New topology in residuated lattices

Definition 2.1

([1]). A residuated lattice is an algebra (L, ∨, ∧, ⊙, →, 0,1) of type (2, 2, 2, 2, 0, 0) such that

(Lr₁) (L, ∨, ∧, 0, 1) is a bounded lattice;

(Lr₂) (L, ⊙, 1) is a commutative monoid;

(Lr₃) ⊙ and → form an adjoint pair, i.e., a ⊙ x ≤ biffx ≤ a → b.

We call them simply residuated lattices. For examples of residuated lattices see [12–17].

In what follows (unless otherwise speciied) we denote by L a residuated lattice. If L is totally ordered, then L is called a chain.

For x ∈ L and n ≥ 1 we define x^* = x → 0, x^** = (x^**)^**, x⁰ = 1 and xⁿ = xⁿ⁻¹ ⊙ x.

We refer to [3–21] for detailed proofs of these well-known results:

Lemma 2.2

LetLbe a residuated lattice. Then for everyx, y, z ∈ L, we have:

(r₁) x → x = 1, x → 1 = 1, 1 → x = x;

(r₂) x ≤ yiffx → y = 1;

(r₃) Ifx ∨ y = 1, thenx ⊙ y = x ∧ y;

(r₄) Ifx ≤ y, thenz ⊙ x ≤ z ⊙ y, z → x ≤ z → y, y → z ≤ x → z;

(r₅) x → (y → z) = (x ⊙ y) → z = y → (x → z);

(r₆) x ⊙ (y ∨ z) = (x ⊙ y) ∨ (x ⊙ z), andx ⊙ (y ∧ z) ≤ (x ⊙ y) ∧ (x ⊙ z);

(r₇) (x → z) ∧ (y → z) = (x ∨ y) → z;

(r₈) (x → z) ∨ (y → z) ≤ (x ∧ y) → z;

(r₉) x → (y ∧ z) = (x → y) ∧ (x → z);

(r₁₀) (x → y) ∨ (x → z) ≤ x → (y ∨ z);

(r₁₁) x ∨ y ≤ ((x → y) → y) ∧ ((y → x) → x);

(r₁₂) (x ∨ y)^* = x^* ∧ y^*, (x ∧ y)^* ≤ x^* ∨ y^*;

(r₁₃) (x → y^**)^** = x → y^**;

(r₁₄) x^** → y^** = y^* → x^* = x → y^** = (x → y^**)^**;

(r₁₅) x ⊙ x^* = 0, 1^* = 0, 0^* = 1, x^*** = x^*;

(r₁₆) x ≤ x^**, x^** ≤ x^* → x, x → y ≤ y^* → x^*;

(r₁₇) x → y ≤ (x → y)^** ≤ x^** → y^**;

(r₁₈) x^** ⊙ y^** ≤ (x ⊙ y)^**, so (x^**)ⁿ ≤ (xⁿ)^**

for every natural number n;

(r₁₉) x^* ⊙ y^* ≤ (x ⊙ y)^*;

(r₂₀) (x ∧ y)^** ≤ x^** ∨ y^** ≤ (x ∨ y)^**.

Following the above mentioned literature, we consider the identities:

(i₁) x ∧ y = x ⊙ (x → y) (divisibility);

(i₂) (x^* ∧ y^*)^* = [x^* ⊙ (x^* → y^*)]^* (semi – divisibility);

(i₃) (x → y) ∨ (y → x) = 1 (pre – linearity);

(i₄) x^* ∨ x^** = 1;

(i₅) x² = x;

(i₆) x = x^**.

Then the residuated lattice L is called:

(i) Divisible if L verifies (i₁);

(ii) Semi-divisible if L verifies (i₂);

(iii) MTL-algebra if L verifies (i₃);

(iv) BL-algebra if L verifies (i₁) and (i₃);

(v) Stonean if L verifies (i₄);

(v) G-algebraif L verifies (i₅);

(vi) Involutiveif L verifies (i₆).

An MV-algebra is an algebra ℒ = (L, ⊕, *, 0) of type (2, 1, 0) satisfying the following equations:

(mv₁) x ⊕ (y ⊕ z) = (x ⊕ y) ⊕ z;

(mv₂) x ⊕ y = y ⊕ x;

(mv₃) x ⊕ 0 = x;

(mv₄) x^** = x;

(mv₅) x ⊕ 0^* = 0^*;

(mv₆) (x^* ⊕ y)^* ⊕ y = (y^* ⊕ x)^* ⊕ x, for all x, y, z ∈ L.

Note that axioms (mv₁) – (mv₃) state that (L, ⊕, *, 0) is a commutative monoid.

Every BL-algebra L with x^** = x for all x ∈ L, is an MV-algebra.

By Lemma 2.2, in the case of divisible residuated lattices we have:

Corollary 2.3

([7]). IfLis a divisible residuated lattice, then for everyx, y ∈ Lwe have:

(r₂₁) (x^** → x)^* = 0;

(r₂₂) (x → y)^** = x^** → y^**;

(r₂₃) (x ∧ y)^** = x^** ∧ y^**;

(r₂₄) x ⊙ (y ∧ z) = (x ⊙ y) ∧ (x ⊙ z);

(r₂₅) x ∧ (y ∨ z) = (x ∧ y) ∨ (x ∧ z).

On any residuated lattice L ([6, 8]) we may define an operator ⊕ by setting for all x, y ∈ L,

By (r₅), the identity (1) is equivalent with

Lemma 2.4

([8]). LetLbe a residuated lattice andx, y, z ∈ L. Then:

(r₂₆)x ⊕ 0 = x^**, x ⊕ 1 = 1, x ⊕ x^* = 1;

(r₂₇) x ⊕ y = y ⊕ x, x, y ≤ x ⊕ y;

(r₂₈) x ⊕ (y ⊕ z) = (x ⊕ y) ⊕ z;

(r₂₉) Ifx ≤ y, thenx ⊕ z ≤ y ⊕ z.

In what follows we will establish other necessary properties of operator ⊕ in a residuated lattice L.

Proposition 2.5

LetLbe a residuated lattice andx, y, z ∈ L. Then:

(r₃₀) (x ⊕ y)^** = x ⊕ y = x^** ⊕ y^**;

(r₃₁) Ifx ∨ y = 1, thenx ⊕ y = 1;

(r₃₂) x ⊕ (y ∨ z)^* = (x ⊕ y^*) ∧ (x ⊕ z^*).

(r₃₃) x ⊕ (y^* ∨ z^*)^* = (x ⊕ y) ∧ (x ⊕ z).

Proof

(r₃₀). We obtain successively (x⊕y)**=(x*⊙y*)***(r15)=(x*⊙y*)*=x⊕y(r15)=[(x**)*⊙(y**)*]*=x**⊕y**..

(r₃₁). Since x ≤ x^**, y ≤ y^** and x^**, y^** ≤ x ⊕ y implies x ∨ y ≤ x ⊕ y. Since 1 = x ∨ y ≤ x ⊕ y, then x ⊕ y = 1.

(r₃₂). By (r₉), (r₁₂), (r₁₅) and (2) we obtain successively x⊕(y∨z)*2=x*→(y∨z)***(r15)=x*→(y∨z)*(r12)=x*→(y*∧z*)(r9)=(x*→y*)∧(x*→z*)(r14)=(y**→x**)∧(z**→x**)(2)=(x⊕y*)∧(x⊕z*).

(r₃₃). By (r₉), (r₁₂), (r₁₅) and (2) we obtain successively x⊕(y*∨z*)*2=x*→(y*∨z*)***(r15)=x*→(y*∨z*)*(r12)=x*→(y**∧z**)(r9)=(x*→y**)∧(x*→z**)(2)=(x⊕y)∧(x⊕z).

Corollary 2.6

LetLbe a residuated lattice andx, y ∈ L. Then:

(r₃₄) (x^* ⊕ y)^* ⊕ y = (y^* ⊕ x)^* ⊕ xiff (x^** → y^**)^** → y^** = (y^** → x^**)^** → x^**.

Proof

(r₃₄). Since (x*⊕y)*⊕y2=(x**→y**)*⊕y2=(x**→y**)**→y** and (y*⊕x)*⊕x2=(y**→x**)*⊕x2=(y**→x**)**→x**. Thus, our claim holds.

Definition 2.7

([2]). An implicative filter (filter, for short) is a nonempty subsetFofLsuch that

(F₁) Ifx ≤ yandx ∈ F, theny ∈ F;

(F₂) Ifx, y ∈ F, thenx ⊙ y ∈ F.

We notice that:

1. F is an implicative filter ([13]) of L iff 1 ∈ F and x, x ͖ y ∈ F, then y ∈ F (that is, F is a deductive system ofL).

2. Every filter is a latticeal filter in the lattice (L, ∧, ∨), but the converse does not hold ([3,14]).

Hence, if we denote by ℱ(L) (ℱ_i(L)) the set of all latticeal filters (implicative filters) of L, then ℱ_i(L) ⊆ ℱ (L).

We have ([13]) ℱ_i(L) = ℱ(L) iff x ⊙ y = x ∧ y for every x, y ∈ L.

We recall ([3, 14]) that for a nonempty subset D of L we denote by 〈D〉 the filter generated by D, and 〈D〉 = {x ∈ L : d₁ ⊙ ... ⊙ d_n ≤ x, for some d₁,..., d_n ∈ D}. If a ∈ L, the filter generated by {a} will be denoted by 〈a〉, and (a) = {x ∈ L : aⁿ ≤ x for some n ≥ 1}. If F ∈ ℱ_i(L) and a ∈ L \ F, then 〈F ∪ {a}〉 will be denoted by F(a), and F(a) = {x ∈ L : aⁿ → x ∈ F for some n ≥ 1}. If F, G ∈ ℱ_i(L), then F ∨ G = F ∨_ℱi.(L)G = 〈F ∪ G〉 = {x ∈ L : b ⊙ c ≤ x for some b ∈ F, c ∈ G}.

We say that P ∈ ℱ_i(L), P ≠ L is a prime filter ([14]) if for x, y ∈ L and x ∨ y ∈ P, then x ∈ P or y ∈ P. We denote by Spec_i(L) the set of all prime filters of L.

We recall that a filter M of L is called maximal if M ≠ L and M is not strictly contained in any proper filter of L.

Every maximal filterM of L is obvious prime because, if there exist two proper filters N, P ∈ ℱ_i(L) such that M = N ∩ P, then M ⊆ N and M ⊆ P, by the maximality of M we deduce that M = N = P, that is, M is an inf-irreducible, so prime element in the lattice of filters (ℱ_i(L), ⊆) of L (by the distributivity of the lattice of filters (ℱ_i(L), ⊆) of L).

So, if we denote by Max_i(L) the set of all maximal filters of L, then Max_i(L) ⊆ Spec_i(L).

Proposition 2.8

([1, 3, 14]). ForM ∈ ℱ_i(L), M ≠ L, the following are equivalent:

(i) M is maximal;

(ii) Ifx ∉ M, then there existsn ≥ 1 such that (xⁿ)^* ∈ M.

Definition 2.9

If L, L′ are residuated lattices, a map f : L → L is called morphism of residuated lattices iff is a morphism of bounded lattices, f (x ⊙ y) = f (x) ⊙ f (y) and f (x → y) = f (x) → f (y) for every x, y ∊ L.

Iff is a bijective map (one-to-one and onto), then we say that L and L′ are isomorphic and we write L ≈ L′.

Remark 2.10

If F ∊ ℱ_i(L′), then f⁻¹(F) ∊ ℱ_i(L). Inparticular f⁻¹ ({1}) = {x ∊ L : f (x) = 1} is denoted by Ker(f) and Ker(f) ∊ ℱ_i (L).

Clearly, f is one-to-one iff Ker(f) = {1}. Also, f (L) is a subalgebra of L′ denoted by Im(f).

Definition 3.1

([11]). Let (X, ≤) be an ordered set. Then we define ↑ : 𝒫(X) → 𝒫(X), by ↑ S = {x ∊ X|a ≤ x, for some a ∊ S}, for any subset S of X. A subset F of X is called an upset if ↑ F = F. We denote by U(X) the set of all upsets of X. Anupset F is called finitely generated if there exists n ∊ N suchthat F =↑ {x₁, x₂, x_n}, for some x₁, x₂, ... , x_n ∊ X.

Remark 3.2

Clearly, if F is a filter of L, then F is an upset of L, but the converse does not hold. Indeed, we consider L = {0, n, a, b, c, d, e, f, m, 1} with 0 < n < a < c < m < 1, 0 < n < a < e < m < 1, 0 < n < b < c < m < 1, 0 < n < b < f < m < 1, 0 < n < d < e < m < 1 and elements {a, b}, {a, d}, {b, d}, {a, f}, {c, d}, {b , e} , {c, e} , {c, f} and {e, f} are pairwise incomparable.

Then ([12]) L becomes a residuated lattice relative to the following operations:

Then ↑ n = {n, a, b, c, d, e, f, m, 1} is an upset, but n ⊙ n = 0, that is ↑ n is nota filter.

Definition 3.3

([11]). Letτ and τbe two topologies on a given set X. Ifτ′ ⊆ τ, then we say thatτis finer thanτ. Let (X, τ) and (Y, τ ) be two topological spaces. A map f : X → Y is called continuous if the inverse image of each open set of Y is open in X. A homeomorphism is a continuous function, bijective and has a continuous inverse.

Lemma 3.4

([11]). Let β and β′ be two bases for topologiesτandτ′ , respectively on X. Then the following are equivalent:

(i) τ′is iner thanτ;

(ii) For any x ∊ X and each basis element B ∊ β containing x, there is a basis element B′ ∊ β such that x ∊ B′ ⊆ B.

Definition 3.5

([11]). Let (A, *) be an algebra of type 2 andτbe a topology on A. Then (A, *, τ) is called:

(i) Right (left) topological algebra, if for all a s A the map * : A → A defined by x ↦ a * x (x ↦ x * a) is continuous;

(ii) Semitopological algebra if A is a right and left topological algebra.

If (A, *) is a commutative algebra, then right and left topological algebras are equivalent.

Definition 3.6

([11]). Let A be a nonempty set, {*_i}_i∊Ibe a family of binary operations on A andτbe a topology on A. Then:

(i) (A, {*_i}_i∊I, τ) is a right (left) topological algebra, if for i s I, (A, *_i, τ) is a right (left) topological algebra;

(ii) (A, {*_i , τ) is a right (left) semitopological algebra, if for i ∊ I, (A, *_i, τ) is a right (left) semitopological algebra.

Definition 3.7

Consider L a residuated lattice and a ∊ L. For any nonempty upset X of L we define the set

where aⁿ ⊖ x = a ⊖ (a^n–1 ⊖ x), for any n ∊ {2, 3, 4,...}.

Lemma 3.8

Let L be a residuated lattice and a, x ∊ L. Then:

(r₃₅) aⁿ ⊖ x = (a**)ⁿ → x**;

(r₃₆) (aⁿ ⊖ x)** = aⁿ ⊖ x, for any n ∊ ℕ.

Proof

(r₃₅). Mathematical induction relative to n.

If n = 2, by (r₅), (r_3O) and identity (4) we obtain successively a2⊖x=a⊖(a⊖x)=id.(4)a**→(a*⊕x)**=(r30)a**→(a*⊕x)=id.(4)a**→(a**→x**)=(r5)(a**⊙a**)→x**=(a**)2→x**.

Suppose a^n–1 ⊖ x = (a**)^n–1 → x**. By (r₅) we obtain successively an⊖x=a⊖(an−1⊖x)=a**→((a**)n−1→x**)=(r5)(a**⊙(a**)n−1)→x**=(a**)n→x**. Therefore, aⁿ ⊖ x = (a**)ⁿ → x**, for any n ∊ ℕ.

(r₃₆). By (r₃₀) we have (an⊖x)**=[a⊖(an−1⊖x)]**=id.(4)[a*⊕(an−1⊖x)]**=(r30)a*⊕(an−1⊖x)=id.(4)a⊖(an−1⊖x)=an⊖x, for any n ∊ ℕ.

Proposition 3.9

Let L be a residuated lattice and a ∊ L. For any nonempty upset X of L theset D_a (X) = {x ∊ L : (a**)ⁿ → x** ∊ X, for some n ∊ ℕ}.

Remark 3.10

If a = 1, then D_a (X) = D₁(X) = {x ∊ L : x** ∊ X} is the set of double complemented elements of X.

Corollary 3.11

If L is a residuated lattice and a, x ∊ L, then:

(rM₃₇) a ⊙ x ≤ a ∧ x ≤ a ∨ x ≤ a** ∨ x** ≤ a ⊕ x;

(r₃₈) x** ≤ a ⊖ x ≤ aⁿ ⊖ x, for any n ∊ ℕ.

Proof

(r₃₇). Indeed, a ⊙ x ≤ a ∧ x ≤ a ∨ x ≤ a** ∨ x**. By (r₂₇) and identity (2) we obtain successively a⊕x=id.(2)a*→x**≥x** and a⊕x=(r27)x⊕a=id.(2)x*→a**≥a**, hence a** ∨ x ≤ a ⊕ x.

(r₃₈). Since an⊖x=(r35)(a**)n→x**≥(r4)a**→x**=a⊖x≥x**, hence x** ≤ a ⊖ x ≤ aⁿ ⊖ x, for any n ∊ ℕ.

Theorem 3.12

If L is a residuated lattice and a, x ∊ L. Consider X, Y two nonempty upsets of L. Then:

(i) D_a(X) is an upset of L;

(ii) 1 ∊ D_a(X), a ∊ D_a(X) and X ⊆ D_a(X);

(iii) a^m ⊖ (aⁿ ⊖ x) = a^m+n ⊖ x, for any m, n ∊ ℕ;

(iv) if X ⊆ Y, then D_a (X) ⊆ D_a (Y);

(v) D_a (D_a (X)) = D_a (X);

(vi) if F is a filter of L, then D_a(F) is a filter of L;

(vii) if a ≤ s, thenD_s (X) ⊆ D_a (X);

(viii) if {X_α|α ∊ I} is a family of upsets of L, then D_a(∪{X_α|α ∊ I}) = ∪(D_a ({X_α|α ∊ I}));

(ix) if {X₁, X₂,..., X_n} is a finite set of upsets of L, then D_a (∩{X_i| i = 1, 2,..., n}) = ∩{D_a (X_i)| i = 1, 2,..., n};

(x) D_a(D_x(X)) = D_x(D_a(X));

(xi) if X₁, X₂, X₃are upsets of L, then D_a(X₁) ∩ [D_a (X₂) ∩ D_a (X₃)] = [D_a (X₁) ∩ D_a (X₂)] ∪ [D_a (X₁) ∩ D_a (X₃)] andD_a(X₁) ∪ [D_a(X₂) ∩ D_a(X₁)] = [D_a(X₁) ∪ D_a(X₂)] ∩ [D_a(X₁) ∪ D_a(X₃)].

Proof

(i). Clearly, D_a(X) ⊆↑ D_a(X). We consider s ∊↑ D_a(X), then there exists x ∊ D_a(X) such that x ≤ s. Since x ∊ D_a(X) we have aⁿ ⊖ x ∊ X, by (r₃₅) we deduce that (a**)ⁿ → x** ∊ X, for some n ∊ ℕ. By (r₄) and (r₁₆), since x ≤ s we obtain successively x** ≤ s**, (a**)ⁿ → x** ≤ (a**)ⁿ → s**. Since X is an upset of L and (a**)ⁿ → x** ∊ X, then (a**)ⁿ → s** ∊ X, hence s ∊ D_a(X) and so ↑ D_a(X) ⊆ D_a(X). We deduce that D_a (X) is an upset of L.

(ii). Since 1 ∊ X, by (r₃₅) we have aⁿ ⊖ 1 = (a**)ⁿ → 1** = (a**)ⁿ → 1 = 1 ∊ X, for any n ∊ ℕ. Hence 1 ∊ D_a (X).

Since 1 ∊ X, by (r₁₈ and (r₃₅) we have aⁿ ⊖ a = (a**)ⁿ → a** = 1 ∊ X, for any n ∊ ℕ. Hence a ∊ D_a(X).

Now, we consider x ∊ X. Then by and (r₃₈) we have x ≤ x** ≤ aⁿ ⊖ x ∊ X, hence x ∊ D_a(X). We deduce that X ⊆ D_a (X).

(iii). Consider m, n ∊ ℕ and a, x ∊ L. By (r₅) and (r₃₆) we obtain successively am⊖(an⊖x)=id.(4)(a**)m→(an⊖x)**=(r36)(a**)m→(an⊖x)=id.(4)(a**)m→((a**)n→x**)=(r5)[(a**)m⊙(a**)n]→x**=(a**)m+n→x**=id.(4)am+n⊖x.

(iv). Consider X ⊆ Y and x ∊ D_a(X). Then there exists n ∊ ℕ such that aⁿ ⊖ x ∊ X ⊆ Y and so x ∊ D_a(Y). Hence D_a(X) ⊆ D_a(Y).

(v). Since X ⊆ D_a (X), by (iv) we have D_a (X) ⊆ D_a (D_a (X)). Consider x ∊ D_a (D_a (X)). Then there exists m ∊ ℕ such that a^m ⊖ x ∊ D_a (X) and so aⁿ ⊖ (a^m ⊖ x) ∊ X, for some n ∊ ℕ. By (iii) we have a^_n+m ⊖ x ∊ X and so x ∊ D_a (X). Hence D_a (X) = D_a (D_a (X)).

(iv). Consider X ⊆ Y and x ∊ D_a(X). Then there exists n ∊ ℕ such that aⁿ ⊖ x ∊ X ⊆ Y and so x ∊ D_a(Y). Hence D_a(X) ⊆ D_a(Y).

(v). Since X ⊆ D_a (X), by (iv) we have D_a (X) ⊆ D_a (D_a (X)). Consider x ∊ D_a (D_a (X)). Then there exists m ∊ ℕ such that a^m ⊖ x ∊ D_a (X) and so aⁿ ⊖ (a^m ⊖ x) ∊ X, for some n ∊ ℕ. By (iii) we have a^_n+m ⊖ x ∊ X and so x ∊ D_a (X). Hence D_a (X) = D_a (D_a (X)).

(vi). Consider F a filter of L. Then F is a nonempty upset and by (ii) we have 1 ∊ D_a(F). Let x, x → y ∊ D_a(F), then there are m, n ∊ ℕ such that (a**)ⁿ → x** ∊ F and (a**)^m → (x → y)** ∊ F. By (r₅) and (r₁₇) we obtain successively ((a**)n→x**)→((a**)n+m→y**)=((a**)n→x**)→((a**)n⊙(a**)m→y**)=(r5)(a**)m→[((a**)n→x**)→((a**)n→y**)]≥(r39)(a**)m→(x**→y**)≥(r17)(a**)m→(x→y)**∈F.

Since F is a filter and (a**)ⁿ → x** ∊ F, then (a**)^n+m → y** ∊ F, hence y ∊ D_a(F). We deduce that D_a (F) is a filter of L.

(vii). Consider a ≤ s and x ∊ D_s(X), then there exists n ∊ ℕ such that (s**)ⁿ → x** ∊ X. By (r₄) we obtain successively a ≤ s, a** ≤ s**, (a**)ⁿ ≤ (s**)ⁿ, (s**)ⁿ → x** ≤ (a**)ⁿ → x**. Since (s**)ⁿ → x** ∊ X, then (a**)ⁿ → x** ∊ X, hence x ∊ D_a(X). We deduce that D_s(X) ⊆ D_a(X).

(viii). Consider x ∊ L. Then we have the equivalences:

x ∊ D_a(∪{X_α|α ∊ I}) iff aⁿ ⊖ x ∊ ∪{X_α|α ∊ I} iff

aⁿ ⊖ x ∊ X_α, for some n ∊ ℕ, and α s I iff

x ∊ D_a(X_α), for some n ∊ ℕ and α ∊ I iff

x ∊ ∪(D_a({X_α|α ∊ I})). Hence D_a(∪{X_α|α ∊ I}) = ∪(D_a({X_α|α ∊ I})).

(ix). Following (iv), since ∩{X_i| i = 1, 2,..., n} ⊆ X_i, for any i ∊ {1, 2,..., n}, then D_a| i = 1, 2, n}) ⊆ D_a(X_i), for any i ∊ {1, 2, n}, hence D_a(∩{X_i| i = 1, 2, n}) ⊆ ∩{X_i| i = 1, 2, n}. Consider x ∊ ∩{D_a(X_i)| i = 1, 2, n}. Then there exist m₁, m₂, m_n ∊ ℕ such that (a**)^m_i → x** ∊ X_i, for any i ∊ { 1 , 2 , ... , n}.

Consider now p = max{m₁, m₂,...,m_n}. By (r₄), since m_i ≤ p, for any i ∊ {1, 2, n}, then (a**)^m_i → x** ≤ (a**)^p → x**. Since (a**)^m_i → x** ∊ X_i, then (a**)^p → x** ∊ X_i, for any i ∊ {1, 2,..., n}. It follows that (a**)^p → x** ∊ ∩{X_i = 1, 2, n}, hence x ∊ D_a(∩{X_i = 1, 2, n}). We deduce that D_a(∩{X_i| i = 1, 2,..., n}) = ∩{D_a (X_i)| i = 1, 2,..., n}.

(x). We have successively:

D_a (D_x (X)) = {t ∊ L| aⁿ ⊖ t ∊ D_x (X), for some n ∊ ℕ} =

= {t ∊ L| x^m ⊖ (aⁿ ⊖ t) ∊ X, for some m, n ∊ ℕ}

=id.(4){t∈L| (x**)m→(an⊖t)**∈X, for some m,n∈ℕ}

=(r36){t∈L| (x**)m→(an⊖t)∈X, for some m,n∈ℕ}

=id.(4){t∈L| (x**)m→[(a**)n→t**]∈X,for some m,n∈ℕ}

=(r5){t∈L| (a**)n→[(x**)m→t**]∈X, for some m,n∈ℕ}

=id.(4){t∈L| an⊖(xm⊖t)∈X, for some m,n∈ℕ}

= {t ∊ L|x^m ⊖ t ∊ D_a (X), for some m ∊ ℕ} = D_x (D_a (X)).

(xi). It follows from (viii) and (ix).

Corollary 3.13

Let a ∊ L. Then the map D_a : U(L) → U(L) is a laticeal monotone modal operator.

Proof

Following Theorem 3.12 (ii), (iv), (v), and (viii) we deduce that the map D_a : U(L) → U(L) is a laticeal monotone modal operator.

Corollary 3.14

For any a ∊ L, the mapD_a : U(L) → U(L) is a closure operator and D_a = D_a**.

Proof

Following Theorem 3.12 (i), (ii), (iv) and (v), we deduce that D_a is a closure operator.

Consider X an upset of L. Following Theorem 3.12 (vii), since by (r₁₆) we have a ≤ a**, then D_a** (X) ⊆ D_a(X). Consider now x ∊ D_a(X). Then there exists n ∊ ℕ such that (a**)ⁿ → x** ∊ X. By (r₁₅) we have a**** = a**, then (a****)ⁿ → x** ∊ X, hence x ∊ D_a**(X). We deduce that D_a(·) = D_a**(·).

Lemma 3.15

Let a ∊ L and F be a filter of L. Then D_a (F) = D₁(F) iff a** ∊ F.

Proof

” ⇐ ”. Consider a ∊ L and F a filter of L such that D_a (F) = D₁(F). Following Theorem 3.12 (ii), we have a ∊ D_a(F) = D₁(F), then a ∊ D₁ (F), and we have successively a ∊ D₁(F), (1**)ⁿ → a** ∊ F, 1ⁿ → a** ∊ F, 1 → a^** ∊ F, a^** ∊ F.

” ⇐ ”. Suppose a** ∊ F. By Theorem 3.12 (vii), since a ≤ 1, then D₁(F) ⊆ D_a(F). Consider now x ∊ D_a (F). Then there exists n ∊ ℕ such that (a**)ⁿ → x** ∊ F. Since a** ∊ F and F is a filter of L, then (a**)ⁿ ∊ F, for any n ∊ ℕ. By (a^**)ⁿ ∊ F, (a^**)ⁿ → x^** ∊ F, we deduce x^** ∊ F. Since x^** = (1^**)ⁿ → x^** ∊ F, then x ∊ D₁(F), hence D_a (F) ⊆ D₁(F). We deduce that D_a (F) = D₁(F).

Corollary 3.16

Let L be an involutive residuated lattice, a ∊ L and F be a filter of L. Then D_a (F) = Fiff a ∊ F.

Proof

” ⇐ ”. Consider a ∊ L and F a filter of L such that D_a(F) = F. Following Theorem 3.12 (ii), we have a ∊ D_a (F) = F, then a ∊ F.

” ⇐ ”. Suppose a ∊ F. By Theorem 3.12 (ii), a ∊ F ⊆ D_a(F). Consider now x ∊ D_a(F). Then there exists n ∊ ℕ such that (a**)ⁿ → x** ∊ F. Since a ≤ a** ∊ F and F is a filter of L, then (a**)ⁿ ∊ F, for any n ∊ ℕ. By (a**)ⁿ ∊ F, (a**)ⁿ → x** ∊ F, we deduce x** ∊ F, hence x ∊ F. Therefore, D_a(F) ⊆ F. We deduce that D_a (F) = F.

Remark 3.17

(i). By Theorem 3.12 (vi), if F is a proper filter, then D_a (F) is a proper filter or D_a (F) = L. We consider the lattice L = {0, a, c, d, m, 1} with 0 < a < m < 1, 0 < c < d < m < 1, but a incomparable with c and d.

Then ([12], page 233) L becomes a residuated lattice relative to the following operations:

We have 〈a〉 = {a, m, 1} is a proper filter of L, D_m (〈a〉) = {x ∊ L | (m**)ⁿ → x** ∊ 〈a〉, for some n ∊ ℕ} = {x ∊ L | 1 → x** ∊ 〈a〉} = 〈a〉 and D_c (〈a〉) = {x ∊ L | (c**)ⁿ → x** ∊ 〈a〉, for some n ∊ ℕ} = {x ∊ L | c → x**} = L.

Moreover, 〈a〉 is a maximal filter of L and D_c (〈a〉) = L ≠ 〈a〉. So, if M is a maximal filter, then D_x (M) = MorD_x(M) = L, for any x ∊ L.

(ii). There are residuated lattices L such that for a proper filter F of L, thereis a ∊ L and D_a (F) = L. Indeed, we consider L = {0, a, b, c, d, e, f, m, 1} with 0 < a < c < m < 1, 0 < a < e < m < 1, 0 < b < c < m < 1, 0 < b < f < m < 1, 0 < d < e < m < iand elements {a, b}, {a, f}, {a, d}, {b, d}, {b, e}, {d, c}, {c, e}, {c, f} and {e, f} are pairwise incomparable.

Then ([12]) L is a residuated lattice with the following operations:

Let F = 〈1〉 = {1}. Since D_m (〈1〉) = {x ∊ L|(m**)ⁿ → x** ∊ 〈1〉, for some n ∊ ℕ} = {x ∊ L|0 → x** ∊ 〈1〉} = {x ∊ L|1 ∊ 〈1〉} = L.

(iii). There are residuated lattices L such that for a prime filter F, there is a ∊ L such that D_a (F) is not prime. Indeed, we consider L = {0, a, b, n, c, d, 1} with 0 < a < n < d < 1, 0 < b < n < c < 1, but (a, b) and (c, d) are pairwise incomparable.

Then ([12], page 191) L becomes a residuated lattice relative to the operations:

Clearly, 〈c〉 = {c, 1} is aprime filter of L. Since D_n (〈c〉) = {x ∊ L|(n**)^m → x** ∊ 〈c〉, for some m ∊ ℕ} = {x ∊ L|1 → x** ∊ 〈c〉} = {x ∊ L|x** ∊ 〈c〉} = {n, c, d, 1}, hence D_n(〈c〉) = {n, c, d, 1}. Since n = a ⋁ b, but a ∉ D_n (〈c〉) and b ∉ D_n (〈c〉), we deduce D_n (〈c〉) is not prime.

(iv). There are residuated lattices L such that for a filter F, there is a ∊ L such that D_a(F) is prime, but F is not prime. Indeed, we consider L = {0, a, b, c, d, e, f, 1} with 0 < a < c < e < 1, 0 < b < e < f < 1 0 < b < c < e < 1, 0 < b < d < e < 1, and elements {a, b}, {a, d}, {a, f}, {c, d}, {c, f}, and {e, f} are pairwise incomparable.

Then [12] L becomes a MTL-algebra with the following operations:

Since 〈e〉 = {e, 1}, with e = c ⋁ d, but c ∉ 〈e〉 and d ∉ 〈e〉, then 〈e〉 is not aprime filter. Since D_d (〈e〉) = {x ∊ L|(d**)ⁿ → x** ∊ 〈e〉, for some n ∊ ℕ} = {x ∊ L|d → x** ∊ 〈e〉} = {d, e, f, 1}. Hence D_d(〈e〉) = {d, e, f, 1}, which is a prime filter of L, but 〈e〉 is not prime.

(v). If M ∊ Max_i (L) is a maximal filter and D_a (M) is a proper filter, then D_a (M) = M. Indeed, by Theorem 3.12 (ii), M ⊆ D_a (M) and D_a (M) is a proper filter. We get that if M is a maximal filter, then M = D_a (M).

Proposition 3.18

Let a ∊ L and F be a filter of L. Then:

(i) if M is a maximal filter of L and a** ∊ M, then D_a (M) = M;

(ii) D₁(F) is a prime filter iff D_a (F) is prime;

(iii) (aⁿ)* ∊ F, for some n ∊ ℕ iff D_a (F) = L;

(iv) if M is a maximal filter of L, then D_a (M) = L iff a ∊ L \ M.

Moreover, if M is a maximal filter of L, then D_a(M) = M iff a ∊ M.

Proof

(i). Let M be a maximal filter of L and a** ∊ M. Following Theorem 3.12 (ii) and (vi) we have D_a(M) is a filter and M ⊆ D_a(M). We must to prove that D_a(M) is a proper filter of L. If 0 ∊ D_a(M), then by (r₅) and (r₁₅) we obtain successively 0 ∊ D_a(M), (a**)ⁿ → 0** ∊ M, (a**)ⁿ → 0 ∊ M, (a**)^n-1 ⊙ a** → 0 ∊ M, (a**)^n-1 → [a** → 0] ∊ M, (a**)^n-1 → a*** ∊ M, (a**)^n-1 → a* ∊ M. Since a** ∊ M and M is a filter of L, then (a**)^n-1 ∊ M. Since (a**)^n-1 ∊ M, (a** )^n-1 → a* ∊ M, then a* ∊ M. Since a*, a** ∊ M and M is a filter of L, then 0 = a* ⊙ a** ∊ M, a contradiction. Hence D_a(M) ≠ L.

By hypothesis, M is a maximal filter and M ⊆ D_a(M) ≠ L, then D_a(M) = M.

(ii). Let F beafilter of L. Following Theorem 3.12 (vi), D₁(F) and D_a (F) are filters of L. Suppose D₁(F) is a prime filter of L and x ∨ y ∊ D₁(F). If x ∨ y ∊ D₁(F), then (1**)ⁿ → (x ∨ y)** = (x ∨ y)** ∊ F, for any n ∊ ℕ. Since F is a filter and (x ∨ y)** ≤ (a**)ⁿ → (x ∨ y)**, then (a**)ⁿ → (x ∨ y)** ∊ F, hence x ∨ y ∊ D_a(F).

Since x ∨ y ∊ D₁(F) and D₁(F) is prime, then x ∊ D₁(F) or y ∊ D₁(F). It follows that x** ∊ F or y** ∊ F. We have successively x** ∊ F, x** ≤ (a**)ⁿ → x** ∊ F or y** ∊ F, y** ≤ (a**)^m → y** ∊ F, then x ∊ D_a(F) or y ∊ D_a(F), for some m, n ∊ ℕ. Hence D_a(F) is prime.

Now, suppose D_a(F) is prime and for a = 1 we deduce D_i(F) is prime, too.

(iii). Suppose (aⁿ)* ∊ F, for some n ∊ ℕ. Following Theorem 3.12 (ii), F ⊆ D_a(F), then (aⁿ)* ∊ D_a(F). By Theorem 3.12 (ii) and (vi), we obtain a ∊ D_a (F) and D_a (F) is a filter, then aⁿ ∊ D_a (F), for any n ∊ ℕ. Since aⁿ ∊ D_a(F), (aⁿ)* ∊ D_a(F) and D_a(F) is a filter, then 0 = aⁿ ⊙ (aⁿ)* ∊ D_a(F). Hence D_a(F) = L.

Now, suppose D_a(F) = L. Then 0 ∊ D_a(F), (a^**)→ 0^** ∊ F, (a^**)ⁿ → 0 ∊ F, for some n ∊ ℕ.

We prove by induction that (a^**)ⁿ⁺¹ → 0 = aⁿ⁺¹ → 0, for n ∊ ℕ. Clear, for n = 0. Suppose (a^**)ⁿ → 0 = aⁿ → 0, for n ∊ ℕ. By (r₅), (a^**)ⁿ⁺¹ → 0 = ((a**)ⁿ ⊙ a^**) → 0 = a^** → ((a∊**)ⁿ → 0) = a^** → (aⁿ → 0) = aⁿ → a∊^*** = aⁿ → (a → 0) = aⁿ⁺¹ → 0.

Since (a^**)ⁿ → 0 ∊ F, then aⁿ → 0 ∊ F, that is (aⁿ)* ∊ F.

(iv) Consider M a maximal filter of L.

”(⇒)”. Consider D_a(M) = L. If a ∊ M, then we get a^** ∊ M, and by (i) we obtain L = D_a(M) = M, a contradiction. We deduce a ∉ M, that is, a ∊ L \ M.

” ⇐ ”. Consider a ∊ L \ M. Since M is a maximal filter, then following Proposition 2.8 there is n ∊ ⊙ such that (aⁿ)^* ∊ M, and by (iii) we obtain D_a(M) = L.

Now, the fact that D_a(M) = M iff a ∊ M is routine.

Georgescu et al. (2015)[15] called Gelfand residuated lattices those residuated lattices in which any prime filter is included in a unique maximal ilter. They are also called normal residuated lattices. Examples of Gelfand residuated lattices are Boolean algebras, BL-algebras and Stonean residuated lattices (see [8]).

Proposition 3.19

Let a ∊ L and P ∊ Spec_i (L) be aprime filter. If P = D_a (P), then L is Gelfand (normal) residuated lattice. The converse does not hold.

Proof

Let P be a prime filter of L such that P = D_a (P). Using Zorn’s Lemma we deduce that P is contained in a maximal filter. Suppose that there are two distinct maximal filters M₁ and M₂ such that P ⊆ M₁ and P ⊆ M₂. Since M₁ ≠ M₂, there is a ∊ M₁ such that a ∌ M₂. By Theorem 3.12 (ii) and (vi) we have that a ∊ D_a(P) and D_a(P) is a filter, then (aⁿ)^** ∊ D_a(P) = P, for any n ∊ ℕ. Hence (aⁿ)^** ∊ P, for any n ∊ ℕ.

Following Proposition 2.8, there is n ≥ 1 such that (aⁿ)^* ∊ M₂. Then (aⁿ)^** ∉ M₂, hence (aⁿ)^** ∉ P, a contradiction.

For the converse we consider the residuated lattice L from Remark 3.17 (v). The prime filter of L are (a) = {a, b, c, d, 1}, 〈b〉 = {b, c, d, 1}, 〈c〉 = {c, 1} and 〈d〉 {d, 1}. The maximal filter of L is 〈a〉, which include the all other prime filters. Hence L is Gelfand, but D_c(〈b〉) = 〈a〉 ≠ 〈b〉.

Proposition 3.20

Let F ∊ ℱ_i (L) and a ∊ F. For x ∊ D_a(F) the following assertions are equivalent:

(i)D_a(F) = F;

(ii)x^** ∊ F, then x ∊ F.

Proof

(i) ⇒ (ii). Consider D_a(F) = F. By hypothesis F is a filter and a ∊ F, then a^** ∊ F, (a^**)ⁿ ∊ F, for every n ∊ ℕ. Since F is a ilter and (a^**)ⁿ ∊ F, (a^**)ⁿ → x ∊ F, then x ∊ F. We obtain successively F = D_a(F) = {x ∊ L | (a^**)ⁿ → x^** ∊ F, for some n ∊ ℕ} = {x ∊ L | x^** ∊ F}, that is, if x^** ∊ F, then x ∊ F, for all x ∊ L.

(ii) ⇒ (i). By hypothesis F is a filter and a ∊ F, then a^** ∊ F, (a^**)ⁿ ∊ F, for every n ∊ ℕ. Since F is a filter and (a^**)ⁿ ∊ F, (a^**)ⁿ → x^** ∊ F, then x^** ∊ F. We obtain successively D_a(F) = {x ∊ L | (a^**)ⁿ → x^** ∊ F, for some n ∊ ℕ} = {x ∊ L | x^** ∊ F} = F.

Theorem 3.21

Let a ∊ L and F be a filter of L. For x ∊ D_a (F) the following assertions are equivalent:

(i)D_a(F) = F

(ii)x^** ∊ F iff x ∊ F.

Proof

(i) ⇒ (ii). Consider D_a(F) = F. By hypothesis F is a filter and x ∊ D_a(F) = F, by (r₁₆) we obtain x ≤ x^** ∊ F, so, if x ∊ F, then x^** ∊ F. Now, we prove that if x^** ∊ F, then x ∊ F. Since F is a filter and x^** ∊ F, x^** ≤ (a^**)ⁿ — x^** ∊ F, for any n ∊ N, then x ∊ D_a(F) = F.

(ii) ⇒ (i). By Theorem 3.12, (ii) we have F ⊆ D_a(F). Now, we consider x ∊ D_a(F) such that x** ∊ F iff x ∊ F. Since x** ∊ F and x** ≤ (a**)ⁿ → x** ∊ F, for any n ∊ ℕ, then we obtain successively D_a(F) = {x ∊ L | (a**)ⁿ → x** ∊ F, for some n ∊ ℕ} ⊆ {x ∊ L | x** ∊ F} ⊆ F. Therefore, D_a(F) = F.

Remark 3.22

Examples of residuated lattices which satisfy the conditions from Proposition 3.20 and Theorem 3.21 are Boolean algebras, MV-algebras and involutive residuated lattices.

Corollary 3.23

The set τ_a = {D_a(X)|X ϵ U(L)} is a topology on L and (L, τ_a) is a topological space.

Proof

Let a ϵ L. Clearly, D_a(ϕ) = ϕ and D_a(L) = L. Following Theorem 3.12 (viii) and (ix) the set τ_a = {D_a(X)|X ϵ U(L)} is a topology on L and (L, τ_a) is a topological space.

Proposition 3.24

Theset β_a = {D_a (↑ x)|x ϵ L)} is a base for the topology τa on L.

Proof

Let Z be an open subset of (L, τ_a). Then there is X ϵ U(L) such that Z = D_a(X). Since X is an upset, then X = u{↑ x|x ϵ X} and by Theorem 3.12 (viii), we deduce that D_a(X) = ∪{D_a (↑ x)| x ϵ X}. Hence β_a is a base for the topology τa on L.

Corollary 3.25

Every open set X relative to the topology τ_a is an upset. But the converse does not hold.

Proof

Following Corollary 3.23 and Proposition 3.24 we have (L, τ_a) is a topological space with β _a a base for the topology τ_a. Since every union of elements of β_a is an upset and every open set X relative to τ_a can be written as a union of elements of β_a, then X is an upset.

For the converse we consider the residuated lattice L from Remark 3.17 (v), where ↑ n = {n, a, b, c, d, 1} is an upset of L. The upsets of L are ↑ 0 = L, ↑ n = {n, a, b, c, d, 1}, ↑ a = {a, b, c, d, 1}, ↑ b = {b, c, d, 1}, ↑ c = {c, 1}, ↑ d = {d, 1} and ↑ 1 = {1}, so U (L) = {↑ 0, ↑ n, ↑ a, ↑ b, ↑ c, ↑ d, ↑ 1}. Since the base of topology τ_p on L is the set β_p = {D_p(↑ x) | p ϵ L and ↑ x ϵ U(L)} = {z ϵ L| (p^**)ⁿ → z^** ϵ↑ x for some n ϵ ℕ}, then it is easy to verify that ↑ n can not be written as a union of elements of βp. Therefore, ↑ n is not an open set relative to the topology τ_p.

Proposition 3.26

If u, v ϵ L such that u ≤ v, then the topology τ_v is finer than topology τ_u.

Proof

We denote by β_u, β_v basis of the topology τ_u, respectively τ_v.

Consider t ϵ L and D_u (↑ x) ⊆ β_u an element of the basis β_u such that t ϵ D_u(↑ x). Then there exists n ϵ N such that (u^**)ⁿ → t^** ϵ↑ x, that is, x ≤ (u^**)ⁿ → t^**.

Following Theorem 3.12 (ii) we have t ϵ D_v(↑ t). Following Theorem 3.12 (vii), since u ≤ v, then D_v (↑ t) ⊆ D_u(↑ t). We prove that D_u (↑ t) ⊆ D_u(↑ x). Now, we consider s ϵ D_u (↑ t). Then there exists m ϵ N such that (u^**)^m → s^** ϵ ↑ t and so t ≤ (u^**)^m → s^**. By (r₄), (r₃₅) and (r₃₆) we obtain successively t ≤ (u^**)^m → s^**, t**≤(r4)[(u**)m→s**]**=(r35),(r36)(u**)m→s**. Since x≤(u**)n→t**≤(u**)n→[(u**)m→s**]=(r5)(u**)m+n→s**, then x ≤ (u^**)^m+n → s^**, hence (u^**)^m+n→ s^** ϵ ↑ x and so s ϵ Du (↑ x).

Since D_u (↑ t) ⊆ D_u (↑ x) and D_v (↑ t) ⊆ D_u (↑ t), it follows that D_v (↑ t) ⊆ D_u (↑ t) ⊆ D_u (↑ x).

Clearly, D_v (↑ t) ⊆ β_v is an element of the basis β_v. We deduce that for any t ϵ D_u (↑ x) ⊆ β_u, there is a basis element D_v (↑ t) ⊆ β_v such that t ϵ D_v (↑ t) ⊆ D_u(↑ t) ⊆ D_u (↑ x). Following Lemma 3.4 we deduce that the topology τv is finer than topology τu.

Lemma 3.27

If X is a nonempty subset of L and a ϵ L, then X is a compact subset of (L, τa) iff X ⊆ D_a (↑ {x_i1, x_i2, x_in}), for some x_i1, x_i2, x_in ϵ X and i ϵ I.

Proof

“ ⇐ “. SupposeX ⊆ D_a(↑ {x_i1, x_i2,..., x_in}), for some x_i1, x_i2,..., x_in ϵ X and {D_a(X_i)|i ϵ I} be a family of open subsets of L whose union contains X. For any j ϵ {1, 2,..., n}, there is i_j ϵ I such that x_ij ϵ D_a (X_ij). Following Theorem 3.12 (i) and (iv), we have D_a(x_ij) ⊆ D_a(X_ij), for any j ϵ {1,2,..., n}. By Theorem 3.12 (viii) weobtain X ⊆ D_a (↑ {xi₁, x_i2x_in}) = D_a (x_i1 ) ∪ D_a (x_i2)∪...∪D_a (x_in) ⊆ D_a (X_i1)∪D_a (X_i2)∪...∪D_a (X_in). Hence X is compact.

“ ⇒ “. Now, suppose X be a compact subset of L. Since X ⊆ ∪{↑ x|x ϵ X}, then by Theorem 3.12(ii), X ⊆ u{D_a(↑ x)|x ϵ X}. Hence {D_a(↑ x)|x ϵ X} is a family of open subsets of L whose union contains X.

Theorem 3.28

The topological space (L, τ_a) is connected.

Proof

Consider X a non-empty subset of L such that is both closed and open relative to the topology T_a. If X is an open set, by Corollary 3.25, then X is an upset. If 0 ∊ X and X is an upset, then X = L. If 0 ∊ L -X, since X is closed, we get L - X is open, by Corollary 3.25, L - X is an upset of L, since 0 ∊ L - X, then L - X = L. Hence X = ∅, a contradiction. We conclude that {∅, L} is the set of all subsets of L which are both closed and open. That is, (L, τ_a) is connected.

Proposition 4.1

Let P, Q ∊ ℱ_i(L) and a ∊ L such that Q ⊆ P. Then D_[a](P/Q) = D_a(P)/Q, where [a] = {a/Q | a ∊ L}.

Proof

Consider x ∊ L. Then

D_[a](P/Q) = {[x] ∊ L/Q/ [a]^m ⊖ [x] ∊ P/Q, for some m ∊ ℕ} = {[x] ∊ L/Q/ [a^m ⊖ x] ∊ P/Q} = {[x] ∊ L/Q\ a^m ⊖ x ∊ P} = D_a(P)/Q.

Following Proposition 3.24 we obtain the following result.

Remark 4.2

(i). Let F ∊ F_i(L) and a ∊ L. Then (L/F, τ_a/_F) is a topological space, where the set τ_a/_F = {D_a/_F (X/F)\ X ∊ U (L)} is a topology on L/F and the set {D_a/_F (↑ x/F)\ x/F ∊ L/F} is a base for the topology Ta/F on L/F.

If π : L → L/F is the canonical morphism defined by π(x) = [x], then π is a continuous map. Indeed, if O is an open set with respect the topology τ_a/_F, then O = ∪{D_a/_F(↑ x_i/F)\for some x_i/F ∊ L/F}, and π^-1(O) = π^-1(∪ (D_a/F(t x_i/F)) = ∪ π^-1 (Da/F(↑ x_i/f)) = ∪ Da(π’¹(↑ x_i/F)), which is an open set on (L, τa).

(ii) Let F ∊ ℱ_i(L) and a ∊ L. Then (L/F,τa−), is a topological space, where the setτa−={Da(X)/F| X∊U(L)}is a topology on L/F and the set {D_a(↑ x)/F | x/F ∊ L/F} is a base for the topologyτa−.

If π : L → L/F is the canonical morphism defined by π(x) = [x], then π is a continuous map. Indeed, if O is an open set with respect the topology τa−, then O = ∪{D_a(↑ x_i)/F| for some x_i/F ∊ L/F}, and π^-1) = π^-1(∪D_a(∧ x_i)/F) = ∪π^-1(D_a(↑ x_i)/F) = ∪D_a(π’¹ (↑x_i/F)), which is an openseton (L, τa).

Theorem 4.3

Let F be a filter of L and a, x ∊ L. Then:

(i) D_a(↑ x)/F ⊆ D_a/F(↑ x/F).

(ii) The topology τ_a/_F is finer than τa−, where τa− is the quotient topology of L/F.

Proof

(i). D_a/F(↑ x/F) = {u/F ∊ L/F\ x/F ≤ ((a/F)**)ⁿ → (u/F)**} = {u/F ∊ L/F| x → [(a**)ⁿ → u**] ∊ F} and D_a(↑ x)/F = {v/F : v ∊ D_a(↑ x)}. Let u/F ∊ D_a(↑ x)/F. Then there exists v ∊ D_a(↑ x) such that u/F = v/F and so x → [(a**)ⁿ → v**] = 1, for some n ∊ ℕ. Since u ≡_F v, then x → [(a**)ⁿ → u**] = x → [(a**)ⁿ → v**] = 1, that is x → [(a**)ⁿ → u**] ∊ F, hence u/F ∊ D_a/F(↑ x/F). Therefore, D_a(↑ x)/F ⊆ D_a/F(↑x/F).

(ii). By Proposition 3.24 the set {D_a/F (↑ x/F): x/F ∊ L/F} is a base for the topology τ_a/F on L/F. Moreover, the set {D_a (↑ x)/F: x ∊ L} is abase for the topology τa−. Now, the proof of (ii) is straightforward by (i).

Let L, L′ be residuated lattices. On L × L′ we consider the relation of order (x, y) ≤ (x′, y′) iff x ≤ x′ and y ≤ y′ and the operations

(x, y) ∧ (x′, y′) = (x ∧x′ , y∧ y′),

(x, y) ∨ (x′, y′) = (x ∨x′, y∨y′),

(x, y) ⊙ (x′ , y′) = (x ⊙ x′, y ⊙ y),

(x, y) → (x′ , y′) = (x → x′ , y→y′) for all x, y ∊ L and x′, y′ ∊ L .

Lemma 4.4

([17]). Let L, L′ be residuated lattices. Then K is a filter of L χ L′ iff there exist P ϵ F_i(L) and Q ϵ F_i (L′) such that K = P × Q.

Proof. “ ⇒ “. If K ϵ F_i(L × L′), we consider P = {x ϵ L : (x, x′) ϵ K for some x′ ϵ L′} and Q = {x′ ϵ L′ : (x, x′) ϵ K for some x ϵ L}.

Clearly, K = P × Q. Since (i, i) ϵ K we get i ϵ P. Let x,y ϵ P. Then there exist x′,y′ ϵ L′ such that (x, x′), (y, y′) ϵ K. Thus, (x, x′) ⊙ (y, y′) = (x ⊙ y, x′ ⊙ y′) ϵ K, so x ⊙ y ϵ P.

Consider x < y and x ϵ P. Then there exists x′ ϵ L′ such that (x, x′) ϵ K. Since (x, x′) < (y, x′), then y ϵ P, hence P ϵ F_i (L). Similarly, Q ϵ F_i(L′).

“ ⇐ ”=. Let K = P × Q for some P ϵ F_i(L′) and Q ϵ F_i(L′). Clearly, K ϵ L × L′. We consider (x, y), (p, q) ϵ K. Then x, p ϵ P and y, q ϵ Q, that is x ⊙ p ϵ P, y ⊙ q ϵ Q. Therefore, (x, y) ⊙ (p, q) = (x ⊙ p, y ⊙ q) ϵ P × Q = K.

Now, we consider (x, y) ϵ K such that (x, y) < (p, q). Then x < p with x ϵ P and y < q with y ϵ Q. Since P ϵ F_i(L) and Q ϵ F_i(L′), then p ϵ P and q ϵ Q, that is (p, q) ϵ K. Hence K ϵ F_i(L × L′).

Lemma 4.5

Let P, Q ϵ F_i(L) and a ϵ L. Then Da(P) × Da(Q) = Da(P × Q).

Proof

By Lemma 4.4, Da(P) × Da(Q) = {x ϵ L|aⁿ θ x ϵ P, for some n ϵ {y s L|a^m θ y ϵ Q, for some m ϵ ℕ} = {x × y ℕ L × L|(a^t θ x, a^t θ y) ϵ P × Q, for some t > max{m, n} ϵ ℕ} = D_a(P × Q).

Theorem 4.6

Let P, Q ϵ F_i (L) and a, b ϵ L. Then there exists a homeomorphism from L×LDa(P×Q) to LDa(P)×LDa(Q)

Proof

We define ϕ:L×L→LDa(P)×LDa(Q), by ϕ(x,y)=(xDa(P),yDa(Q)) Clearly, ϕ is onto. Let (x, y) ϵ L × L. Then (x, y) ϵ ker(ϕ) iff x/D_a (P) = 1/D_a (P) and yϵD_a(Q) = 1/D_a (Q) iff x ϵ D_a (P) and y ϵ D_a (Q). Hence ker(ϕ) = D_a(P) × Da(Q). It follows easily by Lemma 4.5 that D_a(P) × D_a(Q) = D_a(P × Q). Consider the map h:L×LDa(P×Q)→LDa(P)×LDa(Q),, defined by h((x,y)Da(P)×Da(Q))=(x/Da(P),y/Da(Q)) then by the first isomorphism theorem h becomes an isomorphism. If we suppose that X is an open subset of LDa(P)×LDa(Q) then there exist U, Vϵ τ_b open subsets such that X=LDa(P)×LDa(Q) Clearly, h−1=U×VDa(P×Q) is an open subset of L×LDa(P×Q) Hence h is a continuous map. In the same manner we can prove that h^-i is a continuous map. Therefore, h is a homeomorphism.