C om mun.Fac.Sci.U niv.A nk.Series A 1 Volum e 67, N umb er 1, Pages 225–234 (2018) D O I: 10.1501/C om mua1_ 0000000844 ISSN 1303–5991
http://com munications.science.ankara.edu.tr/index.php?series= A 1
ON NEW TYPES OF SETS VIA -OPEN SETS IN
BITOPOLOGICAL SPACES
MERVE ·ILKHAN, MAHMUT AKY·I ¼G·IT, AND EMRAH EVREN KARA
Abstract. In this paper, the concept of (i; j)- -P -open sets in bitopological spaces are introduced and characterizations of their related notions are given.
1. Introduction
The notion of bitopological space (X; 1; 2) which is a nonempty set X endowed
with two topologies 1 and 2 is introduced by Kelly [1]. Also in [1], some basic
results of separation axioms in topological spaces are extended to bitopological spaces. In a bitopological space (X; 1; 2), a set S X is called (i; j)-preopen [2]
if S i-Int( j-Cl(S)), where i; j = 1; 2 and i 6= j. For a subset S of a bitopological
space, i-Int(S) and i-Cl(S) stand for the interior and the closure of S with respect
to the topology i, respectively. A function f : (X; 1; 2) ! (Y; 1; 2) is called
(i; j)-precontinuous [2] if the inverse image of every i-open set in (Y; 1; 2) is (i; j)-preopen in (X; 1; 2).
Ogata [3] de…ned an operation on a topological space (X; ) as a mapping from into the power set P(X) of X such that U (U ) for each U 2 , where (U) denotes the value of at U . By means of this operation, he de…ned the concept of operation-open sets called -open sets. A subset S of X is said to be -open if for each x 2 S, there exists an open set U containing x such that (U) S. It is clear that -open sets are open. A subset of X is called -closed if its complement is -open. -interior of S [4] denoted by -Int(S) is the union of all -open sets of X contained in S and -closure of S [3] denoted by -Cl(S) is the intersection of all -closed sets of X containing S.
In topological spaces, some generalizations of -open sets are introduced and some properties of these sets are investigated. Krishan and Balachandran [5] de…ned -preopen sets. A subset S of a topological space (X; ) is said to be -preopen if S -Int( -Cl(S)). Later, -P -open sets are de…ned and some weak separation
Received by the editors: August 04, 2016; Accepted: January 25, 2017.
2010 Mathematics Subject Classi…cation. Primary 54A05; Secondary 54E55, 54A10. Key words and phrases. Bitopological spaces, -interior, -closure, -P -open sets.
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axioms are introduced by Khalaf and Ibrahim [6]. A subset S of a topological space (X; ) is said to be -P -open if S Int( -Cl(S)). By P O(X; ), we denote the family of all -P -open sets in (X; ).
Operations on bitopological spaces were studied in [7]. An operation on a bitopological space (X; 1; 2) is a mapping : 1[ 2 ! P(X) such that U
(U ) for each U 2 1[ 2. In [8], the author studied some separation axioms in
bitopological spaces by utilizing operations on such spaces.
By i -Int(S) and j -Cl(S), we denote the -interior of S in (X; i) and
-closure of S in (X; j), respectively.
The main purpose of this paper is to extend the concepts of -preopen sets and -P -open sets which are weaker than -open sets to bitopological spaces. We give some properties related to these sets and also introduce some separation axioms in bitopological spaces. Further we de…ne new types of functions in bitopological spaces, namely (i; j)- -pre-continuous and (i; j)- -P -continuous functions.
2. (i; j)- -P -Open Sets
Throughout the paper, we assume that i; j = 1; 2 and i 6= j.
De…nition 1. A subset S of a bitopological space (X; 1; 2) with an operation
on 1[ 2 is said to be:
1. (i; j)- -pre-open if S i -Int( j -Cl(S)).
2. (i; j)- -P -open if S i-Int( j -Cl(S)).
By P O(X; 1; 2), we denote the family of all (i; j)- -P -open sets in (X; 1; 2).
Theorem 1. Every (i; j)- -pre-open set is (i; j)- -P-open.
Proof. Let S be an (i; j)- -pre-open set. Then we have S i -Int( j -Cl(S)) i-Int( j -Cl(S)) and so S is (i; j)- -P-open.
The following example shows that the converse of the above theorem is not true generally.
Example 1. Consider X = fa; b; c; dg with topologies 1= fX; ;; fbg; fdg; fb; dg; fa; b; cgg, 2 = fX; ;; fag; fdg; fa; bg; fa; dg; fa; b; dgg and a subset fa; bg of X. Let be an
operation on 1[ 2 de…ned as follows:
(U ) = U; if U = fdg X; if U 6= fdg Then fa; bg is (1; 2)- -P-open but it is not (1; 2)- -pre-open. Theorem 2. Every (i; j)-pre-open set is (i; j)- -P-open.
Proof. Let S be an (i; j)-pre-open set. Then we have S i-Int( j-Cl(S)) i
-Int( j -Cl(S)) and so S is (i; j)- -P-open.
The following example shows that the converse of the above theorem is not true generally.
Example 2. Consider X = fa; b; cg with topologies 1= fX; ;; fagg, 2= fX; ;; fbgg
and a subset fcg of X. Let be an operation on 1[ 2 de…ned by (U ) = X.
Then fcg is (1; 2)- -P-open but it is not (1; 2)-pre-open.
Conclusion 1. If is identity operator on 1[ 2, then (i; j)-pre-open sets and
(i; j)- -P-open sets coincide.
Lemma 1. There is no relation between (i; j)-pre-open sets and (i; j)- -pre-open sets.
Example 3. Let X, 1, 2 and be as in Example 1.
Then fa; bg is (1; pre-open which is not (1; -pre-open and fc; dg is (1; 2)--pre-open which is not (1; 2)2)--pre-open.
Theorem 3. Every i-open set is (i; j)- -P-open.
It can be seen from the following example that (i; j)- -P-open set is not need to be i-open.
Example 4. Let X, 1, 2 and be as in Example 2.
Then fbg is (1; 2)- -P-open but fbg =2 1.
Corollary 1. Every i-open set is (i; j)- -P-open.
Theorem 4. Let fS : 2 g be a class of (i; j)- -P-open sets. Then [ 2 S is (i; j)- -P-open.
Proof. Since S is an (i; j)- -P-open set, we have S i-Int( j -Cl(S )) for all
2 . Hence it is obtained[ 2 S [ 2 i-Int( j -Cl(S )) i-Int [ 2 j -Cl(S ) ! i-Int j -Cl [ 2 S !! : Therefore [ 2 S is also (i; j)- -P-open.
The intersection of two (i; j)- -P-open sets is not need to be (i; j)- -P-open as shown in the following example.
Example 5. Consider X = fa; b; cg with topologies 1 = fX; ;; fag; fa; bgg and 2= fX; ;; fbg; fa; bgg. Let be an operation on 1[ 2 de…ned as follows:
(U ) = U; if U = fa; bg X; if U 6= fa; bg
Then the sets fa; cg and fb; cg are (1; 2)- -P-open but their intersection fcg is not (1; 2)- -P -open.
Lemma 2. It is obtained that P O(X; 1; 2) 6= P O(X; 1) [ P O(X; 2).
Example 6. Let X = fa; b; cg be given with topologies 1 = fX; ;; fag; fb; cgg, 2= fX; ;; fbgg and be de…ned as follows:
(U ) = U; if U = fag X; if U 6= fag Then it is obtained
P O(X; 1) = fX; ;; fag; fbg; fcg; fa; bg; fb; cg; fa; cgg;
P O(X; 2) = fX; ;; fag; fbg; fcg; fa; bg; fb; cg; fa; cgg
and
P O(X; 1; 2) = fX; ;; fag; fa; bg; fa; cgg:
De…nition 2. A subset F of a bitopological space (X; 1; 2) with an operation on 1[ 2 is said to be (i; j)- -P -closed ((i; j)- -pre-closed) if XnF is (i; j)- -P -open
((i; j)- -pre-open) in X.
By P C(X; 1; 2), we denote the family of all (i; j)- -P -closed sets in (X; 1; 2).
Theorem 5. A subset F is (i; j)- -P -closed in a bitopological space (X; 1; 2) if
and only if i-Cl( j -Int(F )) F .
Proof. Let F be an (i; j)- -P -closed set in X. Then XnF is (i; j)- -P -open that is XnF i-Int( j -Cl(XnF )). It follows that
F Xn i-Int( j -Cl(XnF )) = i-Cl(Xn j -Cl(XnF )) = i-Cl( j -Int(F )): Conversely, we obtain XnF Xn i-Cl( j -Int(F )) = i-Int(Xn j -Int(F )) = i-Int( j -Cl(XnF ))
which completes the proof.
Theorem 6. Let fF : 2 g be a class of (i; j)- -P-closed sets. Then \ 2 F is (i; j)- -P-closed.
Proof. The proof follows from De…nition 2 and Theorem 4.
The union of two (i; j)- -P-closed sets is not need to be (i; j)- -P-closed. By Example 5, the sets fbg and fag are (1; 2)- -P-closed but their union fa; bg is not (1; 2)- -P -closed.
De…nition 3. (i; j)- -P -interior ((i; j)- -pre-interior) point of a subset S of a bitopological space (X; 1; 2) is a point x in X satisfying the inclusion V S for
an (i; j)- -P -open ((i; j)- -pre-open) set V containing x.
By (i; -P -Int(S) ((i; -pre-Int(S)), we denote the (i; -P -interior ((i; j)--pre-interior) of S consisting of all (i; j)- -P -interior ((i; j)- j)--pre-interior) points of S.
Theorem 7. The following properties hold for any subset S of a bitopological space (X; 1; 2):
1. (i; j) P Int(S) is the union of all (i; j) P open sets (the largest (i; j) -P -open set) contained in S.
2. (i; j)- -P -Int(S) is an (i; j)- -P -open set.
3. S is (i; j)- -P -open if and only if S = (i; j)- -P -Int(S). Proof. The proof follows from de…nitions.
Theorem 8. The following properties hold for any subsets S1, S2 and any class of
subsets fS : 2 g of a bitopological space (X; 1; 2):
1. If S1 S2, then (i; j)- -P -Int(S1) (i; j)- -P -Int(S2).
2. S
2
(i; j)- -P -Int(S ) (i; j)- -P -Int(S
2 S ). 3. (i; j)- -P -Int(T 2 S ) T 2 (i; j)- -P -Int(S ).
Proof. The …rst statement is obvious. The second and the third ones follow from 1 which implies (i; j)- -P -Int(S ) (i; j)- -P -Int([ 2 S ) and (i; j)- -P -Int(\ 2 S ) (i; j)- -P -Int(S ) for all 2 . Hence we have the results [ 2 (i; j)- -P -Int(S ) (i; j)- -P -Int([ 2 S ) and (i; j)- -P -Int(\ 2 S ) \ 2 (i; j)- -P -Int(S ),
re-spectively.
The reverse inclusions in 2 and 3 of Theorem 8 may not be applicable as shown in the following examples.
Example 7. Let X, 1, 2 and be as in Example 6. Then we have
fa; bg = (i; j)- -P -Intfa; bg * (i; j)- -P -Intfag [ (i; j)- -P -Intfbg = fag: Example 8. Let X, 1, 2 and be as in Example 5. Then we have
fcg = (i; j)- -P -Intfa; cg \ (i; j)- -P -Intfb; cg * (i; j)- -P -Intfcg = ;: De…nition 4. (i; j)- -P -cluster ((i; j)- -pre-cluster) point of a subset S of a bitopo-logical space (X; 1; 2) is a point x in X satisfying V \ S 6= ; for every (i; j) P
-open ((i; j)- -pre--open) set V containing x.
By (i; -P -Cl(S) ((i; -pre-Cl(S)), we denote the (i; -P -closure ((i; j)--pre-closure) of S consisting of all (i; j)- -P -cluster ((i; j)- -pre-cluster) points of S.
Theorem 9. The following properties hold for any subset S of a bitopological space (X; 1; 2):
1. (i; j)- -P -Cl(S) is the intersection of all (i; j)- -P -closed sets (the smallest (i; j)- -P -closed set) containing S.
2. (i; j)- -P -Cl(S) is an (i; j)- -P -closed set.
3. S is (i; j)- -P -closed if and only if S = (i; j)- -P -Cl(S). Proof. The proof follows from de…nitions.
Theorem 10. The following properties hold for any subsets S1, S2 and any class
of subsets fS : 2 g of a bitopological space (X; 1; 2):
1. If S1 S2, then (i; j)- -P -Cl(S1) (i; j)- -P -Cl(S2).
2. S 2 (i; j)- -P -Cl(S ) (i; j)- -P -Cl( S 2 S ). 3. (i; j)- -P -Cl( T 2 S ) T 2 (i; j)- -P -Cl(S ).
Proof. It can be proved in a similar way with the proof of Theorem 8
The reverse inclusions in 2 and 3 of Theorem 10 may not be applicable as shown in the following examples.
Example 9. Let X, 1, 2 and be as in Example 5. Then it is obtained
P C(X; 1; 2) = fX; ;; fag; fbg; fcg; fb; cg; fa; cgg. Thus we have
X = (i; j)- -P -Clfa; bg * (i; j)- -P -Clfag [ (i; j)- -P -Clfbg = fa; bg: Example 10. Let X, 1, 2 and be as in Example 6. Then it is obtained
P C(X; 1; 2) = fX; ;; fbg; fcg; fb; cgg. Thus we have
fbg = (i; j)- -P -Clfag \ (i; j)- -P -Clfbg * (i; j)- -P -Cl (fag \ fbg) = ;: Theorem 11. The following properties hold for a subset S of a bitopological space (X; 1; 2):
1. (i; j)- -P -Int(XnS) = Xn(i; j)- -P -Cl(S). 2. (i; j)- -P -Cl(XnS) = Xn(i; j)- -P -Int(S). Proof. The proof follows from de…nitions.
De…nition 5. (i; j)- -P neighborhood of a point x in a bitopological space (X; 1; 2)
is a subset Nx of X satisfying the inclusion U Nx for an (i; j)- -P -open set U
containing x.
Theorem 12. A subset of a bitopological space (X; 1; 2) is (i; j)- -P -open if and
only if one of the (i; j)- -P neighborhoods of each points of the subset is itself. Proof. The proof follows from De…nition 5.
3. (i; j)- -P -Tk Spaces
De…nition 6. A bitopological space (X; 1; 2) is said to be (i; j)- -P -T0 if for
every distinct points x and y of X, there exists an (i; j)- -P -open set containing x but not y or vice versa.
Theorem 13. A bitopological space (X; 1; 2) is (i; j)- -P -T0 if and only if for
each distinct points x and y of X (i; j)- -P -Clfxg 6= (i; j)- -P -Clfyg.
Proof. Let x and y be any two distinct points of X. Suppose that U contains x but not y for an (i; j)- -P -open set U of X. Then we have fyg \ U = ; which implies x =2 (i; j)- -P -Clfyg. Hence the result (i; j)- -P -Clfxg 6= (i; j)- -P -Clfyg is clear.
Conversely, suppose that (i; j)- -P -Clfxg 6= (i; j)- -P -Clfyg for every distinct points x and y of X. Let z 2 (i; j)- -P -Clfyg but z =2 (i; j)- -P -Clfxg. Then we have fyg \ U 6= ; for every (i; j)- -P -open set U containing z and fxg \ U = ; for at least one (i; j)- -P -open set U containing z. That is y 2 U and x =2 U for an (i; j)- -P -open set. Thus the space X is (i; j)- -P -T0.
De…nition 7. A bitopological space (X; 1; 2) is said to be (i; j)- -P -T1 if for
every distinct points x and y of X, there exist two (i; j)- -P -open sets which one of them contains x but not y and the other one contains y but not x.
Theorem 14. A bitopological space (X; 1; 2) is (i; j)- -P -T1 if and only if for
each point x of X (i; j)- -P -Clfxg = fxg.
Proof. Let y =2 fxg for x; y in X. Then by hypothesis, there exists an (i; j) P -open set U such that y 2 U but x =2 U. The result follows from fxg \ U = ; which means y =2 (i; j)- -P -Clfxg.
Conversely, let x 6= y for x; y in X. Since x =2 (i; j)- -P -Clfyg and y =2 (i; j) -P -Clfxg, there exist (i; j)- --P -open sets U and V containing x and y, respectively such that fyg \ U = ; and fxg \ V = ;. Hence we have x 2 U, y =2 U and y 2 V , x =2 V . Thus the proof is completed.
De…nition 8. A bitopological space (X; 1; 2) is said to be (i; j)- -P -T2 if for
every distinct points x and y of X, there exist two disjoint (i; j)- -P -open sets which one of them contains x and the other one contains y.
Theorem 15. A bitopological space (X; 1; 2) is (i; j)- -P -T2 if and only if for
each distinct points x and y of X there exists an (i; j)- -P -open set U containing x such that y =2 (i; j)- -P -Cl(U).
Proof. The proof is obvious.
Theorem 16. A bitopological space (X; 1; 2) is (i; j)- -P -T2 if and only the
in-tersection of all (i; j)- -P -closed (i; j)- -P neighborhood of each point of X consists of only that point.
Proof. Let x be any point of X. Suppose that there exist (i; j)- -P -open sets Uy
and Vysuch that x 2 Uy, y 2 Vy and Uy\ Vy = ; for each point y which is distinct
from x. Hence XnVyis an (i; j)- -P -closed (i; j)- -P neighborhood of x which does
not contain y since the inclusion Uy XnVyholds. This means that the equivalence
\fXnVy : y 2 X; y 6= xg = fxg holds which completes the …rst part of the proof.
Let x and y be any two distinct points of X. By our hypothesis, there is an (i; j)- -P -closed (i; j)- -P neighborhood of x in which y is not contained. Then we have y 2 XnU and XnU is (i; j)- -P -open. Also, there exits an (i; j)- -P -open set V containing x such that V U since U is an (i; j)- -P neighborhood of x. It is clear that the sets V and XnU are disjoint. This shows that X is an (i; j)- -P -T2
space.
4. (i; j)- -P -continuous functions
De…nition 9. Let f : (X; 1; 2) ! (Y; 1; 2) be a function and x be any point
of X. f is said to be (i; j)- -P -continuous (resp. (i; j)- -pre-continuous) at x if for every i open set H of Y containing f (x) there exists an (i; j)- -P -open (resp. (i; j)- -pre-open) set G of X containing x such that f (G) H.
Theorem 17. For a function f : (X; 1; 2) ! (Y; 1; 2), the following statements
are equivalent:
1. f is (i; j)- -P -continuous (resp. (i; j)- -pre-continuous).
2. For every i open subset O of Y , f 1(O) is an (i; -P -open (resp. (i;
j)--pre-open) set in X.
3. For every i closed subset C of Y , f 1(C) is an (i; -P -closed (resp. (i;
j)--pre-closed) set in X.
4. For every subset S of X, f ((i; j)- -P -Cl(S)) iCl(f (S)) (resp. f ((i; j) -pre-Cl(S)) i-Cl(f (S))).
5. For every subset T of Y , (i; j)- -P -Cl(f 1(T )) f 1(
i-Cl(T )) (resp. (i;
j)--pre-Cl(f 1(T )) f 1(
i-Cl(T ))).
Proof.
1 ) 2 Let O be i open in Y and x 2 f 1(O). Since f is (i; j)- -P -continuous
on X, there exists an (i; j)- -P -open set G of X containing x such that f (G) O. Hence we have G f 1(O) which means x is an (i; j)- -P -interior point of f 1(O). Thus f 1(O) is an (i; j)- -P -open set in X.
2 ) 1 Let x be any point of X and H be a i open set containing f (x). By the
assumption, f 1(H) is (i; j)- -P -open and x 2 f 1(H). For G = f 1(H), we have
f (G) H which concludes the proof. 2 , 3 It is obvious.
1 ) 4 Let S be a subset of X and f(x) 2 f((i; -P -Cl(S)), where x 2 (i; j)--P -Cl(S). Take any i open set H of Y containing f (x). Since there exists an (i; j)- -P -open set G of X containing x such that f (G) H and also G \ S 6= ;, we have H \ f(S) 6= ;. This means f(x) 2 i-Cl(f (S)). Thus the inclusion f ((i;
4 ) 5 Let T be a subset of Y . Then by the inclusion in 4, we have f((i; j) P -Cl(f 1(T )))
iCl(T ). Taking the preimage of both sides, we obtain (i; j) P
-Cl(f 1(T )) f 1(
i-Cl(T )).
5 ) 3 Let C be iclosed in Y . Then by the inclusion in 5, we have (i; j) P
-Cl(f 1(C)) f 1(C). Hence f 1(C) is (i; j)- -P -closed in X.
Corollary 2.
1. Every (i; j)- -pre-continuous function is (i; j)- -P -continuous. 2. Every (i; j)-pre-continuous function is (i; j)- -P -continuous.
(i; j)- -P -continuous functions may not be (i; j)- -pre-continuous and (i; j)-pre-continuous as seen from the following examples.
Example 11. Let X, 1, 2 and be as in Example 1. De…ne a function f on
X as f (a) = b, f (b) = b, f (c) = d and f (d) = d. Then for fa; b; cg 2 1, the
inverse image f 1(fa; b; cg) = fa; bg is not (1; 2)- -pre-open and so f is not an
(1; 2)- -pre-continuous function. But f is (1; 2)- -P -continuous.
Example 12. Let X, 1, 2 and be as in Example 2. De…ne a function f on
X as f (a) = c, f (b) = b and f (c) = a. Then for fag 2 1, the inverse image
f 1(fag) = fcg is not (1; 2)-pre-open and so f is not an (1; 2)-pre-continuous function. But f is (1; 2)- -P -continuous.
Acknowledgments. The authors would like to thank the anonymous referee for his/her valuable comments.
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Current address : Merve ·Ilkhan: DEPARTMENT OF MATHEMATICS, DUZCE UNIVER-SITY, DÜZCE, TURKEY.
E-mail address : merveilkhan@gmail.com
Current address : Mahmut Akyi¼git: DEPARTMENT OF MATHEMATICS, SAKARYA UNI-VERSITY, SAKARYA, TURKEY.
E-mail address : makyigit@sakarya.edu.tr
Current address : Emrah Evren Kara: DEPARTMENT OF MATHEMATICS, DUZCE UNI-VERSITY, DÜZCE, TURKEY.
