{"id":3685,"date":"2021-05-20T10:47:04","date_gmt":"2021-05-20T08:47:04","guid":{"rendered":"https:\/\/projects.lsv.ens-paris-saclay.fr\/topology\/?page_id=3685"},"modified":"2022-11-19T14:58:28","modified_gmt":"2022-11-19T13:58:28","slug":"the-sorgenfrey-line-is-not-consonant","status":"publish","type":"page","link":"https:\/\/projects.lsv.ens-paris-saclay.fr\/topology\/?page_id=3685","title":{"rendered":"The Sorgenfrey line is not consonant"},"content":{"rendered":"\n

I have already talked about consonant spaces, for example here<\/a>. A space X<\/em> is consonant<\/em> [1] if and only if the compact-open topology and the Isbell topology coincide on the function space [X<\/em> \u2192 Y<\/em>] for every topological space Y<\/em>. It is equivalent to require that they coincide on the single function space [X<\/em> \u2192 S<\/strong>], where S<\/strong> is Sierpi\u0144ski space {0 < 1}. Since a continuous map from X<\/em> to S<\/strong> is the characteristic map of a unique open subset of X<\/em>, X<\/em> is consonant if and only if every Scott-open subset U<\/em><\/strong> of the lattice O<\/strong>X<\/em> of open subsets of X<\/em> is a union of sets of the form \u25a0Q<\/em>\u225d{U<\/em> | Q<\/em> \u2286 U<\/em>}, where the sets Q<\/em> are compact saturated subsets of X<\/em>.<\/p>\n\n\n\n

In Exercise 5.4.12 of the book<\/a>, I ask the reader to prove that neither the space of rationals, Q<\/strong>, nor the Sorgenfrey line, R<\/strong>\u2113<\/sub>, is consonant. That is what I classified as the important blooper #5 in the list of errata<\/a>, because the proof I had originally in mind was too simple-minded to have any chance of succeeding. But showing that R<\/strong>\u2113<\/sub> is not consonant is not that hard, finally. I will explain the argument. This will also be an excuse to explain some additional topological properties of R<\/strong>\u2113<\/sub>.<\/p>\n\n\n\n

The fact that R<\/strong>\u2113<\/sub> is not consonant was proved by Costantini and Watson, as a consequence of a more general result [3]. (Update, April 22nd, 2022: no, Costantini and Watson’s theorem does not apply. Instead, the fact that R<\/strong>\u2113<\/sub> is not consonant is due to Alleche and Calbrix [8]. Sorry for the confusion.) Instead, the route I will take has some similarities with what Bouziad did in [2] in order to prove that Q<\/strong> is not consonant\u2014again, as consequence of a more general result\u2014in the sense that I will rely on a bit of measure theory. (Update, April 22nd, 2022. And indeed, the argument I will develop below is essentially Alleche and Calbrix’s.) If you don’t know anything about measure theory, I will also give an argument that only requires some domain theory, most of which is in appendices, at the end of this post.<\/p>\n\n\n\n

The Sorgenfrey line R<\/strong>\u2113<\/sub><\/h2>\n\n\n\n

Before we start, let me recall what the Sorgenfrey line R<\/strong>\u2113<\/sub> is.<\/p>\n\n\n\n

It is an incredible beast. Its points are just the real numbers, but its topology is generated by the half-open intervals [a<\/em>,b<\/em>[ with a<\/em> < b<\/em>. The topology of R<\/strong>\u2113<\/sub> is finer than the usual metric topology of R<\/strong>. R<\/strong>\u2113<\/sub> is a zero- dimensional, first-countable space (Exercices 4.1.34, 4.7.17 in the book<\/a>). It is paracompact hence T4<\/sub>, and Choquet-complete hence a Baire space (Exercises 6.3.32, 7.6.11). R<\/strong>\u2113<\/sub> is not locally compact, as every compact subset of R<\/strong>\u2113<\/sub> has empty interior (Exercise 4.8.5). Although it is first-countable, R<\/strong>\u2113<\/sub> is not second-countable (Exercise 6.3.10).<\/p>\n\n\n\n

We won’t need any of that! It is enough to remember that R<\/strong>\u2113<\/sub> exhibits a mixture of the nice (paracompact, T4<\/sub>, Choquet-complete, Baire, first-countable) and of the ugly (not locally compact, not second-countable). We will add some new water to the pot of ugliness (non consonant), and also to the jar of niceness (hereditary Lindel\u00f6f, see below).<\/p>\n\n\n\n

R<\/strong>\u2113<\/sub> is not consonant: the short argument<\/h2>\n\n\n\n

We consider the collection U<\/em><\/strong> of open subsets of R<\/strong>\u2113<\/sub> whose Lebesgue measure is > r<\/em>, for some fixed r<\/em>>0. That is a Scott-open subset of O<\/strong>(R<\/strong>\u2113<\/sub>), and a non-empty one, too (since [\u2013r<\/em>, r<\/em>[ is in U<\/em><\/strong>, for example), but we claim that it simply contains no<\/em> subset of the form \u25a0Q<\/em> for any compact (saturated) subset of R<\/strong>\u2113<\/sub>. Hence U<\/em><\/strong> is certainly not a union of sets of the form \u25a0Q<\/em>.<\/p>\n\n\n\n

In order to see this, we realize that every compact subset Q<\/em> of R<\/strong>\u2113<\/sub> is countable. Writing Q<\/em> as {x<\/em>0<\/sub>, x<\/em>1<\/sub>, …, x<\/em>n<\/em><\/sub>, …}, we see that U<\/em> \u225d \u222an<\/em> \u2208 N<\/strong><\/sub> [x<\/em>n<\/em><\/sub>, \u03b5\/2n<\/sup><\/em>[ is an open neighborhood of Q<\/em> for every \u03b5>0, namely that U<\/em> in in \u25a0Q<\/em>. However the Lebesgue measure of U<\/em> is smaller than or equal to \u2211n<\/em> \u2208 N<\/strong><\/sub> \u03b5\/2n<\/sup><\/em> = 2\u03b5, so U<\/em> cannot be in U<\/em><\/strong> as soon as we pick \u03b5\u2264r<\/em>\/2.<\/p>\n\n\n\n

Now, wait! I skipped a number of details here. One is that I have not said why every compact subset Q<\/em> of R<\/strong>\u2113<\/sub> is countable. This is what I am going to show next, and, while we are at it, I will give a complete characterization of the compact subsets of R<\/strong>\u2113<\/sub>. Another missing piece is the reason why U<\/em><\/strong> is Scott-open. That would be the case is the Lebesgue measure of any directed union of open sets were equal to the supremum of their measures, but measures only satisfy this property for countable<\/em> directed unions in general. The measures that satisfy this property for arbitrary unions are called \u03c4-smooth<\/em>. Here a miracle will occur, thanks to a theorem due to Wolfgang Adamski [4], and a magical property called hereditary Lindel\u00f6fness<\/em>, which I will talk about, too. This all uses a bit of measure theory, but, if you have never heard about measure theory, I will give you a purely domain-theoretic construction of U<\/em><\/strong> in Appendix B at the end of this post.<\/p>\n\n\n\n

The compact subsets of the Sorgenfrey line are countable<\/h2>\n\n\n\n

Every compact subset Q<\/em> of R<\/strong>\u2113<\/sub> is countable, and this is easy to see.<\/p>\n\n\n\n

nWe first note that the sets ]\u2212\u221e,r<\/em>\u2013\u03b5[ and [r<\/em>, +\u221e[ are open, for all r<\/em> and \u03b5, because they are unions of basic open sets [r<\/em>\u2013\u03b5\u2013n<\/em>\u20131, r<\/em>\u2013\u03b5\u2013n<\/em>[, n<\/em> \u2208 N<\/strong>, and of [r<\/em>+n<\/em>, r<\/em>+n<\/em>+1[, n<\/em> \u2208 N<\/strong>, respectively.<\/p>\n\n\n\n

Next, given a compact subset Q<\/em> of R<\/strong>\u2113<\/sub>, we realize that, for every r<\/em> in Q<\/em>, the open sets ]\u2212\u221e,r<\/em>\u2013\u03b5[ with \u03b5>0, plus [r<\/em>, +\u221e[, form an open cover of Q<\/em>. (Indeed, of the whole of R<\/strong>\u2113<\/sub>!) We extract a finite subcover: without loss of generality, we may assume that this subcover contains [r<\/em>, +\u221e[, otherwise we add it back; the other elements of the subcover are ]\u2212\u221e,r<\/em>\u2013\u03b51<\/sub>[, …, ]\u2212\u221e,r<\/em>\u2013\u03b5n<\/em><\/sub>[. We ick \u03b5>0 so that \u03b5<\u03b51<\/sub>, …, \u03b5n<\/em><\/sub>, and we see that Q<\/em> is included in ]\u2212\u221e,r<\/em>\u2013\u03b5[ \u222a [r<\/em>, +\u221e[, namely that it does not intersect [r<\/em>\u2013\u03b5, r<\/em>[. In order to display the dependency on r<\/em>, let us write \u03b5r<\/sub><\/em> for that \u03b5.<\/p>\n\n\n\n

Let us pick a rational number qr<\/sub><\/em> in [r<\/em>\u2013\u03b5r<\/sub><\/em>, r<\/em>[, one for each r<\/em> in Q<\/em>. The map r<\/em> \u2208 Q<\/em> \u21a6 qr<\/sub><\/em> is injective: otherwise there would be two distinct elements r<\/em> and s<\/em> of Q<\/em> such that [r<\/em>\u2013\u03b5r<\/sub><\/em>, r<\/em>[ and [s<\/em>\u2013\u03b5s<\/sub><\/em>, s<\/em>[ would overlap (at qr<\/sub><\/em>=qs<\/sub><\/em>). By symmetry, let us assume r<\/em><s<\/em>. Then r<\/em> would be in [s<\/em>\u2013\u03b5s<\/sub><\/em>, s<\/em>[, which is impossible since r<\/em> is in Q<\/em> but [s<\/em>\u2013\u03b5s<\/sub><\/em>, s<\/em>[ does not intersect Q<\/em>.<\/p>\n\n\n\n

We have built an injective map from Q<\/em> to the set of rationals, which is countable. Therefore Q<\/em> is countable.<\/p>\n\n\n\n

That much is enough knowledge for us on the compact subsets of R<\/strong>\u2113<\/sub>. However, it would be foolish to stop here and not investigate the nature of compact subsets of R<\/strong>\u2113<\/sub>, right? Hence, let us embark on characterizing the compact subsets of R<\/strong>\u2113<\/sub> exactly.<\/p>\n\n\n\n

Characterizing the compact subsets of the Sorgenfrey line<\/h2>\n\n\n\n

Here is another property of compact subsets of R<\/strong>\u2113<\/sub> that can be proved in a similar way as above.<\/p>\n\n\n\n

Fact A.<\/strong> Every compact subset Q<\/em> of R<\/strong>\u2113<\/sub> is well-founded with respect to \u2265: there is no infinite chain of the form r<\/em>0<\/sub> < r<\/em>1<\/sub> < … < rn<\/sub><\/em> < … in Q<\/em>.<\/p>\n\n\n\n

Proof.<\/em> Let r<\/em> \u225d supn<\/em>\u2208N<\/strong><\/sub> rn<\/sub><\/em>. Then the open sets ]\u2212\u221e,r<\/em>0<\/sub>[, [r<\/em>0<\/sub>,r<\/em>1<\/sub>[, …, [rn<\/sub><\/em>,rn<\/sub><\/em>+1<\/sub>[, …, together with [r<\/em>,+\u221e[ in case that r<\/em>\u2260+\u221e, would form an open cover of Q<\/em> without a finite subcover. \u2610<\/p>\n\n\n\n

Fact A says that every compact subset of R<\/strong>\u2113<\/sub> is a well-founded<\/em> subset of (R<\/strong>, \u2265), namely of R<\/strong> with the opposite \u2265 of the usual ordering \u2264.<\/p>\n\n\n\n

A directed family in (R<\/strong>, \u2265) is a filtered<\/em> family: a non-empty family in which for any two elements, there is an element less than or equal to both; and “filtered” in not even necessary, since the ordering \u2264 on R<\/strong> is total: in other words, a filtered family is just a non-empty family there.<\/p>\n\n\n\n

Fact B.<\/strong> Every compact subset Q<\/em> of R<\/strong>\u2113<\/sub> is a subdcpo of (R<\/strong>, \u2265): the infimum, taken in R<\/strong>, of every filtered family of elements of Q<\/em> lies in Q<\/em>.<\/p>\n\n\n\n

Proof.<\/em> We first need to note that a net (xi<\/sub><\/em>)i<\/em>\u2208I<\/em>,\u2291<\/sub> converges to x<\/em> in R<\/strong>\u2113<\/sub> if and only if xi<\/sub><\/em> tends to x<\/em> from the right, namely: for every \u03b5 > 0, x<\/em> \u2264 xi<\/sub><\/em> < x<\/em> + \u03b5 for i<\/em> large enough. This is Exercise 4.7.6 in the book<\/a>, but the argument is very short, and I will let you verify this by yourselves.<\/p>\n\n\n\n

Let D<\/em> be any non-empty subset of Q<\/em>, and let r<\/em> \u225d inf D<\/em>. Since the topology of R<\/strong>\u2113<\/sub> is finer than the topology of R<\/strong>, every compact subset of R<\/strong>\u2113<\/sub> is also compact in R<\/strong>, in particular, it is bounded. Therefore r<\/em> is a well-defined real number (not \u2013\u221e).<\/p>\n\n\n\n

By Fact A, Q<\/em> is well-founded, so D<\/em> is isomorphic to a unique ordinal \u03b2, and we can write the elements of D<\/em> as r<\/em>\u03b1<\/sub>, \u03b1<\u03b2, in such a way that for all \u03b1,\u03b1\u2032 <\u03b2, \u03b1\u2264\u03b1\u2032 if and only if r<\/em>\u03b1<\/sub> \u2265 r<\/em>\u03b1’<\/sub>. The net (r<\/em>\u03b1<\/sub>)\u03b1<\u03b2,\u2264<\/sub> then converges to r<\/em> from the right. Since R<\/strong>\u2113<\/sub> is T2<\/sub>, its compact subset Q<\/em> is closed, so r<\/em> is in Q<\/em>. \u2610<\/p>\n\n\n\n

Note that, if the compact set Q<\/em> is non-empty, then in particular inf Q<\/em> lies in Q<\/em>, namely, Q<\/em> has a least element. That could also be derived from the fact that every compact subset of R<\/strong>\u2113<\/sub> is compact in R<\/strong>.<\/p>\n\n\n\n

Here is an example of a compact subset of R<\/strong>\u2113<\/sub>: the set of points of the form 1\/2n<\/sup><\/em>, n<\/em> \u2208 N<\/strong>, plus the point zero. We can index those points by ordinal numbers: point number n<\/em> is 1\/2n<\/sup><\/em>, and there is an additional point numbered \u03c9, which is zero:<\/p>\n\n\n\n

\"\"<\/a>
A compact subset of R<\/strong>\u2113<\/sub>, of order-type \u03c9+1.<\/figcaption><\/figure>\n\n\n\n

Every well-founded subset of (R<\/strong>, \u2265) must be isomorphic to an ordinal (here, \u03c9+1: an ordinal is also the set of all the ordinals strictly below it, so \u03c9+1 = {0, 1, …, n<\/em>, …, \u03c9}.) Hence if Q<\/em> is a compact subset of R<\/strong>\u2113<\/sub>, then its elements must be numbered by ordinal numbers, from right to left, as in the above picture. Additionally, since Q<\/em> is a subdcpo, that enumeration must stop at some ordinal (Q<\/em> has a least element), and points of Q<\/em> must cluster towards points which are numbered with a limit ordinal (from the right), as in the above picture where points cluster towards point number \u03c9, coming from the right.<\/p>\n\n\n\n

There are more complex compact subsets, such as the following one, whose points are indexed by the ordinals from 0 to \u03c92<\/sup>.<\/p>\n\n\n\n

\"\"<\/a>
A compact subset of R\u2113, of order-type \u03c92<\/sup>+1.<\/figcaption><\/figure>\n\n\n\n

How do I know that the latter is compact in R<\/strong>\u2113<\/sub>? Well, here is a complete characterization.<\/p>\n\n\n\n

Proposition C.<\/strong> The compact subsets of the Sorgenfrey line R<\/strong>\u2113<\/sub> are exactly the well-founded subdcpos of (R<\/strong>, \u2265).<\/p>\n\n\n\n

Proof.<\/em> Considering Facts A and B, it suffices to show that every well-founded subdcpo Q<\/em> of (R<\/strong>, \u2265) is compact in R<\/strong>\u2113<\/sub>. We use Alexander’s subbase lemma (Theorem 4.4.29 in the book<\/a>): for every family U<\/em><\/strong> of basic open subsets that covers Q<\/em>, we will extract a finite subcover. We build a (finite or infinite) sequence of elements rn<\/sub><\/em> of Q<\/em> and another sequence of elements [an<\/sub><\/em>,bn<\/sub><\/em>[ of U<\/em><\/strong> by induction on n<\/em> in such a way that:<\/p>\n\n\n\n

    \n
  1. r<\/em>0<\/sub> < r<\/em>1<\/sub> < … < rn<\/sub><\/em> < … (so yes, the sequence will in fact be finite, since Q<\/em> is well-founded in (R<\/strong>, \u2265), but let us not go too fast);<\/li>\n\n\n\n
  2. for each n<\/em>, rn<\/sub><\/em> is in [an<\/sub><\/em>,bn<\/sub><\/em>[;<\/li>\n\n\n\n
  3. for each n<\/em>, the intervals [a<\/em>0<\/sub>,b<\/em>0<\/sub>[, …, [an<\/sub><\/em>,bn<\/sub><\/em>[ cover all the elements of Q<\/em> that are \u2264rn<\/sub><\/em>; in other words, every element of Q<\/em> is in [a<\/em>0<\/sub>,b<\/em>0<\/sub>[ \u222a … \u222a [an<\/sub><\/em>,bn<\/sub><\/em>[ \u222a ]rn<\/sub><\/em>,+\u221e[.<\/li>\n<\/ol>\n\n\n\n

    In the base case (n<\/em>=0), we take r<\/em>0<\/sub>\u225dinf Q<\/em>, which is in Q<\/em> since Q<\/em> is a subdcpo of (R<\/strong>, \u2265). Being in Q<\/em>, r<\/em>0<\/sub> is in some basic open subset [a<\/em>0<\/sub>,b<\/em>0<\/sub>[ \u2208 U<\/em><\/strong>. We then note that every element of Q<\/em> is in [a<\/em>0<\/sub>,b<\/em>0<\/sub>[ \u222a ]r<\/em>0<\/sub>,+\u221e[.<\/p>\n\n\n\n

    In the induction case (n<\/em>\u22651), we assume we already have elements r<\/em>0<\/sub> < r<\/em>1<\/sub> < … < rn<\/sub><\/em>\u20131<\/sub> in Q<\/em>, lying in [a<\/em>0<\/sub>,b<\/em>0<\/sub>[, …, [an<\/sub><\/em>\u20131<\/sub>,bn<\/sub><\/em>\u20131<\/sub>[ respectively, and such that every element of Q<\/em> is in [a<\/em>0<\/sub>,b<\/em>0<\/sub>[ \u222a … \u222a [an<\/sub><\/em>\u20131<\/sub>,bn<\/sub><\/em>\u20131<\/sub>[ or is >rn<\/sub><\/em>\u20131<\/sub>. Let Qn<\/sub><\/em> be Q<\/em> \u2013 ([a<\/em>0<\/sub>,b<\/em>0<\/sub>[ \u222a … \u222a [an<\/sub><\/em>\u20131<\/sub>,bn<\/sub><\/em>\u20131<\/sub>[). If Qn<\/sub><\/em><\/em> is empty, then the sequence stops here, namely at rn<\/sub><\/em>\u20131<\/sub>. Otherwise, Qn<\/sub><\/em> is non-empty hence directed in (R<\/strong>, \u2265). Since Q<\/em> is a subdcpo, inf Qn<\/sub><\/em> exists and is an element of Q<\/em>. We let rn<\/sub><\/em>\u225dinf Qn<\/sub><\/em>. Every element of Qn<\/sub><\/em> is >rn<\/sub><\/em>\u20131<\/sub>, so rn<\/sub><\/em>\u2265rn<\/sub><\/em>\u20131<\/sub>. If rn<\/sub><\/em>=rn<\/sub><\/em>\u20131<\/sub>, then rn<\/sub><\/em> would be in [an<\/sub><\/em>\u20131<\/sub>,bn<\/sub><\/em>\u20131<\/sub>[, in particular an<\/sub><\/em>\u20131<\/sub>\u2264rn<\/sub><\/em>=inf Qn<\/sub><\/em><\/em><bn<\/sub><\/em>\u20131<\/sub>, so some element of Qn<\/sub><\/em><\/em> would be in [an<\/sub><\/em>\u20131<\/sub>,bn<\/sub><\/em>\u20131<\/sub>[, which is impossible. Hence rn<\/sub><\/em>\u20131<\/sub><rn<\/sub><\/em>. As far as item 2 is concerned, we merely use the fact that U<\/em><\/strong> is a cover of Q<\/em>, so rn<\/sub><\/em> must be in some [an<\/sub><\/em>,bn<\/sub><\/em>[ \u2208 U<\/em><\/strong>. Finally, every element x<\/em> of Q<\/em> is in [a<\/em>0<\/sub>,b<\/em>0<\/sub>[ \u222a … \u222a [an<\/sub><\/em>\u20131<\/sub>,bn<\/sub><\/em>\u20131<\/sub>[ or is >rn<\/sub><\/em>\u20131<\/sub>. If x<\/em>>rn<\/sub><\/em>\u20131<\/sub>, then x<\/em> is in Qn<\/sub><\/em><\/em>, hence x<\/em>\u2265rn<\/sub><\/em>, and therefore x<\/em> is in [an<\/sub><\/em>,bn<\/sub><\/em>[ or is strictly larger than rn<\/sub><\/em>.<\/p>\n\n\n\n

    We have built a chain r<\/em>0<\/sub> < r<\/em>1<\/sub> < … < rn<\/sub><\/em> < … of elements of Q<\/em> satisfying properties 1, 2, and 3 above. Since Q<\/em> is well-founded, this chain must be finite: the inductive process described above stops at some rank n<\/em>. In that case Qn<\/sub><\/em> is empty, and therefore Q<\/em> is included in [a<\/em>0<\/sub>,b<\/em>0<\/sub>[ \u222a … \u222a [an<\/sub><\/em>\u20131<\/sub>,bn<\/sub><\/em>\u20131<\/sub>[. \u2610<\/p>\n\n\n\n

    In other words, the compact subsets Q<\/em> of R<\/strong>\u2113<\/sub> are exactly the set of points x<\/em>\u03b1<\/sub> enumerated by all ordinals \u03b1<\u03b2, in such a way that x<\/em>0<\/sub> > x<\/em>1<\/sub> > … > x<\/em>n<\/em><\/sub> > … > x<\/em>\u03c9<\/sub> > x<\/em>\u03c9+1<\/sub> > … > x<\/em>\u03c9.2<\/sub> > …, and so that this forms a subdcpo in the opposite ordering \u2265, namely:<\/p>\n\n\n\n