Exploring Star Filters of Almost Distributive Lattices

Abstract

In an almost distributive lattice (ADL), we have presented and explored the notions of star filters and starlets. Further, we have characterized star filters through their starlets. A set of equivalent conditions is established for a filter in an ADL to become a star filter. Additionally, this paper investigates the topological properties of the prime spectrum associated with star filters in an ADL.

1. Introduction

Boolean algebra, a key branch of abstract algebra, was introduced by George Boole in his notable work “The Mathematical Analysis of Logic” (1847). The emergence of lattice theory initially faced opposition, especially after R. Dedekind’s influential contributions. Dedekind’s two foundational papers, regarded as classics, established the framework for lattice theory and profoundly impacted later studies. In these works, he defined modular and distributive lattices and investigated their essential properties, emphasizing their relevance in various contexts. Among the developments in lattice theory, distributive lattices have been particularly significant. These lattices possess unique and intriguing properties that are not generally observed in other types of lattices, making them essential in both theoretical and practical domains. In [1], Speed studied the properties of annihilator ideals in lattices and Mandelker in [2] established the properties of relative annihilators in lattices. Swamy and Rao [3] introduced an almost distributive lattice (ADL) that established a connection between ideals in ADLs and distributive lattices. Further, they proved that an ADL’s collection of principal ideals, $\mathcal{PI}(\mathcal{R})$, forms a distributive lattice. In [4], the authors derived some additional properties of prime, minimal prime, and annihilator ideals in an ADL. The study of neutrosophic ideals and neutrosophic filters within the context of topology has been explored in the work of Agarwal [5]. Zhang [6] developed the theoretical framework for filters and quotient algebras in inflationary general residuated lattices. Noumen [7] investigated the concept of ideals in triangle algebras and examined their relationship with filters. The introduction of the $\alpha$-multiplier concept for almost distributive lattices, along with an analysis of its properties, was presented by Ying Wang [8]. Additionally, Morton [9] analyzed the relation between prime filter completions and canonical extensions. In [10], the concepts of $\mathcal{E}\text{-}$ideals and prime $\mathcal{E}\text{-}$ideals are introduced in an ADL and their properties are studied. Later, in [11], the concept of $\nu \text{-}$ideals is introduced in an ADL and their properties are discussed. In this paper, we introduce the concepts of star filters and starlets in almost distributive lattices (ADLs). It establishes that the set of star filters in an ADL forms a distributive lattice, and further it proves that the starlets form a sublattice within ADLs. This paper provides an element-wise characterization of star filters and identifies several conditions under which an ideal that is an ADL becomes a star filter. A congruence relation is defined on ADLs, along with conditions that ensure the quotient lattice structure that becomes a Boolean algebra.

Henriksen and Jerison [12] explored the structure of minimal prime ideals within a commutative ring, building on earlier work by Kist [13]. They were able to derive sufficient conditions under which these spaces would exhibit compactness. Inspired by this work, Speed [1,14] turned their attention to the study of minimal prime ideals in distributive lattices containing a zero element. The framework provided by lattice theory allowed Speed [14] to achieve significantly more comprehensive results, including an elegant characterization of the conditions under which the space of minimal prime ideals in a zero-distributive lattice becomes compact. These developments, combined with earlier findings by the authors [15] on the properties of minimal prime ideals, provide a strong foundation for undertaking an in-depth study of the fundamental properties of prime star filters and give equivalent conditions for when the prime spectrum of star filters in an ADL can be considered a Hausdorff space. Finally, a necessary and sufficient condition is given for the prime spectrum to attain the structure of a regular space.

2. Preliminaries

The definitions and important results from [3,16] are collected and given in this section. These concepts will be needed throughout this paper. We use $\mathcal{R}$ to represent an ADL with maximal elements.

Definition 1

([3]). An algebra $(\mathcal{R},\vee ,\wedge ,0)$ of type (2,2,0) satisfying the following properties is an almost distributive lattice (ADL) with zero:

(1) 

$(\theta \vee \vartheta )\wedge \ae =(\theta \wedge \ae )\vee (\vartheta \wedge \ae )$;

(2) 

$\theta \wedge (\vartheta \vee \ae )=(\theta \wedge \vartheta )\vee (\theta \wedge \ae )$;

(3) 

$(\theta \vee \vartheta )\wedge \vartheta =\vartheta$;

(4) 

$(\theta \vee \vartheta )\wedge \theta =\theta$;

(5) 

$\theta \vee (\theta \wedge \vartheta )=\theta$;

(6) 

$0\wedge \theta =0,$   for any $\theta ,\vartheta ,\ae \in \mathcal{R}.$

Let $\mathcal{B}$ be a non-empty set. It is possible to structure $\mathcal{B}$ as an almost distributive lattice (ADL) by selecting an element ${\mu }_0\in \mathcal{B}$ and defining the binary operations ∨ and ∧ on $\mathcal{B}$ by

$$\mu \vee \pi =\left\{\begin{array}{c}\mu \mathrm{if}\mu \ne {\mu }_0\hfill \\ \pi \mathrm{if}\mu ={\mu }_0\hfill \end{array}\right.\mu \wedge \pi =\left\{\begin{array}{c}\pi \mathrm{if}\mu \ne {\mu }_0\hfill \\ {\mu }_0\mathrm{if}\mu ={\mu }_0.\hfill \end{array}\right.$$

The structure $(\mathcal{B},\vee ,\wedge ,{\mu }_0)$, where ${\mu }_0$ represents the zero element, is an example of a discrete ADL. For an ADL $(\mathcal{R},\vee ,\wedge ,0)$, a partial order ≤ can be established on $\mathcal{R}$ by defining $\theta \le \vartheta$ for any $\theta ,\vartheta \in \mathcal{R}$ if and only if $\theta =\theta \wedge \vartheta$ or, equivalently, $\theta \vee \vartheta =\vartheta .$

Theorem 1

([3]). For any $\theta ,\vartheta ,\ae \in \mathcal{R}$, we have the following:

(1) 

$\theta \vee \vartheta =\theta \iff \theta \wedge \vartheta =\vartheta$;

(2) 

$\theta \vee \vartheta =\vartheta \iff \theta \wedge \vartheta =\theta$;

(3) 

$\theta \wedge (\vartheta \wedge \ae )=(\theta \wedge \vartheta )\wedge \ae$;

(4) 

$(\theta \wedge \vartheta )\wedge \ae =(\vartheta \wedge \theta )\wedge \ae$;

(5) 

$(\theta \vee \vartheta )\wedge \ae =(\vartheta \vee \theta )\wedge \ae$;

(6) 

$\theta \wedge \vartheta =0\iff \vartheta \wedge \theta =0$;

(7) 

$\theta \vee (\vartheta \wedge \ae )=(\theta \vee \vartheta )\wedge (\theta \vee \ae )$;

(8) 

$\theta \wedge (\theta \vee \vartheta )=\theta ,(\theta \wedge \vartheta )\vee \vartheta =\vartheta$ and $\theta \vee (\vartheta \wedge \theta )=\theta$;

(9) 

$\theta \le \theta \vee \vartheta$ and $\theta \wedge \vartheta \le \vartheta$;

(10) 

$\theta \wedge \theta =\theta$ and $\theta \vee \theta =\theta$;

(11) 

$0\vee \theta =\theta$ and $\theta \wedge 0=0.$,

(12) 

$\theta \vee \vartheta =(\theta \vee \vartheta )\vee \theta .$

An ADL $\mathcal{R}$ exhibits almost all the characteristics of a distributive lattice, except for the right distributivity of ∨ over $\wedge ,$ the commutativity of $\vee ,$ and the commutativity of $\wedge .$ The satisfaction of any one of these properties ensures that the ADL $\mathcal{R}$ becomes a distributive lattice [17].

Theorem 2

([3]). Let $(\mathcal{R},\vee ,\wedge ,0)$ be an ADL with zero. For any $\theta ,\vartheta ,\ae \in \mathcal{R},$ the following conditions are equivalent:

1. 

$(\mathcal{R},\vee ,\wedge ,0)$ is a distributive lattice;

2. 

$\theta \vee \vartheta =\vartheta \vee \theta$;

3. 

$\theta \wedge \vartheta =\vartheta \wedge \theta$;

4. 

$(\theta \wedge \vartheta )\vee \ae =(\theta \vee \ae )\wedge (\vartheta \vee \ae )$;

5. 

The poset $(\mathcal{R},\le )$ is directed above.

As usual, an element $m\in \mathcal{R}$ is referred to as maximal if it is a maximal element in the partially ordered set $(\mathcal{R},\le ).$ In other words, for any $\theta \in \mathcal{R}$, if $m\le \theta$, then $m=\theta .$ We represent ${\mathcal{M}}_{Max.elts}$ as the collection of all such maximal elements within $\mathcal{R}$.

Theorem 3

([3]). Suppose $\mathcal{R}$ is an ADL and $m\in \mathcal{R}.$ The subsequent statements are equivalent:

1. 

m is maximal with respect to ≤;

2. 

$m\vee \theta =m$, for all $\theta \in \mathcal{R}$;

3. 

$m\wedge \theta =\theta$, for all $\theta \in \mathcal{R}$;

4. 

$\theta \vee m$ is maximal, for all $\theta \in \mathcal{R}$.

In [3] by Swamy, it is worth noticing that an ADL denoted as $\mathcal{R}$ exhibits nearly all features of a distributive lattice, apart from the non-commutativity of ∨ and ∧ and the right distributivity of ∨ over ∧. Either of these properties, if present, would classify $\mathcal{R}$ as a distributive lattice.

A non-empty subset $\mathcal{I}$ of $\mathcal{R}$ is called an ideal (respectively, a filter) of $\mathcal{R}$ if $\theta \vee \vartheta ,\theta \wedge \mu \in \mathcal{I}$ (respectively, $\theta \wedge \vartheta ,\mu \vee \theta \in \mathcal{I}$) for all $\theta ,\vartheta \in \mathcal{I}$ and all $\mu \in \mathcal{R}$. A proper ideal (filter) $\mathcal{A}$ of $\mathcal{R}$ is called a prime ideal (filter) if, for any $\mu ,\pi \in \mathcal{R},$ $\mu \wedge \pi \in \mathcal{A}(\mu \vee \pi \in \mathcal{A})\Rightarrow$ $\mu \in \mathcal{A}$ or $\pi \in \mathcal{A}$. A proper ideal (filter) $\mathcal{X}$ of $\mathcal{R}$ is said to be maximal if it is not properly contained in any proper ideal (filter) of $\mathcal{R}$. It can be observed that every maximal ideal (filter) of $\mathcal{R}$ is a prime ideal (filter). The smallest ideal that contains $\mathcal{G}$ for each subset $\mathcal{G}$ of $\mathcal{R}$ is ($\mathcal{G}$] := $\{({\displaystyle \underset{i=1}{\overset{n}{{\displaystyle \bigvee }}}}{\theta }_i)\wedge \mu \mid {\theta }_i\in \mathcal{G},\mu \in \mathcal{R}\mathrm{and}n\in \mathbb{N}\}.$ The ideal $\mathcal{G}=\{\theta \}$ is written as $(\theta ]$ rather than $(\mathcal{G}]$; this is known as the principal ideal of $\mathcal{R}$. The same way, for each $\mathcal{G}\subseteq \mathcal{R},$ [$\mathcal{G}$) := $\{\mu \vee ({\displaystyle \underset{i=1}{\overset{n}{{\displaystyle \bigwedge }}}}{\theta }_i)\mid {\theta }_i\in \mathcal{G},\mu \in \mathcal{R}\mathrm{and}n\in \mathbb{N}\}.$ A filter like $\mathcal{G}=\{\theta \}$ is written as $[\theta )$ rather than $[\mathcal{G})$; this is known as the principal filter of $\mathcal{R}$. It is easy to see that $(\theta ]\vee (\vartheta ]=(\theta \vee \vartheta ]$ and $(\theta ]\cap (\vartheta ]=(\theta \wedge \vartheta ]$ hold for any $\theta ,\vartheta \in \mathcal{R}$. A sublattice of the distributive lattice $(\mathcal{I}(\mathcal{R}),\vee ,\cap )$ of all ideals of $\mathcal{R}$ is thus the set $(\mathcal{PI}(\mathcal{R}),\vee ,\cap )$ of all principal ideals of $\mathcal{R}$. Moreover, the bounded distributive lattice $(\mathcal{F}(\mathcal{R}),\vee ,\cap )$ containing all filters for $\mathcal{R}$ is defined. In an ADL [16], $\mathcal{R}\setminus \mathcal{A}$ is a prime filter of $\mathcal{R}$ if and only if the prime ideal of $\mathcal{R}$ is $\mathcal{A}.$ Given a non-empty subset $\mathcal{G}$ in $\mathcal{R}$, the set ${\mathcal{G}}^{+}=\{\mu \in \mathcal{R}|\theta \vee \mu \in {\mathcal{M}}_{Max.elts},\mathrm{for}\mathrm{all}\theta \in \mathcal{G}\}$ is a filter of $\mathcal{R}.$ Usually, for every $\theta \in \mathcal{R},{\{\theta \}}^{+}={(\theta )}^{+},$ where $(\theta )=[\theta ).$ For any $\theta \in \mathcal{R}$, we have ${\{\theta \}}^{+}={[\theta )}^{+}$, where $[\theta )$ is the principal filter generated by $\theta$. An element $\theta$ of an ADL $\mathcal{R}$ is called dual dense element if ${[\theta )}^{+}={\mathcal{M}}_{max.elt}$ and the set $\mathcal{E}$ of all dual dense elements in ADL is an ideal if $\mathcal{E}$ is non-empty.

Definition 2

([10]). For any subset $\mathcal{G}$ of an ADL $\mathcal{R}$, define $(\mathcal{G},\mathcal{E})=\{\mu \in \mathcal{R}|\theta \wedge \mu \in \mathcal{E}\mathit{for}\mathit{all}\theta \in \mathcal{G}\}$.

For any $\theta \in \mathcal{R}$, we simply represent $(\{\theta \},\mathcal{E})$ by $(\theta ,\mathcal{E})$. Clearly, $(0,\mathcal{E})=\mathcal{R}.$ It is also obvious that $(m,\mathcal{E})=\mathcal{E}$ for all $m\in {\mathcal{M}}_{Max.elts},$ and $\mathcal{E}\subseteq (\mu ,\mathcal{E})$ for all $\mu \in \mathcal{R}$.

Proposition 1

([10]). Let $\mathcal{R}$ be an ADL. For any $\theta ,\vartheta ,\ae \in \mathcal{R}$, we have the following:

(1)

$\theta \le \vartheta$ implies $(\vartheta ,\mathcal{E})\subseteq (\theta ,\mathcal{E})$;

(2)

$(\theta \vee \vartheta ,\mathcal{E})=(\theta ,\mathcal{E})\cap (\vartheta ,\mathcal{E})$;

(3)

$((\theta \wedge \vartheta ,\mathcal{E}),\mathcal{E})=((\theta ,\mathcal{E}),\mathcal{E})\cap ((\vartheta ,\mathcal{E}),\mathcal{E})$;

(4)

$(\theta ,\mathcal{E})=\mathcal{R}$ if and only if $\theta \in \mathcal{E}$.

3. Star Filters of ADLs

This section presents the concept of star filters and derives a characterization theorem for them. It shows that the collection of all star filters constitutes a complete distributive lattice. Moreover, it establishes a set of equivalent conditions that describe when a filter in an ADL becomes star. Additionally, the concept of starlets in an ADL is introduced. Finally, $\mathcal{E}\text{-}$star quasi-complemented ADLs are defined and characterized in terms of starlets.

Now, we begin with the following definition.

Definition 3.

For any non-empty subset $\mathcal{I}$ of an ADL $\mathcal{R}$, the set ${\mathcal{I}}^{★}$ is defined as ${\mathcal{I}}^{★}=\{\theta \in \mathcal{R}|(\theta ,\mathcal{E})\subseteq (\omega ,\mathcal{E}),\mathit{for}\mathit{some}\omega \in \mathcal{I}\}$.

Lemma 1.

For any non-void subsets $\mathcal{I},\mathcal{U}$ of $\mathcal{R},$ the following statements hold:

(1)

$\mathcal{I}\subseteq {\mathcal{I}}^{★}$;

(2)

$\mathcal{I}\subseteq \mathcal{U}$ implies ${\mathcal{I}}^{★}\subseteq {\mathcal{U}}^{★}$;

(3)

${\mathcal{I}}^{★★}={\mathcal{I}}^{★}$;

(4)

${\mathcal{E}}^{★}=\mathcal{R}.$

Proof. (1) and (2) are clear.

• (3) By (1) and (2), it follows that ${\mathcal{I}}^{★}\subseteq {\mathcal{I}}^{★★}$. Let $\theta \in {\mathcal{I}}^{★★}$. Then, there is $\omega \in {\mathcal{I}}^{★}$ satisfying $(\theta ,\mathcal{E})\subseteq (\omega ,\mathcal{E}).$ Since $\omega \in {\mathcal{I}}^{★}$, there exists $ł\in \mathcal{I}$ such that $(\omega ,\mathcal{E})\subseteq (ł,\mathcal{E}).$ Then, $(\theta ,\mathcal{E})\subseteq (\omega ,\mathcal{E})\subseteq (ł,\mathcal{E})$ and $ł\in \mathcal{I}.$ Therefore, $\theta \in {\mathcal{I}}^{★}$ and hence ${\mathcal{I}}^{★★}\subseteq {\mathcal{I}}^{★}.$ Thus, ${\mathcal{I}}^{★}={\mathcal{I}}^{★★}.$
• (4) Clearly, for any $e\in \mathcal{E},$ we obtain $(e,\mathcal{E})=\mathcal{R}.$ This implies $(\theta ,\mathcal{E})\subseteq \mathcal{R}=(e,\mathcal{E})$ for all $\theta \in \mathcal{R}.$ This gives $\theta \in {\mathcal{E}}^{★},$ for all $\theta \in \mathcal{R}.$ Hence, ${\mathcal{E}}^{★}=\mathcal{R}.$ □

For $\mathcal{I}=\{\omega \}$, we just write ${\{\omega \}}^{★}$ as ${(\omega )}^{★},$ where ${(\omega )}^{★}=\{\theta \in \mathcal{R}|(\theta ,\mathcal{E})\subseteq (\omega ,\mathcal{E})\}$.

Lemma 2.

For any elements $\omega ,ł$ in $\mathcal{R}$ with maximal element $m,$ the following properties hold:

(1)

${(m)}^{★}=\{\theta \in \mathcal{R}|(\theta ,\mathcal{E})=\mathcal{E}\}$;

(2)

${(\omega )}^{★}={([\omega ))}^{★}$;

(3)

${(\omega )}^{★★}={(\omega )}^{★}$;

(4)

$\omega \le ł$ implies ${(ł)}^{★}\subseteq {(\omega )}^{★}$;

(5)

$\omega \in {(ł)}^{★}$ implies ${(\omega )}^{★}\subseteq {(ł)}^{★}$;

(6)

${(\omega \wedge ł)}^{★}={(ł\wedge \omega )}^{★}$ and ${(\omega \vee ł)}^{★}={(ł\vee \omega )}^{★}$;

(7)

${(\omega )}^{★}\cap {(ł)}^{★}={(\omega \vee ł)}^{★}$;

(8)

$(\omega ,\mathcal{E})=(ł,\mathcal{E})$ if and only if ${(\omega )}^{★}={(ł)}^{★}$;

(9)

${(\omega )}^{★}={(ł)}^{★}$ implies ${(\omega \wedge \ae )}^{★}={(ł\wedge \ae )}^{★}$ and ${(\omega \vee \ae )}^{★}={(ł\vee \ae )}^{★},$ for any $\ae \in \mathcal{R}.$

Proof. (1) It is clear.

(2)

Let $\theta \in {(\omega )}^{★}.$ Then, $(\theta ,\mathcal{E})\subseteq (\omega ,\mathcal{E})$. Since $\omega \in [\omega ),$ we obtain that $\theta \in {([\omega ))}^{★}.$ Therefore, ${(\omega )}^{★}\subseteq {([\omega ))}^{★}.$ Let $\theta \in {([\omega ))}^{★}$. Then, there is $\vartheta \in [\omega )$ satisfying $(\theta ,\mathcal{E})\subseteq (\vartheta ,\mathcal{E}).$ Since $\vartheta \in [\omega )$, we obtain $(\vartheta ,\mathcal{E})\subseteq (\omega ,\mathcal{E})$. Hence, $(\theta ,\mathcal{E})\subseteq (\vartheta ,\mathcal{E})\subseteq (\omega ,\mathcal{E})$. Thus, $\theta \in {(\omega )}^{★}$. Therefore, ${([\omega ))}^{★}\subseteq {(\omega )}^{★}$, which gives that ${(\omega )}^{★}={([\omega ))}^{★}$.

(3)

It is clear.

(4)

Assume $\omega \le ł$. Then, $(ł,\mathcal{E})\subseteq (\omega ,\mathcal{E})$. This implies ${(ł)}^{★}\subseteq {(\omega )}^{★}$.

(5)

Let $\omega \in {(ł)}^{★}$. Then, $[\omega )\subseteq {(ł)}^{★}$. By (3), we obtain ${(\omega )}^{★}\subseteq {(ł)}^{★★}={(ł)}^{★}$.

(6)

Since $(\omega \wedge ł,\mathcal{E})=(ł\wedge \omega ,\mathcal{E})$ and $(\omega \vee ł,\mathcal{E})=(ł\vee \omega ,\mathcal{E}),$ we obtain (6).

(7)

For any $\omega ,ł\in \mathcal{R},$ we have ${(\omega \vee ł)}^{★}\subseteq {(\omega )}^{★}\cap {(ł)}^{★}$. Let $\theta \in {(\omega )}^{★}\cap {(ł)}^{★}$. Then, $(\theta ,\mathcal{E})\subseteq (\omega ,\mathcal{E})$ and $(\theta ,\mathcal{E})\subseteq (ł,\mathcal{E})$. This implies $(\theta ,\mathcal{E})\subseteq (\omega ,\mathcal{E})\cap (ł,\mathcal{E})=(\omega \vee ł,\mathcal{E}).$ Therefore, $\theta \in {(\omega \vee ł)}^{★}$ and ${(\omega )}^{★}\cap {(ł)}^{★}\subseteq {(\omega \vee ł)}^{★}$ Thus, ${(\omega )}^{★}\cap {(ł)}^{★}={(\omega \vee ł)}^{★}$.

(8)

Assume $(\omega ,\mathcal{E})=(ł,\mathcal{E})$. Then, clearly ${(\omega )}^{★}={(ł)}^{★}$. Conversely, assume that ${(\omega )}^{★}={(ł)}^{★}$. Since $\omega \in {(\omega )}^{★}={(ł)}^{★}$, we obtain $(\omega ,\mathcal{E})\subseteq (ł,\mathcal{E})$. Similarly, we can obtain that $(ł,\mathcal{E})\subseteq (\omega ,\mathcal{E})$. Therefore, $(\omega ,\mathcal{E})=(ł,\mathcal{E})$.

(9)

Assume that ${(\omega )}^{★}={(ł)}^{★}$. Then, $(\omega ,\mathcal{E})=(ł,\mathcal{E}).$ Let $\ae \in \mathcal{R}.$

Now,

$$\begin{array}{ccc}\hfill {(\omega \vee \ae )}^{★}& =& {(\omega )}^{★}\cap {(\ae )}^{★}\hfill \\ & =& {(ł)}^{★}\cap {(\ae )}^{★}\hfill \\ & =& {(ł\vee \ae )}^{★}.\hfill \end{array}$$

Now,

$$\begin{array}{ccc}\hfill \mu \in {(\omega \wedge \ae )}^{★}\iff (\mu ,\mathcal{E})\subseteq (\omega \wedge \ae ,\mathcal{E})& =& (((\omega \wedge \ae ,\mathcal{E}),\mathcal{E}),\mathcal{E})\hfill \\ & =& (((\omega ,\mathcal{E}),\mathcal{E})\cap ((\ae ,\mathcal{E}),\mathcal{E}),\mathcal{E})\hfill \\ & =& (((ł,\mathcal{E}),\mathcal{E})\cap ((\ae ,\mathcal{E}),\mathcal{E}),\mathcal{E})\hfill \\ & =& (((ł\wedge \ae ,\mathcal{E}),\mathcal{E}),\mathcal{E})\hfill \\ & =& (ł\wedge \ae ,\mathcal{E})\iff \mu \in {(ł\wedge \ae )}^{★}.\hfill \end{array}$$

Therefore, ${(\omega \wedge \ae )}^{★}={(ł\wedge \ae )}^{★}.$ □

Lemma 3.

For each $\mathcal{H}\in \mathcal{F}(\mathcal{R})$, ${\mathcal{H}}^{★}\in \mathcal{F}(\mathcal{R})$ and contains $\mathcal{H}.$

Proof. 

Clearly, we have $\mathcal{H}\subseteq {\mathcal{H}}^{★}$. Let $\theta ,\vartheta \in {\mathcal{H}}^{★}$. Then, there are $\omega ,\ae \in \mathcal{H}$ satisfying $(\theta ,\mathcal{E})\subseteq (\omega ,\mathcal{E})$ and $(\vartheta ,\mathcal{E})\subseteq (\ae ,\mathcal{E}).$ Since $\omega ,\ae \in \mathcal{H},$ we have $\omega \wedge \ae \in \mathcal{H}.$

• Now,

$$\begin{array}{ccc}\hfill (\theta \wedge \vartheta ,\mathcal{E})& =& (((\theta \wedge \vartheta ,\mathcal{E}),\mathcal{E}),\mathcal{E})\hfill \\ & =& ((((\theta ,\mathcal{E}),\mathcal{E})\cap ((\vartheta ,\mathcal{E}),\mathcal{E})),\mathcal{E})\hfill \\ & \subseteq & ((((\omega ,\mathcal{E}),\mathcal{E})\cap ((\ae ,\mathcal{E}),\mathcal{E})),\mathcal{E})\hfill \\ & =& (((\omega \wedge \ae ,\mathcal{E}),\mathcal{E}),\mathcal{E})\hfill \\ & =& (\omega \wedge \ae ,\mathcal{E}).\hfill \end{array}$$

Since $\omega \wedge \ae \in \mathcal{H},$ we obtain that $\theta \wedge \vartheta \in {\mathcal{H}}^{★}.$ Let $\theta \in {\mathcal{H}}^{★}.$ Then, there exists an element $\omega \in \mathcal{H}$ such that $(\theta ,\mathcal{E})\subseteq (\omega ,\mathcal{E}).$ Let $\vartheta \in \mathcal{R}.$

• Then,

$$\begin{array}{ccc}\hfill (\vartheta \vee \theta ,\mathcal{E})& \subseteq & (\vartheta ,\mathcal{E})\cap (\theta ,\mathcal{E})\hfill \\ & \subseteq & (\theta ,\mathcal{E})\hfill \\ & \subseteq & (\omega ,\mathcal{E}).\hfill \end{array}$$

Since $\omega \in \mathcal{H},$ we obtain that $\vartheta \vee \theta \in {\mathcal{H}}^{★}$. Therefore, ${\mathcal{H}}^{★}\in \mathcal{F}(\mathcal{R})$ and contains $\mathcal{H}$. □

Lemma 4.

For any $\mathcal{H},\mathcal{J}\in \mathcal{F}(\mathcal{R}),$ the following properties hold:

(1)

${\mathcal{H}}^{★}\cap {\mathcal{J}}^{★}={(\mathcal{H}\cap \mathcal{J})}^{★}$;

(2)

${(\mathcal{H}\vee \mathcal{J})}^{★}={({\mathcal{H}}^{★}\vee {\mathcal{J}}^{★})}^{★}$.

Proof. (1) It is obvious that ${(\mathcal{H}\cap \mathcal{J})}^{★}\subseteq {\mathcal{H}}^{★}\cap {\mathcal{J}}^{★}$. Let $\theta \in {\mathcal{H}}^{★}\cap {\mathcal{J}}^{★}$. Then, there exist $\omega \in \mathcal{H}$ and $ł\in \mathcal{J}$ such that $(\theta ,\mathcal{E})\subseteq (\omega ,\mathcal{E})$ and $(\theta ,\mathcal{E})\subseteq (ł,\mathcal{E}).$ This implies $(\theta ,\mathcal{E})\subseteq (\omega ,\mathcal{E})\cap (ł,\mathcal{E})=(\omega \vee ł,\mathcal{E}).$ Since $\omega \vee ł\in \mathcal{H}\cap \mathcal{J},$ we obtain $\theta \in {(\mathcal{H}\cap \mathcal{J})}^{★}$. Therefore, ${\mathcal{H}}^{★}\cap {\mathcal{J}}^{★}\subseteq {(\mathcal{H}\cap \mathcal{J})}^{★}$. Hence, ${\mathcal{H}}^{★}\cap {\mathcal{J}}^{★}={(\mathcal{H}\cap \mathcal{J})}^{★}.$

• (2) Clearly, $\mathcal{H}\vee \mathcal{J}\subseteq {\mathcal{H}}^{★}\vee {\mathcal{J}}^{★}$ and hence ${(\mathcal{H}\vee \mathcal{J})}^{★}\subseteq {({\mathcal{H}}^{★}\vee {\mathcal{J}}^{★})}^{★}$. Let $\theta \in {({\mathcal{H}}^{★}\vee {\mathcal{J}}^{★})}^{★}$. There exists $\omega \in {\mathcal{H}}^{★}\vee {\mathcal{J}}^{★}$ such that $(\theta ,\mathcal{E})\subseteq (\omega ,\mathcal{E}).$ Since $\omega \in {\mathcal{H}}^{★}\vee {\mathcal{J}}^{★},$ there are $\vartheta \in {\mathcal{H}}^{★}$ and $\ae \in {\mathcal{J}}^{★}$ satisfying that $\omega =\vartheta \wedge \ae .$ Since $\vartheta \in {\mathcal{H}}^{★}$, there is ${\vartheta }^{\prime }\in \mathcal{H}$ such that $(\vartheta ,\mathcal{E})\subseteq ({\vartheta }^{\prime },\mathcal{E})$. Since $\ae \in {\mathcal{J}}^{★}$, there exists ${\ae }^{\prime }\in \mathcal{J}$ such that $(\ae ,\mathcal{E})\subseteq ({\ae }^{\prime },\mathcal{E})$. Since $(\vartheta ,\mathcal{E})\subseteq ({\vartheta }^{\prime },\mathcal{E})$ and $(\ae ,\mathcal{E})\subseteq ({\ae }^{\prime },\mathcal{E})$, we obtain that $(({\vartheta }^{\prime },\mathcal{E}),\mathcal{E})\subseteq ((\vartheta ,\mathcal{E}),\mathcal{E})$ and $(({\ae }^{\prime },\mathcal{E}),\mathcal{E})\subseteq ((\ae ,\mathcal{E}),\mathcal{E})$. Then, $(({\vartheta }^{\prime }\wedge {\ae }^{\prime },\mathcal{E}),\mathcal{E})=(({\vartheta }^{\prime },\mathcal{E}),\mathcal{E})\cap (({\ae }^{\prime },\mathcal{E}),\mathcal{E})\subseteq ((\vartheta ,\mathcal{E}),\mathcal{E})\cap ((\ae ,\mathcal{E}),\mathcal{E})=((\vartheta \wedge \ae ,\mathcal{E}),\mathcal{E})$. Therefore, $(\theta ,\mathcal{E})\subseteq (\omega ,\mathcal{E})=(\vartheta \wedge \ae ,\mathcal{E})\subseteq ({\vartheta }^{\prime }\wedge {\ae }^{\prime },\mathcal{E}).$ Since ${\vartheta }^{\prime }\wedge {\ae }^{\prime }\in \mathcal{H}\vee \mathcal{J},$ we obtain $\theta \in {(\mathcal{H}\vee \mathcal{J})}^{★}$. Thus, ${({\mathcal{H}}^{★}\vee {\mathcal{J}}^{★})}^{★}\subseteq {(\mathcal{H}\vee \mathcal{J})}^{★}$. □

Proposition 2.

For any $\mathcal{H}\in \mathcal{F}(\mathcal{R}),$ the following conditions are equivalent:

(1)

${\mathcal{H}}^{★}=\mathcal{R}$;

(2)

${\mathcal{H}}^{★}\cap \mathcal{E}\ne \varnothing$;

(3)

$\mathcal{H}\cap \mathcal{E}\ne \varnothing$.

Proof. $(1)\Rightarrow (2)$: Assume ${\mathcal{H}}^{★}=\mathcal{R}$. Then, $0\in {\mathcal{H}}^{★}.$ Since $0\in \mathcal{E},$ we obtain ${\mathcal{H}}^{★}\cap \mathcal{E}\ne \varnothing .$

• $(2)\Rightarrow (3)$: Assume ${\mathcal{H}}^{★}\cap \mathcal{E}\ne \varnothing$. Then, choose $\theta \in {\mathcal{H}}^{★}\cap \mathcal{E}$. Since $\theta \in \mathcal{E}$, by Proposition 1(4), we obtain $(\theta ,\mathcal{E})=\mathcal{R}$. Since $\theta \in {\mathcal{H}}^{★},$ there is $\omega \in \mathcal{H}$ satisfying $\mathcal{R}=(\theta ,\mathcal{E})\subseteq (\omega ,\mathcal{E}).$ Then, $(\omega ,\mathcal{E})=\mathcal{R}$ and hence $\omega \in \mathcal{E}$. Therefore, $\omega \in \mathcal{H}\cap \mathcal{E}$. Thus, $\mathcal{H}\cap \mathcal{E}\ne \varnothing$.
• $(3)\Rightarrow (1)$: Assume $\mathcal{H}\cap \mathcal{E}\ne \varnothing$. Choose $\omega \in \mathcal{H}\cap \mathcal{E}$. Since $\omega \in \mathcal{E},$ we have $(\omega ,\mathcal{E})=\mathcal{R}.$ It gives $(\theta ,\mathcal{E})\subseteq (\omega ,\mathcal{E})=\mathcal{R}$ for all $\theta \in \mathcal{R}.$ Therefore, $\theta \in {\mathcal{H}}^{★},$ for each $\theta \in \mathcal{R}.$ Hence, ${\mathcal{H}}^{★}=\mathcal{R}$. □

The definition of a star filter in an ADL is now presented.

Definition 4.

A filter $\mathcal{H}$ of an ADL $\mathcal{R}$ is called star if $\mathcal{H}={\mathcal{H}}^{★}$.

Example 1.

Let $\mathcal{R}=\{0,1,2,3,4,5,6,7\}$ and define $\vee ,\wedge$ on $\mathcal{R}$ as follows:

$(\mathcal{R},\vee ,\wedge )$ is an ADL. Clearly, we have that $\mathcal{E}=\{0,7\}.$ We have that $(0,\mathcal{E})=(7,\mathcal{E})=\mathcal{R};(1,\mathcal{E})=(2,\mathcal{E})=\mathcal{E};(3,\mathcal{E})=(6,\mathcal{E})=\{0,4,5,7\};(4,\mathcal{E})=(5,\mathcal{E})=\{0,3,,6,7\}.$ Then, clearly ${(0)}^{★}={(7)}^{★}=\mathcal{R};{(1)}^{★}={(2)}^{★}=\mathcal{E};{(3)}^{★}={(6)}^{★}=\{1,2,3,6\}\mathrm{and}{(4)}^{★}={(5)}^{★}=\{1,2,4,5\}.$ Consider the filters ${\mathcal{H}}_1=\{1,2,3,6\}$ and ${\mathcal{H}}_2=\{1,2,6\}.$ Clearly, ${\mathcal{H}}_1$ is a star filter. But ${\mathcal{H}}_2$ is not star, because ${\mathcal{H}}_2^{★}\ne {\mathcal{H}}_2.$

Proposition 3.

For any $\mathcal{W}\in \mathcal{F}(\mathcal{R})$ with $\mathcal{W}\cap \mathcal{E}=\varnothing$, every maximal filter $\mathcal{W}$ is a star filter.

Proof. 

Let $\mathcal{W}\in \mathcal{F}(\mathcal{R})$ with $\mathcal{W}\cap \mathcal{E}=\varnothing$. Suppose $\mathcal{W}$ is maximal. By Proposition 2, it follows that ${\mathcal{W}}^{★}\cap \mathcal{E}=\varnothing$ and ${\mathcal{W}}^{★}\ne \mathcal{R}$. Therefore, $\mathcal{W}\subseteq {\mathcal{W}}^{★}$ and hence $\mathcal{W}={\mathcal{W}}^{★}$. Thus, $\mathcal{W}$ is a star filter of $\mathcal{R}$. □

Let ${\mathfrak{F}}^{★}(\mathcal{R})$ and $Spe{c}^{★}(\mathcal{R})$ represent the set of all star filters and all prime star filters in an ADL $\mathcal{R}$. Although ${\mathfrak{F}}^{★}(\mathcal{R})$ is not necessarily a sublattice of the distributive lattice $\mathfrak{F}(\mathcal{R})$, which consists of all filters of $\mathcal{R}$, we will demonstrate that ${\mathfrak{F}}^{★}(\mathcal{R})$ forms a distributive lattice.

Theorem 4.

For any ADL $\mathcal{R}$, ${\mathfrak{F}}^{★}(\mathcal{R})$ forms a distributive lattice.

Proof. 

For any $\mathcal{H},\mathcal{J}\in {\mathfrak{F}}^{★}(\mathcal{R})$, define ∩ and ⊔ on ${\mathfrak{F}}^{★}(\mathcal{R})$ as in bellow:

$$\mathcal{H}\cap \mathcal{J}={(\mathcal{H}\cap \mathcal{J})}^{★}\mathrm{and}\mathcal{H}\bigsqcup \mathcal{J}={({\mathcal{H}}^{★}\vee {\mathcal{J}}^{★})}^{★}={(\mathcal{H}\vee \mathcal{J})}^{★}$$

Clearly, ${(\mathcal{H}\cap \mathcal{J})}^{★}=inf\{\mathcal{H},\mathcal{J}\}$ in ${\mathfrak{F}}^{★}(\mathcal{R})$ and ${(\mathcal{H}\vee \mathcal{J})}^{★}$ is an upper bound of $\mathcal{H},\mathcal{J}$. By Lemma 4(2), it gives ${({\mathcal{H}}^{★}\vee {\mathcal{J}}^{★})}^{★}={(\mathcal{H}\vee \mathcal{J})}^{★}$. Suppose $\mathcal{Q}\in {\mathfrak{F}}^{★}(\mathcal{R})$ such that $\mathcal{H}\subseteq \mathcal{Q}$ and $\mathcal{J}\subseteq \mathcal{Q}$. Let $\theta \in {(\mathcal{H}\vee \mathcal{J})}^{★}$. Then, there exist $\rho \in \mathcal{H}\subseteq \mathcal{Q}$ and $\kappa \in \mathcal{J}\subseteq \mathcal{Q}$ such that $(\theta ,\mathcal{E})\subseteq (\rho \wedge \kappa ,\mathcal{E}).$ Since $\rho \wedge \kappa \in \mathcal{Q}$, we obtain $\theta \in {\mathcal{Q}}^{★}=\mathcal{Q}$. This leads to ${(\mathcal{H}\vee \mathcal{J})}^{★}=sup\{\mathcal{H},\mathcal{J}\}$ in ${\mathfrak{F}}^{★}(\mathcal{R})$. Consequently, it is straightforward to verify that $({\mathfrak{F}}^{★}(\mathcal{R}),\cap ,\bigsqcup )$ forms a distributive lattice, where $\mathcal{R}$ serves as the greatest element in the lattice $({\mathfrak{F}}^{★}(\mathcal{R}),\cap ,\bigsqcup )$. □

In the following, we characterize the star filter of an ADL element-wise.

Theorem 5.

For any $\mathcal{H}\in \mathcal{F}(\mathcal{R}),$ the following conditions are equivalent:

(1)

$\mathcal{H}$ is star;

(2)

For any $\theta \in \mathcal{R}$, $\theta \in \mathcal{H}$ if and only if ${(\theta )}^{★}\subseteq \mathcal{H}$;

(3)

For any $\theta ,\vartheta \in \mathcal{R}$, $(\theta ,\mathcal{E})=(\vartheta ,\mathcal{E})$ and $\theta \in \mathcal{H}$ imply $\vartheta \in \mathcal{H}$;

(4)

For any $\theta ,\vartheta \in \mathcal{R}$, ${(\theta )}^{★}={(\vartheta )}^{★}$ and $\theta \in \mathcal{H}$ imply $\vartheta \in \mathcal{H}$; $(5)$ $\mathcal{H}={\displaystyle \bigcup _{\theta \in \mathcal{H}}}{(\theta )}^{★}$.

Proof. $(1)\Rightarrow (2)$: Assume (1). Suppose $\theta \in \mathcal{H}.$ Let $\omega \in {(\theta )}^{★}$. Then, $(\omega ,\mathcal{E})\subseteq (\theta ,\mathcal{E}).$ Since $\theta \in \mathcal{H},$ we obtain that $\omega \in {\mathcal{H}}^{★}=\mathcal{H}$. Hence, ${(\theta )}^{★}\subseteq \mathcal{H}$. The converse is clear.

• $(2)\Rightarrow (3)$: Assume (2). Let $\theta ,\vartheta \in \mathcal{R}$ with $(\theta ,\mathcal{E})=(\vartheta ,\mathcal{E})$ and $\theta \in \mathcal{H}$. By our assumption, we obtain ${(\theta )}^{★}\subseteq \mathcal{H}.$ Since $(\theta ,\mathcal{E})=(\vartheta ,\mathcal{E}),$ which gives ${(\theta )}^{★}={(\vartheta )}^{★}$, $\vartheta \in {(\vartheta )}^{★}\subseteq \mathcal{H}$. Therefore, $\vartheta \in \mathcal{H}.$
• $(3)\Rightarrow (4)$: By Lemma 2, (8) is clear.
• $(4)\Rightarrow (5)$: Assume (4). Clearly, we have that $[\theta )\subseteq {(\theta )}^{★},$ for all $\theta \in \mathcal{H}$ and hence $\mathcal{H}={\displaystyle \bigcup _{\theta \in \mathcal{H}}}[\theta )\subseteq {\displaystyle \bigcup _{\theta \in \mathcal{H}}}{(\theta )}^{★}$. Let $\vartheta \in {\displaystyle \bigcup _{\theta \in \mathcal{H}}}{(\theta )}^{★}.$ Then, there is $\mu \in \mathcal{H}$ satisfying $\vartheta \in {(\mu )}^{★}.$ Then, we obtain ${(\vartheta )}^{★}\subseteq {(\mu )}^{★}$. This implies ${(\vartheta )}^{★}={(\vartheta )}^{★}\cap {(\mu )}^{★}={(\mu \vee \vartheta )}^{★}$. Since $\mu \vee \vartheta \in \mathcal{H}$, by condition (4), we obtain $\vartheta \in \mathcal{H}$. Therefore, ${\displaystyle \bigcup _{\theta \in \mathcal{H}}}{(\theta )}^{★}\subseteq \mathcal{H}.$ Hence, ${\displaystyle \bigcup _{\theta \in \mathcal{H}}}{(\theta )}^{★}\subseteq \mathcal{H}$. Thus, $\mathcal{H}={\displaystyle \bigcup _{\theta \in \mathcal{H}}}{(\theta )}^{★}$.
• $(5)\Rightarrow (1)$: Assume (5). It is evident that $\mathcal{H}\subseteq {\mathcal{H}}^{★}$. Let $\theta \in {\mathcal{H}}^{★}$. Then, there is $\omega \in \mathcal{H}$ satisfying $(\theta ,\mathcal{E})\subseteq (\omega ,\mathcal{E}).$ This implies $\theta \in {(\omega )}^{★}.$ Since $\omega \in \mathcal{H},$ we conclude that $\theta \in {\displaystyle \bigcup _{\omega \in \mathcal{H}}}{(\omega )}^{★}=\mathcal{H}$. Therefore, ${\mathcal{H}}^{★}\subseteq \mathcal{H}$ and this shows that $\mathcal{H}$ is star. □

Proposition 4.

Consider $\mathcal{H}$ is a star filter and $\mathcal{A}$ a prime filter in $\mathcal{R}$, where $\mathcal{A}\cap \mathcal{E}=\varnothing$ and $\mathcal{H}\subseteq \mathcal{A}.$ If $\mathcal{A}$ is minimal, it follows that $\mathcal{A}$ is a star filter.

Proof. 

Suppose $\mathcal{A}\notin {\mathfrak{F}}^{★}(\mathcal{R})$. Then, there exists $\theta ,\vartheta \in \mathcal{R}$ satisfying ${(\theta )}^{★}={(\vartheta )}^{★},\theta \in \mathcal{A}$ and $\vartheta \notin \mathcal{A}$. Consider $\mathcal{Q}=(\mathcal{R}\setminus \mathcal{A})\vee (\theta \vee \vartheta ]$. Suppose $\mathcal{H}\cap \mathcal{Q}\ne \varnothing .$ Then, choose an element $\mu \in \mathcal{H}\cap \mathcal{Q}$. Then, $\mu \in \mathcal{H}$ and $\mu \in \mathcal{Q}.$ Since $\mu \in \mathcal{Q},$ there exist $\tau \in \mathcal{R}\setminus \mathcal{A}$ and $\omega \in (\theta \vee \vartheta ]$ such that $\mu =\tau \vee \omega$. Since $\omega \in (\theta \vee \vartheta ],$ it gives $(\theta \vee \vartheta )\wedge \omega =\omega .$ Now, $\mu =\tau \vee \omega =\tau \vee ((\theta \vee \vartheta )\wedge \omega )=(\tau \vee \theta \vee \vartheta )\wedge (\tau \vee \omega ).$ This implies $\tau \vee \theta \vee \vartheta =(\tau \vee \theta \vee \vartheta )\vee (\tau \vee \omega ).$ Since $\mu =\tau \vee \omega \in \mathcal{H},$ we obtain that $\tau \vee \theta \vee \vartheta \in \mathcal{H}.$ Since ${(\theta )}^{★}={(\vartheta )}^{★}$, by Lemma 2(9), we obtain ${(\tau \vee \vartheta )}^{★}={(\tau \vee \theta \vee \vartheta )}^{★}$. Since $\mathcal{H}\in {\mathfrak{F}}^{★}(\mathcal{R})$ and $\tau \vee \theta \vee \vartheta \in \mathcal{H}$, we obtain $\tau \vee \vartheta \in \mathcal{H}\subseteq \mathcal{A}$. This implies $\tau \in \mathcal{A}$ or $\vartheta \in \mathcal{A}$, which leads to a contradiction. Therefore, $\mathcal{H}\cap \mathcal{Q}=\varnothing$. Therefore, a prime filter $\mathcal{X}$ can be found such that $\mathcal{Q}$ and $\mathcal{X}$ are disjoint, and $\mathcal{H}$ is contained within $\mathcal{X}$. As $\mathcal{Q}\cap \mathcal{X}=\varnothing$, it gives $\mathcal{X}\subseteq \mathcal{R}\setminus \mathcal{Q}\subseteq \mathcal{A}$ because of $\mathcal{R}\setminus \mathcal{A}\subseteq \mathcal{Q}$. Since $\theta \vee \vartheta \in \mathcal{Q}$, we obtain $\theta \vee \vartheta \notin \mathcal{X}$. This implies $\theta \vee \vartheta \in \mathcal{A}$ and $\theta \vee \vartheta \notin \mathcal{X}$. Therefore, $\mathcal{H}\subseteq \mathcal{X}\subset \mathcal{A}$. Since $\mathcal{A}\cap \mathcal{E}=\varnothing$, we have $\mathcal{X}\cap \mathcal{E}=\varnothing$. This leads to the conclusion that $\mathcal{A}$ is not the minimal prime ideal among those disjoint from $\mathcal{E}$ and containing $\mathcal{H}$, which leads to a contradiction. Hence, $\mathcal{A}$ is a star filter. □

Corollary 1.

If ${\mathcal{M}}_{max.elt}\in {\mathfrak{F}}^{★}(\mathcal{R}),$ it follows that all minimal prime filters of $\mathcal{R}$ are star filters.

Proof. 

Let $\mathcal{A}$ be a prime filter of $\mathcal{R}$ that is minimal with respect to inclusion. Suppose there exists $\theta \in \mathcal{R}$ such that $\theta \in \mathcal{A}\cap \mathcal{E}.$ Since $\mathcal{A}$ is minimal, there is a non-maximal element $\vartheta \notin \mathcal{A}$ such that $\theta \vee \vartheta \in {\mathcal{M}}_{max.elt}$. This implies $\theta \notin \mathcal{E}$, which leads to a contradiction. It can be concluded that $\mathcal{A}\cap \mathcal{E}=\varnothing$. Since ${\mathcal{M}}_{max.elt}\subseteq \mathcal{A}$, by Proposition 4, $\mathcal{A}$ is star. □

Proposition 5.

The following statements in $\mathcal{R}$ are equivalent:

(1)

For $\theta ,\vartheta \in \mathcal{R}$, $(\theta ,\mathcal{E})=(\vartheta ,\mathcal{E})$ implies $[\theta )=[\vartheta )$;

(2)

For $\theta ,\vartheta \in \mathcal{R}$, ${(\theta )}^{★}={(\vartheta )}^{★}$ implies $[\theta )=[\vartheta )$;

(3)

Each filter $\mathcal{H}$ that does not intersect with $\mathcal{E}$ is star;

(4)

Each prime filter $\mathcal{A}$ that does not intersect with $\mathcal{E}$ is star.

Proof. $(1)\Rightarrow (2)$: It is evident by Lemma 2(6).

• $(2)\Rightarrow (3)$ and $(3)\Rightarrow (4)$ are straightforward.
• $(4)\Rightarrow (1)$: Assume (4). Let $\theta ,\vartheta \in \mathcal{R}$ with $(\theta ,\mathcal{E})=(\vartheta ,\mathcal{E})$. According to Lemma 2(8), it follows that ${(\theta )}^{★}={(\vartheta )}^{★}$. We prove that $[\theta )=[\vartheta ).$ Suppose $[\theta )\ne [\vartheta )$. Assuming without loss of generality that $[\theta )⊈[\vartheta ).$ Then, $\theta \notin [\vartheta )$. Then, there is a prime filter $\mathcal{A}$ of $\mathcal{R}$ satisfying $\vartheta \in [\vartheta )\subseteq \mathcal{A}$ and $\theta \notin \mathcal{A}$. Suppose $\mathcal{A}\cap \mathcal{E}\ne \varnothing .$ Then, ${\mathcal{A}}^{★}=\mathcal{R},$ which gives $\theta \in {\mathcal{A}}^{★}.$ This gives $\mathcal{A}⊈{\mathcal{A}}^{★},$ which leads to a contradiction to $\mathcal{A}\subseteq {\mathcal{A}}^{★}.$ Therefore, $\mathcal{A}\cap \mathcal{E}=\varnothing .$ By our assumption, we obtian that $\mathcal{A}$ is star. Since $\vartheta \in \mathcal{A}$ and ${(\theta )}^{★}={(\vartheta )}^{★},$ we obtain that $\theta \in \mathcal{A},$ which is a contradiction to $\theta \notin \mathcal{A}.$ Hence, $[\theta )=[\vartheta ).$ □

A filter represented as ${(\theta )}^{★}$ is referred to as a starlet of $\mathcal{R}$. Since $(m,\mathcal{E})=\mathcal{E}$ for any maximal element m, it can be noted that ${(m)}^{★}=\{\mu \in \mathcal{R}|(\mu ,\mathcal{E})=\mathcal{E}\}$.

Proposition 6.

For $\theta \in \mathcal{R},$ the following statements in $\mathcal{R}$ are equivalent:

(1)

${(\theta )}^{★}=\mathcal{R}$;

(2)

${(\theta )}^{★}\cap \mathcal{E}\ne \varnothing$;

(3)

$\theta \in \mathcal{E}$.

Lemma 5.

For any $\theta ,\vartheta \in \mathcal{R}$, the properties listed below are valid:

(1)

$\theta \wedge \vartheta =0$ implies ${(\theta )}^{★}\vee {(\vartheta )}^{★}=\mathcal{R}$;

(2)

For any $\theta \notin \mathcal{E},(\theta ,\mathcal{E})\cap {(\theta )}^{★}=\varnothing$;

(3)

${(\theta \wedge \vartheta )}^{★}={({(\theta )}^{★}\vee {(\vartheta )}^{★})}^{★}$.

Proof. (1) Let $\theta ,\vartheta \in \mathcal{R}$ with $\theta \wedge \vartheta =0$. Then, $\mathcal{R}=[\theta \wedge \vartheta )=[\theta )\vee [\vartheta )\subseteq {(\theta )}^{★}\vee {(\vartheta )}^{★}$. This implies ${(\theta )}^{★}\vee {(\vartheta )}^{★}=\mathcal{R}$.

• (2) Let $\theta \notin \mathcal{E}$. Suppose $\omega \in (\theta ,\mathcal{E})\cap {(\theta )}^{★}$. Then, $((\omega ,\mathcal{E}),\mathcal{E})\subseteq (\theta ,\mathcal{E})$ and $(\omega ,\mathcal{E})\subseteq (\theta ,\mathcal{E})$. This implies $\theta \in ((\theta ,\mathcal{E}),\mathcal{E})\subseteq ((\omega ,\mathcal{E}),\mathcal{E})\subseteq (\theta ,\mathcal{E})$. Therefore, $\theta =\theta \wedge \theta \in \mathcal{E}$, which is a contradiction. Hence, $(\theta ,\mathcal{E})\cap {(\theta )}^{★}=\varnothing$.
• (3) This is evident from Lemma 4(2). □

Clearly, every starlet is a star filter. Therefore, for any two starlets ${(\theta )}^{★}$ and ${(\vartheta )}^{★},$ their supremum in ${\mathfrak{F}}^{★}(\mathcal{R})$ can be expressed as

$${(\theta )}^{★}\bigsqcup {(\vartheta )}^{★}={([\theta )\vee [\vartheta ))}^{★}={([\theta \wedge \vartheta ))}^{★}={(\theta \wedge \vartheta )}^{★}$$

Also, their infimum in ${\mathfrak{F}}^{★}(\mathcal{R})$ is ${(\theta )}^{★}\cap {(\vartheta )}^{★}={(\theta \vee \vartheta )}^{★}$.

Theorem 6.

For any ADL $\mathcal{R}$, ${\mathfrak{F}}_{★}(\mathcal{R})$ is a lattice $({\mathfrak{F}}_{★}(\mathcal{R}),\cap ,\bigsqcup )$ and sublattice to the distributive lattice $({\mathfrak{F}}^{★}(\mathcal{R}),\cap ,\bigsqcup ,\mathcal{R})$ of all star filters of $\mathcal{R}$. Moreover, ${\mathfrak{F}}_{★}(\mathcal{R})$ has the same greatest element $\mathcal{R}={(e)}^{★},e\in \mathcal{E}$ as $\mathcal{S}(\mathcal{R})$, while ${\mathfrak{F}}_{★}(\mathcal{R})$ has the smallest element ${(\omega )}^{★}$ if and only if $\mathcal{R}$ has an element ω of the form $(\omega ,\mathcal{E})=\mathcal{E}$.

Proof. 

Clearly, $({\mathfrak{F}}_{★}(\mathcal{R}),\cap ,\bigsqcup )$ is a sublattice to the distributive lattice $({\mathfrak{F}}^{★}(\mathcal{R}),\cap ,\bigsqcup )$. Suppose ${(\omega )}^{★}$ is the smallest element of ${\mathfrak{F}}_{★}(\mathcal{R})$. Let $\theta \in (\omega ,\mathcal{E})$. Then, $\theta \wedge \omega \in \mathcal{E}$. Now, for any $\theta \in \mathcal{R}$ ${(\theta )}^{★}={(\theta )}^{★}\bigsqcup {(\omega )}^{★}={(\theta \wedge \omega )}^{★}=\mathcal{R}$, which gives that $\omega \in \mathcal{E}$. Hence, $(\omega ,\mathcal{E})\subseteq \mathcal{E}$. Therefore, $(\omega ,\mathcal{E})=\mathcal{E}$. Conversely, assume that $\mathcal{R}$ has an element $\omega$ such that $(\omega ,\mathcal{E})=\mathcal{E}$. Let $\theta \in {(\omega )}^{★}$. Then, $(\theta ,\mathcal{E})\subseteq (\omega ,\mathcal{E})=\mathcal{E}\subseteq (ł,\mathcal{E}),$ for all $ł\in \mathcal{R}$. Hence, $\theta \in {(ł)}^{★}$ for all $ł\in \mathcal{R}$. Thus, ${(\omega )}^{★}\subseteq {(ł)}^{★}$ for all $ł\in \mathcal{U}$. Hence, ${(\omega )}^{★}$ is the smallest element of ${\mathfrak{F}}_{★}(\mathcal{R})$. □

In any ADL $\mathcal{R}$, it is a commonly accepted fact that the quotient ADL $\mathcal{R}/\varphi =\{{\left[\theta \right]}_{\varphi }|\theta \in \mathcal{R}\}$, where ${\left[\theta \right]}_{\varphi }$ denotes the congruence class of $\theta$ with respect to $\varphi$, which is a quotient lattice with respect to the operations given by ${\left[\theta \right]}_{\varphi }\cap {\left[\vartheta \right]}_{\varphi }={[\theta \wedge \vartheta ]}_{\varphi }$ and ${\left[\theta \right]}_{\varphi }\vee {\left[\vartheta \right]}_{\varphi }={[\theta \vee \vartheta ]}_{\varphi }$ for all $\theta ,\vartheta \in \mathcal{R}.$

Proposition 7.

Let ϕ be a binary relation defined on $\mathcal{R}$ with maximal element m by

$$(\theta ,\vartheta )\in \varphi \mathit{if} {(\theta )}^{★}={(\vartheta )}^{★}$$

for all $\theta ,\vartheta \in \mathcal{R}$. Then, ϕ is a congruence on $\mathcal{R}$, where ${(m)}^{★}$ is the smallest congruence class and $\mathcal{E}$ is the unit congruence class of $\mathcal{R}/\varphi$. Furthermore, $ker\varphi =\{\theta \in \mathcal{R}|{(\theta )}^{★}={(m)}^{★}$ is a star filter of $\mathcal{R}$.

Proof. 

According to (9) of Lemma 2, $\varphi$ qualifies as a congruence on $\mathcal{R}$. It is evident that ${(m)}^{★}$ is the smallest congruence class in $\mathcal{R}/\varphi$. Let $\theta ,\vartheta \in \mathcal{E}$. By Proposition 6, we conclude that ${(\theta )}^{★}={(\vartheta )}^{★}=\mathcal{R}$. Consequently, $(\theta ,\vartheta )\in \varphi$. This shows that $\mathcal{E}$ represents a congruence class in $\mathcal{R}/\varphi$. Now, consider $\omega \in \mathcal{E}$ and $\theta \in \mathcal{R}$. As $\mathcal{E}$ is an ideal, it follows that $\omega \wedge \theta \in \mathcal{E}$. Since $\mathcal{E}$ is a congruence class under $\varphi$, we have ${\left[\theta \right]}_{\varphi }\vee {\left[\omega \right]}_{\varphi }={[\theta \wedge \omega ]}_{\varphi }=\mathcal{E}$. Therefore, $\mathcal{E}$ is the unit congruence class of $\mathcal{R}/\varphi$. It is noted that $ker\varphi$ is a filter of $\mathcal{R}$. Let $\theta \in ker\varphi$. Then, ${(\theta )}^{★}={(m)}^{★}$. Let $\omega \in {(\theta )}^{★}$. Then, ${(\omega )}^{★}\subseteq {(\theta )}^{★}={(m)}^{★}$. As ${(m)}^{★}\subseteq {(\vartheta )}^{★}$ for any $\vartheta \in \mathcal{R}$, we obtain ${(\omega )}^{★}={(m)}^{★}$, which gives $\omega \in ker\varphi$. It can be concluded that ${(\theta )}^{★}\subseteq ker\varphi$. Thus, $ker\varphi$ is a star filter of $\mathcal{R}$. □

Definition 5.

An ADL $\mathcal{R}$ with maximal element m can be called $\mathcal{E}\text{-}$star quasi-complemented if, for every $\theta \in \mathcal{R}$, there is a $\vartheta \in \mathcal{R}$ such that ${(\theta \vee \vartheta )}^{★}={(m)}^{★}$ and $\theta \wedge \vartheta \in \mathcal{E}$.

From Example 1, it is clearly observed that $\mathcal{R}$ is $\mathcal{E}\text{-}$star quasi-complemented. The $\mathcal{E}\text{-}$star quasi-complemented ADLs are now characterized using the congruence $\varphi$ and starlets.

Theorem 7.

In ADL $\mathcal{R}$ with maximal element m, the conditions given below are equivalent:

(1)

$\mathcal{R}$ is $\mathcal{E}\text{-}$star quasi-complemented;

(2)

${\mathfrak{F}}_{★}(\mathcal{R})$ is a Boolean algebra;

(3)

$\mathcal{R}/\varphi$ is a Boolean algebra.

Proof. $(1)\Rightarrow (2)$: Assume (1). Let ${(\theta )}^{★}\in {\mathfrak{F}}_{★}(\mathcal{R})$. Then, there is ${\theta }^{\prime }\in \mathcal{R}$ satisfying $\theta \vee {\theta }^{\prime }=m$ and $\theta \wedge {\theta }^{\prime }\in \mathcal{E}$. Hence, ${(\theta )}^{★}\cap {({\theta }^{\prime })}^{★}={(\theta \vee {\theta }^{\prime })}^{★}={(m)}^{★}$ and ${(\theta )}^{★}\bigsqcup {({\theta }^{\prime })}^{★}={(\theta \wedge {\theta }^{\prime })}^{★}=\mathcal{R}$. Therefore, ${\mathfrak{F}}_{★}(\mathcal{R})$ is a Boolean algebra.

• $(2)\Rightarrow (3)$: Assume (2). Let ${\left[\theta \right]}_{\varphi }\in \mathcal{R}/\varphi$. Then, ${(\theta )}^{★}\in {\mathfrak{F}}_{★}(\mathcal{R})$. This implies that there is ${(\vartheta )}^{★}\in {\mathfrak{F}}_{★}(\mathcal{R})$ satisfying ${(\theta \vee \vartheta )}^{★}={(\theta )}^{★}\cap {(\vartheta )}^{★}={(m)}^{★}$ and ${(\theta \wedge \vartheta )}^{★}={(\theta )}^{★}\bigsqcup {(\vartheta )}^{★}=\mathcal{R}$. Hence, $\theta \vee \vartheta \in {\left[m\right]}_{\varphi }$ and $\theta \wedge \vartheta \in \mathcal{E}$. Thus, ${\left[\theta \right]}_{\varphi }\cap {\left[\vartheta \right]}_{\varphi }={[\theta \vee \vartheta ]}_{\varphi }={\left[m\right]}_{\varphi }$ and ${\left[\theta \right]}_{\varphi }\vee {\left[\vartheta \right]}_{\varphi }={[\theta \wedge \vartheta ]}_{\varphi }=\mathcal{E}$. Therefore, $\mathcal{R}/\varphi$ is a Boolean algebra.
• $(3)\Rightarrow (1)$: Assume (3). Let $\theta \in \mathcal{R}$. Then, ${\left[\theta \right]}_{\varphi }\in \mathcal{R}/\varphi$. By (3), there exists ${\left[{\theta }^{\prime }\right]}_{\varphi }\in \mathcal{R}/\varphi$ such that ${[\theta \wedge {\theta }^{\prime }]}_{\varphi }={\left[\theta \right]}_{\varphi }\cap {\left[{\theta }^{\prime }\right]}_{\varphi }={\left[m\right]}_{\varphi }$ and ${[\theta \wedge {\theta }^{\prime }]}_{\varphi }={\left[\theta \right]}_{\varphi }\vee {\left[{\theta }^{\prime }\right]}_{\varphi }=\mathcal{E}$. Thus, ${(\theta \vee {\theta }^{\prime })}^{★}={(m)}^{★}$ and $\theta \wedge {\theta }^{\prime }\in \mathcal{E}$. Therefore, $\mathcal{R}$ is $\mathcal{E}\text{-}$star quasi-complemented. □

Theorem 8.

Every ADL $\mathcal{R}$ is an epimorphic image of the lattice $({\mathfrak{F}}_{★}(\mathcal{R}),\bigsqcup ,\cap )$ of starlets.

Proof. 

Define $\psi :\mathcal{R}⟶{\mathfrak{F}}_{★}(\mathcal{R})$ by $\psi (\mu )={(\mu )}^{★}$ for each $\mu \in \mathcal{R}$. It is obvious that $\psi$ is well defined. Let $\theta ,\vartheta \in \mathcal{R}$. Then, $\psi (\theta \vee \vartheta )={(\theta \vee \vartheta )}^{★}={(\theta )}^{★}\cap {(\vartheta )}^{★}=\psi (\theta )\cap \psi (\vartheta )$. By Lemma 5(3), we obtain $\psi (\theta \wedge \vartheta )={(\theta \wedge \vartheta )}^{★}={({(\theta )}^{★}\vee {(\vartheta )}^{★})}^{★}={(\theta )}^{★}\bigsqcup {(\vartheta )}^{★}=\psi (\theta )\bigsqcup \psi (\vartheta )$. This implies that $\psi$ is a homomorphism. We easily obtain that $\psi$ is surjective. □

Proposition 8.

Every maximal star filter of an ADL $\mathcal{R}$ is prime.

Proof. 

Let $\mathcal{W}$ be a maximal star filter of an ADL $\mathcal{R}$. Let $\theta ,\vartheta \in \mathcal{R}$ be such that $\theta \notin \mathcal{W}$ and $\vartheta \notin \mathcal{W}$. Then, $\mathcal{W}\bigsqcup {(\theta )}^{★}=\mathcal{R}$ and $\mathcal{W}\bigsqcup {(\vartheta )}^{★}=\mathcal{R}$. Now, $\mathcal{R}=\mathcal{R}\cap \mathcal{R}=\{\mathcal{W}\bigsqcup {(\theta )}^{★}\}\cap \{\mathcal{W}\bigsqcup {(\vartheta )}^{★}\}=\mathcal{W}\bigsqcup \{{(\theta )}^{★}\cap {(\vartheta )}^{★}\}=\mathcal{W}\bigsqcup {(\theta \vee \vartheta )}^{★}.$

Suppose $\theta \vee \vartheta \in \mathcal{W}$. Since $\mathcal{W}$ is star, we obtain ${(\theta \vee \vartheta )}^{★}\subseteq \mathcal{W}$. Hence, $\mathcal{W}=\mathcal{R}$, which is a contradiction. Therefore, $\mathcal{W}$ is prime. □

Theorem 9.

Let $\mathcal{H}\in {\mathfrak{F}}^{★}(\mathcal{R})$ and $\mathcal{S}\in \mathfrak{I}(R)$ with $\mathcal{S}\cap \mathcal{H}=\varnothing$. Then, there is $\mathcal{A}\in Spe{c}^{★}(\mathcal{R})$ such that $\mathcal{H}\subseteq \mathcal{A}$ and $\mathcal{A}\cap \mathcal{S}=\varnothing$.

Proof. 

Consider $\mathcal{G}=\{\mathcal{F}|\mathcal{F}\mathrm{is}a\mathrm{star}\mathrm{filter},\mathcal{H}\subseteq \mathcal{F}\mathrm{and}\mathcal{F}\cap \mathcal{S}=\varnothing \}$. Clearly, $\mathcal{H}\in \mathcal{G}$. Clearly, $\mathcal{G}$ satisfies the hypothesis of Zorn’s Lemma and hence $\mathcal{G}$ has a maximal element, let it be $\mathcal{W}.$ Suppose $\theta ,\vartheta \in \mathcal{R}$ such that $\theta \notin \mathcal{W}$ and $\vartheta \notin \mathcal{W}$. Then, $\mathcal{W}\subset \mathcal{W}\vee [\theta )\subseteq \mathcal{W}\vee {(\theta )}^{★}$ and $\mathcal{W}\subset \mathcal{W}\vee [\vartheta )\subseteq \mathcal{W}\vee {(\vartheta )}^{★}$. By the maximality of $\mathcal{W}$, we obtain $\{\mathcal{W}\vee {(\theta )}^{★}\}\cap \mathcal{H}\ne \varnothing$ and $\{\mathcal{W}\vee {(\vartheta )}^{★}\}\cap \mathcal{S}\ne \varnothing$. Choose $\omega \in \{\mathcal{W}\vee {(\theta )}^{★}\}\cap \mathcal{S}$ and $ł\in \{\mathcal{W}\vee {(\vartheta )}^{★}\}\cap \mathcal{S}$. Then, $\omega \vee ł\in \mathcal{S}.$ Now, $\omega \vee ł\in \{\mathcal{W}\vee {(\theta )}^{★}\}\cap \{\mathcal{W}\vee {(\vartheta )}^{★}\}=\mathcal{W}\vee \{{(\theta )}^{★}\cap {(\vartheta )}^{★}\}=\mathcal{W}\vee {(\theta \vee \vartheta )}^{★}.$ Suppose $\theta \vee \vartheta \in \mathcal{W}$. Since $\mathcal{W}$ is star, we obtain ${(\theta \vee \vartheta )}^{★}\subseteq \mathcal{W}$. Hence, $\omega \vee ł\in \mathcal{W}$ and thus $\omega \vee ł\in \mathcal{W}\cap \mathcal{S}\ne \varnothing$, which is a contradiction. Therefore, $\mathcal{W}$ is prime. □

Corollary 2.

Let $\mathcal{H}$ be a star filter of an ADL $\mathcal{R}$ and $\mu \notin \mathcal{H}$. Then, there is $\mathcal{A}\in Spe{c}^{★}(\mathcal{R})$ such that $\mathcal{H}\subseteq \mathcal{A}$ and $\mu \notin \mathcal{A}$.

Corollary 3.

For any star filter $\mathcal{H}$ of an ADL $\mathcal{R}$, we have

$$\mathcal{H}=\bigcap \{\mathcal{A}|\mathcal{A}\mathit{is}a\mathit{prime}\mathit{star}\mathit{filter}\mathit{of}\mathcal{R}\mathit{such}\mathit{that}\mathcal{H}\subseteq \mathcal{A}\}$$

Corollary 4.

The intersection of all prime star filters is equal to ${(m)}^{★},$ where $m\in {\mathcal{M}}_{max.elt}$.

Let $\mathcal{H}\in {\mathfrak{F}}^{★}(\mathcal{R})$ and $\mathcal{A}\in Spe{c}^{★}(\mathcal{R})$ with $\mathcal{H}\subseteq \mathcal{A}$. Then, $\mathcal{A}$ is called minimal, belonging to $\mathcal{H}$ if there is $\mathcal{V}\notin Spe{c}^{★}(\mathcal{R})$ such that $\mathcal{H}\subseteq \mathcal{V}\subseteq \mathcal{A}$. A minimal prime star filter belonging to ${({\mathcal{M}}_{max.elt})}^{★}$ is simply referred as a minimal prime star.

The following theorem establishes a necessary and sufficient condition for a prime star filter of an ADL to be minimal.

Theorem 10.

Let $\mathcal{H}\in {\mathfrak{F}}^{★}(\mathcal{R})$ and $\mathcal{A}\in Spe{c}^{★}(\mathcal{R})$, with $\mathcal{H}\subseteq \mathcal{A}$. Then, $\mathcal{A}$ is minimal, belonging to $\mathcal{H}$ if and only if to each $\theta \in \mathcal{A}$ there is $\vartheta \notin \mathcal{A}$ such that $\theta \vee \vartheta \in \mathcal{H}$.

Proof. 

Assume that $\mathcal{A}$ is a minimal prime star filter with $\mathcal{H}\subseteq \mathcal{A}$. By Proposition 2, we obtain $\mathcal{A}\cap \mathcal{E}=\varnothing$. Then, $\mathcal{R}\setminus \mathcal{A}$ is a maximal ideal with respect to the condition that $(\mathcal{R}\setminus \mathcal{A})\cap \mathcal{H}=\varnothing$. Let $\mu \in \mathcal{A}$. Then, $\mathcal{R}\setminus \mathcal{A}\subset (\mathcal{R}\setminus \mathcal{A})\vee (\theta ]$. By the maximality of $\mathcal{R}\setminus \mathcal{A}$, we obtain $\{(\mathcal{R}\setminus \mathcal{A})\vee (\theta ]\}\cap \mathcal{H}\ne \varnothing$. Choose $\omega \in \{(\mathcal{R}\setminus \mathcal{A})\vee (\theta ]\}\cap \mathcal{H}$. This gives $\omega =ł\vee a$ for some $ł\in \mathcal{R}\setminus \mathcal{A}$ and $\omega \in \mathcal{H}$. Therefore, $ł\vee \theta =\omega \in \mathcal{H}$, where $ł\notin \mathcal{A}$. Conversely, let us assume that the specified condition is satisfied. If $\mathcal{A}$ is not a minimal prime star filter associated with $\mathcal{H}$, then there is $\mathcal{Q}\in Spe{c}^{★}(\mathcal{R})$ with $\mathcal{H}\subseteq \mathcal{Q}\subset \mathcal{A}$. Choose an element $\theta \in \mathcal{A}\setminus \mathcal{Q}$. According to the assumed condition, there exists an element $\vartheta \notin \mathcal{A}$ such that $\theta \vee \vartheta \in \mathcal{H}$, which is also a subset of $\mathcal{Q}$. However, since $\theta \notin \mathcal{Q}$, it follows that $\vartheta$ must belong to $\mathcal{Q}$, which is contained within $\mathcal{A}$. This leads to a contradiction. Therefore, we conclude that $\mathcal{A}$ is indeed a minimal prime star filter belonging to $\mathcal{H}$. □

By taking $\{m\},$ where $m\in {\mathcal{M}}_{max.elt}$ in place of $\mathcal{H}$ in Theorem 10, we obtain the following.

Corollary 5.

A prime star filter $\mathcal{A}$ of an ADL with maximal element m is minimal if and only if to each $\theta \in \mathcal{A}$ there exists $\vartheta \notin \mathcal{A}$ such that $\theta \vee \vartheta \in {(m)}^{★}$.

4. Prime Spectrum of Star Filters

In this section, we explore various algebraic properties of prime star filters in an ADL. We provide a set of equivalent conditions under which the space of prime star filters of an ADL becomes a Hausdorff space.

For every $\mathcal{B}\subseteq \mathcal{R}$, let $\mathcal{K}(\mathcal{B})=\{\mathcal{A}\in Spe{c}^{★}(\mathcal{R})|\mathcal{B}⊈\mathcal{A}\}$ and for any $\theta \in \mathcal{R},\mathcal{K}(\theta )=\mathcal{K}(\{\theta \})$.

The following result can be verified directly.

Lemma 6.

Let $\mathcal{R}$ be an ADL with maximal element m. For any $\theta ,\vartheta \in \mathcal{R}$, the following properties hold:

(1)

${\displaystyle \bigcup _{\theta \in \mathcal{R}}}\mathcal{K}(\theta )=Spe{c}^{★}(\mathcal{R})$;

(2)

$\mathcal{K}(\theta )\cap \mathcal{K}(\vartheta )=\mathcal{K}(\theta \vee \vartheta )$;

(3)

$\mathcal{K}(\theta )\cup \mathcal{K}(\vartheta )=\mathcal{K}(\theta \wedge \vartheta )$;

(4)

$\mathcal{K}(\theta )=\varnothing \iff \theta \in {(m)}^{★}$;

(5)

$\mathcal{K}(\theta )=Spe{c}^{★}(\mathcal{R})\iff \theta \in \mathcal{E}$.

From the above lemma, it can be easily observed that the collection $\{\mathcal{K}(\mu )|\mu \in \mathcal{R}\}$ forms a base for a topology on $Spe{c}^{★}(\mathcal{R})$, which is called a hull–kernel topology.

Theorem 11.

In any ADL $\mathcal{R}$, the following properties hold:

(1)

For any $\mu \in \mathcal{R},\mathcal{K}(\mu )$ is compact in $Spe{c}^{★}(\mathcal{R})$,;

(2)

Let $\mathcal{L}$ be a compact open subset of $Spe{c}^{★}(\mathcal{R})$. Then, there exists some $\theta \in \mathcal{R}$ such that $\mathcal{L}=\mathcal{K}(\theta )$;

(3)

$Spe{c}^{★}(\mathcal{R})$ is a $T_0$-space;

(4)

The mapping $\theta \mapsto \mathcal{K}(\theta )$ defines a homomorphism from $\mathcal{R}$ onto the lattice of all compact open subsets of $Spe{c}^{★}(\mathcal{R})$.

Proof. (1) Consider $\theta \in \mathcal{R}$ and $\mathcal{G}\subseteq \mathcal{R}$ with $\mathcal{K}(\theta )\subseteq {\displaystyle \bigcup _{\vartheta \in \mathcal{G}}}\mathcal{K}(\vartheta )$. Take $\mathcal{H}=[\mathcal{G})$. If $\theta \notin {\mathcal{H}}^{★}$, then by Corollary 2, there is $\mathcal{A}\in Spe{c}^{★}(\mathcal{R})$ such that ${\mathcal{H}}^{★}\subseteq \mathcal{A}$ and $\theta \notin \mathcal{A}$. This implies $\mathcal{A}\in \mathcal{K}(\theta )\subseteq {\displaystyle \bigcup _{\vartheta \in \mathcal{G}}}\mathcal{K}(\vartheta )$. Hence, $\vartheta \notin \mathcal{A}$ for some $\vartheta \in \mathcal{G}$, it gives a contradiction to $\vartheta \in \mathcal{G}\subseteq \mathcal{H}\subseteq {\mathcal{H}}^{★}\subseteq \mathcal{A}$. Thus, $\theta \in {\mathcal{H}}^{★}$. This leads to $\theta \in {(\omega )}^{★}$ for some $\omega \in \mathcal{H}$. Since $\mathcal{H}=[\mathcal{G})$ and $\omega \in \mathcal{H},$ there exist $ł\in \mathcal{R}$ and ${\omega }_1,{\omega }_2,\dots ,{\omega }_n\in \mathcal{G}$ such that $\omega =ł\vee ({\displaystyle \underset{i=1}{\overset{n}{\bigwedge }}}{\omega }_i).$ This implies ${(\omega )}^{★}={(({\displaystyle \underset{i=1}{\overset{n}{\bigwedge }}}{\omega }_i)\vee ł)}^{★}\subseteq {({\displaystyle \underset{i=1}{\overset{n}{\bigwedge }}}{\omega }_i)}^{★}.$ This implies $\mathcal{K}(\theta )\subseteq {\displaystyle \bigcup _{i=1}^n}\mathcal{K}({\omega }_i)$, which is a finite subcover of $\mathcal{K}(\theta )$. Hence, $\mathcal{K}(\theta )$ is compact in $Spe{c}^{★}(\mathcal{R})$. Thus, for each $\theta \in \mathcal{R},\mathcal{K}(\theta )$ is a compact open subset of $Spe{c}^{★}(\mathcal{R})$.

• (2) Since $\mathcal{L}$ is open, it gives $\mathcal{L}={\displaystyle \bigcup _{\theta \in \mathcal{G}}}\mathcal{K}(\theta )$ for some $\mathcal{G}\subseteq \mathcal{R}$. As $\mathcal{L}$ is compact, there are ${\theta }_1,{\theta }_2,\dots ,{\theta }_n\in \mathcal{G}$ such that $\mathcal{L}={\displaystyle \bigcup _{i=1}^n}\mathcal{K}({\theta }_i)=\mathcal{K}\left({\displaystyle \underset{i=1}{\overset{n}{\bigwedge }}}{\theta }_i\right).$ Thus, $\mathcal{L}=\mathcal{K}(\mu )$ for some $\mu \in \mathcal{R}$.
• (3) Let $\mathcal{X},\mathcal{W}\in Spe{c}^{★}(\mathcal{R})$ with $\mathcal{X}\ne \mathcal{W}$. Assuming without loss of generality that $\mathcal{X}⊈\mathcal{W}$, let us select an element $\theta \in \mathcal{R}$ such that $\theta$ belongs to $\mathcal{X}$ but not to $\mathcal{W}$. This implies $\mathcal{X}$ is not included in $\mathcal{K}(\theta )$, whereas $\mathcal{W}$ is contained in $\mathcal{K}(\theta )$. Therefore, we can conclude that $Spe{c}^{★}(\mathcal{R})$ is a $T_0$-space.
• (4) This can be derived from statements (2) and (3) of Lemma 6. □

Lemma 7.

The statements given below hold in $\mathcal{R}$:

(1)

$\mathcal{K}(\theta )=\mathcal{K}({(\theta )}^{★})$ for each $\theta \in \mathcal{R}$;

(2)

$\mathcal{K}(\mathcal{H})=\mathcal{K}({\mathcal{H}}^{★})$ for each $\mathcal{H}\in \mathfrak{F}(R)$;

(3)

$\mathcal{K}(\mathcal{H})={\displaystyle \bigcup _{\theta \in \mathcal{H}}}\mathcal{K}({(\theta )}^{★})$ for each $\mathcal{H}\in {\mathfrak{F}}^{★}(\mathcal{R})$.

Proof. (1) Let $\mathcal{A}\in \mathcal{K}(\theta )\cap Spe{c}^{★}(\mathcal{R})$. Then, $\theta \notin \mathcal{A}$. Since $\mathcal{A}\in {\mathfrak{F}}^{★}(\mathcal{R})$, we gain ${(\theta )}^{★}⊈\mathcal{A}$. This gives $\mathcal{A}\in \mathcal{K}({(\theta )}^{★})$. Thus, $\mathcal{K}(\theta )\subseteq \mathcal{K}({(\theta )}^{★})$. In the same way, the other condition is true.

• (2) It is evident that $\mathcal{K}(\mathcal{H})\subseteq \mathcal{K}({\mathcal{H}}^{★})$ for every $\mathcal{H}\subseteq {\mathcal{H}}^{★}$. Let $\mathcal{A}\in \mathcal{K}({\mathcal{H}}^{★})\cap Spe{c}^{★}(\mathcal{R})$. Then, ${\mathcal{H}}^{★}⊈\mathcal{A}$. Choose $\mu \in {\mathcal{H}}^{★}$ and $\mu \notin \mathcal{A}$. Then, $(\theta ,\mathcal{E})\subseteq (\omega ,\mathcal{E}),$ for some $\omega \in \mathcal{H}$. Hence, $\theta \in {(\theta )}^{★}\subseteq {(\omega )}^{★}$. If $\mathcal{A}\notin \mathcal{K}(\mathcal{H})$, then $\omega \in \mathcal{H}\subseteq \mathcal{A}$. Since $\mathcal{A}\in {\mathfrak{F}}^{★}(\mathcal{R})$, which implies $\theta \in {(\theta )}^{★}\subseteq {(\omega )}^{★}\subseteq \mathcal{A}$, it gives a contradiction. It can be concluded that $\mathcal{A}\in \mathcal{K}(\mathcal{H})$. Hence, $\mathcal{K}({\mathcal{H}}^{★})\subseteq \mathcal{K}(\mathcal{H})$.
• (3) Consider $\mathcal{A}\in \mathcal{K}(\mathcal{H})\cap Spe{c}^{★}(\mathcal{R})$. Then, $\mathcal{H}⊈\mathcal{A}$. Choose $\theta \in \mathcal{H}$ such that $\theta \notin \mathcal{A}$. This implies $\mathcal{A}\in \mathcal{K}(\theta )$. As $\theta \in \mathcal{H}$, we obtain $\mathcal{A}\in {\displaystyle \bigcup _{\theta \in \mathcal{H}}}\mathcal{K}(\theta )$. Hence, $\mathcal{K}(\mathcal{H})\subseteq {\displaystyle \bigcup _{\theta \in \mathcal{H}}}\mathcal{K}(\theta )$. Let $\mathcal{A}\in {\displaystyle \bigcup _{\theta \in \mathcal{H}}}\mathcal{K}(\theta )$. Then, $\mathcal{A}\in \mathcal{K}(\theta )$ for some $\theta \in \mathcal{H}$. Then, $\theta \notin \mathcal{A}$ for some $\theta \in \mathcal{H}$. This gives $\mathcal{H}⊈\mathcal{A}$. Therefore, $\mathcal{A}\in \mathcal{K}(\mathcal{H})$. Hence, ${\displaystyle \bigcup _{\theta \in \mathcal{H}}}\mathcal{K}(\theta )\subseteq \mathcal{K}(\mathcal{H})$. □

Theorem 12.

For an ADL $\mathcal{R},$ $(\mathcal{I}(\mathcal{R}),\bigsqcup ,\cap )$ is isomorphic to the lattice of open sets in $Spe{c}^{★}(\mathcal{R})$.

Proof. 

Let the class of all open subsets of the space $Spe{c}^{★}(\mathcal{R})$ be denoted by ℑ. It is clear that $(\Im ,\cap ,\cup )$ forms a lattice. Define the map $\phi :\mathcal{I}(\mathcal{R})\to \Im$ as $\phi ({\mathcal{H}}^{★})=\mathcal{K}(\mathcal{H})$ for every $\mathcal{H}\in \mathcal{I}(\mathcal{R})$. By Lemma 7(2), each open subset of $Spe{c}^{★}(\mathcal{R})$ can be expressed as $\mathcal{K}(\mathcal{H})$ for some $\mathcal{H}\in \mathcal{I}(\mathcal{R})$. Therefore, the mapping $\phi$ is surjective. Now, let $\mathcal{H},\mathcal{J}\in \mathcal{I}(\mathcal{R})$ and assume that $\phi (\mathcal{H})=\phi (\mathcal{J})$. If $\mathcal{H}\ne \mathcal{J}$, then there is some $\alpha \in \mathcal{J}$ such that $\alpha \notin \mathcal{H}$. By Corollary 2, there is a prime star filter $\mathcal{A}$ such that $\mathcal{H}\subseteq \mathcal{A}$ and $\alpha \notin \mathcal{A}$. Hence, $\mathcal{A}\in \mathcal{K}(\alpha )$ for some $\alpha \in \mathcal{J}$. By Lemma 7(3), it follows that $\mathcal{A}\in {\bigcup }_{\alpha \in \mathcal{J}}\mathcal{K}(\alpha )=\mathcal{K}(\mathcal{J})$. Since $\phi (\mathcal{H})=\phi (\mathcal{J})$, we have $\mathcal{K}(\mathcal{H})=\mathcal{K}(\mathcal{J})$. Thus, $\mathcal{A}\in \mathcal{K}(\mathcal{J})=\mathcal{K}(\mathcal{H})$, implying that $\mathcal{H}⊈\mathcal{A}$, which contradicts the choice of $\mathcal{A}$. Hence, we conclude that $\mathcal{H}=\mathcal{J}$, and therefore, $\phi$ is injective.

For each $\mathcal{H},\mathcal{J}\in \mathcal{I}(\mathcal{R})$, we have $\phi (\mathcal{H}\cap \mathcal{J})=\mathcal{K}(\mathcal{H}\cap \mathcal{J})=\mathcal{K}(\mathcal{H})\cap \mathcal{K}(\mathcal{J})=\phi (\mathcal{H})\cap \phi (\mathcal{J})$. Now,

$$\begin{array}{ccc}\hfill \phi (\mathcal{H}\bigsqcup \mathcal{J})& =& \mathcal{K}(\mathcal{H}\bigsqcup \mathcal{J})\hfill \\ & =& \mathcal{K}({(\mathcal{H}\vee \mathcal{J})}^{★})\mathrm{by}\mathrm{Theorem}4\hfill \\ & =& \mathcal{K}(\mathcal{H}\vee \mathcal{J})\mathrm{by}\mathrm{Lemma}7(2)\hfill \\ & =& \mathcal{K}(\mathcal{H})\cup \mathcal{K}(\mathcal{J})\mathrm{by}\mathrm{Lemma}6(3)\hfill \\ & =& \phi (\mathcal{H})\cup \phi (\mathcal{J}).\hfill \end{array}$$

Therefore, $\phi$ is a homomorphism. Hence, $\mathcal{I}(\mathcal{R})\cong \Im$. □

Given any subset $\mathcal{G}\subseteq \mathcal{R}$, let $\mathcal{H}(\mathcal{G})=\{\mathcal{A}\in Spe{c}^{★}(\mathcal{R})\mid \mathcal{G}\subseteq \mathcal{A}\}$. It is evident that $\mathcal{H}(\mathcal{G})=Spe{c}^{★}(\mathcal{R})\setminus \mathcal{K}(\mathcal{G})$. Hence, $\mathcal{H}(\mathcal{G})$ forms a closed set within $Spe{c}^{★}(\mathcal{R})$. Moreover, any closed set in $Spe{c}^{★}(\mathcal{R})$ can be formed as $\mathcal{H}(\mathcal{G})$ for some subset $\mathcal{G}\subseteq \mathcal{R}$. Now, we have the following result.

Theorem 13.

For $\mathcal{B}\subseteq Spe{c}^{★}(\mathcal{R})$, $\overline{\mathcal{G}}=\mathcal{H}({\displaystyle \bigcap _{\mathcal{A}\in \mathcal{G}}}(\mathcal{A}))$.

Proof. 

Let $\mathcal{G}\subseteq Spe{c}^{★}(\mathcal{R})$, and assume $\mathcal{V}\in \mathcal{G}$. Since ${\displaystyle {\displaystyle \bigcap }_{\mathcal{A}\in \mathcal{G}}}\mathcal{A}\subseteq \mathcal{V}$, it follows that $\mathcal{V}\in \mathcal{H}\left({\displaystyle {\displaystyle \bigcap }_{\mathcal{A}\in \mathcal{G}}}\mathcal{A}\right)$. Therefore, the set $\mathcal{H}\left({\displaystyle {\displaystyle \bigcap }_{\mathcal{A}\in \mathcal{G}}}\mathcal{A}\right)$ is a closed set that contains $\mathcal{G}$. Now, let $\mathcal{L}$ be any closed set in $Spe{c}^{★}(\mathcal{R})$. Then, $\mathcal{L}=\mathcal{H}(\mathcal{K})$ for some $\mathcal{K}\subseteq \mathcal{R}$. Since $\mathcal{G}\subseteq \mathcal{L}=\mathcal{H}(\mathcal{K})$, we deduce that $\mathcal{K}\subseteq \mathcal{A}$ for all $\mathcal{A}\in \mathcal{G}$, meaning $\mathcal{K}\subseteq {\displaystyle {\displaystyle \bigcap }_{\mathcal{A}\in \mathcal{G}}}\mathcal{A}$. This clearly gives $\mathcal{H}\left({{\displaystyle \bigcap }}_{\mathcal{A}\in \mathcal{G}}\mathcal{A}\right)\subseteq \mathcal{H}(\mathcal{K})=\mathcal{L}$. Therefore, $\mathcal{H}\left({\displaystyle {\displaystyle \bigcap }_{\mathcal{A}\in \mathcal{G}}}\mathcal{A}\right)$ is the smallest closed set that contains $\mathcal{G}$, which gives us $\overline{\mathcal{G}}=\mathcal{H}\left({\displaystyle {\displaystyle \bigcap }_{\mathcal{A}\in \mathcal{G}}}\mathcal{A}\right)$. □

Theorem 14.

The given conditions in below are equivalent for an ADL $\mathcal{R}$ with maximal element m:

(1)

Every member of $Spe{c}^{★}(\mathcal{R})$ is maximal;

(2)

Every member of $Spe{c}^{★}(\mathcal{R})$ is minimal;

(3)

$Spe{c}^{★}(\mathcal{R})$ is a $T_1$-space;

(4)

$Spe{c}^{★}(\mathcal{R})$ is a $T_2$-space;

(5)

For every $\theta ,\vartheta \in \mathcal{R}$, $\theta \vee \ae \in {(m)}^{★}$ and $\mathcal{K}(b)\cap \{Spe{c}^{★}(R)\setminus \mathcal{K}(a)\}=\mathcal{K}(b\vee c)$ for some $\ae \in \mathcal{R}.$

Proof. $(1)\iff (2)$: As every maximal star filter is prime, this is clear.

• $(2)\Rightarrow (3)$: Assume (2). Let $\mathcal{A},\mathcal{V}\in Spe{c}^{★}(R)$ with $\mathcal{A}\ne \mathcal{V}.$ By (2), we have $\mathcal{A}$ and $\mathcal{V}$ are minimal. This implies that $\mathcal{A}⊈\mathcal{V}$ and $\mathcal{V}⊈\mathcal{A}$. Now, select $\theta \in \mathcal{A}\setminus \mathcal{V}$ and $\vartheta \in \mathcal{V}\setminus \mathcal{A}$. Hence, $\mathcal{V}\in \mathcal{K}(\theta )\setminus \mathcal{K}(\vartheta )$ and $\mathcal{A}\in \mathcal{K}(\vartheta )\setminus \mathcal{K}(\theta )$. Thus, $Spe{c}^{★}(\mathcal{R})$ is a $T_1$-space.
• $(3)\Rightarrow (4)$: Assume (3). Let $\mathcal{A}\in Spe{c}^{★}(\mathcal{R})$. According to Theorem 13, we have $\{\mathcal{A}\}=\overline{\{\mathcal{A}\}}=\{\mathcal{V}\in Spe{c}_{\mathcal{H}}^{★}(\mathcal{R})\mid \mathcal{A}\subseteq \mathcal{V}\}$. This indicates that $\mathcal{A}$ is maximal. Since all maximal star filters are prime, we conclude that every prime star filter is minimal. Let $\mathcal{A},\mathcal{V}\in Spe{c}^{★}(\mathcal{R})$ with $\mathcal{A}\ne \mathcal{V}$. Select an element $\theta \in \mathcal{A}$ such that $\theta \notin \mathcal{V}$. Given that $\mathcal{A}$ is minimal, there exists an element $\vartheta \notin \mathcal{A}$ for which $\theta \vee \vartheta \in {(m)}^{★}$. Therefore, we find that $\mathcal{A}\in \mathcal{K}(\vartheta )$, $\mathcal{V}\in \mathcal{K}(\theta )$, and $\mathcal{K}(\theta )\cap \mathcal{K}(\vartheta )=\mathcal{K}(\theta \vee \vartheta )=\varnothing$. Thus, we conclude that $Spe{c}^{★}(\mathcal{R})$ is a $T_2$-space.
• $(4)\Rightarrow (5)$: Assume that $Spe{c}^{★}(\mathcal{R})$ is a $T_2$-space. Consequently, for every $\theta \in \mathcal{R}$, the set $\mathcal{K}(\theta )$ is a compact subset of $Spe{c}^{★}(\mathcal{R})$. Thus, $\mathcal{K}(\theta )$ is also a clopen subset of $Spe{c}^{★}(\mathcal{R})$. Let $\theta ,\vartheta \in \mathcal{R}$ such that $\theta \ne \vartheta$. Then, the intersection $\mathcal{K}(\vartheta )\cap (Spe{c}^{★}(\mathcal{R})\setminus \mathcal{K}(\theta ))$ forms a compact subset of the compact space $\mathcal{K}(\vartheta )$. Since $\mathcal{K}(\vartheta )$ is open in $Spe{c}^{★}(\mathcal{R})$, it follows that $\mathcal{K}(\vartheta )\cap (Spe{c}^{★}(\mathcal{R})\setminus \mathcal{K}(\theta ))$ is a compact open subset of $Spe{c}^{★}(\mathcal{R})$. Therefore, by Theorem 11(2), there exists an element $\ae \in \mathcal{R}$ such that $\mathcal{K}(\ae )=\mathcal{K}(\vartheta )\cap (Spe{c}^{★}(\mathcal{R})\setminus \mathcal{K}(\theta ))$. Thus, we have that $\mathcal{K}(\vartheta )\cap (Spe{c}^{★}(\mathcal{R})\setminus \mathcal{K}(\theta ))\setminus (\mathcal{K}(\vartheta )\cap \mathcal{K}(\ae ))=\mathcal{K}(\vartheta \vee \ae )$. Moreover, $\mathcal{K}(\theta \vee \ae )=\mathcal{K}(\theta )\cap \mathcal{K}(\ae )=\varnothing$. Hence, it follows that $\theta \vee \ae \in {(m)}^{★}$.
• $(5)\Rightarrow (1)$: Let $\mathcal{A}\in Spe{c}^{★}(\mathcal{R})$. We can choose elements $\theta$ and $\vartheta$ from $\mathcal{R}$ such that $\theta \in \mathcal{A}$ and $\vartheta \notin \mathcal{A}$. By condition (5), there is an element $\ae \in \mathcal{R}$ satisfying $\theta \vee \ae \in {(m)}^{★}$ and the condition that $\mathcal{K}(\vartheta )\cap (Spe{c}^{★}(\mathcal{R})\setminus \mathcal{K}(\theta ))=\mathcal{K}(\vartheta \vee \ae )$. It follows that $\mathcal{A}\in \mathcal{K}(\vartheta )\cap (Spe{c}^{★}(\mathcal{R})\setminus \mathcal{K}(\theta ))=\mathcal{K}(\vartheta \vee \ae )$. If $\ae \in \mathcal{A}$, then $\vartheta \vee \ae \in \mathcal{A}$, which contradicts the fact that $\mathcal{A}\in \mathcal{K}(\vartheta \vee \ae )$. Therefore, we can conclude that $\ae$ cannot be in $\mathcal{A}$. This means that for every $\theta \in \mathcal{A}$, there exists an element $\ae \notin \mathcal{A}$ such that $\theta \vee \ae \in {(m)}^{★}$. Consequently, we can conclude that $\mathcal{A}$ is a minimal prime star filter. □

For any ADL $\mathcal{R}$, it is evident that $\mathcal{Z}(\mathcal{G})=Spe{c}^{★}(\mathcal{R})\setminus \mathcal{K}(\mathcal{G})$, which implies that $\mathcal{Z}(\mathcal{G})$ constitutes a closed set within $Spe{c}^{★}(\mathcal{R})$. The following result establishes a necessary and sufficient condition for the space $Spe{c}^{★}(\mathcal{R})$ to be regular.

Theorem 15.

In an ADL $\mathcal{R},$ $Spe{c}^{★}(\mathcal{R})$ is a regular space if and only if, for every $\mathcal{A}\in Spe{c}^{★}(\mathcal{R})$ and $a\notin \mathcal{A}$, there is a filter $\mathcal{S}$ in $\mathcal{R}$ along with an element $\vartheta \in \mathcal{R}$ such that $\mathcal{A}\in \mathcal{K}(\vartheta )\subseteq \mathcal{Z}(\mathcal{S})\subseteq \mathcal{K}(\theta )$.

Proof. 

Assuming that $Spe{c}^{★}(\mathcal{R})$ is a regular space, let $\mathcal{A}$ be an element of $Spe{c}^{★}(\mathcal{R})$ and consider $\theta \notin \mathcal{A}$ for some $\theta \in \mathcal{R}$. This implies $\mathcal{A}\notin \mathcal{Z}(\theta )$. Since $Spe{c}^{★}(\mathcal{R})$ is regular, there exist two disjoint open sets $\mathcal{J}$ and $\mathcal{F}$ in $Spe{c}^{★}(\mathcal{R})$ such that $\mathcal{A}\in \mathcal{J}$ and $\mathcal{Z}(\theta )\subseteq \mathcal{F}$. Thus, we have $Spe{c}^{★}(\mathcal{R})\setminus \mathcal{F}\subseteq \mathcal{K}(\theta )$. Given that $Spe{c}^{★}(\mathcal{R})\setminus \mathcal{F}$ is a closed set, we can express it as $Spe{c}^{★}(\mathcal{R})\setminus \mathcal{F}=\mathcal{Z}(\mathcal{H})$ for some star filter $\mathcal{H}$ in $\mathcal{R}$. Therefore, it follows that $\mathcal{Z}(\mathcal{H})=Spe{c}^{★}(\mathcal{R})\setminus \mathcal{F}\subseteq \mathcal{K}(\theta )$. Next, since $\mathcal{J}\cap \mathcal{F}=\varnothing$, we conclude that $\mathcal{F}\subseteq Spe{c}^{★}(\mathcal{R})\setminus \mathcal{J}$. The set $Spe{c}^{★}(\mathcal{R})\setminus \mathcal{J}$ is also closed, which allows us to write $Spe{c}^{★}(\mathcal{R})\setminus \mathcal{J}=\mathcal{Z}(\mathcal{D})$ for some star filter $\mathcal{D}$ in $\mathcal{R}$. Since $\mathcal{A}\in \mathcal{J}$, it follows that $\mathcal{A}\notin Spe{c}^{★}(\mathcal{R})\setminus \mathcal{J}=\mathcal{Z}(\mathcal{D})$, indicating that $\mathcal{D}\notin \mathcal{A}$. We can choose an element $\vartheta \in \mathcal{D}$ such that $\vartheta \notin \mathcal{A}$, which implies $\mathcal{A}\in \mathcal{K}(\vartheta )$. Let $\mathcal{U}\in \mathcal{F}$. Because $\mathcal{F}\subseteq \mathcal{Z}(\mathcal{D})$, we have $\mathcal{D}\subseteq \mathcal{U}$. Since $\vartheta \in \mathcal{D}\subseteq \mathcal{U}$, it follows that $\mathcal{U}\in \mathcal{Z}(\vartheta )$. Therefore, we conclude that $\mathcal{F}\subseteq \mathcal{Z}(\vartheta )$. According to (1), we find that $\mathcal{K}(\vartheta )=Spe{c}^{★}(\mathcal{R})\setminus \mathcal{Z}(\vartheta )\subseteq Spe{c}^{★}(\mathcal{R})\setminus \mathcal{F}=\mathcal{Z}(\mathcal{H})$, which leads to the result that $\mathcal{K}(\vartheta )\subseteq \mathcal{Z}(\mathcal{H})$. Thus, for any $\mathcal{A}\in Spe{c}^{★}(\mathcal{R})$ and $\theta \notin \mathcal{A}$, we have a filter $\mathcal{H}$ in $\mathcal{R}$ and an element $\vartheta \in \mathcal{R}$ such that $\mathcal{A}\in \mathcal{K}(\vartheta )\subseteq \mathcal{Z}(\mathcal{H})\subseteq \mathcal{K}(\theta )$. Conversely, assume that for every $\mathcal{A}\in Spe{c}^{★}(\mathcal{R})$ and for any $\theta \notin \mathcal{A}$ there exist a filter $\mathcal{H}$ in $\mathcal{R}$ and an element $\vartheta \in \mathcal{R}$ such that $\mathcal{A}\in \mathcal{K}(\vartheta )\subseteq \mathcal{Z}(\mathcal{H})\subseteq \mathcal{K}(\theta )$. To demonstrate that the space $Spe{c}^{★}(\mathcal{R})$ is regular, let $\mathcal{A}\in Spe{c}^{★}(\mathcal{R})$ and consider any closed set $\mathcal{Z}(\mathcal{Q})$ in $Spe{c}^{★}(\mathcal{R})$ such that $\mathcal{A}\notin \mathcal{Z}(\mathcal{Q})$. This implies that $\mathcal{Q}⊈\mathcal{A}$. Therefore, there exists an element $\theta \in \mathcal{Q}$ such that $\theta \notin \mathcal{A}$. Consequently, we have $\mathcal{A}\in \mathcal{K}(\theta )$. Since $\theta \notin \mathcal{A}$, by our assumption, there exists a filter $\mathcal{H}$ in $\mathcal{R}$ and an element $\vartheta \in \mathcal{R}$ such that $\mathcal{A}\in \mathcal{K}(\vartheta )\subseteq \mathcal{Z}(\mathcal{H})\subseteq \mathcal{K}(\theta )$. Hence, we have $\mathcal{K}(\theta )\cap \mathcal{Z}(\mathcal{Q})=\varnothing$, due to $\mathcal{Q}\in \mathcal{Z}(\theta )$ for $\theta \in \mathcal{Q}$. This leads to the conclusion that $\mathcal{Z}(\mathcal{Q})\subseteq Spe{c}^{★}(\mathcal{R})\setminus \mathcal{K}(\theta )\subseteq Spe{c}^{★}(\mathcal{R})\setminus \mathcal{Z}(\mathcal{H})$. Additionally, we have $\mathcal{K}(\vartheta )\cap \mathcal{K}(\mathcal{H})=\varnothing$. Therefore, we can find two disjoint open sets $\mathcal{K}(\vartheta )$ and $\mathcal{K}(\mathcal{H})$ such that $\mathcal{A}\in \mathcal{K}(\vartheta )$ and $\mathcal{Z}(\mathcal{Q})\subseteq \mathcal{K}(\mathcal{H})$. Thus, we conclude that $Spe{c}^{★}(\mathcal{R})$ is a regular space. □

5. Conclusions

This study introduced the concepts of star filters and starlets in almost distributive lattices (ADLs) and demonstrated that the set of star filters forms a distributive lattice, with starlets as a sublattice. An element-wise characterization of star filters was established, along with conditions for an ideal in an ADL to become a star filter. A congruence relation was defined, ensuring that the quotient lattice structure transforms into a Boolean algebra under specific conditions. Important properties of prime star filters were examined, with equivalent conditions identified for their prime spectrum to be a Hausdorff space. Furthermore, a necessary and sufficient condition for the prime spectrum to form a regular space was provided. Future work can focus on introducing fuzzy star filters in almost distributive lattices (ADLs) to handle uncertainty using fuzzy logic. This includes defining fuzzy star filters, studying their structure, and finding conditions for fuzzy starlets to form a sublattice. Researchers can also explore fuzzy congruences and how they affect quotient lattices, potentially forming fuzzy Boolean algebras. The topological properties of the fuzzy prime spectrum, like being Hausdorff or regular, can be analyzed.