EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 15, No. 3, 2022, 821-829 ISSN 1307-5543 – ejpam.com Published by New York Business Global Semi-Regularization Topological Spaces Dina Abuzaid1, Nouf Alfarsi1,∗, Lutfi Kalantan1 1 King Abdulaziz University, Department of Mathematics, P.O.Box 80203, Jeddah 21589, Saudi Arabia Abstract. If (X , τ ) is a topological space, then the semi-regularization topology τ s on X of τ is the coarser topology on X generated by the family of all open domains of (X , τ ) where a subset U is called an open domain if U = int(U). In this paper, we study the semi-regularity of some generated spaces and some properties of weaker version of normality of the semi-regularization space (X , τ s ) of a space (X , τ ). 2020 Mathematics Subject Classifications: 54A10, 54D10. Key Words and Phrases: Semi-regular, semi-regularization, Alexandroff duplicate, closed ex- tension, discrete extension, open extension, C-normality, epi-normality, submetrizability, scattered. If (X , τ ) is a topological space, then the semi-regularization topology τ s on X of τ is the coarser topology on X generated by the family of all open domains of (X , τ ) where a subset U is called an open domain if U = int(U). In this paper, we study some properties of weaker version of normality of the semi-regularization space (X , τ s ) of a space (X , τ ). Also, we study the semi-regularity of some generated spaces. This paper may considered as a continuation of the study of Mrs̆ević, Reilly and Vamanamurthy in [10]. Throughout this paper, we denote the set of positive integers by N, the rationals by Q, the irrationals by P, and the set of real numbers by R. A T4 space is a T1 normal space and a Tychonoff space (T3 1 2 ) is a T1 completely regular space. We do not assume T2 in the definition of compactness and countable compactness. For a subset A of a space X, intA and A denote the interior and the closure of A, respectively. If two topologies τ and τ ′ on a set X are considered, we denote the interior of A in (X , τ ) by int τA and int τ ′A for the interior of A in (X , τ ′ ). We denote the closure of A in (X , τ ′ ) by A τ ′ and, similarly, A τ denotes the closure of A in (X , τ ). We denote an order pair by ⟨x, y⟩. ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v15i3.4386 Email addresses: dabuzaid@kau.edu.sa, dina.abuzaid@gmail.com (D. Abuzaid), nhamdanalfarsi@stu.kau.edu.sa, noufalfarsi98@gmail.com (N. Alfarsi), lnkalantan@hotmail.com, lkalantan@kau.edu.sa (L. Kalantan) https://www.ejpam.com 821 © 2022 EJPAM All rights reserved. D. Abuzaid, N. Alfarsi, L. Kalantan / Eur. J. Pure Appl. Math, 15 (3) (2022), 821-829 822 1. Semi-Regularity We have to start by recalling some basic definitions. Definition 1. A subset A of a space X is called closed domain [6], called also regular closed, κ-closed [9], if A = intA. A subset A of a space X is called open domain [6], called also regular open, κ-open, if A = int(A ). It is easy to see that a subset is an open domain if and only if it is the interior of a closed set, a subset is a closed domain if and only if it is the closure of an open set, the complement of a closed domain is an open domain and the complement of an open domain is a closed domain [6]. Now, let (X , τ ) be a topological space and let OD denotes the family of all open domains in (X , τ ). Since X is an open domain and an intersection of two open domains is an open domain [6, 1.1.C], then we have the following definition [6]. Definition 2. If (X , τ ) is a topological space, then the semi-regularization topology τ s on X of τ is the coarser topology on X generated by the family of all open domains of (X , τ ). (X , τ ) is called semi-regular if τ= τ s. (X , τ s ) is called the semi-regularization topological space of (X , τ ), see [10]. Since any open domain is an open set, then for any space (X , τ ), we have that τ s is coarser than τ , that is, τ s ⊆ τ . Note that if ∅ ̸= U ⊆ X, then U ∈ τ s if and only if U = ⋃ α∈Λ Vα with Vα is an open domain in (X , τ ) for each α ∈ Λ. Equivalently U ∈ τ s if and only if for each x ∈ U there exists an open domain G in (X , τ ) such that x ∈ G ⊆ U , [11]. If CF is the finite complement topology on an infinite set, then CFs = I, where I is the indiscrete topology. If CC is the countable complement topology on an uncountable set, then CCs = I. If X = { ⟨x, y⟩ : y ≥ 0 }, the closed upper half plan, then the semi- regularization topology of the Half-Disc topology on X [12, Example 78], is the usual metric topology U on X. If X is regular, then it is semi-regular [6, 1.1.8]. The converse is not always true. As an example the simplified Arens square [12, Example 81]. 2. Semi-regularity of generated spaces There are many ways of generating new spaces from old ones. In this section, we study the semi-regularity of the Alexandroff duplicate, the closed extension, the discrete extension, and the open extension of a given space X. Recall that the Alexandroff Duplicate space A(X) of a space X is defined as follows: Let X be any topological space. Let X ′ = X × {1}, so X ′ is just a copy of X. Note that X ∩X ′ = ∅. Let A(X) = X ∪X ′. For simplicity, for an element x ∈ X, we will denote the element ⟨x, 1⟩ in X ′ by x′ and for a subset B ⊆ X, let B′ = {x′ : x ∈ B} = B×{1} ⊆ X ′. For each x′ ∈ X ′, let B(x′) = {{x′}}. For each x ∈ X, let B(x) = {U∪(U ′\E) : U is open in X with x ∈ U and E is a finite subset ofX ′ }. Then B = {B(x) : x ∈ X}∪{B(x′) : x′ ∈ X ′} D. Abuzaid, N. Alfarsi, L. Kalantan / Eur. J. Pure Appl. Math, 15 (3) (2022), 821-829 823 will generate a unique topology on A(X) such that B is its neighborhood system. A(X) with this topology is called the Alexandroff Duplicate of X [2, 5]. Our goal is to show that “if X is semi-regular, then so is its Alexandroff duplicate A(X)”. In order to show this we will follow five steps expressed in the following Lemmas and Theorem 1. As a notation we will call a subset which is closed and open by clopen. Lemma 1. For each x′ ∈ X ′, {x′} is clopen in A(X). Proof. Let x′ ∈ X ′ be arbitrary. We need only to show that {x′} is closed. So, let a ∈ A(X) \ {x′} be arbitrary. If a ∈ X ′, then {a} is an open neighborhood of a in A(X) with {a} ⊂ A(X) \ {x′}. If a ∈ X, pick any open neighborhood U ⊆ X of a. Then U ∪ (U ′ \ {x′}) is an open neighborhood of a in A(X) with U ∪ (U ′ \ {x′}) ⊆ A(X) \ {x′}. Thus A(X) \ {x′} is open in A(X). Therefore, {x′} is closed. Lemma 2. Let (X , τ ) be any topological space. If C is clopen in X and B is an open domain in X , then B \ C is an open domain in X. Proof. we want to show that, int(B \ C) = B \C. We always have B \C ⊆ B \ C , by taking the interior in both sides we get int(B \ C) ⊆ int(B \ C). But int(B \ C) = int(B ∩ (X \C)) = int(B)∩ int(X \C) = B∩ (X \C) = B \C, because C is clopen and B is an open domain so B is an open set, thus B \C is an open set. Thus, B \C ⊆ int(B \ C). Now, we always have int(B \ C ) ⊆ B \ C. Thus int(B \ C) ⊆ B \ C = B ∩ (X \ C) ⊆ B∩ (X \ C) = B ∩ (X \ C), because C is clopen in X. int(B \ C)⊆ B ∩ (X \ C) then, int(int(B \ C)) ⊆ int(B∩ (X \C)) thus, int(B \ C) ⊆ int(B)∩ int(X \C) hence, int(B \ C) ⊆ B∩ (X \C) = B \ C, because B is an open domain and C is clopen in X. Thus, int(B \ C) ⊆ B \ C. Therefore, int(B \ C) = B \ C. Thus, B \ C is an open domain . Notice that, if U is any non-empty open set in X, then U ∪ (U ′ \ ∅) = U ∪U ′ is a basic open neighborhood in A(X) of any x ∈ U . So, we establish that the following lemma. Lemma 3. If U is an open set in X, then U ∪U ′ is an open set in A(X) Theorem 1. If U is an open domain in X, then U ∪U ′ is an open domain in A(X). Proof. Let U be an open domain in X. We show that U ∪ U ′ = intA(X)(U ∪ U ′ A(X) ) first, we show that intA(X)(U ∪ U ′ A(X) ) ⊆ U ∪ U ′. Let x ∈ intA(X)(U ∪ U ′ A(X) ) be arbi- trary, then x ∈ U ∪ U ′ A(X) . There are only two cases. Case 1: x ∈ X ′. So, {x} is an open set in A(X) with x ∈ {x} and {x} ∩ (U ∪ U ′) ̸= ∅, then x ∈ U ′ ⊆ U ∪ U ′. Thus, x ∈ U ∪ U ′. Case 2: x ∈ X. Since x ∈ intA(X)(U ∪ U ′ A(X) ), then there exist an open set V in X with x ∈ V such that x ∈ V ∪ (V ′ \ E) ⊆ U ∪ U ′ A(X) = U A(X) ∪ U ′ A(X) where E is a finite subset of X ′. Thus, x ∈ V ⊆ U A(X) , but U A(X) = U X . So, x ∈ V ⊆ U X , by taking the interior of both sides with respect to X we get, intX(V ) ⊆ intX(U X ) , then V ⊆ U as V is an open set in X and U is an open domain in X. Thus, x ∈ U ⊆ U ∪ U ′. Hence D. Abuzaid, N. Alfarsi, L. Kalantan / Eur. J. Pure Appl. Math, 15 (3) (2022), 821-829 824 intA(X)(U ∪ U ′ A(X) ) ⊆ U ∪ U ′. Now, we show that U ∪ U ′ ⊆ intA(X)(U ∪ U ′ A(X) ). Note that we always have U ∪ U ′ ⊆ U ∪ U ′ A(X) . Since U is an open domain in X, then U is an open set in X , so by Lemma 3, U ∪ U ′ is an open set in A(X). Then by taking the interior of both sides with respect to A(X) we get, U ∪ U ′ = intA(X)(U ∪ U ′) ⊆ intA(X)(U ∪ U ′ A(X) ). Hence, U ∪ U ′ = intA(X)(U ∪ U ′ A(X) ). Therefore, U ∪ U ′ is an open domain in A(X). As an immediate consequence of the Theorem 1 is the following Corollary 1. Corollary 1. If U is an open domain in X, then U ∪(U ′ \E) is an open domain in A(X) where E is a finite subset of X ′ . Proof. Let U be any open domain in X. By Lemma 1, we have any singleton in X ′ is clopen in A(X), then E is clopen in A(X) because finite union of closed sets is closed and arbitrary union of open sets is open. Also, by Theorem 1, we have U ∪ U ′ is an open domain in A(X). But, U ∪(U ′ \E) = (U ∪U ′) \E is an open domain in A(X) by Lemma 2. Theorem 2. If (X , τ ) is semi-regular, then so is its Alexandroff duplicate A(X). Proof. Let (X , τ ) be any semi-regular topological space. Let ∅ ≠ W be any open set in A(X) and let x ∈ W be arbitrary. To show that A(X) is semi-regular, it is enough to exhibit an open domain subset G in A(X) such that x ∈ G ⊆ W . For such an x, we have only two cases. Case 1: x ∈ X ′. Since any clopen subset is an open domain, then by Lemma 1, there exist an open domain G = {x} in A(X) such that x ∈ G ⊆ W . Case 2: x ∈ X. Since W is an open set in A(X) with x ∈ W , then there exists an open set U in X with x ∈ U and U ∪ (U ′ \ E) ⊆ W , where E is a finite subset of X ′. Now, since (X , τ ) is semi-regular, then there exists an open domain V in X such that x ∈ V ⊆ U . Thus, x ∈ (V ∪ (V \E ) ) ⊆ (U ∪ (U \E ) ) ⊆ W By Corollary 1, we get V ∪ (V ′ \E ) = G is an open domain in A(X) such that x ∈ G ⊆ W . Therefore, A(X) is semi-regular. Definition 3. Let (X , τ ) be a topological space and let p be an object not in X, that is, p ̸∈ X. Put Xp = X ∪ { p }. Define a topology τ ⋆ on Xp by τ ⋆ = { ∅ } ∪ {U ∪ { p } : U ∈ τ }. The space (Xp , τ ⋆ ) is called the closed extension space of (X , τ ), see [12, Example 12]. Consider the particular point topology τ p = {W ⊆ Xp : p ∈ W } on Xp, [12, Example 10]. It is easy to see that τ ⋆ is coarser than τ p, that is, τ ⋆ ⊆ τ p. Notice that the closed extension (Xp , τ ⋆ ) of a space (X , τ ) is not semi-regular regardless wither (X , τ ) is semi-regular or not. D. Abuzaid, N. Alfarsi, L. Kalantan / Eur. J. Pure Appl. Math, 15 (3) (2022), 821-829 825 Example 1. Let (X , τ ) be the Simplified Arens Square topological space, [12, Example 81]. So, X = { ⟨0, 0⟩, ⟨1, 0⟩ } ∪ { ⟨x, y⟩ : 0 < x, y < 1 }. The topology τ on X is generated by the following neighborhood system: For each ⟨x, y⟩ ∈ { ⟨x, y⟩ : 0 < x, y < 1 }, let B(⟨x, y⟩) = {Bd(⟨x, y⟩; ϵ) ⊂ S : ϵ > 0 } where d is the usual metric on R2 and Bd(⟨x, y⟩; ϵ) is the open ball centered at ⟨x, y⟩ of radius ϵ > 0 so that ϵ is small enough to make the open ball Bd(⟨x, y⟩; ϵ) is contained in { ⟨x, y⟩ : 0 < x, y < 1 }. Let B(⟨0, 0⟩) = {Un(⟨0, 0⟩) : n ∈ N }, where for each n ∈ N, we have Un(⟨0, 0⟩) = { { ⟨0, 0⟩ } ⋃ { ⟨x, y⟩ ∈ S : 0 < x < 1 2 and 0 < y < 1 n }. Let B(⟨1, 0⟩) = {Un(⟨1, 0⟩) : n ∈ N }, where for each n ∈ N, we have Un(⟨1, 0⟩) = { { ⟨1, 0⟩ } ⋃ { ⟨x, y⟩ ∈ S : 1 2 < x < 1 and 0 < y < 1 n }. In [12, Example 81], it was shown that the Simplified Arens Square space (X , τ ) is semi-regular. Let U be any non-empty proper open subset of X, then U ∪ {p} is an open set in Xp such that U ̸= Xp. Now, U ∪ {p} τ⋆ = U τ⋆ ∪{p} τ⋆ = U τ⋆ ∪Xp = Xp because {p} is dense in (Xp , τ ⋆ ). Hence, int τ⋆(U ∪ {p} τ⋆ ) = int τ⋆ (Xp) = Xp ̸= U ∪ {p}. Thus the only open domains in (Xp , τ ⋆ ) are Xp and ∅, then τ ⋆ s = I on Xp, where I is the indiscrete topology. Therefore, the closed extension topological space (Xp , τ ⋆ ) of the Simplified Arens Square space (X , τ ) is not semi-regular. Definition 4. Let M be a non-empty proper subset of a topological space (X , τ ). Define a new topology τ (M) on X as follows: τ (M) = {U ∪K : U ∈ τ and K ⊆ X \M }. (X , τ (M) ) is called a discrete extension of (X , τ ) and we denote (X , τ (M) ), simply, by XM [1], see also [6, Example 5.1.22]. Observe that if U is an open set in X, then U is also open in XM because we can write U = U ∪ ∅. The space XM has the following neighborhood system: For each x ∈ X \M , let B(x) = {{x}} and for each x ∈ M , let B(x) = {U ∈ τ : x ∈ U }. If X is a semi- regular topological space and ∅ ̸= M ⊂ X, then the discrete extension XM may not be semi-regular as can be shown in the following example. Example 2. Consider, (R , I ) where I is the indiscrete topology. It is clear that (R , I ) is semi-regular. Put M = R\{0}. Then, the discrete extension XM can be describe as follows: B(0) = { {0} } and for each x ̸= 0, B(x) = {R }. XM is not semi-regular because {0} is an open set in XM , but intXM ({0}XM ) = intXM (R)= R ̸= {0}. Thus, {0} is not an open domain in XM . Hence, 0 ∈ {0} with {0} is an open set and there is no open domain G in XM satisfies 0 ∈ G ⊆ {0}. Therefore, XM is not semi-regular. Lemma 4. Let (X , τ ) be a topological space. Let M be any non-empty proper subset of X. Then, for any open domain U in X, U is an open domain in XM . Proof. Let U be any open domain in X, we always have U ⊆ U XM . By taking the interior of both sides with respect to XM we get, intXM (U) ⊆ intXM (U XM ). But since U is an open domain in X, then U is an open set in X. Thus, U is an open set in XM . Hence, intXM (U) = U , therefore U ⊆ intXM (U XM ) ... ⋆. Now, let x ∈ intXM (U XM ) be arbitrary, then x ∈ (U XM ). There are only two cases. Case 1: x ∈ X \M . Since {x} is an open neighborhood of x in XM satisfies {x} ∩U ̸= ∅, D. Abuzaid, N. Alfarsi, L. Kalantan / Eur. J. Pure Appl. Math, 15 (3) (2022), 821-829 826 then x ∈ U . Case 2: x ∈ M . Since x ∈ intXM (U XM ), then there exist an open set V in X such that x ∈ V ⊆ U XM ⊆ U X and the last inclusion is true because the topology on X is coarser than the topology on XM . Therefore, we have x ∈ V ⊆ U X , then by taking the interior of both sides with respect to X we have, x ∈ intXV = V ⊆ intX(U X ) = U because U is an open domain in X and V is an open set in X. Hence, x ∈ U , thus intXM (U XM ) ⊆ U ... ⋆⋆. By ⋆ and ⋆⋆ we get U = intXM (U XM ). Therefore, U is an open domain in XM . In the next theorem, we will use the following fact which was proved in [1]: “If X is T1, then so is XM for any non-empty proper subset M of X”. Theorem 3. If X is T1 and semi-regular, then for any non-empty proper subset M of X, we have that the discrete extension XM of X is semi-regular. Proof. Assume the hypotheses. Let W be an arbitrary non-empty open set in XM . Let x ∈ W be arbitrary. There are only two cases. Case 1: x ∈ X \M . Then we have {x} is an open neighborhood of x in XM . Since X is T1, then XM is also T1. Thus {x} is also closed in XM . Hence {x} is clopen in XM , thus {x} is an open domain in XM such that x ∈ {x} ⊆ W . Case 2: x ∈ M . Since X is semi-regular, then there is a base for X consisting of open domains. Thus, there exists an open domain V in X such that x ∈ V ⊆ W . By Lemma 4, we get V is an open domain in XM . Therefore, XM is semi-regular. Definition 5. Let (X , τ ) be a topological space and let p be an object not in X, that is, p ̸∈ X. Put Xp = X ∪ { p }. Define a topology τ ′ on Xp by τ ′ = {Xp } ∪ {U : U ∈ τ } = {Xp }∪ τ . The space (Xp , τ ′ ) is called the open extension space of (X , τ ), see [12, Example 16]. Observe that (X , τ ) and (Xp , τ ′ ) have the same open sets except for Xp. Also, if U is an open domain in (X , τ ), then U is an open domain in (Xp , τ ′ ) because U Xp = U X ∪ {p} as the only open neighborhood of p in (Xp , τ ′ ) is Xp itself. Thus, intXp( U Xp ) = intXp(U X ∪ {p}) = intX(U X ) = U . It is easy to see that if U is an open domain in (Xp , τ ′ ) such that p ̸∈ U , then U is an open domain in (X , τ ). Theorem 4. (X , τ ) is semi-regular if and only if (Xp , τ ′ ) is semi-regular. Proof. Assume that (X , τ ) is semi-regular. To show that (Xp , τ ′ ) is semi-regular, we only need to prove that τ ′ ⊆ τ ′ s. Let W ∈ τ ′ be an arbitrary such that ∅ ≠ W ̸= Xp, then W ∈ τ . Since (X , τ ) is semi-regular, then τ = τ s. So, the family of all open domains in (X , τ ) is a base for (X , τ ). Thus W can be written as a union of open domains in (X , τ ). So, W can be written as a union of open domains in (Xp , τ ′ ). Thus W ∈ τ ′ s. Thus τ ′ ⊆ τ ′ s. Therefore (Xp , τ ′ ) is semi-regular. Conversely, Assume that (Xp , τ ′ ) is semi-regular, that is, τ ′ = τ ′ s. To show that (X , τ ) is semi-regular, we only need to show that τ ⊆ τ s. Let W ∈ τ be arbitrary, then p /∈ W . But W ∈ τ ′ implies that W can be written as a union of open domains in D. Abuzaid, N. Alfarsi, L. Kalantan / Eur. J. Pure Appl. Math, 15 (3) (2022), 821-829 827 (Xp , τ ′ ). Since any open domain in (Xp , τ ′ ) which does not contain the element p is also an open domain in (X , τ ), then W ∈ τ s, implies that τ ⊆ τ s and hence (X , τ ) is semi-regular. 3. New results about semi-regularization spaces In this section, we study the relationship between a topological space (X , τ ) and its semi-regularization space (X , τ s ) regarding a topological property. We start with the property of scattered. Recall that a space X is scattered if any non-empty subset of X has an isolated point, that is, if ∅ ≠ A ⊆ X, then there exists an element a ∈ A and there exists an open set U such that a ∈ U and U ∩ A = { a }. It is easy to see that if (X , τ s ) is scattered, then so is (X , τ ). This follows from the containment τ s ⊆ τ . But the converse is not always true as can be shown in the following example. Example 3. Consider R with the particular point topology τ 0 which is scattered, see [12, Example 10]. But the semi-regularization of (R ,τ 0 ) is (R , I ) where I is the indiscrete topology which is not scattered. Definition 6. A topological space (X , τ ) is called epi-normal if there exists a coarser topology τ ′ on X such that (X , τ ′ ) is T4, see [3]. Lemma 5. Let (Y , ν ) be a regular space. If f : (X , τ ) −→ (Y , ν ) is continuous, then f : (X , τ s ) −→ (Y , ν ) is continuous, [8]. Theorem 5. (X , τ ) is epi-normal if and only if (X , τ s ) is epi-normal. Proof. Assume that (X , τ ) is epi-normal. Pick a coarser topology τ ′ on X such that (X , τ ′ ) is T4. Consider the identity function idX : (X , τ ) −→ (X , τ ′ ) which is continuous since τ ′ ⊆ τ . Then, by Lemma 5, we have idX : (X , τ s ) −→ (X , τ ′ ) is continuous, hence τ ′ ⊆ τ s . Thus, (X , τ s ) is epi-normal. Conversely, assume that (X , τ s ) is epi-normal. Then there exist a coarser topology τ ′ on X such that (X , τ ′ ) is T4. Since τ s ⊆ τ , then result follows. Definition 7. A topological space X is called submetrizable if there exists a metric d on X such that τ d ⊆ τ , [7]. Similar argument of the proof of Theorem 5 gives the following theorem. Theorem 6. (X , τ ) is submetrizable if and only if (X , τ s ) is submetrizable. Definition 8. A topological space X is called C-normal if there exist a normal space Y and a bijective function f : X −→ Y such that the restriction f |A : A −→ f(A) is a homeomorphism for each compact subspace A ⊆ X, [4]. The following example shows that, if (X , τ s ) is C-normal, then (X , τ ) may not be C-normal. REFERENCES 828 Example 4. Consider R with the particular point topology τ 0 which is not C-normal see [4, Example 1.5]. But the semi-regularization topological space of (R , τ 0 ) is (R , I ), where I is the indiscrete topology, which is a normal space, thus C-normal. Lemma 6. If X is T1 and C-normal, then any witness Y of its C-normality is T4. Proof. Assume that X is T1 and C-normal. Pick a normal space Y and a bijective function f : X −→ Y such that f|A : A −→ f(A) is a homeomorphism for each compact subspace A ⊆ X. Let x, y be any two distinct elements in Y . Since f is bijective, there are unique elements a, b ∈ X such that f(a) = x and f(b) = y such that a ̸= b. Consider {a, b} which is a compact subset ofX. This implies f|{a,b} : {a, b} −→ {x, y} is a homeomorphism. But X is T1, thus {a, b} is a discrete subspace of X, hence {x, y} is a discrete subspace of Y , then there are two open neighborhoods U and V of x and y respectively in Y such that U ∩ {x, y} = {x} and V ∩ {x, y} = {y} where y /∈ U and x /∈ V . Thus Y is T1 and given that Y is normal, thus Y is T4. Recall that a topological space X is called a Fréchet space if for every A ⊆ X and every x ∈ A there exists a sequence (an)n∈N of points of A such that an −→ x, [6]. Lemma 7. If X is Fréchet and C-normal, then any witness function of its C-normality is continuous. Proof. Assume that X is Fréchet and C-normal. Let f : X −→ Y be a witness function of the C-normality of X. Let A ⊂ X and let y ∈ f(A) be arbitrary. Pick the unique element x ∈ X such that f(x) = y. Thus x ∈ A. Since X is a Fréchet space, then there exist a sequence (an) ⊆ A such that (an) converges to x. The subspace B = {x, an : n ∈ N} of X is compact and thus f|B : B −→ f(B) is a homeomorphism. Now, let W ⊆ Y be any open neighborhood of y, then W ∩ f(B) is open in the subspace f(B) containing y. Since f({an : n ∈ N}) ⊆ f(B) ∩ f(A) and W ∩ f(B) ̸= ∅, then W ∩ f(A) ̸= ∅. Hence y ∈ f(A) and thus f(A) ⊆ f(A). Therefore, f is Continuous. Theorem 7. If (X , τ ) is Fréchet, T1 and C-normal, then its semi-regularization topo- logical space (X , τ s ) is C-normal Proof. Assume the hypothesis. Pick a normal topological space (Y , τ ′ ) and a bijective function f : (X , τ ) −→ (Y , τ ′ ) such that f|A : A −→ f(A) is a homeomorphism for any compact subspace A of X. As X is Fréchet, then by Lemma 7, f is continuous and by Lemma 6, we get (Y , τ ′ ) is T4. Pick the same bijection function f : (X , τ s ) −→ (Y , τ ′ ) which is continuous by Lemma 5. Let B be any compact subset of (X , τ s ), then f|B : B −→ f(B) is bijective and continuous, thus by [6, Theorem 3.1.13] f|B is a homeomorphism. Therefore, (X , τ s ) is C-normal. References [1] A Alawadi, L Kalantan, and M M Saeed. On the discrete extension spaces. Journal of Mathematical Analysis, 9(2):150–157., 2018. REFERENCES 829 [2] P S Alexandroff and P S Urysohn. Mémoire sur les espaces topologiques compacts. Verh. Konink. Acad. Wetensch. Amsterdam, 14:1–96., 1929. [3] S AlZahrani and L Kalantan. Epinormality. Journal of Nonlinear Sciences & Appli- cations, 9(9):5398–5402., 2016. [4] S AlZahrani and L Kalantan. C-normal topological property. Filomat, 31(2):407–411., 2017. [5] R Engelking. On the double circumference of alexandroff. Bull. Acad. Pol. Sci. Ser. Astron. Math. Phys., 16(8):629–634., 1968. [6] R Engelking. General Topology. PWN, Warszawa, 1977. [7] G Gruenhage. Generalized metric spaces. Handbook of set-theoretic topology, pages 423–501., 1984. [8] L L Herrington. Characterizations of urysohn-closed spaces. Proceedings of the Amer- ican Mathematical Society, pages 435–439., 1976. [9] L Kalantan. Results about κ-normality. Topology and its Applications, 125(1):47–62., 2002. [10] M Mrs̆ević, I L Reilly, and M K Vamanamurthy. On semi-regularization topologies. Journal of the Australian Mathematical Society, 38(1):40–54., 1985. [11] T. Noiri and V. Popa. On almost b-continuous functions. Acta Math Hungar, 79(4):329–339., 1998. [12] L Steen and J A Seebach. Counterexamples in Topology. Dover Publications INC, USA, 1995.