{"id":2014,"date":"2019-09-24T10:58:57","date_gmt":"2019-09-24T08:58:57","guid":{"rendered":"https:\/\/projects.lsv.ens-paris-saclay.fr\/topology\/?page_id=2014"},"modified":"2022-11-19T15:05:27","modified_gmt":"2022-11-19T14:05:27","slug":"well-filterifications","status":"publish","type":"page","link":"https:\/\/projects.lsv.ens-paris-saclay.fr\/topology\/?page_id=2014","title":{"rendered":"Well-filterifications"},"content":{"rendered":"\n

Xiaodong Jia once asked the following question<\/a>: 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<\/sub> space is in fact sober, and we know that every sober core-compact space is locally compact (Theorem 8.3.10 in the book<\/a>).<\/p>\n\n\n\n

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<\/em> 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.<\/p>\n\n\n\n

The notion of well-filterification is interesting in its own right, and is similar to sobrification, where ‘well-filtered T0<\/sub>‘ 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.<\/p>\n\n\n\n

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.<\/p>\n\n\n\n

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<\/em>, and second, a very interesting topological extension of Rudin’s Lemma due to Heckmann and Keimel [4] (or rather, a further refinement of it).<\/p>\n\n\n\n

Well-filterifications<\/h2>\n\n\n\n

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

A sobrification S<\/strong>(X<\/em>) of a space X<\/em> is a sober space, together with a continuous map \u03b7 : X<\/em> \u2192 S<\/strong>(X<\/em>), with the following universal property: every continuous map f<\/em> from X<\/em> to a sober space Y<\/em> extends along \u03b7 to a unique continuous map from S<\/strong>(X<\/em>) to Y<\/em>. (By ‘extending along \u03b7’, I mean that this map composed with \u03b7 gives you back f<\/em>. Think of \u03b7 as a form of subspace inclusion.)<\/p>\n\n\n\n

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

Does this work if you replace Sob<\/strong> by the larger category Wfil<\/strong>0<\/sub> of well-filtered T0<\/sub> spaces? Namely, is there a well-filterification<\/em> functor Wf<\/strong> that would be left adjoint to the inclusion functor from Wfil<\/strong>0<\/sub> into Top<\/strong>? This is what is proved in [1].<\/p>\n\n\n\n

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<\/em> [3, Definition 6.1]. “WD” means well-filtered determined<\/em>, in case you would wonder.<\/p>\n\n\n\n

WD subsets<\/h2>\n\n\n\n

A WD subset<\/em> of a topological space X<\/em> is a subset A<\/em> such that, for every continuous map f<\/em> from X<\/em> to a well-filtered T0<\/sub> space, there is a (necessarily unique) point y<\/em> in Y<\/em> such that cl(f<\/em>[A<\/em>])=\u2193y<\/em>. (By cl I mean closure, and f<\/em>[A<\/em>] is the image of A<\/em> by f<\/em>.)<\/p>\n\n\n\n

The definition may seem curious, but here is why it is forced on us.<\/p>\n\n\n\n

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

Now imagine that Wf<\/strong>(X<\/em>) exists and is a subspace of S<\/strong>(X<\/em>). Let \u03b7 map x<\/em> to its closure \u2193x<\/em>. Given any continuous map f<\/em> from X<\/em> to a well-filtered T0<\/sub> space Y<\/em>, f<\/em> should extend along \u03b7 to a unique continuous map f’<\/em> from Wf<\/strong>(X<\/em>) to Y<\/em>. For every A<\/em> in Wf<\/strong>(X<\/em>), we inquire what f’<\/em>(A<\/em>) may be. For every open subset V<\/em> of Y<\/em>, f’<\/em>-1<\/sup>(V<\/em>) is an open subset \u2662U<\/em> of Wf<\/strong>(X<\/em>). (We write \u2662U<\/em> for the intersection of the open subset \u2662U<\/em> of S<\/strong>(X<\/em>) with Wf<\/strong>(X<\/em>), by a small abuse of language.) Since f’<\/em> o \u03b7 = f<\/em>, we have the equality f<\/em>-1<\/sup>(V<\/em>) = \u03b7-1<\/sup>(\u2662U<\/em>) = U<\/em>. Hence the open neighborhoods V<\/em> of f’<\/em>(A<\/em>) are those such that f<\/em>-1<\/sup>(V<\/em>) intersects A<\/em>; equivalently, such that f<\/em>[A<\/em>] intersects V<\/em>, or again such that cl(f<\/em>[A<\/em>]) intersects V<\/em>. Since the open neighborhoods V<\/em> of f’<\/em>(A<\/em>) are exactly the open sets V<\/em> that intersect \u2193f’<\/em>(A<\/em>), we have two closed sets, \u2193f’<\/em>(A<\/em>) and cl(f<\/em>[A<\/em>]), which must intersect the same open neighborhoods. Therefore they must be equal. But, for that to make sense, we must require that A<\/em> is a set such that cl(f<\/em>[A<\/em>]) is the downward closure of a point y<\/em> in Y<\/em>, for every continuous map f<\/em> from X<\/em> to a well-filtered T0<\/sub> space Y<\/em>… that means that A<\/em> must be a WD set.<\/p>\n\n\n\n

Hence we should define Wf<\/strong>(X<\/em>) as the set of irreducible closed WD subsets A<\/em> of X<\/em>. However, ‘irreducible’ is redundant here:<\/p>\n\n\n\n

Lemma.<\/strong> Every closed WD subset of X<\/em> is irreducible closed.<\/p>\n\n\n\n

Proof.<\/em> Take Y<\/em>=S<\/strong>(X<\/em>), which is sober, hence in particular well-filtered T0<\/sub>, and consider the continuous map f<\/em>=\u03b7 that sends x<\/em> to its closure \u2193x<\/em>. Since A<\/em> is WD, there is a unique point (a unique irreducible closed set) C<\/em> such that cl(f<\/em> [A<\/em>])=\u2193C<\/em>. For every open subset U<\/em> of X<\/em>, the open subset \u2662U<\/em> of S<\/strong>(X<\/em>) intersects cl(f<\/em> [A<\/em>]) if and only if it intersects f<\/em> [A<\/em>], if and only if there is an x<\/em> in A<\/em> such that \u2193x<\/em> \u2208 \u2662U<\/em> (i.e., x<\/em> \u2208 U<\/em>), if and only if A<\/em> intersects U<\/em>. And \u2662U<\/em> intersects \u2193C<\/em> if and only if C<\/em> intersects U<\/em>. Since A<\/em> is closed, and intersects the same open sets as C<\/em>, A<\/em>=C<\/em>. Therefore A<\/em> is irreducible closed. \u2610<\/p>\n\n\n\n

One can also show that every directed subset of X<\/em> 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<\/strong>(X<\/em>) of closed WD subsets.<\/p>\n\n\n\n

The well-filterification Wf<\/strong>(X<\/em>)<\/h2>\n\n\n\n

Following [3, Section 7], we simply define the well-filterification Wf<\/strong>(X<\/em>) of X<\/em> as the collection of all closed WD subsets of X<\/em>. Since every closed WD subset is irreducible closed, Wf<\/strong>(X<\/em>) is a subset of S<\/strong>(X<\/em>), and we give the former the subspace topology. As before, we will use again the notation \u2662U<\/em> to denote the open subsets of Wf<\/strong>(X<\/em>).<\/p>\n\n\n\n

Let us check that this satisfies the universal property of well-filterifications. There is a map \u03b7 from X<\/em> to Wf<\/strong>(X<\/em>), which maps every x<\/em> to \u2193x<\/em>: \u2193x 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<\/em> from X<\/em> to a well-filtered T0<\/sub> space Y<\/em>, f<\/em> extends along \u03b7 to a unique continuous map f’<\/em> from Wf<\/strong>(X<\/em>) to Y<\/em>.<\/p>\n\n\n\n

If f’<\/em> exists, then for every WD subset A<\/em> of X<\/em>, we have already seen that f’<\/em>(A<\/em>) must be the unique point y<\/em> of Y<\/em> such that cl(f<\/em>[A<\/em>])=\u2193y<\/em>. This shows the uniqueness part. As far as existence is concerned, let us define f’<\/em>(A<\/em>) as the unique point y<\/em> of Y<\/em> such that cl(f<\/em>[A<\/em>])=\u2193y<\/em>. The inverse image of every open subset V<\/em> of Y<\/em> is the set \u2662f<\/em>-1<\/sup>(V<\/em>). (Exercise! You will realize we have already done that check above.) Therefore f’<\/em> is continuous. And for every x<\/em> in X<\/em>, cl(f<\/em>[\u2193x<\/em>])=\u2193f<\/em>(x<\/em>), so f’<\/em> extends f<\/em> along \u03b7.<\/p>\n\n\n\n

Are we finished? No: it remains to check that Wf<\/strong>(X<\/em>) is T0<\/sub> (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<\/em>.<\/p>\n\n\n\n

Heckmann and Keimel’s topological Rudin Lemma<\/h2>\n\n\n\n

Rudin’s Lemma is an incredibly useful lemma. It is mentioned, and proved, as Proposition 5.2.25 in the book<\/a>, and used immediately afterwards.<\/p>\n\n\n\n

Consider the set Q<\/strong>fin<\/sub>(X<\/em>) of all finite non-empty subsets of X<\/em>, where (for now) X<\/em> is any poset. We preorder Q<\/strong>fin<\/sub>(X<\/em>) with the Smyth preordering \u2264#<\/sup>, defined by A<\/em> \u2264#<\/sup> B<\/em> if and only if every element of B<\/em> is above some element of A<\/em>, if and only if \u2191A<\/em> contains \u2191B<\/em> (contains, not “is contained in”).<\/p>\n\n\n\n

What Rudin’s Lemma says is that we can reduce the study of directed families in Q<\/strong>fin<\/sub>(X<\/em>) to (specific) directed families in X<\/em>. Explicitly, given any directed family (E<\/em>i<\/em><\/sub>)i<\/em> \u2208 I<\/em><\/sub> in Q<\/strong>fin<\/sub>(X<\/em>), there is a directed family D<\/em> in X<\/em> such that D<\/em> is included in \u222ai<\/em><\/sub> \u2208 I<\/em><\/sub> E<\/em>i<\/em><\/sub> and every set E<\/em>i<\/em><\/sub> intersects D<\/em>.<\/p>\n\n\n\n

If we replace Q<\/strong>fin<\/sub>(X<\/em>) by Q<\/strong>(X<\/em><\/i><\/span>), the space of all compact saturated subsets of X<\/em>, and replaced ‘directed’ by ‘irreducible’, then we obtain the following [4, Lemma 3.1]. We topologize Q<\/strong>(X<\/em>) with the upper Vietoris topology, whose basic open subsets are of the form \u2610U<\/em> (the set of compact saturated sets that are included in U<\/em>), U<\/em> open in X<\/em>. Its complement are the sets \u2662F<\/em> (F<\/em> closed in X<\/em>) of all compact saturated sets that intersect F<\/em>. We will also say that a set A<\/em> (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<\/em>, contains one element that is already a superset of A<\/em>.<\/p>\n\n\n\n

Proposition (Heckmann and Keimel’s topological Rudin Lemma).<\/strong> Let A<\/strong><\/em> be an irreducible closed subset of Q<\/strong>(X<\/em>). Every closed subset F<\/em> of X<\/em> that intersects every element of A<\/strong><\/em> contains an irreducible closed subset C<\/em> that still intersects every element of A<\/strong><\/em>.<\/p>\n\n\n\n

Before I give its proof, we should notice that this is a generalization of Rudin’s original lemma: if we give X<\/em> the Alexandroff topology of an ordering \u2264, then the elements of Q<\/strong>(X<\/em>) are exactly the sets \u2191A<\/em> with A<\/em> \u2208 Q<\/strong>fin<\/sub>(X<\/em>), the upper Vietoris topology on Q<\/strong>(X<\/em>) 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.<\/p>\n\n\n\n

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<\/em> of F<\/em> that still intersects every element of A<\/strong><\/em>, then we show that every such minimal closed set must be irreducible. Let me decompose this in two steps.<\/p>\n\n\n\n

Lemma A.<\/strong> Let A<\/strong><\/em> be an irreducible closed subset of Q<\/strong>(X<\/em>). Every closed subset F<\/em> of X<\/em> that intersects every element of A<\/strong><\/em> contains an minimal closed subset C<\/em> that still intersects every element of A<\/strong><\/em>.<\/p>\n\n\n\n

Proof.<\/em> Let E<\/strong> be the collection of closed subsets of F<\/em> that intersect every element of A<\/em><\/strong>. E<\/strong> is non-empty, since F<\/em> is in E<\/strong> by assumption. We order E<\/strong> by reverse inclusion, and we observe that this turns E<\/strong> into a dcpo. Indeed, for every directed family (under reverse inclusion, hence filtered under inclusion) (Fi<\/sub><\/em>)i <\/em>\u2208 I<\/em><\/sub> in E<\/strong>, let F<\/em> be the intersection of that family. This is a closed set. For every Q<\/em> in A<\/strong><\/em>, every Fi<\/sub><\/em> intersects Q<\/em>, so F<\/em> also intersects Q<\/em>, because Q<\/em> is compact. Hence F<\/em> intersects every element of A<\/em><\/strong>, and therefore is in E<\/strong>.<\/p>\n\n\n\n

Since E<\/strong> is a dcpo, and contains F<\/em>, by Zorn’s Lemma it contains a maximal element C<\/em> above F<\/em>\u2014maximal under reverse inclusion, hence minimal subset of F<\/em> with respect to inclusion. \u2610<\/p>\n\n\n\n

Lemma B.<\/strong> Let A<\/strong><\/em> be an irreducible closed subset of Q<\/strong>(X<\/em>). Every minimal closed subset C<\/em> that intersects every element of A<\/strong><\/em> is irreducible.<\/p>\n\n\n\n

Proof.<\/em> Imagine that C<\/em> is included in a finite union of closed sets F<\/em>1<\/sub>, …, F<\/em>n<\/em><\/sub>. If C<\/em> is included in no Fi<\/sub><\/em>, then C<\/em> \u2229 Fi<\/sub><\/em> is strictly included in C<\/em> for every i<\/em>, 1\u2264i<\/em>\u2264n<\/em>. By the minimality of C<\/em>, no C<\/em> \u2229 Fi<\/sub><\/em> is in E<\/strong>. Hence, for each i<\/em>, there is an element Qi<\/sub><\/em> of A<\/em><\/strong> that does not intersect C<\/em> \u2229 Fi<\/sub><\/em>. This can be restated by saying that A<\/strong><\/em> is not included in \u2662(C<\/em> \u2229 Fi<\/sub><\/em>). The latter form a finite family of closed subsets of Q<\/strong>(X<\/em>). Since A<\/strong><\/em> is irreducible, if it were included in the finite union \u222ai<\/em>=1<\/sub>n<\/em><\/sup> \u2662(C<\/em> \u2229 Fi<\/sub><\/em>), it would be included in some \u2662(C<\/em> \u2229 Fi<\/sub><\/em>), and that is not the case. Therefore there is an element Q<\/em> of A<\/strong><\/em> which is not in \u222ai<\/em>=1<\/sub>n<\/em><\/sup> \u2662(C<\/em> \u2229 Fi<\/sub><\/em>), namely which is disjoint from every C<\/em> \u2229 Fi<\/sub><\/em>. Then Q<\/em> is disjoint from \u222ai<\/em>=1<\/sub>n<\/em><\/sup> (C<\/em> \u2229 Fi<\/sub><\/em>), which is equal to C<\/em> since C<\/em> is included in the union of F<\/em>1<\/sub>, …, F<\/em>n<\/em><\/sub>. That is impossible since C<\/em> intersects every element of A<\/strong><\/em>. \u2610<\/p>\n\n\n\n

This applies in particular when A<\/strong><\/em> is a filtered family of compact saturated subsets of X<\/em>. However, in that case, Xu, Shen, Xi and Zhao notice that we can say more:<\/p>\n\n\n\n

Lemma C.<\/strong> Let A<\/strong><\/em>=(Ki<\/em><\/sub>)i <\/em>\u2208 I<\/em><\/sub> be a filtered family of compact saturated subsets of X<\/em>. Every minimal closed subset C<\/em> that intersects every element of A<\/strong><\/em> is a closed WD set.<\/p>\n\n\n\n

Proof.<\/em> Let f<\/em> be any continuous map from X<\/em> to a well-filtered T0<\/sub> space Y<\/em>. The trick is to consider the family (\u2191f<\/em>[Ki<\/em><\/sub> \u2229 C<\/em>])i \u2208 I<\/em><\/sub>. (You may be tempted to use (\u2191f<\/em>[Ki<\/em><\/sub>])i \u2208 I<\/em><\/sub>, but that would not work.)<\/p>\n\n\n\n

Note that Ki<\/em><\/sub> \u2229 C<\/em> is compact, being the intersection of a compact set and of a closed set, and non-empty. Therefore (\u2191f<\/em>[Ki<\/em><\/sub> \u2229 C<\/em>])i \u2208 I<\/em><\/sub> is a filtered family of non-empty compact saturated subsets of Y<\/em>. It is easy to see that each set \u2191f<\/em>[Ki<\/em><\/sub> \u2229 C<\/em>] intersects the closed set cl(f<\/em>[C<\/em>]). Since Y<\/em> is well-filtered, \u2229i<\/em><\/sub> \u2208 I<\/em><\/sub> \u2191f<\/em>[Ki<\/em><\/sub> \u2229 C<\/em>] intersects cl(f<\/em>[C<\/em>]), say at y<\/em>.<\/p>\n\n\n\n

Let us look at the closed set f<\/em>-1<\/sup>(\u2193y<\/em>) \u2229 C<\/em> in X<\/em>. For every i<\/em>, since y<\/em> is in \u2191f<\/em>[Ki<\/em><\/sub> \u2229 C<\/em>], there is a point x<\/em> in Ki<\/em><\/sub> \u2229 C<\/em> such that f<\/em>(x<\/em>)\u2264y<\/em>: so x<\/em> is also in f<\/em>-1<\/sup>(\u2193y<\/em>), and that shows that f<\/em>-1<\/sup>(\u2193y<\/em>) \u2229 C<\/em> intersects Ki<\/em><\/sub>. Since f<\/em>-1<\/sup>(\u2193y<\/em>) \u2229 C<\/em> intersects every Ki<\/em><\/sub>, and is included in C<\/em>, the minimality of C<\/em> implies that f<\/em>-1<\/sup>(\u2193y<\/em>) \u2229 C<\/em> = C<\/em>. In other words, for every x<\/em> in C<\/em>, f<\/em>(x<\/em>)\u2264y<\/em>. It follows that f<\/em>[C<\/em>], hence also cl(f<\/em>[C<\/em>]), is included in \u2193y<\/em>. But remember that y<\/em> is in cl(f<\/em>[C<\/em>]), so \u2193y<\/em> is also included in cl(f<\/em>[C<\/em>]). This shows that cl(f<\/em>[C<\/em>])=\u2193y<\/em>, and we are done. \u2610<\/p>\n\n\n\n

Wf(X<\/em>) is well-filtered<\/h2>\n\n\n\n

Let A<\/strong><\/em>=(K<\/strong>i<\/sub>)i <\/em>\u2208 I<\/em><\/sub> be a filtered family of compact saturated subsets of Wf<\/strong>(X<\/em>). (Yes, sorry, this is very<\/em> higher-order: families of subsets of a set of subsets…) We assume that \u2229i<\/em><\/sub> \u2208 I<\/em><\/sub> K<\/strong>i<\/sub><\/em> is included in some open subset of Wf<\/strong>(X<\/em>). That open subset must be of the form \u2662U<\/em> for some open subset of X<\/em>.<\/p>\n\n\n\n

We reason by contradiction, and we assume that no K<\/strong>i<\/sub><\/em> is included in \u2662U<\/em>. Letting F<\/em> be the complement of U<\/em>, this means that K<\/strong>i<\/sub><\/em> intersects the complement \u2610F<\/em> of \u2662U<\/em>. (\u2610F<\/em> is the set of compact saturated sets included in F<\/em>.) Then \u2610F<\/em> is closed and intersects every element K<\/strong>i<\/sub><\/em> of A<\/strong><\/em>. By Lemma A and Lemma C (not just Lemma B), \u2610F<\/em> contains a closed WD subset that still intersects every K<\/strong>i<\/sub><\/em>.<\/p>\n\n\n\n

We remember that every open subset of Wf<\/strong>(X<\/em>) is the diamond of an open subset of X<\/em>, so every closed subset is a box \u2610C<\/em> of some closed subset C<\/em> of X<\/em>. Henceforth, let us write \u2610C<\/em> for the closed subset of \u2610F<\/em><\/i><\/span> that we have just found, and which intersects every K<\/strong>i<\/sub><\/em>.<\/p>\n\n\n\n

We claim that C<\/em> must be a WD subset of X<\/em>. Let f<\/em> be an arbitrary continuous map from X<\/em> to a well-filtered T0<\/sub> space Y<\/em>. We have seen that f<\/em> extends along \u03b7 to a continuous map f’<\/em> from Wf<\/strong>(X<\/em>) to Y<\/em>: for every WD subset A of X<\/em>, f’<\/em>(A<\/em>) is the unique point y<\/em> of Y<\/em> such that cl(f<\/em>[A<\/em>])=\u2193y<\/em>. Since \u2610C<\/em> is a WD subset of Wf<\/strong>(X<\/em>) (don’t lose track of the spaces in which we are working!), there is a unique point y’<\/em> such that cl(f’<\/em>[\u2610C<\/em>])=\u2193y’<\/em>. We claim that \u2193y’<\/em> is equal to cl(f<\/em>[C<\/em>]): this is what we need to show that C<\/em> is a WD set. The open neighborhoods of y’<\/em> are exactly the open subsets V<\/em> of Y<\/em> that intersect cl(f’<\/em>[\u2610C<\/em>]), hence f’<\/em>[\u2610C<\/em>]. Now V<\/em> intersects f’<\/em>[\u2610C<\/em>] if and only if f’<\/em>-1<\/sup>(V<\/em>) intersects \u2610C<\/em>. We have seen earlier that f’<\/em>-1<\/sup>(V<\/em>)=\u2662f<\/em>-1<\/sup>(V<\/em>). And \u2662f<\/em>-1<\/sup>(V<\/em>) intersects \u2610C<\/em> if and only if there is a WD subset A<\/em> of X<\/em> that intersects f<\/em>-1<\/sup>(V<\/em>) and is included in C<\/em>. If such an A<\/em> exists, then f<\/em>-1<\/sup>(V<\/em>) intersects C<\/em>, and conversely, if f<\/em>-1<\/sup>(V<\/em>) intersects C<\/em> at x<\/em>, then \u2193x<\/em> is a WD set that intersects f<\/em>-1<\/sup>(V<\/em>) and is included in C<\/em>. Summing up, the open neighborhoods of y’<\/em> are exactly the open subsets V<\/em> of Y<\/em> such that f<\/em>-1<\/sup>(V<\/em>) intersects C<\/em>. Those are also those such that cl(f<\/em>[C<\/em>]) intersects V<\/em>. We now have two closed sets, cl(f’<\/em>[\u2610C<\/em>])=\u2193y’<\/em> and cl(f<\/em>[C<\/em>]), which intersect exactly the same open sets. Therefore they are equal, and this shows the desired claim cl(f<\/em>[C<\/em>]) =\u2193y’<\/em>.<\/p>\n\n\n\n

What have we got so far? We started with a filtered family A<\/strong><\/em>=(K<\/strong>i<\/sub>)i <\/em>\u2208 I<\/em><\/sub> of compact saturated subsets K<\/strong>i<\/sub><\/em> of Wf<\/strong><\/span>(X<\/em>) such that \u2229i<\/em><\/sub> \u2208 I<\/em><\/sub> K<\/strong>i<\/sub><\/em> is included in \u2662U<\/em>. We have assumed that no K<\/strong>i<\/sub><\/em> is included in \u2662U<\/em>, we have called F<\/em> the complement of U<\/em>, and we have found a WD subset C<\/em> of X<\/em>, included in F<\/em>, such that \u2610C<\/em> intersects every K<\/strong>i<\/sub><\/em>. Since C<\/em> is in \u2610C<\/em> and K<\/strong>i<\/sub><\/em> is upwards-closed, C<\/em> is in every K<\/strong>i<\/sub><\/em>. Therefore C<\/em> is in \u2229i<\/em><\/sub> \u2208 I<\/em><\/sub> K<\/strong>i<\/sub><\/em>, hence in \u2662U<\/em>. That is impossible! … because C<\/em> is included in F<\/em>, which does not intersect U<\/em> by definition.<\/p>\n\n\n\n

Since we have reached a contradiction, it must in fact be the case that some K<\/strong>i<\/sub><\/em> is included in \u2662U<\/em>. This shows that Wf<\/strong>(X<\/em>) is well-filtered, and we are done. (It is T0<\/sub>, as a subspace of S<\/strong>(X<\/em>), which is always T0<\/sub>.)<\/p>\n\n\n\n

Appendix A: every directed set is WD<\/h2>\n\n\n\n

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

(1) We claim that every directed family (yi<\/sub><\/em>)i <\/em>\u2208 I<\/em><\/sub> in Y<\/em> has a supremum. In order to show this, we first observe that the family of upwards-closed sets \u2191yi<\/sub><\/em>, i <\/em>\u2208 I<\/em>, is filtered. The intersection Q<\/em> of the latter family is the collection of upper bounds of (yi<\/sub><\/em>)i <\/em>\u2208 I<\/em><\/sub>. Since Y<\/em> is well-filtered, Q<\/em> is compact saturated and non-empty.<\/p>\n\n\n\n

If (Vj<\/sub><\/em>)j <\/em>\u2208 J<\/em><\/sub> is any family of open sets whose union V<\/em> contains Q<\/em>, then by well-filteredness again some \u2191yi<\/sub><\/em> is in V<\/em>. Hence that yi<\/sub><\/em> is in V<\/em>, and therefore in some Vj<\/sub><\/em>. Since Vj<\/sub><\/em> is upwards-closed, \u2191yi<\/sub><\/em> and therefore also the smaller set Q<\/em> is included in Vj<\/sub><\/em>. This shows that Q<\/em> is supercompact<\/em>: every open cover of Q<\/em> contains an open set that already contains Q<\/em>.<\/p>\n\n\n\n

Now, it is known that every supercompact set Q<\/em> is of the form \u2191y<\/em> for some point y<\/em> of Y<\/em> [4, Fact 2.2]; and that point is unique since Y<\/em> is T0<\/sub>. The proof is easy: since Q<\/em> is supercompact, for every family of closed sets that each intersect Q<\/em>, their intersection must again intersect Q<\/em> (consider the complements of those closed sets, and apply the definition); for every z<\/em> in Q<\/em>, clearly \u2193z<\/em> is closed and intersects Q<\/em>, so the intersection of those closed sets must intersect Q<\/em>, say at y<\/em>; then, by construction, y<\/em> is in Q<\/em> and below every element z<\/em> of Q<\/em><\/span>, which proves the claim.<\/p>\n\n\n\n

(2) We claim that Y<\/em> 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<\/sub> space Y<\/em>.<\/p>\n\n\n\n

This is simply because, for every directed family (yi<\/sub><\/em>)i <\/em>\u2208 I<\/em><\/sub> whose supremum y<\/em> lies in some open subset U<\/em>, then \u2191y<\/em> = \u2229i<\/sub><\/em> \u2191yi<\/sub><\/em> is included in U<\/em>, so by well-filteredness some \u2191yi<\/sub><\/em> is included in U<\/em>.<\/p>\n\n\n\n

All right, back to our problem. We know that f<\/em>[D<\/em>] is directed in Y<\/em>. Since we now know that Y<\/em> is a dcpo in its specialization ordering, f<\/em>[D<\/em>] has a supremum, call it y<\/em>. Then f<\/em>[D<\/em>] is a subset of \u2193y<\/em>, and the latter is closed, so cl (f<\/em>[D<\/em>]) is a subset of \u2193y<\/em>. Since Y<\/em> is monotone convergence, every closed subset is Scott-closed, hence the closure of f<\/em>[D<\/em>] is included in its Scott-closure, which is included in \u2193y<\/em>. \u2610<\/p>\n\n\n\n

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  1. Guohua Wu, Xiaoyong Xi, Xiaoquan Xu, and Dongsheng Zhao.  Existence of well-filterifications of T0<\/sub> topological spaces.  arXiv<\/a> 1906.10832<\/a>, July 2019.  Submitted.<\/li>\n\n\n\n
  2. Jimmie Lawson and Xiaoyong Xi.  Well-filtered spaces, compactness, and the lower topology.  2019.  Submitted.<\/li>\n\n\n\n
  3. Xiaoquan Xu, Chong Shen, Xiaoyong Xi, and Dongsheng Zhao.  On T0<\/sub> spaces determined by well-filtered spaces.  arXiv<\/a> 1909.09303<\/a>, September 2019.  Submitted. (Now Topology and its Applications<\/a>, Volume 282<\/a>, 15 August 2020, https:\/\/doi.org\/10.1016\/j.topol.2020.107323<\/a>.)<\/li>\n\n\n\n
  4. Reinhold Heckmann and Klaus Keimel. Quasicontinuous domains and the Smyth powerdomain. Electronic Notes in Theoretical Computer Science 298 (2013), pp. 215\u2013232.<\/li>\n<\/ol>\n\n\n
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    \"jgl-2011\"<\/figure>\n<\/div>\n\n\n

    \u2014 Jean Goubault-Larrecq<\/a> (September 24th, 2019)<\/p>\n","protected":false},"excerpt":{"rendered":"

    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 … Continue reading →<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"_crdt_document":"","footnotes":""},"class_list":["post-2014","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/projects.lsv.ens-paris-saclay.fr\/topology\/index.php?rest_route=\/wp\/v2\/pages\/2014","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/projects.lsv.ens-paris-saclay.fr\/topology\/index.php?rest_route=\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/projects.lsv.ens-paris-saclay.fr\/topology\/index.php?rest_route=\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/projects.lsv.ens-paris-saclay.fr\/topology\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/projects.lsv.ens-paris-saclay.fr\/topology\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=2014"}],"version-history":[{"count":26,"href":"https:\/\/projects.lsv.ens-paris-saclay.fr\/topology\/index.php?rest_route=\/wp\/v2\/pages\/2014\/revisions"}],"predecessor-version":[{"id":5908,"href":"https:\/\/projects.lsv.ens-paris-saclay.fr\/topology\/index.php?rest_route=\/wp\/v2\/pages\/2014\/revisions\/5908"}],"wp:attachment":[{"href":"https:\/\/projects.lsv.ens-paris-saclay.fr\/topology\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=2014"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}