Some New Generalized Types of J-spaces and Metacompact spaces

Proposition 1.1:

¹The real numbers set $ℝ$ with usual topology is not $J$ -space.

Definition 1.2:

⁶The space $(X,\tau )$ is called metacompact, if for each open cover of the space $(X,\tau )$ it has a point finite open refinement.

Remark 1.3:

⁷Every compact space is metacompact.

Example 1.4:

⁸The usual topology on the real numbers set $ℝ$ is metacompact space but it is not compact.

Theorem 1.5:

⁹The metacompact space $(X,\tau )$ , if it is countably compact, then it is compact.

Theorem 1.6:

¹⁰Let $(X,\tau )$ be a space, if $(X,\tau )$ is metacompact, and $K\subset X$ , with $K$ is a closed in $(X,\tau )$ , then $K$ is metacompact.

Theorem 1.7:

¹¹The disjoint union of metacompact spaces is metacompact.

Definition 1.8:

¹²Let $f:X⟶Y$ be a function, then $f$ is called perfect, if $f$ is a closed, continuous and for all $h\in Y$ , so $f^{-1}\left(h\right)$ is compact.

Definition 1.9:

¹The function $f:(X,\tau )⟶(X^{⸍},{\tau }^{ˊ})$ is called boundary-perfect, when $f$ is a closed and for each $x^{ˊ}\in X^{ˊ}$ , so $\partial \left(f^{-1}\left(x^{ˊ}\right)\right)$ is compact.

Theorem 1.10:

⁸Let $f:(X,\tau )\to (X^{ˊ},{\tau }^{ˊ})$ be a continuous, closed and onto function. If $(X,\tau )$ is metacompact, then $(X^{ˊ},{\tau }^{ˊ})$ is so.

Definition 1.11:

¹³The function $f:(X,\tau )\to (X^{ˊ},{\tau }^{ˊ})$ is known as metacompact-function, when the inverse image for any closed and metacompact set of $(X^{ˊ},{\tau }^{ˊ})$ is metacompact in $(X,\tau )$ .

Theorem 1.12:

¹⁴The function $f:(X,\tau )\to (X^{ˊ},\sigma )$ , if it is a continuous, then $f\left(X\right)$ is connected space for all $X$ is connected.

Definition 1.13:

¹⁰A function $f:(X,\tau )\to (X^{ˊ},{\tau }^{ˊ})$ is called monotone if all fibers $f^{-1}\left(x^{ˊ}\right)$ are connected.

Theorem 1.14:

¹⁰ Suppose that $f:X\to Y$ is a function, if $f$ is monotone such that $f$ is a clopen, so $f^{-1}\left(K\right)$ is connected for each connected subset $K$ of $Y$ .

Lemma 1.15:

¹ The function $f:(X,\tau )\to (X^{ˊ},{\tau }^{ˊ})$ , if it is monotone, and $\{K_1,K_2\}$ is either open or closed cover of $(X,\tau )$ , then $f(K_1\cap K_2)=f\left(K_1\right)\cap f\left(K_2\right)$ .

Definition 2.1:

A space $(X,\tau )$ is said to be $\mathit{MJ}$ -space, if it is achieved for each $\{K_1,K_2\}$ is a closed cover of $(X,\tau )$ with $K_1\cap K_2$ metacompact, so either $K_1$ or $K_2$ is metacompact.

Definition 2.2:

The space $(X,\tau )$ is said to be strong $\mathit{MJ}$ -space, when it is achieved for all $L_1\subset X$ metacompact be included in a closed metacompact $L_2\subset X$ , with $X\backslash L_2$ is connected.

Remark 2.3:

Every metacompact (compact) space is a strong $\mathit{MJ}$ -space, and therefore $\mathit{MJ}$ -space is as explained in Proposition (2.6). For example, the real numbers set $ℝ$ with usual topology.

Proposition 2.4:

Suppose that $(X,\tau )$ is a strong $\mathit{MJ}$ -space, the $(X,\tau )$ is an $\mathit{MJ}$ -space.

Proof:

Assume that $(X,\tau )$ is a strong $\mathit{MJ}$ -space, and let $\{K_1,K_2\}$ be a closed cover of $(X,\tau )$ and $K_1\cap K_2$ metacompact, since $(X,\tau )$ is a strong $\mathit{MJ}$ -space, thus there is metacompact and closed $L\subset X$ , with $K_1\cap K_2\subset L$ and $X\backslash L$ is connected. Thus $\{K_1\cap (X\backslash L),K_2\cap (X\backslash L)\}$ is a closed disjoint cover of $X\backslash L$ . However, $X\backslash L$ is connected, then $X\backslash L$ has to be content in $K_1\cap (X\backslash L)$ or in $K_2\cap (X\backslash L)$ . Consequently, we have $(X\backslash L)\subset K_1$ or $(X\backslash L)\subset K_2$ . By using the complement, we have ${K_1}^c\subset L$ or ${K_2}^c\subset L$ , but $K_1\cap K_2\subset L$ , so $K_1\subset L$ or $K_2\subset L$ . So $K_1$ or $K_2$ is metacompact by Theorem (1.6). Hence $X$ is an $\mathit{MJ}$ -space.

Remark 2.5:

Generally speaking, the inverse of proposition (2.4) cannot be true. For example, we define the co-finite topology on $\mathbb{N}$ , where $\mathbb{N}$ represents the set of natural numbers, where the open sets are $\mathbb{N},\varnothing$ and any subset $A$ of $\mathbb{N},$ $\text{such that}$ the complement is finite. Where the only subsets metacompact of $\mathbb{N}$ are the finite subsets. Consequently, assume that $\{K_1,K_2\}$ is a closed cover $\mathbb{N}$ with $K_1\cap K_2$ metacompact, so it is necessary to have $K_1\cap K_2$ as finite set, where the intersection in that space for any infinite sets has to be an infinite set, thus either $K_1$ or $K_2$ has to be finite, thus $K_1$ or $K_2$ is metacompact. So $\mathbb{N}$ is an $\mathit{MJ}$ -space. However, $\mathbb{N}$ it cannot be a strong $\mathit{MJ}$ -space, because each metacompact subset $L\subset \mathbb{N}$ is finite set, and thus $X\backslash L$ is infinite and each infinite subset of $\mathbb{N}$ it cannot be connected.

Proposition 2.6:

Every metacompact space $(X,\tau )$ is a strong $\mathit{MJ}$ -space.

Proof:

Assume that $(X,\tau )$ is metacompact space, in addition to $L\subset X$ is metacompact, since $(X,\tau )$ is a closed metacompact with $L\subset X$ , and since $X\backslash X=\varnothing$ which is connected. Consequently, $(X,\tau )$ is a strong $\mathit{MJ}$ -space.

Corollary 2.7:

Every $(X,\tau )$ compact space, it is a strong $\mathit{MJ}$ -space.

Proof:

Taken from Proposition (2.6), because each compact space is metacompact space.

Corollary 2.8:

Every metacompact space $(X,\tau )$ is an $\mathit{MJ}$ -space.

Proof:

Resulting from the Proposition (2.6) and the Proposition (2.4).

Remark 2.9:

In most cases the inverse of Corollary (2.8) cannot be true. For example, the topological space $(ℝ,\tau )$ , where $ℝ$ is the real numbers set and $\tau$ is the particular point topology, such that $\tau =\{U\subseteq ℝ:0\in U\mathit{\text{or}}U=\varnothing \}$ , where each closed cover of $ℝ$ has to be contained in $ℝ$ . Assume that $\{ℝ,K\}$ is a closed cover of $ℝ$ such that $ℝ\cap K$ is metacompact, but $ℝ\cap K=K$ , so $K$ is metacompact. Hence $(ℝ,\tau )$ is an $\mathit{MJ}$ -space. But this space is not metacompact, since $C=\{(-n,n):n\in \mathbb{N}\}$ is set of all open sets in $ℝ$ , which is a cover of $ℝ$ , but $C$ has no point finite open refinement.

Lemma 2.10:

Assume that $B$ is a closed subset of $(X,\tau )$ , such that $B$ is non-metacompact and if $C\subset B$ is metacompact, then there is a closed subset $D\subset B$ , such that $D$ is non-metacompact with $D\cap C=\varnothing$ .

Proof:

Assume that $\mathcal{U}$ is a cover of $B$ by open sets, with $\mathcal{U}$ has no point finite subcover of $B$ . Choose a finite $U\subset \mathcal{U}$ is cover of $C$ . So $D=B\backslash \cup U$ is a closed non–metacompact set in $B$ with $D\cap C=\varnothing$ , because if $D$ is metacompact, then by theorem (1.7) $D\cup C$ is metacompact, and thus $B$ is metacompact and this is a contradiction.

Theorem 2.11:

For any space $(X,\tau )$ , the following properties are equivalent.

Proof:

(1) ⟹ (2): Let $K\subset X$ with $\partial \left(K\right)$ be metacompact, (where the symbol $\partial \left(K\right)$ means the boundary of $K$ ), since $\partial \left(K\right)=\overline{K}\cap \overline{(X\backslash K)}$ , in addition to $\{\overline{K},\overline{(X\backslash K)}\}$ is a closed cover of $(X,\tau )$ with $\overline{K}\cap \overline{(X\backslash K)}$ is metacompact. Since $X$ is an $\mathit{MJ}$ -space, so $\overline{K}$ or $\overline{(X\backslash K)}$ is metacompact.

(2) ⟹ (3): Assume that $K_1$ and $K_2$ are disjoint closed subsets of $(X,\tau )$ and also $\partial \left(K_1\right)$ is metacompact. Then by (2) we obtain $\overline{K_1}$ or $\overline{(X\backslash K_1)}$ is metacompact. Since $K_1$ is a closed subset, so $K_1=\overline{K_1}$ . Then $K_1$ or $\overline{(X\backslash K_1)}$ is metacompact, and because $K_2$ is a closed subset of $\overline{(X\backslash K_1)}$ , thus $K_1$ or $K_2$ is metacompact, (in the same way if $\partial \left(K_2\right)$ is metacompact).

(3) ⟹ (1): Assume that $\{K_1,K_2\}$ is a cover of $X$ by closed sets, such that $K_1\cap K_2$ is metacompact. Assume that $K_2$ is non-metacompact, since $K_1\cap K_2\subset K_2$ is metacompact, then by lemma (2.10) there is a closed non-metacompact $D\subset K_2$ such that $D\cap (K_1\cap K_2)=\varnothing$ , it follows that $D\cap K_1=\varnothing$ , so $K_1$ and $D$ are closed subsets disjoint of $X$ , since $\partial \left(K_1\right)$ is metacompact and it is a closed subsets of $K_1\cap K_2$ . So, by (3) $K_1$ or $D$ is metacompact, but $D$ is non-metacompact. Therefore $K_1$ is metacompact, (in the same way if $K_1$ is non-metacompact).

Theorem 2.12:

The following properties are equivalent in topological space.

Proof:

(1) ⟹ (2): Let $L\subset X$ be metacompact, and $\mathcal{U}=\{U_1,U_2\}$ is open cover of $X\backslash L$ with $U_1\cap U_2=\varnothing$ . By (1) there exists $U\in \mathcal{U}$ such that $X\backslash U$ is metacompact, since $\mathcal{U}$ contains only two open sets, so $U$ must be either $U_1$ or $U_2$ , therefore $X\backslash U_1$ or $X\backslash U_2$ is metacompact.

(2) ⟹ (1): Assume that $L\subset X$ is metacompact, and $\mathcal{U}$ is a disjoint open cover of $X\backslash L$ . For some $U\in \mathcal{U}$ , we will use three demarches to show that $X\backslash U$ is metacompact.

First, we will show that, when $A$ is an open subset for $X$ containing $L$ , so ${\mathcal{U}}^{\prime }=\{U\in \mathcal{U}:U⊈A\}$ is point finite, assume that it is not point finite, then $\mathcal{U}=U_1\cup U_2$ with $U_1\cap U_2=\varnothing$ and $U_1\cap {\mathcal{U}}^{\prime }$ and $U_2\cap {\mathcal{U}}^{\prime }$ both point infinite.

Assume that $Z_1=\cup U_1$ and $Z_2=\cup U_2$ , so $Z_1,Z_2$ are two open subsets of $X$ with $Z_1\cap Z_2=\varnothing$ and $X\backslash L=Z_1\cup Z_2$ , so by (2) $X\backslash Z_1$ or $X\backslash Z_2$ is metacompact, but $Z_1\subseteq X\backslash Z_2$ and $Z_2\subseteq X\backslash Z_1$ since $Z_1$ and $Z_2$ are disjoint. It follows that $\overline{Z_1}\subseteq \overline{X\backslash Z_2}=X\backslash Z_2$ and $\overline{Z_2}\subseteq \overline{X\backslash Z_1}=X\backslash Z_1$ , so we obtain $\overline{Z_1}$ or $\overline{Z_2}$ is metacompact by Theorem (1.6). Suppose that $\overline{Z_1}$ is metacompact, then $C=\overline{Z_1}\backslash A$ is metacompact. Now let ${\mathcal{U}}_1^{\prime }=U_1\cap {\mathcal{U}}^{\prime }$ , then ${\mathcal{U}}_1^{\prime }$ covers $C$ and each $U\in {\mathcal{U}}^{\prime }$ intersects $C$ , so $C$ is not metacompact since ${\mathcal{U}}_1^{\prime }$ is disjoint and infinite, and this is a contradiction. Therefore ${\mathcal{U}}^{\prime }$ is point finite.

Second, it will be shown that, when $\overline{U}$ is metacompact, for all $U\in \mathcal{U}$ , and thus $X$ is metacompact. Assume that $\mathcal{Z}$ is set of point finite open subsets covers $X$ , so $\mathcal{Z}$ is a point finite open cover of the space $X$ , where is metacompact, thus $\mathcal{Z}$ contains a point finite subcover $\mathcal{F}$ covers $L$ . Assume that $A=\cup \mathcal{F}$ , by the first step we obtain a point finite family ${\mathcal{U}}^{\prime }=\{U\in \mathcal{U}:U⊈A\}$ , so $\bigcup \{\overline{U}:U\in {\mathcal{U}}^{\prime }\}$ is metacompact and because $\mathcal{Z}$ is an open cover of $\bigcup \{\overline{U}:U\in {\mathcal{U}}^{\prime }\}$ , thus $\bigcup \{\overline{U}:U\in {\mathcal{U}}^{\prime }\}$ is covered by some point finite $\mathcal{E}\subset \mathcal{Z}$ . However, $\bigcup \mathcal{E}\subset \mathcal{Z}$ is point finite and covers $X$ , so $X$ is metacompact.

Finally, we will show that, $X\backslash U$ is metacompact for a few $U\in \mathcal{U}$ . If for any $U\in \mathcal{U}$ , then $\overline{U}$ is metacompact, so $X$ is metacompact through step (2) and because $X\backslash U$ is a closed set in $X$ , then $X\backslash U$ is metacompact. Assume that there is $U_{\circ }\in \mathcal{U}$ with $\overline{U_{\circ }}$ is non-metacompact. Assume that $U^{\ast }=\cup \{U\in \mathcal{U}:U\ne U_{\circ }\}$ , then $\{U_{\circ },U^{\ast }\}$ is an open disjoint cover for $X\backslash L$ , so $X\backslash U_{\circ }$ or $X\backslash U^{\ast }$ is metacompact, by (2). When $X\backslash U^{\ast }$ is metacompact, and because $\overline{U_{\circ }}$ is a closed set in $X\backslash U^{\ast }$ , then $\overline{U_{\circ }}$ is metacompact, and this is a contradiction, then $X\backslash U^{\ast }$ is not metacompact, it follows that is $X\backslash U_{\circ }$ is metacompact.

(2) ⟹ (3): Suppose that $\{K_1,K_2\}$ is a cover of $X$ by closed sets such that $K_1\cap K_2$ is metacompact, so $\{X\backslash K_1,X\backslash K_2\}$ is cover of $X\backslash (K_1\cap K_2)$ by open sets, with $X\backslash K_1\cap X\backslash K_2=\varnothing$ . By (2) $X\backslash (X\backslash K_1)$ or $X\backslash (X\backslash K_2)$ is metacompact, that is $K_1$ or $K_2$ is metacompact. Hence $X$ is an $\mathit{MJ}$ -space.

(3) ⟹ (2): Let $L\subset X$ be a metacompact, and assume that $\mathcal{U}=\{U_1,U_2\}$ is a cover of $X\backslash \mathrm{L}$ by open sets with $U_1\cap U_2=\varnothing$ , so $\{X\backslash U_1,X\backslash U_2\}$ is cover of $X$ by closed sets, with $X\backslash U_1\cap X\backslash U_2=X\backslash (U_1\bigcup U_2)$ which is a closed set in $L$ , since $X\backslash \mathrm{L}\subset U_1\bigcup U_2$ , then by Theorem (1.6) $X\backslash U_1\cap X\backslash U_2$ is metacompact. But $X$ is an $\mathit{MJ}$ -space, thus $X\backslash U_1$ or $X\backslash U_2$ is metacompact.

Theorem 2.13:

Suppose that the closed cover $\{A_1,A_2\}$ of a space $(X,\tau )$ with $A_1\cap A_2$ is metacompact. Then $(X,\tau )$ is an (strong) $\mathit{MJ}$ -space if and only if $A_1$ and $A_2$ are (strong) $\mathit{MJ}$ -space and $A_1$ or $A_2$ is metacompact.

Proof:

(1) $\mathit{MJ}$ -space

Suppose that $(X,\tau )$ is an $\mathit{MJ}$ -space, so $A_1$ or $A_2$ is metacompact from definition of $\mathit{MJ}$ -space. Let $A_1$ be a metacompact, so $A_1$ is an $\mathit{MJ}$ -space. To show that $A_2$ is an $\mathit{MJ}$ -space. Assume that $\{K_1,K_2\}$ is a cover of $(X,\tau )$ by closed sets such that $K_1\cap K_2$ is metacompact, therefore $\{K_1,K_2\cup A_1\}$ is a cover of $(X,\tau )$ be closed sets such that $K_1\cap (K_2\cup A_1)$ is metacompact, so $K_1$ or $K_2\cup A_1$ is metacompact, since $(X,\tau )$ is an $\mathit{MJ}$ -space. However, $K_2$ is a closed subset of $K_2\cup A_1,$ then $K_1$ or $K_2$ is metacompact.

Suppose that $A_1$ and $X_2$ are $\mathit{MJ}$ -space and let $A_2$ be a metacompact, we must demonstrate that $(X,\tau )$ is an $\mathit{MJ}$ -space. Suppose that $\{K,K^{ˊ}\}$ is a closed cover of $(X,\tau )$ such that $K\cap K^{ˊ}$ is metacompact. Now let $K_i=K\cap A_i$ and ${K^{ˊ}}_i=K^{ˊ}\cap A_i$ $(i=1,2),$ consequently $\{K_1,{K^{ˊ}}_1\}$ is a cover of $A_1$ by closed sets, which is an $\mathit{MJ}$ -space, such that $K_1\cap {K^{ˊ}}_1$ is metacompact, so $K_1$ or ${K^{ˊ}}_1$ is metacompact. If $K_1$ is metacompact, then $K=K_1\cup K_2$ is metacompact since $K_2$ is a closed subset of metacompact $A_2$ . In the same manner, when ${K^{ˊ}}_1$ is metacompact, so is $K^{ˊ}.$

(2) Strong $\mathit{MJ}$ -space

Let $(X,\tau )$ be a strong $\mathit{MJ}$ -space, from Proposition (2.4) $(X,\tau )$ is an $\mathit{MJ}$ -space, so $A_1$ or $A_2$ is metacompact. Assume that $A_1$ is metacompact, from Proposition (2.6) $A_1$ is a strong $\mathit{MJ}$ -space. Thus, it is still necessary to demonstrate that $A_2$ is a strong $\mathit{MJ}$ -space. Assume that $L_2\subset A_2$ is metacompact. Define $L=L_2\cup A_1$ , thus $L$ is metacompact subset of $(X,\tau )$ which is a strong $\mathit{MJ}$ -space, then there is a closed metacompact set $L^{ˊ}$ in $(X,\tau )$ with $L\subset L^{ˊ}$ and $X\backslash L^{ˊ}$ is connected. Assume that ${L^{ˊ}}_2=L^{ˊ}\cap A_2,$ so ${L^{ˊ}}_2\subset A_2$ is metacompact since ${L^{ˊ}}_2$ is a closed set in metacompact set $L^{ˊ}$ . In addition to $L_2\subset {L^{ˊ}}_2$ , where $L_2\subseteq L\subset L^{ˊ}$ implies that $L_2\cap A_2\subset L^{ˊ}\cap A_2$ . Also note that $A_1\subset L\subset L^{ˊ}$ which implies $A_2\backslash {L^{ˊ}}_2=X\backslash L^{ˊ}$ , and hens $A_2\backslash {L^{ˊ}}_2$ is connected.

Let $A_1$ and $A_2$ be a strong $\mathit{MJ}$ -space and assume that $A_1$ is metacompact, and assume that $L\subset X$ is metacompact. Define $L_2=(L\cup A_1)\cap A_2$ , so $L_2$ is metacompact, because it is a closed set in metacompact set $L\cup A_1$ , then $L_2$ is metacompact set in the strong $\mathit{MJ}$ -space $A_2$ , thus there is a closed metacompact set ${L^{ˊ}}_2$ in $A_2$ , with $L_2\subset {L^{ˊ}}_2$ and $A_2\backslash {L^{ˊ}}_2$ is connected. And now let $L^{ˊ}={L^{ˊ}}_2\cup A_1$ , so $L^{ˊ}\subset X$ is a closed metacompact and $L\subset L^{ˊ}$ and $X\backslash L^{ˊ}=A_2\backslash {L^{ˊ}}_2$ is connected. Therefore $(X,\tau )$ is a strong $\mathit{MJ}$ -space.

Corollary 2.14:

Assume that $(X,\tau )$ any space, and assume that $K\subset X$ is a closed subset, with $\partial \left(K\right)$ is metacompact. If $X$ is an (strong) $\mathit{MJ}$ -space, then is $K$ .

Proof:

Assume that $K,\overline{(X\backslash K)}$ is two closed subsets of $X$ with $X=K\cup \overline{(X\backslash K)}$ and $K\cap \overline{(X\backslash K)}=\partial \left(K\right)$ which is metacompact. However, $X$ is an (strong) $\mathit{MJ}$ -space by hypothesis, from Theorem (2.13) $K$ is an (strong) $\mathit{MJ}$ -space.

Corollary 2.15:

Assume that $X=A\cup B$ , such that $A$ (strong) $\mathit{MJ}$ -space and $B$ is open with $\overline{B}$ is metacompact. Then $(X,\tau )$ is an (strong) $\mathit{MJ}$ -space.

Proof:

Assume that $C=X\backslash B$ , so $C$ is a closed subset of $A$ such that $\partial \left(C\right)$ is metacompact, since $\partial \left(C\right)=\partial \left(B\right)$ is a closed set in $\overline{B}$ which is metacompact by hypothesis. However, $A$ is an (strong) $\mathit{MJ}$ -space, from Corollary (2.14) which shows that $C$ is a (strong) $\mathit{MJ}$ -space. Thus $\{C,\overline{B}\}$ is a closed cover for $(X,\tau )$ with $C\cap \overline{B}=\partial \left(B\right)$ which is metacompact with $C$ and $\overline{B}$ are (strong) $\mathit{MJ}$ -space, so $X$ is an (strong) $\mathit{MJ}$ -space by Theorem (2.13).

Remark 3.1:

The following example shows that there is an $\mathit{MJ}$ -space but it is not $J$ -space.

Example 3.2:

By Proposition (1.1) the usual topology on the real numbers set $ℝ$ is not $J$ -space. Frome Example (1.4) $ℝ$ is metacompact space, and thus by Corollary (2.8) $ℝ$ is an $\mathit{MJ}$ -space.

Proposition 3.3:

Suppose that $(X,\tau )$ is a space, if $(X,\tau )$ is countably compact, then $(X,\tau )$ is an $\mathit{MJ}$ -space if and only if $(X,\tau )$ is $J$ -space.

Proof:

Assume that $(X,\tau )$ is an $\mathit{MJ}$ -space, to show that $(X,\tau )$ is $J$ -space, suppose that $\{K_1,K_2\}$ is a cover of $(X,\tau )$ by closed sets, with $K_1\cap K_2$ is compact, and so $K_1\cap K_2$ is metacompact, since $(X,\tau )$ is an $\mathit{MJ}$ -space, then $K_1$ or $K_2$ is metacompact, since $K_1$ and $K_2$ are closed subsets of $(X,\tau )$ , so both $K_1,K_2$ are countably compact, and thus $K_1$ or $K_2$ is compact, and then $(X,\tau )$ is $J$ -space.

Assume that $(X,\tau )$ is a $J$ -space, to prove that $(X,\tau )$ is an $\mathit{MJ}$ -space, assume that $\{K_1,K_2\}$ is a closed cover for $(X,\tau )$ such that $K_1\cap K_2$ is metacompact, since $K_1\cap K_2$ is a closed subset of $(X,\tau )$ , so $K_1\cap K_2$ is countably compact, and thus $K_1\cap K_2$ is compact, where $(X,\tau )$ is $J$ -space, so $K_1$ or $K_2$ is compact, and thus $K_1$ or $K_2$ is metacompact, therefore $(X,\tau )$ is an $\mathit{MJ}$ -space.

Proposition 3.4:

Let $(X,\tau )$ be a strong metacompact $J$ -space, if $(X,\tau )$ is countably compact, then $(X,\tau )$ is a strong $J$ -space.

Proof:

Suppose that $(X,\tau )$ is a strong $\mathit{MJ}$ -space, to show that $(X,\tau )$ is a strong $J$ -space, and let $L_1$ be a compact set in $(X,\tau )$ , this means $L_1$ is metacompact subset of $(X,\tau )$ , and because $(X,\tau )$ is a strong $\mathit{MJ}$ -space, thus there exists a closed metacompact set $L_2$ in $(X,\tau )$ , with $L_1$ contained in $L_2$ and $X\backslash L_2$ is connected, sinse $L_2$ is closed subset of $(X,\tau )$ , this leads to $L_2$ is countably compact and metacompact, and thus $L_2$ is compact subset of $(X,\tau )$ .

Definition 4.1:

The function $f:(X,\tau )\to (X^{ˊ},{\tau }^{\prime })$ is said to be meta-perfect, if $f$ is a closed and continuous, and $f^{-1}\left(B\right)\subset X$ is metacompact for all $B\subset X^{ˊ}$ is metacompact. This means $f$ is meta-perfect if $f$ is a continuous, closed and metacompact.

Example 4.2:

Suppose that $X=[0,1]\times [0,1]$ and $Y=[0,1]$ are two topological spaces with usual topology, and let $f:X\to Y$ be a function such that $f\left(\right(x,y\left)\right)=x$ , for all $(x,y)\in X$ , since $f$ is a closed and continuous, and for all $B\subset Y$ is metacompact then $f^{-1}\left(B\right)=B\times [0,1]\subset X$ which is metacompact, because of both $B$ and $[0,1]$ are metacompact, thus $f$ is meta-perfect function.

Remark 4.3:

It is noted that the concepts of meta-perfect function and perfect function, they are independent.

Definition 4.4:

The closed function $f:(X,\tau )\to (Y,{\tau }^{\prime })$ is called boundary-meta-perfect if $\partial \left(f^{-1}\left(y\right)\right)$ is metacompact set in $X$ for all $y\in Y$ .

Example 4.5:

Suppose that $X=[0,2]$ with usual topology and $Y=\{a,b\}$ with discrete topology are two topological spaces, and let $f:X\to Y$ be a function such that $f\left(x\right)=\{\begin{array}{c}a,\mathit{\text{if}}0\le x\le 1\\ b,\mathit{\text{if}}1<x\le 2\end{array}$ , for all $x\in X$ , since $f^{-1}\left(a\right)=[0,1]$ and $f^{-1}\left(b\right)=(1,2]$ , and since $\partial \left(f^{-1}\left(a\right)\right)=\overline{[0,1]}\backslash \mathit{int}\left(\right[0,1\left]\right)=[0,1]\backslash (0,1)=\{0,1\}$ , and $\partial \left(f^{-1}\left(b\right)\right)=\overline{(1,2]}\backslash \mathit{int}\left(\right(1,2\left]\right)=[1,2]\backslash (1,2)=\{1,2\}$ . Thus, both $\partial \left(f^{-1}\left(a\right)\right)$ and $\partial \left(f^{-1}\left(b\right)\right)$ are metacompact, because of both $\{0,1\}$ and $\{1,2\}$ are finite sets and thus metacompact, therefore $f$ is boundary-meta-perfect function.

Remark 4.6:

For each boundary perfect function $f$ , then $f$ is a boundary-meta-perfect, because each compact space is metacompact.

Proposition 4.7:

Assume that $f:(X,\tau )\to (Y,{\tau }^{\prime })$ is a meta-perfect function from $(X,\tau )$ onto $(Y,{\tau }^{\prime })$ , when $(X,\tau )$ is an $\mathit{MJ}$ -space, then $(Y,{\tau }^{\prime })$ is so.

Proof:

Assume that $\{K_1,K_2\}$ is a closed cover of $(Y,{\tau }^{\prime })$ such that $K_1\cap K_2$ is metacompact, so $\{f^{-1}\left(K_1\right),f^{-1}\left(K_2\right)\}$ is a closed cover of $X$ , where $f$ is a continuous and $f^{-1}\left(K_1\right)\cap f^{-1}\left(K_2\right)=f^{-1}(K_1\cap K_2)$ . But $f^{-1}(K_1\cap K_2)$ is metacompact since $f$ is meta-perfect, so $f^{-1}\left(K_1\right)$ or $f^{-1}\left(K_2\right)$ . It follows that $f\left(f^{-1}\left(K_1\right)\right)$ or $f\left(f^{-1}\left(K_2\right)\right)$ is metacompact, since the function $f$ is a continuous and by Theorem (1.10), then $K_1$ or $K_2$ is metacompact because $f$ is onto. Therefore $Y$ is an $\mathit{MJ}$ -space.

Proposition 4.8:

Assume that $f:(X,\tau )\to (Y,{\tau }^{\prime })$ is meta-perfect and monotone function from space $(X,\tau )$ onto $(Y,{\tau }^{\prime })$ , when $(Y,{\tau }^{\prime })$ is an (strong) $\mathit{MJ}$ -space, then $(X,\tau )$ is so.

Proof:

1) Suppose that $Y$ is an $\mathit{MJ}$ -space, and let $\{K_1,K_2\}$ be a closed cover for $X$ such that $K_1\cap K_2$ metacompact, so $\{f\left(K_1\right),f\left(K_2\right)\}$ is a cover of $Y$ by closed sets since $f$ is a closed and continuous, and $f\left(K_1\right)\cap f\left(K_2\right)=f(K_1\cap K_2)$ by Lemma (1.15). But $f(K_1\cap K_2)$ is metacompact by Theorem (1.10), so $f\left(K_1\right)$ or $f\left(K_2\right)$ is metacompact since $Y$ is an $\mathit{MJ}$ -space. Then $f^{-1}\left(f\left(K_1\right)\right)$ or $f^{-1}\left(f\left(K_2\right)\right)$ is metacompact, since $f$ is meta-perfect, it follows by Theorem (1.6), that $K_1$ or $K_2$ is metacompact since $K_1$ and $K_2$ are closed subsets of $f^{-1}\left(f\left(K_1\right)\right)$ and $f^{-1}\left(f\left(K_2\right)\right)$ respectively. Hence $X$ is an $\mathit{MJ}$ -space.

2) Assume that $Y$ is a strong $\mathit{MJ}$ -space, and let $L_1\subset X$ be metacompact, because $f$ is a continuous, onto and closed, then $f\left(L_1\right)\subset Y$ is metacompact, so there is a closed metacompact $L_2\supset f\left(L_1\right)$ in $Y$ , with $Y\backslash L_2$ is connected. Hence $f^{-1}\left(L_2\right)\subset X$ since $f$ is meta-perfect, and thus $f^{-1}\left(L_2\right)\supset L_1$ , and $f^{-1}(Y\backslash L_2)$ is connected because $f$ is monotone and closed. But $X\backslash f^{-1}\left(L_2\right)=f^{-1}(Y\backslash L_2)$ , so $X\backslash f^{-1}\left(L_2\right)$ is connected. Therefore $X$ is a strong $\mathit{MJ}$ -space.

Proposition 4.9:

Suppose that $f:(X,\tau )\to (Y,{\tau }^{\prime })$ is a homeomorphism, if $(X,\tau )$ is an $\mathit{MJ}$ -space, then $(Y,{\tau }^{\prime })$ is so, and vice versa.

Proof:

Suppose that $X$ is an $\mathit{MJ}$ -space, it must be proven that $Y$ is an $\mathit{MJ}$ -space. Suppose that the function $f:X\to Y$ is a homeomorphism, and assume that $K_1$ and $K_2$ are two closed sets in $Y$ , where $Y=K_1\cup K_2$ and $K_1\cap K_2$ metacompact, so $\{f^{-1}\left(K_1\right),f^{-1}\left(K_2\right)\}$ is a closed cover of $X$ , because $f$ is a continuous, since $f^{-1}\left(K_1\right)\cap f^{-1}\left(K_2\right)=f^{-1}(K_1\cap K_2)$ which is metacompact since $f^{-1}$ is a continuous and by Theorem (1.6). It follows by Definition of $\mathit{MJ}$ -space, that $f^{-1}\left(K_1\right)$ or $f^{-1}\left(K_2\right)$ is metacompact. Again, by Theorem (1.6) we can obtain $f\left(f^{-1}\left(K_1\right)\right)$ or $f\left(f^{-1}\left(K_2\right)\right)$ is metacompact since $f$ is a continuous, closed and onto, and thus we have $K_1$ or $K_2$ is metacompact since $f$ is onto. Therefore $Y$ is an $\mathit{MJ}$ -space. Likewise, it can be proven that $X$ is an $\mathit{MJ}$ -space, if $Y$ is an $\mathit{MJ}$ -space.

Proposition 4.10:

The following properties are equivalent, for each space $(X,\tau )$ .

Proof:

$\left(\mathit{a}\right)\mathit{⟹}\left(\mathit{b}\right)$ Clear, because $\{\overline{K},\overline{X\backslash K}\}$ is a cover of $X$ by closed sets and $\overline{K}\cap \overline{X\backslash K}=\partial \left(K\right)$ which is metacompact, since $X$ is an $\mathit{MJ}$ -space, thus $\overline{K}$ or $\overline{X\backslash K}$ is metacompact.

$\left(\mathit{b}\right)\mathit{⟹}\left(\mathit{a}\right)$ Let $\{K_1,K_2\}$ be a closed cover of $X$ with $K_1\cap K_2$ metacompact, since $\partial \left(K_1\right)\subset K_1\cap K_2$ , then by Theorem (1.6) $\partial \left(K_1\right)$ is metacompact, so $K_1$ or $\overline{X\backslash K_1}$ is metacompact by (b). But $K_2=(K_1\cap K_2)\cup \left(\overline{X\backslash K_1}\right)$ , so $K_1$ or $K_2$ is metacompact.

Theorem 4.11:

The following properties are equivalent, for every space $(X,\tau )$ .

Proof:

$\left(\mathit{a}\right)\mathit{⟹}\left(\mathit{b}\right)$ Assume that $f:X⟶Y$ is a boundary-meta-perfect function from $\mathit{MJ}$ -space $X$ onto non-metacompact space $Y$ , and $y\in Y$ , where $f$ is boundary meta-perfect, so $\partial \left(f^{-1}\left(y\right)\right)\subset X$ is metacompact, thus by Proposition (4.10), either $f^{-1}\left(y\right)$ or $\overline{(X\backslash f^{-1}\left(y\right))}$ is metacompact, but $\overline{(X\backslash f^{-1}\left(y\right))}$ is non-metacompact since if $\overline{(X\backslash f^{-1}\left(y\right))}$ is metacompact, since $Y=f\left(\overline{(X\backslash f^{-1}\left(y\right))}\right)\cup \left\{y\right\}$ which is metacompact, and this is a contradiction with the hypothesis, so $f^{-1}\left(y\right)$ is metacompact. Therefore $f$ is meta-perfect.

$\left(\mathit{b}\right)\mathit{⟹}\left(\mathit{a}\right)$ Suppose that if $f$ is a boundary-meta-perfect function from $X$ onto $Y$ which is a non-metacompact, so $f$ is meta-perfect. We need to show that $X$ is an $\mathit{MJ}$ -space, let $K_1$ and $K_2$ be a two closed sets in $X$ , where $X=K_1\cup K_2$ and $K_1\cap K_2$ metacompact, and let $Y=X∕K_2$ and $f:X⟶Y$ be the quotient function and let $y_{\circ }=f\left(K_2\right)$ , so $f$ is boundary-meta-perfect, since $f$ is a closed and $\partial \left(f^{-1}\left(y\right)\right)$ is metacompact for every $y\in Y$ , because if $y=y_{\circ }$ , so $\partial \left(f^{-1}\left(y\right)\right)$ is a closed set in $K_1\cap K_2$ , and if $y\ne y_{\circ }$ , thus $\partial \left(f^{-1}\left(y\right)\right)$ is set containing one element. And now if $Y$ is not metacompact, so $f$ is meta-perfect from hypothesis, and so $K_2=f^{-1}\left(y_{\circ }\right)$ is metacompact. If $Y$ is metacompact, thus $f\left(K_1\right)$ is metacompact since it is closed set in $Y$ . However, ${f|}_{K_1}:K_1⟶f\left(K_1\right)$ is meta-perfect, because $f$ is a closed and its fibers are either is a set containing one element or equal to $K_1\cap K_2$ , so $K_1=f^{-1}\left(f\left(K_1\right)\right)$ is metacompact. Therefor $X$ is an $\mathit{MJ}$ -space.