Strongly compact sets and the double hyperspace construction

The notion of strongly compact set is due to Reinhold Heckmann [1]. A subset Q of a space X is called strongly compact if and only if for every open neighborhood U of Q, there is a finitary compact set ↑E such that Q ⊆ ↑E ⊆ U. (A finitary compact set is the upward closure of a finite set E.)

I have already mentioned there that for every sober space X, the sobrification of Q_fin(X) of finitary compact subsets of X, with the upper Vietoris topology, is the space Q_s(X) of so-called strongly compact saturated subsets of X, not the whole Smyth hyperspace Q(X) as one might expect [2, Proposition 7.33].  Let us explore this in a bit more detail.

First, we need to notice that every strongly compact set is compact. If Q is strongly compact indeed, then every open cover (U_i)_{i ∈ I} of Q is such that Q ⊆ ↑E ⊆ ∪_{i ∈ I} U_i for some finite set E. Then E is included in the union of finite many of the sets U_i, which therefore form a finite subcover of Q.

If X is a continuous dcpo in its Scott topology, or more generally a locally finitary compact space, then, conversely, every compact set Q is strongly compact. Indeed, for every open neighborhood U of Q, we use locally finitary compactness in order to find finitary compact neighborhoods ↑E_x of each point x of Q, and included in U; finitely many cover Q, and the resulting finite union is a finitary compact set that contains Q and is included in U.

But, in more general spaces, there are in general strictly more compact sets than strongly compact sets. For example, in any T₁ space, it is easy to see that the strongly compact sets are exactly the finite sets, and that is very far from exhausting the collection of compact sets.

The sobrification of Q_fin(X)

Let Q(X) be the space of all compact saturated subsets of X, with the upper Vietoris topology: its basic open subsets are of the form ☐U ≝ {Q ∈ Q(X) | Q ⊆ U}, where U ranges over the open subsets of X. Q(X) is a T₀ space, whose specialization ordering is reverse inclusion.

One can say more. First, there is a monad (Q, η, μ) on the category Top₀ of T₀ spaces, whose unit η_X : X → Q(X) maps every point x of X to ↑x, and whose multiplication μ_X : Q(Q(X)) → Q(X) maps every element Q of Q(Q(X)) to ∪Q, the union of all the elements of Q. It is practical to note that η_X^–1(☐U)=U and μ_X^–1(☐☐U)=☐U for every open subset U of X; this is useful in order to show that the unit and the multiplication are continuous, for example.

We will also need to observe that the operator ☐ commutes with finite intersections and with directed unions of open sets; the latter is a consequence of the compactness of elements of Q(X).

Lemma. For every sober space X, Q(X) is sober.

Proof. Let C be an irreducible closed subset of Q(X). We consider the family F of all open subsets U of X such that ☐U intersects C. F is non-empty, upwards-closed and if U and V are any two elements of F, then since C is irreducible, ☐U ∩ ☐V intersects C. Since ☐U ∩ ☐V =☐(U ∩ V), U ∩ V is in F. Hence, F is a filter of open subsets of X. Similarly, but using the fact that ☐ commutes with directed unions of open sets, F is Scott-open. By the Hofmann-Mislove theorem (Theorem 8.3.2 in the book), F is the collection of open neighborhoods of some compact saturated subset Q₀ of X.

If Q₀ were not in C, then it would be in its complement, which is open, so there would be a basic open set ☐U such that Q₀ ∈ ☐U and ☐U does not intersect C. The latter means that U is not in F, hence does not contain Q₀; this contradicts Q₀ ∈ ☐U. Therefore Q₀ is in C.

In order to show that C is the closure of Q₀, it remains to show that every element Q of C is below, namely contains, Q₀. It suffices to show that every open neighborhood U of Q contains Q₀, and that is easy: since Q ⊆ U, ☐U intersects C at Q, so U is in F, and therefore contains Q₀. ☐

One might guess that Q(X) would be the sobrification of its subspace Q_fin(X) of finitary compact subsets, but that is wrong. For example, you can check that, if X is T₂, then Q_fin(X) is already sober, and very different from Q(X). Since that will follow from the next result, I will not bother to show this.

Proposition. For every sober space X, the sobrification of Q_fin(X) is Q_s(X), the subspace of Q(X) consisting of strongly compact saturated subsets of X.

Proof. Since Q(X) is sober, the sobrification of Q_fin(X) is its Skula-closure, as we have seen in this post. (Namely, the closure in the Skula topology. The Skula topology is generated by the open and the closed sets, or equivalently by the sets U ∩ ↓z, where z ranges over the points and U ranges over the open sets in the original topology.)

If Q is in the Skula-closure of Q_fin(X), then for every open neighborhood U of Q, ☐U ∩ ↓_QQ is a Skula-open neighborhood of Q in Q(X); I am writing ↓_Q for downward closure in Q(X) (remember that the ordering there is reverse inclusion!). And that set must therefore intersect Q_fin(X). In other words, there is a finitary compact set ↑E in ☐U ∩ ↓_QQ. We expand this, and we obtain Q ⊆ ↑E ⊆ U. Therefore Q must be strongly compact saturated.

Conversely, for every strongly compact saturated subset Q of X, we claim that Q is in the Skula-closure of Q_fin(X). In other words, for every open neighborhood U of Q in Q(X), we show that U contains some finitary compact set containing Q. Necessarily, U contains a basic open neighborhood ☐U of Q. Since Q is strongly compact, there is a finitary compact set ↑E such that Q ⊆ ↑E ⊆ U; then ↑E is in ☐U, hence in U. ☐

One can then show that the monad (Q, η, μ) cuts down to a monad (Q_s, η, μ), and both monads themselves restrict to monads on the category Sob of sober spaces. As I have said in this post, the algebras of the monad (Q_s, η, μ) on Sob are exactly the deflationary sober semilattices with small semilattices [1, Theorem 7.37], and the proof is a pretty simple extension of what we had done with Q_fin in that same post.

Instead, I would like to mention a funny hyperspace commutation theorem, which I happened to obtain recently by rereading carefully Matthew de Brecht and Tatsuji Kawai’s paper [3].

Hyperspace commutation results

Let H(X) denote the Hoare hyperspace of X: that is the set of closed subsets of X, with the lower Vietoris topology, given by subbasic open sets ♢U ≝ {C ∈ Q(X) | C intersects U}, where U ranges over the open subsets of X. As we have seen here, Schalk proved that H(X) is always a sober space. This also defines a monad (H, η, μ) [let me reuse the same η and μ as before] whose unit η_X : X → H(X) maps every point x of X to ↓x, and whose multiplication μ_X : H(H(X)) → H(X) maps every element C of H(H(X)) to cl(∪C). Also, η_X^–1(♢U)=U and μ_X^–1(♢♢U)=♢U for every open subset U of X.

What de Brecht and Kawai achieved was to show that QHX and HQX are homeomorphic, provided that X is consonant; and that this is actually an if and only if. This is the continuation of a long line of work, showing similar isomorphisms in more restricted categories, mostly of domains. The locale theoretic analogue of this result is due to Steven Vickers and Christopher Townsend [4] shows that the same commutation between the two localic analogues of Q and H (the so-called upper and lower power locale constructions) holds without any assumption on the locale.

Replacing Q by Q_s, I will show that Q_sHX and HQ_sX are homeomorphic for every topological space, consonant or not, sober or not.

Oh, by the way, if you read Heckmann [1], you will see that this commutation requires a property that he calls U_S-conformity, and it may puzzle you that I will not require any assumption on X at all. The main reason for this is that he considers the Scott topology, not the lower Vietoris topology, on HX.

Let us start the proof. This is essentially the same as de Brecht and Kawai’s proof, replacing Q by Q_s. I will simply reorganize it, and mention how each step adapts when we replace Q by Q_s.

The first isomorphisms: QH=O_pO_σ, Q_sH=O_pO_p

The first step consists in showing that there is a homeomorphism between QHX and O_pO_σX, valid for every topological space X [3, Theorem 6.10]. (Well, de Brecht and Kawai only show an order-isomorphism here, not a homeomorphism, but that is not too far.) Here OX is the collection of open subsets of X, and we can topologize it in two ways:

• with the Scott topology of the inclusion ordering, yielding a space that I will write as O_σX;
• or with what I will call the pointwise topology, whose subbasic open sets are [x ∈] ≝ {U ∈ OX | x ∈ U}, where x ranges over the points of X. I will write the resulting space O_pX.

The homeomorphism is defined as follows:

• φ_X maps every Q ∈ QHX to the collection of open subsets U of X that intersect every C ∈ Q;
• ψ_X maps every U ∈ O_pO_σX to the intersection of the sets ♢U, where U ranges over U.

Let us check this.

Proposition. φ_X and ψ_X form a pair of mutually inverse continuous maps between QHX and O_pO_σX. We have φ_X^–1([U ∈]) = ☐♢U, and ψ_X^–1(☐♢U) = [U ∈] for every open subset U of X.

Proof. For every Q ∈ QHX, we first verify that U ≝ φ_X(Q) is Scott-open. It is clearly upwards-closed. Given any directed family (U_i)_i∈I of open subsets of X whose union is in U, by definition of U, every C ∈ Q intersects ∪_{i ∈ I} U_i, hence is in some ♢U_i. Therefore the sets ♢U_i (i ∈ I) form an open cover of Q, which is also directed. Since Q is compact, Q is included in some ♢U_i. Therefore U_i intersects every C ∈ Q, so is in φ_X(Q)=U.

We verify that φ_X^–1([U ∈]) = ☐♢U. For every Q ∈ φ_X^–1([U ∈]), by definition, the collection φ_X(Q) of all open subsets V of X that intersect every C ∈ Q contains U, namely every C ∈ Q intersects U; equivalently, Q ∈ ☐♢U.

This shows, in particular, that φ_X is continuous.

Let us turn to ψ_X . We first show that: (∗) for every upwards-closed family U of open subsets of X, for every open subset V of X, ψ_X(U) ⊆ ♢V if and only if V ∈ U. The if direction is clear. Conversely, we assume ψ_X(U) ⊆ ♢V and we consider the closed set C ≝ X–V. C is not in ♢V, hence not in ψ_X(U). Therefore C fails to intersect some U ∈ U. By definition of C, U is included in V, and therefore V is in U, since U is upwards-closed.

It follows that ψ_X(U) is compact, as we show by using Alexander’s subbase lemma (Theorem 4.4.29 in the book): if ψ_X(U) ⊆ ∪_{i ∈ I} ♢U_i, where (U_i)_i∈I is a directed family of open subsets of X, then ψ_X(U) ⊆ ♢V where V ≝ ∪_{i ∈ I} U_i, since ♢ commutes with arbitrary unions; so V is in U by (∗). Since U is Scott-open, some U_i is in U, and therefore ψ_X(U) ⊆ ♢U_i, using (∗) once again.

It also follows from (∗) that ψ_X^–1(☐♢U) = [U ∈]. In particular, ψ_X is continuous.

For every open subset U of X, (φ_X ◦ ψ_X)⁻¹(☐♢U) = ψ_X−1([U ∈]) = ☐♢U. Hence, for every Q ∈ QHX, Q and φ_X(ψ_X(Q)) belong to the same open sets. Since QHX is T₀, they are equal. We prove that ψ_X ◦ φ_X is the identity map in the same manner. ☐

The following modification (which does not appear in [3]) is concerned with the space Q_sHX of strongly compact saturated subsets of HX. Correspondingly, we replace O_pO_σX with O_pO_pX. (Mind the change from σ to p in indices!) In order to try and avoid some confusion, we write ☐^sU for the collection of strongly compact saturated subsets of an open set U. Notice that Q_sHX is a topological subspace of QHX, and that O_pO_pX is a topological subspace of O_pO_σX: its elements are the pointwise open subsets of OX, all of them are Scott-open, and the topology is indeed the subspace topology.

Proposition. The pair of homeomorphisms φ_X and ψ_X restrict (and corestrict) to homeomorphisms between Q_sHX and O_pO_pX. We have φ_X^–1([U ∈]) = ☐^s♢U, and ψ_X^–1(☐^s♢U) = [U ∈] for every open subset U of X.

Proof. We first show that φ_X maps every strongly compact set Q ∈ Q_sHX to a pointwise open subset U ≝ φ_X(Q). To this end, let U be any element of U; we will find an open subset V of O_pX such that U ∈ V ⊆ U. By definition of φ_X, U intersects every C ∈ Q, so Q is included in ☐^sU. Since Q is strongly compact, there are finitely many closed subsets C₁, …, C_n of X such that Q ⊆ ↑_H{C₁, …, C_n} ⊆ ☐^sU, where ↑_H denotes upward closure in HX. Since ↑_H{C₁, …, C_n} ⊆ ☐^sU, every C_i intersects U, say at x_i. Then U is in V ≝ ∩_i=1ⁿ[x_i ∈]. It remains to show that V is included in U = φ_X(Q). For every V ∈ V, every point x_i is in V, so V intersects every C_i, showing that ↑_H{C₁, …, C_n} ⊆ ♢V. Since Q ⊆ ↑_H{C₁, …, C_n}, every C ∈ Q also intersects V, so V is in φ_X(Q).

We turn to ψ_X. For every closed subset C of X, let us write [C ∩] for the collection of open subsets of X that intersect C. If C is of the form ↓{x₁, …, x_n}, then [C ∩] is the union of the sets [x₁ ∈], …, [x_n ∈]. Every open subset U of O_pX can be written as ∪_i∈I∩_j=1^n_i [x_ij ∈]. By writing the outer union as a directed union of finite unions, then distributing the inner intersections over the just created finite unions, and finally observing that any finite union of sets [x ∈] can be written in the form of [C ∩] (as I mentioned at the beginning of this paragraph), we can write U as a directed union ∪_i∈I∩_j=1ⁿ [C_ij ∩], where each C_ij is closed in X.

A few technical but elementary computations show that ψ_X(U) is then equal to the filtered intersection ∩_i∈I ↑_H{C_i1, …, C_ini}. (Exercise!) Now let us assume that ψ_X(U) is included in some open subset V of HX. Since HX is sober, as I have mentioned earlier, it is well-filtered, so there is an index i in I such that ↑_H{C_i1, …, C_ini} is included in V. Since obviously ψ_X(U) is included in ↑_H{C_i1, …, C_ini}, we have shown that ψ_X(U) is strongly compact. The remaining claims follow from our previous proposition. ☐

The de Brecht-Kawai isomorphism: QH=HQ ⇔ consonance

There are two other maps, defined by de Brecht and Kawai as follows:

• σ_X maps every C ∈ HQX to the collection of closed subsets C of X that intersect every Q ∈ C.
• τ_X maps every Q ∈ QHX to the collection of compact saturated subsets Q of X that intersect every C ∈ Q;

Yes, the definition is very symmetric. I will break this symmetry, and concentrate on σ_X.

Let us study them one after the other. Items 2 and 3 below are Lemma 6.4 of [3]. Item 1 is the reason why the statement QH=HQ is related to consonance, as we will see. Let me write ■Q, for every compact saturated subset Q of X, for the set of open neighborhoods of Q.

Lemma. We have:

1. φ_X o σ_X maps every C ∈ HQX to ∪_Q∈C ■Q;
2. for every open subset U of X, σ_X⁻¹(☐♢U) = ♢☐U;
3. σ_X is a topological embedding.

Proof. 1. For every C ∈ HQX, ∪_Q∈C ■Q is a Scott-open subset of OX. For every closed subset C of X, C ∈ ψ_X (∪_Q∈C ■Q) if and only if C intersects every open set U that contains some Q ∈ C. This is certainly the case if C intersects Q. Conversely, if C does not intersect Q, then U ≝ X–C is an open set that does not intersect C. Therefore ψ_X (∪_Q∈C ■Q) = σ_X(C).

In particular, σ_X(C) is in QHX, showing that σ_X is well-defined, and φ_X(σ_X(C)) = ∪_Q∈C ■Q, since φ_X and ψ_X are mutually inverse.

For every open subset U of X, σ_X⁻¹(☐♢U) = σ_X⁻¹(φ_X⁻¹([U ∈])), and this is is equal to the collection of closed subsets C of QX such that U ∈ ∪_Q∈C ■Q, namely such that Q ⊆ U for some Q ∈ C. Hence σ_X⁻¹(☐♢U) = ♢☐U. In particular, σ_X is continuous, almost-open, and every almost-open map whose domain is a T₀ space is injective, so σ_X is a topological embedding. ☐

Hence the only thing missing for σ_X to be a homeomorphism is surjectivity. A space X is consonant if and only every Scott-open subset of OX is a union of sets of the form ■Q, where each Q is compact saturated in X. We have already encountered the concept a few times. Item 1 of the previous lemma shows that if σ_X is surjective, namely if φ_X o σ_X is surjective, then X is consonant. Conversely, we have:

Lemma. If X is consonant, then σ_X is surjective.

Proof. It suffices to show that φ_X o σ_X is surjective. Let U be a Scott-open subset of OX. Since X is consonant, we can write U as a union of sets of the form ■Q. Let C be the collection of compact saturated subsets Q of X such that ■Q is included in U. We claim that C is closed, in other words that its complement is open in QX. Let Q ∈ QX be outside C. We will show that Q is in some basic open set ☐U that is disjoint from C. Let us assume that this is not the case, and let us consider any open neighborhood U of Q. By assumption, ☐U intersects C, say at Q’. Since Q’ is in C, ■Q’ is included in U. Since Q’ is in ☐U, U is in ■Q’, so U is in U. Hence, we have shown that all the open neighborhoods U of Q are in U. In other words, ■Q is included in U, so Q is in C: this is impossible, since we have assumed Q ∈ QX to be outside C.

In summary, we have built an element C of HQX. By item 1 of the previous lemma, φ_X o σ_X maps it to ∪_Q∈C ■Q, namely to the union of all the sets of the form ■Q (with Q compact saturated in X) included in U. Because of our assumption of consonance, that union contains U, and it is clearly included in U. Therefore (φ_X o σ_X) (C) = U. ☐

We have therefore obtained the following result, due to de Brecht and Kawai [3, Theorem 6.13].

Theorem. For every topological space X, the following are equivalent:

1. X is consonant;
2. σ_X is surjective;
3. σ_X is a homeomorphism of HQX onto QHX.

They also show that the map τ_X is such that τ_X⁻¹(♢☐U) ⊆ ☐♢U, that this inclusion is an equality if and only if any of the items 1–3 above is satisfied; and that, in that case, τ_X is continuous and is the inverse of σ_X.

The isomorphism Q_sH=HQ_s

What happens if we replace Q with Q_s throughout? One may guess—and one would be right—that everything would work the same way, replacing compact saturated sets by strongly compact saturated sets… and also O_pO_σX by O_pO_pX, since now φ_X and ψ_X restrict to homeomorphisms between Q_sHX (instead of QHX) and O_pO_pX (instead of O_pO_σX). The right variant of the notion of consonance is then the following: let us say that X is finitarily consonant if and only if every pointwise open subset U of OX (namely, every open subset of O_pX, not O_σX) is a union of sets of the form ■Q, where Q is strongly compact saturated.

We note that every set ■Q is a pointwise open subset of OX (namely, an open subset of O_pX, or equivalently an element of O_pO_pX), for every strongly compact saturated set Q. Indeed, for every U ∈ ■Q, Q is included in U, and since Q is strongly compact, we can find a finitary compact set ↑E such that Q ⊆ ↑E ⊆ U. Let us write the finite set E as {x₁, …, x_n}. Then U is in ∩_i=1ⁿ[x_i ∈], since each x_i is in U, and ∩_i=1ⁿ[x_i ∈] is included in ■Q, since every open set that contains every x_i must contain ↑E, hence also Q. This finishes to show that ■Q is an open subset of O_pX, for every strongly compact saturated set Q. Hence every union of such sets is also open in O_pX. Finitary consonance requires that those are the only open subsets of O_pX.

We now define σ_X^s just like σ_X: it maps every C ∈ HQ_sX (instead of HQX) to the collection of closed subsets C of X that intersect every Q ∈ C. And we replay the proofs we did above:

Lemma. We have:

1. φ_X o σ_X^s maps every C ∈ HQ_sX to ∪_Q∈C ■Q;
2. for every open subset U of X, (σ_X^s)⁻¹(☐^s♢U) = ♢☐^sU;
3. σ_X^s is a topological embedding.

Proof. Just as before, ψ_X (∪_Q∈C ■Q) = σ_X(C), where now C is a closed subset of Q_sX. Since every Q ∈ C is strongly compact saturated, we have seen that ∪_Q∈C ■Q is in O_pO_pX, so its image by ψ_X is in Q_sHX. This also shows φ_X(σ_X(C)) = ∪_Q∈C ■Q, hence item 1. Items 2 and 3 are proved as before. ☐

The following is also proved just like the analogous result we have proved earlier.

Lemma. If X is finitarily consonant, then σ_X^s is surjective. ☐

However, the new thing is that the notion of finitary consonance is vacuous (I was keeping that for the end!).

Fact. Every topological space is finitary consonant.

Proof. Let U be an open subset of O_pX. By definition, we can write U as ∪_i∈I∩_j=1^n_i [x_ij ∈]. But that is just ∪_i∈I ■Q_i, where Q_i ≝ ↑{x_i1, …, x_ini}, and each Q_i is finitarily compact, hence trivially strongly compact. ☐

Hence we have obtained the final theorem of this post.

Theorem. For every topological space X, the spaces Q_sHX, HQ_sX and O_pO_pX are homeomorphic; the first two, through σ_X^s, and the first and the third through the pair φ_X, ψ_X.

I could say a lot more. For example, the constructions are all natural in X; σ_X^s defines a distributive law, hence Q_sH is really the functor part of a monad; etc. But that’s enough for today!

1. Reinhold Heckmann. An upper power domain construction in terms of strongly compact sets. In: Brookes, S., Main, M., Melton, A., Mislove, M., Schmidt, D. (eds) Mathematical Foundations of Programming Semantics. MFPS 1991. Lecture Notes in Computer Science, vol 598. Springer, Berlin, Heidelberg. https://doi.org/10.1007/3-540-55511-0_14
2. Andrea Schalk. Algebras for Generalized Power Constructions. PhD Thesis, TU Darmstadt, 1993.
3. Matthew de Brecht and Tatsuji Kawai. On the commutativity of the powerspace constructions. Logical Methods in Computer Science, 15(3), 2019.
4. Steven J. Vickers and Christopher F. Townsend. A universal characterization of the double powerlocale. Theoretical Computer Science 316(1-3), 2004, pages 297–321. http://dx.doi.org/10.1016/j.tcs.2004.01.034

— Jean Goubault-Larrecq (October 19th, 2022)