The Sorgenfrey line is not consonant

I have already talked about consonant spaces, for example here. A space X is consonant [1] if and only if the compact-open topology and the Isbell topology coincide on the function space [XY] for every topological space Y. It is equivalent to require that they coincide on the single function space [XS], where S is Sierpiński space {0 < 1}. Since a continuous map from X to S is the characteristic map of a unique open subset of X, X is consonant if and only if every Scott-open subset U of the lattice OX of open subsets of X is a union of sets of the form ■Q≝{U | QU}, where the sets Q are compact saturated subsets of X.

In Exercise 5.4.12 of the book, I ask the reader to prove that neither the space of rationals, Q, nor the Sorgenfrey line, R, is consonant. That is what I classified as the important blooper #5 in the list of errata, because the proof I had originally in mind was too simple-minded to have any chance of succeeding. But showing that R is not consonant is not that hard, finally. I will explain the argument. This will also be an excuse to explain some additional topological properties of R.

The fact that R is not consonant was proved by Costantini and Watson, as a consequence of a more general result [3]. (Update, April 22nd, 2022: no, Costantini and Watson’s theorem does not apply. Instead, the fact that R is not consonant is due to Alleche and Calbrix [8]. Sorry for the confusion.) Instead, the route I will take has some similarities with what Bouziad did in [2] in order to prove that Q is not consonant—again, as consequence of a more general result—in the sense that I will rely on a bit of measure theory. (Update, April 22nd, 2022. And indeed, the argument I will develop below is essentially Alleche and Calbrix’s.) If you don’t know anything about measure theory, I will also give an argument that only requires some domain theory, most of which is in appendices, at the end of this post.

The Sorgenfrey line R

Before we start, let me recall what the Sorgenfrey line R is.

It is an incredible beast. Its points are just the real numbers, but its topology is generated by the half-open intervals [a,b[ with a < b. The topology of R is finer than the usual metric topology of R. R is a zero- dimensional, first-countable space (Exercices 4.1.34, 4.7.17 in the book). It is paracompact hence T4, and Choquet-complete hence a Baire space (Exercises 6.3.32, 7.6.11). R is not locally compact, as every compact subset of R has empty interior (Exercise 4.8.5). Although it is first-countable, R is not second-countable (Exercise 6.3.10).

We won’t need any of that! It is enough to remember that R exhibits a mixture of the nice (paracompact, T4, Choquet-complete, Baire, first-countable) and of the ugly (not locally compact, not second-countable). We will add some new water to the pot of ugliness (non consonant), and also to the jar of niceness (hereditary Lindelöf, see below).

R is not consonant: the short argument

We consider the collection U of open subsets of R whose Lebesgue measure is > r, for some fixed r>0. That is a Scott-open subset of O(R), and a non-empty one, too (since [–r, r[ is in U, for example), but we claim that it simply contains no subset of the form ■Q for any compact (saturated) subset of R. Hence U is certainly not a union of sets of the form ■Q.

In order to see this, we realize that every compact subset Q of R is countable. Writing Q as {x0, x1, …, xn, …}, we see that U ≝ ∪nN [xn, ε/2n[ is an open neighborhood of Q for every ε>0, namely that U in in ■Q. However the Lebesgue measure of U is smaller than or equal to ∑nN ε/2n = 2ε, so U cannot be in U as soon as we pick ε≤r/2.

Now, wait! I skipped a number of details here. One is that I have not said why every compact subset Q of R is countable. This is what I am going to show next, and, while we are at it, I will give a complete characterization of the compact subsets of R. Another missing piece is the reason why U is Scott-open. That would be the case is the Lebesgue measure of any directed union of open sets were equal to the supremum of their measures, but measures only satisfy this property for countable directed unions in general. The measures that satisfy this property for arbitrary unions are called τ-smooth. Here a miracle will occur, thanks to a theorem due to Wolfgang Adamski [4], and a magical property called hereditary Lindelöfness, which I will talk about, too. This all uses a bit of measure theory, but, if you have never heard about measure theory, I will give you a purely domain-theoretic construction of U in Appendix B at the end of this post.

The compact subsets of the Sorgenfrey line are countable

Every compact subset Q of R is countable, and this is easy to see.

nWe first note that the sets ]−∞,r–ε[ and [r, +∞[ are open, for all r and ε, because they are unions of basic open sets [r–ε–n–1, r–ε–n[, nN, and of [r+n, r+n+1[, nN, respectively.

Next, given a compact subset Q of R, we realize that, for every r in Q, the open sets ]−∞,r–ε[ with ε>0, plus [r, +∞[, form an open cover of Q. (Indeed, of the whole of R!) We extract a finite subcover: without loss of generality, we may assume that this subcover contains [r, +∞[, otherwise we add it back; the other elements of the subcover are ]−∞,r–ε1[, …, ]−∞,r–εn[. We ick ε>0 so that ε<ε1, …, εn, and we see that Q is included in ]−∞,r–ε[ ∪ [r, +∞[, namely that it does not intersect [r–ε, r[. In order to display the dependency on r, let us write εr for that ε.

Let us pick a rational number qr in [r–εr, r[, one for each r in Q. The map rQqr is injective: otherwise there would be two distinct elements r and s of Q such that [r–εr, r[ and [s–εs, s[ would overlap (at qr=qs). By symmetry, let us assume r<s. Then r would be in [s–εs, s[, which is impossible since r is in Q but [s–εs, s[ does not intersect Q.

We have built an injective map from Q to the set of rationals, which is countable. Therefore Q is countable.

That much is enough knowledge for us on the compact subsets of R. However, it would be foolish to stop here and not investigate the nature of compact subsets of R, right? Hence, let us embark on characterizing the compact subsets of R exactly.

Characterizing the compact subsets of the Sorgenfrey line

Here is another property of compact subsets of R that can be proved in a similar way as above.

Fact A. Every compact subset Q of R is well-founded with respect to ≥: there is no infinite chain of the form r0 < r1 < … < rn < … in Q.

Proof. Let r ≝ supnN rn. Then the open sets ]−∞,r0[, [r0,r1[, …, [rn,rn+1[, …, together with [r,+∞[ in case that r≠+∞, would form an open cover of Q without a finite subcover. ☐

Fact A says that every compact subset of R is a well-founded subset of (R, ≥), namely of R with the opposite ≥ of the usual ordering ≤.

A directed family in (R, ≥) is a filtered family: a non-empty family in which for any two elements, there is an element less than or equal to both; and “filtered” in not even necessary, since the ordering ≤ on R is total: in other words, a filtered family is just a non-empty family there.

Fact B. Every compact subset Q of R is a subdcpo of (R, ≥): the infimum, taken in R, of every filtered family of elements of Q lies in Q.

Proof. We first need to note that a net (xi)iI,⊑ converges to x in R if and only if xi tends to x from the right, namely: for every ε > 0, xxi < x + ε for i large enough. This is Exercise 4.7.6 in the book, but the argument is very short, and I will let you verify this by yourselves.

Let D be any non-empty subset of Q, and let r ≝ inf D. Since the topology of R is finer than the topology of R, every compact subset of R is also compact in R, in particular, it is bounded. Therefore r is a well-defined real number (not –∞).

By Fact A, Q is well-founded, so D is isomorphic to a unique ordinal β, and we can write the elements of D as rα, α<β, in such a way that for all α,α′ <β, α≤α′ if and only if rαrα’. The net (rα)α<β,≤ then converges to r from the right. Since R is T2, its compact subset Q is closed, so r is in Q. ☐

Note that, if the compact set Q is non-empty, then in particular inf Q lies in Q, namely, Q has a least element. That could also be derived from the fact that every compact subset of R is compact in R.

Here is an example of a compact subset of R: the set of points of the form 1/2n, nN, plus the point zero. We can index those points by ordinal numbers: point number n is 1/2n, and there is an additional point numbered ω, which is zero:

A compact subset of R, of order-type ω+1.

Every well-founded subset of (R, ≥) must be isomorphic to an ordinal (here, ω+1: an ordinal is also the set of all the ordinals strictly below it, so ω+1 = {0, 1, …, n, …, ω}.) Hence if Q is a compact subset of R, then its elements must be numbered by ordinal numbers, from right to left, as in the above picture. Additionally, since Q is a subdcpo, that enumeration must stop at some ordinal (Q has a least element), and points of Q must cluster towards points which are numbered with a limit ordinal (from the right), as in the above picture where points cluster towards point number ω, coming from the right.

There are more complex compact subsets, such as the following one, whose points are indexed by the ordinals from 0 to ω2.

A compact subset of Rℓ, of order-type ω2+1.

How do I know that the latter is compact in R? Well, here is a complete characterization.

Proposition C. The compact subsets of the Sorgenfrey line R are exactly the well-founded subdcpos of (R, ≥).

Proof. Considering Facts A and B, it suffices to show that every well-founded subdcpo Q of (R, ≥) is compact in R. We use Alexander’s subbase lemma (Theorem 4.4.29 in the book): for every family U of basic open subsets that covers Q, we will extract a finite subcover. We build a (finite or infinite) sequence of elements rn of Q and another sequence of elements [an,bn[ of U by induction on n in such a way that:

  1. r0 < r1 < … < rn < … (so yes, the sequence will in fact be finite, since Q is well-founded in (R, ≥), but let us not go too fast);
  2. for each n, rn is in [an,bn[;
  3. for each n, the intervals [a0,b0[, …, [an,bn[ cover all the elements of Q that are ≤rn; in other words, every element of Q is in [a0,b0[ ∪ … ∪ [an,bn[ ∪ ]rn,+∞[.

In the base case (n=0), we take r0≝inf Q, which is in Q since Q is a subdcpo of (R, ≥). Being in Q, r0 is in some basic open subset [a0,b0[ ∈ U. We then note that every element of Q is in [a0,b0[ ∪ ]r0,+∞[.

In the induction case (n≥1), we assume we already have elements r0 < r1 < … < rn–1 in Q, lying in [a0,b0[, …, [an–1,bn–1[ respectively, and such that every element of Q is in [a0,b0[ ∪ … ∪ [an–1,bn–1[ or is >rn–1. Let Qn be Q – ([a0,b0[ ∪ … ∪ [an–1,bn–1[). If Qn is empty, then the sequence stops here, namely at rn–1. Otherwise, Qn is non-empty hence directed in (R, ≥). Since Q is a subdcpo, inf Qn exists and is an element of Q. We let rn≝inf Qn. Every element of Qn is >rn–1, so rnrn–1. If rn=rn–1, then rn would be in [an–1,bn–1[, in particular an–1rn=inf Qn<bn–1, so some element of Qn would be in [an–1,bn–1[, which is impossible. Hence rn–1<rn. As far as item 2 is concerned, we merely use the fact that U is a cover of Q, so rn must be in some [an,bn[ ∈ U. Finally, every element x of Q is in [a0,b0[ ∪ … ∪ [an–1,bn–1[ or is >rn–1. If x>rn–1, then x is in Qn, hence xrn, and therefore x is in [an,bn[ or is strictly larger than rn.

We have built a chain r0 < r1 < … < rn < … of elements of Q satisfying properties 1, 2, and 3 above. Since Q is well-founded, this chain must be finite: the inductive process described above stops at some rank n. In that case Qn is empty, and therefore Q is included in [a0,b0[ ∪ … ∪ [an–1,bn–1[. ☐

In other words, the compact subsets Q of R are exactly the set of points xα enumerated by all ordinals α<β, in such a way that x0 > x1 > … > xn > … > xω > xω+1 > … > xω.2 > …, and so that this forms a subdcpo in the opposite ordering ≥, namely:

  • there is a last (smallest with respect to ≤) point in Q, in other words β is not a limit ordinal (in our previous examples, β was equal to ω+1 and to ω2+1 respectively);
  • for every chain of points xα[i] in Q, iI, infiI xα[i] is again in Q, so must be of the form xα for some ordinal α<β, and it is easy to see that α must in fact be equal to supiI α[i].

The ordinal types of compact subsets of R

We have seen that every compact subset of R is countable. More general, every well-founded subset E of (R, ≥) is countable. (And similarly with the more usual ordering ≤, too, of course.) That is easy to prove, and we don’t need E to be a subdcpo. For every x in E, we look at the interval ]–∞, x[. If ]–∞, x[ ∩ E is empty, then we pick a rational number qx in ]–∞, x[. Otherwise, since E is well-founded with respect to ≥, E contains a largest element y (with respect to ≤) that is inside ]–∞, x[. Then we pick a rational number qx in ]y, x[. The map xEqx is injective: in fact we have y<x if and only if qy<qx. Since there are only countably many rational numbers, E is countable.

In particular, the order-type of any compact subset Q of R, namely the unique ordinal that is order-isomorphic to (Q, ≥), is a countable ordinal. It is also a non-limit ordinal, because those are exactly the ordinals that are (sub)dcpos. For example, the limit ordinal ω={0, 1, …, n, …} is not a dcpo. But ω+1={0, 1, …, n, …, ω} is a dcpo.

One might wonder which countable ordinals are the order-types of compact subsets of R. The answer is simple: all the non-limit countable ordinals. Since I have already spent too much time on compact subsets, I will refer you to Appendix A of this post for details.

The hereditary Lindelöf property

So, enough with the compact subsets of R. I promised we would look at a magical property of Lebesgue measure on R, and that this would have to do with a property called hereditary Lindelöfness.

A topological space X is hereditarily Lindelöf (train yourself to pronounce that… what a tongue twister!) if and only if every family of open subsets of X contains a countable subfamily with the same union. If you already know what a Lindelöf space is, you may realize that a space is hereditarily Lindelöf if and only if all its subspaces are Lindelöf.

We will see that every second-countable space is hereditarily Lindelöf, and that R is an example of a space that is hereditarily Lindelöf… but not second-countable.

Lemma D. Every second-countable space is hereditarily Lindelöf.

Proof. Let X be a second-countable space, and B be a countable base of its topology. Let (Ui)iI be an arbitrary family of open subsets of X, and let U be its union. For every point x of U, there is an index i in I such that x is in Ui, let us call it i[x]. Also, there is an element B[x] of the base B such that xB[x] ⊆ Ui[x]. Then U ⊆ ∪xU B[x]. However, since B is countable, the latter is a countable union. In other words, for each element of the base B that arises as B[x] for some x in U, we pick such a point x; we collect all those points and form a countable subset E of U; then U ⊆ ∪xE B[x]. Now ∪xE B[x] ⊆ ∪xE Ui[x]U, so U = ∪xE B[x] = ∪xE Ui[x]: when x ranges over the set E, the open sets Ui[x] form a countable subfamily of (Ui)iI with the same union, namely U. ☐

The Sorgenfrey line R is not second-countable: you can find the argument in Exercise 6.3.10 of the book. Rather amazingly, we have the following.

Proposition E. R is hereditarily Lindelöf.

Proof. I took the proof from the Math Stack exchange. Let us agree to write Ů for the interior of a set U in the topology of R: not that of R, please bear that in mind.

Let (Ui)iI be any family of open subsets of R, let U be its union and let U’ denote the union ∪iI Ůi of their interiors in the topology of R. Since R is second-countable, hence hereditarily Lindelöf by Lemma D, there is a countable subset J of I such that U’ = ∪iJ Ůi.

We claim that UU’ is countable. For each point xUU’, x is in some Ui, hence in some basic open subset [ax, bx[ ⊆ Ui. Then x is also included in the smaller basic open set [x, bx[. The (fully) open interval ]x, bx[ is included in Ůi, hence in U’. By picking a rational number qx in [x, bx[ for each xUU’, we see that there are only countably many distinct intervals [x, bx[. For any two distinct points x and y of UU’, if the intervals [x, bx[ and [y, by[ overlapped, then either y would be in [x, bx[ or x would be in [y, by[. Let us imagine that y is in [x, bx[. Since yx, y must be in ]x, bx[ ⊆ U’; that is impossible since we have chosen y in UU’. We reason similarly if x is in [y, by[. We have shown that for any two distinct points x and y of UU’, the intervals [x, bx[ and [y, by[ do not overlap; in particular, they are distinct. As there are only countably many such intervals, there are only countably many points in UU’.

For each xUU’, let us pick an index i[x] in I such that x is in Ui[x]. Let K be {i[x] | xUU’}. This is a countable set, and therefore so is JK. The countable subfamily consisting of those sets Ui such that iJK then covers both U’ and UU’, hence it covers the whole of U. It follows that U = ∪iJK Ui. ☐

Hence R once again serves as a counterexample! Namely, as an example that not all hereditarily Lindelöf spaces are second-countable.

Before we proceed, let me mention the following practical fact.

Lemma F. A space X is hereditarily Lindelöf if and only if every directed family (Ui)iI of open sets contains an ascending subsequence Ui[0]Ui[1] ⊆ … ⊆ Ui[n] ⊆ … with the same union.

Proof. If X is hereditary Lindelöf, then we can assume that I is countable, and even that I is just N. We build i[n] by induction on n as follows. We let i[0]≝0. For every n≥1, we define i[n] as any index i such that Ui contains Ui[n–1] and Un. It is easy to see that Ui[0]Ui[1] ⊆ … ⊆ Ui[n] ⊆ … Moreover, for every natural number n, Un is included in Ui[n], so that the union of the sets Ui[n] contains the union of the sets Un, and is therefore equal to it.

Conversely, let us assume that every directed family of open sets contains an ascending subsequence with the same union. Let (Ui)iI be any family of open sets, and U be its union. For every non-empty finite subset J of I, let UJ ≝ ∪iJ Ui: the family (UJ)J finite non-empty ⊆ I is now directed, its union is U, and by assumption we can find an ascending subsequence UJ[0]UJ[1] ⊆ … ⊆ UJ[n] ⊆ … whose union is U. It follows that the union of the sets Ui, where i ranges over the countable family ∪nN J[n], is equal to U. ☐

The Lebesgue measure on R

If you know about Lebesgue measure λ, proving that R is not consonant is now only a few steps away. (If you don’t, then stop reading here and proceed directly to Appendix B.) Realizing that every open subset of R is a countable union of basic open subsets [a,b[ (a simple consequence of the fact that R is hereditarily Lindelöf), and that any subset of the form [a,b[ is a Borel subset of R, we see that R and R have the same Borel σ-algebras. In particular, λ defines a Borel measure on R.

Then we notice that every Borel measure on a hereditarily Lindelöf space is τ-smooth [4]. This is much easier to understand and prove than the intimidating wording would leave us to believe. A measure is τ-smooth if and only if its restriction to the lattice of open sets is Scott-continuous. If X is a hereditarily Lindelöf space, then for every directed family (Ui)iI of open subsets of X, with union U, there is an ascending subsequence Ui[0]Ui[1] ⊆ … ⊆ Ui[n] ⊆ … whose union is also U (see Lemma F). Given any Borel measure μ on X, μ(U) is then equal to μ(∪nN Ui[n]). Since the union ∪nN Ui[n] is a union of a countable chain, the fact that μ is a measure entails that μ(U) = supnN μ(Ui[n]), and one easily checks that this is also equal to supiI μ(Ui). The important point is that measures commute with suprema of countable chains. Going from the countable case to the general case is something we can afford thanks to hereditary Lindelöfness.

By the way, hereditary Lindelöfness is a sufficient and necessary condition: as Adamski proved, a space X is hereditarily Lindelöf if and only if every Borel measure on X is τ-smooth [4].

In any case, Lebesgue measure λ is a Borel measure on R. R is hereditarily Lindelöf by Proposition E, and Adamski’s theorem tells us that λ is τ-smooth, namely that it restricts to a Scott-continuous map from O(R) to R+ ∪ {+∞}. It follows that the set U of all open subsets U of R such that λ(U)>r (for any given r>0) is Scott-open, as promised. That was the only remaining detail we had to fill in the argument we gave at the beginning of this post: R is not consonant.

That’s it! Oh, if you’re not completely exhausted, have a look at Appendix B for a funny purely domain-theoretic of building the same Scott-open family U.

  1. Szymon Dolecki, Gabriele H. Greco, and Alojzy Lechicki, 1995. When do the upper Kuratowski topology (homeomorphically, Scott topology) and the co-compact topology coincide? Transactions of the American Mathematical Society, 347(8), 2869–2884.
  2. Ahmed Bouziad, Borel measures in consonant spaces, Topology and its Applications 70 (1996), 125—138.
  3. Camillo Costantini and Stephen Watson, On the dissonance of some metrizable spaces, Topology and its Applications 84 (1998), 259—268.
  4. Wolfgang Adamski. τ-smooth Borel measures on topological spaces. Mathematische Nachrichten, 78:97–107, 1977.
  5. Claire Jones. Probabilistic Non-Determinism. Ph.D. thesis, University of Edinburgh. Technical Report ECS-LFCS-90-105, 1990.
  6. Nait Saheb-Djahromi. Cpo’s of Measures for Nondeterminism. Theoretical Computer Science, 12, pages 19–37, 1980.
  7. Jimmie Lawson. Valuations in continuous lattices. In: Hoffman, R.-E. (ed.), Continuous Lattices and Related Topics. Mathematik Arbeitspapiere, Universität Bremen 27, 1982.
  8. Boualem Alleche and Jean Calbrix. On the coincidence of the upper Kuratowski topology with the cocompact topology. Topology and its Applications 93(3):207—218, 1999.

Appendix A

We show that every countable non-limit ordinal β+1 is the order-type of some compact subset Qβ of R. (Technically, 0 is another non-limit ordinal, and it is not of the form β+1; but 0 if of course the order-type of the empty compact set.) In fact, we will show that we can even design Qβ so that it is included in any prescribed interval ]a,b], a<b.

In order to show this, we need to know the following interesting little fact: every countable limit ordinal has cofinality ω. This is savant jargon to say that every countable limit ordinal β is the supremum of a monotone sequence of ordinals α0 ≤ α1 ≤ … ≤ αn ≤ … < β indexed by natural numbers. Here is how you prove it. First, β is the supremum of all the ordinals α<β. The family of those ordinals is just β itself, and is therefore countable. This gives us an enumeration of the ordinals α<β by natural numbers; but we are not done yet, because that enumeration may fail to be monotone. So we have to work a bit more. Let f be a bijection between N and β. This is the enumeration we have just talked about. We define αn by induction on n, as being f(m) where m is the first natural number ≥m such that f(m) is strictly larger than all the previously defined ordinals α0, α1, …, αn–1. I’ll let you check by yourselves that the sequence of ordinals αn is well-defined, monotone, and cofinal in the family of all ordinals α<β, so that its supremum is β as well.

We can now build Qβ (compact in R, of order-type β+1, and included in ]a,b], a<b) by induction on β.

  • If β=0, then we take Qβ ≝ {b}.
  • If β is a successor ordinal, say β=γ+1, then by induction hypothesis we have already formed Qγ, and we may assume that Qγ is included in ]a,b]. Then QβQγ ∪ {(a+min Qγ)/2} suits our needs.
  • Finally, let us assume that β is a (countable) limit ordinal. First, we express β as the supremum of a monotone sequence of ordinals α0 ≤ α1 ≤ … ≤ αn ≤ … < β, using the fact that every countable limit ordinal has cofinality ω. We let β[0]≝α0. For every n≥1, the set of ordinals between αn–1 (included) and αn (excluded) is well-founded and total, hence is isomorphic to a unique ordinal β[n] (its order-type). Clearly, β[n] is countable and smaller than or equal to αn, hence strictly smaller than β. We can there apply the induction hypothesis: for each n in N, we find a compact subset Q’n of R or order-type β[n] included in ]c+(bc)/2n+1, c+(bc)/2n], where c is a fixed number in ]a,b], for example (a+b)/2. Let gn be the order-isomorphism between β[n] and Q’n. Then we let Qβ be the (disjoint) union of the sets E’n, for nN, plus {c}.
    It is easy to see that the order-type of Qβ is β+1. We can even describe an explicit order-isomorphism between β+1 and Qβ: this maps any ordinal α < β[0]=α0 to g0(α) the ordinals α0+α with α < β[1] to g1(α), the ordinals α1+α with α < β[2] to g2(α), and so on; and finally, it maps β to c.
    It remains to show that Qβ is a subdcpo of (R, ≥). Let D be a non-empty family of elements of Qβ. If c is in D, then inf D=c, so inf D is in Qβ. In the rest of the argument, we therefore assume that c is not in D, so that D is included in the union of the compact sets Q’n, nN. Let E be the set of natural numbers n such that D contains an element of Q’n. E is non-empty. If E is bounded, then let n be its largest element. Then DQ’n is cofinal in D (with respect to ≥), so inf D = inf (DQ’n), and that lies in Q’n by Fact B, hence in Qβ. If E is unbounded, then there are points of D in intervals ]c+(bc)/2n+1, c+(bc)/2n] for n arbitrarily large. Then inf D=c, and c is in Qβ.

Appendix B

A continuous valuation on a topological space X is a Scott-continuous map ν : OXR+ ∪ {+∞} such that:

  • (strictness) ν(∅)=0
  • (modularity) ν(UV) + ν(UV) = ν(U) + ν(V) for all open subsets U and V of X.

Notions of valuations as strict and modular maps (without Scott-continuity) can be traced back to at least a 1948 paper by Horn and Tarski. Nait Saheb-Djahromi introduced continuous valuations on dcpos in 1980 [6], and Jimmie Lawson proved a fundamental measure extension theorem for continuous valuations in 1982 [7]. Claire Jones’ 1990 PhD thesis [5] popularized the notion among computer scientists, and was probably the first to lay out the general theory of continuous valuations (on dcpos).

Continuous valuations and measures are very close notions. By definition, given any τ-smooth Borel measure μ on X, its restriction to open sets is a continuous valuation. Conversely, there are large classes of spaces on which every continuous valuation extends to a measure on the Borel σ-algebra (Lawson’s paper [7] being the first result of that kind). But enough with measures! I have promised that this section would be purely domain-theoretic.

The collection VX of all continuous valuations on a space X can be ordered by the so-called stochastic ordering: ν≤ν’ if and only if ν(U)≤ν'(U) for every open subset U of X. This turns VX into a dcpo. In order to show the (modularity) law for a directed supremum of continuous valuations, simply observe that + is Scott-continuous on R+ ∪ {+∞}.

There are some very simple continuous valuations: the Dirac valuation δx at x maps every open neighborhood of x to 1, all all other open sets to 0. The finite linear combinations ∑i=1n ai δx[i], where each ai is in R+, are all continuous valuations. The continuous valuations of this form are called the simple valuations. The probabilistic intuition is that such a continuous valuation represents a process by which we draw x[i] at random with probability ai (provided the points x[i] are pairwise distinct and the sum of the coefficients ai equals 1).

Since VX is a dcpo, any directed supremum of simple valuations is a continuous valuation, and I will call the continuous valuations obtained this way quasi-simple. Jones showed that every continuous valuation on a a continuous dcpo is quasi-simple. I am only mentioning this because that is interesting, but we won’t make any use of that here.

We will build a continuous valuation λ on R, which is in fact the restriction of Lebesgue measure to the open subsets of R. No, λ is not quasi-simple (pardon me for omitting the argument), but the following roundabout strategy will work. We will first embed R into a dcpo model, namely a dcpo IR in which R embeds as the space of its maximal elements. We will then build a certain quasi-simple valuation λ on IR, and we will show that λ is supported on its set of maximal elements R (in a sense that I will make precise below). It will follow that λ arises in a certain way from a continuous valuation on R, and this will be λ.

A dcpo model for R

Let IR be the following poset. It contains two kinds of points:

  • The real numbers r, which will form the set of maximal points of R,
  • and the half-open intervals [a,b[, with a<b.

We order them as follows:

  • [a,b[ ⊑ [a‘,b‘[ if and only if a=a’ and b>b’, namely if and only if [a,b[ contains [a‘,b‘[, and has the same left endpoint;
  • [a,b[ ⊑ r for every r ∈ [a,b[;
  • the only element larger than or equal to a real number r is r itself.

If you start from an interval [a,b[, the only way to go up from there is to produce intervals [a,b’[ with smaller and smaller values of b’, leaving a untouched, or to directly go from one such interval [a,b’[ to a real number r in [a,b’[ and stop. For example, look at the picture below. If you start from [a,a+1[, you can either go up, geometrically, following the red arrows, or you can go directly to the top line, following the blue arrows, and landing anywhere between a (included) and a+1 (excluded).

The dcpo IR

It is then easy to see that IR is a dcpo. Given any directed family D,

  • If D contains some real number r already, then r = sup D;
  • Otherwise, D is a family of intervals [a,bi[, iI, with the same left end a, and its supremum is the interval [a, infiI bi[ if a<infiI bi, and the real number a otherwise.

The way-below relation on IR is given by: AB if and only if B is a real number a, and A is an interval [a, b[ (with the same a). In particular, no interval A is way-below any interval [a, b[: to see this, pick any point r in [a,b[ other than a, then r is the supremum of the chain of intervals [r, r+1/2n[, nN, but none of those intervals can be above A. The same argument shows that [a, b[ cannot be way-below any number r in [a,b[ that is different from a. But [a, b[ is way-below a, as one checks easily.

It follows that IR is very far from being a continuous dcpo: no interval B can be obtained as the supremum of any directed family (or any family whatsoever) of intervals way-below B.

Still, we have the following. We equip IR with the Scott topology of ⊑.

Lemma D. IR is a dcpo model of R: R embeds topologically as the subspace of maximal elements of IR.

Proof. Let U be any Scott-open subset of IR. We show that UR is open in R. In order to see this, let r be any point of UR. Then r is the supremum of the chain [r, r+ε[, ε>0, and is in U, so [r, r+ε[ is an element of U for some ε>0. Since U is upwards-closed, every real number in [r, r+ε[ is in U, hence in UR. This shows that UR is an open neighborhood of r, and since r is arbitrary, UR is open in R.

Conversely, we claim that every open subset U of R is also open in the subspace topology induced by the inclusion of R into IR. To this end, it suffices to consider the special case where U is a subbasic open set [a,b[. We consider the set U consisting of all the real numbers in [a,b[, plus all the intervals [a’,b’[ such that aa’<b’<b. It is easy to see that U is Scott-open, and that UR=U. ☐

Building λ

Let us call finite partition (of R) any finite list P ≝ {a[0] < a[1] < … < a[n]} of real numbers, n≥1. We view this as a specification of the finitely many pairwise disjoint intervals [a[0], a[1][, …, [a[n–1], a[n][. The lengths of those intervals are (a[1]–a[0]), …, and (a[n]–a[n–1]). With such a finite partition, we associate the simple valuation λP ≝ ∑i=1n (a[i]–a[i–1]) δ[a[i–1], a[i][ on IR. Explicitly, for every Scott-open subset U of IR, λP(U) is the sum of the lengths (aiai–1) of those intervals [ai–1, ai[ which, seen as elements of IR, belong to U.

For example, the picture below displays a finite partition P ≝{a[0] < a[1] < a[2] < a[3] < a[4]}, and the corresponding intervals [a[0], a[1][, [a[1], a[2][, [a[2], a[3][, and [a[3], a[4][. Those intervals can be read as strips off the top line, R, or as points of IR, shown as the (upside-down) tips at the bottom of the corresponding triangles. The picture also displays a Scott-open subset U of IR, in a light blue shade. [a[1], a[2][ and [a[2], a[3][ are elements of U, but not [a[0], a[1][ or [a[3], a[4][, and therefore λP(U) = a[3]–a[1].

The collection of all finite partitions is directed in the subset ordering ⊆: notably, given any two finite partitions, their union is a finite partition, too.

I claim that the map PλP is monotonic. Let P and Q be two finite partitions, and let us assume that PQ. Hence we can write P as {a[0] < a[1] < … < a[n]}, and for each i with 1≤in, we can write the points of Q that lie between a[i–1] and a[i] as b[i,0]≝a[i–1] < b[i,1] < … < b[i,mi]≝a[i]. For each j with 1≤jmi, the interval [b[i,j–1], b[i,j][ is an element of IR that lies above [a[i–1], a[i][, and the sum of the lengths of those intervals [b[i,j–1], b[i,j][, 1≤jmi, is exactly the length (a[i]–a[i–1]) of [a[i–1], a[i][. For every Scott-open subset U of IR, λP(U) is the sum of the lengths of those intervals [a[i–1], a[i][ that lie in U. Hence λP(U) is also the sum of the lengths of the intervals [b[i,j–1], b[i,j][ such that [a[i–1], a[i][ is in U and 1≤jmi. All those intervals [b[i,j–1], b[i,j][ are in U as well, so λP(U) is less than or equal to the sum of the lengths of all the intervals [b[i,j–1], b[i,j][ that are in U, when both i and j vary. The latter sum is just λQ(U), so λP(U)≤λQ(U).

Since V(IR) is a dcpo, the family of all maps λP has a supremum λ.

λ is supported on the set of maximal elements of IR

I claim that λ is supported on the set R of maximal elements of IR. Intuitively, that means that drawing an element at random with respect to λ will give you an element of R with probability 1. Formally, that really means that for every fixed open subset U of R, and whichever Scott-open subset U of IR we consider whose intersection with R equals U, λ(U) will always give you the same value, and that will be the desired value λ(U) of the Lebesgue valuation λ on U.

The key to this is Proposition 5.2.28 in the book, a consequence of Rudin’s Lemma. Let me restate it. Given any dcpo X, let Qfin(X) be the poset of all non-empty finite subsets of X, preordered by the Smyth preordering ≤#: E # F if and only if every element of F is above some element of E, namely if and only if ↑E ⊇ ↑F. Proposition 5.2.8 says that, given any directed family (Ei)iI in Qfin(X) and any Scott-open subset U of X, if ∩iIEi is included in U, then some Ei is included in U already.

Let us say that the endpoints of a finite partition P ≝ {a[0] < a[1] < … < a[n]} are a[0] and a[n]. Let us also write Part(a,b) for the set of all finite partitions whose endpoints are a and b.

Given any finite partition P ≝ {a[0] < a[1] < … < a[n]}, let us write EP for the set of intervals [a[0], a[1][, …, [a[n–1], a[n][, viewed as elements of the dcpo IR. We observe that, given two finite partitions P and Q in Part (a,b) (hence with the same endpoints a and b), if PQ, then EP# EQ. Part (a,b) is a directed family of finite partitions, since {a, b} is in Part (a,b) and Part (a,b) is closed under binary unions. Hence the collection of elements EP of Qfin(IR) when P varies over Part (a,b) is directed. The intersection of the sets ↑EP, P ∈ Part (a,b), is exactly the interval [a,b[. By Proposition 5.2.28, we obtain: (*) for every Scott-open subset V of IR such that [a,b[ ⊆ V, there is finite partition Q with endpoints a and b such that EQ is included in V.

This is pictured below. The partition Q is {b[0] < b[1] < … < b[10]}.

Let U be any open subset of R, and let U and V be two Scott-open subsets of IR such that UR = VR = U. We wish to show that λ(U)=λ(V). By symmetry, it is enough to show that λ(U)≤λ(V). The situation is exemplified below. Note that U need not be included in V.

In order to show that λ(U)≤λ(V), let r be any positive real number such that r < λ(U). There is a finite partition P such that r < λP(U). (You may imagine that this is the partition {a[0] < a[1] < … < a[4]} showed in the picture above.)

We will show that there is a finite partition Q such that r < λQ(V). We write P as {a[0] < a[1] < … < a[n]}, and we consider any index i with 1≤in such that [a[i–1], a[i][ is an element of U. (For instance, [a[1], a[2][ in the example above.) Every real number in [a[i–1], a[i][ is above the element [a[i–1], a[i][, hence is also in U. Hence the interval [a[i–1], a[i][ is included in UR, namely in U, and therefore [a[i–1], a[i][ is included in VR, hence in V. By (*), proved above, there is a finite partition Qi with endpoints a[i–1] and a[i], say Qi ≝ {b[i,0] < … < b[i,mi]} (with b[i,0]=a[i–1] and b[i,mi]=a[i]) such that the intervals [b[i,j–1], b[i,j][, 1≤jmi, are all in V. In the following picture, we show how [a[1], a[2][ is refined by Q2 ≝ {b[2,0] < … < b[2,10]}.

Let Q be the union of the finite partitions Qi, when i ranges over the indices i with 1≤in such that [a[i–1], a[i][ is an element of U. If we agree to write those indices i as i[0] < i[1] < … i[n], then the elements of Q are b[i[0],0] < … < b[i[0],mi[0]] ≤ b[i[1],0] < … < b[i[1],mi[1]] ≤ … ≤ b[i[n],0] < … < b[i[n],mi[n]], and EQ consists of the intervals [b[i,j–1], b[i,j][ with 1≤jmi, and where i can be equal to i[0], i[1], …, or i[n], plus possibly other intervals [b[i[0],mi[0]], b[i[1],0][, b[i[1],mi[0]] ≤ b[i[2],0], inbetween two consecutive partitions Qi. All the intervals of the former kind are in V, from which we conclude that λQ(V) is larger than or equal to the sum of their lengths b[i,j]–b[i,j–1]. For each i among i[0], i[1], …, i[n], the sum of the lengths b[i,j]–b[i,j–1], 1≤jmi, is equal to b[i,mi]–b[i,0] = a[i]–a[i–1], and summing those quantities over i=i[0], i[1], …, i[n] yields λP(U). Therefore λQ(V)≥λP(U)>r.

Defining the Lebesgue valuation λ on R

We can therefore define λ(U), for every open subset U of R, as λ(U) where U is any Scott-open subset of IR such that UR = U.

I claim that λ is a continuous valuation. For every open subset U of R, there is at least one Scott-open subset U of IR such that UR = U; let me call such a Scott-open set U a representative of U.

  • (strictness) the empty subset of IR is a representative of the empty subset of R, so λ(∅)=λ(∅)=0.
  • (modularity) Let U and V be any two open subsets of R, with respective representatives U and V. Then UV is a representative of UV and UV is a representative of UV, so λ(UV)+λ(UV)=λ(UV)+λ(UV)=λ(U)+λ(V)=λ(U)+λ(V), using the fact that λ is modular in order to justify the middle equation.
  • Scott-continuity. We first show that λ is monotonic. Let U and V be any two open subsets of R, with respective representatives U and V, and let us assume that U is included in V. It may not be the case that U is included in V, but that does not matter. What does matter is that UV is another representative of V: (UV) ∩ R = (UR) ∪ (VR) = UV = V. Therefore λ(V)=λ(UV)≥λ(U)=λ(U), where the middle inequality is by monotonicity of λ.
    Let us consider any directed family (Ui)iI of open subsets of R, and let their union be U. We consider that directed family as a monotone net, by defining ij as UiUj. For each i in I, let Ui be a representative of Ui. Generalizing what we have done in the previous paragraph, we see that the union Vi of all the sets Uj over the indices j in I such that ji is another representative of Ui: ViR is equal to the union of the sets UjR = Uj over the indices ji, which is just Ui. The point of this construction is that the map iIVi is monotonic, so that (Vi)iI is a directed family of open subsets of IR. Additionally, the union V ≝ ∪iI Vi is a representative of U = ∪iI Ui. It follows that λ(U)=λ(V)=supiI λ(Vi)=supiI λ(Ui), using the Scott-continuity of λ.

From the fact that λ is a continuous valuation, we obtain that the collection of open subsets U of R such that λ(U)>r (for any given r>0) is a Scott-open subset of O(R). For that, we did not need to know that λ is strict or modular.

However, in order to complete our proof of consonance, we also needed to know that, given any compact subset Q ≝ {x0, x1, …, xn, …} of R, and for every ε>0, the open set U ≝ ∪nN [xn, ε/2n[ had Lebesgue measure λ(U)≤∑nN ε/2n = 2ε. This requires one to know that λ is a continuous valuation, not just any Scott-continuous map from O(R) to R+ ∪ {+∞}. Indeed, every continuous valuation ν satisfies the following sum bound: ν(∪nN Un)≤∑nN ν(Un). This is proved as follows. By Scott-continuity, ν(∪nN Un)≤supnN ν(∪i=1n Ui). By induction on n, we show that ν(∪i=1n Ui)≤∑i=1n ν(Ui): the essential argument is the inequality ν(UV)≤ν(U)+ν(V), which follows from (modularity). We therefore obtain ν(∪nN Un)≤supnNi=1n ν(Ui)=∑nN ν(Un). And we are done!


Jean Goubault-Larrecq (May 20th, 2021)