Available online at http://ijim.srbiau.ac.ir/

Int. J. Industrial Mathematics (ISSN 2008-5621)

Vol. 9, No. 3, 2017 Article ID IJIM-00976, 12 pages

Research Article

Topological Residuated Lattices

N. Kouhestani ∗, R. A. Borzooei †‡

Received Date: 2016-04-08 Revised Date: 2016-10-24 Accepted Date: 2017-03-04

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Abstract

In this paper, we study the separtion axioms T0, T1, T2 and T5/2 on topological and semitopologicalresiduated lattices and we show that they are equivalent on topological residuated lattices. Thenwe prove that for every infinite cardinal number α, there exists at least one nontrivial Hausdorfftopological residuated lattice of cardinality α. In the follows, we obtain some conditions on (semi)topological residuated lattices under which this spaces will convert into regular and normal spaces.Finally by using of regularity and normality, we convert (semi)topological residuated lattices intometrizable topological residuated lattices.

Keywords : Topological residuated lattice; Filter; Regular space; Normal space; (locally) Compactspace.

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1 Introduction

Residuated lattices have been introduced byM.Ward and R.P. Dilworth [9] as generaliza-

tion of ideal lattices of rings with identity. Theyare a common structure among algebras associ-ated with logical systems. The main examplesof residuated lattices related to logic are MV-algebras and BL-algebras. In recent years somemathematicians have endowed algebraic struc-tures associated with logical systems with a topol-ogy and have found some their propertises. Forexample, Borzooei et.al. in [3] introduced (semi)topological BL-algebras and in [4] and [5] studiedmetrizability and separation axioms on them. In[11] Kouhestani and Borzooei defined the notionof (semi) topological residuated lattices and stud-

∗Fuzzy Systems Research Center, University of Sistanand Baluchestan, Zahedan, Iran.

†Corresponding author. borzooei@sbu.ac.ir, Tel:+989124903982

‡Department of Mathematics, Shahid Beheshti Univer-sity, Tehran, Iran.

ied separation axioms T0, T1 and T2 on them. Fora topological space, metrizability is a highly de-sirable property, for the existence of a such a dis-tance function gives one a valuable tool for prov-ing theorems about the space. In the section 3of this paper, we deal with relations between Ti

spaces, for i = 0, 1, 2, 5/2 and (semi)topologicalresiduated lattices. We will prove that they areequivalent and will show that there are nontrivialHausdorff topological residuated lattices of infi-nite cardinality. Then in section 4 we will findsome conditions under which a (semi)topologicalresiduated lattice convert into a regular or nomalspace. Finally, metrizable topological residuatedlattices will be obtained by using of regularity andnormality.

2 Preliminaries

Recall that a set A with a family U of its subsetsis called a topological space, denoted by (A,U),if A, ∅ ∈ U , the intersection of any finite num-bers of members of U is in U and the arbitrary

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union of members of U is in U . The members ofU are called open sets of A and the complementof U ∈ U , that is A \U , is said to be a closed set.If B is a subset of A, the smallest closed set con-taining B is called the closure of B and denotedby B (or cluB). A subfamily Uα : α ∈ I of Uis said to be a base of U if for each x ∈ U ∈ U ,there exists an α ∈ I such that x ∈ Uα ⊆ U , orequivalently, each U in U is the union of membersof Uα. A subset P of A is said to be a neigh-borhood of x ∈ A, if there exists an open set Usuch that x ∈ U ⊆ P . Let Ux denote the totalityof all neighborhoods of x in A. Then a subfamilyVx of Ux is said to form a fundamental systemof neighborhoods of x, if for each Ux in Ux, thereexists a Vx in Vx such that Vx ⊆ Ux. A directedset I is a partially ordered set such that, for any iand j of I, there is a k ∈ I with k ≥ i and k ≥ j.If I is a directed set, then the subset xi : i ∈ Iof A is called a net. A net xi; i ∈ I convergesto x ∈ A if for each neighborhood U of x, thereexists a j ∈ I such that for all i ≥ j, xi ∈ U. IfB ⊆ A and x ∈ B, then there is a net in B thatis converges to x.Topological space (A,U) is said to be a:(i) T0-space if for each x = y ∈ A, there is atleast one in an open neighborhood excluding theother,(ii) T1-space if for each x = y ∈ A, each has anopen neighborhood not containing the other,(iii) T2-space if for each x = y ∈ A, there twodisjoint open neighborhoods U, V of x and y, re-spectively.A T2-space is also known as a Hausdorff space.[see, [7]]

Definition 2.1 [9] A residuated lattice is analgebra L = (L,∧,∨,⊙,→, 0, 1) of type(2, 2, 2, 2, 0, 0) such that (L,∧,∨, 0, 1) is abounded lattice, (L,⊙, 1) is a commutativemonoid and for any a, b, c ∈ L,

c ≤ a → b ⇔ a⊙ c ≤ b.

A residuated lattice L is divisible if for each a, b ∈L, a⊙ (a → b) = a ∧ b.

Let L be a residuated lattice. We set a′ = a → 0and denote (a′)′ by a′′. We call the map p : L → Lby p(a) = a′, for any a ∈ L, the negation map.Also, for each a ∈ L, we define a0 = 1 and an =an−1 ⊙ a, for each natural numbers n.

Example 2.1 [9](i) Let ⊙ and → on the realunit interval I = [0, 1] be defined as follows:

x⊙ y = minx, y & x → y =

1, x ≤ y,y, x > y.

Then I = (I,min,max,⊙,→, 0, 1) is a residuatedlattice,(ii) Let ⊙ be the usual multiplication of real num-bers on the unit interval I = [0, 1] and x →y = 1 iff, x ≤ y and y/x otherwise. ThenI = (I,min,max,⊙,→, 0, 1) is a residuated lat-tice.(iii) Let L = 0, a, b, c, 1. Define ⊙ and → asfollows :

⊙ 0 a b c 1

0 0 0 0 0 0a 0 a 0 a ab 0 0 b b bc 0 a b c c1 0 a b c 1

→ 0 a b c 1

0 1 1 1 1 1a b 1 b 1 1b a a 1 1 1c 0 a b 1 11 0 a b c 1

Easily we can check that (L,⊙,→, 0, 1) is aresiduated lattice, whose lattice (L,∧,∨, 0, 1) isgiven by the partial order 0 < a < c < 1, 0 < b <c < 1 and x∧ y = minx, y, x∨ y = maxx, y,a ∧ b = 0 and a ∨ b = 1.

Proposition 2.1 [9] Let (L,∧,∨,⊙,→, 0, 1) bea residuated lattice. The following propertieshold.(R1) x → (y → z) = (x⊙ y) → z,(R2) x ≤ y iff, x → y = 1,(R3) 1 ∗ x = x, where ∗ ∈ ∧,⊙,→,(R4) x⊙ 0 = 0, 1′ = 0, 0′ = 1,(R5) x⊙ y ≤ x ∧ y ≤ x, y, and y ≤ (x → y),(R6) (x → y)⊙ x ≤ y,(R7) x ≤ y → (x⊙ y),(R8) x ≤ y implies x ∗ z ≤ y ∗ z, where∗ ∈ ∧,∨,⊙,(R9) x ≤ y implies z → x ≤ z → y andx → z ≥ y → z,(R10) x → y = x → (x ∧ y),(R11) x ≤ y implies x ≤ z → y,(R12) z ⊙ (x ∧ y) ≤ (z ⊙ x) ∧ (z ⊙ y),

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(R13) x → y ≤ (x⊙ z) → (y ⊙ z),(R14) (x → y)⊙ (y → z) ≤ x → z,(R15) x⊙ x′ = 0,(R16) x → y′ = (x⊙ y)′,(R17) x⊙ (y ∨ z) = (x⊙ y) ∨ (x⊙ z),(R18) (x ∨ y) → z = (x → z) ∧ (y → z),(R19) x → (y ∧ z) = (x → y) ∨ (x → z),(R20) (x ∨ y)′ = x′ ∧ y′, and (x ∧ y)′ ≥ x′ ∨ y′,(R21) if x ∨ y = 1, then x → y = y andx⊙ y = x ∨ y,(R22) x

′′′ = x′,(R23) (a → x)⊙(b → y) ≤ (a∗b) → (x∗y), where∗ ∈ ∧,∨.

Let (L,∧,∨,⊙,→, 0, 1) be a residuated lattice.Then a filter of L is a nonempty subset F ⊆ Lwhich satisfies the following conditions:(a) x, y ∈ F implies x⊙ y ∈ F ,(b) if x ∈ F and x ≤ y, then y ∈ F.Let F be a filter of L. Then the relation x ≡F yiff, x → y, y → x ∈ F is a congruence relationon L. Moreovere, if for each x ∈ L, F (x) = y ∈A : y ≡F x, then for each ∗ ∈ ∧,∨,⊙,→,F (x) ∗ F (y) = F (x ∗ y). Thus, the set L/F =F (x) : x ∈ L is a residuated lattice which calledquotient residuated lattice.[See, [9]]

Notation. From now on, in this paper we letL be a residuated lattice and U be a topology onL, unless otherwise state.

3 Hausdorff topological residu-ated lattice

Definition 3.1 [11] Let U be a topology on resid-uated lattice L and ∗ ∈ ∧,∨,⊙,→. Then:(i) the operation ∗ is continuous in the first (sec-ond) variable if for each a ∈ L, the mappingx → x ∗ a(x → a ∗ x) from L into L is con-tinuous.(ii) (L, ∗,U) is semitopological residuated latticeif ∗ is continuous in the first and second variable,(iii) (L, ∗,U) is topological residuated lattice if ∗is continuous,(iv) (L,U) is (semi) topological residuated latticeif for any ∗ ∈ ∧,∨,⊙,→, (L, ∗,U) is (semi)topological residuated lattice.

Proposition 3.1 Let (L, ,U) be a semitopologi-cal residuated lattice.Then (L,U) is T0 iff, it isT1.

Proof. Let (L,U) be T0 and x = y. Then xy = 1or yx = 1. W.L.O.G let xy = 1. Then there existsU ∈ U such that xy ∈ U and 1 ∈ U or xy ∈ U and1 ∈ U. First suppose xy ∈ U and 1 ∈ U. Sinceis continuous in each variable separately, thereare open neighborhoods V and W of x and y,respectively, such that V y ⊆ U and xW ⊆ U. Butx ∈ W because if x ∈ W, then 1 = xx ∈ xW ⊆ Uwhich is a contradiction. Similarly, y ∈ V. Nowlet 1 ∈ U and xy ∈ U. Since xx = yy = 1 ∈ U,there are V,W ∈ U such that x ∈ V, y ∈ W,xV ⊆ U and Wy ⊆ U. If x ∈ W, then xy ∈ Wy ⊆U which is a contradiction. So x ∈ W. Similaly,y ∈ V. Hence (L,U) is a T1 space. Convesely isclear.

Proposition 3.2 Let (L, ,U) be a topologicalresiduated lattice.Then (L,U) is T1 iff, it is Haus-dorff.

Proof. Let (L,U) be T1 and x = y. Then xy = 1or yx = 1. W.L.O.G let xy = 1. Then there existsU, V ∈ U such that xy ∈ U, 1 ∈ V, xy ∈ Vand 1 ∈ U. Since is continuous, there are openneighborhoods V0 and V1 of x and y, respectively,such that V0V1 ⊆ U. We prove that V0 ∩ V1 isempty. Suppose z ∈ V0 ∩ V1, then 1 = zz ∈V0V1 ⊆ U which is a contradiction. So (L,U) isHausdorff. Conversely is clear.

Corollary 3.1 Let (L, ,U) be topological residu-ated lattice. Then (L,U) is T0 iff, it is T1 iff, itis Hausdorff.

Proof. By Propositions 3.1, 3.2, the proof isclear.

Theorem 3.1 Let (L,⊙, ,U) be topologicalresiduated lattice. Then (L,U) is Hausdorff iff,1 or 0 is closed.

Proof. Let 1 be closed. We show that for eacha ∈ L, the set a is closed. Suppose a ∈ L. Since⊙ is continuous, ⊙−1(1) = (1, 1) is closed in L×L. On the other hand, as is continuous, the maph : L → L × L by h(x) = (ax, xa) is continuous.Hence h−1(1, 1) = x : xa = ax = 1 = ais closed in L. This implies that (L,U) is T1. ByProposition 3.2, it is Hausdorff. If 0 is closed,since the negation map p : L → L by p(x) = x′ iscontinuous and 1 = p−1(0), the set 1 is closedand so (L,U) is Hausdorff. Conversely is clear.

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Definition 3.2 Let L be a residuated lattice.Then for each a ∈ L and any V ⊆ L, we define

V (a) = x ∈ L : xa ∈ V, ax ∈ V ,

andV [a] = x ∈ L : xa ∈ V .

Proof. Let (L, ,U) be a semitopological residu-ated lattice. Then for each a ∈ L and V ∈ U thesets V (a) and V [a], both, are open.Proof. Let a ∈ L and V ∈ U . Suppose x ∈ V (a),then xa ∈ V and ax ∈ V. Since is continuous ineach variable, there is W ∈ U such that x is inW and Wa and aW are the subsets of V. Clearly,x ∈ W ⊆ V (a). Similarly, V [a] ∈ U .

Proposition 3.3 Let (L,U) be a topologicalresiduated lattice. Then it is Hausdorff iff, foreach 0 = x ∈ L there is V ∈ U such that 1 ∈ Vand x′ ∈ V.

Proof. First let (L,U) be Hausdorff and x = 0.Then x′ = 1 and so there is an open set V suchthat 1 ∈ V and x′ ∈ V. Conversely, let x ∈ 0.If x = 0, then there is an open set V such that1 ∈ V and x′ ∈ V. By Proposition 3, V (x) is anopen neighborhood of x, so 0 ∈ V (x). This im-plies that x′ ∈ V, a contradiction. Hence 0 isclosed. By Theorem 3.1, (L,U) is Hausdorff. Re-call that a topological space (X,U) is an Uryshonspace if for each x = y ∈ L, there exist two openneighborhoods U and V of x and y, respectively,such that U ∩ V = ϕ. An Uryshon space is alsoknown as a T5/2 space.[See, [7]]

Proposition 3.4 Let (L,→,U) be a topologicalresiduated lattice. Then L is an Uryshon spaceiff, for each x = 1, there exist two open neighbor-hoods U and V of x and 1, respectively, such thatU ∩ V = ϕ.

Proof. If (L,U) is an Uryshon space, the proofis clear. Conversely, let for each x = 1, there ex-ist two open neighborhoods U and V of x and 1,respectively, such that U ∩ V = ϕ. Let x, y ∈ Land x = y. Then x → y = 1 or y → x = 1.W.O.L.G, let x → y = 1 and U and V be twoopen neighborhoods of x → y and 1 such thatU ∩V = ϕ. Since (L,→,U) is a topological resid-uated lattice, there are two open neighborhoodsW1 and W2 of x and y, respectively, such thatW1 → W2 ⊆ U. We prove that W1 and W2 aredisjoint. Let z ∈ W1 ∩W2. Then there exist two

nets xi and yi in W1 and W2, respectively,such that both converge to z. Since the operation→ is continuous, the net xi → yi converges toz → z = 1. Hence 1 ∈ W1 → W2 ⊆ U, which is acontradiction.

Theorem 3.2 Let (L,→,U) be a topologicalresiduated lattice. Then the following statementsare equivalent:(i) (L,U) is a T0 space,(ii) (L,U) is a T1 space,(iii) (L,U) is a Hausdorff space,(iv) (L,U) is an Uryson space,(v)

∩U∈N U = 1, where N is a fundamental sys-

tem of neighborhoods of 1.

Proof. By Corollary 3.1, the statements (i), (ii)and (iiii) are equivalent.(iii) =⇒ (iv) Let (L,U) be a Hausdorff space and1 = x. Then there are two disjoint open sets Uand V such that x ∈ U and 1 ∈ V. Since 1x =x ∈ U, there are two open neighborhoods W andW1 of x and 1, respectively, such that W1W ⊆ U.We prove that W ∩ W1 = ϕ. Let z ∈ W ∩ W1.Then there are nets xi : i ∈ I ⊆ W1 and yi :i ∈ I ⊆ W such that both converges to z. Hencexiyi : i ∈ I is a net in U which converges tozz = 1. This implies that 1 ∈ U. Since 1 ∈ V ∈U , V ∩ U = ϕ, a conteradiction. Therefore, byProposition 3.4, (L,U) is an Uryson space.(iv) =⇒ (v) Let (L,U) be an Uryson space andx = 1. By Proposition 3.4, there are open sets Uand V such that 1 ∈ U, x ∈ V and U ∩ V = ϕ.Hence x ∈ U and so x ∈

∩U∈N U.

(v) =⇒ (ii) Let∩

U∈N U = 1, and 1 = x. Thenthere is a V ∈ N such that x ∈ V. Hence 1 ∈V (x). By [[11], Proposition 4.6], (L,U) is a T1

space.

Proposition 3.5 Let L be a noncountable setand 0, 1 ∈ L. Then there are operations ∧,∨,⊙and on L such that (L,∧,∨,⊙, , 0, 1) is a resid-uated lattice.

Proof. Let L0 = x0 = 0, x1, x2, ... ∪ 1 ⊆ L.Define ,⊕,⊼ and ⊻ on L0 as follows:

xy :=

[0, x], if x ≤ y,(0, y], if x > y = 0,x, if y = 0

1 xi = xi, xi 1 = 1,

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xi xj =

1, if i ≤ j,xj , if otherwise

x⊕ y = minx, y = x ⊼ y, x ⊻ y = maxx, y,

where x0 ≤ x1 ≤ x2 ≤ ... ≤ 1. It is easy to seethat (L0,⊼,⊻,⊕,, 0, 1) is a residuated lattice.Given a ∈ L \ L0 and put L1 = L0 ∪ a. Let⋏,⋎,⊗ and → on the L1 be defined as follows:

x⋏ y =

x ⊼ y, if x, y ∈ L0,minx, y, if otherwise

x⋎ y =

x ⊻ y, if x, y ∈ L0,maxx, y, if otherwise,

x⊗ y =

x⊕ y, if x, y ∈ L0,minx, y, if otherwise,

x → y =

x y, if x, y ∈ L0,1, v x ∈ L1 \ 1, y = a,a, if x = 1, y = a,y, if x = a, y ∈ L0,

where x0 ≤ x1 ≤ x2 ≤ ... ≤ a ≤ 1.Then(L1,⋏,⋎,⊕, →, 0, 1) is a residuated lattice. Nowconsider the operations ∧,∨,⊙ and on L as fol-lows:

x ∧ y = x⊙ y =

minx, y, if x or y ∈ L1,a, if otherwise,

x ∨ y =

maxx, y, if x or y ∈ L1,1, if otherwise,

xy =

x → y, if x, y ∈ L1,1, if x ∈ L1 \ 1, y ∈ L1,y, if x = 1, y ∈ L1,y, if x ∈ L1, y ∈ L1,1, if x = y ∈ L1,y, if x = y ∈ L1,

where x ≤ y if and only if xy = 1. Then it is easyto claim that (L,∧,∨,⊙, , 0, 1) is a residuated lat-tice.

Theorem 3.3 Let L0 and L1 be residuated lat-tices in Proposition 3.5. Then there are twotopologies V and W on L0 and L1, respectively,such that (L0,V) and (L1,W) are Hausdorff topo-logical residuated lattices.

Proof. Let Fk = xk, xk+1, ... ∪ 1, for eachk ∈ 0, 1, 2, 3, .... Then F = Fk : k = 0, 1, 2, ...is a family of filters in L0 which is closed underfinite intersections. Let V = V ⊆ L0 : ∀x ∈V ∃Fk s.t Fk(x) ⊆ V . It is easy to prove that

V is closed under finite intersections and arbi-trary unions. Also, for each x ∈ L and eachFk ∈ F , Fk(x) belongs to V. Hence V is a non-trivial topology on L0. We prove that (L0,V)is a topological resiuated lattice. For this, let∗ ∈ ⊼,⊻,⊕,, x, y ∈ L0 and Fk ∈ F . Ifz1 ∈ Fk(x) and z2 ∈ Fk(y), since ≡F is a con-gruence relation, then z1 ∗ z2 ≡Fk x ∗ y. HenceFk(x)∗Fk(y) ⊆ Fk(x∗y). This proves that the op-eration ∗ is continuous and so (L0,V) is a topolog-ical residuated lattice. Now suppose x ∈ L0 andx = 1. Then there is a Fk ∈ F such that x ∈ Fk.This implies that x ∈ Fk(x) and 1 ∈ Fk(x). By[[11], Proposition 4.4] (L0,V) is a T1 space. ByTheorem 3.2, (L0,V) is a Hausdorff space.In continue we will find a topology W on L1

such that (L1,W) is a Hausdorff topological resid-uated lattice. Let Jk = Fk ∪ a. We showthat Jk is a filter in L1. Let x, y ∈ Jk. Thenx⊗y ∈ x, y, a ⊆ Jk. Suppose x ∈ Jk and x ≤ y.If y = a, then y ∈ Jk. If y = a, then x ∈ Fk,since Fk is a filter, y ∈ Fk ⊆ Jk. Hence Jk isa filter for each k and J = Jk : k = 0, 1, , ...is a family of filters in L1 which is closed un-der finite intersections. In a similar way with theabove paragraph we can prove that W = W ⊆L1 : ∀x ∈ W ∃Jk s.t Jk(x) ⊆ W is a nontrivialtopology on L1 such that (L1,W) is a topologicalresiduated lattice. Since L1 \ 0 = J1 ∈ W, theset 0 is closed in L1. By Theorem 3.1, (L1,W)is Hausdorff.

Theorem 3.4 Let α be an infinite cardinalnumber. Then there is a Hausdorff topologicalresiduated lattice of cardinality α.

Proof. Let L be a set of cardinality α and0, 1 ∈ L. If L is a countable set, then take L = L0,where L0 is the residuated lattice in Proposition3.5. Then by Theorem 3.3, there is a nontrivialtopology V on L such that (L,V) is a Hausdorfftopological residuated lattice. Let L be a non-countable set and L0 = x0 = 0, x1, ....∪1 ⊆ Land L1 = L0∪a, where a ∈ L\L0. As the proofof Proposition 3.5, there are operations ⊼,⊻,⊕on L0 and operations ⋏,⋎,⊗, → on L1 and op-erations ∧,∨,⊙, on L such that (L0,⊼,⊻,⊕ )and (L1,⋏,⋎,⊗, →) and (L,∧,∨,⊙, ) are resid-uated lattices. Suppose for each k ∈ 0, 1, 2, ...,Fk = xk, xk+1, ..., Jk = Fk ∪ a and Hk =Jk ∪ (L \L1), then Fk is a filter in L0 and Jk is afilter in L1. We prove that Hk is a filter in L. Letx, y ∈ Hk, then x⊙ y ∈ a, x, y ⊆ Hk. Let x ≤ y

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and x ∈ Hk. If y ∈ L \ L1, then clearly y ∈ Hk.If y ∈ L1, then x ∈ L1. Since Jk is a filter in L1,y ∈ Jk ⊆ Hk. Since H = Hk : k = 0, 1, 2, ... isclosed under finite intersections, as the proof ofTheorem 3.3, there is a topology U on L such that(L,U) is a topological residuated lattice. SinceL \ 0 = H1 ∈ U , the set 0 is closed in (L,U).By Theorem 3.1, (L,U) is Hausdorff.

4 Regular and normal(semi)topological residuatedlattices

In this section we study regularity and normalityon (semi)topological residuated lattices. First weshow that for each residuated lattice L there ex-ists a topology U such that (L,U) is a regular andnormal topological residuated lattice. In Exam-ple 4.1, we show that there exist residuated lat-tices that are not regular or normal. Then we findsome conditions under which a (semi)topologicalresiduated lattice becomes regular or normal.Recall that a topological space (X,U) is regularif for each x ∈ U ∈ U , there exists an open setH such that x ∈ H ⊆ H ⊆ U. Also, (X,U) isnormal if for each closed set S and each open setU contains S, there is an open set H such thatS ⊆ H ⊆ H ⊆ U. [See, [7]]

Theorem 4.1 Let F be a familly of filters in Lwhich is closed under intersections. Then there isa nontrivial topology U on L such that (L,U) is aregular and normal topological residuated lattice.

Proof. Let U = U ⊆ L : ∀x ∈ U ∃F ∈F s.t F (x) ⊆ U. It is easy to prove that U isclosed under arbitrary intersections and unions.Also, for each x ∈ L and each F ∈ F , F (x) be-longs to U . Hence U is a nontrivial topology. Weprove that (L,U) is a topological resiuated lat-tice. For this, let ∗ ∈ ∧,∨,⊙,→, x, y ∈ L andF ∈ F . If z1 ∈ F (x) and z2 ∈ F (y), since ≡F

is a congruence relation, then z1 ∗ z2 ≡F x ∗ y.Hence F (x) ∗ F (y) ⊆ F (x ∗ y). This proves thatthe operation ∗ is continuous and so (L,U) is atopological residuated lattice. Now we will provethat (L,U) is a regular and normal space. Firstwe show that for each x ∈ L and F ∈ F , theset F (x) is closed. Let x ∈ L, F ∈ F andy ∈ F (x). Since F (y) is an open neighborhoodof y, there is a z ∈ L such that y ≡F z ≡F x.

Hence y ∈ F (x) and so F (x) = F (x). To proveregularity, let x ∈ U ∈ U . Then there is a F ∈ Fsuch that F (x) ⊆ U. Since F (x) is clsoed, weget that x ∈ F (x) ⊆ F (x) ⊆ U, which impliesthat (L,U) is a regular space. To complete theproof, we have to show that (L,U) is a normalspace. Let S be a closed set, U be an open setand S ⊆ U. For each x ∈ S there exits a F ∈ Fsuch that F (x) ⊆ U. Let H =

∪x∈S,F∈F F (x).

Then H is a closed and open set because for eachx ∈ S and F ∈ F , the set F (x) is a closed andopen set. Thus S ⊆ H ⊆ H ⊆ U, which impliesthat (L,U) is a normal space.

Example 4.1 (i) Let L be a nontrivialresiduated lattice and for each a ∈ L,La = x ∈ L : a ≤ x. Then the setB = La : a ∈ L is a base for a topologyU on L. We show that (L,U) is not a reg-ular space. If (L,U) is a regular space and1 ∈ L1 = 1 ∈ U , then there is an open set Hsuch that 1 ∈ H ⊆ H ⊆ L1. Hence 1 = H = H.On the other hand, 1 = L, so L = H = 1,which is a contradiction.(ii) Let L be the residuated lat-tice in Example 2.1(iii). Then U =0, 1, a, 0, 1, 0, 1, b, c, L, ϕ is a topol-ogy on L. Clearly, S = b, c is a closed setand U = 0, 1, b, c is an open set containsb, c which is not a closed set because a ∈ U.Obviously, there is not any open set H such thatS ⊆ H ⊆ H ⊆ U.

Theorem 4.2 For each n ≥ 3 there exist a nor-mal and regular topological residuated lattice oforder n.

Proof. Let L3 = 0, a, 1. Define the operations,⊙,∧ and ∨ on L3 as follows:

⊙ 0 a 1

0 0 0 0a 0 a a1 0 a a

0 a 1

0 1 1 1a 0 1 11 0 a 1

x ∧ y = minx, y x ∨ y = maxx, y

It is easy to prove that (L3,∧,∨,⊙, , 0, 1) is aresiduated lattice and F = a, 1 is a filter inL3. By Theorem 4.1, U = ϕ, 0, 1, a, L3 isa topology on L3 such that (L3,U) is a normaland regular topological residuated lattice. Now

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suppose L4 = L3 ∪ b, where b ∈ L3. Considerthe operations ∧,∨,⊙ and → on L as follows:

x ∧ y = x⊙ y =

minx, y, if x or y ∈ L1,a, if otherwise,

x ∨ y =

maxx, y, if x or y ∈ L1,1, if otherwise,

x → y =

xy, if x, y ∈ L3,1, if x ∈ L4 \ 1, y = b,b, if x = 1, y = b,y, if x = b, y ∈ L3

where x ≤ y if and only if xy = 1. Then it iseasy to claim that (L,∧,∨,⊙, , 0, 1) is a residu-ated lattice. If F = 1, a, b, then F is a filterin L4. By Theorem 4.1, U = ϕ, 0, 1, a, b, L4is a topology on L4 such that (L4,U) is a nor-mal and regular topological residuated lattice oforder 4. In a similar way we can make a normaland regular topological residuated lattice of ordern, for each n ≥ 4.

Theorem 4.3 For every infinte cardinal num-ber α there is a normal, regular and Hausdorfftopological residuated lattice of order α.

Proof. If α is countable cardinal number, thenby Theorem 4.1, Hausdorff topological residuatedlattice (L0,V) in Theorem 3.3, is a normal andregular space. If α is noncuontable cardinal num-ber, then by Theorem 4.1, Hausdorff topologicalresiduated lattice (L,U) in Theorem 3.4, is a nor-mal and regular space.

Theorem 4.4 Let F be a nontrivial filter in L.Then:(i) there exists a nontrivial topology U on L suchthat (L,⊙,U) is a T0 topological residuated latticeand F is a closed and open set,(ii) (L,U) is a regular space iff, for each x ∈ U ∈U , the set F ⊙ x is a closed set,(iii) (L,U) is a normal space iff, for each closedset S, the set

∪x∈S F ⊙ x is a closed set.

Proof. (i) By [[11], Proposition 4.1(ii)] the setU = U ⊆ L : ∀x ∈ U F ⊙ x ⊆ U is a nontrivialtopology on L such that (L,U) is a T0 topologicalresiduated lattice. We prove that F is closed andopen. Since for each x ∈ F, F ⊙ x ⊆ F, we getthat F ∈ U . If x ∈ F, then F ⊙ x∩F = ϕ. Hencethere is a f ∈ F such that f ⊙ x ∈ F. Since F isa filter by (R5), we conclude that x ∈ F. So F isa closed set.

(ii) Let (L,U) be regular and x ∈ U ∈ U . SinceF ⊙ x is an open neighborhood of x, there is anopen set H such that x ∈ H ⊆ H ⊆ F ⊙ x. Sincex ∈ H, the set F ⊙ x is contained in H. HenceF ⊙ x = H, which implies that F ⊙ x is closed.Conversely, let x ∈ U ∈ U . Since F ⊙ x is closed,hence x ∈ F ⊙x = F ⊙ x ⊆ U, which implies that(L,U) is a regular space.(iii) Let (L,U) be normal and S be a closed setin L. Since

∪x∈S F ⊙x is an open set contains S,

there is an open set H such that S ⊆ H ⊆ H ⊆∪x∈S F ⊙ x. Since S ⊆ H, the set

∪x∈S F ⊙ x is

contained inH. Hence∪

x∈S F⊙x = H is a closedset. Conversely, let S be a closed set in L and Ube an open set contains S. Now

∪x∈S F ⊙ x is a

closed and open set such that S ⊆∪

x∈S F ⊙ x =∪x∈S F ⊙ x ⊆ U. Therefore, (L,U) is a normal

space.

Definition 4.1 Let L be a residuated lattice. Anonempty subset V of L is a prefilter if for eachx, y ∈ L, x ≤ y and x ∈ V imply y ∈ V.

Example 4.2 (i) Let I be the residuated latticein Example 2.1(ii). Then for each a ∈ I the set[a, 1] is a prefilter. It is clear that it is not filter.(ii) Let L be residuated lattice in Example2.1(iii). Then V = a, b, c, 1 is a prefilter. Itis not filter because a⊙ b = 0 ∈ V.

Proposition 4.1 Let L be a residuated latticeand V be a prefilter in L. Then:(i) FV = x ∈ V : ∀y ∈ V x⊙ y ∈ V is a filter,(ii) for each x ∈ V, FV (x) ⊆ V (x),(ii) for each x, y ∈ L, FV (x) ∗ FV (y) ⊆ V (x ∗ y),when ∗ ∈ ⊙, ,∧,∨.

Proof. (i) Clearly, 1 ∈ FV . Let x, y ∈ FV andz ∈ V. Then y⊙z ∈ V and so x⊙(y⊙z) ∈ V. Thisimplies that (x ⊙ y) ⊙ z ∈ V. Hence x ⊙ y ∈ FV .Now suppose x ≤ y and x ∈ FV . If z ∈ V, thenby (R8), x⊙z ≤ y⊙z. Since x⊙z ∈ V, y⊙z ∈ V.Hence y ∈ FV .(ii) The proof is easy.(iii) First we prove that FV (x)⊙FV (y) ⊆ V (x⊙y). For this, let a ∈ FV (x) and b ∈ FV (y). Wehave

(xa)((x⊙ y)(a⊙ b))

= (xa)(x(y(a⊙ b)))

= (x⊙ (xa))(y(a⊙ b))

≥ a(y(a⊙ b))

= y(aa⊙ b) ≥ yb

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Since yb and xa, both, belong to FV , by (i), (x⊙y)(a⊙b) ∈ FV ⊆ V. In a similar way, we can showthat (a⊙ b)(x⊙ y) ∈ V. Hence a⊙ b ∈ V (x⊙ y).

This implies that FV (x) ⊙ FV (y) ⊆ V (x ⊙ y).Now we show that FV (x)FV (y) ⊆ V (xy). Let a ∈FV (x) and b ∈ FV (y). We have

(yb)((xy)(ab))

= [(xy)⊙ (yb)](ab)

≥ (xb)(ab)

= a((xb)b) ≥ ax

Since yb and ax, both, belong to FV , (xy)(ab) ∈FV ⊆ V. In a similar way, we can showthat (ab)(xy) ∈ V. Hence ab ∈ V (xy). ThusFV (x)FV (y) ⊆ V (xy). In continue we will showthat FV (x) ∧ FV (y) ⊆ V (x ∧ y). Let a ∈ FV (x)and b ∈ FV (y). By (R5) and (R9), (x ∧ y)a ≥ xaand (x ∧ y)b ≥ yb. Hence (x ∧ y)a and (x ∧ y)b,both, are in FV and so [(x∧y)a]∨ [(x∧y)b] ∈ FV .By (R19),

(x ∧ y)(a ∧ b) = [(x ∧ y)a] ∨ [(x ∧ y)b].

Hence(x ∧ y)(a ∧ b) ∈ FV ⊆ V.

By the similar way, we get that (a∧b)(x∧y) ∈ V.Therefore, a∧b ∈ V (x∧y). To complete the proofwe have to show that FV (x) ∨ FV (y) ⊆ V (x ∨ y).Let a ∈ FV (x) and b ∈ FV (y). By (R9), ax ≤a(x ∨ y) and by ≤ b(x ∨ y). Hence (ax) ∧ (by) isless than

(a(x ∨ y)) ∧ (b(x ∨ y)).

Since ax and by are in filter FV , (a(x∨y))∧(b(x∨y)) ∈ FV . By (R18), (a ∨ b)(x ∨ y) ∈ FV . In asimilar way, (x ∨ y)(a ∨ b) ∈ FV . So a ∨ b ∈ FV .

Theorem 4.5 Suppose Ω is a family of pre-filters in a residuated lattice L such that Ω isclosed under intersections, and for each V ∈ Ω,FV ∈ Ω.

Then there is a topology U on L such that(L,U) is a regular topological residuated lattice.Moreover, Ω is a fundamental system of neigh-borhoods 1.

Proof. Let U = U ⊆ L : ∀x ∈ U ∃V ∈Ω s.t V (x) ⊆ U. Then it is easy to prove thatU is a topology on L. We show that for any x ∈ Land V ∈ Ω, the set V (x) is open and closed. Let

y ∈ V (x). By (ii) of Proposition 4.1, FV (xy) ⊆ Vand FV (yx) ⊆ V. Let z ∈ FV (y). By (R6) and(R1),

yz ≤ (zx)(yx)

andzy ≤ (yx)(zx).

Hence zx ∈ FV (yx) ⊆ V. By the similar way, wecan show that xz ∈ V. Hence z ∈ V (x). This im-plies that y ∈ FV (y) ⊆ V (x), and so V (x) ∈ U .Now let y ∈ V (x), where V ∈ Ω. Then there isa z ∈ FV (y) ∩ V (x). Since zy ∈ FV and xz ∈ V,by Proposition 4.1(i), (xz)⊙ (zy) ∈ V. By (R14),xy ∈ V. Similarly, we can prove that yx ∈ V.Hence y ∈ V (x). Therefore V (x) is closed. Sincefor each x ∈ L and V ∈ Ω, the set V (x) is open,by (iii) of Proposition 4.1, the operations ∧,∨,⊙and are continuous. Hence (L,U) is a topologi-cal residuated lattice. Finally, (L,U) is a regularspace because for any x ∈ U ∈ U , there is a V ∈ Ωsuch that x ∈ V (x) = V (x) ⊆ U. Since for eachV ∈ Ω, V (1) = V, it is easy to see that Ω is afundamental system of neighborhoods of 1.

Definition 4.2 Let L be a residuated lattice.Then we call filter F in L satisfies maximum con-dition if for each x ∈ L, the set F (x) has a max-imum.

If F is a filter in L which satisfies maximum condi-tion, then we also call F is a filter with maximumcondition.

Example 4.3 Let I be the residuated lattices inExample 2.1 (i), and F be a filter in it. Then foreach x ∈ I, F (x) = x or F (x) = F, and so Fsatisfies the maximum condition.

Lemma 4.1 Let F be a filter with maximumcondition in a divisible residuated lattice L. Thenfor each x ∈ L, there exists a x∗ ∈ L such thatF (x) = F ⊙ x∗.

Proof. Let x ∈ L. Since F is a filter with max-imum condition, there exists x∗ ∈ L such thatx∗ = maxF (x). For each f ∈ F, by (R5), sincex∗⊙f ≤ f and f ≤ x∗ → (x∗⊙f), it follows that(x∗ ⊙ f) → f = 1 ∈ F and x∗ → (x∗ ⊙ f) ∈ F.Hence x∗⊙f ≡F x∗. Since x∗ ∈ F (x), we concludethat x∗ ⊙ f ∈ F (x). Therefore, F ⊙ x∗ ⊆ F (x).If z ∈ F (x), since x∗ ∈ F (x), we have z ∈ F (x∗).Since z = x∗∧z = x∗⊙(x∗ → z) and x∗ → z ∈ F,we conclude that z ∈ x∗⊙F. Thus, F (x) ⊆ F⊙x∗.

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Theorem 4.6 Let F be a compact filter in aHausdorff semitopological divisible residuated lat-tice (L,⊙,U) which satisfies maximum condition.If the set L/F is finite, then (L,U) is a normaland regular space.

Proof. First we prove that for each x, y ∈ L,F ⊙x = F ⊙y iff, x = y. If x = y, clearly F ⊙x =F ⊙ y. Let F ⊙ x = F ⊙ y. Since x ∈ F ⊙ x, forsome f ∈ F, x = f⊙y. By (R5), x ≤ y. Similarity,y ≤ x. Hence x = y. Now since the set L/F isfinite, by Lemma 4.1, there exist x1, ..., xn ∈ Lsuch that L = (F ⊙ x1)∪ ...∪ (F ⊙ xn). Since theoperation ⊙ is continuous in the first variable,for each 1 ≤ i ≤ n, F ⊙ xi is compact and soL is compact. Since L is compact and Hausdorffspace, we conclude that L is normal and regularspace.

Theorem 4.7 Let F be a fiter with maximumcondition in topological divisible residuated lattice(L,U). If the mapping ta : L → L by ta(x) = a⊙xis a closed map and F is a normal subspace of(L,U), then (L,U) is a normal space.

Proof. Since for each a ∈ L, the mappingta(x) = a ⊙ x is a closed continuous map of Fonto F ⊙ a, it follows that F ⊙ a is a normalsubspace of (L,U). Also by Lemma 4.1, for eachx ∈ L there is a x∗ ∈ L such that F (x) = F ⊙x∗.Thus L = ∪F ⊙ x∗ is a union of disjoint normalspaces which implies that L is normal.

Proposition 4.2 Let (L, ,U) be a T1 semitopo-logical residuated lattice. Then 1 is closed.

Proof. Let x ∈ 1. If x = 1, then there isan open set U such that 1 ∈ U and x ∈ U. ByProposition 3, U(x) is an open neighborhood ofx. Since x ∈ 1, 1 ∈ U(x). Hence x = 1x ∈ Uwhich is a contradiction.

Proposition 4.3 Let (L,→,U) be a secondcountable semitopological residuated lattice. If Jis an open nontrivial filter in L, then there is anontrivial topology V on L caorser than U suchthat (L,V) is a metrizable topological residuatedlattice.

Proof. Let F be a family of open fiters whichcontains J and is closed under intersection. Itis easy to prove that the set B = F (x) : F ∈F , x ∈ L is a base for the topology V = V ⊆

L : ∀x ∈ V ∃F ∈ F s.t F (x) ⊆ V on L. Since Jis a nontrivial filter and for each x ∈ J, J(x) ⊆ J,hence J ∈ V which implies that V is a nontrivialtopology on L. By Proposition 3, F (x) is in U , forall F ∈ F and x ∈ L. Hence V is coarser than U .As the proof of Proposition 4.1, we can show that(L,V) is a regular topological residuated lattice.By Proposition 4.2, 1 is closed and by Propo-sition 3.1, (L,V) is Hausdorff. Since (L,V) is aT2 second countable regular topological space, itis metrizable.

Proposition 4.4 Let Ω be a family of prefiltersin residuated lattice L which is closed under in-tersection. If(i) for each U ∈ Ω there is a W ∈ Ω such thatW [W ] ⊆ U,(ii) for each x = 0 there is a V ∈ Ω such thatx′ ∈ V,then there is a topology U on L such that(L,⊙, ,U) is a Hausdorff compact topologicalresiduated lattice. Moreover, if (L,U) is secondcountable, it is metrizable.

Proof. First we prove that the set B = V [x] :V ∈ Ω, x ∈ L is a base for a topology U onL. Clearly, L ⊆

∪B. Let V [x] and U [y], both,

be in B and a ∈ V [x] ∩ U [y]. By (i), there is aW ∈ Ω such that W [W ] ⊆ V ∩ U. We show thatW [a] ⊆ V [x] ∩ U [y]. Suppose z ∈ W [a]. By (R5),(zx)(za) ≥ za, so zx ∈ W [za] ⊆ W ]W ] ⊆ V.Similarly, zy ∈ U. Hence W [a] ⊆ V [x] ∩ U [y].Therefore, B is a base for a topology U on L. Nowwe show that V (x) is open, for any V ∈ Ω andx ∈ L. For this, suppose y ∈ V (x) and W [W ] ⊆V, for some W ∈ Ω. Let z ∈ W [y]. By (R5), zx ∈W [zy] ⊆ W [W ] ⊆ V. Similarly, xz ∈ V. Hencey ∈ W [y] ⊆ V (x). This shows that V (x) ∈ U . Letx, y ∈ L, V ∈ Ω. Then there exists a W ∈ Ω suchthat W [W ] ⊆ V. As the proof of Proposition 4.1,we can prove that W [x] ⊙ W [y] ⊆ V [x ⊙ y]. Weshow that W (x)W [y] ⊆ V [xy]. For this, supposea ∈ W (x) and b ∈ W [y]. We have

(by)((ab)(xy))

= (ab)[(by)](xy)]

= (ab)[x((by)y)]

≥ (ab)(xb)

= x((ab)b)

≥ xa

Hence (ab)(xy) ∈ W [by] ⊆ V and soW (x)W [y] ⊆V [xy]. Therefore is continuous. Let Ui : i ∈ I

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be an open cover of L. Then for some V ∈ Ωand i ∈ I, 1 ∈ V [1] ⊆ Ui. It is easy to see thatL = V [1]. Hence L ⊆ Ui, which shows that L iscompact. Now, we show that 0 is closed. Letx ∈ 0. If x = 0, by (ii) there is a V ∈ Ω suchthat x′ ∈ V. Since V (x) is an open neighborhoodof x, 0 ∈ V (x) which implies that x′ ∈ V, a con-tradiction. Hence 0 is closed. By Proposition3.1, (L,U) is Hausdorff. Finally, if (L,U) is sec-ond countable, then it is metrizable because it isHausdorff compact second countable.

Proposition 4.5 Let F be a closed filter withmaximum condition in a second countabletopological Hausdorff divisible residuated lattice(L,U). If F is a normal subspace of (L,U) andthe mapping ta(x) = a ⊙ x is a closed map of Linto L, then (L,U) is metrizable.

Proof. Since (L,U) is a topological residuatedlattice, the mappings la(x) = a → x and ra(x) =x → a of L into L are continuous and so for eachx ∈ L, F (x) = l−1

x (F ) ∩ r−1x (F ) is a closed set in

L. Since L is divisible, by Lemma 4.1, for eachx ∈ L, there is a x∗ ∈ L such that F (x) = F ⊙x∗.On the other hand, for each x ∈ L, the mappingϕ(y) = y ⊙ x∗ is a closed continuous map of Fonto F ⊙ x∗. Hence F (x) = F ⊙ x∗ is a normalsubspace of (L,U). Let C be the family of pairwisedisjoint sets F (x). Since L = ∪C, residuatedlattice L is normal. Since (L,U) is Hausdorff, itis a regular space. Finally, since (L,U) is regularsecond countable, it is metrizable.

Proposition 4.6 Let (L,U) be a second count-able topological residuated lattice and let for eacha ∈ L the mapping ta(x) = a⊙ x be an open andclosed map of L into L. Then (L,U) is metriz-able iff, for each x = 1 and each open neighbor-hood U of 1, there exists an open set V such that1 ∈ V ⊆ V ⊆ U and x ∈ V.

Proof. (⇐) Let for each x = 1 and each openneighborhood U of 1, there exists an open set Vsuch that 1 ∈ V ⊆ V ⊆ U and x ∈ V. Thenby [[11], Proposition 4.10,] (L,U) is a T1 space.By Theorem 3.2, (L,U) is a Hausdorff space. Weprove that it is regular. For this, let x ∈ U ∈ U .Clearly, if x = 1, then there is a V ∈ U such that1 ∈ V ⊆ V ⊆ U. Hence we suppose that x = 1.Since the operation ⊙ is continuous, there is anopen neighborhood W of 1 such that x⊙W ⊆ U.Let V be an open set such that 1 ∈ V ⊆ V ⊆ W.

Since tx is an open and closed map, the set x⊙Vis an open neighborhood of x and x⊙V is a closedset. So we have

x ∈ x⊙ V ⊆ x⊙ V = x⊙ V ⊆ x⊙W ⊆ U

which prove that (L,U) is a regular space. Since(L,U) is second countable, it is metrizable.(⇒) Let (L,U) be metrizable, U be induced topol-ogy by metric d and x = 1. If U is an open neigh-borhood of 1, then there exists a ϵ > 0 such thatx ∈ L : d(x, 1) < ϵ ⊆ U. Let V = x ∈ L :d(x, 1) < ϵ

2. Then clearly, 1 ∈ V ⊆ V ⊆ U andx ∈ V.

Proposition 4.7 Let (L,→,U) be a secondcountable topological residuated lattice and let foreach a ∈ L the mapping ra(x) = x → a be anopen map of L into L. Then (L,U) is metriz-able iff, for each x = 1 and each open neighbor-hood U of 1, there exists an open set V such that1 ∈ V ⊆ V ⊆ U, V is compact and x ∈ V.

Proof. (⇐) Let for each x = 1 and each openneighborhood U of 1, there exists an open set Vsuch that 1 ∈ V ⊆ V ⊆ U, V is compact andx ∈ V. Then by [[11], Proposition 4.11,] (L,U) isa Hausdorff space. We prove that it is regular.For this, let x ∈ U ∈ U . Clearly, if x = 1, thenthere is a V ∈ U such that 1 ∈ V ⊆ V ⊆ U. Hencewe suppose that x = 1. Since the operation → iscontinuous, there is an open neighborhood W of1 such that W → x ⊆ U. Let V be an open setsuch that 1 ∈ V ⊆ V ⊆ W. Since rx is an openand continuous map, the set V → x is an openneighborhood of x and V → x is a compact set.Since (L,U) is Hausdorff, the set V → x is closed.So we have

V → x ⊆ V → x = V → x ⊆ W → x ⊆ U

which prove that (L,U) is a regular space. Since(L,U) is second countable, it is metrizable.(⇒) The proof is similar to the proof of Theorem4.6. Recall, a topological space (X,U) is locallycompact if for each x ∈ X there is an open neigh-borhood U of x such that U is compact. If (X,U)is Hasudorff, then X is locally compact iff, foreach x ∈ U ∈ U there is an open set V such thatx ∈ V ⊆ V ⊆ U and V is compact. It is easy tosee that any compact set is locally compact.[See,[7]]

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Proposition 4.8 Let (L,U) be a second count-able Hausdorff residuated lattice and for eacha ∈ L, ta(x) = a ⊙ x be an open map of L intoL. Then (L,U) is metrizable iff, there is a locallycompact open neighborhood of 1.

Proof. (⇐) Let U be a locally compact openneighborhood of 1. Let a ∈ L. Since ta(x) = x⊙ais an open map of U onto U ⊙a, by [[7], Theorem3.3.15] U⊙a is an open locally compact subset ofL. As L = ∪x∈LU ⊙ x is a union of open locallycompact subspaces of L, then by [[7], Exercise3.3.A] it is a locally compact space. Since (L,U)is Hausdorff and second countable, it is regularand so metrizable.(⇒) Let (L,U) be metrizable and U be topol-ogy induced by metric d. Then clearly, the setx ∈ L : d(x, 1) < r is a locally compact openneighborhood of 1.

Proposition 4.9 Let (L,→,U) be a secondcountable Hausdorff semitopological residuatedlattice. If F is a locally compact closed filter, then(L,U) is metrizable.

Proof. Let for each a ∈ L, ra(x) = x → a andla(x) = a → x be two maps of L into L. Then raand la, both, are continuous. Since F is locallycompact, r−1

a (F ) is locally compact because if x ∈r−1a (F ), then ra(x) ∈ F and so there is an openset V such that ra(x) ∈ V ∩F ⊆ V ∩ F ⊆ F. Now

x ∈ r−1a (V ∩ F ) ⊆ r−1

a (V ∩ F ) ⊆ r−1a (F )

proves that r−1a (F ) is locally compact. Simi-

larity l−1a (F ) is locally compact. Since F (a) =

r−1a (F )∩ l−1

a (F ), and F is locally compact closed,so F (a) is locally compact closed for each a ∈ L.Now since L = ∪a∈LF (a) is a union of locallycompact closed subspaces of L, by [[7], Exercise3.3.B] (L,U) is locally compact. Since (L,U) isHausdorff and second countable, (L,U) is regularand then metrizable.

Definition 4.3 Let U be a topology on a residu-ated lattice L and W ⊆ L. We call:(i) W is a n-quasifilter if x ∈ Wn and x ≤ y im-ply y ∈ Wn,(ii) W is a quasifilter if for each n ≥ 1, W is an-quasifilter,(iii) W is a quasifilter neighborhood of 1 if W isa neighborhood of 1 which is quasifilter,(iv) W is a QN-fundamental system, if W isa fundamental system of open neighborhoods 1which all are quasifilter.

Example 4.4 (i) Every open filter is a quasifil-ter neighborhood.(ii) In Example 2.1, W = 1, a, b, c is a quasifil-ter neighborhood of 1 which is not a filter.(iii) Let I be residuated lattice in Example 2.1(ii). Let 0 = a ∈ I and W = (a, 1]. Then for eachn ≥ 1, Wn = (an, 1]. If U is subspace topologyinduced by real number, then W is a quasifilterneighborhood of 1 which is not a filter. Moreover,W = (a, 1] : a ∈ I is a QN-fundamental sys-tem.

Proposition 4.10 Let W be a QN-fundamentalsystem in the second countable Hausdorff topo-logical residuated lattice (L,U). If for each a ∈ L,ta(x) = a ⊙ x is an open map of L into L, then(L,U) is metrizable provided there exists a com-pact open neighborhood of 1.

Proof. Let U be a compact open neighborhoodof 1 and x ∈ U. Since the operation ⊙ is continu-ous, there exist two open neighborhoods Vx andHx of 1 such that x⊙Vx ⊆ U and Hx⊙Hx ⊆ Vx.Clearly, x ⊙ Hx : x ∈ U is an open coveringof U. Since U is compact, the union of a finitenumber sets xi ⊙Hxi : 1 ≤ i ≤ n covers U. LetV =

∩ni=1Hxi . Then V is an open neighborhood

of 1 such that

U ⊙ V ⊆ (n∪

i=1

xi ⊙Hxi)⊙ V ⊆n∪

i=1

xi ⊙ Vxi ⊆ U.

Since U ∩ V is an open neighborhood of 1, thereis a W ∈ W such that W ⊆ U ∩ V. Thus W 2 ⊆U⊙V ⊆ U. But then by induction, Wn = Wn−1⊙W ⊆ U ⊙ V ⊆ U. Let F = ∪n≥1W

n. If x, y ∈ F,then for some n and m, x ∈ Wn and y ∈ Wm.Hence x ⊙ y ∈ Wn+m ⊆ F. If x ∈ F and x ≤ y,then for some n ≥ 1, x ∈ Wn. Since W is aquasifilter, y ∈ Wn and so y ∈ F. Thus F is afilter. Since for each a ∈ L, ta is an open map,F is an open set. If y ∈ F, then since F ⊙ y isan open neighborhood of y, there is a f ∈ F suchthat f ⊙ y ∈ F. Since F is a filter, y ∈ F. ThusF is a closed set in the compact Hausdorff spaceU and so F is compact. Now by Proposition 4.9,(L,U) is metrizable.

5 Conclusion

In this paper, we studied separation axioms ontopological residuated lattices. In Theorem 3.2

194 N. Kouhestani et al, /IJIM Vol. 9, No. 3 (2017) 183-194

we showed that axioms T0, T1, T2 and T5/2 areequivalence. In Theorem 3.4 we proved that thereis a Hausdorff topological residuated lattice ofeach infinite cardinality. Axioms regularity andnormality were studied in section 4. For the fu-ture we suggest to study metrizability, uniformity,quasi-uniformity on this structure.

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Nader Kouhestani is AssistantProfessor of department of math-ematics of Sistan and Baluches-tan university and Head of FuzzySystems Research Center of Sis-tan and Baluchestan university.His research interests include

topology on Logical algebras such as residuatedlattices, BL-algebras and MV-algebras

Rajab Ali Borzooei is Full Profes-sor of Department of Mathemat-ics of Shahid Beheshti Universityin Iran. He is managing editor ofIranian Journal of Fuzzy Systems.His research interests is Logicalalgebras and Fuzzy Mathematics.