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South Asian Journal of Mathematics
www.sajm-online.com ISSN 2251-1512
2014 , Vol. 4 ( 5 ) : 215 ∼ 224
RESEARCH ARTICLE
A new generalization of closed sets in
bitopology
Z. Duszyñski① ∗ , M. Jeyaraman②, M.S. Joseph③ , O. Ravi④, M.L. Thivagar⑤
①
②
③
④
⑤
Institute of Mathematics, Casimirus the Great University, 85-072 Bydgoszez, Pl. Weyssenhoffa II, Poland, India
Department of Mathematics, Raja Dorai Singam Govt, Arts College, Sivaganga, Tamil Nadu, India
Department of Mathematics, Arul Anandar College, Karumathur, Tamil Nadu, India
Department of Mathematics, P. M. Thevar College, Usilampatti, Madurai District, Tamil Nadu, India
School of Mathematics, Madurai Kamaraj University, Madurai, Tamil Nadu, India
E-mail: [email protected]
Received: June-8-2014; Accepted: Sep-30-2014 *Corresponding author
Abstract In this paper we introduce a new class of sets called (1,2)⋆ -ĝ-closed sets and (1,2)⋆ -αĝ-closed
sets in bitopological spaces. We prove that this (1,2)⋆ -αĝ- closed sets lies between the class of (1,2)⋆ -αclosed sets and the class of (1,2)⋆ -αg-closed sets. Also, we find some basic properties of (1,2)⋆ -αĝ-closed
sets. Applying these sets, we introduce new spaces called (1,2)⋆ -Tαĝ space and (1,2)⋆ -Tĝ space.
Key Words
MSC 2010
1
(1,2)⋆ -ĝ-closed set, (1,2)⋆ -αĝ-closed set, (1,2)⋆ -Tαĝ space, (1,2)⋆ -Tĝ space
54E55, 54E05
Introduction
Levine[6] and Njastad [11] have introduced semi-open sets and α-open sets in topology respectively.
Levine [7] introduced the generalized closed (briefly g-closed) sets and studied their basic properties.
Bhattacharyya and Lahiri[2], Arya and Nour[1] and Maki et al [8],[9] introduced semi-generalized closed
(briefly sg-closed) sets, generalized semi-closed (briefly gs-closed) sets, generalized α-closed (briefly gαclosed) sets and α-generalized closed (briefly αg-closed) sets respectively in topological spaces. Veera
Kumar [22] introduced ĝ-closed sets and Rose Mary et al [21] introduced αĝ-closed sets in topological
spaces. Ravi et al [13] [20], recently, generalized some topological notions of [2], [4], [5], [6], [7], [8], [9].
In the present paper, we introduce a new class of closed sets called (1,2)⋆ -ĝ-closed and (1,2)⋆ -αĝclosed sets in bitopological spaces. We prove that this (1,2)⋆ -αĝ-closed sets lies between the class of (1,2)⋆ α-closed sets and the class of(1,2)⋆ -αg-closed sets. Applying these sets, we introduce new spaces called
(1,2)⋆ -Tαĝ space and (1,2)⋆ -Tĝ space. These generalizations are substantiated with suitable examples at
every stage.
Citation: Z. Duszyñski, M. Jeyaraman, M. Sajan Joseph, O. Ravi, M. Lellis Thivagar, A new generalization of closed sets in
bitopology, South Asian J Math, 2014, 4(5), 215-224.
Z. Duszyñski, et al : A new generalization of closed sets in bitopology
2
Preliminaries
Throughout this paper (X, τ1 , τ2 ) and (Y, σ1 , σ2 ) (or simply X and Y ) denote bitopological spaces.
Definition 2.1. [16] A subset S of X is said to be τ1,2 -open if S = A ∪ B where A ∈ τ1 and B ∈ τ2 .
The complement of τ1,2 -open subset of X is called τ1,2 -closed.
The family of all τ1,2 -open (resp. τ1,2 -closed) subsets of X is denoted by (1,2)⋆ -O(X)[resp.(1,2)⋆ C(X)].
Note 2.2. [16] Recall that τ1,2 -open subsets of X need not necessarily form a topology.
Definition 2.3. [16] Let S be a subset of X. Then
1. The τ1,2 -interior of S in X, denoted by τ1,2 -int(S), is defined as ∪{F/F ⊂ S and F is τ1,2 -open};
2. The τ1,2 -closure of S in X, denoted by τ1,2 -cl(S), is defined as ∩{G/S ⊂ G and G is τ1,2 -closed}.
Definition 2.4. [16] A subset S of X is said to be:
1. (1,2)⋆ -α-open if S⊆τ1,2 -int (τ1,2 -cl(τ1,2 -int(S)));
2. (1,2)⋆ -semi-open if S⊆τ1,2 -cl(τ1,2 -int(S)).
The complement of (1,2)⋆ -α-open (resp.(1,2)⋆ -semi-open) is called (1,2)⋆ -α- closed (resp. (1,2)⋆ semi-closed).
The family of all (1,2)⋆ -α-open (resp.(1,2)⋆ -semi-open, (1,2)⋆ -α-closed and (1,2)⋆ -semi-closed) subsets of X is denoted by (1,2)⋆ -αO(X) (resp. (1,2)⋆ - SO(X), (1,2)⋆ -αC(X) and (1,2)⋆ -SC(X)).
Definition 2.5. [13] A subset S of X is called:
1. (1,2)⋆ -g-closed if τ1,2 -cl(S)⊆U whenever S⊆U and U∈(1,2)⋆ -O(X).
2. (1,2)⋆ -g-open if X\S is (1,2)⋆ -g-closed.
Definition 2.6. [17] The intersection of all (1,2)⋆ -semi-closed sets containing S is called the (1,2)⋆ semi-closure of S and is denoted by (1,2)⋆ -scl(S).
Definition 2.7. A subset S of X is called:
1. (1,2)⋆ -sg-closed [17] if (1,2)⋆ -scl(S)⊆U whenever S⊆U and U∈(1,2)⋆ -SO(X).
2. (1,2)⋆ -sg-open if X\S is (1,2)⋆ -sg-closed.
3. (1,2)⋆ -gs-closed [14] if (1,2)⋆ -scl(S)⊆U whenever S⊆U and U∈(1,2)⋆ -O(X).
4. (1,2)⋆ -gs-open if X\S is (1,2)⋆ -gs-closed.
Definition 2.8. [18] The intersection of all (1,2)⋆ -α-closed sets containing S is called the (1,2)⋆ -αclosure of S and is denoted by (1,2)⋆ -αcl(S). The union of all (1,2)⋆ -α-open sets contained in S is called
the (1,2)⋆ -α-interior of S and is denoted by (1,2)⋆ -αint(S).
Definition 2.9. A subset S of X is called:
1. (1,2)⋆ -αg-closed [20] if (1,2)⋆ -αcl(S)⊆U whenever S⊆U and U∈(1,2)⋆ -O(X).
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2. (1,2)⋆ -αg-open if X\S is (1,2)⋆ -αg-closed.
3. (1,2)⋆ - gα-closed [20] if (1,2)⋆ -αcl(S)⊆U whenever S⊆U and U∈(1,2)⋆ -αO(X).
4. (1,2)⋆ -gα-open if X\S is (1,2)⋆ -gα-closed.
Definition 2.10. Let S be any subset of X .Then
1. (1,2)⋆ -scl(S)=S ∪ τ1,2 -int (τ1,2 -cl(S)); [15]
2. (1,2)⋆ -αcl(S)=S ∩ τ1,2 -cl(τ1,2 -int(τ1,2 -cl (S))); [18]
3. (1,2)⋆ -αint(S)=S ∩ τ1,2 -int(τ1,2 -cl(τ1,2 -int(S))); [18]
Definition 2.11. A bitopological space (X, τ1 , τ2 ) is called:
1. (1,2)⋆ -αTb space [20] if every (1,2)⋆ -αg-closed set in it is τ1,2 -closed.
2. (1,2)⋆ -Tb space [14] if every (1,2)⋆ -gs-closed set in it is τ1,2 -closed.
Result 2.12. Let S be any subset S of X . Then
1. Any τ1,2 -closed set S is (1,2)⋆ -α-closed, but not conversely; [17]
2. Any (1,2)⋆ -α-closed set S is (1,2)⋆ -gα-closed, but not conversely; [20]
3. Any (1,2)⋆ -sg-closed set S is (1,2)⋆ -αg-closed, but not conversely; [20]
4. Any (1,2)⋆ -g-closed set S is (1,2)⋆ -αg-closed, but not conversely; [20]
5. Any (1,2)⋆ -αg-closed set S is (1,2)⋆ -gs-closed, but not conversely; [20]
6. Any (1,2)⋆ -gα-closed set S is (1,2)⋆ -sg-closed, but not conversely; [20].
3
Properties of (1,2)⋆ -ĝ-closed sets
Definition 3.1. A subset S of X is said to be (1,2)⋆ -ĝ-closed set if τ1,2 -cl(S)⊆U whenever S⊆U and
U∈(1,2)⋆ -SO(X).
The complement of (1,2)⋆ -ĝ-closed set is (1,2)⋆ -ĝ-open.
Theorem 3.2. Every τ1,2 -open subset of X is (1,2)⋆ -ĝ-open.
Proof. Let S be an τ1,2 -open subset in X. Then X\S∈(1,2)⋆ -C(X). Therefore τ1,2 -cl(X\S)=X\S⊆X
whenever X\S⊆X and X∈(1,2)⋆ -SO(X). This means X\S is (1,2)⋆ -ĝ-closed. Hence S is (1,2)⋆ -ĝ-open.
Theorem 3.3. Every (1,2)⋆ -ĝ-closed subset of X is (1,2)⋆ -g-closed.
Proof. It is clear by definitions.
Remark 3.4. The converses of Theorem 3.2 and 3.3 are not true as per the following examples.
Example 3.5. Let X = {a, b, c} with τ1 = {∅, X, {a}} and τ2 = {∅, X, {b, c}}. Then (1,2)⋆ -O(X) =
{∅, X, {a}, {b, c}} = (1,2)⋆ -C(X). We consider the set {b} which is (1,2)⋆ -ĝ-open but it is not τ1,2 -open.
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Example 3.6. Let X = {a, b, c} with τ1 = {∅, X, {a}} and τ2 = {∅, X, {a, c}}. Then (1,2)⋆ -O(X) =
{∅, X, {a}, {a, c}} and (1,2)⋆ -C(X)= {∅, X, {b}, {b, c}}. Clearly {a, b} is (1,2)⋆ -g-closed set but it is
not (1,2)⋆ -ĝ-closed.
Proposition 3.7. If a subset S is (1,2)⋆ -ĝ-closed of X, then τ1,2 -cl(S)\S does not contain any non-empty
τ1,2 -closed set.
Proof. Let F be a τ1,2 -closed subset of τ1,2 -cl(S)\S. Then F⊆τ1,2 -cl(S) and F∩S = ∅. Therefore X\F
is τ1,2 -open and hence X\F is (1,2)⋆ -semi-open. Since F∩S = ∅, S⊆X\F. But S is (1,2)⋆ -ĝ-closed, then
τ1,2 -cl(S)⊆ X\F and consequently F⊆X\τ1,2 -cl(S). Therefore F⊆(τ1,2 -cl(S))∩(X\ τ1,2 -cl(S)) and hence F
is empty.
Remark 3.8. The converse of Proposition 3.7 is not true as per the following example.
Example 3.9. Let X = {a, b, c} with τ1 = {∅, X} and τ2 = {∅, X, {a}}. If we consider A = {b}, then
τ1,2 -cl(A)\A = {c} does not contain non-empty τ1,2 -closed set. However A is not (1,2)⋆ -ĝ-closed.
Proposition 3.10. If S is a (1,2)⋆ -ĝ-closed and (1,2)⋆ -semi-open subset of X, then S is τ1,2 -closed.
Proof. Since S is (1,2)⋆ -ĝ-closed and (1,2)⋆ -semi-open, τ1,2 -cl(S)⊆S. Hence S is τ1,2 -closed.
Proposition 3.11. Let A be a (1,2)⋆ -ĝ-closed subset of X. If A⊆B⊆τ1,2 -cl(A), then B is also (1,2)⋆ -ĝclosed in X.
Proof. Let U∈(1,2)⋆ -SO(X) with B⊆U. Then A⊆U. Since A is (1,2)⋆ -ĝ-closed, τ1,2 -cl(A)⊆U. Also,
since B τ1,2 -cl(A), τ1,2 -cl(B)⊆τ1,2 -cl(A)⊆U. Hence B is also (1,2)⋆ -ĝ-closed subset of X.
Remark 3.12. Union of any two (1,2)⋆ -ĝ-closed sets in X need not to be (1,2)⋆ -ĝ- closed as per the
following example.
Example 3.13. Let X = {a, b, c} with τ1 = {∅, X, {a, b}} and τ2 = {∅, X, {b}, {c}, {b, c}, {a, c}}.
Clearly {b} and {c} are (1,2)⋆ -ĝ-closed sets but their union {b, c} is not (1,2)⋆ -ĝ-closed set.
Remark 3.14. Intersection of any two (1,2)⋆ -ĝ-closed sets in X need not be (1,2)⋆ - ĝ-closed as per the
following example.
Example 3.15. Let X = {a, b, c} with τ1 = {∅, X, {a, b}} and τ2 = {∅, X}. Clearly {a, c} and {a, b}
are (1,2)⋆ -ĝ-closed sets but their intersection {a} is not (1,2)⋆ -ĝ-closed set.
Remark 3.16. The following shows that (1,2)⋆ -ĝ-closedness and (1,2)⋆ -α-closedness are independent of
each other.
Example 3.17. The set {b} in Example 3.5 is (1,2)⋆ -ĝ-closed, but it is not (1,2)⋆ -α-closed. The set {b}
in Example 3.9 is (1,2)⋆ -α-closed, but it is not (1,2)⋆ -ĝ- closed.
Corollary 3.18. Every (1,2)⋆ -ĝ-closed subset of X is (1,2)⋆ -g-closed.
Proof. By Theorem 3.3 and Result 2.12(iv).
Corollary 3.19. Every (1,2)⋆ -ĝ-closed subset of X is (1,2)⋆ -gs-closed.
Proof. By Theorem 3.3 and Result 2.12 (iv) and (v).
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Remark 3.20. The converses of Corollary 3.18 and 3.19 are not true in general.
Example 3.21. Let X = {a, b, c} with τ1 = {∅, X, {a}} and τ2 = {∅, X, {a, b}}. Clearly {b} is
(1,2)⋆ -αg-closed and (1,2)⋆ -gs-closed sets but it is not (1,2)⋆ -ĝ-closed.
Theorem 3.22. Every (1,2)⋆ -ĝ-closed subset of X is (1,2)⋆ -sg-closed.
Proof. It is true that (1,2)⋆ -scl(A)⊆τ1,2 -cl(A) for every subset A of X.
Remark 3.23. The converse of Theorem 3.22 is not true as per the following example.
Example 3.24. The set {b} in Example 3.21 is (1,2)⋆ -sg-closed, but it is not (1,2)⋆ - ĝ-closed.
Theorem 3.25. Every (1,2)⋆ -ĝ-closed set of X is (1,2)⋆ -gα-closed.
Proof. We utilize the facts that (1,2)⋆ -scl(A)⊆τ1,2 -cl(A) for every subset A of X and that every
(1,2) -α-open set is (1,2)⋆ -semi-open (not conversely [16] in general).
⋆
Remark 3.26. The converse of Theorem 3.25 is not true in general.
Example 3.27. The set {c} in Example 3.6 is (1,2)⋆ -gα-closed but it is not (1,2)⋆ -ĝ-closed.
4
Properties of (1,2)⋆ -αĝ-closed sets
Definition 4.1. A subset S of X is said to be (1,2)⋆ -αĝ-closed if (1,2)⋆ -αcl(S)⊆U whenever S⊆U and U
is a (1,2)⋆ -ĝ-open in X.
Proposition 4.2. Every (1,2)⋆ -α-closed subset of X is (1,2)⋆ -αĝ-closed.
Proof. Let S be an (1,2)⋆ -α-closed subset in X and U be any (1,2)⋆ -ĝ-open set containing S. Since
S is (1,2)⋆ -α-closed, (1,2)⋆ -αcl(S)=S by ([10], Lemma 2.2(iii)). Therefore (1,2)⋆ -αcl(S)⊆U and hence S is
(1,2)⋆ -αĝ-closed.
Corollary 4.3. Every τ1,2 -closed subset of X is (1,2)⋆ -αĝ-closed.
Proof. It follows from the fact that every τ1,2 -closed subset of X is (1,2)⋆ -α-closed [17].
Remark 4.4. An (1,2)⋆ -αĝ-closed set need not be (1,2)⋆ -α-closed, in general, as shown in the following
example.
Example 4.5. The set {a, c} in Example 3.15 is (1,2)⋆ -αĝ-closed, but it is not (1,2)⋆ -α-closed in X.
Proposition 4.6. Every (1,2)⋆ -αĝ -closed subset of X is (1,2)⋆ -αg-closed.
Proof. Let S be an (1,2)⋆ -αĝ-closed set and U be any τ1,2 -open set containing S in X. Since every
τ1,2 -open subset of X is (1,2)⋆ -ĝ-open, the proof follows immediately.
Remark 4.7. An (1,2)⋆ -αg-closed set need not be (1,2)⋆ -αĝ-closed as shown in the following example.
Example 4.8. The set {b} in Example 3.5 is (1,2)⋆ -αg-closed, but it is not (1,2)⋆ -αĝ-closed in X.
Remark 4.9. From the above results, we note that the class of (1,2)⋆ -αĝ-closed sets lies between the
class of (1,2)⋆ -α-closed sets and the class of (1,2)⋆ -αg-closed sets.
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Proposition 4.10. Every (1,2)⋆ -αĝ-closed subset of X is (1,2)⋆ -gs-closed.
Proof. It is true that (1,2)⋆ -scl(A)⊆(1,2)⋆ -αcl(A) for every subset A of X.
Remark 4.11. A (1,2)⋆ -gs-closed set need not be (1,2)⋆ -αĝ-closed as shown in the following example.
Example 4.12. Let X = {a, b, c} with τ1 = {∅, X, {a}} and τ2 = {∅, X, {b}}. Then the set {a} is
(1,2)⋆ -gs-closed, but it is not (1,2)⋆ -αĝ-closed in X.
Remark 4.13. The following examples show that (1,2)⋆ - ĝ-closedness is independent of (1,2)⋆ -sg-closedness,
(1,2)⋆ -g-closedness, (1,2)⋆ -ĝ-closedness and (1,2)⋆ -gα-closedness.
Example 4.14. The set {a, b} in Example 3.6 is (1,2)⋆ -αĝ-closed, but it is not (1,2)⋆ -sg-closed and
the set {c} in Example 3.6 is an (1,2)⋆ -αĝ-closed but it is neither (1,2)⋆ -g-closed nor (1,2)⋆ -ĝ-closed in
X. Also the set {a} in Example 4.12 is (1,2)⋆ -sg-closed but it is not (1,2)⋆ -αĝ-closed and the set {b}
in Example 3.5 is (1,2)⋆ -g-closed, (1,2)⋆ -ĝ-closed and (1,2)⋆ -gα-closed, but it is not (1,2)⋆ -αĝ-closed.
Moreover, the set {a, b} in Example 3.9 is (1,2)⋆ -αĝ-closed, but it is not (1,2)⋆ -gα-closed.
Remark 4.15. From the above discussion and known results we obtain the following diagram:
τ1,2 -closed
−→
↓
ց
(1,2)⋆ -α-closed
−→
(1,2)⋆ -ĝ-closed
−→
(1,2)⋆ -g-closed
↓
(1,2)⋆ -αĝ-closed
↓
(1,2)⋆ -gα-closed
−→
(1,2)⋆ -sg-closed
−→
(1,2)⋆ -αg-closed
ց
↓
−→
(1,2)⋆ -gs-closed
where A → B (resp. A = B) represents A implies B but not conversely (resp. A and B are independent
of each other).
Remark 4.16. Union of any two (1,2)⋆ -αĝ -closed subsets of X need not to be (1,2)⋆ -αĝ -closed as the
following example shows.
Example 4.17. The sets {b} and {c} in Example 3.13 are (1,2)⋆ -αĝ -closed, but their union {b, c} is
not (1,2)⋆ -αĝ-closed.
Remark 4.18. Intersection of any two (1,2)⋆ -αĝ-open subsets of X need not be (1,2)⋆ -αĝ -open.
Example 4.19. The sets {a, c} and {a, b} in Example 3.13 are (1,2)⋆ -αĝ -open, but their intersection
{a} is not (1,2)⋆ -αĝ-open.
Proposition 4.20. If a subset S is (1,2)⋆ -αĝ-closed in X, then (1,2)⋆ -αcl(S)\S contains no any nonempty τ1,2 -closed set.
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Proof. Let F be a τ1,2 -closed subset of (1,2)⋆ -αcl(S)\S. Then S⊆(X\F), X\F is τ1,2 -open and hence
(1,2)⋆ -ĝ-open. Since S is (1,2)⋆ -αĝ-closed, (1,2)⋆ -αcl(S) ⊆ (X\F). Consequently F⊆(X\(1,2)⋆ -αcl(S)).
Therefore F ⊆ (1,2)⋆ - cl(S) ∩ (X\(1,2)⋆ -cl(S)) and hence F is empty.
Remark 4.21. The converse of Proposition 4.20 is not true by the following example.
Example 4.22. Consider a bitopological space (X, τ1 , τ2 ) as in Example 3.5. If A={b}, then (1,2)⋆ αcl(A)\A={c} does not contain non-empty τ1,2 -closed set. However, A is not an (1,2)⋆ -αĝ-closed subset
of X.
Proposition 4.23. If a subset S is (1,2)⋆ -αĝ-closed in X, then (1,2)⋆ -αcl(S)\S does not contain any
non-empty (1,2)⋆ -ĝ-closed set.
Proof. Suppose that S is (1,2)⋆ -αĝ-closed. Let F be a (1,2)⋆ -ĝ-closed set contained in (1,2)⋆ αcl(S)\S. Since X\F is (1,2)⋆ -ĝ-open set with S⊆(X\F) and S is (1,2)⋆ -αĝ-closed, (1,2)⋆ -αcl(S)⊆(X\F).
Then F⊆(X\(1,2)⋆ -αcl(S)). Therefore F⊆(X\(1,2)⋆ -αcl(S))∩(1,2)⋆ -αcl(S) and hence F is empty.
Proposition 4.24. If S is (1,2)⋆ -ĝ-open and (1,2)⋆ -αĝ-closed subset of X, then S is an (1,2)⋆ -α-closed
subset of X.
Proof. Since S is (1,2)⋆ -ĝ-open and (1,2)⋆ -αĝ-closed, (1,2)⋆ -αcl(S)⊆S. Hence S is (1,2)⋆ -α-closed.
Proposition 4.25. Let A be an (1,2)⋆ -αĝ-closed subset of X. If A⊆B⊆(1,2)⋆ -αcl(A), then B is also an
(1,2)⋆ -αĝ-closed subset of X.
Proof. Let U be a (1,2)⋆ -ĝ-open subset of X such that B⊆U. Then A⊆U. Since A is an (1,2)⋆ -αĝclosed set, (1,2)⋆ -αcl(A)⊆U. Hence B is also an (1,2)⋆ -αĝ-closed subset of X.
Proposition 4.26. For each x ∈ X, either {x} is (1,2)⋆ -ĝ-closed or X\{x} is an (1,2)⋆ -αĝ-closed.
Proof. Suppose that {x} is not (1,2)⋆ -ĝ-closed in X. Then X\{x} is not (1,2)⋆ -ĝ- open and the
only (1,2)⋆ -ĝ-open set containing X\{x} is the space X itself. That is X\{x}⊆X. Therefore (1,2)⋆ αcl(X\{x})⊆X and so X\{x} is (1,2)⋆ -αĝ-closed.
Theorem 4.27. If S is (1,2)⋆ -αĝ-closed and the set (1,2)⋆ -αcl(S)\S is τ1,2 -closed, then (1,2)⋆ -αcl(S) =
S. Moreover, if the family (1,2)⋆ -αO(X) fulfills the condition B [10], then S is (1,2)⋆ -α-closed.
Proof. Since S is (1,2)⋆ -αĝ-closed, by Proposition 4.20, (1,2)⋆ -αcl(S)\S contains no non-empty τ1,2 closed set. But, by assumption (1,2)⋆ -αcl(S)\S is τ1,2 -closed. This implies (1,2)⋆ -αcl(S)\S=∅. That is
(1,2)⋆ -αcl(S)=S. Hence S is (1,2)⋆ -α-closed.
5
Some Properties of (1,2)⋆ -αĝ-open sets
Definition 5.1. A subset S of X is said to be (1,2)⋆ -αĝ-open if and only if X\S is (1,2)⋆ -αĝ-closed.
Lemma 5.2. For a subset A of X, (1,2)⋆ -αcl(X\A)=X\(1,2)⋆ -αint(A).
Proof. Follows by ([10] Lemma 2.2(i)).
Theorem 5.3. Let S be a subset of X. Then S is (1,2)⋆ -αĝ-open in X if and only if F⊆(1,2)⋆ -αint(S)
whenever F is (1,2)⋆ -ĝ-closed and F⊆S.
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Proof. Necessity Let F be an (1,2)⋆ -ĝ-closed such that F⊆S. Then X\S ⊆ X\F where X\F is
(1,2)⋆ -ĝ-open. Since X\S is (1,2)⋆ -αĝ-closed, (1,2)⋆ -αcl(X\S) ⊆ X\F. By Lemma 5.2, X\(1,2)⋆ -αint(S)
⊆ X\F. That is F⊆(1,2)⋆ -αint(S).
Sufficiency Suppose F is (1,2)⋆ -ĝ-closed and F⊆S. Then F⊆(1,2)⋆ -αint(S). Let X\S⊆U where U
is (1,2)⋆ -ĝ-open. Then X\U ⊆ S where X\U is (1,2)⋆ -ĝ-closed. By hypothesis X\U ⊆ (1,2)⋆ -αint(S).
That is X\(1,2)⋆ -αint(S) ⊆ U. By Lemma 5.2, (1,2)⋆ -αcl(X\S) ⊆ U. This implies X\S is (1,2)⋆ -αĝ-closed.
Hence S is (1,2)⋆ -αĝ-open.
Proposition 5.4. If (1,2)⋆ -αint(A) ⊆ B ⊆ A and A is (1,2)⋆ -αĝ-open in X, then B is (1,2)⋆ -αĝ-open.
Proof. (1,2)⋆ -αint(A) ⊆ B ⊆ A implies X\A ⊆ X\B ⊆ X\(1,2)⋆ -αint(A). By Lemma 5.2 X\A ⊆
X\B ⊆ (1,2)⋆ -αcl(X\A). By Proposition 4.25 X\B is (1,2)⋆ -αĝ-closed. Hence B is (1,2)⋆ -αĝ-open.
Remark 5.5. It is true that every τ1,2 -open subset of X is (1,2)⋆ -αĝ-open but the converse may not be
true as shown in the following example.
Example 5.6. The set {b} in Example 3.21 is (1,2)⋆ -αĝ-open, but it is not τ1,2 - open.
6
Applications
We introduce the following type of spaces.
Definition 6.1. A space (X, τ1 , τ2 ) is called a (1,2)⋆ -Tαĝ space if every (1,2)⋆ -αĝ-closed set in it is an
(1,2)⋆ -α-closed.
Definition 6.2. A space (X, τ1 , τ2 ) is called a (1,2)⋆ -Tĝ space if every (1,2)⋆ -ĝ-closed set in it is τ1,2 closed.
Theorem 6.3. For a space (X, τ1 , τ2 ) such that the family (1,2)⋆ -αO(X) fulfills the condition B [11], the
following conditions are equivalent.
1. X is a (1,2)⋆ -Tαĝ space
2. Every singleton set of X is either (1,2)⋆ -ĝ-closed or (1,2)⋆ -α-open.
Proof. i) ⇒ ii) Let x ∈ X. Suppose {x}is not a (1,2)⋆ -ĝ-closed subset of X. Then X\{x} is not a
(1,2)⋆ -ĝ-open subset of X. So X is the only (1,2)⋆ -ĝ-open set of X containing X\{x}. Then X\{x} is an
(1,2)⋆ -αĝ-closed subset of X. Since X is (1,2)⋆ -Tαĝ space, X\{x} ∈ (1,2)⋆ -αC(X) or equivalently {x} ∈
(1,2)⋆ -αO(X).
ii) ⇒ i) Let S be an (1,2)⋆ -αĝ-closed subset of X. Trivially S⊆(1,2)⋆ -αcl(S). Let {x} ∈ (1,2)⋆ -αcl(S).
By (ii), {x} is either (1,2)⋆ -ĝ-closed or (1,2)⋆ -α-open.
Case (a) Suppose that {x} is (1,2)⋆ -ĝ-closed. If x 6∈ S, (1,2)⋆ -αcl(S)\S contains a non-empty
⋆
(1,2) -ĝ-closed set {x}. By Proposition 4.23, we arrive at a contradiction. Thus x ∈ S.
Case (b) Suppose that {x} ∈ (1,2)⋆ -αO(X). Since x ∈ (1,2)⋆ -αcl(S), {x}∩S6= ∅ by ([12], Lemma
3.2). This implies that x ∈ S.
Thus in any case we have x ∈ S, so (1,2)⋆ -αcl(S)⊆S. Therefore (1,2)⋆ -αcl(S)=S and by ([12], Lemma
3.3(3)) S ∈ (1,2)⋆ -αC(X). Hence X is (1,2)⋆ -Tαĝ space.
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Theorem 6.4. Every (1,2)⋆ -Tb space is a (1,2)⋆ -Tαĝ .
Proof. It is true that every (1,2)⋆ -αĝ-closed set is (1,2)⋆ -gs-closed and every τ1,2 -closed set is
(1,2) -α-closed.
⋆
Theorem 6.5. Every (1,2)⋆ -αTb space is a (1,2)⋆ -Tαĝ .
Proof. It is true that every (1,2)⋆ -αĝ-closed set is (1,2)⋆ -αg-closed.
Remark 6.6. The following examples show that the reverse implications in Theorem 6.4 and 6.5 are not
true, in general.
Example 6.7. The space X in Example 4.12 is a (1,2)⋆ -Tαĝ space. However, it is not (1,2)⋆ -Tb space
because {a} is (1,2)⋆ -gs-closed and it is not τ1,2 -closed.
Example 6.8. The space X in Example 3.5 is a (1,2)⋆ -Tαĝ space. However, it is not (1,2)⋆ -αTb space
because {b} is (1,2)⋆ -αg-closed and it is not τ1,2 -closed
Theorem 6.9. For a space (X, τ1 , τ2 ), the following conditions are equivalent.
1. X is a (1,2)⋆ -Tĝ space
2. Every singleton set of X is either (1,2)⋆ -semi-closed or τ1,2 -open.
Proof. i) ⇒ ii) Let x ∈ X. Suppose {x} 6∈ (1,2)⋆ -SC(X). Then X\{x}6∈(1,2)⋆ -SO(X). So X is the
only (1,2)⋆ -semi-open set of X containing X\{x}. Then X\{x} is a (1,2)⋆ -ĝ-closed set of X. Since X is
(1,2)⋆ -Tĝ space, X\{x} is τ1,2 -closed. Hence {x} is τ1,2 -open.
ii) ⇒ i) Let S be a (1,2)⋆ -ĝ-closed subset of X. Trivially S ⊆ τ1,2 -cl(S). Let {x} ∈ τ1,2 -cl(S). By (ii),
{x} is (1,2)⋆ -semi-closed or τ1,2 -open.
Case (a) Suppose that {x} ∈ (1,2)⋆ -SC(X). If x 6∈ S, τ1,2 -cl(S)\S contains non- empty τ1,2 -closed
subset. By Proposition 3.7, we arrive at a contradiction. Thus x ∈ S.
Case (b) Suppose that {x} ∈ (1,2)⋆ -O(X). Since x ∈ τ1,2 -cl(S), {x} ∩ S 6= ∅ ([12], Lemma 3.2). This
implies that x ∈ S. Thus in any case x ∈ S. So, τ1,2 -cl(S)⊆S. Therefore τ1,2 -cl(S)=S or equivalently S is
τ1,2 -closed because the family (1,2)⋆ -O(X) fulfills the condition B [10]. Hence X is a (1,2)⋆ -Tĝ space.
Remark 6.10. (1,2)⋆ -Tαĝ space and (1,2)⋆ -Tĝ spaces are independent of one another as the following
examples show.
Example 6.11. The space X in Example 3.9 is a (1,2)⋆ -Tĝ space but it is not (1,2)⋆ -Tαĝ space because
{a, b} is (1,2)⋆ -αĝ-closed and it is not (1,2)⋆ -α-closed.
Example 6.12. The space X in Example 3.5 is a (1,2)⋆ -Tαĝ space but it is not (1,2)⋆ -Tĝ space because
{a, b} is (1,2)⋆ -ĝ-closed and it is not τ1,2 -closed.
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