Reinhold Heckmann has a very nice paper [1] in which he proves the following: every countably presented locale is spatial. As he himself mentions, the result was first proved by Fourman and Grayson in 1982. The new thing is that he shows how tightly this is connected with the Baire property. This also gives a localic description of Matthew de Brecht’s quasi-Polish spaces, as those whose frame of open sets is countably presented.

## Completely Baire spaces

A topological space is *Baire* if and only if every countable intersection of dense open sets is dense. All complete metric spaces are Baire, but there are more. The following was first proved by John Isbell in a localic context [2]. That is Proposition 8.3.24 in the book.

**Lemma 1.** Every locally compact sober space is Baire.

One can say more. A space is *completely Baire* if and only if all its closed subspaces are Baire. There are Baire spaces that are not completely Baire. For example, take the set **Q** of rational numbers with the usual metric topology. This is not Baire, since the intersection of the open sets **Q**–{*q*}*,* *q* *∈* **Q**, is empty. However, **Q **with a new top element added above all rational numbers, and with the topology whose open sets are the open sets of **Q** union top, plus the empty set, is Baire… because every non-empty open set contains the top element. And **Q **is a closed subspace of that, which is not Baire. (In a previous version, I used **N** instead of **Q**, but that is silly, as Zhenchao Lyu pointed out to me.)

**Lemma 2.** Every locally compact sober space is completely Baire.

Indeed, it is easy to see that every closed subspace of a locally compact sober space is locally compact and sober.

Heckmann calls the completely Baire spaces spaces with the *local Baire* property. I am using the name that Matthew de Brecht uses in his papers.

**Π**^{0}_{2} subsets

Matthew de Brecht does a lot with so-called **Π**^{0}_{2} subsets [3], and so does Heckmann, under a different name.

Let us fix a topological space *X*. Following Heckmann, let us call *UCO subset* of *X* any space that is the union of a closed subset *C* and of an open subset *V* of *X*. (UCO stands for “union of closed and open”.)

If we write *C* as the complement of an open set *U*, then our UCO set can be written as the set of points *x* such that *x* ∈ *U* implies *x* ∈ *V*. (Recall that “*A* implies *B*” means the same thing as “not *A* or *B*“.) Then we can write this set under the convenient notation *U* **imp** *V*, which I will also use even if *U* is not open. (I will not use ⇒ for implication **imp**, because that would be in conflict with the residuation operation on frames which we will use below.)

A **Π**^{0}_{2} subset of *X* is then a countable intersection of UCO sets, namely the subsets of all points satisfying all implications *x* ∈ *U _{n}* implies

*x*∈

*V*, where

_{n}*U*and

_{n}*V*are fixed open subsets of

_{n}*X*,

*n*∈

**N**.

Note that all open subsets, all closed subsets, and all G_{δ} subsets, are **Π**^{0}_{2}.

Heckmann notices that the Baire property extends to **Π**^{0}_{2} subsets.

**Proposition 3.** In a Baire space, every countable intersection of dense **Π**^{0}_{2} subsets is dense.

Proof. Let *X* be a Baire space. If *A* is dense, and is a countable intersection of subsets, then each of these subsets is dense. Moreover, a countable intersection of countable intersections of UCO sets is a countable intersection of UCO sets. With these remarks in mind, we see that it is enough to show that every countable intersection of dense UCO subsets *U _{n}*

**imp**

*V*, is dense.

_{n}Let *O _{n}* be the set cl(

*U*)

_{n}**imp**

*V*, where cl(

_{n}*U*) is the closure of

_{n}*U*. Note that

_{n}*O*is open. We claim that

_{n}*O*is dense. For every non-empty open set

_{n}*U*, we must show that

*U*intersects

*O*. We consider two cases:

_{n}- If
*U*intersects*U*, then_{n}*U*∩*U*is another non-empty open set. It must intersect the dense set_{n}*U*_{n}**imp**V, say at_{n}*x*, and we see that*x*must then be in*V*. In that case_{n}*U*intersects*V*at_{n}*x*, hence also the larger set*O*._{n} - If
*U*does not intersect*U*, then it does not intersect cl(_{n}*U*) either. We pick_{n}*x*in*U*. Since*x*is in the complement of cl(*U*), it is in_{n}*O*. Hence again_{n}*U*intersects*O*._{n}

Now that we know that *O _{n}* is dense for each

*n*, their intersection is dense, by the Baire property. But each

*O*

*is included in*

_{n}*U*

_{n}**imp**V

*, so the intersection of the latter when*

_{n}*n*varies must be dense as well. ☐

## Sublocales defined by relations

Let us now enter the world of frames and locales. Reading one of my previous posts on the subject might be a good idea at this point. For example, this one talks about the various presentations one can give of sublocales, the pointfree analogue of subspaces.

Imagine we start from a frame Ω, and we wish to define a sublocale. One way of doing so is by defining a frame congruence ≡ on Ω, namely an equivalence relation on Ω that is compatible with finite infima and all suprema.

How do we specify such a frame congruence? A popular method is to list a few axioms it should satisfy (infinitely many if you wish), and to consider the least frame congruence satisfying the axioms. The axioms are of the form *u*≡*v*, where *u* and *v* are elements from Ω.

One can even specify axioms of the form *u*⪯*v*, meaning that we would like *u* to be considered as being below *v* in the quotient frame: we take this as an abbreviation for the axiom *u* ≡ *u* ∧ *v*.

Once the frame congruence ≡ has been defined, we obtain a nucleus ν defined by: ν(*u*) is the largest element that is equivalent to *u* with respect to ≡. This now defines a sublocale, the sublocale of all fixed points of ν.

Let us look at a few examples.

## Axiomatizing open sublocales

Fix *v*_{0} in Ω, and consider the frame congruence axiomatized by the single axiom *v*_{0}=⊤. We will see that this defines the open sublocale defined by *v*_{0}.

That axiom defines a frame congruence ≡, and I claim that for all *u* and *v*, *u*≡*v* if and only if *u* ∧ *v*_{0}=*v* ∧ *v*_{0}.

Indeed, let ≡’ be the frame congruence defined by *u*≡’*v* if and only if *u* ∧ *v*_{0}=*v* ∧ *v*_{0}. If *u*≡’*v*, then * u* =

*u*∧ ⊤ ≡

*u*∧

*v*

_{0}=

*v*∧

*v*

_{0}≡

*v*∧ ⊤ =

*v*. Conversely, we have

*v*

_{0}≡’ ⊤, so ≡’ satisfies the axiom. Since ≡ is the least frame congruence satisfying the axiom,

*u*≡

*v*implies

*u*≡’

*v*, for all

*u*and

*v*.

Now the largest element that is ≡-equivalent to *u* is *v*_{0} ⇒ *u*. (Exercise. Use the fact that *v*_{0} ⇒ *u* is the largest *v* such that *v* ∧ *v*_{0} ≤ *u*, hence such that *v* ∧ *v*_{0} ≤ *u* ∧ *v*_{0}.) Hence the associated nucleus is the map that sends every *u* to *v*_{0} ⇒ *u*. This is the *open nucleus* **o**(*v*_{0}), as we have already seen here.

If Ω is the open set lattice **O***X* of a topological space *X*, note that the lattice of open subsets of the open subspace *v*_{0} is exactly the lattice of open subsets *U* of *X* modulo the fact that two open sets that have the same intersection with *v*_{0} must be considered equal. This is exactly our relation ≡. In particular, if Ω is spatial, then the sublocale defined as the set of fixed points of the open nucleus **o**(*v*_{0}) is spatial as well, and is isomorphic to the frame of open subsets of the open subspace *v*_{0}.

## Axiomatizing closed sublocales

Let us now fix an element *u*_{0} of Ω. Consider the axiom *u*_{0}*=*⊥. This should now represent what remains when we erase *u*_{0}, and if *u*_{0} is seen as an open set, that should represent its complement, a closed set.

The associated frame congruence is defined by *u*≡*v* if and only if *u* ⋁ *u*_{0} = *v* ⋁ *u*_{0}. The argument is similar to the previous case of open sublocales. The associated nucleus is the *closed nucleus* **c**(*u*_{0}), which maps every *u* to *u* ⋁ *u*_{0}.

By a similar argument as above, it follows that if Ω is spatial, then the sublocale defined as the set of fixed points of the closed nucleus **c**(*u*_{0}) is spatial as well, and is isomorphic to the frame of open subsets of the closed subspace defined as the complement of the open set *u*_{0}.

## UCO sublocales

We now fix two elements *u*_{0} and *v*_{0} of Ω. We again consider just one axiom, and now this is the inequality *u*_{0}⪯*v*_{0}, i.e., *u*_{0} ≡ *u*_{0} ∧ *v*_{0}. Replacing *v*_{0} by *u*_{0} ∧ *v*_{0}, this can be rewritten as the axiom *u*_{0} ≡ *v*_{0}, where *v*_{0}≤*u*_{0}.

The associated frame congruence is harder to characterize in a synthetic way. I claim that *u*≡*v* if and only if the following two conditions hold:

*u*∧*v*_{0}=*v*∧*v*_{0}*u*⋁*u*_{0}=*v*⋁*u*_{0}.

Note that the first condition looks like the condition we had for open sublocales, and that the second condition looks like the one we had for closed sublocales.

Let us check this. We let *u*≡’*v* if and only if the two conditions above hold. We first claim that if *u*≡’*v*, then *u*≡*v*, where ≡ is axiomatized by the unique axiom *u*_{0} ≡ *v*_{0}. This is done in a series of steps:

- First,
*u*= (*u*⋁*v*_{0}) ∧ (*v*_{0}⇒*u*). This is always true. Indeed, in one direction,*u*is below both*u*⋁*v*_{0}and*v*_{0}⇒*u*, hence also below their infimum. In the other direction, (*u*⋁*v*_{0}) ∧ (*v*_{0}⇒*u*) = (*u*∧ (*v*_{0}⇒*u*)) ⋁ (*v*_{0}∧ (*v*_{0}⇒*u*)) (by distributivity) ≤*u*⋁*u*=*u*. - Similarly,
*v*= (*v*⋁*v*_{0}) ∧ (*v*_{0}⇒*v*). - Next, we have (
*v*_{0}⇒*u*) = (*v*_{0}⇒*v*). We just show that (*v*_{0}⇒*u*) ≤ (*v*_{0}⇒*v*), as the other inequality is similar. (*v*_{0}⇒*u*) ∧*v*_{0}is below*u*by the definition of ⇒, and also below*v*_{0}, hence below*u*∧*v*_{0}. Since the latter is equal to*v*∧*v*_{0}, hence below*v*, (*v*_{0}⇒*u*) ∧*v*_{0}is below*v*, so (*v*_{0}⇒*u*) is below (*v*_{0}⇒*v*). (In fact*u*∧*v*_{0}=*v*∧*v*_{0}is*equivalent*to (*v*_{0}⇒*u*) = (*v*_{0}⇒*v*).) - It follows that
*u*= (*u*⋁*v*_{0}) ∧ (*v*_{0}⇒*u*) = (*u*⋁*v*_{0}) ∧ (*v*_{0}⇒*v*). The latter is ≡-equivalent to (*u*⋁*u*_{0}) ∧ (*v*_{0}⇒*v*) (we at last use the axiom*u*_{0}≡*v*_{0}), which is equal to (*v*⋁*u*_{0}) ∧ (*v*_{0}⇒*v*) (by the second condition), itself ≡-equivalent to (*v*⋁*v*_{0}) ∧ (*v*_{0}⇒*v*) =*v*(by the axiom*u*_{0}≡*v*_{0}).

Conversely, we have *u*_{0} ≡’ *v*_{0}, because *v*_{0}≤*u*_{0}. Therefore ≡’ is one frame congruence satisfy the required axiom, and since ≡ is least, *u*≡*v* implies *u*≡*‘v* for all *u*, *v* in Ω.

This concludes our argument that ≡’ and ≡ are the same relation.

Now assume that Ω is equal to the lattice **O***X* of open subsets of some space *X*. Let *C*_{0 }be the closed subset obtained as the complement of *u*_{0}. Then *C*_{0} and *v*_{0} are disjoint, and *C*_{0} ∪ *v*_{0} is a UCO subset of *X*, which we wrote earlier as *u*_{0} **imp** *v*_{0}.

The open subsets of the subspace *u*_{0} **imp** *v*_{0} are the intersections of open subsets *U* of *X* with *C*_{0} ∪ *v*_{0}. Equivalently, they are the open subsets of *X*, modulo the equivalence that equates two open subsets *U*, *V* if they have the same intersection with *C*_{0} ∪ *v*_{0}. Since *C*_{0} and *v*_{0} are disjoint, *U* and *V* are equivalent in this sense if and only if *U* ∩ *C*_{0} = *V *∩ *C*_{0} and *U* ∩ *v*_{0} = *V *∩ *v*_{0}. I will let you verify that *U* ∩ *C*_{0} = *V *∩ *C*_{0} is equivalent to *U* ∪ *u*_{0} = *V* ∪ *u*_{0}. Hence *U* and *V* are equivalent in this sense if and only if:

*U*∩*v*_{0}=*V*∩*v*_{0}*U*∪*u*_{0}=*V*∪*u*_{0}

But those are exactly our conditions defining the equivalence defined by the single axiom *u*_{0} ≡ *v*_{0}.

To sum up: for two open subsets *u*_{0} and *v*_{0} of a topological space *X*, with *v*_{0}≤*u*_{0}, the sublocale axiomatized by the single equation *u*_{0} ≡ *v*_{0} is spatial, and can be equated with **O**(*u*_{0} **imp** *v*_{0}).

## More general subspaces

At this point, it is tempting to think that every sublocale of a spatial frame would be spatial as well. This is wrong. The main counterargument is based on Isbell’s density theorem: every frame contains a smallest dense sublocale *S*_{0}. (Recall that a sublocale is dense if and only if it contains ⊥.)

Here is the complete argument. Let us fix a topological space *X*. Every subspace *Y* of *X* defines a frame congruence on Ω=**O***X* where two open sets *U* and *V* are equivalent if and only if they have the same intersection with *Y*. This can be axiomatized by a (long) list of axioms *U*≡*V*, where *U*, *V* range over all pairs of equivalent open subsets of *X*. The corresponding nucleus ν* _{Y }* maps each

*U*∈

**O**

*X*to the largest open subset

*V*of

*X*such that

*V*∩

*Y=U*∩

*Y*,

*and the corresponding sublocale is*

*S*

_{Y}*{*

_{=}*U*∈

**O**

*X*|

*U*is the largest open subset of

*X*whose intersection with

*Y*equals

*U*∩

*Y}*. We have already noticed that

*Y*is dense in

*X*if and only if

*S*contains ⊥, namely if

_{Y}*S*is dense as a sublocale of Ω. If the smallest dense sublocale

_{Y}*S*

_{0}were spatial, say equal to

*S*, for some subspace

_{Y}*Y*of

*X*, then

*Y*would be dense as well, and would be included in every dense subspace, too. If

*X*contains no isolated point (e.g., consider the real line

**R**), then

*X*–{

*x*} is dense for every point

*x*of

*X*, so such a space

*Y*would be contained in all of them, hence it would be empty. But, unless

*X*is empty, the empty subspace cannot be dense in

*X*.

Heckmann observes that one can express *S _{Y}* using the notion of

*density*, and this is a key point.

Precisely, he claims that:

**Fact 4.** For every open subset *V* of *X*, *V* is in *S _{Y}* if and only if

*C*∩

*Y*is dense in

*C*, where

*C*is the (closed) complement of

*V*.

*Proof. * If *V* is in *S _{Y}*, then we must show that every non-empty open subset

*U*∩

*C*of

*C*(where

*U*is open in

*X*) intersects

*C*∩

*Y*, or equivalently, just

*Y*. Otherwise,

*U*∩

*C*=

*U*–

*V*would not intersect

*Y*, so

*V*∪ (

*U*–

*V*) would have the same intersection with

*Y*as

*V*, but would be strictly larger, which contradicts the definition of

*S*.

_{Y}Conversely, if *C *∩ *Y* is dense in *C*, consider any open set *U* with the same intersection with *Y* as *V*, and imagine that *U* is strictly larger than *V*. Then *U* ∩ *C* = *U* – *V* is open in *C*, non-empty, hence must intersect *C *∩ *Y*, say at *x*. Then *x* is in *U* ∩ *Y*, and not in *V *∩ *Y* since it is in *C *∩ *Y*: contradiction. ☐

**Π**^{0}_{2} subspaces of Baire spaces

We now come back to **Π**^{0}_{2} subsets. Let *X* be completely Baire, and *Y* be a **Π**^{0}_{2} subset of *X*, defined as the intersection of countably many UCO subsets *U _{n}*

**imp**

*V*. We may assume that

_{n}*V*is included in

_{n}*U*, since we can enforce this by replacing

_{n}*V*by

_{n}*U*∩

_{n}*V*, and this does not change the UCO subset

_{n}*U*

_{n}**imp**

*V*.

_{n}As we have shown in this previous post, the mapping *A *↦ *S _{A }*is monotonic. It follows that

*S*is included in each of the sublocales defined by

_{Y}*U*

_{n}**imp**

*V*, i.e., in each of the sublocales axiomatized by the single equation

_{n}*U*≡

_{n}*V*. Writing the latter as

_{n}*S*(

*U*,

_{n}*V*), S

_{n}*is included in the infimum ⋀*

_{Y}

_{n}*S*(

*U*,

_{n}*V*). The infimum is taken in the coframe of sublocales of

_{n}**O**

*X*.

Since the lattice of frame congruences is isomorphic to the lattice of sublocales, ⋀_{n}*S*(*U _{n}*,

*V*) is also the sublocale axiomatized by the intersection of all the frame congruences axiomatized by just one axiom

_{n}*U*≡

_{n}*V*, and that is also the frame congruence ≡ axiomatized by the countably many equations

_{n}*U*≡

_{n}*V*,

_{n}*n*in

**N**.

We have just shown that *S _{Y}* is included in the sublocale

*S*(≡) corresponding to the frame congruence ≡ axiomatized by the equations

*U*≡

_{n}*V*,

_{n}*n*in

**N**. This does not use the fact that those equations are only countably many, or that

*X*is completely Baire.

However, the converse inclusion holds, and that uses the fact that *X* is completely Baire and that we have only countably many equations. Let *V* be any element of *S*(≡). By definition, *V* is in every *S*(*U _{n}*,

*V*). We use Fact 4:

_{n}*C*∩ (

*U*

_{n}**imp**

*V*) is dense in

_{n}*C*, where

*C*is the (closed) complement of

*V*. Note that

*C*∩ (

*U*

_{n}**imp**

*V*) is a UCO subset of

_{n}*C*, hence in particular a

**Π**

^{0}

_{2}subset of

*C*. Since

*X*is completely Baire,

*C*is Baire, so the intersection of all the sets

*C*∩ (

*U*

_{n}**imp**

*V*),

_{n}*n*in

**N**, is dense in

*C*. Equivalently,

*C*∩

*Y*is dense in

*C*. We use Fact 4 again, now in the opposite direction, and we conclude that

*V*is in

*S*. To sum up:

_{Y}**Theorem 5.** Let *X* be a completely Baire space.

The sublocale *S*(≡), where ≡ is axiomatized by countably many equations *U _{n}* ≡

*V*,

_{n}*n*in

**N**, with

*V*⊆

_{n}*U*, is spatial.

_{n}It is isomorphic to the sublocale

*S*corresponding to the

_{Y}**Π**

^{0}

_{2}subset

*Y*obtained as the intersection of UCO subsets

*U*

_{n}**imp**

*V*,

_{n}*n*in

**N**.

## Countably presented locales

One can describe a frame by generators and relations. Namely, we fix a set *G* of symbols, called the generators, and we form the free frame on *G*. Then, given a set *R* of equations between elements of *G* (the *relations*), we form the sublocale of *G* axiomatized by *R*.

The free frame on *G* is defined in the usual categorical way. Informally, its elements are formal arbitrary suprema of finite infima of generators, and those should obey only the equations that are valid in every frame. It turns out that this construction can be described more synthetically as follows (see Theorem 3.1 of [1]): the free frame on *G* is the lattice of Scott-open subsets of **P**(*G*), where **P**(*G*) is the powerset of *G*, ordered by inclusion.

It follows that the free frame on *G* is spatial. More: since **P**(*G*) is an algebraic complete lattice, in particular it is locally compact and sober, hence completely Baire, by Lemma 2 of this post.

Theorem 5 immediately implies: every frame described by *countably* many relations, on any set of relations, is spatial.

In particular, every countably presented frame, namely every frame described by countably many generators and countably many relations, is spatial. More precisely, it is isomorphic to the frame of open subsets of a **Π**^{0}_{2} subset of **P**(*G*), where *G* is a countable set. Those spaces are exactly de Brecht’s quasi-Polish spaces [3, Corollary 24]. Hence:

*The quasi-Polish spaces are exactly the spaces whose frame of open subsets have a countable presentation.**All countably presented frames are spatial.*

As a final note, remember that every frame described by countably many relations, even on an uncountable set of generators, is spatial, too. They yield the frames of open sets of Ruiyuan Chen’s *countably correlated* spaces [4], a class of spaces that contains quasi-Polish spaces strictly.

- Reinhold Heckmann. Spatiality of countably presentable locales (proved with the Baire category theorem). Mathematical Structures in Computer Science, Volume 25, Special Issue 7 (Computing with Infinite Data: Topological and Logical Foundations Part 1), October 2015 , pages 1607-1625.
- John R. Isbell. Function spaces and adjoints, Mathematica Scandinavica 36 (1975), pp. 317–339.
- Matthew de Brecht. Quasi-Polish spaces. arXiv:1108.1445, Aug 6, 2011.
- Ruiyuan Chen. Notes on quasi-Polish spaces, arXiv:1809.07440v1 [math.LO] (2018).

— Jean Goubault-Larrecq (March 24th, 2019)