Scott’s formula

There is a famous formula in domain theory, called Scott’s formula. If B is a basis of a continuous poset X, and f is a monotonic function from B to some dcpo Y, then one can define a new function f’ from the whole of X to Y by:

f’(x) ≝ supb ∈ B, b ≪ x f(b),

and then f’ will automatically be a continuous function from X to Y. There is no reason why it would coincide with f on B, but it is the best continuous approximation to f on B, in the sense that f’ is below f on B (f’(b)≤f(b) for every b ∈ B) and that it is the largest one with that property (for every continuous function g below f on B, g≤f’).

This is mentioned as Proposition 5.1.60 in the book, although you should definitely read the errata page (important blooper #13).

I would like to explore a few ways that this formula can be extended, in what cases f’ coincides with f on B (i.e., in what cases f’ is a continuous extension of f), and finally in what cases any algebraic laws satisfied by f will still be satisfied by f’.

c-spaces

We can first extend Scott’s formula to the case where X is a c-space, not just a continuous poset. A c-space is a space X in which, for every point x, for every open neighborhood U of x, there is a point y in U such that x is in the interior of ↑y.

It will be practical to write y ≪ x instead of “x is in the interior of ↑y“, and ↟y for the interior of ↑y (not just in c-spaces, but in any topological space). You might be concerned about a possible conflict of notations with the way-below relation on posets. But, in the case where X is a continuous poset, with its Scott topology, the usual way-below relation ≪ is exactly given by y ≪ x if and only if x is in the interior of ↑y, or equivalently, the interior of ↑y is exactly ↟y, defined as the set of elements z such that y is way-below z (this is Proposition 5.1.35 of the book); and I will never consider the way-below relation on posets that are not continuous in this post.

Now let me define a basis of a topological space X as a collection B of points of X such that for every point x in X, for every open neighborhood U of x, there is an element b of B ∩ U such that b ≪ x (i.e., such that x is in the interior of ↑b). Any space with a basis must be c-space, and conversely, and c-space X has a basis, namely X itself. When X is a continuous poset with its Scott topology, we retrieve the usual definition of a basis.

Scott’s formula has an easy generalization to c-spaces and their bases, as we now see. Note that X need not be sober, in particular; if X were sober, then it would be a continuous dcpo with its Scott topology (Proposition 8.3.36 of the book). There is no harm in replacing Y, with its Scott topology, with a monotone convergence space either. All spaces are (pre)ordered with their specialization preorderings; this is what allows us to make sense of the adjective “monotonic” below.

Theorem. Let B be a basis of a c-space X, and f be a monotonic function from B to some monotone convergence space Y. For every x in X, let:

f’(x) ≝ supb ∈ B, b ≪ x f(b)
(Scott’s formula).

Then f’ is a continuous map from X to Y, and is the largest continuous map on X that is below f on B.

Proof. For any given point x, let Fx denote the family of elements b ∈ B such that b ≪ x. We claim that Fx is directed. Once we have proved this, the supremum that defines f’(x) will be a supremum of a directed family, hence will be well-defined, since Y is a monotone convergence space, in particular a dcpo in its specialization ordering. Now every open neighborhood U of x contains an element of Fx, since B is a basis of X. By taking U ≝ X, Fx is non-empty, and for any two elements b1 and b2 of Fx, there is an element b of Fx in int(↑ b1) ∩ int(↑b2); in particular, b1, b2 ≪ b. Hence Fx is directed, as promised.

We claim that f’ is continuous. For every open subset V of Y, for every x ∈ X, f’(x) ∈ V if and only if supb ∈ B, b ≪ x f(b) is in V. Since Y is a monotone convergence space, V is Scott-open, so supb ∈ B, b ≪ x f(b) ∈ V implies that f(b) ∈ V for some b ∈ B such that b ≪ x. Conversely, if f(b) ∈ V for some b ∈ B such that b ≪ x, then f’(x), which is even larger, is in V. Hence f’−1(V) is the union of the sets ↟b, where b ranges over the elements of B such that f(b) ∈ V. In other words, f’−1(V) = ∪b ∈ f−1 (V) int(↑b). Since that is open, f’ is continuous.

For every x ∈ B, we have that every b ∈ B such that b ≪ x (i.e., x ∈ int(↑b)) is such that x is in ↑b, namely b ≤ x; so f(b) ≤ f(x) since f is monotonic. Taking suprema, we obtain that f’(x)≤f(x). Hence f’ is below f on B.

Finally, we consider any continuous map g from X to Y that is below f on B. For every point x of X, for every open neighborhood V of g(x), g−1(V) is an open neighborhood of x, hence contains some element b of B such that b ≪ x, since B is a basis of X. Since g(b) ≤ f(b) and g(b) ∈ V, f(b) is also in V, and therefore f’(x) ≥ f(b) is in V. We have shown that every open neighborhood V of g(x) contains f’(x), so g(x) ≤ f’(x), showing that f’ is the largest continuous map below f on B. ☐

Continuous extensions

Next, we examine when f’ is an extension of f, namely when f’ coincides with f on B. This is dealt with in Corollary 5.1.61 of the book; as with Proposition 5.1.60, you need Y to be a dcpo, not just a bdcpo (see the errata page, important blooper #13; you may also keep Y as a bdcpo, and then you need to require B to be cofinal in X).

The view I have just given of Scott’s formula on c-spaces gives us a complete characterization of when f’ is an extension of f.

Proposition. Let B be a basis of a c-space X, f be a monotonic map from B to a monotone convergence space Y, and let f’ be defined by Scott’s formula. Then f’ extends f if and only if f is continuous from B (with the subspace topology from X) to Y.

Proof. We have seen that f’ is continuous from X to Y. Therefore its restriction to the subspace B is continuous. In particular, if f’ coincides with f on B, f must be continuous from B to Y.

Conversely, let us assume that f is continuous from B to Y. For every open subset V of Y, f−1(V) is open, hence equal to U ∩ B for some open subset U of X. For every x ∈ f−1(V)= U ∩ B, by definition of B there is an element b ∈ B such that b ≪ x and b ∈ U. Hence b is in f −1(V). Equivalently, f(b) ∈ V, and this shows that f’(x) ≥ f(b) is in V. We have shown that for every x ∈ X, every open neighborhood V of f(x) also contains f’(x), so f(x) ≤ f’(x). Since f’ is below f on B, we conclude that f and f’ coincide on B. ☐

When X is a continuous poset with basis B, and Y is a dcpo, the condition that f be continuous from B, with the subspace topology from X, to Y, is equivalent to the fact that f be relatively Scott-continuous on B, which I will define next. There is a pretty subtle point here, as any Scott-continuous map from B to Y will be relatively Scott-continuous on B, but the converse may fail.

Let me call X–supremum of a family D of points of X the least upper bound of D in X, if it exists. Let me call B–supremum of a family D of points of B the least upper bound of D in B, if it exists. Now, given any family D of points of B,

  • If D has an X-supremum x that happens to be in B, then x must be the B-supremum of D as well: x is an upper bound of D, and for every upper bound b of D in B, b is larger than any point of D, so b≤x since x is an X-supremum of D.
  • However, and this is the catch, if D has a B-supremum b, then b will be an upper bound of D in X as well, but not necessarily the least one; in other words, b may well not be the X-supremum of D.
    For an illustration of this, consider the collection of all subsets of a given topological space Y for X, its subcollection of closed subsets for B, both ordered by inclusion. The B-supremum of a collection D≝(Ci)i ∈ I of elements of B (closed sets) is cl(∪i ∈ I Ci), and that is in general different from the X-supremum ∪i ∈ ICi.

Hence, and returning to the case where B is a basis of a continuous poset X, and Y is a dcpo, let me say that a function f from B to Y is relatively Scott-continuous on B if and only if f is monotonic, and for every directed family (bi)i ∈ I in B with an X-supremum (not a B-supremum!) b that happens to be in B, f(b) = supi ∈ I f(bi).

Lemma. Given a basis B of a continuous poset X, and a function f from B to a dcpo Y, f is continuous from B, with the subspace topology induced by the Scott topology on X, if and only if f is relatively Scott-continuous on B.

Proof. Let us first assume that f is continuous from B, with the subspace topology induced by the Scott topology on X. Then f is monotonic, and given any directed family (bi)i ∈ I in B with an X-supremum b that happens to be in B, we claim that f(b) = supi ∈ I f(bi). The inequality f(b) ≥ supi ∈ I f(bi) is a consequence of monotonicity. In order to show the reverse inequality, we consider any (Scott-)open neighborhood V of f(b), and we will show that f(bi) is in V for some i ∈ I. Since f is continuous, f–1(V) is open in the subspace topology induced on B by the Scott topology on X, hence is of the form U ∩ B for some Scott-open subset U of X. Now f–1(V) = U ∩ B contains b, which is the X-supremum of (bi)i ∈ I. Since U is Scott-open (in X), some bi is in U. It is also in B, by definition. Hence bi is in U ∩ B = f–1(V), so that f(bi) is in V, as promised.

That direction of the proof did not use the fact that B is a basis of X, and works perfectly well for any subset B of an arbitrary, not necessarily continuous poset X.

In the converse direction, we will definitely use the fact that B is a basis of X, and therefore that X is a continuous poset. Let us assume that f is relatively Scott-continuous on B. We consider any (Scott-)open subset V of Y, and we claim that f–1(V) can be written as U ∩ B for some Scott-open subset U of X. We define U as the union of all the sets ↟b, where b ranges over f–1(V); to make it clear, ↟b is the collection of points x in X (not B) such that b ≪ x. Then U is (Scott-)open in X. For every b’ ∈ f–1(V), since X is a continuous poset with basis B, b’ is the X-supremum of the directed family of all elements b ≪ b’ with b ∈ B. Since f is relatively Scott-continuous on B, f(b’), which is in V, is the directed supremum of all elements f(b) with b ≪ b’, b ∈ B. Therefore one such element f(b) is in V, since V is Scott-continuous. This shows that b’ ∈ ↟b, where b ∈ f–1(V), and hence that b’ ∈ U ∩ B. Conversely, for every b’ ∈ U ∩ B, there is a b ∈ f–1(V) such that b ≪ b’. Since f is monotonic and V is upwards-closed, b’ ∈ f–1(V). ☐

It follows that:

Fact. A monotonic function f from a basis B of a continuous poset X to a dcpo Y has a Scott-continuous extension f’ from the whole of X to Y if and only if f is relatively Scott-continuous on B.

In Corollary 5.1.61 of the book, the following is stated: if f is Scott-continuous from B to Y, then it has a (unique) continuous extension f’. We have seen that being Scott-continuous from B to Y (=monotonic+mapping existing directed B-suprema to suprema) is different from being relatively Scott-continuous on B. Corollary 5.1.61 of the book is a consequence of the fact we have just stated. Since X-suprema of elements of B that happen to be in B are also B-suprema (not the other way around!), any Scott-continuous map from B to Y is also relatively Scott-continuous on B. It just so happens that Corollary 5.1.61 of the book is not the most general result that one could state.

Perhaps more annoyingly, I have realized recently that Corollary 5.1.61 of the book is almost never what you need in applications. Most of the time, it is much more natural to show that f is relatively Scott-continuous. The reason is that, in usual applications, we have a firm grasp over what directed X-suprema look like in X, but it would require quite some work to characterize B-suprema—and, as we have seen, that is in fact useless.

A sketch of an application

A continuous valuation on a topological space X is a Scott-continuous map ν from the lattice of open sets OX of X to R+ ∪ {∞} that satisfies the following two conditions:

  • strictness: ν(∅)=0;
  • modularity: for all U, V in OX, ν(U ∪ V)+ν(U ∩ V)=ν(U)+ν(V).

I will concentrate on bounded continuous valuations ν, namely those such that ν(X) < ∞.

Continuous valuations are very close to measures: we can integrate with respect to a continuous valuation, and in fact continuous valuations extend to measures on the Borel σ-algebra of X when X is LCS-complete, and measures restrict to continuous valuations on the open sets when X is hereditarily Lindelöf, in particular when X is second-countable. For a space X that is both (typically any quasi-Polish space), this even defines a one-to-one correspondence between bounded continuous valuations and bounded measures. I will not expand on this here.

In [1], Klaus Keimel and I gave pretty simple generalizations of what has been known as Choquet-Kendall-Mathéron theorems. Let me give one example of this.

We consider the Smyth hyperspace QZ of a topological space Z. That is the set of non-empty compact saturated subsets of Z, and we give it the upper Vietoris topology, whose basic open subsets are ☐U ≝ {Q ∈ QZ | Q ⊆ U}, where U ranges over the open subsets of Z.

Given a continuous valuation ν on QZ, we may form a function μ : OZ → R+ ∪ {∞} by letting μ(U) ≝ ν(☐U) for every open subset U of X. Such a function μ has the following properties, which make μ what I call a continuous credibility, and which has been known under the name of a (continuous) totally monotone capacity, or a (continuous) totally convex capacity. Namely, μ is Scott-continuous, strict, and satisfies the following condition of total convexity:

μ(U) ≥ ∑I (-1)|I|+1 μ(∩i∈I Ui)

for all open sets U, U1, …, Un such that U contains U1 ∪ … ∪ Un, and where the summation extends over all non-empty subsets I of {1, …, n}. This is an inequational form of the so-called inclusion-exclusion formula in probability theory, and I have already mentioned this in relation to the baby Groemer theorem.

The point of the (first of three) Choquet-Kendall-Mathéron theorems in [1] is to give a converse to this construction. This way, continuous credibilities are relatively practical representations for continuous valuations on the hyperspace QZ of non-empty compact saturated subsets of Z.

For this, we use that Z is a locally compact space (although a core-compact space would suffice), and we start from a function μ : OZ → R+ (not R+ ∪ {∞}, so μ is, in fact, bounded) that is Scott-continuous, strict, and totally convex in the sense described above. With this as input, we show that there is a (unique, bounded) continuous valuation ν : O(QZ) → R+ such that μ(U) ≝ ν(☐U) for every open subset U of X. Here is how, omitting most of the details, and in the goal of showing where Scott’s formula is used.

Using the baby Groemer theorem, we can show that there is a function ν : EO(QZ) → R+ such that μ(U) ≝ ν(☐U) for every open subset U of Z. Here EO(QZ) denotes the lattice of elementary open subsets of QZ, which we define as the finite unions of basic open sets ☐U. This does not use the fact that Z is locally compact yet. Also, ν is monotonic, strict and modular. The next move is to extend ν to the whole of O(QZ), and for this we use Scott’s formula on the basis B ≝ EO(QZ) of O(QZ); that is indeed a basis because Z is locally compact, and therefore also QZ.

Great: omitting all the details, all that is easy. However, there is still something missing. We would like to show that the extension ν’ of ν that we obtained through Scott’s formula is a continuous valuation. It is continuous, but we also need to verify that it is strict and modular, right? In [1], we proved it by hand, but surely there is a general theorem that would allow us to prove it directly.

This is exactly what I will state. We examine under what conditions any algebraic laws satisfied by ν (or, in general, a continuous map f from B to Y) are still satisfied by its unique continuous extension ν’ (or more generally, f’ : X → Y). In the sequel, I will even look at cases where we can ensure that algebraic laws satisfied by f are also satisfied by f’, even when f’ is not an extension of f, just the largest continuous map below f on B.

Preserving algebraic laws

Given a c-space X with basis B and a monotone convergence space Y, we consider formal inequalities of the form:

g (_(g1 (z)), …, _(gm (z))) ≤ h (_(h1 (z)), …, _(hn (z)))

where g and h are maps from Ym, resp. Yn, to Y, and g1, …, gm, h1, …, hn are maps from Xk to X, for some k ∈ N. The notation z stands for a tuple (z1, …, zk) of k distinct formal variables.

Such a formal inequality is satisfied on X (resp., on B) by a function f : X → Y if and only if

g (f (g1 (u)), …, f(gm (u))) ≤ h (f (h1 (u)), …, f(hn (u)))

for every tuple u = (u1, …, uk) of values in X (resp., in B). For example, a continuous valuation ν on a space X is the same thing as a Scott-continuous map ν: OX → R+ ∪ {∞} that satisfies the inequalities:

  • _(g1()) ≤ h(), where g1 is the 0-ary map with value ∅ and h is the 0-ary map with value 0 (in short, the inequality _(∅) ≤ 0, expressing strictness);
  • _(z1 ∪ z2)+_(z1 ∩ z2) ≤ _(z1)+_(z2);
  • _(z1)+_(z2) ≤ _(z1 ∪ z2)+_(z1 ∩ z2).

Note in particular how we expressed the equality _(z1 ∪ z2)+_(z1 ∩ z2) = _(z1)+_(z2) (modularity) as two inequalities in opposite directions.

Now the answer that I will give below requires the functions g and g1, …, gm to be continuous, but the functions h1, …, hn will need to obey a different assumption. Let me say that a map h : X → Y is quasi-open if and only if for every open subset U of X, ↑h[U] is open in Y. This is the same idea as a notion that Klaus Keimel once used for the addition operation of a semitopological cone, and which he called almost open.

Proposition. Let X be a c-space with basis B, Y be a monotone convergence space, and L be a formal inequality g (_(g1 (z)), …, _(gm (z))) ≤ h (_(h1 (z)), …, _(hn (z))) such that:

  1. g1, …, gm, h1, …, hn applied to elements of B yield values in B;
  2. g is continuous on Y and g1, …, gm are continuous on Xk;
  3. h is monotonic on Y and h1, …, hn are monotonic on Xk.

For every monotonic map f : B → Y, if f satisfies L on B, and if:

  1. either f is continuous on B, with the subspace topology (in which case f’ extends f),
  2. or h1, …, hn are quasi-open on Xk (in which case f’ does not necessarily extend f),

then f’ satisfies L on X.

Proof. For every tuple u ≝ (u1, …, uk) of values in X, we verify that:

g (f’ (g1 (u)), …, f’ (gm (u))) ≤ h (f’ (h1 (u)), …, f’ (hn (u)))

by considering any open neighborhood V of the left-hand side, and showing that it contains the right-hand side. By assumption, (f’ (g1 (u)), …, f’ (gm (u))) is in g−1(V). Since g is continuous, the latter is open, so we can find open neighborhoods U1, …, Um of f’(g1 (u)), …, f’(gm (u)) respectively such that Πi=1m Ui ⊆ g−1(V).

Since Y is a monotone convergence space, each Ui is Scott-open. By definition of f’, it follows that for each i there is an element bi ∈ B such that bi ≪ gi (u) and f(bi) ∈ Ui. In particular, f (b1, …, bm) ∈ V.

Since every gi is continuous, ∩i=1m gi–1(↟bi) is an open neighborhood of u. Since Xk is a c-space, with basis Bk (exercise!), there is a tuple c ≝ (c1, …, ck) of values of B such that c ∈ ∩i=1m gi–1(↟bi) and c ≪ u (namely, such that c1 ≪ u1, …, ck ≪ uk; remember that this means that each ui is in the interior of ↑ci). Then bi ≤ gi (c) for every i, and since f is monotonic, f (b1, …, bm) ≤ f(g1 (c), …, gm (c)). This entails that f(g1 (c), …, gm (c)) ∈ g−1(V), hence that g (f (g1 (c), …, gm (c))) ∈ V.

All the elements g1 (c), …, gm (c), h1 (c), …, hn (c) are in B, by our assumption 1. Since f satisfies L on B,

g (f (g1 (c)), …, f (gm (c))) ≤ h (f (h1 (c)), …, f (hn (c))),

so h (f (h1 (c)), …, f (hn (c))) is in V.

We now consider two cases, depending on whether assumption 4 or 5 is satisfied.

First case: assumption 4 is satisfied. In other words, we assume that f is continuous on B. Then f’ extends f, as we have seen earlier. Then h (f (h1 (c)), …, f (hn (c))) = h (f’ (h1 (c)), …, f’ (hn (c))). Since each ui is in the interior of ↑ci, in particular ci ≤ ui for each i, so, using the fact that f’, h1, …, hn are monotonic, h (f (h1 (c)), …, f (hn (c))) ≤ h (f’ (h1 (u)), …, f’ (hn (u))), and therefore h (f’ (h1 (u)), …, f’ (hn (u))) is in V.

Second case: assumption 5 is satisfied. Then f’ may fail to extend f, but assumption 5 tells us that the maps h1, …, hn are quasi-open. Since each ui is in the interior ↟ci of ↑ci, hi(u) is in the open set ↑hi[↟c1 × … × ↟ck]. The latter is equal to its own interior, and is therefore included in the interior of the larger set ↑hi[↑c1 × … × ↑ck] = ↑hi(c). Hence hi(u) ∈ int(↑hi (c)), in other words hi(c) ≪ hi(u). By definition of f’, f (h1 (c)), …, f (hn (c)) ≤ f’ (h1 (u)), …, f’ (hn (u)). We use the fact that h is monotonic, and we obtain that h (f (h1 (c)), …, f (hn (c))) ≤ h (f’ (h1 (u)), …, f’ (hn (u))), so that h (f’ (h1 (u)), …, f’ (hn (u))) is in V. ☐

Let us see what this yields on our example of inequalities characterizing valuations (strictness, plus two inequalities for modularity):

  • _(∅) ≤ 0;
  • _(z1 ∪ z2)+_(z1 ∩ z2) ≤ _(z1)+_(z2);
  • _(z1)+_(z2) ≤ _(z1 ∪ z2)+_(z1 ∩ z2).

The first inequality is really _(g1()) ≤ h(), where g1 is the 0-ary map with value ∅ and h is the 0-ary map. Considering that the basis B we took in this example was EO(QZ), the set of finite unions of basic open subsets of QZ, condition 1 (mapping Bk to B) is obvious. Condition 2 is that g1 is continuous (and also g, which is the identity map in that case). Condition 3 is that h is monotonic, which is clear as well.

Condition 1 is also clear for the second and third inequalities: namely, EO(QZ) is closed under binary unions, and also under binary intersections (because ☐U ∩ ☐V = ☐(U ∩ V)). As far as conditions 2 and 3 are concerned, both g and h are the + function on R+ ∪ {∞}, and that is continuous (for g; in particular it is monotonic, which is what we want for h). By continuous, I mean jointly continuous, by the way, but separate continuity is enough, by the Banaschewski-Lawson-Ershov observation. The maps gi and hi are: binary union ∪, binary intersection ∩, and first and second projections (this is how you get z1, resp., z2, from z = (z1, z2)). And they are all Scott-continuous, hence separately continuous, hence jointly continuous by the Banaschewski-Lawson-Ershov observation, since this example took place in the setting of a locally compact space, hence one such that OZ is a continuous dcpo.

The setting Klaus Keimel and I were in was where f (the valuation ν on B=EO(QZ) such that μ(U) ≝ ν(☐U) for every open subset U of Z) was already continuous on B, so condition 4 applies, and the extension f’ is now automatically a continuous valuation on the space Z, finishing the proof.

However, let us imagine that we wanted to use condition 5 instead. We would need to examine when binary union, binary intersections, and first and second projections are quasi-open. Let me state how we can simplify the proof of such statements. We will see that satisfying condition 5 would require more from the underlying space Z.

Quasi-open maps and the way-below relation

Condition number 5 (the one that applies in the previous proposition in the case that f is not continuous on B, and therefore does not extend to f’) is about quasi-openness, which may appear as a mysterious property. It is equivalent to the following one, however.

Proposition. For every c-space X, for every topological space Y, for every natural number k, a map h : Xk → X is quasi-open if and only if it preserves ≪, namely if and only if for all x1, …, xk, x’1, …, x’k in X such that x1 ≪ x’1, …, xk ≪ x’k, we have h (x1, …, xk) ≪ h (x’1, …, x’k).

Proof. Let us assume h quasi-open. Then ↑h[↟x1 × … × ↟xk] is open. If x1 ≪ x’1, …, xk ≪ x’k, then h (x’1, …, x’k) is in ↑h[↟x1 × … × ↟xk]. Also, ↑h [↟x1 × … × ↟xk] is included in ↑h [↑x1 × … × ↑xk] = ↑h (x1, …, xk), so h (x’1, …, x’k) is in an open set that is included in ↑h (x1, …, xk). It follows that h (x’1, …, x’k) is in the interior of ↑h (x1, …, xk), which means that h (x1, …, xk) ≪ h (x’1, …, x’k), by our definition of ≪ on topological spaces.

Conversely, let us assume that h preserves ≪. Let U be any open subset of Xk. In order to show that ↑h[U] is open, we show that every element y of ↑h[U] belongs to some open subset V of Y that is included in ↑h[U]. Since y ∈ ↑h[U], there is a tuple (x’1, …, x’k) in U such that h (x’1, …, x’k) ≤ y. Notably, there is an open rectangle U1 × … × Uk included in U and containing (x’1, …, x’k). Since X is a c-space, we can find elements x1 ≪ x’1, …, xk ≪ x’k in U1, …, Uk respectively. In particular, (x1, …, xk) is in U. Since h preserves ≪, h (x1, …, xk) ≪ h (x’1, …, x’k), namely h (x’1, …, x’k) is in the interior V of ↑h (x1, …, xk). Since h (x’1, …, x’k) ≤ y, y is also in V. Finally, V is included in ↑h (x1, …, xk), which is included in ↑h[U] since (x1, …, xk) is in U. ☐

Returning to our example of the inequalities defining modularity, when are binary union, binary intersections, first and second projections quasi-open? We are working on X = OZ where Z is locally compact, hence X is a continuous dcpo and ≪ is the usual notion of way-below relation. It is easy to see that union, first and second projections do preserve ≪. But binary intersection preserves ≪ if and only if Z is core-coherent: this is Proposition 5.2.19 in the book.

In other words, relying on condition 5 dispenses us to show that f is continuous on B (condition 4—that can indeed be a chore!), but requires more from the underlying space X.

Other extension results

Scott’s formula is only one of many possible extension results. I should mention that there is another close one, which requires a lot less on X and a lot more on Y. Given any dense subset B of an arbitrary topological space X, a bc-domain Y, and a monotonic map f from B to Y, there is a largest continuous map f’ from X to Y below f; and if f is continuous, then f’ is a continuous extension of f from B to the whole of X. That is part of Scott’s theorem that the bc-domains are the injective topological spaces (over topological embeddings).

Explicitly, we first extend f to a monotonic map from X to Y by letting f(x) be equal to the infimum of all f(b), where b ranges over the elements of B above x, for every x in X. In a second step, we define f’(x) as sup {y ∈ Y | x ∈ int(f−1(↟y))}. I will not prove this here, maybe another time. (This may be wrong, too. I have just written it mostly from the top of my head, without much verification.)

There is a variant of this where Y is required to be a continuous complete lattice, and we obtain a similar result, where this time B is an arbitrary (not a dense) subset of X.

And there are many other extension results where X is obtained as the sobrification of a space B, or as its well-filterification, or etc. Oh, well.

  1. Jean Goubault-Larrecq and Klaus Keimel.  Choquet-Kendall-Matheron theorems for non-Hausdorff spaces.  Mathematical Structures in Computer Science 21(3), 2011, pages 511-561.

— Jean Goubault-Larrecq (September 20th, 2023)