Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 7s (2024) 349 https://internationalpubls.com Introduction to RG-Closed Type Sets in Topological Ordered Spaces 1 G. Sravani, *2 G. Srinivasa Rao 1 Research Scholar, Department of Mathematics, School of Applied Science & Humanities, VFSTR Deemed to be University, Vadlamudi, Guntur (Dt.), Andhra Pradesh, India. Email:sravanigandikota1998@gmail.com *2 Department of Mathematics, School of Applied Science & Humanities, VFSTR Deemed to be University, Vadlamudi, Guntur (Dt.), Andhra Pradesh, India. Email:gsrinulakshmi77@gmail.com Article History: Received: 01-06-2024 Revised: 03-07-2024 Accepted: 29-07-2024 Abstract: This research paper presents a novel concept in the field of topological ordered spaces, which is the introduction of a new class of sets called "rg-closed sets". This new class of sets is formed by generalizing closed sets using rg-open sets in topological ordered spaces. Notably, rg-closed sets strictly lie between the classes of closed sets and rg-closed collections in topological ordered spaces. Additionally, the article also covers a discussion on rg* closed sets. Keywords: Topological ordered space, rg-closed set (irg, drg, brg-closed sets), r*g*- closed set (ir*g*, dr*g*, br*g*-closed sets). AMS Classification: 55XX22 1. INTRODUCTION The first research on topological ordered spaces was conducted by Leopoldo Nachbin [1]. In 1970, Levine [19] invented a superclass of sets known as rg-closed sets. Later, M. K. R. S. Veera Kumar introduced a novel category of sets [14], in the year 2014 G. Srinivasa Rao et.al [4-6 &19-25] studied and explained g-closed and g * -closed sets in topological ordered space, which should be placed before the rg-closed sets and closed set classes. This new class of sets was not only different but also significant. In 2001, M. K. R. S. Veera Kumar presented research on i-closed, d-closed, and b-closed sets, which were introduced for the first time. A topological ordered space is referred to as a triple ( ) where X is a non-empty set, τ is a topology on X and is a partial order on X. Definition 1.1[5]: For any * ⁄ + will be represented by , -. If P = i(P), where i(P) = ⋃ , - , then a subset P of a topological ordered space ( ) is known to be increasing. Definition 1.2[5]: For any * ⁄ + will be represented by , -. If P = d(P), where d(P) = ⋃ , - , then a subset P of a topological ordered space ( ) is known to be decreasing. An increasing (resp. a decreasing) set is the complement of a decreasing (resp. an increasing) set. C(P) denotes the complement of „P‟ in X. dcl (P) = {F F is a decreasing closed subset of X containing P with F = d(F)}. icl (P) = {F F is an increasing closed subset of X containing P with F = i(F)}. bcl (P) = {F F is a closed subset of X containing P with F=i(F)=d(F)}. IO(X) (resp. DO(X), BO(X)) represents the set of all decreasing (or increasing, both increasing and decreasing) open subsets of a topological ordered space( ). Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 7s (2024) 350 https://internationalpubls.com For a subset P of a space( ), cl(P) (resp. icl(P), bcl(P)) denote the decreasing (resp. increasing, both increasing and decreasing) closure of P. 2. Topological ordered space with rg-closed sets Definition 2.1: A topological space ( ) has a subset P is known as rg-closed [29] set, if cl(P) ⊆ R whenever P ⊆ R and R is regular open in ( ). Definition 2.2: A topological space ( )has a subset P is called r*g*-closed set [29], if rcl(P) ⊆ R whenever P ⊆ R and R is g-open in ( ). Theorem 2.3: Every r*g*-closed set is a rg-closed set. Proof: suppose P ⊆ R and R is regular open. Now, R is regular open R is open. W. k. t every closed set is a g-closed set. So, every open set is a g-open set. Since, R is open we have R is g-open. Therefore, rcl(P) ⊆ R, whenever P ⊆ R and R is g-open. Since P ⊆ R, there exist an open set G we have P ⊆ G ⊆ cl(G) ⊆ R. Since P ⊆ cl(G) cl(P) ⊆ cl(cl(G)) = cl(G) ⊆ R. cl(P) ⊆ R. Therefore cl (P) ⊆ R, whenever P ⊆ R and R is regular open. The following illustration demonstrates that a rg-closed set does not always have to be a r*g*-closed set. Example 2.4: Let X = {p, q, r}, = { , X, {p},{q},{p, q}} and = {(p, p), (q, q),(r, r),(p, q),(q, r), (p, r)}. Clearly, a topological ordered space is ( ). r*g*-closed sets are , X, {r}, {q, r}, {p, r}. rg-closed sets are , X, {r}, {p, q}, {q, r}, {p, r}. Let P = {p, q}. Clearly, A is rg-closed set but not r*g*-closed set. 3. Results between i(r*g*), d(r*g*) and b(r*g*) closed type sets Here are some definitions that we introduce: Definition 3.1: If ircl(P) ⊆ R whenever P ⊆ R and R is g-open in ( ), then a subset P of ( ) is called i(r*g*)-closed set. Definition 3.2: if drcl (P) ⊆ R whenever P ⊆ R and R is g-open in ( ), then a subset P of ( ) is called d(r*g*)-closed set. Definition 3.3: if brcl (P) ⊆ R whenever P ⊆ R and R is g-open in ( ), then a subset P of ( ) is called b(r*g*)-closed set. Theorem 3.4: Every i(r*g*)-closed set is an i(rg)-closed set. Proof: As far as we know, every r*g*-closed set is a rg-closed set. Therefore, every i(r*g*)-closed set is a i(rg)-closed set. In general, the following illustration demonstrates that, a i(rg)-closed set need not be an i(r*g*)-closed set. Example 3.5: Let X = {p, q, r}, = { , X, {p}, {p, q}, {p, r}} and = {(p, p), (q, q), (r, r), (p, q), (q, r)}. Clearly ( ) is a topological ordered space. i(r*g*)-closed sets are , X, {q, r}. i(rg)- closed sets are , X, {r}, {q, r}. Let P = {r}. Clearly P is i(rg)-closed set but not i(r*g*)-closed set. Theorem 3.6: Every d(r*g*)-closed set is an d(rg)-closed set. Proof: We know, every r*g*-closed set is an rg-closed set. Thus, every d(r*g*)-closed set is an d(rg)- closed set. The following illustration demonstrates that, a d(rg)-closed set need not always be an d(r*g*)-closed set. Example 3.7: Let X = {p, q, r}, = { , X, {p, q}} and = {(p, p), (q, q), (r, r), (p, q), (p, r)}. Clearly, a topological ordered space is ( ). , X, {p, r} are d(r*g*)-closed sets. , X, {p}, {p, Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 7s (2024) 351 https://internationalpubls.com q},{p, r} are d(rg)-closed sets. Let P = {p}. Clearly, P is a d(rg)-closed set but not a d(r*g*)-closed set. Theorem 3.8: Every b(r*g*)-closed set is a b(rg)-closed set. Proof: As far as we know, every r*g*-closed set is a rg-closed set. Thus, every b(r*g*)-closed set is a b(rg)-closed set. The following illustration demonstrates that, a b(rg)-closed set does not always have to be a b(r*g*)-closed set. Example 3.9: Let X = {p, q, r}, = { , X, {p}, {p, q}, {p, r}} and = {(p, p), (q, q), (r, r), (p, q), (r, q)}. Clearly, ( ) is a topological ordered space. , X are b(r*g*)-closed sets. , X, {p, q} are b(rg)-closed sets. Let P = {p, q}. Clearly, P is a b(rg)-closed set but not a b(r*g*)-closed set. Theorem 3.10: Every b(r*g*)-closed set is i(r*g*)-closed set. Proof: As far as we know, a balanced set is always an increasing set. Then, all b(r*g*)-closed sets are i(r*g*)-closed sets. The following illustration demonstrates that, an i(r*g*)-closed set need not always be a b(r*g*)-closed set. Example 3.11: Let X = {p, q, r}, = { , X, {p}, {p, r}} and = {(p, p), (q, q), (r, r), (q, p), (r, q), (r, p)}. Clearly, a topological ordered space is ( ). b(r*g*)-closed sets are , X. i(r*g*)-closed sets are , X, {p, q}. Let P = {p, q}. Clearly, P is i(r*g*)-closed set but not be a b(r*g*)-closed set. Theorem 3.12: Every b(r*g*)-closed set is a d(r*g*)-closed set. Proof: As we know, every balanced set is a decreasing set. Any set that is b(r*g*)-closed is also be an d(r*g*)-closed. The following illustration demonstrates that, a d(r*g*)-closed set does not always have to be a b(r*g*)-closed set. Example 3.13: Let X = {p, q, r}, = { , X, {p}, {p, r}} and = {(p, p), (q, q), (r, r), (q, p), (r, q), (r, p)}. Clearly, ( ) is a topological ordered space. b(r*g*)-closed sets are , X. d(r*g*)-closed sets are , X, {q, r}. Let P = {q, r}. Clearly, P is a d(r*g*)-closed set but not a b(r*g*)-closed set. Theorem 3.14: i(r*g*)-closed and d(r*g*)-closed are independent notions. The following example will demonstrate this. Example 3.15: Let X = {p, q, r}, = { , X, {p}, {p, q}, {p, r}} and = {(p, p), (q, q), (r, r), (p, q), (p, r)}. Clearly, ( ) is a topological ordered space. i(r*g*)-closed sets are , X. d(r*g*)-closed sets are , X, {q, r}. Let P = {q, r}. Clearly, P is a d(r*g*)-closed set but not be an i(r*g*)-closed set. Example 3.16: Let X = {p, q, r}, = { , X, {p}, {p, q}, {p, r}} and = {(p, p), (q, q), (r, r), (q, p), (r, q), (r, p)}. Clearly, a topological ordered space is ( ). i(r*g*)-closed sets are , X, {q, r}. d(r*g*)-closed sets are , X. Le t P = {q, r}. Clearly, P is i(r*g*)-closed set but not be a d(r*g*)- closed set. Theorem 3.17: Every b(rg)-closed set is an i(rg)-closed set. Proof: As we know, a balanced set is always an increasing set. Hence, all b(rg)-closed sets are i(rg)- closed sets. Generally the next example demonstrates that, i(rg)-closed sets do not have to be b(rg)- closed sets. Example 3.18: Let X = {p, q, r}, = { , X, {p}, {p, r}} and = {(p, p), (q, q), (r, r), (p, q), (r, q)}. Clearly, ( ) is a topological ordered space. b(rg)-closed sets are , X, {p, q}. i(rg)-closed sets are , X, {q}, {p, q}, {q, r}. Let P = {p}. Clearly, P is i(rg)-closed set but not a b(rg)-closed set. Theorem 3.19: Every b(rg)-closed set is a d(rg)-closed set. Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 7s (2024) 352 https://internationalpubls.com Proof: Every balanced set is a decreasing set, as we are aware. Every b(rg)-closed set is a d(rg)- closed set, hence this is true. The following illustration demonstrates that, a d(rg)-closed set does not always have to be a b(rg)-closed set. Example 3.20: Let X = {p, q, r}, = { , X, {p}, {p, r}} and = {(p, p), (q, q), (r, r), (q, r), (p, r)}. Clearly ( ) is a topological ordered space. b(rg)-closed sets are , X. d(rg)-closed sets are , X, {q}, {p, q}. Let P = {q}. Clearly P is a d(rg)-closed set but not a b(rg)-closed set. Theorem 3.21: i(rg)-closed and d(rg)-closed are independent notions. This will be demonstrated by the example that follows. Example 3.22: Let X = {p, q, r}, = { , X, {p}, {q}, {p, q}} and = {(p, p), (q, q), (r, r), (p, q), (p, r)}. Clearly, a topological ordered space is ( ). i(rg)-closed sets are , X, {r}, {q, r}. d(rg)- closed sets are , X, {p, q}, {p, r}. Let P = {p, r}. Clearly, P is a d(rg)-closed set but not i(rg)-closed set. Let Q = {r}. Clearly, Q is i(rg)-closed set but not a d(rg)-closed set. 4. Topological ordered space with irg-closed type sets Theorem 4.1: A set P Q is irg-closed if P and Q are irg-closed sets. Proof: If P Q ⊆ R and R is regular-open, then P ⊆ R and Q ⊆ R. But P and Q are irg-closed and therefore icl(P) ⊆ R and icl(Q) ⊆ R. Therefore, (icl (P) icl (Q)) ⊆ R icl (P Q) ⊆ R. Hence, P Q is irg-closed. Example 4.2: Let X = {p, q, r}, = { , X, {p}, {q}, {p, q}} and = {(p, p), (q, q), (r, r), (p, q)}. Clearly, a topological ordered space is ( ). Take, P = {r}, Q = {p, q}. If (P Q) = {r} {p, q} = {p, q, r} ⊆ R = X and R is regular-open, then {r}⊆ R and {p, q} ⊆ R. But P and Q are irg-closed and therefore icl(P) ⊆ R and icl(Q) ⊆ R. Therefore, (icl(P) icl (Q)) ⊆ R, and hence icl(P Q) ⊆ R. Hence P Q is irg-closed. Theorem 4.3: Suppose that Q ⊆ P ⊆ X, P is an ig-closed open subset of X and Q is an irg-closed set in relation to P. Then, Q is irg-closed with respect to X. Proof: Let Q ⊆ R and let R be regular-open. We have Q ⊆ (P R). But Q is an irg-closed set relative to P. Hence i ( ) ⊆ (P R). (1) Note that P R is regular-open in P. But i ( ) = icl(Q) P (2) From (1) and (2), (P icl(Q)) ⊆ (P R) Consequently P icl (Q) ⊆ R. Hence, P (icl(Q) C(icl(Q))) ⊆ R C(icl(Q)). That is P X ⊆ (R C(icl(Q))). So P ⊆ (R C(icl(Q))) = G, say (3) But then G is an open set. Since P is ig-closed in X, from (3) we have icl(P) ⊆ (R C(icl(Q))) = G (4) But icl(Q) ⊆ icl(P) (5) From (4) and (5) we have icl(Q) ⊆ (R C(icl(Q))). Hence icl(Q) ⊆ R because icl(Q) C(icl(Q)) = Q is irg-closed relative to X. Corollary 4.4: Let P be an ig-closed, open set. Suppose that Q is an i-closed set. Then P Q is an irg-closed set relative to X. Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 7s (2024) 353 https://internationalpubls.com Proof: We have that P Q is closed in P. Hence icl(P Q) = P Q in P. Let P Q ⊆ R, Where R is regular-open in P. That is icl(P Q) ⊆ R. Hence P Q is an irg-closed set in the ig-closed P. By the theorem [4.3], P Q is an irg-closed set relative to X. Theorem 4.5: If a set P is irg-closed then icl(P)\P contains no nonempty regular-closed set. Proof: Suppose that P is irg-closed. Let S be a regular-closed subset of icl(P)\P. Then S ⊆ (icl(P) C(P)) and so P ⊆ C(S). But P is irg-closed. Therefore icl(P) ⊆ C(S). (1) Consequently S ⊆ C(icl(P)) (2) We have already S ⊆ icl(P) (3) From (2) and (3), S ⊆ (icl(P) C(icl(P))) = Thus S = . Therefore icl(P)\P contains no nonempty regular-closed set. Corollary 4.6: Let P be an irg-closed set. If P is regular-closed then cl(int(P))\P is regular-closed. Proof: Let P be an irg-closed. If P is regular-closed i.e., cl(int(P)) = P. Then cl(int(P))\P = P\P = . But, is always regular-closed. As a result, cl(int(P))\P is regular-closed. On the other hand, imagine that cl(int(P))\P is regular-closed. However, P is irg-closed. Additionally, the regular-closed set cl(int(P))\P is contained in icl(P)\P. By above theorem [4.5], cl(int(P))\P = . Hence cl(int(P)) = P. Therefore P is regular-closed. Theorem 4.7: If P is ig-closed then P is irg-closed. Proof: Suppose that P ⊆ R, Where R is regular-open. Now R regular-open implies that R is open. Thus P ⊆ R and R is open. But P is ig-cosed. Hence icl(P) ⊆ R. Therefore, P is irg-closed. The following illustration demonstrates that an irg-closed set need not always be an ig-closed set. Example 4.8: Let X = {p, q, r}, = { , X, {p}, {q}, {p, q}} and = {(p, p), (q, q), (r, r), (p, q), (r, q)}. Clearly ( ) is a topological ordered space. irg-closed sets are , X, {p, q}, {q, r}. ig- closed sets are , X, {q, r}. Let P = {p, q}. Clearly P is an irg-closed set but not an ig-closed set. Theorem 4.9: If P is irg-closed and P ⊆ Q ⊆ icl(P) then icl(Q)\Q contains no nonempty regular- closed set. Proof: Suppose P is irg-closed and P ⊆ Q ⊆ icl(P). Since P ⊆ Q C(Q) ⊆ C(P) (1) Since Q ⊆ icl(P) icl(Q) ⊆ icl(icl(P)) ⊆ icl(P) (2) That is icl(Q) ⊆ icl(P). From (1) and (2), (icl(Q) C(Q)) ⊆ (icl(P) C(P)). Which implies (icl(Q)\Q) ⊆ (icl(P)\P). Now P is irg-closed. Hence, icl(P)\P has no nonempty regular-closed subsets neither does icl(Q)\Q. Theorem 4.10: Assume that P is irg-closed in X and P ⊆ Y ⊆ X. If Y is open in X, Then P is irg- closed relative to Y. Proof: Assume that R is regular-open in X and that P ⊆ Y R. Therefore, P ⊆ R and hence icl(P) ⊆ R. It follows from this that (Y icl(P)) ⊆ Y R. Therefore P is irg-closed with respect to Y. Theorem 4.11: Let X be a regular space. Prove that every compact subset of X is an irg-closed set. Proof: Assume P R, where R is regular-open. R is open since it is regular-open right now. But in the typical space X, P is compact. Consequently, there exists an open set O in which P ⊆ O ⊆ cl(O) ⊆ R. Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 7s (2024) 354 https://internationalpubls.com Since P ⊆ cl(O) icl(P) ⊆ icl(cl(O)) = cl(cl(O)) = cl(O) ⊆ R. That is icl(P) ⊆ R. Hence P is irg- closed in X. 5. ir*g*- closed type sets in topological ordered spaces: Theorem 5.1: A set P Q is ir*g*-closed if P and Q are ir*g*-closed sets. Proof: If P Q ⊆ R and R is g-open, then P ⊆ R and Q ⊆ R. But P and Q are ir*g*-closed and therefore ircl (P) ⊆ R and ircl (Q) ⊆ R. Therefore, (ircl(P) ircl(Q)) ⊆ R, and hence ircl(P Q) ⊆ R. Hence P Q is ir*g*-closed. Example 5.2: Let X = {p, q, r}, = { , X, {p}, {q}, {p, q}} and = {(p, p), (q, q), (r, r), (p, q), (q, r)}. Clearly ( ) is a topological ordered space. Take P ={r}, Q = {q, r}. If (P Q) ={r} {q, r} = {q, r} ⊆ R = X and R is g-open, then {r} ⊆ R and {p, q} ⊆ R. But P and Q are ir*g*-closed and therefore ircl (P) ⊆ R and ircl (Q) ⊆ R. Therefore, (ircl(P) ircl(Q)) ⊆ R, and hence ircl(P Q) ⊆ R. Hence P Q is ir*g*-closed. Theorem 5.3: If a set P is ir*g*-closed then ircl(P)\P contains no nonempty regular-closed set. Proof: Suppose that P is ir*g*-closed. Let S be a regular-closed subset of ircl(P)\P. Then S ⊆ (ircl(P) C(P)) and so P ⊆ C(S). But P is ir*g*-closed. Therefore ircl(P) ⊆ C(S). (1) Consequently S ⊆ C(ircl(P)) (2) We have already S ⊆ ircl(P) (3). From (2) and (3) S ⊆ (ircl(P) C(ircl(P))) = . Thus S = . Therefore, ircl (P)\P contains no nonempty regular-closed set. Corollary 5.4: If P is an ir*g*-closed set, then P is regular-closed if and only if cl(int (P))\P is regular closed. Proof: Make P an irg-closed. If cl(int(P)) = P, then P is regular-closed. If so, cl(int(P))\P = P\P = . However, is always regular closed. cl(int(P))\P is hence regular-closed. Assume, on the other hand, that cl(int(P))\P is regular-closed. P is, however, irg-closed. The regular-closed set cl(int(P))\P is also contained in ircl(P)\P. The statement "cl(int(P))\P = ." is based on the aforementioned theorem. As a result, cl(int(P)) = P. As a result, P is regular closed. Theorem 5.5: In the event that P is ir*g*-closed and P ⊆ Q ⊆ ircl(P), then ircl(Q)\Q does not contain any nonempty regular-closed sets. Proof: If P is ir*g*-closed and P ⊆ Q ⊆ ircl(P). Since C(P)⊆C(Q) follows from P⊆Q (1) Q ⊆ ircl(P) implies that ircl(Q) ⊆ ircl(ircl(P)) = ircl(P). In this case ircl(Q) ⊆ ircl(P) (2) From (1) & (2) (ircl(Q) C(Q)) ⊆ (ircl(P) C(P)) Implies (ircl(Q)\Q) ⊆ (ircl(P)\P). P is now ir*g*-closed. As a result, neither ircl(P)\P nor ircl(Q)\Q have any nonempty regular-closed subsets. Theorem 5.6: Let P ⊆ Y ⊆ X and suppose that P is ir*g*-closed in X. Then P is ir*g*-closed relative to Y, provided Y is open in X. Proof: Let P ⊆ Y R and suppose that R is g-open in X. Then P ⊆ R and hence ircl(P) ⊆ R. This implies that (Y ircl(P)) ⊆ Y R. 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