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 [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, 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 [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)jgl-2011

  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.

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