{"id":908,"date":"2016-05-11T19:01:22","date_gmt":"2016-05-11T17:01:22","guid":{"rendered":"https:\/\/projects.lsv.ens-paris-saclay.fr\/topology\/?page_id=908"},"modified":"2022-05-17T09:45:56","modified_gmt":"2022-05-17T07:45:56","slug":"locales-sublocales-ii","status":"publish","type":"page","link":"https:\/\/projects.lsv.ens-paris-saclay.fr\/topology\/?page_id=908","title":{"rendered":"Locales, sublocales II: sieves"},"content":{"rendered":"

Last time, I promised you we would explore another way of defining sublocales.\u00a0 We shall again use the naive approach that consists in imagining how we would encode subspaces of a T0<\/sub> topological space X<\/em> by looking at open subsets only, and certainly not at points.<\/p>\n

Hence imagine we know of a T0<\/sub> topological space X<\/em> through its lattice\u00a0\u03a9 of opens, and consider a subset A<\/em> of X<\/em>.\u00a0 I would like to say the most I can say about A<\/em> when I only have access to the open subsets U<\/em> of X<\/em>, not its points.\u00a0 I can say whether U<\/em> intersects A<\/em>, and I can say whether the complement of U<\/em> intersects A<\/em>.<\/p>\n

Crescents, formal crescents<\/h2>\n

I can say something more, in fact.\u00a0 Consider two open subsets U<\/em> and V<\/em> of X<\/em>.\u00a0 Their difference U \u2014 <\/em>V<\/em> is a so-called crescent<\/em>.\u00a0 (This is graphical: imagine U<\/em> is the moon, and V<\/em> is the shadow of the earth; then U<\/em> \u2014 V<\/em> is the visible part of the moon \u2014 a crescent.)\u00a0 We may now test, for each crescent, whether it intersects A<\/em>.\u00a0 This gives us more information than just testing whether U<\/em> alone, or the complement of V<\/em> alone, intersects A<\/em>.<\/p>\n

I will encode A<\/em> as the collection of pairs (U<\/em>, V<\/em>) of open sets such that U \u2014 <\/em>V<\/em> intersects A<\/em>.<\/p>\n

We now have to describe this formally, using a frame\u00a0\u03a9 (supposedly the lattice of open subsets of X<\/em>) only.\u00a0 Let me write u<\/em>, v<\/em>, …, for the elements of \u03a9, to make a parallel with real opens U<\/em>, V<\/em>, …, while stressing that we are now working in a general frame.
\n<\/em><\/p>\n

Call a pair (u<\/em>, v<\/em>) of elements of\u00a0\u03a9 a formal crescent<\/em>.\u00a0 By analogy with actual crescents, we say that (u<\/em>, v<\/em>) is empty<\/em> if and only if u<\/em> \u2264 v<\/em>.\u00a0 Indeed, for actual crescents, U<\/em> \u2014 V<\/em> is empty if and only if U<\/em> \u2286 V<\/em>.<\/p>\n

Again with analogy with actual crescents, we compare formal crescents by: (u<\/em>, v<\/em>) \u2291 (u’<\/em>, v’<\/em>) if and only if u<\/em> \u2264 v \u2228 u’<\/em> and u<\/em> \u2227 v’<\/em> \u2264 v<\/em>.\u00a0 (\u2227 is binary inf, and\u00a0\u2228 is binary sup.)\u00a0 That may not ring a bell, but you may check that, for actual crescents, U<\/em> \u2014 V <\/em>\u2286 U’<\/em> \u2014 V’<\/em> if and only if U<\/em> is included in V<\/em> union U’<\/em> and the intersection of U<\/em> and V’<\/em> is included in V<\/em>.<\/p>\n

The relation \u2291 is a preorder, but in general not an ordering.\u00a0 That is, the equivalence relation \u2263 defined by (u<\/em>, v<\/em>) \u2263 (u’<\/em>, v’<\/em>) if and only if (u<\/em>, v<\/em>) \u2291 (u’<\/em>, v’<\/em>)\u00a0 and (u’<\/em>, v’<\/em>) \u2291 (u<\/em>, v<\/em>), is not the equality relation in general.\u00a0 One may quotient formal crescents by \u2263 and obtain a good locale-theoretic notion of crescent, but we won’t need to do that.<\/p>\n

Sieves<\/h2>\n

Now look at the set E<\/em> of (actual) crescents U<\/em> \u2014 V<\/em> that intersect A<\/em>.\u00a0 It has the following properties:<\/p>\n

    \n
  1. All the crescents in E<\/em> are non-empty.<\/li>\n
  2. E<\/em> is upwards-closed: if U<\/em> \u2014 V<\/em> is a crescent in E<\/em>, then any larger crescent is also in E<\/em>.<\/li>\n
  3. E<\/em> is refinement-closed: if U<\/em> \u2014 V<\/em> is a crescent in E<\/em>, and U’<\/em> is any open subset, then the intersection of (U<\/em> \u2014 V<\/em>) with U’<\/em> or with the complement of U’<\/em> is in E<\/em>.\u00a0 (Given a point in the intersection of U<\/em> \u2014 V<\/em> with A<\/em>, it must be in U’<\/em> or in the complement of U’<\/em>.)<\/li>\n
  4. E<\/em> is accessible: if U<\/em> \u2014 V<\/em> is a crescent in E<\/em>, and if U<\/em> is the union of a family Ui<\/sub><\/em> of opens, i<\/em> in I<\/em>, then Ui<\/sub><\/em> \u2014 V<\/em> is in E<\/em> for some i<\/em> in I<\/em>.\u00a0 (A point in U<\/em> must be in some Ui<\/sub><\/em>.)<\/li>\n<\/ol>\n

    Accordingly, call a sieve<\/em> S<\/em> on a frame\u00a0\u03a9 any set of formal crescents such that:<\/p>\n

      \n
    1. No empty formal crescent is in S<\/em>.<\/li>\n
    2. S<\/em> is upwards-closed<\/em>: if (u<\/em>, v<\/em>) is a formal crescent in S<\/em>, and if (u<\/em>, v<\/em>) \u2291 (u’<\/em>, v’<\/em>), then (u’<\/em>, v’<\/em>) is in S<\/em>.<\/li>\n
    3. S<\/em> is refinement-closed<\/em>: if (u<\/em>, v<\/em>) is in S<\/em>, then for every u’<\/em> in \u03a9, (u<\/em> \u2227 u’<\/em>, v<\/em>) or (u<\/em>, v<\/em>\u00a0\u2228 u’<\/em>) is in S<\/em>.<\/li>\n
    4. S<\/em> is accessible<\/em>: if (u<\/em>, v<\/em>) is in S<\/em> and u<\/em> = \u22c1i \u2208 I<\/sub><\/em> ui<\/sub><\/em> then (ui<\/sub><\/em>, v<\/em>) is in S<\/em> for some i<\/em> in I<\/em>.<\/li>\n<\/ol>\n

      Upward closure (condition 2) in particular implies that if (u<\/em>, v<\/em>) is a formal crescent in S<\/em>, then any formal crescent (u’<\/em>, v’<\/em>) that is equivalent to (u<\/em>, v<\/em>) in the sense that (u<\/em>, v<\/em>) \u2263 (u’<\/em>, v’<\/em>) is also in S<\/em>.\u00a0 This is why we will never need to quotient our formal crescents: every sieve S<\/em> contains all the formal crescents that intuitively denote the same crescent as one already in S.<\/em><\/p>\n

      The poset of sieves<\/h2>\n

      We order sieves by inclusion.\u00a0 This is natural: if A<\/em> and B<\/em> are two subsets of X<\/em>, and if A<\/em> is included in B<\/em>, then all the crescents that intersect A<\/em> also intersect B<\/em>.\u00a0 I doubt that the converse would hold (even assuming that X<\/em> is T0<\/sub>), since that would imply that any two subsets that intersect the same crescents are equal.<\/p>\n

      Let C<\/strong>(\u03a9) denote the poset of all sieves on the frame \u03a9.\u00a0 It is an easy exercise to verify that any union of sieves is a sieve.\u00a0 It follows that C<\/strong>(\u03a9) is a complete lattice.<\/p>\n

      So suprema are simple: they are just unions.\u00a0 It is a general fact that all infima also exist: the inf of a family of sieves is the sup of its set of lower bounds.\u00a0 However, this is less concrete, and hardly usable.\u00a0 We had the same issue with the complete lattice of sublocales, where suprema are simply unions, but infima are more complicated.<\/p>\n

      Let me show that sieves and sublocales form isomorphic lattices.\u00a0 It will be easier to go through frame congruences, and nuclei, though.<\/p>\n

      Given a subset A<\/em> of a space X<\/em>, imagine two opens U<\/em> and V<\/em> of X<\/em> that have different intersections with A<\/em>.\u00a0 There is a point of A<\/em> that is in U<\/em> but not in V<\/em>, or in V<\/em> but not in U<\/em>.\u00a0 That is, U<\/em> \u2014 V<\/em> or V<\/em> \u2014 U<\/em> intersects A<\/em>.\u00a0 The converse direction is easy, too.\u00a0 We obtain that U<\/em> and V<\/em> are in the frame congruence associated with A<\/em> (i.e., U<\/em> \u22c2 A<\/em> = V<\/em> \u22c2 A<\/em>) if and only if the crescents U<\/em> \u2014 V<\/em> and V<\/em> \u2014 U<\/em> are both outside the sieve associated with A<\/em> (i.e., they do not intersect A<\/em>).\u00a0 When U<\/em> is included in V<\/em>, this condition simplifies to: U<\/em> is equivalent to V<\/em> if and only if V<\/em> \u2014 U<\/em> is outside the sieve associated with A<\/em>.\u00a0 In particular, the largest open subset that has the same intersection as U<\/em> with A<\/em> is the union of all the open subsets V<\/em> \u2287 U<\/em> such that V<\/em> \u2014 U<\/em> is outside the sieve associated with A<\/em>.<\/p>\n

      Accordingly, on the localic side, given a sieve S<\/em> in C<\/strong>(\u03a9), let \u03bdS<\/sub><\/em> be the function that maps every u<\/em> in \u03a9 to the union of all the elements v<\/em> \u2265 u<\/em> such that (v<\/em>, u<\/em>) is not in S<\/em>.<\/p>\n

      Lemma.<\/strong> For every sieve S<\/em>, \u03bdS<\/sub><\/em> is a nucleus.<\/p>\n

      Proof.<\/em> We start by checking that \u03bdS<\/sub><\/em> is monotonic.\u00a0 Assume u<\/em> \u2264 u’<\/em>.\u00a0 For every v<\/em> \u2265 u<\/em> such that (v<\/em>, u<\/em>) is not in S<\/em>, I claim that (v<\/em> \u2228 u’<\/em>, u’<\/em>) is not in S<\/em> either.\u00a0 Indeed, we can check that (v<\/em> \u2228 u’<\/em>, u’<\/em>) \u2291 (v<\/em>, u<\/em>), and this then follows from the fact that S<\/em> is upwards-closed.\u00a0 By taking suprema, \u03bdS<\/sub><\/em>(u<\/em>) \u2228 u’<\/em> \u2264 \u03bdS<\/sub><\/em>(u<\/em>‘).\u00a0 This actually shows both monotonicity and the inequality u’<\/em> \u2264 \u03bdS<\/sub><\/em>(u<\/em>‘) at once.<\/p>\n

      To show that it is a closure operator, namely \u03bdS<\/sub><\/em> o \u03bdS<\/sub><\/em> = \u03bdS<\/sub><\/em>, we need to check that \u03bdS<\/sub><\/em>(u<\/em>) is the largest v<\/em> \u2265 u<\/em> such that (v<\/em>, u<\/em>) is not in S<\/em>, i.e., that the supremum of those\u00a0v<\/em>‘s is attained.\u00a0 This amounts to showing that (\u03bdS<\/sub><\/em>(u<\/em>), u<\/em>) is not in S<\/em>.\u00a0 If it were in S<\/em>, since S<\/em> is accessible and\u00a0\u03bdS<\/sub><\/em>(u<\/em>) is the supremum of all elements v<\/em> \u2265 u<\/em> such that (v<\/em>, u<\/em>) is not in S<\/em>, one of those v<\/em>‘s would be such that (v<\/em>, u<\/em>) in S<\/em>.\u00a0 That would be absurd.<\/p>\n

      It remains to show that \u03bdS<\/sub><\/em> preserves binary infima.\u00a0 The inequality \u03bdS<\/sub><\/em>(u<\/em> \u2227 v<\/em>) \u2264 \u03bdS<\/sub><\/em>(u<\/em>) \u2227 \u03bdS<\/sub><\/em>(v<\/em>) follows from monotonicity.\u00a0 In the converse direction, we know that \u03bdS<\/sub><\/em>(u<\/em>) \u2265 u<\/em>, \u03bdS<\/sub><\/em>(v<\/em>) \u2265 v<\/em>,\u00a0(\u03bdS<\/sub><\/em>(u<\/em>), u<\/em>) is not in S<\/em>, and (\u03bdS<\/sub><\/em>(v<\/em>), v<\/em>) is not in S<\/em>.\u00a0 We shall prove that (\u03bdS<\/sub><\/em>(u<\/em>) \u2227 \u03bdS<\/sub><\/em>(v<\/em>), u<\/em> \u2227\u00a0v<\/em>) is not in S<\/em>, which will imply \u03bdS<\/sub><\/em>(u<\/em> \u2227 v<\/em>) \u2265 \u03bdS<\/sub><\/em>(u<\/em>) \u2227 \u03bdS<\/sub><\/em>(v<\/em>).\u00a0 Assume on the contrary that (\u03bdS<\/sub><\/em>(u<\/em>) \u2227 \u03bdS<\/sub><\/em>(v<\/em>), u<\/em> \u2227\u00a0v<\/em>) is in S<\/em>.<\/p>\n

      By refinement-closure, ((\u03bdS<\/sub><\/em>(u<\/em>) \u2227 \u03bdS<\/sub><\/em>(v<\/em>)) \u2227 u<\/em>, u<\/em> \u2227\u00a0v<\/em>) or (\u03bdS<\/sub><\/em>(u<\/em>) \u2227 \u03bdS<\/sub><\/em>(v<\/em>), (u<\/em> \u2227\u00a0v<\/em>) \u2228 u<\/em>) is in S<\/em>.\u00a0 In other words, simplifying all expressions, (u<\/em> \u2227 \u03bdS<\/sub><\/em>(v<\/em>), u<\/em> \u2227\u00a0v<\/em>) or (\u03bdS<\/sub><\/em>(u<\/em>) \u2227 \u03bdS<\/sub><\/em>(v<\/em>), u<\/em>) is in S<\/em>.\u00a0 The second case is impossible since (\u03bdS<\/sub><\/em>(u<\/em>) \u2227 \u03bdS<\/sub><\/em>(v<\/em>), u<\/em>) \u2264 (\u03bdS<\/sub><\/em>(u<\/em>), u<\/em>), which is not in S<\/em>, because S<\/em> is upwards-closed.\u00a0 Hence (u<\/em> \u2227 \u03bdS<\/sub><\/em>(v<\/em>), u<\/em> \u2227\u00a0v<\/em>) is in S<\/em>.<\/p>\n

      By refinement-closure again, ((u<\/em> \u2227 \u03bdS<\/sub><\/em>(v<\/em>)) \u2227 v<\/em>, u<\/em> \u2227\u00a0v<\/em>) or (u<\/em> \u2227 \u03bdS<\/sub><\/em>(v<\/em>), (u<\/em> \u2227\u00a0v<\/em>) \u2228 v<\/em>) is in S<\/em>.\u00a0 Simplifying, (u<\/em> \u2227 v<\/em>, u<\/em> \u2227\u00a0v<\/em>) or (u<\/em> \u2227 \u03bdS<\/sub><\/em>(v<\/em>), v<\/em>) is in S<\/em>. The first case is impossible since no empty formal crescent is in S<\/em>, and the second case is impossible since (u<\/em> \u2227 \u03bdS<\/sub><\/em>(v<\/em>), v<\/em>) \u2264 (\u03bdS<\/sub><\/em>(v<\/em>), v<\/em>), which is not in S.<\/em><\/p>\n

      This concludes our argument that (\u03bdS<\/sub><\/em>(u<\/em>) \u2227 \u03bdS<\/sub><\/em>(v<\/em>), u<\/em> \u2227\u00a0v<\/em>) is not in S<\/em>, hence also the whole proof.\u00a0\u2610<\/p>\n

      The mapping S<\/em> \u27fc \u03bdS<\/sub><\/em> is easily seen to be antitonic.<\/p>\n

      Let us find a mapping in the converse direction.\u00a0 Given a nucleus \u03bd on \u03a9, let S\u03bd<\/sub><\/em> be the set of formal crescents (u<\/em>, v<\/em>) such that u<\/em> \u2270 \u03bd(v<\/em>).<\/p>\n

      Lemma.<\/strong>\u00a0 For every nucleus \u03bd, S\u03bd<\/sub><\/em> is a sieve.<\/p>\n

      Proof.<\/em> 1. If (u<\/em>, v<\/em>) is an empty formal crescent, then u<\/em> \u2264 v<\/em> \u2264 \u03bd(v<\/em>), hence it is not in S\u03bd<\/sub><\/em>.<\/em><\/p>\n

      2. Upward closure.\u00a0 The argument exhibits some subtletly.\u00a0 Assume (u<\/em>, v<\/em>) is in S\u03bd<\/sub><\/em>, and (u<\/em>, v<\/em>) \u2291 (u’<\/em>, v’<\/em>).\u00a0 We must show that (u’<\/em>, v’<\/em>) is in S\u03bd<\/sub><\/em>.\u00a0 Otherwise, u’<\/em> \u2264 \u03bd(v’<\/em>).\u00a0 The inequality (u<\/em>, v<\/em>) \u2291 (u’<\/em>, v’<\/em>) means that u<\/em> \u2264 v \u2228 u’<\/em> and u<\/em> \u2227 v’<\/em> \u2264 v<\/em>.\u00a0 Now u<\/em>\u00a0 = u <\/em>\u2227 u<\/em> \u2264 u<\/em> \u2227 (v \u2228 u’<\/em>) \u2264 u<\/em> \u2227 (v \u2228 \u03bd(v’<\/em><\/em>)) = (u<\/em> \u2227 v<\/em>) \u2228 (u<\/em> \u2227 \u03bd(v’<\/em><\/em>)), the term u<\/em> \u2227 v<\/em> is less than or equal to v<\/em>, the term u<\/em> \u2227 \u03bd(v’<\/em><\/em>) is less than or equal to \u03bd(u<\/em>) \u2227 \u03bd(v’<\/em><\/em>) = \u03bd(u<\/em> \u2227 v’<\/em>) \u2264 \u03bd(v<\/em>); so u <\/em>\u2264 v<\/em> \u2228\u00a0\u03bd(v<\/em>) = \u03bd(v<\/em>).\u00a0 But that contradicts the fact that (u<\/em>, v<\/em>) is in S\u03bd<\/sub><\/em>.<\/p>\n

      3. Refinement closure.\u00a0 Assume (u<\/em>, v<\/em>) is in S\u03bd<\/sub><\/em>, and that for some u’<\/em> in \u03a9, neither (u<\/em> \u2227 u’<\/em>, v<\/em>) nor (u<\/em>, v<\/em>\u00a0\u2228 u’<\/em>) is in S\u03bd<\/sub><\/em>.\u00a0 In other words, u<\/em> \u2227 u’<\/em> \u2264 \u03bd(v<\/em>) and u<\/em> \u2264 \u03bd(v\u00a0\u2228 u’<\/em><\/em>).\u00a0 Again, the argument is slightly subtle.\u00a0 We have u<\/em>\u00a0 = u <\/em>\u2227 u<\/em> \u2264 u <\/em>\u2227 \u03bd(v\u00a0\u2228 u’<\/em><\/em>) \u2264 \u03bd(u<\/em>) \u2227 \u03bd(v\u00a0\u2228 u’<\/em><\/em>) = \u03bd(u<\/em> \u2227 (v\u00a0\u2228 u’<\/em><\/em>)) = \u03bd((u<\/em> \u2227 v<\/em>) \u2228 (u<\/em> \u2227 u’<\/em>)) \u2264 \u03bd((u<\/em> \u2227 v<\/em>) \u2228 \u03bd(v<\/em>)) = \u03bd(\u03bd(v<\/em><\/em>)) [since u<\/em> \u2227 v<\/em> \u2264 \u03bd(v<\/em>)] = \u03bd(v<\/em><\/em>).\u00a0 But u<\/em> \u2264 \u03bd(v<\/em><\/em>) contradicts the assumption that (u<\/em>, v<\/em>) is in S\u03bd<\/sub><\/em>.<\/p>\n

      4. Accessibility.\u00a0 Assume (u<\/em>, v<\/em>) is in\u00a0S\u03bd<\/sub><\/em>, u<\/em> = \u22c1i \u2208 I<\/sub><\/em> ui<\/sub><\/em> but no (ui<\/sub><\/em>, v<\/em>) is in S\u03bd<\/sub><\/em>.\u00a0 In other words, ui<\/sub><\/em> \u2264 \u03bd(v<\/em>) for every i<\/em> in I<\/em>.\u00a0 Clearly, this implies u<\/em> \u2264 \u03bd(v<\/em>), a contradiction.\u00a0\u2610<\/p>\n

      The mapping\u00a0\u03bd \u27fc S\u03bd<\/sub><\/em> is, too, easily seen to be antitonic.<\/p>\n

      Lemma.<\/strong> The mappings \u03bd \u27fc S\u03bd<\/sub><\/em> and S<\/em> \u27fc \u03bdS<\/sub><\/em> are inverse of each other.<\/p>\n

      Proof. Given a nucleus \u03bd, let S<\/em> = S\u03bd<\/sub><\/em>, and let us compute \u03bdS<\/sub><\/em>: \u03bdS<\/sub><\/em>(u<\/em>) is the largest v<\/em> \u2265 u<\/em> such that (v<\/em>, u<\/em>) is not in S\u03bd<\/sub><\/em>, i.e., such that v<\/em> \u2264 \u03bd(u<\/em>).\u00a0 Clearly, this v<\/em> is just \u03bd(u<\/em>), so \u03bdS<\/sub><\/em>=\u03bd.<\/p>\n

      Conversely, given a sieve S<\/em>, let \u03bd=\u03bdS<\/sub><\/em>, and let us compute S\u03bd<\/sub><\/em>.\u00a0 We show that the formal crescents that are not in S<\/em> are exactly those that are not in S\u03bd<\/sub><\/em>, and this will show S<\/em>= S\u03bd<\/sub><\/em>.\u00a0 For every formal crescent (u<\/em>, v<\/em>), if (u<\/em>, v<\/em>) is not in\u00a0S<\/em>, then u<\/em> would be in the family of elements w<\/em> \u2265 v<\/em> such that (w<\/em>, v<\/em>) is not in S<\/em>.\u00a0 The largest element of that family is \u03bdS<\/sub><\/em>(v<\/em>), so we would obtain u<\/em> \u2264 \u03bdS<\/sub><\/em>(v<\/em>), hence (u<\/em>, v<\/em>) is not in S\u03bd<\/sub><\/em>.\u00a0 Conversely, if (u<\/em>, v<\/em>) is not in S\u03bd<\/sub><\/em>, then u<\/em> \u2264 \u03bdS<\/sub><\/em>(v<\/em>).\u00a0 Calling w<\/em> the element \u03bdS<\/sub><\/em>(v<\/em>), w<\/em> is the largest element above v<\/em> such that (w<\/em>, v<\/em>) is not in S<\/em> by definition, and we know that u<\/em> \u2264 w<\/em>.\u00a0 Hence (u<\/em>, v<\/em>) \u2291 (w<\/em>, v<\/em>).\u00a0 Since S<\/em> is upwards-closed, (u<\/em>, v<\/em>) is not in S<\/em>.\u00a0\u2610<\/p>\n

      Since we know that nuclei form a frame, we immediately obtain:<\/p>\n

      Theorem.<\/strong> The lattice of sieves is a co-frame, isomorphic to the opposite of the frame of nuclei (hence to the co-frame of sublocales).<\/p>\n

      Open and closed sieves, crescent sieves<\/h2>\n

      If A<\/em> is an open subset of the T0<\/sub> topological space X<\/em>, then the crescent U<\/em> \u2014 V<\/em> intersects A<\/em> if and only if (U<\/em> \u2014 V<\/em>) \u22c2 A<\/em> = (U\u00a0<\/em>\u22c2 A<\/em>) \u2014\u00a0V<\/em> is non-empty.\u00a0 Note that (U\u00a0<\/em>\u22c2 A<\/em>) \u2014\u00a0V<\/em> is a crescent in this case.<\/p>\n

      We can port this observation to the localic side, and define, for each element\u00a0<\/em>\u03c9 of the frame \u03a9, the open sieve<\/em> o<\/strong>(\u03c9) as the set of pairs (u<\/em>, v<\/em>) such that (u<\/em>\u00a0\u2227 \u03c9,\u00a0v<\/em>) is non-empty.\u00a0\u00a0 I will let you check that this is indeed a sieve.<\/p>\n

      By similar considerations, we can define the closed sieve<\/em> c<\/strong>(\u03c9) as the set of pairs (u<\/em>, v<\/em>) such that (u<\/em>,\u00a0v <\/em>\u2228 \u03c9<\/em>) is non-empty.\u00a0 Thinking of \u03c9 as open, this represents the complement of \u03c9 as a sieve.\u00a0 Indeed, look back at the topological side.\u00a0 If A<\/em> is a closed subset of X<\/em>, then the crescent U<\/em> \u2014 V<\/em> intersects the complement of A<\/em> if and only if the crescent U<\/em> \u2014 (V<\/em> \u22c3 A<\/em>) is non-empty.<\/p>\n

      Generalizing the two constructions, given a formal crescent (u<\/em>0<\/sub>, v<\/em>0<\/sub>), we can define oc<\/strong> (u<\/em>0<\/sub>, v<\/em>0<\/sub>) as the set of pairs (u<\/em>, v<\/em>) such that (u<\/em>\u00a0\u2227 u<\/em>0<\/sub>, v <\/em>\u2228\u00a0v<\/em>0<\/sub>) is non-empty.\u00a0 This intuitively encodes the intersects of the open set\u00a0u<\/em>0<\/sub> and of the complement of the open set v<\/em>0<\/sub>.\u00a0 (That can be checked, too, see below.)<\/p>\n

      Even more generally, given (u<\/em>0<\/sub>, v<\/em>0<\/sub>) and a sieve S<\/em>, define oc<\/strong> (S<\/em>; u<\/em>0<\/sub>, v<\/em>0<\/sub>) as the set of pairs (u<\/em>, v<\/em>) such that (u<\/em>\u00a0\u2227 u<\/em>0<\/sub>, v <\/em>\u2228\u00a0v<\/em>0<\/sub>) is in S<\/em>.\u00a0 We get back oc<\/strong> (u<\/em>0<\/sub>, v<\/em>0<\/sub>) by taking 1<\/strong> for S<\/em>, where 1<\/strong> is the largest sieve (= the set of all non-empty formal crescents).\u00a0 We also obtain o<\/strong>(u<\/em>0<\/sub>) as oc<\/strong> (1<\/strong>; u<\/em>0<\/sub>, \u22a5) and c<\/strong>(v<\/em>0<\/sub>) as oc<\/strong> (1<\/strong>; \u22a4, v<\/em>0<\/sub>).<\/p>\n

      As an exercise, let me suggest that you check the following (without going through nuclei or sublocales):<\/p>\n

        \n
      1. oc<\/strong> (S<\/em>; u<\/em>0<\/sub>, v<\/em>0<\/sub>) is a sieve; in particular o<\/strong>(u<\/em>0<\/sub>) and c<\/strong>(v<\/em>0<\/sub>) are sieves, too.<\/li>\n
      2. oc<\/strong> (S<\/em>; \u22c1i \u2208 I<\/sub><\/em> ui<\/sub><\/em>, v<\/em>0<\/sub>) = \u22c3i \u2208 I<\/sub><\/em> oc<\/strong> (S<\/em>; ui<\/sub><\/em>, v<\/em>0<\/sub>).<\/li>\n
      3. oc<\/strong> (S<\/em>; u<\/em>0<\/sub>, v<\/em>0<\/sub>) is monotonic in S<\/em> and u<\/em>0<\/sub>, and antitonic in v<\/em>0<\/sub>.<\/li>\n
      4. oc<\/strong> (S<\/em>; u<\/em>0<\/sub>, v<\/em>0<\/sub>) is the inf of S<\/em>, o<\/strong>(u<\/em>0<\/sub>) and c<\/strong>(v<\/em>0<\/sub>) in C<\/strong>(\u03a9).\u00a0 (Hint: to check that every sieve S<\/em>‘ that is included in S<\/em>, o<\/strong>(u<\/em>0<\/sub>) and c<\/strong>(v<\/em>0<\/sub>) is also included in oc<\/strong> (S<\/em>; u<\/em>0<\/sub>, v<\/em>0<\/sub>), take an arbitrary formal crescent (u<\/em>, v<\/em>) in S’<\/em>, and use refinement-closure twice to show that (u<\/em>\u00a0\u2227 u0<\/sub>, v<\/em><\/em>), then (u<\/em>\u00a0<\/em>\u2227 u0<\/sub>, v<\/em><\/em>\u00a0\u2228 v<\/em>0<\/sub><\/em><\/em>), is in S’<\/em><\/em>.)
        \n<\/em><\/li>\n
      5. o<\/strong> commutes with finite infima.<\/li>\n
      6. o<\/strong> commutes with arbitrary suprema.<\/li>\n
      7. inf (S<\/em>, o<\/strong>(u<\/em>0<\/sub>)) \u2286 S’<\/em> if and only if S<\/em> \u2286 c<\/strong>(u<\/em>0<\/sub>) \u22c3 S’<\/em>.\u00a0 (Hint: in the \u21d2 direction, let (u<\/em>, v<\/em>) be in S<\/em>, use refinement-closure to obtain that (u<\/em>\u00a0\u2227 u0<\/sub>, v<\/em><\/em>) or (u<\/em><\/em>, v<\/em><\/em>\u00a0\u2228 u<\/em>0<\/sub><\/em><\/em>) is in S<\/em>.\u00a0 In the first case, use item 4 and the assumption to conclude that (u<\/em>, v<\/em>) is in S’<\/em>.\u00a0 In the second case, observe that no element of S<\/em> can be empty and conclude that (u<\/em>, v<\/em>) is in c<\/strong>(u<\/em>0<\/sub>).)<\/li>\n
      8. inf (S<\/em>, c<\/strong>(u<\/em>0<\/sub>)) \u2286 S’<\/em> if and only if S<\/em> \u2286 o<\/strong>(u<\/em>0<\/sub>) \u22c3 S’<\/em>.\u00a0 (A very similar proof.)<\/li>\n
      9. o<\/strong>(u<\/em>0<\/sub>) and c<\/strong>(u<\/em>0<\/sub>) are complements.\u00a0 (Use 7 and 8.)<\/li>\n
      10. For every sieve S<\/em>, for all pairs (u<\/em>0<\/sub>, v<\/em>0<\/sub>) that are not in S<\/em>, S <\/em>\u2286 c<\/strong>(u<\/em>0<\/sub>) \u22c3 o<\/strong>(v<\/em>0<\/sub>).\u00a0 (Use 7, 8, and the fact that S<\/em> is upwards-closed.)<\/li>\n
      11. Conversely, if S <\/em>\u2286 c<\/strong>(u<\/em>0<\/sub>) \u22c3 o<\/strong>(v<\/em>0<\/sub>) then (u<\/em>0<\/sub>, v<\/em>0<\/sub>) cannot be in S<\/em>.<\/li>\n
      12. Every sieve S<\/em> is the intersection (hence the infimum) of the complemented sieves c<\/strong>(u<\/em>0<\/sub>) \u22c3 o<\/strong>(v<\/em>0<\/sub>), when (u<\/em>0<\/sub>, v<\/em>0<\/sub>) ranges over the pairs that are not in S<\/em>.\u00a0 (Use 10 and 11.)<\/li>\n<\/ol>\n

        This way, you will retrieve something that I had announced last month: every sublocale is an infimum of complemented sublocales, each obtained as binary supremum of an open and a closed sublocale.\u00a0 In particular, the co-frame of sieves (or sublocales) is zero-dimensional.<\/p>\n

        Also, items 5 and 6 show that o<\/strong> is a frame homomorphism from \u03a9 to C<\/strong>(\u03a9).\u00a0 It is in fact a frame embedding<\/em>, of \u03a9 into C<\/strong>(\u03a9), since o<\/strong>(u<\/em>) \u2286 o<\/strong>(v<\/em>) is equivalent to u<\/em>\u2264v<\/em>.\u00a0 (o<\/strong> is monotonic since it preserves finite infima for example.\u00a0 Conversely, if u<\/em>\u2270v<\/em> then (u<\/em>, v<\/em>) is in o<\/strong>(u<\/em><\/em>) but not in o<\/strong>(v<\/em><\/em>).)
        \n<\/em><\/p>\n

        I am leaving one question open here.\u00a0 There are several ways of showing that the lattice of nuclei is a frame, or equivalently that the lattice of sublocales is a co-frame, certain simpler than others.\u00a0 Is there a simple, direct way of showing that the lattice of sieves is a co-frame?<\/p>\n

        \u2014 Jean Goubault-Larrecq<\/a>(May 11th, 2016)\"jgl-2011\"<\/p>\n","protected":false},"excerpt":{"rendered":"

        Last time, I promised you we would explore another way of defining sublocales.\u00a0 We shall again use the naive approach that consists in imagining how we would encode subspaces of a T0 topological space X by looking at open subsets … Continue reading →<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"class_list":["post-908","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/projects.lsv.ens-paris-saclay.fr\/topology\/index.php?rest_route=\/wp\/v2\/pages\/908","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/projects.lsv.ens-paris-saclay.fr\/topology\/index.php?rest_route=\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/projects.lsv.ens-paris-saclay.fr\/topology\/index.php?rest_route=\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/projects.lsv.ens-paris-saclay.fr\/topology\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/projects.lsv.ens-paris-saclay.fr\/topology\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=908"}],"version-history":[{"count":22,"href":"https:\/\/projects.lsv.ens-paris-saclay.fr\/topology\/index.php?rest_route=\/wp\/v2\/pages\/908\/revisions"}],"predecessor-version":[{"id":5362,"href":"https:\/\/projects.lsv.ens-paris-saclay.fr\/topology\/index.php?rest_route=\/wp\/v2\/pages\/908\/revisions\/5362"}],"wp:attachment":[{"href":"https:\/\/projects.lsv.ens-paris-saclay.fr\/topology\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=908"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}