EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 16, No. 2, 2023, 819-832
ISSN 1307-5543 – ejpam.com
Published by New York Business Global

F−open and F−closed sets in Topological Spaces

Mesfer H. Alqahtani

Department of Mathematics, University College of Umluj, University of Tabuk, Tabuk,
Saudi Arabia

Abstract. An open (resp., closed) subset A of a topological space (X, T ) is called F -open (resp.,
F -closed) set if cl(A)\A (resp., A\ int(A)) is finite set. In this work, we study the main properties
of these definitions and examine the relationships between F -open and F -closed sets with other
kinds such as regularly open, regularly closed, closed, and open sets. Then, we establish some
operators such as F -interior, F -closure, and F -derived...etc., using F -open and F -closed sets. At
the end of this work, we introduce definitions of F -continuous function, F -compact space, and
other related properties.

2020 Mathematics Subject Classifications: 05C99, 54F15
Key Words and Phrases: Topological space, F -open, open domain, F -interior, F -closure,
F -derived, F -continuous, F -compact

1. Introduction

In the topological space X, a subset B of a space X is said to be a regularly-closed,
called also closed domain if B = cl(int(B)). A subset B of X is said to be a regularly-open,
called also open domain if B = int(cl(B)) [2]. In this work, we are interested in studying
the concepts of F -open and F -closed sets in topological spaces in detail. We organize this
paper as follows. In Section 2, we recall the basic concepts and findings that make this
work self-contained. In Section 3, we introduce definitions of F -open and F -closed sets
and examine the relationships between them and other types such as open domain, closed
domain, closed, and open sets. We also present some important theorems. In Section
4, we establish some operators such as F -interior, F -closure, F -border, F -frontier, F -
exterior, and F -derived using F -open and F -closed sets. Then we introduce some related
theorems. In Section 5, we introduce definitions of F -continuous, F -open, F -closed, and
F -homeomorphism functions. We also provide definitions of F -compact, F -Lindelöf, and
F -countably compact spaces. In addition, we present some related properties. Throughout
this paper, the subset B of a topological space (X, T ), we will denote the complement of B
in (X, T ) by X \B, the set of positive integers by N, the set of integers numbers by Z, the
set of real numbers by R, and usual topology in R by U . Unless or otherwise mentioned,
X stands for the topological space (X, T ).

DOI: https://doi.org/10.29020/nybg.ejpam.v16i2.4583

Email address: m
¯
halqahtani@ut.edu.sa (M. H. Alqahtani)

https://www.ejpam.com 819 © 2023 EJPAM All rights reserved.



M. H. Alqahtani / Eur. J. Pure Appl. Math, 16 (2) (2023), 819-832 820

2. Preliminaries

In this Section, we recall the basic concepts and findings that make this work self-
contained.

Definition 1. [1] Let K be a subset of the topological space (X, T ), then the interior of
K is defined as the union of all open subsets of K (or the largest open set contained in
K) and is denoted by int(K).

Definition 2. [1] Let K be a subset of the topological space (X, T ), then the clouser of K
is defined as the intersection of all closed sets containing K, and is denoted by cl(K).

Definition 3. [1] Let K be a subset of the topological space (X, T ). A point x ∈ X is said
to be limit points (or an accumulation point, or a cluster point) of K if and only if every
open set V containing x, contains at least one point of K different from x. The set of all
limit points of K is called the derived set of K and denoted by D(K).

Definition 4. [1] Let K be a subset of the topological space (X, T ), then the border of K
is defined as Bd(K) = K \ int(K).

Definition 5. [1] Let K be a subset of the topological space (X, T ), then the frontier of
K is defined as Fr(K) = cl(K) \ int(K).

Definition 6. [1] Let K be a subset of the topological space (X, T ), then the exterior of
K is defined as Ext(K) = int(X \K).

Definition 7. [1] A function h from the topological space (X, T ) into the topological space
(Y,P) is said to be continuous if h−1(U) is an open subset in X for every open subsets U
in Y.

Definition 8. [1] A function h from the topological space (X, T ) into the topological space
(Y,P) is said to be open (resp. closed) if h(U) is an open (resp., closed) subset in Y for
every open (resp., closed) subsets U in X.

Definition 9. [1] A bijection function h from the topological space (X, T ) into the topo-
logical space (Y,P) is said to be homeomorphism if and only if h and h−1 are continuous.

Definition 10. [1] Let (X, T ) be a topological space, then (X, T ) is compact (resp.,
Lindelöf) if and only if any open cover of X has a finite (resp., countable) subcover of
open sets.

Definition 11. [1] Let (X, T ) be a topological space, then (X, T ) is countably compact
space if and only if any countable open cover of X has a finite subcover of open sets.



M. H. Alqahtani / Eur. J. Pure Appl. Math, 16 (2) (2023), 819-832 821

3. F -open and F -closed sets

Definition 12. An open subset A of a topological space (X, T ) is called F -open set if
cl(A) \A is finite set. That is, A is an open set and the frontier of A is a finite set.

Definition 13. A closed subset A of a topological space (X, T ) is called F -closed set if
A \ int(A) is finite set. That is, A is a closed set and the frontier of A is a finite set.

Theorem 1. Let (X, T ) be a topological space, then

(i) The complement of any F -open subset of X is a F -closed;

(ii) The complement of any F -closed subset of X is a F -open.

Proof. Let A be any F -open subset in X, then X\A is closed and (X\A)\int(X\A) =
(X \A)\ (X \ cl(A)) = (X \A)∩ cl(A) = cl(A)\A is finite. Therefore, X \A is a F -closed.
Using the same way to prove the complement of any F -closed subset in X is a F -open.

Remark 1.

i) Any clopen (open-and-closed) subset of the topological space (X, T ) is F -open and
F -closed sets;

ii) Any finite closed subset of the topological space (X, T ) is F -closed.

The collection of all F -open (resp., F -closed) subsets of the topological space (X, T )
is denoted by FO(X) (resp., FC(X)).

Example 1. Let (R,U) be a topological space, where U is denoted for the usual topology.
Let A = (2, 100) is an open interval in (R,U), then A is a F -open set, because, (2, 100) ∈ U
and cl(2, 100)\(2, 100) = [2, 100]\(2, 100) = {2, 100} is finite. Any open intervals of (R,U)
is a F -open set. On the other hand, let B = [1, 6] is a closed interval in (R,U), then B
is a F -closed set, because, [1, 6] is closed set and [1, 6] \ int[1, 6] = [1, 6] \ (1, 6) = {1, 6} is
finite. Any closed interval or finite set of (R,U) is a F -closed set.

Definition 14. If (X, T ) is a topological space, x is a point of X, a F -open neighbourhood
of x is a F -open subset U of X, which is containing x.

Lemma 1. Let (X, T ) be a topological space and ∅ ≠ V ⊂ X is F -open set, then for every
x ∈ V there exists a F -open neighborhood Ux of the point x contained in V.

Proof. Assume that V is F -open and pick x ∈ V arbitrary. Let Ux = V . Then V is a
F -open neighborhood of x and x ∈ V ⊆ V.

The converse of the previous Lemma is not true in general. We have the following
example:



M. H. Alqahtani / Eur. J. Pure Appl. Math, 16 (2) (2023), 819-832 822

Example 2. Let An = (n, n + 1) be a subset of (R,U), for all n ∈ Z. Then, An =
(n, n + 1) ∈ U and cl(n, n + 1) \ (n, n + 1) = {n, n + 1} is finite set for all n ∈ Z. Then
An = (n, n + 1) is F -open set for all n ∈ Z. Let B =

⋃
n∈Z(n, n + 1) and let x ∈ B =⋃

n∈Z(n, n+ 1) be arbitrary, then there exists some n1 ∈ Z such that x ∈ (n1, n1 + 1) is a
F -open neighborhood containing x and contained in B, that is mean for any x ∈ B there
exist a F -open neighborhood containing x and contained in B. However, B is not F -open
set, because B =

⋃
n∈Z(n, n+ 1) = R \ Z is open set and cl(B) \ (B) = R \ (R \ Z) = Z is

not finite set. Therefore, B =
⋃

n∈Z(n, n+ 1) is not F -open set.

It is clear by the definitions every F -open and F -closed sets are open and closed,
respectively. However, the converse is not true in general. Here is an example of open
(resp., closed) set which is not F -open (resp., F -closed).

Example 3. There exist A = R\Z is an open subset of (R,U). However, cl(R\Z)\(R\Z) =
R\ (R\Z)) = Z is not finite. Hence A is not F -open set. Also, Z is a closed set in (R,U),
but Z \ int(Z) = Z \ ∅ = Z is not finite. therefore, Z is not F -closed set.

There is an example of F -open (resp., F -closed) set which is not an open domain
(resp., closed domain).

Example 4. Let A = (2, 5) ∪ (5, 9) is a F -open subset in (R,U), because, (2, 5) ∪ (5, 9)
is open and cl((2, 5) ∪ (5, 9)) \ ((2, 5) ∪ ((5, 9)) = [2, 9] \ ((2, 5) ∪ (5, 9)) = {2, 5, 9} is
finite. But A is not open domain, because (2, 5) ∪ (5, 9) ̸= int(cl((2, 5) ∪ (5, 9))) = (2, 9).
Moreover, R \ A is F -closed set, because R \ A is closed and (R \ A) \ int(R \ A) =
((−∞, 2] ∪ [9,∞) ∪ {5}) \ ((−∞, 2) ∪ (9,∞)) = {2, 5, 9} is finite. But R \ A is not closed
domain because the complement of closed domain is open domain or cl(int(R \ A)) =
cl(int((−∞, 2] ∪ [9,∞) ∪ {5})) = cl((−∞, 2) ∪ (9,∞)) = ((−∞, 2] ∪ [9,∞)) ̸= (R \A).

Now, there is an example of an open domain (resp., closed domain) set which is not
F -open (resp., F -closed) set.

Example 5. Let (R, TZ) be the excluded set topological space on R by Z. Let K = (0, 1),
then K ∈ TZ and cl(K) \K = ((0, 1) ∪ Z) \ (0, 1) = Z is not finite, then K is not F -open
set. But K = int(cl(K)) = int(cl(0, 1)) = int((0, 1) ∪ Z) = (0, 1) = K, then K is open
domain. By Theorem 1, the complement of F -closed set is F -open set and the complement
of open domain is closed domain, then X \K = X \ (0, 1) is an example of closed domain
set which is not F -closed.

Theorem 2. Let (X, T ) be a topological space, then

i) Finite union of F -closed subsets in X is a F -closed;

ii) Finite union of F -open subsets in X is a F -open;

iii) Finite intersection of F -open subsets in X is a F -open;

iv) Finite intersection of F -closed subsets in X is a F -closed.



M. H. Alqahtani / Eur. J. Pure Appl. Math, 16 (2) (2023), 819-832 823

Proof.

i) Suppose that Ki be F -closed set for all i ∈ {1, 2, 3, . . . , n}, then Ki is a closed set
and Ki \ int(Ki) is finite for all i. Since

⋃n
i=1Ki is closed, then we need to show the

other condition of F -closed set.
Claim:

n⋃
i=1

Ki \ int(
n⋃

i=1

Ki) ⊆
n⋃

i=1

(Ki \ int(Ki)),

Let x ∈
⋃n

i=1Ki \ int(
⋃n

i=1Ki) be arbitrary. Since
⋃n

i=1 int(Ki) ⊆ int(
⋃n

i=1Ki),
then there exists i1 ∈ {1, 2, 3, . . . , n} such that x ∈ Ki1 and x /∈ int(Ki) for all
i ∈ {1, 2, 3, . . . , n}. Then x ∈ (Ki1) \ int(Ki1), then x ∈

⋃n
i=1(Ki \ int(Ki)). Claim

is proved.
Since the finite union of finite sets is finite. Then,

⋃n
i=1Ki \ int(

⋃n
i=1Ki) is finite.

Therefore,
⋃n

i=1Ki is F -closed.

ii) Suppose that Ki be F -open set for all i ∈ {1, 2, 3, . . . , n}, then Ki is an open set and
cl(Ki) \Ki is finite for all i. Since

⋃n
i=1Ki is open, then we need to show the other

condition of F -open set.
Claim:

cl(
n⋃

i=1

Ki) \
n⋃

i=1

Ki ⊆
n⋃

i=1

(cl(Ki) \Ki),

Let x ∈ cl(
⋃n

i=1Ki) \
⋃n

i=1Ki be arbitrary. Since cl(
⋃n

i=1Ki) =
⋃n

i=1 cl(Ki), then
x ∈

⋃n
i=1 cl(Ki) \

⋃n
i=1Ki, that is mean there exist i1 ∈ {1, 2, 3, . . . , n} such that

x ∈ cl(Ki1) and x /∈ Ki for all i ∈ {1, 2, 3, . . . , n}. Then x ∈ (cl(Ki1) \Ki1), so thus
x ∈

⋃n
i=1(cl(Ki) \Ki). Claim is proved.

Since the finite union of finite sets is finite. Then, cl(
⋃n

i=1Ki) \
⋃n

i=1Ki is finite.
Therefore,

⋃n
i=1Ki is F -open.

iii) Suppose that Ki be F -open set for all i ∈ {1, 2, 3, . . . , n}, then
⋂n

i=1Ki is open and
by Morgan’s Laws, Theorem 1 and Theorem 2 part (i) we have

⋂n
i=1Ki is F -open,

because X \
⋂n

i=1Ki =
⋃n

i=1(X \Ki) is F -closed set.

iv) Suppose that Ki be F -closed set for all i ∈ {1, 2, 3, . . . , n}, then
⋂n

i=1Ki is closed
and by Morgan’s Laws, Theorem 1 and Theorem 2 part (ii) we have

⋂n
i=1Ki is

F -closed, because X \
⋂n

i=1Ki =
⋃n

i=1(X \Ki) is F -open set.

In general, countable union of F -open sets may not be F -open set.

Example 6. Let An = (n, n+1) be a subset of (R,U), for all n ∈ N, then, An = (n, n+1) ∈
U and cl(n, n+1)\ (n, n+1) = {n, n+1} is finite for all n ∈ N. Hence, An = (n, n+1) is
F−open set for all n ∈ N. We have

⋃
n∈N(n, n+1) = [1,∞) \N is an open set. However,

cl(
⋃

n∈N(n, n+1))\
⋃

n∈N(n, n+1) = cl([1,∞)\N)\([1,∞)\N) = [1,∞)\([1,∞)\N) = N
is not finite set. Therefore,

⋃
n∈N(n, n+ 1) is not F -open set.



M. H. Alqahtani / Eur. J. Pure Appl. Math, 16 (2) (2023), 819-832 824

In general countable intersection of F -closed sets may not be F -closed set.

Example 7. By Theorem 1 and Example 6, we have R \ (
⋃

n∈N(n, n + 1)) =
⋂

n∈N(R \
(n, n + 1)) is countable intersection of F -closed sets which is not F -closed set or V =
R \ (n, n+ 1) = (−∞, n] ∪ [n+ 1,∞) is closed for all n ∈ N, and ((−∞, n] ∪ [n+ 1,∞)) \
int((−∞, n]∪[n+1,∞)) = {n, n+1} is finite set for all n ∈ N. Then V is a F -closed set for
all n ∈ N. Since

⋂
n∈N((−∞, n]∪ [n+1,∞)) = (−∞, 1)∪N is closed, then we need to show

the other condition of F -closed set. Let
⋂

n∈N((−∞, n]∪ [n+1,∞)) \ int(
⋂

n∈N((−∞, n]∪
[n + 1,∞))) = ((−∞, 1) ∪ N) \ int((−∞, 1) ∪ N) = ((−∞, 1) ∪ N) \ (−∞, 1) = N is not
finite set. Therefore,

⋂
n∈N(R \ (n, n+ 1)) is not F -closed set.

In general countable union of F -closed sets may not be F -closed set.

Example 8. Let Kn = { 1
n} is a F−closed set in (R,U) for all n ∈ N, because { 1

n} is
closed for all n ∈ N, and { 1

n} \ int({
1
n}) = { 1

n} \ ∅ = { 1
n} is finite for all n ∈ N. However,⋃

n∈N{
1
n} is not a closed set. Hence is not a F -closed set.

In general countable intersection of F -open sets may not be F -open set.

Example 9. We have (−1
n , 1

n) is a F -open set in (R,U) for all n ∈ N. But
⋂

n∈N(
−1
n , 1

n) =
{0} is not open set, then is not F -open set.

4. Some topological properties

Definition 15. Let K be a subset of the topological space (X, T ). Then,

i) The F -interior of K is defined as the union of all F -open subsets of K (or the largest
F -open set contained in K) and is denoted by intF (K).

ii) The F -clouser of K is defined as the intersection of all F -closed sets containing K,
and is denoted by clF (K).

Definition 16. Let x ∈ intF (K) if and only if there exist a F -open set U such that
x ∈ U ⊆ K.

Theorem 3. Let K be a subset of the topological space (X, T ). Then,

i) intF (K) ⊆ int(K) ⊆ K.

ii) K ⊆ cl(K) ⊆ clF (K).

Proof.

i) Let x ∈ intF (K) be arbitrary, then there exist a F -open set U such that x ∈ U ⊆ K.
Since any F -open set is open, then x ∈ int(K). Hence, intF (K) ⊆ int(K). Also by
the interior definition ofK, we have int(K) ⊆ K. Therefore, intF (K) ⊆ int(K) ⊆ K.



M. H. Alqahtani / Eur. J. Pure Appl. Math, 16 (2) (2023), 819-832 825

ii) By the closure definition, we have K ⊆ cl(K). Let cl(K) = F be the smallest closed
set such that K ⊆ F. Since any F -closed set is closed, then F ⊆ clF (K). Therefore,
K ⊆ cl(K) ⊆ clF (K).

In general int(K) ⊈ intF (K) and clF (K) ⊈ cl(K). For example:

Example 10. Let K = R \ Z be a subset of the usual topological space (R,U). Then
int(K) = R \ Z. However, R \ Z is not F -open, then the largest F -open subset of R \ Z
is not equal R \ Z. Therefore, int(K) ⊈ intF (K). Let H = Z be a subset of the usual
topological space (R,U), then H = Z is closed in (R,U), so thus cl(Z) = Z. However, Z is
not F -closed, then Z ⊂ clF (Z). Therefore, clF (H) ⊈ cl(H).

Corollary 1. If U is F -open set and U ∩ V = ∅, then U ∩ clF (V ) = ∅. In particular, If
U and V are disjoint F -open sets, then U ∩ clF (V ) = ∅ = clF (U) ∩ V .

Proof. Since U ∩ V = ∅, then V ⊆ X \ U , so thus clF (V ) ⊆ clF (X \ U). However, U
is F -open, then X \ U is F -closed. Hence, X \ U = clF (X \ U), so thus clF (V ) ⊆ X \ U.
Therefore, U ∩ clF (V ) = ∅.

Definition 17. Let K be a subset of the topological space (X, T ). A point x ∈ X is said
to be F -limit points (or an F -accumulation point, or a F -cluster point) of K if and only
if for any F -open set V containing x, we have (V \ {x}) ∩K ̸= ∅. The set of all F -limit
points of K is called the F -derived set of K and denoted by DF (K).

Theorem 4. Let K and H be subsets of a topological space (X, T ). Then we have the
following properties:

(i) D(K) ⊂ DF (K), where D(K) is the derived set of K.

(ii) If K ⊆ H, then DF (K) ⊆ DF (H).

(iii) DF (K) ∪DF (H) = DF (K ∪H) and DF (K ∩H) ⊂ DF (K) ∩DF (H).

Proof. For (i) it suffices to observe that every F -open is open. For (ii) follow from
Definition 17. For (iii) is a modification of the standard proof for D.

In general DF (K) ⊈ D(K). For example:

Example 11. Let (R, T 1
2
) be a topological space, where T 1

2
is the particular point topology

at 1
2 . Let K = Z, then D(Z) = ∅, because {1

2 , x} for any x ∈ R is an open set containing
x and ({1

2 , x} \ {x}) ∩ Z = ∅. Let V be any F -open subset of R, then we have 1
2 ∈ V, and

cl(V ) \ V is finite. Hence, V = R \H where H is finite. Suppose not, H is infinite, then
cl(V ) \V = R \R \H = H is infinite. Hence, V is not F -open set. let x ∈ R be arbitrary,
then for any F -open set V containing x we have V ∩ Z ̸= ∅. Hence, DF (Z) = R, so thus
DF (Z) ⊈ D(Z). Therefore, DF (K) ⊈ D(K).

In general DF (K ∩H) ⊉ DF (K) ∩DF (H). For example:



M. H. Alqahtani / Eur. J. Pure Appl. Math, 16 (2) (2023), 819-832 826

Example 12. There exists K = (1, 2) and H = (2, 3) are F -open subsets of the usual
topological space (R,U) such that DF (K ∩ H) = DF ((1, 2) ∩ (2, 3)) = DF (∅) = ∅ and
DF (K)∩DF (H) = DF ((1, 2))∩DF ((2, 3)) = [1, 2]∩ [2, 3] = {2}. Therefore, DF (K∩H) ⊉
DF (K) ∩DF (H).

Theorem 5. Let K and H be subsets of a topological space (X, T ). Then we have the
following properties:

(i) intF (X) = X.

(ii) intF (K) ⊆ K.

(iii) If K ⊆ H, then intF (K) ⊆ intF (H).

(iv) intF (intF (K)) = intF (K).

(v) intF (K ∩H) = intF (K) ∩ intF (H).

(vi) intF (K) ∪ intF (H) ⊆ intF (K ∪H).

Proof. The properties (i), (ii), (iii) and (iv) follow from Definitions 12 and Definition
15. To prove (v), by property (ii) we have intF (K) ⊆ K and intF (H) ⊆ H, then intF (K)∩
intF (H) ⊆ K ∩H. As intF (K) ∩ intF (H) is F -open, then we have intF (K) ∩ intF (H) ⊆
intF (K∩H), because intF (K)∩intF (H) is F -open and intF (K∩H) is the largest F -open
set contained in K ∩H. Conversely, (K ∩H) ⊆ K and (K ∩H) ⊆ H, by property (iii)
we have intF (K ∩ H) ⊆ intF (K) and intF (K ∩ H) ⊆ intF (H). Hence, intF (K ∩ H) ⊆
intF (K) ∩ intF (H). Therefore, intF (K ∩H) = intF (K) ∩ intF (H). To prove (vi), since
K ⊆ (K ∪ H) and H ⊆ (K ∪ H), from property (iii) we have intF (K) ⊆ intF (K ∪ H)
and intF (H) ⊆ intF (K ∪H). Therefore, intF (K) ∪ intF (H) ⊆ intF (K ∪H).

In general, intF (K ∪H) ⊈ intF (K) ∪ intF (H). For example:

Example 13. Let K = R\N and H = N be a subsets of the usual topological space (R,U),
where N = {1, 2, 3, . . . }. Then intF (K ∪H) = R. However, R \N is not F -open set, then
intF (R \ N) ⊂ R \ N and intF (H) = ∅. Therefore, intF (K ∪H) ⊈ intF (K) ∪ intF (H).

Theorem 6. Let K and H be subsets of a topological space (X, T ). Then we have the
following properties:

(i) clF (∅) = ∅.

(ii) K ⊆ clF (K).

(iii) If K ⊆ H, then clF (K) ⊆ clF (H).

(iv) clF (K ∪H) = clF (K) ∪ clF (H).

(v) clF (clF (K)) = clF (K).



M. H. Alqahtani / Eur. J. Pure Appl. Math, 16 (2) (2023), 819-832 827

Proof. The properties (i), (ii), (iii) and (v) follow from Definition 13 and Definition
15. The property (iv), follow from property (iii), Definition 12, Definition 15 and using
set theoretic properties.

Theorem 7. Let K be a subset of the topological space (X, T ). Then,

(i) intF (K) = X \ clF (X \K).

(ii) clF (K) = X \ intF (X \K).

Proof.

(i) We have intF (K) = X \ clF (X \K), then X \K ⊆ clF (X \K), thus X \ clF (X \
K) ⊆ K. Since X \ clF (X \ K) is F -open, then X \ clF (X \ K) ⊆ intF (K)...(1).
Now, let V be any F -open set contained in K, i.e., V ⊆ K and V is F -open, then
X \K ⊆ X \V = clF (X \V ). Then clF (X \K) ⊆ X \V , hence V ⊆ X \ clF (X \K).
That is, any F -open set contained in K is contained in X \ clF (X \K), which means
that intF (K) ⊆ X \ clF (X \K) ...(2). From (1) and (2) equality holds.

(ii) Can be proved by replacing K and X \K by X \K and K, respectively in (i) and
using set theoretic properties.

Definition 18. Let K be a subset of the topological space X. Then F -border of K is
defined as BdF (K) = K \ intF (K).

Theorem 8. Let K be a subset of the topological space (X, T ). Then we have the following
properties:

(i) Bd(K) ⊂ BdF (K), where Bd(K) denotes the border of K.

(ii) K = intF (K) ∪BdF (K).

(iii) intF (K) ∩BdF (K) = ∅.

(iv) K is a F−open set if and only if BdF (K) = ∅.

(v) intF (BdF (K)) = ∅.

(vi) BdF (BdF (K)) = BdF (K).

(vii) BdF (K) = K ∩ clF (X \K).

Proof. To prove (i), let x ∈ Bd(K) be arbitrary, then x ∈ K \ int(K), so thus
x /∈ int(K). By Theorem 3 part (i) we have intF (K) ⊆ int(K) ⊆ K, then x /∈ intF (K).
Hence x ∈ (K \ intF (K)) = BdF (K). Therefore, Bd(K) ⊆ BdF (K). The properties
(ii), (iii) and (iv) follow from Definition 18. To prove (v), let x ∈ intF (BdF (K)) be
arbitrary, then x ∈ BdF (K). Since BdF (K) ⊆ K, then x ∈ intF (BdF (K)) ⊆ intF (K),



M. H. Alqahtani / Eur. J. Pure Appl. Math, 16 (2) (2023), 819-832 828

so thus x ∈ intF (K) ∩ BdF (K) which contradicts (iii). For (vi) from Definition 18 and
property (v) we have BdF (BdF (K)) = BdF (K)\intF (BdF (K)) = BdF (K)\∅ = BdF (K).
The property (vii) follow from Theorem 7 part (i) and Definition 18.

In general BdF (K) ⊈ Bd(K). For example:

Example 14. Let K = R \ N be a subset of the usual topological space (R,U), where
N = {1, 2, 3, . . . }. Then Bd(K) = ∅. Since R\N is not F -open set, then intF (R\N) ⊂ R\N.
Hence, ((R \ N) \ intF (R \ N)) ̸= ∅. Therefore, BdF (K) ⊈ Bd(K).

Definition 19. Let K be a subset of the topological space X. Then F -frontier of K is
defined as FrF (K) = clF (K) \ intF (K).

Theorem 9. Let K be a subset of the topological space (X, T ). Then we have the following
properties:

(i) Fr(K) ⊂ FrF (K), where Fr(K) denotes the frontier of K.

(ii) clF (K) = intF (K) ∪ FrF (K).

(iii) intF (K) ∩ FrF (K) = ∅.

(iv) BdF (K) ⊂ FrF (K).

(v) FrF (K) = clF (K) ∩ clF (X \K).

(vi) FrF (K) = FrF (X \K).

(viii) intF (K) = K \ FrF (K).

Proof. The property (i), by Theorem 3 and Definition 19, we have intF (K) ⊆ int(K)
and cl(K) ⊆ clF (K), then Fr(K) = (cl(K) \ int(K)) ⊆ clF (K) \ intF (K) = FrF (K).
The properties (ii) and (iii) follow from Definition 19. For (iv), since K ⊆ clF (K), then
BdF (K) = (K \ intF (K)) ⊆ (clF (K) \ intF (K)) = FrF (K). The property (v) follow from
Theorem 7 and Definition 19. The property (vi) follow from property (v) and Definition
19. The property (vii) follow from Definition 19.

In general FrF (K) ⊈ Fr(K) and FrF (K) ⊈ BdF (K). For example:

Example 15. Let K = N be a subset of the usual topological space (R,U), where N =
{1, 2, 3, . . . }. Then Fr(K) = N. Without loss of generality, we assume that, clF (K) =
{1, 2, 3, . . . , n}∪[n,∞), for some n ∈ N, then FrF (K) = clF (K)\intF (K) = {1, 2, 3, . . . , n}∪
[n,∞) \ ∅ = {1, 2, 3, . . . , n} ∪ [n,∞). Hence, FrF (K) ⊈ Fr(K). For FrF (K) ⊈ BdF (K),
let K = (1, 2] be a subsets in (R,U), then BdF (K) = {2} and FrF (K) = {1, 2}. Hence,
FrF (K) ⊈ BdF (K).

Definition 20. Let K be a subset of the topological space X. Then F -exterior of K is
defined as ExtF (K) = intF (X \K).



M. H. Alqahtani / Eur. J. Pure Appl. Math, 16 (2) (2023), 819-832 829

Theorem 10. Let K be a subset of the topological space (X, T ). Then we have the following
properties:

(i) ExtF (K) ⊂ Ext(K), where Ext(K) denotes the exterior of K.

(ii) ExtF (K) is open.

(iii) ExtF (K) = X \ clF (K).

(iv) ExtF (ExtF (K)) = intF (clF (K)).

(v) If K ⊆ H, then ExtF (H) ⊆ ExtF (K).

(vi) ExtF (X \ ExtF (K)) = ExtF (K).

(vii) intF (K) ⊂ ExtF (ExtF (K)).

(viii) ExtF (K ∪H) ⊂ ExtF (K) ∪ ExtF (H).

(ix) ExtF (K ∩H) ⊃ ExtF (K) ∩ ExtF (H).

(x) X = intF (K) ∪ ExtF (K) ∪ FrF (K).

Proof. The property (i), follow from Definition 20 and Theorem 3. The property (ii),
follow from Definition 15. The property (iii), follow from Theorem 7 and Definition 20.
To prove (iv) ExtF (ExtF (K)) = ExtF (intF (X \ K)) = ExtF (X \ clF (K)) = intF (X \
(X \ clF (K))) = intF (clF (K)). To prove (v) since K ⊆ H, then X \ H ⊆ X \ K, then
intF (X\H) ⊆ intF (X\K), hence, ExtF (H) ⊆ ExtF (K). For (vi), ExtF (X\ExtF (K)) =
ExtF (X\intF (X\K)) = intF (X\(X\intF (X\K))) = intF (intF (X\K)) = intF (X\K) =
ExtF (K). For (vii), intF (K) ⊆ intF (clF (K)) ⊆ intF (X \ intF (X \ K)) = intF (X \
ExtF (K)) = ExtF (ExtF (K)). The property (viii) ExtF (K ∪H) = intF (X \ (K ∪H)) =
(X\clF (K∪H)) = (X\(clF (K)∪clF (H))) ⊂ X\clF (K)∪X\clF (H) = (X\clF (K))∩(X\
clF (H))) = ExtF (K) ∩ ExtF (H) ⊂ ExtF (K) ∪ ExtF (H). The property (ix): ExtF (K ∩
H) = intF (X \ (K ∩H)) = (X \ clF (K ∩H)) ⊃ (X \ (clF (K)∩ clF (H))) = (X \ clF (K))∪
(X \ clF (H))) = ExtF (K) ∪ ExtF (H) ⊃ ExtF (K) ∩ ExtF (H). The property (x) follow
from the Definitions 19 and 20.

In general Ext(K) ⊈ ExtF (K), ExtF (K ∪H) ⊉ ExtF (K)∪ExtF (H) and ExtF (K ∩
H) ⊈ ExtF (K) ∩ ExtF (H). For example:

Example 16. Let K = Z be a subset of the usual topological space (R,U). Then Ext(K) =
int(R\Z) = R\Z. But R\Z is not F -open, then ExtF ((K) = intF ((R\Z) ⊂ R\Z. Hence,
Ext(K) ⊈ ExtF (K). For ExtF (K∪H) ⊉ ExtF (K)∪ExtF (H) , let K = (−∞, 2) and H =
(0,∞), then ExtF (K ∪H) = ExtF (R) = ∅ and ExtF (K)∪ExtF (H) = (2,∞)∪ (−∞, 0).
Hence, ExtF (K ∪H) ⊉ ExtF (K)∪ExtF (H). For ExtF (K ∩H) ⊈ ExtF (K)∩ExtF (H),
let K = (−∞, 2] and H = [2,∞), then ExtF (K ∩ H) = ExtF ({2}) = R \ {2} and
ExtF (K)∩ExtF (H) = (2,∞)∩(−∞, 2) = ∅. Hence, ExtF (K∩H) ⊈ ExtF (K)∩ExtF (H).



M. H. Alqahtani / Eur. J. Pure Appl. Math, 16 (2) (2023), 819-832 830

5. F -continuity and F -compactness

Definition 21. A function h : (X, T ) → (Y,P) is said to be F -continuous if h−1(U) is a
F -open set in X for every open sets U in Y.

Definition 22. A function h : (X, T ) → (Y,P) is said to be F−open if h(U) is F−open
in Y for every open sets U in X.

Definition 23. A function h : (X, T ) → (Y,P) is said to be F -closed if h(U) is F -closed
in Y for every closed sets U in X.

Definition 24. A bijection function h : (X, T ) → (Y,P) is said to be F−hmoeomrphism
if and only if h and h−1 are F -continuous.

Theorem 11. Let h : (X, T ) → (Y,P) be a F -continuous function, then h is continuous
function.

Proof. Let U be any open set in Y , then by F -continuity h−1(U) is F -open set in X.
Since any F -open set is open, then h−1(U) is open set in X.

In general, the converse of the previous theorem is not true. There is an example of
continuous function which is not F−continuous function.

Example 17. The identity function id : (R,U) → (R,U) is a continuous function. How-
ever, there exist (R \ Z) ∈ U and id−1(R \ Z) = R \ Z is not F−open set, because
cl(R \ Z) \ (R \ Z) = Z is not finite.

Definition 25. Let (X, T ) be a topological space. Then (X, T ) is a F -compact (resp., F -
Lindelöf) space if and only if any open cover of X has a finite (resp., countable) subcover
of F -open sets.

Definition 26. Let (X, T ) be a topological space. Then (X, T ) is a F -countably compact
space if and only if any countable open cover of X has a finite subcover of F -open sets.

Theorem 12. Any F -compact space is compact.

Proof. Obvious, because any F -open set is open.

Here an example of compact space which is not F -compact space.

Example 18. Overlapping Interval Topology [3]. On the set X = [−1, 1] we generate a
topology from sets of the form [−1, b) for b > 0 and (a, 1] for a < 0. Then all sets of the form
(a, b) are also open. We have X is a compact space, since in any open covering, the two sets
which include 1 and −1 will cover X. The space X is not F -compact space, because there
exists {[−1, 0.5), (−0.5, 1]} open cover for X has no finite subcover of F -open sets, because
[−1, 0.5) and (−0.5, 1] are not F−open sets (cl[−1, 0.5) \ [−1, 0.5) = [−1, 1] \ [−1, 0.5) =
[0.5, 1] is not finite set and cl(−0.5, 1] \ (−0.5, 1] = [−1, 1] \ (−0.5, 1] = [−1,−0.5] is not
finite set).



M. H. Alqahtani / Eur. J. Pure Appl. Math, 16 (2) (2023), 819-832 831

Corollary 2. Any F -Lindelöf (resp., F -countably compact) space is Lindelöf (resp., count-
ably compact).

Theorem 13. If there exists a F -open F -continuous function h from F -compact space
(X, T ) onto a topological space (Y,P), then (Y,P) is a F -compact space.

Proof. Let {Vα : α ∈ Λ} be any open cover of Y . Since h is F -continuous, then
h−1(Vα) is F -open in X for each α ∈ Λ. Since Y ⊆

⋃
α∈Λ Vα, then X = h−1(Y ) ⊆

h−1(
⋃

α∈Λ Vα) =
⋃

α∈Λ h−1(Vα) ), that is means {h−1(Vα) : α ∈ Λ} is an open cover
of X (because any F -open set is open). Then, by the F -compactness of X, there exist
α1, α2, · · ·αn ∈ Λ such that h−1(Vα1)∪ h−1(Vα2)∪ · · · ∪ h−1(Vαn) = X, then h[h−1(Vα1)∪
h−1(Vα2)∪· · ·∪h−1(Vαn)] = h(X), then h(h−1(Vα1))∪h(h−1(Vα2))∪· · ·∪h(h−1(Vαn)) = Y ,
then Vα1 ∪ Vα2 ∪ · · · ∪ Vαn = Y . Since h is F -open, then {Vα1 ∪ Vα2 ∪ · · · ∪ Vαn} is a finite
subcover of F -open sets for Y . Therefore, (Y,P) is a F -compact space.

A subset B of a space X is F -compact if and only if B is a F -compact topological
space with the subspace topology.

Theorem 14. If there exists a F -continuous function h from F -compact space (X, T )
onto a topological space (Y,P), then (Y,P) is compact.

Proof. Using the same proof of Theorem 13.

Theorem 15. If there exists a F -open F -continuous function h from a F -Lindelöf (resp.,
F -countably compact) space (X, T ) onto a topological space (Y,P), then (Y,P) is F -
Lindelöf (resp., F -countably compact).

Proof. Using the same proof of Theorem 13.

Theorem 16. If there exists a F -continuous function h from a F -Lindelöf space (X, T )
onto a topological space (Y,P), then (Y,P) is Lindelöf.

Proof.
Let {Vα : α ∈ Λ} be any open cover of Y . Since h is F -continuous, then h−1(Vα)

is F -open subsets in X for each α ∈ Λ. Since Y ⊆
⋃

α∈Λ Vα, then X = h−1(Y ) ⊆
h−1(

⋃
α∈Λ Vα) =

⋃
α∈Λ h−1(Vα) ), that is means {h−1(Vα) : α ∈ Λ} is an open cover of X

(because any F -open set is open). Then, by the F -Lindelöfness ofX, there exists α1, α2, · · ·
such that h−1(Vα1)∪ h−1(Vα2)∪ · · · = X, then h[h−1(Vα1)∪ h−1(Vα2)∪ . . . ] = h(X), then
h(h−1(Vα1)) ∪ h(h−1(Vα2)) ∪ · · · = Y , then Vα1 ∪ Vα2 ∪ · · · = Y . Hence, {Vα1 ∪ Vα2 ∪ . . . }
is a countable subcover of open sets for Y . Therefore, (Y,P) is a Lindelöf space.

Theorem 17. If there exists a F -continuous function h from a F -countably compact space
(X, T ) onto a topological space (Y,P), then (Y,P) is countably compact.

Proof. Using the same proof of Theorem 16.



REFERENCES 832

References

[1] R. Engelking. General Topology. PWN, Warszawa, 1977.

[2] C. Kuratowski. Topology I. 4th. ed., in French, Hafner, New york, 1958.

[3] L. Steen and J. Seebach. GCounterexamples in Topology. Dover Publications, INC,
1995.