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* cl_{1}(*A*) as the collection of suprema of directed families *D* included in ↓*A*. In general, cl_{1}(*A*) may fail to be Scott-closed, and we have to build the *two-step closure* cl_{2}(*A*) ≝ cl_{1}(cl_{1}(*A*)), then the three-step closure, and so on, transfinitely.

Wouldn’t it be nice if the one-step closure cl_{1}(*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 [1] 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 [2] 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**_{ℓ}) of the Sorgenfrey line **R**_{ℓ} 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 [1]. 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**_{ℓ}) (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(↓

**). Since**

*A***Q**(

*X*) is first-countable, it is easy to show that there is a descending chain

*U*

_{0}⊇

*U*

_{1}⊇ …

*U*

_{n}⊇ … of open neighborhoods of

*Q*such that the sets ☐

*U*

_{n}form a base of open neighborhoods of

*Q*. Each one of them intersects cl(↓

**) (at**

*A**Q*), hence must intersect ↓

**, say at**

*A**K*

_{n}. As in the proof of the de Brecht-Kawai theorem,

*K’*≝

_{n}*K*

_{n}_{ }∪

*K*

_{n}_{+1 }∪ … ∪

*K*∪ … is compact saturated, so

_{m }*Q*

_{n}≝

*K’*

_{n}_{ }∪

*Q*is compact saturated as well. The sets

*Q*

_{n}form an ascending chain in

**Q**(

*X*), whose supremum is their intersection; since the intersection of the sets

*K’*is included in the intersection of the sets

_{n}*U*

_{n}, which is

*Q*, the intersection of the sets

*Q*

_{n}is exactly

*Q*. Now each

*Q*

_{n}is a superset of, hence is smaller than or equal to,

*K*, which is in ↓

_{n}**by construction; so each**

*A**Q*

_{n}is in ↓

**, and we have obtained**

*A**Q*as a directed supremum of elements of ↓

**. ☐**

*A*Then **R**_{ℓ} is well-filtered since T_{2}, and **Q**(**R**_{ℓ}) 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**_{ℓ} 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 ☐(∪_{x ∈ Q} [*x*,*x*+1/2^{k}[), *k* ∈ **N**. Hence, by the previous lemma, **Q**(**R**_{ℓ}) (with the Scott=upper Vietoris topology) has one-step closure.

However, **Q**(**R**_{ℓ}) 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**_{ℓ} 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 cl_{1}(*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 (*x _{i}*)

_{i ∈ I}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 cl

_{1}(

*A*) where

*A*is the downward closure of {

*x*

_{i}

*| i*∈

*I*}. By assumption cl

_{1}(

*A*) is downwards-closed, so

*x’*is also in cl

_{1}(

*A*). This means that

*x’*is the supremum of some directed family (

*x’*)

_{j}_{j ∈ J}of points that are all in ↓

*A*, namely, in

*A*, since

*A*is downwards-closed. Since

*x’*is in

*U*, some

*x’*is in

_{j}*U*. Also, since

*x’*≤

_{j}*x’*≤

*y*, we obtain that

*x’*is in

_{j}*U*∩ ↓

*y*. Since

*x’*is in

_{j}*A*, by definition of

*A*,

*x’*is below some

_{j}*x*; hence that

_{i}*x*is in ↑(

_{i}*U*∩ ↓

*y*). Therefore ↑(

*U*∩ ↓

*y*) is Scott-open. ☐

In a poset *X* that has one-step closure, cl_{1}(*A*) is the closure of *A*, for every downwards-closed subset *A*. In particular, cl_{1}(*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 [**Z**^{–} → **Z**^{–}] 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 [**Z**^{–} → **Z**^{–}] 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 [**Z**^{–} → **Z**^{–}] of (Scott-)continuous map from **Z**^{–} to **Z**^{–}, up to isomorphism—but not necessarily with the *Scott* topology. I have just said that [**Z**^{–} → **Z**^{–}], with the Scott topology, would not be a c-space, but maybe there is *another* topology that would make [**Z**^{–} → **Z**^{–}] 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 [4], 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 [**Z**^{–} → **Z**^{–}], and we write [**Z**^{–} → **Z**^{–}]_{τ} for [**Z**^{–} → **Z**^{–}] with that topology. It is characterized by the property that [**Z**^{–} → **Z**^{–}]_{τ} is a c-space, and for every c-space *X* and every map *f* : *X* × **Z**^{–} → **Z**^{–}, *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 [**Z**^{–} → **Z**^{–}]_{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 [
**Z**^{–}→**Z**^{–}]_{τ}is the usual pointwise ordering. Indeed, consider Sierpiński space**S**≝ {0 < 1} (certainly a c-space), and two elements*g*,*g’*of [**Z**^{–}→**Z**^{–}]_{τ}. We build the map*f*:*X*×**Z**^{–}→**Z**^{–}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*g*≤*g’*in the pointwise ordering. But Λ(*f*) is continuous if and only if*g*is below*g’*in the specialization ordering of [**Z**^{–}→**Z**^{–}]_{τ}. - [
**Z**^{–}→**Z**^{–}]_{τ}is a monotone convergence space, namely every open subset of [**Z**^{–}→**Z**^{–}]_{τ}is Scott-open (in its specialization ordering, namely the pointwise ordering ≤), and [**Z**^{–}→**Z**^{–}]_{τ}is a dcpo under ≤. The latter is obvious. In order to prove the former, let*W*be any open subset of [**Z**^{–}→**Z**^{–}]_{τ}. It is certainly upwards-closed with respect to ≤. We consider any monotone net (*g*)_{i}_{i ∈ I, ⊑}in [**Z**^{–}→**Z**^{–}]_{τ}, and we assume that its (pointwise) supremum*g*is in*W*.

With the aim of showing that some*g*is already in_{i}*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*i*∈*I*(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*×**Z**^{–}→**Z**^{–}: for every*i*in*I*,*f*(*i*,*n*) ≝(*g*_{i}*n*), and*f*(∞,*n*) ≝*g*(*n*). It is easy to see that*f*is (separately) continuous. Therefore Λ(*f*) is continuous from*Î*to [**Z**^{–}→**Z**^{–}]_{τ}. We compose it with the characteristic map χ: [_{W}**Z**^{–}→**Z**^{–}]_{τ}→**S**, and we obtain a continuous map χo Λ(_{W}*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 χo Λ(_{W}*f*) is of the form*A*∪ {∞} where*A*is non-empty and upwards-closed in*I*. Let*i*be any element of*A*. Then χo Λ(_{W}*f*) maps*i*to 1, and that means thatis in*g*_{i}*W*. - The final nail is the fact that any T
_{0}c-space (such as [**Z**^{–}→**Z**^{–}]_{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 T_{0}-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 [
**Z**^{–}→**Z**^{–}]_{τ}must really be [**Z**^{–}→**Z**^{–}], with the Scott topology, and must be a continuous dcpo. And we are back at step 1: [**Z**^{–}→**Z**^{–}] 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 [4], simply use Lawson’s [5]. (Zh. Lyu, X. Xie and H. Kou show this in a different way, as a consequence of the previous result.)

- Zhiwei Zou, Qingguo Li, and Wengkin Ho, domains via approximation operators, Logical Methods in Computer Science, 14 (2018): 1-17.
- 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. - Xiaoquan Xu and Zhongqiang Yang.
*Coincidence of the upper Vietoris topology and the Scott topology*. Topology and its Applications, 288, 107480, December 2021. - Yuri Leonidovich Ershov.
*The Bounded Complete Hull of an α-Space*. Theoretical Computer Science, 175, 3–13, 1997. - Jimmie D. Lawson.
*T*. Topology and Its Applications, 21(1), 73–76, 1985._{0}-Spaces and Pointwise Convergence

— Jean Goubault-Larrecq (July 20th, 2022)