Fuzzy Closed Filters in Bounded BE-Algebras

ABSTRACT

In this paper we use the method of fuzzification of crisp set to fuzzy set as extension of crisp set to fuzzy set. In particular the concepts of fuzzy closed filters of a bounded BE-algebra are introduced; and also some related properties are investigated. As a result a new concept of fuzzy closed filters of a bounded BE-algebras are discussed. Using the idea of fuzzy closed filters of a bounded BE-algebras, some related properties are investigated. In general characterization theorems that related fuzzy closed filters and fuzzy filter are proved. Furthermore, characterization theorem that relate closed filter and fuzzy closed filter in abounded BE-algebra is investigated.

2010 AMS Classification:

• 06A11
• 06A75
• 08A60
• 08A55

Introduction

Kim and Kim (e-2006) introduced a class of BE-algebras and it was observed that these classes of BE-algebra are a generalization of BCK-algebras. The concept of ideals of BE-algebras was introduced by Ahn and So (2008) and then derived various characterizations of ideals, and introduced the notion of ideals of BE-algebras and proved several characterizations of such ideals (Ahn and So 2009).

The concepts of a fuzzy set and a fuzzy relation on a set were initially defined by Zadeh (1965). Song, Jun, and Lee (2010) discussed the fuzzy ideals in BE- algebras. The concept of closed ideals of B- algebras was introduced by Senapati, Bhowmik and they discussed fuzzy closed ideals of B- algebras with interval valued membership function (Senapati, Bhowmik, and Pal 2013). A. Walendziak (e-2008) studied commutative BE-algebras. Kim (2010) discussed a note on BE-algebras. In 2012 fuzzy completely closed ideal of a BH-algebra was discussed by Abbass and Dahham (2012). Dymek and Walendziak (2013) studied fuzzy filters in a BE-algebra. On a fuzzy completely closed filter with respect of an element in a BH-algebra was introduced by Abdalhussein (2017), and Ciloglue and Ceven (2013) was introduced Commutative and Bounded BE- algebras.

The Concepts of Hilbert Implication algebra and generalized Hilbert Implication algebra were introduced in Dejen (2021).

The main objective of this research paper is introducing new idea by investigating closed filters in a bounded BE-algebra with fuzzyfication to fuzzy closed filter in abounded BE-algebra with additional properties. In particular we dealt with the concepts of fuzzy closed filters of BE- algebras with some other additional properties .

Preliminaries

In this paper we used the method introduced in BE-algebra,bounded BE-algebra, filters and fuzzy filters of BE-Algebra with inclusion of closure property.

BE-Algebras

Definition 2.1 (Kim and Kim e-e-2006) An algebra $(X,\star ,1)$ of type $(2,0)$ is called a BE-algebra if it satisfies the following conditions

1. $x\star x=1$, for all $x\in X$

2. $x\star 1=1$, for all $x\in X$

3. $1\star x=x$, for all $x\in X$

4. $x\star (y\star z)=y\star (x\star z)$, for all $x,y,z\in X$.

A relation $\le$ on a BE-algebra $X$ by $x\le y$ if and only if $x\star y=1$ for all $x,y\in X.$

Theorem 2.2 (Kim 2010) Let (X, $\star$, 1) be a BE-algebra. Then we have the following:

(a) $x\star (y\star x)=1,\mathrm{\forall }x,y\in X.$

(b) $x\star ((x\star y)\star y)=1,\mathrm{\forall }x,y\in X.$

Definition 2.3 (Walendziak e-e-2008) A BE – algebra $X$ is called commutative BE- algebra if $(x\star y)\star y=(y\star x)\star x$, for all $x,y\in X$.

Definition 2.4 (Kim and Kim e-e-2006) A BE- algebra $(X,\circ ,1)$ is said to be self-distributive if $x\circ (y\circ z)=(x\circ y)\circ (x\circ z)$, for all $x,y,z\in X$.

Example 2.5 (Kim and Kim e-e-2006) Let $X=\{1,2,3,4,5\}$ be a set with the following table

Table

Hence $(X,\star ,1)$ is a BE – algebra satisfying self- distributivity.

Definition 2.6 (Ahn and So 2008) A BE -algebra $(X,\star ,1)$ is said to be a transitive BE-algebra if it satisfies the condition $y\star z\le (x\star y)\star (x\star z)$ for all $x,y,z\in X.$

Example 2.7 (Mukkamala 2018) Let $X=\{1,a,b,c\}$ be a set. Define a binary operation $\star$ on $X$ as follows

Table

Hence $(X,\star ,1)$ is a transitive BE- algebra.

Proposition 2.8 (Ahn and So 2008) If X is a self- distributive BE-algebra, then it is transitive, but the converse is not true.

Bounded BE – Algebra

Definition 2.9 (Ciloglue and Ceven 2013) Let X be a BE- algebra. If there exists $0$ satisfying $0\le x$ and $0\star x=1$ for all $x\in X$, then the element $0$ is called a unit of X.

A BE- algebra with unit is called a bounded BE- algebra.

Example 2.10 (Ciloglue and Ceven 2013) Let $X=\{1,p,q,r,s,0\}$ be a set with the following table

Table

Hence $(X,\star ,1)$ is a bounded BE- algebra.

Proposition 2.11 Every bounded BE- algebra is a BE- algebra, but the converse may not be true.

Proof. Suppose $X$ be a bounded BE- algebra. Then we want to show that $X$ is a BE- algebra.

Since $X$ is a bounded BE- algebra implies it satisfies the following properties. (BE1) $x\star x=1$, for all $x\in X$

(BE2) $x\star 1=1$, for all $x\in X$

(BE3) $1\star x=x$, for all $x\in X$

(BE4) $x\star (y\star z)=y\star (x\star z)$, for all $x,y,z\in X$

Hence $X$ is a BE- algebra.

To show the converse, we apply the following counter example.

Let $X=\{1,a,b,0\}$ be a set with the following table

Table

Then $(X,\star ,1)$ is a BE- algebra, but it is not a bounded BE- algebra.

Definition 2.12 (Mukkamala 2018) Let X be a bounded BE – algebra. A non-empty subset S of X is said to be a sub algebra of X if $x\star y\in S$ for all $x,y\in S.$

Filters and Fuzzy Filters of BE – Algebras

Definition 2.13 (Kim and Kim e-e-2006) Let $(X,\circ ,1)$ be a BE-algebra and let $F$ be a non-empty subset of $X$. Then $F$ is said to be a filter of $X$ if

$(F_1)1\in F,$

$(F_2)$ If $x\in F$ and $x\circ y\in F$, then $y\in F$.

Example 2.14 Let $X=\{1,a,b,c,d,0\}$ be a set. Define a binary operation $\circ$ on $X$ as follows

Hence $(X,\circ ,1)$ is a BE- algebra.

Let $F_1=\{1,a,b\}$ and $F_2=\{1,a\}$ be a subset of $X$. Then $F_1$ is a filter of $X$ where as $F_2$ is not a filter of $X$, because $a\circ b=a\in F_2$ but $b\notin F_2$.

Definition 2.15 (Mukkamala 2018) A fuzzy set $\eta$ in a BE-algebra X is called a fuzzy filter of X if it satisfies:

$(F{F}_1)$ $\eta (1)\ge \eta (x),$ f or all $x\in X$,

$(F{F}_1)$ $\eta (y)\ge$ min $\{\eta (x),\eta (x\star y)\}$, $\mathrm{\forall }x,y\in X$.

Main Results

Fuzzy Closed Filter of Bounded BE- Algebra

Definition 3.1 Let X be a bounded BE- algebra. A fuzzy set $\eta$ in a bounded BE- algebra X is called a fuzzy closed filter of $X$ if it satisfies the following conditions for all $x,y\in X$:

1. $\eta (1)\ge \eta (x)$, $\mathrm{\forall }x\in X$.

2. $\eta (y)\ge$ min $\{\eta (x),\eta (x\star y)\}$, $\mathrm{\forall }x,y\in X$.

3. $\eta (0\star x)\ge \eta (x)$, $\mathrm{\forall }x\in X$.

Proposition 3.2 Every fuzzy closed filter of a bounded BE – algebra is a fuzzy filter of a BE – algebra, but the converse is not true.

Proof. Suppose that $\eta$ be a fuzzy closed filter of a bounded BE – algebra X. We need to show that $\eta$ is a fuzzy filter of a BE – algebra X.

Let $x\in X$ and $0\in X\Rightarrow 0\star x\in X$, and $\eta$ is a fuzzy closed filter of a bounded BE – algebra X.

$\Rightarrow$ $\eta (0\star x)\ge \eta (x)$.

$\Rightarrow$ $\eta (1)\ge \eta (x)$, since $0\star x=1$.

Hence,

(3.1) $\eta (1)\ge \eta (x).$(3.1)

Let $x,y\in X$ and $\eta$ is a fuzzy closed filter of a bounded BE – algebra X.

Then

(3.2) $\eta (y)\ge min\{\eta (x),\eta (x\star y)\},\mathrm{\forall }x,y\in X.$(3.2)

Hence $\eta$ is a fuzzy filter of a BE – algebra X.

Conversely, suppose $\eta$ be a fuzzy filter of a BE – algebra X. We want to show that $\eta$ is not a fuzzy closed filter of a bounded BE – algebra X.

Assume $\eta$ be a fuzzy filter of a BE – algebra X and $x\in X$.

Now, $\eta (1)\ge \eta (x)$ $\Rightarrow$ $\eta (0\star x)\ge \eta (x)$, since $1=0\star x,\mathrm{\forall }x\in X$ .

It follows that, $0\le x.$ Since $x\in X$ $\Rightarrow 0\in X$.

Which is a contradiction with $0\notin X$. Hence, $\eta (0\star x)\ge \eta (x)$ does not hold.

Therefore, $\eta$ is not a fuzzy closed filter of X.

Definition 3.3 Let X be a bounded BE – algebra and let $\eta$ be a fuzzy closed filter of X. Then ${\eta }_{\alpha }=\{x\in X:\eta (x)\ge \alpha \}$, where $\alpha \in [0,1]$ is said to be a level set of $X$ or $\alpha$ – cut of X.

Theorem 3.4 Let $\eta$ be a fuzzy closed filter of a bounded BE – algebra X if and only if its non-empty level subset ${\eta }_{\alpha }$ is a closed filter of X, for all $\alpha \in [0,1]$.

Proof. Suppose X be a bounded BE – algebra.

Assume that ${\eta }_{\alpha }$ is a closed filter of X, for each $\alpha \in [0,1]$. We need to show that $\eta$ is a fuzzy closed filter of a bounded BE – algebra X.

Let $1\in {\eta }_{\alpha }$ and $x\in {\eta }_{\alpha }$, for each $\alpha \in [0,1]$.

Now, $\eta (1)\ge \alpha$ and $\eta (x)\ge \alpha$, $\alpha \in [0,1]$.

Since $x\le 1\Rightarrow \eta (1)\ge \eta (x)\ge \alpha$

Hence we have

(3.3) $\eta (1)\ge \eta (x),\mathrm{\forall }x\in X$(3.3)

Let $x,y\in X$ be such that $\eta (x)={\alpha }_1$ and $\eta (x\star y)={\alpha }_2$.

Then $x\in {\eta }_{{\alpha }_1}$ and $x\star y\in {\eta }_{{\alpha }_2}$.

Assume ${\alpha }_1\le {\alpha }_2$, which implies that ${\eta }_{{\alpha }_2}\le {\eta }_{{\alpha }_1}$.

Hence, $x\star y\in {\eta }_{{\alpha }_1}$.

Since ${\eta }_{{\alpha }_1}$ is a closed filter, we have $y\in {\eta }_{{\alpha }_1}$. Hence $\eta (y)\ge {\alpha }_1.$

Now we get $\eta (y)\ge {\alpha }_1$= min $\{{\alpha }_1,{\alpha }_2\}$= min $\{\eta (x),\eta (x\star y)\}$

Hence we have

(3.4) $\eta (y)\ge min\{\eta (x),\eta (x\star y)\}$(3.4)

Let $x\in X$ be such that $\eta (x)\ge {\alpha }_1$. Assume that ${\alpha }_1\le {\alpha }_2$.Then $\eta ({\alpha }_2)\le \eta ({\alpha }_1)$. Hence $0\star x\in {\eta }_{{\alpha }_1}\Rightarrow \eta (0\star x)\ge {\alpha }_1.$

since ${\eta }_{{\alpha }_1}$ is a closed filter, we have $0\in {\eta }_{{\alpha }_1}$.

Now, $\eta (0\star x)=\eta (1)\ge \eta (x)\ge {\alpha }_1$, since $0\star x=1$.

Hence,

(3.5) $\eta (0\star x)\ge \eta (x),\mathrm{\forall }x\in X$(3.5)

Therefore $\eta$ is a fuzzy closed filter of a bounded BE – algebra X.

Conversely, suppose $\eta$ be a fuzzy closed filter of a bounded BE – algebra X. We need to show that ${\eta }_{\alpha }$ is a closed filter of a bounded BE – algebra X.

Let $x\in X$, hence $\eta (1)\ge \eta (x)$

$\Rightarrow \eta (1)\ge \eta (x)\ge \alpha$

$\Rightarrow x\in {\eta }_{\alpha }$ and $1\in {\eta }_{\alpha }$.

Hence we have

(3.6) $1\in {\eta }_{\alpha }$(3.6)

Let $x,x\star y\in {\eta }_{\alpha }$. Then $\eta (x)\ge \alpha$ and $\eta (x\star y)\ge \alpha$.

Since $\eta$ is a fuzzy closed filter of X, we have $\eta (y)\ge min\{\eta (x),\eta (x\star y)\}\ge \alpha$.

$\Rightarrow \eta (y)\ge \alpha$

$\Rightarrow y\in {\eta }_{\alpha }.$

Hence we have

(3.7) $y\in {\eta }_{\alpha }$(3.7)

Let $0,x\in {\eta }_{\alpha }$ implies that $\eta (0)\ge \alpha$ and $\eta (x)\ge \alpha$. Since $\eta$ is a fuzzy closed filter of X, we have $\eta (0\star x)\ge \eta (x)\ge \alpha$.

$\Rightarrow \eta (0\star x)\ge \alpha$

$\Rightarrow 0\star x\in {\eta }_{\alpha }.$

Hence we have

(3.8) $0\star x\in {\eta }_{\alpha }$(3.8)

Therefore ${\eta }_{\alpha }$ is a closed filter of a bounded BE – algebra X.

Definition 3.5 Let X be a bounded BE – algebra X. If $\{{\eta }_i|i\in I\}$ is a family of fuzzy subsets of X, then $({\bigcap }_{\begin{array}{c}i\in I\end{array}}{\eta }_i)(x)=inf\{{\eta }_i(x):i\in I\},\mathrm{\forall }x\in X.$

Proposition 3.6 Let $({\eta }_i{)}_{i\in I}$ be an indexed family of a closed filter of a bounded BE – algebra X. Then ${\bigcap }_{\begin{array}{c}i\in I\end{array}}{\eta }_i$ is a fuzzy closed filter of a bounded BE – algebra X.

Proof. Suppose X be a bounded BE – algebra and let $x\in X$. We want to show that ${\bigcap }_{\begin{array}{c}i\in I\end{array}}{\eta }_i$ is a fuzzy closed filter of a bounded BE – algebra X. Since ${\eta }_i$ is a fuzzy closed filter of X, then we have ${\eta }_i(1)\ge {\eta }_i(x)$, for each $i\in I.$

Now $({\bigcap }_{\begin{array}{c}i\in I\end{array}}{\eta }_i)(1)=inf\{{\eta }_i(1):i\in I\}.$

Since ${\eta }_i(1)\ge {\eta }_i(x)$, implies that $inf\{{\eta }_i(1)\}\ge inf\{{\eta }_i(x)\}:i\in I\},\mathrm{\forall }x\in X$.

$\Rightarrow ({\bigcap }_{\begin{array}{c}i\in I\end{array}}{\eta }_i)(1)\ge inf\{{\eta }_i(x):i\in I\},\mathrm{\forall }x\in X.$

Now we get $inf\{{\eta }_i(x):i\in I\}=({\bigcap }_{\begin{array}{c}i\in I\end{array}}{\eta }_i)(x),\mathrm{\forall }x\in X.$

$\Rightarrow ({\bigcap }_{\begin{array}{c}i\in I\end{array}}{\eta }_i)(1)\ge ({\bigcap }_{\begin{array}{c}i\in I\end{array}}{\eta }_i)(x),\mathrm{\forall }x\in X.$ Hence we have

(3.9) $(\bigcap _{\begin{array}{c}i\in I\end{array}}{\eta }_i)(1)\ge (\bigcap _{\begin{array}{c}i\in I\end{array}}{\eta }_i)(x),\mathrm{\forall }x\in X.$(3.9)

Let $x,y,x\star y\in {\bigcap }_{\begin{array}{c}i\in I\end{array}}{\eta }_i$.

Now we get $({\bigcap }_{\begin{array}{c}i\in I\end{array}}{\eta }_i)(y)=inf\{{\eta }_i(y):i\in I\}.$

Since ${\eta }_i(y)\ge min\{{\eta }_i(x\star y),{\eta }_i(x)\}$, because ${\eta }_i$ is a fuzzy closed filter of X.

Which implies that $inf\{{\eta }_i(y)\}\ge inf\{min\{{\eta }_i(x\star y),{\eta }_i(x)\}:i\in I\}\text{break},\mathrm{\forall }x\in X.$

$\begin{array}{rl}& \Rightarrow ({\bigcap }_{\begin{array}{c}i\in I\end{array}}{\eta }_i)(y)\ge inf\{min\{{\eta }_i(x\star y),{\eta }_i(x)\}:i\in I\},\mathrm{\forall }x\in X.\\ & \Rightarrow ({\bigcap }_{\begin{array}{c}i\in I\end{array}}{\eta }_i)(y)\ge min\{inf\{{\eta }_i(x\star y),{\eta }_i(x)\}:i\in I\},\mathrm{\forall }x\in X.\\ & \Rightarrow ({\bigcap }_{\begin{array}{c}i\in I\end{array}}{\eta }_i)(y)\ge min\{inf\{{\eta }_i(x\star y)\},inf\{{\eta }_i(x)\}:i\in I\},\mathrm{\forall }x\in X.\\ & \Rightarrow ({\bigcap }_{\begin{array}{c}i\in I\end{array}}{\eta }_i)(y)\ge min\{({\bigcap }_{\begin{array}{c}i\in I\end{array}}{\eta }_i)(x\star y),({\bigcap }_{\begin{array}{c}i\in I\end{array}}{\eta }_i)(x)\},\mathrm{\forall }x,y\in X.\end{array}$

Hence we have

(3.10) $(\bigcap _{\begin{array}{c}i\in I\end{array}}{\eta }_i)(y)\ge min\{(\bigcap _{\begin{array}{c}i\in I\end{array}}{\eta }_i)(x\star y),(\bigcap _{\begin{array}{c}i\in I\end{array}}{\eta }_i)(x)\},\mathrm{\forall }x,y\in X.$(3.10)

Let $0,x\in X\Rightarrow 0\star x\in X$, hence closurity holds. Assume that $x\in {\bigcap }_{\begin{array}{c}i\in I\end{array}}{\eta }_i$.

Now we have $({\bigcap }_{\begin{array}{c}i\in I\end{array}}{\eta }_i)(0\star x)=inf\{{\eta }_i(0\star x):i\in I\},\mathrm{\forall }x\in X.$

Since ${\eta }_i(0\star x)\ge {\eta }_i(x)$, because ${\eta }_i$ is a fuzzy closed filter of X.

Which implies that $inf\{{\eta }_i(0\star x)\}\ge inf\{{\eta }_i(x)\}:i\in I\},\mathrm{\forall }x\in X.$

Now we get $({\bigcap }_{\begin{array}{c}i\in I\end{array}}{\eta }_i)(0\star x)\ge inf\{{\eta }_i(x)\}:i\in I\},\mathrm{\forall }x\in X.$

$\Rightarrow ({\bigcap }_{\begin{array}{c}i\in I\end{array}}{\eta }_i)(0\star x)\ge ({\bigcap }_{\begin{array}{c}i\in I\end{array}}{\eta }_i)(x),\mathrm{\forall }x\in X.$

Hence we have

(3.11) $(\bigcap _{\begin{array}{c}i\in I\end{array}}{\eta }_i)(0\star x)\ge (\bigcap _{\begin{array}{c}i\in I\end{array}}{\eta }_i)(x),\mathrm{\forall }x\in X.$(3.11)

Therefore ${\bigcap }_{\begin{array}{c}i\in I\end{array}}{\eta }_i$ is a fuzzy closed filter of a bounded BE – algebra X.

Definition 3.7 Let $\eta$ be a fuzzy subset of a bounded BE – algebra X and $\lambda \in [0,1]$. A fuzzy set ${\eta }_{\lambda }:X\to [0,1]$ is defined by ${\eta }_{\lambda }(x)=\eta (x)\bullet \lambda$, for all $x\in X$; where $\bullet$ indicates the multiplication.

Theorem 3.8 Let $\eta$ be a fuzzy subset of a bounded BE – algebra X. Then $\eta$ is a fuzzy closed filter of X if and only if ${\eta }_{\lambda }$ is a fuzzy closed filter of X for each $\lambda \in [0,1]$.

Proof. Let X be abounded BE – algebra.

Suppose $\eta$ be a fuzzy closed filter of X and let $\lambda \in [0,1]$.

We want to show that ${\eta }_{\lambda }$ is a fuzzy closed filter of X for each $\lambda \in [0,1]$.

Now, let $x\in X$ and $\lambda \in [0,1]$.

${\eta }_{\lambda }(1)=\eta (1)\bullet \lambda$ … … . … . by definition $3.3$.

Since $\eta$ is a closed filter of X, implies that $\eta (1)\ge \eta (x),\mathrm{\forall }x\in X.$

Now we get

${\eta }_{\lambda }(1)=\eta (1)\bullet \lambda$

$\Rightarrow {\eta }_{\lambda }(1)\ge \eta (x)\bullet \lambda ,\mathrm{\forall }x\in X.$

$\Rightarrow {\eta }_{\lambda }(1)\ge {\eta }_{\lambda }(x),\mathrm{\forall }x\in X.$

Hence we have

(3.12) ${\eta }_{\lambda }(1)\ge {\eta }_{\lambda }(x),\mathrm{\forall }x\in X.$(3.12)

Let $x,y\in X$ and $\lambda \in [0,1]$.

Now, ${\eta }_{\lambda }(y)=\eta (y)\bullet \lambda$ … … . … . by definition $\mathrm{4.3.4}$.

Since $\eta$ is a closed filter of X, implies that $\eta (y)\ge$ min $\{\eta (x\star y),\eta (x)\},\mathrm{\forall }x,y\in X.$

Now we get ${\eta }_{\lambda }(y)=\eta (y)\bullet \lambda$

$\Rightarrow {\eta }_{\lambda }(y)\ge$ min $\{\eta (x\star y),\eta (x)\}\bullet \lambda ,\mathrm{\forall }x,y\in X.$

$\Rightarrow {\eta }_{\lambda }(y)\ge$ min $\{\eta (x\star y)\bullet \lambda ,\eta (x)\bullet \lambda \},\mathrm{\forall }x,y\in X.$

$\Rightarrow {\eta }_{\lambda }(y)\ge$ min $\{{\eta }_{\lambda }(x\star y),{\eta }_{\lambda }(x)\},\mathrm{\forall }x,y\in X.$

Hence we have

(3.13) ${\eta }_{\lambda }(y)\ge min\{{\eta }_{\lambda }(x\star y),{\eta }_{\lambda }(x)\},\mathrm{\forall }x,y\in X.$(3.13)

Let $0,x\in X$ and $\lambda \in [0,1]$.

Now we get ${\eta }_{\lambda }(0\star x)=\eta (0\star x)\bullet \lambda$ … … . … . by definition $3.11$.

Since $\eta$ is a closed filter of X, implies that $\eta (0\star x)\ge \eta (x),\mathrm{\forall }x\in X.$

Now we have ${\eta }_{\lambda }(0\star x)=\eta (0\star x)\bullet \lambda$

$\Rightarrow {\eta }_{\lambda }(0\star x)\ge \eta (x)\bullet \lambda ,\mathrm{\forall }x\in X.$

$\Rightarrow {\eta }_{\lambda }(0\star x)\ge {\eta }_{\lambda }(x),\mathrm{\forall }x\in X.$

Hence we have

(3.14) ${\eta }_{\lambda }(0\star x)\ge {\eta }_{\lambda }(x),\mathrm{\forall }x\in X.$(3.14)

Therefore ${\eta }_{\lambda }$ is a fuzzy closed filter of a bounded BE – algebra X.

Conversely, suppose X be a bounded BE – algebra and ${\eta }_{\lambda }$ is a fuzzy closed filter of X, for each $\lambda \in [0,1]$.

We want to show that $\eta$ is a fuzzy closed filter of X.

Let $x\in X$, by definition $3.3$ we have ${\eta }_{\lambda }(1)=\eta (1)\bullet \lambda$ .

Now we have $\eta (1)\bullet \lambda ={\eta }_{\lambda }(1)$, since ${\eta }_{\lambda }$ is a fuzzy closed filter of X, we get ${\eta }_{\lambda }(1)\ge {\eta }_{\lambda }(x).$

Now $\eta (1)\bullet \lambda ={\eta }_{\lambda }(1)$

$\Rightarrow \eta (1)\bullet \lambda \ge {\eta }_{\lambda }(x),\mathrm{\forall }x\in X$, since ${\eta }_{\lambda }$ is a fuzzy closed filter of X .

$\Rightarrow \eta (1)\bullet \lambda \ge \eta (x)\bullet \lambda ,\mathrm{\forall }x\in X$ and for each $\lambda \in [0,1]$.

$\Rightarrow \eta (1)\ge \eta (x),\mathrm{\forall }x\in X.$

Hence we have

(3.15) $\eta (1)\ge \eta (x),\mathrm{\forall }x\in X.$(3.15)

Let $x,y\in X\Rightarrow x\star y\in X$.

Since ${\eta }_{\lambda }$ is a fuzzy closed filter of X, we get ${\eta }_{\lambda }(y)\ge$ min $\{\eta (x\star y)\bullet \lambda ,\eta (x)\bullet \lambda \},\mathrm{\forall }x,y\in X$ and for each $\lambda \in [0,1]$.

Since by definition $3.11$ $\eta (y)\bullet \lambda ={\eta }_{\lambda }(y)$.

Now we have $\eta (y)\bullet \lambda ={\eta }_{\lambda }(y)$

$\Rightarrow \eta (y)\bullet \lambda \ge$ mi$\{{\eta }_{\lambda }(x\star y),{\eta }_{\lambda }(x)\},\mathrm{\forall }x,y\in X$ and for each $\lambda \in [0,1]$.

$\Rightarrow \eta (y)\bullet \lambda \ge$ mi$\{\eta (x\star y)\bullet \lambda ,\eta (x)\bullet \lambda \},\mathrm{\forall }x,y\in X$ and for each $\lambda \in [0,1]$.

$\Rightarrow \eta (y)\bullet \lambda \ge$ mi$\{\eta (x\star y),\eta (x)\}\bullet \lambda ,\mathrm{\forall }x,y\in X$ and for each $\lambda \in [0,1]$.

Now we have $\eta (y)\ge$ min $\{\eta (x\star y),\eta (x)\},\mathrm{\forall }x,y\in X$.

Hence we have

(3.16) $\eta (y)\ge min\{\eta (x\star y),\eta (x)\},\mathrm{\forall }x,y\in X.$(3.16)

Let $0,x\in X\Rightarrow 0\star x\in X$.

Since ${\eta }_{\lambda }$ is a fuzzy closed filter of X, we get ${\eta }_{\lambda }(0\star x)\ge {\eta }_{\lambda }(x),\mathrm{\forall }x\in X$.

Since by definition $3.3$ $\eta (0\star x)\bullet \lambda ={\eta }_{\lambda }(0\star x),\mathrm{\forall }x\in X$ and for each $\lambda \in [0,1]$.

Now we have $\eta (0\star x)\bullet \lambda ={\eta }_{\lambda }(0\star x),\mathrm{\forall }x\in X$ and for each $\lambda \in [0,1]$.

$\Rightarrow \eta (0\star x)\bullet \lambda \ge {\eta }_{\lambda }(x),\mathrm{\forall }x\in X$ and for each $\lambda \in [0,1]$.

$\Rightarrow \eta (0\star x)\bullet \lambda \ge \eta (x)\bullet \lambda ,\mathrm{\forall }x\in X$ and for each $\lambda \in [0,1]$.

Now we have $\eta (0\star x)\ge \eta (x),\mathrm{\forall }x\in X.$

Hence we have

(3.17) $\eta (0\star x)\ge \eta (x),\mathrm{\forall }x\in X.$(3.17)

Therefore $\eta$ is a fuzzy closed filter of a bounded BE – algebra X.

Cartesian Product of Fuzzy Closed Filters of a Bounded BE – Algebras

Definition 3.9 Let X be a bounded BE – algebra, and assume that $\eta$ and $\mu$ be two fuzzy subsets on X. Then the Cartesian product of $\eta$ and $\mu$, is denoted by $\eta \times \mu$, and defined by $(\eta \times \mu )(x,y)=$ min $\{\eta (x),\mu (y)\}$, where $\eta \times \mu :X\times X\to [0,1],\text{break}\mathrm{\forall }x,y\in X.$

Proposition 3.10 Let $\eta$ and $\mu$ be two fuzzy closed filters of a bounded BE- algebra X. Then $\eta \times \mu$ is fuzzy closed filters of a bounded BE- algebra $X\times X$.

Definition 3.11 Let $\eta$ and $v$ be a fuzzy subset in a bounded BE – algebra X. Then the strongest fuzzy relation of $\eta$ on $v$ is a fuzzy relation on a bounded BE – algebra X, is denoted by ${\eta }_v$, is defined by ${\eta }_v(x,y)=$ min $\{v(x),v(y)\},\mathrm{\forall }x,y\in X.$

Theorem 3.12 Let $v$ be a fuzzy subset in a bounded BE – algebra X and ${\eta }_v$ strongest fuzzy relation on a bounded BE – algebra X.

If $v$ is a fuzzy closed filter of a bounded BE – algebra X, then ${\eta }_v$ is a fuzzy closed filter of a bounded BE – algebra X.

Proof. Suppose that X is a bounded BE – algebra and let $v$ is a fuzzy closed filter of X.

We need to show that ${\eta }_v$ is a fuzzy closed filter of $X\times X$.

Let $(x,y)\in X\times X$. Then we have ${\eta }_v(1,1)=$ min $\{v(1),v(1)\}$. … . by definition $3.9$.

Since $v$ is a fuzzy closed filter of X, then we get ${\eta }_v(1,1)\ge$ min $\{v(x),v(y)\},\mathrm{\forall }x,y\in X$.

Now we have min $\{v(x),v(y)\}={\eta }_v(x,y),\mathrm{\forall }x,y\in X.$

Hence we have

(3.18) ${\eta }_v(1,1)\ge {\eta }_v(x,y),\mathrm{\forall }x,y\in X.$(3.18)

Let $(x,y),(z,w)\in X\times X$. Then ${\eta }_v(x,y)=$ min $\{v(x),v(y)\}$, by definition $3.11$.

Now by definition $3.11$ we have ${\eta }_v(x,y)=$ min $\{v(x),v(y)\}$

$\Rightarrow {\eta }_v(x,y)\ge$ min $\{$ min $\{v(x\star z),v(z)\}$, min $\{v(y\star w),v(w)\}\},\mathrm{\forall }x,y,z,\text{break}w\in X$, since $v$ is a fuzzy closed filters of X.

$\Rightarrow {\eta }_v(x,y)\ge$ min $\{$ min $\{v(x\star z),v(y\star w)\}$, min $\{v(z),v(w)\}\},\mathrm{\forall }x,y,z,\text{break}w\in X$, by rearranging the equation.

$\Rightarrow {\eta }_v(x,y)\ge$ min $\{{\eta }_v(x\star z,y\star w),{\eta }_v(z,w)\},\mathrm{\forall }x,y,z,w\in X$, by definition $3.11$ .

$\Rightarrow {\eta }_v(x,y)\ge$ min $\{{\eta }_v((x,y)\star (z,w)),{\eta }_v(z,w)\},\mathrm{\forall }x,y,z,w\in X$, by the property of $\star$.

Hence we have

(3.19) ${\eta }_v(x,y)\ge min\{{\eta }_v((x,y)\star (z,w)),{\eta }_v(z,w)\},\mathrm{\forall }x,y,z,w\in X.$(3.19)

Let $(0,0),(x,y)\in X\times X$. Then $(0\star x,0\star y)\in X\times X$.

Now by definition $\mathrm{3.11.}$ we have ${\eta }_v((0,0)\star (x,y))={\eta }_v(0\star x,0\star y),\text{break}\mathrm{\forall }x,y\in X$.

Now we get ${\eta }_v(0\star x,0\star y)=$ min $\{v(0\star x),v(0\star y)\},\mathrm{\forall }x,y\in X$, by definition $3.11$.

${\eta }_v(0\star x,0\star y)\ge$ min $\{v(x),v(y)\},\mathrm{\forall }x,y\in X$, since $v$ is a fuzzy closed filters of X.

Now we have ${\eta }_v(0\star x,0\star y)\ge {\eta }_v(x,y),\mathrm{\forall }x,y\in X$, by definition $3.9$.

$\Rightarrow {\eta }_v((0,0)\star (x,y))\ge {\eta }_v(x,y),\mathrm{\forall }x,y\in X.$

Hence we have

(3.20) ${\eta }_v((0,0)\star (x,y))\ge {\eta }_v(x,y),\mathrm{\forall }x,y\in X.$(3.20)

Therefore ${\eta }_v$ is a fuzzy closed filters of a bounded BE- algebra $X\times X$.

Conclusion

In this paper the concepts of fuzzy closed filters of a bounded BE- algebras are introduced. Different characterization theorems and properties are investigated. Properties of families of fuzzy closed filters of a bounded BE- algebra are elaborated.

The cartesian product in fuzzy closed filters of a bounded BE- algebra with related properties is investigated. The family of intersection in fuzzy closed filters of a bounded BE- algebras is discussed, and also basic properties of family of intersection fuzzy closed filters are investigated.

As future work we will work on completely closed filters of bounded BE-algebra.

Acknowledgments

The authors of this paper would like to thank the referees for their valuable comments for the improvment of the manuscript. We also indicate there is no conflict of interest on this research paper.

Disclosure Statement

No potential conflict of interest was reported by the author(s).

Additional information

Funding

This work is not supported by any organization in any form except the second author supported by as msc student support from Wollo University.