Role of bacterial biofertilizers in agriculture and forestry

1.   Introduction

Data is a valuable source of knowledge that contains helpful information if exploited effectively ^[1]. One of the challenges facing data researchers is the ambiguity and uncertainty of the data they have access to, which makes it difficult for them to process information. But these challenges are, in a positive sense, opportunities for the development of new techniques and tools, such as they various approaches based on fuzzy set theory ^[2]. The advent of fuzzy theory has prompted extensive work on ideas such as fuzzy sets ^[3], vague sets ^[4], soft sets ^[5], and neutrosophic sets ^[6]. It was originally thought that the development of new theories would eclipse fuzzy theory, but that does not seem to be the case ^[7]. This research field is becoming more and more active, with a number of fundamental contributions to the rapid development of new theories ^[8,9]. One of the most prominent applications is the use of fuzzy set theory in emerging and vibrant fields like machine learning ^[10,11] or topological data analysis ^[12,13].

In recent years, the study of soft sets ^[5] and neutrosophic sets ^[14] has become an attractive research area. Neutrosophic sets recently emerged as a tool for dealing with imprecise, indeterminate, and inconsistent data ^[15]. In contrast, soft sets show potential for dealing with uncertainties that classical methods cannot control ^[16]. Combining these two types of sets results in a unique hybrid structure, a neutrosophic soft set (NS-set) ^[17], for working effectively in uncertain environments. Maji proposed this ^[17,18] in 2013 and it was modified by Deli and Broumi ^[19] in 2015. Furthermore, Karaaslan ^[20] redefined this concept and its operations to be more efficient and complete. Since then, this structure has proved to be quite effective when applied in real life in many fields, such as decision making ^[17], market prediction ^[21], and medical diagnosis ^[22,23].

The topology on NS-sets is one of the issues that needs more attention, alongside neutrosophic topology ^[24,25] and soft topology ^[26]. This issue has emerged recently to help complete the overall picture for NS and aid its practical applications based on topology ^[27,28]. In 2017, Bera and Mahapatra ^[29] gave general operations to construct a topology on NS-sets. They also presented concepts related to topological space such as interior, closure, neighborhood, boundary, regularity, base, subspace, separation axioms, along with specific illustrations and proofs. In 2018, these authors ^[30] continued to develop further studies on connectedness and compactness on NS-topological space. In 2019, Ozturk, Aras, and Bayramov ^[31] introduced a new approach to topology on NS-sets. This approach is quite different from the previous work ^[29], and was further developed by constructing separation axioms ^[32] in the same year, 2018. Recently, the continuum ^[33] or compactness ^[34] on the topological space generated on NS-sets has also been studied with the same properties as the normal space. Many variations ^[35] of the topological space on NS-sets have also attracted the attention of researchers, and most of the related works are inspired by topology on neutrosophic and soft sets with the idea of a hybrid structure ^[36,37].

In this work, we construct the topological space and related concepts on NS-sets through general operations in a way that is very different from the work of Bera and Mahapatra ^[29,30], but more general than the work of Ozturk, Aras, and Bayramov ^[31,32], with our operations based on the generality of min and max operations. This work begins by defining two new operations to create the relationships between NS-sets. These relations are then used as the kernel for forming topology and topological relations on NS-sets. One emphasis shown here is on elucidating the relationship between the topology on NS-sets and the component fuzzy soft topologies. All the ideas in this work are presented convincingly and clearly through definitions, theorems, and their consequences.

In summary, the significant contributions of this study are as follows:

(1) Defining two novel concepts, called $ {\rm{min}}-{\rm{n}}{\rm{o}}{\rm{r}}{\rm{m}} $ and $ {\rm{max}}-{\rm{n}}{\rm{o}}{\rm{r}}{\rm{m}} $, to provide a theoretical foundation for determining operations on NS-sets, including intersection, union, difference, AND, and OR.

(2) Constructing the topology, open set, closed set, interior, closure, and regularity concepts on NS-sets based on just determined operations.

(3) Elucidating the relationship between the topology on NS-sets and the fuzzy soft topologies generated by truth, indeterminacy, falsity degrees by the theorems and counterexamples.

(4) The concepts are well-defined, and the theorems are proved convincingly and logically.

This work is organized as follows: Section 1 presents the motivation and introduces the significant contributions. Section 2 briefly introduces NS-sets and related concepts. The two new ideas, $ {\rm{min}}-{\rm{n}}{\rm{o}}{\rm{r}}{\rm{m}} $ and $ {\rm{max}}-{\rm{n}}{\rm{o}}{\rm{r}}{\rm{m}} $, are provided in Section 3 as a theoretical foundation for determining operations on NS-sets, including intersection, union, difference, AND, and OR. In Section 4, the topology on NS-sets is defined with related concepts such as open set, closed set, interior, closure, and regularity. Furthermore, the relationship between the topology on NS-sets and the fuzzy soft topologies generated by truth, indeterminacy, and falsity functions by theorems and counterexamples in Section 5. The last section presents conclusions and future research trends in this area.

2.   Preliminaries

This section recalls the NS-set proposed in 2013 by Maji ^[17,18], then modified and improved in 2015 by Deli and Broumi ^[19]. This concept is based on combining soft ^[5] and neutrosophic ^[6] sets. Some background related to NS-sets is briefly presented below so that readers can better understand the following sections.

Without loss of generality, we consider $ X $ to be a universal set, $ \mathcal{E} $ to be a parameter set, and $ \mathcal{N}\left(X\right) $ to denote the collection of all neutrosophic sets on $ X $.

Definition 1. (^[18,19]). The pair $ \left(A, \mathcal{E}\right) $ is a NS-set on $ X $ where $ A:\mathcal{E}⟶\mathcal{N}\left(X\right) $ is a set valued function determined by $ e⟼A\left(e\right)≔{A}_{e} $ with

for all $ e\in \mathcal{E} $, and the real function triples $ {T}_{{A}_{e}}, {I}_{{A}_{e}}, {F}_{{A}_{e}}:X⟶{]}^{-}0;{1}^{+}[$ indicate truth, indeterminacy, and falsity degrees, respectively, with no restriction on their sum.

In other words, the NS-set can be described as a set of ordered tuples as follows:

If nothing changes, the symbol $ \mathcal{N}\mathcal{S}\left(X\right) $ indicates the collection of all NS-sets on $ X $. Besides, if the NS-sets consider the same parameter set $ \mathcal{E} $, then it is not mentioned repeatedly in order to simplify the notations. Moreover, because the values of $ T $, $ I $, $ F $ belong to the unit interval $ \left[0;1\right] $, the integral part of the values is almost zero. Typically, it may occur that the integer part is omitted (for example, $.1 $ instead of $ 0.1 $). Therefore, if it does not lead to confusion, this omitted format of a decimal is always used in all the tables used in this paper.

Definition 2. (^[18,19]).

a. $ {\varnothing }_{\mathcal{E}} $ is a null NS-set if

b. $ {\varnothing }_{\widetilde{\mathcal{E}}} $ is a semi-null NS-set if

c. $ {X}_{\mathcal{E}} $ is an absolute NS-set if

d. $ {X}_{\widetilde{\mathcal{E}}} $ is a semi-absolute NS-set if

Definition 3. (^[19,31]). Let $ A $ and $ B $ be two NS-sets on $ X $.

a. $ A $ is a NS-subset of $ B, {\rm{ }}{\rm{w}}{\rm{r}}{\rm{i}}{\rm{t}}{\rm{t}}{\rm{e}}{\rm{n}}{\rm{ }}{\rm{a}}{\rm{s}}A\subseteq B $, if

b. $ A $ is a NS-superset of $ B, {\rm{ }}{\rm{w}}{\rm{r}}{\rm{i}}{\rm{t}}{\rm{t}}{\rm{e}}{\rm{n}}{\rm{ }}{\rm{a}}{\rm{s}}A\supseteq B $, if $ B $ is a NS-subset of $ A $.

c. $ \overline{A} $ is the complement of $ A $ if

Example 1. Let two NS-sets $ M $ and $ N $ be represented in Table 1 as follows:

Based on Eq (9) of Definition 3, $ M\subseteq N $.

Example 2. The NS-set $ P $ and its complement $ \overline{P} $ are represented according to Eq (10) in Table 2 as follows:

Theorem 1. If $ A\in \mathcal{N}\mathcal{S}\left(X\right) $,

(1) $ \overline{\overline{A}} = A, $

(2) $ \overline{{\varnothing }_{\mathcal{E}}} = {X}_{\mathcal{E}}, $

(3) $ \overline{{\varnothing }_{\widetilde{\mathcal{E}}}} = {X}_{\widetilde{\mathcal{E}}}, $

(4) $ \overline{{X}_{\mathcal{E}}} = {\mathcal{\varnothing }}_{\mathcal{E}}, $

(5) $ \overline{{X}_{\widetilde{\mathcal{E}}}} = {\mathcal{\varnothing }}_{\widetilde{\mathcal{E}}}. $

Proof. These properties are directly inferred from the definitions of the null, semi-null, absolute, semi-absolute NS-sets and the complement operation.

3.   Another novel approach for operations on NS-sets

In this section, we focus on defining two novel norms, called $ \mathit{min}-norm $ and $ \mathit{max}-norm $, as the foundations for determining operations on NS-sets in general. Each operation is well-defined along with its well-proven properties.

3.1.   $ \mathit{min}-norm $ and $ \mathit{max}-norm $

Definition 4. A $ {\rm{min}}-{\rm{n}}{\rm{o}}{\rm{r}}{\rm{m}} $ is the binary operation $ \bullet :\left[0;1\right]\times \left[0;1\right]\to \left[{\rm{0, 1}}\right] $ that obeys the conditions as follows:

(a) $ \bullet $ has the commutative and associative properties,

(b) $ \forall x\in \left[{\rm{0, 1}}\right], x \bullet 1 = 1 \bullet x = x, $

(c) $ \forall x\in \left[{\rm{0, 1}}\right], x \bullet 0 = 0 \bullet x = 0, $

(d) $ \forall x, y\in \left[{\rm{0, 1}}\right], x\ge x \bullet y. $

Definition 5. A $ {\rm{max}}-{\rm{n}}{\rm{o}}{\rm{r}}{\rm{m}} $ is the binary operation $ \circ :\left[0;1\right]\times \left[0;1\right]\to \left[{\rm{0, 1}}\right] $ that obeys the following conditions:

(a) $ \circ $ has the commutative and associative properties,

(b) $ \forall x\in \left[{\rm{0, 1}}\right], x\circ 1 = 1\circ x = 1, $

(c) $ \forall x\in \left[{\rm{0, 1}}\right], y\circ 0 = 0\circ x = x, $

(d) $ \forall x, y\in \left[{\rm{0, 1}}\right], x\le x\circ y. $

Definition 6. The $ {\rm{min}}-{\rm{n}}{\rm{o}}{\rm{r}}{\rm{m}} $ $ \bullet $ and $ {\rm{max}}-{\rm{n}}{\rm{o}}{\rm{r}}{\rm{m}} $ $ \circ $ satisfy De Morgan's law if they obey the following conditions:

Some commonly used $ {\rm{min}}-{\rm{n}}{\rm{o}}{\rm{r}}{\rm{m}} $ and $ {\rm{max}}-{\rm{n}}{\rm{o}}{\rm{r}}{\rm{m}} $ are shown in Table 3. On the other hand, all of these norms satisfy De Morgan's law in pairs.

3.2.   Operations on NS-sets

3.2.1.   Intersection

Definition 7. The intersection of the two NS-sets $ A $ and $ B, {\rm{ }}{\rm{w}}{\rm{r}}{\rm{i}}{\rm{t}}{\rm{t}}{\rm{e}}{\rm{n}}{\rm{ }}{\rm{a}}{\rm{s}}A\cap B $, is determined by

Example 3. Let two NS-sets $ A $ and $ B $ be represented in Table 4 as follows:

If using $ {\rm{min}}-{\rm{n}}{\rm{o}}{\rm{r}}{\rm{m}}{\rm{s}} $ $ x \bullet y = {\rm{max}}\left\{x+y-{\rm{1, 0}}\right\} $ and $ {\rm{max}}-{\rm{n}}{\rm{o}}{\rm{r}}{\rm{m}}{\rm{s}} $ $ x\circ y = {\rm{min}}\left\{x+y, 1\right\} $, the intersection $ A\cap B $ of the two above NS-sets is described according to Eq (13) in Table 5 as follows:

Theorem 2. If $ A, B, C\in \mathcal{N}\mathcal{S}\left(X\right) $,

(1) $ A\cap A = A, $

(2) $ A\cap {\varnothing }_{\mathcal{E}} = {\mathcal{\varnothing }}_{\mathcal{E}}, $

(3) $ A\cap {\varnothing }_{\widetilde{\mathcal{E}}} = {\mathcal{\varnothing }}_{\widetilde{\mathcal{E}}}, $

(4) $ A\cap {X}_{\mathcal{E}} = A, $

(5) $ A\cap {X}_{\widetilde{\mathcal{E}}} = A, $

(6) $ A\cap \left(B\cap C\right) = \left(A\cap B\right)\cap C, $

(7) $ A\cap B = B\cap A. $

Proof. These properties are directly inferred from the definitions of norms and intersection operation.

Definition 8. Let $ {\left({A}_{i}\right)}_{i\in I} $ be a collection of NS-sets on $ X $. The intersection of the collection of NS-sets $ {\left({A}_{i}\right)}_{i\in I}, {\rm{ }}{\rm{w}}{\rm{r}}{\rm{i}}{\rm{t}}{\rm{t}}{\rm{e}}{\rm{n}}{\rm{ }}{\rm{a}}{\rm{s}}\bigcap _{i\in I}{A}_{i} $, is determined by

3.2.2.   Union

Definition 9. The union of the two NS-sets $ A $ and $ B, {\rm{ }}{\rm{w}}{\rm{r}}{\rm{i}}{\rm{t}}{\rm{t}}{\rm{e}}{\rm{n}}{\rm{ }}{\rm{a}}{\rm{s}}A\cup B $, is determined by

Example 4. If using $ {\rm{min}}-{\rm{n}}{\rm{o}}{\rm{r}}{\rm{m}}{\rm{s}} $ $ x \bullet y = {\rm{max}}\left\{x+y-{\rm{1, 0}}\right\} $ and $ {\rm{max}}-{\rm{n}}{\rm{o}}{\rm{r}}{\rm{m}}{\rm{s}} $ $ x\circ y = {\rm{min}}\left\{x+y, 1\right\} $, the union $ A\cup B $ of the two above NS-sets $ A $ and $ B $ in Example 3 is described according to Eq (14) in Table 6 as follows:

Theorem 3. If $ A, B, C\in \mathcal{N}\mathcal{S}\left(X\right) $,

(1) $ A\cup A = A, $

(2) $ A\cup {\varnothing }_{\mathcal{E}} = A, $

(3) $ A\cup {\varnothing }_{\widetilde{\mathcal{E}}} = A, $

(4) $ A\cup {X}_{\mathcal{E}} = {X}_{\mathcal{E}}, $

(5) $ A\cup {X}_{\widetilde{\mathcal{E}}} = {X}_{\widetilde{\mathcal{E}}}, $

(6) $ A\cup \left(B\cup C\right) = \left(A\cup B\right)\cup C, $

(7) $ A\cup B = B\cup A. $

Proof. These properties are directly inferred from the definitions of norms and union operation.

Theorem 4. If the $ {\rm{min}}-norm $ and $ {\rm{max}}-norm $ satisfy De Morgan's law, for all $ A, B\in \mathcal{N}\mathcal{S}\left(X\right) $,

(1) $ \overline{A\cap B} = \overline{A}\cup \overline{B}, $

(2) $ \overline{A\cup B} = \overline{A}\cap \overline{B}. $

Proof.

(1) $ \forall e\in \mathcal{E}, \forall x\in X, $

and

Moreover,

due to De Morgan's law of the $ {\rm{min}}-{\rm{n}}{\rm{o}}{\rm{r}}{\rm{m}} $ and $ {\rm{max}}-{\rm{n}}{\rm{o}}{\rm{r}}{\rm{m}} $. Therefore,

(2) $ \forall e\in \mathcal{E}, \forall x\in X, $

and

Moreover,

due to De Morgan's law of the $ {\rm{min}}-{\rm{n}}{\rm{o}}{\rm{r}}{\rm{m}} $ and $ {\rm{max}}-{\rm{n}}{\rm{o}}{\rm{r}}{\rm{m}} $. Therefore,

The distributive properties between intersection and union operations are not satisfied in the case of these general operations. Counterexamples are shown in Example 5.

Example 5. Let the NS-set $ C $ be represented in Table 7 as follows:

If using $ {\rm{min}}-{\rm{n}}{\rm{o}}{\rm{r}}{\rm{m}} $ $ x \bullet y = {\rm{max}}\left\{x+y-{\rm{1, 0}}\right\} $ and $ {\rm{max}}-{\rm{n}}{\rm{o}}{\rm{r}}{\rm{m}} $ $ x\circ y = {\rm{min}}\left\{x+y, 1\right\} $ with the two above NS-sets $ A $ and $ B $ in Example 3, the two NS-sets $ A\cap \left(B\cup C\right) $ and $ \left(A\cap B\right)\cup \left(A\cap C\right) $ can be described in Table 8 as follows:

Therefore, $ A\cap \left(B\cup C\right)\ne \left(A\cap B\right)\cup \left(A\cap C\right) $. Similarly, see Table 9:

Therefore, $ A\cup \left(B\cap C\right)\ne \left(A\cup B\right)\cap \left(A\cup C\right) $.

Definition 10. Let $ {\left({A}_{i}\right)}_{i\in I} $ be a collection of NS-sets on $ X $. The union of the collection of NS-sets $ {\left({A}_{i}\right)}_{i\in I}, {\rm{ }}{\rm{w}}{\rm{r}}{\rm{i}}{\rm{t}}{\rm{t}}{\rm{e}}{\rm{n}}{\rm{ }}{\rm{a}}{\rm{s}}\bigcup _{i\in I}{A}_{i} $, is determined by

3.2.3.   Difference

Definition 11. The difference of the two NS-sets $ A $ and $ B, {\rm{ }}{\rm{w}}{\rm{r}}{\rm{i}}{\rm{t}}{\rm{t}}{\rm{e}}{\rm{n}}{\rm{ }}{\rm{a}}{\rm{s}}A\backslash B $, is determined by $ A\backslash B = A\cap \overline{B} $, i.e.,

Example 6. If using $ {\rm{min}}-{\rm{n}}{\rm{o}}{\rm{r}}{\rm{m}} $ $ x \bullet y = {\rm{max}}\left\{x+y-{\rm{1, 0}}\right\} $ and $ {\rm{max}}-{\rm{n}}{\rm{o}}{\rm{r}}{\rm{m}} $ $ x\circ y = {\rm{min}}\left\{x+y, 1\right\} $, the difference $ A\backslash B $ of the two above NS-sets $ A $ and $ B $ in Example 3 is described according to Eq (25), see Table 10:

Theorem 5. If the $ {\rm{min}}-norm $ and $ {\rm{max}}-norm $ satisfy De Morgan's law, for all $ A, B, C\in \mathcal{N}\mathcal{S}\left(X\right) $,

(1) $ A\backslash B\subseteq A, $

(2) $ \overline{A\backslash B} = \overline{A}\cup B, $

(3) $ \overline{A}\backslash \overline{B} = B\backslash A, $

(4) $ A\backslash \left(B\cup C\right) = \left(A\backslash B\right)\cap \left(A\backslash C\right), $

(5) $ \left(A\cap B\right)\backslash C = \left(A\backslash C\right)\cap \left(B\backslash C\right), $

(6) $ \left(A\backslash B\right)\cap \left(C\backslash D\right) = \left(C\backslash B\right)\cap \left(A\backslash D\right) = \left(A\cap C\right)\backslash \left(B\cup D\right). $

Proof.

(1) $ \forall e\in \mathcal{E}, \forall x\in X, \left\{

$$\begin{array}{c}{T}_{{A\backslash B}_{e}}\left(x\right) = {T}_{{A}_{e}}\left(x\right) \bullet {T}_{{B}_{e}}\left(x\right)\le {T}_{{A}_{e}}\left(x\right)\\ {I}_{{A\backslash B}_{e}}\left(x\right) = {I}_{{A}_{e}}\left(x\right) \bullet \left(1-{I}_{{B}_{e}}\left(x\right)\right)\le {I}_{{A}_{e}}\left(x\right)\\ {F}_{{A\backslash B}_{e}}\left(x\right) = {F}_{{A}_{e}}\left(x\right)\circ {F}_{{B}_{e}}\left(x\right)\ge {F}_{{A}_{e}}\left(x\right)\end{array}$$ \right. $. This implies that $ A\backslash B\subseteq A $.

(2) $ \overline{A\backslash B} = \overline{A\cap \overline{B}} = \overline{A}\cup \overline{\overline {B}} = \overline{A}\cup B $ due to Theorem 1.

(3) $ \overline{A}\backslash \overline{B} = \overline{A}\cap \overline{\overline {B}} = \overline{A}\cap B = B\cap \overline{A} = B\backslash A $.

(4) $ A\backslash \left(B\cup C\right) = A\cap \overline{B\cup C} = A\cap \left(\overline{B}\cap \overline{C}\right) = \left(A\cap \overline{B}\right)\cap \left(A\cap \overline{C}\right) = \left(A\backslash B\right)\cap \left(A\backslash C\right) $ due to Theorems 3 and 4.

(5) $ \left(A\cap B\right)\backslash C = \left(A\cap B\right)\cap \overline{C} = \left(A\cap \overline{C}\right)\cap \left(B\cap \overline{C}\right) = \left(A\backslash C\right)\cap \left(B\backslash C\right) $ due to Theorem 3.

(6) $ \left(A\backslash B\right)\cap \left(C\backslash D\right) = \left(A\cap \overline{B}\right)\cap \left(C\cap \overline{D}\right) = \left(C\cap \overline{B}\right)\cap \left(A\cap \overline{D}\right) = \left(C\backslash B\right)\cap \left(A\backslash D\right) $ due to Theorem 3.

(7) $ \left(A\backslash B\right)\cap \left(C\backslash D\right) = \left(A\cap \overline{B}\right)\cap \left(C\cap \overline{D}\right) = \left(A\cap C\right)\cap \left(\overline{B}\cap \overline{D}\right) = \left(A\cap C\right)\cap \overline{B\cup D}$ $ = \left(A\cap C\right)\backslash \left(B\cup D\right) $ due to Theorems 3 and 4.

3.2.4.   AND and OR

Definition 12. The AND operation of the two NS-sets $ A $ and B with the same parameter set $ \mathcal{E} $, written as $ A\wedge B $, is determined over the same parameter set $ \mathcal{E}\times \mathcal{E} $ by

Definition 13. The OR operation of the two NS-sets $ A $ and B with the same parameter set $ \mathcal{E} $, written as $ A\wedge B $, is determined over the same parameter set $ \mathcal{E}\times \mathcal{E} $ by

Example 7. If using $ {\rm{min}}-{\rm{n}}{\rm{o}}{\rm{r}}{\rm{m}} $ $ x \bullet y = {\rm{max}}\left\{x+y-{\rm{1, 0}}\right\} $ and $ {\rm{max}}-{\rm{n}}{\rm{o}}{\rm{r}}{\rm{m}} $ $ x\circ y = {\rm{min}}\left\{x+y, 1\right\} $, the AND $ A\wedge B $ and OR $ A\vee B $ operations of the two above NS-sets $ A $ and $ B $ in Example 3 is described according to Eqs (26) and (27) in Table 11 as follows:

Theorem 6. If the $ {\rm{min}}-norm $ and $ {\rm{max}}-norm $ satisfy De Morgan's law, for all $ A, B\in \mathcal{N}\mathcal{S}\left(X\right) $,

(1) $ \overline{A\wedge B} = \overline{A}\vee \overline{B}, $

(2) $ \overline{A\vee B} = \overline{A}\wedge \overline{B}. $

Proof.

(1) $ \forall \left({e}_{1}, {e}_{2}\right)\in \mathcal{E}\times \mathcal{E}, \forall x\in X, $

and

Moreover,

due to De Morgan's law of the $ {\rm{min}}-{\rm{n}}{\rm{o}}{\rm{r}}{\rm{m}} $ and $ {\rm{max}}-{\rm{n}}{\rm{o}}{\rm{r}}{\rm{m}} $. Therefore,

(2) $ \forall \left({e}_{1}, {e}_{2}\right)\in \mathcal{E}\times \mathcal{E}, \forall x\in X, $

and

Moreover,

due to De Morgan's law of the $ {\rm{min}}-{\rm{n}}{\rm{o}}{\rm{r}}{\rm{m}} $ and $ {\rm{max}}-{\rm{n}}{\rm{o}}{\rm{r}}{\rm{m}} $. Therefore,

4.   Topology on NS-sets

This section uses the operations just constructed above as the core to build the topology and related concepts on NS-sets. It is important to note that the norms used must satisfy De Morgan's law.

4.1.   NS-topological space

Definition 14. A collection $ \tau \subseteq \mathcal{N}\mathcal{S}\left(X\right) $ is NS-topology on $ X $ if it obeys the following properties:

(a) $ {\varnothing }_{\mathcal{E}} $ and $ {X}_{\mathcal{E}} $ belongs to $ \tau $,

(b) The intersection of any finite collection of $ \tau $'s elements belongs to $ \tau $,

(c) The union of any collection of $ \tau $'s elements belongs to $ \tau $.

Then, the pair $ \left(X, \tau \right) $ is a NS-topological space and each element of $ \tau $ is a NS-open set.

Example 8. Let three NS-sets $ {K}_{1} $, $ {K}_{2} $, $ {K}_{3} $ be represented in Table 12 as follows:

If using the $ {\rm{min}}-{\rm{n}}{\rm{o}}{\rm{r}}{\rm{m}} $ $ x \bullet y = {\rm{min}}\left\{x, y\right\} $, $ {\rm{max}}-{\rm{n}}{\rm{o}}{\rm{r}}{\rm{m}} $ $ x\circ y = {\rm{max}}\left\{x, y\right\} $, the collection $ \tau = \left\{{\varnothing }_{\mathcal{E}}, {X}_{\mathcal{E}}, {K}_{1}, {K}_{2}, {K}_{3}\right\} $ is a NS-topology.

Theorem 7.

(1) $ {\tau }_{0} = \left\{{\varnothing }_{\mathcal{E}}, {X}_{\mathcal{E}}\right\} $ is a NS-topology (anti-discrete).

(2) $ {\tau }_{\infty } = \mathcal{N}\mathcal{S}\left(X\right) $ is a NS-topology (discrete).

(3) If $ {\tau }_{1} $ and $ {\tau }_{2} $ are two NS-topologies, $ {\tau }_{1}\cap {\tau }_{2} $ is a NS-topology.

Proof. This proof focuses on the proof of Property 3 because Properties 1 and 2 are directly inferred.

• $ {\varnothing }_{\mathcal{E}}, {X}_{\mathcal{E}}\in {\tau }_{1};{\varnothing }_{\mathcal{E}}, {X}_{\mathcal{E}}\in {\tau }_{2}\Rightarrow {\varnothing }_{\mathcal{E}}, {X}_{\mathcal{E}}\in {\tau }_{1}\cap {\tau }_{2}. $

• If $ {\left\{{K}_{j}\right\}}_{1}^{n} $ is a finite family of NS-sets in $ {\tau }_{1}\cap {\tau }_{2} $, $ {K}_{i}\in {\tau }_{1} $ and $ {K}_{i}\in {\tau }_{2} $ for all $ i $. So $ \cap {\left\{{K}_{j}\right\}}_{1}^{n}\in {\tau }_{1} $ and $ {\left\{{K}_{j}\right\}}_{1}^{n}\in {\tau }_{2} $. Thus $ \cap {\left\{{K}_{j}\right\}}_{1}^{n}\in {\tau }_{1}\cap {\tau }_{2} $.

• If letting $ \left\{{K}_{i}|i\in I\right\} $ be a family of NS-sets in $ {\tau }_{1}\cap {\tau }_{2} $, $ {K}_{i}\in {\tau }_{1} $ and $ {K}_{i}\in {\tau }_{2} $ for all $ i\in I $. So $ {\cup }_{i\in I}{K}_{i}\in {\tau }_{1} $ and $ {\cup }_{i\in I}{K}_{i}\in {\tau }_{2} $. Therefore, $ {\cup }_{i\in I}{K}_{i}\in {\tau }_{1}\cap {\tau }_{2}. $

It should be noted that if $ {\tau }_{1} $ and $ {\tau }_{2} $ are two NS-topologies, $ {\tau }_{1}\cup {\tau }_{2} $ cannot be a NS-topology. Counterexamples are shown in Example 9.

Example 9. Let three NS-sets $ {H}_{1} $, $ {H}_{2} $, $ {H}_{3} $ be represented in Table 13 as follows:

If using the $ {\rm{min}}-{\rm{n}}{\rm{o}}{\rm{r}}{\rm{m}} $ $ x \bullet y = xy $, $ {\rm{max}}-{\rm{n}}{\rm{o}}{\rm{r}}{\rm{m}} $ $ x\circ y = x+y-xy $ and letting $ {\tau }_{1} = \left\{{\varnothing }_{\mathcal{E}}, {X}_{\mathcal{E}}, {H}_{1}, {H}_{2}\right\} $ and $ {\tau }_{2} = \left\{{\varnothing }_{\mathcal{E}}, {X}_{\mathcal{E}}, {H}_{3}\right\} $ be two NS-topologies, the collection $ {\tau }_{1}\cup {\tau }_{2} = \left\{{\varnothing }_{\mathcal{E}}, {X}_{\mathcal{E}}, {H}_{1}, {H}_{2}, {H}_{3}\right\} $ is not a NS-topology due to $ {H}_{1}\cup {H}_{2}\notin {\tau }_{1}\cup {\tau }_{2} $, see Table 14.

Definition 15. A NS-set $ A\in \mathcal{N}\mathcal{S}\left(X\right) $ is NS-closed set if it has the complement $ \overline{A} $ is a NS-open set. The symbol $ \overline{\tau } $ is denoted as the collection of all NS-closed sets.

Theorem 8.

(1) $ {\varnothing }_{\mathcal{E}} $ and $ {X}_{\mathcal{E}} $ belongs to $ \overline{\tau } $.

(2) The union of any finite collection of $ \overline{\tau } $'s elements belongs to $ \overline{\tau } $.

(3) The intersection of any collection of $ \overline{\tau } $'s elements belongs to $ \overline{\tau } $.

Proof. These properties are directly inferred from the definitions of a NS-closed set and De Morgan's law for intersection and union.

4.2.   NS-interior

Definition 16. The NS-interior of a NS-set $ A, {\rm{ }}{\rm{ }}{\rm{ }}{\rm{ }}{\rm{w}}{\rm{r}}{\rm{i}}{\rm{t}}{\rm{t}}{\rm{e}}{\rm{n}}{\rm{ }}{\rm{a}}{\rm{s}}{\rm{ }}{\rm{i}}{\rm{n}}{\rm{t}}\left(A\right) $, is the union of all NS-open subsets of $ A $. It is considered the biggest NS-open set which is contained by $ A $.

Example 10. Let three NS-sets $ {L}_{1} $, $ {L}_{2} $, $ K $ be represented as follows:

If using the $ {\rm{min}}-{\rm{n}}{\rm{o}}{\rm{r}}{\rm{m}} $ $ x \bullet y = {\rm{max}}\left\{x+y-{\rm{1, 0}}\right\} $, $ {\rm{max}}-{\rm{n}}{\rm{o}}{\rm{r}}{\rm{m}} = {\rm{min}}\left\{x+y, 1\right\} $, the collection $ \tau = \left\{{\varnothing }_{\mathcal{E}}, {X}_{\mathcal{E}}, {L}_{1}, {L}_{2}\right\} $ is the NS-topology. It is easy to see that $ {\varnothing }_{\mathcal{E}}, {L}_{1}\subseteq K $ and $ {\varnothing }_{\mathcal{E}}\cup {L}_{1} = {L}_{1}\subseteq K. $ Therefore, $ {\rm{i}}{\rm{n}}{\rm{t}}\left(K\right) = A. $

Theorem 9. A NS-set $ A $ is a NS-open set if and only if $ A = {\rm{i}}{\rm{n}}{\rm{t}}\left(A\right) $.

Proof. If $ A\in \tau $ then $ A $ is the biggest NS-open set that is contained by $ A $. So $ A = {\rm{i}}{\rm{n}}{\rm{t}}\left(A\right) $. Conversely, $ A = {\rm{i}}{\rm{n}}{\rm{t}}\left(A\right)\in \tau $.

Theorem 10. If $ A, B\in \mathcal{N}\mathcal{S}\left(X\right) $,

(1) $ {\rm{i}}{\rm{n}}{\rm{t}}\left({\rm{i}}{\rm{n}}{\rm{t}}\left(A\right)\right) = {\rm{i}}{\rm{n}}{\rm{t}}\left(A\right) $,

(2) $ {\rm{i}}{\rm{n}}{\rm{t}}\left({\varnothing }_{\mathcal{E}}\right) = {\mathcal{\varnothing }}_{\mathcal{E}} $ and $ {\rm{i}}{\rm{n}}{\rm{t}}\left({X}_{\mathcal{E}}\right) = {X}_{\mathcal{E}} $,

(3) $ A\subseteq B\Rightarrow {\rm{i}}{\rm{n}}{\rm{t}}\left(A\right)\subseteq {\rm{i}}{\rm{n}}{\rm{t}}\left(B\right) $,

(4) $ {\rm{i}}{\rm{n}}{\rm{t}}\left(A\cap B\right) = {\rm{i}}{\rm{n}}{\rm{t}}\left(A\right)\cap {\rm{i}}{\rm{n}}{\rm{t}}\left(B\right) $,

(5) $ {\rm{i}}{\rm{n}}{\rm{t}}\left(A\right)\cup {\rm{i}}{\rm{n}}{\rm{t}}\left(B\right)\subseteq {\rm{i}}{\rm{n}}{\rm{t}}\left(A\cup B\right) $.

Proof.

(1) Due to $ {\rm{i}}{\rm{n}}{\rm{t}}\left(A\right)\in \tau $, $ {\rm{i}}{\rm{n}}{\rm{t}}\left({\rm{i}}{\rm{n}}{\rm{t}}\left(A\right)\right) = {\rm{i}}{\rm{n}}{\rm{t}}\left(A\right). $

(2) $ {\varnothing }_{\mathcal{E}}\in \tau \Rightarrow {\rm{i}}{\rm{n}}{\rm{t}}\left({\varnothing }_{\mathcal{E}}\right) = {\mathcal{\varnothing }}_{\mathcal{E}} $ and $ {X}_{\mathcal{E}}\in \tau \Rightarrow {\rm{i}}{\rm{n}}{\rm{t}}\left({X}_{\mathcal{E}}\right) = {X}_{\mathcal{E}}. $

(3) Due to $ A\subseteq B $, $ {\rm{i}}{\rm{n}}{\rm{t}}\left(A\right)\subseteq A\subseteq B $ and $ {\rm{i}}{\rm{n}}{\rm{t}}\left(B\right)\subseteq B $. Because $ {\rm{i}}{\rm{n}}{\rm{t}}\left(B\right) $ is the biggest NS-open set contained in $ B $, $ {\rm{i}}{\rm{n}}{\rm{t}}\left(A\right)\subseteq {\rm{i}}{\rm{n}}{\rm{t}}\left(B\right). $

(4) Since $ {\rm{i}}{\rm{n}}{\rm{t}}\left(A\right)\in \tau $ and $ {\rm{i}}{\rm{n}}{\rm{t}}\left(B\right)\in \tau $, then $ {\rm{i}}{\rm{n}}{\rm{t}}\left(A\right)\cup {\rm{i}}{\rm{n}}{\rm{t}}\left(B\right)\in \tau $. It is known that $ {\rm{i}}{\rm{n}}{\rm{t}}\left(A\right)\subseteq A $ and $ {\rm{i}}{\rm{n}}{\rm{t}}\left(B\right)\subseteq B, $ so $ {\rm{i}}{\rm{n}}{\rm{t}}\left(A\right)\cup {\rm{i}}{\rm{n}}{\rm{t}}\left(B\right)\subseteq A\cup B. $ Moreover, $ {\rm{i}}{\rm{n}}{\rm{t}}\left(A\cup B\right) $ is the biggest NS-open set contained in $ A\cup B. $ Therefore, $ {\rm{i}}{\rm{n}}{\rm{t}}\left(A\right)\cup {\rm{i}}{\rm{n}}{\rm{t}}\left(B\right)\subseteq {\rm{i}}{\rm{n}}{\rm{t}}\left(A\cup B\right). $

(5) Since $ {\rm{i}}{\rm{n}}{\rm{t}}\left(A\cap B\right)\subseteq A\cap B $, so $ {\rm{i}}{\rm{n}}{\rm{t}}\left(A\cap B\right)\subseteq A $ and $ {\rm{i}}{\rm{n}}{\rm{t}}\left(A\cap B\right)\subseteq B. $ Therefore, $ {\rm{i}}{\rm{n}}{\rm{t}}\left(A\cap B\right)\subseteq {\rm{i}}{\rm{n}}{\rm{t}}\left(A\right) $ and $ {\rm{i}}{\rm{n}}{\rm{t}}\left(A\cap B\right)\subseteq {\rm{i}}{\rm{n}}{\rm{t}}\left(B\right) $ or $ {\rm{i}}{\rm{n}}{\rm{t}}\left(A\cap B\right)\subseteq {\rm{i}}{\rm{n}}{\rm{t}}\left(A\right)\cap {\rm{i}}{\rm{n}}{\rm{t}}\left(B\right). $

Moreover,

and $ {\rm{i}}{\rm{n}}{\rm{t}}\left(A\cap B\right) $ is the biggest NS-open set contained in $ A\cap B $, so

Thus, $ {\rm{i}}{\rm{n}}{\rm{t}}\left(A\cap B\right) = {\rm{i}}{\rm{n}}{\rm{t}}\left(A\right)\cap {\rm{i}}{\rm{n}}{\rm{t}}\left(B\right). $

4.3.   NS-closure

Definition 17. The NS-closure of a NS-set $ A, {\rm{ }}{\rm{ }}{\rm{ }}{\rm{ }}{\rm{w}}{\rm{r}}{\rm{i}}{\rm{t}}{\rm{t}}{\rm{e}}{\rm{n}}{\rm{ }}{\rm{a}}{\rm{s}}{\rm{ }}{\rm{c}}{\rm{l}}\left(A\right) $, is the intersection of all NS-closed supersets of $ A $. The $ {\rm{c}}{\rm{l}}\left(A\right) $ is the smallest NS-closed set which contains $ A $.

Example 11. For the NS-topology $ \tau $ given in Example 10, let NS-set $ H $ be represented in Table 16 as follows:

It is easy to see that $ \overline{{\varnothing }_{\mathcal{E}}} = {X}_{\mathcal{E}} $, $ \overline{{X}_{\mathcal{E}}} = {\mathcal{\varnothing }}_{\mathcal{E}} $. So $ {\mathcal{\varnothing }}_{\mathcal{E}} $, $ {X}_{\mathcal{E}} $, $ \overline{{L}_{1}} $, $ \overline{{L}_{2}} $ are all NS-closed sets. Since $ H\subseteq {X}_{\mathcal{E}} $, $ {\rm{c}}{\rm{l}}\left(H\right) = {L}_{2}. $

Theorem 11. A NS-set $ A $ is a NS-closed set if and only if $ A = {\rm{c}}{\rm{l}}\left(A\right) $.

Proof. Let $ A $ be a NS-closed set. Because $ A\subseteq A $ and $ {\rm{c}}{\rm{l}}\left(A\right) $ is the smallest NS-closed set that contains $ A $, $ {\rm{c}}{\rm{l}}\left(A\right)\subseteq A $. Therefore, $ A = {\rm{c}}{\rm{l}}\left(A\right). $ Conversely, if $ A = {\rm{c}}{\rm{l}}\left(A\right) $ then $ A $ is a NS-closed set.

Theorem 12. If $ A, B\in \mathcal{N}\mathcal{S}\left(X\right) $,

(1) $ {\rm{c}}{\rm{l}}\left({\rm{c}}{\rm{l}}\left(A\right)\right) = {\rm{c}}{\rm{l}}\left(A\right) $,

(2) $ {\rm{c}}{\rm{l}}\left({\varnothing }_{\mathcal{E}}\right) = {\mathcal{\varnothing }}_{\mathcal{E}} $ and $ {\rm{c}}{\rm{l}}\left({X}_{\mathcal{E}}\right) = {X}_{\mathcal{E}} $,

(3) $ A\subseteq B\Rightarrow {\rm{c}}{\rm{l}}\left(A\right)\subseteq {\rm{c}}{\rm{l}}\left(B\right) $,

(4) $ {\rm{c}}{\rm{l}}\left(A\cap B\right)\subseteq {\rm{c}}{\rm{l}}\left(A\right)\cap {\rm{c}}{\rm{l}}\left(B\right) $,

(5) $ {\rm{c}}{\rm{l}}\left(A\cup B\right) = {\rm{c}}{\rm{l}}\left(A\right)\cup {\rm{c}}{\rm{l}}\left(B\right) $.

Proof.

(1) Directly inferring from Theorem 9.

(2) Directly inferring from Definition 14 and Theorem 9.

(3) Since $ A\subseteq B\subseteq {\rm{c}}{\rm{l}}\left(B\right) $ and $ {\rm{c}}{\rm{l}}\left(A\right) $ is the smallest NS-closed set containing $ A $, $ {\rm{c}}{\rm{l}}\left(A\right)\subseteq {\rm{c}}{\rm{l}}\left(B\right). $

(4) Since $ A\cap B\subseteq A\subseteq {\rm{c}}{\rm{l}}\left(A\right) $ and $ A\cap B\subseteq B\subseteq {\rm{c}}{\rm{l}}\left(B\right) $, $ A\cap B\subseteq {\rm{c}}{\rm{l}}\left(A\right)\cap {\rm{c}}{\rm{l}}\left(B\right). $ Therefore, $ {\rm{c}}{\rm{l}}\left(A\cap B\right)\subseteq {\rm{c}}{\rm{l}}\left(A\right)\cap {\rm{c}}{\rm{l}}\left(B\right) $.

(5) It is easy to see that $ A\subseteq A\cup B\subseteq {\rm{c}}{\rm{l}}\left(A\cup B\right) $, $ B\subseteq A\cup B\subseteq {\rm{c}}{\rm{l}}\left(A\cup B\right) $, $ {\rm{c}}{\rm{l}}\left(A\right) $ is the smallest NS-closed set that contains A, and $ {\rm{c}}{\rm{l}}\left(B\right) $ is the smallest NS-closed set that containing $ B $. So $ {\rm{c}}{\rm{l}}\left(A\right)\subseteq {\rm{c}}{\rm{l}}\left(A\cup B\right) $ and $ {\rm{c}}{\rm{l}}\left(B\right)\subseteq {\rm{c}}{\rm{l}}\left(A\cup B\right) $. Therefore, $ {\rm{c}}{\rm{l}}\left(A\right)\cup {\rm{c}}{\rm{l}}\left(B\right)\subseteq {\rm{c}}{\rm{l}}\left(A\cup B\right) $.

Moreover, since $ A\subseteq {\rm{c}}{\rm{l}}\left(A\right) $ and $ B\subseteq {\rm{c}}{\rm{l}}\left(B\right) $, $ A\cup B\subseteq {\rm{c}}{\rm{l}}\left(A\right)\cup {\rm{c}}{\rm{l}}\left(B\right) $. Therefore, $ {\rm{c}}{\rm{l}}\left(A\cup B\right)\subseteq {\rm{c}}{\rm{l}}\left(A\right)\cup {\rm{c}}{\rm{l}}\left(B\right) $.

Thus, $ {\rm{c}}{\rm{l}}\left(A\cup B\right) = {\rm{c}}{\rm{l}}\left(A\right)\cup {\rm{c}}{\rm{l}}\left(B\right) $.

Theorem 13. If $ A, B\in \mathcal{N}\mathcal{S}\left(X\right) $,

(1) $ \overline{{\rm{i}}{\rm{n}}{\rm{t}}\left(A\right)} = {\rm{c}}{\rm{l}}\left(\overline{A}\right) $,

(2) $ \overline{{\rm{c}}{\rm{l}}\left(A\right)} = {\rm{i}}{\rm{n}}{\rm{t}}\left(\overline{A}\right) $.

Proof.

(1) Because

(2) Because

4.4.   NS-boundary

Definition 18. The NS-boundary of a NS-set $ A, \; {\rm{w}}{\rm{r}}{\rm{i}}{\rm{t}}{\rm{t}}{\rm{e}}{\rm{n}}\; {\rm{a}}{\rm{s}}\; \partial A $, is the intersection of the NS-closure of $ A $ and the NS-closure of $ \overline{A} $.

Example 12. For the NS-topology $ \tau $ given in Example 10 and the NS-set $ H $ given in Example 11, the complement of $ H $ is represented in Table 17 as follows:

It is easy to see that $ {\rm{c}}{\rm{l}}\left(H\right) = {X}_{\mathcal{E}} $ and $ {\rm{c}}{\rm{l}}\left(\overline{H}\right) = {X}_{\mathcal{E}} $. So $ \partial H = {X}_{\mathcal{E}}{\cap X}_{\mathcal{E}} = {X}_{\mathcal{E}} $.

Theorem 14. If $ A\in \mathcal{N}\mathcal{S}\left(X\right) $,

(1) $ \partial A = {\rm{c}}{\rm{l}}\left(A\right)\cap \overline{{\rm{i}}{\rm{n}}{\rm{t}}\left(A\right)} $,

(2) $ {\rm{i}}{\rm{n}}{\rm{t}}\left(A\right)\cap \partial A = {\varnothing }_{\mathcal{E}} $,

(3) $ \partial A = {\varnothing }_{\mathcal{E}} $ if and only if $ A $ is a NS-open and NS-closed set.

Proof.

(1) $ \partial A = {\rm{c}}{\rm{l}}\left(A\right)\cap {\rm{c}}{\rm{l}}\left(\overline{A}\right) $ = $ {\rm{c}}{\rm{l}}\left(A\right)\cap \overline{{\rm{i}}{\rm{n}}{\rm{t}}\left(A\right)} $ due to Theorem 13.

(2) It is easy to see that

due to Theorem 13.

(3) Since

${\rm{c}}{\rm{l}}\left(A\right)\cap {\rm{i}}{\rm{n}}{\rm{t}}\left(A\right){\ne \varnothing }_{\mathcal{E}} $. So $ A\subseteq {\rm{c}}{\rm{l}}\left(A\right)\subseteq {\rm{i}}{\rm{n}}{\rm{t}}\left(A\right)\subseteq A $. Therefore, $ A = {\rm{c}}{\rm{l}}\left(A\right) = {\rm{i}}{\rm{n}}{\rm{t}}\left(A\right) $ or $ A $ is a NS-open and NS-closed set.

Conversely, if $ A $ is a NS-open and NS-closed set, $ A = {\rm{i}}{\rm{n}}{\rm{t}}\left(A\right) $ and $ A = {\rm{c}}{\rm{l}}\left(A\right) $. Therefore,

4.5.   Regular property

Definition 19.

a. The NS-open set $ M $ is regular if $ M = {\rm{i}}{\rm{n}}{\rm{t}}\left({\rm{c}}{\rm{l}}\left(M\right)\right) $.

b. The NS-closed set $ M $ is regular if $ M = {\rm{c}}{\rm{l}}\left({\rm{i}}{\rm{n}}{\rm{t}}\left(M\right)\right) $.

Theorem 15. If $ M, N\in \mathcal{N}\mathcal{S}\left(X\right) $,

(1) If $ M $ is a NS-closed set, $ {\rm{i}}{\rm{n}}{\rm{t}}\left(M\right) $ is a regular NS-open set.

(2) If $ M $ is a NS-open set, $ {\rm{c}}{\rm{l}}\left(M\right) $ is a regular NS-closed set.

(3) If $ M $ and $ N $ are two regular NS-open sets, $ M\subseteq N\iff {\rm{c}}{\rm{l}}\left(M\right)\subseteq {\rm{c}}{\rm{l}}\left(N\right) $.

(4) If $ M $ and $ N $ are two regular NS-closed sets, $ M\subseteq N\iff {\rm{i}}{\rm{n}}{\rm{t}}\left(M\right)\subseteq {\rm{i}}{\rm{n}}{\rm{t}}\left(N\right) $.

(5) If $ M $ is a regular NS-closed set, $ \overline{M} $ is a regular NS-open set.

(6) If $ M $ is a regular NS-open set, $ \overline{M} $ is a regular NS-closed set.

Proof.

(1) If $ M $ is a NS-closed set,

Therefore, $ {\rm{i}}{\rm{n}}{\rm{t}}\left(M\right) = {\rm{i}}{\rm{n}}{\rm{t}}\left[{\rm{c}}{\rm{l}}\left({\rm{i}}{\rm{n}}{\rm{t}}\left(M\right)\right)\right] $ or $ {\rm{i}}{\rm{n}}{\rm{t}}\left(M\right) $ is regular.

(2) If $ M $ is a NS-open set,

Because $ {\rm{i}}{\rm{n}}{\rm{t}}\left(M\right) = M $,

Therefore, $ {\rm{c}}{\rm{l}}\left(M\right) = {\rm{c}}{\rm{l}}\left({\rm{i}}{\rm{n}}{\rm{t}}\left({\rm{c}}{\rm{l}}\left(M\right)\right)\right) $ or $ {\rm{c}}{\rm{l}}\left(M\right) $ is regular.

(3) Clearly, $ M\subseteq N \Rightarrow {\rm{c}}{\rm{l}}\left(M\right)\subseteq {\rm{c}}{\rm{l}}\left(N\right) $ and $ {\rm{i}}{\rm{n}}{\rm{t}}\left({\rm{c}}{\rm{l}}\left(M\right)\right) = M $, $ {\rm{i}}{\rm{n}}{\rm{t}}\left({\rm{c}}{\rm{l}}\left(N\right)\right) = N $ due to $ M, N $ are regular. Conversely,

(4) Clearly, $ M\subseteq N \Rightarrow {\rm{i}}{\rm{n}}{\rm{t}}\left(M\right)\subseteq {\rm{i}}{\rm{n}}{\rm{t}}\left(N\right) $ and $ M = {\rm{c}}{\rm{l}}\left({\rm{i}}{\rm{n}}{\rm{t}}\left(M\right)\right) $, $ N = {\rm{c}}{\rm{l}}\left({\rm{i}}{\rm{n}}{\rm{t}}\left(N\right)\right) $ due to $ M, N $ are regular. Conversely,

(5) If $ M $ is a regular NS-open set, $ M = {\rm{i}}{\rm{n}}{\rm{t}}\left({\rm{c}}{\rm{l}}\left(M\right)\right) $. So

Therefore, $ \overline{M} $ is a regular NS-closed set.

(6) Similarly, if $ M $ is a regular NS-closed set, $ {\rm{i}}{\rm{n}}{\rm{t}}\left({\rm{c}}{\rm{l}}\left(\overline{M}\right)\right) = \overline{M} $. So $ \overline{M} $ is a regular NS-open set.

5.   The relationship between NS-topology and fuzzy soft topology

Theorem 16. Let $ \tau = \left\{{K}_{i}:i\in I\right\} $ be NS-topology on $ X $ where

Three collections

are the fuzzy soft topologies on $ X $.

Proof.

• $ {\varnothing }_{\mathcal{E}}\in \tau \Rightarrow \widetilde{\varnothing }\in \mathcal{T}, \widetilde{\mathcal{\varnothing }}\in \mathcal{I}, \widetilde{\mathcal{\varnothing }}\in \mathcal{F}. $

• $ {X}_{\mathcal{E}}\in \tau \Rightarrow \widetilde{X}\in \mathcal{T}; \widetilde{X}\in \mathcal{I}; \widetilde{X}\in \mathcal{F}. $

• Let $ {\left({\mathcal{T}}_{i}\right)}_{i\in I} $ be a family of fuzzy soft sets in $ \mathcal{T} $, $ {\left({\mathcal{I}}_{i}\right)}_{i\in I} $ be a family of fuzzy soft sets in $ \mathcal{I} $, and $ {\left({\mathcal{F}}_{i}\right)}_{i\in I} $ be a family of fuzzy soft sets in $ \mathcal{F} $. They make a family of NS-sets $ \left\{{K}_{i}:i\in I\right\} $ where

So $ {\cup }_{i\in I}{K}_{i}\in \tau $ or

Therefore,

• Let $ {\left\{{\mathcal{T}}_{j}\in \mathcal{T}\right\}}_{1}^{n} $, $ {\left\{{\mathcal{I}}_{j}\in \mathcal{I}\right\}}_{1}^{n}, {\left\{{\mathcal{F}}_{j}\in \mathcal{F}\right\}}_{1}^{n} $ be finite families of fuzzy soft sets on $ X $ and satisfy

So, we have $ {\cap }_{1}^{n}{K}_{j}\in \tau $, i.e.,

Hence,

In the general case, the opposite of Theorem 16 is not true. This is demonstrated through a counterexample, as shown in Example 13.

Example 13. Let four NS-sets $ {H}_{1} $, $ {H}_{2} $, and $ {H}_{3} $be represented in Table 18 as follows:

If using the $ {\rm{min}}-{\rm{n}}{\rm{o}}{\rm{r}}{\rm{m}} $ $ x \bullet y = {\rm{max}}\left\{x+y-{\rm{1, 0}}\right\} $, $ {\rm{max}}-{\rm{n}}{\rm{o}}{\rm{r}}{\rm{m}} = {\rm{min}}\left\{x+y, 1\right\} $, three collections defined in Table 19 as follows are the fuzzy soft topologies on $ X $.

The $ \mathcal{T}, \mathcal{I}, \mathcal{ }\mathcal{F} $ are fuzzy soft topologies on $ X $, but $ \tau = \left\{{\varnothing }_{\mathcal{E}}, {X}_{\mathcal{E}}, {H}_{1}, {H}_{2}, {H}_{3}\right\} $ is not a NS- topology on $ X $ because $ {H}_{1}\cup {H}_{2}\notin \tau. $

Theorem 17. Let three collections

be the fuzzy soft topologies on $ X $. Let $ \tau = \left\{{K}_{i}:i\in I\right\} $ where

If for all $ l, m, n $, we have

Then $ \tau $ is the NS-topology on $ X $.

Proof.

• Obviously, $ {\varnothing }_{\mathcal{E}}, {X}_{\mathcal{E}}\in \tau. $

• Let $ \left\{{K}_{i}:i\in I\right\}\subset \tau $ be a family of NS-sets on $ X. $ We have $ \left\{{\mathcal{T}}_{i}\right\}, \left\{{\mathcal{I}}_{i}\right\}, \left\{{\mathcal{F}}_{i}\right\} $ are families of fuzzy soft sets on $ X. $ So,

Thus, $ {\bigcup }_{i\in I}{K}_{i} = {\mathcal{T}}_{{n}_{0}}\in \tau $.

• Let $ {\left\{{K}_{j}\in \tau \right\}}_{1}^{n} $ be a finite family of NS-sets on $ X. $ We have $ {\left\{{\mathcal{T}}_{j}\right\}}_{1}^{n} $, $ {\left\{{\mathcal{I}}_{j}\right\}}_{1}^{n}, {\left\{{\mathcal{F}}_{j}\right\}}_{1}^{n} $as finite families of fuzzy soft sets on $ X. $ So,

Thus, $ {\cap }_{1}^{n}{K}_{j}\in \tau. $

Theorem 18. Let $ \tau = \left\{{K}_{i}:i\in I\right\} $ be the NS-topology on $ X $ where

For each $ e\in E $, three collections

are the fuzzy topologies on $ X $.

Proof. It can be implied from Theorem 17.

In the general case, the opposite of Theorem 18 is not true. This is demonstrated through the counterexample shown in Example 14.

Example 14. We return to Example 12 with the same hypothesis. Then,

are fuzzy topologies on $ X. $ Similarly,

are also fuzzy topologies, but $ \tau = \left\{{\varnothing }_{\mathcal{E}}, {X}_{\mathcal{E}}, {H}_{1}, {H}_{2}, {H}_{3}\right\} $ is not a NS-topology on $ X $ because $ {K}_{1}\cup {K}_{2}\notin \tau. $

6.   Conclusions

In this paper, two novel norms are proposed to serve as the core for determining operations on NS-sets. These operations are used to construct the topology and related concepts such as open set, closed set, interior, closure, and regularity. Another highlight of this work is demonstrating the relationship between the topologies on NS-sets and fuzzy soft sets. The topology on NS-sets can parameterize the topologies on fuzzy soft sets, but the reverse is not guaranteed. This work's advantage is the structure's logic is presented with well-defined concepts and convincingly proven theorems.

Determining these concepts in a novel way enables a variety of methods for studying NS-sets, and offers a unique opportunity for future research and development in this field. Such research could focus on separation axioms, continuity, compactness, and paracompactness on NS-sets. Moreover, the relationship between topology on hybrid structure, NS-sets, and component structures, neutrosophic sets and soft sets, is also of research interest. In addition, applications of neutrosophic soft topological spaces can be investigated to handle decision-making problems.

Furthermore, we are also turning our interests to building topology on a new type of set, neutrosophic fuzzy sets. We believe these results will be helpful for future studies on neutrosophic fuzzy topology to develop a general framework for practical applications. These issues present opportunities but also challenges for researchers interested in the field of fuzzy theory.

Conflict of interest

The authors declare that they have no competing interests in this paper.