2. Preliminary

Next, we introduce some basic concepts and notations that will be used in the paper. For more details, see Engelking (Reference Engelking1989), Gierz et al. (Reference Gierz, Hofmann, Keimel, Lawson, Mislove and Scott2003), Goubault-Larrecq (2003).

Let P be a poset. A nonempty subset D of P is directed (resp., filtered) if every two elements in D have an upper (resp., lower) bound in D. P is called a directed complete poset or a dcpo for short, if for any directed subset $D\subseteq P$ , the supremum of D, denoted by $\bigvee D$ , exists.

For any subset A of a poset P, we use the following standard notations:

\begin{equation*}\mathord{\uparrow} A=\{y\in P: \exists x\in A, x\leq y\};\ \mathord{\downarrow} A=\{y\in P:\exists x\in A, y\leq x\}.\end{equation*}

In particular, for each $x\in X$ , we write $\mathord{\uparrow} x=\mathord{\uparrow}\{x\}$ and $\mathord{\downarrow} x=\mathord{\downarrow} \{x\}$ . A subset A of P is called a lower (resp., upper) set if $A=\mathord{\downarrow} A$ (resp., $A=\mathord{\uparrow} A$ ).

A subset U of P is Scott open if (i) $U=\mathord{\uparrow}U$ and (ii) for any directed subset D of P for which $\bigvee D$ exists, $\bigvee D\in U$ implies $D\cap U\neq\emptyset$ . All Scott open subsets of P form a topology, called the Scott topology on P, denoted by $\sigma(P)$ . The space $\Sigma P=(P,\sigma(P))$ is called the Scott space of P.

Let X be a $T_0$ space. A subset A of X is called saturated if A equals the intersection of all open sets containing it. The specialization order $\leq $ on X is defined by $x\leq y$ if and only if $x\in \mathrm{cl}(\{y\})$ , where cl is the closure operator. It is important to note that a subset A of X is saturated if and only if $A=\mathord{\uparrow} A$ in the specialization order.

1. (1). Every open (resp., closed) set is an upper (resp., lower) set. In particular, $\mathrm{cl}(\{x\})=\mathord{\downarrow} x$ (Gierz et al. Reference Gierz, Hofmann, Keimel, Lawson, Mislove and Scott2003; Goubault-Larrecq 2003).

2. (2). For each subset K of X, K is compact if and only if $\mathord{\uparrow} K$ is compact (Gierz et al. Reference Gierz, Hofmann, Keimel, Lawson, Mislove and Scott2003; Goubault-Larrecq 2003).

3. (3). It is well-known that a subset K of X is compact saturated if and only if $\min K$ is compact and $K=\mathord{\uparrow} \min K$ , where $\min K$ is the set of all minimal elements of K in the specialization order (see, e.g., Erné Reference Erné2009, pp. 2068).

1. (1). A nonempty subset A of a topological space X is called irreducible if for any closed sets $F_1$ , $F_2$ of X, $A \subseteq F_1\cup F_2$ implies $A \subseteq F_1$ or $A \subseteq F_2$ .

2. (2). A $T_0$ space X is called sober if every irreducible closed subset of X is the closure of a (unique) point.

For a $T_0$ space X, $A\subseteq X$ , and $\mathcal F\subseteq 2^X$ , we use the following notations:

$\begin{array}{ll} \mathcal O(X), & \text{the family of all open subsets of }X;\\ \mathcal C(X), & \text{the family of all closed subsets of }X;\\ \mathcal Q(X), & \text{the family of all compact saturated subsets of }X;\\ \mathcal N(A), &\text{the family } \{U\in\mathcal O(X): A\subseteq U\};\\ \mathcal M(A), & \text{the family } \{U\in\mathcal O(X): A\cap U\neq\emptyset\};\\ \mathfrak{M}(\mathcal F), & \text{the family } \{C\in\mathcal C(X):\forall F\in\mathcal F, C\cap F\neq\emptyset\}; \\ m(\mathcal F), & \text{the family of all minimal members in } (\mathfrak M(\mathcal F),\subseteq). \end{array}$

1. (1). For each compact saturated subset K of X, $\mathcal N(K)\in\mathsf{OF}(X)$ .

2. (2). For each irreducible subset A of X, $\mathcal M(A)\in\mathsf{OF}(X)$ .

3. (3). For each continuous mapping $f:X\longrightarrow Y$ between $T_0$ spaces X and Y, if $\mathcal F\in\mathsf{OF}(X)$ , then $f_*(\mathcal F)=\{V\in\mathcal O(Y):f^{-1}(V)\in\mathcal F\}\in\mathsf{OF}(Y)$ .

The following is a similar result to the Topological Rudin Lemma in Heckmann and Keimel (2013).

1. (1). every element of $m(\mathcal F)$ is irreducible;

2. (2). if $A\in\mathfrak M(\mathcal F)$ , then there is a closed subset C of A such that $C\in m(\mathcal F)$ .

(2) Note that an open set is not in $\mathcal F$ if and only if its complement is in $\mathfrak{M}(\mathcal F)$ . Then, using Proof (i) of Lemma II-1.19 in Gierz et al. (Reference Gierz, Hofmann, Keimel, Lawson, Mislove and Scott2003), condition (2) holds dually.

3. $\Psi$ -Well-Filtered Spaces

For each $K\in\mathcal Q(X)$ , it is trivial that $\mathcal N(K)\in\mathsf{OF}(X)$ and $K=\bigcap\mathcal N(K)$ . Then, the mapping $\mathcal N:(\mathcal Q(X),\supseteq)\longrightarrow (\mathsf{OF}(X),\subseteq)$ , $K\mapsto \mathcal N(K)$ is well-defined and clearly is an order-embedding. In domain theory, the single most important result about sober spaces is the Hofmann-Mislove Theorem (see Gierz et al. Reference Gierz, Hofmann, Keimel, Lawson, Mislove and Scott2003, Theorems II-1.20, II-1.21).

1. (1). X is sober;

2. (2). $\forall\mathcal F\in \mathsf{OF}(X)$ , there is a $K\in\mathcal Q(X)$ such that $\mathcal F=\mathcal N(K)$ ;

3. (3). $\forall \mathcal F\in\mathsf{OF}(X)$ , $\mathcal F=\mathcal N(\bigcap\mathcal F)$ .

As a corollary of the Hofmann-Mislove Theorem, the following result is clear.

1. (1). X is sober;

2. (2). $\mathsf{OF}(X)=\{\mathcal N(K):K\in\mathcal Q(X)\}$ ;

3. (3). $\forall \mathcal F\in \mathsf{OF}(X)$ , $\forall U\in\mathcal O(X)$ , $\bigcap\mathcal F\subseteq U$ implies $U\in\mathcal F$ .

In the following, we would like to provide a unified characterization of Hofmann-Mislove Theorem type for the classes of d-spaces and well-filtered spaces via open filters.

1. (HM1) for each $T_0$ space X, $\{\mathcal N(K):K\in\mathcal Q(X)\}\subseteq\Psi(X)\subseteq \mathsf{OF}(X)$ ;

2. (HM2) for each continuous mapping $f:X\longrightarrow Y$ in ${\bf Top_0}$ , $\Psi(f)(\mathcal F)=f_*(\mathcal F)=\{V\in\mathcal O(Y):f^{-1}(V)\in\mathcal F\}\in\Psi(Y)$ for each $\mathcal F\in\Psi(X)$ .

$\bigcap\mathcal F\subseteq U$ implies $U\in \mathcal F$ .

1. (1). X is $\Psi$ -well-filtered;

2. (2). $\forall \mathcal F\in \Psi(X)$ , $\mathcal F=\mathcal N(\bigcap\mathcal F)$ ;

3. (3). $\Psi(X)= \{\mathcal N(K): K\in\mathcal Q(X)\}$ .

(2) $\Rightarrow$ (3). By the definition of $\Psi$ , it is clear that $ \{\mathcal N(K): K\in\mathcal Q(X)\}\subseteq\Psi(X)$ . Now suppose $\mathcal F\in\Psi(X)$ . By Lemma 3.5, $K_0=\bigcap\mathcal F\in\mathcal Q(X)$ , and by condition (2) $\mathcal F=\mathcal N(K_0)$ . This shows that $\Psi(X)\subseteq \{\mathcal N(K): K\in\mathcal Q(X)\}$ . Thus, condition (3) holds.

(3) $\Rightarrow$ (1). Let $\mathcal F\in\Psi(X)$ and $U\in\mathcal O(X)$ such that $\bigcap\mathcal F\subseteq U$ . By (3), there exists $K\in\mathcal Q(X)$ such that $\mathcal F=\mathcal N(K)$ . Then, $K=\bigcap\mathcal N(K)=\bigcap \mathcal F\subseteq U$ , which implies that $U\in\mathcal N(K)=\mathcal F$ . Therefore, X is $\Psi$ -well-filtered.

The following result shows that the $\Psi$ -well-filteredness is a topological property for each HM-system $\Psi$ .

Proof. Suppose $h:Y\longrightarrow X$ is a homeomorphism. Let $\mathcal F\in\Psi(Y)$ and $W\in\mathcal O(Y)$ such that $\bigcap \mathcal F\subseteq W$ . Since h is a homeomorphism, one can easily obtain that $h_*(\mathcal F)=\{h(U):U\in\mathcal F\}$ , which implies that

\begin{equation*}\bigcap h_*(\mathcal F)=\bigcap\{h(U):U\in\mathcal F\}=h\left(\bigcap\mathcal F\right)\subseteq h(W)\in\mathcal O(X).\end{equation*}

By (HM2), $h_*(\mathcal F)\in\Psi(X)$ , and since X is $\Psi$ -well-filtered, we have that $h(W)\in h_*(\mathcal F)$ , so $W\in\mathcal F$ . Therefore, X is $\Psi$ -well-filtered.

In the following, we will show that all the classes of sober spaces, well-filtered spaces, and d-spaces can be characterized via HM-systems.

Definition 3.8. Define $\Psi_{\mathsf{sob}}, \Psi_{\mathsf{wf}}, \Psi_{\rm d}:{\bf Top_0}\longrightarrow{\bf Set}$ as follows: for each $T_0$ space X,

\begin{align*} \Psi_{\mathsf{sob}}(X)&=\mathsf{OF}(X) ,\\ \Psi_{\mathsf{wf}}(X)&=\left\{\bigcup_{K\in\mathcal G}\mathcal N(K): \mathcal G\text{ is a filtered family of }\mathcal Q(X)\right\},\\ \Psi_{{\rm d}}(X)&=\left\{\bigcup_{x\in D}\mathcal N(\mathord{\uparrow} x):D\text{ is a directed subset of }X\right\}\cup \{\mathcal N(K):K\in\mathcal Q(X)\}. \end{align*}

Then, it is trivial to check that $\Psi_{\mathsf{sob}}, \Psi_{\mathsf{wf}}, \Psi_{\rm d}$ are all HM-systems.

1. (1). X is sober;

2. (2). X is $\Psi_{\mathsf{sob}}$ -well-filtered;

3. (3). $\forall \mathcal F\in \Psi_{\mathsf{sob}}(X)$ , $\mathcal F=\mathcal N(\bigcap\mathcal F)$ ;

4. (4). $\Psi_{\mathsf{sob}}(X)=\{\mathcal N(K): K\in\mathcal Q(X)\}$ .

sober ( $\Psi_{\mathsf{sob}}$ -well-filtered) space $\Longrightarrow$ $\Psi$ -well-filtered space $\Longrightarrow$ $T_0$ ( $\Psi_0$ -well-filtered) space.

In another words, the sober space is the strongest $\Psi$ -well-filtered space, and the $T_0$ space is the weakest one.

1. (1). X is well-filtered;

2. (2). X is $\Psi_{\mathsf{wf}}$ -well-filtered;

3. (3). $\forall \mathcal F\in \Psi_{\mathsf{wf}}(X)$ , $\mathcal F=\mathcal N(\bigcap\mathcal F)$ ;

4. (4). $\Psi_{\mathsf{wf}}(X)=\{\mathcal N(K): K\in\mathcal Q(X)\}$ .

(1) $\Rightarrow$ (2). Suppose X is well-filtered. Let $\mathcal F\in \Psi_{\mathsf{wf}}(X)$ and $U\in\mathcal O(X)$ such that $\bigcap\mathcal F\subseteq U$ . Then, there exists a filtered family $\{K_i:i\in I\}\subseteq\mathcal Q(X)$ such that $\mathcal F=\bigcup_{i\in I}\mathcal N(K_{i})$ . Note that $\bigcap_{i\in I}K_i=\bigcap\mathcal F\subseteq U$ , which implies that $K_{i_0}\subseteq U$ for some $i_0\in I$ because X is well-filtered. It follows that $U\in\mathcal N(K_{i_0})\subseteq \mathcal F$ . Hence, X is $\Psi_{\mathsf{wf}}$ -well-filtered.

(2) $\Rightarrow$ (1). Let $\{K_i:i\in I\}$ be a filtered family of $\mathcal Q(X)$ and $U\in \mathcal O(X)$ such that $\bigcap_{i\in I}K_i\subseteq U$ . Then, $\mathcal F=\bigcup_{i\in I}\mathcal N(K_i)\in\Psi_{\mathsf{wf}}(X)$ . Note that $\bigcap\mathcal F=\bigcap_{i\in I}K_i\subseteq U$ , and since X is $\Psi$ -well-filtered, it follows that $U\in\mathcal F$ . By the definition of $\mathcal F$ , there exists $i_0\in I$ such that $U\in \mathcal N(K_{i_0})$ , that is, $K_{i_0}\subseteq U$ . Hence, X is well-filtered.

1. (1). X is a d-space;

2. (2). X is $\Psi_{\rm d}$ -well-filtered;

3. (3). $\forall \mathcal F\in \Psi_{\rm d}(X)$ , $\mathcal F=\mathcal N(\bigcap\mathcal F)$ ;

4. (4). $\Psi_{\rm d}(X)=\{\mathcal N(K): K\in\mathcal Q(X)\}$ .

(1) $\Rightarrow$ (2). Let $\mathcal F\in\Psi_{\rm d}(X)$ and $U\in\mathcal O(X)$ . If $\mathcal F=\mathcal N(K)$ for some $K\in\mathcal Q(X)$ , then it is clear that $K=\bigcap\mathcal N(K)\subseteq U$ implies $U\in\mathcal N(K)$ . Now assume there is a directed subset D of X such that $\mathcal F=\bigcup_{x\in D}\mathcal N(\mathord{\uparrow} x)$ and $\bigcap\mathcal F\subseteq U$ . Since X is a d-space, $\bigvee D$ exists. We have that $\bigcap \mathcal F=\bigcap_{d\in D}\mathord{\uparrow} d=\mathord{\uparrow} \bigvee D\subseteq U$ , which implies $\bigvee D\in U$ . Since every open set in a d-space is Scott open, $ D\cap U\neq\emptyset$ , and take $x_0\in D\cap U$ . It follows that $U\in \mathcal N(\mathord{\uparrow} x_0)\subseteq \mathcal F$ . Hence, X is $\Psi_{\rm d}$ -well-filtered.

(2) $\Rightarrow $ (1). Let D be a directed subset of X, and $\mathcal F=\bigcup_{x\in D}\mathcal N(\mathord{\uparrow} x)$ . We claim that $\mathrm{cl}(D)\cap\bigcap\mathcal F\neq\emptyset$ . Otherwise, $\bigcap\mathcal F\subseteq X\setminus\mathrm{cl}(D)$ , and since $\mathcal F\in\Psi_{\rm d}(X)$ and X is $\Psi_{\rm d}$ -well-filtered, $X\setminus\mathrm{cl}(D)\in\mathcal F=\bigcup_{x\in D}\mathcal N(\mathord{\uparrow} x)$ . Then, there exists $x_0\in D$ such that $X\setminus\mathrm{cl}(D)\in\mathcal N(\mathord{\uparrow} x_0)$ , which follows that $x_0\in X\setminus\mathrm{cl}(D)$ , contradicting the fact that $x_0\in D$ . Hence, there is $y\in \mathrm{cl}(D)\cap\bigcap\mathcal F\neq\emptyset$ . Then, $y\in\bigcap \mathcal F=\bigcap_{x\in D}\mathord{\uparrow} x$ , so y is an upper bound of D. It follows that $D\subseteq \mathord{\downarrow} y=\mathrm{cl}(\{y\})$ , and since $y\in \mathrm{cl}(D)$ , it follows that $\mathrm{cl}(D)=\mathrm{cl}(\{y\})$ . Hence, by Proposition 2.8 X is a d-space.

sober space $\Longrightarrow$ well-filtered space $\Longrightarrow$ d-space.

4. A Characterization for $\Psi$ -Well-Filtered Spaces

In this section, we will show that a $\Psi$ -well-filtered space X is determined by a class of subsets of X, called $\Psi$ -sets.

1. (1). It is clear that A is a $\Psi$ -set if and only if $\mathrm{cl}(A)$ is a $\Psi$ -set.

2. (2). Every $\Psi$ -set is irreducible by Lemma 2.12.

1. (1). $\mathrm{cl}(A)\cap K\neq\emptyset$ ;

2. (2). $\forall U\in\mathcal N(K)$ , $A\cap U\neq\emptyset$ .

Lemma 4.4. Let X be a $T_0$ space and $K\in\mathcal Q(X)$ . Then,

\begin{equation*}m(\mathcal N(K))=\{\mathrm{cl}(\{x\}):x\in\min K\},\end{equation*}

where $\min K$ is the set of all minimal elements of K in the specialization order of X.

Now assume $A\in m(\mathcal N(K))$ . By Lemma 4.3, there is $a\in A\cap K\neq\emptyset$ . Since $a\in K=\mathord{\uparrow}\min K$ , there exists $x\in\min K$ such that $x\leq a$ , which follows that $x\in \mathrm{cl}(\{a\})\subseteq A$ . Then, $\mathrm{cl}(\{x\})$ is a closed subset of A such that $\mathrm{cl}(\{x\})\in\mathfrak M(\mathcal N(K))$ . By the minimality of A, we have that $A=\mathrm{cl}(\{x\})$ . This shows that $m(\mathcal N(K))\subseteq\{\mathrm{cl}(\{x\}):x\in\min K\}$ , completing the proof.

Using Lemma 4.4, we deduce that $ m(\mathcal N(\mathord{\uparrow} x))=\{\mathrm{cl}(\{x\})\}$ for each point x of a $T_0$ space X, and since $\mathcal N(\mathord{\uparrow} x)\in\Psi(X)$ for each HM-system $\Psi$ , we have the following result.

1. (1). X is $\Psi$ -well-filtered;

2. (2). for each $\Psi$ -set $A\subseteq X$ , there exists $x\in X$ such that $\mathrm{cl}(A)=\mathrm{cl}(\{x\})$ ;

3. (3). for each $\Psi$ -set $A\subseteq X$ and $U\in\mathcal O(X)$ , $\bigcap\mathcal M(A)\subseteq U$ implies $U\in\mathcal M(A)$ ;

4. (4). for each $\Psi$ -set $A\subseteq X$ and $U\in\mathcal O(X)$ , $\bigcap_{a\in A}\mathord{\uparrow} a\subseteq U$ implies $\mathord{\uparrow} a\subseteq U$ for some $a\in A$ .

(2) $\Rightarrow$ (3). It is trivial since $\mathcal M(\mathrm{cl}(\{x\}))=\mathcal N(\{x\})$ for each $x\in X$ .

(3) $\Rightarrow$ (1). Let $\mathcal F\in\Psi(X)$ and $O\in\mathcal O(X)$ such that $\bigcap\mathcal F\subseteq O$ . We need to show that $O\in\mathcal F$ , or equivalently, $U\subseteq O$ for some $U\in\mathcal F$ , because $\mathcal F$ is an upper set. On the contrary, we assume $U\nsubseteq O$ , that is, $U\cap (X\setminus O)\neq\emptyset$ for each $U\in\mathcal F$ . By Lemma 2.12, there exists a closed subset A of $X\setminus O$ such that $A\in m(\mathcal F)$ . Then, A is a $\Psi$ -set in X and note that $\mathcal F\subseteq\mathcal M(A)$ , so $\bigcap\mathcal M(A)\subseteq\bigcap\mathcal F\subseteq O$ . From condition (3), it follows that $O\in\mathcal M(A)$ , that is, $O\cap A\neq\emptyset$ , contradicting the fact that $A\subseteq X\setminus O$ . This implies that $O\in\mathcal F$ . Therefore, X is $\Psi$ -well-filtered.

(2) $\Rightarrow$ (4). Suppose condition (2) is satisfied. Then, there exists $x\in X$ such that $\mathrm{cl}(A)=\mathrm{cl}(\{x\})=\mathord{\downarrow} x$ , which follows that

\begin{equation*}\mathord{\uparrow} x=\bigcap_{a\in\mathord{\downarrow} x}\mathord{\uparrow} a=\bigcap_{a\in\mathrm{cl}(A)}\mathord{\uparrow} a\subseteq\bigcap_{a\in A}\mathord{\uparrow} a\subseteq U.\end{equation*}

This shows that U is an open neighborhood of x, and since $x\in \mathrm{cl}(A)$ , there is $a_0\in U\cap A\neq\emptyset$ , so $\mathord{\uparrow} a_0\subseteq U$ . This gives (4).

(4) $\Rightarrow$ (2). Suppose A is a $\Psi$ -set in X. By Remark 4.2, $\mathrm{cl}(A)$ is also a $\Psi$ -set. Since $\mathord{\uparrow} a\nsubseteq X\setminus\mathrm{cl}(A)$ for each $a\in \mathrm{cl}(A)$ , by condition (4) there exists $x\in \mathrm{cl}(A)\cap\bigcap_{a\in\mathrm{cl}(A)}\mathord{\uparrow} a\neq\emptyset$ . Then, we can easily obtain that $\mathrm{cl}(A)=\mathrm{cl}(\{x\})$ .

1. (1). $\mathcal M(A)\in \mathsf{OF}(X)$ and $\mathrm{cl}(A)\in m(\mathcal M(A))$ .

2. (2). For each $B\in m(\mathcal M(A))$ , $\mathrm{cl}(A)=B$ . Hence, $\mathcal M(B)=\mathcal M(A)$ .

3. (3). If $f:X\longrightarrow Y$ is a continuous mapping, then $f_*(\mathcal M(A))=\mathcal M(f(A))$ , where

   \begin{equation*}f_*(\mathcal M(A))=\{V\in\mathcal O(Y):f^{-1}(V)\in\mathcal M(A)\}.\end{equation*}
   Hence, $\mathrm{cl}_Y(f(A))\in m(f_*(\mathcal M(A)))$ .

(2) Suppose $B\in m(\mathcal M(A))$ . Let $x\in A$ . For each open neighborhood U of x, $U\in\mathcal M(A)$ , hence $B\cap U\neq\emptyset$ . This implies $x\in \mathrm{cl}(B)=B$ . We then deduce that $A\subseteq B$ , so $\mathrm{cl}(A)\subseteq B$ . From the minimality of B, it follows that $\mathrm{cl}(A)=B$ .

(3) For each $V\in\mathcal O(Y)$ , we have that $V\in f_*(\mathcal M(A))$ iff $f^{-1}(V)\cap A\neq\emptyset$ iff $V\cap f(A)\neq\emptyset$ iff $V\in\mathcal M(f(A))$ , which implies that $f_*(\mathcal M(A))=\mathcal M(f(A))$ . Since f is continuous, it follows that f(A) is an irreducible set in Y. Hence, by (1), we have $\mathrm{cl}_Y(f(A))\in m(\mathcal M(f(A)))=m(f_*(\mathcal M(A)))$ .

1. (1). The $\Psi_{\mathsf{sob}}$ -sets are exactly the irreducible sets in X.

2. (2). The $\Psi_{\mathsf{wf}-}$ sets are exactly the KF-sets in X.

3. (3). The closed $\Psi_{\rm d-}$ sets are exactly the closure of directed sets in X.

(2) Suppose $\mathcal F\in \Psi_{\mathsf{wf}}(X)$ . Then, there exists a filtered family $\{K_i:i\in I\}\subseteq\mathcal Q(X)$ such that $\mathcal F=\bigcup_{i\in I}\mathcal N(K_i)$ . By Lemma 4.3, for each subset $A\subseteq X$ , $\mathrm{cl}(A)\in\mathfrak{M}(\mathcal F)$ iff $\mathrm{cl}(A)\cap K_i\neq\emptyset$ for each $i\in I$ . Then, we deduce that the $\Psi_{\mathsf{wf}}$ -sets are exactly the KF-sets.

(3) Suppose A is a closed $\Psi_{\rm d}$ -set in X. Then, there exists a directed subset D of X such that $A\in m(\mathcal M(D))$ . By Lemma 4.7, we have $\mathrm{cl}(D)=A$ . Now suppose E is a directed subset of X. Then, by Lemma 4.7, $\mathrm{cl}(E)\in m(\mathcal M(E))$ . This means that $\mathrm{cl}(E)$ is a closed $\Psi_{\rm d}$ -set in X.

5. Relations between $\Psi$ -Well-Filtered Spaces and H-Sober Spaces

In the paper Xu (Reference Xu2021), Xu introduces the notions of R-subset system and H-sober space, which provides a uniform approach to d-spaces, well-filtered spaces, and sober spaces. In this section, we study the relations between H-sober spaces and $\Psi$ -well-filtered spaces.

1. (H1) for each $T_0$ space X, $\{\{x\} :x\in X\}\subseteq H(X)\subseteq Irr(X)$ ;

2. (H2) For each continuous mapping $f:X\longrightarrow Y$ in ${\bf Top_0}$ and each $A\in H(X)$ , $H(f)(A)=f(A)\in H(Y)$ .

For an R-subset system H and a $T_0$ space X, we call $A\subseteq X$ an H-set if $A\in H(X)$ .

Next, we study the relationship between $\Psi$ -well-filtered spaces and H-sober spaces, the following concept is needed.

1. (1). $f_*(\mathcal N(K))=\mathcal N(f(K))$ .

2. (2). If $A\in m(\mathcal N(K))$ , then $\mathrm{cl}_{Y}(f(A))\in m(\mathcal N (\mathord{\uparrow} f(K\cap A)))$ .

(2) First, by Lemma 4.4 there exists $x\in\min K$ such that $A=\mathrm{cl}_X(\{x\})$ . It follows that $A\cap K=\{x\}$ , so $\mathord{\uparrow} f(K\cap A)=\mathord{\uparrow} f(x)$ . In addition, using Lemma 4.3, one can easily obtain that $\mathrm{cl}_Y(f(A))\in\mathfrak{M}(\mathcal N(\mathord{\uparrow} f(x)))$ . Suppose C is a closed subset of $\mathrm{cl}_Y(f(A))$ such that $C\in\mathfrak{M}(\mathcal N(\mathord{\uparrow} f(x)))$ . By Lemma 4.3, $C\cap \mathord{\uparrow} f(x)\neq\emptyset$ , and since C is a lower set, we have that $f(x)\in C$ , so $A=\mathrm{cl}_X(\{x\})\subseteq f^{-1}(C)$ . It follows that $\mathrm{cl}_Y(f(A))\subseteq C$ . Thus, $C=\mathrm{cl}_Y(f(A))$ . Therefore, $\mathrm{cl}_Y(f(A))\in m(\mathcal N(\mathord{\uparrow} f(K\cap A)))$ .

$\Psi_H(X)=\{\mathcal M(A):A\in H(X)\}\cup\{\mathcal N(K):K\in\mathcal Q(X)\}$ .

for each $T_0$ space X, and for each continuous mapping $f:X\longrightarrow Y$ between $T_0$ spaces X and Y, we define $\Psi_H(f):{\Psi}_H(X)\longrightarrow {\Psi}_H(Y)$ by

\begin{equation*}\Psi_H(f)(\mathcal F)=f_*(\mathcal F)=\{V\in\mathcal O(Y):f^{-1}(V)\in\mathcal F\}\end{equation*}

for each $\mathcal F\in {\Psi}_H(X)$ .

1. (1) $\Psi_H$ is a complete HM-system.

2. (2) A $T_0$ space X is H-sober if and only if it is $\Psi_H$ -well-filtered.

Note that each member of H(X) is irreducible in X, so it is clear that $\Psi_H$ satisfies (H1). To show (HM2), let $f:X\longrightarrow Y$ be a continuous mapping between $T_0$ spaces X and Y, and $A\in H(X)$ , that is, $\mathcal M(A)\in \Psi_H(X)$ . By (H2), $f(A)\in H(Y)$ , and it follows from Lemma 4.7 that $f_*(\mathcal M(A))=\mathcal M(f(A))\in\Psi_H(Y)$ , and since $f_*(\mathcal N(K))=\mathcal N(f(K))\in\Psi_H(Y)$ , (H2) holds. In addition, it is trivial to check that $\Psi_H$ is a covariant functor. Hence, $\Psi_H$ is an HM-system.

Now we prove that $\Psi_H$ is complete. Suppose A is a closed $\Psi_H$ -set in a $T_0$ space X and $f:X\longrightarrow Y$ is a continuous mapping to a $T_0$ space Y. We need to prove that f(A) is a $\Psi_H$ -set in Y. We consider the following cases:

(c1) there exists $K\in\mathcal Q(X)$ such that $A\in m(\mathcal N(K))$ . Note that the intersection $K\cap A$ is compact and since f is continuous, $f (K\cap A)$ is compact in Y, so $\mathord{\uparrow} f(K\cap A)\in\mathcal Q(Y)$ . By Lemma 5.6, $\mathrm{cl}_Y(f(A))\in m(\mathcal N(\mathord{\uparrow} f(K\cap A)))$ . This shows that f(A) is a $\Psi_H$ -set in Y.

(c2) there exists a $B\in H(X)$ such that $A\in m(\mathcal M(B))$ . By Lemma 4.7, $A=\mathrm{cl}_X(B)$ and $\mathrm{cl}_Y(f(A))\in m(\mathcal M(f(A)))$ . Since $f(A)\in H(Y)$ by (H2), it follows that $\mathcal M(f(A))=\mathcal M(f(\mathrm{cl}_X(B)))=\mathcal M(f(B))\in\Psi_H(Y)$ . Thus f(A) is a $\Psi_H$ -set in Y.

By Lemma 5.4, $\Psi$ is a complete HM-system.

(2) ( $\Rightarrow$ ). Assume X is H-sober. Let A be a $\Psi_H$ -set in X. There are two cases:

(c1) $A\in m(\mathcal M(B))$ for some $B\in H(X)$ . By Lemma 4.7, $\mathrm{cl}(B)=\mathrm{cl}(A)$ , and since X is H-sober, $\mathrm{cl}(B)=\mathrm{cl}(A)=\mathrm{cl}(\{x\})$ for some $x\in X$ .

(c2) $A\in m(\mathcal N(K))$ for some $K\in\mathcal Q(X)$ . By Lemma 4.4, $\mathrm{cl}(A)=\mathrm{cl}(\{x\})$ for some $x\in\min K$ .

Then by Theorem 4.6, X is $\Psi_H$ -well-filtered.

( $\Leftarrow$ ). Assume X is a $\Psi_H$ -well-filtered space. Let $A\in H(X)$ . Then by Lemma 4.7, $\mathrm{cl}(A)\in m(\mathcal M(A))\in\Psi_H(X)$ , so A is a $\Psi_H$ -set in X. Since X is $\Psi_H$ -well-filtered, by Theorem 4.6 $\mathrm{cl}(A)=\mathrm{cl}(\{x\})$ for some $x\in X$ . Therefore, X is H-sober.

Theorem 5.8. Let $\Psi:{\bf Top_0}\longrightarrow{\bf Set}$ be a complete HM-system. Define $H_{\Psi}:{\bf Top_0}\longrightarrow{\bf Set}$ by

\begin{equation*}H_{\Psi}(X)=\{A\subseteq X: A \text{ is a }\Psi\text{-set in }X\}\end{equation*}

for each $T_0$ space X, and for each continuous mapping $f:X\longrightarrow Y$ between $T_0$ spaces X and Y, define $H_{\Psi}(f):H_{\Psi}(X)\longrightarrow H_{\Psi}(Y)$ by

\begin{equation*}H_{\Psi}(f)(A)=f(A)\end{equation*}

for each $A\in H_{\Psi}(X)$ .

1. (1). $H_{\Psi}$ is an R-system.

2. (2). A $T_0$ space X is $\Psi$ -well-filtered if and only if it is $H_{\Psi}$ -sober.

(2) It is straightforward by Theorem 4.6.

For an HM-system $\Psi$ , we use $\Psi$ - WF to denote the category of all $\Psi$ -well-filtered spaces with continuous mappings. In Xu (Reference Xu2021), Xu proved that the category of all H-sober spaces with continuous mappings is reflective in the category ${\bf Top_0}$ of $T_0$ spaces. Then by Theorem 5.8, we have the following corollary.

6. The Smyth Power Space of a $\Psi$ -Well-Filtered Space

For a topological space X, the upper Vietoris topology on $\mathcal Q^*(X)=\mathcal Q(X)\setminus\{\emptyset\}$ is generated by the following family (as a base)

\begin{equation*}\Box U=\{K\in\mathcal Q^*(X) : K\subseteq U\},\end{equation*}

where U ranges over the open subsets of X. The resulting space, denoted by $P_s(X)$ , is called the Smyth power space or the upper space.

1. (1). The specialization order of $P_s(X)$ is $\supseteq$ . Hence, for each $\mathcal A\subseteq \mathcal Q^*(X)$ ,

   \begin{equation*}\mathord{\uparrow}_{P_s(X)}\mathcal A=\{K\in\mathcal Q^*(X):K\subseteq G\text{ for some }G\in\mathcal A\}\end{equation*}
   in the specialization order of $P_s(X)$ .

2. (2). Define $\xi:X\longrightarrow P_s(X)$ , $x\mapsto\mathord{\uparrow} x$ . Then $\xi^{-1}(\Box U)=U$ for each $U\in\mathcal O(X)$ , and hence $\xi$ is continuous.

3. (3). If $\mathcal K$ is compact in $P_s(X)$ , then $\bigcup\mathcal K$ is compact in X.

The following theorem strengthens a result in Xu (Reference Xu2021).

1. (1). $P_s(X)$ is H-sober;

2. (2). X is H-sober and H has property Q for X.

From Theorems 5.7 and 5.8, it turns out that there is a one to one correspondence between H-sober spaces and $\Psi$ -well-filtered spaces when $\Psi$ is complete. In other words, the class of $\Psi$ -well-filtered spaces are more general than H-sober spaces. As a generalized result of Theorem 6.4, we have the following result.

1. (1). $P_s(X)$ is $\Psi$ -well-filtered;

2. (2). X is $\Psi$ -well-filtered and $\Psi$ has property Q for X;

3. (3). for each $\Psi$ -set $\mathcal A$ in $P_s(X)$ and each $U\in\mathcal O(X)$ , $\bigcap\mathcal A\subseteq U$ implies $K\subseteq U$ for some $K\in\mathcal A$ .

Proof. (1) $\Rightarrow$ (2). We first prove that X is $\Psi$ -well-filtered. Let $\mathcal F\in\Psi(X)$ and $U\in\mathcal O(X)$ with $\bigcap \mathcal F\subseteq U$ . Consider the continuous mapping $\xi:X\longrightarrow P_s(X)$ , $x\mapsto\mathord{\uparrow} x$ . Then, $\xi_*(\mathcal F)=\{\mathcal U\in\mathcal O(P_s(X)):\xi^{-1}(\mathcal U)\in\mathcal F\}\in\Psi(P_s(X))$ by the definition of $\Psi$ . Let

\begin{equation*}\mathfrak{F}=\xi_*(\mathcal F)\cap\{\Box V:V\in\mathcal O(X)\}.\end{equation*}

For each $V\in\mathcal O(X)$ , observe that $\Box V\in\mathfrak{F}$ iff $\xi^{-1}(\Box V)=V\in\mathcal F$ , which implies that $\mathfrak{F}=\{\Box V:V\in\mathcal F\}$ . If $K\in\bigcap\mathfrak{F}$ , then $K\in\Box V$ (i.e., $K\subseteq V$ ) for all $V\in\mathcal F$ , so $K\subseteq\bigcap\mathcal F\subseteq U$ , which implies $K\in\Box U$ . Thus, $\bigcap\mathfrak{F}\subseteq\Box U$ . Since $P_s(X)$ is $\Psi$ -well-filtered and $\bigcap\xi_*(\mathcal F)\subseteq\bigcap \mathfrak{F}\subseteq\Box U$ , we have that $\Box U\in \xi_*(\mathcal F)$ , so $\xi^{-1}(\Box U)=U\in\mathcal F$ . Hence, X is $\Psi$ -well-filtered. Using a similar method of Theorem 6.4, one can prove that $\Psi$ has property Q for X.

(2) $\Rightarrow$ (3). Assume on the contrary that $K\cap (X\setminus U)\neq\emptyset$ for each $K\in\mathcal A$ . It follows that $X\setminus U\in\mathfrak M(\mathcal A)$ . Since $\Psi$ satisfies property Q for X, there exists a $\Psi$ -set $F\subseteq X\setminus U$ such that $\mathrm{cl}(F)\in\mathfrak M(\mathcal A)$ . Since X is $\Psi$ -well-filtered, by Theorem 4.6 there exists $x\in X$ such that $\mathrm{cl}(F)=\mathrm{cl}(\{x\})$ . For each $K\in \mathcal A$ , since $K=\mathord{\uparrow} K$ and $\mathrm{cl}(F)\cap K=\mathrm{cl}(\{x\})\cap K\neq\emptyset$ , we deduce that $x\in K$ . Thus, $x\in\bigcap\mathcal A\subseteq U$ , contradicting the fact that $x\in \mathrm{cl}(F)\subseteq X\setminus U$ . This shows that $K\subseteq U$ for some $K\in\mathcal A$ , which gives (3).

(3) $\Rightarrow$ (1). Suppose $\mathcal A$ is a $\Psi$ -set in $P_s(X)$ and $\mathcal U$ is an open set in $P_s(X)$ such that $\bigcap_{K\in \mathcal A}\mathord{\uparrow}_{P_s(X)}K\subseteq\mathcal U$ . Then, there exists a family $\{U_i:i\in I\}\subseteq\mathcal O(X)$ such that $\mathcal U=\bigcup_{i\in I}\Box U_i$ . Using a similar proof of Lemma 3.5, one can obtain that $K_0=\bigcap \mathcal A\in\mathcal Q^*(X)$ . Note that $\mathord{\uparrow}_{P_s(X)}K=\{G\in\mathcal Q^*(X):G\subseteq K\}$ for each $K\in\mathcal A$ , so $K_0\in \bigcap_{K\in \mathcal A}\mathord{\uparrow}_{P_s(X)}K\subseteq\bigcup_{i\in I}\Box U_i$ . Then, there exists $i_0\in I$ such that $K_0=\bigcap\mathcal A\in\Box U_{i_0}$ , so $\bigcap\mathcal A\subseteq U_{i_0}$ . By condition (3), there exists $K_1\in\mathcal A$ such that $K_1\subseteq U_{i_0}$ . This implies that $\mathord{\uparrow}_{P_s(X)}K_1\subseteq\Box U_{i_0}\subseteq\mathcal U$ . By Theorem 4.6, we deduce that $P_s(X)$ is $\Psi$ -well-filtered.

Claim: $\bigcup_{K\in\mathcal A}\mathcal N(K)=\bigcup_{i\in I}\mathcal N(\bigcup(\mathcal A\cap \mathcal K_i))$ .

On the one hand, suppose $U\in\bigcup_{i\in I}\mathcal N(\bigcup(\mathcal A\cap \mathcal K_i))$ . Then, there exists $i_0\in I$ such that $\bigcup(\mathcal A\cap\mathcal K_{i_0})\subseteq U$ . Choose one $K_{i_0}\in \mathcal A\cap \mathcal K_{i_0}\neq\emptyset$ . Then, $K_{i_0}\subseteq \bigcup (\mathcal A\cap \mathcal K_{i_0})\subseteq U$ , which implies that $U\in\mathcal N(K_{i_0})\subseteq \bigcup_{K\in\mathcal A}\mathcal N(K)$ . This shows that $\bigcup_{i\in I}\mathcal N(\bigcup(\mathcal A\cap \mathcal K_i))\subseteq\bigcup_{K\in\mathcal A}\mathcal N(K)$ . On the other hand, suppose $U\notin \bigcup_{i\in I}\mathcal N(\bigcup(\mathcal A\cap \mathcal K_i))$ . For each $i\in I$ , $\bigcup(\mathcal A\cap\mathcal K_i)\nsubseteq U$ , so there exists $G_i\in \mathcal A\cap\mathcal K_i$ such that $G_i\nsubseteq U$ ; that is, $G_i\notin \Box U$ . Then, $G_i\in (P_s(X)\setminus \Box U)\cap \mathcal A\cap \mathcal K_i\neq\emptyset$ for all $i\in I$ . Thus, $(P_s(X)\setminus \Box U)\cap \mathcal A$ is a closed subset of $\mathcal A$ such that $(P_s(X)\setminus \Box U)\cap \mathcal A\in\mathfrak{M}(\mathfrak F)$ . Since $\mathcal A\in m(\mathfrak{F})$ , $\mathcal A\cap (P_s(X)\setminus\Box U)=\mathcal A$ , that is, $\mathcal A\subseteq P_s(X)\setminus\Box U$ . It follows that $K\nsubseteq U$ for all $K\in\mathcal A$ ; hence, $U\notin\bigcup_{K\in\mathcal A}\mathcal N(K)$ . Therefore, $\bigcup_{K\in\mathcal A}\mathcal N(K)\subseteq \bigcup_{i\in I}\mathcal N(\bigcup(\mathcal A\cap \mathcal K_i))$ .

Recall that the intersection of a closed set and a compact set is always compact, so $\mathcal A\cap\mathcal K_i$ is compact in $P_s(X)$ for each $i\in I$ . Since $\{\mathcal K_i:i\in I\}$ is a filtered family, by Remark 6.1 $\{\mathord{\uparrow} \bigcup(\mathcal A\cap \mathcal K_i): i\in I\}$ is a filtered family of $\mathcal Q^*(X)$ . Note that for each open subset U of X, $\bigcup(\mathcal A\cap \mathcal K_i)\subseteq U$ if and only if $\mathord{\uparrow} \bigcup(\mathcal A\cap \mathcal K_i)\subseteq U$ , we have that $\bigcup_{K\in\mathcal A}\mathcal N(K)=\bigcup_{i\in I}\mathcal N(\bigcup(\mathcal A\cap \mathcal K_i))=\bigcup_{i\in I}\mathcal N(\mathord{\uparrow}\bigcup\mathcal A\cap \mathcal K_i)\in\Psi_{\mathsf{wf}}(X)$ . Therefore, $\Psi_{\mathsf{wf}}$ satisfies property T.

As a direct consequence of Theorem 6.6, Remark 6.8 and Lemma 6.9, we have the following corollaries.

1. (1). X is sober;

2. (2). $P_s(X)$ is sober.

1. (1). X is well-filtered;

2. (2). $P_s(X)$ is well-filtered.