The category of algebraic L -closure systems

Abstract

In this paper, the concepts of algebraic L-closure system and algebraic L-closure operators are introduced from the fuzzy point of view. It is shown that the category of algebraic L-closure spaces is isomorphic to the category of algebraic L-closure system spaces. It is also proved that the category of algebraic L-closure system spaces is dual to the category algebraic L-lattices.

Keywords

Fuzzy set category

1 Introduction

Closure operators and closure systems play significant roles in several areas of classical mathematics, e.g., topology, algebra, lattice theory, logic, data mining, knowledge representation, and so on. In the classical order theory, they are particularly related to theoretical computer science and have a close relation with Galois connections [6, 7]. Since in any closure system as a complete lattice, join is usually not given by union, algebraic closure systems and algebraic closure operators emerge as the times require. In any algebraic closure system as an algebraic lattice, the join of any directed family is exactly given by set union. This solves the above problem in part.

On the other hand, since the partial order is an important mathematical structure and is useful in many areas, the generalization of the crisp partial order has been studied from the fuzzy point of view. In 1971, the fuzzy orderings was originally introduced by Zadeh [20]. Then, in order to fuzzify the fundamental theorem of concept lattices, the L-ordered set was introduced by B $\overset{ˇ}{e}$ lohl $\overset{´}{a}$ vek [2]. Recently, in order to study quantitative domain theory via fuzzy sets, a kind of L-orders was redefined and studied by Yao [17], where L is a complete residuated lattice. Indeed, it is routine to show that an L-order, in the sense of B $\overset{ˇ}{e}$ lohl $\overset{´}{a}$ vek and that in the sense of Yao are equivalent to each other.

On a set X and for a complete residuated lattice L, a mapping sub_X : L^X × L^X ⟶ L is defined by ${sub}_{X} (A, B) = ⋀_{x \in X} A (x) \to B (x)$

Then sub_X is an L-order on L^X, called the subsethood operator on X and (L^X, sub_X) is a complete L-lattice. In the complete L-lattice (L^X, sub_X), for any $𝔸 \in L^{L^{X}}$ , $⊓ 𝔸 = ⋀_{A \in L^{X}} 𝔸 (A) \to A$ and $⊔ 𝔸 = ⋀_{A \in L^{X}} 𝔸 (A) \land A$ , which are the generalization of set intersection and set union in crisp case.

On this special L-ordered set (L^X, sub_X), L-closure operators and L-closure systems were introduced and studied by B $\overset{ˇ}{e}$ lohl $\overset{´}{a}$ vek [1], which are the fuzzification of the corresponding notions on classical powersets. Moreover, the one-to-one correspondence between L-closure systems and L-closure operators was established. Since in the fuzzy setting, the monotony condition of L-closure operators may take several particular forms, B $\overset{ˇ}{e}$ lohl $\overset{´}{a}$ vek [4] proposed a new notion of L^*- closure operator which satisfies more general monotony condition than that in [1]. In addition, Georgescu and Popescu [8] presented analogous results in the non-commutative fuzzy framework. In another direction, fuzzy closure systems and fuzzy closure L-systems on general L-ordered sets were introduced [9], which can be viewed as the fuzzification of the corresponding notions on classical partially ordered sets and can also be seen as the generalization of the corresponding notions in [1]. Go on this work, fuzzy interior operators and fuzzy interior systems were investigated [3, 18]. It was also shown that the correspondence between fuzzy interior operators and fuzzy interior systems can be established analogously. Recently, Guo, Li and Zhang [10] proposed a new notion of L^∗-closure L-system which can be viewed as an alternative of L^∗-closure systems introduced by B $\overset{ˇ}{e}$ lohl $\overset{´}{a}$ vek [4], and B $\overset{ˇ}{e}$ lohl $\overset{´}{a}$ vek and Konecny [5] presented a decomposition theorem for a general L-closure operator and utilize it for computing generators and bases of the L-closure system.

From the previous works, easily discerned is the problem that the join of L-subset of any L-closure system when performed as a L-complete lattice is not the generalization of set union in crisp case. Then, can establishing a special kind of L-closure system similar to classical partially ordered sets solve the aforementioned problem?

The aim of this paper is to propose a special kind of L-closure operators--algebraic L-closure operators, and a special kind of L-closure system--algebraic L-closure system. Moreover, from category view, discuss the relationship between the category of algebraic L-closure operator spaces and algebraic L-closure system spaces. Additionally, establish the L-subset family representation for algebraic L-lattices by algebraic L-closure systems.

The contents are arranged as follows. Section 2 presents some preliminary concepts and properties. Section 3 propose the concepts of algebraic L-closure systems and algebraic L-closure operators. It is shown that there is a bijective correspondence between algebraic L-closure operators and algebraic L-closure systems and the categories of algebraic L-closure operator spaces and algebraic L-closure system spaces are isomorphic. Section 4, from L-order view, focuses on the relationship between algebraic L-closure systems and algebraic L-lattices. Finally, some conclusions are presented in Section 5.

2 Preliminaries

For the convenience of the reader, in this section, some basic concepts are reviewed. Because of the outstanding properties of the complete Heyting algebra, throughout this paper, it were used as the structures of truth values.

Definition 2.1. [2 , 21] An L-ordered set is a pair (X, e) such that X is a set and e : X × X ⟶ L is a mapping, called an L-order, that satisfies for any x, y, z ∈ X,

e (x, x) =1 (reflexivity);

e (x, y) ∧ e (y, z) ≤ e (x, z) (transitivity);

e (x, y) = e (y, x) =1 implies x = y (antisymmetry).

Given two L-ordered sets (X, e_X) and (Y, e_Y), a mapping f : X ⟶ Y is said to be an L-order preserving mapping if e_X (x, y) ≤ e_Y (f (x) , f (y)) for any x, y ∈ X.

Proposition 2.2. [2 , 21] Let (X, e) be an L-ordered set. Then

x₀ = ⊔ A iff e (x₀, y) = ⋀ _x∈XA (x) → e (x, y) for any y ∈ X;

x₀ = ⊓ A iff e (y, x₀) = ⋀ _x∈XA (x) → e (y, x) for any y ∈ X.

x ∈ X is called the minimal (maximal) element of A ∈ L^X if A (x) =1 and for any y ∈ X, A (y) ≤ e (x, y) (A (y) ≤ e (y, x)).

Let (X, e) be an L-ordered set and x ∈ X. ↓x ∈ L^X and ↑x ∈ L^X are defined as follows: ↓x (y) = e (y, x) , ↑ x (y) = e (x, y) for every y ∈ X. Moreover, define A^l, A^u ∈ Ł ^X as follows: A^l (x) = ⋀ _y∈XA (y) → e (x, y) , A^u (x) = ⋀ _y∈XA (y) → A (y, x) for any x ∈ X.

Definition 2.3. [11, 17] Let (X, e) be an L-ordered set. If an L-subset D of X satisfies

⋁_x∈XD (x) =1;

for any x, y ∈ X, D (x) ∧ D (y) ≤ ⋁ _z∈XD (z) ∧ e (x, z) ∧ e (y, z),

then it is called a directed L-subset in X. In this case, if it is a lower L-subset, then it is called an L-ideal in X. The set of all directed L-ordered subsets in X is denoted by

D (X)

and the set of all L-ideals in X is denoted by

I (X)

An L-ordered set (X, e) is called a directed complete L-ordered set if ⊔D exists for any $D \in D (X)$ . And it is called a complete L-lattice if for any A ∈ L^X, ⊔A exists.

Definition 2.4. [17] Let (X, e_X), (Y, e_Y) be directed complete L-ordered sets. A mapping f : X ⟶ Y is called an L-Scott continuous mapping if it satisfies f (⊔ D) = ⊔ f^→ (D) for any $D \in D (X)$ .

It is noteworthy that every L-Scott continuous mapping is L-order preserving.

Definition 2.5. [11, 17] Let (X, e) be a directed complete L-ordered set. For x, y ∈ X, define ⇓x ∈ L^X as $⇓ x (y) = ⋀_{I \in I (X)} e (x, ⊔ I) \to I (y) .$

Let (X, e) be a directed complete L-ordered set and x ∈ X, if ⇓x (x) =1 (or equivalently, I (x) = e (x, ⊔ I) for any $I \in I (X)$ ), then we call x a compact element in X, and we denote the set of all compact elements in X by $K (X)$ .

Definition 2.6. [14] Let (X, e) be a directed complete L-ordered set and x ∈ X, define K (x) ∈ L^X as $\forall y \in X, K (x) (y) = e (y, x) \land χ_{K (X)} (y) .$

If for any x ∈ X, $K (x) \in D (X)$ and x = ⊔ K (x), then (X, e) is called a algebraic directed complete L-ordered set. If, in addition, (X, e) is a complete L-lattice, then (X, e) is called an algebraic L-lattice.

Remark 2.7. It is noteworthy that an algebra of directed complete L-ordered sets defined in Definition 2.6 is equivalent to the algebra of Ω-categories given by Yao [19], that is, a directed complete L-ordered set (X, e) is algebraic iff for any x ∈ X, $k_{x} = ↓ x |_{K (X)} \in D (K (X))$ and ⊔k_x = x, where $k_{x} (y) = ↓ x |_{K (X)} (y) = e (y, x)$ , for any $y \in K (X)$ .

Let (X, e_X), (Y, e_Y) be L-ordered sets and f : X ⟶ Y be an ordinary mapping. One can define the forward fuzzy powerset operators f^→, f^← : L^X ⟶ L^Y as follows: $f^{\to} (A) (y) = ⋁_{f (x) = y} A (x), f^{\leftarrow} (B) (x) = B (f (x))$ for any A ∈ L^X, B ∈ L^Y and any y ∈ Y, x ∈ X.

Definition 2.8. [11, 16] Let (X, e_X) , (Y, e_Y) be two L-ordered sets and f : (X, e_X) → (Y, e_Y) , g : (Y, e_Y) → (X, e_X) two L-order preserving mappings. The pair (f, g) is called an L-adjunction between X and Y provided that $e_{Y} (f (x), y) = e_{X} (x, g (y))$ for all x ∈ X and all y ∈ Y, where f is called the left L-adjoint of g and dually g the right L-adjoint of f.

Definition 2.9. [12, 13] Let (X, e) be an L-ordered set. Then

(X, e) is tensored in the sense that for any a ∈ L, x ∈ X, there is an element a ⊗ x ∈ X, called the tensor of a with x, such that for any y ∈ X, e (a ⊗ x, y) = a → e (x, y);

(X, e) is cotensored in the sense that for any a ∈ L, x ∈ X, there is an element arightarrowtailx ∈ X, called the cotensor of a with x, such that for any y ∈ X, e (y, arightarrowtailx) = a → e (y, x).

Proposition 2.10. [12, 13] Let (X, e) be a complete L-lattice, then for an L-subset A ∈ L^X, ⊔A = ⋁ _x∈XA (x) ⊗ x and ⊓A = ⋀ _x∈XA (x) rightarrowtailx.

Definition 2.11. [1] An L-closure operator on a set X is a mapping C : L^X ⟶ L^X satisfying

A ≤ C (A) for any A ∈ L^X;

sub_X (A, B) ≤ sub_X (C (A) , C (B)) for any A, B ∈ L^X;

C (A) = C (C (A)) for any A ∈ L^X.

Definition 2.12. [1] Let X be a set. A system $T \subseteq L^{X}$ is called an L-closure system on X if for each A ∈ L^X, $⋀_{B \in T} {sub}_{X} (A, B) \to B \in T .$

Remark 2.13. (1) Let X be a set, $T \subseteq L^{X}$ , then $(T, {sub}_{X})$ is an L-ordered set. Moreover, from the view of L-order, $T \subseteq L^{X}$ is an L-closure system on X is equivalent to that for any $𝔸 \in L^{T}$ , $⊓ i^{\to} (𝔸) \in T$ , where $i : T ⟶ L^{X}$ is a include mapping, i.e., for any $A \in T, i (A) = A$ .

(2) From the definition of the L-closure system, it is easy to check that for any L-closure system $T$ , $1_{X} \in T$ .

3 Algebraic L-closure operators and algebraic L-closure systems

As referred to in the introduction, for any L-closure system as a complete L-lattice, the join of any L-subset of it is not the generalization of set union in crisp case. Therefore, in this section, we introduce the algebraic L-closure operators and the algebraic L-closure systems from L-order view.

Definition 3.1. Let X be a set, $T \subseteq L^{X}$ be an L-closure system on X. Then $T$ is said to be an algebraic L-closure system on X, if for any $𝔻 \in D (T)$ , $⊔ i^{\to} (𝔻) \in T$ .

Remark 3.2. For a set X, it is easy to check that

an algebraic L-closure system $T \subseteq L^{X}$ is also an algebraic closure fuzzy set system given in [15], that is, it is closed under (point-wise) joins of upward directed families of members of $T$ , i.e., for any upward directed families ${A_{i}}_{i \in I} \subseteq T$ , $⋁_{i \in I} A_{i} \in T$ ;

an L-closure system $T$ is algebraic iff for any $𝕀 \in I (T)$ , $⊔ i^{\to} (𝕀) \in T$ .

By definition of algebraic L-closure system, we have that join of any directed L-subset of algebraic L-closure systems is exact the generalization of set union in crisp setting. This partially addresses the question raised in Introduction.

Example 3.3. Let X = {1, 2, 3}, L = {0, 1}.

Obviously, L^X = {φ₁, φ₂, φ₃, φ₄, φ₅, φ₆, φ₇, φ₈}, where φ₁ = (0, 0, 0), φ₂ = (1, 0, 0), φ₃ = (0, 1, 0), φ₄ = (0, 0, 1), φ₅ = (1, 1, 0), φ₆ = (1, 0, 1), φ₇ = (0, 1, 1), φ₈ = (1, 1, 1). Thus sub_X : L^X × L^X ⟶ L as follows: ${sub}_{X} = ({sub}_{X} (φ_{i}, φ_{j})) = [\begin{matrix} 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ 0 & 1 & 0 & 0 & 1 & 1 & 0 & 1 \\ 0 & 0 & 1 & 0 & 1 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 & 0 & 1 & 1 & 1 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \end{matrix}] .$

Let $T = {φ_{1}, φ_{2}, φ_{3}, φ_{5}, φ_{8}}$ , then all L-subsets of $T$ are as follows: Φ₁ = (0, 0, 0, 0, 0), Φ₂ = (1, 0, 0, 0, 0), Φ₃ = (0, 1, 0, 0, 0), Φ₄ = (0, 0, 1, 0, 0), Φ₅ = (0, 0, 0, 1, 0), Φ₆ = (0, 0, 0, 0, 1), Φ₇ = (1, 1, 0, 0, 0), Φ₈ = (1, 0, 1, 0, 0), Φ₉ = (1, 0, 0, 1, 0), Φ₁₀ = (1, 0, 0, 0, 1), Φ₁₁ = (0, 1, 1, 0, 0), Φ₁₂ = (0, 1, 0, 1, 0), Φ₁₃ = (0, 1, 0, 0, 1), Φ₁₄ = (0, 0, 1, 1, 0), Φ₁₅ = (0, 0, 1, 0, 1), Φ₁₆ = (0, 0, 0, 1, 1), Φ₁₇ = (1, 1, 1, 0, 0), Φ₁₈ = (1, 1, 0, 1, 0), Φ₁₉ = (1, 1, 0, 0, 1), Φ₂₀ = (1, 0, 1, 1, 0), Φ₂₁ = (1, 0, 1, 0, 1), Φ₂₂ = (1, 0, 0, 1, 1), Φ₂₃ = (0, 1, 1, 1, 0), Φ₂₄ = (0, 1, 1, 0, 1), Φ₂₅ = (0, 1, 0, 1, 1), Φ₂₆ = (0, 0, 1, 1, 1), Φ₂₇ = (0, 1, 1, 1, 1), Φ₂₈ = (1, 0, 1, 1, 1), Φ₂₉ = (1, 1, 0, 1, 1), Φ₃₀ = (1, 1, 1, 0, 1), Φ₃₁ = (1, 1, 1, 1, 0), Φ₃₂ = (1, 1, 1, 1, 1). It is easy to check that $I (T) = {Φ_{2}, Φ_{7}, Φ_{8}, Φ_{31}, Φ_{32}}$ and $⊔ i^{\to} (Φ_{2}) = φ_{1} \in T$ , $⊔ i^{\to} (Φ_{7}) = φ_{2} \in T$ , $⊔ i^{\to} (Φ_{8}) = φ_{3} \in T$ , $⊔ i^{\to} (Φ_{31}) = φ_{5} \in T$ , $⊔ i^{\to} (Φ_{32}) = φ_{8} \in T$ , which implies that $T$ is an an algebraic L-closure system on X. Moreover, we can verify $(T, {sub}_{X})$ is a complete L-lattice, specific as follows: ⊔Φ₁ = ⊔ Φ₂ = φ₁, ⊔Φ₃ = ⊔ Φ₇ = φ₂, ⊔Φ₄ = ⊔ Φ₈ = φ₃, ⊔Φ₅ = ⊔ Φ₉ = ⊔ Φ₁₁ = ⊔ Φ₁₂ = ⊔ Φ₁₄ = ⊔ Φ₁₇ = ⊔ Φ₁₈ = ⊔ Φ₂₀ = ⊔ Φ₂₃ = ⊔ Φ₃₁ = φ₅, ⊔Φ₆ = ⊔ Φ₁₀ = ⊔ Φ₁₃ = ⊔ Φ₁₅ = ⊔ Φ₁₆ = ⊔ Φ₁₉ = ⊔ Φ₂₁ = ⊔ Φ₂₂ = ⊔ Φ₂₄ = ⊔ Φ₂₅ = ⊔ Φ₂₆ = ⊔ Φ₂₇ = ⊔ Φ₂₈ = ⊔ Φ₂₉ = ⊔ Φ₃₀ = ⊔ Φ₃₂ = φ₈.

Example 3.4. (1) Given an L-ordered set (X, e), then the set of all L-ideals $I (X)$ is an algebraic L-closure system on X.

(2) The set $T$ of all fuzzy equivalence relations on a non-empty set X is an algebraic L-closure system on X × X. A fuzzy equivalence θ on X if and only if θ (x, x) =1, θ (x, y) = θ (y, x) , θ (x, y) ∧ θ (y, Z) ≤ θ (x, z) for all x, y, z ∈ X.

(3) Let (X, e) be an L-ordered set, then the set of all L-Frink ideals, denoted $FI (X)$ , is an algebraic L-closure system on X. An L-subset J ∈ L^X is a Frink L-ideal if for any F ∈ K (L^X), sub_X (F, J) = sub_X (F^ul, J). The detailed proof is as follows:

Proof. Since for any $𝔸 \in L^{FI (X)}$ , $⊓ 𝔸 = ⋀_{A \in FI (X)} 𝔸 (A) \to A,$ and for any $F \in K (L^{X})$ , $\begin{matrix} {sub}_{X} (F, ⊓ 𝔸) \\ = ⋀_{x \in X} F (x) \to ⊓ 𝔸 (x) \\ = ⋀_{x \in X} F (x) \to (⋀_{J \in FI (X)} 𝔸 (J) \to J (x)) \\ = ⋀_{J \in FI (X)} 𝔸 (J) \to {sub}_{X} (F, J) \\ = ⋀_{J \in FI (X)} 𝔸 (J) \to {sub}_{X} (F^{ul}, J) \\ = ⋀_{J \in FI (X)} 𝔸 (J) \to (⋀_{x \in X} F^{ul} (x) \to J (x)) \\ = ⋀_{x \in X} F^{ul} (x) \to (⋀_{J \in FI (X)} 𝔸 (J) \to J (x)) \\ = {sub}_{X} (F^{ul}, ⊓ 𝔸), \end{matrix}$ which implies that $⊓ 𝔸 \in FI (X)$ and thus $(FI (X),$ sub_X) is a complete L-lattice. Moreover, it is easy to check that for any $𝔸 \in L^{FI (X)}$ and any $𝔻 \in D (FI (X))$ , $⊓ 𝔸 = ⊓ i^{\to} (𝔸)$ and $⊔ 𝔻 = ⊔ i^{\to} (𝔻)$ , which implies that $⊓ i^{\to} (𝔸), ⊔ i^{\to} (𝔻) \in FI (X)$ by the completeness of $(FI (X), {sub}_{X})$ . Therefore, $FI (X)$ is an algebraic L-closure system.

Note that for any x ∈ X, (↓ x) ^ul = ↓ x, i.e., $↓ x \in FI (X)$ . Then we have the following embedding theorem.

Theorem 3.5. Let (X, e) be an L-ordered set and define $↓ : X ⟶ FI (X)$ as x ↦ ↓ x for any x ∈ X. Then $FI (X)$ is a completion of X via the mapping ↓.

The above theorems tell us that every L-ordered set can be embedded into an algebraic L-closure system by the Yoneda embedding.

Next, we introduce the definition of algebraic L-closure operator.

Definition 3.6. Let C be an L-closure operator on a set X. Then C is called an algebraic L-closure operator, if for any $𝔻 \in D (L^{X})$ , $C (⊔ 𝔻) = ⊔ C^{\to} (𝔻)$ holds.

Remark 3.7. (1) It is easy to check that C is an algebraic L-closure operator iff for any $𝕀 \in I (L^{X})$ , $C (⊔ 𝕀) = ⊔ C^{\to} (𝕀)$ holds.

(2) If an L-closure operator C is algebraic, then C (A) = ⋁ _B∈L^Xsub_X (B, A) ∧ C (B) for any A ∈ L^X.

Example 3.8. For an L-ordered set (X, e), an operator ↓ : L^X ⟶ L^X defines as follows: for any A ∈ L^X, ↓A (x) = ⋁ _y∈XA (y) ∧ e (x, y) , ∀ x ∈ X, we can easy check that the operator ↓ is an algebraic L-closure operator.

Theorem 3.9. Let C be an algebraic L-closure operator on X and $T$ be an algebraic L-closure system on X. Then $T_{C}$ is an algebraic L-closure system on X and $C_{T}$ is an algebraic L-closure operator on X. Moreover, $C = C_{T_{C}}$ and $T = T_{C_{T}}$ .

Proof. We only need to show that $T_{C}$ is algebraic and $C_{T}$ is algebraic by Theorem 3.4 in [1]. Since for any $𝕀 \in I (T_{C})$ and any x ∈ X, we have $i^{\to} (𝕀) \in I (L^{X})$ and $\begin{matrix} C (⊔ i^{\to} (𝕀)) (x) & = & ⊔ (C \circ i)^{\to} (𝕀)) (x) \\ = & ⋁_{B \in L^{X}} ⋁_{A \in T_{C}, C (A) = B} 𝕀 (A) \land B (x) \\ = & ⋁_{A \in T_{C}} 𝕀 (A) \land C (A) (x) = ⊔ i^{\to} (𝕀) (x), \end{matrix}$ $T_{C}$ is algebraic by Remark 3.7 (1). Next, we show that $C_{T}$ is algebraic, i.e., for any $𝔻 \in D (L^{X})$ , $C_{T} (⊔ 𝔻) = ⊔ C_{T}^{\to} (𝔻)$ . In fact, $\begin{matrix} {sub}_{X} (⊔ C_{T}^{\to} (𝔻), C_{T} (⊔ 𝔻)) \\ = ⋀_{A \in L^{X}} C_{T}^{\to} (𝔻) (A) \to {sub}_{X} (A, C_{T} (⊔ 𝔻)) \\ = ⋀_{A \in L^{X}} 𝔻 (A) \to {sub}_{X} (C_{T} (A), C_{T} (⊔ 𝔻)) \\ \geq ⋀_{A \in L^{X}} 𝔻 (A) \to {sub}_{X} (A, ⊔ 𝔻) \\ = {sub}_{X} (⊔ 𝔻, ⊔ 𝔻) = 1, \end{matrix}$ which implies that $⊔ C_{T}^{\to} (𝔻) \leq C_{T} (⊔ 𝔻)$ . As for the reverse, we first show that $⊔ C_{T}^{\to} (𝔻) \in T$ . Define ${C_{T}}_{o} : L^{X} ⟶ T$ by ${C_{T}}_{o} (A) = C_{T} (A)$ for any A ∈ L^X. Then, we have $C_{T} = i \circ {C_{T}}_{o}$ and so $⊔ C_{T}^{\to} (𝔻) = ⊔ (i \circ {C_{T}}_{o})^{\to} (𝔻) = ⊔ i^{\to} ({C_{T}}_{o}^{\to} (𝔻))$ . As ${C_{T}}_{o}^{\to} (𝔻) \in D (T)$ and thus $⊔ i^{\to} ({C_{T}}_{o}^{\to} (𝔻)) \in D (T)$ , i.e., $⊔ C_{T}^{\to} (𝔻) \in D (T)$ . Moreover, since $\begin{matrix} ⊔ 𝔻 & = & ⋁_{A \in L^{X}} 𝔻 (A) \land A \\ \leq & ⋁_{A \in L^{X}} 𝔻 (A) \land C_{T} (A) = ⊔ C_{T}^{\to} (𝔻), \end{matrix}$ we have $C_{T} (⊔ 𝔻) \leq C_{T} (⊔ C_{T}^{\to} (𝔻)) = ⊔ C_{T}^{\to} (𝔻) .$

Next, two categories are introduced, called the category of algebraic L-closure spaces and the category of algebraic L-closure systems paces, respectively. An algebraic L-closure space is a pair (X, C) consisting of a set X and an algebraic L-closure closure C on X. A mapping f : X ⟶ Y is called an L-continuous mapping between algebraic L-closure spaces (X, C_X) and (Y, C_Y) if f^→ (C_X (A)) ≤ C_Y (f^→ (A)). Alternatively, one may introduce algebraic L-closure system spaces as pairs $(X, T)$ where $T$ is an L-closure system on X. A mapping f : X ⟶ Y is called an L-continuous mapping between algebraic L-closure system spaces $(X, T_{X})$ and $(Y, T_{Y})$ if $f^{\leftarrow} (B) \in T_{X}$ for any $B \in T_{Y}$ . Then we can obtain the following conclusions.

Proposition 3.10. (1) Algebraic L-closure spaces together with L-continuous mappings as morphisms form a category AL - CSP.

(2) Algebraic L-closure system spaces together with L-continuous mappings as morphisms form a category AL - CSSP.

Proof. (1) Let f ∈ Hom_AL-CSP (X, Y) , g ∈ Hom_AL-CSP (Y, Z), and Id_X : X ⟶ X is an identity mapping. Then, for any A ∈ L^X and x ∈ X, $\begin{matrix} (g \circ f)^{\to} (C_{X} (A)) & = & g^{\to} (f^{\to} (C_{X} (A))) \\ \leq & g^{\to} (C_{Y} (f^{\to} (A))) \\ \leq & C_{Z} (g^{\to} (f^{\to} (A))) \\ = & C_{Z} ((g \circ f)^{\to} (A)), \end{matrix}$ which implies g ∘ f ∈ Hom_L-CSP (X, Z) and $\begin{matrix} {Id}_{X}^{\to} (C_{X} (A)) (x) & = & ⋁_{{Id}_{X} (x^{'}) = x} C_{X} (A) (x^{'}) \\ = & C_{X} (A) (x) = C_{X} ({Id}_{X}^{\to} (A)) (x) \end{matrix}$

(2) Let f ∈ Hom_AL-CSSP (X, Y) , g ∈ Hom_AL-CSSP (Y, Z), and Id_X : X ⟶ X is an identity mapping. Then, for any $B \in T_{Z}$ , by (g ∘ f) ^← (B) = f^← (g^← (B)) and the L-continuity of f, g, we have $(g \circ f)^{\leftarrow} (B) \in T_{X}$ , which implies g ∘ f ∈ Hom_AL-CSSP (X, Z). Moreover, for any $A \in T_{X}$ and x ∈ X, we have ${Id}_{X}^{\leftarrow} (A) (x) = A ({Id}_{X} (x)) = A (x)$ .

Proposition 3.11. (1) For all (X, C_X) , (Y, C_Y) ∈ ob (AL - CSP) and f ∈ Hom_AL-CSP (X, Y), let $𝕂 (X) = (X, T_{C_{X}})$ and $𝕂 (f) = f$ . Then $𝕂$ is a functor from the category AL - CSP to the category AL - CSSP, i.e., the following diagram.

(2) For all $(X, T_{X}), (Y, T_{Y}) \in ob (AL - CSSP)$ and f ∈ Hom_AL-CSSP (X, Y), let $ℍ (X) = (X, C_{T_{X}})$ and $ℍ (f) = f$ . Then $ℍ$ is a functor from the category AL - CSSP to the category AL - CSP, i.e., the following diagram.

Proof. (1) By Theorem 3.9 and Proposition 3.10, we have that $T_{C_{X}} \in ob (AL - CSSP)$ and $𝕂$ preserves the composite of morphisms and the identity. Let $B \in T_{C_{Y}}$ and y ∈ Y, then $\begin{matrix} f^{\to} (C_{X} (f^{\leftarrow} (B))) (y) & \leq & C_{Y} (f^{\to} (f^{\leftarrow} (B))) (y) \\ = & C_{Y} (B) (y) = B (y), \end{matrix}$ and so $\begin{matrix} f^{\leftarrow} (B) (x) & = & B (f (x)) \geq f^{\to} (C_{X} (f^{\leftarrow} (B))) (f (x)) \\ = & ⋁_{f (x^{'}) = f (x)} C_{X} (f^{\leftarrow} (B)) (x^{'}) \\ \geq & C_{X} (f^{\leftarrow} (B)) (x) . \end{matrix}$

Therefore, we have f^← (B) = C_X (f^← (B)), i.e., $f^{\leftarrow} (B) \in T_{C_{X}}$ which implies that $𝕂 (f) = f \in {Hom}_{AL - CSSP} (X, Y)$ .

(2) Similar to (1), we only need to show that for any f ∈ Hom_AL-CSSP, $ℍ (f) = f \in {Hom}_{AL - CSP} (X, Y)$ . In fact, for any A ∈ L^X and y ∈ Y, $\begin{matrix} C_{T_{Y}} (f^{\to} (A)) (y) \\ = ⋀_{B \in T_{Y}} {sub}_{Y} (f^{\to} (A), B) \to B (y) \\ = ⋀_{B \in T_{Y}} (⋀_{y^{'} \in Y} f^{\to} (A) (y^{'}) \to B (y^{'})) \to B (y) \\ = ⋀_{B \in T_{Y}} (⋀_{x \in X} A (x) \to B (f (x))) \to B (y) \\ = ⋀_{B \in T_{Y}} {sub}_{X} (A, f^{\leftarrow} (B)) \to B (y) \end{matrix}$ and so $\begin{matrix} f^{\to} (C_{T_{X}} (A)) (y) = ⋁_{f (x) = y} C_{T_{X}} (A) (x) \\ = ⋁_{f (x) = y} ⋀_{A^{'} \in T_{X}} {sub}_{X} (A, A^{'}) \to A^{'} (x) \\ \leq ⋁_{f (x) = y} ⋀_{B \in T_{Y}} {sub}_{X} (A, f^{\leftarrow} (B)) \to f^{\leftarrow} (B) (x) \\ \leq ⋀_{A^{'} \in T_{X}} ⋁_{f (x) = y} {sub}_{X} (A, f^{\leftarrow} (B)) \to f^{\leftarrow} (B) (x) \\ \leq ⋀_{A^{'} \in T_{X}} {sub}_{X} (A, f^{\leftarrow} (B)) \to (⋁_{f (x) = y} f^{\leftarrow} (B) (x)) \\ = ⋀_{B \in T_{Y}} {sub}_{X} (A, f^{\leftarrow} (B)) \to B (y) \\ \leq C_{T_{Y}} (f^{\to} (A)) (y) . \end{matrix}$

By Theorem 3.9 and Proposition 3.11, we obtain the following result.

Theorem 3.12. The categories AL - CSP and AL - CSSP are isomorphic.

It is noteworthy that $(𝕂, ℍ)$ is a adjoint functor, and so there is a Galois correspondence between the category of algebraic L-closure spaces and that of algebraic L-closure system spaces.

4 The relationship between the category of algebraic L-closure system spaces and that of algebraic L-lattices

In order to go to develop the notions of algebraic L-closure system and algebraic L-closure operator and investigate the relationship between the category of algebraic L-closure system spaces and that of algebraic L-lattices, we propose the following lemma.

Lemma 4.1. Let X be a set. Then

$B_{1} \lor B_{2} \in K (L^{X})$ for any $B_{1}, B_{2} \in K (L^{X})$ ;

$0_{X} \in K (L^{X})$ , where 0_X (x) =0 for any x ∈ X;

∀x ∈ X, $1_{x} \in K (L^{X})$ , where 1_x ∈ L^X is defi- ned by 1_x (y) =1 if y = x, and otherwise, 1_x (y) =0;

(L^X, sub_X) is an algebraic L-lattice.

Proof. (1) We only need to show that ${sub}_{X} (B_{1} \lor B_{2}, ⊔ 𝕀) \leq 𝕀 (B_{1} \lor B_{2})$ for any $𝕀 \in I (L^{X})$ . In fact, $\begin{matrix} {sub}_{X} (B_{1} \lor B_{2}, ⊔ 𝕀) \\ = {sub}_{X} (B_{1}, ⊔ 𝕀) \land {sub}_{X} (B_{2}, ⊔ 𝕀) \\ = 𝕀 (B_{1}) \land 𝕀 (B_{2}) = (⋁_{B \in L^{X}} 𝕀 (B) \land {sub}_{X} (B_{1}, B)) \\ \land (⋁_{C \in L^{X}} 𝕀 (C) \land {sub}_{X} (B_{2}, C)) \\ = ⋁_{B, C \in L^{X}} 𝕀 (B) \land 𝕀 (C) \land {sub}_{X} (B_{2}, C) \\ \land {sub}_{X} (B_{1}, B) \\ \leq ⋁_{B, C, D \in L^{X}} 𝕀 (D) \land {sub}_{X} (B, D) \land {sub}_{X} (C, D) \\ \land {sub}_{X} (B_{2}, C) \land {sub}_{X} (B_{1}, B) \\ \leq ⋁_{D \in L^{X}} 𝕀 (D) \land {sub}_{X} (B_{2}, D) \land {sub}_{X} (B_{1}, D) \\ = ⋁_{D \in L^{X}} 𝕀 (D) \land {sub}_{X} (B_{1} \lor B_{2}, D) \\ = 𝕀 (B_{1} \lor B_{2}) . \end{matrix}$

(2) Indeed, for any $𝕀 \in I (L^{X})$ , we have that $\begin{matrix} 𝕀 (0_{X}) & = & ⋁_{B \in L^{X}} 𝕀 (B) \land {sub}_{X} (0_{X}, B) \\ = & 1 = {sub}_{X} (0_{X}, ⊔ 𝕀) . \end{matrix}$

(3) In fact, for any $𝕀 \in I (L^{X})$ , we have that $\begin{matrix} 𝕀 (1_{x}) & = & ⋁_{B \in L^{X}} 𝕀 (B) \land {sub}_{X} (1_{x}, B) \\ = & ⋁_{B \in L^{X}} 𝕀 (B) \land B (x) \\ = & ⊔ 𝕀 (x) = {sub}_{X} (1_{x}, ⊔ 𝕀) . \end{matrix}$

(4) Firstly, we show that for any B ∈ L^X, $K (B) \in D (L^{X})$ . In fact, $\begin{matrix} ⋁_{B^{'} \in L^{X}} K (B) (B^{'}) & = & ⋁_{B^{'} \in K (L^{X})} {sub}_{X} (B^{'}, B) \\ \geq & {sub}_{X} (0_{X}, B) = 1 \end{matrix}$ and by (1), we have, for any B₁, B₂ ∈ K (L^X), $\begin{matrix} {sub}_{X} (B_{1}, B) \land {sub}_{X} (B_{2}, B) \\ = {sub}_{X} (B_{1} \lor B_{2}, B) \\ \land {sub}_{X} (B_{1}, B_{1} \lor B_{2}) \land {sub}_{X} (B_{2}, B_{1} \lor B_{2}) \\ \leq ⋁_{B^{'} \in L^{X}} K (B) (B^{'}) \land {sub}_{X} (B_{1}, B^{'}) \\ \land {sub}_{X} (B_{2}, B^{'}) . \end{matrix}$

Second, we show that ⊔K (B) = B. Indeed, for any x ∈ X, $⊔ K (B) (x) = ⋁_{B^{'} \in K (L^{X})} {sub}_{X} (B^{'}, B) \land B^{'} (x) \leq B (x)$ and on the other hand, $\begin{matrix} ⊔ K (B) (x) & = & ⋁_{B^{'} \in K (L^{X})} {sub}_{X} (B^{'}, B) \land B^{'} (x) \\ \geq & {sub}_{X} (1_{x}, B) \land 1_{x} (x) = B (x) . \end{matrix}$

Theorem 4.2. Let C be an L-closure operator on a set X. Then the following are equivalent:

C is an algebraic L-closure operator;

for any A ∈ L^X and any x ∈ X, C (A) (x) = ⋁ _B∈L^XK (A) (B) ∧ C (B) (x).

Proof. (1)⇒(2): For any A ∈ L^X, $K (A) \in D (L^{X})$ and ⊔K (A) = A by the algebra of (L^X, sub_X). Then, for any x ∈ X, $\begin{matrix} C (A) (x) & = & C (⊔ K (A)) (x) = ⊔ C^{\to} (K (A)) (x) \\ = & ⋁_{B \in L^{X}} C^{\to} (K (A)) (B) \land B (x) \\ = & ⋁_{B \in L^{X}} ⋁_{C (B^{'}) = B} K (A) (B^{'}) \land B (x) \\ = & ⋁_{B \in L^{X}} K (A) (B) \land C (B) (x) . \end{matrix}$

(2)⇒(1): Let $𝕀 \in I (L^{X})$ , then it follows that $\begin{matrix} {sub}_{X} (⊔ C^{\to} (𝕀), C (⊔ 𝕀)) \\ = ⋀_{A \in L^{X}} C^{\to} (𝕀) (A) \to {sub}_{X} (A, C (⊔ 𝕀)) \\ = ⋀_{A \in L^{X}} 𝕀 (A) \to {sub}_{X} (C (A), C (⊔ 𝕀)) \\ \geq ⋀_{A \in L^{X}} 𝕀 (A) \to {sub}_{X} (A, ⊔ 𝕀) = 1, \end{matrix}$ which implies that $⊔ C^{\to} (𝕀) \leq C (⊔ 𝕀)$ . For the reverse inequality, we have that for any x ∈ X, $\begin{matrix} C (⊔ 𝕀) (x) & = & ⋁_{B \in L^{X}} K (⊔ 𝕀) (B) \land C (B) (x) \\ = & ⋁_{B \in K (L^{X})} {sub}_{X} (B, ⊔ 𝕀) \land C (B) (x) \\ = & ⋁_{B \in K (L^{X})} 𝕀 (B) \land C (B) (x) \\ \leq & ⋁_{B \in L^{X}} 𝕀 (B) \land C (B) (x) \\ = & ⊔ C^{\to} (𝕀) (x) . \end{matrix}$

Lemma 4.3. Let C be an algebraic L-closure operator on a set X and let $T_{C}$ be the associated algebraic L-closure system. Then $(T_{C}, {sub}_{X})$ is algebraic L-lattice in which an element A is compact if and only if A = C (B) for some compact element B in (L^X, sub_X).

Proof. As $T_{C}$ is a complete L-lattice, we only need to verify that $(T_{C}, {sub}_{X})$ is algebraic. We first show that A is compact in $(T_{C}, {sub}_{X})$ if and only if A = C (B) for some compact element B in (L^X, sub_X). Let B is a compact element in (L^X, sub_X) and let A = C (B). For any $𝕀 \in I (T_{C})$ , $\begin{matrix} {sub}_{X} (A, ⊔ 𝕀) & = & {sub}_{X} (A, ⊔ i^{\to} (𝕀) \\ = & {sub}_{X} (A, C (⊔ i^{\to} (𝕀)) \\ = & {sub}_{X} (A, ⊔ (C \circ i)^{\to} (𝕀)) \\ \leq & {sub}_{X} (B, ⊔ (C \circ i)^{\to} (𝕀)) \\ = & (C \circ i)^{\to} (𝕀) (B) \\ = & ⋁_{B^{'} \in T_{C}, C (B^{'}) = B} 𝕀 (B^{'}) \\ = & 𝕀 (C (B)), \end{matrix}$ which implies that A is compact element in $T_{C}$ .

Conversely, assume that $A \in T_{C}$ is a compact element. Certainly, $\begin{matrix} 1 & = & {sub}_{X} (A, C (A)) \\ = & {sub}_{X} (A, ⋁_{B \in L^{X}} K (A) (B) \land C (B)) \\ = & {sub}_{X} (A, ⊔ C^{o \to} (K (A)) \\ = & C^{o \to} (K (A)) (A) \\ = & ⋁_{B \in K (L^{X}), C (B) = A} {sub}_{X} (B, A), \end{matrix}$ which implies that there exists $B \in K (L^{X})$ such that C (B) = A.

Next, we show that $(T_{C}, {sub}_{X})$ is algebraic. Indeed, for any $A, B_{1}, B_{2} \in T_{C}$ , $\begin{matrix} ⋁_{B \in T_{C}} K (A) (B) & = & ⋁_{B \in K (T_{C})} {sub}_{X} (B, A) \\ = & ⋁_{B \in K (L^{X})} {sub}_{X} (C (B), C (A)) \\ \geq & ⋁_{B \in K (L^{X})} {sub}_{X} (B, A) = 1, \end{matrix}$ $\begin{matrix} K (A) (B_{1}) \land K (A) (B_{2}) \\ \leq ⋁_{B \in L^{X}} K (A) (B) \land {sub}_{X} (B_{1}, B) \land {sub}_{X} (B_{2}, B) \\ \leq ⋁_{B \in L^{X}} K (A) (C (B)) \land {sub}_{X} (B_{1}, C (B)) \\ \land {sub}_{X} (B_{2}, C (B)) \\ = ⋁_{B \in T_{C}} K (A) (B) \land {sub}_{X} (B_{1}, B) \land {sub}_{X} (B_{2}, B) . \end{matrix}$

Therefore, $K (A) \in D (T_{C})$ . Moreover, by the algebra of (L^X, sub_X), we have, for any $B \in T_{C}$ , $\begin{matrix} {sub}_{X} (A, B) \\ = ⋀_{B^{'} \in L^{X}} K (A) (B^{'}) \to {sub}_{X} (B^{'}, B) \\ \geq ⋀_{B^{'} \in L^{X}} K (A) (C (B^{'})) \to {sub}_{X} (B^{'}, B) \\ \geq ⋀_{B^{'} \in L^{X}} K (A) (C (B^{'})) \to {sub}_{X} (C (B^{'}), B) \\ = ⋁_{B \in T_{C}} K (A) (B^{'}) \land {sub}_{X} (B^{'}, B) \end{matrix}$ and K (A) (B) ≤ sub_X (B, A). Hence, we have ⊔K (A) = A.

Theorem 4.4.(1) Let $T$ be an algebraic L-closure system on a set X. Then $(T, {sub}_{X})$ is an algebraic L-lattice.

(2) Let (X, e) be an algebraic L-lattice and $T = {k_{x} | x \in X}$ , then $T$ is an algebraic L-closure system and $(T, {sub}_{K (X)}) ≅ (X, e)$ .

Proof. Part (1) follows from the preceding lemma. For the part (2), we first show that $T$ is an L-closure system. In fact, let $𝔸 \in L^{T}$ and let $A (x) = 𝔸 (k_{x})$ for any x ∈ X, then A ∈ L^X and ⊓A exists by the completeness of (X, e). Moreover, for any $x \in K (X)$ , $\begin{matrix} ⊓ i^{\to} (𝔸) (x) \\ = ⋀_{B \in L^{K (X)}} i^{\to} (𝔸) (B) \to B (x) \\ = ⋀_{y \in X} 𝔸 (k_{y}) \to k_{y} (x) = ⋀_{y \in X} A (y) \to k_{y} (x) \\ = ⋀_{y \in X} A (y) \to e (x, y) = e (x, ⊓ A) = k_{⊓ A} (x) . \end{matrix}$

Next, we prove that the map f : x ↦ k_x is an isomorphism of (X, e) onto $(T, {sub}_{K (X)})$ and that $T$ is algebraic. Because (X, e) is algebraic, $e (x, y) = {sub}_{K (X)} (k_{x}, k_{y})$ . Therefore, f is an L-isomorphism.

Let $𝕀 \in I (T)$ and let $I (x) = 𝕀 (k_{x})$ , then $I \in I (X)$ . We claim that $⊔ i^{\to} (𝕀) = k_{⊔ I}$ and so belongs to $T$ . Indeed, for any $x \in K (X)$ , it holds that $\begin{matrix} ⊔ i^{\to} (𝕀) (x) & = & ⋁_{B \in L^{K (X)}} i^{\to} (𝕀) (B) \land B (x) \\ = & ⋁_{B \in L^{K (X)}} ⋁_{k_{y} = B} 𝕀 (k_{y}) \land B (x) \\ = & ⋁_{y \in X} 𝕀 (k_{y}) \land k_{y} (x) \\ = & ⋁_{y \in X} I (y) \land e (x, y) \\ = & I (x) = e (x, ⊔ I) = k_{⊔ I} (x) . \end{matrix}$

Next, we illustrate the theorem by Example 3.3.

By Example 3.3, we have $T = {φ_{1}, φ_{2}, φ_{3}, φ_{5}, φ_{8}}$ is an algebraic L-closure system on X and $(T, {sub}_{X})$ is a complete L-lattice, so we only need to check that $(T, {sub}_{X})$ is algebraic. In fact, by $I (T) = {Φ_{2}, Φ_{7}, Φ_{8}, Φ_{31}, Φ_{32}}$ and $⇓ φ_{i} (φ_{i}) = ⋀_{Φ_{j} \in I (T)} {sub}_{X} (φ_{i}, ⊔ Φ_{j}) \to Φ_{j} (φ_{i}),$ we can easy calculate that $K (T) = {φ_{1}, φ_{2}, φ_{3}, φ_{5}, φ_{8}}$ and thus K (φ_i) (φ_j) = sub_X (φ_j, φ_i) for any $φ_{i}, φ_{j} \in T$ , which implies that for any $φ_{i} \in T$ , $K (φ_{i}) \in D (T)$ and φ_i = ⊔ K (φ_i). On the other hand, we define e : X × X ⟶ L as follows: $[\begin{matrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{matrix}] .$

By definition one can easily see that (X, e) be a complete L-lattice and $I (X) = {φ_{2}, φ_{5}, φ_{8}}$ . Then, we have ⇓1 (1) = (e (1, ⊔ φ₂) → φ₂ (1)) ∧ (e (1, ⊔ φ₅) → φ₅ (1)) ∧ (e (1, ⊔ φ₈) → φ₈ (1)) = (e (1, 1) →1) ∧ (e (1, 2) →1) ∧ (e (1, 3) →1) =1, which implies $K (X) = {1, 2, 3} = X$ and thus it is easy to check that (X, e) is algebraic. Now, let $T = {k_{1}, k_{2}, k_{3}} = {↓ 1, ↓ 2, ↓ 3} = {φ_{2}, φ_{5}, φ_{8}},$ then all L-subsets of $T$ are as follows: Φ₁ = (0, 0, 0), Φ₂ = (1, 0, 0), Φ₃ = (0, 1, 0), Φ₄ = (0, 0, 1), Φ₅ = (1, 1, 0), Φ₆ = (1, 0, 1), Φ₇ = (0, 1, 1), Φ₈ = (1, 1, 1). It is easy to check that $I (T) = {Φ_{2}, Φ_{5}, Φ_{8}}$ , and $⊔ i^{\to} (Φ_{2}) = φ_{1} \in T$ , $⊔ i^{\to} (Φ_{5}) = φ_{5} \in T$ , $⊔ i^{\to} (Φ_{8}) = φ_{8} \in T$ , which implies that $T$ is an algebraic L-closure system on X and $(T, {sub}_{K (X)}) ≅ (X, e)$ .

Let L - ALGDOMG denote the category of algebraic L-lattices with L-Scott continuous mappings which having left L-adjoint.

Lemma 4.5. Let g be an L-Scott continuous mapping between algebraic L-lattices (X, e_X) and (Y, e_Y), f be the left L-adjoint of g. Then

f maps compact elements of (Y, e_Y) to compact elements of (X, e_X), that is, $f (K (Y)) \subseteq K (X)$ ;

$H (g) = f |_{K (Y)} : K (Y) ⟶ K (X)$ has the property that for any $I \in I (K (X))$ , ${H (g)}^{\leftarrow} (I) \in I (K (Y))$ .

Proof. (1) First of all, we observe that for any y ∈ Y and $I \in I (X)$ , $\begin{matrix} g^{\to} (I) (y) & = & ⋁_{g (x) = y} I (x) = ⋁_{g (x) = y} I (x) \land e_{Y} (y, g (x)) \\ = & ⋁_{g (x) = y} I (x) \land e_{Y} (f (y), x) \\ \leq & I (f (y)) = f^{\leftarrow} (I) (y) . \end{matrix}$

Then we have $\begin{matrix} ⇓ f (y) (f (y)) & = & ⋀_{I \in I (X)} e_{X} (f (y), ⊔ I) \to I (f (y)) \\ = & ⋀_{I \in I (X)} e_{Y} (y, g (⊔ I)) \to I (f (y)) \\ = & ⋀_{I \in I (X)} e_{Y} (y, ⊔ g^{\to} (I)) \to I (f (y)) \\ \geq & ⋀_{I \in I (X)} e_{Y} (y, ⊔ g^{\to} (I)) \to g^{\to} (I) (y) \\ \geq & ⋀_{J \in I (Y)} e_{Y} (y, ⊔ J) \to J (y) = ⇓ y (y), \end{matrix}$ which implies that for any $y \in K (Y)$ , $f (y) \in K (X)$ , i.e., $f (K (Y)) \subseteq K (X)$ .

(2) Let $I \in I (K (X))$ , we claim that $H (g)^{\leftarrow} (I) = {f |_{K (Y)}}^{\leftarrow} (I)$ is an L-ideal of $K (Y)$ . Note that for any $x \in K (X)$ , I (x) = e_X (x, ⊔ I), then for any $y \in K (Y)$ , it holds that $H (g)^{\leftarrow} (I) (y) = {f |_{K (Y)}}^{\leftarrow} (I) (y) = I (f (y)) = e_{X} (f (y), ⊔ I) = e_{Y} (y, g (⊔ I)) = k_{g (⊔ I)} (y)$ , as (1) and g is the right L-adjoint of f. Thus, H (g) ^← (I) = k_g(⊔I) is an L-ideal of $I (K (Y))$ by the algebra of (Y, e_Y).

Proposition 4.6.(1) For all (X, e_X) , (Y, e_Y) ∈ ob (L - ALGDOMG) and g ∈ Hom_L-ALGDOMG (X, Y), let $𝕂 (X) = (K (X), T_{X})$ and $𝕂 (g) = f |_{K (Y)} : K (Y) ⟶ K (X)$ , where $T_{X} = {k_{x} | x \in X}$ and f is the left L-adjoint of g. Then $𝕂$ is a functor from the category L - ALGDOMG to the category AL - CSSP^op.

(2) For all $(X, T_{X}), (Y, T_{Y}) \in ob (AL - CSSP)$ and f ∈ Hom_AL-CSSP (X, Y), let $ℍ (X) = (T_{X}, {sub}_{X})$ and $ℍ (f) = f^{\leftarrow}$ . Then $ℍ$ is a functor from the category AL - CSSP to the category L - ALGDOMG^op.

Proof. (1) By Theorems 3.9, 4.4 and Proposition 3.10, we have that $(K (X), T_{X}) \in ob (AL - CSSP)$ and $𝕂$ preserves the composite of morphisms and the identity. Moreover, for any x ∈ X, by Lemma 4.5, it holds that ${f |_{K (Y)}}^{\leftarrow} (k_{x}) = k_{g (x)}$ , which implies that for any x ∈ X, ${f |_{K (Y)}}^{\leftarrow} (k_{x}) \in T_{Y}$ .

(2) Similar to (1), we only need to show that for any f ∈ Hom_AL-CSSP (X, Y), $ℍ (f) = f^{\leftarrow} \in {Hom}_{{L - ALGDOMG}^{op}} (X, Y)$ . Firstly, for any $𝕀 \in I (T_{Y})$ and x ∈ X, $\begin{matrix} f^{\leftarrow} (⊔ 𝕀) (x) & = & (⊔ 𝕀) (f (x)) \\ = & ⋁_{B \in T_{Y}} 𝕀 (B) \land B (f (x)) \\ = & ⋁_{A \in T_{X}} ⋁_{f^{\leftarrow} (B) = A} 𝕀 (B) \land A (x) \\ = & ⋁_{A \in T_{X}} (f^{\leftarrow})^{\to} (𝕀) (A) \land A (x) \\ = & ⊔ (f^{\leftarrow})^{\to} (𝕀) (x) . \end{matrix}$

Then, for any $𝔸 \in L^{T_{Y}}$ and any x ∈ X, $\begin{matrix} f^{\leftarrow} (⊓ 𝔸) (x) & = & (⊓ 𝔸) (f (x)) \\ = & ⋀_{B \in T_{Y}} 𝔸 (B) \to B (f (x)) \\ = & ⋁_{A \in T_{X}} ⋁_{f^{\leftarrow} (B) = A} 𝔸 (B) \to A (x) \\ = & ⋁_{A \in T_{X}} (f^{\leftarrow})^{\to} (𝔸) (A) \to A (x) \\ = & ⊓ (f^{\leftarrow})^{\to} (𝔸) (x) . \end{matrix}$

Therefore, by Theorem 3.5 in [16] and the completeness of $(T_{Y}, {sub}_{Y})$ , we have $ℍ (f) = f^{\leftarrow} \in {Hom}_{{L - ALGDOMG}^{op}} (Y, X)$ .

The above proposition can be shown in following diagrams:

Theorem 4.7. Let $η = (X, e_{X}) \mapsto (η_{X} = x \mapsto k_{x} : (X, e_{X}) ⟶ ℍ \circ 𝕂 (X, e_{X}))) : {Id}_{{L - ALGDOMG}^{op}} ⟶ ℍ \circ 𝕂$ is natural isomorphism.

Proof. For any algebraic L-lattice (X, e_X), it is obvious that $η_{X} : (X, e_{X}) ⟶ ℍ \circ 𝕂 (X, e_{X}) = ({k_{x} | x \in X}, {sub}_{K (X)})$ is an isomorphic mapping. Moreover, suppose that (X, e_X) , (Y, e_Y) are algebraic L-lattices and g ∈ Hom_{L-ALGDOMG
^op} (X, Y) with the left adjoint f, then the diagram below is commutative.

In fact, for any x ∈ X, it is easy to check that $\begin{matrix} ℍ \circ 𝕂 \circ η_{X} (x) \\ = {f |_{K (Y)}}^{\leftarrow} (k_{x}) = k_{g (x)} = η_{Y} \circ g (x), \end{matrix}$ i.e., $ℍ \circ 𝕂 \circ η_{X} = η_{Y} \circ g .$

Theorem 4.8. L - ALGDOMG is dual to AL - CSSP.

Proof. The proof can be accomplished in two steps by Proposition 4.6 and Theorem 4.7.

Step 1. We show that the functor $𝕂 : L - ALGDOMG$ ⟶AL - CSSP is faithful. Suppose that (X, e_X) and (Y, e_Y) are all algebraic L-lattice, g₁ ≠ g₂ ∈ Hom_L-ALGDOMG (X, Y), and f₁, f₂ are the left adjoint of them, respectively. we have that $𝕂 (g_{1}) \neq 𝕂 (g_{2})$ . Otherwise, $𝕂 (g_{1}) = 𝕂 (g_{2})$ , then for all x ∈ X, $𝕂 (g_{1})^{\leftarrow} (k_{x}) = 𝕂 (g_{2})^{\leftarrow} (k_{x})$ , i.e., ${f_{1} |_{K (Y)}}^{\leftarrow} (k_{x}) = {f_{2} |_{K (Y)}}^{\leftarrow} (k_{x})$ , and so k_g₁(x) = k_g₂(x), which implies g₁ = g₂ by the algebra of (Y, e_Y). This is contradiction with g₁ ≠ g₂.

Step 2. We show that the functor $𝕂$ is full, i.e., for any Hom_L-ALGDOMG (X, Y), $𝕂$ is surjective from Hom_L-ALGDOMG (X, Y) to ${Hom}_{AL - CSSP} (𝕂 (Y), 𝕂 (X))$ . Suppose that f ∈ Hom_AL-CSSP $(𝕂 (Y), 𝕂 (X))$ , then f is L-order preserving, since e_Y (y₁, y₂) = f^← (f (y₂)) (y₂) ∧e_Y (y₁, y₂) ≤ f^← (f (y₂)) (y₁) = e_X (f (y₁) , f (y₂)) for any $y_{1}, y_{2} \in K (Y)$ . Let g (x) = ⊔ i^→ (f^← (k_x)), then for any $x \in K (X), y \in K (Y)$ , $\begin{matrix} e_{Y} (y, g (x)) = e_{Y} (y, ⊔ i^{\to} (f^{\leftarrow} (k_{x}))) \\ = i^{\to} (f^{\leftarrow} (k_{x})) (y) = ⋁_{i (y^{'}) = y} f^{\leftarrow} (k_{x}) (y^{'}) \\ = f^{\leftarrow} (k_{x}) (y) = e_{X} (f (y), x) . \end{matrix}$

Next, we show that g is an L-Scott continuous mapping between algebraic L-lattices (X, e_X) and (Y, e_Y). In fact, for any y ∈ Y, $\begin{matrix} e_{Y} (⊔ i^{\to} (f^{\leftarrow} (k_{x})), y) \\ = ⋀_{y^{'} \in Y} i^{\to} (f^{\leftarrow} (k_{x})) (y^{'}) \to e_{Y} (y^{'}, y) \\ = ⋀_{y^{'} \in K (Y)} e (f (y^{'}), x) \to e_{Y} (y^{'}, y) \\ = ⋀_{y^{'} \in K (Y)} e (y^{'}, g (x)) \to e_{Y} (y^{'}, y) \\ = e_{Y} (g (x), y) . \end{matrix}$

Therefore, for any $I \in I (X)$ , $\begin{matrix} g (⊔ I) & = & ⊔ i^{\to} (f^{\leftarrow} (k_{⊔ I})) \\ = & ⋁_{y \in Y} i^{\to} (f^{\leftarrow} (k_{⊔ I})) (y) \otimes y \\ = & ⋁_{y \in K (Y)} e_{X} (f (y), ⊔ I) \otimes y \\ = & ⋁_{y \in K (Y)} I (f (y)) \otimes y \\ = & ⋁_{y \in K (Y)} (⋁_{x \in X} I (x) \land e_{X} (f (y), x)) \otimes y \\ = & ⋁_{x \in X} I (x) \otimes (⋁_{y \in K (Y)} e_{X} (f (y), x) \otimes y) \\ = & ⋁_{x \in X} I (x) \otimes (⋁_{y \in Y} i^{\to} (f^{\leftarrow} (k_{x})) (y) \otimes y) \\ = & ⋁_{x \in X} I (x) \otimes ⊔ i^{\to} (f^{\leftarrow} (k_{x})) \\ = & ⋁_{x \in X} I (x) \otimes g (x) \\ = & ⋁_{y \in Y} ⋁_{g (x) = y} I (x) \otimes y \\ = & ⋁_{y \in Y} g^{\to} (I) (y) \otimes y = ⊔ g^{\to} (I) . \end{matrix}$

Therefore, g is L-order preserving and thus $(f, g |_{K (X)})$ is an L-adjunction between $K (Y)$ and $K (X)$ . Finally, let h (y) = ming^← (↑ y) for any y ∈ Y. Then h is the left L-adjoint of g by Theorem 3.4 in [16]. In order to complete the proof, we still show that h (y) = f (y) when $y \in K (Y)$ . In fact, for any $y \in K (Y)$ , e_X (f (y) , ⊓ g^← (↑ y))) ≥ g^← (↑ y) (f (y)) = e_Y (y, g ∘ f (y)) ≥ e_Y (y, y) =1 which implies that ⊓g^← (↑ y) ≥ f (y). For the converse inequality, for any x ∈ X, $\begin{matrix} e_{X} (x, ⊓ g^{\leftarrow} (↑ y)) & = & ⋀_{x^{'} \in X} g^{\leftarrow} (↑ y) (x^{'}) \to e_{X} (x, x^{'}) \\ = & ⋀_{x^{'} \in X} e_{Y} (y, g (x^{'})) \to e_{X} (x, x^{'}) \\ \leq & ⋀_{x^{'} \in K (X)} e_{Y} (y, g (x^{'})) \to e_{X} (x, x^{'}) \\ = & ⋀_{x^{'} \in K (X)} e_{X} (f (y), x^{'}) \to e_{X} (x, x^{'}) \\ = & e_{X} (x, f (y)), \end{matrix}$ which implies that ⊓g^← (↑ y) ≤ f (y). Therefore, by Theorem 2.6 in [16], it obtains that f (y) = ming^← (↑ y) and thus h (y) = f (y) for any $y \in K (Y)$ .

5 Conclusions

Taking the complete Heyting algebra as the structure of truth value, we proposed the notions of the algebraic L-closure system and the algebraic L-closure operator, which generalize the corresponding notions on crisp setting. From L-order view, the algebraic L-closure system has the property that the join of directed L-subset of it is the generalization of set union in crisp case. This partly solves the problem mentioned in Introduction. Moreover, in order to exploit the relationship between the algebraic L-closure systems and the algebraic L-closure operators, we showed that there is a one-to-one correspondence between these two structures and obtained that the category of algebraic L-closure spaces is isomorphic to the category of algebraic L-closure system spaces. As a result of our study, it followed that the category L - ALGDOMG is dual to AL - CSSP.

Acknowledgments

This work is partially supported by the National Natural Science Foundation of China (Nos. 11371130 and 11561002), the National Basic Research Program of China (No. 2011CB311808), the National Social Science Foundation of China (No. 12BJY122), and the Science Foundation of Education Committee of Jiangxi (No. GJJ14481).

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