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AN EXAMPLE OF A JOIN SPACE 
ASSOCIATED WITH A RELATION 

 
Laurenţiu LEOREANU1 

 
Abstract In this paper a join spaces associated with a binary relation is presented. 
 
Keywords Join spaces, Relations 
 
First of all, let us recall what a join space is. 
Let H be a nonemptyset and o : H x H →P* (H), where P* (H) is the set of nonempty 
subsets of H. 
If A ⊂ H, B ⊂ H, then we set A o B =   ∪a∈A ∪b∈B  a o b. 
We denote A≈B if  A∩B ≠ ø. 
If the hyperoperation �o� is associative and ∀ a∈H, we have a o H = H = H o a, then 
(H,o) is a hypergroup. 
Denote a / b={x∈H a∈b o x}, for any (a,b)∈H2. 
A hypergroup (H,o) is called a join space if  �o� is commutative and  

a / b ≈ c / d ⇒ a o d ≈ b o c. 
Join spaces have been introduced by W. Prenowitz and used by himself and J. 
Jantosciak in order to rebuild some branches of non-Euclidian geometries. 
Afterwards, join spaces have also been used in the study of other topics (Graphs and 
Hypergraphs, Lattices, Binary Relations and so on). 
Here, a connection between join spaces and reflexive and symmetric relations is 
presented. 
First, we give an example: 
Let f : H → U be an onto map. 
We define on H the following hyperoperation: 

∀(x,y) ∈ H2, x o x = f-1(f(x)), x o y = x o x ∪ yoy 
(where ∀ Y⊂U, f-1(Y) = {x∈Hf(x) ∈Y}). 
 
Proposition  (H,o) is a join space. 
 
Proof. For any (x,y,z) ∈ H3, we have 
(x o y)o z = x o (y o z) =  f-1(f(x)) ∪ f-1(f(y)) ∪ f-1(f(z)) 
and x o H = ∪a∈H x o a = ∪a∈H f

-1(f(x)) ∪ f-1(f(a)) = H,  
since f is onto. 
So, (H, o) is a commutative hypergroup. 

                                                           
1 Grupul Şcolar �Miron Costin � Roman, România 



 

 

 

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Moreover, any x ∈ H is an identity of H (since ∀ y ∈ H,   y ∈ x o y) and for any 
(x,y)∈H2, x is an inverse of y. 
Let us check now that  a / b ≈ c / d ⇒ a o d ≈ b o c. 
Let x ∈ a / b ∩ c / d that is a ∈ f-1 (f(x)) ∪ f-1 (f(y)) and c ∈ f-1 (f(x)) ∪ f-1 (f(d)).  
It follows f(a) ∈ {f(x), f(b)} and f(c) ∈{f(x), f(d)}. 
We must prove that there is y ∈H, such that y∈ a o d  ∩ b o c, that is  
f(y) ∈ {f(a), f(d)} ∩ {f(b), f(c)}. 
We have the following situations: 

1) if f(a) = f(x) = f(c) then we can choose y = a ; 
2) if f(a) = f(x) and f(c) = f(d), then we choose y = d ; 
3) if f(a) = f(b), then we choose y = a. 

Therefore, (H, o) is a join space. 
 

Now, let us consider R a reflexive and symmetric relation on H. 
Let us consider the following hyperoperation on H 

∀(x,y) ∈ H2 , x oR x = {z(z,x) ∈R}, x oR y = x oR x ∪ y oR y. 
 

Theorem (H, oR) is a join space. 
 
Proof. The associativity is immediate and ∀ x ∈ H, we have   
x oRH = x oRx ∪ ∪a∈H a oRa = H, since 
∪a∈H a oRa = ∪a∈H {z(z,a) ∈ R} = H  (R is reflexive). 
So, (H, oR) is a commutative hypergroup. 
Notice that a∈ a oRa  ⇔ (a,a)∈R. 
Let us check now that a / b ≈ c / d  ⇒  a ∈ a oRd ≈ b oRc. 
Let x∈H, such that a ∈ x oR b and c ∈ x oR d. 
We have a ∈ {t(t,x) ∈ R or (t,b) ∈ R}, whence (a,x)∈R or (a,b)∈R.  
Similarly, (c,x)∈R or (c,d)∈R. 
We must prove that there is y∈H, such that y∈ a oRd and y∈ b oR c, that is  
[(y,a) ∈R or (y,d) ∈R] and [(y,b) ∈R or (y,c) ∈R],  
or equivalently, [(y,a) ∈R and (y,b) ∈R] or [(y,a) ∈R and (y,c) ∈R] 
or [(y,d)∈R and (y,b)∈R] or [(y,d) ∈R and (y,c)∈R]. 
We have the following situations: 
 
1)  (a,x) ∈R and (c,x) ∈R. Since R is symmetric, it follows (x,a) ∈R and (x,c) ∈R. 
In this case, we can choose y=x. 
2)  (a.x) ∈R and (c,d) ∈R. We take y=c and so, (y,d)=(c,d) ∈R and  (y,c)=(c,c) ∈R. 
3)  (a,b) ∈R and [(c,x) ∈R or (c,d) ∈R]. 
4)  We take y=a, so (y,b)=(a,b) ∈R and (y,a)=(a,a) ∈R. 
 
Therefore, (H, oR) is a join space. 
 



 

 

 

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Remark. If R is the relation defined as follows: 
(x,y) ∈R ⇔ f(x)=f(y) 
where f : H →U, then (H, oR ) is the join space presented at the beginning. 
 
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