Topologies on Spaces of Lipschitz Yoneda-continuous Maps

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

The topology of pointwise convergence is the topology induced by the inclusion of Lα(X) into the product [0, ∞]X, namely the space of all (not necessarily continuous) maps from 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 Lα(X) (α>0) are stably compact.

The toned-down version

Now here comes the delicate part.  Lα(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 Lα(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 Lα(X).  By definition, it is determined if and only if any subset U of L(X) whose intersection with Lα(X) is open in Lα(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 Lα(X), with the obvious inclusions. Whence the question:

Let X, d be a complete metric space.  Is the topology of L(X) determined by the topologies of Lα(X), α>0?

If it is not, then under which reasonable conditions is that topology determined by the topologies of Lα(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, Lα(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 Lα(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 Lα(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 Lα(X) (α>0) all have the topology of pointwise convergence, and are stably compact.

Accordingly, the real question is:

Let X, d be a continuous Yoneda-complete metric space.  Is the topology of L(X) determined by the topologies of Lα(X), α>0?

If it is not, then under which reasonable conditions is that topology determined by the topologies of Lα(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 VX = 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 Lα(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 Lα(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 Gδ 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 Lα(X), α>0, there would be a much simpler proof available.

Any ideas?

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

Jean Goubault-Larrecqjgl-2011