Abstract
In the present study, we introduce some new separation axioms for binary topological spaces. This new idea gives the notion of generalized binary (, , , , and spaces) and binary generalized semi (, , , , and spaces) using generalized binary open sets and binary generalized semi open sets to investigate their properties. We also provide adequate examples to assist and understand abstract concepts. In the similar manner, we begin researching the b-sg-, b-sg-, b-sg-, b-sg-, and b-sg- spaces in binary topological spaces. The study on the axioms is done over binary-, binary-, binary-, binary-, and binary- spaces, motivated to do the analysis of the spaces gb(b-gs)-, gb(b-gs)-, gb(b-gs)-, gb(b-gs)-, and gb(b-gs)- as well.
1. Introduction and Preliminaries
Topology is the most advanced area of pure mathematics which studies mathematical structures. Many scholars have recently analyzed the binary topology that was originally developed by Nithyanantha Jothi and Thangavelu [1]. They also investigated topological structures, displaying their many characteristics in relation to binary topological spaces. In 2011, Nithyanantha Jothi and Thangavelu [1, 2] introduced from to . The authors explored the ideas of binary closed, binary closure, binary interior, binary continuity, base, and subbase of a . In 2012, the authors [3] introduced the concept of binary-, binary-, binary-, binary-, and binary- spaces. The binary points are distinct if . In a , a subset is said to be binary semi open [4] if there exists a binary open set so that , in which denotes the binary closure of in . The complement of a binary semi open set is called binary semi closed, and a subset of is said to be generalized binary closed [5] if whenever and is binary open. The complement of generalized binary closed set is called generalized binary open. Izadi et al. and Kosari et al. [6, 7] tried to transform quartic Diophantine equations into cubic elliptic curves in 2021, and Shao et al. [8] introduced some extensions of Fejér-divergences in 2022. Recently, Sathishmohan et al. [9] proposed the idea of b-gs(b-sg)-closed sets in . Consequently, they [10] introduced the concept of bs-, bs-, bs-, bs-, and bs- spaces. This work introduces and identifies the basic features of the gb(b-gs)-, gb(b-gs)-, gb(b-gs)-, gb(b-gs)-, and gb(b-gs)- spaces in . The analysis ended up with b-sg-, b-sg-, b-sg-, b-sg-, and b-sg- spaces in with various illustrations to demonstrate the behaviour of these new classes of functions.
In the present examination, we use the following symbols: , , , , , , , , , , , , , and (topological spaces, binary topological spaces, generalized semi open set, generalized semi closed set semi generalized closed set, semi generalized open set, binary closed set, binary open set, generalized binary open, generalized binary closed, binary generalized semi closed set, binary generalized semi open set, binary semi generalized closed set, and binary semi generalized open set).
2. gb(b-gs)-, gb(b-gs)-, and gb(b-gs)- Spaces
We define the concept of gb(b-gs)-, gb(b-gs)-, and gb(b-gs)- spaces and explore some of their characterizations in the study.
Definition 1. A is called a generalized binary- (briefly, gb-) if for any two jointly distinct points , there exists such that exactly one of the following holds: (i)(ii)
Definition 2. A is called a generalized binary- (briefly, gb-) if for every two jointly distinct points with , there exists and with and such that .
Definition 3. A is called a generalized binary- (briefly, gb-) if for any two jointly distinct points , with , there exists disjoint and such that and .
Definition 4. A is called a binary generalized semi- (briefly, b-gs-) if for any two jointly distinct points , there exists such that exactly one of the following holds: (i)(ii)
Definition 5. A is called a binary generalized semi- (briefly, b-gs-) if for every two jointly distinct points with , there exists and with and such that .
Definition 6. A is called a binary generalized semi- (briefly, b-gs-) if for any two jointly distinct points , with , there exists disjoint and such that and .
Theorem 7. Let be a ; then, for (1)every b- space is gb(b-gs)- space(2)every b- space is gb(b-gs)- space(3)every b- space is gb(b-gs)- space.(4)every gb(b-gs)- space is gb(b-gs)- space(5)every gb(b-gs)- space is gb(b-gs)- space(6)every gb(b-gs)- space is gb(b-gs)- space
Proof. (1)Let () be a b- space and and be a two distinct points of (); as () is b- space, there exists such that and . Even before every is gb(b-gs)-open and ergo is gb(b-gs)-open set such that and , change has occurred () is gb(b-gs)- space(2)Proof of (2) to (6) is obvious
Example 1. Let ={,}, ={,,}. Clearly, is a from to . (1)Let and , and ; there exists gb(b-gs)-open set =({},{}); then, it is gb(b-gs)- space but not b- space(2)Let and . Assume and , and ; then, it is clear that , and , . After that, we can declare it to be gb(b-gs)- space but not b- space(3)Let and . Suppose and , ,() and ; then, it is clear that , and , . Thereafter, we can formally declare it to be gb(b-gs)- space but not b- space(4)Let and . Assume and , and ; then, it is clear that , and , . Once that is done, we can proclaim it to be gb(b-gs)- space but not gb(b-gs)- space(5)Let and . Suppose and , and ; then, it is clear that and . When that is finished, we can declare it to be gb(b-gs)- space but not gb(b-gs)- space(6)Let and . Consider and , and ; then, it is clear that , and , . We may formally proclaim it after it is done to be gb(b-gs)- space but not gb(b-gs)- space
Theorem 8. A is a b-gs- space if and only if every binary point is .
Proof. Consider that is a b-gs-. Let . Let . By demonstrating it, is . It appears likely to depict this. is . Let . This indicates that and . Ergo and . That is, and () are jointly distinct binary points of . Even before is b-gs-, there exists and , and (), such that and (), although . Hence, is a binary neighbourhood of . This implies that is .
Conversely, assume that is , for every (). Let and with , . Therefore, and is . Also, and is . This shows that is a b-gs-.
Theorem 9. If a is a b-gs-, then is gs- and is gs-.
Proof. Even before is a from to , we have as a topology on and as a topology on . Let and with , . Even before is b-gs-, there exists such that either , or , . This implies that either , , , or , , , . This implies that is gs- and is gs-.
Theorem 10. If a is a b-gs-, then the () and are gs-.
Proof. Suppose that is b-gs-. Let and with , . Even before is b-gs-, there exists such that either , or , . This implies that either , , , or , , , . This implies that either , or , and , or , . Even before , we have and . Change has occurred; () and are gs-.
Theorem 11. If a is a b-gs-, then is gs- and is gs-.
Proof. Even before is a from to , we have as a topology on and as a topology on . Let and with , . Even before is b-gs-, there exists with , , such that , . This implies that , and , such that , and , . Hence, is gs- and is gs-.
Theorem 12. A is a (1)gb- space if and only if every binary point is (2)gb-, then is g- and is g-(3)If a is a gb-, then the () and are gb-(4)gb-, then is g- and is g-
Proof. Proof of (1) to (4) follows from Definitions 1, 2, and 3 and Theorems 8, 9, 10, and 11.
3. gb(b-gs)- and gb(b-gs)- Spaces
We use gb(b-gs)-open sets to create gb(b-gs)- and gb(b-gs)- spaces and examine some of their characteristics in this section.
Definition 13. A is called a generalized binary- (briefly, gb-) or generalized binary regular if is gb- and for every () and every such that () , there exists jointly disjoint such that () , .
Definition 14. A is called a generalized binary- (briefly, gb-) or generalized binary normal if is gb- and for every pair of jointly disjoint , there exists jointly disjoint such that and .
Definition 15. A is called a b-gs- or b-gs regular if is b-gs- and for every () and every set such that () , there exists jointly disjoint such that () , .
Definition 16. A is called a b-gs-or b-gs normal if is b-gs- and for every pair of jointly disjoint , there exists jointly disjoint such that and .
Theorem 17. (1)Every binary- is gb(b-gs)-regular space(2)Every gb(b-gs)- is gb(b-gs)- space(3)Every gb(b-gs)- is b-gs- space(4)Every binary- is gb(b-gs)-(5)Every gb(b-gs)- is gb(b-gs)-
Proof. (1)Let () be a binary regular and be a not containing () which implies to be a set not containing (). As () is b-gs regular, there exists jointly disjoint such that () , . Hence, () is b-gs regular(2)Proof of (2) to (5) is obvious
Example 2. From Example 1, (1)let , , , and ; that is why it is gb(b-gs)- space but not b- space(2)let and , and ; there exists and , because of this gb(b-gs)- space but not gb(b-gs)-(3)let and . Let and , ,() and , and and ; then, it is clear that and . After that, we can declare it to be gb(b-gs)- space but not gb(b-gs)- space(4)let , , , and . Otherwise, we might officially declare it to be gb(b-gs)- space but not binary- space(5)let and , , , and . Thereafter, we can formally declare it to be gb(b-gs)- space but not gb(b-gs)- space
Theorem 18. Let () and to be gs- spaces if and only if is called a b-gs-.
Proof. Suppose () and are gs- spaces. Let () and be a (). Therefore, , and , . Even before () is gs-, there exists disjoint , k, and . Also, even before is gs-, there exists disjoint , l, and . This implies that and . Even before and are disjoint , we have . Also even before and are disjoint , we have . Thus, . Ergo and are disjoint . This implies that is b-gs-.
Conversely, assume that is b-gs-. Let and be a subset of (). Let and be a subset of (,). Therefore, and is in . Even before is b-gs-, there exists disjoint and such that () and . Ergo and , and . This consistently shows that () and are gs- spaces.
Theorem 19. A subspace of a b-gs normal space is b-gs normal.
Proof. Let be a subspace of a b-gs normal space. Let and be disjoint subset of . Even before is in (), and () are in (). Even before () is b-gs normal, there exists disjoint and in (), such that and . Even before contains both and , we have , , and . Even before and are in (), and are in . Thus in the subspace , we have disjoint containing and containing . Ergo the subspace is b-gs normal.
Theorem 20. Let () and be gs- spaces iff the is called a b-gs-.
Proof. Suppose () and are gs- spaces. and are disjoint pair of in . Then, are disjoint in () and are disjoint in . Even before () is gs-, there exists disjoint in , , and . Also, even before is gs-, there exists disjoint , , and . This implies that and . Even before and are disjoint , we have . Also even before and are disjoint , we have . Thus, . Hence, and are disjoint . This implies that is a b-gs-.
Conversely, assume that is b-gs-. Let be disjoint in () and be disjoint in . Then, and are in . Even before is b-gs-, there exists disjoint and such that and . That is, , and , . Hence, () and are gs- spaces
Theorem 21. (1)Let () and be g- spaces iff is called a gb-(2)A subspace of a gb-normal space is gb-normal(3)Let () and be g- spaces iff is called a gb-
Proof. Proof of (1) to (3) follows from Definitions 13 and 14 and Theorems 18, 19, and 20.
4. bsg-, bsg-, and bsg- Spaces
The aspects of b-sg-, b-sg-, and b-sg- spaces are established, and some of their corresponding characterizations are studied in the section.
Definition 22. A is called a binary semi generalized- (briefly, bsg-) if for any two jointly distinct points , there exists such that exactly one of the following holds: (i)(ii)
Definition 23. A is called a binary semi generalized- (briefly, bsg-) if for every two jointly distinct points with , there exists and with and such that .
Definition 24. A is called a binary semi generalized- (briefly, bsg-) if for every two jointly distinct points , with , there exists disjoint and such that and .
Theorem 25. Let be a ; then, for (1)every binary- is b-sg-(2)every binary- is b-sg-(3)every binary- is b-sg-(4)every b-sg- is b-sg-(5)every b-sg- is b-sg-(6)every b-sg- is b-sg-
Proof. (1)Let () be a binary- space and and be a two distinct points of (); as () is binary- space, there exists such that and . Even before every is and ergo is such that and , ergo () is b-sg- space(2)Proof of (2) to (6) is obvious
Example 3. Let and . Clearly, is a from to . (1)Let and , and ; there exists b-sg-open set ; then, it is b-sg- space but not b- space(2)Let and . Let and , and ; then, it is clear that , and , . Then, we can say that it is b-sg- space but not binary- space(3)Let and . Let and , and ; then, it is clear that , and , . Then, we might declare it to be b-sg- space but not binary- space(4)Let and . Let and , and ; then, it is clear that , and , . Afterward, we could declare it to be b-sg- space but not b-sg- space(5)Let and . Let and , and ; then, it is clear that and . We might then proclaim it to be b-sg- space but not b-sg- space(6)Let and . Let and , and ; then, it is clear that , and , . Then, we can say that it is b-sg- space but not b-sg- space
Theorem 26. A is a (1)b-sg- space if and only if every binary point is (2)b-sg-, then is sg- and is sg-(3)If a is called a b-sg-, then the () and are sg-(4)b-sg-, then is sg- and is sg-
Proof. Proof of (1) to (4) follows from Definitions 22, 23, and 24 and Theorems 8, 9, 10, and 11.
5. b-sg- and b-sg- Spaces
The initiation of binary semi- and semi- spaces by utilizing b-sg open sets and their properties are examined in this segment.
Definition 27. A is called a b-sg- or b-sg regular if is b-sg- and for every () and every such that () , there exists jointly disjoint such that () , .
Definition 28. A is called a b-sg- or b-sg normal if is b-sg- and for every pair of jointly disjoint , there exists jointly disjoint such that and .
Theorem 29. (1)Every binary is b-sg (2)Every b-sg is b-sg- space(3)Every b-sg is b-sg- space(4)Every binary normal space is b-sg (5)Every b-sg normal space is b-sg
Proof. Proof of (1) to (5) follows from Definitions 27 and 28 and Theorem 17.
Example 4. From Example 3, (1)let , , , and ; then, it is b-sg- space but not b- space(2)let and , and ; there exists and , ; then, it is b-sg- space but not b-sg-(3)let and . Let and , and , and and ; then, it is clear that and . Then, we can say that it is b-sg- space but not b-sg- space(4)let , , , and ; then, it is b-sg- space but not binary- space(5)let and , , , and . Then, we can say that it is b-sg- space but not b-sg- space
Theorem 30. (1)Let () and be sg- spaces if and only if is called a b-sg-(2)A subspace of a b-sg normal space is b-sg normal(3)Let () and be sg- spaces if and only if is called a b-sg-
Proof. Proof of (1) to (3) follows from Definitions 27 and 28 and Theorems 18, 19, and 20.
6. Conclusion
We defined a few separation axioms in binary topological spaces with respect to binary points of a binary topological spaces, compared their characteristics with those of the existing spaces, and established a few theorems in the paper. The separation axioms, namely, g(gs)‐, g(gs)‐, g(gs)‐, g(gs)‐, and g(gs)‐, are extended to . The perceived result is gb(b-gs)- gb(b-gs)- gb(b-gs)- gb(b-gs)- gb(b-gs)-. Eventually, we identified sg-, sg-, sg-, sg-, and sg- spaces extended to . In our future work, we will extend these structures to infer various results such as binary urysohn space and binary tychonoff space in binary topological spaces.
Data Availability
No data were used to support this study.
Conflicts of Interest
The authors declare that they have no conflicts of interest.
Acknowledgments
This work was supported by the Natural Science Foundation of Guangdong Province of China (2022A1515011468) and the Funding by Science and Technology Projects in Guangzhou (202201020237).