Fuzzy closed ideals of bounded BE-algebras

ABSTRACT

In this paper, we introduce the concepts of fuzzy closed ideals of a bounded BE-algebra and prove basic properties. Using the idea of fuzzy closed ideals of a bounded BE-algebras, we discuss some related properties.

KEYWORDS:

• BE-algebra
• closed ideal
• fuzzy closed ideal,
• subalgebra
• bounded BE-Algebra

1. Introduction

Kim & Kim, (2006) introduced a wide class of BE-algebras. The concept of ideals of BE-algebras was introduced by Sun-shin Ahn, So and Keum-Sook and then derived various characterizations of ideals, and the notion of ideals of BE-algebras and proved several characterizations of such ideals (Sun-Shin & Keum-Sook, 2009). The concepts of a fuzzy set and a fuzzy relation on a set were initially defined by Zadeh, (1965). Song et al., (2010) discussed the fuzzy ideals in BE-algebras. The concept of closed ideals of B-algebras was introduced by Senapati et al., (2011), and they discussed fuzzy closed ideals of B-algebras with interval-valued membership function (Senapati et al., 2011).

Walendziak, (2008) was studied on commutative BE-algebras. Fuzzy completely closed ideal of a BH-algebra was discussed by Abbass & Dahham, (2012).

Ciloglue & Ceven, (2013) were introduced Commutative and Bounded BE-algebras.

Gerima et al., (2021) initiated ideals and filters on implication algebras and Tefera & Oli, (2021) introduced the idea of cartesian product with additional property of fuzzy BI-algebras, and Gerima & Mohammed, (2022) discussed fuzzy closed filters with different characterizations. Abughazalah et al., (2022) discuss Bipolar fuzzy closed BCI-positive implicative ideals and bipolar fuzz closed BCI-Commutative ideas of BCI-algebras, and Muhiuddin, G. Al-Kadi, O., Mahboor A. and Amani Aljohani, contribute ideas on interval valued m-polar fuzzy P-ideals, m-polar fuzzy q-ideals and m-polar fuzzy a-ideals in BCI-algebras (Muhiuddin et al., 2021). Abdi Oli and Gerima Tefera (Oli & Tefera, 2023) elaborate basic ideas of left (right) fuzzy derivations of ideals of d-algebra and different characterization of right (left) fuzzy derivation of ideals of d-algebra.

The introduction of closed ideals in B-algebra (Senapati et al., 2011) and bounded BE-algebra (Ciloglue & Ceven, 2013) motivates us to investigate the concepts of fuzzy closed ideals of bounded BE-algebras with some other additional properties.

2. Preliminaries

Definition 2.1.

(Kim & Kim, 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 ≤ 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 & Kim, 2006)

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

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

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

Definition 2.3.

(Walendziak, 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 & Kim, 2006)

A BE- algebra $(X,\star ,1)$ is said to be self distributive if $x\star (y\star z)=(x\star y)\star (x\star z)$, for all $x,y,z\in X$.

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 $x,y,z\in X.$ (Sun-Shin & Keum-Sook, 2009)

Example 2.5.

(Kim & Kim, 2006)

Let $X=\{1,a,b,c\}$. Define a binary operation $\star$ on X as follows

$\begin{array}{l}\begin{array}{lllll}\star & 1& a& b& c\\ 1& 1& a& b& c\\ a& 1& 1& a& a\\ b& 1& 1& 1& a\\ c& 1& 1& a& 1\end{array}\end{array}$

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

If X is a self-distributive BE-algebra, then it is transitive, but the converse is not true (Sun-Shin & Keum-Sook, 2009).

Definition 2.6.

(Ciloglue & 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.

Proposition 2.7.

(Ciloglue & Ceven, 2013)

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

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

$\begin{array}{l}\begin{array}{lllll}\star & 1& a& b& 0\\ 1& 1& a& b& 0\\ a& 1& 1& a& 0\\ b& 1& 1& 1& 0\\ 0& 1& a& b& 1\end{array}\end{array}$

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

Definition 2.8.

(Ciloglue & Ceven, 2013)

Let X and Y be two bounded BE-algebras. A mapping $f:X⟶Y$ is said to be a bounded homomorphism if $f(0)=0$ and $f(x\star y)=f(x)\star f(y)$ for all $x,y\in X.$

If $f:X⟶Y$ is a bounded homomorphism, then it can be easily observed that $f(0)=0$ if and only if $f(x\star 0)=f(x)\star f(0)=f(x)\star 0$ for all $x\in X.$

Definition 2.9.

(Ciloglue & Ceven, 2013)

Let $s:Y⟶[0,1]$ be a map on a bounded BE-algebra X and $f:X⟶Y$ be a bounded homomorphism. Define a map $s^f(x)=(s\circ f)(x)$ for all $x\in X.$

Definition 2.10.

(Ciloglue & Ceven, 2013)

Let X be a bounded BE-algebra. A non-empty subset S of X is said to be a subalgebra of X if $x\star y\in S$ for all $x,y\in S.$

Definition 2.11.

(Kim & Kim, 2006)

A non-empty subset I of a BE-algebra X is said to be an ideal of X if it satisfies:

• $(I_1)\mathrm{\forall }x\in X$ and $\mathrm{\forall }a\in I$ imply $x\star a\in I$, that means $XI\subseteq I$;

• $(I_1)\mathrm{\forall }x\in X$ and $\mathrm{\forall }a,b\in I$ imply $(a\star (b\star x))\star x\in I$.

Example 2.12.

(Sun-Shin & Keum-Sook, 2009)

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

$\begin{array}{l}\begin{array}{lllllll}\star & 1& a& b& c& d& e\\ 1& 1& a& b& c& d& e\\ a& 1& 1& a& c& c& d\\ b& 1& 1& 1& c& c& c\\ c& 1& a& b& 1& a& b\\ d& 1& 1& a& 1& 1& a\\ e& 1& 1& 1& 1& 1& 1\end{array}\end{array}$

Now, we have $(X,\star ,1)$ is a BE-algebra.

Let $I=\{1,a,b\}$ be a subset of X. Then, I is an ideal of X; but let $J=\{1,a\}$ is not an ideal of X, because $(a\star (a\star b))\star b=(a\star a)\star b=1\star b=b\notin J.$

Definition 2.13.

(Ciloglue & Ceven, 2013)

A non-empty subset I of a bounded BE-algebra X is said to be a closed ideal of X if it satisfies:

• $(I_1)\mathrm{\forall }x\in X$ and $\mathrm{\forall }a\in I$ imply $x\star a\in I$, that means $XI\subseteq I$;

• $(I_1)\mathrm{\forall }x\in X$ and $\mathrm{\forall }a,b\in I$ imply $(a\star (b\star x))\star x\in I$.

• $(I_3)0\star x=1\mathrm{\forall },x\in X$.

Definition 2.14.

(Ciloglue & Ceven, 2013)

Let η be a fuzzy subset in a bounded BE-algebra X. Then, η is called a fuzzy subalgebra of X if $\eta (x\star y)\ge$ min $\{eta(x),\eta (y)\}$ for all $x,y\in X.$

Example 2.15.

(Ciloglue & Ceven, 2013)

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

$\begin{array}{l}\begin{array}{lllll}\star & 1& a& b& 0\\ 1& 1& a& b& 0\\ a& 1& 1& a& a\\ b& 1& 1& 1& a\\ 0& 1& 1& 1& 1\end{array}\end{array}$

Now we have X is a bounded BE-algebra.

Define a fuzzy set $\eta :X⟶[0,1]$ by $\eta (0)=0.7,\eta (a)=\eta (b)=\eta (1)=m\in [0,0.7)$ Then η is a fuzzy subalgebra of X.

Lemma 2.16.

(Gerima & Mohammed, 2022)

Let X be a bounded BE-algebra. If η is a fuzzy subalgebra of X, then for all $x\in X$, $\eta (1)\ge \eta (x)$.

3. Main results

3.1. Fuzzy closed ideals of bounded BE-algebra

Definition 3.1.

(Song et al., 2010) A fuzzy set µ in a BE-algebra X is called a fuzzy ideal of a BE-algebra X if it satisfies

• $(F{I}_1)\mathrm{\forall }x,y\in X,\mu (x\star y)\ge \mu (y)$

• $(F{I}_2)\mathrm{\forall }x,y,z\in X,\mu ((x\star (y\star z))\star z)\ge$ min $\{\mu (x),\mu (y)\}$

Definition 3.2.

Let X be a bounded BE-algebra, and let η be a fuzzy subset of a bounded BE-algebra X. Then, η is called a fuzzy closed ideal of X if the following conditions are satisfied.

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

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

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

Example 3.3.

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

$\begin{array}{l}\begin{array}{lllll}\star & 1& p& q& 0\\ 1& 1& p& q& 0\\ p& 1& 1& p& p\\ q& 1& 1& 1& p\\ 0& 1& 1& 1& 1\end{array}\end{array}$

Then $(X,\star ,1)$ is a bounded BE-algebra and assume that $I_1=\{1,p,0\}$ be an ideal of X.

We define a fuzzy closed ideal $\eta :X⟶[0,1]$ of X by $\eta (0)=0.9$ and $\eta (1)=\eta (p)=m$, where $m\in [0,0.9)$.

Therefore, η is a fuzzy closed ideal of X, where X is a bounded BE-algebra.

Theorem 3.4

Let X be a bounded BE-algebra. Then, every fuzzy closed ideal of X is a fuzzy subalgebra of X.

Proof.

Let X be a bounded BE-algebra. Suppose that η is a fuzzy closed ideal of X.

We want to show that η is a subalgebra of X.

Let $x,y,z\in X$ and η is a fuzzy closed ideal of X. We have $\eta (0\star x)\ge \eta (x)$ and $\eta (x)\ge$ min $\{\eta (x\star y),\eta (y)\}$.

Now we have $\eta (x\star y)\ge$ min $\{\eta ((x\star y)\star (0\star y)),\eta (0\star y)\}$, by definition of fuzzy closed ideal.

Since $x\star (y\star z)=y\star (x\star z)$ we get $\eta (x\star y)\ge$ min $\{\eta (0\star (x\star y)\star y),\eta (0\star y)\}$

Since X is a bounded BE-algebra, we get $\eta (0\star (x\star y)\star y)=\eta (1)$

Now we have $\eta (x\star y)\ge$ min $\{\eta (1),\eta (y)\}$, because $\eta (0)\ge \eta (x)$.

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

$\Rightarrow \eta (x\star y)\ge$ min $\{\eta (x),\eta (y)\},\mathrm{\forall }x,y\in X$, this is the definition of fuzzy subalgebra.

Hence, η is a fuzzy subalgebra of X.

Therefore for any bounded BE-algebra X, every fuzzy closed ideal of X is a fuzzy subalgebra of X.

Lemma 3.5.

Let X be a bounded BE-algebra and η be a fuzzy closed ideal of X. If $x\le y$, then $\eta (x)\ge \eta (y)$, for all $x,y\in X.$

Proof.

Let X be a bounded BE-algebra. Suppose η be a fuzzy closed ideal of X and for all $x,y\in X,x\le y$.

We need to show that $\eta (x)\ge \eta (y)$, for all $x,y\in X$.

Let $x,y\in X$ and η be a fuzzy closed ideal of X. Then we have

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

Since $x\le y\Rightarrow x\star y=1$, by properties of bounded BE-algebra.

Now we get $\eta (x)\ge$ min $\{\eta (1),\eta (y)\}$. Since $\eta (1)\ge \eta (y),\mathrm{\forall }y\in X.$

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

Therefore, we have

(3.1) $x\le y\Rightarrow \eta (x)\ge \eta (y),\mathrm{\forall }x,y\in X.$(3.1)

Proposition 3.6.

Let η be a fuzzy subset of a bounded BE-algebra X. If η is a fuzzy closed ideal of X, then $\eta (0)\ge \eta (x)$,for all $x\in X,.$

Lemma 3.7.

Every fuzzy closed ideal of a bounded BE-algebra X is a fuzzy ideal of a bounded BE-algebra X.

Proof.

Let X be a bounded BE-algebra and suppose that η be a fuzzy closed ideal of X.

We need to show that η is a fuzzy ideal of X.

Since η is a fuzzy closed ideal of X, then by definition 3.1 it satisfies

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

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

From the above 1 and 2, they satisfy the definition of fuzzy ideal of X.

Hence η is a fuzzy ideal of X.

Therefore, every fuzzy closed ideal of a bounded BE-algebra X is a fuzzy ideal of a bounded BE-algebra X.

Theorem 3.8

Let X be a bounded BE-algebra and η be a fuzzy subset of X. Then, η is a fuzzy closed ideal of X if and only if ${\eta }^{\prime }(x)=\eta (x)+1-\eta (0),\mathrm{\forall }x\in X,$ is a fuzzy closed ideal of X.

Proof.

Assume X be a bounded BE-algebra.

Suppose η be a fuzzy closed ideal of X.

We want to show that ${\eta }^{\prime }$ is a fuzzy closed ideal of X.

Since η be a fuzzy closed ideal of X, now we have $\eta (0)\ge \eta (x),\mathrm{\forall }x\in X$ .

Now we have

${\eta }^{\prime }(0)=\eta (0)+1-\eta (0)$, by the definition of ${\eta }^{\prime }$.

$\Rightarrow {\eta }^{\prime }(0)\ge \eta (x)+1-\eta (0),\mathrm{\forall }x\in X,$ since $\eta (0)\ge \eta (x).$

$\Rightarrow {\eta }^{\prime }(0)\ge {\eta }^{\prime }(x),\mathrm{\forall }x\in X$, since $\eta (x)+1-\eta (0)={\eta }^{\prime }(x),\mathrm{\forall }x\in X.$

Hence we have

(3.2) ${\eta }^{\prime }(0)\ge {\eta }^{\prime }(x),\mathrm{\forall }x\in X$(3.2)

Let $x\in X$ and since η is a fuzzy closed ideal of X, now we have

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

Now we get ${\eta }^{\prime }(x)=\eta (x)+1-\eta (0)$, by the definition of ${\eta }^{\prime }$.

$\Rightarrow {\eta }^{\prime }(x)\ge$ min $\{\eta (x\star y),\eta (y)\}+1-\eta (0),\mathrm{\forall }x,y\in X$.

$\Rightarrow {\eta }^{\prime }(x)\ge$ min $\{\eta (x\star y)+1-\eta (0),\eta (y)+1-\eta (0)\},\mathrm{\forall }x,y\in X.$

$\Rightarrow {\eta }^{\prime }(x)\ge$ min $\{{\eta }^{\prime }(x\star y),{\eta }^{\prime }(y)\},\mathrm{\forall }x,y\in X$, since $\eta (x\star y)+1-\eta (0)={\eta }^{\prime }(x\star y)$ and $\eta (y)+1-\eta (0)={\eta }^{\prime }(y).$

Hence we have

(3.3) ${\eta }^{\prime }(x)\ge min\{{\eta }^{\prime }(x\star y),{\eta }^{\prime }(y)\},\mathrm{\forall }x,y\in X.$(3.3)

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

Since η is a fuzzy closed ideal of X, now, we have $\eta (0\star x)\ge \eta (x),\mathrm{\forall }x\in X$ .

Now we get ${\eta }^{\prime }(0\star x)=\eta (0\star x)+1-\eta (0)$, by the definition of ${\eta }^{\prime }$.

$\Rightarrow {\eta }^{\prime }(0\star x)\ge \eta (x)+1-\eta (0),\mathrm{\forall }x\in X$, since $\eta (0\star x)\ge \eta (x)$

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

Hence, we have

(3.4) ${\eta }^{\prime }(0\star x)\ge {\eta }^{\prime }(x),\mathrm{\forall }x\in X.$(3.4)

Therefore ${\eta }^{\prime }$ is a fuzzy closed ideal of X.

Conversely, suppose ${\eta }^{\prime }$ be a fuzzy closed ideal of X.

We want to show that η is a fuzzy closed ideal of X.

Since ${\eta }^{\prime }(x)=\eta (x)+1-\eta (0)\Rightarrow \eta (x)={\eta }^{\prime }(x)-1+\eta (0),\mathrm{\forall }x\in X$. Now ${\eta }^{\prime }$ is a fuzzy closed ideal of X, then we have

$\eta (0)={\eta }^{\prime }(0)-1+\eta (0)$, by the definition of η.

$\Rightarrow \eta (0)\ge {\eta }^{\prime }(x)-1+\eta (0),\mathrm{\forall }x\in X,$ since ${\eta }^{\prime }(0)\ge {\eta }^{\prime }(x).$

$\Rightarrow \eta (0)\ge \eta (x),\mathrm{\forall }x\in X$, since ${\eta }^{\prime }(x)-1+\eta (0)=\eta (x),\mathrm{\forall }x\in X.$

Hence, we have

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

Let $x,y\in X$ and since ${\eta }^{\prime }$ be a fuzzy closed ideal of X, now we have

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

Now we get $\eta (x)={\eta }^{\prime }(x)-1+\eta (0)$, by the definition of η.

$\Rightarrow \eta (x)\ge$ min $\{{\eta }^{\prime }(x\star y),{\eta }^{\prime }(y)\}-1+\eta (0),\mathrm{\forall }x,y\in X$.

$\Rightarrow \eta (x)\ge$ min $\{{\eta }^{\prime }(x\star y)-1+\eta (0),{\eta }^{\prime }(y)-1+\eta (0)\},\mathrm{\forall }x,y\in X.$

$\Rightarrow \eta (x)\ge$ min $\{\eta (x\star y),\eta (y)\},\mathrm{\forall }x,y\in X$, since ${\eta }^{\prime }(x\star y)-1+\eta (0)=\eta (x\star y)$ and ${\eta }^{\prime }(y)-1+\eta (0)=\eta (y).$

Hence, we have

(3.6) $\eta (x)\ge min\{\eta (x\star y),\eta (y)\},\mathrm{\forall }x,y\in X.$(3.6)

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

Since ${\eta }^{\prime }$ is a fuzzy closed ideal of X, now we have ${\eta }^{\prime }(0\star x)\ge {\eta }^{\prime }(x),\mathrm{\forall }x\in X$ .

Now we get $\eta (0\star x)={\eta }^{\prime }(0\star x)-1+\eta (0)$, by the definition of η.

$\Rightarrow \eta (0\star x)\ge {\eta }^{\prime }(x)-1+\eta (0),\mathrm{\forall }x\in X$, since ${\eta }^{\prime }(0\star x)\ge {\eta }^{\prime }(x)$

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

Hence, we have

(3.7) $\eta (0\star x)\ge \eta (x),\mathrm{\forall }x\in X.$(3.7)

Therefore, η is a fuzzy closed ideal of X.

Definition 3.9.

Let X be a bounded BE-algebra, and η be a fuzzy closed ideal of X. Then, the set ${\eta }_{\alpha }=\{x\in X:\eta (x)\ge \alpha \}$, for all $\alpha \in [0,1]$, is said to be a level set of X.

Theorem 3.10

η is a fuzzy closed ideal of a bounded BE-algebra X if and only if its non-empty level subset η_α is a closed ideal of X, for all $\alpha \in [0,1]$.

Suppose X be a bounded BE-algebra.

Assume η_α be a closed ideal of X, for each $\alpha \in [0,1]$.

We need to show that η is a fuzzy closed ideal of X.

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

Since $0\le x\Rightarrow \eta (0)\ge \eta (x)\ge \alpha$

Hence we have,

(3.8) $\eta (0)\ge \eta (x),\mathrm{\forall }x\in X$(3.8)

Let $x\star y\in {\eta }_{{\alpha }_1},y\in {\eta }_{{\alpha }_2}$ and $x\in {\eta }_{{\alpha }_3}$ be such that $\eta (x\star y)\ge {\alpha }_1$, $\eta (y)\ge {\alpha }_2$ and $\eta (x)\ge {\alpha }_3$ .

Then, consider the following cases

1. ${\alpha }_1\le {\alpha }_2\le {\alpha }_3$

2. ${\alpha }_2\le {\alpha }_1\le {\alpha }_3$

3. ${\alpha }_3\le {\alpha }_1\le {\alpha }_2$

4. ${\alpha }_3\le {\alpha }_2\le {\alpha }_1$

5. ${\alpha }_1\le {\alpha }_3\le {\alpha }_2$

6. ${\alpha }_2\le {\alpha }_3\le {\alpha }_1$

Now, we apply the above cases, we get the following

1. max$\{{\alpha }_1,{\alpha }_2\}\le {\alpha }_3\le \eta (x)$

$\Rightarrow$ max $\{\eta (x\star y),\eta (y)\}\le \eta (x)$

$\Rightarrow \eta (x)\ge$ max $\{\eta (x\star y),\eta (y)\}$, since max $$\begin{gathered} \{\eta (x\star y), \\ \eta (y)\}\ge \end{gathered}$$ min $\{\eta (x\star y),\eta (y)\}$

Hence we have

(3.9) $\eta (x)\ge min\{\eta (x\star y),\eta (y)\}$(3.9)

2. max$\{{\alpha }_2,{\alpha }_1\}\le {\alpha }_3\le \eta (x)$

$\Rightarrow$ max $\{\eta (y),\eta (x\star y)\}\le \eta (x)$

$\Rightarrow \eta (x)\ge$ max $\{\eta (y),\eta (x\star y)\}$, since max $\{\eta (y),\eta (x\star y)\}\ge$ min $\{\eta (y),\eta (x\star y)\}$

Hence we have

(3.10) $\eta (x)\ge min\{\eta (y),\eta (x\star y)\}$(3.10)

3. max$\{{\alpha }_3,{\alpha }_1\}\le {\alpha }_2\le \eta (y)$

$\Rightarrow$ max $\{\eta (x),\eta (x\star y)\}\le \eta (y)$

$\Rightarrow \eta (y)\ge$ max $\{\eta (x),\eta (x\star y)\}$, since max $\{\eta (x),\eta (x\star y)\}\ge$ min $\{\eta (x),\eta (x\star y)\}$

Hence we have

(3.11) $\eta (y)\ge min\{\eta (x),\eta (x\star y)\}$(3.11)

4. max$\{{\alpha }_3,{\alpha }_2\}\le {\alpha }_1\le \eta (x\star y)$

$\Rightarrow$ max $\{\eta (x),\eta (y)\}\le \eta (x\star y)$

$\Rightarrow \eta (x\star y)\ge$max$\{\eta (x),\eta (y)\}$, since max $$\begin{gathered} \{\eta (x), \\ \eta (y)\}\ge \end{gathered}$$ min $\{\eta (x),\eta (y)\}$

Hence we have

(3.12) $\eta (x\star y)\ge min\{\eta (x),\eta (y)\}$(3.12)

5. max$\{{\alpha }_1,{\alpha }_3\}\le {\alpha }_2\le \eta (y)$

$\Rightarrow$ max $\{\eta (x\star y),\eta (x)\}\le \eta (y)$

$\Rightarrow \eta (y)\ge$ max $\{\eta (x\star y),\eta (x)\}$, since max $\{\eta (x\star y),\eta (x)\}\ge$ min $\{\eta (x\star y),\eta (x)\}$

Hence, we have

(3.13) $\eta (y)\ge min\{\eta (x\star y),\eta (x)\}$(3.13)

6. max$\{{\alpha }_2,{\alpha }_3\}\le {\alpha }_1\le \eta (x\star y)$

$\Rightarrow$ max $\{\eta (y),\eta (x)\}\le \eta (x\star y)$

$\Rightarrow \eta (x\star y)\ge$max$\{\eta (y),\eta (x)\}$, since max $$\begin{gathered} \{\eta (y),\eta (x)\} \\ \ge \end{gathered}$$ min $\{\eta (y),\eta (x)\}$

Hence, we have

(3.14) $\eta (x\star y)\ge min\{\eta (y),\eta (x)\}$(3.14)

Therefore, we get

(3.15) $\eta (x)\ge min\{\eta (x\star y),\eta (y)\}.$(3.15)

Let $0\in {\eta }_{\alpha }$ and $x\in {\eta }_{\alpha }$ 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 ideal, 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,

(3.16) $\eta (0\star x)\ge \eta (x),\mathrm{\forall }x\in X.$(3.16)

Therefore, η is a fuzzy closed ideal of a bounded BE-algebra X.

Conversely, suppose η be a fuzzy closed ideal of a bounded BE-algebra X. We need to show that η_α is a closed ideal of a bounded BE-algebra X.

Let $x\in X$ and since η is a closed ideal of X. Hence we have $\eta (0)\ge \eta (x)$

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

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

Hence, we have

(3.17) $0\in {\eta }_{\alpha }$(3.17)

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

$\Rightarrow \eta (y)\ge {\alpha }_1$ and $\eta (x\star y)\ge {\alpha }_2$.

Assume ${\alpha }_1\le {\alpha }_2$, then we have min $\{\eta (x\star y),\eta (y)\}\ge \alpha$, where $\alpha =$min$\{{\alpha }_1,{\alpha }_2\}$. $\Rightarrow \eta (x)\ge \alpha$, since η be a fuzzy closed ideal of a bounded BE-algebra X.

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

Hence, we have

(3.18) $x\in {\eta }_{\alpha }$(3.18)

Let $0,x\in {\eta }_{\alpha }$ implies that $\eta (0)\ge \alpha$ and $\eta (x)\ge \alpha$. Since η is a fuzzy closed ideal 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

(3.19) $0\star x\in {\eta }_{\alpha }$(3.19)

Therefore, η_α is a closed ideal of a bounded BE-algebra X.

Definition 3.11.

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

Proposition 3.12.

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

Proof.

Suppose X be a bounded BE-algebra.

Assume that $({\eta }_i{)}_{i\in I}$ be an indexed family of a fuzzy closed ideal of X and let $x\in X$.

We want to show that $\bigcap _{\begin{array}{l}i\in I\end{array}}{\eta }_i$ is a fuzzy closed ideal of X.

Since η_i is a fuzzy closed ideal of X, then we have ${\eta }_i(0)\ge {\eta }_i(x)$, for each $i\in I.$

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

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

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

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

Hence, we have

(3.20) $(\bigcap _{\begin{array}{l}i\in I\end{array}}{\eta }_i)(0)\ge (\bigcap _{\begin{array}{l}i\in I\end{array}}{\eta }_i)(x),\mathrm{\forall }x\in X.$(3.20)

Let $x,y\in X$, implies that $x\star y\in X$.

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

Since ${\eta }_i(x)\ge$ min $\{{\eta }_i(x\star y),{\eta }_i(y)\}$, because η_i is a fuzzy closed ideal of X, which implies that inf $\{{\eta }_i(x)\}\ge$ inf $\{$min $\{{\eta }_i(x\star y),{\eta }_i(y)\}:i\in I\},\mathrm{\forall }x\in X.$

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

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

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

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

Hence, we have

(3.21) $\begin{array}{ll}(\bigcap _{\begin{array}{l}i\in I\end{array}}{\eta }_i)(x)\ge & min\{(\bigcap _{\begin{array}{l}i\in I\end{array}}{\eta }_i)(x\star y),(\bigcap _{\begin{array}{l}i\in I\end{array}}{\eta }_i)(y):i\in I\},\\ & \mathrm{\forall }x,y\in X.\end{array}$(3.21)

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

Now, we have $(\bigcap _{\begin{array}{l}i\in I\end{array}}{\eta }_i)(0\star x)=$ 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 η_i is a fuzzy closed ideal 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$(\bigcap _{\begin{array}{l}i\in I\end{array}}{\eta }_i)(0\star x)\ge$ inf $$\begin{gathered} \{{\eta }_i(x)\}:i\in I\}, \\ \mathrm{\forall }x\in X. \end{gathered}$$

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

Hence, we have

(3.22) $(\bigcap _{\begin{array}{l}i\in I\end{array}}{\eta }_i)(0\star x)\ge (\bigcap _{\begin{array}{l}i\in I\end{array}}{\eta }_i)(x),\mathrm{\forall }x\in X.$(3.22)

Therefore, $\bigcap _{\begin{array}{l}i\in I\end{array}}{\eta }_i$ is a fuzzy closed ideal of a bounded BE-algebra X.

3.2. Cartesian product of fuzzy closed ideals of a bounded BE-algebras

Definition 3.13.

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

Theorem 3.14

Let η and µ be two fuzzy closed ideals of a bounded BE-algebra X. Then $\eta \times \mu$ is fuzzy closed ideals of a bounded BE-algebra X × X.

Proof.

Assume X be a bounded BE-algebra and let η and µ be two fuzzy closed ideals of X.

We want to show that $\eta \times \mu$ is fuzzy closed ideals of a bounded BE-algebra X × X.

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

Then, we have

$\begin{array}{ll}(\eta \times \mu )(0,0)=& min\{\eta (0),\mu (0)\}\\ \ge & min\{\eta (x),\mu (y)\},\text{by definition 3.3.1.}\end{array}$

Hence, $(\eta \times \mu )(0,0)\ge$ min $\{\eta (x),\mu (y)\}$ , since η and µ be two fuzzy closed ideals of X.

Now, by definition 3.2. Hence, we have

(3.23) $(\eta \times \mu )(0,0)\ge (\eta \times \mu )(x,y),\mathrm{\forall }x,y\in X.$(3.23)

Let $(x,y),(z,w)\in X\times X$. Then $(\eta \times \mu )(x,y)=$ min $\{\eta (x),\mu (y)\}$, by definition 3.3.1.

Now, by definition 3.3.1 we have $(\eta \times \mu )(x,y)=$ min $\{\eta (x),\mu (y)\}$

$\Rightarrow (\eta \times \mu )(x,y)\ge min\{min\{\eta (x\star z),\eta (z)\},min\{\mu (y\star w),\mu (w)\}\},\mathrm{\forall }x,y,z,w\in X$, since η and µ be two fuzzy closed ideals of X.

$(\eta \times \mu )(x,y)\ge min\{min\{\eta (x\star z),\mu (y\star w)\},min\{\eta (z),\mu (w)\}\},\mathrm{\forall }x,y,z,w\in X$, by rearrange the equation.

$$\begin{gathered} (\eta \times \mu )(x,y)\ge min\{(\eta \times \mu )(x\star z,y\star w),(\eta \times \mu ) \\ (z,w)\},\mathrm{\forall }x,y,z,w\in X \end{gathered}$$, by definition 3.2.

$$\begin{gathered} (\eta \times \mu )(x,y)\ge min\{(\eta \times \mu ) \\ ((x,y)\star (z,w)),(\eta \times \mu )(z,w)\},\mathrm{\forall }x,y,z,w\in X \end{gathered}$$, by the property of $\star$.

Hence, we have

(3.24) $\begin{array}{ll}(\eta \times \mu )(x,y)\ge & min\{(\eta \times \mu )((x,y)\star (z,w)),(\eta \times \mu )(z,w)\},\\ & \mathrm{\forall }x,y,z,w\in X.\end{array}$(3.24)

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

Now, by definition 3.3.1 we have $(\eta \times \mu )((0,0)\star (x,y))=(\eta \times \mu )(0\star x,0\star y),\mathrm{\forall }x,y\in X$.

Now, we get $(\eta \times \mu )(0\star x,0\star y)=min\{\eta (0\star x),\mu (0\star y)\},\mathrm{\forall }x,y\in X$, by definition 3.3.1.

$(\eta \times \mu )(0\star x,0\star y)\ge min\{\eta (x),\mu (y)\},\mathrm{\forall }x,y\in X$, since η and µ be two fuzzy closed ideals of X.

Now, we have $(\eta \times \mu )(0\star x,0\star y)\ge (\eta \times \mu )(x,y),\mathrm{\forall }x,y\in X$, by definition 3.3.1. $\Rightarrow (\eta \times \mu )((0,0)\star (x,y))\ge (\eta \times \mu )(x,y),\mathrm{\forall }x,y\in X$.

Hence, we have

(3.25) $(\eta \times \mu )((0,0)\star (x,y))\ge (\eta \times \mu )(x,y),\mathrm{\forall }x,y\in X.$(3.25)

Therefore, from the above equation (3.3.1), (3.3.2) and (3.3.3), $\eta \times \mu$ is a fuzzy closed ideals of a bounded BE-algebra X × X.

4. Conclusion

In this paper, we introduce the concepts of fuzzy closed ideals of a bounded BE-algebras. Different characterization theorems and properties are investigated. We discuss Properties of families of fuzzy closed ideals of a bounded BE-algebra .

The cartesian product in fuzzy closed ideals of a bounded BE-algebra is investigated with related properties. We also discuss the family of intersection of fuzzy closed ideals of a bounded BE-algebras, and basic properties of family of intersection of fuzzy closed ideals are investigated. As future work, it is possible to work on completely closed ideals of bounded BE-algebra.

Authors Contribution

All authors contributed to the study of concepts and design material preparation, data collection and analysis prepared by Gerima Tefera and Mohammed Adem.

Data statment

This manuscript is our original work, and the material that used to work this research is included by citation. Again, this manuscript does not contain human objects and no data about animal objects involved.

Financial Interest

The authors (1, and 2) have no relevant financial or non-financial interest to disclose. The authors declared that there is no conflict of interest between authors and no other person is involved in doing this research.

Acknowledgements

The authors of this paper would like to thank the referees for their valuable comments for the improvement of the manuscript.

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Additional information

Funding

The authors do not receive fund from any organization in any form but the second author recieved Student support from Wollo University Research and community support as Msc students.