Songklanakarin Journal of Science and Technology SJST R1 Al Ghour. L-open sets and L^*-open sets

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1 L-open sets and L^*-open sets Journal: Manuscript ID Songklanakarin Journal of Science and Technology SJST-0-0.R Manuscript Type: Date Submitted by the Author: Original Article 0-Oct-0 Complete List of Authors: Keyword: Al Ghour, Samer Abufarhan, Khaled; Jordan University of Science and Technology Lindelof, maximal Lindelof, LC space, generalized open set, continuous functions

2 Page of open sets and -open sets Samer Al Ghour and Khaled Abufarhan Department of Mathematics and Statistics Jordan University of Science and Technology Irbid 0, Jordan algore@just.edu.jo Abstract. As new classes of generalized open sets and generalized continuous functions, the notions of -open sets, -open sets, -continuous functions, - continuous functions and weakly -continuous functions are introduced and investigated. As main results it is proved that -open sets forms a topology and decomposition of continuity is obtained. AMS Subject Classification (00): C0. Key Words and Phrases: Lindelof; maximal Lindelof; LC space; generalized open set; continuous functions.. Introduction Let (, ) be a topological space and let. Denote the set of all countable (resp. Lindelof, compact) subsets of by (, ) (resp. (, ), (, )). A point is called a condensation point of if for each with, the set is uncountable. In, Hdeib defined -closed sets and -open sets as follows: is called -closed (Hdeib, ) if it contains all its condensation points. The

3 Page of complement of an -closed set is called -open. Denote the set of all -open sets in (, ) by. Authors in (Al-Zoubi & Al-Nashef, 00), proved that (, ) is a topological space, and { : and (, )} forms a base for. is called co-compact open (notation: coc-open set) (Al Ghour & Samarah, 0) if for every, there exists an open set and (, ) such that. Denote the set of all coc-open sets in (, ) by. Authors in (Al Ghour & Samarah, 0) investigated coc-open sets, proved that (, ) is a topological space, and { : and (, )} forms a base for and gave a decomposition theorem of continuity. It is not difficult to give an example to show that -open sets are not coc-open sets and vice versa, in general. The first main goal of this research is to introduce and investigate -open sets as a new class of sets that is strictly containing both -open sets and coc-open sets. The second main goal of this research is introducing and studying -open sets as a new class of sets that is containing strictly the open sets and give with the L-open sets a decomposition of open sets. The third main goal of this paper is defining three new classes of continuous functions via -open and -open sets. Throughout this paper, R,Q denote the set of real numbers, the set of rational numbers, respectively. For a subset of a topological space (, ), ( ) and ( ) will denote the closure of and the interior of, respectively. Also, we write to denote the relative topology on when is nonempty. For a nonempty set, we will denote the discrete topology on and the indiscrete topology on by τ and τ, respectively. Finally, τ will denote the usual topology on R.

4 Page of open sets Definition.. Let (, ) be a topological space and. A point in is in closure of ( ( )) if ( ) for any and (, ) with. is called -closed if ( )=. The complement of an -closed set is called an -open set. Denote the family of all -open subsets of (, ) by τ. The next result follows directly from Definition.: Proposition.. A subset of a topological space (X,τ) is -open if and only if for every, there exists an open set and (, ) such that. For a topological space (, ) denote the family of -open sets { : and (, )} by B ( ). Theorem.. Let (X,τ) be a topological space. Then the collection τ forms a topology on. Proof. By the definition one has directly that τ. To see that τ, let, take = and =. Then. Let, τ and let. For each =,, we find an open set and (, ) such that. Take = ₁ ₂ and = ₁ ₂. Then is open, (, ), and. It follows that is -open. Let { : } be a collection of -open subsets of (, ) and. Then there exists ₀ such that ₀. Since ₀ is -open, then there exists an open set and (, ), such that ₀. Therefore, we have ₀. Hence, is -open. Remark.. Let (, ) be a topological space. Then

5 Page of (a) The collection B ( ) forms a base for τ. (b) the collection { : (, )} forms a subbase for τ. (c) τ τ τ. Remark.. The inclusion in Remark. (c) is not equality in general, to see this, let =R and =. Then {} τ (τ τ ) and so τ τ τ. The following is an example of a topological space (, ) such that τ =τ and τ τ : Example.. Let =R and ={ } { : }. Then (,τ)={ : } { : and is finite} and (X,τ)={ : } { : and is countable}. Thus, τ = { : and is finite} and τ =τ { : and is countable}. Clearly that τ =τ and τ τ. Definition.. (Mukherji & Sarkar, ) A topological space (, ) is called an LC space if each Lindelof subset is closed. Theorem.. Let (, ) be a topological space. Then the following are equivalent: (a)(, ) is LC. (b) =B ( ). (c) =τ =τ. Proof. (a) (b) B ( ) is obvious. Let B ( ) where and (, ). Since (, ) is LC, then is closed and hence. It follows that B ( ).

6 Page of (b) (c) By Remark. (c), it is sufficient to see that τ. By Remark. (a), B ( ) is a base for τ. Thus by (b), τ. (c) (a) If (, ), then τ and by (c),. Therefore, is closed in. Proposition.. For any topological space (, ), (,τ ) is LC. Proof. Let (,τ ). Since τ, then (,τ ) (, ) and so (, ). Thus, we have τ, and hence is closed in (,τ ). Corollary.0. For any topological space (, ),(τ ) =τ. Proof. This is an immediate consequence of Theorem. and Proposition.. As defined in (Cameron, ) a topological space (, ) is maximal Lindelof if (, ) is Lindelof and there exists no strictly finer Lindelof topology on. Theorem.. (Cameron, ) A Lindelof topological space is maximal Lindelof if and only if it is LC. Corollary.. Let (, ) be a topological space. Then (,τ ) is maximal Lindelof if and only if (,τ ) is Lindelof. Proof. This is an immediate consequence of Proposition. and Theorem.. Corollary.. A Lindelof topological space (, ) is maximal Lindelof if and only if there is a topology on such that =. Proposition.. If (, ) is hereditarily Lindelof, then τ =. Proof. For every, { } is Lindelof and so { }= ( { }) B ( ) τ. It follows that τ =. As a conclusion of Proposition., for any second countable topological space (, ), τ =.

7 Page of The following is an example a nondiscrete topological space which shows that the converse of Proposition. is not true in general: Example.. Let =R and ={ } { R:Q }. It is not difficult to see that (, )={ : Q is countable}. For every, take =Q and =Q { }. Then (, ), and { }=. This shows that τ =. On the other hand, it is clear that (, ) is not hereditarily Lindelof. Theorem.. Let (, ) be a topological space and be a nonempty subset of. Then ( ). Proof. Let ( ) and. Then there exists and a Lindelof set such that. Choose such that =. Now implies that ( ).Hence,. Question.. Let (, ) be a topological space and be a nonempty subset of. Is it true that ( ) =? The following result is a partial answer for Question.: Theorem.. Let (, ) be a topological space and be a nonempty closed set in (X,τ). Then ( ) =. Proof. By Theorem., ( ). Conversely, let and. Choose such that =. Choose and (, ) such that. Thus, we have ( ) ( ), and (, ). It follows that ( ).

8 Page of Theorem.. If :(, ) (, ) is injective, open, and continuous, then :(, ) (, ) is open. Proof. Let = where and (, ) be a basic element for. Since is injective, ( )= ( ) ( ). Since :(, ) (, ) is open, ( ). And since :(, ) (, ) is continuous, ( ) (, ). This ends the proof In Theorem., the condition 'continuous' cannot be dropped as the following example shows: Example.0. Consider :(R, ) (R, ), where as in Example. and is the identity function. Then is bijective and open. On the other hand, since (R, ) is hereditarily Lindelof, we have ( ) =. Since by Example., :(R,( ) ) (R, ) is not open.. -open sets For any topological space (, ) and, it is clear that ( )= ( ). For simplicity, from now on we will use only ( ). Also, it is clear that ( ) ( ). The following example shows that ( ) ( ) in general. Example.. Consider (, )=(, ). By Proposition., =. Therefore, ( )(Q)=Q but (R)=R. The following definition is reasonable: Definition.. Let (, ) be a topological space and.

9 Page 0 of a. is called -closed if ( )= ( ). b. is called -open if is -closed. Theorem.. For a topological space (, ), we have the following a. and are -closed sets in (, ). b. Every closed set in (, ) is -closed set. c. Every finite union of -closed sets in (, ) is -closed set in (, ). Proof. (a) It is obvious. (b) Let be closed in (, ). Then ( ) ( )=. It follows that ( )= ( )= and hence is an -closed set in (, ). (c) Suppose { : } is a finite collection of -closed sets. Then for every, ( )= ( ). Thus, It follows that is -closed. ( ) = ( ) = ( ) = ( ). Corollary.. For a topological space (, ), we have the following a., are -open sets in (, ). b. Every open set in (, ) is -open set. c. Every finite intersection of -open sets in (, ) is -open in (, ).

10 Page of The following theorem characterizes -open sets: Proposition.. A subset of a topological space (, ) is -open if and only if ( )= ( ). Proof. is -open if and only if is -closed if and only if ( )= ( ) if and only if ( )= ( ) if and only if ( )= ( ) if and only if ( )= ( ). The union of two -open sets one of them is open need not to be -open in general, as the following example shows: Example.. Consider the topological space as in Example.. Let =(0, ) and =(,0). Then is open, also since ( )= ( )= {}, then is -open. On the other hand, ( )=R {0,} but ( )=R {0} which implies that is not -open Example. shows also that the converse of Corollary. (b) is not true in general. Let (, ) be a topological space. Denote the family of all -open sets in (, ) by B. By Corollary. (b), B. According to Example., B does not form a topology on in general. By Corollary., B forms a base for some topology on. Denote the topology on which has B as a base by τ.

11 Page of Definition.. (Henriksen & Woods, ) A topological space (, ) is called antilocally Lindelof if any Lindelof subset of has empty interior. If is the Sorgenfrey line and (, ) is the Cartesian product topological space, then (, ) is anti-locally Lindelof (see Corollary. of (Henriksen & Woods, )). Theorem.. Open sets in an anti-locally Lindelof topological space are -closed sets. Proof. Let (, ) be a topological space and. Suppose on the contrary that there is ( ) ( ). Since ( ), there exists such that and =. Choose and (, ) such that. Thus we have. Since ( ), then and hence ( ). This contradicts the assumption that (, ) is anti-locally Lindelof. Recall that a topological space (, ) is locally indiscrete if every open set in (, ) is closed. Corollary.. If (, ) is anti-locally Lindelof such that B =, then (, ) is locally indiscrete. Proof. Let be an open set in (, ). Then by Theorem., is -closed. So is -open and by assumption, is open. Thus, is closed.

12 Page of In Theorem. the condition 'anti-locally Lindelof' cannot be dropped as we can see by the following example: Example.0. Consider (, )=(R, ) as in Example.. Take =(0,), then ( )= but ( )= 0,. The following example shows that the converse of Corollary. (b) is not true in general, even if we add the condition 'anti-locally Lindelof' on the topological space: Example.. Let (, ) be the Cartesian product topological space where is the Sorgenfrey line, then as we mentioned above (, ) is anti-locally Lindelof. If every -open set in (, ) is open, then by Corollary., (, ) is locally indiscrete. We are going to show that (, ) is not locally indiscrete. Let =(0, ) R. Then, on the other hand, if is closed, then = (,0 R and so there exist,,, R such that (0,0), ), ) (,0 R which implies that >0 and (( /),0) (,0 R, which is not true. It follows that (, ) is not locally indiscrete. The following is a decomposition theorem of openness in terms of L-openness and -openness: Theorem.. For any topological space (, ), = B. Proof. By Proposition. (c),, and by Corollary. (b), B. It follows that B. Conversely, let B. Since is -open, then ( )= ( ). Also, since A is -open then ( )=. It follows that ( )= and hence. It follows that B. Corollary.. For any topological space (, ), = τ.

13 Page of Corollary.. If (, ) is a hereditarily Lindelof topological space, then B =. Proof. By Proposition., =. Theorem. ends the proof. Corollary.. If (, ) is a second countable topological space, then B =. Proposition.. If (, ) is an LC topological space, then B =. Proof. Let. By Theorem., we have ( )= ( ) and hence is - open. Corollary.. For any topological space (, ), ( ) =. Proof. By Proposition., (, ) is LC. So by Proposition., we must have ( ) =.. -Continuity and -Continuity Definition.. A function :(, ) (, ) is called -continuous (resp. - continuous, weakly -continuous) at a point, if for every with ( ) there is (resp. B, τ ) such that and ( ). If is continuous (resp. -continuous, weakly -continuous) at each point of, then is said to be -continuous (resp. -continuous, weakly -continuous). The following result follows immediately and its proof is left to the reader: Proposition.. A function :(, ) (, ) is a. -continuous if and only if :(, ) (, ) is continuous. b. -continuous if and only if for each, ¹( ) is -open. c. Weakly -continuous if and only if :(,τ ) (, ) is continuous.

14 Page of Corollary.. Every -continuous function is weakly -continuous Proof. The proof follows directly from parts (b) and (c) of Proposition.. The following example shows that Corollary. is not reversible in general. It is also an example of an -continuous function that is not -continuous. Example.. Let (, ), and be the topological space and the sets as in Example.. Let ={, } and ={,,{ }}. Define :(, ) (, ) by (0)= and ( )= otherwise. As it can be seen from Example., τ B. Since ¹({ })= τ B, then is weakly -continuous that is not -continuous. Also, since ¹({ })=, then is -continuous. The following example shows that -continuity does not imply even weakly - continuity in general: Example.. Let (, )=(R, ) and (, )=(R, ). Since (R, ) is second countable, then by Proposition. (, )=(R, ) and by Corollary., B =. So the identity function :(, ) (, ) is an -continuous function that is not weakly -continuous. The following is an example of an -continuous function that is not -continuous: Example.. Let (, ) be the topological space as in Example., ={, } and ={,,{ }}. Define :(, ) (, ) by ()= and ( )= otherwise. Since

15 Page of ({})= ({})=, then ¹({ })={} B and hence is - continuous. On the other hand, since ({ })={}, then is not continuous. In the following result we list two decomposition theorems for continuity: Proposition.. Let :(, ) (, ) be a function. Then the following are equivalent: a. is continuous. b. is both -continuous and -continuous. c. is both -continuous and weakly -continuous. Proof. (a) (b) The Proof follows immediately from Theorem. and Proposition.. (a) (c) The Proof follows directly from Corollary. and Proposition.. As defined in (Hdeib, ), a function :(, ) (, ) is -continuous if the inverse image of each open set is -open. Any -continuous function is -continuous, however, the function in Example. is -continuous but not -continuous. Remark.. Let :(, ) (, ) be a function for which (, ) is LC. Then the following are equivalent: a. is continuous. b. is -continuous. c. is -continuous. Proof. This is an immediate consequence of Theorem..

16 Page of Remark.. If :(, ) (, ) is any function for which (, ) is LC, then is - continuous. Proof. This is an immediate consequence of Proposition.. Remark.0. Let :(, ) (, ) be a function with (, ) is hereditarily Lindelof (in particular, (, ) is second countable). Then the following are equivalent: a. is continuous. b. is -continuous. Proof. This follows from Corollary.. Theorem.. a. If :(, ) (, ) is -continuous and :(, ) (, ) is continuous, then (, ) (, ) is -continuous. b. If :(, ) (, ) is -continuous and :(, ) (, ) is continuous, then (, ) (, ) is -continuous. c. If :(, ) (, ) is weakly -continuous and :(, ) (, ) is continuous, then (, ) (, ) is weakly -continuous. Proof. (a) Since :(, ) (, ) is -continuous, then by Proposition. (a), :(, ) (, ) is continuous. Therefore, (, ) (, ) is continuous. Again by Proposition. (a), it follows that (, ) (, ) is -continuous. (b) Let. Since :(, ) (, ) is continuous, then ¹( ). Since :(, ) (, ) is -continuous, then ¹( ¹( )) B. Therefore, ( ) ¹( )= ¹( ¹( )) B. (c) Similar to that used in (a). The following Example shows in Theorem. (a) that the continuity condition on cannot be replaced by -continuity:

17 Page of Example.. Let =R, ={,, }, ={, }, be as in Example., =,,{ },{, }, and =,,{ }. Define the function :(, ) (, ) by ( )= if {0,} and ( )= otherwise, and define the function :(, ) (, ) by ( )= ( )= and ( )=. Then and are -continuous functions, but is not -continuous since ( ) ({ })={0,}. Questions.. Is it true that the composition of two -continuous (resp. weakly -continuous) functions is -continuous (resp. weakly -continuous)? Theorem.. Let { (, ) (, ): } be a family of functions. Then the function : (, ) (, ) defined by ( )= ( ) is -continuous (resp. -continuous, weakly -continuous) if and only if for each, is continuous (resp. -continuous, weakly -continuous). Proof. Only we prove it for -continuous, the others are similar. Suppose that is -continuous and let. Then = where :(, ) (, ) is the projection function on. Since is continuous, then by Theorem. (b), is -continuous. Conversely, suppose for each, is - continuous. Let be any subbasic open set of (, ), say = ¹ ( ) for some and. Then ( )= ¹ ( ) = ( )= ¹ ( ). Since by assumption is -continuous, then we have ¹ ( ) B. By Proposition. (b), it follows that is -continuous. Corollary.. A function :(, ) (, ) is -continuous (resp. -continuous, weakly -continuous) if and only if the graph function h:(, ) (, ), given by h( )=(, ( )) for every is -continuous (resp. -continuous, weakly -continuous).

18 Page of Theorem.. The restriction of an -continuous function on a nonempty closed set is -continuous. Proof. Let :(, ) (, ) be -continuous and be a nonempty closed set. Let. Then ( ). So by Theorem., we have ( ) ( )= ( ) ( ). It follows that the restriction function :(, ) (, ) is continuous. Question.. Is it true that the restriction of an -continuous (resp. weakly - continuous) function on a nonempty closed set is -continuous (resp. weakly - continuous)? Proposition.. Let (, ) be a topological space and let and be two -closed sets in (, ) with =. If :(, ) (, ) is a function such that (, ) (, ) and (, ) (, ) are -continuous functions, then is continuous. Proof. It is similar to that used in Theorem. of (Al Ghour & Samarah, 0) and left to the reader. Corollary.. Let (, ) be a topological space, and be two closed sets in (, ) with = and :(, ) (, ) be a function. Then :(, ) (, ) is continuous iff (A, ) (, ) and (A, ) (, ) are -continuous Proof. This is an immediate consequence of Theorem. and Proposition.. Theorem.0. Let :(, ) (, ) be a function. If there is containing such that the restriction of to, (A, ) (, ) is -continuous at, then is -continuous at. Proof. Let with ( ). Since is -continuous at, there is ( ) such that and ( )( )= ( ). By Theorem., ( ) and so. Since, then. This ends the proof.

19 Page 0 of Corollary.. Let :(, ) (, ) be a function. If B is a cover of which consists of nonempty -open sets such that for each B, (A, ) (, ) is -continuous, then is -continuous. Proof. Let. Choose B such that. Then by Theorem.0, is continuous at. References Al Ghour, S., & Samarah S. (0). Cocompact Open Sets and Continuity. Abstract and Applied Analysis, Art. ID, pp. Al-Zoubi, K., & Al-Nashef, B. (00). The Topology of ω-open Subsets. Al- Manarah Journal, (), -. Cameron, D. (). Maximal and Minimal Topologies. Transactions of the American Mathematical Society, 0, -. Hdeib, H., (). ω-closed Mappings. Revista Colombiana de Mathematicas,, -. Hdeib, H., (). ω-continuous Functions. Dirasat Journal,, -. Henriksen, M., & Woods, R. (). Weak P-Spaces and L-Closed Spaces. Questions Answers General Topology,, 0-0. Mukherji, T. & Sarkar, M. (). On a Class of Almost Discrete Spaces. Matematički Vesnik,, -.