Separation axioms via novel operators in the frame of topological spaces and applications

1.   Introduction

Open sets are one of the most important keys to topology. Many researchers have developed different versions of open sets including their weaker and stronger versions. The first development was done by Levine ^[1] in 1963, where he presented the notions of semi-open (closed) sets, and semi-continuity of functions. After that, in 1965, Njåstad ^[2] introduced the concepts of $\alpha$-open sets. The approaches of pre-open sets were presented by Mashhour et al. ^[3] as a tool to study precontinuous and weak precontinuous functions. In 1983, Abd-El-Monsef et al. ^[4] suggested the notions of $\beta$-open sets to investigate $\beta$-continuous functions. The notions of $b$-open or $\gamma$-open sets were studied in detail in ^[5,6] including $\gamma$-continuous functions, and were further developed in ^[7] in terms of the convergence of nets. The somewhat openness of sets was defined by Piotrowski ^[8] to characterize somewhat continuity defined in ^[9]. Recently, Gao and Khalilthe ^[10] introduced more concepts of $\mathfrak{D}\alpha$-Closed sets as a tool for developing many fields of mathematics. Also, the concept of somewhat continuity of function was developed and extended in ^[11,12,13].

In 1968, Veličko ^[14] introduced the concepts of the $\delta$-interior and $\delta$-closure of sets. Several authors have obtained interesting classes of open sets via $\delta$-interior and $\delta$-closure, such as $a$-open ^[15], $Z$-open ^[16], $\delta$-semiopen ^[17], $\delta$-$\theta$-semiopen ^[18], $\delta$-preopen ^[19], $\beta^{\ast}$-open ^[20], $\delta$-semiregular ^[18], and $e^{\ast}$-open ^[21] sets.

This research aims to find a definition called $\aleph$-open, which includes most different classes of open sets, prove its difference from previous similar definitions, and determine its basic properties.

The motive for drafting this paper is as follows: First, we anticipate that by creating a new category of open sets in the frame of topological spaces, we will simplify the path for many future articles on this topic. Second, researchers can investigate alternative concepts, such as covering characteristics and separation axioms via the proposed class of $\aleph$-open and $\aleph$-closed sets. Finally, it reinforces the importance of the concept of classical topology as it is a strong tributary to other modern concepts such as soft and fuzzy topology.

The structure of this work is presented as follows: In Section 2, We recall the fundamental ideas and conclusions that make this work self-contained. In Section 3, we present the main properties of this class and disclose its relationships with many different classes of open sets with the help of appropriate examples and counterexamples. Then, we prove that any $e^{\ast}$-open set is $\aleph$-open, but that the converse may not always be true. The concepts of $\aleph$-interior and $\aleph$-closure operators via $\aleph$-open and $\aleph$-closed sets, respectively, are introduced. In Section 4, we study the concept of $\aleph$-continuity and prove that it does not give continuity and vice versa, and we provide examples of this. After that, we develop several important main theorems about the $\aleph$-continuity of functions via the classes of $\aleph$-open and $\aleph$-closed sets. In Section 5, we study some types of separation axioms via $\aleph$-open and $\aleph$-closed sets and prove their difference from the separation axioms in Euclidean topology, supporting this with several examples. Finally, we present some theorems related to the graph of functions. In Section 6, we end the work with a brief conclusion and potential lines of future research.

2.   Definitions and background

In this paper, spaces imply topological spaces on which no other property is presumed. For a subset $F$ of a space $Z,$ we denote the complement (resp., interior, closure, frontier) of $F$ in $Z$ by $Z\setminus F$ (resp., $F{^{^{\circ}}}$, $\overline{F}$, $fr(F)$). We denote the arbitrary indexing set by $\Theta.$ A subset $F$ of a space $Z$ is called nowhere dense if its closure has empty interior. We also say that $F$ is nowhere dense in $Z$ if $F$ is not dense in any non-empty open subset $V$ of $Z.$ The arbitrary union of nowhere dense sets is a subset of the nowhere dense set of the union. A subset $F$ of a space $Z$ is called $\delta$-open ^[14] if, for each element $z$ in $F,$ there exists a set $K$ such that $z\in\overline{K}^{^{\circ}}\subseteq F.$ Moreover, $Z\setminus F$ is called $\delta$-closed. A point $z\in Z$ is called a $\delta$-accumulation point of $F$ ^[14] if $\overline{K}^{^{\circ}}\cap F\neq \emptyset$ for each open set $K,$ where $z\in K.$ The set of all $\delta$-accumulation points of $F$ is called the $\delta$-closure of $F$ and is denoted by $cl_{\delta}(F)$ or $\overline{F}^{\delta}$. The $\delta$-interior of $F$ is the union of all $\delta$-open sets contained in $F$ and is denoted by $int_{\delta}(F)$ or $F^{^{\circ_{\delta}}}$. A subset $T$ of a space $Z$ is called nowhere dense (resp., semi-open ^[1], somewhere dense ^[2], $\alpha$-open ^[2], pre-open ^[3], $\beta$-open ^[4], $b$-open ^[5] or $sp$-open ^[6] or $\gamma$-open ^[22], $a$-open ^[15], $Z$-open ^[16], $\delta$-semiopen ^[17], $\delta$-preopen ^[19], $\beta^{\ast}$-open ^[20], $e^{\ast}$-open ^[21]), $F$-open ^[23], $C$-open ^[24] sets if $\overline{T}^{^{\circ}} = \emptyset$ (resp., $T\subseteq \overline{T^{^{^{\circ}}}},$ $\overline{T}^{^{\circ}}\neq\emptyset$, $T\subseteq \overline{T^{^{^{\circ}}}}^{^{\circ}},$ $T\subseteq \overline{T}^{^{\circ}},$ $T\subseteq \overline{\overline{T}^{^{^{\circ}}}},$ $T\subseteq(\overline{T}^{^{\circ}}\cup \overline{T^{^{^{\circ}}}}),$ $T\subseteq \overline{T^{^{^{\circ_{\delta}}}}}^{^{\circ}},$ $T\subseteq \overline{T}^{^{\circ}}\cup \overline{T^{^{^{\circ_{\delta}}}}},$ $T\subseteq \overline{T^{^{^{\circ_{\delta}}}}},$ $T\subseteq (\overline{T}^{\delta})^{^{\circ}},$ $T\subseteq \overline{\overline{T}^{^{^{\circ}}}}\cup (\overline{T}^{\delta})^{^{\circ}},$ $T\subseteq \overline{((\overline{T}^{\delta})^{^{^{\circ}}})},$ $T$ is open and $\overline{T}\setminus T$ is finite, $T$ is open and $\overline{T}\setminus T$ is countable).

3.   $\aleph$-open sets and relationships

This section begins by presenting the definitions of $\aleph$-open and $\aleph$-closed sets and the characterizations based on them, which will be applied in what follows.

Definition 1. A subset $F$ of space $Z$ is said to be an $\aleph$-open set when either $F = \emptyset,$ or $F\subseteq \overline{F}^{^{\circ}}\cup fr(F)$ when $F$ is somewhere dense, or $F\subseteq fr(F)$ when $F$ is nowhere dense and $fr(F)$ is infinite.

Remark 2. The complement of an $\aleph$-open subset of space $Z,$ is said to be $\aleph$-closed. Moreover, the collection of all $\aleph$-open (resp. $\aleph$-closed) subsets of space $Z$ is denoted by $\aleph O(Z)$ (resp., $\aleph C(Z)$).

Lemma 3. ^[21] The following are equivalent for a subset $F$ of space $Z$:

(1) $F$ is $e^{\ast}$-open;

(2) There exists a $\delta$-preopen set $V$ such that $V\subseteq\overline{F}^{\delta}\subseteq \overline{V}^{\delta}$;

(3) $\overline{F}^{\delta} = \overline{F^{^{\circ}}}$.

In [20, Remark 2.1], the authors proved that any $e^{\ast}$-open set need not be open (resp., pre-open, semi-open, $\alpha$-open, $\beta$-open, $b$-open, and $\beta^{\ast }$-open) set. In the next proposition, we shall show that, our new notion is generalized to all the above-mentioned concepts.

Proposition 4. Any $e^{\ast}$-open subset $F$ of space $Z$ is $\aleph$-open.

Proof. Suppose that $F$ is an $e^{\ast}$-open subset of $Z.$ If $F = \emptyset,$ then $F$ is $\aleph$-open. Assume that $F\neq \emptyset.$ Now, if $F$ is nowhere dense, then $F$ is not $e^{\ast}$-open, because $F\nsubseteq \overline{((\overline{F}^{\delta})^{^{^{\circ}}})} = \overline{((\overline{F^{\circ}})^{^{^{\circ}}})} = \emptyset.$ Suppose that $F$ is somewhere dense, then $F\subseteq \overline{((\overline{F^{\circ}})^{^{^{\circ}}})}\subseteq \overline{F} = \overline{F}^{^{\circ}}\cup(\overline{F}\setminus F^{\circ}) = \overline{F}^{^{\circ}}\cup fr(F)$ is $\aleph$-open. Therefore, $F$ is $\aleph$-open. □

Here we give an example to explain that the $\aleph$-open set need not be $e^{\ast}$-open.

Example 5. Let $(\mathbb{R}, \mathcal{U})$ be a usual space, and $A = \{1, 2, 3, ...\}\subseteq \mathbb{R}.$ Since $A$ is nowhere dense, $fr(A)$ is infinite, and $A$ is subset of $fr(A),$ then $A$ is $\aleph$-open. However, $A$ is not $e^{\ast}$-open, because $\{1, 2, 3, ...\} \nsubseteq \overline{((\overline{\{1, 2, 3, ...\}}^{\delta})^{^{^{\circ}}})} = \overline{((\{1, 2, 3, ...\})^{^{^{\circ}}})} = \overline{\emptyset} = \emptyset.$

From all the above results the following diagram is obtained:

None of these implications are reversible, as shown from Example 5. Other examples concerning other implications are shown in ^{[1,2,3,4,5,6,11,12,15,16,17,19,20,21,22,23,24]}.

Theorem 6. Suppose that $F_{s}$ is an $\aleph$-open subset of space $Z,$ for each $s \in \Theta.$ Then $\bigcup_{s \in \Theta} F_{s}$ is an $\aleph$-open set.

Proof. Suppose that $F_{s}$ is an $\aleph$-open subset of space $Z,$ for each $s \in \Theta.$ If $F_{s} = \emptyset,$ for each $s \in \Theta,$ then $\bigcup_{s \in \Theta} F_{s} = \emptyset$ is $\aleph$-open. So, suppose that some members are nonempty. However, for the empty set does not affect any union, without loss of generality, we may suppose that $F_{s}\neq \emptyset,$ for each $s \in \Theta.$ If there exists $s_{\gamma} \in \Theta$ such that $F_{s_{\gamma}} = Z,$ then $\bigcup_{s \in \Theta} F_{s} = Z$ is $\aleph$-open. So, assume that $F_{s}\neq Z$ for all $s \in \Theta.$ Now, if $\bigcup_{s \in \Theta}F_{s}$ is nowhere dense and $fr(\bigcup_{s \in \Theta}F_{s})$ is infinite, then by the nowhere denseness of $\bigcup_{s \in \Theta}F_{s},$ we have $fr(\bigcup_{s \in \Theta}F_{s}) = \overline{\bigcup_{s \in \Theta}F_{s}}\setminus (\bigcup_{s \in \Theta}F_{s})^{^{\circ}} = \overline{\bigcup_{s \in \Theta}F_{s}},$ which implies $\bigcup_{s \in \Theta} F_{s} \subseteq fr(\bigcup_{s \in \Theta}F_{s}).$ So, assume that $\bigcup_{s \in \Theta} F_{s}$ is somewhere dense. Since $(\bigcup_{s \in \Theta} F_{s})^{^{\circ}}\subseteq \overline{\bigcup_{s \in \Theta}F_{s}}^{^{\circ}}\subseteq\overline{\bigcup_{s \in \Theta}F_{s}},$ then $\overline{\bigcup_{s \in \Theta}F_{s}}^{^{\circ}}\cup fr(\bigcup_{s \in \Theta} F_{s}) = \overline{\bigcup_{s \in \Theta}F_{s}}^{^{\circ}}\cup(\overline{\bigcup_{s \in \Theta}F_{s}}\setminus (\bigcup_{s \in \Theta} F_{s})^{^{\circ}}) = \overline{ \bigcup_{s \in \Theta} F_{s}},$ hence $\bigcup_{s \in \Theta} F_{s} \subseteq \overline{\bigcup_{s \in \Theta}F_{s}}^{^{\circ}}\cup fr(\bigcup_{s \in \Theta} F_{s}).$ Therefore, $\bigcup_{s \in \Theta} F_{s}$ is an $\aleph$-open set.

□

Theorem 7. Suppose that $F_{s}$ is an $\aleph$-closed subset of space $Z,$ for each $s \in \Theta.$ Then $\bigcap_{s \in \Theta} F_{s}$ is an $\aleph$-closed set.

Proof. It follows from Theorem 6.

□

In general, the finite intersection of $\aleph$-open sets is not $\aleph$-open. For example, $(0, 5]$ and $[5, 7)$ are $\aleph$-open sets in the usual space $\mathbb{R}$, and $(0, 5]\cap [5, 7) = \{5\}$ is not $\aleph$-open, because $\{5\}$ is nowhere dense and has a finite frontier.

Definition 8. Let $F$ be subset of space $Z.$ Then:

$(1)$ $\aleph$-interior $F$ denoted by $F^{^{\circ_{\aleph}}}$ is $F^{^{\circ_{\aleph}}} = \cup\{K\in \aleph O(Z): K\subseteq F\}$;

$(2)$ $\aleph$-closure $F$ denoted by $\overline{F}^{\aleph}$ is the intersection of all $\aleph$-closed sets of $Z$ containing $F.$

The following three propositions are easy to prove and therefore the proofs are omitted, as we need them in the rest of the paper.

Proposition 9. Let $F_{1}$ and $F_{2}$ be subsets of space $Z.$ Then we have the following properties:

(1) $\overline{Z}^{\aleph} = Z$ and $\overline{\emptyset}^{\aleph} = \emptyset$;

(2) $F_{1}\subseteq \overline{F_{1}}^{\aleph}$;

(3) If $F_{1}\subseteq F_{2}$, then $\overline{F_{1}}^{\aleph}\subseteq\overline{F_{2}}^{\aleph}$;

(4) $a\in\overline{F_{1}}^{\aleph}$ if and only if for each an $\aleph$-open set $K$ containing $a,$ $K\cap F_{1}\neq \emptyset$;

(5) $F_{1}$ is $\aleph$-closed set if and only if $F_{1} = \overline{F_{1}}^{\aleph}$;

(6) $\overline{\overline{F_{1}}^{\aleph}}^{\aleph} = \overline{F_{1}}^{\aleph}$;

(7) ($\overline{F_{1}}^{\aleph} \cup\overline{F_{2}}^{\aleph})\subseteq\overline{F_{1} \cup F_{2}}^{\aleph};$

(8) $\overline{F_{1} \cap F_{2}}^{\aleph}\subseteq (\overline{F_{1}}^{\aleph} \cap\overline{F_{2}}^{\aleph}).$

The inclusion relation in parts 7 and 8, cannot be replaced by the equality, for example:

Example 10. Let $Z = \{\alpha, \beta, \gamma, \delta \}$ and $\mathcal{T} = \{\emptyset, \{\alpha\}, \{\beta\}, \{\alpha, \beta\}, Z \}.$ Let $F_{1} = \{\alpha, \gamma \},$ $F_{2} = \{\beta, \gamma \},$ and $F_{3} = \{\alpha, \beta \}.$ Then $\overline{F_{1}}^{\aleph} = F_{1},$ $\overline{F_{2}}^{\aleph} = F_{2},$ and $\overline{F_{1}\cup F_{2}}^{\aleph} = Z.$ Hence, ($\overline{F_{1}}^{\aleph} \cup\overline{F_{2}}^{\aleph})\nsupseteq\overline{F_{1} \cup F_{2}}^{\aleph}.$ Also, $\overline{F_{1}}^{\aleph} = F_{1},$ $\overline{F_{3}}^{\aleph} = Z,$ and $\overline{F_{1}\cap F_{2}}^{\aleph} = \{\alpha\}$. Hence, $\overline{F_{1} \cap F_{2}}^{\aleph}\nsupseteq (\overline{F_{1}}^{\aleph} \cap\overline{F_{2}}^{\aleph}).$

Proposition 11. Let $F_{1},$ $F_{2}$ be subsets of space $Z.$ Then we have the following properties:

$(1)$ $Z^{^{\circ_{\aleph}}} = Z$ and $\emptyset^{^{\circ_{\aleph}}} = \emptyset$;

(2) $F_{1}^{^{\circ_{\aleph}}}\subseteq F_{1}$;

(3) If $F_{1}\subseteq F_{2}$, then $F_{1}^{^{\circ_{\aleph}}}\subseteq F_{2}^{^{\circ_{\aleph}}}$;

(4) $a\in F_{1}^{^{\circ_{\aleph}}}$ if and only if there exists $\aleph$-open $K$ such that $a \in K\subseteq F_{1}$;

(5) $F_{1}$ is an $\aleph$-open set if and only if $F_{1} = F_{1}^{^{\circ_{\aleph}}}$;

(6) $(F_{1}^{^{\circ_{\aleph}}})^{^{\circ_{\aleph}}} = F_{1}^{^{\circ_{\aleph}}}$;

(7) $(F_{1} \cap F_{2})^{^{\circ_{\aleph}}}\subseteq(F_{1}^{^{\circ_{\aleph}}} \cap F_{2}^{^{\circ_{\aleph}}})$;

(8) $F_{1}^{^{\circ_{\aleph}}} \cup F_{2}^{^{\circ_{\aleph}}}\subseteq (F_{1} \cup F_{2})^{^{\circ_{\aleph}}}.$

Proposition 12. For a subset $F$ of space $Z,$ the following statements are true:

(1) $\overline{(Z\setminus F)}^{\aleph} = Z\setminus F^{^{\circ_{\aleph}}};$

(2) $(Z\setminus F)^{^{\circ_{\aleph}}} = Z\setminus \overline{F}^{\aleph}.$

Theorem 13. Let $F$ be subset of space $Z.$ Then,

$F^{^{\circ_{\aleph}}} = \left\{ \begin{array}{ccc} \emptyset & if & \overline{F}^{^{\circ}} = \emptyset\ {{and\ fr(F)\ is\ finite}}\\ F\cap fr(F) & if & \overline{F}^{^{\circ}} = \emptyset\ {{and\ fr(F)\ is\ infinite}} \\ F & if & F\ {{is\ somewhere\ dense}} \end{array} \right.$.

Proof. There are three cases:

Case 1: If $\overline{F}^{^{\circ}} = \emptyset$ and $fr(F)$ is finite. Suppose that there exists a non-empty element $x\in F^{\circ \aleph}.$ Then there exists an $\aleph$-open set $U$ such that $x \in U \subseteq F.$ From the definition of $\aleph$-openness, $x\in U\subseteq (\overline{U}^{^{\circ}}\cup fr(U))\subseteq (\overline{F}^{^{\circ}}\cup fr(F)).$ Since $\overline{F}^{^{\circ}} = \emptyset,$ then $\overline{U}^{^{\circ}} = \emptyset,$ and $x\in U\subseteq fr(U)\subseteq fr(F).$ However, if $fr(U)$ is infinite, then $fr(F)$ is infinite, and thus we have a contradiction. Therefore, $F^{\circ \aleph} = \emptyset.$

Case 2: If $\overline{F}^{^{\circ}} = \emptyset$ and $fr(F)$ is infinite. Let $x\in F^{^{\circ_{\aleph}}}.$ Then there exists an $\aleph$-open set $U$ such that $x \in U \subseteq F.$ From the definition of an $\aleph$-open set, we have $x\in U\subseteq \overline{U}^{^{\circ}}\cup fr(U).$ Since $\overline{F}^{^{\circ}} = \emptyset,$ then $\overline{U}^{^{\circ}} = \emptyset,$ hence $x\in U\subseteq fr(U)\subseteq fr(F).$ Then $x \in F \cap fr(U).$ On the other hand, let $x \in F \cap fr(F).$ Since $F$ is nowhere dense and $fr(F)$ is infinite, then $F \subseteq fr(F),$ and by the definition of $\aleph$-openness, $F$ is $\aleph$-open. From Proposition 11, part 5, we have $x\in \overline{F}^{^{\circ}}.$ Therefore, $F^{^{\circ_{\aleph}}} = F \cap fr(F).$

Case 3: If $F$ is somewhere dense, then

hence $F$ is $\aleph$-open. By Proposition 11 part 5, we have $F^{^{\circ_{\aleph}}} = F.$

□

Example 14. Let $F = \{1, 2, 3\},$ $G = \{\frac{1}{n}:n\in \mathbb{N}\},$ and $H = (0, 5)$ be subsets of the usual space $(\mathbb{R}, \mathcal{U})$. Then:

(1) Since $F$ is nowhere dense, then any non-empty subset $K$ of $F$ is nowhere dense, hence $fr(F)$ and $fr(K)$ are finite. According to the definition of $\aleph$-openness, $\emptyset$ is the only $\aleph$-open subset of $F.$ Hence, $F^{^{\circ_{\aleph}}} = \emptyset;$

(2) $G^{^{\circ_{\aleph}}} = \{\frac{1}{n}:n\in \mathbb{N}\}\cap fr(\{\frac{1}{n}:n\in \mathbb{N}\})$ = $\{\frac{1}{n}:n\in \mathbb{N}\} \cap (\{\frac{1}{n}:n\in \mathbb{N}\}\cup \{0\}) = G$, because $G$ is nowhere dense and $fr(G)$ is infinite, hence $G$ is $\aleph$-open;

(3) $H^{^{\circ_{\aleph}}} = H,$ because $H$ is somewhere dense, then $H\subseteq \overline{H}^{^{\circ}} \cup fr(H),$ hence $H$ is $\aleph$-open.

Theorem 15. Let $F$ be a subset of space $Z.$ Then,

$\overline{F}^{\aleph} = \left\{ \begin{array}{ccc} Z & if & \overline{ Z\setminus F}^{^{\circ}} = \emptyset\ {{and\ fr(Z\setminus F)\ is\ finite}}\\ F & if & \overline{ Z\setminus F}^{^{\circ}} = \emptyset\ {{and\ fr(Z\setminus F)\ is\ infinite}} \\ F & if & Z\setminus F\ {{is\ somewhere\ dense}} \end{array} \right.;$

Proof.

There are three cases:

Case 1: If $\overline{ Z\setminus F}^{^{\circ}} = \emptyset$ and $fr(Z\setminus F)$ is finite. Since $F\subseteq Z,$ then $\overline{F}^{\aleph}\subseteq \overline{Z}^{\aleph} = Z.$ On the other hand, suppose that there exists $x\in Z$ such that $x \notin \overline{F}^{\aleph}.$ Then there exists an $\aleph$-open set $U$ such that $x \in U$ and $U\cap F = \emptyset.$ Since $U$ is $\aleph$-open, then $x\in U\subseteq (\overline{ U}^{^{\circ}}\cup fr(U))$. However, $U\cap F = \emptyset,$ and then $U \subseteq Z\setminus F.$ Since $\overline{ Z\setminus F}^{^{\circ}} = \emptyset,$ then $\overline{ U}^{^{\circ}} = \emptyset,$ hence, from the definition of $\aleph$-openness, $fr(U)$ is infinite. Since $\overline{ Z\setminus F}^{^{\circ}} = \emptyset,$ $\overline{ U}^{^{\circ}} = \emptyset,$ and $U \subseteq Z\setminus F,$ then $fr(U)\subseteq fr(Z\setminus F),$ hence $fr(Z\setminus F)$ is infinite, and thus there is a contradiction. Then we have $\overline{F}^{\aleph}\subseteq Z.$ Therefore, $\overline{F}^{\aleph} = Z.$

Case 2: Let $\overline{ Z\setminus F}^{^{\circ}} = \emptyset$ and $fr(Z\setminus F)$ is infinite, then $(Z\setminus F)\subseteq (\overline{ Z\setminus F}^{^{\circ}}\cup fr(Z\setminus F))\subseteq fr(Z\setminus F),$ hence $Z\setminus F$ is $\aleph$-open. Thus, $F$ is $\aleph$-closed. Therefore, $F = \overline{F}^{\aleph}.$

Case 3: Let $Z\setminus F$ be somewhere dense. Then $(Z\setminus F)\subseteq (\overline{ Z\setminus F}^{^{\circ}}\cup fr(Z\setminus F)),$ hence $Z\setminus F$ is $\aleph$-open. Thus, $F$ is $\aleph$-closed. Therefore, $F = \overline{F}^{\aleph}.$

□

Proposition 16. Any nonempty finite closed nowhere dense set is not $\aleph$-open.

Proof. Let $F$ be a nonempty finite closed set of a space $Z$ and $\overline{F}^{^{\circ}} = \emptyset.$ Since $F$ is closed and $\overline{F}^{^{\circ}} = \emptyset,$ then $fr(F) = F$ is finite. Hence, by the definition of $\aleph$-open sets, we obtain that $F$ is not $\aleph$-open. □

Proposition 17. Any infinite nowhere dense set is $\aleph$-open.

Proof. Obviously, we use the same method that was used in Proposition 16. □

Proposition 18. For space $Z,$ the singleton set $\{z\}$ is either $\aleph$-open or $\overline{\{z\}}^{^{\circ}} = \emptyset.$

Proof. Let $\overline{\{z\}}^{^{\circ}}\neq\emptyset.$ Then $\{z\} \subseteq \overline{\{z\}}^{^{\circ}},$ hence $\{z\} \subseteq(\overline{\{z\}}^{^{\circ}}\cup fr\{z\}),$ which is $\aleph$-open. If $\{z\}$ is not $\aleph$-open, then $\{z\}\nsubseteq (\overline{\{z\}}^{^{\circ}}\cup fr\{z\}),$ that is $\{z\}\nsubseteq \overline{\{z\}}^{^{\circ}}.$ Therefore, $\overline{\{z\}}^{^{\circ}} = \emptyset.$ □

4.   Continuous functions via $\aleph$-open sets

In this section, we study the notion of $\aleph$-continuity of functions, using $\aleph$-open and $\aleph$-closed sets.

Definition 19. A function $h:Z_{1} \to Z_{2}$ is said to be $\aleph$-continuous if $h^{-1}(F)\in \aleph O(Z_{1}),$ for each $F\in \aleph O(Z_{2}).$

In general, $\aleph$-continuity neither implies continuity nor is implied by continuity, as shown in the following examples:

Example 20. Let $(\{1, 2\}, \mathcal{D})$ be a discrete space and $(\mathbb{R}, \mathcal{U})$ be a Euclidean space. Then the function $h:(\mathbb{R}, \mathcal{U})\to (\{1, 2\}, \mathcal{D})$ defined by:

is $\aleph$-continuous, but not continuous, as the inverse image of an open set $\{1\}$ is not open.

Example 21. Let $(\mathbb{R}, \mathcal{I})$ be an indiscrete space and $(\mathbb{R}, \mathcal{U})$ be a usual space. Then the identity function $id:(\mathbb{R}, \mathcal{U})\to (\mathbb{R}, \mathcal{I})$ is continuous, but not $\aleph$-continuous, because $\{1, 2\}$ is an $\aleph$-open set in $(\mathbb{R}, \mathcal{I})$ and $id^{-1}(\{1, 2\})$ is not $\aleph$-open in $(\mathbb{R}, \mathcal{U}).$

Definition 22. Let $(Z, \mathcal{F})$ be a topological space. Then:

(1) A collection $\aleph$-$\mathcal{B} \subseteq \mathcal{F}$ is called an $\aleph$-base for $(Z, \mathcal{F})$ if every non-empty $\aleph$-open subset of $Z$ can be represented as a union of a subfamily of $\aleph$-$\mathcal{B};$

(2) A collection $\aleph$-$\mathcal{P} \subseteq \mathcal{F}$ is called an $\aleph$-subbase for a topological space $(Z, \mathcal{F})$ if the family $\{F_{1}\cap F_{2}\cap...\cap F_{m}:m\in \mathbb{N}, F_{j}\in \aleph\text{-}\mathcal{P}, \forall j\in \{1, 2, ..., m \} \}$ is an $\aleph$-base for $(Z, \mathcal{F})$.

The following theory is easy to prove and therefore it is presented it without proof, as we will need it in the rest of the research.

Theorem 23. Let $Z_{1}$ and $Z_{2}$ be two spaces and $h: Z_{1} \to Z_{2}$ be a function. Then the following statements are equivalent:

(1) $h$ is $\aleph$-continuous;

(2) For each $z_{1}\in Z_{1}$ and $F_{1}\in \aleph O(Z_{2}),$ where $h(z_{1})\in F_{1},$ there exists $F_{2}\in \aleph O(Z_{1})$ where $z_{1}\in F_{2}$ and $h(F_{2})\subseteq F_{1}$;

(3) The inverse images of all $\aleph$-closed subsets of $Z_{2}$ are $\aleph$-closed in $Z_{1};$

(4) The inverse images of all members of an $\aleph$-subbase $\aleph$-$\mathcal{P}$ for $Z_{2}$ are $\aleph$-open in $Z_{1};$

(5) The inverse images of all members of an $\aleph$-base $\aleph$-$\mathcal{B}$ for $Z_{2}$ are $\aleph$-open in $Z_{1};$

(6) For every $F \subseteq Z_{1}$ we have $h(\overline{F}^{\aleph})\subseteq\overline{h(F)}^{\aleph}$;

(7) For every $K\subseteq Z_{2}$ we have $\overline{h^{-1}(K)}^{\aleph} \subseteq h^{-1}(\overline{F}^{\aleph})$;

(8) For every $K\subseteq Z_{2}$ we have $h^{-1}(K^{^{\circ_{\aleph}}})\subseteq (h^{-1}(K))^{^{\circ_{\aleph}}}$.

In the following results, if $Z = \prod_{s\in \Theta}Z_{s}$ is a product space and $F_{s} \subseteq Z_{s}$ for each $s \in \Theta,$ we denote $F_{s_{1}}\times F_{s_{2}}\times... \times F_{s_{n}} \times \prod_{s \notin B}Z_{s},$ by $< F_{s_{1}}, F_{s_{2}}, ..., F_{s_{n}} > ,$ where $B = \{s_{1}, s_{2}, ..., s_{n}\}.$

Lemma 24. Let $Z = \prod_{s = 1}^{n}Z_{s}$ be a finite product space and $F_{s}$ be $\aleph$-open in $Z_{s}$ for each $s\in \{1, 2, ..., n\}.$ Then, $\langle F_{1}, F_{2}, ..., F_{n} \rangle$ is $\aleph$-open in $Z.$

Proof. Let $F_{s}\in \aleph O(Z_{s})$ for each $s\in \{1, 2, ..., n\}.$ If there exists $F_{s_{\gamma}} = \emptyset$ for some $s_{\gamma}\in \{1, 2, ..., n\}$, then $\langle F_{1}, F_{2}, ..., F_{n} \rangle = \emptyset \in \aleph O(Z).$ Suppose that $F_{s}\neq\emptyset$ for each $s\in \{1, 2, ..., n\}.$ If there exists $F_{s_{\gamma}}$ is nowhere dense in $Z_{s_{\gamma}}$ for some $s_{\gamma}\in \{1, 2, ..., n\},$ then $fr(F_{s_{\gamma}})$ is infinite and also $\langle F_{1}, F_{2}, ..., F_{n} \rangle$ is nowhere dense. Hence,

where $fr\langle F_{1}, F_{2}, ..., F_{n} \rangle$ is infinite, because

and, for some $s_{\gamma}\in \{1, 2, ..., n\},$ $fr(F_{s_{\gamma}})$ is infinite. Hence, $\langle F_{1}, F_{2}, ..., F_{n} \rangle\in \aleph O(Z).$ Assume that $F_{s}$ is somewhere dense in $Z_{s}$ for each $s\in \{1, 2, ..., n\}.$ Since $\langle F_{1}, F_{2}, ..., F_{n} \rangle$ is somewhere dense in $Z.$ then, by Remark 2,

Therefore, $\langle F_{1}, F_{2}, ..., F_{n} \rangle\in \aleph O(Z).$ □

If at least one coordinate is nowhere dense or equal to the empty set, then the converse of Lemma 24 is not true. For example:

Example 25. Consider $(\mathbb{R}, \mathcal{U})$, where $\mathcal{U}$ is the usual topology on $\mathbb{R}$ ^[25]. Then $\emptyset \times \{1, 2\} = \emptyset$ is an $\aleph$-open set in $\mathbb{R}^{2}.$ By Theorem 16, $\{1, 2\}$ is not $\aleph$-open in $\mathbb{R}.$ Also, $\mathbb{Z} \times \{1, 2\} \times (0, 1)$ is an $\aleph$-open set in $\mathbb{R}^{3}$ and $\{1, 2\}$ is finite nowhere dense, which is not $\aleph$-open in $\mathbb{R}.$

Corollary 26. Let $Z = \prod_{s = 1}^{n}Z_{s}$ be a finite product space and $F_{s}\subseteq Z_{s}$ for each $s\in \{1, 2, ..., n\}.$ Then

(1) $\overline{\langle F_{1}, F_{2}, ..., F_{n} \rangle}^{\aleph}\subseteq\langle \overline{F_{1}}^{\aleph}, \overline{F_{2}}^{\aleph}, ..., \overline{F_{n}}^{\aleph} \rangle$;

(2) $\langle F_{1}^{^{\circ_{\aleph}}}, F_{2}^{^{\circ_{\aleph}}}, ..., F_{n}^{^{\circ_{\aleph}}} \rangle\subseteq \langle F_{1}, F_{2}, ..., F_{n} \rangle^{^{\circ_{\aleph}}}.$

Since the closure and interior operators are preserved in a finite product, see [26, p. 99–100], then we have the following theorem.

Theorem 27. Let $Z = \prod_{s = 1}^{n}Z_{s}$ be a finite product space and $F_{s}\subseteq Z_{s}$ is not finite nowhere dense in $Z_{s}$ for each $s\in \{1, 2, ..., n\}.$ Then $F_{s}\in \aleph O(Z_{s})$ for each $s\in \{1, 2, ..., n\}$ if and only if $\langle F_{1}, F_{2}, ..., F_{n} \rangle \in \aleph O(Z).$

Proof. Necessity. It follows from Lemma 24.

Sufficiency. Suppose that $F_{s}\subseteq Z_{s}$ is not finite nowhere dense in $Z_{s}$ for each $s\in \{1, 2, ..., n\}$ and $\langle F_{1}, F_{2}, ..., F_{n} \rangle\in \aleph O(Z).$ If $F_{s}$ is somewhere dense in $Z_{s}$ for each $s\in \{1, 2, ..., n\}.$ Then

hence $F_{s}\subseteq \overline{F_{s}} = \overline{F_{s}}^{^{\circ}}\cup fr(F_{s})$ for each $s\in \{1, 2, ..., n\}.$ Then, $F_{s}\in \aleph O(Z_{s})$ for each $s\in \{1, 2, ..., n\}.$ Assume that there exists $F_{s_{\gamma}}$ is infinite nowhere dense in $Z_{s_{\gamma}}$ for some $s_{\gamma}\in \{1, 2, ..., n\}.$ Then

hence $F_{s}\subseteq \overline{F_{s}} = \overline{F_{s}}^{^{\circ}}\cup fr(F_{s})$ for each $s\in \{1, 2, ..., n\}.$ Therefore, $F_{s}\in \aleph O(Z_{s})$ for each $s\in \{1, 2, ..., n\}.$

□

Corollary 28. Let $Z = \prod_{s = 1}^{n}Z_{s}$ be a finite product space and $F_{s}\subseteq Z_{s}$ is not finite nowhere dense in $Z_{s}$ for each $s\in \{1, 2, ..., n\}.$ Then,

(1) $\overline{\langle F_{1}, F_{2}, ..., F_{n} \rangle}^{\aleph} = \langle \overline{F_{1}}^{\aleph}, \overline{F_{2}}^{\aleph}, ..., \overline{F_{n}}^{\aleph} \rangle$;

(2) $\langle F_{1}^{^{\circ_{\aleph}}}, F_{2}^{^{\circ_{\aleph}}}, ..., F_{n}^{^{\circ_{\aleph}}} \rangle = \langle F_{1}, F_{2}, ..., F_{n} \rangle^{^{\circ_{\aleph}}}.$

Lemma 29. The projection maps $\pi_{s}:\prod_{\alpha\in \Theta} Z_{\alpha} \to Z_{s}$ are $\aleph$-continuous.

Proof. For $s\in \Theta,$ let $V_{s}\in \aleph O(Z_{s}).$ Then, $\pi^{-1}_{s}(V_{s}) = V_{s}\times \prod_{s\neq \alpha \in \Theta} Z_{\alpha}.$ By Lemma 24, $\pi^{-1}_{s}(V_{s})\in \aleph O(\prod_{\alpha\in \Theta} Z_{\alpha}).$

□

Theorem 30. For space $Z$ and product space $Y = \prod_{s \in \Theta}Y_{s}$, a function $h: Z \to Y$ is $\aleph$-continuous if and only if $\pi_{s} \circ h$ is $\aleph$-continuous for each $s \in \Theta.$

Proof. Let $h: Z \to Y$ be $\aleph$-continuous. Let $U_{s}$ be arbitrary $\aleph$-open in $Y_{s},$ for each $s \in \Theta.$ By Lemma 29, $\pi^{-1}(U_{s})\in \aleph O(Y),$ for each $s \in \Theta.$ Hence, $(\pi_{s} \circ h)^{-1}(U_{s}) = h^{-1}(\pi^{-1}(U_{s}))$ is $\aleph$-open, for each $s \in \Theta.$ Therefore, $\pi_{s} \circ h$ is $\aleph$-continuous for each $s \in \Theta.$ Conversely, assume that $\pi_{s} \circ h$ is $\aleph$-continuous for each $s\in \Theta.$ Let $V_{s}$ be $\aleph$-open in $Y_{s}$ for each $s \in \Theta.$ Then $\langle V_{s}\rangle$ is an $\aleph$-subbasic $\aleph$-open set in $Y.$ By Theorem 23 and $(\pi_{s} \circ h)^{-1}(V_{s}) = h^{-1}(\pi^{-1}(V_{s})) = h^{-1}(\langle V_{s}\rangle),$ we get that $h^{-1}(\langle V_{s}\rangle)$ is $\aleph$-open in $Z$. Therefore, $h$ is $\aleph$-continuous.

□

Corollary 31. For a space $Z$ and the product space $Y = \prod_{s \in \Theta}Y_{s}$, $h_{s}: Z \to \prod_{s \in \Theta}Y_{s}$ is a function for each $s \in \Theta.$ Let $h: Z \to Y$ be the function defined by $h(z) = < h_{s}(z) > .$ Then $h$ is $\aleph$-continuous if and only if $h_{s}$ is $\aleph$-continuous for each $s \in \Theta.$

Theorem 32. Let $Z = \prod_{s = 1}^{n}Z_{s}$ and $Y = \prod_{s = 1}^{n}Y_{s}$ be finite product spaces. Let $h_{s}: Z_{s} \to Y_{s}$ be an $\aleph$-continuous function for each $s\in \{1, 2, ..., n\}.$ Then the function $h: Z \to Y$ defined by $h(\langle z_{s} \rangle) = \langle h_{s}(z_{s})\rangle$ is $\aleph$-continuous.

Proof. Let $\langle F_{1}, F_{2}, ..., F_{n} \rangle$ be an $\aleph$-basic $\aleph$-open set in $Y.$ Then

Since $h_{s}$ is $\aleph$-continuous for each $s\in \{1, 2, ..., n\},$ then $h_{s}^{-1}(F_{s})$ is $\aleph$-continuous for each $s\in \{1, 2, ..., n\}.$ By Lemma 24, we have, $\langle h^{-1}(F_{1}), h^{-1}(F_{2}), ..., h^{-1}(F_{n}) \rangle$ is $\aleph$-open in $Z.$ Hence, from Theorem 23, $h$ is $\aleph$-continuous. □

5.   Applications on $\aleph$-open sets in separation axioms

In this section, we apply our new notions in separation axioms to explore new types of separation axioms and some theorems related to graphs of functions.

Definition 33. A space $Z$ is said to be an

$(1)$ $\aleph$-$T_{0}$-space if for each distinct pair $z_{1}, z_{2}\in Z,$ then there is either an $\aleph$-open set containing $z_{1}$ but not $z_{2}$ or an $\aleph$-open set containing $z_{2}$ but not $z_{1}.$

$(2)$ $\aleph$-$T_{1}$-space if for each distinct pair $z_{1}, z_{2}\in Z,$ then there are two $\aleph$-open subsets $F_{1}$ and $F_{2}$ of $Z,$ such that $z_{1}\in F_{1}$, $z_{2}\notin F_{1}$, and $z_{1}\notin F_{2}$, $z_{2} \in F_{2}.$

$(3)$ $\aleph$-$T_{2}$-space "$\aleph$-Hausdorff space" if for each distinct pair $z_{1}, z_{2}\in Z,$ then there are two disjoint $\aleph$-open subsets $F_{1}$ and $F_{2}$ of $Z,$ such that $z_{1}\in F_{1}$ and $z_{2} \in F_{2}.$

Theorem 34. Any topological space is $\aleph$-$T_{0}$-space.

Proof. Let $Z$ be a topological space and let $x\neq y$ for $x$ and $y$ in $Z.$ Then $Z\setminus \{x\}$ is either a nowhere dense or somewhere dense set. If $Z\setminus \{x\}$ is nowhere dense, then $\overline{Z\setminus \{x\}}\neq Z,$ hence $Z\setminus \{x\}$ is closed and so $\{x\}$ is an open set. Therefore, $\{x\}\subseteq \overline{\{x\}}^{^{\circ}}\cup fr(\{x\})$ is an $\aleph$-open set containing $x,$ but not $y.$ On the other hand, if $Z\setminus \{x\}$ is somewhere dense, then $Z\setminus \{x\}\subseteq \overline{Z\setminus \{x\}}^{^{\circ}}\cup fr(Z\setminus \{x\})$ is an $\aleph$-open set containing $y,$ but not $x.$ □

Lemma 35. Any infinite subset of a topological space $Z$ is $\aleph$-open.

Proof. Let $K$ be any infinite subset of topological space $Z.$ It follows that $K$ is either a nowhere dense or somewhere dense set. If $K$ is nowhere dense, then $K\subseteq \overline{K}^{^{\circ}}\cup fr(K) = fr(K)$ is $\aleph$-open, because $fr(K)$ is infinite. Suppose that $K$ is somewhere dense. Then $K\subseteq \overline{K }^{^{\circ}}\cup fr(K)$ is $\aleph$-open. □

Theorem 36. Any $T_{1}$ topological space is an $\aleph$-$T_{2}$-space.

Proof. Let $Z$ be a $T_{1}$ topological space and let $x\neq y$ in $Z.$ If $Z$ is finite, then $Z$ is discrete, hence it is an $\aleph$-$T_{2}$-space because any open set is $\aleph$-open. Suppose that $Z$ is infinite. Then there exist two infinite disjoint subsets $K$ and $H$ of $Z$ containing $x$ and $y,$ respectively. By Lemma 35, $K$ and $H$ are $\aleph$-open sets. Therefore, $Z$ is an $\aleph$-$T_{2}$-space. □

Theorem 37. Any infinite topological space is an $\aleph$-$T_{2}$-space.

Proof. This follows from Lemma 35 and the fact that any infinite set contains two disjoint infinite subsets.

□

Theorem 38. Any $\aleph$-$T_{1}$ finite topological space is $\aleph$-$T_{2}$-space.

Proof. Suppose that $Z$ is an $\aleph$-$T_{1}$ finite topological space. If $Z$ contains two or fewer elements, then $Z$ is $\aleph$-$T_{2}$-space. Assume that $Z$ contains more than two elements. Let $x\neq y$ in $Z$ and, by $\aleph$-$T_{1}$-spaceness, there exist two $\aleph$-open sets $U$ and $V$ containing $x$ and $y,$ respectively. If $U\cap V = \emptyset,$ we are done. Assume that $U\cap V\neq \emptyset.$ Since $Z$ is finite, then $U$ and $V$ are somewhere dense.

Claim: $Z\setminus U$ is $\aleph$-open. If $Z\setminus U$ is somewhere dense, we are done. Assume that $Z\setminus U$ is nowhere dense. Since $(V\cap (Z\setminus U))\subseteq(Z\setminus U)$ is nowhere dense, and $V$ is somewhere dense, then $U\cap V$ is somewhere dense. If $U\setminus V$ is somewhere dense, we are done. Assume that $U\setminus V$ is nowhere dense, hence $Z\setminus(U\cap V)$ is nowhere dense, which implies that $U\cap V$ is the only subset of $Z$ such that $(U\cap V)^{\circ}\neq\emptyset.$ If $(U\cap V)^{\circ}$ is an open set, then $Z$ is not $\aleph$-$T_{1},$ hence the interior of $U\cap V$ consists of more than one open set. Since $Z$ consists of more than two elements, then there exists an element $g$ such that $g\neq x\neq y.$ Let $H_{i}$ be any $\aleph$-open set containing $g,$ where $i\in \{1, 2, 3, ..., n\}.$ Since $Z\setminus(U\cap V)$ is nowhere dense, then $H_{i}\cap (U\cap V)\neq \emptyset$ for all $i\in \{1, 2, 3, ..., n\}.$ If $(\bigcap_{i = 1}^{n}H_{i}) \cap (U \cap V)\neq \emptyset,$ then there exists an element $e \in((\bigcap_{i = 1}^{n}H_{i}) \cap (U \cap V))$ such that any $\aleph$-open set containing $x$ or $y$ must contain $e.$ Thus, contradicts the fact that $Z$ is an $\aleph$-$T_{1}$-space. Assume that $(\bigcap_{i = 1}^{n}H_{i})\cap (U \cap V) = \emptyset.$ Then there exist $H_{i_{1}}$ and $H_{i_{2}}$ for some $i_{1}\neq i_{2}\in \{1, 2, 3, ..., n\}$ such that $H_{i_{1}}\cap H_{i_{2}}\cap (U\cap V) = \emptyset,$ hence $H_{i_{1}}\cap (U\cap V)$ and $H_{i_{2}}\cap (U\cap V)$ are $\aleph$-open sets, and $(H_{i_{1}}\cap (U\cap V))\cap (H_{i_{2}}\cap (U\cap V)) = \emptyset.$ Hence, $(U\setminus V)\cup(H_{i_{1}}\cap (U\cap V))$ and $(V\setminus U)\cup(H_{i_{2}}\cap (U\cap V))$ are two disjoint $\aleph$-open sets containing $x$ and $y,$ respectively. Therefore, $Z$ is an $\aleph$-$T_{2}$-space.

□

Remark 39. The following implications hold for a space $Z:$

The converse of these implications are not true in general. Consider the following examples:

Example 40. Let $Z = \{\alpha, \beta \}$ and $\mathcal{T} = \{\emptyset, \{\alpha\}, Z \}.$ Then, $(Z, \mathcal{T})$ is $T_{0}$-space, which implies that $(Z, \mathcal{T})$ is an $\aleph$-$T_{0}$-space, but not an $\aleph$-$T_{1}$-space.

Example 41. Consider $(\mathbb{R}, \mathcal{I})$, where $\mathcal{I}$ is an indiscrete topology on $\mathbb{R}$ ^[25]. Then $(\mathbb{R}, \mathcal{I})$ is an $\aleph$-$T_{2}$-space, but not a $T_{0}$-space.

Let $h: Z_{1}\to Z_{2}$ be a function. Recall that the subset $\{(z_{1}, h(z_{1})): z_{1} \in Z_{1}\}$ of the product space $Z_{1}\times Z_{2}$ is called the graph of $h$ and is denoted by $\Im(h).$

Theorem 42. If $h: Z_{1}\to Z_{2}$ is an $\aleph$-continuous function and $Z_{2}$ is an $\aleph$-$T_{2}$-space, then $\Im(h)$ is $\aleph$-closed in $Z_{1}\times Z_{2}.$

Proof. Let $(z_{1}, z_{2})\in Z_{1}\times Z_{2}$ where $z_{2}\neq h(z_{1}).$ Since $Z_{2}$ is an $\aleph$-$T_{2}$-space, then there exist $F_{1}, F_{2}\in \aleph O(Z_{2})$ where $z_{2}\in F_{1}$, $h(z_{1})\in F_{2}$ and $F_{1}\cap F_{2} = \emptyset.$ Since $h$ is $\aleph$-continuous and, by Theorem 23 there exists an $\aleph$-open set $h^{-1}(F_{2}) = K_{1}$ containing $z_{1},$ by Lemma 24 we have that $K_{1}\times F_{1}$ is an $\aleph$-open set in $Z_{1}\times Z_{2}$ such that $(z_{1}, z_{2})\in K_{1}\times F_{1}.$ Suppose that there exists $(w, h(w))\in K_{1}\times F_{1},$ then $h(w)\in F_{1}.$ However, $w \in K_{1} = h^{-1}(F_{2}),$ which means that $h(w) \in F_{2}.$ Thus, it is contradicts the fact that $F_{1}$ and $F_{2}$ are disjoint. Hence, $Z_{1}\times Z_{2} \setminus \Im(h)$ is $\aleph$-open. Therefore, $\Im(h)$ is $\aleph$-closed in $Z_{1}\times Z_{2}.$ □

Theorem 43. Let $h: Z_{1}\to Z_{2}$ be an $\aleph$-continuous function and $Z_{2}$ be an $\aleph$-$T_{2}$-space. Then the set $M = \{(z_{1}, z_{2})\in Z_{1}\times Z_{1}: h(z_{1}) = h(z_{2})\}$ is $\aleph$-closed in $Z_{1}\times Z_{1}.$

Proof. Let $(z_{1}, z_{2})\notin M,$ then $h(z_{1})\neq h(z_{2})$ in $Z_{1}\times Z_{1}\setminus M.$ Since $Z_{2}$ is an $\aleph$-$T_{2}$-space, there exist $F_{1}, F_{2}\in \aleph O(Z_{2})$ where $h(z_{1})\in F_{1}$, $h(z_{2})\in F_{2},$ and $F_{1}\cap F_{2} = \emptyset.$ However, $h$ is $\aleph$-continuous, and then, from Theorem 23, there exist $\aleph$-open sets $h^{-1}(F_{1}) = K_{1}$ and $h^{-1}(F_{2}) = K_{2}$ in $Z_{1}$ containing $z_{1}$ and $z_{2},$ respectively, such that $h(K_{1})\subseteq F_{1}$ and $h(K_{2})\subseteq F_{2}.$ By Lemma 24 we have that $K_{1}\times K_{2}$ is an $\aleph$-open set in $Z_{1}\times Z_{1}$ such that $(z_{1}, z_{2})\in K_{1}\times K_{2}.$ Claim: $K_{1}\times K_{2}\cap M = \emptyset.$ Suppose that $K_{1}\times K_{2}\cap M\neq \emptyset.$ Then there exists $(a, b)\in K_{1}\times K_{2}\cap M.$ Hence, $a\in K_{1} = h^{-1}(F_{1})$ and $b\in K_{2} = h^{-1}(F_{2}),$ which means that $h(a) \in F_{1}$ and $h(b) \in F_{2}.$ But, $(a, b)\in M$, and so $h(a) = h(b),$ which contradicts the fact that $F_{1}$ and $F_{2}$ are disjoint. Hence, $(z_{1}, z_{2})\notin \overline{M}^{\aleph}.$ Therefore, $M$ is $\aleph$-closed in $Z_{1}\times Z_{1}.$ □

Theorem 44. Let $h_{1}.h_{2}:Z_{1}\to Z_{2}$ be an $\aleph$-continuous functions and $Z_{2}$ be $\aleph$-$T_{2}$-space. Then the set $M = \{(z_{1}, z_{2}): h_{1}(z_{1}) = h_{2}(z_{2})\}$ is $\aleph$-closed in $Z_{1}\times Z_{2}.$

Proof. Let $(z_{1}, z_{2})\notin M.$ Then $h_{1}(z_{1})\neq h_{2}(z_{2}).$ Since $Z_{2}$ is an $\aleph$-$T_{2}$-space, there exist $F_{1}, F_{2}\in \aleph O(Z_{2})$ where $h_{1}(z_{1})\in F_{1}$, $h_{2}(z_{2})\in F_{2},$ and $F_{1}\cap F_{2} = \emptyset.$ Since $h_{1}$ is $\aleph$-continuous, and by Theorem 23, there exists an $\aleph$-open set $K_{1}$ containing $z_{1}$ and $h_{1}(K_{1})\subseteq F_{1},$ then $h_{1}(K_{1})\cap F_{2} = \emptyset.$ Since $h_{2}$ is $\aleph$-continuous, and by Theorem 23 there exists an $\aleph$-open set $K_{2}$ containing $z_{2}$ and $h_{2}(K_{2})\subseteq F_{2},$ then $h_{1}(K_{1})\cap h_{2}(K_{2}) = \emptyset.$ Claim: $(K_{1}\times K_{2})\cap M = \emptyset.$ Suppose that $(K_{1}\times K_{2})\cap F\neq\emptyset.$ Then there exists $a = (a_{1}, a_{2})\in ((K_{1}\times K_{2})\cap M),$ where $h_{1}(a_{1}) = h_{2}(a_{2})$, $a_{1}\in K_{1},$ and $a_{2}\in K_{2}.$ Hence, $h_{1}(a_{1})\in h_{1}(K_{1})\subseteq F_{1}$ and $h_{2}(a_{2})\in h_{2}(K_{2})\subseteq F_{2}.$ Thus, it contradicts the fact that $F_{1}$ and $F_{2}$ are disjoint. Hence from Lemma 24, we have that $K_{1}\times K_{2}$ is an $\aleph$-open set in $Z_{1}\times Z_{2}.$ This implies that, $(z_{1}, z_{2})\notin \overline{M}^{\aleph}.$ Therefore, $M$ is $\aleph$-closed in $Z_{1}\times Z_{2}.$ □

Theorem 45. Let $h_{1}.h_{2}:Z_{1}\to Z_{2}$ be $\aleph$-continuous functions and $Z_{2}$ be an $\aleph$-$T_{2}$-space. Then $\{z: h_{1}(z) = h_{2}(z)\}$ is an $\aleph$-closed subset of $Z_{1}.$

Proof. It is similar to the proof of Theorem 44. □

Theorem 46. Let $(Z_{1}, \mathcal{\rho}_{1})$ and $(Z_{2}, \mathcal{\rho}_{2})$ be two topological spaces and $h:(Z_{1}, \mathcal{\rho}_{1})\to (Z_{2}, \mathcal{\rho}_{2})$ be an $\aleph$-continuous function. Then $\pi_{z_{1}}(M\cap \Im(h))$ is $\aleph$-closed in $Z_{1},$ where $\pi_{z_{1}}$ represents the projection map of $Z_{1}\times Z_{2}$ onto $Z_{1}$ and $M$ is $\aleph$-closed in $Z_{1}\times Z_{2}.$

Proof. Let $z_{1} \in \overline{\pi_{z_{1}}(M\cap \Im(h))}^{\aleph}$ and let $F_{1}$ and $F_{2}$ be two $\aleph$-open sets in $Z_{1}$ and $Z_{2}$, respectively, such that $z_{1}\in F_{1}$ and $h(z_{1})\in F_{2}.$ Since $h$ is $\aleph$-continuous, and by Theorem 23 there exists an $\aleph$-open $h^{-1}(F_{2})$ containing $z_{1},$ and since $F_{2}$ is $\aleph$-open in $Z_{2}$ and $h$ is $\aleph$-continuous, then, by Theorem 23, we have $z_{1}\in h^{-1}(F_{2})\subseteq (h^{-1}(F_{2}))^{^{\circ_{\aleph}}}.$ Then $F_{1}\cap \pi_{z_{1}}(M\cap \Im(h)) \cap (h^{-1}(F_{2}))^{^{\circ_{\aleph}}}$ contains some point $g$ in $Z_{1}$ which implies that $(g, h(g))\in M$ and $h(g)\in F_{2}.$ Thus, we have $(F_{1} \times F_{2})\cap M\neq \emptyset,$ and hence $(z_{1}, h(z_{1}))\in \overline{M}^{\aleph}.$ Since $M$ is $\aleph$-closed, then $(z_{1}, h(z_{1}))\in M\cap \Im(h)$ and $z_{1}\in \pi_{z_{1}}(M\cap \Im(h)).$ Therefore, $\pi_{z_{1}}(M\cap \Im(h))$ is $\aleph$-closed in $Z_{1}.$ □

6.   Concluding remarks and further work

This study contributes to the literature on novel topological qualities by introducing a new class of open sets. The given results explain that, it is possible to gain new examples and qualities that contribute to a broader understanding of topological spaces by generalizing certain topological properties, such as compression, connectivity, and others.

In this paper, we introduced a novel class of open sets named "$\aleph$-open" sets. We have studied the basic properties of these sets, provide the master properties of $\aleph$-open sets, and disclosed their relationships with many different classes of open sets with the support of appropriate counterexamples. Additionally, we provided the $\aleph$-interior and $\aleph$-closure via $\aleph$-open and $\aleph$-closed sets. In detail, we studied the concept of $\aleph$-continuity of product spaces using $\aleph$-open and $\aleph$-closed sets. Finally, we provided some kinds of separation axioms, some theorems related to graph of functions, and $\aleph$-compact spaces via $\aleph$-open and $\aleph$-closed sets.

This work considers auspicious directions for future work First, we expect that by defining a new class of open sets in topological spaces, we will facilitate a path for future research on the subject. Also, one of the main focuses of future work is to study the relationship between $\aleph$-continuous functions and the other types of continuities such as $e^{\ast }$-continuous functions, $\beta^{\ast }$-continuous function, $\alpha$-continuous functions etc. Another direction is to introduce a new framework to be used in the near future to produce soft topological concepts ^[27] and supra soft topological spaces ^[28] such as (supra) soft operators and continuity ^{[29,30,31,32]}, which are inspired by classical topologies. Certainly, researchers can explore other notions like covering properties and separation axioms via the proposed class of $\aleph$-open and $\aleph$-closed sets. Finally, it reinforces the importance of the concept of classical topology as it is a strong tributary to other modern concepts such as (supra) soft and fuzzy topology ^[33] and soft nodec spaces ^[34].

Use of AI tools declaration

The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

Acknowledgments

The authors are appreciative to the anonymous referees for their valuable recommendations, which helped to improve the paper's presentation.

The authors extend their appreciation to the Deanship of Scientific Research at Northern Border University, Arar, KSA for funding this research work through the project number "NBU-FPEJ-2024-2727-04".

Conflict of interest

The authors declare no conflict of interest.