(j-i)sg_κ^*-closed sets and pairwise semi t_ω-axiom in bispaces
aa r X i v : . [ m a t h . GN ] F e b ( j − i ) sg ∗ κ -CLOSED SETS AND PAIRWISE SEMI T ω -AXIOM IN BISPACES JAGANNATH PAL AND AMAR KUMAR BANERJEE
Department of Mathematics, The University of Burdwan, Golapbag, East Burdwan-713104,West Bengal, India. Email: [email protected]; Email: [email protected]. A BSTRACT . Here we have introduced the ideas of ( j − i ) sg ∗ κ -closed sets and a semi gen-eralized closed set in a bispace; i, j = 1 , i = j and then have studied on pairwise semi T -axiom, pairwise semi T -axiom and pairwise semi T ω -axiom. We have investigated someof their topological properties and also established a relation among these axioms under someadditional conditions. Introduction
For many years several generalization on topological spaces are done in a more generalstructure of spaces and obtained interesting results [6, 11, 12, 17]. A. D. Alexandroff [1]generalized a topological space to a σ -space (or simply a space) by weakening union re-quirements where only countable union of open sets were taken to be open. J. C. Kelly [13]turned his attention to generalize topological spaces to a bitopological space. Bitopologicalspae was further generalized to a bispace by Lahiri and Das [15]. Then lot of contributionwere made in this direction by several authors [2, 3, 4, 20]. Simultaneously, closed setsin topological spaces are also generalized in many ways and several topological propertieswere studied on different spaces. N. Levine [16] gave the idea of generalized closed setsin a topological space. In 1987, Bhattacharyya and Lahiri [5] introduced the class of semi-generalized closed sets in topological spaces. P. Das and Rashid [8, 9] defined g ∗ -closed set, sg ∗ -closed set in σ -spaces and studied their various properties. M. S. Sarsak (2011) [19]studied g µ -closed sets on a generalized topological space.In this paper we have studied the notion of semi-generalized closed sets in a bispace ( X, κ , κ ) by introducing ( j − i ) sg ∗ κ -closed sets; i, j = 1 , i = j . We have also introducedpairwise semi T -axiom, pairwise semi T -axiom, pairwise semi T ω -axiom in a bispace andinvestigated some of their various topological properties and established a relation amongthese axioms under certain conditions. AMS Subject Classification (2010): Primary 54A05, 54D10Keywords and phrases: ( j − i ) sg ∗ κ -closed set; i, j = 1 , i = j , pairwise semi T -axiom, pairwisesemi T -axiom, pairwise semi T ω -axiom, pairwise strongly semi-symmetric bispace. PreliminariesDefinition 2.1. [1] . Let P be a collection of subsets of a non empty set X satisfying thefollowing axioms:(1) The intersection of a countable number of sets in P is a set in P ,(2) The union of a finite number of sets in P is a set in P ,(3) The empty set ∅ and the whole set X are sets in P ,then the set X is called an Alexandroff space or a σ -space or simply a space and sets of P are called closed sets, complements are open sets. We may take open sets in lieu of closed setsin the definition subject to the conditions that countable summability, finite intersectability,the whole set X and ∅ should be open. Sometimes we denote the collection of all such opensets by κ and the σ -space by ( X, κ ) . Definition 2.2. (cf. [13] ) . Suppose κ and κ are two collections of subsets of a nonemptyset X such that ( X, κ ) and ( X, κ ) are two σ -spaces, then ( X, κ , κ ) is called a bispace. Definition 2.3. (cf. [14] ). A set B in a bispace ( X, κ , κ ) is said to be semi κ i -open ( sκ i -open in short) if there exists a κ i -open set G in X such that G ⊂ B ⊂ G κ i ; B is said tobe semi κ i -closed ( sκ i -closed in short) if X − B is semi κ i -open; i = 1 , where G κ i is the κ i -closure of G defined parallely as in a topological space. Note that a topological space is a space but in general κ is not a topology. We take thedefinitions of semi-closure, semi-interior, semi-limit point of a set in a σ -space similar as inthe case of a topological space. Closure (resp. s-closure) of a set may not be closed (resp.semi closed). The space ( X, κ ) will simply be denoted by X and sets always mean subsetsof X . The complement of a set A is denoted by A c . Throughout the paper R and Q standrespectively for the set of real numbers and the set of rational numbers.We denote the class of all sκ i -open sets and sκ i -closed sets in a bispace ( X, κ , κ ) re-spectively by κ i - s.o. ( X ) and κ i - s.c. ( X ); i = 1 , . Definition 2.4. (cf. [13] ). A bispace ( X, κ , κ ) is called pairwise semi- T if for any pair ofdistinct points of X , either there is a semi κ -open set U such that x ∈ U, y U or, there isa semi κ -open set V such that y ∈ V, x V . Definition 2.5. (cf. [18] ). A bispace ( X, κ , κ ) is said to be pairwise semi- T if for any pairof distinct points x, y ∈ X there exist semi κ -open set U and semi κ -open set V such that x ∈ U, y U, y ∈ V, x V . If B is a subset of a bispace ( X, κ , κ ) , then we denote semi κ i -closure of B and semi κ i -derived set of B respectively by sB κ i and sB ′ κ i . Theorem 2.1. [10] . If B ⊂ ( X, κ , κ ) , a bispace, then sB κ i = B ∪ sB ′ κ i ; i = 1 , . Definition 2.6. (cf. [18] ). A bispace ( X, κ , κ ) is said to be pairwise semi- R if for everysemi κ i -open set U, x ∈ U implies s { x } κ j ⊂ U ; i, j = 1 , i = j . Definition 2.7. [5] . A topological space is called semi- T if every sg -closed set D is semi-closed (where D is sg -closed if s ( D ) ⊂ O, O is s-open, D ⊂ O ). j − i ) sg ∗ κ -CLOSED SETS AND PAIRWISE SEMI T ω -AXIOM IN BISPACES 3 Definition 2.8. [19] . Let D be a subset in a bispace ( X, κ , κ ) . We define semi-kernel of D with respect to κ i ; i = 1 , denoted by sker i ( D ) = ∩{ U : D ⊂ U ; U is semi κ i -open } . Definition 2.9. (c.f. [7] ) Two sets
E, F in a bispace ( X, κ , κ ) are said to be semi κ i -separated if there are two sκ i -open sets G, H such that E ⊂ G, F ⊂ H and E ∩ H = F ∩ G = ∅ ; i = 1 , . Definition 2.10. (c.f. [7] ) ( X, κ , κ ) is said to be pairwise semi-door bispace if every subsetof it is not sκ i -closed then it is sκ j -open; i, j = 1 , i = j . Theorem 2.2. (c.f. [9] ). Let ( X, κ , κ ) be a bispace, then sκ i -closure of a set is sκ i -closedif and only if arbitrary intersection of κ i -closed sets is sκ i -closed in X ; i = 1 , . ( j − i ) sg ∗ κ -closed sets and pairwise semi- T , pairwise semi- T axioms in bispaces In this section we introduce the ideas of ( j − i ) sg ∗ κ -closed sets and ( j − i ) sg ∗ κ -open sets in abispace to investigate some topological properties of the sets and to corelate with some newseparation axioms like pairwise semi- T , pairwise semi- T and pairwise semi- R axioms. Definition 3.1.
A set B of a bispace ( X, κ , κ ) is said to be semi g ∗ κ i -closed with respect to sκ j denoted by ( j − i ) sg ∗ κ -closed if there is a sκ i -closed set F containing B such that F ⊂ O whenever B ⊂ O and O is sκ j -open. B is called semi g ∗ κ i -open with respect to sκ j denotedby ( j − i ) sg ∗ κ -open if X − B is ( j − i ) sg ∗ κ -closed; i, j = 1 , i = j . Theorem 3.1.
A set B in a bispace ( X, κ , κ ) is ( j − i ) sg ∗ κ -closed if and only if there is a sκ i -closed set F containing B such that F ⊂ sker j ( B ); i, j = 1 , i = j .Proof. Suppose B ⊂ X and B is ( j − i ) sg ∗ κ -closed. Then there is a sκ i -closed set F containing B such that F ⊂ O whenever B ⊂ O and O is sκ j -open. Therefore F ⊂ sker j ( B ); i, j = 1 , i = j .Converse part is obvious. (cid:3) Theorem 3.2.
A set B in a bispace ( X, κ , κ ) is ( j − i ) sg ∗ κ -open if and only if there is a sκ i -open set V contained in B such that F ⊂ V whenever F is sκ j -closed and F ⊂ B ; i, j = 1 , i = j .Proof. Let B be a ( j − i ) sg ∗ κ -open set; i, j = 1 , i = j . Then by Definition 3.1, X − B isa ( j − i ) sg ∗ κ -closed set and hence there is a sκ i -closed set F containing B c such that F ⊂ U whenever B c ⊂ U and U is sκ j -open. Therefore, F c ⊂ B such that F c ⊃ U c whenever B ⊃ U c , where U c is a sκ j -closed set and F c is a sκ i -open set. Hence we can say that B is ( j − i ) sg ∗ κ -open set if and only if there exists a sκ i -open set F c = V, V ⊂ B such that F c = V ⊃ P = U c whenever P ⊂ B and P is a sκ j -closed. (cid:3) Remark 3.1.
Every sκ i -closed set D in a bispace ( X, κ , κ ) is ( j − i ) sg ∗ κ -closed; i, j =1 , i = j but Example 3.1 shows that the converse may not be true; however, it is true if inaddition D is semi κ j -open. Again if a set D = sker j ( D ) then it is ( j − i ) sg ∗ κ -closed if andonly if it is sκ i -closed. JAGANNATH PAL AND AMAR KUMAR BANERJEE
Example 3.1.
Suppose X = R − Q, κ = { X, ∅ , p { } ∪ G i } , κ = { X, ∅ , G i } , G i runsover all countable subsets of X . Then ( X, κ , κ ) is a bispace but not a bitopological space.Incidentally the bispace is pairwise semi- T and also pairwise semi- T . Suppose A is the setof all irrational numbers of (1 , . Then A is not sκ -closed as X − A is not sκ -open (for √ ∈ (1 , ) but A is (2 − sg ∗ κ -closed since X is the only sκ -open set containing A . Theorem 3.3.
If a set B of a bispace ( X, κ , κ ) is ( j − i ) sg ∗ κ -closed; i, j = 1 , i = j thenthere exists a sκ i -closed set F containing B such that F − B does not contain any non-empty sκ j -closed set; i, j = 1 , i = j .Proof. Let B be a ( j − i ) sg ∗ κ -closed set of a bispace ( X, κ , κ ) ; i, j = 1 , i = j . Thenthere exists a sκ i -closed set F containing B such that F ⊂ sker j ( B ) . Let P be a sκ j -closedset such that P ⊂ F − B . Then X − P is sκ j -open and B ⊂ X − P ⇒ F ⊂ X − P ⇒ P ⊂ X − F . So P ⊂ ( X − F ) ∩ F = ∅ . Hence the result follows. (cid:3) But the reverse implication in Theorem 3.3 is not true as seen from Example 3.2 althoughit is true in a space [9].
Example 3.2.
Suppose X = { a, b, c } , κ = {∅ , X, { a } , { a, b }} and κ = {∅ , X, { b } , { b, c }} .Then ( X, κ , κ ) is a bitopological space so a bispace. Consider the set { b } which is con-tained in sκ -closed set { b, c } , so { b, c } − { b } = { c } does not contain any non-empty sκ -closed set. Since { b } itself is a sκ -open set, there is no sκ -closed set containing { b } contained in { b } ; and hence { b } is not (2 − sg ∗ κ -closed. Theorem 3.4.
Suppose D is a ( i − j ) sg ∗ κ -closed set in a bispace ( X, κ , κ ) ; i, j = 1 , i = j ,then D is sκ j -closed if and only if sD κ j and sD κ j − D are sκ j -closed.Proof. Let the ( i − j ) sg ∗ κ -closed set D be a sκ j -closed set in a bispace ( X, κ , κ ) then sD κ j = D is sκ j -closed and sD κ j − D = ∅ is sκ j -closed; i, j = 1 , i = j .Conversely, let sD κ j and sD κ j − D be sκ j -closed. Since D is ( i − j ) sg ∗ κ -closed, then thereis a sκ j -closed set P, P ⊃ D such that P − D does not contain any non-empty sκ i -closedset. So sD κ j − D ⊂ P − D ⇒ sD κ j − D = ∅ ⇒ sD κ j = D ⇒ D is sκ j -closed. (cid:3) Intersection and union two ( j − i ) sg ∗ κ -closed sets in a bispace ( X, κ , κ ) are not in generala ( j − i ) sg ∗ κ -closed; i, j = 1 , i = j as shown by the Examples 3.3 (i) and 3.3 (ii). However,Theorems 3.5and 3.6 show that these are true under certain additional conditions. Example 3.3. (i) : Suppose X = { a, b, c } , κ = {∅ , X, { a, b }} and κ = {∅ , X, { b }} .Then ( Y, κ , κ ) is a bitopological space so a bispace. Now a family of sκ -open sets is {∅ , X, { a, b }} and a family of sκ -open sets is {∅ , X, { b }} . Then clearly the sets { a, b } and { b, c } are (2 − sg ∗ κ -closed sets but { a, b } ∩ { b, c } = { b } is not (2 − sg ∗ κ -closed set.(ii): Suppose X = R − Q and κ = { X, ∅ , G i ∪ {√ } where G i s are countable subsetsof X and κ = { X, ∅ , G i } where G i s are countable subsets of X − {√ } . Then ( X, κ , κ ) is a bispace, but not a bitopological space. Suppose A = X − {√ , √ , √ } and B = X −{√ , √ , √ } . Then X − A is sκ -open since {√ , √ } is a κ -open set and {√ , √ } ⊂ j − i ) sg ∗ κ -CLOSED SETS AND PAIRWISE SEMI T ω -AXIOM IN BISPACES 5 X − A ⊂ {√ , √ } . Similarly, X − B is also sκ -open. Therefore A and B are sκ -closed sets and hence A and B are (1 − sg ∗ κ -closed sets. Again take C = A ∪ B =( X − {√ , √ , √ } ) ∪ ( X − {√ , √ , √ } ) = X − {√ } . Since {√ } is not sκ -open,then X − {√ } is not sκ -closed, so X is the only sκ -closed set containing C . But C is a sκ -open set. Hence by Definition 3.1, C is not a (1 − sg ∗ κ -closed set. Theorem 3.5.
Union of two ( j − i ) sg ∗ κ -closed sets in a bispace ( X, κ , κ ) is ( j − i ) sg ∗ κ -closedif union of two sκ i -closed sets is ( j − i ) sg ∗ κ -closed set; i, j = 1 , i = j .Proof. Let
E, F be two ( j − i ) sg ∗ κ -closed sets in a bispace ( X, κ , κ ); i, j = 1 , i = j .Assume E ∪ F ⊂ G, G is sκ j -open. So E ⊂ G, F ⊂ G and there are sκ i -closed sets P, Q containing respectively
E, F such that P ⊂ G, Q ⊂ G . Hence P ∪ Q ⊃ E ∪ F and P ∪ Q ⊂ G . By our assumption P ∪ Q is ( j − i ) sg ∗ κ -closed and so there is a sκ i -closed set K containing P ∪ Q such that K ⊂ G ⇒ E ∪ F ⊂ P ∪ Q ⊂ K ⊂ G . Hence the resultfollows. (cid:3) Theorem 3.6.
Union of two semi κ j -separated ( j − i ) sg ∗ κ -open sets in a bispace ( X, κ , κ ) is ( j − i ) sg ∗ κ -open; i, j = 1 , i = j .Proof. Suppose E , E are two semi κ j -separated ( j − i ) sg ∗ κ -open sets; i, j = 1 , i = j .Then for semi κ j -separated sets there are sκ j -open sets G , G such that E ⊂ G , E ⊂ G and E ∩ G = E ∩ G = ∅ . Let D = G c , D = G c . Then D , D are sκ j -closed sets and E ⊂ D , E ⊂ D . Again if P n ⊂ E n , n = 1 , P n is sκ j -closed set then for ( j − i ) sg ∗ κ -open sets, there is sκ i -open set U n ⊂ E n such that P n ⊂ U n by Theorem 3.2. Clearly, U ∪ U is a sκ i -open set and U ∪ U ⊂ E ∪ E . Assume D is a sκ j -closed set and D ⊂ E ∪ E .Now D = D ∩ ( E ∪ E ) = ( D ∩ E ) ∪ ( D ∩ E ) ⊂ ( D ∩ D ) ∪ ( D ∩ D ) where ( D ∩ D ) and ( D ∩ D ) are sκ j -closed sets. Further ( D ∩ D ) ⊂ ( E ∪ E ) ∩ D = ( D ∩ E ) ∪ ( D ∩ E ) = ∅ ∪ ( D ∩ E ) ⊂ E and so ( D ∩ D ) ⊂ U . Similarly, ( D ∩ D ) ⊂ U . Then D ⊂ U ∪ U and hence the result follows. (cid:3) Theorem 3.7.
Let ( X, κ , κ ) be a bispace. If C is ( j − i ) sg ∗ κ -closed and C ⊂ D ⊂ sC κ i then D is ( j − i ) sg ∗ κ -closed; i, j = 1 , i = j .Proof. Suppose C is a ( j − i ) sg ∗ κ -closed set in ( X, κ , κ ) , C ⊂ D ⊂ sC κ i and D ⊂ U, U is sκ j -open; i, j = 1 , i = j . Then C ⊂ U and hence there is a sκ i -closed set P containingthe set set C such that P ⊂ U . Now C ⊂ P ⇒ sC κ i ⊂ P ⇒ P ⊃ sC κ i ⊃ D ⇒ D is ( j − i ) sg ∗ κ -closed. (cid:3) Theorem 3.8.
Let ( X, κ , κ ) be a bispace and C ⊂ B where B is sκ j -open and ( j − i ) sg ∗ κ -closed; i, j = 1 , i = j . Then C is ( j − i ) sg ∗ κ -closed if and only if C is ( j − i ) sg ∗ κ -closedrelative to B .Proof. Let ( X, κ , κ ) be a bispace and C ⊂ B where B is sκ j -open and ( j − i ) sg ∗ κ -closed; i, j = 1 , i = j . Then by Remark 3.1, B is sκ i -closed. Let C be ( j − i ) sg ∗ κ -closed. Since B is sκ j -open and C ⊂ B , there is a sκ i -closed set P containing the set C such that P ⊂ B and so P is sκ i -closed in B . Now let C ⊂ G where G is sκ j -open in B . Evidently G is sκ j -open in X and hence P ⊂ G ⇒ C is ( j − i ) sg ∗ κ -closed in B . JAGANNATH PAL AND AMAR KUMAR BANERJEE
Conversely, let C be ( j − i ) sg ∗ κ -closed relative to B . Then there is a sκ i -closed set P in B containing the set C such that P ⊂ G ′ where G ′ is sκ j -open in B containing the set C .As B is sκ i -closed, P is sκ i -closed in X . Let C ⊂ H, H is sκ j -open, then C ⊂ H ∩ B , a sκ j -open set in B . Hence C ⊂ P ⊂ H ∩ B ⇒ C ⊂ P ⊂ H ⇒ C is ( j − i ) sg ∗ κ -closed. (cid:3) Corollary 3.1.
Let B be sκ j -open and ( j − i ) sg ∗ κ -closed set in a bispace ( X, κ , κ ) , then B ∩ C is ( j − i ) sg ∗ κ -closed if C is ( j − i ) sg ∗ κ -closed; i, j = 1 , i = j .Proof. Obviously B is sκ i -closed, so B ∩ C is sκ i -closed in C which implies that B ∩ C is ( j − i ) sg ∗ κ -closed in C . Then by Theorem 3.8, B ∩ C is ( j − i ) sg ∗ κ -closed. (cid:3) Theorem 3.9.
If each subset of a bispace ( X, κ , κ ) is ( j − i ) sg ∗ κ -closed, then κ i - s.c. ( X ) = κ j - s.o. ( X ); i = j ; i, j = 1 , .Proof. Let each subset be ( j − i ) sg ∗ κ -closed and B ∈ κ j - s.o. ( X ) ; i, j = 1 , i = j . Then B is ( j − i ) sg ∗ κ -closed, so there is a sκ i -closed set P containing the set B such that P ⊂ B which implies that P = B ⇒ B is sκ i -closed and so κ j - s.o. ( X ) ⊂ κ i - s.c. ( X ) . Again let E ∈ κ i - s.c. ( X ) ⇒ X − E ∈ κ i - s.o. ( X ) , then by same argument, X − E is sκ j -closed andhence E is sκ j -open. So κ i - s.c. ( X ) ⊂ κ j - s.o. ( X ) ⇒ κ i - s.c. ( X ) = κ j - s.o. ( X ) . (cid:3) The reverse implication of above Theorem is not true as revealed from the Example 3.4.But the converse is true under some additional condition shown in the next Theorem 3.10.
Example 3.4.
Suppose X = R − Q, κ = κ = {∅ , X, G n , D n } , where G n and D n runover respectively all countable sets and cocountable sets of X . Then ( X, κ , κ ) is a bispacebut not a bitopological space. Clearly, we have κ i - s.c. ( X ) = κ j - s.o. ( X ); i = j ; i, j = 1 , .Now consider the set D of all irrational numbers in (0 , . Then D κ i - s.c. ( X ) ⇒ D is not ( j − i ) sg ∗ κ -closed, by Remark 3.1, since D = sker j ( D ) . Theorem 3.10.
Let ( X, κ , κ ) be a bispace with κ i - s.c. ( X ) = κ j - s.o. ( X ); i = j ; i, j = 1 , ,then each subset is ( j − i ) sg ∗ κ -closed if and only if the bispace satisfies the condition (C).(C): Arbitrary intersection of κ i -closed sets is sκ i -closed in X .Proof. Let ( X, κ , κ ) be a bispace with κ i - s.c. ( X ) = κ j - s.o. ( X ); i = j ; i, j = 1 , .Suppose that every subset is ( j − i ) sg ∗ κ -closed and P n is an arbitrary collection of κ i -closedsets and P = T P n . Then P is ( j − i ) sg ∗ κ -closed. So there is a sκ i -closed set P ′ containing P such that P ′ ⊂ V whenever P ⊂ V, V is sκ j -open. Now since each P n ∈ κ i - s.c. ( X ) = κ j - s.o. ( X ) and P ⊂ P n , it follows that P ′ ⊂ P n , for all n i.e. P ′ ⊂ T P n = P . Hence P ′ = P and the result follows.Conversely, let the condition (C) hold and D ⊂ X . So ( D ) κ i is sκ i -closed and henceby Theorem 2.2, s ( D ) κ i = P (say) is sκ i -closed. Let D ⊂ U, U is sκ j -open. Since κ i - s.c. ( X ) = κ j - s.o. ( X ); i = j ; i, j = 1 , U is sκ i -closed. So P = s ( D ) κ i ⊂ s ( U ) κ i = U . So D is ( j − i ) sg ∗ κ -closed. (cid:3) Theorem 3.11.
Suppose ( X, κ , κ ) is a bispace. For each x ∈ X , if { x } is not sκ i -closedthen { x } c is ( i − j ) sg ∗ κ -closed; i, j = 1 , i = j . j − i ) sg ∗ κ -CLOSED SETS AND PAIRWISE SEMI T ω -AXIOM IN BISPACES 7 Proof.
Suppose x ∈ X and { x } is not sκ i -closed, then { x } c is not κ i -open. Therefore X isthe only sκ i -open set which contains { x } c . Also X may be taken as sκ j -closed set containing { x } c . Therefore { x } c is a ( i − j ) sg ∗ κ -closed; i, j = 1 , i = j . (cid:3) Theorem 3.12.
A bispace ( X, κ , κ ) is pairwise semi- T if and only if for any pair of distinctpoints x, y ∈ X , there is a set B which contains only one of them such that B is either sκ i -open or sκ j -closed, i, j = 1 , .Proof. Suppose that the bispace ( X, κ , κ ) is pairwise semi- T and x, y ∈ X, x = y . Sothere exists a sκ i -open set B containing one of x, y say x , but not y . Then it is done. Theremay arise the case that x ∈ B and y B . But B is not a sκ i -open set, then y ∈ X − B, x X − B and X − B is sκ j -open. So B is sκ j -closed.Conversely, suppose the condition holds and B is a sκ i -open set containing x only. Thenclearly the bispace is pairwise semi- T . If x ∈ B but B is sκ j -closed, then X − B is sκ j -openand y ∈ X − B, x X − B . Hence in this case also the bispace is pairwise semi- T . (cid:3) Remark 3.2. If ( X, κ ) or ( X, κ ) are semi- T spaces, then the bispace ( X, κ , κ ) is pair-wise semi- T . Again, it can be easily shown that pairwise semi- T is pairwise semi- T butthe converse may not be true as seen from the Example 3.5 given below. Example 3.5.
Example of pairwise semi- T bispace which is not pairwise semi- T .Suppose X = R − Q, κ = { X, ∅ , {√ , √ } ∪ G i } , κ = { X, ∅ , G i } where G i runsover all countable subsets of X . Then ( X, κ , κ ) is a bispace but not a bitopological space.Clearly, the bispace is pairwise semi- T but it is not pairwise semi- T . For, consider thepair of distinct points √ , √ ∈ X and it will not be possible to find a sκ -open set whichcontains only one of √ , √ . So ( X, κ , κ ) is not pairwise semi- T bispace. Now we are going to find out a condition for a pairwise semi- T bispace to be a pairwisesemi- T . Theorem 3.13.
If a bispace ( X, κ , κ ) is pairwise semi- T then for each pair of distinctpoints p, q ∈ X , either p s { q } κ or q s { p } κ .Proof. Let ( X, κ , κ ) be a pairwise semi- T bispace and p, q ∈ X, p = q . Since X ispairwise semi- T , there exists a sκ -open set U which contains only one of p, q . Supposethat p ∈ U and q U . Then the sκ -open set U has an empty intersection with { q } . Hence p s { q } ′ κ . Since p = q, p s { q } κ . Similarly, there may exist a sκ -open set V such that q ∈ V, p V and we get q s { p } κ . Hence the result follows. (cid:3) Theorem 3.14.
In a pairwise semi- T bispace, p, q ∈ X, p = q implies s { p } κ = s { q } κ .Proof. Let the bispace ( X, κ , κ ) be pairwise semi- T and p, q ∈ X, p = q . Then either p s { q } κ or q s { p } κ . Let p s { q } κ , but p ∈ s { p } κ . Thus s { p } κ = s { q } κ . (cid:3) Theorem 3.15.
Pairwise semi- T and pairwise semi- R bispace is pairwise semi- T .Proof. Let the bispace ( X, κ , κ ) be pairwise semi- T and pairwise semi- R and p, q ∈ X ; p = q . By Theorem 3.13, either p s { q } κ or q s { p } κ . Let q s { p } κ , then JAGANNATH PAL AND AMAR KUMAR BANERJEE there is a sκ -closed set P containing p such that q P . So q ∈ X − P which is a sκ -open set and p X − P . From Definition 2.6 of pairwise semi- R bispace, s { q } κ ⊂ X − P ⇒ s { q } κ ∩ P = ∅ ⇒ s { q } κ ∩ { p } = ∅ ⇒ p s { q } κ . As p = q , then p is nota sκ -limit point of { q } . Then there is a sκ -open set U such that p ∈ U but q U . So p ∈ U, q ∈ X − P, p X − P, q U . Thus the bispace is pairwise semi- T . (cid:3) The condition pairwise semi R in the above Theorem is not necessary for pairwise semi- T as shown by Example 3.6 although pairwise semi- T implies pairwise semi- T . Example 3.6.
Example of pairwise semi- T pairwise semi- R .Let X = { a, b, c } , κ = {∅ , X, { a } , { c } , { a, c }} , κ = {∅ , X, { b } , { a, b }} . Then ( X, κ , κ ) is a bitopological space so a bispace. Here a family of sκ -open sets is {∅ , X, { a } , { c } , { a, c } , { a, b } , { b, c }} and a family of sκ -open sets is {∅ , X, { b } , { a, b } , { b, c }} and the bispace ispairwise semi- T . Now consider b ∈ { a, b } , a sκ -open set, then s { b } κ = X
6⊂ { a, b } which implies that ( X, κ , κ ) is not pairwise semi- R . Definition 3.2.
A bispace ( X, κ , κ ) is said to be pairwise semi-symmetric if for any x, y ∈ X, x ∈ s { y } κ i = ⇒ y ∈ s { x } κ j , i, j = 1 , i = j . Theorem 3.16.
Pairwise semi-symmetric, pairwise semi- T bispace is pairwise semi T .Proof. Let ( X, κ , κ ) be a pairwise semi-symmetric and pairwise semi- T bispace and let a, b ∈ X, a = b . Since the bispace is pairwise semi- T , either a s { b } κ or b s { a } κ .Let a s { b } κ . Then we claim that b s { a } κ . For, if b ∈ s { a } κ then it would implythat a ∈ s { b } κ , since the bispace is pairwise semi-symmetric. But this contradicts that a s { b } κ . Since a s { b } κ , there is a sκ -closed set F such that b ∈ F and a F . So a ∈ X − F , a sκ -open set and b X − F . Again since b s { a } κ , there is a sκ -closedset P such that a ∈ P and b P . So b ∈ X − P , a sκ -open set and a X − P . Hence thebispace is pairwise semi- T . (cid:3) Pairwise semi- T ω bispace In this section we give the idea of pairwise semi- T ω axiom in a bispace and discuss some ofits properties. Though pairwise semi- T ω axiom can not be placed between pairwise semi- T and pairwise semi- T axioms but we try to establish a relation among them. Definition 4.1.
A bispace ( X, κ , κ ) is said to be pairwise semi- T ω if and only if every ( j − i ) sg ∗ κ -closed set is sκ i -closed; i, j = 1 , i = j . A topological space is T [11] if and only if each singleton is either open or closed. For a σ -space, condition is necessary but it is not sufficient [8]. For this we introduce the followingdefinition to find necessary and sufficient conditions for a bispace to be pairwise semi- T ω . Definition 4.2.
For any set D in a bispace ( X, κ , κ ) , we define sD ∗ jgi = T { A : D ⊂ A, A is ( j − i ) sg ∗ κ -closed in X } , then sD ∗ jgi is called ( j − i ) sg ∗ κ -closure of D ; i, j = 1 , i = j .We denote the following sets as G i and G ′ i which will be used in the sequel. j − i ) sg ∗ κ -CLOSED SETS AND PAIRWISE SEMI T ω -AXIOM IN BISPACES 9 (i) G i = { A : s ( X − A ) κ i is sκ i -closed } ; i = 1 , and(ii) G ′ i = { A : s ( X − A ) ∗ jgi is ( j − i ) sg ∗ κ -closed; i, j = 1 , i = j . Theorem 4.1.
Suppose ( X, κ , κ ) is a pairwise semi- T ω bispace then(i): for each x ∈ X, if { x } is is not sκ i -closed, it is sκ j -open; i, j = 1 , i = j ,(ii): κ i - s.o. ( X ) = κ i − j - s.o. ( X ) ∗ where κ i − j - s.o. ( X ) ∗ = { D : D ⊂ X : s ( X − D ) ∗ jgi = X − D ; i = j ; i, j = 1 , .Proof. Let ( X, κ , κ ) be a pairwise semi- T ω bispace.(i): Let x ∈ X, { x } is not sκ i -closed, then by Theorem 3.11, { x } c is ( i − j ) sg ∗ κ -closed, so { x } c is sκ j -closed, i, j = 1 , i = j . This implies that { x } is sκ j -open.(ii): For pairwise semi- T ω bispace, sκ i -closed sets and ( j − i ) sg ∗ κ -closed sets coincideand hence for each D ⊂ X we have s ( D ) κ i = sD ∗ jgi ; i, j = 1 , i = j . Hence the resultfollows. (cid:3) Theorem 4.2.
A bispace ( X, κ , κ ) is pairwise semi- T ω if and only if(a) for each x ∈ X , { x } is not sκ i -closed, then it is sκ j -open; i, j = 1 , i = j and(b) G i = G ′ i where G i and G ′ i are as in Definition 4.2Proof. Suppose ( X, κ , κ ) is a pairwise semi- T ω bispace and x ∈ X, { x } is not sκ i -closed,then by Theorem 3.11, { x } c is ( i − j ) sg ∗ κ -closed and so { x } c is sκ j -closed, i, j = 1 , i = j .This implies that { x } is sκ j -open. Now let D ∈ G i , then s ( X − D ) κ i is sκ i -closed ⇒ s ( X − D ) κ i is ( j − i ) sg ∗ κ -closed ⇒ s ( X − D ) ∗ jgi is ( j − i ) sg ∗ κ -closed ⇒ D ∈ G ′ i ⇒ G i ⊂ G ′ i .Again suppose D ∈ G ′ i , then s ( X − D ) ∗ jgi is ( j − i ) sg ∗ κ -closed ⇒ s ( X − D ) κ i is sκ i -closed ⇒ D ∈ G i ⇒ G ′ i ⊂ G i . Hence G i = G ′ i .Conversely, suppose the conditions hold and D is ( j − i ) sg ∗ κ -closed. Then s ( D ) ∗ jg i = D is ( j − i ) sg ∗ κ -closed ⇒ X − D ∈ G ′ i . Since G i = G ′ i , then X − D ∈ G i ⇒ s ( D ) κ i is sκ i -closed. We claim that s ( D ) κ i = D . If not, then there exists x ∈ s ( D ) κ i − D . Since D is ( j − i ) sg ∗ κ -closed, { x } cannot be sκ j -closed, by Theorem 3.3. Hence by assumption, { x } is sκ i -open and so { x } c is sκ i -closed. But x D ⇒ D ⊂ { x } c ⇒ s ( D ) κ i ⊂ { x } c . Wesee that x ∈ s ( D ) κ i − D and s ( D ) κ i ⊂ { x } c . Both implies that x ∈ { x } c , a contradiction.Hence D is sκ i -closed and the bispace is pairwise semi- T ω . (cid:3) Theorem 4.3.
A bispace ( X, κ , κ ) is pairwise semi- T ω if and only if for i, j = 1 , i = j ;(i) each subset of X is the intersection of all sκ i -closed sets and all sκ j -open sets contain-ing it and(ii) G i = G ′ i where G i and G ′ i are as in Definition 4.2.Proof. Let the bispace ( X, κ , κ ) be pairwise semi- T ω . (i): Let D ⊂ X , then by Theorem4.2, D = T { X − { x } : x ∈ X, x D ; if { x } is not sκ i -closed, then it is sκ j -open; i, j = 1 , i = j } ; so D is the intersection of all sκ i -closed sets and all sκ j -open setscontaining it. (ii) It is proved in Theorem 4.2.Conversely, assume the conditions (i) and (ii) hold and x ∈ X, { x } is not sκ i -closed,then X − { x } is not sκ i -open; i, j = 1 , i = j and X is the only sκ i -open set containing X − { x } . So by assumption, X − { x } is sκ j -closed which implies that { x } is sκ j -open.Hence the result follows by heorem 4.2. (cid:3) Theorem 4.4.
Every semi-door bispace ( X, κ , κ ) is pairwise semi- T ω . The proof is straight forward.
Example 4.1.
Assume X = { p, q } and κ = κ = {∅ , X, { p }} , then ( X, κ , κ ) is a bitopo-logical space so a bispace. Obviously, the bispace ( X, κ , κ ) is pairwise semi- T ω but it isnot pairwise semi- T . Theorem 4.5.
Pairwise semi- T ω bispace ( X, κ , κ ) is pairwise semi- T .Proof. Suppose the bispace ( X, κ , κ ) is pairwise semi- T ω but it is not pairwise semi- T .Then there exist l, m ∈ X ; l = m such that s { l } κ = s { m } κ . We assert that { l } is not sκ -closed. If { l } is sκ -closed, then s { l } κ = { l } 6 = s { m } κ which contradicts the assumption.So by Theorem 3.11, { l } c is (2 − sg ∗ κ -closed. But the bispace is pairwise semi- T ω , so { l } c is sκ -closed. Now we show that { l } c is not sκ -closed. If { l } c is sκ -closed, then m ∈ { l } c ⇒ s { m } κ ⊂ s ( X − { l } ) κ = X − { l } . Therefore s { l } κ = s { m } κ whichcontradicts the assumption again . Hence the result follows. (cid:3) But the converse is not true always as revealed from the Example 3.1.
Remark 4.1.
It is seen that in a topological space, semi- T axiom can be placed betweensemi- T and semi- T axioms [5] . But pairwise semi- T ω axiom in a bispace does not have thisproperty as is evident from Examples 3.1 and 3.5, rather Examples 3.1 and 4.1 show thatpairwise semi- T w and pairwise semi- T axioms in a bispace are independent of each other. Itcan be easily verified that pairwise semi- T bispace is pairwise semi-symmetric but converseis not true as in [16] . In a symmetric topological space, T , T , T axioms are equivalent [16] . In a pairwise semi-symmetric bispace, though pairwise semi- T and pairwise semi- T axioms are equivalent but pairwise semi- T bispace may not imply pairwise semi- T w as canbe seen from the Example 4.2. A topological space is symmetric if and only if each singletonof X is g -closed [16] . There exists bispace which is pairwise semi-symmetric but singletonsare not ( j − i ) sg ∗ κ -closed; i, j = 1 , i = j can be seen from Example 4.2. Definition 4.3.
A bispace ( X, κ , κ ) is said to be pairwise strongly semi-symmetric if eachsingleton of X is ( j − i ) sg ∗ κ -closed, i, j = 1 , i = j . Theorem 4.6.
A pairwise strongly semi-symmetric bispace is pairwise semi-symmetric.Proof.
Suppose a bispace ( X, κ , κ ) is pairwise strongly semi-symmetric and l ∈ s { m } κ i ,but m s { l } κ j for l, m ∈ X, i, j = 1 , i = j . So there is a sκ j -closed set F ⊃ { l } suchthat m F ⇒ { m } ⊂ F c , a sκ j -open set. For pairwise strongly semi-symmetric bispace, { m } is ( j − i ) sg ∗ κ -closed, then there is a sκ i -closed set F ′ ⊃ { m } such that F ′ ⊂ F c . So s { m } κ i ⊂ F ′ and l ∈ s { m } κ i ⊂ F ′ ⊂ F c , a contradiction. Hence the result follows. (cid:3) But the converse may not be true as shown by the following Example. j − i ) sg ∗ κ -CLOSED SETS AND PAIRWISE SEMI T ω -AXIOM IN BISPACES 11 Example 4.2.
Suppose X = R − Q and κ = κ = { X, ∅ , G i } where { G i } are the countablesubsets of X . So ( X, κ , κ ) is a bispace but not a bitopological space. Suppose l, m ∈ X .If l = m then l ∈ s { m } κ i implies that l is a sκ i -limit point of { m } . But { l } is a sκ i -openset containing l which does not intersect { m } . Hence l cannot be a sκ i -limit point of { m } .So we must have l = m and in that case m ∈ s { l } κ j . Hence the bispace is pairwise semi-symmetric. But for each singleton { l } in X, sker j ( l ) = { l } and { l } is not sκ i -closed. So nosingleton is ( j − i ) sg ∗ κ -closed. Hence the result follows. Corollary 4.1.
A pairwise strongly semi-symmetric pairwise semi- T ω bispace is pairwisesemi- T . Proof follows from Theorems 3.16, 4.5, 4.6.
Remark 4.2.
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