# A pointfree description of convex compact subsets

In my thesis I needed to represent compact convex subsets of compact partially-ordered spaces without actually referring to points of the spaces. The motivation behind this is that the points of the Vietoris hyperspace (of compact partially-ordered spaces) are precisely the compact convex subsets and if we want to describe what the points are constructively, we need to be able to axiomatise such subsets without referring to the points of the original space.

Compact partially ordered spaces $(X, \leq, \tau)$, i.e. compact spaces with the order $\leq$ closed in the product $X\times X$, can be equivalently described as compact regular bitopological spaces $(X, \tau_+, \tau_-)$ and that is the description I used in my thesis (the definition of regularity and compactness is given below). The two topologies $\tau_+$ and $\tau_-$ are just the topologies of open upsets and open downsets, respectively.

The key observation is that every compact convex subset $K$ of $X$ can be described in terms of its complement. Take the sets $U_+ = X \setminus {\downarrow K} \text{ and } U_- = X \setminus {\uparrow K}$ Since $\downarrow K$ and $\uparrow K$ are closed subsets we see that $U_+ \in \tau_+$ and $U_- \in \tau_-$. Conversely, for any pair $(U_+, U_-)\in \tau_+\times\tau_-$, the complement of their union $X \setminus (U_+ \cup U_-)$ is compact and convex. However, there might be many pairs $(U_+, U_-)$ representing the same $K$.

Finding a canonical representation of compact convex subsets amounts to trimming down the pairs of opens $(U_+, U_-)$ to the maximal such pairs, that is, whenever $V_+ \cup V_-$ is equal to $U_+ \cup U_-$, for some $(V_+, V_-)\in \tau_+\times\tau_-$, then it must be that $V_+ \subseteq U_+$ and $V_- \subseteq U_+$.

On the other hand, if we are given a pair $(U_+, U_-)$ of opens, we can easily find the canonical representation of the convex compact set $K = X \setminus (U_+ \cup U_-)$ they represent. Just take, the pair $(X \setminus {\downarrow K}, \ \ X \setminus {\uparrow K})$ mentioned earlier.

## The problem

So far we didn’t do anything special. So long we stick to bitopological spaces we can comfortably represent convex compact subsets as described above. The problem comes when we move to the pointfree setting. Instead of a space consisting of two topologies we have a pair of frames $L_+$ and $L_-$ (representing the two topologies) and two relations $\mathrm{con}\subseteq L_+ \times L_-$ and $\mathrm{tot}\subseteq L_+ \times L_-$. The $\mathrm{con}$ relation represents that two (abstract) opens $(a,b)\in L_+\times L_-$ are disjoint from each other and $(a,b)\in \mathrm{tot}$ whenever $a$ and $b$ cover the whole space.

For example, for any bitopological space $(X,\tau_+, \tau_-)$, we have a tuple $(L_+, L_-, \mathrm{con},\mathrm{tot})$ where $L_+ = \tau_+$, $L_- = \tau_-$ and, for $(U_+, U_-)\in \tau_+\times \tau_-$:

$\begin{align*} (U_+, U_-) \in \mathrm{con}&\iff\quad U_+ \cap U_- = \emptyset \\ (U_+, U_-) \in \mathrm{tot}&\iff\quad U_+ \cup U_- = X \end{align*}$

If we only have $\mathrm{con}$ and $\mathrm{tot}$, we cannot directly express that a pair $(U_+, U_-)$ is maximal with the property that $K = X \setminus (U_+ \cup U_-)$. I spent quite a lot time finding the right pointfree axiomatisation of maximal $(U_+, U_-)$ pairs. It can be shown that they are characterised by the following two axioms (see Section 4.4.1 of my thesis [2]):

**(K+)**$\forall W_+\in L_+$ if $(U_+ \cup W_+, U_-) \in \mathrm{tot}$ then $(W_+, U_-)\in \mathrm{tot}$**(K−)**$\forall W_-\in L_-$ if $(U_+, U_- \cup W_-) \in \mathrm{tot}$ then $(U_+, W_-)\in \mathrm{tot}$

Observe that $(U_+ \cup W_+, U_-) \in \mathrm{tot}$ is equivalent to saying that $K \subseteq W_+$ for $K = X \setminus (U_+ \cup U_-)$. So (K+) says that if $W_+$ covers $K$ then it must also cover $\uparrow K$.

These two axioms allowed me to prove everything I needed in my thesis. However, there was one thing bugging me that I wasn’t able to show. Given an arbitrary pair $(U_+, U_-)\in \tau_+\times \tau_-$, I wasn’t able to find a pair of opens satisfying (K+) and (K−) while representing the same compact convex subset. In other words, I didn’t know how to calculate $(X \setminus {\downarrow K}, \ \ X \setminus {\uparrow K})$ for $K = X \setminus (U_+ \cup U_-)$ in terms of $\mathrm{con}$ and $\mathrm{tot}$ alone.

I managed to do it today, three years after submitting my thesis. I would like to thank Guillaume Massas for showing me how to do a special case of this, which prompted me to revisit this topic.

## The solution

The tuples $\mathcal L = (L_+, L_-, \mathrm{con}, \mathrm{tot})$ are called d-frames and are required to satisfy a number of axioms. I will not spell them out here, they are all fairly natural (see my thesis or [1]). As we said earlier our results apply to compact regular bitopological spaces. Compactness and regularity can be expressed entirely in terms of $\mathrm{con}$ and $\mathrm{tot}$:

- For $a\in L_+$, set $a^*$ to be the largest $b\in L_-$ such that $(a,b)\in \mathrm{con}$.
- For $a, b \in L_+$, set $a \triangleleft b$ if $(b,a^*) \in \mathrm{tot}$.

($a^*$ for $a\in L_-$ and $a\triangleleft b$ for for $a, b\in L_-$ are defined symmetrically) - $\mathcal L$ is regular if $a = \bigvee \{ b \mid b \triangleleft a \}$ for every $a\in L_+$ or $a\in L_-$.
- $\mathcal L$ is compact if whenever $\bigsqcup_{i\in I} \ (a_i, b_i) = (\bigvee_i a_i, \bigvee_i b_i)$ is in $\mathrm{tot}$ then $\bigsqcup_{i\in F} \ (a_i, b_i)$ is in $\mathrm{tot}$, for some finite $F\subseteq I$.

Note that $U_+ \triangleleft V_+$, for $U_+, V_+\in \tau_+$ in a bitopological space, precisely whenever the $\tau_-$-closure of $U_+$ is a subset of $V_+$. Details and further explanation of these definitions can be found in Sections 2.1.1 and 2.3.1 of my thesis [2]. In the following we assume that $\mathcal L$ is compact and regular.

By $\mathbb{K}_c$ denote the set of pairs $\alpha = (\alpha_+, \alpha_-)\in L_+ \times L_-$ that satisfy (K+) and (K−). We wish to show the following theorem:

For every $\alpha \in L_+ \times L_-$ there is some $\beta \in \mathbb{K}_c$ such that

- $\alpha_+ \leq \beta_+$ and $\alpha_- \leq \beta_-$
- For every $x\in L_+$ and $y\in L_-$,

$\begin{align*} (\alpha_+ \vee x, \alpha_-)\in \mathrm{tot}&\iff\quad (\beta_+ \vee x, \beta_-)\in \mathrm{tot}\\ (\alpha_+, \alpha_- \vee y)\in \mathrm{tot}&\iff\quad (\beta_+, \beta_- \vee y)\in \mathrm{tot} \end{align*}$

Observe that, if we instantiate $\mathcal L$ with a bitopological space, the two conditions (1) and (2) exactly express that $\alpha$ and $\beta$ represent (the complement of) the same convex compact subset.

Fix $\alpha \in L_+ \times L_-$ and define $\beta = (\beta_+, \beta_-) \in L_+\times L_-$ as follows

$\begin{align*} \beta_+ &= \ \bigvee \{ x \in L_+ \mid (\alpha_+, \alpha_- \vee x^*) \in \mathrm{tot}\} \quad\text{and}\\ \beta_- &= \ \bigvee \{ y \in L_- \mid (\alpha_+\vee y^*, \alpha_-) \in \mathrm{tot}\}. \end{align*}$

Observe that this definition yields precisely the pair $(X \setminus {\downarrow K}, \ \ X \setminus {\uparrow K})$ for $K = X \setminus (U_+ \cup U_-)$ in a compact regular bitopological space $(X,\tau_+, \tau_-)$. Indeed, by the Hofmann-Mislove theorem, the compact downset $\downarrow K$ can be uniquely represented as a Scott-open filter $\mathcal S_K = \{ V_-\in \tau_- \mid \downarrow K \subseteq V_-\} = \{ V_- \mid U_+ \cup U_- \cup V_- = X\}$ on $\tau_-$. Further, $\tau_+$ is order-isomorphic to the poset of Scott-open filters on $\tau_-$ (Lemma 5.3.6 in [2]), where a Scott-open filter $\mathcal S$ is send to the $\tau_+$-open $\bigcup \{ V_-^* \mid V_- \in \mathcal S\}$. Because $V_- \in \mathcal S_K$ implies $V_-^{**} \in \mathcal S_K$ and because $V_-^* = V_-^{***}$, we obtain the expected equality $X \setminus \downarrow K = \bigcup \{ V_- \mid U_+ \cup U_- \cup V_-^* = X\}$.

Next we show that $\beta$ is indeed in $\mathbb{K}_c$ and that (1) and (2) hold. Showing (1) is immediate, let $x \triangleleft\alpha_+$ then, by definition, $(\alpha_+, x^*)\in \mathrm{tot}$ and because $\mathrm{tot}$ is upwards closed also $(\alpha_+, \alpha_- \vee x^*)\in \mathrm{tot}$, giving us $x \leq \beta_+$. Finally, be regularity, $\alpha_+ = \bigvee \{ x \mid x\triangleleft\alpha_+\} \leq \beta_+$. Symmetrically we also obtain $\alpha_- \leq \beta_-$.

We show (2) in two steps.

$(\beta_+ \vee x, \beta_-)\in \mathrm{tot}$ implies $(\beta_+ \vee x, \alpha_-)\in tot$

By compactness and directedness of $\{ y \in L_- \mid (\alpha_+\vee y^*, \alpha_-) \in \mathrm{tot}\}$, there is some $y\in L_-$ such that $(\beta_+ \vee x, y)\in \mathrm{tot}$ and $(\alpha_+\vee y^*, \alpha_-) \in \mathrm{tot}$. Since $(y^*,y)\in \mathrm{con}$ it must be that $y^* \leq \beta_+\vee x$ (by the so-called ($\mathrm{con}$-$\mathrm{tot}$) axiom of d-frames). Lastly, because $\mathrm{tot}$ is upwards closed and $\alpha_+ \leq \beta_+$ by (1), from $(\alpha_+\vee y^*, \alpha_-) \in \mathrm{tot}$ we obtain that $(\beta+\vee x, \alpha_-) \in \mathrm{tot}$.

$(\beta_+ \vee x, \alpha_-)\in \mathrm{tot}$ implies $(\alpha_+ \vee x, \alpha_-)\in \mathrm{tot}$

By compactness $(z \vee x, \alpha_-)\in \mathrm{tot}$ for some $z\in L_+$ such that $(\alpha_+, \alpha_- \vee z^*)\in \mathrm{tot}$. Because $\mathrm{tot}$ is upwards closed both $(z \vee (\alpha_+ \vee x), \alpha_-)$ and $(\alpha_+ \vee x, \alpha_- \vee z^*)$ are in $\mathrm{tot}$. However, $(z, z^*)\in \mathrm{con}$ and by regularity and compactness, $\mathcal L$ satisfies cut rules (see Proposition 6.13 in [1]), hence $(\alpha_+ \vee x, \alpha_-)\in \mathrm{tot}$.

The reverse implication, entailing $(\beta_+ \vee x, \beta_-)\in \mathrm{tot}$ from $(\alpha_+ \vee x, \alpha_-)\in \mathrm{tot}$, follows from (1) and the fact that $\mathrm{tot}$ is upwards closed.

Lastly, we explain why $\beta \in \mathbb{K}_c$. We only check (K+) as (K−) is proved symmetrically. Assume that $(\beta_+ \vee x, \beta_-)\in \mathrm{tot}$. From the previous two lemmas we entail that $(\alpha_+ \vee x, \alpha_-)\in \mathrm{tot}$. By compactness and regularity, there exists an $e\in L_+$ such that $e\triangleleft x$ and $(\alpha_+ \vee e, \alpha_-)\in \mathrm{tot}$. Because $e \leq e^{**}$ and $\mathrm{tot}$ is upwards closed $(\alpha_+ \vee e^{**}, \alpha_-)\in \mathrm{tot}$. Therefore, $e^* \leq \beta_-$. Finally, $e\triangleleft x$ gives us that $(x, e^*)\in \mathrm{tot}$, which implies $(x, \beta_-)\in \mathrm{tot}$ because $\mathrm{tot}$ is upwards closed. This finishes the proof of Theorem 1.

Observe that item 2 of Theorem 1 guarantees that the choice of $\beta\in \mathbb{K}_c$ is unique. Indeed, for any other pair $\gamma\in \mathbb{K}_c$ which satisfies item 2 of Theorem 1, let $a\triangleleft\gamma_+$. From $(\gamma_+, a^*)\in \mathrm{tot}$ we also have $(\gamma_+, \gamma_- \vee a^*)\in \mathrm{tot}$ which, by two applications of item 2 of Theorem 1 and (K−), gives $(\beta_+, a^*)\in \mathrm{tot}$. Hence $a \triangleleft\beta_+$. By regularity $\gamma_+ \leq \beta_+$. The converse direction and also the equivalent statements for the minus side are proved analogously, hence $\beta = \gamma$.

## References

- Achim Jung and M. Andrew Moshier.
*On the bitopological nature of Stone duality.*Technical report, 2006. - Tomáš Jakl.
*d-Frames as algebraic duals of bitopological spaces.*Charles University and University of Birmingham (Ph.D. thesis), 2018.