Last time I mentioned that *S*_{0} 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 *S*_{0} 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 *S*_{0} 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, *U*_{0} be an open subset of *X*, *V*_{0} be an open subset of *Y*, and ** U** be an open cover of

*U*

_{0}×

*V*_{0}by open rectangles. Till Plewe’s game

*G*

_{X,Y}(

*U*

_{0},

*V*

_{0},

**) is played as follows. At each turn**

*U**i*≥1:

- player I picks a point
*x*∈_{i}*U*_{i–1}; - then player II picks an open neighborhood
*U*_{i}of*x*included in_{i}*U*_{i–1}; - player I picks a point
*y*∈_{i}*V*_{i–1}; - then player II picks an open neighborhood
*V*_{i}of*y*included in_{i}*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
*U*_{0}of*X*, for every open subset*V*_{0}of*Y*, for every open coverof*U**U*_{0}×*V*_{0}by open rectangles, player II has a winning strategy in the game*G*_{X,Y}(*U*_{0},*V*_{0},), and then the locale product of*U**X*and of*Y*is spatial; - or there is an open subset
*U*_{0}of*X*, there is an open subset*V*_{0}of*Y*, and there is an open coverof*U**U*_{0}×*V*_{0}by open rectangles such that player I has a winning strategy in the game*G*_{X,Y}(*U*_{0},*V*_{0},), and then the locale product of*U**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 *S*_{0} 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

**. Let me explain.**

*U*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

**any union ∪**

*A*_{i ∈ I}(

*U*

_{i}×

*V*_{i}), where each

*U*

_{i}×

*V*_{i}is in

**, and**

*A**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

**with the same vertical cross-section. Vertical unions are defined similarly; namely, in that case, we require all the open sets**

*A**to be identical instead.*

*U*_{i}The downward closure ↓** U** of an open cover

**is by definition the collection of open rectangles that are included in some element of**

*U***. We define**

*U*

*A*_{0}(

**) as ↓**

*U***. For every ordinal α, if we have already defined**

*U*

*A*_{α}(

**), we define**

*U*

*A*_{α+1}(

**) as the collection of horizontal unions of elements of**

*U*

*A*_{α}(

**), plus all vertical unions of elements of**

*U*

*A*_{α}(

**). (In particular,**

*U*

*A*_{α}(

**) is included in**

*U*

*A*_{α+1}(

**), since one can reobtain any rectangle of**

*U*

*A*_{α}(

**) as, say, a horizontal union of just one rectangle from**

*U*

*A*_{α}(

**).) For a limit ordinal α,**

*U*

*A*_{α}(

**) is the union of the collections**

*U*

*A*_{β}(

**), β<α.**

*U***Lemma.** For every ordinal α, *A*_{α}(** U**) is downward closed.

Proof. By induction on α. Since *A*_{0}(** U**) = ↓

**,**

*U*

*A*_{0}(

**) is downward closed. Assuming**

*U*

*A*_{α}(

**) downward closed, we show that**

*U*

*A*_{α+1}(

**) is downward closed by showing that any open rectangle**

*U**U*×

*included in some horizontal union ∪*

*V*_{i ∈ I}(

*U*

_{i}×

*V*_{i}) of rectangles in

*A*_{α}(

**) is already in**

*U*

*A*_{α+1}(

**) (and similarly for vertical unions; but the argument is symmetrical, hence omitted). This is easy:**

*U**U*×

*is equal to ∪*

*V*_{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}(

**), hence is in**

*U*

*A*_{α}(

**) by induction hypothesis, so**

*U**U*×

*= ∪*

*V*_{i ∈ I}((

*U*

_{i}∩

*U*) × (

*V*_{i}∩

*V*)) is also a horizontal union of elements of

*A*_{α}(

**); therefore**

*U**U*×

*is in*

*V*

*A*_{α+1}(

**). ☐**

*U*Let us write *A*_{∞}(** U**) for the union of the monotone sequence of collections

*A*_{α}(

**).**

*U*

*A*_{∞}(

**) must be obtained as**

*U*

*A*_{α}(

**) for some large enough ordinal α, since otherwise**

*U*

*A*_{∞}(

**) would contain at least as many elements as there are ordinals, namely:**

*U*

*A*_{∞}(

**) would be a proper class. And**

*U*

*A*_{∞}(

**) cannot be a proper class, since it is included in a set—namely the set of all open rectangles on**

*U**X*×

*X*.

It follows that *A*_{∞}(** U**) is closed under both horizontal and vertical unions.

**Lemma.** *A*_{∞}(** U**) is Scott-closed in

**O**

*X*×

**O**

*X*.

Proof. Let (*U*_{i} × *V*_{i})_{i ∈ I} be a directed family of elements of *A*_{∞}(** U**), and let

*U*×

*be its union. Without loss of generality, we may assume that none of the rectangles*

*V**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

*A*_{1}(

**), hence in**

*U*

*A*_{∞}(

**)) or will still be directed. Therefore every**

*U**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*_{∞}(

**), and**

*U*

*A*_{∞}(

**) is downwards closed, so**

*U**U*

_{j}×

*V*_{i}is in

*A*_{∞}(

**), too.**

*U*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*_{∞}(

**), hence is itself in**

*U*

*A*_{∞}(

**). Then**

*U**U*×

*is the union of the sets*

*V**U*×

*, where*

*V*_{i}*i*ranges over

*I*, hence is a vertical union of elements of

*A*_{∞}(

**). It follows that**

*U**U*×

*is also in*

*V*

*A*_{∞}(

**). ☐**

*U*Let us define ** C**(

**) as the collection of open subsets**

*U**U*of

*X*such that

*U*×

*is in*

*U*

*A*_{∞}(

**). The map**

*U**U*↦

*U*×

*is Scott-continuous from*

*U***O**

*X*to

**O**

*X*×

**O**

*X*, so

**(**

*C***) is Scott-closed in**

*U***O**

*X*.

Let ** H**(

**) be the complement of**

*U***(**

*C***) in**

*U***O**

*X*, namely the collection of open subsets

*U*of

*X*such that

*U*×

*is not in*

*U*

*A*_{∞}(

**). We have just proved:**

*U***Fact.** ** H**(

**) is a Scott-open subset of**

*U***O**

*X*.

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 **O***X* 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**(

**); equivalently,**

*U**X*×

*X*is in

*A*_{∞}(

**).**

*U**Proof.* If *X* ∈ ** H**(

**) and**

*U**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***). We aim for a contradiction.**

*U*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

**. Since {**

*U**x*} ×

*Q*is compact, it is included in a finite union ∪

_{y ∈ E(x)}(

*U*

_{xy}×

*V*_{xy}) of open rectangles of

**; notably, the index set**

*U**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

*A*_{0}(

**) = ↓**

*U***. Therefore**

*U**U*

_{x}×

*V*

_{x}is in

*A*_{1}(

**).**

*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 *A*_{1}(** U**). It follows that

*U*×

*V*is in

*A*_{2}(

**).**

*U*Then open rectangle (*U* ∩ *V*) × (*U* ∩ *V*) is included in the latter, hence is also in *A*_{2}(** 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***), or equivalently, (**

*U**U*∩

*V*) × (

*U*∩

*V*) is not in

*A*_{∞}(

**). This directly contradicts the fact that it is in**

*U*

*A*_{2}(

**). ☐**

*U*For every open rectangle *U* × * V* in

*A*_{∞}(

**), there is least ordinal α such that**

*U**U*×

*is in*

*V*

*A*_{α}(

**). Let us call α the**

*U**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*,

**). We start with the open rectangle**

*U**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

*A*_{0}(

**) = ↓**

*U***, in which case player II has won.**

*U*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**S*_{0} is not consonant I: the setup

M. de Brecht uses this to show that *S*_{0} 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

**of**

*U**X*×

*X*(with

*X*≝

*S*

_{0}). In order to do so, we show that player I has a winning strategy in this game.

Let me recap a few things about *S*_{0} from last time. *S*_{0} 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 *S*_{0} is a tree with the root ε at the top:

Let me also recall that the closed subsets of *S*_{0} (hence also the open subsets of *S*_{0}) are in one-to-one correspondence with the *finitely branching* subtrees of *S*_{0}. 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 *S*_{0}. 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 *S*_{0}; 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 *S*_{0}–↓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 *S*_{0}–↓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 *S*_{0}, 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*S*_{0}, 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 *m*_{1}::…::*m _{p}*::ε, and that each

*m*is ≤|σ|+

_{i}*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*::ε of

_{p}*s*. That is not a leaf of

*T*

_{σ,n}, as we have just said: hence

*m*

_{i+1}::…::

*m*::ε is not a suffix of σ, or

_{p}*>|σ|+*

*m*_{i}*n*, or

*m*

_{i}::…::

*m*::ε is a suffix of σ, by definition of the leaves of

_{p}*T*

_{σ,n}. The middle alternative

*>|σ|+*

*m*_{i}*n*is impossible by assumption, so

*m*

_{i+1}::…::

*m*::ε is not a suffix of σ or

_{p}*m*

_{i}::…::

*m*::ε is a suffix of σ. In other words, exploiting that “not a or b” is the same as “if a then b”, if

_{p}*m*

_{i+1}::…::

*m*::ε is a suffix of σ then

_{p}*m*

_{i}::…::

*m*::ε is a suffix of σ. We use this to show that

_{p}*m*

_{i}::…::

*m*::ε is a suffix of σ for every

_{p}*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*=

*m*

_{1}::…::

*m*::ε is a suffix of σ.

_{p}*S*_{0} is not consonant II: playing the game

In order to show that *S*_{0} 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*=

*S*

_{0}.

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

*S*

_{0}), 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

*, and similarly τ is in*

*T*_{σ,|τ|}*, so (σ, τ) is in*

*U*_{τ,|σ|}*×*

*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:

- the elements of
(as a sequence of natural numbers) are smaller than or equal to |*x*_{i}|;*y*_{i} - the elements of
are smaller than or equal to |*y*_{i}|;*x*_{i} - there are two distinct natural numbers
*n*_{i},*n’*_{i}≤|| such that every point below*y*_{i}*n*_{i}::, as well as every point below*x*_{i}*n’*_{i}::, is in*x*_{i}*U*_{i}; - every pair (σ, τ) of finite sequences σ and τ such that (
,*x*_{i}) is in*y*_{i}×*U*_{σ,|τ|}must be below (*U*_{τ,|σ|},*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

*are small enough, rather by making*

*y*_{i}*and*

*x*_{i}*long enough, as sequences of natural numbers. Invariant 3, or at least the existence of*

*y*_{i}*one*natural number

*n*

_{i}≤|

*| such that ↓(*

*y*_{i}*n*

_{i}::

*) is included in*

*x*_{i}*U*

_{i}, will be required to make sure that the points

*we will build are in*

*x*_{i}*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 (

*x*

_{0},

*y*

_{0}) 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

*n*

_{0}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}::is in*x*_{i}*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≝ 0*x*_{i}^{n}::*n*_{i}_{–1}::*x*_{i}_{–1}.

Notice thatis in*x*_{i}*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),
has infinitely many successors whose downward closure is entirely included in*x*_{i}*U*_{i}. We pick*two*of them,*p*::and*x*_{i}*q*::, with*x*_{i}*p*≠*q*and both*p*and*q*larger than or equal to*n*_{i}_{–1}. Player I then decides to play≝ 0*y*_{i}^{p+q}::*n*::*y*_{i}_{–1}. (The number*n*was picked two bullet points above. The numbers*p*and*q*will be theand*n*_{i}needed to establish invariant 3. M. de Brecht chooses*n’*_{i}*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 ||, as we will see below.)*y*_{i}

Notice whyis in*y*_{i}*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:

- The elements of
are those of*x*_{i}*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, we have |*y*_{i}*y*_{i}_{–1}|≤||; 0 is smaller than or equal to any number, and*y*_{i}≤*n*_{i}_{–1}*p*≤||. Therefore all the elements of*y*_{i}are smaller than or equal to |*x*_{i}|.*y*_{i} - The elements of
are those of*y*_{i}*y*_{i}_{–1}, which are all smaller than or equal to |*x*_{i}_{–1}| (hence to ||) by invariant 2 at the end of round*x*_{i}*i*–1, plus*n*and 0. But= 0*x*_{i}^{n}::*n*_{i}_{–1}::*x*_{i}_{–1}, so in particular*n*≤||. It follows that all the elements of*x*_{i}are smaller than or equal to |*y*_{i}|.*x*_{i} - The number
*p*was chosen so that ↓(::*p*) is entirely included in*x*_{i}*U*_{i}. Also,*p*≤||. We can therefore take*y*_{i}≝*n*_{i}*p*. Similarly, we take≝*n’*_{i}*q*, since ↓(::*q*) is entirely included in*x*_{i}*U*_{i}and*q*≤||.*y*_{i} - Let us imagine that (
,*x*_{i}) is in an open rectangle of the form*y*_{i}×*U*_{σ,|τ|}. Since*U*_{τ,|σ|}*x*_{i}_{–1}is aboveand*x*_{i}*y*_{i}_{–1}is above, (*y*_{i}*x*_{i}_{–1},*y*_{i}_{–1}) is also in×*U*_{σ,|τ|}. Hence, by invariant 4 at the end of round*U*_{τ,|σ|}*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 inis*x*_{i}*n*_{i}_{–1}, which is less than or equal to |*y*_{i}_{–1}| as well, by invariant 3. Therefore all the elements ofare less than or equal to |*x*_{i}*y*_{i}_{–1}|, hence to |σ|+|τ|. By property (A), it follows thatis above σ.*x*_{i}

In particular, ||≤|σ|≤|σ|+|τ|. By invariant 2 (at the end of round*x*_{i}*i*this time), all the elements ofare smaller than or equal to |*y*_{i}|, hence to |σ|+|τ|. We can therefore use property (A) once more, and conclude that*x*_{i}is above τ. This allows us to conclude that (*y*_{i},*x*_{i}) is above (σ, τ).*y*_{i}

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}*) is in*

*y*_{i}*U*

_{i}×

*V*

_{i}, it is in

*×*

*U*_{σ,|τ|}*U*

_{τ,|σ|}, and therefore, by invariant 4, (

*,*

*x*_{i}*) is above (σ, τ).*

*y*_{i}This is the place where we need the *two* successors *n*_{i}::* x_{i}* and

*n’*

_{i}::

*of*

*x*_{i}*guaranteed by invariant 3. Invariant 3 tells us that the downward closure of each one is included in*

*x*_{i}*U*

_{i}. In particular, both

*n*

_{i}::

*and*

*x*_{i}*n’*

_{i}::

*are in*

*x*_{i}*U*

_{i}, hence in

*. All the elements of*

*U*_{σ,|τ|}*are smaller than or equal to |*

*x*_{i}*| by invariant 1, hence to |τ|≤|σ|+|τ|, since*

*y*_{i}*is above τ. Invariant 3 also tells us that*

*y*_{i}*n*

_{i},

*n’*

_{i}≤ |

*|, hence they are also both less than or equal to |σ|+|τ|. By property (A), it follows that both*

*y*_{i}*n*

_{i}::

*and*

*x*_{i}*n’*

_{i}::

*are above σ. In other words, they are both suffixes of σ. But that is impossible, since*

*x*_{i}*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

*S*

_{0}is not consonant.

- Matthew de Brecht. Some results on countably based consonant spaces. Recent Developments in General Topology and its Related Fields, RIMS Kôkyûroku No. 2151, 2019.
- 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 - Ruiyuan Chen. Products of Scott topologies and spatiality. Note, last consulted February 19th, 2023.
- Ruiyuan Chen. A geometric proof that
*S*_{0}×*S*_{0}is not spatial. Note, last consulted February 19th, 2023.

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