On countability: the compact completed sequence

A note by Matthew de Brecht

Recently, Matthew de Brecht sent me a clever argument that shows the following:

Theorem (de Brecht, private communication, March 29th, 2019): The poset-theoretic product of two first-countable posets coincides with their topological product.

He insists that the ideas come straight out of a paper by Matthias Schröder [1]. My impression is that this still requires some thinking after reading [1], but certainly M. Schröder has had plenty of clever ideas.

What M. de Brecht’s theorem above means, in detail, is the following. Let X and Y be two posets. You can consider them as topological spaces, written Xσ and Yσ respectively, by giving them their Scott topology. Then you can build their topological product Xσ × Yσ. Or you can first build their product and then take the Scott topology of the result, yielding (X × Y)σ. In general, the two results differ: the Scott topology on X × Y is finer, and in general strictly finer than the product topology on Xσ × Yσ (see Exercise 5.2.16 in the book). What M. de Brecht shows is that there is never such a problem if Xσ and Yσ are first-countable.

This is a pretty surprising result. It is well-known that the problem disappears, namely Xσ × Yσ=(X × Y)σ, if X and Y are continuous posets (Proposition 5.1.54). What we learn is that, for first-countable posets, continuity is not needed.

The completed sequence is compact

The key to that result (apart from other clever arguments, which we will see below) is the following easy observation:

Fact 1: Let X be a topological space, (xn)n ∈ be a sequence of elements of X, and x be any limit of that sequence. Then the set {xn | n ∈ ℕ} ∪ {x} (the completed sequence of the title) is compact.

That is easy to prove. Consider any open cover of that set. One open set in that cover must contain x, and since (xn)n ∈ converges to x, it must contain all the elements xn for n large enough. The finitely many remaining elements can then be covered by finitely many of the remaining open sets from the cover.

It is pretty easy to see that one cannot generalize Fact 1 to nets indexed by anything else than ℕ. That already fails for nets indexed by ω.2, the ordinal with elements 0<1<…<n<… < ω<ω+1<…<ω+n<…, unless you require that xω be a limit of the subsequence of elements xn, n<ω. The situation is worse for chains indexed by ordinals above ω2, and even worse for nets that are not ordinal-indexed chains.

The proof

Let me reproduce Matthew de Brecht’s argument. Since every open set in the product topology is Scott-open, it suffices to show the converse implication. Let W be a Scott-open subset of X × Y. We will show that W is open in the product of the Scott topologies.

We first define f : XOY, where OY is the complete lattice of open subsets of Y, by letting f(x) be the set of points y of Y such that (x, y) is in W. Then f is a Scott-continuous map. This is a trick that is already used in the red book [2, proof of Theorem II-4.13(1)]. The argument used by the authors of that book is rather abstract, but that is also easily proved by hand, hence I will let you do the exercise; don’t forget to verify that f(x) is in OY for every x in X first!

In order to show that W is open in the product topology, we fix an arbitrary pair (x, y) in W, and we wish to show that there is an open rectangle containing (x, y) and included in W.

Let us imagine that this failed for some pair (x, y) in W. Using first-countability, we enumerate a base of open neighborhoods (Un)n ∈ of x, and a base of open neighborhoods (Vn)n ∈ of y. We can assume that those bases form descending sequences: otherwise we replace each Un by the intersection U0 U1 ∩ … ∩ Un for every n, and similarly with Vn.

By assumption, none of the open rectangles Un × Vn is included in W. Therefore, we can find a point (xn, yn) in Un × Vn and outside W.

We note that (xn)n ∈ converges to x: every open neighborhood of x must contain some Un, by the definition of a base of open neighborhoods, and then every xm with mn, is in Um, hence in Un (because we have made sure that the sequence (Um)m is descending). Similarly, (yn)n ∈ converges to y.

Since f(x) is open in Y and y is in f(x), every yn is in f(x) for n large enough, say nn0. We consider the completed sequence K defined as {yn | nn0} ∪ {y}. By Fact 1, K is a compact set, and it is included in the open set f(x).

We now consider the set ■K of all open subsets of Y that contain K. Since K is compact, ■K is Scott-open in OY. Indeed, it is upwards-closed, and every directed family (Ui)i I of open sets whose union is in ■K, i.e., which contains K, must be such that some Ui already contains K, i.e., is in ■K (Proposition 4.4.7 in the book).

Since f is continuous, f-1(■K) is open in X.

We now recall that (xn)n ∈ converges to x. And x is in f-1(■K), since K is included in f(x), as we have observed. It follows that xn is in f-1(■K) for n large enough. Let us pick any such n, in such a way that nn0. We have that xn is in f-1(■K), namely K is included in f(xn). In particular yn is in f(xn), which means that (xn, yn) is in W. But that is impossible, since all such pairs were chosen outside W. ☐

Exercise 5.2.16 in the book

Recall that Exercise 5.2.16 in the book asks you to show that the product and Scott topologies on X × Y do not agree, where X is Johnstone’s dcpo J, and Y is O(J). The idea of the exercise is to show that the graph (∈) = {(x, U) | x in J, U Scott-open in J} of the membership relation is Scott-open. Since Jσ is not core-compact (Exercise 5.2.15), (∈) is not open in the product topology, by Exercise 5.2.7.

Hence J or O(J) is not first-countable in its Scott topology. Which one?

I claim the following (which I have never seen in print, by the way):

Proposition. Johnstone’s dcpo J is not first-countable in its Scott topology.

This may look rather surprising, since J is countable. If you know about the sequential fan, you already know that there are countable topological spaces that are not first-countable. What the proposition above says is that there are even dcpos with this property. The proof is a diagonalization argument similar to the proof that the sequential fan is not first-countable, albeit slightly more complicated.

Let us prove the proposition. We need to understand the structure of open subsets of Jσ. Recall that J is the set of pairs (m, n) where m is a natural number and n is a natural number or ω. We go up in J by increasing n, or by going in one step from (m, n) to (n’, ω) with nn’, in which case we can no longer go up strictly afterwards.

For every open set U, if an element of the form (m, ω) is in U, then it contains elements of the form (m, n) with n large enough, because (m, ω) is the supremum of the chain {(m, n) | n ∈ ℕ}. Let fU(m) denote the smallest n such that (m, n) is in U. If (m, ω) is not in U, then let us define fU(m) as a special symbol ∞, which we place strictly above ω. This way, U is exactly the set of pairs (m, n) where nfU(m).

The function fU is not quite arbitrary: for every natural number m such that fU(m)≠∞, since (m, fU(m))≤(n, ω) for every nfU(m), every such (n, ω) is in U as well, so fU(n)≠∞. In short: if fU(m)≠∞, then for every nfU(m), fU(n)≠∞.

(In particular, if fU(m)≠∞ for some m, then fU(n)≠∞ for all n except finitely many.)

Let us imagine that the point (0, ω) had a countable base of open neighborhoods. The argument is the same for all other points (m, ω). We enumerate those open neighborhoods as U[0], U[1], etc. For each m, U[m] contains (0, ω), so fU[m](0)≠∞, and we have just noticed that this implies that for every nfU[m](0), fU[m](n)≠∞. We can then define an increasing sequence n0<n1<…<nm<… by induction on m in such that a way that nmfU[m](0). This way, fU[m](nm) is different from ∞ for every m.

That being done, we define a function g by g(nm)=fU[m](nm)+1, and g(n)=0 for all natural numbers n that are not in the list n0, n1, , nm,… Let V be the set of points (n, k) such that kg(n).

I claim that V is upwards-closed. If kg(n)and you increase k, the result is still above g(n). The interesting case is when we move from (n, k) to (k’, ω), where k’k: the point is that (k’, ω) is in V for every natural number k’, because we have built g in such a way that g(k’) is always different from ∞.

V is Scott-open, too. In J, the only non-trivial directed sets, that is those that do not contain a largest element, are sets of elements (n, k) with the same n and arbitrarily large values of k. It is easy to see that if the supremum of those k values is above g(n), then one of those k values is already above g(n). Therefore V is Scott-open.

Finally, V is a Scott-open neighborhood of (0, ω) (and in fact of every point of the form (m, ω)). Now, if V contains U[m] for some m, then the point (nm, fU[m](nm)), which is in U[m], must be in V, but that is impossible, by the way we constructed g and V. Therefore V contains no U[m] at all. Hence (0, ω) (and every point of the form (m, ω)) does not have any countable base of open neighborhoods. ☐

  1. Matthias Schröder. A Hofmann-Mislove theorem for Scott-open sets. arXiv:1501.06452, January 26th, 2015.
  2. Gerhard Gierz, Karl Heinrich Hofmann, Klaus Keimel, Jimmie D. Lawson, Michael W. Mislove, and Dana S. Scott. Continuous Lattices and Domains. Number 93 in Encyclopedia of Mathematics and its Applications. Cambridge University Press, Cambridge, 2003.

Jean Goubault-Larrecq (May 20th, 2019)