aa r X i v : . [ m a t h . L O ] J a n Transcendental pairs of generic extensions ∗ Jindˇrich ZapletalUniversity of FloridaJanuary 12, 2021
Abstract
We isolate a new preservation class of Suslin forcings and prove sev-eral associated consistency results in the choiceless theory ZF+DC. Forexample, writing Γ n for the hypergraph on P ( ω ) consisting of n -tupleswhich modulo finite form a partition of ω , it is consistent with ZF+DCthat the chromatic number of Γ is countable, yet the chromatic numberof Γ is not. Following the initial work on geometric set theory [7], I isolate a new class ofbalanced forcings, the transcendental balanced forcings , verify that a numberof partial orders belong to it, and show a number of preservation theorems forthe extension of the Solovay model by forcings in this class. In this way, I ob-tain several independence results in the choiceless ZF+DC set theory regardingchromatic numbers of certain Borel graphs and hypergraphs. Recall that a hy-pergraph
Γ of arity n on a set X is just a subset of [ X ] n , its elements are its hyperedges . A (partial) function c on X is a Γ- coloring if c is not constant onany Γ-hyperedge. The chromatic number of Γ is the smallest cardinal κ suchthat there is a total Γ-coloring c : X → κ . In the absence of the axiom of choice,we only discern between various finite values of the chromatic number and thencountable and uncountable chromatic number. The study of chromatic numbersof Borel hypergraphs has long history [9, 6, 8, 1, 2]; in the choiceless ZF+DCtheory, inequalities between chromatic numbers of such hypergraphs are subjectto many consistency results [11, 7].For a Polish group G one may consider hypergraphs on G associated withvarious equations, and their chromatic numbers. Consider the hypergraph ∆( G )of quadruples which solve the equation g g − g g − = 1. Note that in ZF, thecountable chromatic number of ∆( G ) is inherited by subgroups of G . I prove Theorem 1.1.
Let G be a K σ Polish group. It is consistent relative to an inac-cessible cardinal that ZF+DC holds, the chromatic number of ∆( G ) is countable,yet the chromatic number of ∆( S ∞ ) is uncountable. ∗ S ∞ has (a small strengtheningof) the ample generics property. It seems to be difficult to separate chromaticnumbers of the hypergraphs ∆( G ) for various other Polish groups, such as G = S ∞ and G =the unitary group.For a natural number n ≥ n be the hypergraph on P ( ω ) of arity n consisting of n -tuples of sets which modulo finite form a partition of ω . Inthe presence of a nonprincipal ultrafilter on ω , the chromatic number of eachhypergraph Γ n is two, as the membership in the ultrafilter constitutes a Γ n -coloring. Without an ultrafilter, I learned how to separate chromatic numbersof Γ n ’s: Theorem 1.2.
It is consistent relative to an inaccessible cardinal that ZF+DCholds, the chromatic number of Γ is countable, while the chromatic number of Γ is uncountable. Theorem 1.3.
It is consistent relative to an inaccessible cardinal that ZF+DCholds, the chromatic number of Γ is countable, while the chromatic number of Γ is uncountable. It appears more challenging to separate the chromatic numbers of Γ n for highervalues of n , and I do not know how to do that.For every number n ≥ n be the hypergraph of arity n on P ( ω )consisting of sets d of size n such that T d = 0 and S d = ω , both modulo finite.A membership in a nonprincipal ultrafilter on ω provides a Θ n coloring withtwo colors. However, coloring Θ n without an ultrafilter seems to be a greatchallenge for larger values of n . Θ is locally countable and so can be coloredby rather innocuous posets of [7, Section 6.4]. I do not know how to obtain amodel where the chromatic number of Θ is countable without an ultrafilter.For Θ , I have a preservation theorem. Theorem 1.4.
Let κ be an inaccessible cardinal. In cofinally transcendentallybalanced forcing extensions of the symmetric Solovay model derived from κ , thechromatic number of Θ is uncountable. This rules out nonprincipal ultrafilters on ω in transcendentally balanced ex-tensions. (Diffuse finitely additive probability measures on ω do not exist thereeither for a different chromatic number reason.) The important point here isthat most known balanced Suslin posets which do not add an ultrafilter aretranscendentally balanced. This includes for example the posets adding a tran-scendence basis for R over Q , the coloring posets in [7, Section 8.2], and alsothe coloring posets introduced in the last section of the present paper. I do notknow how to find a model of ZF+DC in which the chromatic number of Θ isneither equal to 2 nor uncountable.In Section 2, I introduce transcendence of pairs of generic extensions, aproperty weaker than mutual genericity, with several properties of mutuallytranscendental pairs of generic extensions. In Section 3 I provide a number ofuseful examples of mutually transcendental pairs. In Section 4, I define thenotion of transcendental balance for Suslin forcing, and examples of Section 42re used to prove a number of preservation theorems for generic extensions ofthe Solovay model obtained with transcendentally balanced forcings. Finally,in Section 5, I show that many known balanced forcings are transcendentallybalanced, and build a new supply of coloring posets which are transcendentallybalanced. These are in turn used to prove the theorems of this introduction.Notation of the paper follows [3], and in matters of geometric set theory [7].In particular, the calculus of virtual conditions in Suslin forcing of Section 4 isestablished in [7, Chapter 5]. The key concept in this paper is a certain perpendicularity notion for pairs ofgeneric extensions, which generalizes mutual genericity.
Definition 2.1.
Let V [ G ] , V [ G ] be two generic extensions in an ambientgeneric extension. Say that V [ G ] is transcendental over V [ G ] if for everyordinal α and every open set O ⊂ α in the model V [ G ], if 2 α ∩ V ⊂ O then2 α ∩ V [ G ] ⊂ O . Say that the models V [ G ], V [ G ] are mutually transcendental if each of them is transcendental over the other one.Here, the space 2 α is equipped with the usual compact product topology. Fora finite partial function h from α to 2 write [ h ] = { x ∈ α : h ⊂ x } Theopen set O ⊂ α is then coded in V [ G ] by a set H ∈ V [ G ] of finite partialfunctions from α to 2 with the understanding that O = S h ∈ H [ h ], and in thisway it is interpreted in V [ G ][ G ]. For the general theory of interpretations oftopological spaces in generic extensions (unnecessary for this paper) see [10]. Itis tempting to deal just with the Cantor space 2 ω instead of its non-metrizablegeneralizations, but the present definition has a number of small advantagesand essentially no disadvantages as compared to the Cantor space treatment.On the other hand, extending the definition to all compact Hausdorff spaces isequivalent to the present form.I first need to show that the notion of transcendence generalizes mutualgenericity and in general behaves well with respect to product forcing. This isthe content of the following proposition. Proposition 2.2.
Let V [ G ] , V [ G ] be generic extensions such that V [ G ] istranscendental over V [ G ] . Let P ∈ V [ G ] and P ∈ V [ G ] be posets. Let H ⊂ P and H ⊂ P be filters mutually generic over the model V [ G ][ G ] .Then V [ G ][ H ] is transcendental over V [ G ][ H ] .Proof. Work in the model V [ G , G ] and consider the product forcing P × P .Let α be an ordinal. Let h p , p i ∈ P × P be a condition, let τ ∈ V [ G ] bea P -name for an open set such that p (cid:13) α ∩ V ⊂ ˙ O , and let η ∈ V [ G ] bea P -name for an element of 2 α . We need to find a stronger condition in theproduct which forces η ∈ τ .First, work in the model V [ G ]. Let A = { h : h is a finite partial functionfrom α to 2 and p (cid:13) ˇ h η } . By a compactness argument, the set S h ∈ A [ h ]3ust not cover the whole space 2 α ; if it did, a finite subset of Q would suffice tocover 2 α and there would be no space left for the point η . Let y ∈ α \ S h ∈ A [ h ]be an arbitrary point. Now work in the model V [ G ] and let A = { h : h isa finite partial function from α to 2 and there is a condition q ≤ p such that q (cid:13) [ h ] ⊂ τ } ; since p (cid:13) α ∩ V ⊂ τ , it must be the case that 2 α ∩ V ⊂ S h ∈ A [ h ].Now, use the transcendence of V [ G ] over V [ G ] to argue that y ∈ Q . Itfollows that there must be a finite partial function h ∈ A such that h ⊂ y . Itfollows that there must be conditions p ′ ≤ p and p ′ ≤ p such that p ′ (cid:13) ˇ h ⊂ η and p ′ (cid:13) [ h ] ⊂ τ . Then the condition h p ′ , p ′ i forces in the product P × P that η ∈ τ as required. Corollary 2.3.
Mutually generic extensions are mutually transcendental.Proof.
Just let V [ G ] = V [ G ] = V in Proposition 2.2.In the remainder of this section, I isolate several properties of mutually tran-scendental extensions which will come handy later. Proposition 2.4.
Let V [ G ] , V [ G ] be mutually transcendental generic exten-sions of V . Then V [ G ] ∩ V [ G ] = V .Proof. It will be enough to show that (2 α ∩ V [ G ]) ∩ (2 α ∩ V [ G ]) = 2 α ∩ V holds for every ordinal α . Let x ∈ α ∩ V [ G ] \ V be an arbitrary point. Theopen set O = 2 α \ { x } in V [ G ] covers 2 α ∩ V . By the mutual transcendence,2 α ∩ V [ G ] ⊂ O must hold as well. In particular, x / ∈ V [ G ] as required. Proposition 2.5.
Let V [ G ] , V [ G ] be mutually transcendental generic exten-sions of V . Let X , X be Polish spaces and C ⊂ X × X be a K σ -set. Let x ∈ X ∩ V [ G ] and x ∈ X ∩ V [ G ] be points such that h x , x i ∈ C . Thenthere is a point x ′ ∈ X ∩ V such that h x ′ , x i ∈ C .Proof. Since C is a countable union of compact sets, there is a compact set K ⊂ C coded in V such that h x , x i ∈ K . Let h : 2 ω → X be a continuousfunction onto the compact projection of the set K to X . Let O ⊂ ω in V [ G ]be the open set of all points y ∈ ω such that h f ( y ) , x i / ∈ K . The set O doesnot cover 2 ω ∩ V [ G ] since h − x ∩ O = 0. By the mutual transcendence, theremust be a point y ∈ ω ∩ V \ O . Let x ′ = h ( y ) and observe that the point x ′ ∈ X works as desired. Corollary 2.6.
Let X be a K σ Polish field. Let p (¯ v , ¯ v ) be a multivariatepolynomial with coefficents in X and variables ¯ v , ¯ v . Let V [ G ] , V [ G ] be mu-tually transcendental generic extensions of V and let ¯ x ∈ V [ G ] , ¯ x ∈ V [ G ] bestrings of elements of X such that p (¯ x , ¯ x ) = 0 . Then there is a string ¯ x ′ ∈ V arbitrarily close to ¯ x such that p (¯ x ′ , ¯ x ) = 0 .Proof. Apply the proposition with the additional insight that the spaces X n forany natural number n are K σ and solutions to a given polynomial form a closedset. 4 orollary 2.7. Let E be a K σ -equivalence relation on a Polish space X . When-ever V [ G ] , V [ G ] are mutually transcendental generic extensions of V and x ∈ X ∩ V [ G ] and x ∈ X ∩ V [ G ] are E -related points, then there is apoint x ∈ X ∩ V E -related to them both.
The last corollary can be generalized to some non- K σ -equivalence relations asin the following proposition. Proposition 2.8.
Let h U n , d n : n ∈ ω i be a sequence of sets and metrics oneach and let X = Q n U n . Let V [ G ] , V [ G ] be mutually transcendental genericextensions of V and x ∈ X ∩ V [ G ] and x ∈ X ∩ V [ G ] be points suchthat lim n d n ( x ( n ) , x ( n )) = 0 . Then there is a point x ∈ X ∩ V such that lim n d n ( x ( n ) , x ( n )) = 0 .Proof. First argue that for every number m ∈ ω there is a point y ∈ X ∩ V such that ∀ n d n ( y ( n ) , x ( n )) ≤ − m . To see this, fix a number k ∈ ω such thatfor all n ≥ k , d n ( x ( n ) , x ( n )) < − m − . Let A = {{ u, v } : ∃ n ≥ k u, v ∈ U n and d n ( u, v ) > − m } , and consider the space Z of all selectors on A , which isnaturally homeomorphic to 2 A . In the model V [ G ], let O = { z ∈ Z : ∃ n ≥ k ∃ v ∈ U n d n ( x ( n ) , v ) > − m and z ( x ( n ) , v ) = v } . This is an open subset ofthe space Z . It does not cover Z ∩ V [ G ] as in the model V [ G ], one can find aselector z ∈ Z such that for all n ≥ k and all { u, v } ∈ Z with u, v ∈ U n , z ( u, v )is one of the points u, v which is not d n -farther from x ( n ) than the other. It isimmediate from the definition of the set O and a triangle inequality argumentthat z / ∈ O . By a mutual transcendence argument, there is a selector z ′ ∈ Z ∩ V such that z ′ / ∈ O holds.Work in V . For each number n ≥ k , let B n = { u ∈ U n : ∀ v ∈ Y n d n ( u, v ) > − m → z ′ ( u, v ) = u } . The set B n contains x ( n ) by the choice of the selector z ′ . Moreover, for any two elements u, v ∈ B n , d n ( u, v ) ≤ − m must hold: inthe opposite case, the selector z ′ could not choose one element from the pair { u, v } without contradicting the definition of the set B n . Now consider anypoint y ∈ X such that for all n < k , y ( n ) = x ( k ) and for all n ≥ k y ( n ) ∈ B n .Then ∀ n d n ( y ( n ) , x ( n )) ≤ − m as desired.Now, let C = ω × ( X ∩ V ) and consider the set B ⊂ C of all pairs h m, y i ∈ A such that lim sup n d n ( y ( n ) , x ( n )) ≤ − m . As written, the set belongs to V [ G ];however, it also belongs to V [ G ] since replacing x in its definition with x results in the same set by the initial assumptions on x , x . By Proposition 2.4, B ∈ V holds. By the work in the previous paragraph, for each m ∈ ω B contains some element whose first coordinate is m . Thus, in V there existsa sequence h y m : m ∈ ω i such that ∀ m h m, y m i ∈ B . By a Mostowski ab-soluteness argument, there must be in V a point x such that for all m ∈ ω ,lim sup n ( y m ( n ) , x ( n )) ≤ − m , since such a point, namely x , exists in V [ G ].A triangular inequality argument then shows that lim n ( x ( n ) , x ( n )) = 0 as de-sired.I do not know whether further generalizations are possible. In particular, thefollowing is open: 5 uestion 2.9. Let E be a pinned Borel equivalence relation on a Polish space X . Let V [ G ] , V [ G ] be mutually transcendental generic extensions of V and x ∈ X ∩ V [ G ] and x ∈ X ∩ V [ G ] are E -related points. Must there be apoint x ∈ X ∩ V E -related to them both?
In this section, I provide several interesting pairs of mutually transcendentalpairs of generic extensions. To set up the notation, for a Polish space X , write P X for the Cohen poset of nonempty open subsets of X ordered by inclusion,with ˙ x gen being its name for a generic element of X . If f : X → Y is a con-tinuous open map then P X forces f ( ˙ x gen ) ∈ Y to be a point generic for P Y [7, Proposition 3.1.1]. The first definition and proposition deal with Cohenelements of Polish spaces. Definition 3.1.
Let
X, Y , Y be compact Polish spaces and f : X → Y and f : X → Y be continuous open maps. Say that f is transcendental over f if for every nonempty open set O ⊂ X there is a point y ∈ Y such that theset f ′′ ( f − { y } ∩ O ) ⊂ Y has nonempty interior for every nonempty open set O ⊂ X . Proposition 3.2.
Suppose that
X, Y , Y are Polish spaces, X is compact, and f : X → Y and f : X → Y are continuous open maps. The following areequivalent:1. f is transcendental over f ;2. P X forces V [ f ( ˙ x gen )] to be transcendental over V [ f ( ˙ x gen )] .Proof. To show that (1) implies (2), let α be an ordinal, let η be a P Y -name foran element of 2 α , and let τ be a P Y -name for an open subset of 2 α which is forcedto contain V ∩ α as a subset. Let O ⊂ X be a nonempty open set. To prove (2),I must find a strengthening O ′ ⊂ O such that O ′ (cid:13) η/f ( ˙ x gen ) ∈ τ /f ( ˙ x gen ).To this end, let y ∈ Y be a point such that the set f ′′ ( f − { y } ∩ O )has nonempty interior, and let O ⊂ Y denote that interior. Use the initialassumption on τ to find, for each z ∈ α , a condition O z ⊂ O and a finitepartial map h z : α → h z ⊂ z and O z (cid:13) [ h z ] ⊂ τ . Use a compactnessargument to find a finite set a ⊂ α such that 2 α = S z ∈ a [ h z ]. The set O = T z ∈ a f ′′ ( O ∩ f − O z ⊂ Y is nonempty as it contains y , and it is open as themaps f , f are continuous and open. Let O ′ ⊂ O be a condition which decidesthe value η ( ˇ β ) for every ordinal β ∈ S z ∈ a dom( h z ). Since S z ∈ a [ h z ] = 0, theremust be a point z ∈ a such that O ′ (cid:13) ˇ h z ⊂ η . The set O ′ = O ∩ f − O ′ ∩ f − O z is nonempty and open, and it forces in P X that η/f ( ˙ x gen ) ∈ [ h z ] and [ h z ] ⊂ τ /f ( ˙ x gen ).The implication (2) → (1) is best proved by a contrapositive. Suppose that(1) fails, as witnessed by some open set O ⊂ X . Let O ′ ⊂ O be some nonempty6pen set whose closure is a subset of O , and let x ∈ O ′ be a point P X -genericover the ground model. Write y = f ( x ) ∈ Y and y = f ( x ) ∈ Y . Let C = f ′′ ( f − { y } ∩ ¯ O ′ ). This is a closed subset of Y coded in V [ y ] whichcontains the point y . For the failure of (3), it is enough to show that C contains no ground model point. Indeed, if y ∈ Y is a point in the groundmodel, then D = f ′′ ( f − { y } ∩ ¯ O ′ ) ⊂ Y is a closed subset of Y coded in theground model which has empty interior by the choice of the set O ; in particular, D ⊂ Y is nowhere dense, and since y ∈ Y is a Cohen generic point, y / ∈ D holds. Comparing the definitions of the sets C and D , it is obvious that y / ∈ C as required. Example 3.3.
Let b be a finite set and let b = a ∪ a be a partition intotwo sets, each of cardinality at least two. Let X be the closed subset of P ( ω ) b consisting of those functions x such that T i ∈ a x ( i ) = 0 and S i ∈ a x ( i ) = ω . Let Y = P ( ω ) a and Y = P ( ω ) a . Let f : X → Y and f : X → Y be theprojection functions. Then f , f are continuous, open, and transcendental overeach other. Proof.
The continuity and openness are left to the reader. To show that f istranscendental over f , let O ⊂ X be a nonempty relatively open set. Thinningthe set O down if necessary, one can find a natural number k ∈ ω and sets c i ⊂ k for i ∈ b such that S i c i = k and T i c i = 0, and O = { x ∈ X : ∀ i ∈ b x ( i ) ∩ k = c i } . Now let y ∈ Y be any point such that ∀ i ∈ a y ( i ) ∩ k = c i and for some i ∈ a y ( i ) ⊂ k , and for another i ∈ a ω \ k ⊂ y ( i ). There issuch a point because | a | ≥ f ′′ ( f − { y } ∩ O ) is exactly the open set of all points y ∈ Y such that ∀ i ∈ a y ( i ) ∩ k = c i } . Example 3.4.
Let b be a finite set and let b = a ∪ a be a partition intononempty sets. Let X, Y , Y be the closed subsets of P ( ω ) b , P ( ω ) a , and P ( ω ) a consisting of tuples of pairwise disjoint subsets of ω respectively. Let f : X → Y and f : X → Y be the projection functions. Then f , f are continuous, open,and mutually transcendental functions. Proof.
The continuity and openness is left to the reader. For the transcendentalpart, I will show that f is transcendental over f . Let O ⊂ X be a relativelyopen nonempty set. Find finite sets c i , d i ⊂ ω for each i ∈ b such that c i ∩ d i = 0and the set {h z i : i ∈ b i ∈ P ( ω ) : ∀ i ∈ b c i ⊂ z i and d i ∩ z i = 0 }∩ X is a nonemptysubset of O . Note that the sets c i for i ∈ b must be pairwise disjoint, and wemay arrange the sets d i so that if i, j ∈ b are distinct elements then c i ⊂ d j .Let y = h c i : i ∈ a i and let O = {h z i : i ∈ a i ∈ P ( ω ) : ∀ i ∈ a c i ⊂ z i and d i ∩ z i = 0 } ∩ Y ; this is a nonempty open subset of Y . It is clear that for eachpoint y ∈ O , h y , y i ∈ O holds and the proof is complete.Another class of examples of transcendental pairs of generic extensions comesfrom actions of Polish groups with dense diagonal orbits [5]. I am going to needa local variant of this notion which appears to be satisfied in all natural actions7ith dense diagonal orbits. Recall that if a group G acts on a set X , then italso acts coordinatewise on the set X n for every natural number n . Definition 3.5.
Let G be a Polish group acting continuously on a Polish space X . The action has1. dense diagonal orbits if for every n ∈ ω there is a point ~x ∈ X n such that { g · ~x : g ∈ G } is dense in X n .2. locally dense diagonal orbits if for every open neighborhood U ⊂ G of theunit and every nonempty open set O ⊂ X there is a nonempty open set O ′ ⊂ O such that for every n ∈ ω there is a point ~x ∈ X n such that { g · ~x : g ∈ U } is dense in ( O ′ ) n . Proposition 3.6.
Let G be a Polish group acting on a Polish space X withlocally dense diagonal orbits. Let Y ⊂ G × X be the closed set of all triples h g, x , x i such that g · x = x . Let P Y be its associated Cohen forcing and h ˙ g, ˙ x , ˙ x i its names for the generic triple. P Y forces the following:1. ˙ g is P G -generic over V ;2. h ˙ x , ˙ x i is P X -generic over V ;3. the model V [ ˙ x , ˙ x ] is transcendental over V [ ˙ g ] .Proof. For the first item, let p ∈ P Y be a condition and D ⊂ G an open denseset. I must find a stronger condition which forces ˙ g into D . There are nonemptyopen neighborhoods U ⊂ G and O ⊂ X such that h g, x, g · x i ∈ p whenever g ∈ U and x ∈ O . Now, just note that the set D ∩ U is nonempty; therefore the set ofall triples h g, x, g · x i where g ∈ U ∩ D and x ∈ O is a nonempty relatively opensubset of Y which forces ˙ g ∈ D as desired.For the second item, suppose first that p ∈ P Y is a condition and D ⊂ X is an open dense set. I must find a stronger condition which forces the pair h ˙ x , ˙ x i into D . There is a point g ∈ G , an open neighborhood U ⊂ G of theunit, and an open set O ⊂ X such that h gh, x , gh · x i ∈ p for all h ∈ U U − and x ∈ O . Use the dense orbit assumption to thin out the set O if necessaryso that for every n ∈ ω there is a point ~x ∈ X n such that { h · ~x : h ∈ U } isdense in O n . Since the set D ⊂ X is open dense, there are open sets P ⊂ O and P ⊂ gO such that P × P ⊂ D . By the choice of the set O ⊂ X , theremust be a point x ∈ P and a point h ∈ U U − such that hx ∈ g − P , in otherwords ghx ∈ P . Now the relatively open set of all triples in p such that theirsecond and third coordinates belong to P and P respectively is nonempty, andit forces h ˙ x , ˙ x i ∈ D as required.For the third item, suppose that α is an ordinal, τ is a P X -name for anopen subset of 2 α which is forced to contain V ∩ α , and η is a P G -name for anelement of 2 α . Suppose that p ∈ P Y is a condition. One can find an element h ∈ G , an open neighborhood U ⊂ G of the unit, and a nonempty open set O ⊂ X such that h gh, x , gh · x i ∈ p for all h ∈ U U − and x ∈ O . Use the8ense orbit assumption to thin out the set O if necessary so that for every n ∈ ω there is a point ~x ∈ X n such that { h · ~x : h ∈ U } is dense in O n .Now, use the initial assumption on the name τ to find, for each z ∈ α , afinite partial map h z : α → O z × O z ⊂ O × gO such that h z ⊂ z and O z × O z (cid:13) [ h z ] ⊂ τ . Use a compactness argument to find afinite set a ⊂ α such that 2 α ⊂ S z ∈ a [ h z ]. Now, the dense orbit assumptionprovides points x z ∈ O z and a point h ∈ U U − such that for each z ∈ a , h · x z ∈ g − O z , or in other words gh · x z ∈ O z . Let U ′ ⊂ U U − be an openneighborhood such that for all k ∈ U ′ and all z ∈ a , gk · x z ∈ O z . Shrinking U ′ if necessary, assume that gU ′ decides the value of η ↾ S z ∈ a dom( h z ). By thechoice of the set a , there must be a point z ∈ a such that gU ′ (cid:13) ˇ h z ⊂ η . Nowthe relatively open set of all triples in p whose coordinates belong to gU ′ , O z and O z respectively is nonempty, and it forces η ∈ τ as desired. Example 3.7.
Let Y be the closed subset of S ∞ consisting of all quadruples h g , g , g , g i such that g g − g g − = 1. The Cohen poset P Y adds a genericquadruple h ˙ g , ˙ g , ˙ g , ˙ g i . It forces V [ ˙ g , ˙ g ] and V [ ˙ g , ˙ g ] to be mutually tran-scendental P S ∞ -generic extensions of the ground model. Proof.
Consider the continuous action of S ∞ on S ∞ given by ( h , h ) · h = h h h − . It has locally dense diagonal orbits: if U ⊂ ( S ∞ ) is an open neigh-borhood of the unit and O ⊂ S ∞ is a nonempty open set, then thinning downone may assume that there is a number n ∈ ω such that U = {h h , h i : h ↾ n = h ↾ n is the identity } and O = { h : [ v ] ⊂ h } for some permutation v of n . Then the action of U on O is naturally homeomorphic to the whole action of S ∞ on S ∞ . That action though has dense diagonal orbits because already theconjugation action of S ∞ on S ∞ has them [5].Now, to show for example that P Y forces V [ ˙ g , ˙ g ] to be transcendental over V [ ˙ x , ˙ x ], consider the self-homeomorphism of S ∞ which takes inverses of thesecond and third coordinates. Note that h g , g , g , g i ∈ Y iff ( g , g − ) · g − = g and apply Proposition 3.6.The last class of examples in this section deals with posets other than the Cohenposet. Definition 3.8.
Let P be a Suslin partial order. Say that P is Suslin- σ -centered if P = S n ∈ ω A n where each set A n ⊂ P is analytic and centered. Proposition 3.9.
Let P be a Suslin poset which is Suslin- σ -centered. Let V [ G ] be an arbitrary generic extension, let H ⊂ P be a filter generic over V [ G ] and let G = H ∩ V . Then the extensions V [ G ] , V [ G ] are mutuallytranscendental.Proof. I use an apparently novel abstract combinatorial property of σ -linkedposets encapsulated in the following claim. Claim 3.10.
Let Q be a σ -linked poset, and B ⊂ [ Q ] < ℵ be a family of finitesubsets of Q such that for every condition q ∈ Q there is a set in B consisting nly of conditions stronger than q . Then there is a countable set C ⊂ B suchthat every condition in Q is compatible with every condition in some set in C .Proof. Let Q = S n D n be a partition of Q into linked subsets. Let M be acountable elementary submodel of a large structure containing this partition andthe set B ; we claim that C = B ∩ M works. Suppose that q ∈ Q is a conditionand find a set b ∈ B which consists solely of conditions stronger than q . Let e = { n ∈ ω : b ∩ D n = 0 } ; this is a finite set and therefore e ∈ M . By elementarityof the model M , there is a set c ∈ C such that e = { n ∈ ω : c ∩ D n = 0 } . Sincethe sets D n for n ∈ ω are linked, it follows that q is compatible with eachelement of c as desired.Let P be the poset generating the V [ G ] extension and work in V . Fix acover P = S n A n consisting of analytic centered sets. To see that V [ G ] istranscendental over V [ G ], work in V and fix and ordinal α . Suppose that τ isa P -name for an open subset of 2 α such that P (cid:13) ω ∩ V ⊂ τ . We will arguethat P ∗ ˙ P (cid:13) ω ∩ V [ ˙ G ] ⊂ τ / ˙ G . To this end, for each condition p ∈ P let O p be the union of all basic open subsets of 2 α which p forces to be subsets of τ ; inparticular, p (cid:13) O p ⊂ τ . Let B ⊂ P < ℵ be the set of all finite sets b ⊂ P suchthat S p ∈ b O p = 2 α . Now, fix a condition q ∈ P . As q (cid:13) α ∩ V ⊂ τ , it mustbe the case that 2 α = S p ≤ q O p holds, and by a compactness argument, thereis a set b ∈ B consisting only of conditions stronger than q . Thus, assumptionsof the claim are satisfied, and there must be a countable set C ⊂ B such thatevery condition in P is compatible with all conditions in some set in C . Notethat this property of the set C persists to the model V [ G ] by the Shoenfieldabsoluteness.Now, move to the model V [ G ], pick a point x ∈ α , and a condition p ∈ P ;we must find a strengthening of p which forces ˇ x ∈ τ / ˙ G . By the work in theprevious paragraph, there is a set c ∈ C such that p is compatible with allconditions in it. Since S { O p : p ∈ c } = 2 α , there is a condition p ∈ c suchthat x ∈ O p . Any common lower bound of p and p forces ˇ x ∈ τ / ˙ G asrequired.To see that V [ G ] is transcendental over V [ G ], let τ be a P -name for anopen subset of 2 α such that P (cid:13) α ∩ V ⊂ ˙ O . Similarly to the previous work,for each condition p ∈ P let O p be the union of all basic open subsets of 2 α which p forces to be subsets of τ and let B ⊂ P < ℵ be the set of all finite sets b ⊂ P such that S p ∈ b O p = 2 α . Thus, for every condition p ∈ P there is aset b ∈ B consisting of conditions stronger than p . Now, let η be a P -namefor an element of 2 α , and let h p , ˙ p i be a condition in the iteration P ∗ ˙ P .We must find a strengthening which forces η ∈ τ / ˙ G . To this end, first use thecenteredness assumption again to strengthen p if necessary to find a number n ∈ ω such that p (cid:13) ˙ p ∈ ˙ A n . Let b ∈ B be a set consisting of conditionsstronger than p , and choose an enumeration b = { p i : i ∈ m } . Let { G i : i ∈ m } be a mutually generic collection of filters on P such that p i ∈ G i holds for all i ∈ m . In the model V [ G i : i ∈ m ], note that the conditions ˙ p /G i ∈ A n havea common lower bound in the poset P ; let H ⊂ P be a filter generic over the10odel V [ G i : i ∈ m ] containing them all. Since S i O p i = 2 α , there must bea number i ∈ m such that η/H ∩ V ∈ τ /G i . This membership relation is astatement of the model V [ G i ][ H ∩ V [ G i ]] which is a P ∗ ˙ P extension of V . Thus,there has to be a condition stronger than h p , ˙ p i forcing it as required. Example 3.11.
Let x ∈ ω ω be a point Hechler-generic over the ground model,and x ∈ ω ω a point Hechler generic over V [ x ]. Then V [ x ] and V [ x ] aremutually transcendental extensions of V . Example 3.12.
One cannot weaken the σ -centeredness assumption in Exam-ple 3.9 to σ -linkedness. To see this, let X be a compact metric space with aBorel probability measure µ . Let V [ G ] be a generic extension of V in whichthere is a compact µ -positive set C ⊂ X containing no points of V ∩ X . Let P be the random forcing associated with µ and let x ∈ C be a point P -genericover V [ G ]. Then V [ G ] is not compactly transcendental to V [ x ] since the com-plement of C is an open set covering X ∩ V but not X ∩ V [ x ]. Note that V [ x ]is transcendental to V [ G ] by the first half of Proposition 3.9, so transcendenceis not a symmetric property of pairs of extensions. As with all similar notions of perpendicularity of generic extensions, transcen-dence gives rise to a natural companion: a preservation property for Suslinforcings.
Definition 4.1.
Let P be a Suslin forcing. We say that a virtual condition ¯ p in P is transcendentally balanced if for every pair of mutually transcendentalgeneric extensions V [ G ] , V [ G ] inside some ambient forcing extension, and forall conditions p ∈ V [ G ] and p ∈ V [ G ] stronger than ¯ p , p and p have acommon lower bound.I now state and several preservation theorems for transcendentally balancedextensions of the symmetric Solovay model. Theorem 4.2.
Let κ be an inaccessible cardinal. In cofinally transcendentallybalanced forcing extensions of the symmetric Solovay model derived from κ ,every nonmeager subset of P ( ω ) contains a collection d of cardinality four suchthat T d = 0 and S d = ω , both modulo finite.Proof. Let P be a Suslin forcing which is cofinally transcendentally balancedbelow κ . Let W be the symmetric Solovay model derived from κ and work in themodel W . Suppose that p ∈ P is a condition, τ is a P -name, and p (cid:13) τ ⊂ P ( ω )is a nonmeager set. I must find a set d ⊂ P ( ω ) of size four such that T d = 0and S d = ω , both modulo finite, and a strengthening of the condition p whichforces ˇ d ⊂ τ .To this end, let z ∈ ω be a point such that p, τ are both definable fromthe parameter z and some parameters in the ground model. Let V [ K ] be an11ntermediate forcing extension obtained by a poset of cardinality less than κ such that z ∈ V [ K ] and V [ K ] | = P is transcendentally balanced. Work in V [ K ]. Let ¯ p ≤ p be a transcendentally balanced virtual condition. Let Q bethe Cohen poset of nonempty open subsets of P ( ω ), adding a single genericpoint ˙ z . There must be a condition q ∈ Q and a poset R of cardinality smallerthan κ and an 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) P ˙ z ∈ τ . Otherwise, in the model W the condition¯ p would force τ to be disjoint from the co-meager set of elements of P ( ω ) whichare Cohen-generic over V [ K ], contradicting the initial assumption on τ .Now, let X = {h x i ∈ P ( ω ) : S i ∈ x ( i ) = ω and T i ∈ x ( i ) = 0 } with thetopology inherited from P ( ω ) . Let x ∈ X be a point generic over V [ K ] forthe Cohen poset with X . By Example 3.3, x (0) , x (1) are mutually Cohen-generic elements of P ( ω ), so are x (2) , x (3), and the models V [ K ][ x (0) , x (1)]and V [ K ][ x (2) , x (3)] are mutually transcendental. Choose finite modifications z i of x i such that z i ∈ q holds for all i ∈
4; each of these points is still Q -generic over V [ K ] and meets the condition q . Let H i : i ∈ R mutually generic over the model V [ K ][ x ] and let p i = σ/z i , H i . By Proposi-tion 2.2, conclude that the models V [ K ][ z ][ H ] and V [ K ][ z ][ H ] are mutu-ally generic extensions of V [ K ], so are V [ K ][ z ][ H ] and V [ K ][ z ][ H ], and themodels V [ K ][ z , z ][ H , H ] and V [ K ][ z , z ][ H , H ] are mutually transcenden-tal extensions of V [ K ]. Now, the balance assumption on the virtual condition ¯ p ,we see that the conditions p , p have a common lower bound p in the model V [ K ][ z , z ][ H , H ], the conditions p and p have a common lower bound p in the model V [ K ][ z , z ][ H , H ], and finally the conditions p and p havea common lower bound as well. The forcing theorem then shows that such alower bound then forces in the model W that ˇ z i ∈ τ holds for all i ∈
4. Theproof is complete.For every number n ≥ n be the hypergraph of arity n on P ( ω ) consistingof sets d of size n such that T d = 0 and S d = ω , both modulo finite. Corollary 4.3.
Let κ be an inaccessible cardinal. In cofinally transcendentallybalanced forcing extensions of the symmetric Solovay model derived from κ , thechromatic number of Θ is uncountable. Theorem 4.4.
Let κ be an inaccessible cardinal. In cofinally transcendentallybalanced forcing extensions of the symmetric Solovay model derived from κ ,every nonmeager subset of S ∞ contains a quadruple of distinct points solvingthe equation g g − g g − = 1 .Proof. Let P be a Suslin forcing which is cofinally transcendentally balancedbelow κ . Let W be the symmetric Solovay model derived from κ and work in themodel W . Suppose that p ∈ P is a condition, τ is a P -name, and p (cid:13) τ ⊂ S ∞ is a nonmeager set. I must find distinct points z , z , z , z ∈ S ∞ such that z z − z z − = 1 and a strengthening of the condition p which forces all four ofthese points into τ .To this end, let z ∈ ω be a point such that p, τ are both definable fromthe parameter z and some parameters in the ground model. Let V [ K ] be an12ntermediate forcing extension obtained by a poset of cardinality less than κ such that z ∈ V [ K ] and V [ K ] | = P is transcendentally balanced. Work in V [ K ]. Let ¯ p ≤ p be a transcendentally balanced virtual condition. Let Q be the Cohen poset of nonempty open subsets of S ∞ , adding a single genericpoint ˙ g . There must be a condition q ∈ Q and a poset R of cardinality smallerthan κ and an 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) P ˙ g ∈ τ . Otherwise, in the model W the condition¯ p would force τ to be disjoint from the co-meager set of elements of S ∞ whichare Cohen-generic over V [ K ], contradicting the initial assumption on τ .Now, let X = { x ∈ S ∞ : x (0) x (1) − x (2) x (3) − = 1 } with the topology in-herited from S ∞ . Consider the nonempty relatively open set O ⊂ X given by O = q ∩ X . Note that the set O is indeed nonempty because any constantquadruple in S ∞ belongs to X . Let h z i : i ∈ i ∈ O be a tuple generic over V [ K ]for the Cohen poset with X . By Example 3.7, z , z are mutually Cohen-genericelements of S ∞ below the condition q , so are z , z , and the models V [ K ][ z , z ]and V [ K ][ z , z ] are mutually transcendental. Let H i : i ∈ R mutually generic over the model V [ K ][ z , z , z , z ] and let p i = σ/g i , H i . ByProposition 2.2, conclude that the models V [ K ][ z ][ H ] and V [ K ][ z ][ H ] aremutually generic extensions of V [ K ], so are V [ K ][ z ][ H ] and V [ K ][ z ][ H ], andthe models V [ K ][ z , z ][ H , H ] and V [ K ][ z , z ][ H , H ] are mutually transcen-dental extensions of V [ K ]. Now, the balance assumption on the virtual condition¯ p , we see that the conditions p , p have a common lower bound p in the model V [ K ][ z , z ][ H , H ], the conditions p and p have a common lower bound p in the model V [ K ][ z , z ][ H , H ], and finally the conditions p and p havea common lower bound as well. The forcing theorem then shows that such alower bound then forces in the model W that ˇ z i ∈ τ holds for all i ∈
4. Theproof is complete.
Corollary 4.5.
Let κ be an inaccessible cardinal. In cofinally transcendentallybalanced forcing extensions of the symmetric Solovay model derived from κ ,the chromatic number of the hypergraph on S ∞ consisting of solutions to theequation g g − g g − = 1 is uncountable. Certain consistency results require amalgamation diagrams with multiple forcingextensions. The following definitions and a theorem show one such possibility.
Definition 4.6.
A finite collection { V [ G i ] : i ∈ a } of generic extensions is mu-tually transcendental if for every index j ∈ a , the models V [ G j ] and V [ G i : i ∈ a, i = j ] are mutually transcendental. The collection is in n -tuples mutuallytranscendental if every subcollection of size n is mutually transcendental. Definition 4.7.
Let m > n be natural numbers. Let P be a Suslin forcing.1. A virtual condition ¯ p in P is m, n - transcendentally balanced if for everytuple h V [ G i ] : i ∈ m i of generic extensions, mutually transcendental in n -tuples, and conditions p i ≤ ¯ p in the respective models V [ G i ], the con-ditions p i for i ∈ m have a common lower bound.13. The poset P is m, n - transcendentally balanced if below every condition p ∈ P there is an m, n -transcendentally balanced virtual condition. Theorem 4.8.
Let κ be an inaccessible cardinal and n ≥ be a number. Incofinally n + 1 , n -transcendentally balanced forcing extensions of the symmetricSolovay model derived from κ , every nonmeager subset of P ( ω ) contains n + 1 many sets which modulo finite form a partition of ω .Proof. Let P be a Suslin forcing which is cofinally n + 1 , n -transcendentallybalanced below κ . Let W be the symmetric Solovay model derived from κ andwork in the model W . Suppose that p ∈ P is a condition, τ is a P -name, and p (cid:13) τ ⊂ P ( ω ) is a nonmeager set. I must find a collection { a i : i ∈ n + 1 } whichis modulo finite a partition of ω and a condition stronger than p which forcesevery element of this collection into τ .To this end, let z ∈ ω be a point such that p, τ are both definable fromthe parameter z and some parameters in the ground model. Let V [ K ] be anintermediate forcing extension obtained by a poset of cardinality less than κ such that z ∈ V [ K ] and V [ K ] | = P is n + 1 , n -transcendentally balanced. Workin V [ K ]. Let ¯ p ≤ p be a n + 1 , n -transcendentally balanced virtual condition.Let Q be the Cohen poset of nonempty open subsets of P ( ω ), adding a singlegeneric point ˙ a . There must be a condition q ∈ Q and a poset R of cardinalitysmaller than κ and an Q × R -name σ for a condition in P stronger than ¯ p suchthat q (cid:13) Q R (cid:13) Coll( ω, < κ ) (cid:13) σ (cid:13) P ˙ a ∈ τ . Otherwise, in the model W thecondition ¯ p would force τ to be disjoint from the co-meager set of elements of P ( ω ) which are Cohen-generic over V [ K ], contradicting the initial assumptionon τ .Let X be the closed subset of P ( ω ) n +1 consisting of tuples of sets whichform a partition of ω . For every set b ⊂ n + 1 of cardinality n , consider theclosed subset Y b of P ( ω ) b consisting of pairwise disjoint sets. It is easy to checkthat the projection map from X to Y b is open. Thus, the poset P X of nonemptyrelatively open subsets of X adds a generic tuple ˙ x , and the restriction ˙ x ↾ b is generic for the poset P Y b of nonempty relatively open subsets of Y by [7,Proposition 3.1.1]. By Example 3.4, P X forces the tuple { V [ K ][ ˙ x ( i )] : i ∈ n + 1 } to be in n -tuples mutually transcendental over V [ K ].Now, move to W and find an n + 1-tuple x which is generic over the model V [ K ] for the poset P X . Let H i ⊂ R for i ∈ n + 1 be a collection of filtersmutually generic over the model V [ K ][ x ]; use Proposition 2.2 and the work inthe previous paragraph to argue that the tuple { V [ K ][ x i ][ H i ] : i ∈ n + 1 } is in n -tuples mutually transcendental. For each i ∈ n + 1, make a finite adjustmentto x ( i ) so that the resulting set a i ⊂ ω meets the condition q ; note that a i is Q -generic over V [ K ]. Let p i = σ/a i , H i . The balance assumption on the virtualcondition ¯ p shows that the conditions p i for i ∈ n + 1 have a common lowerbound. That lower bound forces each set a i for i ∈ ω into τ as required. Corollary 4.9.
Let n ≥ be a natural number and Γ n be the hypergraph on P ( ω ) of n + 1 -tuples which form a modulo finite partition of ω . In cofinally + 1 , n -transcendentally balanced forcing extensions of the symmetric Solovaymodel derived from κ , the chromatic number of Γ n +1 is uncountable. One can also prove a preservation theorem of a different type:
Theorem 4.10.
Let κ be an inaccessible cardinal. In cofinally transcendentallybalanced forcing extensions of the symmetric Solovay model derived from κ ,for every uncountable set A ⊂ ω ω there is a function in ω ω which pointwisedominates uncountably many elements of A .Proof. Let P be a Suslin forcing which is cofinally transcendentally balancedbelow κ . Let W be the symmetric Solovay model derived from κ , let n > W . Suppose that p ∈ P is a conditionand τ is a P -name such that p (cid:13) τ ⊂ ω ω is an uncountable set. I must find afunction h ∈ ω ω and a condition stronger than p which forces that ˇ h dominatesuncountably many elements of τ .To this end, let z ∈ ω be a point such that p, τ are both definable fromthe parameter z and some parameters in the ground model. Let V [ K ] be anintermediate forcing extension obtained by a poset of cardinality less than κ suchthat z ∈ V [ K ] and V [ K ] | = P is transcendentally balanced. Work in V [ K ]. Let¯ p ≤ p be a transcendentally balanced virtual condition. Since the set τ is forcedto be uncountable, there must be a poset Q and Q -names η for an element of ω ω which is not in V [ K ] and σ for a condition in P stronger than ¯ p such that Q (cid:13) Coll( ω, < κ ) (cid:13) σ (cid:13) η ∈ τ . Now, let H ⊂ Q be a filter generic overthe model V [ K ], let p = σ /H and let x = η /H . Let h ∈ ω ω be a functionHechler generic over the model V [ K ][ H ] which pointwise dominates x . ByProposition 3.9, V [ K ][ H ] and V [ K ][ h ] are mutually transcendental extensionsof V [ K ].We claim that in the model W , ¯ p (cid:13) ˇ h pointwise dominates uncountably manyelements of τ . This will complete the proof. Suppose towards contradiction thatthis fails. Work in the model V [ K ][ h ]. There must be a poset Q of cardinalityless than κ , a Q -name η for a countable sequence of elements of ω ω , and a Q -name σ for a condition in P stronger than ¯ p such that Q (cid:13) Coll( ω, < κ ) (cid:13) σ (cid:13) P η enumerates all elements of τ pointwise dominated by ˇ h .Now, in the model W find a filter H ⊂ Q generic over V [ K ][ H ][ h ]. ByProposition 2.2 V [ K ][ H ] and V [ K ][ h ][ H ] are mutually transcendental exten-sions of V [ K ]. Let p = σ /H and y = η /H ; these objects belong to themodel V [ K ][ h ][ H ]. By the mutual transcendenceof the models V [ K ][ H ] and V [ K ][ h ][ H ], x / ∈ V [ K ][ h ][ H ]; in particular, x / ∈ rng( y ). By the mutual tran-scendence and the balance of the condition ¯ p , the conditions p , p are compat-ible in P . Their common lower bound forces ˇ x ∈ τ as well as ˇ y to enumerateall elements of τ pointwise dominated by ˇ h . This is impossible as ˇ x is pointwisedominated by h and does not belong to the range of y .15 Examples II
The whole enterprise in the previous sections would be pointless if there wereno substantial transcendentally balanced posets. In this section, I will produceor point out a number of examples in this direction. At first, I consider posetsor classes of posets known from previous work.
Proposition 5.1.
Every placid Suslin poset is transcendentally balanced.
This class of examples is very broad: it includes among others posets adding aHamel basis for a Polish space over a countable field, posets adding maximalacyclic subsets to Borel graphs, or posets adding a selector to pinned Borelequivalence relations classifiable by countable structures.
Proof.
Recall [7, Definition 9.3.1] that a poset P is placid if below every condi-tion p ∈ P there is a virtual balanced condition ¯ p ≤ p which is placid: whenever V [ G ] and V [ G ] are generic extensions such that V [ G ] ∩ V [ G ] = V and p ∈ V [ G ] and p ∈ V [ G ] are conditions stronger than ¯ p , then p , p arecompatible. Now, if V [ G ] , V [ G ] are mutually transcendental extensions of theground model, then V [ G ] ∩ V [ G ] = V by Proposition 2.4, and therefore a placidvirtual condition also transcendentally balanced. The proposition follows. Proposition 5.2.
Let X be a K σ Polish field with a countable subfield F . Theposet adding a transcendence basis to X over F is transcendentally balanced.Proof. Reviewing the proof of [7, Theorem 6.3.9] it becomes clear that theonly feature of mutually generic extensions V [ G ] and V [ G ] there is that if p is a multivariate polynomial with coefficients in F , ~x ∈ X ∩ V [ G ] and ~x ∈ X ∩ V [ G ] are tuples such that p ( ~x , ~x ) = 0, then there are tuples ~x ′ , ~x ′ inthe ground model such that p ( ~x ′ , ~x ) = p ( ~x , ~x ′ ) = 0. However, this is satisfiedfor mutually transcendental extensions V [ G ] , V [ G ] as well by Corollary 2.6.This completes the proof. Proposition 5.3.
Let E be an equivalence relation on a Polish space of one ofthe following types:1. E is K σ ;2. for some sequence h Y n , d n : n ∈ ω i of countable metric spaces, E is theequivalence relation on X = Q n Y n connecting points x , x if the dis-tances d n ( x ( n ) , x ( n )) tend to zero as n tends to infinity.The poset adding a selector to E is transcendentally balanced.Proof. Note that the equivalence relation E is pinned ([4, Chapter 17], but itfollows directly from Corollary 2.7 or Proposition 2.8) and therefore [7, Theorem6.4.5] applies. The only feature of mutually generic extensions V [ G ] and V [ G ]in the proof of the balance of P is that every E -class represented both in V [ G ]and V [ G ] is represented in V . However, this feature holds true for mutuallytranscendental extensions by Corollary 2.7 or Proposition 2.8.16ow it is time to produce transcendentally balanced posets for some new andmore difficult tasks. I will only look at coloring posets for hypergraphs of acertain type. Definition 5.4.
Let n ≥ X a Polish space, and Γ ahypergraph on X of arity n .1. Γ is redundant if for every set a ⊂ X of cardinality n −
1, the set { x ∈ X : a ∪ { x } ∈ Γ } is countable.If Γ is redundant, then2. a set b ⊂ X is Γ- closed if for every set a ⊂ b of cardinality n − { x ∈ X : a ∪ { x } ∈ Γ } is a subset of b ;3. if a set b ⊂ X is Γ-closed, define the equivalence relation E ( b, Γ) on X \ b as the smallest equivalence containing all pairs { x , x } such that for someset a ⊂ b of cardinality n − a ∪ { x , x } ∈ Γ. Example 5.5.
The hypergraph Γ on R of arity 3 consisting of solutions to theequation x + y + z − xyz = 0 is redundant. Every real closed subfield of R is Γ-closed. Example 5.6.
The hypergraph Γ on R consisting of vertices of equilateraltriangles is redundant. A similar hypergraph on R is not redundant. Example 5.7.
Let n ≥ n on P ( ω ) consist-ing of n -tuples which modulo finite partition ω is redundant. Every Booleansubalgebra of P ( ω ) containing all singletons is Γ n -closed. Example 5.8.
Let G be a Polish group with a countable dense subset d ⊂ G .Let n ≥ G, n ) consisting of all n -tuples whose product belongs to d is redundant. Note that if G is not abelian,then the product depends on the order of the elements, so one must say “theproduct of all elements in some order belongs to d ”. Example 5.9.
Let G be a Polish group and n ≥ G, n ) of all 2 n -tuples whose alternating product g g − g g − . . . in some order is equal to 1 is redundant.In the common case that the set b ⊂ X is countable, the equivalence relation E ( b, Γ) has all classes countable since it is the path connectedness equivalenceof a locally countable graph. If the redundant hypergraph Γ is Borel, then theequivalence relation E ( b, Γ) is Borel as well. The complexity of E ( b, Γ) never en-ters the considerations of this paper; it can be an arbitrarily complex countableBorel equivalence relation, but in the natural examples it is hyperfinite. Forthis paper, the following feature of these equivalences is much more relevant.17 roposition 5.10.
Let X be a K σ Polish space and Γ a redundant F σ hy-pergraph of arity n ≥ on X . Let V [ G ] , V [ G ] be mutually transcendentalgeneric extensions of the ground model. Then on X ∩ V [ G ] , E ( V ∩ X, Γ) = E ( V [ G ] ∩ X, Γ) .Proof. The left-to-right inclusion is obvious as increasing the set b increasesthe equivalence relation E ( b, Γ). The right-to-left inclusion is the heart of thematter. Suppose that x, x ′ ∈ X ∩ V [ G ] are two points in X ∩ V [ G ] whichare E ( X ∩ V [ G ] , Γ) equivalent. Then there must be a number m ∈ ω , points x i ∈ X for i ≤ m and sets y i ∈ [ X ∩ V [ G ]] n − for i < m such that x = x , x ′ = x m , and ∀ i < m y i ∪ { x i , x i +1 } ∈ Γ. The tuple h x i : i ≤ m, y i : i < m i willbe called a walk from x to x ′ .Let K ⊂ X be a compact set coded in the ground model containing all pointsmentioned in the walk. Let d be a complete metric on X and let ε > d -distance at least ε . Let ∆ ⊂ Γ be a ground modelcoded compact set such that all hyperedges in the walk belong to ∆.Now, consider the space Y = {h z i : i ∈ m i : z i ∈ [ K ] n − and distinct pointsin each z i have a distance at least ε } ; this is a compact subspace of ([ K ] n − ) m in the ground model. Consider the set C ⊂ Y consisting of tuples h z i : i ∈ m i which can serve in a walk from x to x ′ which uses only points in K , whosehyperedges belong to ∆, and in which any two distinct points have distance atleast ε . The set C ⊂ Y is compact, as it is a projection of a compact set ofwalks. The set C is coded in V [ G ], and the sequence h y i : i ∈ m i ∈ V [ G ]belongs to it. By the mutual transcendence of the models V [ G ] and V [ G ],the set C contains a ground model element. A review of definitions reveals thatthis means that x, x ′ are E ( X ∩ V, Γ)-related. F σ redundant hypergraphs of arity three or four on K σ spaces can be coloredby a transcendentally balanced Suslin forcing. The following definition andtheorem provide a general treatment. However, in certain special cases it ispossible to find posets which have stronger preservation properties, as is donein [7, Section 8.2]. Arity five (and higher) presents challenges that I do notknow how to overcome in this generality, as the discussion of configurations inthe proof of Claim 5.13 becomes untenable. Definition 5.11.
Let X be a K σ Polish space and Γ be a redundant F σ hyper-graph of arity three or four on X . The coloring poset P Γ consists of all partialΓ colorings p : X → ω × ω whose domain is a countable Γ-closed subset of X .The ordering is defined by p ≤ p if p ⊂ p and for every E (Γ , dom( p ))-class a ⊂ dom( p ), p ′′ a ⊂ ω × ω has all vertical sections finite (and, if the arity of Γis four, the function p ↾ a is an injection). Theorem 5.12.
Let Γ be a redundant F σ hypergraph of arity three on a K σ Polish space X . The poset P Γ is Suslin and σ -closed, and it forces the union ofthe generic filter to be a total Γ -coloring on X . Moreover, . for every total Γ -coloring c : X → ω × ω , the pair h Coll( ω, X ) , ˇ c i is tran-scendentally balanced;2. for every balanced pair h Q, τ i there is a total coloring c : X → ω × ω suchthat the pairs h Q, τ i and h Coll( ω, X ) , ˇ c i are equivalent;3. distinct total Γ -colorings provide inequivalent balanced pairs.In particular, under CH, the poset P Γ is transcendentally balanced.Proof. The easiest part is the σ -closure. If h p i : i ∈ ω i is a decreasing sequenceof conditions in P Γ , then S i p i is their common lower bound. For the propertyof the generic filter, I need to show that for every condition p and a point x ∈ X there is a condition p ≤ p such that x ∈ p . To see that, just find a countableΓ-closed set b ⊂ X such that dom( p ) ∪ { x } ⊂ b , enumerate the set b \ dom( p )by { x n : n ∈ ω } , and define a function p : b → ω × ω by the demands p ⊂ p and p ( x n ) = ( n, n ). It is not difficult to see that p ∈ P Γ and p ≤ p is asrequired.It is clear that the ordering on P Γ is a Borel relation. Borelness of thecompatibility relation of the poset P Γ follows immediately from the followingclaim, which is used later as well. Claim 5.13.
Let p , p ∈ P Γ be conditions. Then p is compatible with p ifand only if the conjunction of the following item occurs:1. p ∪ p is a function;2. for every E (dom( p ) , Γ) -class a , p ′′ a ⊂ ω × ω has all vertical sections finite(and, if the arity of Γ is four, the function p ↾ a is injective);3. the same demand as in (2) except with , interchanged.Proof. The failure of any of the items excludes the existence of the lower commonbound by the definition of the ordering. Now, suppose that the items are satis-fied. Let b ⊂ X be a countable Γ-closed set such that dom( p ) ∪ dom( p ) ⊂ b ,write c = b \ (dom( p ) ∪ dom( p )), enumerate c as { x n : n ∈ ω } and choose afunction p : b → ω × ω so that p ∪ p ⊂ p and for every n ∈ ω , p ( x n ) = ( n, m )so that the point ( n, m ) does not belong to any of the sets p ′′ a where a is the E (dom( p ) , Γ)-class such that x n ∈ a , or p ′′ a where a is the E (dom( p ) , Γ)-classsuch that x n ∈ a . This is possible by items (2) and (3). I will show that p ∈ P Γ and p is a lower bound of p , p .To show that p ∈ P Γ , it is enough to verify that p is a Γ-coloring. Supposethat e is a Γ-hyperedge, and work to show that p ↾ e is not constant. We firstdeal with the case where the arity of Γ is three. Case 1. e ∩ c contains more than one element. Then p ↾ e is not constant since p ↾ c is injective. Case 2. e ∩ dom( p ) contains more than one element. By the Γ-closure ofdom( p ) this means that e ⊂ dom( p ) and so p ↾ e is not constant since p is aΓ-coloring. 19 ase 3. e ∩ dom( p ) contains more than one element. This case is symmetricto Case 2. Case 4.
The last configuration is that the sets c , dom( p ), and dom( p ) con-tain one point of e each. Call these points y , y , y respectively. Then y is E (dom( p ) , Γ)-related to y . By the description of p then, p ( y ) = p ( y ) holds,and p ↾ e is not constant as desired.If the arity of Γ is four, there are more configurations to discuss, and in oneof them the injectivity demand will play a key role. Case 1. e ∩ c contains more than one element. Then p ↾ e is not constant since p ↾ c is injective. Case 2. e ∩ dom( p ) contains more than two elements. By the Γ-closure ofdom( p ) this means that e ⊂ dom( p ) and so p ↾ e is not constant since p is aΓ-coloring. Case 3. e ∩ dom( p ) contains more than two elements. This case is symmetricto Case 2. Case 4. e ∩ dom( p ) and e ∩ dom( p ) both contain exactly two elements.Writing { y , y } = e ∩ dom( p ), it follows that y , y are E (dom( p ) , Γ)-relatedand therefore by item (2) they receive distinct p -colors. Thus p ↾ e is notconstant. Case 5.
Both e ∩ c and e ∩ dom( p ) contain exactly one element. Denotingthese points by y , y respectively, it is clear that they are E (dom( p ) , Γ)-related.Thus, p ( y ) = p ( y ) by the choice of the function p , and p ↾ e is not constant. Case 6.
Both e ∩ c and e ∩ dom( p ) contain exactly one element. This case issymmetric to Case 5.Finally, I have to show that p is a common lower bound of p , p . By sym-metry, it is enough to show that p ≤ p holds. To verify that, note that forevery E (dom( p ) , Γ) class a ⊂ b , p ′′ a has all vertical sections finite by item (2)and moreover, p ′′ ( a \ dom( p )) has all vertical sections of cardinality at mostone by the choice of p . In total, p ′′ a has all vertical sections finite. If the arityof Γ is four, then the function p ↾ a is injective by item (2) and p ↾ a \ dom( p )is an injection and uses no values that p ↾ a uses by the choice of p . In total,the function p ↾ a is an injection. This shows that p ≤ p holds as required.Now, for the first item of the theorem. Suppose that c : X → ω × ω is a totalcoloring. Clearly, Coll( ω, X ) (cid:13) ˇ c ∈ P Γ holds. Now suppose that V [ G ] , V [ G ]are mutually transcendental generic extensions of the ground model and p ∈ V [ G ], p ∈ V [ G ] are conditions stronger than c . I must prove that the con-ditions are compatible. To do this, use Claim 5.13. It is clear that p ∩ p is afunction, since the domains of p and p intersect in X ∩ V , and on that set both p , p are equal to c . To verify the demand (2) of Claim 5.13, just note that E (dom( p ) , Γ) ↾ dom( p ) is equal to E ( X ∩ V, Γ) ↾ dom( p ) by Proposition 5.10and use the fact that p ≤ c holds. Demand (3) of Claim 5.13 is verified in asymmetric way.For the second item of the theorem, let h Q, τ i be a balanced pair. Strength-ening τ if necessary, we may assume that Q (cid:13) X ∩ V ⊂ dom( τ ). A balanceargument then shows that for every ground model point x ∈ X there is a pair20 ( x ) ∈ ω × ω such that Q (cid:13) τ (ˇ x ) = c ( x ). I claim that the Γ-coloring c worksas in (2). It will be enough to show that Q (cid:13) τ ≤ ˇ c . If this failed, then theremust be a condition q ∈ Q which forces that there is some E ( X ∩ V, Γ)-class a ⊂ dom( τ ) such that τ ↾ a has an infinite vertical section. Let G , G ⊂ Q be mutually generic filters containing the condition q and let p = τ /G and p = τ /G . The two conditions p , p should be compatible in P Γ , but thecontradictory assumption together with Claim 5.13 shows that they are not.The third item of the theorem is immediate. For the last sentence, supposethat the continuuum hypothesis holds and let p ∈ P Γ be a condition. I mustfind a total coloring c : X → ω × ω which is stronger than p in the sense of theordering on the poset P Γ . To do this, use the CH assumption to find a continuousincreasing sequence h M α : α ∈ ω i of countable elementary submodels of somelarge structure such that Γ , p ∈ M and X = S α M α . Find a total map c : X → ω × ω such that p ⊂ c and on each of the sets X ∩ M \ dom( p ), X ∩ M α +1 \ M α it is an injection with the range included in the diagonal on ω × ω . Note thateach set X ∩ M α is Γ-closed by elementarity, so any Γ-hyperedge which is nota subset of dom( p ) must have two elements coming from the same set on theabove list; in conclusion, the map c is a Γ-coloring. It is easy to check that c ≤ p as required.In arity four, there are many closed redundant hypergraphs such that theircoloring number being equal to ℵ is equivalent to the continuum hypothesis–for example, the hypergraph of rectangles in R [2]. Thus, in arity four the CHassumption in the theorem is to some extent necessary. I do not know if theassumption is also necessary in arity three in general. There are many specifichypergraphs in which the requisite colorings have been proved to exist in ZFC.The poset P Γ in arity three or four has an important additional preservationproperty. Theorem 5.14. (ZFC+CH) Let X be a K σ Polish space and Γ a redundant F σ -hypergraph on it of arity three (or four). The poset P Γ is , -transcendentallybalanced (or , -transcendentally balanced).Proof. I present the proof for arity three. In view of Theorem 5.12, it will beenough to show the following. If c : X → ω × ω is a total Γ coloring, and { V [ G i ] : i ∈ } is a collection or extensions in triples mutually transcendental, p i ∈ V [ G i ] for i ∈ c , then the conditions { p i : i ∈ } have a common lower bound.To do this, work in the model V [ G i : i ∈ b ⊂ X be a Γ-closed subsetof X such that S i p i ⊂ b . Let d = b \ S i dom( p i ) and enumerate the elementsof d as { x n : n ∈ ω } . Now, define a lower bound p of the conditions { p i : i ∈ } as a function with domain b such that S i p i ⊂ p and for each n ∈ ω , p ( x n ) issome pair ( n, m ) such that for no pair i, j ∈ a forthe E (dom( p i ) , Γ)-class to which x n belongs, ( n, m ) does not belong to p ′′ j a .First of all, observe that the requirements on p can be met. For any pair i, j ∈ V [ G i ] and V [ G j ] are mutually transcendental.By Proposition 5.10, the equivalences E (dom( p i ) , Γ) and E ( X ∩ V, Γ) coincide21n dom( p j ), and since p j ≤ c , it is the case that p ′′ j a has all vertical sectionsfinite. Thus the function p can be found as required.Now, I have to prove that p ∈ P Γ holds and p is a common lower bound ofall conditions p i for i ∈
4. This follows nearly literally the proof of Claim 5.13,with the consideration of one extra (impossible) configuration. Namely, it is im-possible for there to be distinct indices i , i , i and a hyperedge e ⊂ b such thateach set dom( p i k ) for k ∈ e . This is so becausethe redundancy of Γ together with a Mostowski absoluteness argument showsthat e ⊂ V [ G i , G i ] and V [ G i ] ∩ V [ G i , G i ] = V by the mutual transcendenceassumption.Finally, we are in position to prove the theorems in the introduction. Startin the symmetric Solovay model. For Theorem 1.1, let G be a K σ Polish groupand let ∆( G ) be the closed hypergraph of all solutions to g g − g g − = 1. It isimmediate that ∆( G ) is a redundant hypergraph of arity four. The poset P ∆( G ) of Definition 5.11 is transcendentally balanced under CH by Theorem 5.12. ByCorollary 4.5, in the P ∆( G ) extension of the Solovay model, ∆( G ) has countablechromatic number while ∆( S ∞ ) does not.For Theorem 1.2, consider the hypergraph Γ on P ( ω ) of arity three con-sisting of triples which modulo finite form a partition of ω . It is clearly aredundant hypergraph. The poset P Γ of Definition 5.11 is (under CH, but infact also in good old ZFC) 4 , P Γ -extension of the Solovay model, the chromatic numberof Γ is countable while that of Γ is uncountable. Theorem 1.3 is proved in thesame way. Finally, Theorem 1.4 just restates Corollary 4.3. References [1] Jack Ceder. Finite subsets and countable decompositions of Euclideanspaces.
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