C om mun.Fac.Sci.U niv.A nk.Series A 1 Volum e 66, N umb er 2, Pages 323–331 (2017) D O I: 10.1501/C om mua1_ 0000000822 ISSN 1303–5991
http://com munications.science.ankara.edu.tr/index.php?series= A 1
MORE ON -TOPOLOGICAL SPACES
SHYAMAPADA MODAK AND MD. MONIRUL ISLAM
Abstract. The aim of this paper is to introduce a new topology with the help of a-open sets. For this job, we shall de…ne two new types of set and discuss its properties in detail and characterize Njastad’s -open sets and Levine’s semi-open sets through these new types of set.
1. Introduction
The study of ideal in topological space was introduced and studied by Kuratowski [15] and Vaidyanathaswamy [22] but in this study Jankovic and Hamlett gave a new dimension through their paper “New topologies from old via ideals" [14]. Now a days the authors like Navaneethakrishnan et al. [19], Hamlett and Jankovic [12], Arenas et al. [4], Nasef and Mahmoud [18], Mukherjee et al. [17] Dontchev et al. [6] and many others have enriched this study. The authors Al-Omari et al. [1, 2] in their papers “a-local function and its properties in ideal topological spaces" and “The <a operator in ideal topological spaces", have studied Ekici’s [7, 8, 9] a-open
sets in terms of ideals. They have obtained a new topology with the help of two operators viz. <a and ()a , and have shown that this topology is …ner than Ekici’s
a - topology.
In this paper, we have further considered the space which is the joint venture of a-topology and an ideal as like Al-omari et al. have considered in [2, 1]. Through this paper we will solve the question “how much …ner is Noiri’s et al.’s topology than Ekici’s topology?" For solution of this question we have considered Njastad’s
-open sets [20] from literature.
2. Preliminaries
In this section we have discussed some preliminary concepts of literature and introduce some prime results for discussing the paper.
Let A be a subset of a topological space (X; ), then ‘Int(A)’and ‘Cl(A)’will denote ‘interior of A’and ‘closure of A’respectively.
Received by the editors: November 07, 2016, Accepted: February 02, 2017. 2010 Mathematics Subject Classi…cation. 540A, 54C10.
Key words and phrases. set, a-open set, a-local function, <aoperator.
c 2 0 1 7 A n ka ra U n ive rsity C o m m u n ic a tio n s d e la Fa c u lté d e s S c ie n c e s d e l’U n ive rs ité d ’A n ka ra . S é rie s A 1 . M a th e m a t ic s a n d S t a tis tic s .
We de…ne following as a mathematical tool for this research article:
De…nition 1. Let A be a subset of a topological space (X; ). A is said to be regular open [21] (resp. semi-open [16, 11], semi-pre open [3], -open [20]) if A = Int(Cl(A)) (resp. A Cl(Int(A)); A Cl(Int(Cl(A))); A Int(Cl(Int(A)))). De…nition 2. [23] A subset A of a topological space (X; ) is said to be -open if, for each x 2 A, there exists a regular open set G such that x 2 G A.
The complement of -open set is called -closed. Let (X; ) be a topological space, then the point x 2 X is called -cluster point of A if Int(Cl(V )) \ A 6= ;, for each open set V containing x.
The -closure of A is denoted as Cl (A) [23] and it is a set of all -cluster point of A. In this regards, Int (A) [23] is the -interior of A and it is the union of all regular open sets of (X; ) contained in A. If Int (A) = A for a topological space (X; ), then A is -open and conversely [23]. It is remarkable that the collection of all -open sets in a topological space (X; ) forms a topology and it is denoted as
[23].
De…nition 3. [8, 9, 10] A subset A of (X; ) is said to be a-open (resp. a-closed) if A Int(Cl(Int (A))) (resp. Cl(Int(Cl (A))) A).
The family of a-open sets in (X; ) forms a topology on X. This collection is denoted as a[8], and a(x) is denoted as the collection of all a-open sets containing
x.
In this paper we also denote ‘aCl’ by the means of closure operator of Ekici’s a topology [7, 8].
Hereditary class and a-local function are also the mathematical tool for this paper:
De…nition 4. [15] A collection I }(X) is said to be an ideal on X if B A 2 I implies B 2 I and A; B 2 I implies A [ B 2 I.
Let I be an ideal on the topological space (X; ), then (X; ; I) is called an ideal topological space.
According to Al-Omari et al. [2, 1], we give the following:
The a-local function ()a : }(X) ! }(X) for a subset A of an ideal topological
space (X; ; I) is de…ned as (A)a = fx 2 X : U \ A =2 I; for every U 2 a(x)g,
and as like complement operator of ()a , <a: }(X) ! }(X) is de…ned as <a(A) =
X n (X n A)a = fx 2 X : there exists U
x 2 a(x) such that Uxn A 2 Ig.
Due to the operator ()a , we have a topology a [1] whose one of the basis is
(I; ) = fV n I : V 2 a; I 2 Ig [1]. In this respect, we will denote ‘Inta ’and
‘Cla ’as ‘interior’operator and ‘closure’operator of (X; a ) respectively.
Following results help us for repairing the paper:
Theorem 1. [1] Let (X; ; I) be an ideal topological space and U 2 a. Then
Corollary 2. Let A be a subset of an ideal topological space (X; ; I), then aInt(A) <a(A).
Theorem 3. [1] Let A be a subset of an ideal topological space (X; ; I) with a\I =
;. Then <a(A) (A)a .
Corollary 4. Let A be a subset of an ideal topological space (X; ; I) with a\I = ;. Then <a(A) aCl(A).
Lemma 5. Let (X; ; I) be an ideal topological space and O 2 a. Then a\ I = ;
if and only if (O)a = aCl(O).
Proof. Let a\ I = ; and ; 6= O 2 a . Now Oa aCl(O) always. For reverse
inclusion, let x 2 aCl(O). Therefore all neighbourhoods Ux 2 a(x), Ux\ O 6= ;
implies Ux\ O =2 I, since a\ I = ;. Therefore x 2 (O)a . Hence (O)a = aCl(O).
Conversely let O 2 a, (O)a = aCl(O). Then Xa = X and this implies I \ a = ; [2].
Proposition 6. Let (X; ; I) be an ideal topological space with a\ I = ;. Then
following hold:
(1) For A X, <a(A) aInt(aCl(A)).
(2) For a-closed subset A; <a(A) A.
(3) For A X; aInt(aCl(A)) = <a(aInt(aCl(A))).
(4) For any a-regular open subset A; A = <a(A).
(5) For any O 2 a; <
a(O) aInt(aCl(O)) (O)a .
Proof. (1) From Theorem 3, <a(A) (A)a . Then <a(A) aCl(A), and since
<a(A) is open, <a(A) aInt(aCl(A)).
(3) <a(aInt(aCl(A))) (aInt(aCl(A))a = aCl(aInt(aCl(A))) (from Lemma
5) aCl(A). Thus <a(aInt(aCl(A))) aInt(aCl(A)).
Reverse inclusion: aInt(aCl(A)) <a(aInt(aCl(A))) (from Theorem 1).
Thus aInt(aCl(A)) = <a(aInt(aCl(A))).
3. <a aCl sets
De…nition 5. Let (X; ; I) be an ideal topological space and A X, A is said to be a <a aCl set if A aCl(<a(A)).
The collection of all <a aCl sets in (X; ; I) is denoted by <a(X; a).
Note 3.1. Let (X; ; I) be an ideal topological space. If A 2 a, then A 2
<a(X; a).
Later, we shall given the example for the converse of this note.
Theorem 7. Let fAi : i 2 g be a collection of nonempty <a aCl sets in an
Proof. For each i; Ai aCl(<a(Ai)) aCl(<a(
S
i2 Ai)). This implies that
S i2 Ai aCl(<a( S i2 Ai)). Thus S i2 Ai2 <a(X; a).
For intersecting of two <a aCl sets, we give following example:
Example 1. Let X = fe; b; c; dg; = f;; X; feg; fcg; fe; bg; fe; cg; fe; b; cg; fe; c; dgg; I = f;; fbgg. Regular open sets are: ;; X; fcg; fe; bg. Then a= f;; X; fcg; fe; bg;
fe; b; cgg. Therefore <a(fc; dg) = X n (fe; bg)a = X n fe; b; dg = fcg and
aCl(<a(fc; dg)) = fc; dg. Again <a(fe; b; dg) = X n(fcg)a = X nfc; dg = fe; bg and
aCl(<a(fe; b; dg)) = fe; b; dg. Now <a(fdg) = X n (fe; b; cg)a = X n fe; b; c; dg = ;.
Hence we have fc; dg and fe; b; dg are <a aCl sets but they are not a-open sets.
Again their intersection fdg is not a <a aCl set.
We show that the intersecting of a <a aCl set and an - set of a is also a
<a aCl set.
Theorem 8. Let (X; ; I) be an ideal topological space and A 2 <a(X; a). If
U 2 a , then U \ A 2 <a(X; a) ( a denotes the collection of all -open sets in
(X; a)).
Proof. Let G be a-open, and A X, then it is obvious that G \ aCl(A) aCl(G \ A) ... (i).
If V is a-open, then V aInt(aCl(V )) and it is obvious that aCl(aInt(aCl(V ))) aCl(V ). Hence
aCl(V ) = aCl(aInt(aCl(A))) ... (ii). Again for A and B subsets of X,
<a(A \ B) = <a(A) \ <a(B) [1] ... (iii).
Let U 2 a and A 2 <
a(X; a), then we have U \ A aInt(aCl(aInt(U ))) \
aCl(<a(A)) aInt(aCl(<a(U )))\aCl(<a(A)) (Corollary 2). Since aInt(aCl(<a(U )))
is a-open, from (i) we have
U \ A aCl[aInt(aCl(<a(U ))) \ <a(A)] = aCl[aInt[aCl(<a(U )) \ <a(A)]], since
<a(A) is a-open. Now by again from (i), we have U \ A aCl[aInt[aCl(<a(U ) \
<a(A))]]. Since <a(U )\<a(A) is a-open then from (ii), we get U \A aCl(<a(U )\
<a(A)) = aCl(<a(U \ A)) (using (iii)). Therefore, U \ A 2 <a(X; a).
As a a for a topological space (X; ), then we have following corollary:
Corollary 9. Let (X; ; I) be a topological space and A 2 <a(X; a). If U 2 a,
then U \ A 2 <a(X; a).
For next, we recall that, a subset A of an ideal topological space (X; ; I) is said to be Ia-dense [1] if (A)a = X.
Theorem 10. A =2 <a(X; a) if and only if there exists x 2 A such that there is a a - neighbourhood V
Proof. Let A =2 <a(X; a). We are to show that there exists an element x 2 A and
a a - neighbourhood V
x of x satisfying that X n A is relatively Ia - dense in Vx.
Since A * aCl(<a(A)), there exists x 2 X such that x 2 A but x =2 aCl(<a(A)).
Hence there exists a a - neighbourhood V
x of x such that Vx\ <a(A) = ;. This
implies that Vx\(X n(X nA)a = ; and so Vx (X nA)a . Let U be any nonempty
a-open set in Vx. Since Vx (X n A)a , therefore U \ (X n A) =2 I. This implies
that (X n A) is relatively Ia - dense in Vx.
The converse part is obvious by reversing process.
Relations between <a aCl set with generalized open sets:
Theorem 11. Let (X; ; I) be a topological space, then SO(X; a) <
a(X; a)
(SO(X; a) denotes the collection of all semi-open sets in (X; a)).
Proof. For A aCl(aInt(A)), A aCl(aInt(A)) aCl(<a(aInt(A))) aCl(<a(A)).
Thus SO(X; a) <a(X; a).
Theorem 12. Let A be a <a aCl set in a topological space (X; ; I), where a \ I = ;. Then A 2 SP O(X; a) (SP O(X; a) denotes the collection of all
semi-preopen sets in (X; a)).
Let (X; ; I) be an ideal topological space, and In( a) is denoted as the collection
of all nowhere dense subsets of (X; a).
Lemma 13. Let (X; ; I) be an ideal topological space, where I = In( a), then for
A X; <a(A) = aInt(aCl(aInt(A))).
Proof. Proof is obvious from the fact that <a(A) = X n (X n A)a and (A)a =
aInt(aCl(aInt(A))).
Theorem 14. Let (X; ; I) be an ideal topological space, where I = In( a), then
<a(X; a) = SO(X; a).
Proof. Let A 2 <a(X; a), therefore A aCl(<a(A)) = aCl(aInt(aCl(aInt(A))))
(from Lemma 13) = aCl(aInt(A)). Thus A 2 SO(X; a).
Suppose that A 2 SO(X; a). Then A aCl(aInt(A)), so aInt(A) 6= ;. We know that aInt(A) <a(A) by Corollary 2. Thus A aCl(aInt(A))
aCl(<a(A)). Hence the Theorem.
In o.h.i. space two concepts semi-preopen set and <a aCl set are synonymous,
where o.h.i. space is de…ned as follows:
A space (X; ) is said to be resolvable [13] if there is a dense subset D of X such that X n D are dense in (X; ). Otherwise it is said to irresolvable [13]. Real line with usual topology is an example of a resolvable space. A space (X; ) is called open hereditarily irresolvable (in short o.h.i.) [5] if every nonempty open subset of it is irresolvable.
Theorem 15. Let (X; ; I) be an ideal topological space, where (X; a) is an o.h.i.
space, a\ I = ;. Then <
a(X; a) = SP O(X; a).
Proof. We shall prove only the inclusion SP O(X; a) <
a(X; a), reverse inclusion
has already been done. Let A 2 SP O(X; a). Then A aCl(aInt(aCl(A))). Let
x 2 aInt(aCl(A)). Therefore there exists a nonempty a-open set Ox (containing
x) such that Ox aCl(A). Now it is obvious that Ox\ A is dense in Ox. Since the
space is o.h.i., therefore aInt(Ox\ A) is dense in Ox, that is Ox aCl(aInt(A))
and hence x 2 aCl(aInt(A)). Thus aInt(aCl(A)) aCl(aInt(A)). Now A aCl(aInt(aCl(A))) aCl(aInt(A)). But aInt(A) <a(aInt(A)) <a(A), thus
A aCl(<a(A)). Therefore A 2 <a(X; a).
4. - topology of a
De…nition 6. Let (X; ; I) be an ideal topological space and A X. A is said to be a <a - set if A aInt(aCl(<a(A))).
The collection of all <a sets in (X; ; I) is denoted as a
<a
. It is obvious that a a<a
<a(X; a).
Theorem 16. Let (X; ; I) be an ideal topological space with a\ I = ;, then the
collection a<a = fA X : A aInt(aCl(<a(A)))g forms a topology on X.
Proof. We shall prove only …nite intersection property: Let A1; A22 a
<a
. We are to show that A1\ A22 a
<a
. If A1\ A2= ;, we are
done. Let A1\ A26= ;. Let x 2 A1\ A2. Now A1 aInt(aCl(<a(A1))) and A2
aInt(aCl(<a(A2))), implies that x 2 aInt(aCl(<a(A1))) \ aInt(aCl(<a(A2))). So
x 2 aInt[aCl(<a(A1)) \ aCl(<a(A2))]. Therefore there exists an a-open set Vx
containing x such that Vx aCl(<a(A1)) \ aCl(<a(A2)). Let Ux be any
a-neighbourhood of x. Then ; 6= Vx\Ux aCl(<a(A1)) and Vx\Ux aCl(<a(A2)).
Let y 2 Vx\ Ux. Consider any open set Gycontaining y. Without loss of generality
we may suppose that Gy Vx\Ux. So Gy\<a(A1) 6= ;. From de…nition of <a(A1)
there exists a nonempty a-open set U such that U Gyand U nA12 I. Again U
aCl(<a(A2)), so there exists a nonempty a-open set U0 U such that U0n A22 I.
Now U0n(A
1\A2) = (U0nA1)[(U0nA2) (U nA1)[(U0nA2) 2 I (…nite additivity).
Hence from de…nition U0 <
a(A1\ A2). Since U0 Gy; Gy\ <a(A1\ A2) 6= ;,
therefore y 2 aCl(<a(A1\ A2). Since y was any point of Ux\ Vx, it follows that
Ux\ Vx aCl(<a(A1\ A2)), implies that x 2 aInt(aCl(<a(A1\ A2))). Thus
A1\ A2 aInt(aCl(<a(A1\ A2))). Hence A1\ A22 a
<a
.
Theorem 17. Let (X; ; I) be an ideal topological space, where a\ I = ;, then
a a<a
.
Corollary 18. Let (X; ; I) be an ideal topological space, where I = In( a), then a = a<a.
Lemma 19. Let (X; ; I) be an ideal topological space, where a\ I = ;. Then
<a(A) 6= ; if and only if A contains a nonempty a -interior.
Proof. Let <a(A) 6= ;. Therefore <a(A) = [fM : M 2 a; M n A 2 Ig 6= ;,
implies that there exists ; 6= M 2 a such that M n A 2 I. Let M n A = P , where
P 2 I. So M n P A where M n P 6= ;, since a\ I = ;. Since M n P 2 a , so
that A contains a nonempty a - interior.
Conversely suppose that A contains a a - interior M n P (say), where M 2
a
; P 2 I. Thus M n P A, that is M n A P . Hence M n A 2 I. So [fM : M 2 a; M n A 2 Ig 6= ;. This implies that <
a(A) 6= ;.
Corollary 20. Let x 2 X. Then fxg is open in (X; a ) if and only if fxg 2
<a(X; a).
Corollary 21. Let x 2 X, then fxg 2 a<a if and only if fxg 2 <a(X; a).
Theorem 22. a<a is exactly the collection such that A 2 a<a
and B 2 <a(X; a)
implies A \ B 2 <a(X; a).
Proof. Let A 2 a<a and B 2 <
a(X; a). Now we are to show that A \ B 2
<a(X; a). If A \ B = ;, we are done. Let A \ B 6= ;. Let x 2 A \ B. This implies
that x 2 aInt(aCl(<a(A))), therefore aInt(aCl(<a(A))) is a a-neighbourhood of
x. Consider any a-neighbourhood Ux of x, then Ux\ aInt(aCl(<a(A))) is a
a-neighbourhood of x. Since x 2 B aCl(<a(B)), then Ux\ aInt(aCl(<a(A))) \
<a(B) 6= ;. Let V = Ux\ aInt(aCl(<a(A))) \ <a(B), then V aCl(<a(A)).
This implies that Ux\ <a(A) \ <a(B) = V \ <a(A) 6= ;, since <a(A) is a-open.
Therefore x 2 aCl(<a(A) \ <a(B)), that is x 2 aCl(<a(A \ B)). Hence A \ B
aCl(<a(A \ B)), therefore A \ B 2 <a(X; a).
Next we consider a subset A of X such that A \ B 2 <a(X; a) for each
B 2 <a(X; a). We show that A 2 a
<a
, that is A aInt(aCl(<a(A))). If
possible suppose that x 2 A but x =2 aInt(aCl(<a(A))). Therefore x 2 A \ [X n
aInt(aCl(<a(A)))] = A \ aCl(X n aCl(<a(A))) = A \ aCl(G) (say). It is obvious
that G = X n aCl(<a(A)) is a nonempty a open set. Since x 2 aCl(G) then for
all a-open sets Vx containing x, Vx\ G 6= ;. Therefore Vx\ <a(G) 6= ;, since
G <a(G). This implies that
x 2 aCl(<a(G)) aCl(<a(fxg [ G)) ... (i).
Again
G aCl(<a(G)) aCl(<a(fxg [ G)) ...(ii).
From (i) and (ii) fxg [ G aCl(<a(fxg [ G)). Thus fxg [ G 2 <a(X; a). Now
by given condition A \ (fxg [ G) is a <a aCl set.
We shall prove that A \ (fxg [ G) = fxg.
If possible suppose that there exists y 2 X and x 6= y such that y 2 A\(fxg[G). So y 2 A and y 2 G. Now A = A \ X and X 2 <a(X; a), again by given condition
A 2 <a(X; a). Since y 2 A, and y 2 aCl(<a(A)) - a contradiction to the fact that
fxg 2 a<a. So fxg aInt(aCl(<a(fxg))) = aInt(aCl(<a(A \ (fxg [ G))))
aInt(aCl(<a(A))). But x 2 aInt(aCl(<a(A))), that is A 2 a
<a
.
Theorem 23. Let (X; ; I) be an ideal topological space, where a\ I = ;. Then SO(X; a ) = <
a(X; a).
Proof. Let A 2 SO(X; a ). Then A Cla (Inta (A)) = Cla [<
a(A) \ A]
aCl(<a(A) \ A) aCl(<a(A)). Thus A 2 <a(X; a). For reverse inclusion, let
A 2 <a(X; a). We show that A 2 SO(X; a ). Take x 2 A. Consider G12 (I; )
[2] such that x 2 G1. Then G1is of the form G1= G n E, where G 2 a; E 2 I. So
x 2 G. Since A aCl(<a(A)) and G 2 a; G \ (<a(A)) 6= ;. Let y 2 G \ (<a(A)).
Thus there exists Oy 2 a such that Oyn A 2 I by de…nition of <a(A). Consider
; 6= G \ Oy. So (G \ Oy) n A 2 I (by heredity). Let G0 = G \ Oy. Then
G0 6= ;; G0 2 a and G0n A = P say where P 2 I and so G0 n P A. Hence G0n (E [ P ) A where G0n (E [ P ) 6= ;, since a\ I = ;. Write M = G0n (E [ P ). Then ; 6= M 2 a such that M A \ (G n E). Hence A contains a nonempty
a -open set M contained in G n E = G
1. Since x is an arbitrary point of A, we
get A Cla (Inta (A)). Therefore A 2 SO(X; a ).
Corollary 24. Let x 2 X, then fxg 2 SO(X; a ) if and only if fxg 2 a<a. Theorem 25. a<a is exactly the collection such that A 2 a<a and B 2 SO(X; a ) imply A \ B 2 SO(X; a ), where a \ I = ;.
Theorem 26. [20] Let (X; ) be a topological space. consists of exactly those sets A for which A \ B 2 SO(X; ) for all B 2 SO(X; ).
From above Theorem we get the representation of - sets of (X; a):
Theorem 27. Let (X; ; I) be an ideal topological space with a\ I = ;. Then a<a
= a .
Acknowledgement: Authors are thankful to Professor Erdal Ekici for sending reprints.
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E-mail address : spmodak2000@yahoo.co.in
Current address : Md. Monirul Islam: Department of Mathematics, University of Gour Banga P.O. Mokdumpur, Malda - 732103, India
