International Journal of Scientific & Engineering Research, Volume 3, Issue 5, May-2012 1

ISSN 2229-5518

Weak Forms of γ-Open Sets and New Separation

Axioms

Hariwan Z. Ibrahim

Department of Mathematics, Faculty of Science, University of Zakho, Kurdistan- Region, Iraq

E-mail: hariwan_math@yahoo.com

AbstractIn this paper, we introduce some generalizations of γ-open sets and investigate some properties of the sets. Moreover, we use them to obtain new separation axioms.

Index Terms— γ-open, α-γ-open, pre-γ-open, β-γ-open, b-γ-open.

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1 INTRODUCTION

hroughout this paper, (X, τ) and (Y, σ) stand for topo- logical spaces with no separation axioms assumed unless less otherwise stated. For a subset A of X, the closure of A and the interior of A will be denoted by Cl(A) and Int(A), respective- ly.

2 Weak Forms Of γ-Open Sets

Denition 2.1

A subset A of a space X is said to be:
subset A of X with an operation γ on τ is called γ-open [8] if for each
x ∈ A, there exists an open set U such that x ∈ U and γ(U) ⊂ A.
Then, τγ set denotes the of all γ-open set in X. Clearly τγ ⊂ τ. Com-
plements of γ-open sets are called γ-closed. The τγ-interior [7] of A is
denoted by τγ ⊂ τ. Complements of γ-open sets are τγ-Int(A) and
defined to be the union of all γ-open sets of X contained in A. A to-
pological X with an operation γ on τ is said to be γ-regular [8] if for
each x ∈ X and for each open neighborhood V of x, there exists an
open neighborhood U of x such that γ(U) contained in V. It is also to
be noted that τ = τγ if and only if X is a γ-regular space [8].

Denition 1.1

A subset A of a space X is said to be:
1. α-open [6] if A ⊆ Int(Cl(Int(A)));
2. semi-open [4] if A ⊆ Cl(Int(A));
3. pre-open [5] if A ⊆ Int(Cl(A));
4. β-open [1] if A ⊆ Cl(Int(Cl(A)));
5. b-open [2] if A ⊆ Int(Cl(A)) ∪ Cl(Int(A)).
In this paper we introduce and investigate the new notions called α-γ-open sets, pre-γ-open sets, β-γ-open sets and b-γ-open sets which are weaker than γ-open. Moreover, we use these notions to obtain new separation axioms.

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Department of Mathematics, Faculty of Science, University of Zakho, Kurdistan-Region, Iraq.

E-mail: hariwan_math@yahoo.com

4. b-γ-open if A ⊆ τγ-Int(Cl(A)) ∪ Cl(τγ-Int(A)).

Lemma 2.2

Let (X, τ) be a topological space, then the following properties hold: A subset A of a space X is said to be:
1. Every γ-open set is α-γ-open.
2. Every α-γ-open set is pre-γ-open.
3. Every pre-γ-open set is b-γ-open.
4. Every b-γ-open set is β-γ-open.

Proof

1. If A is a γ-open set, then A = τγ-Int(A). Since A ⊆ Cl(A), then A ⊆ Cl(τγ-Int(A)) and A ⊆ τγ-Int(Cl(τγ-Int(A))). There-
fore A is α-γ-open.
2. If A is an α-γ-open set, then A ⊆ τγ-Int(Cl(τγ-Int(A))) ⊆ τγ-
Int(Cl(A)). Therefore A is pre-γ-open.
3. If A is pre-γ-open, then A ⊆ τγ-Int(Cl(A)) ⊆ τγ-Int(Cl(A)) ∪
Cl(τγ-Int(A)). Therefore A is b-γ-open.
4. If A is b-γ-open, then A ⊆ τγ-Int(Cl(A)) ∪ Cl(τγ-Int(A)) ⊆
Cl(τγ-Int(Cl(A))) ∪ Cl(τγ-Int(A)) ⊆ Cl(τγ-Int(Cl(A))).
Therefore A is β-γ-open.
Since every γ-open set is open, then we have the following diagram for properties of subsets.
γ-openα-γ-openpre-γ-openb-γ-openβ-γ-open
    
open  α-open  pre-open  b-open  β-open
The converses need not be true as shown by the following exam- ples.

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Example 2.3

Let X = {a, b, c} and τ = {φ, {a}, {b}, {a, b}, X}. Define an opera- tion γ on τ by γ(A) = A if A = {a, b} and γ(A) = X otherwise. Clearly, τγ = {φ, {a, b}, X}. Then {a} is an open set which is not β-γ-open. Example 2.4
Let X = R with the usual topology τ and γ(A) = A for all A ∈ τ . Let A
= Q ∩ [0, 1]. Then A is a β-γ-open set which is not b-γ-open.

Example 2.5

Let X = {a, b, c} and τ ={φ, {a}, {b}, {a, b}, {b, c}, X}. Define an operation γ on τ by γ(A) = A if A = {b} and γ(A) = X
if A≠ {b}. Clearly, τγ = {φ, {b}, X}. Then {b, c} is a b-γ-open set which is not pre-γ-open.

Example 2.6

Let X = {a, b, c} and τ = {φ, {a}, X}. Define an operation γ on τ by
γ(A) = X for all A ∈ τ. Then {a, c} is a pre-γ-open set which is not α-
γ-open.

Example 2.7

Let X = {a, b, c} and τ = {φ, {a}, {a, b}, X}. Define an operation γ on τ by γ(A) = A if A = {a} and γ(A) = X if A ≠ {a} Clearly, τγ = {φ,
{a}, X}. Then {a, b} is an α-γ-open set which is not γ-open.

Lemma 2.8

If U is an open set, then Cl(U ∩ A) = Cl(U ∩ Cl(A)) and hence U ∩
Cl(A) ⊆ Cl(U ∩ A) for any subset A of a space X [3].

Theorem 2.9

If A is a pre-γ-open subset of a space (X, τ) such that U ⊆ A ⊆ Cl(U)
for a subset U of X, then U is a pre-γ-open set.

Proof

Since A ⊆ τγ-Int(Cl(A)), U ⊆ τγ-Int(Cl(A)). Also Cl(A) ⊆ Cl(U) im-
plies that τγ-Int(Cl(A)) ⊆ τγ-Int(Cl(U)). Thus U ⊆ τγ-Int(Cl(A)) ⊆ τγ-
Int(Cl(U)) and hence U is a pre-γ-open set.

Theorem 2.10

A subset A of a space (X, τ) is semi-open if A is β-γ-open and τγ-
Int(Cl(A)) ⊆ Cl(Int(A)).

Proof

Let A be β-γ-open and τγ-Int(Cl(A)) ⊆ Cl(Int(A)). Then A ⊆ Cl(τγ- Int(Cl(A))) ⊆ Cl(Cl(Int(A))) = Cl(Int(A)). And hence A semi-open.

Proposition 2.11

The intersection of a pre-γ-open set and an open set is pre-open.

Proof

Let A be a pre-γ-open set and U be an open set in X. Then A ⊆ τγ-
Int(Cl(A)) and Int(U ) = U, by Lemma 2.8, we have U ∩ A ⊆ Int(U)
∩ τγ-Int(Cl(A)) ⊆ Int(U) ∩ Int(Cl(A)) = Int(U ∩ Cl(A)) ⊆ Int(Cl(U
∩ A)). Therefore, A ∩ U is pre-open.

Proposition 2.12

The intersection of a β-γ-open set and an open set is β-open.

Proof

Let U be an open set and A a β-γ-open set. Since every γ-open set is open, by Lemma 2.8, we have
U ∩ A ⊆ U ∩ Cl(τγ-Int(Cl(A)))
⊆ U ∩ Cl(Int(Cl(A)))
⊆ Cl(U ∩ Int(Cl(A)))
= Cl(Int(U ) ∩ Int(Cl(A)))
= Cl(Int(U ∩ Cl(A)))
⊆ Cl(Int(Cl(U ∩ A))).
This shows that U ∩ A is β-open.

Proposition 2.13

The intersection of a b-γ-open set and an open set is b-open.

Proof

Let A be b-γ-open and U be open, then A ⊆ τγ-Int(Cl(A)) ∪ Cl(τγ-
Int(A)) and U = Int(U). Then we have
U ∩ A ⊆ U ∩ [τγ-Int(Cl(A)) ∪ Cl(τγ-Int(A))]
= [U ∩ τγ-Int(Cl(A))] ∪ [U ∩ Cl(τγ-Int(A))]
= [Int(U ) ∩ τγ-Int(Cl(A))] ∪ [U ∩ Cl(τγ-Int(A))]
⊆ [Int(U ) ∩ Int(Cl(A))] ∪ [U ∩ Cl(Int(A))]
⊆ [Int(U ∩ Cl(A))] ∪ [Cl(U ∩ Int(A))]
⊆ [Int(Cl(U ∩ A))] ∪ [Cl(Int(U ∩ A))].
This shows that U ∩ A is b-open.
We note that the intersection of two pre-γ-open (resp. b-γ-open, β-
γ-open) sets need not be pre-open (resp. b-γ-open, β-γ-open) as can be seen from the following example:

Example 2.14

Let X = {a, b, c} and τ = {φ, {a, b}, X}. Define an operation γ on τ
by γ (A) = A for all A ∈ τ. Let A = {a, c} and B = {b, c}, then A and
B are pre-γ-open (resp. b-γ-open, β-γ-open), but A ∩ B = {c} which
is not pre-open (resp. b-γ-open, β-γ-open).

Proposition 2.15

The intersection of an α-γ-open set and an open set is α-open.

Theorem 2.16

If {Ak: k ∈ ∆} is a collection of b-γ-open (resp. α-γ-open, pre-γ-open,
β-γ-open) sets of a space (X, τ), then ∪k∈∆Ak is b-γ-open (resp. α-γ-
open, pre-γ-open, β-γ-open).

Proof

We prove only the first case since the other cases are similarly
shown. Since Ak⊆ τγ-Int(Cl(Ak)) ∪ Cl(τγ-Int(Ak)) for every k ∈ ∆, we
have
k∈∆Ak⊆ ∪k∈∆γ-Int(Cl(Ak)) ∪ Cl(τγ-Int(Ak))]
⊆ [∪k∈∆τγ-Int(Cl(Ak))] ∪ [∪k∈∆Cl(τγ-Int(Ak))]
⊆ [τγ-Int(∪k∈∆Cl(Ak))] ∪ [Cl(∪k∈∆τγ-Int(Ak))]
⊆ [τγ-Int(Cl(∪k∈∆Ak))] ∪ [Cl(τγ-Int(∪k∈∆Ak))].
Therefore, ∪k∈∆Ak is b-γ-open.
We note that the intersection of two pre-γ-open (resp. α-γ-open, b-
γ-open, β-γ-open) sets need not be pre-γ-open (resp. α-γ-open, b-γ-
open, β-γ-open) as can be seen from the following example:

Example 2.17

Let X = {a, b, c} and τ = P (X). Define an operation γ on τ by γ(A) = A if A = {a, b} or {a, c} or {b, c} and γ(A) = X otherwise. Clearly, τγ
= {φ, {a, b}, {a, c}, {b, c}, X}. Let A = {a, b} and B = {a, c}, then A
and B are pre-γ-open (resp. α-γ-open, b-γ-open, β-γ-open), but A ∩ B
= {a} which is not pre-γ-open (resp. α-γ-open, b-γ-open, β-γ-open).

Proposition 2.18

Let A be a b-γ-open set such that τγ-Int(A) = φ. Then A is pre-γ-open.
A space (X, τ) is called a door space if every subset of X is
open or closed.

Proposition 2.19

If (X, τ) is a door space and γ-regular, then every pre-γ-open set is γ- open.

Proof

Let A be a pre-γ-open set. If A is open, then A is γ-open. Otherwise,

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A is closed and hence A ⊆ τγ-Int(Cl(A)) = τγ-Int(A) ⊆ A. Therefore,
A = τγ-Int(A) and thus A is a γ-open set.

3 New Separation Axioms

Denition 3.1

A topological space (X, τ) with an operation γ on τ is said to be:
1. pre-γ-T0 (resp α-γ-T0, b-γ-T0, β-γ-T0) if for each pair of dis- tinct points x, y in X, there exists a pre-γ-open (resp. α-γ-
open, b-γ-open, β-γ-open) set U such that either x ∈ U and
y  U or x  U and y ∈ U.
2. pre-γ-T1 (resp. α-γ-T1, b-γ-T1, β-γ-T1) if for each pair of
distinct points x, y in X, there exist two pre-γ-open (resp.
α-γ-open, b-γ-open, β-γ-open) sets U and V such that x ∈
U but y  U and y ∈ V but x  V.
3. pre-γ-T2 (resp. α-γ-T2, b-γ-T2, β-γ-T2) if for each distinct
points x, y in X, there exist two disjoint pre-γ-open (resp.
α-γ-open, b-γ-open, β-γ-open) sets U and V containing x and y respectively.

Remark 3.2

For a topological space (X, τ) with an operation γ on τ, the following
properties hold:
1. If (X, τ) is α-γ-Ti, then it is pre-γ-Ti, for i = 0, 1, 2.
2. If (X, τ) is pre-γ-Ti, then it is b-γ-Ti, for i = 0, 1, 2.
3. If (X, τ) is b-γ-Ti, then it is β-γ-Ti, for i = 0, 1, 2.

Denition 3.3

A subset A of a topological space X is called a pre-γD-set (resp. α- γD-set, b-γD-set, β-γD-set) if there are two pre-γ-open (resp. α-γ- open, b-γ-open, β-γ-open) sets U and V such that U ≠ X and A = U \ V. It is true that every pre-γ-open (resp. α-γ-open, b-γ-open, β-γ-
open) set U different from X is a pre-γD-set (resp. α-γD-set, b-γD-
set, β-γD-set) if A = U and V = φ. So, we can observe the following.

Remark 3.4

Every proper pre-γ-open (resp. α-γ-open, b-γ-open, β-γ-open) set is a pre-γD-set (resp. α-γD-set, b-γD-set, β-γD-set).

Remark 3.5

For a topological space (X, τ) with an operation γ on τ, the following properties hold:
1. Every α-γD-set is pre-γD-set.
2. Every pre-γD-set is b-γD-set.
3. Every b-γD-set is β-γD-set.

Denition 3.6

A topological space (X, τ) with an operation γ on τ is said to be:
1. pre-γ-D0 (resp. α-γ-D0, b-γ-D0, β-γ-D0) if for any pair of dis- tinct points x and y of X there exists a pre-γD-set (resp. α- γD-set, b-γD-set, β-γD-set) of X containing x but not y or a γ-bD-set of X containing y but not x.
2. pre-γ-D1 (resp. α-γ-D1, b-γ-D1, β-γ-D1) if for any pair of distinct points x and y of X there exist two pre-γD-sets (resp. α-γD-sets, b-γD-sets, β-γD-sets) U and V such that x
∈ U but y  U and y ∈ V but x V.
3. pre-γ-D2 (resp. α-γ-D2, b-γ-D2, β-γ-D2) if for any pair of
distinct points x and y of X there exist disjoint pre-γD-sets
(resp. α-γD-sets, b-γD-sets, β-γD-sets) G and E of X con- taining x and y, respectively.

Remark 3.7

For a topological space (X, τ) with an operation γ on τ, the following
properties hold:
1. If (X, τ) is α-γ-Di, then it is pre-γ-Di, for i = 0, 1, 2.
2. If (X, τ) is pre-γ-Di, then it is b-γ-Di, for i = 0, 1, 2.
3. If (X, τ) is b-γ-Di, then it is β-γ-Di, for i = 0, 1, 2.

Remark 3.8

For a topological space (X, τ) with an operation γ on τ, the following
properties hold:
1. If (X, τ) is pre-γ-Ti (resp. α-γ-Ti, b-γ-Ti, β-γ-Ti), then it is pre-γ-Ti−1 (resp. α-γ-Ti−1, b-γ-Ti−1, β-γ-Ti−1), for i = 1, 2.
2. If (X, τ) is pre-γ-Ti (resp. α-γ-Ti, b-γ-Ti, β-γ-Ti), then it is pre-γ-Di (resp. α-γ-Di, b-γ-Di, β-γ-Di), for i = 0, 1, 2.
3. If (X, τ) is pre-γ-Di (resp. α-γ-Di, b-γ-Di, β-γ-Di), then it is pre-γ-D (resp. α-γ-D , b-γ-D ,β-γ-D ), for i = 1, 2.

Theorem 3.9

A space X is pre-γ-D1 (resp. α-γ-D1, b-γ-D1, β-γ-D1) if and only if it is pre-γ-D2 (resp. α-γ-D2, b-γ-D2, β-γ-D2).

Proof

Necessity. Let x, y ∈ X, x≠ y. Then there exist pre-γD-sets (resp. α- γD-sets, b-γD-sets, β-γD-sets) G1, G2 in X such that x ∈ G1, y  G1and y ∈ G2, x  G2 . Let G1= U1\U2 and G2= U3\U4, where U1, U2,
U3 and U4 are pre-γ-open (resp. α-γ-open, b-γ-open, β-γ-open) sets in
X. From xG2, it follows that either x  U3 or x ∈ U3 and x ∈ U4. We
discuss the two cases separately.
i. x  U3. By y  G1 we have two subcases:
(a) y  U1. From x ∈ U1\U2, it follows that x ∈ U1\(U2 ∪ U3), and by
y ∈ U3\ U4 we have y ∈ U3\ (U1∪ U4). Therefore (U1\ (U2∪ U3)) ∩
(U3\ (U1 ∪ U4)) = φ.
(b) y ∈ U1 and y ∈ U2. We have x ∈ U1\U2, and y ∈ U2. Therefore
(U1\ U2) ∩ U2= φ.
ii. x ∈ U3 and x ∈ U4. We have y ∈ U3\U4 and x ∈ U4. Hence
(U3\U4) ∩ U4= φ. Therefore X is pre-γ-D2 (resp. α-γ-D2, b-
γ-D2, β-γ-D2).
Sufficiency. Follows from Remark 3.8 (3).

Theorem 3.10

A space is pre-γ-D0 (resp. α-γ-D0, b-γ-D0, β-γ-D0) if and only if it is pre-γ-T0 (resp. α-γ-T0, b-γ-T0, β-γ-T0).

Proof

Suppose that X is pre-γ-D0 (resp. α-γ-D0, b-γ-D0, β-γ-D0). Then for
each distinct pair x, y ∈ X, at least one of x, y, say x, belongs to a
pre-γD-set (resp. α-γD-set, b-γD-set, β-γD-set) G but y  G. Let G =
U1\ U2 where U1 ≠ X and U1, U2 are two pre-γ-open (resp. α-γ-open,
b-γ-open, β-γ-open) sets. Then x ∈ U1, and for y  G we have two cases: (a) y  U1, (b) y ∈ U1 and y ∈ U2. In case (a), x ∈ U1 but y  U1. In case (b), y ∈ U2 but x  U2. Thus in both the cases, we obtain
that X is pre-γ-T0 (resp. α-γ-T0, b-γ-T0, β-γ-T0).
Conversely, if X is pre-γ-T0 (resp. α-γ-T0, b-γ-T0, β-γ-T0), by Remark
3.8 (2), X is pre-γ-D0 (resp. α-γ-D0, b-γ-D0, β-γ-D0).

Corollary 3.11

If (X, τ) is pre-γ-D1 (resp. α-γ-D1, b-γ-D1, β-γ-D1), then it is pre-γ-T0
(resp. α-γ-T0, b-γ-T0, β-γ-T0).

Proof

Follows from Remark 3.8 (3) and Theorem 3.10.

Denition 3.12

A point x ∈ X which has only X as the pre-γ-neighborhood (resp. α-
γ-neighborhood, b-γ-neighborhood, β-γ neighborhood) is called a
pre-γ-neat point (resp. α-γ-neat point, b-γ-neat point, β-γ-neat point).

Theorem 3.13

For a pre-γ-T0 (resp. α-γ-T0, b-γ-T0, β-γ-T0) topological space (X, τ)
the following are equivalent:

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1. (X, τ) is pre-γ-D1 (resp. α-γ-D1, b-γ-D1, β-γ-D1).
2. (X, τ) has no pre-γ-neat point (resp. α-γ-neat point, b-γ-neat
point, β-γ-neat point).

Proof

1 2. Since (X, τ) is pre-γ-D1 (resp. α-γ-D1, b-γ-D1, β-γ-D1), then
each point x of X is contained in a pre-γD-set (resp. α-γD-set, b-γD-
set, β-γD-set) A = U \ V and thus in U. By definition U≠ X. This im- plies that x is not a pre-γ-neat point (resp. α-γ-neat point, b-γ-neat point, β-γ-neat point).
2 1. If X is pre-γ-T0 (resp. α-γ-T0, b-γ-T0, β-γ-T0), then for each distinct pair of points x, y ∈ X, at least one of them, x (say) has a
pre-γ-neighborhood (resp. α-γ-neighborhood, b-γ-neighborhood, β-γ-
neighborhood) U containing x and not y. Thus which is different
from X is a pre-γD-set (resp. α-γD-set, b-γD-set, β-γD-set). If X has
no pre-γ-neat point (resp. α-γ-neat point, b-γ-neat point, β-γ-neat point), then y is not a pre-γ-neat point (resp. α-γ-neat point, b-γ-neat point, β-γ-neat point). This means that there exists a pre-γ- neighborhood (resp. α-γ-neighborhood, b-γ-neighborhood, β-γ-
neighborhood) V of y such that V≠ X. Thus y ∈ V \ U but not x and
V \ U is a pre-γD-set (resp. α-γD-set, b-γD-set, β-γD-set). Hence X is
pre-γ-D1 (resp. α-γ-D1, b-γ-D1, β-γ-D1).

Corollary 3.14

A pre-γ-T0 (resp. α-γ-T0, b-γ-T0, β-γ-T0) space X is not pre-γ-D1 (resp. α-γ-D1, b-γ-D1, β-γ-D1) if and only if there is a unique pre-γ- neat point (resp. α-γ-neat point, b-γ-neat point, β-γ-neat point) in X. Proof
We only prove the uniqueness of the pre-γ-neat point (resp. α-γ-neat point, b-γ-neat point, β-γ-neat point). If x and y are two pre-γ-neat points (resp. α-γ-neat points, b-γ-neat points, β-γ-neat points) in X, then since X is pre-γ-T0 (resp. α-γ-T0, b-γ-T0, β-γ-T0), at least one of x and y, say x, has a pre-γ-neighborhood (resp. α-γ-neighborhood, b- γ-neighborhood, β-γ-neighborhood) U containing x but not y. Hence U≠ X. Therefore x is not a pre-γ-neat point (resp. α-γ-neat point, b-γ- neat point, β-γ-neat point) which is a contradiction.

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