Near-Generalized Approximations in Groups Based on a Set-Valued Mapping

Dian Winda Setyawati; Bijan Davvaz

doi:10.5391/IJFIS.2024.24.2.125

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International Journal of Fuzzy Logic and Intelligent Systems 2024; 24(2): 125-140

Published online June 25, 2024

https://doi.org/10.5391/IJFIS.2024.24.2.125

Near-Generalized Approximations in Groups Based on a Set-Valued Mapping

Dian Winda Setyawati¹, Subiono¹, and Bijan Davvaz²

¹Department of Mathematics, Institut Teknologi Sepuluh Nopember, Kampus ITS, Sukolilo-Surabaya, Indonesia
²Department of Mathematical Sciences, Yazd University, Yazd, Iran

Correspondence to :
Subiono (subiono2008@matematika.its.ac.id)

Received: June 5, 2023; Revised: August 24, 2023; Accepted: June 5, 2024

This is an Open Access article distributed under the terms of the Creative Commons Attribution Non-Commercial License (http://creativecommons.org/licenses/by-nc/3.0/) which permits unrestricted noncommercial use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

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Abstract
1. Introduction
2. Approximations and Near Approximations in a Group
3. Approximations in a Group based on a Set-Valued Mapping
4. Near-Generalized Approximations in a Group based on a Set-Valued Mapping
5. The Lower and Upper Near Subgroups in a Group based on a Set-Valued Mapping
6. Comparison between Near-Generalized and Generalized Approximations in a Group based on a Set-Valued Mapping
7. Conclusion
Conflict of Interest
References
Biographies

A normal subgroup of a group can partition a group into equivalence classes. Therefore, approximations can be constructed within a group. The near approximations in a group are extensions of the approximations in a group. A set-valued mapping T from group G to the set of all non-empty subsets of group G′ can establish generalized approximations in group G based on the set-valued mapping T. In this study, we introduce the notion of near-generalized approximations in a group based on set-valued mapping, an extension of the concept of generalized approximations in a group based on set-valued mapping and near approximations in a group. We then present some properties of nearby subgroups in a group based on set-valued mapping. Furthermore, we compare these types of near-generalized and generalized approximations in a group based on set-valued mapping.

Keywords: Normal subgroup, Group, Approximations, Near approximations, Generalized approximations, Near-generalized approximations

1. Introduction

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Abstract
1. Introduction
2. Approximations and Near Approximations in a Group
3. Approximations in a Group based on a Set-Valued Mapping
4. Near-Generalized Approximations in a Group based on a Set-Valued Mapping
5. The Lower and Upper Near Subgroups in a Group based on a Set-Valued Mapping
6. Comparison between Near-Generalized and Generalized Approximations in a Group based on a Set-Valued Mapping
7. Conclusion
Conflict of Interest
References
Biographies

In the 1980s, Pawlak [1] introduced rough set theory as a mathematical tool for dealing with the problems of vagueness and uncertainty in decision-making. The theory has been used in various fields, including machine learning, data mining, intelligent systems, pattern recognition, decision analysis, inductive reasoning, mereology, etc. [2–10]. The rough set theory is based on the concept of equivalence relation that is reflexive, symmetric, and transitive. The foundation of this theory is an equivalence class, which is used to construct the lower and upper approximations of a set. The lower approximation of set A is the union of the equivalent classes, subsets of set A, and upper approximation of set A is the union of the equivalent classes that intersect with set A. The concept of lower and upper approximations of a set with respect to the normal subgroups of a group has been extensively studied by many scientists [11–14]. In 2004, Davvaz [15] introduced the concept of the lower and upper approximations of a set with respect to the ideals of a ring. Furthermore, in 2006, Davvaz and Mahdavipour [16] introduced the concept of lower and the upper approximations of a set with respect to the submodules of an R-module. Scientists studied the rough set of the ideals of a ring and the submodules of an R-module [15–22]. In 2022, Salama et al. [23] constructed the lower and upper approximations of a set using dominance relation that is reflexive, antisymmetric, and transitive.

The algebraic structure of rough sets is generalized to approximation mapping. It is generalized to approximation mappings with respect to an equivalence relation of algebraic structures, such as generalized approximation mapping with respect to the normal subgroups of a group [24], the ideals of a ring [25, 26] and the submodules of an R-module [27]. In 2022, Setyawati and Subiono [24] discussed the relationship between the lower and upper approximations of different sets with different normal subgroups of a group based on set-valued mapping.

The near set was introduced by Peters [28], in which the objects were perceived to be close to each other, with a similarity of description to some degree. The basic idea behind near sets is the description and classification of objects, that is, the members of a universal set based on the perceptual information system. Peters and Wasilewski [29] introduced an approach to the foundations of information science considered in the context of near sets. The algebraic properties of near sets are described in [30]. In 2019, Bağirmaz [31] introduced and derived the properties of near approximations in a group. Furthermore, he compared the near approximations and the approximations in a group. The near approximations in a group are an extension of the rough approximations in a group. Near approximations in a group used two or more normal subgroups of a group, whereas approximations in a group used the normal subgroup of a group. Davvaz et al. [32,33] extended the ideas presented by Bağirmaz [31] by using the ideals of a ring and the submodules of an R-module. In 2023, Setyawati et al. [34] applied the concept of near approximations to the Cayley graphs.

In this study, we introduce the notion of near-generalized approximations in a group based on set-valued mapping (an extension of the concept of generalized approximations in a group based on set-valued mapping [24] and near approximations in a group [31]) and investigate their properties. We then present some properties of nearby subgroups in a group based on set-valued mapping. Furthermore, we compare these types of near-generalized and generalized approximations in a group based on set-valued mapping.

The remainder of this paper is organized as follows. After its introduction, Section 2 reviews fundamental definitions of and near approximations in a group. Section 3 reviews fundamental definitions of approximation in a group based on set-valued mapping. Section 4 introduces the notion of near-generalized approximations in a group based on set-valued mapping and proves their properties. Section 5 presents the properties of lower and upper near subgroups in a group based on set-valued mapping. Section 6 compares the near-generalized and generalized approximations in a group based on a set-valued mapping.

2. Approximations and Near Approximations in a Group

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Abstract
1. Introduction
2. Approximations and Near Approximations in a Group
3. Approximations in a Group based on a Set-Valued Mapping
4. Near-Generalized Approximations in a Group based on a Set-Valued Mapping
5. The Lower and Upper Near Subgroups in a Group based on a Set-Valued Mapping
6. Comparison between Near-Generalized and Generalized Approximations in a Group based on a Set-Valued Mapping
7. Conclusion
Conflict of Interest
References
Biographies

In this section, we define near approximations in a group, which is an extended definition of rough approximations in a group. Let K be a finite set of objects and Q be an equivalence relation on K. A pair (K, Q) is called the approximation space. If S is a normal subgroup of G, then pair (G, S) is an approximation space.

Definition 1 [11]

Let G be a group, S a normal subgroup of G, and A a non-empty subset of G. We define the sets

AprS_(A)={x∈G∣xS⊆A},

and

AprS¯(A)={x∈G∣xS∩A≠∅},

called lower and upper approximations, respectively. of set A with respect to the normal subgroup S.

Example 1

Let D₆ = {ρ⁰, ρ¹, ρ², ρ³, ρ⁴, ρ⁵, μ, ρ¹μ, ρ²μ, ρ³μ, ρ⁴μ, ρ⁵μ} be a dihedral group with order 12 and S = {ρ⁰, ρ², ρ⁴} be a normal subgroup of D₆ where ρ⁰ is a neutral element of D₆. Let A = {ρ¹, ρ³, ρ⁵, μρ²}. Then, we obtain Apr_S(A) = {ρ¹, ρ³,ρ⁵} and AprS¯(A)={ρ1,ρ3,ρ5,μ,μρ2,μρ4}.

We denote as the set of all normal subgroups in the group G and ℬ as a subset of . A pair (G, ) is called a nearness approximation space. If (G, ) is a nearness approximation space, then

xℬr=x(∩S∈ℬrS),

where ℬ_r ⊆ ℬ.

Definition 2 [31]

Let (G, ) be a nearness approximation space and A be a non-empty subset of G. We define the sets

Nr(ℬ)_ (A)={x∈G∣xℬr⊆A, for ℬr⊆ℬ},

and

Nr(ℬ)¯ (A)={x∈G∣xℬr∩A≠∅, for ℬr⊆ℬ},

called near lower and near upper approximations of set A with respect to the normal subgroups of G, respectively. Clearly, Nr(ℬ)¯(A)=∪ℬr⊆ℬAprℬr¯(A) and Nr(ℬ)¯(A)=∪ℬr⊆ℬAprℬr¯(A).

Consider the following example. In this example [31], three different near approximations of the subset A of G defined by families N₁(ℬ)(A), N₂(ℬ)(A), N₃(ℬ)(A).

Example 2

Let ℤ₁₂ be a group of integers modulo 12 and ℬ = {S₁, S₂, S₃} where S₁ = {[0]₁₂, [6]₁₂}, S₂ = {[0]₁₂, [4]₁₂, [8]₁₂}, S₃ = {[0]₁₂, [3]₁₂, [6]₁₂, [9]₁₂} be normal subgroups of ℤ₁₈. Let A = {[0]₁₂, [1]₁₂, [3]₁₂, [4]₁₂, [7]₁₂, [8]₁₂, [10]₁₂}. Then, we get

• N1(ℬ)¯(A)=[0]12,[1]12,[4]12,[7]12,[8]12,[10]12,
• N2(ℬ)¯(A)=A,
• N3(ℬ)¯(A)=A,
• N1(ℬ)¯(A)=ℤ12,
• N2(ℬ)¯(A)={[0]12,[1]12,[2]12,[3]12,[4]12,[6]12,[7]12,[8]12,[9]12,[10]12},
• N3(ℬ)¯(A)=A.

From the calculations above, we can conclude that N1(ℬ)_(A)⊂N2(ℬ)_(A)=N3(ℬ)_(A)=N3(ℬ)¯(A)⊂N2(ℬ)¯(A)⊂N1(ℬ)¯(A).

3. Approximations in a Group based on a Set-Valued Mapping

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Abstract
1. Introduction
2. Approximations and Near Approximations in a Group
3. Approximations in a Group based on a Set-Valued Mapping
4. Near-Generalized Approximations in a Group based on a Set-Valued Mapping
5. The Lower and Upper Near Subgroups in a Group based on a Set-Valued Mapping
6. Comparison between Near-Generalized and Generalized Approximations in a Group based on a Set-Valued Mapping
7. Conclusion
Conflict of Interest
References
Biographies

In this study, if N is a non-empty set, then ℘^*(N) denotes the set of all non-empty subsets of N. The following definition is a generalization of Definition 1.

Definition 3 [35]

Let M and N be two non-empty sets and T : M→℘^*(N) be a set-valued mapping. Let A be a non-empty subset of N. We define the sets

T_(A)={x∈M∣T(x)⊆A},

and

T¯(A)={x∈M∣T(x)∩A≠∅},

which are called lower and upper approximations of A under T, respectively.

Example 3

Let M = {p, q, r, s, t}, N = {a, b, c, d, e, f, g, h, i, j, k}, and A = {a, b, c, g, h, k}. Let the set-valued mapping T : M→℘^*(N) be defined as T(p) = {a, c, h}, T(q) = {b}, T(r) = {e, f, g}, T(s) = {d, i, j}, T(t) = {j, k}. Then, T(A) = {p, q} and T̄(A) = {p, q, r, t}.

In [35], the set-valued homomorphism between the two groups is defined.

Definition 4

Let G and G′ be the two groups. A mapping T : G→℘^*(G′) is called a set-valued homomorphism if

(1) T(xy) = T(x)T(y),
(2) (T(x))⁻¹ = {a⁻¹ | a ∈ T(x)} = T (x⁻¹),

for all x, y ∈ G.

Example 4 shows that each group homomorphism is a set-valued homomorphism.

Example 4

Let G and G′ be two groups and f : G → G′ be a group homomorphism. Subsequently, the set-valued mapping T : G→℘^*(G′) defined by T(x) = {f(x)} is a set-valued homomorphism because for every x, y ∈ G we have

(1) T(xy) = {f(xy)} = {f(x)f(y)} = {f(x)} {f(y)} = T(x)T(y),
(2) (T(x))⁻¹ = {(f(x))⁻¹} = {f (x⁻¹)} = T(x⁻¹).

Example 5 [35]

The set ℝ^* of non-zero real numbers is a group. Suppose that D={[100a]|a∈ℝ*}. The set-valued mapping T : ℝ^* → ℘^*(D) defined by T (a)={[100a],[100-a]} is a set-valued homomorphism.

Furthermore, we introduced the generalized lower and upper approximations of a set with respect to the normal subgroup of a group [24]. This definition is a generalization of Definition 3. The following definition is similar to that in [25].

Definition 5

Let G and G′ be two groups and T : G→℘^* (G′) be a set-valued mapping. Let S be a normal subgroup of G′ and A be a non-empty subset of G′. Then,

TS_(A)={x∈G∣T(x)S⊆A},

and

TS¯(A)={x∈G∣(T(x)S)∩A≠∅},

called the generalized lower and upper approximations of A under T with respect to a normal subgroup S of G′, respectively.

Example 6

Let D₆ = {ρ⁰, ρ¹, ρ², ρ³, ρ⁴, ρ⁵, μ, ρ¹μ, ρ²μ, ρ³μ, ρ⁴μ, ρ⁵μ} be a dihedral group with order 12, S = {ρ⁰, ρ², ρ⁴} be a normal subgroup of D₆ where ρ⁰ is a neutral element of D₆. Let T : D₆→℘^* (D₆) be defined as

T(x)={{x}if x∈{ρ0,ρ1,ρ2,ρ3,ρ4,ρ5},{ρi,μ}if x=ρiμ∈{μ,ρ1μ,ρ2μ,ρ3μ,ρ4μ,ρ5μ}.

Let A = {ρ¹, ρ³, ρ⁵, μρ²}. Then, we get

• T_S(A) = {ρ¹, ρ³, ρ⁵},
• TS¯(A)={ρ1,ρ3,ρ5,μ,ρμ,ρ2μ,ρ3μ,ρ4μ,ρ5μ}.

From the above calculations, we conclude that TS_(A)⊂TS¯(A).

4. Near-Generalized Approximations in a Group based on a Set-Valued Mapping

Go to

Abstract
1. Introduction
2. Approximations and Near Approximations in a Group
3. Approximations in a Group based on a Set-Valued Mapping
4. Near-Generalized Approximations in a Group based on a Set-Valued Mapping
5. The Lower and Upper Near Subgroups in a Group based on a Set-Valued Mapping
6. Comparison between Near-Generalized and Generalized Approximations in a Group based on a Set-Valued Mapping
7. Conclusion
Conflict of Interest
References
Biographies

Definitions 2 and 5 motivate to form Definition 6. We denote as the set of all normal subgroups in a group G′ and ℬ as a subset of . A pair (G′, ) is defined as nearness approximation space. If (G′, ) is a nearness approximation space, then

T(x)ℬr=T(x)(∩S∈ℬrS),

where ℬ_r ⊆ ℬ.

Definition 6

Let G and G′ be two groups, (G′, ) be a nearness approximation space, and T : G→℘^*(G′) be a set-valued mapping. Let A be a non-empty subset of G′. We define the sets

TNr(ℬ)_(A)={x∈G∣T(x)ℬr⊆A, for ℬr⊆ℬ},

and

TNr(ℬ)¯(A)={x∈G∣T(x)ℬr∩A≠∅, for ℬr⊆ℬ},

which are called the near-generalized lower and upper approximations of A under T with respect to the normal subgroups ℬ of G′, respectively. Clearly, TNr(ℬ)¯(A)=∪ℬr⊆ℬTℬr¯(A) and TNr(ℬ)¯(A)=∪ℬr⊆ℬTℬr¯(A).

Example 7

Let D₄ = {ρ⁰, ρ¹, ρ², ρ³, μ, ρ¹μ, ρ²μ, ρ³μ} be a dihedral group with order 8, H ⊂ D₄ and H = {ρ⁰, ρ¹, ρ², ρ³} be a group where ρ⁰ is a neutral element of D₄ and H. Let T : ℤ₃₆ × H→℘^*(ℤ₃₆ × D₄) be defined by

T (([x]36, ρi))={([x]36, ρi), ([x+1]36, ρiμ)},

for every ([x]₁₈, ρⁱ) ∈ ℤ₃₆ × H. Let N = {ρ⁰, ρ², μ, ρ²μ}, H be normal subgroups of D₄ and ℬ = {S₁, S₂, S₃}, where S₁ = ⟨[9]₃₆⟩ × H, S₂ = ⟨[6]₃₆⟩ × N, S₃ = ⟨[4]₃₆⟩ × H be normal subgroups of ℤ₃₆ × D₄. Let A ⊂ ℤ₃₆ × D₄ where

A={([a]36,b)∣[a]36∈Z36-{[1]36,[2]36,[3]36,[4]36,[13]36,[14]36,[15]36,[16]36,[19]36,[20]36,[21]36,[25]36,[26]36,[27]36,[28]36} andb∈{ρ0, ρ2, ρ3, μ, ρ2μ, ρ3μ}}.

We determine TNr(ℬ)¯(A) and TNr(ℬ)¯(A).

• The first set of components of A is ℤ₃₆ – {[1]₃₆, [2]₃₆, [3]₃₆, [4]₃₆, [13]₃₆, [14]₃₆, [15]₃₆, [16]₃₆, [19]₃₆, [20]₃₆, [21]₃₆, [25]₃₆, [26]₃₆, [27]₃₆, [28]₃₆}. There are no cosets of ⟨[4]₃₆⟩ contained in the first set of components of A. Cosets of ⟨[6]₃₆⟩ contained in the first set of components of A are [0]₁₈ + ⟨[6]₃₆⟩, [5]₁₈ + ⟨[6]₃₆⟩. There are no cosets of ⟨[9]₃₆⟩ contained in the first set of components of A. Cosets of ⟨[12]₃₆⟩ contained in the first set of components of A are [0]₃₆+⟨[12]₃₆⟩, [5]₃₆+⟨[12]₃₆⟩, [6]₃₆+ ⟨[12]₃₆⟩, [10]₃₆ + ⟨[12]₃₆⟩, [11]₃₆ + ⟨[12]₃₆⟩. Cosets of ⟨[18]₃₆⟩ contained in the first set of components of A are [0]₃₆ + ⟨[18]₃₆⟩, [5]₃₆ + ⟨[18]₃₆⟩, [6]₃₆ + ⟨[18]₃₆⟩, [11]₃₆ + ⟨[18]₃₆⟩, [12]₃₆ + ⟨[18]₃₆⟩, [17]₃₆ + ⟨[18]₃₆⟩.
• The second set of components of A is {ρ⁰, ρ², ρ³, μ, ρ²μ, ρ³μ}. Cosets of {ρ⁰, ρ²} contained in the second set of components of A are {ρ⁰, ρ²}, {μ, ρ²μ}. Cosets of N contained in the second set of components of A are N. There are no cosets of H contained in the second set of components of A.

(1) Determine TN1(ℬ)¯(A).

For r = 1, the subsets ℬ₁ ⊆ ℬ are {S₁}, {S₂}, {S₃}.
- • S₁ = ⟨[9]₃₆⟩×H then ∣(ℤ36×D4)/S1∣=36×84×4=18.
  
  TS1_(A)={([x]36,ρi)∈ℤ36×H∣T (([x]36,ρi)) S1⊆A}={([x]36,ρi)∈ℤ36×H∣{([x]36,ρi),([x+1]36,ρiμ)}(⟨[9]36⟩×H)⊆A}=∅.
- • S₂ = ⟨[6]₃₆⟩×N then ∣(ℤ36×D4)/S2∣=36×86×4=12.
  
  TS2_(A)={([x]36,ρi)∈ℤ36×H∣T (([x]36,ρi)) S2⊆A}={([x]36,ρi)∈ℤ36×H∣{([x]36,ρi),([x+1]36,ρiμ)}(⟨[6]36⟩×N)⊆A}={([x]36,ρi)∈ℤ36×H∣[x]36∈{[5]36,[11]36,[17]36,[23]36,[29]36,[35]36} and ρi∈{ρ0,ρ2}}.
- • S₃ = ⟨[4]₃₆⟩×H then ∣(ℤ36×D4)/S3∣=36×89×4=8.
  
  TS3_(A)={([x]36,ρi)∈ℤ36×H∣T (([x]36,ρi)) S3⊆A}={([x]36,ρi)∈ℤ36×H∣{([x]36,ρi),([x+1]36,ρiμ)}(⟨[4]36⟩×H)⊆A}=∅.
From the above calculations, we obtain

TN1(ℬ)_(A)={([x]36,ρi)∈ℤ36×H∣T (([x]36,ρi)) ℬ1⊆A, for ℬ1⊆ℬ}=TS1_(A)∪TS2_(A)∪TS3_(A)={([x]36,ρi)∈ℤ36×H∣[x]36∈{[5]36,[11]36,[17]36,[23]36,[29]36,[35]36} and ρi∈{ρ0,ρ2}}.
(2) Determine TN2(ℬ)¯(A).

For r = 2, the subsets ℬ₂ ⊆ℬ are {S₁, S₂}, {S₁, S₃}, {S₂, S₃}. By the above explanation, we have T(x)ℬ₂ = T(x)(∩_S∈ℬ₂S.)
- • S₁∩S₂ = ⟨[18]₃₆⟩×{ρ⁰, ρ²} then ∣(ℤ36×D4)/(S1∩S2)∣=36×82×2=72.
  
  TS1∩S2_(A)={([x]36,ρi)∈ℤ36×H∣T (([x]36,ρi)) (S1∩S2)⊆A}={([x]36,ρi)∈ℤ36×H∣{([x]36,ρi),([x+1]36, ρiμ)}(⟨[18]36⟩×{ρ0,ρ2})⊆A}={([x]36,ρi)∈ℤ36×H∣[x]36∈{[5]36,[11]36,[17]36,[23]36,[29]36,[35]36} and ρi∈{ρ0,ρ2}}.
- • S₁ ∩ S₃ = {[0]₃₆} × H then ∣(ℤ36×D4)/(S1∩S3)∣=36×81×4=72.
  
  TS1∩S3_(A)={([x]36,ρi)∈ℤ36×H∣T (([x]36,ρi)) (S1∩S3)⊆A}={([x]36,ρi)∈ℤ36×H∣{([x]36,ρi),([x+1]36, ρiμ)}({[0]36}×H)⊆A}=∅.
- • S₂∩S₃ = ⟨[12]₃₆⟩×{ρ^o, ρ²} then ∣(ℤ36×D4)/(S2∩S3)∣=36×83×2=48.
  
  TS2∩S3_(A)={([x]36,ρi)∈ℤ36×H∣T (([x]36,ρi)) (S2∩S3)⊆A}={([x]36,ρi)∈ℤ36×H∣{([x]36,ρi),([x+1]36, ρiμ)}(⟨[12]36⟩×{ρ0,ρ2})⊆A}={([x]36,ρi)∈ℤ36×H∣[x]36∈{[5]36,[10]36,[11]36,[17]36,[22]36,[23]36,[29]36,[34]36,[35]36} and ρi∈{ρ0,ρ2}}.
From the above calculations, we obtain

TN2(ℬ)_(A)={([x]36,ρi)∈ℤ36×H∣T (([x]36,ρi)) ℬ2⊆A, for ℬ2⊆ℬ}=TS1∩S2_(A)∪TS1∩S3_(A)∪TS2∩S3_(A)={([x]36,ρi)∈ℤ36×H∣[x]36∈{[5]36,[10]36,[11]36,[17]36,[22]36,[23]36,[29]36,[34]36,[35]36} and ρi∈{ρ0,ρ2}}.
(3) Determine TN3(ℬ)¯(A).

For r = 3, the subsets ℬ₃ ⊆ ℬ then {S₁, S₂, S₃}. By the above explanation, we have T(x)ℬ₃ = T(x) (∩_S∈ℬ₃S).
- • S₁ ∩ S₂ ∩ S₃ = {[0]₃₆} × {ρ^o, ρ²} then ∣(ℤ36×D4)/(S1∩S2∩S3)∣=36×81×2=144.
  
  TN3(ℬ)_(A)=TS1∩S2∩S3_(A)={([x]36,ρi)∈ℤ36×H∣T (([x]36,ρi)) (S1∩S2∩S3)⊆A}={([x]36,ρi)∈ℤ36×H∣{([x]36,ρi), ([x+1]36,ρiμ)}({[0]36×{ρ0,ρ2})⊆A}={([x]36,ρi)∈ℤ36×H∣[x]36∈{[5]36,[6]36,[7]36,[8]36,[9]36,[10]36,[11]36,[17]36,[22]36,[23]36,[29]36,[30]36,[31]36,[32]36,[33]36,[34]36,[35]36} and ρi∈{ρ0,ρ2}}.
As a summary, we have TN1(ℬ)¯(A)⊂TN2(ℬ)¯(A)⊂TN3(ℬ)¯(A).
- • The first set of components of A is ℤ₃₆ – {[1]₃₆, [2]₃₆, [3]₃₆, [4]₃₆, [13]₃₆, [14]₃₆, [15]₃₆, [16]₃₆, [19]₃₆, [20]₃₆, [21]₃₆, [25]₃₆, [26]₃₆, [27]₃₆, [28]₃₆}.
  
  Every coset of ⟨[4]₃₆⟩ intersects with the first set of components of A.
  
  Every coset of ⟨[6]₃₆⟩ intersects with the first set of components of A.
  
  Every coset of ⟨[9]₃₆⟩ intersects with the first set of components of A.
  
  Every coset of ⟨[12]₃₆⟩ intersects with the first set of components of A except [1]₃₆ +⟨[12]₃₆⟩, [2]₃₆ + ⟨[12]₃₆⟩, [3]₃₆ + ⟨[12]₃₆⟩, [4]₃₆ + ⟨[12]₃₆⟩.
  
  Every coset of ⟨[18]₃₆⟩ always intersects with the first set of components of A except [1]₃₆ +⟨[18]₃₆⟩, [2]₃₆ + ⟨[18]₃₆⟩, [3]₃₆ + ⟨[18]₃₆⟩.
- • The second set of components of A is {ρ⁰, ρ², ρ³, μ, ρ²μ, ρ³μ}.
  
  Every coset of {ρ⁰, ρ²} intersects with the second set of components of A.
  
  Every coset of N intersects with the second set of components of A.
  
  Every coset of H intersects with the second set of components of A.
(4) Determine TN1(ℬ)(A)¯.

For r = 1, the subsets ℬ₁ ⊆ ℬ are {S₁}, {S₂}, {S₃}.
- • S₁ = ⟨[9]₃₆⟩×H then ∣(ℤ36×D4)/S1∣=36×84×4=18.
  
  TS1¯(A)={([x]36,ρi)∈ℤ36×H∣T (([x]36,ρi)) S1∩A≠∅}={([x]36,ρi)∈ℤ36×H∣({([x]36,ρi),([x+1]36,ρiμ)}(⟨[9]36⟩×H))∩A≠∅}=ℤ36×H.
From the above calculations, we obtain

TN1(ℬ)(A)¯={([x]36,ρi)∈ℤ36×H∣T (([x]36,ρi) ℬ1∩A≠∅, for ℬ1⊆ℬ}=TS1(A)¯∪TS2(A)¯∪TS3(A)¯=ℤ36×H.
(5) Determine TN2(ℬ)(A)¯.

For r = 2, the subsets ℬ₂ ⊆ ℬ are {S₁, S₂}, {S₁, S₃}, {S₂, S₃}. By the above explanation, we have T(x)ℬ₂ = T(x) (∩_S∈ℬ₂S).
- • S₁ ∩ S₂ = ⟨[18]₃₆⟩ × {ρ^o, ρ²} then ∣(ℤ36×D4)/(S1∩S2)∣=36×82×2=72.
  
  TS1∩S2¯(A)={([x]36,ρi)∈ℤ36×H∣T (([x]36,ρi)) (S1∩S2)∩A≠∅}={([x]36,ρi)∈ℤ36×H∣({([x]36,ρi), ([x+1]36,ρiμ)}(⟨[18]36⟩×{ρ0,ρ2}))∩A≠∅}={([x]36,ρi)∈ℤ36×H∣[x]36∈ℤ36-{[1]36,[2]36,[19]36,[20]36,} and ρi∈H}.
- • S₁ ∩ S₃ = {[0]₃₆} × H then ∣(ℤ36×D4)/(S1∩S3)∣=36×81×4=72.
  
  TS1∩S3¯(A)={([x]36,ρi)∈ℤ36×H∣T (([x]36,ρi)) (S1∩S3)∩A≠∅}={([x]36,ρi)∈ℤ36×H∣({([x]36,ρi), ([x+1]36,ρiμ)}({[0]36}×H))∩A≠∅}={([x]36,ρi)∈ℤ36×H∣[x]36∈ℤ36-{[1]36,[2]36,[3]36,[13]36,[14]36,[15]36,[19]36,[20]36,[25]36,[26]36,[27]36} and ρi∈H}.
- • S₂∩S₃ = ⟨[12]₃₆⟩×{ρ^o, ρ²} then ∣(ℤ36×D4)/(S2∩S3)∣=36×83×2=48.
  
  TS2∩S3¯(A)={([x]36,ρi)∈ℤ36×H∣T (([x]36,ρi)) (S1∩S3)∩A≠∅}={([x]36,ρi)∈ℤ36×H∣({([x]36,ρi), ([x+1]36,ρiμ)}(⟨[12]36⟩×{ρo,ρ2}))∩A≠∅}={([x]36,ρi)∈ℤ36×H∣[x]36∈ℤ36-{[1]36,[2]36,[3]36,[13]36,[14]36,[15]36,[25]36,[26]36,[27]36} and ρi∈H}.
From the above calculations, we obtain

TN2(ℬ)(A)¯={([x]36,ρi)∈ℤ36×H∣T (([x]36,ρi)) ℬ2⊆A, for ℬ2⊆ℬ}=TS1∩S2(A)¯∪TS1∩S3(A)¯∪TS2∩S3(A)¯={([x]36,ρi)∈ℤ36×H∣[x]36∈ℤ36-{[1]36,[2]36} and ρi∈H}.
(6) Determine TN3(ℬ)(A)¯.

For r = 3, the subsets ℬ₃ ⊆ ℬ are {S₁, S₂, S₃}. By the above explanation, we have T(x)ℬ₃ = T(x)∩(∩_S∈ℬ₂S).
- • S₁ ∩ S₂ ∩ S₃ = {[0]₃₆} × {ρ⁰, ρ²} then ∣(ℤ36×D4)/(S1∩S2∩S3)∣=36×81×2=144.
  
  TN3(ℬ)(A)¯=TS1∩S2∩S3¯(A)={([x]36,ρi)∈ℤ36×H∣ (([x]36,ρi)) (S1∩S2∩S3)∩A≠∅}={([x]36,ρi)∈ℤ36×H∣({([x]36,ρi), ([x+1]36,ρiμ)}({[0]36}×{ρ0,ρ2}))∩A≠∅}={([x]36,ρi)∈ℤ36×H∣[x]36∈ℤ36-{[1]36,[2]36,[3]36,[13]36,[14]36,[15]36,[19]36,[20]36,[25]36,[26]36,[27]36} and ρi∈H}.

In summary, we have TN3(ℬ)(A)¯⊂TN2(ℬ)(A)¯⊂TN1(ℬ)(A)¯. The relationship between TNr(ℬ)¯(A) and TNr(ℬ)¯(A) is TN1(ℬ)_(A)⊂TN2(ℬ)_(A)⊂TN3(ℬ)_(A)⊂TN3(ℬ)¯(A)⊂TN2(ℬ)¯(A)⊂TN1(ℬ)(A)¯.

Proposition 1

Let G be a group, (G, ) be a nearness approximation space and I : G→℘^*(G) be a set-valued mapping, where I(x) = {x} for each x ∈ G. Let A be a non-empty subset of G. Then,

INr(ℬ)_(A)=Nr(ℬ)(A)_ and INr(ℬ)¯(A)=Nr(ℬ)(A)¯.

Proof

This is clear from Definitions 2 and 6.

Proposition 2

Let G and G′ be two groups, (G′, ) be a nearness approximation space, and T : G→℘^*(G′) be a set-valued mapping. Let A be a non-empty subset of G′. Then,

(1) TNr(ℬ)_(A)⊆TNr(ℬ)¯(A),
(2) TNr(ℬ)_(G′)=G=TNr(ℬ)¯(G′),
(3) If A ⊆ B, then TNr(ℬ)¯(A)⊆TNr(ℬ)¯(B) and TNr(ℬ)¯(A)⊆TNr(ℬ)¯(B),
(4) TNr(ℬ)_(A)⊇(TNr(ℬ)¯(AC))C,
(5) TNr(ℬ)¯(A)⊇(TNr(ℬ)_(AC))C,
(6) TNr(ℬ)¯(A∩B)⊆TNr(ℬ)¯(A)∩TNr(ℬ)¯(B),
(7) TNr(ℬ)¯(A∩B)⊆TNr(ℬ)¯(A)∩TNr(ℬ)¯(B),
(8) TNr(ℬ)¯(A∪B)⊇TNr(ℬ)¯(A)∪TNr(ℬ)¯(B),
(9) TNr(ℬ)¯(A∪B)=TNr(ℬ)¯(A)∪TNr(ℬ)¯(B).

Proof

(1) It is clear from Definition 6.
(2) It is clear from Definition 6.
(3) Let x ∈ TNr(ℬ)¯(A). Subsequently, T(x)ℬ_r ⊆ A ⊆ B, for some ℬ_r ⊆ ℬ so that x ∈ TNr(ℬ)¯(B). Therefore, TNr(ℬ)¯(A)⊆TNr(ℬ)¯(B). Suppose y∈TNr(ℬ)¯(A). Then, T(y)ℬr¯∩A≠∅ for some ℬ_r ⊆ ℬ. Since A ⊆ B then T(y)ℬ_r ∩ B ≠ ∅ for some ℬ_r ⊆ ℬ such that y∈TNr(ℬ)¯(B). Therefore, TNr(ℬ)¯(A)⊆TNr(ℬ)¯(B).
(4) Let x∈(TNr(ℬ)¯(AC))C. Then, x∉(TNr(ℬ)¯(AC) so that T(x) ℬ_r ∩ A^C = ∅ for every ℬ_r ⊆ ℬ, which implies that T(x) ℬ_r ⊆ A for every ℬ_r ⊆ ℬ. Therefore, x ∈ TNr(ℬ)¯(A) so that TNr(ℬ)_(A)⊇(TNr(ℬ)¯(AC))C. The converse is not true in general. Let x ∈ TNr(ℬ)¯(A). Then, T(x) ℬ_r ⊆ A for some ℬ_r ⊆ ℬ but T(x)ℬ_r′ ⊈ A for some ℬ_r′ ⊆ ℬ so that T(x)ℬ_r′ ∩ A^C ≠ ∅, which implies x∈TNr(ℬ)¯(AC). Thus, we have x∉(TNr(ℬ)¯(AC))C so that TNr(ℬ)_(A)⊄(TNr(ℬ)¯(AC))C.
(5) Let x∈TNr(ℬ)¯(AC)C. Then x∉TNr(ℬ)¯(AC) so that T(x)ℬ_r ⊈ A^C for every ℬ_r ⊆ ℬ which implies T(x)ℬ_r ∩ A ≠ ∅ for every ℬ_r ⊆ ℬ. Therefore, x∈TNr(ℬ)¯(A) so that TNr(ℬ)¯(A)⊇(TNr(ℬ)_(AC))C. The converse is not true in general. Let x∈TNr(ℬ)¯(A) Then, T(x)ℬ_r ∩ A ≠ ∅ for some ℬ_r ⊆ ℬ but T(x)ℬ_r′ ∩ A = ∅ for some ℬ_r′ ⊆ ℬ such that T(x)ℬ_r′ ⊆ A^C, which implies x∈TNr(ℬ)¯(AC). Therefore, x∉TNr(ℬ)¯(AC)C so that TNr(ℬ)¯(A)⊄(TNr(ℬ)_(AC))C.
(6) From part (3), we have TNr(ℬ)¯(A∩B)⊆TNr(ℬ)¯(A) and TNr(ℬ)¯(A∩B)⊆TNr(ℬ)¯(B). Then, TNr(ℬ)¯(A∩B)⊆TNr(ℬ)¯(A)∩TNr(ℬ)¯(B).
(7) From part (3), we have TNr(ℬ)¯(A∩B)⊆TNr(ℬ)¯(A) and TNr(ℬ)¯(A∩B)⊆TNr(ℬ)¯(B). Then, TNr(ℬ)¯(A∩B)⊆TNr(ℬ)¯(A)∩TNr(ℬ)¯(B).
(8) From part (3), we have TNr(ℬ)¯(A)⊆TNr(ℬ)¯(A∪B) and TNr(ℬ)¯(B)⊆TNr(ℬ)¯(A∪B). Then, TNr(ℬ)¯(A)∪TNr(ℬ)¯(B)⊆TNr(ℬ)¯(A∪B).
(9) Let x∈TNr(ℬ)¯(A∪B). Then, T(x)ℬ_r ∩ (A ∪ B) ≠ ∅ for some ℬ_r ⊆ ℬ so that T(x)ℬ_r ∩ A ≠ ∅ or T(x)ℬ_r ∩ A ≠ ∅ which implies x∈TNr(ℬ)¯(A) or x∈TNr(ℬ)¯(B). Therefore, TNr(ℬ)¯(A∪B)⊆TNr(ℬ)¯(A)∪TNr(ℬ)¯(B). Conversely, from part (3), we have TNr(ℬ)¯(A)⊆TNr(ℬ)¯(A∪B) and TNr(ℬ)¯(B)⊆TNr(ℬ)¯(A∪B). Then, TNr(ℬ)¯(A)∪TNr(ℬ)¯(B)⊆TNr(ℬ)¯(A∪B).

The following examples show that the converse of Proposition 2 (6,7,8) is generally not true.

Example 8

Let ℤ₁₈ be a group of integers modulo 18, ℬ = {S₁, S₂}, where S₁ = ⟨[9]₁₈⟩ = {[0]₁₈, [9]₁₈}, S₂ = ⟨[6]₁₈⟩ = {[0]₁₈, [6]₁₈, [12]₁₈} be normal subgroups of ℤ₁₈ and T : ℤ₁₈ → ℘^* (ℤ₁₈) be defined as T ([x]₁₈) = {[x + 1]₁₈}, for every [x]₁₈ ∈ ℤ₁₈. Suppose A, B ⊂ ℤ₁₈, where

A=ℤ18-([1]18+⟨[9]18⟩∪[3]18+⟨[9]18⟩∪[2]18+⟨[6]18⟩)=ℤ18-{[1]18,[2]18,[3]18,[8]18,[10]18,[12]18,[14]18},

and

B=ℤ18-([1]18+⟨[9]18⟩∪[3]18+⟨[9]18⟩∪[3]18+⟨[6]18⟩)=ℤ18-{[1]18,[3]18,[4]18,[9]18,[10]18,[13]18,[15]18}.

(1) Determine TN1(ℬ)¯(A) and TN1(ℬ)¯(B).
- • Cosets of S₁ = ⟨[9]₁₈⟩ contained in A are ⟨[9]₁₈⟩, [4]₁₈ + ⟨[9]₁₈⟩, [6]₁₈ + ⟨[9]₁₈⟩, [7]₁₈ + ⟨[9]₁₈⟩.
  
  TS1_(A)={[x]18∈ℤ18∣[x+1]18+⟨[9]18⟩⊆A}={[3]18,[5]18,[6]18,[8]18,[12]18,[14]18,[15]18,[17]18}.
- • Cosets of S₂ = ⟨[6]₁₈⟩ contained in A are [5]₁₈ + ⟨[6]₁₈⟩.
  
  TS2_(A)={[x]18∈ℤ18∣[x+1]18+⟨[6]18⟩⊆A}={[4]18,[10]18,[16]18}.
  
  From the above calculations, we have
  
  TN1(ℬ)_(A)=TS1_(A)∪TS2_(A)={[3]18,[4]18,[5]18,[6]18,[8]18,[10]18,[12]18,[14]18,[15]18,[16]18,[17]18}.
- • Cosets of S₁ = ⟨[9]₁₈⟩ contained in B are [2]₁₈ + ⟨[9]₁₈⟩, [5]₁₈+⟨[9]₁₈⟩, [7]₁₈+⟨[9]₁₈⟩, [8]₁₈+⟨[9]₁₈⟩.
  
  TS1_(B)={[x]18∈ℤ18∣[x+1]18+⟨[9]18⟩⊆B}={[1]18,[4]18,[6]18,[7]18,[10]18,[13]18,[15]18,[16]18}.
- • Cosets of S₂ = ⟨[6]₁₈⟩ contained in B are [0]₁₈ + ⟨[6]₁₈⟩, [2]₁₈ + ⟨[6]₁₈⟩, [5]₁₈ + ⟨[6]₁₈⟩.
  
  TS2_(B)={[x]18∈ℤ18∣[x+1]18+⟨[6]18⟩⊆B}={[1]18,[4]18,[5]18,[7]18,[10]18,[11]18,[13]18,[16]18,[17]18}.
  
  From the above calculations, we have
  
  TN1(ℬ)_(B)=TS1_(B)∪TS2_(B)={[1]18,[4]18,[5]18,[6]18,[7]18,[10]18,[11]18,[13]18,[15]18,[16]18,[17]18}.
  
  As a summary, we have TN1(ℬ)¯(A)∩TN1(ℬ)¯(B)={[4]18,[5]18,[6]18,[10]18,[15]18,[16]18,[17]18}.
(2) Determine TN1(ℬ)¯(A∩B).

A∩B = {[0]₁₈, [5]₁₈, [6]₁₈, [7]₁₈, [11]₁₈, [16]₁₈, [17]₁₈}.
- • Cosets of S₁ = ⟨[9]₁₈⟩ contained in A ∩ B are [7]₁₈ + ⟨[9]₁₈⟩.
  
  TS1_(A∩B)={[x]18∈ℤ18∣[x+1]18+⟨[9]18⟩⊆A∩B}={[6]18,[15]18}.
- • Cosets of S₂ = ⟨[6]₁₈⟩ contained in A ∩ B are [5]₁₈ + ⟨[6]₁₈⟩.
  
  TS2_(A∩B)={[x]18∈ℤ18∣[x+1]18+⟨[6]18⟩⊆A∩B}={[4]18,[10]18,[16]18}.
  
  From the above calculations, we have
  
  TN1(ℬ)_(A∩B)=TS1_(A∩B)∪TS2_(A∩B)={[4]18,[6]18,[10]18,[15]18,[16]18}.
As a summary, we obtain TN1(ℬ)¯(A)∩TN1(ℬ)¯(B)⊈TN1(ℬ)¯(A∩B).
(3) Determine TN1(ℬ)¯(A) and TN1(ℬ)¯(B).
- • Cosets of S₁ = ⟨[9]₁₈⟩ intersected with A are [0]₁₈+⟨[9]₁₈⟩, [2]₁₈+⟨[9]₁₈⟩, [4]₁₈+⟨[9]₁₈⟩, [5]₁₈+ ⟨[9]₁₈⟩, [6]₁₈+⟨[9]₁₈⟩, [7]₁₈+⟨[9]₁₈⟩, [8]₁₈+⟨[9]₁₈⟩.
  
  TS1¯(A)={[x]18∈ℤ18∣([x+1]18+⟨[9]18⟩)∩A≠∅}=ℤ18-{[0]18,[2]18,[9]18,[11]18}.
- • Cosets of S₂ = ⟨[6]₁₈⟩ intersected with A are [0]₁₈+⟨[6]₁₈⟩, [1]₁₈+⟨[6]₁₈⟩, [3]₁₈+⟨[6]₁₈⟩, [4]₁₈+ ⟨[6]₁₈⟩, [5]₁₈ + ⟨[6]₁₈⟩.
  
  TS2¯(A)={[x]18∈ℤ18∣([x+1]18+⟨[6]18⟩)∩A≠∅}=ℤ18-{[1]18,[7]18,[13]18}.
  
  From the above calculations, we obtain
  
  TN1(ℬ)¯(A)=TS1¯(A)∪TS2¯(A)=ℤ18.
- • Cosets of S₁ = ⟨[9]₁₈⟩ intersected with B are [0]₁₈+⟨[9]₁₈⟩, [2]₁₈+⟨[9]₁₈⟩, [3]₁₈+⟨[9]₁₈⟩, [5]₁₈+ ⟨[9]₁₈⟩, [6]₁₈+⟨[9]₁₈⟩, [7]₁₈+⟨[9]₁₈⟩, [8]₁₈+⟨[9]₁₈⟩.
  
  TS1¯(B)={[x]18∈ℤ18∣([x+1]18+⟨[9]18⟩)∩B≠∅}=ℤ18-{[0]18,[3]18,[9]18,[12]18}.
- • Cosets of S₂ = ⟨[6]₁₈⟩ intersected with B are [0]₁₈+⟨[6]₁₈⟩, [1]₁₈+⟨[6]₁₈⟩, [2]₁₈+⟨[6]₁₈⟩, [4]₁₈+ ⟨[6]₁₈⟩, [5]₁₈ + ⟨[6]₁₈⟩.
  
  TS2¯(B)={[x]18∈ℤ18∣([x+1]18+⟨[6]18⟩)∩B≠∅}=ℤ18-{[2]18,[8]18,[14]18}.
  
  From the above calculations, we obtain
  
  TN1(ℬ)¯(B)=TS1¯(B)∪TS2¯(B)=ℤ18.
  
  As a summary, we have TN1(ℬ)¯(A)∩TN1(ℬ)¯(B)=ℤ18.
(4) Determine TN1(ℬ)¯(A∩B).

A∩B = {[0]₁₈, [5]₁₈, [6]₁₈, [7]₁₈, [11]₁₈, [16]₁₈, [17]₁₈}.
- • Cosets of S₁ = ⟨[9]₁₈⟩ intersected with A ∩ B are [0]₁₈ + ⟨[9]₁₈⟩, [2]₁₈ + ⟨[9]₁₈⟩, [5]₁₈ + ⟨[9]₁₈⟩, [6]₁₈ + ⟨[9]₁₈⟩, [7]₁₈ + ⟨[9]₁₈⟩, [8]₁₈ + ⟨[9]₁₈⟩.
  
  TS1¯(A∩B)={[x]18∈ℤ18∣([x+1]18+⟨[9]18⟩)∩(A∩B)≠∅}=ℤ18-{[0]18,[2]18,[3]18,[9]18,[11]18,[12]18}.
- • Cosets of S₂ = ⟨[6]₁₈⟩ intersected with A ∩ B are [0]₁₈ + ⟨[6]₁₈⟩, [1]₁₈ + ⟨[6]₁₈⟩, [4]₁₈ + ⟨[6]₁₈⟩, [5]₁₈ + ⟨[6]₁₈⟩.
  
  TS2¯(A∩B)={[x]18∈ℤ18∣([x+1]18+⟨[6]18⟩)∩(A∩B)≠∅}=ℤ18-{[1]18,[2]18,[7]18,[8]18,[13]18,[14]18}.
  
  From the above calculations, we obtain
  
  TN1(ℬ)¯(A∩B)=TS1¯(A∩B)∪TS2¯(A∩B)=ℤ18-{[2]18}.
As a summary, we obtain TN1(ℬ)¯(A)∩TN1(ℬ)¯(B)⊈TN1(ℬ)¯(A∩B).

Example 9

Let ℤ₁₈ be a group of integers modulo 18, ℬ = {S₁, S₂} where S₁ = ⟨[9]₁₈⟩ = {[0]₁₈, [9]₁₈}, S₂ = ⟨[6]₁₈⟩ = {[0]₁₈, [6]₁₈, [12]₁₈} be normal subgroups of ℤ₁₈ and T : ℤ₁₈ → ℘^* (ℤ₁₈) be defined as

T ([x]18)={[x+1]18},

for every [x]₁₈ ∈ ℤ₁₈. Let A, B ⊂ ℤ₁₈ where A = {[0]₁₈, [5]₁₈, [11]₁₈}, B = {[9]₁₈, [17]₁₈} so A ∪ B = {[0]₁₈, [5]₁₈, [9]₁₈, [11]₁₈, [17]₁₈}.

(1) Determine TN1(ℬ)¯(A) and TN1(ℬ)¯(B).
- • There are no cosets of S₁ = ⟨[9]₁₈⟩ contained in A. T_S₁ (A) = {[x]₁₈ ∈ ℤ₁₈ | [x + 1]₁₈ + ⟨[9]₁₈⟩ ⊆ A} = ∅.
- • There are no cosets of S₂ = ⟨[6]₁₈⟩ contained in A. T_S₂ (A) = {[x]₁₈ ∈ ℤ₁₈ | [x + 1]₁₈ + ⟨[6]₁₈⟩ ⊆ A} = ∅.
  
  From the above calculations, we obtain
  
  TN1(ℬ)_(A)=TS1_(A)∪TS2_(A)=∅.
- • There are no cosets of S₁ = ⟨[9]₁₈⟩ contained in B. T_S₁ (B) = {[x]₁₈ ∈ ℤ₁₈ | [x+1]₁₈+⟨[9]₁₈⟩ ⊆ B} = ∅.
- • There are no cosets of S₂ = ⟨[6]₁₈⟩ contained in B. T_S₂ (B) = {[x]₁₈ ∈ ℤ₁₈ | [x+1]₁₈+⟨[6]₁₈⟩ ⊆ B} = ∅.
  
  From the above calculations, we obtain
  
  TN1(ℬ)_(B)=TS1_(B)∪TS2_(B)=∅.
As a summary, we have TN1(ℬ)¯(A)∪TN1(ℬ)¯(B)=∅
(2) Determine TN1(ℬ)¯(A∪B).
- • Cosets of S₁ = ⟨[9]₁₈⟩ contained in A ∪ B is ⟨[9]₁₈⟩.
  
  TS1_(A∪B)={[x]18∈ℤ18∣[x+1]18+⟨[9]18⟩⊆A∪B}={[8]18,[17]18}.
- • Cosets of S₂ = ⟨[6]₁₈⟩ contained in A∪B is [5]₁₈+ ⟨[6]₁₈⟩.
  
  TS2_(A∪B)={[x]18∈ℤ18∣[x+1]18+⟨[6]18⟩⊆A∪B}={[4]18,[10]18,[16]18}.
  
  From the above calculations, we obtain
  
  TN1(ℬ)_(A∪B)=TS1_(A∪B)∪TS2_(A∪B)={[4]18,[8]18,[10]18,[16]18,[17]18}.
As a summary, we have TN1(ℬ)¯(A∪B)⊈TN1(ℬ)¯(A)∪TN1(ℬ)¯(B).

Proposition 3

Let G and G′ be two groups, (G′, ) be a nearness approximation space, and T : G→℘^*(G′) be a set-valued homomorphism. If A is a non-empty subset of G′ and e′ ∈ A, then e∈TNr(ℬ)¯(A) where e and e′ are neutral elements of G and G′, respectively.

Proof

Since T is a set-valued group homomorphism, it follows that for any a ∈ G, T(e) = T(aa⁻¹) = T(a)T(a⁻¹) = T(a)(T(a))⁻¹. If x ∈ T(a) then e′ = xx⁻¹ ∈ T(a)(T(a))⁻¹ = T(e). Because e′ ∈ A and e′ ∈ T(e) then e′ ∈ T(e)ℬ_r ∩A, for every ℬ_r ⊆ ℬ such that e∈TNr(ℬ)¯(A).

Proposition 4

Let G and G′ be two groups, (G′, ) be a nearness approximation space, and T : G→℘^*(G′) be a set-valued mapping. Then,

(1) TNr(ℬ)¯(A)=∪ℬr⊆ℬT¯(Aℬr),
(2) TN1(ℬ)¯(A)⊆∪ℬr⊆ℬT¯(Aℬr),

where Aℬ_r = A (∩_{S∈ℬ_r}S).

Proof

(1) Similar to the proof of Theorem 4.5 in [26], we have

TNr(ℬ)¯(A)=∪ℬr⊆ℬTℬr¯(A)=∪ℬr⊆ℬT¯(Aℬr).

(2) Similar to the proof of Theorem 4.5 in [26], we have TNr(ℬ)¯(A)=∪ℬr⊆ℬTℬr¯(A)⊆∪ℬr⊆ℬT¯(Aℬr).

The following corollary follows from Proposition 4.

Corollary 1

Let G and G′ be two groups, (G′, ) be a nearness approximation space, and T : G→℘^*(G′) be a set-valued mapping. Let N be a normal subgroup of G′ and A ⊆ G′, A ≠ ∅︀. Then,

(1) TNr(ℬ)¯(A)=∪ℬr⊆ℬTℬr¯(Aℬr),
(2) TNr(ℬ)¯(A)⊆∪ℬr⊆ℬTℬr¯(Aℬr), where Aℬ_r = A (∩_{S∈ℬ_r}S).

Proof

(1) Similar to the proof of Theorem 4.5 in [26], we have Tℬr¯(Aℬr)=T¯(Aℬrℬr)=T¯(Aℬr) so that with Proposition 4, we get TNr(ℬ)¯(A)=∪ℬr⊆ℬTℬr¯(Aℬr).

(2) Since A ⊆ Aℬ_r, it follows that TNr(ℬ)¯(A)=∪ℬr⊆ℬTℬr¯(A)⊆∪ℬr⊆ℬTℬr¯(Aℬr).

Definition 7

Let G and G′ be two groups, (G′, ) be a nearness approximation space, and T : G→℘^*(G′) be a set-valued mapping. Let A be a non-empty subset of G′. We define the sets

TNr(ℬℬ)_(A)={x∈G∣T(x)ℬrℬr′⊆A, for ℬr,ℬr′⊆ℬ},

and

TNr(ℬℬ)¯(A)={x∈G∣T(x)ℬrℬr′∩A≠∅, for ℬr,ℬr′⊆ℬ},

where T(x)ℬrℬr′=T(x) (∩S∈ℬrS) (∩R∈ℬr′R).

Proposition 5

Let G and G′ be two groups, (G′, ) be the nearness approximation space, and T : G→ ℘^*(G′) be a set-valued homomorphism. If A is a non-empty subset of G′, then

(1) TNr(ℬℬ)¯(A)=TNr(ℬ)¯(A),
(2) TNr(ℬ)¯(A)⊆TNr(ℬℬ)¯(A).

Proof

(1) Let x ∈ TNr(ℬℬ)¯(A). Then, T ( x ) ℬ r ℬ r ′ ⊆ A for some ℬ_r, ℬ r ′ ⊆ ℬ. Since T ( x ) ℬ r ⊆ T ( x ) ℬ r ℬ r ′ ⊆ A, we have x ∈ TNr(ℬ)¯(A) so that TNr(ℬℬ)¯(A)⊆TNr(ℬ)¯(A). Let x ∈ TNr(ℬ)¯(A). Subsequently, T(x)ℬ_rℬ_r = T(x)ℬ_r ⊆ A for some ℬ_r ⊆ ℬ so that x ∈ TNr(ℬℬ)¯(A), which implies that TNr(ℬ)¯(A)⊆TNr(ℬℬ)¯(A). Thus, TNr(ℬℬ)¯(A)=TNr(ℬ)¯(A).

(2) Let x∈TNr(ℬ)¯(A). Then, T(x)ℬ_r ∩ A ≠ ∅︀ for some ℬ_r ⊆ ℬ. Since T(x)ℬr⊆T(x)ℬrℬr′ for every ℬr′⊆ℬ, we obtain T(x)ℬrℬr′∩A≠∅ such that x∈TNr(ℬℬ)¯(A). Thus, TNr(ℬ)¯(A)⊆TNr(ℬℬ)¯(A).

The following example shows that the converse of Proposition 5 (2) is not true in general.

Example 10

Let ℤ₂₄ be a group of integers modulo 24, ℬ = {S₁, S₂} where S₁ = ⟨[8]₂₄⟩ = {[0]₂₄, [8]₂₄, [16]₂₄}, S₂ = ⟨[12]₂₄⟩ = {[0]₂₄, [12]₂₄} be normal subgroups of ℤ₂₄ and T : ℤ₂₄ → ℘^*(ℤ₂₄) be defined as

T ([x]24)={(x+1)24},

for every [x]₂₄ ∈ ℤ₂₄. It is easy to see S₁+S₁ = S₁, S₂+S₂ = S₂ and S₁ + S₂ = ⟨[4]₂₄⟩. Let A = {[4]₂₄}. Then,

TS1+S1¯(A)=TS1¯(A)={[3]24,[11]24,[19]24},TS2+S2¯(A)=TS2¯(A)={[3]24,[15]24},TS1+S2¯(A)={[3]24,[7]24,[11]24,[15]24,[19]24,[23]24}.

We have

TN1(ℬ)¯(A)=TS1¯(A)∪TS2¯(A)={[3]24,[11]24,[15]24,[19]24},TN1(ℬ+ℬ)¯(A)=TS1+S1¯(A)∪TS2+S2¯(A)∪TS1+S2¯(A)={[3]24,[7]24,[11]24,[15]24,[19]24,[23]24}.

Thus, we obtain TN1(ℬ+ℬ)(A)¯⊈TN1(ℬ)(A)¯.

Proposition 6

Let G and G′ be two groups, (G′, ) be a nearness approximation space, and T : G→℘^*(G′) be a set-valued homomorphism. If A and B are two non-empty subsets of G′, then

TNr(ℬ)¯(A) TNr(ℬ)¯(B)⊆TNr(ℬℬ)¯(AB).

Proof

Let z∈TNr(ℬ)¯(A) TNr(ℬ)¯(B). Then, there exist x∈TNr(ℬ)¯(A) and y∈TNr(ℬ)¯(B) such that z = xy. Since x∈TNr(ℬ)¯(A) and y∈TNr(ℬ)¯(B) then T(x)ℬ_r ∩ A ≠ ∅︀ for some ℬ_r ⊆ ℬ and T(y)ℬr′∩B≠∅ for some ℬr′⊆ℬ so there exist a, b ∈ G′ where a ∈ T(x)ℬ_r ∩ A and b∈T(y)ℬr′∩B. Since a ∈ T(x)ℬ_r, b∈T(y)ℬr′ and T be a set-valued group homomorphism then ab∈T(x)ℬrT(y)ℬr′=T(x)T(y)ℬrℬr′=T(xy)ℬrℬr′ so that ab∈T(xy)ℬrℬr′∩AB which implies T(xy)ℬrℬr′∩AB≠ϕ. Therefore, z=xy∈TNr(ℬℬ)¯(AB) so that TNr(ℬ)¯(A)TNr(ℬ)¯(B)⊆TNr(ℬℬ)¯(AB).

The following example shows that the converse of Proposition 6 is generally not true.

Example 11

Let ℤ₂₄ be a group of integers modulo 24, ℬ = {S₁, S₂} where S₁ = ⟨[8]₂₄⟩ = {[0]₂₄, [8]₂₄, [16]₂₄}, S₂ = ⟨[12]₂₄⟩ = {[0]₂₄, [12]₂₄}, S₁ + S₂ = ⟨[4]₂₄⟩ = {[0]₂₄, [4]₂₄, [8]₂₄, [12]₂₄, [16]₂₄, [20]₂₄} be normal subgroups of ℤ₂₄ and T : ℤ₂₄→℘^*(ℤ₂₄) be defined as

T ([x]24)={8 [x]24},

for every [x]₂₄ ∈ ℤ₂₄. Notice that T is a set-valued homomorphism. Let A = {[1]₂₄, [5]₂₄}, B = {[3]₂₄}, A + B = {[4]₂₄, [8]₂₄}. We get

• TS1¯(A)=TS2¯(A)=TS1¯(B)=TS2¯(B)=∅,
• TS1+S2¯(A+B)=ℤ24,
• TN1(ℬ)¯(A)=TN1(ℬ)¯(B)=∅,
• TN1(ℬ+ℬ)¯(A+B)=ℤ24.

This shows that TN1(ℬ+ℬ)¯(A+B)⊈TN1(ℬ)¯(A)+TN1(ℬ)¯(B).

Proposition 7

Let G and G′ be two groups, (G′, ) be a nearness approximation space, and T : G→℘^*(G′) be a set-valued homomorphism. If A and B are two non-empty subsets of G′, then

TNr(ℬ)_(A)TNr(ℬ)_(B)⊆TNr(ℬ)_(AB).

Proof

Let z ∈ TNr(ℬ)¯(A)TNr(ℬ)¯(B). Then, there exist x ∈ TNr(ℬ)¯(A) and y ∈ TNr(ℬ)¯(B) so that z = xy. Since x ∈ TNr(ℬ)¯(A) and y ∈ TNr(ℬ)¯(B) then T(x)ℬ_r ⊆ A and T(y)ℬr′⊆B for some ℬ_r, ℬr′⊆ℬ. Since T is a set-valued group homomorphism then T(x)ℬrT(y)ℬr′=T(x)T(y)ℬrℬr′=T(xy)ℬrℬr′⊆AB, which implies that z = xy ∈ TNr(ℬℬ)¯(AB). Hence TNr(ℬ)¯(A)TNr(ℬ)¯(B)⊆TNr(ℬℬ)¯(AB). By Proposition 5(1), we have TNr(ℬ)¯(A)TNr(ℬ)¯(B)⊆TNr(ℬ)¯(AB).

Example 12

Let Q₈ = {1, −1, i, −i, j, −j, k, −k} be a quaternion group, ℬ = {S₁, S₂} where S₁ = {1, −1, i, −i}, S₂ = {1, −1, j, −j} be the normal subgroups of Q₈ and T : Q₈→℘^*(Q₈) be defined as

T(x)={x,-x},

for every x ∈ Q₈. From Example 4.2, in [35], the mapping T is a set-valued homomorphism. Let A = {1, i}, B = {1, −1, −i}, A.B = {1, −1, i, −i} = S₁. Then, we get

• TS1¯(A)=TS2¯(A)=TS1¯(B)=TS2¯(B)=TS2¯(A.B)=∅,
• TS1¯(A.B)={1,−1,i,−i}.

So that

• TN1(ℬ)¯(A)=TN1(ℬ)¯(B)=∅,
• TN1(ℬ)¯(A.B)={1,−1,i,−i}.

This implies that TN1(ℬ)¯(A.B)⊈TN1(ℬ)¯(A).TN1(ℬ)¯(B).

5. The Lower and Upper Near Subgroups in a Group based on a Set-Valued Mapping

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Abstract
1. Introduction
2. Approximations and Near Approximations in a Group
3. Approximations in a Group based on a Set-Valued Mapping
4. Near-Generalized Approximations in a Group based on a Set-Valued Mapping
5. The Lower and Upper Near Subgroups in a Group based on a Set-Valued Mapping
6. Comparison between Near-Generalized and Generalized Approximations in a Group based on a Set-Valued Mapping
7. Conclusion
Conflict of Interest
References
Biographies

Proposition 8

Let G and G′ be two groups, (G′, ) be a nearness approximation space, and T : G→℘^*(G′) be a set-valued homomorphism. Let A be a subgroup of G′, TNr(ℬ)¯(A)=TNr(ℬℬ)¯(A) and TNr(ℬ)¯(A)≠∅. Subsequently, TNr(ℬ)¯(A) is a subgroup of G.

Proof

Let x, y∈TNr(ℬ)¯(A). Then, T(x)ℬ_r ∩ A ≠ ∅︀ and T(y)ℬr′∩A≠∅ for some ℬ_r, ℬr′⊆ℬ. So there exist a ∈ T(x)ℬ^r ∩A and b∈T(y)ℬr′∩A. Since A is a subgroup of G′, we have ab⁻¹ ∈ A. Since T is a set-valued homomorphism, we have ab-1∈T(x)ℬr(T(y)ℬr′)-1=T(x)ℬr(ℬr′)-1(T(y))-1=T(x)ℬrℬr′T(y-1)=T(x)T(y-1)ℬrℬr′=T(xy-1)ℬrℬr′ so that ab-1∈T(xy-1)ℬrℬr′∩A which implies that xy-1∈TNr(ℬℬ)¯(A)=TNr(ℬ)¯(A). Therefore, TNr(ℬ)¯(A) is a subgroup of G.

Proposition 9

Let G and G′ be two groups, (G′, ) be a nearness approximation space, and T : G→℘^* (G′) be a set-valued homomorphism. Let A be a normal subgroup of G′, TNr(ℬ)¯(A)=TNr(ℬℬ)¯(A) and TNr(ℬ)¯(A)≠∅. Subsequently, TNr(ℬ)¯(A) is a normal subgroup of G.

Proof

This is sufficient to demonstrate that TNr(ℬ)¯(A) is normal. Let x∈TNr(ℬ)¯(A) and g ∈ G. Then, T(x)ℬ_r ∩ A ≠ ∅︀ for some ℬ_r ⊆ ℬ and T(g) is a non-empty set, we can select g′ ∈ T(g). Thus, a ∈ T(x)ℬ_r ∩ A. exists and because A is a normal subgroup of G′, we have ag′a⁻¹ ∈ A. Since T is a set-valued homomorphism, we have ag′a⁻¹ ∈ T(x)ℬ_rT(g)(T(x)ℬ_r)⁻¹ = T(x)ℬ_rT(g)ℬ_r⁻¹T(x)⁻¹=T(x) ℬ_rT(g)ℬ_r⁻¹T(x⁻¹)=T(x)T(g)T(x⁻¹)ℬ_rℬ_r⁻¹=T(xgx⁻¹) ℬ_r such that ag′a⁻¹ ∈ T(xgx⁻¹)ℬ_r ∩ A which implies that xgx-1∈TNr(ℬ)¯(A). Therefore, TNr(ℬ)¯(A) is a normal subgroup of G.

Example 13

Let ℤ₂₄ be a group of integers modulo 24, ℬ = {S₁, S₂}, where S₁ = ⟨[8]₂₄⟩ = {[0]₂₄, [8]₂₄, [16]₂₄}, S₂ = ⟨[12]₂₄⟩ = {[0]₂₄, [12]₂₄} be normal subgroups of ℤ₂₄ and T : ℤ₂₄→℘^*(ℤ₂₄) be defined as

T ([x]24)={3 [x]24},

for every [x]₂₄ ∈ ℤ₂₄. Observe that T is a set-valued homomorphism, S₁ + S₁ = S₁, S₂ + S₂ = S₂, and S₁ + S₂ = ⟨[4]₂₄⟩. Let A ⊂ ℤ₂₄ where A = 2ℤ₂₄. Then, A is a normal subgroup of ℤ₂₄. We get

TS1¯(A)=TS2¯(A)=TS1+S1¯(A)=TS2+S2¯(A)=TS1+S2¯(A)=2ℤ24.

so that TN1(ℬ)¯(A)=TN1(ℬ+ℬ)¯(A)=2ℤ24 is a normal subgroup of ℤ₂₄.

Proposition 10

Let G and G′ be two groups, (G′, ) be a nearness approximation space, and T : G→℘^*(G′) be a set-valued homomorphism. Let A be a subgroup of G′ and TNr(ℬ)¯(A) ≠ ∅. Then, TNr(ℬ)¯(A) is a subgroup of G.

Proof

Let x, y ∈ TNr(ℬ)¯(A). Then, T(x)ℬ_r ⊆ A and T(y)ℬr′⊆A for some ℬ_r, ℬr′⊆ℬ. Since A is a subgroup of G′, we have (T(y)ℬr′)-1=T(y-1)ℬr′⊆A. Since T is a set-valued homomorphism, we have T(x)ℬrT(y-1)ℬr′=T(x)T(y-1)ℬrℬr′=T(xy-1)ℬrℬr′⊆A which implies that xy⁻¹ ∈ TNr(ℬB)¯(A). From Proposition 5(1), we have xy⁻¹ ∈ TNr(ℬ)¯(A). Therefore, TNr(ℬ)¯(A) is a subgroup of G.

Proposition 11

Let G and G′ be two groups, (G′, ) be a nearness approximation space, and T : G→℘^*(G′) be a set-valued homomorphism. Let A be a normal subgroup of G′ and TNr(ℬ)¯(A) ≠ ∅. Then, TNr(ℬ)¯(A) is a normal subgroup of G.

Proof

This suffices for demonstrating that TNr(ℬ)¯(A) is normal. Let x ∈ TNr(ℬ)¯(A) and g ∈ G. Then, T(x)ℬ_r ⊆ A for some ℬ_r ⊆ ℬ and T(g) is a non-empty set. Since A is a subgroup of G′, we have (T(x)ℬ_r)⁻¹ = T(x⁻¹)ℬ_r ⊆ A. Since A is a normal subgroup of G′ and T is a set-valued homomorphism, we have T(x)ℬ_rT(g)T(x⁻¹)ℬ_r = T(x)T(g)T(x⁻¹)ℬ_r = T(xgx⁻¹)ℬ_r ⊆ A which implies that xgx⁻¹ ∈ TNr(ℬ)¯(A). Therefore, TNr(ℬ)¯(A) is a normal subgroup of G.

6. Comparison between Near-Generalized and Generalized Approximations in a Group based on a Set-Valued Mapping

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Abstract
1. Introduction
2. Approximations and Near Approximations in a Group
3. Approximations in a Group based on a Set-Valued Mapping
4. Near-Generalized Approximations in a Group based on a Set-Valued Mapping
5. The Lower and Upper Near Subgroups in a Group based on a Set-Valued Mapping
6. Comparison between Near-Generalized and Generalized Approximations in a Group based on a Set-Valued Mapping
7. Conclusion
Conflict of Interest
References
Biographies

Proposition 12

Let G and G′ be two groups, (G′, ) be a nearness approximation space, and T : G→℘^*(G′) be a set-valued mapping. Let A be a non-empty subset of G′. Then,

(1) If |ℬ| = k, then TNk(ℬ)¯(A)=Tℬ¯(A) and TNk(ℬ)¯(A)=Tℬ¯(A).
(2) If 1 ≤ r < k where |ℬ| = k, then Tℬr¯(A)⊆TNr(ℬ)¯(A) and Tℬr¯(A)⊆TNr(ℬ)¯(A).

Proof

The proof results from Definitions 5 and 6

From Proposition 12(1), we observe that the generalized approximations in a group based on set-valued mapping is a special type of near generalized approximation.

Proposition 13

Let G and G′ be two groups, (G′, ) be a nearness approximation space, and T : G→℘^*(G′) be a set-valued mapping and ℬ_i ⊂ ℬ_j ⊆ ℬ, 1 ≤ i < j ≤ |ℬ|. Let A be a non-empty subset of G′. Then

(1) TNr(ℬi)¯(A)⊆TNr(ℬj)¯(A),
(2) TNr(ℬi)¯(A)⊆TNr(ℬj)¯(A).

Proof

(1) From Definition 6, we have TNr(ℬk)¯(A)=∪ℬr⊆ℬkTℬr¯(A), for 1 ≤ k ≤ |ℬ|. Since 1 ≤ i < j ≤ |ℬ|, we have ℬ_i ⊂ ℬ_j ⊆ ℬ then TNr(ℬi)¯(A)⊆TNr(ℬj)¯(A).

(2) From Definition 6, we obtain TNr(ℬk)¯(A)=∪ℬr⊆ℬkTℬr¯(A) for 1 ≤ k ≤ |ℬ|. Since 1 ≤ i < j ≤ |ℬ|, we have ℬ_i ⊂ ℬ_j ⊆ ℬ then TNr(ℬi)¯(A)⊆TNr(ℬj)¯(A).

Proposition 14

Let G and G′ be two groups, (G′, ), be a nearness approximation space, and T : G→℘^*(G′) be a set-valued mapping. Let A be a non-empty subset of G′. Then,

TNi(ℬ)_(A)⊆TNj(ℬ)_(A)⊆TNj(ℬ)¯(A)⊆TNi(ℬ)¯(A),

where 1 ≤ i < j ≤ |ℬ|.

Proof

Let x ∈ TNi(ℬ)¯(A). Subsequently, T(x)ℬ_i ⊆ A for a certain ℬ_i ⊆ ℬ. Since i < j then ℬ_i ⊂ ℬ_j so that T(x)ℬ_j = T(x)∩_{S∈ℬ_j} S ⊆ T(x)∩_{S∈ℬ_i} S = T(x)ℬ_i ⊆ A which yields that x ∈ TNj(ℬ)¯(A). Therefore, TNi(ℬ)¯(A)⊆TNj(ℬ)¯(A). From Proposition 2 (1), we have TNj(ℬ)_(A)⊆TNj(ℬ)¯(A). Let x∈TNj(ℬ)¯(A). Then, T(x)ℬ_j ∩ A ≠ ∅︀ for some ℬ_j ⊆ ℬ. Since i < j then ℬ_i ⊂ ℬ_j so that T(x)ℬ_j = T(x)∩_{S∈ℬ_j}S ⊆ T(x)∩_{S∈ℬ_i}S = T(x)ℬ_i. Thus, we have T(x)ℬ_i ∩ A ≠ ∅︀, which yields x∈TNi(ℬ)¯(A). Therefore, TNi(ℬ)_(A)⊆TNj(ℬ)_(A)⊆TNj(ℬ)¯(A)⊆TNi(ℬ)¯(A).

7. Conclusion

Go to

Abstract
1. Introduction
2. Approximations and Near Approximations in a Group
3. Approximations in a Group based on a Set-Valued Mapping
4. Near-Generalized Approximations in a Group based on a Set-Valued Mapping
5. The Lower and Upper Near Subgroups in a Group based on a Set-Valued Mapping
6. Comparison between Near-Generalized and Generalized Approximations in a Group based on a Set-Valued Mapping
7. Conclusion
Conflict of Interest
References
Biographies

In this study, we showed that near set theory is applicable to group theory. We demonstrated the concepts of near-generalized approximations in a group based on set-valued mapping, an extension of generalized approximations in a group based on set-valued mapping and near approximations in a group. We demonstrate that near-generalized lower and upper approximations of a (normal) subgroup in a group based on set-valued mapping can form a (normal) subgroup of a group. Furthermore, we showed that the generalized approximations in a group based on a set-valued mapping are a special type of near-generalized approximations in a group based on a set-valued mapping. Finally, we found that if the r index is larger, then the near-generalized lower approximations in a group based on set-valued mapping are greater (in terms of the inclusion sets), but the near-generalized upper approximations in a group based on set-valued mapping are smaller sets. Near-generalized lower approximations in a group based on set-valued mapping are subsets of near-generalized upper approximations in a group based on set-valued mapping. As an extension of this study, we are interested in applying this structure in a ring.

Conflict of Interest

Go to

Abstract
1. Introduction
2. Approximations and Near Approximations in a Group
3. Approximations in a Group based on a Set-Valued Mapping
4. Near-Generalized Approximations in a Group based on a Set-Valued Mapping
5. The Lower and Upper Near Subgroups in a Group based on a Set-Valued Mapping
6. Comparison between Near-Generalized and Generalized Approximations in a Group based on a Set-Valued Mapping
7. Conclusion
Conflict of Interest
References
Biographies

No potential conflict of interest relevant to this article was reported.

References

Go to

Abstract
1. Introduction
2. Approximations and Near Approximations in a Group
3. Approximations in a Group based on a Set-Valued Mapping
4. Near-Generalized Approximations in a Group based on a Set-Valued Mapping
5. The Lower and Upper Near Subgroups in a Group based on a Set-Valued Mapping
6. Comparison between Near-Generalized and Generalized Approximations in a Group based on a Set-Valued Mapping
7. Conclusion
Conflict of Interest
References
Biographies

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Kuroki, N, and Wang, PP (1996). The lower and upper approximations in a fuzzy group. Information Sciences. 90, 203-220. https://doi.org/10.1016/0020-0255(95)00282-0
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Glebsky, L (2017). Approximations of groups, characterizations of sofic groups, and equations over groups. Journal of Algebra. 477, 147-162. https://doi.org/10.1016/j.jalgebra.2016.12.012
Mahmood, W, Nazeer, W, and Kang, SM (2018). A comparison between lower and upper approximations in groups with respect to group homomorphisms. Journal of Intelligent & Fuzzy Systems. 35, 693-703. https://doi.org/10.3233/JIFS-171037
Davvaz, B (2004). Roughness in rings. Information Sciences. 164, 147-163. https://doi.org/10.1016/j.ins.2003.10.001
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Davvaz, B (2014). A short note on approximations in a ring by using a neighborhood system as a generalization of Pawlak’s approximations. UPB Scientific Bulletin, Series A: Applied Mathematics and Physics. 76, 77-84.
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Biographies

Go to

Abstract
1. Introduction
2. Approximations and Near Approximations in a Group
3. Approximations in a Group based on a Set-Valued Mapping
4. Near-Generalized Approximations in a Group based on a Set-Valued Mapping
5. The Lower and Upper Near Subgroups in a Group based on a Set-Valued Mapping
6. Comparison between Near-Generalized and Generalized Approximations in a Group based on a Set-Valued Mapping
7. Conclusion
Conflict of Interest
References
Biographies

DianWinda Setyawati is a doctoral student in the Department of Mathematics at Institut Teknologi Sepuluh Nopember, Indonesia. Since 2003, she has worked as a lecturer in the Department of Mathematics at Institut Teknologi Sepuluh Nopember, Indonesia. Her research areas include group theory, ring theory, and rough set theory.

Subiono is a professor in the Department of Mathematics at Institut Teknologi Sepuluh Nopember, Indonesia. He has worked as a lecturer in the Department of Mathematics at Institut Teknologi Sepuluh Nopember, Indonesia, since 1984. His research areas are min-maxpluss algebra and applied algebra.

Bijan Davvaz received his B.Sc. degree in Applied Mathematics from Shiraz University in 1988 and his M.Sc. degree in Pure Mathematics from Tehran University in 1990. In 1998, he received his Ph.D. in mathematics at Tarbiat Modares University. He is a member of the Editorial Boards of 24 mathematical journals. He has also served as the Head of the Department of Mathematics (1998–2002), Chairman of the Faculty of Science (2004–2006), and Vice-President for Research (2006–2008) at Yazd University, Iran. His areas of interest include algebra (groups, rings, and modules), algebraic hyperstructures, rough sets, and fuzzy logic. He has published 12 books on mathematics. He is currently a distinguished professor of Mathematics at Yazd University.

Article

Original Article

International Journal of Fuzzy Logic and Intelligent Systems 2024; 24(2): 125-140

Published online June 25, 2024 https://doi.org/10.5391/IJFIS.2024.24.2.125

Near-Generalized Approximations in Groups Based on a Set-Valued Mapping

Dian Winda Setyawati¹, Subiono¹, and Bijan Davvaz²

¹Department of Mathematics, Institut Teknologi Sepuluh Nopember, Kampus ITS, Sukolilo-Surabaya, Indonesia
²Department of Mathematical Sciences, Yazd University, Yazd, Iran

Correspondence to:Subiono (subiono2008@matematika.its.ac.id)

Received: June 5, 2023; Revised: August 24, 2023; Accepted: June 5, 2024

Abstract

Keywords: Normal subgroup, Group, Approximations, Near approximations, Generalized approximations, Near-generalized approximations

1. Introduction

2. Approximations and Near Approximations in a Group

Definition 1 [11]

Let G be a group, S a normal subgroup of G, and A a non-empty subset of G. We define the sets

\underline{{Apr}_{S}} (A) = {x \in G ∣ x S \subseteq A},

and

\bar{{Apr}_{S}} (A) = {x \in G ∣ x S \cap A \neq \emptyset},

called lower and upper approximations, respectively. of set A with respect to the normal subgroup S.

Example 1

We denote as the set of all normal subgroups in the group G and ℬ as a subset of . A pair (G, ) is called a nearness approximation space. If (G, ) is a nearness approximation space, then

x ℬ_{r} = x (\underset{S \in ℬ_{r}}{\cap} S),

where ℬ_r ⊆ ℬ.

Definition 2 [31]

Let (G, ) be a nearness approximation space and A be a non-empty subset of G. We define the sets

\underline{N_{r} (ℬ)} (A) = {x \in G ∣ x ℬ_{r} \subseteq A, for ℬ_{r} \subseteq ℬ},

and

\bar{N_{r} (ℬ)} (A) = {x \in G ∣ x ℬ_{r} \cap A \neq \emptyset, for ℬ_{r} \subseteq ℬ},

called near lower and near upper approximations of set A with respect to the normal subgroups of G, respectively. Clearly, $\underline{N_{r} (ℬ)} (A) = \cup_{ℬ_{r} \subseteq ℬ} \underline{{Apr}_{ℬ_{r}}} (A)$ and $\bar{N_{r} (ℬ)} (A) = \cup_{ℬ_{r} \subseteq ℬ} \bar{{Apr}_{ℬ_{r}}} (A)$ .

Consider the following example. In this example [31], three different near approximations of the subset A of G defined by families N₁(ℬ)(A), N₂(ℬ)(A), N₃(ℬ)(A).

Example 2

• $\underline{N_{1} (ℬ)} (A) = \{{[0]}_{12}, {[1]}_{12}, {[4]}_{12}, {[7]}_{12}, {[8]}_{12}, {[10]}_{12}\},$
• $\underline{N_{2} (ℬ)} (A) = A,$
• $\underline{N_{3} (ℬ)} (A) = A,$
• $\bar{N_{1} (ℬ)} (A) = ℤ_{12}$ ,
• $\bar{N_{2} (ℬ)} (A) = {{[0]}_{12}, {[1]}_{12}, {[2]}_{12}, {[3]}_{12}, {[4]}_{12}, {[6]}_{12}, {[7]}_{12}, {[8]}_{12}, {[9]}_{12}, {[10]}_{12}}$ ,
• $\bar{N_{3} (ℬ)} (A) = A$ .

From the calculations above, we can conclude that $\underline{N_{1} (ℬ)} (A) \subset \underline{N_{2} (ℬ)} (A) = \underline{N_{3} (ℬ)} (A) = \bar{N_{3} (ℬ)} (A) \subset \bar{N_{2} (ℬ)} (A) \subset \bar{N_{1} (ℬ)} (A)$ .

3. Approximations in a Group based on a Set-Valued Mapping

In this study, if N is a non-empty set, then ℘^*(N) denotes the set of all non-empty subsets of N. The following definition is a generalization of Definition 1.

Definition 3 [35]

Let M and N be two non-empty sets and T : M→℘^*(N) be a set-valued mapping. Let A be a non-empty subset of N. We define the sets

\underline{T} (A) = {x \in M ∣ T (x) \subseteq A},

and

\bar{T} (A) = {x \in M ∣ T (x) \cap A \neq \emptyset},

which are called lower and upper approximations of A under T, respectively.

Example 3

In [35], the set-valued homomorphism between the two groups is defined.

Definition 4

Let G and G′ be the two groups. A mapping T : G→℘^*(G′) is called a set-valued homomorphism if

(1) T(xy) = T(x)T(y),
(2) (T(x))⁻¹ = {a⁻¹ | a ∈ T(x)} = T (x⁻¹),

for all x, y ∈ G.

Example 4 shows that each group homomorphism is a set-valued homomorphism.

Example 4

(1) T(xy) = {f(xy)} = {f(x)f(y)} = {f(x)} {f(y)} = T(x)T(y),
(2) (T(x))⁻¹ = {(f(x))⁻¹} = {f (x⁻¹)} = T(x⁻¹).

Example 5 [35]

The set ℝ^* of non-zero real numbers is a group. Suppose that $D = {[\begin{matrix} 1 & 0 \\ 0 & a \end{matrix}] | a \in ℝ^{*}}$ . The set-valued mapping T : ℝ^* → ℘^*(D) defined by $T (a) = {[\begin{matrix} 1 & 0 \\ 0 & a \end{matrix}], [\begin{matrix} 1 & 0 \\ 0 & - a \end{matrix}]}$ is a set-valued homomorphism.

Definition 5

Let G and G′ be two groups and T : G→℘^* (G′) be a set-valued mapping. Let S be a normal subgroup of G′ and A be a non-empty subset of G′. Then,

\underline{T_{S}} (A) = {x \in G ∣ T (x) S \subseteq A},

and

\bar{T_{S}} (A) = {x \in G ∣ (T (x) S) \cap A \neq \emptyset},

called the generalized lower and upper approximations of A under T with respect to a normal subgroup S of G′, respectively.

Example 6

T (x) = {\begin{array}{l} {x} & if x \in {ρ^{0}, ρ^{1}, ρ^{2}, ρ^{3}, ρ^{4}, ρ^{5}}, \\ {ρ^{i}, μ} & if x = ρ^{i} μ \in {μ, ρ^{1} μ, ρ^{2} μ, ρ^{3} μ, ρ^{4} μ, ρ^{5} μ} . \end{array}

Let A = {ρ¹, ρ³, ρ⁵, μρ²}. Then, we get

• T_S(A) = {ρ¹, ρ³, ρ⁵},
• $\bar{T_{S}} (A) = {ρ^{1}, ρ^{3}, ρ^{5}, μ, ρ μ, ρ^{2} μ, ρ^{3} μ, ρ^{4} μ, ρ^{5} μ}$ .

From the above calculations, we conclude that $\underline{T_{S}} (A) \subset \bar{T_{S}} (A)$ .

4. Near-Generalized Approximations in a Group based on a Set-Valued Mapping

T (x) ℬ_{r} = T (x) (\underset{S \in ℬ_{r}}{\cap} S),

where ℬ_r ⊆ ℬ.

Definition 6

Let G and G′ be two groups, (G′, ) be a nearness approximation space, and T : G→℘^*(G′) be a set-valued mapping. Let A be a non-empty subset of G′. We define the sets

\underline{T_{N_{r} (ℬ)}} (A) = {x \in G ∣ T (x) ℬ_{r} \subseteq A, for ℬ_{r} \subseteq ℬ},

and

\bar{T_{N_{r} (ℬ)}} (A) = {x \in G ∣ T (x) ℬ_{r} \cap A \neq \emptyset, for ℬ_{r} \subseteq ℬ},

which are called the near-generalized lower and upper approximations of A under T with respect to the normal subgroups ℬ of G′, respectively. Clearly, $\underline{T_{N_{r} (ℬ)}} (A) = \cup_{ℬ_{r} \subseteq ℬ} \underline{T_{ℬ_{r}}} (A)$ and $\bar{T_{N_{r} (ℬ)}} (A) = \cup_{ℬ_{r} \subseteq ℬ} \bar{T_{ℬ_{r}}} (A)$ .

Example 7

T (({[x]}_{36}, ρ^{i})) = {({[x]}_{36}, ρ^{i}), ({[x + 1]}_{36}, ρ^{i} μ)},

\begin{array}{l} A = {({[a]}_{36}, b) ∣ {[a]}_{36} \in Z_{36} - {{[1]}_{36}, {[2]}_{36}, {[3]}_{36}, {[4]}_{36}, \\ {[13]}_{36}, {[14]}_{36}, {[15]}_{36}, {[16]}_{36}, {[19]}_{36}, {[20]}_{36}, {[21]}_{36}, \\ {[25]}_{36}, {[26]}_{36}, {[27]}_{36}, {[28]}_{36}} and \\ b \in {ρ^{0}, ρ^{2}, ρ^{3}, μ, ρ^{2} μ, ρ^{3} μ}} . \end{array}

We determine $\underline{T_{N_{r} (ℬ)}} (A)$ and $\bar{T_{N_{r} (ℬ)}} (A)$ .

• The first set of components of A is ℤ₃₆ – {[1]₃₆, [2]₃₆, [3]₃₆, [4]₃₆, [13]₃₆, [14]₃₆, [15]₃₆, [16]₃₆, [19]₃₆, [20]₃₆, [21]₃₆, [25]₃₆, [26]₃₆, [27]₃₆, [28]₃₆}. There are no cosets of ⟨[4]₃₆⟩ contained in the first set of components of A. Cosets of ⟨[6]₃₆⟩ contained in the first set of components of A are [0]₁₈ + ⟨[6]₃₆⟩, [5]₁₈ + ⟨[6]₃₆⟩. There are no cosets of ⟨[9]₃₆⟩ contained in the first set of components of A. Cosets of ⟨[12]₃₆⟩ contained in the first set of components of A are [0]₃₆+⟨[12]₃₆⟩, [5]₃₆+⟨[12]₃₆⟩, [6]₃₆+ ⟨[12]₃₆⟩, [10]₃₆ + ⟨[12]₃₆⟩, [11]₃₆ + ⟨[12]₃₆⟩. Cosets of ⟨[18]₃₆⟩ contained in the first set of components of A are [0]₃₆ + ⟨[18]₃₆⟩, [5]₃₆ + ⟨[18]₃₆⟩, [6]₃₆ + ⟨[18]₃₆⟩, [11]₃₆ + ⟨[18]₃₆⟩, [12]₃₆ + ⟨[18]₃₆⟩, [17]₃₆ + ⟨[18]₃₆⟩.
• The second set of components of A is {ρ⁰, ρ², ρ³, μ, ρ²μ, ρ³μ}. Cosets of {ρ⁰, ρ²} contained in the second set of components of A are {ρ⁰, ρ²}, {μ, ρ²μ}. Cosets of N contained in the second set of components of A are N. There are no cosets of H contained in the second set of components of A.

(1) Determine $\underline{T_{N_{1} (ℬ)}} (A)$ .

For r = 1, the subsets ℬ₁ ⊆ ℬ are {S₁}, {S₂}, {S₃}.
- • S₁ = ⟨[9]₃₆⟩×H then $∣ (ℤ_{36} \times D_{4}) / S_{1} ∣ = \frac{36 \times 8}{4 \times 4} = 18$ .
  
  $\begin{array}{l} \underline{T_{S_{1}}} (A) = {({[x]}_{36}, ρ^{i}) \in ℤ_{36} \times H ∣ T (({[x]}_{36}, ρ^{i})) S_{1} \subseteq A} \\ = {({[x]}_{36}, ρ^{i}) \in ℤ_{36} \times H ∣ {({[x]}_{36}, ρ^{i}), ({[x + 1]}_{36}, ρ^{i} μ)} (⟨ {[9]}_{36} ⟩ \times H) \subseteq A} \\ = \emptyset . \end{array}$
- • S₂ = ⟨[6]₃₆⟩×N then $∣ (ℤ_{36} \times D_{4}) / S_{2} ∣ = \frac{36 \times 8}{6 \times 4} = 12$ .
  
  $\begin{array}{l} \underline{T_{S_{2}}} (A) = {({[x]}_{36}, ρ^{i}) \in ℤ_{36} \times H ∣ T (({[x]}_{36}, ρ^{i})) S_{2} \subseteq A} \\ = {({[x]}_{36}, ρ^{i}) \in ℤ_{36} \times H ∣ {({[x]}_{36}, ρ^{i}), ({[x + 1]}_{36}, ρ^{i} μ)} (⟨ {[6]}_{36} ⟩ \times N) \subseteq A} \\ = {({[x]}_{36}, ρ^{i}) \in ℤ_{36} \times H ∣ {[x]}_{36} \in {{[5]}_{36}, {[11]}_{36}, {[17]}_{36}, {[23]}_{36}, {[29]}_{36}, {[35]}_{36}} and ρ^{i} \in {ρ^{0}, ρ^{2}}} . \end{array}$
- • S₃ = ⟨[4]₃₆⟩×H then $∣ (ℤ_{36} \times D_{4}) / S_{3} ∣ = \frac{36 \times 8}{9 \times 4} = 8$ .
  
  $\begin{array}{l} \underline{T_{S_{3}}} (A) = {({[x]}_{36}, ρ^{i}) \in ℤ_{36} \times H ∣ T (({[x]}_{36}, ρ^{i})) S_{3} \subseteq A} \\ = {({[x]}_{36}, ρ^{i}) \in ℤ_{36} \times H ∣ {({[x]}_{36}, ρ^{i}), ({[x + 1]}_{36}, ρ^{i} μ)} (⟨ {[4]}_{36} ⟩ \times H) \subseteq A} \\ = \emptyset . \end{array}$
From the above calculations, we obtain

$\begin{array}{l} \underline{T_{N_{1} (ℬ)}} (A) = {({[x]}_{36}, ρ^{i}) \in ℤ_{36} \times H ∣ T (({[x]}_{36}, ρ^{i})) ℬ_{1} \subseteq A, for ℬ_{1} \subseteq ℬ} \\ = \underline{T_{S_{1}}} (A) \cup \underline{T_{S_{2}}} (A) \cup \underline{T_{S_{3}}} (A) \\ = {({[x]}_{36}, ρ^{i}) \in ℤ_{36} \times H ∣ {[x]}_{36} \in {{[5]}_{36}, {[11]}_{36}, {[17]}_{36}, {[23]}_{36}, {[29]}_{36}, {[35]}_{36}} and ρ^{i} \in {ρ^{0}, ρ^{2}}} . \end{array}$
(2) Determine $\underline{T_{N_{2} (ℬ)}} (A)$ .

For r = 2, the subsets ℬ₂ ⊆ℬ are {S₁, S₂}, {S₁, S₃}, {S₂, S₃}. By the above explanation, we have T(x)ℬ₂ = T(x)(∩_S∈ℬ₂S.)
- • S₁∩S₂ = ⟨[18]₃₆⟩×{ρ⁰, ρ²} then $∣ (ℤ_{36} \times D_{4}) / (S_{1} \cap S_{2}) ∣ = \frac{36 \times 8}{2 \times 2} = 72$ .
  
  $\begin{array}{l} \underline{T_{S_{1} \cap S_{2}}} (A) = {({[x]}_{36}, ρ^{i}) \in ℤ_{36} \times H ∣ T (({[x]}_{36}, ρ^{i})) (S_{1} \cap S_{2}) \subseteq A} \\ = {({[x]}_{36}, ρ^{i}) \in ℤ_{36} \times H ∣ {({[x]}_{36}, ρ^{i}), ({[x + 1]}_{36}, ρ^{i} μ)} (⟨ {[18]}_{36} ⟩ \times {ρ^{0}, ρ^{2}}) \subseteq A} \\ = {({[x]}_{36}, ρ^{i}) \in ℤ_{36} \times H ∣ {[x]}_{36} \in {{[5]}_{36}, {[11]}_{36}, {[17]}_{36}, {[23]}_{36}, {[29]}_{36}, {[35]}_{36}} and ρ^{i} \in {ρ^{0}, ρ^{2}}} . \end{array}$
- • S₁ ∩ S₃ = {[0]₃₆} × H then $∣ (ℤ_{36} \times D_{4}) / (S_{1} \cap S_{3}) ∣ = \frac{36 \times 8}{1 \times 4} = 72$ .
  
  $\begin{array}{l} \underline{T_{S_{1} \cap S_{3}}} (A) = {({[x]}_{36}, ρ^{i}) \in ℤ_{36} \times H ∣ T (({[x]}_{36}, ρ^{i})) (S_{1} \cap S_{3}) \subseteq A} \\ = {({[x]}_{36}, ρ^{i}) \in ℤ_{36} \times H ∣ {({[x]}_{36}, ρ^{i}), ({[x + 1]}_{36}, ρ^{i} μ)} ({{[0]}_{36}} \times H) \subseteq A} \\ = \emptyset . \end{array}$
- • S₂∩S₃ = ⟨[12]₃₆⟩×{ρ^o, ρ²} then $∣ (ℤ_{36} \times D_{4}) / (S_{2} \cap S_{3}) ∣ = \frac{36 \times 8}{3 \times 2} = 48$ .
  
  $\begin{array}{l} \underline{T_{S_{2} \cap S_{3}}} (A) = {({[x]}_{36}, ρ^{i}) \in ℤ_{36} \times H ∣ T (({[x]}_{36}, ρ^{i})) (S_{2} \cap S_{3}) \subseteq A} \\ = {({[x]}_{36}, ρ^{i}) \in ℤ_{36} \times H ∣ {({[x]}_{36}, ρ^{i}), ({[x + 1]}_{36}, ρ^{i} μ)} (⟨ {[12]}_{36} ⟩ \times {ρ^{0}, ρ^{2}}) \subseteq A} \\ = {({[x]}_{36}, ρ^{i}) \in ℤ_{36} \times H ∣ {[x]}_{36} \in {{[5]}_{36}, {[10]}_{36}, {[11]}_{36}, {[17]}_{36}, {[22]}_{36}, {[23]}_{36}, {[29]}_{36}, {[34]}_{36}, {[35]}_{36}} and ρ^{i} \in {ρ^{0}, ρ^{2}}} . \end{array}$
From the above calculations, we obtain

$\begin{array}{l} \underline{T_{N_{2} (ℬ)}} (A) = {({[x]}_{36}, ρ^{i}) \in ℤ_{36} \times H ∣ T (({[x]}_{36}, ρ^{i})) ℬ_{2} \subseteq A, for ℬ_{2} \subseteq ℬ} \\ = \underline{T_{S_{1} \cap S_{2}}} (A) \cup \underline{T_{S_{1} \cap S_{3}}} (A) \cup \underline{T_{S_{2} \cap S_{3}}} (A) \\ = {({[x]}_{36}, ρ^{i}) \in ℤ_{36} \times H ∣ {[x]}_{36} \in {{[5]}_{36}, {[10]}_{36}, {[11]}_{36}, {[17]}_{36}, {[22]}_{36}, {[23]}_{36}, {[29]}_{36}, {[34]}_{36}, {[35]}_{36}} and ρ^{i} \in {ρ^{0}, ρ^{2}}} . \end{array}$
(3) Determine $\underline{T_{N_{3} (ℬ)}} (A)$ .

For r = 3, the subsets ℬ₃ ⊆ ℬ then {S₁, S₂, S₃}. By the above explanation, we have T(x)ℬ₃ = T(x) (∩_S∈ℬ₃S).
- • S₁ ∩ S₂ ∩ S₃ = {[0]₃₆} × {ρ^o, ρ²} then $∣ (ℤ_{36} \times D_{4}) / (S_{1} \cap S_{2} \cap S_{3}) ∣ = \frac{36 \times 8}{1 \times 2} = 144$ .
  
  $\begin{array}{l} \underline{T_{N_{3} (ℬ)}} (A) = \underline{T_{S_{1} \cap S_{2} \cap S_{3}}} (A) \\ = {({[x]}_{36}, ρ^{i}) \in ℤ_{36} \times H ∣ T (({[x]}_{36}, ρ^{i})) (S_{1} \cap S_{2} \cap S_{3}) \subseteq A} \\ = {({[x]}_{36}, ρ^{i}) \in ℤ_{36} \times H ∣ {({[x]}_{36}, ρ^{i}), ({[x + 1]}_{36}, ρ^{i} μ)} ({{[0]}_{36} \times {ρ^{0}, ρ^{2}}) \subseteq A} \\ = {({[x]}_{36}, ρ^{i}) \in ℤ_{36} \times H ∣ {[x]}_{36} \in {{[5]}_{36}, {[6]}_{36}, {[7]}_{36}, {[8]}_{36}, {[9]}_{36}, {[10]}_{36}, {[11]}_{36}, {[17]}_{36}, {[22]}_{36}, {[23]}_{36}, {[29]}_{36}, {[30]}_{36}, {[31]}_{36}, {[32]}_{36}, {[33]}_{36}, {[34]}_{36}, {[35]}_{36}} and ρ^{i} \in {ρ^{0}, ρ^{2}}} . \end{array}$
As a summary, we have $\underline{T_{N_{1} (ℬ)}} (A) \subset \underline{T_{N_{2} (ℬ)}} (A) \subset \underline{T_{N_{3} (ℬ)}} (A)$ .
- • The first set of components of A is ℤ₃₆ – {[1]₃₆, [2]₃₆, [3]₃₆, [4]₃₆, [13]₃₆, [14]₃₆, [15]₃₆, [16]₃₆, [19]₃₆, [20]₃₆, [21]₃₆, [25]₃₆, [26]₃₆, [27]₃₆, [28]₃₆}.
  
  Every coset of ⟨[4]₃₆⟩ intersects with the first set of components of A.
  
  Every coset of ⟨[6]₃₆⟩ intersects with the first set of components of A.
  
  Every coset of ⟨[9]₃₆⟩ intersects with the first set of components of A.
  
  Every coset of ⟨[12]₃₆⟩ intersects with the first set of components of A except [1]₃₆ +⟨[12]₃₆⟩, [2]₃₆ + ⟨[12]₃₆⟩, [3]₃₆ + ⟨[12]₃₆⟩, [4]₃₆ + ⟨[12]₃₆⟩.
  
  Every coset of ⟨[18]₃₆⟩ always intersects with the first set of components of A except [1]₃₆ +⟨[18]₃₆⟩, [2]₃₆ + ⟨[18]₃₆⟩, [3]₃₆ + ⟨[18]₃₆⟩.
- • The second set of components of A is {ρ⁰, ρ², ρ³, μ, ρ²μ, ρ³μ}.
  
  Every coset of {ρ⁰, ρ²} intersects with the second set of components of A.
  
  Every coset of N intersects with the second set of components of A.
  
  Every coset of H intersects with the second set of components of A.
(4) Determine $\bar{T_{N_{1} (ℬ)} (A)}$ .

For r = 1, the subsets ℬ₁ ⊆ ℬ are {S₁}, {S₂}, {S₃}.
- • S₁ = ⟨[9]₃₆⟩×H then $∣ (ℤ_{36} \times D_{4}) / S_{1} ∣ = \frac{36 \times 8}{4 \times 4} = 18$ .
  
  $\begin{array}{l} \bar{T_{S_{1}}} (A) = {({[x]}_{36}, ρ^{i}) \in ℤ_{36} \times H ∣ T (({[x]}_{36}, ρ^{i})) S_{1} \cap A \neq \emptyset} \\ = {({[x]}_{36}, ρ^{i}) \in ℤ_{36} \times H ∣ ({({[x]}_{36}, ρ^{i}), ({[x + 1]}_{36}, ρ^{i} μ)} ({⟨ [9]}_{36} ⟩ \times H)) \cap A \neq \emptyset} \\ = ℤ_{36} \times H . \end{array}$
From the above calculations, we obtain

$\begin{array}{l} \bar{T_{N_{1} (ℬ)} (A)} = {({[x]}_{36}, ρ^{i}) \in ℤ_{36} \times H ∣ T (({[x]}_{36}, ρ^{i}) ℬ_{1} \cap A \neq \emptyset, for ℬ_{1} \subseteq ℬ} \\ = \bar{T_{S_{1}} (A)} \cup \bar{T_{S_{2}} (A)} \cup \bar{T_{S_{3}} (A)} \\ = ℤ_{36} \times H . \end{array}$
(5) Determine $\bar{T_{N_{2} (ℬ)} (A)}$ .

For r = 2, the subsets ℬ₂ ⊆ ℬ are {S₁, S₂}, {S₁, S₃}, {S₂, S₃}. By the above explanation, we have T(x)ℬ₂ = T(x) (∩_S∈ℬ₂S).
- • S₁ ∩ S₂ = ⟨[18]₃₆⟩ × {ρ^o, ρ²} then $∣ (ℤ_{36} \times D_{4}) / (S_{1} \cap S_{2}) ∣ = \frac{36 \times 8}{2 \times 2} = 72$ .
  
  $\begin{array}{l} \bar{T_{S_{1} \cap S_{2}}} (A) = {({[x]}_{36}, ρ^{i}) \in ℤ_{36} \times H ∣ T (({[x]}_{36}, ρ^{i})) (S_{1} \cap S_{2}) \cap A \neq \emptyset} \\ = {({[x]}_{36}, ρ^{i}) \in ℤ_{36} \times H ∣ ({({[x]}_{36}, ρ^{i}), ({[x + 1]}_{36}, ρ^{i} μ)} (⟨ {[18]}_{36} ⟩ \times {ρ^{0}, ρ^{2}})) \cap A \neq \emptyset} \\ = {({[x]}_{36}, ρ^{i}) \in ℤ_{36} \times H ∣ {[x]}_{36} \in ℤ_{36} - {{[1]}_{36}, {[2]}_{36}, {[19]}_{36}, {[20]}_{36},} and ρ^{i} \in H} . \end{array}$
- • S₁ ∩ S₃ = {[0]₃₆} × H then $∣ (ℤ_{36} \times D_{4}) / (S_{1} \cap S_{3}) ∣ = \frac{36 \times 8}{1 \times 4} = 72$ .
  
  $\begin{array}{l} \bar{T_{S_{1} \cap S_{3}}} (A) = {({[x]}_{36}, ρ^{i}) \in ℤ_{36} \times H ∣ T (({[x]}_{36}, ρ^{i})) (S_{1} \cap S_{3}) \cap A \neq \emptyset} \\ = {({[x]}_{36}, ρ^{i}) \in ℤ_{36} \times H ∣ ({({[x]}_{36}, ρ^{i}), ({[x + 1]}_{36}, ρ^{i} μ)} ({{[0]}_{36}} \times H)) \cap A \neq \emptyset} \\ = {({[x]}_{36}, ρ^{i}) \in ℤ_{36} \times H ∣ {[x]}_{36} \in ℤ_{36} - {{[1]}_{36}, {[2]}_{36}, {[3]}_{36}, {[13]}_{36}, {[14]}_{36}, {[15]}_{36}, {[19]}_{36}, {[20]}_{36}, {[25]}_{36}, {[26]}_{36}, {[27]}_{36}} and ρ^{i} \in H} . \end{array}$
- • S₂∩S₃ = ⟨[12]₃₆⟩×{ρ^o, ρ²} then $∣ (ℤ_{36} \times D_{4}) / (S_{2} \cap S_{3}) ∣ = \frac{36 \times 8}{3 \times 2} = 48$ .
  
  $\begin{array}{l} \bar{T_{S_{2} \cap S_{3}}} (A) = {({[x]}_{36}, ρ^{i}) \in ℤ_{36} \times H ∣ T (({[x]}_{36}, ρ^{i})) (S_{1} \cap S_{3}) \cap A \neq \emptyset} \\ = {({[x]}_{36}, ρ^{i}) \in ℤ_{36} \times H ∣ ({({[x]}_{36}, ρ^{i}), ({[x + 1]}_{36}, ρ^{i} μ)} (⟨ {[12]}_{36} ⟩ \times {ρ^{o}, ρ^{2}})) \cap A \neq \emptyset} \\ = {({[x]}_{36}, ρ^{i}) \in ℤ_{36} \times H ∣ {[x]}_{36} \in ℤ_{36} - {{[1]}_{36}, {[2]}_{36}, {[3]}_{36}, {[13]}_{36}, {[14]}_{36}, {[15]}_{36}, {[25]}_{36}, {[26]}_{36}, {[27]}_{36}} and ρ^{i} \in H} . \end{array}$
From the above calculations, we obtain

$\begin{array}{l} \bar{T_{N_{2} (ℬ)} (A)} = {({[x]}_{36}, ρ^{i}) \in ℤ_{36} \times H ∣ T (({[x]}_{36}, ρ^{i})) ℬ_{2} \subseteq A, for ℬ_{2} \subseteq ℬ} \\ = \bar{T_{S_{1} \cap S_{2}} (A)} \cup \bar{T_{S_{1} \cap S_{3}} (A)} \cup \bar{T_{S_{2} \cap S_{3}} (A)} \\ = {({[x]}_{36}, ρ^{i}) \in ℤ_{36} \times H ∣ {[x]}_{36} \in ℤ_{36} - {{[1]}_{36}, {[2]}_{36}} and ρ^{i} \in H} . \end{array}$
(6) Determine $\bar{T_{N_{3} (ℬ)} (A)}$ .

For r = 3, the subsets ℬ₃ ⊆ ℬ are {S₁, S₂, S₃}. By the above explanation, we have T(x)ℬ₃ = T(x)∩(∩_S∈ℬ₂S).
- • S₁ ∩ S₂ ∩ S₃ = {[0]₃₆} × {ρ⁰, ρ²} then $∣ (ℤ_{36} \times D_{4}) / (S_{1} \cap S_{2} \cap S_{3}) ∣ = \frac{36 \times 8}{1 \times 2} = 144$ .
  
  $\begin{array}{l} \bar{T_{N_{3} (ℬ)} (A)} = \bar{T_{S_{1} \cap S_{2} \cap S_{3}}} (A) \\ = {({[x]}_{36}, ρ^{i}) \in ℤ_{36} \times H ∣ (({[x]}_{36}, ρ^{i})) (S_{1} \cap S_{2} \cap S_{3}) \cap A \neq \emptyset} \\ = {({[x]}_{36}, ρ^{i}) \in ℤ_{36} \times H ∣ ({({[x]}_{36}, ρ^{i}), ({[x + 1]}_{36}, ρ^{i} μ)} ({{[0]}_{36}} \times {ρ^{0}, ρ^{2}})) \cap A \neq \emptyset} \\ = {({[x]}_{36}, ρ^{i}) \in ℤ_{36} \times H ∣ {[x]}_{36} \in ℤ_{36} - {{[1]}_{36}, {[2]}_{36}, {[3]}_{36}, {[13]}_{36}, {[14]}_{36}, {[15]}_{36}, {[19]}_{36}, {[20]}_{36}, {[25]}_{36}, {[26]}_{36}, {[27]}_{36}} and ρ^{i} \in H} . \end{array}$

In summary, we have $\bar{T_{N_{3} (ℬ)} (A)} \subset \bar{T_{N_{2} (ℬ)} (A)} \subset \bar{T_{N_{1} (ℬ)} (A)}$ . The relationship between $\underline{T_{N_{r} (ℬ)}} (A)$ and $\bar{T_{N_{r} (ℬ)}} (A)$ is $\underline{T_{N_{1} (ℬ)}} (A) \subset \underline{T_{N_{2} (ℬ)}} (A) \subset \underline{T_{N_{3} (ℬ)}} (A) \subset \bar{T_{N_{3} (ℬ)}} (A) \subset \bar{T_{N_{2} (ℬ)}} (A) \subset \bar{T_{N_{1} (ℬ)} (A)}$ .

Proposition 1

Let G be a group, (G, ) be a nearness approximation space and I : G→℘^*(G) be a set-valued mapping, where I(x) = {x} for each x ∈ G. Let A be a non-empty subset of G. Then,

\underline{I_{N_{r} (ℬ)}} (A) = \underline{N_{r} (ℬ) (A)} and \bar{I_{N_{r} (ℬ)}} (A) = \bar{N_{r} (ℬ) (A)} .

Proof

This is clear from Definitions 2 and 6.

Proposition 2

Let G and G′ be two groups, (G′, ) be a nearness approximation space, and T : G→℘^*(G′) be a set-valued mapping. Let A be a non-empty subset of G′. Then,

(1) $\underline{T_{N_{r} (ℬ)}} (A) \subseteq \bar{T_{N_{r} (ℬ)}} (A)$ ,
(2) $\underline{T_{N_{r} (ℬ)}} (G^{'}) = G = \bar{T_{N_{r} (ℬ)}} (G^{'})$ ,
(3) If A ⊆ B, then $\underline{T_{N_{r} (ℬ)}} (A) \subseteq \underline{T_{N_{r} (ℬ)}} (B)$ and $\bar{T_{N_{r} (ℬ)}} (A) \subseteq \bar{T_{N_{r} (ℬ)}} (B)$ ,
(4) $\underline{T_{N_{r} (ℬ)}} (A) \supseteq {(\bar{T_{N_{r} (ℬ)}} (A^{C}))}^{C}$ ,
(5) $\bar{T_{N_{r} (ℬ)}} (A) \supseteq {(\underline{T_{N_{r} (ℬ)}} (A^{C}))}^{C}$ ,
(6) $\underline{T_{N_{r} (ℬ)}} (A \cap B) \subseteq \underline{T_{N_{r} (ℬ)}} (A) \cap \underline{T_{N_{r} (ℬ)}} (B),$
(7) $\bar{T_{N_{r} (ℬ)}} (A \cap B) \subseteq \bar{T_{N_{r} (ℬ)}} (A) \cap \bar{T_{N_{r} (ℬ)}} (B)$ ,
(8) $\underline{T_{N_{r} (ℬ)}} (A \cup B) \supseteq \underline{T_{N_{r} (ℬ)}} (A) \cup \underline{T_{N_{r} (ℬ)}} (B),$
(9) $\bar{T_{N_{r} (ℬ)}} (A \cup B) = \bar{T_{N_{r} (ℬ)}} (A) \cup \bar{T_{N_{r} (ℬ)}} (B)$ .

Proof

(1) It is clear from Definition 6.
(2) It is clear from Definition 6.
(3) Let x ∈ $\underline{T_{N_{r} (ℬ)}} (A)$ . Subsequently, T(x)ℬ_r ⊆ A ⊆ B, for some ℬ_r ⊆ ℬ so that x ∈ $\underline{T_{N_{r} (ℬ)}} (B)$ . Therefore, $\underline{T_{N_{r} (ℬ)}} (A) \subseteq \underline{T_{N_{r} (ℬ)}} (B)$ . Suppose $y \in \bar{T_{N_{r} (ℬ)}} (A)$ . Then, $\bar{T (y) ℬ_{r}} \cap A \neq \emptyset$ for some ℬ_r ⊆ ℬ. Since A ⊆ B then T(y)ℬ_r ∩ B ≠ ∅ for some ℬ_r ⊆ ℬ such that $y \in \bar{T_{N_{r} (ℬ)}} (B)$ . Therefore, $\bar{T_{N_{r} (ℬ)}} (A) \subseteq \bar{T_{N_{r} (ℬ)}} (B)$ .
(4) Let $x \in {(\bar{T_{N_{r} (ℬ)}} (A^{C}))}^{C}$ . Then, $x \notin (\bar{T_{N_{r} (ℬ)}} (A^{C})$ so that T(x) ℬ_r ∩ A^C = ∅ for every ℬ_r ⊆ ℬ, which implies that T(x) ℬ_r ⊆ A for every ℬ_r ⊆ ℬ. Therefore, x ∈ $\underline{T_{N_{r} (ℬ)}} (A)$ so that $\underline{T_{N_{r} (ℬ)}} (A) \supseteq {(\bar{T_{N_{r} (ℬ)}} (A^{C}))}^{C}$ . The converse is not true in general. Let x ∈ $\underline{T_{N_{r} (ℬ)}} (A)$ . Then, T(x) ℬ_r ⊆ A for some ℬ_r ⊆ ℬ but T(x)ℬ_r′ ⊈ A for some ℬ_r′ ⊆ ℬ so that T(x)ℬ_r′ ∩ A^C ≠ ∅, which implies $x \in \bar{T_{N_{r} (ℬ)}} (A^{C})$ . Thus, we have $x \notin {(\bar{T_{N_{r} (ℬ)}} (A^{C}))}^{C}$ so that $\underline{T_{N_{r} (ℬ)}} (A) ⊄ {(\bar{T_{N_{r} (ℬ)}} (A^{C}))}^{C}$ .
(5) Let $x \in {(\underline{T_{N_{r} (ℬ)}} (A^{C}))}^{C}$ . Then $x \notin \underline{T_{N_{r} (ℬ)}} (A^{C})$ so that T(x)ℬ_r ⊈ A^C for every ℬ_r ⊆ ℬ which implies T(x)ℬ_r ∩ A ≠ ∅ for every ℬ_r ⊆ ℬ. Therefore, $x \in \bar{T_{N_{r} (ℬ)}} (A)$ so that $\bar{T_{N_{r} (ℬ)}} (A) \supseteq {(\underline{T_{N_{r} (ℬ)}} (A^{C}))}^{C}$ . The converse is not true in general. Let $x \in \bar{T_{N_{r} (ℬ)}} (A)$ Then, T(x)ℬ_r ∩ A ≠ ∅ for some ℬ_r ⊆ ℬ but T(x)ℬ_r′ ∩ A = ∅ for some ℬ_r′ ⊆ ℬ such that T(x)ℬ_r′ ⊆ A^C, which implies $x \in \underline{T_{N_{r} (ℬ)}} (A^{C})$ . Therefore, $x \notin {(\underline{T_{N_{r} (ℬ)}} (A^{C}))}^{C}$ so that $\bar{T_{N_{r} (ℬ)}} (A) ⊄ {(\underline{T_{N_{r} (ℬ)}} (A^{C}))}^{C}$ .
(6) From part (3), we have $\underline{T_{N_{r} (ℬ)}} (A \cap B) \subseteq \underline{T_{N_{r} (ℬ)}} (A)$ and $\underline{T_{N_{r} (ℬ)}} (A \cap B) \subseteq \underline{T_{N_{r} (ℬ)}} (B)$ . Then, $\underline{T_{N_{r} (ℬ)}} (A \cap B) \subseteq \underline{T_{N_{r} (ℬ)}} (A) \cap \underline{T_{N_{r} (ℬ)}} (B)$ .
(7) From part (3), we have $\bar{T_{N_{r} (ℬ)}} (A \cap B) \subseteq \bar{T_{N_{r} (ℬ)}} (A)$ and $\bar{T_{N_{r} (ℬ)}} (A \cap B) \subseteq \bar{T_{N_{r} (ℬ)}} (B)$ . Then, $\bar{T_{N_{r} (ℬ)}} (A \cap B) \subseteq \bar{T_{N_{r} (ℬ)}} (A) \cap \bar{T_{N_{r} (ℬ)}} (B)$ .
(8) From part (3), we have $\underline{T_{N_{r} (ℬ)}} (A) \subseteq \underline{T_{N_{r} (ℬ)}} (A \cup B)$ and $\underline{T_{N_{r} (ℬ)}} (B) \subseteq \underline{T_{N_{r} (ℬ)}} (A \cup B)$ . Then, $\underline{T_{N_{r} (ℬ)}} (A) \cup \underline{T_{N_{r} (ℬ)}} (B) \subseteq \underline{T_{N_{r} (ℬ)}} (A \cup B)$ .
(9) Let $x \in \bar{T_{N_{r} (ℬ)}} (A \cup B)$ . Then, T(x)ℬ_r ∩ (A ∪ B) ≠ ∅ for some ℬ_r ⊆ ℬ so that T(x)ℬ_r ∩ A ≠ ∅ or T(x)ℬ_r ∩ A ≠ ∅ which implies $x \in \bar{T_{N_{r} (ℬ)}} (A)$ or $x \in \bar{T_{N_{r} (ℬ)}} (B)$ . Therefore, $\bar{T_{N_{r} (ℬ)}} (A \cup B) \subseteq \bar{T_{N_{r} (ℬ)}} (A) \cup \bar{T_{N_{r} (ℬ)}} (B)$ . Conversely, from part (3), we have $\bar{T_{N_{r} (ℬ)}} (A) \subseteq \bar{T_{N_{r} (ℬ)}} (A \cup B)$ and $\bar{T_{N_{r} (ℬ)}} (B) \subseteq \bar{T_{N_{r} (ℬ)}} (A \cup B)$ . Then, $\bar{T_{N_{r} (ℬ)}} (A) \cup \bar{T_{N_{r} (ℬ)}} (B) \subseteq \bar{T_{N_{r} (ℬ)}} (A \cup B)$ .

The following examples show that the converse of Proposition 2 (6,7,8) is generally not true.

Example 8

\begin{array}{l} A = ℤ_{18} - ({[1]}_{18} + ⟨ {[9]}_{18} ⟩ \cup {[3]}_{18} + ⟨ {[9]}_{18} ⟩ \cup {[2]}_{18} + ⟨ {[6]}_{18} ⟩) \\ = ℤ_{18} - {{[1]}_{18}, {[2]}_{18}, {[3]}_{18}, {[8]}_{18}, {[10]}_{18}, {[12]}_{18}, {[14]}_{18}}, \end{array}

and

\begin{array}{l} B = ℤ_{18} - ({[1]}_{18} + ⟨ {[9]}_{18} ⟩ \cup {[3]}_{18} + ⟨ {[9]}_{18} ⟩ \cup {[3]}_{18} + ⟨ {[6]}_{18} ⟩) \\ = ℤ_{18} - {{[1]}_{18}, {[3]}_{18}, {[4]}_{18}, {[9]}_{18}, {[10]}_{18}, {[13]}_{18}, {[15]}_{18}} . \end{array}

(1) Determine $\underline{T_{N_{1} (ℬ)}} (A)$ and $\underline{T_{N_{1} (ℬ)}} (B)$ .
- • Cosets of S₁ = ⟨[9]₁₈⟩ contained in A are ⟨[9]₁₈⟩, [4]₁₈ + ⟨[9]₁₈⟩, [6]₁₈ + ⟨[9]₁₈⟩, [7]₁₈ + ⟨[9]₁₈⟩.
  
  $\begin{array}{l} \underline{T_{S_{1}}} (A) = {{[x]}_{18} \in ℤ_{18} ∣ {[x + 1]}_{18} + ⟨ {[9]}_{18} ⟩ \subseteq A} \\ = {{[3]}_{18}, {[5]}_{18}, {[6]}_{18}, {[8]}_{18}, {[12]}_{18}, {[14]}_{18}, {[15]}_{18}, {[17]}_{18}} . \end{array}$
- • Cosets of S₂ = ⟨[6]₁₈⟩ contained in A are [5]₁₈ + ⟨[6]₁₈⟩.
  
  $\begin{array}{l} \underline{T_{S_{2}}} (A) = {{[x]}_{18} \in ℤ_{18} ∣ {[x + 1]}_{18} + ⟨ {[6]}_{18} ⟩ \subseteq A} \\ = {{[4]}_{18}, {[10]}_{18}, {[16]}_{18}} . \end{array}$
  
  From the above calculations, we have
  
  $\begin{array}{l} \underline{T_{N_{1} (ℬ)}} (A) = \underline{T_{S_{1}}} (A) \cup \underline{T_{S_{2}}} (A) \\ = {{[3]}_{18}, {[4]}_{18}, {[5]}_{18}, {[6]}_{18}, {[8]}_{18}, {[10]}_{18}, {[12]}_{18}, {[14]}_{18}, {[15]}_{18}, {[16]}_{18}, {[17]}_{18}} . \end{array}$
- • Cosets of S₁ = ⟨[9]₁₈⟩ contained in B are [2]₁₈ + ⟨[9]₁₈⟩, [5]₁₈+⟨[9]₁₈⟩, [7]₁₈+⟨[9]₁₈⟩, [8]₁₈+⟨[9]₁₈⟩.
  
  $\begin{array}{l} \underline{T_{S_{1}}} (B) = {{[x]}_{18} \in ℤ_{18} ∣ {[x + 1]}_{18} + ⟨ {[9]}_{18} ⟩ \subseteq B} \\ = {{[1]}_{18}, {[4]}_{18}, {[6]}_{18}, {[7]}_{18}, {[10]}_{18}, {[13]}_{18}, {[15]}_{18}, {[16]}_{18}} . \end{array}$
- • Cosets of S₂ = ⟨[6]₁₈⟩ contained in B are [0]₁₈ + ⟨[6]₁₈⟩, [2]₁₈ + ⟨[6]₁₈⟩, [5]₁₈ + ⟨[6]₁₈⟩.
  
  $\begin{array}{l} \underline{T_{S_{2}}} (B) = {{[x]}_{18} \in ℤ_{18} ∣ {[x + 1]}_{18} + ⟨ {[6]}_{18} ⟩ \subseteq B} \\ = {{[1]}_{18}, {[4]}_{18}, {[5]}_{18}, {[7]}_{18}, {[10]}_{18}, {[11]}_{18}, {[13]}_{18}, {[16]}_{18}, {[17]}_{18}} . \end{array}$
  
  From the above calculations, we have
  
  $\begin{array}{l} \underline{T_{N_{1} (ℬ)}} (B) = \underline{T_{S_{1}}} (B) \cup \underline{T_{S_{2}}} (B) \\ = {{[1]}_{18}, {[4]}_{18}, {[5]}_{18}, {[6]}_{18}, {[7]}_{18}, {[10]}_{18}, {[11]}_{18}, {[13]}_{18}, {[15]}_{18}, {[16]}_{18}, {[17]}_{18}} . \end{array}$
  
  As a summary, we have $\underline{T_{N_{1} (ℬ)}} (A) \cap \underline{T_{N_{1} (ℬ)}} (B) = {{[4]}_{18}, {[5]}_{18}, {[6]}_{18}, {[10]}_{18}, {[15]}_{18}, {[16]}_{18}, {[17]}_{18}}$ .
(2) Determine $\underline{T_{N_{1} (ℬ)}} (A \cap B)$ .

A∩B = {[0]₁₈, [5]₁₈, [6]₁₈, [7]₁₈, [11]₁₈, [16]₁₈, [17]₁₈}.
- • Cosets of S₁ = ⟨[9]₁₈⟩ contained in A ∩ B are [7]₁₈ + ⟨[9]₁₈⟩.
  
  $\begin{array}{l} \underline{T_{S_{1}}} (A \cap B) = {{[x]}_{18} \in ℤ_{18} ∣ {[x + 1]}_{18} + ⟨ {[9]}_{18} ⟩ \subseteq A \cap B} \\ = {{[6]}_{18}, {[15]}_{18}} . \end{array}$
- • Cosets of S₂ = ⟨[6]₁₈⟩ contained in A ∩ B are [5]₁₈ + ⟨[6]₁₈⟩.
  
  $\begin{array}{l} \underline{T_{S_{2}}} (A \cap B) = {{[x]}_{18} \in ℤ_{18} ∣ {[x + 1]}_{18} + ⟨ {[6]}_{18} ⟩ \subseteq A \cap B} \\ = {{[4]}_{18}, {[10]}_{18}, {[16]}_{18}} . \end{array}$
  
  From the above calculations, we have
  
  $\begin{array}{l} \underline{T_{N_{1} (ℬ)}} (A \cap B) = \underline{T_{S_{1}}} (A \cap B) \cup \underline{T_{S_{2}}} (A \cap B) \\ = {{[4]}_{18}, {[6]}_{18}, {[10]}_{18}, {[15]}_{18}, {[16]}_{18}} . \end{array}$
As a summary, we obtain $\underline{T_{N_{1} (ℬ)}} (A) \cap \underline{T_{N_{1} (ℬ)}} (B) ⊈ \underline{T_{N_{1} (ℬ)}} (A \cap B)$ .
(3) Determine $\bar{T_{N_{1} (ℬ)}} (A)$ and $\bar{T_{N_{1} (ℬ)}} (B)$ .
- • Cosets of S₁ = ⟨[9]₁₈⟩ intersected with A are [0]₁₈+⟨[9]₁₈⟩, [2]₁₈+⟨[9]₁₈⟩, [4]₁₈+⟨[9]₁₈⟩, [5]₁₈+ ⟨[9]₁₈⟩, [6]₁₈+⟨[9]₁₈⟩, [7]₁₈+⟨[9]₁₈⟩, [8]₁₈+⟨[9]₁₈⟩.
  
  $\begin{array}{l} \bar{T_{S_{1}}} (A) = {{[x]}_{18} \in ℤ_{18} ∣ ({[x + 1]}_{18} + ⟨ {[9]}_{18} ⟩) \cap A \neq \emptyset} \\ = ℤ_{18} - {{[0]}_{18}, {[2]}_{18}, {[9]}_{18}, {[11]}_{18}} . \end{array}$
- • Cosets of S₂ = ⟨[6]₁₈⟩ intersected with A are [0]₁₈+⟨[6]₁₈⟩, [1]₁₈+⟨[6]₁₈⟩, [3]₁₈+⟨[6]₁₈⟩, [4]₁₈+ ⟨[6]₁₈⟩, [5]₁₈ + ⟨[6]₁₈⟩.
  
  $\begin{array}{l} \bar{T_{S_{2}}} (A) = {{[x]}_{18} \in ℤ_{18} ∣ ({[x + 1]}_{18} + ⟨ {[6]}_{18} ⟩) \cap A \neq \emptyset} \\ = ℤ_{18} - {{[1]}_{18}, {[7]}_{18}, {[13]}_{18}} . \end{array}$
  
  From the above calculations, we obtain
  
  $\bar{T_{N_{1} (ℬ)}} (A) = \bar{T_{S_{1}}} (A) \cup \bar{T_{S_{2}}} (A) = ℤ_{18} .$
- • Cosets of S₁ = ⟨[9]₁₈⟩ intersected with B are [0]₁₈+⟨[9]₁₈⟩, [2]₁₈+⟨[9]₁₈⟩, [3]₁₈+⟨[9]₁₈⟩, [5]₁₈+ ⟨[9]₁₈⟩, [6]₁₈+⟨[9]₁₈⟩, [7]₁₈+⟨[9]₁₈⟩, [8]₁₈+⟨[9]₁₈⟩.
  
  $\begin{array}{l} \bar{T_{S_{1}}} (B) = {{[x]}_{18} \in ℤ_{18} ∣ ({[x + 1]}_{18} + ⟨ {[9]}_{18} ⟩) \cap B \neq \emptyset} \\ = ℤ_{18} - {{[0]}_{18}, {[3]}_{18}, {[9]}_{18}, {[12]}_{18}} . \end{array}$
- • Cosets of S₂ = ⟨[6]₁₈⟩ intersected with B are [0]₁₈+⟨[6]₁₈⟩, [1]₁₈+⟨[6]₁₈⟩, [2]₁₈+⟨[6]₁₈⟩, [4]₁₈+ ⟨[6]₁₈⟩, [5]₁₈ + ⟨[6]₁₈⟩.
  
  $\begin{array}{l} \bar{T_{S_{2}}} (B) = {{[x]}_{18} \in ℤ_{18} ∣ ({[x + 1]}_{18} + ⟨ {[6]}_{18} ⟩) \cap B \neq \emptyset} \\ = ℤ_{18} - {{[2]}_{18}, {[8]}_{18}, {[14]}_{18}} . \end{array}$
  
  From the above calculations, we obtain
  
  $\bar{T_{N_{1} (ℬ)}} (B) = \bar{T_{S_{1}}} (B) \cup \bar{T_{S_{2}}} (B) = ℤ_{18} .$
  
  As a summary, we have $\bar{T_{N_{1} (ℬ)}} (A) \cap \bar{T_{N_{1} (ℬ)}} (B) = ℤ_{18}$ .
(4) Determine $\bar{T_{N_{1} (ℬ)}} (A \cap B)$ .

A∩B = {[0]₁₈, [5]₁₈, [6]₁₈, [7]₁₈, [11]₁₈, [16]₁₈, [17]₁₈}.
- • Cosets of S₁ = ⟨[9]₁₈⟩ intersected with A ∩ B are [0]₁₈ + ⟨[9]₁₈⟩, [2]₁₈ + ⟨[9]₁₈⟩, [5]₁₈ + ⟨[9]₁₈⟩, [6]₁₈ + ⟨[9]₁₈⟩, [7]₁₈ + ⟨[9]₁₈⟩, [8]₁₈ + ⟨[9]₁₈⟩.
  
  $\begin{array}{l} \bar{T_{S_{1}}} (A \cap B) = {{[x]}_{18} \in ℤ_{18} ∣ ({[x + 1]}_{18} + ⟨ {[9]}_{18} ⟩) \cap (A \cap B) \neq \emptyset} \\ = ℤ_{18} - {{[0]}_{18}, {[2]}_{18}, {[3]}_{18}, {[9]}_{18}, {[11]}_{18}, {[12]}_{18}} . \end{array}$
- • Cosets of S₂ = ⟨[6]₁₈⟩ intersected with A ∩ B are [0]₁₈ + ⟨[6]₁₈⟩, [1]₁₈ + ⟨[6]₁₈⟩, [4]₁₈ + ⟨[6]₁₈⟩, [5]₁₈ + ⟨[6]₁₈⟩.
  
  $\begin{array}{l} \bar{T_{S_{2}}} (A \cap B) = {{[x]}_{18} \in ℤ_{18} ∣ ({[x + 1]}_{18} + ⟨ {[6]}_{18} ⟩) \cap (A \cap B) \neq \emptyset} \\ = ℤ_{18} - {{[1]}_{18}, {[2]}_{18}, {[7]}_{18}, {[8]}_{18}, {[13]}_{18}, {[14]}_{18}} . \end{array}$
  
  From the above calculations, we obtain
  
  $\begin{array}{l} \bar{T_{N_{1} (ℬ)}} (A \cap B) = \bar{T_{S_{1}}} (A \cap B) \cup \bar{T_{S_{2}}} (A \cap B) \\ = ℤ_{18} - {{[2]}_{18}} . \end{array}$
As a summary, we obtain $\bar{T_{N_{1} (ℬ)}} (A) \cap \bar{T_{N_{1} (ℬ)}} (B) ⊈ \bar{T_{N_{1} (ℬ)}} (A \cap B)$ .

Example 9

T ({[x]}_{18}) = {{[x + 1]}_{18}},

(1) Determine $\underline{T_{N_{1} (ℬ)}} (A)$ and $\underline{T_{N_{1} (ℬ)}} (B)$ .
- • There are no cosets of S₁ = ⟨[9]₁₈⟩ contained in A. T_S₁ (A) = {[x]₁₈ ∈ ℤ₁₈ | [x + 1]₁₈ + ⟨[9]₁₈⟩ ⊆ A} = ∅.
- • There are no cosets of S₂ = ⟨[6]₁₈⟩ contained in A. T_S₂ (A) = {[x]₁₈ ∈ ℤ₁₈ | [x + 1]₁₈ + ⟨[6]₁₈⟩ ⊆ A} = ∅.
  
  From the above calculations, we obtain
  
  $\underline{T_{N_{1} (ℬ)}} (A) = \underline{T_{S_{1}}} (A) \cup \underline{T_{S_{2}}} (A) = \emptyset .$
- • There are no cosets of S₁ = ⟨[9]₁₈⟩ contained in B. T_S₁ (B) = {[x]₁₈ ∈ ℤ₁₈ | [x+1]₁₈+⟨[9]₁₈⟩ ⊆ B} = ∅.
- • There are no cosets of S₂ = ⟨[6]₁₈⟩ contained in B. T_S₂ (B) = {[x]₁₈ ∈ ℤ₁₈ | [x+1]₁₈+⟨[6]₁₈⟩ ⊆ B} = ∅.
  
  From the above calculations, we obtain
  
  $\underline{T_{N_{1} (ℬ)}} (B) = \underline{T_{S_{1}}} (B) \cup \underline{T_{S_{2}}} (B) = \emptyset .$
As a summary, we have $\underline{T_{N_{1} (ℬ)}} (A) \cup \underline{T_{N_{1} (ℬ)}} (B) = \emptyset$
(2) Determine $\underline{T_{N_{1} (ℬ)}} (A \cup B)$ .
- • Cosets of S₁ = ⟨[9]₁₈⟩ contained in A ∪ B is ⟨[9]₁₈⟩.
  
  $\begin{array}{l} \underline{T_{S_{1}}} (A \cup B) = {{[x]}_{18} \in ℤ_{18} ∣ {[x + 1]}_{18} + ⟨ {[9]}_{18} ⟩ \subseteq A \cup B} \\ = {{[8]}_{18}, {[17]}_{18}} . \end{array}$
- • Cosets of S₂ = ⟨[6]₁₈⟩ contained in A∪B is [5]₁₈+ ⟨[6]₁₈⟩.
  
  $\begin{array}{l} \underline{T_{S_{2}}} (A \cup B) = {{[x]}_{18} \in ℤ_{18} ∣ {[x + 1]}_{18} + ⟨ {[6]}_{18} ⟩ \subseteq A \cup B} \\ = {{[4]}_{18}, {[10]}_{18}, {[16]}_{18}} . \end{array}$
  
  From the above calculations, we obtain
  
  $\begin{array}{l} \underline{T_{N_{1} (ℬ)}} (A \cup B) = \underline{T_{S_{1}}} (A \cup B) \cup \underline{T_{S_{2}}} (A \cup B) \\ = {{[4]}_{18}, {[8]}_{18}, {[10]}_{18}, {[16]}_{18}, {[17]}_{18}} . \end{array}$
As a summary, we have $\underline{T_{N_{1} (ℬ)}} (A \cup B) ⊈ \underline{T_{N_{1} (ℬ)}} (A) \cup \underline{T_{N_{1} (ℬ)}} (B)$ .

Proposition 3

Let G and G′ be two groups, (G′, ) be a nearness approximation space, and T : G→℘^*(G′) be a set-valued homomorphism. If A is a non-empty subset of G′ and e′ ∈ A, then $e \in \bar{T_{N_{r} (ℬ)}} (A)$ where e and e′ are neutral elements of G and G′, respectively.

Proof

Proposition 4

Let G and G′ be two groups, (G′, ) be a nearness approximation space, and T : G→℘^*(G′) be a set-valued mapping. Then,

(1) $\bar{T_{N_{r} (ℬ)}} (A) = \cup_{ℬ_{r} \subseteq ℬ} \bar{T} (A ℬ_{r})$ ,
(2) $\underline{T_{N_{1} (ℬ)}} (A) \subseteq \cup_{ℬ_{r} \subseteq ℬ} \underline{T} (A ℬ_{r}),$

where Aℬ_r = A (∩_{S∈ℬ_r}S).

Proof

(1) Similar to the proof of Theorem 4.5 in [26], we have

\bar{T_{N_{r} (ℬ)}} (A) = \underset{ℬ_{r} \subseteq ℬ}{\cup} \bar{T_{ℬ_{r}}} (A) = \underset{ℬ_{r} \subseteq ℬ}{\cup} \bar{T} (A ℬ_{r}) .

(2) Similar to the proof of Theorem 4.5 in [26], we have $\underline{T_{N_{r} (ℬ)}} (A) = \cup_{ℬ_{r} \subseteq ℬ} \underline{T_{ℬ_{r}}} (A) \subseteq \cup_{ℬ_{r} \subseteq ℬ} \underline{T} (A ℬ_{r})$ .

The following corollary follows from Proposition 4.

Corollary 1

Let G and G′ be two groups, (G′, ) be a nearness approximation space, and T : G→℘^*(G′) be a set-valued mapping. Let N be a normal subgroup of G′ and A ⊆ G′, A ≠ ∅︀. Then,

(1) $\bar{T_{N_{r} (ℬ)}} (A) = \cup_{ℬ_{r} \subseteq ℬ} \bar{T_{ℬ_{r}}} (A ℬ_{r})$ ,
(2) $\underline{T_{N_{r} (ℬ)}} (A) \subseteq \cup_{ℬ_{r} \subseteq ℬ} \underline{T_{ℬ_{r}}} (A ℬ_{r}),$ where Aℬ_r = A (∩_{S∈ℬ_r}S).

Proof

(1) Similar to the proof of Theorem 4.5 in [26], we have $\bar{T_{ℬ_{r}}} (A ℬ_{r}) = \bar{T} (A ℬ_{r} ℬ_{r}) = \bar{T} (A ℬ_{r})$ so that with Proposition 4, we get $\bar{T_{N_{r} (ℬ)}} (A) = \cup_{ℬ_{r} \subseteq ℬ} \bar{T_{ℬ_{r}}} (A ℬ_{r})$ .

(2) Since A ⊆ Aℬ_r, it follows that $\underline{T_{N_{r} (ℬ)}} (A) = \cup_{ℬ_{r} \subseteq ℬ} \underline{T_{ℬ_{r}}} (A) \subseteq \cup_{ℬ_{r} \subseteq ℬ} \underline{T_{ℬ_{r}}} (A ℬ_{r})$ .

Definition 7

Let G and G′ be two groups, (G′, ) be a nearness approximation space, and T : G→℘^*(G′) be a set-valued mapping. Let A be a non-empty subset of G′. We define the sets

\underline{T_{N_{r} (ℬ ℬ)}} (A) = {x \in G ∣ T (x) ℬ_{r} ℬ_{r}^{'} \subseteq A, for ℬ_{r}, ℬ_{r}^{'} \subseteq ℬ},

and

\bar{T_{N_{r} (ℬ ℬ)}} (A) = {x \in G ∣ T (x) ℬ_{r} ℬ_{r}^{'} \cap A \neq \emptyset, for ℬ_{r}, ℬ_{r}^{'} \subseteq ℬ},

where $T (x) ℬ_{r} ℬ_{r}^{'} = T (x) (\cap_{S \in ℬ_{r}} S) (\cap_{R \in ℬ_{r}^{'}} R)$ .

Proposition 5

Let G and G′ be two groups, (G′, ) be the nearness approximation space, and T : G→ ℘^*(G′) be a set-valued homomorphism. If A is a non-empty subset of G′, then

(1) $\underline{T_{N_{r} (ℬ ℬ)}} (A) = \underline{T_{N_{r} (ℬ)}} (A)$ ,
(2) $\bar{T_{N_{r} (ℬ)}} (A) \subseteq \bar{T_{N_{r} (ℬ ℬ)}} (A)$ .

Proof

(1) Let x ∈ $\underline{T_{N_{r} (ℬ ℬ)}} (A)$ . Then, $T (x) ℬ_{r} ℬ_{r}^{'} \subseteq A$ for some ℬ_r, $ℬ_{r}^{'} \subseteq ℬ$ . Since $T (x) ℬ_{r} \subseteq T (x) ℬ_{r} ℬ_{r}^{'} \subseteq A$ , we have x ∈ $\underline{T_{N_{r} (ℬ)}} (A)$ so that $\underline{T_{N_{r} (ℬ ℬ)}} (A) \subseteq \underline{T_{N_{r} (ℬ)}} (A)$ . Let x ∈ $\underline{T_{N_{r} (ℬ)}} (A)$ . Subsequently, T(x)ℬ_rℬ_r = T(x)ℬ_r ⊆ A for some ℬ_r ⊆ ℬ so that x ∈ $\underline{T_{N_{r} (ℬ ℬ)}} (A)$ , which implies that $\underline{T_{N_{r} (ℬ)}} (A) \subseteq \underline{T_{N_{r} (ℬ ℬ)}} (A)$ . Thus, $\underline{T_{N_{r} (ℬ ℬ)}} (A) = \underline{T_{N_{r} (ℬ)}} (A)$ .

(2) Let $x \in \bar{T_{N_{r} (ℬ)}} (A)$ . Then, T(x)ℬ_r ∩ A ≠ ∅︀ for some ℬ_r ⊆ ℬ. Since $T (x) ℬ_{r} \subseteq T (x) ℬ_{r} ℬ_{r}^{'}$ for every $ℬ_{r}^{'} \subseteq ℬ$ , we obtain $T (x) ℬ_{r} ℬ_{r}^{'} \cap A \neq \emptyset$ such that $x \in \bar{T_{N_{r} (ℬ ℬ)}} (A)$ . Thus, $\bar{T_{N_{r} (ℬ)}} (A) \subseteq \bar{T_{N_{r} (ℬ ℬ)}} (A)$ .

The following example shows that the converse of Proposition 5 (2) is not true in general.

Example 10

T ({[x]}_{24}) = {{(x + 1)}_{24}},

for every [x]₂₄ ∈ ℤ₂₄. It is easy to see S₁+S₁ = S₁, S₂+S₂ = S₂ and S₁ + S₂ = ⟨[4]₂₄⟩. Let A = {[4]₂₄}. Then,

\begin{array}{l} \bar{T_{S_{1} + S_{1}}} (A) = \bar{T_{S_{1}}} (A) = {{[3]}_{24}, {[11]}_{24}, {[19]}_{24}}, \\ \bar{T_{S_{2} + S_{2}}} (A) = \bar{T_{S_{2}}} (A) = {{[3]}_{24}, {[15]}_{24}}, \\ \bar{T_{S_{1} + S_{2}}} (A) = {{[3]}_{24}, {[7]}_{24}, {[11]}_{24}, {[15]}_{24}, {[19]}_{24}, {[23]}_{24}} . \end{array}

We have

\begin{array}{l} \bar{T_{N_{1} (ℬ)}} (A) = \bar{T_{S_{1}}} (A) \cup \bar{T_{S_{2}}} (A) \\ = {{[3]}_{24}, {[11]}_{24}, {[15]}_{24}, {[19]}_{24}}, \\ \bar{T_{N_{1} (ℬ + ℬ)}} (A) = \bar{T_{S_{1} + S_{1}}} (A) \cup \bar{T_{S_{2} + S_{2}}} (A) \cup \bar{T_{S_{1} + S_{2}}} (A) \\ = {{[3]}_{24}, {[7]}_{24}, {[11]}_{24}, {[15]}_{24}, {[19]}_{24}, {[23]}_{24}} . \end{array}

Thus, we obtain $\bar{T_{N_{1} (ℬ + ℬ)} (A)} ⊈ \bar{T_{N_{1} (ℬ)} (A)}$ .

Proposition 6

Let G and G′ be two groups, (G′, ) be a nearness approximation space, and T : G→℘^*(G′) be a set-valued homomorphism. If A and B are two non-empty subsets of G′, then

\bar{T_{N_{r} (ℬ)}} (A) \bar{T_{N_{r} (ℬ)}} (B) \subseteq \bar{T_{N_{r} (ℬ ℬ)}} (A B) .

Proof

Let $z \in \bar{T_{N_{r} (ℬ)}} (A) \bar{T_{N_{r} (ℬ)}} (B)$ . Then, there exist $x \in \bar{T_{N_{r} (ℬ)}} (A)$ and $y \in \bar{T_{N_{r} (ℬ)}} (B)$ such that z = xy. Since $x \in \bar{T_{N_{r} (ℬ)}} (A)$ and $y \in \bar{T_{N_{r} (ℬ)}} (B)$ then T(x)ℬ_r ∩ A ≠ ∅︀ for some ℬ_r ⊆ ℬ and $T (y) ℬ_{r}^{'} \cap B \neq \emptyset$ for some $ℬ_{r}^{'} \subseteq ℬ$ so there exist a, b ∈ G′ where a ∈ T(x)ℬ_r ∩ A and $b \in T (y) ℬ_{r}^{'} \cap B$ . Since a ∈ T(x)ℬ_r, $b \in T (y) ℬ_{r}^{'}$ and T be a set-valued group homomorphism then $a b \in T (x) ℬ_{r} T (y) ℬ_{r}^{'} = T (x) T (y) ℬ_{r} ℬ_{r}^{'} = T (x y) ℬ_{r} ℬ_{r}^{'}$ so that $a b \in T (x y) ℬ_{r} ℬ_{r}^{'} \cap A B$ which implies $T (x y) ℬ_{r} ℬ_{r}^{'} \cap A B \neq ϕ$ . Therefore, $z = x y \in \bar{T_{N_{r} (ℬ ℬ)}} (A B)$ so that $\bar{T_{N_{r} (ℬ)}} (A) \bar{T_{N_{r} (ℬ)}} (B) \subseteq \bar{T_{N_{r} (ℬ ℬ)}} (A B)$ .

The following example shows that the converse of Proposition 6 is generally not true.

Example 11

T ({[x]}_{24}) = {8 {[x]}_{24}},

for every [x]₂₄ ∈ ℤ₂₄. Notice that T is a set-valued homomorphism. Let A = {[1]₂₄, [5]₂₄}, B = {[3]₂₄}, A + B = {[4]₂₄, [8]₂₄}. We get

• $\bar{T_{S_{1}}} (A) = \bar{T_{S_{2}}} (A) = \bar{T_{S_{1}}} (B) = \bar{T_{S_{2}}} (B) = \emptyset$ ,
• $\bar{T_{S_{1} + S_{2}}} (A + B) = ℤ_{24}$ ,
• $\bar{T_{N_{1} (ℬ)}} (A) = \bar{T_{N_{1} (ℬ)}} (B) = \emptyset$ ,
• $\bar{T_{N_{1} (ℬ + ℬ)}} (A + B) = ℤ_{24}$ .

This shows that $\bar{T_{N_{1} (ℬ + ℬ)}} (A + B) ⊈ \bar{T_{N_{1} (ℬ)}} (A) + \bar{T_{N_{1} (ℬ)}} (B)$ .

Proposition 7

Let G and G′ be two groups, (G′, ) be a nearness approximation space, and T : G→℘^*(G′) be a set-valued homomorphism. If A and B are two non-empty subsets of G′, then

\underline{T_{N_{r} (ℬ)}} (A) \underline{T_{N_{r} (ℬ)}} (B) \subseteq \underline{T_{N_{r} (ℬ)}} (A B) .

Proof

Let z ∈ $\underline{T_{N_{r} (ℬ)}} (A) \underline{T_{N_{r} (ℬ)}} (B)$ . Then, there exist x ∈ $\underline{T_{N_{r} (ℬ)}} (A)$ and y ∈ $\underline{T_{N_{r} (ℬ)}} (B)$ so that z = xy. Since x ∈ $\underline{T_{N_{r} (ℬ)}} (A)$ and y ∈ $\underline{T_{N_{r} (ℬ)}} (B)$ then T(x)ℬ_r ⊆ A and $T (y) ℬ_{r}^{'} \subseteq B$ for some ℬ_r, $ℬ_{r}^{'} \subseteq ℬ$ . Since T is a set-valued group homomorphism then $T (x) ℬ_{r} T (y) ℬ_{r}^{'} = T (x) T (y) ℬ_{r} ℬ_{r}^{'} = T (x y) ℬ_{r} ℬ_{r}^{'} \subseteq A B$ , which implies that z = xy ∈ $\underline{T_{N_{r} (ℬ ℬ)}} (A B)$ . Hence $\underline{T_{N_{r} (ℬ)}} (A) \underline{T_{N_{r} (ℬ)}} (B) \subseteq \underline{T_{N_{r} (ℬ ℬ)}} (A B)$ . By Proposition 5(1), we have $\underline{T_{N_{r} (ℬ)}} (A) \underline{T_{N_{r} (ℬ)}} (B) \subseteq \underline{T_{N_{r} (ℬ)}} (A B)$ .

Example 12

T (x) = {x, - x},

for every x ∈ Q₈. From Example 4.2, in [35], the mapping T is a set-valued homomorphism. Let A = {1, i}, B = {1, −1, −i}, A.B = {1, −1, i, −i} = S₁. Then, we get

• $\underline{T_{S_{1}}} (A) = \underline{T_{S_{2}}} (A) = \underline{T_{S_{1}}} (B) = \underline{T_{S_{2}}} (B) = \underline{T_{S_{2}}} (A . B) = \emptyset$ ,
• $\underline{T_{S_{1}}} (A . B) = {1, - 1, i, - i}$ .

So that

• $\underline{T_{N_{1} (ℬ)}} (A) = \underline{T_{N_{1} (ℬ)}} (B) = \emptyset,$
• $\underline{T_{N_{1} (ℬ)}} (A . B) = {1, - 1, i, - i} .$

This implies that $\underline{T_{N_{1} (ℬ)}} (A . B) ⊈ \underline{T_{N_{1} (ℬ)}} (A) . \underline{T_{N_{1} (ℬ)}} (B) .$

5. The Lower and Upper Near Subgroups in a Group based on a Set-Valued Mapping

Proposition 8

Let G and G′ be two groups, (G′, ) be a nearness approximation space, and T : G→℘^*(G′) be a set-valued homomorphism. Let A be a subgroup of G′, $\bar{T_{N_{r} (ℬ)}} (A) = \bar{T_{N_{r} (ℬ ℬ)}} (A)$ and $\bar{T_{N_{r} (ℬ)}} (A) \neq \emptyset$ . Subsequently, $\bar{T_{N_{r} (ℬ)}} (A)$ is a subgroup of G.

Proof

Let x, $y \in \bar{T_{N_{r} (ℬ)}} (A)$ . Then, T(x)ℬ_r ∩ A ≠ ∅︀ and $T (y) ℬ_{r}^{'} \cap A \neq \emptyset$ for some ℬ_r, $ℬ_{r}^{'} \subseteq ℬ$ . So there exist a ∈ T(x)ℬ^r ∩A and $b \in T (y) ℬ_{r}^{'} \cap A$ . Since A is a subgroup of G′, we have ab⁻¹ ∈ A. Since T is a set-valued homomorphism, we have $a b^{- 1} \in T (x) ℬ_{r} {(T (y) ℬ_{r}^{'})}^{- 1} = T (x) ℬ_{r} {(ℬ_{r}^{'})}^{- 1} {(T (y))}^{- 1} = T (x) ℬ_{r} ℬ_{r}^{'} T (y^{- 1}) = T (x) T (y^{- 1}) ℬ_{r} ℬ_{r}^{'} = T (x y^{- 1}) ℬ_{r} ℬ_{r}^{'}$ so that $a b^{- 1} \in T (x y^{- 1}) ℬ_{r} ℬ_{r}^{'} \cap A$ which implies that $x y^{- 1} \in \bar{T_{N_{r} (ℬ ℬ)}} (A) = \bar{T_{N_{r} (ℬ)}} (A)$ . Therefore, $\bar{T_{N_{r} (ℬ)}} (A)$ is a subgroup of G.

Proposition 9

Let G and G′ be two groups, (G′, ) be a nearness approximation space, and T : G→℘^* (G′) be a set-valued homomorphism. Let A be a normal subgroup of G′, $\bar{T_{N_{r} (ℬ)}} (A) = \bar{T_{N_{r} (ℬ ℬ)}} (A)$ and $\bar{T_{N_{r} (ℬ)}} (A) \neq \emptyset$ . Subsequently, $\bar{T_{N_{r} (ℬ)}} (A)$ is a normal subgroup of G.

Proof

This is sufficient to demonstrate that $\bar{T_{N_{r} (ℬ)}} (A)$ is normal. Let $x \in \bar{T_{N_{r} (ℬ)}} (A)$ and g ∈ G. Then, T(x)ℬ_r ∩ A ≠ ∅︀ for some ℬ_r ⊆ ℬ and T(g) is a non-empty set, we can select g′ ∈ T(g). Thus, a ∈ T(x)ℬ_r ∩ A. exists and because A is a normal subgroup of G′, we have ag′a⁻¹ ∈ A. Since T is a set-valued homomorphism, we have ag′a⁻¹ ∈ T(x)ℬ_rT(g)(T(x)ℬ_r)⁻¹ = T(x)ℬ_rT(g)ℬ_r⁻¹T(x)⁻¹=T(x) ℬ_rT(g)ℬ_r⁻¹T(x⁻¹)=T(x)T(g)T(x⁻¹)ℬ_rℬ_r⁻¹=T(xgx⁻¹) ℬ_r such that ag′a⁻¹ ∈ T(xgx⁻¹)ℬ_r ∩ A which implies that $x g x^{- 1} \in \bar{T_{N_{r} (ℬ)}} (A)$ . Therefore, $\bar{T_{N_{r} (ℬ)}} (A)$ is a normal subgroup of G.

Example 13

T ({[x]}_{24}) = {3 {[x]}_{24}},

\bar{T_{S_{1}}} (A) = \bar{T_{S_{2}}} (A) = \bar{T_{S_{1} + S_{1}}} (A) = \bar{T_{S_{2} + S_{2}}} (A) = \bar{T_{S_{1} + S_{2}}} (A) = 2 ℤ_{24} .

so that $\bar{T_{N_{1} (ℬ)}} (A) = \bar{T_{N_{1} (ℬ + ℬ)}} (A) = 2 ℤ_{24}$ is a normal subgroup of ℤ₂₄.

Proposition 10

Let G and G′ be two groups, (G′, ) be a nearness approximation space, and T : G→℘^*(G′) be a set-valued homomorphism. Let A be a subgroup of G′ and $\underline{T_{N_{r} (ℬ)}} (A)$ ≠ ∅. Then, $\underline{T_{N_{r} (ℬ)}} (A)$ is a subgroup of G.

Proof

Let x, y ∈ $\underline{T_{N_{r} (ℬ)}} (A)$ . Then, T(x)ℬ_r ⊆ A and $T (y) ℬ_{r}^{'} \subseteq A$ for some ℬ_r, $ℬ_{r}^{'} \subseteq ℬ$ . Since A is a subgroup of G′, we have ${(T (y) ℬ_{r}^{'})}^{- 1} = T (y^{- 1}) ℬ_{r}^{'} \subseteq A$ . Since T is a set-valued homomorphism, we have $T (x) ℬ_{r} T (y^{- 1}) ℬ_{r}^{'} = T (x) T (y^{- 1}) ℬ_{r} ℬ_{r}^{'} = T (x y^{- 1}) ℬ_{r} ℬ_{r}^{'} \subseteq A$ which implies that xy⁻¹ ∈ $\underline{T_{N_{r} (ℬ B)}} (A)$ . From Proposition 5(1), we have xy⁻¹ ∈ $\underline{T_{N_{r} (ℬ)}} (A)$ . Therefore, $\underline{T_{N_{r} (ℬ)}} (A)$ is a subgroup of G.

Proposition 11

Let G and G′ be two groups, (G′, ) be a nearness approximation space, and T : G→℘^*(G′) be a set-valued homomorphism. Let A be a normal subgroup of G′ and $\underline{T_{N_{r} (ℬ)}} (A)$ ≠ ∅. Then, $\underline{T_{N_{r} (ℬ)}} (A)$ is a normal subgroup of G.

Proof

This suffices for demonstrating that $\underline{T_{N_{r} (ℬ)}} (A)$ is normal. Let x ∈ $\underline{T_{N_{r} (ℬ)}} (A)$ and g ∈ G. Then, T(x)ℬ_r ⊆ A for some ℬ_r ⊆ ℬ and T(g) is a non-empty set. Since A is a subgroup of G′, we have (T(x)ℬ_r)⁻¹ = T(x⁻¹)ℬ_r ⊆ A. Since A is a normal subgroup of G′ and T is a set-valued homomorphism, we have T(x)ℬ_rT(g)T(x⁻¹)ℬ_r = T(x)T(g)T(x⁻¹)ℬ_r = T(xgx⁻¹)ℬ_r ⊆ A which implies that xgx⁻¹ ∈ $\underline{T_{N_{r} (ℬ)}} (A)$ . Therefore, $\underline{T_{N_{r} (ℬ)}} (A)$ is a normal subgroup of G.

6. Comparison between Near-Generalized and Generalized Approximations in a Group based on a Set-Valued Mapping

Proposition 12

Let G and G′ be two groups, (G′, ) be a nearness approximation space, and T : G→℘^*(G′) be a set-valued mapping. Let A be a non-empty subset of G′. Then,

(1) If |ℬ| = k, then $\underline{T_{N_{k} (ℬ)}} (A) = \underline{T_{ℬ}} (A)$ and $\bar{T_{N_{k} (ℬ)}} (A) = \bar{T_{ℬ}} (A)$ .
(2) If 1 ≤ r < k where |ℬ| = k, then $\underline{T_{ℬ_{r}}} (A) \subseteq \underline{T_{N_{r} (ℬ)}} (A)$ and $\bar{T_{ℬ_{r}}} (A) \subseteq \bar{T_{N_{r} (ℬ)}} (A)$ .

Proof

The proof results from Definitions 5 and 6

From Proposition 12(1), we observe that the generalized approximations in a group based on set-valued mapping is a special type of near generalized approximation.

Proposition 13

(1) $\underline{T_{N_{r} (ℬ_{i})}} (A) \subseteq \underline{T_{N_{r} (ℬ_{j})}} (A),$
(2) $\bar{T_{N_{r} (ℬ_{i})}} (A) \subseteq \bar{T_{N_{r} (ℬ_{j})}} (A)$ .

Proof

(1) From Definition 6, we have $\underline{T_{N_{r} (ℬ_{k})}} (A) = \cup_{ℬ_{r} \subseteq ℬ_{k}} \underline{T_{ℬ_{r}}} (A)$ , for 1 ≤ k ≤ |ℬ|. Since 1 ≤ i < j ≤ |ℬ|, we have ℬ_i ⊂ ℬ_j ⊆ ℬ then $\underline{T_{N_{r} (ℬ_{i})}} (A) \subseteq \underline{T_{N_{r} (ℬ_{j})}} (A)$ .

(2) From Definition 6, we obtain $\bar{T_{N_{r} (ℬ_{k})}} (A) = \cup_{ℬ_{r} \subseteq ℬ_{k}} \bar{T_{ℬ_{r}}} (A)$ for 1 ≤ k ≤ |ℬ|. Since 1 ≤ i < j ≤ |ℬ|, we have ℬ_i ⊂ ℬ_j ⊆ ℬ then $\bar{T_{N_{r} (ℬ_{i})}} (A) \subseteq \bar{T_{N_{r} (ℬ_{j})}} (A)$ .

Proposition 14

Let G and G′ be two groups, (G′, ), be a nearness approximation space, and T : G→℘^*(G′) be a set-valued mapping. Let A be a non-empty subset of G′. Then,

\underline{T_{N_{i} (ℬ)}} (A) \subseteq \underline{T_{N_{j} (ℬ)}} (A) \subseteq \bar{T_{N_{j} (ℬ)}} (A) \subseteq \bar{T_{N_{i} (ℬ)}} (A),

where 1 ≤ i < j ≤ |ℬ|.

Proof

Let x ∈ $\underline{T_{N_{i} (ℬ)}} (A)$ . Subsequently, T(x)ℬ_i ⊆ A for a certain ℬ_i ⊆ ℬ. Since i < j then ℬ_i ⊂ ℬ_j so that T(x)ℬ_j = T(x)∩_{S∈ℬ_j} S ⊆ T(x)∩_{S∈ℬ_i} S = T(x)ℬ_i ⊆ A which yields that x ∈ $\underline{T_{N_{j} (ℬ)}} (A)$ . Therefore, $\underline{T_{N_{i} (ℬ)}} (A) \subseteq \underline{T_{N_{j} (ℬ)}} (A)$ . From Proposition 2 (1), we have $\underline{T_{N_{j} (ℬ)}} (A) \subseteq \bar{T_{N_{j} (ℬ)}} (A)$ . Let $x \in \bar{T_{N_{j} (ℬ)}} (A)$ . Then, T(x)ℬ_j ∩ A ≠ ∅︀ for some ℬ_j ⊆ ℬ. Since i < j then ℬ_i ⊂ ℬ_j so that T(x)ℬ_j = T(x)∩_{S∈ℬ_j}S ⊆ T(x)∩_{S∈ℬ_i}S = T(x)ℬ_i. Thus, we have T(x)ℬ_i ∩ A ≠ ∅︀, which yields $x \in \bar{T_{N_{i} (ℬ)}} (A)$ . Therefore, $\underline{T_{N_{i} (ℬ)}} (A) \subseteq \underline{T_{N_{j} (ℬ)}} (A) \subseteq \bar{T_{N_{j} (ℬ)}} (A) \subseteq \bar{T_{N_{i} (ℬ)}} (A)$ .

7. Conclusion

Abstract
1. Introduction
2. Approximations and Near Approximations in a Group
3. Approximations in a Group based on a Set-Valued Mapping
4. Near-Generalized Approximations in a Group based on a Set-Valued Mapping
5. The Lower and Upper Near Subgroups in a Group based on a Set-Valued Mapping
6. Comparison between Near-Generalized and Generalized Approximations in a Group based on a Set-Valued Mapping
7. Conclusion

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Article Tools

International Journalof Fuzzy Logic andIntelligent Systems

Original Article

Near-Generalized Approximations in Groups Based on a Set-Valued Mapping

Abstract

1. Introduction

2. Approximations and Near Approximations in a Group

Definition 1 [11]

Example 1

Definition 2 [31]

Example 2

3. Approximations in a Group based on a Set-Valued Mapping

Definition 3 [35]

Example 3

Definition 4

Example 4

Example 5 [35]

Definition 5

Example 6

4. Near-Generalized Approximations in a Group based on a Set-Valued Mapping

Definition 6

Example 7

Proposition 1

Proposition 2

Example 8

Example 9

Proposition 3

Proposition 4

Corollary 1

Definition 7

Proposition 5

Example 10

Proposition 6

Example 11

Proposition 7

Example 12

5. The Lower and Upper Near Subgroups in a Group based on a Set-Valued Mapping

Proposition 8

Proposition 9

Example 13

Proposition 10

Proposition 11

6. Comparison between Near-Generalized and Generalized Approximations in a Group based on a Set-Valued Mapping

Proposition 12

Proposition 13

Proposition 14

7. Conclusion

Conflict of Interest

References

Biographies

Article

Original Article

Near-Generalized Approximations in Groups Based on a Set-Valued Mapping

Abstract

1. Introduction

2. Approximations and Near Approximations in a Group

Definition 1 [11]

Example 1

Definition 2 [31]

Example 2

3. Approximations in a Group based on a Set-Valued Mapping

Definition 3 [35]

Example 3

Definition 4

Example 4

Example 5 [35]

Definition 5

Example 6

4. Near-Generalized Approximations in a Group based on a Set-Valued Mapping

Definition 6

Example 7

Proposition 1

Proposition 2

Example 8

Example 9

Proposition 3

Proposition 4

Corollary 1

Definition 7

Proposition 5

Example 10

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