3. Infra Soft Connected Spaces
Through this section, we introduce and characterize the concept of infra soft connected spaces. We demonstrate that the family of infra soft connected sets is closed under a union operator provided that their intersection is non-null. Furthermore, we show that the image of an infra soft connected set is preserved by injective infra soft continuous maps, and prove that the finite product of infra soft connected spaces is infra soft connected. With the aid of examples, we elucidate that there is no relationship between infra soft topology and its parametric infra topologies with respect to possessing the property of infra soft connectedness.
Definition 19. We called and two separated soft subsets of an ISTS provided that and .
It is obvious that the two separated soft sets are disjoint; the next example clarifies that the converse is incorrect.
Example 1. Let be an infra soft topology on with as a set of parameters. Now, the two soft sets and are disjoint. On the other hand, and , . Hence, they are not separated.
Definition 20. An ISTS is said to be infra soft disconnected if can be expressed as a union of two non-null separated soft sets, say, and . Otherwise, is said to be infra soft connected.
We called and disconnection soft sets of .
An ISTS given in Example 1 is infra soft connected. In the following, we present an example of an infra soft disconnected space.
Example 2. Let be an infra soft topology on with as a set of parameters. Now, the two soft sets and are separated and their union is . Hence, is infra soft disconnected.
Proposition 3. The next statements are identical:
- (i)
is infra soft connected;
- (ii)
Two non-null disjoint infra soft closed sets do not exist such that their union is ;
- (iii)
Two non-null disjoint infra soft open sets such that their union is do not exist;
- (iv)
The only infra soft open and infra soft closed subsets are and Φ.
Proof. : Consider and as infra soft closed sets such that and . Then, and . Therefore, and are separated soft sets. This contradicts , which means that is true.
: Consider and as infra soft open sets such that and . Then, and are infra soft closed subsets. However, we obtain a contradiction with . This means that is true.
: Consider as a proper non-null infra soft open and infra soft closed set. Then, and . However, we obtain a contradiction with . This means that is true.
: Suppose that there is an infra soft open and infra soft closed set . Then, and . Therefore, and are separated soft sets such that their union is . Consequently, is infra soft disconnected. Hence, we obtain the desired result. □
Proposition 4. Every extended infra soft topological space is infra soft disconnected.
Proof. Let be an extended infra soft topological space. Without loss of generality, consider . Now, the two soft sets and are non-null infra soft open such that . According to Proposition 3, is infra soft disconnected. □
Proposition 5. Let and be two ISTSs. Then:
- (i)
If is infra soft connected such that , then is infra soft connected;
- (ii)
If is infra soft disconnected such that , then is infra soft disconnected.
Proof. (i): Let be infra soft connected. According to Proposition 3, the only infra soft clopen subsets of are and . Since , we obtain that and are the only infra soft clopen subsets of . Hence, is infra soft connected.
(ii): Let be infra soft disconnected. Suppose that is infra soft connected. Then, the only infra soft clopen subsets of are and . This implies that and are the only infra soft clopen subsets of (because ). This contradicts our assumption. Hence, is infra soft disconnected. □
Definition 21. A subset of is said to be infra soft disconnected if there exist two non-null separated soft sets such that their union is . Otherwise, is said to be infra soft connected.
Proposition 6. Let be a subspace of . Consider and are infra soft closure operators in and , respectively. Then, for each .
Corollary 1. A soft subset of is infra soft connected if is infra soft connected.
Proof. Let
and
be disconnection soft sets of
. Then, the proof comes from the fact that:
□
Theorem 2. If is an infra soft connected set such that , then is infra soft connected.
Proof. Let the given conditions be satisfied. Suppose that is infra soft disconnected. Then, there are disconnection soft sets and of . That is, . Since , and are disjoint such that . It follows from the infra soft connectedness of that or . For example, . This means that ; consequently, . Therefore, . Thus, ; this is a contradiction. Hence, is infra soft connected. □
Definition 22. The difference between infra soft closure points and infra soft interior points of a subset of is called infra soft boundary points. It is denoted by .
Proposition 7. If is a non-null proper subset of an infra soft connected space , then .
Proof. Suppose that . Then, . By hypothesis , then is an infra soft open and infra soft closed set. This is a contradiction with the infra soft connectedness of . Hence, , as required. □
Lemma 1. Consider and as disconnection soft sets of . If is an infra soft connected subset of , then or .
Proof. Since
and
are disconnection soft sets of
,
and
. So:
Note that
. Since
is infra soft connected, we obtain either
or
. This automatically implies that
or
, as required. □
Theorem 3. If is a subset of such that for each , there is an infra soft connected subset of containing , then is infra soft connected.
Proof. Let the given conditions be satisfied. Suppose that is infra soft disconnected. Then, there are disconnection soft sets and of . Therefore, there are two soft points such that and . By hypothesis, there is an infra soft connected set containing such that . It comes from the above lemma that or . This leads to . This is a contradiction because and are disconnection soft sets of . Hence, is infra soft connected. □
Corollary 2. If is a union of infra soft connected sets such that , then is infra soft connected.
Proof. Let the given conditions be satisfied. Suppose that is infra soft disconnected. Then, there are disconnection soft sets and of . By hypothesis , there exists a soft point such that for each i. Now, either or . Say, . Then, . It comes from Theorem 3 that for each i. This means that . This is a contradiction. Hence, is infra soft connected. □
Proposition 8. The injective infra soft continuous image of an infra soft connected set is infra soft connected.
Proof. Consider as an infra soft continuous map and let be an infra soft connected subset of . Suppose that is infra soft disconnected. Then, there are disjoint infra soft open sets and such that . Now, and are two disjoint infra soft open subsets of . Since is injective, . This means that is infra soft disconnected; a contradiction. Hence, is infra soft connected. □
Corollary 3. The property of being an infra soft connected space is an IST property.
Theorem 4. The finite product of infra soft connected spaces is infra soft connected.
Proof. Let and be two infra soft connected spaces. Suppose that is infra soft disconnected. Then, there exist infra soft open subsets and of such that and . Now, , and or . This implies that is infra soft disconnected or is infra soft disconnected. This contradicts the assumption. Hence, is infra soft connected. □
The next examples illustrate that the property of infra soft connected space does not transmit from an ISTS to classical infra topological spaces and vice versa.
Example 3. The family is an infra soft topology on with a set of parameters , where:
;
;
and
.
Obviously, is an infra soft disconnected. On the other hand, and are infra connected.
Example 4. The family is an infra soft topology on with a set of parameters , where:
;
;
and
.
Obviously, is infra soft connected. On the other hand, and are infra disconnected.
Proposition 9. A stable ISTS is infra soft connected if is infra connected.
Proof. Straightforward. □
Remark 1. Note that the property of being an infra soft connected space is not preserved by a subspace; i.e., it fails to be an ISH property. To validate this matter, take a subset of an ISTS given in Example 4. Then, , , . It is clear that and are disconnection infra soft sets of a subspace . Hence, is infra soft disconnected in spite of being infra soft connected.
4. Infra Soft Locally Connected Spaces
This section is allocated to present the concept of infra soft locally connected spaces and explore their main properties. We characterize it and show that the concepts of infra soft connected and infra soft locally connected spaces are independent of each other. Furthermore, we prove that the property of being an infra soft locally connected space is an infra soft open hereditary and IST property. Finally, we show that the finite product of infra soft locally connected spaces is also infra soft locally connected.
Definition 23. The two soft points are called connected in an ISTS if there exists an infra soft connected subset containing them.
Proposition 10. If every two soft points in an ISTS are connected, then is infra soft connected.
Proof. Let be a fixed soft point in an ISTS . Then, for each soft point different than ( if or ), we have an infra soft connected set containing and . Since , it follows from Corollary 2 that is infra soft connected. □
Definition 24. An ISTS is said to be:
- (i)
Infra soft locally connected at if for every infra soft neighborhood of there exists an infra soft neighborhood of such that every two soft points in are connected in .
- (ii)
Infra soft locally connected if it is infra soft locally connected at every soft point in . Otherwise, it is said to be infra soft locally disconnected.
The concepts of infra soft connectedness and infra soft locally connectedness are independent from one another. The next two examples explain this fact.
Example 5. An ISTS given in Example 3 is an infra soft locally connected space, but not infra soft connected.
Example 6. The topologist’s sine curve is an example of an infra soft connected space which is not infra soft locally connected.
We recall that we called an infra soft connected neighborhood of a soft point if it is infra soft connected and infra soft neighborhood of ; i.e., there exists an infra soft open set such that .
Theorem 5. An ISTS is infra soft locally connected at if every infra soft neighborhood of contains an infra soft connected neighborhood of it.
Proof. Necessity: Let be infra soft locally connected at . Consider as an infra soft neighborhood of . Then, there exists an infra soft neighborhood of such that every two soft points in are connected in . For each soft point in , there exists an infra soft connected set containing and . Putting . Now, ; according to Corollary 2, is an infra soft connected neighborhood of . Thus, we obtain the desired result.
Sufficiency: It comes from Definition 24. □
We recall that an ISH property is called an infra soft open hereditary property if it passes from an ISTS to every infra soft open subspace.
Theorem 6. The property of being an infra soft locally connected space is an infra soft open hereditary property.
Proof. Let be an infra soft open subspace of an infra soft locally connected space . Let be an infra soft neighborhood of in . Then, there exists an infra soft open subset of such that . Since is an infra soft open set, is an infra soft open set in ; consequently, is an infra soft neighborhood of in . By hypothesis of the infra soft local connectedness of there exists an infra soft neighborhood of in such that every two soft points in are connected in . Now, is an infra soft neighborhood of in such that every two soft points of it are connected in . Hence, the proof is complete. □
Proposition 11. The property of being an infra soft locally connected space is an IST property.
Proof. Consider as an infra soft homomorphism map such that is an infra soft locally connected space. Let be an infra soft neighborhood of . Then, is a soft point and is an infra soft neighborhood of . By hypothesis of the infra soft locally connectedness of , there exists an infra soft connected neighborhood of . It follows from Proposition 8 that is an infra soft connected neighborhood of such that . Thus, is infra soft locally connected. Hence, we obtain the desired result. □
Theorem 7. The finite product of infra soft locally connected spaces is infra soft locally connected.
Proof. Let and be two infra soft locally connected spaces. Let be an infra soft neighborhood of a soft point in . Now, , and , where and are, respectively, infra soft neighborhoods of soft points and . By hypothesis of the infra soft locally connectedness of and , there exist infra soft connected neighborhoods and of soft points and , respectively, such that and . It follows from Theorem 4 that is an infra soft connected neighborhood of in . Hence, is infra soft locally connected. □
5. Components in the Frame of Infra Soft Topological Spaces
In this part, we explore the concept of components in the frame of ISTSs and elucidate that the family of all components forms a partition for an ISTS. We show some properties of components in soft topology that are invalid in infra soft topology. Furthermore, we determine the conditions under which the number of components is finite or countable as well as discuss under what conditions the infra soft connected subsets are components. Finally, we investigate the image of the components under infra soft homomorphism maps.
Definition 25. A component of corresponding to is the union of all infra soft connected subsets of which contain . It is denoted by . That is:
and is infra soft connected }.
Remark 2. - (i)
According to Corollary 2, every component of a soft point is the largest infra soft connected set containing this soft point;
- (ii)
If is infra soft connected, then is the only component of each soft point.
We present the next example to show the method of calculating a component of each soft point.
Example 7. The family is an infra soft topology on with a set of parameters , where:
;
;
;
,and
.
Obviously, is an infra soft disconnected space. Now, we have the following:
, and .
Theorem 8. The family of all components of any ISTS forms a partition for it.
Proof. Consider as a family of all components of an ISTS . It is clear that . Suppose that there are two distinct soft points such that . According to Corollary 2, is an infra soft connected set. This contradicts that and are the largest infra soft connected sets containing and , respectively. Thus, . Hence, we obtain the desired result. □
The topological result reports that every closed component is one of the results that is invalid in ISTS.
Proposition 12. For any component of an ISTS we have .
Proof. Since is infra soft connected and , it follows from Theorem 2 that is an infra soft connected set as well. However, is the largest infra soft connected set containing ; however, . □
In Example 7, the component corresponding to is an infra soft open set because it is a member of . However, it is not an infra soft closed set because its complement is not infra soft open.
Theorem 9. If the components of an infra soft compact space are infra soft open, then the number of them is finite.
Proof. Let be infra soft compact. Suppose that the number of components of is infinite. Now, every component is infra soft open and forms a cover of . By hypothesis, . This implies that there are some members of the cover that have a non-null intersection; this is a contradiction. Hence, has a finite number of components. □
Corollary 4. If the components of an infra soft Lindelöf space are infra soft open, then the number of them is countable.
Theorem 10. If an ISTS has a finite number of components which are infra soft closed, then they are also infra soft open.
Proof. Consider as an ISTS with a finite number of components which are infra soft closed. As we previously proved that the family of components forms a partition for . Since every component is infra soft closed, the union of any number of them is infra soft closed. Since is infra soft closed, then is infra soft open. This completes the proof. □
Theorem 11. If is a non-null infra soft clopen and infra soft connected subset of , then is a component.
Proof. Let the given condition be satisfied. Suppose that is infra soft connected such that . Now, . Since is a non-null infra soft clopen set, and . Therefore, and are disconnection soft sets of . This contradicts the assumption. Hence, is a component. □
Proposition 13. The infra soft homeomorphism image of a component is also a component.
Proof. Consider as an infra soft homeomorphism map and let be a component of . Suppose that is not a component of . This means, according to Proposition 8, that there is an infra soft connected subset of such that . Since is an infra soft homeomorphism, . Now, is an infra soft connected subset of which contradicts that is a component. Hence, is a component. □