# power structures of directed spaces111Research supported by NSF of China
(Nos. 11871353, 12001385).

Xiaolin Xie<EMAIL_ADDRESS>Yuxu Chen<EMAIL_ADDRESS>Hui Kou
<EMAIL_ADDRESS>Department of Mathematics, Sichuan University,
Chengdu,610064, China

###### Abstract

Powerdomains in domain theory plays an important role in modeling the
semantics of nondeterministic functional programming languages. In this paper,
we extend the notion of powerdomain to the category of directed spaces, which
is equivalent to the notion of the $T_{0}$ monotone-determined space [4]. We
define the notion of upper, lower and convex powerspace of a directed space by
the way of free algebras. We show that the upper, lower and convex powerspace
over any directed space exist and give their concrete structures. Generally,
the upper, lower and convex powerspaces of a directed spaces are different
from the upper, lower and convex powerdomains of a dcpos endowed with the
Scott topology and the observationally-induced upper and lower powerspaces
introduced by Battenfeld and Schöder in 2015.
Keywords: powerdomain, directed lower powerspace of directed spaces, directed
upper powerspace of directed spaces, directed convex powerspace of directed
spaces, observationally-induced lower powerspace, observationally-induced
lower powerspace
Mathematics Subject Classification: 06B35; 54A20; 54B30; 54H10

##  1 Introduction

Powerdomain is one of the most important part of domain theory. Its purpose is
to provide a mathematical model for the semantics of nondeterministic
functional programming languages. There are three classical power structures
in domain theory, they are the lower powerdomain( also known as the Hoaer
powerdomain), the upper powerdomain( also known as the Smyth powerdomain) and
the convex powerdomain( also known as the Plotikin powerdomain). Moreover,
each power structure has a standard topological representation. In recent
years, papers [6, 38, 39, 40] have made a lot of generalizations on these
power structures. Generally speaking, these power structures are free algebras
generated by domain respect to some binary operation. In 2015, I. Batenfeld
and M. Schöder ([2])introduced power structure in general topological spaces,
and in this paper, Boolean algebra 2 is endowed with Sierpinski topology as an
observable structure, and then the upper power structure and lower power
structure are defined, which is called the observationally-induced
upper(lower) powerspace. The work of paper [2] makes it possible for the
category of general topological space to be applied to express the
nondeterministic semantics of functional programming languages. But the method
of power structure induced by Boolean algebra 2 is complicated and difficult
to understand. In this paper, we consider a class of special topological
spaces, the directed space (equivalent to those in [4]). It is proved in [37]
that directed spaces contain the basic objects of domain theory, all directed
complete posets endowed with the Scott topology, and continuous functions,
which forms a cartesian closed category. Thus, the directed space is an
extended framework of domain theory. In this paper, we are going to extend the
structure of powerdomain to the category of directed spaces, and define the
concept of directed upper (lower, convex) powerspace by means of free algebra.
We show that the upper, lower and convex powerspace over any directed space
exist and give their concrete structures. Generally, the upper, lower and
convex powerspaces of a directed spaces are different from the upper, lower
and convex powerdomains of a dcpos endowed with the Scott topology and the
observationally-induced upper and lower powerspaces.

In 1991, Heckmann introduced an algebraic method and it does not rely on any
explicit representations of the powerdomains [15]. In the last part of this
article, we will discuss the commutativity of the directed upper and lower
functors.

##  2 Preliminaries

Now, we introduce the concepts needed in this article. On domain theory,
topology, and category theory, see [1, 7, 23]. Let $P$ be a nonempty set. A
relation $\leq$ on $P$ is called a partial order, if $\leq$ satisfies
reflexivity ($x\leq x$), transitivity ($x\leq y\ \&\ y\leq z\Rightarrow x\leq
z$) and antisymmetry ($x\leq y\ \&\ y\leq x\Rightarrow x=y$). $P$ is called a
partial ordered set(poset) if $\ P\ $is endowed with some partial order
$\leq$. Given $A\subseteq P$, denote $\mathord{\downarrow}A=\\{x\in P:\exists
a\in A,\ \ x\leq a\\}$, $\mathord{\uparrow}A=\\{x\in P:\exists a\in A,\ a\leq
x\\}$. We say $A$ is a lower set (upper set) if $A=\mathord{\downarrow}A$
($A=\mathord{\uparrow}A$). A nonempty set $D\subseteq P$ is called a directed
set if each finite nonempty subset of $D$ has upper bound in $D$.
Particularly, a poset is called a directed complete poset if each directed
subset has a supremum(denoted by $\bigvee D$), abbreviated as dcpo. The subset
$U$ of poset $P$ is called a Scott open set if $U$ is an upper set and for
each directed set $D\subseteq P$, which $\bigvee D$ exists and belongs to $U$,
then $U\cap D\not=\emptyset$. The set of all Scott open sets of poset $P$ is a
toplology on $P$, which is called the Scott topology and denoted by
$\sigma(P)$. Suppose $P,\ E$ are two posets, a fuction $f:P\longrightarrow E$
is called Scott continuous if it is continuous respect to Scott topology
$\sigma(P)$ and $\sigma(E)$.

All topological spaces in this paper are required $T_{0}$ separation. A net of
a topological space $X$ is a map $\xi:J\longrightarrow X$, here $J$ is a
directed set. Thus, each directed subset of a poset can be regarded as a net,
and its index set is itself. Usually, we denote a net by $(x_{j})_{j\in J}$ or
$(x_{j})$. Let $x\in X$, saying $(x_{j})$ converges to $x$, denote by
$(x_{j})\rightarrow x$ or $x\equiv\lim x_{j}$, if $(x_{j})$ is eventually in
every open neighborhood of $x$, that is, for each given open neighborhood $U$
of $x$, there exists $j_{0}\in J$ such that for every $j\in J$, $j\geq
j_{0}\Rightarrow x_{j}\in U$.

Let $X$ be a $T_{0}$ topological space, its topology is denoted by $O(X)$, the
specialization order on $X$ is defined as follows:

$\forall x,\ y\in X,\ x\sqsubseteq y\Leftrightarrow x\in\overline{\\{y\\}}$

here, $\overline{\\{y\\}}$ means the closure of $\\{y\\}$. From now on, the
order of a $T_{0}$ topological space always indicates the specialization order
”$\sqsubseteq$”. Here are some basic properties of specialization order.

######  Proposition 2.1.

[1, 7] For a $T_{0}$ topological space $X$, the following are always true:

  1. 1.

For each open set $U\subseteq X,\ U=\mathord{\uparrow}U;$

  2. 2.

For each closed set $A\subseteq X,\ A=\mathord{\downarrow}A;$

  3. 3.

Suppose $Y$ is another $T_{0}$ topological space, and $f:X\rightarrow Y$ is a
continuous function from $X$ to $Y$. Then for each $x,\ y\in X$, $x\sqsubseteq
y\Rightarrow f(x)\sqsubseteq f(y)$, that is every continuous fuction is
monotone.

Suppose $X$ is a $T_{0}$ space, then every directed set $D\subseteq X$ can be
regarded as a net of $X$, we use $D\rightarrow x$ or $x\equiv\lim D$ to
represent $D$ converges to $x$. Define notation

$D(X)=\\{(D,\ x):x\in X,\ D$ is a directed subset of $X$ and $D\rightarrow
x\\}$.

It is easy to verify that, for each $x,\ y\in X$, $x\sqsubseteq
y\Leftrightarrow\\{y\\}\rightarrow x$. Therefore, if $x\sqsubseteq y$ then
$(\\{y\\},\ x)\in D(X)$. Next, we give the concept of directed space.

######  Definition 2.2.

[37] Let $X$ be a $T_{0}$ space.

  1. (1)

A subset $U$ of $X$ is called a directed open set if $\forall(D,\ x)\in D(X),\
x\in U\Rightarrow D\cap U\neq\emptyset$. Denote all directed open sets of $X$
by $d(X)$.

  2. (2)

$X$ is called a directed space if each directed open set of $X$ is an open
set, that is, $d(X)=O(X)$.

######  Remark 2.3.

  1. (1)

Each open set of a $T_{0}$ space is directed open, but the contrary is not
necessarily true. For example, suppose $Y$ is a non-discrete $T_{1}$
topological space, its specialization order is diagonal, that is, $\forall x,\
y\in Y$, $x\sqsubseteq y\Leftrightarrow x=y$. Thus, all subsets of $Y$ are
directed open. We notice that $Y$ is non-discrete, at least one directed open
set is not an open set.

  2. (2)

The definition of directed space here is equivalent to the $T_{0}$ monotone
determined space defined in [4].

  3. (3)

Every poset endowed with the Scott topology is a directed space [21,
37],besides,each Alexandroff space is a directed space. Thus, the directed
space extends the concept of the Scott topology.

Next, we introduce the directed continuous function.

######  Definition 2.4.

Suppose $X,\ Y$ are two $T_{0}$ spaces. A function $f:X\longrightarrow Y$ is
called directed continuous if it is monotone and preserves all limits of
directed set of $X$; that is, $(D,\ x)\in D(X)\Rightarrow(f(D),\ f(x))\in
D(Y)$.

Here are some characterizations of the directed continuous functions.

######  Proposition 2.5.

[37] Suppose $X,\ Y$ are two $T_{0}$ spaces. $f:X\longrightarrow Y$ is a
function between $X$ and $Y$.

  1. (1)

$f$ is directed continuous if and only if $\forall U\in d(Y),\ f^{-1}(U)\in
d(X)$.

  2. (2)

If $X,\ Y$ are directed spaces, then $f$ is continuous if and only if it is
directed continuous.

Now we introduce the product of directed spaces.

Suppose $X,\ Y$ are two directed spaces. Let $X\times Y$ represents the
cartesian product of $X$ and $Y$, then we have a natural partial order on it:
$\forall(x_{1},\ y_{1}),\ (x_{2},\ y_{2})\in X\times Y$,

$(x_{1},\ y_{1})\leq(x_{2},\ y_{2})\iff x_{1}\sqsubseteq x_{2},\
y_{1}\sqsubseteq y_{2}.$

which is called the pointwise order on $X\times Y$. Now, we define a
topological space $X\otimes Y$ as follows:

  1. 1.

The underlying set of $X\otimes Y$ is $X\times Y$;

  2. 2.

The topology on $X\times Y$ is generated as follows: for each given $\leq$\-
directed set $D\subseteq X\times Y$ and $(x,\ y)\in X\times Y$,

$D\rightarrow(x,\ y)\in X\otimes Y\iff\pi_{1}D\rightarrow x\in X,\
\pi_{2}D\rightarrow y\in Y,\ $

That is, a subset $U\subseteq X\times Y$ is open if and only if for every
directed limit defined as above $D\rightarrow(x,\ y)$, $(x,\ y)\in
U\Rightarrow U\cap D\neq\emptyset$.

######  Theorem 2.6.

[37] Suppose $X$ and $Y$ are two directed spaces.

  1. 1.

The topological space $X\otimes Y$ defined as above is a directed space and
satisfies the following properties: the specialization order on $X\otimes Y$
equals to the pointwise order on $X\times Y$, that is $\sqsubseteq=\leq$.

  2. 2.

Suppose $Z$ is another directed space, then $f:X\otimes Y\longrightarrow Z$ is
continuous if and only if it is continuous in each variable separately.

Denote the category of all directed spaces with continuous functions by ${\bf
Dtop}$. It is proved in [36, 37] that, ${\bf Dtop}$ contains all posets
endowed with the Scott topology and ${\bf Dtop}$ is a cartesian closed
category; specifically, the categorical products of two directed spaces $X$
and $Y$ are isomorphic to $X\otimes Y$. So, the directed space is an extended
framework of Domain Theory.

Let $P$ be a dcpo, and $x,\ y\in P$. We say $x$ way below $y$, if for each
given directed set $D\subseteq P$, $y\leq\bigvee D$ implies there exists some
$d\in D$ such that $x\leq d$. We write
$\mathord{\mbox{\makebox[0.0pt][l]{\raisebox{-1.72218pt}{$\downarrow$}}$\downarrow$}}x=\\{a\in
P:a\ll x\\}$,
$\mathord{\mbox{\makebox[0.0pt][l]{\raisebox{1.72218pt}{$\uparrow$}}$\uparrow$}}x=\\{a\in
P:x\ll a\\}$.

######  Definition 2.7.

A dcpo $P$ is called a continuous domain if for each $x\in P$,
$\mathord{\mbox{\makebox[0.0pt][l]{\raisebox{-1.72218pt}{$\downarrow$}}$\downarrow$}}x$
is directed and
$x=\bigvee\mathord{\mbox{\makebox[0.0pt][l]{\raisebox{-1.72218pt}{$\downarrow$}}$\downarrow$}}x$.

######  Theorem 2.8.

[7] Suppose $P$ is a continuous domain. The followings hold:

  1. (1)

$\forall x,\ y\in P$, $x\ll y\Rightarrow\exists z\in P,\ x\ll z\ll y$.

  2. (2)

$\forall x\in P$,
$\mathord{\mbox{\makebox[0.0pt][l]{\raisebox{1.72218pt}{$\uparrow$}}$\uparrow$}}x$
is a Scott open set. Particularly,
$\\{\mathord{\mbox{\makebox[0.0pt][l]{\raisebox{1.72218pt}{$\uparrow$}}$\uparrow$}}x:x\in
P\\}$ is a base of $(P,\ \sigma(P))$.

##  3 The directed lower powerspaces of directed spaces

As mentioned above, directed space is a extended framework of domain theory,
just like the work done in article [2], extending the powerdomain in the
category of directed space is very meaningful. In this section, we will
construct the directed lower powerspace of the directed space, which is a free
algebra generated by the inflationary operation of the directed space.

######  Definition 3.1.

[34] Let $X$ be a directed space.

  1. (1)

A binary operation $\oplus:X\otimes X\rightarrow X$ on $X$ is called an
inflationary operation if it is continuous and satisfies the following four
conditions: $\forall x,\ y,\ z\in X$,

    1. (a)

$x\oplus x=x$,

    2. (b)

$(x\oplus y)\oplus z=x\oplus(y\oplus z)$,

    3. (c)

$x\oplus y=y\oplus x$,

    4. (d)

$x\oplus y\geq x$.

  2. (2)

If $\oplus$ is a inflationary operation on $X$, then $(X,\ \oplus)$ is called
a directed inflationary semilattice, that is, directed inflationary
semilattices are those directed spaces with inflationary operations.

By Theorem 2.6(2), the operation $\oplus$ on a directed space $X$ is
continuous if and only if it is monotone and for each given $x,\ y\in X$ and
directed set $D\subseteq X$, $\ x\equiv\lim D$ implies $x\oplus
y\equiv\lim(D\oplus y)$. Here, $D\oplus y=\\{d\oplus y:d\in D\\}$.

Here are two examples.

######  Example 3.2.

[34]

  1. (1)

Suppose $P$ is a poset endowed with the Scott topology, and for each $a,\ b\in
P$, the supremum of $a$ and $b$ exists in $P$ (denoted by $a\vee b$ ). Then
$(P,\ \vee)$ is a directed inflationary semilattice.

  2. (2)

Let $I=[0,\ 1]$ (the unit interval), let ${\mathcal{A}}$ denote the topology
generated by $\\{[a,\ 1]:a\in I\\}$, then $\mathcal{A}$ is the Alexandroff
topology on $I$. It is easy to check that, $(I,\ {\mathcal{A}})$ is a directed
space, and $(I,\ max)$ is a directed inflationary semilattice endowed with
$\mathcal{A}$.

######  Definition 3.3.

[34] Suppose $(X,\ \oplus),\ (Y,\ \uplus)$ are two directed inflationary
semilattices, $f:(X,\ \oplus)\rightarrow(Y,\ \uplus)$ is called an
inflationary homomorphism between $X$ and $Y$, if $f$ is continuous and
$f(x\oplus y)=f(x)\uplus f(y)$ holds, $\forall x,\ y\in X$.

Denote the category of all directed inflationary semilattices and inflationary
homomorphisms by ${\bf Disl}$. Then ${\bf Disl}$ is a subcategory of ${\bf
Dtop}$.

######  Lemma 3.4.

[34] Suppose $(X,\ \oplus)$ is a directed inflationary semilattice, then
$\oplus=\vee_{\sqsubseteq}$. Here, $\forall x,\ y\in X,\ x\vee_{\sqsubseteq}y$
means the supremum of $x$ and $y$ respect to the specialization order
$\sqsubseteq$ on $X$ (calling it sup operation). Conversely, suppose $X$ is a
directed space such that for each $x,\ y\in X$, $x\vee_{\sqsubseteq}y$ exists,
and the continuity of $\vee_{\sqsubseteq}$ will naturally imply that $(X,\
\vee_{\sqsubseteq})$ is a directed inflationary semilattice.

Proof By definition 3.1 ,$\forall x,\ y\in X,\ x\oplus y\geq x,\ y$, that is
$x\oplus y$ is a upper bound of $\\{x$, $y\\}$. Suppose $z$ is another
arbitrary upper bound of $\\{x,\ y\\}$, By Theorem 2.6(1), the pointwise order
equals to the specialization order of $X\otimes X$,then $(x,\
y)\sqsubseteq(z,\ z)$. By the continuity and idempotence of the inflationary
operation, we have $x\oplus y\sqsubseteq z\oplus z=z$. That is $x\oplus y$ is
the supremum of $\\{x,\ y\\}$, which means $x\oplus y=x\vee_{\sqsubseteq}y$.
Conversely, continuous sup operation naturally satisfy all conditions in
definition 5.1, thus, we get the conclusion. $\Box$

The above results show that a directed inflationary semilattice $(X,\ \oplus)$
is just a directed space with a continuous sup operation satisfy
$\oplus=\vee_{\sqsubseteq}$. Since the order on the directed space in this
paper is always the specialization order, we will use the symbol $\vee$
instead of $\vee_{\sqsubseteq}$ in the following part. Therefore, a directed
inflationary semilattice is always represented by a tuple of the form $(X,\
\vee)$, here $X$ is a directed space, $\vee$ represents the continuous sup
operation on $X$. Next, we give the definition of directed lower powerspace.

######  Definition 3.5.

[34] Suppose $X$ is a directed space. A directed space $Z$ is called the
directed lower powerspace of $X$ if and only if the following two conditions
are satisfied:

  1. (1)

$Z$ is a directed inflationary semilattice, that is the sup operation $\vee$
on $Z$ exists and which is continuous,

  2. (2)

There is a continuous function $i:X\longrightarrow Z$ satisfying: for an
arbitrary directed inflationary semilattice $(Y,\ \vee)$ and continuous
function $f:X\longrightarrow Y$, there exists an unique inflationary
homomorphism $\bar{f}:(Z,\ \vee)\rightarrow(Y,\ \vee)$ such that
$f=\bar{f}\circ i$.

By the definition above, if directed inflationary semilattices $(Z_{1},\
\vee)$ and $(Z_{2},\ \vee)$ are both the directed lower powerspaces of $X$,
then there exists a topological homomorphism which is also a inflationary
homomorphism $g:Z_{1}\rightarrow Z_{2}$. Therefore, up to of order isomorphism
and topological homomorphism, the directed lower powerspace of a directed
space is unique. Particularly, we denote the directed lower powerspace of each
directed space $X$ by $P_{L}(X)$.

Now, we will prove the existence of the directed lower powerspace of each
directed space $X$ by way of concrete construction.

Let $X$ be a directed space. Set

$LX=\\{\mathord{\downarrow}F:F\subseteq_{fin}X\\},\ $

here, $F\subseteq_{fin}X$ is an arbitrary nonempty finite subset of $X$.
Define an order $\leq_{L}$ on $LX$ as follows:

$\mathord{\downarrow}F_{1}\leq_{L}\mathord{\downarrow}F_{2}\iff\mathord{\downarrow}F_{1}\subseteq\mathord{\downarrow}F_{2}.$

Let $\mathcal{D}\subseteq LX$ be a directed set (respect to order $\leq_{L}$),
$\mathord{\downarrow}F\in LX$. Define a convergence notation
$\mathcal{D}\Rightarrow_{L}\mathord{\downarrow}F$ as follows:

$\mathcal{D}\Rightarrow_{L}\mathord{\downarrow}F\iff\forall a\in F,\ $ there
exists a directed set $D_{a}$ of $X$ satisfying
$D_{a}\subseteq\bigcup\mathcal{D}$ and $D_{a}\rightarrow a$.

A subset $\mathcal{U}\subseteq LX$ is called a $\Rightarrow_{L}$ convergence
open set of $LX$ if and only if for each directed subset $\mathcal{D}$ of $LX$
and $\mathord{\downarrow}F\in LX$,
$\mathcal{D}\Rightarrow_{L}\mathord{\downarrow}F\in\mathcal{U}$ implies
$\mathcal{D}\cap\mathcal{U}\neq\emptyset$. Denote all $\Rightarrow_{L}$
convergence open sets of $LX$ by $O_{\Rightarrow_{L}}(LX)$.

######  Proposition 3.6.

[34] Suppose $X$ is a directed space, the following are true:

  1. (1)

$(LX,\ O_{\Rightarrow_{L}}(LX))$ is a topological space, abbreviated as $LX$.

  2. (2)

The specialization order $\sqsubseteq$ on $(LX,\ O_{\Rightarrow_{L}}(LX))$
equals to $\leq_{L}$.

  3. (3)

$(LX,\ O_{\Rightarrow_{L}}(LX))$ is a directed space, that is
$O_{\Rightarrow_{L}}(LX)=d(LX)$.

Proof (1) Firstly $\emptyset,\ LX\in O_{\Rightarrow_{L}}(LX)$. Suppose
$\mathcal{U}\in O_{\Rightarrow_{L}}(LX)$, $\mathord{\downarrow}F_{1},\
\mathord{\downarrow}F_{2}\in LX$. If
$\mathord{\downarrow}F_{1}\leq\mathord{\downarrow}F_{2}$, then
$\\{\mathord{\downarrow}F_{2}\\}\Rightarrow_{L}\mathord{\downarrow}F_{1}$ is
obviously hold. Therefore $\mathord{\downarrow}F_{1}\in\mathcal{U}$ will imply
$\mathord{\downarrow}F_{2}\in\mathcal{U}$. This means $\mathcal{U}$ is an
upper set respect to order $\leq_{L}$ on $LX$. Suppose we have
$\mathcal{U}_{1},\ \mathcal{U}_{2}\in O_{\Rightarrow_{L}}(LX)$, and
$\mathcal{D}$ is a directed set in $LX$, $\mathord{\downarrow}F\in LX$ and
$\mathcal{D}\Rightarrow\mathord{\downarrow}F\in\mathcal{U}_{1}\cap\mathcal{U}_{2}$,
there will exist $\mathord{\downarrow}G_{1}\in\mathcal{D}\cap\mathcal{U}_{1}$
and $\mathord{\downarrow}G_{2}\in\mathcal{D}\cap\mathcal{U}_{2}$. Since
$\mathcal{D}$ is directed, there exists $\mathord{\downarrow}G\in\mathcal{D}$
such that $\mathord{\downarrow}G_{1},\
\mathord{\downarrow}G_{2}\leq_{L}\mathord{\downarrow}G$, taht is
$\mathord{\downarrow}G\in\mathcal{D}\cap\mathcal{U}_{1}\cap\mathcal{U}_{2}$.
By the same way, $O_{\Rightarrow_{L}}(LX)$ is closed under the arbitrary
union. Thus, $O_{\Rightarrow_{L}}(LX)$ is a topology on $LX$.

(2) Let $\mathord{\downarrow}F_{1},\ \mathord{\downarrow}F_{2}\in LX$. If
$\mathord{\downarrow}F_{1}\leq_{L}\mathord{\downarrow}F_{2}$, according the
proof of (1), every $\Rightarrow_{L}$ convergence open set is an upper set
respect to partial order $\leq_{L}$, then
$\mathord{\downarrow}F_{1}\in\overline{\\{\mathord{\downarrow}F_{2}\\}}$, that
is, $\mathord{\downarrow}F_{1}\sqsubseteq\mathord{\downarrow}F_{2}$. On the
other hand, let
$\mathord{\downarrow}F_{1}\sqsubseteq\mathord{\downarrow}F_{2}$. We want to
prove $\mathord{\downarrow}F_{1}\leq_{L}\mathord{\downarrow}F_{2}$, that is
$\\{\mathord{\downarrow}F\in
LX:\mathord{\downarrow}F\subseteq\mathord{\downarrow}F_{2}\\}$ is a closed set
of $LX$ respect to the topology $O_{\Rightarrow_{L}}(LX)$, since
$\\{\mathord{\downarrow}F\in
LX:\mathord{\downarrow}F\subseteq\mathord{\downarrow}F_{2}\\}\subseteq\overline{\\{\mathord{\downarrow}F_{2}\\}}$,
$\\{\mathord{\downarrow}F\in
LX:\mathord{\downarrow}F\subseteq\mathord{\downarrow}F_{2}\\}$ is a closed set
of $LX$ respect to the topology $O_{\Rightarrow_{L}}(LX)$ will imply
$\\{\mathord{\downarrow}F\in
LX:\mathord{\downarrow}F\subseteq\mathord{\downarrow}F_{2}\\}=\overline{\\{\mathord{\downarrow}F_{2}\\}}$,
that is $\mathord{\downarrow}F_{1}\leq_{L}\mathord{\downarrow}F_{2}$. We are
going to prove that $\\{\mathord{\downarrow}F\in
LX:\mathord{\downarrow}F\subseteq\mathord{\downarrow}F_{2}\\}$ is closed
respcet to topology $O_{\Rightarrow_{L}}(LX)$, that is to show
$\mathcal{U}=LX\setminus\\{\mathord{\downarrow}F\in
LX:\mathord{\downarrow}F\subseteq\mathord{\downarrow}F_{2}\\}$ is a
$\Rightarrow_{L}$ convergence open set in $LX$. Let ${\mathcal{D}}\subseteq
LX$ be a $\leq_{L}$ \- directed set and
$\mathcal{D}\Rightarrow_{L}\mathord{\downarrow}G\in\mathcal{U}$. Let
$G=\\{a_{1},\ a_{2},\ \ldots,\ a_{k}\\}$. By the definition of
$\Rightarrow_{L}$ convergence, we have finitary directed subsets
$D_{i}\subseteq\bigcup{\mathcal{D}}$ in $X$ such that $D_{i}\rightarrow
a_{i}$, $i=1,\ 2,\ \ldots,\ k$. By contradiction, we suppose that
${\mathcal{D}}\cap{\mathcal{U}}=\emptyset$. Then
$\bigcup{\mathcal{D}}\subseteq\mathord{\downarrow}F_{2}$, that is
$D_{i}\subseteq\mathord{\downarrow}F_{2}$, $i=1,\ 2,\ \ldots,\ k$. Since
$\mathord{\downarrow}F_{2}$ is closed in $X$, the limit points of $D_{i}$ are
in $\mathord{\downarrow}F_{2}$, that is $G=\\{a_{1},\ a_{2},\ \ldots,\
a_{k}\\}\subseteq\mathord{\downarrow}F_{2}$, and,
$\mathord{\downarrow}G\leq_{L}\mathord{\downarrow}F_{2}$. This contradicts
with $\mathord{\downarrow}G\in{\mathcal{U}}$. Therefore,
$\mathcal{U}=LX\setminus\\{\mathord{\downarrow}F\in
LX:\mathord{\downarrow}F\subseteq\mathord{\downarrow}F_{2}\\}$ is a
$\Rightarrow_{L}$ convergence open set in $LX$.

(3) For an arbitrary topological space $X$, we have $O(X)\subseteq d(X)$, then
$O_{\Rightarrow_{L}}(LX)\subseteq d(LX)$. On the other hand, according the
definition of $\Rightarrow_{L}$ convergence, if directed set
$\mathcal{D}\Rightarrow_{L}\mathord{\downarrow}F$ in $LX$, according to (2),
$\mathcal{D}$ convergents to $\mathord{\downarrow}F$ respect to the topology
$O_{\Rightarrow_{L}}(LX)$. Then, by the definition of directed open set,
$\mathcal{D}\Rightarrow_{L}\mathord{\downarrow}F\in\mathcal{U}\in d(LX)$, will
imply $\mathcal{U}\cap\mathcal{D}\neq\emptyset$. This means $\mathcal{U}\in
O_{\Rightarrow_{L}}(LX)$, thus $O_{\Rightarrow_{L}}(LX)=d(LX)$, that is,
$(LX,\ O_{\Rightarrow_{L}}(LX))$ is a directed space. $\Box$

######  Proposition 3.7.

[34] Suppose $X,\ Y$ are two directed spaces. Then function $f:LX\rightarrow
Y$ is continuous if and only if for each directed set ${\mathcal{D}}\subseteq
LX$ and $\mathord{\downarrow}F\in LX$,
${\mathcal{D}}\Rightarrow_{L}\mathord{\downarrow}F$ implies
$f(\mathcal{D})\rightarrow f(\mathord{\downarrow}F)$.

Proof Since $\Rightarrow_{L}$ convergence will lead to
$O_{\Rightarrow_{L}}(LX)$ topological convergence, the necessity is obviously.
we are going to prove the sufficiency. First to check that $f$ is monotone. If
$\mathord{\downarrow}F_{1},\ \mathord{\downarrow}F_{2}\in LX$ and
$\mathord{\downarrow}F_{1}\leq_{L}\mathord{\downarrow}F_{2}$, then
$\\{\mathord{\downarrow}F_{2}\\}\Rightarrow_{L}\mathord{\downarrow}F_{1}$, by
the given condition, $\\{f(\mathord{\downarrow}F_{2})\\}\rightarrow
f(\mathord{\downarrow}F_{1})$, thus $f(\mathord{\downarrow}F_{2})\sqsubseteq
f(\mathord{\downarrow}F_{1})$. Suppose $U$ is an open set of $Y$ and the
directed set ${\mathcal{D}}\Rightarrow_{L}\mathord{\downarrow}F\in f^{-1}(U)$,
then $f(\mathcal{D})$ is a directed set of $Y$ and $f(\mathcal{D})\rightarrow
f(\mathord{\downarrow}F)\in U$, there exists
a$\mathord{\downarrow}F\in{\mathcal{D}}$ such that
$f(\mathord{\downarrow}F)\in U$. Thus,
$\mathord{\downarrow}F\in{\mathcal{D}}\cap f^{-1}(U)$. According to the
definition of $\Rightarrow_{L}$ convergence open set, $f^{-1}(U)\in
O_{\Rightarrow_{L}}(LX)$, that is $f$ is continuous. $\Box$

######  Theorem 3.8.

[34] Let $X$ be a directed space. Then $(LX,\ O_{\Rightarrow_{L}}(LX))$
respect to the set union operation $\cup$ is a directed inflationary
semilattice.

Proof According to Proposition 3.6, $(LX,\ O_{\Rightarrow_{L}}(LX))$ is a
directed space. we will prove that $\cup$ is an inflationary operation. For
arbitrary $\mathord{\downarrow}F_{1},\ \mathord{\downarrow}F_{2}\in LX$, then
$\mathord{\downarrow}F_{1}\cup\mathord{\downarrow}F_{2}=\mathord{\downarrow}(F_{1}\cup
F_{2})\in LX$. Obviously, $\cup$ satisfy the conditions (a), (b), (c), (d) in
Definition 3.1, we now prove the continuity of $\cup$. The monotonicity of
$\cup$ is evident. By Theorem 2.6(2) and Proposition 3.7, we only need to
prove that, For each directed set ${\mathcal{D}}\subseteq LX$ and
$\mathord{\downarrow}F,\ \mathord{\downarrow}G\in LX$,
${\mathcal{D}}\Rightarrow_{L}\mathord{\downarrow}F$, will imply
$G\cup{\mathcal{D}}\Rightarrow_{L}(\mathord{\downarrow}G\cup\mathord{\downarrow}F)=\mathord{\downarrow}(G\cup
F)$. Here, $G\cup{\mathcal{D}}=\\{\mathord{\downarrow}(G\cup
F^{\prime}):\mathord{\downarrow}F^{\prime}\in\mathcal{D}\\}$ is a directed
set. For each $a\in G\cup F$, if $a\in G$, then $\\{a\\}\rightarrow a$; if
$a\in F$, since ${\mathcal{D}}\Rightarrow_{L}\mathord{\downarrow}F$,there
exists $D\subseteq\bigcup{\mathcal{D}}$ such that $D\rightarrow a$. According
to the definition of $\Rightarrow_{L}$ convergence, we have
$G\cup{\mathcal{D}}\Rightarrow_{L}\mathord{\downarrow}(G\cup F)$. $\Box$

######  Remark 3.9.

We can directly check that for arbitrary $\mathord{\downarrow}F_{1},\
\mathord{\downarrow}F_{2}\in LX,\
\mathord{\downarrow}F_{1}\cup\mathord{\downarrow}F_{2}=\mathord{\downarrow}(F_{1}\cup
F_{2})$ is well-defined although each $\mathord{\downarrow}F$ may generated by
different $F$.

The following theorem is the main result in this section.

######  Theorem 3.10.

[34] Suppose $X$ is a directed space, then $(LX,\ O_{\Rightarrow_{L}}(LX))$ is
the lower powerspace of $X$, that is, endowed with topology
$O_{\Rightarrow_{L}}(LX)$, $(LX,\ \cup)\cong P_{L}(X)$.

Proof Define function $i:X\rightarrow LX$, $\forall x\in X$,
$i(x)=\mathord{\downarrow}x$. We prove the continuity of $i$. It is evident
that $i$ is monotone. If $D\subseteq X$ and $x\in X$ satisfy $D\rightarrow x$.
Let $\mathcal{D}=\\{\mathord{\downarrow}d:d\in D\\}$, then $\mathcal{D}$ is a
directed set in $LX$ and ${\mathcal{D}}\Rightarrow_{L}\mathord{\downarrow}x$.
But $i(D)=\mathcal{D}$, so $i(D)\Rightarrow_{L}\mathord{\downarrow}x=i(x)$.
According to Proposition 3.7, $i$ is continuous.

Let $(Y,\ \vee)$ be an arbitrary directed inflationary semilattice,
$f:X\rightarrow Y$ is a continuous function. Define $\bar{f}:LX\rightarrow Y$
as follows: $\forall\ \mathord{\downarrow}F\in LX$ (let $F=\\{a_{1},\ a_{2},\
\ldots,\ a_{n}\\}$ ),

$\bar{f}(\mathord{\downarrow}F)=f(a_{1})\vee f(a_{2})\vee\cdots\vee
f(a_{n})=\bigvee\limits_{a\in F}f(a).$

$\bar{f}$ obvious well-defined. Especially, denote
$\bar{f}(\mathord{\downarrow}F)=\vee f(F)$.

(1) $f=\bar{f}\circ i$.

For arbitrary $x\in X$, $(\bar{f}\circ
i)(x)=\bar{f}(i(x))=\bar{f}(\mathord{\downarrow}x)=f(x)$.

(2) $\bar{f}$ is an inflationary homomorphism, that is, $\bar{f}$ is
continuous and for arbitrary $\mathord{\downarrow}F_{1},\
\mathord{\downarrow}F_{2}\in LX$,
$\bar{f}(\mathord{\downarrow}F_{1}\cup\mathord{\downarrow}F_{2})=\bar{f}(\mathord{\downarrow}F_{1})\vee\bar{f}(\mathord{\downarrow}F_{2})$.

First,we prove that $\bar{f}$ preserves the union operation. Let
$\mathord{\downarrow}F_{1},\ \mathord{\downarrow}F_{2}\in LX$. Then
$\bar{f}(\mathord{\downarrow}F_{1}\cup\mathord{\downarrow}F_{2})=\bar{f}(\mathord{\downarrow}(F_{1}\cup
F_{2}))=\vee f(F_{1}\cup F_{2})=(\vee f(F_{1})\vee(\vee
f(F_{2}))=\bar{f}(\mathord{\downarrow}F_{1})\vee\bar{f}(\mathord{\downarrow}F_{2})$.
Next, we prove the continuity of $\bar{f}$. Since $\vee$ is the sup operation,
$\bar{f}$ is monotone. Suppose ${\mathcal{D}}\subseteq LX$ and
$\mathord{\downarrow}G\in LX$ satisfy
${\mathcal{D}}\Rightarrow_{L}\mathord{\downarrow}G$. Let $G=\\{b_{1},\ b_{2},\
\ldots,\ b_{k}\\}$. By the definition of $\Rightarrow_{L}$ convergence, for
each $b_{i}\in G$, there exists a directed set
$D_{i}\subseteq\bigcup{\mathcal{D}}$ such that $D_{i}\rightarrow b_{i}$. By
the continuity of $f$, we have $f(D_{i})\rightarrow f(b_{i})$, $i=1,\ 2,\
\ldots,\ k$. since $\vee$ is continuous,the following convergence hold in Y:

$f(D_{1})\vee f(D_{2})\vee\cdots\vee f(D_{k})\rightarrow f(b_{1})\vee\dots\vee
f(b_{k}).\ \\\ (\ast)$

Here,

$f(D_{1})\vee f(D_{2})\cdots\vee f(D_{k})=\\{f(d_{1})\vee
f(d_{2})\vee\cdots\vee f(d_{k}):(d_{1},\ d_{2},\ \ldots,\
d_{k})\in\prod\limits_{i=1}^{k}D_{i}\\}.\ $

Suppose $(d_{1},\ d_{2},\ \dots,\ d_{k})\in\prod\limits_{i=1}^{k}D_{i}$. For
arbitrary $i\in\\{1,\ 2,\ \ldots,\ k\\}$, there exists
$\mathord{\downarrow}F_{i}\in\mathcal{D}$ such that
$d_{i}\in\mathord{\downarrow}F_{i}$. Since $\mathcal{D}$ is directed, there
exists some $\mathord{\downarrow}F\in\mathcal{D}$ such that
$\mathord{\downarrow}F_{i}\subseteq\mathord{\downarrow}F$. Therefore
$f(d_{1})\vee\dots\vee f(d_{n})\sqsubseteq(\vee f(F_{1}))\vee\dots\vee(\vee
f(F_{k}))\sqsubseteq\vee f(F)$. This means that,

$f(D_{1})\vee\dots\vee f(D_{n})\subseteq\mathord{\downarrow}\\{\vee
f(F):\mathord{\downarrow}F\in\mathcal{D}\\}.\ \\\ (\ast\ast)$

Let $U\subseteq Y$ be an open neighborhood of $f(b_{1})\vee
f(b_{2})\vee\cdots\vee f(b_{k})$. By $(\ast)$, there exists some $(d_{1},\
d_{2}\ldots,\ d_{k})\in\prod\limits_{i=1}^{k}D_{i}$ such that $f(d_{1})\vee
f(d_{2})\vee\cdots\vee f(d_{k})\in U$. For each open set is an upper set, by
$(\ast\ast)$, there exists some $\mathord{\downarrow}F\in{\mathcal{D}}$ such
that $\vee f(F)\in U$. That is
$\bar{f}(\mathcal{D})\rightarrow\bar{f}(\mathord{\downarrow}G)$. Therefore, by
Proposition 3.7, $\bar{f}$ is continuous.

(3) Inflationary homomorphism $\bar{f}$ is unique.

Suppose there is an inflationary homomorphism $g:(LX,\ \cup)\rightarrow(Y,\
\vee)$ satisfy $f=g\circ i$. Then
$g(\mathord{\downarrow}x)=f(x)=\bar{f}(\mathord{\downarrow}x)$. For each
$\mathord{\downarrow}F\in LX$ (let $F=(a_{1},\ \dots,\ a_{n})$ ),

$\displaystyle g(\mathord{\downarrow}F)$ $\displaystyle=$ $\displaystyle
g(\mathord{\downarrow}a_{1}\cup\mathord{\downarrow}a_{2}\cdots\cup\mathord{\downarrow}a_{n})$
$\displaystyle=$ $\displaystyle g(\mathord{\downarrow}a_{1})\vee
g(\mathord{\downarrow}a_{2})\vee\cdots\vee g(\mathord{\downarrow}a_{n})$
$\displaystyle=$
$\displaystyle\bar{f}(\mathord{\downarrow}a_{1})\vee\bar{f}(\mathord{\downarrow}a_{2})\vee\cdots\vee\bar{f}(\mathord{\downarrow}a_{n})$
$\displaystyle=$
$\displaystyle\bar{f}(\mathord{\downarrow}a_{1}\cup\mathord{\downarrow}a_{2}\cdots\cup\mathord{\downarrow}a_{n})$
$\displaystyle=$ $\displaystyle\bar{f}(\mathord{\downarrow}F).$

Thus $\bar{f}$ is unique.

In conclusion, according to Definition 3.5, endowed with topology
$O_{\Rightarrow_{L}}(LX)$, the directed inflationary semilattice $(LX,\ \cup)$
is the lower powerspace of $X$, that is, $P_{L}(X)\cong(LX,\ \cup)$. $\Box$

The lower powerspace is unique in the sense of order isomorphism and
topological homomorphism, so we can directly denote the lower powerspace of
each directed space $X$ by $P_{L}(X)=(LX,\ \cup)$.

Suppose $X,\ Y$ are two directed spaces, $f:X\rightarrow Y$ is a continuous
function. Define map $P_{L}(f):P_{L}(X)\rightarrow P_{L}(Y)$ as follows:
$\forall\mathord{\downarrow}F\in LX$,

$P_{L}(f)(\mathord{\downarrow}F)=\mathord{\downarrow}f(F).$

it is evident that $P_{L}(f)$ is well-defined and order preserving. According
to 3.10, it is easy to check that $P_{L}(f)$ is an inflationary homomorphism
between these two lower powerspaces. If $id_{X}$ is the identity function and
$g:Y\rightarrow Z$ is an arbitrary continuous function from $Y$ to a directed
space $Z$, then, $P_{L}(id_{X})=id_{P_{L}(X)},\ P_{L}(g\circ f)=P_{L}(g)\circ
P_{L}(f)$. Thus, $P_{L}:{\bf Dtop}\rightarrow{\bf Disl}$ is a functor from
${\bf Dtop}$ to ${\bf Disl}$. Let $U:{\bf Disl}\rightarrow{\bf Dtop}$ be the
forgetful functor, by Theorem 3.10, we have the following result.

######  Corollary 3.11.

[34] $P_{L}$ is a left adjoint of the forgetful functor $U$, that is, ${\bf
Disl}$ is a reflective subcategory of ${\bf Dtop}$.

##  4 Relations Between Lower Powerspaces

In this section, we will discuss the relation between the lower powerdomain of
dcpo, observationally-induced lower powerspaces and directed lower
powerspaces.

Suppose $X$ is a topological space (denote the topology by $O(X)$ ). Let
$C(X)$ be the set of all nonempty closed sets of $X$. Obviously, $C(X)$ is
closed under the set operation $\cup$. For each $U\in O(X)$, let

$\langle U\rangle=\\{A\in C(X):A\cap U\not=\emptyset\\}.$

It is easy to check that, $\\{\langle U\rangle:U\in O(X)\\}$ is closed under
finitary intersection and arbitrary union, it forms a topology on $C(X)$,
called the lower Vietoris topology of $C(X)$ and denoted by $V_{L}(C(X))$.
Besides, $C(X)$ is a dcpo respect to the inclusion order of set. Denote the
Scott topology on $C(X)$ by $\sigma(C(X))$. We can check that, the lower
Vietoris topology $V_{L}(C(X))$ is contained in the Scott topology
$\sigma(C(X))$, that is, $V_{L}(C(X))\subseteq\sigma(C(X))$. Particularly,
write $P_{OL}(X)=(C(X),\ \cup)$ endowed with the lower Vietoris topology
$V_{L}(C(X))$, write $H(X)=(C(X),\ \cup)$ endowed with the Scott topology
$\sigma(C(X))$.

######  Theorem 4.1.

[7, Corollary IV-8.6] Let $P$ is a dcpo endowed with the Scott topology
$\sigma(P)$. Then $H(P)$ isomorphic to the lower powerdomian of $P$ (which is
also called Hoare powerdomain ), that is, $H(P)$ is free dcpo sup semilattice
of generated by $P$.

######  Theorem 4.2.

[2, Theorem 3.8] Let $X$ be a topological space. Then $P_{OL}(X)$ is
isomorphic to the observationally-induced lower powerspace over $X$.

Let $P$ be a dcpo, endowed with the Scott topology, then $(P,\ \sigma(P))$ is
also a directed space. By Theorem 3.10, the directed lower powerspace
$P_{L}(P)=(LP,\ \cup)$, endowed with topology $O_{\Rightarrow_{L}}(LP)$,
compared with the lower powerdomain and the observationally-induced lower
powerspace over $P$, in which the operation are both the union of sets $\cup$,
$LP$ is a subset of $C(P)$, and in general, $LP\not=C(P)$. For the topological
structure, $P_{L}(P)$ is neither necessarily a subspace of lower powerdomain
$H(P)$, nor a subspace of the observationally-induced lower powerspace
$P_{OL}(P)$.

The following example tells us that, in general, a dcpo $P$, endowed with
Scott topology, its lower powerdomain is not equal to its observationally-
induced lower powerspace, that is, $H(P)\not=P_{OL}(P)$.

######  Example 4.3.

[34] Let $P=(\mathbb{N}\times\mathbb{N})\cup\\{\infty\\}$, here $\mathbb{N}$
is the set of natural numbers. Define a partial order on $P$ as follows:
$\forall x,\ y\in P$, $x\leq y$ if and only if one of the following conditions
is true:

  * 1.

$y=\infty$,

  * 2.

$\exists n_{0}\in\mathbb{N},\ x=(m,\ n_{0}),\ y=(m^{\prime},\ n_{0})$ and
$m^{\prime}-m\geq 0$.

Evidently, $P$ is a dcpo. And for each nonempty subset $A$ , $A$ is a Scott
closed set of $P$ if and only if $A=P$ or there exists an antichain $B$ of $P$
such that $A=\mathord{\downarrow}B$. Then, we have
$(\mathbb{N}\times\\{n\\})\cup\\{\infty\\}$ is in $C(P)$ but not in $LP$, that
is $LP\not=C(P)$. For each $n\in\mathbb{N}$, let
$D_{n}=\mathbb{N}\times\\{n\\}$, then $D_{n}$ is a directed set and
$\bigvee(\mathbb{N}\times\\{n\\})=\infty$. For each $U\subseteq P$, $U$ is a
Scott open subset if and only if $U=\emptyset$ or $U\cap D_{n}\not=\emptyset$.
So, each nonempty Scott open set of $P$ equals to the upper set of its minimal
elements ${\rm min}U$. Let $\mathcal{U}\subseteq C(P)$, then $\mathcal{U}$ is
a open set respect to the lower Vietoris topology if and only if there exists
some $U\in\sigma(P)$ such that$\ \mathcal{U}=\\{B\in C(P):B\cap{\rm
min}U\not=\emptyset\\}$. Let

$\mathcal{V}=\\{A\in C(P):\exists n\in\mathbb{N},\ (5,n)\in A\\}\cup\\{A\in
C(P):\exists n\in\mathbb{N},\ (4,n),(4,n+1)\in A\\}.$

we can easily check that, $\mathcal{V}$ is a Scott open set of $C(P)$, and for
arbitrary $V\in\sigma(P)$, $\mathcal{V}\not=\langle V\rangle$. Thus,
$\mathcal{V}$ is not an open set in lower Vietoris topology. That is,
$H(P)\not=P_{OL}(P)$.

Next, we are going to consider the relation between the lower powerdomains and
the directed lower powerspaces.

Let $P$ be a dcpo endowed with Scott topology $\sigma(P)$. Write

$\sigma(C(P))|_{LP}=\\{\mathcal{U}\cap LP:\mathcal{U}\in\sigma(C(X))\\}$

to denote the relative topology on $LP$ from the Scott topology on $C(P)$. For
arbitrary $\mathcal{V}\in O_{\Rightarrow_{L}}(LP)$, let

$\uparrow_{C(P)}\mathcal{V}=\\{A\in
C(P):\exists\mathord{\downarrow}F\in\mathcal{V},\
\mathord{\downarrow}F\subseteq A\\}.$

######  Proposition 4.4.

[34] Let $P$ be a dcpo endowed with Scott topology $\sigma(P)$. Then
$\sigma(C(P))|_{LP}\subseteq O_{\Rightarrow_{L}}(LP)$. Particularly,
$\sigma(C(P))|_{LP}=O_{\Rightarrow_{L}}(LP)$ if and only if for each
$\mathcal{V}\in O_{\Rightarrow_{L}}(LP)$,
$\uparrow_{C(P)}\mathcal{V}\in\sigma(C(P))$.

Proof Let $\mathcal{U}\in\sigma(C(P))$, and a directed set
${\mathcal{D}}\subseteq LP$ satisfy
${\mathcal{D}}\Rightarrow_{L}\mathord{\downarrow}F\in\mathcal{U}\cap LP$. Then
for each $a\in F$, there exists a directed set
$D_{a}\subseteq\bigcup\mathcal{D}$ such that $D_{a}\rightarrow a$. Thus,
$\mathord{\downarrow}F\subseteq\overline{\bigcup{\mathcal{D}}}$, Here
$\overline{\bigcup{\mathcal{D}}}$ means the Scott closure of
$\bigcup{\mathcal{D}}$. But $\overline{\bigcup{\mathcal{D}}}$ is just the
supremum of $\bigcup{\mathcal{D}}$ in $C(P)$, Then
$\mathcal{D}\cap\mathcal{U}\cap LP=\mathcal{D}\cap\mathcal{U}\not=\emptyset$.
It follows that $\mathcal{U}\cap LP\in O_{\Rightarrow_{L}}(LP)$, that is,
$\sigma(C(P))|_{LP}\subseteq O_{\Rightarrow_{L}}(LP)$.

On the other hand, for arbitrary Scott closed set $A$ with $A\in\mathcal{U}$,
here $\mathcal{U}$ is a Scott open set of $C(P)$. Let
${\mathcal{F}}(A)=\\{\mathord{\downarrow}F:F\subseteq_{fin}A\\}$, then
${\mathcal{F}}(A)$ is a directed set of $LP$ and $A=\bigcup{\mathcal{F}}(A)$.
We have a nonempty finite set $F$ of $A$ such that
$\mathord{\downarrow}F\in\mathcal{U}$. It means that
$\mathcal{U}=\mathord{\uparrow}_{C(P)}(\mathcal{U}\cap LP)$. Thus,
$\sigma(C(P))|_{LP}=O_{\Rightarrow_{L}}(LP)$ if and only if
$\forall\mathcal{V}\in O_{\Rightarrow_{L}}(LP)$,
$\uparrow_{C(P)}\mathcal{V}\in\sigma(C(P))$. $\Box$

Suppose $A$ is an arbitrary subset of dcpo $P$. Let
$A^{1}=\mathord{\downarrow}\\{x\in P:\exists$ directed set
$D\subseteq\mathord{\downarrow}A,\ \bigvee D=x\\}$.

######  Lemma 4.5.

[6] Let $P$ be a quasi-continuous domain and $A\subseteq P$. Then, the Scott
closure of $A$ equals to $A^{1}$, that is $\overline{A}=A^{1}$.

######  Corollary 4.6.

[34] Let $P$ be a continuous or quasi-continuous domain. Then
$\sigma(C(X))|_{LP}=O_{\Rightarrow_{L}}(LP)$, that is, endowed with the Scott
topology, $P$ is a directed space, the directed lower powerspace $P_{L}(X)$ is
a subspace of the lower powerdomain $H(P)$.

Proof By Proposition 4.4, we only need to prove that, $\forall\mathcal{V}\in
O_{\Rightarrow_{L}}(LP)$, $\uparrow_{C(P)}\mathcal{V}\in\sigma(C(P))$. Suppose
${\mathcal{D}}\subseteq C(P)$ and
$\overline{\bigcup{\mathcal{D}}}\in\uparrow_{C(P)}\mathcal{V}$. Thus we have
some $\mathord{\downarrow}F\in\mathcal{V}$ such that
$\mathord{\downarrow}F\subseteq\overline{\bigcup{\mathcal{D}}^{\prime}}$. Let
${\mathcal{D}}^{\prime}=\\{\mathord{\downarrow}F:F\subseteq_{fin}\bigcup{\mathcal{D}}\\}$,
then ${\mathcal{D}}^{\prime}$ is a directed set of $LP$ and
$\bigcup{\mathcal{D}}^{\prime}=\bigcup{\mathcal{D}}$. By Lemma 4.5, $\forall
a\in F$, $\exists$ a directed set
$D_{a}\subseteq\bigcup{\mathcal{D}}^{\prime}$ such that $D_{a}\rightarrow a$.
Thus,
${\mathcal{D}}^{\prime}\Rightarrow_{L}\mathord{\downarrow}F\in\mathcal{V}$.
Noticed that $\mathcal{V}\in O_{\Rightarrow_{L}}(LP)$, so we have some
$\mathord{\downarrow}G\in{\mathcal{D}}^{\prime}\cap\mathcal{V}$. For
${\mathcal{D}}$ is directed, there exists $A\in{\mathcal{D}}$ such that
$\mathord{\downarrow}G\subseteq A$. Thus $A\in\uparrow_{C(P)}\mathcal{V}$.
That is, $\uparrow_{C(P)}\mathcal{V}\in\sigma(C(P))$. $\Box$

##  5 The Directed Upper Powerspaces of Directed Spaces

In this section, we will construct the directed upper powerspace of the
directed space, which is a free algebra generated by the directed deflationary
operation of the directed space.

######  Definition 5.1.

Let $X$ be a directed space.

  1. (1)

A binary operation $\oplus:X\otimes X\rightarrow X$ on $X$ is called a
deflationary operation if it is continuous and satisfy the following four
conditions: $\forall x,\ y,\ z\in X$,

    1. (a)

$x\oplus x=x$,

    2. (b)

$(x\oplus y)\oplus z=x\oplus(y\oplus z)$,

    3. (c)

$x\oplus y=y\oplus x$,

    4. (d)

$x\oplus y\leq x$.

  2. (2)

If $\oplus$ is a deflationary operation on $X$, then $(X,\ \oplus)$ is called
a directed deflationary semilattice, that is, directed deflationary
semilattices are those directed spaces with deflationary operations.

By Theorem 2.6(2), the operation $\oplus$ on a directed space $X$ is
continuous if and only if it is monotone and for each given $x,\ y\in X$ and
directed set $D\subseteq X$, $D\rightarrow x$ implies $(D\oplus y)\rightarrow
x\oplus y$. Here, $D\oplus y=\\{d\oplus y:d\in D\\}$.

Here are two examples.

######  Example 5.2.

  1. (1)

Suppose $P$ is a poset endowed with the Scott topology, and for each $a,\ b\in
P$, the infimum of $a$ and $b$ exists in $P$ (denote by $a\wedge b$). Then
$(P,\ \wedge)$ is a directed deflationary semilattice.

  2. (2)

Let $I=[0,\ 1]$ (the unit interval), let ${\mathcal{T}}$ denote the topology
generated by $\\{[0,\ a]:a\in I\\}$. It is easy to check that $(I,\
\mathcal{T})$ is a directed space, and $(I,\ min)$ is a directed deflationary
semilattice endowed with $\mathcal{T}$.

######  Definition 5.3.

Suppose $(X,\ \oplus),\ (Y,\ \uplus)\ $ are two directed deflationary
semilattices, function $f:(X,\ \oplus)\rightarrow(Y,\ \uplus)$ is called a
deflationary homomorphism between $X$ and $Y$, if $f$ is continuous and
$f(x\oplus y)=f(x)\uplus f(y)$ holds for $\forall x,\ y\in X$.

Denote the category of all directed deflationary semilattices and deflationary
homomorphisms by ${\bf Ddsl}$. Then ${\bf Ddsl}$ is a subcategory of ${\bf
Dtop}$.

######  Lemma 5.4.

Suppose $(X,\ \oplus)$ is a directed deflationary semilattice, then
$\oplus=\wedge_{\sqsubseteq}$. Here, $\forall x,\ y\in X,\
x\wedge_{\sqsubseteq}y$ means the infimum of $x$ and $y$ respect to the
specialization order $\sqsubseteq$ on $X$ (calling it meet operation).
Conversely, suppose $X$ is a directed space and for each $x,\ y\in X$,
$x\wedge_{\sqsubseteq}y$ exists, the continuity of $\wedge_{\sqsubseteq}$ will
naturally imply that $(X,\ \wedge_{\sqsubseteq})$ is a directed deflationary
semilattice.

Proof By Definition 5.1, $\forall x,\ y\in X,\ x\oplus y\leq x,\ y$, that is
$x\oplus y$ is a lower bound of $\\{x,\ y\\}$. Suppose $z$ is another
arbitrary lower bound of $\\{x,\ y\\}$, by Theorem 2.6(1), the pointwise order
equals to the specialization order of $X\otimes X$, then $(x,\
y)\sqsubseteq(z,\ z)$. By the continuity and idempotence of the deflationary
operation, we have $z\oplus z=z\sqsubseteq x\oplus y$. That is $x\oplus y$ is
the infimum of $\\{x,\ y\\}$, which means $x\oplus y=x\wedge_{\sqsubseteq}y$.
Conversely, continuous meet operation naturally satisfy all conditions in
Definition 5.1, thus, we get the conclusion. $\Box$

The above results show that a directed deflationary semilattice $(X,\ \oplus)$
is just a directed space with a continuous meet operation
$\wedge_{\sqsubseteq}$ satisfying $\oplus=\wedge_{\sqsubseteq}$. Since the
order on the directed space in this paper is always the specialization order,
we will use the symbol $\wedge$ instead of $\wedge_{\sqsubseteq}$ in the
following part. Therefore, a directed deflationary semilattice is always
represented by a tuple of the form $(X,\ \wedge)$, here $X$ is a directed
space, $\wedge$ represents the continuous meet operation on $X$.

Next, we give the definition of directed upper powerspace.

######  Definition 5.5.

Suppose $X$ is a directed space. A directed space $Z$ is called the directed
upper powerspace over $X$ if and only if the following two conditions are
satisfied:

  1. (1)

$Z$ is a directed deflationary semilattice, that is the meet operation
$\wedge$ on $Z$ exists and which is continuous,

  2. (2)

There is a continuous function $i:\ X\longrightarrow Z$ satisfy: for an
arbitrary directed deflationary semilattice $(Y,\ \wedge)$ and continuous
function $f:X\longrightarrow Y$, there exists a unique deflationary
homomorphism $\bar{f}:(Z,\ \wedge)\rightarrow(Y,\ \wedge)$ such that
$f=\bar{f}\circ i$.

By the definition above, if directed deflationary semilattices $(Z_{1},\
\wedge)$ and $(Z_{2},\ \wedge)$ are both the directed upper powerspaces of
$X$, then there exists a topological homomorphism which is also a deflationary
homomorphism $g:Z_{1}\rightarrow Z_{2}$. Therefore, up to order isomorphism
and topological homomorphism, the directed upper powerspace of a directed
space is unique. Particularly, we denote the directed upper powerspace of each
directed space $X$ by $P_{U}(X)$.

Now, we will prove the existence of the directed upper powerspace of each
directed space $X$ by way of concrete construction.

Let $X$ be a directed space. Denote

$UX=\\{\uparrow F:F\subseteq_{fin}X\\},$

here, $F\subseteq_{fin}X$ is an arbitrary nonempty finitary subset of $X$.
Define an order $\leq_{U}$ on $UX$ :

$\uparrow F_{1}\leq_{U}\uparrow F_{2}\iff\uparrow F_{2}\subseteq\uparrow
F_{1}.$

Let $\mathcal{F}\subseteq UX$ be a directed set(respect to order $\leq_{U}$)
and $\mathord{\uparrow}F\in UX$. Define a convergence notation
$\mathcal{F}\Rightarrow_{U}\mathord{\uparrow}F\iff$ there exists finite
directed sets $D_{1},\dots,D_{n}\subseteq X$ such that

  1. 1.

$F\cap\lim D_{i}\neq\emptyset,\ \forall D_{i}$;

  2. 2.

$F\subseteq\bigcup\limits_{i=1}^{n}\lim D_{i}$;

  3. 3.

$\forall(d_{1},\dots,d_{n})\in\prod\limits_{i=1}^{n}D_{i}$, there exists some
$\uparrow F^{{}^{\prime}}\in\mathcal{F}$, such that $\uparrow
F^{{}^{\prime}}\subseteq\bigcup\limits_{i=1}^{n}\uparrow d_{i}$.

A subset $\mathcal{U}\subseteq UX$ is called a $\Rightarrow_{U}$ convergence
open set of $UX$ if and only if for each directed subset $\mathcal{F}$ of $UX$
and $\mathord{\uparrow}F\in UX$, $\mathcal{F}\Rightarrow_{U}\uparrow
F\in\mathcal{U}$ implies $\mathcal{F}\cap\mathcal{U}\neq\emptyset$. Denote all
$\Rightarrow_{U}$ convergence open set of $UX$ by $O_{\Rightarrow_{U}}(UX)$.

######  Proposition 5.6.

Suppose $X$ is a directed space, the following are true:

  1. (1)

$(UX,\ O_{\Rightarrow_{U}}(UX))$ is a topological space, abbreviated as $UX$.

  2. (2)

The specialization order $\sqsubseteq$ of $(UX,\ O_{\Rightarrow_{U}}(UX))$
equals to $\leq_{U}$.

  3. (3)

$(UX,\ O_{\Rightarrow_{U}}(UX))$ is a directed space, that is
$O_{\Rightarrow_{U}}(UX)=d(UX)$.

Proof (1) Obviously we have $\emptyset,\ UX\in O_{\Rightarrow_{U}}(UX)$. If
$\mathcal{U}\in O_{\Rightarrow_{U}}(UX)$, and $\uparrow F_{1}\leq\uparrow
F_{2},\uparrow F_{1}\in\mathcal{U},\ F_{1}=\\{a_{1},\dots,a_{n}\\}$. Then it
is evident that $\\{\uparrow F_{2}\\}\Rightarrow_{U}\uparrow F_{1}$, since we
only need to take $D_{i}=\\{a_{i}\\},\ i=1,\dots,n$. Then, $\\{\uparrow
F_{2}\\}\cap\mathcal{U}\neq\emptyset$, this means $\uparrow
F_{2}\in\mathcal{U}$, and $\mathcal{U}$ is an upper set respect to order
$\leq_{U}$, $\mathcal{U}=\uparrow_{\leq}\mathcal{U}$.

Let $\mathcal{U}_{1},\mathcal{U}_{2}\in O_{\Rightarrow_{U}}(UX)$, and a
directed set $\mathcal{F}\subseteq UX$ with
$\mathcal{F}\Rightarrow_{U}\uparrow F\in\mathcal{U}_{1}\cap\mathcal{U}_{2}$,
then, there exists $\uparrow F_{1}\in\mathcal{F}\cap\mathcal{U}_{1}$ and
$\uparrow F_{2}\in\mathcal{F}\cap\mathcal{U}_{2}$, but $\mathcal{F}$ is
directed, we have $\uparrow F_{3}\in\mathcal{F},\ \uparrow
F_{3}\subset\uparrow F_{1}$ and $\uparrow F_{2}$. Then, $\uparrow
F_{3}\in\mathcal{F}\cap\mathcal{U}_{1}\cap\mathcal{U}_{2}$. By the same way,
we can prove that $O_{\Rightarrow_{U}}(UX)$ is closed under arbitrary union.
It follows that $O_{\Rightarrow_{U}}(UX)$ is a topology.

(2) Let $\mathord{\uparrow}F_{1},\ \mathord{\uparrow}F_{2}\in UX$. If
$\mathord{\uparrow}F_{1}\leq_{L}\mathord{\uparrow}F_{2}$, By the proof of (1),
each $\Rightarrow_{U}$ convergence open set is an upper set respect to
$\leq_{U}$, then
$\mathord{\uparrow}F_{1}\in\overline{\\{\mathord{\uparrow}F_{2}\\}}$, that is,
$\mathord{\uparrow}F_{1}\sqsubseteq\mathord{\uparrow}F_{2}$.

On the other hand, suppose
$\mathord{\uparrow}F_{1}\sqsubseteq\mathord{\uparrow}F_{2}$. We need to prove
that $\mathord{\uparrow}F_{1}\leq_{U}\mathord{\uparrow}F_{2}$, that is to
prove that $\\{\mathord{\uparrow}F\in
UX:\mathord{\uparrow}F_{2}\subseteq\mathord{\uparrow}F\\}$ is a closed set in
$UX$ respect to topology $O_{\Rightarrow_{U}}(UX)$, since
$\\{\mathord{\uparrow}F\in
UX:\mathord{\uparrow}F_{2}\subseteq\mathord{\uparrow}F\\}=\\{\mathord{\uparrow}F:\mathord{\uparrow}F\leq_{U}\mathord{\uparrow}F_{2}\\}\subseteq\\{\mathord{\uparrow}F:\mathord{\uparrow}F\sqsubseteq\mathord{\uparrow}F_{2}\\}=\overline{\\{\mathord{\uparrow}F_{2}\\}}$,
then, $\\{\mathord{\uparrow}F\in
UX:\mathord{\uparrow}F_{2}\subseteq\mathord{\downarrow}F\\}$ is a closed set
in $UX$ respect to $O_{\Rightarrow_{U}}(UX)$ implies
$\\{\mathord{\uparrow}F\in
UX:\mathord{\uparrow}F\subseteq\mathord{\uparrow}F_{2}\\}=\overline{\\{\mathord{\uparrow}F_{2}\\}}$,
that is $\uparrow F_{1}\leq_{U}\uparrow F_{2}$. Now, we prove that
$\\{\mathord{\uparrow}F\in
UX:\mathord{\uparrow}F_{2}\subseteq\mathord{\uparrow}F\\}$ is closed in $UX$
respect to $O_{\Rightarrow_{U}}(UX)$, equivalently,
$\mathcal{U}=UX\setminus\\{\mathord{\uparrow}F\in
UX:\mathord{\uparrow}F_{2}\subseteq\mathord{\uparrow}F\\}$ is a
$\Rightarrow_{U}$ convergence open set in $UX$.

By contradiction, suppose $\mathcal{U}$ is not a $\Rightarrow_{U}$ convergence
open set. Then there exists a directed set $\mathcal{F}$ of $UX$ with
$\mathcal{F}\Rightarrow_{U}\uparrow F\in\mathcal{U}$ but
$\mathcal{U}\cap\mathcal{F}=\emptyset$. According to the definition of
$\Rightarrow_{U}$ convergence, there exists finite directed sets
$D_{1},\dots,\ D_{n}\subseteq X$ such that

  1. 1.

$F\cap\lim D_{i}\neq\emptyset,\ i=1,\ 2,\ldots,n$;

  2. 2.

$F\subseteq\bigcup\limits_{i=1}^{n}\lim D_{i}$;

  3. 3.

$\forall(d_{1},\dots,d_{n})\in\prod\limits_{i=1}^{n}D_{i}$, there exists some
$\uparrow F^{{}^{\prime}}\in\mathcal{F}$, such that $\uparrow
F^{{}^{\prime}}\subseteq\bigcup\limits_{i=1}^{n}\uparrow d_{i}$.

Since $\uparrow F\in\mathcal{U}$, then $\uparrow F_{2}\nsubseteq\uparrow F$,
there exists some $a\in F_{2}$ with $a\not\in\mathord{\uparrow}F$, then
$F\subseteq X\backslash\mathord{\downarrow}a$. According to 1, 2 in the
definition above, for arbitrary $i\in\\{1,\ 2,\ \ldots,\ n\\}$,
$D_{i}\cap(X\backslash\mathord{\downarrow}a)\not=\emptyset$. For each $i$,
pick $d_{i}\in D_{i}\cap(X\backslash\mathord{\downarrow}a)$. Then $(d_{1},\
d_{2},\ \ldots,d_{n})\in\prod\limits_{i=1}^{n}D_{i}$ and
$a\not\in\bigcup\limits_{i=1}^{n}\mathord{\uparrow}d_{i}$. Since
${\mathcal{F}}\cap{\mathcal{U}}=\emptyset$, we have $\forall
F^{\prime}\in{\mathcal{F}}$,
$\mathord{\uparrow}F_{2}\subseteq\mathord{\uparrow}F^{\prime}$, thus
$\mathord{\uparrow}F^{\prime}\not\subseteq\bigcup\limits_{i=1}^{n}\mathord{\uparrow}d_{i}$,
this contradicts with 3 in the definition above. Therefore, $\mathcal{U}$ is a
$\Rightarrow_{U}$ convergence open set in $UX$.

(3) For an arbitrary topological space $X$, $O(X)\subseteq d(X)$ holds, then
$O_{\Rightarrow_{U}}(UX)\subseteq d(UX)$. On the other hand, according to the
definition of $\Rightarrow_{U}$ convergence topology, if directed set
$\mathcal{F}\subseteq UX$ with $\mathcal{F}\Rightarrow_{U}\uparrow F$, then
$\mathcal{F}$ convergents to $\mathord{\uparrow}F$ respect to
$O_{\Rightarrow_{U}}(UX)$. Thus, by the definition of directed open set,
$\mathcal{F}\Rightarrow_{F}\uparrow F\in\mathcal{U}\in d(UX)$ will imply
$\mathcal{U}\cap\mathcal{F}\neq\emptyset$. Then, $\mathcal{U}\in
O_{\Rightarrow_{U}}(UX)$, it follows that $O_{\Rightarrow_{U}}(UX)=d(UX)$,
that is, $(UX,\ O_{\Rightarrow_{U}}(UX))$ is a directed space. $\Box$

######  Proposition 5.7.

Suppose $X,\ Y$ are two directed spaces. Then function $f:(UX,\
O_{\Rightarrow_{U}}(UX))\rightarrow Y$ is continuous if and only if for each
directed set ${\mathcal{F}}\subseteq UX$ and $\mathord{\uparrow}F\in UX$,
${\mathcal{F}}\Rightarrow_{U}\mathord{\uparrow}F$ implies
$f(\mathcal{F})\rightarrow f(\mathord{\uparrow}F)$.

Proof Since $\Rightarrow_{U}$ convergence will lead to
$O_{\Rightarrow_{U}}(UX)$ topological convergence, the necessity is obvious.
We are going to prove the sufficiency. Firstly, we check that $f$ is monotone.
If $\mathord{\uparrow}F_{1},\ \mathord{\uparrow}F_{2}\in UX$ and
$\mathord{\uparrow}F_{1}\leq_{U}\mathord{\uparrow}F_{2}$, then
$\\{\mathord{\uparrow}F_{2}\\}\Rightarrow_{U}\mathord{\uparrow}F_{1}$, by the
hypothesis, $\\{f(\mathord{\uparrow}F_{2})\\}\rightarrow
f(\mathord{\uparrow}F_{1})$, thus $f(\mathord{\uparrow}F_{2})\sqsubseteq
f(\mathord{\uparrow}F_{1})$. Suppose $U$ is an open set of $Y$ and the
directed set ${\mathcal{F}}\Rightarrow_{U}\mathord{\uparrow}F\in f^{-1}(U)$,
then $f(\mathcal{F})$ is a directed set of $Y$ and $f(\mathcal{F})\rightarrow
f(\mathord{\uparrow}F)\in U$, thus
$\exists\mathord{\uparrow}F\in{\mathcal{D}}$ such that
$f(\mathord{\uparrow}F)\in U$. That is,
$\mathord{\uparrow}F\in{\mathcal{F}}\cap f^{-1}(U)$. According to the
definition of $\Rightarrow_{U}$ convergence open set, $f^{-1}(U)\in
O_{\Rightarrow_{U}}(UX)$, that is $f$ is continuous. $\Box$

Define a binary operation $\cup$ on $UX:\forall\mathord{\uparrow}F_{1},\
\mathord{\uparrow}F_{2}\in UX,\
\mathord{\uparrow}F_{1}\cup\mathord{\uparrow}F_{2}=\mathord{\uparrow}(F_{1}\cup
F_{2})$. Suppose we have
$\mathord{\uparrow}{F_{1}}=\mathord{\uparrow}{F_{2}},\
\mathord{\uparrow}{G_{1}}=\mathord{\uparrow}{G_{2}}$ with $F_{1}\neq F_{2},\
G_{1}\neq G_{2}$. $F_{1}\cup G_{1}\subseteq\mathord{\uparrow}(F_{2}\cup
G_{2})$ implies $\mathord{\uparrow}(F_{1}\cup
G_{1})\subseteq\mathord{\uparrow}(F_{2}\cup G_{2})$. Similarly, we have the
opposite containment, thus, $\cup$ is well-defined.

######  Theorem 5.8.

Let $X$ be a directed space. Then $(UX,\ O_{\Rightarrow_{U}}(UX))$ respect to
the set union operation $\cup$ is a directed deflationary semilattice.

Proof By Proposition 5.6, $(UX,\ O_{\Rightarrow_{U}}(UX))$ is a directed
space. We will prove that $\cup$ is a deflationary operation. For arbitrary
$\mathord{\uparrow}F_{1},\ \mathord{\uparrow}F_{2}\in UX$,
$\mathord{\uparrow}F_{1}\cup\mathord{\uparrow}F_{2}=\mathord{\uparrow}(F_{1}\cup
F_{2})\in UX$. Obviously, $\cup$ satisfy the conditions (a), (b), (c), (d) in
Definition 5.1, we now prove the continuity of $\cup$. The monotonicity of
$\cup$ is evident. By Theorem 2.6(2) and Proposition 5.7, we only need to
prove that, for each directed set ${\mathcal{F}}\subseteq UX$ and
$\mathord{\uparrow}F,\ \mathord{\uparrow}G\in UX$,
${\mathcal{F}}\Rightarrow_{U}\mathord{\uparrow}F$ will imply
$G\cup{\mathcal{F}}\Rightarrow_{U}\mathord{\uparrow}G\cup\mathord{\uparrow}F=\mathord{\uparrow}(G\cup
F)$. Here, $G\cup{\mathcal{F}}=\\{\mathord{\uparrow}(G\cup
F^{\prime}):\mathord{\uparrow}F^{\prime}\in\mathcal{F}\\}$ is still a directed
set. According to the definition of $\Rightarrow_{U}$ convergence, there
exists finite directed sets $D_{1},\dots,D_{k}\subseteq X$, satisfy the
conditions such that ${\mathcal{F}}\Rightarrow_{U}\mathord{\uparrow}F$. Let
$G=\\{a_{1},\dots,a_{n}\\}$, and $D_{k+1}=\\{a_{1}\\},\ D_{k+2}=\\{a_{2}\\},\
\ldots,\ D_{k+n}=\\{a_{n}\\}$. It is straightward to verify that, $D_{1},\
D_{2},\ \ldots,\ D_{k},\ D_{k+1},\ \ldots,\ D_{k+n}$ satisfy all the
conditions such that
$G\cup{\mathcal{F}}\Rightarrow_{U}(\mathord{\uparrow}G\cup\mathord{\uparrow}F)$.
It follows that, $(UX,\ \cup)$ is a directed deflationary semilattice. $\Box$

The following theorem is the main result in this paper.

######  Theorem 5.9.

Suppose $X$ is a directed space, then $(UX,\ O_{\Rightarrow_{U}}(UX))$ is the
directed upper powerspace over $X$, that is, endowed with topology
$O_{\Rightarrow_{U}}(UX)$, $(UX,\ \cup)\cong P_{U}(X)$.

Proof Define function $i:X\rightarrow UX$: $\forall x\in X$, $i(x)=\uparrow
x$. We prove the continuity of $i$. It is evident that $i$ is monotone.
Suppose we have directed set $D\subseteq X$ and $x\in X$ satisfy $D\rightarrow
x$. Let $\mathcal{D}=\\{\mathord{\uparrow}d:d\in D\\}$, then $\mathcal{D}$ is
a directed set in $UX$ and ${\mathcal{D}}\Rightarrow_{U}\uparrow x$. Notice
that $i(D)=\mathcal{D}$, so $i(D)\Rightarrow_{U}\mathord{\uparrow}x=i(x)$. By
Proposition 2.5, $i$ is continuous.

Let $(Y,\ \wedge)$ be an arbitrary directed deflationary semilattice,
$f:X\rightarrow Y$ is a continuous function. Define $\bar{f}:UX\rightarrow Y$
as follows: $\forall\mathord{\uparrow}F\in UX$ (let $F=\\{a_{1},\ a_{2},\
\ldots,\ a_{n}\\}$),

$\bar{f}(\mathord{\uparrow}F)=f(a_{1})\wedge f(a_{2})\wedge\cdots\wedge
f(a_{n})=\bigwedge\limits_{a\in F}f(a).$

Particularly, we write $\bar{f}(\mathord{\uparrow}F)=\wedge f(F)$. $\bar{f}$
is well-defined, since if we have $\mathord{\uparrow}F=\mathord{\uparrow}G$
with $F\neq G$, then $f(F)\subseteq\mathord{\uparrow}f(G)$ implies $\wedge
f(G)\leq\wedge f(F)$, that is
$\bar{f}(\mathord{\uparrow}G)\leq\bar{f}(\mathord{\uparrow}F)$. Similarly, we
have $\bar{f}(\mathord{\uparrow}F)\leq\bar{f}(\mathord{\uparrow}G)$. Thus, $f$
is well-defined.

(1) $f=\bar{f}\circ i$.

For arbitrary $x\in X$, $(\bar{f}\circ
i)(x)=\bar{f}(i(x))=\bar{f}(\mathord{\uparrow}x)=f(x)$.

(2) $\bar{f}$ is a deflationary homomorphism, that is, $\bar{f}$ is continuous
and for arbitrary $\mathord{\uparrow}F_{1},\ \mathord{\uparrow}F_{2}\in UX$,
$\bar{f}(\mathord{\uparrow}F_{1}\cup\mathord{\uparrow}F_{2})=\bar{f}(\mathord{\uparrow}F_{1})\wedge\bar{f}(\mathord{\uparrow}F_{2})$.

First, we prove that $\bar{f}$ preserves the union operation. Suppose
$\mathord{\uparrow}F_{1},\ \mathord{\uparrow}F_{2}\in UX$. Then
$\bar{f}(\mathord{\uparrow}F_{1}\cup\mathord{\uparrow}F_{2})=\bar{f}(\mathord{\uparrow}(F_{1}\cup
F_{2}))=\wedge f(F_{1}\cup F_{2})=(\wedge f(F_{1})\wedge(\wedge
f(F_{2}))=\bar{f}(\mathord{\uparrow}F_{1})\wedge\bar{f}(\mathord{\uparrow}F_{2})$.
Next, we prove the continuity of $\bar{f}$. Notice that $\wedge$ is the meet
operation, $\bar{f}$ is evidently monotone. Suppose ${\mathcal{F}}\subseteq
UX$ is a directed set and $\mathord{\uparrow}F\in UX$ satisfy
${\mathcal{F}}\Rightarrow_{U}\mathord{\uparrow}F$. By the definition of
$\Rightarrow_{U}$, there exists finite directed sets
$D_{1},\dots,D_{n}\subseteq X$ such that

  1. 1.

$F\cap\lim D_{i}\neq\emptyset,\ i=1,2,\ldots,n$;

  2. 2.

$F\subseteq\bigcup\limits_{i=1}^{n}\lim D_{i}$;

  3. 3.

$\forall(d_{1},\dots,d_{n})\in\prod\limits_{i=1}^{n}D_{i}$, there exists some
$\uparrow F^{{}^{\prime}}\in\mathcal{F}$, such that $\uparrow
F^{{}^{\prime}}\subseteq\bigcup\limits_{i=1}^{n}\uparrow d_{i}$.

Let $F=\\{b_{1},\ b_{2},\ldots,b_{k}\\}$. By 1, for each $1\leq i\leq n$, we
have some $b_{i}\in F$ such that $D_{i}\rightarrow b_{i}$. If
$F\setminus\\{b_{1},\dots,b_{n}\\}\neq\emptyset$, which is denoted by
$G=\\{a_{1},\ a_{2}\ldots,\ a_{s}\\}$. By 2, For each $a_{j}\in G$, we have
$1\leq i_{j}\leq n$ such that $D_{i_{j}}\rightarrow a_{j}$. By the continuity
of $f$, then $f(D_{i})\rightarrow f(b_{i}),\ i=1,\dots,n$ and,
$f(D_{i_{j}})\rightarrow f(a_{j})$, $j=1,2,\ldots,s$. Since the meet operation
$\wedge$ on $Y$ is continuous, the following convergence holds

$f(D_{1})\wedge\cdots\wedge f(D_{n})\wedge f(D_{i_{1}})\wedge\cdots\wedge
f(D_{i_{s}})\rightarrow f(b_{1})\wedge\dots\wedge f(b_{n})\wedge
f(a_{i_{1}})\wedge\cdots\wedge f(a_{i_{s}}).\ \\\ (\ast)$

Here, $f(D_{1})\wedge\cdots\wedge f(D_{n})\wedge
f(D_{i_{1}})\wedge\cdots\wedge f(D_{i_{s}})=\\{f(d_{1})\wedge\cdots\wedge
f(d_{n})\wedge f(d_{i_{1}})\wedge\cdots\wedge f(d_{i_{s}}):(d_{1},\ \ldots,\
d_{k},\ d_{i_{1}},\ \ldots,\
d_{i_{s}})\in(\prod\limits_{i=1}^{n}D_{i}\\})\times(\prod\limits_{j=1}^{s}D_{i_{j}})\\}$.
Let $U$ be an arbitrary open neighborhood of $\wedge f(F)$, by $(\ast)$, there
exists $(d_{1},\ \ldots,\ d_{n},\ d_{i_{1}},\ \ldots,\
d_{i_{s}})\in(\prod\limits_{i=1}^{n}D_{i}\\})\times(\prod\limits_{j=1}^{s}D_{i_{j}})$
such that $f(d_{1})\wedge\cdots\wedge f(d_{n})\wedge
f(d_{i_{1}})\wedge\cdots\wedge f(d_{i_{s}})\in U$. For each $D_{i_{j}}$
repeats $D_{i}$, and each $D_{i}$ is directed, we have $(d^{{}^{\prime}}_{1},\
d^{{}^{\prime}}_{2},\ \ldots,\
d^{{}^{\prime}}_{n})\in\prod\limits_{i=1}^{n}D_{i}$ such that
$f(d^{{}^{\prime}}_{1})\wedge\cdots\wedge f(d^{{}^{\prime}}_{n})\sqsupseteq
f(d_{1})\wedge\cdots\wedge f(d_{n})\wedge f(d_{i_{1}})\wedge\cdots\wedge
f(d_{i_{s}})$. By 3, there exists some $\uparrow
F^{{}^{\prime}}\in\mathcal{F}$ such that $\uparrow
F^{{}^{\prime}}\subseteq\bigcup\limits_{i=1}^{n}\uparrow d^{{}^{\prime}}_{i}$.
Thus $\bar{f}(\uparrow F^{{}^{\prime}})=\wedge f(F^{{}^{\prime}})\sqsupseteq
f(d^{{}^{\prime}}_{1})\wedge\dots\wedge f(d^{{}^{\prime}}_{n})$. But $U$ is an
upper set, it follows that $\wedge f(F^{{}^{\prime}})=\bar{f}(\uparrow
F^{{}^{\prime}})\in U$, then $\bar{f}({\mathcal{F}})=\\{\wedge
f(F^{{}^{\prime}}):\uparrow F^{{}^{\prime}}\in\mathcal{F}\\}\rightarrow\wedge
f(F)$. By Proposition 5.7, function $\bar{f}$ is continuous.

(3) Homomorphism $\bar{f}$ is unique.

Suppose we have a deflationary homomorphism $g:(UX,\ \cup)\rightarrow(Y,\
\wedge)$ such that $f=g\circ i$, then $g(\uparrow
x)=f(x)=\bar{f}(\mathord{\uparrow}x)$. For each $\mathord{\uparrow}F\in UX$
(let $F=(a_{1},\dots,a_{n})$),

$\displaystyle g(\uparrow F)$ $\displaystyle=$ $\displaystyle
g(\mathord{\uparrow}a_{1}\cup\mathord{\uparrow}a_{2}\cdots\cup\mathord{\uparrow}a_{n})$
$\displaystyle=$ $\displaystyle g(\uparrow a_{1})\wedge
g(\mathord{\uparrow}a_{2})\wedge\cdots\wedge g(\uparrow a_{n})$
$\displaystyle=$ $\displaystyle\bar{f}(\uparrow
a_{1})\wedge\bar{f}(\mathord{\uparrow}a_{2})\wedge\cdots\wedge\bar{f}(\uparrow
a_{n})$ $\displaystyle=$
$\displaystyle\bar{f}(\mathord{\uparrow}a_{1}\cup\mathord{\uparrow}a_{2}\cdots\cup\mathord{\uparrow}a_{n})$
$\displaystyle=$ $\displaystyle\bar{f}(\uparrow F).$

Thus $\bar{f}$ is unique.

In conclusion, according to definition 5.1, endowed with topology
$O_{\Rightarrow_{U}}(UX)$, the directed deflationary semilattice $(UX,\ \cup)$
is the directed upper powerspace of $X$, that is, $P_{U}(X)\cong(UX,\ \cup)$.
$\Box$

The directed upper powerspace is unique in the sense of order isomorphism and
topological homomorphism, so we can directly denote the directed upper
powerspace by $P_{U}(X)=(UX,\ \cup)$ of each directed space $X$.

Suppose $X,\ Y$ are two directed spaces, $f:X\rightarrow Y$ is a continuous
function. Define function $P_{U}(f):P_{U}(X)\rightarrow P_{U}(Y)$ as follows:
$\forall\mathord{\uparrow}F\in UX$,

$P_{U}(f)(\mathord{\uparrow}F)=\mathord{\uparrow}f(F).$

We can check that, $P_{U}(f)$ is well-defined and order preserving. It is easy
to check that, $P_{U}(f)$ is a deflationary homomorphism between these two
lower powerspaces. If $id_{X}$ is the identity function and $g:Y\rightarrow Z$
is an arbitrary continuous function from $Y$ to a directed space $Z$, then,
$P_{U}(id_{X})=id_{P_{U}(X)},\ P_{U}(g\circ f)=P_{U}(g)\circ P_{U}(f)$. Thus,
$P_{U}:{\bf Dtop}\rightarrow{\bf Ddsl}$ is a functor from ${\bf Dtop}$ to
${\bf Ddsl}$. Let $U:{\bf Ddsl}\rightarrow{\bf Dtop}$ be the forgetful
functor. By Theorem 5.9, we have the following result.

######  Proposition 5.10.

$P_{U}$ is a deflationary homomorphism between $P_{U}(X)$ and $P_{U}(Y)$ for
each directed spaces $X$ and $Y$.

Proof It is directed to check that $P_{U}$ preserves deflationary operation
between $P_{U}(X)$ and $P_{U}(Y)$ for each directed spaces $X$ and $Y$. We
only to chek that $P_{U}$ is continuous.

Let $X$ and $Y$ be two directed spaces, let $\mathcal{D}=\\{f_{i}\\}_{i\in I}$
be a directed set in $[X\rightarrow Y]$ and convergent to $f\in[X\rightarrow
Y]$ pointwise. Let $\uparrow F\in P_{U}(X)$ and $F=\\{a_{1},\dots,a_{n}\\}$,
we shall verify that $\\{P_{U}(f_{i})(\uparrow F)\\}_{i\in I}\rightarrow
P_{U}(f)(\uparrow F)$, that is $\\{\uparrow f_{i}(F)\\}_{i\in
I}\rightarrow\uparrow f(F)$ in $P_{U}(Y)$. According to the definition of
$\Rightarrow_{U}$ convergence, firstly, we have $D_{a}=\\{f_{i}(a)\\}_{i\in
I}\rightarrow f(a)$ in $Y$ for each $a\in F$. Secondly, for each
$f_{i_{1}}(a_{1})\in D_{a_{1}},\dots,f_{i_{n}}(a_{n})\in D_{a_{n}}$, since
each $a\in F,\ D_{a}$ is directed, we have some $\uparrow
f_{i_{0}}(F)\in\\{\uparrow f_{i}(F)\\}_{i\in I}$ such that $\uparrow
f_{i_{0}}(F)\subseteq\uparrow(f_{i_{1}}(a_{1}),\dots,f_{i_{n}}(a_{n}))$. Thus
$\\{\uparrow f_{i}(F)\\}_{i\in I}\Rightarrow\uparrow_{U}f(F)$, $P_{U}$ is
continuous.

######  Corollary 5.11.

$P_{U}$ is a left adjoint of the forgetful functor $U$, that is, ${\bf Ddsl}$
is a reflective subcategory of ${\bf Dtop}$.

##  6 Relations Between Upper Powerspace

In this section, we will discuss the relations between the upper powerdomains
of dcpos, the observationally-induced upper powerspaces and the directed upper
powerspaces.

According to the results in the last section, for an arbitrary directed space,
the directed lowerspace is the set of $UX$, endowed with the $\Rightarrow_{U}$
convergence topology, the deflationary operation equals to the union operation
of sets. In general, for an arbitrary topological space and arbitrary dcpo,
although the observationally-induced upper powerspace and the upper
powerdomain exist, their concrete structure cannot be expressed(see article
[2, 13, 14]).

Let $(X,\ O(X))$ be a topological space. We say that an nonempty set
$A\subseteq X$ is a saturated set, if $A=\bigcap\\{U\in O(X):A\subseteq U\\}$.
Denote all nonempty compact saturated sets of $X$ by $Q(X)$. For each $U\in
O(X)$, let

$[U]=\\{K\in Q(X):K\subseteq U\\}.$

Denote ${\mathcal{B}}_{X}=\\{[U]:U\in O(X)\\}$. The upper Vietoris topology of
$Q(X)$ is generated by subbase ${\mathcal{B}_{X}}$, denote by $V_{U}(Q(X))$.
Particularly, for a dcpo, endowed with the Scott topology, the compact
saturated sets are just the compact upper sets.

######  Theorem 6.1.

[2] Suppose $X$ is a sober and locally compact space, $(Q(X),\ V_{U}(Q(X))$ is
order isomorphic and topological homomorphic to the observationally-induced
upper space of $X$ respect to the union operation of sets. Under this
condition, we have $V_{U}(Q(X))=\sigma(Q(X)$. Here, $Q(X)$ with the order
reverse to containment, $\sigma(Q(X)$ denotes the Scott topology.

######  Theorem 6.2.

[7] Let $P$ be a continuous domain. Then $(Q(P),\ \supseteq)$, endowed with
the Scott topology and the order reverse to containment, isomorphic to the
upper powerdomain over $P$ (which is also called Smyth powerdomain ) $S(P)$,
besides, the following holds:

  1. (1)

$(Q(P),\ \supseteq)$ is a continuous meet semilattice,

  2. (2)

$\forall K\in Q(P)$, $K=\bigcap\\{\mathord{\uparrow}F:1\leq|F|<\omega\ \&\
K\subseteq(\mathord{\uparrow}F)^{\circ}\\}$.

According to these two theorems above, for each continuous domain with the
Scott topology, its observationally-induced upper powerspace isomorphic to the
upper powerdomain. Next, we discuss the directed upper powerspace of
continuous domain.

Let $X$ be a continuous domain. Then $(X,\ \sigma(X))$ is a directed space,
and each upper set of nonempty finitary elements is a compact saturated set.
Thus, $UX\subseteq Q(X)$. By $\sigma(Q(X))|_{UX}$ denote the relative topology
from the Scott topology on $Q(X)$.

######  Proposition 6.3.

Suppose $X$ is a continuous domain which is endowed with Scott topology
$\sigma(X)$.

  1. (1)

For each given directed set ${\mathcal{F}}\subseteq UX$ and
$\mathord{\uparrow}F\in UX$, ${\mathcal{F}}\Rightarrow_{U}\mathord{\uparrow}F\
\Leftrightarrow\
\bigcap\\{\mathord{\uparrow}G:\mathord{\uparrow}G\in{\mathcal{F}}\\}\subseteq\mathord{\uparrow}F$.

  2. (2)

$O_{\Rightarrow_{U}}(UX)=\sigma(Q(X))|_{UX}$.

Proof (1) Suppose ${\mathcal{F}}\subseteq UX$ is a directed set of $UX$,
$\mathord{\uparrow}F\in UX$, and
${\mathcal{F}}\Rightarrow_{U}\mathord{\uparrow}F$. By definition, there exist
finitary directed sets $D_{1}\dots,D_{n}\subseteq X$ such that

  1. 1.

$F\cap\lim D_{i}\neq\emptyset,\ i=1,2,\ldots,\ n$;

  2. 2.

$F\subseteq\bigcup\limits_{i=1}^{n}\lim D_{i}$;

  3. 3.

$\forall(d_{1},\dots,d_{n})\in\prod\limits_{i=1}^{n}D_{i}$, there exists some
$\uparrow F^{{}^{\prime}}\in\mathcal{F}$, such that $\uparrow
F^{{}^{\prime}}\subseteq\bigcup\limits_{i=1}^{n}\uparrow d_{i}$.

By contradiction, suppose
$\bigcap\\{\mathord{\uparrow}G:\mathord{\uparrow}G\in{\mathcal{F}}\\}\not\subseteq\mathord{\uparrow}F$.
Then there exists some
$a\in\bigcap\\{\mathord{\uparrow}G:\mathord{\uparrow}G\in{\mathcal{F}}\\}$
such that $a\not\in\mathord{\uparrow}F$. Thus, $F\subseteq
X\backslash\mathord{\downarrow}a$. By 1 and 2, for each $i$, we have $d_{i}\in
D_{i}$ such that $d_{i}\in X\backslash\mathord{\downarrow}a$. By 3, there
exists some $\mathord{\uparrow}F^{{}^{\prime}}\in{\mathcal{F}}$ such that
$\mathord{\uparrow}F^{{}^{\prime}}\subseteq\bigcup\limits_{i=1}^{n}\mathord{\uparrow}d_{i}$,
this contradicts with
$a\not\in\bigcup\limits_{i=1}^{n}\mathord{\uparrow}d_{i}$. That is
$\bigcap\\{\mathord{\uparrow}G:\mathord{\uparrow}G\in{\mathcal{F}}\\}\subseteq\mathord{\uparrow}F$.

On the other hand, suppose
$\bigcap\\{\mathord{\uparrow}G:\mathord{\uparrow}G\in{\mathcal{F}}\\}\subseteq\mathord{\uparrow}F$.
Let $F=\\{a_{1},\ a_{2},\ldots,a_{n}\\}$. For $X$ is a continuous domain, then
each
$\mathord{\mbox{\makebox[0.0pt][l]{\raisebox{-1.72218pt}{$\downarrow$}}$\downarrow$}}a_{i}$
is directed and
$a_{i}=\bigvee\mathord{\mbox{\makebox[0.0pt][l]{\raisebox{-1.72218pt}{$\downarrow$}}$\downarrow$}}a_{i}$,
$i=1,2,\ldots,n$. Let
$D_{i}=\mathord{\mbox{\makebox[0.0pt][l]{\raisebox{-1.72218pt}{$\downarrow$}}$\downarrow$}}a_{i}$,
$D_{i}\rightarrow a_{i}$, then, each $D_{i}$ satisfy 1 and 2 in the definition
of $\Rightarrow_{U}$. For arbitrary $(d_{1},d_{2},\ldots,\
d_{n})\in\prod\limits_{i=1}^{n}D_{i}$, we have
$\mathord{\uparrow}F\subseteq\bigcup\limits_{i=1}^{n}\mathord{\mbox{\makebox[0.0pt][l]{\raisebox{1.72218pt}{$\uparrow$}}$\uparrow$}}d_{i}=(\bigcup\limits_{i=1}^{n}\mathord{\uparrow}d_{i})^{\circ}$.
Since each continuous domain is well-filtered, it follows that there exists
some $\mathord{\uparrow}G\in{\mathcal{F}}$ such that
$\mathord{\uparrow}G\subseteq(\bigcup\limits_{i=1}^{n}\mathord{\uparrow}d_{i})^{\circ}$,
then, 3 in the definition of $\Rightarrow_{U}$ holds. In conclusion,
${\mathcal{F}}\Rightarrow_{U}\mathord{\uparrow}F$.

(2) Suppose ${\mathcal{U}}\in O_{\Rightarrow_{U}}(UX)$. Let
${\mathcal{U}}_{Q}=\\{K\in Q(X):\exists\mathord{\uparrow}F\in{\mathcal{U}},\
K\subseteq\mathord{\uparrow}F\\}$. Obviously,
${\mathcal{U}}={\mathcal{U}}_{Q}\cap Q(X)$. Let ${\mathcal{K}}\subseteq Q(X)$
be a directed set respect to the order reverse to containment and
$\bigcap\\{K:K\in{\mathcal{K}}\\}\in{\mathcal{U}}_{Q}$. There exists some
$\mathord{\uparrow}F\in{\mathcal{U}}$ such that
$\bigcap\\{K:K\in{\mathcal{K}}\\}\subseteq\mathord{\uparrow}F$. Let
$F=\\{a_{1},a_{2},\ldots,\ a_{n}\\}$. Since $X$ is a continuous domain, each
$\mathord{\mbox{\makebox[0.0pt][l]{\raisebox{-1.72218pt}{$\downarrow$}}$\downarrow$}}a_{i}$
is directed and
$a_{i}=\bigvee\mathord{\mbox{\makebox[0.0pt][l]{\raisebox{-1.72218pt}{$\downarrow$}}$\downarrow$}}a_{i}$,
$i=1,2,\ldots,n$. Let
${\mathcal{F}}=\\{\bigcup_{i=1}^{n}\mathord{\uparrow}d_{i}:d_{i}\ll a_{i},\
i=1,2,\ldots,n\\}$. For $X$ is a continuous domain, then $Q(X)$ is a
continuous domain, we have
$\mathord{\uparrow}F\subseteq(\bigcup_{i=1}^{n}\mathord{\uparrow}d_{i})^{\circ}\subseteq\bigcup_{i=1}^{n}\mathord{\uparrow}d_{i}$,
$i=1,2,\ldots,\ n$, besides, $\cap\mathcal{F}=\mathord{\uparrow}F$. By (1),
${\mathcal{F}}\Rightarrow_{U}\mathord{\uparrow}F$. Thus, we have $(d_{1},\
d_{2},\ \ldots d_{n})\in\prod_{1\leq i\leq
n}\mathord{\mbox{\makebox[0.0pt][l]{\raisebox{-1.72218pt}{$\downarrow$}}$\downarrow$}}a_{i}$
such that $\bigcup_{i=1}^{n}\mathord{\uparrow}d_{i}\in{\mathcal{U}}$. Notice
that
$\bigcap\\{K:K\in{\mathcal{K}}\\}\subseteq\mathord{\uparrow}F\subseteq(\bigcup_{i=1}^{n}\mathord{\uparrow}d_{i})^{\circ}$,
there exists some $K\in{\mathcal{K}}$ such that
$K\subseteq(\bigcup_{i=1}^{n}\mathord{\uparrow}d_{i})^{\circ}$. By the
definition of ${\mathcal{U}}_{Q}$, $K\in{\mathcal{U}}$, that is,
${\mathcal{U}}_{Q}$ is a Scott open set in $Q(X)$. Therefore,
$O_{\Rightarrow_{U}}(UX)\subseteq\sigma(Q(X))|_{UX}$.

On the other hand, let ${\mathcal{V}}\in\sigma(Q(X))$, and
${\mathcal{F}}\subseteq UX$ is a directed set with
${\mathcal{F}}\Rightarrow_{U}\mathord{\uparrow}F\in{\mathcal{V}}\cap UX$. By
(1),
$\bigcap\\{\mathord{\uparrow}G:\mathord{\uparrow}G\in{\mathcal{F}}\\}\in{\mathcal{V}}$,
there exists some $\mathord{\uparrow}G\in{\mathcal{F}}\cap{\mathcal{V}}$, that
is, ${\mathcal{F}}\cap{\mathcal{V}}\cap UX\not=\emptyset$. Thus,
$\sigma(Q(X))|_{UX}\subseteq O_{\Rightarrow_{U}}(UX)$. We get the conclusion.
$\Box$

######  Example 6.4.

Let $X={\mathbb{R}}^{n}$ be the $n-$dimensional Euclidean space. Then $X$ is
locally compact $T_{2}$ space, naturally sober and locally compact. Denote the
observationally-induced upper space over $X$ by $P_{O}(X)$. By Theorem 6.1,
$P_{O}(X)=\\{K\subseteq X:K$ is a nonempty compact set of $X$ $\\}$ and the
topology of which is the Scott topology. It is easy to check that, $P_{O}(X)$
is a continuous domain and for each nonempty compact set $K\subseteq X$,
$\\{\\{a\\}:a\in K\\}$ is a compact saturated set of $P_{O}(X)$. It follows
that, for directed space $P_{O}(X)$, its directed upper powerspace
$P_{U}(P_{O}(X))\not=Q(P_{O}(X))$.

##  7 The Directed Convex Powerspace

In this section, we will construct the directed convex powerspace of the
directed space, which is a free algebra generated by the directed semilattice
operation of the directed space.

######  Definition 7.1.

Let $X$ be a directed space.

  1. (1)

A binary operation $+:X\otimes X\rightarrow X$ on $X$ is called a semilattice
operation if it is continuous and satisfy the following three conditions::
$\forall x,\ y,\ z\in X$,

    1. (a)

$x+x=x$,

    2. (b)

$(x+y)+z=x+(y+z)$,

    3. (c)

$x+y=y+x$.

  2. (2)

If $+$ is a directed semilattice operation on $X$, then $(X,\ +)$ is called a
directed semilattice, that is, directed semilattices are those directed spaces
with semilattice operations.

By Theorem 2.6(2), the operation $+$ on $X$ is continuous if and only if it is
monotone and for each given $x,\ y\in X$ and directed set $D\subseteq X$,
$D\rightarrow x$ implies $(D+y)\rightarrow x+y$. Here, $D+y=\\{d+y:d\in D\\}$.

######  Example 7.2.

  1. (1)

Suppose $(P,\ +)$ is a directed inflationary(or deflationary)semilattice,
then, $(P,\ +)$ is a directed semilattice.

  2. (2)

Let $X=\\{0,\ 1,\ 2\\}$ and $0\leq 1\leq 2$, define $0+0=0,\ 1+1=1,\ 2+2=2,\
0+1=1,\ 1+2=1,\ 0+2=1$. Then $(X,\ +\ )$, endowed with the Scott topology, is
a directed semilattice.

######  Definition 7.3.

Suppose $(X,\ \oplus),\ (Y,\ \uplus)$ are two directed semilattice, function
$f:(X,\ \oplus)\rightarrow(Y,\ \uplus)$ is called a directed semilattice
homomorphism between $X$ and $Y$, if $f$ is continuous respect to these two
directed space and $f(x\oplus y)=f(x)\uplus f(y)$ holds, $\forall x,\ y\in X$.

Denote the category of all directed semilattice and directed homomorphism by
${\bf Dsl}$. Then ${\bf Dsl}$ is a subcategory of the category of ${\bf
Dtop}$.

Next, we give the definition of directed convex powerspace.

######  Definition 7.4.

Suppose $X$ is a directed space. A directed space $Z$ is called the directed
convex powerspace over $X$ if and only if the following two conditions are
satisfied:

  1. (1)

$Z$ is a directed semilattice, that is the semilattice operation $+$ on $Z$
exists and which is continuous,

  2. (2)

There is a continuous function $i:X\longrightarrow Z$ satisfy: for an
arbitrary directed semilattice $(Y,\ +)$ and continuous function
$f:X\longrightarrow Y$, there exists an unique directed semilattice
homomorphism $\bar{f}:(Z,\ +)\rightarrow(Y,\ +)$ such that $f=\bar{f}\circ i$.

By the definition above, if directed semilattices $(Z_{1},\ +)$ and $(Z_{2},\
+)$ are both the directed convex powerspaces of $X$, then, there exists a
topological homomorphism which is still a directed semilattice homomorphism
$g:Z_{1}\rightarrow Z_{2}$. Therefore, up to order isomorphism and topological
homomorphism, the directed convex powerspace of a directed space is unique.
Particularly, we denote the directed convex powerspace of each directed space
$X$ by $P_{P}(X)$.

Now, we will prove the existence of the directed convex powerspace of each
directed space $X$ by way of concrete construction.

Let $X$ be a directed space, for each compact set $A\subseteq X$, $\bar{A}$
denotes the closure of $A$, and $sat(A)$ the saturation of $A$, that is
$sat(A)=\cap\\{U\in O(X):A\subseteq U\\}$. Thus, for each finite subset $F$,
it follows that $\bar{F}=\mathord{\downarrow}F,\ sat(F)=\mathord{\uparrow}F$
respect to the specialization order of $X$. Denote

$\widehat{A}=(\bar{A},\ sat(A)),\ \\\ PX=\\{\widehat{F}:F\subseteq_{fin}X\\},\
$

Here, $F\subseteq_{fin}X$ is an arbitrary nonempty finitary subset of $X$.
Define an order $\leq_{P}$ on $PX$ as follows:

$\widehat{F_{1}}\leq_{P}\widehat{F_{2}}\iff\mathord{\downarrow}F_{1}\subseteq\mathord{\downarrow}F_{2}\
\&\uparrow F_{2}\subseteq\uparrow F_{1}.$

Let $\mathcal{D}=\\{\widehat{F_{i}}\\}_{i\in I}\subseteq PX$ be a directed set
(respect to the order $\leq_{P}$), and $\widehat{F}\in PX$. Define a
convergence notation $\mathcal{D}\Rightarrow_{P}\widehat{F}$ as follows:

$\mathcal{D}\Rightarrow_{P}\widehat{F}\iff$ there exists finite directed sets
$D_{1},\dots,D_{k}$ of $X$ satisfy the following four conditions:

  1. (1)

$D_{i}\subseteq\bigcup\limits_{i\in I}\mathord{\downarrow}F_{i},\
i=1,\dots,k$;

  2. (2)

$\forall i=1,\dots,k,\ F\cap\lim D_{i}\neq\emptyset$;

  3. (3)

$F\subseteq\bigcup\limits_{i=1}^{k}\lim D_{i}$;

  4. (4)

$\forall(d_{1},\dots,d_{k})\in\prod\limits_{i=1}^{k}D_{i},\ \exists\
\widehat{F^{{}^{\prime}}}\in\mathcal{D}$, such that
$\mathord{\uparrow}F^{{}^{\prime}}\subseteq\mathord{\uparrow}(d_{1},\dots,d_{k})$.

A subset $\mathcal{U}\subseteq PX$ is called a $\Rightarrow_{P}$ convergence
open set of $PX$ if and only if for each directed subset
$\mathcal{D}=\\{\widehat{F_{i}}\\}_{i\in I}$ of $PX$ and $\widehat{F}\in PX$,
$\mathcal{D}\Rightarrow_{P}\widehat{F}\in\mathcal{U}$ will imply
$\mathcal{D}\cap\mathcal{U}\neq\emptyset$. Denote all $\Rightarrow_{P}$
convergence open set of $PX$ by $O_{\Rightarrow_{P}}(PX)$.

######  Proposition 7.5.

Suppose $X$ is a directed space, the following are true:

  1. (1)

$(PX,\ O_{\Rightarrow_{P}}(PX))$ is a topological space, abbreviated as $PX$.

  2. (2)

The specialization order $\sqsubseteq$ on $(PX,\ O_{\Rightarrow_{P}}(PX))$
equals to $\leq_{P}$.

  3. (3)

$(PX,\ O_{\Rightarrow_{P}}(PX))$ is a directed space, that is
$O_{\Rightarrow_{P}}(PX)=d(PX)$.

Proof (1) Firstly, $\emptyset,\ PX\in O_{\Rightarrow_{P}}(PX)$. Suppose
$\mathcal{U}\in O_{\Rightarrow_{P}}(PX)$, $\widehat{F_{1}},\
\widehat{F_{2}}\in PX$, Let $F_{1}=\\{a_{1},\dots,a_{n}\\}$. If
$\widehat{F_{1}}\leq_{P}\widehat{F_{2}}$, then
$\\{\widehat{F_{2}}\\}\Rightarrow_{P}\widehat{F_{1}}$ holds, since the finite
directed sets $\\{a_{1}\\},\dots,\\{a_{n}\\}$ is sufficient to satisfy the
definition of $\\{\widehat{F_{2}}\\}\Rightarrow_{P}\widehat{F_{1}}$, then
$\widehat{F_{1}}\in\mathcal{U}$ implies $\widehat{F_{2}}\in\mathcal{U}$. It
follows that $\mathcal{U}$ is an upper set of $PX$ respect to $\leq_{P}$.
Suppose $\mathcal{U}_{1},\ \mathcal{U}_{2}\in O_{\Rightarrow_{P}}(PX)$,
$\mathcal{D}$ is a directed set in $PX$, $\widehat{F}\in PX$ and satisfy
$\mathcal{D}\Rightarrow_{P}\widehat{F}\in\mathcal{U}_{1}\cap\mathcal{U}_{2}$,
then there exists some $\widehat{G_{1}}\in\mathcal{D}\cap\mathcal{U}_{1}$ and
$\widehat{G_{2}}\in\mathcal{D}\cap\mathcal{U}_{2}$. Since $\mathcal{D}$ is
directed, we have $\widehat{G}\in\mathcal{D}$ such that $\widehat{G_{1}},\
\widehat{G_{2}}\leq_{P}\widehat{G}$, then
$\widehat{G}\in\mathcal{D}\cap\mathcal{U}_{1}\cap\mathcal{U}_{2}$. By the same
way, we can evidently prove that $O_{\Rightarrow_{P}}(PX)$ is closed under
arbitrary union. That is, $O_{\Rightarrow_{P}}(PX)$ is a topology on $PX$.

(2) Let $\widehat{F_{1}},\widehat{F_{2}}\in PX$. If
$\widehat{F_{1}}\leq_{P}\widehat{F_{2}}$, according the proof of (1), each
$\Rightarrow_{P}$ convergence open set is an upper set respect to partial
order $\leq_{P}$, then $\widehat{F_{1}}\in\overline{\\{\widehat{F_{2}}\\}}$,
that is, $\widehat{F_{1}}\sqsubseteq\widehat{F_{2}}$. On the other hand,
suppose $\widehat{F_{1}}\sqsubseteq\widehat{F_{2}}$. We need to prove that
$\widehat{F_{1}}\leq_{P}\widehat{F_{2}}$, that is to prove that
$\\{\widehat{F}\in PX:\widehat{F}\leq_{P}\widehat{F_{2}}\\}$ is a closed set
in $PX$ respect to $O_{\Rightarrow_{P}}(PX)$, by the proof of (1),
$\\{\widehat{F}\in
PX:\widehat{F}\leq_{P}\widehat{F_{2}}\\}\subseteq\overline{\\{\widehat{F_{2}}\\}}$,
thus, when $\\{\widehat{F}\in PX:\widehat{F}\leq_{P}\widehat{F_{2}}\\}$ is
closed in $PX$ respect to $O_{\Rightarrow_{P}}(PX)$, we have
$\\{\widehat{F}\in
PX:\widehat{F}\leq_{P}\widehat{F_{2}}\\}=\overline{\\{\widehat{F_{2}}\\}}$, it
follows that $\widehat{F_{1}}\leq_{P}\widehat{F_{2}}$. Now, we prove that
$\\{\widehat{F}\in PX:\widehat{F}\leq_{P}\widehat{F_{2}}\\}$ is closed respect
to $O_{\Rightarrow_{P}}(PX)$, equivalently,
$\mathcal{U}=PX\setminus\\{\widehat{F}\in
PX:\widehat{F}\leq_{P}\widehat{F_{2}}\\}$ is a $\Rightarrow_{P}$ convergence
open set. Suppose $\mathcal{D}=\\{\widehat{F_{i}}\\}_{i\in I}\subseteq PX$ is
a $\leq_{P}$ \- directed set and
$\mathcal{D}\Rightarrow_{P}\widehat{G}\in\mathcal{U}$. Let
$G=\\{a_{1},a_{2},\ldots,\ a_{k}\\}$. By the definition of $\Rightarrow_{P}$
convergence, there exists finite directed sets $D_{1},\dots,D_{n}\subseteq X$
such that $D_{1},\dots,D_{n}\subseteq\bigcup_{i\in
I}\\{\mathord{\downarrow}F_{i}\\}$. For each $a_{i}\in G$, there is one of the
above directed set $D_{i}$ such that $D_{i}\rightarrow a_{i}$,
$i=1,2,\ldots,k$. By contradiction, suppose
${\mathcal{D}}\cap{\mathcal{U}}=\emptyset$, then, by the definition of order
$\leq_{P}$, $\forall\widehat{F_{i}}\in\mathcal{D},\
\mathord{\downarrow}F_{i}\subseteq\mathord{\downarrow}F_{2}$, thus
$\bigcup{\mathord{\downarrow}F_{i}}\subseteq\mathord{\downarrow}F_{2}$, it
follows that $D_{i}\subseteq\mathord{\downarrow}F_{2},\ i=1,2,\ldots,\ k$. But
$\mathord{\downarrow}F_{2}$ is a closed set of $X$, then, all limit points of
$D_{i}$ belongs to $\mathord{\downarrow}F_{2}$, it follows that
$G=\\{a_{1},a_{2},\ \ldots,\ a_{k}\\}\subseteq\mathord{\downarrow}F_{2}$.
However, according to the hypothesis, $\widehat{G}\in\mathcal{U}$, that is
$\widehat{G}\nleq_{P}\widehat{F_{2}}$, then, it follows that
$\mathord{\uparrow}F_{2}\nsubseteq\mathord{\uparrow}G$, that is $\exists a\in
F_{2}$ such that $G\subseteq X\setminus\mathord{\downarrow}a$. By the
definition of $\Rightarrow_{P}$ convergence, each $D_{i}$ converges to the
points in $G$, and $X\setminus\mathord{\downarrow}a$ is an open set of $X$,
then $D_{i}\cap(X\setminus\mathord{\downarrow}a)\neq\emptyset$. For each
$i=1,\dots,n$, pick $d_{i}\in D_{i}\cap(X\setminus\mathord{\downarrow}a)$. We
have $(d_{1},\dots,d_{n})\in\prod_{i=1}^{n}D_{i}$ and
$a\notin\mathord{\uparrow}(d_{1},\dots,d_{n})$. By the hypothesis,
$\mathcal{D}\cap\mathcal{U}=\emptyset$, it follows that
$\forall\widehat{F_{i}}\in\mathcal{D},\
\mathord{\uparrow}F_{2}\subseteq\mathord{\uparrow}F_{i}$. Then $\forall i\in
I,\
a\in\mathord{\uparrow}F_{i},\mathord{\uparrow}F_{i}\nsubseteq\mathord{\uparrow}(d_{1},\dots,d_{n})$.
This contradicts with (4) in the definition of $\Rightarrow_{P}$ convergence.
Therefore, $\mathcal{U}$ is a $\Rightarrow_{P}$ convergence open set in $PX$.

(3) For an arbitrary topological space $X$, $O(X)\subseteq d(X)$ holds, then
$O_{\Rightarrow_{P}}(PX)\subseteq d(PX)$. On the other hand, according to the
definition of $\Rightarrow_{P}$ convergence topology, if directed set
$\mathcal{D}\in PX$ with $\mathcal{D}\Rightarrow_{P}\widehat{F}$, then
$\mathcal{D}$ converges to $\widehat{F}$ respect to $O_{\Rightarrow_{P}}(PX)$.
Thus, by the definition of directed open set,
$\mathcal{D}\Rightarrow_{P}\widehat{F}\in\mathcal{U}\in d(PX)$ implies
$\mathcal{U}\cap\mathcal{D}\neq\emptyset$. Then, $\mathcal{U}\in
O_{\Rightarrow_{P}}(PX)$, it follows that $O_{\Rightarrow_{P}}(PX)=d(PX)$,
that is $(PX,\ O_{\Rightarrow_{P}}(PX))$ is a directed space. $\Box$

######  Proposition 7.6.

Suppose $X,\ Y$ are two directed spaces. Then function $f:(PX,\
O_{\Rightarrow_{P}}(PX))\rightarrow Y$ is continuous if and only if for each
directed set ${\mathcal{D}}\subseteq PX$ and $\widehat{F}\in PX$,
${\mathcal{D}}\Rightarrow_{P}\widehat{F}$ implies $f(\mathcal{D})\rightarrow
f(\widehat{F})$.

Proof Since $\Rightarrow_{P}$ convergence will lead to
$O_{\Rightarrow_{P}}(PX)$ topological convergence, the necessity is obvious.
We are going to prove the sufficiency. Firstly, we check that $f$ is monotone.
If $\widehat{F_{1}},\ \widehat{F_{2}}\in PX$ and
$\widehat{F_{1}}\leq_{P}\widehat{F_{2}}$, then
$\\{\widehat{F_{2}}\\}\Rightarrow_{P}\widehat{F_{1}}$. By the hypothesis,
$\\{f(\widehat{F_{2}})\\}\rightarrow f(\widehat{F_{1}})$, it follows that
$f(\widehat{F_{2}})\sqsubseteq f(\widehat{F_{1}})$. Suppose $U$ is an open set
of $Y$ and there is a directed set ${\mathcal{D}}\Rightarrow_{P}\widehat{F}\in
f^{-1}(U)$, then $f(\mathcal{D})$ is a directed set of $Y$ and
$f(\mathcal{D})\rightarrow f(\widehat{F})\in U$, thus, there exists some
$\widehat{F^{\prime}}\in{\mathcal{D}}$ such that $f(\widehat{F^{\prime}})\in
U$, that is, $\widehat{F^{\prime}}\in{\mathcal{D}}\cap f^{-1}(U)$. According
to the definition of $\Rightarrow_{P}$ convergence open set, $f^{-1}(U)\in
O_{\Rightarrow_{P}}(PX)$, that is $f$ is continuous. $\Box$

Define a binary operation $\oplus:PX\times PX\rightarrow PX$ on $PX$
:$\forall\ \widehat{F_{1}},\ \widehat{F_{2}}\in PX,\
\widehat{F_{1}}\oplus\widehat{F_{2}}=\widehat{F_{1}\cup F_{2}}$. For arbitrary
$\widehat{F_{1}}=\widehat{F_{2}},\ \widehat{G_{1}}=\widehat{G_{2}}$ with
$F_{1}\neq F_{2},\ G_{1}\neq G_{2}$. $F_{1}\cup
G_{1}\subseteq\mathord{\downarrow}(F_{2}\cup G_{2})$ and $F_{1}\cup
G_{1}\subseteq\mathord{\uparrow}(F_{2}\cup G_{2})$ imply $\widehat{F_{1}\cup
G_{1}}\leq_{P}\widehat{F_{2}\cup G_{2}}$. Similarly, we have the opposite
inequality, thus, $\oplus$ is well-defined.

######  Theorem 7.7.

Let $X$ be a directed space. Then $(PX,\ \oplus)$ is a directed semilattice.

Proof By Proposition 7.5, $(PX,\ O_{\Rightarrow_{P}}(PX))$ is a directed
space. We will prove that $\oplus$ is a directed semilattice operation. It is
evidently to check that $\oplus$ satisfy (a), (b), (c) in definition 7.1, we
now prove the continuity of $\oplus$. $\oplus$ is obviously monotone. By
Theorem 2.6(2) and proposition 7.6, we only need to prove that, for each
directed set ${\mathcal{D}}=\widehat{F_{i}}\subseteq PX$ and $\widehat{F},\
\widehat{G}\in PX$, ${\mathcal{D}}\Rightarrow_{P}\widehat{F}$ will imply
$\widehat{G}\oplus{\mathcal{D}}\Rightarrow_{P}(\widehat{G}\oplus\widehat{F})=\widehat{(G\cup
F)}$. Here,
$G\cup{\mathcal{D}}=\\{\widehat{G}\oplus\widehat{F_{i}}:\widehat{F_{i}}\in\mathcal{D}\\}$
is also a directed set. By the definition of $\Rightarrow_{P}$ convergence,
there exists finite directed sets $D_{1},\dots,\ D_{n}\subseteq X$ such that
$\mathcal{D}\Rightarrow_{P}\widehat{F}$. Let $G=\\{a_{1},\dots,a_{k}\\}$, take
$D_{n+1}=\\{a_{1}\\},\dots,D_{n+k}=\\{a_{k}\\}$.We can evidently verify that
$D_{1},\dots,D_{n+k}$ is sufficient to satisfy the definition of
$\widehat{G}\oplus\mathcal{D}\Rightarrow_{P}\widehat{G\cup F}$. Therefore,
$(PX,\ \oplus)$ is a directed semilattice. $\Box$

The following theorem is the main result of this section.

######  Theorem 7.8.

Suppose $X$ is a directed space, then $(PX,\ O_{\Rightarrow_{P}}(PX))$ is the
directed convex powerspace over $X$, that is, endowed with topology
$O_{\Rightarrow_{P}}(PX)$, $(PX,\ \oplus)\cong P_{P}(X)$.

Proof Define function $i:X\rightarrow PX$ as follows: $\forall x\in X$,
$i(x)=\widehat{x}$. We prove the continuity of $i$. $i$ is evidently monotone.
Suppose we have directed set $D\subseteq X$ and $x\in X$ satisfy $D\rightarrow
x$. Let $\mathcal{D}=i(D)=\\{\widehat{d}:d\in D\\}$, then $\mathcal{D}$ is a
directed set in $PX$ and ${\mathcal{D}}\Rightarrow_{P}\widehat{x}$. Since $D$
satisfy the definition such that $\mathcal{D}\Rightarrow_{P}\widehat{x}$. By
Proposition 2.5, $i$ is continuous.

Let $(Y,\ +)$ be a directed semilattice, $f:X\rightarrow Y$ is a continuous
function. Define $\bar{f}:PX\rightarrow Y$ as follows: $\forall\widehat{F}\in
PX$ (let $F=\\{a_{1},a_{2},\ldots,a_{n}\\}$),

$\bar{f}(\widehat{F})=f(a_{1})+f(a_{2})+\dots+f(a_{n})=\sum\limits_{a\in
F}f(a)$

Particularly, we write $\bar{f}(\widehat{F})=\Sigma f(F)$.
$\bar{f}$ is well-defined, since if we have $\widehat{F_{1}},\
\widehat{F_{2}}\in PX$ and $\widehat{F_{1}}=\widehat{F_{2}}$ but with
$F_{1}\neq F_{2}$. Let $F_{1}=\\{a_{1},\dots,a_{k}\\},\
F_{2}=\\{b_{1},\dots,b_{n}\\}$. For
$\mathord{\downarrow}F_{1}\subseteq\mathord{\downarrow}F_{2}$, we have
$b_{1},\dots,b_{k}\in F_{2}$ such that $a_{1}\leq b_{1},\dots,a_{k}\leq
b_{k}$. If $F_{2}\setminus\\{b_{1},\dots,b_{k}\\}=\emptyset$, then $\Sigma
f(F_{1})=\bar{f}(\widehat{F_{1}})\leq\Sigma
f(F_{2})=\bar{f}(\widehat{F_{1}})$. If not, suppose
$F_{2}\setminus\\{b_{1},\dots,b_{k}\\}=\\{b_{i1},\dots,b_{is}\\}$, but
$\mathord{\uparrow}F_{2}\subseteq\mathord{\uparrow}F_{1}$, we have
$a_{i1},\dots,a_{is}\in F_{1}$ such that $a_{i1}\leq b_{i1},\ \dots,\
a_{is}\leq b_{is}$, now, we have $a_{1}\leq b_{1},\dots,a_{n}\leq b_{n},\
a_{i1}\leq b_{i1},\dots,a_{is}\leq b_{is}$. Notice that $a_{i1},\dots,a_{is}$
repeat $\\{a_{1},\dots,a_{n}\\}$, it follows that
$\bar{f}(\widehat{F_{1}})=f(a_{1})+\dots+f(a_{n})+f(a_{i1})+\dots+f(a_{is})\leq
f(b_{1})+\dots+f(b_{n})+f(b_{i1})+\dots+f(b_{is})=\bar{f}(\widehat{F_{2}})$.
Similarly, we have $\bar{f}(\widehat{F_{2}})\leq\bar{f}(\widehat{F_{1}})$.
Thus, $\bar{f}$ is well-defined.

(1) $f=\bar{f}\circ i$.

For arbitrary $x\in X$, $(\bar{f}\circ
i)(x)=\bar{f}(i(x))=\bar{f}(\widehat{x})=f(x)$.

(2) $\bar{f}$ is a directed semilattice homomorphism, that is, $\bar{f}$ is
continuous and for arbitrary $\widehat{F_{1}},\ \widehat{F_{2}}\in PX$,
$\bar{f}(\widehat{F_{1}}\oplus\widehat{F_{2}})=\bar{f}(\widehat{F_{1}})+\bar{f}(\widehat{F_{2}})$.

Firstly, we prove that $\bar{f}$ preserves directed semilattice operation.
Suppose $\widehat{F_{1}},\ \widehat{F_{2}}\in PX$. Then
$\bar{f}(\widehat{F_{1}}\oplus\widehat{F_{2}})=\bar{f}(\widehat{F_{1}\cup
F_{2}})=\Sigma f(F_{1}\cup F_{2})=(\Sigma f(F_{1})+(\Sigma
f(F_{2}))=\bar{f}(\widehat{F_{1}})+\bar{f}(\widehat{F_{2}})$. Next, we prove
the continuity of $\bar{f}$. We check that $\bar{f}$ is monotone. Suppose
$\widehat{F_{1}},\ \widehat{F_{2}}\in PX$. Let
$F_{1}=\\{a_{1},\dots,a_{k}\\},\ F_{2}=\\{b_{1},\dots,\ b_{n}\\}$, and
$\widehat{F_{1}}\leq_{P}\widehat{F_{2}}$, that is
$\mathord{\downarrow}F_{1}\subseteq\mathord{\downarrow}F_{2}\
\&\mathord{\uparrow}F_{2}\subseteq\mathord{\uparrow}F_{1}$. For
$\mathord{\downarrow}F_{1}\subseteq\mathord{\downarrow}F_{2}$, we have
$b_{1},\dots,b_{k}\in F_{2}$ such that $a_{1}\leq b_{1},\dots,a_{k}\leq
b_{k}$. If $F_{2}\setminus\\{b_{1},\dots,b_{k}\\}=\emptyset$, then $\Sigma
f(F_{1})=\bar{f}(\widehat{F_{1}})\leq\Sigma
f(F_{2})=\bar{f}(\widehat{F_{1}})$. If not, suppose
$F_{2}\setminus\\{b_{1},\dots,b_{k}\\}=\\{b_{i1},\dots,b_{is}\\}$, but
$\mathord{\uparrow}F_{2}\subseteq\mathord{\uparrow}F_{1}$, we have
$a_{i1},\dots,a_{is}\in F_{1}$ such that $a_{i1}\leq b_{i1},\ \dots,a_{is}\leq
b_{is}$, now, we have $a_{1}\leq b_{1},\dots,a_{n}\leq b_{n},\ a_{i1}\leq
b_{i1},\dots,\ a_{is}\leq b_{is}$. Notice that $a_{i1},\ \dots,\ a_{is}$
repeat $\\{a_{1},\ \dots,\ a_{n}\\}$, it follows that
$\bar{f}(\widehat{F_{1}})=f(a_{1})+\dots+f(a_{n})+f(a_{i1})+\dots+f(a_{is})\leq
f(b_{1})+\dots+f(b_{n})+f(b_{i1})+\dots+f(b_{is})=\bar{f}(\widehat{F_{2}})$.
In conclusion, $\widehat{F_{1}}\leq_{P}\widehat{F_{2}}$ will imply
$\bar{f}(\widehat{F_{1}})\leq\bar{f}(\widehat{F_{2}})$, that is, $\bar{f}$ is
monotone.

Finally, we verify the continuity of $\bar{f}$. Suppose
${\mathcal{D}}\subseteq PX$ is a directed set and
${\mathcal{D}}\Rightarrow_{P}\widehat{F}\in PX$, there exists finite directed
sets $D_{1},\dots,\ D_{n}\subseteq X$ satisfy definition of $\Rightarrow_{P}$
convergence. Let $F=\\{b_{1},b_{2},\ldots,b_{k}\\}$. By (2) in the definition,
for each $D_{i},\ 1\leq i\leq n$, there exists some $b_{i}\in F$ such that
$D_{i}\rightarrow b_{i}$. If $F\setminus\\{b_{1},\dots,b_{n}\\}\neq\emptyset$,
which is denoted by $G=\\{a_{1},a_{2}\ldots,a_{s}\\}$. By (3) of definition,
for each $a_{j}\in G$, there exists $1\leq i_{j}\leq n$ such that
$D_{i_{j}}\rightarrow a_{j}$. By the continuity of $f$, we have
$f(D_{i})\rightarrow f(b_{i}),\ i=1,\dots,n$, and $f(D_{i_{j}})\rightarrow
f(a_{j})$, $j=1,2,\ldots,s$. Since the directed semilattice operation $+$ on
$Y$ is continuous, the following convergence holds

$f(D_{1})+\cdots+f(D_{n})+f(D_{i_{1}})+\cdots+f(D_{i_{s}})\rightarrow
f(b_{1})+\dots+f(b_{n})+f(a_{i_{1}})+\cdots+f(a_{i_{s}}).\ \\\ (\ast)$

Here,
$f(D_{1})+\cdots+f(D_{n})+f(D_{i_{1}})+\cdots+f(D_{i_{s}})=\\{f(d_{1})+\cdots+f(d_{n})+f(d_{i_{1}})+\cdots+f(d_{i_{s}}):(d_{1},\ldots,d_{k},d_{i_{1}},\ldots,\
d_{i_{s}})\in(\prod\limits_{i=1}^{n}D_{i}\\})\times(\prod\limits_{j=1}^{s}D_{i_{j}})\\}$.
Let $U$ be an arbitrary open neighborhood of $\Sigma f(F)$, by $(\ast)$, there
exists some
$(d_{1},\ldots,d_{n},d_{i_{1}},\ldots,d_{i_{s}})\in(\prod\limits_{i=1}^{n}D_{i}\\})\times(\prod\limits_{j=1}^{s}D_{i_{j}})$
such that $f(d_{1})+\cdots+f(d_{n})+f(d_{i_{1}})+\cdots+f(d_{i_{s}})\in U$.
Since each $D_{i_{j}}$ repeats $D_{i}$, there exists
$(d^{{}^{\prime}}_{1},d^{{}^{\prime}}_{2},\ldots,d^{{}^{\prime}}_{n})\in\prod\limits_{i=1}^{n}D_{i}\\}$
such that $f(d^{{}^{\prime}}_{1})+\cdots+f(d^{{}^{\prime}}_{n})\sqsupseteq
f(d_{1})+\cdots+f(d_{n})+f(d_{i_{1}})+\cdots+f(d_{i_{s}})$. By (4) in the
definition, there exists some $\widehat{F_{1}}\in\mathcal{D}$ such that
$\uparrow F_{1}\subseteq\uparrow(d^{{}^{\prime}}_{1},\ \dots,\
d^{{}^{\prime}}_{n})$. On the other hand, by (1) in definition of
$\Rightarrow_{P}$ convergence, for each $d_{i}^{{}^{\prime}}$, there exists
some $\widehat{F_{i}}\in\mathcal{D}$ such that
$d^{{}^{\prime}}_{i}\in\mathord{\downarrow}F_{i}$. Since $\mathcal{D}$ is
directed, we have $\widehat{F_{2}}\in\mathcal{D}$ with
$\widehat{F_{i}}\leq_{P}\widehat{F_{2}},\ i=1,\ \dots,\ n$. Again, by the
directness of $\mathcal{D}$, we obtain $\widehat{F^{\prime}}\in\mathcal{D}$
such that $\widehat{F_{1}},\ \widehat{F_{2}}\leq_{P}\widehat{F^{\prime}}$.
Now, we have
$\mathord{\downarrow}(d^{{}^{\prime}}_{1},\dots,d^{{}^{\prime}}_{n})\subseteq\mathord{\downarrow}F_{2}\subseteq\mathord{\downarrow}F^{{}^{\prime}},\
\mathord{\uparrow}F^{{}^{\prime}}\subseteq\mathord{\uparrow}F_{1}\subseteq\mathord{\uparrow}(d^{{}^{\prime}}_{1},\dots,d^{{}^{\prime}}_{n})$,
that is
$\widehat{\\{d^{{}^{\prime}}_{1},\dots,d^{{}^{\prime}}_{n}\\}}\leq_{P}\widehat{F^{{}^{\prime}}}$,
since $\bar{f}$ is monotone and $U$ is an upper set in $Y\ $, we got
$\bar{f}(\widehat{F^{{}^{\prime}}})\in U$. It follows that
$\bar{f}({\mathcal{D}})=\\{\Sigma
f(F_{i}):\widehat{F_{i}}\in\mathcal{D}\\}\rightarrow\Sigma f(F)$. By
Proposition 7.6, $\bar{f}$ is continuous.

(3) $\bar{f}$ is unique.

Suppose we have a directed semilattice homomorphism $g:(PX,\
\oplus)\rightarrow(Y,\ +)$ such that $f=g\circ i$. Then
$g(\widehat{x})=f(x)=\bar{f}(\widehat{x})$. For each $\widehat{F}\in PX$ (Let
$F=\\{a_{1},\ \dots,\ a_{n}\\}$ ),

$\displaystyle g(\widehat{F})$ $\displaystyle=$ $\displaystyle
g(\widehat{a_{1}}\oplus\widehat{a_{2}}\oplus\cdots\oplus\widehat{a_{n}})$
$\displaystyle=$ $\displaystyle g(\widehat{a_{1}})\oplus
g(\widehat{a_{2}})\oplus\cdots\oplus g(\widehat{a_{n}})$ $\displaystyle=$
$\displaystyle\bar{f}(\widehat{a_{1}})\oplus\bar{f}(\widehat{a_{2}})\oplus\cdots\oplus\bar{f}(\widehat{a_{n}})$
$\displaystyle=$
$\displaystyle\bar{f}(\widehat{a_{1}}\oplus\widehat{a_{2}}\oplus\cdots\oplus\widehat{a_{n}})$
$\displaystyle=$ $\displaystyle\bar{f}(\widehat{F}).$

That is $\bar{f}$ is unique.

In conclusion, by Definition 7.4, endowed with the topology
$O_{\Rightarrow_{P}}(PX)$, the directed semilattice $(PX,\ \oplus)$ is a
directed convex powerspace over $X$, that is, $P_{P}(X)\cong(PX,\ \oplus)$.
$\Box$

The directed convex powerspace is unique in the sense of order isomorphism and
topological homomorphism, so we can denote the directed convex powerspace of
each $X$ by $P_{P}(X)=(PX,\ \oplus)$.

Suppose $X,\ Y$ are two directed spaces, $f:X\rightarrow Y$ is a continuous
function. Define map $P_{P}(f):P_{P}(X)\rightarrow P_{P}(Y)$ as follows:
$\forall\widehat{F}\in PX$,

$P_{P}(f)(\mathord{\downarrow}F)=\widehat{f(F)}.$

It is evident that, $P_{P}(f)$ is well-defined and order preserving. According
to the proof of the theorem above, it is easy to check that $P_{P}(f)$ is a
directed semilattice homomorphism between these two directed convex
powerspaces. If $id_{X}$ is the identity function and $g:Y\rightarrow Z$ is an
arbitrary continuous function from $Y$ to a directed space $Z$, then,
$P_{P}(id_{X})=id_{P_{P}(X)},\ P_{P}(g\circ f)=P_{P}(g)\circ P_{P}(f)$. Thus,
$P_{P}:{\bf Dtop}\rightarrow{\bf Dsl}$ is a functor from ${\bf Dtop}$ to ${\bf
Dsl}$. Let $U:{\bf Dsl}\rightarrow{\bf Dtop}$ be the forgetful functor. By
Theorem 7.8, we have the following results.

######  Corollary 7.9.

$P_{P}$ is a left adjoint of the forgetful functor $U$, that is, ${\bf Dsl}$
is a reflective subcategory of ${\bf Dtop}$.

##  8 Relations Between Convex Powerspaces

In this section, we will discuss the relations between the convex powerdomain
of dcpo and its directed lower powerspace.

Let $L$ be a domain equipped with the Scott topology. A nonempty subset $A$ is
a $lens$ if $A$ can be written as the intersection of a closed set and a
compact saturated set. A lens $A=C\cap K$ has a canonical representation of
the form $\bar{A}\cap\mathord{\uparrow}A$ (since
$A\subseteq\bar{A}\cap\mathord{\uparrow}A\subseteq C\cap K=A$) ;note that
$\mathord{\uparrow}A$ is compact since $A$ is.A pair $(C,\ K)\in C(L)\times
Q(L)$ is called a lens factorization if $C=\bar{A}$ and
$K=\mathord{\uparrow}A$ for some lens $A$.We denote by $LensL$ the set of all
lens.

######  Theorem 8.1.

[7] Let $L$ be a domain equipped with the Scott topology. If $L$ is coherent,
resp. countably based, then the convex powerdomain $P(L)$ (which is also
called the Plotkin powerdomain)may be identified with Lens $L$ equipped with
the Egli-Milner order, resp. topological Egli-Milner order.

######  Proposition 8.2.

Let $X$ be a continuous domain endowed with Scott topology, then
$\sigma(P(X))|_{PX}=O_{\Rightarrow_{P}}(PX)$.

Proof By proposition 4.4 and proposition 6.3(1), it is not hard to check that
$\sigma(P(X))|_{PX}\subseteq O_{\Rightarrow_{P}}(PX)$.

On the other hand, for each $\mathcal{U}\in O_{\Rightarrow_{P}}(PX)$, let

$\mathord{\uparrow}_{P(X)}\mathcal{U}=\\{(C,\ D)\in C(X)\times
Q(X):\exists\widehat{F}\in PX\ such\ that\ \mathord{\downarrow}F\subseteq
C\&D\subseteq\mathord{\uparrow}F\\}.$

Then, it is obviously that
$\mathcal{U}=\mathord{\uparrow}_{P(X)}\mathcal{U}\cap PX$, it is sufficient to
prove that $\mathord{\uparrow}_{P(X)}\mathcal{U}\in\sigma(P(X))$. Let
$\mathcal{D}=\\{(C_{i},\ D_{i})\\}_{i\in I}\subseteq P(X)$ be a directed set
with $\vee\mathcal{D}=(\overline{\cup_{i\in I}C_{i}},\ \cap_{i\in
I}D_{i})\in\mathord{\uparrow}_{P(X)}\mathcal{U}$. By the definition of
$\mathord{\uparrow}_{P(X)}\mathcal{U}$ we have some
$\widehat{F}\in\mathcal{U}$ such that $\mathord{\downarrow}F\subseteq C\ \&\
D\subseteq\mathord{\uparrow}F$. Set $F=\\{a_{1},\ \dots,\ a_{n}\\}$. Let
$D_{i}=\mathord{\mbox{\makebox[0.0pt][l]{\raisebox{-1.72218pt}{$\downarrow$}}$\downarrow$}}a_{i},\
1\leq i\leq n$, $\mathcal{F}=\\{\widehat{\\{d_{1},\ \dots,\ d_{n}\\}}:(d_{1},\
\dots,\ d_{n})\in\prod\limits_{i=1}^{n}D_{i}\\}$. Since $X$ is a continuous
domain and by the proof of proposition 6.3(1), we can directly check that
$\mathcal{F}\Rightarrow_{P}\widehat{F}$, it follows that there exists some
$\widehat{\\{d_{1},\ \dots,\ d_{n}\\}}\in\mathcal{F}\cap\mathcal{U}$. Since
$F\subseteq\overline{\cup_{i\in I}C_{i}}$ and $X$ is continuous, we can infer
that each $D_{i}\subseteq\mathord{\downarrow}\cup_{i\in I}C_{i}=\cup_{i\in
I}C_{i}$. For $\mathcal{D}$ is directed, we have some $(C^{{}^{\prime}},\
D^{{}^{\prime}})\in\mathcal{D}$ such that $\mathord{\downarrow}\\{d_{1},\
\dots,\ d_{n}\\}\subseteq C^{{}^{\prime}}$. Besides, $\cap_{i\in
I}D_{i}\subseteq\mathord{\uparrow}F\subseteq(\cup_{i=1}^{n}\mathord{\uparrow}d_{i})^{\circ}$,
since $X$ is a continuous domain endowed with the Scott topology, hence well-
filtered, we have some $(C^{{}^{\prime\prime}},\
D^{{}^{\prime\prime}})\in\mathcal{D}$ such that
$D^{{}^{\prime\prime}}\subseteq(\cup_{i=1}^{n}\mathord{\uparrow}d_{i})^{\circ}\subseteq\mathord{\uparrow}\\{d_{1},\
\dots,\ d_{n}\\}$. Again, by the directness of $\mathcal{D}$, choose $(C,\
D)\in\mathcal{D}$ with $C^{{}^{\prime}},\ C^{{}^{\prime\prime}}\subseteq C\
\&D\subseteq D^{{}^{\prime}},\ D^{{}^{\prime\prime}}$. Then,
$\mathord{\downarrow}\\{d_{1},\ \dots,\ d_{n}\\}\subseteq
C^{{}^{\prime}}\subseteq C\ \&D\subseteq
D^{{}^{\prime\prime}}\subseteq\mathord{\uparrow}\\{d_{1},\ \dots,\ d_{n}\\}$,
that is $(C,\ D)\in\mathord{\uparrow}_{P(X)}\mathcal{U}$ and
$\mathord{\uparrow}_{P(X)}\mathcal{U}$ is a Scott open set.$\Box$

######  Example 8.3.

Let $X$ be the same as in example 6.4, by example 6.4 we can directly infer
that, $P_{P}(P_{O}(X))\neq P(P_{O}(X))$, that is, in general, the directed
convex powerspace over a dcpo endowed with the Scott topology is not agree
with its convex powerdomain.

##  9 The Commutativity of The Directed Upper and Lower Functors

In domain theory, the Hoare and Smyth powerdomain constructors were firtsly
proved to be commuted under composition in [5] in 1990. In 1991, Heckmann gave
an algebraic methods and does not rely on any explicit representations of the
powerdomains [15]. In this section, we will discuss the commutativity of the
directed upper and lower functors.

In this paper, for convenience, a directed inflationary semilattice is
abbriated as $\vee-$ directed space, a directed deflationary semilattice is
abbriated as $\wedge-$ directed space. A morphism between two directed
deflationary semilattices is called a $\vee-$ morphism, a morphism between two
directed inflationary semilattices is called a $\wedge-$ morphism.

######  Definition 9.1.

  1. 1.

A $\vee-\wedge-$ directed space $X$ is a directed space that is both a $\vee-$
directed space and $\wedge-$ directed space that the distributive law
$a\wedge(b\vee c)=(a\wedge b)\vee(a\wedge c),\ \forall a,b,c\in X$.

  2. 2.

A $\vee-\wedge-$ morphism between two $\vee-\wedge-$ directed spaces that is
both a $\vee-$ morphism and a $\wedge-$ morphism.

Denote the category of all $\vee-\wedge-$ directed spaces together with
$\vee-\wedge-$ morphisms by ${\bf Ddl}$. For all directed spaces $X$, we then
show that both $P_{U}(P_{L}(X))$ and $P_{L}(P_{U}(X))$ are free $\vee-\wedge-$
directed spaces over $X$, whence they are isomorphic.

We will discuss some properties of functor $P_{L}$, and functor $P_{U}$ is
analogously. Theorm 3.10 indicates that for each directed space $X$,
$P_{L}(X)$ is a $\vee-$ directed space. The following proposition tell us that
if $X$ is a $\wedge-$ directed space, then $P_{L}(X)$ is actually a
$\vee-\wedge-$ directed space.

######  Theorem 9.2.

  1. 1.

If $X$ is a $\wedge-$ directed space, then $P_{L}(X)$ is a $\vee-\wedge-$
directed space, and $\eta_{\vee}:X\rightarrow P_{L}(X)$ is a $\wedge-$
morphism, here $\eta_{\wedge}(x)=\mathord{\downarrow}x,\forall x\in X$.

  2. 2.

If $X$ is a $\vee-$ directed space, then $P_{U}(X)$ is a $\vee-\wedge-$
directed space, and $\eta_{\wedge}:X\rightarrow P_{U}(X)$ is a $\vee-$
morphism, here $\eta_{\wedge}(x)=\uparrow x,\forall x\in X$.

Proof We only prove (1). Let $X$ be a $\wedge-$ directed space, then
$P_{L}(X)$ is a $\vee-$ directed space, we have to construct the operation
$\wedge$ in $P_{L}(X)$. To this end, for each
$\mathord{\downarrow}F_{1},\mathord{\downarrow}F_{2}\in P_{L}(X)$, let
$f_{a}(x)=\eta_{\vee}(a\wedge x)$ for each $a\in X$. By Theorem 3.10, there
exist a unique morphism $\bar{f_{a}}:P_{L}(X)\rightarrow P_{L}(X)$ such that
$\bar{f_{a}}(\mathord{\downarrow}F)=\mathord{\downarrow}f_{a}(F)$ for each
$\mathord{\downarrow}F\in P_{L}(X)$. Let
$g(a)=\bar{f_{a}}(\mathord{\downarrow}F_{2})=\mathord{\downarrow}\\{a\wedge
b:b\in F_{2}\\}$, again by Theorem 3.10, there exist a unique morphism
$\bar{g}:P_{L}(X)\rightarrow P_{L}(X)$ such that
$\bar{g}(\mathord{\downarrow}F)=\mathord{\downarrow}\\{a\wedge b:a\in F,b\in
F_{2}\\}$. Thus, we have a continuous operation in $P_{L}(X)$ defined as
follows

$\mathord{\downarrow}F_{1}\sqcap\mathord{\downarrow}F_{2}=\\{\eta_{\wedge}(a\wedge
b):a\in F_{1},b\in F_{2}\\}=\mathord{\downarrow}\\{a\wedge b:a\in F_{1},b\in
F_{2}\\}$

It is straightly to check that $\sqcap$ satisfy all the conditions in
Definition 3.1:

  1. 1.

$\mathord{\downarrow}F\sqcap\mathord{\downarrow}F=\mathord{\downarrow}\\{a\wedge
a:a\in F\\}=\mathord{\downarrow}F,\forall\mathord{\downarrow}F\in P_{L}(X)$.

  2. 2.

$(\mathord{\downarrow}F_{1}\sqcap\mathord{\downarrow}F_{2})\sqcap\mathord{\downarrow}F_{3}=\mathord{\downarrow}\\{a\wedge
b:a\in F_{1},b\in
F_{2}\\}\sqcap\mathord{\downarrow}F_{3}=\mathord{\downarrow}\\{a\wedge b\wedge
c:a\in F_{1},b\in F_{2},c\in
F_{3}\\}=\mathord{\downarrow}F_{1}\sqcap\mathord{\downarrow}\\{b\wedge c:b\in
F_{2},c\in
F_{3}\\}=\mathord{\downarrow}F_{1}\sqcap(\mathord{\downarrow}F_{2}\sqcap\mathord{\downarrow}F_{3}),\forall\mathord{\downarrow}F_{1},\mathord{\downarrow}F_{2},\mathord{\downarrow}F_{3}\in
P_{L}(X)$.

  3. 3.

$\mathord{\downarrow}F_{1}\sqcap\mathord{\downarrow}F_{2}=\mathord{\downarrow}\\{a\wedge
b:a\in F_{1},b\in
F_{2}\\}=\mathord{\downarrow}F_{2}\sqcap\mathord{\downarrow}F_{1},\forall\mathord{\downarrow}F_{1},\mathord{\downarrow}F_{2}\in
P_{L}(X)$.

  4. 4.

$\mathord{\downarrow}F_{1}\sqcap\mathord{\downarrow}F_{2}=\mathord{\downarrow}\\{a\wedge
b:a\in F_{1},b\in
F_{2}\\}\leq\mathord{\downarrow}F_{1},\forall\mathord{\downarrow}F_{1},\mathord{\downarrow}F_{2}\in
P_{L}(X)$.

Next, we shall check that the distributivity law for $\cup$ and $\sqcap$.
Suppose we have $F_{1}=\\{a_{1},\dots,a_{n}\\},\
F_{2}=\\{b_{1},\dots,b_{m}\\},\ F_{3}=\\{c_{1},\dots,c_{k}\\}$.

  1. 1.

$\mathord{\downarrow}F_{1}\sqcap(\mathord{\downarrow}F_{2}\cup\mathord{\downarrow}F_{3})=(\mathord{\downarrow}F_{1}\sqcap\mathord{\downarrow}F_{2})\cup(\mathord{\downarrow}F_{1}\sqcap\mathord{\downarrow}F_{3})$.
$\mathord{\downarrow}F_{1}\sqcap(\mathord{\downarrow}F_{2}\cup\mathord{\downarrow}F_{3})=\mathord{\downarrow}F_{1}\sqcap\mathord{\downarrow}\\{b_{1},\dots,b_{m},c_{1},\dots,b_{k}\\}=\mathord{\downarrow}\\{a\wedge
b,a\wedge c:a\in F_{1},b\in F_{2},c\in F_{3}\\}$.
$(\mathord{\downarrow}F_{1}\sqcap\mathord{\downarrow}F_{2})\cup(\mathord{\downarrow}F_{1}\sqcap\mathord{\downarrow}F_{3})=\mathord{\downarrow}\\{a\wedge
b:a\in F_{1},b\in F_{2}\\}\cup\mathord{\downarrow}\\{a\wedge c:a\in F_{1},c\in
F_{3}\\}=\mathord{\downarrow}\\{a\wedge b,a\wedge c:a\in F_{1},b\in F_{2},c\in
F_{3}\\}$. Thus they are equal.

  2. 2.

$\mathord{\downarrow}F_{1}\cup(\mathord{\downarrow}F_{2}\sqcap\mathord{\downarrow}F_{3})=(\mathord{\downarrow}F_{1}\cup\mathord{\downarrow}F_{2})\sqcap(\mathord{\downarrow}F_{1}\cup\mathord{\downarrow}F_{3})$.
$\mathord{\downarrow}F_{1}\cup(\mathord{\downarrow}F_{2}\sqcap\mathord{\downarrow}F_{3})=\mathord{\downarrow}F_{1}\cup\mathord{\downarrow}\\{b\wedge
c:b\in F_{2},c\in F_{3}\\}=\mathord{\downarrow}\\{a,b\wedge c:a\in F_{1},b\in
F_{2},c\in F_{3}\\}$.
$(\mathord{\downarrow}F_{1}\cup\mathord{\downarrow}F_{2})\sqcap(\mathord{\downarrow}F_{1}\cup\mathord{\downarrow}F_{3})=\mathord{\downarrow}\\{a\wedge
a,a\wedge c,a\wedge b,b\wedge c:a\in F_{1},b\in F_{2},c\in
F_{3}\\}=\mathord{\downarrow}\\{a,b\wedge c:a\in F_{1},b\in F_{2},c\in
F_{3}\\}$. Thus they are equal.

Hence, $(P_{L}(X),\vee,\sqcap)$ is a $\vee-\wedge-$ directed space. Last, we
shall prove that $\eta_{\vee}$ is a $\wedge-$ morphism. For each $x,\ y\in
X,\eta_{\vee}(x\wedge y)=\mathord{\downarrow}(x\wedge
y)=(\mathord{\downarrow}x)\sqcap(\mathord{\downarrow}y)=\eta_{\vee}(x)\sqcap\eta_{\vee}(y)$.
$\Box$

In the pervious part, we showed that the functor $P_{L}$ does not destroy the
$\wedge$ operator, now we prove that $P_{L}$ has the analogous property for
$\wedge-$ morphism.

######  Theorem 9.3.

  1. 1.

Let $X$ be a $\wedge-$ directed space, $Y$ a $\vee-\wedge-$ directed space,
and $f:X\rightarrow Y$ a $\wedge-$ morphism. Then its unique extension
$P_{L}(f):P_{L}(X)\rightarrow Y$ is a $\vee-\wedge-$ morphism.

  2. 2.

Let $X$ be a $\vee-$ directed space, $Y$ a $\vee-\wedge-$ directed space, and
$f:X\rightarrow Y$ a $\vee-$ morphism. Then its unique extension
$P_{U}(f):P_{U}(X)\rightarrow Y$ is a $\vee-\wedge-$ morphism.

Proof We only prove (1). Only need to check that $P_{L}(f)$ is a $\wedge-$
morphism. Suppose we have $F_{1}={a_{1},\dots,a_{n}},\
F_{2}=\\{b_{1},\dots,b_{m}\\}$.

$\displaystyle
P_{L}(f)(\mathord{\downarrow}F_{1}\sqcap\mathord{\downarrow}F_{2})$
$\displaystyle=$ $\displaystyle P_{L}(f)(\mathord{\downarrow}\\{a\wedge b:a\in
F_{1},b\in F_{2}\\})$ $\displaystyle=$ $\displaystyle\bigvee\limits_{a\in
F_{1},b\in F_{2}}f(a\wedge b)$ $\displaystyle=$
$\displaystyle\bigvee\limits_{a\in F_{1},b\in F_{2}}f(a)\wedge f(b)$

On the other hand,

$\displaystyle P_{L}(f)(\mathord{\downarrow}F_{1})\wedge
P_{L}(f)(\mathord{\downarrow}F_{2})$ $\displaystyle=$ $\displaystyle(\vee
f(F_{1}))\wedge(\vee f(F_{2}))$ $\displaystyle=$ $\displaystyle(\vee
f(F_{1})\wedge f(b_{1}))\vee(\vee f(F_{1})\wedge f(b_{2}))\vee\dots\vee(\vee
f(F_{1})\wedge f(b_{m}))$ $\displaystyle=$ $\displaystyle(f(a_{1})\wedge
f(b_{1}))\vee\dots\vee(f(a_{n})\wedge f(b_{1}))\vee\dots\vee(f(a_{n})\wedge
f(b_{m}))$ $\displaystyle=$ $\displaystyle\bigvee\limits_{a\in F_{1},b\in
F_{2}}f(a)\wedge f(b)$

Thus, $P_{L}(f)$ is a $\vee-\wedge-$ morphism. $\Box$ Now, we will show the
main theorem of this section.

######  Theorem 9.4.

  1. 1.

For each directed space $X$, each $\vee-\wedge-$ directed space $Y$, and every
morphism $f:X\rightarrow Y$, there is a unique $\vee-\wedge-$ morphism $F$
from $P_{L}(P_{U}(X))$ to $Y$ such that
$F\circ(\eta_{\vee}\circ\eta_{\wedge})=f$. $F$ is given by $P_{L}(P_{U}(f))$.
Thus, $P_{L}(P_{U}(X))$ is the free $\vee-\wedge-$ directed space over the
directed space $X$.

  2. 2.

For each directed space $X$, each $\vee-\wedge-$ directed space $Y$, and every
morphism $f:X\rightarrow Y$, there is a unique $\vee-\wedge-$ morphism $F$
from $P_{U}(P_{L}(X))$ to $Y$ such that
$F\circ(\eta_{\wedge}\circ\eta_{\vee})=f$. $F$ is given by $P_{U}(P_{L}(f))$.
Thus, $P_{U}(P_{L}(X))$ is the free $\vee-\wedge-$ directed space over the
directed space $X$.

  3. 3.

For each directed space $X$, $P_{L}(P_{U}(X))$ and $P_{U}(P_{L}(X))$ are
isomorphic by a $\vee-\wedge-$ morphism, which maps
$\uparrow(\mathord{\downarrow}x)$ to
$\mathord{\downarrow}(\mathord{\uparrow}x)$.

Proof We prove (1) and (2) is analogous. Suppose $X$ is a directed space, then
$P_{U}(X)$ is a $\wedge-$ direcrted space and $P_{U}(f)$ is a $\wedge-$
morphism from $P_{U}(X)$ to $Y$. By Theorem 9.2 and 9.3, $P_{L}(P_{U}(X))$ is
a $\vee-\wedge-$ directed space and $P_{L}(P_{U}(f))$ is a $\vee-\wedge-$
morphism.
$f=P_{U}(f)\circ\eta_{\wedge}=(P_{L}(P_{U}(f))\circ\eta_{\vee})\circ\eta_{\wedge}=F\circ(\eta_{\vee}\circ\eta_{\wedge})$.
We remain to prove that $F$ is unique. Let $F_{1},F_{2}$ be two such
$\vee-\wedge-$ morphisms, i.e.
$F_{i}\circ\eta_{\vee}\circ\eta_{\wedge}=f,i=1,2$ holds. We have to show
$F_{1}=F_{2}$. Let $F_{i}^{{}^{\prime}}=F_{i}\circ\eta_{\vee},i=1,2$. Since
$F_{i}^{{}^{\prime}}\circ\eta_{\wedge}=f$, by the uniqueness of $P_{U}(f)$,
$F_{1}\circ\eta_{\vee}=F_{2}\circ\eta_{\vee}$. Since the extension of
$P_{L}(P_{U}(X))$ is unique, we have $F_{1}=F_{2}$. Thus, $F$ is unique, and
$P_{L}(P_{U}(X))$ is the free $\vee-\wedge-$ directed space over $X$.
(3) an immediate conclusion by (1) and (2). $\Box$

## data availability statement

Data sharing not applicable to this article as no datasets were generated or
analysed during the current study.

## Conflict of interest

The authors declare that they have no conflict of interest.

## Reference

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