A Study on Hyperatomic Ordered Semihyperrings

Abstract

In this study, we analyze an operator $R_T$ of an ordered semihyperring T with symmetrical hyperoperation ⊕ and show relations with the $L_T$ operator of T. We define the set ${\mathcal{E}}_{R_T\left(K\right)}=\{R_T\left(K\right):\varnothing \ne K\subseteq T\}$. We denote by ${\tau }_{R_T}\left(T\right)$ the topology generated by ${\mathcal{E}}_{R_T\left(K\right)}$. We prove that if $R_T\left(0\right)=T$, then $(T,{\tau }_{R_T}\left(T\right))$ is connected. Some results relating to the hyperatomic ordered semihyperrings and the topology ${\tau }_{R_T}\left(T\right)$ are discussed.

1. Introduction

Marty [1] offered the idea of hypergroup theory as an extension of group theory in 1934. The notion of semihyperring was introduced by Vougiouklis [2] as a generalization of a semiring in 1990. The notion of an ordered semihypergroup was proposed by Heidari and Davvaz [3] in 2011. The concept of pseudoorders in ordered semihypergroups was discussed by Davvaz et al. in [4] and used in constructing ordered semigroups. They constructed an ordered semigroup based on the strongly regular relations on ordered semihypergroups.

Hyperstructures, introduced by Marty, have been of great interest to different fields of science. Hyperrings were introduced by Krasner [5] in connection with his work on valued fields. Researchers are always seeking appropriate and interesting contents. In [6], Jun studied algebraic and geometric aspects of Krasner hyperrings in detail.

Rao et al. [7] utilized derivations in the ordered semihyperring theory. In [8], Kou et al. introduced and studied the concept of the $L_T$ operator and $L_T$ graph in ordered semihyperrings. In continuity of this paper, we study the $R_T$ operator of ordered semihyperrings. A relationship between $L_T$ and $R_T$ operators of ordered semihyperrings and some results relating to the hyperatomic ordered semihyperrings are investigated.

For more details of (ordered) hyperstructures and their related notions, the reader is referred to [9,10,11,12,13,14]. The derivations of hyperrings [9], the left almost hyperideals [10], the almost interior hyperideals [11], the w-pseudo-orders [12], the left k-bi-quasi hyperideals [13] and the right pure (bi-quasi) hyperideals [14] have been investigated. In [15], Rao et al. investigated some properties of hyperatom elements in ordered semihyperrings. Pseudo-atoms of pseudo BE-algebras were studied in [16].

Panganduyon and Canoy [17] discussed a zero divisor graph of a hyper BCI-algebra. Topology is the most advanced area of pure mathematics which studies mathematical (hyper)structures. Many scholars have analyzed the topology that was developed by Panganduyon and Canoy [18]. Panganduyon et al. [19] studied some aspects of the induced topology on hyper BCI-algebras. Jun and Xin [20] investigated some properties of hyperatom elements on hyper BCK-algebras. Jun et al. [21] defined topological BCI-algebras in the year 1999 and investigated the topological ideals and topological homomorphisms. In [22], Al-Tahan and Davvaz presented some applications of ordered hyperstructures in genetics and biological inheritance.

Kou et al. [8] introduced the concept of the $L_T$ operator and proposed the idea of the $L_T$ graph. Now, we analyze an operator

$$R_T\left(K\right):=\{t\in T|k\le t\mathrm{for}\mathrm{all}k\in K\}$$

on an ordered semihyperring T. Inspired by the research performed by Kou et al. [8] on the $L_T$ graph, our paper discusses the topology ${\tau }_{R_T}\left(T\right)$. Considering the ordered semihyperring, a method was suggested to construct ${\tau }_{R_T}\left(T\right)$ topology using the $R_T$ operator. We work on the relationship between $L_T$ and $R_T$ operators of ordered semihyperrings. Some lemmas relating to the hyperatomic ordered semihyperrings are studied. Finally, we show that if ${\tau }_{R_T}\left(T\right)$ is the topology generated by

$${\mathcal{E}}_{R_T\left(K\right)}=\{R_T\left(K\right):\varnothing \ne K\subseteq T\}$$

and $R_T\left(0\right)=T$, then $(T,{\tau }_{R_T}\left(T\right))$ is connected.

2. Preliminaries

Let ${\mathcal{P}}^{*}\left(T\right)$ be the family of all non-empty subsets of a non-empty set T. A mapping $+:T\times T\to {\mathcal{P}}^{*}\left(T\right)$ is said to be a hyperoperation on T. If $\varnothing \ne K,L\subseteq T$ and $t\in T$, then

$$K+L=\bigcup _{\begin{array}{c}\genfrac{}{}{0pt}{}{k\in K}{l\in L}\end{array}}k+l,t+K=\left\{t\right\}+K\mathrm{and}L+t=L+\left\{t\right\}.$$

If $k+(l+t)=(k+l)+t$, for all $k,l,t\in T$, then $(T,+)$ is said to be a semihypergroup.

Let $T,\mathsf{\Gamma }\ne \varnothing$. Then, T is called a $\mathsf{\Gamma }$-semihypergroup if:

(1)

every $\gamma \in \mathsf{\Gamma }$ is a hyperoperation on T;

(2)

for every $\alpha ,\beta \in \mathsf{\Gamma }$ and $a,b,c\in T$, $a\alpha \left(b\beta c\right)=\left(a\alpha b\right)\beta c$.

Note that

$$\begin{array}{c}M\mathsf{\Gamma }N={\displaystyle \bigcup _{\begin{array}{c}\gamma \in \mathsf{\Gamma }\end{array}}}M\gamma N=\bigcup \left\{m\gamma n\right|m\in M,n\in N\mathrm{and}\gamma \in \mathsf{\Gamma }\}.\hfill \end{array}$$

A triple $(T,\mathsf{\Gamma },\le )$ is said to be an ordered $\mathsf{\Gamma }$-semihypergroup [11] if:

(1)

$(T,\mathsf{\Gamma })$ is a $\mathsf{\Gamma }$-semihypergroup;

(2)

$(T,\le )$ is an ordered set;

(3)

for any $a,x,y\in T$ and $\beta \in \mathsf{\Gamma }$, $x\le y$ implies $a\beta x⪯a\beta y$ and $x\beta a⪯y\beta a$.

Note that

$$M⪯N\iff \forall m\in M,\exists n\in N;m\le n.$$

Definition 1

([2]). A semihyperring is a triple $(T,⊞,⊡)$ such that for each $k,l,t\in T$:

(1) 

$(T,⊞)$ is a commutative semihypergroup;

(2) 

$(T,⊡)$ is a semihypergroup;

(3) 

$k⊡(l⊞t)=k⊡l⊞k⊡t$ and $(k⊞l)⊡t=k⊡t⊞l⊡t$;

(4) 

there exists an element $0\in T$ such that $t⊞0=0⊞t=\left\{t\right\}$ and $t⊡0=0⊡t=\left\{0\right\}$ for all t in T.

Definition 2

([23]). A quadruplet $(T,⊞,⊡,\le )$ is called an ordered semihyperring (OS) if for any $q,q^{\prime },x\in R$:

(1) 

$(T,⊞,⊡)$ is a semihyperring;

(2) 

$(T,\le )$ is a partially ordered set;

(3) 

$k\le l\Rightarrow k⊞t⪯l⊞t$;

(4) 

$$k\le l\Rightarrow \left\{\begin{array}{c}k⊡t⪯l⊡t,\hfill \\ \\ t⊡k⪯t⊡l.\hfill \end{array}\right.$$

For every $\varnothing \ne K,L\subseteq T$, $K⪯L$ is defined by $\forall k\in K,\exists l\in L$ such that $k\le l$.

In the remaining part of the paper, let OS be the set of all ordered semihyperrings.

Definition 3

([23]). Let $(T,⊞,⊡,{\le }_T)$ and $(T^{\prime },{⊞}^{\prime },{⊡}^{\prime },{\le }_{T^{\prime }})\in OS$. A function $\psi :T\to T^{\prime }$ is a homomorphism if for all $t,l\in T$:

(1) 

$\psi (t⊞l)\subseteq \psi \left(t\right){⊞}^{\prime }\psi \left(l\right)$;

(2) 

$\psi (t⊡l)\subseteq \psi \left(t\right){⊡}^{\prime }\psi \left(l\right)$;

(3) 

$(t,t^{\prime })\in {\le }_T\Rightarrow (\psi \left(t\right),\psi \left(t^{\prime }\right))\in {\le }_{T^{\prime }}$.

In [15], the authors give the concept of a hyperatom element for an ordered semihyperring T as follows.

Definition 4.

Let $(T,⊞,⊡,\le )\in OS$. An element $t\in T$ is said to be a hyperatom element if:

(1) 

for any $t^{\prime }\in T$, $t^{\prime }\le t\Rightarrow t^{\prime }=0$ or $t^{\prime }=t$;

(2) 

$t⊡t^{\prime }\ne 0\Rightarrow (t⊡t^{\prime })⊡l=t^{\prime }$ for some $l\in T$.

We will use the following notations:

$A\left(T\right)$: the set of all hyperatom elements of $(T,⊞,⊡,\le )$;

$A^{*}\left(T\right):A\left(T\right)\setminus \left\{0\right\}$;

POS: positive ordered semihyperring;

HOS: hyperatomic ordered semihyperring.

Note that $(T,⊞,⊡,\le )\in POS$ if $0\le t$ for any $t\in T$, and $(T,⊞,⊡,\le )\in HOS$ if $A\left(T\right)=T$.

Recently, Kou et al. [8] studied the concept of the $L_T$ operator and the $L_T$ graph in ordered semihyperrings. Following this preceding work [8], in this paper, we introduce an operator $R_T$ on an ordered semihyperring T. The set $L_T\left(K\right)$ [8] is given by

$$L_T\left(K\right):=\{t\in T|t\le k\mathrm{for}\mathrm{all}k\in K\}.$$

Let $\phi :T⟶T^{\prime }$ be a monomorphism. If $W\subseteq T$, then $L_{T^{\prime }}\left(\phi \left(W\right)\right)=\phi \left(L_T\left(W\right)\right)$. If $(T,\oplus ,\odot ,\le )\in HOS$ and $\left|T\right|>2$, then $L_T\left(\{w,w^{\prime }\}\right)=\left\{0\right\}$ or $L_T\left(\{w,w^{\prime }\}\right)=\varnothing$ for all $w,w^{\prime }\in T$ [8]. $\mathsf{{\rm Y}}$ is an $L_T$ graph of a finite OS, T if $V\left(\mathsf{{\rm Y}}\right)=T$ and for all $t,t^{\prime }\in T$ and $t\ne t^{\prime }$, we have

$$t{t}^{\prime }\in E\left(\mathsf{{\rm Y}}\right)\iff L_T\left(\{t,t^{\prime }\}\right)=\left\{0\right\}.$$

Assume that $\mathsf{{\rm Y}}$ is the $L_T$ graph of T and ${\mathsf{{\rm Y}}}^{\prime }$ is the $L_{T^{\prime }}$ graph of $T^{\prime }$, where $T,T^{\prime }\in OS$. If $\psi$ is an isomorphism from T into $T^{\prime }$, then $\mathsf{{\rm Y}}\cong {\mathsf{{\rm Y}}}^{\prime }$ [8].

3. Main Results

We introduce an operator $R_T$ on an ordered semihyperring $(T,⊞,⊡,\le )$ and work on the relationship between $L_T$ and $R_T$ operators of ordered semihyperrings. Some results relating to the hyperatomic ordered semihyperrings are investigated.

Definition 5.

Let $(T,⊞,⊡,\le )\in OS$ and $K\subseteq T$. The set $R_T\left(K\right)$ is given by

$$R_T\left(K\right):=\{t\in T|k\le t\mathrm{for}\mathrm{all}k\in K\}.$$

If $K=\left\{t\right\}$, we write $R_T\left(\left\{t\right\}\right)=R_T\left(t\right)$.

Example 1.

Assume ${\mathbb{N}}_0=\mathbb{N}\cup \left\{0\right\}$, where $\mathbb{N}$ is the set of natural numbers. Consider the semiring $({\mathbb{N}}_0,+,·)$, where + and · are usual addition and multiplication. Define

$$t⊞s=\{t,s\}\mathrm{and}t⊡s=\{ts,qts\},\mathrm{where}q\in {\mathbb{N}}_0.$$

If ≤ is the natural ordering on ${\mathbb{N}}_0$, then $({\mathbb{N}}_0,⊞,⊡,\le )$ is an ordered semihyperring. Let $x\in {\mathbb{N}}_0$. Then, $R_T\left(x\right)=\{x,x+1,x+2,\cdots \}$. Now, let $\varnothing \ne K\subseteq {\mathbb{N}}_0$. If K is bounded and $m=maxK$, then $R_T\left(K\right)={\displaystyle \bigcap _{\begin{array}{c}x\in K\end{array}}}{R}_T\left(x\right)={\displaystyle \bigcap _{\begin{array}{c}x\in K\end{array}}}\{x,x+1,x+2,\cdots \}=\{m,m+1,m+2,\cdots \}$. If K is not bounded and $t\in {\mathbb{N}}_0$, then there exists $x\in K$ such that $t<x$. So, $t\notin {\displaystyle \bigcap _{\begin{array}{c}z\in K\end{array}}}{R}_T\left(z\right)=R_T\left(K\right)$. Thus, $R_T\left(K\right)=\varnothing$.

Lemma 1.

Let $(T,⊞,⊡,\le )\in OS$ and $\varnothing \ne U,V\subseteq T$. Then:

(1) 

$R_T(\varnothing )=T$;

(2) 

if $U\subseteq V$, then $R_T\left(V\right)\subseteq R_T\left(U\right)$;

(3) 

$R_T\left(R_T\left(U\right)\right)\subseteq R_T\left(U\right)$.

Proof. 

(1) Let $R_T(\varnothing )\ne T$. Then,

$$\exists t\in T\mathrm{s}.\mathrm{t}.,t\notin R_T(\varnothing ).$$

So,

$$\exists x\in \varnothing \mathrm{s}.\mathrm{t}.,x\nleqq t,$$

which is a contradiction. Hence, $R_T(\varnothing )=T$.

(2) Let $t\in R_T\left(V\right)$. Then, $v\le t$ for all $v\in V$. As $U\subseteq V$, we obtain

$$u\le t\mathrm{for}\mathrm{all}u\in U.$$

So, $t\in R_T\left(U\right)$. Therefore, $R_T\left(V\right)\subseteq R_T\left(U\right)$.

(3) Let $t\in R_T\left(R_T\left(U\right)\right)$. Then,

$$t^{\prime }\le t\mathrm{for}\mathrm{all}{t}^{\prime }\in R_T\left(U\right).$$

By definition of $R_T\left(U\right)$, we obtain

$$u\le t^{\prime }\mathrm{for}\mathrm{all}u\in U.$$

As $T\in OS$, we have

$$u\le t\mathrm{for}\mathrm{all}u\in U.$$

So, $t\in R_T\left(U\right)$. Therefore, $R_T\left(R_T\left(U\right)\right)\subseteq R_T\left(U\right)$. □

Lemma 2.

Let $(T,⊞,⊡,\le )\in OS$ and $K\subseteq T$. Then:

(1) 

$R_T\left(K\right)={\displaystyle \bigcap _{\begin{array}{c}x\in K\end{array}}}{R}_T\left(x\right)$;

(2) 

$x\in R_T\left(x\right)$ for all $x\in T$.

Proof. 

(1) We have

$$\begin{array}{cc}{R}_T\left(K\right)\hfill & =\{t\in T|k\le t\mathrm{for}\mathrm{all}k\in K\}\hfill \\ \\ & =\{t\in T|t\in R_T\left(x\right),\mathrm{for}\mathrm{all}x\in K\}\hfill \\ \\ & ={\displaystyle \bigcap _{\begin{array}{c}x\in K\end{array}}}{R}_T\left(x\right).\hfill \end{array}$$

(2) By definition,

$$R_T\left(x\right)=\{t\in T|x\le t\}.$$

As $x\le x$ for all $x\in T$, we obtain $x\in R_T\left(x\right)$. □

The set $L_T\left(K\right)$ [8] is given by

$$L_T\left(K\right):=\{t\in T|t\le k\mathrm{for}\mathrm{all}k\in K\}.$$

Now, we compare the $L_T$ and $R_T$ operators of an ordered semihyperring T.

Proposition 1.

Let $(T,⊞,⊡,\le )\in OS$ and $x\subseteq T$. Then,

$$R_T\left(x\right)=\{t\in T|x\in L_T\left(t\right)\}.$$

Proof. 

Let $x^{\prime }\in R_T\left(x\right)$. Then, $x\le x^{\prime }$, i.e., $x\in L_T\left(x^{\prime }\right)$. So,

$$x\in \{t\in T|x\in L_T\left(t\right)\}.$$

Hence, $R_T\left(x\right)\subseteq \{t\in T|x\in L_T\left(t\right)\}$.

Now, let $u\in \{t\in T|x\in L_T\left(t\right)\}$. Thus, $x\in L_T\left(u\right)$ and so $x\le u$. Hence, $u\in R_T\left(x\right)$. Therefore, $\{t\in T|x\in L_T\left(t\right)\}\subseteq R_T\left(x\right)$. □

Proposition 2.

Let $(T,⊞,⊡,\le )\in OS$. Then,

$$R_T\left(0\right)=T\iff 0\in L_T\left(t\right),\mathrm{for}\mathrm{all}t\in T.$$

Proof. 

By Proposition 1,

$$R_T\left(0\right)=\{t\in T|0\in L_T\left(t\right)\}=\{t\in T|0\le t)\}.$$

Hence,

$$R_T\left(0\right)=T\iff 0\in L_T\left(t\right),\mathrm{for}\mathrm{all}t\in T.$$

□

Corollary 1.

Let $(T,⊞,⊡,\le )\in POS$. Then, $R_T\left(0\right)=T$.

Theorem 1.

Let $(T,⊞,⊡,\le )\in OS$ and $\varnothing \ne K\subseteq T$. Then,

$$K\subseteq L_T\left(R_T\left(K\right)\right).$$

Proof. 

Consider the following situations.

Case 1. $R_T\left(K\right)=\varnothing$.

If $R_T\left(K\right)=\varnothing$, then

$$K\subseteq T=L_T(\varnothing )=L_T\left(R_T\left(K\right)\right).$$

Case 2. $R_T\left(K\right)\ne \varnothing$.

Let $t\in R_T\left(K\right)$. Then, $k\le t$ for all $k\in K$. Take any $k^{\prime }\in K$; then, $k^{\prime }\le t$ for any $t\in R_T\left(K\right)$.

Thus, $k^{\prime }\in L_T\left(t\right)$ for any $t\in R_T\left(K\right)$. Hence,

$$k^{\prime }\in \bigcap _{\begin{array}{c}t\in R_T\left(K\right)\end{array}}{L}_T\left(t\right)=L_T\left(R_T\left(K\right)\right).$$

Therefore, $K\subseteq L_T\left(R_T\left(K\right)\right)$. □

Corollary 2.

Let $(T,⊞,⊡,\le )\in OS$. Then,

$$L_T\left(R_T\left(T\right)\right)=T.$$

Theorem 2.

Let $(T,⊞,⊡,\le )\in OS$ and $\left|T\right|\ge 2$. Then,

$$R_T\left(T\right)=\left\{x\right\}\mathrm{for}\mathrm{some}x\in T\setminus \left\{0\right\}\iff L_T\left(x\right)=T.$$

Proof. 

$\Rightarrow )$ If $R_T\left(T\right)=\left\{x\right\}$ for some $x\in T\setminus \left\{0\right\}$, then, by Corollary 5, we obtain

$$T=L_T\left(R_T\left(T\right)\right)=L_T\left(x\right).$$

$\Leftarrow )$ Conversely, let $L_T\left(x\right)=T$ for some $x\in T\setminus \left\{0\right\}$. Then, $t\le x$ for all $t\in T$. It implies that $x\in R_T\left(t\right)$ for all $t\in T$. Thus,

$$x\in \bigcap _{\begin{array}{c}t\in T\end{array}}{R}_T\left(t\right)=R_T\left(T\right).$$

So, $\left\{x\right\}\subseteq R_T\left(T\right)$. Now, let $u\in R_T\left(T\right)\setminus \left\{x\right\}$. Then, $t\le u$ for all $t\in T$. Thus, $x\le u$. As $x\in R_T\left(u\right)$, we obtain $u\le x$. So, $u=x$, which is a contradiction. Hence, $R_T\left(T\right)=\left\{x\right\}$. □

Theorem 3.

Let $(T,⊞,⊡,\le )\in OS$ and $\left\{K_{\lambda }\right|\lambda \in \mathsf{\Lambda }\}$ be a family of subsets of T. Then,

$$\bigcap _{\begin{array}{c}\lambda \in \mathsf{\Lambda }\end{array}}{R}_T\left(K_{\lambda }\right)=R_T\left(\bigcup _{\begin{array}{c}\lambda \in \mathsf{\Lambda }\end{array}}{K}_{\lambda }\right).$$

Proof. 

Let ${\displaystyle \bigcap _{\begin{array}{c}\lambda \in \mathsf{\Lambda }\end{array}}}{R}_T\left(K_{\lambda }\right)=\varnothing$. As $K_{\lambda }\subseteq {\displaystyle \bigcup _{\begin{array}{c}\lambda \in \mathsf{\Lambda }\end{array}}}{K}_{\lambda }$, by Lemma 1, we obtain

$$R_T\left(\bigcup _{\begin{array}{c}\lambda \in \mathsf{\Lambda }\end{array}}{K}_{\lambda }\right)\subseteq R_T\left(K_{\lambda }\right)$$

for all $\lambda \in \mathsf{\Lambda }$. So,

$$R_T\left(\bigcup _{\begin{array}{c}\lambda \in \mathsf{\Lambda }\end{array}}{K}_{\lambda }\right)\subseteq \bigcap _{\begin{array}{c}\lambda \in \mathsf{\Lambda }\end{array}}{R}_T\left(K_{\lambda }\right).$$

Thus, $R_T\left({\displaystyle \bigcup _{\begin{array}{c}\lambda \in \mathsf{\Lambda }\end{array}}}{K}_{\lambda }\right)=\varnothing$.

Now, let ${\displaystyle \bigcap _{\begin{array}{c}\lambda \in \mathsf{\Lambda }\end{array}}}{R}_T\left(K_{\lambda }\right)\ne \varnothing$. Then,

$$\begin{array}{cc}u\in {\displaystyle \bigcap _{\begin{array}{c}\lambda \in \mathsf{\Lambda }\end{array}}}{R}_T\left(K_{\lambda }\right)\hfill & \iff u\in R_T\left(K_{\lambda }\right),\forall \lambda \in \mathsf{\Lambda }\hfill \\ \\ \hfill & \iff v\le u,\forall v\in K_{\lambda },\forall \lambda \in \mathsf{\Lambda }\hfill \\ \\ \hfill & \iff v\le u,\forall v\in {\displaystyle \bigcup _{\begin{array}{c}\lambda \in \mathsf{\Lambda }\end{array}}}{K}_{\lambda }\hfill \\ \\ \hfill & \iff u\in R_T\left({\displaystyle \bigcup _{\begin{array}{c}\lambda \in \mathsf{\Lambda }\end{array}}}{K}_{\lambda }\right)\hfill \end{array}$$

□

Theorem 4.

Let $T,T^{\prime }\in OS$ and $\psi :T⟶T^{\prime }$ be a monomorphism. If $K\subseteq T$, then $R_{T^{\prime }}\left(\psi \left(K\right)\right)=\psi \left(R_T\left(K\right)\right)$.

Proof. 

Let $K\subseteq T$. Then,

$$\begin{array}{cc}x\in R_{T^{\prime }}\left(\psi \left(K\right)\right)\hfill & \iff \psi \left(k\right){\le }_T^{\prime }x,\forall k\in K\hfill \\ \\ \hfill & \iff k{\le }_T{\psi }^{-1}\left(x\right),\forall k\in K\hfill \\ \\ \hfill & \iff {\psi }^{-1}\left(x\right)\in R_T\left(k\right),\forall k\in K\hfill \\ \\ \hfill & \iff {\psi }^{-1}\left(x\right)\in R_T\left(K\right)\hfill \\ \\ \hfill & \iff x\in \psi \left(R_T\left(K\right)\right).\hfill \end{array}$$

Therefore, $R_{T^{\prime }}\left(\psi \left(K\right)\right)=\psi \left(R_T\left(K\right)\right)$. □

Example 2.

Let $T=\{0,t,u,v\}$ and

$$\begin{array}{|ccccc|}\hline ⊞& 0& t& u& v\\ 0& 0& t& u& v\\ t& t& \{t,u\}& u& v\\ u& u& u& \{0,u\}& v\\ v& v& v& v& \{0,v\}\\ \hline\end{array}$$

$$\begin{array}{|ccccc|}\hline ⊡& 0& t& u& v\\ 0& 0& 0& 0& 0\\ t& 0& t& t& t\\ u& 0& u& u& u\\ v& 0& v& v& v\\ \hline\end{array}$$

$$\begin{array}{c}0\le t\le u\le v.\hfill \end{array}$$

Then, $(T,⊞,⊡,\le )\in OS$. We have

$$R_T\left(0\right)=T,$$

$$R_T\left(t\right)=\{t,u,v\},$$

$$R_T\left(u\right)=\{u,v\},$$

$$R_T\left(v\right)=\left\{v\right\},$$

$$R_T\left(\{0,t\}\right)=\{t,u,v\},$$

$$R_T\left(\{0,u\}\right)=\{u,v\},$$

$$R_T\left(\{0,v\}\right)=\left\{v\right\},$$

$$R_T\left(\{t,u\}\right)=\{u,v\},$$

$$R_T\left(\{t,v\}\right)=\left\{v\right\},$$

$$R_T\left(\{u,v\}\right)=\left\{v\right\},$$

$$R_T\left(\{0,t,u\}\right)=\{u,v\},$$

$$R_T\left(\{0,t,v\}\right)=\left\{v\right\},$$

$$R_T\left(\{0,u,v\}\right)=\left\{v\right\},$$

$$R_T\left(\{t,u,v\}\right)=\left\{v\right\},$$

and

$$R_T\left(T\right)=\left\{v\right\}.$$

Clearly, $\mid R_T\left(T\right)\mid =1$ and $L_T\left(R_T\left(T\right)\right)=L_T\left(v\right)=T$.

Theorem 5.

Let $(T,⊞,⊡,\le )\in OS$ and $\mid R_T\left(T\right)\mid \ne 1$. Then, $R_T\left(T\right)=\varnothing$.

Proof. 

Take any $x\in T$; then, by Theorem 2, $L_T\left(x\right)\ne T$. So,

$$\exists a\in T\mathrm{such}\mathrm{that}a\nleqq x.$$

Hence,

$$\exists a\in T\mathrm{such}\mathrm{that}x\notin R_T\left(a\right).$$

Thus,

$$x\notin \bigcap _{\begin{array}{c}b\in T\end{array}}{R}_T\left(b\right)=R_T\left(T\right).$$

Therefore, $R_T\left(T\right)=\varnothing$. □

Example 3.

Let $T=\{0,t,u,v\}$ and

$$\begin{array}{|ccccc|}\hline ⊞& 0& t& u& v\\ 0& 0& t& u& v\\ t& t& t& t& t\\ u& u& t& \{0,u\}& \{0,u,v\}\\ v& v& t& \{0,u,v\}& \{0,v\}\\ \hline\end{array}$$

$$\begin{array}{|ccccc|}\hline ⊡& 0& t& u& v\\ 0& 0& 0& 0& 0\\ t& 0& t& \{0,u\}& 0\\ u& 0& 0& 0& 0\\ v& 0& \{0,v\}& 0& 0\\ \hline\end{array}$$

$$\begin{array}{c}\le :=\left\{\right(x,x):x\in T\}\cup \left\{\right(0,u),(0,v\left)\right\}.\hfill \end{array}$$

Then, $(T,⊞,⊡,\le )\in OS$. We have

$$R_T\left(0\right)=\{0,u,v\},$$

$$R_T\left(t\right)=\left\{t\right\},$$

$$R_T\left(u\right)=\left\{u\right\},$$

$$R_T\left(v\right)=\left\{v\right\},$$

$$R_T\left(\{0,t\}\right)=\varnothing ,$$

$$R_T\left(\{0,u\}\right)=\left\{u\right\},$$

$$R_T\left(\{0,v\}\right)=\left\{v\right\},$$

$$R_T\left(\{t,u\}\right)=\varnothing ,$$

$$R_T\left(\{t,v\}\right)=\varnothing ,$$

$$R_T\left(\{u,v\}\right)=\varnothing ,$$

$$R_T\left(\{0,t,u\}\right)=\varnothing ,$$

$$R_T\left(\{0,t,v\}\right)=\varnothing ,$$

$$R_T\left(\{0,u,v\}\right)=\varnothing ,$$

$$R_T\left(\{t,u,v\}\right)=\varnothing ,$$

and

$$R_T\left(T\right)=\varnothing .$$

Clearly, $L_T\left(R_T\left(T\right)\right)=L_T(\varnothing )=T$.

Let $(T,⊞,⊡,\le )\in OS$ and $\varnothing \ne K\subseteq T$. Define the set

$${\mathcal{E}}_{R_T\left(K\right)}=\{R_T\left(K\right):\varnothing \ne K\subseteq T\}.$$

Clearly, ${\mathcal{E}}_{R_T\left(K\right)}$ is a basis for some topology on T. Indeed, $T={\displaystyle \bigcup _{\begin{array}{c}x\in T\end{array}}}{R}_T\left(x\right)$ and ${\displaystyle \bigcap _{\begin{array}{c}\lambda \in \mathsf{\Lambda }\end{array}}}{R}_T\left(K_{\lambda }\right)=R_T\left({\displaystyle \bigcup _{\begin{array}{c}\lambda \in \mathsf{\Lambda }\end{array}}}{K}_{\lambda }\right)\in {\mathcal{E}}_{R_T\left(K\right)}$.

Denote by ${\tau }_{R_T}\left(T\right)$ the topology generated by ${\mathcal{E}}_{R_T\left(K\right)}$.

Example 4.

In Example 3,

$${\mathcal{E}}_{R_T\left(K\right)}=\{R_T\left(K\right):\varnothing \ne K\subseteq T\}=\{\{0,u,v\},\left\{t\right\},\left\{u\right\},\left\{v\right\},\varnothing \}$$

and

$${\tau }_{R_T}\left(T\right)=\{\varnothing ,T,\{0,u,v\},\left\{t\right\},\left\{u\right\},\left\{v\right\},\{t,u\},\{t,v\},\{u,v\},\{t,u,v\}\}.$$

Thus, $T=\{0,u,v\}\cup \left\{t\right\}$. Therefore, $(T,{\tau }_{R_T}\left(T\right))$ is disconnected.

Theorem 6.

Let $(T,⊞,⊡,\le )\in OS$ and $R_T\left(0\right)=T$. Then, $(T,{\tau }_{R_T}\left(T\right))$ is connected.

Proof. 

For any $x\in T\setminus \left\{0\right\}$, $0\notin R_T\left(x\right)$. Indeed, if $0\in R_T\left(x\right)=\{t\in T\mid x\le t\}=\{t\in T\mid x\in L_T\left(t\right)\}$, then $x\in L_T\left(0\right)=0,$ which is a contradiction. So, if $\left\{0\right\}\ne K\subseteq T$, then $0\notin R_T\left(K\right)$. Now, let $R_T\left(0\right)=T$. Then,

$$\nexists L\in {\tau }_{R_T}\left(T\right)\setminus R_T\left(0\right) \mathrm{such} \mathrm{that} 0\in L.$$

Hence, $(T,{\tau }_{R_T}\left(T\right))$ is connected. □

Corollary 3.

Let $(T,⊞,⊡,\le )\in POS$. Then, $(T,{\tau }_{R_T}\left(T\right))$ is connected.

Example 5.

In Example 2, $(T,{\tau }_{R_T}\left(T\right))$ is connected.

Now, we give some results relating to the hyperatomic ordered semihyperrings (HOS).

Theorem 7.

$(T,⊞,⊡,\le )\in OS$ is hyperatomic if and only if $R_T\left(x\right)=\left\{x\right\}$ for any $x\in T\setminus \left\{0\right\}$.

Proof. 

Necessity. Let $T\in HOS$. Let $x\in A^{*}\left(T\right)$, $R_T\left(x\right)\ne \left\{x\right\}$ and $t\in R_T\left(x\right)\setminus \left\{x\right\}$. Then, $x\le t$. Since $x\in A\left(T\right)$, we obtain $x=0$ or $x=t$, which is a contradiction. Thus, $R_T\left(x\right)=\left\{x\right\}$ for any $x\in T\setminus \left\{0\right\}$.

Sufficiency. Let $R_T\left(x\right)=\left\{x\right\}$ for any $x\in T\setminus \left\{0\right\}$. Let $t\in T$ and $z\le t$. Then, $t\in R_T\left(z\right)$. As $R_T\left(z\right)=\left\{z\right\}$, we have $z\in \{0,t\}$. Thus, $z=0$ or $z=t$. So, $t\in A\left(T\right)$ and hence $T\in HOS$. □

Example 6.

In Example 3, $T\in HOS$ by Theorem 7.

Theorem 8.

Let $(T,⊞,⊡,\le )\in HOS$ and $R_T\left(0\right)=T$. Then:

(1) 

$R_T\left(\{0,t\}\right)=\left\{t\right\}$ for any $t\in T\setminus \left\{0\right\}$;

(2) 

$(T,{\tau }_{R_T}\left(T\right))$ is connected.

Proof. 

(1) Let $t\in T\setminus \left\{0\right\}$. Then,

$$R_T\left(\{0,t\}\right)=R_T\left(0\right)\cap R_T\left(t\right)=T\cap R_T\left(t\right).$$

Since $T\in HOS$, by Theorem 7 we have

$$R_T\left(\{0,t\}\right)=T\cap \left\{t\right\}=\left\{t\right\}.$$

(2) Let $(T,{\tau }_{R_T}\left(T\right))$ be disconnected. Then,

$$\exists \varnothing \ne K,L\subseteq T \mathrm{such} \mathrm{that} K\cap L=\varnothing \mathrm{and} K\cup L=T.$$

Now, let $0\in K$. Then, $R_T\left(0\right)\subseteq K$. Thus, $T\subseteq K$ and so $K=T$. Hence, $L=\varnothing$, a contradiction.

Therefore, $(T,{\tau }_{R_T}\left(T\right))$ is connected. □

Let T be a finite ordered semihyperring. Clearly, $\{R_T\left(x\right)\mid x\in T\}\subseteq {\tau }_{R_T}\left(T\right)$. Let $\varnothing \ne U\in {\mathcal{E}}_{R_T\left(K\right)}$. Then, by Lemma 2, $\varnothing \ne U=R_T\left(K\right)={\displaystyle \bigcap _{\begin{array}{c}x\in K\end{array}}}{R}_T\left(x\right)$. Therefore, $\{R_T\left(x\right)\mid x\in T\}$ is a subbase of ${\tau }_{R_T}\left(T\right)$.

Corollary 4.

Let $(T,⊞,⊡,\le )$ be a finite ordered semihyperring and $T\in HOS$. Then,

$${\mathcal{E}}_{R_T\left(K\right)}=\left\{R_T\left(0\right)\right\}\cup \{\left\{t\right\}\mid t\in T\setminus \left\{0\right\}\}$$

is a subbase of ${\tau }_{R_T}\left(T\right)$.

Proof. 

By Theorem 7, we have

$${\mathcal{E}}_{R_T\left(K\right)}=\left\{R_T\left(0\right)\right\}\cup \{\left\{R_T\left(x\right)\right\}\mid x\in T\setminus \left\{0\right\}\}=\left\{R_T\left(0\right)\right\}\cup \{\left\{x\right\}\mid x\in T\setminus \left\{0\right\}\}.$$

□

Corollary 5.

Let $(T,⊞,⊡,\le )$ be a finite HOS and $K\subseteq T$. Then, $K\in {\tau }_{R_T}\left(T\right)$ if and only if $0\notin K$ or $R_T\left(0\right)\subseteq K$.

Theorem 9.

Let $(T,⊞,⊡,\le )$ be a finite HOS and $K\subseteq T$. Then, K is ${\tau }_{R_T}\left(T\right)$-closed if and only if $0\in K$ or $R_T\left(0\right)\cap K=\varnothing$.

Proof. 

⇒: Let K be ${\tau }_{R_T}\left(T\right)$-closed. Then, $K^c\in {\tau }_{R_T}\left(T\right)$. By Corollary 5, we obtain $0\notin K^c$ or $R_T\left(0\right)\subseteq K^c$. So, $0\in K$ or $R_T\left(0\right)\cap K=\varnothing$.

⇐: Let $0\in K$. Then, $0\notin K^c$. Thus, $\left\{t\right\}\in {\mathcal{E}}_{R_T\left(K^c\right)}$ for all $t\in K^c$. So, $K^c={\displaystyle \bigcup _{\begin{array}{c}t\in K^c\end{array}}}\left\{t\right\}\in {\tau }_{R_T}\left(T\right)$. Hence, K is ${\tau }_{R_T}\left(T\right)$-closed. Now, let $R_T\left(0\right)\cap K=\varnothing$. Then, $R_T\left(0\right)\subseteq K^c$. If $R_T\left(0\right)=K^c$, then clearly $K^c\in {\tau }_{R_T}\left(T\right)$. If $R_T\left(0\right)\ne K^c$, then $\left\{t\right\}\in {\mathcal{E}}_{R_T(K^c\setminus R_T\left(0\right))}$ for all $t\in K^c\setminus R_T\left(0\right)$. Thus, $K^c=\left({\displaystyle \bigcup _{\begin{array}{c}t\in K^c\setminus R_T\left(0\right)\end{array}}}\left\{t\right\}\right)\cup R_T\left(0\right)\in {\tau }_{R_T}\left(T\right)$. Hence, K is ${\tau }_{R_T}\left(T\right)$-closed. □

Theorem 10.

Let $(T,⊞,⊡,\le )$ be a finite HOS. Then, $R_T\left(0\right)=T$ if and only if $(T,{\tau }_{R_T}\left(T\right))$ is connected.

Proof. 

⇒: it follows from Theorem 8.

⇐: Let $(T,{\tau }_{R_T}\left(T\right))$ be connected and $R_T\left(0\right)\ne T$. Then, $T=R_T\left(0\right)\cup (T\setminus R_T\left(0\right))$, which is a contradiction. Thus, $R_T\left(0\right)=T$. □

4. Conclusions

Recently, Kou et al. [8] studied the concept of the $L_T$ operator and $L_T$ graph in ordered semihyperrings. Following this preceding work [8], in this paper, we have introduced an operator $R_T$ on an ordered semihyperring T. We have studied hyperatomic ordered semihyperrings (HOS) in detail. Related properties with respect to $R_T$ and $L_T$ were investigated. Moreover, we proved that if ${\tau }_{R_T}\left(T\right)$ is the topology generated by

$${\mathcal{E}}_{R_T\left(K\right)}=\{R_T\left(K\right):\varnothing \ne K\subseteq T\}$$

and $R_T\left(0\right)=T$, then $(T,{\tau }_{R_T}\left(T\right))$ is connected. We have proved that if $T\in HOS$ and $R_T\left(0\right)=T$, then $(T,{\tau }_{R_T}\left(T\right))$ is connected. Based on the results in this study, we will seek fuzzy hyperatomic ordered semihyperrings in future works. One can further apply these notions on ordered Krasner hyperrings and ordered semihypergroups.