On Induced Topologies by Ideal, Primal, Filter and Grill

1. Introduction

Many topologies with important applications in mathematics have been defined using some additional mathematical structure. For example, if a topological space $(X,\tau )$ and an ideal $\mathcal{I}$ on X are given, a new topology (called an ideal topology) can be obtained by an ideal-associated local function. The ideal topological spaces are extensively studied, see [4,5,7,8,10]. A similar concept was used for a grill [9,13,14,15,16,17], a filter [3,11,12] and a primal [1,2,6].

In general, we can consider four systems, namely an ideal $\mathcal{I}$, a primal $\mathcal{P}$, a filter $\mathcal{F}$ and a grill $\mathcal{G}$ on a topological space $(X,\tau )$, see Definition 1. A derivation of a new topology that is finer than the original topology $\tau$ is as follows: The local function $A_{\mathcal{I}}^{*}$, $A_{\mathcal{P}}^{*}$, $A_{\mathcal{F}}^{*}$, $A_{\mathcal{G}}^{*}$, see Definition 1, derived from $\mathcal{I},\mathcal{P},\mathcal{F},\mathcal{G}$ and $\tau$ defines the Kuratowski closure operator $c{l}_{\mathcal{I}}$, $c{l}_{\mathcal{P}}$, $c{l}_{\mathcal{F}}$, $c{l}_{\mathcal{G}}$, see Definition 3. In the final step, a new topology on X is defined, denoted by ${\tau }_{\mathcal{I}}$, ${\tau }_{\mathcal{P}}$, ${\tau }_{\mathcal{F}}$, ${\tau }_{\mathcal{G}}$, respectively, see Theorem 9. If we look at the achieved results, we can see a striking similarity. In fact, the local functions and topologies generated by this way are equivalent.

The main concept of the article is as follows: Using correspondence between two systems (Theorem 1–6), it is possible to define their equivalence, see Definition 2. Two equivalent systems generate the same topology (Theorem 10), and the results achieved in one topology can be used in a topology determined by an equivalent system. In the last part, we will show the application of this equivalence on examples of compatibility and codense topologies.

Definition 1. 

Let X be a nonempty set. A nonempty system $\mathcal{I}$, $\mathcal{P}$, $\mathcal{F}$, $\mathcal{G}$ of subsets of X is said be an ideal, a primal, a filter, a grill on X if it satisfies the following conditions

respectively. Furthermore, if τ is a topology on X, for $A\subset X$ we define four local functions

$\begin{array}{c}{A}_{\mathcal{I}}^{*}=\{x\in X:A\cap U\notin \mathcal{I}\mathrm{for}\mathrm{any}U\in \tau ,x\in U\},\hfill \\ A_{\mathcal{P}}^{*}=\{x\in X:(X\setminus A)\cup (X\setminus U)\in \mathcal{P}\mathrm{for}\mathrm{any}U\in \tau ,x\in U\},\hfill \\ A_{\mathcal{F}}^{*}=\{x\in X:(X\setminus A)\cup (X\setminus U)\notin \mathcal{F}\mathrm{for}\mathrm{any}U\in \tau ,x\in U\},\hfill \\ A_{\mathcal{G}}^{*}=\{x\in X:A\cap U\in \mathcal{G}\mathrm{for}\mathrm{any}U\in \tau ,x\in U\}.\hfill \end{array}$

An ideal topological space, a primal topological space, a filter topological space, a grill topological space is a topological space $(X,\tau )$ with an ideal $\mathcal{I}$, a primal $\mathcal{P}$, a filter $\mathcal{F}$, a grill $\mathcal{G}$ and it is denoted by $(X,\tau ,\mathcal{I})$, $(X,\tau ,\mathcal{P})$, $(X,\tau ,\mathcal{F})$, $(X,\tau ,\mathcal{G})$, respectively. Let $\mathbb{I},\mathbb{P},\mathbb{F},\mathbb{G}$ be a family of all ideals, primals, filters, grills on X, respectively. Put $\mathbb{X}=\mathbb{I}\cup \mathbb{P}\cup \mathbb{F}\cup \mathbb{G}$. If $\mathcal{Z}\in \mathbb{X}$, then $(X,\tau ,\mathcal{Z})$ is called a $\mathcal{Z}$-topological space.

2. Main Results

In the following two parts we present sixteen operators ${\mathsf{H}}_{{\mathbb{Z}}_2}^{{\mathbb{Z}}_1}$ (Theorem 1–6) between pairs of systems ${\mathbb{Z}}_1,{\mathbb{Z}}_2\in \{\mathbb{I},\mathbb{P},\mathbb{F},\mathbb{G}\}$ and their properties and compositions are studied. Next, the equivalence between ${\mathcal{Z}}_1,{\mathcal{Z}}_2\in \mathbb{X}$ is defined (Definotion 2). The equality of local functions and the equality of generated topologies are proved, provided that ${\mathcal{Z}}_1$ and ${\mathcal{Z}}_2$ and equivalent.

Define four identity operators

$\begin{array}{cc}{\mathsf{H}}_{\mathbb{I}}^{\mathbb{I}}:\mathbb{I}\to \mathbb{I}\mathrm{by}{\mathsf{H}}_{\mathbb{I}}^{\mathbb{I}}\left(\mathcal{I}\right)=\mathcal{I},\hfill & {\mathsf{H}}_{\mathbb{P}}^{\mathbb{P}}:\mathbb{P}\to \mathbb{P}\mathrm{by}{\mathsf{H}}_{\mathbb{P}}^{\mathbb{P}}\left(\mathcal{P}\right)=\mathcal{P},\hfill \\ {\mathsf{H}}_{\mathbb{F}}^{\mathbb{F}}:\mathbb{F}\to \mathbb{F}\mathrm{by}{\mathsf{H}}_{\mathbb{F}}^{\mathbb{F}}\left(\mathcal{F}\right)=\mathcal{F},\hfill & {\mathsf{H}}_{\mathbb{G}}^{\mathbb{G}}:\mathbb{G}\to \mathbb{G}\mathrm{by}{\mathsf{H}}_{\mathbb{G}}^{\mathbb{G}}\left(\mathcal{G}\right)=\mathcal{G}.\hfill \end{array}$

In the next six theorems the proofs of items (1)–(10) are left to the reader. The rest will be proven.

Theorem 1. 

Let $\mathcal{I}$, $\mathcal{P}$ be an ideal, a primal on X, respectively. Define

${\mathsf{H}}_{\mathbb{P}}^{\mathbb{I}}:\mathbb{I}\to \mathbb{P}$ by ${\mathsf{H}}_{\mathbb{P}}^{\mathbb{I}}\left(\mathcal{I}\right)=\{A\subset X:X\setminus A\notin \mathcal{I}\}$,

${\mathsf{H}}_{\mathbb{I}}^{\mathbb{P}}:\mathbb{P}\to \mathbb{I}$ by ${\mathsf{H}}_{\mathbb{I}}^{\mathbb{P}}\left(\mathcal{P}\right)=\{A\subset X:X\setminus A\notin \mathcal{P}\}$.

Then ${\mathsf{H}}_{\mathbb{P}}^{\mathbb{I}}\left(\mathcal{I}\right)$, ${\mathsf{H}}_{\mathbb{I}}^{\mathbb{P}}\left(\mathcal{P}\right)$ is a primal, an ideal on X, respectively and

Proof. 

We prove ${\mathsf{H}}_{\mathbb{P}}^{\mathbb{I}}\left(\mathcal{I}\right)$ is a primal. Since $\varnothing \in \mathcal{I}$, $X\notin {\mathsf{H}}_{\mathbb{P}}^{\mathbb{I}}\left(\mathcal{I}\right)$.

Let $A\in {\mathsf{H}}_{\mathbb{P}}^{\mathbb{I}}\left(\mathcal{I}\right)$ and $B\subset A$. Then $X\setminus A\notin \mathcal{I}$. Since $X\setminus A\subset X\setminus B$ and $\mathcal{I}$ is an ideal, $X\setminus B\notin \mathcal{I}$, so $B\in {\mathsf{H}}_{\mathbb{P}}^{\mathbb{I}}\left(\mathcal{I}\right)$. Let $A\cap B\in {\mathsf{H}}_{\mathbb{P}}^{\mathbb{I}}\left(\mathcal{I}\right)$. Then $(X\setminus A)\cup (X\setminus B)=X\setminus A\cap B\notin \mathcal{I}$. Since $\mathcal{I}$ is an ideal, $X\setminus A$ or $X\setminus B$ is not from $\mathcal{I}$. So $A\in {\mathsf{H}}_{\mathbb{P}}^{\mathbb{I}}\left(\mathcal{I}\right)$ or $B\in {\mathsf{H}}_{\mathbb{P}}^{\mathbb{I}}\left(\mathcal{I}\right)$. That means ${\mathsf{H}}_{\mathbb{P}}^{\mathbb{I}}\left(\mathcal{I}\right)$ is a primal.

We prove ${\mathsf{H}}_{\mathbb{I}}^{\mathbb{P}}\left(\mathcal{P}\right)$ is an ideal. Since $\varnothing \in \mathcal{P}$, $X\notin {\mathsf{H}}_{\mathbb{I}}^{\mathbb{P}}\left(\mathcal{P}\right)$.

Let $A\in {\mathsf{H}}_{\mathbb{I}}^{\mathbb{P}}\left(\mathcal{P}\right)$ and $B\subset A$. Then $X\setminus A\notin \mathcal{P}$. Since $X\setminus A\subset X\setminus B$ and $\mathcal{P}$ is a primal, $X\setminus B\notin \mathcal{P}$, so $B\in {\mathsf{H}}_{\mathbb{I}}^{\mathbb{P}}\left(\mathcal{P}\right)$. Let $A,B\in {\mathsf{H}}_{\mathbb{I}}^{\mathbb{P}}\left(\mathcal{P}\right)$. Then $X\setminus A\notin \mathcal{P}$, $X\setminus B\notin \mathcal{P}$ and $(X\setminus A)\cap (X\setminus B)=X\setminus A\cup B\notin \mathcal{P}$. So $A\cup B\in {\mathsf{H}}_{\mathbb{I}}^{\mathbb{P}}\left(\mathcal{P}\right)$. That means ${\mathsf{H}}_{\mathbb{I}}^{\mathbb{P}}\left(\mathcal{P}\right)$ is an ideal.

(11): $x\in A_{{\mathsf{H}}_{\mathbb{P}}^{\mathbb{I}}\left(\mathcal{I}\right)}^{*}$ if and only if $(X\setminus A)\cup (X\setminus U)=X\setminus A\cap U\in {\mathsf{H}}_{\mathbb{P}}^{\mathbb{I}}\left(\mathcal{I}\right)$ for any nbhd U of x if and only if $A\cap U\notin \mathcal{I}$ if and only if $x\in A_{\mathcal{I}}^{*}$.

(12): $x\in A_{{\mathsf{H}}_{\mathbb{I}}^{\mathbb{P}}\left(\mathcal{P}\right)}^{*}$ if and only if $A\cap U\notin {\mathsf{H}}_{\mathbb{I}}^{\mathbb{P}}\left(\mathcal{P}\right)$ for any nbhd U of x if and only if $X\setminus A\cap U=(X\setminus A)\cup (X\setminus U)\in \mathcal{P}$ if and only if $x\in A_{\mathcal{P}}^{*}$. □

Theorem 2. 

Let $\mathcal{I}$, $\mathcal{G}$ be an ideal, a grill on X, respectively. Define

${\mathsf{H}}_{\mathbb{G}}^{\mathbb{I}}:\mathbb{I}\to \mathbb{G}$ by ${\mathsf{H}}_{\mathbb{G}}^{\mathbb{I}}\left(\mathcal{I}\right)=\{A\subset X:A\notin \mathcal{I}\}$,

${\mathsf{H}}_{\mathbb{I}}^{\mathbb{G}}:\mathbb{G}\to \mathbb{I}$ by ${\mathsf{H}}_{\mathbb{I}}^{\mathbb{G}}\left(\mathcal{G}\right)=\{A\subset X:A\notin \mathcal{G}\}$.

Then ${\mathsf{H}}_{\mathbb{G}}^{\mathbb{I}}\left(\mathcal{I}\right)$, ${\mathsf{H}}_{\mathbb{I}}^{\mathbb{G}}\left(\mathcal{G}\right)$ is a grill, an ideal on X, respectively and

Proof. 

We prove ${\mathsf{H}}_{\mathbb{G}}^{\mathbb{I}}\left(\mathcal{I}\right)$ is a grill. Since $\varnothing \in \mathcal{I}$, $\varnothing \notin {\mathsf{H}}_{\mathbb{G}}^{\mathbb{I}}\left(\mathcal{I}\right)$.

Let $A\in {\mathsf{H}}_{\mathbb{G}}^{\mathbb{I}}\left(\mathcal{I}\right)$ and $A\subset B$. Then $A\notin \mathcal{I}$. Since $\mathcal{I}$ is an ideal, $B\notin \mathcal{I}$, so $B\in {\mathsf{H}}_{\mathbb{G}}^{\mathbb{I}}\left(\mathcal{I}\right)$. Let $A\cup B\in {\mathsf{H}}_{\mathbb{G}}^{\mathbb{I}}\left(\mathcal{I}\right)$. Then $A\cup B\notin \mathcal{I}$. Since $\mathcal{I}$ is an ideal, A or B is not from $\mathcal{I}$. So $A\in {\mathsf{H}}_{\mathbb{G}}^{\mathbb{I}}\left(\mathcal{I}\right)$ or $B\in {\mathsf{H}}_{\mathbb{G}}^{\mathbb{I}}\left(\mathcal{I}\right)$. That means ${\mathsf{H}}_{\mathbb{G}}^{\mathbb{I}}\left(\mathcal{I}\right)$ is a grill.

We prove ${\mathsf{H}}_{\mathbb{I}}^{\mathbb{G}}\left(\mathcal{G}\right)$ is an ideal. Since $X\in \mathcal{G}$, $X\notin {\mathsf{H}}_{\mathbb{I}}^{\mathbb{G}}\left(\mathcal{G}\right)$.

Let $A\in {\mathsf{H}}_{\mathbb{I}}^{\mathbb{G}}\left(\mathcal{G}\right)$ and $B\subset A$. Then $A\notin \mathcal{G}$. Since $\mathcal{G}$ is a grill, $B\notin \mathcal{G}$, so $B\in {\mathsf{H}}_{\mathbb{I}}^{\mathbb{G}}\left(\mathcal{G}\right)$. Let $A,B\in {\mathsf{H}}_{\mathbb{I}}^{\mathbb{G}}\left(\mathcal{G}\right)$. Then $A,B\notin \mathcal{G}$. Since $\mathcal{G}$ is a grill, $A\cup B\notin \mathcal{G}$. So $A\cup B\in {\mathsf{H}}_{\mathbb{I}}^{\mathbb{G}}\left(\mathcal{G}\right)$. That means ${\mathsf{H}}_{\mathbb{I}}^{\mathbb{G}}\left(\mathcal{G}\right)$ is an ideal.

(11): $x\in A_{{\mathsf{H}}_{\mathbb{G}}^{\mathbb{I}}\left(\mathcal{I}\right)}^{*}$ if and only if $A\cap U\in {\mathsf{H}}_{\mathbb{G}}^{\mathbb{I}}\left(\mathcal{I}\right)$ for any nbhd U of x if and only if $A\cap U\notin \mathcal{I}$ if and only if $x\in A_{\mathcal{I}}^{*}$.

(12): $x\in A_{{\mathsf{H}}_{\mathbb{I}}^{\mathbb{G}}\left(\mathcal{G}\right)}^{*}$ if and only if $A\cap U\notin {\mathsf{H}}_{\mathbb{I}}^{\mathbb{G}}\left(\mathcal{G}\right)$ for any nbhd U of x if and only if $A\cap U\in \mathcal{G}$ if and only if $x\in A_{\mathcal{G}}^{*}$. □

Theorem 3. 

Let $\mathcal{F}$, $\mathcal{G}$ be a filter, a grill on X, respectively. Define

${\mathsf{H}}_{\mathbb{G}}^{\mathbb{F}}:\mathbb{F}\to \mathbb{G}$ by ${\mathsf{H}}_{\mathbb{G}}^{\mathbb{F}}\left(\mathcal{F}\right)=\{A\subset X:X\setminus A\notin \mathcal{F}\}$,

${\mathsf{H}}_{\mathbb{F}}^{\mathbb{G}}:\mathbb{G}\to \mathbb{F}$ by ${\mathsf{H}}_{\mathbb{F}}^{\mathbb{G}}\left(\mathcal{G}\right)=\{A\subset X:X\setminus A\notin \mathcal{G}\}$.

Then ${\mathsf{H}}_{\mathbb{G}}^{\mathbb{F}}\left(\mathcal{F}\right)$, ${\mathsf{H}}_{\mathbb{F}}^{\mathbb{G}}\left(\mathcal{G}\right)$ is a grill, a filter on X, respectively and

$\begin{array}{cc}\left(1\right)A\in {\mathsf{H}}_{\mathbb{G}}^{\mathbb{F}}\left(\mathcal{F}\right)\iff X\setminus A\notin \mathcal{F}\hfill & \left(5\right)A\in {\mathsf{H}}_{\mathbb{F}}^{\mathbb{G}}\left(\mathcal{G}\right)\iff X\setminus A\notin \mathcal{G}\hfill \\ \left(2\right)A\notin {\mathsf{H}}_{\mathbb{G}}^{\mathbb{F}}\left(\mathcal{F}\right)\iff X\setminus A\in \mathcal{F}\hfill & \left(6\right)A\notin {\mathsf{H}}_{\mathbb{F}}^{\mathbb{G}}\left(\mathcal{G}\right)\iff X\setminus A\in \mathcal{G}\hfill \\ \left(3\right)X\setminus A\in {\mathsf{H}}_{\mathbb{G}}^{\mathbb{F}}\left(\mathcal{F}\right)\iff A\notin \mathcal{F}\hfill & \left(7\right)X\setminus A\in {\mathsf{H}}_{\mathbb{F}}^{\mathbb{G}}\left(\mathcal{G}\right)\iff A\notin \mathcal{G}\hfill \\ \left(4\right)X\setminus A\notin {\mathsf{H}}_{\mathbb{G}}^{\mathbb{F}}\left(\mathcal{F}\right)\iff A\in \mathcal{F}\hfill & \left(8\right)X\setminus A\notin {\mathsf{H}}_{\mathbb{F}}^{\mathbb{G}}\left(\mathcal{G}\right)\iff A\in \mathcal{G}\hfill \\ \left(9\right){\mathsf{H}}_{\mathbb{G}}^{\mathbb{F}}\left({\mathsf{H}}_{\mathbb{F}}^{\mathbb{G}}\left(\mathcal{G}\right)\right)={\mathsf{H}}_{\mathbb{G}}^{\mathbb{G}}\left(\mathcal{G}\right)=\mathcal{G}\hfill & \left(10\right){\mathsf{H}}_{\mathbb{F}}^{\mathbb{G}}\left({\mathsf{H}}_{\mathbb{G}}^{\mathbb{F}}\left(\mathcal{F}\right)\right)={\mathsf{H}}_{\mathbb{F}}^{\mathbb{F}}\left(\mathcal{F}\right)=\mathcal{F}\hfill \\ \left(11\right)A_{{\mathsf{H}}_{\mathbb{G}}^{\mathbb{F}}\left(\mathcal{F}\right)}^{*}=A_{\mathcal{F}}^{*}\hfill & \left(12\right)A_{{\mathsf{H}}_{\mathbb{F}}^{\mathbb{G}}\left(\mathcal{G}\right)}^{*}=A_{\mathcal{G}}^{*}\hfill \end{array}$

Proof. 

We prove ${\mathsf{H}}_{\mathbb{G}}^{\mathbb{F}}\left(\mathcal{F}\right)$ is a grill. Since $X\in \mathcal{F}$, $\varnothing \notin {\mathsf{H}}_{\mathbb{G}}^{\mathbb{F}}\left(\mathcal{F}\right)$.

Let $A\in {\mathsf{H}}_{\mathbb{G}}^{\mathbb{F}}\left(\mathcal{F}\right)$ and $A\subset B$. Then $X\setminus A\notin \mathcal{F}$. Since $\mathcal{F}$ is a filter and $X\setminus B\subset X\setminus A$, $X\setminus B\notin \mathcal{F}$, so $B\in {\mathsf{H}}_{\mathbb{G}}^{\mathbb{F}}\left(\mathcal{F}\right)$. Let $A\cup B\in {\mathsf{H}}_{\mathbb{G}}^{\mathbb{F}}\left(\mathcal{F}\right)$. Then $X\setminus A\cup B=(X\setminus A)\cap (X\setminus B)\notin \mathcal{F}$. Since $\mathcal{F}$ is an filter, $X\setminus A$ or $X\setminus B$ is not from $\mathcal{F}$. So $A\in {\mathsf{H}}_{\mathbb{G}}^{\mathbb{F}}\left(\mathcal{I}\right)$ or $B\in {\mathsf{H}}_{\mathbb{G}}^{\mathbb{F}}\left(\mathcal{I}\right)$. That means ${\mathsf{H}}_{\mathbb{G}}^{\mathbb{F}}\left(\mathcal{I}\right)$ is a grill.

We prove ${\mathsf{H}}_{\mathbb{F}}^{\mathbb{G}}\left(\mathcal{G}\right)$ is a filter. Since $X\in \mathcal{G}$, $\varnothing \notin {\mathsf{H}}_{\mathbb{F}}^{\mathbb{G}}\left(\mathcal{G}\right)$.

Let $A\in {\mathsf{H}}_{\mathbb{F}}^{\mathbb{G}}\left(\mathcal{G}\right)$ and $A\subset B$. Then $X\setminus A\notin \mathcal{G}$. Since $\mathcal{G}$ is a grill and $X\setminus B\subset X\setminus A$, $X\setminus B\notin \mathcal{G}$, so $B\in {\mathsf{H}}_{\mathbb{F}}^{\mathbb{G}}\left(\mathcal{G}\right)$. Let $A,B\in {\mathsf{H}}_{\mathbb{F}}^{\mathbb{G}}\left(\mathcal{G}\right)$. Then $X\setminus A\notin \mathcal{G}$, $X\setminus B\notin \mathcal{G}$. Then $(X\setminus A)\cup (X\setminus B)=X\setminus A\cap B\notin \mathcal{G}$. So, $A\cap B\in {\mathsf{H}}_{\mathbb{F}}^{\mathbb{G}}\left(\mathcal{G}\right)$. That means ${\mathsf{H}}_{\mathbb{F}}^{\mathbb{G}}\left(\mathcal{G}\right)$ is a filter.

(11): $x\in A_{{\mathsf{H}}_{\mathbb{G}}^{\mathbb{F}}\left(\mathcal{F}\right)}^{*}$ if and only if $A\cap U\in {\mathsf{H}}_{\mathbb{G}}^{\mathbb{F}}\left(\mathcal{F}\right)$ for any nbhd U of x if and only if $X\setminus A\cap U\notin \mathcal{F}$ if and only if $x\in A_{\mathcal{F}}^{*}$.

(12): $x\in A_{{\mathsf{H}}_{\mathbb{F}}^{\mathbb{G}}\left(\mathcal{G}\right)}^{*}$ if and only if $X\setminus A\cap U\notin {\mathsf{H}}_{\mathbb{F}}^{\mathbb{G}}\left(\mathcal{G}\right)$ for any nbhd U of x if and only if $A\cap U\in \mathcal{G}$ if and only if $x\in A_{\mathcal{G}}^{*}$. □

Theorem 4. 

Let $\mathcal{F}$, $\mathcal{P}$ be a filter, a primal on X, respectively. Define

${\mathsf{H}}_{\mathbb{P}}^{\mathbb{F}}:\mathbb{F}\to \mathbb{P}$ by ${\mathsf{H}}_{\mathbb{P}}^{\mathbb{F}}\left(\mathcal{F}\right)=\{A\subset X:A\notin \mathcal{F}\}$,

${\mathsf{H}}_{\mathbb{F}}^{\mathbb{P}}:\mathbb{P}\to \mathbb{F}$ by ${\mathsf{H}}_{\mathbb{F}}^{\mathbb{P}}\left(\mathcal{P}\right)=\{A\subset X:A\notin \mathcal{P}\}$.

Then ${\mathsf{H}}_{\mathbb{P}}^{\mathbb{F}}\left(\mathcal{F}\right)$, ${\mathsf{H}}_{\mathbb{F}}^{\mathbb{P}}\left(\mathcal{P}\right)$ is a primal, a filter on X, respectively and

Proof. 

We prove ${\mathsf{H}}_{\mathbb{P}}^{\mathbb{F}}\left(\mathcal{F}\right)$ is a primal. Since $X\in \mathcal{F}$, $X\notin {\mathsf{H}}_{\mathbb{P}}^{\mathbb{F}}\left(\mathcal{F}\right)$.

Let $A\in {\mathsf{H}}_{\mathbb{P}}^{\mathbb{F}}\left(\mathcal{F}\right)$ and $B\subset A$. Then $A\notin \mathcal{F}$. Since $\mathcal{F}$ is a filter, $B\notin \mathcal{F}$, so $B\in {\mathsf{H}}_{\mathbb{P}}^{\mathbb{F}}\left(\mathcal{F}\right)$. Let $A\cap B\in {\mathsf{H}}_{\mathbb{P}}^{\mathbb{F}}\left(\mathcal{F}\right)$. Then $A\cap B\notin \mathcal{F}$. Since $\mathcal{F}$ is a filter, A or B is not from $\mathcal{F}$. So $A\in {\mathsf{H}}_{\mathbb{P}}^{\mathbb{F}}\left(\mathcal{F}\right)$ or $B\in {\mathsf{H}}_{\mathbb{P}}^{\mathbb{F}}\left(\mathcal{F}\right)$. That means ${\mathsf{H}}_{\mathbb{P}}^{\mathbb{F}}\left(\mathcal{F}\right)$ is a primal.

We prove ${\mathsf{H}}_{\mathbb{F}}^{\mathbb{P}}\left(\mathcal{P}\right)$ is a filter. Since $\varnothing \in \mathcal{P}$, $\varnothing \notin {\mathsf{H}}_{\mathbb{F}}^{\mathbb{P}}\left(\mathcal{P}\right)$.

Let $A\in {\mathsf{H}}_{\mathbb{F}}^{\mathbb{P}}\left(\mathcal{P}\right)$ and $A\subset B$. Then $A\notin \mathcal{P}$. Since $\mathcal{P}$ is a primal, $B\notin \mathcal{P}$, so $B\in {\mathsf{H}}_{\mathbb{F}}^{\mathbb{P}}\left(\mathcal{P}\right)$. Let $A,B\in {\mathsf{H}}_{\mathbb{F}}^{\mathbb{P}}\left(\mathcal{P}\right)$. Then $A,B\notin \mathcal{P}$. Since $\mathcal{P}$ is a primal, $A\cap B\notin \mathcal{P}$. So $A\cap B\in {\mathsf{H}}_{\mathbb{F}}^{\mathbb{P}}\left(\mathcal{P}\right)$. That means ${\mathsf{H}}_{\mathbb{F}}^{\mathbb{P}}\left(\mathcal{P}\right)$ is a filter.

(11): $x\in A_{{\mathsf{H}}_{\mathbb{P}}^{\mathbb{F}}\left(\mathcal{F}\right)}^{*}$ if and only if $(X\setminus A)\cup (X\setminus U)=X\setminus A\cap U\in {\mathsf{H}}_{\mathbb{P}}^{\mathbb{F}}\left(\mathcal{F}\right)$ for any nbhd U of x if and only if $X\setminus A\cap U\notin \mathcal{F}$ if and only if $x\in A_{\mathcal{F}}^{*}$.

(12): $x\in A_{{\mathsf{H}}_{\mathbb{F}}^{\mathbb{P}}\left(\mathcal{P}\right)}^{*}$ if and only if $X\setminus A\cap U\notin {\mathsf{H}}_{\mathbb{F}}^{\mathbb{P}}\left(\mathcal{P}\right)$ for any nbhd U of x if and only if $X\setminus A\cap U=(X\setminus A)\cup (X\setminus U)\in \mathcal{P}$ if and only if $x\in A_{\mathcal{P}}^{*}$. □

Theorem 5. 

Let $\mathcal{G}$, $\mathcal{P}$ be a grill, a primal on X, respectively. Define

${\mathsf{H}}_{\mathbb{P}}^{\mathbb{G}}:\mathbb{G}\to \mathbb{P}$ by ${\mathsf{F}}_{\mathbb{P}}^{\mathbb{G}}\left(\mathcal{G}\right)=\{A\subset X:X\setminus A\in \mathcal{G}\}$,

${\mathsf{H}}_{\mathbb{G}}^{\mathbb{P}}:\mathbb{P}\to \mathbb{G}$ by ${\mathsf{H}}_{\mathbb{G}}^{\mathbb{P}}\left(\mathcal{P}\right)=\{A\subset X:X\setminus A\in \mathcal{P}\}$.

Then ${\mathsf{H}}_{\mathbb{P}}^{\mathbb{G}}\left(\mathcal{G}\right)$, ${\mathsf{H}}_{\mathbb{G}}^{\mathbb{P}}\left(\mathcal{P}\right)$ is a primal, a grill on X, respectively and

Proof. 

We prove ${\mathsf{H}}_{\mathbb{P}}^{\mathbb{G}}\left(\mathcal{G}\right)$ is a primal. Since $\varnothing \notin \mathcal{G}$, $X\notin {\mathsf{H}}_{\mathbb{P}}^{\mathbb{G}}\left(\mathcal{G}\right)$.

Let $A\in {\mathsf{H}}_{\mathbb{P}}^{\mathbb{G}}\left(\mathcal{G}\right)$ and $B\subset A$. Then $X\setminus A\in \mathcal{G}$. Since $\mathcal{G}$ is a grill and $X\setminus A\subset X\setminus B$, $X\setminus B\in \mathcal{G}$, so $B\in {\mathsf{H}}_{\mathbb{P}}^{\mathbb{G}}\left(\mathcal{G}\right)$. Let $A\cap B\in {\mathsf{H}}_{\mathbb{P}}^{\mathbb{G}}\left(\mathcal{G}\right)$. Then $X\setminus A\cap B=(X\setminus A)\cup (X\setminus B)\in \mathcal{G}$. Since $\mathcal{G}$ is a grill, $X\setminus A$ or $X\setminus B$ is from $\mathcal{G}$. So $A\in {\mathsf{H}}_{\mathbb{P}}^{\mathbb{G}}\left(\mathcal{G}\right)$ or $B\in {\mathsf{H}}_{\mathbb{P}}^{\mathbb{G}}\left(\mathcal{G}\right)$. That means ${\mathsf{H}}_{\mathbb{P}}^{\mathbb{G}}\left(\mathcal{G}\right)$ is a primal.

We prove ${\mathsf{H}}_{\mathbb{G}}^{\mathbb{P}}\left(\mathcal{P}\right)$ is a grill. Since $X\notin \mathcal{P}$, $\varnothing \notin {\mathsf{H}}_{\mathbb{G}}^{\mathbb{P}}\left(\mathcal{P}\right)$.

Let $A\in {\mathsf{H}}_{\mathbb{G}}^{\mathbb{P}}\left(\mathcal{P}\right)$ and $A\subset B$. Then $X\setminus A\in \mathcal{P}$. Since $\mathcal{P}$ is a primal and $X\setminus B\subset X\setminus A$, $X\setminus B\in \mathcal{P}$, so $B\in {\mathsf{H}}_{\mathbb{G}}^{\mathbb{P}}\left(\mathcal{P}\right)$. Let $A\cup B\in {\mathsf{H}}_{\mathbb{G}}^{\mathbb{P}}\left(\mathcal{P}\right)$. Then $X\setminus A\cup B=(X\setminus A)\cap (X\setminus B)\in \mathcal{P}$. Since $\mathcal{P}$ is a primal, $X\setminus A$ or $X\setminus B$ is from $\mathcal{P}$. So $A\in {\mathsf{H}}_{\mathbb{G}}^{\mathbb{P}}\left(\mathcal{P}\right)$ or $B\in {\mathsf{H}}_{\mathbb{G}}^{\mathbb{P}}\left(\mathcal{P}\right)$. That means ${\mathsf{H}}_{\mathbb{G}}^{\mathbb{P}}\left(\mathcal{P}\right)$ is a grill.

(11): $x\in A_{{\mathsf{H}}_{\mathbb{P}}^{\mathbb{G}}\left(\mathcal{G}\right)}^{*}$ if and only if $(X\setminus A)\cup (X\setminus U)=X\setminus A\cap U\in {\mathsf{H}}_{\mathbb{P}}^{\mathbb{G}}\left(\mathcal{G}\right)$ for any nbhd U of x if and only if $A\cap U\in \mathcal{G}$ if and only if $x\in A_{\mathcal{G}}^{*}$.

(12): $x\in A_{{\mathsf{H}}_{\mathbb{G}}^{\mathbb{P}}\left(\mathcal{P}\right)}^{*}$ if and only if $A\cap U\in {\mathsf{H}}_{\mathbb{G}}^{\mathbb{P}}\left(\mathcal{P}\right)$ for any nbhd U of x if and only if $X\setminus A\cap U=(X\setminus A)\cup (X\setminus U)\in \mathcal{P}$ if and only if $x\in A_{\mathcal{P}}^{*}$. □

Theorem 6. 

Let $\mathcal{F}$, $\mathcal{I}$ be a filter, a ideal on X, respectively. Define

${\mathsf{H}}_{\mathbb{I}}^{\mathbb{F}}:\mathbb{F}\to \mathbb{I}$ by ${\mathsf{H}}_{\mathbb{I}}^{\mathbb{F}}\left(\mathcal{F}\right)=\{A\subset X:X\setminus A\in \mathcal{F}\}$,

${\mathsf{H}}_{\mathbb{F}}^{\mathbb{I}}:\mathbb{I}\to \mathbb{F}$ by ${\mathsf{H}}_{\mathbb{F}}^{\mathbb{I}}\left(\mathcal{I}\right)=\{A\subset X:X\setminus A\in \mathcal{I}\}$.

Then ${\mathsf{H}}_{\mathbb{I}}^{\mathbb{F}}\left(\mathcal{F}\right)$, ${\mathsf{H}}_{\mathbb{F}}^{\mathbb{I}}\left(\mathcal{I}\right)$ is an ideal, a filter on X, respectively and

Proof. 

We prove ${\mathsf{H}}_{\mathbb{I}}^{\mathbb{F}}\left(\mathcal{F}\right)$ is an ideal. Since $\varnothing \notin \mathcal{F}$, $X\notin {\mathsf{H}}_{\mathbb{I}}^{\mathbb{F}}\left(\mathcal{F}\right)$.

Let $A\in {\mathsf{H}}_{\mathbb{I}}^{\mathbb{F}}\left(\mathcal{F}\right)$ and $B\subset A$. Then $X\setminus A\in \mathcal{F}$. Since $X\setminus A\subset X\setminus B$ and $\mathcal{F}$ is a filter, $X\setminus B\in \mathcal{F}$, so $B\in {\mathsf{H}}_{\mathbb{I}}^{\mathbb{F}}\left(\mathcal{F}\right)$. Let $A,B\in {\mathsf{H}}_{\mathbb{I}}^{\mathbb{F}}\left(\mathcal{F}\right)$. Then $X\setminus A,X\setminus B\in \mathcal{F}$. Since $\mathcal{F}$ is a filter, $(X\setminus A)\cap (X\setminus B)=X\setminus A\cup B\in \mathcal{F}$. So $A\cup B\in {\mathsf{H}}_{\mathbb{I}}^{\mathbb{F}}\left(\mathcal{F}\right)$. That means ${\mathsf{H}}_{\mathbb{I}}^{\mathbb{F}}\left(\mathcal{F}\right)$ is an ideal.

We prove ${\mathsf{H}}_{\mathbb{F}}^{\mathbb{I}}\left(\mathcal{I}\right)$ is a filter. Since $X\notin \mathcal{I}$, $\varnothing \notin {\mathsf{H}}_{\mathbb{F}}^{\mathbb{I}}\left(\mathcal{I}\right)$.

Let $A\in {\mathsf{H}}_{\mathbb{F}}^{\mathbb{I}}\left(\mathcal{I}\right)$ and $A\subset B$. Then $X\setminus A\in \mathcal{I}$. Since $X\setminus B\subset X\setminus A$ and $\mathcal{I}$ is an ideal, $X\setminus B\in \mathcal{I}$, so $B\in {\mathsf{H}}_{\mathbb{F}}^{\mathbb{I}}\left(\mathcal{I}\right)$. Let $A,B\in {\mathsf{H}}_{\mathbb{F}}^{\mathbb{I}}\left(\mathcal{I}\right)$. Then $X\setminus A,X\setminus B\in \mathcal{I}$. Since $\mathcal{I}$ is an ideal, $(X\setminus A)\cup (X\setminus B)=X\setminus A\cap B\in \mathcal{I}$. So $A\cap B\in {\mathsf{H}}_{\mathbb{F}}^{\mathbb{I}}\left(\mathcal{I}\right)$. That means ${\mathsf{H}}_{\mathbb{F}}^{\mathbb{I}}\left(\mathcal{I}\right)$ is a filter.

(11): $x\in A_{{\mathsf{H}}_{\mathbb{I}}^{\mathbb{F}}\left(\mathcal{F}\right)}^{*}$ if and only if $A\cap U\notin {\mathsf{H}}_{\mathbb{I}}^{\mathbb{F}}\left(\mathcal{F}\right)$ for any nbhd U of x if and only if $X\setminus A\cap U\notin \mathcal{F}$ if and only if $x\in A_{\mathcal{F}}^{*}$.

(12): $x\in A_{{\mathsf{H}}_{\mathbb{F}}^{\mathbb{I}}\left(\mathcal{I}\right)}^{*}$ if and only if $X\setminus A\cap U\notin {\mathsf{H}}_{\mathbb{F}}^{\mathbb{I}}\left(\mathcal{I}\right)$ for any nbhd U of x if and only if $A\cap U\notin \mathcal{I}$ if and only if $x\in A_{\mathcal{I}}^{*}$. □

Proposition 1. 

Let ${\mathbb{Z}}_1,{\mathbb{Z}}_2\in \{\mathbb{I},\mathbb{P},\mathbb{F},\mathbb{G}\}$. If ${\mathbb{Z}}_1\ne {\mathbb{Z}}_2$, then each of the twelve operators ${\mathsf{H}}_{{\mathbb{Z}}_2}^{{\mathbb{Z}}_1}$ from Theorem 1-6 is bijective and ${\left({\mathsf{H}}_{{\mathbb{Z}}_2}^{{\mathbb{Z}}_1}\right)}^{-1}={\mathsf{H}}_{{\mathbb{Z}}_1}^{{\mathbb{Z}}_2}$, consequently ${\mathsf{H}}_{{\mathbb{Z}}_2}^{{\mathbb{Z}}_1}$ and ${\mathsf{H}}_{{\mathbb{Z}}_1}^{{\mathbb{Z}}_2}$ are mutually inverse, so ${\mathsf{H}}_{{\mathbb{Z}}_1}^{{\mathbb{Z}}_2}\circ {\mathsf{H}}_{{\mathbb{Z}}_2}^{{\mathbb{Z}}_1}={\mathsf{H}}_{{\mathbb{Z}}_1}^{{\mathbb{Z}}_1}$.

Proof. 

It follows from Theorem 1-6 items (9) and (10). □

Proposition 2. 

${\mathsf{H}}_{\mathbb{F}}^{\mathbb{P}}\circ {\mathsf{H}}_{\mathbb{P}}^{\mathbb{I}}={\mathsf{H}}_{\mathbb{F}}^{\mathbb{I}}$, ${\mathsf{H}}_{\mathbb{I}}^{\mathbb{G}}\circ {\mathsf{H}}_{\mathbb{G}}^{\mathbb{F}}={\mathsf{H}}_{\mathbb{I}}^{\mathbb{F}}$, ${\mathsf{H}}_{\mathbb{G}}^{\mathbb{F}}\circ {\mathsf{H}}_{\mathbb{F}}^{\mathbb{P}}={\mathsf{H}}_{\mathbb{G}}^{\mathbb{P}}$, ${\mathsf{H}}_{\mathbb{P}}^{\mathbb{I}}\circ {\mathsf{H}}_{\mathbb{I}}^{\mathbb{G}}={\mathsf{H}}_{\mathbb{P}}^{\mathbb{G}}$.

Proof. 

$A\in {\mathsf{H}}_{\mathbb{F}}^{\mathbb{P}}\left({\mathsf{H}}_{\mathbb{P}}^{\mathbb{I}}\left(\mathcal{I}\right)\right)$, ${\mathsf{H}}_{\mathbb{I}}^{\mathbb{G}}\left({\mathsf{H}}_{\mathbb{G}}^{\mathbb{F}}\left(\mathcal{F}\right)\right)\iff$$A\notin {\mathsf{H}}_{\mathbb{P}}^{\mathbb{I}}\left(\mathcal{I}\right)$, ${\mathsf{H}}_{\mathbb{G}}^{\mathbb{F}}\left(\mathcal{F}\right)$⇔$X\setminus A\in \mathcal{I},\mathcal{F}\iff A\in {\mathsf{H}}_{\mathbb{F}}^{\mathbb{I}}\left(\mathcal{I}\right),{\mathsf{H}}_{\mathbb{I}}^{\mathbb{F}}\left(\mathcal{F}\right)$, respectively. So, ${\mathsf{H}}_{\mathbb{F}}^{\mathbb{P}}\circ {\mathsf{H}}_{\mathbb{P}}^{\mathbb{I}}={\mathsf{H}}_{\mathbb{F}}^{\mathbb{I}}$, ${\mathsf{H}}_{\mathbb{I}}^{\mathbb{G}}\circ {\mathsf{H}}_{\mathbb{G}}^{\mathbb{F}}={\mathsf{H}}_{\mathbb{I}}^{\mathbb{F}}$.

$A\in {\mathsf{H}}_{\mathbb{G}}^{\mathbb{F}}\left({\mathsf{H}}_{\mathbb{F}}^{\mathbb{P}}\left(\mathcal{P}\right)\right)$, ${\mathsf{H}}_{\mathbb{P}}^{\mathbb{I}}\left({\mathsf{H}}_{\mathbb{I}}^{\mathbb{G}}\left(\mathcal{G}\right)\right)\iff$$X\setminus A\notin {\mathsf{H}}_{\mathbb{F}}^{\mathbb{P}}\left(\mathcal{P}\right)$, ${\mathsf{H}}_{\mathbb{I}}^{\mathbb{G}}\left(\mathcal{G}\right)\iff$$X\setminus A\in \mathcal{P},\mathcal{G}\iff$$A\in {\mathsf{H}}_{\mathbb{G}}^{\mathbb{P}}\left(\mathcal{P}\right),{\mathsf{H}}_{\mathbb{P}}^{\mathbb{G}}\left(\mathcal{G}\right)$, respectively. So, ${\mathsf{H}}_{\mathbb{G}}^{\mathbb{F}}\circ {\mathsf{H}}_{\mathbb{F}}^{\mathbb{P}}={\mathsf{H}}_{\mathbb{G}}^{\mathbb{P}}$, ${\mathsf{H}}_{\mathbb{P}}^{\mathbb{I}}\circ {\mathsf{H}}_{\mathbb{I}}^{\mathbb{G}}={\mathsf{H}}_{\mathbb{P}}^{\mathbb{G}}$. □

Proposition 3. 

Let $\mathbb{Z},{\mathbb{Z}}_1,{\mathbb{Z}}_2\in \{\mathbb{I},\mathbb{P},\mathbb{F},\mathbb{G}\}$. The next conditions are equivalent

(1)

${\mathsf{H}}_{{\mathbb{Z}}_2}^{\mathbb{Z}}\circ {\mathsf{H}}_{\mathbb{Z}}^{{\mathbb{Z}}_1}={\mathsf{H}}_{{\mathbb{Z}}_2}^{{\mathbb{Z}}_1}$ (2) ${\mathsf{H}}_{{\mathbb{Z}}_1}^{\mathbb{Z}}\circ {\mathsf{H}}_{\mathbb{Z}}^{{\mathbb{Z}}_2}={\mathsf{H}}_{{\mathbb{Z}}_1}^{{\mathbb{Z}}_2}$

(3)

${\mathsf{H}}_{\mathbb{Z}}^{{\mathbb{Z}}_2}\circ {\mathsf{H}}_{{\mathbb{Z}}_2}^{{\mathbb{Z}}_1}={\mathsf{H}}_{\mathbb{Z}}^{{\mathbb{Z}}_1}$ (4) ${\mathsf{H}}_{{\mathbb{Z}}_1}^{{\mathbb{Z}}_2}\circ {\mathsf{H}}_{{\mathbb{Z}}_2}^{\mathbb{Z}}={\mathsf{H}}_{{\mathbb{Z}}_1}^{\mathbb{Z}}$

(5)

${\mathsf{H}}_{{\mathbb{Z}}_2}^{{\mathbb{Z}}_1}\circ {\mathsf{H}}_{{\mathbb{Z}}_1}^{\mathbb{Z}}={\mathsf{H}}_{{\mathbb{Z}}_2}^{\mathbb{Z}}$ (6) ${\mathsf{H}}_{\mathbb{Z}}^{{\mathbb{Z}}_1}\circ {\mathsf{H}}_{{\mathbb{Z}}_1}^{{\mathbb{Z}}_2}={\mathsf{H}}_{\mathbb{Z}}^{{\mathbb{Z}}_2}$

Proof.

$\left(1\right)\iff \left(2\right)$:

${\mathsf{H}}_{{\mathbb{Z}}_2}^{\mathbb{Z}}\circ {\mathsf{H}}_{\mathbb{Z}}^{{\mathbb{Z}}_1}={\mathsf{H}}_{{\mathbb{Z}}_2}^{{\mathbb{Z}}_1}$⇔${({\mathsf{H}}_{{\mathbb{Z}}_2}^{\mathbb{Z}}\circ {\mathsf{H}}_{\mathbb{Z}}^{{\mathbb{Z}}_1})}^{-1}={\left({\mathsf{H}}_{{\mathbb{Z}}_2}^{{\mathbb{Z}}_1}\right)}^{-1}$⇔${\mathsf{H}}_{{\mathbb{Z}}_1}^{\mathbb{Z}}\circ {\mathsf{H}}_{\mathbb{Z}}^{{\mathbb{Z}}_2}={\mathsf{H}}_{{\mathbb{Z}}_1}^{{\mathbb{Z}}_2}$

$\left(1\right)\iff \left(3\right)$:

${\mathsf{H}}_{{\mathbb{Z}}_2}^{\mathbb{Z}}\circ {\mathsf{H}}_{\mathbb{Z}}^{{\mathbb{Z}}_1}={\mathsf{H}}_{{\mathbb{Z}}_2}^{{\mathbb{Z}}_1}$⇔${\mathsf{H}}_{\mathbb{Z}}^{{\mathbb{Z}}_1}={\left({\mathsf{H}}_{{\mathbb{Z}}_2}^{\mathbb{Z}}\right)}^{-1}\circ {\mathsf{H}}_{{\mathbb{Z}}_2}^{{\mathbb{Z}}_1}\iff {\mathsf{H}}_{\mathbb{Z}}^{{\mathbb{Z}}_1}={\mathsf{H}}_{\mathbb{Z}}^{{\mathbb{Z}}_2}\circ {\mathsf{H}}_{{\mathbb{Z}}_2}^{{\mathbb{Z}}_1}$

$\left(3\right)\iff \left(4\right)$:

${\mathsf{H}}_{\mathbb{Z}}^{{\mathbb{Z}}_2}\circ {\mathsf{H}}_{{\mathbb{Z}}_2}^{{\mathbb{Z}}_1}={\mathsf{H}}_{\mathbb{Z}}^{{\mathbb{Z}}_1}$⇔${({\mathsf{H}}_{\mathbb{Z}}^{{\mathbb{Z}}_2}\circ {\mathsf{H}}_{{\mathbb{Z}}_2}^{{\mathbb{Z}}_1})}^{-1}={\left({\mathsf{H}}_{\mathbb{Z}}^{{\mathbb{Z}}_1}\right)}^{-1}$⇔${\mathsf{H}}_{{\mathbb{Z}}_1}^{{\mathbb{Z}}_2}\circ {\mathsf{H}}_{{\mathbb{Z}}_2}^{\mathbb{Z}}={\mathsf{H}}_{{\mathbb{Z}}_1}^{\mathbb{Z}}$

$\left(1\right)\iff \left(5\right)$:

${\mathsf{H}}_{{\mathbb{Z}}_2}^{\mathbb{Z}}\circ {\mathsf{H}}_{\mathbb{Z}}^{{\mathbb{Z}}_1}={\mathsf{H}}_{{\mathbb{Z}}_2}^{{\mathbb{Z}}_1}$⇔${\mathsf{H}}_{{\mathbb{Z}}_2}^{\mathbb{Z}}={\mathsf{H}}_{{\mathbb{Z}}_2}^{{\mathbb{Z}}_1}\circ {\left({\mathsf{H}}_{\mathbb{Z}}^{{\mathbb{Z}}_1}\right)}^{-1}\iff {\mathsf{H}}_{{\mathbb{Z}}_2}^{\mathbb{Z}}={\mathsf{H}}_{{\mathbb{Z}}_2}^{{\mathbb{Z}}_1}\circ {\mathsf{H}}_{{\mathbb{Z}}_1}^{\mathbb{Z}}$

$\left(5\right)\iff \left(6\right)$:

${\mathsf{H}}_{{\mathbb{Z}}_2}^{{\mathbb{Z}}_1}\circ {\mathsf{H}}_{{\mathbb{Z}}_1}^{\mathbb{Z}}={\mathsf{H}}_{{\mathbb{Z}}_2}^{\mathbb{Z}}$⇔${({\mathsf{H}}_{{\mathbb{Z}}_2}^{{\mathbb{Z}}_1}\circ {\mathsf{H}}_{{\mathbb{Z}}_1}^{\mathbb{Z}})}^{-1}={\left({\mathsf{H}}_{{\mathbb{Z}}_2}^{\mathbb{Z}}\right)}^{-1}$⇔${\mathsf{H}}_{\mathbb{Z}}^{{\mathbb{Z}}_1}\circ {\mathsf{H}}_{{\mathbb{Z}}_1}^{{\mathbb{Z}}_2}={\mathsf{H}}_{\mathbb{Z}}^{{\mathbb{Z}}_2}$□

Proposition 4. 

Let $\mathbb{Z},{\mathbb{Z}}_1,{\mathbb{Z}}_2\in \{\mathbb{I},\mathbb{P},\mathbb{F},\mathbb{G}\}$. Then for 64 possibilities the equation ${\mathsf{H}}_{{\mathbb{Z}}_2}^{\mathbb{Z}}\circ {\mathsf{H}}_{\mathbb{Z}}^{{\mathbb{Z}}_1}={\mathsf{H}}_{{\mathbb{Z}}_2}^{{\mathbb{Z}}_1}$ holds.

Proof. 

By Proposition 2 and Proposition 3, the equation ${\mathsf{H}}_{{\mathbb{Z}}_2}^{\mathbb{Z}}\circ {\mathsf{H}}_{\mathbb{Z}}^{{\mathbb{Z}}_1}={\mathsf{H}}_{{\mathbb{Z}}_2}^{{\mathbb{Z}}_1}$ holds for 24 possibilities ($\mathbb{Z},{\mathbb{Z}}_1,{\mathbb{Z}}_2$ are mutually different):

Other cases for ${\mathbb{Z}}_1\ne {\mathbb{Z}}_2$ are trivial:

${\mathsf{H}}_{{\mathbb{Z}}_1}^{{\mathbb{Z}}_1}\circ {\mathsf{H}}_{{\mathbb{Z}}_1}^{{\mathbb{Z}}_2}={\mathsf{H}}_{{\mathbb{Z}}_1}^{{\mathbb{Z}}_2}$ (12 possibilities), ${\mathsf{H}}_{{\mathbb{Z}}_2}^{{\mathbb{Z}}_1}\circ {\mathsf{H}}_{{\mathbb{Z}}_1}^{{\mathbb{Z}}_1}={\mathsf{H}}_{{\mathbb{Z}}_2}^{{\mathbb{Z}}_1}$ (12 possibilities),

${\mathsf{H}}_{{\mathbb{Z}}_1}^{{\mathbb{Z}}_2}\circ {\mathsf{H}}_{{\mathbb{Z}}_2}^{{\mathbb{Z}}_1}={\mathsf{H}}_{{\mathbb{Z}}_1}^{{\mathbb{Z}}_1}$ (12 possibilities), ${\mathsf{H}}_{{\mathbb{Z}}_1}^{{\mathbb{Z}}_1}\circ {\mathsf{H}}_{{\mathbb{Z}}_1}^{{\mathbb{Z}}_1}={\mathsf{H}}_{{\mathbb{Z}}_1}^{{\mathbb{Z}}_1}$ (4 possibilities).

□

For a composition of finitely many operators, the domain (the codomain) of the resulting operator is equal to the domain (codomain) of the first (last) operator and the value of local function is independent on the operators as the following theorem states.

Theorem 7. 

Let ${\mathbb{Z}}_k\in \{\mathbb{I},\mathbb{P},\mathbb{F},\mathbb{G}\}$, $k=1,\cdots ,n$ and $\mathcal{Z}\in {\mathbb{Z}}_1$. Then

$${\mathsf{H}}_{{\mathbb{Z}}_n}^{{\mathbb{Z}}_{n-1}}\circ {\mathsf{H}}_{{\mathbb{Z}}_{n-1}}^{{\mathbb{Z}}_{n-2}}\circ \cdots \circ {\mathsf{H}}_{{\mathbb{Z}}_3}^{{\mathbb{Z}}_2}\circ {\mathsf{H}}_{{\mathbb{Z}}_2}^{{\mathbb{Z}}_1}={\mathsf{H}}_{{\mathbb{Z}}_n}^{{\mathbb{Z}}_1}$$

$$A_{\left({\mathsf{H}}_{{\mathbb{Z}}_n}^{{\mathbb{Z}}_{n-1}}\circ {\mathsf{H}}_{{\mathbb{Z}}_{n-1}}^{{\mathbb{Z}}_{n-2}}\circ \cdots \circ {\mathsf{H}}_{{\mathbb{Z}}_3}^{{\mathbb{Z}}_2}\circ {\mathsf{H}}_{{\mathbb{Z}}_2}^{{\mathbb{Z}}_1}\right)\left(\mathcal{Z}\right)}^{*}=A_{{\mathsf{H}}_{{\mathbb{Z}}_n}^{{\mathbb{Z}}_1}\left(\mathcal{Z}\right)}^{*}=A_{\mathcal{Z}}^{*}$$

Consequently,

(1)

$A_{\mathcal{I}}^{*}=A_{{\mathsf{H}}_{\mathbb{I}}^{\mathbb{I}}\left(\mathcal{I}\right)}^{*}=A_{{\mathsf{H}}_{\mathbb{P}}^{\mathbb{I}}\left(\mathcal{I}\right)}^{*}=A_{{\mathsf{H}}_{\mathbb{G}}^{\mathbb{I}}\left(\mathcal{I}\right)}^{*}=A_{{\mathsf{H}}_{\mathbb{F}}^{\mathbb{I}}\left(\mathcal{I}\right)}^{*}$

(2)

$A_{\mathcal{P}}^{*}=A_{{\mathsf{H}}_{\mathbb{P}}^{\mathbb{P}}\left(\mathcal{P}\right)}^{*}=A_{{\mathsf{H}}_{\mathbb{I}}^{\mathbb{P}}\left(\mathcal{P}\right)}^{*}=A_{{\mathsf{H}}_{\mathbb{F}}^{\mathbb{P}}\left(\mathcal{P}\right)}^{*}=A_{{\mathsf{H}}_{\mathbb{G}}^{\mathbb{P}}\left(\mathcal{P}\right)}^{*}$

(3)

$A_{\mathcal{F}}^{*}=A_{{\mathsf{H}}_{\mathbb{F}}^{\mathbb{F}}\left(\mathcal{F}\right)}^{*}=A_{{\mathsf{H}}_{\mathbb{G}}^{\mathbb{F}}\left(\mathcal{F}\right)}^{*}=A_{{\mathsf{H}}_{\mathbb{P}}^{\mathbb{F}}\left(\mathcal{F}\right)}^{*}=A_{{\mathsf{H}}_{\mathbb{I}}^{\mathbb{F}}\left(\mathcal{F}\right)}^{*}$

(4)

$A_{\mathcal{G}}^{*}=A_{{\mathsf{H}}_{\mathbb{G}}^{\mathbb{G}}\left(\mathcal{G}\right)}^{*}=A_{{\mathsf{H}}_{\mathbb{I}}^{\mathbb{G}}\left(\mathcal{G}\right)}^{*}=A_{{\mathsf{H}}_{\mathbb{F}}^{\mathbb{G}}\left(\mathcal{G}\right)}^{*}=A_{{\mathsf{H}}_{\mathbb{P}}^{\mathbb{G}}\left(\mathcal{G}\right)}^{*}$

Proof. 

For the first equation we use the mathematical induction. If $n=3$, it follows from Proposition 4. Suppose the equation holds for $n>3$. Then

${\mathsf{H}}_{{\mathbb{Z}}_{n+1}}^{{\mathbb{Z}}_n}\circ {\mathsf{H}}_{{\mathbb{Z}}_n}^{{\mathbb{Z}}_{n-1}}\circ {\mathsf{H}}_{{\mathbb{Z}}_{n-1}}^{{\mathbb{Z}}_{n-2}}\circ \cdots \circ {\mathsf{H}}_{{\mathbb{Z}}_3}^{{\mathbb{Z}}_2}\circ {\mathsf{H}}_{{\mathbb{Z}}_2}^{{\mathbb{Z}}_1}={\mathsf{H}}_{{\mathbb{Z}}_{n+1}}^{{\mathbb{Z}}_n}\circ {\mathsf{H}}_{{\mathbb{Z}}_n}^{{\mathbb{Z}}_1}={\mathsf{H}}_{{\mathbb{Z}}_{n+1}}^{{\mathbb{Z}}_1}$, by Proposition 4. The second equation follows from the first one and from Theorem 1–6. □

The set $\{{\mathsf{H}}_{{\mathbb{Z}}_2}^{{\mathbb{Z}}_1}:{\mathbb{Z}}_1,{\mathbb{Z}}_2\in \{\mathbb{I},\mathbb{P},\mathbb{F},\mathbb{G}\}\}$ consisting of 16 operators is enclosed under composition. The results can be interpreted by the next diagram.

3. Applications

We have defined 16 operators, which can be expressed by one notation ${\mathsf{H}}_{{\mathbb{Z}}_2}^{{\mathbb{Z}}_1}$ where ${\mathbb{Z}}_1,{\mathbb{Z}}_2\in \{\mathbb{I},\mathbb{P},\mathbb{F},\mathbb{G}\}$. Between the members of ${\mathbb{Z}}_1$ and the members of ${\mathbb{Z}}_2$ we can define an equivalence as the next definition states. Note if $\mathcal{Z}\in \mathbb{X}$, then $\mathcal{Z}\in \mathbb{Z}$ for some $\mathbb{Z}\in \{\mathbb{I},\mathbb{P},\mathbb{F},\mathbb{G}\}$.

Definition 2. 

Let ${\mathcal{Z}}_1,{\mathcal{Z}}_2\in \mathbb{X}$. ${\mathcal{Z}}_1$ is equivalent to ${\mathcal{Z}}_2$ if ${\mathsf{H}}_{{\mathbb{Z}}_2}^{{\mathbb{Z}}_1}\left({\mathcal{Z}}_1\right)={\mathcal{Z}}_2$ and ${\mathsf{H}}_{{\mathbb{Z}}_1}^{{\mathbb{Z}}_2}\left({\mathcal{Z}}_2\right)={\mathcal{Z}}_1$ where ${\mathcal{Z}}_1\in {\mathbb{Z}}_1$, ${\mathcal{Z}}_2\in {\mathbb{Z}}_2$ and ${\mathbb{Z}}_1,{\mathbb{Z}}_2\in \{\mathbb{I},\mathbb{P},\mathbb{F},\mathbb{G}\}$. This relation is denoted by ${\mathcal{Z}}_1\sim {\mathcal{Z}}_2$.

Lemma 1. 

For any $\mathcal{Z}\in \mathbb{X}$, $\mathcal{Z}\sim {\mathsf{H}}_{{\mathbb{Z}}_2}^{{\mathbb{Z}}_1}\left(\mathcal{Z}\right)$ where $\mathcal{Z}\in {\mathbb{Z}}_1$ and ${\mathbb{Z}}_1,{\mathbb{Z}}_2\in \{\mathbb{I},\mathbb{P},\mathbb{F},\mathbb{G}\}$. Moreover, ∼ is an equivalence relation.

Proof. 

Since ${\mathsf{H}}_{{\mathbb{Z}}_2}^{{\mathbb{Z}}_1}\left(\mathcal{Z}\right)={\mathsf{H}}_{{\mathbb{Z}}_2}^{{\mathbb{Z}}_1}\left(\mathcal{Z}\right)$ and $\mathcal{Z}={\mathsf{H}}_{{\mathbb{Z}}_1}^{{\mathbb{Z}}_2}\left({\mathsf{H}}_{{\mathbb{Z}}_2}^{{\mathbb{Z}}_1}\left(\mathcal{Z}\right)\right)$, $\mathcal{Z}\sim {\mathsf{H}}_{{\mathbb{Z}}_2}^{{\mathbb{Z}}_1}\left(\mathcal{Z}\right)$. It is clear ∼ is reflexive and symmetric. Let ${\mathcal{Z}}_1\sim {\mathcal{Z}}_2\sim {\mathcal{Z}}_3$. Then ${\mathcal{Z}}_2={\mathsf{H}}_{{\mathbb{Z}}_2}^{{\mathbb{Z}}_1}\left({\mathcal{Z}}_1\right)$ and ${\mathcal{Z}}_3={\mathsf{H}}_{{\mathbb{Z}}_3}^{{\mathbb{Z}}_2}\left({\mathcal{Z}}_2\right)={\mathsf{H}}_{{\mathbb{Z}}_3}^{{\mathbb{Z}}_2}\left({\mathsf{H}}_{{\mathbb{Z}}_2}^{{\mathbb{Z}}_1}\left({\mathcal{Z}}_1\right)\right)={\mathsf{H}}_{{\mathbb{Z}}_3}^{{\mathbb{Z}}_1}\left({\mathcal{Z}}_1\right)$ and ${\mathcal{Z}}_1={\mathsf{H}}_{{\mathbb{Z}}_1}^{{\mathbb{Z}}_3}\left({\mathcal{Z}}_3\right)$, so ${\mathcal{Z}}_1\sim {\mathcal{Z}}_3$. □

In the next definition we define a dual operator $A_{\mathcal{Z}}^{▹}$ (see [5]) to the operator $A_{\mathcal{Z}}^{*}$ and a closure $c{l}_{\mathcal{Z}}(·)$ and an interior $in{t}_{\mathcal{Z}}(·)$ operator.

Definition 3. 

Let $\mathcal{Z}\in \mathbb{X}$. Define the next operators

$A_{\mathcal{Z}}^{▹}=X\setminus {(X\setminus A)}_{\mathcal{Z}}^{*},$

$c{l}_{\mathcal{Z}}\left(A\right)=A\cup A_{\mathcal{Z}}^{*},$

$in{t}_{\mathcal{Z}}\left(A\right)=A\cap A_{\mathcal{Z}}^{▹}.$

Lemma 2. 

Let ${\mathcal{Z}}_1,{\mathcal{Z}}_2\in \mathbb{X}$. If ${\mathcal{Z}}_1\sim {\mathcal{Z}}_2$, then $A_{{\mathcal{Z}}_2}^{*}=A_{{\mathcal{Z}}_1}^{*}$, $A_{{\mathcal{Z}}_2}^{▹}=A_{{\mathcal{Z}}_1}^{▹}$, $c{l}_{{\mathcal{Z}}_2}\left(A\right)=c{l}_{{\mathcal{Z}}_1}\left(A\right)$, $in{t}_{{\mathcal{Z}}_2}\left(A\right)=in{t}_{{\mathcal{Z}}_1}\left(A\right)$. Consequently, if $\mathcal{I}\sim \mathcal{P}\sim \mathcal{F}\sim \mathcal{G}$, then $A_{\mathcal{I}}^{*}=A_{\mathcal{P}}^{*}=A_{\mathcal{F}}^{*}=A_{\mathcal{G}}^{*}$.

Proof. 

Suppose ${\mathcal{Z}}_1\sim {\mathcal{Z}}_2$. Then ${\mathsf{H}}_{{\mathbb{Z}}_2}^{{\mathbb{Z}}_1}\left({\mathcal{Z}}_1\right)={\mathcal{Z}}_2$. By Theorem 7, $A_{{\mathcal{Z}}_2}^{*}=A_{{\mathsf{H}}_{{\mathbb{Z}}_2}^{{\mathbb{Z}}_1}\left({\mathcal{Z}}_1\right)}^{*}=A_{{\mathcal{Z}}_1}^{*}$. Other equalities are obvious. □

Remark 1. 

It is well known (see for example [7]) that an ideal topology ${\tau }_{\mathcal{I}}$ derived from a topology τ on a set X and an ideal $\mathcal{I}$ on X is defined by a Kuratowski closure operator $c{l}_{\mathcal{I}}\left(A\right)=A\cup A_{\mathcal{I}}^{*}$ and ${\tau }_{\mathcal{I}}$ is finer then τ. A base for ${\tau }_{\mathcal{I}}$ is described as $\beta (\tau ,\mathcal{I})=\{G\setminus A:G\in \tau ,A\in \mathcal{I}\}$ and ${\tau }_{\mathcal{I}}=\{A\subset X:c{l}_{\mathcal{I}}(X\setminus A)=X\setminus A\}$.

In the literature we can find many properties of local functions. Designation of operators is different. For example, $A_{\mathcal{P}}^{*}=A_{\mathcal{P}}^{\diamond }$, $A_{\mathcal{P}}^{▹}=\Psi \left(A\right)$ in [1] or $A_{\mathcal{G}}^{*}={\Phi }_{\mathcal{G}}\left(A\right)$, $A_{\mathcal{G}}^{▹}={\Gamma }_{\mathcal{G}}\left(A\right)$ in [15]. We will list some of them below regardless of what system $\mathcal{Z}\in \mathbb{Z}$ they apply to.

Theorem 8. 

Let $\mathcal{Z}\in \mathbb{X}$. Then

(1) 

${\varnothing }_{\mathcal{Z}}^{*}=\varnothing$,

(2) 

If $A\subset B$, then $A_{\mathcal{Z}}^{*}\subset B_{\mathcal{Z}}^{*}$,

(3) 

${(A\cup B)}_{\mathcal{Z}}^{*}=A_{\mathcal{Z}}^{*}\cup B_{\mathcal{Z}}^{*}$,

(4) 

$A_{\mathcal{Z}}^{*}\setminus B_{\mathcal{Z}}^{*}={(A\setminus B)}_{\mathcal{Z}}^{*}\setminus B_{\mathcal{Z}}^{*}$,

(5) 

${\left(A_{\mathcal{Z}}^{*}\right)}_{\mathcal{Z}}^{*}\subset A_{\mathcal{Z}}^{*}$,

(6) 

If $A\subset B$, then $A_{\mathcal{Z}}^{▹}\subset B_{\mathcal{Z}}^{▹}$,

(7) 

$A_{\mathcal{Z}}^{▹}\subset {\left(A_{\mathcal{Z}}^{▹}\right)}_{\mathcal{Z}}^{▹}$,

(8) 

${(A\cap B)}_{\mathcal{Z}}^{▹}=A_{\mathcal{Z}}^{▹}\cap B_{\mathcal{Z}}^{▹}$.

Proof. 

Let $\mathcal{I}:={\mathsf{H}}_{\mathbb{I}}^{\mathbb{Z}}\left(\mathcal{Z}\right)$. Then $A_{\mathcal{I}}^{*}=A_{{\mathsf{H}}_{\mathbb{I}}^{\mathbb{Z}}\left(\mathcal{Z}\right)}^{*}=A_{\mathcal{Z}}^{*}$, by Theorem 7. Since all items hold for $\mathcal{I}\in \mathbb{I}$ (see for example [7]), they hold for $\mathcal{Z}\in \mathbb{P}\cup \mathbb{F}\cup \mathbb{G}$. □

Similarly, the next theorem holds for $\mathcal{I}\in \mathbb{I}$, so it holds for any $\mathcal{Z}\in \mathbb{X}$.

Theorem 9. 

Let $\mathcal{Z}\in \mathbb{X}$. A family ${\tau }_{\mathcal{Z}}=\{A\subset X:c{l}_{\mathcal{Z}}(X\setminus A)=X\setminus A\}$ is a topology finer then τ and the next conditions are equivalent.

(1) 

$A\in {\tau }_{\mathcal{Z}}$,

(2) 

$in{t}_{\mathcal{Z}}\left(A\right)=A$,

(3) 

$A\subset in{t}_{\mathcal{Z}}\left(A\right)$,

(4) 

$A\subset A_{\mathcal{Z}}^{▹}\left(A\right)$,

(5) 

$c{l}_{\mathcal{Z}}(X\setminus A)=X\setminus A$,

(6) 

$c{l}_{\mathcal{Z}}(X\setminus A)\subset X\setminus A$,

(7) 

${(X\setminus A)}_{\mathcal{Z}}^{*}\subset X\setminus A$.

Theorem 10. 

Let ${\mathcal{Z}}_1,{\mathcal{Z}}_2\in \mathbb{X}$. If ${\mathcal{Z}}_1\sim {\mathcal{Z}}_2$, then ${\tau }_{{\mathcal{Z}}_1}={\tau }_{{\mathcal{Z}}_2}$. Consequently, if $\mathcal{I}\sim \mathcal{P}\sim \mathcal{F}\sim \mathcal{G}$, then ${\tau }_{\mathcal{I}}={\tau }_{\mathcal{P}}={\tau }_{\mathcal{F}}={\tau }_{\mathcal{G}}$. A simple base for the open sets of ${\tau }_{\mathcal{I}},{\tau }_{\mathcal{P}},{\tau }_{\mathcal{F}},{\tau }_{\mathcal{G}}$ is described as follows:

$\beta (\tau ,\mathcal{I})=\{G\setminus A:G\in \tau ,A\in \mathcal{I}\}$,

$\beta (\tau ,\mathcal{P})=\{G\cap A:G\in \tau ,A\notin \mathcal{P}\}$,

$\beta (\tau ,\mathcal{F})=\{G\cap A:G\in \tau ,A\in \mathcal{F}\}$,

$\beta (\tau ,\mathcal{G})=\{G\setminus A:G\in \tau ,A\notin \mathcal{G}\}$, respectively.

Proof. 

The equality ${\tau }_{{\mathcal{Z}}_1}={\tau }_{{\mathcal{Z}}_2}$ follows from Lemma 2.

By Remark 1, $\beta (\tau ,\mathcal{I})=\{G\setminus A:G\in \tau ,A\in \mathcal{I}\}$ is a base for ${\tau }_{\mathcal{I}}$. $H\in \beta (\tau ,\mathcal{I})$ if and only if $H=G\setminus A=G\cap (X\setminus A)$ where $G\in \tau$ and $A\in \mathcal{I}$ if and only if $H=G\cap B$ where $G\in \tau$ and $B:=X\setminus A\notin \mathcal{P}$ (by Theorem 1 (4)), so $H\in \beta (\tau ,\mathcal{P})$ if and only if $H=G\cap B$ where $G\in \tau$ and $B\in \mathcal{F}$ (by Theorem 4 (5)), so $H\in \beta (\tau ,\mathcal{F})$ if and only if $H=G\setminus A=G\setminus (X\setminus B)$ where $G\in \tau$ and $X\setminus B\notin \mathcal{G}$ (by Theorem 3 (4)), so $H\in \beta (\tau ,\mathcal{G})$. □

Definition 4. 

A set A is $\mathcal{I}$-small ($\mathcal{P}$-small, $\mathcal{F}$-small, $\mathcal{G}$-small) if $A\in \mathcal{I}$ ($X\setminus A\notin \mathcal{P}$, $X\setminus A\in \mathcal{F}$, $A\notin \mathcal{G}$) and A is locally $\mathcal{I}$-small ($\mathcal{P}$-small, $\mathcal{F}$-small, $\mathcal{G}$-small) if for any $a\in A$ there is a set $U\in \tau$ containing a such that $A\cap U$ is $\mathcal{I}$-small ($\mathcal{P}$-small, $\mathcal{F}$-small, $\mathcal{G}$-small). Let $\mathcal{Z}\in \mathbb{X}$. $\mathcal{Z}$ is said to be compatible with τ if any locally $\mathcal{Z}$-small set is $\mathcal{Z}$-small, denoted by $\mathcal{Z}\sim \tau$.

Remark 2. 

Let $\mathcal{I}\sim \mathcal{P}\sim \mathcal{F}\sim \mathcal{G}$. Then $\mathcal{I}={\mathsf{H}}_{\mathbb{I}}^{\mathbb{P}}\left(\mathcal{P}\right)$, $\mathcal{F}={\mathsf{H}}_{\mathbb{F}}^{\mathbb{P}}\left(\mathcal{P}\right)$ $\mathcal{G}={\mathsf{H}}_{\mathbb{G}}^{\mathbb{F}}\left(\mathcal{F}\right)$. So, $A\in \mathcal{I}\iff X\setminus A\notin \mathcal{P}\iff X\setminus A\in \mathcal{F}\iff A\notin \mathcal{G}$. That means if ${\mathcal{Z}}_1,{\mathcal{Z}}_2\in \mathbb{X}$ and ${\mathcal{Z}}_1\sim {\mathcal{Z}}_2$, a set A is ${\mathcal{Z}}_1$-small (locally ${\mathcal{Z}}_1$-small) if and only if A is ${\mathcal{Z}}_2$-small (locally ${\mathcal{Z}}_2$-small) and ${\mathcal{Z}}_1\sim \tau$ if and only if ${\mathcal{Z}}_2\sim \tau$. Consequently,

(1) 

If $\mathcal{I}\sim \tau$, then ${\mathsf{H}}_{\mathbb{P}}^{\mathbb{I}}\left(\mathcal{I}\right)\sim \tau ,{\mathsf{H}}_{\mathbb{F}}^{\mathbb{I}}\left(\mathcal{I}\right)\sim \tau ,{\mathsf{H}}_{\mathbb{G}}^{\mathbb{I}}\left(\mathcal{I}\right)\sim \tau$.

(2) 

If $\mathcal{P}\sim \tau$, then ${\mathsf{H}}_{\mathbb{F}}^{\mathbb{P}}\left(\mathcal{P}\right)\sim \tau ,{\mathsf{H}}_{\mathbb{G}}^{\mathbb{P}}\left(\mathcal{P}\right)\sim \tau ,{\mathsf{H}}_{\mathbb{I}}^{\mathbb{P}}\left(\mathcal{P}\right)\sim \tau$.

(3) 

If $\mathcal{F}\sim \tau$, then ${\mathsf{H}}_{\mathbb{G}}^{\mathbb{F}}\left(\mathcal{F}\right)\sim \tau ,{\mathsf{H}}_{\mathbb{I}}^{\mathbb{F}}\left(\mathcal{F}\right)\sim \tau ,{\mathsf{H}}_{\mathbb{P}}^{\mathbb{F}}\left(\mathcal{F}\right)\sim \tau$.

(4) 

If $\mathcal{G}\sim \tau$, then ${\mathsf{H}}_{\mathbb{I}}^{\mathbb{G}}\left(\mathcal{G}\right)\sim \tau ,{\mathsf{H}}_{\mathbb{P}}^{\mathbb{G}}\left(\mathcal{G}\right)\sim \tau ,{\mathsf{H}}_{\mathbb{F}}^{\mathbb{G}}\left(\mathcal{G}\right)\sim \tau$.

In the theory of ideal topological spaces there are several characterizations of compatibility. For $(X,\tau ,\mathcal{I})$ the most common equivalent conditions are as follows (see for example [5,7]).

Theorem 11. 

The next are equivalent

(1) 

$\mathcal{I}\sim \tau$,

(2) 

for every $A\subset X$, if $A\cap A_{\mathcal{I}}^{*}=\varnothing$, then $A\in \mathcal{I}$,

(3) 

for every $A\subset X$, $A\setminus A_{\mathcal{I}}^{*}\in \mathcal{I}$,

(4) 

for every $A\subset X$, if $(X\setminus A)\cup {(X\setminus A)}_{\mathcal{I}}^{▹}=X$, then $A\in \mathcal{I}$,

(5) 

for every $A\subset X$, $A_{\mathcal{I}}^{▹}\setminus A\in \mathcal{I}$.

Regardless of a concept we work in, a compatibility of ${\mathcal{Z}}_1$ can be characterized by another equivalent system ${\mathcal{Z}}_2$ and by operators ${(·)}_{{\mathcal{Z}}_2}^{*}$ and ${(·)}_{{\mathcal{Z}}_2}^{▹}$.

Theorem 12. 

Let ${\mathcal{Z}}_1,{\mathcal{Z}}_2\in \mathbb{X}$. If ${\mathcal{Z}}_1\sim {\mathcal{Z}}_2$, then the next are equivalent

(1) 

${\mathcal{Z}}_1\sim \tau$,

(2) 

for every $A\subset X$, if $A\cap A_{{\mathcal{Z}}_2}^{*}=\varnothing$, then A is ${\mathcal{Z}}_2$-small,

(3) 

for every $A\subset X$, $A\setminus A_{{\mathcal{Z}}_2}^{*}$ is ${\mathcal{Z}}_2$-small,

(4) 

for every $A\subset X$, if $(X\setminus A)\cup {(X\setminus A)}_{{\mathcal{Z}}_2}^{▹}=X$, then A is ${\mathcal{Z}}_2$-small,

(5) 

for every $A\subset X$, $A_{{\mathcal{Z}}_2}^{▹}\setminus A$ is ${\mathcal{Z}}_2$-small.

Proof. 

Let ${\mathcal{I}}_1:={\mathsf{H}}_{\mathbb{I}}^{{\mathbb{Z}}_1}\left({\mathcal{Z}}_1\right)$ where ${\mathcal{Z}}_1\in {\mathbb{Z}}_1\in \{\mathbb{P},\mathbb{F},\mathbb{G}\}$. Since ${\mathcal{I}}_1\sim {\mathcal{Z}}_1\sim {\mathcal{Z}}_2$, ${\mathcal{Z}}_1\sim \tau$ if and only if ${\mathcal{I}}_1\sim \tau$ (by Remark 2), A is ${\mathcal{I}}_1$-small if and only if A is ${\mathcal{Z}}_2$-small (by Remark 2), $A_{{\mathcal{I}}_1}^{*}=A_{{\mathcal{Z}}_1}^{*}=A_{{\mathcal{Z}}_2}^{*}$, $A_{{\mathcal{I}}_1}^{▹}=A_{{\mathcal{Z}}_1}^{▹}=A_{{\mathcal{Z}}_2}^{▹}$ (by Lemma 2). Then the next are equivalent (note $\left(\mathrm{ii}\right)\iff \left(2\right),\left(\mathrm{iii}\right)\iff \left(3\right),\left(\mathrm{iv}\right)\iff \left(4\right),(\mathrm{v})\iff (5)$)

(1)

${\mathcal{Z}}_1\sim \tau$,

(i)

${\mathcal{I}}_1\sim \tau$,

(ii)

for every $A\subset X$, if $A\cap A_{{\mathcal{I}}_1}^{*}=\varnothing$, then A is ${\mathcal{I}}_1$-small,

(iii)

for every $A\subset X$, $A\setminus A_{{\mathcal{I}}_1}^{*}$ is ${\mathcal{I}}_1$-small,

(iv)

for every $A\subset X$, if $(X\setminus A)\cup {(X\setminus A)}_{{\mathcal{I}}_1}^{▹}=X$, then A is ${\mathcal{I}}_1$-small,

(v)

for every $A\subset X$, $A_{{\mathcal{I}}_1}^{▹}\setminus A$ is ${\mathcal{I}}_1$-small.

(2)

for every $A\subset X$, if $A\cap A_{{\mathcal{Z}}_2}^{*}=\varnothing$, then A is ${\mathcal{Z}}_2$-small,

(3)

for every $A\subset X$, $A\setminus A_{{\mathcal{Z}}_2}^{*}$ is ${\mathcal{Z}}_2$-small,

(4)

for every $A\subset X$, if $(X\setminus A)\cup {(X\setminus A)}_{{\mathcal{Z}}_2}^{▹}=X$, then A is ${\mathcal{Z}}_2$-small,

(5)

for every $A\subset X$, $A_{{\mathcal{Z}}_2}^{▹}\setminus A$ is ${\mathcal{Z}}_2$-small.

□

Remark 3. 

Note, in a primal case (see for example [1]) a compatibility is called "a topology suitable for a primal". More precisely, we prove τ is suitable for a primal $\mathcal{P}$ if and only if ${\mathsf{H}}_{\mathbb{I}}^{\mathbb{P}}\left(\mathcal{P}\right)$ is compatible with τ. Proof: τ is suitable for a primal $\mathcal{P}$ if and only if $X\setminus A\notin \mathcal{P}$ whenever $A\cap A_{\mathcal{P}}^{\diamond }=\varnothing$ (see [1] Theorem 4.2) if and only if $X\setminus A\notin \mathcal{P}$ whenever $A\cap A_{\mathcal{P}}^{*}=\varnothing$ if and only if A is $\mathcal{P}$-small whenever $A\cap A_{{\mathsf{H}}_{\mathbb{I}}^{\mathbb{P}}\left(\mathcal{P}\right)}^{*}=\varnothing$ if and only if A is ${\mathsf{H}}_{\mathbb{I}}^{\mathbb{P}}\left(\mathcal{P}\right)$-small whenever $A\cap A_{{\mathsf{H}}_{\mathbb{I}}^{\mathbb{P}}\left(\mathcal{P}\right)}^{*}=\varnothing$ if and only if ${\mathsf{H}}_{\mathbb{I}}^{\mathbb{P}}\left(\mathcal{P}\right)$ is compatible with τ.

Definition 5. 

A set A is $\mathcal{I}$-big ($\mathcal{P}$-big, $\mathcal{F}$-big, $\mathcal{G}$-big) if A is not $\mathcal{I}$-small ($\mathcal{P}$-small, $\mathcal{F}$-small, $\mathcal{G}$-small), equivalently $A\notin \mathcal{I}$ ($X\setminus A\in \mathcal{P}$, $X\setminus A\notin \mathcal{F}$, $A\in \mathcal{G}$). Let $\mathcal{Z}\in \mathbb{Z}$. A topology τ is $\mathcal{Z}$-codense, if any nonempty open set is $\mathcal{Z}$-big.

Remark 4. 

Let $\mathcal{I}\sim \mathcal{P}\sim \mathcal{F}\sim \mathcal{G}$. Then $\mathcal{I}={\mathsf{H}}_{\mathbb{I}}^{\mathbb{P}}\left(\mathcal{P}\right)$, $\mathcal{F}={\mathsf{H}}_{\mathbb{F}}^{\mathbb{P}}\left(\mathcal{P}\right)$, $\mathcal{G}={\mathsf{H}}_{\mathbb{G}}^{\mathbb{F}}\left(\mathcal{F}\right)$. So, $A\notin \mathcal{I}\iff X\setminus A\in \mathcal{P}\iff X\setminus A\notin \mathcal{F}\iff A\in \mathcal{G}$. That means if ${\mathcal{Z}}_1,{\mathcal{Z}}_2\in \mathbb{X}$ and ${\mathcal{Z}}_1\sim {\mathcal{Z}}_2$, a set A is ${\mathcal{Z}}_1$-big if and only if A is ${\mathcal{Z}}_2$-big and τ is ${\mathcal{Z}}_1$-codense if and only if τ is ${\mathcal{Z}}_2$-codense.

From the theory of ideal topological spaces we have the next characterization.

Theorem 13. 

τ is $\mathcal{I}$-codense if and only if $X_{\mathcal{I}}^{*}=X$.

The property of being ${\mathcal{Z}}_1$-codense can be characterized by the operators ${(·)}_{{\mathcal{Z}}_2}^{*}$ and ${(·)}_{{\mathcal{Z}}_2}^{▹}$ with respect to another equivalent system ${\mathcal{Z}}_2$.

Theorem 14. 

Let ${\mathcal{Z}}_1,{\mathcal{Z}}_2\in \mathbb{X}$. If ${\mathcal{Z}}_1\sim {\mathcal{Z}}_2$, then the next are equivalent

(1) 

τ is ${\mathcal{Z}}_1$-codense,

(2) 

$X_{{\mathcal{Z}}_2}^{*}=X$,

(3) 

${\varnothing }_{{\mathcal{Z}}_2}^{▹}=\varnothing$.

Proof. 

Let $\mathcal{I}:={\mathsf{H}}_{\mathbb{I}}^{{\mathbb{Z}}_2}\left({\mathcal{Z}}_2\right)$ where ${\mathcal{Z}}_2\in {\mathbb{Z}}_2\in \{\mathbb{P},\mathbb{F},\mathbb{G}\}$. Then ${\mathcal{Z}}_2\sim \mathcal{I}$. By Remark 4 and Theorem 13, $\tau$ is ${\mathcal{Z}}_1$-codense if and only if $\tau$ is ${\mathcal{Z}}_2$-codense if and only if $\tau$ is $\mathcal{I}$-codense if and only if $X=X_{\mathcal{I}}^{*}=X_{{\mathsf{H}}_{\mathbb{I}}^{{\mathbb{Z}}_2}\left({\mathcal{Z}}_2\right)}^{*}=X_{{\mathcal{Z}}_2}^{*}$ (by Theorem 7), so $\left(1\right)\iff \left(2\right)$. The equivalence $\left(2\right)\iff \left(3\right)$ follows from equation ${\varnothing }_{{\mathcal{Z}}_2}^{▹}=X\setminus X_{{\mathcal{Z}}_2}^{*}$. □

4. Conclusions

The main result of the work is based on the unification of approaches to the creation of new topologies derived from ideals, primals, filters and grills. Each approach has its equivalent counterparts in other approaches, and results valid in one approach can be used in others according to the following scheme: Definitions, theorems and proof methods in a space $(X,\tau ,\mathcal{Z})$, $\mathcal{Z}\in \mathbb{X}$, are valid and usable in the other three spaces $(X,\tau ,{\mathsf{H}}_{{\mathbb{Z}}_1}^{{\mathbb{Z}}_2}\left(\mathcal{Z}\right))$ where $\mathcal{Z}\in {\mathbb{Z}}_2\in \{\mathbb{I},\mathbb{P},\mathbb{F},\mathbb{G}\}$ and ${\mathbb{Z}}_1\in \{\mathbb{I},\mathbb{P},\mathbb{F},\mathbb{G}\}$, ${\mathbb{Z}}_1\ne {\mathbb{Z}}_2$. From this point of view, many proofs do not need to be done, and it is enough to do them only once in one space. On the other hand, such diversity, even if equivalent, can be a stimulus for further research and applications.

This article does not contain any studies with human participants or animals performed by the author.