EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 12, No. 3, 2019, 1082-1095 ISSN 1307-5543 – www.ejpam.com Published by New York Business Global Pre-irresolute functions in closure spaces Halgwrd M. Darwesh1, Sarhad F. Namiq2,∗ 1Department of Mathematics, College of Science, University of Sulaimani, Kurdistan-Region, Iraq 2 Department of Mathematics, College of Education, University of Garmian, Kurdistan-Region, Iraq Abstract. The preopen sets are used to define pre-open functions, pre-closed functions, pre- continuous functions, contra-pre-continuous functions and pre-irresolute functions which are inves- tigated. They are also used to introduce a new type of connectedness and compactness in closure spaces, they called p-connectedness and p-compactness respectively 2010 Mathematics Subject Classifications: 54A05 Key Words and Phrases: Closure operator, closure space, pre-open functions, pre-closed func- tions, contra-pre-continuous functions, pre-irresolute functions, p-connectedness, p-compactness, Tp-spaces 1. Introduction Kazimierz Kuratowski was a Polish mathematician and logician, he defined [14] closure operator by the following: Let X be a set and P (X) its power set. A Kuratowski Closure Operator is a function cl : P (X)→ P (X) with the following properties: (i) cl (φ) = φ (Preservation of Nullary Union) (ii) A ⊆ cl (A) for every subset A ⊆ X (Extensivity) (iii) cl (A ∪B) = cl (A)∪cl (B) for any subsets A,B ⊆ X (Preservation of Binary Union). (iv) cl (cl (A)) = cl (A) for every subset A ⊆ X (Idempotence) If the last axiom(iv), idempotence, is omitted, then the axioms define a preclosure op- erator. A consequence of the third axiom(iii) is: A ⊆ B then cl (A) ⊆ cl (B) (Preservation of Inclusion). Then cl, together with the underlying set X, is called closure space and is ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v12i3.3317 Email addresses: halgwrd.darwesh@univsul.edu.iq (H.M. Darwesh), sarhad1983@gmail.com (S.F. Namiq) http://www.ejpam.com 1082 c© 2019 EJPAM All rights reserved. H.M. Darwesh, S.F. Namiq / Eur. J. Pure Appl. Math, 12 (3) (2019), 1082-1095 1083 denoted by (X, cl). In 1966, Eduard Cech defined closure operator by the following: Let X be a set and P (X) its power set. A function c : P (X) → P (X) with the following properties: (i) c (φ) = φ. (ii) A ⊆ c (A) for every subset A ⊆ X. (iii) c (A ∪B) = c (A) ∪ c (B) for any subsets A,B ⊆ X. Then c, together with the underlying set X, is called a Cech closure space and is denoted by (X, c). If c also satisfies: c (c (A)) = c (A) for every subset A ⊆ X, then (X, c) is a topological space. In 2009, Jeeranunt Khampakdee [15] defined closure operator by the following: A function c : P (X )→ P (X ) defined on the power set P (X ) of a set X is called a closure operator on X and the pair (X, c) is called a closure space if the following axioms are satisfied: (i) c (φ) = φ. (ii) A ⊆ c(A) for every A ⊆ X. (iii) A ⊆ B ⇒ c(A) ⊆ c(B), for all A, B ⊆ X. The concept of closure operator and closure spaces are very usefull material in sev- eral branches of Science, such as Topology [2],[3],[4],[5]Computer Science Theory[18], Biochemistry[6]. The purpose of this paper is to study the concept of preopen sets in closure spaces. Closure spaces were introduced by E.Cech [2] in 1966 and then studied by many math- ematicians, see e.g. [2],[3],[4],[5],[7],[10],[8],[9] and [13]. Closure spaces are sets endowed with a grounded, extensive and monotone closure operator. Mashhure et al[17] introduced the concept of preopen sets and pre-continuous functions. The preopen sets and local dense sets are same in topological space, also the pre-continuity and almost-continuity (in the sense Hussain)[12] are same in topological spaces. Halgwrd M.Darwesh [11] used the teqnique of mashhoury[12] to introduce and study the concept of preopen sets in closure spaces, and then he showed that its differ to local dense sets. However he defined the con- cept of pre-continuous functions in closure spaces and then he showed that the concepts of pre-continuity and almost-continuity (in the sense Hussain) [11] are independent concepts. In this paper, in Section 3, we introduce the notion of pre-open(pre-closed) functions, contra-pre-continuous and study some of their properties. In Section 4, we introduce and discuss pre-irresolute functions in closure spaces. We es- tablish some basic properties of pre-irresolute functions In Section 5, we introduce the notion of p-connectedness and study some of their proper- ties. In Section 6, we introduce the notion of p-compactness and study some of their properties. H.M. Darwesh, S.F. Namiq / Eur. J. Pure Appl. Math, 12 (3) (2019), 1082-1095 1084 2. Preliminaries A function c : P (X ) → P (X ) defined on the power set P (X ) of a set X is called a closure operator on X and the pair (X, c) is called a closure space[15] if the following axioms are satisfied: (i) c (φ) = φ. (ii) A ⊆ c(A) for every A ⊆ X. (iii) A ⊆ B ⇒ c(A) ⊆ c(B), for all A, B ⊆ X. Definition 1. [15] A closure operator c on a set X is called additive if c(A ∪ B) = c(A) ∪ c(B), for all A,B ⊆ X. Definition 2. [15] A closure operator c on a set X is called idempotent if cc(A) = c(A), for all A ⊆ X. Definition 3. [15] A subset A ⊆ X is closed in the closure space (X, c) if c(A) = A. It is called open, if its complement in X is closed. The empty set and the whole space are both open and closed. Definition 4. [11] A subset A of a space (X, c ) is said to be a preopen set, if there exists an open set G such that A ⊆ G ⊆ c(A). The complement of a preopen set is called preclosed. Theorem 1. [11] A subset A of a space(X, c ) is preclosed if and only if there exists a closed set F such that X\c(X\A) ⊆ F ⊆ A. Proposition 1. [15] Let (X, c) be a closure space and {Gα}αεJ be a collection of subsets of X. Then ⋃ αεJ c(Gα) ⊆ c( ⋃ αεJ Gα). Proposition 2. [15] The union(intersection)of any family of open(closed)sets in a closure space (X, c ) is open(closed). Proposition 3. [11] The union(intersection)of any family of preopen(preclosed)sets in a closure space (X, c ) is preopen(preclosed). Definition 5. [11] The interior operator i : P (X) → P (X) corresponding to the closure operator c on X is given by; i (A) = X\c (X\A) . Theorem 2. [11] Let A be a subset of a closure (X, c) . If x ∈ c(A), then G∩A 6= φ, for each open subset G of X containing x. Proposition 4. [11] Let A be a subset of a closure (X, c) and c is idempotent on X, then x ∈ c(A) if and only if G ∩A 6= φ, for each open subset G of X containing x. H.M. Darwesh, S.F. Namiq / Eur. J. Pure Appl. Math, 12 (3) (2019), 1082-1095 1085 Proposition 5. [11] Let c be an idempotent closure operator on a set X. If A is preopen in X and B ⊆ A ⊆ c(B), then B is preopen. Definition 6. [15] A closure space (Y, v), is said to be a subspace of (X, c), if Y ⊆ X and v (A) = c (A) ∩ Y , for each subset A ⊆ Y. Theorem 3. [11] Let A ⊆ Y ⊆ X, where (Y, υ) is a subspace of (X, c). If A is preopen in X, then A is preopen in Y. Proposition 6. [1] The product of a family {(Xα, cα) : α ∈ I} of closure spaces, denoted by ∏ α∈I (Xα, cα) , is the closure space ( ∏ α∈I Xα, c), where ∏ α∈I Xα denotes the cartesian product of sets Xα, α ∈ I and c is the closure operator generated by the projections πα :∏ α∈I (Xα, c) → (Xα, c), α ∈ I, i.e., is defined by c (A) = ∏ α∈I cαπα(A), A ⊆ ∏ α∈I Xα. Then the projection function πα is continuous. Proposition 7. [16] Let {(Xα, cα) : a ∈ J} be a family of closure spaces. Then Fais closed in (Xα, cα), for all a ∈ J if and only if ∏ a∈J Fαis closed in ∏ a∈J (Xα, ca). Proposition 8. [16] Let {(Xα, cα) : a ∈ J} be a collection of closure spaces, G ⊆∏ αεJ Xα. If G is a open in ∏ αεJ (Xα, cα) and πα is a project function, then πα(G) is a open in (Xα, cα). Definition 7. [16] Let (X, c1)and (Y, c2) be closure spaces. A function f : (X, c1) → (Y, c2) is called open (respectively, closed) if the image of every open (respectively, closed) set in (X, c1) is open (respectively, closed) in (Y, c2) . Proposition 9. [16] A function f : (X, c1)→ (Y, c2) is said to be continuous if f(c1(A)) ⊆ c2f(A) for every subset A of X. Proposition 10. [16] Let (X, c1) and (Y, c2) be closure spaces. If f : (X, c1)→ (Y, c2) is a continuous function, then the inverse image under f of each open set in (Y, c2)is open in (X, c1) . Proposition 11. [16] Let (X, c1), (Y, c2) and (Z, c3)be closure spaces, let f : (X, c1) → (Y, c2) and g : (Y, c2)→ (Z, c3) be functions. Then: (i) If f and g are open, then so is gof . (ii) If gof is open and f is a continuous surjection, then g is open. (iii) If gof is open and g is a continuous injection, then f is open. Proposition 12. [16] Let (X, c1) and (Y, c2) be closure spaces and let f : (X, c1)→ (Y, c2) be a function. If f is open, then for every y ∈ Y and every closed subset F of (X, c1) such that f−1 ({y}) ⊆ F , there exists a closed subset K of (Y, c2) such that y ∈ K and f−1 (K) ⊆ F. H.M. Darwesh, S.F. Namiq / Eur. J. Pure Appl. Math, 12 (3) (2019), 1082-1095 1086 Proposition 13. [11] Let (X, c1) and (Y, c2) be closure spaces. A function f : (X, c1)→ (Y, c2) is called pre-continuous if the inverse image of every open set in (Y, c2) is preopen in (X, c1). Proposition 14. [11] Let (X, c1) and (Y, c2) be closure spaces. A function f : (X, c1)→ (Y, c2) is called pre-continuous if and only if the inverse image of every closed set in (Y, c2) is preclosed in (X, c1). Proposition 15. [11] Let (X, c1), (Y, c2) and (Z, c3)be closure spaces. If f : (X, c1) → (Y, c2) and g : (Y, c2)→ (Z, c3) are pre-continuous and continuous respectively.Then gof : (X, c1)→ (Z, c3) is pre-continuous. Proposition 16. [11] Let (X, c1), (Y, c2) and (Z, c3) be closure spaces. Let f : (X, c1)→ (Y, c2) be a surjective open continuous function and g : (Y, c2)→ (Z, c3) is a function such that gof : (X, c1)→ (Z, c3) is pre-continuous. Then g : (Y, c2)→ (Z, c3) is pre-continuous. Definition 8. [16] A closure space (X, c) is said to be connected if φ and X are the only subsets of X which are both closed and open. Definition 9. [16] A collection {Ga}a∈J of sets in a closure space(X, c) is called a cover of a subset Bof X if B ⊆ ⋃ α∈J Ga if holds, and an open cover if Ga is open for each α ∈ J. Furthermore, a cover {Ga}a∈J of a subset B contains a finite subcover, if there exists a finite subset J0 of J such that B ⊆ ⋃ α∈J0 Ga. Definition 10. [16] A subset A of a closure space (X, c) is compact if every open cover of A contains a finite subcover. 3. Pre-Open ( Pre-Closed ) Functions and Contra-Pre-Continuous In the present section, we define and study some properties of pre-open (preclosed)functions and contra-pre-continuous. Definition 11. Let (X, c1) and (Y, c2) be two closure spaces. Let f : (X, c1) → (X, c2) be a function. Then f is pre-open ( or preopen ) if f (G) is preopen in Y, for every open subset G of X. Definition 12. Let (X, c1) and (Y, c2) be two closure spaces. Let f : (X, c1) → (X, c2) be a function. Then f is pre-closed ( or preclosed ) if f (K) is preclosed in Y, for every closed subset K of X. Proposition 17. Let (X ,c1), (Y, c2) and (Z, c3) be closure spaces, let f : (X, c1)→ (Y, c2) and g: (Y, c2)→ (Z, c3) be functions. Then: (i) If f is open and g is preopen, then gof is preopen. (ii) If gof is preopen and f is a continuous surjection, then g is preopen. H.M. Darwesh, S.F. Namiq / Eur. J. Pure Appl. Math, 12 (3) (2019), 1082-1095 1087 Proof. (i) Let G be an open subset of (X, c1). Since f is open, f(G) is open in (Y, c2). Hence g(f(G)) is preopen in (Z, c3). Thus, gof is preopen. (ii) Let G be an open subset of (Y, c2). Since f is a continuous function, f−1(G) is open in (X , c1). Since gof is preopen, gof(f−1(G)) = g(f(f−1(G))) is preopen in (Z, c3). But f is surjection, so that gof(f−1(G)) = g(G). Hence, g(G) is preopen in (Z, c3). Therefore, g is preopen. Proposition 18. Let (X, c1), (Y, c2) and (Z, c3) be closure spaces, let f : (X, c1)→ (Y, c2) and g : (Y, c2) → (Z, c3) be functions. If gof is open and g is a pre-continuous injection, then f is preopen. Proof. Let G be an open subset of (X, c1). Since gof is open, g(f(G)) is open in (Z, c3). As g is pre-continuous, g−1(g(f (G))) is preopen in (Y, c2). But g is injective, so that g−1(g( f (G))) = f (G) is preopen in (Y, c2). Therefore, f is preopen. Proposition 19. Let (X, c1) and (Y, c2) be closure spaces. If f : (X, c1) → (Y, c2) is a bijection, then the following statements are equivalent: (i) The inverse function f−1 : (Y, c2)→ (X, c1) is pre-continuous. (ii) f is a preopen function. (iii) f is a preclosed function. Proof. Obvious Definition 13. A closure space (X, c) is said to be a Tp-space if every preopen set in (X, c) is open. The closure space in the following example is a Tp-space. Example 1. Let X = {1, 2, 3} and defined a closure operator c : P (X)→ P (X) by: c (A) = { A if A ∈ {φ, {1} , {2} , {3}} X Otherwise Clearly (X, c) is a Tp-space, since every preopen set is open set Proposition 20. Let (X, c1) and (Y, c2) be closure spaces and (Y, c2)be a Tp-space. If f : (X, c1)→ (Y, c2) and g : (Y, c2)→ (Z, c3) are pre-continuous, then gof is pre-continuous. Proof. Let H be open in (Z, c3). Since g is pre-continuous, g−1 (H) is preopen in (Y, c2) . But (Y, c2) is a Tp-space, hence g−1(H)is open in (Y, c2). Thus f−1 ( g−1 (H) ) = (gof)−1(H) is preopen in (X, c1). Therefore, gof is pre-continuous H.M. Darwesh, S.F. Namiq / Eur. J. Pure Appl. Math, 12 (3) (2019), 1082-1095 1088 Theorem 4. Let (X, c) be a closure space, {(Yα, cα) : α ∈ J} be a family of closure spaces and f : (X, c)→ ∏ α∈J (Yα, cα) be a function. If f is pre-continuous and πα is a projection function, then παof is pre-continuous for each α ∈ J . Proof. Assume that f : (X, c) → ∏ α∈J (Yα, cα) is pre-continuous for all α ∈ J . Since πα is continuous, παof is pre-continuous for each α ∈ J by Proposition 15. Definition 14. Let (X, c1) and (Y, c2) be closure spaces and let f : (X, c1) → (Y, c2) be a function. Then f is contra-pre-continuous if the inverse image under f of every open subset of (Y, c2) is preclosed in (X, c1) . Proposition 21. Let (X, c1) and (Y, c2) be closure spaces and let f : (X, c1)→ (Y, c2) be a function. Then f is contra-pre-continuous if and only if the inverse image under f of every closed subset of (Y, c2) is preopen in (X, c1) . Proof. Let F be a closed subset in (Y, c2). Then Y/F is open in(Y, c2). Since f is contra-pre-continuous, f−1(Y/F )is preclosed. But f−1(Y/F ) = X/f−1(F ), thus f−1(F ) is preopen in (X, c1) . Conversely, let G be an open subset in (Y, c2) . Then Y/G is closed in (Y, c2). Since the inverse image of each closed subset in (Y, c2)is preopen in (X, c1), f −1(Y/G) is preopen in (X, c1). But f−1(Y/G) = X/f−1(G), thus f−1(G) is preclosed. Therefore, f is contra-pre-continuous. Proposition 22. Let (X, c1), (Y, c2) and (Z, c3) be closure spaces, let f : (X, c1)→ (Y, c2) and g : (Y, c2) → (Z, c3) be functions. If gof is contra-pre-continuous and g is a closed injection, then f is contra-pre-continuous. Proof. Let H be a closed subset of (Y, c2) . Since g is closed, g(H) is closed in (Z, c3). As gof is contra-pre-continuous, (gof)−1 (g (H)) = f−1(g−1 (g (H))) is preopen in (X, c1) by Proposition 21. But g is injective, hence f−1 ( g−1 (g (H)) ) = f−1(H). Therefore, f is contra-pre-continuous. Proposition 23. Let (X, c1) and (Z, c3)be closure spaces and (Y, c2) be a Tp-space. If f : (X, c1) → (Y, c2) and g : (Y, c2) → (Z, c3) are contra-pre-continuous functions, then gof is pre-continuous. Proof. Let H be closed in (Z, c3). Since g is contra-pre-continuous, g−1(H)is pre- open in (Y, c2). But (Y, c2)is a Tp-space, hence g−1(H)is open in (Y, c2) . As f is contra- pre-continuous by Proposition 21, f−1 ( g−1 (H) ) = (gof)−1(H) is preclosed in (X, c1). Therefore, gof is pre-continuous by Proposition 15. The following statement is evident: Proposition 24. Let (X, c1), (Y, c2) and (Z, c3)be closure spaces and let f : (X, c1) → (Y, c2) and g : (Y, c2) → (Z, c3) be functions. If f is contra-pre-continuous and g is continuous, then gof is contra-pre-continuous. H.M. Darwesh, S.F. Namiq / Eur. J. Pure Appl. Math, 12 (3) (2019), 1082-1095 1089 As a direct consequence of Proposition 24, we have: Proposition 25. Let (X, c)be a closure space, {(Yα, cα) : α ∈ J}be a family of closure spaces and f : (X, c)→ ∏ α∈J (Yα, cα) be a function. If f is contra-pre-continuous and πα is a projection function, then παof is contra-pre-continuous for each α ∈ J . 4. Pre-Irresolute Functions In view of the definition of Pre-irresolute Functions, we define Pre-irresolute Functions as: Definition 15. Let (X, c1) and (Y, c2) be closure spaces. A function f : (X, c1)→ (Y, c2) is called pre-irresolute if f−1(G) is preopen in (X, c1) for every preopen set G in (Y, c2). Proposition 26. Let (X, c1) and (Y, c2)be closure spaces and f : (X, c1) → (Y, c2) be a function. Then f is pre-irresolute if and only if f−1(B) is preclosed in (X, c1), whenever B is preclosed in (Y, c2) . Proof. Let B be a preclosed subset of (Y, c2). Then Y/B is preopen in (Y, c2) . Since f : (X, c1) → (Y, c2) is pre-irresolute, f−1(Y/B) is preopen in (X, c1) . But f−1(Y/B) = X /f−1(B), so that f−1(B)is preclosed in (X, c1). Conversely, let A be a preopen subset in (Y, c2). Then Y/A is preclosed in (Y, c2) . By the assumption, f−1(Y/A) is preclosed in (X, c1). But f−1(Y/A) = X/f−1(A). Thus f−1(A) is preopen in (X, c1) . Therefore, f is pre-irresolute. Clearly, every pre-irresolute function is pre-continuous. The converse need not be true as can be seen from the following example. Example 2. Let X = {1, 2, 3} = Y and define a closure operator c1 on X by: c1 (A) =  A if A ∈ {φ, {3}} {1, 2} if A = {1} {2, 3} if A = {2} X Otherwise And also define a closure operator c2 on Y by: c2 (A) =  A if A = φ {1, 3} if A = {1} {2, 3} if A = {2} Y Otherwise Let f : (X, c1)→ (Y, c2) be the function defined by: f (x) =  1 2 3 if x = 1 if x = 2 if x = 3 H.M. Darwesh, S.F. Namiq / Eur. J. Pure Appl. Math, 12 (3) (2019), 1082-1095 1090 The family of all open sets with respect to c1={φ, {1, 2} , X} PO (X, c1) = {φ, {1} , {1, 2} , {1, 3} , {2, 3} , X} . The family of all open sets with respect to c2={φ,X} PO (X, c2) = {φ, {3} , {1, 2} , {1, 3} , {2, 3} , X} . Then f is pre-continuous but not pre-irresolute because {3} is preopen in (Y, c2) but f−1({3}) = {3} is not preopen in (X, c1) . Proposition 27. Let (X, c1), (Y, c2) and (Z, c3) be closure spaces. If f : (X, c1)→ (Y, c2) is a pre-irresolute function and g : (Y, c2)→ (Z, c3) is a pre-continuous function, then the composition gof : (X, c1)→ (Z, c3) is pre-continuous. Proof. Let G be an open subset of (Z, c3). Then g−1(G) is a preopen subset of (Y, c2) as g is pre-continuous. Hence, f−1(g−1(G)) is preopen in (X, c1) because f is pre-irresolute. Thus, gof is pre-continuous. The following statements are evident: Proposition 28. Let (X, c1), (Y, c2) and (Z, c3) be closure spaces. If f : (X, c1)→ (Y, c2) and g : (Y, c2)→ (Z, c3) are pre-irresolute, then gof : (X, c1)→ (Z, c3)is pre-irresolute. Proof. Obvious. Proposition 29. Let (X, c1) and (Z, c3) be closure spaces and (Y, c2) be a TP -space. If f : (X, c1)→ (Y, c2) is a pre-continuous function and g: (Y, c2)→ (Z, c3) is a pre-irresolute function, then the composition gof : (X, c1)→ (Z, c3)is pre-irresolute. Proof. Obvious. Proposition 30. Let (X, c1) and (Y, c2) be closure spaces and f : (X, c1) → (Y, c2) be a bijective function. If f and f−1are continuous, then f and f−1 are pre-irresolute. Proof. Let B be a preopen subset of (Y, c2). Then there exists an open set H in (Y, c2) such that B ⊆ H ⊆ c2(B) , hence f−1(B) ⊆ f−1(H) ⊆ f−1(c2(B)). Since f−1 is continuous, f−1(c2(B)) ⊆ c1f −1 (H). But f is continuous. Thus f−1(H)) is open in (X, c1). Hence, f−1(B) is preopen in (X, c1). Therefore, f is pre-irresolute. Let A be a preopen subset of (X, c1). Then there exists an open set G in (X, c1) such that A ⊆ G ⊆ c1(A). Hence, f (A) ⊆ f (G) ⊆ f (c1 (A)) . As f is continuous, f(c1A) ⊆ c2f(A). Since f−1is continuous and f(G) is the inverse image of G under f−1, f(G) is open in (Y, c2). Thus, f(A) is preopen in (Y, c2). But f(A) is the inverse image of A under f−1, therefore f−1is pre-irresolute. H.M. Darwesh, S.F. Namiq / Eur. J. Pure Appl. Math, 12 (3) (2019), 1082-1095 1091 5. P-Connectedness As another application of preopen sets, a new kind of Connectedness, namely p- Connectedness, is introduced. Definition 16. A closure space (X, c)is said to be p-connected if φ and X are the only subsets of X which are both preopen and preclosed. Clearly, if (X, c) is p-connected, then (X, c) is connected. The converse is not true as can be seen from the following example. Example 3. Let X={1, 2, 3} and define a closure operator c on X by: c (A) =  A if A = φ {1, 3} if A = {1} {2, 3} if A = {2} X Otherwise The family of all open sets ={φ,X} PO (X, c) = {φ, {3} , {1, 2} , {1, 3} , {2, 3} , X} . We have (X, c) is connected, but it is not p-connected, because {1, 2} is preopen set and preclosed set. Proposition 31. Let (X, c) be a closure space. Then the following statements are equiv- alent: (i) X is p-connected. (ii) X cannot be expressed as the union of two disjoint, non-empty, preclosed subsets. (iii) X cannot be expressed as the union of two disjoint, non-empty, preopen subsets. Proof. Statement(i) implies statement(ii): Suppose that X = U ∪ V, where U and V are non-empty, disjoint, preclosed subsets of (X, c). Then U = X/V and U is preopen. Thus, U is a subset of X which is both preopen and preclosed but U is neither X nor φ. Hence, (X, c) is not p-connected. Statement (ii) implies statement (iii): Suppose that X = A ∪ B where A and B are disjoint non-empty preopen subsets of (X, c). Then X/A = B and X/B = A are both complements of preopen sets and hence are preclosed. Thus, X = A ∪B is an expression of X as the union of two disjoint, non-empty, preclosed subset of (X, c), which contradicts (ii). Statement (iii) implies statement (i): Suppose that A is a subset of X which is both preopen and preclosed, but A is neither X nor φ. Then X/ A is also preclosed, preopen and non-empty. Thus, X= (X/A)∪A is the expression of X as the union of two disjoint, non-empty preopen subsets, which contradicts (iii). The following statement is evident: H.M. Darwesh, S.F. Namiq / Eur. J. Pure Appl. Math, 12 (3) (2019), 1082-1095 1092 Proposition 32. Let (X, c) be a Tp-space. Then (X, c) is connected if and only if (X, c) is p-connected. Proof. Obvious. Proposition 33. Let (X, c1) be a closure space and let Y = {0, 1} and c2 be a closure operator on Y defined by: c2 (A) = A, for all subset A of Y. Then the following statements are equivalent: (i) The only contra-pre-continuous functions f : (X, c1) → (Y, c2) are the constant functions. (ii) A closure space (X, c1) is p-connected. Proof. Statement (i) implies statement (ii): Suppose that there is a non-empty subset A of (X, c1) such that A 6= X and A is both preopen and preclosed. Then X/A is both preopen and preclosed in (X, c1). Define a function f : (X, c1)→ (Y, c2) by: f (x) = { 0 x ∈ A 1 x ∈ X/A Consequently, f−1 (φ) = φ , f−1 ({0}) = A , f−1({1}) = X/A = B and f−1(Y ) = X. Since there are only four closed subsets of (Y, c2), namely φ , {0}, {1} and Y , the inverse image under fof any closed subset in (Y, c2) is preopen in (X, c1) . Thus, f is contra-pre- continuous but non-constant, which a contradiction. Therefore, (X, c1) is p-connected. Statement (ii) implies statement (i): Suppose that a contra-pre-continuous function f : (X, c1)→ (Y, c2) is non-constant, where the closure operator c2 on Y is defined by: c (A) = A, for all subset A of Y. Then f−1({0}) and f−1({1}) are non-empty. Further, neither f−1({0}) nor f−1({1}) are equal to X. Since {0} and {1} are closed subset of (Y, c2) and f is contra pre-continuous, f−1({0}) and f−1({1}) are preopen subsets of (X, c1). But f−1 ({0}) = X/f−1({1}). Hence f−1({0}) is both preclosed and preopen. Consequently, X is not p-connected, which a contradiction. Proposition 34. Let (X, c1)and (Y, c2)be closure spaces and f : (X, c1) → (Y, c2) be a function. (i) If f is a contra-pre-continuous function from (X, c1) onto (Y, c2) and (X, c1) is p- connected, then (Y, c2) is connected. (ii) If f is a pre-irresolute function from (X, c1)onto (Y, c2) and (X, c1)is p-connected, then (Y, c2) is p-connected. H.M. Darwesh, S.F. Namiq / Eur. J. Pure Appl. Math, 12 (3) (2019), 1082-1095 1093 Proof. (ii). Suppose that (Y, c2) is not p-connected. Then there is a non-empty subset A of Y, A 6= Y such that A is both preopen and preclosed. Since f is pre-irresolute, the set f−1(A) is both preopen and preclosed. Since f is an onto function and A is a non-empty subset of Y with A 6= Y , it follows that f−1(A) is a non-empty subset of X with f−1(A) 6= X . Hence, (X, c1) is not p-connected which a contradiction. Therefore, (Y, c2) is connected. The proof of (i)is similar to that of (ii). 6. P -Compactness As another application of preopen sets, a new kind of compactness, namely p-compactness, is introduced. Definition 17. A collection {Gα}α∈J of preopen sets in a closure space (X, c1) is called a preopen cover of a subset B of X if B ⊆ ⋃ α∈J Gα holds. Definition 18. A subset A of a closure space (X, c)is p-compact if every preopen cover of A contains a finite subcover. The following statements are evident: Proposition 35. Let (X, c1) be a closure space. If X is p-compact and B is a preclosed subset of X, then B is p-compact Proof. Let {Gα}α∈J be a collection of preopen subsets of X such that B ⊆ ⋃ α∈J Gα. It follows that X= ⋃ α∈J Gα ∪ (X/B). Since B is preclosed, X/B is preopen. Consequently,⋃ α∈J Gα ∪ (X/B) is a preopen cover of X. But X is p-compact, so ⋃ α∈J Gα∪(X/B) contains a finite subcover, i.e. there exits a finite subset J0 of J such that X= ⋃ α∈J0 Gα∪ (X/B). Since B and X/B are disjoint, B ⊆ ⋃ α∈J0 Gα. Thus, any preopen cover {Gα}α∈J of B contains a finite subcover. Therefore, B is p-compact. Proposition 36. Let (X, c1) and (Y, c2) be closure spaces and f : (X, c1) → (Y, c2) be a function. If f is pre-irresolute and a subset B of X is p-compact, then the image f (B) ⊆ Y is p-compact. Proof. Let{Gα}α∈J be a collection of preopen subsets of Y such that f(B) ⊆ ⋃ α∈J Gα it follow that B ⊆ f−1 (f (B)) ⊆ f−1 {⋃ α∈J Gα } = ⋃ α∈J f −1(Gα ), but f is pre-irresolute, so {f−1(Gα)}α∈J is a preopen cover of B. Since B is p-compact, there exists a finite sub- set J0of J such that B ⊆ ⋃ α∈J0 f −1(Gα ). It follow that f(B) ⊆ ⋃ α∈J0 Gα . Thus, any preopen cover {Gα}α∈J of f (B) contains a finite subcover. Therefore, f (B) is p-compact. Proposition 37. Let (X, c1) and (Y, c2) be closure spaces and f : (X, c1) → (Y, c2) be a function. If f is a pre-continuous surjection and X is p-compact, then Y is compact. REFERENCES 1094 Proof. Let {Gα}α∈J be a collection of open subsets of Y such that Y ⊆ ⋃ α∈J Gα . It follows that X =f−1 (Y ) ⊆ f−1 (⋃ α∈J Gα ) = ⋃ α∈J f −1(Gα ). But f is pre-continuous, so {f−1(Gα)}α∈J is a preopen cover of X. Since X is p-compact, there exists a finite subset J0 of J such that X= ⋃ α∈J0 f −1(Gα ). It follow that Y=f (⋃ α∈J0 f −1(Gα ) ) = f ( f−1( ⋃ α∈J0 (Gα )) ) . Since f is a surjection, Y= ⋃ α∈J0 Gα Thus, any open cover {Gα}α∈Jof Y contains a finite subcover. Therefore, Y is compact. Proposition 38. Let (X, c1) and (Y, c2) be closure spaces and f : (X, c1) → (Y, c2) be a function. If f is a preirresolute surjection and X is p-compact, then Y is p-compact. Proof. Let {Aα}α∈J be a collection of preopen subsets of Y such that Y ⊆ ⋃ α∈J Aα. It follow that X=f−1 (Y ) ⊆ f−1 (⋃ α∈J Aα ) = ⋃ α∈J f −1(Aα ). But f is pre-irresolute, hence {f−1(Aα)}α∈J is a preopen cover of X. Since X is p-compact, there exists a fi- nite subset J0 of J such that X = ⋃ α∈J0 f −1(Aα ). Hence, (Y ) =f (⋃ α∈J f −1(Aα ) ) = f(f−1 (⋃ α∈J0 (Aα ) ) . Since f is a surjection Y= ⋃ α∈J0 Aα of Y contains a finite sub- cover. Therefore, Y is p-compact. 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