{"id":7462,"date":"2023-11-20T13:16:31","date_gmt":"2023-11-20T12:16:31","guid":{"rendered":"https:\/\/projects.lsv.ens-paris-saclay.fr\/topology\/?page_id=7462"},"modified":"2023-11-20T13:16:31","modified_gmt":"2023-11-20T12:16:31","slug":"compact-semilattices-without-small-semilattices-ii-gierzs-counterexample","status":"publish","type":"page","link":"https:\/\/projects.lsv.ens-paris-saclay.fr\/topology\/?page_id=7462","title":{"rendered":"Compact semilattices without small semilattices II: Gierz’s counterexample"},"content":{"rendered":"\n

In 1987, Gerhard Gierz produced another example of a compact semilattice without small semilattices [1]. It is a bit simpler than Jimmie Lawson’s counterexample<\/a>, and it has an additional feature: it embeds into the Hilbert cube [0, 1]N<\/sup><\/strong>, the product of a countably infinite number of copies of [0, 1], with its usual Hausdorff topology (although we will not say why here).<\/p>\n\n\n\n

The fact that Gierz’s counterexample embeds into the Hilbert cube may seem paradoxical, as Lawson’s theorem (see Theorem D there<\/a>) states that any compact semilattice that embeds into a power [0, 1]A<\/sup><\/em>, for any set A<\/em>, must have small semilattices, right? Of course, there is a subtle point: any compact semilattice that embeds into [0, 1]A<\/sup><\/em> as a topological semilattice<\/em> will have small semilattices. But Gierz’s counterexample will only embed as a topological space<\/em>. That is, the embedding will be a topological embedding, but it will not<\/em> map sums to sums.<\/p>\n\n\n\n

In any case, let us proceed to the example. This is built in four steps.<\/p>\n\n\n\n

Step 1: The compact semilattice ([0, 1]N<\/sup><\/strong>, \u2293) and the functions fn<\/sub><\/em> and f<\/em> : [0, 1]N<\/sup><\/strong> \u2192 [0, 1]<\/h2>\n\n\n\n

We start from the Hilbert cube [0, 1]N<\/sup><\/strong> itself, and we equip with the \u2293 operation (pointwise min). Gierz uses the pointwise max operation instead. Using mins will make step 2 slightly more complicated, but using maxes would force us to adjust our minds constantly later on, forcing us to juggle with min operations which are really maxes, or to deal with 1<\/strong> being a zero and 0<\/strong> being a one.<\/p>\n\n\n\n

By Lawson’s theorem (see Theorem D there<\/a>), ([0, 1]N<\/sup><\/strong>, \u2293) is a compact lattice with small semilattices. We use the notation x<\/em><\/strong> (in boldface font, for lack of an overhang arrow) to denote elements of [0, 1]N<\/sup><\/strong>, that is, sequences of elements of [0, 1]. For each such sequence x<\/em><\/strong>, we will write xn<\/sub><\/em> (no boldface, but a subscript) to denote its element at position n<\/em>.<\/p>\n\n\n\n

For every number n<\/em>\u22651, we define a function fn<\/sub><\/em> : [0, 1]N<\/sup><\/strong> \u2192 [0, 1], by:<\/p>\n\n\n\n

fn<\/sub><\/em>(x<\/em><\/strong>) \u225d 1 \u2013 log (1+n<\/em>\u2013x<\/em>0<\/sub>\u2013…\u2013xn<\/sub><\/em>\u20131<\/sub>)\/log (1+n<\/em>),<\/p>\n\n\n\n

where log is natural logarithm. In order to fix ideas, fn<\/sub><\/em>(x<\/em><\/strong>) is equal to gn<\/sub><\/em>(x<\/span><\/em><\/strong>n<\/sub><\/em>), where x<\/span><\/em><\/strong>n<\/sub><\/em> is the average of the first n<\/em> entries of the tuple x<\/em><\/strong>, and gn<\/sub><\/em> is the function mapping t<\/em> \u2208 [0, 1] to 1\u2013log(1+n<\/em>\u2013nt<\/em>)\/log(1+n<\/em>) (also in [0, 1]); those functions gn<\/sub><\/em> are depicted in the graph below, for n<\/em>=1, 2, 3, 4, 5, 6, 10, 20, 100 from top to bottom.<\/p>\n\n\n\n

\"\"<\/a><\/figure>\n\n\n\n

The functions fn<\/sub><\/em> are continuous, and monotonic with respect to the pointwise ordering on [0, 1]N<\/sup><\/strong>; if you prefer, they are perfect maps from ([0, 1]\u03c3<\/sub>)N<\/strong><\/sup> to [0, 1]\u03c3<\/sub>. They map any sequence that starts with n<\/em> zeroes to 0, and any sequence that starts with n<\/em> ones to 1.<\/p>\n\n\n\n

Lawson’s counterexample<\/a> also used some funny trickery with logarithms (see Fact (i) there, where we need to observe that for k<\/em> large enough, log2<\/sub>log2<\/sub> (1+k<\/em>) + \u03b5 > log2<\/sub> log2<\/sub> (1+2k<\/em>)). Here we need to observe that fn<\/sub><\/em> is almost a semilattice homomorphism, and looks more and more like one as n<\/em> tends to infinity. In other words, the following inequalities hold.<\/p>\n\n\n\n

Fact A.<\/strong> min (fn<\/sub><\/em>(x<\/em><\/strong>), fn<\/sub><\/em>(y<\/em><\/strong>)) \u2013 log 2\/log (n<\/em>+1) \u2264 fn<\/sub><\/em>(x<\/em><\/strong> \u2293 y<\/em><\/strong>) \u2264 min (fn<\/sub><\/em>(x<\/em><\/strong>), fn<\/sub><\/em>(y<\/em><\/strong>)),
for all x<\/em><\/strong>, y<\/em><\/strong> in [0, 1]N<\/sup><\/strong>.<\/p>\n\n\n\n

Proof.<\/em> The second inequality fn<\/sub><\/em>(x<\/em><\/strong> \u2293 y<\/em><\/strong>) \u2264 min (fn<\/sub><\/em>(x<\/em><\/strong>), fn<\/sub><\/em>(y<\/em><\/strong>)) is because fn<\/sub><\/em> is monotonic in its argument (not in n<\/em>, mind you). The first inequality is more interesting. The proof below is unfortunately slightly more complex than Gierz’s, because of our choice of staying with pointwise mins: Gierz uses maxes instead, and his definition of fn<\/sub><\/em> is slightly simpler as a consequence.<\/p>\n\n\n\n

Without loss of generality, we may assume that x<\/em>0<\/sub>+…+xn<\/sub><\/em>\u20131<\/sub> \u2264 y<\/em>0<\/sub>+…+yn<\/sub><\/em>\u20131<\/sub>, so fn<\/sub><\/em>(x<\/em><\/strong>) \u2264 fn<\/sub><\/em>(y<\/em><\/strong>), and we need to show that fn<\/sub><\/em>(x<\/em><\/strong>) \u2013 log 2\/log (n<\/em>+1) \u2264 fn<\/sub><\/em>(x<\/em><\/strong> \u2293 y<\/em><\/strong>). Let z<\/em><\/strong> \u225d x<\/em><\/strong> \u2293 y<\/em><\/strong>, so that zk<\/sub><\/em> = min (xk<\/sub><\/em>, yk<\/sub><\/em>) for every index k<\/em>.<\/p>\n\n\n\n