The lower and upper approximations in a multiplicative set

Abstract

Let S be a multiplicative subset of a commutative ring R. We considered a R × S as a universal and introduced the notion of lower[upper] approximations in a R × S . Also, we gave some properties of the lower[upper] approximations in a R × S . We obtained some properties of rough ideals

Keywords

Rough set multiplicative set fuzzy ideal approximation space

1 Introduction

The theory of rough sets were introduced by Pawlak [17]. For concepts without sharp boundaries, the concept of Rough sets is a useful mathematical model. The theory of rough sets is an extension of set theory as a subset of a universe is defined by a pair of ordinary sets called the lower and upper approximations. The properties of rough sets can be examined with this description. Pawlak investigated approximate operations on sets, approximate equality of sets, and approximate inclusion of sets. The algebraic structures of rough sets was studied by several authors. In 1994, the notion of rough subgroups was introduced by Biswas and Nanda [2]. Kuroki introduced the rough ideal in a semigroup [12]. Kuroki and Wang examined some properties of the lower and upper approximations with respect to the normal subgroups [13]. In 2004, Davvaz [6] introduced the notion of rough subring (resp. ideal) and examined some properties of the lower and the upper approximations in a ring. Mordeson [15] defined approximation operators on the power set of the given set by using covers of the universal set. Basic properties of the upper approximation operator examined in this paper. Also, he defined a particular cover on the set of ideals of a commutative ring with identity that both the concepts of the (fuzzy) prime spectrum of a ring and rough set theory can simultaneously be brought to bear on the study of (fuzzy) ideals of a ring [15]. Several researchers have studies on rough set theory [1 , 18–23].

In this paper we examined some relations between rough sets and multiplicative subsets of a commutative ring R. Let’s give basic definitions and theorems.

Definition 1.1. A nonempty subset S of a ring R is multiplicative provided that $a, b \in S \Rightarrow ab \in S$

Theorem 1.1. Let S be a multiplicative subset of a commutative ring R. The relation defined on the set R × S by $\begin{matrix} (r, s) \sim (r^{'}, s^{'}) \Leftrightarrow s_{1} ({rs}^{'} - r^{'} s) = 0 \\ for some s_{1} \in S \end{matrix}$ is an equivalence relation. Furthermore if R has no zero divisors and 0 ∉ S, then $(r, s) \sim (r^{'}, s^{'}) \Leftrightarrow {rs}^{'} - r^{'} s = 0$

Definition 1.2. [17] Let U be a non-empty universal set, θ an equivalence relation on U then equivalence classes induced on U by θ . The pair (U, θ) is called an approximation space.

Definition 1.3. [17] A mapping (U, θ, -) : P (U) → P (U) × P (U) defined for every X ∈ P (U) by $(U, θ, X) = ((\underline{U}, θ, X), (\bar{U}, θ, X))$ , where $\begin{matrix} (\underline{U}, θ, X) = {x \in U : {[x]}_{θ} \subseteq X}, \\ (\bar{U}, θ, X) = {x \in U : {[x]}_{θ} \cap X \neg = \emptyset} \end{matrix}$ is called rough approximation in (U, θ) . Let (U, θ) be an approximation space. The sets $(\underline{U}, θ, X)$ and $(\bar{U}, θ, X)$ are called an lower rough approximation and an upper rough approximation of X in (U, θ) , resp.

Given an approximation space (U, θ) , a pair (A, B) ∈ P (U) × P (U) is called a rough set in (U, θ) if and only if (A, B) = (U, θ, X) for some X ∈ P (U) .

Definition 1.4. [17] Let (U, θ, A) and (U, θ, B) be any two rough sets in the approximation space (U, θ). Then $\begin{matrix} 1) (U, θ, A) \cup (U, θ, B) = & ((\underline{U}, θ, A) \cup (\underline{U}, θ, B), \\ (\bar{U}, θ, A) \cup (\bar{U}, θ, B)) \\ 2) (U, θ, A) \cap (U, θ, B) = & ((\underline{U}, θ, A) \cap (\underline{U}, θ, B), \\ (\bar{U}, θ, A) \cap (\bar{U}, θ, B)) \end{matrix}$

3) Then it is called that (U, θ, A) is subset of (U, θ, B) if (U, θ, A) ∩ (U, θ, B) = (U, θ, A) and denoted by (U, θ, A) ⊆ (U, θ, B) $\begin{matrix} (U, θ, A) \subseteq (U, θ, B) if and only if \\ (\underline{U}, θ, A) \subseteq (\underline{U}, θ, B) and (\bar{U}, θ, A) \subseteq (\bar{U}, θ, B) \end{matrix}$

This property of rough inclusion has all the properties of set inclusion. The rough complement of (U, θ, A) denoted by (U, θ, A) ^c is defined by ${(U, θ, A)}^{c} = ({(\bar{U}, θ, A)}^{c}, {(\underline{U}, θ, A)}^{c})$

Also, we can define (U, θ, A) ╲ (U, θ, B) as follows: $\begin{array}{l} (U, θ, A) ╲ (U, θ, B) = (U, θ, A) \cap {(U, θ, B)}^{c} \\ = ((\underline{U}, θ, A) ╲ (\bar{U}, θ, B), (\bar{U}, θ, A) ╲ (\underline{U %}, θ, B)) \end{array}$

The original concept of fuzzy sets was introduced by Zadeh as an extension of crisp sets by enlarging the truth value set to the real unit interval [0, 1] as follows.

Definition 1.5. [25] Let X be a nonempty set. A mapping μ : X → [0, 1] is called a fuzzy set in X. The complement of μ, denoted by μ^c, is the fuzzy set in X given by μ^c (x) = 1 - μ (x) for all x ∈ X.

Some operations on fuzzy sets defined in same paper. Let μ and λ be two fuzzy subsets of X. $\begin{matrix} λ \subseteq μ : \Leftrightarrow λ (x) \leq μ (x) for all x \in X \\ (μ \cap λ) (x) = min {μ (x), λ (x)} for all x \in X \end{matrix}$

Definition 1.6. [26] Let μ and λ be two fuzzy subsets of a ring R. Then the sum μ + λ is defined by $\begin{matrix} (μ + λ) (x) \\ = sup_{x = a + b} {min {μ (a), λ (b)}} for all x \in R \end{matrix}$

This definition is obtained from Zadehs extension principle.

Definition 1.7. [24] A fuzzy subset μ of a ring R is called a fuzzy ideal of R if it has the following properties:

μ (x - y)≥ min { μ (x) , μ (y) } for all x ∈ R

μ (xy)≥ max { μ (x) , μ (y) } for all x ∈ R

Because of each fuzzy set can be uniquely represented by the family of all its t-level subsets, the t-level subsets have very important role in fuzzy set theory.

Definition 1.8. [5] Let μ be a fuzzy ideal of a ring R. For each t ∈ [0, 1], the set $U (μ, t) = {(a, b) \in R \times R : μ (a - b) \geq t}$ is called a t-level relation of μ .

An equivalence relation θ on a ring R is a congruence relation if (a, b) ∈ θ implies (a + x, b + x) ∈ θ and (x + a, x + b) ∈ θ for all x ∈ R .

Lemma 1.2. [5] Let μ be a fuzzy ideal of a ring R, and let t ∈ [0, 1] . Then U (μ, t) is a congruence relation on R.

We denote by [x] _(μ,t) the equivalence class of U (μ, t) containing x of R.

Lemma 1.3. [5] Let μ be a fuzzy ideal of a ring R. If a, b ∈ R and t ∈ [0, 1] , then

[a] _(μ,t) + [b] _(μ,t) = [a + b] _(μ,t)

[- a] _(μ,t) = - ([a] _(μ,t))

Davvaz introduced a definition of the lower and upper approximations of a subset of a ring with respect to a fuzzy ideal as follows. Some properties of the lower and upper approximations were examined in the same paper.

Definition 1.9. [5] Let μ be a fuzzy ideal of a ring R and t ∈ [0, 1] , we know U (μ, t) is an equivalence relation (congruence relation) on R. Therefore, when U = R and θ is the above equivalence relation, then we use (R, μ, t) instead of approximation space (U, θ) .

Let μ be a fuzzy ideal of a ring R and U (μ, t) be a t-level congruence relation of μ on R. Let X be a non-empty subset of R . Then the sets $\begin{matrix} \underline{U} (μ, t, X) = {x \in R : {[x]}_{(μ, t)} \subseteq X}, \\ \bar{U} (μ, t, X) = {x \in R : {[x]}_{(μ, t)} \cap X \neg = \emptyset} \end{matrix}$ are called, respectively, the lower and upper approximations of the set X with respect to U (μ, t) .

Proposition 1.4. [5] For every approximation space (R, μ, t) and every subsets A, B of R, we have:

$\underline{U} (μ, t, A) \subseteq A \subseteq \bar{U} (μ, t, A)$

$\underline{U} (μ, t, \emptyset) = \emptyset = \bar{U} (μ, t, \emptyset)$

$\underline{U} (μ, t, R) = R = \bar{U} (μ, t, R)$

If A ⊆ B, then $\underline{U} (μ, t, A) \subseteq \underline{U} (μ, t, B)$ and $\bar{U} (μ, t, A) \subseteq \bar{U} (μ, t, B)$

$\underline{U} (μ, t, \underline{U} (μ, t, A)) = \underline{U} (μ, t, A)$

$\bar{U} (μ, t, \bar{U} (μ, t, A)) = \bar{U} (μ, t, A)$

$\bar{U} (μ, t, \underline{U} (μ, t, A)) = \underline{U} (μ, t, A)$

$\underline{U} (μ, t, \bar{U} (μ, t, A)) = \bar{U} (μ, t, A)$

$\underline{U} (μ, t, A) = {(\bar{U} (μ, t, A^{c}))}^{c}$

$\bar{U} (μ, t, A) = {(\underline{U} (μ, t, A^{c}))}^{c}$

$\underline{U} (μ, t, A \cap B) = \underline{U} (μ, t, A) \cap \underline{U} (μ, t, B)$

$\bar{U} (μ, t, A \cap B) \subseteq \bar{U} (μ, t, A) \cap \bar{U} (μ, t, B)$

$\underline{U} (μ, t, A \cup B) \supseteq \underline{U} (μ, t, A) \cup \underline{U} (μ, t, B)$

$\bar{U} (μ, t, A \cup B) = \bar{U} (μ, t, A) \cup \bar{U} (μ, t, B)$

$\underline{U} (μ, t, {[x]}_{(μ, t)}) = \bar{U} (μ, t, {[x]}_{(μ, t)})$ for allx ∈ R

Definition 1.10. [5] A non-empty subset A of a ring R is called an upper (lower) rough ideal of R if $\bar{U} (μ, t, A)$ ( $\underline{U} (μ, t, A)$ ) is an ideal of R.

Definition 1.11. [5] Let μ be a fuzzy ideal of a ring R and $(\underline{U} (μ, t, A), \bar{U} (μ, t, A))$ a rough set in the approximation space (R, μ, t). If $\underline{U} (μ, t, A)$ and $\bar{U} (μ, t, A)$ are ideals of R, then we call $(\underline{U} (μ, t, A), \bar{U} (μ, t, A))$ a rough ideal.

2 Main results

In this section, firstly we proved main lemma that we have used to obtain some results for rough ideals. Then we presented some properties of rough ideals and we examined relations between rough sets and multiplicative subsets of a commutative ring R.

Lemma 2.1. Let μ and λ be two fuzzy ideals of a ring R. If a, b ∈ R and t ∈ [0, 1] , then [a] _(μ,t) + [b] _(λ,t) ⊆ [a + b] _(μ+λ,t).

Proof. Suppose x ∈ [a] _(μ,t) + [b] _(λ,t). Then there exist k ∈ [a] _(μ,t) and c ∈ [b] _(λ,t) such that x = k + c . Since (a, k) ∈ U (μ, t) and (b, c) ∈ U (λ, t) , we have μ (a - k) ≥ t, λ (b - c) ≥ t . Then, $\begin{matrix} (μ + λ) (x - a - b) & = & sup_{u + v = x - a - b} {min {μ (u), λ (v)}} \\ \geq & min {μ (a - k), λ (b - c)} \\ \geq & min {t, t} = t \end{matrix}$ and so (x, a + b) ∈ U (μ + λ, t). □

Proposition 2.2. For every approximation space (R, μ, t) and every subsets A,B of R, we have:

$\bar{U} (μ, t, A) ╲ \bar{U} (μ, t, B) \subseteq \bar{U} (μ, t, A ╲ B)$

$\underline{U} (μ, t, A) \cap \bar{U} (μ, t, B) \subseteq \bar{U} (μ, t, A \cap B)$

Proof. (i) We have $\begin{matrix} x & \in & \bar{U} (μ, t, A) ╲ \bar{U} (μ, t, B) \\ \Rightarrow & x \in \bar{U} (μ, t, A) and x \notin \bar{U} (μ, t, B) \\ \Rightarrow & {[x]}_{(μ, t)} \cap A \neg = \emptyset and {[x]}_{(μ, t)} \cap B = \emptyset \\ \Rightarrow & {[x]}_{(μ, t)} \cap A \neg = \emptyset and {[x]}_{(μ, t)} \subseteq B^{c} \\ \Rightarrow & {[x]}_{(μ, t)} \cap (A \cap B^{c}) \neg = \emptyset \\ \Rightarrow & x \in \bar{U} (μ, t, A ╲ B) \end{matrix}$

Therefore $\bar{U} (μ, t, A) ╲ \bar{U} (μ, t, B) \subseteq \bar{U} (μ, t, A ╲ B)$ .

(ii) We have $\begin{matrix} x & \in & \underline{U} (μ, t, A) \cap \bar{U} (μ, t, B) \\ \Rightarrow & x \in \underline{U} (μ, t, A) and x \in \bar{U} (μ, t, B) \\ \Rightarrow & {[x]}_{(μ, t)} \subseteq A and {[x]}_{(μ, t)} \cap B \neg = \emptyset \\ \Rightarrow & {[x]}_{(μ, t)} \cap (A \cap B) \neg = \emptyset \\ \Rightarrow & x \in \bar{U} (μ, t, A \cap B) \end{matrix}$ Therefore $\underline{U} (μ, t, A) \cap \bar{U} (μ, t, B) \subseteq \bar{U} (μ, t, A \cap B) .$ □

The following example shows that the converse of (i) and (ii) in Proposition 2.2 is not true.

Example 1. Let R = Z₃ . Define fuzzy subset μ : Z₃ → [0, 1] by μ (0) = t₀, μ (1) = μ (2) = t₁ where t_i ∈ [0, 1] , 0 ≤ i ≤ 2 and t₁ < t₀ . It is no difficult to see that μ is a fuzzy ideal of R. We have $\begin{matrix} U (μ, t_{1}) \\ = {(a, b) \in Z_{3} \times Z_{3} : μ (a - b) \geq t_{1}} \\ = {(0, 0), (1, 1), (2, 2), (0, 1), (1, 0), (2, 0), \\ (0, 2), (1, 2), (2, 1)} \end{matrix}$

Let A = { 0, 1 } , B = { 1, 2 } . Then $\begin{matrix} \underline{U} (μ, t_{1}, A) = \emptyset, \bar{U} (μ, t_{1}, B) = Z_{3}, \\ \bar{U} (μ, t_{1}, A \cap B) = Z_{3}, \\ \bar{U} (μ, t_{1}, A \cap B^{c}) = Z_{3}, \bar{U} (μ, t_{1}, A) = Z_{3} \end{matrix}$ Therefore $\begin{matrix} \bar{U} (μ, t, A) ╲ \bar{U} (μ, t, B) ⊉ \bar{U} (μ, t, A ╲ B) \\ \underline{U} (μ, t, A) \cap \bar{U} (μ, t, B) ⊉ \bar{U} (μ, t, A \cap B) \end{matrix}$

Proposition 2.3. Let μ be a fuzzy ideal of a ring R and t ∈ [0, 1] . If {I_α : α∈ ∧ } be a nonempty collection of lower rough ideals of R, then $\underset{α \in \land}{\cap} I_{α}$ is a lower rough ideal of R.

Proof. We have $\underline{U} (μ, t, \underset{α \in \land}{\cap} I_{α}) = ⋂_{α \in \land} \underline{U} (μ, t, I_{α})$ Suppose {I_α : α∈ ∧ } be a nonempty collection of lower rough ideals of R . Since $0 \in \underline{U} (μ, t, I_{α})$ for all α, $0 \in ⋂_{α \in \land} \underline{U} (μ, t, I_{α})$ and so $⋂_{α \in \land} \underline{U} (μ, t, I_{α}) \neg = \emptyset .$ Let $a, b \in ⋂_{α \in \land} \underline{U} (μ, t, I_{α}) .$ Then $a, b \in \underline{U} (μ, t, I_{α})$ for all α . Since each $\underline{U} (μ, t, I_{α})$ is ideal, $a - b \in \underline{U} (μ, t, I_{α})$ for all α . Hence, $a - b \in ⋂_{α \in \land} \underline{U} (μ, t, I_{α}) .$ Let r ∈ R . Since each $\underline{U} (μ, t, I_{α})$ is ideal, $ra \in \underline{U} (μ, t, I_{α})$ and $ar \in \underline{U} (μ, t, I_{α})$ for all α and so $ra \in ⋂_{α \in \land} \underline{U} (μ, t, I_{α})$ and $ar \in ⋂_{α \in \land} \underline{U} (μ, t, I_{α}) .$ Thus, $⋂_{α \in \land} \underline{U} (μ, t, I_{α})$ is ideal of R. □

In the above proposition does not hold in general for upper rough ideal.

Example 2. Let R = Z₆ . Define fuzzy subset μ : Z₆ → [0, 1] by $\begin{matrix} μ (0) & = & t_{0}, μ (1) = μ (2) = μ (4) \\ = & μ (5) = t_{1}, μ (3) = t_{2} \end{matrix}$ where t_i ∈ [0, 1] , 0 ≤ i ≤ 2 and t₁ < t₂ < t₀ . It is no difficult to see that μ is a fuzzy ideal of R. We have $\begin{matrix} U (μ, t_{2}) & = & {(a, b) \in Z_{6} \times Z_{6} : μ (a - b) \geq t_{2}} \\ = & {(0, 0), (1, 1), (2, 2), (3, 3), (4, 4), \\ (5, 5), (5, 2), (2, 5), (1, 4), (4, 1), \\ (0, 3), (3, 0)} \end{matrix}$

Let A = { 1, 2, 3 } , B = { 0, 1, 5, 3 } . Then $\begin{matrix} \bar{U} (μ, t_{2}, B) = Z_{6}, \bar{U} (μ, t_{1}, A \cap B) = {0, 1, 3, 4}, \\ \bar{U} (μ, t_{2}, A) = Z_{6} \end{matrix}$

A and B are an upper rough ideal of R but A ∩ B is not an upper rough ideal of R.

Proposition 2.4. 5 Let μ be a fuzzy ideal of a ring R and t ∈ [0, 1] . If { I_α : α∈ ∧ } be a nonempty collection of upper rough ideals of R such that $\forall α, β \in \land, α \neg = β \Rightarrow I_{α} \subseteq I_{β} or I_{β} \subseteq I_{α}$

Then $\underset{α \in \land}{\cup} I_{α}$ is an upper rough ideal R.

Proof. We have $\bar{U} (μ, t, \underset{α \in \land}{\cup} I_{α}) = ⋃_{α \in \land} \bar{U} (μ, t, I_{α})$

Suppose {I_α : α∈ ∧ } be a nonempty collection of upper rough ideals of R such that $\forall α, β \in \land, α \neg = β \Rightarrow I_{α} \subseteq I_{β} or I_{β} \subseteq I_{α}$

Since $0 \in \bar{U} (μ, t, I_{α})$ for all α, $0 \in ⋃_{α \in \land} \bar{U} (μ, t, I_{α})$ and so $⋃_{α \in \land} \bar{U} (μ, t, I_{α}) \neg = \emptyset .$ Let $a, b \in ⋃_{α \in \land} \bar{U} (μ, t, I_{α}) .$ Then there exist α, β∈ ∧ such that $a \in \bar{U} (μ, t, I_{α})$ and $b \in \bar{U} (μ, t, I_{β})$ .

If α = β, then $\begin{matrix} a, b \in \bar{U} (μ, t, I_{α}) & \Rightarrow & a - b \in \bar{U} (μ, t, I_{α}) \\ \Rightarrow & a - b \in \bar{U} (μ, t, \underset{α \in \land}{\cup} I_{α}) \end{matrix}$ Let α ¬ = β . Then $\begin{matrix} I_{α} \subseteq I_{β} & \Rightarrow & \bar{U} (μ, t, I_{α}) \subseteq \bar{U} (μ, t, I_{β}) \\ \Rightarrow & a - b \in \bar{U} (μ, t, I_{β}) \\ \Rightarrow & a - b \in \bar{U} (μ, t, \underset{α \in \land}{\cup} I_{α}) \end{matrix}$

Now, let r ∈ R . Since each $\bar{U} (μ, t, I_{α})$ is a ideal of R, $r . a \in \bar{U} (μ, t, I_{α})$ for all α and so $r . a \in \bar{U} (μ, t, \underset{α \in \land}{\cup} I_{α})$ and $a . r \in \bar{U} (μ, t, \underset{α \in \land}{\cup} I_{α}) .$ Thus, $\bar{U} (μ, t, \underset{α \in \land}{\cup} I_{α})$ is a ideal of R. □

Corollary 2.5. Let μ be a fuzzy ideal of a ring R and t ∈ [0, 1] . If {I_α : α∈ ∧ } be a nonempty collection of rough ideals of R, then $\underset{α \in \land}{\cap} I_{α}$ might not be rough ideal of R .

Proof. This follows from Propositions 2.3 and 2.4, Example 2. □

Let S be a multiplicative subset of a commutative ring R and “∼” the equivalence relation of Theorem 1.1. The equivalence class of (r, s) ∈ R × S will be denoted ${\bar{(r, s)}}_{S} = \frac{r}{s} .$ The set of all equivalence classes of R × S under “∼” will be denoted by S^-1R .

Definition 2.1. Let X be a non-empty subset of R × S. Then sets $\begin{matrix} \underline{S} (X) & : & = {(r, s) \in R \times S : {\bar{(r, s)}}_{S} \subseteq X} \\ \bar{S} (X) & : & = {(r, s) \in R \times S : {\bar{(r, s)}}_{S} \cap X \neg = \emptyset} \end{matrix}$ are called, respectively, the lower and upper approximations of the set X with respect to “∼”.

Proposition 2.6. For every approximation space (R × S, ∼) and every subsets A, B of R × S, we have:

$\underline{S} (A) \subseteq A \subseteq \bar{S} (A)$

$\underline{S} (\emptyset) = \emptyset = \bar{S} (\emptyset)$

$\underline{S} (R \times S) = R \times S = \bar{S} (R \times S)$

If A ⊆ B, then $\underline{S} (A) \subseteq \underline{S} (B)$ and $\bar{S} (A) \subseteq \bar{S} (B)$

$\underline{S} (\underline{S} (A)) = \underline{S} (A)$

$\bar{S} (\bar{S} (A)) = \bar{S} (A)$

$\bar{S} (\underline{S} (A)) = \underline{S} (A)$

$\underline{S} (\bar{S} (A)) = \bar{S} (A)$

$\underline{S} (A) = {(\bar{S} (A^{c}))}^{c}$

$\bar{S} (A) = {(\underline{S} (A^{c}))}^{c}$

$\underline{S} (A \cap B) = \underline{S} (A) \cap \underline{S} (B)$

$\bar{S} (A \cap B) \subseteq \bar{S} (A) \cap \bar{S} (B)$

$\underline{S} (A \cup B) \supseteq \underline{S} (A) \cup \underline{S} (B)$

$\bar{S} (A \cup B) = \bar{S} (A) \cup \bar{S} (B)$

$\bar{S} ({\bar{(r, s)}}_{S}) = \underline{S} ({\bar{(r, s)}}_{S})$ for all (r, s) ∈ R × S

Proof. (1) We have $\begin{matrix} (r, s) \in \underline{S} (A) \Rightarrow {\bar{(r, s)}}_{S} \subseteq A \\ \Rightarrow {\bar{(r, s)}}_{S} \cap A \neg = \emptyset \Rightarrow (r, s) \in \bar{S} (A) \end{matrix}$ (4) Suppose that A ⊆ B . Then $\begin{matrix} (r, s) & \in & \underline{S} (A) \Rightarrow {\bar{(r, s)}}_{S} \subseteq A \subseteq B \Rightarrow (r, s) \in \underline{S} (B) \\ (r, s) & \in & \bar{S} (A) \Rightarrow {\bar{(r, s)}}_{S} \cap A \neg = \emptyset \Rightarrow {\bar{(r, s)}}_{S} \cap B \neg = \emptyset \\ \Rightarrow (r, s) \in \bar{S} (B) \end{matrix}$ (5) We have $(r, s) \in \underline{S} (\underline{S} (A)) \Leftrightarrow {\bar{(r, s)}}_{S} \subseteq \underline{S} (A) \Leftrightarrow (r, s) \in \underline{S} (A)$

The proof of the other statements is similar or is immediate. □

Proposition 2.7. For every approximation space (R × S, ∼) and every subsets A, B of R × S, we have:

$\bar{S} (A) ╲ \bar{S} (B) \subseteq \bar{S} (A ╲ B)$

$\underline{S} (A) \cap \bar{S} (B) \subseteq \bar{S} (A \cap B)$

Proof. (i) We have $\begin{matrix} (r, s) & \in & \bar{S} (A) ╲ \bar{S} (B) \Rightarrow (r, s) \in \bar{S} (A) \\ and (r, s) \notin \bar{S} (B) \\ \Rightarrow & {\bar{(r, s)}}_{S} \cap A \neg = \emptyset and {\bar{(r, s)}}_{S} \cap B = \emptyset \\ \Rightarrow & {\bar{(r, s)}}_{S} \cap A \neg = \emptyset and {\bar{(r, s)}}_{S} \subseteq B^{c} \\ \Rightarrow & {\bar{(r, s)}}_{S} \cap (A \cap B^{c}) \neg = \emptyset \\ \Rightarrow & (r, s) \in \bar{S} (A ╲ B) \end{matrix}$ Therefore $\bar{S} (A) ╲ \bar{S} (B) \subseteq \bar{S} (A ╲ B)$ . (ii) We have $\begin{matrix} (r, s) & \in & \underline{S} (A) \cap \bar{S} (B) \Rightarrow (r, s) \in \underline{S} (A) \\ and (r, s) \in \bar{S} (B) \\ \Rightarrow & {\bar{(r, s)}}_{S} \subseteq A and {\bar{(r, s)}}_{S} \cap B \neg = \emptyset \\ \Rightarrow & {\bar{(r, s)}}_{S} \cap (A \cap B) \neg = \emptyset \\ \Rightarrow & (r, s) \in \bar{S} (A \cap B) \end{matrix}$ Therefore $\underline{S} (A) \cap \bar{S} (B) \subseteq \bar{S} (A \cap B) .$ □

The following example shows that the converse of (12) and (13) in Proposition 2.6 is not true.

Let R = Z₃ and $S = {\bar{1}, \bar{2}} .$ Then S is a multiplicative subset of a commutative ring R. $\begin{matrix} S^{- 1} R & = & {\frac{r}{s} : r \in R s \in S} \\ = & {\frac{r}{s} : r, s \in Z_{3}, s \neg = 0} \\ \frac{0}{1} & = & \frac{0}{2} \Rightarrow \bar{0} = {\frac{0}{1}, \frac{0}{2}}, \\ 2 & = & \frac{2}{1} \Rightarrow \bar{2} = {\frac{2}{1}} \\ \frac{1}{2} & = & {\frac{1}{2}}, \\ \frac{1}{1} & = & \frac{2}{2} \Rightarrow \bar{1} = {\frac{1}{1}, \frac{2}{2}} \end{matrix}$

Let A = { (1, 1) , (2, 2) } , C = { (0, 1) } , D = { (0, 2)} and B = { (1, 1) , (2, 2) } . Then $\begin{matrix} \underline{S} (A) & = & A, \bar{S} (A \cap B) = \emptyset, \\ \bar{S} (B) & = & {(1, 1), (2, 2), (1, 2)} \\ \underline{S} (C) & = & \emptyset, \\ \underline{S} (D) & = & \emptyset, \underline{S} (C \cup D) = {(0, 1), (0, 2)} \end{matrix}$

Therefore $\begin{matrix} \bar{S} (A \cap B) & ⊉ & \bar{S} (A) \cap \bar{S} (B) \\ \underline{S} (C \cup D) & ⊈ & \underline{S} (C) \cup \underline{S} (D) \end{matrix}$

The following example shows that the converse of (i) and (ii) in Proposition 2.7 is not true.

Example 3. Let R = Z₅ and $S = {\bar{1}, \bar{2}, \bar{3}, \bar{4}} .$ Then S is a multiplicative subset of a commutativering R. $\begin{matrix} S^{- 1} R & = & {\frac{r}{s} : r \in R s \in S} \\ = & {\frac{r}{s} : r, s \in Z_{5}, s \neg = 0} \\ \frac{1}{2} & = & \frac{2}{4} \Rightarrow \frac{1}{2} = {\frac{1}{2}, \frac{2}{4}}, \\ \frac{2}{1} & = & \frac{4}{2} \Rightarrow \bar{2} = \frac{2}{1} = {\frac{2}{1}, \frac{4}{2}} \\ \frac{1}{3} & = & {\frac{1}{3}}, \frac{1}{4} = {\frac{1}{4}}, \frac{2}{3} = {\frac{2}{3}}, \\ \frac{3}{1} = \bar{3} = {\frac{3}{1}}, \frac{3}{2} = {\frac{3}{2}} \\ \bar{1} & = & {\frac{1}{1}, \frac{2}{2}, \frac{3}{3}, \frac{4}{4}}, \bar{0} = {\frac{0}{1}, \frac{0}{2}, \frac{0}{3}, \frac{0}{4}}, \\ \frac{4}{3} = {\frac{4}{3}}, \frac{4}{1} = \bar{4} = {4} \end{matrix}$

Let A ={ (1, 4) , (2, 3) } , C = { (0, 1) , (4, 2) } , D = { (2, 1) } and B = { (3, 1) , (2, 3) } . Then $\begin{matrix} \underline{S} (A) & = & \emptyset, \bar{S} (A \cap B) = {(2, 3)}, \\ \bar{S} (B) & = & {(3, 1), (2, 3)}, \\ \bar{S} (C) & = & {(0, 1), (0, 2), (0, 3), (0, 4), (4, 2), (2, 1)}, \\ \bar{S} (D) & = & {(4, 2), (2, 1)} \end{matrix}$

Therefore $\begin{matrix} \bar{S} (C) ╲ \bar{S} (D) & ⊉ & \bar{S} (C ╲ D) \\ \underline{S} (A) \cap \bar{S} (B) & ⊉ & \bar{S} (A \cap B) \end{matrix}$

Lemma 2.8. A, B is a multiplicative subset of a commutative ring R. If A ⊆ B, then ${\bar{(r, s)}}_{A} \subseteq {\bar{(r, s)}}_{B}$

Proof. Suppose A ⊆ B. Then $\begin{matrix} (x, y) & \in & {\bar{(r, s)}}_{A} \Rightarrow z (xs - ry) = 0 \\ for some z \in A \\ \Rightarrow & z (xs - ry) = 0 for some z \in B \\ \Rightarrow & (x, y) \in {\bar{(r, s)}}_{B} \end{matrix}$ Thus ${\bar{(r, s)}}_{A} \subseteq {\bar{(r, s)}}_{B} □$

Proposition 2.9. A, B is a multiplicative subset of a commutative ring R. If A ⊆ B and ∅ ¬ = X ⊆ R × A, then

$\underline{B} (X) \subseteq \underline{A} (X)$

$\bar{A} (X) \subseteq \bar{B} (X)$

Proof. (i) Suppose A ⊆ B and ∅ ¬ = X ⊆ R × B . Then $\begin{matrix} (r, s) \in \underline{B} (X) & \Rightarrow & {\bar{(r, s)}}_{B} \subseteq X \\ \Rightarrow & {\bar{(r, s)}}_{A} \subseteq {\bar{(r, s)}}_{B} \subseteq X \\ \Rightarrow & (r, s) \in \underline{A} (X) \end{matrix}$

Therefore $\begin{matrix} \underline{B} (X) \subseteq \underline{A} (X) \end{matrix}$

(ii) Suppose A ⊆ B and ∅ ¬ = X ⊆ R × B . Then $\begin{matrix} (r, s) \in \bar{A} (X) & \Rightarrow & {\bar{(r, s)}}_{A} \cap X \neg = \emptyset \\ \Rightarrow & {\bar{(r, s)}}_{B} \cap X \neg = \emptyset \\ \Rightarrow & (r, s) \in \bar{B} (X) \end{matrix}$

Therefore $\begin{matrix} \bar{A} (X) \subseteq \bar{B} (X) □ \end{matrix}$

Corollary 2.10. A, B is a multiplicative subset of a commutative ring R. If A∩ B ¬ = ∅ and ∅ ¬ = X ⊆ R × (A ∩ B) , then

$\underline{A} (X) \subseteq \underline{A \cap B} (X)$ and $\underline{B} (X) \subseteq \underline{A \cap B} (X)$

$\bar{A \cap B} (X) \subseteq \bar{B} (X)$ and $\bar{A \cap B} (X) \subseteq \bar{A} (X)$

Proof. (i) Suppose A∩ B ¬ = ∅ and ∅ ¬ = X ⊆ R × (A ∩ B) . Then $\begin{matrix} (r, s) \in \underline{A} (X) & \Rightarrow & {\bar{(r, s)}}_{A} \subseteq X \\ \Rightarrow & {\bar{(r, s)}}_{A \cap B} \subseteq {\bar{(r, s)}}_{A} \subseteq X \\ \Rightarrow & (r, s) \in \underline{A \cap B} (X) \\ (r, s) \in \underline{B} (X) & \Rightarrow & {\bar{(r, s)}}_{B} \subseteq X \\ \Rightarrow & {\bar{(r, s)}}_{A \cap B} \subseteq {\bar{(r, s)}}_{B} \subseteq X \\ \Rightarrow & (r, s) \in \underline{A \cap B} (X) \end{matrix}$ Therefore $\begin{matrix} \underline{A} (X) \subseteq \underline{A \cap B} (X) and \underline{B} (X) \subseteq \underline{A \cap B} (X) \end{matrix}$ (ii) Suppose A∩ B ¬ = ∅ and ∅ ¬ = X ⊆ R × (A ∩ B) . Then $\begin{matrix} (r, s) \in \bar{A \cap B} (X) \Rightarrow {\bar{(r, s)}}_{A \cap B} \cap X \neg = \emptyset \end{matrix}$ Since ${\bar{(r, s)}}_{A \cap B} \subseteq {\bar{(r, s)}}_{A}$ and ${\bar{(r, s)}}_{A \cap B} \subseteq {\bar{(r, s)}}_{B}$ , so $\begin{matrix} {\bar{(r, s)}}_{A} \cap X & \neg = & \emptyset \Rightarrow (r, s) \in \bar{A} (X) \\ {\bar{(r, s)}}_{B} \cap X & \neg = & \emptyset \Rightarrow (r, s) \in \bar{B} (X) □ \end{matrix}$

Corollary 2.11. Let A, B ideals of a commutative ring R. If X is a non-empty subset of R × AB, then

$\underline{A \cap B} (X) \subseteq \underline{AB} (X)$

$\bar{AB} (X) \subseteq \bar{A \cap B} (X)$

Proof. Since A, B ideals of a commutative ring R. Then AB ⊆ A ∩ B . This follows from Proposition 2.9. □

3 Conclusions

Our aim is in this paper to introduce more connections between fuzzy algebraic structures and rough algebraic structures. Some properties of rough ideals were proved. Also, the lower and upper approximations of the set X with respect to “∼” were defined and algebraic properties were examined. The results obtained were supported by examples and counterexamples.

Footnotes

Acknowledgments

This study was supported by the Research Fund of Mersin University in Turkey with Project Number: 2015-TP3-1249.

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