On Till Plewe’s game and Matthew de Brecht’s non-consonance arguments

Last time I mentioned that S₀ is not consonant. I had a preprint by Matthew de Brecht where he proved that, but I could not find the result in published form. Since then, M. de Brecht wrote to me, and told me a lot of interesting things. First, his proof can be found in [1]. His proof uses a topological game first invented by Till Plewe [2] in order to study locale products of spaces, and conditions under which the locale products coincide with the (locales underlying) the topological products, which M. de Brecht had initially used to show that the locale product of S₀ with itself is not spatial. Second, he pointed me towards an intriguing result by Ruiyuan Chen [3], which gives another connection between locale products (of dcpos) and the question of whether the poset-theoretic product of two dcpos with the Scott topology coincides with the topological product. Third, R. Chen has a simpler, alternative proof of the fact that the locale product of S₀ with itself is not spatial [4]. And so on!

Hence I have too much to talk about, so I will have to make a choice. I will concentrate on explaining what [1] is all about. I am afraid that this post will mostly be paraphrasing M. de Brecht’s paper. I have merely added a few pictures, and I have reformulated a few notions; I hope this will contribute to make what he achieved easier to grasp.

Till Plewe’s game

Let X and Y be two topological spaces, U₀ be an open subset of X, V₀ be an open subset of Y, and U be an open cover of U₀ × V₀ by open rectangles. Till Plewe’s game G_X,Y(U₀, V₀, U) is played as follows. At each turn i≥1:

1. player I picks a point x_i ∈ U_i–1;
2. then player II picks an open neighborhood U_i of x_i included in U_i–1;
3. player I picks a point y_i ∈ V_i–1;
4. then player II picks an open neighborhood V_i of y_i included in V_i–1.

So, yes, each player plays twice at each turn.

Player II wins (at round i) if U_i × V_i is included in one of the open rectangles of the cover U. Otherwise, namely if the play goes along without player II even winning, player I wins.

The following in Theorem 1.1 in [2].

Theorem (Plewe). If X and Y are sober spaces, either:

• for every open subset U₀ of X, for every open subset V₀ of Y, for every open cover U of U₀ × V₀ by open rectangles, player II has a winning strategy in the game G_X,Y(U₀, V₀, U), and then the locale product of X and of Y is spatial;
• or there is an open subset U₀ of X, there is an open subset V₀ of Y, and there is an open cover U of U₀ × V₀ by open rectangles such that player I has a winning strategy in the game G_X,Y(U₀, V₀, U), and then the locale product of X and of Y is non-spatial.

And the following is Theorem 4 in [1].

Theorem (de Brecht). If X is a consonant space, then for every open cover U of X × X by open rectangles, player II has a winning strategy in the game G_X,X(X, X, U).

Hence there is an intriguing connection between consonance and the spatiality of products of locales, which seems to me to be largely unexplored. M. de Brecht says “However, the implicit connections we make in this paper are not as strong as we expect they really are” at the end of the introduction of [1]. I don’t quite know what he means by that, except perhaps that he finds them intriguing, and unexplored, too.

I will not prove Plewe’s theorem here, but I will definitely give de Brecht’s proof of his theorem, and I will then explain how he uses it to prove that S₀ is not consonant. He also hints at the fact that this can be used to show that Q is not consonant. We have suffered quite a lot through the latter question, and I should take a stab at it some day, and explain what M. de Brecht had in mind in this respect—but not today.

An ordinal measure of open rectangles

The key to proving the above theorem by M. de Brecht is to realize that given any open cover U of X × X by open rectangles, where X is consonant, one can eventually reach the open rectangle X × X itself by repeating the operations of horizontal unions and vertical unions, transfinitely often, starting from the downward closure of U. Let me explain.

The terms horizontal unions and vertical unions will be specific to this post. If you are given a collection A of open rectangles, then I will call a horizontal union of elements of A any union ∪_{i ∈ I} (U_i × V_i), where each U_i × V_i is in A, and all the open sets V_i are identical. In other words, this is obtained by putting side by side, horizontally, any number of open rectangles from A with the same vertical cross-section. Vertical unions are defined similarly; namely, in that case, we require all the open sets U_i to be identical instead.

The downward closure ↓U of an open cover U is by definition the collection of open rectangles that are included in some element of U. We define A₀(U) as ↓U. For every ordinal α, if we have already defined A_α(U), we define A_α+1(U) as the collection of horizontal unions of elements of A_α(U), plus all vertical unions of elements of A_α(U). (In particular, A_α(U) is included in A_α+1(U), since one can reobtain any rectangle of A_α(U) as, say, a horizontal union of just one rectangle from A_α(U).) For a limit ordinal α, A_α(U) is the union of the collections A_β(U), β<α.

Lemma. For every ordinal α, A_α(U) is downward closed.

Proof. By induction on α. Since A₀(U) = ↓U, A₀(U) is downward closed. Assuming A_α(U) downward closed, we show that A_α+1(U) is downward closed by showing that any open rectangle U × V included in some horizontal union ∪_{i ∈ I} (U_i × V_i) of rectangles in A_α(U) is already in A_α+1(U) (and similarly for vertical unions; but the argument is symmetrical, hence omitted). This is easy: U × V is equal to ∪_{i ∈ I} ((U_i ∩ U) × (V_i ∩ V)), which is a horizontal union of rectangles (U_i ∩ U) × (V_i ∩ V); each such rectangle is a subset of a rectangle in A_α(U), hence is in A_α(U) by induction hypothesis, so U × V = ∪_{i ∈ I} ((U_i ∩ U) × (V_i ∩ V)) is also a horizontal union of elements of A_α(U); therefore U × V is in A_α+1(U). ☐

Let us write A_∞(U) for the union of the monotone sequence of collections A_α(U). A_∞(U) must be obtained as A_α(U) for some large enough ordinal α, since otherwise A_∞(U) would contain at least as many elements as there are ordinals, namely: A_∞(U) would be a proper class. And A_∞(U) cannot be a proper class, since it is included in a set—namely the set of all open rectangles on X × X.

It follows that A_∞(U) is closed under both horizontal and vertical unions.

Lemma. A_∞(U) is Scott-closed in OX × OX.

Proof. Let (U_i × V_i)_{i ∈ I} be a directed family of elements of A_∞(U), and let U × V be its union. Without loss of generality, we may assume that none of the rectangles U_i × V_i is empty; otherwise, remove all of them and what remains will either be an empty family (and the union of the empty family is in A₁(U), hence in A_∞(U)) or will still be directed. Therefore every U_i, and every V_i, will now be considered as non-empty.

The first thing we need to observe is that every rectangle U_j × V_i, where i and j are possibly distinct indices in I, is in A_∞(U). Indeed, since the family is directed, we can find an index k in I such that both U_i × V_i and U_j × V_j are included in U_k × V_k. In particular, and because neither U_j nor V_i is empty, U_j is included in U_k and V_i is included in V_k. But U_k × V_k is in A_∞(U), and A_∞(U) is downwards closed, so U_j × V_i is in A_∞(U), too.

Fixing i ∈ I, U × V_i is the union of the sets U_j × V_i, where j ranges over I. That is a horizontal union of elements of A_∞(U), hence is itself in A_∞(U). Then U × V is the union of the sets U × V_i, where i ranges over I, hence is a vertical union of elements of A_∞(U). It follows that U × V is also in A_∞(U). ☐

Let us define C(U) as the collection of open subsets U of X such that U × U is in A_∞(U). The map U ↦ U × U is Scott-continuous from OX to OX × OX, so C(U) is Scott-closed in OX.

Let H(U) be the complement of C(U) in OX, namely the collection of open subsets U of X such that U × U is not in A_∞(U). We have just proved:

Fact. H(U) is a Scott-open subset of OX.

This is where consonance comes into play. Let me recall (from any of the earlier posts on the topic, namely 1 through 5 on this page) that a space X is consonant if and only if every Scott-open subset of OX is a union of sets of the form ■Q, where Q is compact saturated in X; and ■Q is the collection of open neighborhoods of Q.

Lemma. If X is consonant, then X does not belong to H(U); equivalently, X × X is in A_∞(U).

Proof. If X ∈ H(U) and X is consonant, then there is a compact saturated subset Q of X such that (X is in ■Q, which is obvious, and) ■Q is included in H(U). We aim for a contradiction.

We recall that U is an open cover of X × X by open rectangles. In particular, for every point x in X, {x} × Q is included in the union of the open rectangles of U. Since {x} × Q is compact, it is included in a finite union ∪_{y ∈ E(x)} (U_xy × V_xy) of open rectangles of U; notably, the index set E(x) is finite. Let U_x be the intersection ∩_{y ∈ E(x)} U_xy, and let V_x be the union ∪_{y ∈ E(x)} V_xy. We obtain that {x} × Q is included in U_x × V_x. Note in particular that Q is included in V_x.

Additionally, U_x × V_x is the vertical union of the rectangles U_x × V_xy, y ∈ E(x). Since each U_x × V_xy is included in U_xy × V_xy, which is in U, U_x × V_xy is in A₀(U) = ↓U. Therefore U_x × V_x is in A₁(U).

We use the compactness of Q once again: Q is included in a finite union U ≝ ∪_{x ∈ E} U_x (namely, the index set E is finite). Let V be the intersection ∩_{x ∈ E} V_x. Since each V_x contains Q, V also contains Q. Then U × V is a horizontal union of open rectangles U_x × V. Each of these open rectangles is included in U_x × V_x, hence is in A₁(U). It follows that U × V is in A₂(U).

Then open rectangle (U ∩ V) × (U ∩ V) is included in the latter, hence is also in A₂(U). Additionally, both U and V contain Q, so U ∩ V is in ■Q. Since ■Q is included in H(U), U ∩ V is in H(U), or equivalently, (U ∩ V) × (U ∩ V) is not in A_∞(U). This directly contradicts the fact that it is in A₂(U). ☐

For every open rectangle U × V in A_∞(U), there is least ordinal α such that U × V is in A_α(U). Let us call α the degree of the rectangle U × V.

It is clear that the degree of each rectangle in A_∞(U) is a successor ordinal β+1. We have just proved that if X is consonant, then the degree of X × X is a well-defined notion.

Winning Plewe’s game on consonant spaces

Let me recall that M. de Brecht’s theorem states that if X is a consonant space, then for every open cover U of X × X by open rectangles, player II has a winning strategy in the game G_X,X(X, X, U). We start with the open rectangle X × X, which has a well-defined degree, as we have just seen. Player II’s strategy will consist in finding open sets U_i and V_i, at round i, for each i≥1, in response to player I’s choice of points x_i and y_i, in such a way that the degree of U_i × V_i is strictly smaller than that of U_i–1 × V_i–1, unless the degree of U_i–1 × V_i–1 is already equal to 0. If the latter happens, then U_i–1 × V_i–1 is in A₀(U) = ↓U, in which case player II has won.

Hence, let us assume that we enter round i with an open rectangle U_i–1 × V_i–1 of degree β+1. (We remember that the degree is always a successor ordinal.) Hence U_i–1 × V_i–1 is either a horizontal or vertical union of rectangles of degrees at most β.

• If U_i–1 × V_i–1 is a horizontal union of rectangles U’_j × V_i–1 (j ∈ J) of degrees at most β, we reason as follows. Player I plays a point x_i in U_i–1. Then x_i belongs to some U’_j, and we let player II play U’_j for the next open set U_i. Player I then plays a point y_i in V_i–1, and we let player II play V_i–1 itself for V_i (no change). As promised, the degree of U_i × V_i is at most β, hence strictly below the degree β+1 of U_i–1 × V_i–1.
• If U_i–1 × V_i–1 is a vertical union of rectangles U_i–1 × V’_j (j ∈ J) of degrees at most β, then player I plays x_i in U_i–1, and this time player II simply plays U_i–1 for U_i (no change). Player I plays y_i in V_i–1. Then y_i is in some V’_j, and player II plays that V’_j for V_i. Once again, the degree of U_i × V_i is at most β, hence strictly below the degree β+1 of U_i–1 × V_i–1.

Since the degree of U_i × V_i decreases strictly at each round, and the ordering on ordinals is well-founded, eventually that degree must reach 0. As we have seen above, in that situation, player II has won the game.

Hence we have proved de Brecht’s theorem, which we repeat here:

Theorem (de Brecht). If X is a consonant space, then for every open cover U of X × X by open rectangles, player II has a winning strategy in the game G_X,X(X, X, U).

S₀ is not consonant I: the setup

M. de Brecht uses this to show that S₀ is not consonant [1, Section 4]. We use the contrapositive of the previous theorem: it suffices to show that player II does not have a winning strategy in the game G_X,X(X, X, U), for some open cover U of X × X (with X ≝ S₀). In order to do so, we show that player I has a winning strategy in this game.

Let me recap a few things about S₀ from last time. S₀ is the space of finite sequences of natural numbers, with the upper topology of the (opposite of the) suffix ordering. I write n::s for the list obtained from the list s by adding a number n in front, ε for the empty list, and then S₀ is a tree with the root ε at the top:

Let me also recall that the closed subsets of S₀ (hence also the open subsets of S₀) are in one-to-one correspondence with the finitely branching subtrees of S₀. In other words, a closed subset C is the same thing as the downward closure ↓Min T of the set Min T of leaves of a finitely branching subtree T of S₀. In the following picture, the tree is shown in red, and the closed set in blue. A subtree is simply an upwards-closed subset of S₀; in particular, mind that we accept the empty set as a subtree.

Since the open sets are the complements of the closed sets, the open sets are also characterized through finitely branching subtrees T, namely as the sets S₀–↓Min T of points that are not below any leaf of T.

M. de Brecht defines specific open subsets U_σ,τ, where σ and τ are finite lists of natural numbers, but I will make two modifications to this presentation. First, τ is not really used in the definition except for its length, so I will replace it with a natural number n, and I will write the corresponding open sets as U_σ,n. Second, it is easier to understand what these open sets are through the trees T_σ,n that are used to define them.

T_σ,n is defined as consisting of ε, plus the sequences k::s, where 0≤k≤|σ|+n and s ranges over the suffixes of σ. (|σ| denotes the length of σ.) In other words, draw the path from the root ε to σ, and add the first |σ|+n+1 successors of each node thus obtained. Here is what T_σ,n is for σ = 2::0::ε and n=1. The path σ itself is shown in brown, and the additional successors are obtained by following the additional red edges.

The leaves of T_σ,n are exactly the sequences k::s of natural numbers such that:

• s is a suffix of σ (i.e., above σ);
• 0≤k≤|σ|+n;
• and k::s is not a suffix of σ (otherwise it would be an internal vertex of T_σ,n).

The open set U_σ,n is the open set defined by T_σ,n, namely the set S₀–↓Min T_σ,n of vertices that are below no leaf of T_σ,n.

I originally thought it might be easier to reason on finitely branching subtrees directly, instead of on open subsets of S₀, but that would get us lost. Instead, I will be following M. de Brecht; we make the following observations, which will distill the properties on finitely branching subtrees that we need in the proof to come, in terms of open sets:

• Property (A). For every finite sequence σ of natural numbers, for every natural number n, for every element s of U_σ,n, in order to show that s is above σ, it suffices to show that every element of s (seen as a finite sequence of natural numbers) is ≤|σ|+n.
• Property (B). For every open subset U of S₀, for every σ ∈ U, there are infinitely many successors n::σ of σ such that the complete subtree ↓(n::σ) is included in U.

Those are proved as follows. For property (B), we write U as the set of points that are not below any leaf of some given finitely branching tree T. Hence σ is not below any leaf of T. Since T is finitely branching, there are infinitely many natural numbers such that n::σ is not in T. Let us consider any such n, and any element t of ↓(n::σ). If t were below some leaf x of T, then x would either have n::σ as a suffix or be itself a suffix of σ. The first case would imply that n::σ is in T, since T is upwards closed, and that is impossible, by definition of n. The second case would imply that σ would be below some leaf of T, which is impossible as well. Therefore t is in U.

For property (A), let us imagine that s is the list m₁::…::m_p::ε, and that each m_i is ≤|σ|+n. Since s is in U_σ,n, s does not have a suffix that would be a leaf of T_σ,n. For every i with 1≤i≤p, let us consider the suffix m_i::…::m_p::ε of s. That is not a leaf of T_σ,n, as we have just said: hence m_i+1::…::m_p::ε is not a suffix of σ, or m_i>|σ|+n, or m_i::…::m_p::ε is a suffix of σ, by definition of the leaves of T_σ,n. The middle alternative m_i>|σ|+n is impossible by assumption, so m_i+1::…::m_p::ε is not a suffix of σ or m_i::…::m_p::ε is a suffix of σ. In other words, exploiting that “not a or b” is the same as “if a then b”, if m_i+1::…::m_p::ε is a suffix of σ then m_i::…::m_p::ε is a suffix of σ. We use this to show that m_i::…::m_p::ε is a suffix of σ for every i with 1≤i≤p+1, by induction on p+1–i. The base case (i=p+1: ε is a suffix of σ) is obvious, and the final case i=1 yields that s = m₁::…::m_p::ε is a suffix of σ.

S₀ is not consonant II: playing the game

In order to show that S₀ is not consonant, let me recall that we will show that player I has a winning strategy in some game G_X,X(X, X, U), where X=S₀.

We let U be the open cover of X × X consisting of the sets U_σ,|τ| × U_τ,|σ| where σ and τ range over all finite lists of natural numbers (all vertices of the tree S₀), where |σ| and |τ| denote the lengths of σ and τ respectively.

This is an open cover: σ is in U_σ,|τ|, because σ is not below any leaf of T_σ,|τ|, and similarly τ is in U_τ,|σ|, so (σ, τ) is in U_σ,|τ| × U_τ,|σ|.

M. de Brecht builds player I’s strategy by describing it, and then showing that it has the right properties. The proof is subtler than its short length might suggest. I also find it useful to have an explicit list of the invariants we need to maintain from round i–1 to round i in the game, and here they are. At the end of round i (assuming that those properties already hold with i–1 in place of i at the end of round i–1), we will establish that:

1. the elements of x_i (as a sequence of natural numbers) are smaller than or equal to |y_i|;
2. the elements of y_i are smaller than or equal to |x_i|;
3. there are two distinct natural numbers n_i, n’_i≤|y_i| such that every point below n_i::x_i, as well as every point below n’_i::x_i, is in U_i;
4. every pair (σ, τ) of finite sequences σ and τ such that (x_i, y_i) is in U_σ,|τ| × U_τ,|σ| must be below (x_i, y_i).

Invariants 1 and 2 are they key trick, and will be enforced, not by making sure that the elements of x_i and y_i are small enough, rather by making x_i and y_i long enough, as sequences of natural numbers. Invariant 3, or at least the existence of one natural number n_i≤|y_i| such that ↓(n_i::x_i) is included in U_i, will be required to make sure that the points x_i we will build are in U_i–1 (see below); the existence of a second natural number n’_i will only be important at the very end of the proof. Invariant 4 will be a consequence of invariants 1 and 2, using property (A), and will be an important property in order to show that player I’s strategy is winning.

With all that said, here is how M. de Brecht’s strategy proceeds. At the start of round i (i≥1), player I is given an open rectangle U_i–1 × V_i–1, not included in any rectangle of the form U_σ,|τ| × U_τ,|σ| but containing (x_i–1, y_i–1). The latter pair is well-defined if i≥2, as this is the pair produced by player I at turn number i–1. If i=1, we agree to define (x₀, y₀) as (ε, ε), so that we do not have to make a case distinction later. Also, we are informed that the invariants 1–4 hold at the end of round i–1. (Initially, invariants 1, 2 and 4 are certainly satisfied, pretty vacuously, at the “end of round i–1″ with i=1. Invariant 3 is also satisfied, for any choice of n₀ and of n’₀.) Let us enter round i≥1:

• By invariant 3 at the end of round i–1, there is a natural number n_i–1≤|y_i–1| such that every point below n_i::x_i is in U_i; By (B), y_i–1 has infinitely many successors n::y_i–1 such that ↓(n::y_i–1) is entirely included in V_i–1. We pick one of them. Player I then decides to play x_i ≝ 0ⁿ::n_i–1::x_i–1.
  Notice that x_i is in U_i–1, thanks to invariant 3.
• Now player II plays an open neighborhood U_i of x_i included in U_i–1, over which player I has no control.
• By (B), x_i has infinitely many successors whose downward closure is entirely included in U_i. We pick two of them, p::x_i and q::x_i, with p≠q and both p and q larger than or equal to n_i–1. Player I then decides to play y_i ≝ 0^p+q::n::y_i–1. (The number n was picked two bullet points above. The numbers p and q will be the n_i and n’_i needed to establish invariant 3. M. de Brecht chooses p+q as the exponent, but max(p,q) would be enough: the only requirement here is that p, q should both be smaller than or equal to |y_i|, as we will see below.)
  Notice why y_i is in V_i–1: the way we have picked n was so that ↓(n::y_i–1) is entirely included in V_i–1, so adding any number of zeroes in front of n::y_i–1, however large that may be, will still produce a point in V_i–1.
• Player II plays an open neighborhood V_i of y_i included in V_i–1, over which player I still has no control.

Note how the invariants are maintained:

1. The elements of x_i are those of x_i–1, which are all smaller than or equal to |y_i–1| by invariant 1 at the end of round i–1, plus n_i–1 and 0. Since y_i–1 is a suffix of y_i, we have |y_i–1|≤|y_i|; 0 is smaller than or equal to any number, and n_i–1≤p≤|y_i|. Therefore all the elements of x_i are smaller than or equal to |y_i|.
2. The elements of y_i are those of y_i–1, which are all smaller than or equal to |x_i–1| (hence to |x_i|) by invariant 2 at the end of round i–1, plus n and 0. But x_i = 0ⁿ::n_i–1::x_i–1, so in particular n≤|x_i|. It follows that all the elements of y_i are smaller than or equal to |x_i|.
3. The number p was chosen so that ↓(p::x_i) is entirely included in U_i. Also, p≤|y_i|. We can therefore take n_i ≝ p. Similarly, we take n’_i ≝ q, since ↓(q::x_i) is entirely included in U_i and q≤|y_i|.
4. Let us imagine that (x_i, y_i) is in an open rectangle of the form U_σ,|τ| × U_τ,|σ|. Since x_i–1 is above x_i and y_i–1 is above y_i, (x_i–1, y_i–1) is also in U_σ,|τ| × U_τ,|σ|. Hence, by invariant 4 at the end of round i–1, (x_i–1, y_i–1) is above (σ, τ). In particular (and this is the only thing we will use the latter for), |y_i–1|≤|τ|≤|σ|+|τ|.
   By invariant 1 (at the end of round i–1), the elements of x_i–1 are all less than or equal to |y_i–1|. The only additional non-zero element in x_i is n_i–1, which is less than or equal to |y_i–1| as well, by invariant 3. Therefore all the elements of x_i are less than or equal to |y_i–1|, hence to |σ|+|τ|. By property (A), it follows that x_i is above σ.
   In particular, |x_i|≤|σ|≤|σ|+|τ|. By invariant 2 (at the end of round i this time), all the elements of y_i are smaller than or equal to |x_i|, hence to |σ|+|τ|. We can therefore use property (A) once more, and conclude that y_i is above τ. This allows us to conclude that (x_i, y_i) is above (σ, τ).
   

We now need to show that U_i × V_i cannot be included in any rectangle of the form U_σ,|τ| × U_τ,|σ|. We reason by contradiction. Let us imagine that U_i × V_i ⊆ U_σ,|τ| × U_τ,|σ|, for some finite lists of natural numbers σ and τ. Since (x_i, y_i) is in U_i × V_i, it is in U_σ,|τ| × U_τ,|σ|, and therefore, by invariant 4, (x_i, y_i) is above (σ, τ).

This is the place where we need the two successors n_i::x_i and n’_i::x_i of x_i guaranteed by invariant 3. Invariant 3 tells us that the downward closure of each one is included in U_i. In particular, both n_i::x_i and n’_i::x_i are in U_i, hence in U_σ,|τ|. All the elements of x_i are smaller than or equal to |y_i| by invariant 1, hence to |τ|≤|σ|+|τ|, since y_i is above τ. Invariant 3 also tells us that n_i, n’_i ≤ |y_i|, hence they are also both less than or equal to |σ|+|τ|. By property (A), it follows that both n_i::x_i and n’_i::x_i are above σ. In other words, they are both suffixes of σ. But that is impossible, since n_i≠n’_i.

This allows us to conclude that U_i × V_i cannot be included in any rectangle of the form U_σ,|τ| × U_τ,|σ|, at any round i. Therefore the strategy for player I we have described above is winning, and this completes our argument that S₀ is not consonant.

1. Matthew de Brecht.  Some results on countably based consonant spaces. Recent Developments in General Topology and its Related Fields, RIMS Kôkyûroku No. 2151, 2019.
2. Till Plewe. Localic products of spaces. Proceedings of the London Mathematical Society, Volume s3-73, Issue 3, November 1996, Pages 642–678, https://doi.org/10.1112/plms/s3-73.3.642
3. Ruiyuan Chen. Products of Scott topologies and spatiality. Note, last consulted February 19th, 2023.
4. Ruiyuan Chen. A geometric proof that S₀ × S₀ is not spatial. Note, last consulted February 19th, 2023.

— Jean Goubault-Larrecq (February 20th, 2023)