The relations between Alexandrov L -fuzzy pre-uniformities and approximation operators

Abstract

We introduce the concepts of Alexandrov L-fuzzy pre-uniformities in complete residuated lattices. We obtain Alexandrov L-fuzzy topologies, L-lower approximation operators and L-upper approximation operators induced by Alexandrov L-fuzzy pre-uniformities. Conversely, we obtain Alexandrov L-fuzzy pre-uniformities induced by L-lower approximation operators and L-upper approximation operators.

Keywords

Complete residuated lattices

1 Introduction

Pawlak [12, 13] introduced the rough set theory as a formal tool to deal with imprecision and uncertainty in the data analysis. This theory was extended and applied in many directions [9 , 21].

Ward et al. [19] introduced a complete residuated lattice which is an algebraic structure for many valued logic. It is an important mathematical tool as algebraic structures for many valued logics [1–9 , 14–18]. For an extension of classical rough sets, many researchers [5 , 17] developed L-lower and L-upper approximation operators in complete residuated lattices. By using this concepts, information systems and decision rules were investigated in complete residuated lattices [1 , 9]. Čimoka [2] and Ramadan [15, 16] investigated the relationships between L-fuzzy quasi-uniformities and L-fuzzy topologies in complete residuated lattices.

An interesting and natural research topic in rough set theory is the study of rough set theory and topology. Ma [11] investigated the topological and lattice structures of L-fuzzy rough sets determined by lower and upper sets. Kim [7] studied the relations between L-fuzzy upper and lower approximation spaces and L-fuzzy quasi-uniform spaces in a strictly two-sided, commutative quantale. He [5, 6] investigated the properties of L-lower and L-upper approximate operators and Alexandrov L-fuzzy topologies in complete residuated lattices.

In this paper, we introduce the concepts of Alexandrov L-fuzzy pre-uniformities in complete residuated lattices, which are unified approaches to the following three structures: L-lower approximation operators, L-upper approximation operators and Alexandrov L-fuzzy topologies. We obtain Alexandrov L-fuzzy topologies, L-lower approximation operators and L-upper approximation operators induced by Alexandrov L-fuzzy pre-uniformities (see Theorems 3.7, 3.11 and 3.12). Conversely, we obtain Alexandrov L-fuzzy pre-uniformities induced by L-lower approximation operators and L-upper approximation operators (see Theorems 4.1 and 4.3). Moreover, we investigate their relations and give examples.

2 Preliminaries

Definition 2.1. [1 , 18] An algebra (L, ∧, ∨, ⊙, →, ⊥, ⊤) is called a complete residuated lattice if

(L, ≤, ∨, ∧, ⊥, ⊤) is a complete lattice with the greatest element ⊤ and the least element ⊥,

(L, ⊙, ⊤) is a commutative monoid, and

x ⊙ y ≤ z iff x ≤ y → z for x, y, z ∈ L.

In this paper, we assume that (L, ≤, ⊙, →, ^∗) is a complete residuated lattice with an order reversing involution ^∗ where x^∗ = x→ ⊥.

For α ∈ L, f ∈ L^X, we denote (α → f), (α ⊙ f), α_X ∈ L^X as (α → f) (x) = α → f (x), (α ⊙ f) (x) = α ⊙ f (x), α_X (x) = α, $\begin{matrix} ⊤_{x} (y) & = & {\begin{matrix} ⊤, & if y = x, \\ ⊥, & otherwise, \end{matrix} \\ ⊤_{x}^{*} (y) & = & {\begin{matrix} ⊥, & if y = x, \\ ⊤, & otherwise . \end{matrix} \end{matrix}$

Lemma 2.2. [1 , 18] For each x, y, z, x_i, y_i, w ∈ L, the following hold.

⊤ → x = x, ⊥ ⊙ x = ⊥ .

If y ≤ z, then x ⊙ y ≤ x ⊙ z, x → y ≤ x → z and z → x ≤ y → x.

x ≤ y iff x→ y = ⊤.

x → (⋀ _iy_i) = ⋀ _i (x → y_i).

(⋁ _ix_i) → y = ⋀ _i (x_i → y).

x ⊙ (⋁ _iy_i) = ⋁ _i (x ⊙ y_i).

(x ⊙ y) → z = x → (y → z) = y → (x → z).

x → y ≤ (x ⊙ z) → (y ⊙ z) and (x → y) ⊙ (y → z) ≤ x → z.

x → y = y^* → x^*.

z → x ≤ (x → y) → (z → y) and y → z ≤ (x → y) → (x → z).

⋁_i∈Γx_i → ⋁ _i∈Γy_i ≥ ⋀ _i∈Γ (x_i → y_i) and ⋀_i∈Γx_i → ⋀ _i∈Γy_i ≥ ⋀ _i∈Γ (x_i → y_i).

Definition 2.3. [1, 8] Let X be a set. A map R : X × X → L is called an L-fuzzy preorder if

R (x, x) =⊤ for all x ∈ X (reflexive), and

R (x, y) ⊙ R (y, z) ≤ R (x, z) for all x, y, z ∈ X (transitive).

Lemma 2.4. [1 , 8] For a given set X, define a map S : L^X × L^X → L by $S (f, g) = ⋀_{x \in X} (f (x) \to g (x)) .$

Then, for each f, g ∈ L^X and for each α ∈ L, the following hold.

S is an L-fuzzy preorder on L^X.

f ≤ g iff S (f, g) =⊤.

Definition 2.5. [5] A map $J : L^{X} \to L^{X}$ is called an L-lower approximation operator on X if

$J (⊤_{X}) = ⊤_{X}$ ,

$J (f) \leq f$ for all f ∈ L^X,

$J (⋀_{i \in Γ} f_{i}) = ⋀_{i \in Γ} J (f_{i})$ for all f_i ∈ L^X, and

$J (α \to f) = α \to J (f)$ .

An L-lower approximation operator is called topological if (T) $J (J (f)) = J (f)$ for all f ∈ L^X.

Definition 2.6. [5] A map $H : L^{X} \to L^{X}$ is called an L-upper approximation operator on X if

$H (⊥_{X}) = ⊥_{X}$ ,

$H (f) \geq f$ for all f ∈ L^X,

$H (⋁_{i \in Γ} f_{i}) = ⋁_{i \in Γ} H (f_{i})$ for all f_i ∈ L^X, and

$H (α ⊙ f) = α ⊙ H (f)$ .

An L-upper approximation operator is called topological if (T) $H (H (f)) = H (f)$ , for all f ∈ L^X.

Let $H$ and $J$ be an L-upper and L-lower approximation on X, respectively. The pair $(J (f), H (f))$ is called a fuzzy rough set for f.

Definition 2.7. [5, 8] A map $T : L^{X} \to L$ is called an Alexandrov L-fuzzy topology on X if

$T (⊥_{X}) = T (⊤_{X}) = ⊤$ ,

$T (⋀_{i} f_{i}) \geq ⋀_{i} T (f_{i})$ and $T (⋁_{i} f_{i}) \geq ⋀_{i} T (f_{i})$ for all {f_i} _i∈Γ ⊆ L^X, and

$T (α ⊙ f) \geq T (f)$ and $T (α \to f) \geq T (f)$ for all α ∈ L and f ∈ L^X.

3 Alexandrov L-fuzzy pre-uniformities

Lemma 3.1. For each f, g ∈ L^X, define two maps $u_{[f, g]}, u_{[f, g]}^{- 1} : X \times X \to L$ by u_[f,g] (x, y) = f (x) → g (y) and $u_{[f, g]}^{- 1} (x, y) = u_{[f, g]} (y, x)$ . Then, the following hold.

⊤_X×X = u_{[⊥_X,⊥_X]} = u_{[⊤_X,⊤_X]} .

If f₂ ≤ f₁ and g₁ ≤ g₂, then u_[f₁,g₁] ≤ u_[f₂,g₂] .

For any u_[f,g] ∈ L^X×X and h ∈ L^X, it holds that u_[f,h] ∘ u_[h,g] ≤ u_[f,g] where u_[f,h] (x, y) ∘ u_[h,g] (y, z) = ⋁ _y∈X ((f (x) → h (y)) ⊙ (h (y) → g (z))) .

u_{[⋁_i∈Γf_i,g]} = ⋀ _i∈Γu_{[f_i,g]} and u_{[f,⋀_i∈Γg_i]} = ⋀ _i∈Γu_{[f,g_i]}.

u_[α⊙f,g] = α → u_[f,g] and u_[f,α→g] = α → u_[f,g].

u_{[α⊙f,α⊙g]} ≥ u_[f,g] and u_{[α→f,α→g]} ≥ u_[f,g].

u_[f,g] = ⋀ _z∈X (f (z) → u_{[⊤_z,g]}) and $u_{[f, g]} = ⋀_{z \in X} (g^{*} (z) \to u_{[f, ⊤_{z}^{*}}]) .$

$u_{[f, g]} = ⋀_{y, z \in X} (f (y) \to (g^{*} (z) \to u_{[⊤_{y}, ⊤_{z}^{*}]})) .$

$u_{[f, g]}^{- 1} = u_{[g^{*}, f^{*}]}$ .

Proof. (1) and (2) can be easily proved.

(3) By Lemma 2.2(8), we have

$\begin{matrix} u_{[f, h]} (x, y) \circ u_{[h, g]} (y, z) \\ = ⋁_{y \in X} ((f (x) \to h (y)) ⊙ (h (y) \to g (z))) \\ \leq f (x) \to g (z) = u_{[f, g]} (x, z) . \end{matrix}$

(4) It follows from Lemma 2.2 (4) and (5).

(5) By Lemma 2.2 (7), we have u_[α⊙f,g] (x, y) = α ⊙ f (x) → g (x) = α → (f (x) → g (y)) = α → u_[f,g] (x, y). u_[f,α→g] (x, y) = f (x) → (α → g (y)) = α → (f (x) → g (y)) = α → u_[f,g] (x, y) .

(6) It follows from Lemma 2.2 (8) and (10).

(7) Since f = ⋁ _z∈X (f (z) ⊙ ⊤ _z), we have by (4) and (5) that

$\begin{matrix} u_{[f, g]} & = & u_{[⋁_{z \in X} (f (z) ⊙ ⊤_{z}), g]} \\ = & ⋀_{z \in X} u_{[f (z) ⊙ ⊤_{z}, g]} \\ = & ⋀_{z \in X} (f (z) \to u_{[⊤_{z}, g]}) . \end{matrix}$

Since $g = ⋀_{z \in X} (g^{*} (z) \to ⊤_{z}^{*})$ , we have by (4) and (5) that

$\begin{matrix} u_{[f, g]} & = & u_{[f, ⋀_{z \in X} (g^{*} (z) \to ⊤_{z}^{*})]} \\ = & ⋀_{z \in X} u_{[f, g^{*} (z) \to ⊤_{z}^{*}]} \\ = & ⋀_{z \in X} (g^{*} (z) \to u_{[f, ⊤_{z}^{*}]}) . \end{matrix}$

(8) Since f = ⋁ _z∈X (f (z) ⊙ ⊤ _z) and $g = ⋀_{z \in X} (g^{*} (z) \to ⊤_{z}^{*})$ , we have by (7) that

$\begin{matrix} u_{[f, g]} & = & u_{[⋁_{y \in X} (f (y) ⊙ ⊤_{y}), ⋀_{z \in X} (g^{*} (z) \to ⊤_{z}^{*})]} \\ = & ⋀_{y \in X} (f (y) \to u_{[⊤_{y}, ⋀_{z \in X} (g^{*} (z) \to ⊤_{z}^{*})]}) \\ = & ⋀_{y \in X} (f (y) \to ⋀_{z \in X} (g^{*} (z) \to u_{[⊤_{y}, ⊤_{z}^{*}]})) \\ = & ⋀_{y, z \in X} (f (y) \to (g^{*} (z) \to u_{[⊤_{y}, ⊤_{z}^{*}]})) . \end{matrix}$

(9) By Lemma 2.2(9), we have

$\begin{matrix} u_{[f, g]}^{- 1} (x, y) & = & u_{[f, g]} (y, x) = f (y) \to g (x) \\ = & g^{*} (x) \to f^{*} (y) = u_{[g^{*}, f^{*}]} (x, y) . \end{matrix}$

Definition 3.2. A map $U : L^{X \times X} \to L$ is called an Alexandrov L-fuzzy pre-uniformity on X if

there exists u ∈ L^X×X with $U (u) = ⊤$ ,

if v ≤ u, then $U (v) \leq U (u)$ ,

for every u_i ∈ L^X×X, $U (⋀_{i \in Γ} u_{i}) = ⋀_{i \in Γ} U (u_{i})$ ,

$U (u_{[f, g]}) \leq S (f, g)$ for each f, g ∈ L^X,

$U (α \to u) = α \to U (u)$ for each α ∈ L,

An Alexandrov L-fuzzy pre-uniformity is called an Alexandrov L-fuzzy quasi-uniformity on X if

(AQ) $U (u) \leq ⋁ {U (v) ⊙ U (w) | v \circ w \leq u}$ where v ∘ w (x, z) = ⋁ _y∈X (v (x, y) ⊙ w (y, z)).

Remark 3.3. Let $U$ be an L-fuzzy pre-uniformity on X. Since u ≤ ⊤ _X×X for all u ∈ L^X×X, we have by (AU1) and (AU2) that $U (⊤_{X \times X}) = ⊤$ .

Theorem 3.4. Let $U$ be an Alexandrov L-fuzzy pre-uniformity on X. Define a map $U^{s} : L^{X \times X} \to L$ as

$U^{s} (u) = U (u^{- 1}) .$

Then, the following hold:

$U^{s}$ is an Alexandrov L-fuzzy pre-uniformity on X.

If $U$ is an Alexandrov L-fuzzy quasi-uniformity on X, then $U^{s}$ is an Alexandrov L-fuzzy quasi-uniformity on X.

$U (u_{[f, g]}) = ⋀_{x, y \in X} (U^{*} (u_{[⊤_{x}, ⊤_{y}^{*}]}) \to (f (x) \to g (y)) .$

There exists a reflexive L-fuzzy relation $R_{U} \in L^{X \times X}$ with $U (u_{[f, g]}) = ⋀_{x, y \in X} (R_{U} (x, y) \to (f (x) \to g (y)) .$

There exists a reflexive L-fuzzy relation $R_{U^{s}} = R_{U}^{- 1} \in L^{X \times X}$ such that $U^{s} (u_{[f, g]}) = ⋀_{x, y \in X} (R_{U}^{- 1} (x, y) \to (f (x) \to g (y)) .$

There exists a reflexive L-fuzzy relation $R_{U} \in L^{X \times X}$ with $U (u) = ⋀_{x, y \in X} (R_{U} (x, y) \to u (x, y)) .$

Proof. (1) (AU1), (AU2) and (AU3) can be easily proved.

(AU4) For all f, g ∈ L^X, we have $\begin{matrix} U^{s} (u_{[f, g]}) & = & U (u_{[f, g]}^{- 1}) = U (u_{[g^{*}, f^{*}]}) \\ \leq & S (g^{*}, f^{*}) = S (f, g) . \end{matrix}$

(AU5) For all α ∈ L and u ∈ L^X×X, $\begin{matrix} U^{s} (α \to u) & = & U (α \to u^{- 1}) = α \to U (u^{- 1}) \\ = & α \to U^{s} (u) . \end{matrix}$

Hence $U^{s}$ is an Alexandrov L-fuzzy pre-uniformity on X.

(2) (AQ) For all u ∈ L^X×X, $\begin{matrix} U^{s} (u) & = & U (u^{- 1}) \\ \leq & ⋁ {U (v^{- 1}) ⊙ U (w^{- 1}) | v^{- 1} \circ w^{- 1} \leq u^{- 1}} \\ = & ⋁ {U^{s} (v) ⊙ U^{s} (w) | v \circ w \leq u} . \end{matrix}$

(3) Since f = ⋁ _x∈X (f (x) ⊙ ⊤ _x) and $g = ⋀_{y \in X} (g^{*} (y) \to ⊤_{y}^{*})$ , we have by Lemma 3.1(8) that $u_{[f, g]} = ⋀_{x, y \in X} (f (x) \to (g^{*} (y) \to u_{[⊤_{x}, ⊤_{y}^{*}]}))$ . Now, by (AU3) and (AU5), we have $\begin{matrix} U (u_{[f, g]}) \\ = ⋀_{x, y \in X} (f (x) \to (g^{*} (y) \to U (u_{[⊤_{x}, ⊤_{y}^{*}]}))) \\ = ⋀_{x, y \in X} (f (x) \to (U^{*} (u_{[⊤_{x}, ⊤_{y}^{*}]}) \to g (y))) \\ = ⋀_{x, y \in X} (U^{*} (u_{[⊤_{x}, ⊤_{y}^{*}]}) \to (f (x) \to g (y))) . \end{matrix}$

(4) Let $R_{U} (x, y) = U^{*} (u_{[⊤_{x}, ⊤_{y}^{*}]})$ . By (AU4), $R_{U} (x, x) = U^{*} (u_{[⊤_{x}, ⊤_{x}^{*}]}) \geq S^{*} (⊤_{x}, ⊤_{x}^{*}) = ⊤ .$

(5) By (4), $R_{U^{s}} (x, y) = U^{s *} (u_{[⊤_{x}, ⊤_{y}^{*}]}) = U^{*} (u_{[⊤_{y}, ⊤_{x}^{*}]}) = R_{U}^{- 1} (x, y) .$ Thus, $\begin{matrix} U^{s} (u_{[f, g]}) & = & U (u_{[f, g]}^{- 1}) = U (u_{[g^{*}, f^{*}]}) \\ = & ⋀_{x, y \in X} (R_{U} (x, y) \to (g^{*} (x) \to f^{*} (y))) \\ = & ⋀_{x, y \in X} (R_{U} (x, y) \to (f (y) \to g (x))) \\ = & ⋀_{x, y \in X} (R_{U}^{- 1} (y, x) \to (f (y) \to g (x))) . \end{matrix}$

(6) Since $u (z, w) = ⋀_{x, y \in X} (u^{*} (x, y) \to u_{[⊤_{x}, ⊤_{y}^{*}]} (z, w)) = u^{*} (z, w) \to ⊤$ , we have $u = ⋀_{x, y \in X} (u^{*} (x, y) \to u_{[⊤_{x}, ⊤_{y}^{*}]}) .$ Let $R_{U} (x, y) = U^{*} (u_{[⊤_{x}, ⊤_{y}^{*}]}) .$ Thus, $\begin{matrix} U (u) & = & U (⋀_{x, y \in X} (u^{*} (x, y) \to u_{[⊤_{x}, ⊤_{y}^{*}]})) \\ = & ⋀_{x, y \in X} (u^{*} (x, y) \to U (u_{[⊤_{x}, ⊤_{y}^{*}]})) \\ = & ⋀_{x, y \in X} (U^{*} (u_{[⊤_{x}, ⊤_{y}^{*}]}) \to u (x, y)) \\ = & ⋀_{x, y \in X} (R_{U} (x, y) \to u (x, y)) . \end{matrix}$

Example 3.5. Let R ∈ L^X×X be a reflexive fuzzy relation. Define a map $U : L^{X \times X} \to L$ as $U (u) = ⋀_{x, y \in X} (R (x, y) \to u (x, y)) .$

Then (AU1), (AU2), (AU3) and (AU5) can be easily proved. (AU4) For all f, g ∈ L^X, $\begin{matrix} U (u_{[f, g]}) & = & ⋀_{x, y \in X} (R (x, y) \to u_{[f, g]} (x, y)) \\ = & ⋀_{x, y \in X} (R (x, y) \to (f (x) \to g (y))) \\ \leq & ⋀_{x \in X} (R (x, x) \to (f (x) \to g (x))) \\ = & S (f, g) . \end{matrix}$

Hence $U$ is an Alexandrov L-fuzzy pre-uniformity on X. We obtain $U^{s} : L^{X \times X} \to L$ as $U^{s} (u) = U (u^{- 1}) = ⋀_{x, y \in X} (R (x, y) \to u^{- 1} (x, y)) .$

Theorem 3.6. (1) Let $J : L^{X} \to L^{X}$ be an L-lower approximation operator. Define $H_{J} : L^{X} \to L^{X}$ as $H_{J} (f) = J^{*} (f^{*})$ . Then $H_{J}$ is an L-upper approximation operator.

(2) Let $H : L^{X} \to L^{X}$ be an L-upper approximation operator. Define $J_{H} : L^{X} \to L^{X}$ as $J_{H} (f) = H^{*} (f^{*})$ . Then $J_{H}$ is an L-lower approximation operator.

(3) Let $T : L^{X} \to L$ be an Alexandrov L-fuzzy topology. Define $T^{*} : L^{X} \to L$ as $T^{*} (f) = T (f^{*})$ . Then $T^{*}$ is an Alexandrov L-fuzzy topology.

Proof. (1) For f ∈ L^X and α ∈ L, we have $H_{J} (f) = J^{*} (f^{*}) \geq (f^{*})^{*} = f$ and $H_{J} (α ⊙ f) = J^{*} ((α ⊙ f)^{*}) = J^{*} (α \to f^{*}) = (α \to J (f^{*}))^{*} = α ⊙ J^{*} (f^{*}) = α ⊙ H_{J} (f) .$ The remaining can be proved by a similar way.

Theorem 3.7. Let $U$ be an Alexandrov L-fuzzy pre-uniformity on X. Define a map $T_{U} : L^{X} \to L$ as $T_{U} (f) = U (u_{[f, f]}) .$

Then $T_{U}$ is an Alexandrov L-fuzzy topology with $T_{U}^{*} = T_{U^{s}}$ .

Proof. (AT1) $T_{U} (⊤_{X}) = U (u_{[⊤_{X}, ⊤_{X}]}) = ⊤$ and $T_{U} (⊥_{X}) = U (u_{[⊥_{X}, ⊥_{X}]}) = ⊤$ .

(AT2) By Lemma 3.1(4), we have $T_{U} (⋀_{i} f_{i}) = U (u_{[⋀_{i} f_{i}, ⋀_{i} f_{i}]}) \geq ⋀_{i} U (u_{[f_{i}, f_{i}]}) = ⋀_{i} T_{U} (f_{i})$ and $T_{U} (⋁_{i} f_{i}) = U (u_{[⋁_{i} f_{i}, ⋁_{i} f_{i}]}) \geq ⋀_{i} U (u_{[f_{i}, f_{i}]}) = ⋀_{i} T_{U} (f_{i}) .$

(AT3) By Lemma 3.1(6), we have $T_{U} (α ⊙ f) = U (u_{[α ⊙ f, α ⊙ f]}) \geq U (u_{[f, f]}) = T_{U} (f)$ and $T_{U} (α \to f) = U (u_{[α \to f, α \to f]}) \geq U (u_{[f, f]}) = T_{U} (f) .$ Hence, $T_{U}$ is an Alexandrov L-fuzzy topology on X. Moreover, $T_{U}^{*} (f) = T_{U} (f^{*}) = U (u_{[f^{*}, f^{*}]}) = U^{s} (u_{[f, f]}) = T_{U^{s}} (f) .$

Example 3.8. Consider R and $U$ defined in Example 3.5. By Theorem 3.7, we obtain two Alexandrov L-fuzzy topologies $T_{U}, T_{U^{s}} : L^{X} \to L$ where $T_{U^{s}} (f) = U (u_{[f^{*}, f^{*}]}) = ⋀_{x, y \in X} (R (x, y) \to (f^{*} (x) \to f^{*} (y))) = ⋀_{x, y \in X} (R (x, y) \to (f (y) \to f (x)))$ and $T_{U} (f) = ⋀_{x, y \in X} (R (x, y) \to (f (x) \to f (y))) .$

Theorem 3.9. [5] Let $(X, H)$ be an L-upper approximation space. Define a map $T_{H} : L^{X} \to L$ by: $T_{H} (f) = S (H (f), f) .$

Then $T_{H}$ is an Alexandrov L-fuzzy topology on X with $T_{H}^{*} (f) = S (f, J_{H} (f))$ where $J_{H} (f) = H^{*} (f^{*})$ for all f ∈ L^X.

Corollary 3.10. [5] Let $(X, J)$ be an L-lower approximation space. Define a map $T_{J} : L^{X} \to L$ by: $T_{J} (f) = S (f, J (f)) .$

Then $T_{J}$ is an Alexandrov L-fuzzy topology on X.

Theorem 3.11. Let $U$ be an Alexandrov L-fuzzy preuniformity on X. Define a map $J_{U} : L^{X} \to L^{X}$ as $J_{U} (f) (x) = U (u_{[⊤_{x}, f]}) .$

Then, the following hold:

$J_{U}$ is an L-lower approximation operator on X.

$U^{*} (u_{[⊤_{x}, ⊤_{x}^{*}]}) = ⊤$ .

If $⋁_{y \in X} (U^{*} (u_{[⊤_{x}, ⊤_{y}^{*}]}) ⊙ U^{*} (u_{[⊤_{y}, ⊤_{z}^{*}]})) \leq U^{*} (u_{[⊤_{x}, ⊤_{z}^{*}]}),$ then $J_{U}$ is a topological L-lower approximation operator on X.

$T_{J_{U}} (f) = U (u_{[f, f]}) = T_{U} (f)$ for all f ∈ L^X.

$J_{U^{s}} (f) = U (u_{[f^{*}, ⊤_{x}^{*}]})$ for all f ∈ L^X, and $T_{J_{U}}^{*} = T_{U^{s}} = T_{J_{U^{s}}}$ .

There exists a reflexive L-fuzzy relation $R_{U} \in L^{X \times X}$ with $J_{U} (f) (x) = ⋀_{y \in X} (R_{U} (x, y) \to f (y)) .$ Moreover, there exists a reflexive L-fuzzy relation $R_{U^{s}} = R_{U}^{- 1} \in L^{X \times X}$ such that $J_{U^{s}} (f) (x) = ⋀_{y \in X} (R_{U} (y, x) \to f (y)) .$

$U (u_{[f, g]}) = S (f, J_{U} (g))$ for all f, g ∈ L^X.

Proof. (1) (J1) Since $U (u_{[⊤_{x}, ⊤_{X}]}) \geq U (u_{[⊤_{X}, ⊤_{X}]}) = ⊤$ , we have $J_{U} (⊤_{X}) (x) = U (u_{[⊤_{x}, ⊤_{X}]}) = ⊤ .$

(J2) By (AU4), we have $J_{U} (f) (x) = U (u_{[⊤_{x}, f]}) \leq S (⊤_{x}, f) = f (x) .$

(J3) By Lemma 3.1 (4), we have $\begin{matrix} J_{U} (⋀_{i \in Γ} f_{i}) (x) \\ = U (u_{[⊤_{x}, ⋀_{i \in Γ} f_{i}]}) = U (⋀_{i \in Γ} u_{[⊤_{x}, f_{i}]}) \\ = ⋀_{i \in Γ} U (u_{[⊤_{x}, f_{i}]}) = ⋀_{i \in Γ} J_{U} (f_{i}) (x) . \end{matrix}$

(J4) By Lemma 3.1(5) and (AU5), we have $\begin{matrix} J_{U} (α \to f) (x) \\ = U (u_{[⊤_{x}, α \to f]}) = U (α \to u_{[⊤_{x}, f]}) \\ = α \to U (u_{[⊤_{x}, f]}) = α \to J_{U} (f) . \end{matrix}$

(2) For x ∈ X, $U^{*} (u_{[⊤_{x}, ⊤_{x}^{*}]}) \geq S^{*} (⊤_{x}, ⊤_{x}^{*}) = ⊤ .$

(3) By (2), $⋁_{y \in X} (U^{*} (u_{[⊤_{x}, ⊤_{y}^{*}]}) ⊙ U^{*} (u_{[⊤_{y}, ⊤_{z}^{*}]})) \geq U^{*} (u_{[⊤_{x}, ⊤_{x}^{*}]}) ⊙ U^{*} (u_{[⊤_{x}, ⊤_{z}^{*}]})) = U^{*} (u_{[⊤_{x}, ⊤_{z}^{*}]}) .$ If $⋁_{y \in X} (U^{*} (u_{[⊤_{x}, ⊤_{y}^{*}]}) ⊙ U^{*} (u_{[⊤_{y}, ⊤_{z}^{*}]})) \leq U^{*} (u_{[⊤_{x}, ⊤_{z}^{*}]})$ , then $⋁_{y \in X} (U^{*} (u_{[⊤_{x}, ⊤_{y}^{*}]}) ⊙ U^{*} (u_{[⊤_{y}, ⊤_{z}^{*}]})) = U^{*} (u_{[⊤_{x}, ⊤_{z}^{*}]}) .$ Since $J_{U} (f) = ⋀_{y \in X} (J_{U}^{*} (f) (y) \to ⊤_{y}^{*})$ , we have

$\begin{matrix} J_{U} (J_{U} (f)) (x) = U (u_{[⊤_{x}, J_{U} (f)]}) \\ = U (u_{[⊤_{x}, ⋀_{y \in X} (J_{U}^{*} (f) (y) \to ⊤_{y}^{*})]}) \\ = U (⋀_{y \in X} (J_{U}^{*} (f) (y) \to u_{[⊤_{x}, ⊤_{y}^{*}]})) \\ = ⋀_{y \in X} (J_{U}^{*} (f) (y) \to U (u_{[⊤_{x}, ⊤_{y}^{*}]})) \\ = ⋀_{y \in X} (U^{*} (u_{[⊤_{y}, ⋀_{z \in X} (f^{*} (z) \to ⊤_{z}^{*})]}) \to U (u_{[⊤_{x}, ⊤_{y}^{*}]})) \\ = ⋀_{y \in X} ({(⋀_{z \in X} (f^{*} (z) \to U (u_{[⊤_{y}, ⊤_{z}^{*}]})))}^{*} \to U (u_{[⊤_{x}, ⊤_{y}^{*}]})) \\ = ⋀_{y, z \in X} ((f^{*} (z) ⊙ U^{*} (u_{[⊤_{y}, ⊤_{z}^{*}]})) \to U (u_{[⊤_{x}, ⊤_{y}^{*}]})) \\ = ⋀_{y, z \in X} {((f^{*} (z) ⊙ U^{*} (u_{[⊤_{y}, ⊤_{z}^{*}]})) ⊙ U^{*} (u_{[⊤_{x}, ⊤_{y}^{*}]}))}^{*} \\ = ⋀_{z \in X} {(f^{*} (z) ⊙ ⋁_{y \in X} (U^{*} (u_{[⊤_{y}, ⊤_{z}^{*}]}) ⊙ U^{*} (u_{[⊤_{x}, ⊤_{y}^{*}]})))}^{*} \\ = ⋀_{z \in X} {(f^{*} (z) ⊙ U^{*} (u_{[⊤_{x}, ⊤_{z}^{*}]}))}^{*} \\ = ⋀_{z \in X} (f^{*} (z) \to U (u_{[⊤_{x}, ⊤_{z}^{*}]})) \\ = U (u_{[⊤_{x}, ⋀_{z \in X} (f^{*} (z) \to ⊤_{z}^{*})]}) = U (u_{[⊤_{x}, f]}) = J_{U} (f) (x) . \end{matrix}$

(4) By Corollary 3.10, we have $\begin{matrix} T_{J_{U}} (f) & = & S (f, J_{U} (f)) \\ = & ⋀_{x \in X} (f (x) \to U (u_{[⊤_{x}, f]})) \\ = & ⋀_{x \in X} U (u_{[f (x) ⊙ ⊤_{x}, f]}) \\ = & U (u_{[⋁_{x \in X} (f (x) ⊙ ⊤_{x}), f]}) \\ = & U (u_{[f, f]}) = T_{U} (f) . \end{matrix}$

(5) For f ∈ L^X and x ∈ X, we have $\begin{matrix} J_{U^{s}} (f) (x) & = & U^{s} (u_{[⊤_{x}, f]}) = U (u_{[f^{*}, ⊤_{x}^{*}]}) and \\ T_{J_{U}}^{*} (f) & = & U (u_{[f^{*}, f^{*}]}) = T_{U^{s}} (f) = T_{J_{U^{s}}} (f) . \end{matrix}$

(6) Since $f = ⋀_{y \in X} (f^{*} (y) \to ⊤_{y}^{*})$ , we have by Lemma 3.1(4,5) and (AU3,5) that $\begin{matrix} J_{U} (f) (x) & = & U (u_{[⊤_{x}, f]}) \\ = & U (u_{[⊤_{x}, ⋀_{y \in X} (f^{*} (y) \to ⊤_{y}^{*})]}) \\ = & ⋀_{y \in X} (f^{*} (y) \to U (u_{[⊤_{x}, ⊤_{y}^{*}]})) \\ = & ⋀_{y \in X} (U^{*} (u_{[⊤_{x}, ⊤_{y}^{*}]}) \to f (y)) . \end{matrix}$

Let $R_{U} (x, y) = U^{*} (u_{[⊤_{x}, ⊤_{y}^{*}]})$ . Then, we have by (2) that $R_{U}$ is reflexive and $J_{U} (f) (x) = ⋀_{y \in X} (R_{U} (x, y) \to f (y)) .$

Moreover, $R_{U^{s}} (x, y) = U^{s *} (u_{[⊤_{x}, ⊤_{y}^{*}]}) = U^{*} (u_{[⊤_{y}, ⊤_{x}^{*}]}) = R_{U} (y, x) = R_{U}^{- 1} (x, y)$ such that $\begin{matrix} J_{U^{s}} (f) (x) & = & ⋀_{y \in X} (U^{s *} (u_{[⊤_{x}, ⊤_{y}^{*}]}) \to f (y)) \\ = & ⋀_{y \in X} (U^{*} (u_{[⊤_{y}, ⊤_{x}^{*}]}) \to f (y)) \\ = & ⋀_{y \in X} (R_{U} (y, x) \to f (y)) . \end{matrix}$

(7) By Lemma 3.1(4,5) and (AU3,5), we have $\begin{matrix} S (f, J_{U} (g)) & = & ⋀_{x \in X} (f (x) \to J_{U} (g))) \\ = & ⋀_{x \in X} (f (x) \to U (u_{[⊤_{x}, g]})) \\ = & U (u_{[⋁_{x \in X} (f (x) ⊙ ⊤_{x}), g]}) = U (u_{[f, g]}) . \end{matrix}$

Theorem 3.12. Let $U$ be an Alexandrov L-fuzzy pre-uniformity on X. Define a map $H_{U} : L^{X} \to L^{X}$ as $H_{U} (f) (x) = U^{*} (u_{[f, ⊤_{x}^{*}]}) .$

Then, the following hold:

$H_{U}$ is an L-upper approximation operator on X.

If $⋁_{y \in X} (U^{*} (u_{[⊤_{x}, ⊤_{y}^{*}]}) ⊙ U^{*} (u_{[⊤_{y}, ⊤_{z}^{*}]})) \leq U^{*} (u_{[⊤_{x}, ⊤_{z}^{*}]}),$ then $H_{U}$ is a topological L-upper approximation operator on X.

$H_{U} (f) = J_{U^{s}}^{*} (f^{*})$ for all f ∈ L^X.

$H_{U^{s}} (f) = J_{U}^{*} (f^{*})$ for all f ∈ L^X.

$T_{H_{U}} (f) = U (u_{[f, f]}) = T_{U} (f)$ for all f ∈ L^X.

There exists a reflexive L-fuzzy relation $R_{U} \in L^{X \times X}$ with $H_{U} (f) (x) = ⋁_{y \in X} (R_{U} (y, x) ⊙ f (y)) .$ Moreover, there exists a reflexive L-fuzzy relation $R_{U^{s}} = R_{U}^{- 1} \in L^{X \times X}$ such that $H_{U^{s}} (f) (x) = ⋁_{y \in X} (R_{U} (x, y) ⊙ f (y)) .$

$U (u_{[f, g]}) = S (H_{U} (f), g)$ for all f, g ∈ L^X.

Proof. {(1) (H1) Since $U (u_{[⊥_{X}, ⊤_{x}^{*}]}) \geq U (u_{[⊥_{X}, ⊥_{X}]}) = U (⊤_{X \times X}) = ⊤$ , we have $H_{U} (⊥_{X}) (x) = U (u_{[⊥_{X}, ⊤_{x}^{*}]}) = ⊤ .$

(H2) By (AU4), we have $H_{U} (f) (x) = U^{*} (u_{[f, ⊤_{x}^{*}]}) \geq S^{*} (f, ⊤_{x}^{*}) = (f^{*} (x))^{*} = f (x) .$

(H3) By Lemma 3.1(4), we have $\begin{matrix} H_{U} (⋁_{i \in Γ} f_{i}) (x) = U^{*} (u_{[⋁_{i \in Γ} f_{i}, ⊤_{x}^{*}]}) \\ = U^{*} (⋀_{i \in Γ} u_{[f_{i}, ⊤_{x}^{*}]}) = {(⋀_{i \in Γ} U (u_{[f_{i}, ⊤_{x}^{*}]}))}^{*} \\ = ⋁_{i \in Γ} U^{*} (u_{[f_{i}, ⊤_{x}^{*}]}) = ⋁_{i \in Γ} H_{U} (f_{i}) (x) . \end{matrix}$

(H4) By Lemma 3.1(5) and (AU5), we have $\begin{matrix} H_{U} (α ⊙ f) (x) = U^{*} (u_{[α ⊙ f, ⊤_{x}^{*}]}) \\ = (U (α \to u_{[f, ⊤_{x}^{*}]}))^{*} \\ = (α \to U (u_{[f, ⊤_{x}^{*}]}))^{*} \\ = α ⊙ U^{*} (u_{[f, ⊤_{x}^{*}]}) = α ⊙ H_{U} (f) . \end{matrix}$

Hence $H_{U}$ is an L-upper approximation operator on X.

(2) Since $U^{*} (u_{[⊥_{x}, ⊤_{x}^{*}]}) \geq S^{*} (⊥_{x}, ⊤_{x}^{*}) = ⊤$ , wehave by the assumption that $⋁_{y \in X} (U^{*} (u_{[⊤_{x}, ⊤_{y}^{*}]}) ⊙ U^{*} (u_{[⊤_{y}, ⊤_{z}^{*}]})) = U^{*} (u_{[⊤_{x}, ⊤_{z}^{*}]}) .$ Since $H_{U} (f) = ⋁_{y \in X} (H_{U} (f) (y) ⊙ ⊤_{y})$ , we have $\begin{matrix} H_{U} (H_{U} (f)) (x) = U^{*} (u_{[H_{U} (f), ⊤_{x}^{*}]}) \\ = U^{*} (u_{[⋁_{y \in X} (H_{U} (f) (y) ⊙ ⊤_{y}), ⊤_{x}^{*}]}) \\ = {(U (⋀_{y \in X} (H_{U} (f) (y) \to u_{[⊤_{y}, ⊤_{x}^{*}]})))}^{*} \\ = {(⋀_{y \in X} (H_{U} (f) (y) \to U (u_{[⊤_{y}, ⊤_{x}^{*}]})))}^{*} \\ = {(⋀_{y \in X} (U^{*} (u_{[f, ⊤_{y}^{*}]}) \to U (u_{[⊤_{y}, ⊤_{x}^{*}]})))}^{*} \\ = (⋀_{y \in X} (U^{*} (u_{[⋁_{z \in X} (f (z) ⊙ ⊤_{z}), ⊤_{y}^{*}]}) \\ {\to U (u_{[⊤_{y}, ⊤_{x}^{*}]})))}^{*} \\ = ⋁_{y \in X} (⋁_{z \in X} (f (z) ⊙ U^{*} (u_{[⊤_{z}, ⊤_{y}^{*}]})) \\ ⊙ U^{*} (u_{[⊤_{y}, ⊤_{x}^{*}]})) \\ = ⋁_{z \in X} (f (z) ⊙ ⋁_{y \in X} (U^{*} (u_{[⊤_{z}, ⊤_{y}^{*}]}) \\ ⊙ U^{*} (u_{[⊤_{y}, ⊤_{x}^{*}]}))) \\ = ⋁_{z \in X} (f (z) ⊙ U^{*} (u_{[⊤_{z}, ⊤_{x}^{*}]})) \\ = {(⋀_{z \in X} (f (z) \to U (u_{[⊤_{z}, ⊤_{x}^{*}]})))}^{*} \\ = {(U (u_{[⋁_{z \in X} (f (z) ⊙ ⊤_{z}), ⊤_{x}^{*}]}))}^{*} \\ = U^{*} (u_{[f, ⊤_{x}^{*}]}) = H_{U} (f) (x) . \end{matrix}$

(3) By Lemma 3.1(9), we have $\begin{matrix} H_{U} (f) (x) = U^{*} (u_{[f, ⊤_{x}^{*}]}) = U^{*} (u_{[⊤_{x}, f^{*}]}^{- 1}) \\ = U^{s *} (u_{[⊤_{x}, f^{*}]}) = J_{U^{s}}^{*} (f^{*}) (x) . \end{matrix}$

(4) It can be proved by a similar method as in the proof of (3).

(5) By (AU3) and (AU5), we have $\begin{matrix} T_{H_{U}} (f) & = & S (H_{U} (f), f) \\ = & ⋀_{x \in X} (H_{U} (f) (x) \to f (x)) \\ = & ⋀_{x \in X} (U^{*} (u_{[f, ⊤_{x}^{*}]}) \to f (x)) \\ = & ⋀_{x \in X} (f^{*} (x) \to U (u_{[f, ⊤_{x}^{*}]})) \\ = & U (⋀_{x \in X} (f^{*} (x) \to u_{[f, ⊤_{x}^{*}]})) \\ = & ⋀_{x \in X} U (u_{[f, f^{*} (x) \to ⊤_{x}^{*}]}) \\ = & U (u_{[f, ⋀_{x \in X} (f^{*} (x) \to ⊤_{x})]}) \\ = & U (u_{[f, f]}) = T_{U} (f) . \end{matrix}$

(6) By Lemma 3.1, (AU3) and (AU5), we have $\begin{matrix} H_{U} (f) (x) = U^{*} (u_{[f, ⊤_{x}^{*}]}) \\ = U^{*} (u_{[⋁_{y \in X} (f (y) ⊙ ⊤_{y}), ⊤_{x}^{*}]}) \\ = {(⋀_{y \in X} (f (y) \to U (u_{[⊤_{y}, ⊤_{x}^{*}]}))}^{*} \\ = ⋁_{y \in X} (f (y) ⊙ U^{*} (u_{[⊤_{y}, ⊤_{x}^{*}]})) \end{matrix}$

Let $R_{U} (x, y) = U^{*} (u_{[⊤_{x}, ⊤_{y}^{*}]})$ . Then $H_{U} (f) (x) = ⋁_{y \in X} (f (y) ⊙ R_{U} (y, x)) .$

Moreover, $R_{U^{s}} (x, y) = U^{s *} (u_{[⊤_{x}, ⊤_{y}^{*}]}) = U^{*} (u_{[⊤_{y}, ⊤_{x}^{*}]}) = R_{U} (y, x) = R_{U}^{- 1} (x, y)$ such that $\begin{matrix} H_{U^{s}} (f) (x) = ⋁_{y \in X} (f (y) ⊙ U^{s *} (u_{[⊤_{y}, ⊤_{x}^{*}]})) \\ = ⋁_{y \in X} (f (y) ⊙ U^{*} (u_{[⊤_{x}, ⊤_{y}^{*}]})) \\ = ⋁_{y \in X} (f (y) ⊙ R_{U} (x, y)) . \end{matrix}$

(7) By (AU3) and (AU5), we have $\begin{matrix} S (H_{U} (f), g) & = & ⋀_{y \in X} (H_{U} (f) (y) \to g (y)) \\ = & ⋀_{y \in X} (U^{*} (u_{[f, ⊤_{y}^{*}]}) \to g (y)) \\ = & ⋀_{y \in X} (g^{*} (y) \to U (u_{[f, ⊤_{y}^{*}]})) \\ = & U (u_{[f, ⋀_{y \in X} (g^{*} (y) \to ⊤_{y}^{*})]}) = U (u_{[f, g]}) . \end{matrix}$

Remark 3.13. From Theorems 3.11 and 3.12, a given Alexandrov L-fuzzy pre-uniformity $U$ , we can obtain a fuzzy rough set $(J_{U} (f), H_{U} (f))$ where $J_{U} (f) (x) = U (u_{[⊤_{x}, f]})$ and $H_{U} (f) (x) = U^{*} (u_{[f, ⊤_{x}^{*}]})$ for f ∈ L^X.

Example 3.14. Consider R and $U$ defined in Example 3.5. By Theorem 3.11, we obtain L-lower approximation operators $J_{U}, J_{U^{s}} : L^{X} \to L^{X}$ where $J_{U^{s}} (f) (x) = U (u_{[f^{*}, ⊤_{x}^{*}]}) = ⋀_{z, y \in X} (R (z, y) \to (f^{*} (z) \to ⊤_{x}^{*} (y))) = ⋀_{z \in X} (R (z, x) \to f (z))$ and $J_{U} (f) (x) = ⋀_{y \in X} (R (x, y) \to f (y)) .$ Since $J_{U} (⊤_{y}^{*}) (x) = U^{*} (u_{[⊤_{x}, ⊤_{y}^{*}]}) = R (x, y)$ , we have by Theorem 3.11(3) that if R is transitive, then $J_{U}$ and $J_{U^{s}}$ are topological.

By Theorem 3.12, we obtain L-upper approximation operators $H_{U}, H_{U^{s}} : L^{X} \to L^{X}$ where $H_{U} (f) (x) = U^{*} (u_{[f, ⊤_{x}^{*}]}) = (⋀_{z, y \in X} (R (x, y) \to (f (z) \to ⊤_{x}^{*} (y))))^{*} = ⋁_{y \in X} (R (z, x) ⊙ f (z))$ and $H_{U^{s}} (f) (x) = ⋁_{y \in X} (R (x, y) ⊙ f (y)) .$ Since $H_{U} (⊤_{x}) (y) = U^{*} (u_{[⊤_{x}, ⊤_{y}^{*}]}) = R (x, y)$ , we have by Theorem 3.12(2) that if R is transitive, then $H_{U}$ and $H_{U^{s}}$ are topological.

4 Alexandrov L-fuzzy pre-uniform spaces induced by approximation operators

Theorem 4.1. Let L be a complete residuated lattice where L is a totally order satisfying x → ⋁ _i∈Γy_i = ⋁ _i∈Γ (x → y_i) and x ∧ y = x ⊙ (x → y) for all x, y, y_i ∈ L. Let $(X, J)$ be an L-lower approximation space. Define a map $U_{J} : L^{X \times X} \to L$ by $U_{J} (u) = ⋁ {S (f, J (g)) ∣ u_{[f, g]} \leq u} .$

Then, the following hold:

$U_{J}$ is an Alexandrov L-fuzzy pre-uniformity with $U_{J} (u_{[f, g]}) = ⋀_{x, y \in X} (f (x) \to (g^{*} (y) \to J (⊤_{y}^{*}) (x))) .$ Moreover, $U_{J} (u_{[⊤_{x}, ⊤_{y}^{*}]}) = J (⊤_{y}^{*}) (x)$ with $U_{J} (u_{[⊤_{x}, ⊤_{x}^{*}]})$ =⊥.

$U_{J} (u_{[f, g]}) \geq ⋁_{h \in L^{X}} (U_{J} (u_{[f, h]}) ⊙ U_{J} (u_{[h, g]}))$ . Moreover, the equality holds if $J$ is topological.

If $J$ is topological, then $U_{J}$ is an Alexandrov L-fuzzy quasi-uniformity on X.

$J = J_{U_{J}}$ .

$T_{J} (f) = U_{J} (u_{[f, f]})$ for all f ∈ L^X.

If $U$ is an Alexandrov L-fuzzy pre-uniformity on X, then $U_{J_{U}} (u_{[f, g]}) = U (u_{[f, g]})$ for all f, g ∈ L^X.

Proof. (1) (AU1) and (AU2) are easily proved.

(AU3) Suppose there exists u_i ∈ L^X×X such that $U_{J} (⋀_{i \in Γ} u_{i}) ≱ ⋀_{i \in Γ} U_{J} (u_{i}) .$

For each i ∈ Γ, there exist u_{[f_i,g_i]} ≤ u_i such that $U_{J} (⋀_{i \in Γ} u_{i}) ≱ ⋀_{i \in Γ} S (f_{i}, J (g_{i})) .$

For each i ∈ Γ, there exist j_i, k_i ∈ Γ with (f_{j
_i} = f_i and u_{[f_{j
_i},g_i]} ≤ u_i) or (g_{j
_i} = g_i and u_{[f_i,g_{k
_i}]} ≤ u_i) such that J = {j_i ∣ i ∈ Γ}, K = {k_i ∣ i ∈ Γ}, J ∪ K = Γ and $\begin{matrix} U_{J} (⋀_{i \in Γ} u_{i}) ≱ ⋀_{i \in Γ} S (f_{i}, J (g_{i})) \\ = S (⋁_{j_{i} \in J} f_{j_{i}}, J (⋀_{k_{i} \in K} g_{k_{i}})) . \end{matrix}$

On the other hand, since u_{[⋁_{j_i∈J}f_{j
_i},⋀_{k_i∈K}g_{k
_i}]} = ⋀ _i∈Γu_{[f_i,g_i]} ≤ ⋀ _i∈Γu_i, we have $\begin{matrix} U_{J} (⋀_{i \in Γ} u_{i}) \geq S (⋁_{j_{i} \in J} f_{j_{i}}, J (⋀_{k_{i} \in K} g_{k_{i}})) \\ = S (⋁_{j_{i} \in J} f_{j_{i}}, ⋀_{k_{i} \in K} J (g_{k_{i}})) = ⋀_{i \in Γ} S (f_{i}, J (g_{i})) . \end{matrix}$

This is a contradiction. Hence we have $U_{J} (⋀_{i \in Γ} u_{i}) \geq ⋀_{i \in Γ} U_{J} (u_{i}) .$ Now, by (AU2), we have $U_{J} (⋀_{i \in Γ} u_{i}) \leq ⋀_{i \in Γ} U_{J} (u_{i}) .$

(AU4) Since $J (g) \leq g$ , we have $\begin{matrix} U_{J} (u_{[f, g]}) = ⋀_{x \in X} (f (x) \to J (g) (x)) \\ \leq ⋀_{x \in X} (f (x) \to g (x)) = S (f, g) . \end{matrix}$

(AU5) For u ∈ L^X×X and α ∈ L, we have $\begin{matrix} α \to U_{J} (u) = α \to ⋁ {S (f, J (g)) | u_{[f, g]} \leq u} \\ = ⋁ {α \to S (f, J (g)) | u_{[f, g]} \leq u} \\ \leq ⋁ {S (α ⊙ f, J (g)) | α \to u_{[f, g]} \leq α \to u} \\ \leq ⋁ {S (α ⊙ f, J (g)) | u_{α ⊙ f, g} \leq α \to u} \\ \leq U_{J} (α \to u) . \end{matrix}$

Conversely, first we show $U_{J} (α ⊙ u) \geq α ⊙ U_{J} (u)$ . For u = ⊤ _X×X, since $U_{J} (⊥_{X \times X}) = U_{J} (u_{[⊤_{X}, ⊥_{X}]}) = S (⊤_{X}, J (⊤_{X})) = S (⊤_{X}, ⊥_{X}) = ⊥$ , we have $U_{J} (α \to ⊥_{X \times X}) \geq α \to U_{J} (⊥_{X \times X}) = α^{*}$ . Hence $U_{J} (α ⊙ ⊤_{X \times X}) \geq α ⊙ U_{J} (⊤_{X \times X}) = α .$

If u ≠ ⊤ _X×X, then $\begin{matrix} α ⊙ U_{J} (u) = α ⊙ ⋁ {S (f, J (g)) | u_{[f, g]} \leq u} \\ = ⋁ {α ⊙ S (f, J (g)) | u_{[f, g]} \leq u} \\ = ⋁ {α ⊙ S (α ⊙ f, J (g)) | u_{[α ⊙ f, g]} \leq u} \\ \leq ⋁ {α ⊙ (α \to S (f, J (g))) | u_{[α ⊙ f, g]} \\ \leq α \to α ⊙ u} \\ \leq ⋁ {S (f, J (g)) | α \to u_{[f, g]} \leq α \to α ⊙ u} . \end{matrix}$

Since α → u_[f,g] ≤ α → α ⊙ u ⇔ α ⊙ (α → u_[f,g]) = α ∧ u_[f,g] ≤ α ⊙ u and L is a totally order, we have that if u ≠ ⊤ _X×X, then α ∧ u_[f,g] = u_[f,g]. Thus $\begin{matrix} α ⊙ U_{J} (u) & \leq & ⋁ {S (f, J (g)) | α \to u_{[f, g]} \\ \leq & α \to α ⊙ u} \\ = & ⋁ {S (f, J (g)) | u_{[f, g]} \leq α ⊙ u} \\ = & U_{J} (α ⊙ u) . \end{matrix}$

Since $α ⊙ U_{J} (α \to u) \leq U_{J} (α ⊙ (α \to u)) \leq U_{J} (u)$ , we have $U_{J} (α \to u) \leq α \to U_{J} (u)$ . Thus $U_{J} (α \to u) = α \to U_{J} (u)$ . Hence $U_{J}$ is an Alexandrov L-fuzzy pre-uniformity on X.

For $g = ⋀_{y \in X} (g^{*} (y) \to ⊤_{y}^{*})$ , $\begin{matrix} U_{J} (u_{[f, g]}) = ⋀_{x \in X} (f (x) \to J (g) (x)) \\ = ⋀_{x \in X} (f (x) \to J (⋀_{y \in X} (g^{*} (y) \to ⊤_{y}^{*})) (x)) \\ = ⋀_{x, y \in X} (f (x) \to (g^{*} (y) \to J (⊤_{y}^{*}) (x))) . \end{matrix}$

(2) Note that for any f, g, h ∈ L^X, $\begin{matrix} U_{J} (u_{[f, h]}) ⊙ U_{J} (u_{[h, g]}) \\ = (⋀_{x \in X} (f (x) \to J (h) (x))) \\ ⊙ (⋀_{x \in X} (h (x) \to J (g) (x))) \\ = S (f, J (h)) ⊙ S (h, J (g)) (Since J (h) \leq h,) \\ \leq S (f, h) ⊙ S (h, J (g)) \\ \leq S (f, J (g)) = U_{J} (u_{[f, g]}) . \end{matrix}$

Hence $U_{J} (u_{[f, g]}) \geq ⋁_{h \in L^{X}} (U_{J} (u_{[f, h]}) ⊙ U_{J} (u_{[h, g]}))$ .

If $J$ is topological, then $\begin{matrix} ⋁_{h \in L^{X}} (U_{J} (u_{[f, h]}) ⊙ U_{J} (u_{[h, g]})) \\ = ⋁_{h \in L^{X}} (S (f, J (h)) ⊙ S (h, J (g))) (Put h = J (g),) \\ \geq S (f, J (J (g))) ⊙ S (J (g), J (g)) \\ = S (f, J (g)) = U_{J} (u_{[f, g]}) . \end{matrix}$

(3) Suppose that there exists u ∈ L^X×X such that $U (u) ≰ ⋁ {U (v) ⊙ U (w) | v \circ w \leq u} .$

By the definition of $U (u)$ , there exists u_[f,g] ≤ u such that $U (u_{[f, g]}) ≰ ⋁ {U (v) ⊙ U (w) | v \circ w \leq u} .$ Since $U_{J} (u_{[f, g]}) = ⋁_{h \in L^{X}} (U_{J} (u_{[f, h]}) ⊙ U_{J} (u_{[h, g]}))$ and u_[f,h] ∘ u_[h,g] ≤ u_[f,g] ≤ u, we have by (2) that $\begin{matrix} U (u_{[f, g]}) & = & ⋁_{h \in L^{X}} (U_{J} (u_{[f, h]}) ⊙ U_{J} (u_{[h, g]})) \\ \leq & ⋁ {U (v) ⊙ U (w) | v \circ w \leq u} . \end{matrix}$

This is a contradiction. Hence $U (u) \leq ⋁ {U (v) ⊙ U (w) | v \circ w \leq u} .$

(4) and (5) can be deduced from: $\begin{matrix} J_{U_{J}} (f) (x) & = & U_{J} (u_{[⊤_{x}, f]}) = S (⊤_{x}, J (f)) \\ = & ⋀ (⊤_{x} (y) \to J (f) (y)) = J (f) (x) and \\ T_{J} (f) & = & S (⊤_{x}, J (f)) = U_{J} (u_{[f, f]}) . \end{matrix}$

(6) By (AU3) and (AU4), we have $\begin{matrix} U_{J_{U}} (u_{[f, g]}) & = & ⋀_{x \in X} (f (x) \to J_{U} (g) (x)) \\ = & ⋀_{x \in X} (f (x) \to U (u_{[⊤_{x}, g]})) \\ = & U (u_{[⋁_{x \in X} (f (x) ⊙ ⊤_{x}), g]}) = U (u_{[f, g]}) . \end{matrix}$

Remark 4.2. [18] If L is a continuous BL-algebra and totally order, then L satisfies the condition of Theorem 4.1.

Theorem 4.3. Let L be a complete residuated lattice where L is a totally order satisfying x → ⋁ _i∈Γy_i = ⋁ _i∈Γ (x → y_i) and x ∧ y = x ⊙ (x → y) for all x, y, y_i ∈ L. Let $(X, H)$ be an L-upper approximation space. Define a map $U_{H} : L^{X \times X} \to L$ by $U_{H} (u) = ⋁ {S (H (f), g) ∣ u_{[f, g]} \leq u} .$

Then, the following hold:

$U_{H}$ is an Alexandrov L-fuzzy pre-uniformity such that $U_{H} (u_{[f, g]}) = ⋀_{x, y \in X} (H (⊤_{x}) (y) \to (f (x) \to g (y))) .$

Moreover, $U_{H} (u_{[⊤_{x}, ⊤_{y}^{*}]}) = H^{*} (⊤_{x}) (y)$ with $U_{H} (u_{[⊤_{x}, ⊤_{x}^{*}]}) = H^{*} (⊤_{x}) (x) = ⊥ .$

$U_{H} (u_{[f, g]}) \geq ⋁_{h \in L^{X}} (U_{H} (f, h) ⊙ U_{H} (h, g))$ . The equality holds if $H$ is topological.

If $H$ is topological, then $U_{H}$ is an Alexandrov L-fuzzy quasi-uniformity on X.

$H = H_{U_{H}}$ .

$T_{H} (f) = U_{H} (u_{[f, f]}) = T_{U_{H}} (f)$ for all f ∈ L^X.

Define an operator $J_{H} : L^{X} \to L^{X}$ with $J_{H} (⊤_{x}^{*}) (y) = H^{*} (⊤_{y}) (x)$ as $J_{H} (f) (z) = ⋀_{w \in X} (g^{*} (w) \to J_{H} (⊤_{w}^{*}) (z)) .$

Then $J_{H}$ is an L-lower approximation space with $U_{H} (u_{[f, g]}) = ⋀_{x \in X} (f (x) \to J_{H} (g) (x)) .$

Define an operator $J_{H} : L^{X} \to L^{X}$ as $J_{H} (f) = H^{*} (f^{*}) .$ Then $J_{H}$ is an L-lower approximation space with $U_{H}^{s} (u_{[f, g]}) = ⋀_{x \in X} (f (x) \to J_{H} (g) (x)) .$

If $U$ is an Alexandrov L-fuzzy pre-uniformity on X, then $U_{H_{U}} (u_{[f, g]}) = U (u_{[f, g]}) = U_{J_{U^{s}}}^{s} (u_{[f, g]})$ for all f, g ∈ L^X.

Proof. (1) (AU1) and (AU2) are proved from the definition of $U_{H}$ .

(AU3) Suppose that there exists u_i ∈ L^X×X such that $U_{H} (⋀_{i \in Γ} u_{i}) ⋀_{i \in Γ} U_{H} (u_{i}) .$

For each i ∈ Γ, there exist u_{[f_i,g_i]} ≤ u_i with $U_{H} (⋀_{i \in Γ} u_{i}) ≱ ⋀_{i \in Γ} S (H (f_{i}), g_{i}) .$ For each i ∈ Γ, there exist j_i, k_i ∈ Γ with (f_{j
_i} = f_i and u_{[f_{j
_i},g_i]} ≤ u_i) or (g_{j
_i} = g_i and u_{[f_i,g_{k
_i}]} ≤ u_i) such that J = {j_i ∣ i ∈ Γ}, K = {k_i ∣ i ∈ Γ}, J ∪ K = Γ and $\begin{matrix} U_{H} (⋀_{i \in Γ} u_{i}) ≱ ⋀_{i \in Γ} S (H (f_{i}), g_{i}) \\ = S (H (⋁_{j_{i} \in J} f_{j_{i}}), ⋀_{k_{i} \in K} g_{k_{i}})) \\ = S (⋁_{j_{i} \in J} H (f_{j_{i}}), ⋀_{k_{i} \in K} g_{k_{i}}) . \end{matrix}$

On the other hand, since u_{[⋁_{j_i∈J}f_{j
_i},⋀_{k_i∈K}g_{k
_i}]} = ⋀ _i∈Γu_{[f_i,g_i]} ≤ ⋀ _i∈Γu_i, we have $\begin{matrix} U_{H} (⋀_{i \in Γ} u_{i}) \geq S (H (⋁_{j_{i} \in J} f_{j_{i}}), ⋀_{k_{i} \in K} g_{k_{i}})) \\ = S (⋁_{j_{i} \in J} H (f_{j_{i}}), ⋀_{k_{i} \in K} g_{k_{i}}) = ⋀_{i \in Γ} S (H (f_{i}), g_{i}) . \end{matrix}$

This is a contradiction. Thus $U_{H} (⋀_{i \in Γ} u_{i}) \geq ⋀_{i \in Γ} U_{H} (u_{i}) .$ By (AU2), $U_{H} (⋀_{i \in Γ} u_{i}) \leq ⋀_{i \in Γ} U_{H} (u_{i}) .$

(AU4) Since $H (f) \geq f$ , we have by Lemma 2.2(2) that $\begin{matrix} U_{H} (u_{[f, g]}) & = & ⋀_{x \in X} (H (f) (x) \to g (x)) \\ \leq & ⋀_{x \in X} (f (x) \to g (x)) = S (f, g) . \end{matrix}$

(AU5) For α ∈ L and u ∈ L^X×X, we have $\begin{matrix} α \to U_{H} (u) = α \to ⋁ {S (H (f), g) | u_{[f, g]} \leq u} \\ = ⋁ {α \to S (H (f), g) | u_{[f, g]} \leq u} \\ \leq ⋁ {S (α ⊙ H (f), g) | α \to u_{[f, g]} \leq α \to u} \\ \leq ⋁ {S (H (α ⊙ f), g) | u_{α ⊙ f, g} \leq α \to u} \\ \leq U_{H} (α \to u) . \end{matrix}$

First we show $U_{H} (α ⊙ u) \geq α ⊙ U_{H} (u)$ .

Case 1: u = ⊤ _X×X. Since $U_{H} (⊥_{X \times X}) = U_{H} (u_{[⊤_{X}, ⊥_{X}]}) = S (H (⊤_{X}), ⊥_{X}) = S (⊤_{X}, ⊥_{X}) = ⊥$ , we have $U_{H} (α \to ⊥_{X \times X}) \geq α \to U_{H} (⊥_{X \times X}) = α^{*}$ . Hence $U_{H} (α ⊙ ⊤_{X \times X}) \geq α ⊙ U_{H} (⊤_{X \times X}) = α .$

Case 2: u ≠ ⊤ _X×X. Note that $\begin{matrix} α ⊙ U_{H} (u) = α ⊙ ⋁ {S (H (f), g) | u_{[f, g]} \leq u} \\ = ⋁ {α ⊙ S (H (f), g) | u_{[f, g]} \leq u} \\ = ⋁ {α ⊙ S (H (α ⊙ f), H (g)) | u_{[α ⊙ f, g]} \leq u} \\ \leq ⋁ {α ⊙ (α \to S (H (f), g)) | u_{[α ⊙ f, g]} \\ \leq α \to α ⊙ u} \\ \leq ⋁ {S (H (f), g) | α \to u_{[f, g]} \leq α \to α ⊙ u} . \end{matrix}$

Since α → u_[f,g] ≤ α → α ⊙ u ⇔ α ⊙ (α → u_[f,g]) = α ∧ u_[f,g] ≤ α ⊙ u and L is a totally order, we have that if u ≠ ⊤ _X×X, then α ∧ u_[f,g] = u_[f,g]. Thus, $\begin{matrix} α ⊙ U_{H} (u) & \leq & ⋁ {S (H (f), g) | α \to u_{[f, g]} \\ \leq & α \to α ⊙ u} \\ = & ⋁ {S (H (f), g) | u_{[f, g]} \leq α ⊙ u} \\ = & U_{H} (α ⊙ u) . \end{matrix}$

Since $α ⊙ U_{H} (α \to u) \leq U_{H} (α ⊙ (α \to u)) \leq U_{H} (u),$ we have $U_{H} (α \to u) \leq α \to U_{H} (u)$ . Thus $U_{H} (α \to u) = α \to U_{H} (u)$ . Hence $U_{H}$ is an Alexandrov L-fuzzy pre-uniformity on X.

For f = ⋀ _x∈X (f (x) ⊙ ⊤ _x), $\begin{matrix} U_{H} (u_{[f, g]}) = ⋀_{y \in X} (H (f) (y) \to g (y)) \\ = ⋀_{x \in X} (H (⋁_{x \in X} (f (x) ⊙ ⊤_{x}) (y) \to g (y)) \\ = ⋀_{x, y \in X} (f (x) ⊙ H (⊤_{x}) (y) \to g (y)) \\ = ⋀_{x, y \in X} (H (⊤_{x}) (y) \to (f (x) \to g (y))) . \end{matrix}$

Moreover, $U_{H} (u_{[⊤_{x}, ⊤_{y}^{*}]}) = ⋀_{x, y \in X} (H (⊤_{x}) (y) \to (⊤_{x} (x) \to ⊤_{y}^{*} (y))) = H^{*} (⊤_{x}) (y)$ and $U_{H} (u_{[⊤_{x}, ⊤_{x}^{*}]}) = H^{*} (⊤_{x}) (x) \leq (⊤_{x} (x))^{*} = ⊥ .$

(2) For f, g, h ∈ L^X, $\begin{matrix} U_{H} (u_{[f, h]}) ⊙ U_{H} (u_{[h, g]}) \\ = (⋀_{x \in X} (H (f) (x) \to h (x))) \\ ⊙ (⋀_{x \in X} (H (h) (x) \to g (x))) \\ = S (H (f), h) ⊙ S (H (h), g) (Since H (h) \geq h,) \\ \leq S (H (f), h) ⊙ S (h, g) \leq S (H (f), g) \\ = U_{H} (u_{[f, g]}) . \end{matrix}$

Hence $U_{H} (u_{[f, g]}) \geq ⋁_{h \in L^{X}} (U_{H} (u_{[f, h]}) ⊙ U_{H} (u_{[h, g]}))$ .

If $H$ is topological, then $\begin{matrix} ⋁_{h \in L^{X}} U_{H} (u_{[f, h]}) ⊙ U_{H} (u_{[h, g]}) \\ = S (H (f), h) ⊙ S (H (h), g) (put H (f) = h,) \\ \geq S (H (f), H (f)) ⊙ S (H (H (f)), g) \\ \geq S (H (H (f)), g) = S (H (f), g) = U_{H} (u_{[f, g]}) . \end{matrix}$

(3) It can be proved in the similar way as in the proof of Theorem 4.1 (3).

(4) For f ∈ L^X and x ∈ X, we have $\begin{matrix} H_{U_{H}} (f) = U_{H}^{*} (u_{[f, ⊤_{x}^{*}]}) \\ = {(⋀_{y \in X} (H (f) (y) \to ⊤_{x}^{*} (y)))}^{*} \\ = (H^{*} (f) (x))^{*} = H (f) (x) . \end{matrix}$

(5) $T_{U_{H}} (f) = U_{H} (u_{[f, f]}) = S (H, f) = T_{H} (f) .$

(6) (J1) $J_{H} (⊤_{X}) = ⊤_{X}$ .

(J2) For f ∈ L^X and z ∈ X, we have $\begin{matrix} J_{H} (f) (z) = ⋀_{w \in X} (f^{*} (w) \to J_{H} (⊤_{w}^{*}) (z)) \\ \leq f^{*} (z) \to J_{H} (⊤_{z}^{*}) (z) = f^{*} (z) \to H^{*} (⊤_{z}) (z) \\ \leq f^{*} (z) \to ⊤_{z}^{*} (z) = f (z) . \end{matrix}$

(J3) For f_i ∈ L^X, i ∈ Γ and z ∈ X, we have $\begin{matrix} J_{H} (⋀_{i \in Γ} f_{i}) (z) \\ = ⋀_{w \in X} (⋁_{i \in Γ} f_{i}^{*} (w) \to J_{H} (⊤_{w}^{*}) (z)) \\ = ⋀_{w \in X} ⋀_{i \in Γ} (f_{i}^{*} (w) \to J_{H} (⊤_{w}^{*}) (z)) \\ = ⋀_{i \in Γ} J_{H} (f_{i}) (z) . \end{matrix}$

(J4) For α ∈ L, f ∈ L^X and z ∈ X, we have $\begin{matrix} J_{H} (α \to f) (z) \\ = ⋀_{w \in X} (α ⊙ f^{*} (w) \to J_{H} (⊤_{w}^{*}) (z)) \\ = α \to ⋀_{w \in X} (f^{*} (w) \to J_{H} (⊤_{w}^{*}) (z)) \\ = α \to J_{H} (f) (z) . \end{matrix}$

Hence $J_{H}$ is an L-lower approximation space. Moreover, $\begin{matrix} U_{H} (u_{[f, g]}) \\ = ⋀_{x \in X} (H (f) (x) \to g (x)) \\ = ⋀_{x \in X} (H (⋁_{z \in X} (f (z) ⊙ ⊤_{z})) (x) \to g (x)) \\ = ⋀_{x \in X} (⋁_{z \in X} (f (z) ⊙ H (⊤_{z}) (x)) \to g (x)) \\ = ⋀_{x, z \in X} (f (z) \to (H (⊤_{z}) (x) \to g (x))) \\ = ⋀_{x, z \in X} (f (z) \to (g^{*} (x) \to H^{*} (⊤_{z}) (x))) \\ = ⋀_{x, z \in X} (f (z) \to (g^{*} (x) \to J_{H} (⊤_{x}^{*}) (z))) \\ = ⋀_{z \in X} (f (z) \to J_{H} (⋀_{x \in X} (g^{*} (x) \to ⊤_{x}^{*}) (z))) \\ = ⋀_{z \in X} (f (z) \to J_{H} (g) (z))) \\ = U_{J_{H}} (u_{[f, g]}), \\ U_{H} (u) = ⋁ {S (H (f), g) ∣ u_{[f, g]} \leq u} \\ = ⋁ {S (f, J_{H} (g)) ∣ u_{[f, g]} \leq u} = U_{J_{H}} (u) . \end{matrix}$

(7) For f, g ∈ L^X, we have $\begin{matrix} U_{H}^{s} (u_{[f, g]}) = U_{H} (u_{[g^{*}, f^{*}]}) = S (H (g^{*}), f^{*}) \\ = S (f, H^{*} (g^{*})) = S (f, J_{H} (g)) = U_{J_{H}} (u_{[f, g]}) . \end{matrix}$

(8) For f, g ∈ L^X, we have $\begin{matrix} U_{H_{U}} (u_{[f, g]}) \\ = ⋀_{x \in X} (U^{*} (u_{[f, ⊤_{x}^{*}]}) \to g (x)) \\ = ⋀_{x \in X} (g^{*} (x) \to U (u_{[f, ⊤_{x}^{*}]})) \\ = U (u_{[f, ⋀_{x \in X} (g^{*} (x) \to ⊤_{x}^{*})]}) = U (u_{[f, g]}), \\ U_{H_{U}} (u_{[f, g]}) \\ = S (H_{U} (f), g) = S (g^{*}, H_{U}^{*} (f)) \\ = S (g^{*}, J_{U^{s}} (f^{*})) = U_{J_{U^{s}}} (u_{[g^{*}, f^{*}]}) = U_{J_{U^{s}}}^{s} (u_{[f, g]}) . \end{matrix}$

Example 4.4. Let ([0, 1], ⊙, →, ^*, 0, 1) be a complete residuated lattice (Ref. [1 , 18]) where x ⊙ y = max {0, x + y - 1}, x → y = min {1 - x + y, 1} and x^* = 1 - x .

Let X = {x_i ∣ i = {1, 2, 3}} where x_i (i = 1, 2, 3) are houses. Let f_e, f_b : X → [0, 1] be the two graded functions defined as $\begin{matrix} f_{e} (x_{1}) = 1, f_{e} (x_{2}) = 0.8, f_{e} (x_{3}) = 0.5, \\ f_{b} (x_{1}) = 0.8, f_{b} (x_{2}) = 0.9, f_{b} (x_{3}) = 0.2 . \end{matrix}$

We call e and b as expensive and beautiful house. Then (X, AT = {e, b}, {f_e, f_b}) is an information system where AT is a set of attributes (see [1 , 14]). Define a reflexive relation R ∈ [0, 1] ^X×X as $R (x_{i}, x_{j}) = (f_{e} (x_{i}) \to f_{e} (x_{j})) \land (f_{b} (x_{i}) \to f_{b} (x_{j})) .$

Then R is a [0, 1]-fuzzy preorder as $R = (\begin{matrix} 1 & 0.8 & 0.4 \\ 0.9 & 1 & 0.3 \\ 1 & 1 & 1 \end{matrix}) .$ (1) By Example 3.5, we obtain an Alexandrov [0, 1]-fuzzy pre-uniformity $U : [0, 1]^{X \times X} \to [0, 1]$ as $U (u) = ⋀_{x, y} (R (x, y) \to u (x, y)) = S (R, u),$ where S is an L-fuzzy preorder on [0, 1] ^X×X in Lemma 2.4. If R (x, y) ≤ u (x, y) for all x, y ∈ X, then $U (u) = S (R, u) = 1$ . Define u ∈ [0, 1] ^X×X as $u = (\begin{matrix} 0.5 & 0.9 & 0.8 \\ 0.7 & 0.4 & 0.5 \\ 0.6 & 0.8 & 0.9 \end{matrix}) .$

Then $U (u) = ⋀_{x, y} (R (x, y) \to u (x, y)) = 0.4 .$

(2) For f ∈ [0, 1] ^X with f (x₁) =0.6, f (x₂) =0.8 and f (x₃) =0.3, we have $\begin{matrix} u_{[1_{x_{1}}, f]} = (\begin{matrix} 0.6 & 0.8 & 0.3 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{matrix}), \\ u_{[1_{x_{2}}, f]} = (\begin{matrix} 1 & 1 & 1 \\ 0.6 & 0.8 & 0.3 \\ 1 & 1 & 1 \end{matrix}), \\ u_{[1_{x_{3}}, f]} = (\begin{matrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 0.6 & 0.8 & 0.3 \end{matrix}), \\ u_{[f, 1_{x_{1}}^{*}]} = (\begin{matrix} 0.4 & 0.2 & 0.7 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{matrix}), \\ u_{[f, 1_{x_{2}}^{*}]} = (\begin{matrix} 1 & 1 & 1 \\ 0.4 & 0.2 & 0.7 \\ 1 & 1 & 1 \end{matrix}), \\ u_{[f, 1_{x_{3}}^{*}]} = (\begin{matrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 0.4 & 0.2 & 0.7 \end{matrix}), \\ u_{[f, f]} = (\begin{matrix} 1 & 1 & 0.7 \\ 0.8 & 1 & 0.5 \\ 1 & 1 & 1 \end{matrix}) . \end{matrix}$

We can obtain a fuzzy rough set $(J_{U} (f), H_{U} (f))$ with $J_{U} (f) = (0.6, 0.7, 0.3), H_{U} (f) = (0.6, 0.8, 0.8)$ and an Alexandrov [0, 1]-fuzzy topology $T_{U} (f) = U (u_{[f, f]}) = 0.9$ for f ∈ [0, 1] ^X.

(3) By Theorems 3.11(6) and 3.12(6), $R_{U} (x, y) = U^{*} (u_{[1_{x}, 1_{y}^{*}]}) = R (x, y)$ , and so $\begin{matrix} J_{U} (f) (x_{i}) = ⋀_{x \in X} (R (x_{i}, x) \to f (x)), \\ H_{U} (f) (x_{i}) = ⋁_{x \in X} (R (x, x_{i}) ⊙ f (x)) . \end{matrix}$

Since ([0, 1], ⊙) is a continuous t-norm, we have x → ⋁ _i∈Γy_i = ⋁ _i∈Γ (x → y_i). Moreover, x ∧ y = x ⊙ (x → y) by Theorems 4.1 and 4.3, we obtain $\begin{matrix} U_{J_{U}} (u) = ⋁ {S (f, J_{U} (g)) ∣ u_{[f, g]} \leq u} \\ = ⋁ {⋀_{x \in X} (f (x) \to ⋀_{y \in X} (R (x, y) \to g (y))) \\ ∣ u_{[f, g]} \leq u} \\ = ⋁ {⋀_{y \in X} (⋁_{x \in X} (R (x, y) ⊙ f (x)) \to g (y)) \\ ∣ u_{[f, g]} \leq u} \\ = ⋁ {S (H_{U} (f), g) ∣ u_{[f, g]} \leq u} = U_{H_{U}} (u) \\ = ⋁ {⋀_{x, y \in X} (R (x, y) \to u_{[f, g]} (x, y)) ∣ u_{[f, g]} \leq u} \\ \leq U (u) . \end{matrix}$

5 Conclusion

In this paper, Alexandrov L-fuzzy pre-uniformities in complete residuated lattices are investigated. From a given Alexandrov L-fuzzy pre-uniformity $U$ , we can obtain a fuzzy rough set $(J_{U} (f), H_{U} (f))$ where $J_{U} (f) (x) = U (u_{[⊤_{x}, f]})$ and $H_{U} (f) (x) = U^{*} (u_{[f, ⊤_{x}^{*}]})$ for f ∈ L^X (see Theorems 3.11 and 3.12). Moreover, we obtain an Alexandrov fuzzy topology $T_{U}$ with $T_{U} (f) = U (u_{f, f})$ (see Theorem 3.7). Conversely, for given L-lower approximation operator $J$ and L-upper approximation operator $H$ , we obtain Alexandrov L-fuzzy pre-uniformities $U_{J}$ and $U_{H}$ , respectively (see Theorems 4.1 and 4.3).

It might be interesting to note that the concepts of Alexandrov L-fuzzy pre-uniformities can be applied to formal concept analysis, topological structures and decision-making systems.

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