The nice thing about colleagues is that they sometimes give you a primer on their latest discovery.\u00a0 The irritating thing about colleagues is that they may prefer you not to announce it before this has been published.\u00a0 This is what happened to me with a very nice result due to Jia, Jung and Li [1] a while ago.\u00a0 I guess enough time has passed now.<\/p>\n
The question is one of coherence<\/em>.\u00a0 See Section 5.2.2 in the book<\/a>: a space is coherent if and only if the intersection of any two compact saturated subsets is again compact (and saturated, necessarily).\u00a0 Every T2<\/sub> space is coherent, but that is no longer the case outside the realm of T2<\/sub> space, in particular in dcpos.<\/p>\n A typical counter-example is built as follows. Take N<\/strong>, the set of natural numbers, and order them by equality: i.e., let them all be incomparable.\u00a0 Add two incomparable elements a<\/em> and b<\/em> below all elements of N<\/strong>.\u00a0 Then \u2191a<\/em> is compact saturated, \u2191b<\/em> is compact saturated too, since the upward closure of any finite collection of points is.\u00a0 But their intersection is N<\/strong>, which is not compact: indeed it is covered by the union of the one-point subsets {n<\/em>}, n<\/em> \u2208 N<\/strong>, which are all Scott-open; and certainly that has no finite subcover.<\/p>\n Call that counter-example N<\/strong>a,b<\/sub><\/em>.<\/p>\n How do you prove that a dcpo X<\/em> is not coherent?\u00a0 Of course you have to consider any two compact saturated subsets, and show that their intersection is compact.\u00a0 Among the compact saturated subsets, we have the finitary compacts, and in particular the upward closures of single points \u2191x<\/em>.<\/p>\n Therefore, certainly, if X<\/em> is coherent, then for any two points x<\/em> and y<\/em>, \u2191x <\/em>\u2229 \u2191y <\/em>should be compact.\u00a0 That provides an easy test.\u00a0 If that fails, as in the case of N<\/strong>a,b<\/sub><\/em>, then you know X<\/em> is not coherent.\u00a0 But what if the test succeeds?<\/p>\n There is a particularly simple answer to that question in continuous<\/em> dcpos: a continuous dcpo is coherent if and only if<\/em>\u00a0\u2191x <\/em>\u2229 \u2191y <\/em>is compact for any two points x<\/em> and\u00a0y<\/em> (Proposition 5.2.34 in the book<\/a>).<\/p>\n But what about other dcpos?\u00a0 For example, what about complete lattices?\u00a0 In that case \u2191x <\/em>\u2229 \u2191y <\/em>= \u2191(x<\/em> \u22c1 y<\/em>) is trivially compact, but it was unknown until recently whether there were any non-coherent complete lattices in their Scott topology.\u00a0 (I have got some inside information about that… but I cannot speak about that yet. \u00a0Update [Jan. 10th, 2021]: all complete lattices are coherent in their Scott topology, by a result of Xi and Lawson<\/a>.)<\/p>\n Jia, Jung and Li’s main result is the following:<\/p>\n Thm [1, Lemma 3.1].<\/strong> Let X<\/em> be a well-filtered dcpo. X<\/em> is coherent if and only if \u2191x <\/em>\u2229 \u2191y <\/em>is compact for any two points x<\/em> and\u00a0y<\/em>.<\/p>\n This is a vast generalization of our previous result on continuous dcpos: recall that every continuous dcpo is sober, and that every sober space is well-filtered (Propositions 8.2.12 and 8.3.5 in the book<\/a>).<\/p>\n The proof of that theorem is particularly nifty.<\/p>\n We start with the Smyth powerdomain Q<\/strong>(X<\/em>) of X<\/em>.\u00a0 See Proposition 8.3.25 in the book<\/a>: this is the poset of all non-empty compact saturated subsets of X<\/em> under reverse inclusion.\u00a0 This is a dcpo since X<\/em> is well-filtered, and directed suprema are filtered intersections.<\/p>\n We take a variant of that construction.\u00a0 We keep the same set Q<\/strong>(X<\/em>), but we change the topology.\u00a0 Equip Q<\/strong>(X<\/em>) with the upper Vietoris topology<\/em>, whose basic open sets are of the form \u2610U<\/em>, U<\/em> open in X<\/em>.\u00a0 Here \u2610U<\/em> denotes the set of all non-empty saturated subsets Q<\/em> that are included in U<\/em>.\u00a0 In Proposition 8.3.26 of the book<\/a>, you will find that those sets form a base of the Scott topology, so it may look like the upper Vietoris topology is just another name for the Scott topology of Q<\/strong>(X<\/em>) under reverse inclusion… but that is only the case if X<\/em> is well-filtered and<\/em> locally compact, an assumption which we shall not make.<\/p>\n Q<\/strong>(X<\/em>) has a rich theory, much like H<\/strong>(X<\/em>), the Hoare powerspace with the lower<\/em> Vietoris topology.\u00a0 This was explored in previous posts<\/a>.\u00a0 This is called the Smyth powerspace monad. Oh yes, this was shown to be a monad by Andrea Schalk again [2].\u00a0 In fact, Q<\/strong>(X<\/em>) with the upper Vietoris topology is the free unital de<\/em>flationary sober semi-lattice on X<\/em>.\u00a0 (Oops, that sentence is not true.\u00a0 It looks like it should be that way, but Schalk even has two sections as to why this fails.\u00a0 Still, Q<\/strong> is a monad, and that is the only think we will need.\u00a0 Thanks to Matthew de Brecht and Weng Kin Ho for pointing out this problem\u2014corrected on June 26th, 2018.) Jia, Jung and Li start by showing that a parameterized version of the unit is also continuous.\u00a0 For every compact saturated subset Q<\/em> of X<\/em>, let iQ<\/sub><\/em> be the map x<\/em> \u21a6 Q <\/em>\u2229 \u2191x<\/em>.<\/p>\n When Q<\/em> is of the form \u2191y<\/em>, we know that iQ<\/sub><\/em> maps every point x<\/em> to an element of Q<\/strong>(X<\/em>). In general, let C<\/em> be the subset of Q<\/strong>(X<\/em>) consisting of those non-empty compact saturated subsets Q<\/em> such that Q <\/em>\u2229 \u2191x<\/em> is compact for every point x<\/em> in X<\/em>.\u00a0 The map iQ<\/sub><\/em> is well-defined, as a map from X<\/em> to Q<\/strong>(X<\/em>), exactly when Q<\/em> is in C<\/em>.<\/p>\n Lemma 1.<\/strong>\u00a0 For every Q<\/em> in C<\/em>, iQ<\/sub><\/em> is (well-defined and) continuous.<\/p>\n Proof.<\/em> We compute the inverse image A<\/em> of \u2610U<\/em>, where U<\/em> is open in X<\/em>, and we show that A<\/em> is Scott-open.\u00a0 A<\/em> is upwards-closed, being the inverse image of an upwards-closed subset by a monotonic map (recall that when x<\/em> grows, \u2191x<\/em> becomes smaller under inclusion, hence larger in the specialization ordering of Q<\/strong>(X<\/em>), which is reverse<\/em> inclusion).<\/p>\n To show that A<\/em> is Scott-open, we consider a directed family (xi<\/sub><\/em>)i \u2208 I<\/sub><\/em> of points of X<\/em>, and realize that iQ<\/sub><\/em>(supi \u2208 I<\/sub><\/em> xi<\/sub><\/em>)=Q <\/em>\u2229 \u2191(supi \u2208 I<\/sub><\/em> xi<\/sub><\/em>) = \u2229i \u2208 I<\/sub><\/em> (Q \u2229 \u2191xi<\/sub><\/em>), because the upward closure of\u00a0supi \u2208 I<\/sub><\/em> xi<\/sub><\/em> is the collection of upper bounds of (xi<\/sub><\/em>)i \u2208 I<\/sub><\/em>, and that is the intersection of all the upward closures of each xi<\/sub><\/em>.\u00a0 However, \u2229i \u2208 I<\/sub><\/em> (Q \u2229 \u2191xi<\/sub><\/em>)= \u2229i \u2208 I<\/sub><\/em> iQ<\/sub><\/em>(xi<\/sub><\/em>), and the latter is the supremum of the elements iQ<\/sub><\/em>(xi<\/sub><\/em>): recall that directed suprema in Q<\/strong>(X<\/em>) are exactly filtered intersections, when X<\/em> is well-filtered.\u00a0 \u2610<\/p>\n Lemma 2.<\/strong>\u00a0 For every Q<\/em> in C<\/em>, for every compact saturated set Q’<\/em> of X<\/em>, Q<\/em> \u2229 Q’<\/em> is compact in X<\/em>.<\/p>\n Proof. <\/em>We use what is essentially the multiplication of the Q<\/strong> monad.\u00a0 By Lemma 1, and the fact that the images of compact sets by continuous maps are compact (Proposition 4.4.13 in the book<\/a>), for every Q<\/em> in C<\/em>, the image iQ<\/sub><\/em>[Q’<\/em>] of any given compact saturated subset Q’<\/em> of X<\/em> by\u00a0iQ<\/sub><\/em> is compact in Q<\/strong>(X<\/em>). (It is compact in a space of compact sets!)<\/p>\n Let Q”<\/em> be the union of all the elements of iQ<\/sub><\/em>[Q’<\/em>].\u00a0 We have:<\/p>\nThe main result<\/h2>\n
\nIn particular, the Q<\/strong> monad has a unit, and that is the map that sends x<\/em> to \u2191x<\/em>.\u00a0 But we shall not need that, except to give a hint as what we are going to do.<\/p>\n\n