2.1 Duality theory for closed sets and the Hoare hyperspace

Let X be a topological space. Then, every closed set $C \subseteq X$ can be paired with any open $U \subseteq X$ by defining

(1) \begin{equation} \langle {C} , {U} \rangle \,{:}\,{\raise-1.5pt{=}}\, \begin{cases} 1 & \text{if } C \cap U \ne \varnothing, \\ 0 & \text{if } C \cap U = \varnothing. \end{cases}\end{equation}

We will occasionally say “C hits U” to indicate that $C \cap U \ne \varnothing$ , as commonly done in the literature on hyperspaces (Beer and Tamaki Reference Beer and Tamaki1993).

Intuitively, C is analogous to a measure, and U is analogous to an integrable function, so that the pairing is analogous to integration. To investigate the properties of this pairing further, we identify the set of open subsets $\mathcal{O}(X)$ with the set of maps $X \to S$ , where $S \,{:}\,{\raise-1.5pt{=}}\, \{0,1\}$ is the SierpiŃski space with open sets $$\{ \varnothing ,\{ 1\} ,\{ 0,1\} \} $$ . We denote this set of maps by $S^X$ . Thus, for a closed set C, we have

\begin{equation*} \langle {C} , {-} \rangle \: : \: S^X \to S.\end{equation*}

Intuitively, an S-valued function – which is the same thing as an open set – can be integrated to an element of S – which is the same thing as a truth value, indicating whether C hits the open set or not. Due to the natural bijection $\mathcal{O}(X) \cong S^X$ , we may call the elements of $S^X$ the open subsets of X, may denote them by $U,V \subseteq X$ , and may use them interchangeably as subsets of X and as continuous functions $X \to S$ . Since $\mathcal{O}(X)$ is a complete lattice with respect to the inclusion order, the same holds for $S^X$ , where the partial order is the pointwise order of functions.

We then have the following known characterization of closed sets.

1. (a) Closed subsets of X,

2. (b) Maps $\phi : S^X \to S$ that fulfill the following two conditions:

3. (i) Linearity: For all $U,V\in S^X$ we have

   \begin{equation*} \phi(U \cup V) = \phi(U) \lor \phi(V), \end{equation*}
   and $\phi(\varnothing) = 0$ .

4. (ii) Scott continuity: For any directed net $(U_\lambda)_{\lambda \in \Lambda}$ in $S^X$ , we have

   \begin{equation*} \phi \left( \bigcup_{\lambda \in \Lambda} U_\lambda \right) = \bigvee_{\lambda \in \Lambda} \phi(U_\lambda). \end{equation*}

Under this bijection, the inclusion order on closed sets corresponds exactly to the pointwise order on maps $S^X \to S$ .

We call condition (i) “linearity” since it is analogous to the linearity of integration, where its first subcondition is analogous to additivity and its second to the commutation with scalar multiplication.

Proof. Conditions (i) and (ii) state the preservation of nullary, binary, and directed joins, respectively, of $\phi$ as a map of complete lattices $S^X \to S$ . It is well-known that these are jointly equivalent to the preservation of all joins.

Given a closed set $C \subseteq X$ , the pairing map $\langle {C} , {-} \rangle : S^X \to S$ preserves all joins, since C hits a union of open sets if and only if it hits one of them. Therefore $\langle {C} , {-} \rangle$ satisfies the stated conditions. Conversely given $\phi : S^X \to S$ satisfying the stated conditions, $\phi^{-1}(0)$ is a collection of open sets, and we can consider the closed set

(2) \begin{equation} C \,{:}\,{\raise-1.5pt{=}}\, X \setminus \left( \bigcup \phi^{-1}(0) \right). \end{equation}

We need to prove that these two constructions are inverses of each other.

For a closed set C, $\langle {C} , {-} \rangle^{-1}(0)$ consists of all open sets disjoint from C. Since their union is $X \setminus C$ , the construction (2) recovers C. For given $\phi : S^X \to S$ , the preservation of joins shows that $\phi(U) = 0$ if and only if $U \subseteq \bigcup \phi^{-1}(0)$ . (In other words, $\phi^{-1}(0)$ is an order ideal closed under joins, and therefore principal.) But this condition holds equivalently if U is disjoint from (2), as was to be shown.

It remains to be shown that both constructions are monotone. Given closed sets $C \subseteq D$ , the pointwise $\langle {C} , {-} \rangle \le \langle {D} , {-} \rangle$ is obvious. Conversely, if $\phi, \psi : S^X \to S$ satisfy our conditions and are such that $\phi \le \psi$ pointwise, then $\psi^{-1}(0) \subseteq \phi^{-1}(0)$ , and the containment of the associated closed sets follows from (2).

We will often use this characterization to define a closed set in terms of how it pairs with opens, in which case we need to verify the relevant conditions (i)–(ii), or merely the preservation of all joins.

1. (a) The Hoare hyperspace over X, denoted HX, is the set of closed subsets of X, or, equivalently, the set of maps $S^X \to S$ of Proposition 2.1.

2. (b) We equip HX with the weakest topology which makes the pairing maps

   \begin{equation*} \langle {-} , {U} \rangle \: : \: HX \longrightarrow S \end{equation*}
   continuous for every $U \in S^X$ .

Thus, the subbasic open sets of HX are the subsets of the form $\langle {-} , {U} \rangle^{-1}(1)$ .

Defining HX as a set of maps $S^X \to S$ with this topology is a double dualization procedure and illustrates the sense in which H acquires the structure of a double dualization monad. Following functional-analytic terminology, one could call it the weak topology, as done by Schalk (Reference Schalk1993), Section 1.1.5, ^{Footnote 2} while in the picture of closed sets, the topology on HX is known as the lower Vietoris topology.

Frequently, it will be convenient to pair the closed sets $C \in HX$ not with all open sets, but merely with a specified subset of the open sets. In our framework, a basis of a topological space is a subset $\mathcal{B} \subseteq S^X$ such that every $U \in S^X$ is a join of a subset of $\mathcal{B}$ .

1. (a) For given $C,D \in HX$ , the following three statements are equivalent:

2. (i) For all $U \in \mathcal{B}$ , we have $\langle {C} , {U} \rangle \le \langle {D} , {U} \rangle$ ;

3. (ii) $C \le D$ in the specialization preorder on HX;

4. (iii) $C \subseteq D$ as subsets of X.

5. (b) The topology on HX is the weakest topology which makes all of the pairings

   \begin{equation*} \langle {-} , {U} \rangle \: : \: HX \longrightarrow S \end{equation*}
   continuous for $U \in \mathcal{B}$ .

Proof.

1. a) Since the pairing maps preserve all joins in the second argument, condition (i) holds if and only if it holds with $S^X$ in place of $\mathcal{B}$ . Then, the equivalence of (i) and (ii) is the final part of Proposition 2.1. Replacing $\mathcal{B}$ by $S^X$ , it is also clear that (i) is equivalent to (ii), since (i) then states exactly that every neighborhood of C is a neighborhood of D, which is the definition of the specialization preorder.

2. b) We need to show that, if these maps are continuous, then so is every $\langle {-} , {V} \rangle$ for $V \in S^X$ . But this is an instance of the fact that (arbitrary) suprema of continuous maps to S are continuous.

Remark 2.4 The sets of the form $\langle {-} , {U} \rangle^{-1}(1)$ , which generate the lower Vietoris topology, are sometimes denoted by Hit(U) (Manuel and Tholen 1997). The set of closed subsets of a topological space is more commonly equipped with the full Vietoris topology (Vietoris 1922). The latter has as subbasic open sets not only all the sets Hit(U) for open U, but also the sets

\begin{equation*}\text{Miss}(C) \,{:}\,{\raise-1.5pt{=}}\, \{D\in HX : D \cap C = \varnothing \}\end{equation*}

for closed $C\subseteq X$ . (One can say that “D misses C” if $D\in\text{Miss}(C)$ .) Intuitively, the lower Vietoris topology relates to the full Vietoris topology as the topology of lower semicontinuity on $\mathbb{R}$ , with generating open sets of the form $(a,\infty)$ , relates to the usual topology of $\mathbb{R}$ . We use the lower Vietoris topology instead of the full Vietoris topology for two reasons: firstly, because the lower Vietoris topology of HX parallels exactly the one on the space of continuous valuations VX as constructed in Definition 3.6; secondly, because, as a consequence of this correspondence, taking the support of a continuous valuation results in a continuous map $VX\to HX$ with respect to these topologies (see Section 5.1). Indeed, the full Vietoris topology would not make this map continuous, as Example 5.7 shows, but rather merely lower semicontinuous.

The following observation of Hoffmann (Reference Hoffmann1979), Example 2.3(a) makes the lower Vietoris topology more concrete, but needs to be read with care as it is one of the few statements which have no analogue for continuous valuations. We denote by $\downarrow\!\! \{ C \}$ the principal downset generated by C, the set of closed subsets of C.

Proof. Every $U \in S^X$ is of the form $U = X \setminus C$ for some closed set C, and the open set $\langle {-} , {U} \rangle^{-1}(1)$ associated with U is the complement of $\downarrow\!\! \{C\}$ .

Thus, the lower Vietoris topology is determined by the inclusion order of closed sets and coincides with the upper topology on the set of closed sets Gierz et al. (Reference Gierz2003), Definition O.5.4. It is worth noting that the point $X \in HX$ , being contained in every nonempty open set, is dense in HX. At the other extreme, the only open set which contains the point $\varnothing \in HX$ is the entire space HX, which implies that $\varnothing$ is in the closure of every nonempty set.

The argument is the same in spirit as Heckmann’s proof of sobriety of the space of valuations Heckmann (1995), Proposition 6.1.

Proof. It is a standard fact that limits of sober spaces (in $\mathsf{Top}$ ) are sober Goubault-Larrecq (Reference Goubault-Larrecq2013), Theorem 8.4.13. It is therefore enough to exhibit HX as a limit of sober spaces. A diagram which achieves this can be directly read off the characterization of Proposition 2.1, which characterizes HX as a subset of $S^{S^X}$ equipped with the product topology (which, by the same theorem, is sober). This subset is characterized by the given equations, where imposing each equation amounts to equalizing a pair of continuous maps to S. An inequality $r \le s$ can equivalently be considered as the equation $r \lor s = s$ and $\lor : S \times S \to S$ is continuous. It is therefore enough to prove that each side of each instance of those conditions depends continuously on $\phi \in S^{S^X}$ . This is obvious in all cases, where for $\bigvee_{\lambda \in \Lambda} \phi(U_\lambda)$ one needs to use the fact that $\bigvee : S^I \to S$ is continuous.

Note the difference between $\text{cl}(\{A\})$ , the closure of the singleton $\{A\}$ in HX, and $\text{cl}(A)$ , the closure of A in X.

2.2 Functoriality

The construction of H as a functor $H : \mathsf{Top} \to \mathsf{Top}$ is due to Smyth (Reference Smyth1983). We here treat the functoriality based on our double dualization approach.

Definition 2.7 (Pushforward) Let $f:X\to Y$ be continuous and $C \in HX$ . The pushforward $f_\sharp C \in HY$ is defined through the pairing with any $V \in S^Y$ as ^{Footnote 3}

(3) \begin{equation} \langle {{f_\sharp C}} , {V} \rangle \,{:}\,{\raise-1.5pt{=}}\, \langle {C} , {V \circ f} \rangle.\end{equation}

It is easy to see that $\langle {f_\sharp C} , {-} \rangle : S^Y \to S$ satisfies the conditions of Proposition 2.1. By the analogy with integration, this makes the change of variables formula a definition. In fact, this is quite generally how functoriality works for double dualization functors: the action on morphisms can be defined in terms of the pairing with respect to the first step of dualization, where one simply composes by the given morphism f.

In the picture of closed sets, $f_\sharp C$ is the closure of the image of C,

\begin{equation*} f_\sharp C = \text{cl}\!\big(f(C)\big),\end{equation*}

since an open set V is disjoint from $\text{cl}\!\big(f(C)\big)$ if and only if $f^{-1}(V)$ is disjoint from C.

The definition (3) makes it immediate that $f_\sharp : HX \to HY$ is continuous, a fact that, although simple, is not obvious in the picture of closed sets. A similar comment applies to the following functoriality statement.

Proof. Let $C\in HX$ . Then, we have

\begin{align*} \langle {(g \circ f)_\sharp (C)} , {U} \rangle \;&=\; \langle {C} , {U \circ (g \circ f)} \rangle \;=\; \langle {C} , {(U \circ g) \circ f} \rangle \\ &=\; \langle {f_\sharp C} , {U \circ g} \rangle \;=\; \langle {g_\sharp(f_\sharp C)} , {U} \rangle , \end{align*}

for any $U\in S^Z$ , which implies the claim.

The preservation of identities is trivial. We have therefore obtained a functor $H:\mathsf{Top}\to\mathsf{Top}$ which assigns to each topological space X its Hoare hyperspace HX and to each continuous map $f:X\to Y$ the continuous map $f_\sharp:HX\to HY$ . We call H the Hoare hyperspace functor. The functor H is also a 2-functor in the sense of Section 5.4.

Proof. By Lemma 5.20, we have $V \circ f \le V \circ g$ with respect to the pointwise order on $S^X$ for every $V \in S^Y$ . Therefore,

\begin{equation*}\langle {g_\sharp(f_\sharp C)} , {V} \rangle = \langle {C} , {V \circ f} \rangle \le \langle {C} , {V \circ g} \rangle = \langle {g_\sharp C} , {V} \rangle,\end{equation*}

which, by Definition 5.19 and Lemma 2.3, means $f_\sharp \le g_\sharp$ .

2.3 Monad structure

We equip the functor H with a monad structure, making it a topological analogue of the powerset monad. While doing so, our double dualization perspective will be handy.

2.3.1 Unit

Definition 2.10 Let X be a topological space. We define the map $\sigma : X \to HX$ by

(4) \begin{equation} \langle {\sigma(x)} , {U} \rangle \; \,{:}\,{\raise-1.5pt{=}}\, \; U(x) = \begin{cases} 1 & \textrm{if } x \in U, \\ 0 & \textrm{if } x \not\in U, \end{cases}\end{equation}

for every $U \in S^X$ .

It is obvious that the preservation of joins in the second argument holds, since a point x is contained in a union of open sets if and only if it is contained in one of them. $\sigma$ is continuous by definition, since every $x \mapsto \langle {\sigma(x)} , {U} \rangle$ is continuous as a map $X \to S$ .

As the definition suggests, $\sigma(x)$ is the possibilistic analogue of the Dirac measure at x. In the picture of closed sets, we have

\begin{equation*} \sigma(x) = \text{cl}(\{x\}),\end{equation*}

which is clear since an open set U does not contain x if and only if $\text{cl}(\{x\})$ is disjoint from U. We now turn to naturality.

Lemma 2.11 Let X and Y be topological spaces and let $f:X\to Y$ be continuous. Then the following diagram commutes.

Proof. For $x\in X$ and $U\in S^Y$ , we have

\begin{equation*} \langle {f_\sharp(\sigma(x))} , {U} \rangle \;=\; \langle {\sigma(x)} , {U \circ f} \rangle \;=\; (U \circ f)(x)) \;=\; U(f(x)) \;=\; \langle {\sigma(f(x))} , {U} \rangle .\end{equation*}

Hence, we have a natural transformation $\sigma:id\Rightarrow H$ between endofunctors of $\mathsf{Top}$ .

2.3.2 Topological properties of the unit map

We state conditions on X under which $\sigma:X\to HX$ is a homeomorphism onto its image (a subspace embedding) and consider when $\sigma(X)\subseteq HX \setminus \{\varnothing\}$ is closed. The Kolmogorov quotient of a space X is the quotient space $X/\sim$ where any two points of X that have the same open neighborhoods are identified (Section 5.4).

Proposition 2.12 Let X be a topological space and consider the map $\sigma:X\to HX$ . Then $\sigma(X)$ is the Kolmogorov quotient of X with respect to $\sigma : X \to \sigma(X)$ as the quotient map. In particular, $\sigma$ is a homeomorphism onto its image if and only if X is $T_0$ .

Proof. By definition (4), $\sigma$ is injective if and only if no two distinct points of X have the same open neighborhoods, hence if and only if X is $T_0$ . Thus, the statement holds at the set-theoretical level and we can identify the points of $\sigma(X)$ with the equivalence classes of points of X with respect to topological indistinguishability.

It remains to be shown that the subspace topology on $\sigma(X)$ induced from HX is the quotient topology with respect to $\sigma : X \to \sigma(X)$ as the quotient map. Since taking the initial topology commutes with the passage to subspaces, the topology on $\sigma(X)$ is the weakest topology which makes the maps

\begin{equation*}\sigma(X) \longrightarrow S, \qquad C \longmapsto \langle {C} , {U} \rangle\end{equation*}

for $U \in S^X$ continuous. Since such an open set contains precisely those equivalence classes whose elements are in U, the topology on $\sigma(X)$ is the topology of the Kolmogorov quotient.

It may also be interesting to know whether $\sigma(X) \subseteq HX$ is a closed subspace. Clearly, $\text{cl}(\sigma(X))$ always contains $\varnothing$ , and therefore, $\sigma(X)$ is not closed in HX (unless X is empty). But it is still meaningful to ask when $\text{cl}(\sigma(X)) = \sigma(X) \cup \{\varnothing\}$ , or equivalently when $\sigma(X)$ is closed in $HX \setminus \{\varnothing\}$ . ^{Footnote 4} We next state conditions for this to be the case and note that it is not true in general, even if X is $T_1$ or sober.

Proposition 2.13 Let X be a nonempty topological space and consider a closed set $C\subseteq X$ . Then the following two conditions are equivalent:

1. a) $C \in \text{cl}(\sigma(X))$ ;

2. b) For every finite collection of open sets $U_1,\dots,U_n \subseteq X$ that hit C, their intersection $U_1\cap\dots\cap U_n$ is nonempty (but possibly disjoint from C).

Note that (b) is trivially true for $C = \varnothing$ and when C is the closure of a singleton.

Proof. C is not in the closure of $\sigma(X)$ if and only if this is witnessed by some basic open set. By definition of the topology, this means that there are $U_1, \ldots, U_n \in S^X$ such that

\begin{equation*}\langle {C} , {U_i} \rangle = 1, \quad \forall i=1,\ldots,n,\end{equation*}

but such that for every $x \in X$ there is i with

\begin{equation*}\langle {\sigma(x)} , {U_i} \rangle = U_i(x) = 0.\end{equation*}

The former means exactly that every $U_i$ hits C, and the latter that $U_1 \cap \ldots \cap U_n = \varnothing$ .

Example 2.14 Let X be any nonempty space in which finite intersections of nonempty open sets are nonempty. Then, it follows that $\sigma(X)$ is dense in HX, since condition (b) is always satisfied. For example, X could be any infinite set equipped with the cofinite topology (which is even $T_1$ ). As another example, take any space X which has a dense point, or equivalently a greatest element in the specialization preorder. More concretely, every $X = HY$ for any Y is a sober space with dense point $Y \in HY$ , and therefore, $\sigma(HY)$ is dense in HHY.

Corollary 2.15 Let X be a Hausdorff space. Then $\sigma(X) \cup \{\varnothing\}$ is closed in HX.

Proof. If C contains at least two different points, then C hits disjoint open neighborhoods $U_1$ and $U_2$ which separate these points.

There are non-Hausdorff spaces X for which $\sigma(X) \cup \{\varnothing\}$ is closed. An example is given by the unit interval $X = [0,1]$ equipped with the upper open topology, whose open sets are the intervals of the form (a,1], then we even have $HX = \sigma(X) \cup \{\varnothing\}$ . More interestingly, in the $T_1$ case also the converse to Corollary 2.15 is true.

Proposition 2.16 Let X be a space which is $T_1$ , but not Hausdorff. Then $\sigma(X) \cup \{\varnothing\}$ is not closed in HX.

Proof. Let x,y be distinct points of X such that for any open neighborhoods $U \ni x$ and $V \ni y$ , we have $U \cap V \neq \varnothing$ . We can find such points since X is not Hausdorff. The subset $C\,{:}\,{\raise-1.5pt{=}}\, \{x,y\}$ is closed since X is $T_1$ . Since x and y are distinct and singletons are closed, C is not in the image of $\sigma$ . Let $U_1,\dots, U_n$ be a finite collection of open sets such that C hits all of them. Then, each of them contains x, or y, or both. Reorder the $U_i$ in such a way that $U_1,\dots, U_k$ contain at least x, and $U_{k+1},\dots, U_n$ contain at least y. Then,

\begin{equation*} \bigcap_{i=1}^n U_i = \left(\bigcap_{i=1}^k U_i\right) \cap \left(\bigcap_{j=k+1}^n U_j\right). \end{equation*}

We have that $\cap_{i=1}^k U_i$ is an open neighborhood of x, and $\cap_{j=k+1}^n U_j$ is an open neighborhood of y, and these two neighborhoods have nonempty intersection. By Proposition 2.13, then, C is in the closure of $\sigma(X)$ . Since C is not in $\sigma(X)$ , it follows that $\sigma(X) \cup \{\varnothing\}$ is not closed.

2.3.3 Multiplication

The definition of the monad multiplication is where our double dualization picture is particularly useful, since there is a very simple definition in terms of the pairing map $\langle {-} , {U} \rangle \in S^{HX}$ for $U \in S^X$ .

Definition 2.17 Let X be a topological space. We define the map $\mathcal{U}:HHX\to HX$ on any $\mathcal{C} \in$ HHX by

(5) \begin{equation}\langle {\mathcal{U}\mathcal{C}} , {U} \rangle \,{:}\,{\raise-1.5pt{=}}\, \langle {\mathcal{C}} , {\langle {-} , {U} \rangle} \rangle\end{equation}

for every $U \in S^X$ .

Since the pairing is join-preserving in the second argument, it is clear that (5) satisfies the requirements of Proposition 2.1 which guarantee that $\mathcal{U}(\mathcal{C}) \in HHX$ . In the picture of closed sets, $\mathcal{U}$ assigns to each closed set of closed sets the closure of their union,

\begin{equation*} \mathcal{U} \mathcal{C} \;=\; \text{cl} \left(\bigcup \, \mathcal{C} \right) \;=\; \text{cl} \left( \bigcup_{C\in\mathcal{C}} C \right) .\end{equation*}

This is because, by definition (5), $\mathcal{U}\mathcal{C}$ is disjoint from U if and only if $\mathcal{C}$ is disjoint from the set of all closed sets that hit U.

To show that $\mathcal{U}$ is continuous, we only need to verify that its composition with any $\langle {-} , {U} \rangle$ is continuous, which holds by definition. We turn to naturality.

Proposition 2.18 Let X and Y be topological spaces and let $f:X\to Y$ be continuous. Then the following diagram commutes.

In terms of closed sets, this amounts to the statement that taking images commutes with unions in the case of discrete spaces and to a similar statement involving closures in general.

Proof. Let $\mathcal{C}\in HHX$ and $V \in S^Y$ . We unfold the definition to see

\begin{align*} \langle {\mathcal{U}(f_{\sharp\sharp}\mathcal{C})} , {V} \rangle \;&=\; \langle {f_{\sharp\sharp} \mathcal{C}} , {\langle {-} , {V} \rangle} \rangle \;=\; \langle {{\mathcal{C}}} , {{\langle {-} , {V} \rangle \circ f_\sharp}} \rangle =\; \langle {{\mathcal{C}}} , {{\langle {{f_\sharp(-)}} , {{V}} \rangle}} \rangle \\ &=\; \langle {{\mathcal{C}}} , {{\langle {-} , {{V \circ f}} \rangle}} \rangle \;=\; \langle {{\mathcal{U}\mathcal{C}}} , {{V \circ f}} \rangle \;=\; \langle {{f_\sharp(\mathcal{U}\mathcal{C})}} , {V} \rangle.\end{align*}

2.3.4 Monad axioms

Proposition 2.19 Let X be a topological space. Then the following three diagrams commute.

In terms of closed sets, these diagrams are the topological analogs of basic facts of set theory. For sets, the first two unitality diagrams state that the union over a singleton set of sets is the set itself and that the union of singletons is the set whose elements are the respective singletons. The associativity diagram states that taking unions is an associative operation. In our double dualization frameworks, all three proofs are mere unfoldings of definitions.

Proof of Proposition 2.19. We start with the first diagram, left unitality. Let $C\in HX$ and $U \in S^X$ . Then

\begin{equation*} \langle {{\mathcal{U}\sigma(C)}} , {U} \rangle \;=\; \langle {{\sigma(C)}} , {{\langle {-} , {U} \rangle}} \rangle \;=\; \langle {-} , {U} \rangle(C) \;=\; \langle {C} , {U} \rangle . \end{equation*}

Right unitality works similarly,

\begin{align*} \langle {{\mathcal{U}\sigma_\sharp(C)}} , {U} \rangle \;&=\; \langle {{\sigma_\sharp C}} , {{\langle {-} , {U} \rangle}} \rangle \;=\; \langle {C} , {{\langle {-} , {U} \rangle \circ \sigma}} \rangle \;=\; \langle {C} , {{\langle {{\sigma(-)}} , {U} \rangle} } \rangle\\ &=\; \langle {C} , {{U(-)}} \rangle \;=\; \langle {C} , {U} \rangle. \end{align*}

It remains to consider the associativity diagram. For $\mathcal{K}\in HHHX$ and $U \in S^X$ we get

\begin{align*} \langle {{\mathcal{U}(\mathcal{U}_\sharp \mathcal{K})}} , {U} \rangle \;&=\; \langle {{\mathcal{U}_\sharp \mathcal{K}}} , {{\langle {-} , {U} \rangle}} \rangle \;=\; \langle {{\mathcal{K}}} , {{\langle {-} , {U} \rangle\circ \mathcal{U}} } \rangle\\ &=\; \langle {{\mathcal{K}}} , {{\langle {{\mathcal{U}(-)}} , {U} \rangle}} \rangle \;=\; \langle {{\mathcal{K}}} , {{\langle {-} , {{\langle {-} , {U} \rangle}} \rangle}} \rangle \\ &=\; \langle {{\mathcal{U}\mathcal{K}}} , {{\langle {-} , {U} \rangle}} \rangle \;=\; \langle {{\mathcal{U}(\mathcal{U}\mathcal{K})}} , {U} \rangle.\\[-38pt] \end{align*}

We have proven the following statement.

Theorem 2.20 The triple $(H,\sigma,\mathcal{U})$ is a monad on $\mathsf{Top}$ .

We call $(H,\sigma,\mathcal{U})$ , or just H, the Hoare hyperspace monad. By Lemma 2.9, H is a strict 2-monad for the 2-categorical structure of $\mathsf{Top}$ given in Section 5.4.

As far as we know, this monad was introduced by Schalk (Reference Schalk1993, Section 6.3.1), with the inessential difference that the empty set was excluded from the hyperspace.

2.4 Algebras of H

There is a characterization of the Eilenberg-Moore algebras of the monad H, which is, as far as we know, also due to Schalk (Reference Schalk1993, Section 6.3.1). We review the main results. An additional reference, which also discusses related constructions, is Hoffmann’s earlier article (Hoffmann Reference Hoffmann1979).

Before we begin, it is worth noting that the metric or Lawvere-metric counterpart (Akhvlediani et al. Reference Akhvlediani, Clementino and Tholen2010) of the monad H becomes a Kock-ZÖberlein monad (Kock Reference Kock1995; ZÖberlein Reference ZÖberlein1976) upon considering it as a 2-monad on a strict 2-category. This means that whenever a topological space admits an H-algebra structure, this structure is unique. This phenomenon is a property-like structure (Max and Steve 1997). Nevertheless, we will see that not every morphism of $\mathsf{Top}$ between H-algebras is a morphism of H-algebras.

It is well-known that the algebras of the powerset monad on $\mathsf{Set}$ are the complete join-semilattices. The H-algebras are their topological cousins. An H-algebra is, by definition, a pair (A,a) consisting of a topological space A and a continuous map $a: HA \to A$ such that the following two diagrams commute.

(6)

We refer to these diagrams as the unit diagram and the algebra diagram. A first observation is that every H-algebra A is a $T_0$ space, since $\sigma$ must be injective by the unit diagram, but $\sigma$ identifies topologically indistinguishable points (Proposition 2.12). ^{Footnote 5} In fact, since every HA is sober (Proposition 2.6) and retracts of sober spaces are sober, ^{Footnote 6} the unit diagram shows that every H-algebra is sober.

Since the structure of such a lattice is completely determined by its topology, we may consider these lattices as a particular class of topological spaces. Hoffmann (Reference Hoffmann1979) calls them essentially complete $T_0$ spaces and also $T_0$ topological complete sup-semilattices, while Schalk (Reference Schalk1993) calls them unital inflationary topological semilattices. These spaces admit several equivalent characterizations Hoffmann (Reference Hoffmann1979, Theorem 1.8). They are precisely the H-algebras.

This result has been claimed by Hoffmann (Reference Hoffmann1979, Theorem 2.6) in the form of a monadicity statement, but apparently without explicit proof. As far as we know, the proof is essentially due to Schalk (Reference Schalk1993, Section 6.3), culminating in Theorems 6.9 and 6.10 therein, which state this characterization in the full subcategory of sober spaces. This is not a restriction since only sober spaces can be H-algebras to begin with.

We now present a proof of Theorem 2.22 which is more direct than Schalk’s, starting with some auxiliary results. Since these statements are quite specific to this particular monad, it is more convenient to work in terms of closed sets rather than with the double dualization framework.

The proof only uses the fact that a is a continuous retraction of $\sigma : A \to HA$ .

Proof. Let $C\in HA$ . Since a is continuous, it is monotone for the specialization preorder. For every $x\in C,$ we have $\sigma(x) = \text{cl}(\{x\}) \subseteq C,$ and therefore, by the unit condition for algebras,

\begin{equation*}x \;=\; a(\sigma(x)) \;\le\; a(C) .\end{equation*}

Hence, a(C) is an upper bound for C. On the other hand, let u be any upper bound for C. Then, $C\subseteq \sigma(u)$ , which implies

\begin{equation*}a(C) \;\le\; a(\sigma(u)) = u.\end{equation*}

Therefore, a(C) is a least upper bound for C, as was to be shown.

Proof. A point $x \in X$ is an upper bound for S if and only if $S\subseteq \;\downarrow\!{x}=\text{cl}(\{x\})$ . Since the latter set is closed, $S\subseteq \;\downarrow\!{x}$ if and only if $\text{cl}(S) \subseteq \;\downarrow\!{x}$ . So the set of upper bounds of S is equal to the set of upper bounds of $\text{cl}(S)$ . If this set admits a lowest element, this is the least upper bound of both S and $\text{cl}(S)$ .

The following lemma is somewhat converse to Lemma 2.23.

Proof. We need the continuity of $\bigvee$ to ensure that it is a morphism in $\mathsf{Top}$ . We have to verify the commutativity of the diagrams (6). The unit diagram requires that, for each $x\in A$ , we have $\bigvee \text{cl}(\{x\}) = x$ , which holds by Lemma 2.24. The algebra diagram requires that, for each $\mathcal{C}\in HHA$ , we have $\bigvee \, \mathcal{U}(\mathcal{C}) = \bigvee \left(\bigvee_\sharp \mathcal{C}\right)$ . Using Lemma 2.24 and the lattice-theoretic fact that the join of an arbitrary union of sets is the join of the individual joins, we have that

\begin{align*} \bigvee \, \mathcal{U}(\mathcal{C}) &=\; \bigvee \left(\text{cl} \left( \bigcup \mathcal{C} \right)\right) \;=\; \bigvee \left( \bigcup \mathcal{C}\right) \;=\; \bigvee \left\{ \bigvee C \: : C \in \mathcal{C} \right\} \\ &=\; \bigvee \text{cl} \left( \left\{ \bigvee C \: : C \in \mathcal{C} \right\} \right) \;=\; \bigvee \left( {\bigvee}_\sharp \mathcal{C}\right).\\[-40pt] \end{align*}

Sketch of proof. If $S \subseteq X$ is directed in the specialization order, then $\text{cl}(S)$ is irreducible and therefore the closure of a unique point, which can be seen to be the supremum. This construction of directed joins makes the second statement obvious, since every open set is upward closed.

Proof. As in Remark 2.4, we also write $Hit(U) := \langle {-} , {U} \rangle^{-1}(1)$ for the subbasic open subset of HL associated with an open $U \subseteq L$ , consisting of all those closed sets that hit U.

If binary (and hence finitary) joins exist, then arbitrary joins exist by Lemma 2.26. The converse is trivial. We thus only need to show that the join map $\bigvee : HL \to L$ is continuous if and only if the binary join map $\vee : L \times L \to L$ is.

Suppose that $\bigvee : HL \to L$ is continuous. By Lemma 2.24, it suffices to show that the map $\phi : L \times L \to HL$ defined by $\phi(x,y) = \text{cl}(\{x,y\})$ is continuous. Consider any subbasic open set Hit(U) for open $U \subseteq L$ . It is then enough to prove that the preimage $\phi^{-1}(Hit(U))$ is open. In fact,

\begin{align*}\phi^{-1}(Hit(U)) \;&=\; \{(x,y)\,|\,\text{cl}(\{x,y\}) \cap U \ne \varnothing \} \;=\; \{(x,y)\,|\,\{x,y\} \cap U \ne \varnothing \} \\ &=\; \{(x,y)\,|\,x\in U \} \cup \{(x,y)\,|\,y\in U \} \;=\; (U\times L) \cup (L\times U) ,\end{align*}

which is open in $L\times L$ .

Suppose that the binary join map is continuous. We show that for every open set $U \subseteq L$ , every $C \in \bigvee^{-1}(U)$ has an open neighborhood contained in $\bigvee^{-1}(U)$ . Since U is Scott-open by Lemma 2.26 and $\bigvee C \in U$ , there exists a finite set $\{x_1 , ..., x_n\} \subseteq C$ such that $x_1 \vee \ldots \vee x_n \in U$ . Since the n-ary join map $L^n \to L$ is continuous by continuity of the binary one, there exist open neighborhoods $V_i \ni x_i$ such that for all $y_i \in V_i$ we have $y_1 \vee \ldots \vee y_n \in U$ as well. Consider the open set $W \,{:}\,{\raise-1.5pt{=}}\, \bigcap_{i=1}^n Hit (V_i)$ . We have $C \in W$ by construction, and it is easy to see that $W \subseteq \bigvee^{-1} (U)$ since U is an upper set.

We are now ready to prove the theorem.

Proof of Theorem 2.22. By Lemmas 2.23 and 2.24, we know that for an H-algebra A, every subset of A must have a supremum, that is A is a complete lattice in the specialization preorder. The Hoare hyperspace HA is sober (Proposition 2.6) and the map $\bigvee: HA \to A$ must be continuous. By the unit condition of (6), it follows that A is a retract of a sober space and therefore sober Gierz et al. (Reference Gierz2003, Exercise O.5.16). By Lemma 2.27, the binary join is continuous too. Therefore, A is a topological complete join-semilattice and the algebra map is the join of closed sets.

Conversely, suppose that A is a topological complete join-semilattice. By bothjoins, the join map of closed sets $\bigvee: HA \to A$ is continuous. Using Lemma 2.25, we conclude that $(A,\bigvee)$ is an H-algebra.

To complete the proof, suppose that A and B are H-algebras. A morphism m between them is, by definition, a continuous map such that the following diagram commutes.

Such maps m are precisely those that preserve arbitrary suprema by Lemam 2.24.

We conclude this subsection with a remark. In contradiction to a claim by Schalk (Reference Schalk1993, Sections 6.3 and 6.3.1), not every sober space whose specialization order is a complete lattice is a topological complete join-semilattice. In other words, for a sober space X whose specialization order is a complete lattice, the continuity of the join map of closed sets $\bigvee: HL \to L$ , or equivalently of the binary join map $\vee: L \times L\to L$ , is not guaranteed. A counterexample seems to be given by Hoffmann (Reference Hoffmann1979, Example 5.5 combined with Lemma 1.5); however, it is based on what appears to be an incorrect reference (reference Alvarez-Manilla Reference Alvarez-Manilla2002 therein). We give a concrete counter example, based on Hoffmann’s approach.

Consider the set $X^* \,{:}\,{\raise-1.5pt{=}}\, X \sqcup \{ \bot, \top \}$ with the topology whose nonempty open subsets $W \subseteq X^*$ are those in the form $W = V \cup \{ \top \}$ where V is an open subset of X. The specialization preorder of $X^*$ is a complete lattice where

\begin{equation*} x\vee y = \begin{cases} \bot & \text{if } x = y = \bot, \\ x & \text{if } x = y \in X, \\ \top & \text{otherwise},\end{cases}\end{equation*}

and similarly for arbitrary joins. To see that $X^*$ is sober, let $K \subseteq X^*$ be a nonempty irreducible closed set. If $\top \in K$ , then necessarily $K = X^* = \text{cl}(\{\top\})$ . Hence, we can assume $K = C \cup \{ \bot\}$ for some closed $C \subseteq X$ . Suppose that $C = C_1 \cup C_2$ for closed $C_1,C_2 \subseteq X$ . Then, $$K = ({C_1} \cup \{ \bot \} ) \cup {\rm{ }}({C_2} \cup \{ \bot \} )$$ . Since K is irreducible, we must have $C_1 = C$ or $C_2 = C$ . Therefore, C is also irreducible. Since X is sober, C is the closure of a unique point in X and so must be K in $X^*$ . We conclude that $X^*$ is sober.

Consider the open subset $\{\top\} \subseteq X^*$ . We will show that the preimage

$${ \vee ^{ - 1}}(\{ \top \} ) = \left\{ {(x,y) \in {X^*} \times {X^*}\>|\>x \vee y = \top } \right\}$$

is not open and therefore that $\vee: X^* \times X^* \to X^*$ is not continuous, by exhibiting $$(x,y) \in { \vee ^{ - 1}}(\{ \top \} )$$ which is not in the interior of this preimage. Since X is not $T_2$ , there are distinct points $x, y \in X$ such that any open neighborhoods $U \ni x$ and $V \ni y$ in X intersect. We then have $x \vee y = \top$ since $x \neq y$ . Consider any basic neighborhood $U' \times V'$ of (x,y) in $X^* \times X^*$ . These necessarily have the form $U' = U \cup \{\top\}$ and $V' = V \cup \{\top\}$ for certain open $U,V \subseteq X$ . Since $U \cap V \neq \varnothing$ , we have $z \in X$ such that $(z,z) \in U' \times V'$ , but clearly $(z,z) \not \in \vee^{-1}(\{\top\})$ . Hence (x,y) is not an interior point of $\vee^{-1}(\{\top\})$ .

If $X = \mathbb{N} \cup \{a,b\}$ as above, the failure of continuity of the binary join can be illustrated as follows. The sequence $(1,2,3,\ldots)$ converges to both a and b in $X^*$ , but $(1 \vee 1, 2 \vee 2, 3 \vee 3, \ldots)$ , which is the same sequence $(1,2,3,\ldots)$ , does not converge to $a \vee b = \top$ .

2.5 Products and projections

In this section, we show that H is a commutative monad (Section 5.4), or, equivalently, a symmetric monoidal monad, with respect to the Cartesian monoidal structure on $\mathsf{Top}$ . We start by constructing a strength transformation $X \times HY \to H(X \times Y)$ .

Given an open $W : X \times Y \to S$ and $x \in X$ , we can consider $W(x,-) : Y \to S$ , which is the restriction of W along the continuous map $y \mapsto (x,y)$ . In terms of open subsets, this amounts to starting with an open set W in $X \times Y$ and pulling it back along the inclusion of Y as a slice in $X \times Y$ to an open subset of Y.

Definition 2.29 Let X and Y be topological spaces. We define $s : X \times HY \to H(X \times Y)$ on $x \in X$ and $C \in HY$ such that

(7) \begin{equation} \langle {{s(x,C)}} , {W} \rangle \,{:}\,{\raise-1.5pt{=}}\, \langle {C} , {{W(x,-)}} \rangle \end{equation}

for all $W \in S^{X \times Y}$ .

Since the restriction map $W \mapsto W_x$ preserves joins, the duality of Proposition 2.1 applies, resulting in $s(x,C) \in H(X \times Y)$ . In terms of closed sets, we have

\begin{equation*} s(x,C) = \text{cl}(\{x\} \times C ) = \text{cl}(\{x\}) \times C,\end{equation*}

since $\{x\} \times C$ is disjoint from an open subset $W \subseteq X \times Y$ if and only if C is disjoint from the slice of W at x, and the second equation holds since products of closed sets are closed.

If $W = U \times V$ is a product of open sets, which means $W(x,y) = U(x) V(y)$ in terms of functions, then definition (7) simplifies to

\begin{equation*} \langle {{s(x,C)}} , {{U\times V}} \rangle \;=\; U(x) \, \langle {C} , {V} \rangle .\end{equation*}

The continuity of s follows. Lemma 2.3 and the fact that the products of open sets form a basis imply that it is enough to prove the continuity of the map

\begin{equation*} (x,C) \mapsto U(x) \, \langle {C} , {V} \rangle,\end{equation*}

which is indeed continuous since the multiplication $S \times S \to S$ is continuous and each factor of the product is.

Proposition 2.30 The map s is natural in both arguments: for all continuous functions $f:X\to Z$ and $g:Y\to W$ , the following two diagrams commute.

Proof. Since the products of open sets $U \times V$ form a basis for the product topology, by Lemma 2.3 it is enough in either case to show that both diagrams commute after pairing with a generic product of open sets $U \times V$ . Doing so for the first diagram gives, for $x\in X$ and $C\in HY$ ,

\begin{align*} \langle {{(f\times id)_\sharp s(x,C)}} , {{U\times V}} \rangle \;&=\; \langle {{s(x,C)}} , {{(U \times V) \circ (f \times id)}} \rangle \;=\; \langle {{s(x,C)}} , {{(U \circ f)\times V}} \rangle \\ &=\; U(f(x)) \, \langle {C} , {V} \rangle \;=\; \langle {{s(f(x),C)}} , {{U\times V}} \rangle . \end{align*}

Similarly, for the second diagram,

\begin{align*} \langle {{(id\times g)_\sharp s(x,C)}} , {{U\times V}} \rangle \;&=\; \langle {{s(x,C)}} , {{(U\times V) \circ (id \times g)}} \rangle \;=\; \langle {{s(x,C)}} , {{U \times (V \circ g)}} \rangle \\ &=\; U(x) \, \langle {C} , {V \circ g} \rangle \;=\; U(x) \, \langle {g_\sharp C} , {V} \rangle \;=\; \langle {{s(x,g_\sharp C)}} , {{U\times V}} \rangle.\\[-38pt] \end{align*}

In other words, for all topological spaces X and Y, the following four diagrams commute. The first two involve the unitor u and associator a of the Cartesian monoidal structure of $\mathsf{Top}$ , while the other ones involve the structure maps of the monad.

Proof. For the first diagram, let $C\in HX$ and $U \in S^X$ . Denoting by $\bullet$ the unique point of 1, we have

\begin{equation*} \langle {{u_\sharp (s(\bullet,C))}} , {U} \rangle \;=\; \langle {{s(\bullet,C)}} , {{U \circ u}} \rangle \;=\; \langle {C} , {U} \rangle \;=\; \langle {u(\bullet,C)} , {U} \rangle. \end{equation*}

For the second diagram, we apply Lemma 2.3 to reduce to the evaluation on $U \in S^X$ , $V \in S^Y$ , and $W \in S^Z$ . For each $x\in X$ , $y\in Y$ , and $C\in HZ$ , we have

\begin{align*} \langle {{a_\sharp s((x,y),C)}} , {{U\times (V\times W)}} \rangle \;&=\; \langle {{s((x,y),C)}} , {{(U\times (V\times W)) \circ a}} \rangle \\ &=\; \langle {{s((x,y),C)}} , {{(U\times V)\times W}} \rangle \\ &=\; (U \times V)(x,y) \,\langle {C} , {W} \rangle \\ &=\; U(x) \, V(y) \,\langle {C} , {W} \rangle \\ &=\; U(x) \, \langle {s(y,C)} , {V\times W} \rangle \\ &=\; \langle {s(x,s(y,C))} , {U\times (V\times W)} \rangle . \end{align*}

For the third diagram, we similarly evaluate

\begin{align*} \langle {{s(x,\sigma(y))}} , {{U\times V}} \rangle \;&=\; U(x) \, \langle {{\sigma(y)}} , {V} \rangle \;=\; U(x) \, V(y) \\ &=\; (U \times V)(x,y) \;=\; \langle {{\sigma((x,y))}} , {{U\times V}} \rangle . \end{align*}

And for the last diagram,

\begin{align*} \langle{\mathcal{U}(s_\sharp s(x,\mathcal{C}))}{U\times V}\rangle \;&=\; \langle{s_\sharp s(x,\mathcal{C})},{\langle {-} , {{U\times V}} \rangle}\rangle \;=\; \langle {{s(x,\mathcal{C})}} , {{\langle {{s(-)}} , {{U\times V}} \rangle}} \rangle \\ &=\; \langle {{s(x,\mathcal{C})}} , {{U\times \langle {-} , {V} \rangle}} \rangle \;=\; U(x) \, \langle {{\mathcal{C}}} , {{\langle {-} , {V} \rangle}} \rangle \\ &=\; U(x) \, \langle {{\mathcal{U}\mathcal{C}}} , {V} \rangle \;=\; \langle {{s(x,\mathcal{U}\mathcal{C})}} , {{U\times V}} \rangle.\\[-38pt] \end{align*}

Proposition 2.32 The strength is commutative, in the sense that the following diagram commutes,

where the costrength $t:HX\times Y\to H(X\times Y)$ is obtained from the strength via the braiding. Explicitly,

\begin{equation*} \langle {{t(C,y)}} , {W} \rangle \; = \; \langle {C} , {{W(-,y)}} \rangle\end{equation*}

for all $C\in HX$ , $y\in Y$ and $W \in S^{X \times Y}$ .

Proof. Let $C\in HX$ , $D\in HY$ , $U \in S^X$ , and $V \in S^Y$ . Then

\begin{align*} \langle{\mathcal{U}(t_\sharp s(C,D))}{U\times V}\rangle \;&=\; \langle{t_\sharp s(C,D)}{\langle{-}{U\times V}}\rangle \;=\; \langle {{s(C,D)}} , {{\langle {{t(-)}} , {{U\times V}} \rangle}} \rangle \\ &=\; \langle {{s(C,D)}} , {{\langle {-} , {U} \rangle \times V}} \rangle \;=\; \langle {C} , {U} \rangle \, \langle {D} , {V} \rangle. \end{align*}

An analogous computation shows that $\langle {{\mathcal{U}(s_\sharp t(C,D))}} , {{U\times V}} \rangle = \langle {C} , {U} \rangle \, \langle {D} , {V} \rangle$ as well. This is enough because products of open sets form a basis.

Per Section 5.4, we have the following.

The lax monoidal structure is implemented by the multiplication map $$HX \times HY \to H(X \times Y)$$ given by the product of closed sets $(C,D) \mapsto C\times D$ , as the computation in the proof of Proposition 2.32 shows. Due to the universal property of the product in $\mathsf{Top}$ , H is an oplax monoidal monad as well, even bilax (see Section 5.4), which underlines the analogy between H and probability monads (Fritz and Perrone Reference Fritz and Perrone2018). The comultiplication $H(X\times Y)\to HX\times HY$ projects a closed set in the product space $X\times Y$ to the pair of its projections to X and Y.

3.1 Duality theory for continuous valuations

One can define an integration theory for continuous valuations that is analogous to Lebesgue integration for measures, but the role of measurable functions is played by lower semicontinuous functions (Jung Reference Jung2004; Kirch Reference Kirch1993). This results in a duality between continuous valuations and lower semicontinuous functions, just as closed subsets are dual to open subsets in Proposition 2.1.

We start by recalling the definition of integral of a lower semicontinuous function against a continuous valuation, also known as lower integral. As far as we know, it was first defined in Kirch’s thesis (Kirch Reference Kirch1993), written in German. A reference in English is the slightly later work of Heckmann (1995).

Throughout, we equip the extended nonnegative reals $[0,\infty]$ with the upper topology, whose open sets are the sets of the form $(r,\infty]$ , in addition to $\varnothing$ and the whole space. This topology is also known as the topology of lower semicontinuity, since continuous functions $f : X \to [0,\infty]$ are characterized by the openness of the sets $f^{-1}((r,\infty])$ , a property more commonly known as lower semicontinuity. As we will see, $[0,\infty]$ plays a role for continuous valuations that is perfectly analogous to the one of the SierpiŃski space throughout Section 2.

For every space X, we denote the set of (lower semi-)continuous functions $X \to [0,\infty]$ by $[0,\infty]^X$ . It is an important fact that all joins in $[0,\infty]^X$ exist and are pointwise. As for $S^X$ in Section 2, we equip $[0,\infty]^X$ with the pointwise order and pointwise algebraic structure given by addition and multiplication, using the convention that $\infty \cdot 0 = 0 \cdot \infty = 0$ . There is also an action of every $r \in [0,\infty]$ by “scalar multiplication,” corresponding to multiplication by the constant function $X \to \{r\}$ .

As in Lebesgue integration theory, a lower semicontinuous function $X \to [0,\infty]$ is called simple if it assumes only finitely many values. Every simple lower semicontinuous function $f:X\to[0,\infty]$ can be written as a positive linear combination of indicator functions of open sets, that is in the form

\begin{equation*} f \;=\;\sum_{i=1}^n r_i \unicode{x1D7D9}_{U_i} \end{equation*}

for $r_i\in(0,\infty]$ and $U_i\subseteq X$ open for all i, as can be seen by induction on the number of values that f takes. It is well-known that every lower semicontinuous function $X\to [0,\infty]$ can be expressed as a directed supremum of simple functions (see, e.g., Kirch Reference Kirch1993, Lemma 1.2). The integral is defined such that it is continuous with respect to directed suprema.

Definition 3.3 Let X be a topological space and $\nu : \mathcal{O}(X) \to [0,\infty]$ a continuous valuation. For a simple function

\begin{equation*}f\;=\; \sum_{i=1}^n r_i \unicode{x1D7D9}_{U_i},\end{equation*}

the integral of f with respect to $\nu$ is given by

\begin{equation*}\int f \, d\nu \;:=\; \sum_{i=1}^n r_i\, \nu(U_i).\end{equation*}

For any $g\in [0,\infty]^X$ , the integral of g with respect to $\nu$ is defined as

\begin{equation*}\int g \, d\nu \;:=\; \sup \left\{ \int f \, d\nu \;\bigg|\; f \, \text{is simple and} \, f\le g \right\} .\end{equation*}

It is not immediately obvious that the integral of a simple function is well-defined, since a priori it may depend on the particular representation, but it does not Kirch (Reference Kirch1993, Proposition 1.2 and Lemma 4.1).

Notation 3.4 To emphasize the analogy between the integral and the pairing (1) in the Hoare hyperspace case, we also write

\begin{equation*} \langle{\nu}{g}\rangle \,{:}\,{\raise-1.5pt{=}}\, \int g \, d\nu. \end{equation*}

This pairing notation is also motivated by the following known duality result, which can be seen as a counterpart of Proposition 2.1. Here and in the following, “linear” and “linearity” refers to linear combinations with coefficients in $[0,\infty]$ .

1. a) continuous valuations on X;

2. b) Maps $\phi : [0,\infty]^X \to [0,\infty]$ such that the following two conditions hold:

3. (i) Linearity: For every $f,g \in [0,\infty]^X$ , we have

   \begin{equation*} \phi(f + g) = \phi(f) + \phi(g), \end{equation*}
   and for every $r \in [0,\infty]$ , we have $\phi(rf) = r \phi(f)$ .

4. (ii) Scott continuity: For any directed net $(f_\lambda)_{\lambda \in \Lambda}$ in $S^X$ , we have

   \begin{equation*} \phi \left( \sup_{\lambda \in \Lambda} f_\lambda \right) = \sup_{\lambda \in \Lambda} \phi(f_\lambda). \end{equation*}

Under this bijection, the pointwise order on continuous valuations corresponds exactly to the pointwise order on maps $[0,\infty]^X \to [0,\infty]$ .

Similarly as for Proposition 2.1, “linearity” is intended between semimodules over the semiring $[0,\infty]$ (without negatives).

Note that the Scott continuity of integration can be thought of as an incarnation of the monotone convergence theorem for continuous valuations.

The proof of Theorem 3.5 is fairly straightforward: restricting a given $\phi$ to indicator functions of open sets produces a continuous valuation, and one only needs to argue that this construction is inverse to the formation of the integral. While this is obvious in one direction, the other follows since (i) and (ii) guarantee that $\phi$ is uniquely determined by its values on indicator functions of open sets. This line of argument is used, for example, by Kirch (Reference Kirch1993, Satz 8.1) for the case of core-compact X, where it is worth noting that Kirch’s proof actually works for any topological space X. ^{Footnote 8} A very similar duality result appears in Heckmann’s work Heckmann (1995, Theorem 9.1).

Intuitively, a continuous valuation is a quantitative analogue of a closed set: a closed set may or may not hit a given open set, while a valuation assigns a numerical value to every open set. Closed subsets of X are equivalent to particular maps $S^X \to X$ , and the lower Vietoris topology corresponds to the topology of pointwise convergence of such functionals. Similarly, continuous valuations on X are particular maps $[0,\infty]^X \to [0,\infty]$ by Theorem 3.5. As for H, this double dualization picture will turn out to be useful for the construction of the monad and especially for the treatment of its multiplication.

1. a) The space of continuous valuations over X, denoted VX, is the set of continuous valuations on X, or, equivalently, the set of maps $[0,\infty]^X \to [0,\infty]$ of Theorem 3.5.

2. b) We equip VX with the weakest topology that makes the pairing maps

   \begin{equation*} \langle {-} , {f} \rangle \: : \: HX \longrightarrow [0,\infty] \end{equation*}
   continuous for every $f \in [0,\infty]^X$ .

This weak topology on VX has a well-known analogue for Borel measures (see Section 4.2).

We now develop an analogue of Lemma 2.3, and consider the specialization preorder on VX.

1. (a) For given $\nu,\rho \in VX$ , the following are equivalent:

2. (i) For all $f \in \mathcal{F}$ , we have $\langle {\nu} , {f} \rangle \le \langle {\rho} , {f} \rangle$ .

3. (ii) $\nu \le \rho$ in the specialization preorder on VX.

4. (b) The topology on VX is the weakest topology which makes all of the pairings

   \begin{equation*} \langle {-} , {f} \rangle \: : \: VX \longrightarrow S \end{equation*}
   continuous for $f \in \mathcal{F}$ .

The specialization order is also the canonical order on the space of valuations in the probabilistic powerdomain (Jones and Plotkin Reference Jones and Plotkin1989).

Proof. The proof is analogous to the proof of Lemma 2.3.

1. a) The pairing map is linear and Scott-continuous in the second argument, and therefore, condition (i) holds if and only if it holds with $[0,\infty]^X$ in place of $\mathcal{F}$ . Having replaced $\mathcal{F}$ by $[0,\infty]^X$ , it is also clear that (i) is equivalent to (ii), since (i) states exactly that every neighborhood of $\nu$ is a neighborhood of $\rho$ , which is the definition of the specialization preorder.

2. b) We need to show that, if these maps are continuous, then so is every $\langle {-} , {f} \rangle$ for $f \in [0,\infty]^X$ . This follows since linear combinations of continuous maps to $[0,\infty]$ are continuous, as are directed suprema.

An interesting choice for $\mathcal{F}$ is given by the collection of indicator functions of open sets, for then the pairing maps $\langle {-} , {\unicode{x1D7D9}_U} \rangle$ coincide with the evaluation maps $\nu \mapsto \nu(U)$ , and the fact that every function in $[0,\infty]^X$ is a directed supremum of simple functions implies that these evaluation maps generate the topology as well. In fact, we have the following stronger result of Heckmann, which is one of the few statements whose proof is not analogous to an argument from hyperspace.

Proof. Consider the weakest topology on VX that makes the $\nu \mapsto \nu(U)$ continuous for all $U \in \mathcal{B}$ . It is then enough to show that for every finite sequence $U_1, \ldots, U_n \in \mathcal{B}$ , also the evaluation map $\nu \mapsto \nu(U_1 \cup \ldots \cup U_n)$ is continuous, since the continuity of evaluation on any open set then follows by Scott continuity. By induction, we can extend the modularity equation (8) to the inclusion-exclusion formula

\begin{equation*} \nu(U_1 \cup \ldots \cup U_n) = \sum_{k=1}^n (-1)^{k+1} \sum_{i_1 \le \ldots \le i_k} \nu(U_{i_1} \cap \ldots \cap U_{i_k}). \end{equation*}

This implies the claim since every term on the right-hand side is a continuous function of $\nu$ by the assumption that $\mathcal{B}$ is closed under finite intersections.

Proof. Completely parallel to the proof of Proposition 2.6.

We end this subsection with a small excursion to ordered topological spaces and the stochastic order, which plays an important role in applied probability and economic theory (Shaked and George 2007). This is of independent interest and will play no further role in this paper. It illustrates the utility of working with non-Hausdorff spaces: as we will see, the stochastic order for continuous valuations on a preordered space (say a Hausdorff space) coincides with the above specialization order on a suitable non-Hausdorff space.

A preordered topological space is a topological space X which is also a preordered set $(X,\le)$ such that the set of all ordered pairs $\{(x,y) \mid x \le y\}$ is closed in the product topology of $X \times X$ .

The stochastic order coincides with the specialization preorder of continuous valuations if one replaces the topology by only those open sets that are upward closed with respect to the preorder. This may be a useful thing to do since the pointwise order is conceptually simpler than the stochastic order. Since the topology consisting of the upward closed open sets is typically non-Hausdorff, we find that continuous valuations on non-Hausdorff spaces occur naturally in the context of the stochastic order.

3.2 Functoriality

We now show that V is a functor $\mathsf{Top} \to \mathsf{Top}$ , based on the double dualization approach, as we had done for the Hoare hyperspace monad H in Section 2.2.

Definition 3.12 (Pushforward) Let $f:X\to Y$ be continuous and $\nu\in VX$ . The pushforward $f_*\nu \in VY$ is defined through the pairing with any $g \in [0,\infty]^Y$ by

\begin{equation*}\langle {f_*\nu} , {g} \rangle \; = \; \langle {\nu} , {g\circ f} \rangle . \end{equation*}

It is easy to see that $\langle {{f_*(\nu)}} , {-} \rangle$ satisfies the conditions of Theorem 3.5: for any $g,h\in [0,\infty]^Y$ and $r \in [0,\infty]$ we have, since linear combinations of functions are pointwise,

\begin{align*}&\langle {\nu} , {(g + h)\circ f} \rangle = \langle {\nu} , {(g\circ f) + (h\circ f)} \rangle = \langle {\nu} , {g\circ f} \rangle + \langle {\nu} , {h\circ f} \rangle , \\&\text{and } \langle {{\nu}} , {{(rg) \circ f}} \rangle = \langle {{\nu}} , {{r (g \circ f)}} \rangle = r \langle {{\nu}} , {{g \circ f}} \rangle .\end{align*}

For Scott continuity, consider a directed net $(g_\lambda)_{\lambda\in \Lambda}$ in $[0,\infty]^X$ . Then

\begin{align*} \langle {\nu} , {{\Big(\sup_{\lambda\in \Lambda} g_\lambda\Big)\circ f}} \rangle = \langle {\nu} , {{\sup_{\lambda\in \Lambda} \, (g_\lambda\circ f)}} \rangle = \sup_{\lambda\in \Lambda} \, \langle {\nu} , {{ g_\lambda\circ f}} \rangle ,\end{align*}

where the second step uses that directed suprema in $[0,\infty]^X$ are pointwise.

We have thus turned the change of variables formula into the definition of the pushforward valuation, which also makes it immediate that $f_* : VX \to VY$ is continuous. Plugging in the indicator function of an open set $U \subseteq Y$ for g shows that we could equivalently have defined

\begin{equation*} (f_*\nu)(U)\; = \;\nu(f^{-1}(U)). \end{equation*}

The proof is exactly as in Lemma 2.8. Since $(id_X)* = id_{VX}$ holds trivially, we have a functor $V:\mathsf{Top}\to\mathsf{Top}$ which assigns to a space X its space of continuous valuations VX and to each continuous map $f:X\to Y$ the pushforward map $f_*:VX\to VY$ . In Section 4, it will turn out to be relevant that V preserves subspace embeddings.

Proof. By Proposition 3.8, it is enough to show that the induced map $\mathcal{O}(Y) \to \mathcal{O}(X)$ is surjective, which holds by assumption.

The analogue of Lemma 2.9 also holds, with essentially the same proof.

Proof. By Lemma 5.20, $f \le g$ implies that $h \circ f \le h \circ g$ in $[0,\infty]^X$ for every $h \in [0,\infty]^Y$ . Therefore

\begin{equation*} \langle {{f_*\nu}} , {h} \rangle = \langle {\nu} , {{h \circ f}} \rangle \le \langle {\nu} , {{h \circ g}} \rangle = \langle {{g_*\nu}} , {h} \rangle, \end{equation*}

which, by Definition 5.19 and Lemma 3.7, means $f_* \le g_*$ .

3.3 Monad structure

As we did for H in hyperspace, we equip V with a monad structure using double dualization.

3.3.1 Unit

Definition 3.16 Let X be a topological space. The map $\delta:X\to VX$ is defined in terms of the pairing by

\begin{equation*} \langle {{\delta(x)}} , {f} \rangle \,{:}\,{\raise-1.5pt{=}}\, f(x) \end{equation*}

for every $f \in [0,\infty]^X$ .

It is obvious that linearity and Scott continuity in the second argument indeed hold, again since (directed) suprema in $[0,\infty]^X$ are pointwise. $\delta$ is continuous by definition, since every $x \mapsto \langle {{\delta(x)}} , {f} \rangle$ is continuous as a map $X \to [0,\infty]$ . We turn to showing that $\delta$ is a natural transformation $\delta : id \Rightarrow V$ on $\mathsf{Top}$ .

Lemma 3.17 Let $f: X \to Y$ be continuous. Then the following diagram commutes.

Proof. For $x\in X$ and $g \in [0,\infty]^Y$ ,

\begin{equation*} \langle {{f_*(\delta(x))}} , {g} \rangle \;=\; \langle {{\delta(x)}} , {{g\circ f}} \rangle \;=\; g(f(x)) \;=\; \langle {{\delta(f(x))}} , {g} \rangle .\end{equation*}

Using arguments analogous to the proof of Proposition 2.12, it can be shown that the image $\delta(X) \subseteq VX$ , when equipped with the subspace topology, is homeomorphic to the Kolmogorov quotient of X with respect to $\delta$ as the quotient map. In particular, $\delta$ is injective if and only if X is $T_0$ .

3.3.2 Multiplication

For every $g \in [0,\infty]^X$ , pairing with g can be considered a function $\langle {-} , {g} \rangle \in [0,\infty]^{VX}$ , which, as in Definition 2.17, is central in the definition of the multiplication in terms of double dualization.

Definition 3.18 Let X be a topological space. The map $\mathcal{E}:VVX\to VX$ is defined on any $\xi\in VVX$ by

\begin{equation*}\langle {\mathcal{E}\xi} , {g} \rangle \,{:}\,{\raise-1.5pt{=}}\, \langle {{\xi}} , {{\langle {-} , {g} \rangle}} \rangle \end{equation*}

for every $g \in [0,\infty]^X$ .

Two applications of the linearity and Scott continuity of the pairing in the second argument show that this indeed results in an element $\mathcal{E}\xi$ of VX by Theorem 3.5. It is also obvious that $\mathcal{E} : VVX \to VX$ is continuous. We now turn to naturality, which is proven in the same way as Proposition 2.18.

Proposition 3.19 Let $f: X \to Y$ be continuous. Then the following diagram commutes.

Proof. Let $\xi\in VVX$ and $g \in [0,\infty]^X$ . Then

\begin{align*} \langle {{\mathcal{E}(f_{**}\xi)}} , {g} \rangle &= \langle {{f_{**}\xi}} , {{\langle {-} , {g} \rangle}} \rangle \;=\; \langle {{\xi}} , {{\langle {-} , {g} \rangle \circ f_*}} \rangle \;=\; \langle {{\xi}} , {{\langle {{f_*(-)}} , {g} \rangle}} \rangle \\ &=\; \langle {{\xi}} , {{\langle {-} , {{g\circ f}} \rangle}} \rangle \;=\; \langle {{\mathcal{E}\xi}} , {{g\circ f}} \rangle \;=\; \langle {{f_*(\mathcal{E}\xi)}} , {g} \rangle.\\[-38pt] \end{align*}

3.3.3 Monad axioms

Proposition 3.20 Let X be a topological space. Then the following three diagrams commute.

Proof. The proof is the same as for Proposition 2.19, after the relevant exchange of symbols. We start with the first diagram, left unitality. For every $\nu\in VX$ and $g \in [0,\infty]^X$ ,

\begin{equation*}\langle {{\mathcal{E}\delta(\nu)}} , {g} \rangle \;=\; \langle {{\delta(\nu)}} , {{\langle {-} , {g} \rangle} } \rangle \;=\; \langle {-} , {g} \rangle(\nu) \;=\; \langle {{\nu}} , {g} \rangle .\end{equation*}

Right unitality works similarly,

\begin{equation*}\langle {{\mathcal{E}\delta_*(\nu)}} , {g} \rangle \;=\; \langle {{\delta_*\nu}} , {{\langle {-} , {g} \rangle}} \rangle \;=\; \langle {{\nu}} , {{\langle {-} , {g} \rangle \circ \delta}} \rangle \;=\; \langle {{\nu}} , {{\langle {{\delta(-)}} , {g} \rangle}} \rangle \;=\; \langle {{\nu}} , {{g(-)}} \rangle \;=\; \langle {\nu} , {g} \rangle .\end{equation*}

It remains to consider the associativity diagram. For $\xi\in VVVX$ and $g \in [0,\infty]^X,$ we get

\begin{align*}\langle {{\mathcal{E}(\mathcal{E}_*\xi)}} , {g} \rangle \;&=\; \langle {{\mathcal{E}_*\xi}} , {{\langle {-} , {g} \rangle}} \rangle \;=\; \langle {{\xi}} , {{\langle {-} , {g} \rangle \circ \mathcal{E}}} \rangle \;=\; \langle {{\xi}} , {{\langle {{\mathcal{E}(-)}} , {g} \rangle}} \rangle \\&=\; \langle {{\xi}} , {{\langle {-} , {{\langle {-} , {g} \rangle}} \rangle}} \rangle \;=\; \langle {{\mathcal{E}\xi}} , {{\langle {-} , {g} \rangle}} \rangle \;=\; \langle {{\mathcal{E}(\mathcal{E}\xi)}} , {g} \rangle.\\[-38pt]\end{align*}

We have proven the following statement.

Theorem 3.21 The triple $(V,\delta,\mathcal{E})$ is a monad on $\mathsf{Top}$ .

We call $(V,\delta,\mathcal{E})$ , or just V, the monad of continuous valuations. By Lemma 3.15, V is a strict 2-monad for the 2-categorical structure of $\mathsf{Top}$ given in Section 5.4.

3.4 Algebras of V

Algebras of probability-like monads such as V have some convex structure, in the sense that the algebra map can be interpreted as the formation of the barycenter of a measure or valuation. For example, the algebras of the Radon monad on the category of compact Hausdorff spaces are the compact convex subsets of locally convex topological vector spaces, with the algebra map given by barycenter formation (Keimel Reference Keimel2008; Swirszcz Reference Swirszcz1974). The algebras of the Kantorovich monad on the category of complete metric spaces are the closed convex subsets of Banach spaces, again with respect to barycenter formation (Fritz and Perrone Reference Fritz and Perrone2019). If we drop the normalization of probability, as we do for V, then the algebra map $VA \to A$ can similarly be interpreted as assigning to every continuous valuation the integral of the identity function $A \to A$ ; in particular, such an algebra is a space where positive linear combinations of points can be evaluated to points. In contrast to the Radon and Kantorovich monads, a complete characterization of the category of algebras of V seems to be quite difficult, but partial characterizations are possible. In work concurrent with ours, Goubault-Larrecq and Jia (Reference Goubault-Larrecq and Jia2019) have proven that the algebras of V on the subcategory of $T_0$ spaces are topological cones in the sense of Keimel (Reference Keimel2008) (as far as we know, the definition first appeared in Heckmann 1995, Section 7.1). Topological cones generalize convex cones in topological vector spaces without the requirement that addition is cancellative. We review Keimel’s definition and state some of the results of Goubault-Larrecq and Jia (Reference Goubault-Larrecq and Jia2019). For further details, we refer to the original papers (Goubault-Larrecq and Jia Reference Goubault-Larrecq and Jia2019; Keimel Reference Keimel2008).

1. a) An operation of addition $+:K\times K\to K$ , jointly continuous, with a neutral element $0\in K$ .

2. b) An operation of scalar multiplication $\cdot: [0,\infty) \times K\to K$ , jointly continuous (where $[0,\infty)$ carries the upper topology).

These operations satisfy the axioms of a $[0,\infty)$ -semimodule with respect to the usual semiring structure of $[0,\infty)$ . A morphism of topological cones is a continuous map which preserves the $[0,\infty)$ -semimodule structure.

Proposition 3.23 (Proposition 2 in Cohen et al. 2006) Every V-algebra $e:VA\to A$ admits a canonical topological cone structure given by

\begin{align*} a + b \,{:}\,{\raise-1.5pt{=}}\, e(\delta_a + \delta_b) \text{and} r\,a \,{:}\,{\raise-1.5pt{=}}\, e(r\,\delta_a) , \end{align*}

where the sum and scalar multiplication on the right-hand sides are the pointwise sum and $r \in [0,\infty)$ . Moreover, every morphism of V-algebras is a $[0,\infty)$ -linear, continuous map.

Note that this statement is analogous to the “if” part of Lemma 2.27. Cohen et al. (Reference Cohen, Escardo and Keimel2006) proved this in the category of $T_0$ spaces. Since VX is sober, and retracts of sober spaces are sober, every V-algebra is sober, which is the same reasoning as for H-algebras in Halg. In particular, every V-algebra is $T_0$ , and therefore, the proposition is true on the whole of $\mathsf{Top}$ . Moreover, Goubault-Larrecq and Jia (Reference Goubault-Larrecq and Jia2019, Proposition 4.9) have shown that this cone is weakly locally convex, meaning that for every point $x\in A$ and every open neighborhood $U\ni x$ there exists a convex neighborhood C with $x\in C \subseteq U$ Goubault-Larrecq and Jia (Reference Goubault-Larrecq and Jia2019, Definition 3.9).

Let A be a topological cone. We say that A is cancellative if it is cancellative as a monoid, which means that for all $a,b,c\in A$ ,

\begin{equation*}a + c \;=\; b + c \quad\Longrightarrow\quad a \;=\; b .\end{equation*}

Prominent examples of non-cancellative topological cones are topological complete join-semilattices (Definition 2.21).

3.5 Products and marginals

In applications of measure theory, especially in probability theory, it is crucial to form products of measures and to project measures on product spaces to their marginal measures on the factor spaces. Even the fundamental probabilistic concept of stochastic independence can be understood in these terms: A product measure exhibits independence if it is equal to the product of its marginal measures. This subsection is concerned with the structure of products and marginals for V. We will show that V is a commutative monad (see Section 5.4), in the same way as we have done for H (Section 2.5).

Instead of constructing products of valuations directly, it is easier to equip the monad with the equivalent structure of a commutative strength (see Section 5.4). This simpler approach is known to measure theorists – if not under this name. For example, a map corresponding to the strength has been used by Ressel in his study of products of $\tau$ -smooth Borel measures (Ressel 1977) (See our Corollary 4.18). Conceptually, the use of the strength is as old as the concept of product measures; it is, for example, implicit in Halmos’ treatment of product measures Halmos (Reference Halmos1950), Paragraph 35. The content of this section is mostly a unified treatment of results due to Heckmann (1995).

To construct the strength transformation $s : X \times VY \to V(X \times Y)$ , note that for every $g \in [0,\infty]^{X \times Y}$ and $x \in X$ , we have $g(x,-) \in [0,\infty]^Y$ . We have the following analogue of Definition 2.29.

Definition 3.25 (Proposition 11.1 in Heckmann 1995) Let X and Y be topological spaces. We define $s : X \times VY \to V(X \times Y)$ on $x\in X$ and $\nu\in VY$ such that

\begin{equation*} \langle {{s(x,\nu)}} , {f} \rangle \,{:}\,{\raise-1.5pt{=}}\, \langle {{\nu}} , {{f(x,-)}} \rangle \end{equation*}

for all $f \in [0,\infty]^{X \times Y}$ .

The relevant properties for $s(x,\nu) \in V(X \times Y)$ are straightforward to check. If $f = gh$ is a product of $g \in [0,\infty]^X$ and $h \in [0,\infty]^Y$ , then the definition simplifies to

(9) \begin{equation} \langle {{s(x,\nu)}} , {{gh}} \rangle \;=\; \langle {{\nu}} , {{g(x) h(-)}} \rangle \;=\; g(x) \, \langle {\nu} , {h} \rangle.\end{equation}

The continuity of s follows. By Proposition 3.8, it would even be enough to consider indicator functions of products of open sets, while even on a general g and h the right-hand side of (9) is continuous in x and $\nu$ since the multiplication $[0,\infty] \times [0,\infty] \to [0,\infty]$ is. More generally, we will use the fact that a continuous valuation on a product space is uniquely determined by its pairings with product functions (for any number of factors in the product).

Proposition 3.26 The map s is natural in both arguments: for all continuous functions $f:X\to Z$ and $g:Y\to W$ , the following two diagrams commute.

The proof is straightforward by using (9) and perfectly analogous to the one of Proposition 2.30; hence, we omit it.

In other words, for all topological spaces X and Y, the following four diagrams commute, where u is the unitor and a the associator of the (Cartesian) monoidal structure of $\mathsf{Top}$ .

Since the statement and proof are perfectly analogous to the one of Proposition 2.31, we omit the detailed verification here.

The following proposition defines the product valuations via the diagonal composite map of the diagram. Following Kock (Reference Kock2012, Section 5), this result can be thought of as a version of Fubini’s theorem for continuous valuations (which is how Kock interprets the commutativity of a monad in general).

Proposition 3.28 The strength of V is commutative in the sense that the following diagram commutes,

where the costrength $t:VX\times Y\to V(X\times Y)$ is obtained from the strength via the braiding. Explicitly,

\begin{equation*}\langle {{t(\nu,y)}} , {f} \rangle \;=\; \langle {{\nu}} , {{f(-,y)}} \rangle\end{equation*}

for all $\nu\in VX$ , $y\in Y$ , $f\in[0,\infty]^{X \times Y}$ .

The proof of Proposition 2.32 only needs to be copied and the corresponding symbols replaced.

In particular, we have a natural map $VX \times VY \to V(X \times Y)$ implementing the formation of products of continuous valuations Heckmann (Reference Heckmann1995, Theorem 11.2). The product of $\nu$ and $\rho$ pairs with functions of the form $g\cdot h$ as

\begin{equation*} \langle {{\nu \otimes \rho}} , {{g h}} \rangle \;=\; \langle {{\nu}} , {g} \rangle \, \langle {{\rho}} , {h} \rangle .\end{equation*}

On open sets, the product valuation assigns

\begin{equation*} (\nu \otimes \rho)(U\times V) \;\longmapsto\; \nu(U)\cdot\rho(V) .\end{equation*}

Both formulations correspond to the ordinary product formulas widely used in measure and probability theory.

As for H, the universal property of the product makes $(V,\delta,\mathcal{E})$ also an oplax and therefore even a bilax, monoidal monad. The resulting comultiplication $V(X\times Y)\to VX\times VY$ is the marginalization of valuations. The bilax monoidal structure of measure-like monads such as V plays an important role in probability theory (Fritz and Perrone Reference Fritz and Perrone2018).

4.1 $\tau$ -smooth Borel measures

Definition 4.1 ( $\tau$ -smooth Borel measure) Let X be a topological space. A Borel measure m on X is called $\tau$ -smooth if, for every directed net $(U_\lambda)_{\lambda\in\Lambda}$ of open subsets of X,

\begin{equation*} m \left( \bigcup_{\lambda\in\Lambda} U_\lambda \right) = \sup_\lambda m(U_\lambda) . \end{equation*}

A more common regularity assumption on measures is the Radon property. It is known that, in the category of Hausdorff spaces, every Radon measure is $\tau$ -smooth, and that, in the category of compact Hausdorff spaces, Radon measures and $\tau$ -smooth measures coincide Bogachev (2000, Proposition 7.2.2). Therefore, all statements of this paper apply to Radon measures on Hausdorff spaces. It is clear that every $\tau$ -smooth Borel measure restricts to a continuous valuation.

Since a Borel measure is uniquely determined by its values on open sets, such an extension, if it exists, is unique and $\tau$ -smooth.

Proposition 4.3 Let X be a topological space. Suppose that a continuous valuation $\nu\in VX$ is extendable to a Borel measure m on X. Then, for every lower semicontinuous function $f\in[0,\infty]^X$ , we have

\begin{equation*}\langle {\nu} , {f} \rangle \;=\;\int_X f \, dm.\end{equation*}

Note that all functions in $[0,\infty]^X$ are measurable and nonnegative; therefore, their Lebesgue integral against a probability measure is either a well-defined nonnegative number or $+\infty$ . To prove the statement, we use the following standard result.

Proposition 4.4 (Corollary 414B.a in Fremlin 2006) Let X be a topological space and $\mu$ a $\tau$ -smooth Borel measure on X. Let $(f_\lambda)_{\lambda\in\Lambda}$ be a directed net in $[0,\infty]^X$ . Define $f(x) := \sup_\lambda f_\lambda(x)$ for all $x\in X$ . Then f is lower semicontinuous and

\begin{equation*} \int f \, d\mu = \sup_{\lambda} \int f_\lambda\,d\mu . \end{equation*}

Proof of Proposition 4.3. If f is simple, the two quantities agree by definition of integral of a simple function (both for measures and for valuations), using the fact that $\nu$ and m agree on open sets. If f is not simple, we use the defining supremum of $\langle {\nu} , {f} \rangle$ and Proposition 4.4 to obtain the desired equality.

If X is not sober, a continuous valuation need not be extendable, as the following counterexample illustrates. It is based on the fact that the $\{0,\infty \}$ -valued continuous valuations with $\nu(X) = \infty$ correspond to completely prime filters on $\mathcal{O}(X)$ .

Example 4.5 Let $X \,{:}\,{\raise-1.5pt{=}}\, (0,1)$ carry the upper open topology, hence its open subsets are of the form (a,1) for $a\in[0,1]$ . The Borel $\sigma$ -algebra of X coincides with the Borel $\sigma$ -algebra of (0,1) in the Euclidean topology. Consider the continuous valuation $\nu: \mathcal{O} (X)\to [0,\infty]$ given by

\begin{equation*} \nu(U) \,{:}\,{\raise-1.5pt{=}}\, \begin{cases} 0 & \text{if } U=\varnothing, \\ 1 & \text{otherwise}. \end{cases} \end{equation*}

Suppose that there was a Borel measure m on X which agrees with $\nu$ on the open sets. Consider the measurable set (a,b] for $0<a<b<1$ . We have

\begin{equation*} m\big( (a,b] \big) \;=\; m\big( (a,1) \backslash (b,1) \big) \;=\; m\big( (a,1) \big) - m\big( (b,1) \big) \;=\; 1-1 \;=\; 0 . \end{equation*}

The space X can be expressed as a countable disjoint union,

$$X\; = \;(0,1)\; = \mathop \sqcup \limits_{n = 1}^\infty \left( {1 - {1 \over n}\;,\;1 - {1 \over {n + 1}}} \right].$$

Note that $m(X)=\nu(X)=1$ , while

\begin{equation*} m(X) = \sum_{n=1}^{\infty} m \left( \left( \left. 1-\dfrac{1}{n}\;,\; 1-\dfrac{1}{n+1} \right] \right)\right. \;=\; \sum_{n=1}^{\infty} 0 \;=\; 0. \end{equation*}

Therefore, m is not countably additive.

The interpretation of the above example is that $\nu$ represents a Dirac mass at 1, a point in the sobrification of X. If we sobrify X to (0,1] by including the point 1, then $\nu$ can be extended to $\delta_1$ .

It is known that, on a $T_{3\frac{1}{2}}$ space, every finite continuous valuation extends to a measure (Alvarez-Manilla et al. Reference Alvarez-Manilla, Edalat and Saheb-Djahromi1998). The same holds on spaces that are sober and locally compact (Alvarez-Manilla Reference Alvarez-Manilla2002). In particular, the sets of finite $\tau$ -smooth Borel measures and of finite continuous valuations are in bijection for all locally compact Hausdorff spaces and for all metric spaces. Whether extensions exist for all (finite) continuous valuations on sober spaces seems to be an open question.

4.2 The A-topology

We will construct our monad P as a submonad of the monad V from Section 3.

Definition 4.6 Let X be a topological space. We define the space PX to be the set of $\tau$ -smooth Borel probability measures on X, equipped with the subspace topology inherited from VX, which is the weakest topology which makes the integration maps

\begin{equation*} p \longmapsto \int f \, dp\end{equation*}

continuous for every lower semicontinuous $f : X \to [0,\infty]$ .

By Propositions 3.8 and 4.3, we can equivalently say PX carries the weakest topology which makes the evaluation maps $p \mapsto p(U)$ lower semicontinuous for all $U \in \mathcal{O}(X)$ . This topology is called the A-topology Bogachev (2000), 8.10(iv), named after Pavel Alexandrov. For a thorough study of the A-topology in the case of a Hausdorff space, consult the monographs by TopsØe^{Footnote 9} (Reference TopsØe1970), Part II and Bogachev (2000), Section 8.10 (iv).

Proposition 4.7 (p. 227 of Bogachev 2000) Let X be a $T_{3\frac{1}{2}}$ space. Then the topology of PX is equivalently the weakest topology which makes the integration maps

\begin{equation*} p \longmapsto \int f \, dp \end{equation*}

continuous for all bounded continuous $f : X \to \mathbb{R}$ .

Proof. One direction is obvious, since to every bounded continuous $f : X \to \mathbb{R}$ one can add a constant so as to make it into a lower semicontinuous function with values in $[0,\infty]$ .

For the other direction, we equip PX with the weak topology with respect to bounded continuous functions, for which we prove that the evaluation map $p \mapsto p(U)$ is lower semicontinuous for any $U \in \mathcal{O}(X)$ . But since Proposition 4.3 shows that the integral is Scott-continuous on $[0,\infty]$ -valued lower semicontinuous functions, it is enough to exhibit the indicator function $\unicode{x1D7D9}_U$ as a directed supremum of bounded continuous functions.

To do so, choose for every $x \in U$ a continuous function $f_x : X \to [0,1]$ with $f_x(x) = 1$ and $f_x|_{X \setminus U} = 0$ , which is possible by the $T_{3\frac{1}{2}}$ assumption. For a finite subset $F \subseteq U$ , consider $f_F := \max(f_1,\ldots,f_n)$ . Then, $\unicode{x1D7D9}_U$ is the directed supremum of the $f_F$ by construction, establishing the claim.

The weak topology of Proposition 4.7 is the one with respect to which convergence of measures is usually considered (as in the Portmanteau theorem). This is not the case for general topological spaces, for which the A-topology can be considered to be more well-behaved than the weak topology (see, e.g., the discussion in Bogachev 2000 after Theorem 8.2.3 therein, as well as its Section 8.10.iv). This is why we work with the A-topology. Due to Proposition 4.7, all results stated for the A-topology apply also to the equivalent weak topology in the $T_{3\frac{1}{2}}$ case. This includes all metric spaces and all compact Hausdorff spaces.

4.3 Functoriality

For topological spaces X and Y, every continuous map $f : X \to Y$ is Borel measurable. Therefore, for $p \in PX$ , we have the pushforward measure $f_*p \in PY$ defined by

\begin{equation*} (f_*p)(B) := p(f^{-1}(B))\end{equation*}

for all Borel sets $B \subseteq Y$ . It is easy to see that $f_*(p)$ is also $\tau$ -smooth. Suppose further that $p \in PX$ restricts to $\nu \in VX$ . Then we have, for all $U \in \mathcal{O}(Y)$ ,

$$({f_*}\nu )(U) = \nu {\rm{ }}\left( {{f^{ - 1}}(U)} \right) = p\left( {{f^{ - 1}}(U)} \right) = ({f_*}p)(U),$$

where the first equation holds by Definition 3.12. Hence, $f_*p$ restricts to $f_*\nu$ . We have proven the following proposition.

We can further conclude from Lemma 3.14 that, if $i : Y \hookrightarrow X$ is a subspace embedding, then so is $i_* : PY \hookrightarrow PX$ .

4.4 Monad structure

Here we prove that P can be extended to a submonad of V. Since P is already a subfunctor of V, this monad structure is necessarily unique: we only need to show that both the unit and the multiplication of V restrict to P.

Consider first the unit $\delta: X \to VX$ . Since the Dirac valuation $\delta(x)$ defined by $\delta(x)(U) = \unicode{x1D7D9}_U(x)$ for open $U \subseteq X$ obviously extends to the Dirac measure $\delta_x (B) \,{:}\,{\raise-1.5pt{=}}\, \unicode{x1D7D9}_B (x)$ for Borel sets $B \subseteq X$ , the map $\delta$ factors through the inclusion $PX \hookrightarrow VX$ . We also write $\delta : X \to PX$ , slightly abusing notation. We know from Section 3.3.1 that $\delta$ is a homeomorphism onto its image if X is $T_0$ . For P, the following more precise statement is known.

The remaining issue in showing that P is a submonad of V is to prove that the multiplication $\mathcal{E} : VVX \to VX$ restricts to a map $\text{E} : PPX \to PX$ . To simplify the exposition, we construct $\text{E} : PPX \to PX$ first and then show that E is indeed the restriction of $\mathcal{E}$ .

Proof. If $A\subseteq X$ is open, then $\varepsilon_A$ is lower semicontinuous by definition of the A-topology on PX and therefore also Borel measurable.

In general, we denote by $\Sigma$ the set of Borel measurable $A\subseteq X$ for which $\varepsilon_A:PX\to\mathbb{R}$ is measurable. To see that $\Sigma$ is closed under countable unions, suppose that a measurable set $B\subseteq X$ can be written as a countable union of disjoint measurable subsets,

\begin{equation*} B = \bigcup_{n=1}^\infty A_n, \end{equation*}

such that $A_n\in \Sigma$ for every n. Then, for each $p\in PX$ ,

\begin{equation*} \varepsilon_B(p) \; = \; p(B) \; = \; \sum_{n=1}^\infty p(A_n) \; = \; \sum_{n=1}^\infty \varepsilon_{A_n}(p) . \end{equation*}

Since pointwise suprema of measurable functions are measurable, and every partial sum on the right is measurable in p, we conclude that $\varepsilon_A$ is also measurable in p. Therefore, $\Sigma$ is closed under countable disjoint unions. Consider $A,B\in\Sigma$ with $A\subseteq B$ . Clearly $\varepsilon_{B\backslash A} = \varepsilon_B - \varepsilon_A$ is also measurable. Therefore, $\Sigma$ is closed under relative complements. This makes $\Sigma$ into a Dynkin system. Since it contains the $\pi$ -system of open subsets, the $$\pi - \lambda $$ theorem implies that $\Sigma$ is the $\sigma$ -algebra of Borel sets.

Definition 4.11 Let X be a topological space and $\mu\in PPX$ . Let $A\subseteq X$ be Borel measurable. We define

\begin{equation*} (\text{E} \mu) (A) \,{:}\,{\raise-1.5pt{=}}\, \int_{PX} p(A) \, d\mu(p) . \end{equation*}

The integrand $p\mapsto p(A) \in [0,1]$ is measurable by Proposition 4.10 and bounded; therefore, the integral exists.

Proof. The nontrivial properties to establish are $\sigma$ -additivity and $\tau$ -smoothness. For the former, let $(A_n)_{n\in\mathbb{N}}$ be a countable family of disjoint measurable subsets of X and let A be their union. Then

\begin{align*} (\text{E} \mu)(A) &= \int_{PX} p(A) \, d\mu(p) = \int_{PX} \left( \sum_{n=1}^{\infty} p(A_n) \right) \, d\mu(p) \\ &= \sum_{n=1}^\infty \int_{PX} p(A_n) \, d\mu(p) = \sum_{n=1}^{\infty} \text{E} \mu (A_n) , \end{align*}

where the third step can be thought of as an application of Fubini’s theorem to $PX \times \mathbb{N}$ .

Turning to $\tau$ -smoothness, let $(U_\lambda)_{\lambda\in\Lambda}$ be a directed net of open sets with union U. Then, since every measure $p\in PX$ is $\tau$ -smooth,

\begin{align*} (\text{E} \mu)(U) & = \int_{PX} p(U) \, d\mu(p) = \int_{PX} \sup_{\lambda} p(U_\lambda) \, d\mu(p) \\ & = \sup_{\lambda} \int_{PX} p(U_\lambda) \, d\mu(p) = \sup_{\lambda} \, (\text{E} \mu)(U_\lambda),\end{align*}

where the third step is an application of $\tau$ -smoothness of $\mu$ (in the form of Proposition 4.3 and Scott continuity of integration against a continuous valuation).

Hence, we have a well-defined map $\text{E}: PPX \to PX$ . The next proposition implies that this map is continuous, which we therefore do not prove separately.

The inclusion $\iota: PX \hookrightarrow VX$ is, by definition, a homeomorphism onto its image. Therefore, so is $\iota_* : VPX \hookrightarrow VVX$ by Lemma 3.14. By composing these two embeddings, we can consider PPX as a subspace of VVX.

Proposition 4.13 Let X be a topological space. Then the following diagram commutes.

Proof. Let $\mu \in PPX$ and let $\nu \in VVX$ be its image. We show that $\mathcal{E} \nu$ extends to the measure $\text{E} \mu$ , by showing that they evaluate to the same number on any open set $U \subseteq X$ ,

\begin{align*} (\mathcal{E} \nu) (U) &= \langle {{\mathcal{E} \nu}} , {{\unicode{x1D7D9}_U}} \rangle = \langle {{\nu}} , {{\langle {-} , {\unicode{x1D7D9}_U} \rangle}} \rangle = \int_{VX} \langle {{\rho}} , {{\unicode{x1D7D9}_U} } \rangle\, d \mu(\rho) \\ &= \int_{PX} \varepsilon_U(p) \, d\mu(p) = \int_{PX} p(U) \, d\mu(p) = (\text{E} \mu) (U).\\[-38pt]\end{align*}

Since the unit and the multiplication of V restrict to P, the equations required for P to be a monad follow automatically from those of V. In particular, we do not need to prove the associativity of E. We have proven the following theorem.

Just as V, the monad P is a strict 2-monad if we consider $\mathsf{Top}$ as a 2-category.

Through the submonad embedding, every algebra of V is also an algebra of P in a canonical way, resulting in a faithful functor from V-algebras to P-algebras. The question whether this forgetful functor is full or essentially surjective is related to the extension problem and is, at present, unanswered.

4.5 Product and marginal probabilities

Here, we show that P is a commutative monad by equipping it with a commutative strength. This in particular implies that we have a natural transformation $PX \times PY \to P(X\times Y)$ implementing the formation of product probability measures. This is a non-trivial statement due to the discrepancy between the Borel $\sigma$ -algebra on $X\times Y$ and the product of the Borel $\sigma$ -algebras on X and Y, which leads to known subtleties with the formation of product measures (Bledsoe and Wilks Reference Bledsoe and Wilks1972).

As with the monad multiplication, this strength $X \times PY \to P(X \times Y)$ is inherited from the strength of V. While we only need to prove that the strength of V restricts to P, it is more convenient to first construct the strength map $s : X \times PY \to P(X \times Y)$ explicitly, which we denote by the same symbol as for V.

Lemma 4.15 Let X and Y be topological spaces, and let $A\subseteq X\times Y$ be a Borel subset. Then, for each $x\in X$ , the slice

\begin{equation*} A_x := \{ y \in Y \mid (x,y) \in A\} \end{equation*}

is a Borel subset of Y.

Proof. The map $y \mapsto (x,y)$ is continuous, and hence Borel measurable, and $A_x$ is the preimage of A under this map.

For given $x \in X$ and $p \in PY$ , we define the Borel probability measure s(x,p) on $X \times Y$ on a Borel set $A\subseteq X\times Y$ as

\begin{equation*}s(x,p)(A) \,{:}\,{\raise-1.5pt{=}}\, p(A_x) .\end{equation*}

This is a $\tau$ -smooth probability measure because taking the slice preserves arbitrary unions, intersections, and complements.

Proposition 4.16 Let X and Y be topological spaces. Then the following diagram commutes.

Proof. Suppose that $p\in PY$ and denote $\nu := \iota(p)$ . On a product of open sets $U \times V \subseteq X \times Y$ , we have

\begin{equation*}s(x,p)(U \times V) \;=\; p((U \times V)_x) \;=\; \unicode{x1D7D9}_U(x) \, p(V) \;=\; \unicode{x1D7D9}_U(x) \, \nu(V) \;=\; s(x,\nu)(U \times V) ,\end{equation*}

which is sufficient to compare two continuous valuations by Proposition 3.8.

It is clear that $s : X \times PY \to P(X \times Y)$ inherits the naturality and strength properties from the strength of V. Thus, we have shown the following.

The commutativity of the strength of P is Fubini’s theorem (Kock Reference Kock2012). In particular, P is also a symmetric monoidal monad (Section 5.4), and the operation of forming product measures is a natural transformation $PX \times PY \to P(X \times Y)$ . This recovers the following result of Ressel.

Moreover, since $\iota$ is a morphism of commutative monads by construction, Proposition 5.29 implies that $\iota$ is also a morphism of monoidal monads.

In other words, the following diagram commutes, which implies that the product of two extendable continuous valuations is extendable.

5.1 The support transformation $V \Rightarrow H$

We turn to the case of valuations. In view of the discussion of the support of Borel measures, we give a definition of the support of a continuous valuation. It is phrased in the formalism of hyperspace and uses the function $\text{sgn} : [0,\infty] \to S$ defined as

\begin{equation*} \text{sgn}(r) \,{:}\,{\raise-1.5pt{=}}\, \begin{cases} 1 & \text{if } r > 0, \\ 0 & \text{otherwise}, \end{cases}\end{equation*}

and the function $\text{ngs} : S \to [0,\infty]$ defined as $\text{ngs}(0) \,{:}\,{\raise-1.5pt{=}}\, 0$ and $\text{ngs}(1) \,{:}\,{\raise-1.5pt{=}}\, \infty$ .

Note that both of these maps are continuous and that ngs is right adjoint to sgn. We therefore have a universal way of turning an open $U \in S^X$ into a function $\text{ngs} \circ U \in [0,\infty]^X$ , namely the function that is infinite on U and vanishes elsewhere.

Definition 5.5 (Support of a continuous valuation) Let X be a topological space and consider $\nu \in VX$ . The support of $\nu$ is the closed set $supp(\nu) \in HX$ characterized by

(10) \begin{equation} \langle {{supp(\nu)}} , {U} \rangle \;=\; sgn( \langle {{\nu}} , {{ngs \circ U}} \rangle ) \end{equation}

for every open $U \in S^X$ .

Thus, $\textrm{supp}(\nu)$ is defined such that pairing it with an open set consists of forcing the open set to take values in the right codomain in a universal way, applying the pairing with $\nu$ , and then transporting back in a universal way. As with the double dualization construction of H and V, the abstract definition (10) facilitates proofs by making them into mere unfoldings of definitions.

Phrased more concretely, (10) states that an open set U is disjoint from $\textrm{supp}(\nu)$ if and only if $\langle {{\nu}} , {{\textrm{ngs} \circ U}} \rangle = 0$ . The definition of ngs and the linearity of $\nu$ imply that this happens if and only if $\nu(U) = 0$ . In other words, the support is characterized as the largest closed set with the property $\nu(X \setminus \textrm{supp}(\nu)) = 0$ . In particular, the support of an extendable valuation equals the support of its extension to a Borel measure as defined in Definition 5.1.

We want to record the following intuitive statement, which will not play any further role in this article.

Proposition 5.6 Let X be a topological space, $\nu \in VX$ a continuous valuation, and $f \in [0,\infty]^X$ . Then

(11) \begin{equation} sgn(\langle {\nu} , {f} \rangle) = \langle {{supp(\nu)}} , {{sgn \circ f}} \rangle .\end{equation}

In down-to-earth terms, this means that $\langle {\nu} , {f} \rangle = 0$ is equivalent to $f|_{supp(\nu)} = 0$ , which is a property that one would expect from a well-behaved notion of support.

Proof. By (10), we have to show that

\begin{equation*} \text{sgn}(\langle {\nu} , {f} \rangle) \;=\; \text{sgn}(\langle {{\nu}} , {{\text{ngs} \circ \text{sgn} \circ f}} \rangle), \end{equation*}

which means that the integral of f vanishes if and only if the integral of the function where all positive values are rounded up to $\infty$ vanishes. Since, with respect to scalar multiplication from $[0,\infty]$ , we have $\text{ngs} \circ \text{sgn} \circ f = \infty f,$ this is a consequence of the $[0,\infty]$ -linearity of the pairing

$${\rm{sgn}}\left( {\langle \nu ,{\rm{ngs}} \circ {\rm{sgn}} \circ f\rangle } \right) = \;{\rm{sgn}}\left( {\langle \nu ,\infty f\rangle } \right) = \;{\rm{sgn}}\left( {\infty {\mkern 1mu} \langle \nu ,n{\mkern 1mu} f\rangle } \right) = \;{\rm{sgn}}\left( {\langle \nu ,f\rangle } \right),$$

where the last step can be thought of as using the equation $\text{sgn} \circ \text{ngs} \circ \text{sgn} = \text{sgn}$ .

We now aim towards showing that $supp : V \Rightarrow H$ is a morphism of monads. Its continuity is obvious as the right-hand side of (10) is continuous in $\nu$ . Similar continuity statements for different choices of topologies are known (such as Theorem 17.1 in Charalambos et al. 2006). We want to stress that the continuity of the support map for the lower Vietoris topology on HX can be interpreted as lower semicontiuity rather than continuity, as the following example shows.

Example 5.7 Let X be the two-element set $\{x,y\}$ equipped with the discrete topology. Consider the following sequence of measures (or the corresponding valuations),

\begin{equation*}\nu_n = \dfrac{n-1}{n} \,\delta_x + \dfrac{1}{n} \,\delta_y\end{equation*}

for $n\in \mathbb{N}$ . In VX (or in PX) this sequence tends to $\delta_x$ . We have $supp (\nu_n) = \{x,y\}$ for all $n \in N$ , while $supp(\delta_x)=x$ . Hence,

\begin{equation*} \textrm{supp} \left( \lim_{n\to\infty} \nu_n \right) \subsetneq \lim_{n\to\infty} \textrm{supp}(\nu_n) .\end{equation*}

The support of the limit can be smaller than the limit of the supports, a property that parallels the lower semicontinuity of real functions.

We now turn to naturality, which relies on the Scott continuity of valuations.

Proposition 5.8 Let $f: X \to Y$ be continuous. The following diagram commutes.

Proof. For every $U \in S^Y$ , we have

\begin{align*} \langle {{\textrm{supp}(f_*\nu)}} , {U} \rangle \;&=\; \text{sgn}(\langle {{f_*\nu}} , {{\text{ngs} \circ U}} \rangle) \;=\; \text{sgn}(\langle {{\nu}} , {{\text{ngs} \circ U \circ f}} \rangle) \\ &=\; \langle {{\textrm{supp}(\nu)}} , {{U \circ f}} \rangle \;=\; \langle {{f_\sharp(\textrm{supp}(\nu))}} , {U} \rangle.\\[-38pt] \end{align*}

Therefore, the support induces a natural transformation $\textrm{supp}:V \Rightarrow H$ , which can be restricted to a natural transformation $\textrm{supp} : P \Rightarrow H$ . On the set of all Borel measures, the support as given in Definition 5.1 would not be natural. We give an example using the DieudonnÉ measure of Example 5.3.

The following counterexample shows that the notion of support is also not natural for signed measures, which is why we do not expect a generalization of our results to the signed case.

5.2 The support is a morphism of monads

Proposition 5.11 For any topological space X, the following two diagrams commute.

We refer to these diagrams as the unit and the multiplication diagram, since they express the compatibility of the support map with the respective monad maps. The unit diagram says that the support of the Dirac valuation $\delta(x)$ equals $\text{cl}(\{x\})$ . The multiplication diagram says: If the valuation $\nu\in VX$ is the integral of the valuation $\xi \in VVX$ , then the support of $\nu$ is the closure of the union of the supports of all the valuations in the support of $\xi$ .

To illustrate this statement in the simplest case, suppose that X is finite and discrete and that $\xi \in VVX$ is finitely supported. Then, VX can be identified with the simplex of probability vectors $(p_x)_{x\in X}$ and $\mathcal{E}\xi$ is a finite convex combination of these. The statement is then that the set of nonzero components of $\mathcal{E}\xi$ coincides with the union of the sets of nonzero components in each term contributing to the convex combination. In the discrete case, this statement is straightforward to prove, however, to the best of our knowledge, it has never appeared in a published document. ^{Footnote 10} Our result can be thought of as a generalization to continuous distributions. Despite this, the proof is purely formal and does not require any topological or measure-theoretic input.

Proof of Proposition 5.11. Considering the unit diagram, we have, for any $x\in X$ and any open $U \in S^X$ ,

\begin{equation*}\langle {{\textrm{supp}(\delta(x))}} , {U} \rangle \;=\; \text{sgn}(\langle {{\delta(x)}} , {{\text{ngs} \circ U}} \rangle) \;=\; \text{sgn}( \text{ngs}(U(x)) \;=\; U(x) \;=\; \langle {\sigma(x)} , {U} \rangle .\end{equation*}

Considering the multiplication diagram, let $\xi \in VVX$ . Using (11) in the third and fourth step, we obtain that, for any $U \in S^X$ ,

\begin{align*} \langle {{\textrm{supp}(\mathcal{E}\xi)}} , {U} \rangle \;=\; \text{sgn}(\langle {{\mathcal{E}\xi}} , {{\text{ngs} \circ U}} \rangle) \;=\; \text{sgn}(\langle {{\xi}} , {{\langle {-} , {{\text{ngs} \circ U}} \rangle}} \rangle) \end{align*}

\begin{align*} &=\; \langle {{\textrm{supp}(\xi)}} , {{\text{sgn}(\langle {-} , {{\text{ngs} \circ U}} \rangle)}} \rangle \;=\; \langle {{\textrm{supp}(\xi)}} , {{\langle {{\textrm{supp}(-)}} , {U} \rangle}} \rangle \\ &=\; \langle {{\textrm{supp}_\sharp(\textrm{supp}(\xi))}} , {{\langle {-} , {U} \rangle}} \rangle \;=\; \langle {{\mathcal{U}(\textrm{supp}_\sharp(\textrm{supp}(\xi)))}} , {U} \rangle. \\[-38pt] \end{align*}

We can summarize our results so far as follows.

Corollary 5.12 The formation of the support of a continuous valuation is a morphism of monads

\begin{equation*} supp:(V,\delta,\mathcal{E})\longrightarrow (H,\sigma,\mathcal{U}). \end{equation*}

5.3 Consequences for algebras

It is well-known that a morphism of monads on the same category induces a functor between the respective categories of algebras in the opposite direction.

The above combined with Corollary 5.12 yields the following.

The algebras of H are the complete topological join-semilattices (Halg). Hence, we obtain the following statement.

A very similar result was obtained independently by Goubault-Larrecq and Jia (Reference Goubault-Larrecq and Jia2019), Proposition 4.14.

Since every V-algebra is also a topological cone (Valg), it follows that every topological complete join-semilattice A is a topological cone. Unfolding the definitions shows that its addition is given by the binary join, and its scalar multiplication is

\begin{equation*} r x \;=\; \begin{cases} x &\text{if $r > 0$},\\ \bot & \text{if $r = 0$}, \end{cases} \end{equation*}

which is indeed jointly continuous. This is the same in spirit as the convex spaces of combinatorial type (Fritz Reference Fritz2009).

5.4 The support of products and marginals

The support is not only a morphism of monads, but also respects the strengths (or equivalently the monoidal structures).

Proposition 5.16 Let X and Y be topological spaces. The following diagram commutes.

Proof. Consider $x\in X$ and $\rho\in VY$ . Since the sign function is multiplicative, we have, for every $U\subseteq X$ and $V\subseteq Y$ ,

\begin{align*} \langle {{\textrm{supp}(s(x,\rho))}} , {{U\times V}} \rangle \;&=\; \text{sgn}(\langle {{s(x,\rho)}} , {{\text{ngs} \circ (U\times V)} } \rangle\\ &=\; \text{sgn}(\langle {{s(x,\rho)}} , {{(\text{ngs} \circ U) \times (\text{ngs} \circ V)}} \rangle) \\ &=\; \text{sgn}\left( \text{ngs}(U(x)) \, \langle {{\rho}} , {{\text{ngs} \circ V}} \rangle \right) \\ &=\; \text{sgn}(\text{ngs}(U(x))) \, \text{sgn}(\langle {{\rho}} , {{\unicode{x1D7D9}_V}} \rangle) \\ &=\; U(x) \, \langle {{\textrm{supp}(\rho)}} , {{V}} \rangle \\ &=\; \langle {{s(x,\textrm{supp}(\rho))}} , {{U\times V}} \rangle . \end{align*}

This is enough by Lemma 2.3.

Proof. See Proposition 5.29.

In particular, the following diagram commutes.

In other words, the support of the product distribution is the product of the supports of the factors. For the unitors, we have the trivial statement that the support of the valuation in the image of $u:1\to V1$ is the unique point of 1.

By the universal property of the Cartesian product, we know that the support is also an opmonoidal natural transformation. This means that the supports of the marginals are the projections of the support. Indeed, the product projection $\pi_1:X\times Y\to X$ makes the following diagram commute,

and the same can be said about $\pi_2:X\times Y\to Y$ . The bottom parallelogram of this diagram gives a map $VX\times VY\to HX$ , and similarly, we obtain a map $VX\times VY\to HY$ . By the universal property of the product, there is an induced map $VX\times VY\to HX\times HY$ commuting with the supports, and this is the opmonoidal structure. The commutativity of the square face of the diagram implies that the supports of the marginals are the projections of the support.