**Nota** (added April 2, 2017): my interest in the following questions has faded. As I am saying at the end of this post, the questions I am asking here were a blocking point in trying to solve a problem on quasi-metrics for spaces of previsions. I have now found a way to solve the latter problem [3], and this does not require a complete solution to the questions of this page. Solving the questions on this page may offer a more elegant route, however.

This is a tough question. I’ll have to do a bit of an introduction first. I will then give you a toned-down version of my question. If you manage to solve this one, maybe this will give us a lead into solving the real question. You may also want to look at the real question directly (in the third part). I will briefly say what the application of all that is in the final part of this post.

## Preliminaries

Let *X*, *d* be a metric space. Consider the set *L*(*X*) of all lower semi-continuous maps from *X* to the extended non-negative reals [0, ∞]. This is also the set of all continuous maps from *X* to [0, ∞], where the latter comes with the Scott topology of ≤. This can be given the Scott topology of the pointwise ordering, or the compact-open topology (understanding [0, ∞] with its Scott topology). We have the following:

**Thm 1.** [3, special case of Proposition 7.7] If X, d is a complete metric space, then the Scott topology and the compact-open topology coincide on *L*(*X*).

We can now consider the subsets *L _{α}*(

*X*) of those functions from

*X*to [0, ∞] that are α-Lipschitz Yoneda-continuous, for every α>0. If you wonder about “Yoneda-continuous”, let me say that, because we are considering also the element ∞ in [0, ∞], the topology on [0, ∞], whether Scott or otherwise, is not the open ball topology of any metric or quasi-metric. Hence not all Lipschitz maps are continuous. I know that makes matters more complex. For a definition, see [1, Section 7.4.3]; you may wish to read Section 6 of [2], too.

Finally, let me consider *L _{∞}*(

*X*), the space of all Lipschitz Yoneda-continuous maps. This is the union of all the subspaces

*L*(

_{α}*X*). Give all those subspaces the subspace topology from

*L*(

*X*). As a consequence of Theorem 1, we have:

**Prop.** Let *X*, *d* be a complete metric space. The spaces *L _{∞}*(

*X*) and

*L*(

_{α}*X*) (α>0) all have the compact-open topology.

That can be refined to:

**Thm 2.** [3, special case of Proposition 8.1] Let *X*, *d* be a complete metric space. The spaces *L _{α}*(

*X*) (α>0) all have the topology of pointwise convergence.

The topology of pointwise convergence is the topology induced by the inclusion of *L _{α}*(

*X*) into the product [0, ∞]

*, namely the space of all (not necessarily continuous) maps from*

^{X}*X*to [0, ∞]. For a space of functions to [0, ∞], the topology of pointwise convergence is always coarser than the compact-open topology, which is always coarser than the Scott topology. Here the first two coincide, and coincide with the subspace topology from

*L*(

*X*) with its own Scott topology.

By using the fact that stable compactness under the formation of arbitrary products, under the extraction of closed subspaces, and under retracts, Theorem 2 also implies the following very nice result.

**Thm 3.** [3, special case of Lemma 8.4 (4)] Let *X*, *d* be a complete metric space. Then the spaces *L _{α}*(

*X*) (α>0) are stably compact.

## The toned-down version

Now here comes the delicate part. *L _{α}*(

*X*) is a subspace of

*L*(

_{∞}*X*) for every α>0. I have certain functions

*F*:

*L*(

_{∞}*X*) → [0, ∞] that I would like to show continuous (where [0, ∞] still has the Scott topology), and I can show that their restrictions to

*L*(

_{α}*X*) are continuous for every α>0.

That is *not* enough to show that *F* itself is continuous! The problem is that the topology of *L _{∞}*(

*X*) may fail to be

*determined by*the topologies of

*L*(

_{α}*X*). By definition, it is determined if and only if any subset

*U*of

*L*(

_{∞}*X*) whose intersection with

*L*(

_{α}*X*) is open in

*L*(

_{α}*X*) for each α>0 is open in

*L*(

_{∞}*X*). In categorical terms: it is determined if and only if

*L*(

_{∞}*X*) is a colimit of the diagram formed by the objects

*L*(

_{α}*X*), with the obvious inclusions. Whence the question:

Let

X,dbe a complete metric space. Is the topology ofL(_{∞}X) determined by the topologies ofL(_{α}X), α>0?

If it is not, then under which reasonable conditions is that topology determined by the topologies of *L _{α}*(

*X*), α>0? Those conditions should be as general as possible.

I already know that the topology is indeed determined if *X*, *d* is what I am currently calling *Lipschitz regular* [3, Section 4]. I will give the actual definition later. In the current context, it is perhaps simpler to say that a complete metric space is Lipschitz regular if and only if, writing B(*x*, <*r*) for the open ball of center *x* and radius *r*, the following property holds: for all positive reals *r* and *s* such that *r<s*, B(*x*, <*r*) is relatively compact in B(*x*, <*s*) (i.e., way-below in the lattice of open sets; i.e., every open cover of B(*x*, <*s*) contains a finite subcover of B(*x*, <*r*)).

In case *X*, *d* is Lipschitz regular, *L _{α}*(

*X*) is in fact a retract, by an embedding-projection pair, of

*L*(

_{∞}*X*), where the embedding is just inclusion. That implies that the topology of

*L*(

_{∞}*X*) is determined by the topologies of

*L*(

_{α}*X*), α>0 [3, Proposition 9.2], but is a stronger property.

Lipschitz regularity implies local compactness, and that excludes Baire space, for example. Hence I am not entirely pleased with that assumption. You may also want to show that any complete metric space *X*, *d* such that the topology of *L _{∞}*(

*X*) is determined by the topologies of

*L*(

_{α}*X*), α>0, is in fact Lipschitz regular. That would give me an excuse for the notion.

## The real question

A quasi-metric space is a space *X* with a *quasi-metric* *d*, namely a kind of metric except that *d*(*x*,*y*) is not required to be equal to *d*(*y*,*x*). There is a lot of information about those spaces in [1], chapters 6 and 7. You should also probably read [2] if you are interested.

Theorem 1 is not what I really proved. What I proved is the following more general result:

**Thm 1′.** [3, Proposition 7.7] If X, d is a continuous Yoneda-complete quasi-metric space, then the Scott topology and the compact-open topology coincide on *L*(*X*).

This includes Theorem 1 as a special case, since complete metric spaces are Yoneda-complete, and all metric spaces are continuous.

**Thm 2′ and 3′.** [3, Proposition 8.1, Lemma 8.4] Let *X*, *d* be a complete metric space. Then the spaces *L _{α}*(

*X*) (α>0) all have the topology of pointwise convergence, and are stably compact.

Accordingly, the real question is:

Let

X,dbe a continuous Yoneda-complete metric space. Is the topology ofL(_{∞}X) determined by the topologies ofL(_{α}X), α>0?

If it is not, then under which reasonable conditions is that topology determined by the topologies of *L _{α}*(

*X*), α>0? Those conditions should be as general as possible. Strengthening the assumption from “continuous Yoneda-complete” to “algebraic Yoneda-complete” is perfectly reasonable, and Section 7 of [2] gives a indication how to reduce the continuous case to the algebraic case.

The actual definition of Lipschitz regular is the following. Let me consider that *X* embeds into its space of formal balls **B**(*X*,*d*), through the map *x* ↦ (*x*, 0). In other words, let me equate *x* with (*x*, 0). There is function that maps every *d*-Scott open subset of *U* to the largest Scott-open subset *V* of **B**(*X*,*d*) such that *V* ∩ *X* = *U*. The space *X*, *d* is *Lipschitz regular* if and only if that map is Scott-continuous.

I already know that if *X*, *d* is Lipschitz regular in that sense, then the topology of *L _{∞}*(

*X*) determined by the topologies of

*L*(

_{α}*X*), α>0; in fact

*L*(

_{∞}*X*) is obtained as a limit of a diagram of embedding-projection pairs, which is a stronger property.

## The purpose of all that

There is a famous theorem in measure theory that says that the space of all probability measures on a Polish space is again Polish, due to Prohorov. Concretely, that involves assuming a separable complete metric space *X*, *d*, and exhibiting a metric on the space of probabilities on *X* that makes it separable and complete.

Such metrics include the Lévy-Prohorov metric and the Kantorovitch-Rubinshtein (also called Hutchinson) metric. Completeness is a pretty tough theorem, and is usually proved by arguments on notions of tightness of measures, and of sets of measures.

What I am trying to do is to generalize that to certain classes of *quasi*-metric spaces. It turns out that the space of continuous valuations on *X* (roughly the same as measures) is isomorphic to a certain space of continuous functionals *F : L _{∞}(X)* → [0, ∞]. One gets the functional from the measure by integrating the function given as argument against the measure.

In an attempt to show that that space of functionals is (Yoneda-)complete, I used to be blocked by the fact that I can show that a certain functional *F* is continuous from *L _{α}*(

*X*) to [0, ∞] for every α>0, but that I need it to be continuous on the whole of

*L*(

_{∞}*X*). I have managed to get around this problem by showing that the space of formal balls of

*X*,

*d*, is always Lipschitz regular, and that every continuous Yoneda-complete space embeds as a

*G*δ subset of its space of formal balls, then using measure extension theorems on continuous dcpos… what a mess. As I said, if the topology of

*L*(

_{∞}*X*) is determined by the topologies of

*L*(

_{α}*X*), α>0, there would be a much simpler proof available.

Any ideas?

- Jean Goubault-Larrecq. Non-Hausdorff Topology and Domain Theory — Selected Topics in Point-Set Topology. New Mathematical Monographs 22. Cambridge University Press, 2013.
- Jean Goubault-Larrecq and Kok Min Ng. A few notes on formal balls. Logical Methods in Computer Science 13(4), nov. 28, 2017.
- Jean Goubault-Larrecq. Complete quasi-metrics for hyperspaces, continuous valuations, and previsions. arXiv 1707.03784, version 3, 28 oct. 2017.

— Jean Goubault-Larrecq (February 23rd, 2017)