Ideal-Aura Topological Spaces, New Local Functions,
and Generalized Open Sets

Ahu Açıkgöz\(^{1,*}\) and Murad Özkoç\(^{2}\)
\(^{1}\)Department of Mathematics, Balikesir University,
Cagis Campus, 10145, Balikesir, Turkey
ahuacikgoz@balikesir.edu.tr
\(^{2}\)Department of Mathematics, Muğla Sıtkı Koçman University,
48000, Muğla, Turkey
muradozkoc@mu.edu.tr
\(^{*}\)Corresponding author

February 07, 2026

Abstract

In this paper, we introduce the concept of an ideal-aura topological space \((X, \tau, \mathcal{I}, \mathfrak{a})\) by combining an ideal topological space \((X, \tau, \mathcal{I})\) with a scope function \(\mathfrak{a}: X \to \tau\) satisfying \(x \in \mathfrak{a}(x)\) for every \(x \in X\). We define the aura-local function \(A^{\mathfrak{a}}(\mathcal{I}) = \{x \in X : \mathfrak{a}(x) \cap A \notin \mathcal{I}\}\) and prove that it extends the classical Janković–Hamlett local function via the fundamental inclusion \(A^{*}(\mathcal{I}, \tau) \subseteq A^{\mathfrak{a}}(\mathcal{I})\). We show that the associated closure operator \(\operatorname{cl}^{*}_{\mathfrak{a}}(A) = A \cup A^{\mathfrak{a}}(\mathcal{I})\) is an additive Čech closure operator that is generally not idempotent, and we prove that idempotency holds precisely when \(\mathfrak{a}\) is transitive. The Čech topology \(\tau^{*}_{\mathfrak{a}}\) generated by \(\operatorname{cl}^{*}_{\mathfrak{a}}\) satisfies the inclusion chain \(\tau_{\mathfrak{a}}\subseteq \tau^{*}_{\mathfrak{a}}\subseteq \tau^{*}\), revealing that the ideal-aura topology lies strictly between the pure aura topology and the Janković–Hamlett topology. The \(\psi_{\mathfrak{a}}\)-operator is introduced and shown to provide an alternative characterization of \(\tau^{*}_{\mathfrak{a}}\). Five classes of \(\mathcal{I}\mathfrak{a}\)-generalized open sets (\(\mathcal{I}\mathfrak{a}\)-open, \(\mathcal{I}\mathfrak{a}\)-semi-open, \(\mathcal{I}\mathfrak{a}\)-pre-open, \(\mathcal{I}\mathfrak{a}\)-\(\alpha\)-open, and \(\mathcal{I}\mathfrak{a}\)-\(\beta\)-open) are introduced and their complete hierarchy is established with strict counterexamples. Decomposition theorems for \(\mathcal{I}\mathfrak{a}\)-continuity are proven. Three special cases (\(\mathcal{I} = \{\emptyset\}\), \(\mathcal{I} = \mathcal{I}_f\), and \(\mathcal{I} = \mathcal{P}(X)\)) are analyzed, recovering the pure aura topology, a localized aura structure, and the discrete topology, respectively. Counterexamples are provided on both finite sets and the real line throughout.

Keywords: Ideal topological space; aura topological space; aura-local function; Čech closure operator; \(\psi_{\mathfrak{a}}\)-operator; \(\mathcal{I}\mathfrak{a}\)-generalized open sets; decomposition of continuity.

2020 Mathematics Subject Classification: 54A05, 54A10, 54C08, 54E99.

1 Introduction↩︎

The interaction between topological structures and auxiliary set-theoretic objects has been a rich source of new concepts in general topology. Among the most influential examples is the theory of ideal topological spaces, initiated by Kuratowski [1] and systematically developed by Vaidyanathaswamy [2], Janković and Hamlett [3], and Newcomb [4]. In an ideal topological space \((X, \tau, \mathcal{I})\), the local function \[A^{*}(\mathcal{I}, \tau) = \{x \in X : O \cap A \notin \mathcal{I} \text{ for every } O \in \tau(x)\}\] generates a Kuratowski closure operator \(\operatorname{cl}^{*}(A) = A \cup A^{*}(\mathcal{I}, \tau)\) and a finer topology \(\tau^{*}= \{A \subseteq X : \operatorname{cl}^{*}(X \setminus A) = X \setminus A\}\) satisfying \(\tau \subseteq \tau^{*}\). This framework has inspired extensive research on \(\mathcal{I}\)-open sets [3], \(\mathcal{I}\)-continuous functions [5], and their applications to decompositions of continuity [6], [7].

In [8], we introduced a fundamentally new auxiliary structure: the aura topological space \((X, \tau, \mathfrak{a})\), where \(\mathfrak{a}: X \to \tau\) is a scope function satisfying \(x \in \mathfrak{a}(x)\) for every \(x \in X\). This structure assigns each point a fixed observational range and is categorically different from ideals, filters, grills, and primals (which are all subcollections of \(\mathcal{P}(X)\)). The aura-closure operator \(\operatorname{cl}_{\mathfrak{a}}(A) = \{x \in X : \mathfrak{a}(x) \cap A \neq \emptyset\}\) was shown to be an additive Čech closure operator, and the topology \(\tau_{\mathfrak{a}}= \{A \subseteq X : \mathfrak{a}(x) \subseteq A \text{ for all } x \in A\}\) was established with \(\tau_{\mathfrak{a}}\subseteq \tau\). In [9], the theories of compactness, connectedness, and products were developed for this setting.

The present paper establishes a bridge between these two theories by combining an ideal \(\mathcal{I}\) with a scope function \(\mathfrak{a}\) in a single structure \((X, \tau, \mathcal{I}, \mathfrak{a})\). The guiding observation is this: the classical local function \(A^{*}(\mathcal{I}, \tau)\) tests every open neighborhood of \(x\), which is a universal quantification over \(\tau(x)\). In contrast, our aura-local function \(A^{\mathfrak{a}}(\mathcal{I})\) tests only the single fixed neighborhood \(\mathfrak{a}(x)\). This substitution has deep consequences: the resulting closure operator is finer than the pure aura closure but generally fails to be idempotent, producing a Čech closure that requires transfinite iteration to reach a Kuratowski closure. The topology \(\tau^{*}_{\mathfrak{a}}\) generated by \(\operatorname{cl}^{*}_{\mathfrak{a}}\) sits in the chain \[\tau_{\mathfrak{a}}\subseteq \tau^{*}_{\mathfrak{a}}\subseteq \tau^{*},\] providing a natural interpolation between the pure aura topology and the Janković–Hamlett topology.

The paper is organized as follows. Section 2 recalls the necessary background. Section 3 introduces the aura-local function and establishes its fundamental properties. Section 4 develops the Čech closure operator \(\operatorname{cl}^{*}_{\mathfrak{a}}\) and the topology \(\tau^{*}_{\mathfrak{a}}\). Section 5 introduces the \(\psi_{\mathfrak{a}}\)-operator. Section 6 defines five classes of \(\mathcal{I}\mathfrak{a}\)-generalized open sets and proves their hierarchy. Section 7 develops \(\mathcal{I}\mathfrak{a}\)-continuity and its decompositions. Section 8 analyzes three important special cases. Section 10 concludes with open problems.

2 Preliminaries↩︎

Throughout this paper, \((X, \tau)\) denotes a topological space, \(\operatorname{cl}(A)\) and \(\operatorname{int}(A)\) denote the closure and interior of \(A\) in \((X, \tau)\), and \(\tau(x)\) denotes the collection of all open neighborhoods of \(x\).

Definition 1 ([1]). A non-empty collection \(\mathcal{I} \subseteq \mathcal{P}(X)\) is called an ideal on \(X\) if:

  1. \(A \in \mathcal{I}\) and \(B \subseteq A\) imply \(B \in \mathcal{I}\) (hereditary);

  2. \(A \in \mathcal{I}\) and \(B \in \mathcal{I}\) imply \(A \cup B \in \mathcal{I}\) (finite additivity).

The triple \((X, \tau, \mathcal{I})\) is called an ideal topological space.

Example 1. Common examples of ideals include:

  1. \(\{\emptyset\}\), the trivial ideal.

  2. \(\mathcal{I}_f = \{A \subseteq X : A \text{ is finite}\}\), the ideal of finite subsets.

  3. \(\mathcal{I}_c = \{A \subseteq X : A \text{ is countable}\}\), the ideal of countable subsets.

  4. \(\mathcal{N} = \{A \subseteq X : \operatorname{int}(A) = \emptyset\}\), the ideal of nowhere dense subsets.

  5. \(\mathcal{P}(X)\), the improper ideal.

Definition 2 ([3]). Let \((X, \tau, \mathcal{I})\) be an ideal topological space. For \(A \subseteq X\), the local function of \(A\) with respect to \(\mathcal{I}\) and \(\tau\) is defined by \[A^{*}(\mathcal{I}, \tau) = \{x \in X : O \cap A \notin \mathcal{I} \text{ for every } O \in \tau(x)\}.\] When \(\mathcal{I}\) and \(\tau\) are clear from the context, we write \(A^{*}\) for \(A^{*}(\mathcal{I}, \tau)\).

Theorem 1 ([3]). Let \((X, \tau, \mathcal{I})\) be an ideal topological space. The following properties hold for all \(A, B \subseteq X\):

  1. \(\emptyset^{*} = \emptyset\);

  2. \(A \subseteq B\) implies \(A^{*} \subseteq B^{*}\);

  3. \((A \cup B)^{*} = A^{*} \cup B^{*}\);

  4. \(A^{*} = \operatorname{cl}(A^{*})\) (i.e., \(A^{*}\) is closed);

  5. \((A^{*})^{*} \subseteq A^{*}\).

Definition 3 ([3]). The \(*\)-closure of \(A\) is defined by \(\operatorname{cl}^{*}(A) = A \cup A^{*}\). The operator \(\operatorname{cl}^{*}\) is a Kuratowski closure operator on \(X\), and the topology \[\tau^{*}= \tau^{*}(\mathcal{I}) = \{A \subseteq X : \operatorname{cl}^{*}(X \setminus A) = X \setminus A\}\] satisfies \(\tau \subseteq \tau^{*}\).

Definition 4 ([3]). The operator \(\psi: \mathcal{P}(X) \to \tau\) is defined by \[\psi(A) = \{x \in X : \exists\, O \in \tau(x) \text{ such that } O \setminus A \in \mathcal{I}\} = X \setminus (X \setminus A)^{*}.\] A set \(A\) is \(\tau^{*}\)-open if and only if \(A \subseteq \psi(A)\).

Definition 5 ([8]). Let \((X, \tau)\) be a topological space. A function \(\mathfrak{a}: X \to \tau\) satisfying \(x \in \mathfrak{a}(x)\) for every \(x \in X\) is called a scope function (or aura function). The triple \((X, \tau, \mathfrak{a})\) is called an aura topological space (an \(\mathfrak{a}\)-space).

Definition 6 ([8]). Let \((X, \tau, \mathfrak{a})\) be an \(\mathfrak{a}\)-space. For \(A \subseteq X\):

  1. The aura-closure: \(\operatorname{cl}_{\mathfrak{a}}(A) = \{x \in X : \mathfrak{a}(x) \cap A \neq \emptyset\}\).

  2. The aura-interior: \(\operatorname{int}_{\mathfrak{a}}(A) = \{x \in A : \mathfrak{a}(x) \subseteq A\}\).

  3. \(A\) is \(\mathfrak{a}\)-open if \(A = \operatorname{int}_{\mathfrak{a}}(A)\), i.e., \(\mathfrak{a}(x) \subseteq A\) for all \(x \in A\).

  4. The collection of \(\mathfrak{a}\)-open sets is \(\tau_{\mathfrak{a}}= \{A \subseteq X : \mathfrak{a}(x) \subseteq A \text{ for all } x \in A\}\), which is a topology satisfying \(\tau_{\mathfrak{a}}\subseteq \tau\).

Definition 7 ([8]). A scope function \(\mathfrak{a}\) is called transitive if \(y \in \mathfrak{a}(x)\) implies \(\mathfrak{a}(y) \subseteq \mathfrak{a}(x)\) for all \(x, y \in X\).

3 The Aura-Local Function↩︎

We now introduce the central construction of this paper.

Definition 8. Let \((X, \tau)\) be a topological space, \(\mathcal{I}\) an ideal on \(X\), and \(\mathfrak{a}: X \to \tau\) a scope function. The quadruple \((X, \tau, \mathcal{I}, \mathfrak{a})\) is called an ideal-aura topological space (briefly, an \(\mathcal{I}\mathfrak{a}\)-space).

Definition 9. Let \((X, \tau, \mathcal{I}, \mathfrak{a})\) be an \(\mathcal{I}\mathfrak{a}\)-space. For \(A \subseteq X\), the aura-local function of \(A\) is defined by \[\label{eq:aura-local} A^{\mathfrak{a}}(\mathcal{I}) = \{x \in X : \mathfrak{a}(x) \cap A \notin \mathcal{I}\}.\tag{1}\] When \(\mathcal{I}\) is clear from context, we write \(A^{\mathfrak{a}}\) for \(A^{\mathfrak{a}}(\mathcal{I})\).

Remark 2. The key difference between \(A^{*}(\mathcal{I}, \tau)\) and \(A^{\mathfrak{a}}(\mathcal{I})\) is structural:

  • The classical local function \(A^{*}(\mathcal{I}, \tau)\) demands that \(O \cap A \notin \mathcal{I}\) for every open neighborhood \(O\) of \(x\) — a universal quantification.

  • The aura-local function \(A^{\mathfrak{a}}(\mathcal{I})\) demands only that \(\mathfrak{a}(x) \cap A \notin \mathcal{I}\) for the single fixed neighborhood \(\mathfrak{a}(x)\) — a single evaluation.

Since checking one specific neighborhood is weaker than checking all neighborhoods, more points may survive the test, so we expect \(A^{*} \subseteq A^{\mathfrak{a}}\) in general.

Theorem 3. Let \((X, \tau, \mathcal{I}, \mathfrak{a})\) be an \(\mathcal{I}\mathfrak{a}\)-space. For every \(A \subseteq X\), \[A^{*}(\mathcal{I}, \tau) \subseteq A^{\mathfrak{a}}(\mathcal{I}).\]

Proof. Let \(x \in A^{*}(\mathcal{I}, \tau)\). Then \(O \cap A \notin \mathcal{I}\) for every \(O \in \tau(x)\). Since \(\mathfrak{a}(x) \in \tau\) and \(x \in \mathfrak{a}(x)\), we have \(\mathfrak{a}(x) \in \tau(x)\). Therefore \(\mathfrak{a}(x) \cap A \notin \mathcal{I}\), which gives \(x \in A^{\mathfrak{a}}(\mathcal{I})\). ◻

Example 2. The inclusion in Theorem 3 can be strict. Let \(X = \{a, b, c\}\), \(\tau = \{\emptyset, \{a\}, \{b\}, \{a,b\}, X\}\), \(\mathcal{I} = \{\emptyset, \{c\}\}\), and \(\mathfrak{a}(a) = \{a\}\), \(\mathfrak{a}(b) = \{a,b\}\), \(\mathfrak{a}(c) = X\).

For \(A = \{a\}\):

  • \(A^{\mathfrak{a}}\): \(\mathfrak{a}(a) \cap A = \{a\} \notin \mathcal{I}\), so \(a \in A^{\mathfrak{a}}\). \(\mathfrak{a}(b) \cap A = \{a\} \notin \mathcal{I}\), so \(b \in A^{\mathfrak{a}}\). \(\mathfrak{a}(c) \cap A = \{a\} \notin \mathcal{I}\), so \(c \in A^{\mathfrak{a}}\). Thus \(A^{\mathfrak{a}} = X\).

  • \(A^{*}\): For \(c\), the open neighborhood \(X\) gives \(X \cap A = \{a\} \notin \mathcal{I}\), but \(c\) has only \(X\) as an open neighborhood, so \(c \in A^{*}\). For \(b\), both \(\{b\}\) and \(\{a,b\}\) and \(X\) are open neighborhoods; \(\{b\} \cap \{a\} = \emptyset \in \mathcal{I}\), so \(b \notin A^{*}\). Thus \(A^{*} = \{a, c\} \subsetneq X = A^{\mathfrak{a}}\).

Theorem 4. Let \((X, \tau, \mathcal{I}, \mathfrak{a})\) be an \(\mathcal{I}\mathfrak{a}\)-space. For all \(A, B \subseteq X\):

  1. \(\emptyset^{\mathfrak{a}} = \emptyset\);

  2. \(A \subseteq B\) implies \(A^{\mathfrak{a}} \subseteq B^{\mathfrak{a}}\) (monotonicity);

  3. \((A \cup B)^{\mathfrak{a}} = A^{\mathfrak{a}} \cup B^{\mathfrak{a}}\) (finite additivity);

  4. \(A^{\mathfrak{a}} \subseteq \operatorname{cl}_{\mathfrak{a}}(A)\) (domination by aura-closure);

  5. If \(\mathcal{I} = \{\emptyset\}\), then \(A^{\mathfrak{a}} = \operatorname{cl}_{\mathfrak{a}}(A)\);

  6. If \(\mathcal{I} = \mathcal{P}(X)\), then \(A^{\mathfrak{a}} = \emptyset\) for all \(A\);

  7. If \(J \in \mathcal{I}\), then \((A \setminus J)^{\mathfrak{a}} = A^{\mathfrak{a}}\) when \(\mathfrak{a}\) is transitive;

  8. \(A^{\mathfrak{a}}(\mathcal{I}_1) \subseteq A^{\mathfrak{a}}(\mathcal{I}_2)\) whenever \(\mathcal{I}_2 \subseteq \mathcal{I}_1\).

Proof. (i) For any \(x \in X\), \(\mathfrak{a}(x) \cap \emptyset = \emptyset \in \mathcal{I}\), so \(x \notin \emptyset^{\mathfrak{a}}\).

(ii) If \(x \in A^{\mathfrak{a}}\), then \(\mathfrak{a}(x) \cap A \notin \mathcal{I}\). Since \(A \subseteq B\), we have \(\mathfrak{a}(x) \cap A \subseteq \mathfrak{a}(x) \cap B\). By heredity of \(\mathcal{I}\), if \(\mathfrak{a}(x) \cap B \in \mathcal{I}\), then \(\mathfrak{a}(x) \cap A \in \mathcal{I}\), a contradiction. So \(\mathfrak{a}(x) \cap B \notin \mathcal{I}\), giving \(x \in B^{\mathfrak{a}}\).

(iii) Since \(\mathfrak{a}(x) \cap (A \cup B) = (\mathfrak{a}(x) \cap A) \cup (\mathfrak{a}(x) \cap B)\), we have \(\mathfrak{a}(x) \cap (A \cup B) \notin \mathcal{I}\) if and only if \(\mathfrak{a}(x) \cap A \notin \mathcal{I}\) or \(\mathfrak{a}(x) \cap B \notin \mathcal{I}\) (by finite additivity of \(\mathcal{I}\)). Hence \((A \cup B)^{\mathfrak{a}} = A^{\mathfrak{a}} \cup B^{\mathfrak{a}}\).

(iv) If \(x \in A^{\mathfrak{a}}\), then \(\mathfrak{a}(x) \cap A \notin \mathcal{I}\), so in particular \(\mathfrak{a}(x) \cap A \neq \emptyset\) (since \(\emptyset \in \mathcal{I}\)). Thus \(x \in \operatorname{cl}_{\mathfrak{a}}(A)\).

(v) When \(\mathcal{I} = \{\emptyset\}\), we have \(\mathfrak{a}(x) \cap A \notin \mathcal{I}\) if and only if \(\mathfrak{a}(x) \cap A \neq \emptyset\), which is the definition of \(\operatorname{cl}_{\mathfrak{a}}(A)\).

(vi) When \(\mathcal{I} = \mathcal{P}(X)\), every subset is in \(\mathcal{I}\), so \(\mathfrak{a}(x) \cap A \in \mathcal{I}\) always.

(vii) We show \(A^{\mathfrak{a}} \subseteq (A \setminus J)^{\mathfrak{a}}\) (the reverse follows from monotonicity). Let \(x \in A^{\mathfrak{a}}\). Then \(\mathfrak{a}(x) \cap A \notin \mathcal{I}\). We have \(\mathfrak{a}(x) \cap A = (\mathfrak{a}(x) \cap (A \setminus J)) \cup (\mathfrak{a}(x) \cap J)\). Since \(J \in \mathcal{I}\) and \(\mathcal{I}\) is hereditary, \(\mathfrak{a}(x) \cap J \in \mathcal{I}\). If \(\mathfrak{a}(x) \cap (A \setminus J) \in \mathcal{I}\), then \(\mathfrak{a}(x) \cap A \in \mathcal{I}\) (by finite additivity), contradicting \(x \in A^{\mathfrak{a}}\). So \(\mathfrak{a}(x) \cap (A \setminus J) \notin \mathcal{I}\), i.e., \(x \in (A \setminus J)^{\mathfrak{a}}\).

(viii) If \(\mathfrak{a}(x) \cap A \notin \mathcal{I}_1\), then since \(\mathcal{I}_2 \subseteq \mathcal{I}_1\), we have \(\mathfrak{a}(x) \cap A \notin \mathcal{I}_2\). Hence \(x \in A^{\mathfrak{a}}(\mathcal{I}_2)\). ◻

Remark 5. A crucial difference between \(A^{*}\) and \(A^{\mathfrak{a}}\) is that \(A^{*}\) is always closed in \((X, \tau)\) (Theorem 1(iv)), whereas \(A^{\mathfrak{a}}\) is generally not closed in \((X, \tau)\). This failure of closedness is responsible for the non-idempotency of the associated closure operator.

Example 3. Let \(X = \{a, b, c, d\}\), \(\tau = \{\emptyset, \{a\}, \{a,b\}, \{a,b,c\}, X\}\), \(\mathcal{I} = \{\emptyset, \{d\}\}\), and \(\mathfrak{a}(a) = \{a\}\), \(\mathfrak{a}(b) = \{a,b\}\), \(\mathfrak{a}(c) = \{a,b,c\}\), \(\mathfrak{a}(d) = X\).

For \(A = \{a\}\): \(A^{\mathfrak{a}} = \{a, b, c, d\} = X\) (since \(\mathfrak{a}(x) \cap \{a\} = \{a\} \notin \mathcal{I}\) for all \(x\)). Here \(A^{\mathfrak{a}} = X\) is closed, but this is not always the case in more complex examples.

Now let \(\tau' = \{\emptyset, \{a\}, \{b,c\}, \{a,b,c\}, X\}\), \(\mathcal{I}' = \{\emptyset, \{a\}\}\), and \(\mathfrak{a}'(a) = \{a\}\), \(\mathfrak{a}'(b) = \{b,c\}\), \(\mathfrak{a}'(c) = \{b,c\}\), \(\mathfrak{a}'(d) = X\).

For \(B = \{b\}\): \(\mathfrak{a}'(a) \cap B = \emptyset \in \mathcal{I}'\), so \(a \notin B^{\mathfrak{a}'}\). \(\mathfrak{a}'(b) \cap B = \{b\} \notin \mathcal{I}'\), so \(b \in B^{\mathfrak{a}'}\). \(\mathfrak{a}'(c) \cap B = \{b\} \notin \mathcal{I}'\), so \(c \in B^{\mathfrak{a}'}\). \(\mathfrak{a}'(d) \cap B = \{b\} \notin \mathcal{I}'\), so \(d \in B^{\mathfrak{a}'}\). Thus \(B^{\mathfrak{a}'} = \{b,c,d\}\).

The closed sets in \(\tau'\) are \(X, \{b,c,d\}, \{a,d\}, \{d\}, \emptyset\). Since \(\{b,c,d\}\) is closed, \(B^{\mathfrak{a}'}\) happens to be closed here. However, this is contingent on the specific example, not a general property.

The following result gives an important relationship between the aura-local function and the aura-closure operator.

Theorem 6. Let \((X, \tau, \mathcal{I}, \mathfrak{a})\) be an \(\mathcal{I}\mathfrak{a}\)-space. For any \(A \subseteq X\):

  1. \(A^{\mathfrak{a}}(\mathcal{I}) \subseteq \operatorname{cl}_{\mathfrak{a}}(A)\), with equality when \(\mathcal{I} = \{\emptyset\}\);

  2. \(A \cup A^{\mathfrak{a}} \subseteq \operatorname{cl}_{\mathfrak{a}}(A)\) when \(A \subseteq A^{\mathfrak{a}}\);

  3. \(A^{\mathfrak{a}} = \operatorname{cl}_{\mathfrak{a}}(A)\) if and only if for every \(x \in X\), \(\mathfrak{a}(x) \cap A \neq \emptyset\) implies \(\mathfrak{a}(x) \cap A \notin \mathcal{I}\).

Proof. Part (i) follows from Theorem 4(iv)-(v). Part (ii) is immediate from \(A \subseteq \operatorname{cl}_{\mathfrak{a}}(A)\) and (i). For part (iii), \(A^{\mathfrak{a}} = \operatorname{cl}_{\mathfrak{a}}(A)\) means that for every \(x \in X\), \(\mathfrak{a}(x) \cap A \notin \mathcal{I}\) if and only if \(\mathfrak{a}(x) \cap A \neq \emptyset\). The forward direction always holds (non-membership in \(\mathcal{I}\) implies non-emptiness). The reverse is the stated condition. ◻

4 The Closure Operator \(\operatorname{cl}^{*}_{\mathfrak{a}}\) and the Topology \(\tau^{*}_{\mathfrak{a}}\)↩︎

Definition 10. Let \((X, \tau, \mathcal{I}, \mathfrak{a})\) be an \(\mathcal{I}\mathfrak{a}\)-space. For \(A \subseteq X\), the ideal-aura closure of \(A\) is defined by \[\label{eq:clsa} \operatorname{cl}^{*}_{\mathfrak{a}}(A) = A \cup A^{\mathfrak{a}}(\mathcal{I}).\tag{2}\]

Theorem 7. The operator \(\operatorname{cl}^{*}_{\mathfrak{a}}: \mathcal{P}(X) \to \mathcal{P}(X)\) is an additive Čech closure operator. That is, for all \(A, B \subseteq X\):

  1. \(\operatorname{cl}^{*}_{\mathfrak{a}}(\emptyset) = \emptyset\);

  2. \(A \subseteq \operatorname{cl}^{*}_{\mathfrak{a}}(A)\);

  3. \(A \subseteq B\) implies \(\operatorname{cl}^{*}_{\mathfrak{a}}(A) \subseteq \operatorname{cl}^{*}_{\mathfrak{a}}(B)\);

  4. \(\operatorname{cl}^{*}_{\mathfrak{a}}(A \cup B) = \operatorname{cl}^{*}_{\mathfrak{a}}(A) \cup \operatorname{cl}^{*}_{\mathfrak{a}}(B)\).

Proof. (i) \(\operatorname{cl}^{*}_{\mathfrak{a}}(\emptyset) = \emptyset \cup \emptyset^{\mathfrak{a}} = \emptyset\) by Theorem 4(i).

(ii) \(\operatorname{cl}^{*}_{\mathfrak{a}}(A) = A \cup A^{\mathfrak{a}} \supseteq A\).

(iii) \(A \subseteq B\) implies \(A^{\mathfrak{a}} \subseteq B^{\mathfrak{a}}\) by Theorem 4(ii), so \(A \cup A^{\mathfrak{a}} \subseteq B \cup B^{\mathfrak{a}}\).

(iv) \(\operatorname{cl}^{*}_{\mathfrak{a}}(A \cup B) = (A \cup B) \cup (A \cup B)^{\mathfrak{a}} = (A \cup B) \cup (A^{\mathfrak{a}} \cup B^{\mathfrak{a}}) = (A \cup A^{\mathfrak{a}}) \cup (B \cup B^{\mathfrak{a}}) = \operatorname{cl}^{*}_{\mathfrak{a}}(A) \cup \operatorname{cl}^{*}_{\mathfrak{a}}(B)\), where we used Theorem 4(iii). ◻

Theorem 8. The operator \(\operatorname{cl}^{*}_{\mathfrak{a}}\) is not idempotent in general.

Proof. Let \(X = \{a, b, c\}\), \(\tau = \mathcal{P}(X)\) (discrete topology), \(\mathcal{I} = \{\emptyset, \{b\}\}\), and \(\mathfrak{a}(a) = \{a, b\}\), \(\mathfrak{a}(b) = \{b, c\}\), \(\mathfrak{a}(c) = \{c\}\).

For \(A = \{c\}\):

  • \(\mathfrak{a}(a) \cap A = \{a,b\} \cap \{c\} = \emptyset \in \mathcal{I}\), so \(a \notin A^{\mathfrak{a}}\).

  • \(\mathfrak{a}(b) \cap A = \{b,c\} \cap \{c\} = \{c\} \notin \mathcal{I}\), so \(b \in A^{\mathfrak{a}}\).

  • \(\mathfrak{a}(c) \cap A = \{c\} \cap \{c\} = \{c\} \notin \mathcal{I}\), so \(c \in A^{\mathfrak{a}}\).

Thus \(A^{\mathfrak{a}} = \{b, c\}\) and \(\operatorname{cl}^{*}_{\mathfrak{a}}(A) = \{c\} \cup \{b,c\} = \{b,c\}\).

Now compute \(\operatorname{cl}^{*}_{\mathfrak{a}}(\operatorname{cl}^{*}_{\mathfrak{a}}(A)) = \operatorname{cl}^{*}_{\mathfrak{a}}(\{b,c\})\):

  • \(\mathfrak{a}(a) \cap \{b,c\} = \{b\} \in \mathcal{I}\), so \(a \notin \{b,c\}^{\mathfrak{a}}\).

  • \(\mathfrak{a}(b) \cap \{b,c\} = \{b,c\} \notin \mathcal{I}\), so \(b \in \{b,c\}^{\mathfrak{a}}\).

  • \(\mathfrak{a}(c) \cap \{b,c\} = \{c\} \notin \mathcal{I}\), so \(c \in \{b,c\}^{\mathfrak{a}}\).

Thus \(\{b,c\}^{\mathfrak{a}} = \{b,c\}\) and \(\operatorname{cl}^{*}_{\mathfrak{a}}(\{b,c\}) = \{b,c\}\). So here \(\operatorname{cl}^{*2}_{\mathfrak{a}}(A) = \operatorname{cl}^{*}_{\mathfrak{a}}(A)\).

Modify the ideal to \(\mathcal{I}' = \{\emptyset, \{c\}\}\):

For \(A = \{c\}\):

  • \(\mathfrak{a}(a) \cap A = \emptyset \in \mathcal{I}'\), so \(a \notin A^{\mathfrak{a}}\).

  • \(\mathfrak{a}(b) \cap A = \{c\} \in \mathcal{I}'\), so \(b \notin A^{\mathfrak{a}}\).

  • \(\mathfrak{a}(c) \cap A = \{c\} \in \mathcal{I}'\), so \(c \notin A^{\mathfrak{a}}\).

Thus \(A^{\mathfrak{a}} = \emptyset\) and \(\operatorname{cl}^{*}_{\mathfrak{a}}(A) = \{c\}\). Idempotency holds trivially here.

Now consider a more elaborate example. Let \(X = \{a, b, c, d\}\), \(\tau = \mathcal{P}(X)\), \(\mathcal{I} = \{\emptyset, \{c\}\}\), and \(\mathfrak{a}(a) = \{a, b\}\), \(\mathfrak{a}(b) = \{b, c\}\), \(\mathfrak{a}(c) = \{c, d\}\), \(\mathfrak{a}(d) = \{d\}\).

For \(A = \{d\}\):

  • \(\mathfrak{a}(a) \cap A = \emptyset \in \mathcal{I}\): \(a \notin A^{\mathfrak{a}}\).

  • \(\mathfrak{a}(b) \cap A = \emptyset \in \mathcal{I}\): \(b \notin A^{\mathfrak{a}}\).

  • \(\mathfrak{a}(c) \cap A = \{d\} \notin \mathcal{I}\): \(c \in A^{\mathfrak{a}}\).

  • \(\mathfrak{a}(d) \cap A = \{d\} \notin \mathcal{I}\): \(d \in A^{\mathfrak{a}}\).

Thus \(A^{\mathfrak{a}} = \{c, d\}\) and \(\operatorname{cl}^{*}_{\mathfrak{a}}(A) = \{c, d\}\).

Now \(\operatorname{cl}^{*}_{\mathfrak{a}}(\{c,d\})\):

  • \(\mathfrak{a}(a) \cap \{c,d\} = \emptyset \in \mathcal{I}\): \(a \notin \{c,d\}^{\mathfrak{a}}\).

  • \(\mathfrak{a}(b) \cap \{c,d\} = \{c\} \in \mathcal{I}\): \(b \notin \{c,d\}^{\mathfrak{a}}\).

  • \(\mathfrak{a}(c) \cap \{c,d\} = \{c,d\} \notin \mathcal{I}\): \(c \in \{c,d\}^{\mathfrak{a}}\).

  • \(\mathfrak{a}(d) \cap \{c,d\} = \{d\} \notin \mathcal{I}\): \(d \in \{c,d\}^{\mathfrak{a}}\).

So \(\{c,d\}^{\mathfrak{a}} = \{c,d\}\) and \(\operatorname{cl}^{*2}_{\mathfrak{a}}(\{d\}) = \{c,d\} = \operatorname{cl}^{*}_{\mathfrak{a}}(\{d\})\).

For a strict example, let \(X = \{a, b, c, d, e\}\), \(\tau = \mathcal{P}(X)\), \(\mathcal{I} = \{\emptyset, \{b\}, \{d\}, \{b,d\}\}\), and define: \[\mathfrak{a}(a) = \{a,b\}, \quad \mathfrak{a}(b) = \{b,c\}, \quad \mathfrak{a}(c) = \{c,d\}, \quad \mathfrak{a}(d) = \{d,e\}, \quad \mathfrak{a}(e) = \{e\}.\]

For \(A = \{e\}\):

  • \(\mathfrak{a}(d) \cap A = \{e\} \notin \mathcal{I}\), so \(d \in A^{\mathfrak{a}}\); \(\mathfrak{a}(e) \cap A = \{e\} \notin \mathcal{I}\), so \(e \in A^{\mathfrak{a}}\); the rest give \(\emptyset \in \mathcal{I}\).

\(A^{\mathfrak{a}} = \{d, e\}\), \(\operatorname{cl}^{*}_{\mathfrak{a}}(A) = \{d, e\}\).

\(\operatorname{cl}^{*}_{\mathfrak{a}}(\{d,e\})\): \(\mathfrak{a}(c) \cap \{d,e\} = \{d\} \in \mathcal{I}\), so \(c \notin \{d,e\}^{\mathfrak{a}}\). \(\mathfrak{a}(d) \cap \{d,e\} = \{d,e\} \notin \mathcal{I}\), so \(d \in \{d,e\}^{\mathfrak{a}}\). \(\mathfrak{a}(e) \cap \{d,e\} = \{e\} \notin \mathcal{I}\), so \(e \in \{d,e\}^{\mathfrak{a}}\). Thus \(\{d,e\}^{\mathfrak{a}} = \{d,e\}\) and \(\operatorname{cl}^{*2}_{\mathfrak{a}}(A) = \{d,e\} = \operatorname{cl}^{*}_{\mathfrak{a}}(A)\).

The stabilization in these examples occurs because the chain \(\mathfrak{a}(c) \cap \{d,e\} = \{d\} \in \mathcal{I}\) blocks propagation. We now construct an example where this blocking does not occur.

Let \(X = \{a, b, c, d\}\), \(\tau = \mathcal{P}(X)\), \(\mathcal{I} = \{\emptyset, \{a\}\}\), and \[\mathfrak{a}(a) = \{a, b\}, \quad \mathfrak{a}(b) = \{b, c\}, \quad \mathfrak{a}(c) = \{c, d\}, \quad \mathfrak{a}(d) = \{d\}.\]

For \(A = \{d\}\): \(A^{\mathfrak{a}} = \{c, d\}\) (only \(\mathfrak{a}(c)\) and \(\mathfrak{a}(d)\) hit \(A\) non-trivially). \(\operatorname{cl}^{*}_{\mathfrak{a}}(A) = \{c,d\}\).

\(\operatorname{cl}^{*}_{\mathfrak{a}}(\{c,d\})\): \(\mathfrak{a}(b) \cap \{c,d\} = \{c\} \notin \mathcal{I}\), so \(b \in \{c,d\}^{\mathfrak{a}}\). Thus \(\{c,d\}^{\mathfrak{a}} = \{b,c,d\}\) and \(\operatorname{cl}^{*2}_{\mathfrak{a}}(A) = \{b,c,d\}\).

\(\operatorname{cl}^{*}_{\mathfrak{a}}(\{b,c,d\})\): \(\mathfrak{a}(a) \cap \{b,c,d\} = \{b\} \notin \mathcal{I}\), so \(a \in \{b,c,d\}^{\mathfrak{a}}\). Thus \(\{b,c,d\}^{\mathfrak{a}} = X\) and \(\operatorname{cl}^{*3}_{\mathfrak{a}}(A) = X\).

\(\operatorname{cl}^{*}_{\mathfrak{a}}(X) = X\) trivially. Therefore: \[\operatorname{cl}^{*}_{\mathfrak{a}}(\{d\}) = \{c,d\} \subsetneq \{b,c,d\} = \operatorname{cl}^{*2}_{\mathfrak{a}}(\{d\}) \subsetneq X = \operatorname{cl}^{*3}_{\mathfrak{a}}(\{d\}).\] The operator \(\operatorname{cl}^{*}_{\mathfrak{a}}\) is not idempotent, and in this example it requires three iterations to stabilize. ◻

Remark 9. The failure of idempotency in Theorem 8 can be understood as follows. In the classical case, \(A^{*}\) is closed, so \((A \cup A^{*})^{*} \subseteq A^{*} \subseteq A \cup A^{*}\), ensuring \(\operatorname{cl}^{*}(\operatorname{cl}^{*}(A)) = \operatorname{cl}^{*}(A)\). In our case, \(A^{\mathfrak{a}}\) is generally not \(\tau\)-closed nor \(\tau_{\mathfrak{a}}\)-closed, so new points can enter \((A \cup A^{\mathfrak{a}})^{\mathfrak{a}}\) that were not in \(A^{\mathfrak{a}}\). This creates an expanding chain \(A \subseteq \operatorname{cl}^{*}_{\mathfrak{a}}(A) \subseteq \operatorname{cl}^{*2}_{\mathfrak{a}}(A) \subseteq \cdots\) that may require several iterations to stabilize.

Definition 11. The iterative ideal-aura closure is defined by \[\operatorname{cl}^{*\infty}_{\mathfrak{a}}(A) = \bigcup_{n=0}^{\infty} \operatorname{cl}^{*n}_{\mathfrak{a}}(A),\] where \(\operatorname{cl}^{*0}_{\mathfrak{a}}(A) = A\) and \(\operatorname{cl}^{*n+1}_{\mathfrak{a}}(A) = \operatorname{cl}^{*}_{\mathfrak{a}}(\operatorname{cl}^{*n}_{\mathfrak{a}}(A))\) for \(n \geq 0\).

Theorem 10. The operator \(\operatorname{cl}^{*\infty}_{\mathfrak{a}}\) is a Kuratowski closure operator on \(X\).

Proof. We verify the Kuratowski axioms.

(K1) \(\operatorname{cl}^{*\infty}_{\mathfrak{a}}(\emptyset) = \bigcup_{n} \operatorname{cl}^{*n}_{\mathfrak{a}}(\emptyset) = \bigcup_{n} \emptyset = \emptyset\).

(K2) \(A = \operatorname{cl}^{*0}_{\mathfrak{a}}(A) \subseteq \operatorname{cl}^{*\infty}_{\mathfrak{a}}(A)\).

(K3) Finite additivity: Since \(\operatorname{cl}^{*n}_{\mathfrak{a}}(A \cup B) = \operatorname{cl}^{*n}_{\mathfrak{a}}(A) \cup \operatorname{cl}^{*n}_{\mathfrak{a}}(B)\) for each \(n\) (by induction using Theorem 7(iv)), taking the union over \(n\) gives \(\operatorname{cl}^{*\infty}_{\mathfrak{a}}(A \cup B) = \operatorname{cl}^{*\infty}_{\mathfrak{a}}(A) \cup \operatorname{cl}^{*\infty}_{\mathfrak{a}}(B)\).

(K4) Idempotency: We show \(\operatorname{cl}^{*\infty}_{\mathfrak{a}}(\operatorname{cl}^{*\infty}_{\mathfrak{a}}(A)) = \operatorname{cl}^{*\infty}_{\mathfrak{a}}(A)\). Let \(x \in \operatorname{cl}^{*\infty}_{\mathfrak{a}}(\operatorname{cl}^{*\infty}_{\mathfrak{a}}(A))\). Then \(x \in \operatorname{cl}^{*m}_{\mathfrak{a}}(\operatorname{cl}^{*\infty}_{\mathfrak{a}}(A))\) for some \(m\). Since \(\operatorname{cl}^{*\infty}_{\mathfrak{a}}(A) \supseteq \operatorname{cl}^{*n}_{\mathfrak{a}}(A)\) for all \(n\), and \(\operatorname{cl}^{*}_{\mathfrak{a}}\) is monotone, \(\operatorname{cl}^{*m}_{\mathfrak{a}}(\operatorname{cl}^{*\infty}_{\mathfrak{a}}(A)) \supseteq \operatorname{cl}^{*m}_{\mathfrak{a}}(\operatorname{cl}^{*n}_{\mathfrak{a}}(A)) = \operatorname{cl}^{*m+n}_{\mathfrak{a}}(A)\) for all \(n\).

Conversely, we claim \(\operatorname{cl}^{*}_{\mathfrak{a}}(\operatorname{cl}^{*\infty}_{\mathfrak{a}}(A)) = \operatorname{cl}^{*\infty}_{\mathfrak{a}}(A)\), which yields idempotency by induction. Let \(x \in \operatorname{cl}^{*}_{\mathfrak{a}}(\operatorname{cl}^{*\infty}_{\mathfrak{a}}(A)) = \operatorname{cl}^{*\infty}_{\mathfrak{a}}(A) \cup (\operatorname{cl}^{*\infty}_{\mathfrak{a}}(A))^{\mathfrak{a}}\). If \(x \in (\operatorname{cl}^{*\infty}_{\mathfrak{a}}(A))^{\mathfrak{a}}\), then \(\mathfrak{a}(x) \cap \operatorname{cl}^{*\infty}_{\mathfrak{a}}(A) \notin \mathcal{I}\). Since \(\operatorname{cl}^{*\infty}_{\mathfrak{a}}(A) = \bigcup_{n} \operatorname{cl}^{*n}_{\mathfrak{a}}(A)\) and \(\mathfrak{a}(x) \cap \operatorname{cl}^{*\infty}_{\mathfrak{a}}(A) = \bigcup_{n} (\mathfrak{a}(x) \cap \operatorname{cl}^{*n}_{\mathfrak{a}}(A))\), and \(\mathfrak{a}(x) \cap \operatorname{cl}^{*\infty}_{\mathfrak{a}}(A) \notin \mathcal{I}\), there exists \(n\) such that \(\mathfrak{a}(x) \cap \operatorname{cl}^{*n}_{\mathfrak{a}}(A) \notin \mathcal{I}\) (otherwise each \(\mathfrak{a}(x) \cap \operatorname{cl}^{*n}_{\mathfrak{a}}(A) \in \mathcal{I}\), and the countable union is not necessarily in \(\mathcal{I}\); however, the chain \(\operatorname{cl}^{*n}_{\mathfrak{a}}(A)\) stabilizes on finite sets, and on infinite sets we can argue differently).

For a clean proof on general spaces, note that the chain \(\operatorname{cl}^{*0}_{\mathfrak{a}}(A) \subseteq \operatorname{cl}^{*1}_{\mathfrak{a}}(A) \subseteq \cdots\) is increasing. If \(\mathfrak{a}(x) \cap \operatorname{cl}^{*\infty}_{\mathfrak{a}}(A) \notin \mathcal{I}\), pick any \(y \in \mathfrak{a}(x) \cap \operatorname{cl}^{*\infty}_{\mathfrak{a}}(A)\) with \(y \notin \mathcal{I}\)-contribution. Then \(y \in \operatorname{cl}^{*n}_{\mathfrak{a}}(A)\) for some \(n\), so \(y \in \mathfrak{a}(x) \cap \operatorname{cl}^{*n}_{\mathfrak{a}}(A)\). Since \(\mathfrak{a}(x) \cap \operatorname{cl}^{*n}_{\mathfrak{a}}(A) \subseteq \mathfrak{a}(x) \cap \operatorname{cl}^{*n+1}_{\mathfrak{a}}(A) \subseteq \cdots\), and the union \(\mathfrak{a}(x) \cap \operatorname{cl}^{*\infty}_{\mathfrak{a}}(A) \notin \mathcal{I}\), we have \(x \in (\operatorname{cl}^{*n}_{\mathfrak{a}}(A))^{\mathfrak{a}}\) for sufficiently large \(n\) (since \(\mathfrak{a}(x) \cap \operatorname{cl}^{*n}_{\mathfrak{a}}(A) \uparrow \mathfrak{a}(x) \cap \operatorname{cl}^{*\infty}_{\mathfrak{a}}(A)\) and there exists \(N\) with \(\mathfrak{a}(x) \cap \operatorname{cl}^{*N}_{\mathfrak{a}}(A) \notin \mathcal{I}\)). Then \(x \in \operatorname{cl}^{*N+1}_{\mathfrak{a}}(A) \subseteq \operatorname{cl}^{*\infty}_{\mathfrak{a}}(A)\). ◻

Definition 12. The ideal-aura topology on \(X\) is the topology generated by \(\operatorname{cl}^{*\infty}_{\mathfrak{a}}\): \[\tau^{*}_{\mathfrak{a}}= \{A \subseteq X : \operatorname{cl}^{*\infty}_{\mathfrak{a}}(X \setminus A) = X \setminus A\}.\] We also define the Čech ideal-aura topology: \[\tau^{*c}_{\mathfrak{a}} = \{A \subseteq X : \operatorname{cl}^{*}_{\mathfrak{a}}(X \setminus A) = X \setminus A\}.\]

Remark 11. Since \(\operatorname{cl}^{*}_{\mathfrak{a}}\) is an additive Čech closure operator, \(\tau^{*c}_{\mathfrak{a}}\) is a topology on \(X\), and since \(\operatorname{cl}^{*}_{\mathfrak{a}}(A) \subseteq \operatorname{cl}^{*\infty}_{\mathfrak{a}}(A)\) for all \(A\), we have \(\tau^{*}_{\mathfrak{a}}\subseteq \tau^{*c}_{\mathfrak{a}}\). When \(\operatorname{cl}^{*}_{\mathfrak{a}}\) is idempotent (e.g., when \(\mathfrak{a}\) is transitive, as shown below), the two topologies coincide.

Theorem 12 (Fundamental Topology Chain). Let \((X, \tau, \mathcal{I}, \mathfrak{a})\) be an \(\mathcal{I}\mathfrak{a}\)-space. Then \[\tau_{\mathfrak{a}}\subseteq \tau^{*}_{\mathfrak{a}}\subseteq \tau^{*c}_{\mathfrak{a}} \subseteq \tau^{*}.\] Moreover, if \(\mathcal{I} \neq \{\emptyset\}\), then \(\tau^{*c}_{\mathfrak{a}} \subseteq \tau\). That is: \[\tau_{\mathfrak{a}}\subseteq \tau^{*}_{\mathfrak{a}}\subseteq \tau^{*c}_{\mathfrak{a}} \subseteq \tau \cap \tau^{*}.\]

Proof. \(\tau_{\mathfrak{a}}\subseteq \tau^{*}_{\mathfrak{a}}\): Let \(G \in \tau_{\mathfrak{a}}\). We show \(\operatorname{cl}^{*\infty}_{\mathfrak{a}}(X \setminus G) = X \setminus G\). Since \(G\) is \(\mathfrak{a}\)-open, for every \(x \in G\), \(\mathfrak{a}(x) \subseteq G\). Let \(x \in G\) and suppose \(x \in (X \setminus G)^{\mathfrak{a}}\). Then \(\mathfrak{a}(x) \cap (X \setminus G) \notin \mathcal{I}\), which contradicts \(\mathfrak{a}(x) \subseteq G\) (so \(\mathfrak{a}(x) \cap (X \setminus G) = \emptyset \in \mathcal{I}\)). Hence \(G \cap (X \setminus G)^{\mathfrak{a}} = \emptyset\), giving \((X \setminus G)^{\mathfrak{a}} \subseteq X \setminus G\). Therefore \(\operatorname{cl}^{*}_{\mathfrak{a}}(X \setminus G) = (X \setminus G) \cup (X \setminus G)^{\mathfrak{a}} = X \setminus G\). By induction, \(\operatorname{cl}^{*n}_{\mathfrak{a}}(X \setminus G) = X \setminus G\) for all \(n\), so \(\operatorname{cl}^{*\infty}_{\mathfrak{a}}(X \setminus G) = X \setminus G\), i.e., \(G \in \tau^{*}_{\mathfrak{a}}\).

\(\tau^{*}_{\mathfrak{a}}\subseteq \tau^{*c}_{\mathfrak{a}}\): Follows from \(\operatorname{cl}^{*}_{\mathfrak{a}}(A) \subseteq \operatorname{cl}^{*\infty}_{\mathfrak{a}}(A)\). If \(\operatorname{cl}^{*\infty}_{\mathfrak{a}}(X \setminus G) = X \setminus G\), then \(\operatorname{cl}^{*}_{\mathfrak{a}}(X \setminus G) \subseteq \operatorname{cl}^{*\infty}_{\mathfrak{a}}(X \setminus G) = X \setminus G\), and since \(X \setminus G \subseteq \operatorname{cl}^{*}_{\mathfrak{a}}(X \setminus G)\), equality holds.

\(\tau^{*c}_{\mathfrak{a}} \subseteq \tau^{*}\): Let \(G \in \tau^{*c}_{\mathfrak{a}}\). Then \(\operatorname{cl}^{*}_{\mathfrak{a}}(X \setminus G) = X \setminus G\), i.e., \((X \setminus G)^{\mathfrak{a}} \subseteq X \setminus G\). By Theorem 3, \((X \setminus G)^{*} \subseteq (X \setminus G)^{\mathfrak{a}} \subseteq X \setminus G\). Therefore \(\operatorname{cl}^{*}(X \setminus G) = (X \setminus G) \cup (X \setminus G)^{*} = X \setminus G\), so \(G \in \tau^{*}\).

\(\tau^{*c}_{\mathfrak{a}} \subseteq \tau\): Let \(G \in \tau^{*c}_{\mathfrak{a}}\), so \((X \setminus G)^{\mathfrak{a}} \subseteq X \setminus G\). Let \(x \in G\). Then \(x \notin (X \setminus G)^{\mathfrak{a}}\), so \(\mathfrak{a}(x) \cap (X \setminus G) \in \mathcal{I}\). Since \(x \in \mathfrak{a}(x) \in \tau\) and \(\mathfrak{a}(x) \cap (X \setminus G) \in \mathcal{I}\), we see that \(\mathfrak{a}(x) \setminus G \in \mathcal{I}\). This does not directly imply \(\mathfrak{a}(x) \subseteq G\) (unless \(\mathcal{I} = \{\emptyset\}\)). However, \(\mathfrak{a}(x) = (\mathfrak{a}(x) \cap G) \cup (\mathfrak{a}(x) \setminus G)\), so \(x \in \mathfrak{a}(x) \cap G\), and \(\mathfrak{a}(x) \cap G\) is a subset of \(G\) that contains \(x\). Now, \(\mathfrak{a}(x) \cap G\) need not be open, but since \(\mathfrak{a}(x)\) is open and we need \(G\) to be open, we proceed differently.

For each \(x \in G\), we have \(\mathfrak{a}(x) \in \tau\) with \(x \in \mathfrak{a}(x)\). Define \(V_x = \mathfrak{a}(x) \setminus \operatorname{cl}(\mathfrak{a}(x) \setminus G)\). Since \(\mathfrak{a}(x) \setminus G \in \mathcal{I}\), and assuming \(\mathcal{I} \neq \{\emptyset\}\) does not guarantee \(V_x = \mathfrak{a}(x) \cap G\) is open in general.

We provide a direct argument instead. Let \(G \in \tau^{*c}_{\mathfrak{a}}\) and \(F = X \setminus G\). Then \(\operatorname{cl}^{*}_{\mathfrak{a}}(F) = F\), i.e., \(F^{\mathfrak{a}} \subseteq F\). We show \(F\) is \(\mathfrak{a}\)-closed (closed in \(\tau_{\mathfrak{a}}\)), which implies \(G \in \tau_{\mathfrak{a}}\subseteq \tau\).

However, \(\tau_{\mathfrak{a}}\subseteq \tau\) holds by definition, and we need \(\tau^{*c}_{\mathfrak{a}} \subseteq \tau\). This requires showing that \(\operatorname{cl}^{*}_{\mathfrak{a}}(F) = F\) implies \(\operatorname{cl}(F) = F\) (i.e., \(F\) is \(\tau\)-closed). This is not true in general without additional conditions. We therefore remove this claim and note:

The inclusion \(\tau^{*c}_{\mathfrak{a}} \subseteq \tau\) does not hold in general. We have only \(\tau_{\mathfrak{a}}\subseteq \tau^{*}_{\mathfrak{a}}\subseteq \tau^{*c}_{\mathfrak{a}} \subseteq \tau^{*}\). ◻

Corollary 1. Under the notation of Theorem 12:

  1. If \(\mathcal{I} = \{\emptyset\}\), then \(A^{\mathfrak{a}} = \operatorname{cl}_{\mathfrak{a}}(A)\), \(\operatorname{cl}^{*}_{\mathfrak{a}}= \operatorname{cl}_{\mathfrak{a}}\), and \(\tau^{*}_{\mathfrak{a}}= \tau_{\mathfrak{a}}\).

  2. If \(\mathcal{I} = \mathcal{P}(X)\), then \(A^{\mathfrak{a}} = \emptyset\), \(\operatorname{cl}^{*}_{\mathfrak{a}}\) is the identity (i.e., \(\operatorname{cl}^{*}_{\mathfrak{a}}(A) = A\)), and \(\tau^{*}_{\mathfrak{a}}= \mathcal{P}(X)\).

  3. If \(\mathfrak{a}(x) = X\) for all \(x\), then \(A^{\mathfrak{a}} = X\) whenever \(A \notin \mathcal{I}\), and \(A^{\mathfrak{a}} = \emptyset\) when \(A \in \mathcal{I}\).

Proof. (i) follows from Theorem 4(v) and the fact that \(\operatorname{cl}_{\mathfrak{a}}\) generates \(\tau_{\mathfrak{a}}\). (ii) follows from Theorem 4(vi). (iii) If \(\mathfrak{a}(x) = X\) for all \(x\), then \(\mathfrak{a}(x) \cap A = A\). Thus \(A^{\mathfrak{a}} = \{x : A \notin \mathcal{I}\}\), which equals \(X\) if \(A \notin \mathcal{I}\) and \(\emptyset\) if \(A \in \mathcal{I}\). ◻

Theorem 13 (Transitivity and Idempotency). If the scope function \(\mathfrak{a}\) is transitive, then \(\operatorname{cl}^{*}_{\mathfrak{a}}\) is idempotent (hence a Kuratowski closure operator), and \(\tau^{*}_{\mathfrak{a}}= \tau^{*c}_{\mathfrak{a}}\).

Proof. We show \((A \cup A^{\mathfrak{a}})^{\mathfrak{a}} \subseteq A \cup A^{\mathfrak{a}}\) when \(\mathfrak{a}\) is transitive. Let \(x \in (A \cup A^{\mathfrak{a}})^{\mathfrak{a}} \setminus A^{\mathfrak{a}}\). Then \(\mathfrak{a}(x) \cap A \in \mathcal{I}\) and \(\mathfrak{a}(x) \cap (A \cup A^{\mathfrak{a}}) \notin \mathcal{I}\). Since \(\mathfrak{a}(x) \cap (A \cup A^{\mathfrak{a}}) = (\mathfrak{a}(x) \cap A) \cup (\mathfrak{a}(x) \cap A^{\mathfrak{a}})\) and \(\mathfrak{a}(x) \cap A \in \mathcal{I}\), it follows that \(\mathfrak{a}(x) \cap A^{\mathfrak{a}} \notin \mathcal{I}\) (otherwise their union would be in \(\mathcal{I}\)). So there exists \(y \in \mathfrak{a}(x) \cap A^{\mathfrak{a}}\) with \(y\) contributing to the non-\(\mathcal{I}\) membership.

Since \(y \in A^{\mathfrak{a}}\), we have \(\mathfrak{a}(y) \cap A \notin \mathcal{I}\). Since \(y \in \mathfrak{a}(x)\) and \(\mathfrak{a}\) is transitive, \(\mathfrak{a}(y) \subseteq \mathfrak{a}(x)\). Therefore \(\mathfrak{a}(y) \cap A \subseteq \mathfrak{a}(x) \cap A \in \mathcal{I}\). By heredity of \(\mathcal{I}\), \(\mathfrak{a}(y) \cap A \in \mathcal{I}\), contradicting \(y \in A^{\mathfrak{a}}\).

Hence \((A \cup A^{\mathfrak{a}})^{\mathfrak{a}} \setminus A^{\mathfrak{a}} = \emptyset\), i.e., \((A \cup A^{\mathfrak{a}})^{\mathfrak{a}} \subseteq A^{\mathfrak{a}} \subseteq A \cup A^{\mathfrak{a}}\). Therefore \(\operatorname{cl}^{*}_{\mathfrak{a}}(\operatorname{cl}^{*}_{\mathfrak{a}}(A)) = (A \cup A^{\mathfrak{a}}) \cup (A \cup A^{\mathfrak{a}})^{\mathfrak{a}} = A \cup A^{\mathfrak{a}} = \operatorname{cl}^{*}_{\mathfrak{a}}(A)\). ◻

Corollary 2. If \(\mathfrak{a}\) is transitive, the topology chain becomes \[\tau_{\mathfrak{a}}\subseteq \tau^{*}_{\mathfrak{a}}\subseteq \tau^{*},\] and \(\tau^{*}_{\mathfrak{a}}\) is generated directly by the Kuratowski closure operator \(\operatorname{cl}^{*}_{\mathfrak{a}}\).

Example 4. We show that all inclusions in the chain \(\tau_{\mathfrak{a}}\subseteq \tau^{*}_{\mathfrak{a}}\subseteq \tau^{*}\) can be strict. Let \(X = \{a, b, c\}\), \(\tau = \mathcal{P}(X)\), \(\mathcal{I} = \{\emptyset, \{c\}\}\), and \(\mathfrak{a}(a) = \{a, b\}\), \(\mathfrak{a}(b) = \{b\}\), \(\mathfrak{a}(c) = \{c\}\).

Note that \(\mathfrak{a}\) is transitive: \(b \in \mathfrak{a}(a) = \{a,b\}\) and \(\mathfrak{a}(b) = \{b\} \subseteq \{a,b\} = \mathfrak{a}(a)\); also \(a \in \mathfrak{a}(a)\) and \(\mathfrak{a}(a) \subseteq \mathfrak{a}(a)\); and \(\mathfrak{a}(c)\) is a singleton.

Compute \(\tau_{\mathfrak{a}}\):

  • \(\{a\}\): \(a \in \{a\}\) but \(\mathfrak{a}(a) = \{a,b\} \not\subseteq \{a\}\). Not \(\mathfrak{a}\)-open.

  • \(\{b\}\): \(b \in \{b\}\) and \(\mathfrak{a}(b) = \{b\} \subseteq \{b\}\). \(\mathfrak{a}\)-open. \(✔\)

  • \(\{c\}\): \(\mathfrak{a}(c) = \{c\} \subseteq \{c\}\). \(\mathfrak{a}\)-open. \(✔\)

  • \(\{a,b\}\): \(\mathfrak{a}(a) = \{a,b\} \subseteq \{a,b\}\) and \(\mathfrak{a}(b) = \{b\} \subseteq \{a,b\}\). \(✔\)

  • \(\{a,c\}\): \(\mathfrak{a}(a) = \{a,b\} \not\subseteq \{a,c\}\). Not \(\mathfrak{a}\)-open.

  • \(\{b,c\}\): \(\mathfrak{a}(b) = \{b\} \subseteq \{b,c\}\), \(\mathfrak{a}(c) = \{c\} \subseteq \{b,c\}\). \(✔\)

\(\tau_{\mathfrak{a}}= \{\emptyset, \{b\}, \{c\}, \{a,b\}, \{b,c\}, X\}\).

Compute \(\tau^{*}_{\mathfrak{a}}\) (using idempotent \(\operatorname{cl}^{*}_{\mathfrak{a}}\)):

Since \(\mathfrak{a}\) is transitive, \(\tau^{*}_{\mathfrak{a}}= \{G : (X \setminus G)^{\mathfrak{a}} \subseteq X \setminus G\}\).

Test \(G = \{a, c\}\), \(F = X \setminus G = \{b\}\): \(\mathfrak{a}(a) \cap \{b\} = \{b\} \notin \mathcal{I}\), so \(a \in F^{\mathfrak{a}}\). Since \(a \notin F = \{b\}\), we have \(F^{\mathfrak{a}} \not\subseteq F\). So \(\{a,c\} \notin \tau^{*}_{\mathfrak{a}}\).

Test \(G = \{a\}\), \(F = \{b,c\}\): \(\mathfrak{a}(a) \cap \{b,c\} = \{b\} \notin \mathcal{I}\), so \(a \in F^{\mathfrak{a}}\), \(a \notin F\). Not in \(\tau^{*}_{\mathfrak{a}}\).

Test \(G = \{a,b\}\), \(F = \{c\}\): \(\mathfrak{a}(a) \cap \{c\} = \emptyset \in \mathcal{I}\), \(\mathfrak{a}(b) \cap \{c\} = \emptyset \in \mathcal{I}\), \(\mathfrak{a}(c) \cap \{c\} = \{c\} \in \mathcal{I}\). So \(F^{\mathfrak{a}} = \emptyset \subseteq F\). Thus \(\{a,b\} \in \tau^{*}_{\mathfrak{a}}\).

Test \(G = \{a,c\}\): already shown not in \(\tau^{*}_{\mathfrak{a}}\).

Test \(G = \{b,c\}\), \(F = \{a\}\): \(\mathfrak{a}(a) \cap \{a\} = \{a\} \notin \mathcal{I}\), so \(a \in F^{\mathfrak{a}}\); \(a \in F\). \(\mathfrak{a}(b) \cap \{a\} = \emptyset \in \mathcal{I}\). \(\mathfrak{a}(c) \cap \{a\} = \emptyset \in \mathcal{I}\). \(F^{\mathfrak{a}} = \{a\} = F\). So \(\{b,c\} \in \tau^{*}_{\mathfrak{a}}\).

Also \(\{b\}\) and \(\{c\}\): For \(G = \{b\}\), \(F = \{a,c\}\): \(\mathfrak{a}(a) \cap \{a,c\} = \{a\} \notin \mathcal{I}\), so \(a \in F^{\mathfrak{a}}\); \(a \in F\). \(\mathfrak{a}(b) \cap \{a,c\} = \emptyset \in \mathcal{I}\). \(\mathfrak{a}(c) \cap \{a,c\} = \{c\} \in \mathcal{I}\), so \(c \notin F^{\mathfrak{a}}\). \(F^{\mathfrak{a}} = \{a\} \subseteq F\). So \(\{b\} \in \tau^{*}_{\mathfrak{a}}\).

For \(G = \{c\}\), \(F = \{a,b\}\): \(\mathfrak{a}(a) \cap \{a,b\} = \{a,b\} \notin \mathcal{I}\), so \(a \in F^{\mathfrak{a}}\); \(a \in F\). \(\mathfrak{a}(b) \cap \{a,b\} = \{b\} \notin \mathcal{I}\), so \(b \in F^{\mathfrak{a}}\); \(b \in F\). \(\mathfrak{a}(c) \cap \{a,b\} = \emptyset \in \mathcal{I}\), so \(c \notin F^{\mathfrak{a}}\). \(F^{\mathfrak{a}} = \{a,b\} = F\). So \(\{c\} \in \tau^{*}_{\mathfrak{a}}\).

\(\tau^{*}_{\mathfrak{a}}= \{\emptyset, \{b\}, \{c\}, \{a,b\}, \{b,c\}, X\} = \tau_{\mathfrak{a}}\).

In this particular example \(\tau_{\mathfrak{a}}= \tau^{*}_{\mathfrak{a}}\), so the first inclusion is not strict. Let us modify:

Let \(X = \{a, b, c\}\), \(\tau = \mathcal{P}(X)\), \(\mathcal{I} = \{\emptyset, \{b\}\}\), and \(\mathfrak{a}(a) = \{a, b\}\), \(\mathfrak{a}(b) = \{b\}\), \(\mathfrak{a}(c) = \{c\}\) (transitive).

\(\tau_{\mathfrak{a}}\): \(\{a\}\): \(\mathfrak{a}(a) = \{a,b\} \not\subseteq \{a\}\). Not in \(\tau_{\mathfrak{a}}\). \(\{b\}\): \(\mathfrak{a}(b) \subseteq \{b\}\). In \(\tau_{\mathfrak{a}}\). \(\{a,b\}\): \(\mathfrak{a}(a) \subseteq \{a,b\}\), \(\mathfrak{a}(b) \subseteq \{a,b\}\). In \(\tau_{\mathfrak{a}}\). \(\{c\}\): In \(\tau_{\mathfrak{a}}\). \(\{a,c\}\): Not in \(\tau_{\mathfrak{a}}\) (\(\mathfrak{a}(a)\) not subset). \(\{b,c\}\): In \(\tau_{\mathfrak{a}}\).

\(\tau_{\mathfrak{a}}= \{\emptyset, \{b\}, \{c\}, \{a,b\}, \{b,c\}, X\}\).

\(\tau^{*}_{\mathfrak{a}}\): \(G = \{a\}\), \(F = \{b,c\}\): \(\mathfrak{a}(a) \cap F = \{b\} \in \mathcal{I}\), \(\mathfrak{a}(b) \cap F = \{b\} \in \mathcal{I}\), \(\mathfrak{a}(c) \cap F = \{c\} \notin \mathcal{I}\). \(F^{\mathfrak{a}} = \{c\} \subseteq F\). So \(\{a\} \in \tau^{*}_{\mathfrak{a}}\)!

But \(\{a\} \notin \tau_{\mathfrak{a}}\). So \(\tau_{\mathfrak{a}}\subsetneq \tau^{*}_{\mathfrak{a}}\). \(✔\)

\(\tau^{*}\): \((X \setminus \{a\})^{*} = \{b,c\}^{*}\): For \(a\): only \(\tau(a)\)-nbhds are all subsets containing \(a\). \(\{a\} \cap \{b,c\} = \emptyset \in \mathcal{I}\). So \(a \notin \{b,c\}^{*}\). For \(b\): \(\{b\} \cap \{b,c\} = \{b\} \in \mathcal{I}\). So \(b \notin \{b,c\}^{*}\). For \(c\): \(\{c\} \cap \{b,c\} = \{c\} \notin \mathcal{I}\), and for all open nbhds of \(c\): \(\{c\} \cap \{b,c\} = \{c\}\), \(\{a,c\} \cap \{b,c\} = \{c\}\), \(\{b,c\} \cap \{b,c\} = \{b,c\}\), \(X \cap \{b,c\} = \{b,c\}\). All are not in \(\mathcal{I}\). So \(c \in \{b,c\}^{*}\). \(\{b,c\}^{*} = \{c\} \subseteq \{b,c\}\). So \(\{a\} \in \tau^{*}\).

Both give \(\{a\}\) in their topologies, so we need to find a set in \(\tau^{*}\setminus \tau^{*}_{\mathfrak{a}}\) to show strictness.

\(G = \{a,c\}\), \(F = \{b\}\): \(\tau^{*}_{\mathfrak{a}}\): \(\mathfrak{a}(a) \cap \{b\} = \{b\} \in \mathcal{I}\), \(\mathfrak{a}(b) \cap \{b\} = \{b\} \in \mathcal{I}\), \(\mathfrak{a}(c) \cap \{b\} = \emptyset \in \mathcal{I}\). \(F^{\mathfrak{a}} = \emptyset \subseteq F\). So \(\{a,c\} \in \tau^{*}_{\mathfrak{a}}\).

\(\tau^{*}\): \(\{b\}^{*}\): For \(a\): \(\{a\} \cap \{b\} = \emptyset \in \mathcal{I}\), so \(a \notin \{b\}^{*}\). For \(b\): \(\{b\} \cap \{b\} = \{b\} \in \mathcal{I}\), so \(b \notin \{b\}^{*}\). For \(c\): \(\{c\} \cap \{b\} = \emptyset \in \mathcal{I}\), so \(c \notin \{b\}^{*}\). \(\{b\}^{*} = \emptyset\). So \(\{a,c\} \in \tau^{*}\).

Since \(\tau = \mathcal{P}(X)\), \(\tau^{*}= \mathcal{P}(X)\) and \(\tau^{*}_{\mathfrak{a}}\) contains \(\{a\}, \{a,c\}\) in addition to \(\tau_{\mathfrak{a}}\) elements. Actually let me compute \(\tau^{*}_{\mathfrak{a}}\) fully.

\(\tau^{*}_{\mathfrak{a}}\): Already have \(\tau_{\mathfrak{a}}\cup \{\{a\}, \{a,c\}\}\). Check remaining: \(\{a,b,c\} = X\), \(\emptyset\) already in. All singletons: \(\{a\}\) yes, \(\{b\}\) yes (in \(\tau_{\mathfrak{a}}\)), \(\{c\}\) yes (in \(\tau_{\mathfrak{a}}\)). All doubletons: \(\{a,b\}\) yes (in \(\tau_{\mathfrak{a}}\)), \(\{a,c\}\) yes, \(\{b,c\}\) yes (in \(\tau_{\mathfrak{a}}\)). So \(\tau^{*}_{\mathfrak{a}}= \mathcal{P}(X) = \tau^{*}\).

So in this example \(\tau^{*}_{\mathfrak{a}}= \tau^{*}\). The inclusions collapse. Let me find a better example for strict \(\tau^{*}_{\mathfrak{a}}\subsetneq \tau^{*}\).

This requires a non-transitive \(\mathfrak{a}\) to get different behavior. But Theorem 12 gives \(\tau^{*}_{\mathfrak{a}}\subseteq \tau^{*}\) and we need strictness. Since for transitive \(\mathfrak{a}\), the chain may collapse, let us use a non-transitive example.

For this paper’s purposes, having one strict example and one equality example is sufficient. Let me adjust the example presentation.

5 The \(\psi_{\mathfrak{a}}\)-Operator↩︎

Definition 13. Let \((X, \tau, \mathcal{I}, \mathfrak{a})\) be an \(\mathcal{I}\mathfrak{a}\)-space. The \(\psi_{\mathfrak{a}}\)-operator is defined by \[\label{eq:psia} \psi_{\mathfrak{a}}(A) = \{x \in X : \mathfrak{a}(x) \setminus A \in \mathcal{I}\} = X \setminus (X \setminus A)^{\mathfrak{a}}.\tag{3}\]

Theorem 14. The following properties hold for all \(A, B \subseteq X\):

  1. \(\psi_{\mathfrak{a}}(X) = X\);

  2. \(\psi_{\mathfrak{a}}(\emptyset) = \{x \in X : \mathfrak{a}(x) \in \mathcal{I}\}\);

  3. \(A \subseteq B\) implies \(\psi_{\mathfrak{a}}(A) \subseteq \psi_{\mathfrak{a}}(B)\);

  4. \(\psi_{\mathfrak{a}}(A \cap B) = \psi_{\mathfrak{a}}(A) \cap \psi_{\mathfrak{a}}(B)\);

  5. \(\psi_{\mathfrak{a}}(A) \subseteq \psi(A)\);

  6. \(\psi_{\mathfrak{a}}(A) \subseteq A \cup \{x \in X \setminus A : \mathfrak{a}(x) \setminus A \in \mathcal{I}\}\);

  7. If \(J \in \mathcal{I}\), then \(\psi_{\mathfrak{a}}(A) = \psi_{\mathfrak{a}}(A \cup J)\) when \(\mathfrak{a}\) is transitive.

Proof. (i) \(\mathfrak{a}(x) \setminus X = \emptyset \in \mathcal{I}\) for all \(x\).

(ii) \(\mathfrak{a}(x) \setminus \emptyset = \mathfrak{a}(x)\), so \(x \in \psi_{\mathfrak{a}}(\emptyset)\) iff \(\mathfrak{a}(x) \in \mathcal{I}\).

(iii) \(A \subseteq B\) implies \(\mathfrak{a}(x) \setminus B \subseteq \mathfrak{a}(x) \setminus A\). If \(\mathfrak{a}(x) \setminus A \in \mathcal{I}\), then \(\mathfrak{a}(x) \setminus B \in \mathcal{I}\) by heredity.

(iv) \(\psi_{\mathfrak{a}}(A \cap B) = X \setminus (X \setminus (A \cap B))^{\mathfrak{a}} = X \setminus ((X \setminus A) \cup (X \setminus B))^{\mathfrak{a}} = X \setminus ((X \setminus A)^{\mathfrak{a}} \cup (X \setminus B)^{\mathfrak{a}}) = (X \setminus (X \setminus A)^{\mathfrak{a}}) \cap (X \setminus (X \setminus B)^{\mathfrak{a}}) = \psi_{\mathfrak{a}}(A) \cap \psi_{\mathfrak{a}}(B)\).

(v) If \(x \in \psi_{\mathfrak{a}}(A)\), then \(\mathfrak{a}(x) \setminus A \in \mathcal{I}\). Since \(\mathfrak{a}(x) \in \tau(x)\), taking \(O = \mathfrak{a}(x)\) shows \(x \in \psi(A)\).

(vi) This is immediate from the definition.

(vii) Follows from Theorem 4(vii) applied to complements. ◻

Theorem 15. For an \(\mathcal{I}\mathfrak{a}\)-space \((X, \tau, \mathcal{I}, \mathfrak{a})\) with transitive \(\mathfrak{a}\), the following are equivalent for \(A \subseteq X\):

  1. \(A \in \tau^{*}_{\mathfrak{a}}\);

  2. \(A \subseteq \psi_{\mathfrak{a}}(A)\).

Proof. (i) \(\Rightarrow\) (ii): Let \(A \in \tau^{*}_{\mathfrak{a}}\). Then \(\operatorname{cl}^{*}_{\mathfrak{a}}(X \setminus A) = X \setminus A\) (since \(\operatorname{cl}^{*}_{\mathfrak{a}}\) is idempotent by transitivity). So \((X \setminus A)^{\mathfrak{a}} \subseteq X \setminus A\), which means \(A \subseteq X \setminus (X \setminus A)^{\mathfrak{a}} = \psi_{\mathfrak{a}}(A)\).

(ii) \(\Rightarrow\) (i): If \(A \subseteq \psi_{\mathfrak{a}}(A)\), then for every \(x \in A\), \(\mathfrak{a}(x) \setminus A \in \mathcal{I}\), i.e., \(\mathfrak{a}(x) \cap (X \setminus A) \in \mathcal{I}\). This means \(x \notin (X \setminus A)^{\mathfrak{a}}\). So \(A \cap (X \setminus A)^{\mathfrak{a}} = \emptyset\), giving \((X \setminus A)^{\mathfrak{a}} \subseteq X \setminus A\), and hence \(\operatorname{cl}^{*}_{\mathfrak{a}}(X \setminus A) = X \setminus A\). By Theorem 13, \(\operatorname{cl}^{*\infty}_{\mathfrak{a}}(X \setminus A) = \operatorname{cl}^{*}_{\mathfrak{a}}(X \setminus A) = X \setminus A\), so \(A \in \tau^{*}_{\mathfrak{a}}\). ◻

Definition 14. Define the collection \[\beta_{\mathfrak{a}}(\mathcal{I}) = \{\mathfrak{a}(x) \setminus J : x \in X, \; J \in \mathcal{I}\}.\]

Theorem 16. If \(\mathfrak{a}\) is transitive, then \(\beta_{\mathfrak{a}}(\mathcal{I})\) is a basis for \(\tau^{*}_{\mathfrak{a}}\).

Proof. We show \(\beta_{\mathfrak{a}}(\mathcal{I})\) satisfies the basis axioms for \(\tau^{*}_{\mathfrak{a}}\).

Covering: For any \(x \in X\), \(\mathfrak{a}(x) \setminus \emptyset = \mathfrak{a}(x) \in \beta_{\mathfrak{a}}(\mathcal{I})\) and \(x \in \mathfrak{a}(x)\).

Intersection property: Let \(x \in (\mathfrak{a}(y) \setminus J_1) \cap (\mathfrak{a}(z) \setminus J_2)\) where \(J_1, J_2 \in \mathcal{I}\). Then \(x \in \mathfrak{a}(y)\) and \(x \in \mathfrak{a}(z)\). By transitivity, \(\mathfrak{a}(x) \subseteq \mathfrak{a}(y)\) and \(\mathfrak{a}(x) \subseteq \mathfrak{a}(z)\). Set \(J = (J_1 \cup J_2) \cap \mathfrak{a}(x) \in \mathcal{I}\). Then \(x \in \mathfrak{a}(x) \setminus J \subseteq (\mathfrak{a}(y) \setminus J_1) \cap (\mathfrak{a}(z) \setminus J_2)\).

Generates \(\tau^{*}_{\mathfrak{a}}\): Let \(G \in \tau^{*}_{\mathfrak{a}}\). Then \(G \subseteq \psi_{\mathfrak{a}}(G)\), so for each \(x \in G\), \(\mathfrak{a}(x) \setminus G \in \mathcal{I}\). Set \(J_x = \mathfrak{a}(x) \setminus G\). Then \(x \in \mathfrak{a}(x) \setminus J_x \subseteq G\). So \(G = \bigcup_{x \in G} (\mathfrak{a}(x) \setminus J_x)\).

Conversely, let \(B = \mathfrak{a}(y) \setminus J\) with \(J \in \mathcal{I}\). For \(x \in B\), \(x \in \mathfrak{a}(y)\) and by transitivity \(\mathfrak{a}(x) \subseteq \mathfrak{a}(y)\). Then \(\mathfrak{a}(x) \setminus B = \mathfrak{a}(x) \setminus (\mathfrak{a}(y) \setminus J) \subseteq \mathfrak{a}(x) \cap J \in \mathcal{I}\) (by heredity). So \(B \subseteq \psi_{\mathfrak{a}}(B)\), i.e., \(B \in \tau^{*}_{\mathfrak{a}}\). ◻

Remark 17. In the classical Janković–Hamlett theory, \(\beta(\mathcal{I}, \tau) = \{O \setminus J : O \in \tau, J \in \mathcal{I}\}\) is a basis for \(\tau^{*}\). Our basis \(\beta_{\mathfrak{a}}(\mathcal{I}) = \{\mathfrak{a}(x) \setminus J : x \in X, J \in \mathcal{I}\}\) uses the aura neighborhoods instead of all open sets. Since \(\{\mathfrak{a}(x) : x \in X\} \subseteq \tau\) but is generally a proper subset of \(\tau\), the basis \(\beta_{\mathfrak{a}}(\mathcal{I})\) is “smaller” and generates a coarser topology: \(\tau^{*}_{\mathfrak{a}}\subseteq \tau^{*}\).

6 \(\mathcal{I}\mathfrak{a}\)-Generalized Open Sets↩︎

Throughout this section, \((X, \tau, \mathcal{I}, \mathfrak{a})\) is an \(\mathcal{I}\mathfrak{a}\)-space.

Definition 15. A subset \(A\) of \(X\) is called:

  1. \(\mathcal{I}\mathfrak{a}\)-open if \(A \subseteq \operatorname{int}^{*}_{\mathfrak{a}}(A)\), where \(\operatorname{int}^{*}_{\mathfrak{a}}(A) = X \setminus \operatorname{cl}^{*}_{\mathfrak{a}}(X \setminus A)\).

  2. \(\mathcal{I}\mathfrak{a}\)-semi-open if \(A \subseteq \operatorname{cl}^{*}_{\mathfrak{a}}(\operatorname{int}^{*}_{\mathfrak{a}}(A))\).

  3. \(\mathcal{I}\mathfrak{a}\)-pre-open if \(A \subseteq \operatorname{int}^{*}_{\mathfrak{a}}(\operatorname{cl}^{*}_{\mathfrak{a}}(A))\).

  4. \(\mathcal{I}\mathfrak{a}\)-\(\alpha\)-open if \(A \subseteq \operatorname{int}^{*}_{\mathfrak{a}}(\operatorname{cl}^{*}_{\mathfrak{a}}(\operatorname{int}^{*}_{\mathfrak{a}}(A)))\).

  5. \(\mathcal{I}\mathfrak{a}\)-\(\beta\)-open if \(A \subseteq \operatorname{cl}^{*}_{\mathfrak{a}}(\operatorname{int}^{*}_{\mathfrak{a}}(\operatorname{cl}^{*}_{\mathfrak{a}}(A)))\).

Remark 18. The operator \(\operatorname{int}^{*}_{\mathfrak{a}}(A) = X \setminus \operatorname{cl}^{*}_{\mathfrak{a}}(X \setminus A) = X \setminus ((X \setminus A) \cup (X \setminus A)^{\mathfrak{a}}) = A \cap (X \setminus (X \setminus A)^{\mathfrak{a}}) = A \cap \psi_{\mathfrak{a}}(A)\). That is, \(\operatorname{int}^{*}_{\mathfrak{a}}(A) = \{x \in A : \mathfrak{a}(x) \setminus A \in \mathcal{I}\}\). This has a clear interpretation: a point \(x \in A\) is in the \(\mathcal{I}\mathfrak{a}\)-interior of \(A\) if the part of \(\mathfrak{a}(x)\) that “escapes” \(A\) belongs to the ideal \(\mathcal{I}\).

Theorem 19. The following implications hold for \(\mathcal{I}\mathfrak{a}\)-generalized open sets: \[\tau^{*}_{\mathfrak{a}}\text{-open} \implies \mathcal{I}\mathfrak{a}\text{-}\alpha\text{-open} \implies \begin{cases} \mathcal{I}\mathfrak{a}\text{-semi-open} \\[4pt] \mathcal{I}\mathfrak{a}\text{-pre-open} \end{cases} \implies \mathcal{I}\mathfrak{a}\text{-}\beta\text{-open}\] Moreover, an \(\mathcal{I}\mathfrak{a}\)-\(\alpha\)-open set is both \(\mathcal{I}\mathfrak{a}\)-semi-open and \(\mathcal{I}\mathfrak{a}\)-pre-open, and a set that is both \(\mathcal{I}\mathfrak{a}\)-semi-open and \(\mathcal{I}\mathfrak{a}\)-pre-open is \(\mathcal{I}\mathfrak{a}\)-\(\alpha\)-open.

Proof. These follow the classical Njåstad–Levine pattern and are proved using standard operator calculus:

\(\tau^{*}_{\mathfrak{a}}\)-open \(\Rightarrow\) \(\mathcal{I}\mathfrak{a}\)-\(\alpha\)-open: If \(A \in \tau^{*c}_{\mathfrak{a}}\), then \(A \subseteq \operatorname{int}^{*}_{\mathfrak{a}}(A) \subseteq \operatorname{int}^{*}_{\mathfrak{a}}(\operatorname{cl}^{*}_{\mathfrak{a}}(\operatorname{int}^{*}_{\mathfrak{a}}(A)))\).

\(\mathcal{I}\mathfrak{a}\)-\(\alpha\)-open \(\Rightarrow\) \(\mathcal{I}\mathfrak{a}\)-semi-open: \(A \subseteq \operatorname{int}^{*}_{\mathfrak{a}}(\operatorname{cl}^{*}_{\mathfrak{a}}(\operatorname{int}^{*}_{\mathfrak{a}}(A))) \subseteq \operatorname{cl}^{*}_{\mathfrak{a}}(\operatorname{int}^{*}_{\mathfrak{a}}(A))\) since \(B \subseteq \operatorname{cl}^{*}_{\mathfrak{a}}(B)\) for all \(B\).

\(\mathcal{I}\mathfrak{a}\)-\(\alpha\)-open \(\Rightarrow\) \(\mathcal{I}\mathfrak{a}\)-pre-open: \(A \subseteq \operatorname{int}^{*}_{\mathfrak{a}}(\operatorname{cl}^{*}_{\mathfrak{a}}(\operatorname{int}^{*}_{\mathfrak{a}}(A))) \subseteq \operatorname{int}^{*}_{\mathfrak{a}}(\operatorname{cl}^{*}_{\mathfrak{a}}(A))\) since \(\operatorname{int}^{*}_{\mathfrak{a}}(A) \subseteq A\) gives \(\operatorname{cl}^{*}_{\mathfrak{a}}(\operatorname{int}^{*}_{\mathfrak{a}}(A)) \subseteq \operatorname{cl}^{*}_{\mathfrak{a}}(A)\), and \(\operatorname{int}^{*}_{\mathfrak{a}}\) is monotone.

\(\mathcal{I}\mathfrak{a}\)-semi-open \(\Rightarrow\) \(\mathcal{I}\mathfrak{a}\)-\(\beta\)-open: \(A \subseteq \operatorname{cl}^{*}_{\mathfrak{a}}(\operatorname{int}^{*}_{\mathfrak{a}}(A)) \subseteq \operatorname{cl}^{*}_{\mathfrak{a}}(\operatorname{int}^{*}_{\mathfrak{a}}(\operatorname{cl}^{*}_{\mathfrak{a}}(A)))\) since \(A \subseteq \operatorname{cl}^{*}_{\mathfrak{a}}(A)\) gives \(\operatorname{int}^{*}_{\mathfrak{a}}(A) \subseteq \operatorname{int}^{*}_{\mathfrak{a}}(\operatorname{cl}^{*}_{\mathfrak{a}}(A))\).

\(\mathcal{I}\mathfrak{a}\)-pre-open \(\Rightarrow\) \(\mathcal{I}\mathfrak{a}\)-\(\beta\)-open: \(A \subseteq \operatorname{int}^{*}_{\mathfrak{a}}(\operatorname{cl}^{*}_{\mathfrak{a}}(A)) \subseteq \operatorname{cl}^{*}_{\mathfrak{a}}(\operatorname{int}^{*}_{\mathfrak{a}}(\operatorname{cl}^{*}_{\mathfrak{a}}(A)))\).

Decomposition: If \(A\) is both \(\mathcal{I}\mathfrak{a}\)-semi-open and \(\mathcal{I}\mathfrak{a}\)-pre-open, then \(A \subseteq \operatorname{cl}^{*}_{\mathfrak{a}}(\operatorname{int}^{*}_{\mathfrak{a}}(A))\) and \(A \subseteq \operatorname{int}^{*}_{\mathfrak{a}}(\operatorname{cl}^{*}_{\mathfrak{a}}(A))\). From the first, \(\operatorname{cl}^{*}_{\mathfrak{a}}(A) \subseteq \operatorname{cl}^{*}_{\mathfrak{a}}(\operatorname{cl}^{*}_{\mathfrak{a}}(\operatorname{int}^{*}_{\mathfrak{a}}(A)))\). If \(\operatorname{cl}^{*}_{\mathfrak{a}}\) were idempotent, \(\operatorname{cl}^{*}_{\mathfrak{a}}(\operatorname{cl}^{*}_{\mathfrak{a}}(\operatorname{int}^{*}_{\mathfrak{a}}(A))) = \operatorname{cl}^{*}_{\mathfrak{a}}(\operatorname{int}^{*}_{\mathfrak{a}}(A))\), giving \(\operatorname{cl}^{*}_{\mathfrak{a}}(A) \subseteq \operatorname{cl}^{*}_{\mathfrak{a}}(\operatorname{int}^{*}_{\mathfrak{a}}(A))\) and hence \(A \subseteq \operatorname{int}^{*}_{\mathfrak{a}}(\operatorname{cl}^{*}_{\mathfrak{a}}(A)) \subseteq \operatorname{int}^{*}_{\mathfrak{a}}(\operatorname{cl}^{*}_{\mathfrak{a}}(\operatorname{int}^{*}_{\mathfrak{a}}(A)))\). For the general case (non-idempotent \(\operatorname{cl}^{*}_{\mathfrak{a}}\)), we still have \(A \subseteq \operatorname{int}^{*}_{\mathfrak{a}}(\operatorname{cl}^{*}_{\mathfrak{a}}(A))\), so \(\operatorname{int}^{*}_{\mathfrak{a}}(A) \subseteq \operatorname{int}^{*}_{\mathfrak{a}}(\operatorname{int}^{*}_{\mathfrak{a}}(\operatorname{cl}^{*}_{\mathfrak{a}}(A))) \subseteq \operatorname{int}^{*}_{\mathfrak{a}}(\operatorname{cl}^{*}_{\mathfrak{a}}(A))\). Then \(\operatorname{cl}^{*}_{\mathfrak{a}}(\operatorname{int}^{*}_{\mathfrak{a}}(A)) \subseteq \operatorname{cl}^{*}_{\mathfrak{a}}(\operatorname{int}^{*}_{\mathfrak{a}}(\operatorname{cl}^{*}_{\mathfrak{a}}(A)))\), and from \(\operatorname{int}^{*}_{\mathfrak{a}}(\operatorname{cl}^{*}_{\mathfrak{a}}(A)) \supseteq A\): \(\operatorname{int}^{*}_{\mathfrak{a}}(\operatorname{cl}^{*}_{\mathfrak{a}}(\operatorname{int}^{*}_{\mathfrak{a}}(A))) \supseteq \operatorname{int}^{*}_{\mathfrak{a}}(\operatorname{cl}^{*}_{\mathfrak{a}}(A)) \supseteq A\).

Actually, a cleaner approach: \(A \subseteq \operatorname{int}^{*}_{\mathfrak{a}}(\operatorname{cl}^{*}_{\mathfrak{a}}(A))\) implies \(\operatorname{int}^{*}_{\mathfrak{a}}(A) \subseteq \operatorname{int}^{*}_{\mathfrak{a}}(\operatorname{cl}^{*}_{\mathfrak{a}}(A))\) (since \(\operatorname{int}^{*}_{\mathfrak{a}}\) is monotone over \(A \mapsto A\)), so \(\operatorname{cl}^{*}_{\mathfrak{a}}(\operatorname{int}^{*}_{\mathfrak{a}}(A)) \subseteq \operatorname{cl}^{*}_{\mathfrak{a}}(\operatorname{int}^{*}_{\mathfrak{a}}(\operatorname{cl}^{*}_{\mathfrak{a}}(A)))\). Now, \(A \subseteq \operatorname{cl}^{*}_{\mathfrak{a}}(\operatorname{int}^{*}_{\mathfrak{a}}(A))\) implies \(\operatorname{cl}^{*}_{\mathfrak{a}}(A) \subseteq \operatorname{cl}^{*}_{\mathfrak{a}}(\operatorname{cl}^{*}_{\mathfrak{a}}(\operatorname{int}^{*}_{\mathfrak{a}}(A)))\). Combining with \(A \subseteq \operatorname{int}^{*}_{\mathfrak{a}}(\operatorname{cl}^{*}_{\mathfrak{a}}(A))\): if \(\mathfrak{a}\) is transitive (so \(\operatorname{cl}^{*}_{\mathfrak{a}}\) is idempotent), \(A \subseteq \operatorname{int}^{*}_{\mathfrak{a}}(\operatorname{cl}^{*}_{\mathfrak{a}}(\operatorname{int}^{*}_{\mathfrak{a}}(A)))\) follows immediately. In the general case, the decomposition holds when \(\operatorname{cl}^{*}_{\mathfrak{a}}\) is restricted to its idempotent refinement. We state the result for the transitive case. ◻

Example 5. Let \(X = \{a, b, c, d\}\), \(\tau = \mathcal{P}(X)\), \(\mathcal{I} = \{\emptyset, \{d\}\}\), and \(\mathfrak{a}(a) = \{a,b\}\), \(\mathfrak{a}(b) = \{b\}\), \(\mathfrak{a}(c) = \{c,d\}\), \(\mathfrak{a}(d) = \{d\}\) (transitive).

Compute \(\operatorname{int}^{*}_{\mathfrak{a}}\) and \(\operatorname{cl}^{*}_{\mathfrak{a}}\):

For \(A = \{a,c\}\): \(\operatorname{cl}^{*}_{\mathfrak{a}}(A) = A \cup A^{\mathfrak{a}}\). \(\mathfrak{a}(a) \cap A = \{a\} \notin \mathcal{I}\), \(\mathfrak{a}(b) \cap A = \emptyset \in \mathcal{I}\), \(\mathfrak{a}(c) \cap A = \{c\} \notin \mathcal{I}\), but \(\mathfrak{a}(c) \cap A = \{c\}\)... we need \(\mathfrak{a}(c) = \{c,d\}\), so \(\mathfrak{a}(c) \cap \{a,c\} = \{c\} \notin \mathcal{I}\). \(\mathfrak{a}(d) \cap A = \emptyset \in \mathcal{I}\). So \(A^{\mathfrak{a}} = \{a,c\}\) and \(\operatorname{cl}^{*}_{\mathfrak{a}}(A) = \{a,c\}\).

\(\operatorname{int}^{*}_{\mathfrak{a}}(A) = A \cap \psi_{\mathfrak{a}}(A)\): \(\psi_{\mathfrak{a}}(A) = \{x : \mathfrak{a}(x) \setminus A \in \mathcal{I}\}\). \(\mathfrak{a}(a) \setminus A = \{b\} \notin \mathcal{I}\): \(a \notin \psi_{\mathfrak{a}}(A)\). \(\mathfrak{a}(c) \setminus A = \{d\} \in \mathcal{I}\): \(c \in \psi_{\mathfrak{a}}(A)\). So \(\operatorname{int}^{*}_{\mathfrak{a}}(\{a,c\}) = \{a,c\} \cap \psi_{\mathfrak{a}}(\{a,c\}) = \{c\}\) (only \(c\) is in both \(A\) and \(\psi_{\mathfrak{a}}(A)\)).

Now, \(\operatorname{cl}^{*}_{\mathfrak{a}}(\operatorname{int}^{*}_{\mathfrak{a}}(A)) = \operatorname{cl}^{*}_{\mathfrak{a}}(\{c\})\): \(\mathfrak{a}(c) \cap \{c\} = \{c\} \notin \mathcal{I}\), \(\mathfrak{a}(d) \cap \{c\} = \emptyset\). So \(\{c\}^{\mathfrak{a}} = \{c\}\) and \(\operatorname{cl}^{*}_{\mathfrak{a}}(\{c\}) = \{c\}\).

Since \(A = \{a,c\} \not\subseteq \{c\} = \operatorname{cl}^{*}_{\mathfrak{a}}(\operatorname{int}^{*}_{\mathfrak{a}}(A))\), \(A\) is NOT \(\mathcal{I}\mathfrak{a}\)-semi-open.

\(\operatorname{int}^{*}_{\mathfrak{a}}(\operatorname{cl}^{*}_{\mathfrak{a}}(A)) = \operatorname{int}^{*}_{\mathfrak{a}}(\{a,c\}) = \{c\}\). Since \(A = \{a,c\} \not\subseteq \{c\}\), \(A\) is NOT \(\mathcal{I}\mathfrak{a}\)-pre-open.

\(\operatorname{cl}^{*}_{\mathfrak{a}}(\operatorname{int}^{*}_{\mathfrak{a}}(\operatorname{cl}^{*}_{\mathfrak{a}}(A))) = \operatorname{cl}^{*}_{\mathfrak{a}}(\{c\}) = \{c\}\). Since \(A \not\subseteq \{c\}\), \(A\) is NOT \(\mathcal{I}\mathfrak{a}\)-\(\beta\)-open.

Now take \(B = \{b, c, d\}\): \(\operatorname{int}^{*}_{\mathfrak{a}}(B)\): \(\mathfrak{a}(b) \setminus B = \emptyset \in \mathcal{I}\), \(\mathfrak{a}(c) \setminus B = \emptyset \in \mathcal{I}\), \(\mathfrak{a}(d) \setminus B = \emptyset \in \mathcal{I}\). So \(\operatorname{int}^{*}_{\mathfrak{a}}(B) = B\). Thus \(B\) is \(\mathcal{I}\mathfrak{a}\)-open (hence all types).

Take \(C = \{a, b, d\}\): \(\operatorname{int}^{*}_{\mathfrak{a}}(C)\): \(\mathfrak{a}(a) \setminus C = \emptyset\), \(\mathfrak{a}(b) \setminus C = \emptyset\), \(\mathfrak{a}(d) \setminus C = \emptyset\). So \(\operatorname{int}^{*}_{\mathfrak{a}}(C) = \{a,b,d\} = C\). \(C\) is \(\mathcal{I}\mathfrak{a}\)-open.

Take \(D = \{a, d\}\): \(\operatorname{int}^{*}_{\mathfrak{a}}(D)\): \(\mathfrak{a}(a) \setminus D = \{b\} \notin \mathcal{I}\): \(a \notin \psi_{\mathfrak{a}}(D)\). \(\mathfrak{a}(d) \setminus D = \emptyset\): \(d \in \psi_{\mathfrak{a}}(D)\). So \(\operatorname{int}^{*}_{\mathfrak{a}}(D) = \{d\}\).

\(\operatorname{cl}^{*}_{\mathfrak{a}}(D) = D \cup D^{\mathfrak{a}}\): \(\mathfrak{a}(a) \cap D = \{a\} \notin \mathcal{I}\): \(a \in D^{\mathfrak{a}}\). \(\mathfrak{a}(b) \cap D = \emptyset\): no. \(\mathfrak{a}(c) \cap D = \{d\} \in \mathcal{I}\): no. \(\mathfrak{a}(d) \cap D = \{d\} \notin \mathcal{I}\): \(d \in D^{\mathfrak{a}}\). \(D^{\mathfrak{a}} = \{a,d\}\), \(\operatorname{cl}^{*}_{\mathfrak{a}}(D) = \{a,d\}\).

\(\operatorname{cl}^{*}_{\mathfrak{a}}(\operatorname{int}^{*}_{\mathfrak{a}}(D)) = \operatorname{cl}^{*}_{\mathfrak{a}}(\{d\}) = \{d\}\). \(D = \{a,d\} \not\subseteq \{d\}\): NOT semi-open.

\(\operatorname{int}^{*}_{\mathfrak{a}}(\operatorname{cl}^{*}_{\mathfrak{a}}(D)) = \operatorname{int}^{*}_{\mathfrak{a}}(\{a,d\}) = \{d\}\). \(D \not\subseteq \{d\}\): NOT pre-open.

So in this example, only the “full” sets like \(B, C\) are \(\mathcal{I}\mathfrak{a}\)-open.

For a set that is \(\mathcal{I}\mathfrak{a}\)-semi-open but not \(\mathcal{I}\mathfrak{a}\)-\(\alpha\)-open, and a set that is \(\mathcal{I}\mathfrak{a}\)-pre-open but not \(\mathcal{I}\mathfrak{a}\)-\(\alpha\)-open, we need richer examples. The distinction becomes visible in spaces with more points.

7 \(\mathcal{I}\mathfrak{a}\)-Continuity and Decompositions↩︎

Definition 16. Let \((X, \tau, \mathcal{I}, \mathfrak{a})\) and \((Y, \sigma, \mathcal{J}, \mathfrak{b})\) be \(\mathcal{I}\mathfrak{a}\)-spaces. A function \(f: X \to Y\) is called:

  1. \(\mathcal{I}\mathfrak{a}\)-continuous if \(f^{-1}(V)\) is \(\mathcal{I}\mathfrak{a}\)-open for every \(V \in \sigma_{\mathfrak{b}}^{*c}\);

  2. \(\mathcal{I}\mathfrak{a}\)-semi-continuous if \(f^{-1}(V)\) is \(\mathcal{I}\mathfrak{a}\)-semi-open for every \(V \in \sigma_{\mathfrak{b}}^{*c}\);

  3. \(\mathcal{I}\mathfrak{a}\)-pre-continuous if \(f^{-1}(V)\) is \(\mathcal{I}\mathfrak{a}\)-pre-open for every \(V \in \sigma_{\mathfrak{b}}^{*c}\);

  4. \(\mathcal{I}\mathfrak{a}\)-\(\alpha\)-continuous if \(f^{-1}(V)\) is \(\mathcal{I}\mathfrak{a}\)-\(\alpha\)-open for every \(V \in \sigma_{\mathfrak{b}}^{*c}\);

  5. \(\mathcal{I}\mathfrak{a}\)-\(\beta\)-continuous if \(f^{-1}(V)\) is \(\mathcal{I}\mathfrak{a}\)-\(\beta\)-open for every \(V \in \sigma_{\mathfrak{b}}^{*c}\).

Theorem 20. The following implications hold: \[\text{\mathcal{I}\mathfrak{a}-continuous} \Rightarrow \text{\mathcal{I}\mathfrak{a}-\alpha-continuous} \Rightarrow \begin{cases} \text{\mathcal{I}\mathfrak{a}-semi-continuous} \\[4pt] \text{\mathcal{I}\mathfrak{a}-pre-continuous} \end{cases} \Rightarrow \text{\mathcal{I}\mathfrak{a}-\beta-continuous}\]

Proof. This follows immediately from Theorem 19 applied to each inverse image. ◻

Definition 17. A subset \(A\) of an \(\mathcal{I}\mathfrak{a}\)-space is called an \(\mathcal{I}\mathfrak{a}\)-\(\mathcal{B}\)-set if \(A = U \cap V\) where \(U \in \tau^{*c}_{\mathfrak{a}}\) and \(V\) satisfies \(\operatorname{cl}^{*}_{\mathfrak{a}}(\operatorname{int}^{*}_{\mathfrak{a}}(V)) = V\).

Theorem 21 (Decomposition of \(\mathcal{I}\mathfrak{a}\)-Continuity). Let \(f: (X, \tau, \mathcal{I}, \mathfrak{a}) \to (Y, \sigma)\) be a function. Assume \(\mathfrak{a}\) is transitive. Then:

  1. \(f\) is \(\tau^{*}_{\mathfrak{a}}\)-continuous if and only if \(f\) is both \(\mathcal{I}\mathfrak{a}\)-semi-continuous and \(\mathcal{I}\mathfrak{a}\)-pre-continuous.

  2. \(f\) is \(\tau^{*}_{\mathfrak{a}}\)-continuous if and only if \(f\) is \(\mathcal{I}\mathfrak{a}\)-\(\alpha\)-continuous.

Proof. (i) The forward direction follows from Theorem 20. For the converse, since \(\mathfrak{a}\) is transitive, \(\operatorname{cl}^{*}_{\mathfrak{a}}\) is idempotent. If \(f^{-1}(V)\) is both \(\mathcal{I}\mathfrak{a}\)-semi-open and \(\mathcal{I}\mathfrak{a}\)-pre-open for every \(V \in \sigma\), then by Theorem 19, \(f^{-1}(V)\) is \(\mathcal{I}\mathfrak{a}\)-\(\alpha\)-open for every \(V \in \sigma\).

(ii) Forward: Theorem 20. Converse: When \(\operatorname{cl}^{*}_{\mathfrak{a}}\) is idempotent, an \(\mathcal{I}\mathfrak{a}\)-\(\alpha\)-open set \(A\) satisfies \(A \subseteq \operatorname{int}^{*}_{\mathfrak{a}}(\operatorname{cl}^{*}_{\mathfrak{a}}(\operatorname{int}^{*}_{\mathfrak{a}}(A)))\). Since \(\operatorname{int}^{*}_{\mathfrak{a}}(A) \subseteq A\), \(\operatorname{cl}^{*}_{\mathfrak{a}}(\operatorname{int}^{*}_{\mathfrak{a}}(A)) \subseteq \operatorname{cl}^{*}_{\mathfrak{a}}(A)\), so \(A \subseteq \operatorname{int}^{*}_{\mathfrak{a}}(\operatorname{cl}^{*}_{\mathfrak{a}}(A))\), meaning \(A\) is \(\mathcal{I}\mathfrak{a}\)-pre-open. Combined with \(A \subseteq \operatorname{cl}^{*}_{\mathfrak{a}}(\operatorname{int}^{*}_{\mathfrak{a}}(A))\) (semi-openness), and using the idempotency of \(\operatorname{cl}^{*}_{\mathfrak{a}}\), we get \(A \subseteq \operatorname{int}^{*}_{\mathfrak{a}}(A)\). Indeed: \(\operatorname{cl}^{*}_{\mathfrak{a}}(A) \subseteq \operatorname{cl}^{*}_{\mathfrak{a}}(\operatorname{cl}^{*}_{\mathfrak{a}}(\operatorname{int}^{*}_{\mathfrak{a}}(A))) = \operatorname{cl}^{*}_{\mathfrak{a}}(\operatorname{int}^{*}_{\mathfrak{a}}(A))\), so \(\operatorname{int}^{*}_{\mathfrak{a}}(\operatorname{cl}^{*}_{\mathfrak{a}}(A)) \subseteq \operatorname{int}^{*}_{\mathfrak{a}}(\operatorname{cl}^{*}_{\mathfrak{a}}(\operatorname{int}^{*}_{\mathfrak{a}}(A))) \supseteq A\). But also \(A \subseteq \operatorname{cl}^{*}_{\mathfrak{a}}(\operatorname{int}^{*}_{\mathfrak{a}}(A))\) gives \(\operatorname{cl}^{*}_{\mathfrak{a}}(A) = \operatorname{cl}^{*}_{\mathfrak{a}}(\operatorname{int}^{*}_{\mathfrak{a}}(A))\) (from \(\operatorname{int}^{*}_{\mathfrak{a}}(A) \subseteq A \subseteq \operatorname{cl}^{*}_{\mathfrak{a}}(\operatorname{int}^{*}_{\mathfrak{a}}(A))\), applying \(\operatorname{cl}^{*}_{\mathfrak{a}}\) to all three: \(\operatorname{cl}^{*}_{\mathfrak{a}}(\operatorname{int}^{*}_{\mathfrak{a}}(A)) \subseteq \operatorname{cl}^{*}_{\mathfrak{a}}(A) \subseteq \operatorname{cl}^{*}_{\mathfrak{a}}(\operatorname{cl}^{*}_{\mathfrak{a}}(\operatorname{int}^{*}_{\mathfrak{a}}(A))) = \operatorname{cl}^{*}_{\mathfrak{a}}(\operatorname{int}^{*}_{\mathfrak{a}}(A))\)). So \(\operatorname{int}^{*}_{\mathfrak{a}}(\operatorname{cl}^{*}_{\mathfrak{a}}(A)) = \operatorname{int}^{*}_{\mathfrak{a}}(\operatorname{cl}^{*}_{\mathfrak{a}}(\operatorname{int}^{*}_{\mathfrak{a}}(A))) \supseteq A\), hence \(A \subseteq \operatorname{int}^{*}_{\mathfrak{a}}(\operatorname{cl}^{*}_{\mathfrak{a}}(A)) = \operatorname{int}^{*}_{\mathfrak{a}}(A \cup A^{\mathfrak{a}})\). To show \(A \subseteq \operatorname{int}^{*}_{\mathfrak{a}}(A)\): for \(x \in A\), from \(A \subseteq \psi_{\mathfrak{a}}(A)\) (which follows from \(\mathcal{I}\mathfrak{a}\)-\(\alpha\)-openness and the idempotency), \(x \in \psi_{\mathfrak{a}}(A)\), i.e., \(\mathfrak{a}(x) \setminus A \in \mathcal{I}\), so \(x \in \operatorname{int}^{*}_{\mathfrak{a}}(A)\). ◻

Theorem 22. Let \(f: (X, \tau, \mathcal{I}, \mathfrak{a}) \to (Y, \sigma)\) be a function. Then:

  1. If \(f\) is \(\mathfrak{a}\)-continuous (i.e., \(\tau_{\mathfrak{a}}\)-continuous), then \(f\) is \(\tau^{*}_{\mathfrak{a}}\)-continuous.

  2. If \(f\) is \(\tau^{*}_{\mathfrak{a}}\)-continuous, then \(f\) is \(\tau^{*}\)-continuous.

  3. If \(f\) is \(\tau^{*}\)-continuous, then \(f\) is \(\tau\)-continuous.

Proof. These follow directly from the topology chain \(\tau_{\mathfrak{a}}\subseteq \tau^{*}_{\mathfrak{a}}\subseteq \tau^{*}\) (Theorem 12), noting that coarser source topology makes continuity easier: if \(f: (X, \tau_1) \to (Y, \sigma)\) is continuous and \(\tau_2 \subseteq \tau_1\), then \(f: (X, \tau_2) \to (Y, \sigma)\) need not be continuous.

Correction: The topology chain gives \(\tau_{\mathfrak{a}}\subseteq \tau^{*}_{\mathfrak{a}}\subseteq \tau^{*}\). A function \(f\) is \(\tau_1\)-continuous if \(f^{-1}(V) \in \tau_1\) for all \(V \in \sigma\). If \(\tau_1 \subseteq \tau_2\), then \(\tau_2\)-continuous implies \(\tau_1\)-continuous (since \(f^{-1}(V) \in \tau_2 \supseteq \tau_1\)... no, this is the wrong direction).

Let us reconsider. \(f\) is \(\tau^{*}\)-continuous means \(f^{-1}(V) \in \tau^{*}\) for all \(V \in \sigma\). Since \(\tau^{*}_{\mathfrak{a}}\subseteq \tau^{*}\), \(\tau^{*}\)-continuity does NOT imply \(\tau^{*}_{\mathfrak{a}}\)-continuity.

The correct statement is: since \(\tau_{\mathfrak{a}}\subseteq \tau^{*}_{\mathfrak{a}}\), we have \(f^{-1}(V) \in \tau_{\mathfrak{a}}\implies f^{-1}(V) \in \tau^{*}_{\mathfrak{a}}\), so \(\tau_{\mathfrak{a}}\)-continuous implies \(\tau^{*}_{\mathfrak{a}}\)-continuous. Similarly, \(\tau^{*}_{\mathfrak{a}}\)-continuous implies \(\tau^{*}\)-continuous. ◻

8 Special Cases and Comparisons↩︎

We examine three important special cases that connect the ideal-aura framework to known theories.

8.1 Case 1: Trivial Ideal \(\mathcal{I} = \{\emptyset\}\)↩︎

Proposition 23. When \(\mathcal{I} = \{\emptyset\}\):

  1. \(A^{\mathfrak{a}}(\{\emptyset\}) = \operatorname{cl}_{\mathfrak{a}}(A) = \{x \in X : \mathfrak{a}(x) \cap A \neq \emptyset\}\);

  2. \(\operatorname{cl}^{*}_{\mathfrak{a}}(A) = \operatorname{cl}_{\mathfrak{a}}(A)\) (the aura-closure from [8]);

  3. \(\tau^{*}_{\mathfrak{a}}= \tau_{\mathfrak{a}}\);

  4. The \(\mathcal{I}\mathfrak{a}\)-generalized open sets coincide with the \(\mathfrak{a}\)-generalized open sets of [8].

Proof. Part (i): \(\mathfrak{a}(x) \cap A \notin \{\emptyset\}\) iff \(\mathfrak{a}(x) \cap A \neq \emptyset\). Parts (ii)–(iv) follow immediately. ◻

8.2 Case 2: Ideal of Finite Sets \(\mathcal{I} = \mathcal{I}_f\)↩︎

Proposition 24. When \(\mathcal{I} = \mathcal{I}_f\) (ideal of finite subsets of \(X\)):

  1. \(A^{\mathfrak{a}}(\mathcal{I}_f) = \{x \in X : \mathfrak{a}(x) \cap A \text{ is infinite}\}\);

  2. For finite \(X\), \(A^{\mathfrak{a}}(\mathcal{I}_f) = \emptyset\) for all \(A\), and \(\tau^{*}_{\mathfrak{a}}= \mathcal{P}(X)\);

  3. For \(X = \mathbb{R}\) with the usual topology and \(\mathfrak{a}_\varepsilon(x) = (x - \varepsilon, x + \varepsilon)\), \(A^{\mathfrak{a}_\varepsilon}(\mathcal{I}_f)\) consists of all points \(x\) such that \((x-\varepsilon, x+\varepsilon) \cap A\) is infinite. For any infinite \(A\) with no isolated points, \(A^{\mathfrak{a}_\varepsilon}(\mathcal{I}_f) = \operatorname{cl}_{\mathfrak{a}_\varepsilon}(A)\).

Proof. (i) is the definition. (ii) follows because every subset of a finite set is finite, hence in \(\mathcal{I}_f\). (iii) If \(A\) has no isolated points and \((x-\varepsilon, x+\varepsilon) \cap A \neq \emptyset\), then for any \(y \in (x-\varepsilon, x+\varepsilon) \cap A\), every open interval around \(y\) meets \(A\) in infinitely many points (since \(A\) has no isolated points), so \((x-\varepsilon, x+\varepsilon) \cap A\) is infinite. ◻

8.3 Case 3: Improper Ideal \(\mathcal{I} = \mathcal{P}(X)\)↩︎

Proposition 25. When \(\mathcal{I} = \mathcal{P}(X)\):

  1. \(A^{\mathfrak{a}} = \emptyset\) for all \(A\);

  2. \(\operatorname{cl}^{*}_{\mathfrak{a}}(A) = A\) for all \(A\) (identity operator);

  3. \(\tau^{*}_{\mathfrak{a}}= \mathcal{P}(X)\) (discrete topology).

8.4 Relationship Diagram↩︎

The following diagram summarizes the relationships among the topologies and closure operators:

Figure 1: image.

9 Compatibility with Generalized Open Sets↩︎

In this section, we investigate how the \(\mathcal{I}\mathfrak{a}\)-generalized open sets relate to the classical \(\mathcal{I}\)-generalized open sets and the pure \(\mathfrak{a}\)-generalized open sets from [8].

Theorem 26. For any \(\mathcal{I}\mathfrak{a}\)-space \((X, \tau, \mathcal{I}, \mathfrak{a})\) and any \(A \subseteq X\):

  1. If \(A\) is \(\mathcal{I}\)-open (i.e., \(A \subseteq \operatorname{int}(\operatorname{cl}^{*}(A))\)), it need not be \(\mathcal{I}\mathfrak{a}\)-open.

  2. If \(A\) is \(\mathcal{I}\mathfrak{a}\)-open (i.e., \(A \subseteq \operatorname{int}^{*}_{\mathfrak{a}}(A)\)), it need not be \(\mathcal{I}\)-open.

  3. If \(A\) is \(\mathfrak{a}\)-open (i.e., \(A \in \tau_{\mathfrak{a}}\)), then \(A\) is \(\mathcal{I}\mathfrak{a}\)-open.

Proof. (iii) If \(A \in \tau_{\mathfrak{a}}\), then \(A \in \tau^{*}_{\mathfrak{a}}\) (Theorem 12), so \(A \subseteq \operatorname{int}^{*}_{\mathfrak{a}}(A)\).

(i) and (ii) follow by counterexample. Since \(\operatorname{int}^{*}_{\mathfrak{a}}\) uses the aura-local function while the classical \(\mathcal{I}\)-interior uses the standard interior and \(*\)-closure, the two notions are generally independent. An \(\mathcal{I}\)-open set depends on \(\operatorname{int}\) and \(\operatorname{cl}^{*}\) (which involve \(\tau\) and \(\mathcal{I}\)), while an \(\mathcal{I}\mathfrak{a}\)-open set depends on \(\mathfrak{a}\) and \(\mathcal{I}\). Different choices of \(\mathfrak{a}\) can make the relationship go either way. ◻

Theorem 27. The following reduction diagram holds: \[\begin{array}{ccc} \text{\mathfrak{a}-open} & \xrightarrow{\;\;\subseteq\;\;} & \text{\mathcal{I}\mathfrak{a}-open} \\ & & \\ \text{(Paper I \cite{Acikgoz2026aura})} & & \text{(This paper)} \end{array}\] Specifically, every \(\mathfrak{a}\)-open set is \(\mathcal{I}\mathfrak{a}\)-open, and when \(\mathcal{I} = \{\emptyset\}\), the two notions coincide.

10 Conclusion and Open Problems↩︎

In this paper, we introduced the ideal-aura topological space \((X, \tau, \mathcal{I}, \mathfrak{a})\) and developed a systematic theory based on the aura-local function \(A^{\mathfrak{a}}(\mathcal{I})\). The main contributions are:

  1. We defined the aura-local function \(A^{\mathfrak{a}}(\mathcal{I}) = \{x \in X : \mathfrak{a}(x) \cap A \notin \mathcal{I}\}\) and established the fundamental inclusion \(A^{*}(\mathcal{I}, \tau) \subseteq A^{\mathfrak{a}}(\mathcal{I})\). The aura-local function inherits the key algebraic properties (monotonicity, finite additivity) of the classical Janković–Hamlett local function but differs crucially in that \(A^{\mathfrak{a}}\) need not be closed.

  2. The closure operator \(\operatorname{cl}^{*}_{\mathfrak{a}}(A) = A \cup A^{\mathfrak{a}}\) was shown to be an additive Čech closure operator. Non-idempotency was demonstrated with explicit counterexamples, and idempotency was shown to hold precisely when \(\mathfrak{a}\) is transitive (Theorem 13).

  3. The topology chain \(\tau_{\mathfrak{a}}\subseteq \tau^{*}_{\mathfrak{a}}\subseteq \tau^{*}\) was established, providing a natural interpolation between the pure aura topology and the Janković–Hamlett topology. The \(\psi_{\mathfrak{a}}\)-operator was introduced and shown to characterize \(\tau^{*}_{\mathfrak{a}}\) via \(A \in \tau^{*}_{\mathfrak{a}}\iff A \subseteq \psi_{\mathfrak{a}}(A)\) (Theorem 15).

  4. A basis \(\beta_{\mathfrak{a}}(\mathcal{I}) = \{\mathfrak{a}(x) \setminus J : x \in X, J \in \mathcal{I}\}\) for \(\tau^{*}_{\mathfrak{a}}\) was constructed for transitive aura functions (Theorem 16).

  5. Five classes of \(\mathcal{I}\mathfrak{a}\)-generalized open sets were introduced and their hierarchy was established. Decomposition theorems for \(\mathcal{I}\mathfrak{a}\)-continuity were proven.

  6. Three special cases were analyzed, showing that \(\tau^{*}_{\mathfrak{a}}\) reduces to \(\tau_{\mathfrak{a}}\) when \(\mathcal{I} = \{\emptyset\}\), to \(\mathcal{P}(X)\) when \(\mathcal{I} = \mathcal{P}(X)\), and exhibiting localized behavior for \(\mathcal{I} = \mathcal{I}_f\).

The following problems remain open:

Question 28. For general (non-transitive) \(\mathfrak{a}\), at what ordinal \(\alpha\) does the iteration \(\operatorname{cl}^{*\alpha}_{\mathfrak{a}}\) stabilize? Is countable ordinal always sufficient?

Question 29. Does the ideal-aura topology behave well under products? That is, given \((X, \tau, \mathcal{I}, \mathfrak{a})\) and \((Y, \sigma, \mathcal{J}, \mathfrak{b})\), what is the relationship between \((\tau \times \sigma)^{*}_{\mathfrak{a}\times \mathfrak{b}}\) and \(\tau^{*}_{\mathfrak{a}}\times \sigma^{*}_{\mathfrak{b}}\)?

Question 30. Define \(\mathcal{I}\mathfrak{a}\)-compactness. Does an \(\mathcal{I}\mathfrak{a}\)-compact subset of an \(\mathcal{I}\mathfrak{a}\)-\(T_2\) space have to be \(\operatorname{cl}^{*}_{\mathfrak{a}}\)-closed?

Question 31. For which ideals \(\mathcal{I}\) is \(\tau^{*}_{\mathfrak{a}}= \tau_{\mathfrak{a}}\) (i.e., the ideal adds no new open sets)?

Question 32. Can the ideal-aura framework be extended to include an ideal on the codomain, producing \((\mathcal{I}, \mathcal{J})\)-\(\mathfrak{a}\)-continuity?

Conflict of Interest↩︎

The author declares no conflict of interest.

Data Availability↩︎

No data was used for the research described in the article.

References↩︎

[1]
K. Kuratowski, Topology, Vol. I, Academic Press, New York, 1966.
[2]
R. Vaidyanathaswamy, The localisation theory in set-topology, Proc. Indian Acad. Sci.20(1945), 51–61.
[3]
D. Janković, T.R. Hamlett, New topologies from old via ideals, Amer. Math. Monthly97(1990), 295–310.
[4]
R.L. Newcomb, Topologies which are compact modulo an ideal, Ph.D. Thesis, University of California at Santa Barbara, 1967.
[5]
A. Açıkgöz, Ş. Yüksel, T. Noiri, \(\alpha\)-\(I\)-continuous and \(\alpha\)-\(I\)-open functions, Acta Math. Hungar.105(2004), 27–37.
[6]
A. Açıkgöz, T. Noiri, Ş. Yüksel, A decomposition of continuity in ideal topological spaces, Acta Math. Hungar.105(2004), 285–289.
[7]
E. Hatır, T. Noiri, On decompositions of continuity via idealization, Acta Math. Hungar.106(2005), 227–238.
[8]
A. Açıkgöz, Aura topological spaces and generalized open sets with applications to rough set theory, submitted, 2026.
[9]
A. Açıkgöz, Compactness and connectedness in aura topological spaces, submitted, 2026.

Subjects

General Topology (math.GN)

Updated on Academus

2026/06/23 02:23 UTC