Download REGULAR GENERALIZED – – CLOSED SETS IN TOPOLOGICAL

International Journal of Mathematics and Computing Applications
ISSN: 0976-6790
Vol. 3, Nos. 1-2, January-December 2011, pp. 1-15
© International Science Press
REGULAR GENERALIZED – – CLOSED SETS
IN TOPOLOGICAL SPACES
A.I. EL-Maghrabi1 and A.M. Zahran2
Abstract: In this paper, we introduce and study the notion of regular generalized –γ– closed
sets. Also, some of their properties are investigated. Further, the notion of regular generalized
–γ– open sets and the notion of regular generalized –γ– neighbourhood are discussed.
Moreover, some of its basic properties are introduced. Finally, we give an applications of
rgγ–closed sets, say γ–regular T1 spaces.
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Mathematics Subject Classification: 54C08, 54C10, 54C05.
Keywords: regular generalized –γ– closed sets; regular generalized –γ– open sets; regular
generalized –γ– neighbourhood; regular –γ– kernel.
1. INTRODUCTION AND PRELIMINARIES
N. Levine [15] introduced generalized the notion of closed sets in general topology
as a generalization of closed sets. This notion was found to be useful and many
results in general topology were improved. Many researchers like Balachandran
et.al [4], Arockiarani [2], Dunham [8], Gnanambal [12], Malghan [17], Palaniappan
et.al [19], Arya et.al [3] and Keskin et.al [13] have worked on generalized closed
sets, their generalizations and related concepts in general topology. The purpose of
this paper is to define and study the properties of regular generalized γ– closed
(breifly, rgγ–closed) sets. Also, we define the concept of regular generalized γ– open
(breifly, rgγ–open) sets and obtained some of its properties and results. Further, we
define γ–regular T1 –spaces as the spaces in which every regular generalized
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γ–closed set is γ–closed
and study its properties.
Throughout the paper X and Y denote the topological spaces (X; τ) and (Y, σ)
respectively and on which no separation axioms are assumed unless otherwise
explicitly stated. For any subset A of a space (X, τ), the closure of A, the interior of A,
the gpr–closure of A, the gpr–interior of A, the w–closure of A, the w– interior of A,
1
Department of Mathematics, Faculty of Science, Kafr El-Sheikh University, Kafr El-Sheikh, Egypt.
E-mail: [email protected]
2
Department of Mathematics, Faculty of Science, Al-AZahar University, Assiut, Egypt.
E-mail: [email protected]
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the gpr–closure of A, the gpr–interior of A and the complement of A are denoted by
cl(A), int(A), γcl(A), γ – int(A), ωcl(A), ω–int(A), gpr–cl(A), gpr–int(A) and
Ac(equivalently, X – A or X\A) respectively. (X, τ) will be replaced by X if there is no
chance of confusion.
Let us recall the following definitions as pre-requesties.
Definition 1.1: A subset A of a space X is called:
(i) a preopen set [18] if A ⊆ int(cl(A)) and a preclosed set if cl(int(A)) ⊆ A,
(ii) a semi-open set [14] if A ⊆ cl(int(A)) and a semi-closed set if int(cl(A)) ⊆ A,
(iii) a γ–open [9] (equivlently, b–open[1 ] or sp–open [7 ]) set if A ⊆ int(cl(A)) ∪
cl(int(A)) and a γ–closed ( equivlently, b–closed or sp–closed) set if int(cl(A)) ∩
cl(int(A)) ⊆ A,
(iv) a regular open set [22] if if A = int(cl(A)) and a regular closed set if cl(int(A))
= A,
(v) π–open [23] if A is the finite union of regular open sets and the complement
of π–open set is said to be π–closed.
The intersection of all γ-open subsets of X containing A is called the γ–kernel of
A and is denoted by γ–ker(A). The intersection of all preclosed ( resp. γ–closed )
subsets of X containing A is called the pre-closure (resp. γ–closure ) of A and is
denoted by pcl(A)) ( resp. γcl(A)).
Definition 1.2: A subset A of a space X is called:
(1) generalized closed set (breifly, g–closed ) [15] if cl(A) ⊆ H whenever A ⊆ H
and H is open in X,
(2) γ–generalized closed set (breifly, γg–closed ) [11] if γcl(A) ⊆ H whenever A
⊆ H and H is open in X,
(3) generalized γ– closed set (breifly, gγ–closed )[10 ] if γcl(A) ⊆ H whenever A
⊆ H and H is γ–open in X,
(4) generalized preclosed set (breifly, gp–closed )[16] if pcl(A) ⊆ H whenever A
⊆ H and H is open in X,
(5) regular generalized closed set (breifly, rg–closed )[19] if cl(A) ⊆ H whenever
A ⊆ H and H is regular open in X,
(6) generalized preregular closed set (breifly, gpr–closed )[12] if pcl(A) ⊆ H
whenever A ⊆ H and H is regular open in X,
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(7) π–generalized closed set (breifly, πg–closed )[6] if cl(A) ⊆ H whenever A ⊆
H and H is π–open in X,
(8) weakly closed set, (breifly, ω–closed )[20] if cl(A) ⊆ H whenever A ⊆ H and
H is semi-open in X.
The complements of the above mentioned closed sets are their respectively open
sets.
2. BASIC PROPERTIES OF REGULAR GENERALIZED – – CLOSED SETS
We introduce the following definitions.
Definition 2.1: A subset A of a space X is called a regular–γ–open set ( breifly,
rγ–open ) if there is a regular open set H such that H ⊆ A ⊆ γcl(H). The faimly of all
regular–γ–open sets of X is denoted by RγO(X).
Definition 2.2: A subset A of a space X is called a regular generalized–γ–closed
set (breifly, rgγ–closed ) if γcl(A) ⊆ H whenever A ⊆ H and H is regular–γ– open in
X. The faimly of all regular–gγ–closed sets of X is denoted by RGγC(X).
In the following proposition, we prove that the class of rgγ–closed sets has
properly lies between the class of gγ–closed sets and the class of rg–closed sets.
Proposition 2.3: For a space (X, τ), we have
(i) Every gγ–closed (resp. ω–closed, closed, regular closed, π–closed ) set is
rgγ–closed,
(ii) Every rgγ–closed set is gpr–closed ( resp. rg– closed ).
Proof: The proof follows from definitions and the fact that every regular open
sets are rγ–open.
The converse of the above proposition need not be true, as seen from the following
examples.
Example 2.4: Let X = {a, b, c, d, e} with topology τ = {X, φ, {a}, {d}, {e}, {a,
d}, {a, e}, {d, e}, {a, d, e}}. Then,
(1) the set A = {a, d, e} is rgγ–closed set but not gγ–closed set in X,
(2) the set B = {b} is rgγ–closed set but not ω–closed set in X,
(3) the set D = {a, b} is rg–closed and gpr–closed set but not rgγ–closed set in X,
Remark 2.5: From the above discussions and known results we have the
following implications, in the following diagram by A → B, we mean A implies B
but not conversely and A –  – B means A and B are independent of each other..
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regular closed
↓
π − closed
→ πg − closed
↓
closed → ω − closed
↓
γ − closed
−−
→ rg γ − closed → rg − closed → gpr − closed
↑
→ g γ − closed →
−−
γg − closed
Remark 2.6: The union of two rgγ–closed subsets of X need not be a rgγ–closed
subset of X. Let X = {a, b, c, d, e} with a topology τ = {X, φ, {a}, {b}, {a, b}, {a, b,
c}, {a, b, c, d}, {a, b, c, e}}. Then two subsets {c} and {d} of X are rgγ–closed
subsets but their union {c, d} is not rgγ–closed subset of X.
Theorem 2.7: The arbitrary intersection of any rgγ–closed subsets of X is also
rgγ–closed subset of X.
Proof: Let {Ai : i ∈ I} be any collection of rgγ–closed subsets of X such that ∩iAi
⊆ H and H be rγ–open in X. But Ai is rgγ–closed subset of X, for each i ∈ I, then
γcl(Ai) ⊆ H, for each i ∈ I. Hence ∩iγcl(Ai) ⊆ H, for each i ∈ I. Therefore γcl(∩iAi)
⊆ H. So, ∩iAi a rgγ–closed subset of X.
Theorem 2.8: A subset A of a space (X, τ) is rgγ–closed if and only if for each A
⊆ H and H is rγ–open, there exists a γ–closed set F such that A ⊆ F ⊆ H.
Proof: Suppose that A is a rgγ–closed set, A ⊆ H and H is a rγ–open set. Then
γcl(A) ⊆ H. If we put F = γcl(A), hence A ⊆ F ⊆ H.
Conversely, Assume that A ⊆ H and H is a rγ–open set. Then by hypothesis there
exists a γ–closed set F such that A ⊆ F ⊆ H. So, A ⊆ γ – cl(A) ⊆ F and hence γcl(A)
⊆ H. Therefore A is rgγ–closed.
Theorem 2.9: If A is closed and B is a rgγ–closed subset of a space X, then A ∪
B is also rgγ–closed.
Proof: Suppose that A ∪ B ⊆ H and H is a rγ–open set. Then A ⊆ H and B ⊆ H.
But B is rgγ–closed, then γcl(B) ⊆ H and hence A ∪ B ⊆ A ∪ γcl(B) ⊆ H. But A ∪
γcl(B) a γ–closed set, hence there exists a γ–closed set A ∪ γcl(B) such that A ∪ B ⊆
A ∪ γcl(B) ⊆ H. Thus by Theorem 2.8, A ∪ B is rgγ–closed.
Theorem 2.10: If a subset A of X is a rgγ–closed set in X, then γcl(A)\A does not
contain any non-empty rγ–open set in X.
Proof: Assume that A is a rgγ–closed set in X. Let H be a rγ–open set such that H
⊆ γcl(A)\A and H ≠ φ. New H ⊆ γcl(A)\A. Therefore H ⊆ X\A which implies A ⊆
X\H. Since H is a rγ–open set, then X\H is also rγ–open in X. But A is a rgγ–closed
set in X, then γcl(A) ⊆ X\H. So, H ⊆ X\γcl(A). Also, H ⊆ γcl(A). Hence, H ⊆ (γcl(A)
Regular Generalized – – closed Sets in Topological Spaces
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∩ (X\γcl(A))) = φ. This shows that, H = φ which is a contradiction. Therefore, γcl(A)\A
does not contain any non-empty rγ–open set in X.
The converse of the above theorem need not be true as seen from the following
example.
Example 2.11: In Example 2.4, if A = {a, b}, then γcl(A)\A = {a, b, c}\{a, b} =
{c} does not contain any non-empty rγ–open set in X. But A is not a rgγ–closed set
in X.
Proposition 2.12: If a subset A of X is a rgγ–closed set in X, then γcl(A)\A does
not contain any regular closed (resp.regular open) set in X.
Proof: Follows directly from Theorem 2.10 and the fact that every regular open
set is rγ–open.
Theorem 2.13: For an element p ∈ X, the set X /{p} is rgγ–closed or rγ–open.
Proof: Suppose that X /{p} is not a rγ–open set. Then X is the only rγ–open set
containing X /{p}. This implies that γcl(X\{p}) ⊆ X. Hence X \{p} is a rgγ–closed set
in X.
Theorem 2.14: If A is regular open and rgγ–closed, then A is regular closed and
hence γ–clopen.
Proof: Assume that A is regular open and rgγ–closed. Since every regular open
set is rγ–open and A ⊆ A, we have γcl(A) ⊆ A. But A ⊆ γcl(A). Therefore A = γcl(A),
that is, A is γ–closed. Since A is regular open, A is γ–open. Therefore A is regular
closed and γ–clopen.
Theorem 2.15: If A is a rgγ–closed set of X such that A ⊆ B ⊆ γcl(A), then B is a
rgγ–closed set in X.
Proof: Let H be a rγ–open set of X such that B ⊆ H. Then A ⊆ H. But A is a rgγ–
closed set of X, then γcl(A) ⊆ H. Now, γcl(B) ⊆ γcl(γcl(A)) = γcl(A) ⊆ H. Therefore
B is a rgγ–closed set in X.
Remark 2.16: The converse of the above theorem need not be true in general. In
Example 2.4, if we take a subset A = {b} and B = {b, c}, then A and B are rgγ–closed
set in (X, τ), but A ⊆ B is not subset in γcl(A).
Theorem 2.17: Let A be a rgγ–closed set in X. Then A is γ–closed if and only if
γcl(A)\A is rγ–open.
Proof: Suppose that A is γ–closed in X. Then γcl(A) = A and so γcl(A)\A = φ,
which is rγ–open in X. Conversely, Suppose that γcl(A)\A is a rγ–open set in X. Since
A is rgγ–closed, then by Theorem 2.10, γcl(A)\A does not contain any non-empty
rγ–open set in X. Then γcl(A)\A = φ, hence A is a γ–closed set in X.
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Theorem 2.18: If A is regular open and rg–closed, then A is a rgγ–closed set
in X.
Proof: Let H be any rγ–open set in X such that A ⊆ H. Since A is regular open
and rg–closed, we have γcl(A) ⊆ A. Then γcl(A) ⊆ A ⊆ H. Hence A is a rgγ–closed set
in X.
Theorem 2.19: If a subset A is both rγ–open and rgγ–closed in a topological
space (X, τ), then A is γ–closed.
Proof: Assume that A is both rγ–open and rgγ–closed in a topological space
(X, τ). Then γcl(A) ⊆ A. Hence A is γ–closed.
Theorem 2.20: If A is both rγ–open and rgγ–closed subset in X and F is a closed
set in X, then A ∩ F is a rgγ–closed set in X.
Proof: Let A be rγ–open and rgγ–closed subset in X and F be a closed set in X.
Then by Theorem 2.19, A is γ–closed. So, A ∩ F is γ–closed. Therefore A ∩ F is a
rgγ–closed set in X.
Theorem 2.21: If A is an open set and H is a γ–open set in a topological space
(X, τ), then A ∩ H is a γ–open set in X.
Theorem 2.22: If A is both open and g–closed subset in X, then A is a rgγ–closed
set in X.
Proof: Let A be an open and g–closed subset in X and A ⊆ H, where H is a
rγ–open set in X. Then by hypothesis γcl(A) ⊆ A, that is, γcl(A) ⊆ H. Thus A is a
rgγ–closed set in X.
Remark 2.23: If A is both open and rgγ–closed subset in X, then A need not be
g–closed set in X. In Example 2.4, the subset {a, d, e} is an open and rgγ–closed set
but not g–closed.
Theorem 2.24: For a topological space (X, τ), if RγO(X, φ) = {X, φ}, then every
subset of X is a rgγ–closed subset in X.
Proof: Let (X, τ) be a space and RγO(X, τ) = {X, φ}. Let A be a subset of X. (i) if
A = φ, then A is a rgγ–closed subset in X. (ii) if A ≠ φ, then X is the only a rγ–open set
in X containing A and so γcl(A) ⊆ X. Hence A is a rgγ–closed subset in X.
The converse of Theorem 2.24 need not be true in general as seen from the
following example.
Example 2.25: Let X = {a, b, c, d,} with topology τ = {X, φ, {a, b}, {c, d}}.
Then every subset of (X, τ) is a rgγ–closed subset in X. But RγO(X, τ) = τ.
Theorem 2.26: For a topological space (X, τ), then RγO(X, τ) ⊆ {F ⊆ X : F is
closed} if and only if every subset of X is a rgγ–closed subset in X.
Regular Generalized – – closed Sets in Topological Spaces
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Proof: Suppose that RγO(X, τ) ⊆ {F ⊆ X : F is closed}. Let A be any subet of X
such that A ⊆ H, where H is a rγ–open set in X. Then H ∈ RγO(X, τ) ⊆ {F ⊆ X : F is
closed}. That is H ∈ {F ⊆ X : F is closed}. Thus H is a γ–closed set. Then γcl(H) =
H. Also, γcl(A) γcl(H) ⊆ H. Hence A is a rgγ–closed subset in X.
Conversely, Suppose that every subset of (X, τ) is a rgγ–closed subset in X. Let
H ∈ Rγ (X, τ). Since H ⊆ H and H is rgγ–closed, we have γcl(H) ⊆ H. Thus γcl(H)
= H and H ∈ {F ⊆ X : F is closed}. Therefore RγO(X, τ) ⊆ {F ⊆ X : F is closed}.
Definition 2.27: The intersection of all rγ–open subsets of (X, τ) containing A is
called the regular γ–kernel of A and is denoted by rγ – ker(A).
Lemma 2.28: For any subset A of a toplogical space (X, τ), then A ⊆ rγ – ker(A).
Proof: Follows directly from Definition 2.27.
Lemma 2.29: Let (X, τ) be a topological space and A be a subset of X. If A is a
rγ–open in X, then rγ – ker(A) = A.
Lemma 2.30: Let (X, τ) be a topological space and A be a subset of X. Then γ –
ker(A) ⊆ rγ – ker(A).
Proof: Follows from the implication RγO(X, τ) ⊆ γO(X, τ).
3. REGULAR GENERALIZED –OPEN SETS
In this section, we introduce and study the concept of regular generalized γ– open
(briefly, rgγ–open) sets in topological spaces and obtain some of their properties.
Definiton 3.1: A subset A of a topological space (X, τ) is called a regular
generalized γ– open (breifly, rgγ–open) set in X if Ac is rgγ–closed in X. We denote
the family of all rgγ–open sets in X by RGγO(X).
The following theorem gives equivalent definitions of a rgγ–open set.
Theorem 3.2: Let (X, τ) be a topological space and A ⊆ X. Then the following
statements are equivalent:
(i) A is a rgγ–open set,
(ii) for each rγ–closed set F contained in A, F ⊆ γ – int(A),
(iii) for each rγ–closed set F contained in A, there exists a γ–open set H such that
F ⊆ H ⊆ A.
Proof: (i) ⇒ (ii). Let F ⊆ A and F be a rγ–closed set. Then X\A ⊆ X\F which is
rγ–open of X, hence γcl(X\A) ⊆ X\F. So, F ⊆ γ – int(A).
(ii) ⇒ (iii). Suppose that F ⊆ A and F be a rγ–closed set. Then by hypothesis, F
⊆ γ – int(A). Set H = γ – int(A), hence there exists a γ–open set H such that F ⊆ H
⊆ A.
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(iii) ⇒ (i). Assume that X\A ⊆ V and V is rγ–open of X. Then by hypothesis, there
exists a γ–open set H such that X\V ⊆ H ⊆ A, that is, X\A ⊆ X\H ⊆ V. Therefore, by
Theorem 2.8, X\A is rgγ–closed in X and hence A is rgγ–open in X.
Theorem 3.3: For a topological space (X, τ), then every ω– open set is
rgγ–open.
Proof: Let A be a ω– open set in X. Then Ac is ω–closed in X. Hence by Proposition
2.3, Ac is rgγ–closed in X. Therefore, A is rgγ–open.
The converse of the above theorem need not be true as seen from the following
example.
Example 3.4: In Example 2.4, the subset A = {b} is rgγ–open but not ω– open.
Theorem 3.5: For a topological space (X, τ), then every open ( resp. regularopen) set is rgγ–open but not conversely.
Proof: Follows from S. John [21] ( resp. Stone [22] ) and Theorem 3.3.
Theorem 3.6: Let (X, τ) be a topological space. Then every rgγ–open set of X is
rg–open (resp. gpr–open).
Proof: (i) Let A be a rgγ–open set of X. Then Ac a rgγ–closed set of X. Then by
Proposition 2.3, Ac is a rg–closed set of X. Therfore A is rg–open in X.
(ii) For the case of a gpr–open set is similar to (i).
The converse of the above theorem need not be true as seen from the following
example.
Example 3.7: In Example 2.4, the subset {a, b} is rg–open (resp. gpr–open) but
not a rgγ–open set in X.
Theorem 3.8: The arbitrary union of rgγ–open sets of X is rgγ–open.
Proof: Obvious from Theorem 2.7.
Remark 3.9: The intersection of two rgγ–open subsets of X need not be a
rgγ–open subset of X. In Remark 2.6, two subsets {a, b, d, e} and {a, b, c, e} of X are
rgγ–open subsets but their intersection {a, b, e} is not rgγ–open subset of X.
Theorem 3.10: If A is an open and B is a rgγ–open subsets of a space X, then A
∩ B is also rgγ–open.
Proof: Follows from Theorem 2.9.
Proposition 3.11: If γ – int(A) ⊆ B ⊆ A and A is a rgγ–open set of X, then B is
rgγ–open.
Proposition 3.12: Let A be a rγ–closed and a rgγ–open set of X. Then A is
γ–open.
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Proof: Assume that A is a rγ–closed and a rgγ–open set of X. Then A ⊆ γ – int(A)
and hence A is γ–open.
Theorem 3.13: For a space (X, τ), if A is a rgγ–closed set, then γcl(A)\A is
rgγ–open.
Proof: Suppose that A is a rgγ–closed set and F is a rγ–closed set contained in
γcl(A)\A. Then by Theorem 2.8, F = φ and hence F ⊆ γ – int(γcl(A)\A). Therefore,
γcl(A)\A is rgγ–open.
Theorem 3.14: If a subset A of a space (X, τ) is rgγ–open, then H = X, whenever
H is rγ–open and int(A) ∪ Ac ⊆ H.
Proof: Assume that A is rgγ–open in X. Let H be rγ–open and int(A) ∪ Ac ⊆ H.
This implies that Hc ⊆ (int(A) ∪ Ac)c = (int(A))c ∩ A, that is, Hc ⊆ (int(A))c\Ac. Thus
Hc ⊆ cl(A)c\Ac, since (int(A))c = cl(A)c. Now Hc is rγ–open and Ac is rgγ–closed,
hence by Theorem 2.10, it follows that Hc = φ. Then H = X.
The converse of the above theorem is not true in general as seen from the following
example.
Example 3.15: Let X = {a, b, c, d} with topology τ = {X, φ, {a}, {b}, {a, b}, {b,
c}, {a, b, c}, {a, b, d}}. Then RGγO(X) = {X, φ, {a}, {b}, {c}, {d}, {a, b}, {b, c},
{c, d}, {a, d}, {b, d}, {a, c}, {a, b, c}, {a, b, d}} and RγO(X) = {X, φ, {a}, {b}, {c},
{a, d}, {b, c}, {b, c, d}}. If we take A = {b, c, d}, then A is not rgγ–open. However
int(A) ∪ Ac = {b, c} ∪ {a} = {a, b, c}. So, for some rγ–open H, we have int(A) ∪ Ac
= {a, b, c} ⊆ H, gives H = X, but A is not rgγ–open.
Theorem 3.16: For a topological space (X, τ), then every singleton of X is eiher
rgγ–open or rγ–open.
Proof: Let (X, τ) be a topological space. Let p ∈ X. To prove that {p} is eiher
rgγ–open or rγ–open. That is to prove X\{p} is eiher rgγ–closed or rγ–open, which
follows directly from Theorem 2.13.
4. SOME APPLICATIONS ON rg –CLOSED AND rg –OPEN SETS
In this section, we introduce and study the concept of regular generalized γ–neighbourhood (briefly, rgγ–nbd) in topological spaces by using the concept of rgγ–open sets.
Further, we prove that every neighbourhood of a point p in X is rgγ–nbd of p but
conversely not true.
Definition 4.1: Let (X, τ) be a topological space and let p ∈ X. A subset N of X is
said to be a regular generalized γ–neighbourhood (briefly, rgγ–nbd) of p if and only
if there exists a rγ–open set H such that p ∈ H ⊆ N.
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Definition 4.2: A subset N of a topological space (X, τ) is called a regular
generalized γ–neighbourhood (briefly, rgγ–nbd) of a subset A of a space (X, τ) if and
only if there exists a rγ–open set H such that A ⊆ H ⊆ N.
Remark 4.3: The rgγ–nbd of N of a point p of a space X need not be a rgγ–open
in X.
Example 4.4: Let X = {a, b, c, d} with topology τ = {X, φ, {a}, {b}, {a, b}, {a,
b, c}}. Then RGγO(X) = {X, φ, {a}, {b}, {c}, {d}, {a, b}, {c, d}, {a, b, c}, {a, b,
d}}. Note that {a, c} is not a rgγ–open set, but it is a rgγ–nbd of {a}. Since {a} is a
rgγ–open set such that a ∈ {a} ⊆ {a, c}.
Theorem 4.5: Every nbd N of p ∈ X is a rgγ–nbd) of X.
Proof: Let N be a nbd of point p. To prove that N is a rgγ–nbd of p. Then by
definition of nbd, there exists an open set H such that p ∈ H ⊆ N. Since every open
set is rgγ–open, hence there exists a rgγ–open set H such that p ∈ H ⊆ N. Therefore
N is a rgγ–nbd of p.
The converse of the above theorem is not true as seen from the following example.
Example 4.6: In Example 4.4, the set {a, c} is rgγ–nbd of the point c, since {c}
is a rgγ–open set such that c ∈ {c} ⊆ {a, c}. However the set {a, c} is not a nbd of
the point c, since there is no exists an open set H such that c ∈ H ⊆ {a, c}.
Theorem 4.7: If N is a rgγ–open set of a space (X, τ), then N is a rgγ–nbd of each
of its points.
Proof: Assume that N is a rgγ–open set. Let p ∈ X. We need to prove that N is a
rgγ–nbd of p. For N is a rgγ–open set such that p ∈ N ⊆ N. Since p is an arbitrary
point of N, it follows that N is a rgγ–nbd of each of its points.
The converse of the above theorem is not true in general as seen from the following
example.
Example 4.8: In Example 4.4, the set {a, d} is a rgγ–nbd of the point a, since
{a} is the rgγ–open set such that a ∈ {a} ⊆ {a, d}. Also, the set {a, d} is a rgγ–nbd
of the point d, since {d} is the rgγ–open set such that d ∈ {d} ⊆ {a, d}, that is {a, d}
is a rgγ–nbd of each of its points. However the set {a, d} is not a rgγ–open set in X.
Theorem 4.9: If F is a rgγ–closed set of a space (X, τ) and p ∈ Fc, then there
exists a rgγ–nbd N of p such that N ∩ F = φ.
Proof: Let F be a rgγ–closed subset of a space (X, τ) and p ∈ Fc. Then Fc is a
rgγ–open set of X. So, by Theorem 4.7, Fc contains a rgγ–nbd of each of its points.
Hence, there exists a rgγ–nbd N of p such that N ⊆ Fc, that is, N ∩ F = φ.
Definition 4.10: Let p be a point in a space X. The set of all a rgγ–nbd of p is
called the rgγ–nbd system at p and is denoted by rgγ – N(p).
Regular Generalized – – closed Sets in Topological Spaces
11
Theorem 4.11: Let (X, τ) be a topological space and for each p ∈ X. Let rgγ –
N(p) be the collection of all rgγ–nbds of p. Then the following statements are hold.
(1) rgγ – N(p) ≠ φ, for every p ∈ X,
(2) If N ∈ rgγ – N(p), then p ∈ N,
(3) If N ∈ rgγ – N(p) and N ⊆ M, then M ∈rgγ – N(p),
(4) If N ∈ rgγ – N(p) and M ∈ rgγ – N(p), then N ∩ M ∈ rgγ – N(p),
(5) If N ∈ rgγ – N(p), then there exists M ∈ rgγ – N(p) such that M ⊆ N and M ∈
rgγ – N(q), for every q ∈ M.
Proof: (1) Since X is a rgγ–open set, then X is a rgγ–nbd for every p ∈ X. Hence
there exists at least one rgγ–nbd (namely, X), for each p ∈ X. Therefore rgγ – N(p) ≠
φ, for every p ∈ X.
(2) If N ∈ rgγ – N(p), then N is a rgγ–nbd of p. Hence by definition of rgγ–nbd,
p ∈ N.
(3) Let N ∈ rgγ – N(p) and N ⊆ M. Then there is a rgγ–open set H such that p ∈
H ⊆ N. Since N ⊆ M, then p ∈ H ⊆ M and so M is a rgγ–nbd of p. Hence M ∈ rgγ –
N(p).
(4) Let N ∈ rgγ – N(p) and M ∈ rgγ – N(p). Then by definition of rgγ–nbd, there
exist rgγ–open sets H1 and H2 such that p ∈ H1 ⊆ N and p ∈ H2 ⊆ M. Hence p ∈ H1
∩ H2 ⊆ N ∩ M .... (i). Since H1 ∩ H2 is a rgγ–open set, (being the intersection of two
rgγ–open sets), it follows from (i) that N ∩ M is a rgγ–nbd of p. So, N ∩ M ∈ rgγ –
N(p).
(5) If N ∈ rgγ – N(p), then there exists a rgγ–open set M such that p ∈ M ⊆ N.
Since M is a rgγ–open set, then M is a rgγ–nbd of each of its points. Therefore M ∈
rgγ – N(q), for every q ∈ M.
Theorem 4.12: For each p ∈ X, let rgγ – N(p) be a non empty collection of
subsets of X satisfying the following conditions.
(1) If N ∈ rgγ – N(p) implies that p ∈N,
(2) If N ∈ rgγ – N(p) and M ∈ rgγ – N(p) implies that N∩M ∈ rgγ – N(p).
Let τ consists of the empty set φ and all those non-empty subsets H of X having
the property that p ∈ H implies that there exists an N ∈ rgγ – N(p) such that p ∈ H ⊆
N. Then τ is a topology on X.
Proof: (1) Since φ ∈ τ (given), now we show that X ∈ τ. Let p be an arbitrary
element of X. Since rgγ – N(p) is non empty, then there is an N ∈ rgγ – N(p) and so
by (1), p ∈ N. Since N is a subset of X, then we have p ∈ N ⊆ X. Hence X ∈ τ.
(2) Let H1 ∈ τ and H2 ∈ τ. If p ∈ H1 ∩ H2, then p ∈ H1 and p ∈ H2. Since H1 ∈
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International Journal of Mathematics and Computing Applications
τ and H2 ∈ τ, hence there exists N ∈ rgγ – N(p) and M ∈ rgγ – N(p) such that p ∈ N
⊆ H1 and p ∈ M ⊆ H2. Then p ∈ N ∩ M ⊆ H1 ∩ H2. But by (2), N ∩ M ∈ rgγ – N(p).
Hence H1 ∩ H2 ∈ τ.
(3) Let Hi ∈τ for every i ∈ I. If p ∈ ∪{Hi : i ∈ I}, then p ∈ Hi , for some ip ∈ I.
Since Hip ∈ τ, then there exists an N ∈ rgγ – N(p) such that p ∈ N ⊆ H i and
consequently, p ∈ N ⊆ ∪{Hi : i ∈ I}. Hence ∪{Hi : i ∈ I} ∈ τ. Therefore τ is a
topology on X.
p
p
In the following, we introduce the notion of γ–regular −T1 − spaces.
2
Definition 4.13: A space (X, τ) is called a γ–regular T1 −space if every rg
gγ–
2
closed set is γ–closed.
Example 4.14: Any set with an indiscrete topology is an example for a γ–regular
T1 –space.
2
Remark 4.15: The notions of γ–regular T1 and T1 –spaces are independent of
2
2
each other.
Example 4.16: Let X = {a, b, c} with topologies τ = {X, φ, {a}, {b}, {a, b}} and
σ = {X, φ, {c}, {a, b}}. Then (X, τ) is T12 but not γ–regular T1 where as (X, σ) is γ–
2
regular T1 but not T1 .
2
2
Theorem 4.17: For a topological space (X, τ), the following conditions are
equivalent:
(1) X is γ–regular T12 ,
(2) Every singleton of X is either regular closed or γ–open.
Proof: (1) ⇒ (2). Let p ∈ X and assume that {p} is not regular closed.
Then X\{p} is not regular open and hence X\{p} is rgγ– closed. Hence, by
hypothesis, X\{p} is γ–closed and thus {p} is γ–open.
(2) ⇒ (1). Let A ⊆ X be rgγ– closed and p ∈ γcl(A). We will show that p ∈ A. For
consider the following two cases:
Case (1). The set {p} is regular closed. Then, if p ∉A, then there exists a regular
closed set in γcl(A)\A. Hence by Proposition 2.12, p ∈ A.
Case (2). The set {p} is γ– open. Since p ∈ γcl(A), then {p} ∩ γcl(A) ≠ φ.
Thus p ∈ A. So, in both cases, p ∈ A. This shows that γcl(A) ⊆ A or equivalently
A is γ–closed.
Definition 4.18: A space (X, τ) is called:
Regular Generalized – – closed Sets in Topological Spaces
13
(1) locally indiscrete (equivalently, partition space)[5] if every closed set is regular
closed,
(2) γ– T12 space [11] if every γg–closed set is γ–closed.
Theorem 4.19: For a topological space (X, τ), the following conditions are
equivalent:
(1) X is γ–regular T1 ,
2
(2) X is locally indiscrete and γ – T12 .
Proof: (1) ⇒ (2). Let (X, τ) be a γ–regular T12 space. Then by Theorem 4.17,
every singleton of (X, τ) is closed or γ–open. This implies that (X, τ) is a γ – T12
space.
(2) ⇒ (1). Since the space X is γ – T12 , then every singleton of (X, τ) is closed or
γ–open. But (X, τ) is locally indiscrete, hence all closed singletons are regular closed.
So, by theorem 4.17, the space X is γ–regular T12 .
Remark 4.20: Every γ–regular T12 space is γ – T12 . But the converse is not true as
seen by the following example.
Example 4.21: Let X = {p, q, r} and τ = {φ, X, {p}, {q}, {p, q}}. Then every
singleton of (X, τ) is closed or γ–open. Therefore (X, τ) is γ – T12 , but not γ–regular
T1 , since the rg
gγ– closed set {p, q} is not γ–closed.
2
Proposition 4.22: For a space (X, τ), we have γ (X, τ) ⊆ RGγO(X, τ).
Proof: Let A be a γ–open set. Then Ac is γ– closed and so rgγ– closed. This
implies that A is rgγ– open. Hence γO(X, τ) ⊆ RGγO(X, τ).
Theorem 4.23: For a topological space (X, τ), the following conditions are
equivalent:
(1) X is γ–regular T 12 ,
(2) γO(X, τ) = RGγO(X, τ).
Proof: (1) ⇒ (2). Let (X, τ) be a γ–regular T12 space and let A ∈ RGγO(X, τ).
Then Ac is rgγ– closed. By hypothesis, Ac is γ– closed and thus A is γ–open this
implies that A ∈ γO(X, τ). Hence, γO(X, τ) = RGγO(X, τ).
(2) ⇒ (1). Let γO(X, τ) = RGγO(X, τ) and let A be a rgγ– closed set. Then Ac is
rgγ– open. Hence Ac ∈ γO(X, τ). Thus A is γ– closed thereby implying that (X, τ) is
γ–regular T 12 .
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International Journal of Mathematics and Computing Applications
Theorem 4.24: For a topological space (X, τ), the following conditions are
equivalent:
(1) X is γ–regular T12 ,
(2) Every nowhere dense singleton of X is regular closed,
(3) The only nowhere dense subset of X is the empty set,
(4) Every subset of X is γ–open.
Proof: (1) ⇒ (2). By Theore m 4.17, every singleton is γ–open or regular closed.
Since non-empty nowhere dense set can not be γ–open at the same time, then every
nowhere dense singleton of X is regular closed.
(2) ⇒ (1). Since every singleton is either γ–open or nowhere dense [10], then by
hypothesis, every singleton of X is either γ–open or regular closed. Hence, by Theorem
4.17, X is γ–regular T12 .
(2) ⇒ (3). Let A ⊆ X be a non-empty nowhere dense set. Then, there exists a
nowhere dense singleton in X and hence by hypothesis, this singleton is regular
closed. If the nowhere dense singleton {p} is regular closed, then {p} = cl(int{p}) =
cl(φ) = φ which is the contradiction with this assumption.
(3) ⇒ (4). By hypothesis and the fact that singletons are either γ–open or nowhere
dense, it follows that every singleton of X is γ–open. Since the union of γ–open sets
is γ–open. Hence, every subset of X is γ–open.
(4) ⇒ (1). From the definition of a γ–regular T12 space and by hypothesis, every
subset of X is γ–open and hence γ–closed.
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