Download On Regular b-Open Sets in Topological Spaces

 Int. Journal of Math. Analysis, Vol. 7, 2013, no. 19, 937 - 948
HIKARI Ltd, www.m-hikari.com
On Regular b-Open Sets
in Topological Spaces
A. Narmadha
Hindusthan College of Engineering & Technology
Coimbatore-32, Tamilnadu, South India
[email protected]
N. Nagaveni
Coimbatore Institute of Technology
Coimbatore-14, Tamilnadu, South India
[email protected]
T. Noiri
2949-1 Shiokita-cho, Hinagu
Yatsushiro-shi, Kumamoto-ken, 869-5142 Japan
[email protected]
Copyright © 2013 A. Narmadha et al. This is an open access article distributed
under the Creative Commons Attribution License, which permits unrestricted use,
distribution, and reproduction in any medium, provided the original work is properly
cited.
Abstract. The purpose of this paper is to introduce and study regular bopen sets (briefly rb-open sets) in topological spaces and obtain some of
their properties. We introduce the rb-closure and discuss some basic
properties of the rb-closure. Also we study the new concepts of rb-closed
spaces by means of filter bases.
Mathematics Subject Classification: 54A05, 54A20
Keywords: Topological spaces, rb-open set, rb-closed set, rb-interior, rbclosure, rb-closed space
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1. Introduction
The notion of regular b-closed (briefly rb-closed) sets is
introduced and studied recently by Nagaveni and Narmadha [15],[16].
Regular open sets and b-open sets have been introduced and investigated
by Stone [19] and Andrijevi’c [2], respectively. Levine [11] (resp.
Bhattacharya and Lahiri [4], Palaniappan and Rao [17], Arya and Nour
[3], Maki et al [14], Maki et al [12],[13]) introduced and investigated
generalized closed sets (resp. semi-generalized closed sets, regular
generalized closed sets, generalized semi-closed sets, generalized preclosed sets, generalized α-closed sets and α-generalized closed sets). AlOmari and Noorani [1] investigated the class of generalized b-closed sets
and obtained some of its fundamental properties. We introduce and study
the concepts of rb-open sets and rb-closed spaces.
Throughout this paper, a space means a topological space on
which no separation axioms are assumed unless otherwise mentioned.
For a subset A of a space (X,τ), Cl(A) and Int(A) denote the closure of A
and interior of A, respectively. X-A or Ac denotes the complement of A
in X.
2. Preliminaries
Let us recall the definitions and results which are used in the sequel.
Definition 2.1: A subset A of a space X is said to be regular closed [19]
if A = Cl (Int (A)).
Definition 2.2: A subset A of a space X is said to be b-closed [2] if Int
(Cl (A)) ∩ Cl ( Int (A)) ⊂ A.
Definition 2.3: A subset A of a space X is said to be
1. generalized closed (briefly g-closed) [11] if Cl (A) ⊂ U whenever A
⊂ U and U is open in X,
2. generalized semi-closed (briefly gs-closed) [3] if scl (A) ⊂ U
whenever A ⊂ U and U is open in X,
3. semi-generalized closed (briefly sg-closed) [4] if scl (A) ⊂ U
whenever A ⊂ U and U is semi-open in X,
4. regular-generalized closed (briefly rg-closed) [17] if Cl (A) ⊂ U
whenever A ⊂ U and U is regular-open in X,
5. generalized pre-closed (briefly gp-closed) [14] if pcl (A) ⊂ U
whenever A ⊂ U and U is open in X,
On regular b-open sets in topological spaces
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6. generalized α-closed (briefly gα-closed) [12] if αcl (A) ⊂ U
whenever A ⊂ U and U is α-open in X,
7. α-generalized closed (briefly αg-closed) [13] if αcl (A) ⊂ U
whenever A ⊂ U and U is open in X,
8. generalized b-closed (briefly gb-closed) [1] if bcl (A) ⊂ U whenever
A ⊂ U and U is open in X,
9. δ-generalized closed (briefly δg-closed) [8] if δcl (A) ⊂ U whenever
A ⊂ U and U is open in X.
Definition 2.4: A subset A of a space X is said to be regular b-closed
(briefly rb-closed) [16] if rcl (A) ⊂ U whenever A ⊂ U and U is b-open
in X.
Remark 2.5: The complements of the above mentioned closed sets are
their respective open sets.
Remark 2.6: The intersection of all regular closed sets of X containing
A is called the regular closure, or the δ-closure, of A and is denoted by
rcl (A) [18] or δcl (A) [21]. The regular interior, or the δ–interior, of A is
the union of all regular open sets contained in A and is denoted by rint
(A) or δint (A). The family of all rb-open (respectively rb-closed) sets of
(X,τ) is denoted by RBO(X) (respectively RBC(X)). The family of all rbopen sets of (X,τ) containing a point x ∈ X is denoted by RBO(X, x).
3. Regular b-closed sets (rb-closed sets)
In this section, we obtain the further properties of regular b-closed
sets and some improvements of the results established in [15] , [16].
Proposition 3.1: Every δ-closed set is rb-closed and every rb-closed set
is δg-closed.
Proof: Let A be a δ-closed set and U a b-open set such that A ⊂ U. Then
δcl (A) = A ⊂ U and A is rb-closed. Let A be an rb-closed set and U an
open set such that A ⊂ U. Since every open set is b-open, δcl (A) ⊂ U
and A is δg-closed.
Remark 3.2: The converses of Proposition 3.1 are not always true.
Moreover, δg-closed sets and closed sets are independent as shown by
the following examples.
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Example 3.3: (1) Let X = {a, b, c}. τ = {X, ϕ, {a}, {b}, {a, b}, {b, c}}
and A = {c}. Then A is rb-closed [16, Example 3.6] but it is not δ-closed.
(2) Let X = {a, b, c}. τ = {X,ϕ, {a, b}} and A = {a, c}. Then A is δgclosed [8, Example 3.2]. Since A is b-open and X – Cl(A) ⊂ δcl(A), A is
rb-closed nor closed.
(3) It is shown in Example 3.3 of [8] that there exists a closed set which
is not δg-closed.
Proposition 3.4: For a subset of a space, the following implications hold:
(1) every rb-closed set is gα-closed but not conversely,
(2) every gα-closed set is gp-closed and every gp-closed set is gb-closed,
(3) every δg-closed set is αg-closed and every αg-closed set is gb-closed.
Proof: (1) This is shown in Theorem 3.7 and Example 3.9 in [15].
(2) Suppose that A is gα-closed. Let U be open and A ⊂ U. Since U is
α-open, αcl (A) ⊂ U and hence pcl (A) ⊂ αcl (A) ⊂ U. Therefore A is
gp-closed. Every preopen set is b-open and it is obvious that every gpclosed set is gb-closed.
(3) It is shown in Diagram (p.19) in [8] that every δg-closed set is αgclosed. The other is obvious.
The following diagram is obtained by Propositions 3.1 and 3.4 and the
diagram in [8].
DIAGRAM
regular closed → δ-closed → closed → g-closed
rb-closed → δg-closed → αg-closed
gα-closed → gp-closed → gb-closed
Theorem 3.5: (1) If A and B are rb-closed, then A ∪ B is rb-closed.
(2) The union of countable rb-closed sets is not always rbclosed.
Proof: (1) Let A ∪ B ⊂ U and U be a b-open set. Then, we have
δcl (A ∪ B) = δcl (A) ∪ δcl (B) ⊂ U and hence A ∪ B is rb-closed.
(2) Let X be the real numbers with the usual topology. Since X is regular,
⎧1 ⎫
every singleton is δ-closed and hence rb-closed. Set A = Un∈N ⎨ ⎬ ,
⎩n⎭
where N is the set of all positive integers. There exists a b-open set (0, 1]
On regular b-open sets in topological spaces
941
such that A ⊂ (0, 1] and 0 ∈ δcl (A). Therefore, δcl (A) ⊄ (0, 1]. This
shows that A is not rb-closed.
Theorem 3.6: If A is rb-closed and B is δ-closed, then A ∩ B is rbclosed.
Proof: Let U be a b-open set and A ∩ B ⊂ U. Then U ∪ (X – B) is a bopen set containing A. Since A is rb-closed, we have δcl (A ∩ B) ⊂ δcl
(A) ∩ δcl (B) = δcl (A) ∩ B ⊂ [U ∩ (X – B)] ∩ B = U ∩ B ⊂ U. This
shows that A ∩ B is rb-closed.
Question 3.7: In Theorem 3.6, can we replace the assumption “B is δclosed” with “B is rb-closed”?
It is shown in Theorem 3.10 of [16] that if A is b-open and rbclosed then A is regular closed. However, there is a gap in the proof. The
following theorem improves this result.
Theorem 3.8: A subset A is b-open and rb-closed if and only if it is
regular closed.
Proof: Necessity. Suppose that A is b-open and rb-closed. Then δcl (A)
⊂ A and δcl (A) = A. Therefore, A is δ-closed and closed. Since A is bopen, A ⊂ Int(cl(A)) ∪ Cl(Int(A)) = Int(A) ∪ Cl(Int(A)) = Cl(Int(A)).
Therefore, we obtain A = Cl (A) = Cl(Int(A)) and hence A is regular
closed.
Sufficiency. Suppose that A is regular closed. Every regular closed
set is semi-open and every semi-open set is b-open. Therefore, A is bopen. It is obvious that every regular closed set is rb-closed.
Theorem 3.9: Let A be an rb-closed set in (X,τ). Then A is closed if
Cl A – A is closed.
Proof: Let A be an rb-closed set in (X,τ). We want to show that Cl A – A
is closed. Let F be a closed set such that F ⊂ Cl A – A. Then F ⊂ X – A
implies A ⊂ X – F. The set A is rb-closed and X – F is open. Since every
open set is b-open, X- F is a b-open set. Therefore rcl A ⊂ X – F. Also
since every regular closed set is closed, Cl A ⊂ rcl A. Therefore Cl A ⊂
rcl A ⊂ X – F and hence Cl A ⊂ X – F. That is, F ⊂ X – Cl A. Thus F ⊂
(X – Cl A) ∩ Cl A = ϕ implies Cl A – A = ϕ, which is closed
The converse of the above theorem need not be true as seen from the
following example.
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Example 3.10: Let X = {a, b, c}. τ = { ϕ, X, {a}, {a, b}, {a, c}}. Let A =
{c}. The set A is closed but not rb-closed in (X,τ).
4. Regular b-open sets (rb-open sets)
In this section we introduce a new class of open sets called rb-open
sets.
Definition 4.1: A set A ⊂ X is said to be regular b-open (briefly rb-open)
if its complement is a regular b-closed set.
Theorem 4.2: A subset A ⊂ X is rb-open if and only if F ⊂ rint (A)
whenever F is b-closed and F ⊂ A.
Proof: Necessity. Let A be an rb-open set and suppose F ⊂ A, where F
is b-closed. Then X – A is an rb-closed set contained in the b-open set
X – F . Hence rcl (X – A) ⊂ X – F and X – rint (A) ⊂ X - F. Thus F ⊂
rint (A).
Sufficiency. Let X – A ⊂ U and U be b-open. Then X – U ⊂ A and
X – U is b-closed. By hypothesis, X – U ⊂ rint (A) and hence rcl (X – A)
= X – rint (A) ⊂ U. Hence X – A is an rb-closed set and A is an rb-open
set.
Theorem 4.3: If rint (A) ⊂ B ⊂ A and A is rb-open in X, then B is rbopen in X.
Proof: rint (A) ⊂ B ⊂ A implies X – A ⊂ X – B ⊂ X – rint (A). That is,
X – A ⊂ X – B ⊂ rcl (X –A). Since X – A is rb-closed, by Theorem
3.16 of [16] X – B is rb-closed and hence B is rb-open in X.
Theorem 4.4: A set A is rb-closed in X if and only if rcl (A) – A is rbopen in X.
Proof: Necessity. Let A be an rb-closed set. Let F be a b-closed set such
that F ⊂ rcl (A) – A. Then by Theorem 3.2 in [16], F = ϕ. So F ⊂ rint (
rcl (A) – A ). Thus rcl (A) – A is rb-open.
Sufficiency. Let A ⊂ U and U be a b-open set. Then rcl (A) ∩ U c ⊂
rcl (A) ∩ Ac = rcl (A) – A. Since rcl (A) ∩ U c is b-closed and rcl (A) –
A is rb-open, by Theorem 4.2, rcl (A) ∩ U c ⊂ rint (rcl (A) – A) = ϕ.
Hence rcl (A) ⊂ U and so A is rb-closed in X.
On regular b-open sets in topological spaces
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Theorem 4.5: Let A, B ⊂ X. If B is rb-open and A ⊃ rint (B), then A ∩
B is rb-open.
Proof: Since B is rb-open and A ⊃ rint (B), rint (B) ⊂ A ∩ B ⊂ B. By
Theorem 4.3, A ∩ B is rb-open.
Definition 4.6: For any set A ⊂ X, the rb-interior of A is defined as the
union of all rb-open sets contained in A and is denoted by rbint (A).
Lemma 4.7: For any A ⊂ X, rint (A) ⊂ rbint (A) ⊂ A.
Proof: This follows from Remark 3.40 of [15]
Proposition 4.8: For any two subsets A and B of (X,τ),
(i)
If A ⊂ B, then rbint (A) ⊂ rbint (B)
(ii)
rbint (A ∩ B) = rbint (A) ∩ rbint (B)
(iii) rbint (A) ∪ rbint (B) ⊂ rbint (A ∪ B)
(iv)
rbint (X) = X
(v)
rbint (ϕ) = ϕ
Proof: (i) Let A and B be subsets of X such that A ⊂ B. Let x ∈ rbint
(A). Then there exists an rb-open set U such that x ∈ U ⊂ A. Since A ⊂
B, x ∈ U ⊂ B and hence x ∈ rbint (B). Hence rbint (A) ⊂ rbint (B).
(ii) We know that A ∩ B ⊂ A and A ∩ B ⊂ B. We have by (i), rbint (A
∩ B) ⊂ rbint (A) and rbint (A ∩ B) ⊂ rbint (B). This implies that rbint
(A ∩ B) ⊂ rbint (A) ∩ rbint (B) … (1).
Again, let x ∈ rbint (A) ∩ rbint (B). Then x ∈ rbint (A) and x ∈
rbint (B). There exist rb-open sets U and V such that x ∈ U ⊂ A and x ∈
V ⊂ B. By Theorem 3.5 (1), U ∩ V is an rb-open set such that x ∈ (U ∩
V) ⊂ (A ∩ B). Hence x ∈ rbint (A ∩ B). Thus x ∈ rbint (A) ∩ rbint (B)
implies that x ∈ rbint (A ∩ B). Therefore, rbint (A) ∩ rbint (B) ⊂ rbint
(A ∩ B) … (2).
From (1) and (2), it follows that rbint (A ∩ B) = rbint (A) ∩ rbint (B).
The proofs of (iii), (iv), (v) are obvious.
Remark 4.9: Arbitrary union of rb-open sets is not always rb-open.
Definition 4.10: For every set A ⊂ X, we define the rb-closure of A to be
the intersection of all rb-closed sets containing A and is denoted by rbcl
(A).
If a subset A of a space (X, τ) is rb-closed then A = rbcl (A).
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Lemma 4.11: For an element x ∈ X, x ∈ rbcl (A) if and only if U ∩ A ≠
ϕ for every rb-open set U containing x.
Proof: Necessity. Let x ∈ rbcl (A). Suppose there exists an rb-open set U
containing x such that U ∩ A = ϕ. Then A ⊂ U c. Since U c is an rbclosed set containing A, we have rbcl (A) ⊂ U c, which implies that x ∉
rbcl (A). This is a contradiction.
Sufficiency. Suppose that x ∉ rbcl (A). Then by definition, there
exists an rb-closed set F containing A such that x ∉ F. Then x ∈ F c and
F c is an rb-open set. Also F c ∩ A = ϕ which is contrary to the
hypothesis. Therefore x ∈ rbcl (A).
Proposition 4.12: (i)
(ii)
[rbint (A)] c = rbcl (A c)
[ rbcl (A)] c = [rbint (A c)]
Proof: (i) Let x ∈ [rbint (A)] c. then x ∉ rbint (A). That is, every rbopen set U containing x is such that U ∩ A c ≠ ϕ . By Lemma 4.11, x ∈
rbcl (A c). Therefore, [rbint (A)] c ⊂ rbcl (A c).
Conversely, let x ∈ rbcl (A c). Then by Lemma 4.11, U ∩ A c ≠ ϕ for
every rb-open set U containing x. That is, every rb-open set U containing
x is such that U ⊄ A. This implies x ∉ rbint (A). That is, x ∈ [rbint (A)] c
and so rbcl (A c) ⊂ [rbint (A)] c. Thus [rbint (A)] c = rbcl (A c).
(ii) This follows by replacing A by A c in (i)
5. Regular b-closed spaces (rb-closed spaces)
Definition 5.1: A filter base ℑ = {Aα} rb-converges to a point x0 ∈ X if
for each rb-open set V containing x0 there exists an Aα ∈ ℑ such that Aα
⊂ rbcl (V).
Definition 5.2: A filter base ℑ = {Aα} rb-accumulates to a point x0 ∈ X
if for each rb-open set V containing x0 and Aα ∈ ℑ, Aα ∩ rbcl (V) ≠ ϕ.
Definition 5.3: A topological space X is said to be rb-closed if for every
cover {Vα | α ∈ ∇} of X by rb-open sets of X, there exists a finite subset
∇0 of ∇ such that X = ∪{rbcl (Vα) | α ∈ ∇0}.
Definition 5.4: A subset K of a topological space X is said to be rbclosed relative to X if for any cover {Ui : i ∈ I} of K by rb-open sets,
there exists a finite subset I0 of I such that K ⊂ ∪{rbcl (Ui) : i ∈ I0}
On regular b-open sets in topological spaces
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Remark 5.5: A filter base ℑ is rb-convergent to a point x ∈ X if and
only if ℑ contains the collection {rbcl (U) : U ∈ RBO(X, x)}.
Theorem 5.6: For a topological space, the following properties are
equivalent:
(i)
X is rb-closed.
(ii)
For each family of rb-closed sets {Fα} such that ∩( Fα) = ϕ,
there exists a finite subfamily {Fα i }in=1 such that
I ni=1 rb int( Fα i ) = ϕ
(iii)
(iv)
Each filter base ℑ = {Aα} rb-accumulates to some point x0 ∈
X.
Each maximum filter base ℑ rb-converges.
Proof: (i) ⇒ (iv) Let ℑ = {Aα} be a maximum filter base. Suppose that
ℑ does not rb-converge to any point of X, then ℑ does not rb-accumulate
to any point. This implies that for every x ∈ X, there exists a rb-open set
V(x) containing x and an Aα(x) ∈ ℑ such that Aα(x) ∩ rbcl (V(x)) = ϕ.
Obviously {V(x) | x ∈ X} is an rb-open cover for X and by hypothesis
there exists a finite subfamily such that X = U ni= 1 rbcl (V ( x i )) . Since ℑ is
a filter base, there exists an A0 ∈ ℑ such that A0 ⊂ I ni= 1 Aα ( xi ) . Hence
A0 ∩ rbcl (V(xi)) = ϕ, 1 ≤ i ≤ n, which implies
A0 I (U ni= 1 rbcl (V ( xi )) = A0 I X = ϕ contradicting the essential fact that
A0 ≠ ϕ.
(iv) ⇒ (iii) Let ℑ be a filter base on X and ℑ0 a maximal filter base
such that ℑ ⊂ ℑ0. By (iv), ℑ0 rb-converges to some x ∈ X. For every F
∈ ℑ and every V ∈ RBO(X), there exists F0 ∈ ℑ0 such that F0 ⊂ rbcl
(V). Therefore, we obtain rbcl (V) ∩ F ⊃ F0 ∩ F ≠ ϕ. This shows that ℑ
rb-accumulates at x.
(iii) ⇒ (ii) Let {Fα }α ε Δ be a collection of rb-closed sets such that
Iα ε Fα = ϕ. Suppose that for every finite subfamily γ of Δ
Iα ε γ rb int(Fα ) ≠ ϕ . Let Γ(Δ) be the family of all finite subsets of Δ and
ℑ = {Iα ε γ rb int( Fα ) / γ ε Γ ( Δ )}. Then ℑ is a filter base on X. By
Δ
i
hypothesis, ℑ rb-accumulates to some point x0 ∈ X. This implies that
for every rb-open set V(x0) containing x0, rbint (Fα) ∩ rbcl (V(x0)) ≠ ϕ
for every α ∈ Δ. Since x0 ∉ ∩ Fα, there exists an α0 ∈ Δ such that x0 ∉
Fα 0 . Hence x0 is contained in the rb-open set X − Fα 0 . Therefore,
rb int( Fα 0 ) I rbcl ( X − Fα 0 ) = rb int( Fα 0 ) I ( X − rb int( Fα 0 )) = ϕ
contradicting the fact that ℑ rb-accumulates to x0.
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(ii) ⇒ (i) Let {Vα} be an rb-open covering of X. Then ∩(X - Vα) = ϕ.
By hypothesis, there exists a finite subfamily {X - Vαi} (i = 1, 2,…, n)
such that I ni=1 rb int( X − Vα i ) = I ni=1 ( X − rbcl (Vα i )) = ϕ .
Therefore, U ni=1 rbcl (Vα i ) = X and consequently, X is an rb-closed space.
Theorem 5.7: A subset A of a topological space X is rb-closed relative
to X if and only if every filter base ℑ on X with A ∈ ℑ rb-converges to
a point in A.
Proof: Necessity. Let A be rb-closed relative to X and ℑ a filter base on
X satisfying A ∈ ℑ. Suppose that ℑ does not rb-converge to any a ∈ A.
For each α ∈ A, there corresponds some Uα ∈ RBO(X, a) such that rbcl
(Uα) ∉ ℑ. Now {Uα : α∈ A} is a cover of A by rb-open sets of X. Then
A ⊂ U ni=1 rbcl(U α i ) = U (say) for some positive integer n. Since ℑ is a
filter base, U ∉ ℑ and hence A ∉ ℑ, which is a contradiction.
Sufficiency. Suppose A is not rb-closed relative to X. Then for
some cover U = {Uα : α ∈ Δ} of A by rb-open sets of X,
ℑ = { A \ U α ∈Δ 0 rbcl (U α ) : Δ 0 ε Γ(Δ )} , where Γ(Δ) is the family of all
finite subsets of Δ, is a filter base on X. Then the family ℑ can be
extended to a maximal filter ℑ0 on X. Then ℑ0 is a filter base on X with
A ∈ ℑ0. Now for each x ∈ A, there exists β ∈ Δ such that x ∈ Uβ, as U is
a cover of A. Then for any G ∈ ℑ0, G ∩ (A \ rbcl(Uβ)) ≠ ϕ, so that G ⊄
rbcl (Uβ) for all G ∈ ℑ0. Hence ℑ0 cannot rb-converge to any point of A
which is a contradiction.
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Received: December, 2012