Bipolar soft groups

Abstract

In this study, a group structure on bipolar soft sets called bipolar soft group is constructed and some of its properties are investigated. Also, (α, β)-level set of a bipolar soft set is defined and some of its properties are obtained.

Keywords

Soft set bipolar soft set soft int-group bipolar soft group level set

1 Introduction

In 1999, Molodtsov introduced soft set theory [18] as an alternative approach to fuzzy set theory [22] defined by Zadeh in 1965. After Molodtsov’s study, many researchers have studied on set theoretical approaches and decision making applications of soft sets. For example Maji et al. [17] defined some new operations of soft sets and gave a decision making method based on soft sets. Chen et al. [11] developed a method of parameter reduction on soft set by using knowledge reduction of rough sets. Çağman and Enginoğlu [10] modified the definitions of soft set operations and gave a decision making method called uni-int decision making method. Ali et al. [3] defined some new operations on soft set theory such as extended union and intersection, restriction union and intersection. Sezgin and Atagün [21] studied on soft set operations defined by Ali et al. [3]. Studies related to soft sets have increased rapidly in many fields such as topology and algebra. The soft group, the first algebraic structure of soft sets, was first defined by Aktaş and Çağman [4] in 2007. In 2008, Jun defined soft BCK/BCI-algebras [13] and applied soft sets in ideal theory of BCK/BCI-algebras [14]. Also Jun et al. [15] defined soft ordered semigroup. Then, Çağman et al. [12] defined a new structure called soft int-group and obtained some properties of this new structure. Acar et al. [2] constructed a ring structure on soft sets. Kaygisiz contributed the soft int-group [7] and defined normal soft int-groups [8] and investigated some properties. Concept of a fuzzy soft group structure was defined by Aygünoğlu and Aygün [9] and intuitionistic fuzzy soft groups were introduced by Karaaslan et al. [6].

Bipolar fuzzy set was introduced by Zhang [23, 24] as a generalization of a fuzzy set. Bipolar fuzzy set is an extension of a fuzzy set whose membership degree interval is [-1, 1]. Abdullah et al. [1] introduced notion of bipolar fuzzy soft sets combining soft sets with bipolar fuzzy sets and they also defined operations of bipolar fuzzy soft sets. Naz and Shabir [19] proposed the concept of fuzzy bipolar soft sets and investigated algebraic structures on fuzzy bipolar soft sets. In a soft set, an element of initial universe belongs to the image set related to parameter or not. But, in some cases, an element of universal set may not belong image set and complement of image set related to parameter. In order to express such cases, Shabir and Naz [20] proposed the concept of bipolar soft set and defined some of their set theoretical operations such as union, intersection and complement. But complement of bipolar soft sets defined by Shabir and Naz have not allowed constructing some structures topological and algebraical. Therefore, notions of bipolar soft sets and their operations were redefined by Karaaslan and Karataş [16]. Muhammad et al. [5] defined the concept of bipolar fuzzy soft Γ-subsemigroup and bipolar fuzzy soft Γ-ideals in a Γ-semigroup.

As mentioned above, many studies have been made on group structures and other algebraic structures of soft sets. Since concept of bipolar soft set is novel, there aren’t enough studies on algebraic structures of bipolar soft sets. The main propose of this paper is to introduce a basic version on bipolar soft group. Therefore, we first define a new concept called extending parameter set in order to redefine the concept of bipolar soft sets. After we redefine the concepts of bipolar soft sets, first of all we define some novel concepts such as bipolar soft groupoid, bipolar soft group, full bipolar soft set, e-set of bipolar soft set, bipolar soft subgroup, centralizer of an element of bipolar soft group, level sets of a bipolar soft set, image and pre-image of bipolar soft sets and product of two bipolar soft sets. Then we obtain some properties related to these concepts. Inasmuch as bipolar soft sets are considered as generalization of soft sets, bipolar soft groups may be considered as generalization of soft int-groups.

2 Preliminaries

In this section, we recall basic definitions of soft set theory that are useful for subsequent sections. For more detail see the papers [10 , 20].

Throughout the paper, U refers to an initial universe, E is a set of parameters and P (U) is the power set of U. ⊂ and ⊃ stand for proper subset and superset, respectively.

2.1 Soft sets

Definition 2.1. [18] For any subset A of E, a soft set f_A over U is a set, defined by a function f_A, representing a mapping $f_{A} : E \to P (U)$ A soft set over U can also be represented by the set of ordered pairs $f_{A} = {(x, f_{A} (x)) : x \in E, f_{A} (x) \in P (U)} .$

Note that the set of all soft sets over U will be denoted by $S (U)$ .

Definition 2.2. [10] Let $f, g \in S (U)$ . Then,

If f (e) =∅ for all e ∈ E, f is said to be a null soft set, denoted by Φ.

If f (e) = U for all e ∈ E, f is said to be absolute soft set, denoted by $\hat{U}$ .

f is soft subset of g, denoted by $f \tilde{\subseteq} g$ , if f (e) ⊆ g (e) for all e ∈ E.

f = g, if $f \tilde{\subseteq} g$ and $g \tilde{\subseteq} f$ .

Soft union of f and g, denoted by $f \tilde{\cup} g$ , is a soft set over U and defined by $f \tilde{\cup} g : E \to P (U)$ such that $(f \tilde{\cup} g) (e) = f (e) \cup g (e)$ for all e ∈ E.

Soft intersection of f and g, denoted by $f \tilde{\cap} g$ , is a soft set over U and defined by $f \tilde{\cap} g : E \to P (U)$ such that $(f \tilde{\cap} g) (e) = f (e) \cap g (e)$ for all e ∈ E.

Soft complement of f is denoted by $f^{\tilde{c}}$ and defined by $f^{\tilde{c}} : E \to P (U)$ such that $f^{\tilde{c}} (e) = U \ f (e)$ for all e ∈ E.

Definition 2.3. [17] Let E = {e₁, e₂, e₃, . . . , e_n} be a set of parameters. The NOT set of E denoted by ⌉E is defined by ⌉E = {¬ e₁, ¬ e₂, ¬ e₃, . . . , ¬ e_n} where ¬e_i =not e_i, ∀ i. (It may be noted that ⌉ and ¬ are different operators.)

2.2 Bipolar soft sets

Definition 2.4. [20] A triplet (F, G, A) is called a bipolar soft set over U, where F and G are mappings, given by F : A → P (U) and G : ¬ A → P (U) such that F (e)∩ G (¬ e) = ∅ for all e ∈ A.

Now we will modify concept of bipolar soft set defined by Shabir and Naz [20] and Karaaslan and Karataş [16].

Definition 2.5. Let E be a parameter set, S ⊂ E and f : S → E be an injective function. Then S ∪ f (S) is called extended parameter set of S and denoted by $E_{S}$ .

If S = E, then extended parameter set of S will be denoted by $E$ .

Definition 2.6. Let E be a parameter set, S ⊆ E and $S \cup f (S) = E_{S}$ such that $f : S \to E_{S}$ be an injective function. If $F : S \to P (U)$ and $G : f (S) \to P (U)$ are two mappings such that F (e)∩ G (f (e)) = ∅, then triple (F, G, E) is called bipolar soft set. We can represent a bipolar soft set (F, G, E) defined by a mapping as follows: $f_{S} : E \to P (U) \times P (U)$ such that F (e) =∅and G (f (e)) = Uif e ∈ E \ S and f(e) ∈ E \ ɛ_S.

Also we can write a bipolar soft set f_S as a set of triples following form

$\begin{matrix} G, E) & = & {(e, F (e), G (f (e))) : e \in E and \\ F (e) \cap G (f (e)) = \emptyset} \end{matrix}$

If F (e) =∅ and G (f (e)) = U for e ∈ E, then (e, ∅ , U) will not be appeared in the bipolar soft set f_S.

From now onward we will denote the sets F (e) and G (f (e)) with $f_{S}^{+} (e)$ and $f_{S}^{-} (e)$ , respectively and these sets will be called positive and negative soft sets of bipolar soft set f_S, respectively. Set of all bipolar soft sets over U will be denoted by ${BS}_{E} (U) .$

Note 2.7. Let f_S = ( $f_{S}^{+}$ , $f_{S}^{-}$ , E) be a bipolar soft set over U. We will say that $f_{S} (e) = (f_{S}^{+} (e), f_{S}^{-} (e))$ is image of parameter $e \in E$ .

Example 2.8. Let E = Z₁₂ be a parameter set, S = {0, 3, 6, 9} be a subgroup of Z₁₂, U = {u₁, u₂, . . . , u₁₂} and f : S → Z₁₂ be an injective function such that f (e) = e^-1. Then, $E_{S} = S$ and

$\begin{matrix} F (0) = {u_{1}, u_{2}, u_{3}}, G (f (0)) = G (0) = {u_{4}, u_{5}} \\ F (3) = {u_{4}, u_{5}}, G (f (3)) = G (9) = {u_{1}, u_{6}, u_{8}} \\ F (6) = {u_{1}, u_{5}, u_{7}, u_{8}}, G (f (6)) = G (6) = {u_{2}, u_{6}, u_{10}} \\ F (9) = {u_{2}, u_{7}}, G (f (9)) = G (3) = {u_{1}, u_{4}, u_{9}} . \end{matrix}$

Thus,

$\begin{matrix} f_{S} = {(0, {u_{1}, u_{2}, u_{3}}, {u_{4}, u_{5}}), (3, {u_{4}, u_{5}}, \\ u_{7}, u_{8}}, {u_{2}, u_{6}, u_{10}}), \end{matrix}$

is a bipolar soft set over U.

If parameters are linguistic expressions such as “beautiful”, “expensive”, etc., we use the function f (e) = ¬ e. In this case, our definition is reduced Shabir and Naz’s [20] bipolar soft set definition.

Definition 2.9. [16] Let f_S and $f_{T} \in {BS}_{E} (U)$ . Then,

If $f_{S}^{+} (e) = \emptyset$ and $f_{S}^{-} (e) = U$ for all e ∈ E, f_S is said to be a null bipolar soft set, denoted by $Φ_{b} = (Φ, \hat{U}, E)$ .

If $f_{S}^{+} (e) = U$ and $f_{S}^{-} (e) = Φ$ for all e ∈ E, f_S is said to be absolute bipolar soft set, denoted by ${\hat{U}}_{b} = (\hat{U}, Φ, E)$ .

f_S is bipolar soft subset of f_T, denoted by f_S ⊑ f_T, if $f_{S}^{+} (e) \subseteq f_{T}^{+} (e)$ and $f_{S}^{-} (e) \supseteq f_{T}^{-} (e)$ for all e ∈ E.

f_S = f_T, if f_S ⊑ f_T and f_T ⊑ f_S.

Bipolar soft union of f_S and f_T, denoted by f_S⊔f_T, is a soft set over U and defined by $(f_{S}^{+} \tilde{\cup} f_{T}^{+})$ : S ∪ T → P (U) such that $(f_{S}^{+} \tilde{\cup} f_{T}^{+}) (e)$ = $f_{S}^{+} (e)$ ∪ $f_{T}^{+} (e)$ and $(f_{S}^{-} \tilde{\cap} f_{T}^{-})$ : S ∪ T → P (U) such that $(f_{S}^{-} \tilde{\cap} f_{T}^{-}) (e) = f_{S}^{-} (e) \cap f_{T}^{-} (e)$ for all e ∈ E.

Bipolar soft intersection of f_S and f_T, denoted by f_S ⊓ f_T, is a soft set over U and defined by $(f_{S}^{+} \tilde{\cap} f_{T}^{+})$ : S ∩ T → P (U) such that $(f_{S}^{+} \tilde{\cap} f_{T}^{+}) (e)$ = $f_{S}^{+} (e)$ ∩ $f_{T}^{+} (e)$ and $(f_{S}^{-} \tilde{\cup} f_{T}^{-})$ : S ∩ T → P (U) such that $(f_{S}^{-} \tilde{\cup} f_{T}^{-}) (e)$ = $f_{S}^{-} (e)$ ∪ $f_{T}^{-} (e)$ for all e ∈ E.

Bipolar soft complement of f_S is denoted by $f_{S}^{\tilde{c}}$ and defined by $f_{S}^{\hat{c}} : E \to P (U) \times P (U)$ such that $f_{S}^{\hat{c}} (e) = {(e, f_{S}^{-} (e), f_{S}^{+} (e)) : e \in E}$ .

3 Bipolar soft group

In this section, first of all we give the definition of soft intersection group (soft int-group) defined by Çağman et al. [12]. Then, we define bipolar soft group (BS-group) structure, and investigate some of its properties.

Definition 3.1. [12] Let G be a group and f_G ∈ S (U). Then, f_G is called a soft intersection groupoid over U if f_G (ab) ⊇ f_G (a) ∩ f_G (b) for all a, b ∈ G.

If, for all a ∈ G, the soft intersection groupoid satisfies f_G (a^-1) = f_G (a), then f_G is called a soft intersection group over U.

Definition 3.2. Let G be a group, $f : G \to G$ be injective function. Then, $f_{G} = (f_{G}^{+}, f_{G}^{-}, G) \in {BS}_{G} (U)$ is called a bipolar soft groupoid (BS-groupoid) over U if f_G (ab) ⊒ f_G (a) ⊓ f_G (b) for all $a, b \in G$ .

Here, f_G (ab) ⊒ f_G (a) ⊓ f_G (b) means that $f_{G}^{+} (ab)$ ⊇ $f_{G}^{+} (a)$ ∩ $f_{G}^{+} (b)$ and $f_{G}^{-} (ab)$ ⊆ $f_{G}^{-} (a)$ ∪ $f_{G}^{-} (b)$ .

Definition 3.3. Let f_G be a bipolar soft groupoid over U. If f_G (a^-1) = f_G (a), then f_G is called bipolar soft group (BS-group) and denoted by f_G.

Example 3.4. Let U = {-3, - 2, - 1, 0, 1, 2, 3, 4, 5} be an initial universe and G = S₃ be symmetric group. We define a bipolar soft set using positive and negative soft sets $f_{G}^{+}$ and $f_{G}^{-}$ by $f_{G}^{+} (σ) = {\begin{matrix} {a \in U, a \leq r} & o (σ) = r \neg = 1 \\ U, & o (σ) = 1 \end{matrix}$ and $f_{G}^{-} (σ) = {\begin{matrix} {a \in U, a > r} & o (σ) = r \neg = 1 \\ \emptyset, & o (σ) = 1 \end{matrix},$ respectively. Where o (σ) is the order of σ ∈ S₃. Then, $\begin{matrix} f_{G} = {((1), U, \emptyset), ((12), {- 3, - 2, - 1, 0, 1, 2}, {3, 4, 5}), \\ ((13), {- 3, - 2, - 1, 0, 1, 2}, {3, 4, 5}), \\ ((23), {- 3, - 2, - 1, 0, 1, 2}, {3, 4, 5}), \\ ((123), {- 3, - 2, - 1, 0, 1, 2, 3}, {4, 5}), \\ ((132), {- 3, - 2, - 1, 0, 1, 2, 3}, {4, 5})} . \end{matrix}$ One can easily show that f_G is a bipolar soft group over U.

Example 3.5. Assume that U = {u₁, u₂, u₃, u₄, u₅, u₆, u₇, u₈, u₉, u₁₀, u₁₁, u₁₂, u₁₃, u₁₄} is a universal set and G = Z₅ is the subset of parameters. We define a bipolar soft set f_G by

$\begin{matrix} f_{G}^{+} (0) & = & {u_{1}, u_{2}, u_{3}, u_{4}, u_{5}, u_{6}} \\ f_{G}^{+} (1) & = & {u_{1}, u_{2}, u_{3}, u_{4}, u_{5}} \\ f_{G}^{+} (2) & = & {u_{2}, u_{3}, u_{6}} \\ f_{G}^{+} (3) & = & {u_{2}, u_{3}, u_{6}} \\ f_{G}^{+} (4) & = & {u_{1}, u_{2}, u_{3}, u_{4}, u_{5}} \end{matrix}$

and

$\begin{matrix} f_{G}^{-} (0) & = & {u_{10}, u_{11}} \\ f_{G}^{-} (1) & = & {u_{9}, u_{12}, u_{13}, u_{14}} \\ f_{G}^{-} (2) & = & {u_{4}, u_{5}, u_{10}, u_{11}, u_{12}, u_{14}} \\ f_{G}^{-} (3) & = & {u_{4}, u_{5}, u_{10}, u_{11}, u_{12}, u_{14}} \\ f_{G}^{-} (4) & = & {u_{9}, u_{12}, u_{13}, u_{14}} . \end{matrix}$

Here f_G is not a bipolar soft group over U, because $f_{G}^{-} (1 + 4) \tilde{⊈} f_{G}^{-} (1) \tilde{\cup} f_{G}^{-} (4)$ .

Definition 3.6. Let $f_{G} = (f_{G}^{+}, f_{G}^{-}, G)$ be BS-set over U. If, for all a ∈ G, $f_{G}^{+} (a) \cup f_{G}^{-} (a) = U$ , BS-set $f_{G} = (f_{G}^{+}, f_{G}^{-}, G)$ is called full BS-set.

Example 3.7. In Example 3.5, if we take as $(f_{G}^{+} (a))^{c} = f_{G}^{-} (a)$ for all a ∈ G, then f_G is a full bipolar soft set and so f_G is bipolar soft group.

Theorem 3.8. Let f_G full BS-set over U. Then, f_G is BS-group if and only if $f_{G}^{+}$ is soft int-group.

Proof. The proof is clear from Definition 3.1 and 3.6.

Theorem 3.9.Let f_G be a BS-group over U. Then,

f_G (e) ⊒ f_G (a) for all a ∈ G.

f_G (ab) ⊒ f_G (b) if and only if f_G (a) = f_G (e).

Proof.

Since f_G is a BS-group over U, $f_{G}^{+} ({aa}^{- 1})$ ⊇ $f_{G}^{+} (a)$ ∩ $f_{G}^{+} (a^{- 1})$ = $f_{G}^{+} (a)$ ∩ $f_{G}^{+} (a) = f_{G}^{+} (a)$ and $f_{G}^{-} ({aa}^{- 1})$ ⊆ $f_{G}^{-} (a)$ ∪ $f_{G}^{-} (a^{- 1})$ = $f_{G}^{-} (a)$ ∪ $f_{G}^{-} (a)$ = $f_{G}^{-} (a)$ for all a ∈ G.

Suppose that f_G (ab) ⊒ f_G (b) for all b ∈ G. Then by choosing b = e, we have that $f_{G}^{+} (a) \supseteq f_{G}^{+} (e)$ and $f_{G}^{-} (a) \subseteq f_{G}^{-} (e)$ , so f_G (a) ⊒ f_G (e), by item (1) f_G (a) = f_G (e).

Conversely, suppose that f_G (a) = f_G (e). Then, $f_{G}^{+} (ab)$ ⊇ $f_{G}^{+} (a)$ ∩ $f_{G}^{+} (b)$ = $f_{G}^{+} (e)$ ∩ $f_{G}^{+} (b)$ = $f_{G}^{+} (b)$ and $f_{G}^{-} (ab)$ ⊆ $f_{G}^{-} (a)$ ∪ $f_{G}^{-} (b)$ = $f_{G}^{-} (e)$ ∪ $f_{G}^{-} (b)$ = $f_{G}^{-} (b)$ .

Theorem 3.10. A bipolar soft set f_G over U is a BS-group over U if and only if f_G (ab^-1) ⊒ f_G (a) ⊓ f_G (b) for all a, b ∈ G.

Proof. Assume that f_G be a BS-group over U. Then $f_{G}^{+} ({ab}^{- 1}) \supseteq f_{G}^{+} (a) \cap f_{G}^{+} (b^{- 1}) = f_{G}^{+} (a) \cap f_{G}^{+} (b)$ for all a, b ∈ G and $f_{G}^{-} ({ab}^{- 1}) \subseteq f_{G}^{-} (a) \cup f_{G}^{-} (b^{- 1}) = f_{G}^{-} (a) \cup f_{G}^{-} (b)$ . Therefore f_G (ab^-1) ⊑ f_G (b) ⊓ f_G (b).

Conversely, let f_G (ab^-1) ⊒ f_G (a) ∩ f_G (b) for all a, b ∈ G. If we take a = e, $f_{G}^{+} (b^{- 1})$ ⊇ $f_{G}^{+} (b)$ and $f_{G}^{-} (b^{- 1})$ ⊆ $f_{G}^{-} (b)$ . Hence, $f_{G}^{+} (b)$ = $f_{G}^{+} ((b^{- 1})^{- 1})$ ⊇ $f_{G}^{+} (b^{- 1})$ and $f_{G}^{-} (b)$ = $f_{G}^{-} ((b^{- 1})^{- 1})$ ⊆ $f_{G}^{-} (b^{- 1})$ . Thus, f_G (b) = f_G (b^-1). If a ¬ = e, $f_{G}^{+} (ab)$ = $f_{G}^{+} (a (b^{- 1})^{- 1})$ ⊇ $f_{G}^{+} (a)$ ∩ $f_{G}^{+} (b^{- 1})$ = $f_{G}^{+} (a)$ ∩ $f_{G}^{+} (b)$ and $f_{G}^{-} (ab)$ = $f_{G}^{-} (a (b^{- 1})^{- 1})$ ⊆ $f_{G}^{-} (a)$ ∪ $f_{G}^{-} (b^{- 1})$ = $f_{G}^{-} (a)$ ∪ $f_{G}^{-} (b)$ for all a, b ∈ G. Therefore, f_G is a BS-group.

Theorem 3.11. Let f_G be a BS-group over U. Then, f_G (aⁿ) ⊒ f_G (a) for all a ∈ G where n ∈ N.

Proof. Suppose that f_G is a BS-group over U. Then, $f_{G}^{+} (a^{n}) \supseteq \underset{n}{\underset{︸}{f_{G}^{+} (a) \cap f_{G}^{+} (a) \cap . . . \cap f_{G}^{+} (a)}} = f_{G}^{+} (a)$ and $f_{G}^{-} (a^{n}) \subseteq \underset{n}{\underset{︸}{f_{G}^{-} (a) \cup f_{G}^{-} (a) \cup . . . \cup f_{G}^{-} (a)}} = f_{G}^{-} (a)$ for all a ∈ G. Thus, f_G (aⁿ) ⊒ f_G (a).

Theorem 3.12. Let f_G be a BS-group over U. If, for all a, b ∈ G, $f_{G}^{+} ({ab}^{- 1}) = U$ and $f_{G}^{-} ({ab}^{- 1}) = \emptyset$ , then f_G (a) = f_G (b).

Proof. For any a, b ∈ G

$\begin{matrix} f_{G}^{+} (a) = f_{G}^{+} (({ab}^{- 1}) b) & \supseteq & f_{G}^{+} ({ab}^{- 1}) \cap f_{G}^{+} (b) \\ = & U \cap f_{G}^{+} (b) = f_{G}^{+} (b) \end{matrix}$ and $\begin{matrix} f_{G}^{+} (b) & = & f_{G}^{+} (b^{- 1}) \\ = & f_{G}^{+} (a^{- 1} ({ab}^{- 1})) \\ \supseteq & f_{G}^{+} (a^{- 1}) \cap f_{G}^{+} ({ab}^{- 1}) \\ = & f_{G}^{+} (a^{- 1}) \cap U \\ = & f_{G}^{+} (a) \end{matrix}$ thus,

$f_{G}^{+} (a) = f_{G}^{+} (b)$ (1) Also, $\begin{matrix} f_{G}^{-} (a) = f_{G}^{-} (({ab}^{- 1}) b) & \subseteq & f_{G}^{-} ({ab}^{- 1}) \cup f_{G}^{-} (b) \\ = & \emptyset \cup f_{G}^{-} (b) = f_{G}^{-} (b) \end{matrix}$ and $\begin{matrix} f_{G}^{-} (b) & = & f_{G}^{-} (b^{- 1}) \\ = & f_{G}^{-} (a^{- 1} ({ab}^{- 1})) \\ \subseteq & f_{G}^{-} (a^{- 1}) \cup f_{G}^{-} ({ab}^{- 1}) \\ = & f_{G}^{-} (a^{- 1}) \cup \emptyset \\ = & f_{G}^{-} (a) \end{matrix}$ thus,

$f_{G}^{-} (a) = f_{G}^{-} (b) .$ (2) from equations (1) and (2)f_G (a) = f_G (b) .

Definition 3.13. Let f_G be a BS-set. Then, e-set of f_G, denoted by e_{f
_G}, is defined as $e_{f_{G}} = {a \in G : f_{G} (a) = f_{G} (e)}$

Example 3.14. Let us consider Klein-4 group over G = {e, a, b, c} given as in following Cayley table

and let f_G be a BS-set over U = {u₁, u₂, u₃, u₄, u₅, u₆, u₇} with $f_{G}^{+}$ and $f_{G}^{-}$ given as follows:

$\begin{matrix} f_{G}^{+} (e) & = & {u_{1}, u_{3}, u_{7}} \\ f_{G}^{+} (a) & = & {u_{2}, u_{3}, u_{4}} \\ f_{G}^{+} (b) & = & {u_{1}, u_{3}, u_{7}} \\ f_{G}^{+} (c) & = & {u_{1}, u_{3}, u_{7}} \end{matrix}$ and

$\begin{matrix} f_{G}^{-} (e) & = & {u_{2}, u_{4}, u_{6}} \\ f_{G}^{-} (a) & = & {u_{1}, u_{5}} \\ f_{G}^{-} (b) & = & {u_{2}, u_{4}} \\ f_{G}^{-} (c) & = & {u_{2}, u_{4}, u_{6}} . \end{matrix}$

Then, e_{f
_G} = {e, c} .

Theorem 3.15. Let f_G be a BS-group over U. Then, e_{f
_G} is a subgroup of G.

Proof. From definition of e_{f
_G}, it is obvious that e_{f
_G}¬ = ∅. Let a, b ∈ e_{f
_G}. Then, $f_{G}^{+} (a) = f_{G}^{+} (e)$ and $f_{G}^{-} (b) = f_{G}^{-} (e)$ and so

$\begin{matrix} f_{G}^{+} ({ab}^{- 1}) & \supseteq & f_{G}^{+} (a) \cap f_{G}^{+} (b^{- 1}) \\ = & f_{G}^{+} (a) \cap f_{G}^{+} (b) \\ = & f_{G}^{+} (e) \cap f_{G}^{+} (e) \\ = & f_{G}^{+} (e) \end{matrix}$

from Definition 3.2, we know that $f_{G}^{+} (e) \supseteq f_{G}^{+} ({ab}^{- 1})$ . Thus,

$f_{G}^{+} (e) = f_{G}^{+} ({ab}^{- 1}) .$ (3)

$\begin{matrix} f_{G}^{-} ({ab}^{- 1}) & \subseteq & f_{G}^{-} (a) \cup f_{G}^{-} (b^{- 1}) \\ = & f_{G}^{-} (a) \cup f_{G}^{-} (b) \\ = & f_{G}^{-} (e) \cup f_{G}^{-} (e) \\ = & f_{G}^{-} (e) \end{matrix}$ from Theorem 3.9 we know that $f_{G}^{-} (e) \subseteq f_{G}^{-} ({ab}^{- 1})$ . Therefore,

$f_{G}^{-} (e) = f_{G}^{-} ({ab}^{- 1}) .$ (4) From (3) and (4), f_G (e) = f_G (ab^-1) and ab^-1∈ e_{f
_G}. Hence e_{f
_G} is a subgroup of G.

Example 3.16. Let us consider Klein-4 group over G = {e, a, b, c} given as in following Cayley table

\begin{matrix} * & e & a & b & c \\ e & e & a & b & c \\ a & a & e & c & b \\ b & b & c & e & a \\ c & c & b & a & e \end{matrix}

and let f_G be a BS-group over U = {u₁, u₂, u₃, u₄, u₅, u₆, u₇} with $f_{G}^{+}$ and $f_{G}^{-}$ given as follows,

$\begin{matrix} f_{G}^{+} (e) & = & {u_{1}, u_{2}, u_{3}, u_{4}} \\ f_{G}^{+} (a) & = & {u_{1}, u_{3}} \\ f_{G}^{+} (b) & = & {u_{1}, u_{3}} \\ f_{G}^{+} (c) & = & {u_{1}, u_{2}, u_{3}, u_{4}} \end{matrix}$

and

$\begin{matrix} f_{G}^{-} (e) & = & {u_{5}, u_{7}} \\ f_{G}^{-} (a) & = & {u_{2}, u_{4}, u_{5}, u_{7}} \\ f_{G}^{-} (b) & = & {u_{2}, u_{4}, u_{5}, u_{7}} \\ f_{G}^{-} (c) & = & {u_{5}, u_{7}} . \end{matrix}$

Then, e_{f
_G} = {e, c} .

Theorem 3.17. Let f_G and g_G be two BS-groups over U. Then, f_G ⊓ g_G is a BS-group over U.

Proof. Let a, b ∈ G, then $\begin{matrix} f_{G}^{+} ({ab}^{- 1}) \cap g_{G}^{+} ({ab}^{- 1}) \\ \supseteq (f_{G}^{+} (a) \cap f_{G}^{+} (b)) \cap (g_{G}^{+} (a) \cap g_{G}^{+} (b)) \\ = (f_{G}^{+} (a) \cap g_{G}^{+} (a)) \cap (f_{G}^{+} (b) \cap g_{G}^{+} (b)) \\ = (f_{G}^{+} \tilde{\cap} g_{G}^{+}) (a) \cap (f_{G}^{+} \tilde{\cap} g_{G}^{+}) (b) \end{matrix}$

and $\begin{matrix} (f_{G}^{-} \tilde{\cup} g_{G}^{-}) ({ab}^{- 1}) \subseteq f_{G}^{-} ({ab}^{- 1}) \cup g_{G}^{-} ({ab}^{- 1}) \\ \subseteq & (f_{G}^{-} (a) \cup f_{G}^{-} (b)) \cup (g_{G}^{-} (a) \cup g_{G}^{-} (b)) \\ = & (f_{G}^{-} (a) \cup g_{G}^{-} (a)) \cup (f_{G}^{-} (b) \cup g_{G}^{-} (b)) \\ = & (f_{G}^{+} \tilde{\cup} g_{G}^{-}) (a) \cup (f_{G}^{-} \tilde{\cup} g_{G}^{-}) (b) . \end{matrix}$

This implies that $(f_{G}^{+} \tilde{\cap} g_{G}^{+}) ({ab}^{- 1})$ ⊇ $(f_{G}^{+} \tilde{\cap} g_{G}^{+}) (a)$ ∩ $(f_{G}^{+} \tilde{\cap} g_{G}^{+}) (b)$ and $(f_{G}^{-} \tilde{\cup} g_{G}^{-}) ({ab}^{- 1})$ ⊆ $(f_{G}^{-} \tilde{\cup} g_{G}^{-}) (a)$ ∪ $(f_{G}^{-} \tilde{\cup} g_{G}^{-}) (b)$ . Hence, (f_G ⊓ g_G) (ab^-1) ⊒ f_G (ab^-1) ⊓ g_G (ab^-1) is a BS-group over U.

Note that f_G ⊔ g_G is not a BS-group over U in general.

Example 3.18. Let G = Z₆ be the set of parameters and U = Z be the universal set. If we construct two BS-groups f_G and g_G over U by $\begin{matrix} f_{G}^{+} (0) & = & Z \\ f_{G}^{+} (1) & = & {5, 6, 9, 16} \\ f_{G}^{+} (2) & = & Z \\ f_{G}^{+} (3) & = & {5, 6, 9, 16} \\ f_{G}^{+} (4) & = & Z \\ f_{G}^{+} (5) & = & {5, 6, 9, 16}, \end{matrix}$ $\begin{matrix} f_{G}^{-} (0) & = & \emptyset \\ f_{G}^{-} (1) & = & {0, 1, 4, 12, 13, 14, 19} \\ f_{G}^{-} (2) & = & \emptyset \\ f_{G}^{-} (3) & = & {0, 1, 4, 12, 13, 14, 19} \\ f_{G}^{-} (4) & = & \emptyset \\ f_{G}^{+} (5) & = & {0, 1, 4, 12, 13, 14, 19}, \end{matrix}$ and $\begin{matrix} g_{G}^{+} (0) & = & Z \\ g_{G}^{+} (1) & = & {7, 8, 11, 18} \\ g_{G}^{+} (2) & = & {7, 8, 11, 18} \\ g_{G}^{+} (3) & = & Z \\ g_{G}^{+} (4) & = & {7, 8, 11, 18} \\ g_{G}^{+} (5) & = & {7, 8, 11, 18}, \end{matrix}$

$\begin{matrix} g_{G}^{-} (0) & = & \emptyset \\ g_{G}^{-} (1) & = & {2, 3, 6, 14, 18, 19} \\ g_{G}^{-} (2) & = & {2, 3, 6, 14, 18, 19} \\ g_{G}^{-} (3) & = & \emptyset \\ g_{G}^{-} (4) & = & {2, 3, 6, 14, 18, 19} \\ g_{G}^{-} (5) & = & {2, 3, 6, 14, 18, 19} \end{matrix}$

here

$\begin{matrix} (f_{G} \tilde{\cup} g_{G})^{+} (0) = f_{G}^{+} (0) \cup g_{G}^{+} (0) = Z \\ (f_{G} \tilde{\cup} g_{G})^{+} (1) = f_{G}^{+} (1) \cup g_{G}^{+} (1) = \\ {5, 6, 7, 8, 9, 11, 16, 18} \\ (f_{G} \tilde{\cup} g_{G})^{+} (1) = f_{G}^{+} (1) \cup g_{G}^{+} (1) = \\ {5, 6, 7, 8, 9, 11, 16, 18} \\ (f_{G} \tilde{\cup} g_{G})^{+} (2) = f_{G}^{+} (2) \cup g_{G}^{+} (2) = Z \\ (f_{G} \tilde{\cup} g_{G})^{+} (3) = f_{G}^{+} (3) \cup g_{G}^{+} (3) = Z \\ (f_{G} \tilde{\cup} g_{G})^{+} (4) = f_{G}^{+} (4) \cup g_{G}^{+} (4) = Z \\ (f_{G} \tilde{\cup} g_{G})^{+} (5) = f_{G}^{+} (5) \cup g_{G}^{+} (5) = \\ {5, 6, 7, 8, 9, 11, 16, 18} \end{matrix}$

and

$\begin{matrix} (f_{G} \tilde{\cap} g_{G})^{-} (0) & = & f_{G}^{-} (0) \cap g_{G}^{-} (0) & = & \emptyset \\ (f_{G} \tilde{\cap} g_{G})^{-} (1) & = & f_{G}^{-} (1) \cap g_{G}^{-} (1) & = & {14, 19} \\ (f_{G} \tilde{\cap} g_{G})^{-} (2) & = & f_{G}^{-} (2) \cap g_{G}^{-} (2) & = & \emptyset \\ (f_{G} \tilde{\cap} g_{G})^{-} (3) & = & f_{G}^{-} (3) \cap g_{G}^{-} (3) & = & \emptyset \\ (f_{G} \tilde{\cap} g_{G})^{-} (4) & = & f_{G}^{-} (4) \cap g_{G}^{-} (4) & = & \emptyset \\ (f_{G} \tilde{\cap} g_{G})^{-} (5) & = & f_{G}^{-} (5) \cap g_{G}^{-} (5) & = & {14, 19} . \end{matrix}$

Here since (f_G $\tilde{\cup}$ g_G)⁺(3 +4) ⊉ $(f_{G} \tilde{\cup} g_{G})^{+} (3)$ ∩ $(f_{G} \tilde{\cup} g_{G})^{+} (4)$ and (f_G $\tilde{\cap}$ g_G)^-(3+ 4) ⊈ (f_G $\tilde{\cap}$ g_G)^-(3) ∪ (f_G $\tilde{\cap}$ g_G)^-(4). Hence f_G ⊔ g_G is not a BS-group over U.

Definition 3.19. Let H be a subgroup of G, f_G be a BS-group over U and f_H be a nonempty BS-subset of f_G over U. If f_H is a BS-group over U, then f_H is called a BS-subgroup of f_G over U and denoted by f_H ⪯ f_G.

Example 3.20. Let us consider BS-group f_G over U in Example 3.18, and let H = {0, 2, 4} ≤ G. If we define a BS-set f_H by positive and negative soft sets as follows; $f_{H}^{+} (0) = f_{H}^{+} (2) = f_{H}^{+} (4) = Z$ and $f_{H}^{-} (0) = f_{H}^{-} (2) = f_{H}^{-} (4) = \emptyset .$ Then f_H is a BS-subgroup of f_G over U.

Theorem 3.21. Let f_G be a BS-group over U and f_H, f_N be two BS-subgroups of f_G over U. Then, f_H ⊓ f_N ⪯ f_G over U.

Proof. Let a, b ∈ G. Then, $\begin{matrix} f_{H \cap N}^{+} ({ab}^{- 1}) = f_{H}^{+} ({ab}^{- 1}) \cap f_{N}^{+} ({ab}^{- 1}) \\ \supseteq (f_{H}^{+} (a) \cap f_{H}^{+} (b)) \cap (f_{N}^{+} (a) \cap f_{N}^{+} (b)) \\ = (f_{H}^{+} (a) \cap f_{N}^{+} (a)) \cap (f_{H}^{+} (b) \cap f_{N}^{+} (b)) \\ = f_{H \cap N}^{+} (a) \cap f_{H \cap N}^{+} (b) \end{matrix}$ and $\begin{matrix} f_{H \cap N}^{-} ({ab}^{- 1}) = f_{H}^{-} ({ab}^{- 1}) \cup f_{N}^{-} ({ab}^{- 1}) \\ \subseteq (f_{H}^{-} (a) \cup f_{H}^{-} (b)) \cup (f_{N}^{-} (a) \cup f_{N}^{-} (b)) \\ = (f_{H}^{-} (a) \cup f_{N}^{-} (a)) \cup (f_{H}^{-} (b) \cup f_{N}^{-} (b)) \\ = f_{H \cup N}^{-} (a) \cup f_{H \cup N}^{-} (b) \end{matrix}$ Therefore, f_H ⊓ f_N is a BS-subgroup over U.

Note that f_H ⊔ f_N is not a BS-group over U in general.

Theorem 3.22. Let f_{G
_i} be a family of BS-group over U for all i ∈ I. Then ⊓_i∈If_{G
_i} is a BS-group over U.

Proof. Let a, b ∈ G. Since f_{G
_i} be a BS-group over U. This implies that f_{G
_i} (ab^-1) ⊒ f_{G
_i} (a) ⊓ f_{G
_i} (b) for all i ∈ I. Then, $\begin{matrix} f_{G_{i}}^{+} ({ab}^{- 1}) & \supseteq & f_{G_{i}}^{+} (a) \cap f_{G_{i}}^{+} (b) and \\ ⋂_{i \in I} f_{G_{i}}^{+} ({ab}^{- 1}) & \supseteq & ⋂_{i \in I} (f_{G_{i}}^{+} (a) \cap f_{G_{i}}^{+} (b)) \\ = & (⋂_{i \in I} f_{G_{i}}^{+} (a)) \cap (⋂_{i \in I} f_{G_{i}}^{+} (b)) \end{matrix}$ and $\begin{matrix} f_{G_{i}}^{-} ({ab}^{- 1}) & \subseteq & f_{G_{i}}^{-} (a) \cup f_{G_{i}}^{-} (b) and \\ ⋃_{i \in I} f_{G_{i}}^{-} ({ab}^{- 1}) & \subseteq & ⋃_{i \in I} (f_{G_{i}}^{-} (a) \cup f_{G_{i}}^{-} (b)) \\ = & (⋃_{i \in I} f_{G_{i}}^{-} (a)) \cup (⋃_{i \in I} f_{G_{i}}^{-} (b)) . \end{matrix}$ Thus, ⊓_i∈If_{G
_i} is BS-group over U.

Definition 3.23. Let f_G be a BS-group over U. For any a ∈ G, centralizer of $(a, f_{G}^{+} (a), f_{G}^{-} (a)) \in f_{G}$ defined as follows: $M_{f_{G}} (a) = {(g, f_{G}^{+} (g), g_{G}^{-} (g)) \in G : ag = ga} .$

Example 3.24. Let us consider BS-group f_G in Example 3.4. Then centralizer of $((123), {- 3, - 2, - 1, 0, 1, 2, 3}, {4, 5}) \in f_{G}$ can be obtained as follow: $\begin{matrix} M_{f_{G}} (123) = {((1), U, \emptyset), ((123), {- 3, - 2, - 1, 0, 1, 2, \\ 3}, {4, 5}), ((132), {- 3, - 2, - 1, 0, 1, 2, \\ 3}, {4, 5})} . \end{matrix}$

Theorem 3.25. Let f_G be a BS-group over U and M_{f
_G} (a) be centralizer of a ∈ G. Then, M_{f
_G} (a) is a BS-subgroup of f_G.

Proof. The proof is clear from Definition 3.19.

4 (α, β)-level of bipolar soft set

Definition 4.1. Let f_A be a BS-set over U. Then (α, β)-level of BS-set f_A, denoted by $f_{A}^{(α, β)}$ , is defined as follows: $f_{A}^{(α, β)} = {a \in A : f_{A}^{+} (a) \supseteq α and f_{A}^{-} (a) \subseteq β} .$ Here α∩ β = ∅.

Note that if α =∅ or β = U, then $f_{A}^{(\emptyset, U)} = {a \in A : f_{A}^{+} (a) \neg = \emptyset and f_{A}^{-} (a) \neg = U}$ is called support of f_A, and denoted by Supp (f_A).

Example 4.2. Let U = {u₁, u₂, u₃, u₄, u₅, u₆, u₇} be an initial universe and E = {e₁, e₂, e₃, e₄, e₅} be a parameter set. If we define bipolar soft set as follow: $\begin{matrix} f_{A}^{+} (e_{1}) & = & \emptyset \\ f_{A}^{+} (e_{2}) & = & {u_{2}, u_{3}, u_{4}, u_{5}, u_{6}} \\ f_{A}^{+} (e_{3}) & = & {u_{5}, u_{6}, u_{7}} \\ f_{A}^{+} (e_{4}) & = & \emptyset \\ f_{A}^{+} (e_{5}) & = & {u_{3}, u_{5}, u_{7}} \end{matrix}$ and $\begin{matrix} f_{A}^{-} (e_{1}) & = & {u_{1}, u_{2}, u_{3}} \\ f_{A}^{-} (e_{2}) & = & {u_{1}} \\ f_{A}^{-} (e_{3}) & = & {u_{2}, u_{4}} \\ f_{A}^{-} (e_{4}) & = & U \\ f_{A}^{-} (e_{5}) & = & {u_{2}, u_{4}} . \end{matrix}$

Let α = {u₅, u₇} and β = {u₂, u₄, u₆}. Then $f_{A}^{(α, β)} = {e_{3}, e_{5}} .$

Proposition 4.3. Let f_A and f_B be two BS-sets over U, A, B ⊆ E. Then, the following assertionshold;

$f_{A} ⊑ f_{B} \Rightarrow f_{A}^{(α, β)} \subseteq f_{B}^{(α, β)}$ , for all α, β ⊆ U such that α∩ β = ∅.

If α₁ ⊆ α₂ and β₂ ⊆ β₁, then $f_{A}^{(α_{2}, β_{2})} \subseteq f_{A}^{(α_{1}, β_{1})}$ , for all α₁, α₂, β₁, β₂ ⊆ U such that α₁∩ β₁ = ∅, α₂∩ β₂ = ∅.

$f_{A} = f_{B} \Leftrightarrow f_{A}^{(α, β)} = f_{B}^{(α, β)}$ , for all α, β ⊆ U such that α∩ β = ∅.

Proof. Suppose that f_A and f_B are two BS-sets over U.

Let $a \in f_{A}^{(α, β)}$ , then $f_{A}^{+} (a) \supseteq α and f_{A}^{-} (a) \subseteq β$ Since f_A ⊑ f_B, $α \subseteq f_{A}^{+} (a) \subseteq f_{B}^{+} (a)$ and $f_{A}^{-} (a) \supseteq f_{B}^{-} (a) \supseteq β$ , for all a ∈ G. This implies that $a \in f_{B}^{(α, β)}$ . Hence $f_{A}^{(α, β)} \subseteq f_{B}^{(α, β)}$ .

Let α₁ ⊆ α₂ and β₂ ⊆ β₁ and $a \in f_{A}^{(α_{2}, β_{2})}$ , then $f_{A}^{+} (a) \supseteq α_{2} and f_{A}^{-} (a) \subseteq β_{2}$ since α₁ ⊆ α₂ and β₂ ⊆ β₁, $f_{A}^{+} (a) \supseteq α_{1} and f_{A}^{-} (a) \subseteq β_{1} .$ This implies that $f_{A}^{(α_{2}, β_{2})} \subseteq f_{A}^{(α_{1}, β_{1})}$ .

The proof is clear.

Theorem 4.4. Let f_A and f_B are two BS-sets over U, A, B ⊆ E and α, β ⊆ U such that α∩ β = ∅. Then,

$f_{A}^{(α, β)} \cup f_{B}^{(α, β)} \subseteq (f_{A} ⊔ f_{B})^{(α, β)}$

$f_{A}^{(α, β)} \cap f_{B}^{(α, β)} = (f_{A} ⊓ f_{B})^{(α, β)}$

Proof.

For all a ∈ E, let $a \in f_{A}^{(α, β)} \cup f_{B}^{(α, β)}$ $\begin{matrix} \Rightarrow (f_{A}^{+} (a) \supseteq α and f_{A}^{-} (a) \subseteq β) \\ \lor (f_{B}^{+} (a) \supseteq α and f_{B}^{-} (a) \subseteq β) \\ \Rightarrow (f_{A}^{+} (a) \cup f_{B}^{+} (a) \supseteq α) or (f_{A}^{-} (a) \cap f_{B}^{-} (a) \subseteq β) \\ \Rightarrow a \in (f_{A} ⊔ f_{B})^{(α, β)} . \end{matrix}$ Hence, $f_{A}^{(α, β)} \cup f_{B}^{(α, β)} \subseteq (f_{A} ⊔ f_{B})^{(α, β)} .$

Similar to the proof of 1.

Theorem 4.5. Let I be an index set and f_{A
_i} be a family of BS-sets over U. Then, for any α, β ⊆ U such that α∩ β = ∅,

⋃_(i∈I) (f_{A
_i})^(α, β)⊆ (⊔ _(i∈I)f_{A
_i})^(α, β),

⋂_(i∈I) (f_{A
_i})^(α, β)= (⊓ _(i∈I)f_{A
_i})^(α, β).

Theorem 4.6. Let f_A be a BS-set over U and {α_i : i ∈ I} and {β_j : j ∈ I} be two non-empty family of subsets of U. If $\underline{α} = \cap {α_{i} : i \in I}, \bar{α} = \cup {α_{i} : i \in I}$ , $\underline{β} = \cap {β_{j} : j \in I}$ and $\bar{β} = \cup {β_{j} : j \in I}$ , then the following assertions hold,

∪_(i∈I) $f_{A}^{(α_{i}, β_{j})} \subseteq f_{A}^{(\underline{α}, \bar{β})}$ ,

∩_(i∈I) $f_{A}^{(α_{i}, β_{j})} = f_{A}^{(\bar{α}, \underline{β})}$ .

Proof. The proof is clear from Definition 4.1.

Theorem 4.7. Let f_G be a BS-group over U and α, β ⊆ U such that α∩ β = ∅. Then, $f_{G}^{(α, β)}$ is a subgroup of G whenever it is nonempty.

Proof. It is clear that $f_{G}^{(α, β)} \neg = \emptyset$ . Suppose that $a, b \in f_{G}^{(α, β)}$ , then $f_{G}^{+} (a) \supseteq α, f_{G}^{+} (b) \supseteq α, f_{G}^{-} (a) \subseteq β$ and $f_{G}^{-} (a) \subseteq β$ , $\begin{matrix} f_{G}^{+} ({ab}^{- 1}) & \supseteq & f_{G}^{+} (a) \cap f_{G}^{+} (b^{- 1}) \\ = & f_{G}^{+} (a) \cap f_{G}^{+} (b) \\ \supseteq & α \end{matrix}$ also $\begin{matrix} f_{G}^{-} ({ab}^{- 1}) & \subseteq & f_{G}^{-} (a) \cup f_{G}^{+} (b^{- 1}) \\ = & f_{G}^{-} (a) \cup f_{G}^{-} (b) \\ \subseteq & β . \end{matrix}$ Therefore, ${ab}^{- 1} \in f_{G}^{(α, β)}$ and $f_{G}^{(α, β)}$ is a subgroup of G.

Definition 4.8. Let h be a function from A to B and $f_{A}, f_{B} \in {BS}_{E} (U)$ . Then, bipolar soft image of f_A under h and bipolar soft preimage (or bipolar soft inverse image) of f_B under f are the bipolar soft set h (f_A) and h^-1(f_B) such that $\begin{matrix} f (f_{A}) (b) = \\ {\begin{matrix} {(a, \cup f_{A}^{+} (a), \cap f_{A}^{-} (a)) : a \in A} h (a) = b \\ (\emptyset, U), otherwise \end{matrix} \end{matrix}$

for all b ∈ B and h^-1(f_B) (a) = f_B (h (a)) for all a ∈ A, respectively. Here h (f_A) is called the image of f_A under h and h^-1(f_B) is called preimage (or inverse image) of f_B under h.

Example 4.9. Let U = {u₁, u₂, u₃, u₄, u₅, u₆, u₇} is a universal set. Let A = {-2, - 1, 1, 2} and B = {0, 1, 2, 3, 4} be two subsets of set of parameters, and h : A → B, f (a) = a². We define a BS-set over U by positive and negative soft sets as follows:

$\begin{matrix} f_{A}^{+} (- 2) & = & {u_{1}, u_{3}} \\ f_{A}^{+} (- 1) & = & \emptyset \\ f_{A}^{+} (1) & = & {u_{2}, u_{3}, u_{4}, u_{5}, u_{6}} \\ f_{A}^{+} (2) & = & {u_{5}, u_{6}, u_{7}} \end{matrix}$

and

$\begin{matrix} f_{A}^{-} (- 2) & = & {u_{2}, u_{4}, u_{7}} \\ f_{A}^{-} (- 1) & = & {u_{1}, u_{2}, u_{3}} \\ f_{A}^{-} (1) & = & {u_{1}} \\ f_{A}^{-} (2) & = & {u_{1}, u_{2}} \end{matrix}$

and

$\begin{matrix} f_{B}^{+} (0) & = & {u_{3}, u_{4}, u_{5}, u_{6}} \\ f_{B}^{+} (2) & = & {u_{5}, u_{6}, u_{7}} \\ f_{B}^{+} (3) & = & {u_{5}, u_{6}, u_{7}} \\ f_{B}^{+} (4) & = & {u_{5}, u_{6}, u_{7}} \end{matrix}$

and

$\begin{matrix} f_{B}^{-} (0) & = & {u_{1}, u_{2}, u_{3}} \\ f_{B}^{-} (1) & = & {u_{1}} \\ f_{B}^{-} (2) & = & {u_{1}, u_{2}} \\ f_{B}^{-} (3) & = & {u_{1}} \\ f_{B}^{-} (4) & = & {u_{1}, u_{2}} \end{matrix}$

Then, $\begin{matrix} h (f_{A}) & = & {(0, \emptyset, U), (1, {u_{2}, u_{3}, u_{4}, u_{5}, u_{6}}, {u_{1}}), \\ (2, \emptyset, U), (3, \emptyset, U), (4, {u_{1}, u_{3}, u_{5}, u_{6}, u_{7}}, \\ {u_{2}})}, \end{matrix}$ and $\begin{matrix} h^{- 1} (f_{B}) & = & {(- 1, {u_{2}, u_{3}, u_{4}, u_{5}, u_{6}}, {u_{1}}), \\ (- 2, {u_{5}, u_{6}, u_{7}}, {u_{1}, u_{2}}), \\ (1, {u_{2}, u_{3}, u_{4}, u_{5}, u_{6}}, {u_{1}}), \\ (2, {u_{5}, u_{6}, u_{7}}, {u_{1}, u_{2}})} . \end{matrix}$

Theorem 4.10. Let h be a function from A to B, A_i ⊆ A, B_i ⊆ B and f_{A
_i}, f_{B
_i} be two BS-sets over U for all i ∈ I. Then

h (⊔ _i∈If_{A
_i}) = ⊔ _i∈Ih (f_{A
_i}),

f_A
₁ ⊑ f_A
₂ ⇒ h (f_A
₁) ⊑ h (f_A
₂),

f_B
₁ ⊑ f_B
₂ ⇒ h^-1(f_B
₁) ⊑ h^-1(f_B
₂) .

Proof.

For all i ∈ I, BS-sets f_{A
_i} and b ∈ B $\begin{matrix} h (⊔_{i \in I} f_{A_{i}}) (b) \\ = {(a, \cup_{i \in I} \cup_{a \in A_{i}} f_{A_{i}}^{+} (a), \cap_{i \in I} \cap_{a \in A_{i}} f_{A_{i}}^{-} (a)) : \\ a \in A_{i}, h (a) = b} \\ = ⊔_{i \in I} {(a, \cup_{a \in A_{i}} f_{A_{i}}^{+} (a), \cap_{a \in A_{i}} f_{A_{i}}^{-}) : \\ h (a) = b}, \\ = ⊔_{i \in I} h (f_{A_{i}}) (b) . \end{matrix}$

Let f_A
₁ ⊑ f_A
₂, so A₁ ⊆ A₂, then $\begin{matrix} h (f_{A_{1}}) (b) \\ = {(a, \cup_{a \in A_{1}} f_{A_{1}}^{+} (a), \cap_{a \in A_{1}} f_{A_{1}}^{-} (a) : h (a) = b} \\ ⊑ {(a, \cup_{a \in A_{2}} f_{A_{2}}^{+} (a), \cap_{a \in A_{2}} f_{A_{2}}^{-} (a) : h (a) = b} \\ = h (f_{A_{2}}) (b) . \end{matrix}$

Let f_B
₁ ⊑ f_B
₂ then, for all a ∈ A, h^-1(f_B
₁) (a) = f_B
₁ (h (a)) = ${(a, f_{B_{1}}^{+} (h (a)),$ $f_{B_{1}}^{-} (h (a)))$ : a ∈ A}. Since f_B
₁ ⊑ f_B
₂, $f_{B_{1}}^{+} (h (a))$ ⊆ $f_{B_{2}}^{+} (h (a))$ and $f_{B_{1}}^{-} (h (a))$ ⊇ $f_{B_{2}}^{-} (h (a))$ . Therefore, ${(a, f_{B_{1}}^{+} (h (a)),$ $f_{B_{1}}^{-} (h (a)))$ : a ∈ A} ⊑ ${(a, f_{B_{2}}^{+} (h (a))$ , $f_{B_{2}}^{-} (h (a)))$ : a ∈ A} and f_B
₁ (h (a)) ⊑ f_B
₂ (h (a)) = h^-1(f_B
₂) (a).

Theorem 4.11. Let h be a function from A to B, I be a nonempty index set, B_i ⊆ B and f_{B
_i} be BS-set over U for all i ∈ I. Then,

h^-1(⊔ _i∈If_{B
_i}) = ⊔ _i∈Ih^-1(f_{B
_i})

h^-1(⊓ _i∈If_{B
_i}) = ⊓ _i∈Ih^-1(f_{B
_i}).

Proof. For all a ∈ A,

$\begin{matrix} 1 . h^{- 1} (⊔_{i \in I} f_{B_{i}}) (a) & = & ⊔_{i \in I} f_{B_{i}} (h (a)) \\ = & ⊔_{i \in I} h^{- 1} (f_{B_{i}}) (a) . \end{matrix}$ $\begin{matrix} 2 . h^{- 1} (⊓_{i \in I} f_{B_{i}}) (a) & = & ⊓_{i \in I} f_{B_{i}} (h (a)) \\ = & ⊓_{i \in I} h^{- 1} (f_{B_{i}}) (a) . \end{matrix}$

Theorem 4.12. Let h be a function from A to B. Then, h^-1(h (f_A)) ⊒ f_A for all $f_{A} \in {BS}_{E} (U)$ . In particular, if h is an injective function, then h^-1(h (f_A)) = f_A .

Proof. For all a ∈ A, h^-1(h (f_A)) (a) = h (f_A) (h (a)) = {(a, ∪ $f_{A}^{+} (h (a)),$ ∩ $f_{A}^{-} (h (a)))$ : h (a′) = h (a)} ⊒ f_A. Thus h^-1(h (f_A)) ⊒ f_A.

Corollary 4.13. If h is one to one function, then h (a′) = h (a) implies a′ = a and the last inclusion is reduced to equality.

Theorem 4.14. Let h be a function from A to B. For all $f_{B} \in {BS}_{E} (U)$ , h (h^-1(f_B)) ⊑ f_B . In particular, if f is an surjective function, then h (h^-1(f_B)) = f_B .

Proof. For all a ∈ A,

$\begin{matrix} h (h^{- 1} (f_{B})) (b) & = & ⊔ {h^{- 1} (f_{B}) (a) : a \in A, h (a) = b} \\ = & ⊔ {(f_{B}) (h (a)) : \forall a \in A, h (a) = b} \\ = & {\begin{matrix} f_{B} (b), & if b \in h (A) \\ (\emptyset, U), & otherwise \end{matrix} \\ ⊑ & f_{B} (b) . \end{matrix}$

Therefore h (h^-1(f_B)) ⊑ f_B. If f is an onto function, then b ∈ h (A) for all b ∈ B and so h (h^-1(f_B)) = f_B .

Theorem 4.15. Let h be a function from A to B. Then, h (f_A) ⊑ f_B ⇔ f_A ⊑ h^-1(f_B) for all $f_{A}, f_{B} \in {BS}_{E} (U)$ .

Proof. By theorem 4.10 we know that h (f_A) ⊑ f_B ⇒ h^-1(h (f_A)) ⊑ h^-1(f_B) and from Theorem 4.12, f_A ⊑ h^-1(h (f_A)), so f_A ⊑ h^-1(f_B).

Conversely, assume that f_A ⊑ h^-1(f_B). Then from Theorem 4.10 and 4.14, h (f_A) ⊑ h (h^-1(f_B) ⊑ f_B.

Theorem 4.16. Let h be a function from A to B and g be a function from B to C. Then,
$g (h (f_{A})) = (g \circ h) (f_{A}), for all f_{A} \in {BS}_{E} (U)$ .

$h^{- 1} (g^{- 1} (f_{C})) = (g \circ h)^{- 1} (f_{C}), for all f_{C} \in {BS}_{E} (U)$ .

Proof. Consider any $f_{A} \in {BS}_{E} (U)$ and any c ∈ C, then $\begin{matrix} 1 . & g (h (f_{A})) (c) \\ B, g (b) = c} \\ f_{A}^{+} (a), \cap f_{A}^{-} (a)) : a \in A, h (a) = b} : \\ B, g (b) = c} \\ A, (g \circ h) (a) = c} \\ (goh) (f_{A}) (c) . \end{matrix}$ 2. For any $f_{C} \in {BS}_{E} (U)$ and for all a ∈ A, $\begin{matrix} (g \circ h)^{- 1} (f_{C}) (a) & = & f_{C} (g (h (a))) \\ = & g^{- 1} (f_{C}) (h (a)) \\ = & h^{- 1} (g^{- 1} (f_{C})) (a) . \end{matrix}$

Definition 4.17. Let G be a group and f_G, g_G be two BS-sets over U. Then, product of f_G and g_G is defined as follow, for all a ∈ G, $\begin{matrix} (f_{G} * g_{G}) (a) = ⊔ {f_{G} (b) ⊓ g_{G} (c) : b, c \in \\ G and bc = a} \end{matrix}$ and inverse of f_G is $f_{G}^{- 1} (a) = f_{G} (a^{- 1}) .$

Theorem 4.18. Let f_G, g_G and h_G be three BS-sets over U. Then, $(f_{G} * g_{G}) * h_{G} = f_{G} * (g_{G} * h_{G})$

Proof. Let G be a group and f_G, g_G, h_G∈BS-sets over U. Then,

$\begin{matrix} [(f_{G} * g_{G}) * h_{G}] (a) \\ = ⊔ {(f_{G} * g_{G}) (b) ⊓ h_{G} (c) : bc = a, b, c \in G} \\ = ⊔ {⊔ {(f_{G} (u) ⊓ g_{G} (v)) : uv = b} ⊓ h_{G} (c) : \\ bc = a, b, c \in G} \\ = ⊔ {(f_{G} (u) ⊓ g_{G} (v)) ⊓ h_{G} (c) : uvc = a, u, v, c \in G} \\ = ⊔ {f_{G} (u) ⊓ (g_{G} (v) ⊓ h_{G} (c)) : uvc = a, u, v, c \in G} \\ = ⊔ {f_{G} (u) ⊓ {g_{G} (v) ⊓ h_{G} (c) : vc = t, v, c \in G} : \\ ut = a, u, t \in G} \\ = ⊔ {f_{G} (u) ⊓ (g_{G} * h_{G}) (t) : ut = a, u, t \in G} \\ = [f_{G} * (g_{G} * h_{G})] (a) \end{matrix}$

so it is associative.

Theorem 4.19. Let $f_{G}, g_{G}, h_{i_{G}} \in {BS}_{E} (U)$ for all i ∈ I. Then the following assertions hold,
(f_G ∗ g_G) (a) = ⊔ _b∈G {f_G (b) ⊓ g_G (b^-1a)} = ⊔ _b∈G {f_G (ab^-1) ⊓ g_G (b)}

$[(f_{G}^{- 1})]^{- 1} = f_{G}$

$f_{G} ⊑ f_{G}^{- 1} \Leftrightarrow f_{G}^{- 1} ⊑ f_{G} \Leftrightarrow f_{G}^{- 1} = f_{G}$

$f_{G} ⊑ g_{G} \Leftrightarrow f_{G}^{- 1} ⊑ g_{G}^{- 1}$

$(⊔_{i \in I} h_{i_{G}})^{- 1} = ⊔_{i \in I} h_{i_{G}}^{- 1}$

$(⊓_{i \in I} h_{i_{G}})^{- 1} = ⊓_{i \in I} h_{i_{G}}^{- 1}$

$(f_{G} * g_{G})^{- 1} = g_{G}^{- 1} * f_{G}^{- 1}$ .

Proof.
For all b ∈ G, b (b^-1a) = a or (ab^-1) b = a includes all alternatives of components of a, so equality holds.

For all a ∈ G, $(f_{G}^{- 1})^{- 1} (a) = f_{G}^{- 1} (a^{- 1}) = f_{G} ((a^{- 1})^{- 1}) = f_{G} (a)$

Let $f_{G} ⊑ f_{G}^{- 1}$ . Then, for all a ∈ G, $f_{G} (a) ⊑ f_{G}^{- 1} (a) = f_{G} (a^{- 1})$

In last expression substituting a^-1 instead of a we get $\begin{matrix} f_{G}^{- 1} (a) ⊑ f_{G} ((a^{- 1})^{- 1}) & \Leftrightarrow & f_{G}^{- 1} (a) ⊑ f_{G} (a) \\ \Leftrightarrow & f_{G}^{- 1} ⊑ f_{G} \end{matrix}$

therefore $f_{G}^{- 1} = f_{G}$ .

The proof is direct from (3).

For all a ∈ G, $\begin{matrix} (⊔_{i \in I} h_{i_{G}})^{- 1} (a) & = & (⊔_{i \in I} h_{i_{G}}) (a^{- 1}) \\ = & ⊔_{i \in I} h_{i_{G}} (a^{- 1}) \\ = & ⊔_{i \in I} h_{i_{G}}^{- 1} (a) . \end{matrix}$

The proof is similar to 5.

For all a ∈ G, $\begin{matrix} g_{G})^{- 1} (a) = (f_{G} * g_{G}) (a^{- 1}) \\ = ⊔ {f_{G} (b) ⊓ g_{G} (c) : b, c \in G and \\ bc = a^{- 1}} \\ = ⊔ {f_{G} (b) ⊓ g_{G} (c) : b, c \in G and \\ (bc)^{- 1} = (a^{- 1})^{- 1}} \\ = ⊔ {g_{G} (c^{- 1})^{- 1} ⊓ f_{G} (b^{- 1})^{- 1} : \\ b^{- 1}, c^{- 1} \in G and c^{- 1} b^{- 1} = a} \\ = ⊔ {g_{G}^{- 1} (c^{- 1}) ⊓ f_{G}^{- 1} (b^{- 1}) : b^{- 1}, c^{- 1} \\ \in G and c^{- 1} b^{- 1} = a} \\ = (g_{G}^{- 1} * f_{G}^{- 1}) (a) . \end{matrix}$

5 Conclusion

In this study, some concepts are defined such as bipolar soft group, (α, β)-level set of bipolar soft set, image and pre-image of bipolar soft set. Then, in group theory some properties are extended to bipolar soft groups and some results are obtained about bipolar soft groups and (α, β)-level sets of bipolar soft sets. I hope that researchers may study the properties of bipolar soft groups in other algebraic structures such as normal groups, rings, ideals and fields.

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