1

(119869119872))

= 119891minus1

([119891 (119909))) cap 119891minus1

(1198712 119869119872) cap 119860

= 119891minus1

([119891 (119909)) cap (1198712 119869119872)) cap 119860

= 119891minus1

([119891 (119909)) 119869119872) cap 119860

(30)

By Lemma 21 and Proposition 1 119909 notin 119888119897(119860) Therefore 119888

119897(119860) sube

119888lowast1(119860)

The Scientific World Journal 7

Conversely let 119910 notin 119888119897(119860) By Proposition 1

0 = 119891minus1

([119891 (119909)) 119869119872) cap 119860

= 119891minus1

([119891 (119909))) cap 119860 cap (1198711 119891minus1

(119869119872))

(31)

Thus 119909 notin 119860 119891minus1([119891(119909))) cap 119860 sube 119891minus1(119869119872) and [119891(119909)) cap

119891(119860) sube 119869119872 Then 119869

1= [119891(119909)) cap 119891(119860) isin J Since 119891 is

injective 119891minus1([119891(119909))) cap 119860 = 119891minus1(1198691) By (1) of Lemma 20

119909 notin 119860lowast(119891minus1(J) 119879119897(119879F(1198712)

119891)) Therefore 119909 notin 119888lowast1(119860) and

119888lowast1(119860) sube 119888

119897(119860)

(2) Let 119888lowast2and 119888119903be the closure operators of the left side

and the right side of the equationWe only need to prove 119888lowast2=

119888119903Let 119910 notin 119888lowast

2(119860) Then 119910 notin 119860 and 119910 notin 119860lowast((119891(I)

119879119903(119879F(1198711)

119891))) By (2) of Lemma 20119891([119891minus1(119910)))cap119860 isin 119891(I)Thus there exists 119868 isin I such that 119891([119891minus1(119910))) cap 119860 = 119891(119868)Since119891 is bijective [119891minus1(119910))cap119891minus1(119860) = 119868 By119910 notin 119860119891minus1(119910) notin119891minus1(119860) and so 119891minus1(119910) notin 119868 Since 119868

119872isin I is the greatest

element ofI satisfying119891minus1(119910) notin 119868119872 [119891minus1(119910))cap119891minus1(119860) sube 119868

119872

and [119891minus1(119910)) cap (1198711 119868119872) cap 119891minus1(119860) = 0 By 119891 being bijective

again we have

0 = 119891 (0) = 119891 ([119891minus1

(119910))) cap (1198712 119891 (119868119872)) cap 119860

= 119891 ([119891minus1

(119910)) 119868119872) cap 119860

(32)

By

119891minus1

(119910) isin [119891minus1

(119910)) 119868119872

= 119891minus1

(119891 ([119891minus1

(119910)) 119868119872)) isin 119879

lowast

F (I 119879F(1198711))

(33)

119910 notin 119888119903(119860) Hence 119888lowast

2(119860) sube 119888

119903(119860)

Conversely let 119911 notin 119888119903(119860) Since 119868

119872is the greatest element

of I satisfying 119891minus1(119911) notin 119868119872 by (2) of Lemma 21 [119891minus1(119911))

119868119872

isin 119879lowastF(I 119879F(1198711)) is the smallest neighborhood of 119891minus1(119911)

Thus 119891([119891minus1(119911)) 119868119872) cap 119860 = 0 Since 119891 is bijective

0 = 119891minus1

(0) = ([119891minus1

(119911)) 119868119872) cap 119891minus1

(119860)

= [119891minus1

(119911)) cap (1198711 119868119872) cap 119891minus1

(119860)

(34)

This implies [119891minus1(119911))cap119891minus1(119860) sube 119868119872and [119891minus1(119911))cap119891minus1(119860) isin

I Thus 119891([119891minus1(119910))) cap 119860 isin 119891(I) which implies 119910 isin

119860lowast((119891(I) 119879119903(119879F(1198711)

119891))) Since 119911 notin 119888119903(119860) 119911 notin 119860 Therefore

119911 notin 119888lowast2(119860) and so 119888lowast

2(119860) sube 119888

119903(119860)

Generally if 119891 isin hom(1198711 1198712) is surjective but not bijec-

tive then (2) of Theorem 22 fails

Example 23 Let 1198711

= 01 119886 119887 119888 119889 1

1 be the Hasse lattice

implication algebra of Example 5 Let 1198712= 02 119890 ℎ 1

2 and

010158402= 12 1198901015840 = ℎ ℎ1015840 = 119890 and 11015840

2= 02 The Hasse diagram and

Table 2 The implication operator of 1198712= 02 119890 ℎ 1

2

rarr 02

119890 ℎ 12

02

1 12

12

12

119890 ℎ 12

ℎ 12

ℎ 119890 119890 12

12

12

02

119890 ℎ 12

e

h

12

02

Figure 2 Hasse diagram of 1198712= 02 119890 ℎ 1

2

the implication operator of 1198712are shown by Figure 2 and

Table 2 Then it is clear that

119879lowast

F(1198711)= 0 1

1 119886 1

1 119887 119888 1

1 119886 119887 119888 1

1

119887 119888 119889 11 0 119887 119888 119889 1

1 119886 119887 119888 119889 1

1 119871

119879lowast

F(1198712)= 0 119890 1

2 ℎ 1

2 119890 ℎ 1

2 12 1198712

(35)

A mapping 119891 from 1198711to 1198712is defined as

119891 (01) = 02 119891 (1

1) = 12 119891 (119886) = 119891 (119889) = 119890

119891 (119887) = 119891 (119888) = ℎ(36)

It easy to check 119891 isin hom(1198711 1198712) and 119891 is surjective LetI =

0 119888 ThenI isin I(1198711) and 119891(I) = 0 ℎ isin I(119871

2)

Since ℎ isin 119891(I) by Theorem 4

02 119890 12 = 1198712 ℎ isin 119879

lowast(119891 (I) 119879

119903(119879F(1198711)

119891)) (37)

Observe that 119887 119888lowast(I 119879F(1198711)) = 0

1 119887 119888 119889 sube 119887 119888 We

have

0 119886 119889 1 = 1198711 119887 119888 notin 119879

lowast

F (I 119879F(1198711)) (38)

Moreover by 119891minus1(02 119890 12) = 0

2 119886 119889 1

2 02 119890 12 notin

119879119903(119879lowastF(I 119879F(1198711)

) 119891)In fact we have

119879119903(119879lowast

F (I 119879F(1198711)) 119891) = 0 1

2 1198712

119879lowast(119891 (I) 119879

119903(119879F(1198711)

119891)) = 0 12 02 119890 12 1198712

(39)

Therefore 119879119903(119879lowastF(I 119879F(1198711)

) 119891) = 119879lowast(119891(I) 119879119903(119879F(1198711)

119891))

8 The Scientific World Journal

Corollary 24 Let 1198711and 119871

2be two implication algebras and

let 119891 isin hom(F1 f2) be bijective(1) IfJ isin I(119871

2) then

119879119903(119879lowast(119891minus1

(J) 119879119897(119879F(1198712)

119891)) 119891) = 119879lowast(J 119879F(1198712)

)

(40)

(2) IfI isin I(1198711) then

119879119897(119879lowast(119891 (I) 119879

119903(119879F(1198711)

119891)) 119891) = 119879lowast(I 119879F(1198711)

) (41)

Proof The proof follows fromTheorem 22

Corollary 25 Let 1198711and 119871

2be two implication algebras and

119891 isin hom(f1 f2)(1) If 119891 is injective andJ

1J2isin I(119871

2) then

119879119897(119879lowast

F (J1orJ2 119879F(1198712)

) 119891)

= 119879119897(119879lowast

F (J1 119879F(1198712)

) 119891) or 119879119897(119879lowast

F (J2 119879F(1198712)

) 119891)

(42)

(2) If 119891 is bijective andI1I2isin I(119871

1) then

119879119903(119879lowast

F (I1orI2 119879F(1198711)

) 119891)

= 119879119903(119879lowast

F (I1 119879F(1198711)

) 119891) or 119879119903(119879lowast

F (I2 119879F(1198711)

) 119891)

(43)

Proof (1) Clearly 119891minus1(J1orJ2) = 119891minus1(J

1) or119891minus1(J

2) By (3)

of Corollary 15 andTheorem 22

119879119897(119879lowast

F (J1orJ2 119879F(1198712)

) 119891)

= 119879lowast(119891minus1

(J1orJ2) 119879119897(119879F(1198712)

119891))

= 119879lowast(119891minus1

(J1) 119879119897(119879F(1198712)

119891))

or 119879lowast(119891minus1

(J2) 119879119897(119879F(1198712)

119891))

= 119879119897(119879lowast

F (J1 119879F(1198712)

) 119891) or 119879119897(119879lowast

F (J2 119879F(1198712)

) 119891)

(44)

(2) Is similar to (1)

Corollary 26 Let 1198711and 119871

2be two implication algebras and

let 119891 isin hom(F1 f2) be bijective If J isin I(1198712) and I isin

I(1198711) then

119879lowast(119891minus1

(J) 119879119897(119879lowast

F (J 119879F(1198712)) 119891))

= 119879lowast(119891minus1

(J) 119879119897(119879F(1198712)

119891))

119879lowast(119891 (I) 119879

119903(119879lowast

F (I 119879F(1198711)) 119891))

= 119879lowast(119891 (I) 119879

119903(119879F(1198711)

119891))

(45)

Proof The proof follows from (4) of Corollary 15 andTheorem 22

Conflict of Interests

The authors declare that there is no conflict of interests regar-ding the publication of this paper

Acknowledgments

This work is supported by the National Natural ScienceFoundations of China (no 11471202) and the Natural ScienceFoundation of Guangdong Province (no S2012010008833)

References

[1] J Lukasiewicz ldquoOn 3-valued logicrdquo Ruch Filozoficzny vol 5 pp169ndash170 1920

[2] Y Xu ldquoLattice implication algebrardquo Journal of Southwest Jiao-tong University vol 89 pp 20ndash27 1993

[3] Y Xu and K Y Qin ldquoOn filters of lattice implication algebrasrdquoThe Journal of FuzzyMathematics vol 1 no 2 pp 251ndash260 1993

[4] Y Xu D Ruan K Qin and J Liu Lattice-Valued Logic Springer2003

[5] D Jankovic and T R Hamlett ldquoNew topologies from old viaidealsrdquo The American Mathematical Monthly vol 97 no 4 pp295ndash310 1990

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Stochastic AnalysisInternational Journal of

The Scientific World Journal 3

[119909) cap (119860 cup 119861) = ([119909) cap 119860) cup ([119909) cup 119861) isin I This implies119909 notin (119860 cup 119861)

lowast(I) Therefore (119860 cup 119861)

lowast(I) sube 119860lowast(I) cup 119861lowast(I)

(11) We firstly prove 119860lowast(I) 119861lowast(I) sube (119860 119861)lowast(I)

Assume that 119909 isin (119860lowast(I) 119861lowast(I)) (119860 119861)lowast(I) Then

[119909) cap (119860 119861) isin I and [119909) cap 119861 isin I Thus

[119909) cap 119860 sube [119909) cap ((119860 119861) cup 119861)

= ([119909) cap (119860 119861)) cup ([119909) cap 119861) isin I(5)

This implies 119909 notin 119860lowast(I) which is a contradiction Thus119860lowast(I) 119861lowast(I) sube (119860 119861)

lowast(I) and so119860lowast(I) 119861lowast(I) sube (119860

119861)lowast(I)119861lowast(I) Finally (119860119861)

lowast(I)119861lowast(I) sube 119860lowast(I)119861lowast(I)

follows from (2) Therefore 119860lowast(I) 119861lowast(I) = (119860 119861)lowast(I)

119861lowast(I)(12) Since 1 notin I 1 notin 119868 for each 119868 isin I Assume that

there exists 119909 isin 119871 such that [119909) = 119871 Let 119910 notin 119871[119909)lowast(I)Thus

[119910) cap [119909) isin I Since [119909 or 119910) = [119910) cap [119909) 1 notin [119909 or 119910) which isa contradiction Therefore [119909)lowast(I) = 119871 for each 119909 isin 119871

(13) Assume that there exists 119910 isin 119871 119860lowast(I) Then 1 isin

[119910) cap 119860 isin I which is a contradiction Thus 119860lowast(I) = 119871 and[119860lowast(I)) = 119871 Since 1 isin 119860 119871 = [1)

lowast(I) sube [119860)

lowast(I) follows

from (12) Therefore [119860)lowast(I) = 119871

Proposition 3 Let (119871 119879F) be the filter topological space andI isin I The operator 119888lowastI (briefly 119888lowast) on 2119871 defined by 119888lowast(119860) =

119860 cup 119860lowast for 119860 sube 119871 satisfies the following statements

(1) 119888lowast(0) = 0 119888lowast(119871) = 119871(2) 119888lowast(119888lowast(119860)) = 119888lowast(119860) sube 119888(119860)(3) (119888lowast(119860))

lowast= 119888lowast(119860lowast) = 119860lowast

(4) 119888lowast(119860 cup 119861) = 119888lowast(119860) cup 119888lowast(119861)(5) 119860 isin 119879119888F or 119860 isin I implies 119888lowast(119860) = 119860

Proof (1) 119888lowast(0) = 0 follows from 0lowast = 0 119888lowast(119871) = 119871 is clear(2) By (4) and (9) of Proposition 2

119888lowast(119888lowast

(119860)) = (119860 cup 119860lowast) cup (119860 cup 119860

lowast)lowast

= 119860 cup 119860lowast= 119888lowast

(119860) sube 119888 (119860) (6)

(3) By (5) and (9) of Proposition 2

(119888lowast

(119860))lowast

= (119888lowast

(119860) cup (119888lowast

(119860))lowast

)lowast

= (119860 cup 119860lowastcup (119860 cup 119860

lowast)lowast

)lowast

= (119860 cup 119860lowast)lowast

= 119860lowast

(7)

and 119888lowast(119860lowast) = 119860lowast cup (119860lowast)lowast= 119860lowast

(4) By (10) of Proposition 2

119888lowast

(119860 cup 119861) = (119860 cup 119861) cup (119860 cup 119861)lowast

= (119860 cup 119860lowast) cup (119861 cup 119861

lowast) = 119888lowast

(119860) cup 119888lowast

(119861) (8)

(5) The result follows from (6) and (7) of Proposition 2

Theorem 4 Let (119871 119879F) be the filter topological space andI isin I The operator 119888lowast stated in Proposition 3 is the closure

operator of a new topology which is finer than 119879F and thetopology generated byI119888 (note thatI119888 is not a topology since0 notin I119888 in general case) Such a topology is called a poweridealization filter topology and often denoted by 119879lowastF(I 119879F)119879lowastF(I) or 119879lowastF

Proof Let 119879lowastF = 119860 sube 119871 119888lowast(119871 119860) = 119871 119860 We prove that119879lowastF is a topology on 119871

(1) 0 119871 isin 119879lowast

F follows from (1) of Proposition 3(2) If119860 119861 isin 119879

lowast

F then 119888lowast(119871119860) = 119871119860 and 119888

lowast(119871119861) = 119871119861

Thus

119888lowast

(119871 (119860 cap 119861)) = 119888lowast

((119871 119860) cup (119871 119861))

= 119888lowast

(119871 119860) cup 119888lowast

(119871 119861) = 119871 (119860 cap 119861)

(9)

This implies 119860 cap 119861 isin 119879lowastF(3) Let 119860

119905isin 119879lowastF for 119905 isin 119879 where 119879 is an index set Then

119888lowast(119871 119860119905) = 119871 119860

119905and

119871 (cup119905isin119879

) = cap119905isin119879

(119871 119860119905) sube 119888lowast(cap119905isin119879

(119871 119860119905))

= cap119905isin119879

(119871 119860119905) cup (cap

119905isin119879(119871 119860

119905))lowast

sube cap119905isin119879

((119871 119860119905) cup (119871 119860

119905)lowast

)

= cap119905isin119879

119888lowast(119871 119860

119905) = cap119905isin119879

(119871 119860119905) = 119871 (cup

119905isin119879119860119905)

(10)

Therefore cup119905isin119879

119860119905isin 119879lowastF

Finally by (6) and (7) of Proposition 3119879FI119888 sube 119879lowastF

Example 5 Let 119871 = 0 119886 119887 119888 119889 1 01015840 = 1 1198861015840 = 119888 1198871015840 = 1198891198881015840 = 119886 1198891015840 = 119887 11015840 = 0 and the implication operator rarr bedefined by 119886 rarr 119887 = 1198861015840 or 119887 for 119886 119887 isin 119871 Then (119871 and or 1015840 rarr )

is the Hasse lattice implication algebra (Figure 1 and Table 1)Then

F (119871) = 1 119886 1 119887 119888 1 119886 119887 119888 119889 1 119871

119879F(119871) = 0 1 119886 1 119887 119888 1 119886 119887 119888 1 119886 119887 119888 119889 1 119871

(11)

LetI = 0 0 119886 0 119886 ThenI is a power ideal It is easyto check that

119879lowast

F = 0 1 119886 1 119887 119888 1 119886 119887 119888 1 119887 119888 119889 1

0 119887 119888 119889 1 119886 119887 119888 119889 1 119871 (12)

Clearly 119879F sube 119879lowastF

Proposition 6 Let (119871 119879F) be the filter topological space andIJ isin I Then

(1) 119879lowastF(I0) = 119879F and 119879lowastF(I

119871) = 2119871

(2) ifI sube J then 119879lowast

F(I) sube 119879lowast

F(J)

Proof (1) By (1) of Proposition 2 119860lowast(I0) = 119888(119860) and

119860lowast(I119871) = 0 Thus 119888lowastI0(119860) = 119860 if and only if 119888(119860) = 119860 Simi-

larly 119888lowastI119871(119860) = 119860 for each 119860 sube 119871 Therefore (1) holds

4 The Scientific World Journal

Table 1 The implication operator of 119871 = 0 119886 119887 119888 119889 1

rarr 0 119886 119887 119888 119889 1

0 1 1 1 1 1 1

119886 119888 1 119887 119888 119887 1

119887 119889 119886 1 119887 119886 1

119888 119886 119886 1 1 119886 1

119889 119887 1 1 119887 1 1

1 0 119886 119887 119888 119889 1

ab

c d

1

0

Figure 1 Hasse diagram of 1198712= 0 119886 119887 119888 119889 1

(2) Let I sube J By (3) of Proposition 2 119888lowastJ(119860) sube 119888lowastI(119860)

for 119860 sube 119871 Thus if 119860 isin 119879lowastF(I) then 119860 isin 119879lowastF(J) Therefore119879lowastF(I) sube 119879lowastF(J)

Clearly if I isin I satisfies 119879119888F sube I then 119871 isin I and soI = 2119871 = I

119871 Thus by (1) of Proposition 3 119879lowastF = I119888 = 2119871 If

119879119888F 119871 sube I we have the following proposition

Proposition 7 Let (119871 119879F) be the filter topological space IfI isin I satisfies 119879119888F 119871 sube I then 119879lowastF = I119888 cup 0

Proof I119888 cup 0 sube 119879lowastF follows from Theorem 4 Converselysuppose that 119879lowastF sube I119888 cup0 Then there exists 119861 isin 119879

lowast

F (I119888 cup0) such that (119871 119861)

lowastsube 119871 119861 = 119871 Let 119910 isin (119871 119861) (119871 119861)

lowastThen 119910 notin (119871119861)

lowastThus [119910)cap(119871119861) isin I Put [119910)cap(119871119861) = 119868We have 119871119861 sube 119868cup (119871 [119910)) Since 119879119888F 119871 sube I 119871 [119910) isin IThus 119871119861 sube 119868cup (119871 119861) isin I and so 119871119861 isin I Hence 119861 isin I119888It is a contradiction Therefore 119879lowastF sube I119888 cup 0

Lemma 8 Let (119871 119879F) be the filter topological space andI isin

I If 119860 sube 119871 satisfies 119860 cap 119868 = 0 for each 119868 isin I then 119888lowast(119860) =

119888(119860)

Proof 119888lowast(119860) sube 119888(119860) is clear Conversely if 119909 notin 119888lowast(119860) then119909 notin 119860 and 119909 notin 119860lowastThus 119868 = [119909)cap119860 isin I and119860 sube 119868cup(119871[119909))Since 119860 cap 119868 = 0 119860 sube 119871 [119909) Observe that 119909 notin 119871 [119909) and119871 [119909) is 119879F-closed We have 119909 notin 119888(119860) Thus 119888(119860) sube 119888lowast(119860)Therefore 119888lowast(119860) = 119888(119860)

Proposition 9 IfI isin 119879119888F then 119879lowastF = 119879F

Proof It is clear that 119879F sube 119879lowastF Conversely let 119868119872 be thegreatest element of I and 119860 sube 119871 Thus (119860 119868

119872) cap 119868 = 0

for each 119868 isin I By Lemma 8 119888lowast(119860 119868119872) = 119888(119860 119868

119872)

By (8) of Proposition 2 (119860 119868119872)lowast

= 119860lowast Now notice that119860 cap 119868119872

isin I sube 119879119888F and thus 119888(119860 cap 119868119872) = 119860 cap 119868

119872 We have

119888 (119860) = 119888 ((119860 119868119872) cup (119860 cap 119868

119872))

= 119888 (119860 119868119872) cup 119888 (119860 cap 119868

119872)

= 119888lowast(119860 119868

119872) cup (119860 cap 119868

119872)

= (119860 119868119872) cup (119860 119868

119872)lowast

cup (119860 cap 119868119872)

= (119860 119868119872)lowast

cup 119860 = 119860lowastcup 119860 = 119888

lowast

(119860)

(13)

This implies 119888lowast = 119888 Therefore 119879lowastF = 119879F

Lemma 10 Let (119871 119879F) be the filter topological space andI isin

I If 119868 isin I and 119880 isin 119879F then 119880 119868 isin 119879lowastF

Proof Let 119875 = 119871 119880 Then 119875 isin 119879119888F By (7) and (8) ofProposition 2

119888lowast

(119875 cup 119868) = (119875 cup 119868) cup (119875 cup 119868)lowast= (119875 cup 119868) cup 119875

lowast= 119875 cup 119868

(14)

Thus 119871 (119875 cup 119868) = 119880 cap (119871 119868) = 119880 119868 isin 119879lowastF

Theorem11 Let (119871 119879F) be the filter topological space andI isin

I Then

B119879lowast

F= 119880 119868 119880 isin 119879F 119868 isin I (15)

is a base of 119879lowastF Moreover

B119879lowast

F(119909) = 119881 119868 119881 isin N

119879F(119909) 119909 notin 119868 isin I (16)

is a base of N119879lowast

F(119909) for each 119909 isin 119871 where N

119879lowast

F(119909) is the set

of all 119879lowastF-neighborhoods of 119909 in (119871 119879lowastF) Clearly [119909) 119868119909is

the smallest 119879lowastF-neighborhoods of 119909 where 119868119909is the greatest

element ofI satisfying 119909 notin 119868119909

Proof By Lemma 10B119879lowast

Fsube 119879lowastF Let 119861 sube 119871 Then 119861 isin 119879lowastF hArr

119871 119861 is 119879lowastF-closed hArr (119871 119861)lowastsube (119871 119861) hArr 119861 sube 119871 (119871 119861)

lowastThus 119909 isin 119861 rArr 119909 notin (119871 119861)

lowastrArr there exists 119880 isin N

119879F(119909) such

that (119871119861)cap119880 isin I Let 119868 = (119871119861)cap119880Then 119871119861 sube 119868cup(119871119880)

and

119909 isin 119880 119868 = 119880 cap (119871 119868) = 119871 (119868 cup (119871 119880)) sube 119861 (17)

ThereforeB119879lowast

Fis a base of 119879lowastF

ClearlyB119879lowast

F(119909) sube N

119879lowast

F(119909) Let 119860 isin N

119879lowast

F(119909) and 119910 isin 119860

SinceB119879lowast

Fis a base of 119879lowastF there are 119880119881 isin 119879F and 119868 119869 isin I

such that 119909 isin 119880 119868 sube 119860 and 119910 isin 119881 119869 sube 119860 We can assume

The Scientific World Journal 5

119910 notin 119868 and 119909 notin 119869 (otherwise 119868 and 119869 can be replaced by 119868 119910

and 119869 119909 resp) Then

(119880 119868) cup (119881 119869) = (119880 cap (119871 119868)) cup (119881 cap (119871 119869))

= [(119880 cup 119881) cap ((119871 119868) cup (119871 119869))]

cap [(119880 cup (119871 119869)) cap (119881 cup (119871 119868))]

supe ((119880 cup 119881) (119868 cap 119869)) cap (119871 (119868 cup 119869))

= (119880 cup 119881) (119868 cup 119869)

(18)

Clearly 119910 isin (119880 cup 119881) (119868 cup 119869) isin B119879lowast

F(119909) and

(119880 cup 119881) (119868 cup 119869) sube (119880 119868) cup (119881 119869) sube 119860 (19)

ThereforeB119879lowast

F(119909) is a base ofN

119879lowast

F(119909)

Clearly if 119868119909is the greatest element ofI satisfying 119909 notin 119868

119909

then [119909) 119868119909isin N119879lowast

F(119909) is the smallest 119879lowastF-neighborhoods of

119909

Let (119871 120591) be a topological space andI isin IThe topologythat was generated byB = 119880 119868 119880 isin 120591 119868 isin I is denotedby 119879lowast(I 120591) [5] Clearly 119879lowast(I 119879F) = 119879lowastF(I)

Lemma 12 Let 120595 = 0 119871 be the indiscrete topology on 119871 andI isin I Then 119879lowast(I 120595) = 0 cup 119868119888

Proof By 119879119888 119871 = 0 isin I and Proposition 7 the proof isobvious

Theorem 13 Let (119871 120591) be a topological space and I isin IThen 119879lowast(I 120591) = 120591 or 119879lowast(I 120595) where 120591 or 119879lowast(I 120595) is thetopology generated by the base 119880cap119881 119880 isin 120591 119881 isin 119879lowast(I 120595)

Proof Clearly B119879lowast = 119880 119868 119880 isin 120591 119868 isin I is a base of

119879lowast(I 120591) Since B119879lowast = 119880 cap 119881 119880 isin 120591 119881 isin 119879lowast(I 120595)

B119879lowast is also a base of 120591 or 119879lowast(I 120595) Therefore 119879lowast(I 120591) = 120591 or

119879lowast(I 120595)

Corollary 14 Let (119871 119879F) be the filter topological space andI isin I Then 119879lowastF = 119879F or 119879lowast(I 120595)

Corollary 15 Let (119871 119879F) be the filter topological space andI

J isin I Then

(1) 119879lowast(I orJ 120595) = 119879lowast(I 120595) or 119879lowast(J 120595)(2) 119879lowastF(I orJ) = 119879lowast(I 119879lowastF(J)) = 119879lowast(J 119879lowastF(I))(3) 119879lowastF(I orJ) = 119879lowastF(I) or 119879lowastF(J)(4) 119879lowast(I 119879lowastF(I)) = 119879lowastF(I)

Proof (1) By (2) of Proposition 6 119879lowast(I or J) supe 119879lowast(I) or

119879lowast(J) Conversely let 0 = 119860 isin 119879lowast(I or J) By Theorem 13there exist 119868 isin I and 119869 isin J such that

119860 = 119871 (119868 cup 119868) = (119871 119868) cap (119871 119869) isin 119879lowast

(I) or 119879lowast

(J)

(20)

Thus 119879lowast(I orJ) sube 119879lowast(I) or 119879lowast(J)

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

119879lowast

F (I orJ) = 119879F or 119879lowast(I orJ 120595)

= 119879F or 119879lowast(I 120595) or 119879

lowast(J 120595)

= 119879lowast

F (I) or 119879lowast(J 120595) = 119879

lowast(119879lowast

F (I) J)

(21)

Similarly 119879lowastF(I orJ) = 119879lowast(119879lowastF(J)I)(3) By (1) andTheorem 13

119879lowast

F (I orJ) = 119879F or 119879lowast(I orJ 120595)

= 119879F or 119879lowast(I 120595) or 119879

lowast(J 120595)

= 119879F or 119879lowast(I 120595) or 119879F or 119879

lowast(J 120595)

= 119879lowast

F (I) or 119879lowast

F (J)

(22)

Therefore 119879lowastF(I orJ) = 119879lowastF(I) or 119879lowastF(J)(4) LetI = J Then the proof follows from (2)

Theorem16 Let (119871 119879F) be the filter topological spaceIJ isin

I and 119860 sube 119871 Then

(1) 119860lowast(I capJ 119879F) = 119860lowast(I 119879F) cup 119860lowast(J 119879F)(2) 119860lowast(I orJ 119879F) = 119860lowast(I 119879lowastF(J)) cap 119860lowast(J 119879lowastF(I))

Proof (1) By (3) of Proposition 2 119860lowast(I 119879F) cup 119860lowast(J 119879F) sube

119860lowast(I cap J 119879F) Conversely 119909 notin 119860

lowast(I 119879F) cup 119860

lowast(J 119879F)

Then [119909) cap 119860 isin I and [119909) cap 119860 isin J Let [119909) cap 119860 = 119868 and[119909)cap119860 = 119869 Then119860 sube 119868cup (119871 [119909)) and119860 sube 119869cup (119871 [119909)) Thus

119860 sube (119868 cup (119871 [119909))) cap (119869 cup (119871 [119909))) = (119868 cap 119869) cup (119871 [119909))

(23)

Thus [119909) cap 119860 sube 119868 cap 119869 which implies 119909 notin 119860lowast(I cap J 119879F)

Therefore 119860lowast(I capJ 119879F) sube 119860lowast(I 119879F) cup 119860lowast(J 119879F)(3) Let 119909 notin 119860lowast(I orJ 119879F) Then [119909) cap 119860 isin I orJ Then

there exist 119868 isin I and 119869 isin J such that [119909) cap 119860 = 119868 cup 119869We can assume 119868 cap 119869 = 0 (otherwise 119868 can be replaced by119868 (119868 cap 119869)) Thus 119909 notin 119868 or 119909 notin 119869 (otherwise 119909 isin 119868 cap 119869 whichis a contradiction) Now we take 119909 notin 119868 for example Then

([119909) 119868) cap 119860 = [119909) cap 119860 cap (119871 119868) = 119869 (24)

Since [119909) isin 119879F and 119909 isin [119909) 119868 [119909) 119868 isin B119879lowast

F(I) Thus 119909 notin

119860lowast(J 119879lowastF(I)) and so 119909 notin 119860lowast(J 119879lowastF(I)) cap 119860lowast(I 119879lowastF(J))Hence

119860lowast(J 119879lowast

F (I)) cap 119860lowast(I 119879

lowast

F (J)) sube 119860lowast(I orJ 119879F)

(25)

Conversely let 119909 notin 119860lowast(119879lowastF(I)J) Then there exists 119868 isin

I such that ([119909) 119868) cap 119860 isin J Let ([119909) 119868) cap 119860 = 119869 Then[119909) cap 119860 = 119868 cup 119869 which implies 119909 notin 119860

lowast(I orJ 119879F) Similarlyif 119909 notin 119860lowast(119879lowastF(J)I) then 119909 notin 119860lowast(I or J 119879F) Therefore119860lowast(I orJ 119879F) sube 119860lowast(J 119879lowastF(I)) cap 119860lowast(I 119879lowastF(J))

Corollary 17 Consider 119860lowast(I 119879F) = 119860lowast(I 119879lowastF(I))

Proof LetI = J The proof follows from (2) ofTheorem 16

6 The Scientific World Journal

Corollary 18 Consider 119879lowastF(I capJ) = 119879lowastF(I) cap 119879lowastF(J)

Proof By (2) of Proposition 6119879lowastF(IcapJ) sube 119879lowastF(I)cap119879lowastF(J)Conversely if 119860 notin 119879lowastF(I capJ) then

(119871 119860)lowast(I 119879F) cup (119871 119860)

lowast(J 119879F)

= (119871 119860)lowast(I capJ 119879F) sube (119871 119860)

(26)

Thus (119871 119860)lowast(I 119879F) sube (119871 119860) or (119871 119860)

lowast(J 119879F) sube (119871 119860)

Thus119860 notin 119879lowastF(I) or119860 notin 119879lowastF(J)Therefore119879lowastF(I)cap119879lowastF(J) sube

119879lowastF(I capJ)

3 Power Idealization Filter TopologicalQuotient Spaces

Let (1198711 and or 1015840 rarr

1 01 11) and (119871

2 and or 1015840 rarr

2 02 12) be

two lattice implication algebras A mapping 119891 from 1198711to 1198712

is called lattice implication homomorphism if 119891(119909rarr1119910) =

119891(119909)rarr2119891(119910) for any 119909 119910 isin 119871

1 The set of all lattice

implication homomorphisms from 1198711to 1198712is denoted by

hom(1198711 1198712)

Let 119891 isin hom(1198711 1198712) Then clearly

119879119897(119879F(1198712)

119891) = 119891minus1

(119880) 119880 isin 119879F(1198712)

119879119903(119879F(1198711)

119891) = 119891 (119880) 119880 sube 1198712 119891minus1

(119880) isin 119879F(1198711)

(27)

are topologies [4]

Lemma 19 Let 1198711and 119871

2be two implication algebras and let

119891 isin hom(F1 f2) andI isin 21198711 J isin 21198712 be power ideals

(1) If 119891 is injective then 119891minus1(J) = 119891minus1(119869) 119869 isin J isin

I(1198711)

(2) If 119891 is surjective then 119891(I) = 119891(119868) 119868 isin I isin

I(1198712)

Proof (1) Since 0 isin J then 0 = 119891minus1(0) isin 119891minus1(J)If 1198682isin 119891minus1(J) and 119868

1sube 1198682 then there exist 119869

2isin J such

that 1198682= 119891minus1(119869

2) Thus 119891(119868

1) sube 119891(119868

2) = 1198692and 119891(119868

1) isin J

Since 119891 is injective 1198681= 119891minus1(119891(119868

1)) isin 119891minus1(J)

If 1198681 1198682isin 119891minus1(J) then there exist 119869

1 1198692isin J such that

1198681= 119891minus1(119869

1) and 119868

2= 119891minus1(119869

2) One has 119868

1cup 1198682= 119891minus1(119869

1) cup

119891minus1(1198692) = 119891minus1(119869

1cup 1198692) Since 119869

1cup 1198692isin J 119868

1cup 1198682isin 119891minus1(J)

Therefore 119891minus1(J) is a power ideal(2) By 0 isin I 0 = 119891(0) isin 119891(I)If 1198692isin 119891(J) and 119869

1sube 1198692 then there exists 119868

2isin I such

that 1198692= 119891(119868

2) Let 119868

1= 119891minus1(119869

1) Then 119868

1sube 1198682and 1198681isin I

Since 119891 is surjective 1198691= 119891(1198681) isin 119891(I)

If 1198691 1198692isin 119891(J) then there exist 119868

1 1198682isin I such that 119869

1=

119891(1198681) and 119869

2= 119891(119868

2) Thus 119868

1cup 1198682

isin I Hence 1198691cup 1198692

=

119891(1198681) cup119891(119868

2) = 119891(119868

1cup1198682) isin 119891(I) Therefore 119891(I) is a power

ideal

Lemma 20 Let 1198711and 119871

2be two implication algebras and

119891 isin hom(F1 F2) Then

(1) if 119891 is injective then for each 119909 isin 1198711 119891minus1([119891(119909))) isin

119879119897(119879F(1198712)

119891) is the smallest 119879119897-neighborhood of 119909

(2) if 119891 is bijective then for each 119910 isin 1198712 119891([119891minus1(119910))) isin

119879119903(119879F(1198711)

119891) is the smallest 119879119903-neighborhood of 119910

Proof (1) Clearly [119891(119909)) is the smallest119879F(1198712)-neighborhood

of 119891(119909) Then 119909 isin 119891minus1([119891(119909))) isin 119879119897(119879F(1198712)

119891) Let 119909 isin 119881 isin

119879119897(119879F(1198712)

119891)Then119891(119909) isin 119891(119881) isin 119879F(1198712)and [119891(119909)) sube 119891(119881)

Thus 119891minus1([119891(119909))) sube 119891minus1(119891(119881)) = 119881 Therefore 119891minus1([119891(119909))) isthe smallest one

(2) Clearly [119891minus1(119910)) is the smallest 119879F(1198711)-neighborhood

of 119891minus1(119910) Thus 119910 isin 119891([119891minus1(119910))) isin 119879

119903(119879F(1198711)

119891) Now let119910 isin 119880 isin 119879

119903(119879F(1198711)

119891) Then 119891minus1(119910) isin 119891minus1(119880) isin 119879F(1198711) Thus

[119891minus1(119910)) sube 119891minus1(119880) and so 119891([119891minus1(119910))) sube 119891(119891minus1(119880)) = 119880Therefore (2) holds

Lemma21 Let 1198711and 119871

2be two implication algebras and119891 isin

hom(F1 F2)

(1) If 119891 is injective and J isin I(1198712) then for each 119909 isin

1198711 [119891(119909)) 119869

119872isin 119879lowastF(J 119879F(1198712)

) is the smallestneighborhood of 119891(119909) where 119869

119872is the greatest element

ofJ satisfying 119891(119909) notin 119869119872

(2) If 119891 is bijective and I isin I(1198711) then for each 119910 isin

1198712 [119891minus1(119910)) 119868

119872isin 119879lowastF(I 119879F(1198711)

) is the smallestneighborhood of 119891minus1(119910) where 119868

119872is the greatest

element ofI satisfying 119891minus1(119910) notin 119868119872

Theorem 22 Let 1198711and 119871

2be two implication algebras and

119891 isin hom(F1 F2)

(1) If 119891 is injective andJ isin I(1198712) then

119879lowast(119891minus1

(J) 119879119897(119879F(1198712)

119891)) = 119879119897(119879lowast

F (J 119879F(1198712)) 119891)

(28)

(2) If 119891 is bijective andI isin I(1198711) then

119879lowast(119891 (I) 119879

119903(119879F(1198711)

119891)) = 119879119903(119879lowast

F (I 119879F(1198711)) 119891)

(29)

Proof (1) Let 119888lowast2and 119888119897be the closure operators of the left side

and the right side of the equationWe only need to prove 119888lowast1=

119888119897Let 119860 sube 119871

1and 119909 notin 119888lowast

1(119860) Then 119909 notin 119860 and 119909 notin

119860lowast(119891minus1(J) 119879119897(119879F(1198712)

119891)) By (1) of Lemma 21 119891minus1([119891(119909)))cap119860 isin 119891minus1(J) Thus there exists 119869 isin J such that 119891minus1([119891(119909))) cap119860 = 119891minus1(119869) Since 119909 notin 119860 and 119891 is injective 119891(119909) notin 119869Let 119869119872

isin J be the greatest one satisfying 119891(119909) notin 119869119872 Then

119891minus1([119891(119909))) cap 119860 sube 119891minus1(119869119872) Thus

0 = 119891minus1

([119891 (119909))) cap 119860 cap (1198711 119891minus1

(119869119872))

= 119891minus1

([119891 (119909))) cap 119891minus1

(1198712 119869119872) cap 119860

= 119891minus1

([119891 (119909)) cap (1198712 119869119872)) cap 119860

= 119891minus1

([119891 (119909)) 119869119872) cap 119860

(30)

By Lemma 21 and Proposition 1 119909 notin 119888119897(119860) Therefore 119888

119897(119860) sube

119888lowast1(119860)

The Scientific World Journal 7

Conversely let 119910 notin 119888119897(119860) By Proposition 1

0 = 119891minus1

([119891 (119909)) 119869119872) cap 119860

= 119891minus1

([119891 (119909))) cap 119860 cap (1198711 119891minus1

(119869119872))

(31)

Thus 119909 notin 119860 119891minus1([119891(119909))) cap 119860 sube 119891minus1(119869119872) and [119891(119909)) cap

119891(119860) sube 119869119872 Then 119869

1= [119891(119909)) cap 119891(119860) isin J Since 119891 is

injective 119891minus1([119891(119909))) cap 119860 = 119891minus1(1198691) By (1) of Lemma 20

119909 notin 119860lowast(119891minus1(J) 119879119897(119879F(1198712)

119891)) Therefore 119909 notin 119888lowast1(119860) and

119888lowast1(119860) sube 119888

119897(119860)

(2) Let 119888lowast2and 119888119903be the closure operators of the left side

and the right side of the equationWe only need to prove 119888lowast2=

119888119903Let 119910 notin 119888lowast

2(119860) Then 119910 notin 119860 and 119910 notin 119860lowast((119891(I)

119879119903(119879F(1198711)

119891))) By (2) of Lemma 20119891([119891minus1(119910)))cap119860 isin 119891(I)Thus there exists 119868 isin I such that 119891([119891minus1(119910))) cap 119860 = 119891(119868)Since119891 is bijective [119891minus1(119910))cap119891minus1(119860) = 119868 By119910 notin 119860119891minus1(119910) notin119891minus1(119860) and so 119891minus1(119910) notin 119868 Since 119868

119872isin I is the greatest

element ofI satisfying119891minus1(119910) notin 119868119872 [119891minus1(119910))cap119891minus1(119860) sube 119868

119872

and [119891minus1(119910)) cap (1198711 119868119872) cap 119891minus1(119860) = 0 By 119891 being bijective

again we have

0 = 119891 (0) = 119891 ([119891minus1

(119910))) cap (1198712 119891 (119868119872)) cap 119860

= 119891 ([119891minus1

(119910)) 119868119872) cap 119860

(32)

By

119891minus1

(119910) isin [119891minus1

(119910)) 119868119872

= 119891minus1

(119891 ([119891minus1

(119910)) 119868119872)) isin 119879

lowast

F (I 119879F(1198711))

(33)

119910 notin 119888119903(119860) Hence 119888lowast

2(119860) sube 119888

119903(119860)

Conversely let 119911 notin 119888119903(119860) Since 119868

119872is the greatest element

of I satisfying 119891minus1(119911) notin 119868119872 by (2) of Lemma 21 [119891minus1(119911))

119868119872

isin 119879lowastF(I 119879F(1198711)) is the smallest neighborhood of 119891minus1(119911)

Thus 119891([119891minus1(119911)) 119868119872) cap 119860 = 0 Since 119891 is bijective

0 = 119891minus1

(0) = ([119891minus1

(119911)) 119868119872) cap 119891minus1

(119860)

= [119891minus1

(119911)) cap (1198711 119868119872) cap 119891minus1

(119860)

(34)

This implies [119891minus1(119911))cap119891minus1(119860) sube 119868119872and [119891minus1(119911))cap119891minus1(119860) isin

I Thus 119891([119891minus1(119910))) cap 119860 isin 119891(I) which implies 119910 isin

119860lowast((119891(I) 119879119903(119879F(1198711)

119891))) Since 119911 notin 119888119903(119860) 119911 notin 119860 Therefore

119911 notin 119888lowast2(119860) and so 119888lowast

2(119860) sube 119888

119903(119860)

Generally if 119891 isin hom(1198711 1198712) is surjective but not bijec-

tive then (2) of Theorem 22 fails

Example 23 Let 1198711

= 01 119886 119887 119888 119889 1

1 be the Hasse lattice

implication algebra of Example 5 Let 1198712= 02 119890 ℎ 1

2 and

010158402= 12 1198901015840 = ℎ ℎ1015840 = 119890 and 11015840

2= 02 The Hasse diagram and

Table 2 The implication operator of 1198712= 02 119890 ℎ 1

2

rarr 02

119890 ℎ 12

02

1 12

12

12

119890 ℎ 12

ℎ 12

ℎ 119890 119890 12

12

12

02

119890 ℎ 12

e

h

12

02

Figure 2 Hasse diagram of 1198712= 02 119890 ℎ 1

2

the implication operator of 1198712are shown by Figure 2 and

Table 2 Then it is clear that

119879lowast

F(1198711)= 0 1

1 119886 1

1 119887 119888 1

1 119886 119887 119888 1

1

119887 119888 119889 11 0 119887 119888 119889 1

1 119886 119887 119888 119889 1

1 119871

119879lowast

F(1198712)= 0 119890 1

2 ℎ 1

2 119890 ℎ 1

2 12 1198712

(35)

A mapping 119891 from 1198711to 1198712is defined as

119891 (01) = 02 119891 (1

1) = 12 119891 (119886) = 119891 (119889) = 119890

119891 (119887) = 119891 (119888) = ℎ(36)

It easy to check 119891 isin hom(1198711 1198712) and 119891 is surjective LetI =

0 119888 ThenI isin I(1198711) and 119891(I) = 0 ℎ isin I(119871

2)

Since ℎ isin 119891(I) by Theorem 4

02 119890 12 = 1198712 ℎ isin 119879

lowast(119891 (I) 119879

119903(119879F(1198711)

119891)) (37)

Observe that 119887 119888lowast(I 119879F(1198711)) = 0

1 119887 119888 119889 sube 119887 119888 We

have

0 119886 119889 1 = 1198711 119887 119888 notin 119879

lowast

F (I 119879F(1198711)) (38)

Moreover by 119891minus1(02 119890 12) = 0

2 119886 119889 1

2 02 119890 12 notin

119879119903(119879lowastF(I 119879F(1198711)

) 119891)In fact we have

119879119903(119879lowast

F (I 119879F(1198711)) 119891) = 0 1

2 1198712

119879lowast(119891 (I) 119879

119903(119879F(1198711)

119891)) = 0 12 02 119890 12 1198712

(39)

Therefore 119879119903(119879lowastF(I 119879F(1198711)

) 119891) = 119879lowast(119891(I) 119879119903(119879F(1198711)

119891))

8 The Scientific World Journal

Corollary 24 Let 1198711and 119871

2be two implication algebras and

let 119891 isin hom(F1 f2) be bijective(1) IfJ isin I(119871

2) then

119879119903(119879lowast(119891minus1

(J) 119879119897(119879F(1198712)

119891)) 119891) = 119879lowast(J 119879F(1198712)

)

(40)

(2) IfI isin I(1198711) then

119879119897(119879lowast(119891 (I) 119879

119903(119879F(1198711)

119891)) 119891) = 119879lowast(I 119879F(1198711)

) (41)

Proof The proof follows fromTheorem 22

Corollary 25 Let 1198711and 119871

2be two implication algebras and

119891 isin hom(f1 f2)(1) If 119891 is injective andJ

1J2isin I(119871

2) then

119879119897(119879lowast

F (J1orJ2 119879F(1198712)

) 119891)

= 119879119897(119879lowast

F (J1 119879F(1198712)

) 119891) or 119879119897(119879lowast

F (J2 119879F(1198712)

) 119891)

(42)

(2) If 119891 is bijective andI1I2isin I(119871

1) then

119879119903(119879lowast

F (I1orI2 119879F(1198711)

) 119891)

= 119879119903(119879lowast

F (I1 119879F(1198711)

) 119891) or 119879119903(119879lowast

F (I2 119879F(1198711)

) 119891)

(43)

Proof (1) Clearly 119891minus1(J1orJ2) = 119891minus1(J

1) or119891minus1(J

2) By (3)

of Corollary 15 andTheorem 22

119879119897(119879lowast

F (J1orJ2 119879F(1198712)

) 119891)

= 119879lowast(119891minus1

(J1orJ2) 119879119897(119879F(1198712)

119891))

= 119879lowast(119891minus1

(J1) 119879119897(119879F(1198712)

119891))

or 119879lowast(119891minus1

(J2) 119879119897(119879F(1198712)

119891))

= 119879119897(119879lowast

F (J1 119879F(1198712)

) 119891) or 119879119897(119879lowast

F (J2 119879F(1198712)

) 119891)

(44)

(2) Is similar to (1)

Corollary 26 Let 1198711and 119871

2be two implication algebras and

let 119891 isin hom(F1 f2) be bijective If J isin I(1198712) and I isin

I(1198711) then

119879lowast(119891minus1

(J) 119879119897(119879lowast

F (J 119879F(1198712)) 119891))

= 119879lowast(119891minus1

(J) 119879119897(119879F(1198712)

119891))

119879lowast(119891 (I) 119879

119903(119879lowast

F (I 119879F(1198711)) 119891))

= 119879lowast(119891 (I) 119879

119903(119879F(1198711)

119891))

(45)

Proof The proof follows from (4) of Corollary 15 andTheorem 22

Conflict of Interests

The authors declare that there is no conflict of interests regar-ding the publication of this paper

Acknowledgments

This work is supported by the National Natural ScienceFoundations of China (no 11471202) and the Natural ScienceFoundation of Guangdong Province (no S2012010008833)

References

[1] J Lukasiewicz ldquoOn 3-valued logicrdquo Ruch Filozoficzny vol 5 pp169ndash170 1920

[2] Y Xu ldquoLattice implication algebrardquo Journal of Southwest Jiao-tong University vol 89 pp 20ndash27 1993

[3] Y Xu and K Y Qin ldquoOn filters of lattice implication algebrasrdquoThe Journal of FuzzyMathematics vol 1 no 2 pp 251ndash260 1993

[4] Y Xu D Ruan K Qin and J Liu Lattice-Valued Logic Springer2003

[5] D Jankovic and T R Hamlett ldquoNew topologies from old viaidealsrdquo The American Mathematical Monthly vol 97 no 4 pp295ndash310 1990

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Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Stochastic AnalysisInternational Journal of

4 The Scientific World Journal

Table 1 The implication operator of 119871 = 0 119886 119887 119888 119889 1

rarr 0 119886 119887 119888 119889 1

0 1 1 1 1 1 1

119886 119888 1 119887 119888 119887 1

119887 119889 119886 1 119887 119886 1

119888 119886 119886 1 1 119886 1

119889 119887 1 1 119887 1 1

1 0 119886 119887 119888 119889 1

ab

c d

1

0

Figure 1 Hasse diagram of 1198712= 0 119886 119887 119888 119889 1

(2) Let I sube J By (3) of Proposition 2 119888lowastJ(119860) sube 119888lowastI(119860)

for 119860 sube 119871 Thus if 119860 isin 119879lowastF(I) then 119860 isin 119879lowastF(J) Therefore119879lowastF(I) sube 119879lowastF(J)

Clearly if I isin I satisfies 119879119888F sube I then 119871 isin I and soI = 2119871 = I

119871 Thus by (1) of Proposition 3 119879lowastF = I119888 = 2119871 If

119879119888F 119871 sube I we have the following proposition

Proposition 7 Let (119871 119879F) be the filter topological space IfI isin I satisfies 119879119888F 119871 sube I then 119879lowastF = I119888 cup 0

Proof I119888 cup 0 sube 119879lowastF follows from Theorem 4 Converselysuppose that 119879lowastF sube I119888 cup0 Then there exists 119861 isin 119879

lowast

F (I119888 cup0) such that (119871 119861)

lowastsube 119871 119861 = 119871 Let 119910 isin (119871 119861) (119871 119861)

lowastThen 119910 notin (119871119861)

lowastThus [119910)cap(119871119861) isin I Put [119910)cap(119871119861) = 119868We have 119871119861 sube 119868cup (119871 [119910)) Since 119879119888F 119871 sube I 119871 [119910) isin IThus 119871119861 sube 119868cup (119871 119861) isin I and so 119871119861 isin I Hence 119861 isin I119888It is a contradiction Therefore 119879lowastF sube I119888 cup 0

Lemma 8 Let (119871 119879F) be the filter topological space andI isin

I If 119860 sube 119871 satisfies 119860 cap 119868 = 0 for each 119868 isin I then 119888lowast(119860) =

119888(119860)

Proof 119888lowast(119860) sube 119888(119860) is clear Conversely if 119909 notin 119888lowast(119860) then119909 notin 119860 and 119909 notin 119860lowastThus 119868 = [119909)cap119860 isin I and119860 sube 119868cup(119871[119909))Since 119860 cap 119868 = 0 119860 sube 119871 [119909) Observe that 119909 notin 119871 [119909) and119871 [119909) is 119879F-closed We have 119909 notin 119888(119860) Thus 119888(119860) sube 119888lowast(119860)Therefore 119888lowast(119860) = 119888(119860)

Proposition 9 IfI isin 119879119888F then 119879lowastF = 119879F

Proof It is clear that 119879F sube 119879lowastF Conversely let 119868119872 be thegreatest element of I and 119860 sube 119871 Thus (119860 119868

119872) cap 119868 = 0

for each 119868 isin I By Lemma 8 119888lowast(119860 119868119872) = 119888(119860 119868

119872)

By (8) of Proposition 2 (119860 119868119872)lowast

= 119860lowast Now notice that119860 cap 119868119872

isin I sube 119879119888F and thus 119888(119860 cap 119868119872) = 119860 cap 119868

119872 We have

119888 (119860) = 119888 ((119860 119868119872) cup (119860 cap 119868

119872))

= 119888 (119860 119868119872) cup 119888 (119860 cap 119868

119872)

= 119888lowast(119860 119868

119872) cup (119860 cap 119868

119872)

= (119860 119868119872) cup (119860 119868

119872)lowast

cup (119860 cap 119868119872)

= (119860 119868119872)lowast

cup 119860 = 119860lowastcup 119860 = 119888

lowast

(119860)

(13)

This implies 119888lowast = 119888 Therefore 119879lowastF = 119879F

Lemma 10 Let (119871 119879F) be the filter topological space andI isin

I If 119868 isin I and 119880 isin 119879F then 119880 119868 isin 119879lowastF

Proof Let 119875 = 119871 119880 Then 119875 isin 119879119888F By (7) and (8) ofProposition 2

119888lowast

(119875 cup 119868) = (119875 cup 119868) cup (119875 cup 119868)lowast= (119875 cup 119868) cup 119875

lowast= 119875 cup 119868

(14)

Thus 119871 (119875 cup 119868) = 119880 cap (119871 119868) = 119880 119868 isin 119879lowastF

Theorem11 Let (119871 119879F) be the filter topological space andI isin

I Then

B119879lowast

F= 119880 119868 119880 isin 119879F 119868 isin I (15)

is a base of 119879lowastF Moreover

B119879lowast

F(119909) = 119881 119868 119881 isin N

119879F(119909) 119909 notin 119868 isin I (16)

is a base of N119879lowast

F(119909) for each 119909 isin 119871 where N

119879lowast

F(119909) is the set

of all 119879lowastF-neighborhoods of 119909 in (119871 119879lowastF) Clearly [119909) 119868119909is

the smallest 119879lowastF-neighborhoods of 119909 where 119868119909is the greatest

element ofI satisfying 119909 notin 119868119909

Proof By Lemma 10B119879lowast

Fsube 119879lowastF Let 119861 sube 119871 Then 119861 isin 119879lowastF hArr

119871 119861 is 119879lowastF-closed hArr (119871 119861)lowastsube (119871 119861) hArr 119861 sube 119871 (119871 119861)

lowastThus 119909 isin 119861 rArr 119909 notin (119871 119861)

lowastrArr there exists 119880 isin N

119879F(119909) such

that (119871119861)cap119880 isin I Let 119868 = (119871119861)cap119880Then 119871119861 sube 119868cup(119871119880)

and

119909 isin 119880 119868 = 119880 cap (119871 119868) = 119871 (119868 cup (119871 119880)) sube 119861 (17)

ThereforeB119879lowast

Fis a base of 119879lowastF

ClearlyB119879lowast

F(119909) sube N

119879lowast

F(119909) Let 119860 isin N

119879lowast

F(119909) and 119910 isin 119860

SinceB119879lowast

Fis a base of 119879lowastF there are 119880119881 isin 119879F and 119868 119869 isin I

such that 119909 isin 119880 119868 sube 119860 and 119910 isin 119881 119869 sube 119860 We can assume

The Scientific World Journal 5

119910 notin 119868 and 119909 notin 119869 (otherwise 119868 and 119869 can be replaced by 119868 119910

and 119869 119909 resp) Then

(119880 119868) cup (119881 119869) = (119880 cap (119871 119868)) cup (119881 cap (119871 119869))

= [(119880 cup 119881) cap ((119871 119868) cup (119871 119869))]

cap [(119880 cup (119871 119869)) cap (119881 cup (119871 119868))]

supe ((119880 cup 119881) (119868 cap 119869)) cap (119871 (119868 cup 119869))

= (119880 cup 119881) (119868 cup 119869)

(18)

Clearly 119910 isin (119880 cup 119881) (119868 cup 119869) isin B119879lowast

F(119909) and

(119880 cup 119881) (119868 cup 119869) sube (119880 119868) cup (119881 119869) sube 119860 (19)

ThereforeB119879lowast

F(119909) is a base ofN

119879lowast

F(119909)

Clearly if 119868119909is the greatest element ofI satisfying 119909 notin 119868

119909

then [119909) 119868119909isin N119879lowast

F(119909) is the smallest 119879lowastF-neighborhoods of

119909

Let (119871 120591) be a topological space andI isin IThe topologythat was generated byB = 119880 119868 119880 isin 120591 119868 isin I is denotedby 119879lowast(I 120591) [5] Clearly 119879lowast(I 119879F) = 119879lowastF(I)

Lemma 12 Let 120595 = 0 119871 be the indiscrete topology on 119871 andI isin I Then 119879lowast(I 120595) = 0 cup 119868119888

Proof By 119879119888 119871 = 0 isin I and Proposition 7 the proof isobvious

Theorem 13 Let (119871 120591) be a topological space and I isin IThen 119879lowast(I 120591) = 120591 or 119879lowast(I 120595) where 120591 or 119879lowast(I 120595) is thetopology generated by the base 119880cap119881 119880 isin 120591 119881 isin 119879lowast(I 120595)

Proof Clearly B119879lowast = 119880 119868 119880 isin 120591 119868 isin I is a base of

119879lowast(I 120591) Since B119879lowast = 119880 cap 119881 119880 isin 120591 119881 isin 119879lowast(I 120595)

B119879lowast is also a base of 120591 or 119879lowast(I 120595) Therefore 119879lowast(I 120591) = 120591 or

119879lowast(I 120595)

Corollary 14 Let (119871 119879F) be the filter topological space andI isin I Then 119879lowastF = 119879F or 119879lowast(I 120595)

Corollary 15 Let (119871 119879F) be the filter topological space andI

J isin I Then

(1) 119879lowast(I orJ 120595) = 119879lowast(I 120595) or 119879lowast(J 120595)(2) 119879lowastF(I orJ) = 119879lowast(I 119879lowastF(J)) = 119879lowast(J 119879lowastF(I))(3) 119879lowastF(I orJ) = 119879lowastF(I) or 119879lowastF(J)(4) 119879lowast(I 119879lowastF(I)) = 119879lowastF(I)

Proof (1) By (2) of Proposition 6 119879lowast(I or J) supe 119879lowast(I) or

119879lowast(J) Conversely let 0 = 119860 isin 119879lowast(I or J) By Theorem 13there exist 119868 isin I and 119869 isin J such that

119860 = 119871 (119868 cup 119868) = (119871 119868) cap (119871 119869) isin 119879lowast

(I) or 119879lowast

(J)

(20)

Thus 119879lowast(I orJ) sube 119879lowast(I) or 119879lowast(J)

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

119879lowast

F (I orJ) = 119879F or 119879lowast(I orJ 120595)

= 119879F or 119879lowast(I 120595) or 119879

lowast(J 120595)

= 119879lowast

F (I) or 119879lowast(J 120595) = 119879

lowast(119879lowast

F (I) J)

(21)

Similarly 119879lowastF(I orJ) = 119879lowast(119879lowastF(J)I)(3) By (1) andTheorem 13

119879lowast

F (I orJ) = 119879F or 119879lowast(I orJ 120595)

= 119879F or 119879lowast(I 120595) or 119879

lowast(J 120595)

= 119879F or 119879lowast(I 120595) or 119879F or 119879

lowast(J 120595)

= 119879lowast

F (I) or 119879lowast

F (J)

(22)

Therefore 119879lowastF(I orJ) = 119879lowastF(I) or 119879lowastF(J)(4) LetI = J Then the proof follows from (2)

Theorem16 Let (119871 119879F) be the filter topological spaceIJ isin

I and 119860 sube 119871 Then

(1) 119860lowast(I capJ 119879F) = 119860lowast(I 119879F) cup 119860lowast(J 119879F)(2) 119860lowast(I orJ 119879F) = 119860lowast(I 119879lowastF(J)) cap 119860lowast(J 119879lowastF(I))

Proof (1) By (3) of Proposition 2 119860lowast(I 119879F) cup 119860lowast(J 119879F) sube

119860lowast(I cap J 119879F) Conversely 119909 notin 119860

lowast(I 119879F) cup 119860

lowast(J 119879F)

Then [119909) cap 119860 isin I and [119909) cap 119860 isin J Let [119909) cap 119860 = 119868 and[119909)cap119860 = 119869 Then119860 sube 119868cup (119871 [119909)) and119860 sube 119869cup (119871 [119909)) Thus

119860 sube (119868 cup (119871 [119909))) cap (119869 cup (119871 [119909))) = (119868 cap 119869) cup (119871 [119909))

(23)

Thus [119909) cap 119860 sube 119868 cap 119869 which implies 119909 notin 119860lowast(I cap J 119879F)

Therefore 119860lowast(I capJ 119879F) sube 119860lowast(I 119879F) cup 119860lowast(J 119879F)(3) Let 119909 notin 119860lowast(I orJ 119879F) Then [119909) cap 119860 isin I orJ Then

there exist 119868 isin I and 119869 isin J such that [119909) cap 119860 = 119868 cup 119869We can assume 119868 cap 119869 = 0 (otherwise 119868 can be replaced by119868 (119868 cap 119869)) Thus 119909 notin 119868 or 119909 notin 119869 (otherwise 119909 isin 119868 cap 119869 whichis a contradiction) Now we take 119909 notin 119868 for example Then

([119909) 119868) cap 119860 = [119909) cap 119860 cap (119871 119868) = 119869 (24)

Since [119909) isin 119879F and 119909 isin [119909) 119868 [119909) 119868 isin B119879lowast

F(I) Thus 119909 notin

119860lowast(J 119879lowastF(I)) and so 119909 notin 119860lowast(J 119879lowastF(I)) cap 119860lowast(I 119879lowastF(J))Hence

119860lowast(J 119879lowast

F (I)) cap 119860lowast(I 119879

lowast

F (J)) sube 119860lowast(I orJ 119879F)

(25)

Conversely let 119909 notin 119860lowast(119879lowastF(I)J) Then there exists 119868 isin

I such that ([119909) 119868) cap 119860 isin J Let ([119909) 119868) cap 119860 = 119869 Then[119909) cap 119860 = 119868 cup 119869 which implies 119909 notin 119860

lowast(I orJ 119879F) Similarlyif 119909 notin 119860lowast(119879lowastF(J)I) then 119909 notin 119860lowast(I or J 119879F) Therefore119860lowast(I orJ 119879F) sube 119860lowast(J 119879lowastF(I)) cap 119860lowast(I 119879lowastF(J))

Corollary 17 Consider 119860lowast(I 119879F) = 119860lowast(I 119879lowastF(I))

Proof LetI = J The proof follows from (2) ofTheorem 16

6 The Scientific World Journal

Corollary 18 Consider 119879lowastF(I capJ) = 119879lowastF(I) cap 119879lowastF(J)

Proof By (2) of Proposition 6119879lowastF(IcapJ) sube 119879lowastF(I)cap119879lowastF(J)Conversely if 119860 notin 119879lowastF(I capJ) then

(119871 119860)lowast(I 119879F) cup (119871 119860)

lowast(J 119879F)

= (119871 119860)lowast(I capJ 119879F) sube (119871 119860)

(26)

Thus (119871 119860)lowast(I 119879F) sube (119871 119860) or (119871 119860)

lowast(J 119879F) sube (119871 119860)

Thus119860 notin 119879lowastF(I) or119860 notin 119879lowastF(J)Therefore119879lowastF(I)cap119879lowastF(J) sube

119879lowastF(I capJ)

3 Power Idealization Filter TopologicalQuotient Spaces

Let (1198711 and or 1015840 rarr

1 01 11) and (119871

2 and or 1015840 rarr

2 02 12) be

two lattice implication algebras A mapping 119891 from 1198711to 1198712

is called lattice implication homomorphism if 119891(119909rarr1119910) =

119891(119909)rarr2119891(119910) for any 119909 119910 isin 119871

1 The set of all lattice

implication homomorphisms from 1198711to 1198712is denoted by

hom(1198711 1198712)

Let 119891 isin hom(1198711 1198712) Then clearly

119879119897(119879F(1198712)

119891) = 119891minus1

(119880) 119880 isin 119879F(1198712)

119879119903(119879F(1198711)

119891) = 119891 (119880) 119880 sube 1198712 119891minus1

(119880) isin 119879F(1198711)

(27)

are topologies [4]

Lemma 19 Let 1198711and 119871

2be two implication algebras and let

119891 isin hom(F1 f2) andI isin 21198711 J isin 21198712 be power ideals

(1) If 119891 is injective then 119891minus1(J) = 119891minus1(119869) 119869 isin J isin

I(1198711)

(2) If 119891 is surjective then 119891(I) = 119891(119868) 119868 isin I isin

I(1198712)

Proof (1) Since 0 isin J then 0 = 119891minus1(0) isin 119891minus1(J)If 1198682isin 119891minus1(J) and 119868

1sube 1198682 then there exist 119869

2isin J such

that 1198682= 119891minus1(119869

2) Thus 119891(119868

1) sube 119891(119868

2) = 1198692and 119891(119868

1) isin J

Since 119891 is injective 1198681= 119891minus1(119891(119868

1)) isin 119891minus1(J)

If 1198681 1198682isin 119891minus1(J) then there exist 119869

1 1198692isin J such that

1198681= 119891minus1(119869

1) and 119868

2= 119891minus1(119869

2) One has 119868

1cup 1198682= 119891minus1(119869

1) cup

119891minus1(1198692) = 119891minus1(119869

1cup 1198692) Since 119869

1cup 1198692isin J 119868

1cup 1198682isin 119891minus1(J)

Therefore 119891minus1(J) is a power ideal(2) By 0 isin I 0 = 119891(0) isin 119891(I)If 1198692isin 119891(J) and 119869

1sube 1198692 then there exists 119868

2isin I such

that 1198692= 119891(119868

2) Let 119868

1= 119891minus1(119869

1) Then 119868

1sube 1198682and 1198681isin I

Since 119891 is surjective 1198691= 119891(1198681) isin 119891(I)

If 1198691 1198692isin 119891(J) then there exist 119868

1 1198682isin I such that 119869

1=

119891(1198681) and 119869

2= 119891(119868

2) Thus 119868

1cup 1198682

isin I Hence 1198691cup 1198692

=

119891(1198681) cup119891(119868

2) = 119891(119868

1cup1198682) isin 119891(I) Therefore 119891(I) is a power

ideal

Lemma 20 Let 1198711and 119871

2be two implication algebras and

119891 isin hom(F1 F2) Then

(1) if 119891 is injective then for each 119909 isin 1198711 119891minus1([119891(119909))) isin

119879119897(119879F(1198712)

119891) is the smallest 119879119897-neighborhood of 119909

(2) if 119891 is bijective then for each 119910 isin 1198712 119891([119891minus1(119910))) isin

119879119903(119879F(1198711)

119891) is the smallest 119879119903-neighborhood of 119910

Proof (1) Clearly [119891(119909)) is the smallest119879F(1198712)-neighborhood

of 119891(119909) Then 119909 isin 119891minus1([119891(119909))) isin 119879119897(119879F(1198712)

119891) Let 119909 isin 119881 isin

119879119897(119879F(1198712)

119891)Then119891(119909) isin 119891(119881) isin 119879F(1198712)and [119891(119909)) sube 119891(119881)

Thus 119891minus1([119891(119909))) sube 119891minus1(119891(119881)) = 119881 Therefore 119891minus1([119891(119909))) isthe smallest one

(2) Clearly [119891minus1(119910)) is the smallest 119879F(1198711)-neighborhood

of 119891minus1(119910) Thus 119910 isin 119891([119891minus1(119910))) isin 119879

119903(119879F(1198711)

119891) Now let119910 isin 119880 isin 119879

119903(119879F(1198711)

119891) Then 119891minus1(119910) isin 119891minus1(119880) isin 119879F(1198711) Thus

[119891minus1(119910)) sube 119891minus1(119880) and so 119891([119891minus1(119910))) sube 119891(119891minus1(119880)) = 119880Therefore (2) holds

Lemma21 Let 1198711and 119871

2be two implication algebras and119891 isin

hom(F1 F2)

(1) If 119891 is injective and J isin I(1198712) then for each 119909 isin

1198711 [119891(119909)) 119869

119872isin 119879lowastF(J 119879F(1198712)

) is the smallestneighborhood of 119891(119909) where 119869

119872is the greatest element

ofJ satisfying 119891(119909) notin 119869119872

(2) If 119891 is bijective and I isin I(1198711) then for each 119910 isin

1198712 [119891minus1(119910)) 119868

119872isin 119879lowastF(I 119879F(1198711)

) is the smallestneighborhood of 119891minus1(119910) where 119868

119872is the greatest

element ofI satisfying 119891minus1(119910) notin 119868119872

Theorem 22 Let 1198711and 119871

2be two implication algebras and

119891 isin hom(F1 F2)

(1) If 119891 is injective andJ isin I(1198712) then

119879lowast(119891minus1

(J) 119879119897(119879F(1198712)

119891)) = 119879119897(119879lowast

F (J 119879F(1198712)) 119891)

(28)

(2) If 119891 is bijective andI isin I(1198711) then

119879lowast(119891 (I) 119879

119903(119879F(1198711)

119891)) = 119879119903(119879lowast

F (I 119879F(1198711)) 119891)

(29)

Proof (1) Let 119888lowast2and 119888119897be the closure operators of the left side

and the right side of the equationWe only need to prove 119888lowast1=

119888119897Let 119860 sube 119871

1and 119909 notin 119888lowast

1(119860) Then 119909 notin 119860 and 119909 notin

119860lowast(119891minus1(J) 119879119897(119879F(1198712)

119891)) By (1) of Lemma 21 119891minus1([119891(119909)))cap119860 isin 119891minus1(J) Thus there exists 119869 isin J such that 119891minus1([119891(119909))) cap119860 = 119891minus1(119869) Since 119909 notin 119860 and 119891 is injective 119891(119909) notin 119869Let 119869119872

isin J be the greatest one satisfying 119891(119909) notin 119869119872 Then

119891minus1([119891(119909))) cap 119860 sube 119891minus1(119869119872) Thus

0 = 119891minus1

([119891 (119909))) cap 119860 cap (1198711 119891minus1

(119869119872))

= 119891minus1

([119891 (119909))) cap 119891minus1

(1198712 119869119872) cap 119860

= 119891minus1

([119891 (119909)) cap (1198712 119869119872)) cap 119860

= 119891minus1

([119891 (119909)) 119869119872) cap 119860

(30)

By Lemma 21 and Proposition 1 119909 notin 119888119897(119860) Therefore 119888

119897(119860) sube

119888lowast1(119860)

The Scientific World Journal 7

Conversely let 119910 notin 119888119897(119860) By Proposition 1

0 = 119891minus1

([119891 (119909)) 119869119872) cap 119860

= 119891minus1

([119891 (119909))) cap 119860 cap (1198711 119891minus1

(119869119872))

(31)

Thus 119909 notin 119860 119891minus1([119891(119909))) cap 119860 sube 119891minus1(119869119872) and [119891(119909)) cap

119891(119860) sube 119869119872 Then 119869

1= [119891(119909)) cap 119891(119860) isin J Since 119891 is

injective 119891minus1([119891(119909))) cap 119860 = 119891minus1(1198691) By (1) of Lemma 20

119909 notin 119860lowast(119891minus1(J) 119879119897(119879F(1198712)

119891)) Therefore 119909 notin 119888lowast1(119860) and

119888lowast1(119860) sube 119888

119897(119860)

(2) Let 119888lowast2and 119888119903be the closure operators of the left side

and the right side of the equationWe only need to prove 119888lowast2=

119888119903Let 119910 notin 119888lowast

2(119860) Then 119910 notin 119860 and 119910 notin 119860lowast((119891(I)

119879119903(119879F(1198711)

119891))) By (2) of Lemma 20119891([119891minus1(119910)))cap119860 isin 119891(I)Thus there exists 119868 isin I such that 119891([119891minus1(119910))) cap 119860 = 119891(119868)Since119891 is bijective [119891minus1(119910))cap119891minus1(119860) = 119868 By119910 notin 119860119891minus1(119910) notin119891minus1(119860) and so 119891minus1(119910) notin 119868 Since 119868

119872isin I is the greatest

element ofI satisfying119891minus1(119910) notin 119868119872 [119891minus1(119910))cap119891minus1(119860) sube 119868

119872

and [119891minus1(119910)) cap (1198711 119868119872) cap 119891minus1(119860) = 0 By 119891 being bijective

again we have

0 = 119891 (0) = 119891 ([119891minus1

(119910))) cap (1198712 119891 (119868119872)) cap 119860

= 119891 ([119891minus1

(119910)) 119868119872) cap 119860

(32)

By

119891minus1

(119910) isin [119891minus1

(119910)) 119868119872

= 119891minus1

(119891 ([119891minus1

(119910)) 119868119872)) isin 119879

lowast

F (I 119879F(1198711))

(33)

119910 notin 119888119903(119860) Hence 119888lowast

2(119860) sube 119888

119903(119860)

Conversely let 119911 notin 119888119903(119860) Since 119868

119872is the greatest element

of I satisfying 119891minus1(119911) notin 119868119872 by (2) of Lemma 21 [119891minus1(119911))

119868119872

isin 119879lowastF(I 119879F(1198711)) is the smallest neighborhood of 119891minus1(119911)

Thus 119891([119891minus1(119911)) 119868119872) cap 119860 = 0 Since 119891 is bijective

0 = 119891minus1

(0) = ([119891minus1

(119911)) 119868119872) cap 119891minus1

(119860)

= [119891minus1

(119911)) cap (1198711 119868119872) cap 119891minus1

(119860)

(34)

This implies [119891minus1(119911))cap119891minus1(119860) sube 119868119872and [119891minus1(119911))cap119891minus1(119860) isin

I Thus 119891([119891minus1(119910))) cap 119860 isin 119891(I) which implies 119910 isin

119860lowast((119891(I) 119879119903(119879F(1198711)

119891))) Since 119911 notin 119888119903(119860) 119911 notin 119860 Therefore

119911 notin 119888lowast2(119860) and so 119888lowast

2(119860) sube 119888

119903(119860)

Generally if 119891 isin hom(1198711 1198712) is surjective but not bijec-

tive then (2) of Theorem 22 fails

Example 23 Let 1198711

= 01 119886 119887 119888 119889 1

1 be the Hasse lattice

implication algebra of Example 5 Let 1198712= 02 119890 ℎ 1

2 and

010158402= 12 1198901015840 = ℎ ℎ1015840 = 119890 and 11015840

2= 02 The Hasse diagram and

Table 2 The implication operator of 1198712= 02 119890 ℎ 1

2

rarr 02

119890 ℎ 12

02

1 12

12

12

119890 ℎ 12

ℎ 12

ℎ 119890 119890 12

12

12

02

119890 ℎ 12

e

h

12

02

Figure 2 Hasse diagram of 1198712= 02 119890 ℎ 1

2

the implication operator of 1198712are shown by Figure 2 and

Table 2 Then it is clear that

119879lowast

F(1198711)= 0 1

1 119886 1

1 119887 119888 1

1 119886 119887 119888 1

1

119887 119888 119889 11 0 119887 119888 119889 1

1 119886 119887 119888 119889 1

1 119871

119879lowast

F(1198712)= 0 119890 1

2 ℎ 1

2 119890 ℎ 1

2 12 1198712

(35)

A mapping 119891 from 1198711to 1198712is defined as

119891 (01) = 02 119891 (1

1) = 12 119891 (119886) = 119891 (119889) = 119890

119891 (119887) = 119891 (119888) = ℎ(36)

It easy to check 119891 isin hom(1198711 1198712) and 119891 is surjective LetI =

0 119888 ThenI isin I(1198711) and 119891(I) = 0 ℎ isin I(119871

2)

Since ℎ isin 119891(I) by Theorem 4

02 119890 12 = 1198712 ℎ isin 119879

lowast(119891 (I) 119879

119903(119879F(1198711)

119891)) (37)

Observe that 119887 119888lowast(I 119879F(1198711)) = 0

1 119887 119888 119889 sube 119887 119888 We

have

0 119886 119889 1 = 1198711 119887 119888 notin 119879

lowast

F (I 119879F(1198711)) (38)

Moreover by 119891minus1(02 119890 12) = 0

2 119886 119889 1

2 02 119890 12 notin

119879119903(119879lowastF(I 119879F(1198711)

) 119891)In fact we have

119879119903(119879lowast

F (I 119879F(1198711)) 119891) = 0 1

2 1198712

119879lowast(119891 (I) 119879

119903(119879F(1198711)

119891)) = 0 12 02 119890 12 1198712

(39)

Therefore 119879119903(119879lowastF(I 119879F(1198711)

) 119891) = 119879lowast(119891(I) 119879119903(119879F(1198711)

119891))

8 The Scientific World Journal

Corollary 24 Let 1198711and 119871

2be two implication algebras and

let 119891 isin hom(F1 f2) be bijective(1) IfJ isin I(119871

2) then

119879119903(119879lowast(119891minus1

(J) 119879119897(119879F(1198712)

119891)) 119891) = 119879lowast(J 119879F(1198712)

)

(40)

(2) IfI isin I(1198711) then

119879119897(119879lowast(119891 (I) 119879

119903(119879F(1198711)

119891)) 119891) = 119879lowast(I 119879F(1198711)

) (41)

Proof The proof follows fromTheorem 22

Corollary 25 Let 1198711and 119871

2be two implication algebras and

119891 isin hom(f1 f2)(1) If 119891 is injective andJ

1J2isin I(119871

2) then

119879119897(119879lowast

F (J1orJ2 119879F(1198712)

) 119891)

= 119879119897(119879lowast

F (J1 119879F(1198712)

) 119891) or 119879119897(119879lowast

F (J2 119879F(1198712)

) 119891)

(42)

(2) If 119891 is bijective andI1I2isin I(119871

1) then

119879119903(119879lowast

F (I1orI2 119879F(1198711)

) 119891)

= 119879119903(119879lowast

F (I1 119879F(1198711)

) 119891) or 119879119903(119879lowast

F (I2 119879F(1198711)

) 119891)

(43)

Proof (1) Clearly 119891minus1(J1orJ2) = 119891minus1(J

1) or119891minus1(J

2) By (3)

of Corollary 15 andTheorem 22

119879119897(119879lowast

F (J1orJ2 119879F(1198712)

) 119891)

= 119879lowast(119891minus1

(J1orJ2) 119879119897(119879F(1198712)

119891))

= 119879lowast(119891minus1

(J1) 119879119897(119879F(1198712)

119891))

or 119879lowast(119891minus1

(J2) 119879119897(119879F(1198712)

119891))

= 119879119897(119879lowast

F (J1 119879F(1198712)

) 119891) or 119879119897(119879lowast

F (J2 119879F(1198712)

) 119891)

(44)

(2) Is similar to (1)

Corollary 26 Let 1198711and 119871

2be two implication algebras and

let 119891 isin hom(F1 f2) be bijective If J isin I(1198712) and I isin

I(1198711) then

119879lowast(119891minus1

(J) 119879119897(119879lowast

F (J 119879F(1198712)) 119891))

= 119879lowast(119891minus1

(J) 119879119897(119879F(1198712)

119891))

119879lowast(119891 (I) 119879

119903(119879lowast

F (I 119879F(1198711)) 119891))

= 119879lowast(119891 (I) 119879

119903(119879F(1198711)

119891))

(45)

Proof The proof follows from (4) of Corollary 15 andTheorem 22

Conflict of Interests

The authors declare that there is no conflict of interests regar-ding the publication of this paper

Acknowledgments

This work is supported by the National Natural ScienceFoundations of China (no 11471202) and the Natural ScienceFoundation of Guangdong Province (no S2012010008833)

References

[1] J Lukasiewicz ldquoOn 3-valued logicrdquo Ruch Filozoficzny vol 5 pp169ndash170 1920

[2] Y Xu ldquoLattice implication algebrardquo Journal of Southwest Jiao-tong University vol 89 pp 20ndash27 1993

[3] Y Xu and K Y Qin ldquoOn filters of lattice implication algebrasrdquoThe Journal of FuzzyMathematics vol 1 no 2 pp 251ndash260 1993

[4] Y Xu D Ruan K Qin and J Liu Lattice-Valued Logic Springer2003

[5] D Jankovic and T R Hamlett ldquoNew topologies from old viaidealsrdquo The American Mathematical Monthly vol 97 no 4 pp295ndash310 1990

Submit your manuscripts athttpwwwhindawicom

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Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

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Complex AnalysisJournal of

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OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Operations ResearchAdvances in

Journal of

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

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Algebra

Discrete Dynamics in Nature and Society

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Decision SciencesAdvances in

Discrete MathematicsJournal of

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

The Scientific World Journal 5

119910 notin 119868 and 119909 notin 119869 (otherwise 119868 and 119869 can be replaced by 119868 119910

and 119869 119909 resp) Then

(119880 119868) cup (119881 119869) = (119880 cap (119871 119868)) cup (119881 cap (119871 119869))

= [(119880 cup 119881) cap ((119871 119868) cup (119871 119869))]

cap [(119880 cup (119871 119869)) cap (119881 cup (119871 119868))]

supe ((119880 cup 119881) (119868 cap 119869)) cap (119871 (119868 cup 119869))

= (119880 cup 119881) (119868 cup 119869)

(18)

Clearly 119910 isin (119880 cup 119881) (119868 cup 119869) isin B119879lowast

F(119909) and

(119880 cup 119881) (119868 cup 119869) sube (119880 119868) cup (119881 119869) sube 119860 (19)

ThereforeB119879lowast

F(119909) is a base ofN

119879lowast

F(119909)

Clearly if 119868119909is the greatest element ofI satisfying 119909 notin 119868

119909

then [119909) 119868119909isin N119879lowast

F(119909) is the smallest 119879lowastF-neighborhoods of

119909

Let (119871 120591) be a topological space andI isin IThe topologythat was generated byB = 119880 119868 119880 isin 120591 119868 isin I is denotedby 119879lowast(I 120591) [5] Clearly 119879lowast(I 119879F) = 119879lowastF(I)

Lemma 12 Let 120595 = 0 119871 be the indiscrete topology on 119871 andI isin I Then 119879lowast(I 120595) = 0 cup 119868119888

Proof By 119879119888 119871 = 0 isin I and Proposition 7 the proof isobvious

Theorem 13 Let (119871 120591) be a topological space and I isin IThen 119879lowast(I 120591) = 120591 or 119879lowast(I 120595) where 120591 or 119879lowast(I 120595) is thetopology generated by the base 119880cap119881 119880 isin 120591 119881 isin 119879lowast(I 120595)

Proof Clearly B119879lowast = 119880 119868 119880 isin 120591 119868 isin I is a base of

119879lowast(I 120591) Since B119879lowast = 119880 cap 119881 119880 isin 120591 119881 isin 119879lowast(I 120595)

B119879lowast is also a base of 120591 or 119879lowast(I 120595) Therefore 119879lowast(I 120591) = 120591 or

119879lowast(I 120595)

Corollary 14 Let (119871 119879F) be the filter topological space andI isin I Then 119879lowastF = 119879F or 119879lowast(I 120595)

Corollary 15 Let (119871 119879F) be the filter topological space andI

J isin I Then

(1) 119879lowast(I orJ 120595) = 119879lowast(I 120595) or 119879lowast(J 120595)(2) 119879lowastF(I orJ) = 119879lowast(I 119879lowastF(J)) = 119879lowast(J 119879lowastF(I))(3) 119879lowastF(I orJ) = 119879lowastF(I) or 119879lowastF(J)(4) 119879lowast(I 119879lowastF(I)) = 119879lowastF(I)

Proof (1) By (2) of Proposition 6 119879lowast(I or J) supe 119879lowast(I) or

119879lowast(J) Conversely let 0 = 119860 isin 119879lowast(I or J) By Theorem 13there exist 119868 isin I and 119869 isin J such that

119860 = 119871 (119868 cup 119868) = (119871 119868) cap (119871 119869) isin 119879lowast

(I) or 119879lowast

(J)

(20)

Thus 119879lowast(I orJ) sube 119879lowast(I) or 119879lowast(J)

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

119879lowast

F (I orJ) = 119879F or 119879lowast(I orJ 120595)

= 119879F or 119879lowast(I 120595) or 119879

lowast(J 120595)

= 119879lowast

F (I) or 119879lowast(J 120595) = 119879

lowast(119879lowast

F (I) J)

(21)

Similarly 119879lowastF(I orJ) = 119879lowast(119879lowastF(J)I)(3) By (1) andTheorem 13

119879lowast

F (I orJ) = 119879F or 119879lowast(I orJ 120595)

= 119879F or 119879lowast(I 120595) or 119879

lowast(J 120595)

= 119879F or 119879lowast(I 120595) or 119879F or 119879

lowast(J 120595)

= 119879lowast

F (I) or 119879lowast

F (J)

(22)

Therefore 119879lowastF(I orJ) = 119879lowastF(I) or 119879lowastF(J)(4) LetI = J Then the proof follows from (2)

Theorem16 Let (119871 119879F) be the filter topological spaceIJ isin

I and 119860 sube 119871 Then

(1) 119860lowast(I capJ 119879F) = 119860lowast(I 119879F) cup 119860lowast(J 119879F)(2) 119860lowast(I orJ 119879F) = 119860lowast(I 119879lowastF(J)) cap 119860lowast(J 119879lowastF(I))

Proof (1) By (3) of Proposition 2 119860lowast(I 119879F) cup 119860lowast(J 119879F) sube

119860lowast(I cap J 119879F) Conversely 119909 notin 119860

lowast(I 119879F) cup 119860

lowast(J 119879F)

Then [119909) cap 119860 isin I and [119909) cap 119860 isin J Let [119909) cap 119860 = 119868 and[119909)cap119860 = 119869 Then119860 sube 119868cup (119871 [119909)) and119860 sube 119869cup (119871 [119909)) Thus

119860 sube (119868 cup (119871 [119909))) cap (119869 cup (119871 [119909))) = (119868 cap 119869) cup (119871 [119909))

(23)

Thus [119909) cap 119860 sube 119868 cap 119869 which implies 119909 notin 119860lowast(I cap J 119879F)

Therefore 119860lowast(I capJ 119879F) sube 119860lowast(I 119879F) cup 119860lowast(J 119879F)(3) Let 119909 notin 119860lowast(I orJ 119879F) Then [119909) cap 119860 isin I orJ Then

there exist 119868 isin I and 119869 isin J such that [119909) cap 119860 = 119868 cup 119869We can assume 119868 cap 119869 = 0 (otherwise 119868 can be replaced by119868 (119868 cap 119869)) Thus 119909 notin 119868 or 119909 notin 119869 (otherwise 119909 isin 119868 cap 119869 whichis a contradiction) Now we take 119909 notin 119868 for example Then

([119909) 119868) cap 119860 = [119909) cap 119860 cap (119871 119868) = 119869 (24)

Since [119909) isin 119879F and 119909 isin [119909) 119868 [119909) 119868 isin B119879lowast

F(I) Thus 119909 notin

119860lowast(J 119879lowastF(I)) and so 119909 notin 119860lowast(J 119879lowastF(I)) cap 119860lowast(I 119879lowastF(J))Hence

119860lowast(J 119879lowast

F (I)) cap 119860lowast(I 119879

lowast

F (J)) sube 119860lowast(I orJ 119879F)

(25)

Conversely let 119909 notin 119860lowast(119879lowastF(I)J) Then there exists 119868 isin

I such that ([119909) 119868) cap 119860 isin J Let ([119909) 119868) cap 119860 = 119869 Then[119909) cap 119860 = 119868 cup 119869 which implies 119909 notin 119860

lowast(I orJ 119879F) Similarlyif 119909 notin 119860lowast(119879lowastF(J)I) then 119909 notin 119860lowast(I or J 119879F) Therefore119860lowast(I orJ 119879F) sube 119860lowast(J 119879lowastF(I)) cap 119860lowast(I 119879lowastF(J))

Corollary 17 Consider 119860lowast(I 119879F) = 119860lowast(I 119879lowastF(I))

Proof LetI = J The proof follows from (2) ofTheorem 16

6 The Scientific World Journal

Corollary 18 Consider 119879lowastF(I capJ) = 119879lowastF(I) cap 119879lowastF(J)

Proof By (2) of Proposition 6119879lowastF(IcapJ) sube 119879lowastF(I)cap119879lowastF(J)Conversely if 119860 notin 119879lowastF(I capJ) then

(119871 119860)lowast(I 119879F) cup (119871 119860)

lowast(J 119879F)

= (119871 119860)lowast(I capJ 119879F) sube (119871 119860)

(26)

Thus (119871 119860)lowast(I 119879F) sube (119871 119860) or (119871 119860)

lowast(J 119879F) sube (119871 119860)

Thus119860 notin 119879lowastF(I) or119860 notin 119879lowastF(J)Therefore119879lowastF(I)cap119879lowastF(J) sube

119879lowastF(I capJ)

3 Power Idealization Filter TopologicalQuotient Spaces

Let (1198711 and or 1015840 rarr

1 01 11) and (119871

2 and or 1015840 rarr

2 02 12) be

two lattice implication algebras A mapping 119891 from 1198711to 1198712

is called lattice implication homomorphism if 119891(119909rarr1119910) =

119891(119909)rarr2119891(119910) for any 119909 119910 isin 119871

1 The set of all lattice

implication homomorphisms from 1198711to 1198712is denoted by

hom(1198711 1198712)

Let 119891 isin hom(1198711 1198712) Then clearly

119879119897(119879F(1198712)

119891) = 119891minus1

(119880) 119880 isin 119879F(1198712)

119879119903(119879F(1198711)

119891) = 119891 (119880) 119880 sube 1198712 119891minus1

(119880) isin 119879F(1198711)

(27)

are topologies [4]

Lemma 19 Let 1198711and 119871

2be two implication algebras and let

119891 isin hom(F1 f2) andI isin 21198711 J isin 21198712 be power ideals

(1) If 119891 is injective then 119891minus1(J) = 119891minus1(119869) 119869 isin J isin

I(1198711)

(2) If 119891 is surjective then 119891(I) = 119891(119868) 119868 isin I isin

I(1198712)

Proof (1) Since 0 isin J then 0 = 119891minus1(0) isin 119891minus1(J)If 1198682isin 119891minus1(J) and 119868

1sube 1198682 then there exist 119869

2isin J such

that 1198682= 119891minus1(119869

2) Thus 119891(119868

1) sube 119891(119868

2) = 1198692and 119891(119868

1) isin J

Since 119891 is injective 1198681= 119891minus1(119891(119868

1)) isin 119891minus1(J)

If 1198681 1198682isin 119891minus1(J) then there exist 119869

1 1198692isin J such that

1198681= 119891minus1(119869

1) and 119868

2= 119891minus1(119869

2) One has 119868

1cup 1198682= 119891minus1(119869

1) cup

119891minus1(1198692) = 119891minus1(119869

1cup 1198692) Since 119869

1cup 1198692isin J 119868

1cup 1198682isin 119891minus1(J)

Therefore 119891minus1(J) is a power ideal(2) By 0 isin I 0 = 119891(0) isin 119891(I)If 1198692isin 119891(J) and 119869

1sube 1198692 then there exists 119868

2isin I such

that 1198692= 119891(119868

2) Let 119868

1= 119891minus1(119869

1) Then 119868

1sube 1198682and 1198681isin I

Since 119891 is surjective 1198691= 119891(1198681) isin 119891(I)

If 1198691 1198692isin 119891(J) then there exist 119868

1 1198682isin I such that 119869

1=

119891(1198681) and 119869

2= 119891(119868

2) Thus 119868

1cup 1198682

isin I Hence 1198691cup 1198692

=

119891(1198681) cup119891(119868

2) = 119891(119868

1cup1198682) isin 119891(I) Therefore 119891(I) is a power

ideal

Lemma 20 Let 1198711and 119871

2be two implication algebras and

119891 isin hom(F1 F2) Then

(1) if 119891 is injective then for each 119909 isin 1198711 119891minus1([119891(119909))) isin

119879119897(119879F(1198712)

119891) is the smallest 119879119897-neighborhood of 119909

(2) if 119891 is bijective then for each 119910 isin 1198712 119891([119891minus1(119910))) isin

119879119903(119879F(1198711)

119891) is the smallest 119879119903-neighborhood of 119910

Proof (1) Clearly [119891(119909)) is the smallest119879F(1198712)-neighborhood

of 119891(119909) Then 119909 isin 119891minus1([119891(119909))) isin 119879119897(119879F(1198712)

119891) Let 119909 isin 119881 isin

119879119897(119879F(1198712)

119891)Then119891(119909) isin 119891(119881) isin 119879F(1198712)and [119891(119909)) sube 119891(119881)

Thus 119891minus1([119891(119909))) sube 119891minus1(119891(119881)) = 119881 Therefore 119891minus1([119891(119909))) isthe smallest one

(2) Clearly [119891minus1(119910)) is the smallest 119879F(1198711)-neighborhood

of 119891minus1(119910) Thus 119910 isin 119891([119891minus1(119910))) isin 119879

119903(119879F(1198711)

119891) Now let119910 isin 119880 isin 119879

119903(119879F(1198711)

119891) Then 119891minus1(119910) isin 119891minus1(119880) isin 119879F(1198711) Thus

[119891minus1(119910)) sube 119891minus1(119880) and so 119891([119891minus1(119910))) sube 119891(119891minus1(119880)) = 119880Therefore (2) holds

Lemma21 Let 1198711and 119871

2be two implication algebras and119891 isin

hom(F1 F2)

(1) If 119891 is injective and J isin I(1198712) then for each 119909 isin

1198711 [119891(119909)) 119869

119872isin 119879lowastF(J 119879F(1198712)

) is the smallestneighborhood of 119891(119909) where 119869

119872is the greatest element

ofJ satisfying 119891(119909) notin 119869119872

(2) If 119891 is bijective and I isin I(1198711) then for each 119910 isin

1198712 [119891minus1(119910)) 119868

119872isin 119879lowastF(I 119879F(1198711)

) is the smallestneighborhood of 119891minus1(119910) where 119868

119872is the greatest

element ofI satisfying 119891minus1(119910) notin 119868119872

Theorem 22 Let 1198711and 119871

2be two implication algebras and

119891 isin hom(F1 F2)

(1) If 119891 is injective andJ isin I(1198712) then

119879lowast(119891minus1

(J) 119879119897(119879F(1198712)

119891)) = 119879119897(119879lowast

F (J 119879F(1198712)) 119891)

(28)

(2) If 119891 is bijective andI isin I(1198711) then

119879lowast(119891 (I) 119879

119903(119879F(1198711)

119891)) = 119879119903(119879lowast

F (I 119879F(1198711)) 119891)

(29)

Proof (1) Let 119888lowast2and 119888119897be the closure operators of the left side

and the right side of the equationWe only need to prove 119888lowast1=

119888119897Let 119860 sube 119871

1and 119909 notin 119888lowast

1(119860) Then 119909 notin 119860 and 119909 notin

119860lowast(119891minus1(J) 119879119897(119879F(1198712)

119891)) By (1) of Lemma 21 119891minus1([119891(119909)))cap119860 isin 119891minus1(J) Thus there exists 119869 isin J such that 119891minus1([119891(119909))) cap119860 = 119891minus1(119869) Since 119909 notin 119860 and 119891 is injective 119891(119909) notin 119869Let 119869119872

isin J be the greatest one satisfying 119891(119909) notin 119869119872 Then

119891minus1([119891(119909))) cap 119860 sube 119891minus1(119869119872) Thus

0 = 119891minus1

([119891 (119909))) cap 119860 cap (1198711 119891minus1

(119869119872))

= 119891minus1

([119891 (119909))) cap 119891minus1

(1198712 119869119872) cap 119860

= 119891minus1

([119891 (119909)) cap (1198712 119869119872)) cap 119860

= 119891minus1

([119891 (119909)) 119869119872) cap 119860

(30)

By Lemma 21 and Proposition 1 119909 notin 119888119897(119860) Therefore 119888

119897(119860) sube

119888lowast1(119860)

The Scientific World Journal 7

Conversely let 119910 notin 119888119897(119860) By Proposition 1

0 = 119891minus1

([119891 (119909)) 119869119872) cap 119860

= 119891minus1

([119891 (119909))) cap 119860 cap (1198711 119891minus1

(119869119872))

(31)

Thus 119909 notin 119860 119891minus1([119891(119909))) cap 119860 sube 119891minus1(119869119872) and [119891(119909)) cap

119891(119860) sube 119869119872 Then 119869

1= [119891(119909)) cap 119891(119860) isin J Since 119891 is

injective 119891minus1([119891(119909))) cap 119860 = 119891minus1(1198691) By (1) of Lemma 20

119909 notin 119860lowast(119891minus1(J) 119879119897(119879F(1198712)

119891)) Therefore 119909 notin 119888lowast1(119860) and

119888lowast1(119860) sube 119888

119897(119860)

(2) Let 119888lowast2and 119888119903be the closure operators of the left side

and the right side of the equationWe only need to prove 119888lowast2=

119888119903Let 119910 notin 119888lowast

2(119860) Then 119910 notin 119860 and 119910 notin 119860lowast((119891(I)

119879119903(119879F(1198711)

119891))) By (2) of Lemma 20119891([119891minus1(119910)))cap119860 isin 119891(I)Thus there exists 119868 isin I such that 119891([119891minus1(119910))) cap 119860 = 119891(119868)Since119891 is bijective [119891minus1(119910))cap119891minus1(119860) = 119868 By119910 notin 119860119891minus1(119910) notin119891minus1(119860) and so 119891minus1(119910) notin 119868 Since 119868

119872isin I is the greatest

element ofI satisfying119891minus1(119910) notin 119868119872 [119891minus1(119910))cap119891minus1(119860) sube 119868

119872

and [119891minus1(119910)) cap (1198711 119868119872) cap 119891minus1(119860) = 0 By 119891 being bijective

again we have

0 = 119891 (0) = 119891 ([119891minus1

(119910))) cap (1198712 119891 (119868119872)) cap 119860

= 119891 ([119891minus1

(119910)) 119868119872) cap 119860

(32)

By

119891minus1

(119910) isin [119891minus1

(119910)) 119868119872

= 119891minus1

(119891 ([119891minus1

(119910)) 119868119872)) isin 119879

lowast

F (I 119879F(1198711))

(33)

119910 notin 119888119903(119860) Hence 119888lowast

2(119860) sube 119888

119903(119860)

Conversely let 119911 notin 119888119903(119860) Since 119868

119872is the greatest element

of I satisfying 119891minus1(119911) notin 119868119872 by (2) of Lemma 21 [119891minus1(119911))

119868119872

isin 119879lowastF(I 119879F(1198711)) is the smallest neighborhood of 119891minus1(119911)

Thus 119891([119891minus1(119911)) 119868119872) cap 119860 = 0 Since 119891 is bijective

0 = 119891minus1

(0) = ([119891minus1

(119911)) 119868119872) cap 119891minus1

(119860)

= [119891minus1

(119911)) cap (1198711 119868119872) cap 119891minus1

(119860)

(34)

This implies [119891minus1(119911))cap119891minus1(119860) sube 119868119872and [119891minus1(119911))cap119891minus1(119860) isin

I Thus 119891([119891minus1(119910))) cap 119860 isin 119891(I) which implies 119910 isin

119860lowast((119891(I) 119879119903(119879F(1198711)

119891))) Since 119911 notin 119888119903(119860) 119911 notin 119860 Therefore

119911 notin 119888lowast2(119860) and so 119888lowast

2(119860) sube 119888

119903(119860)

Generally if 119891 isin hom(1198711 1198712) is surjective but not bijec-

tive then (2) of Theorem 22 fails

Example 23 Let 1198711

= 01 119886 119887 119888 119889 1

1 be the Hasse lattice

implication algebra of Example 5 Let 1198712= 02 119890 ℎ 1

2 and

010158402= 12 1198901015840 = ℎ ℎ1015840 = 119890 and 11015840

2= 02 The Hasse diagram and

Table 2 The implication operator of 1198712= 02 119890 ℎ 1

2

rarr 02

119890 ℎ 12

02

1 12

12

12

119890 ℎ 12

ℎ 12

ℎ 119890 119890 12

12

12

02

119890 ℎ 12

e

h

12

02

Figure 2 Hasse diagram of 1198712= 02 119890 ℎ 1

2

the implication operator of 1198712are shown by Figure 2 and

Table 2 Then it is clear that

119879lowast

F(1198711)= 0 1

1 119886 1

1 119887 119888 1

1 119886 119887 119888 1

1

119887 119888 119889 11 0 119887 119888 119889 1

1 119886 119887 119888 119889 1

1 119871

119879lowast

F(1198712)= 0 119890 1

2 ℎ 1

2 119890 ℎ 1

2 12 1198712

(35)

A mapping 119891 from 1198711to 1198712is defined as

119891 (01) = 02 119891 (1

1) = 12 119891 (119886) = 119891 (119889) = 119890

119891 (119887) = 119891 (119888) = ℎ(36)

It easy to check 119891 isin hom(1198711 1198712) and 119891 is surjective LetI =

0 119888 ThenI isin I(1198711) and 119891(I) = 0 ℎ isin I(119871

2)

Since ℎ isin 119891(I) by Theorem 4

02 119890 12 = 1198712 ℎ isin 119879

lowast(119891 (I) 119879

119903(119879F(1198711)

119891)) (37)

Observe that 119887 119888lowast(I 119879F(1198711)) = 0

1 119887 119888 119889 sube 119887 119888 We

have

0 119886 119889 1 = 1198711 119887 119888 notin 119879

lowast

F (I 119879F(1198711)) (38)

Moreover by 119891minus1(02 119890 12) = 0

2 119886 119889 1

2 02 119890 12 notin

119879119903(119879lowastF(I 119879F(1198711)

) 119891)In fact we have

119879119903(119879lowast

F (I 119879F(1198711)) 119891) = 0 1

2 1198712

119879lowast(119891 (I) 119879

119903(119879F(1198711)

119891)) = 0 12 02 119890 12 1198712

(39)

Therefore 119879119903(119879lowastF(I 119879F(1198711)

) 119891) = 119879lowast(119891(I) 119879119903(119879F(1198711)

119891))

8 The Scientific World Journal

Corollary 24 Let 1198711and 119871

2be two implication algebras and

let 119891 isin hom(F1 f2) be bijective(1) IfJ isin I(119871

2) then

119879119903(119879lowast(119891minus1

(J) 119879119897(119879F(1198712)

119891)) 119891) = 119879lowast(J 119879F(1198712)

)

(40)

(2) IfI isin I(1198711) then

119879119897(119879lowast(119891 (I) 119879

119903(119879F(1198711)

119891)) 119891) = 119879lowast(I 119879F(1198711)

) (41)

Proof The proof follows fromTheorem 22

Corollary 25 Let 1198711and 119871

2be two implication algebras and

119891 isin hom(f1 f2)(1) If 119891 is injective andJ

1J2isin I(119871

2) then

119879119897(119879lowast

F (J1orJ2 119879F(1198712)

) 119891)

= 119879119897(119879lowast

F (J1 119879F(1198712)

) 119891) or 119879119897(119879lowast

F (J2 119879F(1198712)

) 119891)

(42)

(2) If 119891 is bijective andI1I2isin I(119871

1) then

119879119903(119879lowast

F (I1orI2 119879F(1198711)

) 119891)

= 119879119903(119879lowast

F (I1 119879F(1198711)

) 119891) or 119879119903(119879lowast

F (I2 119879F(1198711)

) 119891)

(43)

Proof (1) Clearly 119891minus1(J1orJ2) = 119891minus1(J

1) or119891minus1(J

2) By (3)

of Corollary 15 andTheorem 22

119879119897(119879lowast

F (J1orJ2 119879F(1198712)

) 119891)

= 119879lowast(119891minus1

(J1orJ2) 119879119897(119879F(1198712)

119891))

= 119879lowast(119891minus1

(J1) 119879119897(119879F(1198712)

119891))

or 119879lowast(119891minus1

(J2) 119879119897(119879F(1198712)

119891))

= 119879119897(119879lowast

F (J1 119879F(1198712)

) 119891) or 119879119897(119879lowast

F (J2 119879F(1198712)

) 119891)

(44)

(2) Is similar to (1)

Corollary 26 Let 1198711and 119871

2be two implication algebras and

let 119891 isin hom(F1 f2) be bijective If J isin I(1198712) and I isin

I(1198711) then

119879lowast(119891minus1

(J) 119879119897(119879lowast

F (J 119879F(1198712)) 119891))

= 119879lowast(119891minus1

(J) 119879119897(119879F(1198712)

119891))

119879lowast(119891 (I) 119879

119903(119879lowast

F (I 119879F(1198711)) 119891))

= 119879lowast(119891 (I) 119879

119903(119879F(1198711)

119891))

(45)

Proof The proof follows from (4) of Corollary 15 andTheorem 22

Conflict of Interests

The authors declare that there is no conflict of interests regar-ding the publication of this paper

Acknowledgments

This work is supported by the National Natural ScienceFoundations of China (no 11471202) and the Natural ScienceFoundation of Guangdong Province (no S2012010008833)

References

[1] J Lukasiewicz ldquoOn 3-valued logicrdquo Ruch Filozoficzny vol 5 pp169ndash170 1920

[2] Y Xu ldquoLattice implication algebrardquo Journal of Southwest Jiao-tong University vol 89 pp 20ndash27 1993

[3] Y Xu and K Y Qin ldquoOn filters of lattice implication algebrasrdquoThe Journal of FuzzyMathematics vol 1 no 2 pp 251ndash260 1993

[4] Y Xu D Ruan K Qin and J Liu Lattice-Valued Logic Springer2003

[5] D Jankovic and T R Hamlett ldquoNew topologies from old viaidealsrdquo The American Mathematical Monthly vol 97 no 4 pp295ndash310 1990

Submit your manuscripts athttpwwwhindawicom

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MathematicsJournal of

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Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

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Mathematical PhysicsAdvances in

Complex AnalysisJournal of

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OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

6 The Scientific World Journal

Corollary 18 Consider 119879lowastF(I capJ) = 119879lowastF(I) cap 119879lowastF(J)

Proof By (2) of Proposition 6119879lowastF(IcapJ) sube 119879lowastF(I)cap119879lowastF(J)Conversely if 119860 notin 119879lowastF(I capJ) then

(119871 119860)lowast(I 119879F) cup (119871 119860)

lowast(J 119879F)

= (119871 119860)lowast(I capJ 119879F) sube (119871 119860)

(26)

Thus (119871 119860)lowast(I 119879F) sube (119871 119860) or (119871 119860)

lowast(J 119879F) sube (119871 119860)

Thus119860 notin 119879lowastF(I) or119860 notin 119879lowastF(J)Therefore119879lowastF(I)cap119879lowastF(J) sube

119879lowastF(I capJ)

3 Power Idealization Filter TopologicalQuotient Spaces

Let (1198711 and or 1015840 rarr

1 01 11) and (119871

2 and or 1015840 rarr

2 02 12) be

two lattice implication algebras A mapping 119891 from 1198711to 1198712

is called lattice implication homomorphism if 119891(119909rarr1119910) =

119891(119909)rarr2119891(119910) for any 119909 119910 isin 119871

1 The set of all lattice

implication homomorphisms from 1198711to 1198712is denoted by

hom(1198711 1198712)

Let 119891 isin hom(1198711 1198712) Then clearly

119879119897(119879F(1198712)

119891) = 119891minus1

(119880) 119880 isin 119879F(1198712)

119879119903(119879F(1198711)

119891) = 119891 (119880) 119880 sube 1198712 119891minus1

(119880) isin 119879F(1198711)

(27)

are topologies [4]

Lemma 19 Let 1198711and 119871

2be two implication algebras and let

119891 isin hom(F1 f2) andI isin 21198711 J isin 21198712 be power ideals

(1) If 119891 is injective then 119891minus1(J) = 119891minus1(119869) 119869 isin J isin

I(1198711)

(2) If 119891 is surjective then 119891(I) = 119891(119868) 119868 isin I isin

I(1198712)

Proof (1) Since 0 isin J then 0 = 119891minus1(0) isin 119891minus1(J)If 1198682isin 119891minus1(J) and 119868

1sube 1198682 then there exist 119869

2isin J such

that 1198682= 119891minus1(119869

2) Thus 119891(119868

1) sube 119891(119868

2) = 1198692and 119891(119868

1) isin J

Since 119891 is injective 1198681= 119891minus1(119891(119868

1)) isin 119891minus1(J)

If 1198681 1198682isin 119891minus1(J) then there exist 119869

1 1198692isin J such that

1198681= 119891minus1(119869

1) and 119868

2= 119891minus1(119869

2) One has 119868

1cup 1198682= 119891minus1(119869

1) cup

119891minus1(1198692) = 119891minus1(119869

1cup 1198692) Since 119869

1cup 1198692isin J 119868

1cup 1198682isin 119891minus1(J)

Therefore 119891minus1(J) is a power ideal(2) By 0 isin I 0 = 119891(0) isin 119891(I)If 1198692isin 119891(J) and 119869

1sube 1198692 then there exists 119868

2isin I such

that 1198692= 119891(119868

2) Let 119868

1= 119891minus1(119869

1) Then 119868

1sube 1198682and 1198681isin I

Since 119891 is surjective 1198691= 119891(1198681) isin 119891(I)

If 1198691 1198692isin 119891(J) then there exist 119868

1 1198682isin I such that 119869

1=

119891(1198681) and 119869

2= 119891(119868

2) Thus 119868

1cup 1198682

isin I Hence 1198691cup 1198692

=

119891(1198681) cup119891(119868

2) = 119891(119868

1cup1198682) isin 119891(I) Therefore 119891(I) is a power

ideal

Lemma 20 Let 1198711and 119871

2be two implication algebras and

119891 isin hom(F1 F2) Then

(1) if 119891 is injective then for each 119909 isin 1198711 119891minus1([119891(119909))) isin

119879119897(119879F(1198712)

119891) is the smallest 119879119897-neighborhood of 119909

(2) if 119891 is bijective then for each 119910 isin 1198712 119891([119891minus1(119910))) isin

119879119903(119879F(1198711)

119891) is the smallest 119879119903-neighborhood of 119910

Proof (1) Clearly [119891(119909)) is the smallest119879F(1198712)-neighborhood

of 119891(119909) Then 119909 isin 119891minus1([119891(119909))) isin 119879119897(119879F(1198712)

119891) Let 119909 isin 119881 isin

119879119897(119879F(1198712)

119891)Then119891(119909) isin 119891(119881) isin 119879F(1198712)and [119891(119909)) sube 119891(119881)

Thus 119891minus1([119891(119909))) sube 119891minus1(119891(119881)) = 119881 Therefore 119891minus1([119891(119909))) isthe smallest one

(2) Clearly [119891minus1(119910)) is the smallest 119879F(1198711)-neighborhood

of 119891minus1(119910) Thus 119910 isin 119891([119891minus1(119910))) isin 119879

119903(119879F(1198711)

119891) Now let119910 isin 119880 isin 119879

119903(119879F(1198711)

119891) Then 119891minus1(119910) isin 119891minus1(119880) isin 119879F(1198711) Thus

[119891minus1(119910)) sube 119891minus1(119880) and so 119891([119891minus1(119910))) sube 119891(119891minus1(119880)) = 119880Therefore (2) holds

Lemma21 Let 1198711and 119871

2be two implication algebras and119891 isin

hom(F1 F2)

(1) If 119891 is injective and J isin I(1198712) then for each 119909 isin

1198711 [119891(119909)) 119869

119872isin 119879lowastF(J 119879F(1198712)

) is the smallestneighborhood of 119891(119909) where 119869

119872is the greatest element

ofJ satisfying 119891(119909) notin 119869119872

(2) If 119891 is bijective and I isin I(1198711) then for each 119910 isin

1198712 [119891minus1(119910)) 119868

119872isin 119879lowastF(I 119879F(1198711)

) is the smallestneighborhood of 119891minus1(119910) where 119868

119872is the greatest

element ofI satisfying 119891minus1(119910) notin 119868119872

Theorem 22 Let 1198711and 119871

2be two implication algebras and

119891 isin hom(F1 F2)

(1) If 119891 is injective andJ isin I(1198712) then

119879lowast(119891minus1

(J) 119879119897(119879F(1198712)

119891)) = 119879119897(119879lowast

F (J 119879F(1198712)) 119891)

(28)

(2) If 119891 is bijective andI isin I(1198711) then

119879lowast(119891 (I) 119879

119903(119879F(1198711)

119891)) = 119879119903(119879lowast

F (I 119879F(1198711)) 119891)

(29)

Proof (1) Let 119888lowast2and 119888119897be the closure operators of the left side

and the right side of the equationWe only need to prove 119888lowast1=

119888119897Let 119860 sube 119871

1and 119909 notin 119888lowast

1(119860) Then 119909 notin 119860 and 119909 notin

119860lowast(119891minus1(J) 119879119897(119879F(1198712)

119891)) By (1) of Lemma 21 119891minus1([119891(119909)))cap119860 isin 119891minus1(J) Thus there exists 119869 isin J such that 119891minus1([119891(119909))) cap119860 = 119891minus1(119869) Since 119909 notin 119860 and 119891 is injective 119891(119909) notin 119869Let 119869119872

isin J be the greatest one satisfying 119891(119909) notin 119869119872 Then

119891minus1([119891(119909))) cap 119860 sube 119891minus1(119869119872) Thus

0 = 119891minus1

([119891 (119909))) cap 119860 cap (1198711 119891minus1

(119869119872))

= 119891minus1

([119891 (119909))) cap 119891minus1

(1198712 119869119872) cap 119860

= 119891minus1

([119891 (119909)) cap (1198712 119869119872)) cap 119860

= 119891minus1

([119891 (119909)) 119869119872) cap 119860

(30)

By Lemma 21 and Proposition 1 119909 notin 119888119897(119860) Therefore 119888

119897(119860) sube

119888lowast1(119860)

The Scientific World Journal 7

Conversely let 119910 notin 119888119897(119860) By Proposition 1

0 = 119891minus1

([119891 (119909)) 119869119872) cap 119860

= 119891minus1

([119891 (119909))) cap 119860 cap (1198711 119891minus1

(119869119872))

(31)

Thus 119909 notin 119860 119891minus1([119891(119909))) cap 119860 sube 119891minus1(119869119872) and [119891(119909)) cap

119891(119860) sube 119869119872 Then 119869

1= [119891(119909)) cap 119891(119860) isin J Since 119891 is

injective 119891minus1([119891(119909))) cap 119860 = 119891minus1(1198691) By (1) of Lemma 20

119909 notin 119860lowast(119891minus1(J) 119879119897(119879F(1198712)

119891)) Therefore 119909 notin 119888lowast1(119860) and

119888lowast1(119860) sube 119888

119897(119860)

(2) Let 119888lowast2and 119888119903be the closure operators of the left side

and the right side of the equationWe only need to prove 119888lowast2=

119888119903Let 119910 notin 119888lowast

2(119860) Then 119910 notin 119860 and 119910 notin 119860lowast((119891(I)

119879119903(119879F(1198711)

119891))) By (2) of Lemma 20119891([119891minus1(119910)))cap119860 isin 119891(I)Thus there exists 119868 isin I such that 119891([119891minus1(119910))) cap 119860 = 119891(119868)Since119891 is bijective [119891minus1(119910))cap119891minus1(119860) = 119868 By119910 notin 119860119891minus1(119910) notin119891minus1(119860) and so 119891minus1(119910) notin 119868 Since 119868

119872isin I is the greatest

element ofI satisfying119891minus1(119910) notin 119868119872 [119891minus1(119910))cap119891minus1(119860) sube 119868

119872

and [119891minus1(119910)) cap (1198711 119868119872) cap 119891minus1(119860) = 0 By 119891 being bijective

again we have

0 = 119891 (0) = 119891 ([119891minus1

(119910))) cap (1198712 119891 (119868119872)) cap 119860

= 119891 ([119891minus1

(119910)) 119868119872) cap 119860

(32)

By

119891minus1

(119910) isin [119891minus1

(119910)) 119868119872

= 119891minus1

(119891 ([119891minus1

(119910)) 119868119872)) isin 119879

lowast

F (I 119879F(1198711))

(33)

119910 notin 119888119903(119860) Hence 119888lowast

2(119860) sube 119888

119903(119860)

Conversely let 119911 notin 119888119903(119860) Since 119868

119872is the greatest element

of I satisfying 119891minus1(119911) notin 119868119872 by (2) of Lemma 21 [119891minus1(119911))

119868119872

isin 119879lowastF(I 119879F(1198711)) is the smallest neighborhood of 119891minus1(119911)

Thus 119891([119891minus1(119911)) 119868119872) cap 119860 = 0 Since 119891 is bijective

0 = 119891minus1

(0) = ([119891minus1

(119911)) 119868119872) cap 119891minus1

(119860)

= [119891minus1

(119911)) cap (1198711 119868119872) cap 119891minus1

(119860)

(34)

This implies [119891minus1(119911))cap119891minus1(119860) sube 119868119872and [119891minus1(119911))cap119891minus1(119860) isin

I Thus 119891([119891minus1(119910))) cap 119860 isin 119891(I) which implies 119910 isin

119860lowast((119891(I) 119879119903(119879F(1198711)

119891))) Since 119911 notin 119888119903(119860) 119911 notin 119860 Therefore

119911 notin 119888lowast2(119860) and so 119888lowast

2(119860) sube 119888

119903(119860)

Generally if 119891 isin hom(1198711 1198712) is surjective but not bijec-

tive then (2) of Theorem 22 fails

Example 23 Let 1198711

= 01 119886 119887 119888 119889 1

1 be the Hasse lattice

implication algebra of Example 5 Let 1198712= 02 119890 ℎ 1

2 and

010158402= 12 1198901015840 = ℎ ℎ1015840 = 119890 and 11015840

2= 02 The Hasse diagram and

Table 2 The implication operator of 1198712= 02 119890 ℎ 1

2

rarr 02

119890 ℎ 12

02

1 12

12

12

119890 ℎ 12

ℎ 12

ℎ 119890 119890 12

12

12

02

119890 ℎ 12

e

h

12

02

Figure 2 Hasse diagram of 1198712= 02 119890 ℎ 1

2

the implication operator of 1198712are shown by Figure 2 and

Table 2 Then it is clear that

119879lowast

F(1198711)= 0 1

1 119886 1

1 119887 119888 1

1 119886 119887 119888 1

1

119887 119888 119889 11 0 119887 119888 119889 1

1 119886 119887 119888 119889 1

1 119871

119879lowast

F(1198712)= 0 119890 1

2 ℎ 1

2 119890 ℎ 1

2 12 1198712

(35)

A mapping 119891 from 1198711to 1198712is defined as

119891 (01) = 02 119891 (1

1) = 12 119891 (119886) = 119891 (119889) = 119890

119891 (119887) = 119891 (119888) = ℎ(36)

It easy to check 119891 isin hom(1198711 1198712) and 119891 is surjective LetI =

0 119888 ThenI isin I(1198711) and 119891(I) = 0 ℎ isin I(119871

2)

Since ℎ isin 119891(I) by Theorem 4

02 119890 12 = 1198712 ℎ isin 119879

lowast(119891 (I) 119879

119903(119879F(1198711)

119891)) (37)

Observe that 119887 119888lowast(I 119879F(1198711)) = 0

1 119887 119888 119889 sube 119887 119888 We

have

0 119886 119889 1 = 1198711 119887 119888 notin 119879

lowast

F (I 119879F(1198711)) (38)

Moreover by 119891minus1(02 119890 12) = 0

2 119886 119889 1

2 02 119890 12 notin

119879119903(119879lowastF(I 119879F(1198711)

) 119891)In fact we have

119879119903(119879lowast

F (I 119879F(1198711)) 119891) = 0 1

2 1198712

119879lowast(119891 (I) 119879

119903(119879F(1198711)

119891)) = 0 12 02 119890 12 1198712

(39)

Therefore 119879119903(119879lowastF(I 119879F(1198711)

) 119891) = 119879lowast(119891(I) 119879119903(119879F(1198711)

119891))

8 The Scientific World Journal

Corollary 24 Let 1198711and 119871

2be two implication algebras and

let 119891 isin hom(F1 f2) be bijective(1) IfJ isin I(119871

2) then

119879119903(119879lowast(119891minus1

(J) 119879119897(119879F(1198712)

119891)) 119891) = 119879lowast(J 119879F(1198712)

)

(40)

(2) IfI isin I(1198711) then

119879119897(119879lowast(119891 (I) 119879

119903(119879F(1198711)

119891)) 119891) = 119879lowast(I 119879F(1198711)

) (41)

Proof The proof follows fromTheorem 22

Corollary 25 Let 1198711and 119871

2be two implication algebras and

119891 isin hom(f1 f2)(1) If 119891 is injective andJ

1J2isin I(119871

2) then

119879119897(119879lowast

F (J1orJ2 119879F(1198712)

) 119891)

= 119879119897(119879lowast

F (J1 119879F(1198712)

) 119891) or 119879119897(119879lowast

F (J2 119879F(1198712)

) 119891)

(42)

(2) If 119891 is bijective andI1I2isin I(119871

1) then

119879119903(119879lowast

F (I1orI2 119879F(1198711)

) 119891)

= 119879119903(119879lowast

F (I1 119879F(1198711)

) 119891) or 119879119903(119879lowast

F (I2 119879F(1198711)

) 119891)

(43)

Proof (1) Clearly 119891minus1(J1orJ2) = 119891minus1(J

1) or119891minus1(J

2) By (3)

of Corollary 15 andTheorem 22

119879119897(119879lowast

F (J1orJ2 119879F(1198712)

) 119891)

= 119879lowast(119891minus1

(J1orJ2) 119879119897(119879F(1198712)

119891))

= 119879lowast(119891minus1

(J1) 119879119897(119879F(1198712)

119891))

or 119879lowast(119891minus1

(J2) 119879119897(119879F(1198712)

119891))

= 119879119897(119879lowast

F (J1 119879F(1198712)

) 119891) or 119879119897(119879lowast

F (J2 119879F(1198712)

) 119891)

(44)

(2) Is similar to (1)

Corollary 26 Let 1198711and 119871

2be two implication algebras and

let 119891 isin hom(F1 f2) be bijective If J isin I(1198712) and I isin

I(1198711) then

119879lowast(119891minus1

(J) 119879119897(119879lowast

F (J 119879F(1198712)) 119891))

= 119879lowast(119891minus1

(J) 119879119897(119879F(1198712)

119891))

119879lowast(119891 (I) 119879

119903(119879lowast

F (I 119879F(1198711)) 119891))

= 119879lowast(119891 (I) 119879

119903(119879F(1198711)

119891))

(45)

Proof The proof follows from (4) of Corollary 15 andTheorem 22

Conflict of Interests

The authors declare that there is no conflict of interests regar-ding the publication of this paper

Acknowledgments

This work is supported by the National Natural ScienceFoundations of China (no 11471202) and the Natural ScienceFoundation of Guangdong Province (no S2012010008833)

References

[1] J Lukasiewicz ldquoOn 3-valued logicrdquo Ruch Filozoficzny vol 5 pp169ndash170 1920

[2] Y Xu ldquoLattice implication algebrardquo Journal of Southwest Jiao-tong University vol 89 pp 20ndash27 1993

[3] Y Xu and K Y Qin ldquoOn filters of lattice implication algebrasrdquoThe Journal of FuzzyMathematics vol 1 no 2 pp 251ndash260 1993

[4] Y Xu D Ruan K Qin and J Liu Lattice-Valued Logic Springer2003

[5] D Jankovic and T R Hamlett ldquoNew topologies from old viaidealsrdquo The American Mathematical Monthly vol 97 no 4 pp295ndash310 1990

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

The Scientific World Journal 7

Conversely let 119910 notin 119888119897(119860) By Proposition 1

0 = 119891minus1

([119891 (119909)) 119869119872) cap 119860

= 119891minus1

([119891 (119909))) cap 119860 cap (1198711 119891minus1

(119869119872))

(31)

Thus 119909 notin 119860 119891minus1([119891(119909))) cap 119860 sube 119891minus1(119869119872) and [119891(119909)) cap

119891(119860) sube 119869119872 Then 119869

1= [119891(119909)) cap 119891(119860) isin J Since 119891 is

injective 119891minus1([119891(119909))) cap 119860 = 119891minus1(1198691) By (1) of Lemma 20

119909 notin 119860lowast(119891minus1(J) 119879119897(119879F(1198712)

119891)) Therefore 119909 notin 119888lowast1(119860) and

119888lowast1(119860) sube 119888

119897(119860)

(2) Let 119888lowast2and 119888119903be the closure operators of the left side

and the right side of the equationWe only need to prove 119888lowast2=

119888119903Let 119910 notin 119888lowast

2(119860) Then 119910 notin 119860 and 119910 notin 119860lowast((119891(I)

119879119903(119879F(1198711)

119891))) By (2) of Lemma 20119891([119891minus1(119910)))cap119860 isin 119891(I)Thus there exists 119868 isin I such that 119891([119891minus1(119910))) cap 119860 = 119891(119868)Since119891 is bijective [119891minus1(119910))cap119891minus1(119860) = 119868 By119910 notin 119860119891minus1(119910) notin119891minus1(119860) and so 119891minus1(119910) notin 119868 Since 119868

119872isin I is the greatest

element ofI satisfying119891minus1(119910) notin 119868119872 [119891minus1(119910))cap119891minus1(119860) sube 119868

119872

and [119891minus1(119910)) cap (1198711 119868119872) cap 119891minus1(119860) = 0 By 119891 being bijective

again we have

0 = 119891 (0) = 119891 ([119891minus1

(119910))) cap (1198712 119891 (119868119872)) cap 119860

= 119891 ([119891minus1

(119910)) 119868119872) cap 119860

(32)

By

119891minus1

(119910) isin [119891minus1

(119910)) 119868119872

= 119891minus1

(119891 ([119891minus1

(119910)) 119868119872)) isin 119879

lowast

F (I 119879F(1198711))

(33)

119910 notin 119888119903(119860) Hence 119888lowast

2(119860) sube 119888

119903(119860)

Conversely let 119911 notin 119888119903(119860) Since 119868

119872is the greatest element

of I satisfying 119891minus1(119911) notin 119868119872 by (2) of Lemma 21 [119891minus1(119911))

119868119872

isin 119879lowastF(I 119879F(1198711)) is the smallest neighborhood of 119891minus1(119911)

Thus 119891([119891minus1(119911)) 119868119872) cap 119860 = 0 Since 119891 is bijective

0 = 119891minus1

(0) = ([119891minus1

(119911)) 119868119872) cap 119891minus1

(119860)

= [119891minus1

(119911)) cap (1198711 119868119872) cap 119891minus1

(119860)

(34)

This implies [119891minus1(119911))cap119891minus1(119860) sube 119868119872and [119891minus1(119911))cap119891minus1(119860) isin

I Thus 119891([119891minus1(119910))) cap 119860 isin 119891(I) which implies 119910 isin

119860lowast((119891(I) 119879119903(119879F(1198711)

119891))) Since 119911 notin 119888119903(119860) 119911 notin 119860 Therefore

119911 notin 119888lowast2(119860) and so 119888lowast

2(119860) sube 119888

119903(119860)

Generally if 119891 isin hom(1198711 1198712) is surjective but not bijec-

tive then (2) of Theorem 22 fails

Example 23 Let 1198711

= 01 119886 119887 119888 119889 1

1 be the Hasse lattice

implication algebra of Example 5 Let 1198712= 02 119890 ℎ 1

2 and

010158402= 12 1198901015840 = ℎ ℎ1015840 = 119890 and 11015840

2= 02 The Hasse diagram and

Table 2 The implication operator of 1198712= 02 119890 ℎ 1

2

rarr 02

119890 ℎ 12

02

1 12

12

12

119890 ℎ 12

ℎ 12

ℎ 119890 119890 12

12

12

02

119890 ℎ 12

e

h

12

02

Figure 2 Hasse diagram of 1198712= 02 119890 ℎ 1

2

the implication operator of 1198712are shown by Figure 2 and

Table 2 Then it is clear that

119879lowast

F(1198711)= 0 1

1 119886 1

1 119887 119888 1

1 119886 119887 119888 1

1

119887 119888 119889 11 0 119887 119888 119889 1

1 119886 119887 119888 119889 1

1 119871

119879lowast

F(1198712)= 0 119890 1

2 ℎ 1

2 119890 ℎ 1

2 12 1198712

(35)

A mapping 119891 from 1198711to 1198712is defined as

119891 (01) = 02 119891 (1

1) = 12 119891 (119886) = 119891 (119889) = 119890

119891 (119887) = 119891 (119888) = ℎ(36)

It easy to check 119891 isin hom(1198711 1198712) and 119891 is surjective LetI =

0 119888 ThenI isin I(1198711) and 119891(I) = 0 ℎ isin I(119871

2)

Since ℎ isin 119891(I) by Theorem 4

02 119890 12 = 1198712 ℎ isin 119879

lowast(119891 (I) 119879

119903(119879F(1198711)

119891)) (37)

Observe that 119887 119888lowast(I 119879F(1198711)) = 0

1 119887 119888 119889 sube 119887 119888 We

have

0 119886 119889 1 = 1198711 119887 119888 notin 119879

lowast

F (I 119879F(1198711)) (38)

Moreover by 119891minus1(02 119890 12) = 0

2 119886 119889 1

2 02 119890 12 notin

119879119903(119879lowastF(I 119879F(1198711)

) 119891)In fact we have

119879119903(119879lowast

F (I 119879F(1198711)) 119891) = 0 1

2 1198712

119879lowast(119891 (I) 119879

119903(119879F(1198711)

119891)) = 0 12 02 119890 12 1198712

(39)

Therefore 119879119903(119879lowastF(I 119879F(1198711)

) 119891) = 119879lowast(119891(I) 119879119903(119879F(1198711)

119891))

8 The Scientific World Journal

Corollary 24 Let 1198711and 119871

2be two implication algebras and

let 119891 isin hom(F1 f2) be bijective(1) IfJ isin I(119871

2) then

119879119903(119879lowast(119891minus1

(J) 119879119897(119879F(1198712)

119891)) 119891) = 119879lowast(J 119879F(1198712)

)

(40)

(2) IfI isin I(1198711) then

119879119897(119879lowast(119891 (I) 119879

119903(119879F(1198711)

119891)) 119891) = 119879lowast(I 119879F(1198711)

) (41)

Proof The proof follows fromTheorem 22

Corollary 25 Let 1198711and 119871

2be two implication algebras and

119891 isin hom(f1 f2)(1) If 119891 is injective andJ

1J2isin I(119871

2) then

119879119897(119879lowast

F (J1orJ2 119879F(1198712)

) 119891)

= 119879119897(119879lowast

F (J1 119879F(1198712)

) 119891) or 119879119897(119879lowast

F (J2 119879F(1198712)

) 119891)

(42)

(2) If 119891 is bijective andI1I2isin I(119871

1) then

119879119903(119879lowast

F (I1orI2 119879F(1198711)

) 119891)

= 119879119903(119879lowast

F (I1 119879F(1198711)

) 119891) or 119879119903(119879lowast

F (I2 119879F(1198711)

) 119891)

(43)

Proof (1) Clearly 119891minus1(J1orJ2) = 119891minus1(J

1) or119891minus1(J

2) By (3)

of Corollary 15 andTheorem 22

119879119897(119879lowast

F (J1orJ2 119879F(1198712)

) 119891)

= 119879lowast(119891minus1

(J1orJ2) 119879119897(119879F(1198712)

119891))

= 119879lowast(119891minus1

(J1) 119879119897(119879F(1198712)

119891))

or 119879lowast(119891minus1

(J2) 119879119897(119879F(1198712)

119891))

= 119879119897(119879lowast

F (J1 119879F(1198712)

) 119891) or 119879119897(119879lowast

F (J2 119879F(1198712)

) 119891)

(44)

(2) Is similar to (1)

Corollary 26 Let 1198711and 119871

2be two implication algebras and

let 119891 isin hom(F1 f2) be bijective If J isin I(1198712) and I isin

I(1198711) then

119879lowast(119891minus1

(J) 119879119897(119879lowast

F (J 119879F(1198712)) 119891))

= 119879lowast(119891minus1

(J) 119879119897(119879F(1198712)

119891))

119879lowast(119891 (I) 119879

119903(119879lowast

F (I 119879F(1198711)) 119891))

= 119879lowast(119891 (I) 119879

119903(119879F(1198711)

119891))

(45)

Proof The proof follows from (4) of Corollary 15 andTheorem 22

Conflict of Interests

The authors declare that there is no conflict of interests regar-ding the publication of this paper

Acknowledgments

This work is supported by the National Natural ScienceFoundations of China (no 11471202) and the Natural ScienceFoundation of Guangdong Province (no S2012010008833)

References

[1] J Lukasiewicz ldquoOn 3-valued logicrdquo Ruch Filozoficzny vol 5 pp169ndash170 1920

[2] Y Xu ldquoLattice implication algebrardquo Journal of Southwest Jiao-tong University vol 89 pp 20ndash27 1993

[3] Y Xu and K Y Qin ldquoOn filters of lattice implication algebrasrdquoThe Journal of FuzzyMathematics vol 1 no 2 pp 251ndash260 1993

[4] Y Xu D Ruan K Qin and J Liu Lattice-Valued Logic Springer2003

[5] D Jankovic and T R Hamlett ldquoNew topologies from old viaidealsrdquo The American Mathematical Monthly vol 97 no 4 pp295ndash310 1990

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

8 The Scientific World Journal

Corollary 24 Let 1198711and 119871

2be two implication algebras and

let 119891 isin hom(F1 f2) be bijective(1) IfJ isin I(119871

2) then

119879119903(119879lowast(119891minus1

(J) 119879119897(119879F(1198712)

119891)) 119891) = 119879lowast(J 119879F(1198712)

)

(40)

(2) IfI isin I(1198711) then

119879119897(119879lowast(119891 (I) 119879

119903(119879F(1198711)

119891)) 119891) = 119879lowast(I 119879F(1198711)

) (41)

Proof The proof follows fromTheorem 22

Corollary 25 Let 1198711and 119871

2be two implication algebras and

119891 isin hom(f1 f2)(1) If 119891 is injective andJ

1J2isin I(119871

2) then

119879119897(119879lowast

F (J1orJ2 119879F(1198712)

) 119891)

= 119879119897(119879lowast

F (J1 119879F(1198712)

) 119891) or 119879119897(119879lowast

F (J2 119879F(1198712)

) 119891)

(42)

(2) If 119891 is bijective andI1I2isin I(119871

1) then

119879119903(119879lowast

F (I1orI2 119879F(1198711)

) 119891)

= 119879119903(119879lowast

F (I1 119879F(1198711)

) 119891) or 119879119903(119879lowast

F (I2 119879F(1198711)

) 119891)

(43)

Proof (1) Clearly 119891minus1(J1orJ2) = 119891minus1(J

1) or119891minus1(J

2) By (3)

of Corollary 15 andTheorem 22

119879119897(119879lowast

F (J1orJ2 119879F(1198712)

) 119891)

= 119879lowast(119891minus1

(J1orJ2) 119879119897(119879F(1198712)

119891))

= 119879lowast(119891minus1

(J1) 119879119897(119879F(1198712)

119891))

or 119879lowast(119891minus1

(J2) 119879119897(119879F(1198712)

119891))

= 119879119897(119879lowast

F (J1 119879F(1198712)

) 119891) or 119879119897(119879lowast

F (J2 119879F(1198712)

) 119891)

(44)

(2) Is similar to (1)

Corollary 26 Let 1198711and 119871

2be two implication algebras and

let 119891 isin hom(F1 f2) be bijective If J isin I(1198712) and I isin

I(1198711) then

119879lowast(119891minus1

(J) 119879119897(119879lowast

F (J 119879F(1198712)) 119891))

= 119879lowast(119891minus1

(J) 119879119897(119879F(1198712)

119891))

119879lowast(119891 (I) 119879

119903(119879lowast

F (I 119879F(1198711)) 119891))

= 119879lowast(119891 (I) 119879

119903(119879F(1198711)

119891))

(45)

Proof The proof follows from (4) of Corollary 15 andTheorem 22

Conflict of Interests

The authors declare that there is no conflict of interests regar-ding the publication of this paper

Acknowledgments

This work is supported by the National Natural ScienceFoundations of China (no 11471202) and the Natural ScienceFoundation of Guangdong Province (no S2012010008833)

References

[1] J Lukasiewicz ldquoOn 3-valued logicrdquo Ruch Filozoficzny vol 5 pp169ndash170 1920

[2] Y Xu ldquoLattice implication algebrardquo Journal of Southwest Jiao-tong University vol 89 pp 20ndash27 1993

[3] Y Xu and K Y Qin ldquoOn filters of lattice implication algebrasrdquoThe Journal of FuzzyMathematics vol 1 no 2 pp 251ndash260 1993

[4] Y Xu D Ruan K Qin and J Liu Lattice-Valued Logic Springer2003

[5] D Jankovic and T R Hamlett ldquoNew topologies from old viaidealsrdquo The American Mathematical Monthly vol 97 no 4 pp295ndash310 1990

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of