First-countable spaces and their Smyth powerdomain

I have already talked about the Smyth powerdomain Q(X) of a topological space. This is the space of all non-empty compact saturated subsets of X, and it has two natural topologies:

  • the upper Vietoris topology has a base of open sets of the form ☐U ≝ {QQ(X) | QU}, where U is open in X; its specialization ordering is reverse inclusion ⊇;
  • the Scott topology of the reverse inclusion ordering ⊇.

In the book (Proposition 8.3.25), one finds that, when X is well-filtered, Q(X) is a dcpo under ⊇, where directed suprema are filtered intersections; and that if X is not only well-filtered but also locally compact, then Q(X) is a continuous dcpo in which QQ’ if and only if Q’ is included in the interior of Q. It follows that, if X is well-filtered, then the Scott topology is finer than the upper Vietoris topology on Q(X), and if X is well-filtered and locally compact, then those two topologies coincide.

Whenever there is any ambiguity, I will write QV(X) for Q(X) with the upper Vietoris topology, and Qσ(X) with the Scott topology of reverse inclusion. Hence Qσ(X)=QV(X) for every well-filtered, locally compact space X. Are those assumptions on X necessary?

There is a map η : XQ(X), defined by η(x)≝↑x. If X is T0, then η is injective. Noting that η–1(☐U)=U, we obtain that ☐U is a topological embedding of X into QV(X), provided that X is T0. Hence if X is T0, well-filtered and locally compact (equivalently, sober and locally compact), then ☐U is a topological embedding of X into Qσ(X), since in that case Qσ(X)=QV(X). Does this extend to more general cases of spaces X?

The question does not come out of nowhere. Xiaodong Jia and I were in the process of building some counterexamples in the theory of continuous valuations, and we were exploring the structure of Q(R), the Smyth powerdomain of the Sorgenfrey line R. Those questions naturally arose in this context… and I will answer those questions along the way, too. I must thank Xiaodong for telling me (or reminding me, in one case) about the two papers [1] and [2], which I am going to discuss some of the results of.

Embedding X into its Smyth powerdomain using first countability

He, Li, Xi, and Zhao have showed that one can go a long way with well-filtered first-countable spaces [1, Corollary 2.12]. They prove that every T1, well-filtered and first-countable space topologically embeds into Qσ(X) (with the Scott topology). The proof is clever, and reuses the idea of the compact completed sequence. Let me generalize the result slightly, as follows.

Theorem A. For every first-countable space X, η : XQσ(X) is continuous. If X is T0, well-filtered and first-countable, then η : XQσ(X) is a topological embedding.

Note that the Scott and the upper Vietoris topology on Q(X) may or may not coincide under any of those assumptions. We will address the question of the coincidence between those two topologies later in the same post.

Proof. We show that η is continuous. We don’t know much about the shape of (Scott-)open subsets of Qσ(X), but we proceed nonetheless.

Let U be a (Scott-)open set in Qσ(X). We let U ≝ η–1(U), and we wish to show that U is open in X. We reason by contradiction, and we assume that it is not. Hence there is a point x in U such that no open neighborhood V of x is included in U.

We now use the fact that X is first-countable. There is a sequence V0V1 ⊇ … ⊇ Vn ⊇ … of open neighborhoods of X such that every open neighborhood of x contains some Vn. By our assumption, no Vn is included in U, so we can find a point xn in VnU.

The sequence (xn)nN converges to x. Indeed, every open neighborhood of x contains some Vn, and then xn is in Vn, but also xn+1 since xn+1Vn+1Vn, and xn+2, and so on.

It follows that the set Q ≝ ↑{xn | nN} ∪ ↑x is compact (and saturated). This is really a direct consequence of Fact 1 in the previous post on the compact completed sequence. Since the proof is short, let me reproduce it here. Let (Ui)iI be any open cover of Q. Then x must be in Ui for some iI. Since (xn)nN converges to x, xn is in Ui for n large enough, say for every nm. Now x0 is in Ui[0] for some i[0] ∈ I, x1 is in Ui[1] for some i[1] ∈ I, …, and xm–1 is in Ui[m–1] for some i[m–1] ∈ I, and therefore Ui[0], Ui[1], …, Ui[m–1] and Ui form a finite subcover of Q.

The same argument shows that Qm ≝ ↑{xn | nm} ∪ ↑x is compact saturated for every mN.

The sets Qm form a descending sequence (with respect to inclusion). I claim that their intersection is exactly ↑x. In one direction, ↑x is included in every Qm. In the other direction, it suffices to show that every open neighborhood V of ↑x in X contains ∩mN Qm. Such an open neighborhood V must contain some Vn, by definition of the sets Vn. However, Vn contains xn, and xn+1, and xn+2, and so on, as we have already noticed. It also contains x, by definition. Hence Qn is included in Vn, and therefore ∩mN QmQnVnV.

We reformulate the latter as follows: the sequence (Qm)mN is an ascending sequence in Q(X) (with respect to reverse inclusion), whose supremum is ↑x.

Oh, wait, ↑x = η(x), and we had taken x from U ≝ η–1(U), right? So ↑x is in the Scott-open set U. By the definition of Scott-openness, some Qm must already be in U. We have ↑xmQm, namely ↑xm is a larger element of Q(X) than Qm (for reverse inclusion). Since U is upwards closed, ↑xm is in U. In other words, xm is in η–1(U)=U. That is impossible, since we have picked xm from VmU.

So η is continuous. If X is T0, it is injective, and it remains to show that if X is well-filtered, then every open subset U of X is the inverse image of some Scott-open subset of Qσ(X): we just take ☐U, and the fact that it is Scott-open is literally our assumption of well-filteredness. ☐

Application: the Sorgenfrey line (1)

The Sorgenfrey line R is a T2 space, hence it is obviously T0, and it is also sober, hence well-filtered. It is certainly not locally compact, since every compact subset has empty interior (Exercise 4.8.5 in the book). Hence we cannot use the facts we mentioned at the beginning of this post about the Smyth powerdomain of well-filtered, locally compact spaces. But Theorem A applies nicely, and we obtain the following.

Proposition. Q(R) is a dcpo model of R.

Recall that a dcpo model of a topological space X is a dcpo P together with a topological embedding of X into P (with its Scott topology), whose image is exactly the set Max P of maximal points of P. That requires X to be T1, and indeed R is T1, being T2. The maximal points of Q(X) (for any T1 space X) are always its one-point subsets {x}, which are indeed the sets of the form η(x) (since X is T1).

Comparing the Scott and upper Vietoris topologies in the presence of first-countability

Theorem A can be rephrased, once we agree to equate X with a subset of Q(X) through η, by saying that under the relevant assumptions, both the Scott topology and the upper Vietoris topology on Q(X) induce the same topology on the subspace X, and that is the original topology on X.

The next question one might have is whether the Scott topology and the upper Vietoris topology coincide on the whole of Q(X). In general, that is, for spaces X that are not well-filtered and locally compact, it does not [2, Example 3.13]. The case of the Sorgenfrey line can be handled through some recent results by X. Xu and Zh. Yang [2], which, again, were brought to my attention by Xiaodong Jia. Let me cite Corollary 5.9 from [2]:

Theorem. [2, Corollary 5.9] Let X be a first-countable well-filtered space. If the set Min(Q) of minimal elements of Q is countable for every Q ∈ Q(X), then the upper Vietoris topology and the Scott topology on Q(X) coincide.

The condition on the countability of Min(Q) may seem strange, but is necessary. Note that it is true on the Sorgenfrey line, since all its compact subsets are countable anyway, as we have seen last time.

Let me now explain give a proof of that theorem, which I think is simpler and perhaps more illuminating than that of [2]. Before I start, the point of the condition on the countability of Min(Q) is to make sure that Q(X) is itself first-countable in its upper Vietoris topology. For that, it is not enough to assume that X is first-countable: Xu and Yang give a counterexample [2, Example 4.4], which I will skip over.

Proposition. [2, Proposition 4.5] Let X be a first-countable space. If Min(Q) is countable for every Q ∈ Q(X), then QV(X) is first-countable.

Proof. For each point x of X, we pick a countable base of open neighborhoods Vn(x), nN. Given any Q ∈ Q(X), we may form an open neighborhood of Q in X by picking a natural number n(x) for every x ∈ Min(Q), and building the union of the sets Vn(x)(x) where x ranges over Min(Q). Since Q is compact, Q is included in a finite union of sets Vn(x)(x). Let us call special neighborhood of Q any finite union of sets Vn(x)(x), where x ranges over a finite subset E of Min(Q), and n(_) is a map from E to N. Since Min(Q) is countable, there are only countably many finite subsets of Min(Q), and for each, there are only countably many maps n(_). Hence there are only countably many special neighborhoods of Q in X.

It should be clear that every open neighborhood U of Q in X contains some special neighborhood: for every x in Min(Q), x in U hence in Vn(x) for some n in N; we call n(x) that n, and we extract a finite cover from the open cover (Vn(x)(x))x ∈ Min(Q). The union of the sets in that finite cover is a special neighborhood of Q included in U.

It follows that the sets ☐U, where U ranges over the special neighborhoods of Q, form a countable base of open neighborhoods of Q in QV(X). Indeed, for every open subset U of QV(X) that contains Q, U contains a basic open neighborhood ☐V of Q, by definition of the upper Vietoris topology and the fact that ☐ commutes with finite intersections. Then V is an open neighborhood of Q in X, and as we have seen, V must contain a special neighborhood U of Q. It follows that ☐V contains ☐U, and Q is in ☐U. ☐

We will now show that the Scott and upper Vietoris topologies coincide on Q(X) provided that X is well-filtered and QV(X) is first-countable, as shown by Xu and Yang [2, Theorem 5.7]. Their proof is not that long, but is a bit technical, and when one looks at it closely, it uses many similar ingredients as those that we have used in He, Li, Xi, and Zhao’s proof of Theorem A.

When that happens, one should really wonder whether the result is a consequence of Theorem A itself, not just of the techniques used in its proof. And this is exactly what I will demonstrate here.

The first result we need is mentioned in [2] as Lemma 2.4, and is credited to [3]. I have seen it mentioned in many papers.

Lemma B. Let X be any topological space. A directed family (Qi)iI has a supremum Q in Q(X) if and only if ∩iI Qi is compact, and then ∩iI Qi = Q.

Proof. The if direction is clear. In the only if direction, let us imagine that Q is different, hence properly included in ∩iI Qi. There is a point x in ∩iI Qi that is not in Q. Then Q ∪ ↑x is another upper bound of (Qi)iI, which strictly contains (hence is strictly smaller with respect to reverse inclusion than) Q, contradicting the fact that Q is the least upper bound. ☐

There is a map ∪ : Q(QV(X)) → Q(X), which maps every compact set K of compact subsets of QV(X) to ∪K ≝ ∪ {Q | QK}. In order to check that ∪K is indeed compact, we take any directed family (Ui)iI of open subsets of X such that ∪K ⊆ ∪iI Ui, and we show that ∪K is included in some Ui (see Proposition 4.4.7 in the book). For every Q in K, Q ⊆ ∪K ⊆ ∪iI Ui, and since Q is compact in X, Q is included in some Ui. This means that K is included in ∪iIUi. However, K itself is compact, this time in QV(X), and the family (☐Ui)iI is directed, so K is included in some ☐Ui. This means that every Q in K is included in the same open set Ui, so that ∪KUi.

Lemma C. The map ∪ : Q(QV(X)) → Q(X) is continuous from QV(QV(X)) to QV(X), and also from Qσ(QV(X)) to Qσ(X).

Proof. For the first part, the inverse image of ☐U is simply ☐☐U. Since ∪ is continuous from QV(QV(X)) to QV(X), it is monotonic with respect to the underlying specialization orderings, which are both reverse inclusion. In order to show the second part, it remains to show that for every directed family (Ki)iI in Q(QV(X)) with a supremum K, ∪K is the supremum of the directed family (∪Ki)iI. Note that we do not know yet whether the latter family has a supremum. Hence we will show that ∪K is an upper bound first, then the least one.

Since ∪ is monotonic, ∪K is an upper bound of (∪Ki)iI (… in the reverse inclusion ordering, namely ∪K is included in every ∪Ki). We need to show that it is the least one. Let Q be any other upper bound of (∪Ki)iI, in other words, Q is included in every ∪Ki. We wish to show that Q is included in ∪K, and to this end, we consider any point x in Q. For every iI, x is in ∪Ki, so x is in some compact saturated set K in Ki. Since Ki is itself saturated, any element larger than or equal to (namely, included in) any element of Ki is itself in Ki; and since x is in K, ↑x is included in K; so ↑x is in Ki. This holds for every iI, so ↑x is in the intersection of the sets Ki, iI. By Lemma B, that intersection is simply K! Hence ↑x is in K, from which it follows that x is in ∪K. ☐

If QV(X) is first-countable, then η : QV(X) → Qσ(QV(X)) is continuous by Proposition A. We compose it with the continuous map ∪ : Qσ(QV(X)) → Qσ(X) (Lemma C), and we obtain a continuous map from QV(X) to Qσ(X)… which is easily seen to be the identity map. And that means that every Scott-open subset of Q(X) is open in the upper Vietoris topology.

We know that the converse implication holds when X is well-filtered, so we immediately obtain the following result, which we had announced earlier.

Proposition. [2, Theorem 5.7] The Scott and upper Vietoris topologies coincide on Q(X) provided that X is well-filtered and QV(X) is first-countable.

We recall one of our earlier propositions: given a first-countable space X such that Min(Q) is countable for every Q ∈ Q(X), QV(X) is first-countable. We have therefore prove the theorem announced at the beginning of this section: given any first-countable well-filtered space X such that Min(Q) is countable for every Q ∈ Q(X), the upper Vietoris topology and the Scott topology on Q(X) coincide.

Application: the Sorgenfrey line (2)

In the Sorgenfrey line R, every compact (saturated) subset is countable. We have already recalled that R is first-countable, and well-filtered. Hence:

Proposition. The Scott and upper Vietoris topologies agree on Q(R), viz., Qσ(R)=QV(R).

  1. Qingyu He, Gaolin Li, Xiaoyong Xi, and Dongsheng Zhao. Some Results on Poset Models Consisting of Compact Saturated Subsets. Electronic Notes in Theoretical Computer Science 345, August 2019, pages 77-85. Proceedings of ISDT 2019, the 8th International Symposium on Domain Theory and Its Applications, ISDT 2019, Yangzhou, China, June 2019, edited by Achim Jung, Qingguo Li, Luoshan Xu, Guo-Qiang Zhang.
  2. Xiaoquan Xu and Zhongqiang Yang. Coincidence of the upper Vietoris topology and the Scott topology. Topology and its Applications 288, February 2021.
  3. Gerhard Gierz, Karl Heinrich Hofmann, Klaus Keimel, Jimmie D. Lawson, Michael W. Mislove, and Dana S. Scott. Continuous Lattices and Domains. Number 93 in Encyclopedia of Mathematics and its Applications. Cambridge University Press, Cambridge, 2003.

Jean Goubault-Larrecq (June 22nd, 2021)