{"id":1272,"date":"2017-10-30T17:01:02","date_gmt":"2017-10-30T16:01:02","guid":{"rendered":"https:\/\/projects.lsv.ens-paris-saclay.fr\/topology\/?page_id=1272"},"modified":"2023-06-27T14:24:32","modified_gmt":"2023-06-27T12:24:32","slug":"a-characterization-of-fac-spaces","status":"publish","type":"page","link":"https:\/\/projects.lsv.ens-paris-saclay.fr\/topology\/?page_id=1272","title":{"rendered":"A characterization of FAC spaces"},"content":{"rendered":"

In the open problem<\/a> section, I defined a FAC space<\/a> as a topological space in which every closed subspace is a finite union of irreducible closed subspaces.\u00a0 FAC is for “finite antichain property”, since it generalizes the following theorem: a poset has the finite antichain property (namely, all its antichains are finite) if and only if its downwards-closed subsets are finite unions of ideals.<\/p>\n

A bit of history.<\/h2>\n

That theorem (about posets) has a curious history.\u00a0 It has been discovered over and over.\u00a0 I have learned about it by reading M. Kabil and Maurice Pouzet [2], but it had been proved earlier by Jimmie Lawson, Mike Mislove and Hilary Priestley [3], by R. Bonnet [4], and by others.\u00a0 In some of those papers, you will see that the result is credited to Erd\u0151s and Tarski in 1943 [5].\u00a0 The first time I heard about that, I looked at the Erd\u0151s-Tarski paper, and I could not find the result or anything vaguely looking like it.\u00a0 My colleague Alain Finkel once asked Maurice Pouzet about that state of affairs, and apparently the result is implicit in the Erd\u0151s-Tarski paper.<\/p>\n

Characterizing FAC spaces.<\/h2>\n

Anyway, I think I have finally nailed the correct topological generalization of that theorem by Erd\u0151s and Tarski.\u00a0 Here it is.\u00a0 I do not have much merit in proving the characterization theorem below.\u00a0 This is essentially the Lawson-Mislove-Priestley proof [3], suitably modified.<\/p>\n

Definition.<\/strong> A subset A<\/em> of a topological space\u00a0X<\/em> is a topological antichain<\/em> if and only if the subspace topology on A<\/em> is discrete (all its subsets are open).
\n<\/strong><\/p>\n

Theorem.<\/strong> Given a topological space X<\/em>, the following are equivalent.<\/p>\n

    \n
  1. Every closed subspace of X<\/em> is a finite union of irreducible closed subsets;<\/li>\n
  2. Every topological antichain of X<\/em> is finite.<\/li>\n<\/ol>\n

    It therefore really makes sense to call those spaces FAC.\u00a0 The purpose of this post is to give a proof of that theorem.<\/p>\n

    An alternative view on topological antichains.<\/h2>\n

    First, we give an alternative characterization of topological antichains.<\/p>\n

    Lemma 1.<\/strong> Let A<\/em> be a subset of a topological space X<\/em>.\u00a0 A<\/em> is a topological antichain if and only if, for every x<\/em> \u2208 A<\/em>, x<\/em> is not in the closure cl (A<\/em> \u2014 {x<\/em>}).\u00a0 (We take closures in X<\/em>, here.)<\/p>\n

    Proof.<\/em>\u00a0 If A<\/em> is a topological antichain, then for every x<\/em> \u2208 A<\/em>, {x<\/em>} is open in A<\/em>, so A<\/em> \u2014 {x<\/em>} is closed in A<\/em>.\u00a0 That means that A<\/em> \u2014 {x<\/em>} occurs as the intersection of some closed subset of X<\/em> with A<\/em>, and the smallest one is cl (A<\/em> \u2014 {x<\/em>}).\u00a0 That cannot contain x<\/em> since its intersection with A<\/em> is A<\/em> \u2014 {x<\/em>}.<\/p>\n

    Conversely, for every x<\/em> \u2208 A<\/em>, if x<\/em> is not in the closure cl (A<\/em> \u2014 {x<\/em>}), then it lies in some open subset U<\/em> of X<\/em> that does not intersect A<\/em> \u2014 {x<\/em>}.\u00a0 Then U<\/em> \u2229 A<\/em> is open in A<\/em>, but only contains the point x<\/em> from A<\/em>.\u00a0 It follows that every one-element subset of A<\/em> is open in A<\/em>, whence A<\/em> is discrete.\u00a0 \u2610<\/p>\n

    The key lemma.<\/h2>\n

    Second, we prove a useful lemma, generalizing Lemma 3 of [3] to the topological case.<\/p>\n

    Lemma 2.<\/strong> Let X<\/em> be a topological space, and M<\/em> be an infinite family of irreducible closed subsets of X<\/em>, with the property that for every infinite subfamily N<\/em> of M<\/em>, cl (\u222aN<\/em>)=cl (\u222aM<\/em>).\u00a0 (\u222aM<\/em> denotes the union of all the elements of M<\/em>.)\u00a0 Then cl (\u222aM<\/em>) is irreducible closed.<\/p>\n

    If M’<\/em> were finite, then M<\/em> \u2014 M’<\/em> would be infinite, so by assumption cl (\u222a(M<\/em> \u2014 M’<\/em>)) would be equal to cl (\u222aM<\/em>).\u00a0 However, U<\/em> intersects the latter, so it would intersect the former, hence also \u222a(M<\/em> \u2014 M’<\/em>), hence also some I<\/em> \u2208 M<\/em> \u2014 M’<\/em>.\u00a0 That would contradict the definition of M’<\/em>.\u00a0 Therefore M’<\/em> is infinite.<\/p>\n

    Since M’<\/em> is infinite, we use the assumption again and we obtain cl (\u222aM’<\/em>)=cl (\u222aM<\/em>).\u00a0 We now use the fact that V<\/em>, not just U<\/em>, intersects cl (\u222aM<\/em>).\u00a0 Then V<\/em> must also intersect cl (\u222aM’<\/em>), hence \u222aM’<\/em>, and therefore it must intersect some I<\/em> \u2208 M’<\/em>.\u00a0 By definition of M’<\/em>, U<\/em> also intersects I<\/em>.\u00a0 Since I<\/em> is irreducible closed, it must therefore intersect U<\/em> \u2229 V<\/em>.\u00a0 Since I<\/em> is included in cl (\u222aM<\/em>), the latter must also intersect U<\/em> \u2229 V<\/em>.\u00a0 \u2610<\/p>\n

    Proving the hard implication 2 \u21d2 1.<\/h2>\n

    We now prove the difficult implication 2 \u21d2 1 of the theorem.\u00a0 More precisely, we prove its contrapositive, namely: we assume that there is a closed subset C<\/em> of X<\/em> that cannot be written as a finite union of irreducible closed subsets, and we shall build an infinite topological antichain in X<\/em>.<\/p>\n

    Recall that the sobrification S<\/strong>(X<\/em>) of X<\/em> is sober (Corollary 8.2.23 in the book<\/a>), and that every sober space is a dcpo in its specialization ordering (Proposition 8.2.34).\u00a0 The supremum of a directed family of irreducible closed subsets is the closure of their union (it suffices to show that the latter is irreducible: show that if it intersects two opens U<\/em> and V<\/em>, then it intersects U<\/em> \u2229 V<\/em>!).\u00a0 In particular, the supremum of a directed family of irreducible closed subsets included in C<\/em> is again included in C<\/em>.\u00a0 It follows that the poset of all irreducible closed subsets of X<\/em> included in C<\/em> is inductive.\u00a0 Hence we can apply Zorn’s Lemma (Theorem 2.4.2): every irreducible closed subset of X<\/em> included in\u00a0C<\/em> is contained in some maximal irreducible closed subset of\u00a0X<\/em> included in C<\/em>.\u00a0 (Exercise: an irreducible closed subset of X<\/em> included in C<\/em> is the same thing as an irreducible closed subset of the subspace C<\/em>.\u00a0 That simplifies the latter statement, and also allows for a different proof of the latter fact, working in\u00a0S<\/strong>(C<\/em>) instead of S<\/strong>(X<\/em>).)<\/p>\n

    In particular, for every x<\/em> \u2208 C<\/em>, \u2193x<\/em> is such an irreducible closed subset of X<\/em> included in C<\/em>.\u00a0 Let M<\/em>0<\/sub> be the collection of all maximal irreducible closed subsets of X<\/em> included in C<\/em>.\u00a0 We have just shown that C<\/em>=\u222aM<\/em>0<\/sub>.\u00a0 In particular, C<\/em>=cl (\u222aM<\/em>0<\/sub>).<\/p>\n

    Since C<\/em> is not a finite union of irreducible closed subsets, it is in particular not irreducible.\u00a0 The contrapositive of Lemma 2 then implies the existence of an infinite subset\u00a0M<\/em>1<\/sub> of\u00a0M<\/em>0<\/sub> such that cl (\u222aM<\/em>1<\/sub>) is strictly included in cl (\u222aM<\/em>0<\/sub>).<\/p>\n

    If cl (\u222aM<\/em>1<\/sub>) were an irreducible closed subset, since it is included in C<\/em>, it would be included in some maximal irreducible closed subset of X<\/em> included in C<\/em>, namely in some member of M<\/em>0<\/sub>.\u00a0 In particular,\u00a0\u222aM<\/em>1<\/sub> would be included in some member of M<\/em>0<\/sub>.\u00a0 Since all elements of\u00a0M<\/em>0<\/sub> are maximal, hence pairwise incomparable, that would imply that\u00a0M<\/em>1<\/sub> contains only one element.\u00a0 That is impossible, since\u00a0M<\/em>1<\/sub> is infinite.<\/p>\n

    Hence we can reapply Lemma 2: there is an infinite subset M<\/em>2<\/sub> of\u00a0M<\/em>1<\/sub> such that cl (\u222aM<\/em>2<\/sub>) is strictly included in cl (\u222aM<\/em>1<\/sub>).<\/p>\n

    Again cl (\u222aM<\/em>2<\/sub>) cannot be irreducible, hence we can find an infinite subset M<\/em>3<\/sub> of\u00a0M<\/em>2<\/sub> such that cl (\u222aM<\/em>3<\/sub>) is strictly included in cl (\u222aM<\/em>2<\/sub>), and proceed infinitely that way.\u00a0 To make that clear, we have an infinite, antitone sequence of infinite subsets M<\/em>0<\/sub> \u2287 M<\/em>1<\/sub> \u2287 M<\/em>2<\/sub> \u2287 …\u00a0\u2287 Mn<\/sub><\/em> \u2287 … such that cl (\u222aM<\/em>0<\/sub>) \u2283 cl (\u222aM<\/em>1<\/sub>) \u2283 cl (\u222aM<\/em>2<\/sub>) \u2283 … \u2283 cl (\u222aMn<\/sub><\/em>) \u2283 … (all inclusions here are strict; from which we can conclude that the inclusions M<\/em>0<\/sub> \u2287 M<\/em>1<\/sub> \u2287 M<\/em>2<\/sub> \u2287 …\u00a0\u2287 Mn<\/sub><\/em> \u2287 … are strict as well).<\/p>\n

    We can therefore pick an irreducible closed subset In<\/sub><\/em> in Mn<\/sub><\/em> that is not contained in cl (\u222aMn<\/sub><\/em>+1<\/sub>), for each natural number n<\/em>.\u00a0 Indeed, if every element I<\/em> of Mn<\/sub><\/em> were in cl (\u222aMn<\/sub><\/em>+1<\/sub>), then\u00a0\u222aMn<\/sub><\/em> would be included in cl (\u222aMn<\/sub><\/em>+1<\/sub>), so cl (\u222aMn<\/sub><\/em>) would be included in cl (\u222aMn<\/sub><\/em>+1<\/sub>) as well, contradicting cl (\u222aMn<\/sub><\/em>) \u2283 cl (\u222aMn<\/sub><\/em>+1<\/sub>).<\/p>\n

    We now claim that In<\/sub><\/em> cannot be included in cl (\u222a{Im<\/sub><\/em> | m<\/em>\u2260n<\/em>}).\u00a0 Assume the contrary.\u00a0 Since for every m<\/em>>n<\/em>, Im<\/sub><\/em> is in Mm<\/sub><\/em> \u2286 Mn<\/sub><\/em>+1<\/sub>, In<\/sub><\/em> would be included in cl (I<\/em>0<\/sub> \u222a I<\/em>1<\/sub>\u00a0\u222a I<\/em>2<\/sub>\u00a0\u222a …\u00a0\u222a In<\/sub><\/em>-1<\/sub> \u222a (\u222aMn<\/sub><\/em>+1<\/sub>)).\u00a0 Closure commutes with finite unions (not all unions!), so In<\/sub><\/em> would be included in I<\/em>0<\/sub> \u222a I<\/em>1<\/sub>\u00a0\u222a I<\/em>2<\/sub>\u00a0\u222a …\u00a0\u222a In<\/sub><\/em>-1<\/sub> \u222a cl (\u222aMn<\/sub><\/em>+1<\/sub>).\u00a0 (Remember that each one of I<\/em>0<\/sub>, I<\/em>1<\/sub>, I<\/em>2<\/sub>, …, In<\/sub><\/em>-1<\/sub> is already closed.)\u00a0 Since In<\/sub><\/em> is irreducible, it must be included in one term of the union.\u00a0 It cannot be included in cl (\u222aMn<\/sub><\/em>+1<\/sub>), by definition.\u00a0 And it cannot be included in any Ik<\/sub><\/em>, with k<\/em><n<\/em>, since all those irreducible closed subsets, being maximal, are pairwise incomparable.<\/p>\n

    Finally, since In<\/sub><\/em> is not included in cl (\u222a{Im<\/sub><\/em> | m<\/em>\u2260n<\/em>}), we can pick a point xn<\/sub><\/em> of In<\/sub><\/em> that is not in cl (\u222a{Im<\/sub><\/em> | m<\/em>\u2260n<\/em>}), for each natural number n<\/em>.\u00a0 Since xm<\/sub><\/em> is in Im<\/sub><\/em>, for every m<\/em>, cl (\u222a{Im<\/sub><\/em> | m<\/em>\u2260n<\/em>}) contains cl {xm<\/sub><\/em> | m<\/em>\u2260n<\/em>}.\u00a0 It follows that xn<\/sub><\/em> is not in cl {xm<\/sub><\/em> | m<\/em>\u2260n<\/em>}.\u00a0 Lemma 1 states that the family {xn<\/sub><\/em> | n<\/em> \u2208 N<\/strong>} is a (necessarily infinite) topological antichain.<\/p>\n

    The proof of the easier implication 1 \u21d2 2.<\/h2>\n

    For the record, here is the proof of the much easier implication 1 \u21d2 2.\u00a0 We assume an infinite topological antichain A<\/em>.\u00a0 We shall simply show that cl (A<\/em>) cannot be written as a finite union of irreducible closed subsets of X<\/em>.\u00a0 Assume one could write cl (A<\/em>) as the finite union I<\/em>0<\/sub> \u222a I<\/em>1<\/sub>\u00a0\u222a I<\/em>2<\/sub>\u00a0\u222a …\u00a0\u222a In<\/sub><\/em>-1<\/sub>, where each Ik<\/sub><\/em> is irreducible closed in X<\/em>.<\/p>\n

    For each k<\/em>, if Ik<\/sub><\/em> contains some point x<\/em> of A<\/em>, then, writing cl (A<\/em>) as the closure of x<\/em>, namely \u2193x<\/em>, union cl (A<\/em> \u2014 {x<\/em>}) (recall that closures commute with finite unions), and recalling that Ik<\/sub><\/em> is irreducible and included in cl (A<\/em>), we obtain that\u00a0Ik<\/sub><\/em> is included in \u2193x<\/em> or in cl (A<\/em> \u2014 {x<\/em>}).\u00a0 The latter is impossible if Ik<\/sub><\/em> contains x<\/em>, since A<\/em> is a topological antichain: indeed, Lemma 1 states that x<\/em> is not in cl (A<\/em> \u2014 {x<\/em>}).\u00a0 Therefore Ik<\/sub><\/em>=\u2193x<\/em>.<\/p>\n

    That implies that every\u00a0Ik<\/sub><\/em> can contain at most one point x<\/em> from A<\/em>.\u00a0 Explicitly, if it contained two distinct points x<\/em> and y<\/em> of A<\/em>, then Ik<\/sub><\/em> would be equal to \u2193x<\/em>, and also to \u2193y<\/em>.\u00a0 If we had assumed X<\/em> to be T0<\/sub>, we could conclude\u00a0x<\/em>=y<\/em>, but we haven’t made that assumption, and we must therefore work slightly harder.\u00a0 Since A<\/em> is a topological antichain, x<\/em> is not in cl (A<\/em> \u2014 {x<\/em>}), hence neither in the smaller subset cl ({y<\/em>})=\u2193y<\/em>.\u00a0 That shows that x<\/em> is not less than or equal to y<\/em>.\u00a0 That contradicts \u2193x<\/em>=\u2193y<\/em>.<\/p>\n

    We now have infinitely many points in A<\/em>, hence in cl (A<\/em>) = I<\/em>0<\/sub> \u222a I<\/em>1<\/sub>\u00a0\u222a I<\/em>2<\/sub>\u00a0\u222a …\u00a0\u222a In<\/sub><\/em>-1<\/sub>, but each Ik<\/sub><\/em> can contain at most one point from A<\/em>.\u00a0 That is impossible, by the pigeonhole principle (“to put infinitely many pigeons in some pigeonholes, in such a way that each pigeonhole contains at most one pigeon, you need infinitely many pigeonholes”).\u00a0 \u2610<\/p>\n

    To conclude.<\/h2>\n

    Hence we have a full characterization of those topological spaces in which closed subsets are finite unions of irreducible closed subsets, answering our (previously) open question<\/a>.<\/p>\n

    If you want something to think about, remember that I said the following on that page: every topological space X<\/em> such that every closed subspace C<\/em> has a dense Noetherian subspace D<\/em>, is FAC.\u00a0 The following is still open: are the FAC spaces exactly those such that every closed subspace has a dense Noetherian subspace? \u00a0(Added June 27th, 2023: this question was solved in the positive in a 2017 arXiv report<\/a> with Maurice Pouzet.)<\/p>\n

      \n
    1. Jean Goubault-Larrecq. Non-Hausdorff Topology and Domain Theory \u2014 Selected Topics in Point-Set Topology<\/a>. New Mathematical Monographs 22. Cambridge University Press, 2013.<\/li>\n
    2. M. Kabil and M. Pouzet. Une extension d\u2019un th\u00e9or\u00e8me de P. Julien sur les \u00e2ges de mots. Informatique th\u00e9orique et applications, 26(5):449\u2013482, 1992.<\/li>\n
    3. Jimmie D. Lawson, Michael Mislove, and Hilary Priestley.\u00a0 Ordered sets with no infinite antichains<\/a>.\u00a0 Discrete Mathematics 63:225-230, 1987.<\/li>\n
    4. R. Bonnet.\u00a0 On the cardinality of the set of initial intervals of a partially ordered set.\u00a0 Colloquium of the J\u00e1nos Bolyai Mathematical Society (Colloq. Math. Soc. J\u00e1nos Bolyai) 10:189-198, 1973.<\/li>\n
    5. P\u00e1l Erd\u0151s and Alfred Tarski.\u00a0 On families of mutually exclusive sets.\u00a0 Annals of Mathematics 44:315-329, 1943.<\/li>\n<\/ol>\n

      \u2014 Jean Goubault-Larrecq<\/a>(Oct. 30th, 2017)\"jgl-2011\"<\/p>\n","protected":false},"excerpt":{"rendered":"

      In the open problem section, I defined a FAC space as a topological space in which every closed subspace is a finite union of irreducible closed subspaces.\u00a0 FAC is for “finite antichain property”, since it generalizes the following theorem: a … 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":{"_crdt_document":"","footnotes":""},"class_list":["post-1272","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/projects.lsv.ens-paris-saclay.fr\/topology\/index.php?rest_route=\/wp\/v2\/pages\/1272","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=1272"}],"version-history":[{"count":11,"href":"https:\/\/projects.lsv.ens-paris-saclay.fr\/topology\/index.php?rest_route=\/wp\/v2\/pages\/1272\/revisions"}],"predecessor-version":[{"id":7059,"href":"https:\/\/projects.lsv.ens-paris-saclay.fr\/topology\/index.php?rest_route=\/wp\/v2\/pages\/1272\/revisions\/7059"}],"wp:attachment":[{"href":"https:\/\/projects.lsv.ens-paris-saclay.fr\/topology\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=1272"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}

    Proof.<\/em>\u00a0 We shall repeatedly use the fact that if an open set intersects the closure of a set E<\/em>, then it intersects E<\/em>.\u00a0 Let U<\/em> and V<\/em> be two open subsets of X<\/em> that intersect cl (\u222aM<\/em>).\u00a0 It suffices to show that cl (\u222aM<\/em>) intersects U<\/em> \u2229 V<\/em>.\u00a0 (Trick 8.2.3 in the book<\/a>.)\u00a0 Let M’<\/em> be the subfamily of those I<\/em> \u2208 M<\/em> such that U<\/em> intersects I<\/em>.\u00a0 M’<\/em> is non-empty: since U<\/em> intersects cl (\u222aM<\/em>), it intersects \u222aM<\/em>, hence some I<\/em> \u2208 M<\/em>.<\/p>\n