Pacific-Asian Journal of Mathematics, Volume 5, No. 2, July-December 2011 ON BINARY TOPOLOGICAL SPACES S. NITHYANANTHA JOTHI & P. THANGAVELU ABSTRACT: Recently the authors introduced the concept of a binary topology between two sets and investigate its basic properties where a binary topology from X to Y is a binary structure satisfying certain axioms that are analogous to the axioms of topology. In this paper we further study about its base, sub base and their properties. 1. INTRODUCTION Point set topology deals with a nonempty set X (Universal set) together with a collection τ of subsets of X satisfying certain axioms. Such a collection τ is called a topological structure on X. General topologists studied the properties of subsets of X by using the members of τ. That is the information about a subset of X can be known from the information of members of τ. Therefore the study of point set topology can be thought of the study of information. But in the real world situations there may be two or more universal sets. If A is a subset of X and B is subset of Y, the topological structures on X and Y provide little information about the ordered pair (A, B). Our aim is to introduce a single structure which carries the subsets of X as well as the subsets of Y for studying the information about the ordered pair (A, B) of subsets of X and Y. Such a structure is called a binary structure from X to Y. Mathematically a binary structure from X to Y is defined as a set of ordered pairs (A, B) where A X and B Y. The concept of binary topology from X to Y is introduced by the authors [2]. The concepts of binary closed, binary closure and binary interior are also introduced in [2]. 2. PRELIMINARIES Let X and Y be any two non empty sets. The authors defined that a binary topology from X to Y is a binary structure M (X) (Y) that satisfies the following axioms. (i) (, ) and (X, Y ) M, (ii) (A 1 A 2 B 2 ) M whenever (A 1 ) M and (A 2, B 2 ) M, MSC 2010: 54A05, 54A99. Keywords: Binary topology, Binary open, Binary closed, Binary closure, Binary interior, Base, Sub base and Binary continuity.
134 S. Nithyanantha Jothi & P. Thangavelu (iii) If {(A α ): α } is a family of members of M, then Aα M. If M is a binary topology from X to Y then the triplet (X, Y, M) is called a binary topological space and the members of M are called the binary open subsets of the binary topological space (X, Y, M ). The elements of X Y are called the binary points of the binary topological space (X, Y, M). If Y = X then M is called a binary topology on X in which case we write (X, M) as a binary space. The examples of binary topological spaces are given in [2]. Definition 2.1: Let f : Z X Y be a function. Let A X and B Y. We define f 1 (A, B) = {z Z : f (z) = (x, y) (A, B)}. Definition 2.2: Let (X, Y, M ) be a binary topological space and let (Z, τ) be a topological space. Let f : Z X Y be a function. Then f is called binary continuous if f 1 (A, B) is open in Z for every binary open set (A, B) in X Y. Throughout this section (X) and (Y) denote the power sets of X and Y respectively. 3. BASES AND SUB BASES The bases and sub bases of topological spaces play an important role in topology and analysis. In this section, base and sub base of a binary topological space are defined and their basic properties have been discussed. Definition 3. 1: Let (X, Y, M ) be a binary topological space. A sub set B M is called a base for M if for each binary open set (U, V) there is a family {(A α ) : α }. of members of B such that (U, V) = Aα If B is a base for a binary topology M then the members of B are called the basic binary open sets and B completely determines M. Definition 3. 2: Let (X, Y, M ) be a binary topological space. A sub set S M is called a sub base for M if {(U 1 U k V k ): (U i, V i ) S for i = 1,.., k where k > 0 is an integer} is a base for M. If S is a sub base for a binary topology M then the members of S are called sub basic binary open sets and S completely determines M. Example 3.3: Let X = {a, b} and Y = {1, 2, 3}. Let M = {, ), ({a}, {1}), ({b}, {2}), ({a, b}, {1, 2}), (X, Y)}. M is a binary topology from X to Y.
On Binary Topological Spaces 135 Consider B = {({b}, {2}), ({a, b}, {1, 2})}. Clearly B 1 M 1. But B 1 is not a for any family base for M, since the binary open set ({b}, {2}) Aα {(A α ): α } of members of B. However M is a base for M. Example 3.4: Let X = {a, b} and Y = {1, 2, 3}. M = {(, ), (, {1}), ({a}, {1}), ({a}, {1, 2}), ({b}, ), ({b}, {1}), ({b}, {3}), ({b}, {1, 3}), ({a, b}, {1}), ({a, b}, {1, 2}), ({a, b},{1, 3}), (X, Y)}. Then M is binary topologies from X to Y. Consider B = {(, ), (, {1}), ({a}, {1}), ({b}, ), ({b}, {3}), ({a, b}, {1}), ({a, b}, {1, 2})}. Clearly B M. It can be verified that B is a base for M. Let S 1 = {({b}, {1, 3}), ({a, b}, {1}), ({a, b}, {1, 2}), ({a, b}, {1, 3})}. Since {({b}, {1, 3}), ({a, b}, {1}), ({a, b}, {1, 2}), ({a, b}, {1, 3}), ({b}, {1}), ({b}, {1, 3}), ({a, b}, {1})} is not a base for M, S 1 is not a sub base for M. Consider S 2 = {({a}, {1}), ({ a}, {1, 2}), ({ b}, ), ({ b}, {1}), ({ b}, {3}), ({a, b}, {1}), ({a, b}, {1, 2})}. Since {({a}, {1}), ({a}, {1, 2}), ({b}, ), ({b}, {1}), ({b}, {3}), ({a, b}, {1}), ({a, b}, {1, 2}), (, ), (, {1})} is a base for M, This shows that S 2 is a sub base for M. The next proposition characterizes a base for a binary topology. Proposition 3.5: A subset B of (X) (Y) is a base for the binary topological space (X, Y, M ) if and only if (i)b M and (ii) for every binary point (x, y) and for every binary neighborhood (U, V) of (x, y) there are (A, B) and (C, D) in B such that x A C U and y B D V. Proof: Suppose a subset B of (X) (Y) is a base for the binary topological space (X, Y, B ). Then from Definition 3.1, B M. This proves (i). Let (U, V ) be a binary neighborhood of (x, y). Then (U, V ) = Aα for some members (A α α ) B where α. Therefore, x A α and y B α. Choose α and β such that x A α α α and y B β. This proves that x A α A β U and y B α B β V and (A α ) B and (A β, B β ) B. (ii) follows by taking ( A α ) = (A, B) and (A β, B β ) = (C, D). Conversely we assume that (i) and (ii) hold. Let (U, V) be a binary open set. For each (x, y) (U, V ) we choose (A x, B y ) and (C x, D y ) in B such that x A x C x U and y B y D y V. This shows that
136 S. Nithyanantha Jothi & P. Thangavelu that implies by (U, V ) = (),() A x C x B y D y ( x,)( y,)( U,)( V,) x y U V Definition 3.1, B is a base for (X, Y, M ). Proposition 3.6: Any base B of the binary topological space (X, Y, M ) has the following properties. (i) For any (U 1 ), (U 2, V 2 ) B and for every (x, y) (U 1 ) there are (A, B) and (C, D) in B such that x A C U 1 and y B D V 1. (ii) For every ( x, y) X Y there are ( A, B) and ( C, D) in B such that (x, y) (A C, B D). Proof: Let (U 1 ), (U 2, V 2 ) B and (x, y) (U 1 ). Therefore by Proposition 3.7, there are (A, B) and (C, D) in B such that x A C U 1 and y B D V 1. This proves (i). Since (X, Y) M, (ii) follows fromproposition 3.5. The next lemma on functions from a set Z to X Y is very useful in sequel. Lemma 3.7: Let f : Z X Y be a function and {(A α ) : (A α ) X Y, α } be a set ordered pairs of sub sets of X and Y. Then the following hold: (i) (ii) 1 1 α,(,) α = α β α α α, β f A B f A B. 1 1 α,(,) α = α β α α α, β f A B f A B Proof: Straight forward. Proposition 3.8: Let (X, Y, M ) be a binary space. Let B be a base for (X, Y, M ). If f : Z X Y is binary continuous then f 1 (A, B) is open in Z for every (A, B) in B. Proof: We assume that f is binary continuous. Let (A, B) B. We show that f 1 (A, B) is open in Z. Since B M we have (A, B) M. Since f is binary continuous, we have f 1 (A, B) is open in Z. Proposition 3.9: Let (X, Y, M ) be a binary space such that (A, D) and (C, B) are also binary open sets whenever (A, B) and (C, D) are binary open in (X, Y, M ). Let B be a base for (X, Y, M ). The map f : Z X Y is binary continuous if and only if f 1 (A, B), f 1 (C, D), f 1 (A, D), f 1 (C, B) are open in Z for any two elements (A, B), (C, D) in B. Proof: We assume that f is binary continuous. Let (A, B) and (C, D) B.
On Binary Topological Spaces 137 Since B M we have (A, B) and (C, D) M. Therefore, by hypothesis (A, D) and (C, B) M. Since f is binary continuous, we have f 1 (A, B), f 1 (C, D), f 1 (A, D), f 1 (C, B) are open in Z. Conversely, assume that f 1 (A, B), f 1 (C, D), f 1 (A, D), f 1 (C, B) are open in Z for any two elements (A, B), (C, D) in B. Let (U, V) M. Since B is a basis for (X, Y, M ), we have there is a family {(A a, B a ) : α } of members of B such that (U, V) = Aα. Therefore, f 1 (U, V) = f 1 Aα. By Lemma 3.9, f 1 (U, V) = α, β open in Z. This proves the proposition. 1 f ( A,) B α β. This implies that f 1 (U, V) is Proposition 3. 10: Let S be a sub base for (X, Y, M ). If the map f : Z X Y is binary continuous then f 1 (A, B) is open in Z for every (A, B) in S. Proof: We assume that f is binary continuous. Let (A, B) S. We show that f 1 (A, B) is open in Z. Since S M we have (A, B) M. Since f is binary continuous, f 1 (A, B) is open in Z. Proposition 3. 11: Let (X, Y, M ) be a binary space such that (A, D) and (C, B) are also binary open sets whenever (A, B) and (C, D) are binary open in (X, Y, M ). Let S be a sub base for (X, Y, M ). The map f : Z X Y is binary continuous if and only if f 1 (A, B), f 1 (C, D), f 1 (A, D), f 1 (C, B) are open in Z for any two elements (A, B), (C, D) in S. Proof: Suppose f is binary continuous. Let (A, B) and (C, D) S. Since S is a sub base for M, by Definition 3.2, we have (A, B) and (C, D) M that implies by the given conditions, ( A, D) and ( C, B) M. Since f is binary continuous, f 1 (A, B), f 1 (C, D), f 1 (A, D) and f 1 (C, B) are open in Z. Conversely, we assume that f 1 (A, B), f 1 (C, D), f 1 (A, D), f 1 (C, B) are open in Z for any two elements (A, B), (C, D) in S. Since S is a sub base for ( X, Y, M ), we have there is a family B = {(A 1 A 2,, A k B 2,, B k ) : (A i, B i ) S for i = 1, 2,..., k where k > 0 is an integer} is a base for M. Let (A, B) = (A 1 A 2,, A k B 2,, B k ) and (C, D) = (C 1 C 2,, C m, D 1 D 2,, D m ) where (A i, B i ) S for i = 1, 2,..., k and (C j, D j ) S
138 S. Nithyanantha Jothi & P. Thangavelu for j = 1, 2,.., m. Then by using Lemma 3.7(ii) f 1 (A, B) = ( f 1 (A 1 ),, f 1 (A k ), f 1 (B 1 ),, f 1 (B k )), f 1 (A, D) = (f (A 1 ),, f 1 (A k ), f 1 (D 1 ),, f 1 (D k )), f 1 (C, D) = (f 1 (C 1 ),, f 1 (C k ), f 1 (D 1 ),, f 1 (D k )) and f 1 (C, B) = (f 1 (C 1 ),, f 1 (C k ), f 1 (B 1 ),, f 1 (B k )). By our assumption and by using the definition of a binary topology, it follows that f 1 (A, B), f 1 (A, D), f 1 (C, B) and f 1 (C, D) are open in Z. Then by applying Proposition 3.9, f is binary continuous. 4. CONCLUSION A topology between two sets X and Y other than the product topology has been discussed in his paper. Some elementary properties on bases, sub bases and continuity have also been studied. REFERENCES [1] Ryszard Engelking, Generel Topology, Polish Scientific Publishers, Warszawa, (1977). [2] S. Nithyanantha Jothi, and P. Thangavelu, Topology Between Two Sets, (Submitted). S. Nithyanantha Jothi Department of Mathematics, Aditanar College, Tiruchendur-628 216, India. E-mail: nithananthajothi@gmail.com P. Thangavelu Department of Mathematics, Karunya University, Coimbatore-641 114, India. E-mail: ptvelu12@gmail.com, thangavelu@karunya.edu