{"id":2067,"date":"2019-10-20T08:59:40","date_gmt":"2019-10-20T06:59:40","guid":{"rendered":"https:\/\/projects.lsv.ens-paris-saclay.fr\/topology\/?page_id=2067"},"modified":"2022-11-19T15:05:15","modified_gmt":"2022-11-19T14:05:15","slug":"core-compactwell-filtered-t0sober-locally-compact","status":"publish","type":"page","link":"https:\/\/projects.lsv.ens-paris-saclay.fr\/topology\/?page_id=2067","title":{"rendered":"Core-compact+well-filtered T0=sober locally compact"},"content":{"rendered":"\n

Last time<\/a>, I motivated the construction of the well-filterification Wf<\/strong>(X<\/em>) of a space X<\/em> of X. Xu, Ch. Shen, X. Xi and D. Zhao [3] by saying that it was needed to understand their proof of the fact that every core-compact well-filtered T0<\/sub> space is sober, and hence also locally compact. This solves a question<\/a> asked by X. Jia, and first solved positively by J. Lawson and X. Xi [2].<\/p>\n\n\n\n

While polishing up Xu, Shen, Xi, and Zhao’s proof , I realized that the key was a new form of the Hofmann-Mislove theorem for well-filtered spaces, which is interesting in its own right. I will describe it and prove it below.<\/p>\n\n\n\n

Then I realized that one could simplify their proof. Their argument relies on the following theorem [3, Theorem 6.15]: in a core-compact space, every irreducible closed subset is a WD subset<\/a>. The proof of the latter is technical, but from there, it is not hard to show that every core-compact well-filtered T0<\/sub> space is sober. Interestingly, there is a more directed proof: as we will see, every core-compact well-filtered T0<\/sub> space is locally compact, and this will be much simpler. We will then conclude, since every locally compact well-filtered T0<\/sub> space is sober (Proposition 8.3.8 in the book<\/a><\/a>).<\/p>\n\n\n\n

A Hofmann-Mislove theorem for well-filtered spaces<\/h2>\n\n\n\n

Remember the Hofmann-Mislove theorem (Theorem 8.3.2 in the book<\/a>): in a sober space X<\/em>, every Scott-open filter F<\/strong><\/em> of open subsets of X<\/em> is the family of open neighborhoods of a unique compact saturated set Q<\/em>, namely \u2229F<\/strong><\/em>.<\/p>\n\n\n\n

You don’t have that in a well-filtered space. However, I claim that this works if F<\/strong><\/em> is countably<\/em> generated<\/em>, namely if there is a countable descending chain of sets A<\/em>n<\/em><\/sub>, n<\/em> \u2208 \u2115, (i.e.,  A<\/em>0 <\/sub>\u2287 A<\/em>1 <\/sub>\u2287 \u2026 \u2287 An<\/sub><\/em> \u2287 \u2026), such that F<\/strong><\/em> is equal to the collection of open sets U<\/em> that contain some A<\/em>n<\/em><\/sub>. (Note that I am not requiring that the sets A<\/em>n<\/em><\/sub> be open themselves, but that will not be important in the sequel.)<\/p>\n\n\n\n

In the proof I give below, you should recognize arguments similar to the usual proof of the Hofmann-Mislove theorem, combined with arguments similar to those used in the de Brecht-Kawai theorem<\/a> (given any well-filtered second-countable space X<\/em>, the upper Vietoris and Scott topologies on Q<\/strong>(X<\/em>) coincide\u2014Matthew de Brecht insists that the key ideas come from M. Schr\u00f6der’s papers, and although I still don’t quite understand everything that M. Schr\u00f6der says, Matthew must be right), where some pretty funny sets are shown to be compact, and this requires countable index sets. The key argument in the following proof is taken from X. Xu, Ch. Shen, X. Xi and D. Zhao’s proof of [3, Theorem 6.15], where it is somehow hidden.<\/p>\n\n\n\n

Theorem 1 (\u00e0 la Hofmann-Mislove).<\/strong> Let X<\/em> be a well-filtered space. Every countably generated Scott-open filter F<\/strong><\/em> of open subsets of X<\/em> is the family of open neighborhoods of a unique compact saturated set Q<\/em>, namely \u2229F<\/strong><\/em>.<\/p>\n\n\n\n

Proof.<\/em> Let F<\/strong><\/em> be generated by countably many sets A<\/em>n<\/em><\/sub>, n<\/em> \u2208 \u2115, where  A<\/em>0 <\/sub>\u2287 A<\/em>1 <\/sub>\u2287 \u2026 \u2287 An<\/sub><\/em> \u2287 \u2026. We recall that this means that the open sets U<\/em> in F<\/strong><\/em> are exactly those that are supersets of at least one A<\/em>n<\/em><\/sub>.<\/p>\n\n\n\n

We start as in the proof of the Hofmann-Mislove theorem. We let Q<\/em> be the intersection \u2229F<\/strong><\/em> of all the open sets in F<\/strong><\/em>. We claim that F<\/strong><\/em> is exactly the family of open neighborhoods of Q<\/em>. (Please note that we do not know yet that Q<\/em> is compact.)<\/p>\n\n\n\n

Reasoning by contradiction, we assume that there is an open neighborhood of Q<\/em> that is not in F<\/strong><\/em>. The collection E<\/strong><\/em> of open neighborhoods of Q<\/em> that are not in F<\/strong><\/em> is therefore non-empty. With the inclusion ordering, E<\/strong><\/em> is a dcpo, exactly because F<\/strong><\/em> is Scott-open. Hence we can use Zorn’s Lemma: there is a maximal<\/em> open neighborhood U<\/em> of Q<\/em> that is not in F<\/strong><\/em>.<\/p>\n\n\n\n

In the traditional proof of the Hofmann-Mislove theorem, we then use the fact that F<\/strong><\/em> is a filter to obtain that the complement of U<\/em> is an irreducible closed set, and we conclude by sobriety… but we do not have sobriety here.<\/p>\n\n\n\n

This is the point where we have to use that F<\/strong><\/em> is countably generated instead. Since U<\/em> is not in F<\/strong><\/em>, it does not contain any An<\/sub><\/em>, so there is a point xn<\/sub> <\/em>in An<\/sub><\/em> and not in U<\/em>, for every n<\/em> \u2208 \u2115.<\/p>\n\n\n\n

Let Kn<\/sub><\/em> be the set \u2191{x<\/i>m<\/i><\/sub> | m<\/em>\u2265n<\/em>}=\u2191{xn<\/sub><\/em>, xn<\/sub><\/em>+1<\/sub>, …}. As in the de Brecht-Kawai(-Schr\u00f6der) argument, this is the upward closure of an infinite set of points, hence it is not immediately obvious that Kn<\/sub><\/em> is compact, but we claim that it is nonetheless. Let (V<\/em>i<\/em><\/sub>)i<\/em> \u2208 I<\/em><\/sub> be an open cover of Kn<\/sub><\/em>. We wish to extract a finite subcover:<\/p>\n\n\n\n