{"id":5509,"date":"2022-07-20T07:32:38","date_gmt":"2022-07-20T05:32:38","guid":{"rendered":"https:\/\/projects.lsv.ens-paris-saclay.fr\/topology\/?page_id=5509"},"modified":"2022-11-19T14:53:41","modified_gmt":"2022-11-19T13:53:41","slug":"a-report-from-isdt22-one-step-closure-c-spaces-are-not-ccc","status":"publish","type":"page","link":"https:\/\/projects.lsv.ens-paris-saclay.fr\/topology\/?page_id=5509","title":{"rendered":"A report from ISDT’22: one-step closure; c-spaces are not CCC"},"content":{"rendered":"\n

Let me say a few things I heard at the 9th International Symposium on Domain Theory (ISDT’22<\/a>), 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.<\/p>\n\n\n\n

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.<\/p>\n\n\n\n

One-step closure<\/h2>\n\n\n\n

Hualin Miao<\/a> 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!)<\/p>\n\n\n\n

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

Or we can add the missing directed suprema to A<\/em>, 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<\/em> cl1<\/sub>(A<\/em>) as the collection of suprema of directed families D<\/em> included in \u2193A<\/em>. In general, cl1<\/sub>(A<\/em>) may fail to be Scott-closed, and we have to build the two-step closure<\/em> cl2<\/sub>(A<\/em>) \u225d cl1<\/sub>(cl1<\/sub>(A<\/em>)), then the three-step closure, and so on, transfinitely.<\/p>\n\n\n\n

Wouldn’t it be nice if the one-step closure cl1<\/sub>(A<\/em>) of A<\/em> were already its Scott closure, for any subset A<\/em> of X<\/em>?<\/p>\n\n\n\n

A poset X<\/em> with that property is said to have one-step closure<\/em>.<\/p>\n\n\n\n

It is well-known that every continuous poset X<\/em> 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.)<\/p>\n\n\n\n

One of their first results is that there are some posets that have one-step closure but are not continuous: the Smyth powerdomain Q<\/strong>(R<\/strong>\u2113<\/sub>) of the Sorgenfrey line R<\/strong>\u2113<\/sub> is one example, as they demonstrate. I will give their proof below.<\/p>\n\n\n\n

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<\/a> for a discussion of what meet-continuity is, for example. I will also give the short proof below.<\/p>\n\n\n\n

Hence we have the implications:<\/p>\n\n\n\n

continuous poset \u21d2 one-step-closure \u21d2 meet-continuous poset<\/p>\n\n\n\n

and the first implication is strict. The second implication is strict, too, as a consequence of further results of the authors.<\/p>\n\n\n\n

A non-continuous poset that has one-step closure<\/h3>\n\n\n\n

Let me describe how they show that Q<\/strong>(R<\/strong>\u2113<\/sub>) (with its Scott topology) has one-step closure, but is not continuous. They first show the following.<\/p>\n\n\n\n

Lemma.<\/strong> For every well-filtered space X<\/em> whose Smyth hyperspace Q<\/strong>(X<\/em>) (the space of compact saturated subsets of X<\/em>, with the upper Vietoris topology) is first-countable, Q<\/strong>(X<\/em>) has the Scott topology of \u2287, and is a dcpo that has one-step closure.<\/p>\n\n\n\n

Proof.<\/em> The upper Vietoris topology has a base of open sets of the form \u2610U<\/em>, where U<\/em> ranges over the open subsets of X<\/em>; and \u2610U<\/em> denotes the collection of compact saturated subsets of X<\/em> that are included in U<\/em>. The fact that Q<\/strong>(X<\/em>) has the Scott topology of \u2287 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<\/a>, and in fact rests on similar arguments as the ones I am now going to set forth.<\/p>\n\n\n\n

Let A<\/em><\/strong> be any subset of Q<\/strong>(X<\/em>), and let Q<\/em> be any element in the closure cl(\u2193A<\/em><\/strong>). Since Q<\/strong>(X<\/em>) is first-countable, it is easy to show that there is a descending chain U<\/em>0<\/sub> \u2287 U<\/em>1<\/sub> \u2287 … U<\/em>n<\/em><\/sub> \u2287 … of open neighborhoods of Q<\/em> such that the sets \u2610U<\/em>n<\/em><\/sub> form a base of open neighborhoods of Q<\/em>. Each one of them intersects cl(\u2193A<\/em><\/strong>) (at Q<\/em>), hence must intersect \u2193A<\/em><\/strong>, say at K<\/em>n<\/em><\/sub>. As in the proof of the de Brecht-Kawai theorem<\/a>, K\u2019n<\/sub><\/em> \u225d Kn<\/sub><\/em> <\/sub>\u222a Kn<\/sub><\/em>+1 <\/sub>\u222a \u2026 \u222a Km <\/sub><\/em>\u222a \u2026 is compact saturated, so Q<\/em>n<\/em><\/sub> \u225d K’n<\/sub><\/em> <\/sub>\u222a Q<\/em> is compact saturated as well. The sets Q<\/em>n<\/em><\/sub> form an ascending chain in Q<\/strong>(X<\/em>), whose supremum is their intersection; since the intersection of the sets K’n<\/sub><\/em> is included in the intersection of the sets U<\/em>n<\/em><\/sub>, which is Q<\/em>, the intersection of the sets Q<\/em>n<\/em><\/sub> is exactly Q<\/em>. Now each Q<\/em>n<\/em><\/sub> is a superset of, hence is smaller than or equal to, Kn<\/sub><\/em>, which is in \u2193A<\/em><\/strong> by construction; so each Q<\/em>n<\/em><\/sub> is in \u2193A<\/em><\/strong>, and we have obtained Q<\/em> as a directed supremum of elements of \u2193A<\/em><\/strong>. \u2610<\/p>\n\n\n\n

Then R<\/strong>\u2113<\/sub> is well-filtered since T2<\/sub>, and Q<\/strong>(R<\/strong>\u2113<\/sub>) 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<\/strong>\u2113<\/sub> are exactly the well-founded subdcpos of (R<\/strong>, \u2265) (see Proposition C here<\/a>), and that they are all countable. Any such compact subset Q<\/em> has a countable base of open neighborhoods of the form \u2610(\u222ax<\/em> \u2208 Q<\/em><\/sub> [x<\/em>,x<\/em>+1\/2k<\/em><\/sup>[), k<\/em> \u2208 N<\/strong>. Hence, by the previous lemma, Q<\/strong>(R<\/strong>\u2113<\/sub>) (with the Scott=upper Vietoris topology) has one-step closure.<\/p>\n\n\n\n

However, Q<\/strong>(R<\/strong>\u2113<\/sub>) is not a continuous poset. In fact, it is not even core-compact. The reason is the Lyu-Jia theorem<\/a>: a space X<\/em> is locally compact if and only if Q<\/strong>(X<\/em>) is core-compact; and we know that R<\/strong>\u2113<\/sub> is not core-compact (Exercise 4.8.5 in the book<\/a>).<\/p>\n\n\n\n

One-step-closure \u21d2 meet-continuous<\/h3>\n\n\n\n

There are many equivalent definitions of meet-continuity, and we will take the following one from this post<\/a>: a topological space X<\/em> is meet-continuous if and only if for every y<\/em> in X<\/em>, for every open subset U<\/em> of X<\/em>, \u2191(U<\/em> \u2229 \u2193y<\/em>) is open. H. Miao, Q. Li and D. Zhao show the following:<\/p>\n\n\n\n

Lemma.<\/strong> Every poset X<\/em> in which cl1<\/sub>(A<\/em>) is downwards-closed for every downwards-closed subset A<\/em> is meet-continuous in its Scott topology.<\/p>\n\n\n\n

Proof.<\/em> We consider a point y<\/em> of X<\/em>, and a Scott-open subset U<\/em> of x<\/em>. \u2191(U<\/em> \u2229 \u2193y<\/em>) is trivially upwards-closed. We consider any directed family (xi<\/sub><\/em>)i<\/em> \u2208 I<\/em><\/sub> of points with a supremum x<\/em> in \u2191(U<\/em> \u2229 \u2193y<\/em>). Hence there is a point x’<\/em> \u2264 x<\/em> such that x’<\/em> is in U<\/em> \u2229 \u2193y<\/em>. Clearly, x<\/em> is in cl1<\/sub>(A<\/em>) where A<\/em> is the downward closure of {x<\/i>i<\/i><\/sub> | i<\/em> \u2208 I<\/em>}. By assumption cl1<\/sub>(A<\/em>) is downwards-closed, so x’<\/em> is also in cl1<\/sub>(A<\/em>). This means that x’<\/em> is the supremum of some directed family (x’j<\/sub><\/em>)j<\/em> \u2208 J<\/em><\/sub> of points that are all in \u2193A<\/em>, namely, in A<\/em>, since A<\/em> is downwards-closed. Since x’<\/em> is in U<\/em>, some x’j<\/sub><\/em> is in U<\/em>. Also, since x’j<\/sub><\/em> \u2264 x’<\/em> \u2264 y<\/em>, we obtain that x’j<\/sub><\/em> is in U<\/em> \u2229 \u2193y<\/em>. Since x’j<\/sub><\/em> is in A<\/em>, by definition of A<\/em>, x’j<\/sub><\/em> is below some xi<\/sub><\/em>; hence that xi<\/sub><\/em> is in \u2191(U<\/em> \u2229 \u2193y<\/em>). Therefore \u2191(U<\/em> \u2229 \u2193y<\/em>) is Scott-open. \u2610<\/p>\n\n\n\n

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

Further considerations<\/h3>\n\n\n\n

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:<\/p>\n\n\n\n

continuous poset \u21d2 one-step-closure \u21d2 meet-continuous poset<\/p>\n\n\n\n

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<\/a>).<\/p>\n\n\n\n

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<\/em> 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.<\/p>\n\n\n\n

C-spaces are not Cartesian-closed<\/h2>\n\n\n\n

Zhenchao Lyu<\/a> 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.<\/p>\n\n\n\n

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<\/strong>\u2013<\/sup> of non-positive integers, with the usual ordering, and to realize that there is simply no element of [Z<\/strong>\u2013<\/sup> \u2192 Z<\/strong>\u2013<\/sup>] that is way-below the identity map. Now Z<\/strong>\u2013<\/sup> 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<\/strong>\u2013<\/sup> \u2192 Z<\/strong>\u2013<\/sup>] is not a continuous dcpo; hence it is not a c-space, right?<\/p>\n\n\n\n

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<\/strong>\u2013<\/sup> \u2192 Z<\/strong>\u2013<\/sup>] of (Scott-)continuous map from Z<\/strong>\u2013<\/sup> to Z<\/strong>\u2013<\/sup>, up to isomorphism\u2014but not necessarily with the Scott<\/em> topology. I have just said that [Z<\/strong>\u2013<\/sup> \u2192 Z<\/strong>\u2013<\/sup>], with the Scott topology, would not be a c-space, but maybe there is another<\/em> topology that would make [Z<\/strong>\u2013<\/sup> \u2192 Z<\/strong>\u2013<\/sup>] a c-space, and an exponential object in the category of c-spaces.<\/p>\n\n\n\n

The key to that issue is the following wonderful fact [4, Proposition 2]: for any c-space X<\/em>, and any two topological spaces Y<\/em> and Z<\/em>, every separately continuous<\/em> map f<\/em> from X<\/em> \u00d7 Y<\/em> to Z<\/em> is jointly continuous<\/em>. (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<\/em> such that for any two topological spaces Y and Z, every separately continuous map f from X<\/em> \u00d7 Y<\/em> to Z<\/em> is jointly continuous, is exactly<\/em> the class of locally finitary compact spaces\u2014and those include the c-spaces.)<\/p>\n\n\n\n

Now we imagine, by way of contradiction, that the category of c-spaces is Cartesian-closed. There is an exponential topology \u03c4 on [Z<\/strong>\u2013<\/sup> \u2192 Z<\/strong>\u2013<\/sup>], and we write [Z<\/strong>\u2013<\/sup> \u2192 Z<\/strong>\u2013<\/sup>]\u03c4<\/sub> for [Z<\/strong>\u2013<\/sup> \u2192 Z<\/strong>\u2013<\/sup>] with that topology. It is characterized by the property that [Z<\/strong>\u2013<\/sup> \u2192 Z<\/strong>\u2013<\/sup>]\u03c4<\/sub> is a c-space, and for every c-space X<\/em> and every map f<\/em> : X<\/em> \u00d7 Z<\/strong>\u2013<\/sup> \u2192 Z<\/strong>\u2013<\/sup>, f<\/em> is (jointly) continuous if and only if \u039b(f<\/em>) (the function that maps every x<\/em> in X<\/em> to the function that maps every z<\/em> in Z<\/strong>\u2013<\/sup> to f<\/em>(x<\/em>,z<\/em>)) is continuous from X<\/em> to [Z<\/strong>\u2013<\/sup> \u2192 Z<\/strong>\u2013<\/sup>]\u03c4<\/sub>. I have put “jointly” between parentheses simply because we have just seen that f<\/em> is jointly continuous if and only if it is separately continuous. Zh. Lyu, X. Xie and H. Kou now show:<\/p>\n\n\n\n