Johnstone’e Non-Sober Dcpo and Extensions

Abstract

One classic result in domain theory is that the Scott space of every domain (continuous directed complete poset) is sober. Johnstone constructed the first directed complete poset (dcpo for short) whose Scott space is not sober. This non-sober dcpo has been used in many other parts of domain theory and more properties of it have been uncovered. In this survey paper, we first collect and prove the major properties (some of which are new as far as we know) of Johnstone’s dcpo. We then propose a general method of constructing a dcpo from given posets and prove some properties. Some problems are posed for further investigation. This paper can serve as a relatively complete resource on Johnstone’s dcpo.

1. Introduction

Sobriety is one of the most important properties for non-Hausdorff topological spaces. It is a classic result in domain theory that the Scott space of every domain (continuous directed complete posets) is sober. Johnstone [1] constructed the first dcpo whose Scott space is not sober (this dcpo will be called Johnstone’s dcpo). This elegant example inspired many in-depth work on the properties of Scott spaces of posets. For example, Isbell [2] constructed the first complete lattice whose Scott space is not sober. Kou [3] gave the first dcpo whose Scott space is well filtered but not sober (see [4] for another more straightforwardly defined example of such a dcpo), thus answering the problem posed by Heckmann in [5]. Jia [6] found the first countable dcpo with a well-filtered and non-sober Scott space. In [7], Xu, Xi and Zhao constructed a complete Heyting algebra whose Scott space is not sober, thus answering the problem posed by Achim Jung. Applying Johnstone’s dcpo, Ho, Goubault-Larrecq, Jung and Xi [8] constructed a dcpo P whose Scott space is not sober and the sobrification of the Scott space of P is the Scott space of a dcpo Q, showing that there are two non-isomorphic dcpos such that their Scott closed set lattices are isomorphic. In [9], Miao, Li, Xi and Zhao constructed a countable complete lattice with a non-sober Scott space. Recently, Miao, Xi, Jia, Li and Zhao found two sober dcpos such that their order product is not sober [10]. In [11], Xi and Lawson proved that the Scott space of every complete lattice is well filtered; thus, the non-sober complete lattice constructed by Isbell was actually the first example of a well-filtered and non-sober dcpo.

In the current paper, we will collect the major properties of Johnstone’s dcpo, then suggest a general method of constructing a dcpo from the given posets. Some open problems related to Johnstone’s dcpo will be posed for further study.

2. Preliminaries

We quickly recall some basic concepts and results in domain theory that will be used later.

For any subset A of poset P, $\bigvee A$ ($\bigwedge A$, resp.) will denote the supremum (infimum, resp.) of A in P.

A nonempty subset D of a poset is directed if every two elements in D have an upper bound in D (for any $x,y\in D$, there is a $d\in D$ such that $x\le d$ and $y\le d$).

A subset U of a poset $(P,\le )$ is Scott open if (i) U is an upper set ($U=\uparrow U=\{x\in P:y\le x \mathrm{for} \mathrm{some} y\in U\}$), and (ii) for any directed subset $D\subseteq P$, $\bigvee D\in U$ implies $D\cap U\ne \varnothing$ whenever $\bigvee D$ exists. All Scott open sets of a poset P form a topology on P, denoted by $\sigma \left(P\right)$ and called the Scott topology on P. The space $(P,\sigma (P\left)\right)$ is denoted by $\Sigma P$, called the Scott space of P. The complements of Scott open sets are called Scott closed sets.

Remark 1.

(1) For any element x of poset P, the closure $cl\left(\right\{x\left\}\right)$ of $\left\{x\right\}$ with respect to the Scott topology equals

$$\downarrow x=\{y\in P:y\le x\}.$$

(2) 

A mapping $f:P⟶Q$ between two posets is continuous with respect to the Scott topologies on P and Q, respectively, if and only if for any directed subset $D\subseteq P$ with $\bigvee D$ existing, it holds that $f(\bigvee D)=\bigvee f\left(D\right)$. Such a mapping f is said to be Scott continuous.

(3) 

A subset $F\subseteq P$ is Scott closed if it is a lower set ($F=\downarrow F=\{x\in P:x\le y for some y\in F\}$) and for any directed set $D\subseteq F$, it holds that $\bigvee D\in F$ whenever $\bigvee D$ exists.

A poset is called a directed complete poset (dcpo for short) if every directed subset of the poset has a supremum. For more about Scott topology and dcpos, we refer the reader to [12,13].

For two elements x and y in poset P, x is way below y, written as $x\ll y$, if, for any directed set $D\subseteq P$ with $\bigvee D$ existing and $\bigvee D\ge y$, there is some $d\in D$ such that $x\le d$.

An element $x\in P$ is called compact if $x\ll x$. The set of all compact elements of P is denoted by $K\left(P\right)$.

A poset L is called continuous if, for each $x\in L$, $\{z\in L:z\ll x\}$ is directed and $x=\bigvee \{z\in L:z\ll x\}$.

A continuous dcpo is often called a domain.

A poset L is algebraic if, for each $x\in L$, $\{z\in L:z\le x,z\in K(L\left)\right\}$ is directed and $x=\bigvee \{z\in L:z\le x,z\in K(L\left)\right\}$.

A nonempty subset F of topological space X is irreducible if, for any closed subsets $F_1,F_2$ of X, $F\subseteq F_1\cup F_2$ implies $F\subseteq F_1$ or $F\subseteq F_2$.

Clearly, every singleton set is irreducible. Also, the closure of every singleton set is irreducible (see the remark after Definition O-5.5 in [12]). A space X is called sober if every irreducible closed set F of X is the closure of a unique singleton set: there is a unique $x\in X$, such that $F=cl\left(\right\{x\left\}\right)$.

A subset A of a topological space is called saturated if A equals the intersection of all open sets containing it. A $T_0$ space X is well filtered if, for any open set U and filtered family $\mathcal{F}$ of saturated compact subsets of X, $\bigcap \mathcal{F}\subseteq U$ implies $F\subseteq U$ for some $F\in \mathcal{F}$. Every sober space is well filtered (Theorem II-1.21 of [12]).

The specialization order ${\le }_{\tau }$ on a $T_0$ space $(X,\tau )$ is defined by $x{\le }_{\tau }y$ if $x\in cl\left(\right\{y\left\}\right)$ (for $x,y\in X$).

A $T_0$ space X is called bounded -sober if every upper bounded (with respect to the specialization order) irreducible closed set equals the closure of a singleton set [14]. Clearly, every sober space is bounded sober. The converse conclusion is not true.

3. Johnstone’s Dcpo

One of the classic results in domain theory is that the Scott space of every domain is sober (Corollary II-1.13 [12]). However, it is not trivial whether the Scott space of every dcpo is sober. Peter Johnstone constructed the first example of a dcpo whose Scott space is not sober.

Example 1

([1]). Let $\mathbb{N}$ be the set of all positive integers with the ordinary order ≤ of numbers. Let $\mathbb{J}=\mathbb{N}\times (\mathbb{N}\cup \{\infty \left\}\right)$. Extend the order on $\mathbb{N}$ to $\mathbb{N}\cup \{\infty \}$ by letting $n<\infty$ for all $n\in \mathbb{N}$.

Define the order ≤ on $\mathbb{J}$ by $(m,n)\le (m^{\prime },n^{\prime })if$ either

(i) $m=m^{\prime } and n\le n^{\prime };$

(ii) Or $n^{\prime }=\infty$ and $n\le m^{\prime }$.

All the major properties of $\mathbb{J}$ will be numbered as $\left(Jk\right)$ with k a number. Among the nine properties listed below, (J5), (J7), (J8) and (J9) are new (as far as we know).

The properties (J1) and (J2) below are from [1].

(J1) A subset $D\subseteq \mathbb{J}$ is directed if either it has a largest element, or there is an m such that $D=\{(m,d_i):i\in I\}\subseteq \downarrow (m,\infty )$.

In the latter case, $\bigvee D=(m,\bigvee \{d_i:i\in I\})$. Here, either $\bigvee \{d_i:i\in I\}=\infty$ or $\bigvee \{d_i:i\in I\}=n$ for some $n\in \mathbb{N}$.

Thus, in both cases, $\bigvee D$ exists, showing that $\mathbb{J}$ is a dcpo.

(J2) A nonempty Scott closed set F of $\mathbb{J}$ is irreducible if and only if either $F=\downarrow (m,n)$ for some $(m,n)\in \mathbb{J}$ or $F=\mathbb{J}$.

Clearly, $\mathbb{J}\ne \downarrow (m,n)=cl\left(\right\{(m,n)\left\}\right)$ for any $(m,n)\in \mathbb{J}$; thus, the Scott space $(\mathbb{J},\sigma (\mathbb{J}\left)\right)$ is not sober.

However, as $\mathbb{J}$ does not have the top element, we see that $\Sigma \mathbb{J}$ is bounded sober. Hence, $\mathbb{J}$ is a dcpo whose Scott space is bounded sober but not sober.

(J3) $\Sigma \mathbb{J}$ is not well filtered.

To see this, for each $n\in \mathbb{N}$, let $A_n=\{(k,\infty ):k\ge n\}$ and $U=\varnothing$. Then, one can verify that each $A_n$ is a compact saturated set, $\{A_n:n\in \mathbb{N}\}$ is a filtered family and $\bigcap \{A_n:n\in \mathbb{N}\}=\varnothing =U$. Since not one of the $A_n$ is contained in $\varnothing =U$, we see that $\Sigma \mathbb{J}$ is not well filtered.

In the following, we shall show some less trivial properties of $\mathbb{J}$. Some of them have been mentioned or proved elsewhere. However, for the reader’s convenience, we shall provide a brief explanation of them.

An element a of poset P is a maximal element (or maximal point) if, for any $y\in P$, $a\le y$ implies $a=y$. The set of all maximal points of P is denoted by Max(P). In the following, the set Max(P) with the relative Scott topology on P will be called the maximal point space of P. By Zorn’s Lemma, one deduces that for any dcpo P, $P=\downarrow \mathrm{Max}\left(P\right)$; that is, every element of P is below some maximal point(s).

The following result has been known for quite some time; however, we could not find a resource where a complete proof was provided. Here, we give a brief explanation. See the more general Proposition 3 in Section 4.

Proposition 1.

The space Max($\mathbb{J}$) is homeomorphic to $(\mathbb{N},{\tau }_{cof})$, where ${\tau }_{cof}$ is the co-finite topology on $\mathbb{N}$ ($U\in {\tau }_{cof}$ if either $U=\varnothing$ or $\mathbb{N}-U$ is a finite set).

To see this, first note that Max($\mathbb{J}$) = $\left\{\right(m,\infty ):m\in \mathbb{N}\}$.

Let $F\subseteq \mathbb{J}$ be a nonempty Scott closed set. If $F\cap \mathrm{Max}\left(\mathbb{J}\right)$ is an infinite set, then one can deduce that $F=\mathbb{J}$.

Now, for any finite subset $G\subseteq \mathrm{Max}\left(\mathbb{J}\right)$, $\downarrow G$ is a Scott closed set of $\mathbb{J}$ and $G=\downarrow (G\cap \mathrm{Max}(\mathbb{J}\left)\right)$.

It follows that a subset of Max($\mathbb{J}$) is closed (with respect to the relative Scott topology) if and only if it is either a finite set or the whole set of Max($\mathbb{J}$). That is, the relative Scott topology on Max($\mathbb{J}$) is the co-finite topology.

Clearly, there is a bijection between $\mathbb{N}$ and Max($\mathbb{J}$); thus, the space Max($\mathbb{J}$) is homeomorphic to $(\mathbb{N},{\tau }_{cof})$.

A poset model of a $T_1$ space X is a poset P such that X is homeomorphic to Max(P). For any P, Max(P) is $T_1$ [15]. By the above proposition, we have the following.

(J4) The dcpo $\mathbb{J}$ is a poset model of $(\mathbb{N},{\tau }_{cof})$.

Proposition 2.

For any continuous function $f:Max\left(\mathbb{J}\right)⟶Max\left(\mathbb{J}\right)$, there is a continuous function $F:\Sigma P⟶\Sigma P$ such that $F_{Max\left(\mathbb{J}\right)}=f$.

Proof. 

Let $p:\mathbb{J}⟶\mathbb{N}$ be a function defined by $p(m,n)=m$.

Let $f:\mathrm{Max}\left(\mathbb{J}\right)⟶\mathrm{Max}\left(\mathbb{J}\right)$ be a continuous function. For each $n\in \mathbb{N}$, define

$$\alpha \left(n\right)=inf\left\{p\right(f(k,\infty )):n\le k\}.$$

Then, $\alpha :\mathbb{N}⟶\mathbb{N}$ is a function and monotone: $n_1\le n_2$ implies $\alpha \left(n_1\right)\le \alpha \left(n_2\right)$.

(1) If $\alpha$ is a bounded function (there is an l such that $\alpha \left(n\right)\le l$ for all n), then f is a constant function.

In fact, assume that $\alpha \left(n\right)\le l$ holds for all n. Then, for any $n\in \mathbb{N}$, there exists $n^{\prime }\ge n$ such that $f(n^{\prime },\infty )\in B=\{(m,\infty ):m\le l\}$, which is a closed set in $\mathrm{Max}\left(\mathbb{J}\right)$. Thus, $f^{-1}\left(B\right)$ is an infinite and closed set; hence, $f^{-1}\left(B\right)=\mathrm{Max}\left(\mathbb{J}\right)$ since the topology on $\mathrm{Max}\left(\mathbb{J}\right)$ is the co-finite topology. But then $\mathrm{Max}\left(\mathbb{J}\right)=f^{-1}\left(B\right)={\bigcup }_{m\le l}{f}^{-1}\left(\left\{(m,\infty )\right\}\right)$. Note that each $f^{-1}\left(\left\{(m,\infty )\right\}\right)$ is a closed subset of $\mathrm{Max}\left(\mathbb{J}\right)$. Observe that $\mathrm{Max}\left(\mathbb{J}\right)$ can not be expressed as the union of two proper closed subsets (a proper closed subset must be finite); it follows that $\mathrm{Max}\left(\mathbb{J}\right)=f^{-1}\left(\left\{(m,\infty )\right\}\right)$ for some m, showing that f is a constant function.

(2) Assume that f is a constant function and $f(m,\infty )=(m_0,\infty )$ for all m. Define $F:\mathbb{J}⟶\mathbb{J}$ by $F(m,n)=(m_0,n)$. Then, it is easy to see that F is Scott continuous and an extension of f.

(3) Now assume that f is not a constant function. By (1), the function $\alpha$ is not bounded.

Define $F:\mathbb{J}⟶\mathbb{J}$ by

$$F(m,n)=\left\{\begin{array}{cc}f(m,n),\hfill & n=\infty ,\hfill \\ \left(p\right(f(m,\infty )),\alpha (n\left)\right),\hfill & n\ne \infty .\hfill \end{array}\right.$$

Clearly, F is an extension of f. It remains to show that F is Scott continuous. By its definition, one can easily verify that F is monotone.

By Remark 1(2), We need to verify that for any directed set $D\subseteq \mathbb{J}$, $F(\bigvee D)=\bigvee F\left(D\right)$.

If D has the largest element, say $x_0$, then $F(\bigvee D)=F\left(x_0\right)=\bigvee F\left(D\right)$ because F is monotone.

Now assume that D does not have the largest element, then $D=\{(a,n_k):k=1,2,...\}$ for some fixed $a\in \mathbb{N}$ and we can assume that

$$n_1<n_2<\cdots .$$

Now, $F(\bigvee D)=F(a,\infty )=f(a,\infty )=(b,\infty )$ for some $b\in \mathbb{N}$. For any $l\in \mathbb{N}$, as $\alpha$ is not bounded, there is $n^{\prime }$ such that $\alpha \left(n^{\prime }\right)\ge l$. Take one $n_k$ such that $n^{\prime }\le n_k$. Then,

$$l\le \alpha \left(n^{\prime }\right)\le \alpha \left(n_k\right).$$

Hence, $(b,l)\le (b,\alpha \left(n_k\right))=(p\left(f(a,\infty )\right),\alpha \left(n_k\right))=F(a,n_k)\le \bigvee F\left(D\right)$. Therefore, $F(\bigvee D)=(b,\infty )=\bigvee \left\{\right(b,l):l\in \mathbb{N}\}\le \bigvee F\left(D\right)$. Since F is monotone, we have $F(\bigvee D)\ge \bigvee F\left(D\right)$. At last, we have

$$F(\bigvee D)=\bigvee F\left(D\right).$$

The proof is completed.

□

Definition 1.

A poset P is called continuously extendable if, for any continuous mapping $f:Max\left(P\right)⟶Max\left(P\right)$, there is a continuous mapping of $F:\Sigma P⟶\Sigma P$, which is an extension of f: for any $x\in Max\left(P\right)$, $F\left(x\right)=f\left(x\right)$.

Remark 2.

(1) If a poset P has the top element $T_P$, then Max(P) = $\left\{T_P\right\}$. Then, every function $f:Max\left(P\right)\to Max\left(P\right)$ has the continuous extension $\widehat{f}:P⟶P$, which is the constant function $\widehat{f}\left(x\right)=T_P$ for all $x\in P$. Thus, P is continuously extendable. In particular, every complete lattice is continuously extendable.

(2) 

By Theorem 5.7 of [16], for dcpo $P=\mathbf{I}\mathbb{R}$ of all closed intervals of real numbers with the reverse inclusion order, one can only prove that if the continuous function $f:Max\left(\mathbf{I}\mathbb{R}\right)⟶Max\left(\mathbf{I}\mathbb{R}\right)$ satisfies some extra conditions, then it has the continuous extension $\widehat{f}:\mathbf{I}\mathbb{R}⟶\mathbf{I}\mathbb{R}$.

By Proposition 2, we have the following.

(J5) $\mathbb{J}$ is continuously extendable.

By a classic result (Theorem II-1.14 of [12]), a dcpo P is continuous if and only if the lattice $(\sigma (P),\subseteq )$ of all Scott open sets of P with the set inclusion order is a completely distributive lattice (this is equivalent to that the lattice of all Scott closed sets with the set inclusion order is completely distributive) [12]. In addition, the Scott space of every continuous dcpo is sober [12]. Using these facts, one can deduce the following property of continuous dcpo P, where $C_{\sigma }\left(P\right)$ denotes the lattice of all Scott closed sets of P:

$$\mathrm{For} \mathrm{any} \mathrm{dcpo} Q, C_{\sigma }\left(P\right)\cong C_{\sigma }\left(Q\right) \mathrm{implies} P\cong Q,$$

(1)

here, ≅ is the isomorphism relation between posets.

A dcpo P satisfying the property (1) is called $C_{\sigma }$-determined.

For two subsets F and G of a dcpo P, F is way below G, written $F\ll G$, if, for any directed subset D, $\bigvee D\in \uparrow G$ implies $D\cap \uparrow F\ne \varnothing .$

A dcpo P is called quasicontinuous if, for each $x\in P$, the set family

$\{F\subseteq P:F\ll \{x\} \mathrm{and} F \mathrm{is} \mathrm{finite}\}$ is filtered (for any finite $F_1\ll \left\{x\right\},F_2\ll \left\{x\right\}$, there is finite $F\ll \left\{x\right\}$ such that $F\subseteq \uparrow F_1\cap \uparrow F_2$) and

$$\uparrow x=\bigcap \{\uparrow F:F\ll \{x\}\mathrm{and} F \mathrm{is} \mathrm{finite}\}.$$

Every continuous dcpo is quasicontinuous (Proposition III-3.10 of [12]).

A dcpo P is locally quasicontinuous (continuous, resp.) if each subposet $\downarrow x$ ($x\in P$) is quasicontinuous (continuous, resp.) By [17], the dcpo $\mathbb{J}$ is locally quasicontinuous.

Remark 3.

(1) Note that $\mathbb{J}$ is not locally continuous. For instance, the subposet $\downarrow (1,\infty )$ is not continuous.

(2) 

The product of any finite number of quasicontinuous dcpos is quasicontinuous [12]. Hence, for any positive integer n, the Cartesian product ${\mathbb{J}}^n$ is locally quasicontinuous.

The following result was proved in [18].

Theorem 1.

Every locally quasicontinuous dcpo is $C_{\sigma }$-determined.

By the above result and Remark 3 (2), one deduces the following result proved in [18] and [17] using different methods.

(J6) For any $n\in \mathbb{N}$, the dcpo ${\mathbb{J}}^n$ is $C_{\sigma }$-determined.

Given a dcpo P and a Scott closed subset G of P, the set $\downarrow (G\cap \mathrm{Max}(P\left)\right)$ need not be a Scott closed set, although it is a lower set.

Example 2.

Let $P=\mathbb{N}\cup \{a_i:i\in \mathbb{N}\}\cup \{T,b\}$. The partial order ≤ on P is generated by

$$a_i<i, a_i<a_{i+1}<b<T, for each i\in \mathbb{N}.$$

Then, $(P,\le )$ is a dcpo, where Max(P) = $\mathbb{N}\cup \left\{T\right\}$. The set $G=P-\left\{T\right\}$ is a Scott closed set, $\downarrow (G\cap Max(P\left)\right)=P-\{T,b\}$, which is not Scott closed because $D=\{a_i:i\in \mathbb{N}\}$ is a directed subset of $P-\{T,b\}$ and $\bigvee D=b\notin P-\{T,b\}$.

However, for the dcpo $\mathbb{J}$, this is true for all Scott closed sets G of $\mathbb{J}$.

As a matter of fact, let $G\subseteq \mathbb{J}$ be a Scott closed set and $G\cap \mathrm{Max}\left(\mathbb{J}\right)\ne \varnothing$. Assume that $G\cap \mathrm{Max}\left(\mathbb{J}\right)=\{(n_i,\infty ):i\in \mathbb{N}\}$. Then,

$$\downarrow (G\cap \mathrm{Max}\left(\mathbb{J}\right))=\bigcup \{\downarrow (n_i,\infty ):i\in \mathbb{N}\}\subseteq G.$$

Let $D\subseteq \downarrow (G\cap \mathrm{Max}(\mathbb{J}\left)\right)$ be directed. Then, by Remark 1, either D has the largest element or $D=\{(n_{i_0},m_k):k\in \mathbb{N}\}$ for some $i_0$. If D has the largest element, then $\bigvee D\in D\subseteq \downarrow (G\cap \mathrm{Max}(\mathbb{J}\left)\right)$. Otherwise, $\bigvee D=(n_{i_0},t)\le (n_{i_0},\infty )\in \downarrow (G\cap \mathrm{Max}\left(\mathbb{J}\right))$. Since $\downarrow (G\cap \mathbb{J})$ is already a lower set, it is a Scott closed subset of $\mathbb{J}$.

Hence, we have the following property of $\mathbb{J}$.

(J7) For any Scott closed subset G of $\mathbb{J}$, $\downarrow (G\cap \mathrm{Max}(\mathbb{J}\left)\right)$ is a Scott closed subset of $\mathbb{J}$.

A $T_0$ space X is called a $T_D$ space if, for any $x\in X$, $cl\left(\right\{x\left\}\right)-\left\{x\right\}$ is closed. A $T_0$ space X is called a $\mathcal{B}$-space if there is $T_D$ space Y such that $C\left(X\right)\cong C\left(Y\right)$ [19]. This is equivalent to that there is $T_D$ space Y such that the sobrification of X is homeomorphic to the sobrification of Y.

A closed set A of space X is called completely irreducible if for any collection $\{F_i:i\in I\}$ of closed sets $F_i$, $A=cl\left({\bigcup }_{i\in I}\right)F_i$ implies $A=F_i$ for some $i\in I$.

Theorem 2

([19]). A $T_0$ space X is a $\mathcal{B}$-space if and only if every closed set is the supremum of some completely irreducible closed sets: for any closed $F\subset X$, there are completely irreducible closed sets $F_i(i\in I)$ such that

$$F=\bigvee \{F_i:i\in I\}=cl(\bigcup \{F_i:i\in I\}).$$

If a closed set $F\subseteq \Sigma P$, where P is a poset, is completely irreducible, then, as $F=\bigvee \{\downarrow x:x\in F\}$, we have $F=\downarrow x$ for some $x\in F$.

(J8) The space $\Sigma \mathbb{J}$ is a $\mathcal{B}$-space.

For the proof, it is enough to observe that every $\downarrow (m,n)$ ($n\ne \infty$) is a completely irreducible closed set and every closed set is a supremum of such closed sets.

By Cramer [20], a topological space X is called $T^m$, where m is a cardinal, if for any $x\in X$, $cl\left(\right\{x\left\}\right)-\left\{x\right\}$ is the union of at most m closed sets. Thus, $T^1$ spaces are exactly $T_D$ spaces. It is easy to verify the following.

(J9) The space $\Sigma \mathbb{J}$ is a $T^{{\aleph }_0}$ space.

4. Extensions and Problems

In this section, we shall use Johnstone’s method to construct new dcpos and investigate their properties. Some open problems for further study will be listed.

Let Q be a poset, P a dcpo and $\varphi :Q⟶(P\setminus \mathrm{Max}(P\left)\right)$ a monotone mapping. We define the partial order ${\le }_{\varphi }$ on $Q\times P$ by $(q_1,p_1){\le }_{\varphi }(q_2,p_2)$ if

(i) either $q_1=q_2$ and $p_1\le p_2$;

(ii) or $p_2\in \mathrm{Max}\left(P\right)$ and $p_1\le \varphi \left(q_2\right)$.

It is easy to see that ${\le }_{\varphi }$ is a partial order. We shall use $Q\times P_{\varphi }$ to denote the poset $(Q\times P,{\le }_{\varphi })$.

Remark 4.

(1) As in the case of $\mathbb{J}$, a subset $D\subseteq Q\times P$ is directed if either there is a $q\in Q$ such that $D=\{(q,p_i):i\in I\}$ with $\{p_i:i\in I\}$, a directed subset of P, or D has the largest element (if there are $(q_1,p_1),(q_2,p_2)\in D$ with $q_1\ne q_2$). If D has the largest element, it would be the supremum of D. If $D=\{(q,p_i):i\in I\}$ with $\{p_i:i\in I\}$, a directed subset of P, then $\bigvee D=(q,\bigvee \{p_i:i\in I\})$. Hence, $Q\times P_{\varphi }$ is a dcpo.

(2) Max($Q\times P_{\varphi })=Q\times Max\left(P\right)$.

Example 3.

The subsequent examples serve to illustrate this concept.

(1) 

Let $Q=(\mathbb{N},\le )$ be the poset of all positive integers with the ordinary order of numbers and $P=(\mathbb{N}\cup \{\infty \},\le )$ be the extension of Q with the top element ∞ added. Then, $Q\times P_{id}=\mathbb{J}$, where $id:Q⟶P$ is the embedding mapping: $id\left(m\right)=m$ for all $m\in Q$.

(2) 

Let $Q=[0,1)=\{x\in \mathbb{R}:0\le x<1\}$ and $P=[0,1]=\{x\in \mathbb{R}:0\le x\le 1\}$ be the given sets of real numbers with the ordinary order of numbers. Let $id:Q⟶P$ be the embedding mapping. Then, the poset $Q\times P_{id}$ has $[0,1)\times [0,1]$ as underlying set and $(x,y)\le (x^{\prime },y^{\prime })$ holds if and only if either $x=x^{\prime }$ and $y\le y^{\prime }$ or $y^{\prime }=1$ and $y\le x^{\prime }$.

(3) 

In general, let P be a dcpo, $Q=P-Max\left(P\right)$ and $id:Q⟶P$ be the identity mapping: $x\mapsto x(\forall x\in Q)$. Then, we have a natural dcpo, which will be denoted by $J\left(P\right)$. In particular, if ${\mathbb{N}}^T$ is the complete chain $\mathbb{N}\cup \{\infty \}$ with ∞ as the top element, then $\mathbb{J}=J\left({\mathbb{N}}^T\right)$.

(4) 

Let $P={\mathbb{N}}^T$ be the dcpo defined in (3) and $Q=[0,+\infty )$ with the usual order of numbers. Define $\varphi :Q⟶P$ by $\varphi \left(x\right)=\left[x\right]$, the integer part of x for each $x\in Q$. Then, $Q\times P_{\varphi }=\{(x,n):x\in [0,+\infty ),n\in {\mathbb{N}}^T\}$ and $(x,n){\le }_{\varphi }(y,m)$ if and only if either $x=y$ and $n\le m$ or $n\le \left[y\right]$ and $m=\infty$.

Proposition 3.

Let P be a chain that is a dcpo and has the top element $T_P$.

(1) 

A Scott closed subset $F\subseteq J\left(P\right)$ is irreducible if and only if either $F=\downarrow (x,y)$ for some $(x,y)\in J\left(P\right)$ or $F=J\left(P\right)$. Thus, the Scott space of $J\left(P\right)$ is not sober.

(2) 

For any Scott closed set G of $J\left(P\right)$, $\downarrow (G\cap Max(J\left(P\right)\left)\right)$ is a Scott closed subset of $J\left(P\right)$.

(3) 

Max($J\left(P\right)$) is homeomorphic to $(P\setminus \left\{T_P\right\},{\tau }_{cob})$, where $U\in {\tau }_{cob}$ if and only if either $U=\varnothing$ or $P-(U\bigcup \left\{T_P\right\})$ has an upper bound in $P\setminus \left\{T_P\right\}$.

(4) 

$J\left(P\right)$ is $C_{\sigma }$-determined.

Proof. 

The proofs of (1), (2) and (4) are similar to that of $\mathbb{J}$, we skip the details.

We now prove (3). By the definition of ${\tau }_{cob}$, $H\subseteq P-\left\{T_P\right\}$ is closed in $(P,{\tau }_{cob})$ if and only if either $H=P-\left\{T_P\right\}$, or H has an upper bound in $P-\left\{T_P\right\}$.

Note that Max($J\left(P\right))=\{(p,T_P):p\in P-\left\{T_P\right\}\}$.

Let $\varphi :\mathrm{Max}($J(P)$)⟶(P\setminus \left\{T_P\right\})$ be the mapping defined by $\varphi (p,T_P)=p$. Then, clearly, $\varphi$ is a bijection.

(1) Let $F\subseteq \mathrm{Max}\left(J\right(P\left)\right)$ be a closed set in $\mathrm{Max}\left(J\right(P\left)\right)$. Then, there is a Scott closed set $\widehat{F}$ of $J\left(P\right)$, such that $F=\widehat{F}\cap \mathrm{Max}\left(J\left(P\right)\right)$.

If $\varphi \left(F\right)$ is not upper bounded in $P-\left\{T_P\right\}$, then as in the proof of the Proposition 1, one deduces that $\widehat{F}=J\left(P\right)$; thus, $\varphi \left(F\right)=\varphi (J\left(P\right)\cap \mathrm{Max}\left(J\left(P\right)\right))=\varphi \left(J\left(P\right)\right)=P\setminus \left\{T_P\right\}$. Otherwise, $\varphi \left(F\right)$ has an upper bound in $P-\left\{T_P\right\}$. Thus, $\varphi$ sends each closed set of $\mathrm{Max}\left(J\right(P\left)\right)$ to a closed set in $(P-\left\{T_P\right\},{\tau }_{cob})$. Therefore, $\varphi$ is a closed mapping.

(2) Now assume that $H\subseteq P\setminus \left\{T_P\right\}$ is a closed set in $(P\setminus \left\{T_P\right\},{\tau }_{cob})$. If $H=P\setminus \left\{T_P\right\}$, then ${\varphi }^{-1}\left(H\right)=\mathrm{Max}\left(J\left(P\right)\right)$, which is closed in $\mathrm{Max}\left(J\right(P)$. If $H\ne P-\left\{T_P\right\}$, then H has an upper bound, say w in $P\setminus \left\{T_P\right\}$. Hence, $w<T_P$ holds in P.

Let

$$\underline{H}=\{(x,p)\in J\left(P\right):x\notin H \mathrm{and} p\le w\}\cup \{(h,p)\in J\left(P\right):h\in H \mathrm{and} p\in P\}.$$

(i) $\underline{H}$ is a lower set in $J\left(P\right)$

In fact, assume that $(z,u)\le (x,p)$ for some $(x,p)\in \underline{H}$. If $x\notin H$, then $p\le w<T_P$. Hence, $(z,u)\le (x,p)$ implies $z=x,u\le p\le w$. So, $(z,u)\in \underline{H}$. Let $x\in H$. If $z=x$, then $(z,u)=(x,u)\in \underline{H}$. If $z\ne x$, by $(z,u)\le (x,p)$, we have $p=T_p$ and $u\le x\le w$, again implying $(z,u)\in \underline{H}$. Hence, $\underline{H}$ is a lower set.

(ii) Assume that $D\subseteq \underline{H}$ is a directed subset.

If D has the largest element $d_0$, then $\bigvee D=d_0\in D\subseteq \underline{H}$, implying $\bigvee D\in \underline{H}$. If D does not have the largest element, by Remark 4(1), there is $x\in P-\left\{T_P\right\}$ and directed set $\{d_i:i\in I\}\subseteq P$ such that $D=\{(x,d_i):i\in I\}$.

If $x\in H$, then $\bigvee D=(x,\bigvee \{d_i:i\in I\})\in \underline{H}$. If $x\notin H$, then $d_i\le w$ for all $i\in I$, so $\bigvee \{d_i:i\in I\}\le w$ holds in P. Thus, $\bigvee D=(x,\bigvee \{d_i:i\in I\})\in \underline{H}$.

Therefore, $\underline{H}$ is a Scott closed set of $J\left(P\right)$.

Since $w\ne T_P$, we see that $\underline{H}\cap \mathrm{Max}\left(J\left(P\right)\right)=\{(h,T_P):h\in H\}$ is a closed set of $\mathrm{Max}\left(J\right(P\left)\right))$. In addition, ${\varphi }^{-1}\left(H\right)=\underline{H}\cap \mathrm{Max}\left(J\left(P\right)\right)$. Hence, $\varphi :\mathrm{Max}\left(J\left(P\right)\right)⟶P-\left\{T_P\right\}$ is continuous.

The combination of (1) and (2) proves that $\varphi$ is a homeomorphism.

The proof is completed. □

Example 4.

Let $W_1$ be the set of all countable ordinals plus the smallest non-countable ordinal ${\omega }_1$. Then, $W_1$ is a chain with ${\omega }_1$ as the top element. A subset $A\subseteq W_1-\left\{{\omega }_1\right\}$ has an upper bound in $W_1-\left\{{\omega }_1\right\}$ if and only if it is a countable set. Hence, by Proposition 3, $Max\left(J\left(W_1\right)\right)$ is homeomorphic to $(W_1-\left\{{\omega }_1\right\},{\tau }_{cc})$, where ${\tau }_{cc}$ is the co-countable topology: $U\in {\tau }_{cc}$ if and only if either $U=\varnothing$ or $U^c$ is countable. Thus, we deduce the result in [4] as a special case. It was proved in [4] that $J\left(W_1\right)$ is well filtered.

Example 5.

Let $P=[0,1]$ be the set of all real numbers in the unit interval. Then, $1$ is the top element in chain P and $P-\left\{1\right\}=[0,1)$. By Proposition 3, Max($J\left(\right[0,1]$) is homeomorphic to $\left(\right[0,1),\tau )$, where $F\subseteq [0,1)$ is a closed set if and only if either $F=[0,1)$ or $F\subseteq [0,b]$ for some $0\le b<1$. One can check that the Scott space $\Sigma J\left(\right[0,1\left]\right)$ is not well filtered.

Example 6.

Let ${\mathbb{N}}^T=\mathbb{N}\cup \{\infty \}$ be chain $\mathbb{N}$ with the top element ∞ added. Max(${\mathbb{N}}^T)$ is homeomorphic to $(\mathbb{N},\tau )$, where $U\in \tau$ if and only if either $U=\varnothing$ or $\mathbb{N}-U$ is upper bounded, which is equivalent to the notion that $\mathbb{N}-U$ is a finite set. Hence, $\tau ={\tau }_{cof}$. Thus, we re-deduce (J4) as a special case.

The following are some problems related to dcpo $\mathbb{J}$.

Problem 1. Is the countable self-product ${\mathbb{J}}^{\mathbb{N}}$ of $\mathbb{J}$ characterized as a $C_{\sigma }$-determined dcpo?

Problem 2. Does every $T_1$ space have a continuously extendable dcpo model?

Note that it was proved in [15] that every $T_1$ space has a dcpo model.

Problem 3. In which context does dcpo P exhibit the property that its justification $J\left(P\right)$ is well filtered?

Note that $\mathbb{J}$ is not well filtered and $J\left(W_1\right)$ is well filtered.