On Graph Primal Topological Spaces

Abstract

In this paper, we introduce the concept of the “graph primal,” which serves as the dual structure to the “graph grill”. We present several results associated with graph primal operations. Moreover, we introduce two new graph-local functions on graph adjacency topological spaces (graph ATSs). We then explore the basic properties of the proposed graph-local functions and describe a method for generating two new graph ATSs from existing ones via graph primals. In addition, we examine several fundamental properties and connections of the resulting topologies, supported by some counterexamples. Furthermore, we characterize the nature of the open sets of these new topologies in terms of closure operators. Finally, we assess the compatibility of graph ATSs with the graph primal concept.

1. Introduction

General topology, initially defined by Hausdorff in 1914, and graph theory, presented in [1], are two significant areas of mathematics that are intimately connected. Creating topologies on a graph’s collection of vertices and edges is one way that graph theory and general topology are related. Directed and undirected graphs have been used in a number of manuscripts to design various topologies (see [2,3,4,5,6]). The theory of simple undirected graphs, namely the sets of vertices in such graphs, accounts for the majority of these constructs. A key to connecting topological structures with graph theory is a relation on a graph. This relation gives the graph additional kinds of topological structures. The labeled topologies on n points and the labeled transitive directed graph with n points have a one-to-one correspondence, as demonstrated in [7]. The lattice graph of the topologies of transitive directed graphs, as proposed by the authors of [7], was investigated in 1967 in [8]. The relationship between finite topologies and directed graphs was examined in 2010 by [9]. In 2013, the authors of [10] proposed a topology on the vertices of an undirected graph. In 2018, the authors of [11] linked a vertex set of simple graphs without isolated vertices to an incidence topology. In 2018, the authors of [12] developed a novel topology, constructed using the incidence topology on the set of vertices for simple graphs = ($\mathfrak{Q}$(),$\mathfrak{Z}$()) without isolated vertices, which has a subbasis generated by the family of end sets that include only the endpoints of each edge. In [13], the authors used the graphs =($\mathfrak{Q}$(),$\mathfrak{Z}$()) to induce two topologies on the set of its edges $\mathfrak{Z}$(), denoted by compatible edge topology and incompatible edge topology. A relation on graphs was introduced in [14] to produce new styles of topological structures. In [15], the authors explained how to generate a topology using incidence and adjacency relations on the vertex set of graphs. They also examined the closure and interior properties of vertex sets of subgraphs in the graph adjacency topological space (graph ATS). Numerous new ideas have been added to topology, enriching it with a range of recently created fields of study. Closure spaces, proximity spaces, ideals [16], filters [17], grills [18], and primals [19] are a few examples of innovative structures that topologists have developed in an attempt to investigate properties of topological spaces. The new notion of a primal structure was developed and discussed as the dual of the notion of a grill. Recently, many authors have studied the relationships among primal topological spaces, topological spaces, and the soft primal in soft topological spaces and have investigated their basic properties. Primal proximity spaces inspired by primal and proximity notions have had a significant impact on the development of operators in primal topological spaces (see [20,21,22,23]). Additionally, the authors of [24] created graph grills and investigated the characteristics of the generated topologies on the vertex set of graphs. Similarly, graph ideals were developed and investigated in [25] to generate new topologies on the vertex set of graphs. The majority of research to date has focused on ideals and primals in general topological spaces. It is evident that the majority of real-world problems may be represented as graphs and resolved by applying the ideas of graph theory. In the current contribution, we introduce a new link between graph theory and ordinary topological spaces. Based on our new concept of a “graph primal”, we propose new definitions, theorems, methods, applications, and broader generalizations, which were not covered in previous papers (see [24,25]). Moreover, we develop a different methodological approach that provides fresh insights into the interplay between topology and graph theory.

In this paper, we propose the notion of “graph primal” as the dual structure of graph grill. Moreover, we introduce the new graph local function ${(.)}^{•}$ on a graph ATS. Further, we explore the basic facts of the proposed graph local function and describe the method of generating the new operator ${\mathfrak{CL}}^{•}$ via graph primals with the help of ${(.)}^{•}$, generating a unique graph ATS ${\mathcal{T}}_{Ad}^{•}$. In addition, we propose the operator $\Theta$ and its associated topology $\sigma .$ Several fundamental properties and connections of the new topologies $\sigma$ and ${\mathcal{T}}_{Ad}^{•}$ in the graph primal ATSs were examined. This work ends by analyzing the suitability of the graph primal ATS with respect to the graph primal and by characterizing the nature of the open sets of the novel topologies in terms of the closure operators.

1.1. Key Contributions

The key contributions of this paper can be summarized as follows:

• Introduction of Graph Primal Structure: This paper discusses the novel notion of graph primal structure as a dual to existing concepts in topology, enriching the theoretical framework.
• Graph Theory Applications: It highlights how classical graph theory can be integrated with modern topology, potentially leading to new insights and methodologies.
• Development of New Topological Spaces: This research introduces various new topological spaces that can be derived from graph structures.
• Real-World Problem Solving: By representing numerous real-world problems as graphs, this paper underscores the practical implications of these theoretical developments.
• Future Research Opportunities: This paper is significant due to its innovative contributions to the fields of topology and graph theory. By introducing new concepts and exploring their implications, it not only enhances theoretical knowledge but also provides practical tools for addressing real-world problems. The potential for future research stemming from this work underscores its importance in advancing mathematical sciences.

1.2. Comprehensive Analysis

Our study offers original contributions to the understanding of graph theory by proposing the “graph primal”. Graph primal theory introduces innovative concepts, particularly focusing on graph primals and their associated topologies. It provides a framework for analyzing complex problems represented as graphs, utilizing various mathematical structures to enhance understanding and applications in real-world scenarios. Also, this paper offers a novel graph-local function in Definition 10 via our new concept, the “graph primal”, which differs from the corresponding functions given in [24,25]. Neither our new graph-local function given in Definition 10 nor its related properties, theorems, and graphical topological spaces are defined in [24,25]. Furthermore, this analysis synthesizes the essential elements and implications of graph primal theory, showcasing its relevance and potential applications in various domains. The exploration of graph structures through this lens not only furthers mathematical understanding but also enhances practical problem-solving strategies across multiple disciplines.

2. Preliminaries

Some basic definitions and introductions to graph theory and topology can be found in [1,25,26].

A pair ($\mathfrak{Q}$(), $\mathfrak{Z}$()) denotes a graph , where $\mathfrak{Q}$() is a nonempty finite set and $\mathfrak{Z}$() is a set of unordered pairs of elements of $\mathfrak{Q}$(). The vertex set of is the set $\mathfrak{Q}$(), and the edge set of is the set $\mathfrak{Z}$(). The elements of $\mathfrak{Q}$() are called the vertices or the nodes ), and the elements of $\mathfrak{Z}$() are called the edges ). A loop is an edge of $\mathfrak{Z}$() that connects a vertex of $\mathfrak{Z}$() to itself. Multiple or parallel edges are edges that connect the same vertices. Two nodes ${⅁}_1$ and ${⅁}_2$ of are called adjacent to each other if they are connected by an edge $\alpha$ of . In this case, the edge $\alpha$ is said to connect ${⅁}_1$ and ${⅁}_2$. Moreover, the vertices ${⅁}_1$ and ${⅁}_2$ are called the endpoints of this edge. Two vertices ${⅁}_1$ and ${⅁}_2$ of are called non-adjacent to each other if there are no edges between them (they are not adjacent). In a graph , the degree of a vertex $⅁\in \mathfrak{Q}$() is represented by $deg(⅁)$, which is the number of edges that are connected to $⅁$. A regular graph is one in which all of the vertices have the same degree. An “isolated vertex” refers to a vertex of degree 0. For any node $⅁\in \mathfrak{Q}$(, the neighbors of $⅁$ in are the nodes that are adjacent to $⅁.$ The pair = $(\varnothing ,\varnothing )$ stands for the empty graph. If = $(\mathfrak{Q},\mathfrak{Z})$ and ′ = (${\mathfrak{Q}}^{\prime },{\mathfrak{Z}}^{\prime }$), then ∪ ′ = $(\mathfrak{Q}\cup {\mathfrak{Q}}^{\prime },\mathfrak{Z}\cup {\mathfrak{Z}}^{\prime })$ and ∩ ′ = $(\mathfrak{Q}\cap {\mathfrak{Q}}^{\prime },\mathfrak{Z}\cap {\mathfrak{Z}}^{\prime })$. If ∩ ′ = $(\varnothing ,\varnothing ),$ then and ′ are disjoint. If $\mathfrak{Q}\subseteq {\mathfrak{Q}}^{\prime }$ and $\mathfrak{Z}\subseteq {\mathfrak{Z}}^{\prime },$ then ′ is a subgraph of , and is a supergraph of ′, written as ′ ⊆ . A simple graph is a graph that has no loops and no multiple edges. If $\mathfrak{Q}$ can be divided into two distinct subsets, ${\mathfrak{Q}}_1$ and ${\mathfrak{Q}}_2$, such that each edge of connects a vertex of ${\mathfrak{Q}}_1$ to a vertex of ${\mathfrak{Q}}_2$, then the graph is said to be bipartite, and the pair $({\mathfrak{Q}}_1,{\mathfrak{Q}}_2)$ is called a bipartition of the graph .

Definition 1

([10,25]). Let = ($\mathfrak{Q}$(), $\mathfrak{Z}$()) be a graph. For $⅁\in \mathfrak{Q}$(), the neighborhood set $N_{⅁}$ of $⅁$ is defined as $N_{⅁}$ = {${⅁}^{\prime }\in \mathfrak{Q}$(): $⅁{⅁}^{\prime }\in \mathfrak{Z}$()}.

Definition 2

([10,25]). Consider the graph = ($\mathfrak{Q}$(),$\mathfrak{Z}$(), in which no vertex is isolated. The collection of $N_{⅁}$ for all $⅁\in \mathfrak{Q}$() is defined as $S_N$. Stated differently, $S_N$ = {$N_{⅁}:⅁\in \mathfrak{Q}$()}. The graph adjacency topological space (graph ATS) is the pair ($\mathfrak{Q}$(), ${\mathcal{T}}_{Ad}$), and $S_N$ forms a subbase for the topology ${\mathcal{T}}_{Ad}$ on $\mathfrak{Q}$().

Definition 3

([19]). Let E be a nonempty set and $\beth ,\daleth \subseteq E$. A collection $\mathcal{P}\subseteq 2^E$ is said to be a primal on E if it satisfies the following statements:

(1) 

$E\notin \mathcal{P}$;

(2) 

if $\beth \in \mathcal{P}$ and $\daleth \subseteq \beth$, then $\daleth \in \mathcal{P}$;

(3) 

if $\beth \cap \daleth \in \mathcal{P}$, then $\beth \in \mathcal{P}$ or $\daleth \in \mathcal{P}$.

Definition 4

([24]). Consider the graph = ($\mathfrak{Q}$(), $\mathfrak{Z}$()). In the given graph, $P\left(\mathfrak{Q}\right)$ and $P\left(\mathfrak{Z}\right)$ are the power sets of $\mathfrak{Q}$() and $\mathfrak{Z}$(), respectively. The collection $\mathcal{S}$ = {′: ′ = (${\mathfrak{Q}}^{\prime }$, ${\mathfrak{Z}}^{\prime }$), where ${\mathfrak{Q}}^{\prime }\subseteq \mathfrak{Q},{\mathfrak{Z}}^{\prime }\subseteq \mathfrak{Z}$}, is said to be a graph grill on a graph ATS ($\mathfrak{Q}$(), ${\mathcal{T}}_{Ad}$ if it satisfies the following three conditions:

(1) 

$(\varnothing ,\varnothing )\notin \mathcal{S}.$

(2) 

If ′ is a subgraph of ″ for some ′ ∈ $\mathcal{S}$, then ″ ∈ $\mathcal{S}$.

(3) 

If ′ ∪ (″ ∈ $\mathcal{S}$, then ′ ∈ $\mathcal{S}$ or ″ ∈$\mathcal{S}$.

Simple undirected graphs are the ones discussed throughout this study. The symbol refers to a graph. We abbreviate the word “simple graph” to “graph”. The power set of a graph is denoted by $P$() or 2. For any vertex subset $\left(\mathfrak{Q}\right(\beth )$ of a subgraph $\beth$ ⊆ , the closure and interior with respect to the graph primal ATS are denoted by $\mathfrak{CL}\left(\mathfrak{Q}\right(\beth )$ and $\mathfrak{int}\left(\mathfrak{Q}\right(\beth ),$ respectively.

3. Graph Primals and Graph-Local Functions

This section defines a graph primal on a graph ATS, listing specific conditions that must be satisfied. It discusses the implications of these conditions and provides examples to illustrate the concept of graph primals. The introduction of new graph-local functions marks a significant advancement in the field, allowing for the generation of unique graph ATSs and enhancing the understanding of their properties. Several results related to the graph primal ATSs are discussed in detail with the help of some counterexamples. Further, we define another topology on the vertex set of a given graph. The properties and relationships between the newly proposed topologies and existing structures are discussed.

Definition 5.

Consider the graph = ($\mathfrak{Q}$(), $\mathfrak{Z}$()). It is assumed that the defined graph has $P\left(\mathfrak{Q}\right)$ and $P\left(\mathfrak{Z}\right)$ as the power sets of $\mathfrak{Q}$() and $\mathfrak{Z}$(), respectively. The collection $\mathcal{P}$ = {′: ′ = $\left({\mathfrak{Q}}^{\prime },{\mathfrak{Z}}^{\prime }\right)$, where ${\mathfrak{Q}}^{\prime }\subseteq \mathfrak{Q},{\mathfrak{Z}}^{\prime }\subseteq \mathfrak{Z}$} is said to be a graph primal on a graph ATS ($\mathfrak{Q}$(), ${\mathcal{T}}_{Ad}$) if it satisfies the following three statements:

(1) 

∉ $\mathcal{P}$.

(2) 

If ′ is a subgraph of ″ for some ″ ∈$\mathcal{P}$, then ′ ∈$\mathcal{P}$.

(3) 

If ′ ∩″ ∈$\mathcal{P}$, then ′ ∈ $\mathcal{P}$ or ″ ∈ $\mathcal{P}$.

If there is no confusion, we represent the graph primal $\mathcal{P}$ by the union of all of its members. Moreover, the triple ($\mathfrak{Q}$(), ${\mathcal{T}}_{Ad},\mathcal{P}$) is called a graph primal ATS.

Corollary 1.

Consider the graph = ($\mathfrak{Q}$(), $\mathfrak{Z}$()). It is assumed that the defined graph has $P\left(\mathfrak{Q}\right)$ and $P\left(\mathfrak{Z}\right)$ as the power sets of $\mathfrak{Q}$() and $\mathfrak{Z}$(), respectively. The collection $\mathcal{P}$ = {′: ′ = (${\mathfrak{Q}}^{\prime }$, ${\mathfrak{Z}}^{\prime }$, where ${\mathfrak{Q}}^{\prime }\subseteq \mathfrak{Q},{\mathfrak{Z}}^{\prime }\subseteq \mathfrak{Z}$} is said to be a graph primal on a graph ATS ($\mathfrak{Q}$(), ${\mathcal{T}}_{Ad}$ if it satisfies the following three statements:

(1) 

∉ $\mathcal{P}$.

(2) 

If ′ is a subgraph of ″ for some ″ ∉ $\mathcal{P}$, then ′ ∉ $\mathcal{P}$.

(3) 

If ′ ∉ $\mathcal{P}$ for some ″ ∉ $\mathcal{P}$, then ′ ∩ ″ ∉ $\mathcal{P}$.

Example 1.

Let be the graph ($\mathfrak{Q}$(), $\mathfrak{Z}$()), where $\mathfrak{Q}$() = $\{{⅁}_1,{⅁}_2\}$ and $\mathfrak{Z}$() = {α}. A drawing of the graph is shown in Figure 1.

Figure 1. Graph defined in Example 1.

A graph of two vertices has at most one edge, and so all possible graph primals on the above graph are

${\mathcal{P}}_1=\varnothing$, where “∅” is the empty family, which differs from the family containing the empty set “$\{\varnothing \}$”,

${\mathcal{P}}_2=\{(\varnothing ,\varnothing ),(\left\{{⅁}_1\right\},\varnothing )\}$,

${\mathcal{P}}_3=\left\{((\varnothing ,\varnothing ),\left\{{⅁}_2\right\},\varnothing )\right\}$,

${\mathcal{P}}_4=\{(\varnothing ,\varnothing ),(\left\{{⅁}_1\right\},\varnothing ),(\left\{{⅁}_2\right\},\varnothing )\}$.

Note that $\mathcal{P}=\{(\varnothing ,\varnothing ),(\left\{{⅁}_1\right\},\varnothing ),(\left\{{⅁}_2\right\},\varnothing ),(\{{⅁}_1,{⅁}_2\},\left\{\alpha \right\})\}$ is not a graph primal on since = ($\{{⅁}_1,{⅁}_2\},\left\{\alpha \right\}$) ∈ $\mathcal{P}$.

Example 2.

Let be the graph ( $\mathfrak{Q}$(), $\mathfrak{Z}$()), where $\mathfrak{Q}$() = {${⅁}_1,{⅁}_2,{⅁}_3$} and $\mathfrak{Z}$() = {${\alpha }_1,{\alpha }_2,{\alpha }_3$}. A drawing of the graph is shown in Figure 2.

Figure 2. Graph defined in Example 2.

For the above graph, some possible graph primals on are

$$\begin{array}{l}{\mathcal{P}}_1=\varnothing ,\\ {\mathcal{P}}_2=\{(\varnothing ,\varnothing ),(\left\{{⅁}_1\right\},\varnothing ),(\left\{{⅁}_2\right\},\varnothing ),(\left\{{⅁}_3\right\},\varnothing ),(\{{⅁}_1,{⅁}_2\},\varnothing ),(\{{⅁}_2,{⅁}_3\},\varnothing ),(\{{⅁}_1,{⅁}_3\},\varnothing ),(\{{⅁}_1,{⅁}_2,{⅁}_3\},\varnothing ),(\{{⅁}_1,{⅁}_2\},\\ \left\{{\alpha }_1\right\}),(\{{⅁}_1,{⅁}_3\},\left\{{\alpha }_3\right\}),(\{{⅁}_2,{⅁}_3\},\left\{{\alpha }_2\right\}),(\{{⅁}_1,{⅁}_2,{⅁}_3\},\left\{{\alpha }_1\right\}),(\{{⅁}_1,{⅁}_2,{⅁}_3\},\left\{{\alpha }_2\right\}),(\{{⅁}_1,{⅁}_2,{⅁}_3\},\{{\alpha }_1,{\alpha }_2\}\left)\right\},\\ {\mathcal{P}}_3=\{(\varnothing ,\varnothing ),(\left\{{⅁}_1\right\},\varnothing ),(\left\{{⅁}_2\right\},\varnothing ),(\left\{{⅁}_3\right\},\varnothing ),(\{{⅁}_1,{⅁}_2\},\varnothing ),(\{{⅁}_2,{⅁}_3\},\varnothing ),(\{{⅁}_1,{⅁}_3\},\varnothing ),(\{{⅁}_1,{⅁}_2,{⅁}_3\},\varnothing ),(\{{⅁}_1,{⅁}_2\},\\ \left\{{\alpha }_1\right\}),(\{{⅁}_1,{⅁}_3\},\left\{{\alpha }_3\right\}),(\{{⅁}_2,{⅁}_3\},\left\{{\alpha }_2\right\}),(\{{⅁}_1,{⅁}_2,{⅁}_3\},\left\{{\alpha }_1\right\}),(\{{⅁}_1,{⅁}_2,{⅁}_3\},\left\{{\alpha }_3\right\}),(\{{⅁}_1,{⅁}_2,{⅁}_3\},\{{\alpha }_1,{\alpha }_3\}\left)\right\},\\ {\mathcal{P}}_4=\{(\varnothing ,\varnothing ),(\left\{{⅁}_1\right\},\varnothing ),(\left\{{⅁}_2\right\},\varnothing ),(\left\{{⅁}_3\right\},\varnothing ),(\{{⅁}_1,{⅁}_2\},\varnothing ),(\{{⅁}_2,{⅁}_3\},\varnothing ),(\{{⅁}_1,{⅁}_3\},\varnothing ),(\{{⅁}_1,{⅁}_2,{⅁}_3\},\varnothing ),(\{{⅁}_1,{⅁}_2\},\\ \left\{{\alpha }_1\right\}),(\{{⅁}_1,{⅁}_3\},\left\{{\alpha }_3\right\}),(\{{⅁}_2,{⅁}_3\},\left\{{\alpha }_2\right\}),(\{{⅁}_1,{⅁}_2,{⅁}_3\},\left\{{\alpha }_2\right\}),(\{{⅁}_1,{⅁}_2,{⅁}_3\},\left\{{\alpha }_3\right\}),(\{{⅁}_1,{⅁}_2,{⅁}_3\},\{{\alpha }_2,{\alpha }_3\}\left)\right\},\\ {\mathcal{P}}_5=\{(\varnothing ,\varnothing ),(\left\{{⅁}_1\right\},\varnothing ),(\left\{{⅁}_2\right\},\varnothing ),(\left\{{⅁}_3\right\},\varnothing ),(\{{⅁}_1,{⅁}_2\},\varnothing ),(\{{⅁}_2,{⅁}_3\},\varnothing ),(\{{⅁}_1,{⅁}_3\},\varnothing ),(\{{⅁}_1,{⅁}_2,{⅁}_3\},\varnothing ),(\{{⅁}_1,{⅁}_2\},\\ \left\{{\alpha }_1\right\}),(\{{⅁}_1,{⅁}_3\},\left\{{\alpha }_3\right\}),(\{{⅁}_2,{⅁}_3\},\left\{{\alpha }_2\right\}),\left(\right\{{⅁}_1,{⅁}_2,{⅁}_3\},\left\{{\alpha }_1\right\}),(\{{⅁}_1,{⅁}_2,{⅁}_3\},\left\{{\alpha }_2\right\}),(\{{⅁}_1,{⅁}_2,{⅁}_3\},\left\{{\alpha }_3\right\}),\\ \left(\right\{{⅁}_1,{⅁}_2,{⅁}_3\},\{{\alpha }_1,{\alpha }_2\}),(\{{⅁}_1,{⅁}_2,{⅁}_3\},\{{\alpha }_1,{\alpha }_3\})\},\\ {\mathcal{P}}_6=\{(\varnothing ,\varnothing ),(\left\{{⅁}_1\right\},\varnothing ),(\left\{{⅁}_2\right\},\varnothing ),(\left\{{⅁}_3\right\},\varnothing ),(\{{⅁}_1,{⅁}_2\},\varnothing ),(\{{⅁}_2,{⅁}_3\},\varnothing ),(\{{⅁}_1,{⅁}_3\},\varnothing ),(\{{⅁}_1,{⅁}_2,{⅁}_3\},\varnothing ),(\{{⅁}_1,{⅁}_2\},\\ \left\{{\alpha }_1\right\}),(\{{⅁}_1,{⅁}_3\},\left\{{\alpha }_3\right\}),(\{{⅁}_2,{⅁}_3\},\left\{{\alpha }_2\right\}),(\{{⅁}_1,{⅁}_2,{⅁}_3\},\left\{{\alpha }_1\right\}),(\{{⅁}_1,{⅁}_2,{⅁}_3\},\left\{{\alpha }_2\right\}),(\{{⅁}_1,{⅁}_2,{⅁}_3\},\left\{{\alpha }_3\right\}),\\ \left(\right\{{⅁}_1,{⅁}_2,{⅁}_3\},\{{\alpha }_1,{\alpha }_3\}),(\{{⅁}_1,{⅁}_2,{⅁}_3\},\{{\alpha }_2,{\alpha }_3\})\}.\end{array}$$

Note that $\mathcal{P}=\{(\varnothing ,\varnothing ),(\left\{{⅁}_1\right\},\varnothing ),(\left\{{⅁}_2\right\},\varnothing ),(\left\{{⅁}_3\right\},\varnothing ),(\{{⅁}_1,{⅁}_2\},\left\{{\alpha }_1\right\})\}$ is not a graph primal on G since, for example, $G=(\left\{{⅁}_3\right\},\varnothing )\in \mathcal{P}$, but neither $(\{{⅁}_2,{⅁}_3\},\varnothing )\in \mathcal{P}$ nor $(\{{⅁}_1,{⅁}_3\},\varnothing )\in \mathcal{P}.$

Remark 1.

It should be noted that, for a graph = ($\mathfrak{Q}$(), $\mathfrak{Z}\left(G\right)$) with n vertices and no edges ($\mathfrak{Z}$() = $\varnothing$), $2^n$ graph primals can be written on the graph .

Theorem 1.

Let $\mathcal{E}$ be a graph grill on the graph . Then, {″: ″ = (${\mathfrak{Q}}^{\prime \prime }$, ${\mathfrak{Z}}^{\prime \prime }$)|(${\mathfrak{Q}}^{\prime \prime }$)^c = ${\mathfrak{Q}}^{\prime }$, ′ =$({\mathfrak{Q}}^{\prime },{\mathfrak{Z}}^{\prime })\in \mathcal{S}$ where ${\mathfrak{Q}}^{\prime },{\mathfrak{Q}}^{\prime \prime }\subseteq \mathfrak{Q}$ and ${\mathfrak{Z}}^{\prime },{\mathfrak{Z}}^{\prime \prime }\subseteq \mathfrak{Z}$} is a graph primal on .

Proof. 

Let $\mathcal{S}$ be a graph grill of the graph and $\mathcal{P}$ = {″: ″ = (${\mathfrak{Q}}^{\prime \prime },{\mathfrak{Z}}^{\prime \prime }$)|(${\mathfrak{Q}}^{\prime \prime }$)^c = ${\mathfrak{Q}}^{\prime }$, ′ = (${\mathfrak{Q}}^{\prime },{\mathfrak{Z}}^{\prime }$) ∈ $\mathcal{S}$ where ${\mathfrak{Q}}^{\prime },{\mathfrak{Q}}^{\prime \prime }\subseteq \mathfrak{Q}$ and ${\mathfrak{Z}}^{\prime },{\mathfrak{Z}}^{\prime \prime }\subseteq \mathfrak{Z}$}. Then, we show that $\mathcal{P}$ is a graph primal:

(1)

Since $(\varnothing ,\varnothing )\notin \mathcal{S}$, it follows that ∉ $\mathcal{P}$.

(2)

Let ″ =$({\mathfrak{Q}}^{\prime \prime },{\mathfrak{Z}}^{\prime \prime })\in \mathcal{P}$ and ^★★ = $({\mathfrak{Q}}^{★★},{\mathfrak{Z}}^{★★})$ ⊆ ″. Then, ${\mathfrak{Q}}^{\prime }={\left({\mathfrak{Q}}^{\prime \prime }\right)}^c\subseteq {\mathfrak{Q}}^{★}={\left({\mathfrak{Q}}^{★★}\right)}^c$. Since ′ = $({\mathfrak{Q}}^{\prime },{\mathfrak{Z}}^{\prime })\in \mathcal{S}$, it follows that ^★ = $({\mathfrak{Q}}^{★},{\mathfrak{Z}}^{★})\in \mathcal{S}$. Hence, ^★★ = $({\mathfrak{Q}}^{★★},{\mathfrak{Z}}^{★★})\in \mathcal{P}$.

(3)

Let ′ ∩ ″ = $({\mathfrak{Q}}^{\prime },{\mathfrak{Z}}^{\prime })\cap ({\mathfrak{Q}}^{\prime \prime },{\mathfrak{Z}}^{\prime \prime })\in \mathcal{P}$. Then, ${\left({\mathfrak{Q}}^{\prime }\right)}^c\cup {\left({\mathfrak{Q}}^{\prime \prime }\right)}^c={({\mathfrak{Q}}^{\prime }\cap {\mathfrak{Q}}^{\prime \prime })}^c$. Therefore, we get ^★ = $({\left({\mathfrak{Q}}^{\prime }\right)}^c,{\mathfrak{Z}}^{★})\in \mathcal{S}$ or ^★★ = $({\left({\mathfrak{Q}}^{\prime \prime }\right)}^c,{\mathfrak{Z}}^{★★})\in \mathcal{S}$. Thus, ′ ∈ $\mathcal{P}$ or ″ ∈ $\mathcal{P}$. Hence, $\mathcal{P}$ is a graph primal on the graph .

□

Theorem 2.

Let $\mathcal{P}$ and $\mathcal{Q}$ be two graph primals on a graph . Then, $\mathcal{P}\cup \mathcal{Q}$ is a graph primal on the graph .

Proof. 

(1) Let $\mathcal{P}$ and $\mathcal{Q}$ be two graph primals on X. Then, ∉ $\mathcal{P}$ and ∉ $\mathcal{Q}$. Hence, ∉ $\mathcal{P}\cup \mathcal{Q}$.

(2)

Let $\beth \in \mathcal{P}\cup \mathcal{Q}$ and $\daleth \subseteq \beth$. Then, $\beth \in \mathcal{P}$ or $\beth \in \mathcal{Q}$. Therefore, $\daleth \in \mathcal{P}$ or $\daleth \in \mathcal{Q}$. As a result, $\daleth \in \mathcal{P}\cup \mathcal{Q}$.

(3)

Let $\beth \cap \daleth \in \mathcal{P}\cup \mathcal{Q}$. Then, $\beth \cap \daleth \in \mathcal{P}$ or $\beth \cap \daleth \in \mathcal{Q}$. If $\beth \cap \daleth \in \mathcal{P}$, then either $\beth \in \mathcal{P}$ or $\daleth \in \mathcal{P}$. Again, if $\beth \cap \daleth \in \mathcal{Q}$, then either $\daleth \in \mathcal{Q}$ or $\daleth \in \mathcal{Q}$. Then, obviously $\beth \in \mathcal{P}\cup \mathcal{Q}$ or $\daleth \in \mathcal{P}\cup \mathcal{Q}$.

Hence, $\mathcal{P}\cup \mathcal{Q}$ is a graph primal on the graph . □

Remark 2.

The following example shows that the intersection of two graph primals related to a graph need not be a graph primal on .

Example 3.

Let be the graph ($\mathfrak{Q}$(), $\mathfrak{Z}$()), where $\mathfrak{Q}$() = $\{{⅁}_1,{⅁}_2,{⅁}_3\}$ and $\mathfrak{Z}$() = $\{{\alpha }_1,{\alpha }_2\}$. A drawing of the graph is shown in Figure 3.

Figure 3. Graph defined in Example 3.

The representations of the following three graph primals are given in Table 1.

$$\begin{array}{l}{\mathcal{P}}_1=\{(\varnothing ,\varnothing ),(\left\{{⅁}_1\right\},\varnothing ),(\left\{{⅁}_2\right\},\varnothing ),(\left\{{⅁}_3\right\},\varnothing ),(\{{⅁}_1,{⅁}_2\},\varnothing ),\left(\right\{{⅁}_2,{⅁}_3\},\varnothing ),(\{{⅁}_1,{⅁}_3\},\varnothing ),\left(\right\{{⅁}_1,{⅁}_2,{⅁}_3\},\varnothing ),\\ (\{{⅁}_1,{⅁}_2\},\left\{{\alpha }_1\right\}),(\{{⅁}_1,{⅁}_2,{⅁}_3\},\left\{{\alpha }_1\right\})\}.\\ {\mathcal{P}}_2=\{(\varnothing ,\varnothing ),(\left\{{⅁}_1\right\},\varnothing ),(\left\{{⅁}_2\right\},\varnothing ),(\left\{{⅁}_3\right\},\varnothing ),(\{{⅁}_1,{⅁}_2\},\varnothing ),(\{{⅁}_2,{⅁}_3\},\varnothing ),(\{{⅁}_1,{⅁}_3\},\varnothing ),\left(\right\{{⅁}_1,{⅁}_2,{⅁}_3\},\varnothing ),\\ \left(\right\{{⅁}_1,{⅁}_3\},\left\{{\alpha }_2\right\}),(\{{⅁}_1,{⅁}_2,{⅁}_3\},\left\{{\alpha }_2\right\})\}.\end{array}$$

Table 1. Representations of the tree graph primals in Example 3.

The Graph Primal ${\mathcal{P}}_1$	The Graph Primal ${\mathcal{P}}_2$	${\mathcal{P}}_1\cap {\mathcal{P}}_2$

From the computations of the members of ${\mathcal{P}}_1\cap {\mathcal{P}}_2$ in the Table 1, it is clear that ${\mathcal{P}}_1\cap {\mathcal{P}}_2$ does not represent a graph primal on the graph , since ${\mathcal{P}}_1\cap {\mathcal{P}}_2=\{(\varnothing ,\varnothing ),(\left\{{⅁}_1\right\},\varnothing ),(\left\{{⅁}_2\right\},\varnothing )$, $(\left\{{⅁}_3\right\},\varnothing ),(\{{⅁}_1,{⅁}_2\},\varnothing ),$$(\{{⅁}_2,{⅁}_3\},\varnothing ),$$(\{{⅁}_1,{⅁}_3\},\varnothing ),(\{{⅁}_1,{⅁}_2,{⅁}_3\},\varnothing )\}$, and, for example, $\beth =(\left\{{⅁}_3\right\},\varnothing )=(\{{⅁}_1,{⅁}_3\},{\alpha }_1)\cap (\{{⅁}_2,{⅁}_3\},{\alpha }_2)\in {\mathcal{P}}_1\cap {\mathcal{P}}_2$, but neither $(\{{⅁}_1,{⅁}_3\},{\alpha }_1)\in {\mathcal{P}}_1\cap {\mathcal{P}}_2$ nor $(\{{⅁}_2,{⅁}_3\},{\alpha }_2)\in {\mathcal{P}}_1\cap {\mathcal{P}}_2$.

Definition 6.

Let ($\mathfrak{Q}$(), ${\mathcal{T}}_{Ad}$) be a graph ATS and $⅁$ ∈ $\mathfrak{Q}$(). The open neighborhood system at $⅁$, denoted by $N(⅁)$, is given as $N(⅁)$$=\left\{U\in {\mathcal{T}}_{Ad}:⅁\in U\right\}$.

Definition 7.

Let ($\mathfrak{Q}$(), ${\mathcal{T}}_{Ad},\mathcal{P}$) be a graph primal ATS. Let $\daleth$ be a subgraph of . Then, ${\left(\mathfrak{Q}(\daleth )\right)}^{•}\left(\mathcal{P},{\mathcal{T}}_{Ad}\right)={\left(\mathfrak{Q}(\daleth )\right)}^{•}$ = {$⅁$ ∈ $\mathfrak{Q}$(): for every $U\in N(⅁),\mathfrak{Q}{(\daleth )}^c\cup U^c$ = $\mathfrak{Q}$(′) and ′ ∈$\mathcal{P}$} is called the graph-local function of $\mathfrak{Q}(\daleth )$ with respect to $\mathcal{P}$ and ${\mathcal{T}}_{Ad}$.

Example 4.

Let be the graph ($\mathfrak{Q}$(), $\mathfrak{Z}$()), where $\mathfrak{Q}$() = $\{{⅁}_1,{⅁}_2,{⅁}_3,{⅁}_4,{⅁}_5,{⅁}_6\}$ and $\mathfrak{Z}$() = $\{{\alpha }_1,{\alpha }_2,{\alpha }_3,{\alpha }_4\}$. A drawing of the graph is shown in Figure 4.

Figure 4. Graph defined in Example 4.

Now we illustrate some graph-local functions of a subgraph of the above graph.

$$\begin{array}{l}{S}_N=\{\{{⅁}_4,{⅁}_5\},\left\{{⅁}_6\right\},\left\{{⅁}_5\right\},\left\{{⅁}_1\right\},\{{⅁}_1,{⅁}_3\},\left\{{⅁}_2\right\}\}.\beta =\{\varnothing ,\{{⅁}_4,{⅁}_5\},\left\{{⅁}_6\right\},\left\{{⅁}_5\right\},\left\{{⅁}_1\right\},\{{⅁}_1,{⅁}_3\},\left\{{⅁}_2\right\}\}.\\ {\mathcal{T}}_{Ad}=\{\varnothing ,\{{⅁}_4,{⅁}_5\},\left\{{⅁}_6\right\},\left\{{⅁}_5\right\},\left\{{⅁}_1\right\},\{{⅁}_1,{⅁}_3\},\left\{{⅁}_2\right\},\{{⅁}_4,{⅁}_5,{⅁}_6\},\{{⅁}_4,{⅁}_5,{⅁}_1\},\{{⅁}_1,{⅁}_3,{⅁}_4,{⅁}_5\},\{{⅁}_2,{⅁}_4,{⅁}_5\},\\ \{{⅁}_5,{⅁}_6\},\{{⅁}_1,{⅁}_6\},\{{⅁}_1,{⅁}_3,{⅁}_6\},\{{⅁}_2,{⅁}_6\},\{{⅁}_1,{⅁}_5\},\{{⅁}_1,{⅁}_3,{⅁}_5\},\{{⅁}_2,{⅁}_5\},\{{⅁}_1,{⅁}_2\},\{{⅁}_1,{⅁}_5,{⅁}_6\},\{{⅁}_1,{⅁}_4,{⅁}_5\},\\ \{{⅁}_1,{⅁}_4,{⅁}_5,{⅁}_6\},\{{⅁}_1,{⅁}_5,{⅁}_6\},\{{⅁}_1,{⅁}_3,{⅁}_5,{⅁}_6\},\{{⅁}_1,{⅁}_3,{⅁}_4,{⅁}_5,{⅁}_6\},\{{⅁}_2,{⅁}_4,{⅁}_5,{⅁}_6\},\{{⅁}_2,{⅁}_4,{⅁}_5,{⅁}_1\},\\ \{{⅁}_1,{⅁}_2,{⅁}_3,{⅁}_4,{⅁}_5\},\{{⅁}_2,{⅁}_5,{⅁}_6\},\{{⅁}_1,{⅁}_2,{⅁}_6\},\{{⅁}_1,{⅁}_2,{⅁}_3,{⅁}_6\},\{{⅁}_1,{⅁}_2,{⅁}_5\}\{{⅁}_1,{⅁}_2,{⅁}_3,{⅁}_5\},\{{⅁}_1,{⅁}_2,{⅁}_5,{⅁}_6\},\\ \{{⅁}_1,{⅁}_2,{⅁}_4,{⅁}_5\},\{{⅁}_1,{⅁}_2,{⅁}_4,{⅁}_5,{⅁}_6\},\{{⅁}_1,{⅁}_2,{⅁}_5,{⅁}_6\},\{{⅁}_1,{⅁}_2,{⅁}_3,{⅁}_5,{⅁}_6\},\{{⅁}_1,{⅁}_2,{⅁}_3,{⅁}_4,{⅁}_5,{⅁}_6\}\}.\end{array}$$

Let $\mathcal{P}$ = $P$() − $\{(\{{⅁}_2,{⅁}_3,{⅁}_5,{⅁}_6\},\{{\alpha }_3,{\alpha }_4\}),(\{{⅁}_1,{⅁}_2,{⅁}_3,{⅁}_5,{⅁}_6\},\{{\alpha }_3,{\alpha }_4\}),(\{{⅁}_2,{⅁}_3,{⅁}_4,{⅁}_5,{⅁}_6\},\{{\alpha }_3,{\alpha }_4\}),(\{{⅁}_1,{⅁}_2,{⅁}_3,{⅁}_4,{⅁}_5,{⅁}_6\},\{{\alpha }_3,{\alpha }_4\}),(\{{⅁}_1,{⅁}_2,{⅁}_3,{⅁}_5,{⅁}_6\},\{{\alpha }_2,{\alpha }_3,{\alpha }_4\}),(\{{⅁}_1,{⅁}_2,{⅁}_3,{⅁}_4,{⅁}_5,{⅁}_6\},\{{\alpha }_2,{\alpha }_3,{\alpha }_4\}),G\}$ be a graph primal. Let $W=(\{{⅁}_1,{⅁}_2,{⅁}_3\},\varnothing )$ be a subgraph of the given graph. Then, ${\left(\mathfrak{Q}\left(\mathcal{W}\right)\right)}^{•}\left(\mathcal{P},{\mathcal{T}}_{Ad}\right)=$$\{{⅁}_1,{⅁}_2,{⅁}_3,{⅁}_4,{⅁}_5,{⅁}_6\}$ = $\mathfrak{Q}$(). On the other hand, let $W=(\{{⅁}_2,{⅁}_3,{⅁}_5,{⅁}_6\},\{{\alpha }_3,{\alpha }_4\}).$ By computing, ${\left(\mathfrak{Q}\left(\mathcal{W}\right)\right)}^{•}\left(\mathcal{P},{\mathcal{T}}_{Ad}\right)=\varnothing .$ It is clear that ${\left(\mathfrak{Q}\left(\mathcal{W}\right)\right)}^{•}⊈\mathfrak{Q}(\daleth )$ and ${\left(\mathfrak{Q}\left(\mathcal{W}\right)\right)}^{•}⊉\mathfrak{Q}(\daleth ).$

Theorem 3.

Let ($\mathfrak{Q}$(), ${\mathcal{T}}_{Ad},\mathcal{P}$) and ($\mathfrak{Q}$(), ${\mathcal{T}}_{Ad},\mathcal{J}$ be two graph primal ATSs, and let $\beth$, $\daleth$ be two subgraphs of . Then, the graph-local function satisfies the following properties:

(1) 

$\mathfrak{Q}(\beth )\subseteq \mathfrak{Q}(\daleth )\Rightarrow {\left(\mathfrak{Q}(\beth )\right)}^{•}\subseteq {\left(\mathfrak{Q}(\daleth )\right)}^{•}.$

(2) 

$\mathcal{P}\subseteq \mathcal{J}\Rightarrow {\left(\mathfrak{Q}(\daleth )\right)}^{•}\left(\mathcal{J}\right)\subseteq {\left(\mathfrak{Q}(\daleth )\right)}^{•}\left(\mathcal{P}\right)$.

(3) 

${\left(\mathfrak{Q}(\daleth )\right)}^{•}=\mathfrak{CL}\left({\left(\mathfrak{Q}(\daleth )\right)}^{•}\right)$.

(4) 

If $\mathfrak{Q}{(\beth )}^c\in {\mathcal{T}}_{Ad},$ then ${\left(\mathfrak{Q}(\beth )\right)}^{•}\subseteq \mathfrak{Q}(\beth ).$

(5) 

If $\mathfrak{Q}{(\beth )}^c$ = $\mathfrak{Q}$(′) for some ′ ∉ $\mathcal{P}$, then ${\left(\mathfrak{Q}(\beth )\right)}^{•}=\varnothing .$

(6) 

${\left({\left(\mathfrak{Q}(\daleth )\right)}^{•}\right)}^{•}\subseteq {\left(\mathfrak{Q}(\daleth )\right)}^{•}$.

(7) 

$\mathfrak{Q}{(\beth )}^{•}\cup \mathfrak{Q}{(\daleth )}^{•}={(\mathfrak{Q}(\beth )\cup \mathfrak{Q}(\daleth ))}^{•}.$

(8) 

${(\mathfrak{Q}(\beth )\cap \mathfrak{Q}(\daleth ))}^{•}\subseteq \mathfrak{Q}{(\beth )}^{•}\cap \mathfrak{Q}{(\daleth )}^{•}.$

Proof. 

(1) Let $⅁\notin {\left(\mathfrak{Q}(\daleth )\right)}^{•}$. Then, there exists $U\in N(⅁)$ such that $\mathfrak{Q}{(\daleth )}^c\cup U^c$ = $\mathfrak{Q}$(′) for some ′ ∉ $\mathcal{P}$. Since $\mathfrak{Q}(\beth )\subseteq \mathfrak{Q}(\daleth ),\mathfrak{Q}{(\beth )}^c\cup U^c\supseteq \mathfrak{Q}{(\daleth )}^c\cup U^c$. Therefore, there exists $U\in N(⅁)$ such that $\mathfrak{Q}{(\beth )}^c\cup U^c$ = $\mathfrak{Q}$(′), for some ′ ∉ $\mathcal{P}$. So $⅁\notin {\left(\mathfrak{Q}(\beth )\right)}^{•}$. Hence, ${\left(\mathfrak{Q}(\beth )\right)}^{•}\subseteq {\left(\mathfrak{Q}(\daleth )\right)}^{•}$.

(2)

Let $⅁\in {\left(\mathfrak{Q}(\daleth )\right)}^{•}\left(\mathcal{J}\right)$. Then, for every $U\in N(⅁),\mathfrak{Q}{(\daleth )}^c\cup U^c$ = $\mathfrak{Q}$(′) and ′ ∈ $\mathcal{J}$. Since $\mathcal{P}\subseteq \mathcal{J},\mathfrak{Q}{(\daleth )}^c\cup U^c$ = $\mathfrak{Q}$(′) and ′ ∈ $\mathcal{P}$. Thus, for all $U\in N(⅁),\mathfrak{Q}(\daleth )\cap U$ = $\mathfrak{Q}$(′) for some ′ ∈ $\mathcal{P}$. Hence, ${\left(\mathfrak{Q}(\daleth )\right)}^{•}\left(\mathcal{J}\right)\subseteq {\left(\mathfrak{Q}(\daleth )\right)}^{•}\left(\mathcal{P}\right).$

(3)

We always have ${\left(\mathfrak{Q}(\daleth )\right)}^{•}\subseteq \mathfrak{CL}\left({\left(\mathfrak{Q}(\daleth )\right)}^{•}\right).$ Let $⅁\in \mathfrak{CL}\left({\left(\mathfrak{Q}(\daleth )\right)}^{•}\right)$ and $N_{⅁}\in {\mathcal{T}}_A.$ Then, ${\left(\mathfrak{Q}(\daleth )\right)}^{•}\cap N_{⅁}\ne \varnothing .$ Therefore, there exists $\varsigma \in \mathfrak{Q}$() such that $\varsigma \in N_{⅁}$ and $\varsigma \in {\left(\mathfrak{Q}(\daleth )\right)}^{•}.$ Then, $N_{\varsigma }^c\cup \mathfrak{Q}{(\daleth )}^c=\mathfrak{Q}$(′) and ′ ∈ $\mathcal{P}$ for all $N_{\varsigma }\in {\mathcal{T}}_A.$ Therefore, $N_{⅁}^c\cup \mathfrak{Q}{(\daleth )}^c$ = $\mathfrak{Q}$(″) and ″ ∈ $\mathcal{P}$. As a result, $⅁\in {\left(\mathfrak{Q}(\daleth )\right)}^{•}.$ So, ${\left(\mathfrak{Q}(\daleth )\right)}^{•}\supseteq \mathfrak{CL}\left(\mathfrak{Q}(\daleth )\right)$. Hence, ${\left(\mathfrak{Q}(\daleth )\right)}^{•}$ is closed in $G$.

(4)

Let, $\mathfrak{Q}{(\beth )}^c\in {\mathcal{T}}_{Ad}$ and $⅁\in {\left(\mathfrak{Q}(\beth )\right)}^{•}.$ Assume that $⅁\notin \mathfrak{Q}(\beth ).$ Then, $⅁\in \mathfrak{Q}{(\beth )}^c\in {\mathcal{T}}_A.$ Since $⅁\in {\left(\mathfrak{Q}(\beth )\right)}^{•},$$N_{⅁}^c\cup \mathfrak{Q}{(\beth )}^c$ = $\mathfrak{Q}$(′) and ′ ∈ $\mathcal{P}$ for all $N_{⅁}\in {\mathcal{T}}_A.$ Therefore, $\mathfrak{Q}$() = $\mathfrak{Q}(\beth )\cup \mathfrak{Q}{(\beth )}^c={\left(\mathfrak{Q}{(\beth )}^c\right)}^c\cup \mathfrak{Q}{(\beth )}^c$ and $G\in \mathcal{P},$ which is a contradiction with $G\notin \mathcal{P}.$ Hence, ${\left(\mathfrak{Q}(\beth )\right)}^{•}\subseteq \mathfrak{Q}(\beth )$.

(5)

Assume that $⅁\in {\left(\mathfrak{Q}(\beth )\right)}^{•}.$ Then, $N_{⅁}^c\cup \mathfrak{Q}{(\beth )}^c$ = $\mathfrak{Q}$(″) and ″ ∈ $\mathcal{P}$ for all $N_{⅁}\in {\mathcal{T}}_A.$ Since $\mathfrak{Q}{(\beth )}^c$ = $\mathfrak{Q}$(′) for some ′ ∉ $\mathcal{P}$, $N_{⅁}^c\cup \mathfrak{Q}{(\beth )}^c$ = $\mathfrak{Q}$(″) and ″ ∉ $\mathcal{P}$, which is a contradiction. Hence, ${\left(\mathfrak{Q}(\beth )\right)}^{•}=\varnothing .$

(6)

This is straightforward from (3) and (4).

(7)

According to (1), $\mathfrak{Q}{(\beth )}^{•}\subseteq {(\mathfrak{Q}(\beth )\cup \mathfrak{Q}(\daleth ))}^{•}$ and $\mathfrak{Q}{(\daleth )}^{•}\subseteq {(\mathfrak{Q}(\beth )\cup \mathfrak{Q}(\daleth ))}^{•}$. Hence, $\mathfrak{Q}{(\beth )}^{•}\cup \mathfrak{Q}{(\daleth )}^{•}\subseteq {(\mathfrak{Q}(\beth )\cup \mathfrak{Q}(\daleth ))}^{•}.$ Conversely, let $⅁\notin \mathfrak{Q}{(\beth )}^{•}\cup \mathfrak{Q}{(\daleth )}^{•}.$ Then, $⅁\notin \mathfrak{Q}{(\beth )}^{•}$ and $⅁\notin \mathfrak{Q}{(\daleth )}^{•}.$ Then, there exists $\mathfrak{U},\mathfrak{T}\in N_x$ such that $\mathfrak{Q}{(\beth )}^c\cup {\mathfrak{U}}^c$ = $\mathfrak{Q}$(′), for some ′ ∉ $\mathcal{P}$ and $\mathfrak{Q}{(\daleth )}^c\cup {\mathfrak{T}}^c$ = $\mathfrak{Q}$(″), for some ″ ∉ $\mathcal{P}$. Put $\mathfrak{W}=\mathfrak{U}\cap \mathfrak{T}.$ Hence, $\mathfrak{W}\in N_x$ such that $\mathfrak{Q}{(\beth )}^c\cup {\mathfrak{W}}^c$ = $\mathfrak{Q}$(‴), for some ‴ ∉ $\mathcal{P}$ and $\mathfrak{Q}{(\daleth )}^c\cup {\mathfrak{W}}^c$ = $\mathfrak{Q}$(⁗), for some ⁗ ∉ $\mathcal{P}$. Therefore, ${(\mathfrak{Q}(\beth )\cup \mathfrak{Q}(\daleth ))}^c\cup {\mathfrak{W}}^c=(\mathfrak{Q}{(\beth )}^c\cap \mathfrak{Q}{(\daleth )}^c)\cup {\mathfrak{W}}^c=(\mathfrak{Q}{(\beth )}^c\cup {\mathfrak{W}}^c)\cap (\mathfrak{Q}{(\daleth )}^c\cup {\mathfrak{W}}^c)$ = $\mathfrak{Q}$(⁗′), for some ⁗′ ∉ $\mathcal{P}$, since $\mathcal{P}$ is a graph primal. It follows that, $⅁\notin {(\mathfrak{Q}(\beth )\cup \mathfrak{Q}(\daleth ))}^{•}.$ Thus, ${(\mathfrak{Q}(\beth )\cup \mathfrak{Q}(\daleth ))}^{•}\subseteq \mathfrak{Q}{(\beth )}^{•}\cup \mathfrak{Q}{(\daleth )}^{•}.$ Hence, ${(\mathfrak{Q}(\beth )\cup \mathfrak{Q}(\daleth ))}^{•}\subseteq \mathfrak{Q}{(\beth )}^{•}\cup \mathfrak{Q}{(\daleth )}^{•}.$

(8)

This is similar to (7).

□

Remark 3.

The equality in (8) of Theorem 3 need not be true in general, as shown by the following example.

Example 5.

Let be the graph ($\mathfrak{Q}$(), $\mathfrak{Z}$()), where $\mathfrak{Q}$() = $\{{⅁}_1,{⅁}_2,{⅁}_3\}$ and $\mathfrak{Z}$() = $\{{\alpha }_1,{\alpha }_2\}.$ A drawing of the graph is shown in Figure 5.

Figure 5. Graph defined in Example 5.

From the previous results, it follows that $S_N=\{\left\{{⅁}_2\right\},\{{⅁}_1,{⅁}_3\},\left\{{⅁}_2\right\}\}$. So, $\beta =\{\left\{{⅁}_2\right\},\{{⅁}_1,{⅁}_3\},\varnothing \}.$ Thus, ${\mathcal{T}}_A=\{\varnothing ,\{{⅁}_1,{⅁}_2,{⅁}_3\},\left\{{⅁}_2\right\},\{{⅁}_1,{⅁}_3\}\}.$ Define a graph primal as $\mathcal{P}=\{(\varnothing ,\varnothing ),(\left\{{⅁}_1\right\},\varnothing ),(\left\{{⅁}_2\right\},\varnothing )$, $(\left\{{⅁}_3\right\},\varnothing ),(\{{⅁}_1,{⅁}_2\},\varnothing ),(\{{⅁}_2,{⅁}_3\},\varnothing ),$$(\{{⅁}_1,{⅁}_3\},\varnothing ),$$(\{{⅁}_1,{⅁}_2,{⅁}_3\},\varnothing ),$$(\{{⅁}_1,{⅁}_2\},\left\{{\alpha }_1\right\})$, $(\{{⅁}_1,{⅁}_2,{⅁}_3\},\left\{{\alpha }_1\right\})\}.$ Now, if $\mathfrak{Q}(\beth )=\left\{{⅁}_2\right\}$ and $\mathfrak{Q}(\daleth )=\left\{{⅁}_3\right\},$ we have $\mathfrak{Q}{(\beth )}^{•}\cap \mathfrak{Q}{(\daleth )}^{•}$ = $\mathfrak{Q}$() ∩ $\mathfrak{Q}$() = $\mathfrak{Q}$() ≠ ∅ = ${\varnothing }^{•}={(\mathfrak{Q}(\beth )\cap \mathfrak{Q}(\daleth ))}^{•}.$

Theorem 4.

Let ($\mathfrak{Q}$(), ${\mathcal{T}}_{Ad},\mathcal{P}$) be a graph primal ATS and $\beth ,\daleth$ ⊆ . If $\mathfrak{Q}(\beth )$ is open in $\mathfrak{Q}$(), then $\mathfrak{Q}(\beth )\cap \mathfrak{Q}{(\daleth )}^{•}\subseteq {(\mathfrak{Q}(\beth )\cap \mathfrak{Q}(\daleth ))}^{•}.$

Proof. 

Let $\mathfrak{Q}(\beth )\in {\mathcal{T}}_{Ad}$ and $⅁\in \mathfrak{Q}(\beth )\cap \mathfrak{Q}{(\daleth )}^{•}.$ Then, $⅁\in \mathfrak{Q}(\beth )$ and $⅁\in \mathfrak{Q}{(\daleth )}^{•}.$ Therefore, $\mathfrak{Q}{(\daleth )}^c\cup {\mathfrak{U}}^c$ = $\mathfrak{Q}$(′) for some ′ ∈ $\mathcal{P}$ for all $\mathfrak{U}\in {\mathcal{T}}_{Ad}(⅁).$ Since $\mathfrak{Q}(\beth )\in {\mathcal{T}}_{Ad},$ we get ${(\mathfrak{Q}(\beth )\cap \mathfrak{Q}(\daleth ))}^c\cup {\mathfrak{U}}^c=\mathfrak{Q}{(\daleth )}^c\cup {(\mathfrak{Q}(\beth )\cap \mathfrak{U})}^c$ = $\mathfrak{Q}$(′) for some ′ ∈ $\mathcal{P}$ for all $\mathfrak{U}\in {\mathcal{T}}_{Ad}(⅁).$ It follows that $⅁\in {(\mathfrak{Q}(\beth )\cap \mathfrak{Q}(\daleth ))}^{•}.$ Hence, $\mathfrak{Q}(\beth )\cap \mathfrak{Q}{(\daleth )}^{•}\subseteq {(\mathfrak{Q}(\beth )\cap \mathfrak{Q}(\daleth ))}^{•}.$ □

Corollary 2.

Let ($\mathfrak{Q}$(), ${\mathcal{T}}_{Ad},\mathcal{P}$) be a graph primal ATS and $\beth ,\daleth \subseteq$. If $\mathfrak{Q}(\beth )$ is open in $\mathfrak{Q}$(), then $\mathfrak{Q}(\beth )\cap \mathfrak{Q}{(\daleth )}^{•}=\mathfrak{Q}(\beth )\cap {(\mathfrak{Q}(\beth )\cap \mathfrak{Q}(\daleth ))}^{•}\subseteq {(\mathfrak{Q}(\beth )\cap \mathfrak{Q}(\daleth ))}^{•}.$

Proof. 

The proof is straightforward by using Theorem 4. □

Definition 8.

Let ($\mathfrak{Q}$(), ${\mathcal{T}}_{Ad},\mathcal{P}$) be a graph primal ATS. Define a map ${\mathfrak{CL}}^{•}:2^{\mathfrak{Q}}$⁽⁾ → $2^{\mathfrak{Q}}$⁽⁾ as ${\mathfrak{CL}}^{•}\left(\mathfrak{Q}(\beth )\right)=\mathfrak{Q}(\beth )\cup \mathfrak{Q}{(\beth )}^{•}$ where $\mathfrak{Q}(\beth )$ ⊆ $\mathfrak{Q}$().

Theorem 5.

Let ($\mathfrak{Q}$(), ${\mathcal{T}}_{Ad},\mathcal{P}$) be a graph primal ATS and $\beth ,\daleth$ ⊆ . Then, the following statements hold:

(i) 

${\mathfrak{CL}}^{•}(\varnothing )=\varnothing$;

(ii) 

${\mathfrak{CL}}^{•}$($\mathfrak{Q}$()) = $\mathfrak{Q}$();

(iii) 

$\mathfrak{Q}(\beth )\subseteq {\mathfrak{CL}}^{•}\left(\mathfrak{Q}(\beth )\right)$;

(iv) 

If $\mathfrak{Q}(\beth )\subseteq \mathfrak{Q}(\daleth )$, then ${\mathfrak{CL}}^{•}\left(\mathfrak{Q}(\beth )\right)\subseteq {\mathfrak{CL}}^{•}\left(\mathfrak{Q}(\daleth )\right)$;

(v) 

${\mathfrak{CL}}^{•}\left(\mathfrak{Q}(\beth )\right)\cup {\mathfrak{CL}}^{•}\left(\mathfrak{Q}(\daleth )\right)={\mathfrak{CL}}^{•}(\mathfrak{Q}(\beth )\cup \mathfrak{Q}(\daleth ))$;

(vi) 

${\mathfrak{CL}}^{•}\left({\mathfrak{CL}}^{•}\left(\mathfrak{Q}(\beth )\right)\right)={\mathfrak{CL}}^{•}\left(\mathfrak{Q}(\beth )\right).$

Proof. 

Let $\mathfrak{Q}(\beth ),\mathfrak{Q}(\daleth )$ ⊆ $\mathfrak{Q}$()

(i)

Since ${(\varnothing )}^{•}=\varnothing ,$ we have ${\mathfrak{CL}}^{•}(\varnothing )=\varnothing \cup {\varnothing }^{•}=\varnothing .$

(ii)

Since $\mathfrak{Q}$() ∪ $\mathfrak{Q}$()^• = $\mathfrak{Q}$(), we have ${\mathfrak{CL}}^{•}$($\mathfrak{Q}$()) = $\mathfrak{Q}$().

(iii)

Since ${\mathfrak{CL}}^{•}\left(\mathfrak{Q}(\beth )\right)=\mathfrak{Q}(\beth )\cup \mathfrak{Q}{(\beth )}^{•},$ we have $\mathfrak{Q}(\beth )\subseteq {\mathfrak{CL}}^{•}\left(\mathfrak{Q}(\beth )\right)$.

(iv)

Let $\mathfrak{Q}(\beth )\subseteq \mathfrak{Q}(\daleth ).$ According to (1) of Theorem 3, we have $\mathfrak{Q}{(\beth )}^{•}\subseteq \mathfrak{Q}{(\daleth )}^{•}$. Thus, we get $\mathfrak{Q}(\beth )\cup \mathfrak{Q}{(\beth )}^{•}\subseteq \mathfrak{Q}(\daleth )\cup \mathfrak{Q}{(\daleth )}^{•}$, which means that ${\mathfrak{CL}}^{•}\left(\mathfrak{Q}(\beth )\right)\subseteq {\mathfrak{CL}}^{•}\left(\mathfrak{Q}(\daleth )\right)$.

(v)

This is obvious from the definition of the operator ${\mathfrak{CL}}^{•}$ and (1) of Theorem 3.

(vi)

It is obvious from (iii) that ${\mathfrak{CL}}^{•}\left(\mathfrak{Q}(\beth )\right)\subseteq {\mathfrak{CL}}^{•}\left({\mathfrak{CL}}^{•}\left(\mathfrak{Q}(\beth )\right)\right).$ On the other hand, since $\mathfrak{Q}{(\beth )}^{•}$ is closed in $\mathfrak{Q}$(), we have ${\left(\mathfrak{Q}{(\beth )}^{•}\right)}^{•}\subseteq \mathfrak{Q}{(\beth )}^{•}$. Therefore,

$$\begin{array}{ccc}\hfill {\mathfrak{CL}}^{•}\left({\mathfrak{CL}}^{•}\left(\mathfrak{Q}(\beth )\right)\right)& =& {\mathfrak{CL}}^{•}\left(\mathfrak{Q}(\beth )\right)\cup {\left({\mathfrak{CL}}^{•}\left(\mathfrak{Q}(\beth )\right)\right)}^{•}\hfill \\ & =& {\mathfrak{CL}}^{•}\left(\mathfrak{Q}(\beth )\right)\cup {(\mathfrak{Q}(\beth )\cup \mathfrak{Q}{(\beth )}^{•})}^{•}\hfill \\ & =& {\mathfrak{CL}}^{•}\left(\mathfrak{Q}(\beth )\right)\cup \mathfrak{Q}{(\beth )}^{•}\cup {\left(\mathfrak{Q}{(\beth )}^{•}\right)}^{•}\hfill \\ & \subseteq & {\mathfrak{CL}}^{•}\left(\mathfrak{Q}(\beth )\right)\cup \mathfrak{Q}{(\beth )}^{•}\cup \mathfrak{Q}{(\beth )}^{•}\hfill \\ & =& {\mathfrak{CL}}^{•}\left(\mathfrak{Q}(\beth )\right)\hfill \end{array}$$

Hence, we have ${\mathfrak{CL}}^{•}\left({\mathfrak{CL}}^{•}\left(\mathfrak{Q}(\beth )\right)\right)={\mathfrak{CL}}^{•}\left(\mathfrak{Q}(\beth )\right).$ □

Corollary 3.

Let ($\mathfrak{Q}$(), ${\mathcal{T}}_{Ad},\mathcal{P}$) be a graph primal ATS. Then, the function ${\mathfrak{CL}}^{•}:2^{\mathfrak{Q}}$⁽⁾ → $2^{\mathfrak{Q}}$(), denoted by ${\mathfrak{CL}}^{•}\left(\mathfrak{Q}(\beth )\right)=\mathfrak{Q}(\beth )\cup \mathfrak{Q}{(\beth )}^{•}$, where $\beth$ ⊆ , is a Kuratowski’s closure operator.

Definition 9.

Let ($\mathfrak{Q}$(), ${\mathcal{T}}_{Ad},\mathcal{P}$) be a graph primal ATS. Then, the family ${\mathcal{T}}_{Ad}^{•}$ = {$\mathfrak{Q}(\beth )$ ⊆ $\mathfrak{Q}$()|${\mathfrak{CL}}^{•}\left(\mathfrak{Q}{(\beth )}^c\right)=\mathfrak{Q}{(\beth )}^c$} is called the graph primal adjacency topology (graph primal topology, for short) generated by ${\mathfrak{CL}}^{•}$ on $\mathfrak{Q}$() induced by ${\mathcal{T}}_{Ad}$ and a graph primal $\mathcal{P}.$ We can write ${\mathcal{T}}_{\mathcal{P}}^{•}$ instead of ${\mathcal{T}}_{Ad}^{•}$ to specify the graph primal as per our requirements.

Example 6.

Let be the graph ($\mathfrak{Q}$(), $\mathfrak{Z}$()) where $\mathfrak{Q}$() = $\{{⅁}_1,{⅁}_2,{⅁}_3,{⅁}_4\}$ and $\mathfrak{Z}$() = $\{{\alpha }_1,{\alpha }_2,{\alpha }_3,{\alpha }_4\}$. A drawing of the graph is shown in Figure 6.

Figure 6. Graph defined in Example 6.

From the previous results, it follows that $S_N=\{\left\{{⅁}_2\right\},\{{⅁}_1,{⅁}_3,{⅁}_4\},\{{⅁}_2,{⅁}_4\},\{{⅁}_2,{⅁}_3\}\}$. So, $\beta =\{\left\{{⅁}_4\right\},\{{⅁}_2,{⅁}_4\},$$\{{⅁}_2,{⅁}_3\},\left\{{⅁}_2\right\},\{{⅁}_1,{⅁}_3,{⅁}_4\},$ $\left\{{⅁}_3\right\},\varnothing \}.$ Thus, ${\mathcal{T}}_A=\{\varnothing ,\{{⅁}_1,{⅁}_3,{⅁}_4\},$$\left\{{⅁}_2\right\},$$\left\{{⅁}_3\right\},\left\{{⅁}_4\right\},\{{⅁}_2,$${⅁}_4\},\{{⅁}_3,{⅁}_4\},\{{⅁}_2,{⅁}_3\},\{{⅁}_2,{⅁}_3,{⅁}_4\},$$\{{⅁}_1,{⅁}_2,$${⅁}_3,{⅁}_4\left\}\right\}.$ Consider the graph primal $\mathcal{P}$ as $\mathcal{P}$ = P() − $\{(\{{⅁}_1,{⅁}_2,{⅁}_3\},\{{\alpha }_1,{\alpha }_2\}),(\{{⅁}_1,{⅁}_2,{⅁}_3,{⅁}_4,\},\{{\alpha }_1,{\alpha }_2,{\alpha }_3\}),$$(\{{⅁}_1,{⅁}_2,{⅁}_3,{⅁}_4\},\{{\alpha }_1,{\alpha }_2,{\alpha }_4\left\}\right),G\}$. From the ${\mathcal{T}}_A$- open sets, we have $N\left({⅁}_1\right)=\{\{{⅁}_1,{⅁}_3,{⅁}_4\},$$\{{⅁}_1,{⅁}_2,{⅁}_3,{⅁}_4\}\},$$N\left({⅁}_2\right)=\{\left\{{⅁}_2\right\},\{{⅁}_2,{⅁}_3\},\{{⅁}_2,{⅁}_4\},\{{⅁}_2,{⅁}_3,{⅁}_4\},$ $\{{⅁}_1,{⅁}_2,{⅁}_3,{⅁}_4\}\},$$N\left({⅁}_3\right)=\{\left\{{⅁}_3\right\},$$\{{⅁}_2,{⅁}_3\},\{{⅁}_3,{⅁}_4\},$$\{{⅁}_2,{⅁}_3,{⅁}_4\},\{{⅁}_1,{⅁}_2,{⅁}_3,{⅁}_4\}\}$ and $N\left({⅁}_4\right)=\{\left\{{⅁}_4\right\},\{{⅁}_3,{⅁}_4\},$$\{{⅁}_2,{⅁}_4\},$$\{{⅁}_2,{⅁}_3,{⅁}_4\},$$\{{⅁}_1,{⅁}_2,$${⅁}_3,{⅁}_4\left\}\right\}.$ Simple calculations of the graph-local function associated with the defined graph primal are given in Table 2. According to Table 2, ${\mathcal{T}}_{Ad}^{•}={\mathcal{T}}_{Ad}.$

Table 2. Illustration of Theorem 7.

$\mathit{\beth }\subseteq \mathfrak{Q}$()	$\mathfrak{Q}{\left(\mathit{\beth }\right)}^{•}$	$\mathfrak{Q}\left(\mathit{\beth }\right)\cup \mathfrak{Q}{\left(\mathit{\beth }\right)}^{•}$	$\mathfrak{Q}\left(\mathit{\beth }\right)\cup \mathfrak{Q}{\left(\mathit{\beth }\right)}^{•}=\mathfrak{Q}\left(\mathit{\beth }\right)?$
∅	∅	∅	$\mathrm{Yes}$
$\mathfrak{Q}$()	$\{{⅁}_1,{⅁}_2,{⅁}_3\}$	$\{{⅁}_1,{⅁}_2,{⅁}_3,{⅁}_4\}$	$\mathrm{Yes}$
$\left\{{⅁}_1\right\}$	$\left\{{⅁}_1\right\}$	$\left\{{⅁}_1\right\}$	$\mathrm{Yes}$
$\left\{{⅁}_2\right\}$	$\left\{{⅁}_2\right\}$	$\left\{{⅁}_2\right\}$	$\mathrm{Yes}$
$\left\{{⅁}_3\right\}$	$\left\{{⅁}_3\right\}$	$\left\{{⅁}_3\right\}$	$\mathrm{Yes}$
$\left\{{⅁}_4\right\}$	∅	$\left\{{⅁}_4\right\}$	$\mathrm{Yes}$
$\{{⅁}_1,{⅁}_2\}$	$\{{⅁}_1,{⅁}_2\}$	$\{{⅁}_1,{⅁}_2\}$	$\mathrm{Yes}$
$\{{⅁}_1,{⅁}_3\}$	$\{{⅁}_1,{⅁}_3\}$	$\{{⅁}_1,{⅁}_3\}$	$\mathrm{Yes}$
$\{{⅁}_1,{⅁}_4\}$	$\left\{{⅁}_1\right\}$	$\{{⅁}_1,{⅁}_4\}$	$\mathrm{Yes}$
$\{{⅁}_2,{⅁}_3\}$	$\{{⅁}_2,{⅁}_3\}$	$\{{⅁}_2,{⅁}_3\}$	$\mathrm{Yes}$
$\{{⅁}_2,{⅁}_4\}$	$\left\{{⅁}_2\right\}$	$\{{⅁}_2,{⅁}_4\}$	$\mathrm{Yes}$
$\{{⅁}_3,{⅁}_4\}$	$\left\{{⅁}_3\right\}$	$\{{⅁}_3,{⅁}_4\}$	$\mathrm{Yes}$
$\{{⅁}_1,{⅁}_2,{⅁}_3\}$	$\{{⅁}_1,{⅁}_2,{⅁}_3\}$	$\{{⅁}_1,{⅁}_2,{⅁}_3\}$	$\mathrm{Yes}$
$\{{⅁}_1,{⅁}_2,{⅁}_4\}$	$\{{⅁}_1,{⅁}_2\}$	$\{{⅁}_1,{⅁}_2,{⅁}_4\}$	$\mathrm{Yes}$
$\{{⅁}_1,{⅁}_3,{⅁}_4\}$	$\{{⅁}_1,{⅁}_3\}$	$\{{⅁}_1,{⅁}_3,{⅁}_4\}$	$\mathrm{Yes}$
$\{{⅁}_2,{⅁}_3,{⅁}_4\}$	$\{{⅁}_2,{⅁}_3\}$	$\{{⅁}_2,{⅁}_3,{⅁}_4\}$	$\mathrm{Yes}$

Theorem 6.

Let ($\mathfrak{Q}$(), ${\mathcal{T}}_{Ad},\mathcal{P}$) be a graph primal ATS. Then, the graph primal topology ${\mathcal{T}}_{Ad}^{•}$ is finer than ${\mathcal{T}}_{Ad}.$

Proof. 

Let $\mathfrak{Q}(\beth )\in {\mathcal{T}}_{Ad}.$ Then, $\mathfrak{Q}{(\beth )}^c$ is ${\mathcal{T}}_{Ad}$-closed in $\mathfrak{Q}$(). According to (5) of Theorem 3, we get ${\left(\mathfrak{Q}{(\beth )}^c\right)}^{•}\subseteq \mathfrak{Q}{(\beth )}^c.$ So, ${\mathfrak{CL}}^{•}\left(\mathfrak{Q}{(\beth )}^c\right)=\mathfrak{Q}{(\beth )}^c\cup {\left(\mathfrak{Q}{(\beth )}^c\right)}^{•}\subseteq \mathfrak{Q}{(\beth )}^c.$ Since $\mathfrak{Q}{(\beth )}^c\subseteq {\mathfrak{CL}}^{•}\left(\mathfrak{Q}{(\beth )}^c\right)$ is always satisfied for any subset $\mathfrak{Q}(\beth )$ of $\mathfrak{Q}$(), we have ${\mathfrak{CL}}^{•}\left(\mathfrak{Q}{(\beth )}^c\right)=\mathfrak{Q}{(\beth )}^c.$ It follows that $\mathfrak{Q}(\beth )\in {\mathcal{T}}_{Ad}^{•}.$ Hence, we have ${\mathcal{T}}_{Ad}\subseteq {\mathcal{T}}_{Ad}^{•}.$ □

Theorem 7.

Let ($\mathfrak{Q}$(), ${\mathcal{T}}_{Ad},\mathcal{P}$, be a graph primal ATS. Then, the following statements hold:

(i) 

If $\mathcal{P}=\varnothing$, then ${\mathcal{T}}_{Ad}^{•}=2^{\mathfrak{Q}}$⁽⁾;

(ii) 

If $\mathcal{P}$ = 2 ∖ {}, then ${\mathcal{T}}_{Ad}={\mathcal{T}}_{Ad}^{•}.$

Proof. 

(i) We have always ${\mathcal{T}}_{Ad}^{•}\subseteq 2^{\mathfrak{Q}}$⁽⁾. Now, let $\mathfrak{Q}(\beth )\in 2^{\mathfrak{Q}}$⁽⁾. Since $\mathcal{P}=\varnothing ,$ we have $\mathfrak{Q}{(\beth )}^{•}=\varnothing$ for any subset $\mathfrak{Q}(\beth )$ of $\mathfrak{Q}$(). Therefore, ${\mathfrak{CL}}^{•}\left(\mathfrak{Q}{(\beth )}^c\right)=\mathfrak{Q}{(\beth )}^c.$ This means that $\mathfrak{Q}(\beth )\in {\mathcal{T}}_{Ad}^{•}.$ Hence, $2^{\mathfrak{Q}}$⁽⁾ ⊆ ${\mathcal{T}}_{Ad}^{•}$. Thus, we have ${\mathcal{T}}_{Ad}^{•}=2^{\mathfrak{Q}}$().

(ii) We always have ${\mathcal{T}}_{Ad}\subseteq {\mathcal{T}}_{Ad}^{•}$ according to Theorem 6. Now, we prove that ${\mathcal{T}}_{Ad}^{•}\subseteq {\mathcal{T}}_{Ad}$. Let $\mathfrak{Q}(\beth )\in {\mathcal{T}}_{Ad}^{•}$. Then, $\mathfrak{Q}{(\beth )}^c\cup {\left(\mathfrak{Q}{(\beth )}^c\right)}^{•}=\mathfrak{Q}{(\beth )}^c$, which implies that ${\left(\mathfrak{Q}{(\beth )}^c\right)}^{•}\subseteq \mathfrak{Q}{(\beth )}^c$. Now, let $⅁\notin {\left(\mathfrak{Q}{(\beth )}^c\right)}^{•}.$ Then, there exists $\mathfrak{U}\in {\mathcal{T}}_{Ad}(⅁)$ such that ${\mathfrak{U}}^c\cup {\left(\mathfrak{Q}{(\beth )}^c\right)}^c={\mathfrak{U}}^c\cup \mathfrak{Q}(\beth )$ = $\mathfrak{Q}$(′) for some ′ ∉ $\mathcal{P}$. Since $\mathcal{P}$ = 2 ∖ {}, we get ${\mathfrak{U}}^c\cup \mathfrak{Q}(\beth )$ = $\mathfrak{Q}$() and so $\mathfrak{U}\cap \mathfrak{Q}{(\beth )}^c=\varnothing .$ Thus, $⅁\notin \mathfrak{CL}\left(\mathfrak{Q}{(\beth )}^c\right).$ Therefore, we have $\mathfrak{CL}\left(\mathfrak{Q}{(\beth )}^c\right)\subseteq {\left(\mathfrak{Q}{(\beth )}^c\right)}^{•}\subseteq \mathfrak{Q}{(\beth )}^c.$ Hence, $\mathfrak{CL}\left(\mathfrak{Q}{(\beth )}^c\right)=\mathfrak{Q}{(\beth )}^c$, which implies that $\mathfrak{Q}{(\beth )}^c$ is ${\mathcal{T}}_{Ad}$-closed, so $\mathfrak{Q}(\beth )\in {\mathcal{T}}_{Ad}$. As a result, ${\mathcal{T}}_{Ad}^{•}\subseteq {\mathcal{T}}_{Ad}$. Consequently, we have ${\mathcal{T}}_{Ad}={\mathcal{T}}_{Ad}^{•}$. □

Remark 4.

The following example shows that the converse of Theorem 7 need not be true in general.

Example 7.

Let be the graph ($\mathfrak{Q}$(), $\mathfrak{Z}$()), where $\mathfrak{Q}$() = $\{{⅁}_1,{⅁}_2,{⅁}_3\}$ and $\mathfrak{Z}$() = $\{{\alpha }_1,{\alpha }_2,{\alpha }_3\}.$ A drawing of the graph is shown in Figure 7.

Figure 7. Graph defined in Example 7.

From the previous results, it follows that $S_N=\{\{{⅁}_2,{⅁}_3\},\{{⅁}_1,{⅁}_3\},$$\{{⅁}_1,{⅁}_2\}\}$. So, $\beta =\{\left\{{⅁}_1\right\},\{{⅁}_2,{⅁}_3\},\left\{{⅁}_2\right\}$, $\left\{{⅁}_3\right\},\{{⅁}_1,{⅁}_3\},\{{⅁}_1,{⅁}_2\},\varnothing \}.$ Thus, ${\mathcal{T}}_A=\{\varnothing ,\{{⅁}_1,{⅁}_2,{⅁}_3\},$$\left\{{⅁}_1\right\},\left\{{⅁}_2\right\},\left\{{⅁}_3\right\},\{{⅁}_1,{⅁}_2\},\{{⅁}_1,$${⅁}_3\},\{{⅁}_2,{⅁}_3\}\}.$ Define a graph primal as $\mathcal{P}=\{(\varnothing ,\varnothing ),(\left\{{⅁}_1\right\},\varnothing ),(\left\{{⅁}_2\right\},\varnothing ),(\left\{{⅁}_3\right\},\varnothing ),$$(\{{⅁}_1,{⅁}_2\},\varnothing ),(\{{⅁}_2,{⅁}_3\},\varnothing ),$$(\{{⅁}_1,{⅁}_3\},\varnothing ),(\{{⅁}_1,{⅁}_2,{⅁}_3\},\varnothing ),$$(\{{⅁}_1,{⅁}_2\},$$\left\{{\alpha }_1\right\}),(\{{⅁}_1,{⅁}_3\}$, $\left\{{\alpha }_2\right\}),$$(\{{⅁}_2,{⅁}_3\},\left\{{\alpha }_3\right\}),(\{{⅁}_1,{⅁}_2,{⅁}_3\},\left\{{\alpha }_2\right\}),$$(\{{⅁}_1,{⅁}_2,{⅁}_3\},\left\{{\alpha }_3\right\}),\left(\right\{{⅁}_1,{⅁}_2,$${⅁}_3\},\{{\alpha }_2,{\alpha }_3\}\left)\right\}.$ Simple calculations of the graph-local function associated with the defined graph primal are given in Table 3. According to Table 3, ${\mathcal{T}}_{Ad}={\mathcal{T}}_{Ad}^{•}$, but $\mathcal{P}$ is not equal to $2^{\mathfrak{Q}}$⁽⁾ ∖ {$\mathfrak{Q}$()}.

Table 3. Illustration of Theorem 7.

$\mathit{\beth }\subseteq \mathfrak{Q}$()	$\mathfrak{Q}{\left(\mathit{\beth }\right)}^{•}$	$\mathfrak{Q}\left(\mathit{\beth }\right)\cup \mathfrak{Q}{\left(\mathit{\beth }\right)}^{•}$	$\mathit{\beth }\in {\mathcal{T}}_{\mathit{A}\mathit{d}}^{•}?$
∅	∅	∅	$\mathrm{Yes}$
$\mathfrak{Q}$()	$\{{⅁}_1,{⅁}_2\}$	$\{{⅁}_1,{⅁}_2,{⅁}_3\}$	$\mathrm{Yes}$
$\left\{{⅁}_1\right\}$	$\left\{{⅁}_1\right\}$	$\left\{{⅁}_1\right\}$	$\mathrm{Yes}$
$\left\{{⅁}_2\right\}$	$\left\{{⅁}_2\right\}$	$\left\{{⅁}_2\right\}$	$\mathrm{Yes}$
$\left\{{⅁}_3\right\}$	∅	$\left\{{⅁}_3\right\}$	$\mathrm{Yes}$
$\{{⅁}_1,{⅁}_2\}$	$\{{⅁}_1,{⅁}_2\}$	$\{{⅁}_1,{⅁}_2\}$	$\mathrm{Yes}$
$\{{⅁}_1,{⅁}_3\}$	$\left\{{⅁}_1\right\}$	$\{{⅁}_1,{⅁}_3\}$	$\mathrm{Yes}$
$\{{⅁}_2,{⅁}_3\}$	$\left\{{⅁}_2\right\}$	$\{{⅁}_2,{⅁}_3\}$	$\mathrm{Yes}$

Theorem 8.

Let ($\mathfrak{Q}$(), ${\mathcal{T}}_{Ad},\mathcal{P}$) be a graph primal ATS and $\beth$ ⊆ . Then, the following statements hold:

(i) 

$\mathfrak{Q}(\beth )\in {\mathcal{T}}_{Ad}^{•}$ iff for all $⅁$ in $\mathfrak{Q}(\beth ),$ there exists an ${\mathcal{T}}_A$-open set $\mathfrak{U}$ containing $⅁$ such that ${\mathfrak{U}}^c\cup \mathfrak{Q}(\beth )$ = $\mathfrak{Q}$(′) for some ′ ∉ $\mathcal{P}$,

(ii) 

If $\mathfrak{Q}(\beth )$ = $\mathfrak{Q}$(′) for some ′ ∉ $\mathcal{P}$, then $\mathfrak{Q}(\beth )\in {\mathcal{T}}_{Ad}^{•}$.

Proof. 

(i) Let $\mathfrak{Q}(\beth )\in {\mathcal{T}}_{Ad}^{•}.$

(ii) Let $\mathfrak{Q}(\beth )$ = $\mathfrak{Q}$(′), ′ ∉ $\mathcal{P}$ and $⅁\in \mathfrak{Q}(\beth )$. Put $\mathfrak{U}$ = $\mathfrak{Q}$(). Then, $\mathfrak{U}$ is a ${\mathcal{T}}_{Ad}$-open set containing $⅁.$ Since $\mathfrak{Q}(\beth )$ = $\mathfrak{Q}$(′) for some ′ ∉ $\mathcal{P}$ and ${\mathfrak{U}}^c\cup \mathfrak{Q}(\beth )=\mathfrak{Q}(\beth ),$ we have ${\mathfrak{U}}^c\cup \mathfrak{Q}(\beth )$ = $\mathfrak{Q}$(′) for some ′ ∉ $\mathcal{P}$. From (i), we obtain $\mathfrak{Q}(\beth )\in {\mathcal{T}}_{Ad}^{•}$. □

Theorem 9.

Let ($\mathfrak{Q}$(), ${\mathcal{T}}_{Ad},\mathcal{P}$) be a graph primal ATS. Then, the family $\mathfrak{Q}{(\daleth )}_{\mathcal{P}}$ = {$T\cap P|T\in {\mathcal{T}}_{Ad},$$P$ = $\mathfrak{Q}$(′) for some ′ ∉ $\mathcal{P}$} is a base for the graph primal topology ${\mathcal{T}}_{Ad}^{•}$ on $\mathfrak{Q}$().

Proof. 

Let $\mathfrak{Q}(\daleth )\in \mathfrak{Q}{(\daleth )}_{\mathcal{P}}.$ Then, there exists $T\in {\mathcal{T}}_{Ad}$, $P$ = $\mathfrak{Q}$(′) for some ′ ∉ $\mathcal{P}$ such that $\mathfrak{Q}(\daleth )=T\cap P.$ Since ${\mathcal{T}}_{Ad}\subseteq {\mathcal{T}}_{Ad}^{•},$ we get $T\in {\mathcal{T}}_{Ad}^{•}.$ On the other hand, according to Theorem 3 (5), we obtain $P\in {\mathcal{T}}_{Ad}^{•}.$ Therefore, $\mathfrak{Q}(\daleth )\in {\mathcal{T}}_{Ad}^{•}.$ Consequently, $\mathfrak{Q}{(\daleth )}_{\mathcal{P}}\subseteq {\mathcal{T}}_{Ad}^{•}.$ Now, let $\mathfrak{Q}(\beth )\in {\mathcal{T}}_{Ad}^{•}$ and $⅁\in \mathfrak{Q}(\beth ).$ Then, using Theorem 8 (i), there exists $\mathfrak{U}\in {\mathcal{T}}_{Ad}(⅁)$ such that ${\mathfrak{U}}^c\cup \mathfrak{Q}(\beth )$ = $\mathfrak{Q}$(″) for some ″ ∉ $\mathcal{P}$. Now, let $\mathfrak{Q}(\daleth )=\mathfrak{U}\cap ({\mathfrak{U}}^c\cup \mathfrak{Q}(\beth )).$ Hence, we have $\mathfrak{Q}(\daleth )\in \mathfrak{Q}{(\daleth )}_{\mathcal{P}}$ such that $⅁\in \mathfrak{Q}(\daleth )\subseteq \mathfrak{Q}(\beth )$. □

Theorem 10.

Let (, ${\mathcal{T}}_{Ad},\mathcal{P}$) and (, ${\mathcal{T}}_{Ad},\mathcal{Q}$) be two graph primal topological spaces. If $\mathcal{P}\subseteq \mathcal{Q},$ then ${\mathcal{T}}_A^{•}\left(\mathcal{Q}\right)\subseteq {\mathcal{T}}_A^{•}\left(\mathcal{P}\right).$

Proof. 

Let $\mathfrak{Q}(\beth )\in {\mathcal{T}}_A^{•}\left(\mathcal{Q}\right).$ Then, $\mathfrak{Q}{(\beth )}^c\cup {\left(\mathfrak{Q}{(\beth )}^c\right)}_{\mathcal{Q}}^{•}=\mathfrak{Q}{(\beth )}^c$, which means that ${\left(\mathfrak{Q}{(\beth )}^c\right)}_{\mathcal{Q}}^{•}\subseteq \mathfrak{Q}{(\beth )}^c.$ Now, let $⅁\notin \mathfrak{Q}{(\beth )}^c.$ Then, we get $⅁\notin {\left(\mathfrak{Q}{(\beth )}^c\right)}_{\mathcal{Q}}^{•}$, and there exists $\mathfrak{U}\in {\mathcal{T}}_{Ad}(⅁)$ such that ${\mathfrak{U}}^c\cup {\left(\mathfrak{Q}{(\beth )}^c\right)}^c={\mathfrak{U}}^c\cup \mathfrak{Q}(\beth )$ = $\mathfrak{Q}$(′) for some ′ ∉ $\mathcal{Q}$. Since $\mathcal{P}\subseteq \mathcal{Q},$ we have ${\mathfrak{U}}^c\cup \mathfrak{Q}(\beth )$ = $\mathfrak{Q}$(′) for some ′ ∉ $\mathcal{P}$. Thus, $⅁\notin {\left(\mathfrak{Q}{(\beth )}^c\right)}_{\mathcal{P}}^{•}.$ Therefore, ${\left(\mathfrak{Q}{(\beth )}^c\right)}_{\mathcal{P}}^{•}\subseteq \mathfrak{Q}{(\beth )}^c$, so ${\mathfrak{CL}}^{•}\left(\mathfrak{Q}{(\beth )}^c\right)=\mathfrak{Q}{(\beth )}^c\cup {\left(\mathfrak{Q}{(\beth )}^c\right)}_{\mathcal{P}}^{•}=\mathfrak{Q}{(\beth )}^c.$ Hence, $\mathfrak{Q}(\beth )\in {\mathcal{T}}_A^{•}\left(\mathcal{P}\right).$ As a result, we have ${\mathcal{T}}_A^{•}\left(\mathcal{Q}\right)\subseteq {\mathcal{T}}_A^{•}\left(\mathcal{P}\right)$. □

Theorem 11.

Let f: $\mathfrak{Q}$() → $\mathfrak{Q}\left(\mathfrak{Y}\right)$ be a function and $\mathcal{P}$ ⊆ 2. If $\mathcal{P}$ is a graph primal on and f is not surjective, then $\mathcal{Q}$ = {$f\left(P\right)|P$ = $\mathfrak{Q}$(′) for some ′ ∈ $\mathcal{P}$} is a graph primal on $\mathfrak{Y}$.

Proof. 

(1) Assume that $\mathfrak{Y}\in \mathcal{Q}.$ Then, there exists $P$ = $\mathfrak{Q}$(′) for some ′ ∈ $\mathcal{P}$ such that $f\left(P\right)=\mathfrak{Q}(\mathfrak{Y}.$ However, this contradicts the fact that f is not surjective.

(2)

Let $\mathfrak{Q}(\beth )=\mathfrak{Q}\left({\mathfrak{Y}}^{\prime }\right)$ and ${\mathfrak{Y}}^{\prime }\in \mathcal{Q}$ and $\mathfrak{Q}(\daleth )\subseteq \mathfrak{Q}(\beth ).$ Then, there exists $P_1$ = $\mathfrak{Q}$(′) for some ′ ∈ $\mathcal{P}$ such that $\mathfrak{Q}(\beth )=f\left(P_1\right).$ Now, set $P_2=f^{-1}\left(\mathfrak{Q}(\daleth )\right)\cap P_1$. It is obvious that $P_2\subseteq P_1$. Since $\mathcal{P}$ is a graph primal on $\mathfrak{Q}$(), $\mathcal{P}$ is downward closed and we have $P_2$ = $\mathfrak{Q}$(″) for some ″ ∈ $\mathcal{P}$. Also, $\mathfrak{Q}(\daleth )=f\left(P_2\right)$. This means that $\mathfrak{Q}(\daleth )=\mathfrak{Q}\left({\mathfrak{Y}}^{\prime \prime }\right)$ and ${\mathfrak{Y}}^{\prime \prime }\in \mathcal{Q}.$

(3)

Let $\mathfrak{Q}(\beth )\cap \mathfrak{Q}(\daleth )=\mathfrak{Q}\left({\mathfrak{Y}}^{\prime }\right)$ and ${\mathfrak{Y}}^{\prime }\in \mathcal{Q}$. Then, there exists $P_1$ = $\mathfrak{Q}$(′), ′ ∈ $\mathcal{P}$ and $P_2$ = $\mathfrak{Q}$(″), ″ ∈ $\mathcal{P}$ such that $\mathfrak{Q}(\beth )=f\left(P_1\right)$ and $\mathfrak{Q}(\daleth )=f\left(P_2\right)$. Since $\mathcal{P}$ is a graph primal on and $P_1\cap P_2\subseteq P_1,$ we have $P_1\cap P_2$ = $\mathfrak{Q}$(‴) for some ‴ ∈ $\mathcal{P}$. Thus, ′ ∈ $\mathcal{P}$ or ″ ∈ $\mathcal{P}$. Therefore, $\mathfrak{Q}(\beth )=f\left(P_1\right)=\mathfrak{Q}\left({\mathfrak{Y}}^{\prime \prime }\right),$${\mathfrak{Y}}^{\prime \prime }\in \mathcal{Q}$ or $\mathfrak{Q}(\daleth )=f\left(P_2\right)=\mathfrak{Q}\left({\mathfrak{Y}}^{\prime \prime \prime }\right),$${\mathfrak{Y}}^{\prime \prime \prime }\in \mathcal{Q}$.

□

Corollary 4.

Let f: $\mathfrak{Q}$() → $\mathfrak{Q}\left(\mathfrak{Y}\right)$ be a function and $\mathcal{P}$ ⊆ 2. If $\mathcal{P}$ is a graph primal on and f is not surjective, then the property of being graph primal is not a topological property.

Proof. 

This is straightforward by Theorem 11. □

Remark 5.

For a function f:$\mathfrak{Q}$() → $\mathfrak{Q}\left(\mathfrak{Y}\right)$ and a graph primal $\mathcal{Q}$ on $\mathfrak{Y},$ the family $\mathcal{P}=\left\{f^{-1}\left(Q\right)\right|Q=\mathfrak{Q}\left({\mathfrak{Y}}^{\prime }\right),$${\mathfrak{Y}}^{\prime }\in \mathcal{Q}\}$ need not be a graph primal on , as illustrated by the example below.

Example 8.

Let be the graph ($\mathfrak{Q}$(), $\mathfrak{Z}$()), where $\mathfrak{Q}$() = $\{{⅁}_1,{⅁}_2,{⅁}_3,{⅁}_4,{⅁}_5,{⅁}_6\}$ and $\mathfrak{Z}$() = $\{{\alpha }_1,{\alpha }_2,{\alpha }_3,{\alpha }_4,{\alpha }_5,{\alpha }_6\}$. Also, let $\mathfrak{Y}$ be the graph $\left(\mathfrak{Q}\right(\mathfrak{Y}),\mathfrak{Z}(\mathfrak{Y}\left)\right),$ where $\mathfrak{Q}\left(\mathfrak{Y}\right)=\{Z,R\}$ and $\mathfrak{Z}\left(\mathfrak{Y}\right)=\left\{\alpha \right\}$. Drawings of the two graphs and $\mathfrak{Y}$ are shown in Figure 8.

Figure 8. Graphs defined in Example 8.

Define the function f:$\mathfrak{Q}$() → $\mathfrak{Q}\left(\mathfrak{Y}\right)$ by $f\left({⅁}_i\right)=Z,$$i\in \{1,2,3,4,5,6\}.$ Then, $\mathcal{Q}$ is a graph primal on $\mathfrak{Y},$ but $\mathcal{P}=\left\{f^{-1}\left(Q\right)\right|Q=\mathfrak{Q}\left({\mathfrak{Y}}^{\prime }\right),$${\mathfrak{Y}}^{\prime }\in \mathcal{Q}\}$ = {∅, } is not a graph primal on .

Lemma 1.

Let ($\mathfrak{Q}$(), ${\mathcal{T}}_{Ad},\mathcal{P}$) be a graph primal ATS and $\beth ,\daleth$ be two subgraphs of $\mathfrak{Q}$(). Then, $\mathfrak{Q}{(\beth )}^{•}-\mathfrak{Q}{(\daleth )}^{•}={(\mathfrak{Q}(\beth )-\mathfrak{Q}(\daleth ))}^{•}-\mathfrak{Q}{(\daleth )}^{•}$.

Proof. 

According to Theorem 3, we have $\mathfrak{Q}{(\beth )}^{•}={[(\mathfrak{Q}(\beth )-\mathfrak{Q}(\daleth ))\cup (\mathfrak{Q}(\beth )\cap \mathfrak{Q}(\daleth ))]}^{•}={(\mathfrak{Q}(\beth )-\mathfrak{Q}(\daleth ))}^{•}\cup {(\mathfrak{Q}(\beth )\cap \mathfrak{Q}(\daleth ))}^{•}\subseteq {(\mathfrak{Q}(\beth )-\mathfrak{Q}(\daleth ))}^{•}\cup \mathfrak{Q}{(\daleth )}^{•}$. Thus, $\mathfrak{Q}{(\beth )}^{•}-\mathfrak{Q}{(\daleth )}^{•}\subseteq {(\mathfrak{Q}(\beth )-\mathfrak{Q}(\daleth ))}^{•}-\mathfrak{Q}{(\daleth )}^{•}$. Again, according to Theorem 3, ${(\mathfrak{Q}(\beth )-\mathfrak{Q}(\daleth ))}^{•}\subseteq \mathfrak{Q}{(\beth )}^{•}$ and ${(\mathfrak{Q}(\beth )-\mathfrak{Q}(\daleth ))}^{•}-\mathfrak{Q}{(\daleth )}^{•}\subseteq \mathfrak{Q}{(\beth )}^{•}-\mathfrak{Q}{(\daleth )}^{•}$. Hence, $\mathfrak{Q}{(\beth )}^{•}-\mathfrak{Q}{(\daleth )}^{•}={(\mathfrak{Q}(\beth )-\mathfrak{Q}(\daleth ))}^{•}-\mathfrak{Q}{(\daleth )}^{•}$. □

Corollary 5.

Let ($\mathfrak{Q}$(), ${\mathcal{T}}_{Ad},\mathcal{P}$, be a graph primal ATS and $\beth ,\daleth$ be two subgraphs of with $\mathfrak{Q}{(\daleth )}^c$ = $\mathfrak{Q}$(′) for some ′ ∉ $\mathcal{P}$. Then, ${(\mathfrak{Q}(\beth )\cup \mathfrak{Q}(\daleth ))}^{•}=\mathfrak{Q}{(\beth )}^{•}={(\mathfrak{Q}(\beth )-\mathfrak{Q}(\daleth ))}^{•}$.

Proof. 

Since $\mathfrak{Q}{(\daleth )}^c$ = $\mathfrak{Q}$(′) for some ′ ∉ $\mathcal{P}$, it follows that $\mathfrak{Q}{(\daleth )}^{•}=\varnothing$. Again, according to Lemma 1, $\mathfrak{Q}{(\beth )}^{•}={(\mathfrak{Q}(\beth )-\mathfrak{Q}(\daleth ))}^{•}$, and according to Theorem 3, ${(\mathfrak{Q}(\beth )\cup \mathfrak{Q}(\daleth ))}^{•}=\mathfrak{Q}{(\beth )}^{•}\cup \mathfrak{Q}{(\daleth )}^{•}=\mathfrak{Q}{(\beth )}^{•}$. □

Theorem 12.

Let ($\mathfrak{Q}$(), ${\mathcal{T}}_{Ad},\mathcal{P}$) be a graph primal ATS. If $\mathfrak{Q}\left(\mathfrak{F}\right)$ = $\mathfrak{Q}$(^★) for some ^★ ∈ $\mathcal{P}$ for all closed subsets $\mathfrak{Q}\left(\mathfrak{F}\right),\mathfrak{F}$ ≠ , then $\mathfrak{Q}\left(\mathfrak{U}\right)\subseteq \mathfrak{Q}{\left(\mathfrak{U}\right)}^{•}$ for all $\mathfrak{Q}\left(\mathfrak{U}\right)\in {\mathcal{T}}_{Ad},\mathfrak{U}$ ⊆ .

Proof. 

If $\mathfrak{Q}\left(\mathfrak{U}\right)=\varnothing ,$ then $\mathfrak{Q}{\left(\mathfrak{U}\right)}^{•}=\varnothing =\mathfrak{Q}\left(\mathfrak{U}\right).$ Now, if, $\mathfrak{Q}\left(\mathfrak{F}\right)$ = $\mathfrak{Q}$(^★) for some ^★ ∈ $\mathcal{P}$ for all closed subsets $\mathfrak{Q}\left(\mathfrak{F}\right),\mathfrak{F}$ ≠ , then $\mathfrak{Q}$()^• = $\mathfrak{Q}$(). In fact, $⅁$ ∉ $\mathfrak{Q}$()^•. Then, there exists $\mathfrak{Q}\left(\mathfrak{T}\right)\in {\mathcal{T}}_{Ad}(⅁)$ such that $\mathfrak{Q}$()^c ∪ ${\mathfrak{Q}\left(\mathfrak{T}\right)}^c$ = $\mathfrak{Q}$(′) for some ′ ∉ $\mathcal{P}$. Hence, $\mathfrak{Q}{\left(\mathfrak{T}\right)}^c$ = $\mathfrak{Q}$(′) for some ′ ∉ $\mathcal{P}$, which is a contradiction. Now, according to Theorem 4, we have for all $\mathfrak{Q}\left(\mathfrak{U}\right)\in {\mathcal{T}}_{Ad},$$\mathfrak{Q}\left(\mathfrak{U}\right)=\mathfrak{Q}\left(\mathfrak{U}\right)$ ∩ $\mathfrak{Q}$()^• ⊆ ${(\mathfrak{Q}\left(\mathfrak{U}\right)\cap \mathfrak{Q}\left(\mathfrak{Z}\right))}^{•}=\mathfrak{Q}{\left(\mathfrak{U}\right)}^{•}.$ Hence, $\mathfrak{Q}\left(\mathfrak{U}\right)\subseteq \mathfrak{Q}{\left(\mathfrak{U}\right)}^{•}$. □

Definition 10.

Let ($\mathfrak{Q}$(), ${\mathcal{T}}_{Ad},\mathcal{P}$) be a graph primal ATS. An operator $\Theta :2^{\mathfrak{Q}}$⁽⁾ → $2^{\mathfrak{Q}}$() defined by $\Theta \left(\mathfrak{Q}(\beth )\right)$ = { $⅁$ ∈ $\mathfrak{Q}$(): $(\exists \mathfrak{U}\in {\mathcal{T}}_{Ad}(⅁))$(${(\mathfrak{U}-\mathfrak{Q}(\beth ))}^c$ = $\mathfrak{Q}$(′) for some ′ ∉ $\mathcal{P}$)} for every $\mathfrak{Q}(\beth )$ ⊆ $\mathfrak{Q}$(.

Example 9.

Let be the graph ($\mathfrak{Q}$(), $\mathfrak{Z}$()), where $\mathfrak{Q}$() = $\{{⅁}_1,{⅁}_2,{⅁}_3,{⅁}_4\}$ and $\mathfrak{Z}$() = $\{{\alpha }_1,{\alpha }_2,{\alpha }_3,{\alpha }_4\}.$ A drawing of the graph is shown in Figure 9.

Figure 9. Graph defined in Example 9.

From the previous results, it follows that $S_N=\{\{{⅁}_2,{⅁}_4\},\{{⅁}_1,{⅁}_3,{⅁}_4\},\left\{{⅁}_2\right\},\{{⅁}_1,{⅁}_2\}\}$. So, $\beta =\{\left\{{⅁}_1\right\},\left\{{⅁}_2\right\},\left\{{⅁}_4\right\}$, $\{{⅁}_1,{⅁}_2\},\{{⅁}_2,{⅁}_4\},$ $\{{⅁}_1,{⅁}_3,{⅁}_4\},\varnothing \}.$ Thus, ${\mathcal{T}}_A=\{\varnothing ,\{{⅁}_1,{⅁}_3,{⅁}_4\},$$\left\{{⅁}_1\right\},\left\{{⅁}_2\right\},\left\{{⅁}_4\right\},\{{⅁}_1,{⅁}_2\}$, $\{{⅁}_1,{⅁}_4\},\{{⅁}_2,{⅁}_4\},\{{⅁}_1,{⅁}_2,{⅁}_4\},$ $\{{⅁}_1,{⅁}_2,{⅁}_3,{⅁}_4\}\}.$ Define a graph primal as follows: $\mathcal{P}$ = P() − $\{(\{{⅁}_1,{⅁}_2,{⅁}_3\},\{{\alpha }_1,{\alpha }_3\}),(\{{⅁}_1,{⅁}_2,{⅁}_3,{⅁}_4\},\{{\alpha }_1,{\alpha }_2,{\alpha }_3\}),$$(\{{⅁}_1,{⅁}_2,{⅁}_3,{⅁}_4\},\{{\alpha }_1,{\alpha }_3,{\alpha }_4\}),G\}$. According to the open sets of ${\mathcal{T}}_A$, we have the following: $N\left({⅁}_1\right)=\{\left\{{⅁}_1\right\},\{{⅁}_1,{⅁}_2\},\{{⅁}_1,{⅁}_4\},\{{⅁}_1,{⅁}_2,{⅁}_4\}$, $\{{⅁}_1,{⅁}_3,{⅁}_4\}\{{⅁}_1,{⅁}_2,{⅁}_3,{⅁}_4\}\},$$N\left({⅁}_2\right)=\{\left\{{⅁}_2\right\},\{{⅁}_2,{⅁}_4\},$$\{{⅁}_1,{⅁}_2\},\{{⅁}_1,{⅁}_2,{⅁}_4\},\{{⅁}_1,{⅁}_2,{⅁}_3$, ${⅁}_4\left\}\right\},$$N\left({⅁}_3\right)=\{\{{⅁}_1,{⅁}_3,{⅁}_4\},\{{⅁}_1,{⅁}_2,$${⅁}_3,{⅁}_4\left\}\right\}$ and $N\left({⅁}_4\right)=\{\left\{{⅁}_4\right\},\{{⅁}_1,{⅁}_4\},$$\{{⅁}_2,{⅁}_4\},$$\{{⅁}_1,{⅁}_2,{⅁}_4\},\{{⅁}_1,{⅁}_3,{⅁}_4\},\{{⅁}_1,{⅁}_2,{⅁}_3,{⅁}_4\}\}.$ The computations of the graph-local functions $\mathfrak{Q}{(\beth )}^{•}$ and $\Theta \left(\mathfrak{Q}\right(\beth \left)\right)$ of a graph $\beth$ ⊆ associated with the defined graph primal are given in Table 4. According to Table 4, $\Theta \left(\mathfrak{Q}\right(\beth \left)\right)⊈\mathfrak{Q}(\beth )$ and $\Theta \left(\mathfrak{Q}\right(\beth \left)\right)⊉\mathfrak{Q}(\beth ).$

Table 4. Illustration of Definition 11.

$\mathfrak{Q}\left(\mathit{\beth }\right)$ ⊆ $\mathfrak{Q}$()	$\mathfrak{Q}{\left(\mathit{\beth }\right)}^{\mathit{c}}$	$\mathfrak{Q}{\left(\mathit{\beth }\right)}^{•}$	$\mathit{\Theta }\left(\mathfrak{Q}\right(\mathit{\beth }\left)\right)$
∅	$\{{⅁}_1,{⅁}_2,{⅁}_3,{⅁}_4\}$	∅	$\left\{{⅁}_4\right\}$
$\mathfrak{Q}$()	∅	$\{{⅁}_1,{⅁}_2,{⅁}_3\}$	$\{{⅁}_1,{⅁}_2,{⅁}_3,{⅁}_4\}$
$\left\{{⅁}_1\right\}$	$\{{⅁}_2,{⅁}_3,{⅁}_4\}$	$\{{⅁}_1,{⅁}_3\}$	$\{{⅁}_1,{⅁}_4\}$
$\left\{{⅁}_2\right\}$	$\{{⅁}_1,{⅁}_3,{⅁}_4\}$	$\{{⅁}_2,{⅁}_3\}$	$\{{⅁}_2,{⅁}_4\}$
$\left\{{⅁}_3\right\}$	$\{{⅁}_1,{⅁}_2,{⅁}_4\}$	$\left\{{⅁}_3\right\}$	$\{{⅁}_3,{⅁}_4\}$
$\left\{{⅁}_4\right\}$	$\{{⅁}_1,{⅁}_2,{⅁}_3\}$	∅	$\left\{{⅁}_4\right\}$
$\{{⅁}_1,{⅁}_2\}$	$\{{⅁}_3,{⅁}_4\}$	$\{{⅁}_1,{⅁}_2,{⅁}_3\}$	$\{{⅁}_1,{⅁}_2,{⅁}_4\}$
$\{{⅁}_1,{⅁}_3\}$	$\{{⅁}_2,{⅁}_4\}$	$\{{⅁}_1,{⅁}_3\}$	$\{{⅁}_1,{⅁}_3,{⅁}_4\}$
$\{{⅁}_1,{⅁}_4\}$	$\{{⅁}_2,{⅁}_3\}$	$\{{⅁}_1,{⅁}_3\}$	$\{{⅁}_1,{⅁}_4\}$
$\{{⅁}_2,{⅁}_3\}$	$\{{⅁}_1,{⅁}_4\}$	$\{{⅁}_2,{⅁}_3\}$	$\{{⅁}_2,{⅁}_4\}$
$\{{⅁}_2,{⅁}_4\}$	$\{{⅁}_1,{⅁}_3\}$	$\left\{{⅁}_2\right\}$	$\{{⅁}_2,{⅁}_4\}$
$\{{⅁}_3,{⅁}_4\}$	$\{{⅁}_1,{⅁}_2\}$	$\left\{{⅁}_3\right\}$	$\left\{{⅁}_4\right\}$
$\{{⅁}_1,{⅁}_2,{⅁}_3\}$	$\left\{{⅁}_4\right\}$	$\{{⅁}_1,{⅁}_2,{⅁}_3\}$	$\{{⅁}_1,{⅁}_2,{⅁}_3,{⅁}_4\}$
$\{{⅁}_1,{⅁}_2,{⅁}_4\}$	$\left\{{⅁}_3\right\}$	$\{{⅁}_1,{⅁}_2\}$	$\{{⅁}_1,{⅁}_2,{⅁}_4\}$
$\{{⅁}_1,{⅁}_3,{⅁}_4\}$	$\left\{{⅁}_2\right\}$	$\{{⅁}_1,{⅁}_3\}$	$\{{⅁}_1,{⅁}_4\}$
$\{{⅁}_2,{⅁}_3,{⅁}_4\}$	$\left\{{⅁}_1\right\}$	$\{{⅁}_2,{⅁}_3\}$	$\{{⅁}_2,{⅁}_4\}$

Theorem 13.

Let ($\mathfrak{Q}$(), ${\mathcal{T}}_{Ad},\mathcal{P}$) be a graph primal ATS and $\beth ,\daleth$ be two subgraphs of . Then, the following properties hold:

(1) 

If $\mathfrak{Q}(\beth )$ ⊆ $\mathfrak{Q}$(), then $\Theta \left(\mathfrak{Q}(\beth )\right)$ = $\mathfrak{Q}$() − ($\mathfrak{Q}$() − $\mathfrak{Q}(\beth )$)^•;

(2) 

If $\mathfrak{Q}(\beth )$ ⊆ $\mathfrak{Q}$(), then $\Theta \left(\mathfrak{Q}\right(\beth \left)\right)$ is open;

(3) 

If $\mathfrak{Q}(\beth )\subseteq \mathfrak{Q}(\daleth )$, then $\Theta \left(\mathfrak{Q}\right(\beth \left)\right)\subseteq \Theta \left(\mathfrak{Q}\right(\daleth \left)\right)$;

(4) 

If $\mathfrak{Q}(\beth ),\mathfrak{Q}(\daleth )$ ⊆ $\mathfrak{Q}$(), then $\Theta \left(\mathfrak{Q}\right(\beth )\cap \mathfrak{Q}(\daleth \left)\right)=\Theta \left(\mathfrak{Q}\right(\beth \left)\right)\cap \Theta \left(\mathfrak{Q}\right(\daleth \left)\right)$;

(5) 

If $\mathfrak{U}\in {\mathcal{T}}_{Ad}^{•}$, then $\mathfrak{U}\subseteq \Theta \left(\mathfrak{U}\right)$;

(6) 

If $\mathfrak{Q}(\beth )$ ⊆ $\mathfrak{Q}$(), then $\Theta \left(\mathfrak{Q}\right(\beth \left)\right)\subseteq \Theta (\Theta (\mathfrak{Q}(\beth )\left)\right)$;

(7) 

If $\mathfrak{Q}(\beth )$ ⊆ $\mathfrak{Q}$(), then $\Theta \left(\mathfrak{Q}\right(\beth \left)\right)=\Theta (\Theta (\mathfrak{Q}(\beth )\left)\right)$ iff ($\mathfrak{Q}$() − $\mathfrak{Q}(\beth )$)^• = (($\mathfrak{Q}$() − $\mathfrak{Q}(\beth )$)^•)^•;

(8) 

If $\mathfrak{Q}{(\beth )}^c$ = $\mathfrak{Q}$(′) for some ′ ∉ $\mathcal{P}$, then $\Theta \left(\mathfrak{Q}(\beth )\right)$ = $\mathfrak{Q}$() − $\mathfrak{Q}$()^•;

(9) 

If $\mathfrak{Q}(\beth )$ ⊆ $\mathfrak{Q}$(), then $\mathfrak{Q}(\beth )\cap \Theta \left(\mathfrak{Q}(\beth )\right)={\mathfrak{int}}^{•}\left(\mathfrak{Q}(\beth )\right)$;

(10) 

If $\mathfrak{Q}(\beth )$ ⊆ $\mathfrak{Q}$() and $I^c$ = $\mathfrak{Q}$(′) for some ′ ∉ $\mathcal{P}$, then $\Theta \left(\mathfrak{Q}\right(\beth )-I)=\Theta \left(\mathfrak{Q}\right(\beth \left)\right)$;

(11) 

If $\mathfrak{Q}(\beth )$ ⊆ $\mathfrak{Q}$() and $I^c$ = $\mathfrak{Q}$(′) for some ′ ∉ $\mathcal{P}$, then $\Theta \left(\mathfrak{Q}\right(\beth )\cup I)=\Theta \left(\mathfrak{Q}\right(\beth \left)\right)$;

(12) 

If ${[(\mathfrak{Q}(\beth )-\mathfrak{Q}(\daleth ))\cup (\mathfrak{Q}(\daleth )-\mathfrak{Q}(\beth ))]}^c$ = $\mathfrak{Q}$(′) for some ′ ∉ $\mathcal{P}$, then $\Theta \left(\mathfrak{Q}\right(\beth \left)\right)=\Theta \left(\mathfrak{Q}\right(\daleth \left)\right)$.

Proof. 

(1) Let $⅁\in \Theta \left(\mathfrak{Q}\right(\beth \left)\right)$. Then there exists $\mathfrak{U}\in {\mathcal{T}}_{Ad}(⅁)$ such that ${\mathfrak{U}}^c\cup \mathfrak{Q}(\beth )$ = ($\mathfrak{U}$ ∩ ($\mathfrak{Q}$() − $\mathfrak{Q}(\beth )$))^c = ${(\mathfrak{U}-\mathfrak{Q}(\beth ))}^c$ = $\mathfrak{Q}$(′) for some ′ ∉ $\mathcal{P}$. Then, $⅁$ ∉ ($\mathfrak{Q}$() − $\mathfrak{Q}(\beth )$)^• and $⅁$ ∈ $\mathfrak{Q}$() − ($\mathfrak{Q}$() − $\mathfrak{Q}(\beth )$)^•. Conversely, let $⅁$ ∈ $\mathfrak{Q}$() − ($\mathfrak{Q}$() − $\mathfrak{Q}(\beth )$)^•. Then, $⅁$ ∉ ($\mathfrak{Q}$() − $\mathfrak{Q}(\beth )$)^•, and there exists $\mathfrak{U}\in {\mathcal{T}}_{Ad}(⅁)$ such that ${\mathfrak{U}}^c$ ∪ ($\mathfrak{Q}$() − $\mathfrak{Q}(\beth )$)^c = ${(\mathfrak{U}-\mathfrak{Q}(\beth ))}^c$ = $\mathfrak{Q}$(′) for some ′ ∉ $\mathcal{P}$. Hence, $⅁\in \Theta \left(\mathfrak{Q}\right(\beth \left)\right)$ and $\Theta \left(\mathfrak{Q}(\beth )\right)$ = $\mathfrak{Q}$() − ($\mathfrak{Q}$() − $\mathfrak{Q}(\beth )$)^•.

(2)

This is straightforward from (3) of Theorem 3.

(3)

This is straightforward from (1) of Theorem 3.

(4)

It is straightforward from (3) that $\Theta \left(\mathfrak{Q}\right(\beth )\cap \mathfrak{Q}(\daleth \left)\right)\subseteq \Theta \left(\mathfrak{Q}\right(\beth \left)\right)$ and $\Theta \left(\mathfrak{Q}\right(\beth )\cap \mathfrak{Q}(\daleth \left)\right)\subseteq \Theta \left(\mathfrak{Q}\right(\daleth \left)\right)$. Hence, $\Theta \left(\mathfrak{Q}\right(\beth )\cap \mathfrak{Q}(\daleth \left)\right)\subseteq \Theta \left(\mathfrak{Q}\right(\beth \left)\right)\cap \Theta \left(\mathfrak{Q}\right(\daleth \left)\right)$. Now, let $⅁\in \Theta \left(\mathfrak{Q}\right(\beth \left)\right)\cap \Theta \left(\mathfrak{Q}\right(\daleth \left)\right)$. Then, there exists $\mathfrak{U},\mathfrak{T}\in {\mathcal{T}}_{Ad}(⅁)$ such that ${(\mathfrak{U}-\mathfrak{Q}(\beth ))}^c$ = $\mathfrak{Q}$(′) for some ′ ∉ $\mathcal{P}$ and ${(\mathfrak{T}-\mathfrak{Q}(\daleth ))}^c$ = $\mathfrak{Q}$(′) for some ′ ∉$\mathcal{P}$. Let $G=\mathfrak{U}\cap \mathfrak{T}\in {\mathcal{T}}_{Ad}(⅁)$. Then, we have ${(G-\mathfrak{Q}(\beth ))}^c$ = $\mathfrak{Q}$(′) for some ′ ∉ $\mathcal{P}$ and ${(G-\mathfrak{Q}(\daleth ))}^c$ = $\mathfrak{Q}$(′) for some ′ ∉ $\mathcal{P}$ by heredity. Thus, ${[G-(\mathfrak{Q}(\beth )\cap \mathfrak{Q}(\daleth ))]}^c={(G-\mathfrak{Q}(\beth ))}^c\cap {(G-\mathfrak{Q}(\daleth ))}^c$ = $\mathfrak{Q}$(′) for some ′ ∉ $\mathcal{P}$ according to Corollary 1; hence, $⅁\in \Theta \left(\mathfrak{Q}\right(\beth )\cap \mathfrak{Q}(\daleth \left)\right)$. We have shown that $\Theta \left(\mathfrak{Q}\right(\beth \left)\right)\cap \Theta \left(\mathfrak{Q}\right(\daleth \left)\right)\subseteq \Theta \left(\mathfrak{Q}\right(\beth )\cap \mathfrak{Q}(\daleth \left)\right)$, so the proof is completed.

(5)

If $\mathfrak{U}\in {\mathcal{T}}_{Ad}^{•}$, then ($\mathfrak{Q}$() − $\mathfrak{U}$)^• ⊆ $\mathfrak{Q}$() − $\mathfrak{U}$. Hence, $\mathfrak{U}$ ⊆ $\mathfrak{Q}$() − ($\mathfrak{Q}$() − $\mathfrak{U}$)^• = $\Theta \left(\mathfrak{U}\right)$.

(6)

This is straightforward from (2) and (5).

(7)

This follows from the following facts:

(a)

$\Theta \left(\mathfrak{Q}(\beth )\right)$ = $\mathfrak{Q}$() − ($\mathfrak{Q}$() − $\mathfrak{Q}(\beth )$)^•.

(b)

$\Theta (\Theta \left(\mathfrak{Q}(\beth )\right))$ = $\mathfrak{Q}$() − [$\mathfrak{Q}$() − ($\mathfrak{Q}$() − ($\mathfrak{Q}$() − $\mathfrak{Q}(\beth )$)^•)]^• = $\mathfrak{Q}$() − (($\mathfrak{Q}$() − $\mathfrak{Q}(\beth )$)^•)^•.

(8)

By Corollary 5, it follows that ($\mathfrak{Q}$() − $\mathfrak{Q}(\beth )$)^• = $\mathfrak{Q}$()^• if $\mathfrak{Q}{(\beth )}^c$ = $\mathfrak{Q}$(′) for some ′ ∉ $\mathcal{P}$. Then, $\Theta \left(\mathfrak{Q}(\beth )\right)$ = $\mathfrak{Q}$() − ($\mathfrak{Q}$() − $\mathfrak{Q}(\beth )$)^• = $\mathfrak{Q}$() − $\mathfrak{Q}$()^•.

(9)

If $⅁\in \mathfrak{Q}(\beth )\cap \Theta \left(\mathfrak{Q}\right(\beth \left)\right)$, then $⅁\in \mathfrak{Q}(\beth )$, and there exists $U_{⅁}\in {\mathcal{T}}_{Ad}(⅁)$ such that ${(U_{⅁}-\mathfrak{Q}(\beth ))}^c$ = $\mathfrak{Q}$(′) for some ′ ∉ $\mathcal{P}$. Then, according to Theorem 9, $U_{⅁}\cap {(U_{⅁}-\mathfrak{Q}(\beth ))}^c$ is a ${\mathcal{T}}_{Ad}^{•}$-open neighborhood of $⅁$ and $⅁\in {\mathfrak{int}}^{•}\left(\mathfrak{Q}(\beth )\right)$. On the other hand, if $⅁\in {\mathfrak{int}}^{•}\left(\mathfrak{Q}(\beth )\right)$, there exists a basic ${\mathcal{T}}_{Ad}^{•}$-open neighborhood $V_{⅁}\cap I$ of $⅁$, where $V_{⅁}\in {\mathcal{T}}_{Ad}$ and I =$\mathfrak{Q}$(′) for some ′ ∉ $\mathcal{P}$ such that $⅁\in V_{⅁}\cap I\subseteq \mathfrak{Q}(\beth )$, which implies that $I\subseteq {(V_{⅁}-\mathfrak{Q}(\beth ))}^c$, so ${(V_{⅁}-\mathfrak{Q}(\beth ))}^c$ = $\mathfrak{Q}$(′) for some ′ ∉ $\mathcal{P}$. Hence, $⅁\in \mathfrak{Q}(\beth )\cap \Theta \left(\mathfrak{Q}\right(\beth \left)\right)$.

(10)

This follows from Corollary 5 and $\Theta (\mathfrak{Q}(\beth )-I)$ = $\mathfrak{Q}$() − [$\mathfrak{Q}$() − ($\mathfrak{Q}(\beth )-I)$]^• =

$\mathfrak{Q}$() − [($\mathfrak{Q}$() − $\mathfrak{Q}(\beth ))\cup I$]^• = $\mathfrak{Q}$() − ($\mathfrak{Q}$() − $\mathfrak{Q}(\beth )$)^• = $\Theta \left(\mathfrak{Q}(\beth )\right)$.

(11)

This follows from Corollary 5 and $\Theta (\mathfrak{Q}(\beth )\cup I)$ = $\mathfrak{Q}$() − [$\mathfrak{Q}$() − ($\mathfrak{Q}(\beth )\cup I$)]^• =

$\mathfrak{Q}$() − [($\mathfrak{Q}$() − $\mathfrak{Q}(\beth ))-I$]^• = $\mathfrak{Q}$() − ($\mathfrak{Q}$() − $\mathfrak{Q}(\beth )$)^• = $\Theta \left(\mathfrak{Q}(\beth )\right)$.

(12)

Assume that ${[(\mathfrak{Q}(\beth )-\mathfrak{Q}(\daleth ))\cup (\mathfrak{Q}(\daleth )-\mathfrak{Q}(\beth ))]}^c$ = $\mathfrak{Q}$(′) for some ′ ∉ $\mathcal{P}$. Let $\mathfrak{Q}(\beth )-\mathfrak{Q}(\daleth )=I$ and

$\mathfrak{Q}(\daleth )-\mathfrak{Q}(\beth )=J$. Observe that $I^c,J^c$ = $\mathfrak{Q}$(′) for some ′ ∉ $\mathcal{P}$ by heredity. Also, observe that

$\mathfrak{Q}(\daleth )=\left(\mathfrak{Q}\right(\beth )-I)\cup J$. Thus, $\Theta \left(\mathfrak{Q}\right(\beth \left)\right)=\Theta \left(\mathfrak{Q}\right(\beth )-I)=\Theta \left[\right(\mathfrak{Q}(\beth )-I)\cup J]=\Theta \left(\mathfrak{Q}\right(\daleth \left)\right)$ by (10) and (11).

□

Corollary 6.

Let ($\mathfrak{Q}$(), ${\mathcal{T}}_{Ad},\mathcal{P}$) be a graph primal ATS. Then, $\mathfrak{U}\subseteq \Theta \left(\mathfrak{U}\right)$ for every open set $\mathfrak{U}\in {\mathcal{T}}_{Ad}$.

Proof. 

We know that $\Theta \left(\mathfrak{U}\right)$ = $\mathfrak{Q}$() − ($\mathfrak{Q}$() − $\mathfrak{U}$)^•. Now, ($\mathfrak{Q}$() − $\mathfrak{U}$)^• ⊆ $\mathfrak{CL}$($\mathfrak{Q}$() − $\mathfrak{U}$) = $\mathfrak{Q}$() − $\mathfrak{U}$ since $\mathfrak{Q}$() − $\mathfrak{U}$ is closed. As a result, $\mathfrak{U}$ = $\mathfrak{Q}$() − ($\mathfrak{Q}$() − $\mathfrak{U}$) ⊆ $\mathfrak{Q}$() − ($\mathfrak{Q}$() − $\mathfrak{U}$)^• = $\Theta \left(\mathfrak{U}\right)$. □

Theorem 14.

Let ($\mathfrak{Q}$(), ${\mathcal{T}}_{Ad},\mathcal{P}$) be a graph primal ATS and $\mathfrak{Q}(\beth )$ ⊆ $\mathfrak{Q}$(). Then, the following properties hold:

(1) 

$\Theta \left(\mathfrak{Q}(\beth )\right)$ = ⋃{$\mathfrak{U}\in {\mathcal{T}}_{Ad}:{(\mathfrak{U}-\mathfrak{Q}(\beth ))}^c$ = $\mathfrak{Q}$(′) for some ′ ∉ $\mathcal{P}$};

(2) 

$\Theta \left(\mathfrak{Q}(\beth )\right)$ ⊇ ⋃{$\mathfrak{U}\in {\mathcal{T}}_{Ad}:{(\mathfrak{U}-\mathfrak{Q}(\beth ))}^c\cup {(\mathfrak{Q}(\beth )-\mathfrak{U})}^c$ = $\mathfrak{Q}$(′) for some ′ ∉ $\mathcal{P}$}.

Proof. 

(1) This is straightforward from the definition of $\Theta$-operator.

(2)

Since $\mathcal{P}$ is hereditary, ⋃{$\mathfrak{U}\in {\mathcal{T}}_{Ad}:{(\mathfrak{U}-\mathfrak{Q}(\beth ))}^c\cup {(\mathfrak{Q}(\beth )-\mathfrak{U})}^c$ = $\mathfrak{Q}$(′) for some ′ ∉ $\mathcal{P}$} ⊆ ⋃{$\mathfrak{U}\in {\mathcal{T}}_{Ad}:{(\mathfrak{U}-\mathfrak{Q}(\beth ))}^c$ = $\mathfrak{Q}$(″) for some ″ ∉ $\mathcal{P}$} = $\Theta \left(\mathfrak{Q}(\beth )\right)$, for every $\mathfrak{Q}(\beth )$ ⊆ $\mathfrak{Q}$().

□

Theorem 15.

Let ($\mathfrak{Q}$(), ${\mathcal{T}}_{Ad},\mathcal{P}$) be a graph primal ATS. If $\sigma$ = {$\mathfrak{Q}(\beth )$ ⊆ $\mathfrak{Q}$(): $\mathfrak{Q}(\beth )\subseteq \Theta \left(\mathfrak{Q}\right(\beth \left)\right)$}. Then, σ is a topology on $\mathfrak{Q}$() and $\sigma ={\mathcal{T}}_{Ad}^{•}$.

Proof. 

Let $\sigma$ = {$\mathfrak{Q}(\beth )$ ⊆ $\mathfrak{Q}$(): $\mathfrak{Q}(\beth )\subseteq \Theta \left(\mathfrak{Q}\right(\beth \left)\right)$}. First, we show that $\sigma$ is a topology. Note that $\varnothing \subseteq \Theta (\varnothing )$ and $\mathfrak{Q}$() ⊆ Θ($\mathfrak{Q}$()) = $\mathfrak{Q}$(). Thus, ∅ and $\mathfrak{Q}$() ∈ $\sigma$. If $\mathfrak{Q}(\beth ),\mathfrak{Q}(\daleth )\in \sigma$, then $\mathfrak{Q}(\beth )\cap \mathfrak{Q}(\daleth )\subseteq \Theta \left(\mathfrak{Q}\right(\beth \left)\right)\cap \Theta \left(\mathfrak{Q}\right(\daleth \left)\right)=\Theta \left(\mathfrak{Q}\right(\beth )\cap \mathfrak{Q}(\daleth \left)\right)$. As a result, $\mathfrak{Q}(\beth )\cap \mathfrak{Q}(\daleth )\in \sigma$. If $\{A_{\alpha }:\alpha \in \Delta \}\subseteq \sigma$. Then, $A_{\alpha }\subseteq \Theta \left(A_{\alpha }\right)\subseteq \Theta \left({\bigcup }_{\alpha \in \Delta }{A}_{\alpha }\right)$ for every $\alpha \in \Delta$, so $\bigcup A_{\alpha }\subseteq \Theta \left({\bigcup }_{\alpha \in \Delta }{A}_{\alpha }\right)$. Hence, $\sigma$ is a topology on $\mathfrak{Q}$(). Now, if $\mathfrak{U}\in {\mathcal{T}}_{Ad}^{•}$ and $⅁\in \mathfrak{U}$, then according to Theorem 9, there exists $\mathfrak{T}\in {\mathcal{T}}_{Ad}(⅁)$ and I = $\mathfrak{Q}$(′) for some ′ ∉ $\mathcal{P}$ such that $⅁\in \mathfrak{T}\cap I\subseteq \mathfrak{U}$. Clearly, $I\subseteq {(\mathfrak{T}-\mathfrak{U})}^c$, so ${(\mathfrak{T}-\mathfrak{U})}^c$ = $\mathfrak{Q}$(″) for some ″ ∉ $\mathcal{P}$ by heredity, and hence $⅁\in \Theta \left(\mathfrak{U}\right)$. Thus, $\mathfrak{U}\subseteq \Theta \left(\mathfrak{U}\right)$ and ${\mathcal{T}}_{Ad}^{•}\subseteq \sigma$. Now let $\mathfrak{Q}(\beth )\in \sigma$. Then, we have $\mathfrak{Q}(\beth )\subseteq \Theta \left(\mathfrak{Q}\right(\beth \left)\right)$, i.e., $\mathfrak{Q}(\beth )$ ⊆ $\mathfrak{Q}$() − ($\mathfrak{Q}$() − $\mathfrak{Q}(\beth )$)^• and ($\mathfrak{Q}$() − $\mathfrak{Q}(\beth )$)^• ⊆ $\mathfrak{Q}$() − $\mathfrak{Q}(\beth )$. Therefore, $\mathfrak{Q}$() − $\mathfrak{Q}(\beth )$ is ${\mathcal{T}}_{Ad}^{•}$-closed and hence $\mathfrak{Q}(\beth )\in {\mathcal{T}}_{Ad}^{•}$. So, $\sigma \subseteq {\mathcal{T}}_{Ad}^{•}$, and thus $\sigma ={\mathcal{T}}_{Ad}^{•}$. □

4. Suitability of ${\mathcal{T}}_{Ad}$ with $\mathcal{P}$

This section presents the definition of the suitability of the topological space ${\mathcal{T}}_{Ad}$ with the graph primal $\mathcal{P}$. This definition establishes a foundational criterion for evaluating how well the topology interacts with the graph primal structure, indicating the effective linkages and properties between the two notions.

Definition 11.

Let ($\mathfrak{Q}$(), ${\mathcal{T}}_{Ad},\mathcal{P}$) be a graph primal ATS. Then, ${\mathcal{T}}_{Ad}$ is said to be suitable for the graph primal $\mathcal{P}$ if $\mathfrak{Q}{(\beth )}^c\cup \mathfrak{Q}{(\beth )}^{•}$ = $\mathfrak{Q}$(′) for some ′ ∉ $\mathcal{P}$, for any subgraph $\beth$ ⊆ .

Example 10.

Let be the graph ($\mathfrak{Q}$(), $\mathfrak{Z}$()), where $\mathfrak{Q}$() = $\{{⅁}_1,{⅁}_2,{⅁}_3,{⅁}_4\}$ and $\mathfrak{Z}$() = $\{{\alpha }_1,{\alpha }_2,{\alpha }_3,{\alpha }_4\}.$ A drawing of the graph is shown in Figure 10.

Figure 10. Graph defined in Example 10.

By computing $S_N=\{\{{⅁}_2,{⅁}_3\},\{{⅁}_1,{⅁}_3\},\{{⅁}_1,{⅁}_2,{⅁}_4\},\left\{{⅁}_3\right\}\}$. So, $\beta =\{\left\{{⅁}_3\right\},\{{⅁}_2,{⅁}_3\}$, $\{{⅁}_1,{⅁}_3\},\left\{{⅁}_2\right\},\left\{{⅁}_1\right\},\{{⅁}_1,$${⅁}_2,{⅁}_4\},$$\left\{{⅁}_3\right\},\varnothing \}.$ Thus, ${\mathcal{T}}_A=\{\varnothing ,\{{⅁}_1,{⅁}_2,{⅁}_4\},\left\{{⅁}_1\right\},\left\{{⅁}_2\right\},\left\{{⅁}_3\right\}$, $\{{⅁}_1,{⅁}_2\},\{{⅁}_1,{⅁}_3\},\{{⅁}_2,{⅁}_3\},$$\{{⅁}_1,{⅁}_2,$${⅁}_3\},$ $\{{⅁}_1,{⅁}_2,{⅁}_3,{⅁}_4\}\}.$ Define a graph primal as $\mathcal{P}$ = P() − $\{(\{{⅁}_1,{⅁}_2,{⅁}_3\},\{{\alpha }_1,{\alpha }_4\}),(\{{⅁}_1,{⅁}_2,{⅁}_3,{⅁}_4,\},$$\{{\alpha }_1,{\alpha }_2,{\alpha }_4\}),$$(\{{⅁}_1,{⅁}_2,{⅁}_3,{⅁}_4\},\{{\alpha }_1,{\alpha }_3$, ${\alpha }_4\left\}\right),G\}$. According to the open sets of ${\mathcal{T}}_A$, we have $N\left({⅁}_1\right)=\{\left\{{⅁}_1\right\},\{{⅁}_1,{⅁}_2\},\{{⅁}_1,{⅁}_3\},$$\{{⅁}_1,{⅁}_2,{⅁}_4\},\{{⅁}_1,{⅁}_2,{⅁}_3\},$ $\{{⅁}_1,{⅁}_2,{⅁}_3,{⅁}_4\}\},$$N\left({⅁}_2\right)=\{\left\{{⅁}_2\right\},\{{⅁}_2,{⅁}_3\},\{{⅁}_1,$${⅁}_2\},\{{⅁}_1,{⅁}_2,{⅁}_3\}$, $\{{⅁}_1,{⅁}_2,{⅁}_4\},$$\{{⅁}_1,{⅁}_2,{⅁}_3,{⅁}_4\}\},$ $N\left({⅁}_3\right)=\{\left\{{⅁}_3\right\},\{{⅁}_2,{⅁}_3\},\{{⅁}_1,{⅁}_3\},\{{⅁}_1,{⅁}_2,{⅁}_3\},\{{⅁}_1,$${⅁}_2,{⅁}_3$, ${⅁}_4\left\}\right\}$ and $N\left({⅁}_4\right)=\left\{\right\{{⅁}_1,{⅁}_2,$${⅁}_4\},\{{⅁}_1,{⅁}_2,{⅁}_3$, ${⅁}_4\left\}\right\}.$ The computations of the graph-local function associated with the defined graph primal are given in Table 5. According to Table 5, ${\mathcal{T}}_{Ad}$ is suitable for the graph primal $\mathcal{P}$. Clearly, if $\mathcal{P}$ = P () − {}, then ${\mathcal{T}}_{Ad}$ is not suitable for the graph primal $\mathcal{P}.$

Table 5. Illustration of Definition 11.

$\mathit{\beth }$ ⊆ $\mathfrak{Q}$()	$\mathfrak{Q}{\left(\mathit{\beth }\right)}^{\mathit{c}}$	$\mathfrak{Q}{\left(\mathit{\beth }\right)}^{•}$	$\mathfrak{Q}{\left(\mathit{\beth }\right)}^{\mathit{c}}\cup \mathfrak{Q}{\left(\mathit{\beth }\right)}^{•}$	$\mathfrak{Q}{\left(\mathit{\beth }\right)}^{\mathit{c}}\cup \mathfrak{Q}{\left(\mathit{\beth }\right)}^{•}$ = $\mathfrak{Q}$(′) and ′ ∉ $\mathcal{P}$?
∅	$\{{⅁}_1,{⅁}_2,{⅁}_3,{⅁}_4\}$	∅	$\{{⅁}_1,{⅁}_2,{⅁}_3,{⅁}_4\}$	$\mathrm{Yes}$
$\mathfrak{Q}$()	∅	$\{{⅁}_1,{⅁}_2,{⅁}_3,{⅁}_4\}$	$\{{⅁}_1,{⅁}_2,{⅁}_3,{⅁}_4\}$	$\mathrm{Yes}$
$\left\{{⅁}_1\right\}$	$\{{⅁}_2,{⅁}_3,{⅁}_4\}$	$\{{⅁}_1,{⅁}_4\}$	$\{{⅁}_1,{⅁}_2,{⅁}_3,{⅁}_4\}$	$\mathrm{Yes}$
$\left\{{⅁}_2\right\}$	$\{{⅁}_1,{⅁}_3,{⅁}_4\}$	$\{{⅁}_2,{⅁}_4\}$	$\{{⅁}_1,{⅁}_2,{⅁}_3,{⅁}_4\}$	$\mathrm{Yes}$
$\left\{{⅁}_3\right\}$	$\{{⅁}_1,{⅁}_2,{⅁}_4\}$	$\{{⅁}_3,{⅁}_4\}$	$\{{⅁}_1,{⅁}_2,{⅁}_3,{⅁}_4\}$	$\mathrm{Yes}$
$\left\{{⅁}_4\right\}$	$\{{⅁}_1,{⅁}_2,{⅁}_3\}$	∅	$\{{⅁}_1,{⅁}_2,{⅁}_3\}$	$\mathrm{Yes}$
$\{{⅁}_1,{⅁}_2\}$	$\{{⅁}_3,{⅁}_4\}$	$\{{⅁}_1,{⅁}_2,{⅁}_4\}$	$\{{⅁}_1,{⅁}_2,{⅁}_3,{⅁}_4\}$	$\mathrm{Yes}$
$\{{⅁}_1,{⅁}_3\}$	$\{{⅁}_2,{⅁}_4\}$	$\{{⅁}_1,{⅁}_3,{⅁}_4\}$	$\{{⅁}_1,{⅁}_2,{⅁}_3,{⅁}_4\}$	$\mathrm{Yes}$
$\{{⅁}_1,{⅁}_4\}$	$\{{⅁}_2,{⅁}_3\}$	$\{{⅁}_1,{⅁}_4\}$	$\{{⅁}_1,{⅁}_2,{⅁}_3,{⅁}_4\}$	$\mathrm{Yes}$
$\{{⅁}_2,{⅁}_3\}$	$\{{⅁}_1,{⅁}_4\}$	$\{{⅁}_2,{⅁}_3,{⅁}_4\}$	$\{{⅁}_1,{⅁}_2,{⅁}_3,{⅁}_4\}$	$\mathrm{Yes}$
$\{{⅁}_2,{⅁}_4\}$	$\{{⅁}_1,{⅁}_3\}$	$\{{⅁}_2,{⅁}_4\}$	$\{{⅁}_1,{⅁}_2,{⅁}_3,{⅁}_4\}$	$\mathrm{Yes}$
$\{{⅁}_3,{⅁}_4\}$	$\{{⅁}_1,{⅁}_2\}$	$\{{⅁}_3,{⅁}_4\}$	$\{{⅁}_1,{⅁}_2,{⅁}_3,{⅁}_4\}$	$\mathrm{Yes}$
$\{{⅁}_1,{⅁}_2,{⅁}_3\}$	$\left\{{⅁}_4\right\}$	$\{{⅁}_1,{⅁}_2,{⅁}_3,{⅁}_4\}$	$\{{⅁}_1,{⅁}_2,{⅁}_3,{⅁}_4\}$	$\mathrm{Yes}$
$\{{⅁}_1,{⅁}_2,{⅁}_4\}$	$\left\{{⅁}_3\right\}$	$\{{⅁}_1,{⅁}_2,{⅁}_4\}$	$\{{⅁}_1,{⅁}_2,{⅁}_3,{⅁}_4\}$	$\mathrm{Yes}$
$\{{⅁}_1,{⅁}_3,{⅁}_4\}$	$\left\{{⅁}_2\right\}$	$\{{⅁}_1,{⅁}_3,{⅁}_4\}$	$\{{⅁}_1,{⅁}_2,{⅁}_3,{⅁}_4\}$	$\mathrm{Yes}$
$\{{⅁}_2,{⅁}_3,{⅁}_4\}$	$\left\{{⅁}_1\right\}$	$\{{⅁}_2,{⅁}_3,{⅁}_4\}$	$\{{⅁}_1,{⅁}_2,{⅁}_3,{⅁}_4\}$	$\mathrm{Yes}$

We now describe this notion in some analogous terms.

Theorem 16.

For a graph primal ATS ($\mathfrak{Q}$(), ${\mathcal{T}}_{Ad},\mathcal{P}$), the following are equivalent:

(1) 

${\mathcal{T}}_{Ad}$ is suitable for the graph primal $\mathcal{P}$;

(2) 

For any ${\mathcal{T}}_{Ad}^{•}$-closed subset $\mathfrak{Q}(\beth )$ of $\mathfrak{Q}$(), $\mathfrak{Q}{(\beth )}^c\cup \mathfrak{Q}{(\beth )}^{•}$ = $\mathfrak{Q}$(′) for some ′ ∉ $\mathcal{P}$;

(3) 

Whenever for any $\mathfrak{Q}(\beth )$ ⊆ $\mathfrak{Q}$() and each $⅁\in \mathfrak{Q}(\beth )$ there corresponds some $U_x\in {\mathcal{T}}_{Ad}(⅁)$ with $U_{⅁}^c\cup \mathfrak{Q}{(\beth )}^c$ = $\mathfrak{Q}$(′) for some ′ ∉ $\mathcal{P}$, it implies that $\mathfrak{Q}{(\beth )}^c$ = $\mathfrak{Q}$(″) for some ″ ∉ $\mathcal{P}$;

(4) 

For $\mathfrak{Q}(\beth )$ ⊆ $\mathfrak{Q}$() and $\mathfrak{Q}(\beth )\cap \mathfrak{Q}{(\beth )}^{•}=\varnothing$, it follows that $\mathfrak{Q}{(\beth )}^c$ = $\mathfrak{Q}$(′) for some ′ ∉ $\mathcal{P}$.

Proof. 

(1) ⇒ (2): This is straightforward.

(2)

⇒ (3): Let $\mathfrak{Q}(\beth )$ ⊆ $\mathfrak{Q}$(), and suppose that for every $⅁\in \mathfrak{Q}(\beth )$, there exists $\mathfrak{U}\in {\mathcal{T}}_{Ad}(⅁)$ such that ${\mathfrak{U}}^c\cup \mathfrak{Q}{(\beth )}^c$ = $\mathfrak{Q}$(′) for some ′ ∉ $\mathcal{P}$. Then, $⅁\notin \mathfrak{Q}{(\beth )}^{•}$ so that $\mathfrak{Q}(\beth )\cap \mathfrak{Q}{(\beth )}^{•}=\varnothing$. Since $\mathfrak{Q}(\beth )\cup \mathfrak{Q}{(\beth )}^{•}$ is ${\mathcal{T}}_{Ad}^{•}$-closed, by (2), we have ${(\mathfrak{Q}(\beth )\cup \mathfrak{Q}{(\beth )}^{•})}^c\cup {(\mathfrak{Q}(\beth )\cup \mathfrak{Q}{(\beth )}^{•})}^{•}$ = $\mathfrak{Q}$(′) for some ′ ∉ $\mathcal{P}$, i.e., ${(\mathfrak{Q}(\beth )\cup \mathfrak{Q}{(\beth )}^{•})}^c\cup (\mathfrak{Q}{(\beth )}^{•}\cup {\left(\mathfrak{Q}{(\beth )}^{•}\right)}^{•})$ = $\mathfrak{Q}$(′) for some ′ ∉ $\mathcal{P}$ according to Theorem 3, i.e., ${(\mathfrak{Q}(\beth )\cup \mathfrak{Q}{(\beth )}^{•})}^c\cup \mathfrak{Q}{(\beth )}^{•}$ = $\mathfrak{Q}$(^★) for some ^★ ∉ $\mathcal{P}$ according to Theorem 3, i.e., $\mathfrak{Q}{(\beth )}^c$ = $\mathfrak{Q}$(″) for some ″ ∉ $\mathcal{P}$ (as $\mathfrak{Q}(\beth )\cap \mathfrak{Q}{(\beth )}^{•}=\varnothing$).

(3)

⇒ (4): If $\mathfrak{Q}(\beth )$ ⊆ $\mathfrak{Q}$() and $\mathfrak{Q}(\beth )\cap \mathfrak{Q}{(\beth )}^{•}=\varnothing$, then $\mathfrak{Q}(\beth )$ ⊆ $\mathfrak{Q}$() ∖ ${\mathfrak{Q}(\beth )}^{•}$. Let $⅁\in \mathfrak{Q}(\beth )$. Then, $⅁\notin \mathfrak{Q}{(\beth )}^{•}$. So, there exists $\mathfrak{U}\in {\mathcal{T}}_{Ad}(⅁)$ such that ${\mathfrak{U}}^c\cup \mathfrak{Q}{(\beth )}^c$ = $\mathfrak{Q}$(^★) for some ^★ ∉ $\mathcal{P}$. Then, by (3), $\mathfrak{Q}{(\beth )}^c$ = $\mathfrak{Q}$(′) for some ′ ∉ $\mathcal{P}$.

(4)

⇒ (1): Let $\mathfrak{Q}(\beth )$ ⊆ $\mathfrak{Q}$(). We first claim that $(\mathfrak{Q}(\beth )\setminus \mathfrak{Q}{(\beth )}^{•})\cap {(\mathfrak{Q}(\beth )\setminus \mathfrak{Q}{(\beth )}^{•})}^{•}=\varnothing$. In fact, if $⅁\in (\mathfrak{Q}(\beth )\setminus \mathfrak{Q}{(\beth )}^{•})\cap {(\mathfrak{Q}(\beth )\setminus \mathfrak{Q}{(\beth )}^{•})}^{•}$, then $⅁\in \mathfrak{Q}(\beth )\setminus \mathfrak{Q}{(\beth )}^{•}.$ Thus, $⅁\in \mathfrak{Q}(\beth )$ and $⅁\notin \mathfrak{Q}{(\beth )}^{•}.$ Then, there exists $\mathfrak{U}\in {\mathcal{T}}_{Ad}(⅁)$ such that ${\mathfrak{U}}^c\cup \mathfrak{Q}{(\beth )}^c$ = $\mathfrak{Q}$(′) for some ′ ∉ $\mathcal{P}$. Now, ${\mathfrak{U}}^c\cup \mathfrak{Q}{(\beth )}^c\subseteq {\mathfrak{U}}^c\cup {(\mathfrak{Q}(\beth )\setminus \mathfrak{Q}{(\beth )}^{•})}^c$ according to Corollary 1 (2), ${\mathfrak{U}}^c\cup {(\mathfrak{Q}(\beth )\setminus \mathfrak{Q}{(\beth )}^{•})}^c$ = $\mathfrak{Q}$(″) for some ″ ∉ $\mathcal{P}$. Hence, $⅁\notin {(\mathfrak{Q}(\beth )\setminus \mathfrak{Q}{(\beth )}^{•})}^{•}$, which is a contradiction. Hence, by (4), ${(\mathfrak{Q}(\beth )\setminus \mathfrak{Q}{(\beth )}^{•})}^c=\mathfrak{Q}{(\beth )}^c\cup \mathfrak{Q}{(\beth )}^{•}$ = $\mathfrak{Q}$(‴) for some ‴ ∉ $\mathcal{P}$ and ${\mathcal{T}}_{Ad}$ is suitable for the graph primal $\mathcal{P}$.

□

Theorem 17.

For a graph primal ATS ($\mathfrak{Q}$(), ${\mathcal{T}}_{Ad},\mathcal{P}$), the following conditions are equivalent, and any of them is necessary for ${\mathcal{T}}_{Ad}$ to be suitable for the graph primal $\mathcal{P}$:

(1) 

For any $\mathfrak{Q}(\beth )$ ⊆ $\mathfrak{Q}$(), $\mathfrak{Q}(\beth )\cap \mathfrak{Q}{(\beth )}^{•}=\varnothing$, then $\mathfrak{Q}{(\beth )}^{•}=\varnothing$;

(2) 

For any $\mathfrak{Q}(\beth )$ ⊆ $\mathfrak{Q}$(), ${(\mathfrak{Q}(\beth )\setminus \mathfrak{Q}{(\beth )}^{•})}^{•}=\varnothing$;

(3) 

For any $\mathfrak{Q}(\beth )$ ⊆ $\mathfrak{Q}$(), ${(\mathfrak{Q}(\beth )\cap \mathfrak{Q}{(\beth )}^{•})}^{•}=\mathfrak{Q}{(\beth )}^{•}$.

Proof. 

(1) ⇒ (2): It follows by noting that $(\mathfrak{Q}(\beth )\setminus \mathfrak{Q}{(\beth )}^{•})\cap {(\mathfrak{Q}(\beth )\setminus \mathfrak{Q}{(\beth )}^{•})}^{•}=\varnothing$, for all $\mathfrak{Q}(\beth )$ ⊆ $\mathfrak{Q}$().

(2)

⇒ (3): Since $\mathfrak{Q}(\beth )=(\mathfrak{Q}(\beth )\setminus (\mathfrak{Q}(\beth )\cap \mathfrak{Q}{(\beth )}^{•}))\cup (\mathfrak{Q}(\beth )\cap \mathfrak{Q}{(\beth )}^{•}),$ we have $\mathfrak{Q}{(\beth )}^{•}={(\mathfrak{Q}(\beth )\setminus (\mathfrak{Q}(\beth )\cap \mathfrak{Q}{(\beth )}^{•}))}^{•}\cup {(\mathfrak{Q}(\beth )\cap \mathfrak{Q}{(\beth )}^{•})}^{•}={(\mathfrak{Q}(\beth )\setminus \mathfrak{Q}{(\beth )}^{•})}^{•}\cup {(\mathfrak{Q}(\beth )\cap \mathfrak{Q}{(\beth )}^{•})}^{•}={(\mathfrak{Q}(\beth )\cap \mathfrak{Q}{(\beth )}^{•})}^{•}$ by (2).

(3)

⇒ (1): Let $\mathfrak{Q}(\beth )$ ⊆ $\mathfrak{Q}$() and $\mathfrak{Q}(\beth )\cap \mathfrak{Q}{(\beth )}^{•}=\varnothing$. Then, by (3), $\mathfrak{Q}{(\beth )}^{•}={(\mathfrak{Q}(\beth )\cap \mathfrak{Q}{(\beth )}^{•})}^{•}={\varnothing }^{•}=\varnothing$.

□

Corollary 7.

If ($\mathfrak{Q}$(), ${\mathcal{T}}_{Ad},\mathcal{P}$) is a graph primal ATS such that ${\mathcal{T}}_{Ad}$ is suitable for $\mathcal{P}$, then the operator • is an idempotent operator, i.e., $\mathfrak{Q}{(\beth )}^{•}={\left(\mathfrak{Q}{(\beth )}^{•}\right)}^{•}$ for any subgraph $\beth$ ⊆ .

Proof. 

By Theorem 3 (6), we have ${\left(\mathfrak{Q}{(\beth )}^{•}\right)}^{•}\subseteq \mathfrak{Q}{(\beth )}^{•}$. By Theorem 17 and Theorem 3 (1), we get $\mathfrak{Q}{(\beth )}^{•}={(\mathfrak{Q}(\beth )\cap \mathfrak{Q}{(\beth )}^{•})}^{•}\subseteq {\left(\mathfrak{Q}{(\beth )}^{•}\right)}^{•}.$ □

Theorem 18.

Let ($\mathfrak{Q}$(), ${\mathcal{T}}_{Ad},\mathcal{P}$) be a graph primal ATS such that ${\mathcal{T}}_{Ad}$ is suitable for $\mathcal{P}$. Then, a subset $\mathfrak{Q}(\beth )$ of $\mathfrak{Q}$() is ${\mathcal{T}}_{Ad}^{•}$-closed iff it can be expressed as a union of a set $\mathfrak{Q}(\daleth ),$$\daleth$ ⊆ , which is closed in ($\mathfrak{Q}$(), ${\mathcal{T}}_{Ad}$), and $\mathfrak{Q}{(\daleth )}^c$ = $\mathfrak{Q}$(′), ′ ∉ $\mathcal{P}$.

Proof. 

Let $\mathfrak{Q}(\beth )$ be a ${\mathcal{T}}_{Ad}^{•}$-closed subset of $\mathfrak{Q}$(). Then, $\mathfrak{Q}{(\beth )}^{•}\subseteq \mathfrak{Q}(\beth )$. Now, $\mathfrak{Q}(\beth )=\mathfrak{Q}{(\beth )}^{•}\cup (\mathfrak{Q}(\beth )\setminus \mathfrak{Q}{(\beth )}^{•})$. Since ${\mathcal{T}}_{Ad}$ is suitable for $\mathcal{P}$, according to Theorem 16, ${(\mathfrak{Q}(\beth )\setminus \mathfrak{Q}{(\beth )}^{•})}^c$ = $\mathfrak{Q}$(′) for some ′ ∉ $\mathcal{P}$, and according to Theorem 3 (3), $\mathfrak{Q}{(\beth )}^{•}$ is closed.

Conversely, let $\mathfrak{Q}(\beth )=\mathfrak{F}\cup \mathfrak{Q}(\daleth )$, where $\mathfrak{F}$ is closed and $\mathfrak{Q}{(\daleth )}^c$ = $\mathfrak{Q}$(′) for some ′ ∉ $\mathcal{P}$. Then, $\mathfrak{Q}{(\beth )}^{•}={(\mathfrak{F}\cup \mathfrak{Q}(\daleth ))}^{•}={\mathfrak{F}}^{•}$ according to Corollary 5; hence, according to Theorem 3 (3), $\mathfrak{Q}{(\beth )}^{•}={(\mathfrak{F}\cup \mathfrak{Q}(\daleth ))}^{•}={\mathfrak{F}}^{•}=\mathfrak{CL}\left(\mathfrak{F}\right)=\mathfrak{F}\subseteq \mathfrak{Q}(\beth )$. Hence, $\mathfrak{Q}(\beth )$ is ${\mathcal{T}}_{Ad}^{•}$-closed. □

Corollary 8.

Let the topology ${\mathcal{T}}_{Ad}$ on a space $\mathfrak{Q}$() be suitable for a graph primal $\mathcal{P}$ on a graph . Then, $\mathfrak{Q}{(\daleth )}_{\mathcal{P}}$ = {$\mathfrak{U}\cap P:(\mathfrak{U}\in {\mathcal{T}}_{Ad})$(P = $\mathfrak{Q}$(′) for some ′ ∉ $\mathcal{P}$)} is a topology on $\mathfrak{Q}$(); hence, $\mathfrak{Q}{(\daleth )}_{\mathcal{P}}={\mathcal{T}}_{Ad}^{•}$

Proof. 

Let $\mathfrak{U}\in {\mathcal{T}}_{Ad}^{•}$. Then, according to Theorem 18, $\mathfrak{Q}$() ∖ $\mathfrak{U}=\mathfrak{F}\cup \mathfrak{Q}(\daleth )$, where $\mathfrak{F}$ is closed and $\mathfrak{Q}{(\daleth )}^c$ = $\mathfrak{Q}$(′) for some ′ ∉ $\mathcal{P}$. Then, $\mathfrak{U}$ = $\mathfrak{Q}$() ∖ $(\mathfrak{F}\cup \mathfrak{Q}(\daleth \left)\right)$ = ($\mathfrak{Q}$() ∖ $\mathfrak{F}$) ∩ ($\mathfrak{Q}$() ∖ $\mathfrak{Q}(\daleth )$) = $\mathfrak{T}\cap P$, where $\mathfrak{T}={\mathfrak{F}}^c\in {\mathcal{T}}_{Ad}$ and $P=\mathfrak{Q}{(\daleth )}^c$ = $\mathfrak{Q}$(′) for some ′ ∉ $\mathcal{P}$. Thus, every ${\mathcal{T}}_{Ad}^{•}$-open set is of the form $\mathfrak{T}\cap P$, where $\mathfrak{T}\in {\mathcal{T}}_{Ad}$ and P = $\mathfrak{Q}$(′) for some ′ ∉ $\mathcal{P}$. The rest follows from Theorem 9. □

Theorem 19.

Let ($\mathfrak{Q}$(), ${\mathcal{T}}_{Ad},\mathcal{P}$) be a graph primal ATS and ℶ be any subgraph of such that $\mathfrak{Q}(\beth )\subseteq \mathfrak{Q}{(\beth )}^{•}$. Then, $\mathfrak{CL}\left(\mathfrak{Q}(\beth )\right)={\mathfrak{CL}}^{•}\left(\mathfrak{Q}(\beth )\right)=\mathfrak{CL}\left(\mathfrak{Q}{(\beth )}^{•}\right)=\mathfrak{Q}{(\beth )}^{•}$.

Proof. 

Since ${\mathcal{T}}_{Ad}^{•}$ is finer than ${\mathcal{T}}_{Ad}$, then ${\mathfrak{CL}}^{•}\left(\mathfrak{Q}(\beth )\right)\subseteq \mathfrak{CL}\left(\mathfrak{Q}(\beth )\right)$ for any subset $\mathfrak{Q}(\beth )$ of $\mathfrak{Q}$(). Now, $⅁\notin {\mathfrak{CL}}^{•}\left(\mathfrak{Q}(\beth )\right)$, and there exists $\mathfrak{T}\in {\mathcal{T}}_{Ad}$ and $\mathfrak{Q}(\daleth )\in \mathcal{P}$ such that $⅁\in \mathfrak{T}\cap \mathfrak{Q}(\daleth )$ and $(\mathfrak{T}\cap \mathfrak{Q}(\daleth \left)\right)\cap \mathfrak{Q}(\beth )=\varnothing$, and then ${[(\mathfrak{T}\cap \mathfrak{Q}(\daleth ))\cap \mathfrak{Q}(\beth )]}^{•}=\varnothing$. Thus, ${[(\mathfrak{T}\cap \mathfrak{Q}(\beth ))\setminus \mathfrak{Q}{(\daleth )}^c]}^{•}=\varnothing$, and according to Corollary 5, we have ${(\mathfrak{T}\cap \mathfrak{Q}(\beth ))}^{•}=\varnothing$. By Theorem 4, we get $\mathfrak{T}\cap {\left(\mathfrak{Q}(\beth )\right)}^{•}=\varnothing$ and $\mathfrak{T}\cap \mathfrak{Q}(\beth )=\varnothing$ (as $\mathfrak{Q}(\beth )\subseteq \mathfrak{Q}{(\beth )}^{•}$), and then $⅁\notin \mathfrak{CL}\left(\mathfrak{Q}\right(\beth \left)\right)$. Thus, $\mathfrak{CL}\left(\mathfrak{Q}(\beth )\right)={\mathfrak{CL}}^{•}\left(\mathfrak{Q}(\beth )\right)$. Now, according to Theorem 3 (3), $\mathfrak{Q}{(\beth )}^{•}=\mathfrak{CL}\left(\mathfrak{Q}{(\beth )}^{•}\right)$. Now, let $⅁\notin \mathfrak{CL}\left(\mathfrak{Q}\right(\beth \left)\right)$. Then, there exists $\mathfrak{U}\in {\mathcal{T}}_{Ad}(⅁)$ such that $\mathfrak{U}\cap \mathfrak{Q}(\beth )=\varnothing$. Thus, ${(\mathfrak{U}\cap \mathfrak{Q}(\beth ))}^c={\mathfrak{U}}^c\cup \mathfrak{Q}{(\beth )}^c$ = $\mathfrak{Q}$() = $\mathfrak{Q}$(′) for some ′ ∉ $\mathcal{P}$. So, $⅁\notin \mathfrak{Q}{(\beth )}^{•}$ and hence $\mathfrak{Q}{(\beth )}^{•}\subseteq \mathfrak{CL}\left(\mathfrak{Q}(\beth )\right)$. Again, since $\mathfrak{Q}{(\beth )}^{•}\subseteq \mathfrak{CL}\left(\mathfrak{Q}(\beth )\right)$, we have $\mathfrak{CL}\left(\mathfrak{Q}{(\beth )}^{•}\right)\subseteq \mathfrak{CL}\left(\mathfrak{CL}\left(\mathfrak{Q}(\beth )\right)\right)=\mathfrak{CL}\left(\mathfrak{Q}(\beth )\right)$. Also, $\mathfrak{Q}(\beth )\subseteq \mathfrak{Q}{(\beth )}^{•}$, and then $\mathfrak{CL}\left(\mathfrak{Q}(\beth )\right)\subseteq \mathfrak{CL}\left(\mathfrak{Q}{(\beth )}^{•}\right)$. Thus, $\mathfrak{CL}\left(\mathfrak{Q}(\beth )\right)=\mathfrak{CL}\left(\mathfrak{Q}{(\beth )}^{•}\right)=\mathfrak{Q}{(\beth )}^{•}$. □

Theorem 20.

Let ($\mathfrak{Q}$(), ${\mathcal{T}}_{Ad},\mathcal{P}$) be a graph primal ATS such that ${\mathcal{T}}_{Ad}$ is suitable for $\mathcal{P}$ with $\mathfrak{Q}\left(\mathfrak{F}\right)$ = $\mathfrak{Q}$(^★) for some ^★ ∈ $\mathcal{P}$, for all closed subsets $\mathfrak{Q}\left(\mathfrak{F}\right)$, $\mathfrak{F}$ ≠. For $\mathcal{G}$ ⊆ , if $\mathfrak{Q}\left(\mathcal{G}\right)$ is a ${\mathcal{T}}_{Ad}^{•}$-open set such that $\mathfrak{Q}\left(\mathcal{G}\right)=\mathfrak{U}\cap \mathfrak{Q}(\beth )$, where $\mathfrak{U}\in {\mathcal{T}}_{Ad}$ and $\mathfrak{Q}(\beth )$ = $\mathfrak{Q}$(′) for some ′ ∉ $\mathcal{P}$, then $\mathfrak{CL}\left(\mathfrak{Q}\left(\mathcal{G}\right)\right)={\mathfrak{CL}}^{•}\left(\mathfrak{Q}\left(\mathcal{G}\right)\right)=\mathfrak{Q}{\left(\mathcal{G}\right)}^{•}={\mathfrak{U}}^{•}=\mathfrak{CL}\left(\mathfrak{U}\right)={\mathfrak{CL}}^{•}\left(\mathfrak{U}\right)$.

Proof. 

Let $\mathfrak{Q}\left(\mathcal{G}\right)=\mathfrak{U}\cap \mathfrak{Q}(\beth )$, where $\mathfrak{U}\in {\mathcal{T}}_{Ad}$ and $\mathfrak{Q}(\beth )$ = $\mathfrak{Q}$(′) for some ′ ∉ $\mathcal{P}$ (according to Corollary 8, every ${\mathcal{T}}_{Ad}^{•}$-open set $\mathfrak{Q}\left(\mathcal{G}\right)$ is of this form). Since $\mathfrak{Q}\left(\mathfrak{F}\right)$ = $\mathfrak{Q}$(^★) for some ^★ ∈ $\mathcal{P}$ for all closed subsets $\mathfrak{Q}\left(\mathfrak{F}\right)$, $\mathfrak{F}$ ≠, according to Theorem 12, we have $\mathfrak{U}\subseteq {\mathfrak{U}}^{•}$. Hence, according to Theorem 19, we get ${\mathfrak{U}}^{•}=\mathfrak{CL}\left(\mathfrak{U}\right)={\mathfrak{CL}}^{•}\left(\mathfrak{U}\right)$.

Now, let $\mathfrak{Q}\left(\mathcal{G}\right)$ be ${\mathcal{T}}_{Ad}^{•}$-open. We claim that $\mathfrak{Q}\left(\mathcal{G}\right)\subseteq \mathfrak{Q}{\left(\mathcal{G}\right)}^{•}$. In fact, ${\mathfrak{CL}}^{•}$($\mathfrak{Q}$() ∖ $\mathfrak{Q}\left(\mathcal{G}\right)$) = $\mathfrak{Q}$() ∖ $\mathfrak{Q}\left(\mathcal{G}\right)$. Then, ($\mathfrak{Q}$() ∖ $\mathfrak{Q}\left(\mathcal{G}\right)$)^• = $\mathfrak{Q}$() ∖ $\mathfrak{Q}\left(\mathcal{G}\right)$ and $\mathfrak{Q}$()^• ∖ $\mathfrak{Q}{\left(\mathcal{G}\right)}^{•}$ = $\mathfrak{Q}$() ∖ $\mathfrak{Q}\left(\mathcal{G}\right)$ according to Lemma 1, Furthermore, according to Theorem 12, we have $\mathfrak{Q}$() ∖ $\mathfrak{Q}{\left(\mathcal{G}\right)}^{•}$ = $\mathfrak{Q}$() ∖ $\mathfrak{Q}\left(\mathcal{G}\right)$, $\mathfrak{Q}\left(\mathcal{G}\right)\subseteq \mathfrak{Q}{\left(\mathcal{G}\right)}^{•}$. Hence, according to Theorem 19, $\mathfrak{Q}{\left(\mathcal{G}\right)}^{•}=\mathfrak{CL}\left(\mathfrak{Q}\left(\mathcal{G}\right)\right)={\mathfrak{CL}}^{•}\left(\mathfrak{Q}\left(\mathcal{G}\right)\right)$.

Again, $\mathfrak{Q}\left(\mathcal{G}\right)\subseteq \mathfrak{U}$, so $\mathfrak{Q}{\left(\mathcal{G}\right)}^{•}\subseteq {\mathfrak{U}}^{•}$, and $\mathfrak{Q}{\left(\mathcal{G}\right)}^{•}={(\mathfrak{U}\cap \mathfrak{Q}(\beth ))}^{•}={(\mathfrak{U}-\mathfrak{Q}{(\beth )}^c)}^{•}\supseteq {\mathfrak{U}}^{•}-{\left(\mathfrak{Q}{(\beth )}^c\right)}^{•}={\mathfrak{U}}^{•}$ according to Lemma 1 and Theorem 3 (5) as $\mathfrak{Q}(\beth )$ = $\mathfrak{Q}$(′) for some ′ ∉ $\mathcal{P}$. Thus, ${\mathfrak{U}}^{•}=\mathfrak{Q}{\left(\mathcal{G}\right)}^{•}$. Therefore, we have $\mathfrak{CL}\left(\mathfrak{Q}\left(\mathcal{G}\right)\right)={\mathfrak{CL}}^{•}\left(\mathfrak{Q}\left(\mathcal{G}\right)\right)=\mathfrak{Q}{\left(\mathcal{G}\right)}^{•}={\mathfrak{U}}^{•}=\mathfrak{CL}\left(\mathfrak{U}\right)={\mathfrak{CL}}^{•}\left(\mathfrak{U}\right)$. □

Theorem 21.

Let ($\mathfrak{Q}$(), ${\mathcal{T}}_{Ad},\mathcal{P}$) be a graph primal ATS such that ${\mathcal{T}}_{Ad}$ is suitable for $\mathcal{P}$. Then, for every $\mathfrak{Q}(\beth )\in {\mathcal{T}}_{Ad}$ and any $\daleth$ ⊆ , ${(\mathfrak{Q}(\beth )\cap \mathfrak{Q}(\daleth ))}^{•}={(\mathfrak{Q}(\beth )\cap \mathfrak{Q}{(\daleth )}^{•})}^{•}=\mathfrak{CL}(\mathfrak{Q}(\beth )\cap \mathfrak{Q}{(\daleth )}^{•})$.

Proof. 

Let $\mathfrak{Q}(\beth )\in {\mathcal{T}}_{Ad}$. Then, according to Corollary 2, $\mathfrak{Q}(\beth )\cap \mathfrak{Q}{(\daleth )}^{•}=\mathfrak{Q}(\beth )\cap {(\mathfrak{Q}(\beth )\cap \mathfrak{Q}(\daleth ))}^{•}\subseteq {(\mathfrak{Q}(\beth )\cap \mathfrak{Q}(\daleth ))}^{•}$; hence, ${(\mathfrak{Q}(\beth )\cap \mathfrak{Q}{(\daleth )}^{•})}^{•}\subseteq {\left[{(\mathfrak{Q}(\beth )\cap \mathfrak{Q}(\daleth ))}^{•}\right]}^{•}={(\mathfrak{Q}(\beth )\cap \mathfrak{Q}(\daleth ))}^{•}$ according to Corollary 7.

Now, by using Corollary 2 and Theorem 17, we obtain ${[\mathfrak{Q}(\beth )\cap (\mathfrak{Q}(\daleth )\setminus \mathfrak{Q}{(\daleth )}^{•})]}^{•}=\mathfrak{Q}(\beth )\cap {(\mathfrak{Q}(\daleth )\setminus \mathfrak{Q}{(\daleth )}^{•})}^{•}=\mathfrak{Q}(\beth )\cap \varnothing =\varnothing$.

Also, ${(\mathfrak{Q}(\beth )\cap \mathfrak{Q}(\daleth ))}^{•}\setminus {(\mathfrak{Q}(\beth )\cap \mathfrak{Q}{(\daleth )}^{•})}^{•}\subseteq {[(\mathfrak{Q}(\beth )\cap \mathfrak{Q}(\daleth ))\setminus (\mathfrak{Q}(\beth )\cap \mathfrak{Q}{(\daleth )}^{•})]}^{•}={[\mathfrak{Q}(\beth )\cap (\mathfrak{Q}(\daleth )\setminus \mathfrak{Q}{(\daleth )}^{•})]}^{•}=\varnothing$ according to Lemma 1. Hence, ${(\mathfrak{Q}(\beth )\cap \mathfrak{Q}(\daleth ))}^{•}\subseteq {(\mathfrak{Q}(\beth )\cap \mathfrak{Q}{(\daleth )}^{•})}^{•}$, and we get ${(\mathfrak{Q}(\beth )\cap \mathfrak{Q}(\daleth ))}^{•}={(\mathfrak{Q}(\beth )\cap \mathfrak{Q}{(\daleth )}^{•})}^{•}$.

Again, ${(\mathfrak{Q}(\beth )\cap \mathfrak{Q}(\daleth ))}^{•}={(\mathfrak{Q}(\beth )\cap \mathfrak{Q}{(\daleth )}^{•})}^{•}\subseteq \mathfrak{CL}(\mathfrak{Q}(\beth )\cap \mathfrak{Q}{(\daleth )}^{•})$ since ${\mathcal{T}}_{Ad}^{•}$ is finer than ${\mathcal{T}}_{Ad}$. Due to $\mathfrak{Q}(\beth )\cap \mathfrak{Q}{(\daleth )}^{•}\subseteq {(\mathfrak{Q}(\beth )\cap \mathfrak{Q}(\daleth ))}^{•}$, we have $\mathfrak{CL}(\mathfrak{Q}(\beth )\cap \mathfrak{Q}{(\daleth )}^{•})\subseteq \mathfrak{CL}\left({(\mathfrak{Q}(\beth )\cap \mathfrak{Q}(\daleth ))}^{•}\right)={(\mathfrak{Q}(\beth )\cap \mathfrak{Q}(\daleth ))}^{•}$. Hence, ${(\mathfrak{Q}(\beth )\cap \mathfrak{Q}(\daleth ))}^{•}=\mathfrak{CL}(\mathfrak{Q}(\beth )\cap \mathfrak{Q}{(\daleth )}^{•})$. □

Corollary 9.

Let ($\mathfrak{Q}$(), ${\mathcal{T}}_{Ad},\mathcal{P}$) be a graph primal ATS such that ${\mathcal{T}}_{Ad}$ is suitable for $\mathcal{P}$. If $\mathfrak{Q}(\beth )\in {\mathcal{T}}_{Ad}$ and $\mathfrak{Q}{(\beth )}^c$ = $\mathfrak{Q}$(′) for some ′ ∉ $\mathcal{P}$, then $\mathfrak{Q}(\beth )$ ⊆ $\mathfrak{Q}$() ∖ $\mathfrak{Q}$()^•.

Proof. 

Taking $\mathfrak{Q}(\daleth )$ = $\mathfrak{Q}$() in Theorem 21, we get ($\mathfrak{Q}(\beth )$ ∩ $\mathfrak{Q}$())^• = $\mathfrak{CL}$($\mathfrak{Q}(\beth )$ ∩ $\mathfrak{Q}$()^•). Thus, $\mathfrak{Q}{(\beth )}^{•}$ = $\mathfrak{CL}$($\mathfrak{Q}(\beth )$ ∩ $\mathfrak{Q}$()^•), for all $\mathfrak{Q}(\beth )\in {\mathcal{T}}_{Ad}$. Now, if $\mathfrak{Q}{(\beth )}^c$ = $\mathfrak{Q}$(′) for some ′ ∉ $\mathcal{P}$, then $\mathfrak{Q}{(\beth )}^{•}=\varnothing$. Thus, ($\mathfrak{Q}(\beth )$ ∩ $\mathfrak{Q}$()^• = $\mathfrak{CL}$($\mathfrak{Q}(\beth )$ ∩ $\mathfrak{Q}$()^•) = ∅. So, $\mathfrak{Q}(\beth )$ ∩ $\mathfrak{Q}$()^• = ∅ according to Theorem 4; hence, $\mathfrak{Q}(\beth )$ ⊆ $\mathfrak{Q}$() ∖ $\mathfrak{Q}$()^•. □

5. Conclusions

Graph theory and general topology are two dominant areas in discrete mathematics. Graphs can abstractly represent many concepts, making them useful in real-world applications. We have made a new contribution to the field of graph theory by introducing the notion of the “graph primal”, which is the dual of the graph grill. We have studied some basic operations on graph primals. Additionally, in a graph ATS, we present the novel graph-local function ${(.)}^{•}$. We also examine the fundamentals of the proposed graph-local function and explain how to use ${(.)}^{•}$ to create a unique graph ATS ${\mathcal{T}}_{Ad}^{•}$ by defining the new operator ${\mathfrak{CL}}^{•}$ from the older operator via graph primals. The operator $\Theta$ and its corresponding topology $\sigma$ are also proposed. A number of fundamental properties and relationships of the novel topologies $\sigma$ and ${\mathcal{T}}_{Ad}^{•}$ in the graph primal ATSs were investigated with the aid of many counterexamples. We conclude this paper by introducing the notion of topology suitable for a graph primal and obtain its fundamental properties. This is a crucial discovery, as it suggests that these new structures can provide more detailed insights into the nature of graphs and their applications.

This research’s findings are preliminary, and by examining further graph primal ATS features, such as graph primal soft-limit points, separation axioms, compactness, and connectedness, future work may provide additional insights. The outcomes of directed graphs with loops will also be examined. On the other hand, the promising aim of applying the graph primal is to create generalized rough approximation spaces that enhance the accuracy of lower and upper approximations. This methodology is particularly applicable to decision-making problems, where precise approximations are critical for effective outcomes. By utilizing graph theory concepts, the approach seeks to improve the overall decision-making process through refined analytical techniques. It will be interesting work in the future.