I would like to talk about a nifty, recent result due to Yu Chen, Hui Kou, and Zhenchao Lyu [2].
Let H(X) stand for the Hoare hyperspace of a space X. By definition, this is the space of all closed subsets of X; a variant, used in denotational semantics notably, only considers non-empty closed subsets, and most of what I will say below will apply to that other hyperspace.
There are at least two standard topologies on H(X):
- the Scott topology of the inclusion ordering;
- the lower Vietoris topology, defined through a subbase of open sets of the form ◊U ≝ {F ∈ H(X) | F ∩ U ≠ ∅}, where U ranges over the open subsets of X.
It turns out that the complement of ◊U is the collection of closed subsets of X that are included in the complement C of U, namely the downward closure ↓HC of C in H(X). (I am writing ↓HC for the downward closure of C in H(X), namely {F ∈ H(X) | F ⊆ C}, in order to distinguish it from it downward closure in X, which I write as ↓C, and which denotes {x ∈ X | x ≤ y for some y ∈ C}. Not that it matters, since apparently I will not use any of those notations any further in this post.) Hence the closed sets in the lower Vietoris are generated by the downward closures of single points C of H(X). That topology is also known as the upper topology of the poset H(X).
It is easy to see that the Scott topology is always finer than the lower Vietoris topology. One can see this by remembering that the upper topology is the coarsest topology that has inclusion as specialization ordering (Proposition 4.2.12 in the book). Alternatively, one can check that ◊U is Scott-open in H(X), for every open subset U of X. For that, you first need to observe that H(X) is a dcpo under inclusion, whatever X is, and that directed suprema in H(X) are closures of directed unions.
This begs the question: when are those topologies equal?
Scott may be strictly finer than lower Vietoris
We start by observing that there are cases where the Scott topology is strictly finer than the lower Vietoris topology.
The following example is due to de Brecht and Kawai [1, Example 3.2]. We consider the space X≝Ncof of all natural numbers with the cofinite topology. The collection U of all closed subsets of Ncof of cardinality at least 2 is Scott-open, because the only directed families of closed sets of cardinality at most 1 consists of just one set {n} with at most one element, or just the empty set, or just one of each kind. Now {0,1} is in U, and if U were open in the lower Vietoris topology, there would be finitely many open subsets U1, …, Un of Ncof such that {0,1} ∈ ◊U1 ∩ … ∩ ◊Un ⊆ U. No Ui can be empty, so their complements Ci are finite sets, by definition of the cofinite topology. Let m be any number outside C1 ∪ … ∪ Cn. Then {m} is in ◊U1 ∩ … ∩ ◊Un, but has cardinality 1, hence is not in U. Therefore U is a Scott-open subset of H(X) that is not open in the lower Vietoris topology.
De Brecht and Kawai’s example is not Ncof, rather its sobrification S(Ncof). That difference does not matter much: a space and its sobrification always have order-isomorphic lattices of open subsets (Lemma 8.2.26 in the book), hence have order-isomorphic lattices of closed subsets. In turn, those order-isomorphisms are homeomorphisms for the Scott topologies (which are entirely determined by the order), and also for the lower Vietoris topologies (which, being equal to the upper topologies, are also entirely determined by the order). The point in considering the sobrification of Ncof is that S(Ncof) is quasi-Polish: therefore de Brecht and Kawai conclude that requiring that X be quasi-Polish is not enough to guarantee that the Scott and lower Vietoris topologies coincide on H(X).
S(Ncof) is also sober and Noetherian, hence spectral, hence stably compact. Therefore none of these (pretty strong, too!) properties are sufficient to ensure that the Scott and lower Vietoris topologies coincide on H(X) either.
Another source of counterexamples is given by Chen, Kou and Lyu in the paper I cited in the introduction [2, Proposition 3.15]: for every T1 space X whose topology is not discrete, the Scott topology on H(X) is strictly finer than the lower Vietoris topology. This includes our first example, Ncof.
Explicitly, the argument runs as follows. Since the topology of X is not discrete, there is a subset A of X whose closure cl(A) is different from A. Indeed, otherwise, every subset of X would be closed, hence every subset of X would be open, meaning that the topology of X would be discrete. We pick an element x in cl(A)–A, and we consider the collection C of all sets of the form {a} with a in A, plus the empty set. The sets {a} are all in H(X), because X is T1. Additionally, C is Scott-closed: the argument is as for Ncof, and rests on the fact that directed families of elements of C contain at most one set {a}, plus (optionally) the empty set. If C were closed in the lower Vietoris topology, and realizing that {x} is in its complement U, there would be finitely many open subsets U1, …, Un of Ncof such that {x} ∈ ◊U1 ∩ … ∩ ◊Un ⊆ U. Then x would be in every Ui. We let U ≝ U1 ∩ … ∩ Un. Then U intersects cl(A) (at x), hence U also intersects A. Let a be any element in the intersection of U and A. Then {a} is in ◊U1 ∩ … ∩ ◊Un, hence in U. But that is impossible, since {a} is in C, by definition of C.
Let me sum up. None of the following conditions on X is enough to guarantee that the Scott and lower Vietoris topologies agree on H(X):
- quasi-Polish
- Noetherian
- sober Noetherian
- spectral
- stably compact
- T1 non-discrete.
Easy positive answers
We now turn to cases where the Scott and lower Vietoris topologies do agree on H(X).
First, if X has the discrete topology. In that case, H(X) is simply the powerset P(X) of X, a well-known algebraic dcpo. Its Scott topology has a basis of open sets of the form ↑HE, where E ranges over the finite subsets of X, and ↑H denotes upward closure in H(X) (not in X). Writing E as {x1, …, xn}, we see that ↑HE is equal to the collection of (closed) subsets of X that intersect {x1}, …, {xn}, namely to the lower Vietoris open set ◊{x1} ∩ … ∩ ◊{xn}.
Hence and completing the picture we had started to draw at the end of the former section, for a T1 space X, the Scott and lower Vietoris topologies coincide on H(X) if and only if X has the discrete topology (!).
Leaving the realm of T1 spaces, the Scott and lower Vietoris topologies also agree on H(X) if X has the Alexandroff topology of some preordering ≤. (This includes the previous case, since the discrete topology if the Alexandroff topology of =, and the equality relation = is a preordering.) Instead of proving it, we will observe that the Scott and lower Vietoris topologies coincide on H(X) as soon as X is a c-space; and that every space with an Alexandroff topology is a c-space.
A c-space is a space X that satisfies the following strong form of local compactness: for every point x, for every open neighborhood U of x, there is a point y of U such that x is in the interior of ↑y — namely, x ∈ int(↑y) ⊆ ↑y ⊆ U. If X comes with the Alexandroff topology of some preordering ≤, then its open subsets are the upwards-closed subsets with respect to ≤, by definition, and therefore the interior of any set of the form ↑y is ↑y itself. It follows that, in the search for a point y such that x ∈ int(↑y) ⊆ ↑y ⊆ U, we can simply take x for y.
Perhaps the most important class of c-spaces consists of the continuous dcpos, in their Scott topology. In fact, the continuous dcpos are exactly the sober c-spaces (Proposition 8.3.36), and the sobrification of any c-space is a continuous dcpo in its Scott topology (Corollary 8.3.44). It may be interesting to do the following sanity check: for a space X whose topology is the Alexandroff topology of some partial ordering ≤, we have seen that X is a c-space, so its sobrification S(X) should be a c-space; and indeed, in that case, S(X) is the ideal completion of (X, ≤) (Fact 8.2.49), which is even an algebraic dcpo (Proposition 5.1.46).
All right, so I said that the Scott and lower Vietoris topologies coincide on H(X) as soon as X is a c-space… but we have already shown this! See the last proposition in this old post, which shows much more: in that case, H(X) is a continuous dcpo, even a completely distributive complete lattice, and the theorem right before that one establishes that it is also stably compact.
Quite an impressive conclusion! This deserves a formal statement.
Proposition. If X is a c-space, then H(X) is a continuous dcpo, and in fact a completely distributive complete lattice. Its Scott topology coincides with the lower Vietoris topology, and then H(X) is stably compact.
As a side-note, the usual Hoare powerdomain considered in computer science, consisting of non-empty closed subsets, enjoys the same properties, except that it is stably locally compact instead of stably compact.
The Chen-Kou-Lyu property
Can we relax the requirement of being a c-space for X?
For one part, no: X has to be a c-space if you wish H(X) to be a continuous dcpo. I will skip the proof here. This can be deduced from Exercise 8.3.16, Exercise 8.3.18, Lemmata 8.3.41 and 8.3.42 of the book, and working a bit to show that if H(X) is a continuous dcpo, then it is prime-continuous; then O(X) is prime-continuous, and from that we can conclude that X is a c-space. (Lemmata 8.3.41 and 8.3.42 in the book only prove this when X is sober, but it is easy to deduce the general case by showing that X is a c-space if and only if S(X) is, and that H(X) is prime-continuous if and only if H(S(X)) is).
But what about the coincidence of the Scott and lower Vietoris topologies? Chen, Kou and Lyu [2] show that this holds for spaces with rather less stringent requirements than being a c-space… provided we also require them to be posets in their Scott topology. Let me write Xσ for a poset X seen as a topological space with its Scott topology.
Theorem [2, Proposition 3.9, Theorem 3.13]. If X is a poset in its Scott topology, which is either core-compact or first-countable, then the Scott topology coincides with the lower Vietoris topology on H(Xσ).
Proof. The plan of the proof is to show that:
- Under the given assumptions, X has what I will call the Chen-Kou-Lyu property: the Scott topology on Xn coincides with the product topology, for every natural number n.
- For every poset X that has the Chen-Kou-Lyu property, the Scott topology coincides with the lower Vietoris topology on H(X).
As far as point 1 is concerned, let me clarify: given two posets X and Y, one can form two distinct products: the poset product X × Y, which one can then equip with the Scott topology (of the product ordering), yielding a topological space (X × Y)σ; or one can form the topological spaces Xσ and Yσ by giving each poset the Scott topology, and then form the topological product Xσ × Yσ. The spaces (X × Y)σ and Xσ × Yσ, have topologies that I will simply call the Scott and the product topology on X × Y respectively, and while the Scott topology is always finer than the product topology, it can be strictly finer. The Chen-Kou-Lyu property requires that the Scott and product topologies agree on any n-fold product of X with itself.
Let us prove part 1, which I will split into 1(a) core-compact posets and 1(b) first-countable posets.
1(a) We claim that every core-compact poset (in the Scott topology) has the Chen-Kou-Lyu property.
This rests on the following result, which one can find as Theorem II-4.13 of [3], up to minor variations. (Added, June 19th, 2023: another, direct proof can be found as Theorem 2.5.7 in Xiaodong Jia’s PhD thesis: Meet-Continuity and Locally Compact Sober Spaces. PhD thesis, University of Birmingham, 2018.)
Proposition. Let X be a poset. Then Xσ is core-compact if and only if the Scott and product topologies coincide on X × Y for every poset Y (resp., for every complete lattice Y, resp. for the special complete lattice O(Xσ)).
Proof. We use the fact that a space X is core-compact if and only if (∈) ≝ {(x, U) | x ∈ U, U ∈ O(X)} is open in the topological product X × O(X)σ (Exercise 5.2.7 in the book).
Let X be a poset. If Xσ is core-compact, then (∈) is open in the topological product Xσ × O(Xσ)σ. Let Y be any poset, and let W be any Scott-open subset of X × Y, namely any open subset of (X × Y)σ. We form the map f : Y → O(Xσ) defined by f(y) ≝ {x ∈ X | (x, y) ∈ W}. It is easy to see that f(y) is Scott-open for every y in Y, and that f itself is Scott-continuous. Hence idX × f is continuous from Xσ × Yσ (with the product topology) to Xσ × O(Xσ)σ (with the product topology again). The inverse image of (∈) by idX × f is then open in Xσ × Yσ; but that inverse image is just W, so W is open in Xσ × Yσ. Since conversely every open subset of Xσ × Yσ is open in (X × Y)σ, the two topologies agree.
If the Scott and product topologies coincide on X × Y for every poset Y, they certainly again do so for every complete lattice Y, and in particular when Y=O(Xσ). If that is the case, then we note that (∈) is Scott-open (that holds for any poset X), hence open in the product topology. Therefore Xσ is core-compact. ☐
1(b) Next, we claim that for any poset X such that Xσ is first-countable, then X has the Chen-Kou-Lyu property. Indeed, we have seen in an older post that if X and Y are two posets that are first-countable in their Scott topologies, then the Scott and product topologies agree on X × Y. This is due to Matthew de Brecht, who claims that the arguments are due to Matthias Schröder. (The authors of [2] refer to my post for that, but please do not think that the result is mine!)
2. Finally, we show that any poset X that has the Chen-Kou-Lyu property (the Scott and product topologies agree on any finite power of X) is such that the Scott and lower Vietoris topologies coincide on H(X).
It suffices to show that any Scott open subset U of H(X) is open in the lower Vietoris topology. To this end, let C be any element of U. We will show that there is an open neighborhood V of C in the lower Vietoris topology that is included in U, and that will be it.
C is a closed subset of X, and is equal to the directed union of the so-called finitary closed subsets ↓E of C; by finitary closed, I mean the downward closure (in X) of a finite set E. The closure of that directed union of cl(C)=C, so C is also the supremum (the closure of the union) of the directed family of finitary closed subsets ↓E of C.
Since U is Scott-open, one of those finitary closed sets is in U already, say ↓E with E = {x1, …, xn}.
We build a map f : Xn → H(Xσ) by letting f(x’1, …, x’n) ≝ ↓{x’1, …, x’n} for all x’1, …, x’n ∈ X. By definition, f(x1, …, xn) = ↓E is in U, so (x1, …, xn) is in f–1(U).
We claim that the map f is Scott-continuous. Since f(x’1, …, x’n) = ↓x’1 ∪ … ∪ ↓x’n and union is clearly Scott-continuous, it suffices to show that the map x’ ↦ ↓x’ is Scott-continuous from X to H(Xσ). That map is clearly monotonic, and if x’ is the supremum of a family (x’i)i ∈ I, then we need to show that ↓x’ is the supremum of the directed family (↓x’i)i ∈ I in H(Xσ). For that, it is equivalent to show that the upper bounds of ↓x’ in H(Xσ) (namely, the Scott-closed sets that contain x’) are exactly the upper bounds of the family (↓x’i)i ∈ I in H(Xσ) (namely, the Scott-closed sets that contain every x’i). That is clear, since by definition the Scott-closed subsets are closed downwards and closed under directed suprema.
We now use the Chen-Kou-Lyu property: f–1(U) is open in Xn with the Scott topology, hence also with the product topology. Since f–1(U) contains (x1, …, xn), there is an open neighborhood of (x1, …, xn) in Xn, of the form U1 × … × Un where each Ui is open in X, and which is included in f–1(U).
We let V ≝ ◊U1 ∩ … ∩ ◊Un. This is the desired open subset in the lower Vietoris topology. We check that our original closed set C is in V: C contains ↓E, and E = {x1, …, xn}, so C intersects each Ui (at xi). And we check that V is included in U: every closed set C’ in V intersects every Ui, say at x’i; then (x’1, …, x’n) is in f–1(U), so f(x’1, …, x’n) = ↓{x’1, …, x’n} is in U; finally, C’ contains ↓{x’1, …, x’n}, and since U is upwards-closed with respect to inclusion, C’ is in U. ☐
It is slightly puzzling that one would need X to be a poset in its Scott topology (in addition to being core-compact or first-countable) for those nice results to hold, and in particular for the Scott and lower Vietoris topologies to coincide. However, let me recall that without this assumption, the result fails even on the sober, Noetherian, quasi-Polish space S(Ncof), and on all T1, non-discrete spaces.
- Matthew de Brecht and Tatsuji Kawai. On the commutativity of the powerspace constructions. Logical Methods in Computer Science, 15(3), 2019.
- Yu Chen, Hui Kou, and Zhenchao Lyu. Two topologies on the lattice of Scott closed subsets. Topology and its Applications, 306, 107918, 2022.
- 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 (March 20th, 2022)