Topologies induced by pointwise L -fuzzifying quasi-uniformities

Abstract

Based on a completely distributive De Morgan algebra L, the notion of pointwise L-fuzzifying (quasi-)uniformity is introduced. It is shown that a pointwise L-fuzzifying quasi-uniformity $U$ can induce two L-fuzzifying topologies $τ_{U}$ and $η_{U}$ , and the relations between $τ_{U}$ and $η_{U}$ are investigated.

1 Introduction

Uniformity is an important concept close to topology and a convenient tool for an investigation of topology. Since Chang introduced fuzzy set theory to topology, there have been many spectacular and creative works about the theory of uniformities in fuzzy topology [1 , 17]. In the study of fuzzy uniformities, L-(quasi)-uniformity in Hutton’s sense [3] has been accepted by many authors and has attracted wide attention in the literature [1 , 9]. However, Hutton’s fuzzy uniformities can’t directly reflect the characteristic of pointwise fuzzy topology. Motivated by this, Shi [10] introduced the theory of pointwise L-(quasi)-uniformity in L-topological space, which reflects the characteristic of pointwise fuzzy topology. Afterwards, Yue extended both Hutton’s uniformity and Shi’s uniformity in L-topological space to L-fuzzy setting [16, 17].

It is well known that a quasi-uniformity $U$ can induce two topologies $τ_{U}$ and $η_{U}$ , which are not coincident in general. $τ_{U}$ and $η_{U}$ are the same, whenever $U$ is a uniformity. In fuzzy setting, Hutton’s quasi-uniformity [6], Shi’s quasi-uniformity [11] and L-fuzzy quasi-uniformity in the sense of Hutton [18] all follow this rule. In [18], Yue constructed two topological L-fuzzy remote-neighborhoods from a pointwise L-fuzzy quasi-uniformity. However, the relations between these two topological L-fuzzy remote-neighborhoods are uncertain. That is to say, the relations between L-fuzzy topologies induced by a pointwise L-fuzzy quasi-uniformity are not clear. Inspired by this, we will consider the relations between L-fuzzifying topologies induced by a pointwise L-fuzzifying quasi-uniformity.

This paper is organized as follows. In Section 2, some necessary concepts and notions are listed. In Section 3, the characterizations of fuzzifying interior operators and fuzzifying closure operators in [14] are generalized to L-fuzzifying setting. In Section 4, it is shown that a pointwise L-fuzzifying quasi-uniformity $U$ can induce two L-fuzzifying topologies $τ_{U}$ and $η_{U}$ , and the relations between $τ_{U}$ and $η_{U}$ are discussed.

2 Preliminaries

Throughout this paper, (L, ⋁ , ⋀ , ′) denotes a completely distributive De Morgan algebra. The smallest element and the largest element in L are denoted by ⊥ and ⊤, respectively. The set of non-zero coprimes in L is denoted by M (L). For a, b ∈ L, we say “a is wedge below b”, in symbols a ≺ b, if for every subset D ⊆ L, ⋁ D ≥ b implies a ≤ d for some d ∈ D. We denote β (a) = {b ∈ L ∣ b ≺ a} for each a ∈ L. A complete lattice L is completely distributive if and only if a = ⋁ β (a) for each a ∈ L [12]. The wedge below relation in a completely distributive lattice has the interpolation property, i.e., if a ≺ b, then there exists c ∈ L such that a ≺ c ≺ b. Moreover, it is easy to see that a ≺ ⋀ _i∈Ib_i implies a ≺ b_i for every i ∈ I, whereas a ≺ ⋁ _i∈Ib_i is equivalent to a ≺ b_i for some i ∈ I.

Definition 2.1. ([2, 15], for L=[0,1]) An L-fuzzifying topology on X is a mapping τ : 2^X → L which satisfies:

τ (∅) = τ (X) = ⊤ ;

τ (A∩ B) ≥ τ (A) ∧ τ (B) ;

τ (⋃ _j∈JA_j) ≥ ⋀ _j∈Jτ (A_j) .

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

Definition 2.2. ([7], for L=[0,1]) An L-fuzzifying interior operator on a set X is a family of mappings $I = {I_{x} : 2^{X} \to L ∣ x \in X}$ with the following conditions: For all x ∈ X, A, B ∈ 2^X,

$I_{x} (X) = ⊤;$

$I_{x} (A) \neq ⊥ implies x \in A;$

$I_{x} (A \cap B) = I_{x} (A) \land I_{x} (B) .$

The pair $(X, I)$ is called an L-fuzzifying interior space, and it will be called topological if it satisfies moreover: For all x ∈ X, A ∈ 2^X,

$I_{x} (A) = ⋁_{x \in B \subseteq A} ⋀_{y \in B} I_{y} (B) .$

Definition 2.3. ([7], for L=[0,1]) An L-fuzzifying closure operator on a set X is a family of mappings $C = {C_{x} : 2^{X} \to L ∣ x \in X}$ with the following conditions: For all x ∈ X, A, B ∈ 2^X,

$C_{x} (\emptyset) = ⊥;$

$C_{x} (A) \neq ⊤ implies x \notin A;$

$C_{x} (A \lor B) = C_{x} (A) \lor C_{x} (B) .$

The pair $(X, C)$ is called an L-fuzzifying closure space, and it will be called topological if it satisfies moreover: For all x ∈ X, A ∈ 2^X,

$C_{x} (A) = ⋀_{B \supseteq A} (C_{x} (B) \lor ⋁_{y \notin B} C_{y} (B)) .$

Let $D (X)$ be the set of all mappings from X to 2^X such that x ∉ f (x) for all x ∈ X. d₀ is the smallest element of $D (X)$ , i.e., d₀ (x) =∅ for all x ∈ X. For any $f, g \in D (X)$ , we define:

f ≤ g if and only if ∀x ∈ X, f (x) ⊆ g (x) .

(f ∨ g) (x) = f (x) ∨ g (x) .

(f ⋄ g) (x) = ⋂ {f (y) ∣ y ∉ g (x)} .

Then we can prove that $f \lor g, f ⋄ g \in D (X), f ⋄ g \leq f, f ⋄ g \leq g$ and the operations “∨” and “⋄” satisfy the associate law. An element $f \in D (X)$ is said to be symmetric if ∀x, y ∈ X, x ∉ f (y) ⇔ y ∉ f (x), equivalently, x ∈ f (y) ⇔ y ∈ f (x). If f, g are symmetric, then so are f ∨ g and f ⋄ g. Let $D_{s} (X)$ denote the set of all symmetric mappings in $D (X)$ .

Definition 2.4. ([16]) A pointwise L-fuzzifying quasi-uniformity on X is a mapping $U : D (X) \to L$ such that

$U (d_{0}) = ⊤$ ;

$U (f \lor g) = U (f) \land U (g)$ ;

$U (f) = ⋁_{g ⋄ g \geq f} U (g)$ .

U

is a pointwise L-fuzzifying quasi-uniformity on X, the pair

(X, U)

is called a pointwise L-fuzzifying quasi-uniform space.

Definition 2.5. ([16]) A mapping $B : D (X) \to L$ is called a base of one pointwise L-fuzzifying quasi-uniformity if it satisfies:

$B (d_{0}) = ⊤$ ;

$B (f \lor g) \geq B (f) \land B (g)$ ;

$B (f) \leq ⋁_{g ⋄ g \geq f} B (g)$ .

Definition 2.6. ([18]) Let $U$ be a pointwise L-fuzzifying quasi-uniformity on X. $U$ is called a pointwise L-fuzzifying uniformity if there exists a mapping $B : D (X) \to L$ satisfying (FYB1)–(FYB3) and

$B (f) = ⊥$ , $\forall f \in D (X) - D_{s} (X)$ ;

$U (f) = ⋁_{g \geq f} B (g) .$

In this case, we call

B

a symmetric base of

U

3 Characterizations of L-fuzzifying interior operators and L-fuzzifying closure operators

In this section, we will generalize the characterizations of fuzzifying interior operators and fuzzifying closure operators to L-fuzzifying setting. It will be used in the subsequent section.

Theorem 3.1. ([14], for L=[0,1]) Let $I = {I_{x} ∣ x \in X}$ be an L-fuzzifying interior operator. Then the following statements are equivalent for all x ∈ X and A ∈ 2^X,

$I_{x} (A) = ⋁_{x \in B \subseteq A} ⋀_{y \in B} I_{y} (B) .$

$I_{x} (A) = ⋁_{x \in B \subseteq A} (I_{x} (B) \land ⋀_{y \in B} I_{y} (A)) .$

Proof. That (FYI4)⇒(FYI4)^∗ holds trivially. In order to prove (FYI4)^∗⇒(FYI4), suppose (FYI4)^∗ holds, i.e., $I_{x} (A) = ⋁_{x \in B \subseteq A} (I_{x} (B) \land ⋀_{y \in B} I_{y} (A)) .$ Let α ∈ M (L) such that $α ≺ I_{x} (A) = ⋁_{x \in B \subseteq A} (I_{x} (B) \land ⋀_{y \in B} I_{y} (A)) .$ Then there exists some B ⊆ A such that

(1) $α ≺ I_{x} (B)$ ; (2) $\forall y \in B, α ≺ I_{y} (A)$ .

It is clear that the union of subsets of A fulfilling (1) and (2) is still of this kind. So we can define B_α to be the maximal subset of A fulfilling the above two conditions (1) and (2) for α, i.e., $α ≺ I_{x} (B_{α})$ and $\forall y \in B_{α}, α ≺ I_{y} (A)$ . Thus, ∀y ∈ B_α, it follows from $α ≺ I_{y} (A)$ that there exists C_y ⊆ A such that

(3) $α ≺ I_{y} (C_{y})$ ; (4) $\forall z \in C_{y}, α ≺ I_{z} (A)$ .

By the maximality of B_α, it follows that B_α ∪ C_y ⊆ B_α and C_y ⊆ B_α. Hence, ∀y ∈ B_α, $α ≺ I_{y} (C_{y}) \leq I_{y} (B_{α}) .$ Therefore, $α \leq ⋀_{y \in B_{α}} I_{y} (B_{α}) \leq ⋁_{x \in B \subseteq A} ⋀_{y \in B} I_{y} (B) .$ From the arbitrariness of α, we obtain $I_{x} (A) \leq ⋁_{x \in B \subseteq A} ⋀_{y \in B} I_{y} (B) .$ Since $I_{x} (A) \geq ⋁_{x \in B \subseteq A} ⋀_{y \in B} I_{y} (B)$ is obvious, we have $I_{x} (A) = ⋁_{x \in B \subseteq A} ⋀_{y \in B} I_{y} (B),$ as desired. □

Similarly, we have the following theorem to characterize topological L-fuzzifying closure operators.

Theorem 3.2.Let $C = {C_{x} ∣ x \in X}$ be an L-fuzzifying closure operator. Then the following statements are equivalent for all x ∈ X and A ∈ 2^X,

$C_{x} (A) = ⋀_{B \supseteq A} (C_{x} (B) \lor ⋁_{y \notin B} C_{y} (B)) .$

$C_{x} (A) = ⋀_{B \supseteq A} (C_{x} (B) \lor ⋁_{y \notin B} C_{y} (A)) .$

Proof. That (FYC4)⇒(FYC4)^∗ holds trivially. To prove the inverse, suppose (FYC4)^∗ holds, i.e., $C_{x} (A) = ⋀_{B \supseteq A} (C_{x} (B) \lor ⋁_{y \notin B} C_{y} (A)) .$ Now, we prove $C_{x} (A)^{'} \leq ⋁_{B \supseteq A} (C_{x} (B)^{'} \land ⋀_{y \notin B} C_{y} (B)^{'}) .$ Take α ∈ M (L) such that $α ≺ C_{x} (A)^{'} = ⋁_{B \supseteq A} (C_{x} (B)^{'} \land ⋀_{y \notin B} C_{y} (A)^{'}) .$ Then there exists some B ⊇ A such that

(1) $α ≺ C_{x} (B)^{'}$ ; (2) $\forall y \notin B, α ≺ C_{y} (A)^{'}$ .

It is clear that the join of B (containing A) fulfilling (1) and (2) is still of this kind. So we can define B_α to be the minimal set (containing A) fulfilling the above two conditions (1) and (2) for α, i.e., $α ≺ C_{x} (B_{α})^{'}$ and $\forall y \notin B_{α}, α ≺ C_{y} (A)^{'}$ . Thus, ∀y ∉ B_α, by $α ≺ C_{y} (A)^{'} = ⋁_{D \supseteq A} (C_{y} (D)^{'} \land ⋀_{z \notin D} C_{z} (A)^{'}),$

we know there exists D_y ⊇ A such that

(3) $α ≺ C_{y} (D_{y})^{'}$ ; (4) $\forall z \notin D_{y}, α ≺ C_{z} (A)^{'}$ .

By the minimality of B_α, it follows that B_α ⊆ B_α ∩ D_y and B_α ⊆ D_y. Hence, ∀y ∉ B_α, $α ≺ C_{y} (D_{y})^{'} \leq C_{y} (B_{α})^{'} .$ Therefore, $\begin{matrix} α \leq C_{x} (B_{α})^{'} \land ⋀_{y \notin B_{α}} C_{y} (B_{α})^{'} \\ \leq ⋁_{B \supseteq A} (C_{x} (B)^{'} \land ⋀_{y \notin B} C_{y} (B)^{'}) . \end{matrix}$ From the arbitrariness of α, we obtain $C_{x} (A)^{'} \leq ⋁_{B \supseteq A} (C_{x} (B)^{'} \land ⋀_{y \notin B} C_{y} (B)^{'}),$ i.e., $C_{x} (A) \geq ⋀_{B \supseteq A} (C_{x} (B) \lor ⋁_{y \notin B} C_{y} (B)) .$ Since $C_{x} (A) \leq ⋀_{B \supseteq A} (C_{x} (B) \lor ⋁_{y \notin B} C_{y} (B))$ is obvious, we have $C_{x} (A) = ⋀_{B \supseteq A} (C_{x} (B) \lor ⋁_{y \notin B} C_{y} (B)),$ as desired. □

4 Topologies induced by pointwise L-fuzzifying quasi-uniformities

In this section, we will show that a pointwise L-fuzzifying quasi-uniformity $U$ can induce two L-fuzzifying topologies $τ_{U}$ and $η_{U}$ . The relations between $τ_{U}$ and $η_{U}$ are discussed.

Theorem 4.1.Let $U$ be a pointwise L-fuzzifying quasi-uniformity and define $I_{x}^{U} : 2^{X} \to L$ as follows: $\forall A \in 2^{X}, I_{x}^{U} (A) = ⋁_{x \in ⋂_{y \notin A} f (y)} U (f) .$ Then $I^{U} = {I_{x}^{U} ∣ x \in X}$ is a topological L-fuzzifying interior operator.

Proof. We check $I_{x}^{U}$ satisfies (FYI1)–(FYI4)^∗ as follows:

Since ⋂_y∉Xd₀ (y) = ⋂ ∅ = X, it follows that $I_{x}^{U} (X) = ⋁_{x \in ⋂_{y \notin X} f (y)} U (f) \geq U (d_{0}) = ⊤ .$

$\forall x \notin A, I_{x}^{U} (A) = ⋁_{x \in ⋂_{y \notin A} f (y)} U (f) \leq ⋁_{x \in f (x)} U (f) = ⋁ \emptyset = ⊥ .$

From the definition of $I_{x}^{U}$ , we have $I_{x}^{U} (A) \leq I_{x}^{U} (B)$ for A ⊆ B. This implies $I_{x}^{U} (A \cap B) \leq I_{x}^{U} (A) \land I_{x}^{U} (B)$ . The inverse is shown by the following fact: $\begin{matrix} I_{x}^{U} (A) \land I_{x}^{U} (B) & = & ⋁_{x \in ⋂_{y \notin A} f (y)} U (f) \land ⋁_{x \in ⋂_{z \notin B} g (z)} U (g) \\ = & ⋁_{x \in ⋂_{y \notin A} f (y) \cap ⋂_{z \notin B} g (z)} (U (f) \land U (g)) \\ = & ⋁_{x \in ⋂_{y \notin A, z \notin B} f (y) \cap g (z)} U (f \lor g) \\ \leq & ⋁_{x \in ⋂_{y \notin A \cap B} (f \lor g) (y)} U (f \lor g) \\ \leq & ⋁_{x \in ⋂_{y \notin A \cap B} h (y)} U (h) = I_{x}^{U} (A \cap B) . \end{matrix}$

That $I_{x} (A) \geq ⋁_{x \in B \subseteq A} (I_{x} (B) \land ⋀_{y \in B} I_{y} (A))$ trivially holds from $A \subseteq B \Rightarrow I_{x}^{U} (A) \leq I_{x}^{U} (B)$ . In order to prove $I_{x} (A) \leq ⋁_{x \in B \subseteq A} (I_{x} (B) \land ⋀_{y \in B} I_{y} (A))$ , let $α ≺ I_{x}^{U} (A) = ⋁_{x \in ⋂_{y \notin A} f (y)} U (f) = ⋁_{x \in ⋂_{y \notin A} f (y)} ⋁_{g ⋄ g \geq f} U (g) .$

Then there exists $f, g \in D (X)$ such that x ∈ ⋂ _y∉Af (y), g ⋄ g ≥ f and $α \leq U (g)$ . Let B = ⋂ _y∉Ag (y). Then x ∈ B ⊆ A. Further, it follows from g ⋄ g ≥ f that $\begin{matrix} x \in ⋂_{y \notin A} f (y) & \subseteq & ⋂_{y \notin A} g ⋄ g (y) = ⋂_{y \notin A} ⋂_{z \notin g (y)} g (z) \\ = & ⋂_{z \notin ⋂_{y \notin A} g (y)} g (z) = ⋂_{z \notin B} g (z) . \end{matrix}$

Thus, we obtain $I_{x}^{U} (B) = ⋁_{x \in ⋂_{y \notin B} f (y)} U (f) \geq U (g) \geq α$ and $⋀_{y \in B} I_{y}^{U} (A) = ⋀_{y \in B} ⋁_{y \in ⋂_{z \notin A} h (z)} U (h) \geq ⋀_{y \in B} U (g) \geq α .$ Hence, $α \leq I_{x}^{U} (B) \land ⋀_{y \in B} I_{y}^{U} (A) \leq ⋁_{x \in B \subseteq A} (I_{x}^{U} (B) \land ⋀_{y \in B} I_{y}^{U} (A)) .$ From the arbitrariness of α, we have $I_{x}^{U} (A) \leq ⋁_{x \in B \subseteq A} (I_{x}^{U} (B) \land ⋀_{y \in B} I_{y}^{U} (A)),$ as desired. □

Corollary 4.2.Let $U$ be a pointwise L-fuzzifying quasi-uniformity and define $τ_{U} : 2^{X} \to L$ as follows: $\forall A \in 2^{X}, τ_{U} (A) = ⋀_{x \in A} I_{x}^{U} (A) = ⋀_{x \in A} ⋁_{x \in ⋂_{y \notin A} f (y)} U (f) .$ Then $τ_{U}$ is an L-fuzzifying topology on X.

Theorem 4.3.Let $U$ be a pointwise L-fuzzifying quasi-uniformity and define $C_{x}^{U} : 2^{X} \to L$ as follows: $\forall A \in 2^{X}, C_{x}^{U} (A) = ⋀_{A \subseteq f (x)} U (f)^{'} .$

Then $C^{U} = {C_{x}^{U} ∣ x \in X}$ is a topological L-fuzzifying closure operator.

Proof. We check $C_{x}^{U}$ satisfies (FYC1)–(FYC4)^∗ as follows:

$C_{x}^{U} (\emptyset) = ⋀_{\emptyset \subseteq f (x)} U (f)^{'} \leq U (d_{0})^{'} = ⊥ .$

$\forall x \in A, C_{x}^{U} (A) = ⋀_{A \subseteq f (x)} U (f)^{'} = ⋀ \emptyset = ⊤ .$

From the definition of $C_{x}^{U}$ , we obtain $C_{x}^{U} (A) \leq C_{x}^{U} (B)$ for A ⊆ B. This shows that $C_{x}^{U} (A \cup B) \geq C_{x}^{U} (A) \lor C_{x}^{U} (B)$ . Next, we prove the inverse as follows: $\begin{matrix} C_{x}^{U} (A) \lor C_{x}^{U} (B) & = & ⋀_{A \subseteq f (x)} U (f)^{'} \lor ⋀_{B \subseteq g (x)} U (g)^{'} \\ = & ⋀_{A \subseteq f (x), B \subseteq g (x)} U (f)^{'} \lor U (g)^{'} \\ \geq & ⋀_{A, B \subseteq (f \lor g) (x)} U (f \lor g)^{'} \\ \geq & ⋀_{A \cup B \subseteq h (x)} U (h)^{'} \\ = & C_{x}^{U} (A \cup B) . \end{matrix}$

By (FYC3), $C_{x}^{U} (A) \leq ⋀_{B \supseteq A} (C_{x}^{U} (B) \lor ⋁_{y \notin B} C_{y}^{U} (A))$ holds. In order to prove the inverse, it suffices to prove that $C_{x}^{U} (A)^{'} \leq ⋁_{B \supseteq A} (C_{x}^{U} (B)^{'} \land ⋀_{y \notin B} C_{y}^{U} (A)^{'}) .$ Take any α ∈ M (L) such that $α ≺ C_{x}^{U} (A)^{'} = ⋁_{A \subseteq f (x)} U (f) = ⋁_{A \subseteq f (x)} ⋁_{g ⋄ g \geq f} U (g) .$

Then there exist

f, g \in D (X)

such that A ⊆ f (x), g ⋄ g ≥ f and

α \leq U (g) .

Let B = g (x). Then B ⊇ g ⋄ g (x) ⊇ f (x) ⊇ A and

C_{x}^{U} (B)^{'} = ⋁_{B \subseteq h (x)} U (h) \geq U (g) \geq α .

Further, by A ⊆ f (x) ⊆ g ⋄ g (x) = ⋂ {g (y) ∣ y ∉ g (x)}, we have A ⊆ g (y) for all y ∉ g (x) . This implies

⋀_{y \notin B} C_{y}^{U} (A)^{'} = ⋀_{y \notin B} ⋁_{A \subseteq h (y)} U (h) \geq ⋀_{y \notin B} U (g) \geq α .

Thus,

α \leq C_{x}^{U} (B)^{'} \land ⋀_{y \notin B} C_{y}^{U} (A)^{'} \leq ⋁_{B \supseteq A} (C_{x}^{U} (B)^{'} \land ⋀_{y \notin B} C_{y}^{U} (A)^{'}) .

From the arbitrariness of α, we obtain

C_{x}^{U} (A)^{'} \leq ⋁_{B \supseteq A} (C_{x}^{U} (B)^{'} \land ⋀_{y \notin B} C_{y}^{U} (A)^{'}) .

Therefore,

C_{x}^{U} (A) \geq ⋀_{B \supseteq A} (C_{x}^{U} (B) \lor ⋁_{y \notin B} C_{y}^{U} (A))

, as desired. □

Corollary 4.4.Let $U$ be a pointwise L-fuzzifying quasi-uniformity and define $η_{U} : 2^{X} \to L$ as follows: $\begin{matrix} \forall A \in 2^{X}, η_{U} (A) & = & ⋀_{x \in A} C_{x}^{U} (X - A)^{'} \\ = & ⋀_{x \in A} ⋁_{X - A \subseteq f (x)} U (f) . \end{matrix}$ Then $η_{U}$ is an L-fuzzifying topology.

Remark 4.5. When $U$ is a pointwise L-fuzzifying quasi-uniformity, $τ_{U}$ is not necessary coincident with $η_{U}$ . The following example can show this.

Example 4.6. Let X = [0, 1], L = {0, 1}, $D = {(x, y) \in X \times X ∣ 0 \leq x \leq y \leq 1}$ . Define $f_{D} : X \to 2^{X}$ by $f_{D} (x) = {y \in X ∣ (x, y) \notin D}$ . Now define $U : D (X) \to {0, 1}$ as follows: $\begin{matrix} U (f) = {\begin{matrix} 1, & f \leq f_{D}, \\ 0, & otherwise . \end{matrix} \end{matrix}$ Next we verify that $U$ is a pointwise L-fuzzifying quasi-uniformity, i.e., $U$ satisfies (FYU1)–(FYU3).

and (FYU2) hold trivially.

From (FYU2), we obtain $U (f) \geq ⋁_{g ⋄ g \geq f} U (g) .$

To prove the inverse, we first verify

f_{D} ⋄ f_{D} = f_{D}

, i.e.,

f_{D} ⋄ f_{D} (x) = f_{D} (x)

. It suffices to check that

f_{D} (x) \subseteq f_{D} ⋄ f_{D} (x)

. By the definition of

f_{D} ⋄ f_{D}

, we have

\begin{matrix} f_{D} ⋄ f_{D} (x) & = & ⋂ {f_{D} (y) ∣ y \notin f_{D} (x)} \\ = & ⋂ {f_{D} (y) ∣ (x, y) \in D} . \end{matrix}

Thus, we need only prove

f_{D} (x) \subseteq f_{D} (y)

for all

(x, y) \in D

. Take any

z \in f_{D} (x)

, then x > z. Since

(x, y) \in D

, i.e., x ≤ y, we have z < x ≤ y. Thus,

z \in f_{D} (y)

. From the arbitrariness of z, we obtain

f_{D} (x) \subseteq f_{D} ⋄ f_{D} (x)

. Hence,

f_{D} ⋄ f_{D} = f_{D}

Now, we prove $U (f) \leq ⋁_{g ⋄ g \geq f} U (g)$ .

If $U (f) = 0$ , then $U (f) \leq ⋁_{g ⋄ g \geq f} U (g)$ holds obviously.

If $U (f) = 1$ , then $f \leq f_{D} = f_{D} ⋄ f_{D}$ . This implies $⋁_{g ⋄ g \geq f} U (g) \geq U (f_{D}) = 1 = U (f) .$

Next, we show that $τ_{U} \neq η_{U} .$ Since $f_{D} (x) = {y ∣ (x, y) \notin D} = {y ∣ x > y} = [0, x),$ we have $⋂_{y \in (\frac{1}{2}, 1]} f_{D} (y) = ⋂_{y \in (\frac{1}{2}, 1]} [0, y) = [0, \frac{1}{2}] .$ This implies $\begin{matrix} I_{x}^{U} ([0, \frac{1}{2}]) & = & ⋁_{x \in ⋂_{y \notin [0, \frac{1}{2}]} f (y)} U (f) \\ = & ⋁_{x \in ⋂_{y \in (\frac{1}{2}, 1]} f (y)} U (f) \geq U (f_{D}) = 1 \end{matrix}$ for all $x \in [0, \frac{1}{2}] .$ Moreover, $\begin{matrix} C_{x}^{U} {(X - [0, \frac{1}{2}])}^{'} & = & ⋁_{(\frac{1}{2}, 1] \subseteq f (x)} U (f) \\ = & ⋁_{(\frac{1}{2}, 1] \subseteq f (x), f \leq f_{D}} U (f) \\ \leq & ⋁_{(\frac{1}{2}, 1] \subseteq f (x) \subseteq f_{D} (x) = [0, x)} U (f) \\ = & ⋁ \emptyset = 0 \end{matrix}$ for all $x \in [0, \frac{1}{2}] .$ Hence, we have $\begin{matrix} τ_{U} ([0, \frac{1}{2}]) & = & ⋀_{x \in [0, \frac{1}{2}]} I_{x}^{U} ([0, \frac{1}{2}]) = 1 and \\ η_{U} ([0, \frac{1}{2}]) & = & ⋀_{x \in [0, \frac{1}{2}]} C_{x}^{U} {(X - [0, \frac{1}{2}])}^{'} = 0 . \end{matrix}$

As a sequence, we obtain $τ_{U} \neq η_{U} .$

However, for a pointwise L-fuzzifying uniform space $(X, U)$ , the two L-fuzzifying topologies $τ_{U}$ and $η_{U}$ are the same, just as the following theorem shows.

Theorem 4.7. If $(X, U)$ is a pointwise L-fuzzifying uniform space, then $τ_{U} = η_{U}$ .

Proof. It is sufficient to prove that $I_{x}^{U} (A) = C_{x}^{U} (X - A)^{'}$ for all x ∈ X, A ∈ 2^X. Since $U$ is a pointwise L-fuzzifying uniformity, there is a symmetric base $B$ of $U$ such that $\begin{matrix} I_{x}^{U} (A) & = & ⋁_{x \in ⋂_{y \notin A} f (y)} U (f) \\ = & ⋁_{x \in ⋂_{y \notin A} f (y)} ⋁_{f^{*} \geq f, f^{*} \in D_{s} (X)} B (f^{*}) \end{matrix}$ and $\begin{matrix} C_{x}^{U} (X - A)^{'} & = & ⋁_{X - A \subseteq g (x)} U (g) \\ = & ⋁_{X - A \subseteq g (x)} ⋁_{g^{*} \geq g, g^{*} \in D_{s} (X)} B (g^{*}) . \end{matrix}$

Let α ∈ M (L) with $α ≺ I_{x}^{U} (A)$ . Then there exist $f \in D (X), f^{*} \in D_{s} (X)$ such that f^∗ ≥ f, x ∈ ⋂ _y∉Af (y) ⊆ ⋂ _y∉Af^∗ (y) and $α \leq B (f^{*}) .$ Thus, x ∈ f^∗ (z) for all z ∈ X - A. Since f^∗ is symmetric, it follows that z ∈ f^∗ (x). By the arbitrariness of z, we have X - A ⊆ f^∗ (x). This implies $\begin{matrix} C_{x}^{U} (X - A)^{'} & = & ⋁_{X - A \subseteq g (x)} ⋁_{g^{*} \geq g, g^{*} \in D_{s} (X)} B (g^{*}) \\ \geq & B (f^{*}) \geq α . \end{matrix}$ From the arbitrariness of α, we obtain $I_{x}^{U} (A) \leq C_{x}^{U} (X - A)^{'} .$

On the other hand, take any γ ∈ M (L) such that $γ ≺ C_{x}^{U} (X - A)^{'}$ . Then there exist $g \in D (X), g^{*} \in D_{s} (X)$ such that X - A ⊆ g (x), g^∗ ≥ g and $γ \leq B (g^{*})$ . For all y ∉ A, it follows that y ∈ X - A ⊆ g (x) ⊆ g^∗ (x). Since g^∗ is symmetric, it follows that x ∈ g^∗ (y). Further, we have x ∈ ⋂ _y∉Ag^∗ (y). Thus, $\begin{matrix} I_{x}^{U} (A) & = & ⋁_{x \in ⋂_{y \notin A} f (y)} ⋁_{f^{*} \geq f, f^{*} \in D_{s} (X)} B (f^{*}) \\ \geq & B (g^{*}) \geq γ . \end{matrix}$ From the arbitrariness of γ, we obtain $C_{x}^{U} (X - A)^{'} \leq I_{x}^{U} (A) .$

As a result, $C_{x}^{U} (X - A)^{'} = I_{x}^{U} (A)$ for all x ∈ X, A ∈ 2^X. Therefore, $τ_{U} = η_{U}$ . □

Footnotes

Acknowledgments

The authors would like to express their sincere thanks to the anonymous reviewers for their careful reading and constructive comments. This work is supported by National Nature Science Foundation Committee (NSFC) of China (No. 61573119), China Postdoctoral Science Foundation (No. 2015M581434), Fundamental Research Project of Shenzhen (No. JCYJ20120613144110654 and No. JCYJ20140417172417109).

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