3. Semi-Continuous Functions
In this section, we list and prove some properties between kinds of nowhere dense sets and second-category sets in an sGTS. These properties are useful for checking whether a set is -nowhere dense ( -nowhere dense) or not in a strong Baire space. In addition, the meaning of the collections and in hyperconnected spaces are examined. We give the necessary condition to explore whether or not a function is in , . Furthermore, we find new results for these functions in a Baire space. We start by showing that
Lemma 5. If is a sBS, then
Proof. Let . Then, , where Take for some k such that Since each with and is a sBS, is of μ-II category. Since and superset of μ-II category is of μ-II category, we have that Therefore, . □
Example 1 below shows that the hypothesis in Lemma 5 cannot be neglected.
Example 1. Let and Take and Then , but So that, is not a sBS. Here, However,
Moreover, is not closed under finite intersection if is an sBS as shown in the following Example 2.
Example 2. Consider the GTS , whereand . Then is a sBS andknowing that is the collection of subsets of Now, if and then , but Next, we obtain some inclusion between the sets ; and -residual, -residual sets, in an sBS.
Theorem 1. If is a sBS, then:
Proof. 1. Assume that , whereby by Lemma 1 it follows that and so Suppose that , whereby . Since , there is such that
As , by Lemma 5, we have that Since superset of -II category set is of -II category, we have that ; however, this is not possible. Therefore, . Items 2., 3. and 4. are proved in a similar manner. □
Example 3 below shows that the converse part of 1., in Theorem 1 is not true.
Example 3. Consider the GTS , where: Thus, is a sBS and Now, take so and so H is a -nowhere dense set in However, , for that H is not a μ-nowhere dense set in
The same results of Theorem 1 instead of an sBS are obtained in a GTS, but with an additional property.
Theorem 2. Let be a GTS such that , where with Thus, the items of Theorem 1 are given.
Proof. It is enough to prove that Let and suppose that . Note that , and moreover because . Therefore, there exists such that for that where Take for some k such that , whereby and so and by hypothesis, Thus, which is not possible. Therefore,
The rest are tested in a similar manner. □
Example 4 shows that the necessary condition in Theorem 1 cannot be dropped.
Example 4. Consider a GTS , where: Let so , whereby is not a BS. Therefore, is not an sBS. Now, Take so , but This is
Next, we prove other inclusions but for sets , ; , ; , , and -residual, -residual sets, in a Baire space.
Theorem 3. If is a Baire space, then:
Proof. We give the proof for 1.. Let so Assume that . As then so there exists a set such that
As, so is of -II category set, since super set of -II category is of -II category. However, whereby Hence Therefore, The rest are tested in a similar manner. □
Example 5 below shows that the reverse implication of Theorem 3 part 1., is not true.
Example 5. Consider the generalized topological space , where and Then, is a BS and we get that, Let So that Therefore, H is a -nowhere dense set in However, , so that H is not a μ-nowhere dense set in
The following Theorem 4 gives the relations between different types of subsets with respect to the generalized topologies and in a strong Baire space.
Theorem 4. Let be an sBS and sGTS. Then:
Every -residual set is -residual set.
Proof. It is enough to prove 1.. Let By hypothesis and Lemma 3, so that by Theorem 1 and the fact that is a sBS. □
Next, Example 6 shows that the condition “ is an sBS” cannot be omitted in Theorem 4.
Example 6. Here, is not an sBS. Take and so and , but Choose it turns out , so that , but and are -II category sets, which implies Hence,
The below Theorem 5 directly follows from Lemmas 3 and 4.
Theorem 5. Let be a μ-II category space. Then:
Every -meager is μ-meager.
Every -residual is μ-residual.
Every μ-II category set is of -II category.
Now, we obtain some results between the generalized topologies, semi-continuous functions, and the hyperconnected condition.
Theorem 6. If is a hyperconnected space where then J is η-residual for every
Proof. Let . As is a hyperconnected space so Hence, and so since is -closed. Hence, Thus, J is a -residual set in □
The following Example 7 shows the necessity of hyper-connectedness in Theorem 6.
Example 7. Consider the generalized topological space being So:
is not a hyperconnected space. Because , but Take thus and Here is of μ-second category set, so that Thus, W is not μ-residual.
Obviously, is not a hyperconnected space. Take , then and Let , thus and Furthermore, is of -II category, so that is of -II category. Therefore, K is not -residual.
Withwe have that is not a hyperconnected space. Here , but Take then and but is of -II category. This implies is not in , which implies that O is not -residual.
The reverse implication of Theorem 6 is not true as shown by the following Example 8.
Example 8. Let and Here, each is a μ-residual set in Let , so and , but Therefore, H is not a μ-dense set in X and hence is not a hyperconnected space. In the same manner, we can prove that is not a hyperconnected space if each is a η-residual set in X, where .
The following Example 9 shows that if is a BS, then it is not necessarily a hyperconnected space.
Example 9. Consider the generalized topological space where Then is a BS but not a hyperconnected space. Since, if then
Theorem 7. Let be a μ-II category space. If is hyperconnected, then is a BS.
Proof. Let By hypothesis and Theorem 6, Suppose By Lemma 4, Hence which is not possible. Thus, and is a Baire space. □
Next, in the rest of the section with a series of theorems in a space hyperconnected, or
-II category, or strong Baire, the essentials of lower (upper) semi-continuous functions are discussed. Further, the set theory relationship between
and
is analyzed. Finally, we study the notation
defined on [
11]. We show that every continuous function is in
.
The below Theorem 8 is an immediate consequence of Theorem 5 and by the definition of .
Theorem 8. If is a μ-II category space and , then
Remark 1. Since we have Furthermore, and Moreover, Theorem 9. If is a hyperconnected space, then
Proof. Consider a -lower semi-continuous function, and By assumption there is such that Since By hypothesis and Theorem 6, W is a -residual set. Hence h is a -lower semi-continuous function at and hence h is a -lower semi-continuous function on Similar considerations apply to the case of -upper semi-continuous function, we get h is a -upper semi-continuous function on Therefore, □
The proof of the following result is similar to that of Theorem 9.
Theorem 10. If is a hyperconnected space and then
The following Example 10 shows that the condition “ is a hyperconnected space” can not be dropped in Theorem 9.
Example 10. If , then H is not a -dense set in Therefore, is not a hyperconnected space.
Define a function by For any real number for all there exists such that Therefore, f is a μ-lower semi-continuous function. Take and choose for which If and then Now, So, and are -nowhere dense sets in Hence, and are -meager sets in X and A and B are the -residual sets in However, and Therefore, f is not a -lower semi-continuous function.
Define a function by For any real number and for all there exists such that Therefore, g is a μ-upper semi-continuous function. Now, , choose If and then By similar manner, A and B are -residual sets in However, Therefore, g is not a -upper semi-continuous function.
Theorem 11. Let be a sBS and . Then:
Proof. The proof directly follows from the facts that in a strong Baire space, and every -residual set is -residual. □
Theorem 12. Let be μ-II category space. If is hyperconnected, then
Proof. By Theorem 5 part 2., each -residual set is a -residual set. If is a -lower semi-continuous function, and , then there is (with ) such that W is -residual, whereby W is -residual. We deduce that h is a -lower semi-continuous function on Similar considerations apply to the case of -upper semi-continuous function. We conclude that □
The following Example 11 shows the necessity of hyper-connectedness in Theorem 12.
Example 11. Take and g as given in Example 10. Then,and so is not a hyperconnected space. Here, for any real number and for all there exists a set such that Therefore, f is a -lower semi-continuous function. For and , we get that If and then Furthermore, A and B are μ-residual sets. However, Therefore, f is not a -lower semi-continuous function. In the same manner, we can prove that g is a -upper semi-continuous function but not a -upper semi-continuous function. Theorem 13. Let be a sBS. If is hyperconnected, then
Proof. Consider a -lower semi-continuous function and , . So, by definition and Lemma 5, we have a set such that , then by Theorem 6, P is a -residual set. We deduce that h is a -lower semi-continuous function on Similarly, it is shown that h is a -upper semi-continuous function. □
Theorem 14. Let be an sGTS, sBS and Then:
Proof. 1. Let Assume that h is a -lower semi-continuous function. Let and Then, there is being a -residual set such that By hypothesis and Theorem 4 part (4), Q is a -residual set. Thus, h is a -lower semi-continuous function at Since is an arbitrary point of h is a -lower semi-continuous function on Similar way apply to the case of upper semi-continuous function.
2. Assume that, By 1., and the proof is completed. □
Proceeding similarly to the previous demonstration and applying Theorem 3 part 4., we obtain the following result.
Theorem 15. Let be a BS. If , then
In generalized topological space, every -l.(u.)s.c. function is a -l.(u.)s.c. function. The below Theorem 16 shows a fact for the reverse implication of the above statement.
Theorem 16. Let be a μ-II category space. If is hyperconnected, then:
Proof. 1. Let Assume that h is a -lower semi-continuous function. By hypothesis and Theorem 10, h is -lower semi-continuous function. Let and so there is being -residual set, such that . By hypothesis and Theorem 7, it turns out that is a BS, and by Theorem 3 part 4., we have that L is a -residual set. Therefore, h is a -lower semi-continuous function at and hence h is a -lower semi-continuous function on Similar considerations apply to the case of upper semi-continuous function.
2. Assume that so by 1., , as the test is followed. □
The following Example 12 shows that the condition “hyperconnectedness” on can not be dropped in Theorem 16.
Example 12. Consider generalized topological space and the functions , as in Example 10. Then,and is not a hyperconnected space. Now: Clearly, f is a μ-lower semi-continuous. function on X and Choose If and then Now; Therefore, and are -nowhere dense sets in X and so and are -meager sets in Hence, U and V are -residual sets. However, we have that Therefore, f is not a -lower semi-continuous function. In the same manner, we can prove that g is a μ-upper semi-continuous function but not a -upper semi-continuous function.
Choose . If and then Furthermore, U and V are -residual sets. However, Therefore, f is not a -lower semi-continuous function. In the same manner, we can prove that g is a μ-upper semi-continuous function but not a -upper semi-continuous function.
Since it has been proved in a BS, every -residual is -residual. So, the following Theorem 17 gives the relationship between -l.(u.)s.c. functions and -l.(u.)s.c. function in a BS.
Theorem 17. If is a Baire space, then:
If is a sGTS, then
Proof. We give the detailed proof only for 3.. By hypothesis, X is of -II category. Consider and assume that h is a -lower semi-continuous function. Choose is an arbitrary point and , this implies that there is being -residual such that . By Theorem 3 part 4., Q is -residual. Hence h is a -lower semi-continuous function on
The reverse inclusion follows directly from the same above argument and the fact that in a -II-category space, every -residual set is -residual. Apply similar considerations in the case of the upper semi-continuous function. □
For the following theorem, we consider that
Theorem 18. If be a BS, then .
Proof. We present a proof for lower semi-continuous function. Let and consider that h is a -lower semi-continuous function, and Then, there is such that Now, , where for to Since X is a BS, we have that , so that for to , which implies that Therefore, h is a -lower semi-continuous function at , and hence h is a -lower semi-continuous function on □
Corollary 1. Let be a BS, Then the following set inclusions are true.
An interesting result states that every continuous real-valued function is a -l.(u.)s.c. function in a GTS.
Theorem 19. Let be a GTS. If is a continuous map, then
Proof. For and , we have that Since h is continuous on h is continuous at Therefore, there is such that Hence, h is a -lower semi-continuous function at Since is an arbitrary point of h is a -lower semi-continuous function on By similar argument, we can prove h is a -upper semi-continuous function on □
4. Characterizations of Cliquish Functions
In this section, cliquish functions on generalized metric spaces are examined, and several properties of these are obtained. Moreover, we characterize cliquish functions using nowhere dense sets in Baire spaces. With these results, we conclude that the set of all discontinuity points of a real valued function is either a meager set or not, but in a generalized topological space. Finally, we propose some results to examine whether a function is cliquish or not.
The pair
is called a
generalized metric space [
18] (briefly, GMS) if
is a metric on
Denote
, the family of
Ω-open sets [
18] in
more precisely,
if and only if for each
there is
and
such that
, where
and
means domain space of
So, the pair
is a generalized topological spaces.
The following Theorem 20, is a simple method to explore whether or not a set is residual in a generalized metric space, thus reducing the computational complexity.
Theorem 20. Let be a GTS and Q be a η--set, where If then Q is a η-residual set in
Proof. Assume that is a -set. Then, , where for every As, we have that each , whereby Since each , we get that is -closed for each . Thus, each , whereby Therefore, Q is a -residual set. □
The following Example 13 shows that the condition “” cannot be neglected in Theorem 20.
Example 13. Let and , where Then, is a generalized topological space.
Clearly, , but Choose , thus As , we have that Q is a --set, but Q is not -dense in Clearly, contains a -II category set, so that is in Therefore, Q is a not -residual set.
Choose and Then and Furthermore, Since, and , we get Here, , which is of -II category, so that is not a -residual set.
Take , then and Now, and obviously, is of -II category set. Hence, K is not -residual.
In a generalized metric space, we encounter some difficulty in analyzing the meaning of a set of continuity points of a given function. Theorem 21 below easily concludes the nature of this set.
Theorem 21. Let be a GMS. If is a function, then , where for every n in (that is, is a --set in X).
Proof. Let
there is
such that
, whenever
Thus,
is
-open for every
Assume that
h is continuous at
whereby for every
there is
such that
whenever
Take
with
Thus,
since
for every
n in
Therefore,
, and
Conversely, assume that
. Let
, so there is a positive integer
m such that
Since
, we get
This implies that there is
such that
whenever
by definition of
Since
it turns out
Hence,
h is continuous at
y and
Therefore,
From Equations (1) and (2), , where for all □
Theorem 22 gives a shortcut for finding the significance of the set of all discontinuity points of a given function in a generalized metric space.
Theorem 22. Let be a GTS, and be a -cliquish. Then
Proof. We present the proof only for Assume that h is -cliquish. By Theorem 21, it turns out that, and --set in X. Since , so is --set in Thus, is --set in By Theorem 20, is -residual and therefore, □
The following two theorems are considered via a new strategy, such that:
Theorem 23. To easily explore the meaning of a collection of all discontinuity points from a given function.
Theorem 24. To check whether the given function is cliquish from the set of all discontinuity points of that function.
Theorem 23. Let be a BS. If h is a -cliquish function, then is -meager and also -meager.
Proof. Assume that h is -cliquish, so is -dense in X, by Theorem 21 is obtained that is a --set. Since is a BS, it turns out Hence, is --set in By Theorem 20 it follows that is -residual. Thus, is a -meager set.
As and is -dense in X, so is a -dense set and a --set in By Theorem 20, it is obtained that is -residual. Therefore, is a -meager set. □
Theorem 24. Let be a μ-II category GTS, and be a map. If is -meager, then h is -cliquish.
Proof. Assume that is -meager. Let , then K is of -II category. By Theorem 5 part 3., it follows that K is of -II category. So, Therefore, is a -dense in X and h is -cliquish. □
Theorem 25 reduces the complexity for finding whether a given function in a generalized metric space is -cliquish.
Theorem 25. Let be a sGTS, BS. If h is -cliquish, then, h is -cliquish.
Proof. Assume that h is a -cliquish function. By Theorem 22, is a -meager set. Hence, by Theorem 3 part 2., is a -meager set. Since is a sGTS and BS, it turns out that is a GTS of -II category. By Theorem 24, it follows that h is a -cliquish function. □
Definition 9. Let be a GTS and be a map. We denote by:
Next, Theorem 26 gives a characterization of the cliquish functions in terms of nowhere dense sets.
Theorem 26. Let be a GTS and If is a BS, then the following are equivalent.
There is a -cliquish function such that and
the sets are -meager and is -nowhere dense.
Proof. 1. ⇒
2. By hypothesis,
Let
and
Since
, it turns out that
Now;
Thus, Hence, and so is a -meager set, see Theorem 22. Similarly, it is proven that is a -meager set.
On the other hand, assume that
So, there is an element
and we get a set
such that
Since is a BS, G is of -II category and hence is of -II category set, by (3) and the fact that superset of II-category set is of II-category.
Furthermore,
since
Therefore,
is of
-II category set. However,
which implies that it must be
of
-II category set, which is not possible. Therefore,
, and so
is
-nowhere dense in
2. ⇒
1. The hypothesis implies that
, where each
is
-nowhere dense in
Define a function
as:
Thus, and Define , so by Lemma 4 the set is a -meager set. Let and assume that , so for Therefore, or and or So, we have four cases:
Case-1: Assume that and . Then, and there is such that for This implies that and for every . Thus, by (4) is . Since we get Hence, or or Neither of this is possible.
Case-2: If and by (4), . Since h is a constant function on t is a -continuity point of which is not possible.
Case-3: Assume that and . By (4) must be Since h is a constant function on t is a -continuity point of which is a contradiction.
Case-4: Suppose and . Then, , but
Therefore, none of the cases are possible, whereby . So, and is -meager, because is a BS. Hence, h is -cliquish. □
In a Baire space, the existence of a -cliquish function using -nowhere dense sets can be found straightforwardly by Theorem 27.
Theorem 27. Let be a GTS and If is a BS, then the following are equivalent.
There is a -cliquish function such that and
, and is -nowhere dense.
Proof. 1. ⇒ 2. By hypothesis, we have By Theorem 26 it is obtained that is -nowhere dense in Therefore, is -nowhere dense.
2. ⇒
1. Define the sets
,
. Thus,
,
and
By definition of
and
is clear that
and
Hence,
and so
is
-nowhere dense in
Now,
Thus,
is
-nowhere dense. Similarly,
is
-nowhere dense in
Therefore,
and
are
-meager. Furthermore,
Hence, and are -meager. By Theorem 26, there is a -cliquish function such that and Therefore, there is a -cliquish function such that and . □
Theorem 28 provides the easier route to finding the presence of -cliquish function using -meager sets.
Theorem 28. Let be a GTS and If is a BS, then the following are equivalent.
There is a -cliquish function such that and
and the sets are -meager in
Proof. 1. ⇒
2. By hypothesis,
Consider the function
So,
is
-cliquish and
By Theorem 26, it turns out that
is
-meager in
Furthermore,
Therefore, is -meager in With similar considerations we can prove is -meager in
2. ⇒ 1. The hypothesis implies that the set can be expressed as the union of countably many -nowhere dense subsets , . Similarly, where is a family of -nowhere dense subsets of
Define a map
as follows:
So, and Define , where and Thus, is -meager. Let and suppose that . Then, for all Thus, or and or So, we have four cases:
Case-1: Assume that and , Hence, and for which we get four cases.
and
and
and
Thus, all the cases are not possible and our assumption is not true.
Case-2: If and then by (5). Since h is a constant function on t is a -continuity point of which is not possible.
Case-3: Suppose and . Then, by (5). Since h is a constant function on t is a -continuity point of h, but t is a -discontinuity point of
Case-4: Assume that and Consider , thus Now, so In both cases, t is a continuity point of which is not possible.
Thus, all the cases are not possible. Hence, and so , whereby is -meager in Since is a BS, h is a -cliquish function on □
The following Theorem 29 provides an easier way to check the nature of the subsets of a domain space using the cliquish function.
Theorem 29. Let be a GTS and If is a BS and if there is a -cliquish function such that and then and the sets are -meager.
Proof. Let g be a -cliquish function such that and . Thus, By hypothesis and Theorem 28, we have that and are -meager set. Since and we have that and are -meager in □
The two Theorems 30 and 31 below provide shortcuts for finding the existence of -cliquish function in a sBS using -meager sets.
Theorem 30. Let be a sGTS which is sBS, and So, the following are equivalent.
There is a -cliquish function such that and
the sets are -meager and is -nowhere dense in
Proof. 1. ⇒ 2. Note that Take and define the set In the same way as in the demonstration of Theorem 26 part 1. ⇒ 2., we get By hypothesis and Theorem 23, which implies In a similar way we have that .
By hypothesis and Lemma 2, we get is a BS. With similar considerations in the proof of Theorem 26 part (1. ⇒ 2.), we get is in .
2. ⇒ 1. The hypothesis implies that where each
Define a function
as:
Thus, and Define . Thus, is in see Lemma 4. Let and assume that So, we obtain four cases as follows:
Case-1: and
Case-2: and
Case-3: and
Case-4: and
Note that by hypothesis and Lemma 2, is a Baire space. By the same arguments from the proof of Theorem 26 part 2. ⇒ 1., we get Hence Therefore, g is a -cliquish function. □
Theorem 31. Let be a sGTS which is sBS, and So, the following are equivalent.
There is a -cliquish function such that and
and the set is a -nowhere dense set in
Proof. 1. ⇒ 2. By hypothesis, , and by Theorem 30 it turns out Since the subset of a nowhere dense set is nowhere dense, we have that
2. ⇒ 1. With same considerations in the proof of Theorem 26 part 2. ⇒ 1., and using Theorem 30, we get the proof. □
The following Theorem 32 states that in order to investigate the existence of -cliquish function in a sBS, divide the domain into two disjoint sets and verify whether the particular subset of those sets is -meager or not.
Theorem 32. Let be an sGTS, which is sBS, and So, the following are equivalent.
There is a -cliquish function with and
and the sets are -meager in
Proof. 1. ⇒ 2. By hypothesis,
Define a function
Note that
is
-cliquish and
By Theorem 30, it turns out
Obviously,
Therefore, is in . Similarly, we show that is a -meager set in
2. ⇒ 1. Replace by in the proof of Theorem 28 part 2. ⇒ 1., and use the fact that is Baire. □
Theorem 33 states the significance of subsets of domain space, by checking whether the existence of -cliquish function or not.
Theorem 33. Let be a sGTS and If is a sBS and if there is a -cliquish function such that and then and the sets are -meager in
Proof. Let g be a
-cliquish function such that
and
. Thus,
By hypothesis and Theorem 32, it turns out
Here, and By (7) and the fact that the subset of a meager set is meager, the sets are in . □