I have met Szymon Dolecki at the Summer Topology Conference 2016, and he is a charming person. He tried to make me discover the best Sauvignon Blanc in the world, from the Marlborough vineyards in New Zealand; but I don’t drink wine, to his great disappointment.

Apart from that, but still in relation with Szymon, Jimmie Lawson mentioned the Dolecki-Greco-Lechicki Theorem [1, Theorem 4.1], a very nice theorem that attracts my attention more and more. This says that every Čech-complete space is consonant. (I’ll explain below.)

That Theorem is the topic of Exercise 8.3.4 in the book, but I am afraid that, as stated, it is way too hard. My purpose here is to give a complete solution to the exercise.

## Consonance

Consider the lattice **O**(*X*) of all open subsets of a topological space *X*. There are several ways we can topologize this.

### The Scott topology.

One of the most obvious ways, at least to domain theorists, is to give it the Scott topology (of the inclusion ordering).

Szymon made several historical remarks as to who invented this topology first. If we look at convergence with respect to the Scott topology, then on **O**(*X*), or rather on the anti-isomorphic lattice **H**(*X*) of closed subsets of *X*, Kazimierz Kuratowski had invented the same notion of convergence (of closed sets) a long time before Scott, and Szymon told us that Paul Painlevé had invented it even earlier, and that Giuseppe Peano had invented it even earlier again, in 1887 — at a time when Kuratowski was not even born.

He also mentioned what an extraordinary person Painlevé was, managing to be, at the same time, a renowned scientist and a prominent politician. He was one of the leading mathematicians of his time, and worked mostly in the classification of solutions to second-order ordinary differential equations; in 1903, he showed that the equation of fluid mechanics made flying possible, and later completely developed the theory of flight in perfect fluids. He was elected a member of the French Academy of Sciences, and eventually became its president. At the same time, he got interested in politics in the course of the infamous Dreyfus affair, and ended up being several times minister of war, several times “président du conseil” (=prime minister), and several times minister of the (newly created) air force, between the two world wars.

By the way, Peano was also one extraordinary mathematician, and Szymon kindly directed me to his paper [2], with Gabriele Greco: Peano had invented convergence notions for sets, in Euclidean space, well before Kuratowski, Painlevé, but also Hausdorff or Vietoris. I discovered they had a subsequent paper, too, on the “amazing oblivion of Peano’s contributions to mathematics”. Szymon tells me this has evolved into a series of three papers called “*The astonishing oblivion of Peano’s mathematical legacy*“, available from his preprints page, and which should make it into a little book. It is, indeed, amazing how much of modern mathematics Peano had invented, which was later rediscovered by Borel, Lebesgue, Zermelo, Banach, Choquet, and others.

### The compact-open topology.

Anyway, there is at least one other natural topology on **O**(*X*). Since **O**(*X*) can be identified with the space of continuous maps from *X* to Sierpiński space **S**, one can equip it with the compact-open topology. (One can also equip it with the Isbell topology, for example, but in that case the Isbell topology *is* the Scott topology.) This topology is generated by basic open sets of the form ￭*Q*, where *Q* is compact saturated in *X*. By definition, ￭*Q* is the collection of open neighborhoods of *Q*.

In other words, a set * U* of open subsets of

*X*is open in the compact-open topology if and only if, for every

*U*in

*, there is a compact saturated subset*

**U***Q*of

*X*such that

*U*∈ ￭

*Q*⊆

*.*

**U**It is fairly easy to see that ￭*Q* is always Scott-open. Therefore every open in the compact-open topology is also open in the Scott topology.

Dolecki, Greco and Lechicki [1] call *consonant* those spaces *X* for which the converse also holds, i.e., those for which the Scott and the compact-open topologies agree on **O**(*X*). Concretely, those are the spaces *X* such that, if you take any Scott-open subset * U* of

**O**(

*X*), and any

*U*in

*, then there is a compact saturated subset*

**U***Q*of

*X*contained in

*such that every open neighborhood of*

**U***Q*is in

*.*

**U**The following results are part of Exercise 5.4.12 in the book:

- Every locally compact space is consonant.
- A space
*X*is consonant if and only if, for every topological space*Y*, the compact-open and the Isbell topologies agree on the space of continuous maps [*X*→*Y*].

(This exercise also requires you to show that the Sorgenfrey line and the space of rational numbers are not consonant. That is true, but highly non-trivial, and you should probably not try it at home, actually: see the errata page.)

One of the most important theorems from [1], Theorem 4.1, states that every regular Čech-complete space is consonant. In particular, there are non-locally compact consonant spaces, say Baire space. (Baire space is Čech-complete because it is Polish.)

Dolecki et al. [1] have many other theorems, but I will concentrate on that one.

# Čech-completeness

Čech-completeness is a purely topological notion that is meant to capture some essential properties of complete metric spaces, without any reference to the metric itself. By that, I mean that:

- The notion does not refer to any metric whatsoever, and
- The metrizable spaces that are completely metrizable are exactly those that are Čech-complete (Exercise 7.6.21 — they are also exactly those that are Choquet-complete, but this is a different notion).

The definition is pretty obscure: a space *X* is Čech-complete if and only if there is a countable collection of open covers *C _{n}* = (

*U*)

_{ni}*, one for each natural number*

_{i ∈ I[n]}*n*(the index set

*I[n]*may depend on

*n*, and can be arbitrary), with the property that for every filtered family (

*F*)

_{j}*of non-empty closed subsets such that, for every*

_{j ∈ J}*n*, there are indices

*i*and

*j*such that

*F*⊆

_{j}*U*., the intersection of all

_{ni}*F*s is non-empty. Wow.

_{j}The *n*th open cover *C _{n}* in the collection serves as a measure of

*size*. To show that every complete metric space is Čech-complete, one takes all open balls of radius 1/2

^{n}^{+1}for

*C*, say. Then every subset

_{n}*F*of any

_{j}*U*has diameter at most 1/2

_{ni}*. Any family (*

^{n}*F*)

_{j}*as above is then such that the diameters of the*

_{j ∈ J}*F*s tend to 0. This mimicks Cantor’s Theorem (Exercise 7.3.6), which says that in a complete metric space, the intersection of any filtered family of non-empty closed subsets whose diameters tend to 0 is a single point.

_{j}Hence we can rephrase the definition as follows. Given a countable collection of open covers *C _{n}*, say that a set

*has diameter at most 1/2*if and only if it is included in some element

^{n}*U*of

_{ni}*C*. This is an abstract notion of diameter, nothing as concrete as the diameters you get in metric spaces.

_{n}A filtered family (*F _{j}*)

*has*

_{j ∈ J}*vanishing diameters*if and only if one can find

*F*of arbitrary small diameter. That is, if and only if, for every natural number

_{j}*n*, one can find an index

*j*such that

*F*has diameter at most

_{j}*1/2*.

^{n}A space is Čech-complete if and only if it has a countable collection of open covers such that every filtered family (*F _{j}*)

*of non-empty closed subsets with vanishing diameters has non-empty intersection.*

_{j ∈ J}It is usual to require Čech-complete spaces to be regular T_{2} (i.e., T_{3}) as well. We shall keep that a separate property, although we shall require regularity pretty soon.

## Starting the proof

Let *X* be a regular Čech-complete space. We silently assume the notations we used above for the covers *C _{n}*, and we say that a set is

*1/2*

^{n}*-small*if and only if it is included in a finite union of sets of diameter at most

*1/2*.

^{n}The initial hint given in Exercise 8.3.4 is: “Let * U* be a Scott-open subset of

**O**(

*X*). If

*X*is regular and

*U*∈

*, show that one can find open subsets*

**U***U*

_{0},

*U*

_{1}, …,

*U*, … and closed subsets

_{n}*F*

_{0},

*F*

_{1}, …,

*F*, … such that

_{n}*U*=

*U*

_{0}⊇

*F*

_{0}⊇

*U*

_{1}⊇

*F*

_{1}⊇ … ⊇

*U*⊇

_{n}*F*⊇ …, and such that for each

_{n}*n*,

*U*is in

_{n}*and is*

**U***1/2*-small.”

^{n}This is really how Dolecki et al.’s proof starts, as well [1]. Exercice 8.3.4 diverges from their proof at some later point.

Here is how you do it. Start by defining *U*_{0}=*U*, and define all the other open and closed sets by induction on *n*. This means we can assume that we have already defined *U _{n}* (in

*), and we wish to construct*

**U***F*and

_{n}*U*

_{n}_{+1}. For each point

*x*in

*U*, do the following. Since

_{n}*C*

_{n}_{+1}is a cover,

*x*is also in some

*U*

_{(}

_{n}_{+1}

_{)}

*∈*

_{i}*C*

_{n}_{+1}.

*U*⋂

_{n}*U*

_{(}

_{n}_{+1}

_{)}

*is also an open neighborhood of*

_{i}*x*, but additionally, one that has diameter at most

*1/2*

^{n}^{+1}.

By regularity (i.e., by local closure, see Exercise 4.1.21) *x* has a closed neighborhood *F _{nx}* that is included in

*U*⋂

_{n}*U*. Note that

_{ni}*F*again has diameter at most

_{nx}*1/2*

^{n}^{+1}. The family of interiors int(

*F*), when

_{nx}*x*varies over the points of

*U*, is a family of open subsets of

_{n}*U*whose union is the whole of

_{n}*U*, by construction.

_{n}By a familiar trick (Trick 4.4.6 in the book), *U _{n}* is the

*directed*union of the finite unions ∪

_{x }_{in}

*int(*

_{ E}*F*), where

_{nx}*E*ranges over the finite subsets of

*U*. Since

_{n}*U*is in

_{n}*, and*

**U***is Scott-open, one of those directed unions ∪*

**U**

_{x }_{in}

_{ E}_{[}

_{n}_{]}int(

*F*) is in

_{nx}*(where*

**U***E*[

*n*] is some finite set — I am using the notation

*E*[

*n*] instead of the more standard subscript notation

*E*because I will need to write those sets in subscript as well, and the HTML format does not seem to have provision for double subscripts). Hence define:

_{n}*U*_{n}_{+1}as ∪_{x }_{in}_{ E}_{[}_{n}_{]}int(*F*): this is in_{nx}, and is**U***1/2*^{n}^{+1}-small, by construction;*F*as ∪_{n}_{x }_{in }_{E}_{[}_{n}_{]}*F*: this is closed because_{nx}*E*[*n*] is finite, and finite unions of closed sets are closed.

As an additional bonus, note that *F _{n}* itself is

*1/2*

^{n}^{+1}-small.

## Kőnig’s Lemma

The exercise continues by giving you the following instruction: “Show that the family * V* of all open subsets of

*X*that contain some

*F*is a Scott-open filter.”

_{n}That it is a filter is obvious, but the hard part is to show that * V* is Scott-open… and I’m not giving any hint here — I should have.

There are probably several ways of managing to prove this. Dolecki et al. prove directly that the intersection *Q*=∩_{n }*F _{n}* is compact, and that an open set contains

*Q*if and only if it contains some

*U*. But they rely on some results on compactoid filters, and convergence of ultrafilters, which I preferred to avoid. Furthermore, I thought it would be easier to show that

_{n}*is Scott-open and use the Hofmann-Mislove theorem instead. (I was mistaken, as we shall see at the end of this post.) The argument is not that easy to find, although it is a pretty simple argument in the end.*

**V**We use Kőnig‘s Lemma. This says the following: every finitely branching tree whose branches are all finite, is finite. Let me avoid defining what a tree is. There are various formal definitions, which are fine for formal proofs but tend to obscure the intended object. Informally, you build a tree by drawing an initial vertex, called the root of the tree. Then create fresh *successor* vertices (as many as you wish, possibly infinitely many or none at all) to the root vertex. For each of those successor vertices, create fresh successor vertices, and continue with the newly created successors, indefinitely (or until you decide to stop). A tree is *finitely branching* if and only if every vertex has only finitely many successors. (I.e., immediate successors, not successors of successors of successors etc.) A *branch* is any finite or infinite sequence of vertices *v _{n}* starting from the root (i.e.,

*v*

_{0}is the root), and such that

*v*

_{n}_{+1}is a successor of

*v*, for each

_{n}*n*. A tree is

*finite*iff it has only finitely many vertices.

It is customary (at least in computer science) to draw such trees top-down: with the root at the top, and going down as you add successors. Accordingly, I will say that a vertex *v*‘ is* below* a vertex* v* if and only if *v’* is *v*, or a successor of *v*, or a successor of a successor of *v*, etc. It is *strictly below v* if and only if it is below *v* and different from *v*.

The proof of Kőnig’s Lemma is extremely simple. We assume an infinite finitely-branching tree, and we construct an infinite branch. Start by defining *v*_{0} as the root of the tree. Since *v*_{0} has only finitely many successors, and there are infinitely many vertices strictly below *v*_{0}, there must be infinitely many vertices below some successor *v*_{1} of *v*_{0}. (This is an instance of the infinite pigeonhole principle: if you’ve got infinitely many pigeons, and only finitely many pigeonholes to put them in, then some pigeonhole will contain infinitely many pigeons.) By the same argument, there must be infinitely many vertices below some successor *v*_{2} of *v*_{1}, and so on. Collecting *v*_{0}, *v*_{1}, *v*_{2}, … we obtain an infinite branch in the tree.

## Showing that V is Scott-open

To show that * V* is Scott-open, we must take an arbitrary directed family (

*U*)

_{j}*, and assuming that no*

_{j ∈ J}*U*is in

_{j}*, we must show that ∪*

**V**

_{j ∈ J}*U*is not in

_{j}*either.*

**V**Let *F’ _{j}* be the complement of

*U*. Those are closed sets. That

_{j}*U*is not in

_{j}*means that*

**V***F’*intersects

_{j}*F*, for all

_{n}*j*and

*n*. Our goal is to show that ∩

_{j }*F’*intersects

_{j}*F*for every

_{n}*n*, which will prove our claim. We shall in fact show the stronger statement that it intersects ∩

_{n }*F*.

_{n}We build a tree * T* as follows. We shall also label its vertices (except the root), for later convenience. First create a root. Remember that

*F*

_{0}= ∪

_{x }_{in}

_{ E}_{[}

_{0}

_{]}

*F*

_{0x}: create as many successors to the root as there are elements

*x*in

*E*[0], and label each by

*F*

_{0x}. Call those vertices the

*level 0 vertices*. For each level 0 vertex

*v*, do a similar thing with

*F*

_{1}= ∪

_{x }_{in}

_{ E}_{[1]}

*F*

_{1x}: create as many successors to

*v*as there are elements

*x*in

*E*[1], and label each by the intersection of

*F*

_{1x}with

*v*‘s label. The vertices created this way are the

*level 1 vertices*. Continue in the same way, creating level 2 vertices from

*F*

_{2}= ∪

_{x }_{in}

_{ E}_{[2]}

*F*

_{2x}, as successors to all the level 1 vertices we created previously, and so on.

I will write *F _{v}* for the closed set that labels any (non-root) vertex

*v*. Note that, for a level

*n*vertex

*v*,

*F*is an intersection

_{v}*F*

_{0x[0]}⋂

*F*

_{1x[1]}⋂ … ⋂

*F*, for some points

_{nx[n]}*x*[0],

*x*[1], …,

*x*[

*n*] in

*E*[0],

*E*[1], …,

*E*[

*n*] respectively. Some of these sets

*F*may be empty, but this will not stop us: we shall find enough of them that are non-empty. An important point is that

_{v}*F*, being included in

_{v}*F*, is of diameter at most

_{nx[n]}*1/2*

^{n}^{+1}.

Note also that the union of the sets *F _{v}*, when

*v*ranges over all level

*n*vertices (for fixed

*n*), is exactly

*F*. This is proved by an easy induction on

_{n}*n*.

Now recall that *F _{n}* intersects every

*F’*, for every

_{j}*n*. I claim that this implies that:

(*) For every *n*, there is a level *n* vertex *v* such that *F _{v}* intersects every

*F’*.

_{j}Indeed, otherwise, for each level *n* vertex *v*, there would be an index *j*[*v*] (depending on *v*) such that *F _{v}* does not intersect

*F’*. By filteredness, one can find an index

_{j[v]}*j*(independent of

*v*) such that

*F’*is included in every

_{j}*F’*; this is possible because there are only finitely many level

_{j[v]}*n*vertices at all. Then, for each level

*n*vertex

*v*,

*F*does not intersect

_{v}*F’*, and taking unions over all such vertices, we would obtain that

_{j}*F*does not intersect

_{n}*F’*, which is absurd.

_{j}Select all the vertices *v* such that *F _{v}* intersects every

*F’*, plus the root. Those vertices span a subtree

_{j}*of*

**T’***. That just means that if you remove all non-selected vertices from*

**T***, what you get is still a tree. Notably, you won’t disconnect*

**T***into multiple pieces this way. This is because if*

**T***v’*is below

*v*and

*v’*is selected, then

*v*is selected, too. (I.e., if

*v*‘ is below

*v*and

*F*intersects every

_{v’}*F’*, then

_{j}*F*also intersects every

_{v}*F’*: this is because

_{j}*F*contains

_{v}*F*.)

_{v’}By construction, * T* is finitely branching, so

*is finitely branching, too. By claim (*) above, for every*

**T’***n*,

*contains a level*

**T’***n*vertex. In particular,

*is infinite. By Kőnig’s Lemma,*

**T’***must contain an infinite branch: (the root, then)*

**T’***v*[0],

*v*[1],

*v*[2], …

We are almost there. Look at the family of closed sets *F’ _{j}* ⋂

*F*,

_{v[n]}*j*in

*J*,

*n*in

**N**. The whole point in finding our infinite branch is to make sure that this is a

*filtered*family (to show this, recall that

*F*contains

_{v[n]}*F*, notably) of non-empty closed sets (non-empty by the way we selected the vertices of

_{v[n+1]}*from those of*

**T’***), and with vanishing diameters (because*

**T***F*being included in

_{v[n]}*F*, has diameter at most

_{v[n]}*1/2*

^{n}^{+1}, since

*v[n]*is a level

*n*vertex).

We can *at last* use **Čech-completeness**: the intersection ∩* _{j,n}* (

*F’*⋂

_{j}*F*) is non-empty.

_{v[n]}That intersection is the intersection of ∩_{j}*F’ _{j}* with ∩

_{n}*F*⊆ ∩

_{v[n] }

_{n}*F*, so ∩

_{n}

_{j}*F’*intersects the larger set ∩

_{j}

_{n}*F*, as we had claimed.

_{n}## Finishing the proof.

This shows that * V* is Scott-complete, and we can finish the proof of the theorem. Since

*X*is T

_{2}, it is sober (Proposition 8.2.12 in the book), so we can apply the Hofmann-Mislove theorem:

*is the filter of open neighborhoods of a compact saturated set, namely*

**V***Q*=∩

_{n}*F*. Since

_{n}*Q*⊆

*U*

_{0}=

*U*,

*U*is in ￭

*Q*. And every element

*W*of ￭

*Q*is in

*, hence contains some*

**V***F*, and therefore also the smaller set

_{n}*U*; since

_{n+1}*U*is in

_{n+1}*,*

**U***W*is in

*as well, so that ￭*

**U***Q*⊆

**. As we have noticed earlier, this shows that**

*U**X*is consonant.

## Removing the T_{2} axiom.

The Dolecki-Greco-Lechicki theorem is slightly more general: it says that every *regular* (not necessarily T_{2}) Čech-complete space is consonant.

This is an easy consequence of what we have just proved.

Assume *X* is regular Čech-complete, but not necessarily T_{2}. Take its T_{0} quotient (Exercise 4.10.25), and write it as *X’*. The elements of *X’* are equivalence classes [*x*] of points *x* of *X*, with respect to the relation =_{0} defined by *x= _{0}y* if and only if

*x*≤

*y*and

*y*≤

*x*. As usual, ≤ is the specialization quasi-ordering of

*X*.

We first check that *X’* is T_{2}. Assume [*x*]≠[*y*], that is, *x*≠_{0}*y*. There is an open subset *U* that contains one point, say *x*, but not the other (*y*). Let *F* be the complement of *U*: *F* contains *y* but not *x*. Since *X* is regular, there are disjoint open neighborhoods *V* of *x* and *W* of *F* (hence of *y*), and we are done. (This argument is a slight improvement over a classical argument that says that it is enough for a regular space to be T_{0} for it to be T_{2}.)

*X’* is regular, too, because *X* is. It is equally easy to show that *X’* is Čech-complete. Hence *X’* is consonant. Because *X* and *X’* have isomorphic lattices of open sets, and (therefore) also isomorphic lattices of compact saturated sets, the property transfers to *X*: *X* is consonant.

## An alternative end for our proof.

To sum up, we proved the Dolecki-Greco-Lechicki theorem (for regular Čech-complete spaces *X*) in four steps:

- For a fixed
*U*in, find open subsets**U***U*_{0},*U*_{1}, …,*U*, … and closed subsets_{n}*F*_{0},*F*_{1}, …,*F*, … such that_{n}*U*=*U*_{0}⊇*F*_{0}⊇*U*_{1}⊇*F*_{1}⊇ … ⊇*U*⊇_{n}*F*⊇ …, and such that for each_{n}*n*,*U*is in_{n}and is**U***1/2*-small. This used Čech-complete for the definition of^{n}*1/2*-smallness, but otherwise was mostly relying on regularity.^{n} - Define
as the filter of all open subsets that contain some**V***F*, and show that it is Scott-open. This is where we used Kőnig’s Lemma, and where Čech-completeness was required._{n} - In case
*X*is not only regular and Čech-complete, but also T_{2}, use the Hofmann-Mislove theorem to conclude thatis equal to ￭**V***Q*for some compact saturated subset*Q*. - Remove the T
_{2}assumption by taking T_{0}quotients.

I initially liked the use of the Hofmann-Mislove theorem in step 3, and this is why Exercise 8.3.4 is located in Section 8.3 (“The Hofmann-Mislove Theorem”).

However, one can also give a direct argument that replaces steps 3 and 4 above, and does not use the Hofmann-Mislove theorem either.

In step 2, we had actually proved something stronger that what we needed. We had shown the following. Take an arbitrary directed family (*U _{j}*)

*, and let*

_{j ∈ J}*F’*be the complement of

_{j}*U*. Assume that

_{j}*U*is not in

_{j}*, for any*

**V***j*, or equivalently that

*F’*intersects

_{j}*F*, for all

_{n}*j*and

*n*. Then ∩

_{j }*F’*intersects ∩

_{j}

_{n }*F*.

_{n}I claim that this is enough to show that *Q*=∩_{n }*F _{n }*is compact saturated, and that

*is the set of open neighborhoods of*

**V***Q*, without using the Hofmann-Mislove theorem.

*Q*is saturated: because*Q*is also equal to ∩_{n }*U*, an intersection of saturated sets._{n}*Q*is compact: take an arbitrary directed family (*U*)_{j}, and let_{j ∈ J}*F’*be the complement of_{j}*U*. Assuming that_{j}*U*does not contain_{j}*Q*, for any*j*, we must show that their union does not contain*Q*either. Equivalent, assume that*F’*intersects_{j}*Q*for every*j*. In particular,*F’*intersects_{j}*F*, for all_{n}*j*and*n*. We have seen that this implies that ∩_{j }*F’*intersects_{j}*Q*=∩_{n }*F*: this is what we proved in step 2. Therefore ∪_{n}_{j }*U*does not contain_{j }*Q*.- For every open
*U*,*U*is inif and only if**V***U*contains*Q*. The only if direction is clear, so we assume that*U*contains*Q*, and we wish to show that*U*contains some*F*. For that, we take the_{n}*constant*directed family (*U*)_{j}, where_{j ∈ J}*U*=_{j}*U*for every*j*. If*U*contained no*F*, then every_{n}*F’*(= the complement of_{j}*U*) would intersect every*F*. By (a trivial case of) what we proved in step 2, it would intersect_{n}*Q=∩*. That would contradict the fact that_{n }F_{n}*U*contains*Q*.

The purpose of the last step is to show that ￭*Q* is included in * U*. That last step indeed shows that ￭

*Q*=

*, and*

**V***, by construction, is included in*

**V***.*

**U**— Jean Goubault-Larrecq (August 25th, 2016)

- Dolecki, Szymon, Greco, Gabriele H., and Lechicki, Alojzy, 1995.
*When Do the Upper Kuratowski Topology (Homeomorphically, Scott Topology) and the Co-Compact Topology Coincide?*Transactions of the American Mathematical Society, 347(8), 2869–2884. -
Dolecki, Szymon, Greco, Gabriele H, 2010. Towards Historical Roots of Necessary Conditions of Optimality: Regula of Peano. Dedicated to professor Stefan Rolewicz on his 75-th birthday and Commemorating the 150-th anniversary of the birth of Giuseppe Peano.