Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 3 (2025) 879 https://internationalpubls.com Compactness and Rough Isomorphism on Topological Simple Rough Groups P. Tamilarasi [1] And R. Selvi [2] [1] Research Scholar (Reg. No: 22211202092003), Department of Mathematics, Sri Parasakthi College for Women, Courtallam - 627802, Affiliated by Manonmaniam Sundaranar University, Tirunelveli - 627012. E-mail: tamilarasiparamasivan@gmail.com [2] Associate Professor, Department of Mathematics, Sri Parasakthi College for Women, Courtallam - 627802, Affiliated by Manonmaniam Sundaranar University, Tirunelveli - 627012. E-mail: r.selvimuthu@gmail.com Article History: Received: 12-01-2025 Revised: 15-02-2025 Accepted: 01-03-2025 Abstract: In this paper, we study about the compactness on topological simple rough groups. In particular we discuss the open mapping theorems and rough isomorphism theorems in topological simple rough groups. Also, we define a rough double coset space and discuss their role in topological simple rough groups. Further, we examine the relationship between compactness and continuity of quotient maps. Keywords: Rough groups, Rough subgroups, Topological simple rough groups, Compact, Topological rough group homeomorphism, Rough double coset spaces, Quotient spaces. 2020 Mathematics Subject Classification: 20E32, 22C05, 22D05, 54D45. 1.Introduction: The rough set theory was introduced by Pawlak [14] in 1982 which is based on the equivalence relations. After more than 30 years of research, the theory of rough set has been continuously improved and widely expanded in applications. In 1994, Biswas and Nanda [3] introduced the notion of rough groups and rough subgroups, which depends on the upper approximation. Then, Bagirmaz et al. (2016) [13] introduced the concept of topological rough group and extended the notion of a topological group to include algebraic structures of rough groups. In this paper, we discussed compactness and open mapping theorems in topological simple rough groups and we examine the relationship between compactness and continuity of quotient maps. Further, we analysed some results related to topological rough group homeomorphism and also rough isomorphism theorems are discussed. Finally, we defined a rough double coset space and discuss their role in topological simple rough groups. By utilizing techniques from topology and fundamental analysis, we establish criteria for compactness in rough double coset spaces. 2. Preliminaries: Definition 2.1. [4] Let K = (U, R) be an approximation space and ∗ be a binary operation defined on U. A subset G of universe U is called a rough group if the following properties are satisfied: (i) ∀ x, y ∈ G, x∗y ∈ G̅; mailto:tamilarasiparamasivan@gmail.com mailto:r.selvimuthu@gmail.com Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 3 (2025) 880 https://internationalpubls.com (ii) Association property holds in G̅; (iii) ∃ e ∈ G̅ such that ∀ x ∈ G, x∗e = e∗x = x; e is called the rough identity element of rough group G; (iv) ∀ x ∈ G, ∃ y ∈ G such that x∗y = y∗x = e; y is called the rough inverse element of x in G; Theorem 2.2. [4] A necessary and sufficient condition for a subset H of rough group G to be a rough subgroup is that: (i) ∀ x, y ∈ H, x∗y ∈ G̅; (ii) ∀ x ∈ H, x-1 ∈ H. Definition 2.3. [13] A topological rough group is a rough group (G, ∗) together with a topology T on G̅ satisfying the following two properties: (i) the mapping f: G × G → G̅ defined by f(x, y) = xy is continuous with respect to product topology on G × G and the topology TG on G induced by T, (ii) the inverse mapping g: G → G defined by g(x) = x-1 is continuous with respect to the topology TG on G induced by T. Definition 2.4. [4] Let (𝑈1, 𝑅1), (𝑈2, 𝑅2) be two approximation spaces, and ∗, ∗̅ be binary operations over universes 𝑈1, 𝑈2 respectively. Let 𝐺1 ⊂ 𝑈1 and 𝐺2 ⊂ 𝑈2 be rough groups. 𝐺1, 𝐺2 are called rough homomorphism sets if there exists a surjection 𝜑: 𝐺1 ̅̅ ̅ → 𝐺2 ̅̅ ̅ such that ∀ 𝑥, 𝑦 ∈ 𝐺1 ∪ {𝑒}, we have 𝜑(𝑥 ∗ 𝑦) = 𝜑(𝑥) ∗̅ 𝜑(𝑦). If a rough homomorphism is a bijection, then we say that 𝐺1 and 𝐺2 are rough isomorphism. Definition 2.5. [1] A mapping 𝑓: 𝐺1 ̅̅ ̅ → 𝐺2 ̅̅ ̅ is called a topological rough group homomorphism, if 𝑓 is a rough homomorphism and continuous with respect to the topology 𝜏2 on 𝐺2 ̅̅ ̅ inducing 𝜏𝐺2 on 𝐺2 and a topology 𝜏1 on 𝐺1 ̅̅ ̅ inducing 𝜏𝐺1 on 𝐺1. Definition 2.6. [1] Topological rough group homomorphism 𝑓: 𝐺1 ̅̅ ̅ → 𝐺2 ̅̅ ̅ is called a topological rough group homeomorphism, if there exists a topological rough homomorphism 𝑓−1 such that 𝑓−1 ∘ 𝑓 = 1𝐺1 . Definition 2.7. [1] Let Φ: 𝐺1 ̅̅ ̅ → 𝐺2 ̅̅ ̅ be a topological rough group homomorphism and let 𝑒2 be the rough identity element in 𝐺2. Then ker(Φ) = {𝑔 ∈ 𝐺1 ∶ Φ(𝑔) = 𝑒2}. is called the rough kernel associated to the map Φ. Definition 2.8. [12] Let G be a rough group such that G̅ is a group and H is a rough subgroup of G. If H is a normal subgroup in G̅, then G̅ 𝐻⁄ is a rough quotient group. Definition 2.9. [16] A rough group 𝐺ℜ is called a simple rough group if it contains no proper non- trivial rough normal subgroups. That is, 𝐺ℜ has only the rough normal subgroups {e} and 𝐺ℜ. Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 3 (2025) 881 https://internationalpubls.com Definition 2.10. [16] A topological simple rough group is a simple rough group (𝐺ℜ, ∗) together with a topology τ̅ on 𝐺ℜ ̅̅ ̅̅ satisfying the following two properties: (i) The mapping f: 𝐺ℜ × 𝐺ℜ → 𝐺ℜ ̅̅ ̅̅ defined by f(x, y) = xy, x, y ∈ 𝐺ℜ is continuous with respect to the product topology on 𝐺ℜ × 𝐺ℜ and the topology τ on 𝐺ℜ induced by τ̅ (ii) The inverse mapping g: 𝐺ℜ → 𝐺ℜ defined by g(x) = x-1, x ∈ 𝐺ℜ is continuous with respect to the topology τ on 𝐺ℜ induced by τ̅. Proposition 2.11. [16] Let 𝐺ℜ be a topological simple rough group. If U ⊆ 𝐺ℜ ̅̅ ̅̅ is an open set with e ∈ U, then there exists a symmetric open set V of e in 𝐺ℜ such that VV ⊆ U. Lemma 2.12. [17] Let 𝐺ℜ be a topological simple rough group such that 𝐺ℜ is open in 𝐺ℜ ̅̅ ̅̅ and 𝑊 be a neighbourhood of 𝑒 in 𝐺ℜ ̅̅ ̅̅ . Then there is an open set 𝑈 of 𝑒 in 𝐺ℜ such that 𝑈 ⊆ 𝑈𝑛 ⊆ 𝑊, for every 𝑛 ∈ ℕ − {0}. Lemma 2.13. [7] Let Y be a subspace of X. If U is open in Y and Y is open in X, then U is open in X. Theorem 2.14. [5] Every locally compact subspace M of a Hausdorff space X is an open subset of the closure �̅� of the set M in the space X. Remark 2.15. [16] The topological closure of 𝐻ℜ, 𝑐𝑙(𝐻ℜ), in 𝐺ℜ ̅̅ ̅̅ is a topological rough subgroup in 𝐺ℜ ̅̅ ̅̅ . Definition 2.16. [16] A continuous mapping 𝑓: 𝑋 → 𝑌 is perfect if X is a Hausdorff space, 𝑓 is a closed mapping and all fibers 𝑓−1(𝑦) are compact subsets of X. Result 2.17. [5] A 𝑇1- space 𝑋 is a regular space if and only if for every 𝑥 ∈ 𝑋 and every neighbourhood 𝑉 of 𝑥 there exists a neighbourhood 𝑈 of 𝑥 such that 𝑉 ⊆ 𝑈. Theorem 2.18. [5] A continuous mapping 𝑓: 𝑋 → 𝑌 is closed if and only if for every point 𝑦 ∈ 𝑌 and every open set 𝑈 ⊂ 𝑋 which contains 𝑓−1(𝑦), there exists a neighbourhood V of the point y in Y such that 𝑓−1(𝑉) ⊂ 𝑈. Throughout this paper, we consider 𝑋 be the universal set, 𝐺ℜ be a simple rough group with identity 𝑒 and 𝐺ℜ ̅̅ ̅̅ be the upper rough approximation of 𝐺ℜ. Also, the corresponding topologies are denoted by τ̅ for 𝐺ℜ ̅̅ ̅̅ and τ for 𝐺ℜ induced from τ̅. 3. Compactness: Theorem 3.1. Let 𝐺ℜ be a topological simple rough group such that 𝐺ℜ is open in 𝐺ℜ ̅̅ ̅̅ and 𝐴 be a compact subset of 𝐺ℜ ̅̅ ̅̅ . If 𝑀 is a closed subset of 𝐺ℜ ̅̅ ̅̅ with 𝐴 ∩ 𝑀 = ∅ , then there is an open neighbourhood 𝑉 of e in 𝐺ℜ such that 𝐴𝑉 ∩ 𝑀 = ∅ and 𝑉𝐴 ∩ 𝑀 = ∅. Proof: Since the map 𝐿𝑔: 𝐺ℜ → 𝐺ℜ ̅̅ ̅̅ is continuous and 𝑀 is closed, there exists an open neighbourhood 𝑈𝑎 of 𝑒 in 𝐺ℜ ̅̅ ̅̅ such that 𝑎𝑈𝑎 ∩ 𝑀 = ∅, for all 𝑎 ∈ 𝐴. By proposition 2.9, there is a symmetric open set 𝑉𝑎 of e in 𝐺ℜ such that 𝑉𝑎𝑉𝑎 ⊆ 𝑈𝑎. Since 𝐴 is compact, there exists an open cover ⋃ 𝑎𝑎∈𝐴 𝑉𝑎 such that 𝐴 ⊆ ⋃ 𝑎𝑎∈𝐴 𝑉𝑎. Let 𝑉 = ⋂ 𝑉𝑎𝑎∈𝐴 . Suppose there exists an arbitrary element 𝑏 ∈ 𝐴, then 𝑏 ∈ 𝑎𝑉𝑎. Now, 𝑏𝑉 ⊆ 𝑏𝑉𝑎 ⊆ 𝑎𝑉𝑎𝑉𝑎 ⊆ 𝑎𝑈𝑎 which implies 𝑏𝑉 ∩ 𝑀 = ∅. Therefore, 𝐴𝑉 ∩ 𝑀 = ∅. Similarly, we prove 𝑉𝐴 ∩ 𝑀 = ∅. Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 3 (2025) 882 https://internationalpubls.com Theorem 3.2. (Second Closure Lemma) Let 𝐺ℜ be a topological simple rough group such that 𝐺ℜ is open in 𝐺ℜ ̅̅ ̅̅ . Suppose 𝐴 is a compact subset of 𝐺ℜ ̅̅ ̅̅ and 𝑀 is a closed subset of 𝐺ℜ ̅̅ ̅̅ . Then 𝐴𝑀 and 𝑀𝐴 are closed sets in 𝐺ℜ ̅̅ ̅̅ . Proof: Let 𝑎 ∉ 𝐴𝑀. Then 𝐴−1𝑎 ∩ 𝑀 = ∅. Since 𝐴 is a compact subset, 𝐴−1𝑎 is compact. Therefore, by theorem 3.1, there is an open neighbourhood 𝑉 of e in 𝐺ℜ such that 𝐴−1𝑎𝑉 ∩ 𝑀 = ∅ which implies 𝑎𝑉 ∩ 𝐴𝑀 = ∅ and 𝑎𝑉 is an open neighbourhood of 𝑎 in the complement of 𝐴𝑀. Hence 𝐴𝑀 is a closed subset in 𝐺ℜ ̅̅ ̅̅ . Similarly, 𝑀𝐴 is a closed subset in 𝐺ℜ ̅̅ ̅̅ . Theorem 3.3. Let 𝐺ℜ be a topological simple rough group such that 𝐺ℜ is open in 𝐺ℜ ̅̅ ̅̅ . Suppose 𝐴 is a compact subset of 𝐺ℜ ̅̅ ̅̅ . Then there exists an identity neighbourhood 𝑉 ⊆ 𝐺ℜ ̅̅ ̅̅ such that 𝑎𝑉𝑎−1 ⊆ 𝑊, for every open neighbourhood 𝑊 of 𝑒 in 𝐺ℜ ̅̅ ̅̅ and 𝑎 ∈ 𝐴. Proof: Let 𝑊 be a neighbourhood of 𝑒 in 𝐺ℜ ̅̅ ̅̅ . From lemma 2.10, there is an open set 𝑈 of 𝑒 in 𝐺ℜ such that 𝑈 ⊆ 𝑈𝑛 ⊆ 𝑊, for every 𝑛 ∈ ℕ − {0}. Since 𝐴 is compact, there exists an open cover 𝐴 ⊆ 𝑈𝑀 such that 𝑀 is a finite subset of 𝐴. Consider 𝑉 = ⋂ 𝑥−1𝑈𝑥𝑥∈𝑀 . Then e ∈ V is open in 𝐺ℜ. Also, by theorem 2.11, 𝑉 is open in 𝐺ℜ ̅̅ ̅̅ . Now we choose an element 𝑎 ∈ 𝐴 such that 𝑎 = 𝑢𝑥, for some 𝑢 ∈ 𝑈 and 𝑥 ∈ 𝑀. Hence, 𝑎𝑉𝑎−1 = 𝑢𝑥𝑉𝑥−1𝑢−1 ⊆ 𝑢𝑈𝑢−1 ⊆ 𝑈3 ⊆ 𝑊, for any 𝑎 ∈ 𝐴. Theorem 3.4. Let 𝐺ℜ be a topological simple rough group such that 𝐺ℜ ̅̅ ̅̅ is a group and 𝐻ℜ is a subgroup of 𝐺ℜ ̅̅ ̅̅ . If 𝐻ℜ is open in 𝐺ℜ ̅̅ ̅̅ , then 𝐻ℜ is closed in 𝐺ℜ ̅̅ ̅̅ . Proof: The rough quotient space 𝐺ℜ ̅̅ ̅̅ 𝐻ℜ⁄ = {𝑎𝐻ℜ ∶ 𝑎 ∈ 𝐺ℜ ̅̅ ̅̅ }. It is a disjoint open cover of 𝐺ℜ ̅̅ ̅̅ . Since 𝐻ℜ is open, 𝑎𝐻ℜ is also open. Therefore, the complement of 𝐻ℜ = ⋃ 𝑎𝐻ℜ𝑎∉𝐻ℜ is open in 𝐺ℜ ̅̅ ̅̅ . Hence 𝐻ℜ is closed in 𝐺ℜ ̅̅ ̅̅ . Theorem 3.5. Let 𝐺ℜ be a topological simple rough group such that 𝐺ℜ ̅̅ ̅̅ is a group and 𝐺ℜ is open in 𝐺ℜ ̅̅ ̅̅ . If 𝐴 is a compact open neighbourhood of 𝑒 in 𝐺ℜ ̅̅ ̅̅ , there is a compact subgroup 𝐻ℜ of 𝐺ℜ ̅̅ ̅̅ such that 𝐻ℜ ⊆ 𝐴. Proof: Since 𝐴 ⊆ 𝐺ℜ ̅̅ ̅̅ is an open neighbourhood of 𝑒, there exists a symmetric open neighbourhood 𝑉 of 𝑒 in 𝐺ℜ such that 𝑉𝑉 ⊆ 𝐴 and by lemma 2.10, 𝑉 ⊆ 𝑉𝑛 ⊆ 𝐴, for every 𝑛 ∈ ℕ − {0}. Now consider 𝐻ℜ = ⋃ 𝑉𝑛 𝑛∈ℕ−{0} . Then 𝐻ℜ is open in 𝐺ℜ and 𝐻ℜ ⊆ 𝐴. Since 𝐺ℜ is open in 𝐺ℜ ̅̅ ̅̅ , 𝐻ℜ is open in 𝐺ℜ ̅̅ ̅̅ . Let us prove 𝐻ℜ is a subgroup of 𝐺ℜ ̅̅ ̅̅ . Let 𝑎, 𝑏 ∈ 𝐻ℜ. Then 𝑎 ∈ 𝑉𝑛 and 𝑏 ∈ 𝑉𝑚, for some 𝑛, 𝑚 ∈ ℕ − {0} which implies 𝑎𝑏 ∈ 𝑉𝑛+𝑚 ∈ 𝐻ℜ. Also, 𝑎 ∈ 𝐻ℜ implies 𝑎 ∈ 𝑉𝑛 and 𝑎−1 ∈ (𝑉𝑛)−1 = (𝑉−1)𝑛 = 𝑉𝑛 ∈ 𝐻ℜ. Therefore, 𝐻ℜ is a subgroup of 𝐺ℜ ̅̅ ̅̅ . Applying theorem 3.4, 𝐻ℜ is closed in 𝐺ℜ ̅̅ ̅̅ . Hence 𝐻ℜ is a compact subgroup of 𝐺ℜ ̅̅ ̅̅ . Proposition 3.6. Let 𝐺ℜ be a topological simple rough group such that 𝐺ℜ ̅̅ ̅̅ is a group and 𝐻ℜ be a locally compact subgroup of a Hausdorff topological group 𝐺ℜ ̅̅ ̅̅ . Then 𝐻ℜ is closed in 𝐺ℜ ̅̅ ̅̅ . Proof: Since 𝐻ℜ be a locally compact subgroup of a Hausdorff topological group 𝐺ℜ ̅̅ ̅̅ , 𝐻ℜ is open in 𝑐𝑙(𝐻ℜ), closure of 𝐻ℜ and 𝑐𝑙(𝐻ℜ) is a topological rough subgroup in 𝐺ℜ ̅̅ ̅̅ . Therefore, by theorem 3.4, 𝐻ℜ is closed in 𝐺ℜ ̅̅ ̅̅ . Theorem 3.7. Let 𝐺ℜ be a topological simple rough group such that 𝐺ℜ ̅̅ ̅̅ is a Hausdorff topological group and 𝐻ℜ is a compact subgroup 𝐺ℜ ̅̅ ̅̅ . Then the rough quotient mapping 𝜑: 𝐺ℜ ̅̅ ̅̅ → 𝐺ℜ ̅̅ ̅̅ 𝐻ℜ⁄ is perfect. Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 3 (2025) 883 https://internationalpubls.com Proof: Let 𝑀 be a closed subset of 𝐺ℜ ̅̅ ̅̅ . Then by the second closure lemma, 𝑀𝐻ℜ is closed in 𝐺ℜ ̅̅ ̅̅ . That is, 𝜑(𝑀) is closed in rough quotient space 𝐺ℜ ̅̅ ̅̅ 𝐻ℜ⁄ . Therefore, the rough quotient mapping 𝜑 is closed. Let 𝑏 ∈ 𝐺ℜ ̅̅ ̅̅ 𝐻ℜ⁄ and 𝑎 ∈ 𝐺ℜ ̅̅ ̅̅ such that 𝜑(𝑎) = 𝑏. Then 𝜑−1(𝑏) = 𝑎𝐻ℜ, is a compact subset of 𝐺ℜ ̅̅ ̅̅ . Hence, by the definition of 2.14, the rough quotient mapping 𝜑 is perfect. Theorem 3.8. Suppose 𝑀 is a compact subset of a topological simple rough group 𝐺ℜ such that 𝐺ℜ ̅̅ ̅̅ is a group. Then there exists a smallest rough subgroup 𝐻ℜ in 𝐺ℜ ̅̅ ̅̅ containing M such that 𝐻ℜ is 𝜎- compact. Proof: Let 𝐴 = 𝑀 ∪ {e} ∪ 𝑀−1. Since 𝑀 is a compact subset of a topological simple rough group 𝐺ℜ, 𝐴 is compact in 𝐺ℜ. Now define the multiplication mapping 𝑓𝑖: 𝐺ℜ 𝑖 → 𝐺ℜ ̅̅ ̅̅ by 𝑓𝑖(𝑥1, 𝑥2, … , 𝑥𝑖) = 𝑥1𝑥2 … 𝑥𝑖, for 𝑥1, 𝑥2, … , 𝑥𝑖 ∈ 𝐺ℜ and for every 𝑖 ∈ ℕ. Since 𝐺ℜ is a topological simple rough group and continuous image of compact set is compact, the mappings 𝑓𝑖 are continuous which implies 𝑓𝑖(𝐴𝑖) is compact in 𝐺ℜ ̅̅ ̅̅ , for every 𝑖 ∈ ℕ. Therefore, the rough subgroup 𝐻ℜ = ⋃ 𝑓𝑖(𝐴𝑖) 𝑛 𝑖=1 is generated by 𝑀. Hence 𝐻ℜ is 𝜎-compact. Theorem 3.9. (Open Mapping Theorem I) Let 𝐺ℜ and 𝐻ℜ be topological simple rough groups such that 𝐺ℜ and 𝐻ℜ are open in 𝐺ℜ ̅̅ ̅̅ and 𝐻ℜ ̅̅ ̅̅ . Let 𝜋: 𝐺ℜ ̅̅ ̅̅ → 𝐻ℜ ̅̅ ̅̅ be a surjective mapping topological rough group homomorphism. If 𝐺ℜ ̅̅ ̅̅ is a compact space and 𝐻ℜ ̅̅ ̅̅ is a Hausdorff space, then the mapping 𝜋 is open. Proof: Since 𝐺ℜ ̅̅ ̅̅ is a compact space and 𝐻ℜ ̅̅ ̅̅ is a Hausdorff space, 𝜋 is closed. Also, by the continuity of 𝜋, the mapping 𝜋 is a quotient map that means a subset 𝑈 ⊆ 𝐻ℜ ̅̅ ̅̅ is open if and only if 𝜋−1(𝑈) is open in 𝐺ℜ ̅̅ ̅̅ . Let us prove 𝜋 is an open mapping. Let 𝑉 be an open set in 𝐺ℜ ̅̅ ̅̅ . Then 𝜋−1(𝜋(𝑉)) = 𝒦𝜋𝑉 is open in 𝐺ℜ ̅̅ ̅̅ , where 𝒦𝜋 is the rough kernel of 𝜋. Now consider 𝑈 = 𝜋(𝑉) that implies 𝜋−1(𝑈) is open in 𝐺ℜ ̅̅ ̅̅ . Since 𝜋 is a quotient map, 𝑈 = 𝜋(𝑉) is open. Hence the mapping 𝜋 is open. Proposition 3.10. Let 𝐺ℜ be a topological simple rough group and 𝐺ℜ be an open set in 𝐺ℜ ̅̅ ̅̅ . Suppose the subset 𝐴 ⊆ 𝐺ℜ and 𝑖𝑛𝑡(𝑥𝐴 ∩ 𝐺ℜ) ≠ ∅. Then 𝑖𝑛𝑡(𝐴) ≠ ∅, where 𝑖𝑛𝑡 means interior of the set. Proof: Consider an arbitrary element 𝑏 ∈ 𝑖𝑛𝑡(𝑥𝐴 ∩ 𝐺ℜ). Then there is a point 𝑎 ∈ 𝐴 such that 𝑏 = 𝑥𝑎. Since 𝐺ℜ is a topological simple rough group and 𝐺ℜ is an open set in 𝐺ℜ ̅̅ ̅̅ , there is a neighbourhood 𝑈 of 𝑎 in 𝐺ℜ such that 𝑥𝑈 ⊆ 𝑖𝑛𝑡(𝑥𝐴 ∩ 𝐺ℜ) which implies 𝑥𝑈 ⊆ 𝑥𝐴 that is, 𝑈 ⊆ 𝐴. Hence, 𝑖𝑛𝑡(𝐴) ≠ ∅. Theorem 3.11. (Open Mapping Theorem II) Let 𝐺ℜ and 𝐻ℜ be two locally compact Hausdorff topological simple rough groups such that 𝐺ℜ and 𝐻ℜ are open in 𝐺ℜ ̅̅ ̅̅ and 𝐻ℜ ̅̅ ̅̅ . Suppose a surjective mapping 𝜋: 𝐺ℜ → 𝐻ℜ is a continuous rough homomorphism and 𝐺ℜ is a 𝜎-compact space. Then 𝜋 is an open mapping. Proof: Let 𝑈 ⊆ 𝐺ℜ be a symmetric identity neighbourhood. Since 𝐺ℜ is locally compact, there exists a symmetric open neighbourhood 𝑁 of e in 𝐺ℜ such that 𝑐𝑙(𝑁) is compact and 𝑐𝑙(𝑁)𝑐𝑙(𝑁) ⊆ 𝑈, where 𝑐𝑙(𝑁) is the closure of 𝑁. Since 𝑁 is open, 𝑥𝑁 is open, for every 𝑥 ∈ 𝐺ℜ. So, ⋃ 𝑥𝑁𝑥∈𝐺ℜ covers 𝐺ℜ. Therefore, 𝐺ℜ = 𝐺ℜ ∩ ⋃ 𝑥𝑁𝑥∈𝐺ℜ . Since 𝐺ℜ is a 𝜎-compact space, there exists a countable set {𝑥𝑖}𝑖∈ℕ, such that 𝐺ℜ = 𝐺ℜ ∩ ⋃ 𝑥𝑖𝑁𝑖∈ℕ . Since 𝑐𝑙(𝑁) is compact and 𝜋 is continuous, 𝜋(𝐺ℜ) = 𝜋(𝐺ℜ) ∩ ⋃ 𝜋(𝑥𝑖𝑐𝑙(𝑁𝑖∈ℕ )), for every 𝑥𝑖 ∈ 𝐺ℜ implies 𝐻ℜ = 𝐻ℜ ∩ ⋃ 𝜋(𝑥𝑖) 𝜋(𝑐𝑙(𝑁𝑖∈ℕ )) = 𝐻ℜ ∩ Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 3 (2025) 884 https://internationalpubls.com ⋃ 𝑦𝑖 𝜋(𝑐𝑙(𝑁𝑖∈ℕ )), where 𝑦𝑖 = 𝜋(𝑥𝑖) ∈ 𝐻ℜ. Thus, 𝑦𝑖 𝜋(𝑐𝑙(𝑁)) is closed in 𝐻ℜ ̅̅ ̅̅ , for every 𝑖 ∈ ℕ. Therefore, 𝐻ℜ ∩ 𝑦𝑖 𝜋(𝑐𝑙(𝑁)) is closed in 𝐻ℜ. Since 𝐻ℜ is locally compact, 𝑖𝑛𝑡(𝐻ℜ ∩ 𝑦𝑖 𝜋(𝑐𝑙(𝑁))) ≠ ∅, for every 𝑖 ∈ ℕ. By proposition 3.10, we get 𝑖𝑛𝑡(𝜋(𝑐𝑙(𝑁))) ≠ ∅. Then there exists an open set 𝑉 ⊆ 𝐻ℜ such that 𝑉 ⊆ 𝜋(𝑐𝑙(𝑁)). Let 𝑣 ∈ 𝑉. Then there exists a point 𝑛 ∈ 𝑐𝑙(𝑁) such that 𝜋(𝑛) = 𝑣. Therefore, 𝑒′ ∈ 𝑣−1𝑉 ⊆ 𝑣−1𝜋(𝑐𝑙(𝑁)) = 𝜋(𝑛−1)𝜋(𝑐𝑙(𝑁)) ⊆ 𝜋(𝑛−1𝑐𝑙(𝑁)) ⊆ 𝜋(𝑐𝑙(𝑁)𝑐𝑙(𝑁)) ⊆ 𝜋(𝑈). Hence 𝜋 is an open mapping. 4. Topological rough group homomorphism: Proposition 4.1. Let 𝐺ℜ and 𝐻ℜ be topological simple rough groups such that 𝐺ℜ and 𝐻ℜ are open in 𝐺ℜ ̅̅ ̅̅ and 𝐻ℜ ̅̅ ̅̅ . Let the map 𝑓: 𝐺ℜ ̅̅ ̅̅ → 𝐻ℜ ̅̅ ̅̅ be a continuous topological rough group homomorphism. Suppose for every open neighbourhood 𝑁 of 𝑒 in 𝐺ℜ ̅̅ ̅̅ , 𝑓(𝑁) has a non-empty open set in 𝐻ℜ ̅̅ ̅̅ . Then 𝑓 is an open mapping. Proof: Let 𝑈 be an open neighbourhood of 𝑒 in 𝐺ℜ ̅̅ ̅̅ such that 𝑈−1𝑈 ⊆ 𝑁. But by the hypothesis, 𝑓(𝑈) has a non-empty open set in 𝐻ℜ ̅̅ ̅̅ . Consider that open set 𝑉 in 𝐻ℜ ̅̅ ̅̅ . So, 𝑉−1𝑉 is an identity neighbourhood. Therefore, 𝑉−1𝑉 ⊆ 𝑓(𝑈)−1𝑓(𝑈) = 𝑓(𝑈−1𝑈) ⊆ 𝑓(𝑁) which implies 𝑓(𝑁) has an identity in 𝐻ℜ ̅̅ ̅̅ . Let 𝑏 ∈ 𝑓(𝑁). Since 𝑓 is a continuous topological rough group homomorphism, there exists an arbitrary element 𝑎 ∈ N such that 𝑓(𝑎) = 𝑏 and 𝑎𝑈 ⊆ 𝑁. Also, 𝑉 ⊆ 𝑓(𝑈) and 𝑏𝑉 is an open neighbourhood in 𝐻ℜ ̅̅ ̅̅ . Then 𝑏𝑉 ⊆ 𝑓(𝑎𝑈) ⊆ 𝑓(𝑁). Hence the map 𝑓 is open. Proposition 4.2. (i) Let 𝐺ℜ, 𝐻ℜ and 𝐾ℜ be simple rough groups. Let 𝑓: 𝐺ℜ ̅̅ ̅̅ → 𝐻ℜ ̅̅ ̅̅ and 𝑔: 𝐺ℜ ̅̅ ̅̅ → 𝐾ℜ ̅̅ ̅̅ be rough group homomorphisms, where 𝑔(𝐺ℜ ̅̅ ̅̅ ) = 𝐾ℜ ̅̅ ̅̅ and 𝒦𝑓 ⊆ 𝒦𝑔, where 𝒦𝑓 and 𝒦𝑔 represent the rough kernel of 𝑓 and 𝑔. Then ℎ: 𝐾ℜ ̅̅ ̅̅ → 𝐻ℜ ̅̅ ̅̅ is a rough group homomorphism such that 𝑓 = ℎ ∘ 𝑔. (ii) Let 𝐺ℜ, 𝐻ℜ and 𝐾ℜ be topological simple rough groups such that 𝐺ℜ, 𝐻ℜ and 𝐾ℜ are open in 𝐺ℜ ̅̅ ̅̅ , 𝐻ℜ ̅̅ ̅̅ and 𝐾ℜ ̅̅ ̅̅ . Suppose 𝑔−1(𝑈) ⊆ 𝑓−1(𝑉), for every identity neighbourhood 𝑈 in 𝐾ℜ ̅̅ ̅̅ , there exists an identity neighbourhood 𝑉 in 𝐻ℜ ̅̅ ̅̅ , then ℎ: 𝐾ℜ ̅̅ ̅̅ → 𝐻ℜ ̅̅ ̅̅ is continuous. Proof: (i) Since 𝑓 and 𝑔 are rough group homomorphism and 𝑓 = ℎ ∘ 𝑔, ℎ is a rough group homomorphism. (ii) Let 𝑈 be an identity neighbourhood in 𝐻ℜ ̅̅ ̅̅ . By the hypothesis, there exists an identity neighbourhood 𝑉 in 𝐻ℜ ̅̅ ̅̅ such that 𝑔−1(𝑉) ⊆ 𝑓−1(𝑈). Consider 𝑁 = 𝑔−1(𝑉). Therefore, 𝑓(𝑁) ⊆ 𝑈 and ℎ(𝑉) = 𝑓(𝑁) which implies ℎ(𝑉) ⊆ 𝑈. Hence ℎ is continuous at 𝑒 in 𝐾ℜ ̅̅ ̅̅ , which implies ℎ is continuous. Corollary 4.3. Let 𝐺ℜ, 𝐻ℜ and 𝐾ℜ be topological simple rough groups, where 𝐺ℜ, 𝐻ℜ and 𝐾ℜ are open in 𝐺ℜ ̅̅ ̅̅ , 𝐻ℜ ̅̅ ̅̅ and 𝐾ℜ ̅̅ ̅̅ . Let 𝑓: 𝐺ℜ ̅̅ ̅̅ → 𝐻ℜ ̅̅ ̅̅ and 𝑔: 𝐺ℜ ̅̅ ̅̅ → 𝐾ℜ ̅̅ ̅̅ be continuous rough group homomorphisms such that 𝑔(𝐺ℜ ̅̅ ̅̅ ) = 𝐾ℜ ̅̅ ̅̅ and 𝒦𝑓 ⊆ 𝒦𝑔, where 𝒦𝑓 and 𝒦𝑔 represent the rough kernel of 𝑓 and 𝑔. Suppose 𝑔 is open, then there exists a continuous rough group homomorphism ℎ: 𝐾ℜ ̅̅ ̅̅ → 𝐻ℜ ̅̅ ̅̅ such that 𝑓 = ℎ ∘ 𝑔. Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 3 (2025) 885 https://internationalpubls.com Proof: By proposition 4.2 (i), there exists a rough group homomorphism ℎ: 𝐾ℜ ̅̅ ̅̅ → 𝐻ℜ ̅̅ ̅̅ such that 𝑓 = ℎ ∘ 𝑔. Let 𝑈 be an open set in 𝐻ℜ ̅̅ ̅̅ . Then ℎ−1(𝑈) = 𝑔(𝑓−1(𝑈)). Since 𝑓 is continuous and 𝑔 is open, ℎ−1(𝑈) is open. Therefore, ℎ is continuous and hence ℎ is continuous rough group homomorphism. Proposition 4.4. Let 𝐺ℜ and 𝐻ℜ be topological simple rough groups such that 𝐺ℜ ̅̅ ̅̅ and 𝐾ℜ ̅̅ ̅̅ are groups. Let ℒ be a subgroup of 𝐺ℜ and normal subgroup of 𝐺ℜ ̅̅ ̅̅ . Let ℳ = 𝜌(ℒ) be a subgroup of 𝐻ℜ and normal subgroup of 𝐻ℜ ̅̅ ̅̅ . Suppose 𝜌: 𝐺ℜ ̅̅ ̅̅ → 𝐻ℜ ̅̅ ̅̅ is a rough homeomorphism. Then the quotient map 𝛾: 𝐺ℜ ̅̅ ̅̅ ℒ⁄ → 𝐻ℜ ̅̅ ̅̅ ℳ⁄ is a topological rough group homeomorphism which is defined by 𝛾(𝑎ℒ) = 𝑏ℳ, for 𝑎 ∈ 𝐺ℜ ̅̅ ̅̅ and 𝑏 = 𝜌(𝑎). Proof: Consider the quotient maps 𝜇: 𝐺ℜ ̅̅ ̅̅ → 𝐺ℜ ̅̅ ̅̅ ℒ⁄ and 𝜇′: 𝐻ℜ ̅̅ ̅̅ → 𝐻ℜ ̅̅ ̅̅ ℳ⁄ . Then 𝜇′ ∘ 𝜌 = 𝛾 ∘ 𝜇. Since the maps 𝜇′, 𝜌 and 𝜇 are open continuous rough homomorphisms, the map 𝛾 is an open continuous homomorphism. Let 𝑎ℒ ∈ 𝐺ℜ ̅̅ ̅̅ ℒ⁄ and 𝛾(𝑎ℒ) = ℳ. Then 𝜇′(𝑏) = ℳ and 𝑏ℳ = ℳ which implies 𝑏 = 𝜌(𝑎) ∈ ℳ. Since 𝜌 is a rough homeomorphism and ℳ = 𝜌(ℒ), 𝜌(𝑎) = 𝜌(𝑐), for some 𝑐 ∈ ℒ which implies 𝑎 = 𝑐. Therefore, 𝑎 ∈ ℒ implies the kernel of 𝛾 is ℒ that mean the quotient map 𝛾 is injective. Hence the quotient map 𝛾 is a topological rough group homeomorphism. 5. Rough isomorphism: Theorem 5.1. (Rough Isomorphism Theorem - I) Let 𝐺ℜ and 𝐻ℜ be topological simple rough groups such that 𝐺ℜ ̅̅ ̅̅ is a group. Let 𝜌: 𝐺ℜ ̅̅ ̅̅ → 𝐻ℜ ̅̅ ̅̅ be a topological rough group homomorphism and 𝒦𝜌 be the rough kernel of 𝜌. Then the map 𝜑: 𝐺ℜ ̅̅ ̅̅ 𝒦𝜌⁄ → 𝐻ℜ ̅̅ ̅̅ is a continuous rough isomorphism which is defined by φ(𝑎𝒦𝜌) = 𝜌(𝑎), for every 𝑎 ∈ 𝐺ℜ ̅̅ ̅̅ . Also, if 𝜌 is open, the map 𝜑 is a rough homeomorphism. Proof: Let 𝜇: 𝐺ℜ ̅̅ ̅̅ → 𝐺ℜ ̅̅ ̅̅ 𝒦𝜌⁄ be a quotient map. Then 𝜌 = φ ∘ 𝜇. (i) 𝜑 is injective: Since 𝜌 is a topological rough group homomorphism and 𝒦𝜌 be the rough kernel of 𝜌, for some 𝑎, 𝑏 ∈ 𝐺ℜ ̅̅ ̅̅ , 𝑎𝒦𝜌 = 𝑏𝒦𝜌 which implies 𝑎𝑏−1 ∈ 𝒦𝜌. Then 𝜌(𝑎)𝜌(𝑏)−1 = 𝜌(𝑎𝑏−1) = 𝑒′, where 𝑒′ is the identity in 𝐻ℜ ̅̅ ̅̅ . Therefore, 𝜌(𝑎) = 𝜌(𝑏) implies that 𝜑 is one-one. (ii) 𝜑 is surjective: Let 𝑥 ∈ 𝐻ℜ ̅̅ ̅̅ . Then there exists an element 𝑎 ∈ 𝐺ℜ ̅̅ ̅̅ such that 𝜌(𝑎) = 𝑥. Since φ(𝑎𝒦𝜌) = 𝜌(𝑎), for every element of 𝐻ℜ ̅̅ ̅̅ is in the image of 𝜑. Hence 𝜑 is onto. (iii) 𝜑 is rough homomorphism: Let 𝑎, 𝑏 ∈ 𝐺ℜ ̅̅ ̅̅ . Then 𝑎𝒦𝜌, 𝑏𝒦𝜌 ∈ 𝐺ℜ ̅̅ ̅̅ 𝒦𝜌⁄ . Since 𝜌 is a topological rough group homomorphism, φ(𝑎𝑏𝒦𝜌) = 𝜌(𝑎𝑏) = 𝜌(𝑎)𝜌(𝑏) = φ(𝑎𝒦𝜌)φ(𝑏𝒦𝜌). Therefore, 𝜑 is a rough homomorphism. Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 3 (2025) 886 https://internationalpubls.com (iv) 𝜑 is continuous: Since 𝜌 is continuous and 𝜇 is a quotient map, φ = 𝜌 ∘ 𝜇−1 is continuous. Hence the map 𝜑: 𝐺ℜ ̅̅ ̅̅ 𝒦𝜌⁄ → 𝐻ℜ ̅̅ ̅̅ is a continuous rough isomorphism. (v) 𝜑 is a rough homeomorphism if 𝜌 is open: Let 𝑈 be an open neighbourhood in 𝐺ℜ ̅̅ ̅̅ . Since 𝜌 is open, 𝜌(𝑈) is open. Then φ(U) = 𝜌(𝜇−1(𝑈)) is open. Therefore, φ−1 is continuous. Hence 𝜑 is a rough homeomorphism. Theorem 5.2. (Rough Isomorphism Theorem - II) Let 𝐺ℜ and 𝐻ℜ be topological simple rough groups. Let 𝜌: 𝐺ℜ ̅̅ ̅̅ → 𝐻ℜ ̅̅ ̅̅ be a topological rough group homomorphism such that 𝐺ℜ ̅̅ ̅̅ and 𝐻ℜ ̅̅ ̅̅ are groups and ℒ be a normal subgroup of 𝐻ℜ ̅̅ ̅̅ . Define ℳ = 𝜌−1(ℒ) and 𝒦𝜌 be the rough kernel of 𝜌. Then the map 𝜑: (𝐺ℜ ̅̅ ̅̅ 𝒦𝜌⁄ )/(ℳ 𝒦𝜌⁄ ) → 𝐻ℜ ̅̅ ̅̅ ℒ⁄ is a topological rough group homeomorphism. Proof: Consider the rough quotient map 𝜇: 𝐻ℜ ̅̅ ̅̅ → 𝐻ℜ ̅̅ ̅̅ ℒ⁄ which is an open continuous rough homomorphism. Then 𝜇 ∘ 𝜌 ∶ 𝐺ℜ ̅̅ ̅̅ → 𝐻ℜ ̅̅ ̅̅ ℒ⁄ is also a continuous open rough homomorphism. Let 𝜌′ = 𝜇 ∘ 𝜌. Then the rough kernel of 𝜌′, 𝒦𝜌′ = {𝑎 ∈ 𝐺ℜ ̅̅ ̅̅ ∶ 𝜌′(𝑎) = ℒ}. Since 𝜌′(𝑎) = 𝜇(𝜌(𝑎)) = 𝜌(𝑎)ℒ, 𝒦𝜌′ = {𝑎 ∈ 𝐺ℜ ̅̅ ̅̅ ∶ 𝜌(𝑎) ∈ ℒ}. But ℳ = 𝜌−1(ℒ) = {𝑎 ∈ 𝐺ℜ ̅̅ ̅̅ ∶ 𝜌(𝑎) ∈ ℒ}. Therefore, 𝒦𝜌′ = ℳ. By theorem 5.1, 𝐺ℜ ̅̅ ̅̅ ℳ⁄ is topological rough group homeomorphism to 𝐻ℜ ̅̅ ̅̅ ℒ⁄ . In the similar way, define the map 𝜑: 𝐺ℜ ̅̅ ̅̅ 𝒦𝜌⁄ → 𝐻ℜ ̅̅ ̅̅ ℒ⁄ by 𝜑(𝑎𝒦𝜌) = 𝜌(𝑎)ℒ. Then the rough kernel of 𝜑, 𝒦𝜑 = {𝑎𝒦𝜌 ∈ 𝐺ℜ ̅̅ ̅̅ 𝒦𝜌⁄ ∶ 𝜑(𝑎𝒦𝜌) = ℒ} which implies 𝒦𝜑 = {𝑎𝒦𝜌 ∈ 𝐺ℜ ̅̅ ̅̅ 𝒦𝜌⁄ ∶ 𝜌(𝑎) ∈ ℒ}. But ℳ = 𝜌−1(ℒ) = {𝑎 ∈ 𝐺ℜ ̅̅ ̅̅ ∶ 𝜌(𝑎) ∈ ℒ}. That is, the element 𝑎 ∈ ℳ mapped to the identity ℒ in 𝐻ℜ ̅̅ ̅̅ ℒ⁄ . Therefore, the element 𝑎𝒦𝜌 ∈ ℳ 𝒦𝜌⁄ mapped to the identity ℒ in 𝐻ℜ ̅̅ ̅̅ ℒ⁄ which implies the rough kernel of 𝜑, 𝒦𝜑 = ℳ 𝒦𝜌⁄ . By theorem 5.1, (𝐺ℜ ̅̅ ̅̅ 𝒦𝜌⁄ )/(ℳ 𝒦𝜌⁄ ) is topological rough group homeomorphism to 𝐻ℜ ̅̅ ̅̅ ℒ⁄ . Theorem 5.3. (Rough Isomorphism Theorem - III) Let 𝐺ℜ be a topological simple rough group and let ℒ be a normal rough subgroup of 𝐺ℜ ̅̅ ̅̅ . For any topological rough subgroup 𝐻ℜ of 𝐺ℜ, if 𝐺ℜ ̅̅ ̅̅ , 𝐻ℜ ̅̅ ̅̅ and ℒ̅ are groups, where ℒ̅ is the upper approximation of ℒ, then the rough quotient map 𝛾: 𝐻ℜ ̅̅ ̅̅ ℒ ℒ⁄ → 𝜑(𝐻ℜ ̅̅ ̅̅ ) is a topological rough group homeomorphism, where 𝜑: 𝐺ℜ ̅̅ ̅̅ → 𝐺ℜ ̅̅ ̅̅ ℒ⁄ is a rough quotient map and 𝜑(𝐻ℜ ̅̅ ̅̅ ) is a subgroup of 𝐺ℜ ̅̅ ̅̅ ℒ⁄ . Proof: By our assumption, 𝐻ℜ ̅̅ ̅̅ ℒ = 𝜑−1(𝜑(𝐻ℜ ̅̅ ̅̅ )). Let 𝜇: 𝐻ℜ ̅̅ ̅̅ ℒ → 𝜑(𝐻ℜ ̅̅ ̅̅ ) defined by 𝜇(𝑎ℒ) = 𝜑(𝑎). Since 𝜑 is homomorphism, the rough quotient map 𝜇 is homomorphism. Then the rough kernel of 𝜇, 𝒦𝜇 = {𝑎 ∈ 𝐻ℜ ̅̅ ̅̅ ℒ ∶ 𝜇(𝑎ℒ) = 𝑒, is the identity of 𝜑(𝐻ℜ ̅̅ ̅̅ )}. But the identity of 𝐺ℜ ̅̅ ̅̅ ℒ⁄ is ℒ and 𝜑(𝐻ℜ ̅̅ ̅̅ ) is a subgroup of 𝐺ℜ ̅̅ ̅̅ ℒ⁄ . So, 𝒦𝜇 = {𝑎ℒ ∈ 𝐻ℜ ̅̅ ̅̅ ℒ ∶ 𝜑(𝑎) = ℒ} = ℒ. Therefore, by the theorem 5.1, the rough quotient map 𝛾: 𝐻ℜ ̅̅ ̅̅ ℒ ℒ⁄ → 𝜑(𝐻ℜ ̅̅ ̅̅ ) is a topological rough group homeomorphism. 6. Rough double coset spaces: Definition 6.1. Let 𝐺ℜ be a rough group such that 𝐺ℜ ̅̅ ̅̅ is a group and 𝐻ℜ, 𝐾ℜ be rough subgroups in 𝐺ℜ. If 𝐻ℜ, 𝐾ℜ be subgroups in 𝐺ℜ ̅̅ ̅̅ , then 𝐾ℜ\𝐺ℜ ̅̅ ̅̅ /𝐻ℜ = {𝐾ℜ𝑥𝐻ℜ ∶ 𝑥 ∈ 𝐺ℜ ̅̅ ̅̅ } is a rough double coset space (𝒟ℭ). Also, for any 𝑥, 𝑦 ∈ 𝐺ℜ ̅̅ ̅̅ , either 𝐾ℜ𝑥𝐻ℜ = 𝐾ℜ𝑦𝐻ℜ or 𝐾ℜ𝑥𝐻ℜ ∩ 𝐾ℜ𝑦𝐻ℜ = ∅. Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 3 (2025) 887 https://internationalpubls.com Lemma 6.2. Let 𝐺ℜ be a rough group such that 𝐺ℜ ̅̅ ̅̅ is a group and 𝐻ℜ, 𝐾ℜ be rough subgroups in 𝐺ℜ. If 𝐻ℜ, 𝐾ℜ be subgroups in 𝐺ℜ ̅̅ ̅̅ , then all the double cosets form a partition of 𝐺ℜ ̅̅ ̅̅ . Proof: Let 𝑥, 𝑦 ∈ 𝐺ℜ ̅̅ ̅̅ . Then the corresponding cosets are 𝐾ℜ𝑥𝐻ℜ, 𝐾ℜ𝑦𝐻ℜ. These are either disjoint or coincide. Consider an arbitrary element 𝑧 ∈ 𝐾ℜ𝑥𝐻ℜ ∩ 𝐾ℜ𝑦𝐻ℜ, that is, 𝑧 = 𝑘1𝑥ℎ1 = 𝑘2𝑦ℎ2, for some 𝑘1, 𝑘2 ∈ 𝐾ℜ and ℎ1, ℎ2 ∈ 𝐻ℜ. Therefore, 𝑥 ∈ 𝐾ℜ𝑦𝐻ℜ which implies 𝐾ℜ𝑥𝐻ℜ ⊆ 𝐾ℜ𝑦𝐻ℜ. Similarly, we can prove 𝐾ℜ𝑦𝐻ℜ ⊆ 𝐾ℜ𝑥𝐻ℜ. Hence, 𝐾ℜ𝑥𝐻ℜ = 𝐾ℜ𝑦𝐻ℜ. Lemma 6.3. Let 𝐻ℜ and 𝐾ℜ be rough subgroups of a topological simple rough group 𝐺ℜ such that 𝐺ℜ ̅̅ ̅̅ is a group and 𝐻ℜ, 𝐾ℜ are subgroups of 𝐺ℜ ̅̅ ̅̅ . If 𝐴 is a compact subset in 𝒟ℭ and the rough quotient map 𝜑: 𝐺ℜ ̅̅ ̅̅ → 𝒟ℭ is defined by 𝜑(𝑥) = 𝐾ℜ𝑥𝐻ℜ, for 𝑥 ∈ 𝐺ℜ ̅̅ ̅̅ , then there exists a compact subset 𝐵 in 𝐺ℜ ̅̅ ̅̅ such that 𝜑(𝐵) = 𝐴. Proof: Let 𝑈 be an identity neighbourhood in 𝐺ℜ ̅̅ ̅̅ which has a compact closure. Since 𝐴 is compact, there exists a cover ⋃ 𝜑(𝑥𝑖𝑈)𝑛 𝑖=1 , where 𝑥1, 𝑥2, … 𝑥𝑛 ∈ 𝐺ℜ ̅̅ ̅̅ . That is, 𝐴 ⊆ ⋃ 𝜑(𝑥𝑖𝑈)𝑛 𝑖=1 . Now consider 𝐵 = 𝜑−1(𝐴) ∩ ⋃ 𝑥𝑖𝑐𝑙(𝑈)𝑛 𝑖=1 , 𝑐𝑙(𝑈) means closure of 𝑈. Since 𝑐𝑙(𝑈) is compact, 𝜑−1(𝐴) lies in the compact set ⋃ 𝑥𝑖𝑐𝑙(𝑈)𝑛 𝑖=1 . Therefore, 𝐵 is compact and 𝜑(𝐵) = 𝜑(𝜑−1(𝐴) ∩ ⋃ 𝑥𝑖𝑐𝑙(𝑈)𝑛 𝑖=1 ) = 𝐴 ∩ ⋃ 𝜑(𝑥𝑖𝑐𝑙(𝑈))𝑛 𝑖=1 = 𝐴. Lemma 6.4. Let 𝐺ℜ be a topological simple rough group such that 𝐺ℜ ̅̅ ̅̅ is a group and 𝐺ℜ is open in 𝐺ℜ ̅̅ ̅̅ . Suppose 𝐾ℜ and 𝐻ℜ are closed in 𝐺ℜ ̅̅ ̅̅ and 𝐾ℜ is compact in 𝐺ℜ ̅̅ ̅̅ . Then 𝒟ℭ is a closed set in 𝐺ℜ ̅̅ ̅̅ . Proof: Since 𝐻ℜ is closed in 𝐺ℜ ̅̅ ̅̅ , 𝑥𝐻ℜ is closed in 𝐺ℜ ̅̅ ̅̅ . Then using theorem 3.2, 𝐾ℜ𝑥𝐻ℜ is closed in 𝐺ℜ ̅̅ ̅̅ that is, 𝒟ℭ is a closed set in 𝐺ℜ ̅̅ ̅̅ . Proposition 6.5. Let 𝐻ℜ and 𝐾ℜ be rough subgroups of a topological simple rough group 𝐺ℜ such that 𝐺ℜ ̅̅ ̅̅ is a group and 𝐻ℜ, 𝐾ℜ are subgroups of 𝐺ℜ ̅̅ ̅̅ . If 𝐾ℜ is a compact subset in 𝐺ℜ ̅̅ ̅̅ and 𝐻ℜ is a closed subset in 𝐺ℜ ̅̅ ̅̅ , then the rough quotient map 𝜑: 𝐺ℜ ̅̅ ̅̅ → 𝒟ℭ is open. Proof: Let 𝑈 be an open set in 𝐺ℜ ̅̅ ̅̅ . Then 𝜑−1𝜑(𝑈) = 𝐾ℜ𝑈𝐻ℜ is open in 𝐺ℜ ̅̅ ̅̅ which implies 𝜑(𝑈) is open. Therefore, the rough quotient map 𝜑 is open. Proposition 6.6. Let 𝐻ℜ and 𝐾ℜ be rough subgroups of a topological simple rough group 𝐺ℜ such that 𝐺ℜ ̅̅ ̅̅ is a group and 𝐻ℜ, 𝐾ℜ are subgroups of 𝐺ℜ ̅̅ ̅̅ . Then the rough double coset space 𝒟ℭ is regular. Proof: Consider an arbitrary point 𝑐 ∈ 𝒟ℭ and 𝜑(𝑥) = 𝑐, for some 𝑥 ∈ 𝐺ℜ ̅̅ ̅̅ . Then 𝜑−1(𝑐) = 𝐾ℜ𝑥𝐻ℜ. From theorem 4.2, 𝜑−1(𝑐) is closed which implies {𝑐} is closed in 𝒟ℭ. Therefore, for any 𝑐 ∈ 𝒟ℭ, the rough double coset 𝒟ℭ is a 𝑇1- space. Let 𝑈 be an open neighbourhood of 𝑐 in 𝒟ℭ. Then there exist identity neighbourhoods 𝑉 and W in 𝐺ℜ ̅̅ ̅̅ such that 𝜑(𝑉𝑥) ⊆ 𝑈 and WW ⊆ V. Also, by theorem 3.3, there exists a symmetric identity neighbourhood 𝑁 ⊆ 𝑊 in 𝐺ℜ ̅̅ ̅̅ such that 𝑥𝑁𝑥−1 ⊆ 𝑊, for every 𝑥 ∈ 𝐾ℜ which implies 𝑁𝐾ℜ ⊆ 𝐾ℜ𝑊. Since 𝜑 is an open mapping, 𝜑(𝑁𝑥) is an open neighbourhood of 𝜑(𝑥) in 𝒟ℭ and 𝜑(𝑁𝑥) ⊆ 𝑈. Now let us prove the closure of 𝜑(𝑁𝑥) is contained in 𝑈. Let 𝜑(𝑁𝑦) be an open neighbourhood of 𝑦 in 𝐺ℜ ̅̅ ̅̅ and 𝑦 be an accumulation point of 𝜑(𝑁𝑥).That is, 𝑦 in closure of 𝜑(𝑁𝑥). Then 𝜑(𝑁𝑦) ∩ 𝜑(𝑁𝑥) ≠ ∅ implies 𝑁𝑦 ∩ 𝐾ℜ𝑁𝑥𝐻ℜ ≠ ∅. Therefore, 𝑦 ∈ 𝑁𝐾ℜ𝑁𝑥𝐻ℜ ⊆ 𝐾ℜ𝑊𝑁𝑥𝐻ℜ ⊆ 𝐾ℜ𝑊𝑊𝑥𝐻ℜ ⊆ 𝐾ℜ𝑉𝑥𝐻ℜ = 𝜑(𝑉𝑥) ⊆ 𝑈. So, closure of 𝜑(𝑁𝑥) is contained in 𝑈. Hence 𝒟ℭ is a regular space. Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 3 (2025) 888 https://internationalpubls.com Proposition 6.7. Let 𝐻ℜ and 𝐾ℜ be rough subgroups of a topological simple rough group 𝐺ℜ such that 𝐺ℜ ̅̅ ̅̅ is a Hausdorff topological group and 𝐻ℜ, 𝐾ℜ are subgroups of 𝐺ℜ ̅̅ ̅̅ . Then the mapping 𝜌: 𝐺ℜ ̅̅ ̅̅ /𝐻ℜ → 𝒟ℭ defined by ρ(𝑥𝐻ℜ) = 𝐾ℜ𝑥𝐻ℜ, for every 𝑥 ∈ 𝐺ℜ ̅̅ ̅̅ is open and perfect. Proof: Let the mappings 𝜑: 𝐺ℜ ̅̅ ̅̅ → 𝒟ℭ and 𝜑∗: 𝐺ℜ ̅̅ ̅̅ → 𝐺ℜ ̅̅ ̅̅ 𝐻ℜ⁄ defined by 𝜑(𝑥) = 𝐾ℜ𝑥𝐻ℜ and 𝜑∗(𝑥) = 𝑥𝐻ℜ, for 𝑥 ∈ 𝐺ℜ ̅̅ ̅̅ . Then 𝜑 = 𝜌 ∘ 𝜑∗. Since 𝜑 and 𝜑∗ are continuous and open, 𝜌 is an open mapping. Let 𝑐 ∈ 𝒟ℭ such that 𝜑(𝑥) = 𝑐, for some 𝑥 ∈ 𝐺ℜ ̅̅ ̅̅ . Then 𝜑−1(𝑐) = 𝐾ℜ𝑥𝐻ℜ and the preimage 𝜌−1(𝑐) = 𝜑∗(𝐾ℜ𝑥𝐻ℜ) = 𝜑∗(𝐾ℜ𝑥), 𝐾ℜ𝑥 ⊆ 𝐺ℜ ̅̅ ̅̅ . Since continuous image of a compact set is compact, the set of all preimages, 𝜌−1(𝑐) is compact in 𝐺ℜ ̅̅ ̅̅ /𝐻ℜ, for every 𝑐 ∈ 𝒟ℭ. Now let us prove 𝜌 is closed, using the theorem 2.16. Let 𝑁 be an open neighbourhood of 𝜑∗(𝐾ℜ𝑥) in 𝐺ℜ ̅̅ ̅̅ /𝐻ℜ. Then, 𝐾ℜ𝑥 ⊆ 𝜑∗−1(𝑁) which implies there exists an open neighbourhood 𝑈 ⊆ 𝐺ℜ ̅̅ ̅̅ such that 𝐾ℜ𝑥𝑈 ⊆ 𝜑∗−1(𝑁) that is, 𝜑∗(𝐾ℜ𝑥𝑈) ⊆ 𝑁. Let 𝑉 = 𝜑(𝑥𝑈) be an open neighbourhood of 𝑐 ∈ 𝒟ℭ. Therefore, 𝜌−1(𝑉) = 𝜌−1(𝜑(𝑥𝑈)) = 𝜑∗(𝜑−1(𝜑(𝑥𝑈))) = 𝜑∗(𝐾ℜ𝑥𝑈𝐻ℜ) = 𝜑∗(𝐾ℜ𝑥𝑈) ⊆ 𝑁. Hence, 𝜌 is closed which implies 𝜌 is a perfect mapping. References: [1] Alaa Altassan, Nof Alharbi, Hassen Aydi, Cenap Ozel, Rough action on topological rough groups, Appl. Gen. Topol. 21, no. 2(2020), 295-304. [2] Arhangel skii AV, Tkachenko M, Topological groups and related structures, Atlantis Press and World Sci, Paris (2008). [3] Biswas. R, Nanda. S, Rough Groups and Rough Subgroups, Bull. Pol. AC. Math., 42(1994) 251- 254. 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