# Ideal models II

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 .

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, nN, 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, nN, 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 zzn 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, nN.

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 nn‘ 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 nin‘ 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, nni for i large enough.  By restricting to those indices i such that nni, we can therefore assume that nni 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 xixj (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 .  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 ). 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 .

Jean Goubault-Larrecq (January 3rd, 2016) 1. Keye Martin.  Ideal models of spaces.  Theoretical Computer ScienceVolume 305, Issues 1–3, 18 August 2003, Pages 277–297.
2. Hoffmann, R.-E. On the sobrification remainder sXX . Pacific Journal of Mathematics, 83(1), 1979, pages 145–156.