Two Idealizations of h-Open Sets: Interior Operator and Continuity

Abstract

Given an ideal topological space $(X,\tau ,I)$, we consider two opposite ways of idealizing the class of h-open sets. The first, the class of $hI$-open sets introduced by Açıkgöz and Noiri, pushes the ideal inward through the Janković–Hamlett local-function closure: a set A is $hI$-open if $A\subseteq int(A\cup {cl}^{\ast }\left(V\right))$ for every nonempty proper open subset V of X. The second, introduced here as the class of $Ih$-open sets, pushes the ideal outward by tolerating a small defect: a set A is $Ih$-open if $A\setminus int(A\cup V)\in I$ for every nonempty proper open subset V of X. We show that h-openness implies both $hI$-openness and $Ih$-openness, while $hI$-openness and $Ih$-openness are mutually independent. We also show that the two idealizations collapse to h-openness at opposite ends of the ideal lattice: $hI$-openness coincides with h-openness when $I=\mathcal{P}\left(X\right)$, whereas $Ih$-openness coincides with h-openness when $I=\{\varnothing \}$. Replacing the ordinary interior by the $\alpha$-, pre-, semi- and $\beta$-interior operators, we obtain the variants $Ih\alpha$-, $Ihp$-, $Ihs$- and $Ih\beta$-openness, and we describe how they are related. Finally, we study the family ${\tau }^{Ih}$ of all $Ih$-open sets and the induced operator ${int}_{Ih}$. We prove that ${\tau }^{Ih}$ is a topology whenever X is finite or, more generally, whenever I is closed under arbitrary unions of its members. We also give a counterexample showing that, without such a hypothesis, arbitrary unions of $Ih$-open sets need not be $Ih$-open.

1. Introduction

The class of h-open sets, introduced by Abbas [1], is one of the generalized open sets studied in general topology. A subset A of a topological space $(X,\tau )$ is called h-open if

$$A\subseteq int(A\cup V)$$

for every nonempty proper open subset V of X. Since its introduction, this class has been examined and generalized in several directions; in particular, minimal open sets and h-open sets have recently been studied in the wider setting of generalized topological spaces by Castañeda, Salas-Brown and Sanabria [2].

The theory of ideal topological spaces, initiated by Janković and Hamlett [3] and resting on the local function of Kuratowski [4], has been used to study generalized open sets and weak forms of continuity. Motivated by the interaction between ideals and generalized openness, Açıkgöz and Noiri introduced the class of $hI$-open sets in ideal topological spaces [5]. Their definition brings in the ideal through the local-function closure operator and gives an ideal counterpart of h-open sets.

A word about why such idealizations are worth studying may be in order. An ideal singles out a family of subsets that are to be treated as small, and most of the classical small-set families of analysis and topology—the finite sets, the countable sets, the nowhere dense sets, the meager sets, the sets of measure zero—are ideals. From its beginnings, one of the principal services of idealization has been to make constructions and examples available that the purely topological setting cannot provide: since the local function, and with it every idealized notion, changes as the ideal is tuned between the two extremes $\{\varnothing \}$ and $\mathcal{P}\left(X\right)$, a single definition interpolates between a classical concept and a degenerate one, and along this scale one can manufacture examples separating properties that are indistinguishable at the purely topological level [3]. Idealization has also been used to decompose classical notions such as continuity into strictly weaker constituents [6,7], and a result proved once in $(X,\tau ,I)$ specializes simultaneously to all of the small-set families above. The present paper follows this pattern: the ideal supplies the graded scale along which our two idealizations of h-openness move in opposite directions, and the examples that separate the resulting classes, as well as the counterexample concerning arbitrary unions in Section 5, are all obtained by choosing the ideal appropriately; none of them has a counterpart for h-openness alone.

We have two aims here. First, we introduce a second idealization of h-openness, obtained by allowing the failure of the defining inclusion to lie in the ideal; this gives the notion of an $Ih$-open set. In contrast to $hI$-openness, which brings in the ideal by enlarging the auxiliary open set, $Ih$-openness brings it in by relaxing the inclusion itself. Second, we study the family of all $Ih$-open sets and the interior-type operator it induces.

The point of the second idealization is not merely to add one more class. Because $hI$- and $Ih$-openness react to the ideal in opposite directions, together they bracket ${\tau }^h$ from the two sides of the ideal lattice, and the inclusion between them reverses between the trivial and the full ideal; neither notion alone exhibits this reversal. The outer, defect-based side also admits a concrete reading: the induced operators ${int}_{Ih}$ and ${cl}_{Ih}$ approximate an arbitrary set from inside and from outside while treating the members of I as negligible, which links $Ih$-openness to rough set theory and which we make precise in Section 7.

The paper is organized as follows. In Section 2, we recall the necessary definitions concerning ideal topological spaces, h-open sets and $hI$-open sets. Section 3 introduces $Ih$-open sets and establishes their basic properties, including their relationships with h-open and $hI$-open sets and the way these relationships change as the ideal varies. In Section 4, we develop several further $Ih$-type variants based on generalized interior operators, investigate the implication structure among them, and introduce the notion of an $Ih$-neighborhood. Section 5 is devoted to the family ${\tau }^{Ih}$ and the associated interior operator ${int}_{Ih}$, where sufficient conditions are obtained for ${\tau }^{Ih}$ to form a topology. Section 6 introduces $Ih$-continuity and characterizes it through the $Ih$-interior and $Ih$-closure operators. Section 7 gives an application, reading ${int}_{Ih}$ and ${cl}_{Ih}$ as lower and upper approximation operators in the sense of rough set theory. Finally, Section 8 contains concluding remarks and directions for future research.

2. Preliminaries

Definition 1.

Let X be a nonempty set. A nonempty family $I\subseteq \mathcal{P}\left(X\right)$is called an ideal on X if the following two conditions hold:

(i)

if $A\in I$ and $B\subseteq A$, then $B\in I$;

(ii)

if $A,B\in I$, then $A\cup B\in I$.

Let $(X,\tau )$ be a topological space. If I is an ideal on X, then the triple $(X,\tau ,I)$ is called an ideal topological space.

Throughout the paper, ideals are denoted by italic capital letters such as I and J, and every symbol derived from an ideal carries the letter of that ideal: thus $hI$-open, $Ih$-open, ${\tau }^{Ih}$ and ${int}_{Ih}$ refer to the ideal I, while $hJ$-open, $Jh$-open, ${\tau }^{Jh}$ and ${int}_{Jh}$ refer to the ideal J.

Definition 2

([3]). Let $(X,\tau ,I)$ be an ideal topological space. For $A\subseteq X$, the local function of A with respect to I and τ is

$$A^{\ast }=\{x\in X:U\cap A\notin I\mathit{for}\mathit{every}U\in \tau \mathit{with}x\in U\}.$$

The corresponding ∗-closure is denoted by ${cl}^{\ast }\left(A\right)=A\cup A^{\ast }$.

Definition 3

([1]). A subset A of a topological space $(X,\tau )$ is called h-open if, for every nonempty proper open subset V of X,

$$A\subseteq int(A\cup V).$$

The family of all h-open subsets of X is denoted by ${\tau }^h$. It is known that ${\tau }^h$ forms a topology on X that is finer than $\tau$ [1].

Definition 4

([5]). Let $(X,\tau ,I)$ be an ideal topological space. A subset A of X is called $hI$-open if, for every nonempty proper open subset V of X,

$$A\subseteq int(A\cup {cl}^{\ast }\left(V\right)).$$

The complement of an $hI$-open set is called $hI$-closed. The family of all $hI$-open sets of $(X,\tau ,I)$ is denoted by $hIO\left(X\right)$ or ${\tau }^{hI}$. The family ${\tau }^{hI}$ also forms a topology on X, and it is finer than ${\tau }^h$ [5].

Theorem 1

([5]). Let $(X,\tau ,I)$ be an ideal topological space. If A is h-open, then A is $hI$-open. The converse is not true in general.

Theorem 2

([5]). If $I=\mathcal{P}\left(X\right)$, then ${\tau }^h={\tau }^{hI}$.

3. $\mathit{Ih}$-Open Sets

Definition 5.

Let $(X,\tau ,I)$ be an ideal topological space. A subset A of X is called $Ih$-open if, for every nonempty proper open subset V of X,

$$A\setminus int(A\cup V)\in I.$$

The complement of an $Ih$-open set is called $Ih$-closed. The family of all $Ih$-open sets of $(X,\tau ,I)$ is denoted by $IhO\left(X\right)$ or ${\tau }^{Ih}$.

Theorem 3.

Let $(X,\tau ,I)$ be an ideal topological space. If A is h-open, then A is $Ih$-open.

Proof. 

If A is h-open, then $A\subseteq int(A\cup V)$ for every nonempty proper open subset V of X. Hence

$$A\setminus int(A\cup V)=\varnothing \in I,$$

and so A is $Ih$-open. □

Theorem 4.

If $I=\{\varnothing \}$, then ${\tau }^{Ih}={\tau }^h$.

Proof. 

For $I=\{\varnothing \}$, the condition $A\setminus int(A\cup V)\in I$ is equivalent to $A\setminus int(A\cup V)=\varnothing$, that is, to $A\subseteq int(A\cup V)$. Requiring this for every nonempty proper open subset V of X is exactly the condition that A be h-open, so ${\tau }^{Ih}={\tau }^h$. □

Example 1.

The converse of the implication “h-open implies $Ih$-open” is not true in general. Let

$$X=\{a,b,c,d\},\tau =\{\varnothing ,X,\{d\},\{a,c\},\{a,c,d\left\}\right\},$$

and let

$$I=\{\varnothing ,\{a\left\}\right\}.$$

Consider $A=\left\{a\right\}$. For $V=\left\{d\right\}$ we have $A\setminus int(A\cup V)=\left\{a\right\}\in I$. For $V=\{a,c\}$ we have $A\setminus int(A\cup V)=\varnothing \in I$, and for $V=\{a,c,d\}$ we have $A\setminus int(A\cup V)=\varnothing \in I$. Thus A is $Ih$-open. However, for $V=\left\{d\right\}$ we have $int(A\cup V)=\left\{d\right\}$, so $A⊈int(A\cup V)$ and A is not h-open.

The following examples show that the notions of $hI$-openness and $Ih$-openness are independent.

Example 2

(An $hI$-open set that is not $Ih$-open; cf. [5]). Let

$$X=\{a,b,c\},\tau =\{\varnothing ,X,\{b\},\{b,c\left\}\right\},I=\{\varnothing ,\{c\left\}\right\}.$$

The set $A=\{a,b\}$ is $hI$-open, as shown in [5]. However, for $V=\left\{b\right\}$ we have

$$A\setminus int(A\cup V)=\left\{a\right\}\notin I.$$

Therefore A is not $Ih$-open.

Example 3

(An $Ih$-open set that is not $hI$-open). Let

$$X=\{a,b,c\},\tau =\{\varnothing ,X,\{a\left\}\right\},$$

and let

$$I=\{\varnothing ,\{a\},\{b\},\{a,b\left\}\right\}.$$

Consider $B=\left\{b\right\}$. For the only nonempty proper open subset $V=\left\{a\right\}$ of X, we have

$$B\setminus int(B\cup V)=\left\{b\right\}\in I,$$

so B is $Ih$-open. On the other hand, ${cl}^{\ast }\left(V\right)=\left\{a\right\}$, and hence

$$int(B\cup {cl}^{\ast }\left(V\right))=int\left(\{a,b\}\right)=\left\{a\right\}.$$

Thus $B⊈int(B\cup {cl}^{\ast }\left(V\right))$, and B is not $hI$-open.

Putting the two examples together, the three classes are related by

$${\tau }^h\subseteq {\tau }^{hI},{\tau }^h\subseteq {\tau }^{Ih},$$

while, in general, neither ${\tau }^{hI}\subseteq {\tau }^{Ih}$ nor ${\tau }^{Ih}\subseteq {\tau }^{hI}$ holds.

To compare the two classes more closely we let the ideal vary. In accordance with the convention of Section 2, when two ideals $I\subseteq J$ are in play, ${\tau }^{Ih}$ and ${\tau }^{Jh}$ denote the families of $Ih$-open and $Jh$-open sets formed with I and J, respectively, and similarly ${\tau }^{hI}$ and ${\tau }^{hJ}$. The next two lemmas show that the two families move in opposite directions as the ideal grows.

Lemma 1.

Let $(X,\tau )$ be a topological space and let $I\subseteq J$ be ideals on X. Then ${\tau }^{Ih}\subseteq {\tau }^{Jh}$.

Proof. 

Let $A\in {\tau }^{Ih}$ and let V be a nonempty proper open subset of X. Then $A\setminus int(A\cup V)\in I\subseteq J$, so $A\in {\tau }^{Jh}$. □

Lemma 2.

Let $(X,\tau )$ be a topological space and let $I\subseteq J$ be ideals on X. Then ${\tau }^{hJ}\subseteq {\tau }^{hI}$.

Proof. 

Write $V_I^{\ast }$ for the local function of V taken with respect to I. If $x\in V_J^{\ast }$, then $U\cap V\notin J$ for every open U containing x; since $I\subseteq J$, this gives $U\cap V\notin I$ for every such U, so $x\in V_I^{\ast }$. Hence $V_J^{\ast }\subseteq V_I^{\ast }$, and so ${cl}_J^{\ast }\left(V\right)\subseteq {cl}_I^{\ast }\left(V\right)$, which implies that $int(A\cup {cl}_J^{\ast }\left(V\right))\subseteq int(A\cup {cl}_I^{\ast }\left(V\right))$. If $A\in {\tau }^{hJ}$, then for every nonempty proper open subset V of X we have $A\subseteq int(A\cup {cl}_J^{\ast }\left(V\right))\subseteq int(A\cup {cl}_I^{\ast }\left(V\right))$, so $A\in {\tau }^{hI}$. □

Lemmas 1 and 2 tell us that ${\tau }^{Ih}$ can only grow and ${\tau }^{hI}$ can only shrink as the ideal increases from $\{\varnothing \}$ to $\mathcal{P}\left(X\right)$. At the two endpoints the situation is as follows.

Theorem 5.

Let $(X,\tau ,I)$ be an ideal topological space.

(i)

If $I=\{\varnothing \}$, then ${\tau }^{Ih}={\tau }^h\subseteq {\tau }^{hI}$.

(ii)

If $I=\mathcal{P}\left(X\right)$, then ${\tau }^{hI}={\tau }^h\subseteq {\tau }^{Ih}=\mathcal{P}\left(X\right)$.

So at the trivial ideal every $Ih$-open set is $hI$-open, while at the full ideal every $hI$-open set is $Ih$-open. The inclusion between the two classes is reversed at the two ends of the ideal lattice.

Proof. 

(i) The equality ${\tau }^{Ih}={\tau }^h$ is Theorem 4, and ${\tau }^h\subseteq {\tau }^{hI}$ is Theorem 1.

(ii) The equality ${\tau }^{hI}={\tau }^h$ is Theorem 2, and ${\tau }^h\subseteq {\tau }^{Ih}$ is Theorem 3. When $I=\mathcal{P}\left(X\right)$, the set $A\setminus int(A\cup V)$ lies in $\mathcal{P}\left(X\right)$ for every $A\subseteq X$ and every V, so every subset of X is $Ih$-open and ${\tau }^{Ih}=\mathcal{P}\left(X\right)$. □

Once the ideal moves away from an endpoint, the corresponding inclusion can fail, as we demonstrate below using a singleton.

Example 4.

Let $X=\{a,b,c\}$ with $\tau =\{\varnothing ,X,\{a\},\{a,b\left\}\right\}$, and consider the set $\left\{c\right\}$.

Take $I=\{\varnothing \}$. The closed sets of X are $\varnothing ,\left\{c\right\},\{b,c\}$ and X, so ${cl}^{\ast }\left(\left\{a\right\}\right)=cl\left(\left\{a\right\}\right)=X$ and ${cl}^{\ast }\left(\{a,b\}\right)=cl\left(\{a,b\}\right)=X$. Hence $int(\left\{c\right\}\cup {cl}^{\ast }\left(V\right))=int\left(X\right)=X\supseteq \left\{c\right\}$ for both nonempty proper open sets V, so $\left\{c\right\}$ is $hI$-open. It is not $Ih$-open: for $V=\left\{a\right\}$ we have $int\left(\right\{c\}\cup \{a\left\}\right)=int\left(\right\{a,c\left\}\right)=\left\{a\right\}$, hence $\left\{c\right\}\setminus int\left(\right\{c\}\cup \{a\left\}\right)=\left\{c\right\}\notin \{\varnothing \}$.

Now take $I=\mathcal{P}\left(X\right)$. Then $\left\{c\right\}$ is $Ih$-open, because the defect lies in $\mathcal{P}\left(X\right)$ for every V. But it is not $hI$-open: here $V^{\ast }=\varnothing$ for every V, so ${cl}^{\ast }\left(\left\{a\right\}\right)=\left\{a\right\}$ and $int\left(\right\{c\}\cup \{a\left\}\right)=\left\{a\right\}⊉\left\{c\right\}$.

So the same set $\left\{c\right\}$ is $hI$-open but not $Ih$-open at the trivial ideal, and $Ih$-open but not $hI$-open at the full ideal.

Table 1 collects the two idealizations side by side with the original h-open sets, recording how each responds to the ideal.

Table 1. The two idealizations of h-openness. In every row V ranges over the nonempty proper open subsets of X, and I increases from $\{\varnothing \}$ to $\mathcal{P}\left(X\right)$.

	$\mathit{hI}$-Open	$\mathit{Ih}$-Open
Defining condition	$A\subseteq int(A\cup {cl}^{\ast }\left(V\right))$	$A\setminus int(A\cup V)\in I$
How I enters	enlarges V inward via ${cl}^{\ast }(·)$	tolerates the defect outward
Relation to ${\tau }^h$	${\tau }^h\subseteq {\tau }^{hI}$ (Theorem 1)	${\tau }^h\subseteq {\tau }^{Ih}$ (Theorem 3)
Monotonicity in I	shrinks (Lemma 2)	grows (Lemma 1)
At $I=\{\varnothing \}$	$\supseteq {\tau }^h$	$={\tau }^h$ (Theorem 4)
At $I=\mathcal{P}\left(X\right)$	$={\tau }^h$ (Theorem 2)	$=\mathcal{P}\left(X\right)$ (Theorem 5)

4. Further $\mathit{Ih}$-Type Variants

In this section, we present some additional $Ih$-type notions obtained by replacing the ordinary interior in the definition of an $Ih$-open set with standard generalized interior operators. The corresponding $hI$-type notions and their basic implication diagram were considered by Açıkgöz and Noiri [5]; for related ideal versions of generalized open sets and the decompositions of continuity built from them, see, for instance, Dontchev [6] and Hatir and Noiri [7]. The results below are the outward, defect-based counterparts of those notions.

Recall that a subset G of a topological space $(X,\tau )$ is called $\alpha$-open [8], preopen [9], semi-open [10] and $\beta$-open [11], respectively, if

$$G\subseteq int(cl(int\left(G\right)\left)\right),G\subseteq int(cl(G\left)\right),G\subseteq cl(int(G\left)\right),$$

and

$$G\subseteq cl(int(cl\left(G\right)\left)\right).$$

The $\beta$-open sets are also known as semi-preopen sets [12].

For $A\subseteq X$, we denote by $\alpha \mathrm{int}\left(A\right)$, $pint\left(A\right)$, $sint\left(A\right)$ and $\beta \mathrm{int}\left(A\right)$ the unions of all $\alpha$-open, preopen, semi-open and $\beta$-open subsets of A, respectively. Thus

$$int\left(A\right)\subseteq \alpha \mathrm{int}\left(A\right),\alpha \mathrm{int}\left(A\right)\subseteq pint\left(A\right),\alpha \mathrm{int}\left(A\right)\subseteq sint\left(A\right),$$

and

$$pint\left(A\right)\subseteq \beta \mathrm{int}\left(A\right),sint\left(A\right)\subseteq \beta \mathrm{int}\left(A\right).$$

Definition 6.

Let $(X,\tau ,I)$ be an ideal topological space and let $A\subseteq X$.

(i)

A is called $Ih\alpha$-open if

$$A\setminus \alpha \mathrm{int}(A\cup V)\in I$$

for every nonempty proper open subset V of X.

(ii)

A is called $Ihp$-open if

$$A\setminus pint(A\cup V)\in I$$

for every nonempty proper open subset V of X.

(iii)

A is called $Ihs$-open if

$$A\setminus sint(A\cup V)\in I$$

for every nonempty proper open subset V of X.

(iv)

A is called $Ih\beta$-open if

$$A\setminus \beta \mathrm{int}(A\cup V)\in I$$

for every nonempty proper open subset V of X.

The corresponding families are denoted by $Ih\alpha O\left(X\right)$, $IhpO\left(X\right)$, $IhsO\left(X\right)$ and $Ih\beta O\left(X\right)$, respectively.

Theorem 6.

Let $(X,\tau ,I)$ be an ideal topological space. Then the following implications hold:

$$\begin{array}{cccc}\mathit{\text{h-open}}\to \mathit{\text{Ih-open}}\to & \mathit{\text{Ihα-open}}& \to & \mathit{\text{Ihp-open}}\\ & \downarrow & & \downarrow \\ & \mathit{\text{Ihs-open}}& \to & \mathit{\text{Ihβ-open}}\end{array}$$

Proof. 

Let V be a nonempty proper open subset of X. Since

$$int(A\cup V)\subseteq \alpha \mathrm{int}(A\cup V),$$

we have

$$A\setminus \alpha \mathrm{int}(A\cup V)\subseteq A\setminus int(A\cup V).$$

Hence, by heredity of the ideal, every $Ih$-open set is $Ih\alpha$-open.

Similarly, the inclusions

$$\alpha \mathrm{int}(A\cup V)\subseteq pint(A\cup V),\alpha \mathrm{int}(A\cup V)\subseteq sint(A\cup V)$$

imply

$$A\setminus pint(A\cup V)\subseteq A\setminus \alpha \mathrm{int}(A\cup V)$$

and

$$A\setminus sint(A\cup V)\subseteq A\setminus \alpha \mathrm{int}(A\cup V).$$

Thus every $Ih\alpha$-open set is both $Ihp$-open and $Ihs$-open. Finally, since

$$pint(A\cup V)\subseteq \beta \mathrm{int}(A\cup V)\mathrm{and}sint(A\cup V)\subseteq \beta \mathrm{int}(A\cup V),$$

we obtain

$$A\setminus \beta \mathrm{int}(A\cup V)\subseteq A\setminus pint(A\cup V)$$

and

$$A\setminus \beta \mathrm{int}(A\cup V)\subseteq A\setminus sint(A\cup V).$$

Therefore every $Ihp$-open set and every $Ihs$-open set is $Ih\beta$-open. □

The preceding theorem gives the defect-based analog of the $hI$-type implication diagram in [5].

Remark 1.

The reverse implications in the diagram above need not hold. Replacing the ordinary interior by a larger generalized interior operator weakens the defining condition, so a class can be strictly larger than the one above it, as the examples below show.

We now show that the reverse implications in the diagram of Theorem 6 fail in general.

Example 5

(An $Ih\alpha$-open set that is not $Ih$-open). Let

$$X=\{a,b,c,d\},\tau =\{\varnothing ,X,\{a\},\{c\},\{a,c\},\{a,d\},\{a,c,d\left\}\right\},$$

and let

$$I=\{\varnothing ,\{a\left\}\right\}.$$

Consider $A=\{a,b,c\}$. Since $int(cl(int\left(A\right)\left)\right)=int(cl(\{a,c\}\left)\right)=int\left(X\right)=X\supseteq A$, the set A is α-open; hence $A\subseteq \alpha \mathrm{int}(A\cup V)$ for every nonempty proper open subset V of X, so that

$$A\setminus \alpha \mathrm{int}(A\cup V)=\varnothing \in I.$$

Thus A is $Ih\alpha$-open. However, A is not $Ih$-open: for $V=\left\{a\right\}$ we have $int(A\cup V)=int\left(\right\{a,b,c\left\}\right)=\{a,c\}$, and therefore

$$A\setminus int(A\cup V)=\left\{b\right\}\notin I.$$

Example 6.

Let

$$X=\{a,b,c,d\},\tau =\{\varnothing ,X,\{d\},\{a,c\},\{a,c,d\left\}\right\},$$

and let

$$I=\{\varnothing ,\{a\left\}\right\}.$$

The set $B=\left\{b\right\}$ is both $Ihs$-open and $Ih\beta$-open, but neither $Ihp$-open nor $Ih\alpha$-open.

For every nonempty proper open subset V of X, the set $B\cup V$ is semi-open (and hence β-open); consequently $sint(B\cup V)=\beta \mathrm{int}(B\cup V)=B\cup V$ and

$$B\setminus sint(B\cup V)=B\setminus \beta \mathrm{int}(B\cup V)=\varnothing \in I.$$

On the other hand, taking $V=\left\{d\right\}$ gives $B\cup V=\{b,d\}$ with $\alpha \mathrm{int}\left(\right\{b,d\left\}\right)=pint\left(\right\{b,d\left\}\right)=\left\{d\right\}$, so

$$B\setminus \alpha \mathrm{int}(B\cup V)=B\setminus pint(B\cup V)=\left\{b\right\}\notin I.$$

The set $C=\left\{c\right\}$ is both $Ihp$-open and $Ih\beta$-open, but neither $Ih\alpha$-open nor $Ihs$-open.

For every nonempty proper open subset V of X, the set $C\cup V$ is preopen (and hence β-open); consequently $pint(C\cup V)=\beta \mathrm{int}(C\cup V)=C\cup V$ and

$$C\setminus pint(C\cup V)=C\setminus \beta \mathrm{int}(C\cup V)=\varnothing \in I.$$

However, for $V=\left\{d\right\}$ we have $C\cup V=\{c,d\}$ with $\alpha \mathrm{int}\left(\right\{c,d\left\}\right)=sint\left(\right\{c,d\left\}\right)=\left\{d\right\}$, so

$$C\setminus \alpha \mathrm{int}(C\cup V)=C\setminus sint(C\cup V)=\left\{c\right\}\notin I.$$

Definition 7.

Let $(X,\tau ,I)$ be an ideal topological space and let $x\in X$. A subset N of X is called an $Ih$-neighborhood of x if there exists a set $U\in {\tau }^{Ih}$ such that

$$x\in U\subseteq N.$$

Proposition 1.

Every $Ih$-open set containing x is an $Ih$-neighborhood of x. The converse is not true in general.

Proof. 

If N is $Ih$-open and $x\in N$, then the choice $U=N$ shows that N is an $Ih$-neighborhood of x.

For the converse, consider the ideal topological space in Example 1:

$$X=\{a,b,c,d\},\tau =\{\varnothing ,X,\{d\},\{a,c\},\{a,c,d\left\}\right\},$$

with $I=\{\varnothing ,\{a\left\}\right\}$. There we showed that $A=\left\{a\right\}$ is $Ih$-open. Hence $N=\{a,b\}$ is an $Ih$-neighborhood of a, since

$$a\in A=\left\{a\right\}\subseteq N.$$

However, N is not $Ih$-open. Indeed, for $V=\left\{d\right\}$ we have

$$int(N\cup V)=int\left(\right\{a,b,d\left\}\right)=\left\{d\right\},$$

and therefore

$$N\setminus int(N\cup V)=\{a,b\}\notin I.$$

Thus an $Ih$-neighborhood need not be $Ih$-open. □

5. Some Properties of ${\tau }^{\mathit{Ih}}$ and the Induced Interior Operator

We first show that ${\tau }^{Ih}$ need not be a topology in general.

Example 7.

Let $X=\mathbb{N}\cup \{\infty \}$ and let

$$\tau =\{\varnothing ,X,E\},$$

where E is the set of all even positive integers. Let I be the ideal of all finite subsets of X. For each $n\in \mathbb{N}$, put $A_n=\{2n-1\}$. Since

$$A_n\setminus int(A_n\cup E)=\{2n-1\}\in I,$$

each $A_n$ is $Ih$-open. However,

$$\bigcup _{n\in \mathbb{N}}{A}_n$$

is the set of all odd positive integers, and

$$\left(\bigcup _{n\in \mathbb{N}}{A}_n\right)\setminus int\left(\left(\bigcup _{n\in \mathbb{N}}{A}_n\right)\cup E\right)$$

is again the set of all odd positive integers, which is not finite. Hence ${\bigcup }_{n\in \mathbb{N}}{A}_n$ is not $Ih$-open. Therefore ${\tau }^{Ih}$ is not necessarily closed under arbitrary unions.

Theorem 7.

Let $(X,\tau ,I)$ be an ideal topological space. If X is finite or I is closed under arbitrary unions of its members, then ${\tau }^{Ih}$ is a topology on X.

Proof. 

Clearly, ∅ and X are $Ih$-open.

Let A and B be $Ih$-open subsets of X, and let V be a nonempty proper open subset of X. Then

$$A\setminus int(A\cup V)\in I\mathrm{and}B\setminus int(B\cup V)\in I.$$

Since ideals are hereditary,

$$(A\cap B)\setminus int(A\cup V)\in I\mathrm{and}(A\cap B)\setminus int(B\cup V)\in I.$$

By finite additivity of ideals, their union belongs to I. Moreover, because V is open,

$$int(A\cup V)\cap int(B\cup V)=int\left(\right(A\cup V)\cap (B\cup V\left)\right)=int\left(\right(A\cap B)\cup V).$$

Therefore

$$(A\cap B)\setminus int\left(\right(A\cap B)\cup V)\in I,$$

and hence $A\cap B$ is $Ih$-open.

Now let $\{A_{\alpha }:\alpha \in \mathrm{\Lambda }\}$ be a family of $Ih$-open subsets of X, and put $A={\bigcup }_{\alpha \in \mathrm{\Lambda }}{A}_{\alpha }$. For each $\alpha$,

$$A_{\alpha }\setminus int(A\cup V)\subseteq A_{\alpha }\setminus int(A_{\alpha }\cup V)\in I.$$

Thus $A_{\alpha }\setminus int(A\cup V)\in I$ for each $\alpha$. If I is closed under arbitrary unions, then

$$A\setminus int(A\cup V)=\bigcup _{\alpha \in \mathrm{\Lambda }}\left(A_{\alpha }\setminus int(A\cup V)\right)\in I.$$

If X is finite, then X has only finitely many subsets, so the sets $A_{\alpha }\setminus int(A\cup V)$ take only finitely many distinct values; their union is then a finite union of members of I and again lies in I. Hence A is $Ih$-open in either case, and ${\tau }^{Ih}$ is a topology on X. □

Definition 8.

Let $(X,\tau ,I)$ be an ideal topological space and let $A\subseteq X$. The $Ih$-interior of A is

$${int}_{Ih}\left(A\right)=\bigcup \{G\subseteq A:G\in {\tau }^{Ih}\}.$$

Proposition 2.

Let $(X,\tau ,I)$ be an ideal topological space and let $M,N\subseteq X$. Then the following hold:

(i)

if $M\subseteq N$, then ${int}_{Ih}\left(M\right)\subseteq {int}_{Ih}\left(N\right)$;

(ii)

${int}_{Ih}\left(M\right)\subseteq M$;

(iii)

${int}_{Ih}\left(M\right)=M$ whenever M is $Ih$-open; if, in addition, X is finite or I is closed under arbitrary unions, then M is $Ih$-open if and only if ${int}_{Ih}\left(M\right)=M$;

(iv)

${int}_{Ih}\left({int}_{Ih}\left(M\right)\right)\subseteq {int}_{Ih}\left(M\right)$. Equality holds when X is finite or I is closed under arbitrary unions;

(v)

${int}_{Ih}\left(M\right)\cap {int}_{Ih}\left(N\right)={int}_{Ih}(M\cap N)$;

(vi)

${int}_{Ih}\left(M\right)\cup {int}_{Ih}\left(N\right)\subseteq {int}_{Ih}(M\cup N)$.

Proof. 

(i) Let $M\subseteq N$. If $U\in {\tau }^{Ih}$ and $U\subseteq M$, then $U\subseteq N$ as well, so U belongs to the family whose union defines ${int}_{Ih}\left(N\right)$. Taking the union over all such U yields ${int}_{Ih}\left(M\right)\subseteq {int}_{Ih}\left(N\right)$.

(ii) By definition, ${int}_{Ih}\left(M\right)=\bigcup \{U\in {\tau }^{Ih}:U\subseteq M\}$ is a union of subsets of M; hence ${int}_{Ih}\left(M\right)\subseteq M$.

For (iii) and (iv) we first record the following fact. Assume that X is finite or that I is closed under arbitrary unions; call this condition $(★)$. By Theorem 7, the collection ${\tau }^{Ih}$ is then a topology on X and, in particular, is closed under arbitrary unions. Since ${int}_{Ih}\left(M\right)=\cup \{U\in {\tau }^{Ih}:U\subseteq M\}$ is a union of members of ${\tau }^{Ih}$, it follows that ${int}_{Ih}\left(M\right)\in {\tau }^{Ih}$; that is, ${int}_{Ih}\left(M\right)$ is the largest $Ih$-open set contained in M.

(iii) $(\Rightarrow )$ Suppose M is $Ih$-open. Then M is itself an $Ih$-open set contained in M, so $M\subseteq {int}_{Ih}\left(M\right)$; combined with (ii) gives $M={int}_{Ih}\left(M\right)$. (This implication holds without any additional hypothesis).

$(\Leftarrow )$ Suppose $M={int}_{Ih}\left(M\right)$. By $(★)$, ${int}_{Ih}\left(M\right)\in {\tau }^{Ih}$, hence $M\in {\tau }^{Ih}$; that is, M is $Ih$-open.

(iv) By $(★)$, ${int}_{Ih}\left(M\right)$ is $Ih$-open. Applying the implication $(\Rightarrow )$ of (iii) to the $Ih$-open set ${int}_{Ih}\left(M\right)$ gives ${int}_{Ih}\left({int}_{Ih}\left(M\right)\right)={int}_{Ih}\left(M\right)$.

(v) Since $M\cap N\subseteq M$ and $M\cap N\subseteq N$, part (i) yields ${int}_{Ih}(M\cap N)\subseteq {int}_{Ih}\left(M\right)\cap {int}_{Ih}\left(N\right)$. Conversely, let $x\in {int}_{Ih}\left(M\right)\cap {int}_{Ih}\left(N\right)$. Then there exist $U,W\in {\tau }^{Ih}$ with $x\in U\subseteq M$ and $x\in W\subseteq N$. By Theorem 7, the intersection of two $Ih$-open sets is again $Ih$-open, and this needs no hypothesis on X or I; hence $U\cap W\in {\tau }^{Ih}$ with $x\in U\cap W\subseteq M\cap N$. Hence $x\in {int}_{Ih}(M\cap N)$, and equality follows.

(vi) From $M\subseteq M\cup N$ and $N\subseteq M\cup N$, part (i) gives ${int}_{Ih}\left(M\right)\subseteq {int}_{Ih}(M\cup N)$ and ${int}_{Ih}\left(N\right)\subseteq {int}_{Ih}(M\cup N)$; therefore ${int}_{Ih}\left(M\right)\cup {int}_{Ih}\left(N\right)\subseteq {int}_{Ih}(M\cup N)$.

□

6. $\mathit{Ih}$-Continuity

In this section, we introduce the concept of an $Ih$-continuous function and provide some of its characterizations and properties. The notion of $Ih$-openness leads to a corresponding weak form of continuity. Continuity in ideal topological spaces has already been studied in several forms, for instance $\delta$-I-continuity [13], I-continuity [14], and decompositions of continuity obtained by idealization [7]. Throughout this section $(X,\tau ,I)$ is an ideal topological space and $(Y,\sigma )$ is a topological space; ${int}_Y$ and ${cl}_Y$ denote the interior and closure in Y.

Definition 9.

A function $f:(X,\tau ,I)\to (Y,\sigma )$ is called $Ih$-continuous if $f^{-1}\left(V\right)$ is $Ih$-open in $(X,\tau ,I)$ for every open subset V of Y.

Every open subset of X is h-open: if A is open and V is a nonempty proper open subset of X, then $A\cup V$ is open, so $int(A\cup V)=A\cup V\supseteq A$. By Theorem 3, every open set is therefore $Ih$-open, and the comparison with ordinary continuity follows.

Theorem 8.

Every continuous function $f:(X,\tau ,I)\to (Y,\sigma )$ is $Ih$-continuous. The converse is not true in general.

Proof. 

If f is continuous and V is open in Y, then $f^{-1}\left(V\right)$ is open in X, hence $Ih$-open by the remark above, and so f is $Ih$-continuous.

For the converse, take the space of Example 1, namely $X=\{a,b,c,d\}$ with $\tau =\{\varnothing ,X,\{d\},\{a,c\},\{a,c,d\left\}\right\}$ and $I=\{\varnothing ,\{a\left\}\right\}$, in which $\left\{a\right\}$ is $Ih$-open but not open. Let $Y=\{0,1\}$ with $\sigma =\{\varnothing ,Y,\{0\left\}\right\}$, and define f by $f\left(a\right)=0$ and $f\left(b\right)=f\left(c\right)=f\left(d\right)=1$. The preimages of the open sets of Y are $f^{-1}(\varnothing )=\varnothing$, $f^{-1}\left(\left\{0\right\}\right)=\left\{a\right\}$ and $f^{-1}\left(Y\right)=X$, all of which are $Ih$-open, so f is $Ih$-continuous. Since $f^{-1}\left(\left\{0\right\}\right)=\left\{a\right\}$ is not open in X, the function f is not continuous. □

Proposition 3.

Let $f:(X,\tau ,I)\to (Y,\sigma )$ be $Ih$-continuous and let $g:(Y,\sigma )\to (Z,\rho )$ be continuous. Then $g\circ f$ is $Ih$-continuous.

Proof. 

Let W be open in Z. Since g is continuous, $g^{-1}\left(W\right)$ is open in Y; since f is $Ih$-continuous, ${(g\circ f)}^{-1}\left(W\right)=f^{-1}\left(g^{-1}\left(W\right)\right)$ is $Ih$-open in X. Hence $g\circ f$ is $Ih$-continuous. □

The first characterization of an $Ih$-continuous function needs no extra assumption.

Theorem 9.

A function $f:(X,\tau ,I)\to (Y,\sigma )$ is $Ih$-continuous if and only if $f^{-1}\left(F\right)$ is $Ih$-closed for every closed subset F of Y.

Proof. 

A subset of X is $Ih$-open exactly when its complement is $Ih$-closed. If V is open in Y and $F=Y\setminus V$, then $f^{-1}\left(V\right)=X\setminus f^{-1}\left(F\right)$, so $f^{-1}\left(V\right)$ is $Ih$-open if and only if $f^{-1}\left(F\right)$ is $Ih$-closed. As V runs through the open sets of Y, the set F runs through the closed sets, which gives the equivalence. □

For the remaining characterizations we use the $Ih$-interior of Definition 8 together with its dual.

Definition 10.

For $A\subseteq X$, the $Ih$-closure of A is

$${cl}_{Ih}\left(A\right)=X\setminus {int}_{Ih}(X\setminus A).$$

When X is finite or I is closed under arbitrary unions, ${\tau }^{Ih}$ is a topology by Theorem 7, and ${int}_{Ih}$ and ${cl}_{Ih}$ are its interior and closure operators.

Theorem 10.

Suppose that X is finite or I is closed under arbitrary unions. For a function $f:(X,\tau ,I)\to (Y,\sigma )$, the following are equivalent.

(1)

f is $Ih$-continuous.

(2)

$f^{-1}\left(V\right)={int}_{Ih}\left(f^{-1}\left(V\right)\right)$ for every open subset V of Y.

(3)

$f^{-1}\left({int}_Y\left(B\right)\right)\subseteq {int}_{Ih}\left(f^{-1}\left(B\right)\right)$ for every subset B of Y.

(4)

${cl}_{Ih}\left(f^{-1}\left(B\right)\right)\subseteq f^{-1}\left({cl}_Y\left(B\right)\right)$ for every subset B of Y.

(5)

For every $x\in X$ and every open set V with $f\left(x\right)\in V$, there is an $Ih$-open set U with $x\in U\subseteq f^{-1}\left(V\right)$.

Proof. 

Under the stated hypothesis, ${\tau }^{Ih}$ is a topology with interior ${int}_{Ih}$, so a set M is $Ih$-open if and only if ${int}_{Ih}\left(M\right)=M$ by Proposition 2(iii), and ${int}_{Ih}\left(M\right)$ is the largest $Ih$-open set contained in M.

$\left(1\right)\iff \left(2\right)$. For each open V, the set $f^{-1}\left(V\right)$ is $Ih$-open if and only if ${int}_{Ih}\left(f^{-1}\left(V\right)\right)=f^{-1}\left(V\right)$.

$\left(1\right)\Rightarrow \left(3\right)$. Let $B\subseteq Y$. Then ${int}_Y\left(B\right)$ is open, so $f^{-1}\left({int}_Y\left(B\right)\right)$ is $Ih$-open, and it is contained in $f^{-1}\left(B\right)$. Since ${int}_{Ih}\left(f^{-1}\left(B\right)\right)$ is the largest $Ih$-open set inside $f^{-1}\left(B\right)$, we obtain $f^{-1}\left({int}_Y\left(B\right)\right)\subseteq {int}_{Ih}\left(f^{-1}\left(B\right)\right)$.

$\left(3\right)\Rightarrow \left(2\right)$. Apply (3) with an open set V in place of B. Then ${int}_Y\left(V\right)=V$, so $f^{-1}\left(V\right)\subseteq {int}_{Ih}\left(f^{-1}\left(V\right)\right)$. The reverse inclusion holds for every set by Proposition 2(ii), and hence $f^{-1}\left(V\right)={int}_{Ih}\left(f^{-1}\left(V\right)\right)$.

$\left(3\right)\iff \left(4\right)$. Replace B by $Y\setminus B$ in (3). Using ${int}_Y(Y\setminus B)=Y\setminus {cl}_Y\left(B\right)$, the identity ${int}_{Ih}(X\setminus A)=X\setminus {cl}_{Ih}\left(A\right)$, and $f^{-1}(Y\setminus C)=X\setminus f^{-1}\left(C\right)$, the inclusion in (3) becomes $X\setminus f^{-1}\left({cl}_Y\left(B\right)\right)\subseteq X\setminus {cl}_{Ih}\left(f^{-1}\left(B\right)\right)$, which is (4). The steps reverse, so (3) and (4) are equivalent.

$\left(1\right)\Rightarrow \left(5\right)$. If f is $Ih$-continuous and $f\left(x\right)\in V$ with V open, then $U=f^{-1}\left(V\right)$ is $Ih$-open and satisfies $x\in U\subseteq f^{-1}\left(V\right)$.

$\left(5\right)\Rightarrow \left(1\right)$. Let V be open in Y. For each $x\in f^{-1}\left(V\right)$ choose an $Ih$-open set $U_x$ with $x\in U_x\subseteq f^{-1}\left(V\right)$. Then $f^{-1}\left(V\right)={\bigcup }_{x\in f^{-1}\left(V\right)}{U}_x$ is a union of $Ih$-open sets, which is $Ih$-open by Theorem 7. Hence f is $Ih$-continuous. □

We close this section with three further properties of $Ih$-continuity.

For an open subset $X_0$ of X, we write $\tau |X_0=\{U\cap X_0:U\in \tau \}$ for the subspace topology and $I|X_0=\{A\cap X_0:A\in I\}$ for the induced ideal on $X_0$.

Proposition 4.

Let $f:(X,\tau ,I)\to (Y,\sigma )$ be $Ih$-continuous and let $X_0$ be an open subset of X. Then the restriction $f|X_0:(X_0,\tau |X_0,I\left|X_0\right)\to (Y,\sigma )$ is $Ih$-continuous.

Proof. 

Let V be open in Y and put $A=f^{-1}\left(V\right)$, which is $Ih$-open in $(X,\tau ,I)$, then ${\left(f\right|X_0)}^{-1}\left(V\right)=A\cap X_0$. Because $X_0$ is open, the open subsets of $X_0$ are exactly the open subsets of X contained in $X_0$, and ${int}_{X_0}\left(S\right)=int\left(S\right)$ for every $S\subseteq X_0$. Let W be a nonempty proper open subset of $X_0$; then W is also a nonempty proper open subset of X, and $W\subseteq X_0$ gives $(A\cap X_0)\cup W=(A\cup W)\cap X_0$. Since $X_0$ is open,

$${int}_{X_0}\left((A\cap X_0)\cup W\right)=int\left((A\cup W)\cap X_0\right)=int(A\cup W)\cap X_0,$$

and hence

$$(A\cap X_0)\setminus {int}_{X_0}\left((A\cap X_0)\cup W\right)=\left(A\setminus int(A\cup W)\right)\cap X_0.$$

Since A is $Ih$-open, we have $A\setminus int(A\cup W)\in I$, which implies that

$$(A\cap X_0)\setminus {int}_{X_0}\left((A\cap X_0)\cup W\right)=\left(A\setminus int(A\cup W)\right)\cap X_0\in I|X_0.$$

Thus $A\cap X_0$ is $Ih$-open in $(X_0,\tau |X_0,I\left|X_0\right)$, and $f|X_0$ is $Ih$-continuous. □

The next result is the $Ih$-analog of the fact that two continuous functions into a Hausdorff space agree on a closed set.

Theorem 11.

Suppose that X is finite or I is closed under arbitrary unions. If $f,g:(X,\tau ,I)\to (Y,\sigma )$ are $Ih$-continuous and Y is Hausdorff, then the set $\{x\in X:f(x)=g(x\left)\right\}$ is $Ih$-closed.

Proof. 

Under the hypothesis, ${\tau }^{Ih}$ is a topology by Theorem 7. Put $A=\{x\in X:f(x)=g(x\left)\right\}$; we show that $X\setminus A$ is $Ih$-open. Let $x\in X\setminus A$, so $f\left(x\right)\ne g\left(x\right)$. Since Y is Hausdorff, there are disjoint open sets $V_1$ and $V_2$ with $f\left(x\right)\in V_1$ and $g\left(x\right)\in V_2$. As f and g are $Ih$-continuous, $f^{-1}\left(V_1\right)$ and $g^{-1}\left(V_2\right)$ are $Ih$-open, and so is $U_x=f^{-1}\left(V_1\right)\cap g^{-1}\left(V_2\right)$. Then $x\in U_x$, and every $y\in U_x$ satisfies $f\left(y\right)\in V_1$ and $g\left(y\right)\in V_2$, so $f\left(y\right)\ne g\left(y\right)$ and $y\in X\setminus A$. Hence $U_x\subseteq X\setminus A$, and $X\setminus A={\bigcup }_{x\in X\setminus A}{U}_x$ is $Ih$-open. Therefore A is $Ih$-closed. □

$Ih$-continuity also passes to products in the expected way.

Theorem 12.

Let $\{(Y_{\lambda },{\sigma }_{\lambda }):\lambda \in \mathrm{\Lambda }\}$ be a family of topological spaces and let $Y=\prod \{Y_{\lambda }:\lambda \in \mathrm{\Lambda }\}$ carry the product topology σ induced by the projections $p_{\lambda }:Y\to Y_{\lambda }$, $\lambda \in \mathrm{\Lambda }$. If $f:(X,\tau ,I)\to (Y,\sigma )$ is $Ih$-continuous, then every composition $p_{\lambda }\circ f$ is $Ih$-continuous. The converse holds if X is finite or if I is closed under arbitrary unions of its members.

Proof. 

Each projection $p_{\lambda }$ is continuous, so if f is $Ih$-continuous, then $p_{\lambda }\circ f$ is $Ih$-continuous by Proposition 3.

Conversely, suppose that every $p_{\lambda }\circ f$ is $Ih$-continuous and that X is finite or I is closed under arbitrary unions of its members, so that ${\tau }^{Ih}$ is a topology on X by Theorem 7. A basic open subset of Y has the form $B={\bigcap }_{k=1}^n{p}_{{\lambda }_k}^{-1}\left(V_{{\lambda }_k}\right)$, where $n\in \mathbb{N}$ and each $V_{{\lambda }_k}$ is open in $Y_{{\lambda }_k}$. Then

$$f^{-1}\left(B\right)={\displaystyle \bigcap _{k=1}^n}{(p_{{\lambda }_k}\circ f)}^{-1}\left(V_{{\lambda }_k}\right)$$

is a finite intersection of $Ih$-open sets and hence is $Ih$-open. An arbitrary open subset of Y is a union of basic open sets, so its preimage under f is the corresponding union of $Ih$-open sets, which is $Ih$-open because ${\tau }^{Ih}$ is a topology. Hence f is $Ih$-continuous. □

Corollary 1.

Let $\{(Y_{\lambda },{\sigma }_{\lambda }):\lambda \in \mathrm{\Lambda }\}$ be a family of topological spaces and let $f_{\lambda }:(X,\tau ,I)\to (Y_{\lambda },{\sigma }_{\lambda })$ be a function for each $\lambda \in \mathrm{\Lambda }$. Let σ be the product topology on $Y=\prod \{Y_{\lambda }:\lambda \in \mathrm{\Lambda }\}$ and let $f:(X,\tau ,I)\to (Y,\sigma )$ be the function defined by $f\left(x\right)={\left(f_{\lambda }\left(x\right)\right)}_{\lambda \in \mathrm{\Lambda }}$ for each $x\in X$. If f is $Ih$-continuous, then $f_{\lambda }$ is $Ih$-continuous for every $\lambda \in \mathrm{\Lambda }$. The converse holds if X is finite or if I is closed under arbitrary unions of its members.

Proof. 

Since $f_{\lambda }=p_{\lambda }\circ f$ for each $\lambda \in \mathrm{\Lambda }$, both assertions follow from Theorem 12. □

Finally, we record how $Ih$-continuity behaves at the two ends of the ideal lattice.

Proposition 5.

Let $f:(X,\tau ,I)\to (Y,\sigma )$ be a function.

(i)

If $I=\mathcal{P}\left(X\right)$, then f is $Ih$-continuous.

(ii)

If $I=\{\varnothing \}$, then f is $Ih$-continuous if and only if $f^{-1}\left(V\right)$ is h-open for every open subset V of Y.

Proof. 

(i) When $I=\mathcal{P}\left(X\right)$, every subset of X is $Ih$-open by Theorem 5, so every preimage $f^{-1}\left(V\right)$ is $Ih$-open. (ii) When $I=\{\varnothing \}$, $Ih$-openness coincides with h-openness by Theorem 4, so $f^{-1}\left(V\right)$ is $Ih$-open if and only if it is h-open. □

7. An Application to Ideal-Based Approximation

The operator ${int}_{Ih}$ and its dual ${cl}_{Ih}$ approximate an arbitrary subset from inside and from outside, in the spirit of rough set theory [15,16]. In this reading the ideal acts as a tolerance: a set counts as negligible exactly when it belongs to I, and the approximation is permitted to overlook such sets. The word negligible is used here in the same spirit as in measure theory, where the sets of measure zero are the negligible ones, or in Baire category, where the meager sets are negligible;the ideal simply lists the sets that the approximation is allowed to ignore, and no topological smallness is implied. We stress that definability and negligibility are not mutually exclusive. By heredity, every member of I is $Ih$-open—for $A\in I$, the defect $A\setminus int(A\cup V)$ is a subset of A and therefore lies in I—so every negligible set is definable, just as every set of measure zero is measurable. What the ideal contributes is the license to disregard such sets when an arbitrary set is approximated from inside and from outside. Throughout this section we assume that X is finite or that I is closed under arbitrary unions of its members, so that ${\tau }^{Ih}$ is a topology by Theorem 7 and ${int}_{Ih}$, ${cl}_{Ih}$ are its interior and closure operators.

Definition 11.

Let $A\subseteq X$. We call ${int}_{Ih}\left(A\right)$ the lower $Ih$-approximation and ${cl}_{Ih}\left(A\right)$ the upper $Ih$-approximation of A, and we call

$${Bd}_{Ih}\left(A\right)={cl}_{Ih}\left(A\right)\setminus {int}_{Ih}\left(A\right)$$

the $Ih$-boundary of A. The set A is exactly $Ih$-approximable if ${Bd}_{Ih}\left(A\right)=\varnothing$.

Proposition 6.

For all $A,B\subseteq X$ the following hold.

(i)

${int}_{Ih}\left(A\right)\subseteq A\subseteq {cl}_{Ih}\left(A\right)$.

(ii)

If $A\subseteq B$, then ${int}_{Ih}\left(A\right)\subseteq {int}_{Ih}\left(B\right)$ and ${cl}_{Ih}\left(A\right)\subseteq {cl}_{Ih}\left(B\right)$.

(iii)

${int}_{Ih}(A\cap B)={int}_{Ih}\left(A\right)\cap {int}_{Ih}\left(B\right)$ and ${cl}_{Ih}(A\cup B)={cl}_{Ih}\left(A\right)\cup {cl}_{Ih}\left(B\right)$.

(iv)

${int}_{Ih}\left({int}_{Ih}\left(A\right)\right)={int}_{Ih}\left(A\right)$ and ${cl}_{Ih}\left({cl}_{Ih}\left(A\right)\right)={cl}_{Ih}\left(A\right)$.

(v)

${int}_{Ih}\left(A\right)=A$ if and only if A is $Ih$-open, and ${cl}_{Ih}\left(A\right)=A$ if and only if A is $Ih$-closed.

(vi)

${Bd}_{Ih}\left(A\right)={Bd}_{Ih}(X\setminus A)$.

Proof. 

(i) The inclusion ${int}_{Ih}\left(A\right)\subseteq A$ is Proposition 2(ii). Applied to $X\setminus A$ it gives ${int}_{Ih}(X\setminus A)\subseteq X\setminus A$, and complementing yields $A\subseteq X\setminus {int}_{Ih}(X\setminus A)={cl}_{Ih}\left(A\right)$.

(ii) Monotonicity of ${int}_{Ih}$ is Proposition 2(i). If $A\subseteq B$, then $X\setminus B\subseteq X\setminus A$, so ${int}_{Ih}(X\setminus B)\subseteq {int}_{Ih}(X\setminus A)$ and hence ${cl}_{Ih}\left(A\right)\subseteq {cl}_{Ih}\left(B\right)$.

(iii) The first identity is Proposition 2(v). For the second, De Morgan’s laws and Proposition 2(v) give

$${cl}_{Ih}(A\cup B)=X\setminus {int}_{Ih}\left((X\setminus A)\cap (X\setminus B)\right)=X\setminus \left({int}_{Ih}(X\setminus A)\cap {int}_{Ih}(X\setminus B)\right)={cl}_{Ih}\left(A\right)\cup {cl}_{Ih}\left(B\right).$$

(iv) The first equality is Proposition 2(iv); applying it to $X\setminus A$ and complementing gives the second.

(v) The first equivalence is Proposition 2(iii). For the second, ${cl}_{Ih}\left(A\right)=A$ means ${int}_{Ih}(X\setminus A)=X\setminus A$, that is, $X\setminus A$ is $Ih$-open, which is exactly the statement that A is $Ih$-closed.

(vi) Since ${cl}_{Ih}(X\setminus A)=X\setminus {int}_{Ih}\left(A\right)$, we have ${Bd}_{Ih}\left(A\right)={cl}_{Ih}\left(A\right)\setminus {int}_{Ih}\left(A\right)={cl}_{Ih}\left(A\right)\cap {cl}_{Ih}(X\setminus A)$, which is unchanged when A is replaced by $X\setminus A$. □

Thus ${int}_{Ih}$ and ${cl}_{Ih}$ satisfy the standard properties of a pair of lower and upper approximation operators, with the $Ih$-open sets as the definable sets and the members of I as the negligible ones. The next result shows that enlarging the ideal, that is, allowing more sets to be negligible, can only sharpen the approximation.

Proposition 7.

Let $I\subseteq J$ be ideals on X, both satisfying the standing hypothesis. In accordance with the convention of Section 2, we write ${int}_{Jh}$, ${cl}_{Jh}$, ${Bd}_{Jh}$ for the operators formed with J. Then for every $A\subseteq X$,

$${int}_{Ih}\left(A\right)\subseteq {int}_{Jh},{cl}_{Jh}\left(A\right)\subseteq {cl}_{Ih}\left(A\right),{Bd}_{Jh}\left(A\right)\subseteq {Bd}_{Ih}\left(A\right).$$

Proof. 

By Lemma 1, ${\tau }^{Ih}\subseteq {\tau }^{Jh}$. Hence every $Ih$-open subset of A is also $Jh$-open, and taking unions gives ${int}_{Ih}\left(A\right)\subseteq {int}_{Jh}\left(A\right)$. Applying this inclusion to $X\setminus A$ and complementing yields ${cl}_{Jh}\left(A\right)\subseteq {cl}_{Ih}\left(A\right)$. Since ${cl}_{Jh}\left(A\right)\subseteq {cl}_{Ih}\left(A\right)$ and ${int}_{Ih}\left(A\right)\subseteq {int}_{Jh}\left(A\right)$,

$$\begin{array}{c}\hfill {Bd}_{Jh}\left(A\right)={cl}_{Jh}\left(A\right)\setminus {int}_{Jh}\left(A\right)\end{array}\subseteq {cl}_{Ih}\left(A\right)\setminus {int}_{Ih}\left(A\right)={Bd}_{Ih}\left(A\right).$$

□

Example 8.

Take the space of Example 1, with target set $A=\{a,b,c\}$. Table 2 records the two approximations and the boundary of A for the ideal $I=\{\varnothing ,\{a\left\}\right\}$ and for the larger ideal $J=\mathcal{P}\left(\right\{a,b\left\}\right)$.

Table 2. Lower and upper $Ih$-approximations of $A=\{a,b,c\}$ in the space of Example 1, for two nested ideals $I\subseteq J$.

Ideal	Lower Approximation	Upper Approximation	Boundary
$I=\{\varnothing ,\{a\left\}\right\}$	$\{a,c\}$	$\{a,b,c\}$	$\left\{b\right\}$
$J=\mathcal{P}\left(\right\{a,b\left\}\right)$	$\{a,b,c\}$	$\{a,b,c\}$	∅

Under I the point b lies in the boundary, so A is only approximately captured. Passing to J makes more sets $Ih$-open, the boundary collapses to ∅, and A becomes exactly $Ih$-approximable. This is the behavior predicted by the previous proposition: as the tolerance encoded by the ideal grows, the boundary can only shrink.

8. Conclusions and Future Work

This paper studied $Ih$-open sets, a second way of idealizing the h-open sets of an ideal topological space. The construction sits opposite to the earlier class of $hI$-open sets: where $hI$-openness enlarges the auxiliary open set through the local-function closure, $Ih$-openness leaves that set alone and instead tolerates a defect that is negligible with respect to the ideal. Both idealizations contain the h-open sets, so every h-open set is at once $hI$-open and $Ih$-open; past this common core, however, the two notions are independent. At the trivial ideal $\{\varnothing \}$ the defect is forced to vanish, and $Ih$-openness reduces to h-openness. Following the two classes along the ideal lattice made their duality precise: ${\tau }^{Ih}$ can only widen and ${\tau }^{hI}$ can only narrow as the ideal grows, so $hI$-openness is the broader notion near $\{\varnothing \}$ and $Ih$-openness the broader one near the full ideal, with the inclusion between them reversed at the two ends.

The family ${\tau }^{Ih}$ itself need not be a topology. It does become one as soon as X is finite or the ideal is closed under arbitrary unions of its members, and in that case the induced operator ${int}_{Ih}$ is precisely the interior operator of the topology ${\tau }^{Ih}$.

Replacing the ordinary interior operator by the generalized interior operators $\alpha \mathrm{int}$, pint, sint, and $\beta \mathrm{int}$ produced the defect-based variants $Ih\alpha$-open, $Ihp$-open, $Ihs$-open, and $Ih\beta$-open. These fit into a single implication diagram whose reverse arrows all fail, as the counterexamples show, so the outward construction reproduces the $hI$-type hierarchy of [5] class for class.

The same openness notion yields a weak form of continuity for functions into a topological space. Every continuous function is $Ih$-continuous but not conversely, and, under the hypothesis that turns ${\tau }^{Ih}$ into a topology, $Ih$-continuity admits the expected description through the $Ih$-interior and $Ih$-closure operators, in parallel with the classical account of continuity. It also restricts to open subspaces, passes to products and to the components of a function into a product, and any two $Ih$-continuous functions into a Hausdorff space agree on an $Ih$-closed set.

Under the same hypothesis, ${int}_{Ih}$ and ${cl}_{Ih}$ serve as lower and upper approximation operators in the sense of rough set theory, with the ideal acting as a tolerance for negligible sets. The associated boundary can only shrink as the ideal grows, so that a coarser notion of negligibility yields a sharper approximation.

A natural next step is a fuzzy counterpart of this theory. Motivated by the recent study of fuzzy $hI$-open sets and the associated fuzzy $hI$-topology [17], one may introduce fuzzy $Ih$-open sets by allowing the failure of the defining inclusion to lie in a fuzzy ideal. One can then ask whether the resulting class generates a fuzzy topology, and compare the fuzzy $Ih$-topology with the fuzzy $hI$-topology. Further topics include fuzzy $Ih$-continuity, fuzzy $Ih$-irresoluteness, separation axioms, and subspace constructions in fuzzy ideal topological spaces.

Beyond the fuzzy setting, three further directions seem promising. First, h-openness itself has natural analogs outside ordinary topologies; in particular, minimal open sets and h-open sets have recently been investigated in generalized topological spaces [2], and it is natural to ask which of the results of the present paper survive when $\tau$ is replaced by a generalized topology. Secondly, the ideal I could be replaced by a hereditary class, that is, a family of sets closed under subsets but not necessarily under finite unions. Heredity alone already drives several of our arguments (for instance Theorem 3 and the implication diagram of Section 4), whereas finite additivity is what makes ${\tau }^{Ih}$ closed under finite intersections in Theorem 7; separating the two assumptions would show precisely how much of the theory each of them supports. Thirdly, several proofs in Section 7, notably those of Proposition 6, rest on the De Morgan laws relating ${int}_{Ih}$ and ${cl}_{Ih}$. In frameworks such as soft set theory, where union, intersection and complement are sometimes defined in a manner that does not validate the De Morgan laws, the two approximation operators would no longer determine one another, and the lower and upper approximations would have to be developed independently. Investigating the role played by the De Morgan laws, both in general topology and for h-type idealizations, appears to be a worthwhile line of inquiry.