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.
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 , 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 U, there is a compact saturated subset 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  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 U, then there is a compact saturated subset Q of X contained in U such that every open neighborhood of 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 , 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.  have many other theorems, but I will concentrate on that one.
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 Cn = (Uni)i ∈ I[n], one for each natural number n (the index set I[n] may depend on n, and can be arbitrary), with the property that for every filtered family (Fj)j ∈ J of non-empty closed subsets such that, for every n, there are indices i and j such that Fj ⊆ Uni., the intersection of all Fjs is non-empty. Wow.
The nth open cover Cn 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 1/2n+1 for Cn, say. Then every subset Fj of any Uni has diameter at most 1/2n. Any family (Fj)j ∈ J as above is then such that the diameters of the Fjs 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.
Hence we can rephrase the definition as follows. Given a countable collection of open covers Cn, say that a set has diameter at most 1/2n if and only if it is included in some element Uni of Cn. This is an abstract notion of diameter, nothing as concrete as the diameters you get in metric spaces.
A filtered family (Fj)j ∈ J has vanishing diameters if and only if one can find Fj of arbitrary small diameter. That is, if and only if, for every natural number n, one can find an index j such that Fj has diameter at most 1/2n.
A space is Čech-complete if and only if it has a countable collection of open covers such that every filtered family (Fj)j ∈ J of non-empty closed subsets with vanishing diameters has non-empty intersection.
It is usual to require Čech-complete spaces to be regular T2 (i.e., T3) 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 Cn, and we say that a set is 1/2n-small if and only if it is included in a finite union of sets of diameter at most 1/2n.
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 ∈ U, show that one can find open subsets U0, U1, …, Un, … and closed subsets F0, F1, …, Fn, … such that U=U0 ⊇ F0 ⊇ U1 ⊇ F1 ⊇ … ⊇ Un ⊇ Fn ⊇ …, and such that for each n, Un is in U and is 1/2n-small.”
This is really how Dolecki et al.’s proof starts, as well . Exercice 8.3.4 diverges from their proof at some later point.
Here is how you do it. Start by defining U0=U, and define all the other open and closed sets by induction on n. This means we can assume that we have already defined Un (in U), and we wish to construct Fn and Un+1. For each point x in Un, do the following. Since Cn+1 is a cover, x is also in some U(n+1)i ∈ Cn+1. Un ⋂ U(n+1)i is also an open neighborhood of x, but additionally, one that has diameter at most 1/2n+1.
By regularity (i.e., by local closure, see Exercise 4.1.21) x has a closed neighborhood Fnx that is included in Un ⋂ Uni. Note that Fnx again has diameter at most 1/2n+1. The family of interiors int(Fnx), when x varies over the points of Un, is a family of open subsets of Un whose union is the whole of Un, by construction.
By a familiar trick (Trick 4.4.6 in the book), Un is the directed union of the finite unions ∪x in E int(Fnx), where E ranges over the finite subsets of Un. Since Un is in U, and U is Scott-open, one of those directed unions ∪x in E[n] int(Fnx) is in U (where E[n] is some finite set — I am using the notation E[n] instead of the more standard subscript notation En 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:
- Un+1 as ∪x in E[n] int(Fnx): this is in U, and is 1/2n+1-small, by construction;
- Fn as ∪x in E[n] Fnx: this is closed because E[n] is finite, and finite unions of closed sets are closed.
As an additional bonus, note that Fn itself is 1/2n+1-small.
The exercise continues by giving you the following instruction: “Show that the family V of all open subsets of X that contain some Fn is a Scott-open filter.”
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 Fn is compact, and that an open set contains Q if and only if it contains some Un. 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 V 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.
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 vn starting from the root (i.e., v0 is the root), and such that vn+1 is a successor of vn, for each 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 v0 as the root of the tree. Since v0 has only finitely many successors, and there are infinitely many vertices strictly below v0, there must be infinitely many vertices below some successor v1 of v0. (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 v2 of v1, and so on. Collecting v0, v1, v2, … 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 (Uj)j ∈ J, and assuming that no Uj is in V, we must show that ∪j ∈ J Uj is not in V either.
Let F’j be the complement of Uj. Those are closed sets. That Uj is not in V means that F’j intersects Fn, for all j and n. Our goal is to show that ∩j F’j intersects Fn for every n, which will prove our claim. We shall in fact show the stronger statement that it intersects ∩n Fn.
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 F0 = ∪x in E F0x: create as many successors to the root as there are elements x in E, and label each by F0x. Call those vertices the level 0 vertices. For each level 0 vertex v, do a similar thing with F1 = ∪x in E F1x: create as many successors to v as there are elements x in E, and label each by the intersection of F1x with v‘s label. The vertices created this way are the level 1 vertices. Continue in the same way, creating level 2 vertices from F2 = ∪x in E F2x, as successors to all the level 1 vertices we created previously, and so on.
I will write Fv for the closed set that labels any (non-root) vertex v. Note that, for a level n vertex v, Fv is an intersection F0x ⋂ F1x ⋂ … ⋂ Fnx[n], for some points x, x, …, x[n] in E, E, …, E[n] respectively. Some of these sets Fv may be empty, but this will not stop us: we shall find enough of them that are non-empty. An important point is that Fv, being included in Fnx[n], is of diameter at most 1/2n+1.
Note also that the union of the sets Fv, when v ranges over all level n vertices (for fixed n), is exactly Fn. This is proved by an easy induction on n.
Now recally that Fn intersects every F’j, for every n. I claim that this implies that:
(*) For every n, there is a level n vertex v such that Fv intersects every F’j.
Indeed, otherwise, for each level n vertex v, there would be an index j[v] (depending on v) such that Fv does not intersect F’j[v]. By filteredness, one can find an index j (independent of v) such that F’j is included in every F’j[v]; this is possible because there are only finitely many level n vertices at all. Then, for each level n vertex v, Fv does not intersect F’j, and taking unions over all such vertices, we would obtain that Fn does not intersect F’j, which is absurd.
Select all the vertices v such that Fv intersects every F’j, plus the root. Those vertices span a subtree T’ 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 v’ is below v and v’ is selected, then v is selected, too. (I.e., if v‘ is below v and Fv’ intersects every F’j, then Fv also intersects every F’j: this is because Fv contains Fv’.)
By construction, T is finitely branching, so T’ is finitely branching, too. By claim (*) above, for every n, T’ contains a level n vertex. In particular, T’ is infinite. By Kőnig’s Lemma, T’ must contain an infinite branch: (the root, then) v, v, v, …
We are almost there. Look at the family of closed sets F’j ⋂ Fv[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 Fv[n] contains Fv[n+1], notably) of non-empty closed sets (non-empty by the way we selected the vertices of T’ from those of T), and with vanishing diameters (because Fv[n] being included in Fv[n], has diameter at most 1/2n+1, since v[n] is a level n vertex).
We can at last use Čech-completeness: the intersection ∩j,n (F’j ⋂ Fv[n]) is non-empty.
That intersection is the intersection of ∩j F’j with ∩n Fv[n] ⊆ ∩n Fn, so ∩j F’j intersects the larger set ∩n Fn, as we had claimed.
Finishing the proof.
This shows that V is Scott-complete, and we can finish the proof of the theorem. Since X is T2, it is sober (Proposition 8.2.12 in the book), so we can apply the Hofmann-Mislove theorem: V is the filter of open neighborhoods of a compact saturated set, namely Q=∩n Fn. Since Q ⊆ U0=U, U is in ￭Q. And every element W of ￭Q is in V, hence contains some Fn, and therefore also the smaller set Un+1; since Un+1 is in U, W is in U as well, so that ￭Q ⊆ U. As we have noticed earlier, this shows that X is consonant.
Removing the T2 axiom.
The Dolecki-Greco-Lechicki theorem is slightly more general: it says that every regular (not necessarily T2) 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 T2. Take its T0 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=0y if and only if x≤y and y≤x. As usual, ≤ is the specialization quasi-ordering of X.
We first check that X’ is T2. Assume [x]≠[y], that is, x≠0y. 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 T0 for it to be T2.)
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 U, find open subsets U0, U1, …, Un, … and closed subsets F0, F1, …, Fn, … such that U=U0 ⊇ F0 ⊇ U1 ⊇ F1 ⊇ … ⊇ Un ⊇ Fn ⊇ …, and such that for each n, Un is in U and is 1/2n-small. This used Čech-complete for the definition of 1/2n-smallness, but otherwise was mostly relying on regularity.
- Define V as the filter of all open subsets that contain some Fn, and show that it is Scott-open. This is where we used Kőnig’s Lemma, and where Čech-completeness was required.
- In case X is not only regular and Čech-complete, but also T2, use the Hofmann-Mislove theorem to conclude that V is equal to ￭Q for some compact saturated subset Q.
- Remove the T2 assumption by taking T0 quotients.
I initially liked the used of the Hofmann-Mislove theorem in step 3 here, 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 (Uj)j ∈ J, and let F’j be the complement of Uj. Assume that Uj is not in V, for any j, or equivalently that F’j intersects Fn, for all j and n. Then ∩j F’j intersects ∩n Fn.
I claim that this is enough to show that Q=∩n Fn is compact saturated, and that V is the set of open neighborhoods of Q, without using the Hofmann-Mislove theorem.
- Q is saturated: because Q is also equal to ∩n Un, an intersection of saturated sets.
- Q is compact: take an arbitrary directed family (Uj)j ∈ J, and let F’j be the complement of Uj. Assuming that Uj does not contain Q, for any j, we must show that their union does not contain Q either. Equivalent, assume that F’j intersects Q for every j. In particular, F’j intersects Fn, for all j and n. We have seen that this implies that ∩j F’j intersects Q=∩n Fn: this is what we proved in step 2. Therefore ∪j Uj does not contain Q.
- For every open U, U is in V if and only if 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 Fn. For that, we take the constant directed family (Uj)j ∈ J, where Uj=U for every j. If U contained no Fn, then every F’j (= the complement of U) would intersect every Fn. By (a trivial case of) what we proved in step 2, it would intersect Q=∩n Fn. That would contradict the fact that 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=V, and V, by construction, is included in U.
- 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: Regular of Peano. Dedicated to professor Stefan Rolewicz on his 75-th birthday and Commemorating the 150-th anniversary of the birth of Giuseppe Peano.