{"id":1591,"date":"2018-12-25T19:11:34","date_gmt":"2018-12-25T18:11:34","guid":{"rendered":"https:\/\/projects.lsv.ens-paris-saclay.fr\/topology\/?page_id=1591"},"modified":"2022-11-19T15:09:29","modified_gmt":"2022-11-19T14:09:29","slug":"on-countability","status":"publish","type":"page","link":"https:\/\/projects.lsv.ens-paris-saclay.fr\/topology\/?page_id=1591","title":{"rendered":"On countability"},"content":{"rendered":"<p>Everyone has its own prejudices.\u00a0 One of mine is that countability assumptions are usually superfluous.\u00a0 E.g., instead of defining convergence with sequences, and then realizing that closed sets can only be characterized by sequences in <em>first-countable<\/em> spaces, it is better to define convergence using nets, or using filters\u2014that works in every space.<\/p>\n<p>However, there are some places in mathematics where you really need countability assumptions, and some of them are somehow unexpected.<\/p>\n<h2>The expected places<\/h2>\n<p>You would expect countability to be important in questions related to \u211a or \u211d.\u00a0 Metric spaces are first-countable, and that is because metrics take their values in \u211d<sub>+<\/sub> \u222a {\u221e}.\u00a0 Measures, which also take their values in\u00a0\u211d<sub>+<\/sub> \u222a {\u221e}, are \u03c3-additive, i.e., they map countable disjoint unions to sums, for some pretty good reasons, too, related to the existence of non-measurable sets, and results such as the monotone convergence theorem, which applies to\u00a0<em>countable<\/em> sequences of maps to be integrated.<\/p>\n<p>But there are some places where countability is important, and neither \u211a or \u211d is involved.<\/p>\n<h2>Projective limits<\/h2>\n<p>I have spent quite some time on projective limits already.\u00a0 Remember: the projective limit of a projective system of non-empty topological spaces <em>may<\/em> be empty.\u00a0 It is non-empty if all the spaces involved are compact and sober\u2014this is <a href=\"https:\/\/projects.lsv.ens-paris-saclay.fr\/topology\/?page_id=1548\">part II<\/a> of my previous series posts on Steenrod&#8217;s Theorem\u2014or if the system has a <a href=\"https:\/\/projects.lsv.ens-paris-saclay.fr\/topology\/?page_id=1534\">cofinal <em>countable<\/em><\/a> subsystem.\u00a0 And we have seen one counterexample, which indeed has no cofinal countable subsystem.<\/p>\n<p>Note that countability is not <em>required<\/em>. \u00a0What is required is countability\u00a0<em>or<\/em> the fact that all spaces are compact and sober. \u00a0All the examples that we shall see will be of this form: we will need either a pretty strong topological condition (here, compactness and sobriety), or a countability condition.<\/p>\n<h2>The Plotkin powerdomain<\/h2>\n<p>The Plotkin powerdomain was introduced so as to give semantics to non-deterministic choice in semantics. \u00a0I will not describe what this is, because this is a pretty complicated subject (see Section 6.2.1 of [1]), and I will describe a better example, where countability makes wonders, in the next section.<\/p>\n<p>Let me just say that this is one of the first examples I saw where countability (<em>or<\/em> coherence) was required. \u00a0To wit,\u00a0Theorem 6.2.19 of [1] states that the Plotkin powerdomain of a continuous dcpo\u00a0<em>X<\/em>, which is defined there as a certain continuous dcpo (by construction, as the rounded ideal completion of an abstract basis), is isomorphic to a space of so-called lenses over\u00a0<em>X<\/em>, provided that\u00a0<em>X<\/em> is\u00a0\u03c9-continuous, not just continuous. \u00a0This is where the countability condition comes into play.<\/p>\n<p>Again, countability is not really needed. \u00a0Instead, if <em>X<\/em> is not only a continuous dcpo, but also a coherent one, then the Plotkin powerdomain is also isomorphic to the space of lenses (Theorem 6.2.22 of [1]; see also Proposition 4.45 of [3], and its corrigendum, for a complete proof that the space of lenses is a continuous dcpo in this case). \u00a0If we do not wish to require coherence, countability in the form of\u00a0\u03c9-continuity is necessary, as the pretty sophisticated construction of\u00a0Exercise 6.2.23(11) of [1] shows.<\/p>\n<h2>The Smyth powerdomain<\/h2>\n<p>The Smyth powerdomain\u00a0<strong>Q<\/strong>(<em>X<\/em>) is the set of compact saturated subsets of\u00a0<em>X<\/em>. \u00a0(Usually the empty set is excluded. \u00a0I will not do this.) \u00a0It can be equipped with either one of two classical topologies:<\/p>\n<ul>\n<li>the\u00a0<em>upper Vietoris<\/em> topology, whose subbasic open sets\u00a0\u2610<em>U<\/em>, where\u00a0<em>U<\/em> is open in\u00a0<em>X<\/em>, are defined as {<em>Q<\/em>\u00a0\u2208 <strong>Q<\/strong>(<em>X<\/em>) |\u00a0<em>Q<\/em>\u00a0\u2286\u00a0<em>U<\/em>} (those are even <em>basic<\/em> open sets, since \u2610<em>U<\/em>\u00a0\u2229 \u2610<em>V<\/em>=\u00a0\u2610(<em>U<\/em>\u00a0\u2229\u00a0<em>V<\/em>)); its specialization ordering is\u00a0<em>reverse<\/em> inclusion \u2287;<\/li>\n<li>the Scott topology of reverse inclusion\u00a0\u2287.<\/li>\n<\/ul>\n<p>It is easy to see that every set\u00a0\u2610<em>U<\/em>, where\u00a0<em>U<\/em> is open in\u00a0<em>X<\/em>, is Scott-open in\u00a0<strong>Q<\/strong>(<em>X<\/em>), provided that <em>X<\/em> is well-filtered. \u00a0Hence the Scott topology is finer than the upper Vietoris topology.<\/p>\n<p>The following properties are stated in Proposition 8.3.25 of the <a href=\"https:\/\/www.cambridge.org\/gb\/knowledge\/isbn\/item7069109\/Non-Hausdorff%20Topology%20and%20Domain%20Theory\/?site_locale=en_GB\">book<\/a>:<\/p>\n<ul>\n<li>If\u00a0<em>X<\/em> is well-filtered, then\u00a0<strong>Q<\/strong>(<em>X<\/em>) under reverse inclusion is a dcpo, where directed suprema are intersections.<\/li>\n<li>If\u00a0<em>X<\/em> is well-filtered and locally compact (equivalently, sober and locally compact), then\u00a0<strong>Q<\/strong>(<em>X<\/em>) is a continuous dcpo, and\u00a0<em>Q<\/em>\u226a<em>Q&#8217;<\/em> if and only if the interior of\u00a0<em>Q<\/em> contains\u00a0<em>Q&#8217;<\/em>.<\/li>\n<\/ul>\n<p>It follows that, if\u00a0<em>X<\/em> is well-filtered and locally compact, then the upper Vietoris and the Scott topologies coincide on\u00a0<strong>Q<\/strong>(<em>X<\/em>). \u00a0Indeed, a base of the Scott topology is given by the sets\u00a0\u219f<em>Q<\/em>, and those are equal to\u00a0\u2610int(<em>Q<\/em>).<\/p>\n<p>The following result, due to de Brecht and Kawai [2, Proposition 6], is a remarkable example which shows that second countability can be taken as an alternative to local compactness.<\/p>\n<p>(Note added on April 1, 2019: Matthew de Brecht told me that it should be fair to mention Matthias Schr\u00f6der&#8217;s work, from which he claims to have taken some of the techniques.\u00a0 Indeed, the argument that <em>K&#8217;<sub>n<\/sub><\/em> is compact in the proof below is similar to one found in Proposition 2.3 of [4].)<\/p>\n<p><b>Proposition (de Brecht, Ka<\/b><span style=\"font-weight: 600;\">wai)<\/span><b>\u00a0<\/b> \u00a0If\u00a0<em>X<\/em> is well-filtered and second countable, then the upper Vietoris and Scott topologies coincide on\u00a0<strong>Q<\/strong>(<em>X<\/em>).<\/p>\n<p>De Brecht and Kawai&#8217;s argument is a clever one, and I feel it should be described slowly, and in detail. \u00a0They require\u00a0<em>X<\/em> to be sober, but well-filteredness is enough. \u00a0In fact we only need well-filteredness to make sure that the Scott topology is finer than the upper Vietoris topology, and we will only use second countability in the rest of this post.<\/p>\n<p><em>Proof.<\/em> We reason by contradiction, and we assume that there is a Scott-open subset\u00a0<em>H<\/em> of\u00a0<strong>Q<\/strong>(<em>X<\/em>) which is not upper Vietoris open. This implies that there is an element\u00a0<em>K<\/em> of\u00a0<em>H<\/em>\u00a0which has no upper Vietoris open neighborhood\u00a0\u2610<em>U<\/em> included in\u00a0<em>H<\/em>. \u00a0Indeed, otherwise every element\u00a0<em>K<\/em> of\u00a0<em>H<\/em> would have an upper Vietoris open neighborhood\u00a0\u2610<em>U<sub>K<\/sub><\/em> included in\u00a0<em>H<\/em>, and the union of the sets \u2610<em>U<sub>K<\/sub><\/em> would be equal to\u00a0<em>H<\/em> and upper Vietoris open.<\/p>\n<p>Because\u00a0<em>X<\/em> is second countable, we claim that\u00a0<em>K<\/em> has a countable base of open neighborhoods in\u00a0<em>X<\/em>. \u00a0In other words, there is a countable family <em>F<\/em>\u00a0of open neighborhoods <em>U<sub>n<\/sub><\/em>, <em>n<\/em> \u2208 <strong>N<\/strong>, of\u00a0<em>K<\/em> in\u00a0<em>X<\/em> such that every open neighborhood\u00a0<em>U<\/em> of\u00a0<em>K<\/em> contains some <em>U<sub>n<\/sub><\/em>. \u00a0This is built as follows. \u00a0Let (<em>V<sub>n<\/sub><\/em>)<em><sub>n<\/sub><\/em><sub>\u2208<\/sub><sub><strong>N<\/strong><\/sub>\u00a0be a countable base of the topology of\u00a0<em>X<\/em>. \u00a0The family <em>F<\/em>\u00a0consists of all the finite unions of sets <em>V<sub>n\u00a0<\/sub><\/em>that cover\u00a0<em>K<\/em>. \u00a0We check that every open\u00a0neighborhood\u00a0<em>U<\/em> of\u00a0<em>K<\/em> contains some element of\u00a0<em>F<\/em>:\u00a0<em>U<\/em> is a union of sets <em>V<sub>n<\/sub><\/em>, and since\u00a0<em>K<\/em> is compact, we can extract a finite subcover, whose union is then in\u00a0<em>F<\/em> and included in\u00a0<em>U<\/em>.<\/p>\n<p>Furthermore, by replacing each <em>U<sub>n<\/sub><\/em>\u00a0with <em>U<\/em><sub>0\u00a0<\/sub>\u2229\u00a0<em>U<\/em><sub>1\u00a0<\/sub>\u2229 &#8230;\u00a0\u2229\u00a0<em>U<sub>n<\/sub><\/em>, we may assume that\u00a0<em>U<\/em><sub>0\u00a0<\/sub>\u2287\u00a0<em>U<\/em><sub>1\u00a0<\/sub>\u2287 &#8230;\u00a0\u2287\u00a0<em>U<sub>n<\/sub><\/em>.<\/p>\n<p>Using that countable base <em>F<\/em>=(<em>U<sub>n<\/sub><\/em>)<em><sub>n<\/sub><\/em><sub>\u2208<\/sub><sub><strong>N<\/strong><\/sub>\u00a0\u00a0of open neighborhoods of\u00a0<em>K<\/em>, we build a sequence of elements <em>K<sub>n<\/sub><\/em>, <em>n<\/em> \u2208 <strong>N<\/strong>, of elements of\u00a0<strong>Q<\/strong>(<em>X<\/em>) as follows. \u00a0Recall that\u00a0<em>K<\/em> has no upper Vietoris open neighborhood\u00a0\u2610<em>U<\/em> included in\u00a0<em>H<\/em>. \u00a0In particular,\u00a0\u2610<em>U<sub>n<\/sub><\/em>, which is an upper Vietoris open neighborhood of <em>K<\/em>, is\u00a0not included in\u00a0<em>H<\/em>. \u00a0We can therefore find an element <em>K<sub>n<\/sub><\/em>\u00a0of\u00a0<strong>Q<\/strong>(<em>X<\/em>), included in <em>U<sub>n<\/sub><\/em>, but not in\u00a0<em>H<\/em>. \u00a0By replacing <em>K<sub>n<\/sub><\/em>\u00a0by\u00a0<em>K<\/em> \u222a\u00a0<em>K<sub>n<\/sub><\/em>, which is still included in\u00a0<em>U<sub>n<\/sub><\/em>, and still not in\u00a0<em>H<\/em>\u00a0(since\u00a0<em>H<\/em> is upwards-closed in the reverse inclusion ordering), we can also assume that every <em>K<sub>n<\/sub><\/em>\u00a0contains\u00a0<em>K<\/em>.<\/p>\n<p>The sequence\u00a0(<em>K<sub>n<\/sub><\/em>)<em><sub>n<\/sub><\/em><sub>\u2208<\/sub><sub><strong>N<\/strong><\/sub>\u00a0lies entirely outside\u00a0<em>H<\/em>. \u00a0While we will not use it literally, it is interesting to note (as de Brecht and Kawai do) that it converges to\u00a0<em>K<\/em> in the upper Vietoris topology. \u00a0This is by definition: every open neighborhood of\u00a0<em>K<\/em> in the open Vietoris topology must contain a basic open set \u2610<em>U\u00a0<\/em>containing\u00a0<em>K<\/em>. \u00a0Since\u00a0<em>K<\/em> is included in\u00a0<em>U<\/em>, <em>U<sub>n<\/sub><\/em>\u00a0is included in\u00a0<em>U<\/em> for\u00a0<em>n<\/em> large enough, hence\u00a0<em>K<sub>n<\/sub><\/em>\u00a0is included in\u00a0<em>U<\/em> for\u00a0<em>n<\/em> large enough.<\/p>\n<p>Notice also that\u00a0\u2229<em><sub>n<\/sub><\/em><sub>\u2208<\/sub><sub><strong>N\u00a0<\/strong><\/sub><em>K<sub>n<\/sub><\/em>\u00a0= <em>K<\/em>. \u00a0Indeed, we have made sure that every\u00a0<em>K<sub>n<\/sub><\/em>\u00a0contains\u00a0<em>K<\/em>, and conversely\u00a0\u2229<em><sub>n<\/sub><\/em><sub>\u2208<\/sub><sub><strong>N\u00a0<\/strong><\/sub><em>K<sub>n<\/sub><\/em>\u00a0\u2286 \u2229<em><sub>n<\/sub><\/em><sub>\u2208<\/sub><sub><strong>N\u00a0<\/strong><\/sub><em>U<sub>n<\/sub><\/em>\u00a0is the intersection of all the open neighborhoods of\u00a0<em>K<\/em> (since\u00a0<em>F<\/em>=(<em>U<sub>n<\/sub><\/em>)<em><sub>n<\/sub><\/em><sub>\u2208<\/sub><sub><strong>N<\/strong><\/sub>\u00a0is a base of open neighborhoods of\u00a0<em>K<\/em>), which is just\u00a0<em>K<\/em> since\u00a0<em>K<\/em> is saturated.<\/p>\n<p>If the sequence\u00a0(<em>K<sub>n<\/sub><\/em>)<em><sub>n<\/sub><\/em><sub>\u2208<\/sub><sub><strong>N<\/strong><\/sub>\u00a0were monotone, then we could observe that it would have\u00a0<em>K<\/em> as least upper bound. \u00a0Since\u00a0<em>K<\/em> is in the Scott-open set\u00a0<em>H<\/em>, some <em>K<sub>n<\/sub><\/em>\u00a0would be in\u00a0<em>H<\/em>, contradiction.<\/p>\n<p>However, there is no reason why\u00a0(<em>K<sub>n<\/sub><\/em>)<em><sub>n<\/sub><\/em><sub>\u2208<\/sub><sub><strong>N<\/strong><\/sub>\u00a0would be monotone. \u00a0Hence we will build another sequence\u00a0(<em>K&#8217;<sub>n<\/sub><\/em>)<em><sub>n<\/sub><\/em><sub>\u2208<\/sub><sub><strong>N<\/strong><\/sub>\u00a0in <strong>Q<\/strong>(<em>X<\/em>) with\u00a0<em>K<\/em> as least upper bound again, and which will consist of elements outside\u00a0<em>H<\/em> as well, but this one will be monotone.<\/p>\n<p>There are two ways one can build the sets <em>K&#8217;<sub>n<\/sub><\/em>\u00a0a priori. \u00a0The most obvious one, perhaps, would be to define <em>K&#8217;<sub>n<\/sub><\/em>\u00a0as <em>K<\/em><sub>0\u00a0<\/sub>\u2229\u00a0<em>K<\/em><sub>1\u00a0<\/sub>\u2229 &#8230;\u00a0\u2229\u00a0<em>K<sub>n<\/sub><\/em>. \u00a0This would ensure that\u00a0(<em>K&#8217;<sub>n<\/sub><\/em>)<em><sub>n<\/sub><\/em><sub>\u2208<\/sub><sub><strong>N<\/strong><\/sub>\u00a0is monotone indeed, but this suffers from two problems: there is no reason to believe that <em>K&#8217;<sub>n<\/sub><\/em>\u00a0would be compact, unless we assume\u00a0<em>X<\/em> coherent; and <em>K&#8217;<sub>n<\/sub><\/em>\u00a0is\u00a0<em>above\u00a0K<sub>n<\/sub>\u00a0<\/em> in the reverse inclusion ordering, so there is no reason to believe that\u00a0<em>K&#8217;<sub>n<\/sub><\/em>\u00a0would still be outside\u00a0<em>H<\/em>.<\/p>\n<p>The other way, which is what de Brecht and Kawai use, is to define\u00a0<em>K&#8217;<sub>n<\/sub><\/em>\u00a0as <em>K<sub>n<\/sub><\/em><sub>\u00a0<\/sub>\u222a\u00a0<em>K<sub>n<\/sub><\/em><sub>+1\u00a0<\/sub>\u222a &#8230; \u222a\u00a0<em>K<\/em><em><sub>m\u00a0<\/sub><\/em>\u222a &#8230; \u00a0Now it is even less obvious that the sets\u00a0<em>K&#8217;<sub>n<\/sub><\/em>\u00a0would be compact, but if they are, they will indeed form a monotone sequence, and outside\u00a0<em>H<\/em>.<\/p>\n<p>Why is\u00a0<em>K&#8217;<sub>n<\/sub><\/em>\u00a0= <em>K<sub>n<\/sub><\/em><sub>\u00a0<\/sub>\u222a\u00a0<em>K<sub>n<\/sub><\/em><sub>+1\u00a0<\/sub>\u222a &#8230; \u222a\u00a0<em>K<\/em><em><sub>m\u00a0<\/sub><\/em>\u222a &#8230; compact, then? \u00a0Let (<em>V<sub>i<\/sub><\/em>)<em><sub>i<\/sub><\/em><sub>\u2208<em>I<\/em><\/sub>\u00a0be an open cover of <em>K&#8217;<sub>n<\/sub><\/em>, and let\u00a0<em>V<\/em> be the union \u222a<em><sub>i<\/sub><\/em><sub>\u2208<em>I\u00a0<\/em><\/sub><em>V<sub>i<\/sub><\/em>. \u00a0We will even assume that\u00a0(<em>V<sub>i<\/sub><\/em>)<em><sub>i<\/sub><\/em><sub>\u2208<em>I<\/em><\/sub>\u00a0is directed, and our goal is to show that <em>K&#8217;<sub>n<\/sub><\/em>\u00a0is included in some <em>V<sub>i<\/sub><\/em>\u00a0(Proposition 4.4.7 of the <a href=\"https:\/\/www.cambridge.org\/gb\/knowledge\/isbn\/item7069109\/Non-Hausdorff%20Topology%20and%20Domain%20Theory\/?site_locale=en_GB\">book<\/a>).<\/p>\n<p>Since every <em>K<\/em><em><sub>m\u00a0<\/sub><\/em>contains\u00a0<em>K<\/em>,\u00a0<em>V<\/em> also contains\u00a0<em>K<\/em>. \u00a0Since <em>K<\/em> is compact,\u00a0<em>K<\/em> is included in some <em>V<sub>j<\/sub><\/em>, <em>j<\/em> in\u00a0<em>I<\/em>.\u00a0 Let us recall that\u00a0<em>F<\/em>=(<em>U<sub>m<\/sub><\/em>)<em><sub>m<\/sub><\/em><sub>\u2208<\/sub><sub><strong>N<\/strong><\/sub>\u00a0is a base of open neighborhoods of\u00a0<em>K<\/em>: so\u00a0<em>V<sub>j<\/sub><\/em> contains\u00a0<em>U<sub>m<\/sub><\/em>\u00a0for some\u00a0<em>m<\/em> \u2208 <strong>N<\/strong>. \u00a0Then it also contains <em>U<sub>m<\/sub><\/em><sub>+1<\/sub>, <em>U<sub>m<\/sub><\/em><sub>+2<\/sub>, etc., because <em>U<\/em><sub><em>m<\/em>\u00a0<\/sub>\u2287\u00a0<em>U<\/em><sub><em>m<\/em>+1\u00a0<\/sub>\u2287\u00a0<em>U<\/em><sub><em>m<\/em>+2\u00a0<\/sub>\u2287\u00a0&#8230;, and therefore all the sets <em>K<sub>m<\/sub><\/em>,\u00a0<em>K<sub>m<\/sub><\/em><sub>+1<\/sub>, <em>K<sub>m<\/sub><\/em><sub>+2<\/sub>, &#8230;, are in <em>V<sub>j<\/sub><\/em>. \u00a0This was the hard part.<\/p>\n<p>We only have the finitely many sets <em>K<sub>n<\/sub><\/em>,\u00a0<em>K<sub>n<\/sub><\/em><sub>+1<\/sub>, &#8230;,\u00a0<em>K<\/em><em><sub>m<\/sub><\/em><sub>-1<\/sub>, to take care of. \u00a0They are all compact and included in\u00a0<em>V<\/em>, so: <em>K<sub>n<\/sub><\/em>\u00a0is included in <em>V<sub>i<\/sub><\/em><sub>[<\/sub><em><sub>n<\/sub><\/em><sub>]<\/sub>\u00a0for some <em>i<\/em>[<em>n<\/em>] in <em>I<\/em>,\u00a0<em>K<sub>n<\/sub><\/em><sub>+1<\/sub>\u00a0is included in <em>V<sub>i<\/sub><\/em><sub>[<\/sub><em><sub>n<\/sub><\/em><sub>+1<\/sub><sub>]<\/sub>\u00a0for some <em>i<\/em>[<em>n<\/em>+1] in <em>I<\/em>, &#8230;,\u00a0<em>K<sub>m<\/sub><\/em><sub>-1<\/sub>\u00a0is included in <em>V<sub>i<\/sub><\/em><sub>[<\/sub><em><sub>m<\/sub><\/em><sub>-1<\/sub><sub>]<\/sub>\u00a0for some <em>i<\/em>[<em>m<\/em>-1] in <em>I<\/em>. \u00a0By directedness, there is an index\u00a0<em>i<\/em> in\u00a0<em>I<\/em> such that\u00a0<em>V<sub>j<\/sub><\/em>, <em>V<sub>i<\/sub><\/em><sub>[<\/sub><em><sub>n<\/sub><\/em><sub>]<\/sub>, <em>V<sub>i<\/sub><\/em><sub>[<\/sub><em><sub>n<\/sub><\/em><sub>+1<\/sub><sub>]<\/sub>, &#8230;, <em>V<sub>i<\/sub><\/em><sub>[<\/sub><em><sub>m<\/sub><\/em><sub>-1<\/sub><sub>]\u00a0<\/sub>are all included in <em>V<sub>i<\/sub><\/em>, and therefore\u00a0<em>K&#8217;<sub>n<\/sub><\/em>\u00a0is included in <em>V<sub>i<\/sub><\/em>.<\/p>\n<p>Let us resume\u2014and conclude\u2014the proof. \u00a0We have obtained a monotone sequence\u00a0(<em>K&#8217;<sub>n<\/sub><\/em>)<em><sub>n<\/sub><\/em><sub>\u2208<\/sub><sub><strong>N<\/strong><\/sub>\u00a0in <strong>Q<\/strong>(<em>X<\/em>), and since every <em>K&#8217;<sub>n<\/sub><\/em>\u00a0contains (i.e., lies below) <em>K<sub>n<\/sub><\/em>, no <em>K&#8217;<sub>n<\/sub><\/em>\u00a0is in\u00a0<em>H<\/em>. \u00a0The intersection\u00a0\u2229<em><sub>n<\/sub><\/em><sub>\u2208<\/sub><sub><strong>N\u00a0<\/strong><\/sub><em>K&#8217;<sub>n<\/sub><\/em>\u00a0contains\u00a0\u2229<em><sub>n<\/sub><\/em><sub>\u2208<\/sub><sub><strong>N\u00a0<\/strong><\/sub><em>K<sub>n<\/sub><\/em>\u00a0=\u00a0<em>K<\/em>. \u00a0 Conversely, each <em>K&#8217;<sub>n\u00a0<\/sub><\/em>= <em>K<sub>n<\/sub><\/em><sub>\u00a0<\/sub>\u222a\u00a0<em>K<sub>n<\/sub><\/em><sub>+1\u00a0<\/sub>\u222a &#8230; \u222a\u00a0<em>K<\/em><em><sub>m\u00a0<\/sub><\/em>\u222a &#8230; is included in\u00a0<em>U<sub>n<\/sub><\/em><sub>\u00a0<\/sub>\u222a\u00a0<em>U<sub>n<\/sub><\/em><sub>+1\u00a0<\/sub>\u222a &#8230; \u222a\u00a0<em>U<\/em><em><sub>m\u00a0<\/sub><\/em>\u222a &#8230; = <em>U<sub>n<\/sub><\/em>, so \u2229<em><sub>n<\/sub><\/em><sub>\u2208<\/sub><sub><strong>N\u00a0<\/strong><\/sub><em>K&#8217;<sub>n<\/sub><\/em>\u00a0is included in \u2229<em><sub>n<\/sub><\/em><sub>\u2208<\/sub><sub><strong>N\u00a0<\/strong><\/sub><em>U<sub>n\u00a0<\/sub><\/em>=\u00a0<em>K<\/em>. \u00a0Hence \u2229<em><sub>n<\/sub><\/em><sub>\u2208<\/sub><sub><strong>N\u00a0<\/strong><\/sub><em>K&#8217;<sub>n<\/sub><\/em>\u00a0=\u00a0<em>K<\/em>, which shows in particular that\u00a0<em>K<\/em> is the supremum of\u00a0(<em>K&#8217;<sub>n<\/sub><\/em>)<em><sub>n<\/sub><\/em><sub>\u2208<\/sub><sub><strong>N<\/strong><\/sub>\u00a0in <strong>Q<\/strong>(<em>X<\/em>). \u00a0Because\u00a0<em>K<\/em> is in\u00a0<em>H<\/em> and\u00a0<em>H<\/em> is Scott-open, some <em>K&#8217;<sub>n\u00a0<\/sub><\/em>must be in\u00a0<em>H<\/em>: contradiction. \u00a0\u2610<\/p>\n<p>I encourage you to try and remove the second countability assumption in the above proof, and to realize why this fails. \u00a0You would need to replace the countable base\u00a0<em>F<\/em>=(<em>U<sub>m<\/sub><\/em>)<em><sub>m<\/sub><\/em><sub>\u2208<\/sub><sub><strong>N<\/strong><\/sub>\u00a0by some other, not necessarily countable, base <em>F&#8217;\u00a0<\/em>of open neighborhoods of\u00a0<em>K<\/em>, for example the collection of all open neighborhoods of\u00a0<em>K<\/em>. \u00a0Instead of a sequence (<em>K<sub>n<\/sub><\/em>)<em><sub>n<\/sub><\/em><sub>\u2208<\/sub><sub><strong>N<\/strong><\/sub>, you would obtain a net (<em>K<sub>U<\/sub><\/em>)<em><sub>U<\/sub><\/em><sub>\u2208<\/sub><em><sub>F&#8217;<\/sub><\/em><sub>,\u2291<\/sub>\u00a0(and \u2291 would have to be reverse inclusion). \u00a0The key point is in defining the analogue of (<em>K&#8217;<sub>n<\/sub><\/em>)<em><sub>n<\/sub><\/em><sub>\u2208<\/sub><sub><strong>N<\/strong><\/sub>. \u00a0You should soon realize that this only works provided that given any\u00a0<em>U<\/em> in\u00a0<em>F&#8217;<\/em>, there are only finitely many elements\u00a0<em>V<\/em>\u00a0in\u00a0<em>F&#8217;<\/em>\u00a0such that\u00a0<em>U<\/em>\u22e2<em>V<\/em> (replacing the fact that we had only\u00a0finitely many sets <em>K<sub>n<\/sub><\/em>,\u00a0<em>K<sub>n<\/sub><\/em><sub>+1<\/sub>, &#8230;,\u00a0<em>K<\/em><em><sub>m<\/sub><\/em><sub>-1<\/sub>, to take care of in the second half of the argument that showed that <em>K&#8217;<sub>n<\/sub><\/em>\u00a0is compact)\u00a0&#8230; and that forces the family\u00a0<em>F&#8217;<\/em> to be countable\u2014see the Appendix below if you need a proof.<\/p>\n<p style=\"text-align: center;\">Merry Christmas, and a Happy New Year 2019!<\/p>\n<ol>\n<li>Samson Abramsky and Achim Jung. Domain Theory. Pages 1\u2013168 of Abramsky, S., Gabbay, D. M., and Maibaum, T. S. E. (eds.), Handbook of Logic in Computer Science, vol. 3. Oxford University Press, 1994. \u00a0<a href=\"https:\/\/www.cs.bham.ac.uk\/~axj\/pub\/papers\/handy1.pdf\">Corrected and expanded version<\/a>.<\/li>\n<li>Matthew de Brecht and Tatsuji Kawai. \u00a0On the commutativity of the powerspace constructions. \u00a0<a href=\"https:\/\/arxiv.org\/abs\/1709.06226\">arXiv\u00a01709.06226<\/a>, 2017.<\/li>\n<li>Mike Mislove. \u00a0<a href=\"https:\/\/www.sciencedirect.com\/science\/article\/pii\/S0166864197002228\">Topology, domain theory and theoretical computer science<\/a>. \u00a0<a href=\"https:\/\/www.sciencedirect.com\/journal\/topology-and-its-applications\">Topology and its Applications<\/a> <a href=\"https:\/\/www.sciencedirect.com\/journal\/topology-and-its-applications\/vol\/89\/issue\/1\">89(1-2)<\/a>, pages 3-59, 1998. \u00a0See also the <a href=\"https:\/\/www.entcs.org\/mislove\/topandcs.pdf\">ENTCS version<\/a>, and the <a href=\"https:\/\/www.entcs.org\/mislove\/corrig.pdf\">corrigendum<\/a>.<\/li>\n<li>Matthias Schr\u00f6der.\u00a0 A Hofmann-Mislove theorem for Scott-open sets.\u00a0 <a href=\"https:\/\/arxiv.org\/abs\/1501.06452\" data-wplink-edit=\"true\">arXiv 1501.06452<\/a> , 2015.<\/li>\n<\/ol>\n<h2>Appendix<\/h2>\n<p>Here is a proof of the fact that every poset\u00a0<em>F&#8217;<\/em> such that for every\u00a0<em>U<\/em> in\u00a0<em>F&#8217;<\/em>, there are only finitely many elements\u00a0<em>V<\/em>\u00a0in\u00a0<em>F&#8217;<\/em>\u00a0such that\u00a0<em>U<\/em>\u22e2<em>V<\/em>, is countable.\u00a0 For every\u00a0<em>U<\/em> in\u00a0<em>F&#8217;<\/em>, let <em>A<\/em>(<em>U<\/em>)\u00a0be the finite set of elements\u00a0<em>V<\/em>\u00a0in\u00a0<em>F&#8217;<\/em>\u00a0such that\u00a0<em>U<\/em>\u22e2<em>V<\/em>. \u00a0If\u00a0<em>U<\/em>\u228f<em>U&#8217;<\/em> [strictly] in\u00a0<em>F&#8217;<\/em>, then\u00a0<em>A<\/em>(<em>U<\/em>) is included in\u00a0<em>A<\/em>(<em>U&#8217;<\/em>), and strictly so since\u00a0<em>U<\/em>\u00a0is in the latter and not in the former. \u00a0It follows that the set\u00a0<em>B<\/em>(<em>n<\/em>) of those elements\u00a0<em>U<\/em> of\u00a0<em>F&#8217;<\/em> such that card\u00a0<em>A<\/em>(<em>U<\/em>)=<em>n<\/em> is an antichain: it cannot have distinct comparable elements. \u00a0But every antichain of\u00a0<em>F&#8217;<\/em> is finite: if\u00a0<em>U<\/em> is any element of that antichain, the whole antichain must be contained in the finite set\u00a0<em>A<\/em>(<em>U<\/em>). \u00a0Finally,\u00a0<em>F&#8217;<\/em> is the union of the antichains\u00a0<em>B<\/em>(<em>n<\/em>),\u00a0<em>n<\/em> \u2208 <strong>N<\/strong>, and since each of them is finite,\u00a0<em>F&#8217;<\/em> is countable.<\/p>\n<p style=\"text-align: right;\">\u2014 <a href=\"https:\/\/www.lsv.ens-paris-saclay.fr\/~goubault\/?l=en\" rel=\"attachment wp-att-993\">Jean Goubault-Larrecq<\/a>\u00a0(December 25th, 2018)<img loading=\"lazy\" decoding=\"async\" class=\"wp-image-993 alignright\" src=\"https:\/\/projects.lsv.ens-paris-saclay.fr\/topology\/wp-content\/uploads\/2016\/08\/jgl-2011.png\" alt=\"jgl-2011\" width=\"32\" height=\"44\" \/><\/p>\n","protected":false},"excerpt":{"rendered":"<p>Everyone has its own prejudices.\u00a0 One of mine is that countability assumptions are usually superfluous.\u00a0 E.g., instead of defining convergence with sequences, and then realizing that closed sets can only be characterized by sequences in first-countable spaces, it is better &hellip; <a href=\"https:\/\/projects.lsv.ens-paris-saclay.fr\/topology\/?page_id=1591\">Continue reading <span class=\"meta-nav\">&rarr;<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"_crdt_document":"","footnotes":""},"class_list":["post-1591","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/projects.lsv.ens-paris-saclay.fr\/topology\/index.php?rest_route=\/wp\/v2\/pages\/1591","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/projects.lsv.ens-paris-saclay.fr\/topology\/index.php?rest_route=\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/projects.lsv.ens-paris-saclay.fr\/topology\/index.php?rest_route=\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/projects.lsv.ens-paris-saclay.fr\/topology\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/projects.lsv.ens-paris-saclay.fr\/topology\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=1591"}],"version-history":[{"count":16,"href":"https:\/\/projects.lsv.ens-paris-saclay.fr\/topology\/index.php?rest_route=\/wp\/v2\/pages\/1591\/revisions"}],"predecessor-version":[{"id":5919,"href":"https:\/\/projects.lsv.ens-paris-saclay.fr\/topology\/index.php?rest_route=\/wp\/v2\/pages\/1591\/revisions\/5919"}],"wp:attachment":[{"href":"https:\/\/projects.lsv.ens-paris-saclay.fr\/topology\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=1591"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}