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 subsetsAofXthat 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 [*g*_{1}=*g*_{2}] of two continuous maps *g*_{1}=*g*_{2}: *X* → *Y*, where *Y* is any T_{0} space, is sober. (By the way, *Y* has to be T_{0}. 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 χ_{A} 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 χ_{U }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 **Π**^{0}_{2} subspace of a sober space *X* is sober again. A **Π**^{0}_{2} subset is the intersection of countably many sets *U*_{n}* *⇒ *V*_{n}, where each *U*_{n} and each *V*_{n} is open in *X*. (*U*_{n} ⇒ *V*_{n} is the subset of points *x* such that if *x* is in *U*_{n} then *x* is in *V*_{n}.) 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 *g*_{1}(*x*) as the tuple whose *n*th entry is χ_{Un} (*x*), and *g*_{2}(*x*) as the tuple whose *n*th entry is χ_{Un ∩ Vn} (*x*). Then the equalizer of *g*_{1} and of *g*_{2} is the set of points *x* such that *x* is in *U*_{n} if and only if it is in *U*_{n} ∩ *V*_{n}, and that is exactly our **Π**^{0}_{2} subset. (Alternatively, let *Y* be the powerset of **N**, with its Scott topology, define *g*_{1}(*x*) as the set of natural numbers *n* such that *x* is in *U*_{n}, and *g*_{2}(*x*) as the set of natural numbers *n* such that *x* is in *U*_{n} ∩ *V*_{n}.)

In that argument, note that the fact that we are taking a countable intersection of sets *U*_{n} ⇒ *V*_{n} (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 T_{2.})

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 [*g*_{1}=*g*_{2}] of two continuous maps *g*_{1}=*g*_{2}: *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 cl_{s}(*A*). This is a continuous map, and since cl_{s}(*A*) is sober, *i* has a unique continuous extension *j* from **S**(*A*) to cl_{s}(*A*)—namely, *j* o η_{A} = *i*, where the embedding η_{A} : *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*, η_{A}^{-1}(♢(*U* ∩ *A*))=*U* ∩ *A*. Since *i*^{-1}(*U* ∩ cl_{s}(A))=*U* ∩ *A*, and recalling that η_{A}^{-1} is a frame isomorphism, it follows that *j*^{-1}(*U* ∩ cl_{s}(A))=♢(*U* ∩ *A*) for every open subset *U* of *X*.

In the converse direction, given any point *x* in cl_{s}(*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 cl_{s}(*A*) intersects ↓*x* ∩ *U*, if and only if *x* is in *U* (recall that *x* is in cl_{s}(*A*): if cl_{s}(*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 cl_{s}(*A*) intersects ↓*x* ∩ *U* at *x*). To sum up: (**) for every open subset *U* of *X*, for every *x* in cl_{s}(*A*), *f*(*x*) intersects *U* if and only if *x* is in *U*.

Let us fix *x* in cl_{s}(*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* ∩ cl_{s}(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.