aa r X i v : . [ m a t h . L O ] J a n Noetherian spaces in choiceless set theory ∗ Jindˇrich ZapletalUniversity of FloridaJanuary 12, 2021
Abstract
I prove several independence results in the choiceless ZF+DC theorywhich separate algebraic and non-algebraic consequences of the axiom ofchoice. As an example, let E be an equivalence relation resulting froma turbulent Polish group action, let X be a Polish field, and let F bea countable subfield. It is consistent with the choiceless theory ZF+DCthat X has a transcendence basis over F and E does not have a selector. I isolate a novel preservation property, the
Noetherian balance of Suslin forcingsand use it to prove several consistency results in the choiceless theory ZF+DC.On an intuitive level, the preservation property separates applications of Axiomof Choice which are algebraic in nature from those which are not. The techniquedeveloped in this paper has many corollaries; the following theorems are mereillustrative samples.
Theorem 1.1.
Let E be an equivalence relation resulting from a turbulent Pol-ish group action, let X be a K σ Polish field, and let F be a countable subfield.It is consistent relative to an inaccessible cardinal that ZF+DC holds, X has atranscendence basis over F , yet E does not have a selector. To state the following theorem, a hypergraph Γ of arity n ∈ ω is redundant if forevery set a of size n − R which consistsof all quadruples of points which are vertices of a rectangle is redundant; thehypergraph of arity three of isosceles triangles is not. Redundant hypergraphshave countable chromatic number in ZFC+CH, with CH being unnecessary incertain cases.For a natural number n ≥ n on n ω connecting points x , x ∈ n ω if the set { i ∈ ω : x ( i ) = x ( i ) } is finite. Existence of a nonprincipalultrafilter U on ω guarantees that the chromatic number of Ξ n is n , as one can ∗ x to that number j ∈ n such that { i ∈ ω : x ( i ) = j } ∈ U .In the absence of a nonprincipal ultrafilter, coloring the graphs Ξ n for n ≥ Theorem 1.2.
Let Γ be a redundant algebraic hypergraph on a Euclidean space.It is consistent relative to an inaccessible cardinal that ZF+DC holds, the chro-matic number of Γ is countable, yet the chromatic number of Ξ is uncountable. A rather different type of a preservation theorem concerns Lusin-type subsetsof Polish spaces. For the purposes of this introduction, I include the followingspecial case. It should be noted that there are algebraic redundant hypergraphssuch that in ZFC, their countable chromatic number is equivalent to CH (suchas the hypergraph of all rectangles in R [4]), and ZFC+CH proves the negationof the conclusion of the following theorem. Theorem 1.3.
Let Γ be a redundant algebraic hypergraph on a Euclidean space.It is consistent relative to an inaccessible cardinal that ZF+DC holds, the chro-matic number of Γ is countable, and for every uncountable collection of opensubsets of R of Lebesgue measure < ε , a union of some uncountable subcollec-tion has Lebesgue measure < ε . All Noetherian balanced forcings known to me are nearly automatically tran-scendentally balanced as described in [10], and thus the conclusions of the preser-vation theorems in that paper hold for them as well. Noetherian balance canbe stratified by dimension, more specifically the Krull dimension of relevantNoetherian topologies. This will be described in forthcoming work.In Section 2, I introduce analytic Noetherian topologies and several use-ful examples. Section 3 discusses the notion of a mutually Noetherian pair ofgeneric extensions of a model of ZFC and its implications. This is an instrumen-tal weakening of mutual genericity. In Section 4 I produce several interestingmutually Noetherian pairs of generic extensions, notably one induced by a tur-bulent action of a Polish group. Section 5 defines the notion of a Noetherianbalanced Suslin forcing–this is a forcing in which conditions can be successfullyamalgamated across mutually Noetherian pairs of generic extensions. There areseveral attendant preservation theorems for Noetherian balanced Suslin forc-ings. Section 6 lists some Noetherian balanced Suslin forcings and uses themwith the preservation theorems to finally obtain independence results.The paper uses the notation standard of [6]. In matters pertaining to geo-metric set theory, such as the calculus of virtual and balanced virtual conditionsin Suslin posets, it follows the terminology and notation of [8]. The paper usesthe Effros standard Borel space of closed subsets of a Polish space X , oftendenoted by F ( X ) in the literature. The following fact stands at the root of allcomplexity computation of this paper [7, Section 12.C]: Fact 1.4.
Let X be a Polish space. Then1. the membership relation {h x, C i ∈ X × F ( X ) : x ∈ C } ⊂ X × F ( X ) isBorel; . the union relation {h C , C , C i ∈ F ( X ) : C ∪ C = C } ⊂ F ( X ) isBorel, and similarly for unions of any finite number of sets;3. the subset relation {h C , C i ∈ F ( X ) : C ⊂ C } ⊂ F ( X ) is Borel.In addition, if the space X is K σ , then4. the intersection relation {h C , C , C i ∈ F ( X ) : C ∩ C = C } ⊂ F ( X ) is Borel, and similarly for intersections of any finite number of sets;5. the function y C y , whenever Y is a Polish space and C ⊂ Y × X is aclosed set, is Borel. The last two items fail badly for non- K σ spaces, and this is the reason why Irestrict my attention to K σ spaces in many respects. It is in the K σ contextwhere most interesting examples are found, at any rate. The technology of this paper rests on the following rather standard definition.
Definition 2.1.
Let X be a Polish space and let T be a topology on X differentfrom the original Polish one. Say that T is an analytic Noetherian topology if1. T is Noetherian. That is, there is no infinite strictly decreasing sequenceof T -closed sets;2. T is analytic. That is, every T -closed set is closed in the Polish topology,and the collection of T -closed sets is analytic in the Effros Borel space on X .One immediate initial remark is that for a K σ Polish space X , the statement“ T is an analytic Noetherian topology on X ” is Π and therefore absolutethroughout forcing extensions: it simply says that the collection of T -closedsets is closed under finite unions and finite intersections and contains no infinitesequences strictly decreasing under inclusion. Note that the closure under in-tersection cannot be expressed in Π way unless the underlying space X is K σ and intersection is a Borel function from F ( X ) to F ( X ) as per Fact 1.4(3).Analytic Noetherian topologies in this paper are always generated by ananalytic collection of closed sets, which is then closed under finite unions andintersections to obtain all closed sets in the Noetherian topology. There is astandard criterion for this process to work correctly. Proposition 2.2.
Let X be a K σ Polish space and S be an analytic collectionof closed subets of X . The following are equivalent:1. S generates an analytic Noetherian topology on X ; . for every sequence h C n : n ∈ ω i of elements of S there is a number m ∈ ω such that T n ∈ m C n = T n ∈ ω C n .Proof. Clearly, the failure of (2) implies the failure of (1) as it necessitates aninfinite strictly descending sequence of closed sets in the topology generated by S . Now suppose that (2) holds and consider the collection T = { C ⊂ X : forsome finite sets a , a , . . . a n of elements of S , C = T i ∈ n S a i } . It is not difficultto see that the collection T consists of closed sets, it is analytic in the EffrosBorel space by items (1) and (3) of Fact 1.4 (here the K σ assumption is used),and it is closed under finite intersections and unions. To show that it is closedunder arbitrary intersections and is in fact Noetherian topology, it is necessaryand sufficient to show that there is no infinite strictly decreasing sequence ofsets in T . To this end, it is enough to show that for any sequence h a n : n ∈ ω i offinite subsets of S there is a number m ∈ ω such that T n ∈ m S a n = T n ∈ ω S a n .Suppose towards a contradiction that such a number m does not exist. Thenone can by recursion on k ∈ ω pick sets C k ∈ a k such that the intersections T n ∈ m S a n for m ∈ ω do not stabilize below T l ∈ k C l . By the initial assump-tions on S , there is a number k such that T l ∈ k C l = S l ∈ ω C l . But then, theintersections T n ∈ m S a n do stabilize below T l ∈ k C l : namely, for all m ≥ k theyare equal to T l ∈ k C l itself. A contradiction.I include a basic algebraic and a basic combinatorial example of a Noetheriantopology. Example 2.3.
Let T be the topology on F n generated by algebraic sets, where F is a K σ Polish field. Then T is an analytic Noetherian topology. It isgenerated by an analytic collection of closed sets, and the descending chaincondition is verified via a successive use of the Hilbert Basis Theorem. Definition 2.4.
The infinite half-biclique is the graph on ω × h n, i and h m, i if n ≤ m , and contains no other edges. Example 2.5.
Let Γ be a closed graph on a K σ Polish space X . The graphtopology T is the smallest topology on X such that for every point x ∈ X theset C x = { y ∈ x : y = x ∨ y Γ x } is T -closed. If Γ does not contain an injectivehomomorphic image of the infinite half-biclique, then the graph topology isanalytic Noetherian. Proof.
Note first that the set { C x : x ∈ X } ⊂ F ( X ) is analytic as per Fact 1.4(4).In view of Proposition 2.2 it is enough to show that for any sequence h x n : n ∈ ω i of points in X there is a number m ∈ ω such that T n ∈ m C x n = T n ∈ ω C x n .Suppose towards a contradiction that this fails. Then the sets D m = T n ∈ m C x n must be all infinite, and it is possible to choose points y m ∈ D m so that they arepairwise distinct and also y m / ∈ { x n : n ≤ m } . It is then easy to find an infiniteset a ⊂ ω such that all points x n and y n for n ∈ a are pairwise distinct. Thesepoints then form an injective homomorphic image of the infinite half-biclique,contradicting the initial assumptions on Γ.4 Noetherian pairs of generic extensions
The following definition is the key tool for connecting Noetherian topologieswith geometric set theory.
Definition 3.1.
Let V [ G ] and V [ G ] be generic extensions of V inside anambient generic extension. Say that V [ G ] is Noetherian over V [ G ] if forevery K σ Polish space X and an analytic Noetherian topology T on X in theground model V , if O ⊂ X is a T -open set in V [ G ] such that X ∩ V ⊂ O then X ∩ V [ G ] ⊂ O . Say that the extensions V [ G ] , V [ G ] are mutually Noetherian if each is Noetherian over the other.Similar notions of perpendicularity always have a friendly relationship withproduct forcing, as recorded in the following routine proposition. Proposition 3.2.
Let n ≥ be a number. Let V [ G ] , V [ G ] be generic ex-tensions and V [ G ] is Noetherian over V [ G ] . Suppose that P ∈ V [ G ] and P ∈ V [ G ] be posets and H ⊂ P and H ⊂ P be filters mutually generic over V [ G , G ] . Then V [ G ][ H ] is Noetherian over V [ G ][ H ] .Proof. Work in the model V [ G , G ] and consider the poset P × P . Let X be a Polish space and T an analytic Noetherian topology on it, both in V .Let p ∈ P and p ∈ P be conditions and ˙ x ∈ V [ G ] be a P -name for anelement for X and ˙ C ∈ V [ G ] a P -name for a T -closed set. Suppose that h p , p i (cid:13) ˙ x ∈ ˙ C ; I must produce a ground model element y ∈ X such that p (cid:13) ˇ y ∈ ˙ C . Working in V [ G ], let M be a countable elementary submodel ofsome large structure and h ⊂ P ∩ M be a filter generic over M containingthe condition p ; let x = ˙ x/h ∈ X ∩ V [ G ]. Working in the model V [ G ],let M be a countable elementary submodel of some large structure and let { h n : n ∈ ω } be a countable collection of filters on P generic over M , eachcontaining the condition p , and such that every condition in M ∩ P strongerthan p is contained in one of the filters. Let C = T n ˙ C/h n .Still working in the model V [ G ], observe that for each number n ∈ ω the set˙ C/h n is T -closed: this follows from the fact that the collection of T -closed setsis analytic and the Mostowski absoluteness between M [ h n ] and V [ G ]. Thus,also the set C is T -closed. Stepping out of the model V [ G ], observe that x ∈ C must hold: otherwise, there would have to be a basic open set O ⊂ X such that x ∈ O and a number n ∈ ω such that ˙ C/h n ∩ O = 0. By the forcingtheorem applied in M and M respectively, there would have to be conditions p ′ ≤ p in h and p ′ ≤ p in h n such that p ′ (cid:13) ˙ x ∈ O and p ′ (cid:13) ˙ C ∩ O = 0,contradicting the initial assumptions on p and p .Now, since V [ G ] is Noetherian over V [ G ], there must be a point y ∈ X ∩ V such that y ∈ C . I claim that p (cid:13) ˇ y ∈ ˙ C as desired. If this were not the case,there would have to be a basic open set O ⊂ X such that y ∈ O and a condition p ′ ≤ p forcing ˙ C ∩ O = 0. By the elementarity of the model M , one such acondition would exist in M and therefore in a filter h n for some number n ∈ ω .Then y / ∈ ˙ C/h n and y / ∈ C , contradicting the choice of the point y .5n the remainder of this section, I isolate several useful features of mutuallyNoetherian extensions. Proposition 3.3.
Let V [ G ] , V [ G ] be generic mutually Noetherian extensionsof V . Then ω ∩ V [ G ] ∩ V [ G ] = 2 ω ∩ V .Proof. Let T be the topology on 2 ω whose closed sets are exactly the finite setsand 2 ω itself. It is not difficult to see that T is an analytic Noetherian topologyon X . Now suppose that x ∈ ω ∩ V [ G ] \ V is a point. The set 2 ω \ { x } is a T -open set in V [ G ] which covers 2 ω ∩ V . By the mutual Noetherian assumption,it covers 2 ω ∩ V [ G ] as well, and therefore x / ∈ V [ G ] as desired. Proposition 3.4.
Let Γ be a closed graph on a K σ Polish space X which doesnot contain an injective homomorphic image of the infinite half-biclique. Let V [ G ] , V [ G ] be mutually Noetherian extensions and x ∈ X ∩ V [ G ] and x ∈ X ∩ V [ G ] be Γ -connected points. Then x ∈ V or x ∈ V holds.Proof. Towards a contradiction, suppose that x ∈ V [ G ] \ V and x ∈ V [ G ] \ V are Γ-connected points. First argue that for every open neighborhood O ⊂ X of x , there is a point y ∈ ¯ O ∩ V such that { y, x } ∈ Γ. To see that, look atthe Noetherian graph topology T of Example 2.5 restricted to the set ¯ O . Theset of Γ-neighbors of x in ¯ O is a closed set in this topology, containing a point x ∈ V [ G ]. By the Noetherian balance assumption, it has to contain a pointin V as desired.It follows that there is an injective sequence h y n : n ∈ ω i of points in X ∩ V converging to x , all of them Γ-connected with x . Repeatedly using Mostowskiabsoluteness between V and V [ G ], one can recursively find an injective se-quence h z n : n ∈ ω i of points in X ∩ V converging to x such that for each m ∈ ω and each n ≤ m the points y m and z n are Γ-connected. Tails of thesetwo converging sequences must be disjoint, and they form an injective homomor-phic image of the half-biclique, contradicting the initial assumptions on Γ.Noetherian topologies are most common in algebra, and the following featureexploits standard algebraic facts about them. Proposition 3.5.
Let X be a K σ Polish field, and let p (¯ v , ¯ v ) be a multivariatepolynomial with free variables listed and with coefficients in X . Let V [ G ] , V [ G ] be mutually Noetherian extensions and let ¯ x ∈ V [ G ] and ¯ x ∈ V [ G ] be tuplesof elements of A such that p (¯ x , ¯ x ) = 0 . Then there is a tuple ¯ x ′ ∈ V arbitrarilyclose to ¯ x such that p (¯ x ′ , ¯ x ) = 0 .Proof. Let n be the number of variables in ¯ v and n the number of variables in¯ v . The topology T on X n + n generated by algebraic sets is easily checked tobe analytic; it is Noetherian by the Hilbert Basis Theorem. Let O ⊂ X n be anarbitrary open neighborhood of ¯ x . The proposition follows by an applicationof the definitions to the Noetherian topology T ↾ ¯ O .6 Examples I
In this section, I produce several useful pairs of mutually Noetherian extensions.They are all checked with the following proposition.
Proposition 4.1.
Suppose that in some ambient forcing extension V [ H ] thereis a cardinal κ and forcing extensions V [ G ] and V [ G α ] for α ∈ κ such that1. κ is a regular cardinal larger than | ω ∩ V [ G ] | ;2. for disjoint finite sets b , b ⊂ κ , V [ G α : α ∈ b ] ∩ V [ G α : α ∈ b ] = V .Then there is α ∈ κ such that V [ G α ] is Noetherian over V [ G ] .Proof. Suppose towards a contradiction that this fails. Then for every ordinal α ∈ κ , there must be a Polish space X α and an analytic Noetherian topology T α on it in the ground model, and a T α -closed set C α ⊂ X α in the model V [ G α ]which contains no point of X ∩ V but does contain some point x α ∈ X α ∩ V [ G ]. Acounting argument using assumption (1) shows that thinning down the sequenceof forcing extensions if necessary we may assume that all the spaces X α are thesame, equal tosome X , all the topologies T α are the same, equal to some T , andall the points x α ∈ X ∩ V [ G ] are the same, equal to some x .For each ordinal α ∈ κ let D α = T β>α C α . This is a T -closed set, so closedin the Polish topology on X . The sequence h D α : α ∈ κ i is an uncountableincreasing sequence of closed sets, so by a counting argument with a countablebasis of the space X it has to stabilize at some point α with the stable value D . Use the Noetherian property of the topology T to find finite sets b ⊂ κ \ α and b ∈ κ \ max( b ) + 1 such that D = T α ∈ b C α = T α ∈ b C α . The closedset D then belongs to both models V [ G α : α ∈ b ] and V [ G α : α ∈ b ], and byassumption (2), it must be the case that D ∈ V holds. The set D is nonempty in V [ H ], containing the point x . By a Mostowski absoluteness argument betweenthe models V and V [ H ], the set D is nonempty in V , containing some point y ∈ V . Then for any ordinal α > α , y ∈ C α holds, contradicting the choice ofthe set C α . Example 4.2.
Let Γ be a Polish group acting turbulently on a Polish space X . Let γ ∈ Γ and x ∈ X be mutually generic Cohen elements of Γ and X respectively. Then V [ x ] and V [ γ · x ] are mutually Noetherian extensions of V . Proof.
Work in the ground model. Let P X , P Γ denote the Cohen posets ofnonempty open sets ordered by inclusion on X and Γ respectively, with theirattendant names ˙ x gen and ˙ γ gen for the generic points in X and Γ respectively.Recall that P X × P γ (cid:13) ˙ γ gen · ˙ x gen is P X -generic over the ground model.By a symmetry argument, it is enough to show that the model V [ ˙ γ gen · ˙ x gen ] isNoetherian over V [ ˙ x gen ]. Towards a contradiction, suppose that h O, U i ∈ P X × P Γ be a condition forcing the opposite. Let κ = c + and let x ∈ O, γ α ∈ U : α ∈ κ be points which are mutually generic for the finite support product of copiesof P X and P Γ . Denote the resulting model V [ H ] and work in V [ H ]. Write x α = γ α · x for each α ∈ κ . 7 first show that for disjoint finite sets b , b ⊂ κ , V [ x α : α ∈ b ] ∩ V [ x α : α ∈ b ] = V . Without loss, assume that 0 ∈ b holds. We know that V [ x ] ∩ V [ x ] = V holds by the turbulence assumption on the action [8, Theorem 3.2.2]. Now, foreach α ∈ b \ { } , let δ α = γ α · γ − . Since the multiplication by γ − on theright is a self-homeomorphism of the group Γ, it results in an automorphism ofthe poset P Γ . As a result, the tuple h δ α : α ∈ b \ { } , γ α : α ∈ b i is mutually P Γ -generic over the model V [ x ][ γ ]. By a usual mutual genericity argument,the models V [ x ][ γ α : α ∈ b ] and V [ γ · x ][ δ α : α ∈ b \ { } ] still have V as theirintersection. The former model contains V [ γ α · x : α ∈ b ] and the latter contains V [ γ α · x : α ∈ b ]. The equality V [ x α : α ∈ b ] ∩ V [ x α : α ∈ b ] = V follows.Now apply Proposition 4.1 with the models V [ x ] and V [ x α ] for α ∈ κ . Notethat all posets involved are c.c.c. and so κ remains a regular cardinal larger that | ω ∩ V [ x ] | in V [ H ]. The proposition shows that there is α ∈ κ such that V [ x α ]is Noetherian over V [ x ], contradicting the initial assumption on the condition h O, U i ∈ P X × P Γ . Example 4.3.
Let X ⊂ (3 ω ) be the closed set of all pairs x ∈ (3 ω ) such thatfor every i ∈ ω , x (0)( i ) = x (1)( i ). Let P X be the Cohen poset of all nonemptyrelatively open subsets of Y ordered by inclusion, and let x ∈ X be a pair genericover V for P X . Then x (0) , x (1) ∈ ω are points of 3 ω separately Cohen-genericover V and V [ x (0)], V [ x (1)] are mutually Noetherian extensions of V . Proof.
Move back to V . I start with an abstract claim. For a finite set a let Z a ⊂ (3 ω ) a be the closed set of all tuples z ∈ (3 ω ) a such that for all i ∈ ω , { z ( j )( i ) : j ∈ a } 6 = 3 holds. For the following claim, recall the notion ofindependent pairs of continuous open maps of [8, Definition 3.1.3]. Claim 4.4.
Suppose that a = a ∪ a is a partition into nonempty sets. Theprojection functions π : Z a → Z a and π : Z a → Z a are continuous, open, andindependent.Proof. The continuity and openness are left to the reader. For the independence,let O ⊂ Z a be a nonempty relatively open set. Thinning down if necessary, wemay assume that there is a number k ∈ ω and tuples t j ∈ k for j ∈ a such that Q j [ t j ] ∩ Z a = O . It will be enough to show that whenever O r , O l ⊂ π ′′ O arenonempty open sets, then there are points z r , z l ∈ Z a such that π ′′ ( z r ) ∈ O r , π ′′ ( z l ) ∈ O l , and π ( z r ) = π ( z l ).Thinning out if necessary, we may assume that there is a number k ′ > k andtuples s jr , s jl ∈ k ′ for j ∈ a such that t j ⊂ s jr , s jl and Q j s jr ∩ Z a = O r ,and similarly for subscript l . For each number m ∈ k ′ \ k , neither of the sets { s jr ( m ) : j ∈ a } and { s jl ( m ) : j ∈ a } is equal to 3. There is a value u m ∈ { s jr ( m ) : j ∈ a } ∪ { u m } nor { s jl ( m ) : j ∈ a } ∪ { u m } is equalto 3. For each j ∈ a , let s j ∈ k ′ be defined by t j ⊂ s j and s j ( m ) = u m for all m ∈ k ′ ∈ k . Now, consider the points z r and z l ∈ (3 ω ) a defined in the followingway: z r ( j ) extends s jr (if j ∈ a ) or s j (if j ∈ a ) with an infinite sequence ofzeroes, and z l ( j ) extends s jl (if j ∈ a ) or s j (if j ∈ a ) with an infinite sequenceof zeroes. It is immediate that the points z r , z l work as required.8et ˙ x , ˙ x be the P X names for the first and second coordinates of the genericpair. Suppose towards a contradiction that there is a nonempty relatively openset O ⊂ X which forces V [ ˙ x ] not to be Noetherian over V [ ˙ x ]. Thinning out ifnecessary, find a number k ∈ ω and strings t , t ∈ k such that ([ t ] × [ t ]) ∩ X = O . Let x ∈ ω be a point Cohen-generic over V . Working in V [ x ], let Q bethe poset of all strings in 3 <ω which are disjoint from x and extend t . Let κ = c + and force over V [ x ] with the finite support product of κ -many copiesof Q , obtaining points x α for α ∈ κ . It is not difficult to check that for each α ∈ κ , the pair h x , x α i is generic over V for P X , and for each finite set a ⊂ κ , the tuple h x α : α ∈ a i is a generic element of Z a over V . It followsfrom Claim 4.4 and [8, Theorem 3.1.4] that for disjoint finite sets b , b ⊂ κ , V [ x α : α ∈ b ] ∩ V [ x α : α ∈ b ] = V . By Proposition 4.1, there is α ∈ κ suchthat the model V [ x α ] is Noetherian over V [ x ]. This contradicts the forcingtheorem applied to the P X -generic pair h x , x α i . Example 4.5.
Let P be a Suslin c.c.c. poset. Let V [ G ] be any generic exten-sion and let G ⊂ P be a filter generic over V [ G ]. Then V [ G ] and V [ G ∩ V ]are mutually Noetherian extensions. Proof.
Work in the ground model. Let P be the partial order generating theextension V [ G ]. In the iteration P ∗ P , we will denote the name for the genericfilter on the first coordinate by ˙ G and the name for the generic filter on thesecond coordinate by ˙ G .First, work to show that V [ ˙ G ∩ V ] is forced to be Noetherian over V [ ˙ G ].Suppose towards a contradiction that a condition h p , ˙ p i in the iteration forcesthe opposite. Let G ⊂ P be a generic filter. In V [ G ], let κ = c + . Force withthe finite support product of forcings P ↾ ˙ p /G of length κ , introducing filters G α ⊂ P ∩ V [ G ] for α ∈ κ . These filters are in finite tuples mutually genericover V [ G ] and all contain the condition ˙ p /G .Now, consider the filters G α ∩ V for α ∈ κ . Observe that the product Q ofany finite number of copies of P is a Suslin c.c.c. forcing [2, Corollary 3.6.9].By a Shoenfield absoluteness argument, every (countable) maximal antichainof Q in V remains a maximal antichain of Q in V [ G ]. As a result, the filters G α ∩ V for α ∈ κ are in finite tuples mutually generic over V . By the productforcing theorem then, if b , b ⊂ κ are disjoint finite sets, then V [ G α ∩ V : α ∈ b ] ∩ V [ G α ∩ V : α ∈ b ] = V . It follows that the assumptions of Proposition 4.1are satisfied and one of the models V [ G α ∩ V ] for α ∈ κ is Noetherian over V [ G ]. This contradicts the initial assumptions on the condition h p , ˙ p i .Now, work to show that V [ ˙ G ] is forced to be Noetherian over V [ ˙ G ∩ V ].Suppose towards a contradiction that a condition h p , ˙ p i ∈ P ∗ ˙ P forced theopposite. Let κ = max {|P ( P ) | , c } + and let G α : α ∈ κ be filters generic forthe finite support product of κ many copies of P ↾ p . Work in the model V [ G α : α ∈ κ ]. By a κ -c.c. argument, κ is still an uncountable regular cardinal.For each ordinal α ∈ κ , let p α = ˙ p /G α . This is a condition in the Suslin poset P . By the c.c.c. of P , there is a condition p ∈ P which forces in P that κ -many of the conditions p α belong to the generic filter. Let G ⊂ P be a filter9eneric over the model using the Suslin poset P and containing the condition p . Let V [ H ] = V [ G α : α ∈ κ ][ G ].Now, apply Proposition 4.1 with the model V [ H ], the model V [ G ∩ V ] andthe sequence h V [ G α ] : α ∈ κ, p α ∈ G i . Assumption (1) is satisfied by the κ -c.c.of all the forcings concerned. Assumption (2) is satisfied by the product forcingtheorem applied to finite products of copies of the poset P . In conclusion,there is an ordinal α ∈ κ such that V [ G α ] is Noetherian over V [ G ∩ V ]. Thiscontradicts the initial choice of the condition h p , ˙ p i ∈ P ∗ ˙ P . Any notion of perpendicularity similar to Definition 3.1 comes with a naturalnotion of balance for Suslin forcings.
Definition 5.1.
Let P be a Suslin forcing.1. A virtual condition ¯ p in P is Noetherian balanced if for any pair V [ G ], V [ G ] ofmutually Noetherian extensions of the ground model and for everypair p ∈ V [ G ], p ∈ V [ G ] of conditions stronger than ¯ p , the conditions p , p ∈ P have a common lower bound.2. P is Noetherian balanced if for every condition p ∈ P there is a Noetherianbalanced virtual condition ¯ p ≤ p .The supply of mutually Noetherian pairs of extensions provided in the previoussection now makes it possible to prove several preservation theorems. They arestated using the parlance of [8, Convention 1.7.18]. Thus, given an inaccessiblecardinal κ , a Suslin poset P is Noetherian cofinally below κ if for every genericextension V [ K ] generated by poset of cardinality smaller than κ there is a largergeneric extension V [ K ] generated by a poset of cardinality smaller than κ suchthat V κ [ K ] | = P is Noetherian balanced. HC denotes the set of all hereditarilycountable sets. For an equivalence relation E , the symbol | E | stands for thecardinality of the set of all E -classes. Theorem 5.2.
Let E be an orbit equivalence relation of some turbulent Polishgroup action. Let κ be an inaccessible cardinal. Let W be the symmetric Solovaymodel derived from κ . In cofinally Noetherian balanced extensions of W , | E | 6≤| HC | .Proof. Let Γ be a Polish group acting continuously and turbulently on a Polishspace X such that E is the resulting orbit equivalence relation. Let P be a Suslinforcing which is Noetherian balanced cofinally below κ . Work in W . Let p ∈ P be a condition and let τ be a P -name such that p (cid:13) τ is a function from the setof E -classes to hereditarily countable sets. I have to find two E -unrelated points x , x ∈ X and a condition stronger than p which forces τ ([ x ] E ) = τ ([ x ] E ).Both p, τ are definable from some elements of the ground model and anadditional parameter z ∈ ω . Let V [ K ] be an intermediate extension obtained10y a partial order of cardinality less than κ such that z ∈ V [ K ], and such that V [ K ] | = P is Noetherian balanced. Work in V [ K ]. Let ¯ p ≤ p be a virtualNoetherian balanced condition below p . Let Q X be the Cohen poset of allnonempty open subsets of X , adding a point ˙ x gen ∈ X . There must be a poset R of cardinality less than κ , a Q × R -names σ for a condition in P stronger than¯ p and η for a hereditarily countable set such that Q × R (cid:13) Coll( ω, < κ ) (cid:13) σ (cid:13) P τ ([ ˙ x gen ] E ) = η . There are two cases. Case 1.
There is a condition q ∈ Q and a condition r ∈ R and a set a suchthat Q × R (cid:13) η = ˇ a . In such a case, let G ⊂ Q , G ⊂ Q , H ⊂ R , and H ⊂ R be mutually generic filters such that q ∈ G ∩ G and r ∈ H ∩ H .Let x = ˙ x gen /G ∈ X and x = ˙ x gen /G ∈ X , also p = σ/G × H and p = σ/G × H . First, observe that x , x are mutually generic and so x doesnot belong to any meager set coded in V [ K ][ G ]; in particular, x / ∈ [ x ] E and so x , x are E -inequivalent. Second, observe that the extensions V [ K ][ G ][ H ] and V [ K ][ G ][ H ] are mutually generic and so the conditions p , p are compatiblein the poset P by the balance of the virtual condition ¯ p . Their common lowerbound the forces τ ([ x ]) E = τ ([ x ]) E as desired. Case 2.
Case 1 fails. In this case, I actually arrive at a contradiction. Let S be the Cohen partial order of nonempty open subsets of Γ ordered by inclusion.Let x ∈ X and γ ∈ Γ be points mutually Cohen generic over V [ K ] and let x = γ · x . By the turbulence assumption and [8, Theorem 3.2.2] V [ K ][ x ] ∩ V [ K ][ x ] = V [ K ] and moreover, by Example 4.2, the models V [ K ][ x ] and V [ K ][ x ] are mutually Noetherian. Let H , H ⊂ R be filters mutually genericover the model V [ K ][ x , x ]. By Proposition 3.2, the models V [ K ][ x ][ H ] and V [ K ][ x ][ H ] are mutually Noetherian as well; their intersection is equal to V [ K ]again by a mutual genericity argument. Let p = σ/x , H and p = σ/x , H ,and a = η/x , H and a = η/x , H . By the case assumption, a , a / ∈ V [ K ],and by the intersection property of the two models, a = a . By the balanceassumption on the virtual condition ¯ p , p , p are compatible in P . The commonlower bound of these two conditions forces that τ ([ x ] E ) = τ ([ x ] E ). This is animpossibility as x E x holds. Theorem 5.3.
Let κ be an inaccessible cardinal. Let W be the symmetricSolovay model derived from κ . In cofinally Noetherian balanced extensions of W , every nonmeager subset of ω contains points y , y such that the set { i ∈ ω : y ( i ) = y ( i ) } is finite.Proof. Let Γ be a Polish group acting continuously and turbulently on a Polishspace X such that E is the resulting orbit equivalence relation. Let P be aSuslin forcing which is Noetherian balanced cofinally below κ . Work in W . Let p ∈ P be a condition and let τ be a P -name such that p (cid:13) τ ⊂ ω is a nonmeagerset. I have to find two points y , y such that the set { i ∈ ω : y ( i ) = y ( i ) } isfinite a condition stronger than p which forces both to τ .Both p, τ are definable from some elements of the ground model and anadditional parameter z ∈ ω . Let V [ K ] be an intermediate extension obtainedby a partial order of cardinality less than κ such that z ∈ V [ K ], and such11hat V [ K ] | = P is Noetherian balanced. Work in V [ K ]. Let ¯ p ≤ p be avirtual Noetherian balanced condition below p . Let Q be the Cohen poset ofnonempty open subsets of 3 ω ordered by inclusion, adding a Cohen genericpoint ˙ y . There must be a condition q ∈ Q , a poset R of cardinality smallerthan κ , and a Q × R -name σ for a condition in P stronger than ¯ p such that q (cid:13) Q R (cid:13) Coll( ω, < κ ) (cid:13) σ (cid:13) ˙ y ∈ τ . Otherwise, in the model W , the condition¯ p ≤ p would force in P that the comeager set of points Q -generic over V [ K ] tobe disjoint from τ , contradicting the initial assumptions on p and τ .Let X be the closed subset of 3 ω × ω consisting of all pairs x such that x (0)( i ) = x (1)( i ) holds for all i ∈ ω . Move back to the model W . Let x ∈ X be apoint generic over V [ K ] for the Cohen poset with X . By Example 4.3, the points x (0) , x (1) ∈ ω are separately Q -generic over V [ K ] and the models V [ K ][ x (0)], V [ K ][ x ] are mutually Noetherian. Let y ∈ ω be a finite modification of x (0) which belongs to q and let y ∈ ω be a finite modification of x (1) whichbelongs to q . Let H , H ⊂ R be filters mutually generic over V [ K ][ x ] and let p = σ/y , H and p = σ/y , H . These are conditions in P stronger than¯ p , and they are compatible by the balance assumption on ¯ p . By the forcingtheorem applied in the respective models V [ K ][ y ][ H ] and V [ K ][ y ][ H ], thecommon lower bound of these two conditions forces ˇ y , ˇ y ∈ τ as required. Theorem 5.4.
Let ε > be a real number. Let κ be an inaccessible cardinal.Let W be the symmetric Solovay model derived from κ . In cofinally Noetherianbalanced extensions of W , if A is an uncountable collection of open subsets of [0 , of Lebesgue mass < ε , then there is an uncountable subcollection B ⊂ A such that S B has Lebesgue mass < ε .Proof. Let P be a Suslin forcing which is Noetherian balanced cofinally below κ . Work in the model W . Let p ∈ P be a condition and let τ be a P -namesuch that p (cid:13) τ is an uncountable collection of open subsets of [0 , < ε . I must find an open set O ⊂ [0 ,
1] of Lebesgue mass < ε and a condition stronger than p which forces that uncountably many elementsof τ are a subset of O .Both p, τ are definable from some elements of the ground model and anadditional parameter z ∈ ω . Let V [ K ] be an intermediate extension obtainedby a partial order of cardinality less than κ such that z ∈ V [ K ], and suchthat V [ K ] | = P is Noetherian balanced. Work in V [ K ]. Let ¯ p ≤ p be avirtual Noetherian balanced condition below p . Since the set τ is forced to beuncountable, there has to be a a forcing R of cardinality less than κ and an R -name σ for a condition in P stronger than ¯ p and an R -name η for an opensubset of [0 ,
1] of Lebesgue mass < ε which does not belong to V [ K ] such that R (cid:13) Coll( ω, < κ ) (cid:13) σ (cid:13) P (cid:13) η ∈ τ . Let H ⊂ R be a filter generic over V [ K ]and work in V [ K ][ H ]. Let p = σ /H and O = η/H .Fix a rational number δ strictly between the Lebesgue mass of O and ε .Consider the Suslin c.c.c. poset Q of all open subsets of [0 ,
1] of mass < δ ,ordered by reverse inclusion. Thus, O ∈ Q is a condition. Find a filter G ⊂ Q generic over V [ K ][ H ] containing the condition O and consider the model12 [ K ][ G ∩ V [ K ]]. By Proposition 3.2, the models V [ K ][ H ] and V [ K ][ G ∩ V [ K ]]are mutually Noetherian. Work in the model V [ K ][ G ] and let O ⊂ [0 ,
1] be theunion of all basic open subsets of [0 ,
1] which belong to G . Thus, O ⊂ [0 ,
1] isan open set of Lebesgue mass equal to δ and it contains O as a subset.I claim that in the model W , the virtual condition ¯ p forces O to containuncountably many elements of τ as subsets. This will complete the proof. Sup-pose towards a contradiction that this fails. Work in the model V [ K ][ G ∩ V [ K ]].There has to be a partial order R of cardinality less than κ and R -names σ for a condition in P stronger than ¯ p and η for a countable sequence of opensubsets of [0 ,
1] such that R (cid:13) Coll( ω, < κ ) (cid:13) σ (cid:13) η enumerates all subsetsof O which belong to τ .Let H ⊂ P be a filter generic over the model V [ K ][ H ][ G ]. Let p = σ /H and y = η /H . By Example 4.5, the models V [ K ][ H ] and V [ K ][ G ∩ V [ K ][ H ]are mutually Noetherian. It follows that the set O does not belong to the model V [ K ][ G ∩ V [ K ]][ H ] and therefore does not belong to the range of y . Also, bythe balance assumption on the virtual condition ¯ p , the conditions p , p have acommon lower bound in the poset P . That lower bound simultaneously forcesthat y enumerates all subsets of O which belong to τ and O ∈ τ . This isimpossible. In this section, I produce several Noetherian balanced Suslin forcings. This isof course a necessary ingredient for any specific independence result.
Example 6.1.
Let X be a K σ Polish field and F be a countable subfield. Let P be the partial order of countable subsets of X which are algebraically freeover F . The ordering is reverse inclusion. Then P is Noetherian balanced andevery balanced virtual condition is in fact Noetherian balanced.I do not know an example of a G δ matroid such that it would be impossible toforce a basis to it with a Noetherian balanced forcing. Proof. [8, Theorem 6.3.9] shows that balanced virtual conditions are classifiedby transcendence bases of X over F . Let ¯ p be such a basis. Let V [ G ] , V [ G ] bemutually Noetherian generic extensions of V and let p ∈ V [ G ] and p ∈ V [ G ]be conditions such that ¯ p ⊂ p , p holds. I must show that p , p are compatiblein P ; in other words, p ∪ p is algebraically free over F . Suppose towards acontradiction that this fails. Let r be a nonzero multivariate polynomial withcoefficients in F , and let ~x , ~x be tuples from p and p respectively such that r ( ~x , ~x ) = 0 holds. By Proposition 3.5, there must be a tuple ~x ′ in the groundmodel such that r ( ~x ′ , ~x ) holds. Note that all elements of the tuple ~x ′ arealgebraic over ¯ p ; let a ⊂ ¯ p be a finite set such that elements of ~x ′ are algebraicover a . Let x be any element of the tuple ~x and observe that x is algebraicover a and the remainder of ~x , contradicting the assumption that p is analgebraically independent set. 13 xample 6.2. Let X be a K σ Polish space and Γ an F σ graph on it whichdoes not contain an infinite half bi-clique. Let P be the coloring poset of Γ of[8, Definition 8.1.1]. Then P is Noetherian balanced and every balanced virtualcondition is Noetherian balanced.I do not know an example of a Borel graph of countable coloring number whichwould be impossible to color with a Noetherian balanced forcing. Proof.
Note that a graph which does contain an infinite half bi-clique has count-able coloring number [3], [5] (or in the context of analytic graphs, [1]). There-fore, [8, Theorem 8.1.2] applies to show that balanced conditions are classifiedby total Γ-colorings, and in addition below every condition there is a total Γ-coloring. Also, the poset P has a simple dense subset consisting of Γ-colorings p with countable domain such that ∀ y ∈ X \ dom( p ) { x ∈ dom( p ) : { x, y } ∈ Γ } isfinite. The ordering is by inclusion. One advantage of this presentation is thatconditions p , p ∈ P are compatible just in case p ∪ p is a Γ-coloring.Now suppose that c : X → ω is a total Γ-coloring, V [ G ] and V [ G ] aremutually Noetherian generic extensions, and p ∈ V [ G ] and p ∈ V [ G ] areconditions stronger than c . I must show that they are compatible, which is thesame as proving that p ∪ p is a Γ-coloring. To argue for that, first note that p ∪ p is a function because dom( p ) ∩ dom( p ) = X ∩ V by Proposition 3.3, and p ↾ V = p ↾ V = c . Now suppose that x ∈ dom( p ) and x ∈ dom( p ) areΓ-connected points. By Proposition 3.4, one of these points, say x must belongto V . Since p is a Γ-coloring, it follows that x , x receive distinct colors. Example 6.3. (ZF+CH) Let Γ be an algebraic redundant hypergraph on aEuclidean space. Let P be the coloring poset for Γ as defined in [9, Section4]. Then P is Noetherian balanced and every balanced virtual condition isNoetherian balanced.There are non-algebraic redundant hypergraphs which cannot be colored bya Noetherian balanced, or even transcendentally balanced, Suslin forcing [10,Corollary 4.5]. Proof.
Balanced virtual conditions are classified by total Γ-colorings from theEuclidean space X on which Γ lives, to ω × ω by [9, Theorem 3.6]. Let c : X → ω × ω be such a coloring, let V [ G ] , V [ G ] be mutually Noetherian extensionsand let p ∈ V [ G ] and p ∈ V [ G ] be conditions stronger than c . I must showthat p , p are compatible. The proof of compatibility in [9] depends only on thefact that V [ G ] ∩ R and V [ G ] ∩ R are in amalgamation position [9, Definition2.1] which follows immediately by Proposition 3.5.Now it is time to present the proofs of theorems from the introduction. Let κ be an inaccessible cardinal and let W be the symmetric Solovay model derivedfrom it. For Theorem 1.1, consider the extension from Example 6.1. Theo-rem 5.2 then shows that the conclusion of Theorem 1.1 holds in the resultingextension. For Theorem 1.2, use the coloring poset from Example 6.3. Theo-rem 5.3 then shows that the conclusion of Theorem 1.2 holds in the resulting14xtension. Theorem 5.4 then shows that the conclusion of Theorem 1.3 holdsin the resulting extension as well. References [1] Francis Adams and Jindˇrich Zapletal. Cardinal invariants of closed graphs.
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