2. Preliminaries

First, we introduce some basic concepts. For basic knowledge and notations in domain theory, topology, and category theory, we refer to Abramsky and Jung (Reference Abramsky, Jung, Abramsky, Gabbay and Maibaum1994), Gierz et al. (Reference Gierz, Hofmann, Keimel, Lawson, Mislove and Scott2003),MacLane (Reference MacLane1971).

Let P be a partially ordered set (poset for short). Given any subset $A\subseteq P$ , 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 that 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 an upper bound in D. A poset P is called a directed complete poset, abbreviated as a dcpo, if each directed subset D of P has a supremum (denoted by $\bigvee D$ ). A subset U of a poset P is called a Scott open subset if U is an upper set and for each directed subset $D\subseteq P$ such that $\bigvee D$ exists and belongs to U, $U\cap D\not=\emptyset$ . The set of all Scott open subsets of a poset P form a topology on P, which is called the Scott topology and denoted by $\sigma(P)$ . A topological space is called a Scott space if it is a dcpo endowed with the Scott topology. We denote by $\Sigma P = (P,\sigma(P))$ . Suppose that P and E are two posets, a map $f:P\longrightarrow E$ is called Scott continuous if it is continuous with respect to $\sigma(P)$ and $\sigma(E)$ . A poset L is called a complete lattice if any subset has a supremum in L. We say $x\lll y$ in L, if for any set A with $y\leq\bigvee A$ , there exists some $a\in A$ such that $x\leq a$ . A complete lattice L is prime continuous if for any $x\in L$ , $x=\bigvee\{y\in L:y\lll x\}$ .

All topological spaces in this paper are supposed to be $T_0$ spaces. A net in a topological space X is a map $\xi: J\longrightarrow X$ , where J is a directed set. Usually, we denote a net by $(x_j)_{j\in J}$ or $(x_j)$ . We say that $(x_j)$ converges to x, denoted 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 every 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 with its topology denoted by $\mathcal{O}(X)$ . Then, the specialization order on X is defined as follows:

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

where $\overline{\{y\}}$ denotes the closure of $\{y\}$ . From now on, the order of a $T_{0}$ topological space always indicates the specialization order “ $\sqsubseteq$ ”.

Supposing that X is a $T_{0}$ space, then every directed set $D\subseteq X$ can be regarded as a net $(d)_{d \in D}$ in X. We use $D\rightarrow x$ or $x\equiv \lim D$ to represent that D converges to x. Define

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

It is easy to verify that, for every $x, y\in X$ , $x\sqsubseteq y\Longleftrightarrow \{y\}\rightarrow x$ . Therefore, if $x\sqsubseteq y$ then $(\{y\},\ x)\in D(X)$ . Next, we introduce the concept of monotone determined spaces.

Dcpos endowed with the Scott topology are monotone determined spaces and many properties of dcpos can be extended to monotone determined spaces. The topology on a monotone determined space need not be Scott topology; for example, every poset with the Alexanderoff topology is a monotone determined space. The natural numbers with co-finite topology are not a monotone determined space. For more details, we refer to Yu and Kou (Reference Yu and Kou2015).

Monotone determined spaces with continuous maps as morphisms form a Cartesian closed category, denoted by Dtop. In the following, we introduce the representation and some basic properties of categorical product of monotone determined spaces in $\bf Dtop$ .

Suppose that $X,\ Y$ are two monotone determined spaces. Let $X\times Y$ be 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 $X\otimes Y$ is generated as follows: for each given $\leq$ - directed set $D\subseteq X\times Y$ and $(x, y)\in X\times Y$ , define

   \[D\rightarrow_d(x, y)\in X\otimes Y\iff\pi_{1}D\rightarrow x\in X,\ \pi_{2}D\rightarrow y\in Y. \]
   A subset $U\subseteq X\otimes Y$ is open iff for every $D\rightarrow_d (x, y)$ as defined above, $(x, y)\in U\Longrightarrow U\cap D\neq\emptyset$ .

1. (1) The topological space $X\otimes Y$ defined as above is a monotone determined space, and the specialization order on $X\otimes Y$ is equal to the pointwise order on $X\times Y$ .

2. (2) Let Z be a monotone determined space. Then $f:X\otimes Y\longrightarrow Z$ is continuous if and only if it is continuous in each argument separately.

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 monotone determined spaces, then f is continuous if and only if it is directed continuous.

Let P be a dcpo, and $x,\ y\in P$ . We say that x is way-below y, if for each given directed set $D\subseteq P$ , $y\leq\bigvee D$ implies that there exists some $d\in D$ such that $x\leq d$ . We write

A $T_0$ topological space X is a c-space if for each $x\in X$ and each open neighborhood U of x, there exists some $y\in U$ such that $x\in ({\uparrow y})^\circ \subseteq U$ . The c-spaces are classical spaces investigated by many scholars. Continuous domains endowed with the Scott topology are c-spaces, and c-spaces are monotone determined spaces (Erné Reference Erné2009).

The following result gives a characterization of c-space.

3. The Monotone Determined Probabilistic Powerspaces

As mentioned above, monotone determined spaces form a very natural topological extended model of dcpos in Domain theory. Like the work done by Battenfeld and Schöder (1994), extending the power domain to the category of monotone determined spaces is useful. In this section, we will construct the monotone determined probabilistic powerspace of an arbitrary monotone determined space, which is a free algebra generated by the addition and scalar multiplication over the monotone determined space.

1. (1) strictness: $\mu(\emptyset)=0$ ;

2. (2) monotonicity: $\forall V\subseteq U\in\mathcal{O}(X)$ $\implies$ $\mu(V)\leq\mu(U)$ ;

3. (3) modular law: $\mu(U)+\mu(V)=\mu(U\cup V)+\mu(U\cap V), \forall U,V\in\mathcal{O}(X)$ ;

4. (4) continuity: for each directed subset $\mathcal{D}$ , also called a directed family of $\mathcal{O}(X)$ , $\mu(\sup\mathcal{D})=\sup\{\mu(U):U\in\mathcal{D}\}$ .

$\mathcal{V}(X)$ stands for the set of continuous valuations over X.

The following lemma shows that every simple valuation has a unique representation as a linear combination of point valuations.

Topological cones were defined by Keimel (Reference Keimel2008). In the following, we replace the $T_0$ space by monotone determined space to give the definition of monotone determined cone.

1. (1) $x+y=y+x$ , $\forall x,\ y\in X$ ,

2. (2) $(x+y)+z=x+(y+z)$ , $\forall x,\ y,\ z\in X$ ,

3. (3) $0+x=x,\ \forall x\in X$ ,

4. (4) $(kl)\cdot x=k\cdot(l\cdot x),\ \forall k,\ l\in\mathbb{R}^+,\forall x\in X$ ,

5. (5) $(k+l)\cdot x=(k\cdot x)+(l\cdot x)$ , $\forall k,\ l\in\mathbb{R}^+,\ \forall x\in X$ ,

6. (6) $k\cdot(x+y)=(k\cdot x)+(k\cdot y),\ \forall k\in\mathbb{R^+},\ \forall x,\ y\in X$ ,

7. (7) $1\cdot x=x,\ \forall x\in X$ ,

8. (8) $k\cdot 0=0$ , $\forall k\in\mathbb{R}^+$ .

Denote the category of all monotone determined cones and monotone determined cone homomorphisms by Dscone. Then Dscone is a subcategory of Dtop. Now, we define the notion of monotone determined probabilistic powerspaces.

By the definition above, if monotone determined cones $(Z_1, +, \cdot)$ and $(Z_2, \uplus, \ast)$ are both the monotone determined probabilistic powerspaces of X, then there exists a topological homomorphism which is also a monotone determined cone homomorphism $g:Z_1\rightarrow Z_2$ . Therefore, up to isomorphism, the monotone determined probabilistic powerspace of a monotone determined space is unique. Hence, we denote the monotone determined probabilistic powerspace of each monotone determined space X by $P_D(X)$ .

Now, we show the existence of monotone determined probabilistic powerspaces by giving the concrete construction of the monotone determined probabilistic powerspace of any monotone determined space X.

For any monotone determined space X, there is a natural pointwise order $\leq$ on $\mathcal{V}_f(X)$ :

\[\xi\leq\eta\iff \xi(U)\leq\eta(U),\ \forall U\in\mathcal{O}(X).\]

In the following, we will introduce a new convergence on the set of simple valuations.

A subset $\mathcal{U}\subseteq \mathcal{V}_f(X)$ is called a $\Rightarrow_P$ convergence open subset of $\mathcal{V}_f(X)$ if and only if for each directed subset $\mathcal{D}$ of $\mathcal{V}_f(X)$ and $\xi\in \mathcal{V}_f(X)$ , $\mathcal{D}\Rightarrow_{P}\xi\in \mathcal{U}$ implies $\mathcal{D}\cap \mathcal{U}\neq \emptyset$ . Denote the set of all $\Rightarrow_P$ convergence open subsets of $\mathcal{V}_f(X)$ by $O_{\Rightarrow_{P}}(\mathcal{V}_f(X))$ .

1. (1) $(\mathcal{V}_f(X),\ O_{\Rightarrow_P}(\mathcal{V}_f(X)))$ is a topological space, henceforth, denoted as CX.

2. (2) The specialization order $\sqsubseteq $ on CX is equal to the pointwise order.

3. (3) CX is a monotone determined space, that is, $O_{\Rightarrow_P}(\mathcal{V}_f(X))=d(CX)$ .

Let $\mathcal{U}_{1},\mathcal{U}_{2}\in O_{\Rightarrow_P}(\mathcal{V}_f(X))$ , and $\mathcal D$ be a directed subset of $\mathcal{V}_f(X)$ with $\mathcal{D}\Rightarrow_{P} \xi\in \mathcal{U}_{1}\cap\mathcal{U}_{2}$ . Then, there exists some $\xi_{1}\in \mathcal{D}\cap \mathcal{U}_{1}$ and $ \xi_{2}\in \mathcal{D}\cap \mathcal{U}_{2}$ . Since $\mathcal{D}$ is directed, there exists some $\xi_{3}\in \mathcal{D}$ such that $\xi_{3}\geq\xi_{1},\ \xi_{2}$ . Then, $\xi_{3}\in \mathcal{D}\cap \mathcal{U}_{1}\cap\mathcal{U}_{2}$ , and then $\mathcal{U}_{1}\cap\mathcal{U}_{2} \in O_{\Rightarrow_P}(\mathcal{V}_f(X))$ . In the same way, we can show that $O_{\Rightarrow_U}(\mathcal{V}_f(X))$ is closed under arbitrary unions. Hence, $O_{\Rightarrow_U}(\mathcal{V}_f(X))$ is a topology.

(2) By the proof of (1), each $\Rightarrow_P$ convergence open set is an upper set with respect to the pointwise order. We only need to prove that $\downarrow_{\leq}\eta$ is a closed set in CX. Equivalently, we show that $CX\setminus\downarrow_{\leq}\eta$ is a $\Rightarrow_{P}$ convergence open set.

Set $\mathcal{U}=CX\setminus\downarrow_{\leq}\eta$ and $\mathcal{D}\Rightarrow_{P}\xi\in\mathcal{U}$ . To obtain a contradiction, suppose $\mathcal{U}\cap\mathcal{D}=\emptyset$ , that is, $\forall\xi'\in\mathcal{D},\xi'\leq\eta$ . By the definition of $\mathcal{D}\Rightarrow_{P}\xi$ , there is a representation $\xi=\sum_{i=1}^{n}r_{b_{i}}\eta_{b_i}$ , and we have directed sets $D_1,\dots,D_n\subseteq X$ with $D_i\rightarrow b_i,i=1,\dots,n$ , and for any $ (d_1,\dots,d_n)\in\prod_{i=1}^{n}D_i$ and any $r_{b_{i}}'<r_{b_{i}} (1\leq i\leq n)$ , there exists some $\xi'\in\mathcal{D}$ such that $\sum_{i=1}^{n} r_{b_i}'\eta_{d_i}\leq\xi'$ . Now, we claim that $\xi\leq\eta$ , which contradicts $\xi\in\mathcal{U}$ . By the definition of pointwise order $\leq$ , for any $ U\in\mathcal{O}(X)$ , we may assume that $b_1,\dots,b_k\in U,0\leq k\leq n$ . Since $D_i\rightarrow b_i,i=1,\dots,k$ , there exists $(d_1,\dots,d_k)\in\prod_{i=1}^{k}D_i$ such that $d_i\in U$ . For each $r_{b_{i}}'<r_{b_{i}},i=1,\dots,k$ , there exists $\xi'\in\mathcal{D}$ such that $\sum_{i=1}^{k} r_{b_i}'\eta_{d_i}\leq\xi'$ . Note that $k\leq n$ , then,

\[\left(\sum_{i=1}^{k} r_{b_i}'\eta_{b_i}\right)(U)=\sum_{i=1}^{k} r_{b_i}'\leq\xi'(U)\leq\eta(U).\]

But the supremum of the left hand side is $\sum_{i=1}^{k} r_{b_i}=(\sum_{i=1}^{k}r_{b_{i}}\eta_{b_i})(U)=\xi(U)$ .

(3) For an arbitrary topological space X, $\mathcal{O}(X)\subseteq d(X)$ holds. Thus, $O_{\Rightarrow_P}(\mathcal{V}_f(X))\subseteq d(CX)$ . On the other hand, according to the definition of $\Rightarrow_P$ convergence topology, if $\mathcal D\subseteq CX$ is directed and $ \mathcal{D}\Rightarrow_{P}\xi$ , then $\mathcal{D}$ converges to $\xi$ with respect to $O_{\Rightarrow_P}(\mathcal{V}_f(X))$ . By the definition of directed open set, $\mathcal{D}\Rightarrow_{P}\xi \in\mathcal{U}\in d(CX)$ will imply $\mathcal{U}\cap \mathcal{D}\neq \emptyset$ . Then, $\mathcal{U}\in O_{\Rightarrow_P}(\mathcal{V}_f(X))$ , it follows that $O_{\Rightarrow_P}(\mathcal{V}_f(X))= d(CX)$ , that is, CX is a monotone determined space.

The addition operation on CX is defined as follows:

\[\forall \xi, \eta\in CX,\ (\xi +\eta)(U)=\xi(U)+\eta(U),\ \forall U\in\mathcal{O}(X).\]

And the scalar multiplication operation on CX is defined as follows:

\[\forall a\in\mathbb{R}^{+},\ \forall \xi\in CX,\ (a\cdot\xi)(U)=a\xi(U).\]

We now check that the two operations are both continuous, and thus, $(CX, +, \cdot)$ is a monotone determined cone.

For an arbitrary $(d_1,\dots,d_n,c_1,\dots,c_m)\in\prod_{i=1}^{n}D_i\times\prod_{j=1}^{m}\{c_j\}$ , and $\forall r_{b_i}'<r_{b_i},\ s_{c_j}'<s_{c_j},\ i=1,\dots,n,j=1,\dots,m$ , by the definition of $\mathcal{D}\Rightarrow_{P}\xi$ , there exists some $\xi'\in\mathcal{D}$ such that $\sum_{i=1}^{n}r_{b_i}'\eta_{d_{i}}\leq\xi'$ . Since $+$ is monotone,

\[\sum_{i=1}^{n}r_{b_i}'\eta_{d_{i}}+\sum_{j=1}^{m}s_{c_j}'\eta_{c_j}\leq\xi'+\sum_{j=1}^{m}s_{c_j}'\eta_{c_j}\leq \xi'+\eta.\]

Now, we prove continuity of scalar product. For an arbitrary fixed $a\in\mathbb{R}^+$ and a directed set $\mathcal{D}=\{\xi_i\}_{i\in I}\subseteq CX$ with $\mathcal{D}\Rightarrow_{P}\xi$ , we claim that $a\cdot\mathcal{D}=\{a\cdot\xi_i\}_{i\in I}\Rightarrow_{P}(a\cdot\xi)$ . By the hypothesis of $\mathcal{D}\Rightarrow_{P}\xi$ there is a representation of $\xi=\sum_{i=1}^{n}r_{b_i}\eta_{b_i}$ and we have $D_1,\cdots,D_n$ – with $D_i\rightarrow b_i$ for $1\leq i\leq n$ , for $\forall\ (d_1,\dots,d_n)\in\prod_{i=1}^{n}D_i,\ \forall\ r_{b_i}'<r_{b_i},i=1,\dots,n$ , there exists $\xi'\in\mathcal{D}$ , such that $\sum_{i=1}^{n}r_{b_i}'\eta_{d_i}\leq\xi'$ . Hence, we have $\sum_{i=1}^{n}a\cdot r_{b_i}'\eta_{d_i}\leq a\cdot\xi'$ .

For arbitrary $\xi=\sum_{i=1}^{n}r_{b_i}\eta_{b_i}\in CX$ and a directed set $D\subseteq\mathbb{R}^+$ with $D\rightarrow a$ . We claim that $D\cdot\xi=\{d\cdot\xi\}_{d\in D}\Rightarrow_{P}(a\cdot\xi)$ . Let $D_i=\{b_i\},i=1,\dots,n$ , $\forall\ (b_1,\dots,b_n)\in\prod_{i=1}^{n}D_i$ and $\forall\ r_{b_i}'<ar_{b_i},i=1,\dots,n$ , we only need to find $d\in D$ with $r_{b_i}'\leq dr_{b_i},i=1,\dots,n.$ This is possible, because for each $i=1,\dots,n$ , $\dfrac{r_{b_i}'}{r_{b_i}}<a$ and $D\rightarrow a$ .

In conclusion, $(CX, +, \cdot)$ is a monotone determined cone.

We now show that $(CX, +, \cdot)$ is the free monotone determined cone over X. It is the main result of this section. We begin with one useful lemma.

Proposition 18. (Gierz et al. Reference Gierz, Hofmann, Keimel, Lawson, Mislove and Scott2003, Proposition IV-9.18) (Splitting Lemma) For two simple valuations in $\mathcal{V}_f(X)$ , where X is a $T_0$ space, we have $\zeta=\sum_{b\in B}r_b\eta_b\leq\sum_{c\in C}s_c\eta_c=\xi$ if and only if there exist $\{t_{b,c}\in[0,+\infty):b\in B,\ c\in C\}$ such that for each $b\in B,\ c\in C$ ,

\[\sum\limits_{c\in C}t_{b,c}=r_b,\ \sum_{b\in B}t_{b,c}\leq s_c\]

and $t_{b,c}\neq 0$ implies $b\leq c$ .

Let $(Y, \uplus, \ast)$ be a monotone determined cone and $f:X\rightarrow Y$ be a continuous map. Define $\bar{f}:CX\rightarrow Y$ as follows: $\forall \xi=\sum_{i=1}^{n}r_{b_i}\eta_{b_i}\in CX (\text{supp}(\xi)=\{b_1,\cdots,b_n\})$ ,

\[\bar{f}(\xi)=\biguplus\limits_{i=1}^{n}r_{b_i}\ast f(b_i).\]

By Lemma 10, $\bar{f}$ is well-defined.

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

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

1. (2) $\bar{f}$ is a monotone determined cone homomorphism, that is, $\bar{f}$ is continuous and for arbitrary $\sum_{i=1}^{n}r_{b_i}\eta_{b_i}$ , $\bar{f}(\sum_{i=1}^{n}r_{b_i}\eta_{b_i})=\biguplus_{i=1}^{n}r_{b_i}\ast\bar{f}(\eta_{b_i})$ . This equation is evident since $\bar{f}(\eta_{b_i})=f(b_i),\ i=1,\dots,n.$ So, we only need to prove that $\bar{f}$ is continuous.

First, $\bar{f}$ is monotone. Let $\zeta=\sum_{b\in B}r_b\eta_b\leq\sum_{c\in C}s_c\eta_c=\xi$ . By Proposition 18, there exist $\{t_{b,c}\in[0,+\infty):b\in B,\ c\in C\}$ such that for each $b\in B,\ c\in C$ ,

\[\sum\limits_{c\in C}t_{b,c}=r_b,\ \sum_{b\in B}t_{b,c}\leq s_c\]

and $t_{b,c}\neq 0$ implies $b\leq c$ . From the definition of $\bar{f,}$ we have

\[\bar{f}(\zeta)=\biguplus_{b\in B}r_b\ast f(b)=\biguplus\limits_{b\in B}\biguplus\limits_{c\in C}t_{b,c}\ast f(b)\leq\biguplus\limits_{c\in C}\biguplus\limits_{b\in B}t_{b,c}\ast f(c)\leq\biguplus\limits_{c\in C}s_{c}\ast f(c)=\bar{f}(\xi).\]

By Proposition 16, to show the continuity of $\overline{f}$ , we need only to prove that $\bar{f}$ preserves $\Rightarrow_{P}$ convergence class. Suppose that we have a directed set $\mathcal{D}=\{\xi_i\}_{i\in I}\subseteq CX$ and $\xi\in CX$ with $\mathcal{D}\Rightarrow_{P}\xi$ , then, we need to show that $\bar{f}(\mathcal{D})\rightarrow\bar{f}(\xi)$ in Y. There is a representation $\xi=\sum_{i=1}^{n}r_{b_i}\eta_{b_i}$ and let $D_1,\cdots,D_n$ with $D_i\rightarrow b_i (1\leq i\leq n)$ be as in the definition of $\mathcal{D}\Rightarrow_{P}\xi$ . From the continuity of $f,\ \uplus$ and $\ast$ , we have

\[\mathcal{E}=(r_{b_1}'\ast f(D_1))\uplus\dots\uplus (r_{b_n}'\ast f(D_n))\rightarrow\biguplus\limits_{i=1}^nr_{b_i}\ast f(b_i)=\bar{f}(\xi).\]

Notice that $\mathcal{E}=\{(r_{b_1}'\ast f(d_1))\uplus\dots\uplus (r_{b_n}'\ast f(d_n)):(d_1,\dots,d_n)\in\prod_{i=1}^{n}D_i, r_{b_i}'<r_{b_i},i=1,\cdots,n\}$ is a directed set in Y. For an arbitrary open neighborhood U of $\bar{f}(\xi)$ , there exists some $(d_1,\dots,d_n)\in\prod_{i=1}^{n}D_i$ and $r_{b_i}'<r_{b_i},i=1,\cdots,n$ such that $\biguplus_{i=1}^{n}r_{b_i}'\ast f(d_i)\in U$ . For the given $d_1,\dots,d_n$ and $r_{b_i}'<r_{b_i},i=1,\dots,n$ , again by the definition of $\mathcal{D}\Rightarrow_{P}\xi$ , there exists some $\xi'\in\mathcal{D}$ such that $\sum_{i=1}^{n}r_{b_i}'\eta_{d_i}\leq\xi'$ . Then $\bar{f}(\sum_{i=1}^{n}r_{b_i}'\eta_{d_i})=\biguplus_{i=1}^{n}r_{b_i}'\ast f(d_i)\leq\bar{f}(\xi')$ . Since U is an upper set, then $\bar{f}(\xi')$ is included in U. Hence, $\bar{f}$ is continuous.

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

Suppose we have a monotone determined cone homomorphism $g:(CX, +, \cdot)\rightarrow (Y, \uplus, \ast)$ such that $f=g\circ i$ , then $g(\eta_x)=f(x)=\bar{f}(\eta_x)$ . For each $ \xi=\sum_{i=1}^{n}r_{b_i}\eta_{b_i}\in CX$ ,

\begin{eqnarray*} g(\xi)&=& g(r_{b_1}\eta_{b_1}+r_{b_2}\eta_{b_2}\cdots+r_{b_n}\eta_{b_n})\\ &=&g(r_{b_1}\eta_{b_1})\uplus g(r_{b_2}\eta_{b_2})\uplus \cdots \uplus g(r_{b_n}\eta_{b_n})\\ &=&r_{b_1}\ast g(\eta_{b_1})\uplus r_{b_2}\ast g(\eta_{b_2})\uplus \cdots\uplus r_{b_n}\ast g( \eta_{b_n})\\ &=&r_{b_1}\ast f({b_1})\uplus r_{b_2}\ast f({b_2})\uplus \cdots\uplus r_{b_n}\ast f({b_n})\\ &=&\bar{f}(\xi). \end{eqnarray*}

Thus $\bar{f}$ is unique.

In conclusion, the monotone determined cone $(CX, +, \cdot)$ is the monotone determined probabilistic powerspace of X, that is, $P_D(X)\cong (CX, +, \cdot)$ .

The monotone determined probabilistic powerspace is unique up to algebraic isomorphism and topological homemorphism, so we can directly denote the monotone determined probabilistic powerspace by $P_D(X)=(CX, +, \cdot)$ for each monotone determined space X.

Suppose that X and Y are two monotone determined spaces and $f:X\rightarrow Y$ is a continuous map. Define the map $P_D(f):P_D(X)\rightarrow P_D(Y)$ as follows: $\forall \xi=\sum_{i=1}^{n}r_{b_i}\eta_{b_i}\in CX$ ,

\[P_D(f)(\xi)=\sum_{i=1}^{n}r_{b_i}\eta_{f(b_i)}.\]

$P_D(f)$ is well-defined and order preserving. It is also easy to check that, $P_D(f)$ is a monotone determined cone homomorphism between these two extended probabilistic powerspaces. If $id_X$ is the identity map and $g:Y\rightarrow Z$ is an arbitrary continuous map from Y to a monotone determined space Z, then, $P_D(id_X)=id_{P_D(X)},\ P_D(g\circ f)=P_D(g)\circ P_D(f)$ . Thus, $P_D:{\bf Dtop}\rightarrow {\bf Dscone}$ is a functor from Dtop to Dscone. Let $U:{\bf Dscone}\rightarrow {\bf Dtop}$ be the forgetful functor. By Theorem 19, we have the following result.

4. D-Completion and Free dcpo-cone over Any dcpo

Monotone convergence spaces (a.k.a d-spaces) form a very important class of $T_0$ topological spaces in domain theory. Interestingly, every $T_0$ space has a D-completion (See Definition 25). In this section, we will recall some notions about D-completions which are useful for our further discussions.

A subset A of a poset P is called d-closed if for every directed subset D of A that possesses supremum $\bigvee D$ , $\bigvee D$ is included in A. The d-closed sets form the closed sets for a topology, called the d-topology (Keimel and Lawson Reference Keimel and Lawson2009, Lemma 5.1). We denote the closure of A in P with d-topology by $\text{cl}_{d}(A)$ . If $\text{cl}_{d}(A)$ is equal to P, then we say that A is d-dense in P. A map f from a poset P to a poset Q is called d-continuous if f is continuous under the d-topology.

1. (1) A is a monotone convergence space.

2. (2) A is a sub-dcpo.

3. (3) A is d-closed.

Then, (1) implies (2) and (2) implies (3), and all three are equivalent if X is a monotone convergence space.

The following result says that D-completions are universal, and hence, they are unique up to isomorphism.

Next we introduce the concept of Scott completion.

Not every space has a Scott completion (see the following Example 28). An alternative argument is that the category of Scott space with continuous maps is not reflective in the category of $T_0$ spaces, we refer to Sheng et al. (Reference Sheng, Xi and Zhao2023, Theorem 3.6).

1. (1) f is injective.

For any two different numbers n and m, consider the Sierpinski space $\mathbb{S}$ which is the poset $\{\bot,\top\} (\bot<\top)$ equipped with its Scott topology. Define a continuous map $g\colon\mathbb{N}\rightarrow\mathbb{S}$ as follows: $g(n)=\bot, g(x)=\top $ for any $x\neq n$ . Since $(\bar{\mathbb{N}},f)$ is a Scott completion, there is a continuous map $\tilde{g}\colon\bar{\mathbb{N}}\rightarrow\mathbb{S}$ such that $\tilde{g}\circ f=g$ . It follows that $\tilde{g}(f(n))=g(n)=\bot, \tilde{g}(f(m))=g(m)=\top$ . Hence, $f(n)\neq f(m)$ .

1. (2) For any two different numbers n,m of $\mathbb{N}$ , f(n) and f(m) are incomparable in $\bar{\mathbb{N}}$ .

Without loss of generality, we assume that $f(m)<f(n)$ . By the same argument of the first step, $\top=g(m)=\tilde{g}(f(m))\leq\tilde{g}(f(n))=g(n)=\bot$ , a contradiction.

1. (3) For any $x\in\bar{\mathbb{N}}, $ there is some $n\in\mathbb{N}$ such that $f(n)\leq x$ . In other words, ${\downarrow}x\cap f(\mathbb{N})\neq\emptyset$ . To obtain a contradiction, suppose there is some $x_0\in\bar{\mathbb{N}}$ with ${\downarrow}x_0\cap f(\mathbb{N})=\emptyset$ . We take a constant map $g=\lambda n.\top\colon\mathbb{N}\rightarrow\mathbb{S}$ . Clearly, it is continuous. Now let us consider two continuous maps $g_1, g_2\colon\bar{\mathbb{N}}\rightarrow\mathbb{S}$ as follows:

   \[g_1(x)=\top,\forall x;\]

\[g_2(x)= \begin{cases} \top, & x\not\leq x_0,\\ \bot, & x \leq x_0. \end{cases}\]

It is easy to verify that $g=g_1\circ f=g_2\circ f$ , which contradicts the uniqueness of $\tilde{g}$ .

We conclude that $f(\mathbb{N})$ is the set of minimal elements of $\bar{\mathbb{N}}$ from (1), (2) and (3). Notice that every subset A of $f(\mathbb{N})$ is Scott closed in $\bar{\mathbb{N}}$ . Hence, $f(\mathbb{N})$ with its subspace topology of $\bar{\mathbb{N}}$ is a discrete space. It follows that $f^{-1}(A)$ is closed in $\mathbb{N}$ for any $A\subseteq f(X)$ . We see at once that $\mathbb{N}$ is a discrete space as well, a contradiction.

But any poset with Scott topology has a Scott completion (Keimel and Lawson Reference Keimel and Lawson2009) and the following result tells us that every monotone determined space has a Scott completion. Let X be a monotone determined space. The map $i\colon X\rightarrow \Gamma X$ is defined by $i(x)={\downarrow}x$ , where $\Gamma X$ is the lattice of closed subsets of X. We denote the d-closure of $i(X)=\{{\downarrow}x:x\in X\}$ in $\Gamma X$ by $\tilde{X}$ . The map $i\colon X\rightarrow\tilde{X}$ has the universal property: For any Scott space Z and any continuous map $f\colon X\rightarrow Z$ , there is a unique Scott continuous map $\bar{f}\colon\tilde{X}\rightarrow Z$ which is defined by $\bar{f}(A)=\bigvee\text{cl}_{\sigma}(f(A))$ , such that $f=\bar{f}\circ i$ .

Next, we will recall some basic properties of dcpo-cones. As far as we know, there is no representation of free dcpo-cone over general dcpos. In this section, we will construct the free dcpo-cone over any dcpo by using the Scott completion of a monotone determined space.

Definition 32. Let C,D be two dcpo-cones. A map $f\colon C\rightarrow D$ is said to be linear if

\[f(r\cdot a)=rf(a), \text{for all } r\in\mathbb{R^+} \text{and } a\in C\]

and

\[f(a+b)=f(a)+f(b), \text{for all} a,b\in C.\]

Let Dcone denote the category of dcpo-cones with Scott continuous linear maps.

1. (1) $f(\prod_{i=1}^{n}\bar{A_i})\subseteq\bar{B}$ whenever $f(\prod_{i=1}^{n}A_i)\subseteq B$ ;

2. (2) $\overline{f(\prod_{i=1}^{n}\bar{A_i})}=\overline{f(\prod_{i=1}^{n}A_i)}$ ;

3. (3) $f^{\gamma}\colon\prod_{i=1}^{n}\Gamma X_i\rightarrow\Gamma Y$ is separately continuous, hence Scott continuous. Here the definition of $f^{\gamma}$ is as follows:

   \[f^{\gamma}(A_1,\cdots,A_n)=\overline{f\left(\prod_{i=1}^{n}A_i\right)}=\overline{\{f(x_1,\cdots,x_n):x_i\in A_i \text{for} 1\leq i\leq n\}}.\]

Next we will verify that $\tilde{X}$ with these two operations is a dcpo-cone.

Proof. It is easy to see that $X\rightarrow\tilde{X}$ is a linear map. First, we show that $\oplus\colon\tilde{X}\times\tilde{X}\rightarrow\tilde{X}$ is a well-defined Scott continuous map. By Lemma 33(3), $\oplus\colon\Gamma X\times\Gamma X\rightarrow\Gamma X$ is Scott continuous, hence d-continuous. For any $A \in\tilde{X}$ , we consider a map $g\colon\Gamma X\rightarrow\Gamma X, g(F)=A\oplus F, \forall F\in\Gamma X$ . Since g is d-continuous,

\[\forall B\in\tilde{X}: A\oplus B\in g(cl_{d}\{{\downarrow}x:x\in X\})\subseteq cl_{d}(\{A\oplus{\downarrow}x:x\in X\}).\]

Next we consider another map $h\colon\Gamma X\rightarrow\Gamma X, h(F)=F\oplus{\downarrow}y$ for a given $y\in X$ .

Hence, we obtain that $A\oplus B\in\tilde{X}$ .

Second for any $A,B,C\in\Gamma X$ , obviously $A\oplus B=B\oplus A$ , $A\oplus {\downarrow}0=A$ , and:

\begin{align*}\overline{\{a+b+c:a\in A,b\in B,c\in C\}}&\subseteq A\oplus(B\oplus C)\\ &=\overline{\{a+x:a\in A,x\in B\oplus C\}}\\ &=\overline{\bigcup_{a\in A}\{a+x:x\in B\oplus C\}}\\ &\subseteq\overline{\bigcup_{a\in A}cl\{a+(b+c):b\in B,c\in C\}} (\lambda x.a+x\colon X\rightarrow X \text{is continuous})\\ &\subseteq \overline{\{a+b+c:a\in A,b\in B,c\in C\}}\end{align*}

Thus, we have $A\oplus (B\oplus C)=\overline{\{a+b+c:a\in A,b\in B,c\in C\}}$ . For the same reason, $(A\oplus B)\oplus C$ is equal to $\overline{\{a+b+c:a\in A,b\in B,c\in C\}}$ . Therefore, $(A\oplus B)\oplus C=A\oplus (B\oplus C)$ .

For any positive real number r, the map $\lambda x.rx\colon X\rightarrow X$ is a homomorphism. Hence, $r\odot F=\{ra:a\in F\}\in\Gamma X$ for each $F\in\Gamma X$ . Note that the map $f=\lambda F.r\odot F\colon\Gamma X\rightarrow\Gamma X$ is an order isomorphism under the inclusion ordering. For any $A\in\tilde{X}$ ,

Therefore, the map $\lambda A.r\odot A\colon\tilde{X}\rightarrow\tilde{X}$ is well-defined and Scott continuous.

Given any non-negative real numbers r,s and $A,B\in\tilde{X}$ ,learly

\[1\odot A=A, 0\odot A={\downarrow}0, (rs)\odot A=r\odot (s\odot A),\]

and

\[r\odot (A\oplus B)=r\odot\overline{\{a+b:a\in A,b\in B\}}=\overline{\{ra+rb:a\in A,b\in B\}}=(r\odot A)\oplus (r\odot B).\]

Next we show that

\[(r+s)\odot A=(r\odot A)\oplus (s\odot A).\]

If $A={\downarrow}x$ for some $x\in X$ , then $(r+s)\odot{\downarrow}x={\downarrow}(r+s)x$ and $(r\odot{\downarrow}x)\oplus (s\odot{\downarrow}x)$ = $\overline{\{ra+sb:a\leq x,b\leq x\}}={\downarrow}(r+s)x$ . We set $\mathcal{S}:=\{C\in\Gamma X:(r+s)\odot C=(r\odot C)\oplus (s\odot C)\}$ . By the preceding argument, $\{{\downarrow}x:x\in X\}\subseteq\mathcal{S}$ . For any directed family $\{A_i:i\in I\}$ of $\mathcal{S}$ ,

It is easy to see that

\[(r+s)\odot\bigvee_{i\in I}A_i\subseteq(r\odot\bigvee_{i\in I}A_i)\oplus (s\odot\bigvee_{i\in I}A_i).\]

It follows that $\bigvee_{i\in I}A_i\in\mathcal{S}$ . Hence, we have $A\in\tilde{X}\subseteq\mathcal{S}$ .

Finally, for any fixed $A\in\tilde{X}, g=\lambda x.x\odot A\colon\mathbb{R^+}\rightarrow\tilde{X}$ is Scott continuous. Since $\lambda x.xa\colon \mathbb{R^+}\rightarrow X$ is continuous for each $a\in X$ , then g is monotone. Given any bounded directed set $\{x_i:i\in I\}\subseteq\mathbb{R^+}$ , $(\bigvee_{i\in I}^{\uparrow}x_i)\odot A=\{(\bigvee_{i\in I}^{\uparrow}x_i)a:a\in A\}=\{\bigvee_{i\in I}^{\uparrow}x_ia:a\in A\}\subseteq \overline{\bigcup_{i\in I}^{\uparrow}x_i\odot A}=\bigvee_{i\in I}^{\uparrow}(x_i\odot A)$ . It is easily seen that $\bigvee_{i\in I}^{\uparrow}(x_i\odot A)\subseteq(\bigvee_{i\in I}^{\uparrow}x_i)\odot A$ . We conclude that $\bigvee_{i\in I}^{\uparrow}(x_i\odot A)=(\bigvee_{i\in I}^{\uparrow}x_i)\odot A$ . Therefore $(\tilde{X},\oplus,\odot)$ is a dcpo-cone.

Next we give the construction of free dcpo-cone over any dcpo by using the D-completions. It is one of the main results of this paper.

Theorem 36. Let D be a dcpo. The standard Scott completion $\big( (\mathfrak{D},\oplus,\odot),i\big)$ of $P_D(\Sigma D)$ is the free dcpo-cone, that is, given any dcpo-cone $\mathfrak{E}$ and any Scott continuous map $f\colon D\rightarrow\mathfrak{E}$ , if we assign $\xi=i\circ\eta'\colon D\rightarrow$ $\mathfrak{D}$ , where $\forall d\in D:$ $\eta'(d)=\eta_{d}$ . Then, there is a unique Scott continuous linear map $\bar{f}\colon\mathfrak{D}\rightarrow\mathfrak{E}$ such that $\bar{f}\circ\xi=f$ .

Proof. Since $\mathfrak{E}$ is a dcpo-cone, it is also a monotone determined cone with respect to Scott topology. Because $P_D(\Sigma D)$ is the free monotone determined cone over $\Sigma D$ , there exists a unique continuous linear map $\hat{f}\colon P_D(\Sigma D)\rightarrow \mathfrak{E}$ such that $\hat{f}\circ\eta'=f$ . By a similar argument, there is a unique continuous map $\bar{f}\colon\mathfrak{D}\rightarrow \mathfrak{E}$ with $\bar{f}\circ i=\hat{f}$ . We claim that $\bar{f}$ is also a linear map between $\mathfrak{D}$ and $\mathfrak{E}$ . For any element $F\in\mathfrak{D}$ , $\bar{f}(F)=\bigvee\overline{\hat{f}(F)}$ . Given any $\lambda\in\mathbb{R^+}$ , $\bar{f}(\lambda\odot F)=\bigvee\overline{\hat{f}(\lambda\odot F)}=\bigvee\overline{\lambda\odot\hat{f}(F)}=\bigvee(\lambda\overline{\hat{f}(F)})=\lambda\cdot\bigvee(\overline{\hat{f}(F)})=\lambda\cdot\bar{f}(F)$ . For any $A,B\in\mathcal{F}$ , we calculate

Suppose that there is another Scott continuous linear map $g\colon\mathfrak{D}\rightarrow\mathfrak{E}$ such that $g\circ\xi=f$ . Set $S=\{x\in\mathfrak{D}:\bar{f}(x)=g(x)\}$ . Clearly, S is closed under addition and scalar multiplication. Because $\bar{f}$ and g are Scott continuous, S is also a sub-dcpo of $\mathfrak{D}$ . It is easy to see that for any $d\in D$ , $\xi(d)\in S$ . Since i preserves multiplication, $i(r\cdot\eta_{d})=r\odot (i\circ\eta'(d))=r\odot\xi(d)\in S$ . By the additivity of i, $i(a)\in S$ for any $a\in P_D(\Sigma D)$ . Hence, S is equal to $\mathfrak{D}$ because the image of i is d-dense in $\mathfrak{D}$ . Consequently $\bar{f}=g$ .

5. Valuation Power domain and Free dcpo-cone

A well-known result in probabilistic power domain theory is that the space of continuous valuations over $\Sigma D$ is the free dcpo-cone of $\Sigma D$ if D is a continuous domain (Gierz et al. Reference Gierz, Hofmann, Keimel, Lawson, Mislove and Scott2003, Theorem IV-9.24). But it is unknown whether this result still holds for the case of general dcpos. In this section, we will discuss this problem and build some relationships between the space of valuations and the free dcpo-cones for more general dcpos instead of continuous ones.

In the following, we introduce some notations. Let F(D) denote the free dcpo-cone over D if D is a dcpo. For a topological space X, the valuation power domain (also called the extended probabilistic power domain) $\mathcal{V}(X)$ of X is the set of all continuous valuations on $\mathcal{O}(X)$ with the pointwise order, sometimes called the stochastic order: $\mu\leq\nu$ iff $\mu(U)\leq\nu(U)$ for all open sets U.

Let D be a domain. The extension map $\hat{\eta}\colon P_D(\Sigma D)\rightarrow\mathcal{V}(D)$ is a D-completion map. The topology on $P_D(\Sigma D)$ is the subspace topology induced by the Scott topology on $\mathcal{V}(D)$ .

Clearly, the map $\bar{\eta}\colon \Sigma(\mathcal{D})\rightarrow\Sigma(\mathcal{V}(D))$ is also a topological isomorphism. Notice that $i\colon P_D(\Sigma D)\rightarrow\mathcal{D}$ is a D-completion map. It follows that $\hat{\eta}\colon P_D(\Sigma D)\rightarrow\mathcal{V}(D)$ is a D-completion map. Hence, the topology of $P_D(\Sigma D)$ is the subspace topology induced by the Scott topology on $\mathcal{V}(D)$ .

We first show that $\mathcal{V}_{p}(X)$ is closed under addition. Given any two point-continuous valuations $\mu_1,\mu_2$ , for any $r\in \bar{\mathbb{R}}^{+}$ and $U\in\mathcal{O}(X)$ , assume that $\mu_1(U)+\mu_2(U)>r$ . If one of $\mu_1(U)$ and $\mu_2(U)$ is equal to 0, without loss of generality we say $\mu_1(U)=0$ , then $\mu_2(U)>r$ . Since $\mu$ is point-continuous, there is a finite subset F of U such that $\mu_2(V)>r$ for any open set $V\supseteq F$ . It follows that $\mu_1(V)+\mu_2(V)>r$ for any open set $V\supseteq F$ . Another case is that neither of $\mu_1(U)$ nor $\mu_2(U)$ is equal to 0. Take a positive real number $\epsilon$ such that $\epsilon<\mu_1(U), \epsilon<\mu_2(U), \mu_1(U)-\epsilon+\mu_2(U)-\epsilon>r$ . By the point continuity of $\mu_i$ (i=1,2), there is a finite set $F_i\subseteq U$ satisfying $\mu_i(V)>\mu_i(U)-\epsilon$ for any open set $V\supseteq F_i$ . Let F be the union of $F_1$ and $F_2$ , clearly $\mu_1(V)+\mu_2(V)>r$ for any open set $V\supseteq F$ . We conclude that $\mu_1+\mu_2\in\mathcal{V}_{p}(X)$ .

It is easily seen that $\mathcal{V}_{p}(X)$ is closed under scalar multiplication. It is a routine to check that $\cdot\colon\mathbb{R}^{+}\times\mathcal{V}_{p}(X)\rightarrow\mathcal{V}_{p}(X)$ and $+\colon\mathcal{V}_{p}(X)\times\mathcal{V}_{p}(X)\rightarrow\mathcal{V}_{p}(X)$ are Scott continuous. Hence, $\mathcal{V}_{p}(X)$ is a dcpo-cone.

Example 42. Let D be the Smyth power domain ( $\mathcal{Q}(\mathbb{R}_{\ell}))$ of the Sorgenfrey line $\mathbb{R}_{\ell}$ . By the result of Goubault and Jia (Reference Goubault-Larrecq and Jia2021, Theorem 39), the inclusion map $j\colon\mathcal{V}_{p}(D)\rightarrow\mathcal{V}(D)$ is not surjective. We claim that $(\mathcal{V}(D),\eta)$ is not the free dcpo-cone over D. Let $(F(D),\xi)$ be the free dcpo-cone over D and $\eta_1$ the co-restriction of $\eta\colon D\rightarrow\mathcal{V}(D)$ on $\mathcal{V}_{p}(D)$ . Obviously $\eta=j\circ\eta_1$ . By the universal property of F(D), there is a unique Scott continuous linear map f with $\eta_{1}=f\circ\xi$ . Let g denote the map $j\circ f$ , then $g\circ\xi=(j\circ f)\circ\xi=j\circ(f\circ\xi)=j\circ\eta_1=\eta$ . Clearly g is a Scott continuous linear map but not surjective. Hence, $(\mathcal{V}(D),\eta)$ is not a free dcpo-cone over D.