EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 15, No. 4, 2022, 1808-1821
ISSN 1307-5543 – ejpam.com
Published by New York Business Global

Epi-completely regular topological spaces

Ibtesam Alshammari1,∗

1 Department of Mathematics, Faculty of Science, University of Hafr Al Batin, Saudi Arabia

Abstract. The purpose of this work is to introduce and study a new topological property called
epi-complete-regularity. A space (X, T ) is called an epi-completely-regular space if there exists a
topology T ′ on X which is coarser than T such that (X, T ′) is Tychonoff. This new property is
investigated and some examples are presented in this work to illustrate its relationships with other
kinds of normality and complete-regularity.

2020 Mathematics Subject Classifications: 54A10, 54B10, 54C10, 54D10, 54D20, 54D15,
54D70

Key Words and Phrases: Epi-normal, epi-regular, epi-almost normal, epi-quasi normal, epi-
partially normal, completely regular and epi-mildly normal

1. Introduction

The notion of epi-normality was introduced by Arhangel’skii during his visiting to
Department of Mathematics in King Abdulaziz University, Saudi Arabia on 2012. The
notion of epi-normality has been studied by Kalantan and Alzahrani in 2016 [15]. Then,
Alzahrani studied the notion of epi-regularity in 2018 [5]. Kalantan and Alshammari stud-
ied the notion of epi-mild normality in 2018 [18]. At the beginning of 2020, Alshammari
studied the notion of epi-almost normality [3]. Thabit studied the notion of epi-partial
normality in 2021 [32]. At the end of 2021, Thabit and others studied the notion of epi-
quasi normality [31]. The space X means a topological space in whole paper. We need to
recall that: a subset A of a space X is said to be a closed domain subset if it is the closure
of its own interior [20]. The complement of a closed domain subset is called open domain.
A subset A of a space X is called π-closed if it is a finite intersection of closed domain
subsets [33]. The complement of a π-closed subset is called π-open. Two subsets A and
B of a space X are said to be separated if there exist two disjoint open subsets U and V
of X such that A ⊆ U and B ⊆ V [11, 12, 23]. If T and T ′ are two topologies on X such
that T ′ ⊆ T , then T ′ is called a topology coarser than T , and T is called finer [12]. A
T4-space is a T1 normal space, a T3-space is a T1 regular space and a Tychonoff space is a

∗Corresponding author.
DOI: https://doi.org/10.29020/nybg.ejpam.v15i4.4598

Email addresses: iealshamri@hotmail.com, iealshamri@uhb.edu.sa (I. Alshammari)

https://www.ejpam.com 1808 © 2022 EJPAM All rights reserved.



I. Alshammari / Eur. J. Pure Appl. Math, 15 (4) (2022), 1808-1821 1809

T1 completely regular space. A space X is said to be π-normal [14], if any pair of disjoint
closed subsets A and B of X, one of which is π-closed, can be separated. A space X is
said to be almost-normal [14, 28], if any pair of disjoint closed subsets A and B of X, one
of which is closed domain, can be separated. A space X is said to be mildly normal [29], if
any pair of disjoint closed domain subsets A and B of X can be separated. A space X is
said to be partially normal [4], if any pair of disjoint closed subsets A and B of X, one of
which is closed domain and the other is π-closed, can be separated. A space (X, T ) is said
to be epi-normal [15] (resp. epi-mildly normal [18], epi-almost normal [3], epi-regular [5],
epi-quasi normal [31], epi-partially normal [32]), if there exists a topology T ′ on X coarser
than T such that (X, T ′) is a T4 (resp. Hausdorff mildly-normal, Hausdorff almost-normal,
T3, Hausdorff-quasi-normal, Hausdorff partially-normal) space. A space X is said to be
Hausdorff or a T2-space, if for each distinct two points x, y ∈ X there exist two open
subsets U and V of X such that x ∈ U , y ∈ V are U ∩V = ∅ [12]. A space X is said to be
completely Hausdorff or Urysohn [12, 30], if for each distinct two points x, y ∈ X there
exist two open subsets U and V of X such that x ∈ U , y ∈ V and U ∩ V = ∅. A space X
is said to be almost completely-regular if for each x ∈ X and each closed domain subset F
of X such that x ̸∈ F , there exists a continuous function f : X → [0, 1] such that f(x) = 0
and f(F ) = {1} [28]. A space X is said to be almost-regular if for each x ∈ X and each
closed domain subset F of X such that x ̸∈ F , there exist two disjoint open subsets U
and V such that x ∈ U and F ⊆ V [27]. A space X is said to be sub-metrizable [13], if
there exists a metric d on X such that the topology Td on X generated by d is coarser
than T . The topology on X generated by the family of all open domain subsets of X,
denoted by Ts, is coarser than T , and (X, Ts) is called the semi-regularization of X. A
space (X, T ) is called semi-regular if T = Ts [22]. A space X is called H-closed [12], if
X Hausdorff almost-compact [19, 24]. A space X is called C-normal [8] (resp. C-regular
[6], C-Tychonoff [7]) if there exist a normal (resp. regular, Tychonoff) space Y and a
bijective function f : X → Y such that the restriction function f |A : A → f(A) is a home-
omorphism for each compact subspace A ⊆ X. A space X is called L-normal [16] (resp.
CC-normal [17]) if there exist a normal space Y and a bijective function f : X → Y such
that the restriction function f |A : A → f(A) is a homeomorphism for each Lindelöf (resp.
countably compact) subspace A ⊆ X. A space X is called L-regular [6] (resp. L-Tychonoff
[7]) if there exist a regular (resp. Tychonoff) space Y and a bijective function f : X → Y
such that the restriction function f |A : A → f(A) is a homeomorphism for each Lindelöf
subspace A ⊆ X. The basic definitions and any undefined terms in this article can be
found in [31] and [32].

In this paper, I introduce and study a new topological property called epi-complete
regularity. I show that this new property is different from epi-normality, epi-regularity,
epi-mild normality, epi-quasi normality, epi-partial normality and epi-almost normality.
Some properties, counterexample and relationships of this property are investigated. This
paper contains three main sections starting from section 2. In section 2, the definition of
epi-complete regularity is introduced and some examples are presented. Some properties
of epi-complete regularity are studied and given in section 3.



I. Alshammari / Eur. J. Pure Appl. Math, 15 (4) (2022), 1808-1821 1810

2. Preliminaries

First, I present the main definition of this study:

Definition 1. A space (X, T ) is called an epi-completely-regular space if there exists a
topology T ′ on X which is coarser than T such that (X, T ′) is Tychonoff.

From Definition 1, note that: every epi-completely regular space is Hausdorff and
any Tychonoff space is epi-completely-regular, but the converses are not true in general,
for example: the irregular lattice topology, Example 6 is a Hausdorff space which is not
epi-completely regular. The Smirnov’s deleted sequence topology, Example 10, and the
half disc topology, Example 5, are epi-completely regular spaces which are not Tychonoff.
Now, I present the next results:

Theorem 1. Every epi-completely-regular space is Urysohn.

Proof. Let (X, T ) be an epi-completely-regular space. Then, there exists a topology T ′

on X that is coarser than T such that (X, T ′) is T1-completely-regular. Thus, (X, T ′) is
Tychonoff. Hence, (X, T ′) is Uryshon (completely Hausdorff). Since T ′ ⊆ T , we conclude:
(X, T ) is Urysohn.

Observe that: any Urysohn space is not necessary to be epi-completely regular. For
example, the Tychonoff corkscrew topology, Example 9, and the irregular lattice topology,
Example 6, are Urysohn spaces which are not epi-completely-regular. Thus, the converse
of Theorem 1 is not true in general.

Theorem 2. Every epi-completely-regular space is epi-regular.

Proof. Let (X, T ) be an epi-completely-regular space. Then, there exists a topology T ′

on X coarser than T such that (X, T ′) is T1-completely-regular. Since every completely-
regular space is regular [12], we get: (X, T ′) is a T1-regular space. Hence, (X, T ′) is
T3-space. Therefore, (X, T ) is epi-regular.

Note that: the converse of Theorem 2 is not necessarily true in general. For example,
the Tychonoff corkscrew topology, Example 9, is an epi-regular space which is not epi-
completely-regular. Also, complete regularity and epi-complete regularity are different
from each other, for example, the half disc topology, Example 5, is an epi-completely-
regular space, which is not completely-regular and any uncountable indiscrete space is a
completely-regular space which is not epi-completely-regular.

Theorem 3. Every epi-almost-normal space is epi-completely-regular.

Proof. Let (X, T ) be an epi-almost-normal space. Then, there exists a topology T ′

on X which is coarser than T such that (X, T ′) is a Hausdorff almost-normal space.
Since every almost-normal T1-space is almost-regular [27], we have: (X, T ′) is Hausdorff
almost-normal almost-regular. Since every almost-normal almost-regular space is almost-
completely regular [28], we get: (X, T ′) is Hausdorff almost-completely regular. Let the



I. Alshammari / Eur. J. Pure Appl. Math, 15 (4) (2022), 1808-1821 1811

semi regularization of (X, T ′) be (X, T ′
s ). Then, (X, T ′

s ) is a Hausdorff completely-regular
space because the semi regularization of a Hausdorff almost completely regular space is
Hausdorff completely regular [22]. Since T ′

s ⊆ T ′ ⊆ T , we conclude: T ′
s is a topology

on X that is coarser than T such that (X, T ′
s ) is Hausdorff completely-regular and hence

Tychonoff. Therefore, (X, T ) is epi-completely-regular.

Since every epi-completely-regular space is epi-regular (Theorem 2), every sub-metrizable
space is epi-normal and every epi-normal space is epi-almost-normal [3, 15], we obtain:

Corollary 1.

(1) Every sub-metrizable space is epi-completely-regular.

(2) Every epi-normal space is epi-completely-regular.

Thus, we conclude the following implications:

epi-normal =⇒ epi-almost-normal =⇒ epi-completely-regular =⇒ epi-regular

The next example is an epi-completely regular space which is not epi-normal.

Example 1. Consider the Example 10 in [26], let G = Dω1 , where D = {0, 1} with the
discrete topology. Let H be a subspace of G consisting of all points of G with at most
countably many non zero coordinates. Put X = G × H. Raushan Buzyakova proved
that X cannot be mapped onto a normal space Y by a bijective continuous function [9].
It can be observed that: H is a T2-Fréchet space and hence it is a k-space. G is also a
T2-compact space. Hence, X = H × G is a k-space [26]. Since X is Tychonoff, we get
X is epi-completely regular. The space X is not C-normal [26]. Since every C-Tychonoff
Fréchet Lindelöf space is C-normal, we conclude: X is not Lindelöf. Since X is not C-
normal, we obtainX is neither CC-normal, sub-metrizable nor epi-normal. The spaceX is
not a locally compact space as well. Thus, the space X is an epi-completely regular space
which is neither C-normal, CC-normal, epi-normal, sub-metrizable nor locally compact.

Observe that: any C-Tychonoff (resp. C-normal) space is not necessary to be epi-
completely regular. Here is a counterexample:

Example 2. The countable complement topology (R, CC) is both C-Tychonoff and C-
regular space [6, 7], which is neither epi-completely-regular, epi-regular nor epi-mildly
normal because it is not Hausdorff.

The following example is a normal space, which is not epi-completely regular.

Example 3. The left ray topology (R,L), the right ray topology (R,R) [30] are normal
spaces, which are not epi-completely regular because they are not Hausdorff.

Note that: complete regularity (resp. L-regularity) does not imply to epi-complete-
regularity in general as shown by the next example.



I. Alshammari / Eur. J. Pure Appl. Math, 15 (4) (2022), 1808-1821 1812

Example 4. The double pointed reals topology [30, Example 62], is both a regular and
completely regular space [30], which is not epi-completely regular because it is not Haus-
dorff.

The next example is an epi-completely-regular space which is neither Tychonoff nor
completely-regular.

Example 5. The half disc topology [30, Example 78] is not Tychonoff. The semi regular-
ization of X is the closed upper half plane with the Euclidean topology U on R that is a
topology coarser than T and (X,U) is a T4-space. Thus, X is epi-normal. Hence, X is
epi-completely-regular. Since (X, T ) is an almost-completely regular space if and only if
(X, Ts) is completely-regular [22], we get: the half disc topology is almost-completely reg-
ular. Therefore, the half disc topology is an epi-completely-regular space, which is neither
completely-regular, Tychonoff nor almost-normal.

A Urysohn epi-mildly normal Lindelöf space is not necessary to be epi-completely-
regular, for example:

Example 6. The irregular lattice topology [30, Example 79], is a Urysohn Lindelöf space,
which is neither normal, completely regular nor semi-regular [30]. It is also a mildly-normal
space, which is not partially-normal [4]. Hence, it is neither quasi-normal, almost-normal
nor semi-normal. Since every almost-regular Lindelöf space is quasi-normal [21], and X is
a Lindelöf non quasi-normal space, it is not almost-regular. Since (X, T ) is a Hausdorff
mildly-normal space, it is epi-mildly normal. Hence, the irregular lattice topology is a
Urysohn epi-mildly-normal space, which is neither epi-almost-normal, epi-regular nor epi-
completely-regular.

An almost-completely regular space is not necessarily epi-completely-regular. For ex-
ample:

Example 7. The telophase topology [30, Example 73], is a T1-compact, paracompact
space, which is neither Hausdorff, normal nor semi-regular [30]. Clearly that: X is an
almost-regular space. Since it is an almost-regular paracompact space, it is almost-normal.
Since every almost-normal T1 space is almost-completely regular, we have: the telophase
topology is T1-almost-completely regular. Since the telophase topology is not Hausdorff,
it is neither epi-completely-regular, epi-mildly normal nor epi-regular. Therefore, the
telophase topology is an almost-completely regular space, which is neither epi-completely-
regular, epi-mildly-normal nor epi-regular.

An epi-completely-regular space need not be almost-normal nor quasi-normal. Here is
an example:

Example 8. The Thomas’ plank topology [30, Example 93], Let X =
∞⋃
i=0

Li, where L0 =

(0, 1)×{0} and Li = [0, 1)×{1
i } for each i ≥ 1. For each i ≥ 1, each point (x, 1i ) ∈ Li, x ̸=

0, we have {(x, 1i )} is an open subset of X. For each i ≥ 1, the basic open subset of the



I. Alshammari / Eur. J. Pure Appl. Math, 15 (4) (2022), 1808-1821 1813

points (0, 1i ) ∈ Li is a subset Wi of Li such that Li −Wi is finite. The basic open subset
of any point (x, 0) ∈ L0 is of the form Ui(x, 0) = {(x, 0)} ∪ {(x, 1

n) : n > i}. It can be
observed that: each basic open subsets of X is clopen (closed-and-open). Hence, (X, T ) is
a zero-dimensional, Hausdorff, regular, completely-regular, semi-regular, Urysohn, locally-
compact and Tychonoff space, and it is neither normal nor paracompact [30]. Hence, the
Thomas’ plank topology is an almost-regular and almost-completely regular space. Since
it is Hausdorff, we have: the Thomas’ plank topology is epi-completely-regular and epi-
regular space. Since X is Hausdorff locally-compact, we obtain: X is a k-space. Thus, X
is C-normal. It can be observed that: each Li, i ≥ 1 is open because L0 is closed [30].
Also, A = {(0, 1

n) : n ≥ 1} is a closed subset of X [30]. Since A ∩ L0 = ∅, we get: A and
L0 are disjoint closed subsets of X, which cannot be separated [30]. Let U =

⋃
n∈N

L2n and

V =
⋃
n∈N

L2n+1. Then, U and V are disjoint open subsets of X. Thus, U = U ∪ L0 and

V = V ∪ L0. Hence, U and V are closed-domains in X such that U ∩ V = L0. Therefore,
L0 is a π-closed subset of X. Since A and L0 cannot be separated, we obtain: X is not
π-normal.
Claim 1: Any singleton {(x, 0)} is π-closed and any singleton {(0, 1i )}, i ≥ 1 is also
π-closed in X.
Proof of the Claim 1: Let Ux = {(x, 1

2n) : n ∈ N} and Vx = {(x, 1
2n+1) : n ∈ N}. Then, Ux

and Vx are disjoint open subsets of X such that Ux = Ux∪{(x, 0)} and Vx = Vx∪{(x, 0)}.
Therefore, Ux and Vx are closed domain subsets ofX and Ux∩Vx = {(x, 0)}. Thus, {(x, 0)}
is π-closed in X for each x ∈ (0, 1). Now, fix a sequence ⟨(xik,

1
i )⟩ of distinct points of Li.

Consider the two subsequences Ui = {(xi2k,
1
i ) : k ∈ N} and Vi = {(xi2k+1,

1
i ) : k ∈ N}.

Then, Ui and Vi are disjoint open subsets ofX, Ui, Vi ⊂ Li for each i ≥ 1, Ui = Ui∪{(0, 1i )}
and Vi = Vi∪{(0, 1i )}. Since Ui and Vi are closed-domains of X, we get: {(0, 1i )} is π-closed
for each i ≥ 1. Now, let G =

⋃
i≥1

Ui and H =
⋃
i≥1

Vi. Then, G and H are disjoint open

subsets ofX such thatG = G∪A∪{(x2k, 0) : k ∈ N} andH = H∪A∪{(x2k+1, 0) : k ∈ N},
where A = {(0, 1

n) : n ∈ N}. Then, G andH are closed-domains inX such thatG∩H = A.
Hence, A is π-closed. Since A∩L0 = ∅ and they cannot be separated [30], we obtain that:
X is not quasi-normal. It is easy to show that X cannot be semi-normal.
Claim 2: The Thomas’ plank topology is not almost-normal.
Proof of the Claim 2: It can be observed that, A1 = {(0, 1

2n) : n ∈ N} is a closed subset
of X and U =

⋃
n∈N

L2n is an open-domain subset of X such that A1 ⊆ U . Then, for

each open subset W of X such that A1 ⊂ W , we have: A1 ⊆ W ⊆ W ̸⊆ U because
there are some points (x, 0) ∈ W , and (x, 0) ̸∈ U for each (x, 0) ∈ L0. Hence, X is not
almost-normal. Note that: A = {(0, 1

n); n ∈ N} and L0 are disjoint π-closed subsets that
cannot be separated. If U =

⋃
n∈N

Ln is π-open subset of X such that A ⊆ U . For each

open set W of X, we have: A ⊆ W ⊆ W ̸⊆ U and A ⊆ W ⊆ int(W ) ̸⊆ U . Thus, X is
neither quasi-normal nor semi-normal. Therefore, the Thomas’ plank topology is an epi-
completely-regular space, which is neither almost-normal, semi-normal nor quasi-normal.



I. Alshammari / Eur. J. Pure Appl. Math, 15 (4) (2022), 1808-1821 1814

Note that: an epi-regularity does not imply to epi-complete-regularity as shown by the
next example:

Example 9. The Tychonoff corkscrew topology: [30, Example 90], Let X = S ∪{a+, a−},
where (S, T ) is homeomorphic to the deleted Tychonoff plank topology [30]. The basic
open subset U of a+ contains all points ofX which lies above a certain level k. That means:
U = {x ∈ X : L(x) > k+1}. The basic open subset V of a− contains all points of X which
lies below a certain level k. That means: V = {x ∈ X : L(x) < k+1}. The space (X, T ) is
a Hausdorff, regular and semi-regular space, which is neither Tychonoff, Urysohn, locally-
compact, Lindelöf, first-countable, normal nor completely-regular [30]. Since (X, T ) is
a Hausdorff regular space, it is epi-regular. Since every regular almost-normal space is
completely-regular [28], and X is regular non completely-regular, we obtain: (X, T ) is not
almost-normal.
Claim 1: Any Hausdorff topology T ′ onX, which is coarser than T , cannot be completely-
regular.
Proof of the Claim 1: Let T ′ be any Hausdorff topology on X which is coarser than T .
I show (X, T ′) is not a completely-regular space. Let A be any closed subset of (X, T ′)
and a+ ̸∈ A. Then, A is a closed subset of (X, T ) and a+ ̸∈ A. Thus, X \ A is an open
subset of (X, T ) containing a+. But a+ cannot be separated by a continuous function
from a closed subset A of X consisting the complement of the basis neighborhood of a+

[30]. Thus, (X, T ′) is not a completely-regular space. Therefore, any Hausdorff topology
T ′ on X, which is coarser than T cannot be completely-regular. Hence, (X, T ) is not
epi-completely-regular. Hence, X is not epi-almost-normal. Since every T1-semi-regular
almost-completely regular space is epi-completely-regular (Corollary 7), and X is T1-semi-
regular non epi-completely-regular, we obtain that: X is not almost-completely regular.
Therefore, the Tychonoff corkscrew topology is an epi-regular space, which is neither epi-
completely-regular, almost-completely regular nor epi-almost-normal.

Note that: the Mrówka space Ψ(A) [15, Example 2.10], is a Tychonoff, first-countable
and locally compact space, which is neither normal, countably-compact nor epi-normal.
Hence, it is an epi-completely-regular space, which is not epi-normal. The space presented
in [15, Example 3.1], is a sub-metrizable, epi-normal, Tychonoff and C-normal space,
which is not mildly-normal. The space presented in [6, Example 2.8], is an epi-completely-
regular space, which is neither C-normal nor epi-normal. Now, since every Hausdorff
locally compact space is Tychonoff [12], we get:

Corollary 2. Every Hausdorff locally-compact space is epi-completely-regular.

The converse of Corollary 2 cannot be true in general. Here is a counterexample:

Example 10. The Smirnov’s deleted sequence topology [30, Example 64], is a Urysohn
space, which is neither semi-regular, completely-regular, locally-compact nor almost-normal.
Since any closed domain subset of X is just the closed domain in the Euclidean topology
and U ⊆ T [30], we obtain: X is both almost-regular and almost-completely regular. The
Smirnov’s deleted sequence topology is not almost-normal because the closed domain sub-
set B = [−1, 0] is disjoint from the closed subset A = { 1

n : n ∈ N}, and they cannot be



I. Alshammari / Eur. J. Pure Appl. Math, 15 (4) (2022), 1808-1821 1815

separated. Since U ⊆ T , U is the Euclidian topology on R, which is coarser than T , and
(R,U) is a T4-space, we obtain: X is epi-normal (in fact it is sub-metrizable [5]). Since the
Smirnov’s deleted sequence topology is a Lindelöf non regular space, it is not L-regular
[6]. Therefore, the Smirnov’s deleted sequence is an epi-completely-regular space, which is
neither completely-regular, almost-normal, L-regular nor locally-compact. The Niemytzki
plane topology, the sorgenfrey line square and the Michael line are Tychonoff and hence
epi-completely-regular spaces [30], which are not locally-compact.

Example 11. The deleted Tychonoff plank [30, Example 87], is a Tychonoff locally-
compact space. Hence, it is an epi-completely-regular space. The deleted Tychonoff plank
is neither almost-normal nor sub-metrizable [6, 8]. Therefore, the deleted Tychonoff plank
topology is an epi-completely-regular space, which is not sub-metrizable.

Example 12. The odd-even topology [30, Example 6], is a completely regular and normal
space, which is not epi-completely regular being not Hausdorff.

Every Hausdorff semi-regular almost-compact (resp. H-closed) space is not necessary
to be epi-completely regular. Here is a counterexample:

Example 13. The minimal Hausdorff topology [30, Example 100], is a Hausdorff, semi-
regular, second-countable and almost-compact space, which is neither Urysohn, regular,
normal nor compact [30]. Since X is a semi-regular non regular space, we have: X is not
almost-regular. Since X is a T1 non almost-regular space, it is not almost-normal. Hence,
X is a quasi-normal space, which is not semi-normal [31]. Since X is not Urysohn, it is
neither epi-almost-normal, epi-regular, epi-completely-regular nor epi-normal. Therefore,
the minimal Hausdorff topology is a semi-regular, Hausdorff and epi-quasi-normal almost-
compact H-closed space [31], which is neither almost-regular, epi-regular, epi-completely-
regular nor Urysohn.

Observe that: a normal compact space need not be epi-completely regular. For exam-
ple: the excluded point topology [30, Example 15], and the either-or-topology [30, Example
17], are normal compact spaces, which are neither epi-completely-regular, epi-regular nor
epi-normal.

3. Some properties of epi-complete regularity

In this section, I present the following results:

Theorem 4. Epi-complete regularity is a topological property.

Proof. Let (X, T ) ∼= (Y,S) and (X, T ) be an epi-completely-regular space. There are
a homeomorphism f : X → Y and a topology T ′ on X that is coarser than T such that
(X, T ′) is Tychonoff. Define S ′ on Y by S ′ = {f(U) : U ∈ T ′}. Then, S ′ is a topology on
Y , which is coarser than S, and (Y,S ′) is Tychonoff. Thus, (Y,S) is epi-completely-regular.

Theorem 5. Epi-complete regularity is an additive property.



I. Alshammari / Eur. J. Pure Appl. Math, 15 (4) (2022), 1808-1821 1816

Proof. Let Xs be an epi-completely-regular space for each s ∈ S. Then, there exists
a topology Ts′ on Xs, which is coarser than Ts, such that (Xs, Ts′) is a T1-completely-
regular space. Since both T1 and complete-regularity are additive properties, we obtain:
(X, ⊕

s∈S
Ts′) is T1-completely regular (Tychonoff). Since ⊕

s∈S
Ts′ is a topology coarser than

⊕
s∈S

Ts, we get: (X, ⊕
s∈S

Ts) is epi-completely-regular.

Theorem 6. Epi-complete regularity is a hereditary property.

Proof. Let (X, T ) be an epi-completely-regular space, and (M, TM ) be a subspace
of X. Then, there exists a topology T ′ on X that is coarser than T such that (X, T ′)
is T1-completely-regular. To show (M, TM ) is epi-completely-regular, define TM ′ on M
by: TM ′ = {U ∩ M : U ∈ T ′}. Then, TM ′ ⊆ TM . Hence, TM ′ is a topology on M
which is coarser than TM . Since (X, T ′) is a T1-completely-regular space and (M, TM ′)
is a subspace of X, we obtain: (M, TM ′) is a T1-completely-regular subspace. Therefore,
(M, TM ) is epi-completely-regular.

Theorem 7. A product space X = Π
α∈Λ

Xα, Xα ̸= ∅ for each α ∈ Λ, is an epi-completely-

regular space if and only if each factor Xα is epi-completely-regular for each α ∈ Λ.

Proof. Let ( Π
α∈Λ

Xα, T ) be an epi-completely-regular space, Xα ̸= ∅ for each α ∈ Λ.

There exists a topology T ′ which is coarser than T such that ( Π
α∈Λ

Xα, T ′) is T1-completely-

regular. Thus, we have each factor (Xα, T ′
α) is a T1-completely-regular space [12], where Tα′

is a topology coarser than Tα for each α ∈ Λ. Thus, (Xα, Tα) is an epi-completely regular
space for each α ∈ Λ. Conversely, suppose that (Xα, Tα) is an epi-completely-regular
space for each α ∈ Λ. Then, for each α ∈ Λ, there exists a topology Tα′ that is coarser
than Tα such that (Xα, Tα′) is a T1-completely-regular space. Thus, the product space
( Π
α∈Λ

Xα, T ′) is T1-completely-regular, where T ′ is coarser than T . Therefore, ( Π
α∈Λ

Xα, T )

is epi-completely-regular.

Corollary 3. Epi-complete regularity is a multiplicative property.

Theorem 8. Every epi-completely regular nearly-compact (resp. nearly-paracompact)
space is epi-normal.

Proof. Let (X, T ) be an epi-completely-regular nearly-compact (resp. nearly-paracompact)
space. Then, there exists a topology T ′ on X which is coarser than T such that (X, T ′)
is a Tychonoff compact (resp. paracompact) space. Thus, (X, T ′) is a T1-normal space.
Hence, (X, T ′) is a T4-space. Therefore, (X, T ) is epi-normal.

Now, we recall the definition of the Alexandroff duplicate space. For any space X, let
X ′ = X × {1}. Clearly that X ∩X ′ = ∅. Let A(X) = X ∪X ′. For an element x ∈ X, the
element (x, 1) ∈ X ′ and for a subset B ⊆ X, let B×{1} = {(x, 1) : x ∈ B} ⊆ X ′. For each
(x, 1) ∈ X ′, let B((x, 1)) = {{(x, 1)}}. For each x ∈ X, let B(x) = {U∪(U×{1}\{(x, 1)}) :



I. Alshammari / Eur. J. Pure Appl. Math, 15 (4) (2022), 1808-1821 1817

U is open in X with x ∈ U}. Let T denote the unique topology on A(X) which has
{B(x) : x ∈ X} ∪ {B((x, 1)) : (x, 1) ∈ X ′} as its neighborhood system. The space A(X)
with this topology is called the Alexandroff duplicate of X [2].

Theorem 9. The Alexandroff duplicate A(X) of an epi-completely-regular space X is
epi-completely-regular.

Proof. Let (X, T ) be an epi-completely-regular space. Then, there exists a topology
T ′ on X that is coarser than T such that (X, T ′) is T1-completely-regular. Since T1

and complete-regularity are preserved by the Alexandroff duplicate space [2], we obtain:
A(X, T ′) is also a T1-completely-regular space, which is coarser than A(X, T ) by the
topology of the Alexandroff duplicate. Hence, A(X) is epi-completely-regular.

Since every subspace of a cube is completely-regular [12], we get:

Corollary 4. Every T1-subspace of a cube is epi-completely-regular.

Since every C2-paracompact Fréchet space is epi-normal, and any Mrôwka space Ψ(A)
is Tychonoff [18], we obtain:

Corollary 5.

(1) Every C2-paracompact first-countable space is epi-completely-regular.

(2) Any Mrôwka space Ψ(A) is epi-completely-regular.

Note that: a space (X, T ) is an almost-completely regular space if and only if the
semi-regularization (X, Ts) of (X, T ) is completely-regular [22]. Also, complete-regularity
is not a semi-regularization property, but almost-complete regularity is [22]. For example,
the half disc topology (X, T ) is not completely-regular [30], and its semi-regularization
(X, Ts) is the usual topology on the closed upper half plane, which is completely-regular.

Theorem 10. If (X, T ) is an almost-completely regular space such that the semi-regularization
(X, Ts) of (X, T ) is T1, then (X, T ) is epi-completely-regular.

Proof. Let (X, T ) be an almost-completely regular space and the semi-regularization
(X, Ts) of (X, T ) be T1. Since the semi-regularization of an almost-completely regular
space is completely-regular [22], we get: (X, Ts) is T1-completely-regular. Thus, (X, Ts)
is Tychonoff. Since Ts is a topology on X which is coarser than T , we obtain: (X, T ) is
epi-completely-regular.

Since every extremally-disconnected space is T1-π-normal [14], we get: every extremally-
disconnected space is T1-almost-completely regular. Since every extremally-disconnected
semi-regular space is Tychonoff [3], we conclude:

Corollary 6.

(a) Every Hausdorff extremally-disconnected space is epi-completely-regular.



I. Alshammari / Eur. J. Pure Appl. Math, 15 (4) (2022), 1808-1821 1818

(b) Every extremally-disconnected semi-regular space is epi-completely-regular.

In fact, an epi-completely-regular space is not necessary to be extremally-disconnected.
For example: the rational sequence topology [30, Example 65], is a semi-regular epi-
completely-regular space being Tychonoff, which is not extremally disconnected. The
next result is obvious:

Theorem 11. If the semi-regularization space (X, Ts) of a space (X, T ) is an epi-completely-
regular space, then (X, T ) is epi-completely-regular.

Theorem 12. Every Hausdorff almost-completely regular space is epi-completely-regular.

Proof. Let (X, T ) be a Hausdorff almost-completely regular space. Let (X, Ts) be the
semi-regularization of (X, T ). Then, (X, Ts) is a Hausdorff completely regular space be-
cause the semi-regularization of a Hausdorff almost-completely regular space is Hausdorff
completely regular [22]. Thus, (X, Ts) is Tychonoff. Since Ts ⊆ T , we conclude: (X, T ) is
epi-completely regular.

The next results are obvious:

Corollary 7.

(1) Every T1-semi-regular (resp. semi-normal) almost-completely regular space is epi-
completely-regular.

(2) Any nearly-paracompact Hausdorff space is epi-normal.

(3) Every almost-regular Hausdorff Lindelöf space is epi-quasi-normal.

Since every epi-completely-regular space is epi-regular, and every epi-regular space is
C-regular, we get: every epi-completely-regular space is C-regular, but the converse is
not true in general. For example: the odd-even topology, Example 12, is a normal and
completely-regular space [30], which is not T1. Thus, the odd-even topology is a C-regular
space, which is not epi-completely regular.

Theorem 13. Every semi-regular almost-normal T1-space is Tychonoff.

Proof. Let X be a semi-regular T1-almost-normal space. Then, X is almost-regular.
Since every semi-regular almost-regular is regular, we obtain: X is a T1-regular almost-
normal space. Hence, X is a T1-completely-regular space because every regular almost-
normal space is completely-regular [10]. Therefore, X is Tychonoff.

From Theorem 13, we conclude the next corollary:

Corollary 8. Every semi-regular almost-normal T1-space is epi-completely regular.

The next theorem has been presented in [25, Theorem 9 - 1.17, page 306]:

Theorem 14. [25], A space (X, T ) is a T1-space if and only if T contains the finite
complement topology on X. i.e. CF ⊆ T and (X, CF) is the finite complement topology
on X.



REFERENCES 1819

From Theorem 14, we conclude:

Corollary 9. If (X, T ) is a T1-space, then there exists a topology T ′ coarser than T
such that (X, T ′) is T1 almost completely regular (resp. almost regular). We can say that
(X, T ) is epi-almost completely regular (resp. epi-almost regular).

Recall that: any closed extension space (Xp, T ∗) of a given space (X, T ) is always
connected, π-normal, almost normal, separable and it cannot be T1 [1]. Thus, we conclude:

Corollary 10. Every closed extension space (Xp, T ∗) of a given space (X, T ) cannot be
epi-completely regular.

The next problem is still open in this work:

Problem:

• Is there an example of an epi-completely-regular space, which is not epi-mildly-
normal?.

4. Conclusion

A new version of complete regularity called epi-complete regularity has been studied in
this work. I have shown that epi-complete regularity is different from both epi-regularity
and epi-normality. I have proved that epi-complete regularity is a topological, productive,
hereditary and additive property. Some properties, counterexamples and relationships
with some other forms of topological properties have been presented and proved.

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