Green tea catechins and intracellular calcium dynamics in prostate cancer cells

Carla Marchetti; Carla Marchetti

doi:10.3934/molsci.2021001

AIMS Molecular Science

2021, Volume 8, Issue 1: 1-12. doi: 10.3934/molsci.2021001

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Research article

Green tea catechins and intracellular calcium dynamics in prostate cancer cells

Carla Marchetti ^,

Istituto di Biofisica, Consiglio Nazionale delle Ricerche, via De Marini 6, 16149 Genova, Italy

Received: 11 November 2020 Accepted: 22 December 2020 Published: 28 December 2020

Perturbations of internal Ca²⁺ ([Ca²⁺]_i) homeostasis play a key role in several pathologies and in neoplastic transformation, where deregulated cell proliferation, together with the suppression of apoptosis, provides the condition for abnormal tissue growth and invasion. Green tea catechins have been shown to affect cancer development by interference with basic cellular functions, most of which are mediated by [Ca²⁺]_i. Prostate cancer (PCa) is one of the most common malignancy in men in Western countries and the androgen-independent carcinoma is a lethal form for which there is still no effective therapy. Different evidences suggested that consumption of green tea may have beneficial effects against PCa. We have previously described how the main green tea flavonoid, (−)-epigallocatechin-3-gallate (EGCG), inhibited proliferation and induced dose-dependent peaks of [Ca²⁺]_i in metastatic androgen-insensitive DU145 and PC3 PCa cells, by a mechanism that combined Ca²⁺ entry and Ca²⁺-induced Ca²⁺ release. In the present study, we studied the effect of green tea extract (GTE) on the same cell lines. Proliferation, measured by MTT assay, was inhibited by GTE with IC₅₀ close to 60 µg/ml, a value that is higher than that expected by EGCG effect alone. [Ca²⁺]_i, measured in real time by the fluorescent dye Fura-2, was transiently increased by GTE by a mechanism that resembled that described for EGCG, but was largely independent of external Ca²⁺. These observations suggested that other components, acting in synergy with EGCG, were involved in GTE effect, and confirmed the view that the alleged health benefits of green tea for PCa prevention may be related to [Ca²⁺]_i deregulation in malignant cells. These results may be significant to understand the functional mechanisms by which flavonoids exert their beneficial or toxic actions.

Keywords:

androgen-resistant prostate cancer PC3 cells,
DU145 cells,
green tea extract,
internal Ca2+ release

Citation: Carla Marchetti. Green tea catechins and intracellular calcium dynamics in prostate cancer cells[J]. AIMS Molecular Science, 2021, 8(1): 1-12. doi: 10.3934/molsci.2021001

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Abstract

1. Introduction

Henkin and Skolem introduced Hilbert algebras in the fifties for investigations in intuitionistic and other non-classical logics. Diego ^[4] proved that Hilbert algebras form a variety which is locally finite. Bandaru et al. introduced the notion of GE-algebras which is a generalization of Hilbert algebras, and investigated several properties (see ^[1,2,7,8,9]). The notion of interior operator is introduced by Vorster ^[12] in an arbitrary category, and it is used in ^[3] to study the notions of connectedness and disconnectedness in topology. Interior algebras are a certain type of algebraic structure that encodes the idea of the topological interior of a set, and are a generalization of topological spaces defined by means of topological interior operators. Rachůnek and Svoboda ^[6] studied interior operators on bounded residuated lattices, and Svrcek ^[11] studied multiplicative interior operators on GMV-algebras. Lee et al. ^[5] applied the interior operator theory to GE-algebras, and they introduced the concepts of (commutative, transitive, left exchangeable, belligerent, antisymmetric) interior GE-algebras and bordered interior GE-algebras, and investigated their relations and properties. Later, Song et al. ^[10] introduced the notions of an interior GE-filter, a weak interior GE-filter and a belligerent interior GE-filter, and investigate their relations and properties. They provided relations between a belligerent interior GE-filter and an interior GE-filter and conditions for an interior GE-filter to be a belligerent interior GE-filter is considered. Given a subset and an element, they established an interior GE-filter, and they considered conditions for a subset to be a belligerent interior GE-filter. They studied the extensibility of the belligerent interior GE-filter and established relationships between weak interior GE-filter and belligerent interior GE-filter of type 1, type 2 and type 3. Rezaei et al. ^[7] studied prominent GE-filters in GE-algebras. The purpose of this paper is to study by applying interior operator theory to prominent GE-filters in GE-algebras. We introduce the concept of a prominent interior GE-filter, and investigate their properties. We discuss the relationship between a prominent GE-filter and a prominent interior GE-filter and the relationship between an interior GE-filter and a prominent interior GE-filter. We find and provide examples where any interior GE-filter is not a prominent interior GE-filter and any prominent GE-filter is not a prominent interior GE-filter. We provide conditions for an interior GE-filter to be a prominent interior GE-filter. We provide conditions under which an internal GE-filter larger than a given internal GE filter can become a prominent internal GE-filter, and give an example describing it. We also introduce the concept of a prominent interior GE-filter of type 1 and type 2, and investigate their properties. We discuss the relationship between a prominent interior GE-filter and a prominent interior GE-filter of type 1. We give examples to show that A and B are independent of each other, where A and B are:

$(1)$ $\left\{\begin{array}{ll} \text{ A: prominent interior GE-filter of type 1.}\\ \text{ B: prominent interior GE-filter of type 2.} \end{array}\right.$

$(2)$ $\left\{\begin{array}{ll} \text{ A: prominent interior GE-filter.}\\ \text{ B: prominent interior GE-filter of type 2.} \end{array}\right.$

$(3)$ $\left\{\begin{array}{ll} \text{ A: interior GE-filter.}\\ \text{ B: prominent interior GE-filter of type 1.} \end{array}\right.$

$(4)$ $\left\{\begin{array}{ll} \text{ A: interior GE-filter.}\\ \text{ B: prominent interior GE-filter of type 2.} \end{array}\right.$

2. Preliminaries

Definition 2.1. ^[1] By a GE-algebra we mean a non-empty set ${X}$ with a constant $1$ and a binary operation ${*}$ satisfying the following axioms:

(GE1) ${u}{*} {u} = 1$ ,

(GE2) $1{*}{u} = {u}$ ,

(GE3) ${u}{*} ({v}{*} {w}) = {u}{*} ({v}{*} ({u}{*} {w}))$

for all ${u}, {v}, {w}\in {X}$ .

In a GE-algebra ${X}$ , a binary relation " $\le$ " is defined by

$\begin{align} (\forall {x}, {y}\in {X}) \left({x}\le {y} \; \Leftrightarrow \; {x}{*} {y} = 1\right). \end{align}$

(2.1)

Definition 2.2. ^[1,2,8] A GE-algebra ${X}$ is said to be transitive if it satisfies:

$\begin{align} (\forall {x}, {y}, {z}\in {X})\left({x}{*} {y}\le ({z}{*} {x}){*} ({z}{*} {y})\right). \end{align}$

(2.2)

Proposition 2.3. ^[1] Every GE-algebra ${X}$ satisfies the following items:

$\begin{align} & (\forall {u}\in {X}) \left({u}{*} 1 = 1\right). \end{align}$

(2.3)

$\begin{align} & (\forall {u}, {v}\in {X}) \left({u}{*} ({u}{*} {v}) = {u}{*} {v}\right). \end{align}$

(2.4)

$\begin{align} & (\forall {u}, {v}\in {X}) \left({u} \le {v}{*} {u}\right). \end{align}$

(2.5)

$\begin{align} & (\forall {u}, {v}, {w}\in {X}) \left({u}{*} ({v}{*} {w})\le {v}{*} ({u}{*} {w})\right). \end{align}$

(2.6)

$\begin{align} & (\forall {u}\in {X}) \left({1}\le {u} \; \Rightarrow \; {u} = {1}\right). \end{align}$

(2.7)

$\begin{align} & (\forall {u}, {v}\in {X}) \left({u}\le ({v}{*} {u}){*} {u}\right). \end{align}$

(2.8)

$\begin{align} & (\forall {u}, {v}\in {X}) \left({u}\le ({u}{*} {v}){*} {v}\right). \end{align}$

(2.9)

$\begin{align} & (\forall {u}, {v}, {w}\in {X}) \left({u}\le {v}{*} {w} \; \Leftrightarrow \; {v}\le {u}{*} {w}\right). \end{align}$

(2.10)

If ${X}$ is transitive, then

$\begin{align} & (\forall {u}, {v}, {w}\in {X}) \left({u}\le {v} \; \Rightarrow \; {w}{*} {u}\le {w}{*} {v}, \; {v}{*} {w}\le {u}{*} {w}\right). \end{align}$

(2.11)

$\begin{align} & (\forall {u}, {v}, {w}\in {X}) \left({u}{*} {v}\le ({v}{*} {w}){*} ({u}{*} {w})\right). \end{align}$

(2.12)

Lemma 2.4. ^[1] In a GE-algebra ${X}$ , the following facts are equivalent each other.

$\begin{align} & (\forall {x}, {y}, {z}\in {X}) \left({x}{*} {y}\le ({z}{*} {x}){*} ({z}{*} {y})\right). \end{align}$

(2.13)

$\begin{align} & (\forall {x}, {y}, {z}\in {X}) \left({x}{*} {y}\le ({y}{*} {z}){*} ({x}{*} {z})\right). \end{align}$

(2.14)

Definition 2.5. ^[1] A subset ${F}$ of a GE-algebra ${X}$ is called a GE-filter of ${X}$ if it satisfies:

$\begin{align} & 1\in {F}, \end{align}$

(2.15)

$\begin{align} & (\forall {x}, {y}\in {X}) ({x}{*} {y}\in {F}, \; {x}\in {F} \; \Rightarrow \; {y}\in {F}). \end{align}$

(2.16)

Lemma 2.6. ^[1] In a GE-algebra ${X}$ , every filter ${F}$ of ${X}$ satisfies:

$\begin{align} (\forall {x}, {y}\in {X}) \left({x}\le {y}, \; {x}\in {F} \; \Rightarrow \; {y}\in {F}\right). \end{align}$

(2.17)

Definition 2.7. ^[7] A subset ${F}$ of a GE-algebra ${X}$ is called a prominent GE-filter of ${X}$ if it satisfies (2.15) and

$\begin{align} (\forall {x}, {y}, {z}\in {X}) ({x}{*} ({y}{*} {z})\in {F}, \; {x}\in {F} \; \Rightarrow \; (({z}{*} {y}){*} {y}){*} {z}\in {F}). \end{align}$

(2.18)

Note that every prominent GE-filter is a GE-filter in a GE-algebra (see ^[7]).

Definition 2.8. ^[5] By an interior GE-algebra we mean a pair $({X}, f)$ in which ${X}$ is a GE-algebra and $f :{X} \rightarrow {X}$ is a mapping such that

$\begin{align} & (\forall {x}\in {X}) ({x} \le f({x})), \end{align}$

(2.19)

$\begin{align} & (\forall {x}\in {X}) (( f \circ f)({x}) = f({x})), \end{align}$

(2.20)

$\begin{align} & (\forall {x}, {y}\in {X}) ({x}\le {y} \; \Rightarrow \; f({x})\le f({y})). \end{align}$

(2.21)

Definition 2.9. ^[10] Let $({X}, f)$ be an interior GE-algebra. A GE-filter ${F}$ of ${X}$ is said to be interior if it satisfies:

$\begin{align} (\forall {x}\in {X}) ( f({x})\in {F} \; \Rightarrow \; {x}\in {F}). \end{align}$

(2.22)

3. Prominent interior GE-filters

Definition 3.1. Let $({X}, f)$ be an interior GE-algebra. Then a subset ${F}$ of ${X}$ is called a prominent interior GE-filter in $({X}, f)$ if ${F}$ is a prominent GE-filter of ${X}$ which satisfies the condition (2.22).

Example 3.2. Let ${X} = \{{1}, {2}, {3}, {4}, {5}\}$ be a set with the Cayley table which is given in Table 1.

Table 1. Cayley table for the binary operation "

${*}$ ".

${*}$	${1}$	${2}$	${3}$	${4}$	${5}$
${1}$	${1}$	${2}$	${3}$	${4}$	${5}$
${2}$	${1}$	${1}$	${1}$	${4}$	${4}$
${3}$	${1}$	${1}$	${1}$	${5}$	${5}$
${4}$	${1}$	${2}$	${3}$	${1}$	${1}$
${5}$	${1}$	${2}$	${2}$	${1}$	${1}$

| Show Table

DownLoad: CSV

Then ${X}$ is a GE-algebra. If we define a mapping $f$ on ${X}$ as follows:

$\begin{align*} f : {X} \rightarrow {X}, \; {x}\mapsto \left\{\begin{array}{ll} {1} & \text{if } \, {x} \in \{{1}, {4}, {5}\}, \\ {2} & \text{if } \, {x} \in \{{2}, {3}\}, \end{array}\right. \end{align*}$

then $({X}, f)$ is an interior GE-algebra and ${F} = \{{1}, {4}, {5}\}$ is a prominent interior GE-filter in $({X}, f)$ .

It is clear that every prominent interior GE-filter is a prominent GE-filter. But any prominent GE-filter may not be a prominent interior GE-filter in an interior GE-algebra as seen in the following example.

Example 3.3. Let ${X} = \{{1}, {2}, {3}, {4}, {5}\}$ be a set with the Cayley table which is given in Table 2,

Table 2. Cayley table for the binary operation "

${*}$ ".

${*}$	${1}$	${2}$	${3}$	${4}$	${5}$
${1}$	${1}$	${2}$	${3}$	${4}$	${5}$
${2}$	${1}$	${1}$	${3}$	${4}$	${1}$
${3}$	${1}$	${2}$	${1}$	${4}$	${5}$
${4}$	${1}$	${2}$	${3}$	${1}$	${5}$
${5}$	${1}$	${1}$	${3}$	${4}$	${1}$

| Show Table

DownLoad: CSV

and define a mapping $f$ on ${X}$ as follows:

$\begin{align*} f : {X} \rightarrow {X}, \; {x}\mapsto \left\{\begin{array}{ll} {1} & \text{if } \, {x} \in \{{1}, {2}, {3}, {5}\}, \\ {4} & \text{if } \, {x} = {4}. \end{array}\right. \end{align*}$

Then $({X}, f)$ is an interior GE-algebra and ${F} : = \{{1}\}$ is a prominent GE-filter of ${X}$ . But it is not a prominent interior GE-filter in $({X}, f)$ since $f({2}) = {1}\in {F}$ but ${2}\notin {F}$ .

We discuss relationship between interior GE-filter and prominent interior GE-filter.

Theorem 3.4. In an interior GE-algebra, every prominent interior GE-filter is an interior GE-filter.

Proof. It is straightforward because every prominent GE-filter is a GE-filter in a GE-algebra.

In the next example, we can see that any interior GE-filter is not a prominent interior GE-filter in general.

Example 3.5. Let ${X} = \{{1}, {2}, {3}, {4}, {5}\}$ be a set with the Cayley table which is given in Table 3.

Table 3. Cayley table for the binary operation "

${*}$ ".

${*}$	${1}$	${2}$	${3}$	${4}$	${5}$
${1}$	${1}$	${2}$	${3}$	${4}$	${5}$
${2}$	${1}$	${1}$	${1}$	${4}$	${4}$
${3}$	${1}$	${2}$	${1}$	${4}$	${4}$
${4}$	${1}$	${1}$	${3}$	${1}$	${1}$
${5}$	${1}$	${1}$	${1}$	${1}$	${1}$

| Show Table

DownLoad: CSV

Then ${X}$ is a GE-algebra. If we define a mapping $f$ on ${X}$ as follows:

$\begin{align*} f : {X} \rightarrow {X}, \; {x}\mapsto \left\{\begin{array}{ll} {1} & \text{if } \, {x} = {1}, \\ {2} & \text{if } \, {x} \in \{{2}, {4}, {5}\}, \\ {3} & \text{if } \, {x} = {3}, \end{array}\right. \end{align*}$

then $({X}, f)$ is an interior GE-algebra and ${F} = \{{1}\}$ is an interior GE-filter in $({X}, f)$ . But it is not a prominent interior GE-filter in $({X}, f)$ since ${1} {*} ({2}{*} {3}) = {1} \in {F}$ but $(({3}{*} {2}){*} {2}){*} {3} = {3} \notin {F}$ .

Proposition 3.6. Every prominent interior GE-filter ${F}$ in an interior GE-algebra $({X}, f)$ satisfies:

$\begin{align} (\forall {x}, {y}\in {X}) \left( f({x}{*} {y})\in {F} \; \Rightarrow (({y}{*} {x}){*} {x}){*} {y} \in {F}\right). \end{align}$

(3.1)

Proof. Let ${F}$ be a prominent interior GE-filter in $({X}, f)$ . Let ${x}, {y}\in {X}$ be such that $f({x}{*} {y})\in {F}$ . Then ${x}{*} {y}\in {F}$ by (2.22), and so $1{*} ({x}{*} {y}) = {x}{*} {y}\in {F}$ by (GE2). Since $1\in {F}$ , it follows from (2.18) that $(({y}{*} {x}){*} {x}){*} {y} \in {F}$ .

Corollary 3.7. Every prominent interior GE-filter ${F}$ in an interior GE-algebra $({X}, f)$ satisfies:

$\begin{align} (\forall {x}, {y}\in {X}) \left({x}{*} {y}\in {F} \; \Rightarrow (({y}{*} {x}){*} {x}){*} {y} \in {F}\right). \end{align}$

(3.2)

Proof. Let ${F}$ be a prominent interior GE-filter in $({X}, f)$ . Then ${F}$ is an interior GE-filter in $({X}, f)$ by Theorem 3.4. Let ${x}, {y}\in {X}$ be such that ${x}{*} {y}\in {F}$ . Since ${x}{*} {y} \le f({x}{*} {y})$ by (2.19), it follows from Lemma 2.6 that $f({x}{*} {y})\in {F}$ . Hence $(({y}{*} {x}){*} {x}){*} {y} \in {F}$ by Proposition 3.6.

Corollary 3.8. Every prominent interior GE-filter ${F}$ in an interior GE-algebra $({X}, f)$ satisfies:

$\begin{align*} (\forall {x}, {y}\in {X}) \left({x}{*} {y}\in {F} \; \Rightarrow \; f((({y}{*} {x}){*} {x}){*} {y}) \in {F}\right). \end{align*}$

Proof. Straightforward.

Corollary 3.9. Every prominent interior GE-filter ${F}$ in an interior GE-algebra $({X}, f)$ satisfies:

$\begin{align*} (\forall {x}, {y}\in {X}) \left( f({x}{*} {y})\in {F} \; \Rightarrow \; f((({y}{*} {x}){*} {x}){*} {y}) \in {F}\right). \end{align*}$

Proof. Straightforward.

In the following example, we can see that any interior GE-filter ${F}$ in an interior GE-algebra $({X}, f)$ does not satisfy the conditions (3.1) and (3.2).

Example 3.10. Consider the interior GE-algebra $({X}, f)$ in Example 3.4. The interior GE-filter ${F} : = \{{1}\}$ does not satisfy conditions (3.1) and (3.2) since $f({2}{*} {3}) = f({1}) = {1}\in {F}$ and ${2}{*} {3} = {1}\in {F}$ but $(({3}{*} {2}){*} {2}){*} {3} = {3} \notin {F}$ .

We provide conditions for an interior GE-filter to be a prominent interior GE-filter.

Theorem 3.11. If an interior GE-filter ${F}$ in an interior GE-algebra $({X}, f)$ satisfies the condition $(3.1)$ , then ${F}$ is a prominent interior GE-filter in $({X}, f)$ .

Proof. Let ${F}$ be an interior GE-filter in $({X}, f)$ that satisfies the condition (3.1). Let ${x}, {y}, {z}\in {X}$ be such that ${x}{*} ({y}{*} {z})\in {F}$ and ${x}\in {F}$ . Then ${y}{*} {z}\in {F}$ . Since ${y}{*} {z} \le f({y}{*} {z})$ by (2.19), it follows from Lemma 2.6 that $f({y}{*} {z})\in {F}$ . Hence $(({z}{*} {y}){*} {y}){*} {z} \in {F}$ by (3.1), and therefore ${F}$ is a prominent interior GE-filter in $({X}, f)$ .

Lemma 3.12. ^[10] In an interior GE-algebra, the intersection of interior GE-filters is also an interior GE-filter.

Theorem 3.13. In an interior GE-algebra, the intersection of prominent interior GE-filters is also a prominent interior GE-filter.

Proof. Let $\{{F}_i \mid i\in \Lambda \}$ be a set of prominent interior GE-filters in an interior GE-algebra $({X}, f)$ where $\Lambda$ is an index set. Then $\{{F}_i \mid i\in \Lambda \}$ is a set of interior GE-filters in $({X}, f)$ , and so $\cap \{{F}_i \mid i\in \Lambda \}$ is an interior GE-filter in $({X}, f)$ by Lemma 3.12. Let ${x}, {y}\in {X}$ be such that $f({x}{*} {y})\in \cap \{{F}_i \mid i\in \Lambda \}$ . Then $f({x}{*} {y})\in {F}_i$ for all $i\in \Lambda$ . It follows from Proposition 3.6 that $(({y}{*} {x}){*} {x}){*} {y}\in {F}_i$ for all $i\in \Lambda$ . Hence $(({y}{*} {x}){*} {x}){*} {y}\in \cap \{{F}_i \mid i\in \Lambda \}$ and therefore $\cap \{{F}_i \mid i\in \Lambda \}$ is a prominent interior GE-filter in $({X}, f)$ by Theorem 3.11.

Theorem 3.14. If an interior GE-filter ${F}$ in an interior GE-algebra $({X}, f)$ satisfies the condition $(3.2)$ , then ${F}$ is a prominent interior GE-filter in $({X}, f)$ .

Proof. Let ${F}$ be an interior GE-filter in $({X}, f)$ that satisfies the condition (3.2). Let ${x}, {y}, {z}\in {X}$ be such that ${x}{*} ({y}{*} {z})\in {F}$ and ${x}\in {F}$ . Then ${y}{*} {z}\in {F}$ and thus $(({z}{*} {y}){*} {y}){*} {z} \in {F}$ . Therefore ${F}$ is a prominent interior GE-filter in $({X}, f)$ .

Given an interior GE-filter ${F}$ in an interior GE-algebra $({X}, f)$ , we consider an interior GE-filter ${G}$ which is greater than ${F}$ in $({X}, f)$ , that is, we take two interior GE-filters ${F}$ and ${G}$ such that ${F} \subseteq {G}$ in an interior GE-algebra $({X}, f)$ . We are now trying to find the condition that ${G}$ can be a prominent interior GE-filter in $({X}, f)$ .

Theorem 3.15. Let $({X}, f)$ be an interior GE-algebra in which ${X}$ is transitive. Let ${F}$ and ${G}$ be interior GE-filters in $({X}, f)$ . If ${F} \subseteq {G}$ and ${F}$ is a prominent interior GE-filter in $({X}, f)$ , then ${G}$ is also a prominent interior GE-filter in $({X}, f)$ .

Proof. Assume that ${F}$ is a prominent interior GE-filter in $({X}, f)$ . Then it is an interior GE-filter in $({X}, f)$ by Theorem 3.4. Let ${x}, {y}\in {X}$ be such that $f({x}{*} {y})\in {G}$ . Then ${x}{*} {y}\in {G}$ by (2.22), and so $1 = ({x}{*} {y}){*} ({x}{*} {y})\le {x}{*} (({x}{*} {y}){*} {y})$ by (GE1) and (2.6). Since $1\in {F}$ , it follows from Lemma 2.6 that ${x}{*} (({x}{*} {y}){*} {y})\in {F}$ . Hence $(((({x}{*} {y}){*} {y}){*} {x}){*} {x}){*} (({x}{*} {y}){*} {y})\in {F} \subseteq {G}$ by Corollary 3.7. Since

$\begin{align*} (((({x}{*} {y}){*} {y}){*} {x}){*} {x}){*} (({x}{*} {y}){*} {y})\le ({x}{*} {y}){*} ((((({x}{*} {y}){*} {y}){*} {x}){*} {x}) {*} {y}) \end{align*}$

by (2.6), we have $({x}{*} {y}){*} ((((({x}{*} {y}){*} {y}){*} {x}){*} {x}) {*} {y})\in {G}$ by Lemma 2.6. Hence

$(((({x}{*} {y}){*} {y}){*} {x}){*} {x}) {*} {y}\in {G}.$

Since ${y}\le ({x}{*} {y}){*} {y}$ , it follows from (2.11) that

$\begin{align*} (((({x}{*} {y}){*} {y}){*} {x}){*} {x}){*} {y}\le (({y}{*} {x}){*} {x}){*} {y}. \end{align*}$

Thus $(({y}{*} {x}){*} {x}){*} {y}\in {G}$ by Lemma 2.6. Therefore ${G}$ is a prominent interior GE-filter in $({X}, f)$ . by Theorem 3.11.

The following example describes Theorem 3.15.

Example 3.16. Let ${X} = \{{1}, {2}, {3}, {4}, {5}\}$ be a set with the Cayley table which is given in Table 4,

Table 4. Cayley table for the binary operation "

${*}$ ".

${*}$	${1}$	${2}$	${3}$	${4}$	${5}$
${1}$	${1}$	${2}$	${3}$	${4}$	${5}$
${2}$	${1}$	${1}$	${1}$	${5}$	${5}$
${3}$	${1}$	${1}$	${1}$	${5}$	${5}$
${4}$	${1}$	${3}$	${3}$	${1}$	${1}$
${5}$	${1}$	${3}$	${3}$	${1}$	${1}$

| Show Table

DownLoad: CSV

and define a mapping $f$ on ${X}$ as follows:

$\begin{align*} f : {X} \rightarrow {X}, \; {x}\mapsto \left\{\begin{array}{ll} {1} & \text{if } \, {x} = {1}, \\ {3} & \text{if } \, {x} \in \{{2}, {3}\}, \\ {5} & \text{if } \, {x} \in \{{4}, {5}\}. \end{array}\right. \end{align*}$

Then $({X}, f)$ is an interior GE-algebra in which ${X}$ is transitive, and ${F} : = \{{1}\}$ and ${G} : = \{{1}, {4}, {5}\}$ are interior GE-filters in $({X}, f)$ with ${F}\subseteq {G}$ . Also we can observe that ${F}$ and ${G}$ are prominent interior GE-filters in $({X}, f)$ .

In Theorem 3.15, if ${F}$ is an interior GE-filter which is not prominent, then ${G}$ is also not a prominent interior GE-filter in $({X}, f)$ as shown in the next example.

Example 3.17. Let ${X} = \{{1}, {2}, {3}, {4}, {5}\}$ be a set with the Cayley table which is given in Table 5,

Table 5. Cayley table for the binary operation "

${*}$ ".

${*}$	${1}$	${2}$	${3}$	${4}$	${5}$
${1}$	${1}$	${2}$	${3}$	${4}$	${5}$
${2}$	${1}$	${1}$	${1}$	${4}$	${1}$
${3}$	${1}$	${5}$	${1}$	${4}$	${5}$
${4}$	${1}$	${1}$	${1}$	${1}$	${1}$
${5}$	${1}$	${1}$	${1}$	${4}$	${1}$

| Show Table

DownLoad: CSV

and define a mapping $f$ on ${X}$ as follows:

$\begin{align*} f : {X} \rightarrow {X}, \; {x}\mapsto \left\{\begin{array}{ll} {1} & \text{if } \, {x} = {1}, \\ {3} & \text{if } \, {x} = {3}, \\ {4} & \text{if } \, {x} = {4}, \\ {2} & \text{if } \, {x} \in \{{2}, {5}\}. \end{array}\right. \end{align*}$

Then $({X}, f)$ is an interior GE-algebra in which ${X}$ is transitive, and ${F} : = \{{1}\}$ and ${G} : = \{{1}, {3}\}$ are interior GE-filters in $({X}, f)$ with ${F}\subseteq {G}$ . We can observe that ${F}$ and ${G}$ are not prominent interior GE-filters in $({X}, f)$ since ${2} {*} {3} = 1\in {F}$ but $(({3} {*} {2}){*} {2}) {*} {3} = ({5} {*} {2}) {*} {3} = {1} {*} {3} = {3}\notin {F}$ , and ${4} {*} {2} = {1} \in {G}$ but $(({2} {*} {4}){*} {4}) {*} {2} = ({4} {*} {4}) {*} {2} = {1} {*} {2} = {2}\notin {G}.$

In Theorem 3.15, if ${X}$ is not transitive, then Theorem 3.15 is false as seen in the following example.

Example 3.18. Let ${X} = \{{1}, {2}, {3}, {4}, {5}, {6}\}$ be a set with the Cayley table which is given in Table 6.

Table 6. Cayley table for the binary operation "

${*}$ ".

${*}$	${1}$	${2}$	${3}$	${4}$	${5}$	${6}$
${1}$	${1}$	${2}$	${3}$	${4}$	${5}$	${6}$
${2}$	${1}$	${1}$	${1}$	${6}$	${6}$	${6}$
${3}$	${1}$	${1}$	${1}$	${5}$	${5}$	${5}$
${4}$	${1}$	${1}$	${3}$	${1}$	${1}$	${1}$
${5}$	${1}$	${2}$	${3}$	${2}$	${1}$	${1}$
${6}$	${1}$	${2}$	${3}$	${2}$	${1}$	${1}$

| Show Table

DownLoad: CSV

If we define a mapping $f$ on ${X}$ as follows:

$\begin{align*} f : {X} \rightarrow {X}, \; {x}\mapsto \left\{\begin{array}{ll} {1} & \text{if } \, {x} = {1}, \\ {4} & \text{if } \, {x} = {4}, \\ {5} & \text{if } \, {x} = {5}, \\ {6} & \text{if } \, {x} = {6}, \\ {2} & \text{if } \, {x} \in \{{2}, {3}\}, \\ \end{array}\right. \end{align*}$

then $({X}, f)$ is an interior GE-algebra in which ${X}$ is not transitive. Let ${F} : = \{{1}\}$ and ${G} : = \{{1}, {5}, {6}\}$ . Then ${F}$ is a prominent interior GE-filter in $({X}, f)$ and ${G}$ is an interior GE-filter in $({X}, f)$ with ${F}\subseteq {G}$ . But ${G}$ is not prominent interior GE-filter since ${5} {*} ({3} {*} {4}) = {5} {*} {5} = {1}\in {G}$ and ${5}\in {G}$ but $(({4} {*} {3}) {*} {3}){*} {4} = ({3}{*} {3}) {*} {4} = {1} {*} {4} = {4} \notin {G}$ .

Definition 3.19. Let $({X}, f)$ be an interior GE-algebra and let ${F}$ be a subset of ${X}$ which satisfies (2.15). Then ${F}$ is called:

$\bullet$ A prominent interior GE-filter of type 1 in $({X}, f)$ if it satisfies:

$\begin{align} (\forall {x}, {y}, {z}\in {X}) \left({x}{*} ({y}{*} f({z}))\in {F}, \, f({x})\in {F} \; \Rightarrow (( f({z}){*} {y}){*} {y}){*} f({z})\in {F}\right). \end{align}$

(3.3)

$\bullet$ A prominent interior GE-filter of type 2 in $({X}, f)$ if it satisfies:

$\begin{align} (\forall {x}, {y}, {z}\in {X}) \left({x}{*} ({y}{*} f({z}))\in {F}, \, f({x})\in {F} \; \Rightarrow (({z}{*} f({y})){*} f({y})){*} {z}\in {F}\right). \end{align}$

(3.4)

Example 3.20. (1). Let ${X} = \{{1}, {2}, {3}, {4}, {5}\}$ be a set with the Cayley table which is given in Table 7,

Table 7. Cayley table for the binary operation "

${*}$ ".

${*}$	${1}$	${2}$	${3}$	${4}$	${5}$
${1}$	${1}$	${2}$	${3}$	${4}$	${5}$
${2}$	${1}$	${1}$	${1}$	${1}$	${1}$
${3}$	${1}$	${2}$	${1}$	${2}$	${2}$
${4}$	${1}$	${1}$	${1}$	${1}$	${1}$
${5}$	${1}$	${1}$	${1}$	${1}$	${1}$

| Show Table

DownLoad: CSV

and define a mapping $f$ on ${X}$ as follows:

$\begin{align*} f : {X} \rightarrow {X}, \; {x}\mapsto \left\{\begin{array}{ll} {1} & \text{if } \, {x} \in \{{1}, {3}\} \\ {2} & \text{if } \, {x} = {2}, \\ {4} & \text{if } \, {x} = {4}, \\ {5} & \text{if } \, {x} = {5}. \\ \end{array}\right. \end{align*}$

Then $({X}, f)$ is an interior GE-algebra and ${F} : = \{{1}, {3}\}$ is a prominent interior GE-filter of type 1 in $({X}, f)$ .

(2). Let ${X} = \{{1}, {2}, {3}, {4}, {5}\}$ be a set with the Cayley table which is given in Table 8,

Table 8. Cayley table for the binary operation "

${*}$ ".

${*}$	${1}$	${2}$	${3}$	${4}$	${5}$
${1}$	${1}$	${2}$	${3}$	${4}$	${5}$
${2}$	${1}$	${1}$	${1}$	${1}$	${1}$
${3}$	${1}$	${1}$	${1}$	${4}$	${1}$
${4}$	${1}$	${1}$	${1}$	${1}$	${5}$
${5}$	${1}$	${1}$	${3}$	${4}$	${1}$

| Show Table

DownLoad: CSV

and define a mapping $f$ on ${X}$ as follows:

$\begin{align*} f : {X} \rightarrow {X}, \; {x}\mapsto \left\{\begin{array}{ll} {1} & \text{if } \, {x} = {1}, \\ {2} & \text{if } \, {x} \in \{{2}, {3}, {4}, {5}\}. \\ \end{array}\right. \end{align*}$

Then $({X}, f)$ is an interior GE-algebra and ${F} : = \{{1}, {3}\}$ is a prominent interior GE-filter of type 2 in $({X}, f)$ .

Theorem 3.21. In an interior GE-algebra, every prominent interior GE-filter is of type 1.

Proof. Let ${F}$ be a prominent interior GE-filter in an interior GE-algebra $({X}, f)$ . Let ${x}, {y}, {z}\in {X}$ be such that ${x}{*} ({y}{*} f({z}))\in {F}$ and $f({x})\in {F}$ . Then ${x}\in {F}$ by (2.22). It follows from (2.18) that $((f({z}){*} {y}){*} {y}){*} f({z})\in {F}$ . Hence ${F}$ is a prominent interior GE-filter of type 1 in $({X}, f)$ .

The following example shows that the converse of Theorem 3.21 may not be true.

Example 3.22. Let ${X} = \{{1}, {2}, {3}, {4}, {5}\}$ be a set with the Cayley table which is given in Table 9,

Table 9. Cayley table for the binary operation "

${*}$ ".

${*}$	${1}$	${2}$	${3}$	${4}$	${5}$
${1}$	${1}$	${2}$	${3}$	${4}$	${5}$
${2}$	${1}$	${1}$	${1}$	${1}$	${1}$
${3}$	${1}$	${1}$	${1}$	${1}$	${5}$
${4}$	${1}$	${1}$	${3}$	${1}$	${1}$
${5}$	${1}$	${1}$	${1}$	${1}$	${1}$

| Show Table

DownLoad: CSV

and define a mapping $f$ on ${X}$ as follows:

$\begin{align*} f : {X} \rightarrow {X}, \; {x}\mapsto \left\{\begin{array}{ll} {1} & \text{if } \, {x} = {1}, \\ {2} & \text{if } \, {x} \in \{{2}, {3}\}, \\ {5} & \text{if } \, {x} \in \{{4}, {5}\}. \\ \end{array}\right. \end{align*}$

Then $({X}, f)$ is an interior GE-algebra and ${F} : = \{{1}\}$ is a prominent interior GE-filter of type 1 in $({X}, f)$ . But it is not a prominent interior GE-filter in $({X}, f)$ since ${1}{*} ({3}{*} {4}) = {1}\in {F}$ but $({4}{*} {3}){*} {3}){*} {4} = {4} \notin {F}$ .

The following example shows that prominent interior GE-filter and prominent interior GE-filter of type 2 are independent of each other, i.e., prominent interior GE-filter is not prominent interior GE-filter of type 2 and neither is the inverse.

Example 3.23. (1). Let ${X} = \{{1}, {2}, {3}, {4}, {5}\}$ be a set with the Cayley table which is given in the following Table 10,

Table 10. Cayley table for the binary operation "

${*}$ ".

${*}$	${1}$	${2}$	${3}$	${4}$	${5}$
${1}$	${1}$	${2}$	${3}$	${4}$	${5}$
${2}$	${1}$	${1}$	${1}$	${1}$	${1}$
${3}$	${1}$	${5}$	${1}$	${1}$	${5}$
${4}$	${1}$	${1}$	${1}$	${1}$	${1}$
${5}$	${1}$	${3}$	${3}$	${1}$	${1}$

| Show Table

DownLoad: CSV

and define a mapping $f$ on ${X}$ as follows:

$\begin{align*} f : {X} \rightarrow {X}, \; {x}\mapsto \left\{\begin{array}{ll} {1} & \text{if } \, {x} = {1}, \\ {4} & \text{if } \, {x} \in \{{3}, {4}\} \\ {5} & \text{if } \, {x} \in \{{2}, {5}\}. \end{array}\right. \end{align*}$

Then $({X}, f)$ is an interior GE-algebra and ${F} : = \{{1}\}$ ${F}$ is a prominent interior GE-filter in $({X}, f)$ . But it is not a prominent interior GE-filter of type 2 since ${1}{*} ({5} {*} f({2})) = {5} {*} {5} = {1}\in {F}$ and $f({1}) = {1} \in {F}$ but $(({2} {*} f({5})){*} f(5)) {*} {2} = (({2} {*} {5}) {*} {5}) {*} {2} = ({1} {*} {5}) {*} {2} = {5} {*} {2} = {3}\notin {F}$ .

(2). Let ${X} = \{{1}, {2}, {3}, {4}, {5}\}$ be a set with the Cayley table which is given in the following Table 11,

Table 11. Cayley table for the binary operation "

${*}$ ".

${*}$	${1}$	${2}$	${3}$	${4}$	${5}$
${1}$	${1}$	${2}$	${3}$	${4}$	${5}$
${2}$	${1}$	${1}$	${1}$	${1}$	${1}$
${3}$	${1}$	${2}$	${1}$	${1}$	${1}$
${4}$	${1}$	${1}$	${1}$	${1}$	${1}$
${5}$	${1}$	${1}$	${1}$	${1}$	${1}$

| Show Table

DownLoad: CSV

and define a mapping $f$ on ${X}$ as follows:

$\begin{align*} f : {X} \rightarrow {X}, \; {x}\mapsto \left\{\begin{array}{ll} {1} & \text{if } \, {x} = {1}, \\ {5} & \text{if } \, {x} \in \{{2}, {3}, {4}, {5}\}. \end{array}\right. \end{align*}$

Then $({X}, f)$ is an interior GE-algebra and ${F} : = \{{1}\}$ is a prominent interior GE-filter of type2 in $({X}, f)$ . But it is not a prominent interior GE-filter in $({X}, f)$ since ${1}{*} ({2} {*} {3}) = {1} {*} {1} = {1}\in {F}$ and ${1}\in {F}$ but $(({3} {*} {2}){*} {2}) {*} {3} = ({2} {*} {2}) {*} {3} = {1} {*}{3} = {3}\notin {F}$ .

The following example shows that prominent interior GE-filter of type 1 and prominent interior GE-filter of type 2 are independent of each other.

Example 3.24. (1). Let ${X} = \{{1}, {2}, {3}, {4}, {5}\}$ be a set with the Cayley table which is given in the following Table 12,

Table 12. Cayley table for the binary operation "

${*}$ ".

${*}$	${1}$	${2}$	${3}$	${4}$	${5}$
${1}$	${1}$	${2}$	${3}$	${4}$	${5}$
${2}$	${1}$	${1}$	${1}$	${5}$	${5}$
${3}$	${1}$	${1}$	${1}$	${1}$	${1}$
${4}$	${1}$	${1}$	${1}$	${1}$	${1}$
${5}$	${1}$	${1}$	${1}$	${1}$	${1}$

| Show Table

DownLoad: CSV

and define a mapping $f$ on ${X}$ as follows:

Then $({X}, f)$ is an interior GE-algebra and ${F} : = \{{1}, {2}, {4}\}$ is a prominent interior GE-filter of type 1 in $({X}, f)$ . But it is not a prominent interior GE-filter of type 2 since ${1}{*} ({5} {*} f({2})) = {1} {*} ({5} {*} {3}) = {1} {*} {1} = {1}\in {F}$ and $f({1}) = {1} \in {F}$ but $(({2} {*} f({5})){*} f(5)) {*} {2} = (({2} {*} {5}) {*} {5}) {*} {2} = ({5} {*} {5}) {*} {2} = {1} {*} {2} = {2}\notin {F}$ .

(2). Let ${X} = \{{1}, {2}, {3}, {4}, {5}\}$ be a set with the Cayley table which is given in the following Table 13,

Table 13. Cayley table for the binary operation "

${*}$ ".

${*}$	${1}$	${2}$	${3}$	${4}$	${5}$
${1}$	${1}$	${2}$	${3}$	${4}$	${5}$
${2}$	${1}$	${1}$	${4}$	${4}$	${5}$
${3}$	${1}$	${1}$	${1}$	${1}$	${1}$
${4}$	${1}$	${2}$	${2}$	${1}$	${5}$
${5}$	${1}$	${1}$	${1}$	${1}$	${1}$

| Show Table

DownLoad: CSV

and define a mapping $f$ on ${X}$ as follows:

$\begin{align*} f : {X} \rightarrow {X}, \; {x}\mapsto \left\{\begin{array}{ll} {1} & \text{if } \, {x} = {1}, \\ {2} & \text{if } \, {x} = {2}, \\ {4} & \text{if } \, {x} = {4}, \\ {3} & \text{if } \, {x} \in \{{3}, {5}\}. \end{array}\right. \end{align*}$

Then $({X}, f)$ is an interior GE-algebra and ${F} : = \{{1}\}$ is a prominent interior GE-filter of type 2 in $({X}, f)$ . But it is not a prominent interior GE-filter of type 1 in $({X}, f)$ since ${1}{*} ({5} {*} f(2)) = {1} {*} ({5} {*} {2}) = {1} {*} {1} = {1}\in {F}$ and $f({1})\in {F}$ but $((f(2) {*} {5}){*} {5}) {*} f(2) = (({2} {*} {5}){*} {5}) {*} {2} = ({5} {*} {5}) {*} {2} = {1} {*} {2} = {2}\notin {F}$ .

The following example shows that interior GE-filter and prominent interior GE-filter of type 1 are independent of each other.

Example 3.25. (1). Let ${X} = \{{1}, {2}, {3}, {4}, {5}\}$ be a set with the Cayley table which is given in the following Table 14,

Table 14. Cayley table for the binary operation "

${*}$ ".

${*}$	${1}$	${2}$	${3}$	${4}$	${5}$
${1}$	${1}$	${2}$	${3}$	${4}$	${5}$
${2}$	${1}$	${1}$	${5}$	${5}$	${5}$
${3}$	${1}$	${1}$	${1}$	${1}$	${1}$
${4}$	${1}$	${1}$	${1}$	${1}$	${1}$
${5}$	${1}$	${1}$	${1}$	${1}$	${1}$

| Show Table

DownLoad: CSV

and define a mapping $f$ on ${X}$ as follows:

$\begin{align*} f : {X} \rightarrow {X}, \; {x}\mapsto \left\{\begin{array}{ll} {1} & \text{if } \, {x} = {1}, \\ {2} & \text{if } \, {x} = {2}, \\ {5} & \text{if } \, {x} \in \{{3}, {4}, {5}\}. \end{array}\right. \end{align*}$

Then $({X}, f)$ is an interior GE-algebra and ${F} : = \{{1}\}$ is an interior GE-filter in $({X}, f)$ . But ${F}$ is not prominent interior GE-filter of type 1 since ${1}{*} ({5} {*} f({2})) = {1} {*} ({5} {*} {2}) = {1} {*} {1} = {1}\in {F}$ and $f({1}) = {1} \in {F}$ but $((f({2}) {*} {5}){*} {5}) {*} {2} = (({2} {*} {5}) {*} {5}) {*} {2} = ({5} {*} {5}) {*} {2} = {1} {*} {2} = {2}\notin {F}$ .

(2). Let ${X} = \{{1}, {2}, {3}, {4}, {5}\}$ be a set with the Cayley table which is given in the following Table 15,

Table 15. Cayley table for the binary operation "

${*}$ ".

${*}$	${1}$	${2}$	${3}$	${4}$	${5}$
${1}$	${1}$	${2}$	${3}$	${4}$	${5}$
${2}$	${1}$	${1}$	${5}$	${1}$	${5}$
${3}$	${1}$	${2}$	${1}$	${1}$	${1}$
${4}$	${1}$	${1}$	${3}$	${1}$	${5}$
${5}$	${1}$	${1}$	${1}$	${1}$	${1}$

| Show Table

DownLoad: CSV

and define a mapping $f$ on ${X}$ as follows:

$\begin{align*} f : {X} \rightarrow {X}, \; {x}\mapsto \left\{\begin{array}{ll} {1} & \text{if } \, {x} \in \{{1}, {2}, {4}\}, \\ {5} & \text{if } \, {x} \in \{{3}, {5}\}. \end{array}\right. \end{align*}$

Then $({X}, f)$ is an interior GE-algebra and ${F} : = \{{1}, {2} \}$ is a prominent interior GE-filter of type 1 in $({X}, f)$ . But it is not an interior GE-filter in $({X}, f)$ since ${2} {*} {4} = {1}$ and ${2} \in {F}$ but ${4} \notin {F}$ .

The following example shows that interior GE-filter and prominent interior GE-filter of type 2 are independent of each other.

Example 3.26. (1). Let ${X} = \{{1}, {2}, {3}, {4}, {5}\}$ be a set with the Cayley table which is given in the following Table 16,

Table 16. Cayley table for the binary operation "

${*}$ ".

${*}$	${1}$	${2}$	${3}$	${4}$	${5}$
${1}$	${1}$	${2}$	${3}$	${4}$	${5}$
${2}$	${1}$	${1}$	${1}$	${1}$	${1}$
${3}$	${1}$	${2}$	${1}$	${1}$	${2}$
${4}$	${1}$	${2}$	${3}$	${1}$	${5}$
${5}$	${1}$	${1}$	${1}$	${1}$	${1}$

| Show Table

DownLoad: CSV

and define a mapping $f$ on ${X}$ as follows:

$\begin{align*} f : {X} \rightarrow {X}, \; {x}\mapsto \left\{\begin{array}{ll} {1} & \text{if } \, {x} \in \{{1}, {4}\} \\ {2} & \text{if } \, {x} = {2}, \\ {3} & \text{if } \, {x} = {3}, \\ {5} & \text{if } \, {x} = {5}. \end{array}\right. \end{align*}$

Then $({X}, f)$ is an interior GE-algebra and ${F} : = \{{1}, {4}\}$ is an interior GE-filter in $({X}, f)$ . But ${F}$ is not prominent interior GE-filter of type 2 since ${4}{*} ({2} {*} f({3})) = {4} {*} ({2} {*} {3}) = {4} {*} {1} = {1}\in {F}$ and $f({4}) = {1} \in {F}$ but $((3 {*} f({2})) {*} f(2)){*} {3} = (({3} {*} {2}) {*} {2}) {*} {3} = ({2} {*} {2}) {*} {3} = {1} {*} {3} = {3}\notin {F}$ .

(2). Let ${X} = \{{1}, {2}, {3}, {4}, {5}\}$ be a set with the Cayley table which is given in the following Table 17,

Table 17. Cayley table for the binary operation "

${*}$ ".

${*}$	${1}$	${2}$	${3}$	${4}$	${5}$
${1}$	${1}$	${2}$	${3}$	${4}$	${5}$
${2}$	${1}$	${1}$	${1}$	${1}$	${5}$
${3}$	${1}$	${1}$	${1}$	${1}$	${1}$
${4}$	${1}$	${1}$	${1}$	${1}$	${5}$
${5}$	${1}$	${1}$	${1}$	${1}$	${1}$

| Show Table

DownLoad: CSV

and define a mapping $f$ on ${X}$ as follows:

$\begin{align*} f : {X} \rightarrow {X}, \; {x}\mapsto \left\{\begin{array}{ll} {1} & \text{if } \, {x} = {1}, \\ {3} & \text{if } \, {x} \in \{{2}, {3}, {4}, {5}\}. \end{array}\right. \end{align*}$

Then $({X}, f)$ is an interior GE-algebra and ${F} : = \{{1}, {2}, {5} \}$ is a prominent interior GE-filter of type 2 in $({X}, f)$ . But it is not an interior GE-filter in $({X}, f)$ since ${5} {*} {4} = {1}$ and ${5} \in {F}$ but ${4} \notin {F}$ .

Before we conclude this paper, we raise the following question.

Question. Let $({X}, f)$ be an interior GE-algebra. Let ${F}$ and ${G}$ be interior GE-filters in $({X}, f)$ . If ${F} \subseteq {G}$ and ${F}$ is a prominent interior GE-filter of type 1 (resp., type 2) in $({X}, f)$ , then is ${G}$ also a prominent interior GE-filter of type 1 (resp., type 2) in $({X}, f)$ ?

4. Conclusions

We have introduced the concept of a prominent interior GE-filter (of type 1 and type 2), and have investigated their properties. We have discussed the relationship between a prominent GE-filter and a prominent interior GE-filter and the relationship between an interior GE-filter and a prominent interior GE-filter. We have found and provide examples where any interior GE-filter is not a prominent interior GE-filter and any prominent GE-filter is not a prominent interior GE-filter. We have provided conditions for an interior GE-filter to be a prominent interior GE-filter. We have given conditions under which an internal GE-filter larger than a given internal GE filter can become a prominent internal GE-filter, and have provided an example describing it. We have considered the relationship between a prominent interior GE-filter and a prominent interior GE-filter of type 1. We have found and provide examples to verify that a prominent interior GE-filter of type 1 and a prominent interior GE-filter of type 2, a prominent interior GE-filter and a prominent interior GE-filter of type 2, an interior GE-filter and a prominent interior GE-filter of type 1, and an interior GE-filter and a prominent interior GE-filter of type 2 are independent each other. In future, we will study the prime and maximal prominent interior GE-filters and their topological properties. Moreover, based on the ideas and results obtained in this paper, we will study the interior operator theory in related algebraic systems such as MV-algebra, BL-algebra, EQ-algebra, etc. It will also be used for pseudo algebra systems and further to conduct research related to the very true operator theory.

Acknowledgments

This research was supported by Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education (NRF-2016R1D1A1B02006812).

The authors wish to thank the anonymous reviewers for their valuable suggestions.

Conflict of interest

All authors declare no conflicts of interest in this paper.

Funding sources

This research was supported by National Research Council (CNR Italy) and did not receive any specific grant from funding agencies in the public, commercial, or not-for-profit sectors.

Conflict of interest

The author declares no conflicts of interest in this paper.

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AIMS Molecular Science

Green tea catechins and intracellular calcium dynamics in prostate cancer cells

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Green tea catechins and intracellular calcium dynamics in prostate cancer cells

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