Last time, I motivated the construction of the well-filterification Wf(X) of a space X of X. Xu, Ch. Shen, X. Xi and D. Zhao [3] by saying that it was needed to understand their proof of the fact that every core-compact well-filtered T0 space is sober, and hence also locally compact. This solves a question asked by X. Jia, and first solved positively by J. Lawson and X. Xi [2].
While polishing up Xu, Shen, Xi, and Zhao’s proof , I realized that the key was a new form of the Hofmann-Mislove theorem for well-filtered spaces, which is interesting in its own right. I will describe it and prove it below.
Then I realized that one could simplify their proof. Their argument relies on the following theorem [3, Theorem 6.15]: in a core-compact space, every irreducible closed subset is a WD subset. The proof of the latter is technical, but from there, it is not hard to show that every core-compact well-filtered T0 space is sober. Interestingly, there is a more directed proof: as we will see, every core-compact well-filtered T0 space is locally compact, and this will be much simpler. We will then conclude, since every locally compact well-filtered T0 space is sober (Proposition 8.3.8 in the book).
A Hofmann-Mislove theorem for well-filtered spaces
Remember the Hofmann-Mislove theorem (Theorem 8.3.2 in the book): in a sober space X, every Scott-open filter F of open subsets of X is the family of open neighborhoods of a unique compact saturated set Q, namely ∩F.
You don’t have that in a well-filtered space. However, I claim that this works if F is countably generated, namely if there is a countable descending chain of sets An, n ∈ ℕ, (i.e., A0 ⊇ A1 ⊇ … ⊇ An ⊇ …), such that F is equal to the collection of open sets U that contain some An. (Note that I am not requiring that the sets An be open themselves, but that will not be important in the sequel.)
In the proof I give below, you should recognize arguments similar to the usual proof of the Hofmann-Mislove theorem, combined with arguments similar to those used in the de Brecht-Kawai theorem (given any well-filtered second-countable space X, the upper Vietoris and Scott topologies on Q(X) coincide—Matthew de Brecht insists that the key ideas come from M. Schröder’s papers, and although I still don’t quite understand everything that M. Schröder says, Matthew must be right), where some pretty funny sets are shown to be compact, and this requires countable index sets. The key argument in the following proof is taken from X. Xu, Ch. Shen, X. Xi and D. Zhao’s proof of [3, Theorem 6.15], where it is somehow hidden.
Theorem 1 (à la Hofmann-Mislove). Let X be a well-filtered space. Every countably generated Scott-open filter F of open subsets of X is the family of open neighborhoods of a unique compact saturated set Q, namely ∩F.
Proof. Let F be generated by countably many sets An, n ∈ ℕ, where A0 ⊇ A1 ⊇ … ⊇ An ⊇ …. We recall that this means that the open sets U in F are exactly those that are supersets of at least one An.
We start as in the proof of the Hofmann-Mislove theorem. We let Q be the intersection ∩F of all the open sets in F. We claim that F is exactly the family of open neighborhoods of Q. (Please note that we do not know yet that Q is compact.)
Reasoning by contradiction, we assume that there is an open neighborhood of Q that is not in F. The collection E of open neighborhoods of Q that are not in F is therefore non-empty. With the inclusion ordering, E is a dcpo, exactly because F is Scott-open. Hence we can use Zorn’s Lemma: there is a maximal open neighborhood U of Q that is not in F.
In the traditional proof of the Hofmann-Mislove theorem, we then use the fact that F is a filter to obtain that the complement of U is an irreducible closed set, and we conclude by sobriety… but we do not have sobriety here.
This is the point where we have to use that F is countably generated instead. Since U is not in F, it does not contain any An, so there is a point xn in An and not in U, for every n ∈ ℕ.
Let Kn be the set ↑{xm | m≥n}=↑{xn, xn+1, …}. As in the de Brecht-Kawai(-Schröder) argument, this is the upward closure of an infinite set of points, hence it is not immediately obvious that Kn is compact, but we claim that it is nonetheless. Let (Vi)i ∈ I be an open cover of Kn. We wish to extract a finite subcover:
- First, we imagine that for every index i ∈ I, there are infinitely many numbers m≥n such that Vi does not contain xm. We will soon see that this is impossible.
For every index i, for every number k, Ak is not included in U ∪ Vi: indeed, picking some number m above both n and k such that Vi does not contain xm, we have that xm is in Am hence also in the larger set Ak, while xm is not in U and not in Vi by construction.
Since the elements of F are exactly the open supersets of some Ak, we have just shown that for every index i ∈ I, U ∪ Vi is not in F. However, remember that U was chosen maximal among the open sets (containing Q) that are not in F. Therefore, U ∪ Vi = U. Equivalently, Vi is included in U, and this is true for every i ∈ I.
It follows that Kn ⊆ ∪i ∈ I Vi ⊆ U, which is impossible since xn (for example) is in Kn but not in U. - Since the previous case is impossible, there is an index j, and there is a number k≥n such that Vj contains xm for every m≥k. Then we can extract a finite subcover as follows: xn is in ∪i ∈ I Vi hence in some Vi[n] (I write i[n] instead of in, because HTML does not know about double indices), similarly xn+1 is in some Vi[n+1], …, xm–1 is in some Vi[m–1], and the remaining points are all in Vj; therefore Vi[n], Vi[n+1], …, Vi[m–1], and Vj form a finite open cover of Kn.
We have just shown that every open cover (Vi)i ∈ I of Kn contains a finite sub cover. Therefore Kn is compact, as promised. It is of course saturated as well.
Note that, as in the de Brecht-Kawai(-Schröder) argument, the fact we can index the sets An by natural numbers, and not by an arbitrary directed preordered set, is crucial. We can afford this, because F is countably generated.
Now we use that X is well-filtered at last. The family (Kn)n ∈ ℕ is filtered (in fact a descending chain), so K=∩n ∈ ℕ Kn is compact saturated. Also, no Kn is included in U, since xn is in Kn but not in U, so well-filteredness tells us that K is not included in U either.
We also have that Kn is included in An for every number n. Indeed, every point xm with m≥n is in Am hence in An. It follows that K is included in every element U of our original countably generated Scott-open filter F: for every such U, U is a superset of some An, which then contains K. In turn, this implies that K is included in ∩F = Q. Now recall that U contains Q: U was built as a maximal open neighborhood of Q that is not in F. Therefore K is included in U. This contradicts the conclusion of the previous paragraph!
We have reached a contradiction, so our initial assumption was wrong: F is exactly the family of open neighborhoods of Q.
It remains to show that Q is compact saturated. The saturated part is obvious. For compactness, we proceed as in the final steps of the usual proof of the Hofmann-Mislove theorem: Let (Vi)i ∈ I be a directed open cover of Q, so ∪i ∈ I Vi contains Q, and by the result we have just shown, ∪i ∈ I Vi is in F; since F is Scott-open, some Vi is in F, so Vi is a superset of Q. ☐
Remark. As Xiaoquan Xi mentioned to me (Thursday, September 26th, 2019), the proof above never uses the full power of well-filteredness. Instead, call a space ω-well-filtered if and only if, given any descending sequence (Kn)n ∈ ℕ of compact saturated sets, and any open set U such that ∩n ∈ ℕ Kn is included in U, some Kn is already included in U. Then Theorem 1 still holds under the weaker assumption that X is ω-well-filtered, not well-filtered.
Scott-open filters of open sets
The second important property that we will use—or rather, we will use a refinement of it—is that the Scott topology on the dcpo O(X) of open subsets of a core-compact space X has a base of Scott-open filters. Indeed, since X is core-compact, O(X) is a continuous dcpo, and the Scott topology of every continuous dcpo (even poset) has a base of Scott-open filtered sets: this is Proposition 5.1.19 in the book.
This means that, for every open subset U of X, for every Scott-open subset U of O(X) such that U ∈ U, there is a Scott-open filter F such that U ∈ F ⊆ U. The way F is built is clever, and is a construction of J. Lawson’s: since X is core-compact, O(X) is a continuous dcpo (and I usually write ⋐ for its way-below relation); so we can find an element U0 ⋐ U of U, then another element U1 ⋐ U0, then U2⋐ U1, etc., all in U. We define F as the collection of open sets V such that Un ⋐ V for some n (showing that F is Scott-open), or equivalently such that Un is included in V for some n (which shows that the intersection of any two element of F is still in F).
Fantastically enough, that F we have just built is countably generated. Hence we have:
Lemma 2. For every core-compact space X, the Scott topology on O(X) has a base of countably generated Scott-open filters of open subsets of X.
There is nothing special with O(X) here, and the same argument shows the following more general property, refining Proposition 5.1.19 in the book:
Fact 3. The Scott topology on a continuous poset has a base of countably generated Scott-open filters.
However, this allows us to give a simple proof of:
Proposition. Every core-compact (ω-)well-filtered space is locally compact.
Proof. Let X be a core-compact (ω-)well-filtered space, let x be a point in X, and let U be an open neighborhood of x in X. Since X is core-compact, we can find another open neighborhood V of x such that V ⋐ U. The collection ↟V={W ∈ O(X) | V ⋐ W} is an open neighborhood of U in O(X), so by Lemma 2 there is a countably generated Scott-open filter F included in ↟V and which contains U. We now use Theorem 1 (see also the subsequent Remark in the ω-well-filtered case): the set Q=∩F is compact saturated. Clearly Q is a superset of V, hence an open neighborhood of x, and Q is included in U. ☐
I have already said that every locally compact well-filtered T0 space is sober (Proposition 8.3.8 in the book). Hence we obtain the announced theorem:
Theorem. Every core-compact well-filtered T0 space is sober. ☐
Conclusion
There are two remarkable points to be made here.
The first one is that we can summarize the whole situation as follows. Since every core-compact sober space is locally compact, and since every locally compact well-filtered space is sober, we have the following situation. There is an array of implications:
- sober ⇒ well-filtered
- locally compact ⇒ core-compact,
and any pair of properties, one taken from each line, implies all of them.
The second remarkable thing is that countability entered the picture in a rather unexpected way here—again. That still amazes me.
- Guohua Wu, Xiaoyong Xi, Xiaoquan Xu, and Dongsheng Zhao. Existence of well-filterifications of T0 topological spaces. arXiv 1906.10832, July 2019. Submitted.
- Jimmie Lawson and Xiaoyong Xi. Well-filtered spaces, compactness, and the lower topology. 2019. Submitted.
- Xiaoquan Xu, Chong Shen, Xiaoyong Xi, and Dongsheng Zhao. On T0 spaces determined by well-filtered spaces. arXiv 1909.09303, September 2019. Submitted.
— Jean Goubault-Larrecq (October 20th, 2019)