# Strongly compact sets and the double hyperspace construction

The notion of strongly compact set is due to Reinhold Heckmann . 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 ⊆ ↑EU. (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 Qfin(X) of finitary compact subsets of X, with the upper Vietoris topology, is the space Qs(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 (Ui)iI of Q is such that Q ⊆ ↑E ⊆ ∪iI Ui for some finite set E. Then E is included in the union of finite many of the sets Ui, 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 ↑Ex 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 Qfin(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 ≝ {QQ(X) | QU}, 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 ηX : X → Q(X) maps every point x of X to ↑x, and whose multiplication μX : 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 ηX–1(☐U)=U and μX–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 =☐(UV), UV 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 QU, ☐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 Qfin(X) of finitary compact subsets, but that is wrong. For example, you can check that, if X is T2, then Qfin(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 Qfin(X) is Qs(X), the subspace of Q(X) consisting of strongly compact saturated subsets of X.

Proof. Since Q(X) is sober, the sobrification of Qfin(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 Qfin(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 Qfin(X). In other words, there is a finitary compact set ↑E in ☐U ∩ ↓QQ. We expand this, and we obtain Q ⊆ ↑EU. 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 Qfin(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 ⊆ ↑EU; then ↑E is in ☐U, hence in U. ☐

One can then show that the monad (Q, η, μ) cuts down to a monad (Qs, 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 (Qs, 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 Qfin 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 .

## 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 ≝ {CQ(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 ηX : X → H(X) maps every point x of X to ↓x, and whose multiplication μX : H(H(X)) → H(X) maps every element C of H(H(X)) to cl(∪C). Also, ηX–1(♢U)=U and μX–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  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 Qs, I will show that QsHX and HQsX are homeomorphic for every topological space, consonant or not, sober or not.

Oh, by the way, if you read Heckmann , you will see that this commutation requires a property that he calls US-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 Qs. I will simply reorganize it, and mention how each step adapts when we replace Q by Qs.

## The first isomorphisms: QH=OpOσ, QsH=OpOp

The first step consists in showing that there is a homeomorphism between QHX and OpOσX, 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 OσX;
• or with what I will call the pointwise topology, whose subbasic open sets are [x ∈] ≝ {UOX | xU}, where x ranges over the points of X. I will write the resulting space OpX.

The homeomorphism is defined as follows:

• φX maps every QQHX to the collection of open subsets U of X that intersect every CQ;
• ψX maps every UOpOσX to the intersection of the sets ♢U, where U ranges over U.

Let us check this.

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

Proof. For every QQHX, we first verify that U ≝ φX(Q) is Scott-open. It is clearly upwards-closed. Given any directed family (Ui)iI of open subsets of X whose union is in U, by definition of U, every CQ intersects ∪iI Ui, hence is in some ♢Ui. Therefore the sets ♢Ui (iI) form an open cover of Q, which is also directed. Since Q is compact, Q is included in some ♢Ui. Therefore Ui intersects every CQ, so is in φX(Q)=U.

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

This shows, in particular, that φX is continuous.

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

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

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

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

The following modification (which does not appear in ) is concerned with the space QsHX of strongly compact saturated subsets of HX. Correspondingly, we replace OpOσX with OpOpX. (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 QsHX is a topological subspace of QHX, and that OpOpX is a topological subspace of OpOσX: 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 φX and ψX restrict (and corestrict) to homeomorphisms between QsHX and OpOpX. We have φX–1([U ∈]) = ☐sU, and ψX–1(☐sU) = [U ∈] for every open subset U of X.

Proof. We first show that φX maps every strongly compact set QQsHX to a pointwise open subset U ≝ φX(Q). To this end, let U be any element of U; we will find an open subset V of OpX such that UVU. By definition of φX, U intersects every CQ, so Q is included in ☐sU. Since Q is strongly compact, there are finitely many closed subsets C1, …, Cn of X such that Q ⊆ ↑H{C1, …, Cn} ⊆ ☐sU, where ↑H denotes upward closure in HX. Since ↑H{C1, …, Cn} ⊆ ☐sU, every Ci intersects U, say at xi. Then U is in V ≝ ∩i=1n[xi ∈]. It remains to show that V is included in U = φX(Q). For every VV, every point xi is in V, so V intersects every Ci, showing that ↑H{C1, …, Cn} ⊆ ♢V. Since Q ⊆ ↑H{C1, …, Cn}, every CQ also intersects V, so V is in φX(Q).

We turn to ψX. 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, …, xn}, then [C ∩] is the union of the sets [x1 ∈], …, [xn ∈]. Every open subset U of OpX can be written as ∪iIj=1ni [xij ∈]. 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 ∪iIj=1n [Cij ∩], where each Cij is closed in X.

A few technical but elementary computations show that ψX(U) is then equal to the filtered intersection ∩iIH{Ci1, …, Cini}. (Exercise!) Now let us assume that ψX(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{Ci1, …, Cini} is included in V. Since obviously ψX(U) is included in ↑H{Ci1, …, Cini}, we have shown that ψX(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:

• σX maps every CHQX to the collection of closed subsets C of X that intersect every QC.
• τX maps every QQHX to the collection of compact saturated subsets Q of X that intersect every CQ;

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

Let us study them one after the other. Items 2 and 3 below are Lemma 6.4 of . 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. φX o σX maps every CHQX to ∪QCQ;
2. for every open subset U of X, σX−1(☐♢U) = ♢☐U;
3. σX is a topological embedding.

Proof. 1. For every CHQX, ∪QCQ is a Scott-open subset of OX. For every closed subset C of X, C ∈ ψX (∪QCQ) if and only if C intersects every open set U that contains some QC. This is certainly the case if C intersects Q. Conversely, if C does not intersect Q, then UXC is an open set that does not intersect C. Therefore ψX (∪QCQ) = σX(C).

In particular, σX(C) is in QHX, showing that σX is well-defined, and φXX(C)) = ∪QCQ, since φX and ψX are mutually inverse.

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

Hence the only thing missing for σX 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 σX is surjective, namely if φX o σX is surjective, then X is consonant. Conversely, we have:

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

Proof. It suffices to show that φX o σX 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 QQX 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 QQX to be outside C.

In summary, we have built an element C of HQX. By item 1 of the previous lemma, φX o σX maps it to ∪QCQ, 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 (φX o σX) (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. σX is surjective;
3. σX is a homeomorphism of HQX onto QHX.

They also show that the map τX is such that τX−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, τX is continuous and is the inverse of σX.

## The isomorphism QsH=HQs

What happens if we replace Q with Qs 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 OpOσX by OpOpX, since now φX and ψX restrict to homeomorphisms between QsHX (instead of QHX) and OpOpX (instead of OpOσX). 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 OpX, not OσX) 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 OpX, or equivalently an element of OpOpX), 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 ⊆ ↑EU. Let us write the finite set E as {x1, …, xn}. Then U is in ∩i=1n[xi ∈], since each xi is in U, and ∩i=1n[xi ∈] is included in ■Q, since every open set that contains every xi must contain ↑E, hence also Q. This finishes to show that ■Q is an open subset of OpX, for every strongly compact saturated set Q. Hence every union of such sets is also open in OpX. Finitary consonance requires that those are the only open subsets of OpX.

We now define σXs just like σX: it maps every CHQsX (instead of HQX) to the collection of closed subsets C of X that intersect every QC. And we replay the proofs we did above:

Lemma. We have:

1. φX o σXs maps every CHQsX to ∪QCQ;
2. for every open subset U of X, (σXs)−1(☐sU) = ♢☐sU;
3. σXs is a topological embedding.

Proof. Just as before, ψX (∪QCQ) = σX(C), where now C is a closed subset of QsX. Since every QC is strongly compact saturated, we have seen that ∪QCQ is in OpOpX, so its image by ψX is in QsHX. This also shows φXX(C)) = ∪QCQ, 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 σXs 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 OpX. By definition, we can write U as ∪iIj=1ni [xij ∈]. But that is just ∪iIQi, where Qi ≝ ↑{xi1, …, xini}, and each Qi 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 QsHX, HQsX and OpOpX are homeomorphic; the first two, through σXs, and the first and the third through the pair φX, ψX.

I could say a lot more. For example, the constructions are all natural in X; σXs defines a distributive law, hence QsH 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)