# A report from ISDT’22: one-step closure; c-spaces are not CCC

Let me say a few things I heard at the 9th International Symposium on Domain Theory (ISDT’22), which took place online, July 4-6, 2022, in Singapore. I will not give a lot of details; that will be quite a change from some of my recent posts, and hopefully will give us all a bit of rest.

The selection of topics I will talk about below does not imply any specific notion of ranking. In other words, if you gave a talk at ISDT’22 and I do not speak about it here, that does not mean that I didn’t appreciate your talk. In fact, I am only going to report on two of the contributed talks.

## One-step closure

Hualin Miao talked about one-step closure, in a joint paper with Qingguo Li and Dongsheng Zhao. (Update, July 22nd, 2022: I am learning that H. Miao has been awarded the Best Student Paper Award at ISDT’22. Congratulations!)

Given a subset A of a poset X, we can form its closure cl(A) in the Scott topology in at least two ways. We can take the intersection of all the Scott-closed subsets of X that contain A; that is perfect, except it gives you no idea what elements are in cl(A) and what elements are not.

Or we can add the missing directed suprema to A, iteratively. This is a more concrete process, but is somewhat complex. Here is how you do it. Since a Scott-closed set must be downwards-closed and closed under directed suprema, we form the one-step closure cl1(A) as the collection of suprema of directed families D included in ↓A. In general, cl1(A) may fail to be Scott-closed, and we have to build the two-step closure cl2(A) ≝ cl1(cl1(A)), then the three-step closure, and so on, transfinitely.

Wouldn’t it be nice if the one-step closure cl1(A) of A were already its Scott closure, for any subset A of X?

A poset X with that property is said to have one-step closure.

It is well-known that every continuous poset X has one-step closure. They cite a 2018 paper by Zhiwei Zou, Qingguo Li, and Wengkin Ho  for that, but my impression was that this had been well-known for much longer. (This shows a personal bias: this result was mentioned in a 2009 paper of mine [2, Proposition 3.5] with Alain Finkel; see Appendix A in  for a proof, but note that I am not claiming I was the first to prove it: I was pretty sure already at that time that this was well-known, although I would be incapable of giving a precise reference.)

One of their first results is that there are some posets that have one-step closure but are not continuous: the Smyth powerdomain Q(R) of the Sorgenfrey line R is one example, as they demonstrate. I will give their proof below.

Then they remind us that every poset that has one-step closure must be meet-continuous. That had been shown by Zh. Zou, Q. Li, and W. Ho . See here for a discussion of what meet-continuity is, for example. I will also give the short proof below.

Hence we have the implications:

continuous poset ⇒ one-step-closure ⇒ meet-continuous poset

and the first implication is strict. The second implication is strict, too, as a consequence of further results of the authors.

### A non-continuous poset that has one-step closure

Let me describe how they show that Q(R) (with its Scott topology) has one-step closure, but is not continuous. They first show the following.

Lemma. For every well-filtered space X whose Smyth hyperspace Q(X) (the space of compact saturated subsets of X, with the upper Vietoris topology) is first-countable, Q(X) has the Scott topology of ⊇, and is a dcpo that has one-step closure.

Proof. The upper Vietoris topology has a base of open sets of the form ☐U, where U ranges over the open subsets of X; and ☐U denotes the collection of compact saturated subsets of X that are included in U. The fact that Q(X) has the Scott topology of ⊇ is a result due to X. Xu and Zh. Yang [3, Theorem 5.7], and I will not explain how this is proved here. The proof is rather similar to that of the de Brecht-Kawai theorem, and in fact rests on similar arguments as the ones I am now going to set forth.

Let A be any subset of Q(X), and let Q be any element in the closure cl(↓A). Since Q(X) is first-countable, it is easy to show that there is a descending chain U0U1 ⊇ … Un ⊇ … of open neighborhoods of Q such that the sets ☐Un form a base of open neighborhoods of Q. Each one of them intersects cl(↓A) (at Q), hence must intersect ↓A, say at Kn. As in the proof of the de Brecht-Kawai theorem, K’n ≝ Kn ∪ Kn+1 ∪ … ∪ K∪ … is compact saturated, so Qn ≝ K’n ∪ Q is compact saturated as well. The sets Qn form an ascending chain in Q(X), whose supremum is their intersection; since the intersection of the sets K’n is included in the intersection of the sets Un, which is Q, the intersection of the sets Qn is exactly Q. Now each Qn is a superset of, hence is smaller than or equal to, Kn, which is in ↓A by construction; so each Qn is in ↓A, and we have obtained Q as a directed supremum of elements of ↓A. ☐

Then R is well-filtered since T2, and Q(R) is first-countable, as shown by X. Xu and Zh. Yang [3, Example 5.14 (2)]. The latter can be proved as follows. We remember that the compact subsets of R are exactly the well-founded subdcpos of (R, ≥) (see Proposition C here), and that they are all countable. Any such compact subset Q has a countable base of open neighborhoods of the form ☐(∪xQ [x,x+1/2k[), kN. Hence, by the previous lemma, Q(R) (with the Scott=upper Vietoris topology) has one-step closure.

However, Q(R) is not a continuous poset. In fact, it is not even core-compact. The reason is the Lyu-Jia theorem: a space X is locally compact if and only if Q(X) is core-compact; and we know that R is not core-compact (Exercise 4.8.5 in the book).

### One-step-closure ⇒ meet-continuous

There are many equivalent definitions of meet-continuity, and we will take the following one from this post: a topological space X is meet-continuous if and only if for every y in X, for every open subset U of X, ↑(U ∩ ↓y) is open. H. Miao, Q. Li and D. Zhao show the following:

Lemma. Every poset X in which cl1(A) is downwards-closed for every downwards-closed subset A is meet-continuous in its Scott topology.

Proof. We consider a point y of X, and a Scott-open subset U of x. ↑(U ∩ ↓y) is trivially upwards-closed. We consider any directed family (xi)iI of points with a supremum x in ↑(U ∩ ↓y). Hence there is a point x’x such that x’ is in U ∩ ↓y. Clearly, x is in cl1(A) where A is the downward closure of {xi | iI}. By assumption cl1(A) is downwards-closed, so x’ is also in cl1(A). This means that x’ is the supremum of some directed family (x’j)jJ of points that are all in ↓A, namely, in A, since A is downwards-closed. Since x’ is in U, some x’j is in U. Also, since x’j ≤ x’y, we obtain that x’j is in U ∩ ↓y. Since x’j is in A, by definition of A, x’j is below some xi; hence that xi is in ↑(U ∩ ↓y). Therefore ↑(U ∩ ↓y) is Scott-open. ☐

In a poset X that has one-step closure, cl1(A) is the closure of A, for every downwards-closed subset A. In particular, cl1(A) is downwards-closed for every such A. The Lemma above then implies that X is meet-continuous in its Scott topology.

### Further considerations

I have only scratched the surface of what H. Miao, Q. Li, and D. Zhao show, and I will not say much more. Still, I must mention that they looked at the string of implications:

continuous poset ⇒ one-step-closure ⇒ meet-continuous poset

and compared it with the fact that the continuous posets are exactly those that are both quasi-continuous and meet-continuous (a fact I have touched upon here).

This raises the question of the relationships between having one-step closure and all those notions. For example, they show that a poset has one-step closure if and only if it has the weaker property of having weak one-step closure and is meet-continuous. They also show that every quasi-continuous poset has weak one-step closure. But many questions remain open.

## C-spaces are not Cartesian-closed

Zhenchao Lyu gave a proof that the category of c-spaces is not Cartesian-closed, with Xiaolin Xie and Hui Kou. This is a pretty funny result. My first impression was that it was trivial. My second impression was that there was a well-hidden (also, well-known) difficulty. My third impression was that, if you knew enough about domain theory (and Zhenchao certainly knows enough), that difficulty is easily dealt with. Hence, in a sense, it is both easy and clever.

I initially thought that it was trivial for the following reason. It is well-known that the category of continuous dcpos (or of continuous posets) is not Cartesian-closed. The argument consists in looking at the poset Z of non-positive integers, with the usual ordering, and to realize that there is simply no element of [ZZ] that is way-below the identity map. Now Z is a continuous dcpo (in fact even an algebraic dcpo), and in particular certainly a c-space in its Scott topology; I have just argued that [ZZ] is not a continuous dcpo; hence it is not a c-space, right?

Well, yes, but that does not solve the question. Here is the difficulty. If the category of c-spaces is Cartesian-closed, then certainly there is an exponential object, which will be the set [ZZ] of (Scott-)continuous map from Z to Z, up to isomorphism—but not necessarily with the Scott topology. I have just said that [ZZ], with the Scott topology, would not be a c-space, but maybe there is another topology that would make [ZZ] a c-space, and an exponential object in the category of c-spaces.

The key to that issue is the following wonderful fact [4, Proposition 2]: for any c-space X, and any two topological spaces Y and Z, every separately continuous map f from X × Y to Z is jointly continuous. (Yuri Ershov published this in 1997 , but Jimmie Lawson had proved an even more general theorem twelves years earlier [5, Theorem 2]: the class of spaces X such that for any two topological spaces Y and Z, every separately continuous map f from X × Y to Z is jointly continuous, is exactly the class of locally finitary compact spaces—and those include the c-spaces.)

Now we imagine, by way of contradiction, that the category of c-spaces is Cartesian-closed. There is an exponential topology τ on [ZZ], and we write [ZZ]τ for [ZZ] with that topology. It is characterized by the property that [ZZ]τ is a c-space, and for every c-space X and every map f : X × ZZ, f is (jointly) continuous if and only if Λ(f) (the function that maps every x in X to the function that maps every z in Z to f(x,z)) is continuous from X to [ZZ]τ. I have put “jointly” between parentheses simply because we have just seen that f is jointly continuous if and only if it is separately continuous. Zh. Lyu, X. Xie and H. Kou now show:

• The specialization ordering of [ZZ]τ is the usual pointwise ordering. Indeed, consider Sierpiński space S ≝ {0 < 1} (certainly a c-space), and two elements g, g’ of [ZZ]τ. We build the map f : X × ZZ defined by f(0,n) ≝ g(n) and f(1,n) ≝ g‘(n). The map f is (separately) continuous if and only if it is monotonic in its first argument, if and only if gg’ in the pointwise ordering. But Λ(f) is continuous if and only if g is below g’ in the specialization ordering of [ZZ]τ.
• [ZZ]τ is a monotone convergence space, namely every open subset of [ZZ]τ is Scott-open (in its specialization ordering, namely the pointwise ordering ≤), and [ZZ]τ is a dcpo under ≤. The latter is obvious. In order to prove the former, let W be any open subset of [ZZ]τ. It is certainly upwards-closed with respect to ≤. We consider any monotone net (gi)iI, ⊑ in [ZZ]τ, and we assume that its (pointwise) supremum g is in W.
With the aim of showing that some gi is already in W, we form the following c-space Î. Î is just I itself, preordered by ⊑, plus a fresh element ∞ on top of all others. Then we give Î the topology whose non-empty open sets are exactly the sets of the form A ∪ {∞} where A is non-empty and upwards-closed in I. Î is a c-space: in fact the upward closures of points iI (namely ↑i ∪ {∞} where ↑i denotes upward closure in I) are compact and open, and form a base of the topology.
Let us build the following function f : Î × ZZ: for every i in I, f(i,n) ≝ gi(n), and f(∞,n) ≝ g(n). It is easy to see that f is (separately) continuous. Therefore Λ(f) is continuous from Î to [ZZ]τ. We compose it with the characteristic map χW : [ZZ]τS, and we obtain a continuous map χW o Λ(f) from Î to S. That continuous function maps ∞ to 1, since g is in W. Hence, by the definition of the topology on Î, the inverse image of {1} by χW o Λ(f) is of the form A ∪ {∞} where A is non-empty and upwards-closed in I. Let i be any element of A. Then χW o Λ(f) maps i to 1, and that means that gi is in W.
• The final nail is the fact that any T0 c-space (such as [ZZ]τ) that is at the same time a monotone convergence space is in fact a continuous dcpo, with the Scott topology. This is what the proof of Proposition 8.3.36 in the book really shows (although it is only stated for sober c-spaces: sobriety is only used to obtain T0-ness and monotone convergence; update, July 21st, 2022: you can also find that result in the book Continuous lattices and domains, Theorem II-3.16, equivalence of (1) and (5), and you have to notice that (5) is equivalent to X being a c-space). Zh. Lyu, X. Xie and H. Kou do this step a bit differently, by the way, using the theory of directed spaces (which I really ought to talk about some day!).
• Hence [ZZ]τ must really be [ZZ], with the Scott topology, and must be a continuous dcpo. And we are back at step 1: [ZZ] is not a continuous dcpo, because no continuous function is way-below the identity map on Z. We have reached a contradiction. Hence the category of c-spaces is not Cartesian-closed.

The exact same proof shows that the category of locally finitary compact spaces is not Cartesian-closed either. Instead of using Ershov’s result , simply use Lawson’s . (Zh. Lyu, X. Xie and H. Kou show this in a different way, as a consequence of the previous result.)

1. Zhiwei Zou, Qingguo Li, and Wengkin Ho, domains via approximation operators, Logical Methods in Computer Science, 14 (2018): 1-17.
2. Alain Finkel and Jean Goubault-Larrecq. Forward Analysis for WSTS, Part I: Completions. In Proceedings of the 26th Annual Symposium on Theoretical Aspects of Computer Science (STACS’09), volume 3 of Leibniz International Proceedings in Informatics, pages 433–444, Freiburg, Germany, February 2009. Leibniz-Zentrum für Informatik.
3. Xiaoquan Xu and Zhongqiang Yang. Coincidence of the upper Vietoris topology and the Scott topology. Topology and its Applications, 288, 107480, December 2021.
4. Yuri Leonidovich Ershov. The Bounded Complete Hull of an α-Space. Theoretical Computer Science, 175, 3–13, 1997.
5. Jimmie D. Lawson. T0-Spaces and Pointwise Convergence. Topology and Its Applications, 21(1), 73–76, 1985.

Jean Goubault-Larrecq (July 20th, 2022)