Choquet-Wilker spaces

There is a natural property of topological spaces, which I initially learned from [1, Discussion before Lemma 6.2], and which Klaus Keimel and Jimmie Lawson call Wilker’s condition, after Peter Wilker [2, Theorem 3]. Before I explain what it is, let me note that the property had been known for much longer [3, Definition 17.3]. Gustave Choquet says that a family K of subsets of a topological space X is rich if and only if:

for all open subsets U₁ and U₂ of X, for every Q ∈ K such that Q ⊆ U₁ ∪ U₂,
there are elements Q₁, Q₂ ∈ K such that
Q₁ ⊆ U₁, Q₂ ⊆ U₂ and Q ⊆ Q₁ ∪ Q₂.

He then usually applies this condition to the family of compact subsets of X. This is a very natural condition that is used to build measures on topological spaces: for example, see [5, Section 53]. Wilker’s condition is simply that the collection QX of all compact saturated subsets of X be a rich family. Hence let me call spaces satisfying this condition Choquet-Wilker spaces.

Definition. A Choquet-Wilker space is a space X such that the collection QX is rich.

My purpose is not to explain how to build measures on Choquet-Wilker spaces, rather to show how large the class of Choquet-Wilker spaces is. Hence I will exhibit large families of spaces that are Choquet-Wilker, and I will give one example of a space that is not. I will end with a neat characterization of those consonant spaces that are Choquet-Wilker, due to Matthew de Brecht and Klaus Keimel.

Locally compact spaces

Proposition A. Every locally compact space X is Choquet-Wilker. X even satisfies the following stronger condition:

for all open subsets U₁ and U₂ of X, for every Q ∈ K such that Q ⊆ U₁ ∪ U₂,
there are elements Q₁, Q₂ ∈ K such that
Q₁ ⊆ U₁, Q₂ ⊆ U₂ and Q ⊆ int(Q₁) ∪ int(Q₂),

where int( ) denotes interior.

Proof. Let Q be a compact saturated subset of X included in a union U₁ ∪ U₂ of two open subsets of X. For every x ∈ Q, if x ∈ U₁, we pick a compact saturated neighborhood Q_x of x included in U₁; otherwise, x is in U₂, and we pick a compact saturated neighborhood, which we also call Q_x, of x included in U₂. The open sets int(Q_x) cover Q as x ranges over Q.

Since Q is compact, we can extract a finite subcover (int(Q_x))_x∈E, where E is a finite subset of Q. Let E₁ ≝ E ∩ U₁ and E₂ ≝ E–E₁ consist of the remaining elements. Let also Q₁ ≝ ∪_x∈E1 Q_x and Q₂ ≝ ∪_x∈E2 Q_x. Then:

• Q₁ ∪ Q₂ = ∪_x∈E Q_x contains Q. In fact, we claim that the required stronger condition Q ⊆ int(Q₁) ∪ int(Q₂) holds. Every point of Q is in int(Q_x) for some x ∈ E; if x ∈ E₁ then int(Q_x) ⊆ int(∪_x∈E1 Q_x) = int(Q₁), and otherwise, int(Q_x) ⊆ int(∪_x∈E2 Q_x) = int(Q₂).
• For every x ∈ E₁, x is in U₁, so Q_x ⊆ U₁ by definition. Hence Q₁ ≝ ∪_x∈E1 Q_x ⊆ U₁.
• For every x ∈ E₂, x is in U₂, so Q_x ⊆ U₂ by definition. Hence Q₂ ≝ ∪_x∈E2 Q_x ⊆ U₂. ☐

LCS-complete spaces

I have already mentioned LCS-complete spaces in the August 2025 post and in the November 2024 post. An LCS-complete space is a topological space X that is homeomorphic to a G_δ subspace of a locally compact sober space Y. Every locally compact sober space is LCS-complete, but the class is larger. Most notably, every quasi-Polish space is LCS-complete, and in particular every Polish space (even not locally compact, such as Baire space N^N) is LCS-complete.

The following is Theorem 8.2 in [4].

Proposition B. Every LCS-complete space X is Choquet-Wilker.

Proof. Up to isomorphism, we can consider that X is a G_δ subspace of a locally compact sober space Y. Being G_δ, we can write X as the intersection ∩_n∈N W_n of a descending sequence of open subsets W₀ ⊇ W₁ ⊇ … ⊇ W_n ⊇ … of Y.

Let Q be a compact saturated of X included in a union U₁ ∪ U₂ of two open subsets of X. We write U₁ as Û₁ ∩ X and U₂ as Û₂ ∩ X, where Û₁ and Û₂ are open in Y.

The notion of compactness is absolute, in the sense that a subset of X is compact in X (with the subspace topology), if and only if it is compact in Y. This is an easy exercise. Also, since X is an intersection of open sets, hence of upwards-closed sets (in the specialization ordering of Y), X is upwards-closed in Y. The specialization ordering of X is the restriction of that of Y to X (another easy exercise), so a subset of X is saturated (=upwards-closed) in X if and only if it is saturated in Y. Therefore the compact saturated subsets of X are exactly the compact saturated subsets of Y that are included in X. We will use this several times.

We use it right away: Q is a compact saturated subset of Y. By Proposition A, and specifically, not just the fact that Y is a Choquet-Wilker space, but rather that it satisfies the stronger property stated in Proposition A, and since Q ⊆ U₁ ∪ U₂ hence Q ⊆ (Û₁ ∩ W₀) ∪ (Û₂ ∩ W₀), there are two compact saturated subsets Q₀₁ and Q₀₂ of Y such that Q ⊆ int(Q₀₁) ∪ int(Q₀₂), Q₀₁ ⊆ Û₁ ∩ W₀, and Q₀₂ ⊆ Û₂ ∩ W₀.

We reuse that stronger property on the inclusion Q ⊆ int(Q₀₁) ∪ int(Q₀₂), or rather on the inclusion Q ⊆ (int(Q₀₁) ∩ W₁) ∪ (int(Q₀₂) ∩ W₂), which is deduced from the former and the inclusion Q ⊆ X ⊆ W₁. We obtain that there are two compact saturated subsets Q₁₁ and Q₁₂ of Y such that Q ⊆ int(Q₁₁) ∪ int(Q₁₂), Q₁₁ ⊆ int(Q₀₁) ∩ W₁, and Q₁₂ ⊆ int(Q₀₂) ∩ W₁.

And we continue: there are two compact saturated subsets Q₂₁ and Q₂₂ of Y such that Q ⊆ int(Q₂₁) ∪ int(Q₂₂), Q₂₁ ⊆ int(Q₁₁) ∩ W₂, and Q₂₂ ⊆ int(Q₂₂) ∩ W₂, and in general, for every n ∈ N, we obtain two compact saturated subsets Q₂₁ and Q₂₂ of Y such that Q ⊆ int(Q_(n+1)1) ∪ int(Q_(n+1)2), Q_(n+1)1 ⊆ int(Q_n1) ∩ W_n+1, and Q_(n+1)2 ⊆ int(Q_n2) ∩ W_n+1.

We define Q₁ as the intersection ∩_n∈N Q_n1, and Q₂ as the intersection ∩_n∈N Q_n2. Those are both filtered intersections of compact saturated subsets of Y; in that they are intersections of descending families of compact saturated set. Since Y is sober, it is well-filtered (Proposition 8.3.5 in the book), and in a well-filtered space every filtered intersection of compact saturated subsets is compact saturated (Proposition 8.3.6 in the book). Therefore Q₁ and Q₂ are compact saturated subsets of Y.

They are also included in every W_n, since Q_(n+1)1 ⊆ int(Q_n1) ∩ W_n+1 and Q_(n+1)2 ⊆ int(Q_n2) ∩ W_n+1, so they are included in ∩_n∈N W_n, namely in X. Being compact saturated in Y and included in X, Q₁ and Q₂ are compact saturated subsets of X.

We note that Q is included in Q₁ ∪ Q₂. Otherwise, there would be a point x in Q and outside both Q₁ and Q₂, hence outside int(Q_m1) for some m ∈ N and outside int(Q_n2) for some n ∈ N. Hence x would be outside int(Q_k1) ∪ int(Q_k2) where k ≝ max(m, n). That is impossible since Q ⊆ int(Q_k1) ∪ int(Q_k2).

Finally, Q₁ is included in U₁ because Q₁ ⊆ Q₀₁ ∩ X ⊆ (Û₁ ∩ W₀) ∩ X = U₁, and similarly Q₂ ⊆ U₂. ☐

KT₄ spaces, Hausdorff spaces, regular spaces

A topological space is regular if and only if given any closed subset C and any point x outside of C, there are an open neighborhood U of x and an open neighborhood V of C such that U and V are disjoint (Definition 4.1.19 in the book).

A topological space is normal if and only if for any two disjoint closed subsets C₁ and C₂ have disjoint open neighborhoods (Definition 4.1.22 in the book). A space is KT₄ if all its compact subspaces are normal. The following was observed by Wilker [2, Theorem 8].

Proposition C. Every KT₄ space is Choquet-Wilker. Every Hausdorff space, every regular space is KT₄, hence Choquet-Wilker.

Proof. We start with the first claim, and we consider a KT₄ space X, a compact saturated subset Q of X, and two open subsets U₁ and U₂ of X whose union contains Q. Since X is KT₄, Q as a subspace is normal. Now Q–U₁ and Q–U₂ are disjoint closed subsets of Q. Therefore there are disjoint open subsets of Q, which we write as Q ∩ V₁ and Q ∩ V₂ respectively, where V₁ and V₂ are open in X, such that Q–U₁ ⊆ Q ∩ V₁ and Q–U₂ ⊆ Q ∩ V₂. Taking complements in Q, Q–V₁ ⊆ Q ∩ U₁ and Q–V₂ ⊆ Q ∩ U₂.

But Q–V₁ is the intersection of Q with the complement of V₁ in X (a closed set), hence it is compact (Corollary 4.4.10 in the book). Its upward closure in the specialization preordering of X, ↑(Q–V₁), is then compact saturated in X (Proposition 4.4.14 in the book). Since Q–V₁ ⊆ U₁ and U₁ is upwards-closed, ↑(Q–V₁) is included in U₁. We let Q₁ ≝ ↑(Q–V₁). Similarly, Q₂ ≝ ↑(Q–V₂) is compact saturated and included in U₂.

It remains to show that Q ⊆ Q₁ ∪ Q₂. We have (Q–V₁) ∪ (Q–V₂) = Q–(V₁ ∩ V₂) = Q, since V₁ and V₂ are disjoint. In particular, Q = (Q–V₁) ∪ (Q–V₂) ⊆ Q₁ ∪ Q₂. This concludes the proof that the KT₄ space X is Choquet-Wilker.

We proceed by showing that every Hausdorff space X is KT₄. Every compact subspace of X is compact and Hausdorff, since Hausdorffness is preserved by taking subspaces. Every compact Hausdorff space is normal (Proposition 4.4.17 in the book), so every compact subspace of X is normal: X is KT₄.

We finish this proof by showing that every regular space X is KT₄. We first claim that every subspace Q of X is regular, too. Given any point x ∈ Q and any closed subset C ∩ Q (with C closed in X) of Q that does not contain x, C cannot contain x either. Since X is regular, there are two disjoint open subsets U and V of X, with x ∈ U and C ⊆ V. Then U ∩ Q and V ∩ Q are two disjoint open subsets of the subspace Q, U ∩ Q contains x and C ∩ Q ⊆ V ∩ Q.

Second, given any compact subspace Q of X, Q is both compact and regular. It remains to see that every compact regular space is normal. Curiously, this seems to be absent from the book. Let C₁ and C₂ be two disjoint closed subsets of Q. For each x ∈ C₁, there is an open neighborhood U_x of x in Q and an open neighborhood V_x of C₂ in Q such that U_x and V_x are disjoint. Since C₁ is closed in a compact space, it is compact (Proposition 4.4.15 in the book), and therefore we can find a finite subcover (U_x)_x∈E of C₁, for some finite subset E of C₁. Let U ≝ ∪_x∈E U_x, V ≝ ∩_x∈E V_x. Then U and V are disjoint, and open, U contains C₁, and V contains C₂. We are done. ☐

A non-Choquet-Wilker space

At this point, we have seen that so many families of topological spaces are Choquet-Wilker, and we might wonder whether there is any space that is not Choquet-Wilker at all. And indeed, that exists, as we will demonstrate; and therefore the notion of a Choquet-Wilker space is not a trivial one.

We start with a technical lemma, which relies on the notion of a KC-space, namely a space in which every compact subset is closed. I have already mentioned such spaces in the January 2024 post, where we have observed that every Hausdorff space is a KC-space, that every KC-space is T₁ and coherent (in the sense that the intersection of any two compact (necessarily saturated) sets is compact), and also at the end of the October 2024 post.

Lemma D. Every compact Choquet-Wilker KC-space X is normal.

Proof. We consider two disjoint closed subsets C₁ and C₂ of X, and we let Q ≝ X, U₁ ≝ X–C₁ and U₂ ≝ X–C₂. Q is compact saturated by assumption, U₁ and U₂ are open, and U₁ ∪ U₂ = X–(C₁ ⋂ C₂) = X, since C₁ and C₂ are disjoint. In particular, Q ⊆ U₁ ∪ U₂. Since X is Choquet-Wilker, there are compact saturated subsets Q₁, Q₂ of X such that Q₁ ⊆ U₁, Q₂ ⊆ U₂ and Q ⊆ Q₁ ∪ Q₂. Since X is a KC-space, Q₁ and Q₂ are closed, so their complements are open. We have C₁ = X–U₁ ⊆ X–Q₁, C₂ = X–U₂ ⊆ X–Q₂, and it remains to see that X–Q₁ and X–Q₂ are disjoint. Their intersection is X–(Q₁ ∪ Q₂), and X = Q ⊆ Q₁ ∪ Q₂, so that intersection is empty, as desired. ☐

The following was stated right after Definition 4.1.22 in the book.

Fact E. Every T₁ normal space is Hausdorff (in fact T₄, hence T₃, hence T₂).

Our counterexample will be the Alexandroff (one-point) compactification α(Q) of the space Q of rational numbers. Alexandroff compactifications are defined in Exercise 4.8.9 of the book for locally compact Hausdorff spaces only, but they make sense for any Hausdorff space X—except that α(X) will not be compact Hausdorff in general.

For every Hausdorff topological space X, α(X) is obtained as follows:

• The points of α(X) are those of X, plus an additional point which we write as ∞.
• The open subsets of α(X) are the open subsets of X, plus the sets α(X)–K, where K ranges over the compact subsets of X.  I will let you check that this forms a topology, and this depends on X being Hausdorff.

We equip Q with the subspace topology from R, itself with its usual metric topology. We note that Q is Hausdorff. We will also need to know that every compact subset of Q has empty interior (see Exercise 4.8.4 in the book).

Lemma F. The space α(Q) is T₁ but not Hausdorff.

Proof. Given any two distinct points x and y of α(Q), either both x and y are in Q, and then they have disjoint open neighborhoods in Q, hence in α(Q); or one of them is equal to ∞, say y, in which case α(Q)–{x} is an open neighborhood of y that does not contain x. Therefore α(Q) is T₁.

In order to show that α(Q) is not Hausdorff, we fix any x ∈ Q, and we will show that we cannot separate x from ∞ by any pair of disjoint open neighborhoods. For every open neighborhood U of x, for every open neighborhood V of ∞, necessarily of the form α(Q)–K where K is compact in Q, U ∩ V is equal to U–K. That is empty if and only if U ⊆ K. But K is compact, hence has empty interior, so U ⊆ K would entail that U is empty, which is impossible since U contains x. Hence U ∩ V cannot be empty. This shows that α(Q) is not Hausdorff. ☐

Fact G. The topology of Q is the subspace topology induced by the inclusion in α(Q).

Indeed, in one direction every open subset U of Q appears as an open subset of α(Q), hence as a set of the form U ∩ Q for some open subset U of α(Q). In the other direction, we show that the intersection of any open subset of α(Q) with Q is open in Q. This is clear for open subsets U that are already open in Q, and for open subsets of the form α(Q)–K with K compact in Q, (α(Q)–K) ∩ Q = Q–K is open since K, being compact in a Hausdorff space, is closed (Proposition 4.4.15 in the book).

Proposition H. The compact subsets Q of α(Q) that contain ∞ are exactly those of the form C ∪ {∞} where C is a closed subset of Q. In particular,

1. α(Q) is a compact space;
2. α(Q) is a KC-space.

Proof. Let us see right away why the first part of the Proposition entails items 1 and 2.

1. Let Q ≝ α(Q). We can write Q as C ∪ {∞}, where C ≝ Q is closed in Q, so Q is compact.
2. Given any compact subset Q of α(Q),

• either Q contains ∞, then Q is equal to C ∪ {∞} where C is a closed subset of Q, by the first part of the Proposition; hence its complement is equal to α(Q)–(C ∪ {∞}) = Q–C, which is open in Q, hence in α(Q) by Fact G: so Q is closed in α(Q);
• or Q does not contain ∞, hence is a compact subset of α(Q) that is included in the subspace Q (it is a subspace by Fact G); therefore Q is a compact subset of Q, and since Q is Hausdorff, Q is closed in Q; by Fact G, it follows that Q is closed in α(Q).

We now prove the first part of the Proposition. Let Q be a compact subset of α(Q) containing ∞, and let C ≝ Q ∩ Q = Q–{∞}.

For every compact saturated subset K of Q, we show that C ∩ K is compact in Q by considering any open cover (U_i)_i∈I of C ∩ K, and realizing that C ∩ K ⊆ ∪_i∈I U_i is equivalent to Q ∩ K ⊆ ∪_i∈I U_i, hence to Q ⊆ (α(Q)–K) ∪ ∪_i∈I U_i. Since Q is compact in α(Q), we can extract a finite subcover. That finite subcover must contain α(Q)–K, since ∞ ∈ Q but ∞ is none of the sets U_i, which are open in Q; so there is a finite subset I′ of I such that Q ⊆ (α(Q)–K) ∪ ∪_i∈I’ U_i.. Equivalently, C ∩ K ⊆ ∪_i∈I’ U_i. Therefore, C ∩ K is compact in Q.

For every sequence (x_n)_n∈N of elements of C, converging to some x ∈ Q, the set K ≝ {x_n | n ∈ N} ∪ {x} is compact in Q. This is Fact 1 in the May 2019 post. In the previous paragraph, we have shown that C ∩ K is compact in Q. Since Q is Hausdorff, C ∩ K is closed in Q. Since every point x_n is in C ∩ K, so is the limit x. In particular, x ∈ C. This shows that C is sequentially closed. But all sequentially closed subsets of first-countable spaces are closed (Exercise 4.7.14 in the book), and Q is first-countable, since the sets ]x–1/2ⁿ, x+1/2ⁿ[ with n ∈ N form a countable base of open neighborhoods of any point x (in fact Q is even second-countable). Therefore C is closed in Q.

Since C = Q–{∞} and Q contains ∞, Q must be equal to C ∪ {∞}.

Conversely, let C be a closed subset of Q, and let Q ≝ C ∪ {∞}. We consider any open cover of Q, necessarily given as the union of two families of open sets: a family of open subsets U_i of Q, where i ranges over some index set I, and a family of sets of the form α(Q)–K_j, where j ranges over some index set J and each K_j is compact in Q. We note that J is non-empty, since ∞ ∈ Q. Then, Q ⊆ ∪_i∈I U_i ∪ ∪_j∈J (α(Q)–K_j), namely Q ∩ ∩_j∈J K_j ⊆ ∪_i∈I U_i. Since J is non-empty, Q ∩ ∩_j∈J K_j = C ∩ ∩_j∈J K_j.

In a Hausdorff space such as Q, every non-empty intersection of compact sets such as ∩_j∈J K_j is compact, because they are all closed (Proposition 4.4.15 in the book) and included in one of the sets K_j, which is compact. Taking the intersection with the closed set C then yields a compact set again. Hence C ∩ ∩_j∈J K_j is compact in Q.

Since Q ∩ ∩_j∈J K_j = C ∩ ∩_j∈J K_j is included in ∪_i∈I U_i, there is a finite subset I′ of I such that C ∩ ∩_j∈J K_j ⊆ ∪_i∈I‘ U_i. We note that C ∩ ∩_j∈J K_j = ∩_j∈J (C ∩ K_j). Since Q is Hausdorff, it is sober (Proposition 8.2.12 in the book) hence well-filtered (Proposition 8.3.5 in the book), so there is a finite subset J′ of J such that ∩_j∈J‘ (C ∩ K_j) ⊆ ∪_i∈I‘ U_i. Unfolding back, we have obtained a finite subcover of Q, by the sets U_i with i ∈ I′ and the sets α(Q)–K_j with j ∈J′. ☐

We can now conclude.

Theorem I. α(Q) is not a Choquet-Wilker space.

Proof. By Proposition H, α(Q) is a compact KC-space. If it were Choquet-Wilker, it would be normal by Lemma D. Since it is T₁ (being a KC-space, or by Lemma F), Lemma E tells us that it would be Hausdorff. But Lemma F tells us that this is not the case. ☐

Theorem I even shows that there is a compact KC-space (α(Q), of course) that is not Choquet-Wilker.

The proof of Theorem I is pretty abstract, and here is a more concrete proof, where we will exhibit an actual compact saturated subset Q of α(Q), included in the union U₁ ∪ U₂ of two actual open subsets, and such that cannot find compact saturated subsets Q₁ and Q₂ such that Q₁ ⊆ U₁, Q₂ ⊆ U₂ and Q ⊆ Q₁ ∪ Q₂.

We let Q ≝ α(Q), U₁ ≝ α(Q)–{x} for some arbitrary rational number x, and U₂ ≝ Q. (In other words, we just remove one rational point from α(Q) in order to obtain U₁, and we instead remove just the point ∞ from α(Q) in order to obtain U₂.) By Proposition H, Q is compact (and saturated since α(Q) is T₁ by Lemma F). U₁ is open in α(Q) because {x} is compact in Q, and U₂ is open because it is an open subset of Q.

If α(Q) were Choquet-Wilker, there would be compact (saturated) subsets Q₁ and Q₂ of α(Q) such that Q₁ ⊆ U₁, Q₂ ⊆ U₂ and Q ⊆ Q₁ ∪ Q₂. Since U₂ does not contain ∞, neither does Q₂, so Q₂ is a compact subset of α(Q) included in Q, hence a compact subset of Q, with the subspace topology (by Fact G). Since ∞ ∈ Q ⊆ Q₁ ∪ Q₂, ∞ must be in Q₁. By Proposition H, Q₁ must be of the form C ∪ {∞} for some closed subset C of Q. From Q ⊆ Q₁ ∪ Q₂, we obtain that (C ∪ {∞}) ∪ Q₂ = α(Q), and in particular C ∪ Q₂ = Q. This implies that the complement of C in Q is included in Q₂; but every compact subset of Q has empty interior, so the complement of C in Q must be empty, or equivalently, C = Q. This is impossible since Q₁ = C ∪ {∞}is included in U₁ = α(Q)–{x}.

Consonant Choquet-Wilker spaces

A while back, Matthew de Brecht sent me an unpublished note, and the very first lemma was a nice observation about consonant Choquet-Wilker spaces, which he says he discovered during a conversation with Klaus Keimel. This will conclude this month’s note.

I have already devoted quite a few blog entries on the notion of consonance. Let me give a quick refresher. For every compact saturated subset Q of a space X, let us write ■Q ≝ {U | Q ⊆ U}: this is a a Scott-open subset of the complete lattice OX of open subsets of X, and even a Scott-open filter of open subsets. The space X is consonant if and only if every Scott-open subset of OX is a union of such sets ■Q. In other words, if and only if for every Scott-open subset U of OX, for every U ∈ U, there is a compact saturated subset Q of X such that Q ⊆ U (i.e., U ∈ ■Q) and ■Q is included in U.

The point of consonance is that X is consonant if and only if the compact-open and Isbell topologies coincide on the space [X → Y] of continuous maps from X to Y, for every topological space Y (or just when Y is Sierpiński space S), see Exercise 5.4.12 in the book; and every locally compact space is consonant, a fact you can learn from the same exercise. Every LCS-complete space is consonant, too: this is the first theorem in the October 2016 post.

Proposition J (M. de Brecht 2025, from a conversation with K. Keimel). For a consonant space X, the following are equivalent:

1. X is a Choquet-Wilker space;
2. the union map ∪ : OX × OX → OX is continuous, where OX is given the Scott topology of inclusion and OX × OX has the product topology.

Proof. 1⇒2. Let U be a Scott-open subset of OX, and let U₁ and U₂ be two open subsets of X that are mapped by the union map to an element of U. In other words, we assume that U₁ ∪ U₂ is in U. We wish to show that there is an open neighborhood of (U₁, U₂), in OX × OX with the product topology, which is mapped into U by the union map.

Since X is consonant, there is a compact saturated subset Q of X such that Q ⊆ U₁ ∪ U₂. Since X is Choquet-Wilker, there are compact saturated subsets Q₁ and Q₂ of X such that Q₁ ⊆ U₁, Q₂ ⊆ U₂ and Q ⊆ Q₁ ∪ Q₂. Then ■Q₁ is a Scott-open neighborhood of U in OX, ■Q₂ is a Scott-open neighborhood of V in OX, and it remains to observe that the image of ■Q₁ × ■Q₂ under the union map is included in U. To this end, let V₁ be any element of ■Q₁, V₂ be any element of ■Q₂; then Q ⊆ Q₁ ∪ Q₂ ⊆ V₁ ∪ V₂, so V₁ ∪ V₂ ∈ ■Q ⊆ U.

2⇒1. Here we assume that X is consonant and that ∪ is continuous. Let Q be a compact saturated subset of X that is included in the union U₁ ∪ U₂ of two open subsets of X. Then U₁ ∪ U₂ is in the Scott-open set ■Q. Since ∪ is continuous, there are two open subsets U₁ and U₂ of OX, with U₁ ∈ U₁ and U₂ ∈ U₂, and such that the union of any element of U₁ with any element of U₂ is in ■Q. Since X is consonant, there is a compact saturated subset Q₁ of X such that Q₁ ⊆ U₁ and ■Q₁ ⊆ U₁, and there is a compact saturated subset Q₂ of X such that Q₂ ⊆ U₂ and ■Q₂ ⊆ U₂.

It remains to show that Q ⊆ Q₁ ∪ Q₂. For every open neighborhood V₁ of Q₁, for every open neighborhood V₂ of Q₂, V₁ ∈ ■Q₁ and V₂ ∈ ■Q₂, hence V₁ ∈ U₁ and V₂ ∈ U₂, so V₁ ∪ V₂ ∈ ■Q; the latter means that Q ⊆ V₁ ∪ V₂. But Q₁ and Q₂ are saturated, so Q₁ = ∩_{V1 ∈ ■Q1} V₁ and Q₂ = ∩_{V2 ∈ ■Q2} V₂. Therefore Q₁ ∪ Q₂ = ∩_{V1 ∈ ■Q1, V2 ∈ ■Q2} (V₁ ∪ V₂) ⊇ Q. ☐

In particular, ∪ : OX × OX → OX is continuous if X is locally compact or LCS-complete.

Property 2 of Proposition J looks a lot like meet-continuity, a.k.a. Erné’s web spaces: a meet-continuous space (=a web space) is a space X such that the intersection map ⋂ : HX × HX → HX is Scott-continuous, where HX is the lattice of closed subsets of X. Since closed sets are complements of open sets, and complements turn intersections into unions, the meet-continuous spaces are also characterized by a continuity property for the union map on open subsets. However, in that case, OX would have to be given the Scott topology of the reverse inclusion ordering ⊇. Another less visible difference is that the union map would have to be Scott-continuous, namely separately continuous, while Proposition J requires joint continuity. This is important: the union map is always separately continuous from OX × OX to OX, for every space X.

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— Jean Goubault-Larrecq (April 20th, 2026)