Xiaodong Jia once asked the following question: is every core-compact, well-filtered space automatically locally compact? The question was solved positively this year by J. Lawson and X. Xi [2]. They show that every core-compact, well-filtered T0 space is in fact sober, and we know that every sober core-compact space is locally compact (Theorem 8.3.10 in the book).

This is not at all trivial, and, according to somebody I know who had the opportunity to read an early version of [2], this rests heavily on a construction that Wu, Xi, Xu and Zhao obtained earlier: the well-filterification of a space [1]. I will explain what this is below. I will not explain how you can build on this and show the solution to X. Jia’s question: this will be for next time.

The notion of well-filterification is interesting in its own right, and is similar to sobrification, where ‘well-filtered T0‘ replaces ‘sober’. However, the construction of the well-filterification proposed by Wu, Xi, Xu and Zhao is rather abstract: they show that it exists, as a kind of smallest well-filtered subspace of the sobrification.

Even more recently, in fact a few days ago, X. Xu, Ch. Shen, X. Xi and D. Zhao came up with a much more concrete description of the well-filterification, and with a new proof that core-compactness and well-filteredness imply sobriety [3]. This time, and next time, I wil describe the problem and their solution. Since they also offer a simpler way of understanding well-filterifications, I will start by explaining it today.

Some parts will be technical, by necessity, but there are at least two very interesting constructions involved in the process: first, a new notion of so-called WD sets, and second, a very interesting topological extension of Rudin’s Lemma due to Heckmann and Keimel [4] (or rather, a further refinement of it).


Well, first, if you manage to pronounce that (“well-filterification”), congratulations! I am only starting to be able to do so…

A sobrification S(X) of a space X is a sober space, together with a continuous map η : XS(X), with the following universal property: every continuous map f from X to a sober space Y extends along η to a unique continuous map from S(X) to Y. (By ‘extending along η’, I mean that this map composed with η gives you back f. Think of η as a form of subspace inclusion.)

If you are into abstract thinking: in categorical terms, the sobrification functor is left-adjoint to the inclusion functor from the category Sob of sober spaces into the category Top of topological spaces.

Does this work if you replace Sob by the larger category Wfil0 of well-filtered T0 spaces? Namely, is there a well-filterification functor Wf that would be left adjoint to the inclusion functor from Wfil0 into Top? This is what is proved in [1].

The original solution is rather abstract, but the more recent description of it by Xu, Shen, Xi, and Zhao [3] is pretty simple. It all rests on the new notion of WD subset [3, Definition 6.1]. “WD” means well-filtered determined, in case you would wonder.

WD subsets

A WD subset of a topological space X is a subset A such that, for every continuous map f from X to a well-filtered T0 space, there is a (necessarily unique) point y in Y such that cl(f[A])=↓y. (By cl I mean closure, and f[A] is the image of A by f.)

The definition may seem curious, but here is why it is forced on us.

If X has a well-filterification Wf(X), then the inclusion of X into its sobrification S(X) must have a unique extension as a map from Wf(X) to S(X). One can check that this extension must be a topological embedding. This shows that, up to homeomorphism, Wf(X) must be a subspace of S(X).

Now imagine that Wf(X) exists and is a subspace of S(X). Let η map x to its closure ↓x. Given any continuous map f from X to a well-filtered T0 space Y, f should extend along η to a unique continuous map f’ from Wf(X) to Y. For every A in Wf(X), we inquire what f’(A) may be. For every open subset V of Y, f’-1(V) is an open subset ♢U of Wf(X). (We write ♢U for the intersection of the open subset ♢U of S(X) with Wf(X), by a small abuse of language.) Since f’ o η = f, we have the equality f-1(V) = η-1(♢U) = U. Hence the open neighborhoods V of f’(A) are those such that f-1(V) intersects A; equivalently, such that f[A] intersects V, or again such that cl(f[A]) intersects V. Since the open neighborhoods V of f’(A) are exactly the open sets V that intersect ↓f’(A), we have two closed sets, ↓f’(A) and cl(f[A]), which must intersect the same open neighborhoods. Therefore they must be equal. But, for that to make sense, we must require that A is a set such that cl(f[A]) is the downward closure of a point y in Y, for every continuous map f from X to a well-filtered T0 space Y… that means that A must be a WD set.

Hence we should define Wf(X) as the set of irreducible closed WD subsets A of X. However, ‘irreducible’ is redundant here:

Lemma. Every closed WD subset of X is irreducible closed.

Proof. Take Y=S(X), which is sober, hence in particular well-filtered T0, and consider the continuous map f=η that sends x to its closure ↓x. Since A is WD, there is a unique point (a unique irreducible closed set) C such that cl(f [A])=↓C. For every open subset U of X, the open subset ♢U of S(X) intersects cl(f [A]) if and only if it intersects f [A], if and only if there is an x in A such that ↓x ∈ ♢U (i.e., xU), if and only if A intersects U. And ♢U intersects ↓C if and only if C intersects U. Since A is closed, and intersects the same open sets as C, A=C. Therefore A is irreducible closed. ☐

One can also show that every directed subset of X is a WD subset. See Appendix A if you are curious about the proof. I would like to proceed directly to the study of the set Wf(X) of closed WD subsets.

The well-filterification Wf(X)

Following [3, Section 7], we simply define the well-filterification Wf(X) of X as the collection of all closed WD subsets of X. Since every closed WD subset is irreducible closed, Wf(X) is a subset of S(X), and we give the former the subspace topology. As before, we will use again the notation ♢U to denote the open subsets of Wf(X).

Let us check that this satisfies the universal property of well-filterifications. There is a map η from X to Wf(X), which maps every x to ↓x: ↓x is WD, for example because it is directed, but it is easy to verify this directly. It is continuous, and we need to show that given any continuous map f from X to a well-filtered T0 space Y, f extends along η to a unique continuous map f’ from Wf(X) to Y.

If f’ exists, then for every WD subset A of X, we have already seen that f’(A) must be the unique point y of Y such that cl(f[A])=↓y. This shows the uniqueness part. As far as existence is concerned, let us define f’(A) as the unique point y of Y such that cl(f[A])=↓y. The inverse image of every open subset V of Y is the set ♢f-1(V). (Exercise! You will realize we have already done that check above.) Therefore f’ is continuous. And for every x in X, cl(f[↓x])=↓f(x), so f’ extends f along η.

Are we finished? No: it remains to check that Wf(X) is T0 (which is easy), and well-filtered… and that is technical. We must first go through a very useful result, due to R. Heckmann and K. Keimel [4], and nowadays called the topological Rudin Lemma.

Heckmann and Keimel’s topological Rudin Lemma

Rudin’s Lemma is an incredibly useful lemma. It is mentioned, and proved, as Proposition 5.2.25 in the book, and used immediately afterwards.

Consider the set Qfin(X) of all finite non-empty subsets of X, where (for now) X is any poset. We preorder Qfin(X) with the Smyth preordering ≤#, defined by A# B if and only if every element of B is above some element of A, if and only if ↑A contains ↑B (contains, not “is contained in”).

What Rudin’s Lemma says is that we can reduce the study of directed families in Qfin(X) to (specific) directed families in X. Explicitly, given any directed family (Ei)iI in Qfin(X), there is a directed family D in X such that D is included in ∪i I Ei and every set Ei intersects D.

If we replace Qfin(X) by Q(X), the space of all compact saturated subsets of X, and replaced ‘directed’ by ‘irreducible’, then we obtain the following [4, Lemma 3.1]. We topologize Q(X) with the upper Vietoris topology, whose basic open subsets are of the form ☐U (the set of compact saturated sets that are included in U), U open in X. Its complement are the sets ♢F (F closed in X) of all compact saturated sets that intersect F. We will also say that a set A (not necessarily closed) is irreducible if and only if its closure is, namely, if and only if every finite collection of closed sets whose union is a superset of A, contains one element that is already a superset of A.

Proposition (Heckmann and Keimel’s topological Rudin Lemma). Let A be an irreducible closed subset of Q(X). Every closed subset F of X that intersects every element of A contains an irreducible closed subset C that still intersects every element of A.

Before I give its proof, we should notice that this is a generalization of Rudin’s original lemma: if we give X the Alexandroff topology of an ordering ≤, then the elements of Q(X) are exactly the sets ↑A with AQfin(X), the upper Vietoris topology on Q(X) is the Alexandroff topology of reverse inclusion, and the irreducible subsets of a space with an Alexandroff topology are exactly the directed sets; a wee bit of work then shows that we retrieve Rudin’s Lemma as a corollary.

The proof of the topological Rudin Lemma is very similar to the proof of Rudin’s original lemma. We use Zorn’s Lemma to show that there is a minimal closed subset C of F that still intersects every element of A, then we show that every such minimal closed set must be irreducible. Let me decompose this in two steps.

Lemma A. Let A be an irreducible closed subset of Q(X). Every closed subset F of X that intersects every element of A contains an minimal closed subset C that still intersects every element of A.

Proof. Let E be the collection of closed subsets of F that intersect every element of A. E is non-empty, since F is in E by assumption. We order E by reverse inclusion, and we observe that this turns E into a dcpo. Indeed, for every directed family (under reverse inclusion, hence filtered under inclusion) (Fi)i I in E, let F be the intersection of that family. This is a closed set. For every Q in A, every Fi intersects Q, so F also intersects Q, because Q is compact. Hence F intersects every element of A, and therefore is in E.

Since E is a dcpo, and contains F, by Zorn’s Lemma it contains a maximal element C above F—maximal under reverse inclusion, hence minimal subset of F with respect to inclusion. ☐

Lemma B. Let A be an irreducible closed subset of Q(X). Every minimal closed subset C that intersects every element of A is irreducible.

Proof. Imagine that C is included in a finite union of closed sets F1, …, Fn. If C is included in no Fi, then CFi is strictly included in C for every i, 1≤in. By the minimality of C, no CFi is in E. Hence, for each i, there is an element Qi of A that does not intersect CFi. This can be restated by saying that A is not included in ♢(CFi). The latter form a finite family of closed subsets of Q(X). Since A is irreducible, if it were included in the finite union ∪i=1n ♢(CFi), it would be included in some ♢(CFi), and that is not the case. Therefore there is an element Q of A which is not in ∪i=1n ♢(CFi), namely which is disjoint from every CFi. Then Q is disjoint from ∪i=1n (CFi), which is equal to C since C is included in the union of F1, …, Fn. That is impossible since C intersects every element of A. ☐

This applies in particular when A is a filtered family of compact saturated subsets of X. However, in that case, Xu, Shen, Xi and Zhao notice that we can say more:

Lemma C. Let A=(Ki)i I be a filtered family of compact saturated subsets of X. Every minimal closed subset C that intersects every element of A is a closed WD set.

Proof. Let f be any continuous map from X to a well-filtered T0 space Y. The trick is to consider the family (↑f[KiC])i ∈ I. (You may be tempted to use (↑f[Ki])i ∈ I, but that would not work.)

Note that KiC is compact, being the intersection of a compact set and of a closed set, and non-empty. Therefore (↑f[KiC])i ∈ I is a filtered family of non-empty compact saturated subsets of Y. It is easy to see that each set ↑f[KiC] intersects the closed set cl(f[C]). Since Y is well-filtered, ∩i ∈ If[KiC] intersects cl(f[C]), say at y.

Let us look at the closed set f-1(↓y) ∩ C in X. For every i, since y is in ↑f[KiC], there is a point x in KiC such that f(x)≤y: so x is also in f-1(↓y), and that shows that f-1(↓y) ∩ C intersects Ki. Since f-1(↓y) ∩ C intersects every Ki, and is included in C, the minimality of C implies that f-1(↓y) ∩ C = C. In other words, for every x in C, f(x)≤y. It follows that f[C], hence also cl(f[C]), is included in ↓y. But remember that y is in cl(f[C]), so ↓y is also included in cl(f[C]). This shows that cl(f[C])=↓y, and we are done. ☐

Wf(X) is well-filtered

Let A=(Ki)i I be a filtered family of compact saturated subsets of Wf(X). (Yes, sorry, this is very higher-order: families of subsets of a set of subsets…) We assume that ∩i I Ki is included in some open subset of Wf(X). That open subset must be of the form ♢U for some open subset of X.

We reason by contradiction, and we assume that no Ki is included in ♢U. Letting F be the complement of U, this means that Ki intersects the complement ☐F of ♢U. (☐F is the set of compact saturated sets included in F.) Then ☐F is closed and intersects every element Ki of A. By Lemma A and Lemma C (not just Lemma B), ☐F contains a closed WD subset that still intersects every Ki.

We remember that every open subset of Wf(X) is the diamond of an open subset of X, so every closed subset is a box ☐C of some closed subset C of X. Henceforth, let us write ☐C for the closed subset of ☐F that we have just found, and which intersects every Ki.

We claim that C must be a WD subset of X. Let f be an arbitrary continuous map from X to a well-filtered T0 space Y. We have seen that f extends along η to a continuous map f’ from Wf(X) to Y: for every WD subset A of X, f’(A) is the unique point y of Y such that cl(f[A])=↓y. Since ☐C is a WD subset of Wf(X) (don’t lose track of the spaces in which we are working!), there is a unique point y’ such that cl(f’[☐C])=↓y’. We claim that ↓y’ is equal to cl(f[C]): this is what we need to show that C is a WD set. The open neighborhoods of y’ are exactly the open subsets V of Y that intersect cl(f’[☐C]), hence f’[☐C]. Now V intersects f’[☐C] if and only if f’-1(V) intersects ☐C. We have seen earlier that f’-1(V)=♢f-1(V). And ♢f-1(V) intersects ☐C if and only if there is a WD subset A of X that intersects f-1(V) and is included in C. If such an A exists, then f-1(V) intersects C, and conversely, if f-1(V) intersects C at x, then ↓x is a WD set that intersects f-1(V) and is included in C. Summing up, the open neighborhoods of y’ are exactly the open subsets V of Y such that f-1(V) intersects C. Those are also those such that cl(f[C]) intersects V. We now have two closed sets, cl(f’[☐C])=↓y’ and cl(f[C]), which intersect exactly the same open sets. Therefore they are equal, and this shows the desired claim cl(f[C]) =↓y’.

What have we got so far? We started with a filtered family A=(Ki)i I of compact saturated subsets Ki of Wf(X) such that ∩i I Ki is included in ♢U. We have assumed that no Ki is included in ♢U, we have called F the complement of U, and we have found a WD subset C of X, included in F, such that ☐C intersects every Ki. Since C is in ☐C and Ki is upwards-closed, C is in every Ki. Therefore C is in ∩i I Ki, hence in ♢U. That is impossible! … because C is included in F, which does not intersect U by definition.

Since we have reached a contradiction, it must in fact be the case that some Ki is included in ♢U. This shows that Wf(X) is well-filtered, and we are done. (It is T0, as a subspace of S(X), which is always T0.)

Appendix A: every directed set is WD

We consider a directed subset D of X, and a continuous map f from X to some well-filtered T0 space Y. Since f is monotonic with respect to the respective specialization preorderings, f[D] is directed in Y.

(1) We claim that every directed family (yi)i I in Y has a supremum. In order to show this, we first observe that the family of upwards-closed sets ↑yi, i I, is filtered. The intersection Q of the latter family is the collection of upper bounds of (yi)i I. Since Y is well-filtered, Q is compact saturated and non-empty.

If (Vj)j J is any family of open sets whose union V contains Q, then by well-filteredness again some ↑yi is in V. Hence that yi is in V, and therefore in some Vj. Since Vj is upwards-closed, ↑yi and therefore also the smaller set Q is included in Vj. This shows that Q is supercompact: every open cover of Q contains an open set that already contains Q.

Now, it is known that every supercompact set Q is of the form ↑y for some point y of Y [4, Fact 2.2]; and that point is unique since Y is T0. The proof is easy: since Q is supercompact, for every family of closed sets that each intersect Q, their intersection must again intersect Q (consider the complements of those closed sets, and apply the definition); for every z in Q, clearly ↓z is closed and intersects Q, so the intersection of those closed sets must intersect Q, say at y; then, by construction, y is in Q and below every element z of Q, which proves the claim.

(2) We claim that Y is a monotone convergence space, namely: every directed family has a supremum, and every open subset is Scott-open. This holds for every well-filtered T0 space Y.

This is simply because, for every directed family (yi)i I whose supremum y lies in some open subset U, then ↑y = ∩iyi is included in U, so by well-filteredness some ↑yi is included in U.

All right, back to our problem. We know that f[D] is directed in Y. Since we now know that Y is a dcpo in its specialization ordering, f[D] has a supremum, call it y. Then f[D] is a subset of ↓y, and the latter is closed, so cl (f[D]) is a subset of ↓y. Since Y is monotone convergence, every closed subset is Scott-closed, hence the closure of f[D] is included in its Scott-closure, which is included in ↓y. ☐

  1. Guohua Wu, Xiaoyong Xi, Xiaoquan Xu, and Dongsheng Zhao.  Existence of well-filterifications of T0 topological spaces.  arXiv 1906.10832, July 2019.  Submitted.
  2. Jimmie Lawson and Xiaoyong Xi.  Well-filtered spaces, compactness, and the lower topology.  2019.  Submitted.
  3. Xiaoquan Xu, Chong Shen, Xiaoyong Xi, and Dongsheng Zhao.  On T0 spaces determined by well-filtered spaces.  arXiv 1909.09303, September 2019.  Submitted.
  4. Reinhold Heckmann and Klaus Keimel. Quasicontinuous domains and the Smyth powerdomain. Electronic Notes in Theoretical Computer Science 298 (2013), pp. 215–232.

Jean Goubault-Larrecq