Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 10s (2025) 1358 https://internationalpubls.com Decomposition of (Gζ, Ξ) -Continuity R. Ramesh1, K. Rajupillai2,∗ and R. Uma 3 1 Department of Mathematics, Dr. Mahalingam College of Engineering and Technology, Pollachi, Tamil Nadu, India. 2∗ Department of Mathematics, Government College of Engineering, Thanjavur, Tamil Nadu, India. 3 Department of Mathematics, Sree Saraswathi Thyagaraja College, Tamil Nadu, India. ∗-Correspondance: rajupillai@gct.ac. Article History: Received: 12-01-2025 Revised: 15-02-2025 Accepted: 01-03-2025 Abstract: In this work, we introduce and investigate a new type of open sets. We also present a decomposition of both (gζ, ξ) - c and decomposition of (ζ, ξ)-c. Keywords: hereditary generalized topology, α-Hg-O, σ-Hg-O and π-Hg-O sets, β-Hg-O sets. 1. Introduction In 2002, generalized topology and generalized continuity introduced by Csaszar in [1]. In 2005, Csaszar introduced and studied generalized open sets (ζ-α-O, ζ-σ-O, ζ-π-O, ζ-β-O)[2]. The notion ζ-b-O introduced by Sarsak in [11]. A space Z is called a C0 -space [12], if C0=Z, where C0 is the set of all representative elements of sets of ζ and x is called a represent element of u ∈ ζ if u ⊂ v for each v ∈ ζ(X). A subset A of generalized topological space (X, ζ) is said to be gζ -closed [4] (resp. ωζ -closed [7]), if c(A)⊆M whenever A⊆M and M is ζ-O (resp. ζ-σ-O) in X. The complement of ωζ -closed (resp. gζ-O ) is ωζ-O[7] (resp. gζ-O [4]). The gζ-interior (resp. ωζ-interior) is the largest gζ-O (resp. ωζ-O) set contained in A and is denoted by ig(A) (resp. iω(A)). In 2005, Csaszar introduced hereditary class in [3]. In this work hereditary generalized topological space (Z, ζ, H) is denoted by HGTS. Definition 1.1. [3] The set ψ is said to be α-H-O (resp. σ-H-O, π-H-O, β-H-O, β∗-H-O, ζ∗ -closed), if ψ⊆ic∗(ψ) (resp. ψ⊆c∗i(ψ), ψ⊆ic∗(ψ), ψ⊆cic∗(ψ), ψ⊆c∗ic∗(ψ), c∗(ψ)⊂ψ). Definition 1.2. A set ψ is said to be b-H-O [8], if ψ ⊆ ic∗(ψ)∪c∗i (ψ). Definition 1.3. [10] A set ψ is said to be 1. α-Hg-O, if ψ ⊆ igc∗ig(ψ). 2. σ-Hg-O, if ψ ⊆ c∗ig(ψ). 3. π-Hg-O, if ψ ⊆ igc∗(ψ). 4. β-Hg-O, if ψ ⊆ cigc∗(ψ). 5. S-β-Hg- O, if ψ ⊆ c∗igc∗(ψ). mailto:-Correspondance:%20rajupillai@gct.ac. Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 10s (2025) 1359 https://internationalpubls.com 2. b-Hg-open set Definition 2.1. The set ψ⊂(Z,ζ,H) is called as b-Hg-open (b-Hg-O), if ψ⊆igc∗(ψ)∪c∗ig(ψ). Proposition 2.2. In HGT S, every ζ-O set is b-Hg-O. Proof. A subset ψ⊂Z is ζ-O. Then ψ = i(ψ). Now ψ ⊆ i(ψ) ⊆ ig(ψ) ⊆igc∗(ψ) ∪ c∗ig(ψ). Hence ψ is b- Hg-O. Remark 2.3. The converse of Proposition 2.2 need not be true from the following example. Example 2.4. Assume Z = {1, 2, 3, 4}, ζ={∅, {1}, {2}, {1, 2}, {2, 3, 4}, Z}, H = {∅, {1}, {3}}. Then ψ = {1, 2, 3} is b-Hg-O but not ζ-O. Proposition 2.5. In HGTS (Z, ζ, H), every gζ-O set is b-Hg-O but not conversely. Proof. Assume a subset ψ of HGTS (Z, ζ, H) is gζ- O. Then ψ = ig(ψ). Now ψ ⊆ ig(ψ) ⊆ igc∗(ψ) ∪ c∗ig(ψ). Hence ψ is b-Hg-O. Proposition 2.6. In HGTS, every ωζ-O set is b-Hg-O but not conversely. Proof. Assume a subset ψ of HGT S (Z, ζ, H) is ωζ-O. Then ψ = iω(ψ). Now ψ⊆iω(ψ) ⊆ ig(ψ) ⊆ igc∗(ψ) ∪ c∗ig(ψ). Hence ψ is b-Hg-O. Example 2.7. Assume Z = {1, 2, 3, 4}, ζ={∅, {1}, {2}, {1, 2}, {2, 3, 4}, Z}, H = {∅, {1}, {3}}. Then ψ = {1, 2, 3} is b - Hg - O but not gζ-O. Remark 2.8. The notions of b-Hg-O and ζ-b-O are independent. Example 2.9. Assume Z = {1, 2, 3, 4}, ζ={∅, {1}, {2}, {1, 2}, {2, 3, 4}, Z}, H = {∅, {1}, {3}}. Then ψ = {4} is b-Hg-O but not ζ-b-O. Example 2.10. Assume Z = {1, 2, 3, 4}, ζ = {∅, {1}, {1, 2, 3}, {3, 4}, Z}, H ={∅, {1}, {3}}. Then M = {1, 4} is ζ- b- O but not b- Hg-O. Proposition 2.11. In HGTS (Z, ζ, H) every b-H-O is b-Hg-O. Proof. Assume ψ be a b-H-O ψ⊆ic∗(ψ)∪c∗i(ψ)⊆igc∗(ψ)∪c∗ig(ψ). Hence ψ is b-Hg-O.Remark 2.12. The converse of Proposition 2.11 need not be correct from the following examples. Example 2.13. Assume Z = {1, 2, 3, 4}, ζ = {∅, {1, 3}, {2, 3}, {1, 2, 3}, {1, 4}, {1, 3, 4}, Z}, H = {∅, {1, 2}}. Then ψ = {1} is b-Hg-O but not b-H- O . Proposition 2.14. Every α-Hg-O (resp. σ-Hg-O, π-Hg-O) is b-Hg-O but not conversely. Proof. 1. Assume ψ be α-Hg-O. Then ψ⊆ igc∗ig(ψ)⊆c∗ig(ψ)∪igc∗(ψ). Which implies ψ is b-Hg-O. 2. Assume ψ is σ- Hg-O. Then ψ ⊆ c∗ig(ψ) ⊆ c∗ig(ψ)∪igc∗(ψ). Which implies ψ is b-Hg-O. 3. Assume ψ be π-Hg-O. Then ψ⊆igc∗(ψ) ⊆ c∗ig(ψ)∪ igc∗(ψ). Which implies ψ is b- Hg-O. Example 2.15. Assume Z = {1, 2, 3, 4}, ζ={∅, {1, 3}, {2, 3}, {1, 2, 3}, {1, 4}, {1, 3, 4}, Z}, H = {∅, {1}, {2}}. Then ψ={2} is b-Hg-O but not α-Hg-O (resp. σ-Hg- O, π-Hg-O). Theorem 2.16. If ψ is b-Hg-O and ζ-σ-O, then it is β-H- O. Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 10s (2025) 1360 https://internationalpubls.com ζ Proof. Let ψ is b-Hg-O and ζ-σ- O. Then ψ ⊆ igc∗(ψ) ∪ c∗ig(ψ) and ψ ⊆ ci (ψ). Now ψ ⊆ igc∗(ψ)∪c∗ig(ψ)⊆c∗(ψ), which implies ci (ψ) ⊆ cic∗(ψ). So ψ⊆ci(ψ)⊆ci∗(ψ). Hence ψ is β-H-O. Theorem 2.17. If ψ is b - Hg - O and ζ∗ -closed, then it is σ-Hg-O. Proof. Let ψ is b-Hg-O and ζ∗-closed. Then ψ⊆ igc∗(ψ)∪c∗ig(ψ) and c∗(ψ) ⊆ψ. Now ψ ⊆ igc∗(ψ)∪c∗ig(ψ) ⊆ c∗ig(ψ)∪ig(ψ) = c∗ig(ψ). Hence ψ is σ-Hg-O. Theorem 2.18. If ψ is b-Hg-O and ζ-closed, then it is σ-Hg-O. Proof. Let ψ is b-Hg-O and ζ-closed. Then ψ⊆igc∗(ψ)∪c∗ig(ψ) and c∗(ψ)⊆ψ by Proposition 2.9 of [6]. Which implies igc∗(ψ) ⊆ ig(ψ). Now ψ⊆igc∗(ψ)∪c∗ig(ψ)⊆c∗ig(ψ)∪ig(ψ) = c∗ig(ψ). Hence σ- Hg-O. Theorem 2.19. If ψ is b-Hg-O such that ig(ψ) = ∅, then it is π-Hg-O. Proof. Let ψ be a b-Hg-O and ig(ψ)=∅. Then ψ⊆igc∗(ψ)∪c∗ig(ψ) = igc∗(ψ). Hence ψ is π-Hg-O. Theorem 2.20. If ψ⊂Z is b-Hg-O and ψ∈H, then it is σ-Hg-O. Proof. Let ψ is b-Hg-O and ψ∈ H. Then ψ⊆ igc∗(ψ)∪c∗ig(ψ) and c∗(ψ) = ψ by Remark 2.10 of [6]. Now ψ⊆igc∗(ψ)∪c∗ig(ψ) = ig(ψ)∪c∗ig(ψ)=c∗ig(ψ). Hence ψ is σ-Hg-O. Theorem 2.21. If ψ⊂Z is b-Hg-O and H=P (Z) then it is σ-Hg-O. Proof. Let ψ is b-Hg-O and H = P (Z) Then ψ ⊆ igc∗(ψ)∪c∗ig(ψ) and c∗(ψ)=ψ by Remark 2.10 of [6]. Now ψ⊆igc∗(ψ)∪c∗ig(ψ)=ig(ψ)∪c∗ig(ψ)=c∗ig(ψ). Hence ψ is σ-Hg-O. Definition 2.22. For ψ⊂Z, ibHg(ψ) is the largest b-Hg-O set contained in ψ. Definition 2.23. A subset ψ of HGT S (Z, ζ, H) is called Db(c, Hg)-s, if ig(ψ)=ibHg (ψ). Theorem 2.24. For a subset ψ of HGT S (Z, ζ, H), the following conditions are equivalent. 1. ψ is gζ-O, 2. ψ is b-Hg-O and Db(c, Hg)-s Proof. (1)⇒(2) Let ψ is gζ-O. Then ψ is b-Hg-O. So ψ = ig(ψ) and ψ = ibH(ψ). Therefore ig(ψ) = ibH(ψ). Hence ψ is Db(c, Hg)-s (2) ⇒ (1) Let ψ is b-Hg-O and Db(c, Hg)-s Then ψ = ibH (ψ) and ig(ψ) =ibH(ψ). Therefore ig(ψ)=ψ. Hence ψ is gζ-O. Remark 2.25. The notions of ψ is b-Hg-O and Db(c, Hg)-s are independent. Example 2.26. Assume Z={1, 2, 3, 4}, ζ={∅, {1, 3}, {2, 3}, {1, 2, 3}, {1, 4}, {1, 3, 4}, Z}, H={∅, {1}, {2}}. Then ψ = {2} is b-Hg-O but not Db(c, Hg)-s and M = {4} is Db(c, Hg)-s but not b-Hg-O set. 3. New types of Sets Definition 3.1. A subset ψ⊂Z is called 1. α∗-Hg-s, if igc∗ig(ψ) = i(ψ). 2. σ∗-Hg-s, if c∗ig(ψ)=i(ψ). Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 10s (2025) 1361 https://internationalpubls.com 3. π∗-Hg-s, if igc∗(ψ)=i(ψ). 4. b∗-Hg-s, if c∗ig(ψ)∪igc∗(ψ)=i(ψ). 5. β∗-Hg-s (β∗-Hg-s), if cigc∗(ψ) = i(ψ). Remark 3.2. The notions of α∗-Hg-s (resp. σ∗-Hg-s, π∗-Hg-s, b∗-Hg-s, β∗-Hg-s) and α-g-O (resp. σ- Hg-O, π-Hg-O, b-Hg-O, β-Hg-O) are independent. Example 3.3. Assume Z={1, 2, 3, 4}, ζ={∅, {1, 3}, {2, 3}, {1, 2, 3},{1, 4}, {1, 3, 4}, Z}, H = {∅, {1}, {2}}. Then ψ={1} is α-Hg-O (resp. σ-Hg-O, π-Hg-O, b-Hg-O, β-Hg-O) but not α∗-Hg-s (resp. σ∗- Hg-s , π∗-Hg-s, b∗-Hg-s , β∗-Hg-s) and M={2} is α∗-Hg-s (resp. σ∗- Hg-s , π∗-Hg-s, b∗-Hg-s, β∗-Hg- s) but not α-Hg-O (resp. σ-Hg O, π-Hg-O, b-Hg-O, β-Hg-O). Definition 3.4. The subset ψ ⊂ Z of is called 1. α∗-B-Hg-s ( α∗-B-Hg-s), if ψ=U∩V, where U is ζ-O and V is α∗-Hg-s. 2. σ∗-B-Hg-s (σ∗-B-Hg-s), if ψ=U∩V, where U is ζ-O and V is σ∗-Hg-s. 3. π∗-B-Hg-s (π∗-B-Hg-s), if ψ=U∩V, where U is ζ-O and V is π∗-Hg-s. 4. b∗-B-Hg-s (b∗-B-Hg-s), if ψ=U∩V, where U is ζ-O and V is b∗-Hg-s. 5. β∗-B-Hg-s (β∗-B-Hg-s), if ψ=U∩V, where U is ζ-O and V is β∗-Hg-s. Theorem 3.5. If ψ⊂Z is b-Hg-O and π∗-Hg-s, then it is σ-Hg-O. Proof. Let ψ is b-Hg-O and π∗-Hg-s. Then ψ⊆igc∗(ψ)∪c∗ig(ψ) and igc∗(ψ)=ig(ψ). Now ψ ⊆ igc∗(ψ)∪c∗ig(ψ)⊆c∗ig(ψ)∪ig(ψ) = c∗ig(ψ). Hence ψ is σ-Hg-O. Theorem 3.6. If ψ⊂Z is b-Hg-O and σ∗-Hg-s, then it is π-Hg-O. Proof. Let ψ is b-Hg-O and σ∗-Hg-s. Then ψ⊆igc∗(ψ) ∪ c∗ig(ψ) and c∗ig(ψ) = ig(ψ). Now ψ⊆igc∗(ψ)∪c∗ig(ψ)⊆igc∗(ψ)∪ig(ψ)=igc∗(ψ). Hence ψ is π-Hg-O. Proposition 3.7. Let (Z, ζ, H) be a strong HGTS and ψ⊂Z. Then the following holds: 1. If ψ is α∗-Hg-s, then ψ is α∗-B-Hg-s, 2. If ψ is σ∗-Hg-s, then ψ is σ∗-B-Hg-s. 3. If ψ is π∗-Hg-s, then ψ is π∗-B-Hg-s, 4. If ψ is b∗-Hg-s, then ψ is b∗-B-Hg-s, 5. If ψ is β∗-Hg-s, then ψ is β∗-B-Hg-s. Proof. Let ψ be a π∗-Hg-s. If we take M=Z∈ζ, then ψ=M∩ψ and hence ψ is a π∗-B-Hg-s. Proof of (2), (3), (4), (5) are similar of Proof of (1). Proposition 3.8. For a subset ψ a HGTS (Z, ζ, H), the following properties are hold: 1. If ψ is an σ∗-Hg-s and gζ-O, then ψ is α∗-Hg-s. 2. If ψ is an π∗-Hg-s and gζ-O, then ψ is α∗-Hg-s. 3. If ψ is an b∗-Hg-s, then ψ is π∗-Hg-s. 4. If ψ is an b∗-Hg-s, then ψ is σ∗-Hg-s. Proof. (1). Let ψ is σ∗-Hg-s and gζ-O. Then igc∗ig(ψ)⊂c∗ig(ψ) = i(A). Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 10s (2025) 1362 https://internationalpubls.com ζ ζ ζ Therefore igc∗ig(ψ)=i(A). Hence ψ is α∗-Hg-s. (2). Let ψ is π∗-Hg-s and gζ-O. Then igc∗ig(ψ)⊂igc∗(ψ)=i (A). Therefore igc∗ig(ψ)=i(A). Hence ψ is α∗-Hg-s. (3). Let ψ b∗-Hg-s. Then igc∗(ψ)⊂igc∗(ψ)∪c∗ig(ψ)=i(ψ). Therefore igc∗(ψ)= i(ψ). Hence ψ is π∗-Hg- s. (4). Let ψ b∗-Hg-s. Then c∗ig(ψ)⊂igc∗(ψ)∪c∗ig(ψ)=i(ψ). Therefore c∗ig(ψ)=i(ψ). Hence ψ is σ∗-H-s. Theorem 3.9. Let (Z, ζ, H) be a strong HGT S where Z is C0 -space and ψ ⊂ Z. Then the following conditions are equivalent. 1. ψ is ζ-O, 2. ψ is α-Hg-O and α∗-B-Hg-s, 3. ψ is σ-Hg-O and σ∗-B-Hg-s. 4. ψ is π-Hg-O and π∗-B-Hg-s, 5. ψ is β-Hg-O and β∗-B-Hg-s. Proof. (1) ⇒ (2), (1) ⇒ (3), (1) ⇒ (4), are obvious. (2)⇒(1). Let ψ is both α-Hg-O and α∗-B-Hg-s. Then ψ⊆igc∗ig(ψ)=igc∗ig(M ∩N), where M∈ζ and N is α∗-Hg-s. Hence ψ⊆igc∗ig(M)∩igc∗ig(N). Now ψ⊆M∩ψ⊆M∩[igc∗ig(M)∩i(N )]=M∩i(N)=i(ψ). Hence ψ is ζ-O. (3)⇒(1). Let ψ is both σ-Hg-O and σ∗-B-Hg-s. Then ψ⊆c∗ig(ψ) = c∗ig(M∩N), where M∈ζ and N is σ∗-Hg-s. Hence ψ⊆c∗ig(M)∩c∗ig(N). Now ψ⊆M∩ψ⊆M∩[c∗ig(M )∩i(N)]=M∩i(N )=i(ψ). Hence ψ is ζ-O. (4)⇒ (1). Let ψ is both π-Hg-O and π∗-B-Hg-s. Then ψ⊆igc∗(ψ)=igc∗(M∩N), where M∈ζ and N is π∗-Hg-s. Hence ψ⊆igc∗(M)∩igc∗(N). Now ψ⊆M∩ψ⊆M∩[igc∗(M)∩i (N )]=M∩i (N ) = i(ψ). Hence ψ is ζ-O. (5)⇒ (1). Let ψ is both β-Hg-O and β∗-B-Hg-s. Then ψ⊆cigc∗(ψ)=cigc∗(M∩N), where M∈ζ and N is β∗-Hg-s. Hence ψ⊆cigc∗(M)∩cigc∗(N). Now ψ⊆M∩ψ⊆M∩[cigc∗(M)∩i (N)]=M∩i(N) = i(ψ). Hence ψ is ζ-O. Remark 3.10. The notions of α-Hg-O (resp. σ-Hg-O, π-Hg-O, β-Hg-O) and α∗-B-Hg-s (resp. σ∗-B - Hg-s, π∗-B-Hg-s, β∗-B-Hg -s) are independent. Example 3.11. Assume Z = {1, 2, 3, 4}, ζ={∅, {1, 3}, {2, 3}, {1, 2, 3},{1, 4}, {1, 3, 4}, Z}, H={∅, {1}, {2}}. Then ψ={1} is α-Hg-O (resp. σ-Hg-O, π-Hg-O, β-Hg-O) but not α∗- B-Hg-s (rep. σ∗-B- Hg-s, π∗-B-Hg-s, β∗-B-Hg-s) and M={2} is α∗-B-Hg-s (rep. σ∗-B-Hg -s, π∗-B-Hg-s, β∗-B-Hg-s) but not α-Hg-O (resp. σ-Hg-O, π-Hg-O, β-Hg-O). Theorem 3.12. Let (Z, ζ, H) be a strong HGT S where Z is C0 -space and ψ⊂Z. Then the following conditions are equivalent. 1. ψ is ζ-O, 2. ψ is σ-Hg-O and b∗-B-Hg-s Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 10s (2025) 1363 https://internationalpubls.com 3. ψ is π-Hg-O and b∗-B-Hg-s 4. ψ is b-Hg-O and b∗-B-Hg-s Proof. (1) ⇒ (2) ⇒ (4) and (1) ⇒ (3) ⇒ (4) are obvious, since Z is b∗-B-Hg-s (4) ⇒ (1). Let ψ is b-Hg-O and b∗-B-Hg-s. Then ψ⊆igc∗(ψ)∪c∗ig(ψ) =igc∗(M∩N)∪c∗ig(M∩N), where ψ=M∩N, M∈ζ and V is b∗-Hg-s Hence ψ⊆M∩ψ⊆M∩ [igc∗(M∩N)∪c∗ig(M∩N)]⊆[M∩igc∗(M)∩igc∗(N)]∪[M∩c∗ig(M)∩c∗ig(N)]⊆[M∩igc∗(N)]∪[M∩c∗ig (N)]=M∩[igc∗(N)∪c∗ig(N)]=M∩i(V)=i(ψ). Remark 3.13. The notions of σ-Hg-O (resp. π-Hg-O, b-Hg-O) and b∗-B-Hg-s are independent. Example 3.14. Assume Z = {1, 2, 3, 4}, ζ={∅, {1, 3}, {2, 3}, {1, 2, 3},{1, 4}, {1, 3, 4}, Z}, H={∅, {1}, {2}}. Then ψ={1} is σ-Hg-O (resp. π-Hg-O, b-Hg-O) but not b∗-B-Hg-s and M={2} is b∗-B- Hg-s but not σ-Hg-O (resp. π-Hg-O, b-Hg-O). Theorem 3.15. Let (Z, ζ, H) be a strong HGTS, where Z is C0 -space and ψ⊂Z. Then the following conditions are equivalent. 1. ψ is ζ-O, 2. ψ is α-Hg-O and σ∗-B-Hg- s, 3. ψ is σ-Hg-O and σ∗-B-Hg-s . Proof. (1) ⇒ (2). Let a subset ψ of Z is ζ-O. Then it is α-Hg-O and σ∗-B-Hg-s. (2)⇒(3). Let a subset ψ of Z is both α-Hg-O and σ∗-B-Hg-s. Then it is both σ-Hg-O and σ∗-B-Hg-s. (3)⇒(1). This is from Theorem 3.9. Remark 3.16. The notions of α-Hg-O and σ∗-B-Hg-s are independent. Example 3.17. Assume Z={1, 2, 3, 4}, ζ={∅, {1, 3}, {2, 3}, {1, 2, 3},{1, 4}, {1, 3, 4}, Z}, H={∅, {1}, {2}}. Then ψ={1} is α-Hg-O but not σ∗-B-Hg-s and M={2} is σ∗-B-Hg-s but not α-Hg-O . Theorem 3.18. Let (Z, ζ, H) be a strong HGTS where Z is C0 -space and ψ⊂Z. Then the following conditions are equivalent. 1. ψ is ζ-O, 2. ψ is α-Hg-O and π∗-B-Hg-s, 3. ψ is π-Hg-O and π∗-B-Hg-s. Proof. (1)⇒(2). Let a subset ψ of Z is ζ-O. Then it is α-Hg-O and π∗-B-Hg-s. (2)⇒(3). Let a subset ψ of Z is both α-Hg-O and π∗-B-Hg-s. Then it is both π-Hg-O and π∗-B-Hg-s. (3)⇒(1). This is from Theorem 3.9. Theorem 3.19. Let (Z, ζ, H) be a strong HGTS where Z is C0 -space and ψ⊂Z. Then the following conditions are equivalent. 1. ψ is ζ-O, 2. ψ is α-Hg-O and β∗-B-Hg-s, 3. ψ is β-Hg-O and β∗-B-Hg-s. Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 10s (2025) 1364 https://internationalpubls.com ζ Proof. (1)⇒(2). Let a subset ψ of Z is ζ-O. Then it is α-Hg-O and β∗-B-Hg-s. (2)⇒(3). Let a subset ψ of Z is both α-Hg-O and β∗-B-Hg-s. Then it is both β-Hg-O and β∗-B-Hg-s. (3)⇒(1). This is from Theorem 3.9. Remark 3.20. The notions of α-Hg-O and π∗-B-Hg-s are independent. Example 3.21. Assume Z={1, 2, 3, 4}, ζ={∅, {1, 3}, {2, 3}, {1, 2, 3},{1, 4}, {1, 3, 4}, Z}, H={∅, {1}, {2}}. Then ψ={1} is α-Hg-O but not π∗-B-Hg-s (resp. β∗-B-Hg-s ) and M={2} is π∗-B-Hg-s (resp. β∗-B-Hg-s ) but not α-Hg-O. Theorem 3.22. Let (Z, ζ, H) be a strong HGTS where Z is C0 -space and ψ⊂Z. Then the following conditions are equivalent. 1. ψ is ζ-O, 2. ψ is σ-Hg-O and β∗-B-Hg-s, 3. ψ is β-Hg-O and β∗-B-Hg-s . Proof. (1) ⇒ (2). Let a subset ψ of Z is ζ-O. Then it is σ-Hg-O and β∗-B-Hg-s. (2)⇒ (3). Let a subset ψ of Z is both σ-Hg-O and β∗-B-Hg-s. Then it is both β-Hg-O and β∗-B-Hg-s. (3)⇒ (1). This is from Theorem 3.9. Remark 3.23. The notions of σ-Hg-O and β∗-B-Hg-s are independent. Example 3.24. Assume Z={1, 2, 3, 4}, ζ={∅, {1, 3}, {2, 3}, {1, 2, 3}, {1, 4}, {1, 3, 4}, Z}, H={∅, {1}, {2}}. Then ψ={1} is σ-Hg-O but not β∗-B-Hg-s and M={2} is β∗-B-Hg-s but not σ-Hg-O. Theorem 3.25. Let (Z, ζ, H) be a strong HGTS, where Z is C0 -space and ψ⊂Z. Then the following conditions are equivalent. 1. ψ is ζ-O, 2. ψ is π-Hg-O and β∗-B-Hg-s, 3. ψ is β-Hg-O and β∗-B-Hg-s. Proof. (1) ⇒ (2). Let a subset ψ of Z is ζ-O. Then it is π-Hg-O and β∗-B-Hg-s. (2)⇒(3). Let a subset ψ of Z is both π-Hg-O and β∗-B-Hg-s. Then it is both β-Hg-O and β∗-B-Hg-s. (3)⇒(1). This is from Theorem 3.9. Remark 3.26. The notions of π-Hg-O and β∗-B-Hg-s are independent. Example 3.27. Assume Z = {1, 2, 3, 4}, ζ={∅, {1, 3}, {2, 3}, {1, 2, 3},{1, 4}, {1, 3, 4}, Z}, H={∅, {1}, {2}}. Then ψ={1} is π-Hg-O but not β∗-B-Hg-s and M={2} is β∗-B-Hg-s but not π-Hg-O. Definition 3.28. A subset ψ of a HGTS (Z, ζ, H) is called 1. ξ∗-Hg-s, igc∗ig(ψ)=ig(ψ). 2. σ∗-Hg-s, if c∗ig(ψ)=ig(ψ). 3. π∗-Hg-s, if igc∗(ψ)=ig(ψ). 4. Φ∗-Hg-s, if c∗ig(ψ)∪igc∗(ψ)=ig(ψ). Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 10s (2025) 1365 https://internationalpubls.com ζ ζ 5. ∆∗-Hg-s, if cigc∗(ψ)=ig(ψ). Definition 3.29. A subset ψ of HGT S (Z, ζ, H) is called 1. ξ∗-B-Hg-s, if ψ=U∩V, where U is gζ-O and V is ξ∗-Hg-s 2. σ∗-B-Hg-s, if ψ=U∩V, where U is gζ-O and V is σ∗-Hg-s. 3. π∗-B-Hg-s, if ψ=U∩V, where U is gζ-O and V is π∗-Hg-s. 4. Φ∗-B-Hg-s, if ψ=U∩V, where U is gζ-O and V is Φ∗-Hg-s 5. ∆∗-B-Hg-s , if ψ=U∩V, where U is gζ-O and V is ∆∗-Hg-s Theorem 3.30. If ψ⊂Z is both π-Hg-O and ζ∗-closed, then it is π∗-Hg-s. Proof. Let ψ is both π-Hg-O and ζ∗-closed. Then ψ⊆igc∗(ψ) and c∗(ψ)⊂ψ. Now igc∗(ψ)⊂ c∗(ψ)⊂ψ. So, ψ=igc∗(ψ). Thus ig(ψ) = igc∗(ψ). Hence ψ is π∗-Hg-s. Theorem 3.31. If ψ⊂Z is both σ-Hg-O and ζ∗-closed, then it is ξ∗-Hg-s Proof. Let ψ is both σ-Hg-O and ζ∗-closed. Then ψ⊆c∗ig(ψ) and c∗(ψ)⊂ψ. Now c∗ig(ψ)⊂c∗(ψ)⊂ψ. So, ψ=c∗ig(ψ). Thus ig(ψ) = igc∗ig(ψ). Hence ψ is ξ∗-Hg-s Theorem 3.32. Let (Z, ζ, H) be a strong HGTS where Z is C0 -space and L⊂Z. Then the following conditions are equivalent. 1. ψ is gζ-O, 2. ψ is α-Hg-O and ξ∗-B-Hg-s, 3. ψ is σ-Hg-O and σ∗-B-Hg-s, 4. ψ is π-Hg-O and π∗-B-Hg-s. 5. ψ is β-Hg-O and ∆∗-B-Hg-s Proof. (1) ⇒ (2), (1) ⇒ (3), (1) ⇒ (4), are obvious. (2)⇒ (1). Let ψ is both α-Hg-O and ξ∗-B-Hg-s Then ψ⊆igc∗ig(ψ)=igc∗ig(U∩V), where U is gζ-O and V is ξ∗-Hg-s Hence ψ⊆igc∗ig(U)∩igc∗ig(V).Now ψ⊆U∩ψ⊆U∩[igc∗ig(U)∩ig(V)] = ig(U)∩ig(V)=ig(ψ). Hence ψ is gζ-O. (3) ⇒ (1). Let ψ is both σ-Hg-O and σ∗-B-Hg-s. Then ψ⊆c∗ig(ψ)=c∗ig(U∩V), where U is gζ-O and V is σ∗-Hg-s. Hence ψ⊆c∗ig(U)∩c∗ig(V). Now ψ⊆U∩ψ⊆U∩[c∗ig(U)∩ig(V)] = ig(U)∩ig(V)=ig(ψ). Hence ψ is gζ-O. (4)⇒ (1). Let ψ is both π-Hg-O and π∗-B-Hg-s. Then ψ⊆igc∗(ψ) =igc∗(U∩V), where U is gζ-O and V is π∗-Hg-s. Hence ψ⊆igc∗(U)∩igc∗(V). Now ψ⊆U∩ψ ⊆U∩[igc∗(U)∩ig(V)] = ig(U)∩ig(V)=ig(ψ). Hence ψ is gζ-O. (5)⇒ (1). Let ψ is both β-Hg-O and ∆∗-B-Hg-s Then ψ⊆cigc∗(ψ) =cigc∗(U∩V), where U is gζ-O and V is ∆∗-Hg-s Hence ψ ⊆ cigc∗(U)∩cigc∗(V). Now ψ⊆U∩ψ⊆U∩[cigc∗(U)∩ig(V)]=ig(U)∩ig(V)=ig(ψ). Hence ψ is gζ-O. Remark 3.33. The notions of α-Hg-O (resp. σ-Hg-O, π-Hg-O, β-Hg-O) and ξ∗-B-Hg-s (resp. σ∗-B- Hg-s, π∗-B-Hg-s, ∆∗-B-Hg-s) are independent. Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 10s (2025) 1366 https://internationalpubls.com Example 3.34. Assume Z={1, 2, 3, 4}, ζ={∅, {1, 3}, {2, 3}, {1, 2, 3},{1, 4}, {1, 3, 4}, Z}, H={∅, {1}, {2}}. Then ψ={3, 4} is σ-Hg-O (resp. π-Hg-O, bπ-Hg-O) but not Φ∗-B-Hg-s and M={2} is Φ∗-B- Hg-s but not σ-Hg-O (resp. π-Hg-O, b-Hg-O). Theorem 3.35. Let (Z, ζ, H) be a strong HGTS where Z is C0 -space and ψ⊂Z. Then the following conditions are equivalent. 1. ψ is gζ-O, 2. ψ is σ-Hg-O and Φ∗-B-Hg-s 3. ψ is π-Hg-O and Φ∗-B-Hg-s 4. ψ is b-Hg-O and Φ∗-B-Hg-s. Proof. (1)⇒(2)⇒(4) and (1)⇒(3)⇒(4) are obvious, since Z is Φ∗-B-Hg-s (4)⇒(1). Let ψ is b-Hg-O and Φ∗-B-Hg-s Then ψ⊆igc∗(ψ)∪c∗ig(ψ)=igc∗(M∩N)∪c∗ig(M∩ N), where ψ=M∩N, M is gζ-O and V is Φ∗-Hg-s. Hence ψ⊆M∩ψ⊆M∩[igc∗(M∩N)∪c∗ig(M∩N)]⊆[M∩igc∗(M)∩igc∗(N)]∪[M∩c∗ig(M)∩c∗ig(N)] ⊆[M∩igc∗(N)]∪[M∩c∗ig(N)]=M∩[igc∗(N)∪c∗ig(N )]=M∩ig(V)=ig(ψ). Remark 3.36. The notions of σ-Hg-O (resp. π-Hg-O, b-Hg-O) and Φ∗-B-Hg-s are independent. Theorem 3.37. Let (Z, ζ, H) be a strong HGTS, where Z is C0 -space and ψ⊂Z. Then the following conditions are equivalent. 1. ψ is gζ-O. 2. ψ is α-Hg-O and π∗-B-Hg-s. 3. ψ is π-Hg-O and π∗-B-Hg-s. Proof. (1)⇒(2). Let a subset ψ of Z is gζ-O. Then it is α-Hg-O and π∗-B-Hg-s. (2)⇒(3). Let a subset ψ of Z is both α-Hg-O and π∗-B-Hg-s. Then it is both ψ is π-Hg-O and π∗-B- Hg-s. (3)⇒(1). This is from Theorem 3.32. Theorem 3.38. Let (Z, ζ, H) be a strong HGTS where Z is C0 -space and L⊂Z. Then the following conditions are equivalent. 1. ψ is gζ-O, 2. ψ is α-Hg-O and σ∗-B-Hg-s, 3. ψ is σ-Hg-O and σ∗-B-Hg-s. Proof. (1)⇒(2). Let a subset ψ of Z is gζ-O. Then it is α-Hg-O and σ∗-B-Hg-s. (2) ⇒ (3). Let a subset ψ of Z is both α-Hg-O and σ∗-B-Hg-s. Then it is both σ-Hg-O and σ∗-B-Hg-s. (3) ⇒ (1). This is from Theorem 3.32. Theorem 3.39. Let (Z, ζ, H) be a strong HGTS, where Z is C0 -space and L⊂Z. Then the following conditions are equivalent. Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 10s (2025) 1367 https://internationalpubls.com 1. ψ is gζ-O, 2. ψ is α-Hg-O and ∆∗-B-Hg-s, 3. ψ is β-Hg-O and ∆∗-B-Hg-s. Proof. (1) ⇒ (2). Let a subset ψ of Z is gζ-O. Then it is α-Hg-O and ∆∗-B-Hg-s. (2) ⇒ (3). Let a subset ψ of Z is both α-Hg-O and ∆∗-B-Hg-s Then it is both β-Hg-O and ∆∗-B-Hg-s. (3) ⇒ (1). This is from Theorem 3.32. Theorem 3.40. Let (Z, ζ, H) be a strong HGTS, where Z is C0 -space and L⊂Z. Then the following conditions are equivalent. 1. ψ is gζ-O, 2. ψ is σ-Hg-O and ∆∗-B-Hg-s, 3. ψ is β-Hg-O and ∆∗-B-Hg- s. Proof. (1) ⇒ (2). Let a subset ψ of Z is gζ-O. Then it is σ-Hg-O and ∆∗-B-Hg-s (2) ⇒ (3). Let a subset ψ of Z is both σ-Hg-O and ∆∗-B-Hg-s. Then it is both β-Hg-O and ∆∗-B-Hg- s. (3) ⇒ (1). This is from Theorem 3.32. Theorem 3.41. Let (Z, ζ, H) be a strong HGTS, where Z is C0 -space and ψ⊂Z. Then the following conditions are equivalent. 1. ψ is gζ-O, 2. ψ is π-Hg-O and ∆∗-B-Hg-s, 3. ψ is β-Hg-O and ∆∗-B-Hg-s. Proof. (1) ⇒ (2). Let a subset ψ of Z is gζ-O. Then it is π-Hg-O and ∆∗-B-Hg-s (2)⇒(3). Let a subset ψ of Z is both π-Hg-O and ∆∗-B-Hg- s. Then it is both β-Hg-O and ∆∗-B-Hg-s. (3) ⇒ (1). This is from Theorem 3.32. 4. Decomposition of (gζ, ξ) -Continuity Definition 4.1. A map ν:(Z, ζ, H)→ (W, ξ) is (b- Hg, ξ)-c, if j−1(V) is b-Hg-O for each ξ-O set V in (W, ξ). Definition 4.2. A map ν:(Z, ζ, H)→(W, ξ) is (R∗g, ξ)-c ( (R∗g, ξ)-c), (resp. ((Db(c, Hg), ξ)- c), if ν−1(V) is R∗g set (resp. (Db(c, Hg)-s for each ξ-O set V in (W, ξ). Definition 4.3. A function ν:(Z, ζ, H)→(W, ξ) is said to be (α∗-B-Hg, ξ)-c (resp. (π∗-B-Hg, ξ)-c, ( σ∗- B-Hg, ξ)-c, (b∗-B-Hg, ξ)-c, (β∗-B-Hg, ξ)-c), if ν−1(V ) is α∗-B-Hg-s (resp. π∗-B-Hg-s, σ∗-B-Hg-s, b∗- B-Hg-s, β∗-B-Hg-s) for each ξ-O set V in (W, ξ). Definition 4.4. A function ν:(Z, ζ, H)→(W, ξ) is said to be (ξ∗-B-Hg, ξ)-c (resp. (Π∗-B-Hg, ξ)-c, ( Σ∗- B-Hg, ξ)-c, ( Φ∗-B-Hg, ξ)-c, ( ∆∗-B-Hg, ξ)-c ), if ν−1(V ) is ξ∗-B-Hg-s (resp. π∗-B-Hg-s, ∆∗-B-Hg-s ) for each ξ - O set V in (W, ξ). Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 10s (2025) 1368 https://internationalpubls.com Theorem 4.5. For a map ν:(Z, ζ, H)→(W, ξ) where Z is C0 -space, the following results are equivalent. 1. ν is (gζ, ξ)-c, 2. ν is (b-Hg, ξ)- c and (Db(c, Hg), ξ)-c. Proof. The proof is clear by Theorem 2.53. Theorem 4.6. For a map ν:(Z, ζ, H)→(W, ξ) where Z is C0 -space, the following results are equivalent. 1. ν is (ζ, ξ)-c, 2. ν is (α-Hg, ξ)-c and (α∗-B-Hg, ξ)-c, 3. ν is (σ-Hg, ξ)-c and (σ∗-B-Hg, ξ)-c, 4. ν is (π-Hg, ξ)-c and (π∗-B-Hg, ξ)-c, 5. ν is (β-Hg, ξ)-c and (β∗-B-Hg, ξ)- c. Proof. The proof is clear by Theorem 3.9. Theorem 4.7. For a map ν:(Z, ζ, H)→(W, ξ) where Z is C0 -space, the following results are equivalent. 1. ν is (ζ, ξ)-c, 2. ν is (σ-Hg, ξ)-c and ( b∗ - B - Hg, ξ) - c, 3. ν is (π-Hg, ξ)-c and ( b∗ - B - Hg, ξ) - c, 4. ν is (b - Hg, ξ) - c and (b∗ - B - Hg, ξ) - c. Proof. The proof is clear by Theorem 3.12. Theorem 4.8. Let (Z, ζ, H) be a strong HGTS for a function ν:(Z, ζ, H) →(W, ξ), Z is C0 -space. Then the following conditions are equivalent. 1. ν is (ζ, ξ)-c, 2. ν is (α-Hg, ξ)-c and (σ∗-B-Hg, ξ)- c, 3. ν is (σ- Hg, ξ)-c and ( σ∗-B- Hg, ξ)-c . Proof. The proof is clear by Theorem 3.15. Theorem 4.9. Let (Z, ζ, H) be a strong HGTS for a function ν:(Z, ζ, H)→(W, ξ), Z is C0 -space. Then the following conditions are equivalent. 1. ν is (ζ, ξ)-c , 2. ν is (α - Hg, ξ)-c and (π∗-B-Hg, ξ)-c, 3. ν is (π-Hg, ξ)-c and ( π∗-B-Hg, ξ)-c. Proof. The proof is clear by Theorem 3.18. Theorem 4.10. Let (Z, ζ, H) be a strong HGTS for a function ν:(Z, ζ, H)→(W, ξ), where Z is C0 - space. Then the following conditions are equivalent. 1. ν is (ζ, ξ)-c, 2. ν is (α-Hg, ξ)-c and (β∗-B-Hg, ξ)-c, 3. ν is (β-Hg, ξ)-c and (β∗-B-Hg, ξ)-c. Proof. The proof is clear by Theorem 3.19. Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 10s (2025) 1369 https://internationalpubls.com Theorem 4.11. Let (Z, ζ, H) be a strong HGTS for a function ν:(Z, ζ, H)→(W, ξ), where Z is C0 - space. Then the following conditions are equivalent. 1. ν is (ζ, ξ)-c, 2. ν is (σ-Hg, ξ)-c and (β∗-B-Hg, ξ)-c, 3. ν is (β-Hg, ξ)-c and ( β∗-B- Hg, ξ)-c. Proof. The proof is clear by Theorem 3.22. Theorem 4.12. Let (Z, ζ, H) be a strong HGTS for a function ν:(Z, ζ, H)→(W, ξ), where Z is C0 - space. Then the following conditions are equivalent. 1. ν is (ζ, ξ)-c, 2. ν is (π-Hg, ξ)-c and (β∗-B- Hg, ξ)- c, 3. ν is (β-Hg, ξ)-c and ( β∗-B- Hg, ξ)-c. Proof. The proof is clear by Theorem 3.25. Theorem 4.13. Let (Z, ζ, H) be a strong HGTS, where Z is c0 space for a function ν:(Z, ζ,H)→(W, ξ), where Z is C0 -space. Then the following conditions are equivalent. 1. ν is (gζ, ξ)-c, 2. ν is (α-Hg, ξ)-c and (ξ∗-B-Hg, ξ)-c, 3. ν is (σ-Hg, ξ)-c and (Σ∗-B-Hg, ξ)-c, 4. ν is (π- Hg, ξ)-c and (Π∗-B-Hg, ξ)-c, 5. ν is (β-Hg, ξ)-c and (∆∗-B-Hg, ξ)-c. Proof. The proof is clear by Theorem 3.32. Theorem 4.14. Let (Z, ζ, H) be a strong HGTS, where Z is c0 space for a function ν:(Z, ζ, H)→(W, ξ), where Z is C0 -space. Then the following conditions are equivalent. 1. ν is (gζ, ξ)-c, 2. ν is (σ-Hg, ξ)-c and (Φ∗-B-Hg, ξ)-c, 3. ν is (π-Hg, ξ)-c and (Φ∗-B- Hg, ξ)-c, 4. ν is (b-Hg, ξ)-c and (Φ∗-B- Hg, ξ)-c. Proof. The proof is clear by Theorem 3.35. Theorem 4.15. Let (Z, ζ, H) be a strong HGTS, where Z is c0 space for a function ν:(Z, ζ, H)→(W, ξ), where Z is C0 -space. Then the following conditions are equivalent. 1. ν is (gζ, ξ)-c, 2. ν is (α-Hg, ξ)-c and (Π∗-B-Hg, ξ)-c, 3. ν is (π- Hg, ξ)-c and (Π∗-B-Hg, ξ)-c. Proof. The proof is clear by Theorem 3.37. Theorem 4.16. Let (Z, ζ, H) be a strong HGTS, where Z is c0 space for a function ν:(Z, ζ, H)→(W, ξ). Then the following conditions are equivalent. 1. ν is (gζ, ξ)-c, Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 10s (2025) 1370 https://internationalpubls.com 2. ν is (α- Hg, ξ)-c and (Σ∗-B-Hg, ξ)-c, 3. ν is (σ-Hg, ξ)-c and (Σ∗-B-Hg, ξ)-c. Proof. The proof is clear by Theorem 3.38. Theorem 4.17. Let (Z, ζ, H) be a strong HGTS, where Z is c0 space for a function ν:(Z, ζ, H)→(W, ξ). Then the following conditions are equivalent. 1. ν is (gζ, ξ)-c, 2. ν is (α-Hg, ξ)-c and (∆∗-B-Hg, ξ)-c, 3. ν is (β-Hg, ξ)-c and (∆∗-B- Hg, ξ)-c. Proof. The proof is clear by Theorem 3.39. Theorem 4.18. Let (Z, ζ, H) be a strong HGTS, where Z is c0 space for a function ν:(Z, ζ, H)→(W, ξ). Then the following conditions are equivalent. 1. ν is (gζ, ξ)-c, 2. ν is (σ-Hg, ξ)-c and (∆∗-B-Hg, ξ)-c, 3. ν is (β-Hg, ξ)-c and (∆∗-B-Hg, ξ)-c. Proof. The proof is clear by Theorem 3.40. Theorem 4.19. Let (Z, ζ, H) be a strong HGTS, where Z is c0 space for a function ν:(Z, ζ, H)→(W, ξ). Then the following conditions are equivalent. 1. ν is (gζ, ξ)-c, 2. ν is (π-Hg, ξ)-c and (∆∗-B-Hg, ξ)-c, 3. ν is (β-Hg, ξ)-c and (∆∗-B- Hg, ξ)-c. Proof. The proof is clear by Theorem 3.41. Reference [1] A.Csaszar, Generalized topology generalized continuity Acta Mathematica Hungarica 96 (2002), 351-357. [2] A.Csaszar, Generalized open sets in generalized topologies Acta Mathematica Hungarica 106 (2005), 53-56. [3] A.Csaszar, Modification of generalized topologies via hereditary classes Acta Mathematica Hungarica 115(2007), 29-36. [4] S.Maragathavalli, M. Sheik John and D. Sivaraj On g-closed sets in generalized topological spaces, Journal of Advanced Research in Pure Mathematics (2)(2010), 3:24-33. [5] W. K. Min, Generalized continuous maps defined by generalized open sets on generalized topological spaces Acta Mathematica Hungarica 128(4) (2010)pp 299-306. [6] M. Rajamani, V. Inthumathi and R. Ramesh, Some new generalized topologies via hereditary classes Bol. Soc. Paran. Mat. 30(2)(2012), 71-77. [7] M. Rajamani, V. Inthumathi and R. Ramesh, (ωµ, λ) -continuity in generalized topological spaces, International Journal of Mathematical Archive, 3(10)(2012), 3696-3703. Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 10s (2025) 1371 https://internationalpubls.com [8] R. Ramesh and R. Mariappan Generalized open sets in hereditary generalized topological spaces, J. Math. Comput. Sci., 5(2) (2015), 149-159. [9] R. Ramesh and Ahmad Al-Omari, b - Hσ -open sets in HGTS, Poincare Journal of Analysis and Applications 9(1) (2022), 31-40. [10] R. Ramesh and Ahmad Al-Omari, Decomposition of (α-Hg, λ) -continuity, Poincare Journal of Analysis and Applications, 10(1) (2023), 155-163. [11] M.S. Sarsak, On some properties of Generalized open sets in Generalized topological spaces, Demonstratio Math. (2013). [12] GE Xun and GE Ying, µ -Separations in generalized topological spaces, Appl. Math. J. Chinese Univ., 25(2)(2010), 243-252.