Properties of [γ, γ ]-Preopen Sets

International Journal of Applied Engineering Research ISSN 09734562 Volume 13, Number 22 (2018) pp. 1551915529 Properties of [γ, γ ]Preopen Sets Dr. S. Kousalya Devi 1 and P.Komalavalli 2 1 Principal, Government Arts Colleges, Kulithalai, Tamilnadu, India. 2 Assistant professor, Department of mathematics, Thanthai Hans Roever College,Perambalur, TamilNadu, India. Abstract: In this paper, we investigate some more properties of bioperationpreopen sets in bioperationtopological space. Keywords:Topological spaces, open set, preopen set. 2000 Mathematics Subject Classification. :54A05, 54A10. 1. INTRODUCTION Generalized open sets play a very important role in General Topology and they are now the research topics of many topologists worldwide. Indeed a significant theme in General Topology and Real Analysis concerns the various modified forms of continuity, separation axioms etc. by utilizing generalized open sets. Kasahara [1] defined the concept of an operation on topological spaces and introduce the concept of closed graphs of a function. Maki and Noiri [4] introduced the notion of which is the collection of all open sets in a topological space (X, τ ). In this paper, we investigate some more properties of bioperationpreopen sets in bioperationtopological space. 2. PRELIMINARIES Definition 2.1. Let (X, τ) be a topological space. An operation [, ] [1] on the topology τ is function from τ on to power set P(X) of X such that V V γ for each V τ, where V γ denotes the value of at V. It is denoted by γ : τ P(X). Definition 2.2. Let (X, τ) be a topological space. An operation on τ is said to be regular [1] if for every open neighborhood U and V of each x X, there exists an open neighborhood W of x such that U γ V γ W γ. Definition 2.3. A subset A of a topological space (X, τ ) is said to be open set [4] if for each x A there exist open neighbourhoods Uand V of x such that U γ V γ A. The complement of openset is called closed. denotes set of all open sets in(x, τ ). Definition 2.4. [4] Let A be subset of a topological space (X, τ) and, be operations on τ. Then (i) the closure of A is defined as intersection of all closed sets containing A. That is Cl(A) = {F : F is closed and A F}. [, ] (ii) the interior of A is defined as union of all open sets contained in A. That is, Int(A) = {U : U is open and U A}. [, ] Definition 2.5. A subset A of a topological space (X, τ ) is said to be (i) [, ] regular open [3] if Int( Cl(A)) = A. (ii) [, ] (iii) [, ] preopen [2] if A Int( Cl(A)). dense [2] if Cl(A)) = X. The complement of a preopen set is called a preclosed set. The familyof all preopen (resp. preclosed) sets of (X, τ ) is denoted by PO(X) (resp. PC(X)). The family of all preopen sets of (X, τ ) containing the point x is denoted by PO(X, x). Definition 2.6. [2] Let A be subset of a topological space (X, τ ) and, be operations on τ. Then (i) the preclosure of A is defined as intersection of all preclosed sets containing A. That is, p Cl(A) = {F : F is preclosed and A F}. (ii) the preinterior of A is defined as union of all 15519

International Journal of Applied Engineering Research ISSN 09734562 Volume 13, Number 22 (2018) pp. 1551915529 preopen sets contained in A. That is, p Int(A) = {U : U is preopen and U A}. Theorem 2.7. [2] Let A be subset of a topological space (X, τ) and, be operations on τ. Then (i) A is [, ] preopen if and only if A = p Int(A). (ii) A point x τ γ Cl(A) if and only if U A ᴓ for every U PO(X, x). (iii) pcl(a) is the smallest [, ] subset of X containing A. (iv) A is [, ] pcl(a). (v) preclosed preclosed if and only if A = pint (X \ A) = X \ pcl(a). (vi) pcl (X \ A) = X \ pint (A). Definition 2.8. [2] Let (X, τ ) be a topological space and, be operations on τ. Then a subset A of X is said to be pre g.closed (written as pg.closed) set if pcl(a) U whenever A U and U is preopen. (iii) S is the intersection of a [, ] and a dense set. (iv) S is the intersection of a [, ] dense set. regular open set open set and a Proof. (i) (ii): Let S PO(X). Then S Int( Cl(S)). Let G = Int(S). Then G is regular open with S G and S = G. (ii) (iii): Let D = S (X \ G). Then D is dense and S = G D. (iii) (iv): This is trivial. (iv) (i): Suppose S =G D with G is open and D dense. Then S = G, hence S G Cl(G) = S. Theorem 3.2. If every subset of X is either open or closed, then every preopen set in X is open. Proof.Let A be a preopen in X. If A is not open, then A is closed by hypothesis. Hence A = Cl(A), and Int( Cl(A)) = [, ] Int(A) is a proper subset of A. Thus, A Int( [, ] Cl( A)),so that Ais not preopen, contradiction. Theorem 2.9. [2] Let (X, τ ) be a topological space and, be operations on τ. Then subset A of X is pg.closed,then pcl(a) \ A does not contain any nonempty preclosed set. 3. PROPERTIES OF PREOPENSETS Throught this paper, the operators and are defined on (X, τ ) and the operators and are defined on (Y, σ). Theorem 3.1. For any subset of a space (X, τ ) the following are equivalent: (i) S [, ] PO(X). (ii) There is a [, ] regular open set G X such that S G and Cl(S) = Cl(G). Theorem 3.3. Let (X, τ ) be a topological space in which every preopen set in X is open. Then each singleton in X is either open or closed. Proof.Let x X, and suppose that {x} is not open. Then {x} is not preopen. Hence {x} Int( [, ] Cl({x}), so that Int( Cl({x}) =. We have that Int( Cl(X \ {x}) Int( Cl(X \ ( Cl({x}))) = Int(X \ ( Int( Cl({x})))) = X X \ {x}. Thus, X \ {x} is preopen and hence open. Therefore, {x} is closed. Theorem 3.4. For a topological space (X, τ ) and regular op erations on τ, the following are equivalent: (i) Every [, ] preopen set is open., be 15520

International Journal of Applied Engineering Research ISSN 09734562 Volume 13, Number 22 (2018) pp. 1551915529 (ii) Every [, ] dense set is open. Proof.(i) (ii): Let A be a dense subset of X. Then Int( Cl(A) = X, so that A Int( Cl(A)) and A is preopen. Hence A is open. (ii) (i): Let B be a preopen subset of X, so that B Int( Cl(B)) = G, say.hen Cl(B) = Cl(G), so that Cl(X \ G) B) = Cl(X \ G) Cl(B) = (X \ G) [, ] Cl(G) = X, and thus (X \ G) B is [, ] dense in X. Thus, (X \ G) B is open. Now, B = ((X \ G) B) G, the intersection of two open sets is open ([4], Proposition 2.7), so that B is open. Theorem 3.5. (X, τ ) is a topological space in which every subset is preopen if and only if every open set in (X, τ ) is closed. Proof. Let G be open. Then X \ G = Cl(X \ G) which is preopen, so that Cl(X \ G) [, ] Int( Cl( Cl(X \ G))) = Int( Cl(X \ G)) = Int(X \ G). Thus, X \ G = Int(X \ G), so that X \ G is open, and G is closed. Conversely, let A be any subset of X. Then X \ Cl(A) is open, and hence closed. Thus, X \ Cl(A) = Cl(X \ Cl(A)) = X \ [, ] Int( Cl(A)), so that A ( [, ] Cl(A) = Int Cl(A)), and hence A is preopen. Theorem 3.6. Let (X, τ ) be a topological space, G be a open subset of X and b be a point of Cl(G) \ G. Then {b} is not preopen in (X, τ ). Proof. Suppose {b} is preopen, so that {b} Int( Cl({b})). Thus, G Int( Cl({b})). Let c G Int( Cl({b})), so c Cl({b}) and hence {b} (G Int( Cl({b})). This contradicts the fact that {b} G =. Hence {b} is not preopen. Theorem 3.7. Let (X, τ ) be a topological space, G be a regular open subset of X and b be a point of Cl(G) \ G. Then G {b} is not preopen in (X, τ ). Proof. We have, since Cl({b}) Cl(G), so that [, ] Int( Cl(G {b})) = Int( Cl(G) [, ] Cl({b})) = Int( Cl(G) = G, and thus G {b} Int( Cl(G {b})). Hence G {b} is not preopen. Theorem 3.8. Let (X, τ ) be a topological space. Then every singleton of X is either open or preclosed. Proof. If x} is not open, then Thus, Cl( Int({x}) =. Int({x} )) = ; hence {x} is preclosed. The proof of the second part is straightforward. Theorem 3.9. If A is a preopen and pg.closed subset of (X, τ ), then A is preclosed. Proof. Since A is preopen and pg.closed, pcl(a) A and hence [, ] pcl(a) = A. This implies that A is preclosed by Theorem 2.7 (iv). Theorem 3.10. If A is a pg.closed subset of (X, τ ) such that A B pcl(a), then B is also pg.closed subset of (X, τ ). Proof. Let U be a preopen set in (X, τ ) such that B U. Then A U. Since A is [, ] pcl(a) U. Now, since [, ] preclosed, [, ] pcl(b) pg.closed, then pcl(a) is pcl( pcl(a)) = pcl(a) U. Therefore, B is also a [, ] pg.closed. Theorem 3.11. A set A in a topological space (X, τ ) is pg.open if and only if F p Int(A) whenever F is preclosed in (X, τ ) and F A. Proof. Let A be pg.open. Let F be 15521

International Journal of Applied Engineering Research ISSN 09734562 Volume 13, Number 22 (2018) pp. 1551915529 preclosed and F A. Then X \ A X \ F, where X \ F is preopen. pg.closedness of X \ A implies pcl(x \ A) X \ F. By Theorem 2.7, X \ p [, ] Int(A) X \ F. That is, F p Int(A). Conversely, Suppose if F is preclosed and F A implies F p Int(A). Let X \ A U where U is preopen. Then X \ U A where X \ U is preclosed. By supposition, X \ U p Int(A). That is, X \ p Int(A) U. By Theorem 2.7, pcl(x \ A) U. This implies X \ A is pg.closed and hence A is pg.open. Theorem 3.12. If p Int(A) B A and A is pg.open, then B is pg.open. Proof. Easily follows from Theorems 2.7 and 3.10. Proof. Let X \ (A B) = (X \ A) (X \ B) U, where U is preopen. Then X \ A U and X \ B U. Since A and B are pg.open, pcl(x \ A) U and pcl(x \ B) U. By hypothesis, pcl((x \ A) (X \ B)) = pcl(x \ A) pcl(x \ B) U. That is, pcl(x \ (A B)) U. This shows that A B is pg.open. Theorem 3.16. If A X is pg.closed, then pcl(a) \ A is pg.open. Proof. Let A be pg.closed. Let F be a preclosed set such that F pcl(a) \ A. Then by Theorem 2.9 F =. So, F pint( pcl(a) \ A). This shows pcl(a)\a is pg.open. Theorem 3.13. If a set A is pg.open in a topological space (X, τ ), then G = X whenever G is preopen in (X, τ ) and p Int(A) X \ A G. Proof. Suppose that G is preopen and pint(a) X \ A G. Now X \ G pcl(x \ A) A = p Cl(X \ A) \ X \ A. Since X \ G is preclosed and X \ A is pg.closed, by Theorem 2.9, X \ G = and hence G = X. Proposition 3.14. Let (X, τ ) be a topological space and A, B X. If B is pg.open and if A pint(b), then A B is pg.open. Proof. Since B is pg.open and A pint(b), [, ] p Int(B) A B B. By Theorem 3.12, A B is pg.open. Proposition 3.15. Let the family PO(X) of all preopensubsets of (X, τ) be closed under finite intersections i.e., let PO(X) be the topology on X. If A and B are pg.open in (X, τ ), then A B is pg.open. Definition 3.17. A topological space (X, τ) with operations and on τ is called [, ] preregular if for each preclosed set F of X not containing x, there exists disjoint preopen sets U and V such that x U and F V. Theorem 3.18. The following are equivalent for a topological space (X, τ) with operations and on τ: (i) X is [, ] preregular. (ii) For each x X and each U [, ] PO(X, x), there exists a V PO(X, x) such that x V pcl(v ) U. [, ] (iii) For each [, ] preclosed set F of X, { p Cl(V ) : F V,V PO(X) } = F (iv) For each A subset of X and each U [, ] PO(X) with A U, there exists a V PO(X) such that A U and pcl(v ) U. (v) For each nonempty subset A of X and each preclosed subset F of X with A F =, there exists V, W PO(X) such that A V, 15522

International Journal of Applied Engineering Research ISSN 09734562 Volume 13, Number 22 (2018) pp. 1551915529 F W and W V = (vi) For each [, ] preclosed set F and x F, there exists U PO(X) and a pg.open set V such that x U, F V and U V =. (vii) For each A X and each [, ] preclosed set F with A F =, there exists U PO(X) and a pg.open set V such that A U, F V and U V =. (viii) For each [, ] preclosed set F of X, F = { pcl(v ) : F V, V is pg.open} Proof. (i) (ii) Let x X \ U, where U PO(X, x). Thenthere exists G, V PO(X) such that (X \ U ) G, x V and G V =. Therefore V (X \ G) and so x V pcl(v ) (X \ G) U. (ii) (iii) Let (X \ F) PO(X, x). Then by(2) there exists an U PO(X, x) such that x U pcl(u ) (X \ F ). So, F X \ pcl(u ) = V, V PO(X) and V U =.Then by Theorem 2.7, x pcl(v ). Thus F { pcl(v): F V, V PO(X)}. (iii) (iv) Let U PO(X) with x U A. Then x (X \ U ) and hence by(iii) there exists a preopen set W such that (X \ U ) W and x pcl(w ). We put V = X \ pcl(w ), which is a preopen set containing x and hence V U. Now V (X \ W ) and so pcl(v ) (X \ W ) U (iv) (v) Let F be a set as in hypothesis of (v). Then (X \ F) is preopen and (X \ F) A. Then there exists V PO(X) such that A V and pcl(v ) (X \ F ). If we put W = X \ pcl(v ), then F W and W V =. (v) (i) Let F be a preclosed set not containing x. Then by (v), there exist W, V PO(X) such that F W and x V and W V =. (i) (vi) Obvious. (vi) (vii) For a A, a F and hence by (vi) there exists U PO(X) anda pg.open set V such that a U, F V and U V =. So, A U. (vii) (i) Let xf, where F is pg.closed. Since {x} F =, by (vii) there exists U PO(X) and pg.open set W such that x U,F W and U W =. Now put V = pint(w). Using definition of pg.open sets we get F V and V U =. (iii) (viii) We have F { pcl(v ): F V and V is pg.open} { pcl(v ): F V and V is preopen}= F. (viii) (i) Let F be a preclosed set in X not containing x. Then by (viii) there exists a pg.open set W such that F W and x X \ pcl(w ). Since F is preclosed and W is pg.open, F pint(w ). Take V = pint(w ). Then F V, x U = X \ pcl(v ) and U V =. Definition 3.19. A topological space (X, τ ) with operations and on τ is called [, ] prenormal if for any pair of disjoint preclosed sets A and B of X, there exist disjoint preopen sets U and V such that A U and F V. Theorem 3.20. For a topological space (X, τ) with operations and on τ, the following are equivalent: (i) X is [, ] prenormal. (ii) For each pair of disjoint [, ] preclosed sets A and B of X, there exist disjoint pg.open sets U and V such that A U and B V. (iii) For each [, ] preclosed set A and any preopen set V containing A, there exists a pg.open set U such that A U pcl(u ) V. (iv) For each [, ] preclosed set A and any pg.open set B containing A, there exists a pg.open set U such that A U pcl(u ) [, ] pint(b) (v) For each [, ] preclosed set A and any pg.open set B containing A, there exists a preopen set G such that A G pcl(g) [, ] p Int(B). (vi) For each [, ] pg.closed set A and any 15523

International Journal of Applied Engineering Research ISSN 09734562 Volume 13, Number 22 (2018) pp. 1551915529 preopen set B containing A, there exists a preopen set U such that pcl(a) U pcl(u ) B. [, ] (vii) For each [, ] pg.closed set A and any preopen set B containing A, there exists a pg.open set G such that pcl(a) G pcl(g) B. [, ] Proof. ( i) ( ii) : Follows from the fact that every preopen set is pg.open. ( ii) ( iii) : let A be a closed set and V any preopen set containing A. Since A and (X \ V ) are disjoint preclosed sets, there exist pg.open sets U and W such that U, (X \ V ) W and U W =. By Theorem 3.11, we get (X \ V ) pint(w ). Since U pint(w ) =, we have pcl(u ) pint(w ) =, and hence p Cl(U ) X \ pint(w ) V. Therefore, A U pcl(u ) V. ( iii) ( i) : Let A and B be any disjoint preclosed sets of X. Since (X \ B) is an preopen set containing A, there exists a pg.open set G such that A G p Cl(G) X \ B. Since G is a pg.open set, [, ] using Theorem 3.11, we have A p Int(G). Taking U = p Int(G) and V = X \ p Cl(G), we have two disjoint preopen sets U and V such that A U and B V. Hence X is prenormal. ( v) ( iii) : let A be a closed set and V any preopen set containing A. Since every preopen set is pg.open, there exists a preopen set G such that A G pcl(g) [, ] pint(v ). Also. we have a pg.open set G such that A G pcl(g) pint(v ) V. ( iii) ( v) : Let A be a closed set and B aby pg.open set containing A. Using Theorem 3.11 of a pg.open set we get A pint(b) = V, say. Then applying (iii), we get a pg.open set U such that A = pcl(a) U pcl(u ) V.Again, using the same Theorem 3.11 we get A pint(u ), and hence A pint(u ) U pcl(u ) V ; which implies A pint(u ) pcl( pint(u )) [, ] pcl(u ) V,that is, A G pcl(g) pint(b), where G = pint(u ). ( iii) ( vii) : Let A be a pg.closed set abd B any preopen set containing A. Since A is a pg.closed set, we have pcl(a) B, therefore, we can find a pg.open set U such that pcl(a) U pcl(u ) B. ( vii) ( vi) : [, ] Let A be a [, ] pg.closed set and B any [, ] preopen set containing A, then by (vii) there exists a [, ] pg.open set G such that τ β Cl(A) G pcl(g) B. Since G is a [, ] we get [, ] pg.open set, then by Theorem 3.11, pcl(a) U = pint(g), the proof follows. pint(g). If we take Definition 3.21. Let (X, τ ) be a topological space and A X. (i) The set denoted by for every [, ] pd(a) and defined by {x : preopen set U containing x, U ( A \ {x}) } is called the prederived set of A. (ii) The prefrontier of A, denoted by pfr(a) is defined as pcl(a) pcl(x \ A). Proposition 3.22. Let (X, τ ) be a topological space and A X. Then the following propeties hold: (i) (ii) Proof. Clear. pint(a) = A \ ( pd(x \ A). pcl(a) = A \ pd(a). 15524

International Journal of Applied Engineering Research ISSN 09734562 Volume 13, Number 22 (2018) pp. 1551915529 4. [, ] [, ] PRECONTINUOUS FUNCTIONS Definition 4.1. A function f : (X, τ ) (Y, σ) is said to be [, ] precontinuous [2] at a point xx if for each [, ] preopen subset V in Y containing f (x), there exists a preopen subset U of X containing x such that f (U ) V. ( p Int(f 1 ( [, ] pint (f 1 (V ))). p Int(V )))) pcl( Corollary 4.4. Let f : (X, τ ) (Y, σ) be a [, ] [, ] precontinuous function. Then for each subset V of Y, pint ( pcl (f 1 (V ))) f 1 ( [ pcl (V )). Theorem 4.2. For a function f : (X, τ ) (Y, σ), the following statements are equivalent: (i) f is [, ] precontinuous; (ii) The inverse image of each [, ] is [, ] preopen in X; (iii) For each subset B of Y, f 1 ( [ pcl(b)); (iv) For each subset A of X, f( [ pcl(f (A)); (v) For each subset A of X, f( [ pcl(f (A)); (vi) For any subset B of Y, f 1 ( pint(f 1 (A)); (vii) For each subset C of Y, f 1 ( [ pfr(b)). Proof. Easy proof and hence omitted. preopen set in Y pcl(f 1 (B)) pcl(a)) pd(a)) pint(b)) Theorem 4.3. Let f : (X, τ ) (Y, σ) be a [, ] pfr(f 1 (C)) [, ] precontinuous function. Then for each subset V of Y, f 1 ( [ pint(v )) pcl( pint(f 1 (V ))). Proof. Let V be any subset of Y. Then [ p Int(V ) is [, ] preopen in Y and so f 1 ( pint(v )) is preopen in X. Hence f 1 ( [, ] pint(v )) pcl Theorem 4.5. Let f : (X, τ ) (Y, σ) be a bijection. Then f is [, ] precontinuous if and only if [ pint (f (U)) f ( pint (U)) for each subset U of X. Proof. Let U be any subset of X. Then by Theorem 4.2, f 1 ( [ pint (f (U ))) pint (f 1 (f (U ))). Since f is bijection, [ pint (f (U )) = f (f 1 ( pint (f (U ))) f ( pint (U )). Conversely, let V be any subset of Y. Then [ pint (f (f 1 (V ))) f ( pint (f 1 (V ))). Since f is bijection, [ pint (V ) = [ pint (f (f 1 (V ))) f ( pint (f 1 (V ))); hence f 1 ( [ pint (V ) pint (f 1 (V )). Therefore, by Theorem 4.2, f is precontinuous. [, ] Theorem 4.6. Let f : (X, τ ) (Y, σ) be a function. Then X\ pc(f ) = { pfr (f 1 (V )) : V [, ] [, ] PO(Y), f (x)v, xx },where pc(f ) denotes the set of points at which f is [, ] precontinuous. Proof. Let x X \ V [, ] pc(f ). Then there exists PO(Y ) containing f (x) such that f (U ) is not a subset of V, for every preopen set U containing x. Hence U (X \ f 1 (V )) for every preopen set U containing x. Then, x pcl (X \ f 1 (V )). Then x f 1 (V ) pcl (X \ f 1 (V )) pf r (f 1 (V )). So,X \ pc(f ) { [, ] pfr(f 1 (V )) : V [, ] PO(Y ), f (x) V, x X }.Conversely, let x X \ 15525

International Journal of Applied Engineering Research ISSN 09734562 Volume 13, Number 22 (2018) pp. 1551915529 pc(f ). Then for each V [, ] f 1 (V ) is a [, ] PO(Y) containing f (x), preopen set U containing x. Thus, x pint (f 1 (V )) and hence x pf r(f 1 (V )), for every V βso(y ) containing f (x). Therefore, X \ pc(f ) { [, ] V, x X}. pfr(f 1 (V )) : V [, ] Theorem 4.7. Let f : (X, τ ) (Y, σ) be a [, ] precontinuous, [, ] PO(Y ), f (x) [, ] [, ] preclosed, [, ] preopen surjective function on (X, τ ). If X is [, ] preregular, then Y is [, ] preregular. Proof. Let K be a [, ] Since f is [, ] preclosed in Y and y K. [, ] precontinuous and X is preregular for each point x f 1 (y), there exist disjoint V, W PO(X) such that x V and f 1 (K) W. Now, since f is [, ] preclosed, there exists a [, ] preopen set U containing K such that preopen function, f 1 (U ) W. As f is a [, ] we have y = f (x)f (V ) and f (V ) is [, ] preopen in Y. Now, f 1 (U ) V = ; hence U f (V ) =. Therefore, Y is [, ] preregular. Theorem 4.8. If f : (X, τ ) (Y, σ) is a [, ] precontinuous, [, ] function and X is [, ] prenormal. [, ] preclosed surjective prenormal, then Y is [, ] Proof. Let A and B be two disjoint [, ] preclosed sets in Y. Since f is [, ] precontinuous, f 1 (A) and f 1 (B) are disjoint preclosed sets in X. Now as X is prenormal, there exist disjoint preopen sets V and W such that f 1 (A) V and f 1 (B) W. Since f is [, ] [, ] preclosed, there exist [, ] preopen sets M and N such that A M, B N, f 1 (M ) V and f 1 (N ) W. Since V W =, we have M N = ; hence Y is [, ] prenormal. Definition 4.9. A function f : (X, τ ) (Y, σ) is said to be: (i) [, ] preopen if f (U) is a [, ] preopen set of Y for every preopen set U of X. (ii) [, ] preclosed [2] if f (U ) is a [, ] preclosed set of Y for every preclosed set U of X. Theorem 4.10. For a bijective function f : (X, τ ) (Y, σ), the following statements are equivalent: (i) f is [, ] (ii) f is [, ] (iii) f is [, ] Proof. The proof is clear. preopen; preclosed; precontinuous. Theorem 4.11. For a function f : (X, τ ) (Y, σ), the followingstatements are equivalent: (i) f is [, ] (ii) f ( (iii) preopen; pint (U )) (, ) pint (f (U )) for each subset U of X; pint (f 1 (V )) f 1 ( (, ) pint (V )) for each subset V of Y. Proof. ( i) ( ii) : Let U be any subset of X. Then pint (U) is a preopen set of X. Then [, ] pint (U )) is a [, ] f ( [, ] preopen set of Y. Since f ( pint (U )) f (U ), f ( pint (U )) = [ pint (f ( pint (U ))) [ pint (f (U )). (ii) (iii): Let V be any subset of Y. Then f 1 (V ) is a subset of X. Hence f ( pint (f 1 (V ))) [ pint (f (f 1 (V ))) [ pint (V )). Then pint (f 1 (V )) f 1 (f ( pint (f 1 (V )))) f 1 ( [ pint (V)).(iii) (i): Let U be any preopen set of 15526

International Journal of Applied Engineering Research ISSN 09734562 Volume 13, Number 22 (2018) pp. 1551915529 X.Then pint (U ) = U and f (U ) is a subset of Y. Now, V = pint (V ) pint (f 1 (f (V ))) f 1 ( [ pint (f (V ))). Then f (V) f (f 1 ( [ pint (f (V )))) [ pint (f (V )) and [ pint (f (V )) f (V ). Hence f (V ) is a [, ] preopen set of Y ; hence f is preopen. [, ] Corollary 4.12. Let f : (X, τ ) (Y, σ) be a function. Then f is [, ] preclosed and [, ] precontinuous if and only if f ( pcl (V )) = (, ) p Cl(f (V ))) for evey subset V of X. Corollary 4.13. Let f : (X, τ ) (Y, σ) be a function. Then f is [, ] preopen and [, ] precontinuous if and only if f 1 ( (, ) pcl (V ))) = pcl (f 1 ((V ))) for evey subset V of Y. Theorem 4.14. Let f : (X, τ ) (Y, σ) be a function. Then f is a [, ] preclosed function if and only if for each subset V of X, [ pcl (f (V )) f ( pcl (V )). Proof. Let f be a [, ] [, ] preclosed function and V any subset of X.Then f (V ) f ( pcl(v )) and pcl(v )) is a [, ] preclosed set of Y. f ( [, ] We have [, ] pcl (f (V )) pcl (f ( pcl (V ))) = f ( pcl (V )). Conversely, let V be a preopen set of X. Then f (V ) f ( [, ] pcl (f (V )) pcl (V )) = f (V ); hence f (V ) is a [, ] preclosed subset of Y. Therefore, f is a [, ] [, ] preclosed function. Theorem 4.15. Let f : (X, τ ) (Y, σ) be a function. Then f is a [, ] preclosed function if and only if for each subset V of Y, f 1 ( [ pcl (V )) pcl (f 1 (V )). Proof. Let V be any subset of Y. Then by Theorem 4.14, [ pcl (V) f ( pcl (f 1 (V ))). Since f is bijection, f 1 ( [ pcl (V)) = f 1 ( [ pcl (f (f 1 (V )))) f 1 (f ( pcl (f 1 (V)))) = [ pcl (f 1 (V)). Conversely, let U be any subset of X. Since f is bijection, [ pcl (f (U )) = f (f 1 ( [ pcl (f (U ))) f ( pcl (f 1 (f (U )))) = f ( pcl (U )). Therefore, by Theorem 4.14, f is a [, ] preclosed function. Theorem 4.16. Let f : (X, τ ) (Y, σ) be a [, ] preopen function. If V is a subset of Y and U is a preclosed subset of X containing f 1 (V ), then there exists a [, ] preclosed set F of Y containing V such that f 1 (F ) U. Proof. Let V be any subset of Y and U a [, ] preclosed subset of X containing f 1 (V ), and let F = Y \ (f (X \ V )). Then f (X \ V ) f (f 1 (X \ V )) X \ V and X \ U is a preopen set of X. Since f is [, ] preopen, f (X \ U ) is a [, ] preopen set of Y. Hence F is a [, ] (f (X \ U )) U. preclosed set of Y and f 1 (F ) = f 1 (Y \ Theorem 4.17. Let f : (X, τ ) (Y, σ) be a [, ] [, ] preclosed function. If V is a subset of Y and U is a preopen subset of X containing f 1 (V ), then there exists [, ] preopen set F of Y containing V such that f 1 (F ) U. Proof. The proof is similar to the Theorem 4.16. Theorem 4.18. Let f : (X, τ ) (Y, σ) be a [, ] [, ] preopen function. Then for each subset V of Y, f 1 ( [ pint ( [ pcl (V )) pcl (f 1 (V ). 15527

International Journal of Applied Engineering Research ISSN 09734562 Volume 13, Number 22 (2018) pp. 1551915529 Proof. Let V be any subset of Y. Then pcl (f 1 (V ) is a preclosed set of X containing f 1 (V ). Since f is [, ] preopen, by Theorem 4.16, there is a [, ] preopen set F of Y containing V such that f 1 ( [ pint ( [, ] (F )) f 1 (F ) pcl (V )) pcl (f 1 (V )). pint ( pcl Theorem 4.19. Let f : (X, τ ) (Y, σ) be a bijection such that foreach subset V of Y, f 1 ( [ pint ( [ pcl (V )) [, ] preopen function pcl (f 1 (V)). Then f is a [, ] Proof. Let U be a [, ] preopen subset of X. Then f (X \ U ) is a subset of Y and f 1 ( [ pint ( [ pcl (f (X \ U )))) pcl (f 1 (f (X \ U ))) = pcl (X \ U ) = X \ U, and so [ pint ( [ pcl (f (X \ U ))) preclosed set f (X \ U ). Hence f (X \ U ) is a [, ] of Y and f (U ) = X \ (f (X \ U )) is a [, ] preopen set of Y. Therefore, f is a [, ] preopen function. Definition 4.20. [2] A function f : (X, τ ) (Y, σ) is said to be [, ] prehomeomorphism if f and f 1 are [, ] precontinuous. (v) f ( pcl (V )) = [ pcl (f (V )) for each subset V of X; (vi) f ( pint (V )) = [ pint (f (V )) for each subset V of X; (vii) f 1 ( (viii) pint (V )) = pint (f 1 (V )) for each subset V of Y ; pcl (f 1 (V )) = f 1 ( [ pcl (V )) for each subset V of Y. Proof. (i) (ii): It follows immediately from the definition of a [, ] prehomeomorphism. (ii) (iii) (iv): It follows from Theorem 4.10. (iv) (v): It follows from Theorem 4.15 and Corollary 4.12. (v) (vi): Let U be a subset of X. Then by Theorem 2.7, f ( pint (U )) = X \ f ( pcl (X \ U )) = X \ [ pcl (f (X \ U )) = [ pint (f (U )). (vi) (vii): Let V be a subset of Y. Then f ( pint (f 1 (V ))) = [ pint (f (f 1 (V ))) = [ pint (f (V )). Hence f 1 (f( pint (f 1 (V)))) = f 1 ( [ pint (V )). Therefore, f 1 ( [ pint (V )) = pint (f 1 (V )). (vii) (viii): Let V be a subset of Y. Then by Theorem 2.7, pcl (f 1 (V )) = X \ (f 1 ( [ pint (Y \ V ))) = X \ ( pint (f 1 ((X \ V )))) = f 1 ( [ pcl (V )).(viii) (i): It follows from Theorem 4.15 and Corollary 4.13. Theorem 4.21. Let f : (X, τ ) (Y, σ) be a bijection.then the following statements are equivalent: (i) f is [, ] (ii) f 1 is [, ] prehomeomorphism; prehomeomorphism; (iii) f and f 1 are [, ] preopen ( [, ] preclosed); (iv) f is [, ] precontinuous and [, ] preopen ( [, ] preclosed); Theorem 4.22. Every topological space (X, τ ) with operations and on τ is pret 1/2. Proof. Let x X. Wr prove (X, τ ) is pret 1/2, it is sufficient to show that {x} is preopen or open, then it is preclosed. Now, if {x} is [, ] obviously [, ] preopen. If {x} is not [, ] open, then Int({x}) = ; hence Cl Int({x})) = {x}. Therefore, {x} is preclosed. 15528

International Journal of Applied Engineering Research ISSN 09734562 Volume 13, Number 22 (2018) pp. 1551915529 REFERENCES [1] S.Kasahara, Operationcompact spaces, Math. Japonica 24 (1979), 97 105. [2] C. Carpintero, N. Rajesh and E. Rosas, On [, ] preopen sets, Demon. Math., XLVI (3) (2013), 617 629. [3] S. Kousalyadevi and P. Komalavalli, Bioperationvia Regular open sets (sub mited). [4] H. Maki and T. Noiri, Bioperations and some separation axioms, Scientiae Math. Japonicae, 53(1)(2001), 165180. 15529