2 Preliminaries

In this section, we briefly recall some standard definitions and notations that will be used in the paper. Some basic properties of irreducible sets and compact saturated sets are presented.

For a set X, $|X|$ will denote the cardinality of X. Let $\mathbb{N}$ denote the set of all natural numbers and $\omega=|\mathbb N|$ . As a poset (indeed a chain), $\mathbb{N}$ is always endowed with the usual order if no other explanation is given. The set of all subsets of X is denoted by $2^X$ . Let $X^{(<\omega)}=\{F\subseteq X : F \mbox{~is a finite set}\}$ and $X^{(\leq\omega)}=\{F\subseteq X : F \mbox{~is a countable set}\}$ .

For a poset P and $A\subseteq P$ , let $\mathord{\downarrow}A=\{x\in P: x\leq a \mbox{ for some }a\in A\}$ and $\mathord{\uparrow}A=\{x\in P: x\geq a \mbox{ for some } a\in A\}$ . For $x\in P$ , we write $\mathord{\downarrow}x$ for $\mathord{\downarrow}\{x\}$ and $\mathord{\uparrow}x$ for $\mathord{\uparrow}\{x\}$ . A subset A is called a lower set (resp., an upper set) if $A=\mathord{\downarrow}A$ (resp., $A=\mathord{\uparrow}A$ ). Let $\mathbf{Fin} P=\{\uparrow\! F : F\in P^{(<\omega)}\}$ . For a nonempty subset A of P, define $\mathrm{min} (A)=\{u\in A : u$ is a minimal element of $A\}$ and $\mathrm{max} (A)=\{v\in A : v$ is a maximal element of $A\}$ .

A nonempty subset D of a poset P is directed if every two elements in D have an upper bound in D. The set of all directed sets of P is denoted by $\mathcal D(P)$ . $I\subseteq P$ is called an ideal of P if I is a directed lower subset of P. Let $\mathrm{Id} (P)$ be the poset (with the order of set inclusion) of all ideals of P. Dually, we define the notion of filters and denote the poset of all filters of P by $\mathrm{Filt}(P)$ . The poset P is called a directed complete poset, or dcpo for short, if for any $D\in \mathcal D(P)$ , $\vee D$ exists in P.

The poset P is said to be Noetherian if it satisfies the ascending chain condition (ACC for short): every ascending chain has a greatest member. Clearly, P is Noetherian iff every directed set of P has a largest element or, equivalently, every ideal of P is principal (cf. Zhao et al. 2015).

As in Erné (Reference Erné2018), a topological space X is locally hypercompact if for each $x\in X$ and each open neighborhood U of x, there is $\mathord{\uparrow} F\in \mathbf{Fin}X$ such that $x\in{\rm int}\,\mathord{\uparrow} F\subseteq\mathord{\uparrow} F\subseteq U$ . The space X is called a c-space if for each $x\in X$ and each open neighborhood U of x, there is $u\in X$ such that $x\in{\rm int}\,\mathord{\uparrow} u\subseteq\mathord{\uparrow} u\subseteq U$ ). A set K of X is called supercompact if for any family $\{U_i : i\in I\}$ of open sets of X, $K\subseteq \bigcup_{i\in I} U_i$ implies $K\subseteq U$ for some $i\in I$ . It is easy to verify that the nonempty supercompact saturated sets of X are exactly the sets $\mathord{\uparrow} x$ with $x \in X$ (see Heckmann and Keimel Reference Heckmann and Keimel2013, Fact 2.2). It is well-known that X is a c-space iff $\mathcal O(X)$ is a completely distributive lattice (cf. Erné Reference Erné2009).

The category of all $T_0$ spaces and continuous mappings is denoted by $\mathbf{Top}_0$ . For $X\in ob(\mathbf{Top}_0)$ , we use $\leq_X$ to denote the specialization order of X: $x\leq_X y$ iff $x\in \overline{\{y\}}$ ). In the following, when a $T_0$ space X is considered as a poset, the order always refers to the specialization order if no other explanation is given. Let $\mathcal O(X)$ (resp., $\mathcal C(X)$ ) be the set of all open subsets (resp., closed subsets) of X, and let $\mathcal S^u(X)=\{\mathord{\uparrow} x : x\in X\}$ . Define $\mathcal S_c(X)=\{\overline{{\{x\}}} : x\in X\}$ and $\mathcal D_c(X)=\{\overline{D} : D\in \mathcal D(X)\}$ .

Let X be a topological space and $A, B\subseteq X$ . Then,

1. (1) $\overline{A}=\overline{B}$ if and only if for any $U\in \mathcal O(X)$ , $A\cap U\neq \emptyset$ iff $B\cap U\neq\emptyset$ .

2. (2) If $\tau_1, \tau_2$ are two topologies on the set X and $\tau_1\subseteq \tau_2$ , then ${\rm cl}_{\tau_2}A={\rm cl}_{\tau_2}B$ implies ${\rm cl}_{\tau_1}A={\rm cl}_{\tau_1}B$ .

For a $T_0$ space X and a nonempty subset A of X, A is irreducible if for any $\{F_1, F_2\}\subseteq \mathcal C(X)$ , $A \subseteq F_1\cup F_2$ implies $A \subseteq F_1$ or $A \subseteq F_2$ . Denote by ${\mathsf{Irr}}(X)$ (resp., ${\mathsf{Irr}}_c(X)$ ) the set of all irreducible (resp., irreducible closed) subsets of X. Clearly, every subset of X that is directed under $\leq_X$ is irreducible.

The following lemma is well-known and can be easily verified.

For any $T_0$ space X, the lower Vietoris topology on ${\mathsf{Irr}}_c(X)$ is the topology $\{\Diamond U : U\in \mathcal O(X)\}$ , where $\Diamond U=\{A\in {\mathsf{Irr}}_c(X): A\cap U\neq\emptyset\}$ . The resulting space, denoted by $X^s$ , with the canonical mapping $\eta_{X}: X\longrightarrow X^s, x\mapsto\overline {\{x\}}$ , is the sobrification of X (cf. Gierz et al. Reference Gierz, Hofmann, Keimel, Lawson, Mislove and Scott2003; Goubault 2013).

A subset A of a space X is called saturated if A equals the intersection of all open sets containing it (or equivalently, A is an upper set in the specialization order). We shall use $\mathsf{K}(X)$ to denote the set of all nonempty compact saturated subsets of X and endow it with the Smyth preorder $\sqsubseteq$ , that is, for $K_1,K_2\in \mathord{\mathsf{K}}(X)$ , $K_1\sqsubseteq K_2$ iff $K_2\subseteq K_1$ . The poset $(\mathord{\mathsf{K}}(X), \sqsubseteq)$ (shortly $\mathsf{K}(X)$ ) will be called the Smyth power poset of X. The upper Vietoris topology on $\mathsf{K}(X)$ is the topology that has $\{\Box U : U\in \mathcal O(X)\}$ as a base, where $\Box U=\{K\in \mathsf{K}(X) : K\subseteq U\}$ , and the resulting space is called the Smyth power space or upper space of X and is denoted by $P_S(X)$ (cf. Heckmann Reference Heckmann1992; Schalk Reference Schalk1993). For $A\subseteq X$ , define $\Diamond A=\{K\in \mathsf{K}(X) : K\cap A\neq\emptyset\}$ .

1. (1) The specialization order on $P_S(X)$ is the Smyth order (that is, $\leq_{P_S(X)}=\sqsubseteq$ ).

2. (2) The canonical mapping $\xi_X: X\longrightarrow P_S(X)$ , $x\mapsto\mathord{\uparrow} x$ , is an order and topological embedding Heckmann Reference Heckmann1992; Heckmann and Keimel Reference Heckmann and Keimel2013; Schalk Reference Schalk1993).

3. (3) X is homeomorphic to the subspace $S^u(X)$ of $P_S(X)$ by means of $\xi_X$ .

Proof. Clearly, $\Diamond \overline{A}=\mathsf{K}(X)\setminus \Box (X\setminus \overline{A})$ is closed in $P_S(X),$ and hence, ${\rm cl}_{\mathcal O(P_S(X))}\xi_X(A)\subseteq\Diamond \overline{A}$ . Since $\{\Diamond C : C\in \mathcal C(X)\}$ is a (closed) base of $P_S(X)$ , there is a family $\{C_i : i\in I\}\subseteq \mathcal C(X)$ such that ${\rm cl}_{\mathcal O(P_S(X))}\xi_X(A)=\bigcap_{i\in I} \Diamond C_i$ . Then for each $i\in I$ , $\xi_X(A)\subseteq \Diamond C_i$ , and consequently, $\mathord{\uparrow} a\cap C_i\neq \emptyset$ for each $a\in A$ ; whence, for each $a\in A$ , $a\in C_i$ as $C_i=\mathord{\downarrow} C_i$ . It follow that $\overline{A}\subseteq C_i$ for each $i\in I,$ and hence, $\Diamond \overline{A}\subseteq \bigcap_{i\in I}\Diamond C_i={\rm cl}_{\mathcal O(P_S(X))}\xi_X(A)$ . Thus, ${\rm cl}_{\mathcal O(P_S(X))}\xi_X(A)=\Diamond \overline{A}$ .

For a nonempty subset C of a $T_0$ space X, it is easy to see that C is compact iff $\mathord{\uparrow} C\in \mathord{\mathsf{K}} (X)$ . Furthermore, we have the following useful result (see, e.g., Erné Reference Erné2009, pp. 2068).

Proof. Suppose that $\{K_i : i\in I\}\subseteq \mathord{\mathsf{K}} (X)$ is a nonempty family and $\bigvee_{i\in I} K_i$ exists in $\mathord{\mathsf{K}} (X)$ . Let $K=\bigvee_{i\in I} K_i$ . Then, $K\subseteq K_i$ for all $i\in I$ , and hence, $K\subseteq \bigcap_{i\in I} K_i$ . For any $x\in \bigcap_{i\in I} K_i$ , $\mathord{\uparrow} x$ is a upper bound of $\{K_i : i\in I\}\subseteq \mathord{\mathsf{K}} (X)$ , whence $K\sqsubseteq \mathord{\uparrow} x$ or, equivalently, $\mathord{\uparrow} x \subseteq K$ . Therefore, $\bigcap_{i\in I} K_i\subseteq K$ . Thus, $\bigcap_{i\in I} K_i=K\in \mathord{\mathsf{K}} (X)$ .

Conversely, if $\bigcap_{i\in I} K_i\in \mathord{\mathsf{K}} (X)$ , then $\bigcap_{i\in I} K_i$ is an upper bound of $\{K_i : i\in I\}$ in $\mathord{\mathsf{K}} (X)$ . Let $G\in \mathord{\mathsf{K}} (X)$ be another upper bound of $\{K_i : i\in I\}$ , then $G\subseteq K_i$ for all $i\in I$ , and hence, $G\subseteq \bigcap_{i\in I} K_i$ , that is, $\bigcap_{i\in I} K_i\sqsubseteq G$ , proving that $\bigvee_{i\in I} K_i=\bigcap_{i\in I} K_i$ .

Similarly, we have the following.

1. (1) If $\mathcal K\in\mathord{\mathsf{K}}(P_S(X))$ , then $\bigcup \mathcal K\in\mathord{\mathsf{K}}(X)$ .

2. (2) The mapping $\bigcup : P_S(P_S(X)) \longrightarrow P_S(X)$ , $\mathcal K\mapsto \bigcup \mathcal K$ , is continuous.

3. Scott Topology and Continuous Domains

For a poset P, a subset U of P is Scott open if (i) $U=\mathord{\uparrow}U$ and (ii) for any directed subset D with $\vee D$ existing, $\vee D\in U$ implies $D\cap U\neq\emptyset$ . All Scott open subsets of P form a topology, called the Scott topology on P and denoted by $\sigma(P)$ . The space $\Sigma~\!\! P=(P,\sigma(P))$ is called the Scott space of P. For the chain $2=\{0, 1\}$ (with the order $0<1$ ), we have $\sigma(2)=\{\emptyset, \{1\}, \{0,1\}\}$ . The space $\Sigma~\!\!2$ is well-known under the name of Sierpiński space. The upper topology on P, generated by the complements of the principal ideals of P, is denoted by $\upsilon (P)$ . The upper sets of P form the (upper) Alexandroff topology $\alpha (P)$ .

1. (1) $f : \Sigma~\!\! P \longrightarrow \Sigma~\!\! Q$ is continuous.

2. (2) For each $D\in \mathcal D(P)$ for which $\vee D$ exists in P, $\vee f(D)$ exists in Q and $f(\vee D)=\vee f(D)$ .

A mapping $f : P \longrightarrow Q$ satisfying the equivalent conditions (1) and (2) of Lemma 12 is said to be Scott continuous. Let $\mathbf{DCPO}$ denote the category of all dcpos and Scott continuous mappings.

For a dcpo P and $A, B\subseteq P$ , we say A is way below B, written $A\ll B$ , if for each $D\in \mathcal D(P)$ , $\vee D\in \mathord{\uparrow} B$ implies $D\cap \mathord{\uparrow} A\neq \emptyset$ . For $B=\{x\}$ , a singleton, $A\ll B$ is written $A\ll x$ for short. For $x\in P$ , let $w(x)=\{F\in P^{(<\omega)} : F\ll x\}$ , $\Downarrow x = \{u\in P : u\ll x\}$ and $K(P)=\{k\in P : k\ll k\}$ . Points in K(P) are called compact elements of P.

For the following definition and related conceptions, please refer to Gierz et al. (Reference Gierz, Hofmann, Keimel, Lawson, Mislove and Scott2003).

1. (1) P is called a continuous domain, if for each $x\in P$ , $\Downarrow x$ is directed and $x=\vee\Downarrow x$ . When a complete lattice L is continuous, we call L a continuous lattice.

2. (2) P is called an algebraic domain, if for each $x\in P$ , $\{k\in K(P) : k\leq x\}$ is directed and $x=\vee \{k\in K(P) : k\leq x\}$ . When a complete lattice L is algebraic, we call L an algebraic lattice.

3. (3) P is called a quasicontinuous domain, if for each $x\in P$ , $\{\mathord{\uparrow} F : F\in w(x)\}$ is filtered and $\mathord{\uparrow} x=\bigcap\{\mathord{\uparrow} F : F\in w(x)\}$ .

4. (4) X is called core-compact if $\mathcal O(X)$ is a continuous lattice.

A topological space X is said to be a Noetherian space if every open subset is compact (see Goubault 2013, Definition 9.7.1) or, equivalently, if $U\ll U$ in $\mathcal O(X)$ for any open subset U of X. Clearly, if X is a Noetherian space, then $\mathcal O(X)$ is an algebraic lattice.

It is well-known that if a topological space X is locally compact, then it is core-compact (see, e.g., Gierz et al. Reference Gierz, Hofmann, Keimel, Lawson, Mislove and Scott2003, Examples I-1.7). (1978, Section 7) (see also Gierz et al. Reference Gierz, Hofmann, Keimel, Lawson, Mislove and Scott2003, Exercise V-5.25), Hofmann and Lawson gave a second-countable core-compact $T_0$ space X in which every compact subset of X has empty interior, and hence, it is not locally compact.

The following result is well-known (see Gierz et al. Reference Gierz, Hofmann, Keimel, Lawson, Mislove and Scott2003).

1. (1) If P is algebraic, then it is continuous.

2. (2) If P is continuous, then it is quasicontinuous.

3. (3) P is continuous iff $\Sigma ~\!\! P$ is a c-space.

4. (4) P is quasicontinuous iff $\Sigma ~\!\! P$ is locally hypercompact.

4. d-spaces, Well-Filtered Spaces, and Sober Spaces

A $T_0$ space X is called a d-space (or monotone convergence space) if X (with the specialization order) is a dcpo and $\mathcal O(X) \subseteq \sigma(X)$ (cf. Gierz et al. Reference Gierz, Hofmann, Keimel, Lawson, Mislove and Scott2003; Wyler Reference Wyler1981).

It is easy to verify the following result (cf. Gierz et al. 2003; Xu et al. 2020b).

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

2. (2) $\mathcal D_c(X)=\mathcal S_c(X)$ .

3. (3) X is a dcpo, and $\overline{D}=\overline{\{\vee D\}}$ for any $D\in \mathcal D(X)$ .

4. (4) For any $D\in \mathcal D(X)$ and $U\in \mathcal O(X)$ , $\bigcap\limits\limits_{d\in D}\mathord{\uparrow} d\subseteq U$ implies $\mathord{\uparrow} d \subseteq U$ (i.e., $d\in U$ ) for some $d\in D$ .

A topological space X is called sober, if for any $F\in{\mathsf{Irr}}_c(X)$ , there is a unique point $a\in X$ such that $F=\overline{\{a\}}$ . Hausdorff spaces are always sober (see, e.g., Goubault 2013, Proposition 8.2.12), and sober spaces are always $T_0$ since $\overline{\{x\}}=\overline{\{y\}}$ always implies $x=y$ . The Sierpinski space $\Sigma~\!\!2$ is sober but not $T_1,$ and an infinite set with the co-finite topology is $T_1$ but not sober (see Example 57).

The following conclusion is well-known (see, e.g., Gierz et al. Reference Gierz, Hofmann, Keimel, Lawson, Mislove and Scott2003, Reference Gierz, Lawson and Stralka1983; Heckmann Reference Heckmann1992).

For the sobriety of the Smyth power spaces, we have the following well-known result.

1. (1) X is sober.

2. (2) For any $\mathcal A\in {\mathsf{Irr}}(P_S(X))$ and $U\in \mathcal O(X)$ , $\bigcap\mathcal A\subseteq U$ implies $K \subseteq U$ for some $K\in \mathcal A$ .

3. (3) $P_S(X)$ is sober.

A $T_0$ space X is called well-filtered if for any filtered family $\mathcal{K}\subseteq \mathord{\mathsf{K}}(X)$ and open set U, $\bigcap\mathcal{K}{\subseteq} U$ implies $K{\subseteq} U$ for some $K{\in}\mathcal{K}$ . The full subcategory of $\mathbf{Top}_0$ of well-filtered spaces is denoted by $\mathbf{Top}_w$ .

In Xu et al. (2020b), Xu et al. obtained the following equational characterization of well-filtered spaces.

1. (1) X is well-filtered.

2. (2) For every continuous mapping $f : X\longrightarrow Y$ from X to a $T_0$ space Y and a filtered family $\mathcal K\subseteq \mathord{\mathsf{K}}(X)$ , $\mathord{\uparrow} f\left(\bigcap\mathcal K\right)=\bigcap\limits_{K\in\mathcal K}\mathord{\uparrow} f(K)$ .

3. (3) For every continuous mapping $f : X\longrightarrow Y$ from X to a well-filtered space Y and a filtered family $\mathcal K\subseteq \mathord{\mathsf{K}}(X)$ , $\mathord{\uparrow} f\left(\bigcap\mathcal K\right)=\bigcap\limits_{K\in\mathcal K}\mathord{\uparrow} f(K)$ .

4. (4) For every continuous mapping $f : X\longrightarrow Y$ from X to a sober space Y and a filtered family $\mathcal K\subseteq \mathord{\mathsf{K}}(X)$ , $\mathord{\uparrow} f\left(\bigcap\mathcal K\right)=\bigcap\limits_{K\in\mathcal K}\mathord{\uparrow} f(K)$ .

In Xi et al. (2017), Hofmann et al. (1978), and Kou (Reference Kou2001), the following two useful results were given.

1. (1) X locally compact and sober.

2. (2) X is locally compact and well-filtered.

3. (3) X is core-compact and sober.

For the well-filteredness of topological spaces, a similar result to Theorem 19 was proved in Xu et al. (2021b) (see also Xu et al. 2020b).

1. (1) X is well-filtered.

2. (2) $P_S(X)$ is a d-space.

3. (3) $P_S(X)$ is well-filtered.

By Theorem 24 and Corollary 25, we know that for a $T_0$ space X, if $P_S(X)$ is a d-space (equivalently, X is a well-filtered space), then $\Sigma~\!\!\mathsf{K}(X)$ is a d-space. Example 68 below shows that $\Sigma~\!\!\mathsf{K}(X)$ is a sober space does not imply that X is well-filtered (i.e., $P_S(X)$ is a d-space) in general.

The following example shows that there is a $T_0$ space X such that $\mathsf{K}(X)$ (with the Smyth order) is a dcpo but X is not well-filtered.

Figure 1. Johnstone’s dcpo $\mathbb{J}$ .

The set $\mathbb{J}_{max}=\{(n, \infty) : n\in\mathbb N \}$ is the set of all maximal elements of $\mathbb{J}$ . Adding top $\top$ to $\mathbb{J}$ yields a dcpo $\mathbb{J}_{\top}=\mathbb{J}\cup \{\top\}$ ( $x\leq \top$ for any $x\in \mathbb{J}$ ). Then, $\top$ is the largest element of $\mathbb{J}_{\top}$ and $\{\top\}\in \sigma (\mathbb{J}_{\top})$ . The following three conclusions about $\Sigma~\!\!\mathbb{J}$ are known (see, e.g., Lu et al. 2017, Example 3.1 and Miao et al. Reference Miao, Li and Zhao2021, Lemma 3.1):

1. (i) ${\mathsf{Irr}}_c (\Sigma~\!\!\mathbb{J})=\{\overline{\{x\}}=\mathord{\downarrow}_{\mathbb{J}} x : x\in \mathbb{J}\}\cup \{\mathbb{J}\}$ .

2. (ii) $\mathsf{K}(\Sigma~\!\!\mathbb{J})=(2^{\mathbb{J}_{max}} \setminus \{\emptyset\})\bigcup \mathbf{Fin}~\!\mathbb{J}$ .

3. (iii) $\Sigma~\!\!\mathbb{J}$ is not well-filtered.

Hence, we have

1. ${\mathsf{Irr}}_c (\Sigma~\!\!\mathbb{J}_\top)=\{\overline{\{x\}}=\mathord{\downarrow}_{\mathbb{J}_{\top}} x : x\in \mathbb{J}_\top\}\cup \{\mathbb{J}\}$ by (i).

2. $\mathsf{K}(\Sigma~\!\!\mathbb{J}_\top)=\{\mathord{\uparrow}_{\mathbb{J}_{\top}} G : G \mbox{~is nonempty and~} G\subseteq \mathbb{J}_{max}\cup\{\top\}\}\bigcup \mathbf{Fin}~\!\mathbb{J}_\top$ by (ii).

3. $\mathsf{K}(\Sigma~\!\!\mathbb{J})$ is not a dcpo.

Let $\mathcal G=\{\mathbb{J}_{max}\setminus F : F\in (\mathbb{J}_{max})^{(<\omega)}\}$ . Then by (ii), $\mathcal G\subseteq \mathsf{K}(\Sigma~\!\!\mathbb{J}_\top)$ is a filtered family and $\bigcap\mathcal{G}=\bigcap_{F\in (\mathbb{J}_{max})^{(<\omega)}} (\mathbb{J}_{max}\setminus F)=\mathbb{J}_{max}\setminus \bigcup (\mathbb{J}_{max})^{(<\omega)}=\emptyset$ , whence by Lemma 9 $\mathcal G$ has no least upper bound in $\mathsf{K}(X)$ . Thus $\mathsf{K}(\Sigma~\!\!\mathbb{J})$ is not a dcpo.

4. $\mathsf{K}(\Sigma~\!\!\mathbb{J}_\top)$ is a dcpo.

Suppose that $\{K_d : d\in D\}$ is directed in $\mathsf{K}(\Sigma~\!\!\mathbb{J}_\top)$ (with the Smyth order). Then, $\top \in\bigcap_{d\in D}K_d,$ and hence, $\bigcap_{d\in D}K_d\neq\emptyset$ . Now we show that $\bigcap_{d\in D}K_d\in \mathsf{K}(\Sigma~\!\!\mathbb{J}_\top)$ . If $\bigcap_{d\in D}K_d=\{\top\}$ , then obviously $\bigcap_{d\in D}K_d\in \mathsf{K}(\Sigma~\!\!\mathbb{J}_\top)$ . Now we assume $\bigcap_{d\in D}K_d\neq \{\top\}$ and $\{V_i : i\in I\}\subseteq \mathcal \sigma(\mathbb{J}_\top))$ is an open cover of $\bigcap_{d\in D}K_d$ . For each $d\in D$ and $i\in I$ , let $H_d=K_d\setminus \{\top\}$ and $U_i=V_i\setminus \{\top\}$ . Then, $H_d\in \mathsf{K}(\Sigma~\!\!\mathbb{J})$ ( $d\in D$ ), $U_i\in \sigma(\mathbb{J})$ ( $i\in I$ ) and $\emptyset\neq\bigcap_{d\in D}H_d=\bigcap_{d\in D}K_{d_0}\setminus \{\top\}\subseteq \bigcup_{i\in I}V_i\setminus \{\top\}=\bigcup_{i\in I}U_i$ . By Lu et al. (2017, Example 3.1), there is $d_0\in D$ such that $H_{d_0}\in \bigcup_{i\in I}U_i$ , and consequently, there is $J\in I^{(<\omega)}$ such that $H_d\subseteq \bigcup_{i\in J}U_i$ . It follows that $\bigcap_{d\in D}K_d\subseteq K_{d_0}\subseteq \bigcup_{i\in J}V_i$ . Thus $,\bigcap_{d\in D}K_d\in \mathsf{K}(\Sigma~\!\!\mathbb{J}_\top)$ . By Lemma 9, $\mathsf{K}(\Sigma~\!\!\mathbb{J}_\top)$ is a dcpo.

5. $\Sigma~\!\!\mathbb{J}_\top$ is not well-filtered.

Indeed, let $\mathcal K=\{\mathord{\uparrow}_{\mathbb{J}_\top} (\mathbb{J}_{max}\setminus F) : F\in (\mathbb{J}_{max})^{(<\omega)}\}$ . Then by (b), $\mathcal K\subseteq \mathsf{K}(\Sigma~\!\!\mathbb{J}_\top)$ is a filtered family and $\bigcap\mathcal{K}=\bigcap_{F\in (\mathbb{J}_{max})^{(<\omega)}} \mathord{\uparrow}_{\mathbb{J}_\top} (\mathbb{J}_{max}\setminus F)=\bigcap_{F\in (\mathbb{J}_{max})^{(<\omega)}} ((\mathbb{J}_{max}\setminus F)\cup\{\top\})=\{\top\}\cup(\mathbb{J}_{max}\setminus \bigcup (\mathbb{J}_{max})^{(<\omega)})=\{\top\}\in \sigma (\mathbb{J}_\top)$ , but there is no $F\in (\mathbb{J}_{max})^{(<\omega)}$ with $\mathord{\uparrow}_{\mathbb{J}_\top} (\mathbb{J}_{max}\setminus F)\subseteq \{\top\}$ . Therefore, $\Sigma~\!\!\mathbb{J}_\top$ is not well-filtered.

5. Topological Rudin Lemma, Rudin Spaces, and Well-Filtered Determined Spaces

In Section 5, we recall some concepts and results about the topological Rudin Lemma, Rudin spaces, $\omega$ -Rudin spaces, well-filtered determined spaces, and $\omega$ -well-filtered determined spaces that will be used in the next four sections.

Rudin’s Lemma is a useful tool in non-Hausdorff topology and plays a crucial role in domain theory (see Gierz et al. Reference Gierz, Hofmann, Keimel, Lawson, Mislove and Scott2003, Reference Gierz, Lawson and Stralka1983; Heckmann Reference Heckmann1992). Rudin (1980) proved her lemma by transfinite methods, using the Axiom of Choice. Heckmann and Keimel (Reference Heckmann and Keimel2013) presented the following topological variant of Rudin’s Lemma.

Applying Lemma 27 to the Alexandroff topology on a poset P, one obtains the original Rudin’s Lemma.

For a $T_0$ space X and $\mathcal{K}\subseteq \mathord{\mathsf{K}}(X)$ , let $M(\mathcal{K})=\{A\in \mathcal C(X) : K\bigcap A\neq\emptyset \mbox{~for all~} K\in \mathcal{K}\}$ (that is, $\mathcal K\subseteq \Diamond A$ ) and $m(\mathcal{K})=\{A\in \mathcal C(X) : A \mbox{~is a minimal member of~} M(\mathcal{K})\}$ .

In Shen et al. (Reference Shen, Xi, Xu and Zhao2019) and Reference Xu, Shen, Xi and ZhaoXu et al. (2020b ), based on topological Rudin’s Lemma, Rudin spaces and well-filtered determined spaces ( $\mathord{\mathsf{WD}}$ spaces for short) were introduced and investigated. These two spaces are closely related to sober spaces and well-filtered spaces (see Shen et al. Reference Shen, Xi, Xu and Zhao2019; Reference Xu, Shen, Xi and ZhaoXu et al. 2020b ).

1. (1) A nonempty subset A of X is said to have the Rudin property , if there exists a filtered family $\mathcal K\subseteq \mathord{\mathsf{K}}(X)$ such that $\overline{A}\in m(\mathcal K)$ (i.e., $\overline{A}$ is a minimal closed set that intersects all members of $\mathcal K$ ). Let $\mathsf{RD}(X)=\{A\in \mathcal C(X) : A\mbox{~has Rudin property}\}$ . The sets in $\mathsf{RD}(X)$ will also be called Rudin sets.

2. (2) X is called a Rudin space, $\mathsf{RD}$ space for short, if ${\mathsf{Irr}}_c(X)=\mathsf{RD}(X)$ , that is, all irreducible closed sets of X are Rudin sets.

The Rudin property is called the compactly filtered property in Shen et al. (Reference Shen, Xi, Xu and Zhao2019). In order to emphasize its origin from (topological) Rudin’s Lemma, such a property was called the Rudin property in Reference Xu, Shen, Xi and ZhaoXu et al. (2020b). Clearly, A has Rudin property iff $\overline{A}$ has Rudin property (that is, $\overline{A}$ is a Rudin set).

Motivated by Proposition 30, the following concept was introduced in Xu et al. (Reference Xu, Shen, Xi and Zhao2020b).

1. (1) A subset A of X is called a well-filtered determined set, $\mathord{\mathsf{WD}}$ set for short, if for any continuous mapping $ f:X\longrightarrow Y$ to a well-filtered space Y, there exists a unique $y_A\in Y$ such that $\overline{f(A)}=\overline{\{y_A\}}$ . Denote by $\mathsf{WD}(X)$ the set of all closed well-filtered determined subsets of X.

2. (2) X is called a well-filtered determined space, $\mathsf{WD}$ space for short, if all irreducible closed subsets of X are well-filtered determined, that is, ${\mathsf{Irr}}_c(X)=\mathord{\mathsf{WD}} (X)$ .

Obviously, a subset A of a space X is well-filtered determined iff $\overline{A}$ is well-filtered determined.

1. (1) X is well-filtered.

2. (2) $\mathsf{RD}(X)=\mathcal S_c(X)$ .

3. (3) $\mathord{\mathsf{WD}} (X)=\mathcal S_c(X)$ .

1. (1) X is sober.

2. (2) X is a $\mathsf{DC}$ d-space.

3. (3) X is a well-filtered $\mathsf{DC}$ space.

4. (4) X is a well-filtered Rudin space.

5. (5) X is a well-filtered $\mathsf{WD}$ space.

1. (1) If X is locally compact, then X is a Rudin space.

2. (2) If X is core-compact, then X is a $\mathsf{WD}$ space.

It is still not known whether every core-compact $T_0$ space is a Rudin space (see Xu et al. 2021d, Question 5.14).

Question 39. For a core-compact $T_0$ space X, is the Smyth power space $P_S(X)$ a $\mathsf{WD}$ space? Is the Scott space $\Sigma~\!\!\mathord{\mathsf{K}} (X)$ a $\mathsf{WD}$ space?

From Theorem 36 and Proposition 38, one can immediately get the following result, which was first proved by Lawson et al. (2020) using a different method.

By Corollary 40, Theorem23 can be strengthened into the following one.

1. (1) X locally compact and sober.

2. (2) X is locally compact and well-filtered.

3. (3) X is core-compact and sober.

4. (4) X is core-compact and well-filtered.

Figure 2. Certain relations among some kinds of spaces.

Fig. 2 shows certain relations among some kinds of spaces.

In order to emphasize the Scott topology, we introduce the following notions.

Clearly, a sober dcpo is a well-filtered dcpo. For Isbell’s lattice L constructed in Isbell (Reference Isbell1982), $\Sigma~\!\! L$ is non-sober, namely L is not a sober dcpo, and by Proposition 22, L is well-filtered. The Johnstone’s dcpo $\mathbb{J}$ (see Example 26) is not well-filtered.

1. (1) P is said to be a $\mathsf{DC}$ poset if $\Sigma~\!\!P$ is a $\mathsf{DC}$ space.

2. (2) P is said to be a Rudin poset if $\Sigma~\!\! P$ is a Rudin space.

3. (3) P is said to be a well-filtered determined poset, a $\mathord{\mathsf{WD }}$ poset for short, if $\Sigma~\!\! P$ is a well-filtered determined space.

4. (4) When a dcpo P is a Rudin poset (resp., a well-filtered determined poset), we will call P a Rudin dcpo (resp., a well-filtered determined dcpo).

The following corollary follows directly from Theorem 36.

1. (1) P is a sober dcpo.

2. (2) P is a $\mathsf{DC}$ dcpo.

3. (3) P is a $\mathsf{DC}$ well-filtered dcpo.

4. (4) P is a Rudin well-filtered dcpo.

5. (5) P is a $\mathord{\mathsf{WD}}$ well-filtered dcpo.

In Xu et al. (Reference Xu, Shen, Xi and Zhao2020a), the following countable versions of Rudin spaces and $\mathsf{WD}$ spaces were introduced and studied.

1. (a) The set A is said to be an $\omega$ -Rudin set, if there exists a countable filtered family $\mathcal K\subseteq \mathord{\mathsf{K}}(X)$ such that $\overline{A}\in m(\mathcal K)$ . Let $\mathsf{RD}_\omega(X)$ denote the set of all closed $\omega$ -Rudin sets of X.

2. (b) The space X is called $\omega$ -Rudin space, if ${\mathsf{Irr}}_c(X)=\mathsf{RD}_\omega(X)$ or, equivalently, all irreducible (closed) subsets of X are $\omega$ -Rudin sets.

1. (a) The set A is called an $\omega$ -well-filtered determined set, $\omega$ - $\mathord{\mathsf{WD}}$ set for short, if for any continuous mapping $ f:X\longrightarrow Y$ to an $\omega$ -well-filtered space Y, there exists a (unique) $y_A\in Y$ such that $\overline{f(A)}=\overline{\{y_A\}}$ . Denote by $\mathsf{WD}_\omega(X)$ the set of all closed $\omega$ -well-filtered determined subsets of X.

2. (b) The space X is called $\omega$ -well-filtered determined, $\omega$ - $\mathsf{WD}$ space for short, if ${\mathsf{Irr}}_c(X)=\mathsf{WD}_\omega(X)$ or, equivalently, all irreducible (closed) subsets of X are $\omega$ -well-filtered determined.

For a $T_0$ space X, it was proved in Reference Xu, Shen, Xi and ZhaoXu et al. (2020a , Proposition 5.5) that $\mathcal{S}_c(X)\subseteq \mathsf{RD}_\omega(X)\subseteq\mathsf{WD}_\omega(X)\subseteq{\mathsf{Irr}}_c(X)$ . Therefore, every $\omega$ -Rudin space is $\omega$ -well-filtered determined.

The following result is a countable version of Theorem 36.

1. (1) X is sober.

2. (2) X is an $\omega$ -Rudin and $\omega$ -well-filtered space.

3. (3) X is an $\omega$ -well-filtered determined and $\omega$ -well-filtered space.

If the sobrification $X^s$ of X is first-countable, then X is an $\omega$ -Rudin space.

1. (1) If X is first-countable, then X is a $\mathsf{WD}$ space.

From Theorems 36 and 49, we immediately deduce the following result.

It is still not known whether a first-countable $T_0$ space is a Rudin space (see Reference Xu, Shen, Xi and ZhaoXu et al. 2021a , Problem 6.15). Since the first-countability is a hereditary property, from Remark 4 and Theorem 49 we know that if the Smyth power space $P_S(X)$ of a $T_0$ space X is first-countable, then X is a $\mathsf{WD}$ space.

So naturally, we ask the following question.

Question 51. Is a $T_0$ space with a first-countable Smyth power space a Rudin space?

In Example 67, a $T_0$ space X is given for which the Scott space $\Sigma~\!\!\mathsf{K}(X)$ is a first-countable sober c-space but X is not a $\mathsf{WD}$ space (and hence not a Rudin space).

By Proposition 48 and Theorem 49, we have the following result.

In Theorem 49 and Corollary 52, the first-countability of $X^s$ cannot be weakened to that of X as shown in the following example. It also shows that the first-countability of a $T_0$ space X does not imply the first-countability of $X^s$ in general.

1. (a) $\mathcal C(\Sigma P)=\{\mathord{\downarrow} t : t\in P\}\cup\{\emptyset, P\}$ .

2. (b) $\Sigma P$ is first-countable and compact (since P has a least element 0).

3. (c) $(\Sigma P)^s$ is not first-countable.

   In fact, it is easy to verify that $(\Sigma P)^s$ is homeomorphic to $\Sigma [0, \omega_1]$ . Since sup of a countable family of countable ordinal numbers is still a countable ordinal number, $\Sigma [0, \omega_1]$ has no countable base at the point $\omega_1$ .

4. (d) $\mathsf{K}(\Sigma P)=\{\mathord{\uparrow} x : x\in P\}$ and $\Sigma P$ is not an $\omega$ -Rudin space.

For $K\in \mathsf{K}(\Sigma P)$ , we have $\mathrm{inf}~K\in K$ , and hence $K=\mathord{\uparrow} \mathrm{inf}~K$ . So $\mathsf{K}(\Sigma P)=\{\mathord{\uparrow} x : x\in P\}$ . Now we show that the irreducible closed set P is not an $\omega$ -Rudin set. For any countable filtered family $\{\mathord{\uparrow} \alpha_n : n\in\mathbb N \}\subseteq \mathsf{K}(\Sigma P)$ , let $\beta=\mathrm{sup}\{\alpha_n : n\in\mathbb N\}$ . Then, $\beta$ is still a countable ordinal number. Clearly, $\mathord{\downarrow} \beta \in M(\{\mathord{\uparrow} \alpha_n : n\in \mathbb N\})$ and $P\neq \mathord{\downarrow} \beta$ . Therefore, $P\notin m(\{\mathord{\uparrow} \alpha_n : n\in \mathbb N\})$ . Thus, P is not an $\omega$ -Rudin set, and hence, $\Sigma P$ is not an $\omega$ -Rudin space.

1. (e) $\Sigma P$ is a Rudin space.

It is easy to check that ${\mathsf{Irr}}_c(\Sigma P)=\{\downarrow x : x\in P\}\cup \{P\}$ . Clearly, $\downarrow x$ is a Rudin set for each $x\in P$ . Now we show that P is a Rudin set. First, $\{\mathord{\uparrow} s : s\in P\}$ is filtered. Second, $P\in M(\{\mathord{\uparrow} s : s\in P\})$ . For a closed subset B of $\Sigma P$ , if $B\neq P$ , then $B=\mathord{\downarrow} t$ for some $t\in P$ , and hence, $\mathord{\uparrow} (t+1)\cap\mathord{\downarrow} t=\emptyset$ . Thus, $B\notin M(\{\mathord{\uparrow} s : s\in P\})$ , proving that P is a Rudin set.

1. (f) P is not a dcpo (note that P is directed and $\vee P$ does not exist). So $\Sigma P$ is not a d-space, and hence, $\Sigma P$ is neither well-filtered nor sober (see Remark 20).

2. (g) $\Sigma P$ is $\omega$ -well-filtered.

If $\{\mathord{\uparrow} x_n : n\in \mathbb N\}\subseteq \mathsf{K}(\Sigma P)$ is countable filtered family and $U\in \sigma (P)$ with $\bigcap_{n\in \mathbb N}\mathord{\uparrow} x_n\subseteq U$ , then $\{x_n : i\in\mathbb N\}$ is a countable subset of $P=[0, \omega_1)$ . Since sup of a countable family of countable ordinal numbers is still a countable ordinal number, we have $\beta=\mathrm{sup}\{x_n : n\in \mathbb N\}\in P$ , and hence, $\mathord{\uparrow} \beta=\bigcap_{n\in \mathbb N}\mathord{\uparrow} x_n\subseteq U$ . Therefore, $\beta\in U$ , and consequently, $x_n\in U$ for some $n\in \mathbb N$ or, equivalently, $\mathord{\uparrow} x_n\subseteq U$ , proving that $\Sigma P$ is $\omega$ -well-filtered.

6 Well-Filteredness of Scott Topology on Smyth Power Posets

In this section, we mainly discuss the following two questions:

Question 1. Is the Scott space $\Sigma~\!\! \mathord{\mathsf{K}}(X)$ of a d-space X a d-space?

Question 2. Is the Scott space $\Sigma~\!\! \mathord{\mathsf{K}}(X)$ of a well-filtered space X well-filtered?

First, Example 57 below shows that there is a second-countable Noetherian d-space X for which $\mathord{\mathsf{K}} (X)$ is not a dcpo, and hence, neither the Smyth power space $P_S(X)$ nor the Scott space $\Sigma~\!\!\mathord{\mathsf{K}} (X)$ is a d-space, which gives a negative answer to Question 1.

In order to present the example, we need the following lemma.

Proof. It was proved by Schalk (Reference Schalk1993, the proof of Lemma 7.26). We present a more direct proof here.

Suppose that $\mathcal U\in \sigma(\mathsf{K}(X))$ and $K\in \mathcal U$ . Let $\mathcal K=\{G\in \mathsf{K}(X) : K\subseteq {\rm int} ~\!G\}$ . Now we show that $\mathcal K$ is filtered and $G=\bigcap \mathcal K$ .

$1^{\circ}$ For each $U\in \mathcal O(X)$ with $K\subseteq U$ , there is $G_U\in \mathcal K$ with $G_U\subseteq U$ .

If $U\in \mathcal O(X)$ for which $K\subseteq U$ , then for each $x\in K$ , there is $K_x\in \mathsf{K}(X)$ such that $x\in {\rm int}~\!K_x\subseteq K_x\subseteq U$ since X is locally compact. By the compactness of K, there is $\{x_1, x_2, ..., x_n\}\subseteq K$ such that $K\subseteq \bigcup\limits_{i=1}^{n} {\rm int}~\!K_{x_i}$ . Let $G_U=\bigcup\limits_{i=1}^{n} K_{x_i}$ . Then, $K\subseteq {\rm int}~\!G_U\subseteq G_U\subseteq U$ , whence $G_U\in \mathcal K$ and $G_U\subseteq U$ .

$2^{\circ}$ $\mathcal K$ is filtered.

Suppose that $G_1, G_2\in \mathcal K$ . Then, $K\subseteq {\rm int}~\!G_1\cap {\rm int}~\!G_2$ . Hence, by what was shown above, there is $G_3\in \mathcal K$ with $G_3\subseteq {\rm int}~\!G_1\cap {\rm int}~\!G_2\subseteq G_1\cap G_2$ , proving the filteredness of $\mathcal K$ .

By $1^{\circ}$ and $2^{\circ}$ , $K\subseteq \bigcap \mathcal K\subseteq \{U\in \mathcal O(X) : K\subseteq U\}=K$ , whence $K=\bigcap \mathcal K=\bigvee_{\mathsf{K}(X)}\mathcal K$ by Lemma 9. Since $K\in \mathcal U\in \sigma(\mathsf{K}(X))$ , $G\in \mathcal U$ for some $G\in \mathcal K$ . Hence, $K\in \Box_{\mathsf{K}(X)}{\rm int}~\!G\subseteq \mathcal U$ . Thus, $\mathcal U\in \mathcal O(P_S(X))$ .

By Corollary 25 and Lemma 54, we get the following corollary.

Considering Remark 14 and Lemma 54, we have the following question.

Question 56. For a core-compact $T_0$ space X, is the Scott topology coarser than the upper Vietoris topology on $\mathsf{K}(X)$ ?

1. (a) $\mathcal C(X_{cof})=\{\emptyset, X\}\cup X^{(<\omega)}$ , $X_{cof}$ is $T_1$ and hence a d-space.

2. (b) ${\mathsf{Irr}}_c (X_{cof})=\{\{x\} : x\in X\}\cup \{X\}$ .

3. (c) $\mathord{\mathsf{K}} (X_{cof})=2^X\setminus \{\emptyset\}$ .

4. (d) $X_{cof}$ is second-countable.

Clearly, $\mathcal O(X_{cof})$ is countable, and hence, $X_{cof}$ is second-countable.

1. (e) $X_{cof}$ is Noetherian and hence locally compact.

Since every subset of X is compact in $X_{cof}$ , the space $X_{cof}$ is a Noetherian space and hence a locally compact space.

1. (f) $X_{cof}$ is a Rudin space.

By (e) and Proposition 38 (or by (d) and Corollary 108 below), $X_{cof}$ is a Rudin space.

1. (g) $\mathord{\mathsf{K}} (X_{cof})$ is not a dcpo, and hence, $X_{cof}$ is neither well-filtered nor sober.

$\mathcal{K}=\{X\setminus F : F\in X^{(<\omega)}\}\subseteq\mathord{\mathsf{K}} (X_{cof})$ is countable filtered and $\bigcap \mathcal{K}_X=X\setminus \bigcup X^{(<\omega)}=X\setminus X=\emptyset$ , whence $\bigvee \mathcal K$ does not exist in $\mathord{\mathsf{K}} (X_{cof})$ by Lemma 9. Thus, $\mathord{\mathsf{K}} (X_{cof})$ is not a dcpo, whence by Remark 20 and Theorem 24, $X_{cof}$ is neither well-filtered nor sober.

1. (h) The upper Vietoris topology and the Scott topology on $\mathord{\mathsf{K}} (X_{cof})$ agree.

By the local compactness of $X_{cof}$ and Lemma 54, we have $\sigma (\mathord{\mathsf{K}} (X_{cof})\subseteq \mathcal O(P_S(X_{cof}))$ . Now we show that $\Box U\in \sigma (\mathsf{K}(X_{cof})$ for each $U\in \mathcal O(X_{cof})\setminus \{\emptyset\}$ . Clearly, $\Box U=\mathord{\uparrow}_{\mathsf{K}(X_{cof})}\Box U$ . Suppose that $\mathcal K_D=\{K_d : d\in D\}\in \mathcal D(\mathord{\mathsf{K}} (X_{cof})$ and $\bigvee_{\mathord{\mathsf{K}} (X_{cof})} \mathcal K_D\in \Box U$ . Then by Lemma 9 $\bigcap_{d\in D}K_d=\bigvee_{\mathord{\mathsf{K}} (X_{cof})} \mathcal K_D\subseteq U$ or, equivalently, $X\setminus U\subseteq \bigcup_{d\in D}(X\setminus K_d)$ . Since $X\setminus U$ is finite and $ \{X\setminus K_d : d\in D\}$ is directed, there is $d_0\in D$ with $X\setminus U\subseteq X\setminus K_{d_0}$ , whence $K_{d_0}\subseteq U$ . Thus, $\Box U\in \sigma (\mathsf{K}(X_{cof})$ . Therefore, $\mathcal O(P_S(X_{cof}))\subseteq\sigma (\mathord{\mathsf{K}} (X_{cof}),$ and hence, $\sigma (\mathord{\mathsf{K}} (X_{cof})=\mathcal O(P_S(X_{cof}))$ .

1. (i) $\Sigma~\!\!\mathord{\mathsf{K}} (X_{cof})$ is not a d-space, and hence, it is neither a well-filtered space nor a sober space.

Since $\mathord{\mathsf{K}} (X_{cof})$ is not a dcpo, $\Sigma~\!\!\mathord{\mathsf{K}} (X_{cof})$ is not a d-space. By Remark 20, $\Sigma~\!\!\mathord{\mathsf{K}} (X_{cof})$ is neither a well-filtered space nor a sober space.

Now we investigate Question 2. First, as one of the main results of this paper, we have the following conclusion.

Proof. By Corollary 25, $\mathord{\mathsf{K}} (X)$ is a dcpo and $\mathcal O(P_S(X))\subseteq \sigma (\mathord{\mathsf{K}} (X))$ (i.e., $\Box U\in \sigma (\mathord{\mathsf{K}} (X))$ for all $U\in O(X)$ ). Suppose that $\{\mathcal K_d : d\in D\}\subseteq \mathord{\mathsf{K}}(\Sigma~\!\! \mathord{\mathsf{K}} (X))$ is filtered, $\mathcal U\in \sigma (\mathord{\mathsf{K}} (X))$ and $\bigcap\limits_{d\in D} \mathcal K_d \subseteq \mathcal U$ . If $\mathcal K_d\not\subseteq \mathcal U$ for each $d\in D$ , that is, $\mathcal K_d\bigcap (\mathord{\mathsf{K}} (X)\setminus \mathcal U)\neq\emptyset$ , then $\{\mathcal K_d : d\in D\}\in {\mathsf{Irr}} (P_S(\mathord{\mathsf{K}}(\Sigma~\!\! \mathord{\mathsf{K}} (X)))),$ and hence by Lemma 27, $\mathord{\mathsf{K}} (X)\setminus \mathcal U$ contains a minimal irreducible closed subset $\mathcal A$ that still meets all members $\mathcal K_d$ . For each $d\in D$ , let $K_d=\bigcup \mathord{\uparrow}_{\mathsf{K}(X)} (\mathcal K_d\cap \mathcal A)$ .

Claim 1: For each $d\in D$ , $K_d\in \mathord{\mathsf{K}} (X)$ and $K_d\in \mathcal A$ .

By $\mathcal K_d\in \mathord{\mathsf{K}}(\Sigma~\!\! \mathord{\mathsf{K}} (X))$ and $\mathcal A\in \mathcal C(\Sigma~\!\! \mathord{\mathsf{K}} (X))$ , we have that $\mathord{\uparrow}_{\mathsf{K}(X)} (\mathcal K_d\cap \mathcal A)\in \mathord{\mathsf{K}}(\Sigma~\!\! \mathord{\mathsf{K}} (X))$ , and hence, $\mathord{\uparrow}_{\mathsf{K}(X)} (\mathcal K_d\cap \mathcal A)\in \mathord{\mathsf{K}} (P_s(X))$ by $\mathcal O(P_S(X))\subseteq \sigma (\mathord{\mathsf{K}} (X))$ . By Lemma 11, $K_d=\bigcup \mathord{\uparrow}_{\mathsf{K}(X)} (\mathcal K_d\cap \mathcal A)=\bigcup (\mathcal K_d\cap \mathcal A)\in \mathord{\mathsf{K}} (X)$ . Since $\mathcal A=\mathord{\downarrow}_{\mathord{\mathsf{K}} (X)}\mathcal A$ and $\mathcal K_d\cap \mathcal A\neq\emptyset$ , we have $K_d\in\mathcal A$ .

Claim 2: $\{K_d : d\in D\}\subseteq \mathord{\mathsf{K}} (X)$ is filtered (by Claim 1 and the filteredness of $\{\mathcal K_d : d\in D\}$ ).

Claim 3: $K=\bigcap_{d\in D}K_d\in \mathord{\mathsf{K}} (X)$ and $K\in \mathcal A$ .

By the well-filteredness of X, $K=\bigcap_{d\in D}K_d\in \mathord{\mathsf{K}} (X)$ . By Claims 1, 2 and Lemma 9, $K=\bigvee_{\mathord{\mathsf{K}} (X)} \{K_d : d\in D\}\in \mathcal A$ since $\mathcal A\in \mathcal C(\Sigma \mathord{\mathsf{K}} (X))$ .

Claim 4: For each $k\in K$ , $\mathcal A\subseteq \Diamond_{\mathord{\mathsf{K}} (X)}\overline{\{k\}}$ .

For each $d\in D$ , we have $k\in K\subseteq K_d=\bigcup (\mathcal K_d\cap \mathcal A)$ , whence there is $G_d\in \mathcal K_d\cap \mathcal A$ such that $k\in G_d$ , and consequently, $G_d\in \mathcal K_d\cap \mathcal A\cap \Diamond_{\mathord{\mathsf{K}} (K)}\overline{\{k\}}$ . Therefore, $\mathcal A\cap \Diamond_{\mathord{\mathsf{K}} (K)}\overline{\{k\}}\in M(\{\mathcal K_d : d\in D\})$ . By the minimality of $\mathcal A$ and $\Diamond_{\mathord{\mathsf{K}} (K)}\overline{\{k\}}\in \mathcal C(P_S(X))\subseteq \mathcal C(\Sigma~\!\!\mathord{\mathsf{K}} (X))$ , we have $\mathcal A=\Diamond_{\mathord{\mathsf{K}} (K)}\overline{\{k\}}\bigcap\mathcal A$ , that is, $\mathcal A\subseteq \Diamond_{\mathord{\mathsf{K}} (K)}\overline{\{k\}}$ .

Claim 5: $\mathcal A=\mathord{\downarrow}_{\mathord{\mathsf{K}} (X)}K$ .

By Claims 3 and 4, $\mathord{\downarrow}_{\mathord{\mathsf{K}} (X)}K\subseteq\mathcal A\subseteq \bigcap_{k\in K} \Diamond_{\mathord{\mathsf{K}} (K)}\overline{\{k\}}$ . Clearly,

This implies that $\bigcap_{k\in K} \Diamond_{\mathord{\mathsf{K}} (K)}\overline{\{k\}}=\mathord{\downarrow}_{\mathord{\mathsf{K}} (X)}K$ , and hence, $\mathcal A=\mathord{\downarrow}_{\mathord{\mathsf{K}} (X)}K$ .

Claim 6: $K\in \bigcap_{d\in D}\mathcal K_d$ .

For each $d\in D$ , by $\mathcal K_d\bigcap \mathcal A\neq\emptyset$ , $\mathcal K_d=\mathord{\uparrow}_{\mathord{\mathsf{K}} (X)}\mathcal K_d$ and $\mathcal A=\mathord{\downarrow}_{\mathord{\mathsf{K}} (X)}K$ , we have $K\in \mathcal K_d$ , whence $K\in \bigcap_{d\in D}\mathcal K_d\subseteq \mathcal U$ , being a contradiction with $K\in \mathcal A\subseteq \mathord{\mathsf{K}} (X)\setminus \mathcal U$ .

Therefore, there is $d_0\in D$ such that $\mathcal K_{d_0}\subseteq\mathcal U$ , proving that $\Sigma~\!\! \mathord{\mathsf{K}} (X)$ is well-filtered.

Proof. Let $f : X \longrightarrow Y$ be a morphism in $\mathbf{Top}_w$ .

Claim 1: $\Phi(f) : \Phi(X) \longrightarrow \Phi(Y)$ is well-defined.

By Theorem 58, $\Phi(X)$ and $\Phi(Y)$ are well-filtered dcpos. For any $ K\in\Phi(X)$ , since $f : X \longrightarrow Y$ is continuous, $\Phi(f)(K)=\mathord{\uparrow} f(K)\in \Phi(Y)$ . Thus, $\Phi(f) : \Phi(X) \longrightarrow \Phi(Y)$ is well-defined.

Claim 2: $\Phi(f) : \Phi(X) \longrightarrow \Phi(Y)$ is Scott continuous.

Let $\{K_d : d\in D\}\in \mathcal{D}(\Phi(X))$ (note that $\Phi(X)$ is endowed with the Smyth order). Then by Lemma 9 and Proposition 21, we have that $\bigvee_{\Phi(X)} \{K_d : d\in D\}=\bigcap_{d\in D}K_d$ and

It follows that $\Phi(f) : \Phi(X) \longrightarrow \Phi(Y)$ is Scott continuous by Lemma 12.

Claim 3: $\Phi(id_X)=id_{\Phi(X)}$

For each $K\in \Phi(X)$ , $P_S(id_X)(K)=\mathord{\uparrow} id_X(K)=\mathord{\uparrow} K=K$ .

Claim 4: For any morphism $g : Y \longrightarrow Z$ in $\mathbf{Top}_w$ , $\Phi(g\circ f)=\Phi(g)\circ \Phi(f)$ .

For any $K\in \Phi(X)$ , $\Phi(g\circ f)(K)=\mathord{\uparrow} g\circ f(K)=\mathord{\uparrow} g(f(K))=\mathord{\uparrow} g(\mathord{\uparrow} f(K))=\Phi(g)\circ \Phi(f)(K)$ . So $\Phi(g\circ f)=\Phi(g)\circ \Phi(f)$ .

Thus, $\Phi : \mathbf{Top}_w \longrightarrow \mathbf{DCPO}_w$ is a covariant functor.

Example 68 shows that unlike Smyth power spaces (see Theorem 24), the converse of Theorem 58 does not hold.

From Theorems 41 and 58, we deduce the following result.

1. (1) $\Sigma~\!\!\mathord{\mathsf{K}} (X)$ is core-compact.

2. (2) $\Sigma~\!\!\mathord{\mathsf{K}} (X)$ is locally compact.

By Theorems 36 and 58, we have the following corollary.

1. (1) $\Sigma~\!\!\mathord{\mathsf{K}} (X)$ is sober.

2. (2) $\Sigma~\!\!\mathord{\mathsf{K}} (X)$ is Rudin.

3. (3) $\Sigma~\!\!\mathord{\mathsf{K}} (X)$ is well-filtered determined.

1. (1) If $\mathcal K\in\mathord{\mathsf{K}}(\Sigma~\!\!\mathord{\mathsf{K}} (X))$ , then $\bigcup \mathcal K\in\mathord{\mathsf{K}}(X)$ .

2. (2) The mapping $\bigcup : \Sigma~\!\!\mathord{\mathsf{K}}(\Sigma~\!\!\mathord{\mathsf{K}} (X)) \longrightarrow \Sigma~\!\!\mathord{\mathsf{K}} (X)$ , $\mathcal K\mapsto \bigcup \mathcal K$ , is continuous.

Proof. (1): By Corollary 25, $\mathcal O(P_S(X))\subseteq \sigma (\mathord{\mathsf{K}} (X))$ . For $\mathcal K\in \mathord{\mathsf{K}}(\Sigma~\!\!\mathord{\mathsf{K}} (X))$ , we have $\mathcal K\in \mathord{\mathsf{K}}(P_S(X))$ since $\mathcal O(P_S(X))\subseteq \sigma (\mathord{\mathsf{K}} (X))$ . Then by Lemma 11, $\bigcup \mathcal K\in\mathord{\mathsf{K}}(X)$ .

(2): Suppose that $\{\mathcal K_d : d\in D\}\subseteq \mathord{\mathsf{K}}(\Sigma~\!\!\mathord{\mathsf{K}} (X))$ is directed (with the Smyth order) for which $\bigvee_{\mathord{\mathsf{K}}(\Sigma~\!\!\mathord{\mathsf{K}} (X))}\{\mathcal K_d : d\in D\}$ exists. Then by Lemma 9, $\bigvee_{\mathord{\mathsf{K}}(\Sigma~\!\!\mathord{\mathsf{K}} (X))}\{\mathcal K_d : d\in D\}=\bigcap_{d\in D}\mathcal K_d$ . It follows that $\bigcup \bigvee_{\mathord{\mathsf{K}}(\Sigma~\!\!\mathord{\mathsf{K}} (X))}\{\mathcal K_d : d\in D\}=\bigcup \bigcap_{d\in D}\mathcal K_d$ and $\bigvee_{d\in D} \bigcup \mathcal K_d=\bigcap_{d\in D}\bigcup \mathcal K_d=\bigcup_{\varphi\in \prod_{d\in D}\mathcal K_d}\bigcap_{d\in D}\varphi(d)$ by Lemma 9. For each $K\in \bigcap_{d\in D}\mathcal K_d$ , define $\varphi_{K}\in \prod_{d\in D}\mathcal K_d$ by $\varphi_K(d)\equiv K$ for all $d\in D$ . Then, $\bigcap_{d\in D}\varphi_K(d)=K$ . Hence $\bigcup \bigcap_{d\in D}\mathcal K_d\subseteq \bigcup_{\varphi\in \prod_{d\in D}\mathcal K_d}\bigcap_{d\in D}\varphi(d)$ .

Conversely, for each $\varphi\in \prod_{d\in D}\mathcal K_d$ , $x\in \bigcap_{d\in D}\varphi(d),$ and $d^{\prime}\in D$ , we have that $\mathord{\uparrow} x\sqsupseteq\varphi (d^{\prime})\in \mathcal K_{d^{\prime}}=\mathord{\uparrow}_{\mathord{\mathsf{K}}(\Sigma~\!\!\mathord{\mathsf{K}} (X))}\mathcal K_d^{\prime}$ , and consequently, $\mathord{\uparrow} x\in \bigcap_{d^{\prime}\in D}\mathcal K_d^{\prime}$ and hence, $\mathord{\uparrow} x\subseteq \bigcup \bigcap_{d\in D}\mathcal K_d$ . It follows that $\bigcap_{d\in D}\varphi(d)\subseteq \bigcup \bigcap_{d\in D}\mathcal K_d$ . Therefore, $\bigcup_{\varphi\in \prod_{d\in D}\mathcal K_d}\bigcap_{d\in D}\varphi(d)\subseteq \bigcup \bigcap_{d\in D}\mathcal K_d$ .

Thus, $\bigcup \bigvee_{\mathord{\mathsf{K}}(\Sigma~\!\!\mathord{\mathsf{K}} (X))}\{\mathcal K_d : d\in D\}=\bigvee_{d\in D} \bigcup \mathcal K_d$ . By Lemma 12, $\bigcup : \Sigma~\!\!\mathord{\mathsf{K}}(\Sigma~\!\!\mathord{\mathsf{K}} (X)) \longrightarrow \Sigma~\!\!\mathord{\mathsf{K}} (X)$ is continuous.

Proof. Suppose that $\{K_d : d\in D\}\subseteq \mathsf{K}(X)$ is filtered and $U\in \mathcal O(X)$ with $\bigcap_{d\in D}K_d\subseteq U$ . Then, $\{\mathord{\uparrow}_{\mathsf{K}(X)}K_d : d\in D\}\subseteq \mathsf{K}(\Sigma~\!\! \mathsf{K}(X))$ is filtered, $\Box U\in \mathcal O(P_S(X))\subseteq \sigma (\mathsf{K}(X))$ and $\bigcap_{d\in D}\mathord{\uparrow}_{\mathsf{K}(X)}K_d\subseteq \Box U$ . By the well-filteredness of $\Sigma~\!\! \mathsf{K}(X)$ , there is $d\in D$ such that $\mathord{\uparrow}_{\mathsf{K}(X)}K_d\subseteq \Box U$ , and hence, $K_d\subseteq U$ . Thus, X is well-filtered.

Example 68 below shows that when X lacks the condition of $\mathcal O(P_S(X))\subseteq \sigma (\mathsf{K}(X))$ , Proposition 65 may not hold.

1. (1) X is well-filtered.

2. (2) The upper Vietoris topology is coarser than the Scott topology on $\mathsf{K}(X)$ , and $\Sigma~\!\! \mathsf{K}(X)$ is well-filtered.

3. (3) The upper Vietoris topology is coarser than the Scott topology on $\mathsf{K}(X)$ , and $\Sigma~\!\! \mathsf{K}(X)$ is a d-space.

4. (4) $\mathsf{K}(X)$ is a dcpo, and the upper Vietoris topology is coarser than the Scott topology on $\mathsf{K}(X)$ .

Proof. (1) $\Rightarrow$ (2): By Corollary 25 and Theorem 58.

(2) $\Rightarrow$ (3): By Remark 20.

(3) $\Rightarrow$ (4): Trivial.

(4) $\Rightarrow$ (1): By (4), $P_S(X)$ is a d-space, whence X is well-filtered by Theorem 24.

7. Non-sobriety of Scott Topology on Smyth Power Poset of a Sober Space

In this section, we investigate the following question:

Question 3. Is the Scott space $\Sigma~\!\! \mathord{\mathsf{K}}(X)$ of a sober space X sober?

First, the following example shows that there is a well-filtered space X for which the Scott space $\Sigma~\!\!\mathord{\mathsf{K}} (X)$ is a first-countable sober c-space, but X is not sober although the Scott space $\Sigma~\!\! \mathord{\mathsf{K}} (X)$ is sober by Corollary 94. Hence, by Corollaries 40 and 50, X is neither core-compact nor first-countable. So the sobriety of the Scott space $\Sigma~\!\!\mathord{\mathsf{K}} (X)$ of a $T_0$ space X does not imply the sobriety of X in general.

(a) $\mathcal C(X_{coc})=\{\emptyset, X\}\bigcup X^{(\leq\omega)}$ , $X_{coc}$ is $T_1$ and hence a d-space, and the specialization order on $X_{coc}$ is the discrete order.

(b) Neither $X_{coc}$ nor $P_S(X_{coc})$ is first-countable.

For a point $x\in X$ , suppose that there is a countable base $\{X\setminus C_n : n\in \mathbb{N}, C_n\in X^{(\leq\omega)}\}$ at x in $X_{coc}$ . Let $C=\bigcup_{n\in \mathbb{N}}C_n$ . Then, $C\in X^{(\leq\omega)}$ . Select $t\in X\setminus (C\cup\{x\})$ and let $U=X\setminus \{t\}$ . Then, $x\in U\in \mathcal O(X_{coc})$ . But $X\setminus C_n\not\subseteq U$ for every $n\in \mathbb{N}$ , a contradiction. Thus, $X_{coc}$ is not first-countable. Since the first-countability is a hereditary property and $X_{coc}$ is homeomorphic to the subspace $\mathcal S^u(X_{coc})$ of $P_S(X_{coc})$ (see Remark 4 or Proposition 102 below), $P_S(X_{coc})$ is not first-countable.

(c) ${\mathsf{Irr}}_c(X_{coc})=\{\overline{\{x\}} : x\in X\}\cup\{X\}=\{\{x\} : x\in X\}\cup\{X\}$ . Therefore, $X_{coc}$ is not sober.

(d) $\mathord{\mathsf{K}} (X_{coc})=X^{(<\omega)}\setminus \{\emptyset\}$ and ${\rm int}\ K=\emptyset$ for all $K\in \mathord{\mathsf{K}} (X_{coc})$ , and hence, $X_{coc}$ is not locally compact.

Clearly, every finite subset is compact. Conversely, if $C\subseteq X$ is infinite, then C has an infinite countable subset $\{c_n : n\in\mathbb N\}$ . Let $C_0=\{c_n : n\in\mathbb N\}$ and $U_m=(X\setminus C_0)\cup \{c_m\}$ for each $m\in \mathbb N$ . Then, $\{U_n : n\in\mathbb N\}$ is an open cover of C, but has no finite subcover. Hence, C is not compact. Thus, $\mathord{\mathsf{K}} (X_{coc})=X^{(<\omega)}\setminus \{\emptyset\}$ . Clearly, ${\rm int}~\!K=\emptyset$ for all $K\in \mathord{\mathsf{K}} (X_{coc})$ . Hence, $X_{coc}$ is not locally compact.

(e) $X_{coc}$ is well-filtered and not core-compact.

Suppose that $\{F_d : d\in D\}\subseteq \mathord{\mathsf{K}} (X_{coc})$ is a filtered family and $U\in \mathcal O(X_{coc})$ with $\bigcap_{d\in D}F_d\subseteq U$ . As $\{F_d : d\in D\}$ is filtered and all $F_d$ are finite, $\{F_d : d\in D\}$ has a least element $F_{d_0}$ , and hence, $F_{d_0}=\bigcap_{d\in D}F_d\subseteq U$ , proving that $X_{coc}$ is well-filtered. By (d) and Theorem 41, $X_{coc}$ is not core-compact.

(f) $\mathord{\mathsf{K}} (X_{coc})$ is a Noetherian dcpo, and hence, $\Sigma~\!\!\mathord{\mathsf{K}} (X_{coc})=(\mathord{\mathsf{K}} (X_{coc}), \alpha(\mathord{\mathsf{K}} (X_{coc}))$ is first-countable.

Clearly, $\mathord{\mathsf{K}} (X_{coc})=X^{(<\omega)}\setminus \{\emptyset\}$ (with the Smyth order) is a Noetherian dcpo and $\sigma(\mathord{\mathsf{K}} (X_{coc}))=\alpha(\mathord{\mathsf{K}} (X_{coc}))$ . For any $F\in \mathord{\mathsf{K}} (X_{coc})=X^{(<\omega)}\setminus \{\emptyset\}$ , $\{\mathord{\uparrow}_{\mathord{\mathsf{K}} (X_{coc})}F\}$ is a base at F in $\Sigma~\!\! \mathord{\mathsf{K}}(X_{coc})$ . Hence $\Sigma~\!\! \mathord{\mathsf{K}}(X_{coc})$ is first-countable.

(g) The upper Vietoris topology and the Scott topology on $\mathord{\mathsf{K}} (X_{coc})$ do not agree.

By (e) and Corollary 25, $\mathcal O(P_S(X_{coc}))\subseteq \sigma (\mathord{\mathsf{K}} (X_{coc})$ . For $F\in X^{(<\omega)}\setminus \{\emptyset\}$ , $\mathord{\uparrow}_{\mathord{\mathsf{K}} (X_{coc})}F\in \alpha(\mathord{\mathsf{K}} (X_{coc}))=\sigma(\mathord{\mathsf{K}} (X_{coc}))$ but $\mathord{\uparrow}_{\mathord{\mathsf{K}} (X_{coc})}F\not\in \mathcal O(P_S(X_{coc}))$ since there is no $G\in X^{(<\omega)}$ with $F\in \Box (X\setminus G)=(X\setminus G)^{(<\omega)}\setminus \{\emptyset\}\subseteq \mathord{\uparrow}_{\mathord{\mathsf{K}} (X_{coc})}F$ . Thus, $\sigma(\mathord{\mathsf{K}} (X_{coc}))\not\subseteq\mathcal O(P_S(X_{coc}))$ .

(h) The Scott space $\Sigma~\!\! \mathord{\mathsf{K}}(X_{coc})$ is a sober c-space. So it is Rudin and well-filtered determined.

$\mathord{\mathsf{K}} (X_{coc})=X^{(<\omega)}\setminus \{\emptyset\}$ (with the Smyth order) is a Noetherian dcpo, and hence, it is an algebraic domain. By Theorem 15 and Proposition 18, $\Sigma~\!\! \mathord{\mathsf{K}}(X)$ is a sober c-space. Hence by Theorem 36, $\Sigma~\!\! \mathord{\mathsf{K}}(X)$ is Rudin and well-filtered determined.

(i) $X_{coc}$ is neither a Rudin space nor a $\mathsf{WD}$ space.

By (c), (e) and Theorem 36, $X_{coc}$ is neither a Rudin space nor a $\mathsf{WD}$ space.

(j) The Smyth power space $P_S(X_{coc})$ is well-filtered but non-sober. Hence, it is neither a Rudin space nor a $\mathsf{WD}$ space.

By (c), (e), Theorems 19 and 24, $P_S(X_{coc})$ is well-filtered and non-sober. Hence, $P_S(X_{coc})$ is neither a Rudin space nor a $\mathsf{WD}$ space by Theorem 36.

(k) $P_S(X_{coc})$ is not core-compact.

By (j) and Corollary 40 (or Theorem 41), $P_S(X_{coc})$ is not core-compact.

The following example shows there is even a second-countable Noetherian $T_0$ space X such that the Scott space $\Sigma ~\!\!\mathord{\mathsf{K}} (X)$ is a second-countable sober space but X is not well-filtered (and hence not sober).

Figure 3. The poset P.

Let $\tau=\{(\mathbb{N}\setminus F)\cup \{\infty\} : F\in \mathbb{N}^{(<\omega)}\}\cup\{\emptyset, P\}\cup \{\{\infty\}\}$ . It is straightforward to verify that $\tau$ is a $T_0$ topology on P and the specialization order of $(P, \tau)$ agrees with the original order on P. Now we have

1. $\mathcal C((P, \tau))=\mathbb{N}^{(<\omega)}\cup \{\emptyset, P\}\cup \{\mathbb{N}\}$ .

2. ${\mathsf{Irr}}_c((P, \tau))=\{\overline{\{n\}}=\{n\} : n\in \mathbb{N}\}\cup\{\overline{\{\infty\}}=P\}\cup\{\mathbb{N}\}$ and hence $(P, \tau)$ is not sober.

3. $\mathord{\mathsf{K}} ((P, \tau))=\{A\cup \{\infty\} : A\subseteq \mathbb{N}\}$ .

4. $(P, \tau)$ is not well-filtered.

Let $\mathcal K=\{(\mathbb{N}\setminus F)\cup\{\infty\} : F\in \mathbb{N}^{(<\omega)}\}$ . Then, $\mathcal K\subseteq \mathord{\mathsf{K}} ((P, \tau))$ is a filtered family and $\bigcap \mathcal K=\{\infty\}\in \tau$ . But there is no $F\in \mathbb{N}^{(<\omega)}$ with $(\mathbb{N}\setminus F)\cup\{\infty\}=\{\infty\}$ . Thus, $(P, \tau)$ is not well-filtered. In fact, $(P, \tau)$ is not weak well-filtered in the sense of Lu et al. (2017).

5. $(P, \tau)$ is Noetherian and second-countable, and hence, it is a Rudin space.

Since $|\tau|=\omega$ , $(P, \tau)$ is second-countable. As every subset of P is compact in $(P, \tau)$ , the space $(P, \tau)$ is a Noetherian space (and hence a locally compact space). Hence by Proposition 38, $(P, \tau)$ is a Rudin space.

6. $\Sigma~\!\!\mathord{\mathsf{K}} ((P, \tau))$ is a second-countable sober space.

Clearly, $\mathord{\mathsf{K}} ((P, \tau))$ is isomorphic with the algebraic lattice $2^{\mathbb{N}}$ (with the order of set inclusion) via the poset isomorphism $\varphi : \mathord{\mathsf{K}} ((P, \tau))\rightarrow 2^{\mathbb{N}}$ defined by $\varphi (A\cup \{\infty\})=\mathbb{N}\setminus A$ for each $A\in 2^{\mathbb{N}}$ (note that the order on $\mathord{\mathsf{K}} ((P, \tau))$ is the Smyth order). Hence $\Sigma~\!\!\mathord{\mathsf{K}} ((P, \tau))\cong \Sigma~\!\!2^{\mathbb{N}}$ . Clearly, $2^{\mathbb{N}}$ is an algebraic lattice, whence by Theorem 15 and Proposition 18, $\Sigma~\!\!2^{\mathbb{N}}$ is sober and hence $\Sigma~\!\!\mathord{\mathsf{K}} ((P, \tau))$ is sober. Clearly, $\Sigma~\!\!2^{\mathbb{N}}$ is second-countable since $\{\mathord{\uparrow}_{2^{\mathbb{N}}} F : F\in (2^{\mathbb{N}})^{(<\omega)}\}$ is a countable base of $\Sigma~\!\!2^{\mathbb{N}}$ . So $\Sigma~\!\!\mathord{\mathsf{K}} ((P, \tau))$ is second-countable.

7. $P_S((P, \tau))$ is second-countable.

Clearly, $\{\Box U : U\in \tau\}$ is a countable base of $P_S((P, \tau))$ (note that $|\tau|=\omega$ ). Hence $P_S((P, \tau))$ is second-countable.

8. $\sigma(\mathsf{K}((P, \tau))\subseteq \mathcal O(P_S((P, \tau)))$ but $\mathcal O(P_S((P, \tau))\not\subseteq \sigma(\mathsf{K}((P, \tau)))$ .

Since $(P, \tau)$ is locally compact, $\sigma(\mathsf{K}((P, \tau))\subseteq \mathcal O(P_S((P, \tau)))$ by Lemma 54. Clearly, $\Box \{\infty\}=\{\{\infty\}\}\in \mathcal O(P_S((P, \tau))$ . Now we show that $\Box \{\infty\}\not\in \sigma(\mathsf{K}((P, \tau)))$ . By Lemma 9, $\bigvee \{F\cup\{\infty\} : F\in (\mathbb{N})^{(<\omega)}\setminus \{\emptyset\}\}=\bigcap \{F\cup\{\infty\} : F\in (\mathbb{N})^{(<\omega)}\setminus \{\emptyset\}\}=\{\infty\}\in \Box \{\infty\}$ , but there is no $F\in (\mathbb{N})^{(<\omega)}\setminus \{\emptyset\}$ with $F\cup\{\infty\}\in \Box \{\infty\}=\{\{\infty\}\}$ . Thus, $\Box \{\infty\}\not\in \sigma(\mathsf{K}((P, \tau)))$ .

In the following, we will construct a sober space X for which the Scott space $\Sigma~\!\!\mathsf{K}(X)$ is non-sober (see Theorem 91 below).

Let $\mathcal{L}=\mathbb{N}\times \mathbb{N}\times (\mathbb{N}\cup \{\infty\})$ , where $\mathbb{N}$ is the set of natural numbers with the usual order. Define an order $\leq$ on $\mathcal L$ as follows:

$(i_1, j_1, k_1)\leq (i_2, j_2, k_2)$ if and only if:

(1) $i_1=i_2, j_1=j_2, k_1\leq k_2\leq \infty$ ; or

(2) $i_2=i_1+1, k_1\leq j_2, k_2=\infty$ .

$\mathcal L$ is a known dcpo constructed by Jia in Jia (Reference Jia2018, Example 2.6.1). It can be easily depicted as in Fig. 4 taken from Jia (Reference Jia2018).

Figure 4. A non-sober well-filtered dcpo $\mathcal L$ .

For each $(n, m)\in \mathbb{N}\times \mathbb{N}$ , let

$\mathcal L_{(n,m)}=\{(n, , l) : l\in \mathbb{N}\cup \{\infty\}$ (the (n, m)th line of $\mathcal L$ ),

$\mathcal L_n=\bigcup_{j\in \mathbb{N}}\mathcal L_{(n,j)}$ (the nth row of $\mathcal L$ ) and $\mathcal L_0=\emptyset$ ,

$\mathcal L_n^{\infty}=\{(n, j, \infty) : j\in \mathbb{N}\}$ (the set of all maximal elements of $\mathcal L_n$ ),

$\mathcal L_n^{<\infty}=\mathcal L_n\setminus \mathcal L_n^{\infty}$ (the set of all finite height elements of $\mathcal L_n$ ),

$\mathcal L^{\infty}=\bigcup_{n\in \mathbb{N}}\mathcal L_n^{\infty}=\{(i, j, \infty) : (i, j)\in \mathbb{N}\times\mathbb{N}\}$ (the set of all maximal elements of $\mathcal L$ ),

$\mathcal L^{<\infty}=\mathcal L\setminus \mathcal L^{\infty}$ (the set of all elements of finite height),

$\mathcal L_{(\leq n)}=\bigcup\limits_{i=1}^{n}\mathcal L_i=\{(i, j, l) : i\leq n, j\in \mathbb{N}, l\in \mathbb{N}\cup\{\infty\}\}$ ,

$\mathcal L_{(\geq n)}=\bigcup\limits_{i\geq n}\mathcal L_i=\{(i, j, l) : i\geq n, j\in \mathbb{N}, l\in \mathbb{N}\cup\{\infty\}\}$ ,

$\mathcal L_{(< n+1)}^{\infty}=\mathcal L_{(\leq n)}^{\infty}=\bigcup\limits_{i=1}^{n}\mathcal L_i^{\infty}=\{(i, j, \infty) : i\leq n, j\in \mathbb{N}\}$ ,

$\mathcal L_{(> n)}^{\infty}=\mathcal L_{(\geq n+1)}^{\infty}=\bigcup\limits_{i\geq n}\mathcal L_i^{\infty}=\{(i, j, \infty) : i> n, j\in \mathbb{N}\}$ , and

$\mathcal L_{(n, \geq m)}^{\infty}=\{(n, j, \infty) : j\geq m\}$ .

For simplicity, let $h(\emptyset)=0$ .

Proof. Since $U=\uparrow U$ and $U\cap \mathcal L_i\neq\emptyset$ , we can assume that $(i, j(i), \infty)\in U$ for some $j(i)\in \mathbb{N}$ . As $\bigvee_{l\in \mathbb{N}}(i, j(i), l)=(i, j(i), \infty)\in U\in\sigma (\mathcal L)$ , there is $l(i)\in \mathbb{N}$ such that $(i, j(i), l(i))\in U$ . Let $j^{\prime}=l(i)$ . Then, $\mathcal L_{(i+1, \geq j^{\prime})}^{\infty}\subseteq \uparrow (i, j(i), l(i))\subseteq U$ .

1. (i) $(i_D, j_D, \infty)\not\in D$ ,

2. (ii) $D\subseteq \{(i_D, j_D, l) : l\in \mathbb{N}\}$ , and

3. (iii) $(i_D, j_D, \infty)=\bigvee_{\mathcal L} D$ .

Hence, $\mathcal L$ is a dcpo.

Proof. If there is $(i_0, j_0, \infty)\in D\cap \mathcal L^{\infty}$ , then for each $d=(i_d, j_d, l_d)\in D$ , there is $d^*=(i_{d^*}, j_{d^*}, l_{d^*})\in D$ such that $(i_0, j_0, \infty)\leq d^*=(i_{d^*}, j_{d^*}, l_{d^*})$ and $d=(i_d, j_d, l_d)\leq d^*=(i_{d^*}, j_{d^*}, l_{d^*})$ , whence $l_{d^*}=\infty, i_{d^*}=i_0, j_{d^*}=j_0$ (i.e., $d^*=(i_0, j_0, \infty)$ ) and $d\leq d^*=(i_0, j_0, \infty)$ . Hence, $(i_D, j_D, \infty)=(i_0, j_0, \infty)$ is the (unique) largest element of D.

Now suppose that $D\cap \mathcal L^{\infty}=\emptyset$ , that is, $D\subseteq \mathbb{N}\times \mathbb{N}\times \mathbb{N}$ . Select a $d_1=(i_{d_1}, j_{d_1}, l_{d_1})\in D$ . Then for each $d=(i_d, j_d, l_d)\in D$ , by the directedness of D, there is $d^{\prime}=(i_{d^{\prime}}, j_{d^{\prime}}, l_{d^{\prime}})\in D$ such that $d_1=(i_{d_1}, j_{d_1}, l_{d_1})\leq d^{\prime}=(i_{d^{\prime}}, j_{d^{\prime}}, l_{d^{\prime}})$ and $d=(i_d, j_d, l_d)\leq d^{\prime}=(i_{d^{\prime}}, j_{d^{\prime}}, l_{d^{\prime}})$ . Hence, $i_{d^{\prime}}=i_d=i_{d_0}, j_{d^{\prime}}=j_d=j_{d_0}$ and $l_{d_0}\leq l_{d^{\prime}}, l_d\leq l_{d^{\prime}}$ . Let $i_D=i_{d_1}$ and $j_D=j_{d_1}$ . Then, $D\subseteq \{(i_D, j_D, l) : l\in \mathbb{N}\}$ . Clearly, $(i_D, j_D, \infty)$ is the unique element of $\mathcal L^{\infty}$ satisfying Conditions (i)–(iii).

For any $D\in \mathcal D(\mathcal L)$ , if D has a largest element s, then $s=\vee D$ ; if D has no largest element, then D is infinite directed subset of $\mathcal L$ , whence there is a unique $(i, j, \infty)\in \mathcal L^{\infty}$ such that Conditions (i)–(iii) are satisfied. Then, $(i, j, \infty)=\bigvee_{\mathcal L} D$ . Hence, $\mathcal L$ is a dcpo.

Proof. Since D has no largest element, by Lemma 71, there is a unique $(i_D, j_D, \infty)\in \mathcal L^{\infty}$ such that the following three conditions are satisfied:

1. (i) $(i_D, j_D, \infty)\not\in D$ ,

2. (ii) $D\subseteq \{(i_D, j_D, l) : l\in \mathbb{N}\}$ , and

3. (iii) $(i_D, j_D, \infty)=\bigvee_{\mathcal L} D$ .

Suppose that $(i_D, j_D, \infty)\not\in \mathrm{max}(A)$ . We will show that $\mathrm{max}(A)\cap \mathcal L_{i_D+1}^{\infty}$ is infinite. For each $d=(i_D, j_D, l_d)\in D$ , by $A=\mathord{\downarrow} \mathrm{max}(A)$ , there is $(i(d), j(d), l(d))\in \mathrm{max}(A)$ with $d=(i_D, j_D, l_d)\leq (i(d), j(d), l(d))$ . If $i(d)=i_D$ , then $j(d)=j_D$ and $l_d\leq l(d)$ . Since $D\subseteq A\cap \{(i_D, j_D, l) : l\in \mathbb{N}\}$ is infinite, $\{l_{d^{\prime}} : d^{\prime}=(i_D, j_D, l_{d^{\prime}})\in D\}\subseteq \mathbb{N}$ is infinite and $(i_D, j_D, l(d))=(i(d), j(d), l(d))\in \mathrm{max}(A)$ , we have that $l(d)=\infty$ , which is in contradiction with $(i_D, j_D, \infty)\not\in \mathrm{max}(A)$ . Therefore, $i(d)=i_D+1$ and hence $l(d)=\infty$ and $l_d\leq j(d)$ by $(i_D, j_D, l_d)\leq (i(d), j(d), l(d))=(i_D+1, j(d), l(d))$ . Since $\{l_d : d=(i_D, j_D, l_d)\in D\}\subseteq \mathbb{N}$ is infinite, $\{(i(d), j(d), l(d))=(i_D+1, j(d), \infty) : d\in D\}\subseteq \mathrm{max}(A)\cap \mathcal L^{\infty}$ is infinite (note that $l_d\leq j(d)$ for each $d=(i_D, j_D, l_d)\in D$ ). Thus, $\mathrm{max}(A)\cap \mathcal L_{i_D+1}^{\infty}$ is infinite.

1. (1) $A\subseteq \mathcal L^{<\infty}$ (or equivalently, $\mathrm{max}(A)\subseteq \mathcal L^{<\infty}$ ).

2. (2) $A\cap \mathcal L^{\infty}\neq \emptyset$ and $|A\cap \mathcal L_n^{\infty}|<\omega$ for each $n\in \mathbb{N}$ .

3. (3) $i(A)=\mathrm{max}\{i\in \mathbb{N} : |A\cap L_i^{\infty}|=\omega \}$ exists and $\mathcal L_{n}\subseteq A$ for each $n\leq i(A)-1$ .

Proof. Suppose that A is Scott closed. Then, $A=\mathord{\downarrow} \mathrm{max} (A)$ by Lemma 17. Now we show that A satisfies one of Conditions (1)–(3). If neither Condition (1) nor Condition (2) holds, then there is some $i_0\in \mathbb{N}$ such that $A\cap L_{i_0}^{\infty}$ is infinite. We first show that A satisfies the following property Q:

(Q) For $n\in \mathbb{N}$ , if $A\cap L_{n}^{\infty}$ is infinite, then $\mathcal L_{i}\subseteq A$ for each $i\leq n-1$ .

If $n=1$ , then $\mathcal L_{n-1}=\mathcal L_0=\emptyset\subseteq A$ . Now we assume $2\leq n$ . For each $(j, l)\in \mathbb{N}\times \mathbb{N}$ , since $A\cap L_{n}^{\infty}$ is infinite (i.e., $\{j^{\prime}\in \mathbb{N} : (n, j^{\prime}, \infty)\in A\}$ is infinite), there is $j^{\prime}\in \mathbb{N}$ such that $(n, j^{\prime}, \infty)\in A$ and $l\leq j^{\prime}$ , whence $(n-1, j, l)\leq (n, j^{\prime}, \infty)$ . Thus, $\{(n-1, j, l) : (j, l)\in \mathbb{N}\times \mathbb{N}\}\subseteq \mathord{\downarrow} (A\cap L_{n}^{\infty})\subseteq \mathord{\downarrow} A=A$ . For each $j\in \mathbb{N}$ , since $(n-1, j, \infty)=\bigvee_{l\in \mathbb{N}}(n-1, j, l)$ and A is Scott closed, we have $(n-1, j, \infty)\in A$ . Hence, $\mathcal L_{n-1}\subseteq A$ . In particular, $\mathcal L_{n-1}^{\infty}\subseteq A$ . Then by induction, we get that $\mathcal L_i\subseteq A$ for any $1\leq i\leq n-1$ .

By property Q, if $\{i\in \mathbb{N} : |A\cap L_i^{\infty}|=\omega \}$ is infinite, then for each $n\in \mathbb{N}$ , $\mathcal L_{n-1}\subseteq A$ . Hence, $\mathcal L=\bigcup_{n\in \mathbb{N}}\mathcal L_n\subseteq A$ , which contradicts $A\neq \mathcal L$ . Hence $\{i\in \mathbb{N} : |A\cap L_i^{\infty}|=\omega \}$ is a nonempty finite subset of $\mathbb{N}$ , and consequently, $i(A)=\mathrm{max}\{i\in \mathbb{N} : |A\cap L_i^{\infty}|=\omega \}$ exists. By property Q, we have $\mathcal L_{i}\subseteq A$ for each $i\leq i(A)-1$ . This completes the proof of Condition (3).

Conversely, assume that $A=\mathord{\downarrow} \mathrm{max}(A)$ and one of Conditions (1)–(3) is satisfied. We will show that A is Scott closed.

Suppose that D is a directed subset of A. If D has a largest element s, then $\vee D=s\in A$ . Now suppose that D has no largest element. Then, D is infinite, whence by Lemma 71, there is a unique $(i_D, j_D, \infty)\in \mathcal L^{\infty}$ such that the following three conditions are satisfied:

1. (i) $(i_D, j_D, \infty)\not\in D$ ,

2. (ii) $D\subseteq \{(i_D, j_D, l) : l\in \mathbb{N}\}$ , and

3. (iii) $(i_D, j_D, \infty)=\bigvee_{\mathcal L} D$ .

If $(i_D, j_D, \infty)\not\in A$ , then by Corollary 72, $\mathrm{max}(A)\cap \mathcal L_{i_D+1}^{\infty}$ is infinite, which is a contradiction with Condition (1) or Condition (2). So $\bigvee_{\mathcal L} D=(i_D, j_D, \infty)\in A$ . Thus, A is Scott closed.

Case 2. $i(A)=\mathrm{max}\{i\in \mathbb{N} : |A\cap L_i^{\infty}|=\omega \}$ exists and $\mathcal L_{n}\subseteq A$ for each $n\leq i(A)-1$ .

Suppose that D is a directed subset of A. When D has a largest element, we clearly have $\bigvee_{\mathcal L} D\in A$ . Now we assume that D has no largest element. Then, D is infinite, whence by Lemma71, there is a unique $(i_D, j_, \infty)\in \mathcal L^{\infty}$ satisfying Conditions (i)–(iii) in Lemma 71. If $(i_D, j_D, \infty)\in A$ or $i_D\leq i(A)-1$ , then $\bigvee_{\mathcal L} D=(i_D, j_D, \infty)\in A$ or $\bigvee_{\mathcal L} D=(i_D, j_D, \infty)\in \mathcal L_{i_D}\subseteq A$ . If $i_D\geq i(A)$ and $(i_D, j_D, \infty)\not\in A$ , then by Corollary 72, $\mathrm{max}(A)\cap \mathcal L_{i_D+1}^{\infty}$ is infinite. It follows that $i_D+1\leq i(A)$ , which is a contradiction with $i_D\geq i(A)$ . So $A\in \mathcal C(\Sigma~\!\! \mathcal L)$ .

1. (1) A is Scott closed.

2. (2) For each $(i, j)\in \mathbb{N}\times \mathbb{N}$ , if $(i, j, \infty)\not\in A$ , then $A\cap\{(i, j, l) : l\in \mathbb{N}\}$ is finite.

3. (3) For each $(i, j)\in \mathbb{N}\times \mathbb{N}$ , $\mathcal L_{(i, j)}\subseteq A$ or $(i, j, \infty)\not\in A$ and $A\cap\{(i, j, l) : l\in \mathbb{N}\}$ is finite.

4. (4) For each $(i, j)\in \mathbb{N}\times \mathbb{N}$ , $\mathcal L_{(i, j)}\subseteq A$ or $A\cap\mathcal L_{(i, j)}$ is finite.

Proof. (1) $\Rightarrow$ (2): Trivial. (2) $\Leftrightarrow$ (3) $\Leftrightarrow$ (4): Trivial (since A is a lower set).

(2) $\Rightarrow$ (1): Suppose that D is a directed subset of A. When D has a largest element, then $\bigvee_{\mathcal L} D\in A$ . Now assume that D has no largest element. Then, D is infinite, whence by Lemma 71, there is a unique $(i_D, j_D, \infty)\in \mathcal L^{\infty}$ satisfying Conditions (i)-(iii) in Lemma 4. Since $D\subseteq A\cap \{(i_D, j_D, l) : l\in \mathbb{N}\}$ is infinite, by (2), $\bigvee_{\mathcal L}D=(i_D, j_D, \infty)\in A$ . Thus, A is Scott closed.

Dually, as a direct corollary of Proposition 77, we have the following.

1. (1) U is Scott open.

2. (2) For each $(i, j)\in \mathbb{N}\times \mathbb{N}$ , if $(i, j, \infty)\in U$ , then $U\cap \mathcal L_{(i, j)}$ is infinite.

3. (3) For each $(i, j)\in \mathbb{N}\times \mathbb{N}$ , $U\cap\mathcal L_{(i, j)}=\emptyset$ or $(i, j, \infty)\in U$ and $U\cap \mathcal L_{(i, j)}$ is infinite.

4. (4) For each $(i, j)\in \mathbb{N}\times \mathbb{N}$ , $U\cap\mathcal L_{(i, j)}=\emptyset$ or $U\cap \mathcal L_{(i, j)}$ is infinite.

By Lemma 75 and Proposition 77, we get the following.

1. (1) $A\subseteq \mathcal L^{<\infty}$ (or equivalently, $\mathrm{max}(A)\subseteq \mathcal L^{<\infty}$ ) and $A\cap \{(i, j, l) : l\in \mathbb{N}\}$ is finite for each $(i, j)\in \mathbb{N}\times \mathbb{N}$ .

2. (2) $A\cap \mathcal L^{\infty}\neq \emptyset$ , $|A\cap \mathcal L_i^{\infty}|<\omega$ for each $i\in \mathbb{N}$ and $A\cap \{(i, j, l) : l\in \mathbb{N}\}$ is finite for each $(i, j)\in \mathbb{N}\times \mathbb{N}$ with $(i, j, \infty)\not\in A$ .

3. (3) $i(A)=\mathrm{max}\{i\in \mathbb{N} : |A\cap L_i^{\infty}|=\omega \}$ exists, $\mathcal L_{n}\subseteq A$ for each $n\leq i(A)-1$ , and $A\cap \{(i, j, l) : l\in \mathbb{N}\}$ is finite for each $(i, j)\in \mathbb{N}$ with $(i, j, \infty)\not\in A$ .

Lemma 80. ${\mathsf{Irr}}_c(\Sigma~\!\!\mathcal L)=\{\overline{\{x\}}=\mathord{\downarrow} x : x\in \mathcal L\}\cup\{\mathcal L\}$ .

Proof. Clearly, $\{\overline{\{x\}}=\mathord{\downarrow} x : x\in \mathcal L\}\subseteq {\mathsf{Irr}}_c(\Sigma~\!\!\mathcal L)$ . It was shown in Jia (Reference Jia2018, Example 2.6.1) that $\mathcal L\in {\mathsf{Irr}}_c(\Sigma~\!\!\mathcal L)$ . This can be easily proved by Lemma 75. Suppose that $B, C\in \mathcal C(\Sigma~\!\!\mathcal L)$ and $\mathcal L=B\cup C$ . Let $\mathbb{N}_B=\{n\in \mathbb{N} : |B\cap L_n^{\infty}|=\omega\}$ and $\mathbb{N}_C=\{n\in \mathbb{N} : |C\cap L_n^{\infty}|=\omega\}$ . As $\mathcal L=B\cup C$ , we have $\mathbb{N}=\mathbb{N}_B\cup \mathbb{N}_C,$ and hence, at least one of $\mathbb{N}_B$ and $\mathbb{N}_C$ is infinite. Without loss of generality, assume $\mathbb{N}_B$ is infinite. Then by property Q in the proof of Lemma 75, we have $B=\mathcal L$ . Thus, $\mathcal L\in {\mathsf{Irr}}_c(\Sigma~\!\!\mathcal L)$ .

Conversely, suppose that $A\in {\mathsf{Irr}}_c(\Sigma~\!\!\mathcal L)$ and $A\neq \mathcal L$ .

Case 1. $\mathrm{max}(A)\cap \mathcal L^{<\infty}\neq \emptyset$ (i.e., A has a maximal point of finite height).

Select an $(i, j, l)\in \mathrm{max}(A)\cap \mathcal L^{<\infty}$ . Then, $l\in \mathbb{N}$ . Let $B=\downarrow (\mathrm{max}(A)\setminus \{(i, j, l)\})$ (B may be the empty set). Then, $A=B\cup \downarrow (i, j, l)$ . Clearly, $\downarrow (i, j, l)$ is Scott closed. By Lemma 75 or Proposition 77, it is easy to verify that B is Scott closed. Indeed, we can give a direct proof of the Scott closedness of B. Suppose that D is a directed subset of B. When D has a largest element, we clearly have $\bigvee_{\mathcal L} D\in D\subseteq B$ . Now we assume that D has no largest element. Then, D is infinite, whence by Lemma 71, there is a unique $(i_D, j_, \infty)\in \mathcal L^{\infty}$ satisfying Conditions (i)-(iii) in Lemma 71. Since A is Scott closed, $(i_D, j_D, \infty)=\bigvee_{\mathcal L} D\in A$ . As $l\in \mathbb{N}$ , $(i_D, j_D, \infty)\not\in \downarrow (i, j, l)$ , whence $(i_D, j_D, \infty)\in \mathrm{max}(A)\setminus \{(i, j, l)\}\subseteq B$ . Thus, B is Scott closed. By the irreducibility of A, we have $A=B$ or $A=\downarrow (i, j, l)$ . If $A=B$ , then there is $(i^{\ast}, j^{\ast}, l^{\ast})\in \mathrm{max}(A)\setminus \{(i, j, l)\}$ with $(i, j, l)\leq (i^{\ast}, j^{\ast}, l^{\ast})$ . Hence, $(i, j, l)=(i^{\ast}, j^{\ast}, l^{\ast})$ by $(i, j, l)\in \mathrm{max}(A)$ , a contradiction. Therefore, $A=\downarrow (i, j, l)={\rm cl}_{\sigma (\mathcal L)}\{(i, j, l)\}$ .

Case 2. $\mathrm{max}(A)\subseteq \mathcal L^{\infty}$ (i.e., every maximal point of A is a point of infinite height).

Since $A\neq \emptyset$ , $A=\downarrow \mathrm{max}(A)$ and $\mathrm{max}(A)\subseteq \mathcal L^{\infty}$ , we have that $\mathrm{max}(A)\cap\mathcal L^{\infty}\neq\emptyset$ . If $|A\cap L_n^{\infty}|<\omega$ for all $n\in \mathbb{N}$ , then select any point $(i, j, \infty)\in \mathrm{max}(A)$ and let $n(A)=i$ . We clearly have that $|A\cap L_{n(A)+1}^{\infty}|<\omega$ . If $|A\cap L_m^{\infty}|=\omega$ for some $m\in \mathbb{N}$ , then by property Q in the proof of Lemma 75, the set $\{n\in\mathbb{N} : |A\cap L_n^{\infty}|=\omega\}$ is a nonempty finite set, whence $n(A)=\mathrm{max}\{n\in\mathbb{N} : |A\cap L_n^{\infty}|=\omega\}$ exists. Then, $A\cap L_{n(A)+1}^{\infty}$ is finite. Select an $(n(A), j, \infty)\in A\cap L_{n(A)}^{\infty}$ . Let $B=\downarrow (\mathrm{max}(A)\setminus \{(n(A), j, \infty)\})$ . Then, $A=B\cup \downarrow (n(A), j, \infty)$ . Clearly, $\downarrow (n(A), j, \infty)$ is Scott closed. Now we show that B is Scott closed. Suppose that D is a directed subset of B. If D has a largest element s, then $\bigvee_{\mathcal L} D=s\in D\subseteq B$ ; if D has no largest element, then D is infinite, whence by Lemma 71, there is a unique $(i_D, j_D, \infty)\in \mathcal L^{\infty}$ such that the following three conditions are satisfied:

1. (i) $(i_D, j_D, \infty)\not\in D$ ,

2. (ii) $D\subseteq \{(i_D, j_D, l) : l\in \mathbb{N}\}$ , and

3. (iii) $(i_D, j_D, \infty)=\bigvee_{\mathcal L} D$ .

Since A is Scott closed, $(i_D, j_D, \infty)=\bigvee_{\mathcal L} D\in A$ , whence $(i_D, j_D, \infty)\in \mathrm{max}(A)$ . Now we show that $(i_D, j_D, \infty)\neq (n(A), j, \infty)$ . Suppose, on the contrary, that $(i_D, j_D, \infty)=(n(A), j, \infty)$ . For each $d=(i_D, j_D, l_d)\in D$ , there is an $(i_d, j_d, \infty)\in \mathrm{max}(A)\setminus \{(n(A), j, \infty)\}$ such that $d=(i_D, j_D, l_d)\leq (i_d, j_d, \infty)$ . Since $(i_d, j_d, \infty)\neq (n(A), j, \infty)=(i_D, j_D, \infty)$ , we have that $i_d=i_D+1=n(A)+1$ and $l_d\leq j_d$ . As $D\subseteq \{(i_D, j_D, l) : l\in \mathbb{N}\}$ is infinite, the set $\{(n(A)+1, l_d, \infty)=(i_D+1, l_d, \infty) : d\in D\}$ is infinite, and consequently, $A\cap L_{n(A)+1}^{\infty}$ is infinite, a contradiction. So $(i_D, j_D, \infty)\neq (n(A), j, \infty)$ and hence $\bigvee_{\mathcal L} D=(i_D, j_D, \infty)\in \mathrm{max}(A)\setminus \{(n(A), j, \infty)\}\subseteq B$ . Thus, B is Scott closed.

By the irreducibility of A, we have $A=B$ or $A=\downarrow (n(A), j, \infty)$ . It follows from $(n(A), j, \infty)\not\in B$ that $A=\downarrow (n(A), j, \infty)={\rm cl}_{\sigma (\mathcal L)}\{(n(A), j, \infty)\}$ .

All these together deduce that ${\mathsf{Irr}}_c(\Sigma~\!\!\mathcal L)=\{\overline{\{x\}}=\mathord{\downarrow} x : x\in \mathcal L\}\cup\{\mathcal L\}$ .

1. (1) $\mathrm{min}(K)\cap \mathcal L^{<\infty}$ is finite,

2. (2) $\{n\in \mathbb{N} : \mathcal L_n^{\infty}\cap K\neq\emptyset\}$ is finite, and

3. (3) there is a unique $(i_K, j_K)\in \mathbb{N}\times \mathbb{N}$ such that $K\cap (\mathcal L_{(<{i_K})}^{\infty}\cup \mathcal L_{(i_K, \geq j_K)}^{\infty})=\{(i_K, j_K, \infty)\}$ (or equivalently, $K\cap (\mathcal L_{(<{i_K})}\cup \mathcal L_{(i_K, \geq j_K)}^{\infty})=\{(i_K, j_K, \infty)\}$ ).

Proof. We can assume that D is directed and $K_{d_2}\subseteq K_{d_1}$ iff $d_1\leq d_2$ (indeed, D can be defined an order by $d_1\leq d_2$ iff $K_{d_2}\subseteq K_{d_1}$ ). For each $d\in D$ , by Lemma 81, there exists a unique $(i_d, j_d)\in \mathbb{N}\times \mathbb{N}$ such that $(\mathcal L_{(<i_d)}^{\infty}\cup \mathcal L_{(i_d, \geq j_d)}^{\infty})\cap K_d=\{(i_d, j_d, \infty)\}$ and the set $\mathbb{N}_d=\{n\in \mathbb{N} : \mathcal L_n^{\infty}\cap K_d\neq\emptyset\}$ is finite. Select a $d_0\in D$ . Then for each $d\in D$ with $d_0\leq d$ (whence $K_d\subseteq K_{d_0}$ ), we have that $i_d\in \mathbb{N}_d\subseteq \mathbb{N}_{d_0}$ and $i_{d_0}\leq i_d$ (otherwise, $i_{d_0}> i_d$ would imply that $(i_d, j_d, \infty)\in (\mathcal L_{(<i_d)}^{\infty}\cup \mathcal L_{(i_d, \geq j_d)}^{\infty})\cap K_d\subseteq L_{(<i_{d_0})}^{\infty}\cap K_{d_0}$ , which contradicts $L_{(<i_{d_0})}^{\infty}\cap K_{d_0}=\emptyset$ ). Let $D_{d_0}=\{d\in D : d_0\leq d\}$ and $D_i=\{d\in D_{d_0} : i_d=i\}$ for each $i\in \mathbb{N}_{d_0}$ . Since $\mathbb{N}_{d_0}$ is finite, $D_{d_0}$ is directed and $D_{d_0}=\bigcup_{i\in\mathbb{N}_{d_0}} D_i$ , there is $i_0\in \mathbb{N}_{d_0}$ such that $D_{i_0}$ is a cofinal subset of $D_{d_0}$ and hence a cofinal subset of D, more precisely, for each $d\in D$ , there is $d^*\in D$ such that $d^*\in \mathord{\uparrow} d_0\cap \mathord{\uparrow} d$ and $i_{d^*}=i_0$ .

Clearly, $D_{i_0}$ is also directed. Select a $d_1\in D_{i_0}$ . Then, $D_{d_1}=\{d\in D_{i_0} : d_1\leq d\}$ is a directed and cofinal subset of $D_{i_0}$ and hence a directed and cofinal subset of D. For each $d\in D_{d_1}$ (note that $K_d\subseteq K_{d_1}$ ), we have that $i_{d_1}=i_d=i_0$ , $(i_d=i_0, j_d, \infty)\in K_d\subseteq K_{d_1}$ and $\mathcal L_{(i_0, \geq j_{d_1})}^{\infty}\cap K_{d_1}=\{(i_0, j_{d_1}, \infty)\}$ . It follows that $j_d\leq j_{d_1}$ . For each $1\leq j\leq j_{d_1}$ , let $\widetilde{D}_{j}=\{d\in D_{d_1} : j_d=j\}$ . Since $\{1, 2, ..., j_{d_1}\}$ is finite, $D_{d_1}$ is directed and $D_{d_1}=\bigcup_{i\in\mathbb{N}_{d_0}} \widetilde{D}_{j}$ , there is $1\leq j_0\leq j_{d_1}$ such that $\widetilde{D}_{j_0}$ is a cofinal subset of $D_{d_1}$ and hence a cofinal subset of D; more precisely, for each $d\in D$ , there is $d^{\prime}\in D$ such that $d^{\prime}\in \mathord{\uparrow} d_0\cap \mathord{\uparrow} d_1\cap \mathord{\uparrow} d$ , $i_{d^{\prime}}=i_0$ and $j_{d^{\prime}}=j_0$ . It follows that $(i_0, j_0, \infty)\in \bigcap_{d\in D}K_d$ .

Proof. By Lemma 80, $\mathcal L$ is an irreducible closed subset of $\Sigma~\!\!\mathcal L$ but has no largest element, so $\Sigma~\!\!\mathcal L$ is non-sober.

The well-filteredness of $\Sigma~\!\!\mathcal L$ was proved in Jia (Reference Jia2018) (see Jia Reference Jia2018, the proof of Claim 2.6.4). Using Topological Rudin Lemma, Lemma 80 and Corollary 82, we can give a short proof of the well-filteredness of $\Sigma~\!\!\mathcal L$ . Suppose that $\{K_d : d\in D\}\subseteq \mathord{\mathsf{K}} (\Sigma~\!\!\mathcal L)$ is a filtered family and $U\in \sigma (\mathcal L)$ with $\bigcap_{d\in D}K_d\subseteq U$ . Assume, on the contrary, that $K_d\not\subseteq U$ for each $d\in D$ (whence $U\neq \mathcal L$ ). Then by Lemma 27, $\mathcal L\setminus U$ contains a minimal irreducible closed subset A that still meets all members $K_d$ . By Corollary 82, $\bigcap_{d\in D}K_d\neq\emptyset$ , whence $U\neq\emptyset$ and $A\neq \mathcal L$ . It follows from Lemma 80 that $A=\overline{\{x\}}$ for some $x\in \mathcal L$ . Then, $x\in \bigcap_{d\in D}K_d\subseteq U$ , which contradicts $x\in A\subseteq \mathcal L \setminus U$ . Thus $\Sigma~\!\!\mathcal L$ is well-filtered.

As X is $T_0$ , $X_{\top}$ is also $T_0$ and $\overline{\{\top\}}=X\cup\{\top\}$ in $X_{\top}$ . Hence, $\top$ is a largest element of $X_{\top}$ and for $x, y\in X$ , $x\leq_X y$ iff $x\leq y$ in $X_{\top}$ . It is worthy noting that the set $\{\top\}$ is not open in $X_{\top}$ .

Proof. Clearly, $\mathcal O(X_{\top})=\{U\cup\{\top\} : U\in \mathcal O(X)\setminus \{\emptyset\}\}\cup\{\emptyset\}$ .

First, if $K\in \mathsf{K}(X_{\top})\setminus \{\{\top\}\}$ , then $G=K\setminus \{\top\}$ is a nonempty saturated subset of X. Now we verify that G is a compact subset of X. Suppose that $\{U_i : i\in I\}\subseteq \mathcal O(X)\setminus \{\emptyset\}$ is an open cover of G. Then, $\{U_i\cup\{\top\} : i\in I\}\subseteq \mathcal O(X_{\top})$ is an open cover of $K=G\cup \{\top\}$ . By the compactness of K, there is $I_0\in I^{(<\omega)}$ such that $K\subseteq \bigcup_{i\in I_0}U_i\cup\{\top\}$ , whence $G=K\setminus \{\top\}\subseteq \bigcup_{i\in I_0}U_i$ , proving that $G\in \mathsf{K}(X)$ .

Conversely, assume that $G\in \mathsf{K}(X)$ and $\{W_j : j\in J\}\subseteq \mathcal O(X_{\top})\setminus \{\emptyset\}$ is an open cover of $K=G\cup \{\top\}$ . Then, K is saturated and for each $j\in J$ , there is $V_j\in \mathcal O(X)$ such that $W_j=V_j\cup\{\top\}$ . Hence, $\{V_j : j\in J\}\subseteq \mathcal O(X)$ is an open cover of G. By the compactness of G, there is $J_0\in J^{(<\omega)}$ such that $G\subseteq \bigcup_{j\in J_0}V_j$ , whence $K=G\cup\{\top\}\subseteq \bigcup_{j\in J_0}W_j$ . So $K\in \mathsf{K}(X_{\top})$ .

Thus, $\mathsf{K}(X_{\top})=\{G\cup\{\top\} : G\in \mathsf{K}(X)\}\cup\{\{\top\}\}$ .

Proof. Since X is non-sober and ${\mathsf{Irr}}_c(X)=\{\overline{\{x\}} : x\in X\}\bigcup\{X\}$ , $X\neq\overline{\{x\}}$ for every $x\in X$ . It is well-known that the space $X^s$ with the canonical mapping $\eta_{X}: X\longrightarrow X^s$ , $\eta_{X}(x)=x$ , is a sobrification of X (see, e.g., Gierz et al. Reference Gierz, Hofmann, Keimel, Lawson, Mislove and Scott2003, Exercise V-4.9). For $C\in \mathcal C(X)$ , we have

Define a mapping $f : X^s \rightarrow X_{\top}$ by

For each $C\in (\mathcal C(X)\setminus \{X\})\bigcup \{X_{\top}\}$ and $B\in \mathcal C(X)$ , we have $f^{-1}(C)=\Box_{{\mathsf{Irr}}_c(X)} C$ and

Thus, f is a homeomorphism.

So $\langle X_{\top},\zeta_{X}=f\circ\eta_X\rangle$ is a sobrification of X.

The following corollary is straightforward from Lemma 80, Proposition 83, and Lemma 87.

Corollary 88. $\langle (\Sigma~\!\!\mathcal L)_{\top},\zeta_{\mathcal L} \rangle$ is a sobrification of $\Sigma~\!\!\mathcal L$ , where $\zeta_{\mathcal L}: \Sigma~\!\!\mathcal L\rightarrow (\Sigma~\!\!\mathcal L)_{\top}$ is defined by $\zeta_{\mathcal L}(x)=x$ for each $x\in \mathcal L$ .

Note that although the set $\{\top\}$ is open in $\Sigma~\!\!\mathcal L_{\top}$ (or equivalently, $\top$ is a compact element of the dcpo $\mathcal L\cup\{\top\}$ ), it is not open in $(\Sigma~\!\!\mathcal L)_{\top}$ .

By Lemmas 80 and 86, we get the following.

Corollary 89. $\mathsf{K}((\Sigma~\!\!\mathcal L)_{\top})=\{G\cup\{\top\} : G\in \mathsf{K}(\Sigma~\!\!\mathcal L))\}\cup\{\{\top\}\}$ .

Lemma 90. $\{\top\}$ is a compact element in the dcpo $\mathsf{K}((\Sigma~\!\!\mathcal L)_{\top})$ . Hence, $\{\{\top\}\}$ is open in the Scott space $\Sigma~\!\!\mathsf{K}((\Sigma~\!\!\mathcal L)_{\top})$ .

Proof. By Proposition 83, $\Sigma~\!\!\mathcal L$ is well-filtered, whence $\mathsf{K}(\Sigma~\!\!\mathcal L)$ (with the Smyth order) is a dcpo. So by Corollary 89, $\mathsf{K}((\Sigma~\!\!\mathcal L)_{\top})$ is a dcpo. Now we show that $\{\top\}\ll \{\top\}$ in $\mathsf{K}((\Sigma~\!\!\mathcal L)_{\top})$ . Suppose that $\{K_d : d\in D\}$ is a directed subset of $\mathsf{K}((\Sigma~\!\!\mathcal L)_{\top})$ and $\{\top\}\sqsubseteq \bigvee_{d\in D}K_d$ . Then by Lemma 9, $\bigvee_{d\in D}K_d=\bigcap_{d\in D}K_d,$ and hence, $\bigcap_{d\in D}K_d\subseteq \{\top\}$ . It follows that $\bigcap_{d\in D}(K_d\setminus \{\top\})=\emptyset$ . By Corollaries 82 and 89, there is $d\in D$ such that $K_d\setminus \{\top\}=\emptyset$ , that is, $K_d=\{\top\}$ . Thus, $\{\top\}$ is a compact element in $\mathsf{K}((\Sigma~\!\!\mathcal L)_{\top})$ . Hence, $\{\{\top\}\}\in \sigma(\mathsf{K}((\Sigma~\!\!\mathcal L)_{\top}))$ .

Proof. For simplicity, let $\mathcal A=\{G\cup\{\top\} : G\in \mathsf{K}(\Sigma~\!\!\mathcal L)\}$ .

Claim 1: $\mathcal A$ is a closed subset of $\Sigma~\!\!\mathsf{K}((\Sigma~\!\!\mathcal L)_{\top})$ .

By Corollary 89 and Lemma 90, $\mathcal A$ is Scott closed.

Claim 2: $\mathcal A$ is irreducible.

Suppose that $\mathcal U, \mathcal V\in \sigma(\mathsf{K}((\Sigma \mathcal L)_{\top}))$ and $\mathcal A\bigcap \mathcal U\neq \emptyset \neq \mathcal A\bigcap \mathcal V$ . Then by Corollary 89, there are some $G_1, G_2\in \mathsf{K}(\Sigma~\!\!\mathcal L))$ such that $G_1\cup\{\top\}\in \mathcal A\bigcap \mathcal U$ and $G_2\cup\{\top\}\in \mathcal A\bigcap \mathcal V$ , whence $(i_1, j_1, \infty)\in G_1$ and $(i_2, j_2, \infty)\in G_2$ for some $(i_1, j_1), (i_2, j_2)\in \mathbb{N}\times \mathbb{N}$ . Hence by Corollary 89, $\mathord{\uparrow} (i_1, j_1, \infty)\cup \{\top\}\in \mathcal A\bigcap \mathcal U$ and $\mathord{\uparrow} (i_2, j_2, \infty)\cup \{\top\}\in \mathcal A\bigcap \mathcal V$ (note that $\mathcal U, \mathcal V$ are upper sets and $G_1\cup\{\top\}\sqsubseteq \mathord{\uparrow} (i_1, j_1, \infty)\cup \{\top\}, G_2\cup\{\top\}\sqsubseteq \mathord{\uparrow} (i_2, j_2, \infty)\cup \{\top\}$ ). Without loss of generality, we assume $i_1\leq i_2$ . Since $\bigvee_{l\in \mathbb{N}}(\mathord{\uparrow} (i_1, j_1, l)\cup \{\top\})=\bigcap_{l\in \mathbb{N}}(\mathord{\uparrow} (i_1, j_1, l)\cup \{\top\})=\mathord{\uparrow} (i_1, j_1, \infty)\cup \{\top\}\in \mathcal U\in \sigma(\mathsf{K}((\Sigma~\!\!\mathcal L)_{\top}))$ , we have some $l_1\in\mathbb{N}$ such that $\mathord{\uparrow} (i_1, j_1, l_1)\in \mathcal U$ . So by Corollary 89, $\mathord{\uparrow} (i_1+1, l_1, \infty)\cup\{\top\}\in \mathcal U$ since $(i_1, j_1, l_1)\leq (i_1+1, l_1, \infty)$ and $\mathcal U$ is an upper set. Then by induction, we have $\mathord{\uparrow} (i_2, j^{\prime}, \infty)\cup\{\top\}\in \mathcal U$ for some $j^{\prime}\in \mathbb{N}$ . Again, since $\bigvee_{l\in \mathbb{N}}(\mathord{\uparrow} (i_2, j^{\prime}, l)\cup \{\top\})=\bigcap_{l\in \mathbb{N}}(\mathord{\uparrow} (i_2, j^{\prime}, l)\cup \{\top\})=\mathord{\uparrow} (i_2, j^{\prime}, l)\cup \{\top\}\in \mathcal U\in \sigma(\mathsf{K}((\Sigma~\!\!\mathcal L)_{\top}))$ and $\bigvee_{l\in \mathbb{N}}(\mathord{\uparrow} (i_2, j_2, l)\cup \{\top\})=\bigcap_{l\in \mathbb{N}}(\mathord{\uparrow} (i_2, j_2, l)\cup \{\top\})=\mathord{\uparrow} (i_2, j_2, \infty)\cup \{\top\}\in \mathcal V\in \sigma(\mathsf{K}((\Sigma~\!\!\mathcal L)_{\top}))$ , we have some $k_1, k_2\in\mathbb{N}$ such that $\mathord{\uparrow} (i_2, j^{\prime}, k_1)\in \mathcal U$ and $\mathord{\uparrow} (i_2, j_2, k_2)\in \mathcal V$ . Take $m=\mathrm{max}\{k_1, k_2\}$ . Then, $\mathord{\uparrow} (i_2, m, \infty)\cup\{\top\}\in \mathcal A\bigcap \mathcal U\bigcap \mathcal V$ . Thus, $\mathcal A\in {\mathsf{Irr}}_c(\Sigma~\!\!\mathsf{K}((\Sigma~\!\!\mathcal L)_{\top}))$ .

Claim 3: $\mathcal A$ has no largest element.

Clearly, $\{\mathord{\uparrow} (i, j, \infty)\cup \{\top\} : i, j\in\mathbb{N}\}$ is the set of all maximal elements of $\mathcal A,$ and hence, $\mathcal A$ has no largest element.

By Claims 1–3, the Scott space $\Sigma~\!\!\mathsf{K}((\Sigma~\!\!\mathcal L)_{\top})$ is non-sober.

Question 92. For a dcpo (especially, a complete lattice) P with the sober Scott topology, is the Scott space $\Sigma~\!\!\mathsf{K}(\Sigma~\!\!P)$ sober?

We know that every $T_2$ space is sober, and hence, the Scott space $\Sigma~\!\!\mathsf{K}(X)$ is well-filtered by Theorem 58. In the next section, we will show that for a locally compact (especially, compact) $T_2$ space X, the Soctt space $\Sigma~\!\!\mathsf{K}(X)$ is sober (see Corollary 99 below).

By Theorem 19, Corollary 25, Theorem 91, and Corollary 99 below, we naturally pose the following question.

Question 93. For a $T_2$ space X, is the Scott space $\Sigma~\!\!\mathord{\mathsf{K}} (X)$ sober?