J King Saud Univ, , VoL 21, Science (Special Issue), pp, 243-250, Riyadh (2009/1430H.)
Applications of O-Open
Raja Mohammad Latif
King Fahd University of Petroleum and Minerals
Departmenl and Statistics
Dhahran 31261, Saudi Arabia [email protected]
semi-open set, interior, open set, derived set, border, frontier, exterior, theta- theta-exterior, theta-saturated, theta-compact, theta-kernel, net, filter, theta- convergence.
Abstract. In 1986 Velicko [29] introduced the notion of theta-closed and theta-closure subsets in a space for the purpose of studying the imp0l1ant of H-c1osed in terms of arbitrary An element x in a X is called a theta -clusture point of a subset of X if the intesection of closure of U and S is nonempty every open set U of X containing x, The set of all theta -closure pomts of S is called the theta-closure of S, A subset S is called theta-closed if the theta c10sue of S is equal to S, The complement of a theta-closed set is said to be a theta-open set .The family of all theta-open subsets of (X ,T) fonns a tau-theta topology on X, The objective of this paper is to introduce and study topological properties of theta- derived, theta-border, theta-frontier and theta-exterior of a set using the concept of theta-open sets, Moreover, we investigate and study some furthuer topological properties of the we ii-known notions theta -closure and theta- interior of a set in a topological space,
1 Introduction
The notions of 8 - open subsets, 8 - closed subsets and 8 closure were introduced by Velicko [29] for the purpose of studying the important class of H - closed spaces in tenus of arbitrary filterbases.
Dickman and Porter [5], [6], Joseph[ 15] and [21] continued the work of Velicko. Recently Noiri and Jafari [23] and Jafari [12] have also obtained several new and interesting results related to these sets. For these sets, we introduce the notions of
8 -derived, 8 -border, 8 -frontier and 8- exterior
of a set and show that some of their are analogous to those for open sets, we
additional of 8 closure and 8 - interior of a set due to [29]. Throughout this paper, (X, 1) (simply X) always mean space on which no axioms are assumed unless
stated. We denote the interior and the closure of a subset A of X by Int (A) and CI (A) , respectively. A point x 5 X is called a 8 adherent point of A [29].
if A CI (V ) =6 <p for every open set V contain- x, The set of all 8 adherent points of A is called the 8 -closure of A and is denoted Clo
243
(A) . A subset A of X is called 8-closed if A C I a (A) . Dontchev and Maki [[7], Lemma 3.9] have shown that if A and B are subsets of a space (X, T) , then Clo (A l\ B) Cia (A) l\ Clo (B) and Clo (A _ B)
=
Clo (A) _ Clo (B) . Dickman and Porter [5] proved that a compact subspace of a Hausdorff space is a-closed, Moreover, they showed that a 8- closed subspace of a Hausdorff space is closed.Jankovic [13] proved that a space is Hausdorff if and only if every compact set is 8 closed. The complement of a 8 - closed set is called a 8 - open set. The 8 interior of set A in X, written I n
to
(AJ ' consists of those pointsx
of A such that for some open set U containingx,
CI (U) C A. A set A is 0 - open if and only if A IntO (A), or equivalently, (X - A) is 0 - closed, The family of all 0 - open sets fonus a topology on X and is denotedteo
This topology is weaker than t and it is well-known that a space (X, t) is regular if and only if t =. to. It is also obvious that a set A ise
-closed in t) if and only if it is closed in to), The collection of all e closed subsets of a topological space (X, t) is denoted byce
Velicko [29] introduced the notions of 8 - closure and 8 interior operations. A subset S of X is called a
244 R.M. Latif: Applications ore-Open Sets
semi-open set [20] ifS CI [Int (S)]. The complement of a semi-open set is called a semi-closed set. The family of all semi-open sets in a topological space (X, T) will be de-noted by SO (X). A subset M (x) of a space X is called a semi-neighbourhood of a point x 5 X if there exists a semi-open set S such that x 5 S M (x). In [18] Latif introduced the notion of semi- convergence of filters and investigated some characterizations related to semi-open continuous function. A point x 5 X is called the 8 - cluster point of A X if A _ Int [Cl (U)] 6= <p for every open set U of X containing x. The set of all 8 - cluster points of A is called the
a -
closure of A, denoted byCia
(A).A subset A X is called 8 - closed if A
=
CI8 (A). The complement of a 8 - closed set is called 8 - open. The collection of all 8 - open sets in a topological space (X, T) forms a topology T8 on X, called the semiregularization topology of T, weaker than T and the class of all regular open sets in T forms an open basis for T8. We observe that for any topological space (X, T) the relation T8 T8 T always holds. We also have A CI (A) CI8 (A) CI8 (A), for any subset A ofX.2 Applications of 8-0pen Sets
Definition 2. I. Let A be a subset of a space X. A point x 5 A is said to be a 8 - limit point of A if for each 8 - open set U containing x, U_ (A - {x}) =6 <po The set of all 8 - limit points of A is called a 8 - derived set of A and is denoted by D8 (A).
Theorem 2.2. For subsets A, B ofa space X, the following statements hold:
(I) D (A) D8 (A) , where D(A) is the derived set of A;
(2) if A B, then D8 (A) D8(B);
(3) D8(AY' D8(B) = D8(N' B)andD8(A_ B) D8(A) _ D8(B);
(4) [D8 (D8(A)) - A] D8(A);
(5) D8 [N' D8(A)] N' D8(A).
Proof. ( 1) It suffices to observe that every 8 - open set is open.
(2) Obvious.
(3) D8(A) /\ D8(B) = D8(A /\ B) is a modification of the standard proof for D, where open sets are replaced by 8 -open sets.
D8(A _ B) D8(A) _ D8(B) follows by (2).
(4) Ifx 5 [D8 (D8(A)) - A] and U is a 8 - open set containing x, then U _ [D8 (A) - {x}] =6 <po Let y 5 U _ [D8 (A) - {x}]. Then, since y 5 D8 (A) and y 5 U, so U_ [A-{y}] =6 <p.Letz 5 U_ [A-{y}].Then,z
=6 x for z 5 A and x 51 A. Hence, U _ [A - {x}] =6
<po Therefore, x 5 D8 (A).
(5) Let x 5 D8 [A/\ D8 (A)]. Ifx 5 A, the result is obvious. So, let x 5 [D8 (A /\ De (A)) -A] ,
then, for 8-open set Ucontainingx, U_[A/\ D8 (A) - {x}] =6 <po Thus, U _ (A - {x}) =6 <p or U _ [D8 (A) -{x}] =6 <po
Now, it follows similarly from (4) that U _ [A -{x}]
=6
<p. Hence, x 5 Do (A) . Therefore, in any case, Do [A/\ Do (A)]C [A/\ Do (A)].
Theorem 2.3. For any subset A of a space X, C I o( A) = A 1\ D o( A ) .
Proof. Since Do (A) C Clo (A). A /\ Do (A) C C 10 (A) . On the other hand, let x 5 C I 0 (A) . If x 5 A, then the proof is complete. If x 51 A, each 8 - open set U containing x intersects A at a point distinct from x; so
x5 Do (A). Thus,Clo (A)C[A/\ Do (A)] , which completes the proof.
CoroUary 2.4. A subset A of X is 8 - closed if and only if it contains the set of its 8 -limit points.
Definition 2.5. A point x 5 X is said to be a e - inferior point of A if there exists a 8 - open set U containing x such that U C A. The set of all 8 - interior points of A is said to be the 8 - interior of A [15] and is denoted by In to (A).
Theorem 2.6. For subsets A, B of a space X, the fol- lowing statements are true:
(1)
r
n to (A) is the largest 8 - open set contained in A;(2) A is 8 - open if and only if A = I n t 0 (A);
(3) Into [Into (A)]=lnto (A);
(4)lnto (A)=[A-Do (X-A)];
(5)[X-Into (A)] =Clo (X - A);
(6) [X -C I 0 (A)] = In to (X - A);
(7)ACB,thenlnto (A)Clnto (B);
(8) In to (A) /\ In to (B) C I n to (A 1\ B);
(9) Into (A _ B)=Into(A)_ Into (B). Proof.
(4)lfx5 [A-Do (X-A)],thenx
x5 Into (A),thatis,[A-Do (X-A)]Clnto (A).
On the other hand, if x 5
r
n t 0 (A) . then x 51 Do (X -A) since In to (A) is 8 -0 pen and [I n to (AL (x- A)] = <p. Hence, In to (A) = A -Do (X -A) . ( 5) X - I n t 0 (A) = X - [A - D o( X - A)]= (X - A) /\ D 9( X - A) = C I o( X - A) . Definition 2.7. B d 0 (A) = [A - In to (A)] is said to be the 8 - border of A.
J King Saud Univ., Vol. 21, Science (Special Issue), Riyadh (2009114JOH.) 245
Theorem 2.8. For a subset A of a space X, the follow- ing statements hold:
(1) Bd (A) C B d 0 (A) where Bd (A) denotes the bor- der of A;
(2) A = I n to (A)" B d 0 (A) ; ( 3 ) f n to (A) _ B d a (A) = <p ;
(4) Aisae - opensetifandonlyifBdo (A) =
<p;
( 5 ) B d 0 [I n to (A)]
=
<p ;( 6 ) In t 0 [B d 0 (A)] = <p ; (7) B do [B d e (A)]
=
B d a (A) , (8) B d 0 (A) = A _ [C 10 (X -A)] , (9) B d 0 (A) = D a (X -A) .Proof. (6) Ifx 5 Into [Bdo (A)], then x 5 Bdo (A) . On the other hand, since B d 0 (A) C A, x 5 fnto [Bdo (A)]CTnto (A). Hence, x 5 lnto (A) _ B d a (A), which contradicts (3) . Thus, In to [B d 0
(A)] =
<p. (8) B d a (A) = A - In to (A) = A - [X -
C I o( X - A) J = A _ C I o( X - A) .
(9)Bdo (A)=A-Into (A)=A-[A-Do (X-A)J
= Do (X - A)
In general the equality of Theorem 2.8 (I) does not hold as it is verified by the next example.
Example 2.9. Let X = {I, 2, 3} with T
{<p, {I} , {2} , {I, 2} , X} . Then itcan be easily
verified that for A
=
{2} , we obtain Bdo (A)*
Bd (A) Definition 2.10. Fro (A)=
[Cia (A)-Into (A)] is said to be thee -
frontier [12J of A.Theorem 2.11. For a subset A of a space X, the fol- lowing statements hold:
(1) F r (A) C F ro (A) where F r (A) denotes the frontier of A;
(2) Cia (A) = lnto (A) /\ F ro (A), (3)Tnto (A)_ F ro (A) = <p;
(4) Bdo (A) c F ro (A) ,
(5) Fro (A) = Bdo (A) " Do (A) ,
(6) A is a
e -
open set if and only if F ro (A) = Do (A) ;(7) Fro (A) = Cia (A) _ Cia (X-A);
(8) Fro (A) = Fro (X-A);
(9) Fro (A) is e - closed;
(10) F ro [F ro (A)] C F ro (A);
(1/) F ro [Into (A)] C F ro (A);
(12) F ro [Cia (A)] c F ro (AJ;
(/3) Into (A) = A -Fro (A).
Proof. (2) Into (A)" F ro (A) = f n to (AJ" [C 10 (A) - Into (A)] = Cia (A)
(3) fn to (A) _ F 1'0 (A) fnto (A) _ [Cia (A) -Into (A)]
=
<po(5) Since I n to (A)"F ro (A)
=
Into (A)"Bdo (A)/\Do (A),
F ro (A) = Bdo (A) /\ Do (A) .
(7) F ro (A) = Cia (A) -Into (A) = Cia (A) _ Cia (X- A).
(9) Cia [Fro (A)] = Cia [CIa (A) _ Cia (X-A)]
c Cia [Cia (A)] _ Cia [Cia (X -A)] = Cia (A)_
Cia (X -A)
=
F ro (A) . Hence, F ro (A) is e - closed(10) Fro [F ry (A)] Cia [F ro
r)L
Cia [X - Fro (A)]
c Cia [F ro (,4)] = F ro (A).
(12) Fro (Clo (,4)) = Cia [Clo (A)] -Into [Clo (A)] = Cia (A) -Into [Clo (A)] c [Cia (A) -In to (A)]
=
F ro (A) .(/3) A -F ro (A) = A -[C 10 (A) -Into (A)] = I n to (A) .
In general, the equalities in (I) and (4) of the Theorem 2.11 do not hold as it is shown by the following example.
Example 2.12. Consider the topological space (X, T) given in Example 2.9. If A = {2} . Then F ro (A) = {2, 3}
*
(3) = F r (A) and also F ro (A) = {2, 3}*
(2) = Bdo (A).
Remark 2.13. Let A and B be subsets of X Then A C B does not imply that either F ro (B) C F ro (A) or F ro (A) C F ro (B) .
Definition 214. Ex to (A) = I n to (X - A) is said to be a
e -
exleriorof A.Theorem 2.15. For a subset A of a space X, the fol- lowing statements hold:
(I) Ex to (A) C Ext (A) where £'(1 (A) denotes the exterior of A;
(2) Exto (A) is
e -
open;(3) Exto (A) = Into (X - A)
=
[X -Cia (A)];(4) Exto [Exto (AJ] = Into [Cia (A)], (5) If A C B, then Exto (A) Exto (B);
(6) Exto (A " B) = Exto (A) _ Exto (B);
(7) Exto (A) __ Exto (B) C Exto (A _ B); (8) Exto (Aj = <p;
(9) Ext e (<p) =
x,
(10) Exte (,4)= Exte [X-Exte (AJ);
(1/) In t a (A) C Ext e [E x t e (A)],
(/2) X= Into (A) " Exte (A) " F ro (AJ;
246 R.M. Latif: Applications ofe-Open Sets (13) Exte (A)" Exte CExte (A B).
(14) Proof. (4) Exte [Exte (A)]= Exte Cle
=
Into {X- Cle (A))] = Into [Cle (10) Ext e Ext e (AJIExt e I n t 0 (X-A)]
Into Into
I n to (A) C I n t 0 [C I e (A)]
In to [X -Ext e = Ext e [E x t e (AJ/.
(13)Exte "Exte (B) = Into (X-A)"
Definition 2.16. Let X be a topological space. A set A C Xis said to be
e
saturatediffor every x 5 A it follows C I e CA. The set of alle
satw"ated sets in X we denote by B 0Theorem 2.17. Let X be a B o i s a complete Boolean set
space. Then
Proof. We will prove that all the unions and comple- ments of elements of B 0 (X) are members of B 0 (X).
Obviously, only the proof the
is not trivial. Let A 5 B 0 (X) and suppose that C 1 e
'"
(X - A) for some x 5 (X A). Then there exists y 5 A such that y 5 C I e It follows that x, y have no disjoint neighbourhoods. Then x 5 C I e . But this is a contradiction, because by the definition of B a (X) we have
C Ie CA. Hence, C 1 e x 5 (X - A) , which
A) for every Bo
Corollary 2.18. Be (X) contains every union and every intersection ofa closedand a open sets in X Definition 2.19. [28] A space X is said to be S Hausdorff iffor every x=6 y 5 X, there exist
e -
open sets U U, V vsuch that x 5 U~, Y 5 V vand U U_V, <p.
Theorem 2.20. The four properties are ( 1 ) X is S - T 2;
Let x 5 X. For each X; there exists a
e -
ope n set U such that x 5 U and y 5 / C I e (V);
(3) For each x 5 X, _ {Cle(U) IU 5 wand x 5 U} = {x} ;
diagonal A x) I x 5 is
e -
closed inX x X.Proof. (2) . Let x 5 X and y=6 x Then there are disjoint
e -
open sets U and V such that x 5 Uand y 5 V. Clearly, V c is
e -
C I e(L?
eVe, y 5/ V c and therefore / C Ie(2) ==> . Ify=6 x, there exists a
e
open set U such that x 5 U and y 5 / C I e (Uj. So yto{ D\Ie( <U 6¥~0)]r
lMra:!
~ -54M}E.x t erj'l)t±ldX '(4Jl.twb'PrMe that Ac is a - open. Let (x, y) 5/ A. Theny=6 xand since_ {Cte (L?IU 5 . 0
and x 5 = {x} there is some U5 . 0 with x5 Uand y 51 C I e . Since C Ie
(L?!
<p, U x [C 1 e is a 8 - open set such that (x, y) 5 U x [C I e CDc.= => (1). x, then (x; y) 5/ A and thus there exist a - open sets U and V such that (t, yJ 5 U x V and V ) _ A <p Clearly, for the
e -
Uand V
We have: x 5 5
Definition 2.21. [28] A subset A of a space X IS
said to be theta-compact (briefly, 8 - compact) if every cover of a - open sets has a finite subcover.
It is well-known that every closed subset of a compact space is The next theorem
this result for a - compactness.
Theorem 2.22. A
e -
compact subset of an S- Hausdorff space ise
closed.Proof. Let A be a a compact subset of an S- Hausdorff space X. We wili show that A) is
e -
open. Let x 5 (X - A) then for each a 5 A there exist
e -
open sets UX,a and Va such that x 5 UX,a and a 5 Va and Ux. a Va = <po The collection {Va:a 5 is a
e
open cover of A. Therefore, there exists a finite subcollection Va 1, Va 2, . . . • Va n that covers A. LetUx = Ux ,a 1 _ Ux , a . . . _ Ux , an. Then x 5 Ux, Ux is
a- open and Ux = <po This proves that A is 0- closed.
Theorem 2.23. A
e
closed subset of ae -
compactspace is
e
and let A be a a-closed cover of A. Then
r
=A} is a cover ofX. Since Xis a- compact, this collection
r
has a finite collection AJ. King Saud Univ., Vol. 21, Science (Special Issue), Riyadh (2009/1430H.) 247 that covers X. But then
r
ha a finite subcollection 1\1\ - - A} that covers A as we need.
Definition 2.24. Let A be a subset of a topological space X. Then 0 -kernel of A, denoted KerO (A)
5 TO OJ.
Definition 2.25. Let x be a point of a tODOJo:~nca
space
X. Then O-kernel of x, denotedby KerO ({x}) is defined to be the set KerO ({x}) {O 5 TOlx 5 OJ.
Lemma 2.26. Let x 5 X. ThenKerO
<p}
Proof. Let x 5 KerO and CIO ( ) A = <po Hence x 5/ KerO )] which is a 0 open set contamlng This is impossible, since x 5 KerO
Consequently, KerO ( ) _ A =6 <po Let CIO ( A 6= <p and x 5/ KerO Then there exists a 0 - open set D containing A and x 5/ D. Let y 5 CIO ({ )_ A. Hence, D is a 0 open of y with x 5/ D. By this x 5 KerO (A) and the claim.
Definition 2.27. A topological space (X, T) is said to be a 0 RO space if every 0 - open set contains the 0 - closure of each of its
Lemma 2.28. Let (X, T) be a topological space and x 5 X. Then y 5 KerO ({x}) ifand only if x 5 KerEl ( ).
Proof. Suppose that y 5/ KerO ({x}). Then there ex- ists a 0 - open set V containing x such that y 5/ V.
There-fore we have x 5/ CIS ( ). The of the converse case can be done similarly.
Lemma 2.29. The statements are equivalent for any points x and y in a
(I)KerS ({x}) =6 KerEl ( (2)CIS ( ) =6 CIS ( ).
space (X, T) :
Proof. (I) (2) : Suppose that KerO ({ x}) =6 KerEl ( }). Then there exists a point z in X such that z 5 KerO ({ x}) and z 51 KerO ({y}). It follows from z 5 KerO ( }) that {x} _ CIEl ({x}) =6 <po This implies that x 5 CIO ({ z}). z 5/ KerO ({y}), we have _C19 ({z}) <p. Since x 5 CIS ( ) and CIS ({x}) CIO ( ). Hence CIS ({x}) <po CIS ( ) =6 C19 ( ).
==> (l) : that C I e =6 C I e Then there exists a point z 5 X such that z 5 C Ie ({x)) and z 5/ C I e ({y}). there exists a 0 open set containing z and therefore x but not y, i.e., y 51 Kere ({x)). Hence Kere ((x)) =6 Kere
Theorem 2.30. A topological space T) is a 0 - R 0 space if and only if for every x and y in X, C I e ({x))
=6 Cle implies Cle _ Cle ({y}) <po Proof. Necessity. that (X, T) is
e -
RO andx, y 5 X such that C I e =6 C I e . Then, there exists z 5 C I e such that z 5/ C I e
z 5 C I e such that z 51 C I e There exists V 5 TO such that y 5/ Vand z 5 V ; hence x 5 V.
we have x 5/ Cle ([v)) . Thus x 5 C Ie ({y))] 5 TO, which implies C I e ({x:)) C {X -C I e
and Cle ((x)) Cle <po The prooffor otherwise is similar.
Sufficiency. Let V 5 TO and let x 5 V. We will show that C I e C V. Let y 5/ V, i.e., y 5 V ) . Then
x=6yandx5/ Cle ({y)) ThisshowsthatCle
=6 Cle ({y)). assumption, Cle Cle ((y))
= <po Hence y 5/ CJe and therefore Cle
CV.
Theorem 2.31.A topological space T) is a 0 R 0 space if and if for any points x and y in
=6 Kere implies Kere Kere
Proof. Suppose that r) is a S - RO space. Thus by Lemma for any x and y in X if Kere
=6 Kere then Cle =6 Cle Now we prove that Kere ({ x}
Assume that z 5 Kere
Kere and that x 5
Kere ({z}).
Since x 5 K ere ({'C)) , by Theorem C I e Cle ({z}). Si milarly, we have Cle (Lv)) Cle
C I e ({x}) . This is a contradiction.
we have Kere ({x)) _ Kere ((y)) <po
Conversely, let T) be a topological space such that for any x and y in X, Kere =6 Kere Kere _ Kere ([v)) <po IfCle
=6 C Ie , then by Lemma Kere
=6 Kere ([v}). Hence Kere ({x}) Kere ({y)) <p which implies C I e C Ie ({y)) (j). Because z 5 Kere implies that x 5 Kere ({z)). Therefore Kere ({x)) Kere =6 <p By hypothesis, we have Kere = Kere . Then 5 Cle
Cle implies that Kere ({t)) Kere K ere . This is a contradiction. C I e
248 R.M. Latif: Applications ofe-Open Sets
_ C I e ({y}) = <p. By Theorem 2.30 (X, 1) is a a - RO space.
Theorem 2.32. For a topological space (X, 1) , the following properties are equivalent:
(1) (X, 1) is a a - RO space;
(2) ForanyA=6<p andG5 10 such that A _ G
=6 <p, there exists F 5 C B (X, 1) such that A _ F =6
<p and FCG;
( 3 ) Any G 5 10, G =" {F 5 C 9( X, 1)
IF
CG} ;
(4) Any F 5 C e (X, T) , F = _ {G 5 19
IF
C G};(5) For any x 5 X, C I e ({x}) C K ere ({x}).
Proof. (I) ==> (2) : Let A be a nonempty subset of X and G 5 10 such that A _ G =6 <p. There exists x 5 A _ G. Since x 5 G 5 TO, Cle ({x}) C G. Set F = Cle ({x}) . Then F is a a - closed subset of X such that F C G and A _ F =6 <p .
(2) ==> (3) : Let G5 10. Then
" {F 5 CB (X, 1)
IF
C G} C G. Let x be any point of G. There exists F 5 CB (X, 1) such that x 5 F and F CG.Therefore,wehavex5 FC" (F5 CB(X, 1)IF
CG}andhenceG =" {F 5 C9(X, 1) IF CG}(3) ==> (4) : This is obvious.
(4) ==> (5) : Let x be any point of X and y 51
Kero ({x}). There exists V 5 TO such that x 5 Vand y51 V; henceClo ({x})_ V= <po By(4)C {G5 10lCIe ({y}) C G}) _ V = <po There exists G 5 10
such that x 5 I G and C I 0 ({y}) C G. Therefore C I 0 ({x}) _ G
=
<p and y 5 I C I 0 ({x}). Consequently, we obtainClo ({x})CKero ({x}).(5) ==> (I) : Let G 5 TO and x 5 G. Suppose y 5 Kero ({x}) . Then x 5 Clo ({y)) and y 5 G. This implies that Clo ({x}) C Kero ({x)) C G. Therefore, (X, 1) is a a - R 0 space.
Corollary 2.33. For a topological space (X, 1), the following properties are equivalent:
(I) (X, 1) is a a - RO space;
(2) Clo ({x}) = Kero ({x)) for all x 5 X
Proof. (I) ==> (2) : Suppose that (X, 1) is a a - RO space. By Theorem 2.32, Clo ({x}) C Kero ({x}) for eachx5 XLety5 Kero ({x}). Thenx5 Clo ({y}) and so C 10 ({x}) = C I 0 ({y}) . Therefore, y 5 Clo ({x}) and hence Kero ({J;)) C Clo ({x)). This shows thatClo ({x)) = Kero ((x}).
(3) ==> (I) : This is obvious by Theorem 2.32.
(4) Theorem 2.34.For a topological space (X, 1) , the following properties are equivalent:
(1) (X, 1) is a a - RO space;
(2) x 5 C 10 ({y}) if and only if y 5 C I 0 ({x}) , for any points x and y in X
Proof. (I) ==> (2) : Assume that X is a - RD. Let x 5 C I 0 ({y}) and 0 be any a - open set such that y
5 D. Now by hypothesis, x 5 D. Therefore, every a - open set containing y contains x. Hence y 5 C I 0 ({x}) . (2) ==> (I) : Let U be a a - open set and x 5 U. If Y 51 U,thenx51 Cle({y}) andhencey51 Cle({x}). This implies thatClo ((x))C U. Hence (x,l)isa-RO.
Theorem 2.35. For a topological space (X, 1) , the following properties are equivalent:
( I) (X, 1) is a a - RO space;
(2) IfFisa - closed,thenF= Kero(F);
(3)If Fis a -closed and x5 F , then Kero (F ) C
F;
(4) Ifx5 X, then Kero ({x})CClo ({x}).
Proof. (I) ==> (2) : Let F be a - closed and x 5 I F.
Thus (X-F)isa-open containsx. Since (X; 1) a-RO.
C I 0 ({x}) C (X -F ). Thus C 10 ({x)) _ F = <p and by Lemma 2.26 x 51 Kero (F ) . Therefore Kero (F )
=F
(2) ==> (3) : In general, A C B implies Kero (A) C C I 0 (B) . Therefore, it follows from (2) that K e r 0
({x})CKero (F) = F
(3) ==> (4): Sincex5 Clo ({x})andClo ({x})isa- closed, by (3) , Kero ({xJ~ C Clo ({xJ~'
(4) ==> (I) : We show the implication by using The- orem 2.34. Let x 5 C I 0 ({y}). Then by Lemma 2.28, y 5 Kero ({x)). Since x 5 Clo ({x}) and Clo ({xJ~ is a - closed, by (4) we obtainy5 Kero ({x)) CClo ({x}) . Therefore x 5 C I 0 ({y}) implies y 5 C I 0 ((x}) . The con-verse is obvious and (X, 1) is a - RD.
Theorem 2.36. Let (X, 1) be a topological space.
Then _ {C Ie ({x)) Ix 5 X} = <p if and only if Kero ({x}) =6 X for every x 5 X
Proof. Necessity. Suppose that _ {Cle ({x}) Ix 5 X}
= <po Assume that there is a point y in X such that Kere ({y)) = X. Then y 51 0, where 0 is some proper a - open subset of X. This implies that y 5 _
{Cle ({x}) Ix 5 X} . But this is a contradiction.
Sufficiency. Assume that Kere ((x)) =6 X for every
J King Saud Un;"., Vol. 21, Science (Special Issue), Riyadh (200911430H.) 249
x 5 X. If there exists a point y 5 X such that y 5 _ {Cle ((x)) Ix 5
Xl ,
then every a - open set containing y must contain every point of X. This implies that the space X is the unique a - open set containing y. Hence Kere ({x)) = X which is a contradiction. Therefore, _ {Cle ((x)) Ix 5 X) =cp.Definition 2.37. A filter base z is called a - convergent to a point x in X, if for any a - open set U of
X containing x, there exists B in z such that B is a subset ofU.
Lemma 2.38. Let (X, r) be a topological space and x and y be any two points in X such that every net in X a - converging to y a-converges to x. Then x 5 Cle ((y)) .
Proof. Suppose that xa = y for a 5 1. Then {xa : a 5 I) is a net in Cle ({y)) Since {xa : a 5 I) a - converges to y, so {xa : a 5 1) a -converges to
x and this implies thatx 5 Cle ((y)) .
Theorem 2.39. For a topological space (X, r) , the following statements are equivalent:
(I) (X, 1) is a - RO space;
(2) Ifx, y 5 X, theny 5 Cle ({x)) ifand only if every net in X a -converging to y a - converges to x.
Proof. (I) ==> (2): Let x, y 5 X such thaty 5 Cle ((x)) . Suppose that {xa : a 5 1) is a net in X such that this net a-converges to y. Since y 5 Cle ((x}) so by Theorem 2.30 we have Cle ((x)) = Cle ((y)) Therefore
x 5 Cle ((y)) . This means that the net {xa : a 5 I) a -converges to x.
Conversely, let x, y 5 X such that every net in X a - converging to y a - converges to x. Then x 5 Cle ({y)) by Lemma 2.38. By Theorem 2.30, we have Cle ((x)) = Cle ((y)). Thereforey 5 Cle ((x)).
(2) ==> (I) : Assume that x and yare any two points of X such that Cle ((x)) _ CJe ({y)) =6 cp. Let z 5 Cle ((x)) _ Cle ({y)) . So there exists a net {xa : a 5 1) in Cle ((x)) such that {xa : a 5 I) a - converges toz. Sincez 5 Cle ((y)). So by hypothesis {xa: a 5 1) a - converges to y. It follows that y 5 Cle ((x}) . Similarly we obtain x 5 Cle ((y)) Therefore Cle ((x)) = Cle ((y)) and by Theorem 2.30, (X, 1) is a - RO.
A.cknowledgement. The author tS highly and gratefully indebted to the King Fahd University of
Petroleum and Minerals, Dhahran, Saudi Arabia, for providing necessary research facilities during the preparation of this paper.
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