EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
2025, Vol. 18, Issue 2, Article Number 5394
ISSN 1307-5543 – ejpam.com
Published by New York Business Global

Topologies on Hyper BCK-Algebra [0, 1]

Erneta Payla1,∗, Luzviminda Ranara2

1 General Education, Caraga State University, 8605 Cabadbaran City, Philippines
2 Department of Mathematics, Mindanao State University, 9700 Marawi City, Philippines

Abstract. In this paper, we introduce the definition of a hyper operation ∗ on the set [0,1], and
with this hyper operation, we will show that [0,1] is a hyper BCK-algebra. We also investigate
the topologies that will be formulated with BR([0, 1]) and BL([0, 1]) and show some topology of
R[0, 1](A) and L[0, 1](A). Furthermore, we investigate a basis for the intersection of topologies
τR(H) and τL(H).

2020 Mathematics Subject Classifications: 06F35, 54A10, 08A72, 03G25
Key Words and Phrases: Hyper BCK-algebra, hyper order, bases, hyperoperation, topology

1. Introduction

In 1966, Imai and Isèki [1] introduced the concept of BCK-algebra as a generalization
of the concept of set-theoretic difference and propositional calculi. The study of algebraic
hyperstructure theory (or multialgebras) was introduced in 1934 by F. Marty [2] at the
8th Congress of Scandinavian Mathematics. Since then it becomes the interest of many
researchers. Recently, Jun, et al.,[4] proposed hyperstructure theory on BCK-algebras and
they were able to prove that a hyper BCK-algebra is a generalization of a BCK-algebra.

In the paper of Patangan and Canoy [3, 4], they defined the sets RH(A) = {x ∈ H :
a ≪ x,∀a ∈ A} = {x ∈ H : 0 ∈ a ∗ x,∀a ∈ A} and LH(A) = {x ∈ H : x ≪ a,∀a ∈ A} =
{x ∈ H : 0 ∈ x ∗ a,∀a ∈ A} by the right applications of hyperorder on H, respectively.
They showed that BR(H) consisting of the sets RH(A), is a basis for some topology τR(H)
on a hyper BCK-algebra via right application of hyperorder. Also, BL(H) consisting of
the sets LH(A), is a basis for some topology τL(H) on a hyper BCK-algebra via left
application of hyperorder.

This paper is motivated by the work of Patangan and Canoy [3] on A Topology on
a Hyper BCK-Algebra, as published in JP Journal of Algebra, Number Theory and Ap-
plications. In this paper, we define a hyperoperation ∗ on the set [0, 1], and with this
operation, we will show that [0, 1] is a hyper BCK-algebra. We investigate the basis for
intersection of topologies τR(H) and τL(H). We also investigate the topologies that will
be formulated or generated with bases BR([0, 1]) and BL([0, 1]) and their intersection.

∗Corresponding author.
DOI: https://doi.org/10.29020/nybg.ejpam.v18i2.5394

Email addresses: eipayla@csucc.edu.ph (E. Payla), luzviminda.ranara@msumain.edu.ph (L. Ranara)

https://www.ejpam.com 1 Copyright: © 2025 The Author(s). (CC BY-NC 4.0)



E. Payla, L. Ranara / Eur. J. Pure Appl. Math, 18 (2) (2025), 5394 2 of 10

2. Known Results

Definition 1. [5] Let P(H) be the power set of a nonempty set H. Consider P(H)∗ =
P(H) \ {∅}. A hyperoperation on a nonempty set H is a function ∗: H ×H → P(H)∗.
The image of (x, y) ∈ H×H under ∗ is denoted by x∗y. If x ∈ H and A,B are nonempty
subsets of H, then we define

(i) A ∗B =
⋃

a∈A, b∈B
a ∗ b;

(ii) A ∗ x = A ∗ {x}; and,

(iii) x ∗B = {x} ∗B.

Definition 2. [5] Let x, y ∈ H and A,B ⊆ H. Then

(i) x ≪ y if and only if 0 ∈ x ∗ y; and

(ii) A ≪ B if and only if for any a ∈ A, there exist b ∈ B such that a ≪ b. We call
“ ≪” a hyperorder on H.

Definition 3. [6] A hyper BCK-algebra is a nonempty set H endowed with a hyperoper-
ation “∗” and a constant 0 satisfying the following axioms: for all x, y, z ∈ H,

(i) (x ∗ z) ∗ (y ∗ z) ≪ x ∗ y,

(ii) (x ∗ y) ∗ z = (x ∗ z) ∗ y,

(iii) x ∗H ≪ x,

(iv) x ≪ y and y ≪ x imply x = y.

Proposition 1. [6] In a hyper BCK-algebra (H, ∗, 0), the condition (iii) of Definition 3
is equivalent to the condition x ∗ y ≪ {x} for all x, y ∈ H.

Theorem 1. [7] Let B ⊆ τ . The following two properties of B are equivalent:

(1) B is a basis for τ .

(2) For each G ∈ τ and each x ∈ G there is a U ∈ B with x ∈ U ⊆ G.

Theorem 2. [7] Let X ̸= ∅. A class of subsets B of X is a basis for some topology τ on
X if it satisfies the following:

(i) B covers X, and

(ii) For each x ∈ Uα∩Uβ, there exists U ∈ B such that x ∈ U ⊆ Uα∩Uβ where Uα,Uβ ∈ B.

Definition 4. [7] A space X is said to be connected if X is not the union of two disjoint
open sets. Otherwise, it is said to be disconnected.



E. Payla, L. Ranara / Eur. J. Pure Appl. Math, 18 (2) (2025), 5394 3 of 10

Definition 5. [3] Let H be a hyper BCK-algebra and A ⊆ H. Then the set RH(A) is
defined as RH(A) = {x ∈ H : a ≪ x for all a ∈ A} = {x ∈ H : 0 ∈ a ∗ x for all a ∈ A}. If
A = {a}, then we write RH({a}) = RH(a).

Definition 6. [3] An element a of H of a hyper BCK-algebra H is called a hyperatom
if x ≪ a implies x = 0 or x = a for all x ∈ H. Denote A(H) the set of all hyperatoms of
H and A∗(H) = A(H) \ {0}. Obviously, 0 ∈ A(H). If each element of H is a hyperatom,
then H is said to be hyperatomic, that is, A∗(H) = H \ {0}. A hyper BCK-algebra H is
called ordered if the hyperorder “≪” is transitive.

Proposition 2. [3] Let A and B be subsets of H. the the following hold:

(i) RH(∅) = H.

(ii) If A ⊆ B, then RH(B) ⊆ RH(A).

(iii) If H is an ordered hyper BCK-algebra, then RH(RH(A)) ⊆ RH(A).

Theorem 3. [3] Let H be a hyper BCK-algebra then BR(H) = {RH(A) : A ⊆ H} is a
basis for some topology in H.

Remark 1. [3] Let A and B be nonempty subsets of H. RH(A) ∩RH(B) = RH(A ∪B).

Definition 7. [4] Let H be a hyper BCK-algebra and A ⊆ H. Then the set LH(A) is
defined as LH(A) = {x ∈ H : x ≪ a, for all a ∈ A} = {x ∈ H : 0 ∈ x ∗ a, for all a ∈ A}.
If A = {a}, the we write LH({a}) = LH(a).

Theorem 4. [8] The set [0, 1], together with the binary operation “∗”, is a BCK-algebra.

3. A Hyperoperation on [0, 1]

On the set [0, 1], we define ”∗” as follows: x ∗ y = {x− y}, if x > y and x ∗ y = {0} if
x ≤ y.

Lemma 1. For each x, y, z ∈ [0, 1], (x ∗ z) ∗ (y ∗ z) ≪ x ∗ y.

Proof. Case 1: x ≥ y
If y ≤ z ≤ x then x− z ≥ 0 so that

(x ∗ z) ∗ (y ∗ z) = {x− z} ∗ {0}
= {x− z − 0}
= {x− z}

while x∗y = {x−y} so that (x∗z)∗(y∗z) = {x−z} ≪ {x−y} for 0 ∈ {x−z}∗{x−y} =
{0}, since x− z ≤ x− y.
If z ≤ y ≤ x then x− y ≤ x− z so that



E. Payla, L. Ranara / Eur. J. Pure Appl. Math, 18 (2) (2025), 5394 4 of 10

(x ∗ z) ∗ (y ∗ z) = {x− z} ∗ {y − z}
= {x− z − (y − z)} for x− z ≥ y − z

= {x− y}
= x ∗ y

Thus, (x ∗ z) ∗ (y ∗ z) = {x− y} ≪ x ∗ y for 0 ∈ {x− y} ∗ {x− y} = {0}.
If y ≤ x ≤ z then (x∗z)∗(y∗z) = {0}∗{0} = {0}, x∗y = {x−y} and 0 ∈ 0∗{x−y} = {0}.
Thus, (x ∗ z) ∗ (y ∗ z) = {0} ≪ {x− y} = x ∗ y.
Case 2: If x ≤ y then x ∗ y = {0}.
If x ≤ z ≤ y then (x ∗ z) ∗ (y ∗ z) = {0} ∗ {y − z} = {0} ≪ {0} = x ∗ y for 0 ∈ 0 ∗ 0 = {0}.
If z ≤ x ≤ y then (a ∗ z) ∗ (y ∗ z) = {x − z} ∗ {y − z} = {0} for x − z ≤ y − z so
that(x ∗ z) ∗ (y ∗ z) = {0} ≪ {0} = x ∗ y for 0 ∈ 0 ∗ 0 = {0}.
If x ≤ y ≤ z then (x ∗ z) ∗ (y ∗ z) = {0} ∗ {0} = {0} ≪ {0} = x ∗ y.

■

Lemma 2. For all x, y, z ∈ [0, 1], (x ∗ y) ∗ z = (x ∗ z) ∗ y.

Proof. Case 1: If x ≥ y.

If y ≤ z ≤ x, (x ∗ y) ∗ z = {x− y} ∗ z =

{
{0}, if x− y ≤ z,

{x− y − z}, if x− y > z
while

(x ∗ z) ∗ y = {x− z} ∗ y =

{
{x− z − y}, if x− z ≥ y,

{0}, if x− z ≤ y

Note that when the result of (x ∗ y) ∗ z is {0}, it happens when x − y < z and it is just
the same as x− z < y. Also if the result is {x− y − z}, it happens when x− y > z and it
is just the same as x− z > y. Thus, (x ∗ y) ∗ z = (x ∗ z) ∗ y.
If z ≤ y ≤ x then

(x ∗ y) ∗ z = {x− y} ∗ z =

{
{x− y − z}, if x− y ≥ z

{0}, if x− y < z

(x ∗ z) ∗ y = {x− z} ∗ y

=

{
{x− z − y}, if x− z ≥ y

{0}, if x− z < y

=

{
{x− y − z}, if x− y ≥ z

{0}, if x− y < z

= (x ∗ y) ∗ z

If y ≤ x ≤ z then (x ∗ y) ∗ z = {x− y} ∗ z = {0} for x− y < z and



E. Payla, L. Ranara / Eur. J. Pure Appl. Math, 18 (2) (2025), 5394 5 of 10

(x ∗ z) ∗ y = {0} ∗ y
= {0} for 0 ≤ y

= (x ∗ y) ∗ z

Case 2: If x < y
If x ≤ z < y, then (x ∗ y) ∗ z = {0} ∗ z = {0} and (x ∗ z) ∗ y = {0} ∗ y = {0} = (x ∗ y) ∗ z.
If z ≤ x ≤ y, then (x ∗ y) ∗ z = {0} ∗ z = {0} and

(x ∗ z) ∗ y = {x− z} ∗ y
= {0} for x− z < y

= (x ∗ y) ∗ z

If x ≤ y ≤ z then (x ∗ y) ∗ z = {0} ∗ z = {0} and (x ∗ z) ∗ y = {0} ∗ y = {0} = (x ∗ y) ∗ z.
Thus (x ∗ y) ∗ z = (x ∗ z) ∗ y.

■

Lemma 3. For each x ∈ [0, 1], x ∗ [0, 1] ≪ {x}.

Proof. : Case 1: If x ≥ y, then x ∗ y = {x− y} and that x− y < x so that x− y ≪ x
for 0 ∈ x− y ∗ x = {0}.
Case 2: If x < y, then x ∗ y = {0} ≪ {x} for 0 ∈ 0 ∗ x = {0}.
In any case, for all y ∈ H, x ∗ y ≪ {x}. Thus x ∗ [0, 1] ≪ {x}.

■

Lemma 4. For all x, y ∈ [0, 1] x ≪ y and y ≪ x implies x = y.

Proof. : Suppose x ≪ y and y ≪ x. Then

0 ∈ x ∗ y =

{
{x− y}, if x > y

{0}, if x ≤ y

and

0 ∈ y ∗ x =

{
{y − x}, if y > x

{0}, if y ≤ x

This implies that x ∗ y = {0} = y ∗ x. Hence, x ≤ y and y ≤ x implies x = y.

■

Theorem 5. ([0, 1] , ∗, 0) is a hyper BCK-algebra.

Proof. : It follows immediately from Lemmas 1, 2, 3 and 4.

■



E. Payla, L. Ranara / Eur. J. Pure Appl. Math, 18 (2) (2025), 5394 6 of 10

4. The topology τR[0, 1]

By Definition 5, RH(A) = {x ∈ H : a ≪ x, ∀ a ∈ A} = {x ∈ H : 0 ∈ a ∗ x, ∀ a ∈ A}
and considering the hyper BCK-algebra [0, 1]. Note that from now on H = [0, 1], we have
the following example.

Example 1. Consider A =

{
1

2
,
1

3

}
. Then

RH(A) = {x ∈ [0, 1] : 0 ∈ a ∗ x, ∀ a ∈ A}

=

{
x ∈ [0, 1] : 0 ∈ a ∗ x, ∀ a ∈

{
1

2
,
1

3

}}
=

{
x ∈ [0, 1] : 0 ∈ 1

2
∗ x and 0 ∈ 1

3
∗ x

}

Note that for 0 to be in 1
2 ∗ x, 1

2 ∗ x = {0} this implies x ≥ 1
2 or

[
1
2 , 1

]
. Also for

0 ∈ 1
3 ∗ x, x ≥ 1

3 or
[
1
3 , 1

]
. Since RH(A) =

⋂
a∈ARH(a) this implies that RH({1

2 ,
1
3}) =

[12 , 1] ∩ [13 , 1] = [12 , 1]. Therefore, RH

({
1
2 ,

1
3

})
=

[
1
2 , 1

]
.

Theorem 6. Let A ⊆ H, RH(A) = [sup A, 1] ∩H. In particular,

RH(A) =

{
H if A = ∅
[sup A, 1] if A ̸= ∅.

.

Proof. : Note that x ∈ RH(A) if and only if 0 ∈ a ∗x for all a ∈ A. Now, 0 ∈ a ∗x for
all a ∈ A if and only if a ≤ x for all a ∈ A, that is, x ∈ H is an upperbound of A. Hence,
x ∈ RH(A) if and only if x ∈ [sup A, 1] ∩ H. This shows that RH(A) = [sup A, 1] ∩ H.
If A = ∅, then sup A = −∞ and so RH(A) = H. Otherwise, sup A ∈ H implying that
RH(A) = [sup A, 1].

■

Example 2. Consider A = (18 ,
1
2). Then the sup A = 1

2 . Thus, RH(A) = [12 , 1].

The next result follows from Theorem 6.

Corollary 1. Let a ∈ [0, 1]. Then R[0,1](a) = [a, 1].

Proposition 3. (H, ∗, 0) is not hyperatomic. In particular, A(H) = 0.

Proof. Clearly, 0 ∈ H. Let a ∈ H \{0}, i.e., a ∈ (0, 1]. Since
a

2
< a,

a

2
∗a = {0}. Here,

we find that 0 ∈ a

2
but

a

2
/∈ {0, a}. Thus a /∈ A(H), showing that A(H) = {0}. Therefore,

(H, ∗, 0) is not hyperatomic.

■



E. Payla, L. Ranara / Eur. J. Pure Appl. Math, 18 (2) (2025), 5394 7 of 10

Corollary 2. BR([0, 1]) = {[r, 1] : r ∈ [0, 1]}.

Let τR(H) be the topology generated by BR(H). That is, for each G ∈ τR(H), G =⋃
Bi , i ∈ K ⊆ BR(H). Note that for any a ∈ [0, 1], the set (a, 1] is open for (a, 1] =

∞⋃
n=1

[
a+ 1

n , 1
]
.

Example 3. The following are open set in [0, 1]:

• Sets of the form (r, 1] for (r, 1] =
∞⋃
n=1

[
r + 1

n , 1
]

• {1} is open for {1} = [1, 1]

Example 4. The following are closed sets in [0, 1]:

• [0, r) for [0, r) = [r, 1]c

• [0, r] for [0, r] = (r, 1]c, r ∈ [0, 1]

Thus, the next theorem follows:

Theorem 7. In a hyper BCK-algebra [0, 1], τR ([0, 1]) = {∅, [0, 1] ,
(r, 1] , [r, 1] : r ∈ [0, 1]}.

Proof. Let G ∈ τR ([0, 1]), and G ̸= ∅. Then G =
⋃

i∈K⊆BR([0,1])

Bi. Thus, Bi are of

the form [ri, 1], G =
⋃

[ri, 1] = [r, 1] , r0 = inf{ri} and (r, 1] =
∞⋃
n=1

[
r + 1

n , 1
]
. Thus,

τR ([0, 1]) = {∅, [0, 1] , (r, 1] , [r, 1] : r ∈ [0, 1]}.

■

Theorem 8. [0, 1] with topology τR([0, 1]) is connected.

Proof. Suppose [0, 1] is disconnected. Then, there exist disjoint open sets A,B such
that [0, 1] = A ∪ B. Since A is open, A = [r, 1] or (r, 1]. If A = [r, 1] then B = [0, r). If
A = (r, 1] then B = [0, r]. Whether B = [0, r) or [0, r], B is not open. This contradicts
the statement that A and B are open sets. Therefore, by Definition 4, [0, 1] is connected.

■

5. The topology τL[0, 1]

Example 5. Consider the hyper BCK-algebra [0, 1] and A = {1
2 ,

1
3}.

LH(A) = {x ∈ [0, 1] : 0 ∈ x ∗ a, ∀ a ∈ A}



E. Payla, L. Ranara / Eur. J. Pure Appl. Math, 18 (2) (2025), 5394 8 of 10

=

{
x ∈ [0, 1] : 0 ∈ x ∗ a, ∀ a ∈ {1

2
,
1

3
}
}

=

{
x ∈ [0, 1] : 0 ∈ x ∗ 1

2
and 0 ∈ x ∗ 1

3

}

Note that for 0 to be in x ∗ 1
2 , x ∗ 1

2 = {0} implies x ≤ 1
2 or

[
0, 12

]
. Also for 0 ∈ x ∗ 1

3 ,
x ∗ 1

3 = {0} which implies x ≤ 1
3 or

[
0, 13

]
. For these two to hold, Therefore LH({1

2 ,
1
3}) =[

0, 13
]
.

The proof of the following theorems are analogous to that topology τR[0, 1].

Theorem 9. In the hyper BCK-algebra [0, 1], LH(a) = [0, a].

Theorem 10. Let A ⊆ [0, 1]. L[0,1](A) = [0, r] where r = infA.

Theorem 11. In the hyper BCK-algebra [0, 1], BL([0, 1]) = {[0, r] : r ∈ [0, 1]}.

The following examples are open and closed sets in [0, 1].

Example 6. The following are open set in [0, 1]:

• [0, a) for [0, a) =
∞⋃
n=1

[
0, a− 1

n

]
• {0} for {0} = [0, 0]

Example 7. The following are closed set in [0, 1]:

• (a, 1] for (a, 1]c = [0, a] is open

• [a, 1] for [a, 1]c = [0, a) is open

• (0, 1] for (0, 1] = {0}c

Thus we have the following result.

Theorem 12. In a hyper BCK-algebra [0, 1], τL([0, 1]) = {∅, [0, 1] , [0, r] , [0, r) : r ∈
[0, 1]}.

Proof. Let G ∈ τL ([0, 1]), and G ̸= ∅. Then G =
⋃

i∈K⊆BL([0,1])

Bi. Thus, Bi are of the

form [0, r], G =
⋃
[0, ri] = [0, r] , r0 = inf{ri}. Thus, τL([0, 1]) = {∅, [0, 1] , [0, r] , [0, r) :

r ∈ [0, 1]}.

■

Corollary 3. τL([0, 1]) is not a subspace of R with the usual topology.

Theorem 13. [0, 1] with topology τL([0, 1]) is connected.



E. Payla, L. Ranara / Eur. J. Pure Appl. Math, 18 (2) (2025), 5394 9 of 10

Proof. Suppose [0, 1] is disconnected. Then, there exist disjoint open sets A,B such
that [0, 1] = A ∪ B. Since A is open, A = [0, r] or [0, r). If A = [0, r] then B = [r, 1]. If
A = [0, r) then B = [r, 1]. Whether B = [r, 1] or [r, 1], B is not open. This contradicts
the statement that A and B are open sets. Therefore, by Definition 4, [0, 1] is connected.

■

Theorem 14. In a hyper BCK-algebra [0, 1], τR([0, 1])∩τL([0, 1]) is the indiscrete topology
on [0, 1].

Proof. In [0, 1], BR(H) = {[r, 1] : r ∈ [0, 1]} and BL(H) = {[0, p] : p ∈ [0, 1]}. Suppose
there exixts G ∈ [τR(H) ∩ τL(H)] \ {∅, H}. Let x ∈ G. Then there exist a ∈ (0, 1] and
b ∈ [0, 1) such that x ∈ [a, 1] ⊆ G and x ∈ [0, b] ⊆ G. This implies that a ≤ b; hence,
[a, 1] ∪ [0, b] = [0, 1] ⊆ G. This is a contradiction to our assumption of set G. Therefore,
τR(H) ∩ τL(H) is the indiscrete topology on H.

■

6. Conclusion

The paper explores the structure and topology of a hyper BCK-algebra [0, 1]. Two
topologies, τR([0, 1]) and τL([0, 1]), are constructed based on right and left applications of
hyperorder, respectively. The paper demonstrates the properties of these topologies, in-
cluding the formation of bases, open and closed sets, and connectedness. The intersection
of these topologies results in the indiscrete topology on [0, 1], highlighting a unique conver-
gence of the two structures. This study builds on previous work in algebraic hyperstruc-
tures and provides a foundational exploration of topologies within hyper BCK-algebras.

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