Relations among ( L,M )-fuzzy convex structures,( L,M )-fuzzy closure systems and ( L,M )-fuzzy Alexandrov topologies in a degree sense

Abstract

In this paper, by means of implication operator → in a completely distributive lattice M, we define degrees of (L, M)-fuzzy convex structures, (L, M)-fuzzy closure systems and (L, M)-fuzzy Alexandrov topologies. Moreover, We discuss some properties of the degree of (L, M)-fuzzy convex spaces, the relations between the degree of (L, M)-fuzzy convex structures and the degree of (L, M)-fuzzy closure systems, and the relations between the degree of (L, M)-fuzzy convex structures and the degree of (L, M)-fuzzy Alexandrov topologies.

Keywords

degree

1. Introduction

Axiomatic convexity theory is a branch of mathematics dealing with set-theoretic structures satisfying axioms similar to that convex sets in real linear spaces fulfill. The axiomatic convexity (convex structure) exists in many mathematical structures, such as convexities in metric spaces and graphs [15, 29], convexities in lattices and in Boolean algebras [36]. Also, axiomatic convexity appeared naturally in topology, especially in the theory of supercompact spaces [17]. Some more details about axiomatic convexity can be found in [36].

Fuzzy subsets have been applied to various branches of mathematics, such as fuzzy topology [1 , 34] and fuzzy convergence [10–13 , 19]. In 1994, Rosa [28] first applied fuzzy subsets to convex structures and introduced a fuzzy convex structure $(X, C)$ , which was defined as a subset of [0, 1] ^X satisfying certain axioms. In 2009, based on a completely distributive lattice L, Maruyama [16] gave a notion of L-convex structures, which was a subset of L^X. In recent years, this kind of fuzzy convex structures were widely studied by many researchers [8 , 37].

In 2014, Shi and Xiu [33] introduced a new approach to the fuzzification of convex structures–an M-fuzzifying convex structure, which is a mapping $C : 2^{X} ⟶ M$ satisfying three axioms. As a continuation of the work, there are more and more research on this M-fuzzifying convex structures [30 , 38–40].

Based on the idea of fuzzy topologies in the sense of Kubiak [9] and Šostak [34], Shi and Xiu [32] introduced the notion of (L, M)-fuzzy convex structures, which were the generalization of M-fuzzifying convex structures and L-convex structures, and discussed the categorical relationships between M-fuzzifying convex structures and (L, M)-fuzzy convex structures. Late, Li [12] studied the categorical properties of enriched (L, M)-fuzzy convex spaces.

In a completely distributive lattice M, there exists an implication operation → as the right adjoint for the meet operation ∧, i.e., p ∧ q ≤ r ⇔ q ≤ p → r. The operation ∧ can be interpreted as the logic connective conjunction and the operation → as the logic connective implication. Based on these interpretations, basic notions and results of classical logic can be extended to the many valued case. In this paper, based on the implication operation → in a completely distributive lattice M, we define the degree of (L, M)-fuzzy convex structures, (L, M)-fuzzy closure systems and (L, M)-fuzzy Alexandrov topologies and also discuss some properties of the degree of (L, M)-fuzzy convex structures and the relations between the degree of (L, M)-fuzzy Alexandrov topologies and the degree of (L, M)-fuzzy convex structures, and the relations between the degree of (L, M)-fuzzy convex structures and the degree of (L, M)-fuzzy closure systems.

2. Preliminaries

Let L (resp. M) be a complete lattice with the smallest element ⊥_L (resp. ⊥_M) and the largest element ⊤_L (resp. ⊤_M). The relation ≺ in M is defined as follows: for a, b ∈ M, a ≺ b if for every subset D ⊆ M, the relation b ≤ ⋁ D always implies the existence of d ∈ D with a ≤ d [12]. A complete lattice M is completely distributive if and only if b = ⋁ {a ∈ M : a ≺ b} for each b ∈ M. For any b ∈ M, define β (b) = {a ∈ L : a ≺ b}. An element a in M is called co-prime if a ≤ b ∨ c implies a ≤ b or a ≤ c. The set of non-zero co-prime elements in M is denoted by J (M).

For a, b ∈ L, we say that a is way below b in L (in symbols, a ⪡ b) if for all directed subsets D ⊆ L, b ⩽ ⋁ D always implies that a ⩽ d for some d ∈ D. A complete lattice L is said to be continuous if for all x ∈ L, ⇓x is directed and x = ⋁ ⇓ x, where ⇓x = {y ∈ L ∣ y ⪡ x} ([5]). For a nonempty set X, L^X denotes the set of all L-subsets on X. The operators on L can be translated onto L^X in a pointwise way. In this case, L^X is also a complete lattice. The smallest element and the largest element in L^X are denoted by ${\underline{⊥}}_{X}$ and ${\underline{⊤}}_{X}$ , respectively. {B_k : k ∈ K} ⊆ L^X means {B_k : k ∈ K} is an up-directed subset of L^X and $⋁_{k \in K}^{dir} B_{k}$ means the union of {B_k : k ∈ K}.

In a completely distributive lattice M, there exists an implication operation → : M × M ⟶ M as the right adjoint for the meet operation ∧ by $a \to b = ⋁ {c \in M ∣ a \land c \leq b} .$

For the completely distributive lattice M, the following properties of → hold: for all a, b, c ∈ M, {a_j} _j∈J, {b_j} _j∈J ⊆ M,

(a → b) ≥ c ⇔ a ∧ c ≤ b;

a → b = ⊤ ⇔ a ≤ b;

a → (⋀ _ib_i) = ⋀ _i (a → b_i);

(⋁ _ia_i) → b = ⋀ _i (a_i → b);

(a → c) ∧ (c → b) ≤ a → b;

a ≤ b ⇒ c → a ≤ c → b.

a ≤ b ⇒ b → c ≤ a → c.

(a → b) ∧ (c → d) ≤ a ∧ c → b ∧ d.

Let f : X → Y be a mapping. Define $f_{L}^{\to} : L^{X} \to L^{Y}$ and $f_{L}^{\leftarrow} : L^{Y} \to L^{X}$ by $f_{L}^{\to} (A) (y) = ⋁_{f (x) = y} A (x)$ for A ∈ L^X and y ∈ Y, and $f_{L}^{\leftarrow} (B) = B \circ f$ for B ∈ L^Y, respectively.

Throughout this paper, L denotes a continuous lattice and M denotes a completely distributive lattice, unless otherwise stated.

Lemma 2.1. [5] Let L be a continuous lattice and let {a_j,k ∣ j ∈ J, k ∈ K (j)} be a nonempty family of elements in L such that {a_j,k ∣ k ∈ K (j)} is directed for all j ∈ J. Then the following identity holds: $(DD) ⋀_{j \in J} ⋁_{k \in K (j)}^{dir} a_{j, k} = ⋁_{h \in N}^{dir} ⋀_{j \in J} a_{j, h (j)},$ where N is the set of all choice functions h : J ⟶ ⋃ _j∈JK (j) with h (j) ∈ K (j) for all j ∈ J.

Definition 2.2. [33] A mapping $C : L^{X} ⟶ M$ is an (L, M)-fuzzy convex structure on X if it satisfies the following conditions:

(LMC1) $C ({\underline{⊥}}_{X}) = C ({\underline{⊤}}_{X}) = ⊤_{M}$ ;

(LMC2) if {A_i : i ∈ Ω} ⊆ L^X is nonempty, then $C (⋀_{i \in Ω} A_{i}) \geq ⋀_{i \in Ω} C (A_{i})$ ;

(LMC3) if {A_i : i ∈ Ω} ⊆ L^X is an up-directed set, then $C (⋁_{i \in Ω} A_{i}) \geq ⋀_{i \in Ω} C (A_{i})$ .

If $C$ is an (L, M)-fuzzy convex structure on X, then the pair $(X, C)$ is an (L, M)-fuzzy convex space.

Definition 2.3. [4] A mapping $S : L^{X} ⟶ M$ is an (L, M)-fuzzy closure system on X if it satisfies:

(LMS1) $S ({\underline{⊥}}_{X}) = S ({\underline{⊤}}_{X}) = ⊤_{M}$ ;

(LMS2) $S (⋀_{i \in Ω} A_{i}) ⩾ ⋀_{i \in Ω} S (A_{i})$ .

For an (L, M)-fuzzy closure system $S$ on X, the pair $(X, S)$ is called an (L, M)-fuzzy closure space.

Definition 2.4. [3 , 35] A mapping τ : L^X ⟶ M is an (L, M)-fuzzy topology on X if it satisfies:

(LMFT1) $τ ({\underline{⊤}}_{X}) = τ ({\underline{⊥}}_{X}) = ⊤_{M}$ ;

(LMFT2) τ (A₂ ∧ A₂) ≥ τ (A₁) ∧ τ (A₂);

(LMFT3) τ (⋁ _i∈ΩA_i) ≥ ⋀ _i∈Ωτ (A_i).

For an (L, M)-fuzzy topology τ on X, the pair (X, τ) is called an (L, M)-fuzzy topological space.

Definition 2.5. An (L, M)-fuzzy topology τ is called an (L, M)-fuzzy Alexandrov topology if, in addition, τ satisfies:

(LMFTs) τ (⋀ _i∈ΩA_i) ≥ ⋀ _i∈Ωτ (A_i).

For an (L, M)-fuzzy Alexandrov topology τ on X, the pair (X, τ) is called an (L, M)-fuzzy Alexandrov topological space.

Theorem 2.6. [8] Suppose that $A = ⋁_{k \in K}^{dir} A_{k} = ⋁_{k \in K}^{dir} ⋁_{j \in J_{k}}^{dir} B_{k, j} .$

Let $L_{fin}^{A} = {F | F ⊑ A, | F | is finite}$ , where F ⊑ A means F (x) ⪡ A (x) for each x ∈ X and |F| means the set of {x ∈ X|F (x) ≠ ⊥ _L}. Now define a mapping $σ : L_{fin}^{A} ⟶ L^{X}$ as follows: $\forall F \in L_{fin}^{A}, σ (F) = ⋀ {B_{k, j} ∣ F \leq B_{k, j}} .$

Then σ is definable, the set ${σ (F) | F \in L_{fin}^{A}}$ is up-directed and $A = ⋁_{F \in L_{fin}^{A}} σ (F)$ .

3. The degree of (L, M)-fuzzy convex structures

In this section, we will equip each mapping from L^X to M with some degrees to become an (L, M)-fuzzy convex structure. Then we will give some degree representations for the properties of (L, M)-fuzzy convex structures. For convenience, we first give the following notations.

For each $C \in P_{M} (L^{X}) = {C | C : L^{X} ⟶ M}$ ,

$D_{T} (X, C) = C ({\underline{⊥}}_{X}) \land C ({\underline{⊤}}_{X})$ .

$D_{\land} (X, C) = ⋀_{{A_{k} : k \in K} \subseteq L^{X}} (⋀_{k \in K} C (A_{k}) \to C (⋀_{k \in K} A_{k}))$ .

$D_{\lor^{d}} (X, C) = ⋀_{{A_{k} : k \in K} \overset{dir}{\subseteq} L^{X}} (⋀_{k \in K} C (A_{k}) \to C (⋁_{k \in K}^{dir} A_{k}))$ .

In order to make clear the above notations, we give some explanations. If $D_{\land} (X, C) = ⊤_{M}$ , then $C (⋀_{k \in K} A_{k}) \geq ⋀_{k \in K} C (A_{k})$ for each {A_k : k ∈ K} ⊆ L^X. This means $D_{\land} (X, C)$ is a logical extension of the axiom (LMC2). Similarly, $D_{\lor^{d}} (X, C)$ presents the logical extension of the axiom (LMC3).

Next let us give the main definition of this section.

Definition 3.1. For each $C \in P_{M} (L^{X})$ , define $D^{con} (X, C)$ as follows: $D^{con} (X, C) = D_{T} (X, C) \land D_{\land} (X, C) \land D_{\lor^{d}} (X, C) .$ Then $D^{con} (X, C)$ is called the degree to which $C$ is an (L, M)-fuzzy convex structure on X. Obviously, $C$ is an (L, M)-fuzzy convex structure on X if and only if $D^{con} (X, C) = ⊤_{M}$ .

The above definition allows us to talk on the degree to which an arbitrary mapping $C : L^{X} \to M$ becomes an (L, M)-fuzzy convex structure on X even $C$ is not. The degree $D^{con} (X, C)$ is a natural measure to which $(X, C)$ is an (L, M)-fuzzy convex structure on X. In the sequel, we will show the degree $D^{con} (X, C)$ naturally suggests many-valued logical extensions of properties that the classical convex structure possesses.

Proposition 3.2. Let ${(X, C_{t})}_{t \in T}$ be a family of mappings from L^X to M and define $⋀_{t \in T} C_{t} : L^{X} ⟶ M$ by $\forall A \in L^{X}, (⋀_{t \in T} C_{t}) (A) = ⋀_{t \in T} C_{t} (A) .$

Then $D^{con} (X, ⋀_{t \in T} C_{t}) \geq ⋀_{t \in T} D^{con} (X, C_{t})$ .

Proof. We first verify the following three inequalities.

(1) $D_{T} (X, ⋀_{t \in T} C_{t}) \geq ⋀_{t \in T} D_{T} (X, C_{t})$ . This can be shown by $\begin{matrix} D_{T} (X, ⋀_{t \in T} C_{t}) \\ = & (⋀_{t \in T} C_{t}) ({\underline{⊥}}_{X}) \land (⋀_{t \in T} C_{t}) ({\underline{⊤}}_{X}) \\ = & ⋀_{t \in T} (C_{t} ({\underline{⊥}}_{X}) \land C_{t} (\underline{⊤})_{X}) = ⋀_{t \in T} D_{T} (X, C_{t}) . \end{matrix}$

(2) $D_{\land} (X, ⋀_{t \in T} C_{t}) \geq ⋀_{t \in T} D_{\land} (X, C_{t})$ . This can be shown by $\begin{matrix} D_{\land} (X, ⋀_{t \in T} C_{t}) \\ = & ⋀_{{A_{k} : k \in K} \subseteq L^{X}} (⋀_{k \in K} (⋀_{t \in T} C_{t}) (A_{k}) \to (⋀_{t \in T} C_{t}) (⋀_{k \in K} A_{k})) \\ = & ⋀_{{A_{k} : k \in K} \subseteq L^{X}} (⋀_{t \in T} ⋀_{k \in K} C_{t} (A_{k}) \to ⋀_{t \in T} C_{t} (⋀_{k \in K} A_{k})) \\ \geq & ⋀_{{A_{k} : k \in K} \subseteq L^{X}} ⋀_{t \in T} (⋀_{k \in K} C_{t} (A_{k}) \to C_{t} (⋀_{k \in K} A_{k})) \end{matrix}$

$\begin{matrix} \geq & ⋀_{{A_{k} : k \in K} \subseteq L^{X}} ⋀_{t \in T} D_{\land} (X, C_{t}) \\ = & ⋀_{t \in T} D_{\land} (X, C_{t}) . \end{matrix}$

(3) The proof of $D_{\lor^{d}} (X, ⋀_{t \in T} C_{t}) \geq ⋀_{t \in T} D_{\lor^{d}} (X, C_{t})$ is similar to (2).

Therefore, $\begin{matrix} D^{con} (X, ⋀_{t \in T} C_{t}) \\ = & D_{T} (X, ⋀_{t \in T} C_{t}) \land D_{\land} (X, ⋀_{t \in T} C_{t}) \land D_{\lor^{d}} (X, ⋀_{t \in T} C_{t}) \\ \geq & ⋀_{t \in T} D_{T} (X, C_{t}) \land ⋀_{t \in T} D_{\land} (X, C_{t}) \land ⋀_{t \in T} D_{\lor^{d}} (X, C_{t}) \\ = & ⋀_{t \in T} D^{con} (X, C_{t}) . □ \end{matrix}$

In the above proposition, if $⋀_{t \in T} D^{con} (X, C_{t}) = ⊤_{M}$ , then $D^{con} (X, ⋀_{t \in T} C_{t}) = ⊤_{M}$ . This implies that if $D^{con} (X, C_{t}) = ⊤_{M}$ for each t ∈ T, then $D^{con} (X, ⋀_{t \in T} C_{t}) = ⊤_{M}$ . It is exactly the many-valued extension of the following conclusion with respect to (L, M)-fuzzy convex structures: if ${C_{t} : t \in T}$ is a family of (L, M)-fuzzy convexities on X, so is $⋀_{t \in T} C_{t}$ .

In [33], Shi and Xiu provided a method of constructing a new (L, M)-fuzzy convex structure in the following way.

Proposition 3.3. Let $(Y, D)$ be an (L, M)-fuzzy convex structure and f : X ⟶ Y be a surjective function. Define a mapping $f^{- 1} (D) : L^{X} ⟶ M$ by $\forall A \in L^{X}, f^{- 1} (D) (A) = ⋁ {D (B) : f_{L}^{\leftarrow} (B) = A} .$

Then $(X, f^{- 1} (D))$ is an (L, M)-fuzzy convex structure.

Now let us give a degree description of this result.

Proposition 3.4. Let f : X ⟶ Y be a surjective function and $D \in P_{M} (L^{Y})$ . Define $f^{- 1} (D) : L^{X} ⟶ M$ by $\forall A \in L^{X}, f^{- 1} (D) (A) = ⋁ {D (B) : f_{L}^{\leftarrow} (B) = A} .$

Then $D^{con} (X, f^{- 1} (D)) \geq D^{con} (Y, D)$ .

Proof. We first verify the following three inequalities.

(1) $D_{T} (X, f^{- 1} (D)) \geq D_{T} (Y, D)$ . This can be shown by

$\begin{matrix} D_{T} (X, f^{- 1} (D)) \\ = & f^{- 1} (D) ({\underline{⊤}}_{X}) \land f^{- 1} (D) ({\underline{⊥}}_{X}) \\ = & ⋁_{f_{L}^{\leftarrow} (B) = {\underline{⊤}}_{X}} D (B) \land ⋁_{f_{L}^{\leftarrow} (B) = {\underline{⊥}}_{X}} D (B) \\ \geq & D ({\underline{⊤}}_{Y}) \land D ({\underline{⊥}}_{Y}) = D_{T} (Y, D) . \end{matrix}$

(2) $D_{\land} (X, f^{- 1} (D)) \geq D_{\land} (Y, D)$ . That is, $\begin{matrix} ⋀_{{A_{k} : k \in K} \subseteq L^{X}} (⋀_{k \in K} f^{- 1} (D) (A_{k}) \\ \to f^{- 1} (D) (⋀_{k \in K} A_{k})) \\ \geq & ⋀_{{B_{k} : k \in K} \subseteq L^{Y}} (⋀_{k \in K} D (B_{k}) \to D (⋀_{k \in K} B_{k})) . \end{matrix}$

Take any α ∈ M such that $\begin{matrix} α \leq ⋀_{{B_{k} : k \in K} \subseteq L^{Y}} (⋀_{k \in K} D (B_{k}) \to D (⋀_{k \in K} B_{k})) . \end{matrix}$

Then for each {B_k : k ∈ K} ⊆ L^Y, it follows that $\begin{matrix} α \land ⋀_{k \in K} D (B_{k}) \leq D (⋀_{k \in K} B_{k}) . \end{matrix}$

In order to show $\begin{matrix} α \leq ⋀_{{A_{k} : k \in K} \subseteq L^{X}} (⋀_{k \in K} f^{- 1} (D) (A_{k}) \to \\ f^{- 1} (D) (⋀_{k \in K} A_{k})), \end{matrix}$ we need only show that for each {A_k : k ∈ K} ⊆ L^X, it follows that

$α \land ⋀_{k \in K} f^{- 1} (D) (A_{k}) \leq f^{- 1} (D) (⋀_{k \in K} A_{k}),$

i.e., $\begin{matrix} α \land ⋀_{k \in K} ⋁_{f_{L}^{\leftarrow} (B) = A_{k}} D (B) \leq ⋁_{f_{L}^{\leftarrow} (B) = ⋀_{k \in K} A_{k}} D (B) . \end{matrix}$ Take each β ∈ J (M) such that $β ≺ α \land ⋀_{k \in K} ⋁_{f_{L}^{\leftarrow} (B) = A_{k}} D (B) .$

Then β ≤ α and for each k ∈ K, there exists C_k ∈ L^Y such that $f_{L}^{\leftarrow} (C_{k}) = A_{k}$ and $D (C_{k}) \geq β$ . This implies $⋀_{k \in K} D (C_{k}) \geq β$ . Put B_k = C_k for each k ∈ K and C = ⋀ _k∈KB_k. Then $f_{L}^{\leftarrow} (C) = f_{L}^{\leftarrow} (⋀_{k \in K} B_{k}) = ⋀_{k \in K} f_{L}^{\leftarrow} (B_{k}) = ⋀_{k \in K} A_{k} .$

Furthermore, it follows that $\begin{matrix} ⋁_{f_{L}^{\leftarrow} (B) = ⋀_{k \in K} A_{k}} D (B) & \geq & D (C) = D (⋀_{k \in K} B_{k}) \\ \geq & ⋀_{k \in K} D (B_{k}) ⋀ α \geq β . \end{matrix}$

By the arbitrariness of β, we obtain that for each {A_k : k ∈ K} ⊆ L^X, $α \land ⋀_{k \in K} f^{- 1} (D) (A_{k}) \leq f^{- 1} (D) (⋀_{k \in K} A_{k}) .$ This means that $\begin{matrix} α \leq ⋀_{{A_{k} : k \in K} \subseteq L^{X}} (⋀_{k \in K} f^{- 1} (D) (A_{k} \\ \to f^{- 1} (D) (⋀_{k \in K} A_{k})) . \end{matrix}$

By the arbitrariness of α, we obtain that $\begin{matrix} ⋀_{{A_{k} : k \in K} \subseteq L^{X}} (⋀_{k \in K} f^{- 1} (D) (A_{k}) \\ \to f^{- 1} (D) (⋀_{k \in K} A_{k})) \\ \geq & ⋀_{{B_{k} : k \in K} \subseteq L^{Y}} (⋀_{k \in K} D (B_{k}) \to D (⋀_{k \in K} B_{k})), \end{matrix}$ as desired.

(3) The proof of $D_{\lor^{d}} (X, f^{- 1} (D)) \geq D_{\lor^{d}} (Y, D)$ is similar to (2).

As a result, we get $\begin{matrix} D^{con} (X, f^{- 1} (D)) \\ = & D_{T} (X, f^{- 1} (D)) \land D_{\land} (X, f^{- 1} (D)) \\ \land D_{\lor^{d}} (X, f^{- 1} (D)) \\ \geq & D_{T} (Y, D) \land D_{\land} (Y, D) \land D_{\lor^{d}} (Y, D) \\ = & D^{con} (Y, D) . \end{matrix}$

□

Subspaces and quotient spaces are two kinds of important concepts in (L, M)-fuzzy convex spaces. Next we will give the corresponding degree with respect to these two concepts. For this, we first present their definitions, respectively.

Definition 3.5. [33] Let $(X, C)$ be an (L, M)-fuzzy convex space, ∅ ≠ Y ⊆ X, and $B | Y = B \land {\underline{⊤}}_{Y}$ . Then $(Y, C |_{Y})$ is an (L, M)-fuzzy convex space on Y, where ∀A ∈ L^Y, $(C |_{Y}) (A) = ⋁ {C (B) : B \in L^{X}, B | Y = A} .$

We call $(Y, C |_{Y})$ a subspace of $(X, C)$ .

Definition 3.6.[33] Let $(X, C)$ be an (L, M)-fuzzy convex space and f : X ⟶ Y be a surjective function. Define a mapping $D_{f} : L^{Y} ⟶ M$ by $\forall B \in L^{Y}, D_{f} (B) = C (f_{L}^{\leftarrow} (B)) .$

Then $(Y, D_{f})$ is an (L, M)-fuzzy convex space and we call $D_{f}$ a quotient (L, M)-fuzzy convex structure on X with respect to f and $C$ .

Now let us equip with these concepts with some degrees, respectively.

Proposition 3.7. Let $C \in P_{M} (L^{X})$ and ∅ ≠ Y ⊆ X. Define $(C |_{Y})$ by ∀A ∈ L^Y, $(C |_{Y}) (A) = ⋁ {C (B) : B \in L^{X}, B | Y = A} .$ Then $D^{con} (Y, C |_{Y}) \geq D^{con} (X, C)$ .

Proof. It is enough to prove the following three inequalities.

(1) $D_{T} (Y, C |_{Y}) \geq D_{T} (X, C)$ . By the definition of $C |_{Y}$ , it follows that $\begin{matrix} D_{T} (Y, C |_{Y}) \\ = & C |_{Y} ({\underline{⊤}}_{Y}) \land C |_{Y} ({\underline{⊥}}_{Y}) \\ = & ⋁_{B \in L^{X}, B | Y = {\underline{⊤}}_{Y}} C (B) \land ⋁_{B \in L^{X}, B | Y = {\underline{⊥}}_{Y}} C (B) \\ \geq & C ({\underline{⊤}}_{X}) \land C ({\underline{⊥}}_{X}) \\ = & D_{T} (X, C) . \end{matrix}$

(2) $D_{\land} (Y, C |_{Y}) \geq D_{T} (X, C)$ . That is, $\begin{matrix} ⋀_{{A_{k} : k \in K} \subseteq L^{Y}} (⋀_{k \in K} C |_{Y} (A_{k}) \to C |_{Y} (⋀_{k \in K} A_{k})) \\ \geq & ⋀_{{B_{k} : k \in K} \subseteq L^{X}} (⋀_{k \in K} C (B_{k}) \to C (⋀_{k \in K} B_{k})) . \end{matrix}$

Take any α ∈ M such that $\begin{matrix} α \leq ⋀_{{B_{k} : k \in K} \subseteq L^{X}} (⋀_{k \in K} C (B_{k}) \to C (⋀_{k \in K} B_{k})) . \end{matrix}$ Then for each {B_k : k ∈ K} ⊆ L^X, it follows that $α \land ⋀_{k \in K} C (B_{k}) \leq C (⋀_{k \in K} B_{k})$ . In order to show $α \leq ⋀_{{A_{k} : k \in K} \subseteq L^{Y}} (⋀_{k \in K} C |_{Y} (A_{k}) \to C |_{Y} (⋀_{k \in K} A_{k}))$ , we need only show that for each {A_k : k ∈ K} ⊆ L^Y, it follows that $α \land ⋀_{k \in K} C |_{Y} (A_{k}) \leq C |_{Y} (⋀_{k \in K} A_{k}),$

i.e., $\begin{matrix} α \land ⋀_{k \in K} ⋁_{B \in L^{X}, B | Y = A_{k}} C (B) \\ \leq & ⋁_{B \in L^{X}, B | Y = ⋀_{k \in K} A_{k}} C (B) . \end{matrix}$

Take each β ∈ J (M) such that $β ≺ α \land ⋀_{k \in K} ⋁_{B \in L^{X}, B | Y = A_{k}} C (B) .$

Then β ≤ α and for each k ∈ K, there exists C_k ∈ L^X such that C_k|Y = A_k and $C (C_{k}) \geq β$ . This implies $⋀_{k \in K} C (C_{k}) \geq β$ . Put B_k = C_k for each k ∈ K and C = ⋀ _k∈KB_k. Then $C | Y = (⋀_{k \in K} B_{k}) | Y = ⋀_{k \in K} (B_{k} | Y) = ⋀_{k \in K} A_{k} .$ Furthermore, it follows that $\begin{matrix} ⋁_{B \in L^{X}, B | Y = ⋀_{k \in K} A_{k}} C (B) & \geq & C (C) = C (⋀_{k \in K} B_{k}) \\ \geq & α \land ⋀_{k \in K} C (B_{k}) \\ \geq & β . \end{matrix}$

By the arbitrariness of β, we obtain that for each {A_k : k ∈ K} ⊆ L^X, $α \land ⋀_{k \in K} C |_{Y} (A_{k}) \leq C |_{Y} (⋀_{k \in K} A_{k}) .$

This means that $α \leq ⋀_{{A_{k} : k \in K} \subseteq L^{X}} (⋀_{k \in K} C |_{Y} (A_{k}) \to C |_{Y} (⋀_{k \in K} A_{k})) .$

By the arbitrariness of α, we obtain that $\begin{matrix} ⋀_{{A_{k} : k \in K} \subseteq L^{Y}} (⋀_{k \in K} C |_{Y} (A_{k}) \to C |_{Y} (⋀_{k \in K} A_{k})) \\ \geq & ⋀_{{B_{k} : k \in K} \subseteq L^{X}} (⋀_{k \in K} C (B_{k}) \to C (⋀_{k \in K} B_{k})), \end{matrix}$ as desired.

(3) The proof of $D_{\lor^{d}} (Y, C |_{Y}) \geq D_{\lor^{d}} (X, C)$ is similar to (2).

As a result, we get $\begin{matrix} D^{con} (Y, C |_{Y}) \\ = & D_{T} (Y, C |_{Y}) \land D_{\land} (Y, C |_{Y}) \land D_{\lor^{d}} (Y, C |_{Y}) \\ \geq & D_{T} (X, C) \land D_{\land} (X, C) \land D_{\lor^{d}} (X, C) \\ = & D^{con} (X, C) . \end{matrix}$

□

Proposition 3.8. Let $C_{X} \in P_{M} (L^{X})$ and f : X ⟶ Y be a surjective function. Define $C_{Y} : L^{Y} ⟶ M$ by $\forall B \in L^{Y}, C_{Y} (B) = C_{X} (f_{L}^{\leftarrow} (B)) .$

Then $D^{con} (Y, C_{Y}) \geq D^{con} (X, C_{X})$ .

Proof. It is enough to prove the following three inequalities.

(1) $D_{T} (Y, C_{Y}) \geq D_{T} (X, C_{X})$ . By the definition of $C_{Y}$ , it follows that $\begin{matrix} D_{T} (Y, C_{Y}) \\ = & C_{Y} ({\underline{⊤}}_{Y}) \land C_{Y} ({\underline{⊥}}_{Y}) \\ = & C_{X} (f_{L}^{\leftarrow} ({\underline{⊤}}_{Y})) \land C_{X} (f_{L}^{\leftarrow} ({\underline{⊥}}_{Y})) \\ \geq & C_{X} (X {\underline{⊤}}_{X}) \land C_{X} ({\underline{⊥}}_{X}) \\ = & D_{T} (X, C_{X}) . \end{matrix}$

(2) $D_{\land} (Y, C_{Y}) \geq D_{T} (X, C_{X})$ . It can be checked as follows: $\begin{matrix} D_{\land} (Y, C_{Y}) \\ = & ⋀_{{B_{k} : k \in K} \subseteq L^{Y}} (⋀_{k \in K} C_{Y} (B_{k}) \to C_{Y} (⋀_{k \in K} B_{k})) \\ = & ⋀_{{B_{k} : k \in K} \subseteq L^{Y}} (⋀_{k \in K} C_{X} (f_{L}^{\leftarrow} (B_{k})) \to \\ C_{X} (f_{L}^{\leftarrow} (⋀_{k \in K} B_{k})) \\ = & ⋀_{{B_{k} : k \in K} \subseteq L^{Y}} (⋀_{k \in K} C_{X} (f_{L}^{\leftarrow} (B_{k})) \to \\ C_{X} (⋀_{k \in K} f_{L}^{\leftarrow} (B_{k})) \\ \geq & ⋀_{{A_{k} : k \in K} \subseteq L^{X}} (⋀_{k \in K} C_{X} (A_{k}) \to C_{X} (⋀_{k \in K} A_{k})) \\ = & D_{\land} (X, C_{X}) . \end{matrix}$

(3) The proof of $D_{\lor^{d}} (Y, C_{Y}) \geq D_{\lor^{d}} (X, C_{X})$ is similar to (2).

As a result, we get $\begin{matrix} D^{con} (Y, C_{Y}) \\ = & D_{T} (Y, C_{Y}) \land D_{\land} (Y, C_{Y}) \land D_{\lor^{d}} (Y, C_{Y}) \\ \geq & D_{T} (X, C_{X}) \land D_{\land} (X, C_{X}) \land D_{\lor^{d}} (X, C_{X}) \\ = & D^{con} (X, C_{X}) . \end{matrix}$

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4. The degree of (L, M)-fuzzy closure systems and (L, M)-fuzzy Alexandrov topologies

In this section, we will apply the degree approach to (L, M)-fuzzy closure systems and (L, M)-fuzzy Alexandrov topologies. Then we will study its connection with the degree of (L, M)-fuzzy convex structures.

Adopting the notations D_T and D_∧. We first give the following definition.

Definition 4.1. For each $C \in P_{M} (L^{X})$ , define $D^{clo} (X, C)$ as follows: $D^{clo} (X, C) = D_{T} (X, C) \land D_{\land} (X, C) .$

Then $D^{clo} (X, C)$ is called the degree to which $C$ is an (L, M)-fuzzy closure system on X. Obviously, $C$ is an (L, M)-fuzzy closure system on X if and only if $D^{clo} (X, C) = ⊤_{M}$ .

If $D^{clo} (X, C) = ⊤_{M}$ , then $D_{T} (X, C) = ⊤_{M}$ and $D_{\land} (X, C) = ⊤_{M}$ . This implies (LMS1) and (LMS2). Conversely, For each $C \in P_{M} (L^{X})$ , if it satisfies (LMS1) and (LMS2), then $D^{clo} (X, C) = ⊤_{M}$ .

Proposition 4.2. For each $C \in P_{M} (L^{X})$ , $C$ is an (L, M)-fuzzy closure system on X if and only if $D^{clo} (X, C) = ⊤_{M}$ .

Proposition 4.3. For each $C \in P_{M} (L^{X})$ , $\begin{matrix} D^{con} (X, C) \leq D^{clo} (X, C) . \end{matrix}$

Actually, there are close relations between (L, M)-fuzzy closure systems and (L, M)-fuzzy convex structures. In [40], Pang and Xiu provided a transforming method from (L, M)-fuzzy closure systems to (L, M)-fuzzy convex structures in the following way.

Proposition 4.4. Let $(X, C)$ be an (L, M)-fuzzy closure space. Define a mapping $C^{*} : L^{X} ⟶ M$ by $\forall A \in L^{X}, C^{*} (A) = ⋁_{⋁_{λ \in Λ}^{dir} B_{λ} = A} ⋀_{λ \in Λ} C (B_{λ}) .$

Then $C^{*}$ is an (L, M)-fuzzy convex structure on X.

Now let’s give a degree description for this.

Proposition 4.5. Let’s $C \in P_{M} (L^{X})$ and define $C^{*} : L^{X} ⟶ M$ by $\forall A \in L^{X}, C^{*} (A) = ⋁_{⋁_{λ \in Λ}^{dir} B_{λ} = A} ⋀_{λ \in Λ} C (B_{λ}) .$

Then $D^{con} (X, C^{*}) \geq D^{clo} (X, C)$

Proof. We first verify three inequalities in the following.

(1) $D_{T} (X, C^{*}) \geq D_{T} (X, C)$ . By the definition of $C^{*}$ , we have $\begin{matrix} D_{T} (X, C^{*}) \\ = & C^{*} ({\underline{⊤}}_{X}) \land C^{*} ({\underline{⊥}}_{X}) \\ = & ⋁_{⋁_{λ \in Λ}^{dir} B_{λ} = {\underline{⊤}}_{X}} ⋀_{λ \in Λ} C (B_{λ}) \land ⋁_{⋁_{λ \in Λ}^{dir} B_{λ} = {\underline{⊥}}_{X}} ⋀_{λ \in Λ} C (B_{λ}) \\ \geq & C ({\underline{⊤}}_{X}) \land C ({\underline{⊥}}_{X}) \\ = & D_{T} (X, C) . \end{matrix}$

(2) $D_{\land} (X, C^{*}) \geq D_{\land} (X, C)$ . It suffices to show the following inequality. $\begin{matrix} ⋀_{{A_{k} : k \in K} \subseteq L^{X}} (⋀_{k \in K} ⋁_{⋁_{j \in J_{k}}^{dir} B_{k, j} = A_{k}} ⋀_{j \in J_{k}} C (B_{k, j}) \to \\ ⋁_{⋁_{t \in T}^{dir} B_{t} = ⋀_{k \in K} A_{k}} ⋀_{t \in T} C (B_{t})) \\ \geq & ⋀_{{B_{k} : k \in K} \subseteq L^{X}} (⋀_{k \in K} C (B_{k}) \to C (⋀_{k \in K} B_{k})) . \end{matrix}$

Take any α ∈ M such that $α \leq ⋀_{{B_{k} : k \in K} \subseteq L^{X}} (⋀_{k \in K} C (B_{k}) \to C (⋀_{k \in K} B_{k})) .$

Then for each {B_k : k ∈ K} ⊆ L^X, it follows that $α \land ⋀_{k \in K} C (B_{k}) \leq C (⋀_{k \in K} B_{k})$ . Now we need only show that for each {A_k : k ∈ K} ⊆ L^X, $\begin{matrix} α \land ⋀_{k \in K} ⋁_{⋁_{j \in J_{k}}^{dir} B_{k, j} = A_{k}} ⋀_{j \in J_{k}} C (B_{k, j}) \\ \leq & ⋁_{⋁_{t \in T}^{dir} B_{t} = ⋀_{k \in K} A_{k}} ⋀_{t \in T} C (B_{t}) . \end{matrix}$

Take each β ∈ J (M) such that $\begin{matrix} β ≺ α \land ⋀_{k \in K} ⋁_{⋁_{j \in J_{k}}^{dir} B_{k, j} = A_{k}} ⋀_{j \in J_{k}} C (B_{k, j}) . \end{matrix}$

Then β ≤ α and for each k ∈ K, there exists an up-directed set {B_k,j : j ∈ J_k} such that $⋁_{j \in J_{k}}^{dir} B_{k, j} = A_{k}$ and for each j ∈ J_k, $C (B_{k, j}) \geq β$ . By DD, it follows that $⋀_{k \in K} A_{k} = ⋀_{k \in K} ⋁_{j \in J_{k}}^{dir} B_{k, j} = ⋁_{f \in \prod_{k \in K} J_{k}} ⋀_{k \in K} B_{k, f (k)}$ . Put C_f = ⋀ _k∈KB_k,f(k) for each $f \in \prod_{k \in K} J_{k}$ . Since {B_k,j : j ∈ J_k} is up-directed, it is trivial to verify that ${C_{f} : f \in \prod_{k \in K} J_{k}}$ is up-directed. Then for each $f \in \prod_{k \in K} J_{k}$ , it follow that

$C (C_{f}) = C (⋀_{k \in K} B_{k, f (k)}) \geq α \land ⋀_{k \in K} C (B_{k, f (k)}) \geq β .$

This implies $⋀_{f \in \prod_{k \in K} J_{k}} C (C_{f}) \geq β$ . Since ${C_{f} : f \in \prod_{k \in K} J_{k}}$ is up-directed and $⋀_{k \in K} A_{k} = ⋁_{f \in \prod_{k \in K} J_{k}}^{dir} C_{f}$ , it follows that

$⋁_{⋁_{t \in T}^{dir} B_{t} = ⋀_{k \in K} A_{k}} ⋀_{t \in T} C (B_{t}) \geq ⋀_{f \in \prod_{k \in K} J_{k}} C (C_{f}) \geq β .$ By the arbitrariness of β, we obtain that for each {A_k : k ∈ K} ⊆ L^X,

$\begin{matrix} α \land ⋀_{k \in K} ⋁_{⋁_{j \in J_{k}}^{dir} B_{k, j} = A_{k}} ⋀_{j \in J_{k}} C (B_{k, j}) \\ \leq & ⋁_{⋁_{t \in T}^{dir} B_{t} = ⋀_{k \in K} A_{k}} ⋀_{t \in T} C (B_{t}) . \end{matrix}$ This means that

$α \leq ⋀_{{A_{k} : k \in K} \subseteq L^{X}} (⋀_{k \in K} C^{*} (A_{k}) \to C^{*} (⋀_{k \in K} A_{k})) .$

By the arbitrariness of α, we obtain that

$\begin{matrix} ⋀_{{A_{k} : k \in K} \subseteq L^{X}} (⋀_{k \in K} C^{*} (A_{k}) \to C^{*} (⋀_{k \in K} A_{k})) \\ \geq & ⋀_{{B_{k} : k \in K} \subseteq L^{X}} (⋀_{k \in K} C (B_{k}) \to C (⋀_{k \in K} B_{k})), \end{matrix}$ as desired.

(3) $D_{\lor^{d}} (X, C^{*}) \geq D_{\land} (X, C)$ . It suffices to show the following inequality.

$\begin{matrix} ⋀_{{A_{k} : k \in K} \overset{dir}{\subseteq} L^{X}} (⋀_{k \in K} ⋁_{⋁_{j \in J_{k}}^{dir} B_{k, j} = A_{k}} ⋀_{j \in J_{k}} C (B_{k, j}) \to \\ ⋁_{⋁_{t \in T}^{dir} B_{t} = ⋁_{k \in K}^{dir} A_{k}} ⋀_{t \in T} C (B_{t})) \\ \geq & ⋀_{{B_{k} : k \in K} \subseteq L^{X}} (⋀_{k \in K} C (B_{k}) \to C (⋀_{k \in K} B_{k})) . \end{matrix}$

Take any α ∈ M such that

$\begin{matrix} α \leq ⋀_{{B_{k} : k \in K} \subseteq L^{X}} (⋀_{k \in K} C (B_{k}) \to C (⋀_{k \in K} B_{k})) . \end{matrix}$

Then for each {B_k : k ∈ K} ⊆ L^X, it follows that $α \land ⋀_{k \in K} C (B_{k}) \leq C (⋀_{k \in K} B_{k})$ . Now we need only show that for each ${A_{k} : k \in K} \overset{dir}{\subseteq} L^{X}$ , $\begin{matrix} α \land ⋀_{k \in K} ⋁_{⋁_{j \in J_{k}}^{dir} B_{k, j} = A_{k}} ⋀_{j \in J_{k}} C (B_{k, j}) \\ \leq & ⋁_{⋁_{t \in T}^{dir} B_{t} = ⋁_{k \in K}^{dir} A_{k}} ⋀_{t \in T} C (B_{t}) . \end{matrix}$

Take each β ∈ J (M) such that $β ≺ α \land ⋀_{k \in K} ⋁_{⋁_{j \in J_{k}}^{dir} B_{k, j} = A_{k}} ⋀_{j \in J_{k}} C (B_{k, j}) .$

Then β ≤ α and for each k ∈ K, there exists an up-directed set {B_k,j : j ∈ J_k} such that $⋁_{j \in J_{k}}^{dir} B_{k, j} = A_{k}$ and for each j ∈ J_k, $C (B_{k, j}) \geq β$ . Let $A = ⋁_{k \in K}^{dir} A_{k} = ⋁_{k \in K}^{dir} ⋁_{j \in J_{k}}^{dir} B_{k, j}$ . Define a mapping $σ : L_{fin}^{A} ⟶ L^{X}$ as follows: $\forall F \in L_{fin}^{A}, σ (F) = ⋀ {B_{k, j} ∣ F \leq B_{k, j}} .$

By Theorem 2.6, we know that σ is definable, the set ${σ (F) | F \in L_{fin}^{A}}$ is up-directed and $A = ⋁_{F \in L_{fin}^{A}} σ (F)$ . Then $A = ⋁_{F \in L_{fin}^{A}} σ (F) = ⋁_{F \in L_{fin}^{A}} ⋀_{F \leq B_{k, j}} B_{k, j} .$

Now for each $F \in L_{fin}^{A}$ , Put ${B_{t} | t \in T} = {B_{k, j} | F \leq B_{k, j}} .$

Then $⋀_{t \in T} C (B_{t}) = ⋀_{F \leq B_{k, j}} C (B_{k, j}) \geq β .$

This implies $C (σ (F)) = C (⋀_{F \leq B_{k, j}} B_{k, j}) = ⋀_{t \in T} C (B_{t}) ⋀ α \geq β .$

By the arbitrariness of F, we obtain $⋀_{F \in L_{fin}^{A}} C (σ (F)) \geq β$ . Since $A = ⋁_{k \in K}^{dir} A_{k} = ⋁_{F \in L_{fin}^{A}}^{dir} σ (F)$ , it follows that $⋁_{⋁_{t \in T}^{dir} B_{t} = ⋀_{k \in K} A_{k}} ⋀_{t \in T} C (B_{t}) \geq ⋀_{F \in L_{fin}^{A}} C (σ (F)) \geq β .$

By the arbitrariness of β, we obtain that for each ${A_{k} : k \in K} \overset{dir}{\subseteq} L^{X}$ , $\begin{matrix} α \land ⋀_{k \in K} ⋁_{⋁_{j \in J_{k}}^{dir} B_{k, j} = A_{k}} ⋀_{j \in J_{k}} C (B_{k, j}) \\ \leq & ⋁_{⋁_{t \in T}^{dir} B_{t} = ⋁_{k \in K}^{dir} A_{k}} ⋀_{t \in T} C (B_{t}) . \end{matrix}$

This means that $\begin{matrix} α \leq ⋀_{{A_{k} : k \in K} \overset{dir}{\subseteq} L^{X}} (⋀_{k \in K} C^{*} (A_{k}) \to C^{*} (⋁_{k \in K}^{dir} A_{k})) . \end{matrix}$

By the arbitrariness of α, we obtain that $\begin{matrix} ⋀_{{A_{k} : k \in K} \overset{dir}{\subseteq} L^{X}} (⋀_{k \in K} C^{*} (A_{k}) \to C^{*} (⋁_{k \in K}^{dir} A_{k})) \\ \geq & ⋀_{{B_{k} : k \in K} \subseteq L^{X}} (⋀_{k \in K} C (B_{k}) \to C (⋀_{k \in K} B_{k})), \end{matrix}$ as desired.

By (1), (2) and (3), we have $\begin{matrix} D^{con} (X, C^{*}) \\ = & D_{T} (X, C^{*}) \land D_{\land} (X, C^{*}) \land D_{\lor^{d}} (X, C^{*}) \\ \geq & D_{T} (X, C) \land D_{\land} (X, C) \land D_{\land} (X, C) \\ = & D^{clo} (X, C) . \end{matrix}$

□

In order to introduce the degree of (L, M)-fuzzy Alexandrov topologies, we first give the following notation. For each $C \in P_{M} (L^{X})$ , define $D_{\lor} (X, C) = ⋀_{{A_{k} : k \in K} \subseteq L^{X}} (⋀_{k \in K} C (A_{k}) \to C (⋁_{k \in K} A_{k})) .$

Actually, this definition offers a degree description to which $C$ preserve arbitrary unions.

By using this, let’s equip each $C \in P_{M} (L^{X})$ with some degree to which $C$ becomes an (L, M)-fuzzy Alexandrov topology.

Definition 4.6. For $C \in P_{M} (L^{X})$ , define $D^{top} (X, C)$ as follows: $D^{top} (X, C) = D_{T} (X, C) \land D_{\land} (X, C) \land D_{\lor} (X, C) .$

Then $D^{top} (X, C)$ is called the degree to which $C$ becomes an (L, M)-fuzzy Alexandrov topology on X.

Remark 4.7. For $C \in P_{M} (L^{X})$ , $C$ is an (L, M)-fuzzy Alexandrov topology if and only if $D^{top} (X, C) = ⊤_{M}$ .

Proposition 4.8. Let’s $C \in P_{M} (L^{X})$ . Then $D^{top} (X, C) \leq D^{con} (X, C) .$

Proposition 4.9. Suppose that L is a completely distributive lattice. Let’s $C \in P_{M} (L^{X})$ and define $\bar{C} : L^{X} ⟶ M$ by $\begin{matrix} \forall A \in L^{X}, \bar{C} (A) = ⋁_{⋁_{j \in J} B_{j} = A} ⋀_{j \in J} C (B_{j}) . \end{matrix}$

Then $D^{top} (X, \bar{C}) \geq D^{con} (X, C)$ .

Proof. We prove it in the following steps.

(1) $\begin{matrix} D_{T} (X, \bar{C}) \\ = & \bar{C} ({\underline{⊤}}_{X}) \land \bar{C} ({\underline{⊥}}_{X}) \\ = & ⋁_{⋁_{j \in J} B_{j} = {\underline{⊤}}_{X}} ⋀_{j \in J} C (B_{j}) \land ⋁_{⋁_{j \in J} B_{j} = {\underline{⊥}}_{X}} ⋀_{j \in J} C (B_{j}) \\ \geq & C ({\underline{⊤}}_{X}) \land C ({\underline{⊥}}_{X}) \\ = & D_{T} (X, C) . \end{matrix}$

(2) $D_{\land} (X, \bar{C}) \geq D_{\land} (X, C)$ . That is, $\begin{matrix} ⋀_{{A_{k} : k \in K} \subseteq L^{X}} (⋀_{k \in K} \bar{C} (A_{k}) \to \bar{C} (⋀_{k \in K} A_{k})) \\ \geq & ⋀_{{B_{k} : k \in K} \subseteq L^{X}} (⋀_{k \in K} C (B_{k}) \to C (⋀_{k \in K} B_{k})) . \end{matrix}$

By the definition of $\bar{C}$ , it suffices to show $\begin{matrix} ⋀_{{A_{k} : k \in K} \subseteq L^{X}} (⋀_{k \in K} ⋁_{⋁_{j \in J_{k}} B_{k, j} = A_{k}} ⋀_{j \in J_{k}} C (B_{k, j}) \\ \to ⋁_{⋁_{j \in J} B_{j} = ⋀_{k \in K} A_{k}} ⋀_{j \in J} C (B_{j})) \\ \geq & ⋀_{{B_{k} : k \in K} \subseteq L^{X}} (⋀_{k \in K} C (B_{k}) \to C (⋀_{k \in K} B_{k})) . \end{matrix}$

Take any α ∈ M such that $α \leq ⋀_{{B_{k} : k \in K} \subseteq L^{X}} (⋀_{k \in K} C (B_{k}) \to C (⋀_{k \in K} B_{k})) .$

Then for each {B_k : k ∈ K} ⊆ L^X, it follows that $α \land ⋀_{k \in K} C (B_{k}) \leq C (⋀_{k \in K} B_{k})$ . In order to show $α \leq ⋀_{{A_{k} : k \in K} \subseteq L^{X}} (⋀_{k \in K} \bar{C} (B_{k}) \to \bar{C} (⋀_{k \in K} B_{k}))$ , we need only show that for each {A_k : k ∈ K} ⊆ L^X, it follows that $α \land ⋀_{k \in K} \bar{C} (B_{k}) \leq \bar{C} (⋀_{k \in K} B_{k})$ , i.e., $\begin{matrix} α \land ⋀_{k \in K} ⋁_{⋁_{j \in J_{k}} B_{k, j} = A_{k}} ⋀_{j \in J_{k}} C (B_{k, j}) \\ \leq & ⋁_{⋁_{j \in J} B_{j} = ⋀_{k \in K} A_{k}} ⋀_{j \in J} C (B_{j})) . \end{matrix}$

Take each β ∈ J (M) such that $\begin{matrix} β ≺ α \land ⋀_{k \in K} ⋁_{⋁_{j \in J_{k}} B_{k, j} = A_{k}} ⋀_{j \in J_{k}} C (B_{k, j}) . \end{matrix}$

Then β ≤ α and for each k ∈ K, there exists a set {B_k,j : j ∈ J_k} such that ⋁_{j∈J_k}B_k,j = A_k and for each j ∈ J_k, $C (B_{k, j}) \geq β$ . By the completely distributive law, it follows that $\begin{matrix} ⋀_{k \in K} A_{k} = ⋀_{k \in K} ⋁_{j \in J_{k}} B_{k, j} = ⋁_{f \in \prod_{k \in K} J_{k}} ⋀_{k \in K} B_{k, f (k)} . \end{matrix}$

Put D_f = ⋀ _k∈KB_k,f(k) for each $f \in \prod_{k \in K} J_{k}$ . Then for each $f \in \prod_{k \in K} J_{k}$ , it follow that $\begin{matrix} C (D_{f}) = C (⋀_{k \in K} B_{k, f (k)}) \geq α \land ⋀_{k \in K} C (B_{k, f (k)}) \geq β . \end{matrix}$

This implies $⋀_{f \in \prod_{k \in K} J_{k}} C (D_{f}) \geq β$ . By the arbitrariness of β, we obtain that for each {A_k : k ∈ K} ⊆ L^X, $\begin{matrix} α \land ⋀_{k \in K} ⋁_{⋁_{j \in J_{k}} B_{k, j} = A_{k}} ⋀_{j \in J_{k}} C (B_{k, j}) \\ \leq & ⋁_{⋁_{j \in J} B_{j} = ⋀_{k \in K} A_{k}} ⋀_{j \in J} C (B_{j}) . \end{matrix}$

This means that $α \leq ⋀_{{A_{k} : k \in K} \subseteq L^{X}} (⋀_{k \in K} \bar{C} (A_{k}) \to \bar{C} (⋀_{k \in K} A_{k})) .$ By the arbitrariness of α, we obtain that $\begin{matrix} ⋀_{{A_{k} : k \in K} \subseteq L^{X}} (⋀_{k \in K} \bar{C} (A_{k}) \to \bar{C} (⋀_{k \in K} A_{k})) \\ \geq & ⋀_{{B_{k} : k \in K} \subseteq L^{X}} (⋀_{k \in K} C (B_{k}) \to C (⋀_{k \in K} B_{k})), \end{matrix}$ as desired.

(3) $D_{\lor} (X, \bar{C}) = ⊤_{M}$ . It suffices to show $\begin{matrix} ⋀_{k \in K} ⋁_{⋁_{j \in J_{k}} B_{k, j} = A_{k}} ⋀_{j \in J_{k}} C (B_{k, j}) \\ \leq & ⋁_{⋁_{j \in J} B_{j} = ⋁_{k \in K} A_{k}} ⋀_{j \in J} C (B_{j}) . \end{matrix}$ Take each α ∈ J (M) such that $α ≺ ⋀_{k \in K} ⋁_{⋁_{j \in J_{k}} B_{k, j} = A_{k}} ⋀_{j \in J_{k}} C (B_{k, j}) .$

Then for each k ∈ K, there exists a set {B_k,j : j ∈ J_k} such that ⋁_{j∈J_k}B_k,j = A_k and for each j ∈ J_k, $C (B_{k, j}) \geq α$ . Put {C_t : t ∈ T} = {B_k,j : k ∈ K, j ∈ J_k}. Then ⋁_k∈KA_k = ⋁ _k∈K ⋁ _{j∈J_k}B_k,j = ⋁ _t∈TC_t and $⋀_{t \in T} C (C_{t}) \geq ⋀_{k \in K} ⋀_{j \in J_{k}} C (B_{k, j}) \geq α$ . This implies that $⋁_{⋁_{j \in J} B_{j} = ⋁_{k \in K} A_{k}} ⋀_{j \in J} C (B_{j}) \geq ⋀_{t \in T} C (C_{t}) \geq α .$

By the arbitrariness of α, $\begin{matrix} ⋀_{k \in K} ⋁_{⋁_{j \in J_{k}} B_{k, j} = A_{k}} ⋀_{j \in J_{k}} C (B_{k, j}) \\ \leq & ⋁_{⋁_{j \in J} B_{j} = ⋁_{k \in K} A_{k}} ⋀_{j \in J} C (B_{j}), \end{matrix}$

i.e., $\begin{matrix} ⋀_{k \in K} ⋁_{⋁_{j \in J_{k}} B_{k, j} = A_{k}} ⋀_{j \in J_{k}} C (B_{k, j}) \\ \to & ⋁_{⋁_{j \in J} B_{j} = ⋁_{k \in K} A_{k}} ⋀_{j \in J} C (B_{j}) = ⊤_{M} . \end{matrix}$ This means $D_{\lor} (X, \bar{C}) = ⊤_{M}$ .

By (1), (2) and (3), we have $\begin{matrix} D^{top} (X, \bar{C}) & = & D_{T} (X, \bar{C}) \land D_{\land} (X, \bar{C}) \land D_{\lor} (X, \bar{C}) \\ \geq & D_{T} (X, C) \land D_{\land} (X, C) ⋀ D_{\lor^{d}} (X, C) \\ = & D^{con} (X, C) . \end{matrix}$

□

Corollary 4.10. Suppose that L is a completely distributive lattice. Let $(X, C)$ be an (L, M)-fuzzy convex space and define $\bar{C} : L^{X} ⟶ M$ by $\forall A \in L^{X}, \bar{C} (A) = ⋁_{⋁_{j \in J} B_{j} = A} ⋀_{j \in J} C (B_{j}) .$

Then $\bar{C}$ becomes an (L, M)-fuzzy Alexandrov topology on X.

5. Conclusions

In this paper, we mainly applied an degree approach to (L, M)-fuzzy convex structures, (L, M)-fuzzy closure systems and (L, M)-fuzzy Alexandrov topologies. In this way, we proposed the degrees of (L, M)-fuzzy convex structures, (L, M)-fuzzy closure systems and (L, M)-fuzzy Alexandrov topologies. From a logical viewpoint, we represented some properties of (L, M)-fuzzy convex structures and investigated the relations between the degree of (L, M)-fuzzy convex structures and the degree of (L, M)-fuzzy closure systems, and the relations between the degree of (L, M)-fuzzy convex structures and the degree of (L, M)-fuzzy Alexandrov topologies.

In the future, we will consider the degree of CP-mappings between (L, M)-fuzzy convex spaces and combine it with the degree of (L, M)-fuzzy convex structure. In other words, we will apply the degree method to study (L, M)-fuzzy convex spaces.

Footnotes

Acknowledgments

The authors would like to express their sincere thanks to the anonymous reviewers and the area editor for their careful reading and constructive comments. This work is supported by Projects (11871097, 11771134) funded by the Natural Science Foundation of China, the Project (2017M622563) funded by China Postdoctoral Science Foundation and Projects (KYTZ201631, CRF201611, 2017Z056) funded by the Scientific Research Foundation of CUIT.

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