On the compactness via primal topological spaces

1.   Introduction

Numerous topologies of significant applications have been characterized through the incorporation of some mathematical structures. For instance, Choquet developed the concept of a grill structure with topological spaces in ^[1]. Moreover, several topological concepts were presented, such as the ideal ^[2,3] and the filter ^[4]. The concept of primal topological space $\mathcal{PS}$ was introduced by S. Acharjee et al. in ^[5]. Then, several papers discussed the topological properties in $\mathcal{PS}$, such as ^[6], which presented definitions of $\mathfrak{P}$-regularity, $\mathfrak{P}$-Hausdorff, and $\mathfrak{P}$-normality. Additionally, Al-Omari and Alqahtani provided definitions of new closure operators using a primal structure in ^[7]. Then, Alghamdi et al. introduced novel operators by leveraging the primal structure in ^[8]. Additional primal operators were defined in ^[9]. Moreover, Al-Saadi and Al-Malki discussed various categories of open sets within the framework of generalized topological spaces, thereby utilizing the primal structure ^[10]. In this paper, we introduce some properties concerning compactness in $\mathcal{PS}$. These properties are named $\mathfrak{P}$-compactness, strongly $\mathfrak{P}$-compactness, and super $\mathfrak{P}$-compactness. We provide some results and examples which connect these concepts together. Throughout this paper, $(\mathcal{T}, \mu, \mathfrak{P})$ represents a primal topological space $\mathcal{PS}$ such that $\mu$ is a topology on $\mathcal{T}$. Moreover, we use the symbol $\mathcal{CL}(A)$ for the closure of a set $A\subset \mathcal{T}$ and $\mathfrak{H}$ for an index set. Furthermore, we use the symbol $2^ \mathcal{T}$ for the power set of the set $\mathcal{T}$.

Definition 1.1. $($^[5]$)$ For a nonempty set $\mathcal{T}$, we define a primal collection $\mathfrak{P}\subseteq 2^ \mathcal{T}$ on $\mathcal{T}$ as follows:

(1) $\mathcal{T} \notin \mathfrak{P}$,

(2) if $R\in \mathfrak{P}$ and $T\subseteq R$, then $T\in \mathfrak{P}$,

(3) if $R\cap T\in \mathfrak{P}$, then either $R\in \mathfrak{P}$ or $T\in \mathfrak{P}$.

Corollary 1.1. $($^[5]$)$ If $\mathcal{T}\not = \emptyset$, then $\mathfrak{P}\subseteq 2^\mathcal{T}$ is a primal collection on $\mathcal{T}$ if and only if:

(1) $\mathcal{T} \notin \mathfrak{P}$,

(2) if $T\notin \mathfrak{P}$ and $T\subseteq R$, then $R\notin \mathfrak{P}$,

(3) if $R\notin \mathfrak{P}$ and $T\notin \mathfrak{P},$ then $R\cap T\notin \mathfrak{P}.$

Definition 1.2. $($^[5]$)$ A topological space $(\mathcal{T}, \nu)$ with a primal collection $\mathfrak{P}$ on $\mathcal{T}$ is called a primal topological space $\mathcal{PS}$ and is denoted by $(\mathcal{T}, \nu, \mathfrak{P}).$

2.   $\mathfrak{P}$-compact spaces

Definition 2.1. Let $(\mathcal{T}, \rho, \mathfrak{P})$ be a $\mathcal{PS}$. We say that $(\mathcal{T}, \rho, \mathfrak{P})$ is a primal compact space ($\mathfrak{P}$-compact space) if for every open cover $\{V_\eta \}_{\eta\in \mathfrak{H}}$ of $\mathcal{T}$, there exists a finite set $\mathfrak{H}_0\subseteq \mathfrak{H}$ with $\bigcup_{\eta\in \mathfrak{H}_0} V_\eta\notin \mathfrak{P}$. Let $N\subseteq \mathcal{T}$. Then, $N$ is called a $\mathfrak{P}$-compact subspace of $\mathcal{T}$ if for every open cover $\{W_\eta \}_{\eta\in \mathfrak{H}}$ of $N$, there exists a finite set $\mathfrak{H}_0\subseteq \mathfrak{H}$ such that $\mathcal{T}\setminus [N\setminus\bigcup_{\eta\in \mathfrak{H}_0} W_\eta]\notin \mathfrak{P}$.

Theorem 2.1. Let $(\mathcal{T}, \rho, \mathfrak{P})$ be a $\mathcal{PS}$ and $B\subseteq \mathcal{T}$. If $B$ is a compact subspace of $\mathcal{T}$, then $B$ is a $\mathfrak{P}$-compact subspace of $\mathcal{T}$.

Proof. Let $\{V_\eta\}_{\eta\in \mathfrak{H}}$ be an open cover of $B$. Then, since $B$ is a compact subspace of $\mathcal{T}$, there exists a finite set $\mathfrak{H}_0\subseteq \mathfrak{H}$ such that $B\subseteq\bigcup_{\eta\in \mathfrak{H}_0}V_\eta.$ Hence,

Therefore, $B$ is a $\mathfrak{P}$-compact subspace of $\mathcal{T}$. □

The converse of Theorem 2.1 is not necessarily true as considered in the following example.

Example 2.1. Let $(\mathbb{R}, \tau_1, \mathfrak{P}_1)$ be defined as follows: $U\in\tau_1$ if and only if either $U = \emptyset$ or $1\in U$, see Example 10 in ^[11]. Let $\mathfrak{P}_1$ be defined on $\mathbb{R}$ as follows: $U\in \mathfrak{P}_1$ if and only if $1\notin U$. Then, $(\mathbb{R}, \tau_1, \mathfrak{P}_1)$ is a $\mathcal{PS}$. Let $\mathbb{N}$ be the set of natural numbers and let $\{V_\eta\}_{\eta\in \mathfrak{H}}$ be any open cover of $\mathbb{N}$ such that $V_\eta\not = \emptyset$ for every $\eta\in \mathfrak{H}$. Let $\mathfrak{H}_0 = \{V_i\}_{i = 1}^{n}\subseteq \{V_\eta\}_{\eta\in \mathfrak{H}}$. Then, $1\in\mathbb{R}\setminus [\mathbb{N}\setminus\bigcup_{i = 1}^{n} V_i]$, which means that $\mathbb{R}\setminus [\mathbb{N}\setminus\bigcup_{i = 1}^{n} V_i]\notin \mathfrak{P}_1.$ Hence, $\mathbb{N}$ is a $\mathfrak{P}$-compact subspace of $\mathcal{T}$. Note that $\mathbb{N}$ is not compact. Indeed, $\{j, 1\}_{j\in\mathbb{N}}$ is an open cover of $\mathbb{N}$, which has no finite subcover.

Example 2.2. Let $(\mathbb{R}, \mathcal{D}, \mathfrak{P})$ be a $\mathcal{PS}$ defined as follows: $U\in \mathfrak{P}$ if and only if $\mathbb{R}\setminus U$ is an infinite subset of $\mathbb{R}$. Moreover, $V\in\mathcal{D}$ if and only if $V\subseteq \mathbb{R}$ (the discrete topological space on $\mathbb{R},$ see Example 3 in ^[11]). Then, $\Lambda = \{r\}_{r\in\mathbb{R}}$ is an open cover of $\mathbb{R}$. If $\{V_1, V_2, ..., V_n\}$ is an arbitrary finite subfamily of $\Lambda$, then $\bigcup_{i = 1}^{n} V_i = \{r_1, ..., r_n\}\in \mathfrak{P}.$ Thus, $\mathbb{R}$ is not a $\mathfrak{P}$-compact space.

Theorem 2.2. $\mathfrak{P}$-compactness is hereditarily defined with respect to closed subspaces.

Proof. Assume that $(\mathcal{T}, \rho, \mathfrak{P})$ is a $\mathfrak{P}$-compact space and $M\subseteq \mathcal{T}$ is any closed subspace. Suppose that $\mathcal{Q} = \{V_\eta\}_{\eta\in \mathfrak{H}}$ is an open cover of $M$. Then, $\{V_\eta\}_{\eta\in \mathfrak{H}} \bigcup (\mathcal{T}\setminus M)$ is an open cover of $\mathcal{T}$. Hence, there exists a finite set $\mathfrak{H}_0 = \{V_1, V_2, ..., V_n\}\subseteq \{\mathcal{T}\setminus M\}\bigcup \{V_\eta:\eta\in \mathfrak{H}\}$ such that $\bigcup_{i = 1}^{n} V_i\notin \mathfrak{P}.$ Thus, $\mathcal{T}\setminus \left[M\setminus\bigcup_{i = 1}^{n} V_i\right]\notin \mathfrak{P}$, which implies that $M$ is a $\mathfrak{P}$-compact subspace of $\mathcal{T}$. □

The following example shows that if the subspace of $\mathcal{T}$ is not closed, then it may not be a $\mathfrak{P}$-compact subspace.

Example 2.3. Let $(\mathbb{R}, \mathcal{F}, \mathfrak{P})$ be a $\mathcal{PS}$ defined as follows:

$U\in \mathcal{F}$ if and only if either $\sqrt{2}\in\mathbb{R}\setminus U$ or $\mathbb{R}\setminus U$ is a finite subset of $\mathbb{R}$, see Example 24 in ^[11].

Let $\mathfrak{P}$ be defined as in Example 2.2. Let $\mathcal{Q} = \{O_\eta\}_{\eta\in \mathfrak{H}}$ be an open cover of $\mathbb{R}$. Then, there exists $\lambda\in \mathfrak{H}$ such that $\sqrt{2}\in O_\lambda$. Hence, $\mathbb{R}\setminus O_\lambda$ is a finite subset of $\mathbb{R}$. Let $\mathcal{Q}_0 = \{O_\lambda\}\subseteq \mathcal{Q}$. Then, since $O_\lambda\notin \mathfrak{P}$, $\mathbb{R}$ is a $\mathfrak{P}$-compact space. Now, consider the subspace $\mathbb{R}\setminus\{\sqrt{2}\}$. Claim that $\mathbb{R}\setminus\{\sqrt{2}\}$ is not a $\mathfrak{P}$-compact subspace. Indeed, if $\mathcal{Q}_0$ is any finite subfamily of $\mathcal{Q} = \{t\}_{t\in\mathbb{R}\setminus \{\sqrt{2}\}}$, then $\bigcup_{O\in\mathcal{Q}_0}O\in \mathfrak{P}$. Observe that $\mathbb{R}\setminus\{\sqrt{2}\}$ is a discrete subspace of $\mathbb{R}$ that is not closed.

Theorem 2.3. Let $(\mathcal{T}, \nu, \mathfrak{P})$ be a $\mathcal{PS}$. For a subset $K$ of $\mathcal{T}$, the following properties are equivalent:

(1) $K$ is a $\mathfrak{P}$-compact subspace; and

(2) for every family $\{ L_\delta \}_{\delta\in \mathfrak{H}}$ of closed sets such that $K \cap \left(\bigcap_{\delta\in \mathfrak{H}} L_\delta\right) = \emptyset$, there exists a finite subset $\mathfrak{H}_0$ of $\mathfrak{H}$ such that

Proof. (1) $\Rightarrow$ (2): Let $\{ L_\delta \}_{\delta \in \mathfrak{H}}$ be a collection of closed sets in $\mathcal{T}$ such that $K \cap \left(\bigcap_{ \delta \in \mathfrak{H}} L_\delta \right) = \emptyset$. Then, we have the following:

Since $\mathcal{T} \setminus L_\delta$ is open for each $\delta \in \mathfrak{H}$ and $K$ is a $\mathfrak{P}$-compact subspace of $\mathcal{T}$, then there exists a finite subset $\mathfrak{H}_0$ of $\mathfrak{H}$ such that

Now, we have the following:

(2) $\Rightarrow$ (1): Let $\{ V_\delta \}_{\delta \in \mathfrak{H}}$ be any cover of $K$ which consists of open sets in $\mathcal{T}$. Then, $K \cap \left(\mathcal{T} \setminus \bigcup_{\delta \in \mathfrak{H}} V_\delta \right) = K\cap \left[\bigcap_{\delta \in \mathfrak{H}} \left(\mathcal{T} \setminus V_\delta\right) \right] = \emptyset$.

Since $\mathcal{T} \setminus V_\delta$ is closed for each $\delta \in \mathfrak{H}$, then by (2), there exists a finite subset $\mathfrak{H}_0$ of $\mathfrak{H}$ such that

Therefore, we have the following:

Hence, $K$ is a $\mathfrak{P}$-compact subspace of $\mathcal{T}$. □

Corollary 2.1. If $(\mathcal{T}, \nu, \mathfrak{P})$ is a $\mathcal{PS}$ and $\{ L_\delta\}_{\delta \in \mathfrak{H}}$ is a family of closed sets in $\mathcal{T}$ such that $\bigcap_{\delta \in \mathfrak{H}} L_\delta = \emptyset$, then $(\mathcal{T}, \nu, \mathfrak{P})$ is a $\mathfrak{P}$-compact space if and only if there exists a finite set $\mathfrak{H}_0\subseteq \mathfrak{H}$ such that $\bigcup_{\delta \in \mathfrak{H}_0 } \left(\mathcal{T}\setminus L_\delta\right) \notin \mathfrak{P}$.

Theorem 2.4. Let $(\mathcal{T}, \rho, \mathfrak{P})$ be a $\mathcal{PS}$. If $R, T \subseteq \mathcal{T}$ are both $\mathfrak{P}$-compact subspaces of $\mathcal{T},$ then $R \cup T$ is a $\mathfrak{P}$-compact subspace of $\mathcal{T}$.

Proof. Let $\{O_\delta \}_{\delta\in \mathfrak{H}}$ be an open cover of $R\cup T$. Since both $R$ and $T$ are $\mathfrak{P}$-compact subspaces of $\mathcal{T}$, then there are two finite subsets of $\mathfrak{H}$, namely $\mathfrak{H}_0$ and $\mathfrak{H}_1$, such that $\mathcal{T}\setminus \left(R\setminus \bigcup_{\delta\in \mathfrak{H}_0} O_\delta\right)\notin \mathfrak{P}$ and $\mathcal{T}\setminus \left(T\setminus \bigcup_{\delta\in \mathfrak{H}_1} O_\delta\right)\notin \mathfrak{P}$. Hence, $\mathcal{T}\setminus \left[(R\cup T) \setminus \bigcup_{\delta\in \mathfrak{H}_0\cup \mathfrak{H}_1} O_\delta\right]\notin \mathfrak{P}$. Thus, $R\cup T$ is a $\mathfrak{P}$-compact subspace of $\mathcal{T}$. □

Theorem 2.5. Let $(\mathcal{T}, \nu, \mathfrak{P})$ be a $\mathcal{PS}$ and let $R, S$ be any subsets of $\mathcal{T}$. If $R$ is a $\mathfrak{P}$-compact subspace of $\mathcal{T}$ and $S$ is a closed set, then $R \cap S$ is a $\mathfrak{P}$-compact subspace of $\mathcal{T}$.

Proof. Let $\{O_\delta\}_{\delta\in \mathfrak{H}}$ be an open cover of $R\cap S$. Then, $\mathcal{Q} = \{O_\delta\}_{\delta\in \mathfrak{H}} \cup (\mathcal{T}\setminus S)$ is an open cover of $R$. Hence, there exists a finite subset of $\mathcal{Q}$, namely $\mathcal{Q}_0$, such that $\mathcal{T}\setminus \left[R\setminus \left(\bigcup_{O\in\mathcal{Q}_0} O \right)\right]\notin \mathfrak{P}$. Since $\mathcal{T}\setminus \left[R\setminus \left(\bigcup_{O\in\mathcal{Q}_0} O \right)\right]\subseteq \mathcal{T}\setminus \left[(R\cap T)\setminus \left(\bigcup_{O\in\mathcal{Q}_0} O \right)\right]$, then $\mathcal{T}\setminus \left[(R\cap T)\setminus \left(\bigcup_{O\in\mathcal{Q}_0} O \right)\right]\notin \mathfrak{P}$, which implies that $R\cap T$ is a $\mathfrak{P}$-compact subspace of $\mathcal{T}$. □

Lemma 2.1. Let $f : (\mathcal{T}, \rho) \rightarrow (\mathcal{Y}, \nu)$ be a function. Then, the following properties hold:

(1) If $f$ is a bijective function and $\mathfrak{P}$ is a primal collection on $\mathcal{T}$, then $f(\mathfrak{P}) = \{ f(V) : V \in \mathfrak{P} \}$ is a primal collection on $\mathcal{Y}$; and

(2) If $f$ is a bijective function and $\mathcal{J}$ is a primal collection on $\mathcal{Y}$, then $f^{-1}(\mathcal{J}) = \{ f^{-1}(B) : B \in \mathcal{J} \}$ is a primal collection on $\mathcal{T}$.

Proof. (1) Since $f$ is surjective, then $f(\mathcal{T}) = \mathcal{Y}\notin f(\mathfrak{P})$. Let $W\in f(\mathfrak{P})$ and let $Q\subseteq W$. Since $W\in f(\mathfrak{P})$, then $\exists M\in \mathfrak{P}$ such that $W = f(M)\Rightarrow f^{-1} (W) = M$. Hence, $f^{-1}(Q)\subseteq f^{-1}(W)$; then, $f^{-1}(Q)\in \mathfrak{P}$, which implies that $Q\in f(\mathfrak{P}).$ Now, let $W\cap Q\in f(\mathfrak{P})$. Then, there exists $R\in \mathfrak{P}$ such that $W\cap Q = f(R)$. Thus, $f^{-1}(W\cap Q) = f^{-1}(W)\cap f^{-1}(Q) = R$. Hence, either $f^{-1}(W)\in \mathfrak{P}$ or $f^{-1}(Q)\in \mathfrak{P}$. Then, either $W\in f(\mathfrak{P})$ or $Q\in f(\mathfrak{P})$. Therefore, $f(\mathfrak{P})$ is a primal collection on $\mathcal{Y}$.

(2) We know that $f^{-1} (\mathcal{Y}) = \mathcal{T}$; since $\mathcal{Y}\notin \mathcal{J}$, then $f^{-1}(\mathcal{Y}) = \mathcal{T}\notin f^{-1}(\mathcal{J})$. Let $A\in f^{-1}(\mathcal{J})$ and let $B\subseteq A$. Then, $\exists C\in \mathcal{J}$ such that $A = f^{-1}(C)$. Hence, $f(A) = f(f^{-1}(C)) = C$. As $f(B)\subseteq f(A) = C$, then $f(B)\in \mathcal{J}$, which implies that $B\in f^{-1}(\mathcal{J}).$ Now, suppose that $A\cap C\in f^{-1}(\mathcal{J})$. Then, $\exists R\in \mathcal{J}$ such that $A\cap C = f^{-1}(R)$. Then, $f(A\cap C) = f(f^{-1}(R)) = R$. Thus, $f(A)\cap f(C) = R\in \mathcal{J}$ implies that either $f(A)\in \mathcal{J}$ or $f(C)\in \mathcal{J}$. Therefore, either $A\in f^{-1}(\mathcal{J})$ or $C\in f^{-1}(\mathcal{J})$. □

Lemma 2.2. Let $(\mathcal{T}, \rho, \mathfrak{P})$ be a $\mathcal{PS}$. If $f : (\mathcal{T}, \rho, \mathfrak{P}) \rightarrow (\mathcal{Y}, \nu)$ is a function and $\mathcal{J}_ \mathfrak{P} = \{ B \subset \mathcal{Y}: f^{-1}(B) \in \mathfrak{P}\}$, then the following hold:

(1) $\mathcal{J}_ \mathfrak{P}$ is a primal collection on $\mathcal{Y}$;

(2) if f is injective, then $\mathfrak{P} \subseteq f^{-1}(\mathcal{J}_ \mathfrak{P})$;

(3) if f is surjective, then $\mathcal{J}_ \mathfrak{P} \subseteq f(\mathfrak{P})$; and

(4) if f is bijective, then $\mathcal{J}_ \mathfrak{P} = f(\mathfrak{P})$.

Proof. (1) We know that $f^{-1}(\mathcal{Y}) = \mathcal{T}\notin \mathfrak{P}$. Then, $\mathcal{Y}\notin \mathcal{J}_ \mathfrak{P}$. Let $A\in \mathcal{J}_ \mathfrak{P}$ and let $B\subseteq A$. Then, $A\subset \mathcal{Y}$ and $f^{-1}(A)\in \mathfrak{P}$. Since $f^{-1}(B)\subseteq f^{-1} (A)$, then $f^{-1}(B)\in \mathfrak{P}$; hence $B\in \mathcal{J}_ \mathfrak{P}$. Now, suppose that $A\cap B\in \mathcal{J}_ \mathfrak{P}.$ Then, $f^{-1}(A\cap B)\in \mathfrak{P}$, which implies that $f^{-1}(A)\cap f^{-1}(B)\in \mathfrak{P}$. Hence, either $f^{-1}(A)\in \mathfrak{P}$ or $f^{-1}(B)\in \mathfrak{P}$. Therefore, either $A\in \mathcal{J}_ \mathfrak{P}$ or $B\in \mathcal{J}_ \mathfrak{P}$.

(2) Let $A\in \mathfrak{P}$ and suppose that $f$ is an injective function. Then, $f(A)\subset Y$ and $f^{-1}(f(A)) = A\in \mathfrak{P}.$ Hence, $f(A)\in \mathcal{J}_ \mathfrak{P}$, which implies that $A\in f^{-1}(\mathcal{J}_ \mathfrak{P})$. Then, $\mathfrak{P}\subseteq f^{-1}(\mathcal{J}_ \mathfrak{P})$.

(3) Suppose that $A\in \mathcal{J}_ \mathfrak{P}$. Then, $f^{-1}(A)\in \mathfrak{P}$; hence, $f(f^{-1}(A)) = A\in f(\mathfrak{P}).$

(4) From (2) and (3), we have $\mathcal{J}_ \mathfrak{P} = f(\mathfrak{P}).$ □

Theorem 2.6. If $f : (\mathcal{T}, \Gamma, \mathfrak{P}) \rightarrow (L, \nu, f(\mathfrak{P}))$ is a surjective continuous function and $W$ is a $\mathfrak{P}$-compact subspace of $\mathcal{T},$ then $f(W)$ is a $\mathfrak{P}$-compact subspace of $L$.

Proof. Let $\{O_\delta\}_{\delta\in \mathfrak{H}}$ be an open cover of $f(W)$. Since $f$ is a continuous function, then $\{f^{-1}(O_\delta) \}_{\delta\in \mathfrak{H}}$ is an open cover of $f^{-1} (f(W)).$ As $W\subseteq f^{-1} (f(W))$, then $\{f^{-1}(O_\delta) \}_{\delta\in \mathfrak{H}}$ is an open cover of $W$. Since $W$ is a $\mathfrak{P}$-compact space, then there exists a finite set $\mathfrak{H}_0\subseteq \mathfrak{H}$ such that $\mathcal{T}\setminus \left[W \setminus \bigcup_{\delta\in \mathfrak{H}_0} f^{-1}(O_\delta)\right]\notin \mathfrak{P}$. Then, $f(\mathcal{T})\setminus \left[f(W) \setminus f(f^{-1} (\bigcup_{\delta\in \mathfrak{H}_0} O_\delta))\right]\notin f(\mathfrak{P}).$ Hence, $L\setminus \left[f(W) \setminus \bigcup_{\delta\in \mathfrak{H}_0}O_\delta\right]\notin f(\mathfrak{P})$, since $f$ is a surjective function. Then, $f(W)$ is a $\mathfrak{P}$-compact subspace of $L$. □

Corollary 2.2. If $f : (\mathcal{T}, \Gamma, \mathfrak{P}) \rightarrow (L, \nu, f(\mathfrak{P}))$ is a surjective continuous function and $(\mathcal{T}, \rho, \mathfrak{P})$ is a $\mathfrak{P}$-compact space, then $(L, \nu, f(\mathfrak{P}))$ is a $\mathfrak{P}$-compact space.

Definition 2.2. Let $(\mathcal{T}, \rho, \mathfrak{P})$ be a $\mathcal{PS}$. A subset $A$ of $\mathcal{T}$ is said to be as follows:

(1) $\mathfrak{P} g$-closed if $\mathcal{CL}(A) \subseteq U$ whenever $\mathcal{T}\setminus (A \setminus U) = (\mathcal{T}\setminus A)\cup U \notin \mathfrak{P}$ and $U$ is open; and

(2) $g$-closed if $\mathcal{CL}(A) \subseteq U$ whenever $A \subset U$ and $U$ is open.

From the definition above, we have the following remark.

Remark 2.1.

(1) Every closed set is a $g$-closed set, but the converse is not true in general.

(2) The concept of $\mathfrak{P} g$-closed depends on the definition of the primal space.

To illustrate Remark 2.1, we present the following examples.

Example 2.4. Let $\mathcal{T} = \{r, d, b\}$ and let $\rho = \{ \mathcal{T}, \emptyset, \{r\}\}$. Consider the set $H = \{d\}$. Then, $H\subseteq U\in\rho$ if and only if $U = \mathcal{T}$; hence, $H$ is $g$-closed but it is not a closed set since $\mathcal{CL}(H) = \{d, b\}\not = H$.

Example 2.5. Let $(\mathcal{T}, \rho)$ and $H$ be defined as in Example 2.4. If $\mathfrak{P} = \{\emptyset\}$, then $H$ is not a $\mathfrak{P} g$-closed since $\mathcal{CL}(H)\not\subseteq \{r\}$, although $(\mathcal{T}\setminus H) \cup \{r\} = \{r, b\}\notin \mathfrak{P}$.

Now, let $\mathfrak{P} = 2^ \mathcal{T}\setminus\{ \mathcal{T}\}$. Then, $H$ is $\mathfrak{P} g$-closed since $(\mathcal{T}\setminus H)\cup U\notin \mathfrak{P}$ if and only if $U = \mathcal{T}$.

Theorem 2.7. Let $(\mathcal{T}, \rho, \mathfrak{P})$ be a $\mathcal{PS}$ and let $A, B$ be subsets of $\mathcal{T}$ such that $A \subseteq B \subseteq \mathcal{CL}(A)$. Then, the following properties hold:

(1) If $A$ is a $\mathfrak{P}$-compact subspace of $\mathcal{T}$ and $\mathfrak{P} g$-closed, then $B$ is a compact subspace of $\mathcal{T}$; and

(2) If $B$ is a $\mathfrak{P}$-compact subspace of $\mathcal{T}$ and $A$ is $g$-closed, then $A$ is a $\mathfrak{P}$-compact subspace of $\mathcal{T}$.

Proof. (1) Suppose that $A$ is a $\mathfrak{P}$-compact subspace of $\mathcal{T}$ and $\mathfrak{P} g$-closed. Let $\{ O_\delta \}_{\delta\in \mathfrak{H}}$ be any open cover of $B$. Then, $\{ O_\delta \}_{\delta\in \mathfrak{H}}$ is an open cover of $A$. Since $A$ is a $\mathfrak{P}$-compact subspace of $\mathcal{T},$ then there exists a finite set $\mathfrak{H}_0\subseteq \mathfrak{H}$ such that $\mathcal{T}\setminus \left[A \setminus \bigcup_{\delta\in \mathfrak{H}_0} O_\delta \right] \notin \mathfrak{P}$. Since $A$ is $\mathfrak{P} g$-closed, then $\mathcal{CL}(A) \subseteq \bigcup_{\delta \in \mathfrak{H}_0} O_\delta$. Then, $B \subseteq \bigcup_{\delta \in \mathfrak{H}_0} O_\delta$. Therefore, $B$ is a compact subspace of $\mathcal{T}$.

(2) Suppose that $B$ is a $\mathfrak{P}$-compact subspace of $\mathcal{T}$ and $A$ is $g$-closed. Let $\{ O_\delta \}_{\delta\in \mathfrak{H}}$ be any open cover of $A$. Now, since $B\subseteq \mathcal{CL}(A)$ and $A$ is a $g$-closed, then $B\subseteq \mathcal{CL}(A) \subseteq \bigcup_{\delta\in \mathfrak{H}} O_\delta$. Hence, there exists a finite set $\mathfrak{H}_0\subseteq \mathfrak{H}$ such that $\mathcal{T}\setminus \left[B \setminus \bigcup_{\delta\in \mathfrak{H}_0} O_\delta \right] \notin \mathfrak{P}$ because $B$ is a $\mathfrak{P}$-compact subspace of $\mathcal{T}$. Then, $\mathcal{T}\setminus \left[A \setminus \bigcup_{\delta\in \mathfrak{H}_0} O_\delta \right] \notin \mathfrak{P}$ since $A\subseteq B$. Therefore, $A$ is a $\mathfrak{P}$-compact subspace of $\mathcal{T}$. □

Corollary 2.3. Let $(\mathcal{T}, \rho, \mathfrak{P})$ be a $\mathcal{PS}$. If $A$ is $\mathfrak{P} g$-closed and $A \subseteq B \subseteq \mathcal{CL} (A)$, then $A$ is a $\mathfrak{P}$-compact subspace of $\mathcal{T} \Leftrightarrow B$ is a $\mathfrak{P}$-compact subspace of $\mathcal{T}$.

3.   Strongly $\mathfrak{P}$-compact spaces

Definition 3.1. Let $(\mathcal{T}, \rho, \mathfrak{P})$ be a $\mathcal{PS}$. We say that $\mathcal{T}$ is a strongly $\mathfrak{P}$-compact space (S$\mathfrak{P}$-compact space) if for every family of open sets $\{O_\delta\}_{\delta\in \mathfrak{H}}$ such that $\bigcup_{\delta\in \mathfrak{H}} O_{\delta}\notin \mathfrak{P}$, then there exists a finite set $\mathfrak{H}_0\subseteq \mathfrak{H}$ such that $\bigcup_{\delta\in \mathfrak{H}_0} O_{\delta}\notin \mathfrak{P}$. A subset $K$ of $\mathcal{T}$ is said to be an S$\mathfrak{P}$-compact subspace of $\mathcal{T}$ if for every family $\{O_\delta\}_{\delta\in \mathfrak{H}}$ of open sets of $\mathcal{T}$ such that $\mathcal{T}\setminus \left[K \setminus \bigcup_{\delta \in \mathfrak{H} } O_\delta \right] \notin \mathfrak{P}$, then there exists a finite set $\mathfrak{H}_0 \subseteq \mathfrak{H}$ such that $\mathcal{T}\setminus \left[K \setminus \bigcup_{\delta \in \mathfrak{H}_0 } O_\delta \right] \notin \mathfrak{P}$.

Example 3.1. Let $(\mathbb{R}, \tau_1, \mathfrak{P}_1)$ be a $\mathcal{PS}$ defined in Example 2.1. Let $\{O_\delta\}_{\delta\in \mathfrak{H}}$ be any family of open sets. Then,

Case 1. $O_\delta = \emptyset$ for every $\delta\in \mathfrak{H}$. Then, since $\mathbb{R}\setminus\left[\mathbb{N}\setminus\bigcup_{\delta\in \mathfrak{H}}O_\delta\right]\in \mathfrak{P}_1$, there is nothing to prove.

Case 2. $\exists\; \lambda\in \mathfrak{H}$ such that $O_\lambda\not = \emptyset$. Then, $\mathbb{R}\setminus\left[\mathbb{N}\setminus\bigcup_{\delta\in \mathfrak{H}}O_\delta\right]\notin \mathfrak{P}_1$. Pick a finite set $\mathfrak{H}_0\subseteq \mathfrak{H}$ such that $\lambda\in \mathfrak{H}_0$. Hence, $\mathbb{R}\setminus \left[ \mathbb{N}\setminus\bigcup_{\delta\in \mathfrak{H}_0} O_\delta\right]\notin \mathfrak{P}_1$. Thus, $\mathbb{N}$ is an S$\mathfrak{P}$-compact subspace of $\mathbb{R}$.

From the definition, it is clear that every S$\mathfrak{P}$-compact is a $\mathfrak{P}$-compact subspace of $\mathcal{T}$. However, this relation is not reversible, which is proven in next example.

Example 3.2. Let $(\mathbb{R}, \mathcal{F}, \mathfrak{P})$ be as defined in Example 2.3. Consider the family $\mathcal{M} = \{\{x\} : x\in\mathbb{R}\; \; \text{and}\; \; x\not = \sqrt{2}\}$. Then, $\bigcup_{x\in \mathbb{R}\setminus \{\sqrt{2}\}} \{x\} = \mathbb{R}\setminus\{\sqrt{2}\}\notin \mathfrak{P}.$ Now, let $\{M_i : i\in\{1, ..., n\}\}$ be an arbitrary finite subfamily of $\mathcal{M}$. Then, $\bigcup_{i = 1}^{n} M_i\in \mathfrak{P}$. Hence, $\mathbb{R}$ is not an S$\mathfrak{P}$-compact space. Observe that $\mathbb{R}$ is a $\mathfrak{P}$-compact space.

Example 3.3. Let $H = \mathbb{R\times (R^{+}}\cup\{0\})$. For $(n, m)\in H$ and $r > 0$. Define the set $M_{r}(n, m)$ as follows:

Let $\mathfrak{B} = \{M_{r}(n, m)\}$ be a base for the topology $\mu$ on the set $H$. Then, $(H, \mu, \mathfrak{P})$, where $\mathfrak{P} = \{\emptyset\}$ is a $\mathcal{PS}$. Hence,

(1) $(H, \mu, \mathfrak{P})$ is not a compact subspace of $H$. To show that, consider the family $\mathcal{Q} = \{M_{1}(n, 0)\}\cup \{M_{1}(n, m): m\geq 1\}$. Then, $\mathcal{Q}$ is an open cover of $H$. Since $(t, 0)\notin \{M_{1}(n, m): m\geq 1\}$ and $(t, 0)\in \{M_{1}(n, 0)\}$ if and only if $n = t$, then the above open cover has no finite subcover. Thus, $H$ is not compact.

(2) $(H, \mu, \mathfrak{P})$ is an S$\mathfrak{P}$-compact subspace of $H$ since $\mathfrak{P} = \emptyset$.

Theorem 3.1. Let $(\mathcal{T}, \Gamma, \mathfrak{P})$ be a $\mathcal{PS}$ and let $K\subseteq\mathcal{T}$. Consider the family of closed sets $\{C_\delta\}_{\delta\in \mathfrak{H}}$ such that $(\mathcal{T}\setminus K) \bigcup \left[\bigcup_{\delta\in \mathfrak{H}} (\mathcal{T}\setminus C_\delta) \right] \notin \mathfrak{P}$. Then, $K$ is an S$\mathfrak{P}$-compact subspace of $\mathcal{T}$ if and only if there exists a finite set $\mathfrak{H}_0\subseteq \mathfrak{H}$ such that $(\mathcal{T}\setminus K) \bigcup \left[\bigcup_{\delta \in \mathfrak{H}_0} (\mathcal{T}\setminus C_\delta) \right] \notin \mathfrak{P}$.

Proof. Suppose that $K$ is an S$\mathfrak{P}$-compact subspace of $\mathcal{T}$ and let $\{ C_\delta \}_{\delta\in \mathfrak{H}}$ be a family of closed sets such that $(\mathcal{T}\setminus K) \bigcup \left[\bigcup_{\delta \in \mathfrak{H}} (\mathcal{T}\setminus C_\delta) \right] \notin \mathfrak{P}$. Then,

Since $\mathcal{T} \setminus C_\delta$ is an open set for each $\delta \in \mathfrak{H}$ and $K$ is an S$\mathfrak{P}$-compact subspace of $\mathcal{T}$, then there exists a finite set $\mathfrak{H}_0\subseteq \mathfrak{H}$ such that

Then,

Now, suppose that the condition in the theorem holds and let $\{ O_\delta \}_{\delta\in \mathfrak{H}}$ be a family of open sets such that $\mathcal{T}\setminus \left[ K \setminus \bigcup_{\delta \in \mathfrak{H}} O_\delta \right] \notin \mathfrak{P}$. Then, $\{(\mathcal{T} \setminus O_\delta) \}_{\delta \in \mathfrak{H}}$ is a family of closed sets. Now, we have the following:

Thus, there is a finite set $\mathfrak{H}_0\subseteq \mathfrak{H}$ such that

Therefore, we have the following:

This shows that $K$ is an S$\mathfrak{P}$-compact subspace of $\mathcal{T}$. □

Corollary 3.1. Let $(\mathcal{T}, \rho, \mathfrak{P})$ be a $\mathcal{PS}$ and let $\{ H_\eta\}_{ \eta \in \mathfrak{H}}$ be a collection of closed sets such that $\bigcup_{\eta\in \mathfrak{H}} (\mathcal{T}\setminus H_\eta) \notin \mathfrak{P}$. Then, $(\mathcal{T}, \Gamma, \mathfrak{P})$ is an S$\mathfrak{P}$-compact space if and only if there exists a finite set $\mathfrak{H}_0\subseteq \mathfrak{H}$ such that $\ \bigcup_{\eta \in \mathfrak{H}_0} (\mathcal{T}\setminus H_\eta)\notin \mathfrak{P}$.

Theorem 3.2. Let $(\mathcal{T}, \rho, \mathfrak{P})$ be a $\mathcal{PS}$. If $A$ is $\mathfrak{P} g$-closed and $A \subseteq B \subseteq \mathcal{CL} (A)$, then $A$ is an S$\mathfrak{P}$-compact subspace of $\mathcal{T}$ if and only if $B$ is an S$\mathfrak{P}$-compact subspace of $\mathcal{T}$.

Proof. (1) Let $A$ be an S$\mathfrak{P}$-compact subspace of $\mathcal{T}$ and let $\{O_\delta\}_{\delta\in \mathfrak{H}}$ be a family of open sets such that $\mathcal{T}\setminus \left[B\setminus\bigcup_{\delta\in \mathfrak{H}} O_\delta\right]\notin \mathfrak{P}$. Then, since $A\subseteq B$, we have $\mathcal{T}\setminus \left[A\setminus\bigcup_{\delta\in \mathfrak{H}} O_\delta\right]\notin \mathfrak{P}$; then, there exists a finite set $\mathfrak{H}_0\subseteq \mathfrak{H}$ such that $\mathcal{T}\setminus\left[ A\setminus\bigcup_{\delta\in \mathfrak{H}_0} O_\delta\right]\notin \mathfrak{P}$ because $A$ is an S$\mathfrak{P}$-compact subspace. Now, as $A$ is $\mathfrak{P} g$-closed, we have $\mathcal{CL}(A)\subseteq \bigcup_{\delta\in \mathfrak{H}_0} O_\delta$. Then, $\mathcal{T}\setminus\left[ B\setminus \bigcup_{\delta\in \mathfrak{H}_0} O_\delta\right] = \mathcal{T}\notin \mathfrak{P}$. Hence, $B$ is an S$\mathfrak{P}$-compact subspace.

(2) Let $B$ be an S$\mathfrak{P}$-compact subspace of $\mathcal{T}$ and let $\{O_\delta\}_{\delta\in \mathfrak{H}}$ be a family of open sets such that $\mathcal{T}\setminus \left[A\setminus\bigcup_{\delta\in \mathfrak{H}} O_\delta\right]\notin \mathfrak{P}$. Since $A$ is $\mathfrak{P} g$-closed, then $\mathcal{CL}(A)\subseteq \bigcup_{\delta\in \mathfrak{H}} O_\delta.$ As $A\subseteq B\subseteq \mathcal{CL}(A)$, then $B\subseteq \bigcup_{\delta\in \mathfrak{H}} O_\delta$, which implies that $\mathcal{T}\setminus \left[B\setminus \bigcup_{\delta\in \mathfrak{H}} O_\delta\right] \notin \mathfrak{P}$. Since $B$ is an S$\mathfrak{P}$-compact space, then there exists a finite set $\mathfrak{H}_0 \subseteq \mathfrak{H}$ such that $\mathcal{T}\setminus \left[B\setminus \bigcup_{\delta\in \mathfrak{H}_0} O_\delta\right] \notin \mathfrak{P}$. Therefore, $\mathcal{T}\setminus\left[ A\setminus \bigcup_{\delta\in \mathfrak{H}_0} O_\delta\right]\notin \mathfrak{P}$, which implies that $A$ is an S$\mathfrak{P}$-compact subspace of $\mathcal{T}$. □

Theorem 3.3. Let $(\mathcal{T}, \Gamma, \mathfrak{P})$ be a $\mathcal{PS}$. If $R, K\subseteq\mathcal{T}$ are both S$\mathfrak{P}$-compact subspaces of $\mathcal{T}$, then $R \cup K$ is an S$\mathfrak{P}$-compact subspace of $\mathcal{T}$.

Proof. Let $\{O_\delta\}_{\delta\in \mathfrak{H}}$ be a family of open sets such that

Then, $\mathcal{T}\setminus \left[R\setminus \bigcup_{\delta\in \mathfrak{H}} O_\delta\right]\notin \mathfrak{P}$ and $\mathcal{T}\setminus \left[K\setminus \bigcup_{\delta\in \mathfrak{H}} O_\delta\right]\notin \mathfrak{P}$. Since $R$ and $K$ are both S$\mathfrak{P}$-compact, then there exist two finite sets $\mathfrak{H}_0\subseteq \mathfrak{H}$ and $\mathfrak{H}_1\subseteq \mathfrak{H}$ such that $\mathcal{T}\setminus [R\setminus \bigcup_{\delta\in \mathfrak{H}_0} O_\delta]\notin \mathfrak{P}$ and $\mathcal{T}\setminus [K\setminus \bigcup_{\delta\in \mathfrak{H}_1} O_\delta ]\notin \mathfrak{P}$, respectively. Hence, $[\mathcal{T}\setminus (R\setminus \bigcup_{\delta\in \mathfrak{H}_0} O_\delta)]\bigcap [ \mathcal{T}\setminus (K\setminus \bigcup_{\delta\in \mathfrak{H}_1} O_\delta)]\notin \mathfrak{P}.$ Thus, $\mathcal{T}\setminus \left[(R\cup K)\setminus \bigcup_{\delta\in \mathfrak{H}_0\cup \mathfrak{H}_1} O_\delta\right]\notin \mathfrak{P}$, which implies that $R\cup T$ is an S$\mathfrak{P}$-compact space. □

Theorem 3.4. Let $(\mathcal{T}, \Gamma, \mathfrak{P})$ be a $\mathcal{PS}$ and $R, K$ be subsets of $\mathcal{T}$. If $R$ is an S$\mathfrak{P}$-compact subspace of $\mathcal{T}$ and $K$ is a closed set, then $R\cap K$ is an S$\mathfrak{P}$-compact subspace of $\mathcal{T}$.

Proof. Let $\{O_\delta\}_{\delta\in \mathfrak{H}}$ be a family of open sets such that

Then, $[ \mathcal{T}\setminus (R\setminus \bigcup_{\delta\in \mathfrak{H}} O_\delta)] \bigcup [ \mathcal{T}\setminus (K\setminus \bigcup_{\delta\in \mathfrak{H}} O_\delta)]\notin \mathfrak{P}.$ Let $G = \mathcal{T}\setminus \left[K\setminus \bigcup_{\delta\in \mathfrak{H}} O_\delta\right].$ Then, $G$ is an open set. Since $\mathcal{T}\setminus [R\setminus (\bigcup_{\delta\in \mathfrak{H}} O_\delta \cup G)]\notin \mathfrak{P}$ and $R$ is an S$\mathfrak{P}$-compact subspace of $\mathcal{T}$, then there exists a finite set $\{O_i\}_{i = 1}^n\subseteq \{G, O_\delta :\delta\in \mathfrak{H}\}$ such that $\mathcal{T}\setminus [R\setminus \bigcup_{i = 1}^{n} O_i] \notin \mathfrak{P}.$ Now, since $\mathcal{T}\setminus [R\setminus \bigcup_{i = 1}^{n} O_i] \subseteq \mathcal{T}\setminus [(R\cap K)\setminus \bigcup_{i = 1}^{n} O_i],$ then $\mathcal{T}\setminus [(R\cap K)\setminus \bigcup_{i = 1}^{n} O_i]\notin \mathfrak{P}$, which implies that $R\cap K$ is an S$\mathfrak{P}$-compact subspace of $\mathcal{T}$. □

Corollary 3.2. Let $(\mathcal{T}, \Gamma, \mathfrak{P})$ be an S$\mathfrak{P}$-compact space and $B$ be a closed set. Then, $B$ is an S$\mathfrak{P}$-compact subspace of $\mathcal{T}$.

Theorem 3.5. If $h : (\mathcal{T}, \Gamma, \mathfrak{P}) \rightarrow (L, \nu, h(\mathfrak{P}))$ is a bijective continuous function and $Q$ is an S$\mathfrak{P}$-compact subspace of $\mathcal{T}$, then $h(Q)$ is an S$\mathfrak{P}$-compact subspace of $L$.

Proof. Suppose that $\{W_\eta\}_{\eta\in \mathfrak{H}}$ is a family of open sets such that

Then, $h^{-1}(L)\setminus [h^{-1}(h(Q))\setminus \bigcup_{\eta\in \mathfrak{H}} h^{-1}(W_\eta)]\notin \mathfrak{P}.$ Hence, $\mathcal{T}\setminus [Q\setminus \bigcup_{\eta\in \mathfrak{H}}h^{-1}(W_\eta)]\notin \mathfrak{P}$, and $\{h^{-1}(W_\eta) \}_{\eta\in \mathfrak{H}}$ is a family of open sets in $\mathcal{T}$ since $h$ is a continuous function. Therefore, there exists a finite set $\mathfrak{H}_0\subseteq \mathfrak{H}$ such that $\mathcal{T}\setminus [Q\setminus \bigcup_{\eta\in \mathfrak{H}_0} h^{-1}(W_\eta)]\notin \mathfrak{P}$, which implies that $L\setminus [h(Q)\setminus \bigcup_{\eta\in \mathfrak{H}_0} W_\eta]\notin h(\mathfrak{P})$. Hence, $h(Q)$ is an S$\mathfrak{P}$-compact subspace of $L$. □

Corollary 3.3. If $d : (\mathcal{T}, \Gamma, \mathfrak{P}) \rightarrow (L, \nu, d(\mathfrak{P}))$ is a bijective continuous function and $\mathcal{T}$ is an S$\mathfrak{P}$-compact space, then $(L, \nu, d(\mathfrak{P}))$ is an S$\mathfrak{P}$-compact space.

Theorem 3.6. If $\hbar : (\mathcal{T}, \Gamma, \mathfrak{P}) \rightarrow (L, \nu, \mathcal{J}_ \mathfrak{P})$ is a continuous bijective function and $Q$ is an S$\mathfrak{P}$-compact subspace of $\mathcal{T}$, then $\hbar(Q)$ is an S$\mathfrak{P}$-compact subspace of $L$.

Proof. Let $\{O_\delta\}_{\delta\in \mathfrak{H}}$ be a family of open sets such that

Then, $\hbar^{-1}\left(L\setminus \left[\hbar(Q)\setminus \bigcup_{\delta\in \mathfrak{H}} O_\delta\right]\right)\notin \mathfrak{P}.$ Therefore, $\mathcal{T}\setminus [Q \setminus \bigcup_{\delta\in \mathfrak{H}} \hbar^{-1}(O_\delta)]\notin \mathfrak{P}.$ Since $Q$ is an S$\mathfrak{P}$-compact subspace, then there exists a finite set $\mathfrak{H}_0\subseteq \mathfrak{H}$ such that $\mathcal{T}\setminus [Q \setminus \bigcup_{\delta\in \mathfrak{H}_0} \hbar^{-1}(O_\delta)]\notin \mathfrak{P}.$ Hence,

□

Corollary 3.4. If $\hbar : (\mathcal{T}, \Gamma, \mathfrak{P}) \rightarrow (\mathcal{R}, \nu, \mathcal{J}_ \mathfrak{P})$ is a bijective continuous function and $\mathcal{T}$ is an S$\mathfrak{P}$-compact space, then $(\mathcal{R}, \nu, \mathcal{J}_ \mathfrak{P})$ is an S$\mathfrak{P}$-compact space.

4.   Super $\mathfrak{P}$-compact spaces

Definition 4.1. Let $(\mathcal{T}, \rho, \mathfrak{P})$ be a $\mathcal{PS}$. We say that $(\mathcal{T}, \rho, \mathfrak{P})$ is a super $\mathfrak{P}$-compact space (SU$\mathfrak{P}$-compact space) if for every family of open sets $\{V_\eta\}_{\eta\in \mathfrak{H}}$ such that $\bigcup_{\eta\in \mathfrak{H}}V_\eta\notin \mathfrak{P}$, then there exists a finite set $\mathfrak{H}_0\subseteq \mathfrak{H}$ such that $\mathcal{T}\subseteq \bigcup_{\eta\in \mathfrak{H}_0}V_\eta$. Let $A\subseteq \mathcal{T}$. Then, $A$ is an SU$\mathfrak{P}$-compact subspace of $\mathcal{T}$ if for every family of open sets $\{V_\eta\}_{\eta\in \mathfrak{H}}$ such that $\mathcal{T}\setminus\left[A\setminus\bigcup_{\eta\in \mathfrak{H}} V_\eta\right]\notin \mathfrak{P}$, then there exists a finite set $\mathfrak{H}_0\subseteq \mathfrak{H}$ such that $A\subseteq\bigcup_{\eta\in \mathfrak{H}_0}V_\eta$.

Example 4.1. Let $(\mathbb{R}, \Gamma_\mathbb{P}, \mathfrak{P})$, where $\mathbb{P}$ is the set of irrational numbers, be defined as follows:

$U\in\Gamma_\mathbb{P}$ if and only if either $U\cap\mathbb{P} = \emptyset$ or $U = \mathbb{R}$ and $U\in \mathfrak{P}$ if and only if $\sqrt{2}\notin U$. Let $\{W_\eta\}_{\eta\in \mathfrak{H}}$ be any family of open sets such that $\bigcup_{\eta\in \mathfrak{H}}W_\eta \notin \mathfrak{P}$. Then, $\sqrt{2}\in \bigcup_{\eta\in \mathfrak{H}}W_\eta$, which implies that $\exists\; \gamma\in \mathfrak{H}$ such that $W_\gamma = \mathbb{R}$. Therefore, $(\mathbb{R}, \Gamma_\mathbb{P}, \mathfrak{P})$ is an SU$\mathfrak{P}$-compact space.

Remark 4.1. From the Definition 4.1, it is obvious that every SU$\mathfrak{P}$-compact subspace of $\mathcal{T}$ is a compact subspace. Indeed, let $(\mathcal{T}, \rho, \mathfrak{P})$ be a $\mathcal{PS}$ and let $A\subseteq \mathcal{T}$ be an SU$\mathfrak{P}$-compact subspace of $\mathcal{T}$. Assume that $\{W_\eta\}_{\eta\in \mathfrak{H}}$ is an open cover of $A\subseteq \mathcal{T}$. Then, $\mathcal{T}\setminus [A\setminus \bigcup_{\eta\in \mathfrak{H}} W_\eta] = \mathcal{T}\notin \mathfrak{P}$. Hence, there exists a finite set $\mathfrak{H}_0\subseteq \mathfrak{H}$ such that $A\subseteq \bigcup_{\eta\in \mathfrak{H}_0} W_\eta.$

The following example shows that not every compact space is an SU$\mathfrak{P}$-compact space.

Example 4.2. Let $(\mathbb{R}, \rho_0, \mathfrak{P})$ be defined as follows:

$U\in\rho_0$ if and only if either $0\notin U$ or $U = \mathbb{R},$ and let $\mathfrak{P}$ be defined as in Example 2.2. Then, $\mathcal{V} = \{\{x\} : x\in \mathbb{R} \; \text{and}\; x\not = 0\}$ is a family of open sets such that $\bigcup_{x\in\mathbb{R}\setminus\{0\}} \{x\} = \mathbb{R}\setminus\{0\}\notin \mathfrak{P}.$ However, if $\mathcal{V}_0$ is any finite subfamily of $\mathcal{V}$, then $\mathbb{R}\not\subseteq\bigcup_{V\in\mathcal{V}_0} V$. Hence, $(\mathbb{R}, \rho_0, \mathfrak{P})$ is an example of a compact space that is not an SU$\mathfrak{P}$-compact space.

On the other hand, every SU$\mathfrak{P}$-compact space is an S$\mathfrak{P}$-compact space. However, not every S$\mathfrak{P}$-compact space is an SU$\mathfrak{P}$-compact space, as shown in the following example.

Example 4.3. Consider $(\mathbb{R}, \tau_1, \mathfrak{P}_1)$ that is defined in Example 2.1. In Example 3.1, we proved that $(\mathbb{R}, \tau_1, \mathfrak{P}_1)$ is an S$\mathfrak{P}$-compact space. Consider the family of open sets $\mathcal{V} = \{V_t = \{1, t\} : t\in\mathbb{N}\}$. Let $\mathcal{V}_0$ be any finite subfamily of $\mathcal{V}$. Then, $\bigcup_{V\in\mathcal{V}_0} V = \{1, t_1, t_2, ..., t_k\}$ for some $k\in\mathbb{N}$ and $\mathbb{N}\not\subseteq\bigcup_{V\in\mathcal{V}_0}V.$ Hence, $\mathbb{N}$ is not an SU$\mathfrak{P}$-compact space.

Theorem 4.1. Let $(\mathcal{T}, \rho, \mathfrak{P})$ be a $\mathcal{PS}$ and let $K\subseteq \mathcal{T}$. Suppose that $\{ E_\eta \}_{\eta \in \mathfrak{H}}$ is a collection of closed sets such that $(\mathcal{T}\setminus K) \bigcup \left[\bigcup_{\eta \in \mathfrak{H} } (\mathcal{T}\setminus E_\eta) \right] \notin \mathfrak{P}$. Then, $K$ is an SU$\mathfrak{P}$-compact subspace of $\mathcal{T}$ if and only if there exists a finite subset $\mathfrak{H}_0\subseteq \mathfrak{H}$ such that $K \cap \left[\bigcap_{\eta\in \mathfrak{H}_0} E_\eta\right] = \emptyset$.

Proof. First: Suppose that $K$ is an SU$\mathfrak{P}$-compact space. Let $\{ E_\eta \}_{\eta\in \mathfrak{H}}$ be a collection of closed sets of $\mathcal{T}$ such that

Since $K$ is an SU$\mathfrak{P}$-compact subspace and $\{ \mathcal{T} \setminus E_\eta \}_{\eta \in \mathfrak{H} }$ is a family of open sets, then $K \subseteq \bigcup_{\eta \in \mathfrak{H}_0} \left(\mathcal{T} \setminus E_\eta\right)$. Hence, $K \cap \left(\bigcap_{\eta \in \mathfrak{H}_0} E_\eta\right) = \emptyset$.

Second: Suppose that the condition in the theorem holds and let $\{ W_\eta \}_{\eta \in \mathfrak{H} }$ be a family of open sets such that $\mathcal{T}\setminus \left[K \setminus \bigcup_{\eta \in \mathfrak{H} } W_\eta \right] \notin \mathfrak{P}.$ Then, $\{ \mathcal{T} \setminus W_\eta\}_{\eta \in \mathfrak{H}}$ is a family of closed sets; hence,

Thus, there exists a finite set $\mathfrak{H}_0\subseteq \mathfrak{H}$ such that

Hence, $K\subseteq \bigcup_{ \eta \in \mathfrak{H}_0} W_\eta$. This shows that $(\mathcal{T}, \rho, \mathfrak{P})$ is an SU$\mathfrak{P}$-compact space. □

Corollary 4.1. Let $(\mathcal{T}, \rho, \mathfrak{P})$ be a $\mathcal{PS}$ and $\{E_\eta\}_{\eta\in \mathfrak{H}}$ be a collection of closed sets such that $\bigcup_{\eta \in \mathfrak{H}} (\mathcal{T}\setminus E_\eta) \notin \mathfrak{P}$. Then, $(\mathcal{T}, \rho, \mathfrak{P})$ is an SU$\mathfrak{P}$-compact space if and only if there exists a finite subset $\mathfrak{H}_0\subseteq \mathfrak{H}$ such that $\bigcap_{\eta \in \mathfrak{H}_0 } E_\eta = \emptyset$.

Theorem 4.2. Let $(\mathcal{T}, \rho, \mathfrak{P})$ be a $\mathcal{PS}$ and $A, B\subseteq \mathcal{T}$ such that $A \subseteq B \subseteq \mathcal{CL}(A)$. Then, the following properties hold:

(1) If $A$ is an SU$\mathfrak{P}$-compact subspace and $g$-closed, then $B$ is an SU$\mathfrak{P}$-compact subspace.

(2) If $A$ is an S$\mathfrak{P}$-compact subspace and $\mathfrak{P} g$-closed, then $B$ is an SU$\mathfrak{P}$-compact subspace.

(3) If $B$ is a compact subspace and $A$ is $\mathfrak{P} g$-closed, then $A$ is an SU$\mathfrak{P}$-compact subspace.

Proof. (1) Suppose that $A$ is an SU$\mathfrak{P}$-compact subspace of $\mathcal{T}$ and $g$-closed. Let $\{ V_\eta\}_{\eta \in \mathfrak{H} }$ be a family of open sets such that $\mathcal{T}\setminus \left[B \setminus \bigcup_{\eta \in \mathfrak{H}} V_\eta\right]\notin \mathfrak{P}$. Then, $\mathcal{T}\setminus \left[A \setminus \bigcup_{\eta \in \mathfrak{H}} V_\eta\right] \notin \mathfrak{P}$. Since $A$ is an SU$\mathfrak{P}$-compact subspace of $\mathcal{T}$, then there exists a finite subset $\mathfrak{H}_0\subseteq \mathfrak{H}$ such that $A \subseteq \bigcup_{\eta \in \mathfrak{H}_0} V_\eta$. Since $A$ is $g$-closed, then $\mathcal{CL}(A) \subseteq \bigcup_{\eta \in \mathfrak{H}_0} V_\eta$. Hence, $B \subseteq \bigcup_{ \eta \in \mathfrak{H}_0} V_\eta$. Therefore, $B$ is an SU$\mathfrak{P}$-compact subspace of $\mathcal{T}$.

(2) Suppose that $A$ is an S$\mathfrak{P}$-compact subspace of $\mathcal{T}$ and $\mathfrak{P} g$-closed. Let $\{ V_\eta \}_{ \eta \in \mathfrak{H}}$ be a family of open sets such that $\mathcal{T}\setminus \left[B \setminus \bigcup_{ \eta \in \mathfrak{H} } V_\eta \right] \notin \mathfrak{P}$. Then, $\mathcal{T}\setminus \left[A \setminus \bigcup_{\eta \in \mathfrak{H}} V_\eta \right] \notin \mathfrak{P}$. Since $A$ is an S$\mathfrak{P}$-compact subspace of $\mathcal{T}$, then there exists a finite set $\mathfrak{H}_0\subseteq \mathfrak{H}$ such that $\mathcal{T}\setminus \left[A \setminus \bigcup_{\eta \in \mathfrak{H}_0} V_\eta \right] \notin \mathfrak{P}$. Therefore, $\mathcal{CL}(A) \subseteq \bigcup_{\eta \in \mathfrak{H}_0} V_\eta$ because $A$ is $\mathfrak{P} g$-closed. Thus, $B \subseteq \bigcup_{\eta \in \mathfrak{H}_0} V_\eta$. Hence, $B$ is an SU$\mathfrak{P}$-compact subspace of $\mathcal{T}$.

(3) Suppose that $B$ is a compact subspace of $\mathcal{T}$ and $A$ is $\mathfrak{P} g$-closed. Let $\{ V_\eta\}_{\eta \in \mathfrak{H}}$ be any family of open sets such that $\mathcal{T}\setminus \left[A \setminus \bigcup_{\eta\in \mathfrak{H}} V_\eta\right] \notin \mathfrak{P}$. Since $A$ is $\mathfrak{P} g$-closed, then we have $B \subseteq \mathcal{CL}(A) \subseteq \bigcup_{\eta \in \mathfrak{H} } V_\eta$. Hence, there exists a finite set $\mathfrak{H}_0\subseteq \mathfrak{H}$ such that $B \subseteq \bigcup_{ \eta \in \mathfrak{H}_0} V_\eta$. Then, $A \subseteq \bigcup_{ \eta \in \mathfrak{H}_0} V_\eta$, which implies that $A$ is an SU$\mathfrak{P}$-compact subspace of $\mathcal{T}$. □

Corollary 4.2. Let $(\mathcal{T}, \rho, \mathfrak{P})$ be a $\mathcal{PS}$ and let $A$ be $\mathfrak{P} g$-closed such that $A \subseteq B \subseteq \mathcal{CL} (A)$. Then, $A$ is an SU$\mathfrak{P}$-compact subspace of $\mathcal{T}$ if and only if $B$ is an SU$\mathfrak{P}$-compact subspace of $\mathcal{T}$.

Theorem 4.3. Let $(\mathcal{T}, \rho, \mathfrak{P})$ be a $\mathcal{PS}$ and let $A, B\subseteq \mathcal{T}$ both be SU$\mathfrak{P}$-compact subspaces of $\mathcal{T}$. Then, $A \cup B$ is an SU$\mathfrak{P}$-compact subspace of $\mathcal{T}$.

Proof. Let $\{ O_\eta\}_{\eta \in \mathfrak{H} }$ be any family of open sets such that

Then, $\mathcal{T}\setminus \left[A \setminus \bigcup_{ \eta \in \mathfrak{H}} O_\eta \right]\notin \mathfrak{P}$ and $\mathcal{T}\setminus \left[B \setminus \bigcup_{\eta \in \mathfrak{H}} O_\eta \right] \notin \mathfrak{P}$. Since $A$ and $B$ are both SU$\mathfrak{P}$-compact subspaces of $\mathcal{T}$, then there exist finite subsets of $\mathfrak{H}$, namely $\mathfrak{H}_A$ and $\mathfrak{H}_B$, such that $A \subseteq \bigcup_{\eta \in \mathfrak{H}_A } O_\eta$ and $B \subseteq \bigcup_{\eta \in \mathfrak{H}_B} O_\eta$. Hence, $A \cup B \subseteq \bigcup_{\eta \in \mathfrak{H}_A \cup \mathfrak{H}_B} O_\eta.$ This shows that $A \cup B$ is an SU$\mathfrak{P}$-compact subspace of $\mathcal{T}$. □

Theorem 4.4. Let $(\mathcal{T}, \rho, \mathfrak{P})$ be a $\mathcal{PS}$ and let $A, B\subseteq \mathcal{T}$. If $A$ is an SU$\mathfrak{P}$-compact subspace of $\mathcal{T}$ and $B$ is closed, then $A \cap B$ is an SU$\mathfrak{P}$-compact subspace of $\mathcal{T}$.

Proof. Let $\{W_{\delta}\}_{ \delta \in \mathfrak{H}}$ be a family of open sets such that

Then, $\{W_{\delta}\}_{ \delta \in \mathfrak{H}}\cup \{ \mathcal{T} \setminus B\}$ is a family of open sets such that

Since $A$ is an SU$\mathfrak{P}$-compact subspace of $\mathcal{T}$, then there exists a finite subfamily $\mathcal{W} = \{W_i\}_{i = 1}^{n}\subseteq\{W_{\delta}:\delta \in \mathfrak{H}\} \cup \{ \mathcal{T} \setminus B\}$ such that $A \subseteq \bigcup_{i = 1}^{n} W_i$. Then, $A \cap B \subseteq \bigcup_{i = 1}^{n} W_i.$ This shows that $A \cap B$ is an SU$\mathfrak{P}$-compact subspace of $\mathcal{T}$. □

Corollary 4.3. If $(\mathcal{T}, \rho, \mathfrak{P})$ is an SU$\mathfrak{P}$-compact space and $B\subseteq \mathcal{T}$ is closed, then $B$ is an SU$\mathfrak{P}$-compact subspace of $\mathcal{T}$.

Theorem 4.5. If $\hbar: (\mathcal{T}, \Lambda, \mathfrak{P}) \rightarrow (L, \Gamma, \hbar(\mathfrak{P}))$ is a bijective continuous function and $Q$ is an SU$\mathfrak{P}$-compact subspace of $\mathcal{T}$, then $\hbar(Q)$ is an SU$\mathfrak{P}$-compact subspace of $L$.

Proof. Let $\{V_\lambda\}_{\lambda\in \mathfrak{H}}$ be a family of open sets such that

Then, $\mathcal{T}\setminus [Q\setminus\bigcup_{\lambda\in \mathfrak{H}} \hbar^{-1}(V_\lambda)]\notin \mathfrak{P}$. Hence, $Q\subseteq \bigcup_{\lambda\in \mathfrak{H}_0} \hbar^{-1} (V_\lambda)$ for a finite set $\mathfrak{H}_0\subseteq \mathfrak{H}.$ Thus, $\hbar(Q)\subseteq\bigcup_{\lambda\in \mathfrak{H}_0} V_\lambda$, which implies that $\hbar(Q)$ is an SU$\mathfrak{P}$-compact subspace of $L$. □

Corollary 4.4. If $\hbar : (\mathcal{T}, \Lambda, \mathfrak{P}) \rightarrow (L, \Gamma, \hbar(\mathfrak{P}))$ is a bijective continuous function and $(\mathcal{T}, \Lambda, \mathfrak{P})$ is an SU$\mathfrak{P}$-compact space, then $(L, \Gamma, \hbar(\mathfrak{P}))$ is an SU$\mathfrak{P}$-compact space.

Theorem 4.6. If $\hbar : (\mathcal{T}, \Lambda, \mathfrak{P}) \rightarrow (L, \Gamma, \mathcal{J}_ \mathfrak{P})$ is a surjective continuous function and $Q$ is an SU$\mathfrak{P}$-compact subspace of $\mathcal{T}$, then $\hbar(Q)$ is an SU$\mathfrak{P}$-compact subspace of $L$.

Proof. Suppose that $\{V_\delta\}_{\delta\in \mathfrak{H}}$ is a family of open sets such that

Then, $\mathcal{T}\setminus [Q\setminus\bigcup_{\delta\in \mathfrak{H}} \hbar^{-1}(V_\delta)]\notin \mathfrak{P}.$ Hence, $Q\subseteq \bigcup_{\delta\in \mathfrak{H}_0} \hbar^{-1}(V_\delta)$ for a finite set $\mathfrak{H}_0\subseteq \mathfrak{H}.$ Therefore, $\hbar(Q)\subseteq \bigcup_{\delta\in \mathfrak{H}_0} V_\delta$, which implies that $\hbar(Q)$ is an SU$\mathfrak{P}$-compact subspace. □

Corollary 4.5. If $f : (\mathcal{T}, \rho, \mathfrak{P}) \rightarrow (L, \nu, \mathcal{J}_ \mathfrak{P})$ is a surjective continuous function and $(\mathcal{T}, \rho, \mathfrak{P})$ is an SU$\mathfrak{P}$-compact space, then $(L, \nu, \mathcal{J}_ \mathfrak{P})$ is an SU$\mathfrak{P}$-compact space.

Example 4.4. Let $(\mathbb{R}, \mathcal{U}, \mathfrak{P})$ be defined as follows:

see Example 28 ^[11]. If $\{V_\delta\}_{\delta\in \mathfrak{H}}$ is a family of open sets, then we have the following two cases:

Case 1. $0\notin V_\delta$ for every $\delta\in \mathfrak{H}$. Then, there is nothing to prove since $\bigcup_{\delta\in \mathfrak{H}} V_\delta\in \mathfrak{P}$.

Case 2. There exists $\lambda\in \mathfrak{H}$ such that $0\in\ V_\lambda$. Then, $V_\lambda\notin \mathfrak{P}$. Hence, $(\mathbb{R}, \mathcal{U}, \mathfrak{P})$ is an S$\mathfrak{P}$-compact space, which implies that $(\mathbb{R}, \mathcal{U}, \mathfrak{P})$ is a $\mathfrak{P}$-compact space.

Consider the family $\mathcal{V} = \{V_n = (-n, n) : n\in\mathbb{N}\}$. Then, $\bigcup_{n\in\mathbb{N}} V_n = \mathbb{R}\notin \mathfrak{P}$. Let $\mathcal{V}_0 = \{V_k = (-k, k): k\leq m, k\in\mathbb{N}\} \subseteq\mathcal{V}$ for some $m\in\mathbb{N}$. Then, since $\mathbb{R}\not\subseteq \bigcup_{k\leq m} V_k,$ $(\mathbb{R}, \mathcal{U}, \mathfrak{P})$ is not an SU$\mathfrak{P}$-compact space.

Remark 4.2. We have the following relationships:

5.   Conclusions

In this work, we introduced new notions using a primal structure. We started by providing a definition of $\mathfrak{P}$-compactness. Then, we proposed a definition of another concept called strongly $\mathfrak{P}$-compactness (S$\mathfrak{P}$-compactness) and observed that every S$\mathfrak{P}$-compact space is a $\mathfrak{P}$-compact space. A counterexample was discussed to show the converse of that relation is not necessary true. Furthermore, we defined super $\mathfrak{P}$-compact spaces (SU$\mathfrak{P}$-compact spaces). Additionally, more counterexamples and results were presented to illustrate the relations between SU$\mathfrak{P}$-compactness, S$\mathfrak{P}$-compactness, $\mathfrak{P}$-compactness, and compactness. It is worth noting that the primal structure was considered in both fuzzy and soft theories, as discussed in ^[12,13]. In future work, we aim to define the concepts of $\mathfrak{P}$-compactness, S$\mathfrak{P}$-compactness, and SU$\mathfrak{P}$-compactness within the framework of a fuzzy primal structure.

Acknowledgments

The author expresses gratitude to the editors and reviewers for their valuable time and insightful comments.

Conflict of interest

The author declares that they have no conflict of interest to report regarding the publication of this article.