On some types of functions and a form of compactness via $\omega _{s}$-open sets

1.   Introduction

Let $B$ be a subset of a topological space $\left(Y, \sigma \right)$. The set of condensation points of $B$ is denoted by $Cond\left(B\right)$ and defined by $Cond\left(B\right) = \left\{ y\in Y\text{: for every }V\in \sigma \text{ with }y\in V\text{, }V\cap B\text{ is uncountable}\right\}$. For the purpose of characterizing Lindelöf topological spaces and improving some known mapping theorems, the author in ^[1] defined $\omega$-closed sets as follows: $B$ is called an $\omega$-closed set in $\left(Y, \sigma \right)$ if $Cond\left(B\right) \subseteq B$. $B$ is called an $\omega$-open set in $\left(Y, \sigma \right)$ ^[1] if $Y-B$ is $\omega$-closed. It is well known that $B$ is $\omega$-open in $\left(Y, \sigma \right)$ if and only if for every $b\in B$, there are $V\in \sigma$ and a countable subset $F\subseteq Y$ with $b\in V-F\subseteq B$. The family of all $\omega$-open sets in $\left(Y, \sigma \right)$ is denoted by $\sigma _{\omega }$. It is well known that $\sigma _{\omega }$ is a topology on $Y$ that is finer than $\sigma$. Many research papers related to $\omega$-open sets have appeared in ^{[2,3,4,5,6,7,8]} and others. Authors in ^[9,10,11] included $\omega$-openness in both soft and fuzzy topological spaces. In this paper, we will denote the closure of $B$ in $\left(Y, \sigma \right)$, the closure of $B$ in $\left(Y, \sigma _{\omega }\right)$, the interior of $B$ in $\left(Y, \sigma \right)$, and the interior of $B$ in $\left(Y, \sigma _{\omega }\right)$ by $\overline{B}$, $\overline{B}^{\omega }$, $Int\left(B\right)$, and $Int_{\omega }\left(B\right)$, respectively. $B$ is called a semi-open set in $\left(Y, \sigma \right)$ ^[12] if there exists $V\in \sigma$ such that $V\subseteq B\subseteq \overline{V}$. Complements of semi-open sets are called semi-closed sets. The family of all semi-open sets in $\left(Y, \sigma \right)$ will be denoted by $SO\left(Y, \sigma \right)$. Authors in ^{[12,13,14,15]} have used semi-open sets to define semi-continuity, semi-openness, irresoluteness, pre-semi-openness, and slight semi-continuity. The area of research related to semi-open sets is still hot ^{[16,17,18,19,20,21,22,23,24,25,26,27,28]}. Authors in ^[29] have defined $\omega _{s}$-open sets as a strong form of semi-open sets as follows: $B$ is called an $\omega _{s}$-open set in $\left(Y, \sigma \right)$ if there exists $O\in \sigma$ such that $O\subseteq B\subseteq \overline{O}^{\omega }$. Complements of $\omega _{s}$-open sets are called $\omega _{s}$-closed sets. The family of all $\omega _{s}$-open sets in $\left(Y, \sigma \right)$ will be denoted by $\omega _{s}\left(Y, \sigma \right)$. The intersection of all $\omega _{s}$-closed sets in $\left(Y, \sigma \right)$ which contains $B$ will be denoted by $\overline{B} ^{\omega _{s}}$, and the union of all $\omega _{s}$-open sets in $\left(Y, \sigma \right)$ which contained in $B$ will be denoted by $Int_{\omega _{s}}\left(B\right)$. Authors in ^[29] have defined and investigated the class of $\omega _{s}$-continuity which lies strictly between the classes of continuity and semi-continuity.

In this paper, $\omega _{s}$-irresoluteness as a strong form of $\omega _{s}$-continuity is introduced. It is proved that $\omega _{s}$-irresoluteness is independent of each of continuity and $\omega _{s}$-irresoluteness. Also, $\omega _{s}$-openness which lies strictly between openness and semi-openness is introduced and investigated, and pre-$\omega _{s}$-openness which is a strong form of $\omega _{s}$ -openness and independent of openness is introduced and investigated. Moreover, slight $\omega _{s}$-continuity as a new class of functions which lies between slight continuity and slightly semi-continuity is introduced and investigated. In addition to these, $\omega _{s}$-compactness as a new class of topological spaces that lies strictly between compactness and semi-compactness is introduced. Several implications, examples, counter-examples, characterizations, and mapping theorems are introduced. In particular, several sufficient conditions for the equivalence between our new concepts and other related concepts are given. We hope that this will open the door for a number of future related studies such as $\omega _{s}$ -separation axioms and $\omega _{s}$-connectedness.

Throughout this paper, the usual topology on $\mathbb{R}$ will be denoted by $\tau _{u}$.

Recall that a topological space $(Y, \sigma)$ is called locally countable ^[30] (resp. anti-locally countable ^[31]) if $(Y, \sigma)$ has a base consisting of countable sets (all non-empty open sets are uncountable sets).

The following results will be used in the sequel:

Proposition 1.1. ^[29] Let $(Y, \sigma)$ be a topological space. Then

(a) $\sigma \subseteq \omega _{s}(Y, \sigma)\subseteq SO(Y, \sigma)$.

(b) If $(Y, \sigma)$ is locally countable, then $\sigma = \omega _{s}(Y, \sigma)$.

(c) If $(Y, \sigma)$ is anti-locally countable, then $\omega _{s}(Y, \sigma) = SO(Y, \sigma)$.

Proposition 1.2. ^[29] Let $(Y, \sigma)$ be a topological space and let $B, C\subseteq Y$. Then we have the following:

(a) If $B\subseteq C\subseteq \; \overline{B}^{\omega }$ and $B \; \in \; \omega _{s}(Y, \sigma)$, then $C \; \in \omega _{s}(Y, \sigma)$.

(b) If $B\in \sigma$ and $C \; \in \omega _{s}(Y, \sigma)$, then $B \; \cap \; C\in \omega _{s}(Y, \sigma)$.

Proposition 1.3. ^[31] Let $(Y, \sigma)$ be anti-locally countable. Then for every $B \; \in \sigma _{\omega }$, we have $\overline{B}^{\omega } = \overline{B}$.

2.   ${ \omega }_{{ s}}$-Irresolute functions

Definition 2.1. A function $g:(Y, \sigma)\longrightarrow \left(Z, \gamma \right)$ is called irresolute ^[14] (resp. $\omega _{s}$ -continuous ^[29]), if for every $V\in SO\left(Z, \gamma \right)$ (resp. $V\in \gamma$), $g^{-1}(V)\in SO(Y, \sigma)$ (resp. $g^{-1}(V)\in \omega _{s}(Y, \sigma)$).

Definition 2.2. A function $g:(Y, \sigma)\longrightarrow \left(Z, \gamma \right)$ is called $\omega _{s}$-irresolute, if for every $V\in \omega _{s}\left(Z, \gamma \right)$, $g^{-1}(V)\in \omega _{s}(Y, \sigma)$.

The following two examples will show that irresoluteness and $\omega _{s}$ -irresoluteness are independent concepts:

Example 2.3. Let $X = \mathbb{R}$, $Y = \left\{ a, b\right\}$, $\tau = \left\{ \emptyset, \mathbb{R}, \mathbb{N}, \mathbb{Q} ^{c}, \mathbb{N} \cup \mathbb{Q} ^{c}\right\}$, and $\sigma = \left\{ \emptyset, Y, \left\{ a\right\}, \left\{ b\right\} \right\}$. Define $f:(X, \tau)\longrightarrow (Y, \sigma)$ by

Since $f^{-1}\left(\left\{ a\right\} \right) = \mathbb{Q} ^{c}\in \tau \subseteq SO(X, \tau)$ and $f^{-1}\left(\left\{ b\right\} \right) = \mathbb{Q} \in SO(X, \tau)-\omega _{s}(X, \tau)$, then $f$ is irresolute but not $\omega _{s}$-irresolute.

Example 2.4. Consider the topology $\sigma = \left\{ \emptyset, \mathbb{N}, \left\{ 1\right\}, \left\{ 2\right\}, \left\{ 1, 2\right\} \right\}$ on $\mathbb{N}$. Define $g:(\mathbb{N}, \sigma)\longrightarrow (\mathbb{N}, \sigma)$ by

Since $(\mathbb{N}, \sigma)$ is locally countable, then by Proposition 1.1 (b) we have $\omega _{s}(\mathbb{N}, \sigma) = \sigma$. Since $f^{-1}\left(\left\{ 1\right\} \right) = \left\{ 1, 2\right\} \in \sigma$ and $f^{-1}\left(\left\{ 2\right\} \right) = \emptyset \in \sigma$, then $f$ is $\omega _{s}$-irresolute. However, $f$ is not irresolute because there is $\overline{\left\{ 2\right\} } = \mathbb{N} -\left\{ 1\right\} \in SO(\mathbb{N}, \sigma)$ such that $f^{-1}\left(\mathbb{N} -\left\{ 1\right\} \right) = \mathbb{N} -\left\{ 1, 2\right\} \notin SO(\mathbb{N}, \sigma)$.

Theorem 2.5. Let $\left(Y, \sigma \right)$ and $(Z, \gamma)$ be two anti-locally countable topological spaces. Then for any function $g:\left(Y, \sigma \right) \longrightarrow (Z, \gamma)$ the followings are equivalent:

(a) $g$ is irresolute.

(b) $g$ is $\omega _{s}$-irresolute.

Proof. Follows from the definitions and Proposition 1.1 (c).

The following two examples will show that continuity and $\omega _{s}$ -irresoluteness are independent concepts:

Example 2.6. Let $X = \mathbb{R} \ $and $\tau = \left\{ \emptyset, \mathbb{R}, \left(3, \infty \right), \left\{ 2\right\}, \left\{ 2\right\} \cup \left(3, \infty \right) \right\}$. Define $f:(X, \tau)\longrightarrow (X, \tau)$ by

Since $f^{-1}\left(\left(3, \infty \right) \right) = \emptyset \in \tau$ and $f^{-1}\left(\left\{ 2\right\} \right) = \left\{ 2\right\} \cup \left(3, \infty \right) \in \tau$, then $f$ is continuous. Since $\overline{\left(3, \infty \right) }^{\omega } = \mathbb{R} -\left\{ 2\right\}$, then $\mathbb{R} -\left\{ 2\right\} \in \omega _{s}(X, \tau)$. Since $f^{-1}\left(\mathbb{R} -\left\{ 2\right\} \right) = \mathbb{R} -\left(\left\{ 2\right\} \cup \left(3, \infty \right) \right) \notin \omega _{s}(X, \tau)$, then $f$ is not $\omega _{s}$-irresolute.

Example 2.7. Let $\tau _{disc}$ be the discrete topology on $\left\{ a, b\right\}$. Define $f:(\mathbb{R}, \tau _{u})\longrightarrow (\left\{ a, b\right\}, \tau _{disc})$ by

Clearly that $\omega _{s}(\left\{ a, b\right\}, \tau _{disc}) = \tau _{disc}$. Since $f^{-1}\left(\left\{ a\right\} \right) = \left(-\infty, 0\right) \in \tau _{u}\subseteq \omega _{s}(\mathbb{R}, \tau _{u})$ and $f^{-1}\left(\left\{ b\right\} \right) = \left[0, \infty \right) \in \omega _{s}(\mathbb{R}, \tau _{u})$, then $f$ is $\omega _{s}$-irresolute. However, $f$ is not continuous because there is $\left\{ b\right\} \in \tau _{disc}$ such that $f^{-1}\left(\left\{ b\right\} \right) = \left[0, \infty \right) \notin \tau _{u}$.

Theorem 2.8. Let $\left(Y, \sigma \right)$ and $(Z, \gamma)$ be two locally countable topological spaces. Then for any function $g:\left(Y, \sigma \right) \longrightarrow (Z, \gamma)$ the followings are equivalent:

(a) $g$ is continuous.

(b) $g$ is $\omega _{s}$-irresolute.

Proof. Follows from the definitions and Proposition 1.1 (b).

Theorem 2.9. If $g:\left(Y, \sigma \right) \longrightarrow (Z, \gamma)$ is a continuous function such that $g:\left(Y, \sigma _{\omega }\right) \longrightarrow (Z, \gamma _{\omega })$ is an open function, then $g$ is $\omega _{s}$-irresolute.

Proof. Let $H\in \omega _{s}(Z, \gamma)$. Then there exists $W\in \gamma$ such that $W\subseteq H\subseteq \overline{W}^{\omega }$ and so $g^{-1}\left(W\right) \subseteq g^{-1}\left(H\right) \subseteq g^{-1}\left(\overline{W}^{\omega }\right)$. Since $g:\left(Y, \sigma \right) \longrightarrow (Z, \gamma)$ is continuous, then $g^{-1}(W)\in \sigma$. Since $g:\left(Y, \sigma _{\omega }\right) \longrightarrow (Z, \gamma _{\omega })$ is open, then $g^{-1}\left(\overline{W}^{\omega }\right) \subseteq \overline{g^{-1}\left(W\right) }^{\omega }$. Thus, $g^{-1}\left(W\right) \subseteq g^{-1}\left(H\right) \subseteq \overline{g^{-1}\left(W\right) }^{\omega }$, and hence $g^{-1}\left(H\right) \in \omega _{s}\left(Y, \sigma \right)$. Therefore, $g$ is $\omega _{s}$-irresolute.

Theorem 2.10. Every $\omega _{s}$-irresolute function is $\omega _{s}$-continuous.

Proof. Let $f:(X, \tau)\longrightarrow (Y, \sigma)$ be $\omega _{s}$ -irresolute. Let $V\in \sigma$. Then by Proposition 1.1 (a), $V\in \omega _{s}(Y, \sigma)$. Since $f$ is $\omega _{s}$-irresolute, then $f^{-1}\left(V\right) \in \omega _{s}(X, \tau)$. Therefore, $f$ is $\omega _{s}$ -irresolute.

The function in Example 2.6 is continuous and hence $\omega _{s}$ -continuous. Therefore, the converse of Theorem 2.10 is not true, in general.

Theorem 2.11. A function $g:\left(Y, \sigma \right) \longrightarrow (Z, \gamma)$ is $\omega _{s}$-irresolute if and only if for every $\omega _{s}$-closed subset $B$ of $(Z, \gamma)$, $g^{-1}\left(B\right)$ is $\omega _{s}$-closed in $\left(Y, \sigma \right)$.

Proof. Necessity. Assume that $g$ is $\omega _{s}$-irresolute. Let $B$ be an $\omega _{s}$-closed set in $(Z, \gamma)$. Then $Z-B\in \omega _{s}(Z, \gamma)$. Since $g$ is $\omega _{s}$-irresolute, then $g^{-1}\left(Z-B\right) = Y-g^{-1}\left(B\right) \in \omega _{s}\left(Y, \sigma \right)$. Hence, $g^{-1}\left(B\right)$ is $\omega _{s}$-closed in $\left(Y, \sigma \right)$.

Sufficiency. Suppose that for every $\omega _{s}$-closed subset $B$ of $(Z, \gamma)$, $g^{-1}\left(B\right)$ is $\omega _{s}$ -closed in $\left(Y, \sigma \right)$. Let $W\in \omega _{s}(Z, \gamma)$. Then $Z-W$ is $\omega _{s}$-closed in $(Z, \gamma)$. By assumption, $g^{-1}\left(Z-W\right) = Y-g^{-1}\left(W\right)$ is $\omega _{s}$-closed in $\left(Y, \sigma \right)$, and so $g^{-1}\left(W\right) \in \omega _{s}\left(Y, \sigma \right)$. Therefore, $g$ is $\omega _{s}$-irresolute.

Theorem 2.12. A function $g:\left(Y, \sigma \right) \longrightarrow (Z, \gamma)$ is $\omega _{s}$-irresolute if and only if for every $H\subseteq Y$, $g\left(\overline{H}^{\omega _{s}}\right) \subseteq \overline{g\left(H\right) }^{\omega _{s}}$.

Proof. Necessity. Suppose that $g$ is $\omega _{s}$-irresolute and let $H\subseteq Y$. Then $\overline{g\left(H\right) } ^{\omega _{s}}$ is $\omega _{s}$-closed in $(Z, \gamma)$, and by Theorem 2.11, $g^{-1}\left(\overline{g\left(H\right) }^{\omega _{s}}\right)$ is $\omega _{s}$-closed in $\left(Y, \sigma \right)$. Since $H\subseteq g^{-1}\left(\overline{g\left(H\right) }^{\omega _{s}}\right)$, then $\overline{H}^{\omega _{s}}\subseteq g^{-1}\left(\overline{g\left(A\right) } ^{\omega _{s}}\right)$. Thus,

Sufficiency. Suppose that for every subset $H\subseteq Y$, $g\left(\overline{H}^{\omega _{s}}\right) \subseteq \overline{g\left(H\right) } ^{\omega _{s}}$. We will apply Theorem 2.11 to show that $g$ is $\omega _{s}$ -irresolute. Let $B$ be an $\omega _{s}$-closed subset of $(Z, \gamma)$. Then by assumption we have $g\left(\overline{g^{-1}(B)}^{\omega _{s}}\right) \subseteq \overline{g\left(g^{-1}(B\right))}^{\omega _{s}}\subseteq \overline{B}^{\omega _{s}} = B$, and so

Therefore, $\overline{g^{-1}(B)}^{\omega _{s}} = g^{-1}\left(B\right)$, and hence $g^{-1}\left(B\right)$ is $\omega _{s}$-closed in $\left(Y, \sigma \right)$. This shows that $g$ is $\omega _{s}$-irresolute.

Theorem 2.13. A function $g:\left(Y, \sigma \right) \longrightarrow (Z, \gamma)$ is $\omega _{s}$-irresolute if and only if for every $H\subseteq Z$, $\overline{g^{-1}\left(H\right) }^{\omega _{s}}\subseteq g^{-1}\left(\overline{H}^{\omega _{s}}\right)$.

Proof. Necessity. Suppose that $g$ is $\omega _{s}$-irresolute and let $H\subseteq Z$. Then by Theorem 2.11, $g^{-1}\left(\overline{H}^{\omega _{s}}\right)$ is $\omega _{s}$-closed in $\left(Y, \sigma \right)$. Since $g^{-1}\left(H\right) \subseteq g^{-1}\left(\overline{H}^{\omega _{s}}\right)$, then $\overline{g^{-1}\left(H\right) } ^{\omega _{s}}\subseteq g^{-1}\left(\overline{H}^{\omega _{s}}\right)$.

Sufficiency. Suppose that for every $H\subseteq Z$, $\overline{ g^{-1}\left(H\right) }^{\omega _{s}}\subseteq g^{-1}\left(\overline{H} ^{\omega _{s}}\right)$. We will apply Theorem 2.11 to show that $g$ is $\omega _{s}$-irresolute. Let $B$ be an $\omega _{s}$-closed subset of $(Z, \gamma)$. Then $\overline{B}^{\omega _{s}} = B$. So by assumption, $\overline{g^{-1}(B)}^{\omega _{s}}\subseteq g^{-1}\left(B\right)$, and so $\overline{g^{-1}(B)}^{\omega _{s}} = g^{-1}\left(B\right)$. Therefore, $g^{-1}\left(B\right)$ is $\omega _{s}$-closed in $\left(Y, \sigma \right)$.

Theorem 2.14. The composition of two $\omega _{s}$ -irresolute functions is $\omega _{s}$-irresolute.

Proof. Let $g:(Y, \sigma)\longrightarrow (Z, \lambda)$ and $h:(Z, \lambda)\longrightarrow (M, \gamma)$ be $\omega _{s}$-irresolute functions. Let $C\in \omega _{s}(M, \gamma)$. Since $h$ is $\omega _{s}$ -irresolute, then $h^{-1}(C)\in \omega _{s}(Z, \lambda)$. Since $g$ is $\omega _{s}$-irresolute, then $(h\circ g)^{-1}(C) = g^{-1}\left(h^{-1}(C)\right) \in \omega _{s}(Y, \sigma)$. Therefore, $h\circ g$ is $\omega _{s}$-irresolute.

3.   ${ \omega }_{{ s}}$-open and pre-${ \omega }_{{ s}}$-open functions

Definition 3.1. A function $g:(Y, \sigma)\longrightarrow (Z, \lambda)$ is called

(a) $\omega _{s}$-open (resp. semi-open ^[13]) if for each $U\in \sigma$, $f(U)\in \omega _{s}(Z, \lambda)$ (resp. $f(U)\in SO(Z, \lambda)$).

(b) pre-semi-open ^[14] if for each $U\in SO(Y, \sigma)$, $f(U)\in SO(Z, \lambda)$.

Theorem 3.2. Let $g:(Y, \sigma)\longrightarrow (Z, \lambda)$ be a function. If for a base $\mathcal{B}$ of $(Y, \sigma)$, $g(B)\in \lambda$ for all $B\in \mathcal{B}$, then $g$ is $\omega _{s}$-open.

Proof. Suppose that for a base $\mathcal{B}$ of $(Y, \sigma)$, $g(B)\in \omega _{s}(Z, \lambda)$ for all $B\in \mathcal{B}$. Let $V\in \sigma -\left\{ \emptyset \right\}$. Choose $\mathcal{B}_{1}\subseteq \mathcal{B}$ such that $V = \cup \left\{ B:B\in \mathcal{B}_{1}\right\}$. Then

Since by assumption, $g\left(B\right) \in \omega _{s}(Z, \lambda)$ for all $B\in \mathcal{B}_{1}$, then $g\left(V\right) \in \omega _{s}(Z, \lambda)$.

Theorem 3.3. Every open function is $\omega _{s}$-open.

Proof. Let $g:\left(Y, \sigma \right) \longrightarrow (Z, \gamma)$ be an open function and let $V\in \sigma$. Since $g$ is open, then $g(V)\in \gamma \subseteq \omega _{s}(Z, \gamma)$.

The converse of Theorem 3.3 is not true as shown in the next example:

Example 3.4. Consider the function $g:(\mathbb{R}, \tau _{u})\longrightarrow (\mathbb{R}, \tau _{u})$ defined by $g\left(y\right) = y^{2}$. We apply Theorem 3.2 to show that $g$ is $\omega _{s}$-open. Consider the base $\left\{ \left(c, d\right) :c, d\in \mathbb{R} \text{ and }c < d\right\}$ for $(\mathbb{R}, \tau _{u})$. Then for all $c, d\in \mathbb{R}$ with $c < d$ we have

and so $g\left(\left(c, d\right) \right) \in \omega _{s}(\mathbb{R}, \tau _{u})$. Therefore, $g$ is $\omega _{s}$-open. On the other hand, since $\mathbb{R} \in \tau _{u}$ but $g\left(\mathbb{R} \right) = \left[0, \infty \right) \notin \tau _{u}$, then $g$ is not open.

Theorem 3.5. If $g:\left(Y, \sigma \right) \longrightarrow (Z, \gamma)$ is an $\omega _{s}$-open function such that $(Z, \gamma)$ is locally countable, then $g$ is open.

Proof. Follows from the definitions and Proposition 1.1 (b).

Theorem 3.6. Every $\omega _{s}$-open function is semi-open.

Proof. Let $g:\left(Y, \sigma \right) \longrightarrow (Z, \gamma)$ be an $\omega _{s}$-open function and let $V\in \sigma$. Since $g$ is $\omega _{s}$-open, then $g(V)\in \omega _{s}(Z, \gamma)\subseteq SO(Z, \gamma)$.

The converse of Theorem 3.6 is not true as shown in the next example:

Example 3.7. Consider $(\mathbb{R}, \sigma)$ where $\sigma = \left\{ \emptyset, \mathbb{N}, \mathbb{R} \right\}$. Define $g:(\mathbb{R}, \sigma)\longrightarrow (\mathbb{R}, \sigma)$ by $g\left(y\right) = y-1$. Since $g\left(\mathbb{N} \right) = \left\{ 0\right\} \cup \mathbb{N} \subseteq \overline{ \mathbb{N} } = \mathbb{R}$, then, $g\left(\mathbb{N} \right) \in SO\left(\mathbb{R}, \sigma \right)$. Also, $g\left(\emptyset \right) = \emptyset \in SO\left(\mathbb{R}, \sigma \right) \ $and $g\left(\mathbb{R} \right) = \mathbb{R} \in SO\left(\mathbb{R}, \sigma \right)$. Therefore, $g$ is semi-open. Conversely, $g$ is not $\omega _{s}$-open since there is $\mathbb{N} \in \sigma$ such that $g\left(\mathbb{N} \right) = \left\{ 0\right\} \cup \mathbb{N} \notin \omega _{s}\left(\mathbb{R}, \sigma \right)$.

Theorem 3.8. If $g:\left(Y, \sigma \right) \longrightarrow (Z, \gamma)$ is a semi-open function such that $(Z, \gamma)$ is anti-locally countable, then $g$ is $\omega _{s}$-open.

Proof. Follows from the definitions and Proposition 1.1 (c).

Definition 3.9. A function $g:\left(Y, \sigma \right) \longrightarrow (Z, \gamma)$ is called pre-$\omega _{s}$-open, if for every $A\in \omega _{s}\left(Y, \sigma \right)$, $g(A)\in \omega _{s}(Z, \gamma)$.

The following two examples will show that openness and pre-$\omega _{s}$ -openness are independent concepts:

Example 3.10. Let $X = \mathbb{R}$, $Y = \left\{ a, b, c\right\}$, $\tau = \left\{ \emptyset, \left(-\infty, 1\right), X\right\}$, and $\sigma = \left\{ \emptyset, \left\{ a\right\}, Y\right\}$. Define $f:(X, \tau)\longrightarrow (Y, \sigma)$ by

Since $f\left(\emptyset \right) = \emptyset \in \sigma$, $f\left(\left(-\infty, 1\right) \right) = \left\{ a\right\} \in \sigma$, and $f\left(X\right) = Y\in \sigma$, then $f$ is open. Since $(X, \tau)$ is anti-locally countable, then by Proposition 1.3, $\overline{\left(-\infty, 1\right) } ^{\omega } = \overline{\left(-\infty, 1\right) } = \mathbb{R}$. So, we have $\left(-\infty, 1\right] \in \omega _{s}\left(X, \tau \right)$ but $f\left(\left(-\infty, 1\right] \right) = \left\{ a, b\right\} \notin \omega _{s}(Y, \sigma) = \sigma$. This shows that $f$ is not pre-$\omega _{s}$-open.

Example 3.11. Consider $\left(\mathbb{R}, \tau \right)$ where $\tau = \left\{ \emptyset, \mathbb{R}, \left[1, \infty \right) \right\}$. Define $f:\left(\mathbb{R}, \tau \right) \longrightarrow \left(\mathbb{R}, \tau \right)$ by $f\left(x\right) = x-1$. Since $\left(\mathbb{R}, \tau \right)$ is anti-locally countable, then by Proposition 1.1 (c), $\omega _{s}\left(\mathbb{R}, \tau \right) = SO\left(\mathbb{R}, \tau \right) = \left\{ \emptyset \right\} \cup \left\{ H:\left[1, \infty \right) \subseteq H\right\}$. To see that $f$ is pre-$\omega _{s}$-open, let $H\in \omega _{s}\left(\mathbb{R}, \tau \right) -\left\{ \emptyset \right\}$. Then $\left[0, \infty \right) = f\left(\left[1, \infty \right) \right) \subseteq f\left(H\right)$, and thus $\left[1, \infty \right) \subseteq f\left(H\right)$. Hence, $f\left(H\right) \in \omega _{s}\left(\mathbb{R}, \tau \right)$. Therefore, $f$ is pre-$\omega _{s}$-open. Conversely, $f$ is not open since there is $\left[1, \infty \right) \in \tau$ such that $f\left(\left[1, \infty \right) \right) = \left[0, \infty \right) \notin \tau$.

Theorem 3.12. Let $\left(Y, \sigma \right)$ and $(Z, \gamma)$ be locally countable, and let $g:\left(Y, \sigma \right) \longrightarrow (Z, \gamma)$ be a function. Then the followings are equivalent:

(a) $g$ is open.

(b) $g$ is pre-$\omega _{s}$-open.

Proof. Follows from the definitions and Proposition 1.1 (b).

The following two examples will show that pre-semi-openness and pre-$\omega _{s}$-openness are independent concepts:

Example 3.13. Consider the function $g$ as in Example 3.7. It is not difficult to see that $SO\left(\mathbb{R}, \sigma \right) = \left\{ \emptyset \right\} \cup \left\{ H: \mathbb{N} \subseteq H\right\}$ and $\omega _{s}\left(\mathbb{R}, \sigma \right) = \sigma$. To see that $g$ is pre-semi-open, let $H\in SO\left(\mathbb{R}, \sigma \right) -\left\{ \emptyset \right\}$. Then $\mathbb{N} \subseteq H$ and so $\mathbb{N} \subseteq \mathbb{N} \cup \left\{ 0\right\} = g\left(\mathbb{N} \right)$. Thus, $g\left(\mathbb{N} \right) \in SO\left(\mathbb{R}, \sigma \right)$. This shows that $g$ is pre-semi-open. Conversely, $g$ is not pre-$\omega _{s}$-open since there is $\mathbb{N} \in \omega _{s}\left(\mathbb{R}, \sigma \right)$ such that $g\left(\mathbb{N} \right) = \left\{ 0\right\} \cup \mathbb{N} \notin \omega _{s}\left(\mathbb{R}, \sigma \right)$.

Example 3.14. Let $X = Y = \{1, 2, 3, 4\}$, $\tau = \{\emptyset, X, \{1, 2\}, \{1\}, \{2\}\}$, and $\sigma = \{\emptyset, Y, \{2, 3, 4\}, \{1, 2\}, \{1\}, \{2\}\}$. Let $f:\left(X, \tau \right) \longrightarrow (Y, \sigma)$ be the identity function. It is clear that $f$ is open. Since $\left(X, \tau \right)$ and $(Y, \sigma)$ are locally countable, then by Theorem 3.12, $f$ is pre-$\omega _{s}$-open. Conversely, $f$ is not pre-semi-open because there is $\left\{ 1, 3\right\} \in \; SO\left(X, \tau \right)$ such that $f\left(\left\{ 1, 3\right\} \right) = \left\{ 1, 3\right\} \notin SO\left(Y, \sigma \right)$.

Theorem 3.15. Let $\left(X, \tau \right)$ and $(Y, \sigma)$ be anti-locally countable topological spaces and let $f:\left(X, \tau \right) \longrightarrow (Y, \sigma)$ be a function. Then $f$ is pre-semi-open if and only if $f$ is pre-$\omega _{s}$-open.

Proof. Follows from definitions and Proposition 1.1 (c).

Theorem 3.16. Every pre-$\omega _{s}$-open function is $\omega _{s}$ -open. Proof. Let $f:\left(X, \tau \right) \longrightarrow (Y, \sigma)$ be a pre-$\omega _{s}$-open function and let $U\in \tau \subseteq \omega _{s}\left(X, \tau \right)$. Since $f$ is pre-$\omega _{s}$-open, then $f(U)\in \omega _{s}(Y, \sigma)$.

The function $f$ in Example 3.10 is open but not pre-$\omega _{s}$-open, and by Theorem 3.3, $f$ is $\omega _{s}$-open. Therefore, the converse of the implication in Theorem 3.16 is not true, in general.

Theorem 3.17. For a function $g:(Y, \sigma)\longrightarrow (Z, \lambda)$, the followings are equivalent:

(a) $g$ is $\omega _{s}$-open.

(b) $g^{-1}\left(\overline{B}^{\omega _{s}}\right) \subseteq \overline{ g^{-1}\left(B\right) }$ for every $B\subseteq Z$.

(c) $Int\left(g^{-1}\left(B\right) \right) \subseteq g^{-1}\left(Int_{\omega _{s}}\left(B\right) \right)$ for every $B\subseteq Z$.

Proof. (a) $\Longrightarrow$ (b): Suppose that $g$ is $\omega _{s}$-open and let $B\subseteq Z$. Let $y\in g^{-1}\left(\overline{B}^{\omega _{s}}\right)$. To show that $y\in \overline{ g^{-1}\left(B\right) }$, let $V\in \sigma$ such that $y\in V$. Then $g\left(y\right) \in g\left(V\right)$. Since $g$ is $\omega _{s}$-open, then $g\left(V\right) \in \omega _{s}(Z, \lambda)$. Since $g\left(y\right) \in g\left(V\right) \cap \overline{B}^{\omega _{s}}$, then $g\left(V\right) \cap B\neq \emptyset$. Choose $t\in V$ such that $g\left(t\right) \in B$. Then $t\in V\cap g^{-1}\left(B\right)$, and hence $V\cap g^{-1}\left(B\right) \neq \emptyset$. It follows that $y\in \overline{ g^{-1}\left(B\right) }$.

(b) $\Longrightarrow$ (a): Suppose that $g^{-1}\left(\overline{B}^{\omega _{s}}\right) \subseteq \overline{g^{-1}\left(B\right) }$ for every $B\subseteq Z$, and suppose to the contrary that $g$ is not $\omega _{s}$-open. Then there exists $V\in \sigma$ such that $g\left(V\right) \notin \omega _{s}(Z, \lambda)$ and so, $Z-g\left(V\right)$ is not $\omega _{s}$-closed. Thus, there exists $z\in g\left(V\right) \cap \overline{Z-g\left(V\right) }^{\omega _{s}}$. Choose $y\in V$ such that $z = g\left(y\right)$. Then $y\in g^{-1}\left(\overline{Z-g\left(V\right) } ^{\omega _{s}}\right)$. By assumption, we have

Therefore, $y\in Y-V$. But $y\in V$, a contradiction.

(b) $\Longrightarrow$ (c): Suppose that $g^{-1}\left(\overline{B} ^{\omega _{s}}\right) \subseteq \overline{g^{-1}\left(B\right) }$ for every $B\subseteq Z$. Let $B\subseteq Z$. Then by (b),

(c) $\Longrightarrow$ (b): Suppose that $Int\left(g^{-1}\left(B\right) \right) \subseteq g^{-1}\left(Int_{\omega _{s}}\left(B\right) \right)$ for every $B\subseteq Z$. Let $B\subseteq Z$. Then by (c),

Theorem 3.18. For a function $g:(Y, \sigma)\longrightarrow (Z, \lambda)$, the followings are equivalent:

(a) $g$ is pre-$\omega _{s}$-open.

(b) $g^{-1}\left(\overline{B}^{\omega _{s}}\right) \subseteq \overline{ g^{-1}\left(B\right) }^{\omega _{s}}$ for every $B\subseteq Z$.

(c) $Int_{\omega _{s}}\left(g^{-1}\left(B\right) \right) \subseteq g^{-1}\left(Int_{\omega _{s}}\left(B\right) \right)$ for every $B\subseteq Z$.

Proof. (a) $\Longrightarrow$ (b): Suppose that $g$ is pre-$\omega _{s}$-open and let $B\subseteq Z$. Let $y\in g^{-1}\left(\overline{B}^{\omega _{s}}\right)$. To show that $y\in \overline{ g^{-1}\left(B\right) }^{\omega _{s}}$, let $C\in \omega _{s}(Y, \sigma)$ such that $y\in C$. Then $g\left(y\right) \in g\left(C\right)$. Since $g$ is pre-$\omega _{s}$-open, then $g\left(C\right) \in \omega _{s}(Z, \lambda)$. Since $g\left(y\right) \in g\left(C\right) \cap \overline{B}^{\omega _{s}}$, then $g\left(C\right) \cap B\neq \emptyset$. Choose $t\in C$ such that $g\left(t\right) \in B$. Then $t\in C\cap g^{-1}\left(B\right)$ and hence $C\cap g^{-1}\left(B\right) \neq \emptyset$. It follows that $y\in \overline{g^{-1}\left(B\right) }^{\omega _{s}}$.

(b) $\Longrightarrow$ (a): Suppose that $g^{-1}\left(\overline{B}^{\omega _{s}}\right) \subseteq \overline{ g^{-1}\left(B\right) }^{\omega _{s}}$ for every $B\subseteq Z$, and suppose to the contrary that $g$ is not pre-$\omega _{s}$-open. Then there is $C\in \omega _{s}(Y, \sigma)$ such that $g\left(C\right) \notin \omega _{s}(Z, \lambda)$ and so $Z-g\left(C\right)$ is not $\omega _{s}$-closed. And so there exists $z\in g\left(C\right) \cap \overline{Z-g\left(C\right) }^{\omega _{s}}$. Choose $y\in C$ such that $z = g\left(y\right)$. Then $y\in g^{-1}\left(\overline{Z-g\left(C\right) }^{\omega _{s}}\right)$. By assumption we have

Therefore, $y\in Y-C$ but $y\in C$, a contradiction.

(b) $\Longrightarrow$ (c): Suppose that $g^{-1}\left(\overline{B} ^{\omega _{s}}\right) \subseteq \overline{g^{-1}\left(B\right) }^{\omega _{s}}$ for every $B\subseteq Z$. Let $B\subseteq Z$. Then by (b),

(c) $\Longrightarrow$ (b): Suppose that $Int_{\omega _{s}}\left(g^{-1}\left(B\right) \right) \subseteq g^{-1}\left(Int_{\omega _{s}}\left(B\right) \right)$ for every $B\subseteq Z$. Let $B\subseteq Z$. Then by (c),

Theorem 3.19. If $g:(Y, \sigma)\longrightarrow (Z, \lambda)$ is $\omega _{s}$-continuous such that $g:(Y, \sigma _{\omega })\longrightarrow (Z, \lambda _{\omega })$ is open, then $g:(Y, \sigma)\longrightarrow (Z, \lambda)$ is $\omega _{s}$-irresolute.

Proof. Suppose that $g:(Y, \sigma)\longrightarrow (Z, \lambda)$ is $\omega _{s}$-continuous with $g:(Y, \sigma _{\omega })\longrightarrow (Z, \lambda _{\omega })$ is open. Let $G\in \omega _{s}(Z, \lambda)$, choose $W\in \lambda$ such that $W\subseteq G\subseteq \overline{W}^{\omega }$. Thus, we have $g^{-1}\left(W\right) \subseteq g^{-1}\left(G\right) \subseteq g^{-1}\left(\overline{W}^{\omega }\right)$. Since $g:(Y, \sigma)\longrightarrow (Z, \lambda)$ is $\omega _{s}$ -continuous, then $g^{-1}\left(W\right) \in \omega _{s}(Y, \sigma)$. Since $g:(Y, \sigma _{\omega })\longrightarrow (Z, \lambda _{\omega })$ is open, then $g^{-1}\left(\overline{W}^{\omega }\right) \subseteq \overline{g^{-1}\left(W\right) }^{\omega }$. Since we have $g^{-1}\left(W\right) \subseteq g^{-1}\left(G\right) \subseteq \overline{g^{-1}\left(W\right) }^{\omega }$ with $g^{-1}\left(W\right) \in \omega _{s}(Y, \sigma)$, then by Proposition 1.2 (a), $g^{-1}\left(G\right) \in \omega _{s}(Y, \sigma)$. This ends the proof.

Theorem 3.20. If $g:(Y, \sigma)\longrightarrow (Z, \lambda)$ is an $\omega _{s}$-open function such that $g:(Y, \sigma _{\omega })\longrightarrow (Z, \lambda _{\omega })$ is a continuous function, then $g:(Y, \sigma)\longrightarrow (Z, \lambda)$ is pre-$\omega _{s}$-open.

Proof. Suppose that $g:(Y, \sigma)\longrightarrow (Z, \lambda)$ is $\omega _{s}$-open with $g:(Y, \sigma _{\omega })\longrightarrow (Z, \lambda _{\omega })$ is continuous. Let $H\in \omega _{s}(Y, \sigma)$, then we find $V\in \sigma$ such that $V\subseteq H\subseteq \overline{V}^{\omega }$. Thus, we have $g\left(V\right) \subseteq g\left(H\right) \subseteq g\left(\overline{V}^{\omega }\right)$. Since $g:(Y, \sigma)\longrightarrow (Z, \lambda)$ is $\omega _{s}$-open, then $g\left(V\right) \in \omega _{s}(Z, \lambda)$. By continuity of $g:(Y, \sigma _{\omega })\longrightarrow (Z, \lambda _{\omega })$, $g\left(\overline{V}^{\omega }\right) \subseteq \overline{g\left(V\right) }^{\omega }$. Since we have $g\left(V\right) \subseteq g\left(H\right) \subseteq \overline{g\left(V\right) }^{\omega }$ with $g\left(V\right) \in \omega _{s}(Z, \lambda)$, then by Proposition 1.2 (a), $g\left(V\right) \in \omega _{s}(Z, \lambda)$. This ends the proof.

As defined in ^[32], a function $g:(Y, \sigma)\longrightarrow (Z, \lambda)$ is $\omega$-continuous if for each $W\in \lambda$, $g^{-1}(W)\in \sigma _{\omega }$.

Theorem 3.21. Let $g:(Y, \sigma)\longrightarrow (Z, \lambda)$ be pre-$\omega _{s}$-open and $\omega _{s}$-irresolute such that $(Z, \lambda)$ is semi-regular and dense in itself, then $g$ is $\omega$-continuous.

Proof. Suppose to the contrary that $g$ is not $\omega$ -continuous. Then there is $W\in \lambda$ such that $g^{-1}\left(W\right) \notin \sigma _{\omega }$. So, there exists $y\in g^{-1}\left(W\right) -Int_{\omega }\left(g^{-1}\left(W\right) \right)$. Since $g\left(y\right) \in W$ and $(Z, \lambda)$ is semi-regular, then there is a regular open set $M$ of $(Z, \lambda)$ such that $g\left(y\right) \in M\subseteq W$. Since $Int\left(\overline{M}^{\omega }\right) \subseteq Int\left(\overline{M}\right) = M$, then by Theorem 2.16 of ^[29], $M$ is $\omega _{s}$ -closed. Since $g$ is $\omega _{s}$-irresolute, then by Theorem 2.11, $g^{-1}\left(M\right)$ is $\omega _{s}$-closed, and so $Y-g^{-1}\left(M\right) \in \omega _{s}(Y, \sigma)$. Since $g^{-1}\left(M\right) \subseteq g^{-1}\left(W\right)$, then $Int_{\omega }\left(g^{-1}\left(M\right) \right) \subseteq Int_{\omega }\left(g^{-1}\left(W\right) \right)$. Since $y\notin Int_{\omega }\left(g^{-1}\left(W\right) \right)$, then $y\notin Int_{\omega }\left(g^{-1}\left(M\right) \right)$, and so

Thus, by Proposition 1.2 (a), we have $\left(Y-g^{-1}\left(M\right) \right) \cup \left\{ y\right\} \in \omega _{s}(Y, \sigma)$ with $y\notin Y-g^{-1}\left(M\right)$. Put $S = g\left(\left(Y-g^{-1}\left(M\right) \right) \cup \left\{ y\right\} \right) = g\left(Y-g^{-1}\left(M\right) \right))\cup \left\{ g\left(y\right) \right\}$. Since $g$ is pre-$\omega _{s}$-open, then $S\in \omega _{s}(Z, \lambda)$. So by Proposition 1.2 (b), we have $S\cap M\in \omega _{s}(Z, \lambda)$. Since $g\left(Y-g^{-1}\left(M\right) \right))\subseteq Z-M$, then we have

and thus $S\cap M = \left\{ g\left(y\right) \right\}$. Therefore, $\left\{ g\left(y\right) \right\} \in \omega _{s}(Z, \lambda)$. Thus, there exists $O\in \lambda$ such that $O\subseteq \left\{ g\left(y\right) \right\} \subseteq \overline{O}^{\omega }$, and hence $O = \left\{ g\left(y\right) \right\}$. This implies that $\left\{ g\left(y\right) \right\} \in \lambda$. But by assumption $(Z, \lambda)$ is dense in itself, a contradiction.

The condition '$(Z, \lambda)$ is dense in itself' in Theorem 3.21 cannot be dropped as our next example shows:

Example 3.22. Take $f$ as in Example 2.7. Then $f$ is $\omega _{s}$ -irresolute and pre-$\omega _{s}$-open. Also, $(\left\{ a, b\right\}, \tau _{disc})$ is semi-regular. On the other hand, since $\left\{ b\right\} \in \tau _{disc}$ but $f^{-1}\left(\left\{ b\right\} \right) = \left[0, \infty \right) \notin \left(\tau _{u}\right) _{\omega }$, then $f$ is not $\omega$ -continuous.

Theorem 3.23. Let $g:(Y, \sigma)\longrightarrow (Z, \lambda)$ be injective, pre-$\omega _{s}$-open, and $\omega _{s}$ -irresolute such that $(Z, \lambda)$ is semi-regular, then $g$ is $\omega$ -continuous.

Proof. Suppose to the contrary that $g$ is not $\omega$ -continuous. Then there is $W\in \lambda$ such that $g^{-1}\left(W\right) \notin \sigma _{\omega }$. So, there exists $y\in g^{-1}\left(W\right) -Int_{\omega }\left(g^{-1}\left(W\right) \right)$. Since $g\left(y\right) \in W$ and $(Z, \lambda)$ is semi-regular, then there is a regular open set $M$ of $(Z, \lambda)$ such that $g\left(y\right) \in M\subseteq W$. Since $Int\left(\overline{M}^{\omega }\right) \subseteq Int\left(\overline{M}\right) = M$, then by Theorem 2.16 of ^[29], $M$ is $\omega _{s}$ -closed. Since $g$ is $\omega _{s}$-irresolute, then by Theorem 2.11, $g^{-1}\left(M\right)$ is $\omega _{s}$-closed, and so $Y-g^{-1}\left(M\right) \in \omega _{s}\left(Y, \sigma \right)$. Since $g^{-1}\left(M\right) \subseteq g^{-1}\left(W\right)$, then $Int_{\omega }\left(g^{-1}\left(M\right) \right) \subseteq Int_{\omega }\left(g^{-1}\left(W\right) \right)$. Since $y\notin Int_{\omega }\left(g^{-1}\left(W\right) \right)$, then $y\notin Int_{\omega }\left(g^{-1}\left(M\right) \right)$, and so

Thus, by Proposition 1.2 (a), we have $\left(Y-g^{-1}\left(M\right) \right) \cup \left\{ y\right\} \in \omega _{s}\left(Y, \sigma \right)$ with $y\notin Y-g^{-1}\left(M\right)$. Put $S = g\left(\left(Y-g^{-1}\left(M\right) \right) \cup \left\{ y\right\} \right) = g\left(Y-g^{-1}\left(M\right) \right))\cup \left\{ g\left(y\right) \right\}$. Since $g$ is pre- $\omega _{s}$-open, then $S\in \omega _{s}(Z, \lambda)$, and by Proposition 1.2 (b) we have $S\cap M\in \omega _{s}(Z, \lambda)$. Since $g\left(Y-g^{-1}\left(M\right) \right))\subseteq Z-M$, then we have

and thus $S\cap M = \left\{ g\left(y\right) \right\}$. Therefore, $\left\{ g\left(y\right) \right\} \in \omega _{s}(Z, \lambda)$. Since $g$ is $\omega _{s}$-irresolute and injective, then $g^{-1}\left(\left\{ g\left(y\right) \right\} \right) = \left\{ y\right\} \in \omega _{s}\left(Y, \sigma \right)$. So, there exists $V\in \sigma$ such that $V\subseteq \left\{ y\right\} \subseteq \overline{V}^{\omega }$, and hence $\left\{ y\right\} \in \sigma$. Since $y\in g^{-1}\left(W\right)$ and $\left\{ y\right\} \in \sigma \subseteq \sigma _{\omega }$, then $y\in Int_{\omega }\left(g^{-1}\left(W\right) \right)$, a contradiction.

Example 3.23 shows that the condition 'injective' in Theorem 3.23 cannot be dropped.

Definition 3.24. A function $g:(Y, \sigma)\longrightarrow (Z, \lambda)$ is called$\ \omega _{s}$-closed if for each $\omega _{s}$ -closed set $C$ of $(Y, \sigma)$, $g(C)$ is $\omega _{s}$-closed set in $(Z, \lambda)$.

As defined in ^[33] a topological space $(X, \tau)$ is called $\omega$ -regular if for each closed set $F$ in $(X, \tau)$ and $x\in X-F$, there exist $U \; \in \; \tau$ and $V\in \; \tau _{\omega }$ such that $x\in U$, $F\subseteq V$ and $U\cap V = \emptyset$. As defined in ^[31], a function $g:(Y, \sigma)\longrightarrow (Z, \lambda)$ is $\omega$-open if for each $V\in \sigma$, $g(V)\in \lambda _{\omega }$.

Theorem 3.25. If $g:(Y, \sigma)\longrightarrow (Z, \lambda)$ is $\omega _{s}$-closed, pre-$\omega _{s}$-open, and $\omega _{s}$ -irresolute such that $(Y, \sigma)$ is $\omega$-regular, then $g$ is $\omega$-open.

Proof. Suppose to the contrary that there exists $V\in \sigma$ such that $g\left(V\right) \notin \lambda _{\omega }$. Then we find $y\in V$ such that $g\left(y\right) \in g\left(V\right) -Int_{\omega }\left(g\left(V\right) \right)$. By $\omega$-regularity of $(Y, \sigma)$, we find $M\in \sigma$ such that $y\in M\subseteq \overline{M}^{\omega }\subseteq V$. Since $\overline{M}^{\omega }$ is $\omega _{s}$-closed and $g$ is $\omega _{s}$-closed, then $Z-g\left(\overline{M}^{\omega }\right) \in \omega _{s}(Z, \lambda)$ with $g\left(y\right) \notin Z-g\left(\overline{M} ^{\omega }\right)$. Since $g\left(y\right) \notin Int_{\omega }\left(g\left(V\right) \right)$, then $g\left(y\right) \notin Int_{\omega }\left(g\left(\overline{M}^{\omega }\right) \right)$. Thus,

So by Proposition 1.2 (a), $\left(Z-g\left(\overline{M}^{\omega }\right) \right) \cup \left\{ g\left(y\right) \right\} \in \omega _{s}(Z, \lambda)$. Set $B = g^{-1}\left(\left(Z-g\left(\overline{M}^{\omega }\right) \right) \cup \left\{ g\left(y\right) \right\} \right)$. Since $g$ is $\omega _{s}$ -irresolute, then $B\in \omega _{s}(Y, \sigma)$ and by Proposition 1.2 (b), $M\cap B\in \omega _{s}(Y, \sigma)$. Since $g$ is pre-$\omega _{s}$-open, then $g\left(M\cap B\right) \in \omega _{s}(Z, \lambda)$. Since

then $\left\{ g\left(y\right) \right\} \in \omega _{s}(Z, \lambda)$. Thus, there is $K\in \lambda$ such that $K\subseteq \left\{ g\left(y\right) \right\} \subseteq \overline{K}^{\omega }$. Hence, $\left\{ g\left(y\right) \right\} \in \lambda$. Since $g\left(y\right) \in g\left(V\right)$, then $g\left(y\right) \in Int\left(g\left(V\right) \right) \subseteq Int_{\omega }\left(g\left(V\right) \right)$, a contradiction.

4.   Slightly $\omega _{s}$-continuous functions

Let $(Y, \sigma)$ be a topological space and let $B$ be a subset of $Y$. Then $B$ is called clopen (resp. semi-clopen, $\omega _{s}$-clopen) in $(Y, \sigma)$ if $B$ both open and closed (resp. semi-open and semi-closed, $\omega _{s}$-open and $\omega _{s}$-closed) in $(Y, \sigma)$. Throughout this section, the family of all clopen (resp., semi-clopen, $\omega _{s}$ -clopen) subsets of the topological space $(Y, \sigma)$ will be denoted by $CO(Y, \sigma)$ (resp. $SCO(Y, \sigma)$, $\omega _{s}CO(Y, \sigma)$).

Definition 4.1. A function $g:(Y, \sigma)\longrightarrow (Z, \lambda)$ is called slightly continuous ^[34] (resp. slightly semi-continuous ^[15], slightly $\omega _{s}$-continuous), if for every $y \; \in \; Y$ and every $W\in CO(Z, \lambda)$ with $g\left(y\right) \in W$, there exists $V\in \sigma$ (resp. $V\in SO(Y, \sigma)$, $V\in \omega _{s}(Y, \sigma)$) such that $y\in V$ and $g(V) \; \subseteq \; W$.

As an example of a slightly continuous function that is not continuous, take the function $g:(\mathbb{R}, \tau _{u})\rightarrow (\mathbb{R}, \tau _{u})$ defined by $g(y) = [y]$, where $[y]$ is the greatest integer of $y$.

Theorem 4.2. For a function $g:(Y, \sigma)\longrightarrow (Z, \lambda)$, the followings are equivalent:

(a) $g$ is slightly $\omega _{s}$-continuous.

(b) For all $W\in CO(Z, \lambda)$, $g^{-1}\left(W\right) \in \omega _{s}(Y, \sigma)$.

(c) For all $W\in CO(Z, \lambda)$, $g^{-1}\left(W\right) \in \omega _{s}CO(Y, \sigma)$.

Proof. (a) $\Longrightarrow$ (b): Let $W\in CO(Z, \lambda)$. Then for each $y\in g \; ^{-1}(W)$, $g(y)\in W$ and by (a), there exists $V_{y}\in \omega _{s}(Y, \sigma)$ such that $y \; \in \; V_{y}$ and $g(V_{y}) \; \subseteq \; W$. Thus,

Therefore, $g^{-1}(W)$ is a union of $\omega _{s}$-open sets, and hence $g^{-1}(W)$ is $\omega _{s}$-open.

(b) $\Longrightarrow$ (c): Let $W\in CO(Z, \lambda)$. Then $Z-W \; \in CO(Z, \lambda)$. Thus, by (b), $g^{-1}(W)\in \omega _{s}(Y, \sigma)$ and $g^{-1}(Z \; -W) = Y-g^{-1}(W)\in \omega _{s}(Y, \sigma)$. Therefore, $g^{-1}(W)\in \omega _{s}CO(Y, \sigma)$.

(c) $\Longrightarrow$ (a): Let $y\in Y$ and $W\in CO(Z, \lambda)$ with $g(y) \; \in \; W$. By (c), $g^{-1}(W)\in \omega _{s}CO(Y, \sigma)\subseteq \omega _{s}(Y, \sigma)$. Put $V \; = \; g^{-1}(W)$. Then $V\in \omega _{s}(Y, \sigma)$, $y\in V$, and $g(V) = g(g \; ^{-1}(W))\subseteq W$. This shows that $g$ is slightly $\omega _{s}$-continuous.

Theorem 4.3. Every slightly continuous function is slightly $\omega _{s}$-continuous.

Proof. Assume that $g:(Y, \sigma)\longrightarrow (Z, \lambda)$ is slightly continuous. Let $W\in CO(Z, \lambda)$. Since $g$ is slightly continuous, then $g^{-1}(W)\in \sigma$. So by Proposition 1.1 (a), $g^{-1}(W)\in \omega _{s}(Y, \sigma)$. Therefore, by Theorem 4.2, it follows that $g$ is slightly $\omega _{s}$-continuous.

Our next example shows that the converse of Theorem 4.3 is not true, in general:

Example 4.4. Let $X = Y = \mathbb{R}$, $\tau = \left\{ \emptyset, \mathbb{R}, \mathbb{N}, \mathbb{Q} ^{c}, \mathbb{N} \cup \mathbb{Q} ^{c}\right\}$, and $\sigma \; = \{\emptyset, \mathbb{R}, \mathbb{N}, \mathbb{R} - \mathbb{N} \}$. Define $f\ :(\mathbb{R}, \tau) \; \rightarrow \; (\mathbb{R}, \sigma)$ by $f(x) = x$. Note that $CO(\mathbb{R}, \sigma) = \sigma$. Since $f^{-1}(\mathbb{N}) = \mathbb{N} \in \tau \subseteq \omega _{s}(X, \tau)$ and $f^{-1}(\mathbb{R} - \mathbb{N}) = \mathbb{R} - \mathbb{N} \in \; \omega _{s}(X, \tau)$, then $f$ is$\ $slightly $\omega _{s}$ -continuous. On the other hand, since $\mathbb{R} - \mathbb{N} \in CO(\mathbb{R}, \sigma)$ but $f^{-1}(\mathbb{R} - \mathbb{N}) \; = \; \mathbb{R} \; - \; \mathbb{N} \; \notin \; \tau$, then $f$ is not slightly continuous.

Theorem 4.5. Every slightly $\omega _{s}$-continuous function is slightly semi-continuous.

Proof. Assume that $g:(Y, \sigma)\longrightarrow (Z, \lambda)$ is slightly $\omega _{s}$-continuous. Let $W\in CO(Z, \lambda)$. Since $g$ is slightly $\omega _{s}$-continuous, then $g^{-1}(W)\in \omega _{s}(Y, \sigma)$. So by Proposition 1.1 (a), $g^{-1}(W)\in SO(Y, \sigma)$. Therefore, $g$ is slightly semi-continuous.

The converse of Theorem 4.5 is not true in general as the following example shows:

Example 4.6. Let $X = Y = \mathbb{R}$, $\ \tau = \left\{ \emptyset, \mathbb{R}, \mathbb{N}, \mathbb{Q} ^{c}, \mathbb{N} \cup \mathbb{Q} ^{c}\right\}$, and $\sigma \; = \{\emptyset, \mathbb{R}, \mathbb{Q}, \mathbb{R} - \mathbb{Q} \}$. Define $f\ :(\mathbb{R}, \tau) \; \rightarrow \; (\mathbb{R}, \sigma)$ by $f(x) = x$. Note that $CO(\mathbb{R}, \sigma) = \; \sigma$. Since $f^{-1}(\mathbb{Q}) = \mathbb{Q} \in SO(X, \tau)$ and $f^{-1}(\mathbb{R} - \mathbb{Q}) = \mathbb{R} - \mathbb{Q} \in \; \tau \subseteq SO(X, \tau)$, then $f$ is$\ $slightly semi-continuous. On the other hand, since $\mathbb{Q} \in CO(\mathbb{R}, \sigma)$ but $f^{-1}(\mathbb{Q}) \; = \; \mathbb{Q} \; \notin \; \omega _{s}(\mathbb{R}, \tau)$, then $f$ is not slightly $\omega _{s}$-continuous.

Theorem 4.7.(a) If $g:(Y, \sigma)\longrightarrow (Z, \lambda)$ is slightly $\omega _{s}$-continuous such that $(Y, \sigma)$ is locally countable, then $g$ is continuous.

(b) If $g:(Y, \sigma)\longrightarrow (Z, \lambda)$ is slightly semi-continuous with $(Y, \sigma)$ is anti-locally countable, then $g$ is $\omega _{s}$-continuous.

Proof. (a) Follows from the definitions and Proposition 1.1 (c).

(b) Follows from the definitions and Proposition 1.1 (b).

Theorem 4.8. Let $g:(Y, \sigma)\longrightarrow (Z, \lambda)$ be a function and let $\gamma$ be the product topology of $(Y, \sigma)$ and $(Z, \lambda)$. Let $h:(Y, \sigma)\longrightarrow \left(Y\times Z, \gamma \right)$, where $h(y) = (y, g(y))$ be the graph of $g$. Then $g$ is slightly $\omega _{s}$-continuous if and only if $h$ is slightly $\omega _{s}$-continuous.

Proof. Let $y \; \in \; Y$ and let $M\in CO\left(Y\times Z, \gamma \right)$ such that $h(y) = (y, g(y))\in M$. Then $M\cap \left(\{y\}\times Z\right)$ is a clopen set in $\{y\}\times Z$ which contains $h(y) = (y, g(y))$. Since $\{y\}\times Z$ is homeomorphic to $Z$, then $\{z\in Z:(y, z)\in M\}\in CO(Z, \lambda)$. Since $g$ is slightly $\omega _{s}$-continuous, then $\cup \{g^{-1}(z):(y, z)\in M\}\in \omega _{s}(Y, \sigma)$. Moreover, $y\in \cup \; \{g^{-1}(z):(y, z)\in M\}\subseteq h^{-1}(M)$. Hence, $h^{-1}(M)\in \omega _{s}(Y, \sigma)$. It follows that $h$ is slightly $\omega _{s}$-continuous.

Conversely, let $H\in CO(Z, \lambda)$. Then $Y\times H\in CO\left(Y\times Z, \gamma \right)$. By slight $\omega _{s}$-continuity of $h$, $h^{-1}(Y\times H)\in \omega _{s}CO(Y, \sigma)$. Also, $h^{-1}(Y\times H) \; = g^{-1}(H)$. It follows that $g$ is slightly $\omega _{s}$-continuous.

Theorem 4.9. If $g:(Y, \sigma)\longrightarrow (Z, \lambda)$ is slightly $\omega _{s}$-continuous and $h \; :(Z, \lambda)\longrightarrow \left(W, \delta \right)$ is slightly continuous, then $h\circ g:(Y, \sigma)\longrightarrow \left(W, \delta \right)$ is slightly $\omega _{s}$ -continuous.

Proof. Let $M\in CO\left(W, \delta \right)$. By slight continuity of $h$, $g^{-1}(M)\in CO(Z, \lambda)$. By slight $\omega _{s}$-continuity of $g$, $g^{-1}(h^{-1}(M)) = (h\circ g)^{-1}(M)\in \omega _{s}CO(Y, \sigma)$. Hence, $h\circ g$ is slightly $\omega _{s}$-continuous.

As defined a topological space $\left(Y, \sigma \right)$ is called semi-connected if $SCO\left(Y, \sigma \right) = \left\{ \emptyset, Y\right\}$.

Definition 4.10. A topological space $\left(Y, \sigma \right)$ is called $\omega _{s}$-connected if $\omega _{s}CO\left(Y, \sigma \right) = \left\{ \emptyset, Y\right\}$.

Theorem 4.11. Every $\omega _{s}$-connected topological space is connected.

Proof. Let $\left(Y, \sigma \right)$ be $\omega _{s}$-connected. Then $\omega _{s}CO\left(Y, \sigma \right) = \left\{ \emptyset, Y\right\}$. Thus, by Proposition 1.1 (a), we have $\left\{ \emptyset, Y\right\} \subseteq CO\left(Y, \sigma \right) \subseteq \omega _{s}CO(Y, \sigma) = \left\{ \emptyset, Y\right\}$. Hence, $CO(Y, \sigma) = \left\{ \emptyset, Y\right\}$. Therefore, $\left(Y, \sigma \right)$ is connected.

The following example will show that the converse of Theorem 4.11 is not true, in general:

Example 4.12. Consider the topological space $\left(\mathbb{R}, \tau _{u}\right)$. To see that $\left(\mathbb{R}, \tau _{u}\right)$ is not $\omega _{s}$-connected, let $M = \left(-\infty, 0\right)$, then $M\in \tau _{u}\subseteq \omega _{s}\left(\mathbb{R}, \tau _{u}\right)$. Since $\left(0, \infty \right) \in \tau _{u}$ and $\overline{\left(0, \infty \right) }^{\omega } = \overline{\left(0, \infty \right) } = \left[0, \infty \right) = \mathbb{R} -M$, then $\mathbb{R} -M\in \omega _{s}\left(\mathbb{R}, \tau _{u}\right)$. Therefore, $M\in \omega _{s}CO\left(\mathbb{R}, \tau _{u}\right) -\left\{ \emptyset, \mathbb{R} \right\}$, and hence $\left(\mathbb{R}, \tau _{u}\right)$ is not $\omega _{s}$-connected. On the other hand, $\left(\mathbb{R}, \tau _{u}\right)$ is connected.

Theorem 4.13. Every connected locally countable topological space is $\omega _{s}$-connected.

Proof. Follows from the definitions and Proposition 1.1 (b).

Theorem 4.14. Every semi-connected topological space is $\omega _{s}$-connected.

Proof. Let $\left(Y, \sigma \right)$ be semi-connected. Then $SCO\left(Y, \sigma \right) = \left\{ \emptyset, Y\right\}$. Thus, by Proposition 1.1 (a), we have $\left\{ \emptyset, Y\right\} \subseteq \omega _{s}CO\left(Y, \sigma \right) \subseteq SCO(Y, \sigma) = \left\{ \emptyset, Y\right\}$. Hence, $\omega _{s}CO(Y, \sigma) = \left\{ \emptyset, Y\right\}$. Therefore, $\left(Y, \sigma \right)$ is $\omega _{s}$-connected.

Question 4.15. Is it true that $\omega _{s}$-connected topological spaces are semi-connected?

The following theorem answers Question 4.15 partially:

Theorem 4.16. Every anti-locally countable $\omega _{s}$-connected topological space is semi-connected.

Proof. Follows from the definitions and Proposition 1.1 (c).

Theorem 4.17. A slightly $\omega _{s}$-continuous image of an $\omega _{s}$-connected space is connected.

Proof. Let $g:(Y, \sigma)\longrightarrow (Z, \lambda)$ be surjective and slightly $\omega _{s}$-continuous, where $(Y, \sigma)$ is $\omega _{s}$-connected. Suppose that $(Z, \lambda)$ is not connected. Then there exists $M\in CO(Z, \lambda)-\left\{ \emptyset, Z\right\}$. By Theorem 4.2, $g^{-1}(M)\in \omega _{s}CO(Y, \sigma)$. Since $\emptyset \neq M\neq Z$ and $g$ is surjective, then $\emptyset \neq g^{-1}\left(M\right) \neq Z$. Therefore, $g^{-1}\left(M\right) \in \omega _{s}CO(Y, \sigma)-\left\{ \emptyset, Y\right\}$ which contradicts the assumption that $(Y, \sigma)$ is $\omega _{s}$-connected.

5.   ${ \omega }_{{ s}}$-Compact topological spaces

Definition 5.1. A topological space $\left(Y, \sigma \right)$ is called $\omega _{s}$-compact (resp. semi-compact ^[36]) if for any cover $\mathcal{A}$ of $Y$ with $\mathcal{A}\subseteq \omega _{s}\left(Y, \sigma \right)$ (resp. $\mathcal{A}\subseteq SO\left(Y, \sigma \right)$), there is a finite subfamily $\mathcal{B}\subseteq \mathcal{A}$ such that $\mathcal{ B}$ is also a cover of $Y$.

Theorem 5.2. Every $\omega _{s}$-compact topological space is compact.

Proof. Let $\left(Y, \sigma \right)$ be $\omega _{s}$ -compact and let $\mathcal{A}$ be a cover of $Y$ with $\mathcal{A}\subseteq \sigma$. Then by Proposition 1.1 (a), $\mathcal{A}\subseteq \omega _{s}\left(Y, \sigma \right)$. Since $\left(Y, \sigma \right)$ is $\omega _{s}$-compact, then there exists a finite subfamily $\mathcal{B}\subseteq \mathcal{A}$ such that $\mathcal{B}$ is also a cover of $Y$. This shows that $\left(Y, \sigma \right)$ is compact.

The following example will show that the converse of Theorem 5.2 is not true, in general:

Example 5.3. Consider $(\left[0, \infty \right), \sigma)$, where $\sigma = \left\{ \emptyset, \left[0, \infty \right) \right\} \cup \left\{ \left(a, \infty \right) :a\geq 0\right\}$. To see that $(\left[0, \infty \right), \sigma)$ is compact, let $\mathcal{A}$ be a cover of $\left[0, \infty \right)$ with $\mathcal{A}\subseteq \sigma$. Then $\left[0, \infty \right) \in \mathcal{A}$. Choose $\mathcal{B} = \left\{ \left[0, \infty \right) \right\}$. Then $\mathcal{B}$ is a finite subfamily of $\mathcal{A}$ such that $\mathcal{B}$ is also a cover of $\left[0, \infty \right)$. This shows that $(\left[0, \infty \right), \sigma)$ is compact. Let $\mathcal{A} = \left\{ \left(1, \infty \right) \cup \left\{ x\right\} :x\in \left[0, 1\right] \right\}$. Then $\mathcal{A}$ is a cover of $\left[0, \infty \right)$. Since $\left(1, \infty \right) \in \sigma \subseteq \omega _{s}\left(\left[0, \infty \right), \sigma \right)$ and $\overline{ \left(1, \infty \right) }^{\omega } = \overline{\left(1, \infty \right) } = \; \left[0, \infty \right)$, then by Proposition 1.2 (a), we have $\mathcal{A} \subseteq \omega _{s}\left(\left[0, \infty \right), \sigma \right)$. On the other hand, if $\mathcal{B}$ is a finite subfamily of $\mathcal{A}$, then $\mathcal{B}$ is not a cover of $\left[0, \infty \right)$. This shows that $(\left[0, \infty \right), \sigma)$ is not $\omega _{s}$-compact.

Theorem 5.4. Let $\left(Y, \sigma \right)$ be a locally countable topological space. Then $\left(Y, \sigma \right)$ is $\omega _{s}$ -compact if and only if $\left(Y, \sigma \right)$ is compact.

Proof. Necessity. Follows from Theorem 5.2.

Sufficiency. Suppose that $\left(Y, \sigma \right)$ is compact and let $\mathcal{A}$ be a cover of $Y$ such that $\mathcal{A} \subseteq \omega _{s}\left(Y, \sigma \right)$. Since $\left(Y, \sigma \right)$ is locally countable, then by Proposition 1.1 (b), $\mathcal{A} \subseteq \sigma$. Since $\left(Y, \sigma \right)$ is compact, then there exists a finite subfamily $\mathcal{B}\subseteq \mathcal{A}$ such that $\mathcal{B}$ is also a cover of $Y$. This shows that $\left(Y, \sigma \right)$ is $\omega _{s}$-compact.

Theorem 5.5. Every semi-compact topological space is $\omega _{s}$ -compact.

Proof. Let $\left(Y, \sigma \right)$ be semi-compact and let $\mathcal{A}$ be a cover of $Y$ such that $\mathcal{A}\subseteq \omega _{s}\left(Y, \sigma \right)$. Then by Proposition 1.1 (a), $\mathcal{A} \subseteq SO\left(Y, \sigma \right)$. Since $\left(Y, \sigma \right)$ is semi-compact, then there exists a finite subfamily $\mathcal{B}\subseteq \mathcal{A}$ such that $\mathcal{B}$ is also a cover of $Y$. This shows that $\left(Y, \sigma \right)$ is $\omega _{s}$-compact.

The following example will show that the converse of Theorem 5.5 is not true, in general:

Example 5.6. Consider $(\mathbb{N}, \sigma)$, where $\sigma = \left\{ \emptyset, \mathbb{N}, \left\{ 1\right\}, \left\{ 2\right\}, \left\{ 1, 2\right\} \right\}$. Since $\left\{ 1\right\}$, $\left\{ 2\right\}$, and $\left\{ 1, 2\right\}$ are countable sets, then $\overline{\left\{ 1\right\} }^{\omega } = \left\{ 1\right\}$, $\overline{\left\{ 2\right\} }^{\omega }$, and $\overline{ \left\{ 1, 2\right\} }^{\omega } = \; \left\{ 1, 2\right\}$. Thus, $\sigma = \omega _{s}\left(Y, \sigma \right)$. This shows that $(\mathbb{N}, \sigma)$ is $\omega _{s}$-compact. Let $\mathcal{A} = \left\{ \left\{ 2\right\} \right\} \cup \left\{ \left\{ 1, x\right\} :x\in \mathbb{N} -\left\{ 1, 2\right\} \right\}$. Then $\mathcal{A}$ is a cover of $\mathbb{N}$. Since $\overline{\left\{ 1\right\} } = \mathbb{N} -\left\{ 2\right\}$, then $\mathcal{A}\subseteq SO\left(Y, \sigma \right)$. If $\mathcal{B}\ $is a finite subfamily of $\mathcal{A}$, then $\bigcup \mathcal{B}$ is a finite subset of $\mathbb{N}$. This shows that $(\mathbb{N}, \sigma)$ is not semi-compact.

Theorem 5.7. Let $\left(Y, \sigma \right)$ be an anti-locally countable topological space. Then $\left(Y, \sigma \right)$ is semi-compact if and only if $\left(Y, \sigma \right)$ is $\omega _{s}$ -compact.

Proof. Necessity. Follows from Theorem 2.5.

Sufficiency. Suppose that $\left(Y, \sigma \right)$ is $\omega _{s}$-compact and let $\mathcal{A}$ be a cover of $Y$ such that $\mathcal{A}\subseteq SO\left(Y, \sigma \right)$. Since $\left(Y, \sigma \right)$ is anti-locally countable, then by Proposition 1.1 (c), $\mathcal{A }\subseteq \omega _{s}\left(Y, \sigma \right)$. Since $\left(Y, \sigma \right)$ is $\omega _{s}$-compact, then there is a finite subfamily $\mathcal{B}\subseteq \mathcal{A}$ such that $\mathcal{B}$ is also a cover of $Y$. This shows that $\left(Y, \sigma \right)$ is semi-compact.

Theorem 5.8. A topological space $\left(Y, \sigma \right)$ is $\omega _{s}$-compact if and only if every family of $\omega _{s}$ -closed sets which has the finite intersection property must have non-empty intersection.

Proof. Necessity. Suppose that $\left(Y, \sigma \right)$ is $\omega _{s}$-compact, and suppose to the contrary that there exists a family $\mathcal{H}$ of $\omega _{s}$-closed such that $\mathcal{H}$ has the finite intersection property and $\bigcap \mathcal{H} = \emptyset$. Let $\mathcal{A} = \left\{ Y-H:H\in \mathcal{H}\right\}$. Then $\mathcal{A}$ is a cover of $Y$ and $\mathcal{A}\subseteq \omega _{s}\left(Y, \sigma \right)$. Since $\left(Y, \sigma \right)$ is $\omega _{s}$-compact, then there is a finite subfamily $\mathcal{A}_{1}\subseteq \mathcal{A}$ such that $\mathcal{A }_{1}$ is also a cover of $Y$. Let $\mathcal{H}_{1}\mathcal{ = }\left\{ Y-A:A\in \mathcal{A}_{1}\right\}$. Then $\mathcal{H}_{1}$ is a finite subcollection of $\mathcal{H}$ such that

This contradicts the assumption that $\mathcal{H}$ has the finite intersection property.

Sufficiency. Suppose that every family of $\omega _{s}$-closed sets which has the finite intersection property must have non-empty intersection, and suppose to the contrary that $\left(Y, \sigma \right)$ is not $\omega _{s}$-compact. Then there is a cover $\mathcal{A}$ of $Y$ such that $\mathcal{A}\subseteq \omega _{s}\left(Y, \sigma \right)$ and any finite subcollection of $\mathcal{A}$ is not a cover of $Y$. Let $\mathcal{H} = \left\{ Y-A:A\in \mathcal{A}\right\}$. Then $\mathcal{H}$ is a family of $\omega _{s}$-closed sets and $\mathcal{H}$ has the finite intersection property. So, by assumption $\bigcap \mathcal{H} \; \neq \; \emptyset$, and thus $Y-\bigcap \mathcal{H}\neq Y$. But

a contradiction.

Definition 5.9. Let $\left(Y, \sigma \right)$ be a topological space and let $\left(x_{d}\right) _{d\in D}$ be a net in $\left(Y, \sigma \right)$. A point $y\in Y$ is called an $\omega _{s}$ -cluster point of $\left(y_{d}\right) _{d\in D}$ in $\left(Y, \sigma \right)$ if for every $V\in \omega _{s}\left(Y, \sigma \right)$ with $y\in V$ and every $d\in D$, there is $d_{0}\in D$ such that $d\leq d_{0}$ and $y_{d_{0}}\in V$.

Theorem 5.10. A topological space $\left(Y, \sigma \right)$ is $\omega _{s}$-compact if and only if every net in $\left(Y, \sigma \right)$ has an $\omega _{s}$-cluster point.

Proof. Necessity. Suppose that $\left(Y, \sigma \right)$ is $\omega _{s}$-compact and let $\left(y_{d}\right) _{d\in D}$ be a net in $\left(Y, \sigma \right)$. For each $d\in D$, let $T_{d} = \left\{ y_{d^{\prime }}:d^{\prime }\in D\text{ and }d\leq d^{\prime }\right\}$. Let $\mathcal{A} = \left\{ \overline{T_{d}}^{\omega _{s}}:d\in D\right\}$. Then $\mathcal{A}$ is a family of $\omega _{s}$-closed sets.

Claim 1. $\mathcal{A}$ has the finite intersection property.

Proof of Claim 1. Let $d_{1}, d_{2}, ..., d_{n}\in D$. Choose $d_{0}\in D$ such that $d_{i}\leq d_{0}$ for all $i = 1, 2, ..., n$. Then $y_{d_{0}}\in \bigcap\limits_{i = 1}^{n}T_{d_{i}}\subseteq \bigcap\limits_{i = 1}^{n}\overline{T_{d}}_{i}^{\omega _{s}}$. This ends the proof that $\mathcal{A}$ has the finite intersection property.

Since $\left(Y, \sigma \right)$ is $\omega _{s}$-compact, then by Claim 1 and Theorem 5.8, there exists $y\in \bigcap\limits_{d\in D}\overline{T_{d}} ^{\omega _{s}}$.

Claim 2. $y$ is an $\omega _{s}$-cluster point of $\left(y_{d}\right) _{d\in D}$ in $\left(Y, \sigma \right)$.

Proof of Claim 2. Let $V\in \omega _{s}\left(Y, \sigma \right)$ such that $y\in V$, and let $d\in D$. Since $y\in V\cap \overline{T_{d}} ^{\omega _{s}}$, then $V\cap T_{d}\neq \emptyset$, and so there exists $d^{\prime }\in D$ such that $d\leq d^{\prime }$ and $x_{d^{\prime }}\in V$. This shows that $y$ is an $\omega _{s}$-cluster point of $\left(y_{d}\right) _{d\in D}$ in $\left(Y, \sigma \right)$.

Sufficiency. Suppose that every net in $\left(Y, \sigma \right)$ has an $\omega _{s}$-cluster point. We will apply Theorem 5.8. Let $\mathcal{A}$ be a family of $\omega _{s}$-closed sets which has the finite intersection property. Let $D$ be the family of all finite intersections of members of $\mathcal{A}$. Define the relation $\leq$ on $D$ as follows:

Then $\left(D, \leq \right)$ is a directed set. For every $d\in D$, choose $y_{d}\in d$. By assumption, there is an $\omega _{s}$-cluster point $y$ of $\left(y_{d}\right) _{d\in D}$.

Claim 3. $y\in \overline{A}^{\omega _{s}}$ for all $A\in \mathcal{A}$, and hence $y\in \bigcap\limits_{A\in \mathcal{A}}\overline{A}^{\omega _{s}} = \bigcap\limits_{A\in \mathcal{A}}A$.

Proof of Claim 3. Let $A\in \mathcal{A}$ and $V\in \omega _{s}\left(Y, \sigma \right)$ with $y\in V$. Let $d = A$, then $d\in D$. Since $y$ is an $\omega _{s}$-cluster point of $\left(y_{d}\right) _{d\in D}$, then there is $d^{\prime }\in D$ such that $d\leq d^{\prime }$ and $y_{d^{\prime }}\in V$, say $d^{\prime } = F$. Then $F\subseteq A$, and hence $y_{d^{\prime }}\in V\cap A$. Therefore, $y\in \overline{A}^{\omega _{s}}$.

Theorem 5.11. Let $g:\left(Y, \sigma \right) \longrightarrow (Z, \gamma)$ be $\omega _{s}$-irresolute and surjective. If $\left(Y, \sigma \right)$ is $\omega _{s}$-compact, then $(Z, \gamma)$ is $\omega _{s}$-compact.

Proof. Suppose that $\left(Y, \sigma \right)$ is $\omega _{s}$-compact and let $\mathcal{H}$ be a cover of $Z$ such that $\mathcal{H} \subseteq \omega _{s}(Z, \gamma)$. Let $\mathcal{M} = \left\{ g^{-1}\left(H\right) :H\in \mathcal{H}\right\}$. Then $\mathcal{M}$ is a cover of $Y$. Also, by $\omega _{s}$-irresoluteness of $g$, we have $\mathcal{M}\subseteq \omega _{s}\left(Y, \sigma \right)$. Since $\left(Y, \sigma \right)$ is $\omega _{s}$-compact, then there exist $H_{1}, H_{2}, ..., H_{n}\in \mathcal{H}$ such that $\bigcup\limits_{i = 1}^{n}g^{-1}\left(H_{i}\right) = Y$, and so $\bigcup\limits_{i = 1}^{n}H_{i}\subseteq g\left(g^{-1}\left(\bigcup\limits_{i = 1}^{n}H_{i}\right) \right) = g\left(Y\right)$. Since $g$ is surjective, then $g\left(Y\right) = Z$. Therefore, $Z = \bigcup \limits_{i = 1}^{n}H_{i}$. Hence, $(Z, \gamma)$ is $\omega _{s}$-compact.

6.   Conclusions

In this paper, we introduce $\omega _{s}$-irresoluteness, $\omega _{s}$ -openness, pre-$\omega _{s}$-openness, and slight $\omega _{s}$-continuity as new classes of functions. And, we define $\omega _{s}$-compactness as a new class of topological spaces which lies between the classes compactness and semi-compactness. Several implications, examples, counter-examples, characterizations, and mapping theorems are introduced. The following topics could be considered in future studies: (1) To define $\omega _{s}$-open separation axioms; (2) To define $\omega _{s}$-connectedness; (3) To improve some known topological results.

Conflict of interest

We declare no conflicts of interest in this paper.