{"id":1785,"date":"2019-04-21T16:06:35","date_gmt":"2019-04-21T14:06:35","guid":{"rendered":"https:\/\/projects.lsv.ens-paris-saclay.fr\/topology\/?page_id=1785"},"modified":"2022-11-19T15:07:34","modified_gmt":"2022-11-19T14:07:34","slug":"isbells-non-sober-complete-lattice","status":"publish","type":"page","link":"https:\/\/projects.lsv.ens-paris-saclay.fr\/topology\/?page_id=1785","title":{"rendered":"Isbell’s non sober complete lattice"},"content":{"rendered":"\n

Every continuous dcpo is sober in its Scott topology, and every sober space is a dcpo in its specialization ordering. For some time, it was open whether every dcpo, not necessarily continuous, was sober, until Peter Johnstone found a counterexample [1]. This is explored in Exercises 5.2.15 (J<\/strong> is not core-compact), 8.2.14 (J<\/strong> is not sober), 8.3.9 (J<\/strong> is not well-filtered) in the book<\/a>.<\/p>\n\n\n\n

Now is every complete lattice sober in its Scott topology? The answer is again no, as shown by Isbell [2], but the non sober complete lattice he constructs is much more complicated. I aim to explain that construction today. I will only make one small change to Isbell’s original construction, and that will not change much really.<\/p>\n\n\n\n

Let me say that I benefited from Xiaodong Jia and Zhenchao Lyu’s explanations, during some of our Wednesday meetings. Xiaodong already knew the ins and outs of the construction, and Zhenchao took the time to read Isbell’s paper [2] carefully, and to explain it. Additionally, Zhenchao explained the contents of the recent paper by Xu, Xi and Zhao [5] to us both, and I will conclude this post with an excerpt of their results.<\/p>\n\n\n\n

Before we start, let me remind you that complete lattices are more regular than dcpos. In particular, they are all well-filtered, by a result of Lawson and Xi<\/a> [3], and hence coherent, by a result of Jia, Jung, and Li<\/a> [4].<\/p>\n\n\n\n

How to build a non sober complete lattice<\/h2>\n\n\n\n

We wish to find a non sober complete lattice L<\/em>. Hence there should be a closed subset C<\/em> of L<\/em> with the following properties:<\/p>\n\n\n\n

    \n
  1. C<\/em> should be irreducible, meaning that (C<\/em> is non-empty and) for any two open sets U<\/em> and V<\/em> that intersect C<\/em>, their intersection U<\/em> \u2229 V<\/em> should also intersect C<\/em>.<\/li>\n\n\n\n
  2. C<\/em> should not be the downward closure \u2193x<\/em> of any point x<\/em>, namely it should not have a largest element.<\/li>\n<\/ol>\n\n\n\n

    If we did not require L<\/em> to be a complete lattice, then we could just take C<\/em>=L<\/em>. This is impossible here because any complete lattice has a largest element, contradicting requirement 2.<\/p>\n\n\n\n

    Hence we opt for the next possible option. We will define C<\/em> as being the whole of L<\/em>, minus its top element \u22a4. We should still make sure that C<\/em> is closed, though, and that forces \u22a4 to be compact.<\/p>\n\n\n\n

    Next, any closed subset C<\/em> of L<\/em> is a dcpo in the induced ordering. By Zorn’s Lemma, every element of C<\/em> must therefore lie below some maximal element of C<\/em>. Hence C<\/em> will be equal to \u2193Max C<\/em>, where Max C<\/em> denotes the set of maximal elements of C<\/em>. Requirement 1 then implies that Max C<\/em> must be infinite. Indeed, assume that Max C<\/em> is a finite set {x<\/em>1<\/sub>, …, xn<\/sub><\/em>}, where x<\/em>1<\/sub>, …, xn<\/sub><\/em> are incomparable. Note that n<\/em>\u22652 by requirement 2. Then C<\/em> intersects U<\/em>=L<\/em>\u2013\u2193x<\/em>1<\/sub> (at x<\/em>2<\/sub>,<\/em> say) and V<\/em>=L<\/em>\u2013\u2193{x<\/em>2<\/sub>,<\/em> …, xn<\/sub><\/em>} (at x<\/em>1<\/sub>) but not their intersection.<\/p>\n\n\n\n

    Hence we should build L<\/em> with infinitely many coatoms<\/em>, namely with infinitely many incomparable elements right below \u22a4. And since \u22a4 should be compact, the only directed families whose supremum is \u22a4 are those that contain \u22a4 already.<\/p>\n\n\n\n

    By the way, although the title of Isbell’s paper [2] seems to indicate that his construction would be some form of completion of Johnstone space J<\/strong>, that is entirely misleading. It will be a completion of another non-sober dcpo P<\/em>. We will discover near the end of this post that it is important that P<\/em> is not just non-sober, but also that it is well-filtered, contrarily to J<\/strong>.<\/p>\n\n\n\n

    Note (added by JGL, July 28th, 2019): Zhenchao Lyu told me that the completion process we will apply is exactly the same as the one used here<\/a>\u2014namely this is the so-called Dedekind-MacNeille completion.<\/p>\n\n\n\n

    The gadget<\/h2>\n\n\n\n

    Isbell’s lattice is constructed in several steps, and the first step is the following gadget. We will need to assemble many copies of this gadget, and then we will add all missing suprema and infima.<\/p>\n\n\n\n

    \"\"
    Figure 1. The basic gadget in Isbell’s construction<\/figcaption><\/figure>\n\n\n\n

    The gadget is obtained as follows. First, we take copies of N<\/strong>, with its usual ordering, one per natural number, and we put them side by side. This is the left part of the gadget, shown on an orange background. I will write mk<\/sub><\/em> for element number m<\/em> in column number k<\/em>.<\/p>\n\n\n\n

    On top of each orange column k<\/em>, we add an element \u03c9k<\/sub><\/em>, and we order them by \u03c91<\/sub> <\/sub><\/em>\u2264 \u03c92<\/sub> <\/sub><\/em>\u2264 … \u2264 \u03c9k<\/sub><\/em> \u2264 … On top of that, we add a top element \u03a9.<\/p>\n\n\n\n

    Next, we build another set of copies of N<\/strong> (on the right part of the gadget, shown on a green background). Those copies are indexed by maps f<\/em> : N<\/strong> \u2192 N<\/strong>, and they all have \u03a9 as supremum. (Yes, there are uncountably many green columns.) I will write nf <\/sub><\/em> for the element number n<\/em> in the green column f.<\/em><\/p>\n\n\n\n

    This gadget, per se, is not a complete lattice. It is a dcpo, though. One can check that it is sober, too. In particular, non-sobriety will not arise from the structure of the gadget.<\/p>\n\n\n\n

    Assembling the gadgets<\/h2>\n\n\n\n

    Remember we wish to build a complete lattice L<\/em> with infinitely many coatoms. To that end, we will assemble together infinitely many copies of the gadget, in such a way that their elements \u03a9 are<\/em> the desired coatoms. The point in having those funny orange and green parts in the gadgets is to organize the new ordering relations which we shall require between elements of different copies of the gadget.<\/p>\n\n\n\n

    First, we put c<\/strong> many copies of Isbell’s gadget side by side, where c<\/strong> is the cardinality of R<\/strong>. Those copies will be indexed by positive real numbers \u03b1, \u03b2. (Isbell uses ordinal numbers < c<\/strong>, but I feel that using real numbers will be easier. It will certainly allow me to give a completely explicit definition of the map i below. This is the announced modification to Isbell’s construction, and that is entirely benign.)<\/p>\n\n\n\n

    We consider the elements of gadget number \u03b1 and those of gadget number \u03b2 (\u03b1\u2260\u03b2) as incomparable, except for the following new relations. Whenever \u03b1\u2260\u03b2:<\/p>\n\n\n\n