{"id":6249,"date":"2023-02-20T19:06:33","date_gmt":"2023-02-20T18:06:33","guid":{"rendered":"https:\/\/projects.lsv.ens-paris-saclay.fr\/topology\/?page_id=6249"},"modified":"2023-03-16T18:10:07","modified_gmt":"2023-03-16T17:10:07","slug":"on-till-plewes-game-and-matthew-de-brechts-non-consonance-arguments","status":"publish","type":"page","link":"https:\/\/projects.lsv.ens-paris-saclay.fr\/topology\/?page_id=6249","title":{"rendered":"On Till Plewe’s game and Matthew de Brecht’s non-consonance arguments"},"content":{"rendered":"\n

Last time<\/a> I mentioned that S<\/em>0<\/sub> is not consonant. I had a preprint by Matthew de Brecht where he proved that, but I could not find the result in published form. Since then, M. de Brecht wrote to me, and told me a lot of interesting things. First, his proof can be found in [1]. His proof uses a topological game first invented by Till Plewe [2] in order to study locale products of spaces, and conditions under which the locale products coincide with the (locales underlying) the topological products, which M. de Brecht had initially used to show that the locale product of S<\/em>0<\/sub> with itself is not spatial. Second, he pointed me towards an intriguing result by Ruiyuan Chen [3], which gives another connection between locale products (of dcpos) and the question of whether the poset-theoretic product of two dcpos with the Scott topology coincides with the topological product. Third, R. Chen has a simpler, alternative proof of the fact that the locale product of S<\/em>0<\/sub> with itself is not spatial [4]. And so on!<\/p>\n\n\n\n

Hence I have too much to talk about, so I will have to make a choice. I will concentrate on explaining what [1] is all about. I am afraid that this post will mostly be paraphrasing M. de Brecht’s paper. I have merely added a few pictures, and I have reformulated a few notions; I hope this will contribute to make what he achieved easier to grasp.<\/p>\n\n\n\n

Till Plewe’s game<\/h2>\n\n\n\n

Let X<\/em> and Y<\/em> be two topological spaces, U<\/em>0<\/sub> be an open subset of X<\/em>, V<\/em>0<\/sub> be an open subset of Y<\/em>, and U<\/em><\/strong> be an open cover of U<\/em>0<\/sub> \u00d7 V<\/em><\/em>0<\/sub> by open rectangles. Till Plewe’s game G<\/em>X<\/em>,Y<\/em><\/sub>(U<\/em>0<\/sub>, V<\/em>0<\/sub>, U<\/em><\/strong>) is played as follows. At each turn i<\/em>\u22651:<\/p>\n\n\n\n

    \n
  1. player I picks a point xi<\/sub><\/em> \u2208 U<\/em>i<\/em>\u20131<\/sub>;<\/li>\n\n\n\n
  2. then player II picks an open neighborhood U<\/em>i<\/em><\/sub> of xi<\/sub><\/em> included in U<\/em>i<\/em>\u20131<\/sub>;<\/li>\n\n\n\n
  3. player I picks a point yi<\/sub><\/em> \u2208 V<\/em>i<\/em>\u20131<\/sub>;<\/li>\n\n\n\n
  4. then player II picks an open neighborhood V<\/em>i<\/em><\/sub> of yi<\/sub><\/em> included in V<\/em>i<\/em>\u20131<\/sub>.<\/li>\n<\/ol>\n\n\n\n

    So, yes, each player plays twice at each turn.<\/p>\n\n\n\n

    Player II wins (at round i<\/em>) if U<\/em>i<\/em><\/sub> \u00d7 V<\/em>i<\/em><\/sub> is included in one of the open rectangles of the cover U<\/em><\/strong>. Otherwise, namely if the play goes along without player II even winning, player I wins.<\/p>\n\n\n\n

    The following in Theorem 1.1 in [2].<\/p>\n\n\n\n

    Theorem (Plewe).<\/strong> If X<\/em> and Y<\/em> are sober spaces, either:<\/p>\n\n\n\n