#### Abstract

We introduce the different notions of a new class of continuous functions called generalized semi Lambda (gs) continuous function in topological spaces. Its properties and characterization are also discussed.

#### 1. Introduction

In 1986, Maki  continued the work of Levine and Dunham on generalized closed sets and closure operators by introducing the notion of -sets in topological spaces. A -set is a set which is equal to its kernel (= saturated set), that is, to the intersection of all open supersets of . Arenas et al.  introduced and investigated the notion of -closed sets and -open sets by involving -sets.

In 2008 Caldas et al.  introduced generalized closed sets (g, -g, g) and their properties. They also studied the concept of closed maps. In 2007, Caldas et al.  introduced the concept of irresolute maps.

In this paper, we establish a new class of maps called gs continuous function and study its properties and characteristics.

Throughout this paper, (), (), and () (or simply , and will always denote topological spaces on which no separation axioms are assumed unless explicitly stated. , , , , gs, and gs denote the interior of , closure of , lambda interior of , lambda closure of , gs Lambda closure of and gs Lambda interior of , respectively.

#### 2. Preliminary Definitions

Definition 1. A subset of a space () is called (1)a semiopen set  if , (2)a preopen set  if ,(3)a regular open set  if .
The complement sets of semi open (resp., preopen and regular open) are called semi closed sets (resp., preclosed and regular closed). The semiclosure (resp., preclosure) of a subset of denoted by sCl(), (pCl()) is the intersection of all semi closed sets (pre closed sets) containing .

A topological space () is said to be (1)a generalized closed  if Cl() , whenever and is open in (),(2)a closed  if Cl() , whenever and is g-open in (,), (3)semigeneralized closed (denoted by sg-closed)  if sCl(, whenever and is semi open in (), (4)generalized semiclosed (denoted by gs-closed)  if sCl(, whenever and is open in (), (5)a subset of a space () is called -closed  if , where is a -set and is a closed set, (6)a subset of () is said to be a g closed set  if Cl( whenever , where is open in (), (7)a subset of () is said to be a -g closed set  if whenever , where is open in (), (8)a subset of () is said to be a g closed set  if whenever , where is open in (), (9)a subset of () is said to be a gs closed set  if whenever , where is semi open in .

The complement of the above closed sets are called its respective open sets.

Definition 2. A function ) is called (1)semicontinuous  if is semi open in () for every open set in (), (2)semi open (semiclosed)  if is semi open (semi closed) in () for any open (closed) set in (),(3)pre-semiopen (pre-semiclosed)  if is semi open (semi closed) in (, ) for every semi open (semi closed) set in (), (4)irresolute  if for any semi open set of (), is semi open in (), (5) continuous  if () is open in () for every open set in (), (6) continuous  if () is open in () for every open set in (), (7)contracontinuous  if () is closed in () for every open set in (), (8) continuous [2, 14] if () is open ( closed) in () for every open (closed) set in , (9) closed  if is closed in (, ) for every closed set of (), (10) irresolute  if the inverse image of open sets in are open in (), (11)sg continuous  if () is sg open (sg closed) in for every open (closed) set in (),(12)gs closed map (gs open map)  if the image of each closed (open) set in   is gs closed (gs open) in .(13)M.gs closed map (gs open map)  if the image of each gs closed (gs open) set in is (gs open) gs closed in .

Lemma 3 (see ). If is continuous and closed and if B is closed (or open) subset of , then is closed (or open) in .

Definition 4. A space is called(i)  a space if every closed subset of is closed in , (ii)   a space if every closed subset of is closed in ,(iii)   a space if every closed subset of is closed in .

Proposition 5 (see ). In a topological space (, ), the following properties hold. (1)Every closed set is gs closed. (2)Every open set is gs closed. (3)Every closed set is gs closed. (4)In space every gs closed set is closed. (5)In space every g closed set is gs closed. (6)In partition space every gs closed set is closed and closed. (7)In a door space every subset is gs closed. (8)In space every subset is gs closed.

Definition 6. Let () be the topological space and . We define gs closure of (briefly gs ) to be the intersection of all gs closed sets containing , gs interior of (briefly gs ) to be the union of all gs open sets contained in .

Lemma 7 (see ). Let A and B be subsets of a topological space (). The following properties hold. (1)gsCl(A) is the smallest gs closed set containing . (2)If A is gs closed then A = gs. Converse need not be true. (3)A gs.(4)If , then gs gs(5)gs gs) = gs.(6)gs is the largest gs open set contained in . (7)If is gs open then gs. Converse need not be true.

Proofs are obvious from the definition and properties of gs closed sets and gs open sets.

#### 3. gs Continuous Function

Definition 8. A map is called gs continuous function if the inverse image of each closed set in () is gs closed in ().

Theorem 9. Every continuous function is gs continuous function.

Proof. Let be a closed set in () and a function be a continuous function. Hence, is closed in (). As every closed set is gs closed set by Proposition 5, we have    is gs closed in . Thus, is a gs continuous function.

Converse need not be true as seen from the following example.

Example 10. Let and , . The identity function is a gs continuous function but not continuous function as is closed in () but   is not closed in ().

Theorem 11. Every continuous function is gs continuous function.

Proof. Let be a closed set in () and a function be a continuous function. Hence, is closed in (). As every closed set is gs closed set by Proposition 5, we have is gs closed in X. Thus, is a gs continuous function.

Converse need not be true as seen from the following example.

Example 12. Let and , . The identity function is a gs continuous function but not continuous function as is closed in () but is not closed in ().

Theorem 13. If () is a space then every gs continuous function is continuous function.

Proof. The proof is simple to the readers as in space every gs closed set is closed.

Theorem 14. Every continuous function is gs continuous function.

Proof. As every closed set is gs closed set  is a gs continuous function.

Converse need not be true as seen from the following example.

Example 15. Let and ,,,, ,,, = ,,,,,,,,,. The identity function is a gs continuous function but not continuous function as is closed in () but is not closed in ().

Theorem 16. If () is a partition space, then every gs continuous function is continuous function.

Proof. The proof is simple as in a partition space every gs closed set is closed .

Theorem 17. Every contra continuous function is gs continuous function.

Proof. Let be a closed set in () and a function be a contra continuous function. Hence, () is open in (). As every open set is gs closed set by Proposition 5, we have () is gs closed in . Thus, is a gs continuous function.

Converse need not be true as seen from the following example.

Example 18. Let and , , = ,,,,,,,,,. The identity function is a gs continuous function but not contra continuous function as is closed in () but is not open in .

Theorem 19. Every irresolute function is gs continuous function.

Proof. Let be a closed set in (). As every closed set is closed set, is a closed set in . Since is a irresolute function, () is closed in (). As every closed set is gs closed set by Proposition 5, we have () is gs closed in (). Thus, is a gs continuous function.

Converse need not be true as seen from the following example.

Example 20. Let and ,,,,,,,,,,,,,,,,,,,,,,,, ,,,,,,,. The identity function is a gs continuous function but not irresolute function as is closed in but is not closed in .

Theorem 21. If () is a partition space, then every gs continuous function is continuous function.

Proof. Let be a closed set in . Since a function is a gs continuous function is gs closed in . Since is a partition space by Proposition 5, we have is closed in . Thus, is a continuous function.

Theorem 22. Any function , where the domain is a (door) space is a gs continuous function.

Proof. Since in (door) space every subset is gs closed set , the proof is clear to the readers.

Theorem 23. Let a bijective map is a irresolute and pre-semiopen function. If is gs closed in then () is gs closed in .

Proof. Let be a gs closed set of (). Let be a semi open set of containing . Then , where is semi open in (), as is a pre semi open function. Since is a gs closed set of , we have . Thus, we have (, where is closed in (). Thus, we have (. Hence, is gs closed in .

Lemma 24 (see ). If is semi open and gs closed then is closed.

Lemma 25 (see ). A function is contra irresolute if and only if the inverse image of each semi closed set in is semi open in .

Theorem 26. If a function is a gs continuous function and contra irresolute function then is a continuous function.

Proof. Let be a closed set of . As every closed set is semi closed, is a semi closed set in . Since is a gs continuous function and contra irresolute function, is gs closed and semi open in . Now, by Lemma 24 the proof is very clear.

Lemma 27 (see ). Let , where is open in . If is gs closed in , then is gs closed in .

Theorem 28. If a function is a continuous function and is an open subset of , then the restriction is also gs continuous.

Proof. Let be a closed set of and be an open subset of . As every open set is closed, is closed in and since is continuous function is closed in . Hence, we have is closed in , which is also gs closed in . Since where is an open subset of , by Lemma 27 is gs closed in . Thus, the restriction is also gs continuous.

Definition 29. (i) Let be a point of and be a subset of . Then is called a gs neighbourhood of in , if there exist a gs open set of (), such that .
(ii) The intersection of all gs closed sets each containing a set in a topological space is called the gs closure of and is denoted by gs.

Theorem 30. Let be a subset of . Then if and only if for any gs neighbourhood of in .

Proof. Necessity. Assume that gs. Suppose that there exist a gs neighbourhood of such that . Since is a gs neighbourhood of in by definition, there exists a gs open set such that . Since is such that , we have we have . Since is a gs closed set containing , we have gs) and therefore gs Cl(), which is a contradiction.
Sufficiency. Assume that for each gs neighbourhood of in , . Suppose gs). Then there exist a gs closed set of (), such that and . Thus, and is a gs open set in . But which is a contradiction. Thus, gs.

Theorem 31. Let be a function.Then the following are equivalent. (i)The function is gs continuous. (ii)The inverse of each open set in is gs open in . (iii)For each in , the inverse of every neighbourhood of is a gs neighbourhood of .(iv)For each in , and each neighbourhood of , there is a gs neighbourhood of , such that .(v)For each subset of , . (vi)For each subset of , gs.

Proof. (i)(ii). Let be a gs continuous function and be an open set in . Then is closed set in . Hence, by the definition of gs continuous function is gs closed set in . Thus, is a gs open set in . The converse is analogous.
(ii)(iii). For , let be the neighbourhood of . Then there exist an open set in such that . Consequently, is gs open set in and . Thus, is a gs neighbourhood in . Converse is very simple to the reader.
(iii)(iv). Let and be the neighbourhood of . Then by assumption is a gs neighbourhood of and . Converse is clear.
(iv)(v). Suppose that (iv) holds and let gs) and let be the neighbourhood of . Then there exist and a gs neighbourhood of , such that , gs and . Since gs, by Theorem 30   and hence . Hence, .
Conversely suppose that (v) holds and let , be the neighbourhood of . Let . Since , we have gs. Since gs Cl(), there exist a gs neighbourhood of such that and hence .
(v)(vi). Suppose that (v) holds and be any subset of (). Then replacing by () in (), we obtain gs. That is gs.
Conversely, suppose that (vi) holds, and let , where is the subset of . Then we have gs) gs and so (gs.
This completes the proof of the theorem.

Theorem 32. Let be any family of topological spaces. If is a gs continuous function, then of is gs continuous for each , where is the projection of on .

Proof. We will consider a fixed . Suppose is an arbitrary open set in . Then is open in . Since is a gs continuous function, we have is gs open in . Hence, of is a gs continuous function.

Definition 33. Let be a subset of . mapping is called a gs continuous retraction if is gs continuous and the restriction is the identity mapping on .

Definition 34. A topological space () is called a gs Hausdroff if for each pair of distinct points of , there exist gs neighbourhoods and of and , respectively, that are disjoint.

Theorem 35. Let be a subset of and be a gs continuous retraction. If is gs Hausdroff, then is a gs closed set of .

Proof. Suppose that is not gs closed. Then there exist a point in such that gsCl() but . It follows that because is gs continuous retraction. Since is gs Hausdroff there exist disjoint gs open sets and in such that and . Now let be an arbitrary gs neighbourhood of . Then is a gs neighbourhood of . Since gs, by Theorem 30, we have . Therefore, there exist a point . Since , we have and hence . This implies that because . This is a contrary to gs continuity of . Consequently, is a gs closed set of .

#### 4. On Composite Functions

Theorem 36. Composition of continuous functions is a gs continuous function.

Proof. Since every closed set is a gs closed set, the proof is simple to prove.

Theorem 37. Composition of irresolute functions is a gs continuous function.

Proof. Let and are irresolute functions and let be a closed set of (), which is also closed set by . Then is a closed set in () and is a closed set in . Since every closed set is gs closed set by Proposition 5, we get is a gs continuous function.

Theorem 38. Composition of contra continuous functions is a gs continuous function.

Proof. Proof is clear by definition.

Remark 39. But composition of gs continuous function is not a gs continuous function.

Example 40. Let X = Y = Z = {a,b,c,d,e} and (X,) = ,X,{a},{b},{a,b},{b,c},{a,b,c},{b,c,d},{a,b,c,d},{b,c,d,e , , = ,Y,{a},{b,c,d},{a,b,c,d , and , = ,Z,{a},{b},{a,b},{b,e},{c,d},{b,c,d},{a,b,e},{a,c,d},{a,b,c,d},{b,c,d,e . The identity function and is a gs continuous functions but () is not gs continuous function as is closed in but is not gs closed in .

Theorem 41. (i) If is a gs continuous function and is a continuous function, then is a gs continuous function.
(ii) If is a continuous function and is a continuous function, then is a gs continuous function.
(iii) If is a irresolute function and is a continuous function, then is a gs continuous function.
(iv) If is a continuous function and is a continuous function, then is a gs continuous function.
(v) If is a contra continuous function and is a continuous function, then is a gs continuous function.

Proof. (i) Let is a gs continuous function and is a continuous function and let be a closed set of (). Then () is a closed set in () and (() is a gs closed set in as is a gs continuous function. Thus is a gs continuous function.
(ii) It is clear that is a continuous function and hence gs continuous function by Theorem 11.
(iii) Since every closed set is closed and every closed set is gs closed set (Proposition 5), is a gs continuous function.
(iv) Let is a continuous function and is a continuous function and let be a closed set of . Then () is a closed set in and is a closed set in as is a continuous function. Since every continuous function is gs continuous function by Theorem 14, we get is a gs continuous function.
(v) As every closed set is gs open by Proposition 5 the proof is very clear.

Theorem 42. (i) Let is a gs continuous function and is a continuous function, then is a g continuous ( continuous) function if is a partition space.
(ii) Let is a gs continuous function and is a continuous function, then is a continuous function if is a space.

Proof. (i) In partition space every gs closed set is closed ( closed) set by Proposition 5, the theorem is proved.
(ii) The proof is clear as in a space every gs closed set is closed by Proposition 5.

Lemma 43 (see ). If a map is irresolute and a closed map then for every gs closed set of is gs closed set in .

Theorem 44. Let and are bijective.
(i) If is a continuous function and is gs closed map then is a gs continuous function.
(ii) If is a continuous function and is closed map then is a gs continuous function.
(iii) If is a irresolute function and is closed map then is a gs continuous function.
(iv) If is a continuous function and is closed map then is a gs continuous function.
(v) If is a gs continuous function and is irresolute and a closed map then is a gs continuous function.
(vi) If is a gs continuous function and is M.gs closed map then is a gs continuous function.

Proof. (i) Let be a closed set in . Since is a continuous function ( is closed in (). Now since is a gs closed map is gs closed set in (). Thus is a gs continuous function.
(ii) Let be a closed set in (). Since is a continuous function () = (()) is closed in (). Now since is a closed map is closed which is also gs closed set in . Thus, is a gs continuous function.
(iii) Let be a closed set in which is also closed in . Since is a irresolute function is closed in . Now, since is a closed map is closed which is also gs closed set in () . Thus, is a gs continuous function.
(iv) Let be a closed set in (, ). Since is a continuous function is closed in , which is also closed in . Now, since is a closed map is closed which is also gs closed set in . Thus, is a gs continuous function.
(v) Let be a closed set in . Since is a gs continuous function is gs closed in . By Lemma 43   is gs closed set in . Thus is a gs continuous function.
(vi) Let be a closed set in . Since is a gs continuous function is gs closed in . Now, since is a M.gs closed map is gs closed. Thus, is a gs continuous function.

Theorem 45. For any bijection , the following are equivalent:(i) is gs continuous,(ii) is gs open, (iii) is gs closed.

Proof. (i)(ii) Let be an open set in (). By assumption is gs open in and hence is gs open map.
(ii)(iii) Let be a closed set in . Then is open in . By assumption is gs open in and therefore is gs closed in . Hence, is gs closed.
(iii)(i) Let be a closed set in (). By assumption is gs closed in . But and therefore is is gs continuous.