{"id":808,"date":"2016-01-03T15:54:05","date_gmt":"2016-01-03T14:54:05","guid":{"rendered":"https:\/\/projects.lsv.ens-paris-saclay.fr\/topology\/?page_id=808"},"modified":"2022-11-19T15:25:56","modified_gmt":"2022-11-19T14:25:56","slug":"ideal-models-ii","status":"publish","type":"page","link":"https:\/\/projects.lsv.ens-paris-saclay.fr\/topology\/?page_id=808","title":{"rendered":"Ideal models II"},"content":{"rendered":"

Recall that an ideal domain is a dcpo where every non-finite element is maximal.\u00a0 All ideal domains are algebraic and first-countable.\u00a0 The notion, and all the results presented here, are due to Keye Martin [1].<\/p>\n

Last time, we have seen that every completely metrizable space X<\/em> has an ideal model, that is, that X<\/em> can be embedded into an ideal domain Y <\/em>in such a way that we can equate X<\/em> with the subspace of maximal elements of Y.<\/p>\n

We have also seen the converse to that: if X<\/em> is a metrizable space with an ideal model, then X<\/em> is completely metrizable.\u00a0 The proof made use of Choquet-completeness, and eventually relied on the fact that, if X<\/em> has an ideal model Y<\/em>, then it has one in which it arises as a G\u03b4<\/sub> subset of Y<\/em>.\u00a0 Keye Martin shows that the set of maximal elements of an ideal domain Y<\/em> is always<\/em> a G\u03b4<\/sub> subset of Y<\/em>.\u00a0 The proof is pretty technical.\u00a0 We shall be happy to replace Y<\/em> by another ideal domain Y<\/em>‘ whose set of maximal elements will again be\u00a0X<\/em>, but will be more easily seen to be a G\u03b4<\/sub> subset of Y<\/em>‘.<\/p>\n

First-countable continuous posets<\/h2>\n

Recall that Y<\/em>, being an ideal domain, is first-countable.\u00a0 We claim that this implies that every element x<\/em> of X<\/em> = Max Y<\/em> (the set of maximal elements of Y<\/em>) is the supremum of a countable chain of finite elements of Y<\/em>.\u00a0 In fact, we have the more general result.<\/p>\n

Lemma<\/strong>. A continuous poset Z<\/em> is first-countable in its Scott topology if and only if every element is the supremum of a countable chain<\/em> of elements way-below it.<\/p>\n

This is similar to Norberg’s Lemma (Lemma 7.7.13 in the book<\/a>) that a continuous poset is second-countable in its Scott topology if and only if it has a countable basis.<\/p>\n

Proof.<\/strong> If every element z<\/em> is the supremum of a countable chain (zn<\/sub><\/em>)n \u2208 N<\/strong><\/sub><\/em> of elements way-below z, then the family of open subsets \u219fzn<\/sub><\/em>, n<\/em> \u2208 N<\/strong>, is a countable basis of open neighborhoods of z<\/em>: for every open neighborhood U<\/em> of z<\/em>, some\u00a0zn<\/sub><\/em> is in U<\/em>, and then \u219fzn<\/sub><\/em> is an open subset of U that contains z<\/em>.<\/p>\n

Conversely, if Z<\/em> is first-countable in its Scott topology, then every element z<\/em> has a countable basis of open neighborhoods Un<\/sub><\/em>, n<\/em> \u2208 N<\/strong>, and we can even assume that they form a decreasing chain (Exercise 4.7.14). Since Z<\/em> is a continuous poset, z<\/em> is the supremum of a directed family of elements way-below z<\/em>, and one of them, call it z<\/em>0<\/sub>, is in U<\/em>0<\/sub>.\u00a0 The open subset \u219fz<\/em>0<\/sub> contains z<\/em>, and by the defining property of a basis of open neighborhoods, it must therefore contain some Un<\/sub><\/em>.\u00a0 Up to some reindexing, imagine that it contains U<\/em>1<\/sub>.\u00a0 We repeat the argument, and find an element\u00a0z<\/em>1<\/sub> way-below z<\/em>, and (up to some reindexing again) \u219fz<\/em>1<\/sub> contains U<\/em>2<\/sub>.\u00a0 Continuing this way, we build z<\/em>2<\/sub>, …, zn<\/sub><\/em>, … , and so on.\u00a0 They are all way-below\u00a0z<\/em>,\u00a0zn<\/sub><\/em> is in Un<\/sub><\/em>, and \u219fzn<\/sub><\/em> contains Un+1<\/sub><\/em>.\u00a0 To show that the supremum of that chain equals z<\/em>, it suffices to show that every open neighborhood of z<\/em> contains some zn<\/sub><\/em>.\u00a0 That is clear, since that open neighborhood must contain some Un<\/sub><\/em>, by the defining property of bases of open neighborhoods.\u00a0 \u2610<\/p>\n

By similar arguments, we also obtain:<\/p>\n

Lemma<\/strong>. An algebraic poset Z<\/em> is first-countable in its Scott topology if and only if every element is the supremum of a countable chain<\/em> of finite elements below it.<\/p>\n

Tweaking an ideal domain into a convenient one<\/h2>\n

Let us return to the ideal domain Y<\/em>, with set of maximal elements X<\/em>.\u00a0 As a consequence of the above lemma, every element x<\/em> of X<\/em> is the supremum of a countable chain of finite elements\u00a0x[n<\/em>] way-below x<\/em>, n<\/em> \u2208 N<\/strong>.<\/p>\n

We use the Axiom of Choice implicitly here: for each x<\/em> in X<\/em>, we fix a countable chain of finite elements x[n<\/em>] whose supremum equals x<\/em>.<\/p>\n

We extend the notation x<\/em>[n<\/em>] to the case where n<\/em>=+\u221e, and let x<\/em>[+\u221e] = x<\/em>.<\/p>\n

Define a new poset Y<\/em>‘ as follows.\u00a0 The elements of Y<\/em>‘ are pairs (x<\/em>, n<\/em>) where x<\/em> is in X<\/em> and n<\/em> is in N<\/strong> U {+\u221e}.\u00a0 Reserve the notation \u2264 for the ordering on Y<\/em> (or for the ordering on natural numbers n<\/em>).\u00a0 The ordering \u2264’ on Y<\/em>‘ is given by:<\/p>\n