EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 15, No. 2, 2022, 589-601 ISSN 1307-5543 – ejpam.com Published by New York Business Global Some Properties of Weak Separation Axioms in Coc-Compact Sets Fuad A. Abushaheen Basic Science Department, Middle East University, Amman, Jordan Abstract. In this paper, we introduce some separation axioms in coc-compact set, namely coc-T0- space, coc-T 1 4 - space, coc-T 3 8 - space, coc-T 1 2 - space, coc-T 5 8 - space, coc-Di- space, coc-Ri- space for i = 0, 1, weak coc-D1- space and weak coc-R0- space, and we study some relations between them, also we prove that some of these separation axioms have ”hereditary property”. 2020 Mathematics Subject Classifications: 54B05, 54B10, 54D20. Key Words and Phrases: coc-Ti- space for i = 0, 1 4 , 1 2 , 5 8 , 3 4 , coc-Di- space, coc-Ri- space for i = 0, 1, weak coc-D1- space and weak coc-R0- space 1. Introduction and Preliminaries In [4], the authors defined a new type of open sets called coc-compact set as a gener- alizations of open sets. After this paper many papers in this concept were appeared, see [1–3]. Also many authors studied weak separation axioms in different types of open sets, for example [5, 8, 9] . Definition 1. [4] A subset A of a topological space (X, τ) is called co-compact open set (notation: coc-open) if for every x ∈ A, there exists an open set U ⊆ X and a compact subset K of X such that x ∈ U −K ⊆ A. The complement of a coc-open subset is called coc-closed. The family of all coc-open subsets of a topological space X will be denoted by τk. Theorem 1. [4] Let (X, τ) be a topological space. Then (i) The collection τk forms a topology on X with τ ⊆ τk. (ii) The set {U −K : U ∈ τ and K is compact in X} forms a base for τk . Lemma 1. [4] Let (X, τ) be a topological space and A be a closed subset of X. Then( τ |A )k = τk |A . DOI: https://doi.org/10.29020/nybg.ejpam.v15i2.4298 Email address: fshaheen@meu.edu.jo (F.A. Abushaheen) https://www.ejpam.com 589 © 2022 EJPAM All rights reserved. F.A. Abushaheen / Eur. J. Pure Appl. Math, 15 (2) (2022), 589-601 590 Throughout this paper, we use R,Q and N to denote the set of real numbers, rational numbers and natural numbers, respectively. The coc-closure of A and the coc- interior of A will be denoted by A cocand intcoc(A), respectively. Terms and notations not explained in this paper are taken from [4, 7]. 2. Coc-T -spaces and Coc-D-spaces Definition 2. A space (X, τ) is called co-compact-T0-space (coc-T0-space) if for all x 6= y ∈ X, there exists a coc-open set U contains one point but not other. Definition 3. A subset A of a topological space (X, τ) is called coc-D-set if A = U − V , for some U, V ∈ τk. Definition 4. A space (X, τ) is called co-compact-D0-space (coc-D0-space) if for all x 6= y ∈ X, there exists a coc-D-set U contains one point but not other. Theorem 2. A coc-closed subspace of a coc-D0-space (X, τ) is coc-D0-space. Proof. Let A be a coc-closed subset X and let x 6= y ∈ A. So there exists a coc-D-set D = U − V with U, V ∈ τk such that x ∈ D and y /∈ D. Now x ∈ D ∩A = (U − V )∩A = (A∩U)− (A∩ V ), so by Lemma 1 we have A∩U and A∩ V ∈ (τ |A)k = τk |A, hence the result. Theorem 3. A space (X, τ) is coc-T0-space if and only if it is coc-D0-space. Proof. (⇒) It is clear since every proper coc-open subset of X is coc-D-set. (⇐) Let x 6= y ∈ X, so there exists a coc-D0- set U contains x with U = U1 − U2 where U1, U2 ∈ τk i.e. x ∈ U1 and x /∈ U2. For y, we have the following cases:(1) If y /∈ U1, we are done. (2) If y ∈ U1 and y ∈ U2, so U2 contains y but not x. Theorem 4. A space (X, τ) is coc-T0-space if and only if for all x 6= y ∈ X, we have {x}coc = {y}coc. Proof. (⇒) Let x 6= y ∈ X, there exists a coc- open set U contains one point but not other, say x ∈ U and y /∈ U . Then X−U is a coc-closed set contains y and {y}coc ⊆ X−U, so x /∈ {y}coc, hence {x}coc 6= {y}coc. (⇐) Let x 6= y ∈ X. Then it is clear that X − {y}coc is coc-open set contains x but not y, hence X is coc-T0-space. Definition 5. A space (X, τ) is called co-compact -T1-space (coc-T1-space) if for all x 6= y ∈ X, there exist coc-open sets Ux, Vy with {Ux, Vy}∩ τ 6= φ such that x ∈ Ux, y ∈ Vy and y /∈ Ux, x /∈ Vy. Definition 6. [3] A space (X, τ) is called co-compact -T2-space (coc-T2-space) if for all x 6= y ∈ X, there exist coc-open sets Ux, Vy with {Ux, Vy}∩ τ 6= φ such that x ∈ Ux, y ∈ Vy and Ux ∩ Vy = φ. F.A. Abushaheen / Eur. J. Pure Appl. Math, 15 (2) (2022), 589-601 591 It is clear that if (X, τ) is coc-T1-space, then (X, τk) is T1-space. And every T1-space is coc-T1-space, but the converse need not be true, consider the following example. Example 1. Let X = R and τ = {φ} ∪ {U ⊆ R, 0 ∈ U}. Proof. A space (X, τ) is coc-T1- space, to prove this let x 6= y ∈ X, so we have the following cases : (i) For x = 0, y 6= 0, let U = {0} and V = {y, 0}−{0}, then U, V ∈ τk and {U, V }∩τ = {V } with x /∈ V and y /∈ U . (ii) For y = 0, x 6= 0, same as (i). (iii) For x 6= 0, y 6= 0, let U = {x, 0}, V = {y, 0} − {0}, then U, V ∈ τk and {U, V } ∩ τ = {V }, then x /∈ V and y /∈ U . But (X, τ) is not T1- space, for instance take x = 0, y = 1, then there is no open set contains y but not x. Theorem 5. A space (X, τ) is coc-T1-space if and only if every singleton is coc-closed. Definition 7. A space (X, τ) is called co-compact-D1-space (coc-D1-space) if for all x 6= y ∈ X, there exist coc-D-sets Ux, Vy such that x ∈ Ux , y ∈ Vy and y /∈ Ux, x /∈ Vy. Theorem 6. A coc-closed subspace of a coc-D1-space (X, τ) is coc-D1-space. Definition 8. A space (X, τ) is called co-compact-D2-space (coc-D2-space) if for all x 6= y ∈ X, there exist disjoint coc-D-sets Ux, Vy such that x ∈ Ux, y ∈ Vy and y /∈ Ux, x /∈ Vy. Theorem 7. Let (X, τ) be a topological space. Then: (i) If X is coc-Ti-space, then X is coc-Ti−1-space for i = 1, 2. (ii) If X is coc-Ti-space, then X is coc-Di-space for i = 1, 2. (iii) If X is coc-Di-space, then X is coc-Di−1-space for i = 1, 2. (iv) If X is coc-D1-space, then X is coc-T0-space. (v) X is coc-D1-space if and only if X is coc-D2-space . Proof. We will prove (v) only. (⇐) Obvious. (⇒) For x 6= y ∈ X, there exist coc-D-sets U1, U2 with x ∈ U1, y /∈ U1 and y ∈ U2, x /∈ U1, assume U1 = V1−W1, U2 = V2−W2 where V1,W1, V2,W2 ∈ τk. Then for x /∈ U2, we have the following cases: (1) x /∈ V2 (2) x ∈ V2 and x ∈ W2. For (1) If x /∈ V2, we have: (i) If y /∈ V1, x ∈ V1 − W1, then x ∈ V1 − (V2 ∪ W1) and y ∈ V2 − W2, so y ∈ V2 − (V1 ∪ W2) and F.A. Abushaheen / Eur. J. Pure Appl. Math, 15 (2) (2022), 589-601 592( V1−(V2∪W1) ) ∩ ( V2−(V1∪W2) ) = φ. (ii) If y ∈ V1 and y ∈ W1, we have x ∈ U1−U2, y ∈ U2 and ( U1 − U2 ) ∩ U2 = φ. For (2) If y ∈ U2 = V2 −W2 , then x ∈ W2 and ( V2 −W2 ) ∩W2 = φ. From (1) and (2), X is coc-D2-space . The following theorem gives improvement of Theorem 7(iv). Theorem 8. A space (X, τ) is coc-D1-space if and only if X is coc-T0-space and intcoc(Ax) 6= X for all x ∈ Ax ⊆ X. Proof. (⇒) For x ∈ X, there exists a coc-D-set Ox = U − V with U, V ∈ τk and x ∈ Ox, but U 6= X, so intcoc(U) 6= X, hence the result. (⇐) For x 6= y ∈ X, with out loss of generality there exists a coc-open set U contains x but not y and there exists coc-open set V contains y and intcoc(V ) 6= X, hence y ∈ V −U , therefore X is coc-D1-space. 3. Coc-R0 and Coc-R1-spaces Definition 9. A space (X, τ) is called co-compact-R0-space (coc-R0-space) if every coc- open set contains the coc-closure of its singletons, i.e. for each coc-open set O we have {x}coc ⊆ O for all x ∈ O. Definition 10. A space (X, τ) is called co-compact-R1-space (coc-R1-space) if for x 6= y ∈ X with {x}coc 6= {y}coc, then there exist disjoint coc-open sets U, V with {x}coc ⊆ U , {y}coc ⊆ V . The following theorem is obvious. Theorem 9. Let (X, τ) be a topological space. Then: (i) A coc-closed subspace of a coc-R0-space X is coc-R0-space. (ii) A coc-closed subspace of a coc-R1-space X is coc-R1-space. Theorem 10. Every coc-R1-space (X, τ) is coc-R0-space. Proof. Let U be a coc-open set in X with x ∈ U . For y /∈ U , we have x /∈ {y}coc, thus {x}coc 6= {y}coc, but X is coc-R1-space, so there exits a coc-open set Vy contains y such that {y}coc ⊆ Vy and x /∈ Vy, hence {x}coc ⊆ U , thus X is coc-R0-space. Theorem 11. A space (X, τ) is coc-T1-space if and only if it is coc-T0-space and coc-R0- space. Proof. (⇒) Notes that {x} is coc-closed subset of X for all x ∈ X. (⇐) Let x 6= y ∈ X, with out loss of generality there exists a coc-open set O with x ∈ O ⊆ X − {y}. Thus x /∈ {y}coc, so y /∈ {x}coc, hence X − {x}coc is coc-open set contains y but not x. F.A. Abushaheen / Eur. J. Pure Appl. Math, 15 (2) (2022), 589-601 593 Corollary 1. Let (X, τ) be a coc-R0-space. Then the following are equivalent: (i) X is coc-T2-space, (ii) X is coc-T1-space, (iii) X is coc-T0-space. Definition 11. Let (X, τ) be a topological space and A ⊆ X. Then the coc-Kernal of A define by: coc-ker(A) = ∩{U ∈ τk : A ⊆ U}, if there no coc-open set contains A, then coc-ker(A) = X. Lemma 2. If (X, τ) is a topological space and A is a subset of X, then coc-ker(A) = {x ∈ X : {x}coc ∩A 6= φ}. Proof. For x /∈ coc-ker(A), there exists a coc-open set U contains A and x /∈ U , then {x}coc ∩ U = φ. For {x}coc ∩ U = φ, we have x /∈ X − {x}coc , thus x /∈ coc-ker(A). Lemma 3. Let (X, τ) be a topological space and x ∈ X. Then y ∈ coc-ker({x}) if and only if x ∈ {y}coc. Theorem 12. Let (X, τ) be a topological space and x 6= y ∈ X. Then coc-ker({x}) 6= coc-ker({y}) if and only if {x}coc 6= {y}coc. Proof. (⇒) Let w ∈ coc-ker({x}) and w /∈ coc-ker({y}). Then {w}coc ∩ {x} 6= φ and {w}coc ∩ {y} = φ, so x ∈ {w}coc, and hence {x}coc ⊆ {w}coc, therefore {w}coc ∩ {y} = φ and hence y /∈ {x}coc. (⇐) Since coc-ker({x}) 6= coc-ker({y}), there is z ∈ {x}coc and z /∈ {y}coc, hence there exists a coc-open set Uz with x ∈ Uz and y /∈ Uz, so y /∈ coc-ker({x}). Theorem 13. A space (X, τ) is coc-R0-space if and only if for x 6= y ∈ X, {x}coc 6= {y}coc gives {x}coc ∩ {y}coc = φ. Proof. (⇐) Let x ∈ Ox ∈ τk and assume that y /∈ Ox. Then x /∈ {y}coc, hence {x}coc 6= {y}coc, so {x}coc ∩ {y}coc = φ, therefore y /∈ {x}coc and {x}coc ⊆ Ox, so X is coc-R0-space. (⇒) Let x 6= y ∈ X with {x}coc 6= {y}coc. So there exists z ∈ {x}coc and z /∈ {y}coc, then z ∈ X−{y}coc, so there exists a coc-open set U contains z but not y, but z ∈ {x}coc, so x ∈ U and x /∈ {y}coc, hence {x}coc ⊆ X − {y}coc, therefore {x}coc ∩ {y}coc = φ. Theorem 14. A space (X, τ) is coc-R0-space if and only if for x 6= y ∈ X, coc-ker({x}) 6= coc-ker({y}) gives coc-ker({x}) ∩ coc-ker({y}) = φ. F.A. Abushaheen / Eur. J. Pure Appl. Math, 15 (2) (2022), 589-601 594 Proof. (⇒) Let X be a coc-R0-space and for x 6= y ∈ X with coc-ker({x}) 6= coc-ker({y}). Let w ∈ coc-ker({x})∩coc-ker({y}). Then w ∈ coc-ker({x}) so x ∈ {w}coc, and then by Lemma 3 {x}coc = {w}coc, in same method we have {y}coc = {w}coc, and this is a contradiction which completes the proof. (⇐) Assume {x}coc 6= {y}coc, then coc-ker({x}) 6= coc-ker({y}), so coc-ker({x}) ∩ coc-ker({y}) = φ. If z ∈ {x}coc, then x ∈ coc-ker({z}) and coc-ker({x})∩ coc-ker({z}) = φ, so coc-ker({x}) = coc-ker({z}). Now for z ∈ {x}coc ∩ {y}coc , we have coc-ker({x}) = coc-ker({y}) = coc-ker({z}), and this is a contradiction, hence {x}coc ∩ {y}coc = φ. Theorem 15. For a topological space (X, τ). The following are equivalent: (i) X is a coc-R0-space, (ii) For a subset A of X and G coc-open set of X such that A ∩ G 6= φ, there exists a coc-closed subset F of X such that A ∩ F 6= φ and F ⊆ G, (iii) For any coc-open set G of X, G = ∪{F : F is coc-closed subset with F ⊆ G}, (iv) For any coc-closed subset F of X, F = coc-ker(F ), (v) For any x ∈ X, {x}coc ⊆ coc-ker({x}). Proof. (iii) ⇒ (iv), (v) ⇒ (i) Obvious. (i) ⇒ (ii) Let A ⊆ X and G is a coc-open set and let x ∈ A ∩ G. Then the needed coc-closed subset F is {x}coc. (ii) ⇒ (iii) For a coc-open set G ⊇ ∪{F : F is a coc-closed with F ⊆ G}, let x ∈ G, then there exists a coc-closed set F such that x ∈ F and F ⊆ G, so x ∈ F ⊆ ∪{F : F is a coc-closed, F ⊆ G}, hence the result. (iv) ⇒ (v) Let x ∈ X and y /∈ coc-ker({x}), there exists a coc-open set Ux contains x with y /∈ U , so {y}coc ∩ U = φ and hence coc-ker ( {y}coc ) ∩ U = φ, therefore there exists a coc-open set Oy such that x /∈ Oy and {y}coc ⊆ Oy, so {x}coc ∩Oy = φ and y /∈ {x}coc, hence the result. Lemma 4. A topological space (X, τ) is coc-R0-space if and only if for each x 6= y ∈ X with x ∈ {y}coc gives y ∈ {x}coc Proof. (⇒) Let X be a coc-R0-space and x ∈ {y}coc. If U is any coc-open set with y ∈ U , then x ∈ U and any coc-open set contains y must contains x, hence y ∈ {x}coc. (⇐) Let U be a coc-open set with x ∈ U . For x ∈ {y}coc, we have y ∈ {x}coc, therefore {x}coc ⊆ U , hence X is coc-R0-space. Theorem 16. For a topological space (X, τ). The following are equivalent: (i) X is a coc-R0-space, (ii) If F is a coc-closed subset of X with x ∈ F , then coc-ker({x}) ⊆ F , F.A. Abushaheen / Eur. J. Pure Appl. Math, 15 (2) (2022), 589-601 595 (iii) If x ∈ X, then coc-ker({x}) ⊆ {x}coc. Proof. (ii) ⇒ (iii) Obvious . (i) ⇒ (ii) Let F be a coc-closed and x ∈ F . So coc-ker({x}) ⊆ coc-ker(F ), then by Theorem 15 we have coc-ker({x}) ⊆ F . (iii) ⇒ (i) Let x ∈ {y}coc. So y ∈ coc-ker({x}), therefore by (iii) y ∈ {x}coc, and the result comes from Lemma 4. Corollary 2. A topological space (X, τ) is coc-R0-space if and only if coc-ker({x}) = {x}coc for all x ∈ X. 4. Coc-T 1 2 -space, Coc-T 3 8 -space and Coc-T 1 4 -space In this section we define more weak separation axioms in coc-open set, but before this we need some definitions and lemmas. Definition 12. Let A be a subset of a topological space (X, τ). Then A is called coc-g- closed if {A}coc ⊆ U , whenever A ⊆ U and U is coc-open set. A is called coc-g-open if X −A is coc-g-closed. Clearly, A is a coc-g-closed of (X, τ) if F ⊆ intcoc(A), whenever F ⊆ A and F is coc-closed set of X. Definition 13. Let A be a subset of a topological space (X, τ). Then coc-A∨ = ∪{F : X − F ∈ τk : F ⊆ A}, if there is no coc-closed set contains in A, then coc-A∨ = φ. Lemma 5. Let A be a subset of a topological space (X, τ). Then A is coc-g-closed (coc- g-open) if and only if {A}coc ⊆ coc-ker(A) (coc-A∨ ⊆ intcoc(A)). Definition 14. Let A be a subset of a topological space (X, τ) . Then A is called coc-∧- set (coc-∨-set) if A = coc-ker(A)(A = coc-A∨), or equivalently, A is the intersection of coc-open sets or A = X(A is the union of coc-closed sets or A = φ). Lemma 6. Let A,B are subsets of a topological space (X, τ). Then : (i) coc-ker{φ} = φ, coc-φ∨ = φ, coc-ker{X} = X, coc-X∨ = X. (ii) A ⊆ coc-ker(A), coc-A∨ ⊆ A. (iii) coc-ker(coc-ker(A))=coc-ker(A), coc- ( coc-A∨)∨ = coc-A∨. (iv) If A ⊆ B, then coc-ker(A) ⊆ coc-ker(B). (v) If A ⊆ B, then coc-A∨ ⊆ coc-B∨. Lemma 7. Let (X, τ) be a topological space. Then the following are hold : (i) If A is coc-∧-set (coc-A∨-set), then A is coc-g-closed (coc-g-open) if and only if A is coc-closed (coc-open). F.A. Abushaheen / Eur. J. Pure Appl. Math, 15 (2) (2022), 589-601 596 (ii) For A ⊆ X, if coc-ker(A) is coc-g-closed set (coc-A∨ is coc-g-open set), then A is coc-g-closed (coc-g-open). Proof. (i) Obvious. (ii) From Lemma 5 and Lemma 6. The following definition gives a weaker form of coc-∧-set. Definition 15. A subset A of a space (X, τ) is called generalized coc-kernal set (g-coc-∧- set) if coc-ker(A) ⊆ {A}coc, or equivalently coc-ker(A) ⊆ F , whenever A ⊆ F and F is coc-closed. A subset A of a space (X, τ) is called generalized coc-∨-set (g-coc-∨-set) if X −A is g-coc-∧-set, or equivalently intcoc(A) ⊆ coc−A∨. Lemma 8. Let A be subset of a topological space (X, τ). If A is coc-∧-set (coc-A∨-set), then it is g-coc−∧-set (g-coc-∨-set). Theorem 17. Let (X, τ) be a topological space. Then for x ∈ X, {x} is either coc-open or g-coc-∨-set. Proof. Let x ∈ X and {x} is not coc-open subset of X. Hence X − {x} is not coc- closed subset of X and {X − {x}}coc = X, so coc-ker (X−{x}) ⊆ {X − {x}}coc, therefore X − {x} is g-coc-∧-set, i.e. {x} is g-coc-∨-set. Definition 16. A topological space (X, τ) is called coc-T 1 2 -space if every coc-g-closed subset of X is coc-closed. Lemma 9. Let (X, τ) be a topological space and A ⊆ X. Then A is coc-g-closed subset if and only if Acoc −A contains no coc-closed subset of X. Proof. (⇐) Obvious. (⇒) Let A be coc-g-closed and assume there exists a coc-closed subset F with A ⊆ X−F . Since A is coc-g-closed set, we have A coc ⊆ X − F , hence F ⊆ X − A coc and this is a contradiction which completes the proof. Theorem 18. A topological space (X, τ) is coc-T 1 2 -space if and only if every singleton of X is coc-open or coc-closed. Proof. (⇒) Let x ∈ X and {x} is not coc-closed set. Hence X − {x} is not coc-open, therefore X is the only coc-open set with X − {x} ⊆ X, that is mean X − {x} is coc-g- closed, so X − {x} is coc-closed, i.e. {x} is coc-open. (⇐) Let x ∈ X and A is coc-g-closed subset of X with x ∈ A coc. If {x} is a coc-open set, then {x}∩A 6= φ and hence x ∈ A. If {x} is a coc-closed, then by Lemma 9, x /∈ A coc−A, hence x ∈ A and A = A coc, therefore X is coc-T 1 2 -space. Corollary 3. Every coc-T1-space is coc-T 1 2 -space . Theorem 19. For a topological space (X, τ). The following are equivalent: F.A. Abushaheen / Eur. J. Pure Appl. Math, 15 (2) (2022), 589-601 597 (i) X is coc-T 1 2 -space, (ii) Every g-coc-∧- set is coc-∧-set, (iii) Every g-coc-∨-set is coc-∨-set. Proof. (iii) ⇒ (ii) Obvious. (ii) ⇒ (i) Let x ∈ X. If {x} is not coc-open, then X − {x} is not coc-closed, so the only coc-open set contains X − {x} is X, but X − {x} is g-coc-∧-set, so X − {x} is coc-∧-set, therefore X − {x} is coc-open, hence {x} is coc-closed set, that’s complete the proof. (i) ⇒ (ii) Assume that a subset A of X is g-coc-∧-set which is not coc-∧-set, then coc-ker(A) * A, so there exists x ∈ coc-ker(A) and x /∈ A, but X is a coc-T 1 2 -space, so {x} is a coc-open or coc-closed set, we need to discuss the following two cases: (1) If {x} is a coc-closed, then X−{x} is a coc-open set contains A, but x ∈ coc−ker(A), so x ∈ X − {x} and this is a contradiction. (2) If {x} is coc-open set, then X − {x} is a coc-open set contains A, by assumption coc-ker (A) ⊆ X − {x}, i.e. x /∈coc-ker(A) and this is a contradiction, hence A is coc-∧-set. Definition 17. A subset A of a topological space (X, τ) is called coc-λ-closed if A = L∩F, where L is coc-∧-set and F is coc-closed set. A subset A is coc-λ-open if X −A is coc-λ- closed. Lemma 10. For a subset A of (X, τ). The following are equivalent : (i) A is coc-λ-closed, (ii) A = L ∩A coc, where L is coc-∧-set, (iii) A=coc-ker(A) ∩A coc. Theorem 20. A topological space (X, τ) is coc-T 1 2 -space if and only if every subset of X is coc-λ-closed. Proof. (⇐) Let x ∈ X. Assume that {x} is not coc-open, then A = X − {x} is not coc-closed, but A is coc-λ-closed, so A is coc-∧-set, thus A is coc-open set, then A is coc-open, that is {x} is coc-closed, which is complete the proof. (⇒) Let A ⊆ X and x ∈ X − A. Then {x} is coc-open or coc-closed subset of X. Define B = {x ∈ X − A, {x} ∈ τk}, C = {x ∈ X − A,X − {x} ∈ τk}. Also define F = ⋂ x∈B ( X − {x} ) = X − B, and L = ⋂ x∈C ( X − {x} ) = X − C, then F is coc-closed set and L is coc-∧-set with L ∩ F = A, hence A is coc-λ-set. Definition 18. A topological space (X, τ) is called coc-T 1 4 -space if every finite subset F of X and every y ∈ X −F , there exists a set Ay with F ⊆ Ay such that {y} ∩Ay = φ and Ay is either coc-open or coc-closed. F.A. Abushaheen / Eur. J. Pure Appl. Math, 15 (2) (2022), 589-601 598 Theorem 21. A topological space (X, τ) is coc-T 1 4 -space if and only if every finite subset of X is coc-λ-closed. Proof. (⇒) Let F be any finite subset of X and y ∈ X−F . So there exist a set Ay such that Ay∩{x} = φ, and Ay is either coc-open or coc-closed. Let C be the intersection of all coc-open sets Ay and let L be the intersection of all coc-closed sets Ay, clearly F = C ∩L, C is coc-∧-set and L is coc-closed set, hence F is coc-λ-closed set. (⇐) Let F = L ∩ C and y ∈ X − F where C is coc-∧-set and L coc-closed set. If y /∈ C, we are done. If y ∈ C, then y /∈ L, so there exists a coc-open set Uy with y ∈ Uy, hence X is coc-T 1 4 -space. Definition 19. A topological space (X, τ) is called coc-T 3 8 -space if every countable subset F of X and every y ∈ X − F , there exists a set Ay with F ⊆ Ay such that {y} ∩ Ay = φ and Ay is coc-open or coc-closed. Clearly every coc-T 1 2 -space is coc-T 3 8 -space and hence coc-T 1 4 -space. Theorem 22. A topological space (X, τ) is coc-T 3 8 -space if and only if every countable subset of X is coc-λ-closed. Proof. Same as Theorem 21. In the end of this section, we give weak forms of coc-D1-space and coc-R0-space. Definition 20. A topological space (X, τ) is called weak coc-D1-space if ⋂ x∈X {x}coc = φ. Theorem 23. A coc-closed subspace of weak coc-D1-space (X, τ) is weak coc-D1-space. Theorem 24. A topological space (X, τ) is weak coc-D1-space if and only if intcoc(Ax) 6= X for all x ∈ Ax ⊆ X. Proof. (⇒) Assume that there exists y ∈ X with intcoc({Ay}) = X, then y ∈ {x}coc for each x ∈ X, this is a contradiction, hence the result. (⇐) Let y ∈ ⋂ x∈X {x}coc, then the coc-open set contains y must be X, so intcoc({Ay}) = X, this is a contradiction, hence the result. Corollary 4. A topological space (X, τ) is coc-D1-space if and only if (X, τ) is coc-T0-space and weak coc-D1-space. Theorem 25. A topological space (X, τ) is weak coc-D1-space if and only if coc-ker({x}) 6= X for all x ∈ X. Proof. (⇒) Obvious. (⇐) From Theorem 24. F.A. Abushaheen / Eur. J. Pure Appl. Math, 15 (2) (2022), 589-601 599 Definition 21. A topological space (X, τ) is called weak coc-R0-space if every coc-λ-closed singleton is a coc-∧-set. Theorem 26. Every coc-R0-space (X, τ) is weak coc-R0-space. Proof. Let x ∈ X with {x} is coc−λ−closed. By Lemma 10 {x} = coc-ker({x}) ∩ {x}coc. If {x} is not coc-ker-set, then there exists y ∈ coc−ker({x})−{x} with y /∈ {x}coc, but X is coc-R0-space, so {x}coc ∩ {y}coc = φ and x ∈ {y}coc, therefore there exists a coc- open set Ux contains x but not y, thus y /∈ coc − ker({x}), and this is a contradiction which completes the proof. The following theorems are easily to prove. Theorem 27. For a topological space (X, τ). The following are equivalent: (i) X is coc-T1-space, (ii) Every subset of X is coc-∧-set, (iii) Every singleton of X is coc-∧-set. Theorem 28. For a topological space (X, τ). The following are equivalent: (i) X is coc-T1-space, (ii) X is coc-T0-space and coc-R0-space, (iii) X is coc-T0-space and weak coc-R0-space. Corollary 5. For a weak coc-R0-space (X, τ). The following are equivalent: (i) X is coc-T0-space, (ii) X is coc-T 1 4 -space, (iii) X is coc-T 3 8 -space, (iv) X is coc-T 1 2 -space, (v) X is coc-T1-space. 5. Hereditary Property for Weak Coc-compact Separation Axioms In this section, we discuss the known problem that appeared by Arenas [6] “If every subspace of a topological space X has a property, then the space X has this property” in weak separation axioms via coc-open sets. Theorem 29. If every proper subspace of a topological space (X, τ) is coc-T 1 2 -space, then X is coc-T 1 2 -space with |X| ≥ 4. F.A. Abushaheen / Eur. J. Pure Appl. Math, 15 (2) (2022), 589-601 600 Proof. Let x ∈ X and let z1 6= z2 6= z3 ∈ X − {x} and Zi = X − {zi} for i = 1, 2, 3. So {x} is either coc-open or coc-closed in Zi, therefore either {x} is coc-open in at least two of Z1, Z2, Z3, and hence {x} is coc-open in X, or {x} is coc-closed in at least two of Z1, Z2, Z3, and hence {x} is coc-closed in X, hence the result. Theorem 30. Let (X, τ) be infinite topological space. If every proper subspace of a topological space X is coc-T 1 4 -space, then X is coc-T 1 4 -space. Proof. Let F be a finite set and y /∈ F and let z ∈ X − ( F ∪ {y} ) . So there exists a set A contains F and y /∈ A which is either coc-open or coc-closed in X − {z}, therefore there exists a set B which is either coc-open or coc-closed in X with A = B ∩ (X − {x}), hence X is coc-T 1 4 -space. Theorem 31. Let (X, τ) be infinite topological space. If every proper subspace of a topological space X is coc-T 3 8 -space, then X is coc-T 3 8 -space. Proof. Same as Theorem 30. Theorem 32. If every proper subspace of a topological space (X, τ) is coc-R0-space, then X is coc-R0-space with |X| ≥ 3. Proof. Assume that all proper subspaces of X are coc-R0-space. Let U be coc-open subset of X. If X = U we are done, so we may assume X 6= U . Let x /∈ U and p ∈ U with y ∈ X − {p, x}. So we have the following cases : (1) If y ∈ U , so X − {y} is coc-R0-space, so by Theorem 15 (iii) there is a coc-closed set Gy in X − {y} such that p ∈ Gy ⊆ U − {y} and also there exists a coc-closed set G in X such that Gy = G ∩ (X − {y}), then p ∈ G ⊆ Gy ∪ {y} ⊆ (U − {y}) ∪ {y} = U , hence X is coc-R0-space. (2) If y /∈ U , then X − {x} and X − {y} are proper subspaces of X, so there exist coc- closed subsets Gx, Gy in X − {x} and X − {y}, respectively such that p ∈ Gx ⊆ U and p ∈ Gy ⊆ U , also there exist coc-closed sets G1, G2 in X such that Gx = G1 ∩ (X − {x}) and Gy = G2 ∩ (X − {y}). Define G = G1 ∩G2, so p ∈ G ⊆ (Gx ∪ {x})∩ (Gy ∪ {y}) ⊆ U , hence X is coc-R0-space. Theorem 33. If every proper subspace of a topological space (X, τ) is coc-T1-space, then X is coc-T1-space with |X| ≥ 3. Proof. Suppose that X is not a coc-T1-space, so there exists x ∈ X such that {x} is not coc-closed in X. Let z ∈ X − {x}. 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