# 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) ≝ supbB, bx 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 bB) and that it is the largest one with that property (for every continuous function g below f on B, gf’).

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 yx 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 yx 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 BU such that bx (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) ≝ supbB, bx 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 bB such that bx. 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 UX, 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, b2b. Hence Fx is directed, as promised.

We claim that f’ is continuous. For every open subset V of Y, for every xX, f’(x) ∈ V if and only if supbB, bx f(b) is in V. Since Y is a monotone convergence space, V is Scott-open, so supbB, bx f(b) ∈ V implies that f(b) ∈ V for some bB such that bx. Conversely, if f(b) ∈ V for some bB such that bx, 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) = ∪bf−1 (V) int(↑b). Since that is open, f’ is continuous.

For every xB, we have that every bB such that bx (i.e., x ∈ int(↑b)) is such that x is in ↑b, namely bx; 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 bx, 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 UB for some open subset U of X. For every xf−1(V)= UB, by definition of B there is an element bB such that bx and bU. 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 xX, 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 Xsupremum of a family D of points of X the least upper bound of D in X, if it exists. Let me call Bsupremum 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 bx 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)iI of elements of B (closed sets) is cl(∪iI Ci), and that is in general different from the X-supremum ∪iICi.

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)iI in B with an X-supremum (not a B-supremum!) b that happens to be in B, f(b) = supiI 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)iI in B with an X-supremum b that happens to be in B, we claim that f(b) = supiI f(bi). The inequality f(b) ≥ supiI 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 iI. 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 UB for some Scott-open subset U of X. Now f–1(V) = UB contains b, which is the X-supremum of (bi)iI. Since U is Scott-open (in X), some bi is in U. It is also in B, by definition. Hence bi is in UB = 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 UB 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 bx. 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 bb’ with bB. Since f is relatively Scott-continuous on B, f(b’), which is in V, is the directed supremum of all elements f(b) with bb’, bB. Therefore one such element f(b) is in V, since V is Scott-continuous. This shows that b’ ∈ ↟b, where bf–1(V), and hence that b’UB. Conversely, for every b’UB, there is a bf–1(V) such that bb’. 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 UV 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 ≝ {QQZ | QU}, where U ranges over the open subsets of Z.

Given a continuous valuation ν on QZ, we may form a function μ : OZR+ ∪ {∞} 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(∩iI Ui)

for all open sets UU1, …, 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 : OZR+ (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 BEO(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’ : XY). 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 kN. 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 : XY 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: OXR+ ∪ {∞} 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);
• _(z1z2)+_(z1z2) ≤ _(z1)+_(z2);
• _(z1)+_(z2) ≤ _(z1z2)+_(z1z2).

Note in particular how we expressed the equality _(z1z2)+_(z1z2) = _(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 : XY 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 : BY, 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 Uig−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 biB such that bigi (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 cu (namely, such that c1u1, …, ckuk; remember that this means that each ui is in the interior of ↑ci). Then bigi (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 ciui 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;
• _(z1z2)+_(z1z2) ≤ _(z1)+_(z2);
• _(z1)+_(z2) ≤ _(z1z2)+_(z1z2).

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 = ☐(UV)). 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 x1x’1, …, xkx’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 x1x’1, …, xkx’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 x1x’1, …, xkx’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 {yY | 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)