Linear fuzzifying uniformities

Abstract

The main purpose of this paper is to study Hutton type fuzzifying uniformities on linear spaces. Firstly, we show that if a base of a fuzzifying uniformity defined over a linear space is translation-invariant, balanced and absorbed, then it generates a linear fuzzifying topology. From this linear fuzzifying topology, we can construct a new linear fuzzifying uniformity (i.e., a fuzzifying uniformity compatible with the linear structure) which is equivalent to the original fuzzifying uniformity. Secondly, the Hausdorff separation and complete boundedness in linear fuzzifying uniformities are investigated. In addition, as an example, the linear fuzzifying uniformity induced by a fuzzy norm is also discussed.

Keywords

Linear fuzzifying uniformity linear fuzzifying topology complete boundedness Hausdorff separation axiom fuzzy norm

1 Introduction

It is well known that the theory of uniform structures plays an important role in the area of topology as a bridge linking metrics and topological spaces. Hutton [13], Höhle [10 –12] and Lowen [17] studied uniformities in the fuzzy context. After that, Ying [23] introduced the concept of fuzzifying uniform spaces in the framework of fuzzifying topology and established some fundamental properties of fuzzifying uniform spaces. Gutiérrez García et al. [9, 32] discussed different uniformities in the fuzzy context and gave an essentially more general concept of an L-valued uniformity, specially L = [0, 1], by using a filter approach. With a different approach, fuzzifying uniform spaces were considered in [4] and it was proved that they constitute a category isomorphic to fuzzy uniform spaces in the sense of Lowen. Khedr et al. [15] considered the concept of a strong fuzzifying uniformity in Ying’s sense of a fuzzifying uniformity in 2003. Some relations between strong fuzzifying uniformities and corresponding fuzzifying topologies are also discussed in [15]. In 2006, Yue and Fang [25] studied the Kubiak-Šostak extension of Hutton’s quasi-uniformities and gave the concept of fuzzifying uniformities. Additionally, Ramadan et al. [30] investigated the properties of L-fuzzy topologies induced by L-quasi-uniformity in complete residuated lattices.

In the last decade, many researchers worked on studying fuzzifying uniformities generated by fuzzy pseudo-metrics. For example, Mardones-Pérez and de Prada Vicente [19] explored the fuzzifying uniformity generated by Kramosil and Michalek fuzzy pseudo-metric. Yue and Shi [26] discussed the fuzzifying uniformity generated by a George and Veeramani’s fuzzy pseudo-metrics which can be characterized by a family of compatible ordinary pseudo-metric. Meanwhile, it was also shown that the fuzzy pseudo-metric and its induced fuzzifying uniformity generate the same fuzzifying topology. In particular, Yue and Gu [27] endowed fuzzy partial pseudo-metric with fuzzifying topology and partial fuzzifying uniformity. In [28], Yue and Fang have constructed an I-uniformity generated by the Hutton quasi-uniformity of [7] in a fuzzy metric space in the sense of Kramosil and Michalek. In 2018, Gutiérrez García et al. [8] established a relationship between the categories of probabilistic uniformities and Hutton [0, 1](-quasi)-uniformities indeced by a fuzzy metric space in the sense of George and Veeramani with the category of classical uniformities respectively. Wu et al. [34] investigated the relations among (L, M)-fuzzy pseudo-metrics, pointwise pseudo-metric chains and pointwise (L, M)-fuzzy uniformities. Particularly, Kim et al. [16] showed that the category of Alexandrov L-fuzzy pre-uniform spaces and the category of L-lower (resp. upper) approximation spaces are isomorphic. Besides the above mentioned application of fuzzifying uniformity structure to study fuzzy pseudo-metrics and fuzzifying topologies, Khalaf and Mohammed [14] studied (E, L)-fuzzifying matroid spaces by using fuzzifying uniformity and solve combinatorial optimizations problem via (E, L)-fuzzifying matroid space.

It is worthy noting that Wu and Ma [33] has put forward that a (QL) fuzzy linear topological space can be induced by a unique translation-invariant fuzzy uniformity. Motivated by [26] and [33], we intend to study the Hutton type fuzzifying uniformity on linear spaces which are compatible with the linear structure. Meanwhile, we also discuss the linear fuzzifying uniformity induced by a fuzzy norm as an example. Besides, to our knowledge, there is no paper involved in the Hausdorff separation axiom in fuzzifying uniformities up to now, and we will discuss it as an important content.

The paper is organized as follows. Firstly, we show that if a base of a fuzzifying uniformity defined over a linear space is translation-invariant, balanced and absorbed, then it generates a linear fuzzifying topology. From this linear fuzzifying topology, we can construct a new linear fuzzifying uniformity which is equivalent to the original uniformity. Secondly, we investigate the Hausdorff separation and complete boundedness in linear fuzzifying uniform spaces. Furthermore, we prove that the degree to which linear fuzzifying uniform spaces and fuzzifying topological linear spaces satisfy Hausdorff separation axiom is equivalent. The same conclusion is also proved on complete boundedness. Finally, we study the linear fuzzifying uniformity generated by a fuzzy norm and the continuity of the mappings between fuzzifying uniform spaces and fuzzifying topological linear spaces is also discussed. These results enrich the theory of uniform structures and extend its applications. Our discussion may provide some assistance to establish a more systematic theory of fuzzifying uniformity.

2 Preliminaries

In this section, we recall some necessary notions and fundamental results which are used in this paper.

Throughout this paper, X always denotes a universe of discourse. 2^X and $F (X)$ denote the classes of all crisp and fuzzy subsets, respectively, of X. For any mapping F : X → Y, the notation F^→ : 2^X → 2^Y is defined by F^→ (U) ={ F (x) ∈ Y|x ∈ U } for U ∈ 2^X and F^← : 2^Y → 2^X is defined by F^← (V) ={ x ∈ X|F (x) ∈ V } for V ∈ 2^X, respectively (see [3, 31]). X \ W means the complement set of W in X. For every x ∈ X, $\dot{x}$ is the filter generated by x, that is, $\dot{x} = {U \subseteq X | x \in U}$ . The notation 2^(X) is the set of all non-empty finite subsets of X. $K$ and $ℝ$ represent the field of real or complex numbers and the field of real numbers, respectively. Moreover, the symbol 0 denotes the neutral element of a linear space.

Definition 2.1 (Ying [22]) A fuzzifying topology is a mapping τ : 2^X → [0, 1] such that

(FY1) τ (X) = τ (∅) = 1;

(FY2) τ (U ∩ V) ≥ τ (U) ∧ τ (V) for all U, V ∈ 2^X;

(FY3) $τ (⋃_{j \in J} U_{j}) \geq \underset{j \in J}{⋀} τ (U_{j})$ for every family {U_j|j ∈ J } ⊆ 2^X.

F : (X, τ) → (Y, δ) is called continuous with respect to the two fuzzifying topologies τ and δ if δ (V) ≤ τ (F^← (V)) holds for all V ∈ 2^Y.

Definition 2.2 (Ying [22]) Let (X, τ) be a fuzzifying topological space. For any x ∈ X, $N_{x} \in F (2^{X})$ , called a fuzzifying neighborhood system of x, is defined as follows. For any A ∈ 2^X, $N_{x} (A) = ⋁_{x \in B \subseteq A} τ (B)$ .

Definition 2.3 (Ying [24]) Let (X, τ) and (Y, δ) be two fuzzifying topological spaces, and $N_{x}$ a fuzzifying neighborhood system of x. For any mapping f : X → Y, the degree to which f is continuous at x is defined by $[C (f)] = \underset{U \subseteq Y}{⋀} min {1, 1 - N_{f (x)}^{Y} (U) + N_{x}^{X} (f^{\leftarrow} (U))} .$ The degree to which f is continuous is defined by [C (f)] = ⋀ _x∈X [C_x (f)]. Specially, f is called continuous whenever [C (f)] =1.

Definition 2.4 (Ying [22]) Let (X, τ) be a fuzzifying topological space. The degree to which (X, τ) satisfies T₂ separation axiom is defined by $[T_{2} (X, τ)] = \underset{x \neq y}{⋀} \underset{A \cap B = \emptyset}{⋁} N_{x} (A) \land N_{y} (B) .$

Definition 2.5 (Qiu [29]) Let (X, τ) be a fuzzifying topological space. Suppose that X is a linear space over $K$ . Then (X, τ) is called a fuzzifying topological linear space if it fulfills the following conditions:

(A1) For any x, y ∈ X and any V ⊆ X with x + y ∈ V, $\begin{matrix} ⊨ V \in τ \to (\exists V_{x}, V_{y} \in 2^{X}) ((V_{x} + V_{y} \subseteq V) \\ \land (V_{x} \in τ \land V_{y} \in τ)) . \end{matrix}$ (A2) For each $s \in K$ , any x ∈ X, and any V ∈ 2^X with sx ∈ V, $\begin{matrix} ⊨ V \in τ \to (\exists δ > 0) (\exists V_{x} \in 2^{X}) ((V_{x} \in τ) \\ \land (\forall t \in K) (| t - s | < δ \to t V_{x} \subseteq V)) . \end{matrix}$

Theorem 2.6 (Yan [21]) Let (X, τ) be a fuzzifying topological linear space on $K$ and $N_{0} (\cdot)$ be its corresponding fuzzifying neighborhood system of the neutral element. Then it has the following properties:

(P1) $N_{0} (X) = 1$ ;

(P2) $\forall U \subseteq X, N_{0} (U) > 0 \Rightarrow 0 \in U$ ;

(P3) $\forall U, V \subseteq X, N_{0} (U \cap V) = N_{0} (U) \land N_{0} (V)$ ;

(P4) $\forall W \subseteq X, N_{0} (W) \leq \underset{U + V \subseteq W}{⋁} N_{0} (U) \land N_{0} (V)$ ;

(P5) $\forall U \subseteq X, x \in X, N_{0} (U) > 0 \Rightarrow \exists ɛ > 0$ such that kx ∈ U for all |k| < ɛ;

(P6) $\forall U \subseteq X, N_{0} (U) > a$ implies there exists a balanced set V ⊆ U such that $N_{0} (V) > a$ .

Conversely, let X be a linear space over $K$ and consider a set-valued function $N_{0} (\cdot) : 2^{X} \to [0, 1]$ which satisfies the conditions (P1)-(P6). Then there exists a fuzzifying topology $τ_{N}$ on X such that $(X, τ_{N})$ is a fuzzifying topological linear space and $N_{0} (\cdot)$ is a fuzzifying neighborhood system of the neutral element.

Definition 2.7 (Qiu [29]) Let (X, τ) be a fuzzifying topological linear space. Then the unary fuzzy predicate $BD \in F (2^{X})$ , called fuzzy complete boundedness, is defined as follows: $BD (A) \overset{Δ}{=} (\forall V \in 2^{X}) (V \in N_{0} \to V \in ω (A)),$ for any A ∈ 2^X, where $ω (A) : = {V \in 2^{X} | \exists z_{1}, z_{2}, . . ., z_{n} \in A, s . t . A \subseteq ⋃_{i = 1}^{n} (z_{i} + V)}$ .

Intuitively, the degree to which A is complete bounded is $[BD (A)] = \underset{U \subseteq X}{⋀} {1 - N_{0} (U) | \forall F \in 2^{(A)}, A ⊈ F + U} .$

Let $D (X)$ denote the family of all mappings f : 2^X → 2^X such that:

(D1) U ⊆ f (U) for all U ⊆ X;

(D2) f (⋃ _j∈JU_j) = ⋃ _j∈Jf (U_j) for {U_j|j ∈ J } ⊆ 2^X.

The partial order “≤” is defined as follows: f ≤ g, $f, g \in D (X)$ if and only if f (A) ≤ g (A) for all A ∈ 2^X. Clearly, f₁ denotes the biggest element of $D (X)$ , i.e., f₁ (U) =∅ when U =∅ and f₁ (U) = X otherwise. For $f, g \in D (X)$ , we know $f \cap g \in D (X)$ , where $\begin{matrix} f \cap g (U) = ⋃_{B \cup C = U} f (B) \cap g (C) and \\ f \circ g (U) = f (g (U)) . \end{matrix}$

For $f \in D (X)$ , define f⁾ : 2^X → 2^X by f⁾ (B) =⋂ { C ∈ 2^X|f (X \ C) ⊆ X \ B }. Then $f^{)} \in D (X)$ . Suppose F : X → Y is a mapping, $f \in D (Y)$ and $g \in D (X)$ . Then define F^← (f) : 2^X → 2^X by F^← (f) (U) = (F^← ∘ f ∘ F^→) (U) for U ∈ 2^X and F^⇒ (g) (V) = (F^→ ∘ f ∘ F^←) (V) ∪ V for V ∈ 2^Y, then $F^{\leftarrow} (f) \in D (X)$ and $F^{\Rightarrow} (g) \in D (Y)$ (see [35]).

Definition 2.8 (Yue and Fang [25]) A fuzzifying uniformity on a non-empty set X is a mapping $U : D (X) \to [0, 1]$ such that

(FQU1) $U (f_{1}) = 1$ ;

(FQU2) $U (f \cap g) = U (f) \land U (g)$ for all $f, g \in D (X)$ ;

(FQU3) $U (f) = \underset{g \circ g \leq f}{⋁} U (g)$ for all $f \in D (X)$ ;

(FQU4) $U (f) = U (f^{)})$ for all $f \in D (X)$ .

Definition 2.9 (Yue and Shi [26]) Let $B : D (X) \to [0, 1]$ be a mapping. If $U$ , defined by $U (f) = \underset{g \leq f}{⋁} B (g)$ , is a fuzzifying uniformity, then $B$ is called a base of $U$ .

Lemma 2.10 (Yue and Fang [25], Yue and Shi [26]) Let $(X, U)$ be a fuzzifying uniform space and define $N_{x}^{U} : 2^{X} \to [0, 1]$ by $N_{x}^{U} (U) = \underset{x \notin f (X \ U)}{⋁} U (f)$ for $f \in D (X)$ and U ∈ 2^X. Then $N^{U} = {N_{x}^{U} | x \in X}$ is a fuzzifying neighborhood system.

Definition 2.11 (Yue and Fang [25]) Let $F : (X, U) \to (Y, U_{1})$ be a mapping. The degree to which F is uniformly continuous is defined as follows: $[UC (F)] = \underset{g \in D (Y)}{⋀} min {1, 1 - U (F^{\leftarrow} (g)) + U_{1} (g)} .$

Specially, F is called uniformly continuous if [UC (F)] =1. Clearly, F is uniformly continuous if and only if $U (F^{\leftarrow} (g)) \geq U_{1} (g)$ holds for all $g \in D (Y)$ .

Definition 2.12 (Bag and Samanta [2]) Let X be a linear space over a field $K$ . A fuzzy subset N of $X \times ℝ$ is called a fuzzy norm on X iff for all x, u ∈ X and $c \in K$ ,

(N1) for all $t \in ℝ$ with t ≤ 0, N (x, t) = 0;

(N2) (for all $t \in ℝ$ , t > 0, N (x, t) = 1) iff x = 0;

(N3) for all $t \in ℝ$ , t > 0, $N (cx, t) = N (x, \frac{t}{| c |})$ if c ≠ 0;

(N4) for all $s, t \in ℝ$ , x, u ∈ X, N (x+ u, s + t) ≥ min { N (x, s) , N (u, t) };

(N5) N (x, ·) is a non-decreasing function of $ℝ$ and $lim_{t \to \infty} N (x, t) = 1$ .

The pair (X, N) is called a fuzzy normed linear space.

Definition 2.13 (Bag and Samanta [2]) A mapping T : (X, N₁) → (Y, N₂) is said to be strongly fuzzy continuous at x₀ ∈ X, if for each ɛ > 0, there exists δ > 0 such that for all x ∈ X, $N_{2} (T (x) - T (x_{0}), ɛ) \geq N_{1} (x - x_{0}, δ) .$ T is said to be strongly fuzzy continuous on X if T is strongly fuzzy continuous at each point on X.

3 Linear fuzzifying uniformities

The aim of this section is to study linear fuzzifying uniformities. We will prove that if a base of a fuzzifying uniformity defined over a linear space is translation-invariant, balanced and absorbed, then it generates a linear fuzzifying topology. From this linear fuzzifying topology, we can construct a new linear fuzzifying uniformity which is equivalent to the original fuzzifying uniformity. The Hausdorff separation and complete boundedness in linear fuzzifying uniformitis are also investigated.

Theorem 3.1. Let X be a linear space over $K$ and $U$ a fuzzifying uniformity on X. If $B$ , a base of $U$ , satisfies the following properties.

(C1) For all $t < B (f)$ , there exists g ≤ f such that $t < B (g)$ and g ({ 0 }) + x = g ({ x }) for all x ∈ X;

(C2) For all $0 < B (f)$ ⇒ λf ≤ f for |λ|≤1, $f \in D (X)$ ;

(C3) For all x ∈ X, there exists k > 0, such that kx ∈ f ({ 0 }) for $0 < B (f)$ , $f \in D (X)$ .

Then there exists a fuzzifying topology $τ^{U}$ such that $(X, τ^{U})$ is a fuzzifying topological linear space. (For short, $τ^{U}$ is called a linear fuzzifying topology.)

Proof. We need to prove that the set function $N_{0}^{U} (\cdot)$ which appears in Lemma 2.10 satisfies conditions (P1)–(P6) in Theorem 2.6. (P1)-(P3) are trivial. We only prove (P4)-(P6).

(P4): We only need to show $N_{0}^{U} (W) \leq \underset{V + V \subseteq W}{⋁} N_{0}^{U} (V)$ for all W ⊆ X. From Lemma 2.10 and Definition 2.9, this is equivalent to $\underset{0 \notin f (X \ W)}{⋁} U (f) \leq \underset{V + V \subseteq W}{⋁} \underset{0 \notin f (X \ V)}{⋁} U (f) .$

Let $\underset{0 \notin f (X \ W)}{⋁} U (f) > t$ , there exists $f \in D (X)$ such that $U (f) > t$ with 0 ∉ f (X \ W). From the condition (C1), there exists $h \in D (X)$ with h ≤ f such that $B (h) > t$ and h ({0}) + x = h ({x}) for all x ∈ X. By (FQU3), there exists $g \in D (X)$ satisfing g ∘ g ≤ f such that $U (g) > t$ . From Definition 2.9 and condition (C1), we know that there exists g₁ ≤ g such that $t < B (g_{1})$ with g₁ ({0}) + x = g₁ ({x}) for all x ∈ X. Define V_{g
₁} = g₁ ({ 0 }). For all x, y ∈ V_{g
₁}. We get $\begin{matrix} x + y \in g_{1} ({0}) + y = g_{1} ({y}) \subseteq g_{1} \circ g_{1} ({0}) \\ \subseteq g \circ g ({0}) \subseteq f ({0}) . \end{matrix}$

For all x ∉ W, from the fact 0 ∉ f (X \ W) ⊇ h (X \ W) = ⋃ _y∈X\Wh ({ y }). We have 0 ∉ h ({y}) whenever y ∉ W. Then 0 ∉ h ({x}) = x + h ({0}). That is to say -x ∉ h ({0}). By the condition (C2), we have x ∉ h ({0}). Thus, we get h ({ 0 }) ⊆ W. Furthermore, V_{g
₁} + V_{g
₁} ⊆ W.

Finally, we show that 0 ∉ g₁ (X \ V_{g
₁}). If 0 ∈ g₁ (X \ V_{g
₁}), then 0 ∈ ⋃ _{y∈X\V_{g
₁}}g₁ ({ y }). On one hand, there exists y ∈ X \ V_{g
₁} such that y ∉ g₁ ({ 0 }). On the other hand, 0 ∈ g₁ ({ y }) = g₁ ({ 0 }) + y, so -y ∈ g₁ ({ 0 }). From (C2), it deduces a contradiction. Then, $\begin{matrix} t < B (g_{1}) \leq U (g_{1}) \leq \underset{0 \notin f (X \ V_{g_{1}})}{⋁} U (f) \leq \underset{V + V \subseteq W}{⋁} \\ \underset{0 \notin f (X \ V)}{⋁} U (f) . \end{matrix}$

(P5): For any fixed U ⊆ X with $N_{0}^{U} (U) > 0$ , there exist $g, f \in D (X)$ such that $B (g) > 0$ with 0 ∉ f (X \ U) and g ≤ f. By (C3), we know that for all x ∈ X, there exists t > 0 such that tx ∈ g ({0}) ⊆ f ({0}). Taking the same method used in the proof of (P4), we get tx ∈ f ({0}) ⊆ U for some t > 0.

(P6): From $N_{0}^{U} (U) > t$ , we get $\underset{0 \notin f (X \ U)}{⋁} \underset{h \leq f}{⋁} B (h) > t$ , i.e., there exists h ≤ f such that $B (h) > t$ for 0 ∉ f (X \ U). Similarly to the proof of (P4), we can also construct V_{g
₁} = g₁ ({ 0 }), where g₁ ≤ g, $t < B (g_{1})$ and g ∘ g ≤ h. By (C2), it is obvious that V_{g
₁} is balanced. From (C1) and the proof of (P4), we obtain 0 ∉ g (X \ V_{g
₁}) and there exists $k \in D (X)$ with k ≤ g₁ such that $t < B (k)$ . Thus, $\underset{0 \notin g_{1} (X \ V_{g_{1}})}{⋁} \underset{k \leq g_{1}}{⋁} B (k) > t$ , i.e., $N_{0}^{U} (V_{g_{1}}) > t$ .

Therefore, there exists a fuzzifying topology $τ^{U}$ such that $(X, τ^{U})$ is a fuzzifying topological linear space and $N_{0}^{U}$ is a fuzzifying neighborhood system of the neutral element. This completes the proof.

Definition 3.2. Let X be a linear space on $K$ and $U$ be a fuzzifying uniformity. If $B$ , a base of $U$ , satisfies (C1)-(C3), then we call $U$ a linear fuzzifying uniformity.

Theorem 3.3. Suppose that τ is a linear fuzzifying topology on X. Let $U^{τ} : D (X) \to [0, 1]$ the mapping defined as follows: $U^{τ} (f) = \underset{k \leq f}{⋁} \underset{g_{V} \leq k}{⋁} N_{0} (V)$ for all $f \in D (X)$ , where g_V (A) =⋃ _y∈A { x ∈ X : x - y ∈ V } for all nonempty A ∈ 2^X and g_V (∅) = ∅, where $k \in D (X)$ . Then $U^{τ}$ is a linear fuzzifying uniformity. Moreover, $τ = τ^{U^{τ}}$ .

Proof. First, Let $B (k) = \underset{g_{V} \leq k}{⋁} N_{0} (V) .$ We will prove that $U^{τ}$ satisfies (C1)–(C3) in Theorem 3.1 at first. (C1) Let $B (k) > t$ , then there exists g_V ≤ k such that $N_{0} (V) > t$ . Thus, $\begin{matrix} z \in g_{V} ({0}) + x \Leftrightarrow z - x \in g_{V} ({0}) \Leftrightarrow z \in x \\ + V \Leftrightarrow z \in g_{V} ({x}) . \end{matrix}$ It is obvious that $B (g_{V}) > t$ . That means (C1) holds. By the linearity of τ, it is easy to see that (C2) and (C3) hold.

Next, we prove that $U^{τ}$ satisfies the conditions (FQU1)–(FQU4) in Definition 2.8.

(FQU1) and (FQU2) are trivial. In the following, the conditions (FQU3) and (FQU4) will be verified.

(FQU3): From (FQU2), $U^{τ} (f) \geq \underset{h \circ h \leq f}{⋁} U^{τ} (h)$ is obvious. We need to show $U^{τ} (f) \leq \underset{h \circ h \leq f}{⋁} U^{τ} (h)$ . That is $\begin{matrix} \underset{k \leq f}{⋁} \underset{g_{V} \leq k}{⋁} N_{0} (V) \leq \underset{h \circ h \leq f}{⋁} \underset{k \leq h}{⋁} \underset{g_{V} \leq k}{⋁} N_{0} (V) . \end{matrix}$

Let $t < \underset{k \leq f}{⋁} \underset{g_{V} \leq k}{⋁} N_{0} (V)$ . There exist k ≤ f and g_V ≤ k such that $N_{0} (V) > t$ . By the linearity of τ, we have $N_{0} (V) \leq \underset{W + W \subseteq V}{⋁} N_{0} (W)$ , i.e., there exists W ∈ 2^X such that $N_{0} (W) > t$ with W + W ⊆ V. We define a mapping g_W : 2^X → 2^X given by g_W (A) =⋃ _y∈A { x ∈ X : x - y ∈ W } for all nonempty A ∈ 2^X and g_W (∅) = ∅. $g_{W} \in D (X)$ is obvious and g_W ({ 0 }) = W. For all x ∈ g_W ∘ g_W ({ 0 }), there exists y ∈ g_W ({ 0 }) such that x ∈ g_W ({ y }), then x - y ∈ W. Furthermore, x ∈ W + W ⊆ V. Thus, g_W ∘ g_W ≤ g_V ≤ f. And this implies $t < N_{0} (W) \leq ⋁_{h \circ h \leq f} \underset{k \leq h}{⋁} \underset{g_{W} \leq k}{⋁} N_{0} (W)$ , i.e., $U^{τ} (f) \leq \underset{h \circ h \leq f}{⋁} U^{τ} (h)$ .

(FQU4): We need to prove $U^{τ} (f) = U^{τ} (f^{)})$ , i.e., $\begin{matrix} \underset{k \leq f}{⋁} \underset{g_{V} \leq k}{⋁} N_{0} (V) = \underset{k \leq f^{)}}{⋁} \underset{g_{V} \leq k}{⋁} N_{0} (V) . \end{matrix}$

First, we prove $\begin{matrix} \underset{k \leq f}{⋁} \underset{g_{V} \leq k}{⋁} N_{0} (V) \leq \underset{k \leq f^{)}}{⋁} \underset{g_{V} \leq k}{⋁} N_{0} (V) . \end{matrix}$ Let $t < \underset{k \leq f}{⋁} \underset{g_{V} \leq k}{⋁} N_{0} (V)$ . Then there exist k ≤ f and g_V ≤ k such that $t < N_{0} (V)$ . For all $y \notin g_{V}^{)} ({0}) = ⋂ {C \in 2^{X} | g_{V} (X \ C) \subseteq X \ {0}}$ , there is some C ∈ 2^X with g_V (X\ C) ⊆ X \ { 0 } such that y ∉ C. Since 0 ∉ g_V (X \ C), we obtain 0 ∉ g_V ({ y }). Moreover, we get y ∉ g_V ({ 0 }) as g_V is balanced. Hence, we have $g_{V}^{)} ({0}) \supseteq g_{V} ({0})$ . This implies $g_{V} \leq g_{V}^{)} \leq f^{)}$ . Thus, we get $t < \underset{k \leq f^{)}}{⋁} \underset{g_{V} \leq k}{⋁} N_{0} (V)$ .

Similarly, we obtain $\underset{k \leq f}{⋁} \underset{g_{V} \leq k}{⋁} N_{0} (V) \geq \underset{k \leq f^{)}}{⋁} \underset{g_{V} \leq k}{⋁} N_{0} (V)$ .

Thus, $U^{τ}$ is a linear fuzzifying uniformity.

Finally, we need to prove that for every U ⊆ X, $\begin{matrix} N_{0} (U) = \underset{0 \notin f (X \ U)}{⋁} \underset{k \leq f}{⋁} \underset{g_{V} \leq k}{⋁} N_{0} (V) . \end{matrix}$

The inequality $N_{0} (U) \geq \underset{0 \notin f (X \ U)}{⋁} \underset{k \leq f}{⋁} \underset{g_{V} \leq k}{⋁} N_{0} (V)$ is obvious. We only have to show $\begin{matrix} N_{0} (U) \leq \underset{0 \notin f (X \ U)}{⋁} \underset{k \leq f}{⋁} \underset{g_{V} \leq k}{⋁} N_{0} (V) . \end{matrix}$ Let $N_{0} (U) > t$ , from (P4) in Theorem 2.6, we get $N_{0} (U) \leq \underset{V + V \subseteq U}{⋁} N_{0} (V)$ . Then there exists V ∈ 2^X such that $N_{0} (V) > t$ for V + V ⊆ U. Define the mappings g_U, g_V : 2^X → 2^X as previous part. From the properties of fuzzifying topological linear spaces, we may suppose the set U is balanced, this deduces that 0 ∉ g_U (X \ U). Moreover, g_V ≤ g_U. Thus, we have $N_{0} (U) \leq \underset{0 \notin h (X \ U)}{⋁} \underset{k \leq h}{⋁} \underset{g_{V} \leq k}{⋁} N_{0} (V)$ , as desired.

Theorem 3.4. Let $U$ be a linear fuzzifying uniformity on X. Then $U = U^{τ^{U}}$ , where $τ^{U}$ as given in Theorem 3.1.

Proof. We need to show $U = U^{τ^{U}}$ , i.e., $\begin{matrix} U (f) = \underset{k \leq f}{⋁} \underset{g_{V} \leq k}{⋁} \underset{0 \notin g (X \ V)}{⋁} U (g) \end{matrix}$ for all $f \in D (X)$ . The inequality $U (f) \geq \underset{k \leq f}{⋁} \underset{g_{V} \leq k}{⋁} \underset{0 \notin g (X \ V)}{⋁} U (g)$ is obvious. We only need to prove $\begin{matrix} U (f) \leq \underset{k \leq f}{⋁} \underset{g_{V} \leq k}{⋁} \underset{0 \notin g (X \ V)}{⋁} U (g) . \end{matrix}$

Since $U$ be a linear fuzzifying uniformity on X, by Theorem 3.1 and Theorem 3.3, there exists a base $B$ of $U$ which satisfies conditions (C1)-(C3). Then for any $t < U (f)$ , there exists $g \in D (X)$ such that g ≤ f and $t < B (g)$ . Let V = g ({ 0 }). It is obvious that 0 ∉ g (X \ V) and g = g_V. Thus $t < \underset{k \leq f}{⋁} \underset{g_{V} \leq k}{⋁} \underset{0 \notin g (X \ V)}{⋁} U (g)$ . This completes the proof.

Theorem 3.5. Let τ_i (i = 1, 2) be linear fuzzifying topologies. If a linear mapping F : (X, τ₁) → (Y, τ₂) is continuous, then $F : (X, U^{τ_{1}}) \to (Y, U^{τ_{2}})$ is uniformly continuous.

Proof. From Definition 2.8 and Theorem 3.3, we need to obtain $\begin{matrix} \underset{k \leq F^{\leftarrow} (g)}{⋁} \underset{h_{V} \leq k}{⋁} N_{0}^{X} (V) \geq \underset{k \leq g}{⋁} \underset{h_{V} \leq k}{⋁} N_{0}^{Y} (V), \end{matrix}$ where $k, g, h_{V} \in D (X)$ .

Put $\underset{k \leq g}{⋁} \underset{h_{V} \leq k}{⋁} N_{0}^{Y} (V) > t$ . Then, there exist k ≤ g and h_V ≤ k such that $N_{0}^{Y} (V) > t$ . Since F : (X, τ₁) → (Y, τ₂) is continuous, we have $N_{0}^{X} (F^{\leftarrow} (V)) \geq N_{0}^{Y} (V)$ . Then $N_{0}^{X} (F^{\leftarrow} (V)) > t$ . Moreover, $\begin{matrix} F^{\leftarrow} (V) & = {x | x \in F^{\leftarrow} \circ h_{V} ({0})} \\ = {x | x \in F^{\leftarrow} (h_{V}) ({0})} . \end{matrix}$

Hence, by h_V ≤ k ≤ g, we assert F^← (h_V) ≤ F^← (k) ≤ F^← (g). Let k₁ = F^← (k), then $\begin{matrix} \underset{k_{1} \leq F^{\leftarrow} (g)}{⋁} \underset{F^{\leftarrow} (h_{V}) \leq k_{1}}{⋁} N_{0}^{X} (F^{\leftarrow} (V)) > t . \end{matrix}$

It is well known that the notion of T₂ separation axiom plays an important role in the theory of fuzzy topologies, and it is generalized to the theory of soft set and fuzzy soft setting [1 , 20]. However, to the best of our knowledge, there is no paper describing the T₂ separation axiom in fuzzifying uniformities up to now. Thus, we will discuss it as an important part in the following content.

Definition 3.6. Let Σ be the family of all fuzzifying uniform spaces. A unary predicate $T_{2} \in F (Σ)$ is called T₂-separation predicate if it satisfies the following condition: $\begin{matrix} T_{2} (X, U) : = \forall x \forall y ((x \in X) \land (y \in X) \land (x \neq y) \\ \to (\exists f \in D (X)) (f \in U) \land (x \notin f ({y}))) . \end{matrix}$

The degree to which $(X, U)$ satisfies T₂ separation axiom is defined by $\begin{matrix} [T_{2} (X, U)] = \underset{x \neq y}{⋀} \underset{x \notin f ({y})}{⋁} U (f) . \end{matrix}$

Example 3.7. Let $(X, U)$ be a fuzzifying uniform space, X ={ x, y }, $D (X) = {f_{0}, f_{x}, f_{y}, f_{1}}$ , where $\begin{matrix} f_{0} (U) = {\begin{matrix} \emptyset, U = \emptyset, \\ {x}, U = {x}, \\ {y}, U = {y}, \\ X, U = X, \end{matrix} f_{x} (U) = {\begin{matrix} \emptyset, U = \emptyset, \\ X, U = {x}, \\ {y}, U = {y}, \\ X, U = X, \end{matrix} \end{matrix}$ $\begin{matrix} f_{y} (U) = {\begin{matrix} \emptyset, U = \emptyset, \\ {x}, U = {x}, \\ X, U = {y}, \\ X, U = X, \end{matrix} f_{1} (U) = {\begin{matrix} \emptyset, U = \emptyset, \\ X, U = {x}, \\ X, U = {y}, \\ X, U = X . \end{matrix} \end{matrix}$

And $U : D (X) \to [0, 1]$ defined as follows: $\begin{matrix} U (f) = {\begin{matrix} 0, f = f_{0}, \\ \frac{1}{2}, f = f_{x}, \\ \frac{1}{2}, f = f_{y}, \\ 1, f = f_{1} . \end{matrix} \end{matrix}$

Then it is easy to check that $(X, U)$ is a fuzzifying uniform space and $[T_{2} (X, U)] = \frac{1}{2}$ .

If we define $\begin{matrix} U_{1} (f) = {\begin{matrix} 0, f = f_{0}, \\ 1, f = f_{x}, \\ 1, f = f_{y}, \\ 1, f = f_{1} . \end{matrix} \end{matrix}$

Then $[T_{2} (X, U_{1})] = 1$ and $(X, U_{1})$ satisfies T₂ separation axiom.

Theorem 3.8. Let $(X, U)$ be a fuzzifying uniform space and $U_{r} = {f \in D (X) | U (f) > r}$ . Then $[T_{2} (X, U)] = λ$ (λ ∈ [0, 1]) if and only if for each r < λ, $(X, U_{r})$ is Hausdorff and $(X, U_{r})$ is not Hausdorff for each r > λ.

Proof. Necessity. From Ref. [26, Theorem 5.7], we know that $U_{r}$ is a fuzzifying uniformity. For each r < λ and x ≠ y, since $[T_{2} (X, U)] = λ = \underset{x \neq y}{⋀} \underset{x \notin f ({y})}{⋁} U (f)$ , there is a $f \in D (X)$ such that $U (f) > r$ for x ∉ f ({ y }). By (FQU3), there exists $g \in D (X)$ such that $U (g) > r$ for g ∘ g ≤ f. Obviously, $g \in U_{r}$ and x ∉ g ({ y }). It also means that $⋂_{g \in U_{r}} g ({y}) \subseteq {y}$ . Hence, $⋂_{g \in U_{r}} g ({y}) = {y}$ . From the definition of classical Hutton type uniformity (see [10]), we know that the T₂ separation axiom in classical Hutton type uniform space $(X, U_{r})$ can be described as $⋂_{f \in U_{r}} f ({x}) = {x}$ . This follows that $(X, U_{r})$ is Hausdorff. On the other hand, if there exists r₀ > λ such that $(X, U_{r_{0}})$ is Hausdorff, we can find a contradiction with the hypothesis for $[T_{2} (X, U)] \geq r_{0} > λ$ .

Sufficiency. We easily get that $[T_{2} (X, U)] \geq λ$ since $(X, U_{r})$ is Hausdorff. It suffices to prove $[T_{2} (X, U)] \leq λ$ . Otherwise, if $[T_{2} (X, U)] > λ$ , from the proof of necessity, we can obtain a contradiction. Hence, $[T_{2} (X, U)] = λ$ . This completes the proof.

Theorem 3.9. Let $(X, U)$ be a linear fuzzifying uniform space and $(X, τ^{U})$ a fuzzifying topological linear space. Then $\begin{matrix} [T_{2} (X, τ^{U})] = [T_{2} (X, U)] . \end{matrix}$

Proof. We need to obtain $\begin{matrix} \underset{x \neq 0}{⋀} \underset{U \notin \dot{x}}{⋁} {N_{0}}^{U} (U) = \underset{x \neq 0}{⋀} \underset{x \notin f ({0})}{⋁} U (f) . \end{matrix}$

First, we prove $\underset{x \neq 0}{⋀} \underset{U \notin \dot{x}}{⋁} {N_{0}}^{U} (U) \leq \underset{x \neq 0}{⋀} \underset{x \notin f ({0})}{⋁} U (f)$ . Let $t < \underset{x \neq 0}{⋀} \underset{U \notin \dot{x}}{⋁} {N_{0}}^{U} (U)$ . Then there exists $U \notin \dot{x}$ such that ${N_{0}}^{U} (U) > t$ for all x ≠ 0. By ${N_{0}}^{U} (U) = \underset{0 \notin g (X \ U)}{⋁} U (g)$ , we obtain $\underset{x \neq 0}{⋀} \underset{x \notin g ({0})}{⋁} U (g) > t$ , i.e., $\underset{x \neq 0}{⋀} \underset{U \notin \dot{x}}{⋁} {N_{0}}^{U} (U) \leq \underset{x \neq 0}{⋀} \underset{x \notin g ({0})}{⋁} U (g)$ .

Next, we show $\underset{x \neq 0}{⋀} \underset{U \notin \dot{x}}{⋁} {N_{0}}^{U} (U) \geq \underset{x \neq 0}{⋀} \underset{x \notin f ({0})}{⋁} U (f)$ . Let $t < \underset{x \neq 0}{⋀} \underset{x \notin f ({0})}{⋁} U (f)$ , there exists x ∉ f ({ 0 }) such that $U (f) > t$ for all x ≠ 0. Define U_f = f ({ 0 }). Then $U_{f} \notin \dot{x}$ . Similarly, a mapping generated by U_f can be defined as: f_{U
_f} : 2^X → 2^X, f_{U
_f} (A) =⋃ _y∈A { x ∈ X : x - y ∈ U_f } for all nonempty A ∈ 2^X and f_{U
_f} (∅) = ∅. It is obvious that $f_{U_{f}} \in D (X)$ and f_{U
_f} ({ 0 }) = U_f. Since $\begin{matrix} y \in f_{U_{f}} (A) \Leftrightarrow \exists x \in A, y - x \in f ({0}) \Leftrightarrow \exists x \in A, \\ y \in f ({x}) \Leftrightarrow y \in f (A), \end{matrix}$ we obtain f_{U
_f} = f. From ${N_{0}}^{U} (U_{f}) = \underset{0 \notin f_{U_{f}} (X \ U_{f})}{⋁} U (f_{U_{f}})$ , we have $\underset{x \neq 0}{⋀} \underset{U_{f} \notin \dot{x}}{⋁} {N_{0}}^{U} (U_{f}) \geq \underset{x \neq 0}{⋀} \underset{x \notin f ({0})}{⋁} U (f)$ . This completes the proof.

Definition 3.10. Let $(X, U)$ be a linear fuzzifying uniform space. Then the unary fuzzy predicates ${BD}_{U} \in F (2^{X})$ , called fuzzy complete boundedness, is defined as follows: $\begin{matrix} A \in {BD}_{U} : = \forall (f \in D (X)) ((f \in U) \to \exists \\ (F \in 2^{(A)}) \land (A \subseteq F + f {(0)})) . \end{matrix}$

And the degree to which A is complete bounded is $\begin{matrix} [{BD}_{U} (A)] = \underset{f \in D (X)}{⋀} {1 - U (f) | \forall F \in 2^{(A)}, \\ A ⊈ F + f ({0})} . \end{matrix}$

Theorem 3.11. Let $(X, U)$ be a linear fuzzifying uniform space and $(X, τ^{U})$ a fuzzifying topological linear space. Then $[{BD}_{U} (A)] = [BD (A)]$ for all A ⊆ X.

Proof. We need to show $\begin{matrix} \underset{f \in D (X)}{⋀} {1 - U (f) | \forall F \in 2^{(A)}, A ⊈ F + f ({0})} \\ = \underset{U \subseteq X}{⋀} {1 - {N_{0}}^{U} (U) | \forall F \in 2^{(A)}, A ⊈ F + U} . \end{matrix}$

First, we show $\begin{matrix} \underset{f \in D (X)}{⋀} {1 - U (f) | \forall F \in 2^{(A)}, A ⊈ F + f ({0})} \\ \geq \underset{U \subseteq X}{⋀} {1 - {N_{0}}^{U} (U) | \forall F \in 2^{(A)}, A ⊈ F + U} . \end{matrix}$

Let $\underset{U \subseteq X}{⋀} {1 - {N_{0}}^{U} (U) | \forall F \in 2^{(A)}, A ⊈ F + U} > t$ , that is $1 - {N_{0}}^{U} (U) > t$ for A ⊈ F + U. Define a mapping f : 2^X → 2^X given by f (A) =⋃ _y∈A { x ∈ X : x - y ∈ U } for all nonempty A ∈ 2^X and f (∅) = ∅. It is obvious that $f \in D (X)$ and f ({ 0 }) = U. Then we have A ⊈ F + f ({ 0 }) and ${N_{0}}^{U} (U) = \underset{0 \notin f (X \ U)}{⋁} U (f) < 1 - t$ . Thus, $1 - U (f) > t$ , as desired.

Similarly, $\begin{matrix} \underset{f \in D (X)}{⋀} {1 - U (f) | \forall F \in 2^{(A)}, A ⊈ F + f ({0})} \\ \leq \underset{U \subseteq X}{⋀} {1 - {N_{0}}^{U} (U) | \forall F \in 2^{(A)}, A ⊈ F + U} . \end{matrix}$

This completes the proof.

4 Linear fuzzifying uniformities induced by fuzzy norms

In this section, we will discuss the linear fuzzifying uniformities induced by fuzzy norms. We now study the continuity of the mappings between linear fuzzifying uniform spaces and fuzzifying topological linear spaces.

Theorem 4.1. Let N be a fuzzy norm on a linear space X and define $N_{0}^{N} : 2^{X} \to [0, 1]$ as follows: $\begin{matrix} N_{0}^{N} (U) = \underset{ɛ > 0}{⋁} \underset{z \notin U}{⋀} (1 - N (z, ɛ)) \end{matrix}$ for all U ∈ 2^X. Then $N_{0}^{N} (\cdot)$ is a fuzzifying neighborhood system of the neutral element.

Proof. We only prove (P4)-(P6) and leave the other conditions to the reader.

(P4). For each U ∈ 2^X, what we need to show is $\begin{matrix} \underset{ɛ > 0}{⋁} \underset{z \notin U}{⋀} (1 - N (z, ɛ)) \leq \underset{V + V \subseteq U}{⋁} \underset{ɛ > 0}{⋁} \underset{z \notin V}{⋀} (1 - N (z, ɛ)) . \end{matrix}$

Let $t < \underset{ɛ > 0}{⋁} \underset{z \notin U}{⋀} (1 - N (z, ɛ))$ . Then there exists δ > 0 such that t < 1 - N (z, δ) for all z ∉ U. Hence $W = {x | N (x, \frac{δ}{2}) > 1 - t} \subseteq {x \in X | N (x, δ) > 1 - t} \subseteq U$ . Moreover, given x, y ∈ W then $N (x + y, δ) \geq N (x, \frac{δ}{2}) \land N (y, \frac{δ}{2}) > 1 - t$ , so x + y ∈ U. Consequently, W + W ⊆ U. Hence $N_{0} (W) = \underset{ɛ > 0}{⋁} \underset{z \notin W}{⋀} (1 - N (z, ɛ)) \geq \underset{z \notin W}{⋀} (1 - N (z, \frac{δ}{2})) > t$ . Therefore, $t < N_{0} (W) \leq \underset{V + V \subseteq U}{⋁} \underset{ɛ > 0}{⋁} \underset{z \notin V}{⋀} (1 - N (z, ɛ))$ , i.e., $\begin{matrix} N_{0}^{N} (U) \leq \underset{V + V \subseteq U}{⋁} N_{0}^{N} (V) . \end{matrix}$

(P5). For all U ∈ 2^X, x ∈ X with $N_{0}^{N} (U) = a > 0$ , there exists δ > 0 such that $N (z, δ) > 1 - \frac{a}{2}$ for all z ∉ U. Thus ${y \in X | N (y, δ) > 1 - \frac{a}{2}} \subseteq U$ . By the condition of fuzzy norm, there exists s₀ > 0 such that $N (x, s_{0}) > 1 - \frac{a}{2}$ . Put $t_{0} = \frac{δ}{s_{0}} > 0$ , then for all |t| < t₀, if t ≠ 0, $N (tx, δ) = N (x, \frac{δ}{| t |}) \geq N (x, s_{0}) > 1 - \frac{a}{2}$ . In the case t = 0, $N (tx, δ) = N (0, δ) = 1 > 1 - \frac{a}{2}$ . This means that ta ∈ U.

(P6). For each U ∈ 2^X with $N_{0}^{N} (U) > a > 0$ , there exists b > a such that $N_{0}^{N} (U) > b$ . Then there exists δ > 0 such that b < 1 - N (z, δ) for all z ∉ U. Hence $W = {x | N (x, \frac{δ}{2}) > 1 - b} \subseteq {x \in X | N (x, δ) > 1 - b} \subseteq U$ . This implies $N_{0}^{N} (W) \geq b > a$ . It is easy to find W is a balanced set.

Therefore, $N_{0}^{N} (\cdot)$ is a fuzzifying neighborhood system of neutral element.

Remark 4.2. By Theorem 2.6 and Theorem 4.1, for every fuzzy normed linear space (X, N), there is a fuzzifying topology on X such that this fuzzifying topology is compatible to linear operations. It will be denoted τ_N.

Similarly to the proof of Theorem 5.1 in [26], we can prove the following Theorem 4.3.

Theorem 4.3. Let N be a fuzzy norm on a linear space X, define $U^{N} : D (X) \to [0, 1]$ as follows: $\begin{matrix} U^{N} (f) = \underset{ɛ > 0}{⋁} \underset{x \in X}{⋀} \underset{y \notin f ({x})}{⋀} (1 - N (x - y, ɛ)) \end{matrix}$ for all $f \in D (X)$ . Then $U^{N}$ is a linear fuzzifying uniformity.

Proof. (FQU1) and (FQU2) are trivial. We prove (FQU3) and (FQU4).

(FQU3). From (FQU2), $U^{N} (f) \geq \underset{g \circ g \leq f}{⋁} U^{N} (g)$ is obvious. We need to show that $U^{N} (f) \leq \underset{g \circ g \leq f}{⋁} U^{N} (g)$ . Let $t < U^{N} (f)$ . Then there exists ɛ₀ > 0 such that 1 - N (x - y, ɛ₀) > t for all x ∈ X and y ∉ f ({ x }). Define g : 2^X → 2^X given by $\begin{matrix} g (A) = ⋃_{z \in A} {w \in X | N (z - w, \frac{ɛ_{0}}{2}) > 1 - t} \end{matrix}$ for all nonempty A ∈ 2^X and g (∅) = ∅. It is easy to check that $g \in D (X)$ . We can assert that g ∘ g ≤ f. In fact, let z ∈ g ∘ g ({ x }) = ⋃ _y∈g({x})g ({ y }). Then, there exists y ∈ g ({ x }) such that z ∈ g ({ y }). Therefore, we have $N (x - y, \frac{ɛ_{0}}{2}) > 1 - t$ and $N (y - z, \frac{ɛ_{0}}{2}) > 1 - t$ . Moreover, $\begin{matrix} N (x - z, ɛ_{0}) \geq N (x - y, \frac{ɛ_{0}}{2}) \\ \land N (y - z, \frac{ɛ_{0}}{2}) > 1 - t . \end{matrix}$

Then z ∈ f ({ x }). That is g ∘ g ({ x }) ⊆ f ({ x }), i.e., g ∘ g ≤ f. Hence, $\begin{matrix} U^{N} (g) & = & \underset{ɛ > 0}{⋁} \underset{x \in X}{⋀} \underset{y \notin g ({x})}{⋀} (1 - N (x - y, ɛ)) \\ \geq & \underset{x \in X}{⋀} \underset{y \notin g ({x})}{⋀} (1 - N (x - y, \frac{ɛ_{0}}{2})) \\ \geq & t . \end{matrix}$ Since t is arbitrary, we have $U^{N} (f) \leq \underset{g \circ g \leq f}{⋁} U^{N} (g)$ , as desired.

(FQU4). We need to prove $U^{N} (f) = U^{N} (f^{)})$ for all $f \in D (X)$ , i.e., $\begin{matrix} \underset{ɛ > 0}{⋁} \underset{x \in X}{⋀} \underset{y \notin f ({x})}{⋀} (1 - N (x - y, ɛ)) \\ = \underset{ɛ > 0}{⋁} \underset{x \in X}{⋀} \underset{y \notin f^{)} ({x})}{⋀} (1 - N (x - y, ɛ)) . \end{matrix}$ Let $t < \underset{ɛ > 0}{⋁} \underset{x \in X}{⋀} \underset{y \notin f ({x})}{⋀} (1 - N (x - y, ɛ))$ , then there exists ɛ₀ > 0 such that 1 - N (x - y, ɛ₀) > t for all x ∈ X and y ∉ f ({ x }). For all x ∈ X and $\begin{matrix} y \notin f^{)} ({x}) = ⋂ {C \in 2^{X} | f (X \ C) \leq X \ {x}}, \end{matrix}$ there exists C ∈ 2^X such that y ∉ C and x ∉ f (X \ C). Hence, x ∉ f ({ y }). Therefore, 1 - N (y - x, ɛ₀) = 1 - N (x - y, ɛ₀) > t. Thus, $\underset{ɛ > 0}{⋁} \underset{x \in X}{⋀} \underset{y \notin f^{)} ({x})}{⋀} (1 - N (x - y, ɛ)) > t$ . So, $\begin{matrix} \underset{ɛ > 0}{⋁} \underset{x \in X}{⋀} \underset{y \notin f ({x})}{⋀} (1 - N (x - y, ɛ)) \\ \leq \underset{ɛ > 0}{⋁} \underset{x \in X}{⋀} \underset{y \notin f^{)} ({x})}{⋀} (1 - N (x - y, ɛ)) . \end{matrix}$

Similarly, we have $\begin{matrix} \underset{ɛ > 0}{⋁} \underset{x \in X}{⋀} \underset{y \notin f ({x})}{⋀} (1 - N (x - y, ɛ)) \\ \geq \underset{ɛ > 0}{⋁} \underset{x \in X}{⋀} \underset{y \notin f^{)} ({x})}{⋀} (1 - N (x - y, ɛ)) . \end{matrix}$

To prove that $U^{N}$ is linear, we need to show that $N_{0}^{U^{N}}$ is equivalent to $N_{0}^{N}$ as defined in Theorem 4.1, i.e., $\begin{matrix} \underset{ɛ > 0}{⋁} \underset{z \notin U}{⋀} (1 - N (z, ɛ)) = \underset{0 \in f (X \ U)}{⋁} \underset{ɛ > 0}{⋁} \\ \underset{z \notin f ({0})}{⋀} (1 - N (z, ɛ)) . \end{matrix}$ Let $t < \underset{ɛ > 0}{⋁} \underset{z \notin U}{⋀} (1 - N (z, ɛ))$ , then there exists ɛ > 0 such that 1 - N (z, ɛ) > t for all z ∉ U. Let V = {x|N (x, ɛ) >1 - t}, clearly, V ⊆ U and V is balanced. Define a mapping f_V : 2^X → 2^X given by f_V (A) =⋃ _y∈A { x ∈ X : x - y ∈ V } for all nonempty A ∈ 2^X and f_V (∅) = ∅. Obviously, $f_{V} \in D (X)$ and f_V ({ 0 }) = V. Taking the same method as the proof in Theorem 3.1, we have 0 ∉ f_V (X \ V) ⊇ f_V (X \ U). Then for all z ∉ f_V ({ 0 }) = V, 1 - N (x, ɛ) ≥ t. Then $\begin{matrix} \underset{ɛ > 0}{⋁} \underset{z \notin U}{⋀} (1 - N (z, ɛ)) \leq \underset{0 \in f (X \ U)}{⋀} \\ \underset{ɛ > 0}{⋁} \underset{z \notin f ({0})}{⋀} (1 - N (z, ɛ)), \end{matrix}$ as desired. Similarly, we obtain $\begin{matrix} \underset{ɛ > 0}{⋁} \underset{z \notin U}{⋀} (1 - N (z, ɛ)) \geq \underset{0 \in f (X \ U)}{⋀} \\ \underset{ɛ > 0}{⋁} \underset{z \notin f ({0})}{⋀} (1 - N (z, ɛ)) . \end{matrix}$

Thus, we can assert that $U^{N}$ is a linear fuzzifying uniformity. The proof is complete.

Corollary 4.4. Let N be a fuzzy norm and $U^{N}$ be a linear fuzzifying uniformity induced by N. Then they generate the same linear fuzzifying topology.

Theorem 4.5. Let (X, N₁) and (Y, N₂) be fuzzy normed linear spaces and F : (X, N₁) → (Y, N₂) be a strongly fuzzy continuous mapping. Then $F : (X, U^{N_{1}}) \to (Y, U^{N_{2}})$ is uniformly continuous.

Proof. For each $g \in D (Y)$ , we need to show $\begin{matrix} \underset{ɛ > 0}{⋁} \underset{x_{1} \in X}{⋀} \underset{x_{2} \notin F^{\leftarrow} (g) ({x_{1}})}{⋀} (1 - N_{1} (x_{1} - x_{2}, ɛ)) \\ \geq \underset{ɛ > 0}{⋁} \underset{y_{1} \in Y}{⋀} \underset{y_{2} \notin g ({y_{1}})}{⋀} (1 - N_{2} (y_{1} - y_{2}, ɛ)) . \end{matrix}$

By Theorem 3.3 and Theorem 4.3, we may suppose $B_{2}$ is a base of $U^{N_{2}}$ . Then for any $t < \underset{ɛ > 0}{⋁} \underset{y_{1} \in Y}{⋀} \underset{y_{2} \notin g ({y_{1}})}{⋀} (1 - N_{2} (y_{1} - y_{2}, ɛ))$ , there exists g₁ ≤ g with $B_{2} (g_{1}) > t$ and g₁ ({y}) = y + g₁ ({0}) for all y ∈ Y. Thus $t < U^{N_{2}} (g_{1}) = \underset{ɛ > 0}{⋁} \underset{y_{1} \in Y}{⋀} \underset{y_{2} \notin g_{1} ({y_{1}})}{⋀} (1 - N_{2} (y_{1} - y_{2}, ɛ))$ . Hence there exists ɛ₀ > 0 such that N₂ (y₁ - y₂, ɛ₀) < 1 - t for all y₁ ∈ Y, y₂ ∉ g₁ ({ y₁ }). Since F : (X, N₁) → (Y, N₂) is strongly continuous, then for above ɛ₀ > 0, and each x₁ ∈ X, x₂ ∉ F^← (g₁) ({ x₁ }), there exists δ_{ɛ
₀} > 0 such that N₂ (F (x₁) - F (x₂) , ɛ₀) ≥ N₁ (x₁ - x₂, δ_{ɛ
₀}). Since x₂ ∉ F^← (g₁) ({ x₁ }), it implies F (x₂) ∉ g₁ ({F (x₁)}). Then N₂ (F (x₁) - F (x₂) , ɛ₀) < 1 - t. So N₁ (x₁ - x₂, δ_{ɛ
₀}) <1 - t. This implies $U^{N_{1}} (F^{\leftarrow} (g_{1})) \geq t$ . Furthermore, $U^{N_{1}} (F^{\leftarrow} (g)) \geq U^{N_{1}} (F^{\leftarrow} (g_{1})) \geq t$ . This completes the proof $U^{N_{2}} (g) \leq U^{N_{1}} (F^{\leftarrow} (g))$ . Hence $F : (X, U^{N_{1}}) \to (Y, U^{N_{2}})$ is uniformly continuous.

Theorem 4.6. Let (X, N₁) and (Y, N₂) be fuzzy normed linear spaces and $F : (X, U^{N_{1}}) \to (Y, U^{N_{2}})$ be a mapping. Then [UC (F)] ≤ [C (F)].

Proof. From the Definition 2.3 and Definition 2.11, for all x ∈ X, $g \in D (Y)$ , U ∈ 2^Y, taking the notation as Theorem 4.1, $N_{x}^{X} (\cdot), N_{F (x)}^{Y} (\cdot)$ denote the fuzzifying neighborhood system of x and F (x) respectively. We need to prove the following inequality: $\begin{matrix} min {1, 1 - U^{N_{2}} (g) + U^{N_{1}} (F^{\leftarrow} (g))} \\ \leq min {1, 1 - N_{F (x)}^{Y} (U) + N_{x}^{X} (F^{\leftarrow} (U))} . \end{matrix}$

Case 1. $N_{F (x)}^{Y} (U) \leq N_{x}^{X} (F^{\leftarrow} (U))$ . This case is obvious.

Case 2. $N_{F (x)}^{Y} (U) > N_{x}^{X} (F^{\leftarrow} (U))$ . From Lemma 2.10, we have $N_{F (x)}^{Y} (U) = \underset{F (x) \notin g (Y \ U)}{⋁} U^{N_{2}} (g) and$ $\begin{matrix} N_{x}^{X} (F^{\leftarrow} (U)) = \underset{x \notin F (X \ F^{\leftarrow} (U))}{⋁} U^{N_{2}} (g) \\ \geq \underset{x \notin F^{\leftarrow} (g) (X \ F^{\leftarrow} (U))}{⋁} U^{N_{1}} (F^{\leftarrow} (g)) . \end{matrix}$ Then, $\begin{matrix} N_{F (x)}^{Y} (U) - N_{x}^{X} (F^{\leftarrow} (U)) \\ = \underset{F (x) \notin g (Y \ U)}{⋁} U^{N_{2}} (g) - \underset{x \notin F^{\leftarrow} (g) (X \ F^{\leftarrow} (U))}{⋁} U^{N_{1}} (F^{\leftarrow} (g)) \\ = \underset{F (x) \notin g (Y \ U)}{⋁} U^{N_{2}} (g) - \underset{F (x) \notin g \circ F (X \ F^{\leftarrow} (U))}{⋁} U^{N_{1}} (F^{\leftarrow} (g)) . \end{matrix}$ Next, we show g (Y \ U) ⊇ g ∘ F (X \ F^← (U)), i.e., Y \ U ⊇ F (X \ F^← (U)).

If F (x) = y ∈ U, then F (x) ∉ Y \ U. On the other hand, x ∈ F^← (U). Hence, F (x) ∉ F (X \ F^← (U)) . Then we get g (Y \ U) ⊇ g ∘ F (X \ F^← (U)). Therefore, $\begin{matrix} N_{F (x)}^{Y} (U) - N_{x}^{X} (F^{\leftarrow} (U)) \leq \underset{F (x) \notin g (Y \ U)}{⋁} \\ (U^{N_{2}} (g) - U^{N_{1}} (F^{\leftarrow} (g))) . \end{matrix}$

Then, $1 - U^{N_{2}} (g) + U^{N_{1}} (F^{\leftarrow} (g)) \leq 1 - N_{F (x)}^{Y} (U) + N_{x}^{X} (F^{\leftarrow} (U)),$ as desired. The proof is completed.

As a directly consequence of Theorem 3.5, we have the following corollary.

Corollary 4.7. Let (X, N₁) and (Y, N₂) be fuzzy normed linear spaces and a linear mapping F : (X, τ_{N
₁}) → (Y, τ_{N
₂}) be continuous. Then $F : (X, U^{N_{1}}) \to (Y, U^{N_{2}})$ is uniformly continuous.

5 Conclusion

In the present paper, we have discussed the condition which a fuzzifying uniformity is compatible with the linear structure. Also, we studied the Hausdorff separation and complete boundedness in linear fuzzifying uniform spaces. As an example, the linear fuzzifying uniformity induced by a fuzzy norm was discussed. In fact, the linear uniformity constitutes an efficient tool to discuss the metrizable of topological linear spaces. It is a natural way to study the fuzzy metrizable of fuzzifying topological linear spaces and fuzzifying bornological linear spaces based on linear fuzzifying uniformites, it is left for further work.

Footnotes

Acknowledgments

The authors are grateful to the referees and Editorial Board for their valuable comments and helpful suggestions in modifying this paper. We also acknowledge the support of National Natural Science Foundation of China under Grant No. 11571006, No. 12071225 and Priority Academic Program Development of Jiangsu Higher Education Institutions.

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