It often happens that one wishes to show that a certain subspace A of a given sober space X is sober. The following is a pearl due to Keimel and Lawson [1, Corollary 3.5], which was mentioned to me by Zhenchao Lyu in July:
Given a sober space X, the subsets A of X that are sober as subspaces are exactly its Skula-closed subsets.
My goal today is to comment on that, and to (re)prove it.
In search of sober subspaces
It is not hard to show that if A is a closed subspace of a sober space X, then it is sober. This is also true if A is open.
I have regularly used the following practical result (Lemma 8.4.12 in the book): every subspace A of a sober space X that arises as the equalizer [g1=g2] of two continuous maps g1=g2: X → Y, where Y is any T0 space, is sober. (By the way, Y has to be T0. This is one of the mistakes in the book.)
In particular, if A is open in X, then it arises as the equalizer of its characteristic map χA and the constant map equal to 1 from X to Sierpiński space S (the space {0, 1} with the Alexandroff topology of 0<1). And if A is closed, then it arises as the equalizer of the characteristic map χU of its complement U and the constant map equal to 0 from X to S.
As another application of Lemma 8.4.12 of the book, one can show that every Π02 subspace of a sober space X is sober again. A Π02 subset is the intersection of countably many sets Un ⇒ Vn, where each Un and each Vn is open in X. (Un ⇒ Vn is the subset of points x such that if x is in Un then x is in Vn.) That result was used several times by Matthew de Brecht in the study of quasi-Polish spaces.
How do you prove that? Well, take Y to be the product of countably many copies of S. Define g1(x) as the tuple whose nth entry is χUn (x), and g2(x) as the tuple whose nth entry is χUn ∩ Vn (x). Then the equalizer of g1 and of g2 is the set of points x such that x is in Un if and only if it is in Un ∩ Vn, and that is exactly our Π02 subset. (Alternatively, let Y be the powerset of N, with its Scott topology, define g1(x) as the set of natural numbers n such that x is in Un, and g2(x) as the set of natural numbers n such that x is in Un ∩ Vn.)
In that argument, note that the fact that we are taking a countable intersection of sets Un ⇒ Vn (so-called UCO sets) is entirely irrelevant. Any intersection of UCO subsets of a sober space X, of whatever cardinality, is sober.
The Skula topology
The Skula topology (also called the strong topology) on a topological space X is the topology generated by the open subsets and the closed subsets of X. It is equivalent to define it as the topology generated by the open subsets and the downwards-closed subsets of X, because every downwards-closed subset of X is a union of closed sets (namely the downward closures of single points).
This is a rather remarkable topology. For example, a theorem by R.-E. Hoffmann states that X is Skula-compact (i.e., compact in its Skula topology) if and only if X is sober Noetherian (Exercise 9.7.16 in the book; note that the Skula topology is always T2.)
A subset of X is Skula-closed (i.e., closed in the Skula topology) if and only if it is a (generally infinite) intersection of unions V ∪ C of an open subset V and a closed subset C of X. If you write U for the complement of C, you will realize that such a union V ∪ C is nothing but the UCO subset U ⇒ V. But recall that any intersection of UCO subsets of a sober space is sober in the subspace topology of X. Thus we obtain one half of Keimel and Lawson’s Lemma: every Skula-closed subset A of a sober space X is sober in the subspace topology from X.
In the converse direction, let A be a sober subspace of X, and assume that A is not Skula-closed. Hence its complement is not Skula-open, and that implies that there is a point x in X–A whose Skula-open neighborhoods all intersect A. (Reason by contradiction: otherwise every point in X–A would have a Skula-open neighborhood included in X–A.)
In particular, for every open neighborhood U of x (in the original topology of X), U ∩ ↓x is a Skula-open neighborhood of x, so it must intersect A. Said in another way: (*) every open neighborhood U of x intersects ↓x ∩ A.
Note that ↓x ∩ A is a closed subset of A. We claim that it is irreducible in A.
Using (*) with U=X, we obtain that ↓x ∩ A is non-empty. Given two open subsets U and V of X that intersect ↓x ∩ A, both U and V must contain x, so U ∩ V also contains x, and using (*) we obtain that U ∩ V intersects ↓x ∩ A. It follows that ↓x ∩ A is irreducible closed in A.
Since A is sober, ↓x ∩ A is the closure of some unique point y in A.
In particular, y is in ↓x ∩ A, so y≤x. Since x is in X–A, and y is in A, we have x≠y, so x is not below y. We use (*) with U equal to the complement of ↓y: there is a point in ↓x ∩ A and in U; since ↓x ∩ A is the closure of y in A, and since an open set intersects the closure of a set B if and only if it intersects B itself, y must be in U—but that contradicts the definition of U.
We have proved the promised statement:
Proposition [1, Corollary 3.5]. Given a sober space X, the subsets A of X that are sober as subspaces are exactly its Skula-closed subsets.
This also shows that Lemma 8.4.12 of the book is in a sense optimal: given any sober subspace A of a sober space X, A is the equalizer [g1=g2] of two continuous maps g1=g2: X → Y. We can even take Y to be a power of S (equivalently, a powerset of some set, in its Scott topology). Indeed, A is Skula-closed, hence an intersection of some family of UCO subsets, and we have already seen that any such intersection can be realized as such an equalizer.
Sobrifications
Corollary 3.5 of [1] is—as the name indicates—a corollary of a more general result (I am using the notion of sobrification from the book):
Theorem [1, Proposition 3.4]. Let X be a sober space, and A be a subset of X. The sobrification S(A) of the subspace A is homeomorphic to its Skula-closure in X.
Proof. We look at the inclusion map i from A into its Skula-closure cls(A). This is a continuous map, and since cls(A) is sober, i has a unique continuous extension j from S(A) to cls(A)—namely, j o ηA = i, where the embedding ηA : A → S(A) maps x to its downward closure in A, namely ↓x ∩ A.
Note that, for every open subset of S(A), namely for every set of the form ♢(U ∩ A) where U is open in X, ηA-1(♢(U ∩ A))=U ∩ A. Since i-1(U ∩ cls(A))=U ∩ A, and recalling that ηA-1 is a frame isomorphism, it follows that j-1(U ∩ cls(A))=♢(U ∩ A) for every open subset U of X.
In the converse direction, given any point x in cls(A), we let f(x)=↓x ∩ A. For every open subset U of X, f(x) intersects U if and only if U intersects ↓x ∩ A, if and only if A intersects the Skula-open set ↓x ∩ U, if and only if cls(A) intersects ↓x ∩ U, if and only if x is in U (recall that x is in cls(A): if cls(A) intersects ↓x ∩ U, say at y, then y≤x and y is in U, so x is in U, and conversely if x is in U, then cls(A) intersects ↓x ∩ U at x). To sum up: (**) for every open subset U of X, for every x in cls(A), f(x) intersects U if and only if x is in U.
Let us fix x in cls(A). Using (**) with U empty, we obtain that f(x) is not empty; given any two open sets U and V that each intersect f(x), by (**) we obtain that both U and V contain x, so U ∩ V also contains x, and using (**) again, f(x) intersects U ∩ V. This shows that f(x) is irreducible (closed) in A, hence an element of S(A).
We can now rephrase (**) as: for every open subset U of X, f-1(♢(U ∩ A))=U ∩ cls(A). In particular, f is continuous. It also follows that f-1 is a frame homomorphism that is inverse to the frame homomorphism j-1. Since f and j are continuous maps between sober spaces, they are uniquely determined by the frame homomorphisms f-1 and j-1. It follows that f and j are mutual inverses. ☐
- Klaus Keimel and Jimmie D. Lawson. D-completions and the d-topology. Annals of Pure and Applied Logic 159(3), June 2009, pages 292-306.
— Jean Goubault-Larrecq (August 22nd, 2019)