5. Product of Soft Separation Axioms and Soft (First and Second) Coordinate Spaces
In this section, the engagement of section two and section three with some important results with respect to crisp points under soft semiopen sets is addressed. The product of two soft axioms with respect crisp points with almost all possibilities, the product of two soft axioms with respect crisp points with almost all possibilities and product of two soft axioms with respect to crisp points with almost all possibilities in soft bitopological spaces relative to semiopen sets are introduced. In addition to this, soft (countability, base, subbase, finite intersection property, continuity) are addressed with respect to semiopen sets in soft bitopological spaces. Finally, the product of soft first and second coordinate spaces are addressed with respect to semiopen sets in soft bitopological spaces.
Theorem 1. If and be two over the crisp sets and respectively such that these are two soft . Then their product is soft .
Proof. Let be the soft product space. We prove that it is soft . Suppose are two distinct soft points in then different cases are or
Case If being soft , corresponding to this pair of distinct soft points there exists a soft -open set such that and Thus, is soft -open set catching but bay passing
Case If being soft , corresponding to this pair of distinct soft points there exists a soft -open set such that and Thus, is soft -open set catching but bay passing
Case If being soft , corresponding to this pair of distinct points there exists a –soft -open set such that and Thus, is soft -open set arresting and bypassing Thus for any two distinct points of there exists a soft -open set arresting one of them and bypassing the other. Hence is soft .
Case If being soft , corresponding to this pair of distinct points there exists a soft -open set such that and Thus, is soft -open set arresting and bypassing Thus for any two distinct points of there exists a soft -open set arresting one of them and bypassing the other. Hence is soft space. □
Theorem 2. If and be two over the crisp sets and respectively such that these are two soft . Then their product is soft .
Proof. Let and be two such that they are soft respectively. Then we have to show that their product that is is also soft . For this proof, it is sufficient to show that each soft subset of consisting of exactly one soft point is a soft -closed set. Let so that and .
Case Now, being soft , is soft closed and therefore, is soft - Also, being soft . So that is and hence, is soft closed. Hence is soft .
Case Now, being soft , is soft closed and therefore, is soft - Also, being soft . So that is and hence, is soft -closed. Hence is soft . □
Theorem 3. If and be two over the crisp sets and respectively such that these are two soft -Hausdorff spaces. Then their product is soft -Hausdorff space.
Proof. Let be the product space. We prove that it is a soft Hausdorff space. Suppose and are two distinct points in
Case When then let therefore By the soft Hausdorff space property, given a pair of elements such that there are disjoint soft -open sets such that such that , that . Then and ) are disjoint soft -pen sets in . For, implies that , implies that this whole situation leads us to the conclusion that is a soft Hausdorff space.
Case If let then too By the soft Hausdorff space property, given a pair of elements such that there are disjoint -open sets in such that such that , that . Then and are disjoint soft -open sets in . For, implies that implies that This whole situation leads us to the conclusion that is a soft -Hausdorff space. □
Theorem 4. Let and be two second-countable then their product, that is is also soft second-countable .
Proof. To prove is second-countable . Our assumption implies that there are countable soft s-bases and for and respectively. such that and are soft -open such that is a soft for soft product topology . If we write is soft-countable, this implies that is soft-countable. By definition of soft base and implies that there exists a soft -open sets such that such that implies this implies that there exists such that this implies that . By definition this proves that is a soft base for the soft product topology In addition, has been shown to be soft-countable. Hence, is soft second-countable relative to soft -open set. □
Theorem 5. Let and be two on the crisp sets and respectively. the collection is a for some product soft topologies , where and are soft -open sets in their corresponding .
Proof. Let and be two on the crisp sets and respectively. Suppose be the soft product topology.
where is soft -open and is soft-open We need to prove is a t se for some soft tgy on . To prove that . Clearly, this implies that . Next, let where is soft -open -open and suppose . To prove that there exists -open -open such that such that that is this implies This implies that implies that , implies that , . On taking =,= this implies . , this implies that , this implies that , This implies that , this implies that Thus we have shown that there exists such that . Remaining to prove that Let be arbitrary distinct points. implies that implies that implies that this implies that , this implies that . This implies . Thus is a base. □
Theorem 6. Let and be two on the crisp sets and respectively. Let and be soft bases for and respectively. Let be the soft product space.
Then is a soft base for the product soft topology relative to soft -open sets.
Proof. Then is a soft base for the topology We have to prove that is a soft base for . According to the soft base, for any where is soft -open relative implies there exists a soft -open sets and soft-open set such that such that . Again implies that Applying the definition of a soft base, this implies that there exists such that
implies that there exists such that . Now this implies that . Now maxing (1) and (2) implies that there exists such that or . By definition, this proves that is a soft base for relative to soft -open sets. □
Theorem 7. Let and be two on the crisp set and respectively. Suppose and be soft subbases for and respectively. Then the coon of all soft sets of the rm and is a soft subbas for the product soft here
Proof. In order to prove that is a soft subbase for we have to prove that the soft collection of -open sets of the finite intersection of soft members of form a soft base for Since the intersection of empty soft subcollection of and so . Next, we suppose that
be a non-empty infinite soft subcollection of . This soft intersection of these elements belong to , by the construction of This elements of is .
= . We supposed that is a soft base for generated by the elements of and is a soft base for generated by the elements of . Since the finite soft intersection of soft subbase form, the soft base for that soft topology. In view of the above statements,, . Hence of is expressible as . Then is soft -open such that is a soft base for But is obtained from the finite soft intersection of members of . It follows that is a soft subbase for □
Theorem 8. Let and be two on the crisp set respectively. Let be the product space. Then, the soft projection maps and are soft continuous, and soft -open.
Proof. Suppose be a product space of and Then Define soft maps such that such that for all Then and both are called soft projection maps on the soft first and second coordinate spaces, respectively.
Step (1) First, we show that is soft continuous. Let be arbitrary soft -open set. soft -open set in This implies is soft continuous. The proof runs on similar lines to show that is soft continuous.
Step (2) To prove that the soft projection maps are soft -open maps. Let be an arbitrary soft s-open set. Let [ be arbitrary. This means that there exists such that this implies that because Now. Let be the base for the soft topology Then by definition of a soft base, implies that there exists a soft s-open set and soft s-open set such that such that implies that for This proves that is a soft interior point of . But is an arbitrary soft point of. Therefore, every point of is a soft interior point. This proves that is soft -open in This proves that the soft map is the soft -open map. A proof can be written on the same lines to show that is a soft -open map. Consequently, projection maps are soft -open maps. This finishes the proof. □
Theorem 9. Let and be two on the crisp set and respectively. be the product space, , be the soft projections maps on the first and second coordinate spaces, respectively. Let be another soft bitopological space such that . Then is soft continuous and are soft continuous maps.
Proof. Let be the soft product topological space. Let be another soft bitopological space. Let be the soft base for the soft product topology Now , are also soft maps. Let be soft continuous. Now and are continuous soft maps. Conversely, suppose that and are continuous soft maps. To show that is continuous. Let be arbitrary soft -open set. If we prove that is soft - in , the result will follow. Let be an arbitrary, then therefore, is an element of and hence it can be taken as by definition of the soft base, there exists a soft s-open set and soft - set and such that such that implies that
and this implies that and . For Similarly. ) = Similarity,) =. Thus, ) = (y) =. From the above this implies that and implies that . Therefore, are given to be soft continuous and hence and are soft -open in this implies that is soft -open in On taking ,. We have are soft -open in by the above result, (say). Any, implies that and implies that implies that implies that and implies that ] and ] implies that and Therefore, any implies that . This implies that . Thus we have shown that any implies that there exists a soft -open set such that This implies that is a soft interior point of and hence every point of is a soft interior point, showing thereby is soft -open in □
6. Attachment of Separation Axioms with Soft Connectedness and Soft Coordinate Spaces
In this section, the characterization of soft separation axioms with soft connectedness is addressed with respect to semiopen sets in soft bitopological spaces. In addition to this, the product of two soft topological spaces is ( space if each coordinate space is soft space, a product of two sot topological spaces is (S-regular and S-C regular) space if each coordinate space is (S-regular and S-C regular), a product of two soft topological spaces is connected if each coordinate space is soft connected and the product of two soft topological +spaces is (first-countable, second-countable) if each coordinate space is (first-countable, second-countable).
Theorem 10. Let and be two on the crisp set and respectively. be the soft product space, then the product space is soft -connected if both and are soft -connected.
Proof. Suppose be a product space of soft bitopological spaces and If the soft product is soft -connected then we have to prove that and are soft -connected. Define soft maps such that such that then and are called soft projection maps. Also Now is soft continuous and is soft -connected.
This implies that is soft -connected that is is soft S-connected. Again is continuous and is soft -connected. This implies For what we have done, it follows that is soft -connected sets. Conversely, let be the product space of soft -connected spaces We have to prove that is soft -connected. Pick any point and consider the soft sets and . Define the soft maps by requiring that for all and by writing for all . Now and are continuous soft maps so that and are soft -connected sets that are and are soft -connected sets. Write Then and are soft -connected sets. So that being a finite soft union of soft -connected sets having non-empty intersection. is soft -connected that is, is soft -connected. Consider the family of soft -connected sets that is this implies that . Finally, If we select fixed members of the above family, then their soft intersection is not soft null. Hence, is soft -connected. □
Theorem 11. Let and be two on the crisp set and respectively. be the soft product space, then the product space is soft each soft coordinate space is soft .
Proof. Suppose each coordinate space is soft and let be an element of . Then, for each . Since is soft space, it follows that is soft closed for each Now, each soft projection mapping being soft continuous, it follows that is soft closed is for every . Consequently, So, every singleton soft subspace of is soft closed. Hence, is soft . Conversely, let is soft and let be an arbitrary soft coordinate space of Let and be any two soft distinct points of Choose and in whose - coordinate is and respectively. Since we have or or or But being soft , corresponding to the soft distinct points and of there exists soft -open sets and in such that but and but . So, there exists basic soft -open sets and such that and . Clearly, and . Thus, is soft -open set containing but not ; and, is soft -open set containing but not . This shows that is soft . □
Theorem 12. Let and be two such that they are soft on the crisp set and respectively. be the soft product space, then the product space is soft each soft coordinate space is soft .
Proof. Suppose each soft coordinate space is soft and let and be two points of such that Then,