# Countably presented locales

Reinhold Heckmann has a very nice paper  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 . 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.

## Π02 subsets

Matthew de Brecht does a lot with so-called Π02 subsets , 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 xU implies xV. (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 Π02 subset of X is then a countable intersection of UCO sets, namely the subsets of all points satisfying all implications xUn implies xVn, where Un and Vn are fixed open subsets of X, nN.

Note that all open subsets, all closed subsets, and all Gδ subsets, are Π02.

Heckmann notices that the Baire property extends to Π02 subsets.

Proposition 3. In a Baire space, every countable intersection of dense Π02 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 Un imp Vn, is dense.

Let On be the set cl(Un) imp Vn, where cl(Un) is the closure of Un. Note that On is open. We claim that On is dense. For every non-empty open set U, we must show that U intersects On. We consider two cases:

• If U intersects Un, then UUn is another non-empty open set. It must intersect the dense set Un imp Vn, say at x, and we see that x must then be in Vn. In that case U intersects Vn at x, hence also the larger set On.
• If U does not intersect Un, then it does not intersect cl(Un) either. We pick x in U. Since x is in the complement of cl(Un), it is in On. Hence again U intersects On.

Now that we know that On is dense for each n, their intersection is dense, by the Baire property. But each On is included in Un imp Vn, so the intersection of the latter when 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 uv, where u and v are elements from Ω.

One can even specify axioms of the form uv, 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 uuv.

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 v0 in Ω, and consider the frame congruence axiomatized by the single axiom v0=⊤. We will see that this defines the open sublocale defined by v0.

That axiom defines a frame congruence ≡, and I claim that for all u and v, uv if and only if uv0=vv0.

Indeed, let ≡’ be the frame congruence defined by u≡’v if and only if uv0=vv0. If u≡’v, then u = u ∧ ⊤ ≡ uv0 = vv0v ∧ ⊤ = v. Conversely, we have v0 ≡’ ⊤, so ≡’ satisfies the axiom. Since ≡ is the least frame congruence satisfying the axiom, uv implies u≡’v, for all u and v.

Now the largest element that is ≡-equivalent to u is v0u. (Exercise. Use the fact that v0u is the largest v such that vv0u, hence such that vv0uv0.) Hence the associated nucleus is the map that sends every u to v0u. This is the open nucleus o(v0), as we have already seen here.

If Ω is the open set lattice OX of a topological space X, note that the lattice of open subsets of the open subspace v0 is exactly the lattice of open subsets U of X modulo the fact that two open sets that have the same intersection with v0 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(v0) is spatial as well, and is isomorphic to the frame of open subsets of the open subspace v0.

## Axiomatizing closed sublocales

Let us now fix an element u0 of Ω. Consider the axiom u0=⊥. This should now represent what remains when we erase u0, and if u0 is seen as an open set, that should represent its complement, a closed set.

The associated frame congruence is defined by uv if and only if uu0 = vu0. The argument is similar to the previous case of open sublocales. The associated nucleus is the closed nucleus c(u0), which maps every u to uu0.

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(u0) 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 u0.

## UCO sublocales

We now fix two elements u0 and v0 of Ω. We again consider just one axiom, and now this is the inequality u0v0, i.e., u0u0v0. Replacing v0 by u0v0, this can be rewritten as the axiom u0v0, where v0u0.

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

• u v0 = v v0
• uu0 = vu0.

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 uv, where ≡ is axiomatized by the unique axiom u0v0. This is done in a series of steps:

• First, u = (uv0) ∧ (v0u). This is always true. Indeed, in one direction, u is below both uv0 and v0u, hence also below their infimum. In the other direction, (uv0) ∧ (v0u) = (u ∧ (v0u)) ⋁ (v0 ∧ (v0u)) (by distributivity) ≤ uu = u.
• Similarly, v = (vv0) ∧ (v0v).
• Next, we have (v0u) = (v0v). We just show that (v0u) ≤ (v0v), as the other inequality is similar. (v0u) ∧ v0 is below u by the definition of ⇒, and also below v0, hence below u v0. Since the latter is equal to v v0, hence below v, (v0u) ∧ v0 is below v, so (v0u) is below (v0v). (In fact u v0 = v v0 is equivalent to (v0u) = (v0v).)
• It follows that u = (uv0) ∧ (v0u) = (uv0) ∧ (v0v). The latter is ≡-equivalent to (uu0) ∧ (v0v) (we at last use the axiom u0v0), which is equal to (vu0) ∧ (v0v) (by the second condition), itself ≡-equivalent to (vv0) ∧ (v0v) = v (by the axiom u0v0).

Conversely, we have u0 ≡’ v0, because v0u0. Therefore ≡’ is one frame congruence satisfy the required axiom, and since ≡ is least, uv 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 OX of open subsets of some space X. Let C0 be the closed subset obtained as the complement of u0. Then C0 and v0 are disjoint, and C0v0 is a UCO subset of X, which we wrote earlier as u0 imp v0.

The open subsets of the subspace u0 imp v0 are the intersections of open subsets U of X with C0v0. 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 C0v0. Since C0 and v0 are disjoint, U and V are equivalent in this sense if and only if UC0 = V C0 and Uv0 = V v0. I will let you verify that UC0 = V C0 is equivalent to Uu0 = Vu0. Hence U and V are equivalent in this sense if and only if:

• Uv0 = V v0
• Uu0 = Vu0

But those are exactly our conditions defining the equivalence defined by the single axiom u0v0.

To sum up: for two open subsets u0 and v0 of a topological space X, with v0u0, the sublocale axiomatized by the single equation u0v0 is spatial, and can be equated with O(u0 imp v0).

## 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 S0. (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=OX 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 UV, where U, V range over all pairs of equivalent open subsets of X. The corresponding nucleus ν maps each U ∈ OX to the largest open subset V of X such that V ∩ Y=U ∩ Y, and the corresponding sublocale is SY = {U ∈ OXU is the largest open subset of X whose intersection with equals U ∩ Y}. We have already noticed that Y is dense in X if and only if SY contains ⊥, namely if SY is dense as a sublocale of Ω. If the smallest dense sublocale S0 were spatial, say equal to SY, for some subspace 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 SY 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 SY if and only if C Y is dense in C, where C is the (closed) complement of V.

Proof. If V is in SY, then we must show that every non-empty open subset UC of C (where U is open in X) intersects C Y, or equivalently, just Y. Otherwise, UC = UV would not intersect Y, so V ∪ (UV) would have the same intersection with Y as V, but would be strictly larger, which contradicts the definition of SY.

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 UC = UV is open in C, non-empty, hence must intersect C Y, say at x. Then x is in UY, and not in V Y since it is in C Y: contradiction. ☐

## Π02 subspaces of Baire spaces

We now come back to Π02 subsets. Let X be completely Baire, and Y be a Π02 subset of X, defined as the intersection of countably many UCO subsets Un imp Vn. We may assume that Vn is included in Un, since we can enforce this by replacing Vn by UnVn, and this does not change the UCO subset Un imp Vn.

As we have shown in this previous post, the mapping ↦ Sis monotonic. It follows that SY is included in each of the sublocales defined by Un imp Vn, i.e., in each of the sublocales axiomatized by the single equation UnVn. Writing the latter as S(Un, Vn), SY is included in the infimum ⋀n S(Un, Vn). The infimum is taken in the coframe of sublocales of OX.

Since the lattice of frame congruences is isomorphic to the lattice of sublocales, ⋀n S(Un, Vn) is also the sublocale axiomatized by the intersection of all the frame congruences axiomatized by just one axiom UnVn, and that is also the frame congruence ≡ axiomatized by the countably many equations UnVn, n in N.

We have just shown that SY is included in the sublocale S(≡) corresponding to the frame congruence ≡ axiomatized by the equations UnVn, 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(Un, Vn). We use Fact 4: C ∩ (Un imp Vn) is dense in C, where C is the (closed) complement of V. Note that C ∩ (Un imp Vn) is a UCO subset of C, hence in particular a Π02 subset of C. Since X is completely Baire, C is Baire, so the intersection of all the sets C ∩ (Un imp Vn), 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 SY. To sum up:

Theorem 5. Let X be a completely Baire space.
The sublocale S(≡), where ≡ is axiomatized by countably many equations UnVn, n in N, with VnUn, is spatial.
It is isomorphic to the sublocale SY corresponding to the Π02 subset Y obtained as the intersection of UCO subsets Un imp Vn, 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 ): 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 Π02 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 , a class of spaces that contains quasi-Polish spaces strictly.

1. 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.
2. John R. Isbell. Function spaces and adjoints, Mathematica Scandinavica 36 (1975), pp. 317–339.
3. Matthew de Brecht.  Quasi-Polish spaces.  arXiv:1108.1445, Aug 6, 2011.
4. Riuyuan Chen. Notes on quasi-Polish spaces, arXiv:1809.07440v1 [math.LO] (2018).

Jean Goubault-Larrecq 