Noetherian spaces and quantifier elimination Matthew de Brecht 1 CiNet, NICT, Osaka, Japan Dagstuhl Seminar: 2016, January 17 - 22 1 This work was supported by JSPS Core-to-Core Program, A. Advanced Research Networks and by JSPS KAKENHI Grant Number 15K15940. 1 / 20
Introduction A space is Noetherian iff every strictly ascending chain of open subsets if finite. Noetherian spaces naturally arise in algebraic geometry as the spectrum of Noetherian rings. J. Goubault-Larrecq has also demonstrated that they can be applied to verification for various classes of transition systems that are not based on well-quasi-orders (wqos). We prove a quantifier elimination result for countably based sober Noetherian (i.e., quasi-Polish Noetherian) spaces: Assume X and Y are quasi-Polish Noetherian spaces, and let P ⊆ X × Y be constructible. Then ∃ X P := { y ∈ Y | ( ∃ x ∈ X ) P ( x, y ) } , and ∀ X P := { y ∈ Y | ( ∀ x ∈ X ) P ( x, y ) } are constructible subsets of Y . This is a weak version of a scheme-theoretic result by Chevellay from algebraic geometry, which is related to quantifier elimination for the theory of algebraically closed fields. 2 / 20
Table of Contents 1 Noetherian spaces 2 Countably based sober Noetherian spaces 3 / 20
Table of Contents 1 Noetherian spaces 2 Countably based sober Noetherian spaces 4 / 20
Noetherian spaces Definition A topological space is Noetherian iff every open subset is compact. Equivalently, a space is Noetherian iff its open set lattice satisfies the ascending chain condition, iff its lattice of closed sets is well-founded. (Recall that the Alexandroff topology on a quasi-order P is generated by the sets ↑ x for x ∈ P ). Thereom (see J. Goubault-Larrecq, 2007) A quasi-order P is a well-quasi-order iff the Alexandroff topology on P is Noetherian. However, the specialization order of a Noetherian space is not always a wqo. 5 / 20
Noetherian Constructions J. Goubault-Larrecq proved that Noetherian spaces are closed under the following: Subspaces Disjoint unions ( X + Y ) The lower power space construction ( A ( X ) ) This is the set of closed subsets of X , ordered by inclusion, with the upper topology (the topology generated by complements of sets of the form ↓ x ). Topological products ( X × Y ) → A ( X ) × A ( Y ) ∼ Proof: X × Y ֒ = A ( X + Y ) Continuous images (in particular, quotients) and several others important constructions... In particular, the category of Noetherian spaces is finitely complete and finitely co-complete, but it is not cartesian closed. 6 / 20
Sober Noetherian spaces A space is sober iff every irreducible closed set equals the closure of a single point. A non-empty closed set is irreducible iff it does not equal the union of two closed proper subsets. Sober Noetherian spaces are completely characterized by their specialization order: Theorem (J. Goubault-Larrecq, 2007) X is a sober Noetherian space iff The topology on X is the upper topology of a well-founded partial order, and For any finite F ⊆ X there is finite G ⊆ X such that � x ∈ F ↓ x = � y ∈ G ↓ y . Furthermore, a space is Noetherian iff its soberification is Noetherian, so we don’t lose much by assuming sobriety. 7 / 20
Table of Contents 1 Noetherian spaces 2 Countably based sober Noetherian spaces 8 / 20
Preliminaries (1) The second level of the Borel hierarchy for general spaces (V. Selivanov): A subset A of a topological space X is a Π 0 2 -set iff there is a sequence U i , V i ( i ∈ ω ) of open subsets of X such that x ∈ A ⇐ ⇒ ( ∀ i ∈ ω )[ x ∈ U i ⇒ x ∈ V i ] for each x ∈ X . If X is metrizable then the Π 0 2 -sets are exactly the G δ -sets. The complement of a Π 0 2 -set is called a Σ 0 2 -set. A set which is both Π 0 2 and Σ 0 2 is called a ∆ 0 2 -set. We will also use the following terminology (common in algebraic geometry): A subset of a space is constructible if it is equal to a finite boolean combination of open subsets. 9 / 20
Quantification Σ 0 2 -sets correspond to existential quantification over a constructible predicate with a variable from a countable (discrete) space: If P ⊆ X × Y is constructible (or even Σ 0 2 ) and X is countable then ∃ X P := { y ∈ Y | ( ∃ x ∈ X ) P ( x, y ) } is a Σ 0 2 -subset of Y . Conversely, if A is a Σ 0 2 -subset of Y , then there is a constructible P ⊆ ω × Y such that A = ∃ ω P . Π 0 2 -sets correspond to universal quantification in a similar sense. Quantification over more complicated predicates and spaces result in higher levels of the Borel and (hyper-) projective hierarchies. 10 / 20
Preliminaries (2) A space X is quasi-Polish iff it is countably based and the topology is generated by a (Smyth-) complete quasi-metric. Equivalently, iff X is homeomorphic to a Π 0 2 -subset of P ( ω ) (the powerset of the natural numbers with the Scott-topology) (And there are several other characterizations...) Some basic facts: Every quasi-Polish space is sober. A metrizable space is quasi-Polish iff it is Polish. Every countably based locally compact sober space is quasi-Polish A subspace of a quasi-Polish space is quasi-Polish iff it is a Π 0 2 -subspace. 11 / 20
Preliminaries (3) Every quasi-Polish space X satisfies a generalized Baire category theorem (see R. Heckmann, 2012 and V. Becker & S. Grigorieff, 2012) : Proposition i ∈ ω A i , with each A i a Σ 0 If X = � 2 -set, then some A i has non-empty interior. Every quasi-Polish space X satisfies the Hausdorff-Kuratowski theorm: Proposition If A is a ∆ 0 2 -subset of X , then there is a countable ordinal α and an increasing sequence { U β } β<α of open subsets of X such that � � A = { U β \ U γ | β < α, r ( β ) � = r ( α ) } , γ<β where r ( α ) = 0 is α is even, and r ( α ) = 1 if α is odd. 12 / 20
Quasi-Polish Noetherian spaces (1) Theorem The following are equivalent for a sober Noetherian space X : 1 X has countably many points. 2 X is countably based. 3 X is quasi-Polish. Proof: (1 ⇒ 2) : By J. Goubault-Larrecq’s characterization of sober Noetherian spaces, X has the upper topology, so countability implies a countable basis. (2 ⇒ 1) : Every open is compact, hence equal to a finite union of basic opens, so the topology is countable. Sobriety then implies countability. (2 ⇒ 3) holds because sober Noetherian spaces are locally compact sober spaces, and (3 ⇒ 2) holds by definition. Corollary A subspace of a countable sober Noetherian space is sober iff it is a Π 0 2 -subspace. 13 / 20
Quasi-Polish Noetherian spaces (2) Lemma Every covering of a quasi-Polish Noetherian space X by ∆ 0 2 -sets admits a finite subcovering. (Equivalently, the topology generated by the ∆ 0 2 -sets is compact, and clearly Hausdorff) Proof: Since X is countable we can assume the covering is countable. By the Baire category theorem, there is a ∆ 0 2 -set A 0 in the covering such that its interior, U 0 , is non-empty. For n ≥ 0 , if X � = U n , then we repeat the same argument with respect to X \ U n to get a ∆ 0 2 -set A n +1 in the covering with non-empty interior relative to X \ U n . Define U n +1 to be the union of U n and the relative interior of A n +1 . Then U n +1 is an open subset of X which strictly contains U n . Since X is Noetherian, eventually X = U n , and A 0 , . . . , A n will yield a finite subcovering of X . 14 / 20
Quasi-Polish Noetherian spaces (3) Corollary Every infinite quasi-Polish Noetherian space X contains a singleton set which is not a ∆ 0 2 -set. (In other words, every quasi-Polish Noetherian T D -space is finite) Proof: Assume for a contradiction that every singleton is ∆ 0 2 . x ∈ X { x } is a covering of X by ∆ 0 Then � 2 -sets with no finite subcover. Contradiction. (Note: Infinite non-sober Noetherian T D -spaces do exist) 15 / 20
Quasi-Polish Noetherian spaces (4) Lemma Every ∆ 0 2 -subset A of a quasi-Polish Noetherian space X is constructible. Proof: By the Hausdorff-Kuratowski theorem, � � A = { U β \ U γ | β < α, r ( β ) � = r ( α ) } γ<β for some α < ω 1 and increasing sequence { U β } β<α of open subsets of X . The Noetherian assumption implies we can take α to be finite. (Note: This does not hold for non-sober Noetherian spaces) 16 / 20
Closed sets Theorem (J. Goubault-Larrecq, 2007) Every closed subset of a sober Noetherian space is finitely generated (equal to the closure of a finite set). Proof: (J. Goubault-Larrecq gave a general and elegant proof using de Groot duality. We just prove the quasi-Polish case) It suffices to show that each Noetherian quasi-Polish space X equals the closure of a finite set. x ∈ X cl ( { x } ) . Since closed sets are ∆ 0 Clearly, X = � 2 , from our previous lemma there is finite F ⊆ X such that X = � x ∈ F cl ( { x } ) . Therefore, X = cl ( F ) . 17 / 20
Recommend
More recommend
