# Characterizations of Hyperideals and Interior Hyperideals in OrderedΓ-Semihypergroups

Authors: Yongsheng Rao; Qixin Zhou; Ruiqi Cai; Aysha Khan
Journal: Mathematical Problems in Engineering
 (2022)

---

## Abstract

We give some conditions on orderedΓ-semihypergroups under which their interior hyperideal is equal to the hyperideal. In this paper, it is shown that in regular (resp., intraregular, semisimple) ordered Γ-semihypergroups, the hyperideals and the interior hyperideals coincide. To show the importance of these results, some examples and conclusions are provided.


---

## Body

## 1. Introduction and Preliminaries

Heidari and Davvaz [1] gave the idea of an ordered semihypergroup in 2011. Connection between ordered semihypergroups was studied by Tang et al. [2]. For some works on ordered Γ-semihypergroups, we may refer to Ref. [3].The general structure of factorizable ordered hypergroupoids is studied in Ref. [4]. Tang et al. [5] and Tipachot and Pibaljommee [6] combined the fuzzy set with ordered hyperstructures and proposed the concept of fuzzy interior hyperideal and proved some results. The notion of hypergroups was initially founded by F. Marty [7] in 1934.Recently, many authors, for example, those in Refs. [8, 9], have investigated on ordered hyperstructures. The paper given in Ref. [8] is a detailed study of interior hyperfilters in ordered Γ-semihypergroups. In Ref. [9], w-pseudo-orders in ordered (semi)-hyperrings were defined, and some important properties are investigated.The notion of uni-soft interiorΓ-hyperideals is investigated in Ref. [10]. Motivated by these studies, this note investigates the ordered Γ-semihypergroups that their interior hyperideal is equal to the hyperideal. We prove that in regular (resp., intraregular, semisimple) ordered Γ-semihypergroups, the concepts of interior Γ-hyperideals and Γ-hyperideals coincide.Definition 1 (see [11]).
LetH and Γ be two nonempty sets. Then, H is called a Γ-semihypergroup if every γ∈Γ is a hyperoperation on H, i.e., xγy⊆H for every x,y∈H, and for every α,β∈Γ and x,y,z∈H, we have xαyβz=xαyβz.
LetA and B be two nonempty subsets of H. We define(1)AΓB=∪aγb|a∈A,b∈Bandγ∈Γ=∪γ∈ΓAγB.Definition 2.
An orderedΓ-semihypergroup H,Γ,≤ is a Γ-semihypergroup H,Γ together with a partial order relation ≤ such that for any h,h′,x∈H and α∈Γ, we have(2)h≤h′⇒xαh⪯¯xαh′,hαx⪯¯h′αx.
Here,C⪯¯D means that for any c∈C, there exists d∈D such that c≤d, where ∅≠C,D⊆H.
Now, let
K≔h∈H|h≤kforsomek∈K.
Then,H,Γ,≤ can be called as follows:(1)
Regular (resp., intraregular) ifK⊆KΓHΓK (resp., K⊆HΓKΓKΓH) for every K⊆H(2)
H,Γ,≤ is called semisimple if K⊆HΓKΓHΓKΓH for every K⊆H
A nonempty subsetK of H is called a Γ-hyperideal of H if(1)
HΓK⊆K and KΓH⊆K(2)
K⊆KDefinition 3 (see [5]).
A subΓ-semihypergroup K of H is called an interior Γ-hyperideal (briefly, I-Γ-hyperideal) if(1)
HΓKΓH⊆K(2)
K⊆KRemark 1.
Note that each hyperideal of an ordered hyperstructureH is an I-Γ-hyperideal, but an I-Γ-hyperideal need not be hyperideal.Example 1.
LetH=a,b,c,d and Γ=γ. Define the hyperoperation γ (as shown in Table 1) and (partial) order relation ≤ on H as follows:(3)≤≔a,a,a,b,a,c,a,d,b,b,c,c,d,d.
Here,A=a,c is an I-Γ-hyperideal of ordered Γ-semihypergroup H but not a Γ-hyperideal of H. Indeed, as cγd=a,b and b∉A, A is not a Γ-hyperideal of H.
In this note, we investigate on the orderedΓ-semihypergroups that their interior hyperideal is equal to the hyperideal.Table 1
Table ofγ for Example 1.
γabcdaaaaabaaaacaaaa,bdaaa,ba,b,c


## 2. Main Results

This section aims to outline sufficient conditions for anI-Γ-hyperideal to be a Γ-hyperideal. We continue our study with the characterization of regular (resp., Intraregular, semisimple) ordered Γ-semihypergroup in terms of I-Γ-hyperideals.Theorem 1.
LetH,Γ,≤ be regular. Then, every I-Γ-hyperideal of H is a Γ-hyperideal.Proof.
Assume thatK is an I-Γ-hyperideal of S and a∈K. By hypothesis, there exist h∈H and μ,λ∈Γ such that a⪯¯aμhλa. It means that K⊆KΓHΓK. If x∈H and γ∈Γ, then(4)aγx⪯¯aμhλaγx,=aμhλaγx,⊆KΓHΓKΓH,=KΓHΓKΓH,⊆KΓK,⊆K.
Thus,KΓH⊆K=K. Similarly, HΓK⊆K.Example 2.
Consider theΓ-semihypergroup H,Γ [12] (see Tables 2 and 3).
Now, we set(5)≤:=a,a,b,a,b,b,b,c,c,c,d,d,d,e.
Clearly,H,Γ,≤ is regular. The only I-Γ-hyperideals of H are K1=d,e and K2=H. Both the I-Γ-hyperideals are Γ-hyperideal.Table 2
Table ofγ for Example 2.
γabcdeaa,cba,cd,eebbbbd,eeca,cbcd,eedd,ed,ed,edeeeeeeeTable 3
Table ofβ for Example 2.
βabcdeaaba,cd,eebbbbd,eeca,cba,cd,eedd,ed,ed,edeeeeeeeTheorem 2.
LetH,Γ,≤ be intraregular. Then, we get those as follows:(1)
EveryI-Γ-hyperideal of H is a Γ-hyperideal(2)
EveryI-Γ-hyperideal of H is idempotentProof.
(1)
Assume thatK is an I-Γ-hyperideal of H and a∈K. By hypothesis, there exist h,h′∈H and μ,λ,δ∈Γ such that a⪯¯hμaλaδh′. It means that K⊆HΓKΓKΓH. If x∈H and γ∈Γ, then(6)aγx⪯¯hμaλaδh′γx,=hμaλaδh′γx,⊆HΓKΓKΓH,⊆HΓKΓH,⊆K.So,KΓH⊆K=K. Similarly, HΓK⊆K.(2)
Assume thatK is an I-Γ-hyperideal of H. Then, we have(7)K⊆HΓKΓKΓH,⊆HΓHΓKΓKΓHΓHΓKΓKΓHΓH,⊆HΓHΓKΓKΓHΓHΓKΓKΓHΓH,⊆HΓKΓHΓHΓKΓH,⊆KΓK.
Now, leta∈​KΓK​. Then, a⪯¯kγk′ for some k,k′∈K and γ∈Γ. By hypothesis, there exist h,h′∈H and μ,λ,δ∈Γ such that a⪯¯hμaλaδh′. We have(8)a⪯¯hμaλaδh′,⪯¯hμkγk′λkγk′δh′,⊆HΓKΓH,⊆K.
Thus,a∈K and so KΓK⊆K.Example 3.
Consider theΓ-semihypergroup H,Γ [13] (see Tables 4 and 5).
Now, we set(9)≤:=a,a,a,b,a,c,b,b,b,c,c,c,d,d,e,ecs.
Clearly,H,Γ,≤ is an intraregular ordered Γ-semihypergroup. The only I-Γ-hyperideals of H are K1=d,e and K2=H. Both the I-Γ-hyperideals are Γ-hyperideal and idempotent.Table 4
Table ofγ for Example 3.
γabcdeaa,bb,ccd,eebb,cccd,eeccccd,eedd,ed,ed,edeeeeeeeTable 5
Table ofβ for Example 3.
βabcdeab,cccd,eebcccd,eeccccd,eedd,ed,ed,edeeeeeeeTheorem 3.
LetH,Γ,≤ be a semisimple ordered Γ-semihypergroup. Then, every I-Γ-hyperideal of H is a Γ-hyperideal.Proof.
Assume thatK is an I-Γ-hyperideal of H and a∈K. By hypothesis, there exist x,y,z∈H and α,β,γ,δ∈Γ such that a⪯¯xαaβyγaδz. It means that K⊆HΓKΓHΓKΓH. If h∈H and λ∈Γ, then(10)aλh⪯¯xαaβyγaδzλh,=xαaβyγaδzλh,⊆HΓKΓHΓKΓHΓH,=HΓKΓHΓKΓHΓH,⊆HΓKΓKΓH,⊆HΓKΓH,⊆K.
So,KΓH⊆K=K. Similarly, HΓK⊆K.Theorem 4.
H,Γ,≤ is semisimple if and only if every Γ-hyperideal of H is idempotent.Proof.
(Necessity). Let K be a Γ-hyperideal of H. By hypothesis, we have(11)K⊆HΓKΓHΓKΓH,=HΓKΓHΓKΓH,⊆KΓHΓK,⊆KΓK.
Also,(12)KΓK⊆SΓK,⊆K,=K.
So,K=KΓK, and it completes the proof. Sufficiency. Let a∈H. We denote by IHa the Γ-hyperideal of H generated by a. Then, we get IHa=a∪HΓa∪aΓH∪HΓaΓH.
By hypothesis, we have(13)a∈IHa=IHaΓIHa,=a∪HΓa∪aΓH∪HΓaΓHΓa∪HΓa∪aΓH∪HΓaΓH,⊆HΓaΓHΓaΓH.
Therefore,H is semisimple.Example 4.
In Example2,(14)≤:=a,a,a,b,b,b,c,b,c,c,d,d,d,e,e,e,is a partial order relation. Clearly, H,Γ,≤ is semisimple. The only I-Γ-hyperideals of H are K1=d,e and K2=H. Both the I-Γ-hyperideals are Γ-hyperideal and idempotent.


## 3. Conclusions

This paper gives some conditions under which theI-Γ-hyperideals are Γ-hyperideals. By Theorems 1–3, we prove that in a regular (resp., intraregular, semisimple) ordered hyperstructure H, every interior hyperideal of H is a hyperideal. By Theorems 3 and 4, H is a semisimple ordered hyperstructure if and only if every interior hyperideal of H is idempotent. Our future work will concentrate on some results which are related with the fuzzy interior hyperideals of ordered hyperstructures.



---

*Source: 2292712-2022-11-01.xml*