Stone duality leads naturally to the idea of locale theory. Quickly said, the idea is that, instead of reasoning with topological spaces, we reason with frames. The two concepts are not completely interchangeable, but the O ⊣ pt adjunction shows that they are close. More: sober spaces form a category that is equivalent to the opposite of that of spatial frames.
Topology without points
It has come to the mind of several mathematicians, from John Isbell to Peter Johnstone, and including Bernhard Banaschewski and others, that one might redo most of topology by looking at frames instead of topological spaces.
The result is sometimes known as pointless topology (a horrible pun), because frames that are not necessarily spatial can be thought of as sober spaces, except that they might lack a few points. In fact, some of those frames, such as the frame of regular open sets of [0, 1], have no point at all, and despite that, have a very rich structure.
If you wish to learn about the theory of locales, let me recommend Picado and Pultr’s excellent book . They do a really great job of presenting the theory as simply as possible, and have proofs that are probably the simplest possible.
Let me also remind you that the adjunction O ⊣ pt between topological spaces and frames is actually an adjunction between the category Top of topological spaces and the opposite of the category Frm of frames and frame homomorphisms. To obtain the right pointless equivalent of Top, we must therefore move to Frmop, the opposite of Frm. That opposite category Loc=Frmop, is called the category of locales.
Hence a locale is just a frame, but a morphism f : L → L’ in Loc should be thought the other way around, as a frame homomorphism from L’ to L.
This inversion of the direction of morphisms is the source of immense confusion. We shall instead reason with frames. Picado and Pultr do the same, and call that dealing with locale theory “mostly covariantly”.
Things that go well with locales
There are many things that work well in the realm of locales. You may have heard of Johnstone’s result that “Tychonoff’s theorem can be proved in the realm of locales without any recourse to the axiom of choice”. This is usually the first thing I hear mentioned when a discussion comes to the matter of locales.
This is true, but, as I will recall later, locale products are not the same as topological products. Also, it is a pretty complicated theorem. But it certainly is a neat result.
Let us look at something simpler: coproducts of locales, i.e., products of frames. They always exist, and the lattice of open sets of a coproduct of topological spaces is exactly the locale coproduct, i.e., the frame product, of the various lattices of open sets involved. In short: O(coproduct of spaces Xi) = frame product of O(Xi). For binary coproducts, this mostly says that an open set of X+Y is just a pair of an open set from X and one from Y.
In general, Loc has all colimits, and O preserves them. The latter part is because O is a right adjoint. The former must be checked by hand. Products of frames are computed componentwise, and equalizers are obtained as subframes, i.e., as subsets of frames that are closed under finite infima and arbitrary suprema: there is no difficulty in doing that verification.
Constructions that are more painful: locale limits
The situation is more painful with limits. Section 8.4.4 of the book is devoted to the case of binary products. This is already complicated: binary products of locales, that is, binary coproducts of frames, are described as a frame of Galois connections. (More general frame coproducts are described by so-called C-ideals: for frames Li, i in I, their coproduct is the set of relations, i.e., of subsets of ∏ Li, that are downwards-closed and separately Scott-closed in each subscript i, with the componentwise ordering.)
Worse is that the O functor does not preserve binary products, that is, O(X x Y) may fail to be the frame coproduct of O(X) and of O(Y). For that, X and Y must not be core-compact, as O(X x Y) = O(X) + O(Y) when X or Y is core-compact (Exercise 8.4.23 in the book). I may one day brace myself and try to explain Johnstone’s counterexample showing that, in general, O(X x Y) and O(X) + O(Y) are not isomorphic.
No, what I would like to consider today is the case of subspaces. A topological subspace (together with the canonical inclusion map) is a special case of limit: it is an equalizer, and an equalizer is a special case of limit. Conversely, all limits can be built as equalizers of products, so products and equalizers are really the only kinds of limits we have to consider.
To show that topological subspaces are exactly equalizers in Top, we must show that for every topological subspace A of a topological space X, the inclusion map m : A → X equalizes a pair of arrows from X to some space Y. There is a general recipe to show that. Form the cokernel pair of m, i.e., the pushout of a diagram made of twice the map m starting from apex A. This is the space Y defined as the quotient of X+X (whose elements are (0, x) and (1, x) for x in X) by the equivalence relation that equates (0, a) with (1, a) for each a in A. We can now check that m is the equalizer of the two maps x ⟼ equivalence class of (0, x) and x ⟼ equivalence class of (1, x).
Hence finding the right notion of sublocale — the equivalent of subspaces for locales — is just a matter of understanding what equalizers are in Loc, or equivalently what coequalizers are in Frm. I will not take the categorical route, however. Just as for products, I will try to indicate what sublocales should be by analogy to topological subspaces. And just as for products, this analogy will eventually fail very badly, although it is useful for starters.
Sublocales, nuclei, and congruences
There are three possible, equivalent, ways of defining sublocales: as… something called sublocales, first; as nuclei; and as frame congruences. Here are the raw definitions.
- A sublocale of a frame Ω is a subset L of Ω that is closed under arbitrary infima (taken in Ω), and such that ω ⟹ x is in L for every x in L and every ω in Ω. Here ⟹ is residuation (a.k.a., intuitionistic implication): a ⟹ b is the largest c such that inf (a, c) ≤ b. Note that a sublocale is not a subframe, which would be a subset of Ω that is closed under finite infima and arbitrary suprema, as I have already said.
- A nucleus on Ω is a closure operator on Ω that preserves binary infima. A closure operator is by definition a monotonic map ν : Ω → Ω such that ν(ω) ≥ ω for every ω in Ω, and ν(ν(ω)) = ν(ω) for every ω in Ω. A nucleus additionally satisfies ν(inf(ω, ω’)) = inf (ν(ω), ν(ω’)). It automatically satisfies ν(⊤) = ⊤, because of the law ν(ω) ≥ ω, so ν in fact preserves all finite infima.
- A frame congruence on Ω is an equivalence relation ≡ that is compatible with finite infima and arbitrary suprema: if ω1≡ω’1 and ω2≡ω’2 then inf (ω1, ω2)≡inf(ω’1, ω’2), and if ωi≡ω’i for every i in I, then sup ωi ≡ sup ω’i.
This all looked difficult to understand the first time I saw those definitions. So let me try to rederive those definitions, as we did in the book with products and frame coproducts, by imagining naively that sublocales are a pointfree encoding of topological subspaces. In other words, imagine A is a topological subspace of X, and let me try to give a description of how the open subsets of A can be encoded from the knowledge of the lattice Ω of the open subsets of X, and without mentioning points.
The easiest thing to explain is the frame congruence encoding. An open subset of A is an intersection U ⋂ A, and we may try to encode it as U. But then, such an open subset of A may have many encodings. However, all those encodings are related by the equivalence relation ≣ (on Ω) defined by U ≣ V if and only if U ⋂ A = V ⋂ A. We can check that this is a frame congruence. From a localic point of view, giving an congruence on Ω is enough to describe the lattice of open subsets of A: we take the quotient of Ω by ≣, this gives you a frame isomorphic to the frame O(A) of open subsets of A, and the quotient map q : Ω → Ω/≣ is the localic encoding of the inclusion map m : A → X, in the sense that q = m-1 : Ω = O(X) → Ω/≣ = O(A).
Instead of encoding A by a congruence, we can pick a distinguished representative in each equivalence class. There is an obvious choice for that: define ν(U) as the largest open subset of X that is equivalent with U, that is, that has the same intersection with A as U. More precisely, the union of all the open subsets that are equivalent to U is again equivalent to U, and is the largest such equivalent open subset. I’ll let you check that ν is a nucleus.
Any nucleus, in fact, any closure operator ν is entirely determined by its set L of fixed points. Indeed, ν(ω) is the least fixed point of ν above ω. You may also note that L is nothing but the image of ν, as well.
Hence we can encode the subspace A (together with its inclusion map) as the set of open subsets U of X that are largest in the class of all open subsets of X that have a given intersection with A. Checking directly that it is a sublocale is not entirely obvious, and it is as easy (or as difficult) to show directly that sublocales are adequate encoding of nuclei, in general.
Let me do so now. To deal with residuation ⟹, I will use the following equivalence: (*) inf (a, b) ≤ c if and only if a ≤ [b ⟹ c]. This holds not only in every frame (= complete Heyting algebra), but more generally in every Heyting algebra. Note that, by taking a = [b ⟹ c], this entails the inequality: (**) inf ([b ⟹ c], b) ≤ c.
Lemma. For every nucleus ν, its set L of fixed points is a sublocale.
Proof. Because ν commutes with finite infima, L is closed under binary infima. It is in fact closed under arbitrary infima, for the following reason. Given a family of elements xi of L, ν(inf xi) ≤ inf ν (xi) by monotonicity, and since each xi is a fixed point, ν(inf xi) ≤ inf xi. Because ν is a closure operator, ν(inf xi) ≥ inf xi, whence the equality follows.
The second property of sublocales will follow from the general identity ν (a ⟹ ν (b)) = a ⟹ ν (b). Indeed, if x is a fixed point of ν, then ν (ω ⟹ x) = ν (ω ⟹ ν (x)) = ω ⟹ ν (x) = ω ⟹ x, showing that ω ⟹ x is also a fixed point of ν.
We show that identity as follows. Let c = ν (a ⟹ ν (b)). The infimum of c and a is below inf (c, ν (a)), and the latter is equal to ν (inf (a ⟹ ν (b), a)) because ν commutes with binary infima. Using the inequality (**), inf (a ⟹ ν (b), a) ≤ ν (b), so inf (c, a) ≤ ν (ν (b)) = ν (b). By (*), this implies c ≤ [a ⟹ ν (b)]. The converse inequality is because ν(ω) ≥ ω for every ω in Ω. ☐
Lemma. For every sublocale L, the map ν that sends ω to the smallest element of L above ω is a nucleus.
Proof. That smallest element is the infimum of the elements of L above ω, which is in L because L is closed under arbitrary infima. The fact that ν is a closure operator is clear. Since ν is monotonic, ν(inf(ω, ω’)) ≤ inf (ν(ω), ν(ω’)) for all ω, ω’.
We note that inf(ω, ω’) ≤ ν(inf(ω, ω’)), so ω ≤ [ω’ ⟹ ν(inf(ω, ω’))] by (*). Note that ν(inf(ω, ω’)) is in L by definition, so ω’ ⟹ ν(inf(ω, ω’)) is also in L, using the second clause in the definition of sublocales. By definition of ν(ω) as the smallest element of L above ω, it follows that ν(ω) ≤ [ω’ ⟹ ν(inf(ω, ω’))]. By (*) read in the other direction, inf (ν(ω), ω’) is below ν(inf(ω, ω’)). Permute the two arguments of the first inf: inf (ω’, ν(ω)) is below ν(inf(ω, ω’)), so ω’ ≤ [ν(ω) ⟹ ν(inf(ω, ω’))], using (*) again. By the same argument as above, ν(ω’) ≤ [ν(ω) ⟹ ν(inf(ω, ω’))], so inf (ν(ω), ν(ω’)) ≤ ν(inf(ω, ω’)). ☐
Moreover, the two constructions, from a nucleus to a sublocale and conversely, are inverses of each other.
Subspaces and sublocales
Despite the fact that I explained those constructions by imitating the construction of the lattice of open subsets of a topological subspace, sublocales (resp., nuclei, resp. congruences) are very different from topological subspaces.
For example, even spatial locales have sublocales that are not spatial. This is the dark side of a coin, whose bright side is Isbell’s density theorem: every locale contains a least dense sublocale [1, 8.3]. The latter is, in general, not spatial. The topological counterpart, which would say that every space contains a least dense topological subspace, is completely wrong.
The most curious result that shows how different sublocales are subspaces are is probably the following. The poset P(X) of topological subspaces of a given topological space X is a complete atomic Boolean lattice. In particular every subspace A of X is a complemented element of P(X): there is another subspace X—A, whose infimum (=intersection) with A is the bottom element (the empty subspace) and whose supremum (=union) with A is the top element (X itself). On the contrary, the poset of all sublocales is in general only a coframe, that is, its opposite is a frame, but not all sublocales are complemented. In fact, all complemented sublocales are spatial, and I have already mentioned that not all sublocales of a locale are spatial in general.
The correspondence between sublocales and nuclei reverses the ordering (inclusion of sublocales becomes pointwise ≥ on nuclei), and the former can therefore be rephrased as: nuclei form a frame, in which not all elements are complemented.
It is a zero-dimensional frame, that is, every element of the frame is the (directed) supremum of complemented elements. This is shown as follows. For each u in Ω (which we may think encodes an open subset of a topological space X), there is a so-called open nucleus o(u), which maps every ω to u ⟹ ω (which intuitively encodes the subset u as a subspace), and there is a so-called closed nucleus c(u), which maps every ω to sup (u, ω) (and intuitively encodes the complement of u as a subspace). Those particular nuclei are complemented, and as one might expect, o(u) is the complement of c(u). Then every nucleus ν can be written as the supremum of the (complemented) nuclei inf (c(ν(u)), o(u)), u in Ω. (There is a short, but tricky proof.) That shows that the frame of nuclei is zero-dimensional, although not all nuclei are complemented.
The analogous result on the topological side would say something like: “every subspace is an intersection of special subsets of the form U ⋃ C, where U is open and C is closed”, which would be completely wrong again. (I realize I don’t know how to prove this, but that would seem strange anyway.)
The facts that nuclei form a frame, and that certain, so-called open and closed nuclei are complemented and generate the frame of all nuclei would probably deserve another post. Instead, next time, I may choose to describe a fourth way of “representing” subspaces, and to show that it is another equivalent encoding of sublocales / nuclei / congruences.
 Jorge Picado and Aleš Pultr. Frames and locales — topology without points. Birkhäuser, 2010.