{"id":5736,"date":"2022-10-19T16:25:59","date_gmt":"2022-10-19T14:25:59","guid":{"rendered":"https:\/\/projects.lsv.ens-paris-saclay.fr\/topology\/?page_id=5736"},"modified":"2022-11-19T14:52:41","modified_gmt":"2022-11-19T13:52:41","slug":"strongly-compact-sets-and-the-double-hyperspace-construction","status":"publish","type":"page","link":"https:\/\/projects.lsv.ens-paris-saclay.fr\/topology\/?page_id=5736","title":{"rendered":"Strongly compact sets and the double hyperspace construction"},"content":{"rendered":"\n

The notion of strongly compact set is due to Reinhold Heckmann [1]. A subset Q<\/em> of a space X<\/em> is called strongly compact<\/em> if and only if for every open neighborhood U<\/em> of Q<\/em>, there is a finitary compact set \u2191E<\/em> such that Q<\/em> \u2286 \u2191E<\/em> \u2286 U<\/em>. (A finitary compact set is the upward closure of a finite set E<\/em>.)<\/p>\n\n\n\n

I have already mentioned there<\/a> that for every sober space X<\/em>, the sobrification of Q<\/strong>fin<\/sub>(X<\/em>) of finitary compact subsets of X<\/em>, with the upper Vietoris topology, is the space Q<\/strong>s<\/sub>(X<\/em>) of so-called strongly<\/em> compact saturated subsets of X<\/em>, not the whole Smyth hyperspace Q<\/strong>(X<\/em>) as one might expect [2, Proposition 7.33].  Let us explore this in a bit more detail.<\/p>\n\n\n\n

First, we need to notice that every strongly compact set is compact. If Q<\/em> is strongly compact indeed, then every open cover (Ui<\/sub><\/em>)i<\/em> \u2208 I<\/em><\/sub> of Q<\/em> is such that Q<\/em> \u2286 \u2191E<\/em> \u2286 \u222ai<\/em> \u2208 I<\/em><\/sub> Ui<\/sub><\/em> for some finite set E<\/em>. Then E<\/em> is included in the union of finite many of the sets Ui<\/sub><\/em>, which therefore form a finite subcover of Q<\/em>.<\/p>\n\n\n\n

If X<\/em> is a continuous dcpo in its Scott topology, or more generally a locally finitary compact space, then, conversely, every compact set Q<\/em> is strongly compact. Indeed, for every open neighborhood U<\/em> of Q<\/em>, we use locally finitary compactness in order to find finitary compact neighborhoods \u2191Ex<\/sub><\/em> of each point x<\/em> of Q<\/em>, and included in U<\/em>; finitely many cover Q<\/em>, and the resulting finite union is a finitary compact set that contains Q<\/em> and is included in U<\/em>.<\/p>\n\n\n\n

But, in more general spaces, there are in general strictly more compact sets than strongly compact sets. For example, in any T1<\/sub> space, it is easy to see that the strongly compact sets are exactly the finite sets, and that is very far from exhausting the collection of compact sets.<\/p>\n\n\n\n

The sobrification of Q<\/strong>fin<\/sub>(X<\/em>)<\/h2>\n\n\n\n

Let Q<\/strong>(X<\/em>) be the space of all compact saturated subsets of X<\/em>, with the upper Vietoris topology: its basic open subsets are of the form \u2610U<\/em> \u225d {Q<\/em> \u2208 Q<\/strong>(X<\/em>) | Q<\/em> \u2286 U<\/em>}, where U<\/em> ranges over the open subsets of X<\/em>. Q<\/strong>(X<\/em>) is a T0<\/sub> space, whose specialization ordering is reverse<\/em> inclusion.<\/p>\n\n\n\n

One can say more. First, there is a monad<\/a> (Q<\/strong>, \u03b7, \u03bc) on the category Top<\/strong>0<\/sub> of T0<\/sub> spaces, whose unit \u03b7X<\/sub><\/em> : X<\/em> \u2192 Q<\/strong>(X<\/em>) maps every point x<\/em> of X<\/em> to \u2191x<\/em>, and whose multiplication \u03bcX<\/sub><\/em> : Q<\/strong>(Q<\/strong>(X<\/em>)) \u2192 Q<\/strong>(X<\/em>) maps every element Q<\/em><\/strong> of Q<\/strong>(Q<\/strong>(X<\/em>)) to \u222aQ<\/em><\/strong>, the union of all the elements of Q<\/em><\/strong>. It is practical to note that \u03b7X<\/sub><\/em>\u20131<\/sup>(\u2610U<\/em>)=U<\/em> and \u03bcX<\/sub><\/em>\u20131<\/sup>(\u2610\u2610U<\/em>)=\u2610U<\/em> for every open subset U<\/em> of X<\/em>; this is useful in order to show that the unit and the multiplication are continuous, for example.<\/p>\n\n\n\n

We will also need to observe that the operator \u2610 commutes with finite intersections and with directed unions of open sets; the latter is a consequence of the compactness of elements of Q<\/strong>(X<\/em>).<\/p>\n\n\n\n

Lemma.<\/strong> For every sober space X<\/em>, Q<\/strong>(X<\/em>) is sober.<\/p>\n\n\n\n

Proof.<\/em> Let C<\/strong> be an irreducible closed subset of Q<\/strong>(X<\/em>). We consider the family F<\/strong> of all open subsets U<\/em> of X<\/em> such that \u2610U<\/em> intersects C<\/strong>. F<\/strong> is non-empty, upwards-closed and if U<\/em> and V<\/em> are any two elements of F<\/strong>, then since C<\/strong> is irreducible, \u2610U<\/em> \u2229 \u2610V<\/em> intersects C<\/strong>. Since \u2610U<\/em> \u2229 \u2610V<\/em> =\u2610(U<\/em> \u2229 V<\/em>), U<\/em> \u2229 V<\/em> is in F<\/strong>. Hence, F<\/strong> is a filter of open subsets of X<\/em>. Similarly, but using the fact that \u2610 commutes with directed unions of open sets, F<\/strong> is Scott-open. By the Hofmann-Mislove theorem (Theorem 8.3.2 in the <\/a>book<\/a>), F<\/strong> is the collection of open neighborhoods of some compact saturated subset Q<\/em>0<\/sub> of X<\/em>.<\/p>\n\n\n\n

If Q<\/em>0<\/sub> were not in C<\/strong>, then it would be in its complement, which is open, so there would be a basic open set \u2610U<\/em> such that Q<\/em>0<\/sub> \u2208 \u2610U<\/em> and \u2610U<\/em> does not intersect C<\/strong>. The latter means that U<\/em> is not in F<\/strong>, hence does not contain Q<\/em>0<\/sub>; this contradicts Q<\/em>0<\/sub> \u2208 \u2610U<\/em>. Therefore Q<\/em>0<\/sub> is in C<\/strong>.<\/p>\n\n\n\n

In order to show that C<\/strong> is the closure of Q<\/em>0<\/sub>, it remains to show that every element Q<\/em> of C<\/strong> is below, namely contains, Q<\/em>0<\/sub>. It suffices to show that every open neighborhood U<\/em> of Q<\/em> contains Q<\/em>0<\/sub>, and that is easy: since Q<\/em> \u2286 U<\/em>, \u2610U<\/em> intersects C<\/strong> at Q<\/em>, so U<\/em> is in F<\/strong>, and therefore contains Q<\/em>0<\/sub>. \u2610<\/p>\n\n\n\n

One might guess that Q<\/strong>(X<\/em>) would be the sobrification of its subspace Q<\/strong>fin<\/sub>(X<\/em>) of finitary compact subsets, but that is wrong. For example, you can check that, if X<\/em> is T2<\/sub>, then Q<\/strong>fin<\/sub>(X<\/em>) is already sober, and very different from Q<\/strong>(X<\/em>). Since that will follow from the next result, I will not bother to show this.<\/p>\n\n\n\n

Proposition.<\/strong> For every sober space X<\/em>, the sobrification of Q<\/strong>fin<\/sub>(X<\/em>) is Q<\/strong>s<\/sub>(X<\/em>), the subspace of Q<\/strong>(X<\/em>) consisting of strongly compact saturated subsets of X<\/em>.<\/p>\n\n\n\n

Proof.<\/em> Since Q<\/strong>(X<\/em>) is sober, the sobrification of Q<\/strong>fin<\/sub>(X<\/em>) is its Skula-closure, as we have seen in this post<\/a>. (Namely, the closure in the Skula topology. The Skula topology is generated by the open and<\/em> the closed sets, or equivalently by the sets U<\/em> \u2229 \u2193z<\/em>, where z<\/em> ranges over the points and U<\/em> ranges over the open sets in the original topology.)<\/p>\n\n\n\n

If Q<\/em> is in the Skula-closure of Q<\/strong>fin<\/sub>(X<\/em>), then for every open neighborhood U<\/em> of Q<\/em>, \u2610U<\/em> \u2229 \u2193Q<\/sub><\/strong>Q<\/em> is a Skula-open neighborhood of Q<\/em> in Q<\/strong>(X<\/em>); I am writing \u2193Q<\/sub><\/strong> for downward closure in Q<\/strong>(X<\/em>) (remember that the ordering there is reverse<\/em> inclusion!). And that set must therefore intersect Q<\/strong>fin<\/sub>(X<\/em>). In other words, there is a finitary compact set \u2191E<\/em> in \u2610U<\/em> \u2229 \u2193Q<\/sub><\/strong>Q<\/em>. We expand this, and we obtain Q<\/em> \u2286 \u2191E<\/em> \u2286 U<\/em>. Therefore Q<\/em> must be strongly compact saturated.<\/p>\n\n\n\n

Conversely, for every strongly compact saturated subset Q<\/em> of X<\/em>, we claim that Q<\/em> is in the Skula-closure of Q<\/strong>fin<\/sub>(X<\/em>). In other words, for every open neighborhood U<\/strong> of Q<\/em> in Q<\/strong>(X<\/em>), we show that U<\/strong> contains some finitary compact set containing Q<\/em>. Necessarily, U<\/strong> contains a basic open neighborhood \u2610U<\/em> of Q<\/em>. Since Q<\/em> is strongly compact, there is a finitary compact set \u2191E<\/em> such that Q<\/em> \u2286 \u2191E<\/em> \u2286 U<\/em>; then \u2191E<\/em> is in \u2610U<\/em>, hence in U<\/strong>. \u2610<\/p>\n\n\n\n

One can then show that the monad (Q<\/strong>, \u03b7, \u03bc) cuts down to a monad (Q<\/strong>s<\/sub>, \u03b7, \u03bc), and both monads themselves restrict to monads on the category Sob<\/strong> of sober spaces. As I have said in this post<\/a>, the algebras of the monad (Q<\/strong>s<\/sub>, \u03b7, \u03bc) on Sob<\/strong> are exactly the deflationary sober semilattices with small semilattices<\/em> [1, Theorem 7.37], and the proof is a pretty simple extension of what we had done with Q<\/strong>fin<\/sub> in that same post.<\/p>\n\n\n\n

Instead, I would like to mention a funny hyperspace commutation theorem, which I happened to obtain recently by rereading carefully Matthew de Brecht and Tatsuji Kawai’s paper [3].<\/p>\n\n\n\n

Hyperspace commutation results<\/h2>\n\n\n\n

Let H<\/strong>(X<\/em>) denote the Hoare hyperspace of X<\/em>: that is the set of closed subsets of X<\/em>, with the lower Vietoris topology, given by subbasic open sets \u2662U<\/em> \u225d {C<\/em> \u2208 Q<\/strong>(X<\/em>) | C<\/em> intersects U<\/em>}, where U<\/em> ranges over the open subsets of X<\/em>. As we have seen here<\/a>, Schalk proved that H<\/strong>(X<\/em>) is always<\/em> a sober space. This also defines a monad (H<\/strong>, \u03b7, \u03bc) [let me reuse the same \u03b7 and \u03bc as before] whose unit \u03b7X<\/sub><\/em> : X<\/em> \u2192 H<\/strong>(X<\/em>) maps every point x<\/em> of X<\/em> to \u2193x<\/em>, and whose multiplication \u03bcX<\/sub><\/em> : H<\/strong>(H<\/strong>(X<\/em>)) \u2192 H<\/strong>(X<\/em>) maps every element C<\/em><\/strong> of H<\/strong>(H<\/strong>(X<\/em>)) to cl(\u222aC<\/em><\/strong>). Also, \u03b7X<\/sub><\/em>\u20131<\/sup>(\u2662U<\/em>)=U<\/em> and \u03bcX<\/sub><\/em>\u20131<\/sup>(\u2662\u2662U<\/em>)=\u2662U<\/em> for every open subset U<\/em> of X<\/em>.<\/p>\n\n\n\n

What de Brecht and Kawai achieved was to show that QH<\/strong>X<\/em> and HQ<\/strong>X<\/em> are homeomorphic, provided that X<\/em> is consonant; and that this is actually an if and only if. This is the continuation of a long line of work, showing similar isomorphisms in more restricted categories, mostly of domains. The locale theoretic analogue of this result is due to Steven Vickers and Christopher Townsend [4] shows that the same commutation between the two localic analogues of Q<\/strong> and H<\/strong> (the so-called upper and lower power locale constructions) holds without<\/em> any assumption on the locale.<\/p>\n\n\n\n

Replacing Q<\/strong> by Q<\/strong><\/strong>s<\/sub>, I will show that Q<\/strong><\/strong>s<\/sub>H<\/strong>X<\/em> and HQ<\/strong><\/strong>s<\/sub>X<\/em> are homeomorphic for every<\/em> topological space, consonant or not, sober or not.<\/p>\n\n\n\n

Oh, by the way, if you read Heckmann [1], you will see that this commutation requires a property that he calls US<\/sub>-conformity, and it may puzzle you that I will not require any assumption on X<\/em> at all. The main reason for this is that he considers the Scott topology, not the lower Vietoris topology, on H<\/strong>X<\/em>.<\/p>\n\n\n\n

Let us start the proof. This is essentially the same as de Brecht and Kawai’s proof, replacing Q<\/strong> by Q<\/strong>s<\/sub>. I will simply reorganize it, and mention how each step adapts when we replace Q<\/strong> by Q<\/strong>s<\/sub>.<\/p>\n\n\n\n

The first isomorphisms: QH<\/strong>=O<\/strong>p<\/sub>O<\/strong>\u03c3<\/sub>, Q<\/strong>s<\/sub>H<\/strong>=O<\/strong>p<\/sub>O<\/strong>p<\/sub><\/h2>\n\n\n\n

The first step consists in showing that there is a homeomorphism between QH<\/strong>X<\/em> and O<\/strong>p<\/sub>O<\/strong>\u03c3<\/sub>X<\/em>, valid for every topological space X<\/em> [3, Theorem 6.10]. (Well, de Brecht and Kawai only show an order-isomorphism here, not a homeomorphism, but that is not too far.) Here O<\/strong>X<\/em> is the collection of open subsets of X<\/em>, and we can topologize it in two ways:<\/p>\n\n\n\n