Dominated Ideals and Baire Ideals in Complete Domains

Abstract

We consider an operator $G_{i,I}$ generated by a given interior operator i and an ideal I. Among the properties of the operator $G_{i,I}$, two properties denoted by ${\mathcal{P}}_i$ and ${\mathcal{Q}}_i$ are remarkable. The ideals with the property ${\mathcal{P}}_i$ are called dominated ideals, and the ideals with the property ${\mathcal{Q}}_i$ are called Baire ideals. We present methods of constructing dominated and Baire ideals, providing several examples and (topological) results in the framework of domain theory.

1. Introduction

McKinsey and Tarski initiated the study of interior algebras in [1], where the authors developed a structure called ’closure algebra’ using set-theoretic topology in a natural way. An interior algebra is a Boolean algebra enriched with an interior operator; such an algebra is an algebraic generalization of topological spaces. Essentially, an interior algebra is an algebraic structure that encodes the idea of the topological interior of a set (the interior of a subset S of a topological space X is the union of all subsets of S that are open in X). Their approach emphasized the interior operators, since the theory of closure operators [2] can be easily translated into the language of interior operators (by replacing sets with their complements).

In general, a topology on a set X creates the mathematical framework for defining the concept of convergence and approximation, fundamental notions of mathematical analysis. A topology is a family of subsets of a given set that satisfy certain conditions: the empty set and the entire set are both included, arbitrary unions of subsets are included, and the intersection of any two subsets is also included. Every topology uniquely generates an interior operator, and every interior operator uniquely generates a topology. In a topological space, we define an interior operator as a function that maps any subset to the set of its interior points; we call it the topological interior operator. Conversely, given an interior operator i over the set of subsets $\mathcal{P}\left(X\right)$, the set of its fixed points is a topology, and the associated topological interior operator is even i. Also, the topology generated by the interior operator is uniquely determined by the interior operator (it cannot be derived from other methods). Therefore, there is a bijection between the set of topologies on a non-empty set X and the set of interior operators on the set of subsets $\mathcal{P}\left(X\right)$. Thus, it is enough to define an interior operator on the set $\mathcal{P}\left(X\right)$ to be able to define the other topological notions: open and closed set, neighbourhood, compact set, axioms of separation, separability, convergence, etc.

Domain theory is a mathematical framework to study specific partially ordered sets called domains [3]. This theory is used in denotational semantics by formalizing the ideas of approximation and convergence [4]. It is also closely related to topology.

Starting from these aspects, the authors of [5] began to study a special class of interior operators $G_{i,I}$ generated by a given interior operator i and an ideal I. The framework was given by the complete domains (which generalizes the set of subsets of a set endowed with the inclusion relation). Point sets are represented by minimal elements, and the fact that a set is represented as the union of its point subsets is ensured by the property that the set of minimal elements forms a basis. The role of the complement is played by the function c. An interior operator is a kernel operator which preserves the infimum for finite sets. We were interested in the representation of the image $G_{i,I}\left(X\right)$ depending on the image $i\left(X\right)$. Some properties of the operator $G_{i,I}$ are studied; among them, two properties (denoted by ${\mathcal{P}}_i$ and ${\mathcal{Q}}_i$) were distinctive. Thus, if the ideal I has the property ${\mathcal{P}}_i$, then the image $G_{i,I}\left(X\right)$ has a simple representation $G_{i,I}\left(X\right)=\{u\sqcap cy|u\in i\left(X\right),y\in I\}$, showing a tight connection with the image $i\left(X\right)$. Ideals with the property ${\mathcal{P}}_i$ are also called dominated ideals because each element of such an ideal is dominated by its image via the function $g_{i,I}$. Ideals with the property ${\mathcal{Q}}_i$ are called Baire ideals because an important example is given by the ideal of meagre sets in a Baire space.

In this article, we present methods of constructing ideals with the properties ${\mathcal{P}}_i$ and ${\mathcal{Q}}_i$. Theorem 3 provides a method of constructing the ideals with the property ${\mathcal{P}}_i$. By customizing the functions f, g and h, and later the ideal J, several examples are presented. Next, we construct ideals with the property ${\mathcal{Q}}_i$; for this, we construct an interior operator by means of a topological interior operator on the set of minimal elements, and we present some topological results.

Many of the notations used in this article are taken from [4].

For a non-empty set X, we denote the set of all subsets of X by $\mathcal{P}\left(X\right)$. Let $(X,\le )$ be a partially ordered set (poset), and let $A\subseteq X$. We denote the upper set of A by $\uparrow A$, i.e., $\{x\in X|\exists a\in A,a\le x\}$. Also, the set $\downarrow A$ is the lower set of A, i.e., $\{x\in X|\exists a\in A,x\le a\}$. For $x\in X$, the set $\uparrow \left\{x\right\}$ is denoted by $\uparrow x$, and the set $\downarrow \left\{x\right\}$ is denoted by $\downarrow x$. A subset A of X is an upper (lower) set if $\uparrow A=A$ ($\downarrow A=A$). Also, $ub\left(A\right)$ denotes the upper bound set of A, i.e., $\{x\in X|\forall a\in A,a\le x\}$, and $lb\left(A\right)$ denotes Sthe lower bounds set of A, i.e., $\{x\in X|\forall a\in A,x\le a\}$.

If there exists the smallest element of X, it is denoted by ⊥ (called bottom). If there exists the biggest element of X, it is denoted by ⊤ (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 $\downarrow x=\{x,\perp \}$. $min\left(X\right)$ (or shortly, $min$) denotes 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; 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 $x\sqcap y$.

For a cardinal number $\alpha \ge 2$, the set A is called $\alpha$-directed if $\forall B\subseteq A$ with $\left|B\right|\le \alpha$, $A\cap ub\left(B\right)\ne \varnothing$. The 2-directed sets are called directed sets. It is obvious that a set is a directed set iff it is an $\alpha$-directed set for any finite cardinal $\alpha \ge 2$. A set A is called an $\alpha$-ideal if it is a lower set and $\alpha$-directed; A is called 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. Let ${\mathcal{I}}_{\alpha }\left(X\right)$ be the set of $\alpha$-ideals of $(X,\le )$, and $\mathcal{I}\left(X\right)$ be 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 in $\mathcal{L}$.

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

Let $(X,\le )$ be a pointed poset, $\mathcal{L}$ be a non-empty family of subsets of X, and $g:X⟶X$ be a function. For $x,z\in X$, we say 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 $z\le g\left(a\right)$. In this case, we denote $z{\ll }_{(g,\mathcal{L})}x$. We say that x is $(g,\mathcal{L})$-compact if it approximates itself relative to $(g,\mathcal{L})$. For $\mathcal{L}=\mathcal{P}\left(X\right)$ and $g=1_X$, $z{\ll }_{(g,\mathcal{L})}x$ is denoted by $z\ll x$, and we say z approximates x. Also, x is called 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 is denoted by $K\left(X\right)$. For $x\in X$, $↡x$ denotes the set $\{z\in X\setminus \{\perp \left\}\right|z\ll x\}$. Then, $↡x\subseteq \downarrow x$ for all $x\in X$.

A set $B\subseteq X$ is called a basis if there exists $\bigsqcup (B\cap ↡x)=x$ for all $x\in X\setminus \{\perp \}$. A domain is called continuous if it has a basis, and it is called algebraic if it has a basis of compact elements.

Let $f:X⟶X$ be a function. $fix\left(f\right)$ denotes the set of fixed points of f, i.e., $\{x\in X|f\left(x\right)=x\}$, and $f\left(X\right)$ denotes the set $\left\{f\right(x\left)\right|x\in X\}$. The function f is called isotone (antitone) if $x\le y\Rightarrow f\left(x\right)\le f\left(y\right)(\mathrm{respectively},f\left(y\right)\le f\left(x\right))$. The function 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 $f:X⟶X$ is called $(g,\mathcal{L})$-Scott-continuous if f is isotone and $\forall A\in \mathcal{L}$ such that $\exists \bigsqcup A$, there exists $\bigsqcup f\left(A\right)$ and $g(\bigsqcup f(A\left)\right)=gf(\bigsqcup A)$. Also, f is called Scott-continuous if f is $(1_X,\mathcal{P}\left(X\right))$-Scott-continuous.

2. Interior Operators on Complete Domains

We give the definition of the operator $G_{i,I}$ introduced in [5], and present some results; the results of this section are proved in [5].

Definition 1. 

A poset $(X,\le )$ is a complete domain if X is complete and $min$ is a basis.

Let $(X,\le )$ be a complete domain, and $c:X\to X$ be the function $c\left(x\right)=\bigsqcup \{y\in X|y\sqcap x=\perp \}$ for all $x\in X$. To simplify the notations, we use $cx$ instead of $c\left(x\right)$.

Proposition 1. 

The function c acts as a complement, and has the following properties:

• $cx=\bigsqcup \{y\in min|y\nleqq x\}$ for all $x\in X$;
• if $x\in min$, then $cx=\bigsqcup (min\setminus \{x\left\}\right)$;
• if $x\le y$, then $cy\le cx$;
• for all $x\in X,x\sqcap cx=\perp$ and $x\bigsqcup cx=\top$;
• if $cx\le x\bigsqcup y$, then $cx\le y$;
• $cx=cy\Rightarrow x=y$;
• $c^2=1_X$;
• for all $A\subseteq X,c(\bigsqcup A)=\sqcap \left\{ca\right|a\in A\}$ and $c(\sqcap A)=\bigsqcup \left\{ca\right|a\in A\}$.

A complete domain generalizes the family of subsets of a non-empty set, endowed with the inclusion relation. Point sets are represented by minimal elements. The fact that a set is represented as the union of its point subsets is ensured by the property that the set of minimal elements forms a basis. The role of the complement is played by the function c. Moreover, any complete domain is an algebraic domain (see [5]). Not every complete poset is a complete domain. A simple example is the set $X=\{\perp ,a,b,c,\top \}$ with the partial order $\le =\left\{\right(\perp ,\perp ),(\perp ,a),(\perp ,b),(\perp ,c),(a,a),(b,b),(c,c),(a,\top ),(b,\top ),(c,\top ),(\top ,\top \left)\right\}$. It is clear that $(X,\le )$ is a complete poset; however, it is not a complete domain.

Definition 2. 

A function $i:X⟶X$ is called an interior operator (on X) if

• $i\le 1_X$,
• $i^2=i$,
• $i(\sqcap A)=\sqcap i\left(A\right)$ for every finite subset A of X.

From the third property, if $A=\varnothing$, then we have $i(\top )=i(\sqcap \varnothing )=\sqcap i(\varnothing )=\sqcap \varnothing =\top$. In fact, the third property is equivalent to $i(\top )=\top$ and $i(x\sqcap y)=i\left(x\right)\sqcap i\left(y\right)$ for all $x,y\in X$. Also, if $x,y\in X$ such that $x\le y$, then $i\left(x\right)=i(x\sqcap y)=i\left(x\right)\sqcap i\left(y\right)\le i\left(y\right)$, namely i is isotone. In conclusion, an interior operator is a kernel operator ([3], p. 26) which preserves the infimum for finite sets. Since $i^2=i$, it results that $i\left(X\right)=fix\left(i\right)$.

Let i be an interior operator, and let $I\in \mathcal{I}\left(X\right)$.

Definition 3. 

Let $x,y\in X$ with $x\ne y$, and let $A\subseteq X$ be a subset with at least two elements.

• We say that x is $(i,I)$-separated from y if $\exists u_x\in i\left(X\right)$ such that $x\le u_x$ and $u_x\sqcap y\in I$.
• We say that x and y are i-separated if $\exists u_x,u_y\in i\left(X\right)$ such that $x\le u_x$, $y\le u_y$, and $u_x\sqcap u_y=\perp$.
• We say that A is i-separated if for all $x,y\in A$ with $x\ne y$, x and y are i-separated.

If x and y are i-separated, then x is $(i,I)$-separated from y, and y is $(i,I)$-separated from x.

Proposition 2. 

Let $S_i$ the family of i-separated subsets of X, and $f:X⟶X$ be an isotone function such that $fi\le i$ (particularly, $f\le i$ or $f\le 1_X$). Then, $f\left(S_i\right)\subseteq S_i$.

For each $x\in X$, we define the set

$$A_{i,I}\left(x\right)=\{m\in min|m\mathit{is}(i,I)-\mathit{separated}\mathit{from}x\}.$$

Then, the function $g_{i,I}:X⟶X$ is well defined:

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

Proposition 3. 

The function $g_{i,I}$ has the following properties:

• $g_{i,I}$ is antitone;
• $g_{i,I}(x\bigsqcup y)=g_{i,I}\left(x\right)\sqcap g_{i,I}\left(y\right)$ for all $x,y\in X$;
• $g_{i,I}\left(X\right)\subseteq i\left(X\right)$;
• $g_{i,I}\le g_{i,I}c{g}_{i,I}$;
• $i\le g_{i,I}c$;
• $I\subseteq g_{i,I}^{-1}(\top )$;
• for all $x\in X$ and $y\in I,g_{i,I}(x\bigsqcup y)=g_{i,I}(x\sqcap cy)=g_{i,I}\left(x\right)$.

These properties are proved in [5] (in Propositions 12–18).

Because $I\subseteq g_{i,I}^{-1}(\top )$, if $x\in I$, then $x\le g_{i,I}\left(x\right)$. If the reverse is true, we say that I has the property ${\mathcal{P}}_i$. Thus, ${\mathcal{P}}_i=\{I\in \mathcal{I}\left(X\right)|x\le g_{i,I}\left(x\right)\Rightarrow x\in I\}$, or ${\mathcal{P}}_i=\{I\in \mathcal{I}\left(X\right)|I=\{x\in X|x\le g_{i,I}\left(x\right)\}\}$. An ideal with the property ${\mathcal{P}}_i$ is called a dominated ideal (relative to the operator i).

Proposition 4 

([5], Proposition 20). $I\in {\mathcal{P}}_i$ iff $x\sqcap g_{i,I}\left(x\right)\in I$ for all $x\in X$.

We say that I has the property ${\mathcal{Q}}_i$ if $g_{i,I}(\top )=\perp$, and denote ${\mathcal{Q}}_i=\{I\in \mathcal{I}\left(X\right)|g_{i,I}(\top )=\perp \}$. An ideal with the property ${\mathcal{Q}}_i$ is called a Baire ideal (relative to the operator i).

Proposition 5 

([5], Proposition 21). $I\in {\mathcal{Q}}_i$ iff $i\left(X\right)\cap I=\{\perp \}$.

If $I\in {\mathcal{Q}}_i$, then $g_{i,I}=ic$ on $i\left(X\right)$. $G_{i,I}$ denotes the function $1_X\sqcap g_{i,I}c$. In [5], namely in Theorem 1 and Propositions 23–25, we proved the following properties of the function $G_{i,I}$.

Theorem 1. 

$G_{i,I}$ has the following properties:

• $G_{i,I}$ is an interior operator on X;
• $i\le G_{i,I}$;
• $i\left(X\right)\subseteq G_{i,I}\left(X\right)$;
• $G_{i,I}i=i{G}_{i,I}=i$;
• If $I\in {\mathcal{P}}_i$, then $G_{i,I}\left(X\right)=\{u\sqcap cy|u\in i\left(X\right),y\in I\}$;
• If $I\in {\mathcal{P}}_i$, then $I\in {\mathcal{P}}_{G_{i,I}}$;
• If $I\in {\mathcal{P}}_i\cap {\mathcal{Q}}_i$, then $I\in {\mathcal{P}}_{G_{i,I}}\cap {\mathcal{Q}}_{G_{i,I}}$.

Thus, any interior operator i on a complete domain X together with any ideal I generates a new interior operator $G_{i,I}$, closely related to the initial operator i. An important property is provided by item 5, which expresses the image $G_{i,I}$ as a function depending on the image $i\left(X\right)$. More properties of $G_{i,I}$ can be found in [5].

3. Dominated Ideals: The $\alpha$-Ideals with Property ${\mathcal{P}}_i$

The dominated ideals relative to an interior operator i are ideals with the property ${\mathcal{P}}_i$, namely ideals $I\in \mathcal{I}\left(X\right)$ with the elements $x\in X$ for which $x\le g_{i,I}\left(x\right)$.

Let $(X,\le )$ be a complete domain, $\alpha \ge 2$ be a cardinal number, and $A\subseteq X$. We denote $S_{\alpha }\left(A\right)=\{\bigsqcup M|M\subseteq A,\left|M\right|\le \alpha \}$ and $S\left(A\right)=\{\bigsqcup M|M\subseteq A,M\mathrm{finite}\}$.

Remark 1. 

• If A is a lower set, then $S_{\alpha }\left(A\right)$ and $S\left(A\right)$ are lower sets.
• If $Z\subseteq S_{\alpha }\left(A\right)$ with $\left|Z\right|\le \beta$, then $\bigsqcup Z\in S_{\alpha \beta }\left(A\right)$.

Consequently, if A is a lower set, then $S\left(A\right)\in \mathcal{I}\left(X\right)$, and $S_{\alpha }\left(A\right)\in {\mathcal{I}}_{\alpha }\left(X\right)$ for each transfinite cardinal $\alpha$ with ${\alpha }^2=\alpha$.

Lemma 1. 

Let i be an interior operator on X, and $f,h:X⟶X$ be two functions such that $f(\perp )=\perp$, h is isotone, and $h\le f{\left(ic\right)}^2$. Let $J\subseteq X$ be a lower set and $I=S_{\alpha }\left(h^{-1}\left(J\right)\right)$. Then, for all $x\in X$ with $x\le g_{i,I}\left(x\right)$, there exist $y,z\in X$ such that $x=y\bigsqcup z$, $h\left(y\right)=\perp$ and $z=\bigsqcup Z$, where $\left|Z\right|\le \alpha$ and for all $v\in Z$, there exists an i-separated set $A_v\subseteq h^{-1}\left(J\right)$ such that $v=\bigsqcup A_v$.

Proof. 

Let $x\in X$ such that $x\le g_{i,I}\left(x\right)$. For all $m\in min$ with $m\le x$, we have $m\in A_{i,I}\left(x\right)$. Hence, for all $m\in min$ with $m\le x$, there is $u_m\in i\left(X\right)$ such that $m\le u_m$ and $u_m\sqcap x\in I$. We denote $u_m\sqcap x$ with $x_m$. Let $U=\{u\in i(X\left)\right|u\sqcap x\in I\}$ and $\mathcal{V}=\{V\subseteq U|v_1\sqcap v_2=\perp$ for all $v_1,v_2\in V\}$. Since every chain in $(\mathcal{V},\subseteq )$ has an upper bound, according to Zorn’s lemma, there is a maximal element $V_0$ in $(\mathcal{V},\subseteq )$. Let $v_0=\bigsqcup V_0$. Since $V_0\subseteq U\subseteq i\left(X\right)$, then $v_0\in i\left(X\right)$. We have $x=x\sqcap \top =x\sqcap (c{v}_0\bigsqcup v_0)=y\bigsqcup z$, where $y=x\sqcap c{v}_0$ and $z=x\sqcap v_0$. We show that y and z satisfy the conditions of the lemma.

Let $u_0=\left(ic\right)\left(v_0\right)$, and assume that $u_0\ne \perp$. Since $u_0\le c{v}_0$ and $v_0\sqcap c{v}_0=\perp$, we have $u_0\sqcap v_0=\perp$. Then, $u_0\sqcap v=\perp$ for all $v\in V_0$. Since $u_0\in i\left(X\right)$, if $u_0\sqcap x\in I$ then $u_0\in U$. Therefore, $V_0\cup \left\{u_0\right\}\in \mathcal{V}$ and $V_0{V}_0\cup \left\{u_0\right\}$, which contradicts the maximality of $V_0$. Then, $u_0\sqcap x\notin I$, and so $u_0\sqcap x\ne \perp$. Since $x=x\sqcap x=x\sqcap \bigsqcup \{m\in min|m\le x\}\le x\sqcap \bigsqcup \left\{u_m\right|m\in min,m\le x\}=\bigsqcup \{u_m\sqcap x|m\in min,m\le x\}=\bigsqcup \left\{x_m\right|m\in min,m\le x\}\le x$, we have $x=\bigsqcup \left\{x_m\right|m\in min,m\le x\}$, and so $u_0\sqcap x=\bigsqcup \{u_0\sqcap x_m|m\in min,m\le x\}$. Since $u_0\sqcap x\ne \perp$, there exists $m\in min$ such that $m\le x$ and $u_0\sqcap x_m\ne \perp$. Since $x_m=u_m\sqcap x=i\left(u_m\right)\sqcap x=i(u_m\sqcap (x\bigsqcup cx))\sqcap x\le i(x_m\bigsqcup cx)\sqcap x\le (x_m\bigsqcup cx)\sqcap x\le x_m$, we have $x_m=i(x_m\bigsqcup cx)\sqcap x$. Let $u_1=i(x_m\bigsqcup cx)\sqcap u_0$. Since $u_0\in i\left(X\right)$ and $i^2=i$, $i\left(u_1\right)=i^2(x_m\bigsqcup cx)\sqcap i\left(u_0\right)=u_1$, and then $u_1\in i\left(X\right)$. Moreover, $u_1\sqcap x\le (x_m\bigsqcup cx)\sqcap u_0\sqcap x=x_m\sqcap u_0\sqcap x\le x_m$; since $x_m\in I$ and I is a lower set, it results that $u_1\sqcap x\in I$. Then, $u_1\in U$. Also, from $\perp \ne u_0\sqcap x_m=u_1\sqcap x\le u_1$, it follows that $u_1\ne \perp$, and from $u_1\le u_0\le c{v}_0$, it results that $u_1\sqcap v_0=\perp$. Thus, $u_1\sqcap v=\perp$ for all $v\in V_0$. Therefore, $V_0\cup \left\{u_1\right\}\in \mathcal{V}$ and $V_0{V}_0\cup \left\{u_1\right\}$, which contradicts the maximality of $V_0$. In conclusion, $u_0=\perp$, i.e., $\left(ic\right)\left(v_0\right)=\perp$. Then, we obtain $\perp \le h\left(y\right)=h(x\sqcap c{v}_0)\le h\left(c{v}_0\right)\le f{\left(ic\right)}^2\left(c{v}_0\right)=fic\left(i\left(v_0\right)\right)=fic\left(v_0\right)=f\left(ic\left(v_0\right)\right)=f(\perp )=\perp$, and so $h\left(y\right)=\perp$.

On the other hand, since $V_0\subseteq U$, we obtain $v\sqcap x\in I$ for all $v\in V_0$. It follows that for all $v\in V_0$, there exists $M_v\subseteq h^{-1}\left(J\right)$ with $|M_v|\le \alpha$ and $v\sqcap x=\bigsqcup M_v$. Let $v\in V_0$, and let K be a arbitrary set with $\left|K\right|=\alpha$. Since $|M_v|\le |K|$, there exists a surjective function $f_v:K⟶M_v$. Then, $z=x\sqcap v_0=x\sqcap \bigsqcup V_0={\bigsqcup }_{v\in V_0}(x\sqcap v)={\bigsqcup }_{v\in V_0}(\bigsqcup M_v)={\bigsqcup }_{v\in V_0}(\bigsqcup f_v\left(K\right))={\bigsqcup }_{v\in V_0}{\bigsqcup }_{k\in K}{f}_v\left(k\right)={\bigsqcup }_{k\in K}{\bigsqcup }_{v\in V_0}{f}_v\left(k\right)=\bigsqcup Z$, where $Z=\{\bigsqcup A_k|k\in K\}$ and $A_k=\left\{f_v\left(k\right)\right|v\in V_0\}$ for all $k\in K$.

We have $\left|Z\right|\le \left|K\right|=\alpha$, and show that each $A_k$ is i-separated. Let $k\in K$; then, for all $v\in V_0$, we have $v\in i\left(X\right)$; since $f_v\left(k\right)\in M_v$, it results that $f_v\left(k\right)\le \bigsqcup M_v=v\sqcap x\le v$. Let $a_1,a_2\in A_k$ with $a_1\ne a_2$. Then, there exist $v_1,v_2\in V_0$ such that $v_1\ne v_2$, $a_1=f_{v_1}\left(k\right)$ and $a_2=f_{v_2}\left(k\right)$. From $v_1,v_2\in V_0$ with $v_1\ne v_2$ and $V_0\in \mathcal{V}$, it follows that $v_1\sqcap v_2=\perp$. In conclusion, $A_k$ is i-separated. □

We recall that the ideals with the property ${\mathcal{P}}_i$ are ideals $I\in \mathcal{I}\left(X\right)$ with the elements $x\in X$ for which $x\le g_{i,I}\left(x\right)$.

Theorem 2. 

Let $\alpha \ge 2$ be a cardinal number with ${\alpha }^2=\alpha$, $J\subseteq X$ be a lower set, and $f,h:X⟶X$ be two functions such that $f(\perp )=\perp$, h is isotone, $h\le f{\left(ic\right)}^2$, and $\bigsqcup A\in h^{-1}\left(J\right)$ for all i-separated sets $A\subseteq h^{-1}\left(J\right)$. Then, $S_{\alpha }\left(h^{-1}\left(J\right)\right)\in {\mathcal{I}}_{\alpha }\left(X\right)\cap {\mathcal{P}}_i$.

Proof. 

Since J is a lower set and h is isotone, $h^{-1}\left(J\right)$ is a lower set. Then, $S_{\alpha }\left(h^{-1}\left(J\right)\right)$ is a lower set. Since ${\alpha }^2=\alpha$, it results that $S_{\alpha }\left(h^{-1}\left(J\right)\right)$ is $\alpha$-directed. Consequently, $S_{\alpha }\left(h^{-1}\left(J\right)\right)$ is an $\alpha$-ideal. We prove that $S_{\alpha }\left(h^{-1}\left(J\right)\right)\in {\mathcal{P}}_i$. I denotes the set $S_{\alpha }\left(h^{-1}\left(J\right)\right)$, and we consider $x\in X$ such that $x\le g_{i,I}\left(x\right)$. According to Lemma 1, there exist $y,z\in X$ such that $x=y\bigsqcup z$, $h\left(y\right)=\perp$ and $z=\bigsqcup Z$, where $\left|Z\right|\le \alpha$ and for all $v\in Z$, there exists an i-separated set $A_v\subseteq h^{-1}\left(J\right)$ such that $v=\bigsqcup A_v$. Since $h\left(y\right)=\perp \in J$ implies that $y\in h^{-1}\left(J\right)\subseteq I$, we have $y\in I$. Let $v\in Z$. Since $A_v\subseteq h^{-1}\left(J\right)$ and $A_v$ is an i-separated set, from the hypothesis, we obtain $\bigsqcup A_v\in h^{-1}\left(J\right)$. Thus, $z=\bigsqcup Z$, where $\left|Z\right|\le \alpha$ and $Z\subseteq h^{-1}\left(J\right)$, i.e., $z\in I$. It follows that $x\in I$. In conclusion, $I\in {\mathcal{P}}_i$. □

Theorem 3. 

Let $\alpha \ge 2$ be a cardinal number with ${\alpha }^2=\alpha$, $J\subseteq X$ be a complete ideal, and $f,g,h:X⟶X$ be three functions such that

• $f(\perp )=\perp$ and $h\le f{\left(ic\right)}^2$,
• $1_X\le g$ and $g\left(J\right)\subseteq J$,
• h is $(g,S_i)$-Scott-continuous.

Then $S_{\alpha }\left(h^{-1}\left(J\right)\right)\in {\mathcal{I}}_{\alpha }\left(X\right)\cap {\mathcal{P}}_i$.

Proof. 

Let $A\subseteq h^{-1}\left(J\right)$ be an i-separated set. Then, $h\left(A\right)\subseteq J$, and since J is a complete ideal, we obtain $\bigsqcup h\left(A\right)\in J$. Since $g\left(J\right)\subseteq J$, $g(\bigsqcup h(A\left)\right)\in J$. Since h is $(g,S_i)$-Scott-continuous, $gh(\bigsqcup A)=g(\bigsqcup h(A\left)\right)$. It results that $gh(\bigsqcup A)\in J$. Because $1_X\le g$, it follows that $h(\bigsqcup A)\le gh(\bigsqcup A)$, and so $h(\bigsqcup A)\in J$. Then, $\bigsqcup A\in h^{-1}\left(J\right)$. The desired conclusion follows from Theorem 2. □

We present some properties of the $(g,\mathcal{L})$-Scott-continuous functions, and some examples.

Remark 2. 

Let $\mathcal{L}$ be a non-empty family of subsets of X, and $f,g:X⟶X$ be two functions such that f is $(g,\mathcal{L})$-Scott-continuous. Then, f is $(hg,{\mathcal{L}}^{\prime })$-Scott-continuous for any function $h:X⟶X$ and any non-empty subset ${\mathcal{L}}^{\prime }\subseteq \mathcal{L}$.

Proposition 6. 

Let $\mathcal{L},{\mathcal{L}}^{\prime }$ be two non-empty families of subsets of X, and $f,g,h:X⟶X$ be three functions such that $f\left(\mathcal{L}\right)\subseteq {\mathcal{L}}^{\prime }$, f is $(g,\mathcal{L})$-Scott-continuous, and g is $(h,{\mathcal{L}}^{\prime })$-Scott-continuous. Then, $gf$ is $(h,\mathcal{L})$-Scott-continuous.

Proof. 

Since f and g are isotone, then $gf$ is isotone. For $A\in \mathcal{L}$, since f is $(g,\mathcal{L})$-Scott-continuous, we have $gf(\bigsqcup A)=g(\bigsqcup f(A\left)\right)$. Since $f\left(A\right)\in {\mathcal{L}}^{\prime }$ and g is $(h,{\mathcal{L}}^{\prime })$-Scott-continuous, we have $hg(\bigsqcup f(A\left)\right)=h(\bigsqcup gf(A\left)\right)$. Then, $hgf(\bigsqcup A)=hg(\bigsqcup f(A\left)\right)=h(\bigsqcup gf(A\left)\right)$. In conclusion, $gf$ is $(h,\mathcal{L})$-Scott-continuous. □

Proposition 7. 

Let $f,g,h:X⟶X$ be three functions such that $fi\le i$, f is $(g,S_i)$-Scott-continuous, and g is $(h,S_i)$-Scott-continuous. Then, $gf$ is $(h,S_i)$-Scott-continuous.

Corollary 1. 

Let $f:X⟶X$ be a function such that $f\le 1_X$ and f is $(f,S_i)$-Scott-continuous. Then, $f^n$ is $(f,S_i)$-Scott-continuous for all $n\in {\mathbb{N}}^{*}$.

Proposition 8. 

The interior operator i is $(1_X,S_i)$-Scott-continuous.

Proof. 

Since i is an interior operator, i is isotone. Let $A\subseteq X$ an i-separated set, and $z=\bigsqcup A$. Since A is i-separated, for all $x\in A$ there is $u_x\in i\left(X\right)$ such that $x\le u_x$, and $u_x\sqcap u_y=\perp$ for all $x,y\in A$ with $x\ne y$. Let $x\in A$. Since $x\le u_x$, then $x=x\sqcap u_x$. Since for all $y\in A$ with $y\ne x$ we have $u_y\sqcap u_x=\perp$ and $y\le u_y$, we obtain $y\sqcap u_x=\perp$. Then, $z\sqcap u_x={\bigsqcup }_{y\in A}(y\sqcap u_x)=x\sqcap u_x=x$. It follows that $z\sqcap cx=z\sqcap c(z\sqcap u_x)=z\sqcap (cz\bigsqcup c{u}_x)=z\sqcap c{u}_x\le c{u}_x$, and so $u_x\le c(z\sqcap cx)=cz\bigsqcup x$. Then, $x\le u_x=i\left(u_x\right)\le i(cz\bigsqcup x)$, and so $z=\bigsqcup A\le {\bigsqcup }_{x\in A}i(cz\bigsqcup x)$. It follows that $i\left(z\right)=i\left(z\right)\sqcap z\le i\left(z\right)\sqcap {\bigsqcup }_{x\in A}i(cz\bigsqcup x)={\bigsqcup }_{x\in A}(i\left(z\right)\sqcap i(cz\bigsqcup x))={\bigsqcup }_{x\in A}i(z\sqcap (cz\bigsqcup x))={\bigsqcup }_{x\in A}i(z\sqcap x)\le {\bigsqcup }_{x\in A}i\left(x\right)\le i\left(z\right)$. Therefore, $i\left(z\right)={\bigsqcup }_{x\in A}i\left(x\right)$, i.e., $i(\bigsqcup A)=\bigsqcup i\left(A\right)$. □

Proposition 9. 

The function $cic$ is $(ic,\mathcal{P}(X\left)\right)$-Scott-continuous.

Proof. 

Since i is isotone and c is antitone, it follows that $cic$ is isotone. Let $A\subseteq X$ and $x\in A$. Since $i(\sqcap cA)\le \sqcap cA\le cx$, we have $i(\sqcap cA)=i^2(\sqcap cA)\le ic\left(x\right)$. Then, $i(\sqcap cA)\le \sqcap ic\left(A\right)$, and so $i(\sqcap cA)\le i(\sqcap ic(A\left)\right)$. On the other hand, $\sqcap ic\left(A\right)\le \sqcap c\left(A\right)$, and then $i(\sqcap ic(A\left)\right)\le i(\sqcap c(A\left)\right)$. Then, $i(\sqcap cA)=i(\sqcap ic(A\left)\right)$. It follows that $ic(\bigsqcup A)=i(\sqcap ccic(A\left)\right)$, namely $ic(\bigsqcup A)=ic(\bigsqcup cic(A\left)\right)$, namely $ic\left(cic\right)(\bigsqcup A)=ic(\bigsqcup cic(A\left)\right)$. In conclusion, $cic$ is $(ic,\mathcal{P}(X\left)\right)$-Scott-continuous. □

Proposition 10. 

The functions ${\left(ic\right)}^2$, ${\left(ci\right)}^2$, $i{\left(ci\right)}^2$, $c{\left(ic\right)}^3$, ${\left(ci\right)}^4$ are $(ic,S_i)$-Scott-continuous.

Proof. 

Since i is isotone and c is antitone, it follows that all these functions are isotone. Let $A\in S_i$ and let $z=\bigsqcup A$. Then, $z\le {\bigsqcup }_{x\in A}i(cz\bigsqcup x)$ (see the demonstration of Proposition 8).

We show that ${\left(ic\right)}^2$ is $(ic,S_i)$-Scott-continuous.

${\left(ic\right)}^2\left(z\right)={\left(ic\right)}^2\left(z\right)\sqcap cic\left(z\right)\le {\left(ic\right)}^2\left(z\right)\sqcap cic(z\bigsqcup c{\left(ic\right)}^2\left(z\right))={\left(ic\right)}^2\left(z\right)\sqcap cic((z\sqcap {\left(ic\right)}^2\left(z\right))\bigsqcup c{\left(ic\right)}^2\left(z\right))=[{\left(ic\right)}^2\left(z\right)\sqcap cic(z\sqcap {\left(ic\right)}^2\left(z\right))]\bigsqcup [{\left(ic\right)}^2\left(z\right)\sqcap cic\left(c{\left(ic\right)}^2\left(z\right)\right)]\le cic(z\sqcap {\left(ic\right)}^2\left(z\right))\bigsqcup [{\left(ic\right)}^2\left(z\right)\sqcap c{\left(ic\right)}^2\left(z\right)]=cic(z\sqcap {\left(ic\right)}^2\left(z\right))\le cic(\left({\bigsqcup }_{x\in A}i(cz\bigsqcup x)\right)\sqcap {\left(ic\right)}^2\left(z\right))=cic\left({\bigsqcup }_{x\in A}(i(cz\bigsqcup x)\sqcap {\left(ic\right)}^2\left(z\right))\right)$. $(\ast )$

Let $x\in A$. Since $x\le z$, $ic\left(z\right)=ic(x\bigsqcup z\sqcap cx)=i(cx\sqcap c(z\sqcap cx\left)\right)=ic\left(x\right)\sqcap ic(z\sqcap cx)=ic\left(x\right)\sqcap i(cz\bigsqcup x)$. Then ${\left(ic\right)}^2\left(z\right)=\left(ic\right)\left(ic\right)\left(z\right)\le c\left(ic\right)\left(z\right)=c(ic\left(x\right)\sqcap i(cz\bigsqcup x))=cic\left(x\right)\bigsqcup ci(cz\bigsqcup x)$. It follows that $i(cz\bigsqcup x)\sqcap {\left(ic\right)}^2\left(z\right)\le i(cz\bigsqcup x)\sqcap (cic\left(x\right)\bigsqcup ci(cz\bigsqcup x))=[i(cz\bigsqcup x)\sqcap cic\left(x\right)]\bigsqcup [i(cz\bigsqcup x)\sqcap ci(cz\bigsqcup x)]\le cic\left(x\right)$, and then $i(cz\bigsqcup x)\sqcap {\left(ic\right)}^2\left(z\right)=i(i(cz\bigsqcup x)\sqcap {\left(ic\right)}^2\left(z\right))\le {\left(ic\right)}^2\left(x\right)$. From $(\ast )$, it results that ${\left(ic\right)}^2\left(z\right)\le cic\left({\bigsqcup }_{x\in A}(i(cz\bigsqcup x)\sqcap {\left(ic\right)}^2\left(z\right))\right)\le cic\left({\bigsqcup }_{x\in A}{\left(ic\right)}^2\left(x\right)\right)=cic(\bigsqcup {\left(ic\right)}^2\left(A\right))$. It follows that $ic(\bigsqcup {\left(ic\right)}^2\left(A\right))\le c{\left(ic\right)}^2\left(z\right)$ and $ic(\bigsqcup {\left(ic\right)}^2\left(A\right))\le ic{\left(ic\right)}^2\left(z\right)$. On the other hand, $x\le z$ for all $x\in A$, and so ${\left(ic\right)}^2\left(x\right)\le {\left(ic\right)}^2\left(z\right)$. Hence, $\bigsqcup {\left(ic\right)}^2\left(A\right)\le {\left(ic\right)}^2\left(z\right)$. It follows that $ic{\left(ic\right)}^2\left(z\right)\le ic(\bigsqcup {\left(ic\right)}^2\left(A\right))$. In conclusion, $ic{\left(ic\right)}^2\left(z\right)=ic(\bigsqcup {\left(ic\right)}^2\left(A\right))$, and so ${\left(ic\right)}^2$ is $(ic,S_i)$-Scott-continuous.

In Proposition 7, we consider $f=i$ and $g=cic$. Since f is $(1_X,S_i)$-Scott-continuous and g is $(ic,\mathcal{P}(X\left)\right)$-Scott-continuous, from Remark 2, it results that f is $(g,S_i)$-Scott-continuous, and g is $(ic,S_i)$-Scott-continuous. Then, ${\left(ci\right)}^2=gf$ is $(ic,S_i)$-Scott-continuous.

In Proposition 7, we take $f=i$ and $g={\left(ic\right)}^2$. Since f is $(1_X,S_i)$-Scott-continuous, f is $(g,S_i)$-Scott-continuous. Since g is $(ic,S_i)$-Scott-continuous, it results that $i{\left(ci\right)}^2=gf$ is $(ic,S_i)$-Scott-continuous.

In Proposition 7, we take $f={\left(ic\right)}^2$ and $g=cic$. Since f is $(ic,S_i)$-Scott-continuous, f is $(g,S_i)$-Scott-continuous. Since g is $(ic,\mathcal{P}(X\left)\right)$-Scott-continuous, it results that $c{\left(ic\right)}^3=gf$ is $(ic,S_i)$-Scott-continuous.

In Proposition 7, we take $f=i$ and $g=c{\left(ic\right)}^3$. Since f is $(g,S_i)$-Scott-continuous, and g is $(ic,S_i)$-Scott-continuous, it results that ${\left(ci\right)}^4=gf$ is $(ic,S_i)$-Scott-continuous. □

Theorem 4. 

Let $\alpha \ge 2$ be a cardinal number with ${\alpha }^2=\alpha$, and $J\subseteq X$ be a complete ideal.

Then, $S_{\alpha }\left(i^{-1}\left(J\right)\right)\in {\mathcal{I}}_{\alpha }\left(X\right)\cap {\mathcal{P}}_i$.

Proof. 

We consider $f=g=1_X$ and $h=i$ in Theorem 3. It results that $f(\perp )=\perp$, $1_X\le g$, $g\left(J\right)=J$, and h is $(1_X,S_i)$-Scott-continuous. Since $i\le 1_x$ implies that $ic\le c$, it follows that $1_X\le cic$ and $i\le {\left(ic\right)}^2$; thus, we obtain $h\le f{\left(ic\right)}^2$. Then, $S_{\alpha }\left(h^{-1}\left(J\right)\right)\in {\mathcal{I}}_{\alpha }\left(X\right)\cap {\mathcal{P}}_i$. □

Corollary 2. 

Let $\alpha \ge 2$ be a cardinal number with ${\alpha }^2=\alpha$. Then, $S_{\alpha }\left(i^{-1}(\perp )\right)\in {\mathcal{I}}_{\alpha }\left(X\right)\cap {\mathcal{P}}_i$.

Proof. 

Just consider $J=\{\perp \}$ in Theorem 4. □

Theorem 5. 

Let $\alpha \ge 2$ be a cardinal number with ${\alpha }^2=\alpha$, and $J\subseteq X$ be a complete ideal such that $cic\left(J\right)\subseteq J$. If $h\in \{{\left(ic\right)}^2,{\left(ci\right)}^2,i{\left(ci\right)}^2,c{\left(ic\right)}^3,{\left(ci\right)}^4\}$, then $S_{\alpha }\left(h^{-1}\left(J\right)\right)\in {\mathcal{I}}_{\alpha }\left(X\right)\cap {\mathcal{P}}_i$ .

Proof. 

Let $g=cic$. Then, $1_X\le g$ and $g\left(J\right)\subseteq J$. According to Proposition 10, function h is $(ic,S_i)$-Scott-continuous, and so h is $(g,S_i)$-Scott-continuous. If $h={\left(ic\right)}^2$, we take $f=1_X$. If $h={\left(ci\right)}^2$, we take $f=g$. If $h=i{\left(ci\right)}^2$, we take $f={\left(ic\right)}^2$. If $h=c{\left(ic\right)}^3$, we take $f=g$. If $h={\left(ci\right)}^4$, we take $f=g$. In all these situations, we have $f(\perp )=\perp$ and $h\le f{\left(ic\right)}^2$. Consequently, according to Theorem 3, it follows that $S_{\alpha }\left(h^{-1}\left(J\right)\right)\in {\mathcal{I}}_{\alpha }\left(X\right)\cap {\mathcal{P}}_i$. □

Corollary 3. 

Let $\alpha \ge 2$ be a cardinal number with ${\alpha }^2=\alpha$, and $h\in \{{\left(ic\right)}^2,{\left(ci\right)}^2,i{\left(ci\right)}^2\}$.

Then, $S_{\alpha }\left(h^{-1}(\perp )\right)\in {\mathcal{I}}_{\alpha }\left(X\right)\cap {\mathcal{P}}_i$.

Proof. 

Just consider $J=\{\perp \}$ in Theorem 5. □

4. Baire Ideals: The $\alpha$-Ideals with Property ${\mathcal{Q}}_i$

The Baire ideals relative to an interior operator i are ideals with the property ${\mathcal{Q}}_i$, namely ideals $I\in \mathcal{I}\left(X\right)$ for which $g_{i,I}(\top )=\perp$.

Let $(X,\le )$ be a complete domain, and $\tau \subseteq \mathcal{P}\left(min\right)$ be a topology. For $A\subseteq min$, $in{t}_{\tau }A$ denotes the interior of A relative to $\tau$. Let $i:X\to X$ be the function defined by $i\left(x\right)=\bigsqcup in{t}_{\tau }(min\cap \downarrow x)$.

Proposition 11. 

The function i is an interior operator.

Proof. 

We prove the following:

1. 

$i(\perp )=\perp$,

2. 

$i\le 1_X$,

3. 

$x\le y$ implies $i\left(x\right)\le i\left(y\right)$,

4. 

$i(x\sqcap y)=i\left(x\right)\sqcap i\left(y\right)$ for all $x,y\in X$,

5. 

$i^2=i$.

1. $i(\perp )=\bigsqcup in{t}_{\tau }(min\cap \downarrow \perp )=\bigsqcup in{t}_{\tau }\varnothing =\bigsqcup \varnothing =\perp$.

2. For all $x\in X,in{t}_{\tau }(min\cap \downarrow x)\subseteq min\cap \downarrow x\subseteq \downarrow x$. Then, $i\left(x\right)\le x$, and so $i\le 1_X$.

3. $x\le y$ implies $\downarrow x\subseteq \downarrow y$; it follows that $min\cap \downarrow x\subseteq min\cap \downarrow y$ and $in{t}_{\tau }(min\cap \downarrow x)\subseteq in{t}_{\tau }(min\cap \downarrow y)$ Thus, $i\left(x\right)\le i\left(y\right)$.

4. Let $x,y\in X$. $x\sqcap y\le x,y$ implies $i(x\sqcap y)\le i\left(x\right),i\left(y\right)$, meaning that $i(x\sqcap y)\le i\left(x\right)\sqcap i\left(y\right)$. Let $m\in min$ such that $m\le i\left(x\right)\sqcap i\left(y\right)$. Since $m\le i\left(x\right)=\bigsqcup in{t}_{\tau }(min\cap \downarrow x)$, then $m\in in{t}_{\tau }(min\cap \downarrow x)$. Also, $m\le i\left(y\right)$; it follows that $m\in in{t}_{\tau }(min\cap \downarrow y)$. Then, $m\in in{t}_{\tau }(min\cap \downarrow x)\cap in{t}_{\tau }(min\cap \downarrow y)=in{t}_{\tau }(min\cap \downarrow x\cap \downarrow y)=in{t}_{\tau }(min\cap \downarrow (x\sqcap y))$. Therefore, $m\le i(x\sqcap y)$. Consequently, $i\left(x\right)\sqcap i\left(y\right)\le i(x\sqcap y)$. Thus, $i(x\sqcap y)=i\left(x\right)\sqcap i\left(y\right)$.

5. If $m\in min\cap \downarrow i\left(x\right)$, then $m\in min$ and $m\le i\left(x\right)$. This means that $m\in in{t}_{\tau }(min\cap \downarrow x)$. Therefore, $min\cap \downarrow i\left(x\right)=in{t}_{\tau }(min\cap \downarrow x)$. Then, $in{t}_{\tau }(min\cap \downarrow i\left(x\right))=in{t}_{\tau }(min\cap \downarrow x)$, and so $i\left(i\right(x\left)\right)=i\left(x\right)$. □

For the following results, let $\alpha \ge 2$ be a cardinal number, $\mathcal{I}\subseteq \mathcal{P}\left(min\right)$ an $\alpha$-ideal relative to inclusion, and $I=\{\bigsqcup A|A\in \mathcal{I}\}$.

Proposition 12. 

$I\in {\mathcal{I}}_{\alpha }\left(X\right)$.

Proof. 

Let $J\subseteq I$ such that $\left|J\right|\le \alpha$. For each $j\in J$, there exists $A_j\in \mathcal{I}$ such that $j=\bigsqcup A_j$. Since $\mathcal{I}$ is $\alpha$-directed relative to inclusion, ${\cup }_{j\in J}{A}_j\in \mathcal{I}$. Then, $\bigsqcup A_k\le \bigsqcup \left({\cup }_{j\in J}{A}_j\right)$ for all $k\in J$. Hence, I is $\alpha$-directed. Now, let $A\in \mathcal{I}$ and $x\le \bigsqcup A$. Let $m\in min\cap \downarrow x$. Since $m\le x\le \bigsqcup A$ and $m\ll m$, there exists $a\in A$ such that $m\le a$. Since $a\in min$, it results that $m=a$. Thus, $min\cap \downarrow x\subseteq A$, and so $min\cap \downarrow x\in \mathcal{I}$. Then, $x=\bigsqcup (min\cap \downarrow x)\in I$. Consequently, I is a lower set, and so I is an $\alpha$-ideal. □

Proposition 13. 

$i\left(x\right)\in I$ iff $in{t}_{\tau }(min\cap \downarrow x)\in \mathcal{I}$.

Proof. 

If $x\in X$ and $i\left(x\right)\in I$, then $A\in \mathcal{I}$ exists such that $i\left(x\right)=\bigsqcup A$. Then, for all $m\in in{t}_{\tau }(min\cap \downarrow x)$, we have $m\le i\left(x\right)$, implying $m\le \bigsqcup A$ and $m\in A$ because $m\ll m$ and $A\subseteq min$. Hence, $in{t}_{\tau }(min\cap \downarrow x)\subseteq A$, and so $in{t}_{\tau }(min\cap \downarrow x)\in \mathcal{I}$.

The reverse of this implication is obvious. □

Corollary 4. 

$i\left(X\right)\cap I=\{\perp \}$ iff $\tau \cap \mathcal{I}=\{\varnothing \}$.

Proof. 

We assume $i\left(X\right)\cap I=\{\perp \}$ and $\tau \cap \mathcal{I}\ne \{\varnothing \}$. Then, there is $D\in \tau \cap \mathcal{I}$ such that $D\ne \varnothing$. Let $x=\bigsqcup D$. Since $x=\bigsqcup (min\cap \downarrow x)$ and $D\subseteq min$, it results that $min\cap \downarrow x=D$. Then, $in{t}_{\tau }(min\cap \downarrow x)=D\in \mathcal{I}$, and so $i\left(x\right)\in I$. From the hypothesis, $i\left(x\right)=\perp$; hence, $in{t}_{\tau }(min\cap \downarrow x)=\varnothing$. Then, $D=\varnothing$; contradiction!

Now, let us assume that $\tau \cap \mathcal{I}=\{\varnothing \}$ and $i\left(X\right)\cap I\ne \{\perp \}$. Thus, there is $x\in X$ such that $i\left(x\right)\in I$ and $i\left(x\right)\ne \perp$. Then, $in{t}_{\tau }(min\cap \downarrow x)\in \mathcal{I}$. However, $in{t}_{\tau }(min\cap \downarrow x)\in \tau$, and from the hypothesis, we obtain $in{t}_{\tau }(min\cap \downarrow x)=\varnothing$. Then, $i\left(x\right)=\bigsqcup \varnothing =\perp$; a contradiction! □

The following result is a consequence of Proposition 5 (claiming $I\in {\mathcal{Q}}_i$ iff $i\left(X\right)\cap I=\{\perp \}$).

Corollary 5. 

$I\in {\mathcal{Q}}_i$ iff $\tau \cap \mathcal{I}=\{\varnothing \}$.

Let us consider ideals $\mathcal{I}$ with the property $\tau \cap \mathcal{I}=\{\varnothing \}$. The Baire spaces have this property relative to the ideal of meagre sets [6]; according to Baire’s theorem (see [6], p. 296), each complete metric space and each compact Hausdorff space is a Baire space. A meagre set (also called first-category set) is a subset of a topological space that is small (negligible) in a precise sense. A subset $A\subseteq min$ is $\tau$-nowhere-dense if its closure has empty interior; a subset $A\subseteq min$ is said to be of the $\tau$-first-category if it is the union of a countable collection of $\tau$-nowhere-dense sets. From Corollary 5 and Baire’s theorem, we obtain the following result.

Theorem 6. 

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

A more interesting case is when $min$ can be organized as an algebraic domain.

Theorem 7. 

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

Proof. 

We consider the Scott topology ${\sigma }_{min}$ on $(min,⊑)$, and the essential topology ${\tau }_e=\{A\subseteq min|↡A\subseteq A\}$ on $(min,⊑)$ (see [7]).

Let $\rho$ be the smallest common refinement of Scott topology and essential topology; this $\rho$ is called the density topology on $(min,⊑)$, and it is generated by the basis $\{D\cap G|D\in {\sigma }_{min},G\in {\tau }_e\}$ (see [7]). According to [7], since $(min,⊑)$ is an algebraic domain, we have that the union of every family of non-empty $\rho$-nowhere dense sets is not a $\rho$-open set. Therefore, $\rho \cap \mathcal{I}=\{\varnothing \}$, and so $I\in {\mathcal{I}}_{\alpha }\left(X\right)\cap {\mathcal{Q}}_i$. □

5. Conclusions

Domain theory [3,4] offers interesting opportunities to explore and develop new approaches of construction ideals having specific properties, and relating algebra (ideals) to topology (interior operators). In the framework of domain theory, we introduced recently the interior operators generated by ideals in complete domains [5]. To the best of our knowledge, the idea of interior operators generated by ideals on complete domains is new, and no other authors treated it.

Considering complete domains, we started from an interior operator $G_{i,I}$ generated by a given interior operator i and an ideal I. Among the properties of the operator $G_{i,I}$, two properties denoted by ${\mathcal{P}}_i$ and ${\mathcal{Q}}_i$ were remarkable. To build ideals with either ${\mathcal{P}}_i$ (dominated ideals) or ${\mathcal{Q}}_i$ property (Baire ideals) is not an easy task. In this paper, we described general methods for the construction of dominated ideals and Baire ideals. By varying the functions f, g, h, as well as the ideal J in Theorem 3, various dominated ideals were obtained. Also, by varying the topology in the Theorems 6 and 7, various Baire ideals were obtained. These particular ideals can lead to additional properties of the induced interior operators. These properties will be the subject of a future study.