IOSR Journal of Mathematics (IOSR-JM) e-ISSN: 2278-5728, p-ISSN: 2319-765X. Volume 17, Issue 3 Ser. II (May – June 2021), PP 41-48 www.iosrjournals.org

On Epi 휶 -Regular Spaces

NADIA GHEITH1 and ALAMIN ABUSBAIHA2 1 Mathematics Department/University of Gharyan, Gharyan, Libya

2Mathematics Department/University of Gharyan, Mezda, Libya

Abstract: A (푋, 휏) is said to be epi 훼-regular if a coarser 휏′ on 푋 exists such that ′ 푋, 휏 is 푇1, 훼-regular. In this article we introduce and implement this property and give some examples to show the relationships between epi 훼 -regular, epiregular, epinormal, submetrizable semiregular and Almost 훼- normal (almost 훽-normal). Keywords: Epiregular, Epi 훼-regular, Epinormal, Semiregular, Submetrizable, and Almost 훼-normal (almost 훽-normal). Mathematics Subject Classification: 54퐷15 − 54퐵10. ------Date of Submission: 30-05-2021 Date of Acceptance: 13-06-2021 ------

I. Introduction A topological space (푋, 휏) is said to be epiregular [4] if a coarser topology 휏 on 푋 exists such that ′ 푋, 휏 is 푇1, regular . A topological space (푋, 휏) is said to be 훼 -regular [10], [30] if for every closed subset 퐹 of 푋 and 푥 ∈ 푋 such that 푥 ∉ 퐹 there exist disjoint open sets 푈 and 푉 such that 푥 ∈ 푈 and 푉 ∩ 퐹 is dense in 퐹. We use these definitions to introduce another new as a simultaneous generalization called epi 훼 -regularity. The intent of this article is to implement this property. We show the relationship between epi 훼 - , 훼 -normal, 훼 -regular, epinormal, epiregular, semiregular and Almost α-normal (almost β-normal) spaces. Also we show that every epi 훼 -regular space is Hausdorff. We prove that submetrizability or 푇1, 훼 - regularity imply epi 훼 -regularity but the converse is not correct in general. We give some examples to show that epi 훼 -regularity, 훼 -regularity and semiregularity are not necessarily related.

II. Epi 휶 -Regularity Definition 1.1. A topological space (푋, 휏) is said to be epi 훼 -regular if a coarser topology 휏′ on 푋 exists such ′ that 푋, 휏 is 푇1, 훼 -regular. Note that if we necessarily let 푋, 휏′ to be just 훼 -regular in the above definition, then any space will be epi 훼 -regular since the indiscrete topology will satisfy the property. ′ Observe that if for any topological space 푋, 휏 which is 푇푖, 푖 ∈ {0,1,2} then any larger topology 휏 on 푋 so is, and since every 훼 -regular 푇1 is Hausdorff [10], [30], then we can conclude the following. Theorem 1.2. Every epi 훼 -regular space is Hausdorff.■ We note that if 푋 is not 푇푖, where 푖 ∈ {0,1,2}, then 푋 is not epi 훼 regular. For example, Sierpinski space and The closed extension topology see [9], are not Hausdorff, then they cannot be epi 훼 -regular. Since every regular space is 훼 -regular, then the next theorem is true. Theorem 1.3. Every epiregular space is epi 훼 -regular.■ The opposite direction of the above statement is not always true, but we still have the following correct. Theorem 1.4. If (푋, 휏) is an epi 훼 -regular space, and the witnesses of epi 훼 -regularity 푋, 휏′ is first countable, then (푋, 휏) is epiregular.■ Before proofing the above theorem, we need the following proposition which is proved by a similar argument found in [29]. Proposition 1.5. [30] Every first countable 훼 -regular is regular. Proof. Using a contradiction, we suppose that 푋 is a first countable, Hausdorff and non regular space. Then there is an 푥 ∈ 푋 and a closed subset 퐴 of 푋 such that 푥 ∉ 퐴 where there are no disjoint open sets that separate them. Let 푈푛 : 푛 ∈ 휔 be an open in 푥 such that 푈푛+1 ⊂ 푈푛 for all 푛 ∈ 휔. Let 퐻 = 푥푛 : 푥푛 ∈ 푈 푛 ∩ 퐴, 푛 ∈ 휔 . Note that 푥푛 was chosen inductively and because the space 푋 is Hausdorff, we can also suppose at each step of the induction that 푥푛 ∉ 푈 푛 + 1 , it follows that 푥푛 ∈ 푈 푚 if and only if 푚 ≤ 푛.

DOI: 10.9790/5728-1703024148 www.iosrjournals.org 41 | Page On Epi 훼 -Regular Spaces

∘ ∘ The set 퐻 is closed. Indeed, if 푦 ∉ 퐻 , then 푋 ∖ 푈 푛 ∩ 퐴 is an containing 푦 and not intersecting 퐻 which implies that 푋 ∖ 푈 푛 ∩ 퐴 is a neighborhood open set containing 푦 and not intersecting 퐻. Therefore 푦 ∉ 퐻. Note that 푥 ∉ 퐻. Since 푥 and 퐻 can not be separated, so 푋 is not an 훼-regular space. ■ Proof of theorem (ퟏ. ퟒ): It is straightforward by proposition 1.5 and theorem 1.3.■ Theorem 1.6. If (푋, 휏) is an epi 훼-regular space, and the witnesses of epi 훼 -regularity 푋, 휏′ is first countable, then (푋, 휏) is completely Hausdorff. Proof. Let (푋, 휏) be any epi 훼-regular space, and let 푥, 푦 be any distinct points in 푋, then one can find a coarser ′ ′ ′ topology 휏 on 푋 such that 푋, 휏 is 푇1, 훼 -regular, and then 푋, 휏 is Hausdorff [10]. It follows that there exist two disjoint open sets 퐺, 퐻 ∈ 휏′ such that 푥 ∈ 퐺, 푦 ∈ 퐻,. Now since 푋, 휏′ is first countable then by proposition ′ ′ 1.5 푋, 휏′ is regular, so there exist 푈, 푉 ∈ 휏′ such that 푥 ∈ 푈 ⊆ 푈 휏 ⊆ 퐺 and 푦 ∈ 푉 ⊆ 푉 휏 ⊆ 퐻, where ′ ′ ′ 푈 휏 = 푥 ∈ 푋: 푊 ∩ 푈 ≠ ∅, ∀ open 푊 in 휏 ′ , 푥 ∈ 푊 similarly 푉 휏 . Since 퐴 휏 ⊆ 퐴 휏 , for any 퐴 ⊆ 푋, this implies ′ ′ ′ That 푈휏 ⊆ 푈 휏 . 퐴s 푈 휏 ∩ 푉 휏 = 0. Thus (푋, 휏) is completely Hansdorff. ■ Thus any space (푋, 휏) which is not completely Hausdorff, such that any coarser topology of it is 푇2 first countable, cannot be epi 훼 -regular. Since any 훽 -normal or 훼 -normal [26] satisfying 푇1 axiom is 훼-regular [30], [10], then we end to the following theorem Theorem 1.7. Every epi 훽 -normal (epi 훼 -normal) space is epi 훼 -regular. ■ As every second countable 푇3 space is metrizable, [[8],4.2.9], and since every second countable is first countable then by proposition 1.5 we have the following corollary. Corollary 1.8. If (푋, 휏) is epi 훼 -regular and the witness of epi 훼-regularity 푋, 휏′ is second countable, then (푋, 휏) is submetrizable. ■

Note that corollary 1.8 is not correct in general. For example, the Tychonoft Plank 휔1 + 1 × 휔0 + 1 ∖ 휔1, 휔0 is Tychonoff being Hausdorff locally compact, and hence it is epi 훼 -regular, but it is not submetrizable, because if it was, then 휔1 + 1 × {0} ⊆ 휔1 + 1 × 휔0 + 1 ∖ 휔1, 휔0 is submetrizable, because submetrizablity is hereditary, but 휔1 + 1 × {0} ≅ 휔1 + 1 and 휔1 + 1 is not submetrizable. It is well known that 푇2 is 푇4, then we have the following result proved. Corollary 1.9. If (푋, 휏) is epi 훼-regular and the witness of epi 훼 -regularity 푋, 휏′ is paracompact, then 푋, 휏 is 푇4. ■ Also, we remind that any 푇2 is 푇4, and we conclude. Corollary 1.10. Any epi 훼-regular compact space is 푇4. ■ A Hausdorff space 푋 is said to be 퐻-closed if 푋 is a closed subspace of every Hausdorff space in which it is contained [[8],3.12.5]. Since a regular space is 퐻 -closed if and only if it is compact [[8],3.12.5]. Then we can prove a similar argument for epi 훼 -regularity. Corollary 1.11. If (푋, 휏) is epi 훼 -regular compact space, then the witness of epi 훼 -regularity 푋, 휏′ is 퐻- closed. ■ Theorem 1.12. If (푋, 휏) is an epi 훼 -regular space, then for every compact subset 퐹 of 푋 and every 푥 ∈ 푋 such that 푥 ∉ 퐹, there exist disjoint open sets 푈, 푊 such that 퐹 ∩ 푈 = 퐹 and 푥 ∈ 푊. Proof. Let (푋, 휏) be an epi 훼 -regular space, then a coarser topology 휏′ on 푋 exists such that 푋, 휏′ is 훼 - ′ regular, 푇1. Let 퐹 be any compact set in (푋, 휏) and let 푥 ∉ 퐹, hence 퐹 is closed in 푋, 휏 and 푥 ∉ 퐹, by 훼- regularity of 푋, 휏′ , there exist 푈, 푊 ∈ 휏′ such that 퐹 ∩ 푈 = 퐹, 푥 ∈ 푊 and 푈 ∩ 푊 = ∅. ■ Corollary 1.13. If 퐹 and 퐸 disjoint compact sets in an epi 훼-regular space 푋, then there exist disjoint open sets 푈 and 푊 such that 퐹 ∩ 푈 = 퐹, 퐸 ∩ 푊 = 퐸. Proof. Let (푋, 휏) be an epi 훼 -regular space, then there exists a coarser topology 휏′ on 푋 such that 푋, 휏′ is 훼 - ′ regular, 푇1. Let 퐹, 퐸 be any disjoint compact subsets of (푋, 휏), hence they are disjoint compact subsets of 푋, 휏 and by theorem 1.12 for each 푎 ∈ 퐹 and compact set 퐸, there exist open sets 푈푎 , 푊푎 such that 푎 ∈ 푈푎 , 퐸 ∩ 푊 푎 = 퐸 and 푈푎 ∩ 푊푎 = ∅. Now consider 퐹 is an arbitrary compact set disjoint from 퐸. For each 푎 in 퐹, by theorem

1.12 gives disjoint open sets 푈푎푖 containing 푎 and 퐸 ∩ 푊푎푖 = 퐸 and 푈푎푖 ∩ 푊푎푖 = 0. The family 푈푎푖 : 푖 ∈ 퐼 is an open cover of 퐹, since 퐹 is compact, there is a finite subfamily 푈푎1 , … , 푈푎푛 which covers 퐹 and the 푛 corresponding 푊푎1 , … , 푊푎푛 is a closed cover of 퐸. So that 푈 =∪푖=1 푈푎푖 is an open set containing 퐹 and 푛 disjoint from 푊 = ⋂푖=1 푊푎푖 which is an open set. such that 퐹 ∩ 푈 = 퐹, 퐸 ∩ 푊 = 퐸. Indeed, it is obvious that 퐸 ∩ 푊 ⊆ 퐸. On the other hand, let 푥 ∈ 퐸, and 퐺 is an open set containing 푥, we need to show 퐺 ∩ 퐸 ∩ 푊 ≠ 0.

Let 퐺 ∩ 퐸 ∩ 푊 = 0, then there is 1 ≤ 푗 ≤ 푛 such that 퐺 ∩ 퐸 ∩ 푊푎푗 = 0, since 퐺 is open then 푥 ∉ 퐸 ∩ 푊푎푗 which is a contradiction. Therefore 퐺 ∩ 퐸 ∩ 푊 ≠ 0 which implies that 푥 ∈ 퐸 ∩ 푊 . Hence 퐸 ∩ 푊 = 퐸, and we are done.■

DOI: 10.9790/5728-1703024148 www.iosrjournals.org 42 | Page On Epi 훼 -Regular Spaces

III. Properties of Epi 휶-Regularity Theorem 2.1. [10] Let 푋 be an 훼-regular space, 푓: 푋 → 푌 is an onto, continuous, open, and closed function. Then 푌 is 훼-regular. Proof. Let 푋 be an 훼-regular space, 퐴 be a closed subset of 푌 and 푦 ∈ 푌 such that 푦 ∉ 퐴. Then 푓−1(퐴) is a closed subset of 푋 and there exists 푥 ∈ 푋 such that 푓(푥) = 푦 and 푥 ∉ 푓−1(퐴). Since 푋 is an 훼-regular space, there exist disjoint open subsets 퐺 and 퐻 of 푋 such that 푥 ∈ 퐻 and 푓−1(퐴) ∩ 퐺 = 푓−1(퐴), and so 푥 ∉ 퐺 . Since 푥 ∉ 퐺 , then 푦 ∉ 푓(퐺 ). It is clear that 푓(퐺 ) is a containing the open set 푓(퐺), 푓 ( 퐺 ) ⊆ 푓(퐺 ). Thus 푦 ∉ 푓 ( 퐺 ) which implies 푦 ∈ 푓(퐻) and 푓(퐺) ∩ 푓(퐻) = 0. Now it is sufficient to show that 퐴 ∩ 푓 ( 퐺 ) = 퐴. Let 푧 ∈ 퐴 and 푊 is an open set containing 푧, then 푓−1(푧) ⊆ 푓−1(퐴) ∩ 푓−1(푊). Since 푓 − 1 ( 퐴 ) ∩ 퐺 = 푓−1(퐴),푓−1(퐴) ∩ 퐺 ∩ 푓−1(푊) ≠ ∅. Hence by surjectivity of 푓, 퐴 ∩ 푓(퐺) ∩ 푊 = 푓 푓−1(퐴) ∩ 푓(퐺) ∩ 푓(푓−1(푊)) ⊇ 푓 푓−1(퐴) ∩ 퐺 ∩ 푓−1(푊) ≠ 0 as required.■ Corollary 2.2. Let (푋, 휏) be an epi 훼 -regular space, 푓: (푋, 휏) → (푌, 𝒮) is an onto, continuous, open, and closed function. Then 푌 is epi 훼 -reqular. ′ ′ Proof. Let (푋, 휏) be any epi 훼 -regular space, let 휏 be a coarser topology on 푋 such that 푋, 휏 is 훼 -regular, 푇1. Since 푓: 푋 → 푌 is an onto, continuous, open, and closed function then by theorem 2.13 푌, 𝒮′ , where 𝒮′ = ′ 푓{푈}: 푈 ∈ 휏 , is 훼 -regular, and it is obviously 푇1. Hence (푌, 𝒮) is epi 훼 -regular.■ Corollary 2.3. Epi 훼 -regularity is a topological property.■ The proof of the following theorem is due to Murtinová. Theorem 2.4. [30], [10] Every subspace of an 훼 -regular space is 훼 -regular.■ Proof. Let 푋 be an 훼 -regular space and 퐴 is a subspace 푋, 푦 ∈ 퐴 and 푦 ∉ 퐹 ⊂ 퐴, 퐹 ∩ 퐴 = 퐹 where 퐹 refers to the of 퐹 in 푋. Then 푦 ∉ 퐹 and 푋 is 훼 -regular, hence there are disjoint open sets 푈, 푉 in 푋 such that 푦 ∈ 푈 and 퐹 ∩ 푉 = 퐹. The sets 푈 ∩ 퐴 and 푉 ∩ 퐴 are the sets witnessing 훼 -regularity of 퐴. Indeed, they are disjoint, open in 퐴, 푦 ∈ 푈 ∩ 퐴. It remains to show that 퐹 ∩ 푉(∩ 퐴) is dense in 퐹 in the space 퐴. The 퐴-closure of 퐹 ∩ 푉 is 퐹 ∩ 푉 ∩ 퐴 ⊂ F ∩ 퐴 = 퐹. On the other hand, let 푥 ∈ 퐹, 푊 is an open subset in 푋, 푥 ∈ 푊. We have to prove that 푊 ∩ 퐹 ∩ 푉 ≠ ∅. Suppose for contradiction that 푊 ∩ 퐹 ∩ 푉 = ∅. Since 푊 ∩ 푉 is open, 푊 ∩ 퐹 ∩ 푉 = ∅ as well. And since 푊 is open, ∅ = 푊 ∩ 퐹 ∩ 푉 = 푊 ∩ 퐹 . But 푥 ∈ 푊 ∩ 퐹 which is a contradiction.■ Corollary 2.5. Epi 훼-regularity is a hereditary property. ′ Proof. Let (푋, 휏) be an epi 훼 -regular space and let 퐴, 휏퐴 be a subspace of (푋, 휏). Let 휏 be a coarser topology ′ ′ on 푋 such that 푋, 휏 is 훼 -regular, 푇1. The subspace 퐴, 휏퐴 is 훼 -regular, 푇1 as 훼 -regular [30], [10], 푇1 is ′ hereditary 2.5, and 휏퐴 ⊆ 휏퐴, therefore 퐴, 휏퐴 is epi 훼 -regular.■ Theorem 2.6. 훼-regularity are additive properties. Proof. Let 푋훼 훼∈퐴 be a family of 훼 -regular spaces, and 퐴 be a closed subset of the sum ⨁훼∈훬 푋훼 , 푥 ∈ ⨁훼∈Λ 푋훼 such that 푥 ∉ 퐴. By proposition 2.2.1 in[8] the intersections 퐴 ∩ 푋푎 is closed in 푋훼 for every 훼 ∈ Λ and 푥 ∉ 퐴 ∩ 푋훼 From 훼 -regularity of 푋α it follows that there are two open sets 푈α and 푉α in 푋훼 and such that 퐴 ∩ 푋 훼 ∩ 푈 α = 퐴 ∩ 푋α , 푥 ∈ 푉훼 and 푈 ∩ 퐴 = ∅ Let 푈 =∪훼∈Λ 푈훼 and 푉 =∪훼∈Λ 푉훼 , then clearly 퐴 ∩ 푈 = 푈 퐴 ∪ 푈 =∪ 퐴 = 퐴 = 퐴, 푥 ∈ 푉 훼∈훬 훼 훼∈Λ 푈 ∩ V =∪훼∈훬 푈훼 ∩∪훼∈훬 푉훼 =∪훼∈Λ 푈훼 ∩ 푉훼 = 0 Since 푈 and 푉 are open in ⨁훼∈훬 푋훼 , the sum ⨁훼∈훬 푋훼 is 훼 -regular.■ Theorem 2.7. Epi 훼-regularity is an additive property. ′ Proof. Let 푋훼 , 휏훼 be an epi 훼 -regular space for each 훼 ∈ Λ For each 훼 ∈ Λ, let 휏훼 be a topology on 푋푎 coarser ′ than 휏훼 such that 푋훼 , 휏훼 is 훼 -regular, 푇1. since 푇1 is additive see [[8],2.2.7] and 훼 -regularity is also additive ′ by theorem 2.6. Then ⊕훼∈훬 푋훼 , 휏훼 is 훼 -regular, 푇1, and its topology is coarser than the topology on ⊕훼∈퐴 푋푎 , 휏훼 .■ Theorem 2.8. Let 푋훼 , 휏훼 : 훼 ∈ Λ be a family of epi-regular spaces, and let 푋 = ∏훼∈Λ 푋훼 . Then (푋, 휏) is epi 훼 -regular, where 휏 is the Tychonoff , if only if 푋훼 , 휏훼 is epi 훼 -regular for each 훼 ∈ Λ. Proof. Let (푋, 휏) be an epi 훼 -regular space, and let 훽 ∈ Λ, by Theorem 2.5, every subspace of (푋, 휏) is epi 훼 - regular. By [[31],2.39], there is a subspace of (푋, 휏) that is homeomorphic to 푋훽 . since epi 훼-regularity is a topological property then 푋훽 , 휏훽 is epi 훼 -regular. ′ Now let 푋훼 , 휏훼 be epiregular, epi 훼 -regular space for each 훼 ∈ Λ. For each 훼 ∈ Λ, let 휏α be a topology on 푋훼 , ′ coarser than 휏훼 such that 푋훼 , 휏훼 is 푇3. since 푇3 is multiplicative [[8],2.3.11]. Then ∏훼∈Λ 푋훼 is 푇3 with respect ′′ of the product topology of 휏훼 s, which implies that ∏훼∈Λ 푋훼 is 훼 -regular 푇1 with respect of the product topology ′′ of 휏훼 s and its topology is coarser than the topology on ∏훼∈Λ 푋훼 , 휏훼 .■

DOI: 10.9790/5728-1703024148 www.iosrjournals.org 43 | Page On Epi 훼 -Regular Spaces

Let ℝ be the real line. Let ℙ be the set of all irrational numbers and ℚ be the rational numbers. Let 푈 be the usual topology of the real line ℝ. The real line with the topology generated by ℬ = {(푥 − 휀, 푥 + 휀): 푥 ∈ ℚ} ∪ {{푥}: 푥 ∈ ℙ} is called the Michael line and is denoted by 푀. And 푀 × 푃, where ℙ has the usual topology, is called the Michael product [8]. As the Michael line is 훼 -regular, 푇1, hence we have the following corollary. Corollary 2.9. The Michael line is epi 훼 -regular space.■ The space 필 × ℙ is 훼 -regular, 푇1 space being product of two (regular) 훼 regular, 푇1 spaces, so we have the following theorem. Theorem 2.10. The Michael product is an epi 훼 -regular space. ■ Note that epiregularity is invariant under products, however, this is not the case for 푎 -regularity as Murtinová in [30] proved that 훼 -regularity is not preserved under products. Regarding Murtinová result in [30], the following theorem proves that epi 훼 -regularity is not preserved by products and at the same time we construct a non epi 훼 -regular space from a non epiregular space. Theorem 2.11. Let 퐴(휅) is the one-point compactification of a discrete set of cardinality 휅. Then for every non- epiregular 푇1 space 푋 there is 휅 ≤ 휒(푋) such that 푋 × 퐴(휅) is not epi 훼 -regular. Proof. By a similar argument used in theorem [7] in [30].■ It follows that product of an epi 훼 -regular space and a compact zero dimensional space may fail to be epi 훼- regular. In particular it means that epi 훼-regularity is not preserved by products. There are many ways of producing a new topological space from an old one. In 1929 , Alexandroff introduced his method by constructing the Double Circumference Space [1]. In 1968, R. Engelking generalized this construction to an arbitrary space as follows: Let 푋 be any topological space. Let 푋′ = 푋 × {1}. Note that 푋 ∩ 푋′ = ∅. Let 퐴(푋) = 푋 ∪ 푋′ . For simplicity, for an element 푥 ∈ 푋, we will denote the element (푥, 1) in 푋′ by 푥′ and for a subset 퐵 ⊆ 푋 let 퐵′ = 푥′ : 푥 ∈ 퐵 = 퐵 × {1} ⊆ 푋′ . For each 푥′ ∈ 푋′ , let ℬ 푥′ = 푥′ For each 푥 ∈ 푋, let ℬ(푥) = 푈 ∪ 푈′ \ 푥′ : 푈 is open in 푋 with 푥 ∈ 푈}. Let 휏 denote the unique topology on 퐴(푋) which has {ℬ(푥): 푥 ∈ 푋} ∪ ℬ 푥′ : 푥′ ∈ 푋′ as its neighborhood system. 퐴(푋) with this topology is called the Alexandroff Duplicate of 푋 [9]. The following is easy to prove. Lemma 2.12. If 푋 is 푇1 then its Alexandroff Duplicate 퐴(푋) is also 푇1.■ Theorem 2.13. [10] If 푋 is 훼-regular satisfying 푇1 axiom, then its Alexandroff Duplicate 퐴(푋) is also 훼-regular. Proof. Let 퐸 be aclosed set in 퐴(푋) and 푥 ∈ 퐴(푋) such that 푥 ∉ 퐸. Write 퐸 = 퐸1 ∪ 퐸2, where 퐸1 = 퐸 ∩ 푋, 퐸2 = ′ ′ 퐸 ∩ 푋 . So 푥 ∉ 퐸1 in 푋 and 푥 = (푥, 1) ∉ 퐸2. By 훼 -regularity of 푋, there exist two disjoint open sets 푈 and 푉 of ′ 푋 such that 퐸1 ∩ 푈 is dense in 퐸1 and 푥 ∈ 푉. Since 푋 is 푇1 we can choose 푊1 = 푈 ∪ 푈 ∪ 퐸2 ∖ {푥} and ′ ′ ′ 푊2 = 푉 ∪ 푉 ∪ 푥 ∖ 퐸. Then 푊1 and 푊2are disjoint open sets in 퐴(푋), and 푥 ∈ 푊2. Now, we prove 푊1 ∩ 퐸 is dense in 퐸. Note that 푊1 ∩ 퐸 = 푊1 ∩ 퐸1 ∪ 푊1 ∩ 퐸2 = 푈 ∩ 퐸1 ∪ 퐸2, so 푊1 ∩ 퐸 = 푈 ∩ 퐸1 ∪ 퐸2 =

푈 ∩ 퐸 1 ∪ 퐸 2 ) ⊃ 퐸1 ∪ 퐸 2 ⊃ 퐸. Therefore, 푊1 ∩ 퐸 is dense in 퐸. Then 퐴(푋) is 훼-regular. Hence 훼-regular is preserved by the Alexandroff Duplicate space.■ Theorem 2.14. If (푋, 풦) is epi 훼-regular, then so is its Alexandroff Duplicate (퐴(푋), 휏). ′ ′ Proof. Let (푋, 풦) be an epi 훼 -regualr space, then a coarser topology 풦 on 푋 exists such that 푋, 풦 is 푇1, 훼 - regular. Let 퐴(푋), 휏′ be the Alexandroff Duplicate of 푋, 풦′ . Since by theorem 2.13 훼-regularity is ′ preserved by the Alexandroff Duplicate space and also 푇1, then 퐴(푋), 휏 is also 푇1, 훼 -regular, and it is obviously coarser than (퐴(푋), 휏) by the topology of the Alexandroff Duplicate. Hence, 퐴(푋) is epi 훼 -regualr.■ In 1951, Bing [5] and Hanner [14] introduced a new topological space by generating it from an old topological space. This new space is called discrete extension. Definition 2.15. Let 푀 be a non-empty proper subset of a topological space (푋, 휏). Define a new topology 휏푀 = {푈 ∪ 퐾: 푈 ∈ 휏 and 퐾 ⊆ 푋 ∖ 푀}. The space 푋, 휏푀 is called discrete extension, and donated by 푋푀, see [8], [21]. In [21], properties such as countable tightness, Fréchet, and weaker types of normality were investigated for discrete extension. Here we study the relationship between a space 푋 and a discrete extension 푋푀 of 푋 according to epi 훼 -regularity. ′ ′ For any epi 훼 -regular space (푋, 휏) we have 휏 ⊆ 휏 ⊆ 휏푀 where 휏 is Tychonoff, so we have the following proved. Theorem 2.16. If (푋, 휏) is an epi 훼 -regular space, then also is 푋푀.■ Since any Hausdorff locally compact is Tychonoff and hence epi 훼 -regular, then by theorem 2.16 the following is easy to prove Corollary 2.17. If (푋, 휏) is a Hausdorff , then 푋푀 is epi 훼 -regular.■

IV. Epi 훼 -Regularity And Some Other Separation Axioms A topological space (푋, 휏) is called submetrizable if there exists a metric 푑 on 푋 such that the topology 휏푑 on 푋 generated by 푑 is coarser than 휏, i.e., 휏푑 ⊆ 휏, see [13], since, by definitions, any submetrizable space is epi 훼-regular. The converse of the last statement is not true in general. For example, 휔1 + 1 is epi 훼-regular being 푇2 compact, hence 훼-regular 푇1 and therefore epi 훼-regular. But it is not submetrizable, because if 휔1 + 1 DOI: 10.9790/5728-1703024148 www.iosrjournals.org 44 | Page On Epi 훼 -Regular Spaces was submetrizable, then there would be a metric 푑 on 휔1 + 1 such that the topology 휏푑 on 휔1 + 1 generated by 푑 is coarser than the usual ordered topology. This means that 휔1 + 1, 휏푑 is perfectly normal. So, the closed set 휔1 is a 퐺훿 −set in 휔1 + 1, 휏푑 . i.e., 휔1 =∩푛∈푁 푈푛 , where 푈푛 ∈ 휏푑 , hence 푈푛 is open in the usual ordered topology on 휔1 + 1, which is a contradiction. ′ Obviously, any 훼-regular, 푇1 space is epi 푎-regular, just by taking 휏 = 휏. However epi 훼 -regularity and 훼 -regularity do not imply each other. For example, the Half-Disc space [33] is epi 훼-regular which is not 훼-regular by proposition 1.5, since the space is Hausdorrf first countable not regular. Similarly, Deleted Diameter topology [32] is epi 훼 -regular being submetrizable, but it is not 훼 -regular. Any indiscrete space which has more than one element is an example of an 훼 -regular space which is not epi 훼 -regular. Semiregularization were studied in [27], a is 푇2 space in which the regular open sets form a basis for the topology [33]. Epi 훼 -regularity and semiregularity are independent, for example the Half-Disc space [33], is epi 훼-regular but not semiregular, It is epi 훼 -regular because it is submetrizable. and any indiscrete space which has more than one element is an example of a semiregular space which is not epi 훼-regular. Recall that a topological space (푋, 휏) is called extremally disconnected if it is 푇1 and the closure of any open set is open [18]. Since every 훼 -regular, extremely disconnected space is regular [10], then we have the following correct. Corollary 3.1. If 푋 is an epi 훼-regular space and the attested of epi 훼-regularity is extremely disconnected, then 푋 is epiregular.■ Recall that a topological space (푋, 휏) is called Zero-dimensional if it is a non-empty 푇1 space and has a base consisting of open-and-closed sets [8]. Clearly, every zero-dimensional space is , and hence 푇3, so we conclude. Corollary 3.2. Any zero-dimensional space is epi 훼-regular.■ The converse of the above result is not always correct. For example, The Euclidean topology on the set of real numbers is epi 훼 -regular since it is 푇3 but not zero dimensional. The following example [22] is a modified example of Mysior’s example from [28]. Example 3.3. Let 퐴 ⊆ ℝ be such that the intersection 퐴푘 = 퐴 ∩ [푘, 푘 + 1) is uncountable for every integer 푘 ∈ ℤ. Let Δ = {(푎, 푎): 푎 ∈ 퐴} be the diagonal of 푋 = 퐴2 and define the following sets 푈푘 = {(푎, 푏) ∈ 푋: 푎 > 푘} for 푘 ∈ ℤ Γ푎 = {(푎 + 휀, 푎) ∈ 푋: 휀 ∈ [0,3]} ∪ {(푎, 푎 − 휀) ∈ 푋: 휀 ∈ [0,3]} for 푎 ∈ A. Consider a topology 휏 on 푋 = 퐴2 generated by a basis consisting of all singletons {푥} with 푥 ∈ 푋 ∖ Δ and all sets Γ푎 ∖ 퐹, where 푎 ∈ 퐴 and 퐹 is finite. Clearly 푋 is Hausdorff and zero-dimensional, and so is epi 훼- regular. The following example is constructed by Murtinová in [28] as she showed that it is an example of an 훼 - normal Hausdorff, hence 훼 -regular, non regular space. Example 3.4.[29] Let 푋 = 휔1 + 1 and define a topology 휏 such that: 휔1 with the ordinal topology is an open subspace and a base in the point 휔1 will be the collection: 푈퐶 = 휔1 ∪ {훼 + 1: 훼 ∈ 퐶} where 퐶 is a closed unbounded subset of 휔1 (Club). The topology 휏 is Hausdorff since it is stronger than the onder topology on 휔1 + 1. This space is epi 훼 -regular since it is 훼 -regular Hausdorff and it is epinegular since it is stronger that the order topology on 휔1 + 1 but it is not regular nor first countable. Note that the right order topology defined on the set of real numbers ℝ[33] is an example of 훽 -normal, 훼- normal since there are no disjoint closed sets on it and it is not epi 훼-regular since it is not Hansdorff. Recall that a topological space (푋, 휏) is called epicompletely regular [12] if there is a coarser topology 휏′ on 푋 such that 푋, 휏′ is Tychonoff. Note that if a topological space 푋 is epicompletely regular, then the space is epi 훼-regular. But the converse of the above statement is not always true. however, the following theorem is correct since epiregularity implies epicompletely regularity. Corollary 3.5. If (푋, 휏) is an epi 훼-regular space, and the witnesses of epi 훼-regularity 푋, 휏′ is first countable, then (푋, 휏) is epicompletely regular.■ It is well known that every compact second countable topological space satisfying 푇2 axiom is metrizable, [[ 8],4.2.8 and this induces another result. Corollary 3.6. If a topological space 푋 is epi 훼-regular, compact, and the attested of epi 훼-regular is second countable then the space is submetrizable.■ Remind that a topological space (푋, 휏) is 퐶2-paracompact if there is a 푇2, paracompact space (푌, 훿) and a bijective map 푓: (푋, 휏) ⟶ (푌, 훿) such that the restriction 푓∣퐴: 퐴 ⟶ 푓(퐴) is a homeomorphism for every

DOI: 10.9790/5728-1703024148 www.iosrjournals.org 45 | Page On Epi 훼 -Regular Spaces compact subspace 퐴 ⊆ 푋. For more details see [15]. A space X is called Fréchet if for every 퐴 ⊆ 푋 and every 푥 ∈ 퐴 there exists a sequence (푥푛 )푛∈ℕ of points of 퐴 such that 푥푛 → 푥, see [8]. Theorem 3.7. Let (푋, 휏) be a 퐶2 -paracompact and Fréchet, then (푋, 휏) is epi 훼-regular. Proof. Let (푋, 휏) be a 퐶2 -paracompact and Fréchet, then (푋, 휏) is epinormal by theorem 2.16 in [15], then it is epi 훼 -normal. Hence (푋, 휏) is epi 훼 regular.■ Theorem 3.8. If (푋, 휏) is Lindelöff epi 훼-regular space and the attasted of epi 훼-regularity 푋, 휏′ is first countable, then (푋, 휏) is 퐶2 -paracompact. Proof. Let (푋, 휏) be a Lindelöff epi 훼-regular space, then there exists a coarser topological space 푋, 휏′ that is 푇1, 훼-regular first countable, and hence is regular by propostion 1.5. since (푋, 휏) is a Lindelöff space, then ′ ′ 푋, 휏 is also Lindelöff and regular which implies that 푋, 휏 is 푇2 and paracompact, and therefore the identity ′ map id : (푋, 휏) ⟶ 푋, 휏 is the required map to have our space (푋, 휏) to be 퐶2 -paracompact.■ Since any regular Lindelöff space is normal, then this is not hard to show Corollary 3.9. Let (푋, 휏) be an epi 훼-regular Lindeloff space, and the attested of epi 훼-regularity is first countable, then (푋, 휏) is epinormal.■ Remind that a topological space (푋, 휏) is called nearly compact [23] if every open cover of 푋 has a finite subfamily the interiors of the closures of whose members covers 푋. Theorem 3.10. If (푋, 휏) is a Hausdorff nearly compact space, then (푋, 휏) is epi 훼-regular. Proof. Let (푋, 휏) be a Hausdorff nearly compact space, and let 휏퐴 be the semi regularization of 휏, then 휏푠 is a Hausdorff nearly compact space. Therefore 휏퐴 is 푇4, and hence 푇2, 훼-regular. Therefore (푋, 휏) is epi 훼-regular.■ Remind that a topological space (푋, 휏) is called partially normal [18] if for any two disjoint subsets 퐴 and 퐵 of 푋, where 퐴 is regularly closed and 퐵 is 휋-closed, there exist two disjoint open subsets 푈 and 푉 of 푋 containing 퐴 and 퐵 respectively. Theorem 3.11. If (푋, 휏) is a semi regular partial and 휏푠 is 푇1, then (푋, 휏) is epi 훼-regular. Proof. It is enough to show that 푋, 휏푠 is 훼-regular. Let 푈 be any open set containing 푥 in 푋, 휏푠 . By semiregularity, there is an open set 푊 such that 푥 ⊆ 푊 ⊆ int⁡(W ) ⊆ 푈. Since int (W ) is regularly open and using the same idea of theorem 2.11 in [2] there exists an open set 푉 in 푋, 휏푠 such that 푥 ⊆ 푉 ⊆ 푉 ⊆ int⁡(W ) ⊆ 푈. Therefore 퐴 ∩ 푉 ⊆ 푉 ⊆ int⁡(푈 ) ⊆ 퐵. Hence 푋, 휏푠 is 훼-regular, and then (푋, 휏) is epi 훼- regular.■ Epi 훼-regularity and 훼-normality do not imply each other. For example, the Dieudonné topology and The deleted Tychonoff Plank, see [26] and [33], are not normal space nor 훼 -normal, but they are epi 훼-regular because they are zero dimensional. Remind that a space 푋, 휏 is called almost 훼-normal [11] if for any two disjoint closed subsets 퐴 and 퐵 of 푋 one of which is regularly closed there exist disjoint open subsets 푈 and 푉 of 푋 such that 퐴 ∩ 푈 is dense in 퐴 and 퐵 ∩ 푉 is dense in 퐵. That is, 퐴 ∩ 푈 = 퐴 and 퐵 ∩ 푉 = 퐵. and a space 푋, 휏 is called almost 훽-normal [11] if for any two disjoint closed subsets 퐴 and 퐵 of 푋 one of which is regularly closed there exist disjoint open subsets 푈 and 푉 of 푋 such that 퐴 ∩ 푈 is dense in 퐴 and 퐵 ∩ 푉 is dense in 퐵. That is, 퐴 ∩ 푈 = 퐴 , 퐵 ∩ 푉 = 퐵 and 푈 ∩ 푉 = ∅. Note that almost α-normality (almost β-normality) and epi 훼-regularity are not related to each other. For example, ℝ with the particular point topology 휏푝, see [8], [33], where the particular point is 푝 ∈ ℝ, is not normal nor 훽-normal nor 훼-normal. But the space is almost 훽-normal and almost 훼-normal since the only regularly closed sets are ℝ and ∅. However this space is not Hausdorff and then it is epi 훼-regular. Conversely, Any indiscrete space which has more than one element is an example of an almost α-normal (almost β-normal) space which is not epi 훼-regular. However, every almost α normal extremely disconnected space is epi 훼- regular. A 훽-normal epi 훼 -regular non normal space example found in [29] which as follows: Example 3.12. Let 푆 = 훼 < 휔2: 푐푓(훼) = 휔1 , and consider the set 푋 = (훼, 훽): 훽 ≤ 훼 ≤ 휔2, (훼, 훽) ≠ 휔2,휔2 and its partition into 퐴 = (훼, 훼): 훼 < 휔2 퐵 = 휔2, 훽 : 훽 < 휔2

퐷 = 훼, 훽): 훽 < 훼 < 휔2 Topologize 푋 as follow: Let each (훼, 훽) ∈ 퐷 be isolated, and let an open base in (훼, 훼) ∈ 퐴 consists of all sets of type (훾, 훾): 훼0 < 훾 ≤ 훼 ∪ ⋃ {훾} × 퐶훾 : 훼0 < 훾 ≤ 훼, 훾 ∈ 푆 where 훼0 < 훼 and every 퐶훾 is a closed and unbounded (club) subset of 훾, and let an open base in 휔2, 훽 ∈ 퐵 consists of all sets

(훼, 훾): 훽0 < 훾 ≤ 훽, 훼훾 < 훾 ≤ 휔2

DOI: 10.9790/5728-1703024148 www.iosrjournals.org 46 | Page On Epi 훼 -Regular Spaces where 훽0 < 훽, 훽 ≤ 훼훾 < 휔2. All above defined basic open neighborhoods are closed. That is, 푋 is zero dimensional hence it is epi 훼 -regular. Murtinová in [29] proved that this space is 훽 -normal non normal. Remind that a toplogical space (푋, 휏) is called epi-mildly normal [17] if there exists a coarser topology 휏′ on 푋 such that 푋, 휏′ is Hausdorff, mildly normal. The following theorem induced by theorem 2.4 in [26] shows a relationship between epi-mildly normality, 훽 -normal and epi 훼 -regular. Theorem 3.13. If a topological space is epi-mildly normal and the witness of epi-mildly normality is 훽-normal then (푋, 휏) is epi 훼-regular. Proof. Let (푋, 휏) be a topological epi-mildly normal space and the attested of epi-mildly normal epinormal is 훽 - normal, then there is a coarser topological space 푋, 휏′ that is Hausdorff, mildly normal and 훽-normal, so by theorem (2.4) in [26] then 푋, 휏′ is Hausdorrf and normal, and therefore it is Hansdorff 훼 -normal, and so ′ 푋, 휏 is 푇1, 훼 -regular [10]. Hence (푋, 휏) is epi 훼 -regular.■ Epi 훼 -regularity does not imply mildly normality. Example 3.14. [3] Let ℙ denote the irrationals and ℚ denote the rationals. For each 푝 ∈ ℙ and 푛 ∈ ℕ, let 푝 = 푝, 1 ∈ ℝ2. For each 푝 ∈ ℙ, choose a sequence 푝∗ of rationals such that 푝′ = 푝∗ , 0 → (푝, 0) 푛 푛 푛 푛∈N 푛 푛 2 where the convergence is taken in ℝ with its usual topology 풰. For each 푝 ∈ ℙ and 푛 ∈ ℕ, let 퐴푛 ((푝, 0)) = ′ 푝푘 : 푘 ≥ 푛 and 퐵푛 ((푝, 0)) = 푝푘 : 푘 ≥ 푛 . Now,for each 푝 ∈ ℙ and 푛 ∈ ℕ, let 푈푛 ((푝, 0)) = {(푝, 0)} ∪ 퐴 ((푝, 0)) ∪ 퐵 ((푝, 0)). Let 푋 = (푥; 0) ∈ ℝ2: 푥 ∈ ℝ ∪ {푝 = 푝, 1 ∈ ℝ2: 푝 ∈ ℙ and 푛 ∈ ℕ . For each 푞 ∈ 푛 푛 푛 푛 ℚ, let 퐵((푞, 0)) = {{(푞, 0)}}. For each 푝 ∈ ℙ, let 퐵((푝, 0)) = 푈푛 ((푝, 0)): 푛 ∈ ℕ . For each 푝 ∈ ℙ and each

푛 ∈ ℕ, let 퐵 푝푛 = 푝푛 . Denote by 휏 the unique topology on 푋 that has 퐵((푥, 0)), 퐵 푝푛 : 푥 ∈ ℝ, 푝 ∈ Pandn ∈ ℕ} as its neighbomood system. Let 푍 = {(푥, 0): 푥 ∈ ℝ}. That is, 푍 is the 푥 -axis. Then (푍, 휏) ≅ (ℝ, ℛ𝒮), where ℛ𝒮 is the Rational Sequence Topology, see [33]. Since 푍 is closed in 푋 and (ℝ, ℛ𝒮) is not normal, then 푋 is not normal, but, since any basic open set is closed-and-open and 푋 is 푇1, then 푋 is zero-dimensional, hence epi 훼-regular. Now, Let 퐴 ⊆ ℙ and 퐵 ⊆ ℙ be closed disjoint subsets that cannot be separated in (ℝ, ℛ𝒮). Let 퐺 =∪ 퐵1((푝, 0)): 푝 ∈ 퐴 and 퐻 =∪ 퐵1((푝, 0)): 푝 ∈ 퐵 . Then 퐺 and 퐻are both open in (푋, 휏) and 퐺 and 퐻 are disjoint closed domains that cannot be separated, hence 푋 is not mildly normal. Epi 훼-regularity does not imply epinormality, and here is an example. Example 3.15. Let 퐺 = 퐷ω1 where 퐷 = {1,2} with the discrete topology. Let 퐻 be the subspace of 퐺 consisting of all points of 퐺 with at most countabl many zero coondinates. Put 푋 = 퐺 × 퐻. Raushan Buzyakova proved that 푋 cannot be mapped onto a normal space 푌 be a bijective continuous function [7]. Using Buzyakova’s and the fact that 푋 is 푘 -space [[8],3.3.27], then this implies that 푋 is Tychonoff and so is epi 훼 -regular and it cannot be 퐶-normal see 3], and since epinormality implies 퐶-normailty, then 푋 cannot be epinormal.

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