The O functor does not preserve binary products

In Exercise 8.4.23 of the book, I said:

Exercise 8.4.21 may give you the false impression that the O functor preserves binary products. This is wrong, although an explicit counterexample seems too complicated to study here: see Johnstone (1982, 2.14).

My purpose here is to show that that is not that complicated after all.

My initial plan was to follow Isbell [1], Theorem 2.  The proof is only 5 lines, so that should be doable… or so I thought.  Isbell wrote in such a terse style that an incredible amount of mathematics would have to be called in just to explain those 5 lines.  That seems to have been part of his personality.  I’ll let you judge.  The following is what Isbell writes:

Theorem 2. If two subspaces of a Hausdorff space have no common point but a nonempty locale intersection, then their product locale is not a space.

Proof. Let A and B be such subspaces of X. The traces on A × B of the open rectangles Uα × Vα of X × X disjoint from the diagonal D cover the points of A × B. But if I is the intersection locale A B, the diagonal of I × I is contained in A × B and (in D) disjoint from ⋁Uα × Vα=W.  Thus W ∩ (A × B) is an open proper part of A × B containing all the points.

If you understand it, you certainly do not need to read the following.

What Isbell calls a ‘part’ is a sublocale.  He silently equates any topological space with the corresponding locale and every subspace with corresponding sublocales.  The so-called intersection locale is quite probably defined as a suitable pullback.  I do not know what D stands for.  (I think I have seen it defined in one of this other papers, but cannot find it back.)

Instead, I will give you an elementary proof.  This is quite probably the same as Isbell’s proof, and I was certainly guided by it.  I would be unable to say whether they are exactly the same at the moment.  Notably, I need the open subsets Uα and Vα above (which I’ll rename Ui and Vi, because I would like to reserve α for the first component of Galois connections) to be regular opens, although Isbell does not seem to make such an assumption.  That won’t cost me anything, as I will claim in the ‘separation by regular open subsets’ section below.

Accordingly, we shall spend some time studying the regular open subsets of a topological space, in relation with its dense subsets.  That will be needed in building certain Galois connections later—recall that the frame of Galois connections between two frames is their coproduct in Frm, hence their product in the category of locales.

Regular open subsets

A regular open subset of a topological space X is a subset U such that U=int(cl(U)) (see Exercise 8.1.8 in the book).  Of course every regular open is open: the interior operation int always gives you an open set as result.  But not all open subsets are regular.  For example, the union of open intervals (0, 1) ∪ (1, 2) in R is not regular: its closure is [0, 2], and the interior of the latter is (0, 2).

An alternative description of regular opens is as the regular elements of the frame O(X).  Those were introduced in Exercise 8.1.7.  A frame Ω is a complete Heyting algebra, meaning that there is an implication (or residuation) operation ⇒, with the defining property that uv is the largest element w such that uwv.  Equivalently, for all u, v, w in Ω, uwv if and only  if wuv.

When v = ⊥, we obtain intuitionistic negation ¬u. When Ω is the frame of opens O(X) of a topological space X, ¬U is the interior of the complement of U — not the complement of U, which would in general fail to be open.

The regular elements of a frame Ω are the elements u that are equal to their double negation ¬¬u.  Equivalently, those are exactly the elements that can be expressed as the negation ¬u of some u ∈ Ω.  That is so, because intuitionistic negation has the property that ¬¬¬u=¬u for every u (although ¬¬u≠u in general).

It turns out that the regular elements of O(X) are exactly the regular open subsets of X, as Exercise 8.1.8 asks you to check.  The idea is simply that ¬¬U=int(cl(U)).

Owing to the definition of intuitionistic negation (through implication), we shall use the following equivalence: for all u, v in Ω, uv = ⊥ if and only  if v ≤ ¬u, if and only if u ≤ ¬v.  When Ω=O(X), this translates to: for two open subsets U and V, U and V are disjoint if and only if V ⊆ ¬U, if and only if U ⊆ ¬V.

Separation by regular open subsets

Recall that a topological space X is Hausdorff, or T2, if and only if for every pair of distinct points xy, there are disjoint open subsets U and V such that U contains x and V contains y.

I claim that you can take both U and V regular open here.  In other words,

Lemma. A topological space X is T2 if and only if, for every pair of distinct points xy, there are disjoint regular open subsets U and V such that U contains x and V contains y.

Proof. The if direction is obvious.  In the only if direction, imagine we have disjoint open subsets U and V such that U contains x and V contains y.  Since U and V are disjoint, V ⊆ ¬U, and since ¬U=¬¬¬U, V ⊆ ¬¬¬U.  It follows that V and ¬¬U are disjoint.  Since ¬¬U and V are disjoint, ¬¬U ¬V = ¬¬¬V, so ¬¬U and ¬¬V are disjoint.

We have U ⊆ ¬¬U, so ¬¬U is again an open neighborhood of x.  Similarly, ¬¬V is an open neighborhood of y.  They are disjoint, and regular open.  ☐

Dense subspaces and regular opens

Fix a given dense subset A of a topological space X.

For every open subset U of X, UA is an open subset of A, viewed as a subspace of X.  Can we retrieve U from the data of UA?  I.e., is the map UUA injective?

That would seem so, because A is dense, but that is wrong in general.  For example, if you take the real line R for X, with its usual metric topology, and the subset Q of rationals for A, then (0, 5) and (0, π) ∪ (π, 5) have exactly the same intersection with A.

But that works with regular opens:

Lemma. Let A be a dense subset of a topological space X.  There is at most one regular open subset U of X such that UA is equal to a given open subset V of A, and that is int(cl(V)).  (Interior and closure are taken in X.)
Equivalently, for every regular open subset U of X, U=int(cl(UA)).

Proof. We first claim that if U is any open subset of X, then U is included in int(cl(UA)).  To prove this, let x be an arbitrary element of U.  For every open neighborhood U’ of xUU’ is open, non-empty (it contains x), so its intersects A, because A is dense: (UU’) ∩ A is non-empty, or equivalently, U’ ∩ (U ∩ A) is non-empty.  This shows that every open neighborhood U’ of x intersects U ∩ A, hence that x is in cl(UA).  (Otherwise, taken the complement of cl(UA) for U’.)  Since x is an arbitrary element of U, U is included in cl(UA).  Because U is open, it follows that U is included in the largest open included in cl(UA), namely int(cl(UA)).

Conversely, int(cl(UA)) is always included in int(cl(U)), and that is equal to U if U is regular open.

This show that if U is regular open and UA = V, then U = int(cl(V)), showing that such a U is unique.  ☐

Corollary. Let A be a dense subset of a topological space X.  For two regular open subsets U and V of X, U V if and only if UAVA.

Proof. If UAVA, then (U ∪ V) ∩ A = (UA) ∪ (VA) = VA.  The previous Lemma implies that U ∪ V = V, that is, U V.  ☐

Locale products

Let us start moving towards our goal.

Given two frames Ω1 and Ω2, the frame Gal(Ω1, Ω2) of Galois connections between Ω1 and Ω2 is their coproduct in the category Frm of frames.  (See Exercise 8.4.28 in the book.)

Among those Galois connections, the so-called (u1 × u2)-rectangles play a fundamental role.  (Definition 8.4.20.)  Those are the Galois connections (α, γ) such that α maps:

  • ⊥ to ⊤,
  • every u different from ⊥ and ≤ u1 to u2,
  • every u not below u1 to ⊥.

The other component γ is determined uniquely from α.

The canonical injections are ι1 : Ω1 → Gal(Ω1, Ω2) and ι2 : Ω2 → Gal(Ω1, Ω2), defined as follows:

  • for every u1 in Ω1, ι1(u1) is the  (u1 × ⊤)-rectangle;
  • for every u2 in Ω2, ι2(u2) is the  (⊤ × u2)-rectangle.

We now consider the case where Ω1=O(A) and Ω2=O(B), the frames of open subsets of two topological spaces A and B.  In that case, we also have frame homomorphisms π1-1 : Ω1O(A × B) and π2-1 : Ω2O(A × B), obtained as the inverse image maps by the two canonical projections.

By the universal property of the frame coproduct Gal(O(A), O(B)), there is a unique frame homomorphism ρ : Gal(O(A), O(B)) → O(A × B) such that ρ o ι1 = π1-1 and ρ o ι2 = π2-1.

Explicitly, ρ applied to any (U × ⊤)-rectangle, with U open in A, yields U × B, and ρ applied to any (⊤ × V)-rectangle, with V open in B, yields A × V.  By using Lemma 8.4.24, the (U × V)-rectangle is the inf of the (U × ⊤)-rectangle and of the (⊤ × V)-rectangle.  Since ρ preserves finite infima, ρ must map the (U × V)-rectangle to the open rectangle U × V.

We now use the fact that every Galois connection (α, γ) is the (pointwise) supremum of the (U × V)-rectangles below (α, γ), that is, where V ⊆ α(U) (Lemma 8.4.22).  So:

  • ρ : Gal(O(A), O(B)) → O(A × B) maps every (α, γ) to the union of the open rectangles U × α(U), where U ranges over the open subsets of A.

The question whether O preserves binary products is whether ρ is an order-isomorphism.  It certainly is when A or B is core-compact (Exercise 8.4.23).  In general, we shall see that it is not.  Precisely, we shall exhibit a counterexample where ρ is not injective.

Meanwhile, we observe that there is a map θ that really looks like an inverse to ρ.  (We are so close to showing that ρ is an order-isomorphism!)  That map, from O(A × B) to Gal(O(A), O(B)), is described in Proposition 8.4.18 of the book.  Explicitly:

  • θ : O(A × B) → Gal(O(A), O(B)) maps every open subset W of A × B to the Galois connection (α, γ) where α(U) is the largest open subset V of B such that U × VW.

We have:

Lemma. ρ o θ is the identity on O(A × B).

Proof. For every open subset W of A × B, ρ(θ(W)) is the union of the open rectangles U × α(U), where U ranges over the open subsets of A and α(U) is the largest open subset V of B such that U × VW.  In particular, ρ(θ(W)) is included in W.  Conversely, any open rectangle U × V included in W is included in U × α(U), hence in ρ(θ(W)).  ☐

In particular, ρ is surjective.

Building a counterexample

We take a non-empty T2 topological space X with two disjoint, dense subsets A and BX=R, with the set of rationals for A and the set of irrationals for B, is a natural example.

Look at the collection of pairs (U, V) of disjoint regular open subsets of X.  For clarity, organize them as a family (Ui, Vi), iI.  We observe:

Fact. The family (UiA, ViB), iI, is an open cover of A × B.

Indeed, for every element (x, y) of A × B, x and y are distinct.  Since X is T2, we can separate them by disjoint open subsets U and V, and by the first Lemma of this post, we can take both U and V regular open.  Hence U=Ui and V=Vi for some iI.

For each i ∈ I, we can build the ((UiA) × ( ViB))-rectangle (Definition 8.4.20, Lemma 8.4.22).  That is a particular Galois connection, which I will simply write (αi, γi).

Let me recall part of the definition: for each open subset W of A, αi(W) is:

  • the whole of B if W is empty,
  • ViB if W is non-empty but included in UiA,
  • and empty if W is not included in UiA.

I will not need γi, which is anyway uniquely determined from αi.

Since Gal(O(A), O(B)) is a frame, the family (αi, γi), i ∈ I, has a supremum which we denote by (α, γ).  That may not be the pointwise supremum of (αi, γi), i ∈ I, but we do not mind.  Incidentally, that is the supremum ⋁Uα × Vα=W Isbell is talking about, except for the fact that I require each Ui and each Vi to be regular open; we shall need that below.

The Galois connection (α, γ) = supi ∈ Ii, γi) is the localic analogue of the union of rectangles (UiA, ViB), iI.  By the Fact given earlier, that union is simply the whole product space A × B, and therefore one might think that (α, γ) is the (A × B)-rectangle.  We will start to understand that something is fishy when we realize that this is wrong (see the Proposition below).

The way we show this is with some little help from the following Galois connection.  This is (or should be) the explicit form of ‘the diagonal of I × I‘, where ‘I is the intersection locale A B‘ mentioned by Isbell.

Fact. Let α¬(W)=B—cl(W), γ¬(W’)=A—cl(W’): (α¬, γ¬) is an element of Gal(O(A), O(B)).
Moreover, for every regular open subset U of X, α¬(U ∩ A) = ¬U ∩ B.

Proof. Clearly α¬ and γ¬ are antitonic.  We have W’ ⊆ α¬(W) if and only if W’ and cl(W) are disjoint, if and only if W’ and W are disjoint.  (Because an open set intersects the closure of a set W if and only if it intersects W itself, see Corollary 4.1.28 in the book.)  Similarly, W ⊆ γ¬(W’) if and only if W and cl(W’) are disjoint, if and only if W and W’ are disjoint.

For the second part, let U be a regular open subset of X.  Clearly, cl(U ∩ A) is included in cl(U).  Conversely, since U is regular open, cl(U)=cl(int(cl(U ∩ A))) by the second Lemma in this post.  Since int(E) is included in E for any E and closure is idempotent, cl(U) is included in cl(U ∩ A).  This implies that cl(U)=cl(U ∩ A).  Then α¬(U ∩ A) = B—cl(U ∩ A) = B—cl(U) = ¬U ∩ B.  ☐

Isbell claims that ‘the diagonal of I × I is contained in A × B and (in D) disjoint from ⋁Uα × Vα=W‘.  In explicit form, that would be the following.

Lemma., γ) = supi ∈ Ii, γi) is below (α¬, γ¬) in Gal(O(A), O(B)).

Proof.  We have to show that for every i ∈ I, (αi, γi) is below (α¬, γ¬) in Gal(O(A), O(B)), namely that αi ≤ α¬.  Explicitly, this means showing that, for every open subset W of A, αi(W) is included in α¬(W).

If W is empty, then αi(W)=α¬(W)=B, and if W is not included in UiA, then αi(W) is empty.  It remains to examine the case where W is non-empty and included in UiA, in which case αi(W)=ViB.  Since W is included in UiA, int(cl(W)) is included in int(cl(UiA)), namely in UiA since Ui is regular open.  (Recall our second Lemma.)  We now remember that Ui and Vi are disjoint, so W and Vi must be disjoint, too.  Since Vi is open, cl(W) and Vi must be disjoint.  It follows that every element of αi(W)=ViB must be in B—cl(W)=α¬(W).  ☐

Proposition., γ) is not equal to the (A × B)-rectangle.

Proof. The (A × B)-rectangle is (αB, γA), where αB maps every open subset of A to B, and γA maps every open subset of B to A.  (See again Definition 8.4.20 in the book.)  For every regular open subset U of X, α(U ∩ A) ⊆ α¬(U ∩ A) = ¬U B, using the previous Lemma.  In particular, that is different from B as soon as U is non-empty: in that case, since B is dense, U ∩ B is non-empty; and any element b of U ∩ B will be outside ¬U ∩ B.  To conclude, we must show that there are non-empty regular open subsets U of X.  We may just take U=X.  ☐

Recall that θ : O(A × B) → Gal(O(A), O(B)) maps every open subset W of A × B to the Galois connection (α, γ) where α(U‘) is the largest open subset V’ of B such that U’ × V’W.  In particular, θ maps every non-empty open rectangle U × V to the (U × V)-rectangle.

Corollary. θ : O(A × B) → Gal(O(A), O(B)) does not preserve suprema.

Proof. By the first Fact we stated in this section, the family (UiA, ViB), iI, is an open cover of A × B.  In other words, the open rectangle A × B is the supremum of the family of open rectangles (UiA) × (ViB), iI.  By the previous Proposition, θ(A × B), the (A × B)-rectangle, is not equal to (α, γ), which is the supremum of the Galois connections (αi, γi)=θ((UiA) × (ViB)), iI.  ☐

Corollary. ρ is not injective.

Proof. Recall that ρ o θ = id.  Since ρ is a frame homomorphism (contrarily to θ, as we have just seen), it preserves suprema.  So A × B, which is the union of the open rectangles (UiA) × (ViB) = ρ(θ((UiA) × (ViB)))=ρ(αi, γi), iI, is also equal to ρ(α, γ).  However, A × B is also equal to ρ(θ(A × B)).  So ρ maps both (α, γ) and the (A × B)-rectangle θ(A × B) to the same open subset (A × B itself), of A × B.  We have seen that (α, γ) is different from the (A × B)-rectangle, and this allows us to conclude.  ☐

Since ρ is not injective, it certainly cannot be an order-isomorphism: we have reached our goal.  However, Isbell goes further, and shows that Gal(O(A), O(B)) is not even spatial.

The points of Gal(O(A), O(B))

This is the final sentence in Isbell’s proof: ‘Thus W ∩ (A × B) is an open proper part of A × B containing all the points.’  But what points?  Those of the topological space A × B, or those of the topological space pt(Gal(O(A), O(B)))?

It turns out that those are the same, up to homeomorphism.  To see that, you should refrain from trying to characterize the points of the frame Gal(O(A), O(B)).  There is a much simpler route.  Recall that Gal(O(A), O(B)) is the frame coproduct of O(A) and O(B), hence their locale product.  The functor pt, being right adjoint, preserves all limits (p.176 in the book), hence all products.  It follows that pt(Gal(O(A), O(B))) is a product of the spaces pt(O(A)) and pt(O(B)).  Now pt(O(Z)) is (homeomorphic to) the sobrification of the space Z (Proposition 8.2.22), the sobrification of a sober space Z is homeomorphic to Z itself (Fact 8.2.24 (c)), and every T2 space is sober (Proposition 8.2.12 (a)).  Since A and B are subspaces of the T2 space X, they are T2 as well, so pt(Gal(O(A), O(B))) is homeomorphic to A × B.

Hence, if Gal(O(A), O(B)) were spatial, then it would be order-isomorphic to O(A × B), and a little extra work shows that the isomorphism would have to be ρ.  (You need to remember that limits, in category theory, are not just spaces, but cones, that is, a space plus morphisms ending on that space, and you need to reason on those, too.)  We have seen that ρ is not even injective.

Hence we have proved Isbell’s Theorem:

Theorem. Let X be a non-empty T2 topological space with two disjoint, dense subsets A and B.  The frame Gal(O(A), O(B)) is not spatial. The frame homomorphism ρ : Gal(O(A), O(B)) → O(A × B) is surjective but not injective.

However, this took us a bit more than 5 lines!

  • John Isbell. Product spaces in locales. Proceedings of the American Mathematical Society, 81(1), January 1981.
  • Jorge Picado and Aleš Pultr. Frames and locales — topology without points.  Birkhäuser, 2010.

Jean Goubault-Larrecqjgl-2011