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 T_{0} 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 T_{0} space is sober. Interestingly, there is a more directed proof: as we will see, every core-compact well-filtered T_{0} space is locally compact, and this will be much simpler. We will then conclude, since every locally compact well-filtered T_{0} 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

*A*

_{n},

*n*∈ ℕ, (i.e.,

*A*

_{0 }⊇

*A*

_{1 }⊇ … ⊇

*A*⊇ …), such that

_{n}*is equal to the collection of open sets*

**F***U*that contain some

*A*

_{n}. (Note that I am not requiring that the sets

*A*

_{n}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

*A*

_{n},

*n*∈ ℕ, where

*A*

_{0 }⊇

*A*

_{1 }⊇ … ⊇

*A*⊇ …. We recall that this means that the open sets

_{n}*U*in

*are exactly those that are supersets of at least one*

**F***A*

_{n}.

We start as in the proof of the Hofmann-Mislove theorem. We let *Q* be the intersection ∩* F* of all the open sets in

*. We claim that*

**F***is exactly the family of open neighborhoods of*

**F***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

*of open neighborhoods of*

**E***Q*that are not in

*is therefore non-empty. With the inclusion ordering,*

**F***is a dcpo, exactly because*

**E***is Scott-open. Hence we can use Zorn’s Lemma: there is a*

**F***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

*, it does not contain any*

**F***A*, so there is a point

_{n}*x*in

_{n}*A*and not in

_{n}*U*, for every

*n*∈ ℕ.

Let *K _{n}* be the set ↑{

*x*

_{m}|

*m*≥

*n*}=↑{

*x*,

_{n}*x*

_{n}_{+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

*K*is compact, but we claim that it is nonetheless. Let (

_{n}*V*

_{i})

_{i ∈ I}be an open cover of

*K*. We wish to extract a finite subcover:

_{n}- First, we imagine that for every index
*i*∈*I*, there are infinitely many numbers*m*≥*n*such that*V*_{i}does not contain*x*_{m}. We will soon see that this is impossible.

For every index*i*, for every number*k*,*A*_{k}is not included in*U*∪*V*_{i}: indeed, picking some number*m*above both*n*and*k*such that*V*_{i}does not contain*x*_{m}, we have that*x*_{m}is in*A*hence also in the larger set_{m}*A*while_{k,}*x*_{m}is not in*U*and not in*V*_{i}by construction.

Since the elements ofare exactly the open supersets of some**F***A*_{k}, we have just shown that for every index*i*∈*I*,*U*∪*V*_{i}is not in. However, remember that**F***U*was chosen maximal among the open sets (containing*Q*) that are not in. Therefore,**F***U*∪*V*_{i}=*U*. Equivalently,*V*_{i}is included in*U*, and this is true for every*i*∈*I*.

It follows that*K*⊆ ∪_{n}_{i}_{∈ I}*V*_{i}⊆*U*, which is impossible since*x*(for example) is in_{n}*K*but not in_{n}*U*. - Since the previous case is impossible, there is an index
*j*, and there is a number*k*≥*n*such that*V*_{j}*x*_{m}for every*m*≥*k*. Then we can extract a finite subcover as follows:*x*is in ∪_{n}_{i}_{∈ I}*V*_{i}hence in some*V*_{i[n] }(I write*i*[*n*] instead of*i*_{n}*,*because HTML does not know about double indices), similarly*x*_{n}_{+1}is in some*V*_{i[n+1]}, …,*x*_{m–}_{1}is in some_{ }*V*_{i[m–1]}, and the remaining points are all in*V*_{j}; therefore*V*_{i[n]},*V*_{i[n+1]}, …,*V*_{i[m–1]}, and*V*_{j}form a finite open cover of*K*._{n}

We have just shown that every open cover (*V*_{i})_{i ∈ I} of *K _{n}* contains a finite sub cover. Therefore

*K*is compact, as promised. It is of course saturated as well.

_{n}Note that, as in the de Brecht-Kawai(-Schröder) argument, the fact we can index the sets *A _{n}* by natural numbers, and not by an arbitrary directed preordered set, is crucial. We can afford this, because

*is countably generated.*

**F**Now we use that *X* is well-filtered at last. The family (*K*_{n})_{n ∈ ℕ} is filtered (in fact a descending chain), so *K*=∩_{n} _{∈ ℕ} *K*_{n} is compact saturated. Also, no *K*_{n} is included in *U*, since *x _{n}* is in

*K*

_{n}but not in

*U*, so well-filteredness tells us that

*K*is not included in

*U*either.

We also have that *K*_{n} is included in *A _{n}* for every number

*n*. Indeed, every point

*x*

_{m}with

*m*≥

*n*is in

*A*hence in

_{m}*A*. It follows that

_{n}*K*is included in every element

*U*of our original countably generated Scott-open filter

*: for every such*

**F***U*,

*U*is a superset of some

*A*, which then contains

_{n}*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

*. Therefore*

**F***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 (*V*_{i})_{i ∈ I} be a directed open cover of *Q*, so ∪_{i} _{∈ I} *V*_{i} contains *Q*, and by the result we have just shown, ∪_{i} _{∈ I} *V*_{i} is in * F*; since

*is Scott-open, some*

**F***V*

_{i}is in

*, so*

**F***V*

_{i}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* (*K*_{n})_{n ∈ ℕ} of compact saturated sets, and any open set *U* such that ∩_{n} _{∈ ℕ} *K*_{n} is included in *U*, some *K*_{n} 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*∈

**, there is a Scott-open filter**

*U**such that*

**F***U*∈

*⊆*

**F****. The way**

*U**is built is clever, and is a construction of J. Lawson’s: since*

**F***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

*U*

_{0}⋐

*U*of

**, then another element**

*U**U*

_{1}

*⋐*

*U*

_{0}, then

*U*

_{2}⋐

*U*

_{1}

*,*etc., all in

*. We define*

**U***as the collection of open sets*

**F***V*such that

*U*

_{n}⋐

*V*for some

*n*(showing that

*is Scott-open), or equivalently such that*

**F***U*

_{n}is included in

*V*for some

*n*(which shows that the intersection of any two element of

*is still in*

**F***).*

**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*=∩

*is compact saturated. Clearly*

**F***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 T_{0} space is sober (Proposition 8.3.8 in the book). Hence we obtain the announced theorem:

**Theorem.** Every core-compact well-filtered T_{0} 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 T
_{0}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 T
_{0}spaces determined by well-filtered spaces. arXiv 1909.09303, September 2019. Submitted.

— Jean Goubault-Larrecq (October 20th, 2019)