Recently, Matthew de Brecht sent me a clever argument that shows the following:<\/p>\n\n\n\n
Theorem (de Brecht, private communication, March 29th, 2019):<\/strong> The poset-theoretic product of two first-countable posets coincides with their topological product.<\/p>\n\n\n\n He insists that the ideas come straight out of a paper by Matthias Schr\u00f6der [1]. My impression is that this still requires some thinking after reading [1], but certainly M. Schr\u00f6der has had plenty of clever ideas.<\/p>\n\n\n\n What M. de Brecht’s theorem above means, in detail, is the following. Let X<\/em> and Y<\/em> be two posets. You can consider them as topological spaces, written X<\/em>\u03c3<\/sub> and Y<\/em>\u03c3<\/sub> respectively, by giving them their Scott topology. Then you can build their topological product X<\/em>\u03c3<\/sub> \u00d7 Y<\/em>\u03c3<\/sub>. Or you can first build their product and then take the Scott topology of the result, yielding (X<\/em> \u00d7 Y<\/em>)\u03c3.<\/sub> In general, the two results differ: the Scott topology on X<\/em> \u00d7 Y<\/em> is finer, and in general strictly finer than the product topology on X<\/em>\u03c3<\/sub> \u00d7 Y<\/em>\u03c3<\/sub> (see Exercise 5.2.16 in the book<\/a>). What M. de Brecht shows is that there is never such a problem if X<\/em>\u03c3<\/sub> and Y<\/em>\u03c3<\/sub> are first-countable.<\/p>\n\n\n\n This is a pretty surprising result. It is well-known that the problem disappears, namely X<\/em>\u03c3<\/sub> \u00d7 Y<\/em>\u03c3<\/sub>=(X<\/em> \u00d7 Y<\/em>)\u03c3<\/sub>, if X<\/em> and Y<\/em> are continuous posets (Proposition 5.1.54). What we learn is that, for first-countable posets, continuity is not needed<\/em>.<\/p>\n\n\n\n The key to that result (apart from other clever arguments, which we will see below) is the following easy observation:<\/p>\n\n\n\n Fact 1:<\/strong> Let X<\/em> be a topological space, (xn<\/sub><\/em>)n \u2208 <\/sub><\/em>\u2115<\/sub> be a sequence of elements of X<\/em>, and x<\/em> be any limit of that sequence. Then the set {xn<\/sub><\/em> | n<\/em> \u2208 \u2115} \u222a {x<\/em>} (the completed sequence<\/em> of the title) is compact.<\/p>\n\n\n\n That is easy to prove. Consider any open cover of that set. One open set in that cover must contain x<\/em>, and since (xn<\/sub><\/em>)n \u2208 <\/sub><\/em>\u2115<\/sub> converges to x<\/em>, it must contain all the elements xn<\/sub><\/em> for n<\/em> large enough. The finitely many remaining elements can then be covered by finitely many of the remaining open sets from the cover.<\/p>\n\n\n\n It is pretty easy to see that one cannot generalize Fact 1 to nets indexed by anything else than \u2115. That already fails for nets indexed by \u03c9.2, the ordinal with elements 0<1<…<n<\/em><… < \u03c9<\u03c9+1<…<\u03c9+n<\/em><…, unless you require that x\u03c9<\/sub><\/em> be a limit of the subsequence of elements xn<\/sub><\/em>, n<\/em><\u03c9. The situation is worse for chains indexed by ordinals above \u03c92<\/sup>, and even worse for nets that are not ordinal-indexed chains.<\/p>\n\n\n\n Let me reproduce Matthew de Brecht’s argument. Since every open set in the product topology is Scott-open, it suffices to show the converse implication. Let W<\/em> be a Scott-open subset of X<\/em> \u00d7 Y<\/em>. We will show that W<\/em> is open in the product of the Scott topologies.<\/p>\n\n\n\n We first define f<\/em> : X<\/em> \u2192 O<\/strong>Y<\/em>, where O<\/strong>Y<\/em> is the complete lattice of open subsets of Y<\/em>, by letting f<\/em>(x<\/em>) be the set of points y<\/em> of Y<\/em> such that (x<\/em>, y<\/em>) is in W<\/em>. Then f<\/em><\/i><\/span> is a Scott-continuous map. This is a trick that is already used in the red book [2, proof of Theorem II-4.13(1)]. The argument used by the authors of that book is rather abstract, but that is also easily proved by hand, hence I will let you do the exercise; don’t forget to verify that f<\/em>(x<\/em>) is in O<\/strong>Y<\/em> for every x<\/em> in X<\/em> first!<\/p>\n\n\n\n In order to show that W<\/em> is open in the product topology, we fix an arbitrary pair (x<\/em>, y<\/em>) in W<\/em>, and we wish to show that there is an open rectangle containing (x<\/em>, y<\/em>) and included in W<\/em>.<\/p>\n\n\n\n Let us imagine that this failed for some pair (x<\/em>, y<\/em>) in W<\/em>. Using first-countability, we enumerate a base of open neighborhoods (Un<\/sub><\/em>)n \u2208 <\/sub><\/em>\u2115<\/sub> of x<\/em>, and a base of open neighborhoods (Vn<\/sub><\/em>)n \u2208 <\/sub><\/em>\u2115<\/sub> of y<\/em>. We can assume that those bases form descending sequences: otherwise we replace each Un<\/sub><\/em> by the intersection U<\/em>0<\/sub> <\/sub><\/em>\u2229 U<\/em>1<\/sub> <\/sub><\/em>\u2229 … \u2229 Un<\/sub><\/em> for every n<\/em>, and similarly with Vn<\/sub><\/em>.<\/p>\n\n\n\n By assumption, none of the open rectangles Un<\/sub><\/em> \u00d7 Vn<\/sub><\/em> is included in W<\/em>. Therefore, we can find a point (xn<\/sub><\/em>, yn<\/sub><\/em>) in Un<\/sub><\/em> \u00d7 Vn<\/sub><\/em> and outside W<\/em>.<\/p>\n\n\n\n We note that (xn<\/sub><\/em>)n \u2208 <\/sub><\/em>\u2115<\/sub> converges to x<\/em>: every open neighborhood of x<\/em> must contain some Un<\/sub><\/em>, by the definition of a base of open neighborhoods, and then every xm <\/sub><\/em>with m<\/em>\u2265n<\/em>, is in Um<\/sub>, <\/sub><\/em>hence in Un<\/sub><\/em> (because we have made sure that the sequence (Um)<\/sub><\/em>m<\/sub><\/em> <\/sub><\/em>\u2208 <\/sub><\/em>\u2115<\/sub> is descending). Similarly, (yn<\/sub><\/em>)n \u2208 <\/sub><\/em>\u2115<\/sub> converges to y<\/em>.<\/p>\n\n\n\n Since f<\/em>(x<\/em>) is open in Y<\/em> and y<\/em> is in f<\/em>(x<\/em>), every yn<\/sub><\/em> is in f<\/em>(x<\/em>) for n<\/em> large enough, say n<\/em>\u2265n<\/em>0<\/sub>. We consider the completed sequence K<\/em> defined as {yn<\/sub> <\/em>| n<\/em>\u2265n<\/em>0<\/sub>} \u222a {y<\/em>}. By Fact 1, K<\/em> is a compact set, and it is included in the open set f<\/em>(x<\/em>).<\/p>\n\n\n\n We now consider the set \u25a0K<\/em> of all open subsets of Y<\/em> that contain K<\/em>. Since K<\/em> is compact, \u25a0K<\/em> is Scott-open in O<\/strong>Y<\/em>. Indeed, it is upwards-closed, and every directed family (U<\/em>i<\/em><\/sub>)i <\/em>\u2208<\/sub> I<\/sub> <\/em>of open sets whose union is in \u25a0K<\/em>, i.e., which contains K<\/em>, must be such that some U<\/em>i<\/em><\/sub> already contains K<\/em>, i.e., is in \u25a0K<\/em> (Proposition 4.4.7 in the book<\/a>).<\/p>\n\n\n\n Since f<\/em> is continuous, f<\/em>-1<\/sup>(\u25a0K<\/em>) is open in X<\/em>.<\/p>\n\n\n\n We now recall that (xn<\/sub><\/em>)n \u2208 <\/sub><\/em>\u2115<\/sub> converges to x<\/em>. And x<\/em> is in f<\/em>-1<\/sup>(\u25a0The completed sequence is compact<\/h2>\n\n\n\n
The proof<\/h2>\n\n\n\n