The category of well-filtered dcpo’s is not $\Gamma$ -faithful

Hualin Miao, Huijun Hou, Xiaodong Jia¹¹1Corresponding author, Email: jiaxiaodong@hnu.edu.cn and Qingguo Li
School of Mathematics, Hunan University, Changsha, Hunan 410082, China

Abstract

The Ho-Zhao problem asks whether any two dcpo’s with isomorphic Scott closed set lattices are themselves isomorphic, that is, whether the category $\mathbf{DCPO}$ of dcpo’s and Scott-continuous maps is $\Gamma$ -faithful. In 2018, Ho, Goubault-Larrecq, Jung and Xi answered this question in the negative, and they introduced the category $\mathbf{DOMI}$ of dominated dcpo’s and proved that it is $\Gamma$ -faithful. Dominated dcpo’s subsume many familiar families of dcpo’s in domain theory, such as the category of bounded-complete dcpo’s and that of sober dcpo’s, among others. However, it is unknown whether the category of dominated dcpo’s dominates all well-filtered dcpo’s, a class strictly larger than that of bounded-complete lattices and that of sober dcpo’s. In this paper, we address this very natural question and show that the category $\mathbf{WF}$ of well-filtered dcpo’s is not $\Gamma$ -faithful, and as a result of it, well-filtered dcpo’s need not be dominated in general. Since not all dcpo’s are well-filtered, our work refines the results of Ho, Goubault-Larrecq, Jung and Xi.

As a second contribution, we confirm that the Lawson’s category of $\Omega^{*}$ -compact dcpo’s is $\Gamma$ -faithful. Moreover, we locate a class of dcpo’s which we call weakly dominated dcpo’s, and show that this class is $\Gamma$ -faithful and strictly larger than $\mathbf{DOMI}$ .

Keywords: Ho-Zhao Problem, $\Gamma$ -faithfulness, well-filtered dcpo’s, weakly dominated dcpo’s.

1 Introduction

One of the most important topologies on posets is the so-called Scott topology, which consists of upper subsets that are inaccessible by suprema of directed subsets in given posets, and has been playing prominent rôle in domain theory, non-Hausdorff topology and denotational semantics, among others. As can be seen from definition, Scott topologies are uniquely determined by the order structure on posets. A more intriguing question is the converse: does the Scott topology of a poset determine the order of the poset? In order to form this question more rigorously, we let $\Gamma P$ denote the set of all Scott closed subsets of a poset $P$ , ordered by inclusion. The poset $\Gamma P$ is a complete lattice for each $P$ . The aforementioned question is then formalized as:

If $P$ and $Q$ are posets with $\Gamma P$ isomorphic to $\Gamma Q$ ( $\Gamma P\cong\Gamma Q$ , in symbols), is it true that $P\cong Q$ ?

This question was perfectly answered by Zhao and Fan in [15], where they showed that each poset $P$ admits a so-called dcpo-completion $E(P)$ , with the property that $E(P)$ is a directly-complete poset (dcpo for short) and $\Gamma P\cong\Gamma E(P)$ . Hence, any poset which fails to be a dcpo, together with its dcpo-completion $E(P)$ , fails the above question. This observation has led Ho and Zhao narrow this question to the category $\mathbf{DCPO}$ of dcpo’s and Scott-continuous maps [7], and they call a full subcategory $\mathbf{C}$ of $\mathbf{DCPO}$ $\Gamma$ -faithful if for every pair $P$ and $Q$ in $\mathbf{C}$ , $\Gamma P\cong\Gamma Q$ implies $P\cong Q$ . In [7], Ho and Zhao identified the category of bounded complete dcpo’s as one of $\Gamma$ -faithful subcategories of $\mathbf{DCPO}$ , and asked whether the category $\mathbf{DCPO}$ itself is $\Gamma$ -faithful. The question was later dubbed the Ho-Zhao problem. In 2016, Ho, Goubault-Larrecq, Jung and Xi exhibited a dcpo $\mathcal{H}$ which is not sober in the Scott topology, and showed that $\mathcal{H}$ and its sobrification $\hat{\mathcal{H}}$ serve as instances for the fact that $\mathbf{DCPO}$ fails to be $\Gamma$ -faithful. In addition, they gave a $\Gamma$ -faithful full subcategory $\mathbf{DOMI}$ of $\mathbf{DCPO}$ , which consists of the so-called dominated dcpo’s. Notably, the category $\mathbf{DOMI}$ of dominated dcpo’s subsumes many known $\Gamma$ -faithful subcategories of dcpo’s, for example, the category of sober dcpo’s and that of bounded-complete lattices, to name a few. In Section $3$ , we show that the category $\mathbf{DOMI}$ of dominated dcpo’s also includes the category of $\Omega^{*}$ -compact dcpo’s.

Generalizing both sober dcpo’s and $\Omega^{*}$ -compact dcpo’s, well-filtered dcpo’s are ones that are well-filtered in the Scott topology, and the category $\mathbf{WF}$ of well-filtered dcpo’s is a strictly smaller subclass of $\mathbf{DCPO}$ . A natural question arises as whether well-filtered dcpo’s also are subsumed under dominated dcpo’s, or whether the category $\mathbf{WF}$ is $\Gamma$ -faithful? In this paper, we will mainly investigate this natural question and give concrete examples to deduce that $\mathbf{WF}$ is not $\Gamma$ -faithful. Hence, well-filtered dcpo’s may fail to be dominated. Our examples make use of an example $Y$ given by Zhao and Xi in [16], where they showed that $Y$ is a well-filtered dcpo, but not sober in its Scott topology.

While the paper itself displays a negative result, it refines the results of Ho, Goubault-Larrecq, Jung and Xi, and it also reveals the distinction between the category $\mathbf{WF}$ and the newly found $\mathbf{DOMI}$ , where the later class $\mathbf{DOMI}$ definitely needs more clarification. Finally, we introduce the category of weak dominated dcpo’s, and show that it is a $\Gamma$ -faithful category that is strictly larger than the class of dominated dcpo’s.

2 Preliminaries

In this section, we introduce some basic concepts and notations that will be used in this paper.

Let $P$ be a partially ordered set (poset, for short), $D\subseteq P$ is directed (resp., filtered) if $D$ is nonempty and for any finite subset $F\subseteq D$ , there is a $d\in D$ such that $d$ is an upper bound (resp., a lower bound) of $F$ . A poset $P$ is called directed complete (dcpo, for short) if every directed subset $D$ of $P$ has a least upper bound, which we denote by $\sup D$ , or $\bigvee D$ . For any subset $A\subseteq P$ , let $\uparrow$ $A$ = $\{x\in P:x\geq$ a for some $a\in A\}$ and $\downarrow$ $A$ = $\{x\in P:x\leq a$ for some $a\in A\}$ . Specifically, we write $\uparrow$ $x$ = $\uparrow$ $\{x\}$ and $\downarrow$ $x$ = $\downarrow$ $\{x\}$ . We will call $A\subseteq P$ an upper set (resp., a lower set) if $A$ = $\uparrow$ $A$ (resp., $A$ = $\downarrow$ $A$ ).

A subset $U$ of $P$ is Scott open if $U$ = $\uparrow$ $U$ and for any directed subset $D$ for which sup $D$ exists, sup $D$ $\in U$ implies $D\cap U\neq\emptyset$ . Accordingly, $A\subseteq P$ is Scott closed if $A$ = $\downarrow$ $A$ and for any directed subset $D$ of $P$ with sup $D$ existing, $D\subseteq A$ implies sup $D$ $\in A$ . The set of all Scott open sets of $P$ forms the Scott topology on $P$ , which is denoted by $\sigma P$ , and the set of all Scott closed sets of $P$ is denoted by $\Gamma P$ . Furthermore, for a subset $A$ of $P$ , we will use $\overline{A}$ or $\mathrm{cl}(A)$ to denote the closure of $A$ with respect to the Scott topology on $P$ . The space $(P,\sigma P)$ also is denoted by $\Sigma P$ , some authors refer such a space, a poset endowed with the Scott topology, a Scott space. Recall that the lower topology $\omega(L)$ on $L$ is defined to have the principal filters $\mathord{\uparrow}a$ for $a\in L$ as subbasic closed sets.

For a $T_{0}$ space $X$ , the partial order $\leq$ _X, defined by $x\leq_{X}y$ if and only if $x$ is in the closure of $y$ , is called the specialization order on $X$ . Naturally, the closure of a single point $x$ is $\downarrow$ $x$ , the order considered here is, of course, the specialization order. The specialization order on a Scott space $\Sigma P$ coincides with the original order on $P$ . A subset $K$ of $X$ is compact if every open cover of $K$ admits a finite subcover, and the saturation of a subset $A$ is the intersection of all open sets that contain it. A subset $K$ is called to be saturated if it equals its saturation. Besides, for every subset $A$ of $X$ , the saturation of $A$ coincides with $\uparrow$ $A$ under the specialization order. Every compact saturated subset $K$ in a Scott space is of the form $\mathord{\uparrow}\mathrm{min}K$ , where $\mathrm{min}K$ is the set of minimal elements of $K$ [8]. And usually, we use $Q(X)$ to denote the set of all non-empty compact saturated subsets of $X$ .

A $T_{0}$ space $X$ is sober if every irreducible closed subset $C$ of $X$ is the closure of some unique singleton set $\{c\}$ , where $C$ is called irreducible if $C\subseteq A\cup B$ for closed subsets $A$ and $B$ implies that $C\subseteq A$ or $C\subseteq B$ . We denote the set of all closed irreducible subsets of $X$ by $IRR(X)$ . Every sober space $X$ is a well-filtered space, in the sense that for every filtered family (in the conclusion order) of compact saturated subsets $K_{i},i\in I$ and every open subset $U$ of $X$ , $\bigcap_{i\in I}K_{i}\subseteq U$ implies that there is already some $K_{i}$ with $K_{i}\subseteq U$ . Well-filtered spaces can also be characterized by the so-called KF-sets. In a $T_{0}$ space $X$ , a nonempty subset $A$ of $X$ is said to have the compactly filtered property ( $KF$ property), if there exists a filtered family $\mathcal{K}\subseteq_{\mathrm{flt}}Q(X)$ such that $\mathrm{cl}(A)$ is a minimal closed set that intersects all members of $\mathcal{K}$ . We call such a set a $KF$ -set and denote by $KF(X)$ the set of all closed $KF$ -sets of $X$ . It is shown in [14] that a $T_{0}$ space is well-filtered if and only if every $KF$ -set $K$ of $X$ is the closure of some unique singleton set $\{k\}$ . Finally, we remark that every $KF$ -set is irreducible, i.e., $KF(X)\subseteq IRR(X)$ for each $T_{0}$ space $X$ [14].

3 $\Omega^{*}$ -compact dcpo’s are dominated

In this section, we deduce that the category $\mathbf{DOMI}$ of dominated dcpo’s subsumes the category of $\Omega^{*}$ -compact dcpo’s.

Definition 3.1.

[9] Let $X$ be a $T_{0}$ space and $\leq$ be its order of specialization. We equip X also with its $\omega$ -topology defined from the order of specialization. We say that $X$ is $\Omega^{*}$ -compact if every closed subset of $X$ is compact in the $\omega$ -topology.

Definition 3.2.

[6] Given $A,B\in IRR(L)$ , we write $A\lhd B$ if there is $b\in B$ such that $A\subseteq\mathord{\downarrow}b$ . We write $\nabla B$ for the set $\{A\in IRR(L)\mid A\lhd B\}$ . A dcpo $L$ is called dominated if for every closed irreducible subset $A$ of $L$ , the collection $\nabla A$ is Scott closed in $IRR(L)$ .

Lemma 3.3.

Let $L$ be an $\Omega^{*}$ -compact dcpo. Then $L$ is a dominated dcpo.

Proof.

From the definition of dominated dcpo’s, it suffices to prove that $\nabla A$ is a Scott closed subset of $IRR(L)$ for any $A\in IRR(L)$ . To this end, let $(A_{i})_{i\in I}$ be a directed subset of $\nabla A$ . Then there exists $a_{i}\in A$ such that $A_{i}\subseteq\mathord{\downarrow}a_{i}$ for any $i\in I$ . This means that $(\bigcap_{x\in A_{i}}\mathord{\uparrow}x)\cap A\neq\emptyset$ for any $i\in I$ . Note that $\bigcap_{x\in A_{i}}\mathord{\uparrow}x$ is closed in the lower topology and $(\bigcap_{x\in A_{i}}\mathord{\uparrow}x)_{i\in I}$ is filtered. Then the fact that $L$ is $\Omega^{*}$ -compact reveals that $A\cap\bigcap_{x\in\bigcup_{i\in I}A_{i}}\mathord{\uparrow}x\neq\emptyset$ . Since $\bigcap_{x\in cl(\bigcup_{i\in I}A_{i})}\mathord{\uparrow}x=\bigcap_{x\in\bigcup_{i\in I}A_{i}}\mathord{\uparrow}x$ , we have $A\cap(\bigcap_{x\in cl(\bigcup_{i\in I}A_{i})}\mathord{\uparrow}x)\neq\emptyset$ . Hence, $\sup_{i\in I}A_{i}=cl(\bigcup_{i\in I}A_{i})\in\nabla L$ . ∎

From the above lemma, we can arrive at the following theorem immediately.

Theorem 3.4.

The category of $\Omega^{*}$ -compact dcpo’s is $\Gamma$ -faithful.

We know that $\Omega^{*}$ -compact dcpo’s are well-filtered dcpo’s. It is nature to ask whether the category $\mathbf{WF}$ of well-filtered dcpo’s is $\Gamma$ -faithful. Next, we study the problem.

4 The category of well-filtered dcpo’s is not $\Gamma$ -faithful

In this section, we first give the definition of a dcpo $\mathcal{Z}$ , which will be shown to be well-filtered but not sober in the Scott topology. Moreover, we will see that $\Gamma\mathcal{Z}$ is isomorphic to $\Gamma\hat{\mathcal{Z}}$ , where $\hat{\mathcal{Z}}$ is the poset of all irreducible closed subsets of $\mathcal{Z}$ in the set inclusion order, and $\hat{\mathcal{Z}}$ is a sober dcpo. Since $\mathcal{Z}$ is not sober, it cannot be isomorphic to $\hat{\mathcal{Z}}$ . Hence, the pair of well-filtered dcpo’s $\mathcal{Z}$ and $\hat{\mathcal{Z}}$ illustrates that the category of well-filtered dcpo’s is not $\Gamma$ -faithful.

4.1 The definition of the counterexample $\mathcal{Z}$

Let $\omega_{1}$ be the first non-countable ordinal and $\mathbb{W}=[0,\omega_{1})$ be the set of all ordinals strictly less than $\omega_{1}$ . Then $\mathbb{W}$ consists of all finite and infinite countable ordinals.

Remark 4.1.

[3] The following results about $\mathbb{W}$ are well-known:

1.

$|\mathbb{W}|=\aleph_{1}$ , where $|\mathbb{W}|$ denotes the cardinality of $\mathbb{W}$ .
2.

$\mathbb{W}$ is sequentially complete. That is, for every countable subsequence $D\subseteq\mathbb{W}$ , $\sup D\in\mathbb{W}$ , here the $\sup D$ is taken with respect to the usual linear order on ordinals.
3.

For any $\alpha\in\mathbb{W}$ , $\{\beta:\beta\leq\alpha\}$ is a finite or countably infinite subset of $\mathbb{W}$ .

Now let us review a poset $Y$ given by Zhao and Xi in [16]. We will use this poset as building blocks for our dcpo $\mathcal{Z}$ . Let $\mathbb{W}_{1}=[0,\omega_{1}]=\mathbb{W}\cup\{\omega_{1}\}$ and $Y=\mathbb{W}\times\mathbb{W}_{1}$ . The order on $Y$ is given by $(m,u)\leq(m^{\prime},u^{\prime})$ if and only if:

•

$m=m^{\prime}$ and $u\leq u^{\prime}$ , or
•

$u^{\prime}=\omega_{1}$ and if $u\leq m^{\prime}$ ,

and the order structure of $Y$ can be easily depicted, as in Figure $1$ .

Figure 1: A non-sober well-filtered dcpo $Y$ .

We gather some known results about $Y$ . The reader can find details in [16].

Lemma 4.2.

1.

The poset $Y$ is a dcpo;
2.

$Y$ is well-filtered, but not sober in the Scott topology;
3.

The only closed irreducible subset of $Y$ that is not a principal ideal is $Y$ itself.

For each $u\in[0,\omega_{1})$ , we use the convention $L_{u}$ to denote the set of elements on the $u$ -th level of $Y$ . That is, $L_{u}=\{(x,u)\in Y\mid x\in[0,\omega_{1})\}$ . We call $L_{\omega_{1}}$ the maximal level in $Y$ . Similarly, we say that elements of the form $(u,x),x\in[0,\omega_{1}]$ , are in the $u$ -th column of $Y$ .

The process of constructing $\mathcal{Z}$ is similar to that of $\mathcal{H}$ in [6]. An informal description of this construction is as follows: We begin with $Y$ and for each level $L_{u}$ of $Y$ , we add a copy $Y_{u}$ of $Y$ below $L_{u}$ and identify maximal points of $Y_{u}$ with elements in $L_{u}$ in the canonical way – identifying $(x,\omega_{1})$ in $Y_{u}$ with $(x,u)$ in $L_{u}$ . No order relation between the non-maximal elements of two different $Y_{u},Y_{u^{\prime}}$ is introduced. Now, we repeat this copy-and-identify process and add copies of $Y$ below non-maximal levels of $Y_{u},u\in[0,\omega_{1})$ , copies of $Y$ below non-maximal levels of these copies of $Y$ which are already added in the previous step, and proceed infinitely countable many times.

In order to keep all copies of $Y$ in right place, we use strings on $\mathbb{W}=[0,\omega_{1})$ to index them. To start with, we let $\mathbb{W}^{*}$ be the set of finite strings of elements in $[0,\omega_{1})$ , and for any $s,t\in\mathbb{W}^{*},u\in\mathbb{W}$ , $u.s$ is such that adding the element $u$ to the front of $s$ , and $ts$ is the concatenation of $t$ and $s$ . Now, we use $Y_{\varepsilon}$ to denote the original $Y$ , where $\varepsilon$ is the empty string; for strings $s$ of length $1$ , i.e., $s\in[0,\omega_{1})$ , $Y_{s}$ denotes the copy of $Y$ that is attached below $L_{s}$ of $Y_{\varepsilon}$ , the $s$ -th level of $Y_{\varepsilon}$ ; for strings $s$ of length larger than or equal to $2$ , say $s=x_{1}.s^{\prime}$ with $x_{1}\in[0,\omega_{1})$ and $s^{\prime}$ being the obvious remaining substring of $s$ , $Y_{s}$ denotes the copy of $Y$ that is attached below the $x_{1}$ -level of $Y_{s^{\prime}}$ which is already well placed by induction. Then, the union $\bigcup_{s\in\mathbb{W}^{*}}Y_{s}$ will be the underlying set of our poset $\mathcal{Z}$ . However, there are two issues to be settled. First, we need to denote elements in this big union. This is easy, as we can use $(m,u,s)$ with $(m,u,s)\in[0,\omega_{1})\times[0,\omega_{1}]\times\mathbb{W}^{*}$ to denote the element $(m,u)$ in $Y_{s}$ in the canonical way. Second, we need to take the identifying process into the consideration. In this scenario, we will identify $(m,u,s),u\neq\omega_{1},s\neq\varepsilon$ , the element $(m,u)$ in $Y_{s}$ , with $(m,\omega_{1},u.s)$ , the maximal point $(m,\omega_{1})$ in $Y_{u.s}$ . What we get after this identification is the correct underlying set of our poset $\mathcal{Z}$ .

Note that we equate the elements $(m,u,s)$ of $Y_{s}$ with $(m,\omega_{1},u.s)$ of $Y_{u.s}$ , each element of $\mathcal{Z}$ can be marked as the form of $(m,\omega_{1},s)$ , $m\in\mathbb{W},s\in\mathbb{W}^{*}$ , and $s$ may be $\varepsilon$ . As $\omega_{1}$ appears as the second coordinate of each element in $\mathcal{Z}$ , we simply omit it. By doing so it enables us to use $\{(m,s):(m,s)\in\mathbb{W}\times\mathbb{W}^{*}\}$ to label all elements of $\mathcal{Z}$ . With this all in mind, we now give the rigorous definition of $\mathcal{Z}$ .

Let $\mathcal{Z}$ be $\mathbb{W}\times\mathbb{W}^{*}$ . We define the following relations on $\mathcal{Z}$ (where $m,m^{\prime},u,u^{\prime}\in\mathbb{W},s,t\in\mathbb{W}^{*}$ ), which is reminiscent of the order on $\mathcal{H}$ in [6].

•

$(m,u.s)<_{1}(m,u^{\prime}.s)$ if $u<u^{\prime}$ ;
•

$(m,ts)<_{2}(m,s)$ if $t\neq\varepsilon$ ;
•

$(m,ts)<_{3}(m^{\prime},s)$ if $t\neq\varepsilon$ and $\mathrm{min}(t)\leq m^{\prime}$ .

In the third dotted item, $\mathrm{min}(t)$ denotes the least ordinal appearing in the string $t$ , hence $\mathrm{min}(t)\leq m^{\prime}$ makes sense. The definitions of these three kinds of relations are similar to that on $\mathcal{H}$ given by Ho et al. in [6], and we can obtain a partial order through the following results, where we use $``;"$ to mean composition of relations.

Proposition 4.3.

1.

$<_{1},<_{2},<_{3}$ are transitive and irreflexive, respectively.
2.

$<_{1};<_{2}\;=\;<_{2}$
3.

$<_{1};<_{3}\;\subseteq\;<_{3}$
4.

$<_{2};<_{3}\;\subseteq\;<_{3}$
5.

$<_{3};<_{2}\;\subseteq\;<_{3}$
6.

$<\;:=\;<_{1}\cup<_{2}\cup<_{3}\cup\;(<_{2};<_{1})\cup(<_{3};<_{1})$ is transitive and irreflexive.
7.

$\leq\;:=(<\cup=)$ is a partial order relation on $\mathcal{Z}$ .

Definition 4.4.

We define $\mathcal{Z}$ to be the poset which consists of the set $\mathbb{W}\times\mathbb{W}^{*}$ and the order relation $\leq$ in 4.3.

In the rest of this section, we will prove in steps that $\mathcal{Z}$ is a well-filtered dcpo and that the pair $\mathcal{Z}$ and its sobrification $\hat{\mathcal{Z}}$ witness the fact that the category of well-filtered $\mathrm{dcpo^{\prime}s}$ is not $\Gamma$ -faithful.

4.2 $\mathcal{Z}$ is a dcpo

Sketch of the proof. We prove that $\mathcal{Z}$ is a dcpo by showing that every chain $C$ in $\mathcal{Z}$ has a supremum [10, Corollary 2]. Since $\sup C$ always exists when $C$ itself has a largest element, we only need to take care of the case where $C$ is a strictly increasing chain without a largest element. And of course, in the latter case $C$ must be infinite and we call such chains non-trivial. Similar to Proposition 5.3 in [6], we will be able to show that every non-trivial chain in $\mathcal{Z}$ contains a cofinal chain of the form $(m,u.s)_{u\in W}$ , where $m\in\mathbb{W}$ and $s\in\mathbb{W}^{*}$ are fixed, and $W$ is an infinite subset of $\mathbb{W}$ . This means that every non-trivial chain in $\mathcal{Z}$ eventually stays in the $m$ -column of copy $Y_{s}$ for some string $s$ . Hence, in order to prove that $\mathcal{Z}$ is a dcpo, we only need to verify the existence of suprema of non-trivial chains in that particular form, which is then doable.

Proposition 4.5.

$\mathcal{Z}$ is a $dcpo$ .

Proof.

Let $C$ be a non-trivial chain in $\mathcal{Z}$ .

$\mathbf{Claim~{}1}$ : $C$ contains a cofinal chain of the form $(m,u.s)_{u\in W}$ , with $m$ and $s$ fixed.

Let $(m_{1},s_{1})<(m_{2},s_{2})<\cdots$ be a non-trivial chain. Each relationship $(m_{i},s_{i})<(m_{i+1},s_{i+1})$ has to be one of the five types listed in item (6) of Proposition 4.3. All of these, except $<_{1}$ , strictly reduce the length of the string $s_{i}$ , so they can occur only finitely often along the chain. Therefore, from some index $i_{0}$ onward, the connecting relationship must always be $<_{1}$ which implies that the shape of the entries $(m_{i},s_{i})$ , $i\geq i_{0}$ , is as stated.

$\mathbf{Claim~{}2}$ : If $W$ is uncountable, then $\sup_{u\in W}(m,u.s)=(m,s)$ .

Since $(m,u.s)<_{2}(m,s)$ for any $u\in W$ , $(m,s)$ is an upper bound of $\{(m,u.s):u\in W\}$ . It remains to confirm that $(m,s)$ is the least upper bound of $\{(m,u.s):u\in W\}$ . Assume that $(m^{\prime},s^{\prime})$ is another upper bound of $\{(m,u.s):u\in W\}$ . Then each $(m,u.s)$ must be related to it by one of the five types listed in Proposition 4.3. We set

$A_{<_{r}}=\{(m,u.s):u\in W\ \mathrm{and}\ (m,u.s)<_{r}(m^{\prime},s^{\prime})\}$ ,

where $<_{r}\in\{<_{1},\ <_{2},\ <_{3},\ <_{2};<_{1},\ <_{3};<_{1}\}$ . It follows that at least one of the five sets is uncountable from the uncountability of $W$ .

Case 1, $A_{<_{1}}$ is uncountable. For any $(m,u.s)\in A_{<_{1}}$ , $(m,u.s)<_{1}(m^{\prime},s^{\prime})$ , then $m=m^{\prime}$ and $s^{\prime}$ can be written as $u^{\prime}.s$ for some fixed $u^{\prime}\in\mathbb{W}$ . It is obvious that $\downarrow\!u^{\prime}$ is countable, which contradicts the fact that $A_{<_{1}}$ is uncountable.

Case 2, $A_{<_{2}}$ is uncountable. For any $(m,u.s)\in A_{<_{2}}$ , $(m,u.s)<_{2}(m^{\prime},s^{\prime})$ , so we have $m=m^{\prime}$ and $s^{\prime}\subseteq s$ . Assume that $s=ts^{\prime}$ , where $t\in\mathbb{W^{*}}$ . Then $(m,s)=(m^{\prime},s^{\prime})$ when $t=\varepsilon$ or $(m,s)<_{2}(m^{\prime},s^{\prime})$ when $t\neq\varepsilon$ .

Case 3, $A_{<_{2};<_{1}}$ is uncountable. For any $(m,u.s)\in A_{<_{2};<_{1}}$ , $(m,u.s)<_{2};<_{1}(m^{\prime},s^{\prime})$ , that is, there exists $(m_{u},s_{u})\in\mathcal{Z}$ such that $(m,u.s)<_{2}(m_{u},s_{u})<_{1}(m^{\prime},s^{\prime})$ . Then $m=m_{u}=m^{\prime}$ and $s_{u}$ can be fixed since all of these $s_{u}$ have the same lengths as $s^{\prime}$ , and $s_{u}\subseteq s$ . Set $s_{u}=s^{*}$ which has been fixed. So we have $(m,s)=(m^{\prime},s^{*})<_{1}(m^{\prime},s^{\prime})$ or $(m,s)<_{2}(m^{\prime},s^{*})<_{1}(m^{\prime},s^{\prime})$ .

Case 4, $A_{<_{3}}$ is uncountable. For any $(m,u.s)\in A_{<_{3}}$ , $(m,u.s)<_{3}(m^{\prime},s^{\prime})$ , then $s=ts^{\prime}$ for some $t\in\mathbb{W^{*}}$ . If $t=\varepsilon$ , then $u\leq m^{\prime}$ for all $u\in W$ , which will result in a contradiction as $\downarrow\!m^{\prime}$ is countable, but $W$ is uncountable. Thus $t\neq\varepsilon$ and $\min(ut)\leq m^{\prime}$ for any $u\in W$ . The fact that $\min t\in\mathbb{W}$ is a fixed countable ordinal guarantees the existence of $u_{0}$ with $u_{0}\geq\min t$ . It follows that $\min t\leq m^{\prime}$ according to $(m,u_{0}.s)<_{3}(m^{\prime},s^{\prime})$ . Hence, it turns out that $(m,s)=(m,ts^{\prime})<_{3}(m^{\prime},s^{\prime})$ .

Case 5, $A_{<_{3};<_{1}}$ is uncountable. For any $(m,u.s)\in A_{<_{3};<_{1}}$ , $(m,u.s)<_{3};<_{1}(m^{\prime},s^{\prime})$ , that is, there exists $(m_{u},s_{u})\in\mathcal{Z}$ such that $(m,u.s)<_{3}(m_{u},s_{u})<_{1}(m^{\prime},s^{\prime})$ for any $u\in W$ . Then $m_{u}=m^{\prime}$ and $s_{u}$ can be fixed because all of these $s_{u}$ have the same lengths as $s^{\prime}$ , and $s_{u}\subseteq s$ . Set $s_{u}=s^{*}$ which has been fixed. Thus $(m,u.s)<_{3}(m^{\prime},s^{*})<_{1}(m^{\prime},s^{\prime})$ . Similar to the proof of Case $4$ , we can get that $(m,s)<_{3}(m^{\prime},s^{*})<_{1}(m^{\prime},s^{\prime})$ .

$\mathbf{Claim~{}3}$ : If $W$ is countably infinite, then $\sup_{u\in W}(m,u.s)=(m,u_{0}.s)$ , where $u_{0}=\sup W$ .

It is evident that $(m,u_{0}.s)$ is an upper bound of $\{(m,u.s):u\in W\}$ because $(m,u.s)<_{1}(m,u_{0}.s)$ for each $u\in W$ . Assume that $(m^{\prime},s^{\prime})$ is another upper bound. Now we set the same set ${A_{<_{r}}}$ as which in Claim $2$ . Then there must exist at least one of the five sets being the cofinal subset of $\{(m,u.s):u\in W\}$ . One sees directly that $\sup A_{<_{r}}\!=\sup_{u\in W}(m,u.s)$ if $A_{<_{r}}$ is a cofinal subset of $\{(m,u.s):u\in W\}$ .

Case 1, $A_{<_{1}}$ is cofinal in $\{(m,u.s):u\in W\}$ . For each $(m,u.s)\in A_{<_{1}}$ , $(m,u.s)<_{1}(m^{\prime},s^{\prime})$ , this means $m=m^{\prime}$ and $u<u^{\prime}$ if we set $s^{\prime}=u^{\prime}.s$ for some $u^{\prime}\in\mathbb{W}$ fixed. Then $u_{0}=\sup W=\sup\{u:(m,u.s)\in A_{<_{1}}\}\leq u^{\prime}$ . Thus $(m,u_{0}.s)=(m,u^{\prime}.s)=(m^{\prime},s^{\prime})$ or $(m,u_{0}.s)<_{1}(m,u^{\prime}.s)=(m^{\prime},s^{\prime})$ .

Case 2, $A_{<_{2}}$ is cofinal in $\{(m,u.s):u\in W\}$ . For each $(m,u.s)\in A_{<_{2}}$ , $(m,u.s)<_{2}(m^{\prime},s^{\prime})$ , then $m=m^{\prime}$ and $s^{\prime}\subseteq s$ , which yields that $(m,u_{0}.s)<_{2}(m^{\prime},s^{\prime})$ .

Case 3, $A_{<_{2};<_{1}}$ is cofinal in $\{(m,u.s):u\in W\}$ . Then for each $(m,u.s)\in A_{<_{2};<_{1}}$ , $(m,u.s)<_{2};<_{1}(m^{\prime},s^{\prime})$ , that is, there is $(m_{u},s_{u})\in\mathcal{Z}$ such that $(m,u.s)<_{2}(m_{u},s_{u})<_{1}(m^{\prime},s^{\prime})$ . This implies that $m=m_{u}=m^{\prime}$ and $s_{u}$ can be fixed because all of these $s_{u}$ have the same lengths as $s^{\prime}$ , and $s_{u}\subseteq s$ . Set $s_{u}=s^{*}$ which has been fixed. Hence, $(m,u_{0}.s)<_{2}(m^{\prime},s^{*})<_{1}(m^{\prime},s^{\prime})$ holds.

Case 4, $A_{<_{3}}$ is cofinal in $\{(m,u.s):u\in W\}$ . Then for each $(m,u.s)\in A_{<_{3}}$ , $(m,u.s)<_{3}(m^{\prime},s^{\prime})$ , so $s=ts^{\prime}$ for some $t\in\mathbb{W}^{*}$ and $\min(ut)\leq m^{\prime}$ . If $t=\varepsilon$ or $u\leq\min t$ for all $u$ , then $\min(ut)=u\leq m^{\prime}$ for each $u\in W$ . This manifests that $u_{0}\leq m^{\prime}$ . Note that $u_{0}.s=u_{0}.ts^{\prime}$ and $\min(u_{0}.t)=u_{0}\leq m^{\prime}$ . As a result, $(m,u_{0}.s)<_{3}(m^{\prime},s^{\prime})$ . If $t\neq\varepsilon$ and there is a $u_{1}>\min t$ for some $u_{1}\in\{u\in W:(m,u.s)\in A_{<_{3}}\}$ , then $\min t\leq m^{\prime}$ because of $(m,u_{1}.s)<_{3}(m^{\prime},s^{\prime})$ . It follows that $\min(u_{0}t)=\min t$ . So we can gain that $(m,u_{0}.s)<_{3}(m^{\prime},s^{\prime})$ .

Case 5, $A_{<_{3};<_{1}}$ is cofinal in $\{(m,u.s):u\in W\}$ . For each $(m,u.s)\in A_{<_{3};<_{1}}$ , we know $(m,u.s)<_{3};<_{1}(m^{\prime},s^{\prime})$ . This means that there exists $(m_{u},s_{u})\in\mathcal{Z}$ satisfying $(m,u.s)<_{3}(m_{u},s_{u})$ $<_{1}(m^{\prime},s^{\prime})$ for each $u\in\{u\in W:(m,u.s)\in A_{<_{3};<_{1}}\}$ . So $m_{u}=m^{\prime}$ and $s_{u}$ can be fixed as $s^{*}$ since the length of each $s_{u}$ is same as $s^{\prime}$ , and $s_{u}\subseteq s$ . This suggests that $(m,u.s)<_{3}(m^{\prime},s^{*})<_{1}(m^{\prime},s^{\prime})$ . By using similar analysis of Case $4$ , we can obtain that $(m,u_{0}.s)<_{3}(m^{\prime},s^{*})<_{1}(m^{\prime},s^{\prime})$ .

This covers all cases to be considered and we conclude that $\mathcal{Z}$ is a $\mathrm{dcpo}$ . ∎

4.3 The Scott closed irreducible subsets of $\mathcal{Z}$

Next, we shall characterize the Scott closed irreducible subsets of $\mathcal{Z}$ and make a conclusion that $I\!R\!R(\mathcal{Z})=\{\downarrow\!(m,s):(m,s)\in\mathcal{Z}\}\bigcup\;\{\downarrow\!L_{s}:s\in\mathbb{W}^{*}\}$ , where $L_{s}=\{(m,s)\in\mathcal{Z}:m\in\mathbb{W}\}$ .

The following lemma is immediate and we omit the proof.

Lemma 4.6.

Let $A$ be a subset of $\mathcal{Z}$ . Then $A$ is Scott closed if it is downward closed and contains the sups of every non-trivial chain of the form $(m,u.s)_{u\in W}$ contained in $A$ , where $W\subseteq\mathbb{W}$ .

Lemma 4.7.

Let $A\subseteq\mathcal{Z}$ be a Scott closed subset. If $A$ contains uncountable number of elements of $L_{s}$ for some $s\in\mathbb{W}^{*}$ , then $L_{s}\subseteq A$ .

Proof.

Let $\{(m,s):m\in W\}$ be a subset of $A$ with an uncountable set $W$ . For any $m\in W$ , $L_{m.s}\subseteq A$ since each element of $L_{m.s}$ is related to $(m,s)$ by the relation $<_{3}$ and $A$ is a downward closed set. Thus for any $a\in\mathbb{W}$ , $\sup_{m\in W}(a,m.s)=(a,s)\in A$ , which implies that $L_{s}\subseteq A$ . ∎

Proposition 4.8.

For any $s\in\mathbb{W}^{*}$ , $\downarrow\!L_{s}$ is Scott closed and irreducible.

Proof.

$\mathbf{Claim~{}1}$ : $\downarrow\!L_{s}$ is Scott closed.

Obviously, $\mathord{\downarrow}L_{s}$ is a lower set. Now it remains to confirm that $\mathord{\downarrow}L_{s}$ contains the supremum of every non-trivial chain of the form $(m,u.s)_{u\in W}$ contained in $\mathord{\downarrow}L_{s}$ by Lemma 4.6. To this end, let $(m,u.s_{0})_{u\in W}\subseteq\ \downarrow\!L_{s}$ be a non-trivial chain. We set

$A_{<_{r}}=\{(m,u.s_{0}):u\in W,\ \exists\ (m_{u},s)\in L_{s}\ s.t.\ (m,u.s_{0})<_{r}(m_{u},s)\}$ ,

where $<_{r}\in\!\{<_{1},\ <_{2},\ <_{3},\ <_{2};<_{1},\ <_{3};<_{1}\}$ . Then there exists at least one of the five sets being the cofinal subset of $\{(m,u.s_{0}):u\in W\}$ whether $W$ is countable infinite or uncountable.

Case 1, $A_{<_{1}}$ is cofinal in $\{(m,u.s_{0}):u\in W\}$ . For any $(m,u.s_{0})\in A_{<_{1}}$ , $(m,u.s_{0})<_{1}(m_{u},s)$ for some $m_{u}\in\mathbb{W}$ . Then we have $m_{u}=m,u<u^{\prime}$ if we set $s$ as $u^{\prime}.s_{0}$ for some $u^{\prime}\in\mathbb{W}$ . Under this condition, $W$ is countable and $\sup W\leq u^{\prime}$ . So $\sup_{u\in W}(m,u.s_{0})=(m,\sup W.s)=(m,u^{\prime}.s_{0})=(m,s)$ or $\sup_{u\in W}(m,u.s_{0})=(m,\sup W.s)<_{1}(m,u^{\prime}.s_{0})=(m,s)$ .

Case 2, $A_{<_{2}}$ is cofinal in $\{(m,u.s_{0}):u\in W\}$ . For any $(m,u.s_{0})\in A_{<_{2}}$ , $(m,u.s_{0})<_{2}(m_{u},s)$ for some $m_{u}\in\mathbb{W}$ . We can get that $m_{u}=m$ and $s\subseteq s_{0}$ , so $\sup_{u\in W}(m,u.s_{0})=(m,s)$ or $\sup_{u\in W}(m,u.s_{0})<_{2}(m,s)$ .

Case 3, $A_{<_{2};<_{1}}$ is cofinal in $\{(m,u.s_{0}):u\in W\}$ . For any $(m,u.s_{0})\in A_{<_{2};<_{1}}$ , there is $(m_{u},s_{u})\in\mathcal{Z}$ such that $(m,u.s_{0})<_{2}(m_{u},s_{u})<_{1}(m_{u},s)$ . Then we have $m_{u}=m$ , $s_{u}\subseteq s_{0}$ and $s_{u}$ has the same length as $s$ , which leads to the result that $s_{u}$ can be fixed as $s^{*}$ . Thus $\sup_{u\in W}(m,u.s_{0})=(m,s^{*})<_{1}(m,s)$ or $\sup_{u\in W}(m,u.s_{0})<_{2}(m,s^{*})<_{1}(m,s)$ .

Case 4, $A_{<_{3}}$ is cofinal in $\{(m,u.s_{0}):u\in W\}$ . For any $(m,u.s_{0})\in A_{<_{3}}$ , $(m,u.s_{0})<_{3}(m_{u},s)$ for some $m_{u}\in\mathbb{W}$ . Then $s\subseteq s_{0}$ and hence, $\sup_{u\in W}(m,u.s_{0})=(m,s)$ or $\sup_{u\in W}(m,u.s_{0})<_{2}(m,s)$ .

Case 5, $A_{<_{3};<_{1}}$ is cofinal in $\{(m,u.s_{0}):u\in W\}$ . For any $(m,u.s_{0})\in A_{<_{3};<_{1}}$ , $(m,u.s_{0})<_{3};<_{1}(m_{u},s)$ for some $m_{u}\in\mathbb{W}$ , which implies that there exists $s_{u}\in\mathbb{W}^{*}$ such that $(m,u.s_{0})<_{3}(m_{u},s_{u})<_{1}(m_{u},s)$ . Then $s_{u}\subseteq s_{0}$ and $s_{u}$ has the same length as $s$ . This yields that $s_{u}$ can be fixed as $s^{*}$ . It follows that $\sup_{u\in W}(m,u.s_{0})=(m,s^{*})<_{1}(m,s)$ or $\sup_{u\in W}(m,u.s_{0})<_{2}(m,s^{*})<_{1}(m,s)$ .

Now we can gain our desired result that $\mathord{\downarrow}L_{s}$ is Scott closed.

$\mathbf{Claim~{}2}$ : $\downarrow\!L_{s}$ is irreducible.

It suffices to deduce that $L_{s}$ is irreducible. Let $U_{1},U_{2}$ be two Scott open subsets of $\mathcal{Z}$ with $L_{s}\cap U_{1}\neq\emptyset,\ L_{s}\cap U_{2}\neq\emptyset$ . Then choose $(m_{1},s)\in L_{s}\cap U_{1}$ and $(m_{2},s)\in L_{s}\cap U_{2}$ . Because $(m_{1},s)=\sup_{u\in\mathbb{W}}(m_{1},u.s),\ (m_{2},s)=\sup_{u\in\mathbb{W}}(m_{2},u.s)$ , we can find $(m_{1},u_{1}.s)\in U_{1}$ and $(m_{2},u_{2}.s)\in U_{2}$ for some $u_{i}\in\mathbb{W},\ i=1,2$ from the Scott openness of $U_{1}$ and $U_{2}$ . Without loss of generality, suppose that $u_{1}\leq u_{2}$ , then $(m,u_{i}.s)<_{3}(u_{2},s),\ i=1,2$ , which result in the conclusion that $(u_{2},s)\in L_{s}\cap U_{1}\cap U_{2}$ . Therefore, $L_{s}$ is irreducible. ∎

From the above proposition, we know that $\mathord{\downarrow}L_{s}$ must be Scott irreducible closed sets. Conversely, to prove that $IRR(\mathcal{Z})\subseteq\{\mathord{\downarrow}L_{s}:s\in\mathbb{W}^{*}\}$ , we need to consider $\mathbb{W}^{*}$ endowed with the order $\sqsubseteq$ at first, which is defined in the following.

Based on the observation of the definition of the order relation on $\mathcal{Z}$ , we can find that the two relations $<_{1},<_{2}$ concern only the strings component of points $(m,s)$ . We use them to define an order $\sqsubseteq$ on $\mathbb{W}^{*}$ (where $u,u^{\prime}\in\mathbb{W}$ , $s,t\in\mathbb{W}^{*}$ ):

•

$u.s\sqsubset_{1}u^{\prime}.s$ if $u<u^{\prime}$
•

$ts\sqsubset_{2}s$ if $t\neq\varepsilon$ .

Then $\sqsubset\ :=\ \sqsubset_{1}\cup\sqsubset_{2}\cup\ (\sqsubset_{2};\sqsubset_{1})$ is transitive and irreflexive, hence, $\sqsubseteq\ :=(\sqsubset\cup=)$ is an order relation on $\mathbb{W}^{*}$ , and $\mathbb{W}^{*}$ with the order relation $\sqsubseteq$ is a poset. We denote it by $\mathcal{T}$ and describe parts of $\mathcal{T}$ in Figure $2$ .

Remark 4.9.

Combine with the order on $\mathcal{Z}$ , we have:

1.

If $s\sqsubseteq s^{\prime}$ , then $(m,s)\leq(m,s^{\prime})$ ;
2.

If $(m,s)\leq(m^{\prime},s^{\prime})$ , then $s\sqsubseteq s^{\prime}$ ,

where $m,m^{\prime}\in\mathbb{W}$ , $s,s^{\prime}\in\mathbb{W}^{*}$ .

We state and derive an analogous result for $\mathcal{T}$ with the aid of [6, Proposition 5.9] and Proposition 4.5.

Proposition 4.10.

1.

For any $s\in\mathbb{W}^{*}$ , $\uparrow\!s$ is a linear ordered subset of $\mathcal{T}$ .
2.

The poset $\mathcal{T}$ endowed with the Scott topology is sober.
3.

The map $f:\mathcal{Z}\rightarrow\mathcal{T}$ defined by $f(m,s)=s$ is Scott continuous.

Figure 2: The order $\sqsubseteq$ on $\mathbb{W}^{*}$ .

We remark that for any chain $C\subseteq\mathbb{W}^{*}$ (or $C^{\prime}\subseteq\mathcal{Z}$ ), $\sup C\ (\sup C^{\prime})$ exists and $\sup C=\sup_{u\in W_{1}}u.s$ (or $\sup C^{\prime}=\sup_{u\in W_{2}}(m,u.s)$ ) since there exists $\{u.s:u\in W_{1}\}$ (or $\{(m,u.s):u\in W_{2}\}$ ) being a cofinal subset of $C$ (or $C^{\prime}$ ), where $W_{1}$ (or $W_{2}$ ) is the subset of $\mathbb{W}$ . Thus in the following context, we can regard any non-trivial chain in $\mathcal{T}$ (or in $\mathcal{Z}$ ) as the form of $\{u.s:u\in W_{0}\}$ (or $\{(m,u.s):u\in W^{\prime}\}$ ) with some $W_{0}$ (or $W^{\prime}$ ) $\subseteq\mathbb{W}$ . In addition, a straightforward verification establishes that $\sup_{u\in W_{0}}u.s=s$ when $W_{0}$ is uncountable, and $\sup_{u\in W_{0}}u.s=\sup W_{0}.s$ when $W_{0}$ is countable.

Lemma 4.11.

Let $B$ be a subset of $\mathbb{W}^{*}$ . If $B$ is closed for the $\mathrm{sups}$ of any uncountable non-trivial chain, then we have $\overline{B}=B^{\prime}$ , where

$B^{\prime}=\bigcup\;\{\downarrow\!\sup C:C\subseteq\ \downarrow\!B\ \mathrm{is}\ \mathrm{a}\ \mathrm{countable}\ \mathrm{chain}\}$ .

Proof.

One sees immediately that $B^{\prime}\subseteq\overline{B}$ and $B\subseteq B^{\prime}$ , then it remains to prove that $B^{\prime}$ is Scott closed.

Claim 1: $\downarrow\!B$ is closed for the $\mathrm{sups}$ of any uncountable chain contained in $\downarrow\!B$ .

Assume that $\{u.s^{\prime}:u\in W\}\subseteq\;\downarrow\!B$ be an uncountable chain. Then there exists $s_{u}\in B$ such that $u.s^{\prime}\sqsubseteq s_{u}$ for each $u\in W$ . If for any $u\in W$ , $u.s^{\prime}=s_{u}$ or $u.s^{\prime}\sqsubset_{1}s_{u}$ , then $\{s_{u}:u\in W\}$ is an uncountable chain of $B$ , so $\sup_{u\in W}s_{u}\in B$ by the assumption that $B$ is closed for the $\mathrm{sups}$ of any uncountable non-trivial chain. This yields that $\sup_{u\in W}u.s^{\prime}=s^{\prime}\sqsubseteq\sup_{u\in W}s_{u}$ . If there exists $u^{\prime}.s^{\prime}\sqsubset_{2}s_{u^{\prime}}$ , then $u.s^{\prime}\sqsubset_{2}s_{u^{\prime}}$ for all $u\in W$ . Thus $s^{\prime}=s_{u^{\prime}}$ or $s^{\prime}\sqsubset_{2}s_{u^{\prime}}$ . The case of which there exists $u^{\prime}.s^{\prime}\sqsubset_{2};\sqsubset_{1}s_{u^{\prime}}$ is similar to the former. So $\sup_{u\in W}u.s^{\prime}=s^{\prime}\in\downarrow\!B$ .

Claim 2: $B^{\prime}$ is Scott closed.

Let $\{v.s:v\in W_{1}\}$ be a non-trivial chain of $B^{\prime}$ with $W_{1}\subseteq\mathbb{W}$ being uncountable. Then there exists $C_{v}\subseteq B$ being a countable chain such that $v.s\sqsubseteq\sup C_{v}$ for any $v\in W_{1}$ . We distinguish the following two cases:

Case 1, For each $v\in W_{1}$ , $v.s=\sup C_{v}$ or $v.s\sqsubset_{1}\sup C_{v}$ . If $\{\sup C_{v}:v\in W_{1}\}$ is uncountable, then there must exist $W_{2}\subseteq\mathbb{W}$ being uncountable such that $\{a.s:a\in W_{2}\}$ is a cofinal subset of $\bigcup_{v\in W_{1}}C_{v}$ . By Claim $1$ , we have $s=\sup_{a\in W_{2}}a.s\in\ \downarrow\!B$ . Thus $s\in B^{\prime}$ . If $\{\sup C_{v}:v\in W_{1}\}$ is countable, then there exists a $v^{\prime}\in W_{1}$ such that the set $\{v.s:v\in W_{1},v.s\sqsubset_{1}\sup C_{v^{\prime}}\}$ is uncountable, which implies that $\{v.s:v\in W_{1},v.s\sqsubset_{1}\sup C_{v^{\prime}}\}$ is a cofinal subset of $\{v.s:v\in W_{1}\}$ . It follows that $v\leq v_{0}$ if we set $\sup C_{v^{\prime}}$ as $v_{0}.s$ for some $v_{0}\in\mathbb{W}$ . This means that $W_{1}\subseteq\{a\in\mathbb{W}:a\leq v_{0}\}$ . So we can obtain that $W_{1}$ is a countable set, which is impossible.

Case 2, There exists $v_{0}\in W_{1}$ such that $v_{0}.s\sqsubset_{2}\sup C_{v_{0}}$ or $v_{0}.s\sqsubset_{2};\sqsubset_{1}\sup C_{v_{0}}$ (since the proof of the two cases are similar, we omit that of the latter). Then we have $v.s\sqsubset_{2}\sup C_{v_{0}}$ for all $v\in W_{1}$ . Hence, $\sup_{v\in W_{1}}v.s=s=\sup C_{v_{0}}$ or $\sup_{v\in W_{1}}v.s=s\sqsubset_{2}\sup C_{v_{0}}$ , so $s\in B^{\prime}$ .

So we have proved that $B^{\prime}$ is closed for the $\mathrm{sups}$ of any uncountable chain. Ones sees obviously that $B^{\prime}$ is closed for the $\mathrm{sups}$ of any countable chain and $B^{\prime}$ is a lower set. We can conclude that $B^{\prime}$ is Scott closed. ∎

Lemma 4.12.

Let $A$ be a Scott closed irreducible subset of $\mathcal{Z}$ . Then $\sup f(A)$ exists and $\sup f(A)\in f(A)$ , where $f$ is the Scott continuous map defined in Proposition 4.10.

Proof.

By reason that $A$ is irreducible in $\mathcal{Z}$ and $f$ is Scott continuous, we have $\overline{f(A)}\in I\!R\!R(\mathcal{T})$ . Then there exists $s_{0}\in\mathbb{W}^{*}$ such that $\overline{f(A)}=\overline{\{s_{0}\}}=\ \downarrow\!\!s_{0}$ from the sobriety of $\mathcal{T}$ , which results in $\sup f(A)=s_{0}$ . It suffices to show that $s_{0}\in f(A)$ .

$\mathbf{Claim~{}1}$ : $f(A)$ is closed for the $\mathrm{sups}$ of any uncountable non-trivial chain, and a lower set.

Due to the item (2) in Remark 4.9, the result that $f(A)$ is a lower set follows immediately. Let $\{u.s:u\in W\}$ be an uncountable non-trivial chain in $f(A)$ . If for any $u\in W$ , $A_{u.s}$ is uncountable, where $A_{u.s}=\{m\in\mathbb{W}:(m,u.s)\in A\}$ , then $L_{u.s}\subseteq A$ by Lemma 4.7, so for each $m\in\mathbb{W},\{(m,u.s):u\in W\}\subseteq A$ . Because $W$ is uncountable and $A$ is Scott closed, $\sup_{u\in W}(m,u.s)=(m,s)\in A$ for any $m\in\mathbb{W}$ . It follows that $s\in f(A)$ . If $A_{u^{\prime}.s}$ is countable for some $u^{\prime}\in W$ , then the fact that $A$ is a lower set implies that $A_{u.s}\subseteq A_{u^{\prime}.s}$ for any $u\geq u^{\prime}$ . This means that $A_{u.s}$ is countable for any $u\geq u^{\prime}$ , which guarantees the existence of $u_{1}\geq u^{\prime}$ with the property $A_{u.s}=A_{u_{1}.s}$ for all $u\geq u_{1}$ , owing to the uncountability of $W$ . Thus for each $m\in A_{u_{1}.s},\{(m,u.s):u\geq u_{1}\}\subseteq A$ . Therefore, we have $\sup_{u\geq u_{1}}(m,u.s)=(m,s)\in A$ for any $m\in A_{u_{1}.s}$ as $A$ is Scott closed, that is, $s\in f(A)$ .

Assume for the sake of a contradiction that $s_{0}\notin f(A)$ . By Claim 1, we know $\overline{f(A)}=f(A)^{\prime}$ . This infers that $s_{0}\in\overline{f(A)}\setminus f(A)=f(A)^{\prime}\setminus f(A)$ , where

$f(A)^{\prime}=\bigcup\{\downarrow\!\sup C:C\subseteq\;\downarrow\!f(A)=f(A)\ \mathrm{is}\ \mathrm{a}\ \mathrm{countable}\ \mathrm{chain}\}$ .

This implies that there exists a countable chain $C\subseteq f(A)$ satisfying $s_{0}=\sup C$ . Assume that $\sup C=\sup_{u\in D}u.s^{\prime}=\sup D.s^{\prime}$ , where $D\subseteq\mathbb{W}$ is a countable subset, that is, $s_{0}=\sup D.s^{\prime}$ . Then there must exist $u^{\prime}\in D$ such that $A_{u^{\prime}.s^{\prime}}$ is countable. If not, $A_{u.s^{\prime}}$ is uncountable for all $u\in D$ , then $L_{u.s^{\prime}}\subseteq A$ by Lemma 4.7, and for each $m\in\mathbb{W}$ , $\{(m,u.s^{\prime}):u\in D\}\subseteq A$ , which reveals that $(m,s_{0})=(m,\sup_{u\in D}u.s^{\prime})\in A$ , that is, $s_{0}\in f(A)$ , a contradiction. Thus $A_{u^{\prime}.s^{\prime}}$ is countable for some $u^{\prime}\in D$ . This guarantees the existence of $\sup A_{u^{\prime}.s^{\prime}}$ . Now we let $m_{0}=\sup A_{u^{\prime}.s^{\prime}}$ , $B=\{s\in f(A):s\geq u^{\prime}.s^{\prime}\}$ . Set

$P=\bigcup_{s\in B}\bigcup_{m\in A_{s}}\downarrow\!(m,s)\bigcup\downarrow\!(m_{0},s_{0})$ .

$\mathbf{Claim~{}2}$ : $P$ is Scott closed.

Let $\{(a,u.s):u\in D^{\prime}\}$ be a non-trivial chain in $P$ . If $\{(a,u.s):u\in D^{\prime}\}\bigcap\downarrow\!(m_{0},s_{0})$ is cofinal in $\{(a,u.s):u\in D^{\prime}\}$ , then obviously, $\sup_{u\in D^{\prime}}(a,u.s)$ is less than or equal to $(m_{0},s_{0})$ . Therefore, it belongs to $P$ . If $\{(a,u.s):u\in D^{\prime}\}\bigcap\,(\bigcup_{s\in B}\bigcup_{m\in A_{s}}\downarrow\!(m,s))$ is cofinal in $\{(a,u.s):u\in D^{\prime}\}$ , then for any $u\in D^{\prime}$ , there exist $s_{u}\in B$ and $m_{u}\in A_{s_{u}}$ such that $(a,u.s)<(m_{u},s_{u})$ . Let

$A_{<_{r}}=\{(a,u.s):u\in D^{\prime}\ \mathrm{and}\ (a,u.s)<_{r}(m_{u},s_{u})\ \mathrm{for}\ \mathrm{some}\ s_{u}\in B,\ m_{u}\in A_{s_{u}}\}$ ,

and

$U_{<_{r}}=\{u\in D^{\prime}:(a,u.s)\in A_{<_{r}}\}$ ,

where $<_{r}\in\{<_{1},\;<_{2},\;<_{3},\;<_{2};<_{1},\;<_{3};<_{1}\}$ . Now we need to distinguish the following cases.

Case 1, $A_{<_{1}}$ is cofinal in $\{(a,u.s):u\in D^{\prime}\}$ . For any $(a,u.s)\in A_{<_{1}}$ , there exist $s_{u}\in B$ and $m_{u}\in A_{s_{u}}$ such that $(a,u.s)<_{1}(m_{u},s_{u})$ . Then $m_{u}=a$ and $\{s_{u}:u\in U_{<_{1}}\}$ is a chain contained in $B$ , so $\sup_{u\in U_{<_{1}}}s_{u}$ exists and $(a,\sup_{u\in U_{<_{1}}}s_{u})\in A$ . This means that $\sup_{u\in U_{<_{1}}}s_{u}\in B$ and $\sup_{u\in D^{\prime}}(a,u.s)\leq(a,\sup_{u\in U_{<_{1}}}s_{u})$ . We conclude that $\sup_{u\in D^{\prime}}(a,u.s)\in P$ .

Case 2, $A_{<_{2}}$ or $A_{<_{2};<_{1}}$ is cofinal in $\{(a,u.s):u\in D^{\prime}\}$ . It is easy to verify that $\sup_{u\in D^{\prime}}(a,u.s)\in P$ , so we omit the proof here.

Case 3, $A_{<_{3}}$ is cofinal in $\{(a,u.s):u\in D^{\prime}\}$ . For any $(a,u.s)\in A_{<_{3}}$ , there exist $s_{u}\in B$ and $m_{u}\in A_{s_{u}}$ such that $(a,u.s)<_{3}(m_{u},s_{u})$ , so $s_{u}\subseteq s$ . The finiteness of the length of $s$ ensures the existence of $s_{u_{0}}\in B$ such that $\{(a,u.s):u\in U_{<_{3}},(a,u.s)<_{3}(m_{u},s_{u_{0}})\}$ is cofinal in $A_{<_{3}}$ . Assume that $s=ts_{u_{0}}$ , then $\min(ut)\leq m_{u}$ . If $t=\varepsilon$ or $u\leq\min t$ for all $u\in U_{<_{3}}$ , then $u\leq m_{u}$ for any $u\in U_{<_{3}}$ and $D^{\prime}$ is countable under this condition. It follows that $A_{s_{u_{0}}}\subseteq A_{u^{\prime}.s^{\prime}}$ as $s_{u_{0}}\sqsupseteq u^{\prime}.s^{\prime}$ . This indicates that $\sup D^{\prime}=\sup U_{<_{3}}\leq\sup_{u\in U_{<_{3}}}m_{u}\leq m_{0}$ . Thus that $\sup_{u\in D^{\prime}}(a,u.s)<_{3}(m_{0},s_{u_{0}})<(m_{0},s_{0})$ holds. Else, $t\neq\varepsilon$ and there is a $u_{1}\in D^{\prime}$ such that $u_{1}\!>\min t$ . Then we have $\min t\leq m_{u_{1}}$ according to $(a,u_{1}.s)<_{3}(m_{u_{1}},s_{u_{0}})$ , which leads that $\sup_{u\in D^{\prime}}(a,u.s)<_{3}(m_{u_{1}},s_{u_{0}})$ . Hence $\sup_{u\in D^{\prime}}(a,u.s)\in P$ .

Case 4, $A_{<_{3};<_{1}}$ is cofinal in $\{(a,u.s):u\in D^{\prime}\}$ . For any $(a,u.s)\in A_{<_{3};<_{1}}$ , there exist $s_{u}\in B$ and $m_{u}\in A_{s_{u}}$ such that $(a,u.s)<_{3};<_{1}(m_{u},s_{u})$ , that is, $(a,u.s)<_{3}(m_{u},s_{u}^{\prime})<_{1}(m_{u},s_{u})$ for some $s_{u}^{\prime}\in\mathbb{W}^{*}$ . Then there exists $s^{*}\in\mathbb{W}^{*}$ such that $\{(a,u.s):u\in U_{<_{3};<_{1}}\ \mathrm{and}\ (a,u.s)<_{3}(m_{u},s^{*})<_{1}(m_{u},s_{u})\}$ is cofinal in $A_{<_{3};<_{1}}$ as the length of $s$ is finite and $s_{u}^{\prime}\subseteq s$ . The rest of the proof is similar to Case $3$ . Thus we have $\sup_{u\in D^{\prime}}(a,u.s)\in P$ .

Therefore, $P$ is Scott closed.

Note that $A\subseteq P\cup\overline{(A\setminus\!P)}$ and $A$ is irreducible. Then $A\subseteq P$ or $A\subseteq\overline{A\setminus\!P}$ .

Claim 3: $A\subseteq P$ .

Suppose $A\not\subseteq P$ . Then $A\subseteq\overline{A\setminus\!P}$ . The continuity of $f$ results in $\overline{f(A)}\subseteq\overline{f(A\setminus\!P)}$ . It follows that $\overline{f(A)}\subseteq f(A\setminus\!P)^{\prime}$ as $f(A\setminus\!P)$ is closed for the sups of any uncountable non-trivial chain (the proof is similar to that of $f(A)$ ). Then $s_{0}\in f(A\setminus\!P)^{\prime}$ . This means $s_{0}=\sup C_{0}$ for some countable chain $C_{0}\subseteq\ \downarrow\!\!f(A\setminus\!P)$ . Since $s_{0}=\sup_{u\in D}u.s^{\prime}$ , we can obtain that $C_{0}$ is a cofinal subset of $\{u.s^{\prime}\mid u\in D\}$ , which yields that there exists a $s^{\prime\prime}\in f(A\setminus\!P)$ such that $u^{\prime}.s^{\prime}\sqsubseteq s^{\prime\prime}$ . So we can find $m^{\prime}\in\mathbb{W}$ such that $(m^{\prime},s^{\prime\prime})\in A\setminus\!P$ . However, according to the construction of $P$ , the result that $(m^{\prime},s^{\prime\prime})\in P$ follows immediately, which is a contradiction to the fact that $(m^{\prime},s^{\prime\prime})\in A\backslash P$ . Hence, $A\subseteq P$ .

Now we claim that for any $u\in D$ , there is a $u^{*}\geq u$ such that $A_{u^{*}.s^{\prime}}\subsetneqq A_{u.s^{\prime}}$ . Suppose not, if there is a $u^{*}\in D$ such that for any $u\in D$ with $u\geq u^{*}$ , $A_{u.s^{\prime}}=A_{u^{*}.s^{\prime}}$ , then for each $m\in A_{u^{*}.s^{\prime}}$ , $\{(m,u.s^{\prime}):u\in D,u\geq u^{*}\}\subseteq A$ , and we know that $(m,s_{0})=(m,\sup_{u\in\uparrow\!u^{*}\cap D}u.s^{\prime})\in A$ , which contradicts the assumption that $s_{0}\notin f(A)$ .

Thus for any $u\in D$ , there is a $u^{*}\geq u$ such that $A_{u^{*}.s^{\prime}}\subsetneqq A_{u.s^{\prime}}$ . It follows that for $u^{\prime}\in D$ , we can find $u^{\prime\prime}\in D$ such that $u^{\prime\prime}>u^{\prime}$ , $A_{u^{\prime\prime}.s^{\prime}}\subsetneqq A_{u^{\prime}.s^{\prime}}$ . The assumption that $s_{0}\notin f(A)$ reveals that $m_{0}\notin A_{u_{1}.s^{\prime}}$ for some $u_{1}>u^{\prime\prime}$ . It turns out that there is $u_{2}>u_{1}$ in $D$ satisfying that $A_{u_{2}.s^{\prime}}\subsetneqq A_{u_{1}.s^{\prime}}$ . Write $E=\{u\in D:A_{u.s^{\prime}}=A_{u_{1}.s^{\prime}}\}$ . The Scott closedness of $A$ indicates that $\sup E\in E$ . Let $\hat{u}=\sup E$ . Then $A_{\hat{u}+1.s^{\prime}}\subsetneqq A_{\hat{u}.s^{\prime}}$ as $\hat{u}+1\notin E$ and $\hat{u}+1\leq u_{2}$ , which yields $A_{u_{2}.s^{\prime}}\subseteq A_{\hat{u}+1.s^{\prime}}$ . Note that $A_{u_{2}.s^{\prime}}$ is nonempty by reason of $u_{2}.s^{\prime}\in f(A)$ . Then $A_{\hat{u}+1.s^{\prime}}$ is nonempty. Pick $m_{1}\in A_{\hat{u}.s^{\prime}}\setminus A_{\hat{u}+1.s^{\prime}}$ . We can get that $m_{0}\neq m_{1}$ from $A_{\hat{u}.s^{\prime}}=A_{u_{1}.s^{\prime}}$ .

Now we set $Q=\downarrow\!(P\setminus\downarrow\!(m_{1},\hat{u}.s^{\prime}))$ , then $P=\downarrow\!(m_{1},\hat{u}.s^{\prime})\cup Q$ .

$\mathbf{Claim~{}4}$ : $Q$ is Scott closed.

By the construction of $P$ , we know

$Q=\downarrow(\bigcup_{s\in B}\bigcup_{m\in A_{s}}\downarrow\!(m,s)\setminus\downarrow\!(m_{1},\hat{u}.s^{\prime}))\bigcup\downarrow(\downarrow\!(m_{0},s_{0})\setminus\downarrow\!(m_{1},\hat{u}.s^{\prime}))$ .

The set $Q$ is obviously a lower set. So it remains to prove that $Q$ is closed for the $\mathrm{sups}$ of any non-trivial chain. To begin with, we first notice that $(m_{0},s_{0})\notin\downarrow\!(m_{1},\hat{u}.s^{\prime})$ . Suppose $(m_{0},s_{0})\leq(m_{1},\hat{u}.s^{\prime})$ , then $(m_{0},s_{0})\in A$ because of the fact that $(m_{1},\hat{u}.s^{\prime})\in A$ , which is a contradiction to the assumption that $s_{0}\notin f(A)$ . It follows that $\downarrow(\downarrow\!(m_{0},s_{0})\setminus\downarrow\!(m_{1},\hat{u}.s^{\prime}))=\mathord{\downarrow}(m_{0},s_{0})$ . In other words,

$Q=\downarrow(\bigcup_{s\in B}\bigcup_{m\in A_{s}}\downarrow\!(m,s)\setminus\downarrow\!(m_{1},\hat{u}.s^{\prime}))\bigcup\downarrow\!(m_{0},s_{0})$ .

Let $\{(a,u.s):u\in D_{0}\}$ be a non-trivial chain in $Q$ . If $\{(a,u.s):u\in D_{0}\}\bigcap\downarrow(m_{0},s_{0})$ is cofinal in $\{(a,u.s):u\in D_{0}\}$ , then $\sup_{u\in D_{0}}(a,u.s)\leq(m_{0},s_{0})$ . Else, $\{(a,u.s):u\in D_{0}\}\bigcap\downarrow(\bigcup_{s\in B}\bigcup_{m\in A_{s}}\downarrow\!(m,s)\setminus\downarrow\!(m_{1},\hat{u}.s^{\prime}))$ is cofinal in $\{(a,u.s):u\in D_{0}\}$ . Then, for any $u\in D_{0}$ , there exists $s_{u}\in B,m_{u}\in A_{s_{u}}$ such that $(a,u.s)<(m_{u},s_{u})$ and $(m_{u},s_{u})\nleq(m_{1},\hat{u}.s^{\prime})$ . We set

$A_{<_{r}}=\{(a,u.s):u\in D_{0},\ \exists s_{u}\in B,m_{u}\in A_{s_{u}}\ \mathrm{with}\ (m_{u},s_{u})\nleq(m_{1},\hat{u}.s^{\prime})\ \mathrm{s.t.}\ (a,u.s)<_{r}(m_{u},s_{u})\}$ ,

and

$U_{<_{r}}=\{u\in D_{0}:(a,u.s)\in A_{<_{r}}\}$ ,

where $<_{r}\in\{<_{1},\ <_{2},\ <_{3},\ <_{2};<_{1},\ <_{3};<_{1}\}$ . Now we need to distinguish the following cases.

Case 1, $A_{<_{1}}$ is cofinal in $\{(a,u.s):u\in D_{0}\}$ . For any $(a,u.s)\in A_{<_{1}}$ , there exist $s_{u}\in B$ and $m_{u}\in A_{s_{u}}$ with $(m_{u},s_{u})\nleq(m_{1},\hat{u}.s^{\prime})$ such that $(a,u.s)<_{1}(m_{u},s_{u})$ . Then we have $m_{u}=a$ and $\{s_{u}:u\in U_{<_{1}}\}$ is a chain in $B$ . It follows that $(a,\sup_{u\in U_{<_{1}}}s_{u})\in A$ from the Scott closedness of $A$ . Furthermore, $\sup_{u\in D_{0}}(a,u.s)\leq(a,\sup_{u\in U_{<_{1}}}s_{u})$ . It is obvious that $(a,\sup_{u\in U_{<_{1}}}s_{u})\nleqslant(m_{1},\hat{u}.s^{\prime})$ since $(a,s_{u})\nleqslant(m_{1},\hat{u}.s^{\prime})$ for each $u\in U_{<_{1}}$ . So we have $\sup_{u\in D_{0}}(a,u.s)\in Q$ .

Case 2, $A_{<_{2}}$ or $A_{<_{2};<_{1}}$ is cofinal in $\{(a,u.s):u\in D_{0}\}$ . Clearly, $\sup_{u\in D_{0}}(a,u.s)\in Q$ in this case.

Case 3, $A_{<_{3}}$ is cofinal in $\{(a,u.s):u\in D_{0}\}$ . For any $(a,u.s)\in A_{<_{3}}$ , there exist $s_{u}\in B$ and $m_{u}\in A_{s_{u}}$ with $(m_{u},s_{u})\nleq(m_{1},\hat{u}.s^{\prime})$ such that $(a,u.s)<_{3}(m_{u},s_{u})$ . Then $s_{u}\subseteq s$ . Since the length of $s$ is finite, there exists $s_{u_{0}}\in B$ such that $\{(a,u.s):u\in U_{<_{3}},(a,u.s)<_{3}(m_{u},s_{u_{0}})\}$ is cofinal in $A_{<_{3}}$ . Let $s=ts_{u_{0}}$ . Then $\min(ut)\leq m_{u}$ and $m_{u}\in A_{s_{u_{0}}}\subseteq A_{u^{\prime}.s^{\prime}}$ for all $u$ . If $t=\varepsilon$ or $u\leq\min t$ for each $u$ , then $u=\min(ut)\leq m_{u}$ , which implies that $\sup U_{<_{3}}\leq m_{0},\sup U_{<_{3}}\leq\min t$ and $D_{0}$ is countable. So $\sup_{u\in D_{0}}(a,u.s)<_{3}(m_{0},s_{u_{0}})\leq(m_{0},s_{0})$ . Thus $\sup_{u\in D_{0}}(a,u.s)\in Q$ . If $t\neq\varepsilon$ and $u_{1}>\min t$ for some $u_{1}\in U_{<_{3}}$ , then $\min t\leq m_{u_{1}}$ , which yields that $\sup_{u\in D_{0}}(a,u.s)<_{3}(m_{u_{1}},s_{u_{0}})$ . This means that $\sup_{u\in D_{0}}(a,u.s)\in Q$ by the fact that $(m_{u_{1}},s_{u_{0}})\nleq(m_{1},\hat{u}.s^{\prime})$ .

Case 4, $A_{<_{3};<_{1}}$ is cofinal in $\{(a,u.s):u\in D_{0}\}$ . For any $(a,u.s)\in A_{<_{3};<_{1}}$ , there exist $s_{u}\in B$ and $m_{u}\in A_{s_{u}}$ with $(m_{u},s_{u})\nleq(m_{1},\hat{u}.s^{\prime})$ such that $(a,u.s)<_{3};<_{1}(m_{u},s_{u})$ , that is, there is a $s_{u}^{\prime}$ for each $u$ such that $(a,u.s)<_{3}(m_{u},s_{u}^{\prime})<_{1}(m_{u},s_{u})$ . We can also find $s^{*}\in\mathbb{W}^{*}$ to make $\{(a,u.s):u\in U_{<_{3};<_{1}},(a,u.s)<_{3}(m_{u},s^{*})<_{1}(m_{u},s_{u})\}$ cofinal in $A_{<_{3};<_{1}}$ . Then with a process similar to that in Case $3$ , we conclude that $\sup_{u\in D_{0}}(a,u.s)\in Q$ .

Therefore, $Q$ is Scott closed. Since $A\subseteq P$ is irreducible in $\mathcal{Z}$ , we have $A\subseteq\ \downarrow\!(m_{1},\hat{u}.s^{\prime})$ or $A\subseteq Q=\ \downarrow\!(P\setminus\downarrow\!(m_{1},\hat{u}.s^{\prime}))$ . If $A\subseteq\ \downarrow\!(m_{1},\hat{u}.s^{\prime})$ , then for any $m\in A_{\hat{u}+1.s^{\prime}}$ , $(m,\hat{u}+1.s^{\prime})\in A\subseteq\ \downarrow\!(m_{1},\hat{u}.s^{\prime})$ . This means that $\hat{u}+1.s^{\prime}\leq\hat{u}.s^{\prime}$ , a contradiction. Thus $A\subseteq Q=\ \downarrow\!(P\setminus\downarrow\!(m_{1},\hat{u}.s^{\prime}))$ . It follows that $(m_{1},\hat{u}.s^{\prime})\in\ \downarrow\!(P\setminus\downarrow\!(m_{1},\hat{u}.s^{\prime}))$ , that is, there exists $(m_{2},s_{2})\in P\setminus\downarrow\!(m_{1},\hat{u}.s^{\prime})$ such that $(m_{1},\hat{u}.s^{\prime})\leq(m_{2},s_{2})$ . Now we need to distinguish the following two cases.

Case $1$ , $(m_{2},s_{2})\in(\bigcup_{s\in B}\bigcup_{m\in A_{s}}\downarrow\!(m,s)\setminus\downarrow\!(m_{1},\hat{u}.s^{\prime}))$ . Then $(m_{2},s_{2})\leq(m_{3},s_{3})$ , where $s_{3}\in B$ , $m_{3}\in A_{s_{3}}$ and $(m_{3},s_{3})\nleq(m_{1},\hat{u}.s^{\prime})$ . Via $(m_{1},\hat{u}.s^{\prime})\leq(m_{2},s_{2})\leq(m_{3},s_{3})$ , we have $\hat{u}.s^{\prime}\sqsubseteq s_{2}\sqsubseteq s_{3}\sqsubset s_{0}=\sup_{u\in D}u.s^{\prime}$ . So $m_{1}=m_{2}=m_{3}$ . The assumption that $(m_{3},s_{3})\nleqslant(m_{1},\hat{u}.s^{\prime})$ indicates that $(m_{1},\hat{u}.s^{\prime})<_{1}(m_{3},s_{3})$ . Then $\hat{u}.s^{\prime}\sqsubset_{1}s_{3}$ . This implies $(\hat{u}\!+\!1).s^{\prime}\sqsubset_{1}s_{3}$ or $(\hat{u}\!+\!1).s^{\prime}=s_{3}$ . Hence, it turns out that $(\hat{u}\!\!+\!\!1).s^{\prime}\sqsubseteq s_{3}$ , which yields that $A_{s_{3}}\subseteq A_{\hat{u}+1.s^{\prime}}$ and $m_{1}\in A_{\hat{u}+1.s^{\prime}}$ by reason of the fact that $m_{1}=m_{3}\in A_{s_{3}}$ . This is a contradiction to the assumption that $m_{1}\notin A_{\hat{u}+1.s^{\prime}}$ .

Case $2$ , $(m_{2},s_{2})\in\downarrow\!(m_{0},s_{0})\setminus\downarrow\!(m_{1},\hat{u}.s^{\prime})$ . Then $(m_{1},\hat{u}.s^{\prime})<(m_{2},s_{2})<(m_{0},s_{0})$ . This means that $\hat{u}.s^{\prime}\sqsubset_{1}s_{0}$ , which leads that $m_{1}=m_{0}$ . It violates the assumption that $m_{1}\neq m_{0}$ .

In a conclusion, $s_{0}\in f(A)$ . ∎

Theorem 4.13.

The irreducible closed subsets of $\mathcal{Z}$ are just closures of single elements and the closures of levels.

Proof.

Let $A$ be an irreducible closed subset of $\mathcal{Z}$ and $s_{0}=\sup f(A)$ . Then we have $s_{0}\in f(A)$ by Lemma 4.12, which implies that $A_{s_{0}}\neq\emptyset$ and $f(A)\subseteq\ \downarrow\!s_{0}$ , it is equal to saying that $A\subseteq\ \downarrow\!L_{s_{0}}$ . Next we want to show that $A=\downarrow\!L_{s_{0}}$ or $A=\downarrow\!(m_{0},s_{0})$ for some $m_{0}\in\mathbb{W}$ . If $A_{s_{0}}$ is uncountable, then $L_{s_{0}}\subseteq A$ and so $A=\downarrow\!L_{s_{0}}$ holds. Else, the set $A_{s_{0}}$ is countable, pick $m_{0}\in\mathbb{W}$ such that $(m_{0},s_{0})\in A$ . Let $M=\downarrow\!(A\setminus\downarrow\!(m_{0},s_{0}))$ . Then $A\subseteq\ \downarrow\!(m_{0},s_{0})\bigcup M$ . Now we construct a Scott closed set $W$ containing $M$ . For $M$ , we know $\max M$ exists and $\max M=\max(A\setminus\downarrow\!(m_{0},s_{0}))$ . Let

$E=\{s\in f(\max M):M_{s}\ \mathrm{is}\ \mathrm{countable}\}$ , $B=\{s\in E:\sup M_{s}\notin M_{s}\}$ ,

where $M_{s}=\{m\in\mathbb{W}:(m,s)\in M\}$ . And for any $s\in B$ , we set

$T_{s}=\{(\vee M_{s}).t.p:p=s\ \mathrm{or}\ p<_{1}s,\ \vee M_{s}\leq\min t\ \mathrm{or}\ t=\varepsilon\}$ .

Now let $W=\bigcup_{s\in f(\max M)}W_{s}$ , where $\{W_{s}:s\in f(\max M)\}$ are defined as follows:

•

$W_{s}=\downarrow\!L_{s}$ if $s\in f(\max M)\setminus E$
•

$W_{s}=\bigcup_{a\in M_{s}}\downarrow\!(a,s)$ if $s\in E\setminus B$
•

$W_{s}=\bigcup_{a\in M_{s}}\downarrow\!(a,s)\bigcup\,(\bigcup\{\downarrow\!L_{\vee C}:C\subseteq\bigcup_{s\in B}T_{s}\ \mathrm{is\ a\ chain}\})$ if $s\in B$ .

$\mathbf{Claim~{}1}$ : If $C\subseteq\bigcup_{s\in B}T_{s}$ is a non-trivial chain, then $C$ is countable.

Assume that there exists an uncountable non-trivial chain $C=\{u.s:u\in D\}$ contained in $\bigcup_{s\in B}T_{s}$ . Then for any $u.s\in C$ , there is a $s_{u}\in B$ such that $u.s\in T_{s_{u}}$ , that is, $u.s=(\vee M_{s_{u}}).t_{u}.p_{u}$ with $p_{u}=s_{u}\ \mathrm{or}\ p_{u}\sqsubset_{1}s_{u},\ \vee M_{s_{u}}\leq\min t_{u}\ \mathrm{or}\ t_{u}=\varepsilon$ . It is easy to verify that $s_{u^{\prime}}\neq s_{u^{\prime\prime}}$ and $M_{s_{u^{\prime}}}\neq M_{s_{u^{\prime\prime}}}$ if $u^{\prime}\neq u^{\prime\prime}$ for any $u^{\prime},u^{\prime\prime}\in D$ . In addition, note that $t_{u}.p_{u}=s$ for each $u\in D$ , which means that $s\sqsubseteq p_{u}$ . The fact that $p_{u}\sqsubseteq s_{u}$ for any $u\in D$ suggests that $\{s_{u}:u\in D\}\subseteq\ \uparrow\!s$ . This reveals that $\{s_{u}\mid u\in D\}$ is a chain by Proposition 4.10, which ensures that we can find a cofinal subset $\{s_{u}:u\in D^{\prime}\}$ of $\{s_{u}:u\in D\}$ in which there is only the relation $\sqsubset_{1}$ among all elements. For any $u_{1},u_{2}\in D^{\prime}$ with $u_{1}<u_{2}$ , it turns out that $\vee M_{s_{u_{1}}}<\vee M_{s_{u_{2}}}$ . Note that for any $s_{1},s_{2}\in\mathbb{W}^{*}$ with $s_{1}\sqsubseteq s_{2}$ , we can conclude that $M_{s_{2}}\subseteq M_{s_{1}}$ because $(m,s_{1})\leq(m,s_{2})$ for any $m\in\mathbb{W}$ and $M$ is a lower set. Then the fact that $\{s_{u}:u\in D^{\prime}\}$ is a chain infers that $M_{s_{u^{\prime}}}\subsetneqq M_{s_{u^{\prime\prime}}}$ or $M_{s_{u^{\prime\prime}}}\subsetneqq M_{s_{u^{\prime}}}$ for any $u^{\prime},u^{\prime\prime}\in D^{\prime}$ with $u^{\prime}\neq u^{\prime\prime}$ . Due to the assumption that $\vee M_{s_{u_{1}}}<\vee M_{s_{u_{2}}}$ , we can deduce that $M_{s_{u_{1}}}\subsetneqq M_{s_{u_{2}}}$ , so $s_{u_{1}}\!\sqsupseteq\!s_{u_{2}}$ . Now for a fixed $u_{0}\in D^{\prime}$ , we have $\{s_{u}:u\in D^{\prime},u>u_{0}\}\subseteq\{s\in\mathbb{W}^{*}:s\sqsubset_{1}s_{u_{0}}\}$ . Because $D$ is uncountable and $u=\vee M_{s_{u}}$ for any $u\in D$ , the set $\{s_{u}:u\in D\}$ is uncountable. It follows that $\{s_{u}:u\in D^{\prime}\}$ is uncountable. Furthermore, $\{s_{u}:u\in D^{\prime},u>u_{0}\}$ is uncountable, which contradicts the fact that $\{s\in\mathbb{W}^{*}:s\sqsubset_{1}s_{u_{0}}\}$ is countable.

$\mathbf{Claim~{}2}$ : $W=\bigcup_{s\in f(\max M)}W_{s}$ is Scott closed.

One sees clearly that $W$ is a lower set. So it remains to confirm that $W$ is closed for the $\mathrm{sups}$ of any non-trivial chain.

Let $\{(m,u.s):u\in D_{0}\}$ be a non-trivial chain in $W$ , where $D_{0}\subseteq\mathbb{W}$ . Now we need to distinguish the following cases.

Case 1, $\{(m,u.s):u\in D_{0}\}\bigcap\ (\bigcup_{s\in E}\bigcup_{a\in M_{s}}\downarrow\!(a,s))$ is cofinal in $\{(m,u.s):u\in D_{0}\}$ . For any $u\in D_{0}$ , there are $s_{u}\in E$ and $a_{u}\in M_{s_{u}}$ such that $(m,u.s)<(a_{u},s_{u})$ , without loss of generality, assume that $(a_{u},s_{u})\in\max M$ . We set

$A_{<_{r}}=\{(m,u.s):u\in D_{0},\exists(a_{u},s_{u})\in\max M\ \mathrm{with}\ s_{u}\in E\ s.t.\ (m,u.s)<_{r}(a_{u},s_{u})\}$

and

$U_{<_{r}}=\{u\in D_{0}:(m,u.s)\in A_{<_{r}}\}$ ,

where $<_{r}\in\{<_{1},\ <_{2},\ <_{3},\ <_{2};<_{1},\ <_{3};<_{1}\}$ .

Case 1.1, $A_{<_{1}}$ is cofinal in $\{(m,u.s):u\in D_{0}\}$ . For any $(m,u.s)\in A_{<_{1}}$ , there exists $(a_{u},s_{u})\in\max M$ with $s_{u}\in E$ such that $(m,u.s)<_{1}(a_{u},s_{u})$ . Then $a_{u}=m$ and there is a unique $(m,s_{u_{0}})$ such that $(m,u.s)<_{1}(m,s_{u_{0}})$ since $(a_{u},s_{u})\in\max M$ for each $u\in U_{<_{1}}$ . Thus $\sup_{u\in D_{0}}(m,u.s)\leq(m,s_{u_{0}})\in W$ .

Case 1.2, $A_{<_{2}}$ or $A_{<_{2};<_{1}}$ is cofinal in $\{(m,u.s):u\in D_{0}\}$ . It can be easily verified that $\sup_{u\in D_{0}}(a,u.s)\in W$ in both cases, so we omit the proof here.

Case 1.3, $A_{<_{3}}$ is cofinal in $\{(m,u.s):u\in D_{0}\}$ . For any $(m,u.s)\in A_{<_{3}}$ , there exists $(a_{u},s_{u})\in\max M$ with $s_{u}\in E$ such that $(m,u.s)<_{3}(a_{u},s_{u})$ . We can find $s_{u_{0}}$ such that $\{(m,u.s):u\in U_{<_{3}},(m,u.s)<_{3}(a_{u},s_{u_{0}})\}$ is cofinal in $A_{<_{3}}$ by reason that $s_{u}\subseteq s$ for each $u\in U_{<_{3}}$ and the length of $s$ is finite. If $s_{u_{0}}\in E\setminus B$ , that is, $\vee M_{s_{u_{0}}}\!\in M_{s_{u_{0}}}$ , then $a_{u}\leq\vee M_{s_{u_{0}}}$ . It follows that $(m,u.s)<_{3}(\vee M_{s_{u_{0}}},s_{u_{0}})$ for each $u$ , which implies that $\sup_{u\in D_{0}}(m,u.s)\leq(\vee M_{s_{u_{0}}},s_{u_{0}})\in M$ . Hence, $\sup_{u\in D_{0}}(m,u.s)\in W$ . Suppose that $s_{u_{0}}\in B$ , that is, $\vee M_{s_{u_{0}}}\!\!\notin M_{s_{u_{0}}}$ . Under this condition, we assume $s=ts_{u_{0}}$ , where $t=\varepsilon$ or $t\neq\varepsilon$ . In the case that $t\neq\varepsilon$ and there exists a $u_{1}>\min t$ , then $\min t\leq a_{u_{1}}$ , and so $\sup_{u\in D_{0}}(m,u.s)<_{3}(a_{u_{1}},s_{u_{0}})$ . In the case that $t=\varepsilon$ or $u\leq\min t$ for all $u$ , we can get that $u\leq a_{u}$ for each $u$ . Since $a_{u}\in M_{s_{u_{0}}}$ and $M_{s_{u_{0}}}$ is countable, $D_{0}$ is countable under each case and $\sup U_{<_{3}}\leq\vee M_{s_{u_{0}}}$ . Thus we have $\sup_{u\in D_{0}}(m,u.s)=(m,\vee U_{<_{3}}.t.s_{u_{0}})\leq(m,\vee M_{s_{u_{0}}}.t.s_{u_{0}})$ . Note that $\vee M_{s_{u_{0}}}.t.s_{u_{0}}\in T_{s_{u_{0}}}$ . Hence, $\sup_{u\in D_{0}}(m,u.s)\in W$ .

Case 1.4, $A_{<_{3};<_{1}}$ is cofinal in $\{(m,u.s):u\in D_{0}\}$ . The argument of this case is similar to the above case.

Case 2, $\{(m,u.s):u\in D_{0}\}\bigcap\ (\bigcup_{s\in f(\max M)\setminus E}\downarrow\!L_{s})$ is cofinal in $\{(m,u.s):u\in D_{0}\}$ . For any $u\in D_{0}$ , there are $s_{u}\in f(\max M)\setminus E,a_{u}\in\mathbb{W}$ such that $(m,u.s)<(a_{u},s_{u})$ . We set

$A_{<_{r}}=\{(m,u.s):u\in D_{0},\exists s_{u}\in f(\max M)\setminus E,a_{u}\in\mathbb{W}\ s.t.\ (m,u.s)<_{r}(a_{u},s_{u})\}$

and

$U_{<_{r}}=\{u\in D_{0}:(m,u.s)\in A_{<_{r}}\}$ ,

where $<_{r}\in\{<_{1},\ <_{2},\ <_{3},\ <_{2};<_{1},\ <_{3};<_{1}\}$ .

Case 2.1, $A_{<_{1}}$ is cofinal in $\{(m,u.s):u\in D_{0}\}$ . For any $(m,u.s)\in A_{<_{1}}$ , there exist $s_{u}\in f(\max M)\setminus E$ and $a_{u}\in\mathbb{W}$ such that $(m,u.s)<_{1}(a_{u},s_{u})$ . Then $a_{u}=m,u.s\sqsubset_{1}s_{u}$ . It turns out that $s_{u^{\prime}}=s_{u^{\prime\prime}}$ or $s_{u^{\prime}}\sqsubset_{1}s_{u^{\prime\prime}}$ . As $s_{u}\in f(\max M)\setminus E$ , that is, $M_{s_{u}}$ is uncountable for any $u\in D_{0}$ . Next, we claim that $M_{s_{u}}=\mathbb{W}\setminus\{m_{0}\}$ . Assume that there is a $m\in M_{s_{u}}$ such that $m=m_{0}$ . By $(m,s_{u})\in M=\downarrow\!(A\setminus\downarrow\!(m_{0},s_{0}))$ , we have $(m,s_{u})\leq(m_{1},s_{1})$ for some $(m_{1},s_{1})\in A\setminus\downarrow\!(m_{0},s_{0})$ . This indicates that $m_{1}\neq m_{0}$ , or else, $(m_{1},s_{1})\leq(m_{0},s_{0})$ since $s_{1}\in f(A)$ and $s_{0}=\sup f(A)$ , a contradiction. Thus $(m,s_{u})<_{3}(m_{1},s_{1})$ or $(m,s_{u})<_{3};<_{1}(m_{1},s_{1})$ . The assumption that $s_{u}\in f(\max M)$ guarantees the existence of $n_{u}\in\mathbb{W}$ such that $(n_{u},s_{u})\in\max M$ . It follows that $(n_{u},s_{u})<_{3}(m_{1},s_{1})$ or $(n_{u},s_{u})<_{3};<_{1}(m_{1},s_{1})$ , which will result in a contradiction since there would be a maximal element of $M$ being strictly less than an element of $M$ . Therefore, $M_{s_{u}}\subseteq\mathbb{W}\setminus\{m_{0}\}$ . For the converse, By reason that $A_{s_{u}}$ contains a uncountable set $M_{s_{u}}$ , we know that $L_{s_{u}}\subseteq A$ . Let $m\in\mathbb{W}\setminus\{m_{0}\}$ . If $(m,s_{u})\leq(m_{0},s_{0})$ , then $(m,s_{u})<_{3}(m_{0},s_{0})$ or $(m,s_{u})<_{3};<_{1}(m_{0},s_{0})$ since $m\neq m_{0}$ . By the assumption that $s_{u}\in f(\max M)$ again, we can find $c_{u}\in\mathbb{W}$ satisfying $(c_{u},s_{u})\in\max M$ . Then $(c_{u},s_{u})<_{3}(m_{0},s_{0})$ or $(c_{u},s_{u})<_{3};<_{1}(m_{0},s_{0})$ , but this will contradict to the fact that $(c_{u},s_{u})\in\max M\subseteq A\setminus\downarrow\!(m_{0},s_{0})$ . Hence, we can obtain $\mathbb{W}\setminus\{m_{0}\}=M_{s_{u}}$ . Now suppose that there are $u_{1},u_{2}\in D_{0}$ such that $s_{u_{1}}\sqsubset_{1}s_{u_{2}}$ . From $s_{u_{1}}\in f(\max M)$ , we can identify $(n_{1},s_{u_{1}})\in\max M$ for some $n_{1}\in\mathbb{W}\setminus\{m_{0}\}$ . Furthermore, $n_{1}\in M_{s_{u_{2}}}$ as $M_{s_{u}}=\mathbb{W}\setminus\{m_{0}\}$ for any $u\in D_{0}$ , this is to say $(n_{1},s_{u_{2}})\in M$ . Note that we have assumed that $s_{u_{1}}\sqsubset_{1}s_{u_{2}}$ , we have $(n_{1},s_{u_{1}})<_{1}(n_{1},s_{u_{2}})$ , a contradiction. So for any $u\in D_{0}$ , there exists a unique $s^{*}\in f(\max M)\setminus E$ such that $(m,u.s)<_{1}(m,s^{*})$ , which leads to the result that $\sup_{u\in D_{0}}(m,u.s)\leq(m,s^{*})$ , and we can conclude that $\sup_{u\in D_{0}}(m,u.s)\in W$ .

Case 2.2, $A_{<_{2}}$ or $A_{<_{2};<_{1}}$ is cofinal in $\{(m,u.s):u\in D_{0}\}$ . One sees immediately that $\sup_{u\in D_{0}}(m,u.s)\in W$ in the case.

Case 2.3, $A_{<_{3}}$ is cofinal in $\{(m,u.s):u\in D_{0}\}$ . For any $(m,u.s)\in A_{<_{3}}$ , there exist $s_{u}\in f(\max M)\setminus E$ and $\ a_{u}\in\mathbb{W}$ such that $(m,u.s)<_{3}(a_{u},s_{u})$ . Then we have $s_{u}\subseteq s$ for any $u\in U_{<_{3}}$ , which implies $s\sqsubseteq s_{u}$ . Thus for any given $u^{\prime}\in U_{<_{3}}$ , $\sup_{u\in D_{0}}(m,u.s)=\sup_{u\in U_{<_{3}}}(m,u.s)\leq(m,s_{u^{\prime}})\in L_{s_{u^{\prime}}}\subseteq W$ .

Case 2.4, $A_{<_{3};<_{1}}$ is cofinal in $\{(m,u.s):u\in D_{0}\}$ . The proof is similar to Case $2.3$ .

Case 3, $\{(m,u.s):u\in D_{0}\}\bigcap\ (\bigcup\{\downarrow\!L_{\vee C}:C\subseteq\bigcup_{s\in B}T_{s}\ \mathrm{is}\ \mathrm{a}\ \mathrm{chain}\})$ is cofinal in $\{(m,u.s):u\in D_{0}\}$ . For any $u\in D_{0}$ , there exist $C_{u}\subseteq\bigcup_{s\in B}T_{s}\ \mathrm{and}\ m_{u}\in\mathbb{W}$ such that $(m,u.s)<(m_{u},\vee C_{u})$ . Set

$A_{<_{r}}=\{(m,u.s):u\in D_{0},\exists\ C_{u}\subseteq\bigcup_{s\in B}T_{s}\ \mathrm{and}\ m_{u}\in\mathbb{W}\ s.t.\ (m,u.s)<_{r}(m_{u},\vee C_{u})\}$

and

$U_{<_{r}}=\{u\in D_{0}:(m,u.s)\in A_{<_{r}}\}$ ,

where $<_{r}\in\{<_{1},\ <_{2},\ <_{3},\ <_{2};<_{1},\ <_{3};<_{1}\}$ .

Case 3.1, $A_{<_{1}}$ is cofinal in $\{(m,u.s):u\in D_{0}\}$ . For any $(m,u.s)\in A_{<_{1}}$ , there are $C_{u}\subseteq\bigcup_{s\in B}T_{s}\ \mathrm{and}\ m_{u}\in\mathbb{W}$ such that $(m,u.s)<_{1}(m_{u},\vee C_{u})$ . Then $m_{u}=m$ and $u.s<_{1}\vee C_{u}$ . We know each $C_{u}$ is countable by Claim 1, so there exists a chain $E_{u}$ of the form $\{a.s:a\in D_{u}\}$ is cofinal in $C_{u}$ . It follows that $\bigcup_{u\in U_{<_{1}}}E_{u}$ is a chain in $\bigcup_{s\in B}T_{s}$ . Again applying Claim 1, we know that $\bigcup_{u\in U_{<_{1}}}E_{u}$ is a countable chain. So there exists a chain $C^{\prime}$ of the form $\{a.s:a\in D^{\prime}\}$ contained in $\bigcup_{u\in U_{<_{1}}}C_{u}$ as a cofinal subset, and $C^{\prime}\subseteq\bigcup_{s\in B}T_{s}$ . Hence, we can get that $\sup_{u\in D^{\prime}}(m,u.s)\leq(m,\vee C^{\prime})$ , and hence $\sup_{u\in D_{0}}(m,u.s)\in W$ .

The proof of the residual cases are similar to that of the case 2.2, case 2.3, case 2.4, respectively. Therefore, $W$ is Scott closed. Because $A\subseteq\downarrow\!(m_{0},s_{0})\bigcup M$ and $M\subseteq W$ , $A\subseteq\ \downarrow\!(m_{0},s_{0})\bigcup W$ . So our proof will be complete if we can illustrate that $A$ cannot be contained in $W$ . Suppose not, $A\subseteq W$ , then $(m_{0},s_{0})\in W$ .

•

If $(m_{0},s_{0})\in\bigcup_{s\in E}\bigcup_{a\in M_{s}}\downarrow\!(a,s)$ , then there is $(a_{1},s_{1})\in\max M$ with $s_{1}\in E$ such that $(m_{0},s_{0})\leq(a_{1},s_{1})$ , so $s_{0}\sqsubseteq s_{1}$ . Since $s_{1}\in f(\max M)\subseteq f(A)$ , $s_{1}\sqsubseteq s_{0}$ , that is, $s_{1}=s_{0}$ . Thus $m_{0}=a_{1}$ and so $(m_{0},s_{0})=(a_{1},s_{1})$ , but this contradicts to the fact that $(a_{1},s_{1})\in\max M\subseteq A\setminus\downarrow\!(m_{0},s_{0})$ .
•

If $(m_{0},s_{0})\in\bigcup_{s\in f(\max M)\setminus E}\downarrow\!L_{s}$ , then there are $m^{\prime}\in\mathbb{W},\ s^{\prime}\in f(\max M)\setminus E$ such that $(m_{0},s_{0})\leq(m^{\prime},s^{\prime})$ . This yields that $s_{0}\leq s^{\prime}$ . As $M_{s^{\prime}}$ is uncountable, so is $A_{s^{\prime}}$ . This means that $s^{\prime}\in f(A)$ , which induces $s^{\prime}\sqsubseteq s_{0}$ . As a result, $s^{\prime}=s_{0}$ . It follows that $s_{0}\in f(\max M)\setminus E$ , so we know that $M_{s_{0}}$ is uncountable, which indicates that $A_{s_{0}}$ is uncountable, but this violates the assumption that $A_{s_{0}}$ is countable.
•

If $(m_{0},s_{0})\in\bigcup\{\downarrow\!L_{\vee C}:C\subseteq\bigcup_{s\in B}T_{s}\ \mathrm{is\ a\ chain}\}$ , then there are $C^{\prime}\subseteq\bigcup_{s\in B}T_{s}$ being a countable chain and $m^{\prime}\in\mathbb{W}$ with $(m_{0},s_{0})\leq(m^{\prime},\vee C^{\prime})$ . For any $c\in C^{\prime}$ , there is a $s_{c}\in B$ such that $c=\vee M_{s_{c}}.t_{c}.p_{c}$ , where $\vee M_{s_{c}}\leq\min t_{c}$ , or $t=\varepsilon$ and $\ p_{c}=s_{c}$ or $p_{c}\sqsubset_{1}s_{c}$ . Then there must exist $C^{*}$ being a cofinal subset of $C^{\prime}$ , where $c=\vee M_{s_{c}}.t_{c}.p_{c}=\vee M_{s_{c}}.s^{*}$ for any $c\in C^{*}$ . Because for any $c\in C^{*}$ , we have $s_{c}\in B\subseteq f(\max M)\subseteq f(A)$ , $s_{c}\sqsubseteq s_{0}$ . This implies that $c\sqsubseteq s_{c}\sqsubseteq s_{0}$ for each $c\in C^{*}$ , thus $\vee C^{*}\sqsubseteq s_{0}$ . As $(m_{0},s_{0})\leq(m^{\prime},\vee C^{\prime})$ , $s_{0}\sqsubseteq\vee C^{\prime}=\vee C^{*}$ . We can conclude that $s_{0}=\vee C^{*}=\bigvee_{c\in C^{*}}\vee M_{s_{c}}.s^{*}$ . This means that $|p_{c}|\leq|s^{*}|<|s_{0}|$ , where $|s|$ denotes the string length of $s$ for any $s\in\mathbb{W}^{*}$ . But from $p_{c}\sqsubseteq s_{c}\sqsubseteq s_{0}$ , we know $|p_{c}|\geq|s_{0}|$ , a contradiction. Hence, $A\not\subseteq W$ , and so $A\subseteq\ \downarrow\!(m_{0},s_{0})$ .

In a conclusion, for every irreducible closed subset $A$ of $\mathcal{Z}$ , we have proved that either $A={\downarrow}(m_{0},s_{0})$ or $A=\downarrow\!L_{s_{0}}$ . ∎

4.4 $\mathcal{Z}$ is well-filtered

In this subsection, we confirm that $\mathcal{Z}$ is well-filtered. First, we consider the compact saturated subsets of $\mathcal{Z}$ .

Lemma 4.14.

If $K$ is a compact saturated subset of $\mathcal{T}$ , then $\min K$ is finite, that is, $K=\uparrow\!F$ for some finite subset $F\subseteq\!\mathbb{W}^{*}$ .

Proof.

Assume for the sake of a contradiction that $\min K$ is infinite. Then for any finite subset $F$ contained in $\min K$ , we set $B_{F}=\ \downarrow\!(\min K\setminus F)$ . We claim that $B_{F}$ is Scott closed. To this end, let $\{u.s:u\in D\}$ be a non-trivial chain contained in $B_{F}$ . For any $u\in D$ , there is a $s_{u}\in\min K\setminus F$ such that $u.s\sqsubset s_{u}$ . Assume that $u_{1}<u_{2}$ for some $u_{1},u_{2}\in D$ , then $\{s_{u_{1}},s_{u_{2}}\}\subseteq\ \uparrow\!\!u_{1}.s$ . So $s_{u_{1}}\sqsubseteq s_{u_{2}}$ or $s_{u_{2}}\sqsubseteq s_{u_{1}}$ from Proposition 4.10. Since $\{s_{u_{1}},s_{u_{2}}\}\subseteq\min K$ , $s_{u_{1}}=s_{u_{2}}$ . Thus for any $u\in D$ , there exists a unique $s_{0}\in\min K\setminus F$ such that $u.s\sqsubset s_{0}$ . It follows that $\sup_{u\in D}u.s\sqsubseteq s_{0}$ , which yields that $\sup_{u\in D}u.s\in B_{F}$ . The set $B_{F}$ is obviously a lower set, and therefore, $B_{F}$ is Scott closed. Note that $\{B_{F}:F\subseteq_{f}\min K\}$ is filtered and $B_{F}\cap\min K\neq\emptyset$ , where $F\subseteq_{f}\min K$ denotes $F$ is a finite subset of $\min K$ . Then we arrive at $\bigcap_{F\subseteq_{f}\min K}B_{F}\bigcap\min K\neq\emptyset$ . Pick $s\in\bigcap_{F\subseteq_{f}\min K}B_{F}\bigcap\min K$ , which means that $s\in B_{\{s\}}=\mathord{\downarrow}(\min K\setminus\{s\})$ , which is absurd. ∎

Lemma 4.15.

If $K$ is a compact saturated subset of $\mathcal{Z}$ , then for any $s\in f(\min K)$ , $K_{s}$ is countable, where $K_{s}=\{m\in\mathbb{W}:(m,s)\in\min K\}$ .

Proof.

Assume that there is a $s_{0}\in f(\min K)$ such that $K_{s_{0}}$ is uncountable. Then pick $M=\{m_{n}:n\in\mathbb{N}\}\subseteq K_{s_{0}}$ , where $\mathbb{N}$ denotes the set of all positive integers. We know that $M$ is countable, which ensures the existence of $\vee M$ . Now for each $n\in\mathbb{N}$ , we set

$B_{n}=\bigcup_{i\in\mathbb{N}\setminus\{1,2,...,n-1\}}\downarrow\!(m_{i},s_{0})\ \bigcup\ (\bigcup\{\downarrow\!L_{s}:s\in T\})$ ,

where $T=\{\vee M.t.a:t=\varepsilon\ \mathrm{or}\vee M\leq\min t,a=s_{0}\ \mathrm{or}\ a\sqsubset_{1}s_{0}\}$ .

$\mathbf{Claim}$ : For each $n\in\mathbb{N}$ , $B_{n}$ is Scott closed.

Let $\{(a,u.s):u\in D\}$ be a non-trivial chain contained in $B_{n}$ .

Case 1, $\{(a,u.s):u\in D\}\bigcap\,(\bigcup_{i\in\mathbb{N}\setminus\{1,2,...,n-1\}}\downarrow\!(m_{i},s_{0}))$ is cofinal in $\{(a,u.s):u\in D\}$ . Then for any $u\in D$ , there is $m_{u}\in M\setminus\{m_{1},m_{2},m_{3},...,m_{n-1}\}$ such that $(a,u.s)<(m_{u},s_{0})$ . For convenience we set

$A_{<_{r}}=\{(a,u.s):u\in D,\exists m_{u}\in M\setminus\{m_{1},m_{2},m_{3},...,m_{n-1}\}\ s.t.\ (a,u.s)<_{r}(m_{u},s_{0})\}$ ,

where $<_{r}\in\{<_{1},\ <_{2},\ <_{3},\ <_{2};<_{1},\ <_{3};<_{1}\}$ .

Case 1.1, $A_{<_{1}}$ is cofinal in $\{(a,u.s):u\in D\}$ . For any $(a,u.s)\in A_{<_{1}}$ , $(a,u.s)<_{1}(m_{u},s_{0})$ for some $m_{u}\in M\setminus\{m_{1},m_{2},m_{3},...,m_{n-1}\}$ , which implies that $m_{u}=a$ , and $(a,u.s)<_{1}(a,s_{0})$ . So $\sup_{u\in D}(a,u.s)\leq(a,s_{0})\in B_{n}$ .

Case 1.2, $A_{<_{2}}$ or $A_{<_{2};<_{1}}$ is cofinal in $\{(a,u.s):u\in D\}$ . In this case, for any $u\in D$ , $(a,u.s)$ is less than a fixed point $(m_{0},s_{0})$ , where $a=m_{0}\in M\setminus\{m_{1},m_{2},m_{3},...,m_{n-1}\}$ . Hence $\sup_{u\in D}(a,u.s)\in B_{n}$ .

Case 1.3, $A_{<_{3}}$ is cofinal in $\{(a,u.s):u\in D\}$ . For any $(a,u.s)\in A_{<_{3}}$ , $(a,u.s)<_{3}(m_{u},s_{0})$ for some $m_{u}\in M\setminus\{m_{1},m_{2},m_{3},...,m_{n-1}\}$ . Then $s=ts_{0}$ for some $t\in\mathbb{W}^{*}$ . If $t\neq\varepsilon$ and there is a $u_{0}>\min t$ , then $\min t\leq m_{u_{0}}$ and we can get that $\sup_{u\in D}(a,u.s)<_{3}(m_{u_{0}},s_{0})$ . If $t=\varepsilon$ or $u\leq\min t$ for all $u$ , then $D$ is countable and $u\leq m_{u}$ . So we have $\sup D\leq\vee M$ and $\sup D\leq\min t$ or $t=\varepsilon$ . Now we need to distinguish the following two cases.

Case 1.3.1, $\sup D=\vee M$ . Then $\vee M\leq\min t$ or $t=\varepsilon$ . This implies that $\vee M.t.s_{0}\in T$ . One sees immediately that $\sup_{u\in D}(a,u.s)\leq(a,\vee M.t.s_{0})\in B_{n}$ . Hence, $\sup_{u\in D}(a,u.s)\in B_{n}$ .

Case 1.3.2, $\sup D<\vee M$ . Then we can find $m_{n_{1}}\in M\setminus\{m_{1},m_{2},m_{3},...,m_{n-1}\}$ such that $\sup D\leq m_{n_{1}}$ . This means that $\sup_{u\in D}(a,u.s)<_{3}(m_{n_{1}},s_{0})$ , and we also have that $\sup_{u\in D}(a,u.s)\in B_{n}$ .

Case 1.4, $A_{<_{3};<_{1}}$ is cofinal in $\{(a,u.s):u\in D\}$ . For any $(a,u.s)\in A_{<_{3};<_{1}}$ , $(a,u.s)<_{3};<_{1}(m_{u},s_{0})$ for some $m_{u}\in M\setminus\{m_{1},m_{2},m_{3},...,m_{n-1}\}$ . That is, there is an $s_{u}\in\mathbb{W}^{*}$ , such that $(a,u.s)<_{3}(m_{u},s_{u})<_{1}(m_{u},s_{0})$ . Then there is a fixed $s^{*}\in\mathbb{W}^{*}$ such that $\{(a,u.s):\exists\ m_{u}\in M\setminus\{m_{1},m_{2},m_{3},...,m_{n-1}\}\ s.t.\ (a,u.s)<_{3}(m_{u},s^{*})<_{1}(m_{u},s_{0})\}$ being a cofinal subset of $A_{<_{3};<_{1}}$ since each $s_{u}\subseteq s$ and the length of $s$ is finite. Then $s=ts^{*}$ and the residual analysis is similar to that of the above case.

Case 2, $\{(a,u.s):u\in D\}\bigcap\ (\bigcup\{\downarrow\!L_{s}:s\in T\})$ is cofinal in $\{(a,u.s):u\in D\}$ . That is, for any $u\in D$ , there exists $m_{u}\in\mathbb{W}$ and $s_{u}=\vee M.t_{u}.p_{u}\in T$ such that $(a,u.s)<(m_{u},s_{u})$ , with $t=\varepsilon\ \mathrm{or}\vee M\leq\min t,p_{u}=s_{0}\ \mathrm{or}\ p_{u}\sqsubset_{1}s_{0}$ . Similar to the above case, we set

$A^{\prime}_{<_{r}}=\{(a,u.s):u\in D,\exists m_{u}\in\mathbb{W},s_{u}=\vee M.t_{u}.p_{u}\in T\ s.t.\ (a,u.s)<_{r}(m_{u},s_{u})\}$ ,

where $<_{r}\in\{<_{1},\ <_{2},\ <_{3},\ <_{2};<_{1},\ <_{3};<_{1}\}$ .

Case 2.1, $A^{\prime}_{<_{1}}$ is cofinal in $\{(a,u.s):u\in D\}$ . For any $(a,u.s)\in A^{\prime}_{<_{1}}$ , there exist $m_{u}\in\mathbb{W}$ and $s_{u}=\vee M.t_{u}.p_{u}\in T$ such that $(a,u.s)<_{1}(m_{u},s_{u})$ . Then $m_{u}=a,\ u.s\sqsubset_{1}s_{u}=\vee M.t_{u}.p_{u}$ . Thus $t_{u}.p_{u}=s$ for each $u$ . It follows that $(m_{u},s_{u})=(a,\vee M.s)$ for any $u\in D$ . This means that $\sup_{u\in D}(a,u.s)\leq(a,\vee M.s)$ . Hence, $\sup_{u\in D}(a,u.s)\in B_{n}$ .

Case 2.2, $A^{\prime}_{<_{2}}$ or $A^{\prime}_{<_{2};<_{1}}$ is cofinal in $\{(a,u.s):u\in D\}$ . For the case, it is easy to verify that $\sup_{u\in D}(a,u.s)\in B_{n}$ .

Case 2.3, $A^{\prime}_{<_{3}}$ is cofinal in $\{(a,u.s):u\in D\}$ . For any $(a,u.s)\in A^{\prime}_{<_{3}}$ , there exist $m_{u}\in\mathbb{W}$ and $s_{u}=\vee M.t_{u}.p_{u}\in T$ such that $(a,u.s)<_{3}(m_{u},s_{u})$ . Then we have $s_{u}\subseteq s$ , that is $s\sqsubseteq s_{u}$ . Thus for any fixed $(a,u_{0}.s)\in A^{\prime}_{<_{3}}$ , we have $\sup_{u\in D}(a,u.s)\leq(a,s_{u_{0}})\in B_{n}$ . Therefore, $\sup_{u\in D}(a,u.s)\in B_{n}$ .

Case 2.4, $A^{\prime}_{<_{3};<_{1}}$ is cofinal in $\{(a,u.s):u\in D\}$ . The proof is similar to Case $2.3$ .

One sees immediately that $B_{n}$ is a lower set for any $n\in\mathbb{N}$ . Thus $B_{n}$ is Scott closed for each $n\in\mathbb{N}$ .

Since $\{B_{n}:n\in\mathbb{N}\}$ is filtered and $B_{n}\cap\min K\neq\emptyset$ , $\bigcap_{n\in\mathbb{N}}B_{n}\bigcap\min K\neq\emptyset$ . Choose $(m,s)\in\bigcap_{n\in\mathbb{N}}B_{n}\bigcap\min K$ . We claim that $(m,s)\notin\downarrow\!L_{t}$ for any $t\in T$ . Suppose not, there are $m^{\prime}\in\mathbb{W},s^{\prime}\in T$ such that $(m,s)\leq(m^{\prime},s^{\prime})$ . Assume that $s^{\prime}=\vee M.t^{\prime}.p^{\prime}$ with $\vee M\leq\min t^{\prime}$ or $t=\varepsilon$ , and $p^{\prime}=s_{0}$ or $p^{\prime}\sqsubset_{1}s_{0}$ , then we can obtain that $\mathord{\uparrow}\vee M\cap K_{s_{0}}\neq\emptyset$ from the uncountability of $K_{s_{0}}$ . It follows that

$(m,s)\leq(m^{\prime},s^{\prime})=(m^{\prime},\vee M.t^{\prime}.p^{\prime})<_{3}(m_{1},p^{\prime})\leq(m_{1},s_{0})$

for some $m_{1}\in\mathord{\uparrow}\vee M\cap K_{s_{0}}$ , which implies contradiction as $(m_{1},s_{0})\in\min K$ . Now we suppose that $(m,s)\in B_{n_{0}}$ for some $n_{0}\in\mathbb{N}$ , then $(m,s)\in\bigcup_{i\in\mathbb{N}\setminus\{1,2,...,n_{0}-1\}}\downarrow\!(m_{i},s_{0})$ . This means that there is an $i_{0}\geq n_{0}$ such that $(m,s)\leq(m_{i_{0}},s_{0})$ , and hence, $(m,s)=(m_{i_{0}},s_{0})$ since both of them are minimal elements of $K$ . The fact that $(m,s)\in\bigcap_{n\in\mathbb{N}}B_{n}$ indicates that $(m,s)\in B_{i_{0}+1}$ , that is, $(m,s)\in\bigcup_{i\in\mathbb{N}\setminus\{1,2,...,i_{0}\}}\downarrow\!(m_{i},s_{0})$ . However, this is impossible. Therefore, the original hypothesis doesn’t hold. So $K_{s}$ is countable for any $s\in f(\min K)$ . ∎

Lemma 4.16.

If $K$ is a compact saturated subset of $\mathcal{Z}$ , then for any $s\in\mathbb{W}^{*}$ , $(\bigcup_{u\in\mathbb{W}}L_{u.s})\bigcap\min K$ is countable, where $L_{u.s}=\{(m,u.s):m\in\mathbb{W}\}$ .

Proof.

Suppose that there exists $s_{0}\in\mathbb{W}^{*}$ such that $(\bigcup_{u\in\mathbb{W}}L_{u.s_{0}})\bigcap\min K$ is uncountable. Let

$A=\{u.s_{0}:\exists\ m_{u}\in\mathbb{W}\ s.t.\ (m_{u},u.s_{0})\in\min K\}$

By Lemma 4.15, we have $L_{u.s_{0}}\cap\min K$ is countable for any $u\in\mathbb{W}$ , which implies that $A$ must be uncountable. Then pick $\{u_{n}.s_{0}:n\in\mathbb{N}\}$ as a countable subset of $A$ , besides, for each $n\in\mathbb{N}$ , we extract $m_{n}\in\mathbb{W}$ so that $(m_{n},u_{n}.s_{0})\in\min K$ , where $\mathbb{N}$ denotes all positive integers. Then set

$G=\bigcup_{n\in\mathbb{N}}\downarrow\!(m_{n},u_{n}.s_{0})$

and

$E_{n}=G_{u_{n}.s_{0}}=\{a\in\mathbb{W}:(a,u_{n}.s_{0})\in G\}$ .

Note that each $E_{n}$ is countable and $E_{n_{1}}\subseteq E_{n_{2}}$ if $n_{1}\geq n_{2}$ . Now we construct $B_{n}\ (\forall n\in\mathbb{N})$ as follows:

$B_{n}=\bigcup_{i\in\mathbb{N}\setminus\{1,2,...,n-1\}}\bigcup_{a\in E_{i}}\downarrow\!(a,u_{i}.s_{0})\bigcup\ (\bigcup\{\downarrow L_{\vee C}:C\subseteq\bigcup_{n\in\mathbb{N}}T_{u_{n}.s_{0}}\ \mathrm{is\ a\ chain}\})$ ,

where $T_{u_{n}.s_{0}}=\{\vee E_{n}.t.p:t=\varepsilon\ \mathrm{or}\ \vee E_{n}\leq\min t,\ p=u_{n}.s_{0}\ \mathrm{or}\ p\sqsubset_{1}u_{n}.s_{0}\}$ .

$\mathbf{Claim~{}1}$ : For any chain $C\subseteq\bigcup_{n\in\mathbb{N}}T_{u_{n}.s_{0}}$ , $C$ is countable.

Suppose that there is a chain $C\subseteq\bigcup_{n\in\mathbb{N}}T_{u_{n}.s_{0}}$ being uncountable. Then there must exist an uncountable chain $C_{0}=\{v.s:v\in D_{0}\}\subseteq C$ contained in $T_{u_{n_{0}}.s_{0}}$ for some $n_{0}\in\mathbb{N}$ . That is to say, for any $v.s\in C_{0}$ , $v.s=\vee E_{n_{0}}.t_{c}.p_{c}$ and so $t_{c}.p_{c}$ is equal to $s$ . Then $C_{0}=\{\vee E_{n_{0}}.s\}$ is a single set, which contradicts the uncountability of $C_{0}$ .

$\mathbf{Claim~{}2}$ : For any $n\in\mathbb{N}$ , $B_{n}$ is Scott closed.

To this end, let $\{(m,b.s):b\in D\}$ be a non-trivial chain contained in $B_{n}$ .

Case 1, $\{(m,b.s):b\in D\}\bigcap(\bigcup_{i\in\mathbb{N}\setminus\{1,2,...,n-1\}}\bigcup_{a\in E_{i}}\downarrow\!(a,u_{i}.s_{0}))$ is cofinal in $\{(m,b.s):b\in D\}$ . For any $b\in D$ , there are $i_{b}\in\mathbb{N}\setminus\{1,2,...,n-1\},a_{b}\in E_{i_{b}}$ such that $(m,b.s)<(a_{b},u_{i_{b}}.s_{0})$ , where each $(a_{b},u_{i_{b}}.s_{0})$ can be assumed to be the minimal element of $K$ by the constructions of $G$ and $\{E_{n}:n\in\mathbb{N}\}$ . We set

$A_{<_{r}}=\{(m,b.s):b\in D,\exists\ i_{b}\in\mathbb{N}\setminus\{1,2,...,n-1\},\ a_{b}\in E_{i_{b}}\ s.t.\ (m,b.s)<_{r}(a_{b},u_{i_{b}}.s_{0})\}$ ,

where $<_{r}\ \in\{<_{1},\ <_{2},\ <_{3},\ <_{2};<_{1},\ <_{3};<_{1}\}$ .

Case 1.1, $A_{<_{1}}$ is cofinal in $\{(m,b.s):b\in D\}$ . For any $(m,b.s)\in\!A_{<_{1}}$ , there are some $i_{b}\in\mathbb{N}\setminus\{1,2,...,n-1\},\ a_{b}\in E_{i_{b}}$ such that $(m,b.s)<_{1}(a_{b},u_{i_{b}}.s_{0})$ . Then $a_{b}=m$ and there is an $i_{0}\in\mathbb{N}\setminus\{1,2,...,n-1\}$ such that $(m,b.s)<_{1}(m,u_{i_{0}}.s_{0})$ from the minimality in $K$ of $(a_{b},u_{i_{b}}.s_{0})$ for any $(m,b.s)\in A_{<_{1}}$ . Thus $\sup_{b\in D}(m,b.s)\leq(m,u_{i_{0}}.s_{0})$ , which means $\sup_{b\in D}(m,b.s)\in B_{n}$ .

Case 1.2, $A_{<_{2}}$ or $A_{<_{2};<_{1}}$ is cofinal in $\{(m,b.s):b\in D\}$ . It is easy to demonstrate that $\sup_{b\in D}(m,b.s)\in B_{n}$ in the case.

Case 1.3, $A_{<_{3}}$ is cofinal in $\{(m,b.s):b\in D\}$ . For any $(m,b.s)\in A_{<_{3}}$ , there are some $i_{b}\in\mathbb{N}\setminus\{1,2,...,n-1\},\ a_{b}\in E_{i_{b}}$ such that $(m,b.s)<_{3}(a_{b},u_{i_{b}}.s_{0})$ . Then $u_{i_{b}}.s_{0}\subseteq s$ , which yields that $u_{i_{b}}.s_{0}$ is unique at the moment, denoted by $u_{i^{\prime}}.s_{0}$ , that is, for any $(m,b.s)\in A_{<_{3}}$ , there exists $a_{b}\in E_{i^{\prime}}$ such that $(m,b.s)<_{3}(a_{b},u_{i^{\prime}}.s_{0})$ . Because $(a_{b},u_{i^{\prime}}.s_{0})\in\min K$ , and we pick only a minimal element of $K$ on each level $L_{u_{n}.s_{0}}$ for any $n\in\mathbb{N}$ , $(a_{b},u_{i^{\prime}}.s_{0})$ is a fixed point denoted by $(a_{0},u_{i^{\prime}}.s_{0})$ . Thus $\sup_{b\in D}(m,b.s)\leq(a_{0},u_{i^{\prime}}.s_{0})$ and so $\sup_{b\in D}(m,b.s)\in B_{n}$ .

Case 1.4, $A_{<_{3};<_{1}}$ is cofinal in $\{(m,b.s):b\in D\}$ . For any $(m,b.s)\in A_{<_{3};<_{1}}$ , there are some $i_{b}\in\mathbb{N}\setminus\{1,2,...,n-1\},\ a_{b}\in E_{i_{b}}$ such that $(m,b.s)<_{3};<_{1}(a_{b},u_{i_{b}}.s_{0})$ , that is, $(m,b.s)<_{3}(a_{b},s_{b})<_{1}(a_{b},u_{i_{b}}.s_{0})$ for some $s_{b}\in\mathbb{W}^{*}$ . Then $s_{b}\subseteq s$ is unique since the length of each $s_{b}$ is same as that of $u_{i_{b}}.s_{0}$ , written as $s^{*}$ . Assume $s=t.s^{*}$ for some $t\in\mathbb{W}^{*}$ . It follows that $\min(bt)\leq a_{b}$ in light of $(m,b.s)<_{3}(a_{b},s^{*})$ , where $t$ may be $\varepsilon$ .

Let $i_{b_{0}}=\inf R$ , where $R=\{u_{i}\mid s^{*}\sqsubset_{1}u_{i}.s_{0}\}$ . Then $E_{i_{b}}\subseteq E_{i_{b_{0}}}$ for any $b\in D$ with $(m,b.s)\in A_{<_{3};<_{1}}$ . Therefore, $a_{b}\in E_{i_{b_{0}}}$ . If $t\neq\varepsilon$ and there is $b_{1}>\min t$ , then $\min t=\min(b_{1}.t)\leq a_{b_{1}}$ . So we have $\sup_{b\in D}(a,b.s)<_{3}(a_{b_{1}},s^{*})<_{1}(a_{b_{1}},u_{i_{b_{1}}}.s_{0})$ . If $t=\varepsilon$ or $b\leq\min t$ for each $b\in D$ , then $\sup D\leq\min t$ or $t=\varepsilon$ . The fact that $\min(b.t)\leq a_{b}$ suggests that $\min(b.t)=b\leq a_{b}$ , which leads to that $\sup D\leq\vee E_{i_{b_{0}}}$ . Now we need to distinguish the following two cases.

Case 1.4.1, $\sup D=\vee E_{i_{b_{0}}}$ . Then $\sup_{b\in D}(m,b.s)=(m,\vee E_{i_{b_{0}}}.s)=(m,\vee E_{i_{b_{0}}}.t.s^{*})$ . Note that $\vee E_{i_{b_{0}}}=\sup D\leq\min t$ or $t=\varepsilon$ , which yields that $\vee E_{i_{b_{0}}}.t.s^{*}\in T_{i_{b_{0}}.s_{0}}$ .

Case 1.4.2, $\sup D<\vee E_{i_{b_{0}}}$ . Then there exists $a^{\prime}\in E_{i_{b_{0}}}$ such that $\sup D\leq a^{\prime}$ , which results in the conclusion that $\sup_{b\in D}(m,b.s)<_{3}(a^{\prime},s^{*})<_{1}(a^{\prime},u_{i_{b_{0}}}.s_{0})\in B_{n}$ .

Case 2, $\{(m,b.s):b\in D\}\bigcap\;(\bigcup\{\downarrow L_{\vee C}:C\subseteq\bigcup_{n\in\mathbb{N}}T_{u_{n}.s_{0}}\ \mathrm{is\ a\ chain}\})$ is cofinal in $\{(m,b.s):b\in D\}$ . That is to say, for any $b\in D$ , there are some $m_{b}\in\mathbb{W}$ and a chain $C_{b}\subseteq\bigcup_{n\in\mathbb{N}}T_{u_{n}.s_{0}}$ such that $(m,b.s)<(m_{b},\vee C_{b})$ . Now we set

$A^{\prime}_{<_{r}}=\{(m,b.s):b\in D,\exists\ m_{b}\in\mathbb{W},\ C_{b}\subseteq\bigcup_{n\in\mathbb{N}}T_{u_{n}.s_{0}}\ s.t.\ (m,b.s)<_{r}(m_{b},\vee C_{b})\}$ ,

and

$U^{\prime}_{<_{r}}=\{u\in D_{0}:(m,u.s)\in A^{\prime}_{<_{r}}\}$ ,

where $<_{r}\ \in\{<_{1},\ <_{2},\ <_{3},\ <_{2};<_{1},\ <_{3};<_{1}\}$ .

Case 2.1, $A^{\prime}_{<_{1}}$ is cofinal in $\{(m,b.s):b\in D\}$ . For any $(m,b.s)\in A^{\prime}_{<_{1}}$ , there are $m_{b}\in\mathbb{W}$ and $C_{b}\subseteq\bigcup_{n\in\mathbb{N}}T_{u_{n}.s_{0}}$ such that $(m,b.s)<_{1}(m_{b},\vee C_{b})$ . Then $m_{b}=m$ . By Claim $1$ , we know that $C_{b}$ must be countable, which ensures the existence of the countable chain $C_{0}\subseteq\bigcup_{b\in U^{\prime}_{<_{1}}}C_{b}$ being the cofinal subset of $\bigcup_{b\in U^{\prime}_{<_{1}}}C_{b}$ . It follows that $C_{0}\subseteq\bigcup_{b\in U^{\prime}_{<_{1}}}C_{b}\subseteq\bigcup_{n\in\mathbb{N}}T_{u_{n}.s_{0}}$ . This implies that $(m,b.s)<_{1}(m,\vee C_{0})$ for any $b\in D$ . So $\sup_{b\in D}(m,b.s)\leq(m,\vee C_{0})\in B_{n}$ .

Case 2.2, $A^{\prime}_{<_{2}}$ or $A^{\prime}_{<_{3}}$ is cofinal in $\{(m,b.s):b\in D\}$ . For any $(m,b.s)\in A^{\prime}_{<_{2}}$ or $A^{\prime}_{<_{3}}$ , there are some $m_{b}\in\mathbb{W}$ and $C_{b}\subseteq\bigcup_{n\in\mathbb{N}}T_{u_{n}.s_{0}}$ such that $(m,b.s)<_{2}(m_{b},\vee C_{b})$ or $(m,b.s)<_{3}(m_{b},\vee C_{b})$ . Then $\vee C_{b}\subseteq s$ , that is, $s\sqsubseteq\vee C_{b}$ . Hence, $\sup_{b\in D}(m,b.s)\leq(m,\vee C_{b_{0}})$ for any given $b_{0}\in U^{\prime}_{<_{2}}$ or $b_{0}\in U^{\prime}_{<_{3}}$ , which leads to that $\sup_{b\in D}(m,b.s)\in B_{n}$ .

Case 2.3, $A^{\prime}_{<_{2};<_{1}}$ or $A^{\prime}_{<_{3};<_{1}}$ is cofinal in $\{(m,b.s):b\in D\}$ . The analysis of this case is similar to Case $2.2$ .

As $B_{n}$ is a lower set for each $n\in\mathbb{N}$ , $B_{n}$ is Scott closed. Because $\{B_{n}:n\in\mathbb{N}\}$ is filtered and $\min K\bigcap B_{n}\neq\emptyset$ for any $n\in\mathbb{N}$ , the set $\bigcap_{n\in\mathbb{N}}B_{n}$ meets $\min K$ . Choose $(m,s)\in\bigcap_{n\in\mathbb{N}}B_{n}\bigcap\min K$ . The uncountability of $A$ guarantees the existence of $u^{*}\in\mathord{\uparrow}\vee_{n\in\mathbb{N}}u_{n}$ with $u^{*}.s_{0}\in A$ . Now we claim that there exist $u^{\prime}>u^{*}$ , $m^{\prime}\in\mathord{\uparrow}\vee(\bigcup_{n\in\mathbb{N}}E_{n})=\mathord{\uparrow}\vee E_{1}$ such that $(m^{\prime},u^{\prime}.s_{0})\in\min K$ . Suppose not, for any $u>u^{*}$ , any $m\in\mathbb{W}$ with $(m,u.s_{0})\in\min K$ , $m\notin{E_{1}}^{u}$ . Then we could find $a_{m}\in E_{1}$ such that $a_{m}\nleqslant m$ , i.e., $a_{m}>m$ . It follows that

$\{m\in\mathbb{W}:(m,u.s_{0})\in\min K\ \mathrm{with}\ u>u^{*}\}\subseteq\;\downarrow\!E_{1}$ ,

which is a contradiction since the left side of the inclusion is uncountable but the right is countable. Hence, there exists $u^{\prime}>u^{*}$ , $m^{\prime}\in\mathord{\uparrow}\vee(\bigcup_{n\in\mathbb{N}}E_{n})=\mathord{\uparrow}\vee E_{1}$ such that $(m^{\prime},u^{\prime}.s_{0})\in\min K$ . Now based on the above discussion, we shall know that $(m,s)\notin\ \downarrow\!L_{\vee C}$ for any chain $C\subseteq\bigcup_{n\in\mathbb{N}}T_{u_{n}.s_{0}}$ . Suppose not, $(m,s)\leq(m_{0},\vee C_{0})$ for some $m_{0}\in\mathbb{W}$ and $C_{0}\subseteq\bigcup_{n\in\mathbb{N}}T_{u_{n}.s_{0}}$ . For each $c\in C_{0}$ , $c$ can be written as $\vee E_{n_{c}}.t_{c}.p_{c}$ with $t_{c}=\varepsilon$ or $\vee E_{n_{c}}\leq\min t_{c},\ p_{c}=u_{n_{c}}.s_{0}$ or $p_{c}\sqsubset_{1}u_{n_{c}}.s_{0}$ . Note that there is a cofinal subset $C^{\prime}_{0}$ of $C_{0}$ of the form $\{m_{c}.\hat{s}:c\in C^{\prime}_{0}\}$ , in which $m_{c}=\vee E_{n_{c}},\ \hat{s}=t_{c}.p_{c}\triangleq t.p$ for each $c\in C^{\prime}_{0}$ , $\bigvee(\bigcup_{c\in C^{\prime}_{0}}E_{n_{c}})\leq\min t$ or $t=\varepsilon$ . According to $\bigvee(\bigcup_{c\in C^{\prime}_{0}}E_{n_{c}})\leq\vee E_{1}\leq m^{\prime}$ , we can get that

$(m,s)\leq(m_{0},\vee C_{0})=(m_{0},\vee C^{\prime}_{0})\leq(m_{0},\bigvee(\bigcup_{c\in C^{\prime}_{0}}E_{n_{c}}).t.p)<_{3}(m^{\prime},p)<_{1}(m^{\prime},u^{\prime}.s_{0})$ .

This violates the assumption that both $(m,s)$ and $(m^{\prime},u^{\prime}.s_{0})$ are minimal elements of $K$ . As a result, for $(m,s)\in\bigcap_{n\in\mathbb{N}}B_{n}\bigcap\min K$ , $(m,s)\in\bigcap_{n\in\mathbb{N}}(\bigcup_{i\in\mathbb{N}\setminus\{1,2,...,n-1\}}\bigcup_{a\in E_{i}}\downarrow\!(a,u_{i}.s_{0}))$ . But this is impossible with a similar reason in the proof of Lemma 4.15. ∎

Lemma 4.17.

If $K$ is a compact saturated subset of $\mathcal{Z}$ , then $\min K$ is countable.

Proof.

By way of contradiction, suppose that $\min K$ is uncountable. In virtue of Lemma 4.15, we have $\min K\cap L_{s}$ is countable for any $s\in\min K$ . The fact that $\min K=\bigcup_{s\in f(\min K)}(\min K\cap L_{s})$ induces that $f(\min K)$ is uncountable, where the map $f:\mathcal{Z}\rightarrow\mathcal{T}$ is defined by $f(m,s)=s$ . In addition, the continuity of $f$ reveals that $f(\min K)$ is a compact subset of $\mathcal{T}$ . Then $\min f(\min K)$ is finite due to Lemma 4.14, which yields that there is $s\in\min f(\min K)$ such that $\uparrow\!s\cap f(\min K)$ is uncountable. Since the length of $s$ is finite, there exists $s_{0}\geq s$ such that $\min K\bigcap(\bigcup_{u\in\mathbb{W}}L_{u.s_{0}})$ is uncountable. This contradicts the result of Lemma 4.16. Hence, $\min K$ is countable. ∎

Theorem 4.18.

$\Sigma\mathcal{Z}$ is well-filtered.

Proof.

Recall that a $T_{0}$ space if well-filtered if and only if all its $K\!F$ -sets are closures of singletons. We assume that $\Sigma\mathcal{Z}$ is not well-filtered. Then there exists a $K\!F$ -set $\downarrow\!L_{s_{0}}$ for some $s_{0}\in\mathbb{W}^{*}$ by Theorem 4.13 and that all $K\!F$ -sets are irreducible. This implies that there is a filtered family $\{Q_{i}:i\in I\}$ contained in $Q(\mathcal{Z})$ such that $\downarrow\!\!L_{s_{0}}$ is a minimal closed set that intersects each $Q_{i}$ . Set $K_{i}=\uparrow\!(Q_{i}\ \cap\downarrow\!L_{s_{0}})$ , then a straightforward verification establishes that $\{K_{i}:i\in I\}\subseteq_{flt}Q(\mathcal{Z})$ and $\downarrow\!L_{s_{0}}$ is still a minimal closed set that intersects each $K_{i}$ . Note that $\min K_{i}\subseteq\ \downarrow\!L_{s_{0}}$ for each $i\in I$ . Now we fix $j_{1}\in I$ and let $J=\uparrow\!j_{1}\cap I$ , then $J$ is a cofinal subset of $I$ and $\downarrow\!L_{s_{0}}$ is a minimal closed set that intersects each $K_{j},j\in J$ . The assumption that $\downarrow\!L_{s_{0}}\cap K_{j}\neq\emptyset$ implies that $\downarrow\!L_{s_{0}}\cap\min K_{j}\neq\emptyset$ for all $j\in J$ . Pick $(m_{j},s_{j})\in\ \downarrow\!L_{s_{0}}\cap\min K_{j}$ for each $j\in J$ .

$\mathbf{Claim~{}1}$ : There exists $j_{0}\in J$ such that $s_{j}=s_{0}$ for any $j\geq j_{0}$ .

Assume for the sake of a contradiction that for any $j\in J$ , there is $i\geq j$ such that $s_{i}\neq s_{0}$ , that is, $s_{i}\sqsubset s_{0}$ . Let $J_{1}=\{j\in J:s_{j}\sqsubset s_{0}\}$ . Then $J_{1}$ is cofinal in $J$ and so $\downarrow\!L_{s_{0}}$ is a minimal closed set that intersects each $K_{j},j\in J_{1}$ . The minimality of $\mathord{\downarrow}L_{s_{0}}$ infers that $\overline{\{(m_{j},s_{j}):j\in J_{1}\}}=\downarrow\!L_{s_{0}}$ . This indicates that $\{(m_{j},s_{j}):j\in J_{1}\}$ is uncountable. If not, $\{(m_{j},s_{j}):j\in J_{1}\}$ is countable, we set $B=\bigcup_{j\in J_{1}}\downarrow\!(m_{j},s_{0})\bigcup\ \downarrow\!(\vee_{j\in J_{1}}m_{j},s_{0})$ . Then it is easy to verify that $B$ is Scott closed and $B\cap K_{j}\neq\emptyset,j\in J_{1}$ , but $B\subsetneqq\downarrow\!L_{s_{0}}$ , which is a contradiction to the minimality of $\downarrow\!L_{s_{0}}$ . The above discussion also reveals that $\{m_{j}:j\in J_{1}\}$ is uncountable. Now we let

$B_{j}=\{(m_{i},s_{i}):i\geq j,(m_{i},s_{i})\in\min K_{j}\},\forall j\in J_{1}$ ,

then it follows that each $B_{j}$ is countable by Lemma 4.17. Note that for any given $j\in J_{1}$ , $I_{j}=\{i\in J_{i}:i\geq j\}$ is a cofinal subset of $I$ . This means that $\overline{\{(m_{r},s_{r}):r\in I_{j}\}}=\downarrow\!L_{s_{0}}$ , which yields that $\{m_{r}:r\in I_{j}\}$ is uncountable for a similar reason as $J_{1}$ . Therefore,

$B^{\prime}_{j}=\{(m_{i},s_{i}):i\geq j,(m_{i},s_{i})\notin\min K_{j}\}$

is uncountable for all $j\in J_{1}$ , and $\{m_{i}:(m_{i},s_{i})\in B^{\prime}_{j}\}$ is uncountable. It turns out that $\overline{B^{\prime}_{j}}=\mathord{\downarrow}L_{s_{0}}$ . Given $j\in J_{1}$ , for any $(m_{i},s_{i})\in B^{\prime}_{j}$ , there is $(n_{i_{k}},a_{i_{k}})\in\min K_{j}$ such that $(n_{i_{k}},a_{i_{k}})<(m_{i},s_{i})$ . The countability of $\min K_{j}$ ensures the existence of $(n_{i_{0}},a_{i_{0}})\in\min K_{j}$ with $\uparrow\!\!(n_{i_{0}},a_{i_{0}})\bigcap B^{\prime}_{j}$ and $E_{i_{0}}=\{m_{i}:(m_{i},s_{i})\in\ \uparrow\!\!(n_{i_{0}},a_{i_{0}})\bigcap B^{\prime}_{j}\}$ being uncountable, by the aids of the equation $\{m_{i}:(m_{i},s_{i})\in B^{\prime}_{j}\}=\bigcup_{i\geq j}\{m_{i}:(m_{i},s_{i})\in\ \uparrow\!(n_{i_{k}},a_{i_{k}})\bigcap B^{\prime}_{j}\}$ and the uncountability of the set $\{m_{i}:(m_{i},s_{i})\in B^{\prime}_{j}\}$ . Thus there is a $m_{i}\in E_{i_{0}}$ with $m_{i}\neq n_{i_{0}}$ . In a conclusion, we can find $(m_{i_{j}},s_{i_{j}})\in B^{\prime}_{j}$ and $(n_{i_{j}},a_{i_{j}})\in\min K_{j}$ satisfying $(n_{i_{j}},a_{i_{j}})<(m_{i_{j}},s_{i_{j}})$ , with $n_{i_{j}}\neq m_{i_{j}}$ for any $j\in J_{1}$ , which means that $a_{i_{j}}\sqsubset_{2}s_{i_{j}}$ or $a_{i_{j}}\sqsubset_{2};\sqsubset_{1}s_{i_{j}}$ . The fact that $\overline{\{(n_{i_{j}},a_{i_{j}}):j\in J_{1}\}}\bigcap K_{j}\neq\emptyset$ , for any $j\in J_{1}$ suggests that $\overline{\{(n_{i_{j}},a_{i_{j}}):j\in J_{1}\}}=\downarrow\!L_{s_{0}}$ from the minimality of $\mathord{\downarrow}L_{s_{0}}$ . This implies that $\overline{\{a_{i_{j}}:j\in J_{1}\}}=\overline{\{s_{0}\}}=\ \downarrow\!\!s_{0}$ . It follows that $s_{0}=\vee_{j\in J_{1}}a_{i_{j}}$ . The fact that $a_{i_{j}}\in f(\min K_{j})\subseteq\ f(K_{j_{1}})\subseteq\ \uparrow\!\min f(\min K_{j_{1}})=\ \uparrow\!F$ indicates that $\{a_{i_{j}}:j\in J_{1}\}=\bigcup_{a\in F}(\uparrow\!a\cap\{a_{i_{j}}:j\in J_{1}\})$ , where $F=\min f(\min K_{j_{1}})$ . Then $F$ is finite by Lemma 4.14, which leads to

$\overline{\{a_{i_{j}}:j\in J_{1}\}}\subseteq\bigcup_{a\in F}\downarrow\!\vee(\uparrow\!a\cap\{a_{i_{j}}:j\in J_{1}\})$ .

It follows that $s_{0}\in\downarrow\!\vee(\uparrow\!a_{0}\cap\{a_{i_{j}}:j\in J_{1}\})$ for some $a_{0}\in F$ , that is, $s_{0}\leq\vee(\uparrow\!a_{0}\cap\{a_{i_{j}}:j\in J_{1}\})$ . Then $s_{0}=\vee(\uparrow\!a_{0}\cap\{a_{i_{j}}:j\in J_{1}\})$ according to $s_{0}=\vee_{j\in J_{1}}a_{i_{j}}$ . It turns out that there is a chain of the form $\{v.s:v\in D\}$ being a cofinal subset of $\uparrow\!a_{0}\cap\{a_{i_{j}}:j\in J_{1}\}$ enjoying $s_{0}=\sup_{v\in D}v.s$ . This means that $|s_{0}|=|v.s|$ or $|s_{0}|+1=|v.s|$ . Note that $v.s\in(\uparrow\!a_{0}\cap\{a_{i_{j}}:j\in J_{1}\})$ for any $v\in D$ . Then assume $v.s=a_{i_{j_{v}}}$ . We notice that $v.s\sqsubset_{2}s_{i_{j_{v}}}$ or $v.s\sqsubset_{2};\sqsubset_{1}s_{i_{j_{v}}}$ , and conclude that $|v.s|>|s_{i_{j_{v}}}|$ . We claim that $|s_{i_{j_{v}}}|=|s_{0}|$ . If not, $|s_{i_{j_{v}}}|\neq|s_{0}|$ . The fact that $s_{i_{j_{v}}}\sqsubset s_{0}$ deduces $|s_{i_{j_{v}}}|>|s_{0}|$ , then $|v.s|>|s_{i_{j_{v}}}|>|s_{0}|$ , that is, $|v.s|>|s_{0}|+1$ , a contradiction. Hence, $|s_{i_{j_{v}}}|=|s_{0}|$ , so $|v.s|>|s_{0}|$ . This implies that $D$ is uncountable because $s_{0}=\sup_{v\in D}v.s$ , which in turn implies that $s_{0}=s$ , and it means $v.s_{0}\sqsubset_{2}s_{i_{j_{v}}}$ or $v.s_{0}\sqsubset_{2};\sqsubset_{1}s_{i_{j_{v}}}$ . Then $s_{0}\sqsubseteq s_{i_{j_{v}}}$ . But this is a contradiction to the fact that $s_{i_{j_{v}}}\sqsubset s_{0}$ . Therefore, Claim $1$ holds.

$\mathbf{Claim~{}2}$ : There exists $j_{2}\in J$ such that $\min K_{j_{2}}\subseteq L_{s_{0}}$ .

Assume that for any $j\in J$ , $\min K_{j}$ is not contained in $L_{s_{0}}$ . Then $\min K_{j}\bigcap\ (\downarrow\!L_{s_{0}}\!\setminus L_{s_{0}})\neq\emptyset$ from the fact that $\min K_{j}\subseteq\downarrow\!L_{s_{0}}$ for any $j\in J$ . We pick $(m_{j},s_{j})\in\min K_{j}\bigcap(\downarrow\!L_{s_{0}}\setminus\!L_{s_{0}})$ , then $s_{j}\sqsubset s_{0}$ for all $j\in J$ . This contradicts Claim $1$ .

$\mathbf{Claim~{}3}$ : $\min K_{j_{2}}$ is finite.

We proceed by contradiction. Suppose that $\min K_{j_{2}}$ is infinite. By Lemma 4.17, we know that $\min K_{j_{2}}$ is countable. Let $\min K_{j_{2}}=\{(m_{n},s_{0}):n\in\mathbb{N}\}$ , where $\mathbb{N}$ denotes all positive natural numbers. Then we can identify $m^{\prime}\in\mathbb{W}$ , which is an upper bound of $\{m_{n}:n\in\mathbb{N}\}$ .

For any $n\in\mathbb{N}$ , set

$B_{n}=\bigcup_{i\in\mathbb{N}\setminus\{1,2,...,n-1\}}\downarrow\!(m_{i},s_{0})\bigcup\downarrow\!(m^{\prime},s_{0})$ .

It is easy to check that $B_{n}$ is Scott closed for any $n\in\mathbb{N}$ . By the construction of $B_{n}$ , we have $\min K_{j_{2}}\cap B_{n}\neq\emptyset$ for each $n\in\mathbb{N}$ . Then the compactness of $\min K_{j_{2}}$ implies that $\bigcap_{n\in\mathbb{N}}B_{n}\bigcap\min K_{j_{2}}\neq\emptyset$ . Choose $(m_{n_{0}},s_{0})\in\bigcap_{n\in\mathbb{N}}B_{n}\bigcap\min K_{j_{2}}$ . It follows that $(m_{n_{0}},s_{0})\in B_{n_{0}+1}$ . Now the remaining arguments are similar to that in Lemma 4.15.

$\mathbf{Claim~{}4}$ : For any $j\geq j_{2},\min K_{j}\subseteq\min K_{j_{2}}$ .

By way of contradiction, assume that there is a $j^{\prime}\geq j_{2}$ such that $\min K_{j^{\prime}}\not\subseteq\min K_{j_{2}}$ , that is, there exists $(m,s)\in\min K_{j^{\prime}}\setminus\min K_{j_{2}}$ . It follows that there is $(m_{1},s_{1})\in\min K_{j_{2}}$ such that $(m,s)>(m_{1},s_{1})$ , which implies $s_{1}\sqsubset s$ . The fact that $\min K_{j_{2}}\subseteq L_{s_{0}}$ suggests that $s_{1}=s_{0}$ . It indicates that $s_{0}\sqsubset s$ , which contradicts $s\sqsubseteq s_{0}$ as $\min K_{j^{\prime}}$ is contained in $\downarrow\!L_{s_{0}}$ .

Now let $J^{\prime}=\{j\in J:j\geq j_{2}\}$ , which is a cofinal subset of $J$ , and so $\downarrow\!L_{s_{0}}$ is a minimal closed set intersects $K_{j}$ , for any $j\in J^{\prime}$ . Assume $\min K_{j_{2}}=\{(a_{i},s_{0}):i=1,2,...,n\}$ and set $F=\bigcup_{i\in\{1,2,...,n\}}\downarrow\!(a_{i},s_{0})$ . Then $F$ is Scott closed. Note that $F\cap K_{j}\neq\emptyset$ for any $j\in J^{\prime}$ from Cliam 4. But $F\subsetneqq\ \downarrow\!L_{s_{0}}$ . This violates the minimality of $\downarrow\!L_{s_{0}}$ . So our assumption that $\Sigma\mathcal{Z}$ is not well-filtered must have been wrong. ∎

4.5 $\mathbf{WF}$ is not $\Gamma$ -faithful

In this subsection, we will show that $\mathbf{WF}$ is not $\Gamma$ -faithful by using the well-filtered dcpo $\mathcal{Z}$ constructed above. It is trivial to check that $\Sigma\mathcal{Z}$ and $\Sigma(I\!R\!R(\mathcal{Z}))\triangleq\Sigma\hat{\mathcal{Z}}$ are not homeomorphism. Thus if we derive $\Gamma\mathcal{Z}\cong\Gamma\hat{\mathcal{Z}}$ and $\Sigma\hat{\mathcal{Z}}$ is well-filtered, then this suffices to show that the category $\mathbf{WF}$ is not $\Gamma$ -faithful.

Theorem 4.19.

$\Gamma\mathcal{Z}\cong\Gamma\hat{\mathcal{Z}}$ , that is, the family of Scott closed subsets of $\mathcal{Z}$ and $\hat{\mathcal{Z}}$ are isomorphic.

Proof.

One sees clearly that the Scott topology of $\mathcal{Z}$ is isomorphic to the lower Vietoris topology of $\hat{\mathcal{Z}}$ , which is the family $\tau=\{\lozenge U:U\in\sigma(\mathcal{Z})\}$ , where $\lozenge U=\{A\in\hat{\mathcal{Z}}:A\cap U\neq\emptyset\}$ . Note that $\hat{\mathcal{Z}}$ endowed with the lower Vietoris topology is well-filtered. So it remain to prove that the Scott topology of $\hat{\mathcal{Z}}$ and the lower Vietoris topology of $\hat{\mathcal{Z}}$ coincide. It is clear to see that each closed set in the lower Vietoris topology is Scott closed. Now we show the converse. To this end, for each $A\in\Gamma(\mathcal{Z})$ , we write $\Box A=\{B\in\hat{\mathcal{Z}}:B\subseteq A\}$ and choose $\mathcal{A}\in\Gamma(\hat{\mathcal{Z}})$ . Then it suffices to show that $\bigcup\mathcal{A}$ is Scott closed and $\mathcal{A}=\Box(\bigcup\mathcal{A})$ .

First, by Proposition 2.2 (4) in [6], we know that for the dcpo $\mathcal{Z}$ , $\bigcup\mathcal{A}$ is Scott closed. Next, we confirm that $\mathcal{A}=\Box(\bigcup\mathcal{A})$ . That $\mathcal{A}\subseteq\Box(\bigcup\mathcal{A})$ is trivial. Conversely, let $A\in\Box(\bigcup\mathcal{A})$ . Then $A\subseteq\bigcup\mathcal{A}$ . By Theorem 4.13, we know that $A\in\{\mathord{\downarrow}L_{s}:s\in\mathbb{W}^{*}\}\cup\{\mathord{\downarrow}(m,s):(m,s)\in\mathbb{W}\times\mathbb{W}^{*}\}$ . If $A$ is the form $\downarrow\!(m,s)$ for some $(m,s)\in\mathbb{W}\times\mathbb{W}^{*}$ , then $A\in\mathcal{A}$ evidently. Else, $A=\ \downarrow\!L_{s}$ for some $s\in\mathbb{W}^{*}$ . It follows that for each $m\in\mathbb{W}$ , there is $A_{m}\in\mathcal{A}$ such that $(m,s)\in A_{m}$ . It turns out that $\downarrow\!(m,s)\in\mathcal{A}$ since $\mathcal{A}$ is a lower set. Furthermore, we get that $\downarrow\!L_{m.s}\in\mathcal{A}$ for each $m\in\mathbb{W}$ as $\downarrow\!L_{m.s}\subseteq\ \downarrow\!(m,s)$ . Note that $\{\downarrow\!L_{m.s}:m\in\mathbb{W}\}$ is a directed family contained in $\mathcal{A}$ and $\mathcal{A}$ is Scott closed. Then $A=\downarrow\!L_{s}=\sup_{m\in\mathbb{W}}\downarrow\!L_{m.s}\in\mathcal{A}$ . So $\Box(\bigcup\mathcal{A})\subseteq\mathcal{A}$ . Indeed, we have proved that $\Gamma\mathcal{Z}\cong\Gamma\hat{\mathcal{Z}}$ . ∎

5 The category of weak dominated dcpo’s is $\Gamma$ -faithful

Finally, we give a $\Gamma$ -faithful category, that of weakly dominated dcpo’s, which is strictly larger than the category of dominated dcpo’s.

Definition 5.1.

[7] Let $P$ be a poset and $x,y\in P$ . We say that $x$ is beneath y, denoted by $x\prec^{*}y$ , if for every nonempty Scott closed set $C\subseteq P$ for which $\sup C$ exists, the relation $y\leq\sup C$ always implies that $x\in C$ . An element $x$ of a poset $P$ is called $C$ -compact if $x\prec^{*}x$ . If $P=\Gamma L$ for some poset $L$ , then the set of all $C$ -compact elements of $\Gamma L$ is denoted by $C(\Gamma L)$ .

The following proposition is a corollary of [7, Proposition 3.4].

Proposition 5.2.

$C(\Gamma L)$ is a subdcpo of $\Gamma L$ .

Corollary 5.3.

[7] The family of sets $\{\mathord{\downarrow}x:x\in L\}$ is a subset of $C(\Gamma L)$ .

The following results is crucial for further discussion.

Lemma 5.4.

Let $L$ be a poset. Then $\bigcup\mathcal{A}$ is a Scott closed subset of $L$ for each $\mathcal{A}\in\Gamma(C(\Gamma L))$ .

Proof.

One sees directly that $\bigcup\mathcal{A}$ is a lower set since each element in $\mathcal{A}$ is a lower set. Let $D$ be a directed subset of $\bigcup\mathcal{A}$ . Then there exists $A_{d}\in\mathcal{A}$ such that $d\in A_{d}$ for any $d\in D$ . It follows that $\mathord{\downarrow}d\subseteq A_{d}$ because $A_{d}$ is a lower set. From Corollary 5.3, we know that $\mathord{\downarrow}d\in C(\Gamma L)$ for any $d\in D$ . The fact that $\mathcal{A}$ is a lower set in $C(\Gamma L)$ implies that $\mathord{\downarrow}d\in\mathcal{A}$ . This means that $\{\mathord{\downarrow}d:d\in D\}$ is a directed subset of $\mathcal{A}$ , which yields that $\sup_{d\in D}\mathord{\downarrow}d=\mathord{\downarrow}\sup D\in\mathcal{A}$ . Hence, $\sup D\in\bigcup\mathcal{A}$ . ∎

Definition 5.5.

Let $L$ be a poset. A set $B$ is said to be a $C$ -compact set, if $cl(B)$ is a $C$ -compact element of $\Gamma L$ .

C-compact sets are preserved by Scott-continuous functions.

Lemma 5.6.

Let $L,P$ be two posets, if the function $f:L\rightarrow P$ is Scott continuous, then $f(A)$ is a $C$ -compact set of $P$ for any $A\in C(\Gamma L)$ .

Proof.

We prove that $cl(f(A))\prec^{*}cl(f(A))$ . To this end, let $\mathcal{A}\in\Gamma(\Gamma P)$ with $cl(f(A))\subseteq\bigcup\mathcal{A}$ . Then $A\subseteq\bigcup_{B\in\mathcal{A}}f^{-1}(B)$ . We write $\mathcal{B}=\mathord{\downarrow}\{f^{-1}(B):B\in\mathcal{A}\}$ .

We claim that $\mathcal{B}$ is a Scott closed set of $\Gamma L$ . Obviously, $\mathcal{B}$ is a lower set. Let $(A_{i})_{i\in I}$ be a directed subset of $\mathcal{B}$ . Then we know that there is $B_{i}\in\mathcal{A}$ with $A_{i}\subseteq f^{-1}(B_{i})$ for each $i\in I$ . It follows that $f(A_{i})\subseteq B_{i}$ , which means that $cl(f(A_{i}))\subseteq B_{i}\in\mathcal{A}$ for any $i\in I$ . The fact that $\mathcal{A}$ is a lower set in $\Gamma P$ tells us that $cl(f(A_{i}))\in\mathcal{A}$ for each $i\in I$ . Then we have $cl(f(A_{i}))_{i\in I}$ is a directed subset of $\mathcal{A}$ . As $\mathcal{A}$ is Scott closed, it follows that $\sup_{i\in I}cl(f(A_{i}))=cl(f(\bigcup_{i\in I}A_{i}))\in\mathcal{A}$ . Thus, $\sup_{i\in I}A_{i}=cl(\bigcup_{i\in I}A_{i})\in\mathord{\downarrow}\{f^{-1}(B)\mid B\in\mathcal{A}\}$ .

It is easy to see that $A\subseteq\sup\mathcal{B}=\bigcup\mathcal{B}$ . Then the assumption that $A$ is a $C$ -compact element of $\Gamma L$ guarantees the existence of $B\in\mathcal{A}$ such that $A\subseteq f^{-1}(B)$ . So we conclude $cl(f(A))\subseteq B\in\mathcal{A}$ . Again, as $\mathcal{A}$ is Scott closed, hence a lower set, we obtain that $cl(f(A))\in\mathcal{A}$ . As a result, $f(A)$ is a $C$ -compact set of $P$ . ∎

Lemma 5.7.

Let $L$ be a poset and $\mathcal{B}\in\Gamma(C(\Gamma L))$ . If $\mathcal{B}$ is closed under suprema of $C$ -compact sets, then $\mathcal{B}=\Box\bigcup\mathcal{B}=\{A\in C(\Gamma L):A\subseteq\bigcup\mathcal{B}\}$ .

Proof.

That $\mathcal{B}\subseteq\Box\bigcup\mathcal{B}$ is trivial. Conversely, for any $A\in\Box\bigcup\mathcal{B}$ , similar to the proof of Lemma 5.4, we know that $\{\mathord{\downarrow}x:x\in A\}$ is a subset of $\bigcup\mathcal{B}$ . Then the function $\eta=(x\mapsto\mathord{\downarrow}x):L\rightarrow C(\Gamma L)$ is well-defined and Scott continuous. Then $\eta(A)=\{\mathord{\downarrow}x:x\in A\}$ is a $C$ -compact set of $C(\Gamma L)$ by Lemma 5.6. The assumption that $\mathcal{B}$ is closed under suprema of $C$ -compact sets ensures that $A=\sup_{x\in A}\mathord{\downarrow}x=\sup\eta(A)\in\mathcal{B}$ . ∎

By applying Lemma 5.4 and Lemma 5.7, we have the following lemma.

Lemma 5.8.

For arbitrary dcpo’s D and E, $C(\Gamma D)\cong C(\Gamma E)$ if and only if $\Gamma D\cong\Gamma E$ .

Proof.

Assume that $g\colon C(\Gamma D)\to C(\Gamma E)$ is an isomorphism. Then the map $g^{\prime}\colon\Gamma D\to\Gamma E$ that sends a Scott closed subset $A$ of $D$ to $\bigcup\{g(K)\mid K\in C(\Gamma D)~{}\text{and}~{}K\subseteq A\}$ witnesses an isomorphism between $\Gamma D$ and $\Gamma E$ . ∎

Definition 5.9.

Given $A,B\in C(\Gamma P)$ , we write $A\lhd B$ if there is $x\in B$ such that $A\subseteq\mathord{\downarrow}x$ . We write $\nabla B$ for the set $\{A\in C(\Gamma P):A\lhd B\}$ .

Lemma 5.10.

Let $L$ be a dcpo and $A\in C(\Gamma L)$ .

$(1)$ $A=\sup\nabla A$ .

$(2)$ $\nabla A$ is a $C$ -compact set of $C(\Gamma L)$ .

Proof.

The first statement is trivial because $\mathord{\downarrow}x\lhd A$ for any $x\in A$ .

For the second statement, it suffices to prove that $cl(\nabla A)\prec^{*}cl(\nabla A)$ . To this end, let $\mathcal{A}^{2}\in\Gamma(\Gamma(C(\Gamma L)))$ with $cl(\nabla A)\subseteq\bigcup\mathcal{A}^{2}$ .

$\mathbf{Claim~{}1}$ : $A\subseteq\bigcup_{\mathcal{A}\in\mathcal{A}^{2}}\bigcup\mathcal{A}$ .

For any $x\in A$ , $\mathord{\downarrow}x\in\nabla A\subseteq\bigcup\mathcal{A}^{2}$ . Then there is $\mathcal{A}_{x}\in\mathcal{A}^{2}$ such that $\mathord{\downarrow}x\in\mathcal{A}_{x}$ for each $x\in A$ . It follows that $x\in\bigcup\mathcal{A}_{x}$ , and we have $A\subseteq\bigcup_{\mathcal{A}\in\mathcal{A}^{2}}\bigcup\mathcal{A}$ .

$\mathbf{Claim~{}2}$ : $\mathcal{B}=\mathord{\downarrow}\{\bigcup\mathcal{A}\mid\mathcal{A}\in\mathcal{A}^{2}\}\in\Gamma(\Gamma L)$ .

It is obvious that $\mathcal{B}$ is a lower set. Let $(B_{i})_{i\in I}$ be a directed subset of $\mathcal{B}$ . Then there exists $\mathcal{A}_{i}\in\mathcal{A}^{2}$ with $B_{i}\subseteq\bigcup\mathcal{A}_{i}$ for every $i\in I$ . By Corollary 5.3, we know that $\{\mathord{\downarrow}x\mid x\in B_{i}\}\subseteq\mathcal{A}_{i}$ for every $i\in I$ . It follows that the Scott closure $\mathcal{B}_{i}=cl(\{\mathord{\downarrow}x\mid x\in B_{i}\})$ , taken inside $C(\Gamma L)$ , is contained in $\mathcal{A}_{i}$ . Then $\mathcal{B}_{i}\in\mathcal{A}^{2}$ as $\mathcal{A}^{2}$ is a lower set. This deduces that $\sup_{i\in I}\mathcal{B}_{i}=cl(\bigcup_{i\in I}\mathcal{B}_{i})\in\mathcal{A}^{2}$ since $(\mathcal{B}_{i})_{i\in I}$ is a directed subset of $\mathcal{A}^{2}$ . It turns out that $cl(\bigcup_{i\in I}B_{i})\subseteq\bigcup cl(\bigcup_{i\in I}\mathcal{B}_{i})$ . Therefore, $\sup_{i\in I}B_{i}=cl(\bigcup_{i\in I}B_{i})\in\mathcal{B}$ . In a conclusion, $\mathcal{B}$ is Scott closed in $\Gamma L$ .

Since $A$ is a $C$ -compact element of $\Gamma L$ and $A\subseteq\bigcup\mathcal{B}$ , that $A\in\mathcal{B}$ holds. This means that there exists $\mathcal{A}\in\mathcal{A}^{2}$ such that $A\subseteq\bigcup\mathcal{A}$ . It turns out that $\{\mathord{\downarrow}x\mid x\in A\}\subseteq\mathcal{A}$ , which yields that $\nabla A\subseteq\mathcal{A}$ . So $cl(\nabla A)\in\mathcal{A}^{2}$ . ∎

Definition 5.11.

A dcpo $L$ is called $C$ -compactly complete if every $C$ -compact subset of $L$ has a supremum.

Proposition 5.12.

Let $L$ be a dcpo. Then $(C(\Gamma L),\subseteq)$ is $C$ -compactly complete. Especially, $\sup\mathcal{A}=\bigcup\mathcal{A}$ for any $\mathcal{A}\in C(\Gamma(C(\Gamma L)))$ .

Proof.

It suffices to prove that $\bigcup\mathcal{A}\in C(\Gamma L)$ . To this end, let $\mathcal{B}\in\Gamma(\Gamma L)$ with $\bigcup\mathcal{A}\subseteq\bigcup\mathcal{B}$ . Then $A\subseteq\bigcup\mathcal{B}$ for any $A\in\mathcal{A}$ . The fact that $A$ is a $C$ -compact element of $\Gamma L$ implies that $A\in\mathcal{B}$ . This means that $\mathcal{A}\subseteq\bigcup_{B\in\mathcal{B}}\Box B$ , where $\Box B=\{D\in C(\Gamma L)\mid D\subseteq B\}$ . Ones sees obviously that $\Box B$ is a Scott closed set of $C(\Gamma L)$ .

Now we claim that $\mathcal{B}^{2}=\mathord{\downarrow}\{\Box B\mid B\in\mathcal{B}\}$ is a Scott closed set of $\Gamma(C(\Gamma L))$ . Clearly, $\mathcal{B}^{2}$ is a lower set. Let $(\mathcal{A}_{i})_{i\in I}$ be a directed subset of $\mathcal{B}^{2}$ . Then there exists $B_{i}\in\mathcal{B}$ with $\mathcal{A}_{i}\subseteq\Box B_{i}$ for any $i\in I$ . This means that $\bigcup\mathcal{A}_{i}\subseteq B_{i}$ for any $i\in{I}$ . By Lemma 5.4, we know that the set $\bigcup\mathcal{A}_{i}$ is Scott closed for $i\in I$ , which leads to that $\bigcup\mathcal{A}_{i}\in\mathcal{B}$ , as $\mathcal{B}$ is Scott closed. It follows that $(\bigcup\mathcal{A}_{i})_{i\in I}$ is a directed subset of $\mathcal{B}$ . So $\sup_{i\in I}\bigcup\mathcal{A}_{i}=cl(\bigcup_{i\in I}\bigcup\mathcal{A}_{i})\in\mathcal{B}$ . This implies that $\sup_{i\in I}\mathcal{A}_{i}\subseteq\Box(cl(\bigcup_{i\in I}\bigcup\mathcal{A}_{i}))$ . So, $\sup_{i\in I}\mathcal{A}_{i}\in\mathcal{B}^{2}$ .

Note that $\mathcal{A}\subseteq\bigcup\mathcal{B}^{2}$ . Then $\mathcal{A}\in\mathcal{B}^{2}$ from the fact that $\mathcal{A}\in C(\Gamma(C(\Gamma L)))$ . So there exists $B\in\mathcal{B}$ with $\mathcal{A}\subseteq\Box B$ , and then $\bigcup\mathcal{A}\subseteq B$ . Thus $\bigcup\mathcal{A}\in\mathcal{B}$ by Scott closedness of $\mathcal{B}$ . ∎

Finally, we introduce the category of all weakly dominated dcpo’s which serves as a $\Gamma$ -faithful category.

Definition 5.13.

A dcpo $L$ is called weakly dominated if for every $A\in C(\Gamma L)$ , the collection $\nabla A$ is Scott closed in $C(\Gamma L)$ .

Definition 5.14.

Let $L$ be a $C$ -compactly complete dcpo and $x,y\in L$ . We write $x\prec y$ if for all closed $C$ -compact subsets $A$ , $y\leq\sup A$ implies that $x\in A$ . We say that $x\in L$ is $\prec$ -compact if $x\prec x$ , and denote the set of all $\prec$ -compact elements by $K(L)$ .

Proposition 5.15.

$A\lhd B$ implies that $A\prec B$ for any $A,B\in C(\Gamma L)$ .

Proof.

One sees immediately by Proposition 5.12. ∎

Lemma 5.16.

A dcpo $L$ is weakly dominated, if and only if $B\prec A$ implies $B\lhd A$ for all $A,B\in C(\Gamma L)$ .

Proof.

If $L$ is weakly dominated, then $\nabla A\in C(\Gamma(C\Gamma(L)))$ by Lemma 5.10. Assume $B\prec A$ , then $B\prec A\subseteq\bigcup\nabla A$ . This means that $B\in\nabla A$ .

Conversely, if $B\prec A$ implies $B\lhd A$ for all $A,B\in C(\Gamma L)$ . We need to check that $\nabla A$ is Scott closed in $C(\Gamma L)$ . The set $\nabla A$ is obviously a lower set. Let $(B_{i})_{i\in I}$ be a directed subset of $\nabla A$ . Then it remains to show that $\sup_{i\in I}B_{i}\prec A$ by the assumption that $B\prec A$ implies $B\lhd A$ . For any $\mathcal{A}\in C(\Gamma(C(\Gamma L)))$ , if $A\subseteq\bigcup\mathcal{A}$ , then hat $B_{i}\prec A$ implies that $B_{i}\in\mathcal{A}$ . Note that $\mathcal{A}$ is a Scott closed set of $C(\Gamma L)$ and $(B_{i})_{i\in I}$ is a directed subset of $\mathcal{A}$ . We conclude that $\sup_{i\in I}B_{i}=cl(\bigcup_{i\in I}B_{i})\in\mathcal{A}$ . ∎

By observing the above results, we can get the following corollaries.

Proposition 5.17.

For a weakly dominated dcpo $L$ , the only $\prec$ -compact elements of $C(\Gamma L)$ are exactly the principal ideals $\mathord{\downarrow}x$ , $x\in L$ .

Theorem 5.18.

Let $L$ and $P$ be weakly dominated dcpo’s. The following are equivalent:

1.

$L\cong P$ ;
2.

$\Gamma L\cong\Gamma P$ ;
3.

$C(\Gamma L)\cong C(\Gamma P)$ . $\Box$

The following theorem follows immediately.

Theorem 5.19.

The category of all weakly dominated dcpo’s is $\Gamma$ -faithful. $\Box$

6 Weakly dominated dcpo’s may fail to be dominated

From the definitions of dominated dcpo’s and weakly dominated dcpo’s and the fact that $C(\Gamma L)$ is contained in $IRR(L)$ for each dcpo $L$ , we could easily see that dominated dcpo’s are weakly dominated. We show in this section, by giving a concrete example, that weak dominated dcpo’s form a strictly larger class than that of dominated ones. As we will see, our example is given in Example 6.4, which relies on two steps of preliminary constructions that are based on Isbell’s complete lattice and Johnstone’s dcpo.

Example 6.1.

Let $\mathbb{N}^{\mathbb{N}}$ denote the maps from $\mathbb{N}$ to $\mathbb{N}$ , $S=\mathbb{N}\times\mathbb{N}$ , $T=\mathbb{N}^{\mathbb{N}}\times\mathbb{N}$ , $L=S\cup T\cup\{\top\}$ , where $\mathbb{N}$ is the set of positive natural numbers. An order $\leq$ on $L$ is defined as follows:

$(m_{1},n_{1})\leq(m_{2},n_{2})$ if and only if:

•

$(m_{2},n_{2})=\top$ , $(m_{1},n_{1})\in L$ ;
•

$m_{1}=m_{2}$ , $n_{1}\leq n_{2}$ .

Next, let $P=L\times\mathbb{R}$ , where $\mathbb{R}$ is the set of real numbers. Then there exists an injection $i:\mathbb{R}\times\mathbb{R}\times\mathbb{N}^{\mathbb{N}}\times\mathbb{N}\rightarrow\mathbb{R}$ such that $i(s,t,f,k)>s,t$ for any $(s,t,f,k)\in\mathbb{R}\times\mathbb{R}\times\mathbb{N}^{\mathbb{N}}\times\mathbb{N}$ with $t>s$ . We write $((n,j),k)$ by $(n,j,k)$ for any $((n,j),k)\in P$ and $\{(n,i,r):(n,i)\in L\}$ by $L_{r}$ for any $r\in\mathbb{R}$ . An order $\leq$ on $P$ is defined as: $(n_{1},i_{1},j_{1})\leq(n_{2},i_{2},j_{2})$ if and only if:

•

$j_{1}=j_{2}$ , $(n_{1},i_{1})\leq(n_{2},i_{2})$ in $L_{j_{1}}$ ;
•

$(n_{2},i_{2})=\top$ . If $(n_{1},i_{1})\in S$ , then there exists $t_{1}\in\mathbb{R}$ , $f\in\mathbb{N}^{\mathbb{N}},k\in\mathbb{N}$ such that $j_{2}=i(j_{1},t_{1},f,k)$ , $t_{1}>j_{1}$ and $(n_{1},i_{1})=(k,f(k))$ . Else, $(n_{1},i_{1})\in T$ , then there exists $t_{2}\in\mathbb{R}$ such that $j_{2}=i(t_{2},j_{1},n_{1},i_{1})$ and $j_{1}>t_{2}$ .

Lemma 6.2.

The poset $P$ is a dcpo with the fact that there are only finitely many minimal upper bounds for each pair of elements in $P$ .

Proof.

The proof is similar to [11, Fact 4.1]. ∎

Example 6.3.

Let $\mathbb{J}_{\infty}=\mathbb{N}\times(\mathbb{N}\times(\mathbb{N}\cup\{\omega\}))$ , where $\mathbb{N}$ is the set of all positive natural numbers. We write $(k,(n,j))$ by $(k,n,j)$ for any $(k,(n,j))\in\mathbb{J}_{\infty}$ and $\{(k,n,i):(n,i)\in\mathbb{N}\times(\mathbb{N}\cup\{\omega\})\}$ by $\mathbb{J}_{k}$ for any $k\in\mathbb{N}$ . An order $\leq$ on $\mathbb{J}_{\infty}$ is defined as: $(k,m,n)\leq(c,a,b)$ if and only if:

•

$k=c$ , $m=a$ , $n\leq b$ ;
•

$k\leq c$ , $n\leq a$ , $b=\omega$ .

Then $C(\Gamma\mathbb{J}_{\infty})=\{\mathord{\downarrow}x:x\in\mathbb{J}_{\infty}\}\cup\{\bigcup_{i=1}^{n}\mathbb{J}_{i}:n\in\mathbb{N}\}\cup\{\mathbb{J}_{\infty}\}$ (We depict $\mathbb{J}_{\infty}$ as in Figure $3$ ).

Proof.

$\mathbf{Claim~{}1}$ : rhs $\subseteq$ lhs.

By Corollary 5.3, we know that $\{\mathord{\downarrow}x:x\in\mathbb{J}_{\infty}\}$ is contained in $C(\Gamma\mathbb{J}_{\infty})$ . For any $n\in\mathbb{N}$ , $\bigcup_{i=1}^{n}\mathbb{J}_{i}$ is Scott closed. It remains to prove that $\bigcup_{i=1}^{n}\mathbb{J}_{i}$ and $\mathbb{J}_{\infty}$ are $C$ -compact elements of $\Gamma\mathbb{J}_{\infty}$ . To this end, let $\mathcal{A}\in\Gamma(\Gamma\mathbb{J}_{\infty})$ with $\bigcup_{i=1}^{n}\mathbb{J}_{i}\subseteq\bigcup\mathcal{A}$ . For any $(n,m,\omega)\in\mathbb{J}_{n}$ , the result $\mathord{\downarrow}(n,m,\omega)\in\mathcal{A}$ follows obviously. Observing the order on $\mathbb{J}_{\infty}$ , we get that the Scott closed set $A_{m}=\{(c,a,b)\in\bigcup_{i=1}^{n}\mathbb{J}_{i}:c\leq n,a\in\mathbb{N},b\leq m\}\subseteq\mathord{\downarrow}(n,m,\omega)\in\mathcal{A}$ . Then $A_{m}$ belongs to $\mathcal{A}$ for all $m\in\mathbb{N}$ . Note that $(A_{m})_{m\in\mathbb{N}}$ is a directed subset of $\mathcal{A}$ . Then $\sup_{m\in\mathbb{N}}A_{m}=\bigcup_{i=1}^{n}\mathbb{J}_{i}\in\mathcal{A}$ as $\mathcal{A}$ is Scott closed. Therefore, $\bigcup_{i=1}^{n}\mathbb{J}_{i}$ is a $C$ -compact element of $\Gamma\mathbb{J}_{\infty}$ . The proof of that $\mathbb{J}_{\infty}\in C(\Gamma\mathbb{J}_{\infty})$ runs similarly.

$\mathbf{Cliam~{}2}$ : lhs $\subseteq$ rhs.

Let $A\in C(\Gamma\mathbb{J}_{\infty})$ . For the sake of a contradiction we assume that $A\notin\{\mathord{\downarrow}x:x\in\mathbb{J}_{\infty}\}\cup\{\bigcup_{i=1}^{n}\mathbb{J}_{i}:n\in\mathbb{N}\}\cup\{\mathbb{J}_{\infty}\}$ .

$\mathbf{Claim~{}2.1}$ : $\max A\subseteq\max\mathbb{J}_{\infty}$ .

Assume that there is $a\in\max A\backslash\max\mathbb{J}_{\infty}$ . Then $A=\mathord{\downarrow}\max A\subseteq\mathord{\downarrow}a\cup\mathord{\downarrow}(\max A\backslash\{a\})$ . One sees immediately that $\mathord{\downarrow}\max A\backslash\{a\}$ is Scott closed from the order of $\mathbb{J}_{\infty}$ . Because $C(\Gamma\mathbb{J}_{\infty})$ is contained in $IRR(\mathbb{J}_{\infty})$ , $A$ is irreducible. It follows that $A=\mathord{\downarrow}a$ or $A=\mathord{\downarrow}(\max A\backslash\{a\})$ . The assumption that $A\notin\{\mathord{\downarrow}x:x\in\mathbb{J}_{\infty}\}$ implies that $A=\mathord{\downarrow}(\max A\backslash\{a\})$ . This means that $a\in\max A\subseteq\mathord{\downarrow}(\max A\backslash\{a\})$ , which yields that $a\in\max A\backslash\{a\}$ . A contradiction.
$\mathbf{Claim~{}2.2}$ : Define $A_{\mathbb{N}}=\{n\in\mathbb{N}:A\cap\mathbb{J}_{n}\neq\emptyset\}$ . Then $A_{\mathbb{N}}$ is infinite.

Suppose $A_{\mathbb{N}}$ is finite. Let $n_{0}=\max A_{\mathbb{N}}$ . According to $A\notin\{\bigcup_{i=1}^{n}\mathbb{J}_{i}:n\in\mathbb{N}\}$ , we have that $\max A\cap\mathbb{J}_{n_{0}}$ is finite. Then $A\subseteq\mathord{\downarrow}(\max A\cap\mathbb{J}_{{}_{n_{0}}})\cup\bigcup_{i=1}^{n_{0}-1}\mathbb{J}_{i}$ . The set $\mathord{\downarrow}(\max A\cap\mathbb{J}_{n_{0}})$ is Scott closed as $\max A\cap\mathbb{J}_{n_{0}}$ is finite. It is easy to see that $\bigcup_{i=1}^{n_{0}-1}\mathbb{J}_{i}$ is Scott closed. From the irreducibility of $A$ and the fact that $n_{0}$ belongs to $A_{\mathbb{N}}$ , we know that $A=\mathord{\downarrow}(\max A\cap\mathbb{J}_{n_{0}})=\bigcup_{a\in\max A\cap\mathbb{J}_{n_{0}}}\mathord{\downarrow}a$ . This indicates that there exists $a\in\max A\cap\mathbb{J}_{n_{0}}$ such that $A=\mathord{\downarrow}a$ , which is a contradiction. Hence, $A_{\mathbb{N}}$ is infinite.

The assumption that $A\neq\mathbb{J}_{\infty}$ implies that there exists a minimum natural number $k_{0}\in\mathbb{N}$ such that $\max A\cap\mathbb{J}_{k}$ is finite for any $k\geq k_{0}$ . If not, for any $n\in\mathbb{N}$ , there exists $k_{n}\geq n$ such that $\max A\cap\mathbb{J}_{k_{n}}$ is infinite. As $A$ is Scott closed, we have that $\mathbb{J}_{n}\subseteq\bigcup_{i\leq k_{n}}\mathbb{J}_{i}\subseteq A$ . This yields $\mathbb{J}_{\infty}=A$ , a contradiction. We write $B=\{m\in\mathbb{N}:\exists(n,m,\omega)\in A,n\geq k_{0}\}$ , and distinguish the following two cases for $B$ .

Case $1$ : $B$ is finite. Then $A\subseteq\mathord{\downarrow}(\max A\cap\mathbb{J}_{k_{0}})\cup\mathord{\downarrow}(\max A\backslash\mathbb{J}_{k_{0}})$ . Since $\max A\cap\mathbb{J}_{k}$ is finite for any $k\geq k_{0}$ , we have $\mathord{\downarrow}(\max A\cap\mathbb{J}_{k_{0}})$ is Scott closed. The finiteness of $B$ ensures that $\mathord{\downarrow}(\max A\backslash\mathbb{J}_{k_{0}})$ is Scott closed. Similar to the proof of Claim 2.2, we know that $A=\mathord{\downarrow}x$ for some $x\in\mathbb{J}_{\infty}$ . A contradiction.

Case $2$ : $B$ is infinite. Note that $\max A\cap\mathbb{J}_{i}$ is infinite for any $i\leq k_{0}-1$ . Then $\bigcup_{i=1}^{k_{0}-1}\mathbb{J}_{i}\subseteq A$ as $A$ is Scott closed. It follows that for any $(c,a,b)\in\mathbb{J}_{\infty}\backslash\max\mathbb{J}_{\infty}$ with $c\geq k_{0}$ , the set $C=\{m\in B:m\geq b\}$ is infinite from the infiniteness of $B$ . Define $E=\{k\geq k_{0}:\exists(k,m,\omega)\in A,m\in C\}$ . If E is finite, then there is $k\in E$ such that $\max A\cap\mathbb{J}_{k}$ is infinite. It follows that $k_{0}\geq k+1$ as $k_{0}$ is the minimum natural number with the property that $\max A\cap\mathbb{J}_{k}$ is finite for any $k\geq k_{0}$ . This contradicts that $k\in E$ . Else, $E$ is infinite. Then we could find $k\in E$ with $k\geq c$ . This guarantees the existence of $(k,m,\omega)\in A$ such that $(c,a,b)\leq(k,m,\omega)$ , which yields that $(c,a,b)\in A$ for any $(c,a,b)\in\mathbb{J}_{\infty}\backslash\max\mathbb{J}_{\infty}$ with $c\geq k_{0}$ . Then we would know that $A=\mathbb{J}_{\infty}$ , again from the Scott closedness of $A$ . But that violates the assumption that $A\neq\mathbb{J}_{\infty}$ . In summary, we have that Claim 2 holds.

In a conclusion, $C(\Gamma\mathbb{J}_{\infty})=\{\mathord{\downarrow}x:x\in\mathbb{J}_{\infty}\}\cup\{\bigcup_{i=1}^{n}\mathbb{J}_{i}:n\in\mathbb{N}\}\cup\{\mathbb{J}_{\infty}\}$ . ∎

Figure 3: The order between different blocks on $\mathbb{J}_{\infty}$ .

Now we construct a weakly dominated dcpo $M$ based the above dcpo’s $P$ and $\mathbb{J}_{\infty}$ , which fails to be dominated.

Example 6.4.

Let $M=P\cup\mathbb{J}_{\infty}$ . Fix $\{a_{\mathbb{N}}:n\in\mathbb{N}\}\subseteq\mathbb{R}\backslash i(\mathbb{R}\times\mathbb{R}\times\mathbb{N}^{\mathbb{N}}\times\mathbb{N})$ , where the function $i$ is defined as in Example 6.1. An order $\leq$ on $M$ is defined as $x\leq y$ if and only if:

•

$x\leq y$ in $P$ ;
•

$x\leq y$ in $\mathbb{J}_{\infty}$ ;
•

$y=(a_{n},\top)\in P$ , $x=(c,a,b)\in\mathbb{J}_{\infty}$ , $c\leq n$ .

Then $(M,\leq)$ is weak dominated, but not dominated.

Proof.

$\mathbf{Claim~{}1}$ : $M$ is not dominated.

Similar to the argument of [11, Remark 4.1], $M$ is an irreducible closed set. Set $\nabla^{*}M=\{B\in IRR(M):\exists x\in M,s.t.~{}B\subseteq\mathord{\downarrow}x\}$ . Notice that $\{\bigcup_{i=1}^{n}\mathbb{J}_{i}:n\in\mathbb{N}\}$ is a directed subset of $\nabla^{*}M$ , but $\sup_{n\in\mathbb{N}}\bigcup_{i=1}^{n}\mathbb{J}_{i}=\mathbb{J}_{\infty}\notin\nabla^{*}M$ . Hence, by definition, $M$ is not dominated.

$\mathbf{Claim~{}2}$ : $C(\Gamma M)=\{\mathord{\downarrow}x:x\in M\}\cup\{\bigcup_{i=1}^{n}\mathbb{J}_{i}:n\in\mathbb{N}\}\cup\{\mathbb{J}_{\infty}\}$ .

The right side of the equation is obviously contained in the left side by Example 6.3. Conversely, let $A\in C(\Gamma M)$ . By way of contradiction we assume that $A\notin\{\mathord{\downarrow}x:x\in M\}\cup\{\bigcup_{i=1}^{n}\mathbb{J}_{i}:n\in\mathbb{N}\}\cup\{\mathbb{J}_{\infty}\}$ .

We distinguish the cases whether $A$ is contained in $\mathbb{J}_{\infty}$ :

Case $1$ : $A\subseteq\mathbb{J}_{\infty}$ . In this case we claim that $A$ is a $C$ -compact element of $\Gamma\mathbb{J}_{\infty}$ . To this end, let $\mathcal{A}\in\Gamma(\Gamma\mathbb{J}_{\infty})$ with $A\subseteq\bigcup\mathcal{A}$ . Then the fact that $\mathbb{J}_{\infty}$ is a Scott closed set of $M$ indicates that $\mathcal{A}$ is a Scott closed subset of $\Gamma M$ . It follows that $A\in\mathcal{A}$ as $A\in C(\Gamma M)$ . So we have $A$ is a $C$ -compact element of $\Gamma\mathbb{J}_{\infty}$ , that is, $A\in C(\Gamma\mathbb{J}_{\infty})=\{\mathord{\downarrow}x:x\in\mathbb{J_{\infty}}\}\cup\{\bigcup_{i=1}^{n}\mathbb{J}_{i}:n\in\mathbb{N}\}\cup\{\mathbb{J}_{\infty}\}$ . That is a contradiction.

Case $2$ : $A\not\subseteq\mathbb{J}_{\infty}$ . Then $A\cap P\neq\emptyset$ . We claim that $A=\mathord{\downarrow}(A\cap P)$ . Note that $A\subseteq\mathord{\downarrow}(A\cap P)\cup\mathbb{J_{\infty}}$ . By the definition of $M$ , one sees immediately that $\mathord{\downarrow}(A\cap P)$ is a Scott closed set of $M$ . Since $A$ is a $C$ -compact element of $\Gamma(M)$ , we know that $A$ is irreducible. This means that $A\subseteq\mathord{\downarrow}(A\cap P)$ or $A\subseteq\mathbb{J}_{\infty}$ . The assumption that $A\not\subseteq\mathbb{J}_{\infty}$ implies that $A\subseteq\mathord{\downarrow}(A\cap P)$ . In other words, $A=\mathord{\downarrow}(A\cap P)$ . Similar to the proof of Example 6.3, we have $\max A\subseteq\max P$ . Define $\mathcal{A}=\mathord{\downarrow}(\{\mathord{\downarrow}x:x\in\max A\})\cup\mathord{\downarrow}\mathbb{J}_{\infty}$ . It is obvious that $\mathcal{A}$ is a lower set of $\Gamma M$ . We want to show that $\mathcal{A}$ is Scott closed. To this end, let $(A_{i})_{i\in I}$ be a non-trivial chain of $\mathcal{A}$ . Now we distinguish the following two cases for $(A_{i})_{i\in I}$ .

Case $2.1$ : $(A_{i})_{i\in I}\subseteq\mathord{\downarrow}\mathbb{J_{\infty}}$ . Then the fact that $\mathbb{J}_{\infty}$ is Scott closed concludes that $\sup_{i\in I}A_{i}=cl(\bigcup_{i\in I}A_{i})\in\mathord{\downarrow}\mathbb{J_{\infty}}\subseteq\mathcal{A}$ .

Case $2.2$ : $(A_{i})_{i\in I}\not\subseteq\mathord{\downarrow}\mathbb{J_{\infty}}$ . Then there exists $i_{0}\in I$ such that $A_{i}\cap P\neq\emptyset$ for any $i\geq i_{0}$ . This means that there exists $b_{i}\in\mathbb{R}$ such that $A_{i}\subseteq\mathord{\downarrow}(b_{i},\top)$ for all $i\geq i_{0}$ . Now we need to further distinguish the following two cases.

Case $2.2.1$ : $A_{i}\subseteq P$ for any $i\geq i_{0}$ . If $\{b_{i}\mid i\geq i_{0}\}$ is a finite set, then there exists $b_{i}$ such that $A_{i}\subseteq\mathord{\downarrow}(b_{i},\top)$ for any $i\in I$ . It follows that $\sup_{i\in I}A_{i}\in\mathord{\downarrow}(\mathord{\downarrow}(b_{i},\top))\subseteq\mathcal{A}$ . Otherwise, $\{b_{i}\mid i\geq i_{0}\}$ is an infinite set. Then $A_{i}\subseteq\bigcap_{j\geq i}\mathord{\downarrow}(b_{j},\top)$ for any $i\geq i_{0}$ . In light of Example 6.1, we have that $P$ is a dcpo with the condition that there are only finitely many minimal upper bounds of each pair of elements in $P$ , which yields that for any $i\geq i_{0}$ , $A_{i}$ must be the form of $\{(k,m,n):n\leq n_{i}\}$ , for some fixed $n_{i}\in\mathbb{N},m\in\mathbb{N}\cup\mathbb{N}^{\mathbb{N}}$ and $k\in\mathbb{R}$ , and this reveals that $\{(k,m,n):n\in\mathbb{N}\}$ is a directed subset of $A$ since $A$ is a lower set. Scott closedness of $A$ then implies that $(k,m,\omega)\in A$ . Hence, $\mathord{\downarrow}(k,m,\omega)=cl(\bigcup_{i\in I}A_{i})=\sup_{i\in I}A_{i}\in\mathcal{A}$ .

Case $2.2.2$ : there exists $i_{1}\geq i_{0}$ such that $A_{i_{1}}\cap\mathbb{J}_{\infty}\neq\emptyset$ . Then $b_{i}\in\mathbb{R}\backslash i(\mathbb{R}\times\mathbb{R}\times\mathbb{N}^{\mathbb{N}}\times\mathbb{N})$ for any $i\geq i_{1}$ . For any $i,j\geq i_{1}$ with $j\geq i$ , we know $\emptyset\neq A_{i}\cap P\subseteq A_{j}\cap P$ . This implies that $b_{i}=b_{j}$ by the order defined on $M$ , which yields that $b_{i}=b_{0}$ for any $i\geq i_{1}$ . It turns out that $A_{i}\subseteq\mathord{\downarrow}(b_{0},\top)$ for any $i\in I$ . Thus, $\sup_{i\in I}A_{i}\subseteq\mathord{\downarrow}(b_{0},\top)\in\mathcal{A}$ .

In the both sub-cases, hence in the Case 2.2, $\mathcal{A}$ is Scott closed.

Now in the Case 2, we then know that $A\in\mathcal{A}$ since $A\in C(\Gamma M)$ . This gives that $A=\mathord{\downarrow}x$ for some $x\in A$ or $A\subseteq\mathbb{J}_{\infty}$ . A contradiction. Hence we finish proving Claim 2.

Now it is trivial to check that $\nabla A$ is Scott closed for $A\in C(\Gamma M)$ . So Indeed, $(M,\leq)$ is weak dominated, but not dominated. ∎

Acknowledgement

This work is supported by the National Natural Science Foundation of China (No.12401596, No.12231007 and No.12371457).

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The category of well-filtered dcpo’s is not Γ\Gamma-faithful

Abstract

1 Introduction

2 Preliminaries

3 Ω∗\Omega^{*}-compact dcpo’s are dominated

Definition 3.1.

Definition 3.2.

Lemma 3.3.

Proof.

Theorem 3.4.

4 The category of well-filtered dcpo’s is not Γ\Gamma-faithful

4.1 The definition of the counterexample 𝒵\mathcal{Z}

Remark 4.1.

Lemma 4.2.

Proposition 4.3.

Definition 4.4.

4.2 𝒵\mathcal{Z} is a dcpo

Proposition 4.5.

Proof.

4.3 The Scott closed irreducible subsets of 𝒵\mathcal{Z}

Lemma 4.6.

Lemma 4.7.

Proof.

Proposition 4.8.

Proof.

Remark 4.9.

Proposition 4.10.

Lemma 4.11.

Proof.

Lemma 4.12.

Proof.

Theorem 4.13.

Proof.

4.4 𝒵\mathcal{Z} is well-filtered

Lemma 4.14.

Proof.

Lemma 4.15.

Proof.

Lemma 4.16.

Proof.

Lemma 4.17.

Proof.

Theorem 4.18.

Proof.

4.5 𝐖𝐅\mathbf{WF} is not Γ\Gamma-faithful

Theorem 4.19.

Proof.

5 The category of weak dominated dcpo’s is Γ\Gamma-faithful

Definition 5.1.

Proposition 5.2.

Corollary 5.3.

Lemma 5.4.

Proof.

Definition 5.5.

Lemma 5.6.

Proof.

Lemma 5.7.

Proof.

Lemma 5.8.

Proof.

Definition 5.9.

Lemma 5.10.

Proof.

Definition 5.11.

Proposition 5.12.

Proof.

Definition 5.13.

Definition 5.14.

Proposition 5.15.

Proof.

Lemma 5.16.

Proof.

Proposition 5.17.

Theorem 5.18.

Theorem 5.19.

6 Weakly dominated dcpo’s may fail to be dominated

Example 6.1.

Lemma 6.2.

Proof.

Example 6.3.

The category of well-filtered dcpo’s is not $\Gamma$ -faithful

3 $\Omega^{*}$ -compact dcpo’s are dominated

4 The category of well-filtered dcpo’s is not $\Gamma$ -faithful

4.1 The definition of the counterexample $\mathcal{Z}$

4.2 $\mathcal{Z}$ is a dcpo

4.3 The Scott closed irreducible subsets of $\mathcal{Z}$

4.4 $\mathcal{Z}$ is well-filtered

4.5 $\mathbf{WF}$ is not $\Gamma$ -faithful

5 The category of weak dominated dcpo’s is $\Gamma$ -faithful