Interior Operators Generated by Ideals in Complete Domains

Abstract

This article presents some properties of a special class of interior operators generated by ideals. The mathematical framework is given by complete domains, namely complete posets in which the set of minimal elements is a basis. The first part of the paper presents some preliminary results; in the second part we present the novel interior operator denoted by $G(i,I)$, an operator built starting from an interior operator i and an ideal I. Various properties of this operator are presented; in particular, the connection between the properties of the ideal I and the properties of the operator $G(i,I)$. Two such properties (denoted by ${\mathcal{P}}_i$ and ${\mathcal{Q}}_i$) are extensively analyzed and characterized. Additionally, some characterizations for compact elements are presented.

1. Introduction

In potential theory and in the study of harmonic functions, the fine topology is the natural topology for subharmonic functions [1]. A good example of a fine topology is given by the density topology on the real line. The density topology is related to the interior and closure (topological) operators. The interior of a subset S of a topological space X can be defined as the union of all subsets of S that are open in X; it can also be defined as the largest open subset of X contained (as a subset) in S. Interior and closure are dual notions, the interior of S being the complement of the closure of the complement of S.

The density topology is a good example of a topology generated by ideals. The ideals used to generate the density topology is given by the negligible Lebesgue sets; more details are presented in [1]. In other approaches, the ideal of negligible Lebesgue sets is replaced by the ideal of meagre sets (also called sets of first category, namely sets that are small or negligible as a subset of a topological space). The advantage of the meagre sets is that the whole construction could be used in any topological space, and does not depend on the properties of the measure. After these approaches, new ideals were considered to generate specific topologies. In these topologies, the adherence could be expressed by using the elements of the ideal; thus, the interior could be expressed by using the ideal. However, the topological studies do not insist on these aspects (adherence and interior).

Moving to the algebraic theory of domains [2,3], it is worth noting that certain operators with properties similar to that of interior operators appear in Galois connections. Moreover, if we obtain such an operator, we can derive such a Galois connections. Inspired by the developments in potential and topology theory, we investigate the notion of interior generated by an ideal in the framework of domain theory. This article presents the results obtained using this novel approach.

2. Preliminaries

Considering a nonempty set X, by $\mathcal{P}\left(X\right)$ we denote the set of all subsets of X. If $(X,\le )$ is a partially ordered set (poset) and $A\subseteq X$, then by $\uparrow A$ is denoted the upper set of A, namely $\uparrow A=\{x\in X|\exists a\in A$ with $a\le x\}$. The set $\downarrow A$ is the lower set of A, namely $\downarrow A=\{x\in X|\exists a\in A$ with $x\le a\}$. Given $x\in X$, the notation $\uparrow \left\{x\right\}$ is abbreviated by $\uparrow x$, and $\downarrow \left\{x\right\}$ is abbreviated by $\downarrow x$. A subset $A\subseteq X$ is an upper (lower) set whenever $\uparrow A=A$ ($\downarrow A=A$). In addition, the upper bound set of A is denoted by $ub\left(A\right)$ and the lower bound set of A is denoted by $lb\left(A\right)$; this means that $ub\left(A\right)=\{x\in X|a\le x$ for all $a\in A\}$ and $lb\left(A\right)=\{x\in X|x\le a$ for all $a\in A\}$.

For a number $\alpha \ge 2$, the set A is called $\alpha$-directed if for all $B\subseteq A$ with $\left|B\right|\le \alpha$ we have $A\cap ub\left(B\right)\ne \varnothing$. The 2-directed sets are simply called directed sets. The set A is an $\alpha$-ideal if it is a lower set and $\alpha$-directed; the set A is a complete ideal if it is a lower set and $\alpha$-directed for any cardinal $\alpha \ge 2$. If $\alpha$ is a finite cardinal, an $\alpha$-ideal is simply called ideal. We denote by ${\mathcal{I}}_{\alpha }\left(X\right)$ the set of $\alpha$-ideals of $(X,\le )$, and by $\mathcal{I}\left(X\right)$ the set of ideals of $(X,\le )$. If $\mathcal{L}\subseteq \mathcal{P}\left(X\right)$, we denote by ${\mathcal{L}}_d$ the family of directed sets of $\mathcal{L}$.

If there exists the smallest element of X, it is denoted by ⊥ and is called bottom. If there exists the biggest element of X, it is denoted by ⊤ and is called top. A poset X is called pointed if it has the smallest element. If $(X,\le )$ is a pointed poset, an element $x\in X$ is called minimal if $x\ne \perp$ and $\downarrow x=\{x,\perp \}$. We denote by $min\left(X\right)$ the set of minimal elements of $(X,\le )$. If $ub\left(A\right)$ has the smallest element z, then z is called the supremum of A and it is denoted by $\bigsqcup A$. If $lb\left(A\right)$ has the biggest element z, then z is called the infimum of A and it is denoted by $\sqcap A$. For $A=\{x,y\}$, we denote $\bigsqcup A$ with $x\bigsqcup y$, and $\sqcap A$ with $x\sqcap y$.

A poset $(X,\le )$ is called a poset or dcpo (directed complete partial order set) if any directed subset of X has supremum; a poset $(X,\le )$ is called complete if each subset of X has supremum and infimum.

Let $(X,\le )$ be a pointed poset, $\mathcal{L}$ a nonempty family of subsets of X, and there be a function $g:X\to X$. For $x,z\in X$ we say either that z approximates x relative to $(g,\mathcal{L})$ or z is essential part of x relative to $(g,\mathcal{L})$ if for each set $A\in \mathcal{L}$ such that there exists $\bigsqcup A$ and $x\le \bigsqcup A$, there is $a\in A$ such that $z\le g\left(a\right)$; this is denoted by $z{\ll }_{(g,\mathcal{L})}x$. We say that x is $(g,\mathcal{L})$-compact if it approximates itself relative to $(g,\mathcal{L})$. When $\mathcal{L}=\mathcal{P}\left(X\right)$ and $g=1_X$, $z{\ll }_{(g,\mathcal{L})}x$ is denoted by $z\ll x$, and we say that z approximates x. In addition, x is called simply compact if x is $(1_X,\mathcal{P}\left(X\right))$-compact. The set of $(g,\mathcal{L})$-compact elements is denoted by $K_{g,\mathcal{L}}\left(X\right)$, and the set of compact elements by $K\left(X\right)$. More details can be found in [2].

For $x\in X$, the set $\{z\in X\backslash \{\perp \left\}\right|z\ll x\}$ is denoted by $↡x$. Obviously, $↡x\subseteq \downarrow x$ for all $x\in X$. A set $B\subseteq X$ is called a basis if for all $x\in X\backslash \{\perp \}$ there is $\bigsqcup (B\cap ↡x)=x$. A domain is called continuous if it has a basis, and a domain is called algebraic if it has a basis given by compact elements.

Let $f:X\to X$ be a function. We denote by $fix\left(f\right)$ the set of fixed points of f, i.e., $\{x\in X|f\left(x\right)=x\}$, and by $f\left(X\right)$ the set $\left\{f\right(x\left)\right|x\in X\}$. A function f is called isotone (antitone) if $x\le y\Rightarrow f\left(x\right)\le f\left(y\right)$ (respectively, $f\left(y\right)\le f\left(x\right)$). The composition $f\circ g$ is denoted by $fg$, and by $f\le g$ we mean $f\left(x\right)\le g\left(x\right)$ for all $x\in X$. The function $f:X\to X$ is called $(g,\mathcal{L})$-Scott-continuous if f is isotone and for each $A\in \mathcal{L}$ there exists $\bigsqcup A$, then there also exist $\bigsqcup f\left(A\right)$ and $g(\bigsqcup f(A\left)\right)=gf(\bigsqcup A)$; f is simply called Scott-continuous if f is $(1_X,\mathcal{P}\left(X\right))$-Scott-continuous [3].

3. Complete Domains

A poset $(X,\le )$ is a complete domain if X is complete and $min\left(X\right)$ is a basis. For instance, $\left(\mathcal{P}\right(Y),\subseteq )$ is a complete domain (for any nonempty set Y). From now on, we consider that $(X,\le )$ is a complete domain.

Proposition 1.

$K\left(X\right)\backslash \{\perp \}=min\left(X\right)$.

Proof. 

Let us consider $x\in K\left(X\right)\backslash \{\perp \}$. Since $min\left(X\right)$ is a basis, $x=\bigsqcup \left(min\right(X)\cap ↡x)$. Since $x\ll x$, there exists $m\in min\left(X\right)\cap ↡x$ such that $x\le m$. Thus, $x=m\in min\left(X\right)$. Now let $m\in min\left(X\right)$. Since $m=\bigsqcup \left(min\right(X)\cap ↡m)$, it follows that $↡m\ne \varnothing$, and there exists $x\in ↡m$. Since $x\ne \perp$, $x\ll m\le m$ and $m\in min\left(X\right)$, it follows that $x=m$, and so $m\ll m$. In conclusion, $m\in K\left(X\right)\backslash \{\perp \}$. □

Corollary 1.

Any complete domain is an algebraic domain.

Corollary 2.

Let $m\in min\left(X\right)$ and $A\subseteq min\left(X\right)$. Then $m\le \bigsqcup A$ if and only if $m\in A$.

Proof. 

Let $m\le \bigsqcup A$. Since $m\in K\left(X\right)\backslash \{\perp \}$, we have $m\ll m\le \bigsqcup A$, and so there is $a\in A$ such that $m\le a$. Since $a\in min\left(X\right)$, then $m=a\in A$. □

Let $c:X\to X$ be the function defined by $c\left(x\right)=\bigsqcup \{y\in X|y\sqcap x=\perp \}$ for all $x\in X$. Just to simplify the notation, we use $cx$ instead of $c\left(x\right)$.

Proposition 2.

$cx=\bigsqcup \{y\in min(X\left)\right|y\nleqq x\}$ for all $x\in X$.

Proof. 

Let $x\in X$ and $y\in min\left(X\right)$ such that $y\nleqq x$. If $y\sqcap x\ne \perp$, since $y\in min\left(X\right)$ and $y\sqcap x\le y$, we have $y=y\sqcap x\le x$, which is a contradiction! Therefore, $y\sqcap x=\perp$. Then $\{y\in min(X\left)\right|y\nleqq x\}\subseteq \{y\in X|y\sqcap x=\perp \}$, and $\bigsqcup \{y\in min(X\left)\right|y\nleqq x\}\le cx$.

On the other hand, $cx=\bigsqcup \{y\in X|y\sqcap x$ = $\perp \}=\bigsqcup \{\bigsqcup \left(min\right(X)\cap ↡y)|y\sqcap x$ = $\perp \}$ $=\bigsqcup {\cup }_{y\sqcap x=\perp }(min\left(X\right)\cap ↡y)$. However, ${\cup }_{y\sqcap x=\perp }(min\left(X\right)\cap ↡y)=min\left(X\right)\cap {\cup }_{y\sqcap x=\perp }↡y\subseteq min\left(X\right)\cap {\cup }_{y\sqcap x=\perp }\downarrow y=\{y\in min\left(X\right)|y\sqcap x$ = $\perp \}\subseteq \{y\in min\left(X\right)|y\nleqq x\}$, and so $\bigsqcup {\cup }_{y\sqcap x=\perp }(min\left(X\right)\cap ↡y)\le \bigsqcup \{y\in min\left(X\right)|y\nleqq x\}$. Thus, $cx\le \bigsqcup \{y\in min(X\left)\right|y\nleqq x\}$. □

If $x\in min\left(X\right)$, then $\{y\in min(X\left)\right|y\nleqq x\}$ = $\{y\in min(X\left)\right|y\ne x\}$, and $cx=\bigsqcup \left(min\right(X)\backslash \{x\left\}\right)$.

Proposition 3.

If $x\le y$ then $cy\le cx$.

Proof. 

If $x\le y$ then $\{z\in X|z\sqcap y=\perp \}\subseteq \{z\in X|z\sqcap x=\perp \}$, meaning that $cy\le cx$. □

Proposition 4.

$x\sqcap cx=\perp$ for all $x\in X$.

Proof. 

If we assume $x\sqcap cx\ne \perp$, there is $y\in min\left(X\right)$ such that $y\le x\sqcap cx$. Since $y\in min\left(X\right)$, we have $y\in K\left(X\right)$, and because $y\le cx=\bigsqcup \{z\in min(X\left)\right|z\nleqq x\}$, there exists $z\in min\left(X\right)$ such that $z\nleqq x$ and $y\le z$. Then $z=y\le x$, which is a contradiction. □

Proposition 5.

$x\bigsqcup cx=\top$ for all $x\in X$.

Proof. 

Let $x\in X$. Then $x\bigsqcup cx=(\bigsqcup \{y\in min\left(X\right)|y\le x\})\bigsqcup$ $(\bigsqcup \{y\in min\left(X\right)|y\nleqq x\})=$ $\bigsqcup min\left(X\right)=\bigsqcup \{z\in min(X\left)\right|z\le \top \}=\top$. □

Proposition 6.

If $m\in min\left(X\right)$ and $m\le x\bigsqcup y$, then either $m\le x$ or $m\le y$.

Proof. 

$x\bigsqcup y=\bigsqcup \left(min\right(X)\cap ↡x)\bigsqcup \bigsqcup \left(min\right(X)\cap ↡y)=\bigsqcup \left(min\right(X)\cap (↡x\cup ↡y\left)\right)$. Since $m\in min\left(X\right)$, we have $m\le x\bigsqcup y$ if and only if $m\in min\left(X\right)\cap (↡x\cup ↡y)$ if and only if $m\in min\left(X\right)\cap ↡x$ or $m\in min\left(X\right)\cap ↡x$ if and only if $m\le x$ or $m\le y$. □

Corollary 3.

If $cx\le x\bigsqcup y$, then $cx\le y$.

Proof. 

Let $m\in min\left(X\right)$ such that $m\le cx$. Since $m\le x\bigsqcup y$, it follows that $m\le x$ or $m\le y$. Since $m\le cx$, then $m\nleqq x$, and so $m\le y$. Thus, $cx\le y$. □

Proposition 7.

$x\sqcap (y\bigsqcup z)=(x\sqcap y)\bigsqcup (x\sqcap z)$ for all $x,y,z\in X$.

Proof. 

Let $x,y,z\in X$. We have $x\sqcap y\le x\sqcap (y\bigsqcup z)$ and $x\sqcap z\le x\sqcap (y\bigsqcup z)$. Let $u\in X$ such that $x\sqcap y\le u$ and $x\sqcap z\le u$, and let $m\in min\left(X\right)$ such that $m\le x\sqcap (y\bigsqcup z)$. Since $m\le y\bigsqcup z$ and $m\in K\left(X\right)$, then either $m\le y$ or $m\le z$. Since $m\le x$, we have $m\le x\sqcap y$ or $m\le x\sqcap z$, and so $m\le u$. Thus, $x\sqcap (y\bigsqcup z)=\bigsqcup \{m\in min(X\left)\right|m\le x\sqcap (y\bigsqcup z)\}\le u$, and $x\sqcap (y\bigsqcup z)=(x\sqcap y)\bigsqcup (x\sqcap z)$. □

Proposition 8.

$c(\bigsqcup A)=\sqcap \left\{ca\right|a\in A\}$ for all $A\subseteq X$.

Proof. 

Let $A\subseteq X$. Then $c(\bigsqcup A)=\bigsqcup \{y\in min(X\left)\right|y\nleqq \bigsqcup A\}=\bigsqcup \{y\in min\left(X\right)|y\nleqq a$ for all $a\in A\}\le \bigsqcup \{y\in min\left(X\right)|y\nleqq a\}$ for all $a\in A$. Let $v\in X$ such that $v\le \bigsqcup \{y\in min(X\left)\right|y\nleqq a\}$ for all $a\in A$, and let $u\in min\left(X\right)$ such that $u\le v$. Then, for all $a\in A$, there exists $y_a\in min\left(X\right)$ such that $y_a\nleqq a$ and $u\le y_a$. Since $u,y_a\in min\left(X\right)$ and $u\le y_a$, it follows that $u=y_a$. Then $u\nleqq a$; hence, $u\nleqq a$ for all $a\in A$, and so $u\nleqq \bigsqcup A$. Thus, $u\le c(\bigsqcup A)$. It follows that $v\le c(\bigsqcup A)$. Thus, $c(\bigsqcup A)={\sqcap }_{a\in A}\bigsqcup \{y\in min\left(X\right)|y\nleqq a\}=\sqcap \left\{ca\right|a\in A\}$. □

Proposition 9.

If $cx=cy$ then $x=y$.

Proof. 

We have $cx=\bigsqcup \{u\in min(X\left)\right|u\nleqq x\}$ and $cy=\bigsqcup \{v\in min(X\left)\right|v\nleqq y\}$. Let $u\in min\left(X\right)$ with $u\nleqq x$. Since $u\le cx=cy$, there exists $v\in min\left(X\right)$ such that $v\nleqq y$ and $u\le v$. Then $u=v$, and $u\nleqq y$. It follows that $\{u\in min(X\left)\right|u\nleqq x\}=\{v\in min\left(X\right)|v\nleqq y\}$, which is equivalent to $\{u\in min(X\left)\right|u\le x\}=\{v\in min\left(X\right)|v\le y\}$. Then $x=\bigsqcup \{u\in min(X\left)\right|u\le x\}=\bigsqcup \{v\in min\left(X\right)|v\le y\}=y$. □

Proposition 10.

$c^2=1_X$.

Proof. 

Let $x\in X$. Since $cx=\bigsqcup \{z\in min(X\left)\right|z\nleqq x\}$, we have $c^{2}x=c\left(cx\right)=c(\bigsqcup \{z\in min\left(X\right)|z\nleqq x\})=\sqcap \left\{cz\right|z\in min\left(X\right),z\nleqq x\}$. Let $z\in min\left(X\right)$ with $z\nleqq x$, and let $u\in min\left(X\right)$ with $u\le x$. Since $u\ne z$, $u\in min\left(X\right)\backslash \left\{z\right\}$. Then $u\le \bigsqcup \left(min\right(X)\backslash \{z\left\}\right)=cz$ (because $z\in min\left(X\right)$). It follows that $x\le cz$. Thus, for each $z\in min\left(X\right)$ with $z\nleqq x$ we have $x\le cz$, and so $x\le \sqcap \left\{cz\right|z\in min\left(X\right),z\nleqq x\}=c^{2}x$. Thus, $x\le c^{2}x$ for all $x\in X$.

On the other hand, if $x\le c^{2}x$ then $cx\ge c^{3}x=c^2\left(cx\right)\ge cx$. Then $c^{3}x=cx$. Since $c\left(c^{2}x\right)=cx$, we obtain $c^{2}x=x$. □

According to these results, the mapping c is a bijection.

Proposition 11.

For all $A\subseteq X$ we have $c(\sqcap A)=\bigsqcup \left\{ca\right|a\in A\}$.

Proof. 

Let $A\subseteq X$. Then $c(\bigsqcup \{ca|a\in A\})=\sqcap \{cca|a\in A\}==\sqcap \left\{a\right|a\in A\}=\sqcap A$. Thus, $\bigsqcup \left\{ca\right|a\in A\}=cc(\bigsqcup \left\{ca\right|a\in A\left\}\right)=c(\sqcap A)$. □

4. Interior Operators over Complete Domains

Definition 1.

Let $(X,\le )$ be a complete domain.

A function $i:X\to X$ is called an interior operator (over X) if:

1.

$i\le 1_X$;

2.

$i^2=i$;

3.

$i(\sqcap A)=\sqcap i\left(A\right)$ for all finite subsets A of X.

Let i be an interior operator. From the third condition applied for $A=\varnothing$, we have $i(\top )=i(\sqcap \varnothing )=\sqcap i(\varnothing )=\sqcap \varnothing =\top$. In fact, the third condition is equivalent to $i(\top )=\top$ and $i(x\sqcap y)=i\left(x\right)\sqcap i\left(y\right)$ for all $x,y\in X$. In addition, if $x,y\in X$ such that $x\le y$, then $i\left(x\right)=i(x\sqcap y)=i\left(x\right)\sqcap i\left(y\right)\le i\left(y\right)$. Therefore, i is isotone. Since $i^2=i$, it follows that $i\left(X\right)=fix\left(i\right)$.

Let I be a nonempty subset of X. For each $x\in X$, we define the set

$$A_{i,I}\left(x\right)=\{m\in min\left(X\right)|\exists u\in i\left(X\right)\mathrm{such}\mathrm{that}m\le u\mathrm{and}u\sqcap x\in I\}.$$

Then, the function $g_{i,I}:X\to X$ given by

$$g_{i,I}\left(x\right)=\left\{\begin{array}{c}\bigsqcup A_{i,I}\left(x\right),A_{i,I}\left(x\right)\ne \varnothing \\ \perp ,A_{i,I}\left(x\right)=\varnothing \end{array}\right.$$

is well-defined.

Proposition 12.

If I is a lower subset, then $g_{i,I}$ is antitone.

Proof. 

Given $x,y\in X$ such that $x\le y$, if $A_{i,I}\left(y\right)=\varnothing$ then $g_{i,I}\left(y\right)=\perp \le g_{i,I}\left(x\right)$. We suppose that $A_{i,I}\left(y\right)\ne \varnothing$, and $m\in A_{i,I}\left(y\right)$. Then there exists $u\in i\left(X\right)$ such that $m\le u$ and $u\sqcap y\in I$. If $x\le y$ then $u\sqcap x\le u\sqcap y$, and so $u\sqcap x\in I$. Thus, $m\in A_{i,I}\left(x\right)$ and $m\le g_{i,I}\left(x\right)$. Hence, $g_{i,I}\left(y\right)\le g_{i,I}\left(x\right)$. □

In what follows, we assume that I is an ideal of $(X,\le )$, i.e., $I\in \mathcal{I}\left(X\right)$.

Proposition 13.

$g_{i,I}(x\bigsqcup y)=g_{i,I}\left(x\right)\sqcap g_{i,I}\left(y\right)$ for all $x,y\in X$.

Proof. 

Let $x,y\in X$. Since $x,y\le x\bigsqcup y$, we have $g_{i,I}(x\bigsqcup y)\le g_{i,I}\left(x\right),g_{i,I}\left(y\right)$ and so $g_{i,I}(x\bigsqcup y)\le g_{i,I}\left(x\right)\sqcap g_{i,I}\left(y\right)$. If $g_{i,I}\left(x\right)\sqcap g_{i,I}\left(y\right)=\perp$, then $g_{i,I}(x\bigsqcup y)=\perp$, and so $g_{i,I}(x\bigsqcup y)=g_{i,I}\left(x\right)\sqcap g_{i,I}\left(y\right)$.

We suppose that $g_{i,I}\left(x\right)\sqcap g_{i,I}\left(y\right)\ne \perp$; let $m\in min\left(X\right)$ such that $m\le g_{i,I}\left(x\right)\sqcap g_{i,I}\left(y\right)$. Since $m\le g_{i,I}\left(x\right)=\bigsqcup A_{i,I}\left(x\right)$ and $A_{i,I}\left(x\right)\subseteq min\left(X\right)$, from Corollary 2 we have that $m\in A_{i,I}\left(x\right)$, and so there exists $u_1\in i\left(X\right)$ such that $m\le u_1$ and $u_1\sqcap x\in I$. Since $m\le g_{i,I}\left(y\right)=\bigsqcup A_{i,I}\left(x\right)$, there is $u_2\in i\left(X\right)$ such that $m\le u_2$ and $u_2\sqcap y\in I$. Let $u=u_1\sqcap u_2$; then $i\left(u\right)=i(u_1\sqcap u_2)=i\left(u_1\right)\sqcap i\left(u_2\right)=u_1\sqcap u_2=u$, and so $u\in i\left(X\right)$. In addition, $m\le u$. Since $u\sqcap x\le u_1\sqcap x\in I$ and $u\sqcap y\le u_2\sqcap y\in I$, we have $u\sqcap (x\bigsqcup y)=(u\sqcap x)\bigsqcup (u\sqcap y)\in I$. Consequently, $m\in A_{i,I}(x\bigsqcup y)$ and $m\le g_{i,I}(x\bigsqcup y)$. Thus, $g_{i,I}\left(x\right)\sqcap g_{i,I}\left(y\right)=\bigsqcup \{m\in min\left(X\right)|m\le g_{i,I}\left(x\right)\sqcap g_{i,I}\left(y\right)\}\le g_{i,I}(x\bigsqcup y)$. □

Proposition 14.

$g_{i,I}\left(X\right)\subseteq i\left(X\right)$.

Proof. 

Let $x\in X$ and $m\in A_{i,I}\left(x\right)$. Then there is $u\in i\left(X\right)$ such that $m\le u$ and $u\sqcap x\in I$. If $u\nleqq g_{i,I}\left(x\right)$, there exists $m_1\in min\left(X\right)$ such that $m_1\le u$ and $m_1\nleqq g_{i,I}\left(x\right)$. However, $m_1\nleqq g_{i,I}\left(x\right)$ if and only if $m_1\notin A_{i,I}\left(x\right)$, hence for all $v\in i\left(X\right)$ such that $m_1\le v$, we have $v\sqcap x\notin I$. Since $u\in i\left(X\right)$ such that $m_1\le u$, then $u\sqcap x\notin I$, which is a contradiction. Hence $u\le g_{i,I}\left(x\right)$, and then $m\le u=i\left(u\right)\le i\left(g_{i,I}\left(x\right)\right)$. Therefore, $g_{i,I}\left(x\right)\le i\left(g_{i,I}\left(x\right)\right)$. Since $i\left(g_{i,I}\left(x\right)\right)\le g_{i,I}\left(x\right)$, we have $i\left(g_{i,I}\left(x\right)\right)=g_{i,I}\left(x\right)$, i.e., $g_{i,I}\left(x\right)\in i\left(X\right)$. □

Proposition 15.

$g_{i,I}\le g_{i,I}c{g}_{i,I}$.

Proof. 

Let $x\in X$ and we suppose that there exists $m\in min\left(X\right)$ such that $m\le g_{i,I}\left(x\right)$ and $m\nleqq \left(g_{i,I}c{g}_{i,I}\right)\left(x\right)$. Since $m\le g_{i,I}\left(x\right)$ if and only if $m\in A_{i,I}\left(x\right)$, there is $u\in i\left(X\right)$ such that $m\le u$ and $u\sqcap x\in I$. Since $m\nleqq g_{i,I}\left(c{g}_{i,I}\left(x\right)\right)$ if and only if $m\notin A_{i,I}\left(c{g}_{i,I}\left(x\right)\right)$, it follows that for all $v\in i\left(X\right)$ with $m\le v$ we have $v\sqcap c{g}_{i,I}\left(x\right)\notin I$. Then $u\sqcap c{g}_{i,I}\left(x\right)\notin I$, and so there exists $m_1\in min\left(X\right)$ such that $m_1\le u\sqcap c{g}_{i,I}\left(x\right)$. Since $m_1\le c{g}_{i,I}\left(x\right)$, we have $m_1\nleqq g_{i,I}\left(x\right)$ and $m_1\notin A_{i,I}\left(x\right)$. Because $m_1\in min\left(X\right)$ and $m_1\le u$, it follows that $u\sqcap x\notin I$, which is a contradiction. To conclude, for all $m\in min\left(X\right)$ such that $m\le g_{i,I}\left(x\right)$, we obtain $m\le \left(g_{i,I}c{g}_{i,I}\right)\left(x\right)$. Thus, $g_{i,I}\left(x\right)=\bigsqcup (min\left(X\right)\cap \downarrow g_{i,I}\left(x\right))\le \left(g_{i,I}c{g}_{i,I}\right)\left(x\right)$. □

Proposition 16.

$i\le g_{i,I}c$.

Proof. 

Let $x\in X$, and assume that there is $m\in min\left(X\right)$ such that $m\le i\left(x\right)$ and $m\nleqq g_{i,I}\left(cx\right)$. Since $m\le g_{i,I}\left(cx\right)$ if and only if $m\in A_{i,I}\left(x\right)$, we have $m\notin A_{i,I}\left(x\right)$. Thus, for all $u\in i\left(X\right)$ with $m\le u$, we have $u\sqcap cx\notin I$. Since $m\le i\left(x\right)$ and $i\left(x\right)\in i\left(X\right)$, we obtain $i\left(x\right)\sqcap cx\notin I$. On the other hand, if $i\left(x\right)\le x$ then $i\left(x\right)\sqcap cx=\perp \in I$, which is a contradiction. Therefore, for all $m\in min\left(X\right)$ such that $m\le i\left(x\right)$ we have $m\le g_{i,I}\left(cx\right)$. Thus, $i\left(x\right)\le g_{i,I}\left(cx\right)$. □

Corollary 4.

We have $i\le g_{i,I}ci$ and $i\le i{g}_{i,I}c$.

Proof. 

Since $i\le g_{i,I}c$, it follows that $i=i^2\le g_{i,I}ci$. Since i is isotone, it follows that $i=i^2\le i{g}_{i,I}c$. □

From the previous result, it follows that $\top =i(\top )\le g_{i,I}c(\top )=g_{i,I}(\perp )\le \top$, and so $g_{i,I}(\perp )=\top$. More generally, we have the following result.

Proposition 17.

$I\subseteq g_{i,I}^{-1}(\top )$.

Proof. 

Let $x\in I$. For all $m\in min\left(X\right)$ there is $u=\top \in i\left(X\right)$ such that $m\le \top$ and $u\sqcap x=x\in I$. Then $min\left(X\right)\subseteq A_{i,I}\left(x\right)$, and so $A_{i,I}\left(x\right)=min\left(X\right)$. It follows that $g_{i,I}\left(x\right)=\bigsqcup min\left(X\right)=\top$. □

Proposition 18.

$g_{i,I}(x\bigsqcup y)=g_{i,I}(x\sqcap cy)=g_{i,I}\left(x\right)$ for all $x\in X$ and $y\in I$.

Proof. 

If $x\in X$ and $y\in I$, then $g_{i,I}(x\bigsqcup y)=g_{i,I}\left(x\right)\sqcap g_{i,I}\left(y\right)=g_{i,I}\left(x\right)\sqcap \top =g_{i,I}\left(x\right)$. Since $x\sqcap cy\le x$, $g_{i,I}\left(x\right)\le g_{i,I}(x\sqcap cy)$. Let $m\in min\left(X\right)$ such that $m\le g_{i,I}(x\sqcap cy)$. Then $m\in A_{i,I}(x\sqcap cy)$, and so there is $u\in i\left(X\right)$ such that $m\le u$ and $u\sqcap x\sqcap cy\in I$. Since $u\sqcap x\le (u\sqcap x\sqcap cy)\bigsqcup y$ and $y\in I$, we obtain $u\sqcap x\in I$. Thus, $m\in A_{i,I}\left(x\right)$ and so $g_{i,I}(x\sqcap cy)\le g_{i,I}\left(x\right)$. □

Proposition 19.

Let $U\subseteq X$, $\tilde{U}=\{u\sqcap cy|u\in U,y\in I\}$, $\mathcal{L}=\mathcal{P}\left(U\right)$, and $\tilde{\mathcal{L}}=\mathcal{P}\left(\tilde{U}\right)$.

For every function $h:X\to X$ we have:

1.

$K_{h{g}_{i,I},\mathcal{L}}\left(X\right)=K_{h{g}_{i,I},\tilde{\mathcal{L}}}\left(X\right)$;

2.

For all $u_1,u_2\in U$ with $u_1\bigsqcup u_2\in U$, we have $x\in K_{c{g}_{i,I},{\mathcal{L}}_d}\left(X\right)$ if and only if for all $A\in \mathcal{L}$ with $x\le \bigsqcup A$, there exist $a_1,\dots ,a_n\in A$ such that $x\le c{g}_{i,I}\left(a_1\right)\bigsqcup \dots \bigsqcup c{g}_{i,I}\left(a_n\right)$.

Proof. 

1. Since $\perp \in I$, we have $U\subseteq \tilde{U}$ and $\mathcal{L}\subseteq \tilde{\mathcal{L}}$. Therefore, $K_{h{g}_{i,I},\tilde{\mathcal{L}}}\left(X\right)\subseteq K_{h{g}_{i,I},\mathcal{L}}\left(X\right)$. Let $x\in K_{h{g}_{i,I},\mathcal{L}}\left(X\right)$ and $A\in \tilde{\mathcal{L}}$ such that $x\le \bigsqcup A$. If $A\subseteq \tilde{U}$, then for all $a\in A$ there exist $u_a\in U$ and $y_a\in I$ such that $a=u_a\sqcap c{y}_a$. Let $B=\left\{u_a\right|a\in A\}$. Then $B\in \mathcal{L}$ and $x\le \bigsqcup A\le \bigsqcup B$. Since $x\in K_{h{g}_{i,I},\mathcal{L}}\left(X\right)$, there is $a\in A$ such that $x\le h{g}_{i,I}\left(u_a\right)$. Thus, $g_{i,I}\left(u_a\right)=g_{i,I}(u_a\sqcap c{y}_a)$, and so $x\le h{g}_{i,I}\left(a\right)$. Therefore, $x\in K_{h{g}_{i,I},\tilde{\mathcal{L}}}\left(X\right)$.

2. “⇒”: Let $x\in K_{c{g}_{i,I},{\mathcal{L}}_d}\left(X\right)$ and $A\in \mathcal{L}$ such that $x\le \bigsqcup A$. Let $B=\{\bigsqcup M|M\subseteq A,M\mathrm{finite}\}$. Then $B\subseteq U$ and B is directed. Hence, $B\in {\mathcal{L}}_d$. Since $x\le \bigsqcup A=\bigsqcup B$, there is $b\in B$ such that $x\le c{g}_{i,I}\left(b\right)$. Let us suppose that $b=\bigsqcup \{a_1,\dots ,a_n\}$, where $a_1,\dots ,a_n\in A$. Then $c{g}_{i,I}\left(b\right)=c{g}_{i,I}(a_1\bigsqcup \dots \bigsqcup a_n)=c(g_{i,I}\left(a_1\right)\sqcap \dots \sqcap g_{i,I}\left(a_n\right))=c{g}_{i,I}\left(a_1\right)\bigsqcup \dots \bigsqcup c{g}_{i,I}\left(a_n\right)$, and so $x\le c{g}_{i,I}\left(a_1\right)\bigsqcup \dots \bigsqcup c{g}_{i,I}\left(a_n\right)$.

“⇐”: Let $x\in X$ and $A\in {\mathcal{L}}_d$ such that $x\le \bigsqcup A$. From the hypothesis, it follows that there exist $a_1,\dots ,a_n\in A$ such that $x\le c{g}_{i,I}\left(a_1\right)\bigsqcup \dots \bigsqcup c{g}_{i,I}\left(a_n\right)$. Since A is directed, there is $a\in A$ such that $a_1\bigsqcup \dots \bigsqcup a_n\le a$. Then $c{g}_{i,I}\left(a_1\right)\bigsqcup \dots \bigsqcup c{g}_{i,I}\left(a_n\right)=c{g}_{i,I}(a_1\bigsqcup \dots \bigsqcup a_n)\le c{g}_{i,I}\left(a\right)$. Therefore, $x\in K_{c{g}_{i,I},{\mathcal{L}}_d}\left(X\right)$. □

Since $I\subseteq g_{i,I}^{-1}(\top )$, from $x\in I$ it follows that $x\le g_{i,I}\left(x\right)$. If the converse is true, we say that I has the property ${\mathcal{P}}_i$. Therefore, ${\mathcal{P}}_i=\{I\in \mathcal{I}\left(X\right)|x\le g_{i,I}\left(x\right)$ implies $x\in I\}$; also, ${\mathcal{P}}_i=\{I\in \mathcal{I}\left(X\right)|I=\{x\in X|x\le g_{i,I}\left(x\right)\}\}$.

Proposition 20.

The following statements are equivalent:

1.

$I\in {\mathcal{P}}_i$,

2.

$x\sqcap g_{i,I}\left(x\right)\in I$ for all $x\in X$.

Proof. 

We suppose that $I\in {\mathcal{P}}_i$, and consider $x\in X$. Since $x\sqcap g_{i,I}\left(x\right)\le g_{i,I}\left(x\right)\le g_{i,I}\left(x\right)\bigsqcup g_{i,I}\left(g_{i,I}\left(x\right)\right)=g_{i,I}(x\sqcap g_{i,I}\left(x\right))$, then $x\sqcap g_{i,I}\left(x\right)\in I$.

Conversely, suppose that $x\sqcap g_{i,I}\left(x\right)\in I$ for all $x\in X$. If $x\le g_{i,I}\left(x\right)$, then $x=x\sqcap g_{i,I}\left(x\right)\in I$. Therefore, $I\in {\mathcal{P}}_i$. □

Remark 1.

If $I\in {\mathcal{P}}_i$, then $I=g_{i,I}^{-1}(\top )$.

Since $g_{i,I}$ is antitone, $\perp \le g_{i,I}(\top )\le g_{i,I}\left(x\right)$ for every $x\in X$. A special case is when $g_{i,I}(\top )=\perp$. We say that I has the property ${\mathcal{Q}}_i$ if $g_{i,I}(\top )=\perp$. Thus, we have ${\mathcal{Q}}_i=\{I\in \mathcal{I}\left(X\right)|g_{i,I}(\top )=\perp \}$.

Proposition 21.

The following statements are equivalent:

1.

$I\in {\mathcal{Q}}_i$;

2.

$i\left(X\right)\cap I=\{\perp \}$.

Proof. 

$I\in {\mathcal{Q}}_i$ if and only if $g_{i,I}(\top )=\perp$ if and only if $A_{i,I}(\top )=\varnothing$ if and only if for all $m\in min\left(X\right)$ and $u\in i\left(X\right)$ such that $m\le u$ we have $u\sqcap \top \notin I$ if and only if for all $u\in i\left(X\right)$ such that $u\ne \perp ,u\notin I$ we have $i\left(X\right)\cap I=\{\perp \}$. □

Proposition 22.

If $I\in {\mathcal{Q}}_i$, then $g_{i,I}=ic$ over $i\left(X\right)$.

Proof. 

Let $x\in i\left(X\right)$ and $m\in min\left(X\right)$ such that $m\le g_{i,I}\left(x\right)$. Then $m\in A_{i,I}\left(x\right)$, and so there is $u\in i\left(X\right)$ such that $m\le u$ and $u\sqcap x\in I$. Since $I\in {\mathcal{Q}}_i$ and $u\sqcap x\in i\left(X\right)$, we have $u\sqcap x=\perp$. Then, since $m\le u$, it follows that $m\nleqq x$ and $m\le cx$. Thus, $g_{i,I}\left(x\right)\le cx$ and so $i\left(g_{i,I}\left(x\right)\right)\le i\left(cx\right)$. Since $g_{i,I}\left(x\right)\in i\left(X\right)$, then $g_{i,I}\left(x\right)=i\left(g_{i,I}\left(x\right)\right)\le i\left(cx\right)$. According to Proposition 16, it follows that $i\left(cx\right)\le g_{i,I}\left(x\right)$. Therefore, $g_{i,I}\left(x\right)=i\left(cx\right)$. □

Theorem 1.

$G_{i,I}=1_X\sqcap g_{i,I}c$ is an interior operator on X such that $i\le G_{i,I}$.

Proof. 

By definition, $G_{i,I}\le 1_X$. Let $x\in X$. $G_{i,I}^2\left(x\right)=G_{i,I}\left(G_{i,I}\left(x\right)\right)=G_{i,I}(x\sqcap g_{i,I}\left(cx\right))=(x\sqcap g_{i,I}\left(cx\right))\sqcap g_{i,I}\left(c(x\sqcap g_{i,I}\left(cx\right))\right)=x\sqcap g_{i,I}\left(cx\right)\sqcap g_{i,I}(cx\bigsqcup c{g}_{i,I}\left(cx\right))=x\sqcap g_{i,I}\left(cx\right)\sqcap g_{i,I}\left(cx\right)\sqcap g_{i,I}\left(c{g}_{i,I}\left(cx\right)\right)=x\sqcap g_{i,I}\left(cx\right)$, since $g_{i,I}\left(cx\right)\le \left(g_{i,I}c{g}_{i,I}\right)\left(cx\right)$ (according to Proposition 15). Hence, $G_{i,I}^2\left(x\right)=G_{i,I}\left(x\right)$.

Let $x,y\in X$. Then $G_{i,I}(x\sqcap y)=(x\sqcap y)\sqcap g_{i,I}\left(c(x\sqcap y)\right)=(x\sqcap y)\sqcap g_{i,I}(cx\bigsqcup cy)=(x\sqcap y)\sqcap (g_{i,I}\left(cx\right)\sqcap g_{i,I}\left(cy\right))=G_{i,I}\left(x\right)\sqcap G_{i,I}\left(y\right)$. In addition, $G_{i,I}(\top )=g_{i,I}(\perp )=\top$, and so $G_{i,I}$ is an interior operator. Since $i\le 1_X$ and $i\le g_{i,I}c$, it follows that $i\le G_{i,I}$. □

To simplify the notation, the operator $G_{i,I}$ is denoted also by $G(i,I)$.

Proposition 23.

$i\left(X\right)\subseteq G_{i,I}\left(X\right)$.

Proof. 

Since $i\le G_{i,I}\le 1_X$, $x=i\left(x\right)\le G_{i,I}\left(x\right)\le x$ for all $x\in i\left(X\right)$. Thus, $x=G_{i,I}\left(x\right)\in G_{i,I}\left(X\right)$. □

Proposition 24.

$G_{i,I}i=i{G}_{i,I}=i$.

Proof. 

From Corollary 4, we have $i\le g_{i,I}ci$ and $i\le i{g}_{i,I}c$. Thus, $G_{i,I}i=i\sqcap g_{i,I}ci=i$. In addition, $i{G}_{i,I}=i(1_X\sqcap g_{i,I}c)=i{1}_X\sqcap i{g}_{i,I}c=i\sqcap i{g}_{i,I}c=i$. □

Proposition 25.

1.

$G_{i,I}\left(X\right)\subseteq \{u\sqcap c(v\sqcap g_{i,I}\left(v\right))|u\in i\left(X\right),v\in X\}$;

2.

$G_{i,I}\left(X\right)=\{u\sqcap c(v\sqcap g_{i,I}\left(v\right))|u\in i\left(X\right),v\in X\}$ if and only if $g_{i,I}(v\sqcap g_{i,I}\left(v\right))=\top$ for all $v\in X$.

Proof. 

1. Let $x\in G_{i,I}\left(X\right)$. Then $G_{i,I}\left(x\right)=x\sqcap g_{i,I}\left(cx\right)=x$ if and only if $x\le g_{i,I}\left(cx\right)$ if and only if $c{g}_{i,I}\left(cx\right)\le cx$. Therefore, $cx=cx\sqcap \top =cx\sqcap (c{g}_{i,I}\left(cx\right)\bigsqcup cc{g}_{i,I}\left(cx\right))=c{g}_{i,I}\left(cx\right)\bigsqcup (cx\sqcap g_{i,I}\left(cx\right))$, and so $x=g_{i,I}\left(cx\right)\sqcap c(cx\sqcap g_{i,I}\left(cx\right))=u\sqcap c(v\sqcap g_{i,I}\left(v\right))$, where $u=g_{i,I}\left(cx\right)\in i\left(X\right)$ and $v=cx$.

2. Let $u\in i\left(X\right)$ and $y\in X$. Then $u\sqcap cy\in G_{i,I}\left(X\right)$ if and only if $G_{i,I}(u\sqcap cy)=u\sqcap cy$ if and only if $G_{i,I}\left(u\right)\sqcap G_{i,I}\left(cy\right)=u\sqcap cy$ if and only if $u\sqcap G_{i,I}\left(cy\right)=u\sqcap cy$ if and only if $(u\sqcap cy)\sqcap g_{i,I}\left(y\right)=u\sqcap cy$ if and only if $u\sqcap cy\le g_{i,I}\left(y\right)$. Thus, $u\sqcap cy\in G_{i,I}\left(X\right)$ if and only if $u\sqcap cy\le g_{i,I}\left(y\right)$ $(\ast )$

“⇒”: Let $v\in X$. Since $\top \in i\left(X\right)$, it follows from the hypothesis that $c(v\sqcap g_{i,I}\left(v\right))=\top \sqcap c(v\sqcap g_{i,I}\left(v\right))\in G_{i,I}\left(X\right)$. Then, according to $(\ast )$, we have $c(v\sqcap g_{i,I}\left(v\right))\le g_{i,I}(v\sqcap g_{i,I}\left(v\right))$. However, $c(v\sqcap g_{i,I}\left(v\right))\le g_{i,I}(v\sqcap g_{i,I}\left(v\right))$ that implies $cv\bigsqcup c{g}_{i,I}\left(v\right)\le g_{i,I}\left(v\right)\bigsqcup g_{i,I}\left(g_{i,I}\left(v\right)\right)$ that implies $c{g}_{i,I}\left(v\right)\le g_{i,I}\left(g_{i,I}\left(v\right)\right)$. Then we obtain $\top =g_{i,I}\left(v\right)\bigsqcup c{g}_{i,I}\left(v\right)\le g_{i,I}\left(v\right)\bigsqcup g_{i,I}\left(g_{i,I}\left(v\right)\right)=g_{i,I}(v\sqcap g_{i,I}\left(v\right))\le \top$. Thus, $g_{i,I}(v\sqcap g_{i,I}\left(v\right))=\top$.

“⇐”: Let $u\in i\left(X\right)$ and $v\in X$. From the hypothesis we have $g_{i,I}(v\sqcap g_{i,I}\left(v\right))=\top$, and then $u\sqcap c(v\sqcap g_{i,I}\left(v\right))\le \top =g_{i,I}(v\sqcap g_{i,I}\left(v\right))$. From $(\ast )$ it follows that $u\sqcap c(v\sqcap g_{i,I}\left(v\right))\in G_{i,I}\left(X\right)$. □

Corollary 5.

If $I\in {\mathcal{P}}_i$, then $G_{i,I}\left(X\right)=\{u\sqcap cy|u\in i\left(X\right),y\in I\}$.

From Corollary 5, Propositions 19 and 22, the next result follows.

Proposition 26.

Let $\mathcal{L}=\mathcal{P}\left(i\right(X\left)\right)$, $\tilde{\mathcal{L}}=\mathcal{P}\left(G_{i,I}\left(X\right)\right)$ and $h:X\to X$ be a mapping.

1.

If $I\in {\mathcal{P}}_i$, then $K_{h{g}_{i,I},\mathcal{L}}\left(X\right)=K_{h{g}_{i,I},\tilde{\mathcal{L}}}\left(X\right)$.

2.

If $I\in {\mathcal{P}}_i\cap {\mathcal{Q}}_i$, then $K_{1_X,\mathcal{L}}\left(X\right)\subseteq K_{cic,\mathcal{L}}\left(X\right)=K_{cic,\tilde{\mathcal{L}}}\left(X\right)$.

Let $i_1,i_2$ be two interior operators over X, and $I_1,I_2\in \mathcal{I}\left(X\right)$.

Remark 2.

If $i_1\le i_2$ and $I_1\subseteq I_2$, then $A_{i_1,I_1}\left(x\right)\subseteq A_{i_2,I_2}\left(x\right)$ for all $x\in X$.

Moreover, $g_{i_1,I_1}\le g_{i_2,I_2}$ and $G(i_1,I_1)\le G(i_2,I_2)$.

Proposition 27.

If $I_1\in {\mathcal{P}}_{i_1}$ and $I_1\subseteq I_2$, then $g_{G(i_1,I_1),I_2}\le g_{i_1,I_2}$.

Proof. 

Let $x\in X$ and $m\in A_{G(i_1,I_1),I_2}\left(x\right)$. Then $m\in min\left(X\right)$ and there is $u\in G(i_1,I_1)\left(X\right)$ such that $m\le u$ and $u\sqcap x\in I_2$. From $I_1\in {\mathcal{P}}_{i_1}$, it follows that $G(i_1,I_1)\left(X\right)=\{v\sqcap cy|v\in i_1\left(X\right),y\in I_1\}$. Therefore, there is $v\in i_1\left(X\right)$ and $y\in I_1$ such that $u=v\sqcap cy$, and $m\le v$ and $v\sqcap x=v\sqcap (y\bigsqcup cy)\sqcap x\le y\bigsqcup (u\sqcap x)$. Since $I_1\subseteq I_2$, $y\in I_1$ and $u\sqcap x\in I_2$, then $y\bigsqcup (u\sqcap x)\in I_2$, and so $v\sqcap x\in I_2$. Hence, $m\in A_{i_1,I_2}$. It follows that $A_{G(i_1,I_1),I_2}\left(x\right)\subseteq A_{i_1,I_2}$, and so $g_{G(i_1,I_1),I_2}\le g_{i_1,I_2}$. □

Corollary 6.

If $i_1\le i_2$, $I_1\subseteq I_2$ and $I_1\in {\mathcal{P}}_{i_1}$, then$G(i_1,I_1)\le G(G(i_1,I_1),I_2)\le G(i_2,I_2)$.

Proof. 

From Theorem 1, the inequality $G(i_1,I_1)\le G(G(i_1,I_1),I_2)$ is generally true (without the conditions of the hypothesis). Since $g_{G(i_1,I_1),I_2}\le g_{i_1,I_2}\le g_{i_2,I_2}$, it follows also that $G(G(i_1,I_1),I_2)\le G(i_2,I_2)$. □

Corollary 7.

If $I\in {\mathcal{P}}_i$, then $G\left(G\right(i,I),I)=G(i,I)$.

Corollary 8.

If $I_1\subseteq I_2$ and $I_1,I_2\in {\mathcal{P}}_{i_1}$, then $I_2\in {\mathcal{P}}_{G(i_1,I_1)}$.

Corollary 9.

If $I\in {\mathcal{P}}_i$, then $I\in {\mathcal{P}}_{G(i,I)}$.

Corollary 10.

If $I_1\subseteq I_2$, $I_1\in {\mathcal{P}}_{i_1}$, and $I_2\in {\mathcal{Q}}_{i_1}$, then $I_2\in {\mathcal{Q}}_{G(i_1,I_1)}$.

Corollary 11.

If $I\in {\mathcal{P}}_i\cap {\mathcal{Q}}_i$, then $I\in {\mathcal{P}}_{G(i,I)}\cap {\mathcal{Q}}_{G(i,I)}$.

We present some examples of $\alpha$-ideals having the ${\mathcal{P}}_i$ property.

Let us consider $(X,\le )$ a complete domain, $\alpha \ge 2$ a cardinal number, and $A\subseteq X$. We define $S_{\alpha }\left(A\right)$ by $S_{\alpha }\left(A\right)=\{\bigsqcup M|M\subseteq A,\left|M\right|\le \alpha \}$. If $\alpha \ge 2$ satisfies ${\alpha }^2=\alpha$ and $J\subseteq X$ is a complete ideal, then $S_{\alpha }\left(i^{-1}\left(J\right)\right)\in {\mathcal{I}}_{\alpha }\left(X\right)\cap {\mathcal{P}}_i$. Consequently, if $\alpha \ge 2$ with ${\alpha }^2=\alpha$, then $S_{\alpha }\left(i^{-1}(\perp )\right)\in {\mathcal{I}}_{\alpha }\left(X\right)\cap {\mathcal{P}}_i$. In addition, if $\alpha \ge 2$ is a cardinal number with ${\alpha }^2=\alpha$, $J\subseteq X$ is a complete ideal such that $cic\left(J\right)\subseteq J$ and $h\in \{{\left(ic\right)}^2,{\left(ci\right)}^2,i{\left(ci\right)}^2,c{\left(ic\right)}^3,{\left(ci\right)}^4\}$, then $S_{\alpha }\left(h^{-1}\left(J\right)\right)\in {\mathcal{I}}_{\alpha }\left(X\right)\cap {\mathcal{P}}_i$. Thus, if $\alpha \ge 2$ is a cardinal number with ${\alpha }^2=\alpha$ and $h\in \{{\left(ic\right)}^2,{\left(ci\right)}^2,i{\left(ci\right)}^2\}$, then $S_{\alpha }\left(h^{-1}(\perp )\right)\in {\mathcal{I}}_{\alpha }\left(X\right)\cap {\mathcal{P}}_i$.

We also present some examples of $\alpha$-ideals with the ${\mathcal{Q}}_i$ property.

Let $(X,\le )$ be a complete domain, and $\tau \subseteq \mathcal{P}\left(min\right(X\left)\right)$ be a topology. For $A\subseteq min\left(X\right)$, we denote by $in{t}_{\tau }A$ the interior of A relative to $\tau$. The function $i:X\to X$ defined by $i\left(x\right)=\bigsqcup in{t}_{\tau }(min\left(X\right)\cap \downarrow x)$ is an interior operator on X. Let $\alpha \ge 2$ be a cardinal number, $\mathcal{I}\subseteq \mathcal{P}\left(min\right(X\left)\right)$ be an $\alpha$-ideal relative to inclusion and $I=\{\bigsqcup A|A\in \mathcal{I}\}$. Then $I\in {\mathcal{I}}_{\alpha }\left(X\right)$ and $I\in {\mathcal{Q}}_i$ if and only if $\tau \cap \mathcal{I}=\{\varnothing \}$.

Let us consider ideals $\mathcal{I}$ with the property $\tau \cap \mathcal{I}=\{\varnothing \}$. From Baire’s theorem, each complete metric space is a Baire space [4]. We get the following result.

Theorem 2.

Let d be a metric on $min\left(X\right)$ such that $\left(min\right(X),d)$ is a complete metric space, let τ be the metric topology and $\mathcal{I}$ an ${\aleph }_0$-ideal of meagre sets. Then $I\in {\mathcal{I}}_{{\aleph }_0}\left(X\right)\cap {\mathcal{Q}}_i$.

An interesting situation is when $min\left(X\right)$ can be organized as an algebraic domain. In such a case, based on the results presented in [5], we obtain the following result.

Theorem 3.

Let ⊑ be a partial order on $min\left(X\right)$ such that $\left(min\right(X),⊑)$ is an algebraic domain, τ be the density topology (according to [5]), $\alpha \ge 2$ be a cardinal number and $\mathcal{I}$ be the ideal of all α-unions of τ-nowhere dense subsets of $min\left(X\right)$. Then $I\in {\mathcal{I}}_{\alpha }\left(X\right)\cap {\mathcal{Q}}_i$.

5. Conclusions

Inspired by the developments in topology and potential theory, we defined and studied an interior operator $G(i,I)$ generated by a given interior operator i and an ideal I in a complete domain X. We presented various properties of this interior operator, and emphasized the connection between the properties of the ideal I and the properties of the operator $G(i,I)$. Two specific properties denoted by ${\mathcal{P}}_i$ and ${\mathcal{Q}}_i$ were investigated. Considering $l:G(i,I)\left(X\right)\to X$ defined by $l\left(x\right)=x$ for every $x\in G(i,I)\left(X\right)$, we obtained a Galois connection between $G(i,I)\left(X\right)$ and X. As future work, we plan to study the properties of this connection, especially in the cases $I\in {\mathcal{P}}_i$ and $I\in {\mathcal{Q}}_i$, in order to find new properties and a new class of Galois connections. On the other hand, if we consider a Galois connection $(l,u)$ between the posets X and Y (with Y being a complete domain), we may expect some conditions such that the function $l\circ u$ could become an interior operator over Y having the form $G(i,I)$, eventually having the properties ${\mathcal{P}}_i$ and ${\mathcal{Q}}_i$.