Recall that an ideal domain is a dcpo where every non-finite element is maximal. All ideal domains are algebraic and first-countable. The notion, and all the results presented here, are due to Keye Martin [1].
Last time, we have seen that every completely metrizable space X has an ideal model, that is, that X can be embedded into an ideal domain Y in such a way that we can equate X with the subspace of maximal elements of Y.
We have also seen the converse to that: if X is a metrizable space with an ideal model, then X is completely metrizable. The proof made use of Choquet-completeness, and eventually relied on the fact that, if X has an ideal model Y, then it has one in which it arises as a Gδ subset of Y. Keye Martin shows that the set of maximal elements of an ideal domain Y is always a Gδ subset of Y. The proof is pretty technical. We shall be happy to replace Y by another ideal domain Y‘ whose set of maximal elements will again be X, but will be more easily seen to be a Gδ subset of Y‘.
First-countable continuous posets
Recall that Y, being an ideal domain, is first-countable. We claim that this implies that every element x of X = Max Y (the set of maximal elements of Y) is the supremum of a countable chain of finite elements of Y. In fact, we have the more general result.
Lemma. A continuous poset Z is first-countable in its Scott topology if and only if every element is the supremum of a countable chain of elements way-below it.
This is similar to Norberg’s Lemma (Lemma 7.7.13 in the book) that a continuous poset is second-countable in its Scott topology if and only if it has a countable basis.
Proof. If every element z is the supremum of a countable chain (zn)n ∈ N of elements way-below z, then the family of open subsets ↟zn, n ∈ N, is a countable basis of open neighborhoods of z: for every open neighborhood U of z, some zn is in U, and then ↟zn is an open subset of U that contains z.
Conversely, if Z is first-countable in its Scott topology, then every element z has a countable basis of open neighborhoods Un, n ∈ N, and we can even assume that they form a decreasing chain (Exercise 4.7.14). Since Z is a continuous poset, z is the supremum of a directed family of elements way-below z, and one of them, call it z0, is in U0. The open subset ↟z0 contains z, and by the defining property of a basis of open neighborhoods, it must therefore contain some Un. Up to some reindexing, imagine that it contains U1. We repeat the argument, and find an element z1 way-below z, and (up to some reindexing again) ↟z1 contains U2. Continuing this way, we build z2, …, zn, … , and so on. They are all way-below z, zn is in Un, and ↟zn contains Un+1. To show that the supremum of that chain equals z, it suffices to show that every open neighborhood of z contains some zn. That is clear, since that open neighborhood must contain some Un, by the defining property of bases of open neighborhoods. ☐
By similar arguments, we also obtain:
Lemma. An algebraic poset Z is first-countable in its Scott topology if and only if every element is the supremum of a countable chain of finite elements below it.
Tweaking an ideal domain into a convenient one
Let us return to the ideal domain Y, with set of maximal elements X. As a consequence of the above lemma, every element x of X is the supremum of a countable chain of finite elements x[n] way-below x, n ∈ N.
We use the Axiom of Choice implicitly here: for each x in X, we fix a countable chain of finite elements x[n] whose supremum equals x.
We extend the notation x[n] to the case where n=+∞, and let x[+∞] = x.
Define a new poset Y‘ as follows. The elements of Y‘ are pairs (x, n) where x is in X and n is in N U {+∞}. Reserve the notation ≤ for the ordering on Y (or for the ordering on natural numbers n). The ordering ≤’ on Y‘ is given by:
- for all n, n‘ in N U {+∞}, (x, n) ≤’ (x‘, n‘) if and only if n≤n‘ and x[n] ≤ x’[n‘].
In other words, we compare elements (x, n) as though they really denote x[n], except that we first compare their level n: an element (x, n) can only be below (x‘, n‘) if its level n is below the level n‘ of (x‘, n‘). We naturally equate x in X with the element (x, +∞).
Lemma. Every directed family (xi, ni)i ∈ I in Y‘ has a supremum, and it is equal to (supi ∈ I xi[ni], supi ∈ I ni).
Proof. Let (x, n) = (supi ∈ I xi[ni], supi ∈ I ni). This is clearly an upper bound of the family, and if (x‘, n‘) is any other upper bound, then ni≤n‘ for every i, so n=supi ∈ I ni≤n‘, and xi[ni]≤x‘ for every i, so x=supi ∈ I xi[ni]≤x‘. ☐
Lemma. For every n in N, for every x in X, (x, n) is a finite element of Y‘.
Proof. Assume (x, n) is below some directed supremum (supi ∈ I xi[ni], supi ∈ I ni), i.e., for some monotone net (xi, ni)i ∈ I in Y‘. Since n is finite, n≤ni for i large enough. By restricting to those indices i such that n≤ni, we can therefore assume that n≤ni for every i in I.
If supi ∈ I ni=+∞, then (x, n) ≤’ (supi ∈ I xi[ni], supi ∈ I ni) is equivalent to x[n] ≤ supi ∈ I xi[ni], in which case x[n] ≤ xi[ni] for i large enough, since x[n] is finite in Y. Then (x, n) ≤’ (xi, ni).
If supi ∈ I ni<+∞, then let n‘=supi ∈ I ni. For i large enough, ni=n‘. By reindexing again, we may assume that ni=n‘ for every i. Observe that (xi, ni)≤’ (xj, nj) if and only if xi ≤ xj (since ni=nj=n‘) if and only if xi = xj (since the elements of X are maximal in Y, hence incomparable or equal). Since (xi, ni)i ∈ I is directed (and ni=n‘ for every i in I), all its elements are equal, hence also equal to their supremum. The fact that x[n] ≤ xi[ni] for some i in I (in fact, all), is then obvious. ☐
Lemma. For every x in X, (x, +∞) is a maximal element of Y‘, and is not finite.
Proof. It is maximal: if (x, +∞) ≤’ (x‘, n‘), then n‘=+∞ and x=x[+∞] ≤ x‘[+∞]=x‘, from which we obtain x=x’, since any two comparable elements in X, being maximal in Y, must be equal.
By definition, (x, +∞) is the supremum of the directed family (x, n), n in N. Since (x, +∞) ≤’ (x, n) for no n in N, (x, +∞) is not finite. ☐
Finding X as a Gδ subset of Y‘
We are almost through. With Y‘ constructed as above, for every n in N, the set Un defined as the union of the sets ↑(x, n), x in X, is a union of open sets, hence is open. This is the set of elements “at level n or higher”.
Every element x of X, equated with (x, +∞), is in every Un. Conversely, any element that is in the intersection of the sets Un must have a level larger than any natural number, hence be of the form (x, +∞).
This shows that X, the set of maximal elements of Y‘, is a countable intersection of open subsets, namely:
Proposition. X is a Gδ subset of Y‘.
We have already seen last time that this was the final touch to the following theorem, due to Keye Martin [1]. The key is that Y‘, as an algebraic, hence continuous dcpo, is Choquet-complete, that a Gδ subset of a Choquet-complete space is itself Choquet-complete, and that a Choquet-complete metrizable space is completely metrizable.
Theorem (Martin [1]). The metrizable spaces that have an ideal model are exactly the completely metrizable spaces.
Next time, if all goes well, I’ll explain the connection there is to so-called remainders [2].
— Jean Goubault-Larrecq (January 3rd, 2016)
- Keye Martin. Ideal models of spaces. Theoretical Computer Science, Volume 305, Issues 1–3, 18 August 2003, Pages 277–297.
- Hoffmann, R.-E. On the sobrification remainder sX − X . Pacific Journal of Mathematics, 83(1), 1979, pages 145–156.