Topological structures and distance functions in complete co-residuated lattices

Abstract

Most of the research in fuzzy rough sets and fuzzy topological structures have been studied on the basis of fuzzy partially ordered sets. Instead of fuzzy partially ordered sets, the concept of distance functions in complete co-residuated lattices is introduced. Using distance functions, we define Alexandrov pretopology, Alexandrov precotopology and fuzzy interior (fuzzy closure) operators in complete co-residuated lattices, and we investigate their properties. Moreover, we prove that there exist isomorphic categories and Galois correspondence between topological categories.

Keywords

Complete co-residuated lattice,distance spaces,Alexandrov pretopologies,Alexandrov precotopology,fuzzy interior (fuzzy closure) operators,Galois correspondence

1 Introduction

Alexandrov [2] introduced an Alexandrov topology in which the intersection and union of any family of open sets is open. Given a preordered set (X, ≤), we can define Alexandrov topologies τ_≤, τ_≤^-1 on X by choosing the open sets to be the upper sets: $τ_{\leq} = {U \subseteq X ∣ \forall x, y \in X, (x \in U) \land (x \leq y) \to y \in U},$ (1) and by choosing the open sets to be the lower sets: $τ_{\leq^{- 1}} = {L \subseteq X ∣ \forall x, y \in X, (y \in L) \land (x \leq y) \to x \in L} .$ (2)

Given an Alexandrov topology (X, τ), we can define a preorder ≤_τ on X by $x \leq_{τ} y iff x \leq \bar{y} = ⋂ {F ∣ {y} \subset F, F^{c} \in τ} .$ (3)

Ward et al. [31] introduced the complete residuated lattice as an important mathematical tool for studying algebraic structures for many valued logics [3 , 36]. Bělohlávek [3] investigated information systems and decision rules on complete residuated lattices. Höhle [9, 10] introduced L-fuzzy topologies with algebraic structure L (cqm, quantales, MV-algebra). For an extension of Pawlak’s rough sets [18, 19], many researchers [3 , 27–29] developed fuzzy rough sets, L-lower and L-upper approximation operators in complete residuated lattices.

Lai et al. [16] and Ma et al. [17] investigated the Alexandrov L-topology and lattice structures on L-fuzzy rough sets determined by lower and upper sets in complete residuated lattice (L, ∨ , ∧ , ⊙ , → , ⊥ , ⊤). Given a fuzzy preordered set (X, e_X), Alexandrov topologies τ_{e
_X} and $τ_{e_{X}^{- 1}}$ on X are defined by choosing the open sets to be the upper sets: $τ_{e_{X}} = {A \in L^{X} ∣ \forall x, y \in X, A (x) ⊙ e_{X} (x, y) \leq A (y)},$ (4) and by choosing the open sets to be the lower sets: $τ_{e_{X}^{- 1}} = {A \in L^{X} ∣ \forall x, y \in X, A (y) ⊙ e_{X} (x, y) \leq A (x)} .$ (5)

Given an Alexandrov topology (X, τ), we can define a fuzzy preorder on X by $e_{τ} (x, y) = ⋀_{A \in τ} (A (x) \to A (y)), \forall x, y \in X .$ (6)

Fang [4] studied the relationship between L-fuzzy closure systems and L-fuzzy topological spaces from a category viewpoint on a complete residuated lattice L. Kim et al. [12 –14] studied the properties of fuzzy join and meet completeness, L-fuzzy upper and lower approximation spaces and Alexandrov L-topologies with fuzzy partially ordered spaces on complete residuated lattices. Fuzzy topologies, L-lower and L-upper approximation operators are studied on complete residuated lattices [3–6 , 32–35].

Zheng et al. [36] introduced a complete co-residuated lattice as the generalization of t-conorm. Junsheng et al. [11] investigated (⊙ , ⊕)-generalized fuzzy rough set on (L, ∨ , ∧ , ⊙ , ⊕ , ⊥ , ⊤) where (L, ∨ , ∧ , ⊙ , ⊥ , ⊤) is a complete residuated lattice and (L, ∨ , ∧ , ⊕ , ⊥ , ⊤) is complete co-residuated lattice in a sense [36]. Ko et al. [15] studied distance spaces instead of fuzzy partially ordered spaces in complete co-residuated lattices.

It is well known that fuzzy partially ordered sets (resp. equivalence relations) plays an important role in fuzzy rough sets and fuzzy topological structures. In this paper, we introduce the concept of the distance functions (resp. trivial metrics) instead of fuzzy partially ordered sets (resp. equivalence relations). Using distance functions, we define Alexandrov pretopology, Alexandrov precotopology and fuzzy interior (fuzzy closure) operators on complete co-residuated lattices, and we investigate their properties and relations.

This paper is organized as follows. In Section 2, we recall the definitions of complete co-residuated lattices and distance spaces. Moreover, we give their examples. In section 3, given a distance space (X, d_X) on the complete co-residuated lattice (L, ∨ , ∧ , ⊕ , ⊖ , ⊥ , ⊤), we can define Alexandrov topologies $τ_{d_{X}}, τ_{d_{X}^{- 1}}$ on X by $τ_{d_{X}} = {A \in L^{X} ∣ \forall x, y \in X, A (x) \oplus d_{X} (x, y) \geq A (y)}$ (7) and $τ_{d_{X}^{- 1}} = {A \in L^{X} ∣ \forall x, y \in X, A (y) \oplus d_{X} (x, y) \geq A (x)} .$ (8)

Given an Alexandrov topology (X, τ), we can define a distance function on X by $d_{τ} (x, y) = ⋀_{A \in τ} (A (y) ⊖ A (x)), \forall x, y \in X .$ (9)

Moreover, we investigated the properties of Alexandrov pretopology, Alexandrov precotopology and fuzzy interior (fuzzy closure) operators on complete co-residuated lattices. Furthermore, their relations and examples are studied.

In section 4, we prove that there exist isomorphic categories and Galois correspondence between the category of topological categories.

2 Preliminaries

Definition 2.1. [11, 36] An algebra (L, ∧ , ∨ , ⊕ , ⊥ , ⊤) is called a complete co-residuated lattice if it satisfies the following conditions:

(C1) L = (L, ≤ , ∨ , ∧ , ⊥ , ⊤) is a complete lattice where ⊥ is the bottom element and ⊤ is the top element;

(C2) a = a⊕ ⊥, a ⊕ b = b ⊕ a and a ⊕ (b ⊕ c) = (a ⊕ b) ⊕ c for all a, b, c ∈ L;

(C3) (⋀ _i∈Γa_i) ⊕ b = ⋀ _i∈Γ (a_i ⊕ b) .

Let (L, ≤ , ⊕) be a complete co-residuated lattice. For all x, y ∈ L, define $x ⊖ y = ⋀ {z \in L ∣ y \oplus z \geq x} .$ (10) Then (x ⊕ y) ≥ z if and only if x ≥ (z ⊖ y) .

Let n (x) = ⊤ ⊖ x. The condition n (n (x)) = x for all x ∈ L is called a double negative law. For ⊤_x, ⊥ _x ∈ L^X, we define $⊤_{x} (y) = {\begin{matrix} ⊤, & if y = x, \\ ⊥, & otherwise, \end{matrix} ⊥_{x} (y) = {\begin{matrix} ⊥, & if y = x, \\ ⊤, & otherwise . \end{matrix}$

For all α ∈ L and A ∈ L^X, define $\begin{matrix} (A ⊖ α) (x) & = A (x) ⊖ α, \\ (α \oplus A) (x) & = α \oplus A (x), \\ α_{X} (x) & = α . \end{matrix}$ (11) Then (α ⊖ A) , (α ⊕ A) , α_X ∈ L^X.

Remark 2.2. (1) An infinitely distributive lattice (L, ≤ , ∨ , ∧ , ⊕ = ∨ , ⊥ , ⊤) is a complete co-residuated lattice. In particular, the unit interval $([0, 1], \leq, \lor, \land, \oplus = \lor, 0, 1)$ is a complete co-residuated lattice. Define $x ⊖ y = ⋀ {z \in L ∣ y \lor z \geq x} .$ (12) Then $x ⊖ y = {\begin{matrix} 0, & if y \geq x, \\ x, & if y ⪈ x . \end{matrix}$ (13)

Define $n (x) = {\begin{matrix} 1 ⊖ x = 1, & if x \neq 1, \\ 0, & if x = 1 . \end{matrix}$ (14) Then $n (n (x)) = {\begin{matrix} 0, & if x \neq 1, \\ 1, & if x = 1 . \end{matrix}$ (15) Hence n does not satisfy a double negative law.

(2) The unit interval ([0, 1] , ≤ , ⊕) is a complete co-residuated lattice where ⊕ is a right-continuous t-conorm(see [26]).

(3) The algebra ([1, ∞] , ≤ , ∨ , ⊕ = · , ∧ , 1, ∞) is a complete co-residuated lattice where ∞· a = a · ∞ = ∞ for all a ∈ [1, ∞]. Define $x ⊖ y = ⋀ {z \in [1, \infty] ∣ yz \geq x} .$ (16) Then $x ⊖ y = {\begin{matrix} 1, & if y \geq x, \\ \frac{x}{y}, & if y ⪈ x \end{matrix}$ (17) and ∞ ⊖ ∞ =1. Define $n (x) = {\begin{matrix} \infty ⊖ x = \infty, & if x \neq \infty, \\ 1, & if x = \infty . \end{matrix}$ (18) Then $n (n (x)) = {\begin{matrix} 1, & if x \neq \infty, \\ \infty, & if x = \infty . \end{matrix}$ (19) Hence n does not satisfy a double negative law.

(4) The algebra ([0, ∞] , ≤ , ∨ , ⊕ = + , ∧ , 0, ∞) is a complete co-residuated lattice where ∞+ a = a + ∞ = ∞ for all a ∈ [0, ∞]. Define $y ⊖ x = ⋀ {z \in [0, \infty] ∣ x + z \geq y} .$ (20) Then $\begin{matrix} y ⊖ x & = ⋀ {z \in [0, \infty] ∣ z \geq - x + y} \\ = (y - x) \lor 0 \end{matrix}$ (21) and ∞ ⊖ ∞ =0. Define $n (x) = {\begin{matrix} \infty ⊖ x = \infty, & if x \neq \infty, \\ 0, & if x = \infty . \end{matrix}$ (22) Then $n (n (x)) = {\begin{matrix} 0, & if x \neq \infty, \\ \infty, & if x = \infty . \end{matrix}$ (23) Hence n does not satisfy a double negative law.

(5) The algebra ([0, 1] , ≤ , ∨ , ⊕ , ∧ , 0, 1) is a complete co-residuated lattice where $x \oplus y = (x^{p} + y^{p})^{\frac{1}{p}} \land 1$ and 1≤ p < ∞(see [26]). Define $x ⊖ y = ⋀ {z \in [0, 1] ∣ (z^{p} + y^{p})^{\frac{1}{p}} \geq x} .$ (24) Then $\begin{matrix} x ⊖ y & = ⋀ {z \in [0, 1] ∣ z \geq (x^{p} - y^{p})^{\frac{1}{p}}} \\ = (x^{p} - y^{p})^{\frac{1}{p}} \lor 0 . \end{matrix}$ (25) Define $n (x) = 1 ⊖ x = (1 - x^{p})^{\frac{1}{p}}$ for 1≤ p < ∞. Then n (n (x)) = x for all x ∈ [0, 1]. Hence n satisfies a double negative law.

(6) Let P (X) be the collection of all subsets of X. Then (P (X) , ⊂ , ∪ , ∩ , ⊕ = ∪ , ∅ , X) is a complete co-residuated lattice. Define $A ⊖ B = ⋀ {C \in P (X) ∣ B \cup C \supset A} .$ (26) Then $A ⊖ B = A \cap B^{c} = A - B .$ (27) Define n (A) = X ⊖ A = A^c for all A ⊂ X. Then n (n (A)) = A. Hence n satisfies a double negative law.

Lemma 2.3. [15] Let (L, ∧ , ∨ , ⊕ , ⊖ , ⊥ , ⊤) be a complete co-residuated lattice. Let x, y, z, x_i, y_i ∈ L. Then the following hold.

(1) If y ≤ z, then x ⊕ y ≤ x ⊕ z, y ⊖ x ≤ z ⊖ x and x ⊖ z ≤ x ⊖ y.

(2) (⋁ _i∈Γx_i) ⊖ y = ⋁ _i∈Γ (x_i ⊖ y) and x ⊖ (⋀ _i∈Γy_i) = ⋁ _i∈Γ (x ⊖ y_i) .

(3) (⋀ _i∈Γx_i) ⊖ y ≤ ⋀ _i∈Γ (x_i ⊖ y) .

(4) x ⊖ (⋁ _i∈Γy_i) ≤ ⋀ _i∈Γ (x ⊖ y_i) .

(5) x⊖ x = ⊥, x ⊖ ⊥ = x and ⊥⊖ x = ⊥. Moreover, x⊖ y = ⊥ if and only if x ≤ y.

(6) y ⊕ (x ⊖ y) ≥ x, y ≥ x ⊖ (x ⊖ y), (x ⊖ y) ⊕ (y ⊖ z) ≥ x ⊖ z and x ⊖ y ≥ (x ⊖ z) ⊖ (y ⊖ z).

(7) x ⊖ (y ⊕ z) = (x ⊖ y) ⊖ z = (x ⊖ z) ⊖ y .

(8) x ⊖ y ≥ (x ⊕ z) ⊖ (y ⊕ z), y ⊖ x ≥ (z ⊖ x) ⊖ (z ⊖ y) and (x ⊕ y) ⊖ (z ⊕ w) ≤ (x ⊖ z) ⊕ (y ⊖ w).

(9) x⊕ y = ⊥ if and only if x =⊥ and y =⊥.

(10) (x ⊕ y) ⊖ z ≤ x ⊕ (y ⊖ z) and (x ⊖ y) ⊕ z ≥ x ⊖ (y ⊖ z).

(11) If L satisfies a double negative law and n (x) = ⊤ ⊖ x, then n (x ⊕ y) = n (x) ⊖ y = n (y) ⊖ x, x ⊖ y = n (y) ⊖ n (x), n (⋀ _i∈Γx_i) = ⋁ _i∈Γn (x_i) and n (⋁ _i∈Γx_i) = ⋀ _i∈Γn (x_i).

Definition 2.4. [15] Let (L, ∧ , ∨ , ⊕ , ⊖ , ⊥ , ⊤) be a complete co-residuated lattice. Let X be a set. A function d_X : X × X → L is called a distance function if it satisfies the following conditions:

(D1) d_X (x, x) =⊥ for all x ∈ X;

(D2) d_X (x, y) ⊕ d_X (y, z) ≥ d_X (x, z) for all x, y, z ∈ X.

The pair (X, d_X) is called a distance space.

Remark 2.5. (1) Define a distance function d_X : X × X → [0, ∞]. Then (X, d_X) is called a pseudo-quasi-metric space.

(2) Let (L, ∧ , ∨ , ⊕ , ⊖ , ⊥ , ⊤) be a complete co-residuated lattice. Define a function d_L : L × L → L by d_L (x, y) = x ⊖ y . By Lemma 2.3 (5) and (6), (L, d_L) is a distance space. Define a function d_{L
^X} : L^X × L^X → L by d_{L
^X} (A, B) = ⋁ _x∈X (A (x) ⊖ B (x)) . Then (L^X, d_{L
^X}) is a distance space.

3 Topological structures in complete co-residuated lattices

In this section, we assume that (L, ∧ , ∨ , ⊕ , ⊖ , ⊥ , ⊤) is a complete co-residuated lattice, unless otherwise stated specifically.

Definition 3.1. (1) A subset τ ⊂ L^X is called an Alexandrov pretopology on X if it satisfies the following conditions:

(O1) α_X ∈ τ;

(O2) If A_i ∈ τ for all i ∈ I, then ⋁_i∈IA_i ∈ τ;

(O3) If A ∈ τ and α ∈ L, then A ⊖ α ∈ τ.

(2) A subset η ⊂ L^X is called an Alexandrov precotopology on X if it satisfies the following conditions:

(CO1) α_X ∈ η;

(CO2) If A_i ∈ η for all i ∈ I, then ⋀_i∈IA_i ∈ η;

(CO3) If A ∈ η and α ∈ L, then α ⊕ A ∈ η.

A subset τ ⊂ L^X is called an Alexandrov topology on X if it is both Alexandrov pretopology and Alexandrov precotopology on X.

Theorem 3.2. Let d_X be a distance function on X.

(1) Define $τ_{d_{X}} = {A \in L^{X} ∣ A (x) \oplus d_{X} (x, y) \geq A (y)} .$ (28)Then τ_{d
_X} is an Alexandrov topology on X.

(2) If $d_{X}^{- 1} (x, y) = d_{X} (y, x)$ for all x, y ∈ X, then $τ_{d_{X}^{- 1}} = {A \in L^{X} ∣ A (x) \oplus d_{X} (y, x) \geq A (y)} .$ (29)Moreover, if L satisfies a double negative law, then $τ_{d_{X}^{- 1}} = η_{τ_{d_{X}}}$ where η_{τ
_{d
_X}} = {n (A) ∈ L^X ∣ A ∈ τ_{d
_X}}.

Proof. (1) (O1, CO1) ⊥_X, ⊤ _X ∈ τ_{d
_X}.

(O2,CO2) If A_i ∈ τ_{d
_X} for all i ∈ I, then $\begin{matrix} (⋁_{i \in I} A_{i} (x)) \oplus d_{X} (x, y) & \geq ⋁_{i \in I} (A_{i} (x) \oplus d_{X} (x, y)) \\ \geq ⋁_{i \in I} A_{i} (y), \\ (⋀_{i \in I} A_{i} (x)) \oplus d_{X} (x, y) & = ⋀_{i \in I} (A_{i} (x) \oplus d_{X} (x, y)) \\ \geq ⋀_{i \in I} A_{i} (y) . \end{matrix}$

(O3, CO3) Let A ∈ τ_{d
_X} and α ∈ L. Since $\begin{matrix} (α \oplus A (x)) \oplus d_{X} (x, y) & = α \oplus (A (x) \oplus d_{X} (x, y)) \\ \geq α \oplus A (y), \end{matrix}$ we have that $\begin{matrix} α \oplus (A ⊖ α) (x) \oplus d_{X} (x, y) & \geq A (x) \oplus d_{X} (x, y) \\ \geq A (y) \end{matrix}$ if and only if $(A ⊖ α) (x) \oplus d_{X} (x, y) \geq (A ⊖ α) (y) .$ Thus α ⊕ A, A ⊖ α ∈ τ_{d
_X}. Therefore, τ_{d
_X} is an Alexandrov topology on X.

(2) Since η_{τ
_{d
_X}} = {A ∈ L^X ∣ n (A) ∈ τ_{d
_X}}, we have by Lemma 2.3(11) that n (A) (x) ⊕ d_X (x, y) ≥ n (A) (y) if and only if A (y) ≥ A (x) ⊖ d_X (x, y) if and only if A (y) ⊕ d_X (x, y) ≥ A (x) if and only if $A \in τ_{d_{X}^{- 1}}$ . Hence the result holds.□

Theorem 3.3. (1) Let (X, τ) be an Alexandrov pretopological space. Define $d_{τ} (x, y) = ⋁_{A \in τ} (A (y) ⊖ A (x)) .$ (30)Then d_τ is a distance function with τ ⊂ τ_{d
_τ}. Moreover, if τ is an Alexandrov topology on X, then τ = τ_{d
_τ}.

(2) Let (X, η) be an Alexandrov precotopological space. Define $d_{η} (x, y) = ⋁_{A \in η} (A (y) ⊖ A (x)) .$ (31)Then d_η is a distance function with η ⊂ τ_{d
_η} where $τ_{d_{η}} = {B \in L^{X} ∣ B (x) \oplus d_{η} (x, y) \geq B (y)} .$ (32)

Proof. (1) (D1) d_η (x, x) =⋁ _A∈η (A (x) ⊖ A (x)) = ⊥ for all x ∈ X;

(D2) For all x, y, z ∈ X, $\begin{matrix} d_{τ} & (x, y) \oplus d_{τ} (y, z) \\ = ⋁_{A \in τ} (A (y) ⊖ A (x)) \oplus ⋁_{A \in τ} (A (z) ⊖ A (y)) \\ \geq ⋁_{A \in τ} (A (y) ⊖ A (x)) \oplus (A (z) ⊖ A (y)) \\ \geq ⋁_{A \in τ} (A (z) ⊖ A (x)) = d_{X} (x, z) . \end{matrix}$ Hence d_τ is a distance function. Let B ∈ τ. Then $\begin{matrix} B (x) \oplus ⋁_{A \in τ} (A (y) ⊖ A (x)) \\ \geq B (x) \oplus (B (y) ⊖ B (x)) \geq B (y) . \end{matrix}$ Hence B ∈ τ_{d
_τ}. Assume that τ is an Alexandrov topology on X. For all B ∈ τ_{d
_τ}, we have $B = B (x) \oplus ⋁_{A \in τ} (A (-) ⊖ A (x)) \in τ,$ and so B ∈ τ. Hence τ = τ_{d
_τ}.

(2) Let B ∈ η. Then $\begin{matrix} B (x) \oplus ⋁_{A \in η} (A (y) ⊖ A (x)) \\ \geq B (x) \oplus (B (y) ⊖ B (x)) \geq B (y) . \end{matrix}$ Thus B ∈ τ_{d
_η} and τ_{d
_η} is an Alexandrov topology by Theorem 3.2(1). If η is an Alexandrov topology on X, then B (x) ⊕ ⋁ _A∈η (A (-) ⊖ A (x)) ∈ η. For all B ∈ τ_{d
_η}, $⋀_{x \in X} (B (x) \oplus ⋁_{A \in η} (A (y) ⊖ A (x))) \geq B (y)$ and $\begin{matrix} ⋀_{x \in X} (B (x) \oplus ⋁_{A \in η} (A (y) ⊖ A (x))) \\ \leq B (y) \oplus ⋁_{A \in η} (A (y) ⊖ A (y)) = B (y) . \end{matrix}$ Thus $B = ⋀_{x \in X} (B (x) \oplus ⋁_{A \in η} (A (-) ⊖ A (x))) \in η .$ Hence η = τ_{d
_η}.□

Definition 3.4. A map $C : L^{X} \to L^{X}$ is called a fuzzy closure operator if it satisfies the following conditions:

(C1) $C (⊥_{X}) = ⊥_{X}$ ;

(C2) $A \leq C (A)$ and $C (A) = C (C (A))$ for all A ∈ L^X;

(C3) $d_{L^{X}} (A, B) \geq d_{L^{X}} (C (A), C (B))$ for all A, B ∈ L^X.

The pair $(X, C)$ is called a fuzzy closure space.

Definition 3.5. A map $I : L^{X} \to L^{X}$ is called a fuzzy interior operator if it satisfies the following conditions:

(I1) $I (⊤_{X}) = ⊤_{X}$ ;

(I2) $I (A) \leq A$ and $I (A) = I (I (A))$ for all A ∈ L^X;

(I3) $d_{L^{X}} (A, B) \geq d_{L^{X}} (I (A), I (B))$ for all A, B ∈ L^X.

The pair $(X, I)$ is called a fuzzy interior space.

Let $(X, I)$ be a fuzzy interior space and let $(X, C)$ be a fuzzy closure space. As a generalization of fuzzy rough set (I, C), the pair $(I (A), C (A))$ is called a fuzzy rough set for A ∈ L^X. The map α : L^X → L by $α (A) = ⋁_{x \in X} (C (A) (x) ⊖ I (A) (x))$ (33) is called a fuzzy accuracy measure.

Theorem 3.6. Let $D : L^{X} \to L^{X}$ be a map. The following three statements are equivalent.

(1) $d_{L^{X}} (A, B) \geq d_{L^{X}} (D (A), D (B))$ for all A, B ∈ L^X.

(2) $α \oplus D (A) \geq D (α \oplus A)$ for all α ∈ L and A ∈ L^X, and $D (A) \leq D (B)$ for A ≤ B.

(3) $D (A ⊖ α) \geq D (A) ⊖ α$ for all α ∈ L and A ∈ L^X, and $D (A) \leq D (B)$ for A ≤ B.

Proof. (1) ⇒ (2). Let A ≤ B. By Lemma 2.3(5), we have d_{L
^X} (A, B) =⊥ and $d_{L^{X}} (D (A), D (B)) = ⊥$ , and so $D (A) \leq D (B)$ .

Since α ≥ (α ⊕ A (x)) ⊖ A (x), we have $α \geq d_{L^{X}} (α \oplus A, A) \geq d_{L^{X}} (D (α \oplus A), D (A)),$ and so $α \oplus D (A) \geq D (α \oplus A)$ .

(2)⇒ (1). Since d_{L
^X} (A, B) ⊕ B ≥ A, we have $D (d_{L^{X}} (A, B) \oplus B) \geq D (A) .$ Let α = d_{L
^X} (A, B). Since $d_{L^{X}} (A, B) \oplus D (B) \geq D (d_{L^{X}} (A, B) \oplus B) \geq D (A),$ we have $d_{L^{X}} (A, B) \geq d_{L^{X}} (D (A), D (B))$ .

(2) ⇒ (3). If A ≤ B, then $D (A) \leq D (B)$ . Since $α \oplus D (A ⊖ α) \geq D (α \oplus (A ⊖ α)) \geq D (A),$ we have $D (A ⊖ α) \geq D (A) ⊖ α$ .

(3)⇒ (1). Let α = d_{L
^X} (A, B). Since $D (B) \geq D (A ⊖ d_{L^{X}} (A, B)) \geq D (A) ⊖ d_{L^{X}} (A, B),$ we have $d_{L^{X}} (A, B) \geq d_{L^{X}} (D (A), D (B))$ .□

Lemma 3.7. Let (L, ∧ , ∨ , ⊕ , ⊖ , ⊥ , ⊤) be a complete co-residuated lattice. Let x_i, y_i ∈ L. Then the following hold.

(1) (⋁ _i∈Γx_i) ⊖ (⋁ _i∈Γy_i) ≤ ⋁ _i∈Γ (x_i ⊖ y_i).

(2) (⋀ _i∈Γx_i) ⊖ (⋀ _i∈Γy_i) ≤ ⋁ _i∈Γ (x_i ⊖ y_i).

Proof. (1) Since $x_{i} \leq (x_{i} ⊖ y_{i}) \oplus y_{i} \leq ⋁_{i \in Γ} (x_{i} ⊖ y_{i}) \oplus (⋁_{i \in Γ} y_{i}),$ we have ⋁_i∈Γx_i ≤ ⋁ _i∈Γ (x_i ⊖ y_i) ⊕ (⋁ _i∈Γy_i). Hence $(⋁_{i \in Γ} x_{i}) ⊖ (⋁_{i \in Γ} y_{i}) \leq ⋁_{i \in Γ} (x_{i} ⊖ y_{i}) .$

(2) Since x_i ≤ (x_i ⊖ y_i) ⊕ y_i, we have $\begin{matrix} ⋀_{i \in Γ} x_{i} \leq (x_{i} ⊖ y_{i}) \oplus y_{i} if and only if \\ x_{i} ⊖ y_{i} \geq ⋀_{i \in Γ} x_{i} ⊖ y_{i} . \end{matrix}$ Thus $\begin{matrix} ⋁_{i \in Γ} (x_{i} ⊖ y_{i}) & \geq ⋁_{i \in Γ} (⋀_{i \in Γ} x_{i} ⊖ y_{i}) \\ = (⋀_{i \in Γ} x_{i}) ⊖ (⋀_{i \in Γ} y_{i}) . \end{matrix}$ □

Theorem 3.8. Let d_X be a distance function on X.

(1) Define $C_{d_{X}}, I_{d_{X}} : L^{X} \to L^{X}$ by $\begin{matrix} C_{d_{X}} (A) (x) & = ⋁_{y \in X} (A (y) ⊖ d_{X} (x, y)), \\ I_{d_{X}} (A) (y) & = ⋀_{x \in X} (d_{X} (x, y) \oplus A (x)) . \end{matrix}$ (34)Then $C_{d_{X}}$ is a fuzzy closure operator and $I_{d_{X}}$ is a fuzzy interior operator. Moreover, $\begin{matrix} C_{d_{X}} (A ⊖ α) & = C_{d_{X}} (A) ⊖ α, \\ C_{d_{X}} (⋁_{i \in Γ} A_{i}) & = ⋁_{i \in Γ} C_{d_{X}} (A_{i}), \\ I_{d_{X}} (⋀_{i \in Γ} A_{i}) & = ⋀_{i \in Γ} I_{d_{X}} (A_{i}), \\ I_{d_{X}} (A \oplus α) & = I_{d_{X}} (A) \oplus α \end{matrix}$ (35)where A, A_i ∈ L^X and α ∈ L.

(2) $τ_{d_{X}} = {I_{d_{X}} (A) = ⋀_{x \in X} (A (x) \oplus d_{X} (x, -)) ∣ A \in L^{X}}$ .

(3) $τ_{d_{X}} = {C_{d_{X}} (A) = ⋁_{y \in X} (A (y) ⊖ d_{X} (-, y)) ∣ A \in L^{X}}$ .

(4) Assume that L satisfies a double negative law and n (x) = ⊤ ⊖ x. Then $C_{d_{X}^{- 1}} (A) = n (I_{d_{X}} (n (A)))$ for all A ∈ L^X where $d_{X}^{- 1} (x, y) = d_{X} (y, x)$ for all x, y ∈ X.

Proof. (1) (C1) $C_{d_{X}} (⊥_{X}) (y) = ⋁_{x \in X} (⊥_{X} (x) ⊖ d_{X} (y, x)) = ⊥ .$

(C2) For all A ∈ L^X, we have $C_{d_{X}} (A) (y) \geq (A (y) ⊖ d_{X} (y, y)) = A (y)$ and $\begin{matrix} C_{d_{X}} (C_{d_{X}} (A)) (z) \\ = ⋁_{x \in X} (C_{d_{X}} (A) (x) ⊖ d_{X} (z, x)) \\ = ⋁_{x \in X} (⋁_{y \in X} (A (y) ⊖ d_{X} (x, y)) ⊖ d_{X} (z, x)) \\ = ⋁_{x \in X} ⋁_{y \in X} (A (y) ⊖ (d_{X} (x, y) \oplus d_{X} (z, x))) \\ (by Lemma 2.3 (7) and (12)) \\ = ⋁_{y \in X} (A (y) ⊖ ⋀_{x \in X} (d_{X} (x, y) \oplus d_{X} (z, x))) \\ = ⋁_{y \in X} (A (y) ⊖ d_{X} (z, y)) = C_{d_{X}} (A) (z) . \end{matrix}$ (C3) For all A, B ∈ L^X, $\begin{matrix} d_{L^{X}} (C_{d_{X}} (A), C_{d_{X}} (B)) \\ = ⋁_{x \in X} (C_{d_{X}} (A) (x) ⊖ C_{d_{X}} (B) (x)) \\ = ⋁_{x \in X} (⋁_{y \in X} (A (y) ⊖ d_{X} (x, y)) \\ ⊖ ⋁_{z \in X} (B (z) ⊖ d_{X} (x, z))) \\ \leq ⋁_{x \in X} (⋁_{y \in X} ((A (y) ⊖ d_{X} (x, y)) ⊖ \\ (B (y) ⊖ d_{X} (x, y)))) \\ (by Lemma 3.7 (1)) \\ \leq ⋁_{y \in X} (A (y) ⊖ B (y)) = d_{L^{X}} (A, B) \\ (by Lemma 2.3 (6)) . \end{matrix}$ (I1) $I_{d_{X}} (⊤_{X}) (y) = ⋀_{x \in X} (⊤_{X} (x) \oplus d_{X} (x, y)) = ⊤$ .

(I2) For all A ∈ L^X, $I_{d_{X}} (A) (y) \leq (A (y) \oplus d_{X} (y, y)) = A (y)$ and $\begin{matrix} I_{d_{X}} (I_{d_{X}} (A)) (z) \\ = ⋀_{x \in X} (I_{d_{X}} (A) (x) \oplus d_{X} (x, z)) \\ = ⋀_{x \in X} (⋀_{y \in X} (A (y) \oplus d_{X} (y, x)) \oplus d_{X} (x, z)) \\ = ⋀_{y \in X} (A (y) \oplus ⋀_{x \in X} (d_{X} (y, x) \oplus d_{X} (x, z))) \\ (by Lemma 2.3 (7) and (12)) \\ = ⋀_{y \in X} (A (y) \oplus d_{X} (y, z)) = I_{d_{X}} (A) (z) . \end{matrix}$ (I3) For all A, B ∈ L^X, $\begin{matrix} d_{L^{X}} (I_{d_{X}} (A), I_{d_{X}} (B)) \\ = ⋁_{x \in X} (I_{d_{X}} (A) (x) ⊖ I_{d_{X}} (B) (x)) \\ = ⋁_{x \in X} (⋀_{y \in X} (A (y) \oplus d_{X} (y, x)) \\ ⊖ ⋀_{z \in X} (B (z) \oplus d_{X} (z, x))) \\ \leq ⋁_{x \in X} (⋁_{y \in X} ((A (y) \oplus d_{X} (y, x)) \\ ⊖ (B (y) \oplus d_{X} (y, x)))) (by Lemma 3.7 (2)) \\ \leq ⋁_{y \in X} (A (y) ⊖ B (y)) = d_{L^{X}} (A, B) \\ (by Lemma 2.3 (8)) . \end{matrix}$ For all A, A_i ∈ L^X and α ∈ L, we have $\begin{matrix} C_{d_{X}} (⋁_{i \in Γ} A_{i}) (z) & = ⋁_{x \in X} (⋁_{i \in Γ} A_{i} (x) ⊖ d_{X} (z, x)) \\ = ⋁_{i \in Γ} ⋁_{x \in X} (A_{i} (x) ⊖ d_{X} (z, x)) \\ (by Lemma 2.3 (2)) \\ = ⋁_{i \in Γ} C_{d_{X}} (A_{i}) (z), \end{matrix}$ $\begin{matrix} C_{d_{X}} (A ⊖ α) (z) & = ⋁_{x \in X} ((A ⊖ α) (x) ⊖ d_{X} (z, x)) \\ = ⋁_{x \in X} (A (x) ⊖ d_{X} (z, x)) ⊖ α \\ (by Lemma 2.3 (2, 7)) \\ = C_{d_{X}} (A) (z) ⊖ α \end{matrix}$ and $I_{d_{X}} (⋀_{i \in Γ} A_{i}) = ⋀_{i \in Γ} I_{d_{X}} (A_{i})$ and $I_{d_{X}} (A \oplus α) = I_{d_{X}} (A) \oplus α$ .

(2) Since d_X (x, y) ⊕ d_X (y, z) ≥ d_X (x, z), we have d_X (x, -) ∈ τ_{d
_X} and ⋀_x∈X (A (x) ⊕ d_X (x, -)) ∈ τ_{d
_X} for all A ∈ L^X. Hence ${⋁_{x \in X} (A (x) \oplus d_{X} (x, -)) ∣ A \in L^{X}} \subset τ_{d_{X}} .$

Let A ∈ τ_{d
_X}. Then $⋀_{x \in X} (A (x) \oplus d_{X} (x, y)) \geq A (y) .$ Moreover, $A (y) = A (y) \oplus d_{X} (y, y) \geq ⋀_{x \in X} (A (x) \oplus d_{X} (x, y)) .$ Hence A = ⋀ _x∈X (A (x) ⊕ d_X (x, -)).

(3) Since $\begin{matrix} (A (y) ⊖ d_{X} (x, y)) \oplus d_{X} (x, z) \oplus d_{X} (z, y) \\ \geq (A (y) ⊖ d_{X} (x, y)) \oplus d_{X} (x, y) \geq A (y), \end{matrix}$ we have $(A (y) ⊖ d_{X} (x, y)) \oplus d_{X} (x, z) \geq (A (y) ⊖ d_{X} (z, y)) .$ Thus (A (y) ⊖ d_X (- , y)) ∈ τ_{d
_X}. Hence ⋁_y∈X (A (y) ⊖ d_X (- , y)) ∈ τ_{d
_X}.

Let A ∈ τ_{d
_X}. Then A (x) ⊕ d_X (x, y) ≥ A (y) if and only if A (x) ≥ A (y) ⊖ d_X (x, y) . Thus $\begin{matrix} A (x) & \geq ⋁_{y \in X} (A (y) ⊖ d_{X} (x, y)) \\ \geq A (y) ⊖ d_{X} (x, x) = A (x) . \end{matrix}$ So ⋁_y∈X (A (y) ⊖ d_X (- , y)) ∈ τ_{d
_X}. Hence $τ_{d_{X}} = {⋁_{y \in X} (A (y) ⊖ d_{X} (-, y)) ∣ A \in L^{X}} .$

(4) For all A ∈ L^X, we have $\begin{matrix} n (I_{d_{X}} & (n (A))) (y) \\ = n (⋀_{x \in X} (d_{X} (x, y) \oplus n (A) (x))) \\ = ⋁_{x \in X} (⊤ ⊖ (d_{X} (x, y) \oplus n (A) (x))) \\ (by Lemma 2.3 (7, 11)) \\ = ⋁_{x \in X} (A (x) ⊖ d_{X}^{- 1} (y, x)) = C_{d_{X}^{- 1}} (A) (y) . \end{matrix}$ □

Theorem 3.9. Let $(X, I)$ be a fuzzy interior space.

(1) If L satisfies a double negative law, then $τ_{I} = {A \in L^{X} ∣ A = I (A)}$ is an Alexandrov pretopology on X. Moreover, if $I (⋀_{i \in Γ} A_{i}) = ⋀_{i \in Γ} I (A_{i})$ and $I (α \oplus A) = α \oplus I (A)$ for all A_i, A ∈ L^X, then $τ_{I}$ is an Alexandrov topology on X.

(2) Define $d_{I} (x, y) = I (⊥_{x}) (y)$ for all x, y ∈ X. Then $d_{I}$ is a distance function with $I_{d_{I}} (A) \geq I (A)$ and $τ_{I} \subset τ_{I_{d_{I}}}$ .

(3) Define $d_{I}^{1} (x, y) = ⋁_{A \in L^{X}} (I (A) (y) ⊖ I (A) (x))$ . Then $d_{I}^{1}$ is a distance function with $I_{d_{I}^{1}} (A) \geq I (A)$ and $τ_{I} \subset τ_{I_{d_{I}^{1}}}$ where $I_{d_{I}^{1}} (A) (y) = ⋀_{x \in X} (A (x) \oplus d_{I}^{1} (x, y)) .$ (36)Moreover, $d_{I}^{1} \geq d_{I}$ .

(4) If $I (⋀_{i \in Γ} A_{i}) = ⋀_{i \in Γ} I (A_{i})$ and $I (A \oplus α) = I (A) \oplus α$ for all A_i, A ∈ L^X, then $I_{d_{I}} (A) = I (A)$ , $d_{I}^{1} = d_{I}$ and $τ_{I} = τ_{d_{I}}$ .

(5) If d_X is a distance function, then $d_{I_{d_{X}}} = d_{X}$ .

Proof. (1) (O1) Since $⊤_{X} ⊖ α = I (⊤_{X}) ⊖ α \leq I (⊤_{X} ⊖ α) \leq ⊤_{X} ⊖ α$ by Theorem 3.6(3), we have $I (⊤_{X} ⊖ α_{X}) = ⊤_{X} ⊖ α_{X}$ . Since L satisfies a double negative law, we have $α_{X} \in τ_{I}$ .

(O2) and (O3) Let $A, A_{i} \in τ_{I}$ . By Lemma 2.3(11) and Theorem 3.6(3), we have $\begin{matrix} I (⋁_{i} A_{i}) \geq ⋁_{i} I (A_{i}) = ⋁_{i} A_{i}, \\ I (A ⊖ α) \geq I (A) ⊖ α = A ⊖ α . \end{matrix}$ Thus $⋁_{i} A_{i}, A ⊖ α \in τ_{I}$ . Hence $τ_{I}$ is an Alexandrov pretopology on X.

Moreover, if $I (⋀_{i \in Γ} A_{i}) = ⋀_{i \in Γ} I (A_{i})$ and $I (α \oplus A) = α \oplus I (A)$ for all A_i, A ∈ L^X, then one can see that $τ_{I}$ is an Alexandrov topology on X.

(2) Since $I (⊥_{x}) = ⋀_{y \in X} (I (⊥_{x}) (y) \oplus ⊥_{y})$ , we have by Theorem 3.6(2) that $\begin{matrix} d_{I} (x, z) & = I (⊥_{x}) (z) = I (I (⊥_{x})) (z) \\ = I (⋀_{y \in X} I (⊥_{x}) (y) \oplus ⊥_{y}) (z) \\ \leq ⋀_{y \in X} (I (⊥_{x}) (y) \oplus I (⊥_{y}) (z)) \\ = ⋀_{y \in X} (d_{I} (x, y) \oplus d_{I} (y, z)) . \end{matrix}$ Since $α \oplus I (A) \geq I (α \oplus A)$ by Theorem 3.6(2), we have $\begin{matrix} I_{d_{I}} (A) (y) & = ⋀_{x \in X} (d_{I} (y, x) \oplus A (y)) \\ = ⋀_{x \in X} (I (⊥_{y}) (x) \oplus A (y)) \\ \geq I (⋀_{y \in X} (A (y) \oplus ⊥_{y})) (x) \\ = I (A) (x) . \end{matrix}$ For $A \in τ_{I}$ , we have $A = I (A) \leq I_{d_{I}} (A)$ . Then by (1), we have $A \in τ_{I_{d_{I}}}$ .

(3) It is easy to see that $d_{I}^{1}$ is a distance function. Since $d_{I}^{1}$ is a distance function, we have $\begin{matrix} I_{d_{I}^{1}} (A) (x) = ⋀_{y \in X} (d_{I}^{1} (y, x) \oplus A (y)) \\ \geq ⋀_{y \in X} ((I (A) (x) ⊖ I (A) (y)) \oplus I (A) (y) \\ \geq I (A) (x) . \end{matrix}$ Let $A \in τ_{I}$ . Since $A = I (A) \leq I_{d_{I}^{1}} (A)$ , we have by (1) that $A \in τ_{I_{d_{I}^{1}}}$ . Moreover, $\begin{matrix} d_{I}^{1} (x, y) = ⋁_{A \in L^{X}} (I (A) (y) ⊖ I (A) (x)) \\ \geq ⋁_{z \in X} (I (⊥_{z}) (y) ⊖ I (⊥_{z}) (x)) \\ = ⋁_{z \in X} (d_{I} (z, y) ⊖ d_{I} (z, x)) = d_{I} (x, y) . \end{matrix}$

(4) If $I (⋀_{i \in Γ} A_{i}) = ⋀_{i \in Γ} I (A_{i})$ and $I (α ⊖ A) = α ⊖ I (A)$ for all A_i, A ∈ L^X, then

$\begin{matrix} I_{d_{I}} (A) (y) = ⋀_{x \in X} (d_{I} (y, x) \oplus A (y)) \\ = ⋀_{x \in X} (I (⊥_{y}) (x) \oplus A (y)) \\ = I (⋀_{y \in X} (A (y) \oplus ⊥_{y})) (x) = I (A) (x) . \end{matrix}$ Since A = ⋀ _z∈X (A (z) ⊕ ⊥ _z), we have $\begin{matrix} d_{I}^{1} (x, y) = ⋁_{A \in L^{X}} (I (A) (y) ⊖ I (A) (x)) \\ = ⋁_{A \in L^{X}} I (⋀_{z \in X} (A (z) \oplus ⊥_{z}) (y) \\ ⊖ I (⋀_{z \in X} (A (z) \oplus ⊥_{z}) (x)) \\ = ⋁_{A \in L^{X}} (⋀_{z \in X} (A (z) \oplus I (⊥_{z}) (y) \\ ⊖ ⋀_{z \in X} (A (z) \oplus I (⊥_{z}) (x))) \\ \leq ⋁_{A \in L^{X}} ⋁_{z \in X} ((A (z) \oplus I (⊥_{z}) (y) \\ ⊖ (A (z) \oplus I (⊥_{z}) (x))) (by Lemma 3.7 (2)) \\ \leq ⋁_{z \in X} (I (⊥_{z}) (y) ⊖ I (⊥_{z}) (x)) (by Lemma 2.3 (8)) \\ = ⋁_{z \in X} (d_{I} (z, y) ⊖ d_{I} (z, x)) = d_{I} (x, y) . \end{matrix}$

(5) Let x, y ∈ X. Then $\begin{matrix} d_{I_{d_{X}}} (x, y) & = I_{d_{X}} (⊥_{x}) (y) \\ = ⋀_{z \in X} (⊥_{x} (z) \oplus d_{X} (z, y)) \\ = d_{X} (x, y) . \end{matrix}$ □

Theorem 3.10. Let $(X, C)$ be a fuzzy closure space. The the following hold.

(1) $η_{C} = {A \in L^{X} ∣ C (A) = A}$ is an Alexandrov precotopology on X. Moreover, if $C (⋁_{i \in Γ} A_{i}) = ⋁_{i \in Γ} C (A_{i})$ and $C (A ⊖ α) = C (A) ⊖ α$ for all A_i, A ∈ L^X, then $η_{C}$ is an Alexandrov topology on X.

(2) If L satisfies a double negative law and $d_{C} (x, y) = n (C (n (⊥_{y}))) (x)$ for all x, y ∈ X, then $d_{C}$ is a distance function with $C_{d_{C}} (A) \leq C (A)$ and $η_{C} \subset η_{C_{d_{C}}}$ .

(3) Let $d_{C}^{1} (x, y) = ⋁_{A \in L^{X}} (C (A) (y) ⊖ C (A) (x))$ . Then $d_{C}^{1}$ is a distance function.

(4) If L satisfies a double negative law, then $d_{C}^{1} \leq d_{C}$ , $C_{d_{C}^{1}} (A) \leq C (A)$ and $η_{C} \subset η_{d_{C}^{1}}$ where $C_{d_{C}^{1}} (A) (y) = ⋁_{x \in X} (A (x) ⊖ d_{C}^{1} (y, x))$ . Moreover, if $C (⋁_{i \in Γ} A_{i}) = ⋁_{i \in Γ} C (A_{i})$ and $C (A ⊖ α) = C (A) ⊖ α$ for all A_i, A ∈ L^X, then $d_{C}^{1} = d_{C}$ and $C_{d_{C}^{1}} (A) = C (A)$ .

(5) If d_X is a distance function, then $d_{C_{d_{X}}} = d_{X}$ , $C_{d_{X}} (⋁_{i \in Γ} A_{i}) = ⋁_{i \in Γ} C_{d_{X}} (A_{i})$ and $C_{d_{X}} (A ⊖ α) = C_{d_{X}} (A) ⊖ α$ for all A_i, A ∈ L^X.

(6) Assume that L satisfies a double negative law and n (x) = ⊤ ⊖ x. Define $I (A) = n (C (n (A)))$ for all A ∈ L^X. Then $I$ is a fuzzy interior operator with $A \in τ_{I}$ if and only if $n (A) \in η_{C}$ .

Proof. (1) It can be proved by a similar method used in the proof of Theorem 3.9.

(2) (D1) $d_{C} (x, x) = n (C (n (⊥_{x}))) (x) \leq ⊥_{x} (x) = ⊥$ .

(D2) Since $C (A ⊖ α) \geq C (A) ⊖ α$ by Theorem 3.6(3) and $\begin{matrix} n (C (n (⊥_{x})) & = ⋀_{y \in X} (n (C (n (⊥_{x})) (y) \oplus ⊥_{y}), \\ C (n (⊥_{x})) & = ⋁_{y \in X} (n (⊥_{y}) ⊖ n (C (n (⊥_{x}))) (y)), \end{matrix}$ we have $\begin{matrix} C (n (⊥_{x})) (z) = C (C (n (⊥_{x}))) (z) \\ = C (⋁_{y \in X} (n (⊥_{y}) ⊖ n (C (n (⊥_{x}))) (y))) (z) \\ \geq ⋁_{y \in X} (C (n (⊥_{y})) (z) ⊖ n (C (n (⊥_{x}))) (y)) . \end{matrix}$ Thus by Lemma 2.3(11), we have $\begin{matrix} d_{C} (z, x) = n C (n (⊥_{x})) (z) \\ \leq ⋀_{y \in X} (n (C (n (⊥_{y}))) (z) \oplus n (C (n (⊥_{x}))) (y)) \\ = ⋁_{y \in X} (d_{C} (z, y) \oplus d_{C} (y, x)) . \end{matrix}$

Since $C (A ⊖ α) \geq C (A) ⊖ α$ , we have $\begin{matrix} C_{d_{C}} (A) (y) = ⋁_{x \in X} (A (x) ⊖ d_{C} (y, x)) \\ = ⋁_{x \in X} (A (x) ⊖ n (C (n (⊥_{x}))) (y)) \\ = ⋁_{x \in X} (C (n (⊥_{x})) (y) ⊖ n (A) (x)) \\ \leq C (⋁_{x \in X} (n (⊥_{x}) ⊖ n (A) (x))) (y) = C (A) (y) . \end{matrix}$

(3) It can be easily proved.

(4) By Lemma 2.3(11), we have $\begin{matrix} d_{C}^{1} (x, y) = ⋁_{A \in L^{X}} (C (A) (y) ⊖ C (A) (x)) \\ \geq ⋁_{z \in X} (C (n (⊥_{z})) (y) ⊖ C (n (⊥_{z})) (x)) \\ = ⋁_{z \in X} (n (C (n (⊥_{z}))) (x) ⊖ n (C (n (⊥_{z}))) (y)) \\ = ⋁_{z \in X} (d_{C} (x, z) ⊖ d_{C} (y, z)) = d_{C} (x, y) . \end{matrix}$ Since $d_{C}^{1}$ is a distance function, we have $\begin{matrix} C_{d_{C}^{1}} (A) (y) = ⋁_{x \in X} (A (x) ⊖ d_{C}^{1} (y, x)) \\ \leq ⋁_{x \in X} (C (A) (x) ⊖ (C (A) (x) ⊖ C (A) (y))) \\ \leq C (A) (y) . \end{matrix}$ Thus $η_{C} \subset η_{d_{C}^{1}}$ .

If $C (⋁_{i \in Γ} A_{i}) = ⋁_{i \in Γ} C (A_{i})$ and $C (α \oplus A) = α \oplus C (A)$ for all A_i, A ∈ L^X, then $\begin{matrix} C_{d_{C}} (A) (y) & = ⋁_{x \in X} (A (x) ⊖ d_{C} (y, x)) \\ = ⋁_{x \in X} (A (x) ⊖ n (C (n (⊥_{x}))) (y)) \\ = ⋁_{x \in X} (C (n (⊥_{x})) (y) ⊖ n (A) (x)) \\ = C (⋁_{x \in X} (n (⊥_{x}) ⊖ n (A) (x))) (y) \\ = C (A) (y) . \end{matrix}$ Since A = ⋁ _z∈X (n (⊥ _z) ⊖ n (A) (z)), we have $\begin{matrix} d_{C}^{1} (x, y) = ⋁_{A \in L^{X}} (C (A) (y) ⊖ C (A) (x)) \\ = ⋁_{A \in L^{X}} C (⋁_{z \in X} (n (⊥_{z}) ⊖ n (A) (z)) (y) \\ ⊖ C (⋁_{z \in X} (n (⊥_{z}) ⊖ n (A) (z)) (x)) \\ = ⋁_{A \in L^{X}} ⋁_{z \in X} (C (n (⊥_{z})) (y) ⊖ n (A) (z)) \\ ⊖ ⋁_{z \in X} (C (n (⊥_{z})) (x) ⊖ n (A) (z)) \\ (by Lemmas 3.7 (2) and 2.3 (6)) \\ \leq ⋁_{z \in X} (C (n (⊥_{z})) (y) ⊖ C (n (⊥_{z})) (x)) \\ = ⋁_{z \in X} (n (C (n (⊥_{z})) (x)) ⊖ n (C (n (⊥_{z}))) (y)) \\ (by Lemma 2.3 (11)) \\ = ⋁_{z \in X} (d_{C} (x, z) ⊖ d_{C} (y, z)) = d_{C} (x, y) . \end{matrix}$

(5) For all x, y ∈ X, we have $\begin{matrix} d_{C_{d_{X}}} (y, x) & = n (C_{d_{X}} (n (⊥_{x}))) (y) \\ = n (⋁_{z \in X} (n (⊥_{x}) (z) ⊖ d_{X} (y, z))) \\ = ⋀_{z \in X} (⊥_{x} (z) \oplus d_{X} (y, z)) \\ = d_{X} (y, x) . \end{matrix}$

(6) (I1) $I (⊤_{X}) = n (C (n (⊤_{X}))) = ⊤_{X}$ .

(I2) For all A ∈ L^X, $I (A) = n (C (n (A))) \leq n (n (A)) = A$ and $I (I (A)) = I (n (C (n (A)))) = n (C (C (n (A)))) = A$ .

(I3) For all A, B ∈ L^X, $\begin{matrix} d_{L^{X}} (I (A), I (B)) \\ = d_{L^{X}} (n (C (n (A))), n (C (n (B)))) \\ = d_{L^{X}} (C (n (B)), C (n (A))) (by Lemma 2.3 (11)) \\ \leq d_{L^{X}} (n (B), n (A)) = d_{L^{X}} (A, B) . \end{matrix}$ Moreover, $A \in τ_{I}$ if and only if $I (A) = A$ if and only if $C (n (A)) = n (A)$ if and only if $n (A) \in η_{C}$ . □

Theorem 3.11. Let (X, τ) be an Alexandrov pretopological space. Define $I_{τ} : L^{X} \to L^{X}$ by $I_{τ} (A) = ⋁_{C \in τ} (C ⊖ d_{L^{X}} (C, A))$ . Then the following hold.

(1) $I_{τ} (A) = ⋁_{i \in Γ} {A_{i} ∣ A_{i} \leq A, A_{i} \in τ} .$

(2) $I_{τ}$ is a fuzzy interior operator on X.

(3) $τ_{I_{τ}} = τ$ where $τ_{I_{τ}} = {A \in L^{X} ∣ A = I_{τ} (A)}$ .

(4) If $I$ is a fuzzy interior operator on X, then $I_{τ_{I}} = I$ .

(5) $I_{τ_{d_{τ}}} \geq I_{τ}$ where τ_{d
_τ} = {B ∈ L^X ∣ B (x) ⊕ d_τ (x, y) ≥ B (y)}. Moreover, if τ is an Alexandrov topology on X, then $I_{τ_{d_{τ}}} = I_{τ}$ .

Proof. (1) Let $I_{τ} (A) = ⋁_{A_{i} \in τ} (A_{i} ⊖ d_{L^{X}} (A_{i}, A))$ and I₁ (A) = ⋁ _i∈Γ {A_i ∣ A_i ≤ A, A_i ∈ τ}. Since A_i ≤ A ⊕ d_{L
^X} (A_i, A) if and only if A_i ⊖ d_{L
^X} (A_i, A) ≤ A, we have ⋁_{A_i∈τ} (A_i ⊖ d_{L
^X} (A_i, A)) ≤ A. Since ⋁_{A_i∈τ} (A_i ⊖ d_{L
^X} (A_i, A)) ∈ τ, we have $I_{τ} (A) \leq I_{1} (A)$ . Since I₁ (A) ∈ τ, we have $I_{τ} (A) \geq I_{1} (A) ⊖ d_{L^{X}} (I_{1} (A), A) = I_{1} (A) .$ Hence $I_{τ} = I_{1}$ .

(2) (I1) Let x ∈ X. Then $\begin{matrix} I_{τ} (⊤_{X}) (x) & = ⋁_{C \in τ} (C (x) ⊖ d_{L^{X}} (C, ⊤_{X})) \\ \geq ⊤_{X} (x) ⊖ d_{L^{X}} (⊤_{X}, ⊤_{X}) = ⊤ . \end{matrix}$

(I2) Let A ∈ L^X. Then B ⊖ d_{L
^X} (B, A) ≤ A. Hence $I_{τ} (A) \leq A$ . Since $I_{τ} (A) \in τ$ , we have $\begin{matrix} I_{τ} (I_{τ} (A)) & = ⋁_{C \in τ} (d_{L^{X}} (C, I_{τ} (A)) \oplus C) \\ \geq d_{L^{X}} (I_{τ} (A), I_{τ} (A)) \oplus I_{τ} (A) \\ = I_{τ} (A) . \end{matrix}$

(I3) Let A, C ∈ L^X. Then $\begin{matrix} \begin{matrix} d_{L^{X}} (I_{τ} (A), I_{τ} (C)) \\ = ⋁_{x \in X} (I_{τ} (A) (x) ⊖ I_{τ} (C) (x)) \end{matrix} \end{matrix}$ $\begin{matrix} \begin{matrix} = ⋁_{x \in X} (⋁_{B \in τ} (B (x) ⊖ d_{L^{X}} (B, A)) \\ ⊖ ⋁_{D \in τ} (B (x) ⊖ d_{L^{X}} (D, C))) \\ \leq ⋁_{x \in X} ⋁_{B \in τ} ((B (x) ⊖ d_{L^{X}} (B, A)) \\ ⊖ (B (x) ⊖ d_{L^{X}} (B, C))) (by Lemma 3.7) \\ \leq ⋁_{B \in τ} (d_{L^{X}} (B, C) ⊖ d_{L^{X}} (B, A)) = d_{L^{X}} (A, C) \\ (by Lemma 2.3 (8)) . \end{matrix} \end{matrix}$

(3) Let A ∈ τ. Then $A = I_{τ} (A)$ , and so $A \in τ_{I_{τ}}$ .

Let $A \in τ_{I_{τ}}$ . Then $A = I_{τ} (A) \in τ$ , and so A ∈ τ.

(4) Let A ∈ L^X. Then $\begin{matrix} I_{τ_{I}} (A) (x) & = ⋁_{C \in τ_{I}} (C (x) ⊖ d_{L^{X}} (C, A)) \\ \leq ⋁_{C \in τ_{I}} (C (x) ⊖ d_{L^{X}} (I (C), I (A))) \\ \leq I (A) (x) . \end{matrix}$

Since $I (I (A)) = I (A)$ for all A ∈ L^X, we have $I (A) \in τ_{I}$ and $\begin{matrix} I_{τ_{I}} (A) (x) & = ⋁_{C \in τ_{I}} (C (x) ⊖ d_{L^{X}} (C, A)) \\ \geq I (A) (x) ⊖ d_{L^{X}} (I (A), A) = I (A) (x) . \end{matrix}$

(5) $I_{τ} (A) = ⋁_{i \in Γ} {A_{i} ∣ A_{i} \leq A, A_{i} \in τ} \leq I_{τ_{d_{τ}}} (A) .$ Moreover, $I_{τ_{d_{τ}}} (A) = I_{τ} (A)$ .□

Theorem 3.12. Let (X, η) be an Alexandrov precotopological space. Define $C_{η} : L^{X} \to L^{X}$ by $C_{η} (A) = ⋀_{B \in η} (d_{L^{X}} (A, B) \oplus B) .$ (37)Then the following hold.

(1) $C_{η} (A) = ⋀ {B \in η ∣ A \leq B} .$

(2) $C_{η}$ is a fuzzy closure space on X.

(3) Assume that L satisfies a double negative law. Define τ_η ⊂ L^X by A ∈ τ_η if and only if n (A) ∈ η. Then τ_η is an Alexandrov pretopology.

(4) $η_{C_{η}} = η$ where $η_{C_{η}} = {A \in L^{X} ∣ A = C_{η} (A)}$ .

(5) Assume that L satisfies a double negative law. Then $C_{η} (A) = n (I_{τ_{η}} (n (A)))$ for all A ∈ L^X and $A \in τ_{I_{τ_{η}}}$ if and only if $n (A) \in η_{C_{η}}$ .

(6) If $C$ is a fuzzy closure space on X, then $C_{η_{C}} = C$ .

(7) $C_{τ_{d_{η}}} \leq C_{η}$ . If η is an Alexandrov topology on X, then $C_{τ_{d_{η}}} = C_{η}$ and $I_{τ_{d_{η}}} = I_{η}$ . Moreover, the pair $(I_{η}, C_{η})$ is a fuzzy rough set for A.

Proof. (1) Let $C_{1} (A) = ⋀_{i \in Γ} {A_{i} ∣ A \leq A_{i}, A_{i} \in η} .$ Since A ≤ (d_{L
^X} (A, A_i) ⊕ A_i, we have A ≤ ⋀ _{A_i∈η} (d_{L
^X} (A, A_i) ⊕ A_i) and ⋀_{A_i∈η} (d_{L
^X} (A, A_i) ⊕ A_i) ∈ η. Hence $C_{1} (A) \leq C_{η} (A)$ .

Since C₁ (A) ∈ η, we have $C_{η} (A) \leq d_{L^{X}} (A, C_{1} (A)) \oplus C_{1} (A) = C_{1} (A) .$

(2) (C1) Let x ∈ X. Then $⊥_{X} \leq C_{η} (⊥_{X})$ and $\begin{matrix} C_{η} (⊥_{X}) (x) = ⋀_{B \in L^{X}} (d_{L^{X}} (⊥_{X}, B) \oplus B (x)) \\ \leq d_{L^{X}} (⊥_{X}, ⊥_{X}) \oplus ⊥_{X} (x) = ⊥_{X} (x) . \end{matrix}$

(C2) Note that A ≤ d_{L
^X} (A, B) ⊕ B. Since $C_{η} (A) \in η$ , we have $\begin{matrix} C_{η} (C_{η} (A)) \leq d_{L^{X}} (C_{η} (A), C_{η} (A)) \oplus C_{η} (A) = C_{η} (A) . \end{matrix}$ Thus $C_{η} (A) = C_{η} (C_{η} (A))$ .

(C3) Let A, C ∈ L^X. Then $\begin{matrix} d_{L^{X}} (C_{η} (A), C_{η} (C)) \\ = ⋁_{x \in X} (C_{η} (A) (x) ⊖ C_{η} (C) (x)) \\ = ⋁_{x \in X} (⋀_{B \in η} (d_{L^{X}} (A, B) \oplus B (x)) \\ ⊖ ⋀_{D \in η} (d_{L^{X}} (C, D) \oplus D (x)))) \\ (by Lemma 3.7 (2)) \\ \leq ⋁_{x \in X} ⋁_{B \in η} (d_{L^{X}} (A, B) \oplus B (x)) \\ ⊖ (d_{L^{X}} (C, B) \oplus B (x))) \\ \leq ⋁_{B \in η} (d_{L^{X}} (A, B) ⊖ d_{L^{X}} (C, B)) \\ (by Lemma 2.3 (8)) \\ \leq ⋁_{B \in η} ⋁_{x \in X} ((A (x) ⊖ B (x)) ⊖ (C (x) ⊖ B (x))) \\ \leq d_{L^{X}} (A, C) . \end{matrix}$

(3) Since n (α_X) = n (α) _X ∈ η, we have α_X ∈ τ_η. If A ∈ τ_η, then n (A ⊖ α) = n (A) ⊕ α ∈ η. Thus A ⊖ α ∈ τ_η.

If A_i ∈ τ_η for all i ∈ Γ, then n (⋁ _i∈ΓA_i) = ⋀ _i∈Γn (A_i) ∈ η. Hence ⋁_i∈ΓA_i ∈ τ_η.

(4) It can be similarly proved as in that of Theorem 3.11(3).

(5) Let A ∈ L^X. Then $\begin{matrix} n (I_{τ_{η}} (n (A))) & = n (⋁_{B \in τ_{η}} (B ⊖ d_{L^{X}} (B, n (A))) \\ = ⋀_{B \in τ_{η}} (d_{L^{X}} (A, n (B)) \oplus n (B)) \\ = ⋀_{C \in η} (d_{L^{X}} (A, C) \oplus C) \\ = C_{η} (A) . \end{matrix}$ Moreover, $A \in τ_{I_{τ_{η}}}$ if and only if $A = I_{τ_{η}} (A)$ if and only if $C_{η} (n (A)) = n (A)$ if and only if $n (A) \in η_{C_{η}}$ .

(6) Let A ∈ L^X. Then $\begin{matrix} C_{η_{C}} (A) & = ⋀_{B \in η_{C}} (d_{L^{X}} (A, B) \oplus B) \\ \geq ⋀_{B \in η_{C}} ((d_{L^{X}} (C (A), C (B))) \oplus B) \\ \geq C (A) . \end{matrix}$

Since $C (C (A)) = C (A)$ for all A ∈ L^X, we have $C (A) \in η_{C}$ . Thus $\begin{matrix} C_{η_{C}} (A) (x) & = ⋀_{B \in η_{C}} (d_{L^{X}} (A, B) \oplus B (x)) \\ \leq d_{L^{X}} (A, C (A)) \oplus C (A) (x) \\ = C (A) (x) . \end{matrix}$

(7) $C_{η} (A) = ⋀_{i \in Γ} {A_{i} ∣ A \leq A_{i}, A_{i} \in η} \geq C_{τ_{d_{η}}} (A) .$ If η is an Alexandrov topology on X, then B = B (x) ⊕ ⋁ _A∈η (A (y) ⊖ A (-)) ∈ η for all B ∈ τ_{d
_η}. Thus B ∈ η. Hence η = τ_{d
_η}, $C_{τ_{d_{η}}} = C_{η}$ and $I_{τ_{d_{η}}} = I_{η}$ .□

Theorem 3.13. Let d_X be a distance function and let τ_{d
_X} = {A ∈ L^X ∣ A (x) ⊕ d_X (x, y) ≥ A (y)} be an Alexandrov topology on X. Then the following hold.

(1) $I_{τ_{d_{X}}} (A) = ⋁_{C \in τ_{d_{X}}} (C ⊖ d_{L^{X}} (C, A)) = ⋁ {B \in τ_{d_{X}} ∣ B \leq A} = ⋀_{y \in X} (d_{X} (y, -) \oplus A (y)) = I_{d_{X}} (A) .$

(2) $C_{τ_{d_{X}}} (A) = ⋀_{B \in τ_{d_{X}}} (d_{L^{X}} (A, B) \oplus B) = ⋀ {B \in τ_{d_{X}} ∣ A \leq B} = ⋁_{x \in X} (A (x) ⊖ d_{X} (-, x)) = C_{d_{X}} (A) .$

(3) The pair $(I_{τ_{d_{X}}} (A), C_{τ_{d_{X}}} (A))$ is a fuzzy rough set for A.

(4) d_{τ
_{d
_X}} = d_X where d_{τ
_{d
_X}} (x, y) = ⋁ _{A∈τ_{d
_X}} (A (y) ⊖ A (x)) for all x, y ∈ X.

Proof. (1) By Theorem 3.11(1), we have $I_{τ_{d_{X}}} (A) = ⋁ {B \in τ_{d_{X}} ∣ B \leq A} .$ Since ⋀_y∈X (d_X (y, -) ⊕ A (y)) ∈ τ_{d
_X} by Theorem 3.8(2) and ⋀_y∈X (d_X (y, -) ⊕ A (y)) ≤ A, we have $⋀_{y \in X} (d_{X} (y, -) \oplus A (y)) \leq I_{τ_{d_{X}}} (A) .$ Since $I_{τ_{d_{X}}} (A) \leq A$ and $I_{τ_{d_{X}}} (A) \in τ_{d_{X}}$ , we have $I_{τ_{d_{X}}} (A) (x) \oplus d_{X} (x, y) \geq I_{τ_{d_{X}}} (A) (y)$ , and so $\begin{matrix} I_{τ_{d_{X}}} (A) & = ⋀_{y \in X} (d_{X} (y, -) \oplus I_{τ_{d_{X}}} (A) (y)) \\ \leq ⋀_{y \in X} (d_{X} (y, -) \oplus A (y)) . \end{matrix}$ Hence $I_{τ_{d_{X}}} (A) = ⋀_{y \in X} (d_{X} (y, -) \oplus A (y)) .$

(2) By Theorem 3.12(1), we have $C_{τ_{d_{X}}} (A) = ⋀ {B \in τ_{d_{X}} ∣ A \leq B} .$ Since ⋁_x∈X (A (x) ⊖ d_X (- , x)) ∈ τ_{d
_X} by Theorem 3.8(3) and A ≤ ⋁ _x∈X (A (x) ⊖ d_X (- , x)), we have $C_{τ_{d_{X}}} (A) \leq ⋁_{x \in X} (A (x) ⊖ d_{X} (-, x)) .$ Since $A \leq C_{τ_{d_{X}}} (A)$ and $C_{τ_{d_{X}}} (A) \in τ_{d_{X}}$ , we have $\begin{matrix} ⋁_{x \in X} (A (x) ⊖ d_{X} (-, x)) \\ \leq ⋁_{x \in X} (C_{τ_{d_{X}}} (A) (x) ⊖ d_{X} (-, x)) \\ = C_{τ_{d_{X}}} (A) . \end{matrix}$ Thus $C_{τ_{d_{X}}} (A) = ⋁_{x \in X} (A (x) ⊖ d_{X} (-, x)) .$

(3) It follows from (1) and (2).

(4) Since d_X (z, -) ∈ τ_{d
_X} and $⋀_{z \in X} (A (z) \oplus d_{X} (z, x)) = A (x)$ for all A ∈ τ_{d
_X}, we have $\begin{matrix} d_{τ_{d_{X}}} (x, y) = ⋁_{A \in τ_{d_{X}}} (A (y) ⊖ A (x)) \\ = ⋁_{A \in τ_{d_{X}}} (⋀_{z \in X} (A (z) \oplus d_{X} (z, y)) \\ ⊖ ⋀_{z \in X} (A (z) \oplus d_{X} (z, x))) \\ \leq ⋁_{A \in τ_{d_{X}}} (⋁_{z \in X} ((A (z) \oplus d_{X} (z, y)) \\ ⊖ (A (z) \oplus d_{X} (z, x)))) (by Lemma 3.7 (2)) \\ \leq ⋁_{z \in X} (d_{X} (z, y) ⊖ d_{X} (z, x)) = d_{X} (x, y) \\ (by Lemma 2.3 (8)) \end{matrix}$ and $\begin{matrix} d_{τ_{d_{X}}} (x, y) & = ⋁_{A \in τ_{d_{X}}} (A (y) ⊖ A (x)) \\ \geq ⋁_{z \in X} (d_{X} (z, y) ⊖ d_{X} (z, x)) \\ = d_{X} (x, y) . \end{matrix}$ □

Example 3.14. Let X = {x, y, z} and let ([0, ∞] , ≤ , ∨ , ⊕ = + , ∧ , 0, ∞) be a complete co-residuated lattice in Remark 2.2(4). Let A ∈ [0, ∞] ^X with A (x) =8, A (y) =3 and A (z) =9.

(1) Define an Alexandrov pretopology τ_X = {(A ⊖ α) ∨ β_X ∣ α, β ∈ L} . Since ((A ⊖ α) ∨ β_X) ⊖ γ = (A ⊖ (α ⊕ γ)) ∨ (β_X ⊖ γ), we have ((A ⊖ α) ∨ β_X) ⊖ γ ∈ τ_X.

Since (A (x) ⊖ α) ⊕ (A (y) ⊖ A (x)) ≥ (A (x) ⊕ (A (y) ⊖ A (x))) ⊖ α ≥ (A (y) ⊖ α) by Lemma 2.3(10) and (A (x) ∨ β_X) ⊕ (A (y) ⊖ A (x)) ≥ (A (y) ∨ β_X), we have d_{τ
_X} (x, y) = ⋁ _{C∈τ_X} (C (y) ⊖ C (x)) = A (y) ⊖ A (x) where $d_{τ_{X}} = (\begin{matrix} 0 & 0 & 1 \\ 5 & 0 & 6 \\ 0 & 0 & 0 \end{matrix}) .$ By Theorem 3.2, we obtain an Alexandrov topology τ_{d
_{τ
_X}} such that 6 ⊕ A ∈ τ_{d
_{τ
_X}}, but 6 ⊕ A ∉ τ_X.

By Theorem 3.11(3), we have $τ_{I_{τ_{X}}} = τ_{X}$ where $τ_{I_{τ_{X}}} = {A \in [0, \infty]^{X} ∣ A = I_{τ_{X}} (A)}$ . Since $6 \oplus A = (14, 9, 15) \notin τ_{X} = τ_{I_{τ_{X}}}$ , we have that $τ_{X} = τ_{I_{τ_{X}}}$ is not an Alexandrov precotopology.

Let B = (2, 4, 3) ∈ [0, ∞] ^X. Then $\begin{matrix} I_{τ_{X}} (B) & = ⋁ {A_{i} \in τ_{X} ∣ A_{i} \leq B} \\ = (A ⊖ 6) \lor 2_{X} \\ = (2, 0, 3) \lor 2_{X} \\ = (2, 2, 3) . \end{matrix}$ Note that $I_{τ_{X}} (6 \oplus A) = 9_{X} < I_{τ_{d_{τ_{X}}}} (6 \oplus A) = I_{d_{τ_{X}}} (6 \oplus A) = 6 \oplus A$ .

By Theorem 3.8(2) and (3), we obtain an Alexandrov topology $\begin{matrix} τ_{d_{τ_{X}}} & = {((A ⊖ α) \lor β_{X}), (α \oplus A) \land β_{X} ∣ α, β \in L} \\ = {⋀_{x \in X} (C (x) \oplus d_{τ_{X}} (x, -)) ∣ C \in [0, \infty]^{X}} \\ = {⋁_{x \in X} (D (x) ⊖ d_{τ_{X}} (-, x)) ∣ D \in [0, \infty]^{X}} \end{matrix}$ where $\begin{matrix} I_{d_{τ_{X}}} (C) & = ⋀_{x \in X} (C (x) \oplus d_{τ_{X}} (x, -)) \\ = (\begin{matrix} C (x) \land (C (y) \oplus 5) \land C (z) \\ C (x) \land C (y) \land C (z) \\ (C (x) \oplus 1) \land (C (y) \oplus 6) \land C (z) \end{matrix}) \end{matrix}$ and $\begin{matrix} C_{d_{τ_{X}}} (D) & = ⋁_{x \in X} (D (x) ⊖ d_{τ_{X}} (-, x)) \\ = (\begin{matrix} D (x) \lor D (y) \lor (D (z) - 1) \\ (D (x) - 5) \lor D (y) \lor (D (z) - 6) \\ D (x) \lor D (y) \lor D (z) \end{matrix}) . \end{matrix}$ The pair $(I_{d_{τ_{X}}} (D) = (3, 3, 4), C_{d_{τ_{X}}} (D) = (7, 4, 8))$ is a fuzzy rough set of D = (3, 4, 8). The fuzzy accuracy measure α (D) of D is $α (D) = ⋁_{x \in X} (C_{d_{τ_{X}}} (D) (x) ⊖ I_{d_{τ_{X}}} (D) (x)) = 4 .$ (38)

(2) Define an Alexandrov precotopology η_X = {(α ⊕ A) ∧ β_X ∣ α, β ∈ L} . Let B = (2, 4, 3) ∈ [0, ∞] ^X. Then $\begin{matrix} C_{η_{X}} (B) & = ⋀ {A_{i} \in η_{X} ∣ B \leq A_{i}} \\ = (1 \oplus A) \land 4_{X} = (3, 4, 4) . \end{matrix}$ Since $\begin{matrix} ((α \oplus A (x)) \land β_{X}) \oplus (A (y) ⊖ A (x)) \\ \geq (α \oplus A (y)) \land β_{X}, \end{matrix}$ we obtain a distance function d_{η
_X} with $\begin{matrix} d_{η_{X}} (x, y) & = ⋁_{B \in η_{X}} (B (y) ⊖ B (x)) \\ = A (y) ⊖ A (x) = d_{τ_{X}} (x, y) \end{matrix}$ where τ_X is defined in (1). By Theorem 3.8(2) and (3), we obtain an Alexandrov topology $\begin{matrix} τ_{d_{η_{X}}} & = {((A ⊖ α) \lor β_{X}), (α \oplus A) \land β_{X} ∣ α, β \in L} \\ = {⋀_{x \in X} (C (x) \oplus d_{η_{X}} (x, -)) ∣ C \in [0, \infty]^{X}} \\ = {⋁_{x \in X} (D (x) ⊖ d_{η_{X}} (-, x)) ∣ D \in [0, \infty]^{X}} . \end{matrix}$

The pair $(I_{d_{η_{X}}} (B) = (2, 2, 3), C_{d_{η_{X}}} (B) = (3, 4, 4))$ is a fuzzy rough set of B = (2, 4, 3) ∈ [0, ∞] ^X and $α (B) = ⋁_{x \in X} (C_{d_{η_{X}}} (B) (x) ⊖ I_{d_{η_{X}}} (B) (x)) = 2 .$

Example 3.15. Let X = {h_i ∣ i = {1, 2, 3}} and Y = {e, b, w, c, i} be sets with h_i=house and e=expensive,b= beautiful, w=wooden, c= creative, i=in the green surroundings. Let ([0, 1] , ⊕ , ⊖ , 0, 1) be a complete co-residuated lattice defined in Remark 2.2(5) with p = 1 where $x \oplus y = min {x + y, 1}, x ⊖ y = max {x - y, 0} .$ (39) Define n (x) =1 ⊖ x = 1 - x. Then n (n (x)) = x. Let R ∈ [0, 1] ^X×Y be a fuzzy information by $\begin{matrix} R & e & b & w & c & i \\ h_{1} & 0.3 & 0.6 & 0.5 & 0.8 & 0.2 \\ h_{2} & 0.6 & 0.4 & 0.4 & 0.3 & 0.5 \\ h_{3} & 0.4 & 0.5 & 0.6 & 0.5 & 0.4 \end{matrix}$ (40) Define two distance functions $d_{X}^{Y}, d_{X}^{{e, b}} \in [0, 1]^{X \times X}$ by $\begin{matrix} d_{X}^{Y} (h_{i}, h_{j}) & = ⋁_{y \in Y} (R (h_{i}, y) ⊖ R (h_{j}, y)), \\ d_{X}^{{e, b}} (h_{i}, h_{j}) & = ⋁_{y \in {e, b}} (R (h_{i}, y) ⊖ R (h_{j}, y)), \end{matrix}$ $d_{X}^{Y} = (\begin{matrix} 0 & 0.5 & 0.3 \\ 0.3 & 0 & 0.2 \\ 0.2 & 0.2 & 0 \end{matrix}), d_{X}^{{e, b}} = (\begin{matrix} 0 & 0.2 & 0.1 \\ 0.3 & 0 & 0.2 \\ 0.1 & 0.1 & 0 \end{matrix}) .$ Then $d_{X}^{Y} (h_{1}, -) = (0, 0.5, 0.3) \in τ_{d_{X}^{Y}}$ and $d_{X}^{Y} (h_{1}, -) \notin τ_{d_{X}^{{e, b}}}$ because $d_{X}^{Y} (h_{1}, h_{1}) \oplus d_{X}^{{e, b}} (h_{1}, h_{2}) = 0.2 ⪈ d_{X}^{Y} (h_{1}, h_{2}) = 0.5 .$ By Theorem 3.8(2) and (3), we obtain an Alexandrov [0, 1]-topology $\begin{matrix} τ_{d_{X}^{Y}} & = {⋀_{x \in X} (A (x) \oplus d_{X}^{Y} (x, -)) ∣ A \in [0, 1]^{X}} \\ = {⋁_{x \in X} (A (x) ⊖ d_{X}^{Y} (-, x)) ∣ A \in [0, 1]^{X}} \end{matrix}$ where $\begin{matrix} I_{d_{X}^{Y}} & (A) = ⋀_{x \in X} (A (x) \oplus d_{X}^{Y} (x, -)) \\ = (\begin{matrix} A (x) \land (A (y) + 0.3) \land (A (z) + 0.2) \\ (A (x) + 0.5) \land A (y) \land (A (z) + 0.2) \\ (A (x) + 0.3) \land (A (y) + 0.2) \land A (z) \end{matrix}), \\ C_{d_{X}^{Y}} & (A) = ⋁_{x \in X} (A (x) ⊖ d_{X}^{Y} (-, x)) \\ = (\begin{matrix} A (x) \lor (A (y) - 0.5) \lor (A (z) - 0.3) \\ (A (x) - 0.3) \lor A (y) \lor (A (z) - 0.2) \\ (A (x) - 0.2) \lor (A (y) - 0.2) \lor A (z) \end{matrix}) . \end{matrix}$ The pair $(I_{d_{X}^{Y}} (A), C_{d_{X}^{Y}} (A))$ is a fuzzy rough set for A. By Theorem 3.8, we easily show that $C_{(d_{X}^{Y})^{- 1}} (A) = n (I_{d_{X}^{Y}} (n (A)))$ and $I_{(d_{X}^{Y})^{- 1}} (A) = n (C_{d_{X}^{Y}} (n (A)))$ .

Moreover, we obtain an Alexandrov [0, 1]-topology $\begin{matrix} τ_{d_{X}^{{e, b}}} & = {⋀_{x \in X} (A (x) \oplus d_{X}^{{e, b}} (x, -)) ∣ A \in L^{X}} \\ = {⋁_{x \in X} (A (x) ⊖ d_{X}^{{e, b}} (-, x)) ∣ A \in L^{X}} \end{matrix}$ where $\begin{matrix} I_{d_{X}^{{e, b}}} & (A) = ⋀_{x \in X} (A (x) \oplus d_{X}^{{e, b}} (x, -)) \\ = (\begin{matrix} A (x) \land (A (y) + 0.3) \land (A (z) + 0.1) \\ (A (x) + 0.2) \land A (y) \land (A (z) + 0.1) \\ (A (x) + 0.1) \land (A (y) + 0.2) \land A (z) \end{matrix}), \\ C_{d_{X}^{{e, b}}} & (A) = ⋁_{x \in X} (A (x) ⊖ d_{X}^{{e, b}} (-, x)) \\ = (\begin{matrix} A (x) \lor (A (y) - 0.2) \lor (A (z) - 0.1) \\ (A (x) - 0.3) \lor A (y) \lor (A (z) - 0.2) \\ (A (x) - 0.1) \lor (A (y) - 0.1) \lor A (z) \end{matrix}) . \end{matrix}$ The pair $(I_{d_{X}^{{e, b}}} (A), C_{d_{X}^{{e, b}}} (A))$ is a fuzzy rough set for A. By Theorem 3.8, one can easily see that $C_{(d_{X}^{{e, b}})^{- 1}} (A) = n (I_{d_{X}^{{e, b}}} (n (A)))$ and $I_{(d_{X}^{{e, b}})^{- 1}} (A) = n (C_{d_{X}^{{e, b}}} (n (A))) .$

4 The categorical relation between topological structures

Definition 4.1. [1, 8] Let $F : D \to C$ and $G : C \to D$ be functors.

(1) $C$ and $D$ are isomorphic if $G \circ F = {id}_{D}$ and $F \circ G = {id}_{C}$ .

(2) The pair (F, G) is called a Galois correspondence between $C$ and $D$ if for all $Y \in C,$ id_Y : F ∘ G (Y) → Y is a $C$ -morphism, and for all $X \in D$ , id_X : X → G ∘ F (X) is a $D$ -morphism.

If (F, G) is a Galois correspondence, then it is easy to check that F is a left adjoint of G, or equivalently that G is a right adjoint of F .

Let DIS be a category with object (X, d_X) where d_X is a distance function with a D-morphism f : (X, d_X) → (Y, d_Y) such that d_Y (f (x) , f (y)) ≤ d_X (x, y) for all x, y ∈ X.

Let ATOP(resp. APT, PCT) be a category with object (X, τ_X) where τ_X is an Alexandrov topology (resp. pretopolgy, precotopolgy) with an A-morphism (resp. P-morphism, PC-morphism) f : (X, τ_X) → (Y, τ_Y) such that f^← (B) ∈ τ_X for all B ∈ τ_Y.

Theorem 4.2. (1) Δ : DIS → APT is a left adjoint of Ω : APT → DIS, i.e., (Δ, Ω) is a Galois correspondence. Moreover, Ω is a left inverse of Δ, i.e., Ω ∘ Δ (X, d_X) = (X, d_X) for all (X, d_X) ∈ DIS.

(2) Π : DIS → PCT is a left adjoint of Σ : PCT → DIS, i.e., (Π, Σ) is a Galois correspondence. Moreover, Σ is a left inverse of Π, i.e., Σ ∘ Π (X, d_X) = (X, d_X) for all (X, d_X) ∈ DIS.

(3) Two categories ATOP and DIS are isomorphic.

Proof. (1) Define Δ : DIS → APT by Δ (X, d_X) = (X, τ_{d
_X}) where τ_{d
_X} is an Alexandrov topology on X in Theorem 3.2.

Let d_X (x, z) ≥ d_Y (f (x) , f (z)). For all B ∈ τ_{d
_X}, we have $\begin{matrix} f^{\leftarrow} (B) (x) \oplus d_{X} (x, z) & \geq B (f (x)) \oplus d_{Y} (f (x), f (z)) \\ \geq B (f (z)) \\ = f^{\leftarrow} (B) (z) . \end{matrix}$ Hence Δ is a functor.

Define Ω : APT → DIS by Ω (X, τ_X) = (X, d_{τ
_X}) where d_{τ
_X} (x, y) = ⋁ _{A∈τ_X} (A (y) ⊖ A (x)).

Let f^← (B) ∈ τ_X for all B ∈ τ_Y. Then $\begin{matrix} d_{τ_{Y}} (f (x), f (y)) & = ⋁_{B \in τ_{Y}} (B (f (y)) ⊖ B (f (x))) \\ = ⋁_{B \in τ_{Y}} (f^{\leftarrow} (B) (y) ⊖ f^{\leftarrow} (B) (x)) \\ \leq ⋁_{A \in τ_{X}} (A (y) ⊖ A (x)) \\ = d_{τ_{X}} (x, y) . \end{matrix}$ Hence Ω is a functor.

By Theorem 3.13(4), we have Ω (Δ (X, d_X)) = Ω (X, τ_{d
_X}) = (X, d_{τ
_{d
_X}}).

Since τ_X ⊂ τ_{d
_{τ
_X}} by Theorem 3.3(1), we have id_X : Δ (Ω (X, τ_X)) = Δ (X, d_{τ
_X}) = (X, τ_{d
_{τ
_X}}) → (X, τ_X) is a P-morphism.

(2) Define Π : DIS → PCT by Π (X, d_X) = (X, τ_{d
_X}) where τ_{d
_X} is an Alexandrov topology on X in Theorem 3.2. Define Σ : PCT → DIS by Σ (X, η_X) = (X, d_{η
_X}) where $d_{η_{X}} (x, y) = ⋁_{A \in η_{X}} (A (y) ⊖ A (x)) .$ Then Π and Σ are functors. By Theorem 3.13(4), we have Σ (Π (X, d_X)) = Σ (X, τ_{d
_X}) = (X, d_{τ
_{d
_X}}).

Since η_X ⊂ τ_{d
_{η
_X}} by Theorem 3.3(2), we have id_X : Π (Σ (X, η_X)) = Π (X, d_{η
_X}) = (X, τ_{d
_{η
_X}}) → (X, η_X) is a PC-morphism.

(3) It follows by (1) and Theorem 3.3(1).□

Let FI be a category with object $(X, I_{X})$ where $I_{X}$ is a fuzzy interior operator with an I-morphism $f : (X, I_{X}) \to (Y, I_{Y})$ such that $f^{\leftarrow} (I_{Y} (B)) \leq I_{X} (f^{\leftarrow} (B))$ for all B ∈ L^Y.

Theorem 4.3. (1) Θ : DIS → FI is a left adjoint of Φ : FI → DIS, i.e., (Θ, Φ) is a Galois correspondence. Moreover, Φ is a left inverse of Θ, i.e., Φ ∘ Θ (X, d_X) = (X, d_X) for all (X, d_X) ∈ DIS.

(2) If L satisfies a double negation law, then two categories APT and FI are isomorphic.

Proof. (1) Define Θ : DIS → FI by $Θ (X, d_{X}) = (X, I_{d_{X}})$ where $I_{d_{X}}$ is a fuzzy interior operator defined in Theorem 3.8(1). Let d_Y (f (x) , f (z)) ≤ d_X (x, z). Since $\begin{matrix} f^{\leftarrow} (I_{d_{Y}} (B)) & (x) = ⋀_{w \in Y} (B (w) \oplus d_{Y} (w, f (x))) \\ \leq ⋀_{z \in X} (B (f (z)) \oplus d_{Y} (f (z), f (x))) \\ \leq ⋀_{z \in X} (f^{\leftarrow} (B) (z) \oplus d_{X} (z, x)) \\ = I_{d_{X}} (f^{\leftarrow} (B)) (x), \end{matrix}$ $f : (X, I_{d_{X}}) \to (Y, I_{d_{Y}})$ is an I-morphism. Hence Θ is a functor.

Define a functor Φ : FI → DIS by $G (X, I_{X}) = (X, d_{I_{X}})$ where $d_{I_{X}}$ is the distance function in Theorem 3.9(2). Let $f^{\leftarrow} (I_{Y} (B)) \leq I_{X} (f^{\to} (B))$ . Then $\begin{matrix} d_{I_{Y}} (f (x), f (z)) & = I_{Y} (⊥_{f (x)}) (f (y)) \\ = f^{\leftarrow} (I_{Y} (⊥_{f (x)})) (y) \\ \leq I_{X} (f^{\leftarrow} (⊥_{f (x)})) (y) \\ \leq I_{X} (⊥_{x}) (y) \\ = d_{I_{X}} (x, y) . \end{matrix}$ Hence Φ is a functor. Moreover, by Theorem 3.9(5), we have $Φ (Θ (X, d_{X})) = Φ (X, I_{d_{X}}) = (X, d_{I_{d_{X}}}) = (X, d_{X}) .$ By Theorem 3.9(2), we have $Θ (Φ (X, I_{X})) = Θ (X, d_{I_{X}}) = (X, I_{d_{I_{X}}})$ and ${id}_{X} : Θ (Φ (X, I_{X})) = (X, I_{d_{I_{X}}}) \to (X, I_{X})$ is an I-morphism because $I_{X} \leq I_{d_{I_{X}}}$ .

(2) Define Ξ : APT → FI by $Ξ (X, τ_{X}) = (X, I_{τ_{X}})$ where $I_{τ_{X}} (A) = ⋁ {A_{i} \in τ_{X} ∣ A_{i} \leq A}$ in Theorem 3.11(1). Let f^← (B) ∈ τ_X for each B ∈ τ_Y. Since $\begin{matrix} f^{\leftarrow} & (I_{τ_{Y}} (B)) = f^{\leftarrow} (⋁ {B_{i} \in τ_{Y} ∣ B_{i} \leq B}) \\ \leq ⋁ {f^{\leftarrow} (B_{i}) \in τ_{X} ∣ f^{\leftarrow} (B_{i}) \leq f^{\leftarrow} (B)}) \\ \leq ⋁ {A \in τ_{X} ∣ A \leq f^{\leftarrow} (B)}) \\ = I_{τ_{X}} (f^{\leftarrow} (B)), \end{matrix}$ $f : (X, I_{τ_{X}}) \to (Y, I_{τ_{Y}})$ is an I-morphism. Hence Ξ is a functor.

Define a functor Γ : FI → APT by $Γ (X, I_{X}) = (X, τ_{I_{X}})$ where $τ_{I_{X}}$ is an Alexandrov pretopology in Theorem 3.9(1). Let $f^{\leftarrow} (I_{Y} (B)) \leq I_{X} (f^{\to} (B))$ . For all $B \in τ_{I_{Y}}$ , $f^{\leftarrow} (B) = f^{\leftarrow} (I_{Y} (B)) \leq I_{X} (f^{\leftarrow} (B)),$ and so, $f^{\leftarrow} (B) \in τ_{I_{X}}$ . Hence Γ is a functor. Moreover, by Theorem 3.11(3), we have $Γ (Ξ (X, τ_{X})) = Γ (X, I_{τ_{X}}) = (X, τ_{I_{τ_{X}}}) = (X, τ_{X}) .$ By Theorem 3.11 (4), we have $Ξ (Γ (X, I_{X})) = Ξ (X, τ_{I_{X}}) = (X, I_{τ_{I_{X}}}) = I_{X} .$ □

Let FC be a category with object $(X, C_{X})$ where $C_{X}$ is a fuzzy closure operator with a C-morphism $f : (X, C_{X}) \to (Y, C_{Y})$ such that $f^{\leftarrow} (C_{Y} (B)) \geq C_{X} (f^{\leftarrow} (B))$ for all B ∈ L^Y.

Theorem 4.4. (1) If L satisfies a double negation law, then Λ : DIS → FC is a left adjoint of Ψ : FC → DIS, i.e., (Λ, Ψ) is a Galois correspondence. Moreover, Ψ is a left inverse of Λ, i.e., Ψ ∘ Λ (X, d_X) = (X, d_X) for any (X, d_X) ∈ DIS.

(2) Two categories PCT and FC are isomorphic.

Proof. (1) Define Λ : DIS → FC by $Λ (X, d_{X}) = (X, C_{d_{X}})$ where $C_{d_{X}}$ is a fuzzy closure operator in Theorem 3.8(1).

Let d_X (x, z) ≥ d_Y (f (x) , f (z)). Since $\begin{matrix} f^{\leftarrow} (C_{d_{Y}} (B)) & (x) = C_{d_{Y}} (B) (f (x)) \\ = ⋁_{y \in Y} (B (y) ⊖ d_{Y} (f (x), y)) \\ \geq ⋁_{x \in X} (B (f (z)) ⊖ d_{Y} (f (x), f (z))) \\ \geq ⋁_{x \in X} (f^{\leftarrow} (B) (z) ⊖ d_{X} (x, z)) \\ = C_{d_{X}} (f^{\leftarrow} (B)) (x), \end{matrix}$ $f : (X, C_{X}) \to (Y, C_{Y})$ is a C-morphism. Hence Λ is a functor.

Define Ψ : FC → DIS by $Ψ (X, C_{X}) = (X, d_{C_{X}})$ where $d_{C_{X}}$ is a distance function in Theorem 3.10(2).

Let $f^{\leftarrow} (C_{Y} (B)) \geq C_{X} (f^{\leftarrow} (B))$ for all B ∈ L^Y. Let B = n (⊥ _f(x)). Since $f^{\leftarrow} (n (⊥_{f (x)})) (y) = n (⊥_{f (x)}) (f (y)) \geq n (⊥_{x}) (y),$ we have $\begin{matrix} f^{\leftarrow} (C_{Y} (n (⊥_{f (x)}))) (y) & \geq C_{X} (f^{\leftarrow} (n (⊥_{f (x)}))) (y) \\ \geq C_{X} (n (⊥_{x})) (y), \end{matrix}$ and so $C_{Y} (n (⊥_{f (x)})) (f (y)) \geq C_{X} (n (⊥_{x})) (y)$ . Thus $\begin{matrix} d_{C_{X}} (x, y) & = n (C_{X} (n (⊥_{x}))) (y) \\ \geq n (C_{Y} (n (⊥_{f (x)}))) (f (y)) \\ = d_{C_{Y}} (f (x), f (y)) . \end{matrix}$ Hence Ψ is a functor.

Note by Theorem 3.10(5) that $Ψ (Λ (X, d_{X})) = Ψ (X, C_{d_{X}}) = (X, d_{C_{d_{X}}}) = (X, d_{X}) .$

Since $C_{d_{C_{X}}} \leq C_{X}$ by Theorem 3.10(2), We have $Λ (Ψ (X, C_{X})) = Λ (X, d_{C_{X}}) = (X, C_{d_{C_{X}}})$ and ${id}_{X} : Λ (Ψ (X, C_{X})) = (X, C_{d_{C_{X}}}) \to (X, C_{X})$ is a C-morphism.

(2) Define ϒ : PCT→ FC by $ϒ (X, η_{X}) = (X, C_{η_{X}})$ where $C_{η_{X}} (A) = ⋀ {A_{i} \in η_{X} ∣ A \leq A_{i}}$ in Theorem 3.12(1). Let f^← (B) ∈ η_X for all B ∈ η_Y. Since $\begin{matrix} f^{\leftarrow} (C_{η_{Y}} & (B)) = f^{\leftarrow} (⋀ {B_{i} \in η_{Y} ∣ B \leq B_{i}}) \\ \geq ⋀ {f^{\leftarrow} (B_{i}) \in η_{X} ∣ f^{\leftarrow} (B) \leq f^{\leftarrow} (B_{i})}) \\ \geq ⋁ {A \in η_{X} ∣ f^{\leftarrow} (B) \leq A}) \\ = C_{η_{X}} (f^{\leftarrow} (B)), \end{matrix}$ $f : (X, C_{η_{X}}) \to (Y, C_{η_{Y}})$ is a C-morphism. Hence ϒ is a functor.

Define a functor ∇ : FC → PCT by $\nabla (X, C_{X}) = (X, η_{C_{X}})$ where $η_{C_{X}}$ is an Alexandrov pretopology in Theorem 3.10(1). Let $f^{\leftarrow} (C_{Y} (B)) \geq C_{X} (f^{\to} (B))$ . For all $B \in η_{C_{Y}}$ , $f^{\leftarrow} (B) = f^{\leftarrow} (C_{Y} (B)) \geq C_{X} (f^{\leftarrow} (B)),$ and so, $f^{\leftarrow} (B) \in η_{C_{X}}$ . Hence ∇ is a functor. Moreover, by Theorem 3.12 (4), we have $\nabla (ϒ (X, η_{X})) = \nabla (X, C_{η_{X}}) = (X, η_{C_{η_{X}}}) = (X, η_{X}) .$ By Theorem 3.12 (6), we have $ϒ (\nabla (X, C_{X})) = ϒ (X, η_{C_{X}}) = (X, C_{η_{C_{X}}}) = C_{X}$ . Hence PCT and FC are isomorphic.□

5 Conclusion

Using distance functions, we have discussed the topological structures in complete co-residuated lattices. As a main result, there is a Galois correspondence between DIS and APT (resp. PCT, FI, FC). Moreover, DIS and ATOP (resp. APT and FI, PCT and FC) are isomorphic.

In the future, we plan to investigate fuzzy rough sets, information systems and decision rules by using the concepts of distance spaces in complete co-residuated lattices.

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