Let us continue our exploration of quasi-uniform spaces. I was recently reading a paper by Jimmie Lawson [2] about stably compact spaces, and which, I must confess, I did not know about until a few months ago. It contains a wealth of information on stably compact spaces, giving a lucid and comprehensive view of the area. If I could rewrite Chapter 9 of the book, I would definitely take example on this paper.
And, in Section 3.5 of [2], surprise! Jimmie mentions a neat result about the quasi-uniformities that induce the topology of a stably compact space, which he attributes to Künzi and Brümmer [3]. I immediately decided to talk about it here.
Let me say what this is about. In the setting of uniform spaces, it has been well-known for a long time that the topology of a compact Hausdorff space is induced by a unique uniformity [1, §4, 1, théorème 1]. Jimmie says that this result extends to stably compact spaces, and I tried to understand [3] in order to explain why in a hopefully simpler way. Unfortunately, the expected result, which would be that there is a unique quasi-uniformity that induces any given stably compact topology, is wrong, as I will demonstrate below. We will see what goes wrong, and progressively make our way towards a theorem that indeed generalizes the situation on compact Hausdorff spaces.
Is the topology of stably compact space induced by a unique quasi-uniformity?
That is the natural conjecture we might have that would generalize the aforementioned theorem on compact Hausdorff spaces. A quick reading of [2] or [3] might convince you that this what Künzi and Brümmer somehow proved, but that would be wrong.
Here is a simple counter-example—in fact almost the same counter-example as last time. Consider the interval [0, 1] with the Scott topology of ≤, whose open intervals are the empty set, the whole interval, and the half-open intervals ]a,1], with 0<a<1. This is a stably compact space. But there are at least two quasi-uniformities that induce the Scott topology:
- the Pervin quasi-uniformity, generated by the entourages RU ≝ {(x,y) ∈ [0,1] × [0,1] | x ∈ U implies y ∈ U}, where U ranges over the Scott-open subsets of [0, 1], and
- the quasi-metric quasi-uniformity, namely the quasi-uniformity generated by the entourages [<r] ≝ {(x,y) ∈ [0,1] × [0,1] | x<y+r}, where r is a (non-zero) positive real number, in other words the quasi-uniformity induced by the quasi-metric dR on [0,1].
Now consider RU, where U ≝ ]1/2, 1]. That cannot contain any finite intersection [<r1] ∩ … ∩ [<rn] of entourages of the second quasi-uniformity. We argue as follows. First, if RU contained such an intersection, then it would included the smaller entourage [<r], where r ≝ min (r1, …, rn, 3/2). Note that r is non zero. Second, the pair (1/2+r/3, 1/2–r/3) is in [<r], but not in RU. (The reason of the “3/2” term in the definition of r is to make sure that the two components of this pair are in the interval [0,1].) Hence RU is not in the quasi-metric quasi-uniformity on [0, 1]. This shows that the Pervin quasi-uniformity differs from the quasi-metric quasi-uniformity. (It in fact contains it strictly.)
The specialization preordering
Given any quasi-uniformity U on X, there is a binary relation ∩U, which is simply the intersection of all the entourages in U: (x,y) is in ∩U if and only if, for every entourage R in U, (x,y) is in R. It turns out that this is a preordering, and a familiar one.
Lemma A. For every quasi-uniformity U on X, ∩U is the specialization preordering of the topology induced by U.
Proof. Let ≤ be the specialization preordering of the topology induced by U. If x ≤ y, then by definition every neighborhood of x contains y. This applies to every neighborhood of the form R[x] with R an entourage in U, so y is in R[x] for every R ∈ U. In other words, (x,y) is in R for every R ∈ U, namely, (x,y) is in ∩U. Conversely, if (x,y) is in ∩U, then for every open neighborhood U of x in the induced topology, by definition there is an entourage R such that R[x] is included in U. Since (x,y) is in ∩U, in particular (x,y) is in R, so y is in R[x], and therefore in U. It follows that x≤y. ☐
The dual quasi-uniformity
Given a quasi-uniformity U on a set X, there is a dual quasi-uniformity U–1, whose entourages are the opposites R–1 (also written as Rop) of entourages R in U—by definition, (x,y) is in R–1 if and only if (y,x) is in R.
The quasi-uniformity U induces a topology on X, and from now on I will consider X as a topological space with that topology.
The dual quasi-uniformity U–1 also induces a topology on X, but what can we say about it? Without any further assumption, we can say at least two things. The first one has to do with specialization preorderings. Let me write X–1 for X with the topology induced by U–1.
Lemma B. Let X be a set with a quasi-uniformity U, and let ≤ be the specialization preordering of X. The specialization preordering of X–1 is ≤–1 (which I will also write as ≥).
Proof. By Lemma A, the specialization preordering of X–1 is ∩(U–1), the intersection of all the relations R–1, when R ranges over the entourages of U. This is also equal to (∩U)–1, namely ≥, by Lemma A again. ☐
The second thing has to do with compact saturated subsets.
Lemma C. Let X be a set with a quasi-uniformity U. Every compact saturated subset of X is closed in X–1.
Proof. Let Q be a compact saturated subset of X, and O be its complement. We wish to show that O is open in X–1, and that means that for every point x outside Q, we would like to find an entourage R ∈ U such that R–1[x] is included in O, in other words does not intersect Q.
We fix such a point x outside Q. Since Q is saturated, it is the intersection of its open neighborhoods. Hence there is an open neighborhood U of Q that does not contain x. By definition of the topology of X, for every point y of Q (which is in U), there is an entourage Sy ∈ U such that Sy[y] ⊆ U.
Using the property of quasi-uniformities which we called (relaxed transitivity) last time, for each such y ∈ Q, there is a further entourage Ry ∈ U such that Ry o Ry ⊆ Sy.
Each set Ry[y] is a neighborhood of y. Hence the collection of all interiors of sets Ry[y], y ∈ Q, forms an open cover of Q. Since Q is compact, we can extract a finite subcover. In other words, there is a finite subset E of Q such that Q is included in ∪y ∈ E Ry[y]. Let R be the intersection of the entourages Ry, y ∈ E. Since E is finite, this is again an entourage, that is, it is again in U. We claim that R is the desired entourage, namely that R–1[x] does not intersect Q.
In order to prove this, we will show that R–1[x] does not intersect the larger set ∪y ∈ E Ry[y]. If it did, then R–1[x] would intersect Ry[y] for some y ∈ E, say at z. Since z ∈ Ry[y], we would have (y,z) ∈ Ry. Since z ∈ R–1[x], we would have (z,x) ∈ R, hence (z,x) ∈ Ry. Therefore (y,x) would be in Ry o Ry, hence in Sy. In turn, this would mean that x would be in Sy[y], and therefore in U. But U does not contain x by definition. We have reached a contradiction, and that terminates the proof. ☐
This is promising! If X is stably compact, then we would expect its de Groot dual Xd to coincide with X–1, and the results above are natural steps in that direction.
I do no think that we can say much more without requiring some further properties from X. We now look at the case where X is core-compact, then we will specialize to the locally compact case, and then to the stably compact case.
Core-compact induced topologies
Last time, we had noticed that the Pervin quasi-uniformity was in general neither the smallest nor the largest quasi-uniformity inducing a given topology. If that topology is core-compact, the situation changes, as we see now.
Let therefore X be a core-compact space, and let us write ⋐ for the way-below relation on OX. For every pair of open subsets U, V such that U ⋐ V, let us write [U ⇒ V] for the binary relation consisting of all the pairs of points (x,y) such that x ∈ U implies y ∈ V (namely, such that x is not in U, or y is in V). We have:
Proposition D. Let X be a core-compact space. The finite intersections of relations [U ⇒ V], where U, V range over the pairs of open subsets such that U ⋐ V, form a base for a quasi-uniformity U0. The quasi-uniformity U0 is the smallest compatible quasi-uniformity, namely the smallest uniformity that induces the given topology on X.
(Note added on December 12th, 2020: Proposition D is essentially Lemma 5 of [3].)
Proof. In order to verify that U0 is a quasi-uniformity, or rather that the given finite intersections form a base of a quasi-uniformity, let me recall that we need to verify the following, where B0 is the collection of intersections [U1 ⇒ V1] ∩ … ∩ [Un ⇒ Vn], n ∈ N, where for each i, Ui and Vi are open and Ui ⋐ Vi:
- (reflexivity) for every relation R in B0, for every x in X, (x,x) is in R;
- (filter 3) for all R, S in B0, there is a relation T in B0 such that T ⊆ R ∩ S;
- (relaxed transitivity) for every S in B0, there is a relation R in B0 such that R o R ⊆ S.
Only the latter requires an argument. Given any S ≝ [U1 ⇒ V1] ∩ … ∩ [Un ⇒ Vn] in B0, we use interpolation (remember that X core-compact means that OX is a continuous dcpo) and we find open subsets Wi such that Ui ⋐ Wi ⋐ Vi for each i. Let R be the intersection of the 2n relations [Ui ⇒ Wi] and [Wi ⇒ Vi], 1≤i≤n. Every pair of points (x,y) in R o R is such that there is a point z with (x,z) and (z,y) in R. In particular, if x is in Ui then z is in Wi; then y is in Vi. As this holds for every i, (x,y) is in S.
In order to show that U0 is the smallest compatible quasi-uniformity, let U be an arbitrary compatible quasi-uniformity. For every pair of open subsets U and V of X such that U ⋐ V, we claim that [U ⇒ V] is in U. Since U is closed under finite intersections, this will show that U0 is included in U.
For every point x of V, there is an entourage Rx in U such that Rx[x] is included in V, since the topology of X is induced by U. Using (relaxed reflexivity), we find an entourage Sx in U such that Sx o Sx ⊆ Rx.
The interiors of the sets Sx[x], when x varies in V, form an open cover of V. Since U ⋐ V, we can extract a finite subcover of U. In other words, there is a finite subset E of V such that U is included in the union of the interiors of the sets Sx[x], x ∈ E. Let R be the intersection of the finitely many entourages Sx, x ∈ E. This is again in U. Now every pair of points (y,z) in R is in [U ⇒ V]: if y is in U, then y is in Sx[x] for some x ∈ E; since (y,z) in R, it is also in Sx, so z is in Sx[Sx[x]], hence in Rx[x] (since Sx o Sx ⊆ Rx), and therefore in V.
We have shown that R is included in [U ⇒ V]. Since R is in U, so is [U ⇒ V]. This shows that every subbasic entourage [U ⇒ V] of U0 is in U, hence that U0 is included in U. ☐
Locally compact induced topologies
We can prove a similar result on locally compact spaces, using a slightly different subbase of entourages in order to define U0. For every compact saturated subset Q of X and every open neighborhood V of Q, let us write [Q ⇒ V] for the binary relation consisting of all the pairs of points (x,y) such that x ∈ Q implies y ∈ V.
If X is locally compact, then we claim that those entourages form a subbase for the same quasi-uniformity U0. In one direction, given any pair of open sets U ⋐ V, there must be a compact saturated subset Q such that U ⊆ Q ⊆ V (Theorem 5.2.9 in the book). Then [U ⇒ V] contains [Q ⇒ V]. In the other direction, given any compact saturated subset Q of X and every open neighborhood V of Q, then by interpolation (Proposition 4.8.14), there is a compact saturated set Q’ included in V and whose interior U contains Q. We see that [Q ⇒ V] contains [U ⇒ V]. Notice that we also have U ⋐ V. We can rephrase all this as follows.
Proposition D’. Let X be a locally compact space. The finite intersections of relations [Q ⇒ V], where Q ranges over the compact saturated subsets of X and V ranges over the open neighborhoods of Q, form a base for the smallest compatible quasi-uniformity U0.
To sum up, and still assuming X locally compact, a base of entourages of U0 is given by the finite intersections [Q1 ⇒ U1] ∩ … ∩ [Qn ⇒ Un], n ∈ N, where each Qi is compact saturated, each Ui is open, and Qi ⊆ Ui. Those basic entourages can be rewritten in another form.
We rely on a bit of Boolean gymnastics. A pair (x,y) is in [Q1 ⇒ U1] ∩ … ∩ [Qn ⇒ Un] if and only if the conjunction (over all i, 1≤i≤n) of the formulae ‘x ∈ X–Qi or y ∈ Ui‘ holds. Distributing ors over ands, that is equivalent to the disjunction, over all subsets I of {1, …, n}, of ‘x ∈ X–QI and y ∈ UI‘, where UI is the intersection of the sets Ui, i ∈ I, and QI is the union of the sets Qi, i ∈ I. (If I is empty, UI is equal to the whole of X.)
Hence we have shown that if X is locally compact, any basic entourage [Q1 ⇒ U1] ∩ … ∩ [Qn ⇒ Un] of U0 is equal to a finite union of products (X–QI) × UI, where each QI is compact saturated and each UI is open.
Moreover, that union contains (≤) ≝ {(x,y) ∈ X × X | x≤y}, the graph of the specialization preordering ≤. Indeed, for any pair of points x and y such that x≤y, by Lemma A, (x,y) is in every entourage of U0, in particular in [Q1 ⇒ U1] ∩ … ∩ [Qn ⇒ Un] = ∪I ⊆ {1, …, n} (X–QI) × UI.
In other words, if X is locally compact, then any basic entourage [Q1 ⇒ U1] ∩ … ∩ [Qn ⇒ Un] of U0 is an open neighborhood of (≤) in Xd × X. It follows that every entourage of U0, which contains such a basic entourage by definition, is also a neighborhood of (≤) in Xd × X.
Oh yes, I am using the notation Xd for the de Groot dual of X, even when X is not stably compact. Its topology, the cocompact topology, is the topology generated by the complements of the compact saturated subsets of X. The closed subsets of Xd are the intersections of compact saturated subsets of X, and with compactness and coherence, those are also the filtered intersections of compact saturated subsets of X. When X is well-filtered, those filtered intersections are themselves compact saturated, and we retrieve the definition of the de Groot dual of the book.
Stably compact spaces
Proposition E. If X is stably compact, then U0 is exactly the collection of neighborhoods of (≤) in Xd × X.
Proof. Given what we have just argued, it is enough to show that every neighborhood R of (≤) in Xd × X is in U0. We will use the fact that (Xpatch,≤) is a compact pospace. (See Section 9.1 of the book for the relation between compact pospaces and stably compact spaces.)
For every x ∈ X, (x,x) is in (≤), so there is an open neighborhood X–Qx of x in Xd (namely, Qx is compact saturated in X and x is not in Qx) and an open neighborhood Ux of x in X such that (X–Qx) × Ux ⊆ R. The set (X–Qx) ∩ Ux = Ux – Qx is open in the patch topology of X, which is by definition the coarsest topology containing both the open subsets of X and the complements of compact saturated subsets of X. Therefore the sets Ux – Qx, x ∈ E, forms an open cover of Xpatch.
Since Xpatch is compact, there is a finite subset E of X such that the sets Ux – Qx, x ∈ E, already forms an open cover of Xpatch. Let us consider S ≝ ∪x ∈ E (X–Qx) × Ux. We observe that S is included in R: this is because (X–Qx) × Ux ⊆ R for every x ∈ X.
For each subset I of E, let QI be the intersection of the sets Qx with x ∈ I, and let UI be the union of the sets Ux with x ∈ I. The pairs (y,z) that are in S are exactly those that satisfy the disjunction (over all x in E) of the formulae ‘y ∉ Qx and z ∈ Ux‘. By distributing ands over ors, they are those that satisfy the conjunction, over all subsets I of E, of the formulae ‘ for some x in I, y is not in Qx, or for some x in E–I, z is in Ux‘, equivalently, ‘y ∉ QI or z ∈ UE–I‘. In other words, S is equal to ∩I ⊆ E [QI ⇒ UE–I].
This (almost) proves that S is in U0, and therefore that the larger relation R is in U0. We still need to check that QI is included in UE–I for every subset I of E! And that holds because the union of the sets Ux – Qx, x ∈ E, is the whole of X. Indeed, the latter implies that the union of the larger sets Ux (instead of Ux – Qx) with x ∈ E–I, and of the larger sets X – Qx (instead of Ux – Qx) with x ∈ E–I, is also the whole of X. But that new union is UE–I ∪ (X – QI), and the fact that it is equal to X is exactly equivalent to the inclusion QI ⊆ UE–I. ☐
We can also rewrite this as follows. The diagonal is the set of pairs (x,x). This is the graph (=) of the equality relation =.
Proposition E’. If X is stably compact, then U0 is exactly the collection of neighborhoods of the diagonal (=) in Xd × X.
Indeed, any neighborhood R of (=) in Xd × X must contain ∩U0=(≤), by Lemma A.
Duality
I have already said that, given a set X with a quasi-uniformity U, seen with the induced topology, every compact saturated subset of X is closed in X–1. This means that the cocompact topology on X is coarser than the topology of X–1. When U is U0, the minimal compatible quasi-uniformity (see Proposition D), those two topologies coincide, as we now argue.
Proposition F. Let X be a locally compact topological space, and U be the minimal compatible quasi-uniformity U0. The topology induced by the dual quasi-uniformity U–1 on X coincides with the cocompact topology.
Proof. Let O be an open subset of X in the topology induced by U–1. By definition, for every x ∈ O, there is a basic entourage R≝[Q1 ⇒ U1] ∩ … ∩ [Qn ⇒ Un] (where each Qi is compact saturated, each Ui is open, and Qi ⊆ Ui) such that R–1[x] ⊆ O.
We expand the definition of R: R–1[x] is the set of points y such that for every i, if y ∈ Qi then x ∈ Ui; equivalently, such that for every i, if x ∉ Ui then x ∉ Qi; in other words, it is the complement of QI, where QI is the union of the sets Qi, i ∈ I, and I is the collection of indices i such that x ∉ Ui. QI is compact saturated, so its complement R–1[x] is open in the cocompact topology. This complement contains x (R–1[x] always contains x), and is included in O. This shows that O is an open neighborhood, in the cocompact topology, of each of its points, so that O is open in the cocompact topology.
Conversely, let O be any open subset of X in the cocompact topology. Its complement Q is compact saturated in X. By Lemma C, O is open in the topology induced by U–1. ☐
We finally reach the result promised at the beginning of this post.
Theorem. Let X be a stably compact topological space. There is a unique quasi-uniformity U that induces the topology of X and such that the dual quasi-uniformity U–1 induces the cocompact topology, and this is the minimal compatible quasi-uniformity U0.
Proof. Existence is by Proposition F. In order to show uniqueness, we fix a quasi-uniformity U that induces the topology of X and such that the dual quasi-uniformity U–1 induces the cocompact topology. By Proposition D’, U contains U0, so we concentrate on showing the reverse inclusion.
Let R be any entourage of U. There is an entourage S in U such that S o S ⊆ R. For every x ∈ X, it follows that S–1[x] × S[x] is included in R: every pair (y,z) in S–1[x] × S[x] is such that (y,x) ∈ S and (x,z) ∈ S, so (y,z) ∈ R. S–1[x] is an open neighborhood of x in Xd since U–1 induces the cocompact topology on X, and S[x] is an open neighborhood of x in X since U induces the original topology on X. Therefore R is an open neighborhood of (x,x) in Xd × X, for every x ∈ X. In other words, R is an open neighborhood of (=) in Xd × X. By Proposition E’, R in U0, and this finishes the proof. ☐
An extremely short glimpse at the bitopological view
The proper way of stating the latter theorem is by appealing to the notion of a bitopology. (Yet another concept I have never talked about, at least until now!) A bitopology on a set X is simply a pair of topologies on X, and a bitopological space is simply a set with a bitopology. Most concepts in the theory of bitopological space rely on the interaction of the two topologies, and I really should say more about this some time. (Tomáš Jakl, notably, drew my attention to this subject again pretty recently.)
Every stably compact space has a canonical bitopology, which consists of its original topology, and its cocompact topology. If, as above, we define the cocompact topology as being generated by the complements of compact saturated sets, this even makes sense for arbitrary spaces. I do not know of any standard name for this bitopology; let me call it the cocompact-open bitopology on X.
Every quasi-uniformity U on a space X induces a bitopology on X, too, namely the one formed by the topology induced by U and the topology induced by U–1. Lemma C states that this bitopology is pairwise finer than the cocompact-open bitopology (with the obvious meaning of “pairwise finer”). Proposition F states that the bitopology induced by the minimal quasi-uniformity U0 on a locally compact space coincides with the cocompact-open bitopology.
Using this bit of vocabulary, we can rephrase our findings as follows, when X is stably compact:
- there is no unique quasi-uniformity that induces the topology of X,
- but there is a minimal one U0,
- U0 is also the unique quasi-uniformity that induces the cocompact-open bitopology on X,
- and is simply the set of open neighborhoods of (≤), or equivalently of the diagonal (=), in Xd × X.
Let me conclude by saying how this generalizes the result mentioned at the very beginning of this post: that the topology of a compact Hausdorff space is induced by a unique uniformity [1, §4, 1, théorème 1]. We argue as follows. If X is compact Hausdorff, then its cocompact and its open topologies coincide. By what we have seen, there is a unique quasi-uniformity U on X such that both U and U–1 induce the topology of X. That is, of course, U0. Since Xd = X when X is compact Hausdorff, U0 is simply the set of open neighborhoods of (=), and that is a symmetric quasi-uniformity (i.e., U=U–1), in other words it is a uniformity.
- Nicolas Bourbaki. Topologie générale (éléments de mathématique), chapitre 2. Springer, 2007. (Many other, previous editions, too.)
- Jimmie Lawson. Stably compact spaces. Mathematical Structures in Computer Science 21(1):125-169, Feb. 2011.
- Hans-Peter Albert Künzi and Guillaume C. L. Brümmer. Sobrification and bicompletion of totally bounded quasi-uniform spaces, Mathematical Proceedings of the Cambridge Philosophical Society 101(2):237–247, 1987.
— Jean Goubault-Larrecq (November 21st, 2020)