{"id":1998,"date":"2019-08-22T10:58:48","date_gmt":"2019-08-22T08:58:48","guid":{"rendered":"https:\/\/projects.lsv.ens-paris-saclay.fr\/topology\/?page_id=1998"},"modified":"2022-11-19T15:05:49","modified_gmt":"2022-11-19T14:05:49","slug":"sober-subspaces-and-the-skula-topology","status":"publish","type":"page","link":"https:\/\/projects.lsv.ens-paris-saclay.fr\/topology\/?page_id=1998","title":{"rendered":"Sober subspaces and the Skula topology"},"content":{"rendered":"\n

It often happens that one wishes to show that a certain subspace A<\/em> of a given sober space X<\/em> is sober. The following is a pearl due to Keimel and Lawson [1, Corollary 3.5], which was mentioned to me by Zhenchao Lyu in July:<\/p>\n\n\n\n

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Given a sober space X<\/em>, the subsets A<\/em> of X<\/em> that are sober as subspaces are exactly its Skula-closed subsets.<\/p>\n<\/blockquote>\n\n\n\n

My goal today is to comment on that, and to (re)prove it.<\/p>\n\n\n\n

In search of sober subspaces<\/h2>\n\n\n\n

It is not hard to show that if A<\/em> is a closed subspace of a sober space X<\/em>, then it is sober. This is also true if A<\/em> is open.<\/p>\n\n\n\n

I have regularly used the following practical result (Lemma 8.4.12 in the book<\/a>): every subspace A<\/em> of a sober space X<\/em> that arises as the equalizer [g<\/em>1<\/sub>=g<\/em>2<\/sub>] of two continuous maps g<\/em>1<\/sub>=g<\/em>2<\/sub>: X<\/em> \u2192 Y<\/em>, where Y<\/em> is any T0<\/sub> space, is sober. (By the way, Y<\/em> has to be T0<\/sub>. This is one of the mistakes<\/a> in the book<\/a>.)<\/p>\n\n\n\n

In particular, if A<\/em> is open in X<\/em>, then it arises as the equalizer of its characteristic map \u03c7A<\/em><\/sub> and the constant map equal to 1 from X<\/em> to Sierpi\u0144ski space S<\/strong> (the space {0, 1} with the Alexandroff topology of 0<1). And if A<\/em> is closed, then it arises as the equalizer of the characteristic map \u03c7U <\/em><\/sub>of its complement U<\/em> and the constant map equal to 0 from X<\/em> to S<\/strong>.<\/p>\n\n\n\n

As another application of Lemma 8.4.12 of the book<\/a>, one can show that every \u03a0<\/strong>0<\/sup>2<\/sub> subspace of a sober space X<\/em> is sober again. A \u03a0<\/strong>0<\/sup>2<\/sub> subset<\/a> is the intersection of countably many sets U<\/em>n<\/em><\/sub> <\/em>\u21d2 V<\/em>n<\/em><\/sub>, where each U<\/em>n<\/em><\/sub> and each V<\/em>n<\/em><\/sub> is open in X<\/em>. (U<\/em>n<\/em><\/sub> \u21d2 V<\/em>n<\/em><\/sub> is the subset of points x<\/em> such that if x<\/em> is in U<\/em>n<\/em><\/sub> then x<\/em> is in V<\/em>n<\/em><\/sub>.) That result was used several times by Matthew de Brecht in the study of quasi-Polish spaces.<\/p>\n\n\n\n

How do you prove that? Well, take Y<\/em> to be the product of countably many copies of S<\/strong>. Define g<\/em>1<\/sub>(x<\/em>) as the tuple whose n<\/em>th entry is \u03c7Un<\/em><\/sub> (x<\/em>), and g<\/em>2<\/sub>(x<\/em>) as the tuple whose n<\/em>th entry is \u03c7Un<\/em> \u2229 Vn<\/em><\/sub> (x<\/em>). Then the equalizer of g<\/em>1<\/sub> and of g<\/em>2<\/sub> is the set of points x<\/em> such that x<\/em> is in U<\/em>n<\/em><\/sub> if and only if it is in U<\/em>n<\/em><\/sub> \u2229 V<\/em>n<\/em><\/sub>, and that is exactly our \u03a0<\/strong>0<\/sup>2<\/sub> subset. (Alternatively, let Y<\/em> be the powerset of N<\/strong>, with its Scott topology, define g<\/em>1<\/sub>(x<\/em>) as the set of natural numbers n<\/em> such that x<\/em> is in U<\/em>n<\/em><\/sub>, and g<\/em>2<\/sub>(x<\/em>) as the set of natural numbers n<\/em> such that x<\/em> is in U<\/em>n<\/em><\/sub> \u2229 V<\/em>n<\/em><\/sub>.)<\/p>\n\n\n\n

In that argument, note that the fact that we are taking a countable intersection of sets U<\/em>n<\/em><\/sub> \u21d2 V<\/em>n<\/em><\/sub> (so-called UCO sets<\/em>) is entirely irrelevant. Any intersection of UCO subsets of a sober space X<\/em>, of whatever cardinality, is sober.<\/p>\n\n\n\n

The Skula topology<\/h2>\n\n\n\n

The Skula topology<\/em> (also called the strong topology<\/em>) on a topological space X<\/em> is the topology generated by the open subsets and<\/em> the closed subsets of X<\/em>. It is equivalent to define it as the topology generated by the open subsets and the downwards-closed subsets of X<\/em>, because every downwards-closed subset of X<\/em> is a union of closed sets (namely the downward closures of single points).<\/p>\n\n\n\n

A subset of X<\/em> is Skula-closed (i.e., closed in the Skula topology) if and only if it is a (generally infinite) intersection of unions V<\/em> \u222a C<\/em> of an open subset V<\/em> and a closed subset C<\/em> of X<\/em>. If you write U<\/em> for the complement of C<\/em>, you will realize that such a union V<\/em> \u222a C<\/em> is nothing but the UCO subset U<\/em> \u21d2 V<\/em>. But recall that any intersection of UCO subsets of a sober space is sober in the subspace topology of X<\/em>. Thus we obtain one half of Keimel and Lawson’s Lemma: every Skula-closed subset A<\/em> of a sober space X<\/em> is sober in the subspace topology from X<\/em>.<\/p>\n\n\n\n

In the converse direction, let A<\/em> be a sober subspace of X<\/em>, and assume that A<\/em> is not Skula-closed. Hence its complement is not Skula-open, and that implies that there is a point x<\/em> in X<\/em>\u2013A<\/em> whose Skula-open neighborhoods all intersect A<\/em>. (Reason by contradiction: otherwise every point in X<\/em>\u2013A<\/em> would have a Skula-open neighborhood included in X<\/em>\u2013A<\/em>.)<\/p>\n\n\n\n

In particular, for every open neighborhood U<\/em> of x<\/em> (in the original topology of X<\/em>), U<\/em> \u2229 \u2193x<\/em> is a Skula-open neighborhood of x<\/em>, so it must intersect A<\/em>. Said in another way: (*) every open neighborhood U<\/em> of x<\/em> intersects \u2193x<\/em> \u2229 A<\/em>.<\/p>\n\n\n\n

Note that \u2193x<\/em> \u2229 A<\/em> is a closed subset of A<\/em>. We claim that it is irreducible in A<\/em>.<\/p>\n\n\n\n

Using (*) with U<\/em>=X<\/em>, we obtain that \u2193x<\/em> \u2229 A<\/em> is non-empty. Given two open subsets U<\/em> and V<\/em> of X<\/em> that intersect \u2193x<\/em> \u2229 A<\/em>, both U<\/em> and V<\/em> must contain x<\/em>, so U<\/em> \u2229 V<\/em> also contains x<\/em>, and using (*) we obtain that U<\/em> \u2229 V<\/em> intersects \u2193x<\/em> \u2229 A<\/em>. It follows that \u2193x<\/em> \u2229 A<\/em> is irreducible closed in A<\/em>.<\/p>\n\n\n\n

Since A<\/em> is sober, \u2193x<\/em> \u2229 A<\/em> is the closure of some unique point y<\/em> in A<\/em>.<\/p>\n\n\n\n

In particular, y<\/em> is in \u2193x<\/em> \u2229 A<\/em>, so y<\/em>\u2264x<\/em>. Since x<\/em> is in X<\/em>\u2013A<\/em>, and y<\/em> is in A<\/em>, we have x<\/em>\u2260y<\/em>, so x<\/em> is not below y<\/em>. We use (*) with U<\/em> equal to the complement of \u2193y<\/em>: there is a point in \u2193x<\/em> \u2229 A<\/em> and in U<\/em>; since \u2193x<\/em> \u2229 A<\/em> is the closure of y<\/em> in A<\/em>, and since an open set intersects the closure of a set B<\/em> if and only if it intersects B<\/em> itself, y<\/em> must be in U<\/em>\u2014but that contradicts the definition of U<\/em>.<\/p>\n\n\n\n

We have proved the promised statement:<\/p>\n\n\n\n

Proposition [1, Corollary 3.5]. <\/strong>Given a sober space X<\/em>, the subsets A<\/em> of X<\/em> that are sober as subspaces are exactly its Skula-closed subsets.<\/p>\n\n\n\n

This also shows that Lemma 8.4.12 of the book<\/a> is in a sense optimal: given any sober subspace A<\/em> of a sober space X<\/em>, A<\/em> is the equalizer [g<\/em>1<\/sub>=g<\/em>2<\/sub>] of two continuous maps g<\/em>1<\/sub>=g<\/em>2<\/sub>: X<\/em> \u2192 Y<\/em>. We can even take Y<\/em> to be a power of S<\/strong> (equivalently, a powerset of some set, in its Scott topology). Indeed, A<\/em> is Skula-closed, hence an intersection of some family of UCO subsets, and we have already seen that any such intersection can be realized as such an equalizer.<\/p>\n\n\n\n

Sobrifications<\/h2>\n\n\n\n

Corollary 3.5 of [1] is\u2014as the name indicates\u2014a corollary of a more general result (I am using the notion of sobrification from the book<\/a>):<\/p>\n\n\n\n

Theorem [1, Proposition 3.4].<\/strong> Let X<\/em> be a sober space, and A<\/em> be a subset of X<\/em>. The sobrification S<\/strong>(A<\/em>) of the subspace A<\/em> is homeomorphic to its Skula-closure in X<\/em>.<\/p>\n\n\n\n

Proof. We look at the inclusion map i<\/em> from A<\/em> into its Skula-closure cls<\/sub>(A<\/em>). This is a continuous map, and since cls<\/sub>(A<\/em>) is sober, i<\/em> has a unique continuous extension j<\/em> from S<\/strong>(A<\/em>) to cls<\/sub>(A<\/em>)\u2014namely, j<\/em> o \u03b7A<\/em><\/sub> = i<\/em>, where the embedding \u03b7A<\/em><\/sub> : A<\/em> \u2192 S<\/strong>(A<\/em>) maps x<\/em> to its downward closure in A<\/em>, namely \u2193x<\/em> \u2229 A<\/em>.<\/p>\n\n\n\n

Note that, for every open subset of S<\/strong>(A<\/em>), namely for every set of the form \u2662(U<\/em> \u2229 A<\/em>) where U<\/em> is open in X<\/em>, \u03b7A<\/em><\/sub>-1<\/sup>(\u2662(U<\/em> \u2229 A<\/em>))=U<\/em> \u2229 A<\/em>. Since i<\/em>-1<\/sup>(U<\/em> \u2229 cls<\/sub>(A))=U<\/em> \u2229 A<\/em>, and recalling that \u03b7A<\/em><\/sub>-1<\/sup> is a frame isomorphism, it follows that j<\/em>-1<\/sup>(U<\/em> \u2229 cls<\/sub>(A))=\u2662(U<\/em> \u2229 A<\/em>) for every open subset U<\/em> of X<\/em>.<\/p>\n\n\n\n

In the converse direction, given any point x<\/em> in cls<\/sub>(A<\/em>), we let f<\/em>(x<\/em>)=\u2193x<\/em> \u2229 A<\/em>. For every open subset U<\/em> of X<\/em>, f<\/em>(x<\/em>) intersects U<\/em> if and only if U<\/em> intersects \u2193x<\/em> \u2229 A<\/em>, if and only if A<\/em> intersects the Skula-open set \u2193x<\/em> \u2229 U<\/em>, if and only if cls<\/sub>(A<\/em>) intersects \u2193x<\/em> \u2229 U<\/em>, if and only if x<\/em> is in U<\/em> (recall that x<\/em> is in cls<\/sub>(A<\/em>): if cls<\/sub>(A<\/em>) intersects \u2193x<\/em> \u2229 U<\/em>, say at y<\/em>, then y<\/em>\u2264x<\/em> and y<\/em> is in U<\/em>, so x<\/em> is in U<\/em>, and conversely if x<\/em> is in U<\/em>, then cls<\/sub>(A<\/em>) intersects \u2193x<\/em> \u2229 U<\/em> at x<\/em>). To sum up: (**) for every open subset U<\/em> of X<\/em>, for every x<\/em> in cls<\/sub>(A<\/em>), f<\/em>(x<\/em>) intersects U<\/em> if and only if x<\/em> is in U<\/em>.<\/p>\n\n\n\n

Let us fix x<\/em> in cls<\/sub>(A<\/em>). Using (**) with U<\/em> empty, we obtain that f<\/em>(x<\/em>) is not empty; given any two open sets U<\/em> and V<\/em> that each intersect f<\/em>(x<\/em>), by (**) we obtain that both U<\/em> and V<\/em> contain x<\/em>, so U<\/em> \u2229 V<\/em> also contains x<\/em>, and using (**) again, f<\/em>(x<\/em>) intersects U<\/em> \u2229 V<\/em>. This shows that f<\/em>(x<\/em>) is irreducible (closed) in A<\/em>, hence an element of S<\/strong>(A<\/em>).<\/p>\n\n\n\n

We can now rephrase (**) as: for every open subset U<\/em> of X<\/em>, f<\/em>-1<\/sup>(\u2662(U<\/em> \u2229 A<\/em>))=U<\/em> \u2229 cls<\/sub>(A). In particular, f<\/em> is continuous. It also follows that f<\/em>-1<\/sup> is a frame homomorphism that is inverse to the frame homomorphism j<\/em>-1<\/sup>. Since f<\/em> and j<\/em> are continuous maps between sober spaces, they are uniquely determined by the frame homomorphisms f<\/em>-1<\/sup> and j<\/em>-1<\/sup>. It follows that f<\/em> and j<\/em> are mutual inverses. \u2610<\/p>\n\n\n

    \n
  1. Klaus Keimel and Jimmie D. Lawson.  D-completions and the d-topology<\/em><\/a>.  Annals of Pure and Applied Logic 159(3), June 2009, pages 292-306.<\/li>\n<\/ol>\n

    \u2014 Jean Goubault-Larrecq<\/a> (August 22nd, 2019)\"jgl-2011\"<\/p>\n\n","protected":false},"excerpt":{"rendered":"

    It often happens that one wishes to show that a certain subspace A of a given sober space X is sober. The following is a pearl due to Keimel and Lawson [1, Corollary 3.5], which was mentioned to me by … 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-1998","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/projects.lsv.ens-paris-saclay.fr\/topology\/index.php?rest_route=\/wp\/v2\/pages\/1998","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=1998"}],"version-history":[{"count":10,"href":"https:\/\/projects.lsv.ens-paris-saclay.fr\/topology\/index.php?rest_route=\/wp\/v2\/pages\/1998\/revisions"}],"predecessor-version":[{"id":5909,"href":"https:\/\/projects.lsv.ens-paris-saclay.fr\/topology\/index.php?rest_route=\/wp\/v2\/pages\/1998\/revisions\/5909"}],"wp:attachment":[{"href":"https:\/\/projects.lsv.ens-paris-saclay.fr\/topology\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=1998"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}

This is a rather remarkable topology. For example, a theorem by R.-E. Hoffmann states that X<\/em> is Skula-compact (i.e., compact in its Skula topology) if and only if X<\/em> is sober Noetherian (Exercise 9.7.16 in the book<\/a>; note that the Skula topology is always T2.<\/sub>)<\/p>\n\n\n\n