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,,τ)(X, \leq, \tau), i.e. compact spaces with the order \leq closed in the product X×XX\times X, can be equivalently described as compact regular bitopological spaces (X,τ+,τ)(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 τ+\tau_+ and τ\tau_- are just the topologies of open upsets and open downsets, respectively.

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

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

On the other hand, if we are given a pair (U+,U)(U_+, U_-) of opens, we can easily find the canonical representation of the convex compact set K=X\(U+U)K = X \setminus (U_+ \cup U_-) they represent. Just take, the pair (X\K,X\K)(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+L_+ and LL_- (representing the two topologies) and two relations conL+×L\mathrm{con}\subseteq L_+ \times L_- and totL+×L\mathrm{tot}\subseteq L_+ \times L_-. The con\mathrm{con} relation represents that two (abstract) opens (a,b)L+×L(a,b)\in L_+\times L_- are disjoint from each other and (a,b)tot(a,b)\in \mathrm{tot} whenever aa and bb cover the whole space.

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

(U+,U)conU+U=(U+,U)totU+U=X\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 con\mathrm{con} and tot\mathrm{tot}, we cannot directly express that a pair (U+,U)(U_+, U_-) is maximal with the property that K=X\(U+U)K = X \setminus (U_+ \cup U_-). I spent quite a lot time finding the right pointfree axiomatisation of maximal (U+,U)(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+) W+L+\forall W_+\in L_+ if (U+W+,U)tot(U_+ \cup W_+, U_-) \in \mathrm{tot} then (W+,U)tot(W_+, U_-)\in \mathrm{tot}
  • (K−) WL\forall W_-\in L_- if (U+,UW)tot(U_+, U_- \cup W_-) \in \mathrm{tot} then (U+,W)tot(U_+, W_-)\in \mathrm{tot}

Observe that (U+W+,U)tot(U_+ \cup W_+, U_-) \in \mathrm{tot} is equivalent to saying that KW+K \subseteq W_+ for K=X\(U+U)K = X \setminus (U_+ \cup U_-). So (K+) says that if W+W_+ covers KK then it must also cover K\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)τ+×τ(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\K,X\K)(X \setminus {\downarrow K}, \ \ X \setminus {\uparrow K}) for K=X\(U+U)K = X \setminus (U_+ \cup U_-) in terms of con\mathrm{con} and tot\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 =(L+,L,con,tot)\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 con\mathrm{con} and tot\mathrm{tot}:

  1. For aL+a\in L_+, set a*a^* to be the largest bLb\in L_- such that (a,b)con(a,b)\in \mathrm{con}.
  2. For a,bL+a, b \in L_+, set aba \triangleleft b if (b,a*)tot(b,a^*) \in \mathrm{tot}.
    (a*a^* for aLa\in L_- and aba\triangleleft b for for a,bLa, b\in L_- are defined symmetrically)
  3. \mathcal L is regular if a={bba}a = \bigvee \{ b \mid b \triangleleft a \} for every aL+a\in L_+ or aLa\in L_-.
  4. \mathcal L is compact if whenever iI(ai,bi)=(iai,ibi)\bigsqcup_{i\in I} \ (a_i, b_i) = (\bigvee_i a_i, \bigvee_i b_i) is in tot\mathrm{tot} then iF(ai,bi)\bigsqcup_{i\in F} \ (a_i, b_i) is in tot\mathrm{tot}, for some finite FIF\subseteq I.

Note that U+V+U_+ \triangleleft V_+, for U+,V+τ+U_+, V_+\in \tau_+ in a bitopological space, precisely whenever the τ\tau_--closure of U+U_+ is a subset of V+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 𝕂c\mathbb{K}_c denote the set of pairs α=(α+,α)L+×L\alpha = (\alpha_+, \alpha_-)\in L_+ \times L_- that satisfy (K+) and (K−). We wish to show the following theorem:

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

  1. α+β+\alpha_+ \leq \beta_+ and αβ\alpha_- \leq \beta_-
  2. For every xL+x\in L_+ and yLy\in L_-,

(α+x,α)tot(β+x,β)tot(α+,αy)tot(β+,βy)tot\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 α\alpha and β\beta represent (the complement of) the same convex compact subset.

Fix αL+×L\alpha \in L_+ \times L_- and define β=(β+,β)L+×L\beta = (\beta_+, \beta_-) \in L_+\times L_- as follows

β+={xL+(α+,αx*)tot}andβ={yL(α+y*,α)tot}.\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\K,X\K)(X \setminus {\downarrow K}, \ \ X \setminus {\uparrow K}) for K=X\(U+U)K = X \setminus (U_+ \cup U_-) in a compact regular bitopological space (X,τ+,τ)(X,\tau_+, \tau_-). Indeed, by the Hofmann-Mislove theorem, the compact downset K\downarrow K can be uniquely represented as a Scott-open filter 𝒮K={VτKV}={VU+UV=X}\mathcal S_K = \{ V_-\in \tau_- \mid \downarrow K \subseteq V_-\} = \{ V_- \mid U_+ \cup U_- \cup V_- = X\} on τ\tau_-. Further, τ+\tau_+ is order-isomorphic to the poset of Scott-open filters on τ\tau_- (Lemma 5.3.6 in [2]), where a Scott-open filter 𝒮\mathcal S is send to the τ+\tau_+-open {V*V𝒮}\bigcup \{ V_-^* \mid V_- \in \mathcal S\}. Because V𝒮KV_- \in \mathcal S_K implies V**𝒮KV_-^{**} \in \mathcal S_K and because V*=V***V_-^* = V_-^{***}, we obtain the expected equality X\K={VU+UV*=X}X \setminus \downarrow K = \bigcup \{ V_- \mid U_+ \cup U_- \cup V_-^* = X\}.

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

We show (2) in two steps.

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

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

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

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

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

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

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

References

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

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