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 G_{δ} subset of *Y*. Keye Martin shows that the set of maximal elements of an ideal domain *Y* is *always* a G_{δ} 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 G_{δ} 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 (*z _{n}*)

*of elements way-below z, then the family of open subsets ↟*

_{n ∈ N}*z*,

_{n}*n*∈

**N**, is a countable basis of open neighborhoods of

*z*: for every open neighborhood

*U*of

*z*, some

*z*is in

_{n}*U*, and then ↟

*z*is an open subset of U that contains

_{n}*z*.

Conversely, if *Z* is first-countable in its Scott topology, then every element *z* has a countable basis of open neighborhoods *U _{n}*,

*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

*z*

_{0}, is in

*U*

_{0}. The open subset ↟

*z*

_{0}contains

*z*, and by the defining property of a basis of open neighborhoods, it must therefore contain some

*U*. Up to some reindexing, imagine that it contains

_{n}*U*

_{1}. We repeat the argument, and find an element

*z*

_{1}way-below

*z*, and (up to some reindexing again) ↟

*z*

_{1}contains

*U*

_{2}. Continuing this way, we build

*z*

_{2}, …,

*z*, … , and so on. They are all way-below

_{n}*z*,

*z*is in

_{n}*U*, and ↟

_{n}*z*contains

_{n}*U*. To show that the supremum of that chain equals

_{n+1}*z*, it suffices to show that every open neighborhood of

*z*contains some

*z*. That is clear, since that open neighborhood must contain some

_{n}*U*, by the defining property of bases of open neighborhoods. ☐

_{n}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 (*x _{i}*,

*n*)

_{i}*in*

_{i ∈ I}*Y*‘ has a supremum, and it is equal to (sup

_{i ∈ I}*x*[

_{i}*n*], sup

_{i}

_{i ∈ I}*n*).

_{i}**Proof.** Let (*x*, *n*) = (sup_{i ∈ I}*x _{i}*[

*n*], sup

_{i}

_{i ∈ I}*n*). This is clearly an upper bound of the family, and if (

_{i}*x*‘,

*n*‘) is any other upper bound, then

*n*≤

_{i}*n*‘ for every

*i*, so

*n*=sup

_{i}

*‘, and*

_{ ∈ I}*n*≤n_{i}*x*[

_{i}*n*]≤

_{i}*x*‘ for every

*i*, so

*x*=sup

_{i ∈ I}*x*[

_{i}*n*]≤

_{i}*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 (sup_{i ∈ I}*x _{i}*[

*n*], sup

_{i}

_{i ∈ I}*n*), i.e., for some monotone net (

_{i}*x*,

_{i}*n*)

_{i}*in*

_{i ∈ I}*Y*‘. Since

*n*is finite,

*n*≤

*n*for

_{i}*i*large enough. By restricting to those indices

*i*such that

*n*≤

*n*, we can therefore assume that

_{i}*n*≤

*n*for every

_{i}*i*in

*I*.

If sup_{i ∈ I}*n _{i}*=+∞, then (

*x*,

*n*) ≤’ (sup

_{i ∈ I}*x*[

_{i}*n*], sup

_{i}

_{i ∈ I}*n*) is equivalent to

_{i}*x*[

*n*] ≤ sup

_{i ∈ I}*x*[

_{i}*n*], in which case

_{i}*x*[

*n*] ≤

*x*[

_{i}*n*] for

_{i}*i*large enough, since

*x*[

*n*] is finite in

*Y*. Then (

*x*,

*n*) ≤’ (

*x*,

_{i}*n*).

_{i}If sup_{i ∈ I}*n _{i}*<+∞, then let

*n*‘=sup

_{i ∈ I}*n*. For

_{i}*i*large enough,

*n*=

_{i}*n*‘. By reindexing again, we may assume that

*n*=

_{i}*n*‘ for every

*i*. Observe that (

*x*,

_{i}*n*)≤’ (

_{i}*x*,

_{j}*n*) if and only if

_{j}*x*≤

_{i}*x*(since

_{j}*n*=

_{i}*n*=

_{j}*n*‘) if and only if

*x*=

_{i}*x*(since the elements of

_{j}*X*are maximal in

*Y*, hence incomparable or equal). Since (

*x*,

_{i}*n*)

_{i}*is directed (and*

_{i ∈ I}*n*=

_{i}*n*‘ for every

*i*in

*I*), all its elements are equal, hence also equal to their supremum. The fact that

*x*[

*n*] ≤

*x*[

_{i}*n*] for some

_{i}*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 G_{δ} subset of *Y*‘

We are almost through. With *Y*‘ constructed as above, for every *n* in **N**, the set *U _{n}* 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 *U _{n}*. Conversely, any element that is in the intersection of the sets

*U*must have a level larger than any natural number, hence be of the form (

_{n}*x*,

*+∞*).

This shows that *X*, the set of maximal elements of *Y*‘, is a countable intersection of open subsets, namely:

**Proposition.** *X* is a G_{δ} 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 G_{δ} 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
^{s}*X*−*X*. Pacific Journal of Mathematics, 83(1), 1979, pages 145–156.