# Localic products and Till Plewe’s game

I have talked about Till Plewe’s topological game  last time, showing how Matthew de Brecht used it to show that the space S0 is not consonant . However, the original purpose of that game was to show that certain products are computed differently in the category of topological spaces and in the category of locales.

## Localic products, topological products

Let me recall how products are built in the category Top of topological spaces and continuous maps. The product X × Y of two topological spaces has open subsets W built as arbitrary unions of open rectangles U × V, where U ranges over OX and V ranges over OY.

Any such open set W gives rise to a Galois connection (αW, γW) between the frames OX and OY: αW(U) is defined as the largest open subset V of Y such that U × VW, for every UOX, and γW(V) is the largest open subset U of X such that U × VW, for every VOY. It is easy to see that αW and γW are antitonic, and that V ⊆ αW(U) if and only if U ⊆ γW(V) (if and only if U × VW); those are the properties defining a Galois connection.

In general, given two frames Ω1 and Ω2 (taking the place of OX and of OY respectively), a Galois connection between them is a pair of antitonic maps (α, γ), from Ω1 to Ω2 and from Ω2 to Ω1 respectively, satisfying the equivalence v ≤ α(u) if and only if u ≤ γ(v). Those Galois connections form a frame Gal(Ω1, Ω2); the ordering consists in comparing the α parts using the pointwise ordering, or equivalently the γ parts using the pointwise ordering. It does not matter whether we use α or γ, because any one of them determines the other one uniquely, and if one grows, the other grows as well.

Gal(Ω1, Ω2) is the coproduct of Ω1 and Ω2 in the category Frm of frames and frame homomorphisms. Therefore it is their product in the opposite category Loc=Frmop of locales.

This is how I introduce localic products in Section 8.4.4 of the book.

A funny thing is that the elements of Gal(Ω1, Ω2) are suprema of rectangles, just like in the case of topological spaces. Those rectangles have a strange definition, although it is exactly what you would expect if you look at the Galois connections defined from regular open rectangles U × V in the product of two topological spaces, and abstract away from spaces in order to obtain a formula that makes sense on all frames. Explicitly, given u ∈ Ω1 and v ∈ Ω2, the (u × v)-rectangle uv is the Galois connection (αuv, γuv) where αuv maps ⊥ to ⊤, every element below u and different from ⊥ to v, and all other elements to ⊥. (The γuv part is obtained in a unique way from α, and—as you might guess—by a symmetric formula. In the book, αuv is written as αu×v and γuv is written as γu×v, but I now find this small × sign visually disturbing. Note that I am also using the notation uv, just like Plewe ; so uv is the same thing as the Galois connection (αuv, γuv).)

The point is that every element (α, γ) of Gal(Ω1, Ω2) is the supremum of all the rectangles below it. Additionally, this supremum is pointwise: for every u’, α(u’) is the supremum of the family αuv(u’) where (αuv, γuv) ranges over the rectangles below (α, γ) (and similarly for γ(v’)). This is remarkable, since general suprema of Galois connections are not computed pointwise; only infima are computed pointwise.

We have seen that every open set W in the topological product X × Y defines a Galois connection between OX and OY. In other words, O(X × Y) sits inside Gal(OX, OY), through the mapping that sends W to (αW, γW). Let me call θ this mapping. Explicitly:

Definition. θ : O(X × Y) → Gal(OX, OY) maps every open subset W of X × Y to (αW, γW), where:

• αW(U) is defined as the largest open subset V of Y such that U × VW, for every UOX;
• γW(V) is the largest open subset U of X such that U × VW, for every VOY.

One may wonder whether the two frames are isomorphic, and, if you do not know the answer to that riddle yet, you might surmise that they are indeed isomorphic. But you would be wrong. We have O(X × Y) ≅ Gal(OX, OY) through θ if X or Y is core-compact, but the equality fails in general. I have already given John Isbell’s proof that this isomorphism fails when X and Y are two disjoint dense subsets of a non-empty T2 space , in a 2017 post.

My objective is to give another way of understanding how the isomorphism O(X × Y) ≅ Gal(OX, OY) may succeed or fail, following Till Plewe .

## Till Plewe’s game

I have already introduced Till Plewe’s game last time, but let me repeat it here. 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 GX,Y(U0, V0, U) is played as follows. At each round i≥1:

1. player I picks a point xiUi–1;
2. then player II picks an open neighborhood Ui of xi included in Ui–1;
3. player I picks a point yiVi–1;
4. then player II picks an open neighborhood Vi of yi included in Vi–1.

(Each player plays twice at each round; this is deliberate.)

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

Plewe’s Theorem 1.1  roughly states that Gal(OX, OY) is spatial if and only if player II has a winning strategy in this game, and otherwise player I has a winning strategy. He requires X and Y to be sober, but we will see that we can dispense with that assumption. The core of this post will consist in proving the following very similar result. That is an (apparently) weaker result, since we are asking when θ is an order-isomorphism, not whether there is a (possibly different) order-isomorphism between O(X × Y) and Gal(OX, OY). We will see later that the two results are, in fact, equivalent.

Theorem. For any two topological spaces X and Y, exactly one of the following mutually exclusive alternatives much occur:

• θ is an order-isomorphism of O(X × Y) onto Gal(OX, OY), and 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 GX,Y(U0, V0, U);
• or θ is not an order-isomorphism—in fact, θ is not surjective from O(X × Y) to Gal(OX, OY)—and there are an open subset U0 of X, an open subset V0 of Y, and an open cover U of U0 × V0 by open rectangles such that player I has a winning strategy in the game GX,Y(U0, V0, U).

Plewe really considers arbitrary products, but binary products will be enough for one post.

## C-ideals

The key to Plewe’s Theorem is the following alternate characterization of suprema (not pointwise suprema) in the localic product (=frame coproduct) of two frames.

A C-ideal on a pair of frames Ω1, Ω2 is a downwards closed set D of (localic) rectangles that is closed under C-suprema (for “suprema taken componentwise”), namely such that:

• for every u ∈ Ω1, for every family of elements (vi)iI with some supremum v in Ω2, such that each rectangle uvi is in D, then uv is also in D, and
• for every v ∈ Ω2, for every family of elements (ui)iI with some supremum u in Ω1, such that each rectangle uiv is in D, then uv is also in D.

Let me recall that uv stands for the Galois connection (αuv, γuv). The family (vi)iI can be taken to be empty in the first of the two conditions given above; therefore u ⊗ ⊥ is in D for every u ∈ Ω1; similarly, ⊥ ⊗ v is in D for every v ∈ Ω2.

There is an isomorphism of frames between Gal(Ω1, Ω2) and the poset of C-ideals on Ω1, Ω2 ordered by inclusion. The characterization of localic products as the space of ideals is from Dowker and Strauss ; the isomorphism with Gal(Ω1, Ω2) is due to Wigner , and is treated in Exercise 8.4.27 of the book.

• Given any element (α, γ) in Gal(Ω1, Ω2), the collection of open rectangles uv smaller than or equal to (α, γ) is a C-ideal; let me call that C-ideal Down(α, γ). The reason why this is a C-ideal is that uv ≤ (α, γ) if and only if v ≤ α(u) if and only if u ≤ γ(v) (Lemma 8.4.22 of the book); then the set of elements v that are below α(u) (for u fixed) is downwards closed and closed under all suprema, while the set of elements u that are below γ(v) (for v fixed) is also downwards closed and closed under all suprema.
• Conversely, given any C-ideal D, we can form a Galois connection Sup(D) ≝ (α, γ) as the pointwise supremum of all the rectangles in D. In other words, we define α(u’), for every u’ ∈ Ω1, as the supremum of {αuv(u’) | uvD}, and γ(v’), for every v’ ∈ Ω2, as the supremum of {γuv(v’) | uvD}. In order to see that this is a Galois connection, and therefore that it is the supremum of all the rectangles in D, we first simplify these expressions.
We recall that, for all u and v, αuv(u’) equals ⊤ if u’=⊥, v if u’≠⊥ and u’u, and ⊥ otherwise; therefore α(u’) is equal to ⊤ if u’=⊥, and otherwise to the supremum of all the elements v such that u’u for some rectangle uvD. Since D is downwards-closed, it is equivalent to require that u’u for some rectangle uvD, or to require that u’v itself be in D. Hence, if u’≠⊥, then α(u’) is equal to the supremum of all the elements v such that u’vD. Since D is closed under C-suprema, u’ ⊗ α(u’) is in D; in other words, α(u’) is the largest element v ∈ Ω2 such that u’vD. (This continues to hold if u’=⊥, since every element v is such that ⊥ ⊗ vD, so that their supremum is ⊤.)
Symmetrically, for every v’ ∈ Ω2, γ(v’) is the largest element u ∈ Ω1 such that uv’D.
The statements u’ ≤ γ(v’), v’ ≤ α(u’), and u’v’D are then equivalent; in particular, (α, γ) is a Galois connection, as desired.
• Given Sup(D) = (α, γ), as we have just defined it, Down(Sup(D)) is the collection of rectangles u’v’ below (α, γ), equivalently such that v’ ≤ α(u’), equivalently such that u’v’D. Therefore Down(Sup(D)) = D.
• For every Galois connection (α, γ), letting (α’, γ’) ≝ Sup(Down(α, γ)), we have that for every u’ ∈ Ω1, α'(u’) is the largest element v ∈ Ω2 such that u’v ∈ Down(α, γ), namely such that u’v ≤ (α, γ). But every Galois connection is the supremum of the rectangles below it, so Sup(Down(α, γ)) = (α, γ).
• Finally, it is clear that the map Down is monotonic. Since Sup is an inverse to Down, it is also monotonic. Therefore Down and Sup form a pair of mutually inverse order-isomorphisms.

## Localic suprema of rectangles

Plewe’s theorem is really about how one would compute suprema of rectangles in Gal(OX, OY). It is equivalent to consider suprema of rectangles in the space of C-ideals on OX, OY, by the correspondence between C-ideals and Galois connections that we have just seen (or recalled).

Hence let us fix a a family U of (topological) open rectangles Ui × Vi, iI. Each one defines a localic rectangle UiVi, and we wish to compute their localic supremum, namely their supremum in Gal(OX, OY).

We work in the isomorphic frame of C-ideals, and our problem is, equivalently, to find the supremum of the C-ideals Down(UiVi). That is, by definition, the smallest C-ideal D containing Down(UiVi) for every iI. We start off the construction by defining D0(U) as the union of the sets Down(UiVi), iI. This is downwards-closed, but is quite probably not closed under C-suprema. We repair this by defining D1(U) as all C-suprema of elements of D0(U). One can see that D1(U) will still be downwards-closed (we will see that below), but it may still not be closed under C-suprema, so we iterate the process for transfinitely long.

Explicitly, we define collections Dβ(U) of rectangles by induction on the ordinal β as follows. We have already defined D0(U). Assuming that we have built Dβ(U), and that it is downwards-closed, we let Dβ+1(U) be the collection of rectangles UV where either V=∪iI Vi for open sets Vi such that every rectangle UVi is in Dβ(U), or U=∪iI Ui where every Ui is open in X and every rectangle UiV is in Dβ(U). (Dβ+1(U) will still be downwards-closed. For example, if UV is obtained by taking V=∪iI Vi where UVi is in Dβ(U) for every i, and if U’V’UV, then U’ is included in U and V’ is included in V [or U’ or V’ is empty]; since Dβ(U) is downwards-closed, U’ ⊗ (ViV’) is in Dβ(U) for every i, hence therefore also U’ ⊗ ∪iI (ViV’) = U’ ⊗ (VV’) = U’V’.) When β is a limit ordinal, we define Dβ(U) as the union of Dβ’(U), where β'<β.

The process must stabilize at some ordinal β, namely all the sets Dβ(U) should be equal for β large enough. Otherwise the elements added at each stage would form a proper class of rectangles in the end; but the rectangles on OX and OY only form a set, not a proper class.

I will write D(U) for the union of all the sets Dβ(U). We have just argued that D(U) = Dβ(U) for β large enough. The fact that Dβ(U) = Dβ+1(U) means that Dβ(U) is closed under C-suprema. We have argued earlier that Dβ(U) was downwards-closed. Hence Dβ(U) is a C-ideal. It contains every rectangle UiVi, since D0(U) is included in Dβ(U). Also, every C-ideal D containing every rectangle UiVi must contain D0(U) and, by the fact that D is closed under C-suprema and by ordinal induction, Dβ(U) is included in D for every ordinal β. Therefore:

Fact A. For every family U of (topological) open rectangles Ui × Vi, D(U) is the supremum of the C-ideals Down(UiVi) in the space of C-ideals on the frames OX, OY. Therefore Sup(D(U)) is the supremum of the rectangles UiVi in Gal(OX, OY).

## When is θ an isomorphism?

We recall that θ maps every open subset W of X × Y to the Galois connection (αW, γW), where for every UOX, αW(U) is the largest open subset V of Y such that U × VW, and for every VOY, γW(V) is the largest open subset U of X such that U × VW.

The map θ is an order-embedding, meaning that WW’ if and only if αW≤αW’ (or equivalently, γW≤γW’), for all open subsets W and W’ of X × Y. The left-to-right direction follows from the fact that if U × VW then U × VW’, while the right-to-left direction is because, if αW≤αW’, then for every open rectangle U × VW, we have V ⊆ αW(U), hence V ⊆ αW’(U), hence U × VW‘; we use the fact that, by definition of αW, U × VW if and only if V ⊆ αW(U), and similarly with W’.

Hence θ is an isomorphism if and only if it is surjective.

Lemma B. The map θ is surjective (equivalently, an order-isomorphism) if and only if 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, D(U) contains U0V0.

Proof. Given any collection U of open rectangles in X × Y and any open subset W of X × Y, we claim that Sup(D(U)) = θ(W) if and only if: (*) for every open subset U0 of X, for every open subset V0 of Y, U0V0D(U) if and only if U0 × V0W.

Indeed, Sup(D(U)) = θ(W) if and only if D(U) = Down(θ(W)), using the fact that G and Down are mutual inverses. Now U0V0 ∈ Down(θ(W)) if and only if U0V0 is below θ(W) = (αW, γW). Using Lemma 8.4.22 in the book (a result we have already used several times in this post), U0V0 is below (αW, γW) if and only if V0 ⊆ αW(U0), if and only if U0 × V0W.

If θ is surjective, then for every family U of open rectangles, Sup(D(U)) = θ(W) for some open subset W of X × Y. Hence by (*), the open rectangles U0 × V0 included in W are exactly those such that U0V0 is in D(U). Every open rectangle U0 × V0 in U is such that U0V0 is in D0(U), hence in D(U), so U0 × V0 must be included in W. Therefore the union of the open rectangles in U is included in W; or equivalently, U is a cover of W by open rectangles. It follows that for every open subset U0 of X, for every open subset V0 of Y, if U0 × V0 is in U, in particular U0 × V0 is included in W, so by (*), used a second time, U0V0D(U). This shows the “only if” (left to right) direction of the Lemma.

Conversely, let us assume that the part of the Lemma that follows “if and only if” holds, namely that for every collection U of open rectangles in X × Y, for every open subset U0 of X and for every open subset V0 of Y such that U0 × V0 is included in the union W of the open rectangles in U, D(U) contains U0V0. By (*), this means that for every collection U of open rectangles in X × Y, Sup(D(U)) = θ(W), where W is the union of the open rectangles in U.

Let (α, γ) be any element of Gal(OX, OY). We recall that (α, γ) = Sup(Down(α, γ)), since Sup and Down are mutually inverse. Let us write Down(α, γ) as a collection of rectangles UiVi, iI. Let U be the corresponding collection of open rectangles Ui × Vi, iI. By Fact A, Sup(D(U)) is the supremum of the rectangles UiVi in Gal(OX, OY). But that supremum is also the supremum of Down(α, γ), namely Sup(Down(α, γ)), equivalently, (α, γ). Therefore (α, γ) = Sup(D(U)) = θ(W), showing that θ is surjective. ☐

Plewe’s theorem is now merely a game-theoretic rephrasing of Lemma B, as we now demonstrate.

## Player II wins when θ : O(X × Y) ≅ Gal(OX, OY)

Let us show the first half of the promised theorem: that if θ is an order-isomorphism from O(X × Y) onto Gal(OX, OY), then player II has a winning strategy in the game GX,Y(U0, V0, U), for every open subset U0 of X, for every open subset V0 of Y, and for every open cover U of U0 × V0 by open rectangles.

We fix U0, V0, and an open cover U of U0 × V0 by open rectangles as above. Let me recall how this game is played. At each round i≥1:

1. player I picks a point xiUi–1;
2. then player II picks an open neighborhood Ui of xi included in Ui–1;
3. player I picks a point yiVi–1;
4. then player II picks an open neighborhood Vi of yi included in Vi–1.

Player II wins at round i if Ui × Vi is included in one of the open rectangles of the cover U. Otherwise, player I wins.

If θ is an order-isomorphism, then by Lemma B, D(U) contains U0V0. Since D(U) is equal to the union of the sets Dβ(U), where β ranges over the ordinals, every rectangle in D(U) has a rank, which is by definition the least ordinal β such that the rectangle belongs to Dβ(U). We let β0 be the rank of U0V0. We will let player II play in such a way that after round i, UiVi has rank βi, where β0 > β1 > … > βi.

If βi=0, then UiVi is in D0(U), which is by definition the union of the sets Down(UV) where U × V ranges over the open rectangles of U, and therefore player II wins.

Let us assume that i≥1 and that player II has not won yet, that is, βi–1>0. Ui–1Vi–1 is in Dβi–1(U), and we argue that βi–1 cannot be a limit ordinal; otherwise, since Dβi–1(U) is the union of the sets Dβ(U) with β<βi–1, Ui–1Vi–1 would be in Dβ(U) for some β<βi–1, contradicting the fact that βi–1 is the least ordinal β such that Ui–1Vi–1 is in Dβ(U). Therefore βi–1 is a successor ordinal, say β+1, and Ui–1Vi–1 is obtained as a C-supremum of rectangles in Dβ(U). There are two cases: either Ui–1 is a union of open sets U’j (jJ) such that each rectangle U’jVi–1 is in Dβ(U), or Vi–1 is a union of open sets V’j (jJ) such that each rectangle Ui–1V’j is in Dβ(U).

• In the first case, once player I has picked a point xiUi–1, player II will play some U’j that contains xi for Ui; player I picks a point yiVi–1, and player II simply plays Vi–1 for Vi. Since UiVi = U’jVi–1 is in Dβ(U), it has rank βi≤β<βi–1.
• In the second case, player I picks a point xiUi–1, player II plays Ui–1 for Ui, then player I picks a point yiVi–1, and player II plays some V’j that contains yi for Vi. Since UiVi = Ui–1V’j is in Dβ(U), it has rank βi≤β<βi–1.

Since there is no infinite strictly decreasing sequence of ordinals, player II will eventually win.

## Player I wins if θ : O(X × Y) ≇ Gal(OX, OY)

If θ is not an order-isomorphism, namely if it is not surjective, then by Lemma B, there is a way of choosing U0, V0, and an open cover U of U0 × V0 in such a way that D(U) does not contain U0V0.

We verify that player I wins, whatever strategy player II uses. To this end, we will maintain the invariant that, at round i, UiVi is not in D(U). This is certainly true when i=0.

When i≥1, the invariant at round i–1 tells us that Ui–1Vi–1 is not in D(U).

• If we could find an open neighborhood U’x of every point x in Ui–1 such that U’xVi–1 were in D(U), then Ui–1Vi–1 would be a C-supremum of the rectangles U’xVi–1 where x ranges over Ui–1, hence would be in D(U). Therefore there is a point x in Ui–1—we pick one, call it xi, and this is the point that player I plays—such that whatever open neighborhood Ui of xi included in Ui–1 is played next by player II, UiVi–1 is still not in D(U).
• By a similar argument, if we could find an open neighborhood V’y of every point y in Vi–1 such that UiV’y were in D(U), then UiVi–1 would be a C-supremum of the rectangles UiV’y where y ranges over Vi–1, hence would be in D(U). Therefore there is a point y in Vi–1—we pick one, call it yi, and this is the point that player I plays—such that whatever open neighborhood Vi of yi included in Vi–1 is played next by player II, Ui–1Vi–1 is still not in D(U).

Since the game goes on forever, player I wins the game.

This concludes the proof of the theorem, which we now recall. As we have seen, the proof is really a rephrasing in game-theoretic terms of a construction of the suprema of rectangles in Gal(OX, OY) by ordinal induction.

Theorem. (Repeated from earlier in this post.) For any two topological spaces X and Y, exactly one of the following mutually exclusive alternatives much occur:

• θ is an order-isomorphism of O(X × Y) onto Gal(OX, OY), and 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 GX,Y(U0, V0, U);
• or θ is not an order-isomorphism—in fact, θ is not surjective from O(X × Y) to Gal(OX, OY)—and there are an open subset U0 of X, an open subset V0 of Y, and an open cover U of U0 × V0 by open rectangles such that player I has a winning strategy in the game GX,Y(U0, V0, U).

## Spatiality of products

Before I give a few applications of this result, let me notice that Plewe’s theorem is not about θ being an order-isomorphism or not, and rather is about whether Gal(OX, OY) is a spatial frame or not; in other words whether there exists an order-isomorphism between O(X × Y) and Gal(OX, OY) at all.

But that is equivalent. One can find the argument (for arbitrary products of sober spaces) in [6, Proposition IV-5.4.2]. I will give a proof below; it is somewhat tricky, and I will stress the tricky parts. The point is not so much in showing that if Gal(OX, OY) is spatial then it must be order-isomorphic to O(X × Y), but to show that the order-isomorphism in question must be θ. By the way, the assumption that X and Y are sober is useless (as we will show during the course of the proof), so we will drop it.

Lemma C. For all topological spaces X and Y, Gal(OX, OY) is spatial if and only if θ is an order-isomorphism, if and only if θ is surjective.

Proof. If θ is surjective, or equivalently an order-isomorphism, then O(X × Y) ≅ Gal(OX, OY), so Gal(OX, OY) is spatial.

Conversely, let us assume that L ≝ Gal(OX, OY) is spatial. We can assume that X and Y are sober, otherwise we replace X with its sobrification S(X), and similarly with Y. Indeed, X and S(X) have isomorphic lattices of open sets (Lemma 8.2.26 in the book) and S(X) is sober (Corollary 8.2.23), and similarly with Y; also, S(X × Y) is isomorphic to S(X) × S(Y) (Theorem 8.4.8). Hence, if we prove that θ is surjective from O(X × Y) to Gal(OX, OY) for all sober spaces X and Y, then the corresponding map from O(S(X × Y)) to Gal(O(S(X)), O(S(Y))) will be surjective for all spaces X and Y (not necessarily sober), and therefore also θ itself, from O(X × Y) to Gal(OX, OY). In any case, we will therefore assume, without loss of generality, that X and Y are sober.

There is an adjunction Opt (Theorem 8.1.26 in the book) between Top and Frmop, and with unit ηStone and counit εStone; we will not need to know how they are defined. By Proposition 8.1.17 of the book, the fact that L is spatial is equivalent to the fact that εLStone : LO(pt(L)) is an order-isomorphism.

We use the fact that L=Gal(OX, OY) is the coproduct of OX and OY in Frm, with canonical injections ιX : OXL mapping every open subset U of X to UY and ιY : OYL mapping every open subset V of Y to XV (Exercise 8.4.28), and we will use this to show that pt(L) is (homeomorphic to) the product of X and Y in Top. We need to make the homeomorphism explicit, too: we will use it to show that O applied to that homeomorphism (and composed with the inverse of εStone) is exactly θ, and that is how we will obtain that θ is an order-isomorphism.

Passing to opposite categories, L=Gal(OX, OY) is the product of OX and OY in Frmop. Since right adjoints preserve all limits, hence all products, pt(L) is a product of pt(O(X)) and pt(O(Y)) in Top, with canonical projections ptX) : pt(L) → pt(O(X)) and ptY) : pt(L) → pt(O(Y)).

Since X is sober, the map ηXStone : Xpt(O(X)) is a homeomorphism (Proposition 8.2.22), and similarly with Y. Therefore pt(L) is also a product of X and Y in Top, with canonical projections (ηXStone)–1 o ptX) : pt(L) → X and (ηYStone)–1 o ptY) : pt(L) → Y. But we have another product, namely X × Y, so the pairing 〈(ηXStone)–1 o ptX), (ηYStone)–1 o ptY)〉 : pt(L) → X × Y is a homeomorphism.

We apply the O functor, and we obtain that O 〈(ηXStone)–1 o ptX), (ηYStone)–1 o ptY)〉 : O(X × Y) → O(pt(L)) is an order-isomorphism. Therefore (εLStone)–1 o O 〈(ηXStone)–1 o ptX), (ηYStone)–1 o ptY)〉 : O(X × Y) → L is also an order-isomorphism. We claim that this order-isomorphism—call it ϴ for short—is nothing else than θ.

In order to see this, one would like to apply both ϴ and θ to open rectangles U × V, and to check whether we get the same values. One would be tempted to say that this is enough because every element of O(X × Y) is a union of open rectangles, and because both ϴ and θ are frame homomorphisms, but the latter claim is incorrect. ϴ is a frame homomorphism, but we do not know that much about θ (that is largely what we are trying to prove, in fact), and in particular we do not know whether θ preserves suprema.

Still, it is a good idea to apply ϴ to open rectangles U × V, and to see what we get. Let us write pX : X × YX and pY : X × YY for the canonical projections. Then U × V = pX–1(U) ∩ pY–1(V), and ϴ(U × V) = ϴ(pX–1(U)) ∩ ϴ(pY–1(V)). We concentrate on the computation of ϴ(pX–1(U)), since ϴ(pY–1(V)) will be obtained in a symmetric way. We have:

• ϴ(pX–1(U)) = ϴ(O(pX)(U))
• = (εLStone)–1 [ O 〈(ηXStone)–1 o ptX), (ηYStone)–1 o ptY)〉 [O(pX)(U)] ] by definition of ϴ
• = (εLStone)–1 [ O (pX o 〈(ηXStone)–1 o ptX), (ηYStone)–1 o ptY)〉) [U] ] since O is a (contravariant) functor
• = (εLStone)–1 [ O ((ηXStone)–1 o ptX)) (U) ] using the law pX o 〈a, b〉 = a.

One of the coherence diagrams for adjunctions (see Section 5.5.2 in the book, and taking into account the fact that O is contravariant) is that OXStone) o εOXStone = idOX, so O ((ηXStone)–1) = εOXStone. Therefore O ((ηXStone)–1 o ptX)) = O(ptX)) o O ((ηXStone)–1) = O(ptX)) o εOXStone, and that is equal to εLStone o ιX, by the naturality of εStone. Hence O ((ηXStone)–1 o ptX)) (U) = (εLStone o ιX) (U) = εLStone (UY) (by definition of ιX).

We apply (εLStone)–1, and therefore ϴ(pX–1(U)) = UY. Symmetrically, ϴ(pY–1(V)) = XV, so ϴ(U × V) = ϴ(pX–1(U)) ∩ ϴ(pY–1(V)) = (UY) ∩ (XV), and that is equal to (UX) ⊗ (YV) by Lemma 8.4.24 in the book, hence to UV.

For every open subset W of X × Y, we remember that θ(W) = (αW, γW), just like any element of L=Gal(OX, OY), is the (pointwise) supremum of all the rectangles UV below it. Now UV is below (αW, γW) if and only if V ⊆ αW(U), if and only if U × VW. Therefore θ(W) is the supremum of all the rectangles UV = ϴ(U × V), where U × V ranges over the open rectangles included in W. Since ϴ preserves suprema, it follows that θ(W)=ϴ(W). (Notice how we never assumed that θ preserves suprema. We have just recalled that every Galois connection is a supremum of rectangles, and we have looked at what it means for a rectangle to be below a Galois connection of the form (αW, γW).)

At any rate, we have just proved that θ=ϴ. Since ϴ is an order-isomorphism, so is θ. (And, yes, now at last we know that θ preserves suprema!) ☐

Hence we obtain the final form of Plewe’s theorem.

Theorem (Plewe). For any two topological spaces X and Y, exactly one of the following mutually exclusive alternatives much occur:

• Gal(OX, OY) is spatial, θ is an order-isomorphism of O(X × Y) onto Gal(OX, OY), and 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 GX,Y(U0, V0, U);
• or Gal(OX, OY) is not spatial, θ is not an order-isomorphism—in fact, θ is not surjective from O(X × Y) to Gal(OX, OY)—and there are an open subset U0 of X, an open subset V0 of Y, and an open cover U of U0 × V0 by open rectangles such that player I has a winning strategy in the game GX,Y(U0, V0, U).

## Applications

### 1. The localic product of S0 with itself is not spatial

A first, direct application is Matthew de Brecht’s proof that the localic product of S0 with itself is not spatial . In other words, using Lemma C, that θ is not an order-isomorphism O(S0 × S0) onto Gal(OS0, OS0). Indeed, last time, we have seen that S0 is not consonant, and more specifically that player I has a winning strategy in the game GX,X(X, X, U), for some open cover U of X × X (with XS0). Plewe’s theorem tells us immediately that θ is not an order-isomorphism, and therefore the localic product of S0 by itself is not spatial.

### 2. The localic product of Q with R–Q is not spatial

In a 2017 post, I had argued that, given any non-empty T2 topological space Z with two disjoint, dense subsets X and Y, Gal(OX, OY) is not spatial (and θ does not preserve suprema). This is due to J. Isbell . Plewe says that this can be seen using his game (Section of , final paragraph); and it ought to, since the game entirely characterizes the cases where Gal(OX, OY) is spatial.

One trick that makes this work is the following equivalence. I feel that it ought to be written explicitly, otherwise Plewe’s description of the game in this setting remains somewhat mysterious.

Fact D. For any two subsets U and V of a topological space Z, U intersects int(cl(V)) if and only if V intersects int(cl(U)).

Here int and cl are interior and closure in Z, respectively. (U and V will be open subsets in X and in Y respectively below, but are not open in Z.)

Proof. Let us write ¬ for the “interior of complement” operator. The point is that OZ is a frame, hence a complete Heyting algebra, and ¬A stands for A ⇒ ⊥, where ⇒ is the residuation operator (intuitionistic implication) and ⊥ is the bottom element (the empty set); see Exercise 8.1.6 in the book. This allows us to reason in intuitionistic logic, with ⋁ (“or”) being union and ⋀ (“and”) being intersection, while ≤ (“consequence”) is inclusion. The double negation operator ¬¬ coincides with int(cl(_)), and therefore the claim amounts to the fact that U ⋀ ¬¬V ≠ ⊥ if and only if V ⋀ ¬¬U ≠ ⊥. By taking negations, and realizing that being equal to ⊥ is equivalent to being ≤ ⊥, we must show that U ⋀ ¬¬V ≤ ⊥ if and only if V ⋀ ¬¬U ≤ ⊥. By symmetry, it suffices to show that U ⋀ ¬¬V ≤ ⊥ implies V ⋀ ¬¬U ≤ ⊥. If U ⋀ ¬¬V ≤ ⊥, then ¬¬V ≤ ¬U, by definition of residuation. Since negation is antitonic, it follows that ¬¬U ≤ ¬¬¬V. But ¬¬¬V = ¬V (Exercise 8.1.6, item (iv)), so ¬¬U ≤ ¬V. By definition of residuation, V ⋀ ¬¬U ≤ ⊥. ☐

We come back to the case where X and Y are two disjoint dense subsets of a non-empty T2 topological space Z. We recall that a regular open subset of Z is an open subset U’ such that U’ = int(cl(U’)), namely such that U’ = ¬¬U’.

We play GX,Y(U0, V0, U) where U0X, V0Y, and U is the open cover of X × Y consisting of all the open rectangles (U’X) × (V’Y), where U’ and V’ range over pairs of disjoint regular open subsets of Z. (Plewe seems to have forgotten to say “regular” here.) This is a cover, because any two distinct points of a T2 space, in particular a point in X and a point in Y, can be separated by two disjoint regular open neighborhoods. (We have seen this in the 2017 post mentioned above.)

We will maintain the invariant that Ui intersects int(cl(Vi)), or equivalently (by Fact D) that Vi intersects int(cl(Ui)). That is true for i=0, since by density int(cl(V0)) = int(Z)=Z and U0=X, so U0 ∩ int(cl(V0)) = X is non-empty, since it is a dense subset of a non-empty space.

This invariant has the consequence that player II cannot win at round i. Indeed, if Ui × Vi were included in (U’X) × (V’Y) where U’ and V’ are regular open and disjoint, then we would have UiU’X and ViV’Y (because neither Ui nor Vi is empty, since Ui intersects int(cl(Vi))); hence int(cl(Vi)) ⊆ int(cl(V’)) = V’ (this is where we need V’ to be regular), and that implies that the non-empty set Ui ∩ int(cl(Vi)) would be included in U’V’, which is empty by definition.

In any case, at round i≥1, we assume that Ui–1 and Vi–1 satisfy the invariant. We let player I pick xiUi–1 ∩ int(cl(Vi–1)). This is possible since the invariant is precisely that Ui–1 ∩ int(cl(Vi–1)) is non-empty. Now let player II play Ui, where Ui is an arbitrary open neighborhood of xi in X included in Ui–1. Then Ui ∩ int(cl(Vi–1)) is non-empty, since it contains xi. By Fact D, Vi–1 ∩ int(cl(Ui)) is non-empty, so we let player I pick a point yiVi–1 ∩ int(cl(Ui)). Player II plays an arbitrary open neighborhood Vi of yi in Y included in Vi–1. Then Vi ∩ int(cl(Ui)) is non-empty, since it contains yi. By Fact D, this is equivalent to the fact that Ui ∩ int(cl(Vi)) is non-empty, therefore the invariant is satisfied of Ui and Vi.

We have obtained the following.

Proposition E. Let X and Y are two disjoint dense subsets of a non-empty T2 topological space Z. Player I has a winning strategy in Plewe’s game GX,Y(X, Y, U), where U is the open cover of X × Y consisting of all the open rectangles (U’X) × (V’Y), where U’ and V’ range over pairs of disjoint regular open subsets of Z.

We note that this strategy is stationary (or positional), meaning that player I only has to look at the current values of the open sets Ui–1 and Vi–1 (or Ui and Vi–1, the second time she or he plays during the same round), and not at the open sets or at the points played earlier in the game.

By Plewe’s Theorem, we have obtained a new proof of the following result, already described in the 2017 post mentioned above, and due to J. Isbell .

Theorem. Let X and Y are two disjoint dense subsets of a non-empty T2 topological space Z. Then Gal(OX, OY) is not spatial.

This applies notably to the case where X is the set of rational numbers, Y is the set of irrational numbers, and Z is R.

### 3. The localic product of Q with itself is not spatial… and Q is not consonant

But let’s look at another case. That will be the last topic I will deal with this time. Let us imagine that X and Y are, once again, two disjoint dense subsets of a non-empty T2 topological space Z, but that, additionally, they are isomorphic. Let f : XY be the isomorphism. This is the case, for example, if X is the set Q of rational numbers, and Y is an isomorphic but disjoint copy of it sitting inside R, for example Q+√2.

By Proposition E, player I has a winning strategy in Plewe’s game GX,Y(X, Y, U) for some open cover U of X × Y by open rectangles. This implies that player I also has a winning strategy in GX,X(X, X, U’), where U’ ≝ {U’ × f–1(V’) | U’ × V’U’}. Explicitly, player I on X × X simulates player I on X × Y as follows. In order to make things clearer, let me put primes on the names of players (I’, I”) and on the points (x’i, y’i) and the open sets (U’i, V’i) played on X × X, while keeping the unprimed notations for the original game on X × Y. Initially, U’0 = U0 = X and V’0 = f[V0] = Y. At every round i≥1, we are given U’i–1 and V’i–1, and then player I plays as follows.

• We form Ui–1U’i–1 and Vi–1f–1(V’i–1), and we let player I (in the game on X × Y) play a point xi in Ui–1. (Player I only needs Ui–1 and Vi–1 to do so, because her or his strategy is stationary; but that is not crucial, and only serves to simplify the exposition.) Player I’ simulates that move by playing x’ixi.
• Now player II’ plays an open neighborhood U’i of x’i in X included in U’i–1. (We are not assuming that player II’ plays with a stationary strategy, and she or he may look at all the previous points and open sets played during the game.)
• We let UiU’i, and we let player I play a point yi in Vi–1, given its knowledge of Ui and Vi–1. Player I’ simulates that move that playing y’if(yi).
• Player II’ then plays an open neighborhood V’i of y’i in Y included in V’i–1.

Finally, player II’ wins if for some i the open rectangle U’i × V’i is included in one of the open rectangles in U’, or equivalently if some open rectangle Ui × Vi is included in one of the open rectangles in U; otherwise, player I’ wins. In particular, since player I has a winning strategy in GX,Y(X, Y, U), player I’ has a winning strategy in GX,X(X, X, U’).

Now, the fact that player I’ has a winning strategy in GX,X(X, X, U’) means two things. By Plewe’s theorem, Gal(OX, OX) is not spatial; and by Matthew de Brecht’s theorem [1, Theorem 4], seen last time, X is not consonant. We sum this up as follows.

Theorem. Let X be a space that has two isomorphic, disjoint copies as dense subsets of a non-empty T2 topological space Z. Then Gal(OX, OX) is not spatial and X is not consonant. In particular, the localic product Gal(OQ, OQ) is not spatial, and Q is not consonant.

That proof of the non-consonance of Q is hopefully quite a lot simpler than the one I described a few months ago! It was hinted in the following terms at the end of the introduction of , and must therefore be credited to Matthew de Brecht:

The connection with C-ideals should make it an easy exercise for the reader to convert P. Johnstone’s proof in  of the non-spatiality of the localic product Q ×loc Q into a winning strategy for the first player in G(Q), thereby obtaining a new proof that Q is dissonant.

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. John Isbell. Product spaces in locales. Proceedings of the American Mathematical Society, 81(1), January 1981.
4. Clifford Hugh Dowker and Dona Strauss. (1976). Products and sums in the category of frames. In: Binz, E., Herrlich, H. (eds) Categorical Topology. Lecture Notes in Mathematics, vol 540. Springer, Berlin, Heidelberg, 1976, pages 208–219. https://doi.org/10.1007/BFb0080860
5. David Wigner. Two notes on frames. Journal of the Australian Mathematical Society (Series A), Volume 28, 1979, page 257–268.
6. Jorge Picado and Aleš Pultr. Frames and locales — topology without points. Birkhäuser, 2010.
7. Matthew de Brecht.  A note on the spatiality of localic products of countably based spaces. Workshop on Computability, Continuity, Constructivity – from Logic to Algorithms (CCC 2019), Ljubljana, Slovenia, 2019.

Jean Goubault-Larrecq (March 20th, 2023)