{"id":5109,"date":"2022-06-20T11:21:05","date_gmt":"2022-06-20T09:21:05","guid":{"rendered":"https:\/\/projects.lsv.ens-paris-saclay.fr\/topology\/?page_id=5109"},"modified":"2022-11-19T14:54:02","modified_gmt":"2022-11-19T13:54:02","slug":"q-is-not-consonant-the-costantini-watson-argument","status":"publish","type":"page","link":"https:\/\/projects.lsv.ens-paris-saclay.fr\/topology\/?page_id=5109","title":{"rendered":"Q is not consonant: the Costantini-Watson argument"},"content":{"rendered":"\n

I have already given an argument for the non-consonance of the Sorgenfrey line R<\/strong>\u2113<\/sub> here<\/a>. As I have said last time<\/a>, I would like to deal with the case of the space Q<\/strong> of rational numbers, and explain an argument due to Costantini and Watson [4]. Before we start, let me make an apology. This is probably the longest post I will ever have made, and it is pretty technical, too. Nonetheless, I have made my best to make the argument readable. Please forgive me if you find that all that is too long!<\/p>\n\n\n\n

A bit of history<\/h2>\n\n\n\n

That Q<\/strong> is not consonant was first proved by A. Bouziad [2]. His argument is a pretty simple measure theoretic argument, which shows that every consonant space is a Prohorov space, namely a space X<\/em> such that the compact subsets of P<\/em>(X<\/em>) are exactly the uniformly tight sets\u2014P<\/em>(X<\/em>) is the space of Radon probability measures on X<\/em>, with the weak topology. As you may have guessed, this requires some knowledge of topological measure theory, which I will not get into. But this is not so difficult to acquire. The real difficulty is the second step: you need to argue that Q<\/strong> is not a Prohorov space, and that is a deep and technically demanding result due to D. Preiss [3].<\/p>\n\n\n\n

Another argument was proposed by C. Costantini and S. Watson [4] a few years later. This one is much easier to explain. It is an elementary argument, one which builds fairly strange objects, but at least one that one can describe explicitly. Note that by elementary<\/em>, I mean an argument that does not require elaborate notions or a lot of prior knowledge; an elementary argument need not be simple<\/em>.<\/p>\n\n\n\n

Consonant spaces<\/h2>\n\n\n\n

As promised, let me recall that what a consonant<\/em> space is. I have already spoken about them, for example here<\/a>, here<\/a>, or much more recently, here<\/a>, but I guess that a refresher is never inappropriate.<\/p>\n\n\n\n

Given any topological space X<\/em>, you can form its lattice of open sets O<\/strong>X<\/em>, and then consider it with the Scott topology of inclusion. Among the Scott-open subsets of O<\/strong>X<\/em>, you can find the sets of the form \u25a0Q<\/em>, where Q<\/em> is compact saturated in X<\/em>: \u25a0Q<\/em> is the collection of open neighborhoods of Q<\/em>, and the fact that Q<\/em> is compact entails that \u25a0Q<\/em> is Scott-open in O<\/strong>X<\/em>. In fact, if X<\/em> is sober, those sets \u25a0Q<\/em> are the only Scott-open filters<\/em> that exist, by the Hofmann-Mislove theorem (Theorem 8.3.2 in the book<\/a>). Any union of sets of the form \u25a0Q<\/em> is Scott-open in O<\/strong>X<\/em>, but does the converse hold? A space X<\/em> is consonant<\/em> precisely when this is the case, namely precisely when every Scott-open subset of O<\/strong>X<\/em> is a union of sets of the form \u25a0Q<\/em> with Q<\/em> compact saturated.<\/p>\n\n\n\n

The notion, introduced in [1], crops up regularly.<\/p>\n\n\n\n

The consonant spaces are exactly those spaces X<\/em> such that the Isbell and compact-open topologies coincide, on any space of continuous maps from X<\/em> to any topological space Y<\/em> (or just for Y<\/em>=S<\/strong>, the Sierpi\u0144ski space).<\/p>\n\n\n\n

Every locally compact space is consonant (Exercise 5.4.12), and every T3<\/sub> \u010cech-complete space is consonant (Exercise 8.3.4, namely the Dolecki-Greco-Lechicki theorem: Theorem 4.1 in [1]). Hence in particular Baire space NN<\/sup><\/strong> is consonant, although it is not locally compact. Since then, we have learned that all quasi-Polish spaces, and even all LCS-complete spaces are consonant [5, Proposition 12.1].<\/p>\n\n\n\n

In an important paper [6], M. de Brecht and T. Kawai have considered the question of the commutativity between the Hoare and Smyth hyperspace constructions. (It had been well known that the Hoare and Smyth powerlocales<\/em> commute [7], without any assumption. But power locales and hyperspaces do not always coincide.) Let me not describe the question fully, and let me remain elusive about a few details. There is a Hoare hyperspace<\/a> H<\/strong>(X<\/em>) consisting of the closed subsets of X<\/em>, with the so-called lower Vietoris topology. There is also a Smyth hyperspace Q<\/strong>(X<\/em>) consisting of the compact saturated subsets of X<\/em>, with the upper Vietoris topology. Now there is a continuous map \u03c3 : H<\/strong>(Q<\/strong>(X<\/em>)) \u2192 Q<\/strong>(H<\/strong>(X<\/em>)), which maps every closed subset C<\/strong> of Q<\/strong>(X<\/em>) to {C<\/em> \u2208 H<\/strong>(X<\/em>) | C<\/em> intersects every compact subset of X<\/em> that is in C<\/strong>}. De Brecht and Kawai show that \u03c3 is a homeomorphism if and only if X<\/em> is consonant.<\/p>\n\n\n\n

All right! Let us start attacking the question of the failure of consonance of Q<\/strong> for good.<\/p>\n\n\n\n

Q<\/strong> is zero-dimensional<\/h2>\n\n\n\n

Q<\/strong>, as a subspace of R<\/strong>, is second-countable. Contrarily to R<\/strong>, it is zero-dimensional<\/em>, meaning that its topology has a base of clopen sets. (A set is clopen<\/em> if and only if it is both open and closed.)<\/p>\n\n\n\n

Explicitly, for every irrational number z<\/em>, the set Q<\/strong> \u2229 ]\u2013\u221e, z<\/em>[ is open. Its complement in Q<\/strong> is the set Q<\/strong> \u2229 ]z<\/em>, +\u221e[, since z<\/em> is irrational, and that complement is open, too. Hence both Q<\/strong> \u2229 ]\u2013\u221e, z<\/em>[ and Q<\/strong> \u2229 ]z<\/em>, +\u221e[ are clopen. It follows that, for any two irrational numbers z<\/em><z’<\/em>, Q<\/strong> \u2229 ]z<\/em>, z’<\/em>[ is also clopen, as the intersection of Q<\/strong> \u2229 ]\u2013\u221e, z’<\/em>[ and of Q<\/strong> \u2229 ]z<\/em>, +\u221e[. Now, given any point x<\/em> of Q<\/strong> and any open neighborhood Q<\/strong> \u2229 U<\/em> of x<\/em> in Q<\/strong> (where U<\/em> is open in R<\/strong>), U<\/em> contains an open interval ]a<\/em>, b<\/em>[ that contains x<\/em>, and then there are irrational numbers z<\/em> and z’<\/em> such that a<\/em><z<\/em><x<\/em> and x<\/em><z’<\/em><b<\/em>. Hence the clopen set Q<\/strong> \u2229 ]z<\/em>, z’<\/em>[ is an open neighborhood of x<\/em> included in Q<\/strong> \u2229 U<\/em>. This shows that the sets Q<\/strong> \u2229 ]z<\/em>, z’<\/em>[ with z<\/em><z’<\/em> irrational form a base of clopen sets.<\/p>\n\n\n\n

This base is uncountable, but there must be a countable one, by the following observation, and the fact that Q<\/strong> is second-countable (its topology is generated by sets Q<\/strong> \u2229 ]a<\/em>, b<\/em>[ where a<\/em><b<\/em> are both rational<\/em>).<\/p>\n\n\n\n

Lemma A.<\/strong> Every second-countable, zero-dimensional space has a countable<\/em> base of clopen sets.<\/p>\n\n\n\n

Proof.<\/em> Let Vn<\/sub><\/em>, n<\/em> \u2208 N<\/strong>, enumerate the elements of a given countable base of the topology of a second-countable, zero-dimensional space X<\/em>. Let me say that the pair of natural numbers (m<\/em>, n<\/em>) is good<\/em> if and only if not only Vm<\/sub><\/em> is included in Vn<\/sub><\/em>, but there is even a clopen set O<\/em> such that Vm<\/sub><\/em> \u2286 O<\/em> \u2286 Vn<\/sub><\/em>. For every good pair (m<\/em>, n<\/em>), we pick exactly one clopen set Omn<\/sub><\/em> such that Vm<\/sub><\/em> \u2286 Omn<\/sub><\/em> \u2286 Vn<\/sub><\/em>. (Yes, we use to axiom of (countable) choice to do so.) I claim that the sets Omn<\/sub><\/em> form a countable base of clopen sets.<\/p>\n\n\n\n

To this end, it suffices to show that they form a base. Let x<\/em> be any point of X<\/em>, and U<\/em> be any open neighborhood of x<\/em> in X<\/em>. Since the sets Vn<\/sub><\/em> form a base of the topology, we can find one such that x<\/em> \u2208 Vn<\/sub><\/em> \u2286 U<\/em>. Since X<\/em> is zero-dimensional, there is a clopen set O<\/em> such that x<\/em> \u2208 O<\/em> \u2286 Vn<\/sub><\/em>. Since the sets Vm<\/sub><\/em>, m<\/em> \u2208 N<\/strong>, form a base, we can find one such that x<\/em> \u2208 Vm<\/sub><\/em> \u2286 O<\/em>. To sum up, we have x<\/em> \u2208 Vm<\/sub><\/em> \u2286 O<\/em> \u2286 Vn<\/sub><\/em> \u2286 U<\/em>. In particular, (m<\/em>, n<\/em>) is a good pair. By definition, we have x<\/em> \u2208 Vm<\/sub><\/em> \u2286 Omn<\/sub><\/em><\/em> \u2286 Vn<\/sub><\/em> \u2286 U<\/em>, in particular x<\/em> \u2208 Omn<\/sub><\/em><\/em> \u2286 U<\/em>, and this concludes the proof. \u2610<\/p>\n\n\n\n

The sketch of Costantini and Watson’s argument<\/h2>\n\n\n\n

Costantini and Watson’s argument applies to any non-empty space that is:<\/p>\n\n\n\n

    \n
  1. second-countable, zero-dimensional (we have just seen that Q<\/strong> fits)<\/li>\n\n\n\n
  2. in which every compact subset is scattered (this is the case of Q<\/strong>, as we have seen in Lemma A last time<\/a>)<\/li>\n\n\n\n
  3. and with no non-empty compact open subset.<\/li>\n<\/ol>\n\n\n\n

    This is not exactly what Theorem 1 of [4] assumes. There, Costantini and Watson assume a zero-dimensional separable metrizable space in which every compact subset is scattered, and with no isolated point. (I guess they implicitly assumed it would be non-empty as well, otherwise their theorem would be trivially wrong!)<\/p>\n\n\n\n

    However, for a non-empty space X<\/em>, Costantini and Watson’s assumptions imply ours. Indeed, first, any separable metrizable space is second-countable. Second, if every compact subset is scattered and the space has no isolated point, then it has no non-empty compact open set. (Imagine K<\/em> is non-empty, compact, and open in such a space X<\/em>. By assumption, K<\/em> is scattered, hence it has an isolated point x<\/em>. Namely, there is an open subset U<\/em> of X<\/em> such that U<\/em> \u2229 K<\/em> = {x<\/em>}. Since K<\/em> is open, so is U<\/em> \u2229 K<\/em>, and therefore x<\/em> is isolated in X<\/em>, which is impossible.)<\/p>\n\n\n\n

    Q<\/strong> has no isolated point: for any rational number x<\/em>, every open neighborhood of x<\/em> must contain a basic open set Q<\/strong> \u2229 ]a<\/em>, b<\/em>[ where a<\/em><x<\/em><b<\/em>, and that open neighborhood therefore cannot be just {x<\/em>}. It follows that, indeed, Q<\/strong> contains no non-empty compact open subset. Alternatively, remember that the interior of every compact subset of Q<\/strong> is empty (see Exercise 4.8.4 in the book<\/a>), so every compact open set coincides with its interior, hence is empty.<\/p>\n\n\n\n

    Hence Q<\/strong> satisfies all the assumptions 1\u20133 given above. And those are all the assumptions we will need.<\/p>\n\n\n\n

    I am saying that as though we were going to prove a more general result than Costantini and Watson’s, but that is certainly not the case. Our assumptions are in fact equivalent to theirs. Let us see why. Our assumption 2 implies that X<\/em> is T0<\/sub>: if x<\/em> and y<\/em> are two points that have the same open neighborhoods, then {x<\/em>,y<\/em>} has no isolated point, hence is dense-in-itself; it also compact, hence scattered by 2, and therefore contains no non-empty dense-in-itself subset, which is impossible. Together with assumption 1, X<\/em> is T2<\/sub>: if x<\/em>\u2260y<\/em> then some open neighborhood U<\/em> of x<\/em>, say, fails to contain y<\/em>, and then we may assume that U<\/em> is clopen. By a similar argument, X<\/em> is even T3<\/sub>. Now any space that is second-countable and T3<\/sub> is separable metrizable, by the Urysohn-Tychonoff metrization theorem (Theorem 6.3.44 in the book<\/a>). Finally, if there is no non-empty compact open set, then no point can be isolated: if x<\/em> were an isolated point, then {x<\/em>} would be non-empty, compact, and open. Hence Costantini and Watson’s assumptions are equivalent to ours.<\/p>\n\n\n\n

    Costantini and Watson’s construction proceeds as follows. Let X<\/em> be a space satisfying conditions 1\u20133. We form a collection of clopen sets (which they write as \u03c8(n<\/em>,f<\/em>), where n<\/em> ranges over the natural numbers and f<\/em> ranges over certain functions\u2014in fact when (n<\/em>, f<\/em>) ranges over so-called reachable configurations<\/em>, as we will call them below), and we form the collection U<\/em><\/strong> of all open subsets of X<\/em> that intersect every \u03c8(n<\/em>,f<\/em>).<\/p>\n\n\n\n

    The point of the construction is that, by devising the sets \u03c8(n<\/em>,f<\/em>) in a suitably clever fashion,<\/p>\n\n\n\n