A while back (in March 2019, to be precise), Tomáš Jakl told me that he had a nice, short proof of the fact that the categories of stably compact spaces (and perfect maps) and compact pospaces (and continuous order-preserving maps) are equivalent . He uses an approach through bitopological spaces, and this will give me an opportunity to talk about them.
By the way, I am not claiming, and he is not claiming either, that he is the inventor of that proof. All the ingredients had been there for a long time. But it is worth to present them in a standalone fashion, as he did, with just as little of the theory of bitopological spaces as needed.
So what is a bitopological space?
Well, simply a set with two topologies on it. Nothing more. Rather disappointing, at first sight. As simple as it may seem, the whole beauty of the notion is in the interaction between the two topologies. The main interaction we will study here is order separation; this will be defined below.
Bitopological spaces were introduced by J. C. Kelly in 1963 , and further studied by a number of authors. Let me mention Ralph Kopperman’s comprehensive paper on the subject .
Examples of bitopological spaces
Let me write (X, τ+, τ–) for a bitopological space: a set X, and two topologies τ+ and τ–. Trivially, (X, τ–, τ+) is also a bitopological space, called the dual of (X, τ+, τ–). And the bidual, namely the dual of the dual, is the original bitopological space. Nothing to be amazed at, really, of course.
Here are a few classical examples of (classes of) bitopological spaces.
- (The bitopological space of a hemi-metric space) Let X,d be a hemi-metric space. There is an opposite hemi-metric dop on X, defined by dop(x,y) ≝ d(y,x). Then X with the open ball topology of d for τ+ and with the open ball topology of dop for τ– is a bitopological space.
- (The bitopological space of a quasi-uniform space) Generalizing the previous example, let X be a quasi-uniform space, with quasi-uniformity U. As we have seen in Part II of the posts on quasi-uniform spaces, there is a dual quasi-uniformity U–1, whose entourages are the opposites R–1 (also written as Rop) of entourages R in U. We may take the induced topology of U for τ+ and the induced topology of U–1 for τ–.
- (The bitopological space of a pospace) Let (X,⪯) be a pospace. We obtain a bitopological space by letting τ+ be its upward topology (whose open sets are the open subsets of X that are upwards closed with respect to ⪯) and τ– be its downward topology (the open subsets that are downwards closed with respect to ⪯). We will be especially interested in the case where X is compact.
- (The Lawson bitopological space of a dcpo) Let X be a dcpo. We take τ+ to be the Scott topology of X, and τ– to be its lower topology (whose open sets are unions of sets of the form ↑E, with E finite). That is particularly interesting when X is a continuous dcpo, or more generally a quasi-continuous dcpo.
Of course, those examples are far from random. In each case, the join τ+ ∨ τ– of the two topologies, namely the coarsest topology that contains both τ+ and τ–, is an interesting topology in its own right.
- In case 1, this is the open ball topology of the symmetrized metric dsym, defined by dsym(x,y) ≝ max (d(x,y), dop(x,y)). (Proposition 6.1.19 in the book.)
- In case 2, this is the induced topology of the symmetrized uniformity Usym, which we have already talked about in Part III of the posts of quasi-uniform spaces. (I will justify this claim near the end of this post, when we return to that case for the last time.)
- In case 4, this is the so-called Lawson topology on the dcpo X.
- Finally, in case 3, this is the patch topology, which turns out to be the original topology on X provided that (X,⪯) is a compact pospace (Proposition 9.1.34).
We will redo the theory of compact pospaces and stably compact spaces in the perspective of the bitopological spaces. Ideally, I would have liked to redo it completely, not assuming any particular result of Section 9.1 of the book, but that would be inconvenient. But we will manage to avoid using some of the more technical results, and in particular the dreaded Lemma 9.1.29.
Specialization preorderings and order separation
Given a bitopological space (X, τ+, τ–), we have two specialization preorderings: let me write ≤+ for the specialization preordering of τ+, and ≤– for the specialization preordering of τ–.
In all the examples that I have considered above, ≤– is simply the opposite of ≤+: x≤–y if and only if y≤+x. (We will reprove it in case 3.) In general, there need not be any relationship between ≤+ and ≤–, in the same way as there need be no relationship between the topologies τ+ and τ–. But be reassured: in all situations that will be of concern to us, ≤– will be the opposite of ≤+.
Let ≤ be the preordering defined by: x≤y if and only if x≤+y and y≤–x. This is the associated preordering of (X, τ+, τ–). More succinctly, this is the intersection of ≤+ and of the opposite ≥– of ≤–. In the examples above, this is just ≤+. But note that the definition also makes sense when ≤– is not the opposite of ≤+.
Definition [1, Definition 2.1.5]. We say that (X, τ+, τ–) is order-separated if and only if the following two conditions hold:
- ≤ is antisymmetric, namely it is a partial ordering;
- For any two points x and y such that x≰y, we can find a τ+-open neighborhood U+ of x and a τ–-open neighborhood U– of y such that U+ and U– are disjoint.
This is really a bitopological variant of the notion of T2 space.
Let us review that definition in light of our examples. In all four cases, ≤ is simply ≤+ (= ≥–).
- In case 1, where X,d is a quasi-metric (not just hemi-metric) space, τ+ is the open ball topology of d and τ– is the open ball topology of dop. Since d is a quasi-metric, ≤ (viz., ≤+) is a partial ordering, not just a preordering.
Now assume that x≰y, namely that d(x,y)≠0. Let r be any positive real number smaller than d(x,y). Let U+ be the open ball centered at x with radius r/2, with respect to d. Let U– be the open ball centered at y with radius r/2, with respect to dop. We claim that U+ and U– are disjoint. Any point z in their intersection must satisfy d(x,z)<r/2 and dop(y,z)<r/2, that is, d(z,y)<r/2. By the triangular inequality, d(x,y)<r, which is impossible.
Hence the bitopological space of a quasi-metric space is order-separated. In fact, the bitopological space of a hemi-metric space X,d is order-separated if and only if X,d is T0.
- In case 2, where X,U is a quasi-uniform space. That generalizes case 1. In order to be order-separated, the induced topology must be T0, and I will call a quasi-uniform space T0 if and only if it is T0 in its induced topology, if and only if its specialization ordering (which is also the intersection of all entourages, see Lemma A in Part II of the series on quasi-uniform spaces) is antisymmetric.
Now assume that x≰y. Since ≤ is the intersection of all entourages, there is an entourage R in U such that (x,y) is not in R. Let ½R be a splitter for R (using a notation from Part IV), namely an entourage such that ½R o ½R ⊆ R. In the induced topology of U, ½R[x] is a neighborhood of x, and we let U+ be its interior. Symmetrically, in the induced topology of U–1, (½R)–1[y] is a neighborhood of x, and we let U– be its interior (still in the induced topology of U–1). Any point z is the intersection of U+ and U– would be both in ½R[x] and in (½R)–1[y], meaning that (x,z) would be in ½R and that (z,y) would be in ½R, so that (x,y) would in R, and that is impossible.
Hence the bitopological space of a quasi-uniform space X,U is order-separated if and only if X,U is T0.
- Ah, Case 3! Let (X, ⪯) be a compact pospace. Then ≤=≤+=≥–=⪯ (Lemma 9.1.18 in the book), in particular ≤ is antisymmetric. Lemma 9.1.15 states that given any two points x and y such that x⋠y, we can find a τ+-open neighborhood U+ of x and a τ–-open neighborhood U– of y such that U+ and U– are disjoint. This is exactly the property of order separation.
(You may also wish to look at the corresponding proof in [1, Lemma 2.1.6]. However, be warned that, as far as I can tell, there is a gap in that proof: although the sets U+ and U– built there are τ+-open and τ–-open respectively, and are disjoint, the argument that x is in U+ and that y is in U– is missing; and I doubt that you can make that construction work without modification.)
- Case 4: X is a dcpo, τ+ is its Scott topology, and τ– is the lower topology. Let me assume that X is a continuous dcpo. We have ≤=≤+=≥–, and that is the original ordering on X. Now pick any two points x and y such that x≰y. Since X is a continuous dcpo and x is in the Scott-open set X–↓y, there is a point x’≪x in X–↓y. Then U+ ≝ ↟x’ is a τ+-open neighborhood of x. Now since x’ is in X–↓y, we have x’≰y, so y in the τ–-open set U– ≝ X–↑x’. Moreover, U+ and U– are disjoint: any point z in the intersection would have to satisfy x’≪z and x‘≰z, which is impossible.
This also works when X is assumed to be quasi-continuous instead of continuous. As above, if x≰y, then x is in the Scott-open set X–↓y. Now, X is locally finitary compact (Exercise 5.2.31) in its Scott topology, so there is a finite set E included in X–↓y such that ↑E is a neighborhood of x. We let U+ be the interior of ↑E. Since E ⊆ X–↓y, no point of E is below y, so y is in the τ–-open set U– ≝ X–↑E. Finally, U+ and U– are disjoint: any point in the intersection would have to be both in (the interior of) ↑E and in X–↑E, which is impossible.
We summarize case 3 as follows.
Lemma A. The bitopological space (X, τ+, τ–) of a compact pospace (X,⪯) is order-separated, and ≤=≤+=≥–=⪯.
I have said earlier that order separation is a bitopological form of the T2 separation axiom. We can make this a bit more precise as follows. Let me write τ+ ∨ τ– for the join of τ+ and τ–, namely the coarsest topology finer than both.
Lemma B. For every order-separated bitopological space (X, τ+, τ–), τ+ ∨ τ– is a T2 topology.
Proof. Let x and y be two distinct points in X. Since ≤ is antisymmetric, x≰y or y≰x. By symmetry, let us assume that x≰y. By order separation, there are a τ+-open neighborhood U+ of x and a τ–-open neighborhood U– of y such that U+ and U– are disjoint. But U+ and U– are clearly open in τ+ ∨ τ–. ☐
A bitopological space (X, τ+, τ–) is called d-compact if and only if X, with the topology τ+ ∨ τ–, is a compact topological space. This is equivalent to Jakl’s Definition 2.1.11 , using Alexander’s subbase lemma.
In the literature, one finds a very closely related property: (X, τ+, τ–) is joincompact if and only if X, with the topology τ+ ∨ τ–, is compact and T0. It would not make a difference to consider joincompactness or d-compactness here, since in the presence of order separation, the T0 property (even T2) is automatic (see Lemma B).
There is a map, which Jakl calls bi, that sends every compact pospace (X,⪯) to its associated bitopological space (X, τ+, τ–). We have seen that (X, τ+, τ–) is then order-separated and that ≤=≤+=≥–=⪯ in that case.
Lemma C. For every compact pospace (X,⪯), bi (X,⪯) is an order-separated, d-compact bitopological space (X, τ+, τ–). Moreover, τ+ ∨ τ– is the original topology τ on X, and ≤=≤+=≥–=⪯.
Proof. The topology τ+ ∨ τ– is evidently coarser than τ. It is T2 by Lemma B, since (X, τ+, τ–) is order-separated by Lemma A. But τ is a compact topology, and any T2 topology coarser than a compact topology must coincide with it (Theorem 4.4.27 in the book). In other words, τ+ ∨ τ– = τ, and therefore (X, τ+, τ–) is d-compact. ☐
In the reverse direction, let bi–1 map every bitopological space (X, τ+, τ–) to the pair (X, ≤), where X is equipped with the join topology τ+ ∨ τ–, and ≤ is the associated preordering ≤+ ∩ ≥–. Lemma C implies that, for every compact pospace (X,⪯), bi–1(bi (X,⪯))=(X,⪯). This is Proposition 2.2.2 in , and is credited to Achim Jung and Drew Moshier (2006).
We will turn to the other equality bi (bi–1 (X, τ+, τ–))=(X, τ+, τ–) shortly. For that, we need the following lemma.
Lemma D [1, Lemma 2.2.3]. For every order-separated, d-compact bitopological space (X, τ+, τ–), ≤=≤+=≥–.
Proof. Let x≰+y. In particular, x≰y. Order separation implies that there are a τ+-open neighborhood U+ of x and a τ–-open neighborhood U– of y such that U+ and U– are disjoint. In particular, y is in U– but x is not, so y≰–x. By contraposition, we have shown that ≥– is included in ≤+. Similarly, x≰–y implies y≰+x, so by contraposition, ≤+ is included in ≥–. Hence ≥– and ≤+ are the same relation. ☐
Proposition E. For every order-separated, d-compact bitopological space (X, τ+, τ–), bi–1 (X, τ+, τ–) is a compact pospace (X, ≤), and bi (bi–1 (X, τ+, τ–))=(X, τ+, τ–).
Proof. That is a bit harder than the previous results. Most of it appears as Proposition 2.2.4 of . Let (X, ≤) be bi–1 (X, τ+, τ–). Hence the topology on X is τ+ ∨ τ–, and d-compactness means that with that topology, X is compact. By Lemma D, ≤ is both equal to ≤+ and to ≥–. In order to show that (X, ≤) is a compact pospace, it remains to show that the graph (≤) of ≤ is closed in X × X, or equivalently that its complement W is open.
To this end, let (x,y) be any pair in W, namely, x≰y. By order separation, there are a τ+-open neighborhood U+ of x and a τ–-open neighborhood U– of y such that U+ and U– are disjoint. The open rectangle U+ × U– is an open neighborhood of (x,y), and we merely have to show that it is included in W. If that were not the case, there would be a pair (x’,y’) in U+ × U– such that x’≤y’. In particular, x’≤+y’ (and x’≥–y’). Since U+ is ≤+-upwards closed by definition of ≤+ as the specialization preordering of τ+, and since x’ is in U+, y’ is in U+, too. But y’ is also in U–, which contradicts the fact that U+ and U– are disjoint. We have shown that W is an open neighborhood of any of its elements, so it is open. It follows that (X, ≤) is indeed a (compact) pospace.
Let us write the bitopological space bi (bi–1 (X, τ+, τ–)) as (X, τ+2, τ–2). We aim to show that τ+2=τ+ and that τ–2=τ–. The two cases are symmetric, so we concentrate on showing τ+2=τ+. Let me recall that τ+2 is the collection of ≤-upwards-closed (τ+ ∨ τ–)-open subsets of X.
Every τ+-open set is clearly ≤+-upwards closed, hence ≤-upwards-closed (since ≤=≤+), and (τ+ ∨ τ–)-open. In other words, every τ+-open is τ+2-open. We need to show the converse.
Let U be any ≤-upwards closed (τ+ ∨ τ–)-open subset of X. We will show that U is a τ+-open neighborhood of each one of its points x. For this, we fix an arbitrary point x of U. For each point y in the complement X–U of U, we must have x≰y: otherwise x≤y, and since U is ≤-upwards closed, y would be in U. By order separation, we can therefore find a τ+-open neighborhood U+y of x and a τ–-open neighborhood U–y of y such that U+y and U–y are disjoint.
The sets U–y, where y ranges over X–U, together with U itself, form a cover of X. They are all (τ+ ∨ τ–)-open: this is by assumption for U, and this is because each U–y is even τ–-open. Since (X, τ+, τ–) is d-compact, X is compact in the topology τ+ ∨ τ–. It follows that we can extract a finite subcover from that cover. In other words, there is a finite subset E of X–U such that X ⊆ U ∪ ∪y ∈ E U–y.
Since E is finite, the set V ≝ ∩y ∈ E U+y is τ+-open, and it is easy to see that it contains x. It remains to verify that V is included in U: for every z in V, since X ⊆ U ∪ ∪y ∈ E U–y, either z is in U and we are done; or z is in U–y for some y ∈ E, but then the fact that z is also in V, hence in U+y, contradicts the fact that U+y and U–y are disjoint.
This finishes to prove that U is τ+-open. Since U is arbitrary, every ≤-upwards closed (τ+ ∨ τ–)-open subset of X is τ+-open, meaning that every τ+2-open set is τ+-open. Therefore τ+2=τ+. ☐
There is a category BiTop of bitopological spaces, whose morphisms are the bicontinuous maps f : (X, τ+, τ–) → (Y, τ’+, τ’–), namely the maps from X to Y that are continuous both from X with the τ+ topology to Y with the τ‘+ topology, and from X with the τ– topology to Y with the τ‘– topology.
Every such bicontinuous map f is automatically continuous from X with the join topology τ+ ∨ τ– to Y with the topology τ’+ ∨ τ’–. Being continuous with respect to the + topologies, it is monotonic with respect to the + specialization preorderings; similarly for the – specialization preorderings; and therefore f is monotonic with respect to the respective associated preorderings. In summary, f is continuous with respect to the join topologies and preserves the associated preorderings.
Let CompSepBiTop be the full subcategory of BiTop whose objects are the d-compact order-separated bitopological spaces. Let also CompPoSpace be the category of compact pospaces and continuous order-preserving maps.
The map bi extends to a functor from CompSepBiTop to CompPoSpace. Its action on objects was defined at (and before) Lemma C. On morphisms, bi maps each morphism f to itself.
In the reverse direction, we claim that bi–1, which maps compact pospaces to their bitopological spaces, defines a functor from CompPoSpace to CompSepBiTop. On objects, this is by the first part of Proposition E. On morphisms, we let bi–1 map every continuous order-preserving map f : (X,⪯) → (Y,⪯’) to f itself, seen as a map between the associated bitopological spaces (X, τ+, τ–) ≝ bi–1 (X,⪯) and (Y, τ’+, τ’–) ≝ bi–1 (Y,⪯’). Since τ+ is just the collection of ⪯-upwards closed open subsets of X, and similarly for the other topologies, it is immediate that f is bicontinuous.
We obtain the following.
Theorem [1, Theorem 2.2.5]. The functors bi and bi–1 define an isomorphism (not just an equivalence) of categories between CompPoSpace and CompSepBiTop. ☐
One should notice how short, and pretty elementary, the whole proof of that fact is. Hence Tomáš Jakl was right when he told me that he had a nice, short proof of the equivalence.
But this is cheating. We have not proved the equivalence between CompPoSpace and the category of stably compact spaces and perfect maps, as I had announced at the beginning of this post. In fact, we have not talked about stably compact spaces at all yet!
Stably compact spaces
The key in solving that issue is the following proposition. Let me recall that the cocompact topology τd on any topological space X (with topology τ) is the coarsest topology that contains the complements of τ-compact saturated subsets of X. The patch topology is the join τd ⋁ τ.
Let us say that a dual topology τ* on X is a topology whose specialization preordering is the opposite of that of X. It is separating if and only if (X, τ, τ*) is order-separated. The following is from [4, Definition VI-6.17, Proposition VI-6.18].
Proposition F. Let X be a topological space, and τ be its topology.
- The cocompact topology τd is always dual.
- If X is locally compact, then it is separating dual and is coarser than the patch topology.
- If X is stably compact, then the only separating dual topology coarser than the patch topology on X is the cocompact topology τd.
Proof. 1. The closure of any point x in the cocompact topology τd is ↑x (the upward closure of x in the specialization preordering ≤ of X), because ↑x is clearly the smallest compact saturated subset of X that contains x. Its specialization preordering ≤d is defined by x≤dy if and only if x is in the τd-closure of y, namely in ↑y. This shows that τd is dual.
2. Let us assume that X is locally compact. Let x and y be two points such that x≰y. Then x is in the open set X–↓y. Since X is locally compact, there are a compact saturated subset Q and an open subset U of X such that x ∈ U ⊆ Q ⊆ X–↓y. U is in τ, X–Q is in τd and contains y, and U and X–Q are disjoint since U ⊆ Q. Therefore (X, τ, τd) is order-separated, namely τd is separating.
3. Finally, let us assume that X is stably compact, and let τ* be a separating dual topology coarser than the patch topology τd ⋁ τ. We have the following.
- (a) X with the patch topology τd ⋁ τ is compact: this is shown in Proposition 9.1.27 of the book, and is an easy application of Alexander’s subbase lemma. Let us do it explicitly. We consider an open cover of X by subbasic (τd ⋁ τ)-open sets: they come in two kinds, a family of τ-open subsets Ui, i ∈ I, and a family of complements X–Qj, j ∈ J, of compact saturated subsets Qj. By definition of a cover, X ⊆ ∪i ∈ I Ui ∪ ∪j ∈ J (X–Qj). We rewrite this as ∩j ∈ J Qj ⊆ ∪i ∈ I Ui. We notice that ∩j ∈ J Qj can be written as the filtered intersection of finite intersections of sets Qj. Since X is well-filtered, there is a finite subset J’ of J such that ∩j ∈ J’ Qj ⊆ ∪i ∈ I Ui. Now, since X is compact and coherent, ∩j ∈ J’ Qj is compact saturated (compactness is needed in case J’ is empty), so there is a finite subset I’ of I such that ∩j ∈ J’ Qj ⊆ ∪i ∈ I’ Ui. We rewrite this is X ⊆ ∪i ∈ I’ Ui ∪ ∪j ∈ J’ (X–Qj), and this makes us realize that the finite collection of sets Ui, i ∈ I’, and X–Qj, j ∈ J’, is a finite subcover of our original cover.
- (b) It follows that (X, τ, τd) is d-compact. Since X is locally compact, item 2 of the current proposition shows that (X, τ, τd) is order-separated. Proposition E then tells us that bi–1 (X, τ, τd) is a compact pospace (X, ≤), and bi (bi–1 (X, τ, τd))=(X, τ, τd). Explicitly, that means that τ is the collection of ≤-upwards closed (τd ⋁ τ)-open subsets of X, and that τd is the collection of ≤-downwards closed (τd ⋁ τ)-open subsets of X. We retrieve Theorem 9.1.32 of the book… without going through the dreaded Lemma 9.1.29.
- (c) By Lemma B, since (X, τ, τ*) is order-separated, the topology τ* ⋁ τ is T2. Since τ* is coarser than τd ⋁ τ, τ* ⋁ τ is also coarser than τd ⋁ τ. But We now use the fact that any T2 topology coarser than a compact topology must coincide with it (Theorem 4.4.27). It follows that τ* ⋁ τ = τd ⋁ τ.
- (d) By (c), τ* ⋁ τ = τd ⋁ τ is a compact topology, so (X, τ, τ*) is d-compact. It is order-separated by assumption. We reason as in (b), and we obtain that τ* is the collection of ≤-downwards closed (τ* ⋁ τ)-open (equivalently, (τd ⋁ τ)-open) subsets of X.
- Combining (b) and (d), we obtain that τ*=τd. ☐
We obtain the following easily now.
Theorem G. The category SComp of stably compact spaces and perfect maps is equivalent to CompSepBiTop, and therefore also to CompPoSpace.
Proof. There is a functor F : CompSepBiTop → SComp which maps every d-compact order-separated bitopological space (X, τ+, τ–) to X with the topology τ+, and is the identity on morphisms. This is justified as follows. By Proposition E, bi–1 (X, τ+, τ–) is a compact pospace (X, ≤), and bi (bi–1 (X, τ+, τ–))=(X, τ+, τ–). In particular, τ+ is the upward topology of (X, ≤).
This turns X into a stably compact space (Proposition 9.1.20 in the book; you may also want to reprove it by yourself). By Proposition F, τ– must be the cocompact topology (τ+)d.
Since τ+ turns X into a stably compact space, Lemma 9.1.25 implies that the open subsets of τ–=(τ+)d are exactly the complements of compact saturated subsets; there are no other. (Again, you might want to reprove it by yourself: realize that well-filteredness implies that any filtered intersection of compact saturated sets is compact saturated; coherence and compactness then imply that any intersection whatsoever of compact saturated sets is compact saturated.)
Given any bicontinuous map f : (X, τ+, τ–) → (Y, τ’+, τ’–) between d-compact order-separated bitopological spaces, it follows from the above considerations that f is continuous from X with the τ+ topology to Y with the τ’+ topology. It is also continuous from X with the τ– topology to Y with the τ’– topology, meaning that the inverse image of a τ’–-closed subset of Y (a compact saturated subset of Y in the τ’+ topology) is τ–-closed (compact saturated). In other words, f is perfect.
In the reverse direction, every stably compact space X (with topology τ) induces a d-compact order-separated bitopological space G(X), defined as (X, τ, τd), by Proposition F. (That in fact holds when X is merely locally compact.) We extend G to a functor by letting it map every morphism f (a perfect map between stably compact spaces X and Y) to itself. It is clear that f is bicontinuous from G(X) to G(Y).
Finally, the two functors F and G are inverse of each other. This is trivial on morphisms. On objects, for every stably compact space X (with topology τ), F(G(X)) = F (X, τ, τd) is X, with τ as topology, by definition. For every d-compact order-separated bitopological space (X, τ+, τ–), G(F (X, τ+, τ–)) is equal to the application of G to the stably compact space X, with the τ+ topology, hence to (X, τ+, (τ+)d). We conclude by recalling that (τ+)d=τ–, as shown in Proposition F. ☐
Finally, we have shown the equivalence between CompPoSpace and Scomp, as promised, and without using the dreaded Lemma 9.1.29. Is the proof short? I would not say so, especially if we take into account that we had to reuse many of the results of Section 9.1 of the book. However, it is nice indeed. Notably, it exhibits the notion of order separation as the key ingredient.
I would like to finish this post by mentioning a few low hanging fruit.
Remember that the bitopological spaces of a quasi-metric space X,d is always order-separated. We therefore obtain the following, almost for free. Just as the previous results in this post, none of those results are new. For instance, the following is Theorem 9.1.39 in the book.
Theorem. For a quasi-metric space X,d, the following are equivalent:
- X,d is symcompact;
- X is stably compact in the open ball topology τ+ of d, and the cocompact topology (τ+)d coincides with the open ball topology τ– of dop.
Proof. X,d is symcompact if and only if (X, τ+, τ–) is d-compact, where τ+ and τ– are the open ball topologies defined above. Indeed, τ+ ∨ τ– is the open ball topology of the symmetrized metric (Proposition 6.1.19). Theorem G (with Proposition F) implies that, if X,d is symcompact, then X is stably compact in the τ+ topology, and that τ–=(τ+)d. Therefore 1 implies 2. Conversely, if 2 holds, then τ–=(τ+)d, and we know that (X, τ+, (τ+)d) is d-compact, so 1 holds. ☐
Similarly, and generalizing the latter, we have the following. We recall from Part III of the series on quasi-uniform spaces that the symmetrized uniformity Usym has a base of entourages of the form R ∩ R–1, where R ranges over the entourages in U; alternatively, a base of entourages of the form R ∩ S–1, where R and S range over the entourages in U. The following fact was announced near the beginning of this post.
Lemma. Let U be a quasi-uniformity on X, let τ+ be the induced topology of U, and τ– be that of U–1. The induced topology of Usym is τ+ ∨ τ–.
Proof. Let U be open in the induced topology of Usym. For every point x of U, there is an entourage R in U such that (R ∩ R–1) [x] is included in U. But (R ∩ R–1) [x] is the set of points y such that (x,y) and (y,x) are both in R, or equivalently the intersection R[x] ∩ R–1[x]. The latter is a neighborhood of x in τ+ ∨ τ–, because it contains the intersection of the τ+-interior of R[x] and of the τ–-interior of R–1[x]. Hence U is (τ+ ∨ τ–)-open.
Conversely, let U be (τ+ ∨ τ–)-open. For every point x of U, x is contained in an intersection U+ ∩ U– of a τ+-open set U+ and of a τ–-open set U–, and that intersection is included in U. By definition of induced topologies, there are two entourages R and S in U such that R[x] ⊆ U+ and S–1[x] ⊆ U+. Then (R ∩ S–1)[x] ⊆ R[x] ∩ S–1[x] ⊆ U+ ∩ U– ⊆ U, showing that U is open in the induced topology of Usym. ☐
The same proof as in the quasi-metric case then establishes the following.
Theorem. For a quasi-uniform space X,U, the following are equivalent:
- X,U is symcompact;
- X is stably compact in the induced topology τ+ of U, and the cocompact topology (τ+)d coincides with the induced topology τ– of U–1.
Finally, let X be a continuous dcpo, or more generally a quasi-continuous dcpo. We had observed that (X, τ+, τ–) is order-separated in that case, where τ+ is the Scott topology on X and τ– is the lower topology, and that τ+ ∨ τ– is the Lawson topology.
Theorem. For a quasi-continuous dcpo X, the following are equivalent:
- X is Lawson-compact, that is, X is compact in its Lawson topology;
- X is stably compact in its Scott topology, and the cocompact topology coincides with the lower topology.
Together with Theorem G, we obtain that if any of those two properties is satisfied (still assuming that X is a continuous dcpo), then the Lawson topology coincides with the patch topology. Thus, we retrieve the results of Exercise 9.1.36.
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