EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 17, No. 3, 2024, 1463-1470 ISSN 1307-5543 – ejpam.com Published by New York Business Global Separation Axioms via F -open Sets M. Baloush1,∗, T. Noiri2, M. Talafha3, S. Che dzul-Kifli3 1 Department of Basic Sciences, National University College of Technology, Amman, Jordan 2 2949-1 Shiokita-cho, Hinagu, Yatsushiro-shi, Kumamoto-ken, 869-5142, Japan 3 School of Mathematical Sciences, National University of Malaysia, Kuala Lumpur, Malaysia Abstract. In this paper, we introduce some types of separation axioms via F -open sets, namely FTi (i = 0, 1, 2, 3, 4), F -regular and F -normal spaces, and investigate their properties, relationships and characterizations. We show that every FTi space is a Ti space for i = 0, 1, 2, 3, 4. However, the converse is true whenever X is finite. 2020 Mathematics Subject Classifications: 54D40 Key Words and Phrases: F -open, FTi (i = 0, 1, 2, 3, 4), F -regular, F -normal 1. Introduction In 2012, Alias et al.[3] introduced some types of separation axioms via ω−open sets, namely ω−regular, completely ω−regular and ω−normal space and investigated their fun- damental properties. After that, in 2024, Alqahtani and Abd El-latif [1] interduced some kinds of separation axioms via ℵ−poen sets, namley ℵ − T0−space, ℵ − T1−space and ℵ − T2-space. Quite recently, Alqahtani [2] has introduced the notion of F -open sets in a topological space and obtained the fundamental properties of F -open sets. Furthermore, several notions such as F -continuous functions, F -compact spaces and related properties are defined and investigated. Motivated by these works, and to simplify the path for many future articles on this topic, in this paper, we introduce some types of separation axioms via F -open sets, namely, FTi (i = 0, 1, 2, 3, 4), F -regular and F -normal spaces, and their relationships and charac- terizations are obtained. Moreover, it is shown that 1) an F -compact FT2 space is FT4, 2) F -normal spaces are preserved under F -closed preserving and continuous surjections. First, we recall some notions defind by Alqahtani[2]. Let (X, τ) be a topological space and A be a subset of X. The closure of A and interior of A are denoted by cl(A) and ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v17i3.5223 Email addresses: malohblosh@yahoo.com [M. Baloush), t.noiri@nifty.com (T. Noiri), mitalafha@gmail.com (M.Talafha), syahida@ukm.edu.my (S. Che dzul-Kifli) https://www.ejpam.com 1463 © 2024 EJPAM All rights reserved. M. Baloush et al. / Eur. J. Pure Appl. Math, 17 (3) (2024), 1463-1470 1464 int(A), respectively. Definition 1. [2] An open subset A of a topological space (X, τ) is called an F−open set if cl(A)\ A is a finite set. Definition 2. [2] A closed subset A of a topological space (X, τ) is called an F−closed set if A\int(A) is a finite set. Definition 3. [2] Let A be a subset of a topological space (X, τ). The F−closure of A is defined as the intersection of all F−closed sets containing A, and is denoted by clF (A). Definition 4. [2] Let (X, τ) be a topological space. Then (X, τ) is called an F−compact space if any open cover of X has a finite subcover of F−open sets. 2. Separation axioms via F -open sets For any topological space (X, τ), by τF we denote the collection of all F− open subsets of X. Example 1. 1) For any discrete topology τ on X, τF = τ since every A ⊆ X is clopen, and therefore cl(A)\A = ∅ is finite. 2) For any indiscrete topology τ on X, we have τF = τ . The only open sets in X are X and ∅. Since both are clopen, so cl(X)\X = ∅ and cl(∅)\∅ = ∅. 3) The odd-even topology on N induced by its bases P = {{2k − 1, 2k} , k∈ N} = {{1, 2} , {3, 4} , {5, 6} , · · · } satisfies τ = τF . 4) Also, the deleted integer topology defined by letting X = ⋃ n∈N (n− 1, n)⊂ R and P = {(0, 1), (1, 2), (2, 3), . . .} satisfies τ = τF since every open set is clopen. 5) For any metrizable space, the topology τ induced by the metric also satisfies τ = τF . This is because every singleton is closed and therefore every open set is clopen. Remark 1. The converse of 5 of Example 1 is not true by the following counter examples. Example 2. The indiscrete topology is not metrizable but yet satisfies τ = τF . Example 3. The cofinite topology on R which is defined for a nonempty set X by τcof = {U ⊆ X : X\U isfinite} ∪ {∅} i.e., U⊂ R is open whenever R \U is finite is not metrizable since R is infinite. M. Baloush et al. / Eur. J. Pure Appl. Math, 17 (3) (2024), 1463-1470 1465 However, for any open set, cl(U)\U⊆ R\U , so it is finite and therefore U is F−open. Therefore, τF = τ . In general, if every open set is clopen, then τF = τ. We can also see that if X is finite, then τF = τ. Note that τF is not necessarily another topology on X, since countable union of F−open set is not necessarily an F−open set as shown by the following Example. Example 4. [2] Let An = (n, n+ 1) be a subset of (R,U) for all n ∈ N, then, An = (n, n+1) ∈ U and cl(n, n+1) \ (n, n+1) = {n, n+1} is finite for all n ∈ N. Hence, An = (n, n+1) is F−open for all n ∈ N. Now, we have ⋃ n∈N (n, n+ 1) = [1,∞) \ N is an open set. However, cl( ⋃ n∈N (n, n+ 1))\ ⋃ n∈N (n, n+ 1) = cl([1,∞)\N)\([1,∞)\ N) = [1,∞)\([1,∞)\ N) = N is not a finite set. Therefore, ⋃ n∈N (n, n+ 1) is not F−open. Definition 5. A topological space (X, τ) is called an FT0 − space if for any two distinct points x, y ∈ X there is an F−open set U in X containing x but not y or an F− open set U containing y but not x. Example 5. Let (R,U) be a topological space, where U is the usual topology. Let (a,b) be any open interval in (R,U), since cl(a, b)\(a, b) = [a, b]\(a, b) = {a, b} is finte, then any open interval in (R,U) is F -open. So, for any two distinct points x, y ∈ R there is an open interval (a, b)containing x but not y or an open interval (a, b) containing y but not x.Therefore, (R,U) is an FT0 − space. Definition 6. A topological space (X, τ) is called an FT1 − space if for any two distinct points x, y ∈ X there is an F−open set U in X containing x but not y and an F− open set V in X containing y but not x. Example 6. Let (R,U) be the usual topology for R. Let x, y ∈ R, y > x, and y − x = c. Then U = ( x− c 3 , x+ c 3 ) and V = ( y − c 3 , y + c 3 ) are two F−open sets where x ∈ U, y /∈ U and y ∈ V, x /∈ V .Therefore, (R,U) is an FT1 − space. Theorem 1. An FT1 − space is an FT0 − space, but the converse is not true. Proof. Let (X, τ) be an FT1 − space. Then for any distinct points x, y ∈ X, there exist F−open sets U, V ∈ τF such that: x ∈ U, y /∈ U and y ∈ V, x /∈ V Hence, there exists an F−open set U ∈ τF such that: x ∈ U and y /∈ U or y ∈ U and x /∈ U Therefore,(X, τ) is an FT0 − space. For the converse, let us consider the Sierpinski space, where S = { 0 , 1 } and τ = {∅, {1} , {0, 1}} . It is clear that the Sierpinski space is FTo but not FT1. M. Baloush et al. / Eur. J. Pure Appl. Math, 17 (3) (2024), 1463-1470 1466 Theorem 2. A topological space (X, τ) is an FT1 space if and only if for each x ∈ X, the singleton set {x} is F−closed. Proof. Let (X, τ) be an FT1-space and let x ∈ X be any point. We want to show that {x} is F−closed, i.e., {x} is closed and {x}\int({x}) is a finite set. Since X is an FT1 space, then it is a T1-space and every singleton {x} is closed. Now {x} \int ({x}) = {x} \∅ = {x} is finite. Therefore, {x} is F−closed. Conversely, let every singleton {x} be F−closed, and let x ̸= y be two points in X, then since {x} is F−closed, X\{x} is F− open and contains y but not x. Similarly, X\{y} is F− open and contains x but not y. Therefore, (X, τ) is an FT1- space. Definition 7. A topological space (X, τ) is called an FT2 − space if for any two distinct points x, y ∈ X there exist F−open sets U and V in X, such that x ∈ U and y ∈ V , and U ∩ V = ∅. Example 7. Let us consider the usual topology for R as in Example 5.Then U = ( x− c 3 , x+ c 3 ) and V = ( y − c 3 , y + c 3 ) are two F−open sets where x ∈ U, y ∈ V, x and U ∩ V = ∅. Therefore, (R,U) is an FT2 − space. Theorem 3. An FT2 − space is an FT1 − space, but the converse is not true. Proof. Let (X, τ) be an FT2− space. Then ∀x, y ∈ X : x ̸= y, there exist F−open sets U, V such that: x ∈ U, y ∈ V and U ∩ V = ∅ Therefore, we have U ∈ τF : x ∈ U, y /∈ U and V ∈ τF : y ∈ V, x /∈ V This shows that (X, τ) is an FT1 − space. For the converse, let us consider the cofinite topology τcof on an infinite set X. Then (X, τcof) is an FT1−space but is not FT2. Let x ∈ X be any point, since {x} \int ({x}) = {x} \∅ = {x} is finite, then the singleton {x} is F−closed and hence, by Theorem 2, X is an FT1 − space. To show that X is not an FT2−space, let U and V be any two open subsets of X such that U ∩ V = ∅. Since, U is open and cl(U)\U⊆ R\U , then cl(U)\U is finite and then U is F−open. Similarly, V is also an F−open set. Now, since U ∩ V = ∅, then M. Baloush et al. / Eur. J. Pure Appl. Math, 17 (3) (2024), 1463-1470 1467 (X\U) ∪ (X\V ) = X This is a contradiction since X is infinite. Hence, there is no two distinct points inX belong to two disjoint F−open sets. Therefore, X is not an FT2−space. Definition 8. A topological space (X, τ) is called an F−regular space (briefly Fr − space) if for each closed subset W ⊂ X and each x /∈ W, there exist F−open sets U and V in X, such that x ∈ U , W ⊆ V , and U ∩ V = ∅. An F− regular FT1 − space is called an FT3 − space. Example 8. Let us consider the usual topology for R as in Example 5. It’s easy to show that (R,U) is FT3 − space. Theorem 4. For a topological space (X, τ), the following properties are equivalent: (1) (X, τ) is F -regular; (2) For any x ∈ X and any open set U containing x, there exists an F -open set V such that x ∈ V ⊆ clF (V ) ⊆ U ; (3) For any x ∈ X and any closed set B such that x /∈ B, there exists an F -open set U such that x ∈ U and B ∩ clF (U) = ∅. Proof. (1) ⇒ (2): Let X be an F -regular space and U be any open set containing x. Let K = X \ U , then K is a closed set not containing x. Since (X, τ) is F -regular, then there exist F -open sets V,W of X such that x ∈ V,K ⊂ W and V ∩W = ∅. Hence clF (V ) ∩W = ∅. Therefore, we have x ∈ V ⊆ clF (V ) ⊆ U . (2) ⇒ (3): Let x be any point of X and B be any closed set in X not containing x. Then X \ B is an open set containing x. By (2), there exists an F -open set U such that x ∈ U ⊆ clF (U) ⊆ (X \B). Therefore, there exists an F -open set U such that x ∈ U and B ∩ clF (U) = ∅. (3) ⇒ (1): Let K be a closed set not containing x. Then x ∈ X \K and X \K is an open set. Hence there exists an F -open set U such that x ∈ U and K ∩ clF (U) = ∅. Let V = X \ clF (U). Then V is F -open, K ⊆ V and U ∩ V = U ∩ (X \ cl F (U)) = ∅. This shows that (X, τ) is F -regular. Definition 9. A topological space (X, τ) is called an F−normal space (briefly Fn−space) if for each pair of disjoint closed subsets W1 and W2 of X, there exist F−open sets U and V in X, such that W1 ⊆ U , W2 ⊆ V , and U∩V = ∅. An F− normal FT1−space is called a FT4 − space. Example 9. The usual topology for R as in Example 5 is an Fn and FT1 − space and hence it is an FT4 − space. Remark 2. Every F−normal space in not necessarily F−regular. M. Baloush et al. / Eur. J. Pure Appl. Math, 17 (3) (2024), 1463-1470 1468 Example 10. Let us consider the Sierpinski space, where S = { 0 , 1 } and τ = {∅, {1} , {0, 1}} . It is clear that the Sierpinski space is F−normal since the only closed subsets of S are {0} and ∅, but S is not F-regular, because 1 and {0} which is closed set cannot be sepa- rated by two disjoint F-open sets, i.e., 1 /∈ {0} which is closed sets, 1 ∈ {1}, {0} ⊆ {0, 1} where {1} , {0, 1} are F−open sets and {1} ∩ {0, 1} = {1} ≠ ∅ Theorem 5. Every FT4 − space is an FT3 − space. Proof. Let (X, τ) be an FT4 − space. Then (X, τ) is an FT1 space, hence we need only show that X is an Fr− space. For that, let K ⊆ X be an F − closed set such that x /∈ K. Since X is an FT1 space, then {x} is an F -closed set such that {x} ∩ K = ∅. Since X is an Fn−space, there exist F−open sets U and V in X, such that {x} ⊆ U , K ⊆ V , and U ∩ V = ∅. Hence, for each closed subset K ⊂ X and each x /∈ K, there exist F−open sets U and V in X such that x ∈ U , K ⊆ V , and U ∩ V = ∅. Therefore, X is Fr− regular. Theorem 6. For a topological space (X, τ), the following properties are equivalent: (1) (X, τ) is F -normal; (2) For any closed set F and any open set U containing F , there exists an F -open set V such that F ⊆ V ⊆ clF (V ) ⊆ U ; (3) For any disjoint closed sets A,B, there exists an F -open set U such that A ⊆ U and B ∩ clF (U) = ∅. Proof. The proof is similar with Theorem 4. Lemma 1. Let (X, τ) be a topological space and Y be a subset of X. If U is an F -open set of X, then U ∩ Y is an F -open set of the subspace Y . Proof. Since U is open in X, U ∩ Y is open in Y . clY (U ∩ Y ) \ (U ∩ Y ) = clX(U ∩ Y ) ∩ Y \ (U ∩ Y ) ⊆ clX(U) ∩ Y \ (U ∩ Y ) ⊂ clX(U) \ U . Since U is an F -open set of X, clX(U) \ U is a finite set. Therefore, U ∩ Y is F -open in the subspace Y . Theorem 7. If X is an FTi − space, then any subspace Y of X is an FTi − space for i = 0, 1, 2, 3, 4. Proof. We consider only the cases i = 1, 3 and the other cases can be proved by the same argument. 1) Let i = 1. Let x, y be any distinct points of a subspace Y of an FT1-space X. There exists an F -open set U of X such that x ∈ U and y /∈ U . By Lemma 1, U ∩ Y is an F -open set of Y such that x ∈ U ∩ Y and y /∈ U ∩ Y . Similarly, we can find an F - open set V in Y containing y but not x. Therefore, the subspace Y is an FT1−space. M. Baloush et al. / Eur. J. Pure Appl. Math, 17 (3) (2024), 1463-1470 1469 2) Let i = 3. Let K be any closed set in Y and y ∈ Y \ K. Then there exists a closed set KX in X such that K = KX ∩ Y , where y /∈ KX . Since X is FT3, there exist disjoint F -open sets UX , VX in X such that y ∈ VX and KX ⊂ UX . Now, let U = UX ∩ Y, V = VX ∩ Y , then by Lemma 1, U, V are F -open sets in Y and K = KX ∩Y ⊂ UX ∩Y = U and y ∈ VX ∩Y = V . Moreover, U ∩V ⊂ UX ∩VX = ∅. Therefore, Y is FT3. Lemma 2. [2] Let (X, τ) be a topological space. For F -open sets of X, the following properties hold: (1) The finite union of F -open sets is F -open, (2) The finite intersection of F -open sets is F -open. Theorem 8. Let (X, τ) be a topological space. Then every FTi space is a Ti space for i = 0, 1, 2, 3, 4. However, the converse is true whenever X is finite. Proof. Easy, since F−open sets are open sets. For the converse, let X be a finite set, then every open set is a F−open set and then every Ti space is a FTi space for i = 0, 1, 2, 3, 4. Theorem 9. If (X, τ) is F -compact and FT2, then (X, τ) is FT4. Proof. Let K,L be any disjoint closed sets. Then they are F -compact. Let k be any point of K. For any pointx ∈ L, x /∈ K and x ̸= k. Since (X, τ) is FT2, there exist disjoint F -open sets Ux(k) and Vk(x) such that k ∈ Ux(k) and x ∈ Vk(x). Now, fix x, then ∪{Ux(k) : k ∈ K} is F -open cover of K and there exists a finite subset K0 of K such that K ⊂ ∪{Ux(k) : k ∈ K0}. Now, put U(x) = ∪{Ux(k) : k ∈ K0} and VK0(x) = ∩{Vk(x) : k ∈ K0}. Then U(x) ∩ VK0(x) = ∅ for each x ∈ L. Since ∪{VK0(x) : x ∈ L} is an F -open cover of L and L is F -compact, there exists a finite subset L0 of L such that L ⊂ ∪{VK0(x) : x ∈ L0}. Now, put VL =⊂ ∪{VK0(x) : x ∈ L0} and UK = ∩{U(x) : x ∈ L0}. Then, by Lemma 2, UK , VL are disjoint F -open sets such that L ⊂ VL,K ⊂ Uk. Therefore, (X, τ) is FT4. Definition 10. A function f : (X, τ) → (Y, σ) is said to be F -closed preserving if for any F -closed set K of X, f(K) is F -closed in Y . Theorem 10. A surjective function f : (X, τ) → (Y, σ) is F -closed preserving if and only if for any subset S of Y and any F -open set U such that f−1(S) ⊆ U , there exists an F -open set V in Y such that S ⊂ V and f−1(V ) ⊆ U . Proof. (⇒) Let S be any subset of Y and U be any F -open set such that f−1(S) ⊆ U . Since X \U is F -closed, f(X \U) is F -closed in Y . Since f−1(S) ⊆ U , X \U ⊆ X \f−1(S). Let V = Y \ f(X \ U). Then we have S ⊆ Y \ f(X \ U) = V and f−1(V ) ⊆ U . (⇐) Let K be any F -closed in X. We show that f(K) is F -closed in Y . Let S = Y \f(K). Then f−1(S) = f−1(Y \f(K)) ⊆ X \K. Since X \K is F -open, there exists an F -open set V such that Y \ f(K) ⊂ V and f−1(V ) ⊂ X \K. Hence K ⊂ X \ f−1(V ) = f−1(Y \ V ) and f(K) ⊆ Y \ V . On the other hand, we have Y \ V ⊆ f(K). Therefore, Y \ V = f(K) and f(K) is F -closed in Y . REFERENCES 1470 Theorem 11. Let f : (X, τ) → (Y, σ) be an F -closed preserving continuous surjection. If (X, τ) is F -normal, then (Y, σ) is F -normal. Proof. Let (X, τ) be F -normal. For any disjoint closed sets K1 and K2 in Y , f−1(K1) and f−1(K2) are disjoint closed sets in X since f is cotinuous. Since (X, τ) is F -normal, there exist F -open sets U1, U2 in X such that f−1(Ki) ⊆ Ui (i = 1, 2) and U1∩U2 = ∅. By Theorem 10, there exist F -open sets Vi such that Ki ⊂ Vi, f −1Vi) ⊆ Ui (i = 1, 2). Since U1 ∩ U2 = ∅ and f is surjective, V1 ∩ V2 = ∅. Therefore, (Y, σ) is F -normal. Corollary 1. Let f : (X, τ) → (Y, σ) be an F -closed preserving continuous surjection. If (X, τ) is FT4, then (Y, σ) is FT4. Proof. It is necessary to show that if (X, τ) is FT1, then (Y, σ) is FT1. For any point y ∈ Y , there exists x ∈ X such that f(x) = y. Since X is FT1, by Theorem ??, {x} is F -closed in X and f({x}) = {y} is F -closed in Y . Conclusion. By using F -open sets in a topological space, we defined separation ax- ioms FTi(i = 0, 1, 2, 3, 4) and obtained their properties, characterization and relationships. Furthermore, it was shown that (1) F -compact FT2-spaces are FT4, (2) F -normality is preserved under F -closed preserving continuous surjections, (3) for FTi(i = 0, 1, 2, 3, 4) the following DIAGRAM holds: FT4 ⇒ FT3 ⇒ FT2 ⇒ FT1 ⇒ FT0 ⇓ ⇓ ⇓ ⇓ ⇓ T4 ⇒ T3 ⇒ T2 ⇒ T1 ⇒ T0 Acknowledgements The authors extend their appreciation to the National University of Malaysia for sup- porting this work under the research grant : GP-2021-K014049 Ganjaran Penerbitan. References [1] M. H. Alqahtani and A.M. Abd El-latif. Separation axioms via novel operators in the frame of topological spaces and applications. AIMS Mathematicsl, 9(6):14213–14227, 2024. [2] M.H. Alqahtani. F-open and f-closed sets in topological spaces. Eur. J. Pure Appl. Math., 6(2):819–832, 2023. [3] A.B. Khalaf, H.M. Darwesh, and K. Kannan. Some Types of Separation Axioms in Topological Spaces. Tamsui Oxford Journal of Information and Mathematical Sciences, 28(3):303–326, 2012.