2. Preliminaries

In this section, we recall some basic definitions and results to be used later. For more details on them, we refer the reader to Gierz et al. (Reference Gierz, Hofmann, Keimel, Lawson, Mislove and Scott2003) and Goubault-Larrecq (Reference Goubault-Larrecq2013).

Let P be a poset. A nonempty subset D of P is directed if every two elements of D have an upper bound in D. If D is also a lower set ( $D={\downarrow} D=\{x\in P: x\le d \mbox{ for some } d\in D\}$ ), then D is called an ideal. A poset is called a directed complete poset (dcpo, for short) if its every directed subset has a supremum. A complete lattice is a poset in which every subset has a supremum and an infimum. A subset U of a poset P is Scott open if (i) it is an upper set ( $U=\mathord{\uparrow} U=\{x\in P: u\le x \mbox{ for some } u\in U\}$ ) and (ii) for every directed subset D of P with $\sup D$ existing and $\sup D\in U$ , it follows that $D\cap U\not=\emptyset$ . The complements of Scott open sets are called Scott closed sets. The collection of all Scott open subsets of P forms a topology on P, called the Scott topology of P, and is denoted by $\sigma(P)$ . The collection of all Scott closed subsets of P is denoted by $\Gamma (P)$ . The space $(P,\sigma(P))$ called the Scott space of P is written as $\Sigma P$ .

For two elements x, y in a poset P, x is way-below y, denoted by $x \ll y$ , if for any directed subset D of P for which $\sup D$ exists, $y \leq \sup D$ implies $D \cap \mathord{\uparrow} x \neq \emptyset$ . A poset P is continuous if for each is directed and A continuous dcpo is usually called a domain.

An element x of P is compact if $x\ll x$ . The set of all compact elements of P is denoted by $\textit{K}(P)$ . A poset P is algebraic if for every $x\in P$ , the set $\textit{K}(P)\cap \mathord{\downarrow} x$ is directed and $x=\sup (\mbox{K}(P)\cap \mathord{\downarrow} x).$ For any compact element $x\in P$ , $\mathord{\uparrow} x\in \sigma(P)$ . Every algebraic poset is continuous. If L is a complete lattice such that $\mbox{K}(L)=L$ (all elements are compact), then L is algebraic.

A subset K of a topological space X is compact if every open cover of K has a finite subcover. A set K of a topological space is called saturated if it is the intersection of its open neighborhood ( $K=\mathord{\uparrow} K$ in its specialization order). The saturation satA of a set A is the intersection of all its open neighborhoods.

Lemma 2.3. Assume that L is a complete lattice such that $\mbox{K}(L)=L$ . Let

$$L^*=\{\mathord{\uparrow} F: F\subseteq L \mbox{ is finite}\}.$$

Then $(L^{*}, \subseteq)$ is a distributive complete lattice.

For any $\mathord{\uparrow} F, \mathord{\uparrow} G\in L^{*}$ , it is easy to verify that in $L^*$ ,

$$\sup\{\mathord{\uparrow} F, \mathord{\uparrow} G\}=\mathord{\uparrow} (F\cup G), \inf\{\mathord{\uparrow} F, \mathord{\uparrow} G\}=\mathord{\uparrow} F\wedge \mathord{\uparrow} G=(\mathord{\uparrow} F)\cap (\mathord{\uparrow} G)=\mathord{\uparrow}\{x\vee y: x\in F, y\in G\}. $$

Thus, $L^{*}$ is a lattice.

1. (1) $L^{*}$ is a distributive lattice. For any two elements $\mathord{\uparrow} F, \mathord{\uparrow} G $ in $L^{*}$ , it is easy to see that $\inf\{\mathord{\uparrow} F, \mathord{\uparrow} G\}=\mathord{\uparrow} F\cap \mathord{\uparrow} G$ and $\sup\{\mathord{\uparrow} F,\mathord{\uparrow} G\}=\mathord{\uparrow} F\cup \mathord{\uparrow} G$ . Hence, the finite sup (inf) in $L^{*}$ is the set union (intersection), which means that $L^{*}$ is a sublattice of the distributive powerset lattice $\mathcal{P}(L)$ ; thus, $L^{*}$ is distributive.

2. (2) $L^{*}$ is a complete lattice

Since $L^{*}$ is a lattice and has a top element $\mathord{\uparrow} 0_{L}$ , in order to prove $L^{*}$ is a complete lattice, it remains to show that every filtered subset of $L^{*}$ has an infimum. Let $\mathcal{D}=\{\mathord{\uparrow} F_i: i\in I\}$ be a filtered subset of $L^{*}$ . Then each $\mathord{\uparrow} F_{i}$ is a compact saturated subset of $\Sigma L$ . Hence, by Remarks 2.2(3), the intersection $A=\bigcap \{\mathord{\uparrow} F_i: i\in I\}$ is a compact saturated subset of $\Sigma L$ . For each $x\in A$ , x is compact, so $\mathord{\uparrow} x\in\sigma(L)$ .

Now $A=\bigcup\{\mathord{\uparrow} x: x\in A\}$ (note that A is an upper set). As A is compact, there is a finite subset $G\subseteq A$ , such that $A=\bigcup\{\mathord{\uparrow} y: y\in G\}=\mathord{\uparrow} G$ , which is in $L^{*}$ . Clearly $\inf \mathcal{D}=\mathord{\uparrow} G$ .

It follows that $(L^{*}, \subseteq)$ is a complete lattice.

An ideal I of a poset P is incremental if I is not a principal ideal ( $I\neq \mathord{\downarrow} x$ for any $x\in P$ ). We use Id(P) to denote the set of all incremental ideals of a poset P.

Let $\mathbb{N}$ be the set of all nonnegative integers. Then, $\mathbb{N}$ is a poset with the ordinary order $\le$ of numbers. Let $\mathbb{N}^{<\mathbb{N}}$ be the set of all nonempty finite words (or, finite strings) over $\mathbb{N}$ . The prefix order “ $\le$ ” on $\mathbb{N}^{<\mathbb{N}}$ is defined as follows:

For any $x=a_1a_2\cdots a_n, y= b_1b_2\cdots b_m$ in $\mathbb{N}^{<\mathbb{N}}$ ,

$$x\le y \mbox{ if and only if } n\le m \mbox{ and } a_i=b_i \mbox{ for all } 1\le i\le n.$$

The poset $\mathbb{N}^{<\mathbb{N}}$ is countable and does not have infinite decreasing sequences. One can arrange all the elements from $\mathbb{N}^{<\mathbb{N}}$ in a sequence such that larger elements appear later. Thus, we can define a monotone injective function from $\mathbb{N}^{<\mathbb{N}}$ to $\mathbb{N}$ .

We shall make use of the following result, for its proof, see Remark 6.1.

For example, let $E_{1}=\{2m: m\in \mathbb{N}\}$ . Assume that we have defined disjoint subsets $E_{k} (k=1, 2, \cdots, m)$ such that each $E_{i}$ is an infinite subset of $\mathbb{N} - \bigcup\{E_j: j\le i-1\}$ and $\mathbb{N} - \bigcup\{E_j: j\le i\}$ is infinite. Then choose $E_{m+1}$ to be an infinite subset of $\mathbb{N} - \bigcup\{E_i: i\le m\}$ such that $\mathbb{N} - \bigcup\{E_i: i\le m+1\}$ is infinite.

By induction, we can have a sequence $\{E_k\}_{k\in\mathbb{N}}$ of disjoint infinite subsets of $\mathbb{N}$ .

Since $\mathbb{N}\times \mathbb{N}$ is a countable set, there exists a bijection $h: \mathbb{N}\times \mathbb{N}\rightarrow \mathbb{N}$ . Using h and the above remarks, we see that there is an injective function $i: \mathbb{N}\times \mathbb{N}\rightarrow \mathcal{P}(\mathbb{N})$ such that each i(m, n) is an infinite set and $i(m_1, n_1)\cap i(m_2, n_2)=\emptyset$ whenever $(m_1, n_1)\not=(m_2, n_2)$ . In addition, by subtracting $\{1, 2, \cdots, m\}$ from i(m, n), we can guarantee that all numbers in i(m, n) are strictly greater than m.

Now for each $(m, n)\in \mathbb{N}\times\mathbb{N}$ , there is a monotone injection $h_{m, n}: \mathbb{N}\rightarrow i(m, n)$ . Let $f_{m,n}=h_{m,n}\circ f$ . Then, $f_{m,n}$ is a monotone injective function from $\mathbb{N}^{<\mathbb{N}}$ into i(m, n).

3. A Countable Complete Distributive Lattice Whose Scott Space is Non-sober

We now construct a countable complete distributive lattice whose Scott space is non-sober.

Let $L=\mathbb{N}\cup \mathbb{N}^{<\mathbb{N}}\cup \{\top\}$ with the order $\leq $ such that both $\mathbb{N}^{<\mathbb{N}}$ and $\mathbb{N}$ are sub posets and T is the top element.

The poset L can be depicted as Fig. 1.

Figure 1.

The basic gadget of P.

Next, let $P=\mathbb{N}\times L$ .

Let $L_{n}=\{(n, x)\in P: x\in L\}$ . In this section, for $s\in \mathbb{N}^{<\mathbb{N}}$ with length equaling 1 sometimes is considered as a natural number s. We first define the relations $<_{1}, <_{2}, <_{3}$ and $<_{4}$ on P as follows:

• $(n, x)<_{1}(m,y)$ if $n=m$ and $x< y$ holds in L;

• $(n, x)<_{2}(m,y)$ if $y=\top$ , $x\in \mathbb{N}^{<\mathbb{N}}$ and there exists $k\in \mathbb{N}$ with $k>n$ such that $m\in i(n,k)$ and $m=f_{n,k}(x)$ . (In other words, $(n, x)<_{2}(f_{n,k}(x),\top)$ for all $n<k$ ).

• $(n, x)<_{3}(m,y)$ if $y=\top$ , $x\in \mathbb{N}$ and there exists $d\in \mathbb{N}$ with $d<n$ such that $m\in i(d, n)$ and $m=f_{d,n}(x)$ . (In other words, $(n, x)<_{3}(f_{d, n}(x),\top)$ for all $d<n$ ).

Here $(n, x)\in \mathbb{N}\times\mathbb{N}$ , and in the definition of $f_{d, n}(x)$ , $x\in\mathbb{N}$ is taken as an element of $\mathbb{N}^{<\mathbb{N}}$ with length equaling 1.

• $(n,x)<_{4}(m,y)$ if $y=\top$ , $x\in \mathbb{N}$ and there exists $a,b\in \mathbb{N}$ , $s\in\mathbb{N}$ with $a<b$ such that $f_{a,b}(s)=n$ and $f_{a,b}(s.x)=m$ . (In other words, $(f_{a,b}(s),x)<_{4}(f_{a,b}(s.x),\top)$ for all $(a,b)\in \mathbb{N}\times \mathbb{N}$ with $a<b$ ).

By the above definitions, it is clear that as subsets of $P\times P$ , $<_1$ , $<_2$ , $<_3$ and $<_4$ are disjoint. Here we explain, in particular, why $<_3$ and $<_4$ are disjoint. In fact, suppose that $((f_{a,b}(s),x),(f_{a,b}(s.x), \top))$ is in $<_4$ . Let $n=f_{a,b}(s)$ . Then by the definition of $f_{a,b}$ , $n>b$ , hence, $f_{a,b}(s.x)\neq f_{d,n}(x)$ for any $d \in \mathbb{N}$ with $d<n$ . Hence, $((f_{a,b}(s),x),(f_{a,b}(s.x), \top))$ is not in $<_3$ .

Now let $\le$ be the partial order on P generated by $<_1\cup <_2\cup <_3\cup <_4$ (the smallest partial order relation containing all $<_i (i=1, 2, 3, 4)$ ). Note that the partial order $\leq $ coincides with $\leq_{1}\cup (\leq_{1}; <_{2})\cup (\leq _{1}; <_{3})\cup (\leq_{1}; <_4)$ , where $\leq_{1}$ is the reflexive closure of $<_{1}$ and $(R; S)$ denotes the composition of two relations R and S, defined by $x(R; S)y$ iff $\exists z : xRzSy$ .

If $n=m$ , then the strict order $<_1$ is depicted as in Fig. 1. For $n\neq m$ , the strict orders $<_2, <_3, <_4$ of P are depicted in the following figures, respectively.

When $k \gt n$ , the red lines in Fig. 2 illustrate three specific cases: $(n,1)<_{2}(f_{n,k}(1),\top)$ , $(n,2)<_{2}(f_{n,k}(2),\top)$ and $(n,x)<_{2}(f_{n,k}(x),\top)$ .

Figure 2.

The strict order $<_{3}$ .

The red lines in Fig. 3 illustrate the cases: $(n,1)<_{3}(f_{d,n}(1),\top)$ , $(n,2)<_{3}(f_{d,n}(2),\top)$ , and $(n,x)<_{3}(f_{d,n}(x),\top)$ .

Figure 3.

The strict order $<_{3}$ .

The red lines in Fig. 4 illustrate the cases: $(f_{a,b}(1),1)<_{4}(f_{a,b}(1.1),\top)$ and $(f_{a,b}(1),x)<_{4}(f_{a,b}(1.x),\top)$ .

Figure 4.

The strict order $<_{4}$ .

The red lines in Fig. 5 illustrate the cases: $(f_{a,b}(1),x)<_{4}(f_{a,b}(1.x),\top)$ and $(f_{a,b}(1.x),y)<_{4}(f_{a,b}(1.x.y),\top)$ .

Figure 5.

The strict order $<_{4}$ .

The red lines in Fig. 6 illustrate the cases: $(b,y)<_{3}(f_{a,b}(y),\top)$ ; $(f_{a,b}(y),z)<_{4}(f_{a,b}(y.z),\top)$ ; $(f_{a,b}(y.z),u)<_{4}(f_{a,b}(y.z.u),\top)$ .

Figure 6.

Assembling the strict orders $<_{3}$ and $<_{4}$ .

In Fig. 7, the red lines are the same as Fig. 6 and the blue lines add the cases of $<_{2}$ : $(a,y)<_{2}(f_{a,b}(y),\top)$ , $(a,y.z)<_{2}(f_{a,b}(y.z),\top)$ , and $(a,y.z.u)<_{2}(f_{a,b}(y.z.u),\top)$ .

Figure 7.

Assembling the strict orders $<_{2}$ , $<_{3}$ and $<_{4}$ .

Since U and V are nonempty Scott open sets, there exist $(n_{0}, \top)\in U$ and $(m_{0}, \top)\in V$ .

If $n_{0}=m_{0}$ , then $(n_{0},\top)\in U\cap V$ .

Next, assume that $n_{0} \not= m_{0}$ . Without loss of generality, we just consider the case $n_{0}<m_{0}$ .

Since V is Scott open and $\bigvee\{(m_0, k): k\in\mathbb{N}\}=(m_0, \top)\in V$ , there exists $a_{1}\in \mathbb{N}$ such that $(m_{0}, a_{1})\in V$ .

From the definition of $<_{3}$ , it follows that $(m_{0},a_{1})<_{3}(f_{n_{0}, m_{0}}(a_{1}),\top)$ . Whence, $(f_{n_{0},m_{0}}(a_{1}),\top)\in V$ . By the similar reason for the existence of $a_1$ , there is $a_2\in \mathbb{N}$ such that $(f_{n_{0},m_{0}}(a_{1}), a_{2})\in V$ .

By the definition of $<_{4}$ , we have that $(f_{n_{0}, m_{0}}(a_{1}), a_{2})<_{4}(f_{n_{0}, m_{0}}(a_{1}.a_{2}),\top)$ . It follows that $(f_{n_{0},m_{0}}(a_{1}.a_{2}),\top)\in V$ .

By induction on $\mathbb{N}$ , for any $n\in \mathbb{N}$ , there exists $(f_{n_{0},m_{0}}(a_{1}.a_{2}.\cdots .a_{n}),\top)\in V$ . Note that $\{(n_0, a_{1}.a_{2}.\cdots .a_{k})): k\in\mathbb{N}\}$ is an increasing sequence in P and

$$\bigvee \{(n_0, a_{1}.a_{2}.\cdots .a_{k})): k\in\mathbb{N}\}=(n_0, \top)\in U.$$

Thus, there exists $k\in \mathbb{N}$ such that $(n_{0}, a_{1}.\cdots.a_{k})\in U$ . By the definition of $<_{2}$ , we have that $(n_{0}, a_{1}.\cdots.a_{k}) <_{2} (f_{n_{0},m_{0}}(a_{1}.\cdots.a_{k}),\top)$ (see Fig. 8 for the process of the proof).

Figure 8.

The proof of Lemma 3.2 ( $n<m$ ).

Hence, $(f_{n_{0},m_{0}}(a_{1}.\cdots.a_{k}),\top)\in U$ , implying that $(f_{n_{0},m_{0}}(a_{1}.\cdots.a_{k}),\top)\in U\cap V$ .

now show that M is a dcpo. Let $B=\{(n,m)\in \mathbb{N}\times \mathbb{N}: n<m\}$ . In order to determine what the intersections of two principal[-350pt][0pt]Figures 8–15 have not been mentioned in the text. Please cite the figure in the relevant place in the text. ideals of P are, we first list all types of principal ideals $\mathord{\downarrow} x$ of P.

Figure 9.

The Type I ideals.

Figure 10.

The Type II ideals.

Figure 11.

The Type III ideals.

Figure 12.

The Type IV ideals.

Figure 13.

The Type V ideals.

Figure 14.

The Type $\mathrm{I}\cup \mathrm{II}^{1}$ sets.

Figure 15.

The Type $\mathrm{I}\cup \mathrm{II}^{2}$ sets.

Types I : $\mathord{\downarrow} (m_{0}, s_0)=\{(m_{0}, s): s\leq s_{0}\}$ for some $m_{0}\in \mathbb{N},s_{0}\in \mathbb{N}^{<\mathbb{N}}$ (see Fig. 9 for Type I ideals).

Types II : $\mathord{\downarrow} (m_{0}, n_0)=\{(m_{0}, n): n\leq n_{0}\}$ for some $m_{0}, n_{0}\in \mathbb{N}$ (see Fig. 10 for Type II ideals).

Types III. : $\mathord{\downarrow} (n_0, \top)=L_{n_{0}}$ for some $n_{0}\in \mathbb{N}\backslash \bigcup_{(n,m)\in B}i(n,m)$ (see Fig. 11 for Type III ideals).

Types IV : $\mathord{\downarrow}(f_{m_{0}, n_{0}}(s_{0}),\top)=L_{f_{m_{0},n_{0}}(s_{0})}\cup \{(m_{0},s_{0})\}\cup\{(n_{0},n): n\leq s_{0}\}$ for some $(m_{0},n_{0})\in B, s_{0}\in \mathbb{N}^{<\mathbb{N}}$ with $ {\mid}s_{0}{\mid}=1$ (see Fig. 12 for Type IV ideals).

Types V : $\mathord{\downarrow}(f_{m_{0},n_{0}}(s_{0}),\top)=L_{f_{m_{0},n_{0}}(s_{0})}\cup \{(m_{0},s): s\leq s_{0}\}\cup\{(f_{m_{0},n_{0}}(s_{0}^{*}),n): n\leq n_{0}^{*}\}$ for some $(m_{0},n_{0})\in B, s_{0}=s_{0}^{*}.n_{0}^{*}\in \mathbb{N}^{<\mathbb{N}}$ with $s_{0}^{*}\in \mathbb{N}^{<\mathbb{N}},n_{0}^{*}\in \mathbb{N}$ (see Fig. 13 for Type V ideals).

The Types I, II, and III are easily understood.

For the Type IV principle ideal $\mathord{\downarrow}(f_{m_{0}, n_{0}}(s_{0}),\top)$ , clearly it contains the whole $L_{f_{m_{0},n_{0}}(s_{0})}$ . Also $(m_0, s_0) <_{2} (f_{m_{0}, n_{0}}(s_{0}),\top)$ , where $(m_0, s_0)$ is taken as an element in $\mathbb{N}\times \mathbb{N}^{<\mathbb{N}}$ . Thus, $(m_0, s_0)\in \mathord{\downarrow}(f_{m_{0},n_{0}}(s_{0}),\top)$ . Note that this $(m_0, s_0)$ is a minimal element of P. Next, $(n_0, s_0) <_{3} (f_{m_{0}, n_{0}}(s_{0}),\top)$ where the $s_0$ in $(n_0, s_0)$ is a member of $\mathbb{N}$ , thus we have $\mathord{\downarrow} (n_0, s_0)= \{(n_{0},n): n\leq s_{0}\}\subseteq \mathord{\downarrow}(f_{m_{0}, n_{0}}(s_{0}),\top)$ .

Now consider the Type V principle ideal $\mathord{\downarrow}(f_{m_{0},n_{0}}(s_{0}),\top)$ , where $s_0\in \mathbb{N}^{<\mathbb{N}}$ . Trivially, it contains the whole $L_{f_{m_{0}, n_{0}}(s_{0})}$ . Next, $(m_0, s_0) <_{2} (f_{m_{0},n_{0}}(s_{0}),\top)$ , thus $\mathord{\downarrow} (m_0, s_0)=\{(m_{0},s): s\leq s_{0}\}\subseteq \mathord{\downarrow}(f_{m_{0},n_{0}}(s_{0}),\top)$ . Furthermore, if $s_0=s_{0}^{*}.n_{0}^{*}$ with $n_{0}^{*}\in\mathbb{N}$ , then $(f_{m_{0},n_{0}}(s_{0}^{*}),n_{0}^{*}) <_{4} (f_{m_{0},n_{0}}(s_{0}^{*}. n_{0}^{*}), \top)=(f_{m_{0},n_{0}}(s_{0}), \top)$ , hence $\mathord{\downarrow} (f_{m_{0},n_{0}}(s_{0}^{*}),n_{0}^{*})=\{(f_{m_{0},n_{0}}(s_{0}^{*}),n): n\leq n_{0}^{*}\}\subseteq \mathord{\downarrow}(f_{m_{0},n_{0}}(s_{0}),\top)$ . And these are all the elements in $\mathord{\downarrow}(f_{m_{0},n_{0}}(s_{0}),\top)$ .

These types of principle ideas are depicted as below (as the blue regions).

We now list all the subsets of P which are the intersections of two principal ideals in the following table.

In the aforementioned table, Type I $\cup \mathrm{II}^{1}$ sets are of the form $\{(m_{0},s_{0})\}\cup\{(n_{0},n): n\leq k_{0}\}$ for some $(m_{0},n_{0})\in B, s_{0}\in \mathbb{N}^{<\mathbb{N}},k_{0}\in \mathbb{N}$ with $ {\mid}s_{0}{\mid}=1$ , $k_{0}\leq s_{0}$ (see Fig. 14 for Type I $\cup \mathrm{II}^{1}$ ideals).

Type I $\cup \mathrm{II}^{2}$ sets are of the form $\{(m_{0},s)\mid s\leq s_{0}\}\cup\{(f_{m_{0},n_{0}}(s_{0}),n): n\leq k_{0}\}$ for some $(m_{0},n_{0})\in B, s_{0}\in \mathbb{N}^{<\mathbb{N}},k_{0}\in \mathbb{N}$ (see Fig. 15 for Type I $\cup \mathrm{II}^{2}$ ideals).

The two new types of subsets are depicted in Fig. 14, as the blue regions.

For the intersections with Type I, Type II, and Type III principle ideals, the results are easily seen. We now explain the intersections of two Type IV principle ideas, one Type IV and one Type V principle ideals and two Type V principle ideals.

Intersections of two Type V principle ideals

Next, we show that the intersection of two Type V ideals has the form I/II/V/I $\cup$ $ \mathrm{II}^{2}$ / $\emptyset$ , that is, either a Type I ideal, a Type II ideal, a Type V ideal, and a Type $\mathrm{I}\cup \mathrm{II}^{2}$ set or the empty set.

Let $I_{1}, I_{2}$ be two Type V ideals, where

$I_{1}=\mathord{\downarrow}(f_{m_{1},n_{1}}(s_{1}),\top)=L_{f_{m_{1},n_{1}}(s_{1})}\cup \{(m_{1},s): s\leq s_{1}\}\cup\{(f_{m_{1},n_{1}}(s_{1}^{*}),n): n\leq n_{1}^{*}\}$ for some $(m_{1},n_{1})\in B, s_{1}=s_{1}^{*}.n_{1}^{*}\in \mathbb{N}^{<\mathbb{N}}$ with $s_{1}^{*}\in \mathbb{N}^{<\mathbb{N}},n_{1}^{*}\in \mathbb{N}$ , and

$I_{2}=\mathord{\downarrow}(f_{m_{2},n_{2}}(s_{2}),\top)=L_{f_{m_{2},n_{2}}(s_{2})}\cup \{(m_{2},s): s\leq s_{2}\}\cup\{(f_{m_{2},n_{2}}(s_{2}^{*}),n): n\leq n_{2}^{*}\}$ for some $(m_{2},n_{2})\in B, s_{2}=s_{2}^{*}.n_{2}^{*}\in \mathbb{N}^{<\mathbb{N}}$ with $s_{2}^{*}\in \mathbb{N}^{<\mathbb{N}},n_{2}^{*}\in \mathbb{N}$ .

We prove for the case $f_{m_{2},n_{2}}(s_{2})\le f_{m_{1},n_{1}}(s_{1})$ . The proof for the case $f_{m_{1},n_{1}}(s_{1})\le f_{m_{2},n_{2}}(s_{2})$ is similar.

1. (1) $f_{m_{2},n_{2}}(s_{2})<m_{1}$ Then $I_{1}\cap I_{2}=\emptyset$ .

2. (2) $f_{m_{2},n_{2}}(s_{2})=m_{1}$ Then $I_{1}\cap I_{2}=\{(m_{1},s): s\leq s_{1}\}$ , which is a Type I ideal.

3. (3) $m_{1}<f_{m_{2},n_{2}}(s_{2})<f_{m_{1},n_{1}}(s_{1}^{*})$

   1. (3.1) $m_{2}\neq m_{1}$ Then $I_{1}\cap I_{2}=\emptyset$ .

   2. (3.2) $m_{2}=m_{1}$ If $\mathord{\downarrow} s_{1}\cap \mathord{\downarrow} s_{2}=\emptyset$ , then $I_{1}\cap I_{2}=\emptyset$ . If $\mathord{\downarrow} s_{1}\cap \mathord{\downarrow} s_{2}=\emptyset$ , then $I_{1}\cap I_{2}=\{(m_{1}, s): s\leq\inf\{s_{1},s_{2}\}$ , which is a Type I ideal.

4. (4) $f_{m_{2},n_{2}}(s_{2})=f_{m_{1},n_{1}}(s_{1}^{*})$ Then $(m_{1},n_{1})=(m_{2},n_{2}), s_{1}^{*}=s_{2}$ by the property of i and $f_{m_{1},n_{1}}$ . Note that $s_2 < s_{1}$ in this case. We have

   $$I_{1}\cap I_{2}=\{(m_{2},s): s\leq s_{2})\}\cup\{(f_{m_{2},n_{2}}(s_{2}), n): n\leq n_{1}^{*}\}$$
   which is a Type $\mathrm{I}\cup \mathrm{II}^{2}$ set.

5. (5) $f_{m_{1},n_{1}}(s_{1}^{*})<f_{m_{2},n_{2}}(s_{2})<f_{m_{1},n_{1}}(s_{1})$

   1. (5.1) $m_{1}\neq m_{2}, f_{m_{1},n_{1}}(s_{1}^{*})\neq f_{m_{2},n_{2}}(s_{2}^{*})$ Then $I_{1}\cap I_{2}=\emptyset$ .

   2. (5.2) $m_{1}=m_{2}, f_{m_{1},n_{1}}(s_{1}^{*})\neq f_{m_{2},n_{2}}(s_{2}^{*})$ If $\mathord{\downarrow} s_{1}\cap \mathord{\downarrow} s_{2}=\emptyset$ , then $I_{1}\cap I_{2}=\emptyset$ . If $\mathord{\downarrow} s_{1}\cap \mathord{\downarrow} s_{2}\neq \emptyset$ , then $ I_{1}\cap I_{2}=\{(m_{1}, s): s\leq \inf\{s_{1}, s_{2}\}\}$ , which is a Type I ideal.

   3. (5.3) $m_{1}=m_{2}, f_{m_{1},n_{1}}(s_{1}^{*})= f_{m_{2},n_{2}}(s_{2}^{*})$ Then $(m_{1},n_{1})=(m_{2},n_{2}), s_{1}^{*}=s_{2}^{*}$ by the property of i and $f_{m_{1},n_{1}}$ . Thus, $I_{1}\cap I_{2}=\{(m_{1},s): s\leq s_{1}^{*})\}\cup\{(f_{m_{1},n_{1}}(s_{1}^{*}),n): n\leq \min\{n_{1}^{*},n_{2}^{*}\}\}$ , which is a Type $\mathrm{I}\cup \mathrm{II}^{2}$ set.

   4. (5.4) $m_{1}\neq m_{2}, f_{m_{1},n_{1}}(s_{1}^{*})= f_{m_{2},n_{2}}(s_{2}^{*})$ The second equality implies $m_{1}=m_{2}$ , contradicting the first inequality $m_{1}\neq m_{2}$ . Thus, this case does not exist.

6. (6) $f_{m_{2},n_{2}}(s_{2})=f_{m_{1}, n_{1}}(s_{1})$ Then $(m_{1},n_{1})=(m_{2},n_{2}),s_{1}=s_{2}$ by the property of i and $f_{m_{1},n_{1}}$ . This reveals that $I_{1}\cap I_{2}=I_{1}$ , which is a type V ideal.

This covers all possible cases, and we have confirmed that the intersections of two Type V ideals can be a Type I ideal, a Type II ideal, a Type V ideal, Type $\mathrm{I}\cup \mathrm{II}^{2}$ set, or the empty set.

Next, we consider the intersections of Type I $\cup$ $ \mathrm{II}^{1}$ (Type I $\cup$ $ \mathrm{II}^{2}$ , resp.) sets with Type I $\cup$ $ \mathrm{II}^{1}$ , I $\cup$ $ \mathrm{II}^{2}$ sets, Type I, Type II, Type III, Type IV, and Type V ideals.

The results are shown in the following table.

We now explain the aforementioned table.

We only consider the following nontrivial cases.

Intersections of a Type IV ideal and a Type I $\cup \mathrm{II}^{1}$ set

Let $I_{1}=\{(m_{1},s_{1})\}\cup\{(n_{1},n): n\leq k_{1}\}$ for some $(m_{1},n_{1})\in B, s_{1}\in \mathbb{N}^{<\mathbb{N}},k_{1}\in \mathbb{N}$ with $ {\mid}s_{1}{\mid}=1$ , $k_{1}\leq s_{1}$ , and

$I_{2}=\mathord{\downarrow}(f_{m_{2},n_{2}}(s_{2}),\top)=L_{f_{m_{2},n_{2}}(s_{2})}\cup \{(m_{2},s_{2})\}\cup\{(n_{2},n): n\leq s_{2}\}$ for some $(m_{2},n_{2})\in B, s_{2}\in \mathbb{N}^{<\mathbb{N}}$ with $ {\mid}s_{2}{\mid}=1$ .

Then $I_{1}$ is a Type I $\cup \mathrm{II}^{1}$ set and $I_{2}$ is a type IV ideal.

We prove this by considering the following cases for $f_{m_{2},n_{2}}(s_{2})$ .

1. (1) $f_{m_{2},n_{2}}(s_{2})<m_{1}$ Then $I_{1}\cap I_{2}=\emptyset$ .

2. (2) $f_{m_{2},n_{2}}(s_{2})=m_{1}$ Then $I_{1}\cap I_{2}=\{(m_{1},s_{1})\}$ , which is a Type I ideal.

3. (3) $m_{1}<f_{m_{2},n_{2}}(s_{2})<n_{1}$

   1. (3.1) $m_{2}\neq m_{1}$ Then $I_{1}\cap I_{2}=\emptyset$ .

   2. (3.2) $m_{2}= m_{1}$ If $s_{2}\neq s_{1}$ , then $I_{1}\cap I_{2}=\emptyset$ . If $s_{2}=s_{1}$ , then $I_{1}\cap I_{2}=\{(m_{1},s_{1})\}$ , which is a Type I ideal.

4. (4) $f_{m_{2},n_{2}}(s_{2})=n_{1}$

   1. (4.1) $m_{2}\neq m_{1}$ Then $I_{1}\cap I_{2}=\{(n_{1},n): n\leq k_{1}\}$ , which is a Type II ideal.

   2. (4.2) $m_{2}=m_{1}$ If $s_{2}\neq s_{1}$ , then we have the same result as in case $m_{2}\neq m_{1}$ .If $s_{2}=s_{1}$ , then $I_{1}\cap I_{2}=I_{1}$ , which is a Type I $\cup$ $\mathrm{II}^{1}$ set.

5. (5) $f_{m_{2},n_{2}}(s_{2})>n_{1}$ If $m_{2}\neq m_{1}$ , $n_{2}\neq n_{1}$ , then $I_{1}\cap I_{2}=\emptyset$ .If $m_{2}\neq m_{1}$ , $n_{2}=n_{1}$ , then $I_{1}\cap I_{2}=\{(n_{2},n): n\leq \min\{k_{1},s_{2}\}\}$ , which is a Type II ideal. If $m_{2}=m_{1}$ , $n_{2}\neq n_{1}$ , and $s_{2}\neq s_{1}$ , then $I_{1}\cap I_{2}=\emptyset$ .If $m_{2}=m_{1}$ , $n_{2}\neq n_{1}$ , and $s_{2}= s_{1}$ , then $I_{1}\cap I_{2}=\{(m_{2},s_{2})\}$ , which is a Type I ideal. If $m_{2}=m_{1}$ , $n_{2}=n_{1}$ , and $s_{2}\neq s_{1}$ , then $I_{1}\cap I_{2}=\{(n_{2},n):n\leq \min\{k_{1},s_{2}\}\}$ , which is a Type II ideal. If $m_{2}=m_{1}$ , $n_{2}=n_{1}$ , and $s_{2}= s_{1}$ , then $I_{1}\cap I_{2}=\{(m_{2},s_{2})\}\cup \{(n_{2},n): n\leq \min\{k_{1},s_{2}\}\}$ , which is a Type I $\cup$ $\mathrm{II}^{1}$ set.

These cover all possible cases and we are done.

Intersections of a Type V ideal and a Type I $\cup \mathrm{II}^{1}$ set

Let $I_{1}=\{(m_{1},s_{1})\}\cup\{(n_{1},n): n\leq k_{1}\}$ for some $(m_{1},n_{1})\in B, s_{1}\in \mathbb{N}^{<\mathbb{N}},k_{1}\in \mathbb{N}$ with $ {\mid}s_{1}{\mid}=1$ , $k_{1}\leq s_{1}$ , and

$I_{2}=\mathord{\downarrow}(f_{m_{2},n_{2}}(s_{2}),\top)=L_{f_{m_{2},n_{2}}(s_{2})}\cup \{(m_{2},s): s\leq s_{2}\}\cup\{(f_{m_{2},n_{2}}(s_{2}^{*}),n): n\leq n_{2}^{*}\}$ for some $(m_{2},n_{2})\in B, s_{2}=s_{2}^{*}.n_{2}^{*}\in \mathbb{N}^{<\mathbb{N}}$ with $s_{2}^{*}\in \mathbb{N}^{<\mathbb{N}},n_{2}^{*}\in \mathbb{N}$ .

Then $I_{1}$ is a Type I $\cup \mathrm{II}^{1}$ set and $I_{2}$ is a type V ideal.

Note that in this case, $s_{1}\neq s_{2}$ because $|s_1|=1$ and $|s_2| >1$ .

We prove this by considering the following cases for $f_{m_{2},n_{2}}(s_{2})$ .

1. (1) $f_{m_{2},n_{2}}(s_{2})<m_{1}$ Then $I_{1}\cap I_{2}=\emptyset$ .

2. (2) $f_{m_{2},n_{2}}(s_{2})=m_{1}$ Then $I_{1}\cap I_{2}=\{(m_{1},s_{1})\}$ , which is a Type I ideal.

3. (3) $m_{1}<f_{m_{2},n_{2}}(s_{2})<n_{1}$

   1. (3.1) $m_{2}\neq m_{1}$ Then $I_{1}\cap I_{2}=\emptyset$ .

   2. (3.2) $m_{2}= m_{1}$ If $s_{1}\nleq s_{2}$ , then $I_{1}\cap I_{2}=\emptyset$ .If $s_{1}\leq s_{2}$ , then $I_{1}\cap I_{2}=\{(m_{1}, s_{1})\}$ , which is a Type I ideal.

4. (4) $f_{m_{2},n_{2}}(s_{2})=n_{1}$

   1. (4.1) $m_{2}\neq m_{1}$ Then $I_{1}\cap I_{2}=\{(n_{1},n): n\leq k_{1}\}$ , which is a Type II ideal.

   2. (4.2) $m_{2}=m_{1}$ If $s_{1}\nleq s_{2}$ , then we have the same result as in case $m_{2}\neq m_{1}$ . If $s_{1}\leq s_{2}$ , then $I_{1}\cap I_{2}=I_{1}$ , which is a Type I $\cup$ $\mathrm{II}^{1}$ set.

5. (5) $f_{m_{2},n_{2}}(s_{2})>n_{1}$

   1. (5.1) $m_{2}\neq m_{1}$ If $f_{m_{2},n_{2}}(s_{2}^{*})\neq n_{1}$ , then $I_{1}\cap I_{2}=\emptyset$ .If $f_{m_{2},n_{2}}(s_{2}^{*})=n_{1}$ , then $I_{1}\cap I_{2}=\{(f_{m_{2},n_{2}}(s_{2}^{*}),n): n\leq \min\{k_{1},n_{2}^{*}\}\}$ , which is a Type II ideal.

   2. (5.2) $m_{2}=m_{1}$ If $f_{m_{2},n_{2}}(s_{2}^{*})\neq n_{1}$ and $s_{1}\nleq s_{2}$ , then $I_{1}\cap I_{2}=\emptyset$ .If $f_{m_{2},n_{2}}(s_{2}^{*})\neq n_{1}$ and $s_{1}\leq s_{2}$ , then $I_{1}\cap I_{2}=\{(m_{1},s_{1})\}$ , which is a Type I ideal. $f_{m_{2},n_{2}}(s_{2}^{*})=n_{1}$ and $s_{1}\nleq s_{2}$ , then $I_{1}\cap I_{2}=\{(f_{m_{2},n_{2}}(s_{2}^{*}),n): n\leq \min\{k_{1},n_{2}^{*}\}\}$ , which is a Type II ideal. If $f_{m_{2},n_{2}}(s_{2}^{*})=n_{1}$ and $s_{1}\leq s_{2}$ , then $I_{1}\cap I_{2}=\{(m_{1},s_{1})\}\cup \{(f_{m_{2},n_{2}}(s_{2}^{*}),n): n\leq \min\{k_{1},n_{2}^{*}\}\}$ , which is a Type I $\cup$ $\mathrm{II}^{1}$ set.

These cover all possible cases and we are done.

Intersections of two Type I $\cup \mathrm{II}^{1}$ sets

Let $I_{1}=\{(m_{1},s_{1})\}\cup\{(n_{1},n): n\leq k_{1}\}$ for some $(m_{1},n_{1})\in B, s_{1}\in \mathbb{N}^{<\mathbb{N}},k_{1}\in \mathbb{N}$ with $ {\mid}s_{1}{\mid}=1$ , $k_{1}\leq s_{1}$ , and

$I_{2}=\{(m_{2},s_{2})\}\cup\{(n_{2},n): n\leq k_{2}\}$ for some $(m_{2},n_{2})\in B, s_{2}\in \mathbb{N}^{<\mathbb{N}},k_{2}\in \mathbb{N}$ with $ {\mid}s_{2}{\mid}=1$ , $k_{2}\leq s_{2}$ .

Then $I_{1}, I_{2}$ are both Type I $\cup \mathrm{II}^{1}$ sets.

We prove this by considering the following cases.

1. (1) $m_{2}\neq m_{1}$ , $n_{2}\neq n_{1}$ Then $I_{1}\cap I_{2}=\emptyset$ .

2. (2) $m_{2}\neq m_{1}$ , $n_{2}=n_{1}$ Then $I_{1}\cap I_{2}=\{(n_{2},n): n\leq \min\{k_{1},k_{2}\}\}$ , which is a Type II ideal.

3. (3) $m_{2}=m_{1}$ , $n_{2}\neq n_{1}$ , and $s_{2}\neq s_{1}$ Then $I_{1}\cap I_{2}=\emptyset$ .

4. (4) $m_{2}=m_{1}$ , $n_{2}\neq n_{1}$ , and $s_{2}= s_{1}$ Then $I_{1}\cap I_{2}=\{(m_{2},s_{2})\}$ , which is a Type I ideal.

5. (5) $m_{2}=m_{1}$ , $n_{2}=n_{1}$ , and $s_{2}\neq s_{1}$ then $I_{1}\cap I_{2}=\{(n_{2},n): n\leq \min\{k_{1},k_{2}\}\}$ , which is a Type II ideal.

6. (6) $m_{2}=m_{1}$ , $n_{2}=n_{1}$ , and $s_{2}= s_{1}$ then $I_{1}\cap I_{2}=\{(m_{2},s_{2})\}\cup \{(n_{2},n): n\leq \min\{k_{1},k_{2}\}\}$ , which is a Type I $\cup$ $\mathrm{II}^{1}$ set.

These cover all possible cases and we are done.

Intersections of a Type I $\cup \mathrm{II}^{1}$ set and a I $\cup \mathrm{II}^{2}$ set

Let $I_{1}=\{(m_{1},s_{1})\}\cup\{(n_{1},n): n\leq k_{1}\}$ for some $(m_{1},n_{1})\in B, s_{1}\in \mathbb{N}^{<\mathbb{N}},k_{1}\in \mathbb{N}$ with $ {\mid}s_{1}{\mid}=1$ , $k_{1}\leq s_{1}$ , and

$I_{2}=\{(m_{2},s):s\leq s_{2}\}\cup\{(f_{m_{2},n_{2}}(s_{2}),n): n\leq k_{2}\}$ for some $(m_{2},n_{2})\in B, s_{2}\in \mathbb{N}^{<\mathbb{N}},k_{2}\in \mathbb{N}$ .

Then $I_{1}$ is a Type I $\cup \mathrm{II}^{1}$ set and $I_{2}$ is a Type I $\cup \mathrm{II}^{2}$ set.

We prove this by considering the following cases.

1. (1) $m_{2}\neq m_{1}$ If $f_{m_{2},n_{2}}(s_{2})\neq n_{1}$ , then $I_{1}\cap I_{2}=\emptyset$ .If $f_{m_{2},n_{2}}(s_{2})=n_{1}$ , then $I_{1}\cap I_{2}=\{(f_{m_{2},n_{2}}(s_{2}),n): n\leq \min\{k_{1},k_{2}\}\}$ , which is a Type II ideal.

2. (2) $m_{2}=m_{1}$

   1. (2.1) $f_{m_{2},n_{2}}(s_{2})\neq n_{1}$ If $s_{1}\nleq s_{2}$ , then $I_{1}\cap I_{2}=\emptyset$ . If $s_{1}\leq s_{2}$ , then $I_{1}\cap I_{2}=\{(m_{1},s_{1})\}$ , which is a Type I ideal.

   2. (2.2) $f_{m_{2},n_{2}}(s_{2})=n_{1}$ If $s_{1}\nleq s_{2}$ , then $I_{1}\cap I_{2}=\{(f_{m_{2},n_{2}}(s_{2}),n): n\leq \min\{k_{1},k_{2}\}\}$ , which is a Type II ideal. If $s_{1}\leq s_{2}$ , then $I_{1}\cap I_{2}=\{(m_{1},s_{1})\}\cup \{(n_{1},n): n\leq \min\{k_{1},k_{2}\}\}$ , which is a Type I $\cup$ $\mathrm{II}^{1}$ set.

These cover all possible cases and we are done.

Intersections of a Type IV ideal and a Type I $\cup \mathrm{II}^{2}$ set

Let $I_{1}=\{(m_{1},s): s\leq s_{1})\}\cup\{(f_{m_{1},n_{1}}(s_{1}),n): n\leq k_{1}\}$ for some $(m_{1},n_{1})\in B, s_{1}\in \mathbb{N}^{<\mathbb{N}}$ with $k_{1}\in \mathbb{N}$ , and

$I_{2}=\mathord{\downarrow}(f_{m_{2},n_{2}}(s_{2}),\top)=L_{f_{m_{2},n_{2}}(s_{2})}\cup \{(m_{2},s_{2})\}\cup\{(n_{2},n): n\leq s_{2}\}$ for some $(m_{2},n_{2})\in B, s_{2}\in \mathbb{N}^{<\mathbb{N}}$ with $ {\mid}s_{2}{\mid}=1$ .

Then $I_{1}$ is a Type I $\cup \mathrm{II}^{2}$ set and $I_{2}$ is a Type IV ideal.

We prove this by considering the following cases for $f_{m_{2},n_{2}}(s_{2})$ .

1. (1) $f_{m_{2},n_{2}}(s_{2})<m_{1}$ Then $I_{1}\cap I_{2}=\emptyset$ .

2. (2) $f_{m_{2},n_{2}}(s_{2})=m_{1}$ Then $I_{1}\cap I_{2}=\{(m_{1},s):s\leq s_{1}\}$ , which is a Type I ideal.

3. (3) $m_{1}<f_{m_{2},n_{2}}(s_{2})<f_{m_{1},n_{1}}(s_{1})$

   1. (3.1) $m_{2}\neq m_{1}$ Then $I_{1}\cap I_{2}=\emptyset$ .

   2. (3.2) $m_{2}= m_{1}$ If $s_{2}\nleq s_{1}$ , then $I_{1}\cap I_{2}=\emptyset$ . If $s_{2}\leq s_{1}$ , then $I_{1}\cap I_{2}=\{(m_{2},s_{2})\}$ , which is a Type I ideal.

4. (4) $f_{m_{2},n_{2}}(s_{2})=f_{m_{1},n_{1}}(s_{1})$ Then $(m_{2},n_{2})=(m_{1},n_{1})$ , $s_{1}=s_{2}$ by the property of i and $f_{m_{1},n_{1}}$ .So $I_{1}\cap I_{2}=I_{1}$ , which is a Type I $\cup$ $\mathrm{II}^{2}$ set.

5. (5) $f_{m_{2},n_{2}}(s_{2})>f_{m_{1},n_{1}}(s_{1})$

   1. (5.1) $m_{2}\neq m_{1}$ If $n_{2}\neq f_{m_{1},n_{1}}(s_{1})$ , then $I_{1}\cap I_{2}=\emptyset$ . If $n_{2}=f_{m_{1},n_{1}}(s_{1})$ , then $I_{1}\cap I_{2}=\{(n_{2},n): n\leq \min\{k_{1},s_{2}\}\}$ , which is a Type II ideal.

   2. (5.2) $m_{2}=m_{1}$ If $n_{2}\neq f_{m_{1},n_{1}}(s_{1})$ and $s_{2}\nleq s_{1}$ , then $I_{1}\cap I_{2}=\emptyset$ .If $n_{2}\neq f_{m_{1},n_{1}}(s_{1})$ and $s_{2}\leq s_{1}$ , then $I_{1}\cap I_{2}=\{(m_{2},s_{2})\}$ , which is a Type I ideal. If $n_{2}=f_{m_{1},n_{1}}(s_{1})$ and $s_{2}\nleq s_{1}$ , then $I_{1}\cap I_{2}=\{(n_{2},n): n\leq \min\{k_{1},s_{2}\}\}$ , which is a Type II ideal. If $n_{2}=f_{m_{1},n_{1}}(s_{1})$ , and $s_{2}\leq s_{1}$ , then $I_{1}\cap I_{2}=\{(m_{2},s_{2})\}\cup \{(n_{2},n): n\leq \min\{k_{1},s_{2}\}\}$ , which is a Type I $\cup$ $\mathrm{II}^{1}$ set.

These cover all possible cases and we are done.

Intersections of a Type V ideal and a Type I $\cup \mathrm{II}^{2}$ set

Let $I_{1}=\{(m_{1},s): s\leq s_{1})\}\cup\{(f_{m_{1},n_{1}}(s_{1}),n): n\leq k_{1}\}$ for some $(m_{1},n_{1})\in B, s_{1}\in \mathbb{N}^{<\mathbb{N}}$ with $k_{1}\in \mathbb{N}$ , and

$I_{2}=\mathord{\downarrow}(f_{m_{2},n_{2}}(s_{2}),\top)=L_{f_{m_{2},n_{2}}(s_{2})}\cup \{(m_{2},s): s\leq s_{2}\}\cup\{(f_{m_{2},n_{2}}(s_{2}^{*}),n): n\leq n_{2}^{*}\}$ for some $(m_{2},n_{2})\in B, s_{2}=s_{2}^{*}.n_{2}^{*}\in \mathbb{N}^{<\mathbb{N}}$ with $s_{2}^{*}\in \mathbb{N}^{<\mathbb{N}},n_{2}^{*}\in \mathbb{N}$ .

Then $I_{1}$ is a Type I $\cup \mathrm{II}^{2}$ set and $I_{2}$ is a Type V ideal.

We prove this by considering the following cases for $f_{m_{2},n_{2}}(s_{2})$ .

1. (1) $f_{m_{2},n_{2}}(s_{2})<m_{1}$ Then $I_{1}\cap I_{2}=\emptyset$ .

2. (2) $f_{m_{2},n_{2}}(s_{2})=m_{1}$ Then $I_{1}\cap I_{2}=\{(m_{1},s): s\leq s_{1}\}$ , which is a Type I ideal.

3. (3) $m_{1}<f_{m_{2},n_{2}}(s_{2})<f_{m_{1},n_{1}}(s_{1})$

   1. (3.1) $m_{2}\neq m_{1}$ Then $I_{1}\cap I_{2}=\emptyset$ .

   2. (3.2) $m_{2}= m_{1}$ If $\mathord{\downarrow} s_{1}\cap \mathord{\downarrow} s_{2}=\emptyset$ , then $I_{1}\cap I_{2}=\emptyset$ .If $\mathord{\downarrow} s_{1}\cap \mathord{\downarrow} s_{2}\neq \emptyset$ , then $I_{1}\cap I_{2}=\{(m_{1},s): s\leq \inf\{s_{1},s_{2}\}\}$ , which is a Type I ideal.

4. (4) $f_{m_{2},n_{2}}(s_{2})=f_{m_{1},n_{1}}(s_{1})$ Then $(m_{2},n_{2})=(m_{1},n_{1})$ , $s_{1}=s_{2}$ by the property of i and $f_{m_{1},n_{1}}$ . Thus $I_{1}\cap I_{2}=I_{1}$ , which is a Type I $\cup$ $\mathrm{II}^{2}$ set.

5. (5) $f_{m_{2},n_{2}}(s_{2})>f_{m_{1},n_{1}}(s_{1})$

   1. (5.1) $m_{2}\neq m_{1}$ If $f_{m_{2},n_{2}}(s_{2}^{*})\neq f_{m_{1},n_{1}}(s_{1})$ , then $I_{1}\cap I_{2}=\emptyset$ .If $f_{m_{2},n_{2}}(s_{2}^{*})=f_{m_{1},n_{1}}(s_{1})$ , then $(m_{2},n_{2})=(m_{1},n_{1})$ , $s_{2}^{*}=s_{1}$ by the property of i and $f_{m_{1},n_{1}}$ , which contradicts $m_{2}\neq m_{1}$ . Thus, this case does not occur.

   2. (5.2) $m_{2}=m_{1}$ If $f_{m_{2},n_{2}}(s_{2}^{*})\neq f_{m_{1},n_{1}}(s_{1})$ and $\mathord{\downarrow} s_{1}\cap \mathord{\downarrow} s_{2}=\emptyset$ , then $I_{1}\cap I_{2}=\emptyset$ .If $f_{m_{2},n_{2}}(s_{2}^{*})\neq f_{m_{1},n_{1}}(s_{1})$ and $\mathord{\downarrow} s_{1}\cap \mathord{\downarrow} s_{2}\neq \emptyset$ , then $I_{1}\cap I_{2}=\{(m_{1},s): s\leq \inf\{s_{1},s_{2}\}\}$ , which is a Type I ideal. If $f_{m_{2},n_{2}}(s_{2}^{*})=f_{m_{1},n_{1}}(s_{1})$ , then $(m_{2},n_{2})=(m_{1},n_{1})$ , $s_{2}^{*}=s_{1}$ . It follows that $I_{1}\cap I_{2}=\{(m_{1},s): s\leq s_{2}^{*}\}\cup \{(f_{m_{2},n_{2}}(s_{2}^{*}),n): n\leq \min\{k_{1},n_{2}^{*}\}\}$ , which is a Type I $\cup$ $\mathrm{II}^{2}$ set.

These cover all possible cases and we are done.

Intersections of two Type I $\cup \mathrm{II}^{2}$ sets

Let $I_{1}=\{(m_{1},s): s\leq s_{1}\}\cup\{(f_{m_{1},n_{1}}(s_{1}),n): n\leq k_{1}\}$ for some $(m_{1},n_{1})\in B, s_{1}\in \mathbb{N}^{<\mathbb{N}}$ with $k_{1}\in \mathbb{N}$ , and

$I_{2}=\{(m_{2},s): s\leq s_{2}\}\cup\{(f_{m_{2},n_{2}}(s_{2}),n): n\leq k_{2}\}$ for some $(m_{2},n_{2})\in B, s_{2}\in \mathbb{N}^{<\mathbb{N}}$ with $k_{2}\in \mathbb{N}$ .

Then $I_{1}, I_{2}$ are both Type I $\cup \mathrm{II}^{2}$ sets. We prove this by considering the following cases.

1. (1) $m_{2}\neq m_{1}$ , $f_{m_{2},n_{2}}(s_{2})\neq f_{m_{1},n_{1}}(s_{1})$ Then $I_{1}\cap I_{2}=\emptyset$ .

2. (2) $m_{2}\neq m_{1}$ , $f_{m_{2},n_{2}}(s_{2})= f_{m_{1},n_{1}}(s_{1})$ Then $(m_{2},n_{2})=(m_{1},n_{1})$ , $s_{2}=s_{1}$ , which contradicts the assumption that $m_{2}\neq m_{1}$ . Hence, this case does not occur.

3. (3) $m_{2}=m_{1}$ , $f_{m_{2},n_{2}}(s_{2})\neq f_{m_{1},n_{1}}(s_{1})$ , and $\mathord{\downarrow} s_{2}\cap \mathord{\downarrow} s_{1}= \emptyset$ Then $I_{1}\cap I_{2}=\emptyset$ .

4. (4) $m_{2}=m_{1}$ , $f_{m_{2},n_{2}}(s_{2})\neq f_{m_{1},n_{1}}(s_{1})$ , and $\mathord{\downarrow} s_{2}\cap \mathord{\downarrow} s_{1}\neq \emptyset$ Then $I_{1}\cap I_{2}=\{(m_{2},s):s\leq \min\{s_{1},s_{2}\}\}$ , which is a Type I ideal.

5. (5) $m_{2}=m_{1}$ , $f_{m_{2},n_{2}}(s_{2})= f_{m_{1},n_{1}}(s_{1})$ , then $(m_{2},n_{2})=(m_{1},n_{1})$ , $s_{2}=s_{1}$ . It follows that $I_{1}\cap I_{2}=\{(m_{2},s): s\leq s_{2}\}\cup \{(f_{m_{2},n_{2}}(s_{2}),n): n\leq \min\{k_{1},k_{2}\}\}$ , which is a Type I $\cup$ $\mathrm{II}^{2}$ set.

These cover all possible cases and we are done.

It follows that the intersection of finite number of principle ideals of P is either a principle ideal, or a Type I $\cup \mathrm{II}^{1}$ set, or a Type I $\cup \mathrm{II}^{2}$ set, or the empty set.

Let

$$\Psi=\{\mbox{all principle ideals of} ~P\}\cup\{\mbox{all Type} ~\mathrm{I}\cup \mathrm{II}^{1}~ \mbox{sets of } P\}\cup\{\mbox{all Type} ~\mathrm{I}\cup \mathrm{II}^{2}~ \mbox{sets of } P\}\cup\{\emptyset\}.$$

By the above arguments, we deduce that $\Psi$ is closed under finite non-nullary intersections. In addition, since all elements of $\Psi$ are principal ideals or finite subsets of P, and P has no infinite decreasing chains, we have that $\Psi$ does not contain an infinite decreasing chain.

For any nonempty $E\subseteq P$ ,

$$\bigcap\{\mathord{\downarrow} x: x\in E\} =\bigcap\{\bigcap\{\mathord{\downarrow} x: x\in F\}: F \mbox{ is a finite subset of } E\}.$$

Each $\bigcap\{\mathord{\downarrow} x: x\in F\}$ is a member of $\Psi$ , so $\{\bigcap\{\mathord{\downarrow} x: x\in F\}: F \mbox{ is a finite subset of } E\}$ is a filter of members of $\Psi$ . Since $\Psi$ does not have infinite decreasing chains, this family must have a smallest member, which equals $\bigcap\{\mathord{\downarrow} x: x\in E\}$ and is in $\Psi$ . Therefore, $M=\Psi$ .

We now show that every directed subset of M has a supremum in M.

Let $\mathcal{D}=\{I_{i}:i\in I\}$ be a directed subset of M. If $\mathcal{D}$ has a largest member, then its supremum exists and equals the largest member. Now we assume that $\mathcal{D}$ does not have a largest member. Then $\mathcal{D}$ contains an infinite chain. Since Type III ideals, Type IV ideals and Type V ideals are maximal elements of M, we deduce that none of the $I_i$ is a Type III ideal, Type IV ideal, and Type V ideal.

We consider the following two remaining cases.

Case 1: All members of $\{I_{i}: i\in I\}$ are Type I ideals or Type II ideals.

Then, as a Type I ideal and a Type II ideal are not comparable, it follows that either all $I_{i}$ are Type I ideals or they are all Type II ideals. Thus, $\sup_{i\in I}I_{i}$ exists because P is a dcpo.

Case 2: There exists $i_{0}\in I$ such that $I_{i_{0}}$ is a Type $\mathrm{I}\cup \mathrm{II}^{1}$ set or a Type $\mathrm{I}\cup \mathrm{II}^{2}$ set.

Then all members of $\{I_{i}:I_{i}\geq I_{i_{0}}\}$ are either Type $\mathrm{I}\cup \mathrm{II}^{1}$ sets or Type $\mathrm{I}\cup \mathrm{II}^{2}$ sets. Note that there exist no infinite increasing chains consisting of Type $\mathrm{I}\cup \mathrm{II}^{1}$ sets. Hence, there exists $i_{1}$ such that $\mathcal{D}_{1}=\{I_{i}: I_{i}\ge I_{i_{1}}\}$ consists of only Type $\mathrm{I}\cup \mathrm{II}^{2}$ sets. In order to prove $\sup\mathcal{D}$ exists, it is enough to prove $\sup\mathcal{D}_1$ exists.

Let $\mathcal{D}_{1}=\{I_{j}: j\in J\}$ , where J is a subset of I.

For each $j\in J$ , let $I_{j}=\{(m_{j}, s): s\leq s_{j}\}\cup\{(f_{m_{j},n_{j}}(s_{j}),n): n\leq k_{j}\}$ for some $(m_{j},n_{j})\in B, s_{j}\in \mathbb{N}^{<\mathbb{N}},k_{j}\in \mathbb{N}$ .

Then $f_{m_{i}, n_{i}}(s_{i})=f_{m_{j},n_{j}}(s_{j})$ for all $i, j\in J$ , which implies that $(m_{i}, n_{i})=(m_{j}, n_{j})$ and $ s_{i}=s_{j}$ . Let $m_{j}=m_{0}$ , $n_{j}=n_{0}$ and $s_{j}=s_{0} ~(j\in J)$ . Then, $I_{j}=\{(m_{0}, s): s\leq s_{0}\}\cup\{(f_{m_{0}, n_{0}}(s_{0}),n): n\leq k_{j}\}$ . Since $\mathcal{D}$ does not have a largest member, $\mathcal{D}_{1}$ has no a largest member either. It follows that the set $\{k_j: j\in J\}$ does not have a largest element.

We claim that $\sup \mathcal{D}_1=\mathord{\downarrow} (f_{m_{0},n_{0}}(s_{0}),\top)$ .

Clearly, $\mathord{\downarrow} (f_{m_{0},n_{0}}(s_{0}),\top)$ is an upper bound of $\mathcal{D}_1$ . For any upper bound I of $\mathcal{D}_1$ in M, we have that $\bigcup\{\{(f_{m_{0},n_{0}}(s_{0}),n): n\leq k_{j}\}: j\in J\}\subseteq I$ . Note that I is a Scott closed set of P, and $\bigcup\{\{(f_{m_{0},n_{0}}(s_{0}),n): n\leq k_{j}\}: j\in J\}$ is a directed subset of P. Therefore, $\sup \bigcup\{\{(f_{m_{0},n_{0}}(s_{0}),n): n\leq k_{j}\}: j\in J\}=(f_{m_{0},n_{0}}(s_{0}),\top)\in I$ .

All the above together show that $\sup \mathcal{D}=\sup\mathcal{D}_1=\mathord{\downarrow} (f_{m_{0},n_{0}}(s_{0}),\top)$ . Thus, M is a dcpo.

Define the mapping $g: P\rightarrow M$ by $g(x)=\mathord{\downarrow} x$ for all $x\in P$ . By the last part of the proof of Lemma 3.3, we see easily that g is Scott continuous. By Lemma 3.2, P is irreducible; thus, g(P) is irreducible. Clearly, the closure of g(P) equals M, so M is irreducible (with respect to the Scott topology). As M does not have a top element, it is not the closure of a singleton set. It follows that $\Sigma M$ is not sober.

Let $\hat{M}=M\cup\{\mbox{T}\}$ be the poset obtained by adding a top element $\mbox{T}$ to M. Then $\hat{M}$ is a countable non-sober complete lattice.

Note that the non-sober complete lattice $\hat{M}$ constructed above is not distributive. Thus, it remains to know whether there is a distributive countable non-sober complete lattice.

By applying the Lemma 2.3, we give a negative answer to the above problem.

Define $g: \hat{M} \rightarrow \mathcal{F}$ by $g(x) = \mathord{\downarrow} x$ and $f : \mathcal{F} \rightarrow \hat{M}$ by $f(A) = \sup A$ for any $A \in \mathcal{F}$ . Clearly, both f and g are monotone. In addition, for any $x\in \hat{M}$ and $\mathord{\downarrow} F\in\mathcal{F}$ , we have

$$f(\mathord{\downarrow} F)\leq x \mbox{ iff } \bigvee F\le x \mbox{ iff } \mathord{\downarrow} F\subset \mathord{\downarrow} x=g(x).$$

Hence, f is the left adjoint of g. Then f preserves all suprema, in particular the suprema of directed subsets. Hence, it is Scott continuous. Also by the structure of $\hat{M}$ and $\mathcal{F}$ , we easily see that g also preserves the suprema of directed subsets; hence, it is also Scott continuous. Note that $f\circ g=id_{\hat{M}}$ . It thus follows that $\Sigma \hat{M}$ is a retraction of $\Sigma \mathcal{F}$ . Since $\Sigma \hat{M}$ is non-sober, thus $\Sigma \mathcal{F}$ is non-sober.

4. A Sufficient Condition for Complete Lattices to be Sober

In this section, we prove some positive results on the sobriety of Scott spaces of dcpos. One immediate corollary is that every complete lattice L with Id(L) countable has a sober Scott space.

From the above section, we know that a countable complete lattice may not be sober in the Scott topology. In this section, we show that the Scott space of a complete lattice with Id(P) countable is sober.

The following lemma is critical for our later discussions.

For $n=1$ , $A_{1}=\{a_{1}\},B_{1}=\{b_{1}\}$ .

For $n=2$ , we define $A_{2}$ and $B_{2}$ below:

If $\sup I_{1}^{P}\in \mathord{\uparrow} A_{1}$ , then $(\sup I_{1}^{P},b_{1})\in U$ . It follows that there exists $d_{1}^{P}\in I_{1}^{P}$ such that $(d_{1}^{P},b_{1})\in U$ by the Scott openness of U. Let $A_{2}=\{d_{1}^{P}\}$ in this case, and $A_{2}=\emptyset$ otherwise. Note that $(A_{1}\cup A_{2})\times B_{1}\subseteq U$ .

If $\sup I_{1}^{Q}\in \mathord{\uparrow} B_{1}$ , we have $(A_{1}\cup A_{2})\times \{\sup I_{1}^{Q}\}\subseteq U$ . For each $a\in A_{1}\cup A_{2}$ , we can choose a $d_a$ form $I_1^Q$ satisfying $(a,d_a)\in U$ by the Scott openness of U. Since $A_1\cup A_2$ is finite and $I_1^Q$ is directed, there exists $d_{1}^{Q}\in I_{1}^{Q}$ such that $(A_{1}\cup A_{2})\times \{d_{1}^{Q}\}\subseteq U$ . Let $B_{2}=\{d_{1}^{Q}\}$ in this case, and $B_{2}=\emptyset$ otherwise. We conclude that $(A_{1}\cup A_{2})\times (B_{1}\cup B_{2})\subseteq U$ .

For $n=3$ , we first consider the two index sets:

\begin{align*} E_{1} & =\Big\{i\in\{1\}\mid \sup I_{i}^{P}\notin \mathord{\uparrow} A_{1} ~ {\rm and}~ \sup I_{i}^{P}\in \mathord{\uparrow} A_{2}\Big\}\cup\Big\{i\in \{2\}\mid \sup I_{i}^{P}\in \mathord{\uparrow} (A_{1}\cup A_{2})\Big\}~ {\rm and} \\F_{1} & =\Big\{i\in\{1\}\mid \sup I_{i}^{Q} \notin \mathord{\uparrow} B_{1} ~ {\rm and}~ \sup I_{i}^{Q}\in \mathord{\uparrow} B_{2}\Big\}\cup\Big\{i\in \{2\}\mid \sup I_{i}^{Q}\in \mathord{\uparrow} (B_{1}\cup B_{2})\Big\}.\end{align*}

However, if $\sup I_{1}^{P} \notin \mathord{\uparrow} A_{1}$ and $\sup I_{1}^{Q} \notin \mathord{\uparrow} B_{1}$ , then $A_2=\emptyset$ and $B_2=\emptyset$ from the above step. In this way, $\Big\{i\in\{1\}\mid \sup I_{i}^{P} \notin \mathord{\uparrow} A_{1} $ and $ \sup I_{i}^{P}\in \mathord{\uparrow} A_{2}\Big\}$ and $\Big\{i\in\{1\}\mid \sup I_{i}^{Q} \notin \mathord{\uparrow} B_{1} $ and $\sup I_{i}^{Q}\in \mathord{\uparrow} B_{2}\Big\}$ must be empty.

Next, we define $A_{3}$ and $B_{3}$ in the similar way as before.

If $E_1\neq \emptyset$ , then $E_1=\{2\}$ . We have $\{\sup I_{2}^{P}\}\times (B_{1}\cup B_{2})\subseteq U$ . Through the similar discussion process, we can deduce that there exists $d_{2}^{P}\in I_{2}^{P}$ such that $\{d_{2}^{P}\}\times (B_{1}\cup B_{2})\subseteq U$ because $B_{1}\cup B_{2}$ is finite and $I_{2}^{P}$ is directed. Let $A_{3}=\{d_{2}^{P}\}$ in this case, and $A_{3}=\emptyset$ otherwise. Note that $(A_{1}\cup A_{2}\cup A_{3})\times (B_{1}\cup B_{2})\subseteq U$ .

If $F_{1}\neq \emptyset$ , then $F_{1}=\{2\}$ . Thus, $(A_{1}\cup A_{2}\cup A_{3})\times \{\sup I_{2}^{Q}\}\subseteq U$ . Note that $A_{1}\cup A_{2}\cup A_{3}$ is a finite set. It follows that there exists $d_{2}^{Q}\in I_{2}^{Q}$ such that $(A_{1}\cup A_{2}\cup A_{3})\times \{d_{2}^{Q}\}\subseteq U$ . Let $B_{3}=\{d_{2}^{Q}\}$ in this case, and $B_{3}=\emptyset$ otherwise. We conclude that $(A_{1}\cup A_{2}\cup A_{3})\times (B_{1}\cup B_{2}\cup B_{3})\subseteq U$ .

For $n=4$ , we also consider the two index sets:

\begin{align*} E_{2} & =\Big\{i\in\{1,2\}\mid \sup I_{i}^{P}\notin \bigcup_{k=1}^{i}\mathord{\uparrow} A_{k} ~ {\rm and}~ \sup I_{i}^{P}\in \bigcup_{k=i+1}^{3}\mathord{\uparrow} A_{k}\Big\}\cup\Big\{i\in \{3\}\mid \sup I_{i}^{P}\in \bigcup_{k=1}^{3}\mathord{\uparrow} A_{k}\Big\}, \\ F_{2} & =\Big\{i\in\{1,2\}\mid \sup I_{i}^{Q} \notin \bigcup_{k=1}^{i}\mathord{\uparrow} B_{k}~ {\rm and}~ \sup I_{i}^{Q}\in \bigcup_{k=i+1}^{3}\mathord{\uparrow} B_{k}\Big\}\cup\Big\{i\in \{3\}\mid \sup I_{i}^{Q}\in \bigcup_{k=1}^{3}\mathord{\uparrow} B_{k}\Big\}. \end{align*}

Next, we define $A_{4}$ and $B_{4}$ in the following:

If $E_2\neq \emptyset$ , then $i\in \{1,2\}$ implies $\sup I_{i}^{P}\in \bigcup_{k=i+1}^{3}\mathord{\uparrow} A_{k}\subseteq \bigcup_{k=1}^{3}\mathord{\uparrow} A_{k}$ , and $i=3$ implies $\sup I_{i}^{P}\in \bigcup_{k=1}^{3}\mathord{\uparrow} A_{k}$ . Thus, $\sup I_{i}^{P}\in \bigcup_{k=1}^{3}\mathord{\uparrow} A_{k}$ for all $i\in E_2$ . So for each $i\in E_2$ , $\{\sup I_{i}^{P}\}\times (B_{1}\cup B_{2}\cup B_3)\subseteq U$ implies that there exists $d_{i}^{P}\in I_{i}^{P}$ such that $\{d_{i}^{P}\}\times (B_{1}\cup B_{2}\cup B_3)\subseteq U$ because $B_{1}\cup B_{2}\cup B_3$ is finite and $I_{i}^{P}$ is directed. Let $A_{4}=\{d_{i}^{P}\mid i\in E_2\}$ in this case, and $A_{4}=\emptyset$ otherwise. Note that

$$\Big(\bigcup_{k=1}^{4}\mathord{\uparrow} A_k\Big)\times \Big(\bigcup_{k=1}^{3}\mathord{\uparrow} B_k\Big)\subseteq U.$$

If $F_{2}\neq \emptyset$ , then $i\in \{1,2\}$ implies $\sup I_{i}^{Q}\in \bigcup_{k=i+1}^{3}\mathord{\uparrow} B_{k}\subseteq \bigcup_{k=1}^{3}\mathord{\uparrow} B_{k}$ , and $i=3$ implies $\sup I_{i}^{Q}\in \bigcup_{k=1}^{3}\mathord{\uparrow} B_{k}$ . Thus, $\sup I_{i}^{Q}\in \bigcup_{k=1}^{3}\mathord{\uparrow} B_{k}$ for all $i\in F_2$ . So for each $i\in F_2$ , $(\bigcup_{k=1}^{4}A_{k})\times \{\sup I_{i}^{Q}\}\subseteq U$ implies that there exists $d_{i}^{Q}\in I_{i}^{Q}$ such that $(\bigcup_{k=1}^{4}A_{k})\times \{d_{i}^{Q}\}\subseteq U$ since $\bigcup_{k=1}^{4}A_{k}$ is a finite set and $I_{i}^{Q}$ is directed. Let $B_{4}=\{d_{i}^{Q}\mid i\in F_2\}$ in this case, and $B_{4}=\emptyset$ otherwise. We conclude that

$$\Big(\bigcup_{k=1}^{4}A_{k}\Big)\times \Big(\bigcup_{k=1}^{4}B_{k}\Big)\subseteq U.$$

For $n\geq 4$ , we assume that

$$\Big(\bigcup_{k=1}^{n-1}A_k \Big)\times \Big(\bigcup_{k=1}^{n-1} B_k \Big)\subseteq U.$$

Then, we define $A_{n}$ and $B_{n}$ inductively.

We first consider the following two index sets:

\begin{align*} E_{n-2} = &\Big\{i\in\{1,\ldots, n-2\}\mid \sup I_{i}^{P} \notin \bigcup_{k=1}^{i}\mathord{\uparrow} A_{k} ~{\rm and}~\sup I_{i}^{P}\in \bigcup_{k=i+1}^{n-1}\mathord{\uparrow} A_{k}\Big\}\\ &\cup\Big\{i\in \{n-1\}\mid \sup I_{i}^{P}\in \bigcup_{k=1}^{n-1}\mathord{\uparrow} A_{k}\Big\},\\ F_{n-2} =& \Big\{i \in \{1,\ldots, n-2\}\mid \sup I_{i}^{Q} \notin \bigcup_{k=1}^{i} \mathord{\uparrow} B_{k}~{\rm and}~\sup I_{i}^{Q}\in \bigcup_{k=i+1}^{n-1} \mathord{\uparrow} B_{k}\Big\}\\ &\cup\Big\{i\in \{n-1\}\mid \sup I_{i}^{Q}\in \bigcup_{k=1}^{n-1} \mathord{\uparrow} B_{k}\Big\}. \end{align*}

Note that $\Big\{i\in\{1,\ldots, n-2\}\mid \sup I_{i}^{P} \notin \bigcup_{k=1}^{i}\mathord{\uparrow} A_{k} ~{\rm and}~\sup I_{i}^{P}\in \bigcup_{k=i+1}^{n-1}\mathord{\uparrow} A_{k}\Big\}$ and $\Big\{i \in \{1,\ldots, n-2\}\mid \sup I_{i}^{Q} \notin \bigcup_{k=1}^{i} \mathord{\uparrow} B_{k}~{\rm and}~\sup I_{i}^{Q}\in \bigcup_{k=i+1}^{n-1} \mathord{\uparrow} B_{k}\Big\}$ may not be empty.

If $E_{n-2}\neq \emptyset$ , similarly, we can deduce { $\sup I_{i}^{P}\}\times \big(\bigcup_{k=1}^{n-1} B_{k}\big)\subseteq U$ for any $i\in E_{n-2}$ . Note that $\bigcup_{k=1}^{n-1} B_{k}$ is a finite set and each $I_{i}^{P}$ is directed. Thus, there exists $d_{i}^{P}\in I_{i}^{P}$ such that $\{d_{i}^{P}\}\times \big(\bigcup_{k=1}^{n-1} B_{k}\big)\subseteq U$ for any $i\in E_{n-2}$ . Let $A_{n}=\{d_{i}^{P}{\mid} i\in E_{n-2}\}$ in this case, and $A_{n}=\emptyset$ otherwise. It follows that

$$\Big(\bigcup_{k=1}^{n} A_{k}\Big)\times \Big(\bigcup_{k=1}^{n-1} B_{k}\Big)\subseteq U.$$

If $F_{n-2}\neq \emptyset$ , then $\big(\bigcup_{k=1}^{n} A_{k}\big)\times \{\sup I_{i}^{Q}\}\subseteq U$ for any $i\in F_{n-2}$ . Note that $\bigcup_{k=1}^{n} A_{k}$ is a finite set. This means that there exists $d_{i}^{Q}\in I_{i}^{Q}$ such that $\big(\bigcup_{k=1}^{n} A_{k}\big)\times \{d_{i}^{Q}\}\subseteq U$ for any $i\in F_{n-2}$ . Let $B_{n}=\{d_{i}^{Q}\mid i\in F_{n-2}\}$ in this case, and $B_{n}=\emptyset$ otherwise. We conclude that

$$\Big(\bigcup_{k=1}^{n} A_{k}\Big)\times \Big(\bigcup_{k=1}^{n} B_{k}\Big)\subseteq U.$$

Let $A=\bigcup_{n\in \mathbb{N}}A_{n}$ and $B=\bigcup_{n\in \mathbb{N}}B_{n}$ . It is easy to see that $(a_{1},b_{1})\in A_1\times B_1 \subseteq\mathord{\uparrow} A\times \mathord{\uparrow} B \subseteq U$ . It suffices to prove that $\mathord{\uparrow} A,\mathord{\uparrow} B$ are both Scott open.

Let D be a directed subset of P with $\sup D\in \mathord{\uparrow} A$ . If $\sup D \in D$ , then $D\cap \mathord{\uparrow} A\neq \emptyset$ . If $\sup D \notin D$ , i.e., D contains no maximal element, then $\mathord{\downarrow} D \in Id(P)$ . Thus, there exists $n_0\in \mathbb{N}$ such that $\mathord{\downarrow} D=I_{n_{0}}^{P}$ .

Therefore, $\sup D\in \mathord{\uparrow} A$ can imply that $\sup I_{n_{0}}^{P}\in \mathord{\uparrow} A$ . Let $n_{1}=\inf\{n\in \mathbb{N}\mid \sup I_{n_{0}}^{P}\in \mathord{\uparrow} A_{n}\}$ . Then $\sup I_{n_{0}}^{P}\in \mathord{\uparrow} A_{n_{1}}$ . Now we need to distinguish between the following two cases for $n_{0},n_{1}$ .

Case 1, $n_{0}<n_{1}$ . If $n_0=1,n_1=2$ , then $\sup I_1^P \notin \mathord{\uparrow} A_1$ implies $A_2= \emptyset$ , which contradicts the condition $\sup I_1^P \in \mathord{\uparrow} A_2$ . So $n_1 \geqslant 3$ . The fact that $\sup I_{n_{0}}^{P}\notin \bigcup_{k=1}^{n_{0}}\mathord{\uparrow} A_{k}$ and $\sup I_{n_{0}}^{P}\in \bigcup_{k=n_{0}+1}^{n_{1}}\mathord{\uparrow} A_{k}$ can imply $n_0 \in E_{n_1-1}$ . This means that $I_{n_{0}}^{P}\cap A_{n_{1}+1}\neq \emptyset$ . Hence, $D\cap \mathord{\uparrow} A\neq \emptyset$ .

Case 2, $n_{0}\geq n_{1}$ . If $n_0=n_1=1$ , then $\sup I_1^P \in \mathord{\uparrow} A_1$ implies $I_1^P \cap A_2 \neq \emptyset$ . If $n_0 \geqslant 2$ , then $\sup I_{n_{0}}^{P}\in \mathord{\uparrow} A_{n_1}\subseteq \bigcup_{k=1}^{n_{0}}\mathord{\uparrow} A_{k}$ , which implies $n_0 \in E_{n_0-1}$ . It follows that $I_{n_{0}}^{P}\cap A_{n_{0}+1}\neq \emptyset$ . Therefore, $D\cap \mathord{\uparrow} A\neq \emptyset$ .

Hence, $\mathord{\uparrow} A$ is Scott open, and $\mathord{\uparrow} B$ is Scott open by the similar proof.

Write $B=\{(a,b)\in L\times L\mid \mathord{\uparrow} a\cap \mathord{\uparrow} b\subseteq L\backslash A\}$ . We claim that B is Scott open in $L\times L$ . Obviously, B is an upper set. Let $(x_{i},y_{i})_{i\in I}$ be a directed subset of $L\times L$ with $\sup_{i\in I}(x_{i},y_{i})\in B$ . Then $(\sup_{i\in I}x_{i},\sup_{i\in I}y_{i})\in B$ , which is equivalent to saying that $\mathord{\uparrow} \sup_{i\in I}x_{i}\cap \mathord{\uparrow} \sup_{i\in I}y_{i}\subseteq L\backslash A$ . It follows that $\bigcap_{i\in I} (\mathord{\uparrow} x_{i}\cap \mathord{\uparrow} y_{i})\subseteq L\backslash A$ . Since $\Sigma L$ is coherent and well filtered, we can find some index $i\in I$ such that $\mathord{\uparrow} x_{i}\cap \mathord{\uparrow} y_{i}\subseteq L\backslash A$ . This implies that $(x_{i},y_{i})\in B$ . Thus, B is Scott open.

It is worth noting that Id(L) is countable. $\Sigma(L\times L)=\Sigma L\times \Sigma L$ from Lemma 4.1. For the sake of contradiction, we assume that there are $x,y\in A$ such that $\mathord{\uparrow} x\cap \mathord{\uparrow} y\cap A=\emptyset$ . The fact that $(x,y)\in B\subseteq \sigma(L\times L)$ implies that we can find $U_{x},U_{y}\in \sigma (L)$ such that $(x,y)\in U_{x}\times U_{y}\subseteq B$ . Note that $x\in U_{x}\cap A$ and $y\in U_{y}\cap A$ . By the irreducibility of A, we have $A\cap U_{x}\cap U_{y}\neq \emptyset$ . Pick $a\in A\cap U_{x}\cap U_{y}$ . Then $(a,a)\in U_{x}\times U_{y}\subseteq B$ , that is, $a\in \mathord{\uparrow} a \cap \mathord{\uparrow} a\subseteq L\backslash A$ . It contradicts the assumption that $a\in A$ . Hence, A is directed and $\sup A\in A$ . So $A=\mathord{\downarrow} \sup A$ .