When is ℒ(X) locally compact?

Let us write ℒ(X) for the dcpo of lower semicontinuous maps from a topological space X to R+ ∪ {∞}. Under what conditions on X is it locally compact?

The question occurred while I was looking for a space X such that ℒ(X) is not locally compact, aiming at a counterexample. I have tried to take X ≝ Q, the space of rational numbers with its usual metric topology, and I have not reached a conclusion. But, if X is a poset with its Scott topology, we will see that there is a complete answer to the question. The proof is particularly short and clever, and is due to Xiaodong Jia. If X is an ⊙-consonant second countable space (I will give definitions below), we will also see that there is a complete answer to the question. This is essentially a consequence of results of Matthew de Brecht and Tatsuji Kawai [2], and that is way more complicated.

The space ℒ(X)

My definition of ℒ(X) above is a bit sketchy. Let us give it in full.

I will consider the dcpo R+ ∪ {∞}, obtained by adding a new element ∞ above all non-negative reals numbers. Some would write +∞, but Klaus Keimel once convinced me that if one never has to talk about –∞, one might as well simply write ∞.

We give R+ ∪ {∞} the Scott topology of ≤. Its open subsets, apart from the empty set and the whole space, are the intervals ]t, ∞], where t ranges over R+. A lower semicontinuous map from a space X to R+ ∪ {∞} is then simply a continuous map from X to R+ ∪ {∞}, where the latter has the Scott topology.

We equip ℒ(X) itself with the Scott topology of an ordering, and that ordering is the pointwise ordering: f≤g if and only if f(x)≤g(x) for every x ∈ X. ℒ(X) is a very nicely behaved space provided that X is core-compact. By this, I mean:

  • ℒ(X) is a complete lattice: suprema of arbitrary families are computed pointwise, and once you have suprema of arbitrary families, then you also have infima of arbitrary families (the infimum of a family of lower semicontinuous maps is the supremum of the family of its lower bounds).
  • If X is core-compact, namely if its lattice O(X) of open subsets is a continuous dcpo (Definition 5.2.3 in the book), then ℒ(X) is a continuous complete lattice. In fact, the dcpo [X → Y] of all continuous maps from a core-compact space X to a continuous complete lattice (such as R+ ∪ {∞}) is itself a continuous complete lattice. This can be found as Theorem II-2.12 of the red book [1], for example.
  • Every continuous complete lattice is stably compact in its Scott topology (see Fact 9.1.6 in the book). Hence ℒ(X) is stably compact in its Scott topology, for every core-compact space X, and is in particular, locally compact, hence core-compact.

What if ℒ(X) is locally compact?

I was looking at an example of a space X for which ℒ(X) would be badly behaved. Because of what I have just mentioned, we should look for a non-core-compact space. Instead of looking for a simple example, we will show that, if X is a poset with its Scott topology, then ℒ(X) is locally compact if and only if X is core-compact. Hence any non-core-compact poset X will fit.

We first reduce the question to the study of the dcpo O(X) of open subsets of X. I will also write [X → L] for the dcpo of all continuous maps from a space X to an arbitrary dcpo L, where L is given its Scott topology, so that ℒ(X) = [X → R+ ∪ {∞}]. Finally, I will write Pσ for P with its Scott topology, for every poset P.

Lemma. For every space X, O(X)σ occurs as a retract of ℒ(X)σ. More generally, for every poset L with at least two elements a<b, O(X)σ occurs as a retract of [X → L]σ.

Proof. The first claim follows from the second by taking L ≝ R+ ∪ {∞}, which clearly contains at least two elements a and b with a<b. In order to show the second claim, we pick two elements a and b from L such that a<b. Since the Scott topology is T0 and since ≤ is its specialization ordering (Proposition 4.2.18 in the book), and since b≰a, there is a Scott-open subset V of L that contains b but not a.

For every open subset U of X, let (U ↦ b else a) denote the function from X to L that maps every element of U to b, and all others to a. This is in [X → L], because the inverse image of an arbitrary Scott-open subset W of L is empty if W does not contain b (and therefore does not contain a, since a≤b), U if W contains b but not a, and the whole of X if W contains a (and therefore also b).

We define the section s : O(X)σ → [X → L]σ as mapping every open subset U of X to the function (U ↦ b else a). It is easy to see that this is a (Scott-)continuous map.

We define the retraction r : [X → L]σ → O(X)σ as mapping every f ∈ [X → L] to f–1(V); remember, we have chosen V as a fixed open set that contains b but not a. This is well-defined: f–1(V) is open in X for every f ∈ [X → L], because V is open and every such f is continuous by definition. The function r is also Scott-continuous. Notably, if (fi)i ∈ I is any directed family in [X → L], with (pointwise) supremum f, we show that r(f) = supi ∈ I r(fi) as follows. The supremum on the right-hand side is a union, and specifically, ∪i ∈ I fi–1(V). The left-hand side if f–1(V). Now a point x of X is in f–1(V) if and only if f(x) = supi ∈ I fi(x) is in V, and since V is Scott-open (this is where we use that L is a poset with the Scott topology), that is equivalent to the existence of an index i in I such that fi(x) is in V, namely to x ∈ ∪i ∈ I fi–1(V); and this completes the argument that r is (Scott-)continuous.

For every open subset U of X, r(s(U)) = r(U ↦ b else a) = (U ↦ b else a)–1(V). Since V contains b but not a, this is equal to U. Hence r and s form a retraction.

Retractions preserve many topological properties, including local compactness (Proposition 4.10.33 or Proposition 9.2.3 in the book). ☐

Corollary. For every space X, if ℒ(X)σ is locally compact, then so is O(X)σ.

The following is well-known, but let me include a proof.

Lemma. Every retract of a core-compact space is core-compact.

Proof. Let r : Y → X be a retraction with associated section s : X → Y, and let us assume Y core-compact. For every point x of X and every open neighborhood U of X, r–1(U) is an open neighborhood of s(x) in Y. Since Y is core-compact, there is an open neighborhood V of s(x) in Y such that V ⋐ r–1(U), where ⋐ is the way-below relation between open sets. Then s–1(V) is an open neighborhood of x in X, and it remains to observe that s–1(V) ⋐ U. For every directed family (Ui)i ∈ I of open subsets of X whose union contains U, (r–1(Ui))i ∈ I is a directed family of open subsets of Y whose union contains r–1(U). Since V ⋐ r–1(U), there is an index i in I such that V ⊆ r–1(Ui). Then s–1(V) ⊆ s–1(r–1(Ui)) = Ui, using the fact that r o s is the identity map. ☐

Corollary. For every space X, if ℒ(X)σ is core-compact, then so is O(X)σ.

Local compactness and core-compactness

I made a subtle distinction above between the cases where ℒ(X)σ (resp., O(X)σ) is locally compact and those where it is core-compact. But those are exactly the same situations!

Indeed, both ℒ(X) and O(X) are complete lattices. In the June 2017 post, we have seen that every complete lattice is well-filtered in its Scott topology (the Xi-Lawson theorem), and in the October 2019 post, we have seen that every core-compact well-filtered space is locally compact (a theorem due to Xu, Shen, Xi and Zhao… and, independently, to Lawson and Xi). Hence any core-compact complete lattice is locally compact. Since, conversely, every locally compact space is core-compact (Theorem 5.2.9 in the book), we obtain the following.

Fact. Every complete lattice, and notably ℒ(X) and O(X) for any space X, is core-compact if and only if it is locally compact (in its Scott topology).

What if O(X) is locally compact? The case of posets

I don’t know whether the implication “O(X)σ locally compact ⇒ X core-compact” or the implication “O(X)σ core-compact ⇒ X core-compact” holds in general. Any of these implications would be very surprising. For example, if the first one holds, then we remember that the fact that X is core-compact means that O(X) is a continuous dcpo. But it is much stronger to require a dcpo to be continuous than to require it to be locally compact in its Scott topology. Nevertheless, those implications do hold if X is a poset with its Scott topology.

The clever (and short!) proof is due to Xiaodong Jia.

Theorem A. Given a poset P, the following are equivalent:

  1. Pσ is core-compact;
  2. O(Pσ) is a continuous dcpo;
  3. O(Pσ)σ is locally compact;
  4. O(Pσ)σ is core-compact.

Proof. By definition of core-compactness (Definition 5.2.3 in the book), 1 and 2 are equivalent. 2 implies 3, since every continuous dcpo is locally compact in its Scott topology (Corollary 5.1.36 in the book) and 3 implies 4, since every locally compact space is core-compact (Theorem 5.2.9 in the book). We have just seen that 3 is in fact equivalent to 4, but it is equally easy to show that 4 implies 1, directly. We do this by using the following result, which I am quoting from the March 20th, 2022 post:

Proposition. Let X be a poset. Then Xσ is core-compact if and only if the Scott and product topologies coincide on X × Y for every poset Y (resp., for every complete lattice Y, resp. for the special complete lattice O(Xσ)).

Hence, if O(Pσ)σ is core-compact, then the Scott and product topologies coincide on O(Pσ) × Y for every poset Y. In particular, they coincide on O(Pσ) × P. By permuting the two sides of ×, the Scott and product topologies coincide on P × O(Pσ). But the above proposition (or rather, the last “resp.”—which can also be found as Exercise 5.2.7 in the book) says that, then, Pσ is core-compact. ☐

Putting everything together, we know that:

  • For every space X, if X is core-compact then ℒ(X)σ is locally compact (even stably compact).
  • For every space X, if ℒ(X)σ is locally compact (resp. core-compact) then so is O(X)σ.
  • For every poset P, if O(Pσ)σ is core-compact, then so is Pσ.

Hence:

Theorem B. For every poset P, the following are equivalent:

  1. Pσ is core-compact;
  2. O(Pσ)σ is core-compact (equivalently, locally compact);
  3. ℒ(X)σ is core-compact (equivalently, locally compact);
  4. ℒ(X)σ is stably compact.

In particular, for P one may take Johnstone space J. This is a dcpo that is not core-compact in its Scott topology (Exercise 5.2.15 in the book). Hence ℒ(Jσ)σ is not locally compact and not core-compact.

What if O(X) is locally compact? The case of ⊙-consonant, second-countable spaces

By putting together a few results from [2] (a brilliant paper full of clever ideas), we will show that if X is ⊙-consonant and second-countable, then the fact that O(X)σ is locally compact implies that X is locally compact. This goes through an amazing list of detours. We will need to talk about the following (think of it as a table of contents):

  1. ⊙-consonant spaces and the Hoare hyperspace;
  2. co-consonant spaces;
  3. compact subsets of co-consonant spaces;
  4. the intersection map;
  5. retracting Q(OσX) onto OσX.

1. ⊙-consonant spaces and the Hoare hyperspace

Well, I will start by something that is not in [2], but I promise that everything else below is from [2]. M. de Brecht and T. Kawai work on quasi-Polish spaces. Then, for every quasi-Polish space X, H(X) is quasi-Polish (I will not prove it), hence consonant. We will show, by a direct proof, that H(X) is consonant for every ⊙-consonant space. But I am going too fast, and I haven’t defined H(X) or ⊙-consonance; I haven’t even recalled what consonance is.

Let me write ■Q, for every compact saturated subset Q of a space X, for the set of open neighborhoods of Q. 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.

It turns out that the coproduct of two consonant spaces may fail to be consonant. A notion I have used now and then that fixes this issue in applications is ⊙-consonance. For every space X and every natural number n, let n ⊙ X denote the nth copower of X, namely the coproduct of n copies of X. A space is ⊙-consonant if and only if all its copowers n ⊙ X are consonant. Every LCS-complete space is ⊙-consonant [3, Lemma 13.2]; an LCS-complete space is a space that is homeomorphic to a Gδ-subspace of a locally compact sober space [3], and that includes all of M. de Brecht’s quasi-Polish spaces. (I have not yet really talked about LCS-complete spaces here, and you will have to read [3] if you want to understand more, sorry.)

Let us consider the Hoare hyperspace H(X) of a topological space. I have already talked about this in the March 2015 post and a few others. This is the space of all closed subsets of X, with the lower Vietoris topology, generated by subbasic open sets ♢U ≝ {C closed subset of X | C intersects U}, where U ranges over the open subsets of X.

Let us also use the notation ♢U even when U is not open. For example, for every compact saturated subset Q of X, ♢Q is the collection of closed subsets of X that intersect Q.

Lemma C. Let X be a topological space. For all compact saturated subsets Q1, …, Qn of X, ♢Q1 ∩ … ∩ ♢Qn is compact saturated in H(X).

Proof. The set ♢Q1 ∩ … ∩ ♢Qn is clearly saturated. We use Alexander’s subbase lemma (Theorem 4.4.29 in the book) in order to show that it is compact. Let (♢Uj)j∈J be an open cover of ♢Q1 ∩ … ∩ ♢Qn by subbasic open sets ♢Uj. Let C0 be the complement of ∪j∈J Uj. C0 is a closed set which is not in ∪j∈J ♢Uj, so C0 is not in ♢Q1 ∩ … ∩ ♢Qn either. Hence C0 does not intersect Qi, for some i ∈ {1, …, n}. That Qi must then be included in the complement of C0, which is ∪j∈J Uj. Since Qi is compact, there is a finite subset J′ of J such that Qi ⊆ ∪j∈J’ Uj. Every element C of ♢Q1 ∩ … ∩ ♢Qn intersects Qi, hence Uj for some j ∈ J′. Therefore ♢Q1 ∩ … ∩ ♢Qn is included in ∪j∈J’ ♢Uj. ☐

Lemma D. Let X be a topological space. For all subsets (compact, open, or not) Q1, …, Qn, and for all open subsets V1, …, Vm of X, ♢Q1 ∩ … ∩ ♢Qn ⊆ ♢V1 ∩ … ∩ ♢Vm if and only if for every j ∈ {1, …, m}, there is an i ∈ {1, …, n} such that Qi ⊆ Vj.

Proof. I will not repeat the domain of variation of the indices j and i. If for every j, there is an i such that Qi ⊆ Vj, then ♢Qi ⊆ ♢Vj (the ♢ operator is monotonic), so ♢Q1 ∩ … ∩ ♢Qn ⊆ ♢Vj, and since that holds for every j, ♢Q1 ∩ … ∩ ♢Qn ⊆ ♢V1 ∩ … ∩ ♢Vm. Conversely, we assume the latter inclusion. For every j, we have ♢Q1 ∩ … ∩ ♢Qn ⊆ ♢Vj. Let C be the complement of Vj. This is a closed subset of X that is not in ♢Vj, hence not in ♢Q1 ∩ … ∩ ♢Qn. Hence there is an i such that C is not in ♢Qi, namely such that C does not intersect Qi, or equivalently, Qi ⊆ Vj. ☐

The proof of the following uses a classical trick in domain theory, which already used by Chen, Kou and Lyu, for example (see the March 2022 post).

Proposition E. For every ⊙-consonant space X, H(X) is consonant.

Proof. Let W be a Scott-open collection of open subsets of H(X). (Yes, that is quite a tower of sets: a Scott-open collection of open subsets of a space of closed sets.) Let also U be an open subset of H(X) that lies in W. We wish to find a compact saturated subset Q of H(X) such that U ∈ ■Q ⊆ W.

We first write U as a union ∪i∈I ∩j=1ni ♢Uij of finite intersections of subbasic open sets. We can always write a union as a directed union of finite unions (Trick 4.4.6 in the book), and since U is in W, which is Scott-open, a finite union ∪i∈I‘ ∩j=1ni ♢Uij (namely, with I’ finite and included in I) is already in W. We distribute unions over intersections, and realizing that the ♢ operator commutes with unions, we realize that we can rewrite the latter as just a finite intersection of subbasic open sets. Hence, without loss of generality, we may assume that U is a finite intersection ♢U1 ∩ … ∩ ♢Un of subbasic open subsets of H(X).

Since X is ⊙-consonant, n ⊙ X is consonant. Its open subsets are exactly the sets V1 + … + Vn, where each Vi ranges over the open subsets of X, and the notation V1 + … + Vn denotes the disjoint union of V1, …, Vn; formally, the collection of pairs (i, x) with 1≤i≤n and x ∈ Vi.

Let f : O(n ⊙ X) → O(H(X)) be the function that maps every open subset V1 + … + Vn of n ⊙ X to ♢V1 ∩ … ∩ ♢Vn. Using the fact that ♢ commutes with directed unions (in fact, all unions), and that ∩ does, too, we see that f is Scott-continuous in each of its arguments separately; and therefore, Scott-continuous.

Let W’ ≝ f–1(W). Since f is Scott-continuous, W’ is a Scott-open family of open subsets of n ⊙ X. Additionally, U1 + … + Un is in W’. Since n ⊙ X is consonant, there is a compact saturated subset Q of n ⊙ X such that Q ⊆ U1 + … + Un and such that every open neighborhood of Q is in W’.

I claim that Q is of the form Q1 + … + Qn where each Qi is compact saturated in X. Each copy of X in n ⊙ X is open. It is also closed, because it is the complement of the union of all the other copies. Hence the intersection of Q with each copy of X is compact, too. Let us call Qi the intersection of Q with copy number i of X, and let us equate it with the corresponding compact saturated subset of X. (Explicitly, there is a map x ↦ (i, x) that embeds X into n ⊙ X as its copy number i. This is a topological embedding, hence a homeomorphism onto its image, and therefore Qi, which is the image of Q by the inverse map, is compact.) It is easy to see that Qi is upwards-closed, and that Q = Q1 + … + Qn.

By Lemma C, ♢Q1 ∩ … ∩ ♢Qn is compact saturated in H(X). We let Q ≝ ♢Q1 ∩ … ∩ ♢Qn. Since Q = Q1 + … + Qn ⊆ U1 + … + Un, each Qi is included in the corresponding Ui, so (using the fact that ♢ is monotonic) ♢Q1 ∩ … ∩ ♢Qn ⊆ ♢U1 ∩ … ∩ ♢Un = U. In other words, U ∈ ■Q.

It remains to show that ■Q ⊆ W. For every V in ■Q, namely for every open subset V of H(X) that contains Q, we write V as a union of finite intersections of subbasic open sets, hence as a directed union of finite unions of finite intersections of subbasic open sets (as we did with U). As with U, we can rewrite finite unions of finite intersections of subbasic open sets as just finite intersections of subbasic open sets. Since Q is compact, we obtain that there is a finite intersection ♢V1 ∩ … ∩ ♢Vm of subbasic open sets such that Q ⊆ ♢V1 ∩ … ∩ ♢Vm ⊆ V. Since Q ⊆ ♢V1 ∩ … ∩ ♢Vm, namely ♢Q1 ∩ … ∩ ♢Qn ⊆ ♢V1 ∩ … ∩ ♢Vm, we obtain that for every j ∈ {1, …, m}, there is an i ∈ {1, …, n} such that Qi ⊆ Vj, by Lemma D. For each j, we pick such an i and call it g(j). Hence Qg(j) ⊆ Vj for every j. For every i, let V’i be the intersection of all the open sets Vj such that g(j)=i. Hence Qi ⊆ V’i for every i. It follows that Q = Q1 + … + Qn ⊆ V’1 + … + V’n. Every open neighborhood of Q is in W’, so V’1 + … + V’n is in W’. By definition of W’, ♢V’1 ∩ … ∩ ♢V’n is in W. But, for each i, ♢V’i = ♢(∩j / g(j)=i Vj) ⊆ ∩j / g(j)=i ♢Vj: any closed set that intersects an intersection such as ∩j / g(j)=i Vj must intersect every Vj. Since W is upwards-closed (being Scott-open), ∩j / g(j)=1 ♢Vj ∩ … ∩ ∩j / g(j)=n ♢Vj is in W. In other words, ♢V1 ∩ … ∩ ♢Vm is in W, and therefore the even larger set V is in W. Since V is arbitrary in ■Q, we have shown the desired inclusion ■Q ⊆ W. ☐

2. Co-consonant spaces

Let OσX abbreviate O(X)σ. We also consider OpX, which is also the space of open subsets of X, but with the pointwise topology, generated by subbasic open sets [x ∈] ≝ {U ∈ O(X) | x ∈ U}, where x ranges over X.

After de Brecht and Kawai [2, Definition 7.1], let us say that a space Y is co-consonant if and only if the Scott and pointwise topologies coincide on O(Y), namely if OpY = OσY.

Proposition F. For every second-countable space X, O(X)σ is co-consonant.

Proof. This is one half of Proposition 7.2 of [2], which also says that O(X)σ is consonant; but we will not need that.

For every space Y, let Q(Y) stands for the collection of compact saturated subsets of a space Y, with the upper Vietoris topology on Q(Y). This topology is generated by basic open sets ☐U ≝ {Q ∈ Q(Y) | Q ⊆ U}, where U ranges over the open subsets of Y. Its specialization ordering is reverse inclusion ⊇. In the December 2018 post, we have stated one of de Brecht and Kawai’s theorems as follows (changing the name of the arbitrary space to Y in order to avoid confusion).

Proposition (de Brecht, Kawai)  If Y is well-filtered and second countable, then the upper Vietoris and Scott topologies coincide on Q(Y).

The same conclusion would hold if we assumed Y well-filtered and locally compact; but we wish to set Y ≝ H(X), and we don’t know whether H(X) is locally compact, only O(X)σ.

Now, since X is second-countable, then Y ≝ H(X) is also second-countable in its lower Vietoris topology. Indeed, let B be a countable base of the topology of X. Since the ♢ operator commutes with arbitrary unions, every subbasic open set ♢U can be written as a union of sets ♢V where V is in B, so that those (countably many) sets ♢V form a subbase of the lower Vietoris topology. H(X) is also sober in its lower Vietoris topology: this is a theorem due to Andrea Schalk, which we have seen in the March 2015 post. It is in particular well-filtered, so the above proposition by de Brecht and Kawai tells us that the upper Vietoris and Scott topologies coincide on Q(H(X)) (where the inner H(X) is given its lower Vietoris topology).

We now use the following result from the October 2022 post, which also comes from [2], and applies to any topological space X at all:

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

The function φX maps every Q ∈ Q(H(X)) to the collection of open subsets U of X that intersect every C ∈ Q; ψX maps every U ∈ OpOσX to the intersection of the sets ♢U, where U ranges over U.

The important point is not the definition of φX and ψX, rather that Q(H(X)) (with its upper Vietoris topology) and OpOσX are homeomorphic through φX and ψX. In particular, Q(H(X)) and O(OσX) are isomorphic as posets. Therefore they are homeomorphic when given their Scott topologies, still through φX and ψX. But the Scott and upper Vietoris topologies coincide on Q(H(X)), as we have seen above, so the Scott and pointwise topologies coincide on O(OσX): OpOσX = OσOσX. And that means precisely that OσX is co-consonant. ☐

3. Compact subsets of co-consonant spaces

A subset Q of a space Y 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.) The October 2022 post was about strongly compact sets, and we have seen there that every strongly compact subset is compact. We wrote Qs(X) for the subspace of Q(X) consisting of strongly compact (saturated) subsets of X, and we have noticed that if X is locally finitary compact, then every compact saturated subset of X is strongly compact, namely Q(X) = Qs(X). Here is another case.

Lemma G [2, Proposition 7.4]. Every compact subset Q of a co-consonant space Y is strongly compact.

Proof. Let U be any open neighborhood of the compact set Q. Then U is in ■Q, by definition of ■Q. Now U is in O(Y), and ■Q is Scott-open in O(Y), hence is an element of O(OσY). Since Y is co-consonant, by definition, OσY = OpY. Therefore, there is a basic open neighborhood [x1 ∈] ∩ … ∩ [xn ∈] of U in the pointwise topology that is included in ■Q.

Let E ≝ {x1, …, xn}. We verify that Q ⊆ ↑E ⊆ U. The second inclusion is easy: every point xi of E is in U, because U ∈ [xi ∈], so E  ⊆ U, and therefore ↑E ⊆ U, since U is upwards closed. For the first inclusion, we consider an arbitrary point x in Q. The set V ≝ Y–↓x is open, and not in ■Q: indeed, being in ■Q would mean Q ⊆ Y–↓x, which is impossible since x is in the left-hand side but not in the right-hand side. But [x1 ∈] ∩ … ∩ [xn ∈] ⊆ ■Q, so V is not in [x1 ∈] ∩ … ∩ [xn ∈] either. Hence, for some i with 1≤i≤n, V is not in [xi ∈], that is to say, xi is not in V. Since V = Y–↓x, xi is in ↓x, namely, xi≤x. We have just shown that x is in ↑E. As x is arbitrary in Q, Q ⊆ ↑E. ☐

4. The intersection map

Let X be any topological space. For every Q ∈ Q(OσX), we write ∩Q for the intersection of all the elements U of Q, which are all open subsets of X.

Proposition H. For every ⊙-consonant second countable space X, there is a continuous function ∩ : Q(OσX) → OσX that maps every Q ∈ Q(OσX) to ∩Q. (The domain Q(OσX) has the upper Vietoris topology, the codomain OσX has the Scott topology, as the σ subscript indicates.)

Proof. This is Proposition 9.6 of [2], except that de Brecht and Kawai assume X to be quasi-Polish. But their proof only uses second countability and the fact that H(X) is consonant, which we will deduce from Proposition E.

We first show that, for every Q ∈ Q(OσX), γ(Q) ≝ {C ∈ H(X) | C intersects every element U of Q} is open in H(X) (with the lower Vietoris topology). Let C be any element of γ(Q). Since C ∈ γ(Q), every element U of Q intersects C, hence is in the subset V ≝ ∪x ∈ C [x ∈] of O(X). V is not just open in the pointwise topology, but also in the Scott topology, because the Scott topology is finer than the pointwise topology. (We are not using co-consonance here.) We have just shown that Q ⊆ V. By Lemma C, OσX is co-consonant, so by Lemma G with Y ≝ OσX, Q is not just compact, but even strongly compact in OσX. Therefore Q ⊆ ↑{U1, …, Un} ⊆ V for some finite family of open subsets U1, …, Un of X (points of Y). Then:

  • C is in ♢U1 ∩ … ∩ ♢Un, which is open in H(X). Indeed, each Ui is in ↑{U1, …, Un}, hence in V, and since V = ∪x ∈ C [x ∈], there must be a point x of C such that Ui ∈ [x ∈], namely that is also in Ui.
  • ♢U1 ∩ … ∩ ♢Un ⊆ γ(Q). Indeed, let us consider any closed subset C’ of X that is in ♢U1 ∩ … ∩ ♢Un, namely which intersects every Ui. In order to show that C’ is in γ(Q), we consider any element U of Q, and we will show that C’ and U intersect. Since Q ⊆ ↑{U1, …, Un}, and U is in Q, U must be above some Ui, namely Ui ⊆ U. But C’ intersects Ui, so it intersects U.

This shows that γ(Q) is an open neighborhood of each of its elements C, hence that it is open.

This defines a map γ : Q(OσX) → Oσ(H(X)), and we claim that it is continuous. Since X is ⊙-consonant, H(X) is consonant, by Proposition E. Hence every Scott-open subset of Oσ(H(X))—every Scott-open family of open subsets of H(X)—is a union of sets of the form ■K, where K is compact saturated in H(X). Therefore, it suffices to show that γ–1(■K) is open in the upper Vietoris topology on Q(OσX). Now the elements of γ–1(■K) are exactly the compact saturated subsets Q of OσX such that K ⊆ γ(Q), namely such that every C in K intersects every U in Q. Do you remember φX? We have used it in the proof of Proposition F: φX(K) is the collection of open subsets U of X that intersect every C in K, and φX is a (continuous map) from Q(H(X)) to OpOσX. Well, γ–1(■K) is the collection of elements Q of Q(OσX) such that every U in Q intersects every C in K, namely such that Q ⊆ φX(K). Equivalently, γ–1(■K) = ☐φX(K), which is therefore open in Q(OσX).

Having shown this, we realize that the map η : X → H(X) which sends every point x to ↓x is continuous: the inverse image of the subbasic open set ♢U is U. (This is the unit of the H monad.) Hence there is a function η–1 : O(H(X)) → O(X) that maps every open subset of H(X) to its inverse image under η. It is a frame homomorphism, and is in particular Scott-continuous, but I will not use that (de Brecht and Kawai do).

For every Q in Q(OσX), η–1(γ(Q)) is in O(X) by construction, and we compute what it is as follows: η–1(γ(Q)) is the collection of points x in X such that ↓x intersects every element U of Q, equivalently such that x is in every element U of Q. Therefore η–1(γ(Q)) = ∩Q, which shows that ∩Q is open in X, as desired. Additionally, ∩ is the composition of the two continuous maps η–1 and γ, and is therefore continuous. ☐

5. Retracting Q(OσX) onto OσX

It turns out that γ has an inverse if X is quasi-Polish, but let me diverge slightly from [2] and conclude slightly more quickly.

Theorem I. For every ⊙-consonant second-countable space X, the intersection map ∩ : Q(OσX) → OσX is a retraction, with section s mapping every open subset U of X to ↑OσXU.

Proof. The map s is really the unit of the Q monad. Explicitly, for every space Y (here, OσX), this unit is a map from Y to Q(Y), which maps every point y of Y to ↑y. The inverse image of any basic open set ☐V of Q(Y) (where V is open in Y) is just V. Hence s is well-defined and continuous. Proposition H tells us that ∩ is also well-defined and continuous, and for every open subset U of X, ∩(s(U)) is the intersection of all the open subsets of X that contain U, which is U itself. Hence ∩ o s = id. ☐

Theorem I has some family resemblance with the Hofmann-Mislove theorem. Instead of considering a Scott-open filter of open sets, we consider a compact saturated set of open sets.

Let us return to our initial problem: what happens if OσX is locally compact?

Well, we know from Proposition 8.3.25 and Lemma 8.3.26 in the book that, for every locally compact well-filtered space Y, Q(Y) has the Scott topology of the reverse inclusion ordering ⊇, and is a continuous dcpo. We note that OσX is well-filtered, because it is a complete lattice in its Scott topology (we have already mentioned the Xi-Lawson theorem; see the June 2017 post). Hence, if OσX is locally compact, then Q(OσX) is a continuous dcpo and has the Scott topology of ⊇.

By Corollary 8.3.37 in the book, every retract of a continuous dcpo (with its Scott topology) is a continuous dcpo (with the Scott topology). Hence Theorem I immediately implies that a ⊙-consonant second-countable space X such that OσX is locally compact is such that OX is a continuous dcpo; in other words, X is core-compact. This shows the implication 2 ⇒ 1 of the following proposition. The other implications were established at the beginning of this post.

Proposition J. For every ⊙-consonant second-countable space X, the following are equivalent:

  1. X is core-compact;
  2. O(X)σ is core-compact (equivalently, locally compact);
  3. ℒ(X)σ is core-compact (equivalently, locally compact);
  4. ℒ(X)σ is stably compact.

As an application, we consider Baire space NN. This is a consonant space (Exercise 8.3.4 in the book) that is not core-compact (Example 5.2.14 in the book). It is even ⊙-consonant, which we see as follows. NN is Polish (Exercise 7.7.6 in the book), hence so is any finite copower of NN (by Lemma 7.7.4 in the book). Every Polish space is second countable, T3 and Čech-complete (Exercise 7.7.10 in the book), and by the Dolecki-Greco-Lechicki theorem (Exercise 8.3.4 in the book) every T3 Čech-complete space is consonant; so every finite copower of NN is consonant, showing that NN itself is ⊙-consonant. Since NN is Polish, it is also second countable. Hence Proposition J applies. Since NN is not core-compact, we obtain:

Corollary K. O(NN)σ, ℒ(NN)σ are not core-compact.

I will stop here. It is frustrating that neither Proposition J nor Theorem B (on posets) allows us to conclude about whether O(X)σ or ℒ(X)σ is core-compact when X is the Sorgenfrey line Rℓ or the space of rational numbers Q:

  • These two spaces are Hausdorff, hence sober (Proposition 8.2.12 in the book), and for a sober space core-compactness and local compactness are equivalent (Theorem 8.3.10 in the book). Neither Rℓ nor Q is locally compact (see Exercise 4.8.5, resp. Exercise 4.8.4 in the book), hence none is core-compact.
  • But Theorem B does not allow us to conclude anything, since none of them is a poset in its Scott topology, and Proposition J does not apply either, because neither Rℓ nor Q is consonant: see the March 2021 post for Rℓ, and the February 2023 post for Q (or the June 2022 post for a more complex argument).
  • And Rℓ is not even second countable, see Exercise 6.3.10 in the book.
  1. Gerhard Gierz, Karl Heinrich Hofmann, Klaus Keimel, Jimmie D. Lawson, Michael W. Mislove, and Dana S. Scott. Continuous Lattices and Domains. Number 93 in Encyclopedia of Mathematics and its Applications. Cambridge University Press, Cambridge, 2003.
  2. Matthew de Brecht and Tatsuji Kawai. On the commutativity of the powerspace constructions. Logical Methods in Computer Science, 15(3), 2019.
  3. Matthew de Brecht, Jean Goubault-Larrecq, Xiaodong Jia, Zhenchao Lyu. Domain-complete and LCS-complete spaces. Proceedings of the 8th ISDT, Electronic Notes in Theoretical Computer Science 345:3–35, 2019.
jgl-2011

— Jean Goubault-Larrecq (November 21st, 2024)