aa r X i v : . [ m a t h . L O ] J a n Coloring redundant algebraic hypergraphs ∗ Jindˇrich ZapletalUniversity of FloridaJanuary 12, 2021
Abstract
We prove several consistency results in choiceless set theory ZF+DCregarding countable chromatic numbers of various algebraic hypergraphson Euclidean spaces.
This paper presents a major first step in a research program which compares col-orings of algebraic hypergraphs with other complicated subsets of Polish spacesin the context of choiceless set theory ZF+DC. A parallel project in the contextof ZFC theory has been completed successfully by Schmerl [7]. Schmerl showedthat the chromatic number χ (Γ) of any algebraic hypergraph Γ on a Euclideanspace satisfies an interesting multichotomy: either ZFC proves χ (Γ) is count-able, or ZFC proves that χ (Γ) is uncountable, or there is a natural number n ∈ ω such that ZFC proves that χ (Γ) is countable if and only if 2 ℵ ≤ ℵ n . Inaddition, there is a computer program which determines, for a given algebraicequation, which item of the multichotomy is satisfied for the hypergraph givenby the equation.A parallel program in ZF+DC can hardly have such a satisfactory resolu-tion. Implications between countable chromatic numbers of various algebraichypergraphs are fewer, apparently just consequences of primitive manipulationsof hypergraphs in question. It is also difficult to find implications betweencountable colorings and other types of “paradoxical” objects obtained using theaxiom of choice. However, intuitive insights of this type seem to be difficultto confirm with clear consistency results, owing in part to the multitude andcomplexity of algebraic hypergraphs.In Section 4, we provide the first satisfactory general result for coloringalgebraic hypergraphs in the choiceless context. We deal with the followingclass of hypergraphs: Definition 1.1.
A hypergraph Γ on a Polish space X is redundant if for anyfinite set a ⊂ X , the set { x ∈ X : a ∪ { x } ∈ Γ } is countable. ∗ △ of equilateral triangles in R is count-ably chromatic in ZFC [1], and the redundant hypergraph Γ of rectangles in R is countably chromatic if and only if CH holds [2]. There are many algebraichypergraphs which are countably chromatic in ZFC yet are not redundant, suchas the hypergraph Γ of isosceles triangles in R [6]. We prove: Theorem 1.2.
Let Γ be a redundant algebraic hypergraph of finite arity on aEuclidean space. Then it is consistent relative to an inaccessible cardinal thatZF+DC holds, the chromatic number of Γ is countable, yet there is no sequenceof pairwise distinct Borel sets of bounded complexity and no discontinuous ho-momorphism between Polish groups. The theorem allows for generalization in several natural directions which are notfully spelled out. For one, one can replace R with any Polish real closed fieldor algebraically closed field. Also, in the spirit of Schmerl’s work [7], for eachdimension k ∈ ω , there a balanced forcing which adds a master coloring of R k :no monochromatic set in the coloring is a hyperedge in any redundant algebraichypergraph on R k . The theorem also somewhat understates the understandingof the model we construct, and the conclusion can be strengthened using thetheorems of [4, Sections 9.1, 14.1, 14.2] to include all standard features of bal-anced extensions of the symmetric Solovay model. In forthcoming work [8] [9]we prove many more preservation features of the coloring posets of this paper.The assumption that the hypergraphs are algebraic is used at every turn: theamalgamation problems associated with the balance of the coloring forcing areresolved using quantifier elimination for real closed fields. Thus, the followingquestion remains open. Question 1.3.
Let Γ be a closed redundant hypergraph on a Polish space X .Is it consistent with ZF+DC that the chromatic number of Γ is countable whilethere is no discontinuous homomorphism of Polish groups?For separating chromatic numbers of different redundant hypergraphs, the posetisolated in Definition 4.1 turns out to be too rough a tool, and it will have to besharpened. As an illustration of this procedure, in Section 5 we consider threeredundant hypergraphs on the Euclidean plane, Γ △ of arity three consisting ofall (vertices of) equilateral triangles, Γ of arity four consisting of all (verticesof) parallelograms, and Γ of arity four consisting of all (vertices of) rectangles. Theorem 1.4.
It is consistent relative to an inaccessible cardinal that ZF+DCholds, the chromatic number of Γ is countable, yet the chromatic numbers of Γ and Γ △ are both uncountable. A brief discussion of the opposite implication in ZF is in order. Since Γ ⊂ Γ ,it is clear that if the chromatic number of Γ is countable, then so is that ofΓ . A similar implication for Γ △ fails and the reason is visible in the theoryZFC+ ¬ CH. There, the chromatic number of Γ △ is countable [1], while the2hromatic number of Γ is uncountable [2]. The proof of Theorem 1.4 is of sucha nature that it leaves the following question entirely open. Question 1.5.
Is it consistent with ZF+DC that the chromatic number of Γis countable, yet the chromatic number of Γ △ is uncountable?There is also a simple delimitative result. There are natural algebraic hyper-graphs whose chromatic number is countable with ZFC+CH and which areexceedingly difficult to color without providing an object close to well-orderingof the reals. This makes them impossible to study in balanced extensions ofthe symmetric Solovay model. As an example, let Γ be the hypergraph on R of arity three consisting of those sets a whose projections to both coordinateaxes have cardinality two. This is a sub-hypergraph of the hypergraph of righttriangles in the plane. In the ZFC context, Γ belongs to a rather large class ofhypergraphs whose countable chromatic number is equivalent to the ContinuumHypothesis. We have the following. Theorem 1.6. (ZF)
If the chromatic number of Γ is countable then there isa countable-to-one map from R to ω . To understand the significance of this result, recall that in cofinally balancedextensions of the Solovay model there are no uncountable sequences of pair-wise distinct Borel sets of bounded complexity, in particular no uncountablesequences of pairwise disjoint countable sets [4, Section 9.1]. It follows thenfrom Theorem 1.6 that in all cofinally balanced extensions of the symmetricSolovay model, the chromatic number of Γ is uncountable. In consequence,it is impossible to generalize Theorem 1.2 to all algebraic hypergraphs whichunder ZFC+CH have countable chromatic number.Section 2 isolates and briefly investigates a natural notion of amalgamationposition of pairs of real closed subfields of R . Section 3 builds on that discussionand defines a notion of Suslin forcing with algebraic amalgamation of conditions;such forcings are balanced. Section 4 provides a general coloring poset forredundant algebraic hypergraphs. The main point is the verification of thealgebraic amalgamation property for such posets. The net outcome of thissection is Theorem 1.2. Section 5 discusses a finer coloring poset for the specialcase of the rectangle hypergraph, which allows the proof of Theorem 1.4. Finally,Section 6 proves the delimitative Theorem 1.6.As for the notation, we follow the set theoretic standard of [3]. The short-hand CH denotes the Continuum Hypothesis. An algebraic hypergraph is asubset of [ R k ] n defined by a polynomial equation in kn unknowns; throughout,we neglect the real parameters used in the polynomial. We apply the methodol-ogy and terminology of geometric set theory [4] throughout. All of our models ofZF+DC are obtained as extensions of the symmetric Solovay model by σ -closedSuslin forcings. For the natural but somewhat verbose definitions of virtualconditions and balanced virtual conditions the reader is referred to [4].3 Fields in amalgamation position
Mutual genericity is at the heart of the balanced forcing technology. In thissection, we discuss an algebraic counterpart to it which will be sufficient for allpurposes of this paper. It has the advantage that it removes the forcing relationfrom many considerations. To begin, recall that h R , + , · , ≤i is a real closed field:its ordering is invariant under addition and multiplication by positive elements,and every polynomial of odd degree has a root. Model theory of real closed fieldsis well-known and treated in [5, Section 3.3]. In particular, real closed fields havequantifier elimination [5, Theorem 3.3.15] and they are model complete: everyreal closed subfield is in fact an elementary submodel of R . Definition 2.1.
Let F , F ⊂ R be real closed subfields of R . We say that F , F are in amalgamation position if for every polynomial p (¯ x , ¯ x ) with integercoefficients and all variables listed, and every pair of tuples ¯ r ∈ F and ¯ r ∈ F such that p (¯ r , ¯ r ) = 0 holds, and all open neighborhoods O and O of¯ r , ¯ r respectively, there are tuples ¯ r ′ ∈ O and ¯ r ′ ∈ O in F ∩ F such that p (¯ r ′ , ¯ r ) = p (¯ r , ¯ r ′ ) = 0.In view of elimination of quantifiers for real closed fields, the amalgamationposition yields seemingly stronger consequences for the two fields than statedin the definition. Proposition 2.2.
Let F , F ⊂ R be in amalgamation position. Let φ (¯ x , ¯ x ) be a formula of real closed fields with all free variables listed. For every pair oftuples ¯ r ∈ F and ¯ r ∈ F such that φ (¯ r , ¯ r ) holds there are tuples ¯ r ′ , ¯ r ′ in F ∩ F such that φ (¯ r ′ , ¯ r ) and φ (¯ r , ¯ r ′ ) both hold.Proof. By the quantifier elimination for real closed fields, we may and will as-sume that φ is quantifier free. We also assume that all atomic formulas ap-pearing in it are of the form p (¯ x , ¯ x ) = 0 or p (¯ x , ¯ x ) > p . Let { p i : i ∈ n } be a list of all polynomials appearing in φ . Let a = { i ∈ n : p i (¯ r , ¯ r ) = 0 } . Let O , O be open neighborhoods of ¯ r , ¯ r respec-tively such that for all i ∈ a , the polynomials p i have constant sign on O × O .Use the amalgamation position assumption to find tuples ¯ r ′ ∈ O , ¯ r ′ ∈ O in F ∩ F such that Σ i ∈ n \ a p i (¯ r ′ , ¯ r ) = Σ i ∈ n \ a p i (¯ r , ¯ r ′ ) = 0. It is immediate thatthe tuples ¯ r ′ , ¯ r ′ work as required. Proposition 2.3.
Let F , F ⊂ R be in amalgamation position and let F ⊃ F be a field of finite transcendence degree over F . Then F ∩ F has finitetranscendence degree over F ∩ F .Proof. Let u ⊂ F ∩ F be an inclusion maximal set which is algebraically freeover F . Then u is finite and has cardinality at most that of the transcendencedegree of F over F . It will be enough to show that every element of F ∩ F isalgebraic over F ∩ F and u .Indeed, if x ∈ F ∩ F is any element, then there is a nontrivial polynomial p with coefficients in F such that p ( u, x ) = 0. By the amalgamation position4ssumption and Proposition 2.2, there is a nontrivial polynomial q with coef-ficients in F ∩ F such that q ( u, x ) = 0. In other words, x is algebraic over F ∩ F and u .The following two propositions use the Hilbert basis theorem [5, Theorem 3.2.5]in addition to the amalgamation position assumption. Proposition 2.4.
Let F , F ⊂ R be in amalgamation position. Let p (¯ x , ¯ x ) be a polynomial with integer coefficients and all variables listed. Suppose that ¯ r ∈ F , ¯ r ∈ F are tuples such that p (¯ r , ¯ r ) = 0 holds. Then there is apolynomial q (¯ y, ¯ x ) and a tuple ¯ s ∈ F ∩ F such that1. q (¯ s, ¯ r ) = 0 ;2. { ¯ x : q (¯ s, ¯ x ) = 0 } ⊂ { ¯ x : p (¯ r , ¯ x ) = 0 } . In other words, if a tuple in F belongs to an algebraic set defined from a tuplein F , then it belongs to an algebraic subset of it definable from a tuple in F ∩ F . Proof.
Let A be the set of all tuples ¯ r ′ in F ∩ F such that p (¯ r ′ , ¯ r ) = 0. Theset A is nonempty by the amalgamation position assumption. By the Hilbertbasis theorem, there is a finite set a ⊂ A such that { ¯ x : ∀ ¯ r ′ ∈ a p (¯ r ′ , x ) =0 } = { ¯ x : ∀ ¯ r ′ ∈ A p (¯ r ′ , ¯ x ) = 0 } . Let ¯ s be the concatenation of all tuples ¯ r ′ for r ′ ∈ a and let q (¯ s, ¯ x ) = Σ ¯ r ′ ∈ a p (¯ r ′ , ¯ x ). We claim that this polynomial works.To see this, observe first that (1) follows immediately by the definition ofthe set A . For (2), suppose towards a contradiction that ¯ t is a tuple such that p (¯ r , ¯ t ) = 0 and q (¯ s, ¯ t ) = 0. Since F is an elementary submodel of R , such atuple can be found in F . Let O be an open neighborhood of ¯ r such that thepolynomial p does not change sign on O × { ¯ t } . Use the amalgamation positionassumption to find a tuple r ′ ∈ O in F ∩ F such that p (¯ r ′ , ¯ r ) = 0. Then¯ r ′ ∈ A and ¯ r ′ contradicts the choice of the set a , since adding it to the set a makes the zero set smaller, removing the tuple ¯ t . Proposition 2.5.
Let F , F ⊂ R be in amalgamation position. Let p (¯ x , ¯ x , ¯ x ) be a polynomial with integer coefficients and all variables listed. Suppose that ¯ r , ¯ r are tuples in F and ¯ r is a tuple in F such that ¯ r is the unique tuple in F satisfying p (¯ r , ¯ r , ¯ r ) = 0 . Then all entries of ¯ r are algebraic over F ∩ F and ¯ r .Proof. Let A be the set of all tuples ¯ r ′ in F ∩ F such that p (¯ r , ¯ r ′ , ¯ r ) = 0. Theset A is nonempty by the amalgamation position assumption. By the Hilbertbasis theorem, there is a finite set a ⊂ A such that { ¯ x : ∀ ¯ r ′ ∈ a p (¯ r , ¯ r ′ , ¯ x ) =0 } = { ¯ x : ∀ ¯ r ′ ∈ A p (¯ r , ¯ r ′ , ¯ x ) = 0 } . Let ¯ s be the concatenation of all tuples¯ r ′ for r ′ ∈ a and let q (¯ r , ¯ s, ¯ x ) = Σ ¯ r ′ ∈ a p (¯ r , ¯ r ′ , ¯ x ). It will be enough to showthat ¯ r is the unique solution to the equation q (¯ r , ¯ s, ¯ x ) = 0.Suppose this is not the case. Since F is an elementary submodel of R , thereis a solution ¯ t in F which is distinct from ¯ r . By the initial assumption on ¯ r ,5t must be the case that p (¯ r , ¯ r , ¯ t ) = 0. Let O be a basic open neighborhoodof ¯ r such that p ↾ { ¯ r } × O × { ¯ t } has constant sign. Use the amalgamationposition assumption to find a tuple ¯ r ′ in F ∩ F such that p (¯ r , ¯ r ′ , ¯ r ) = 0.Then ¯ r ′ ∈ A and ¯ r ′ contradicts the choice of the set a , since adding to the set a makes the zero set smaller, removing the tuple ¯ t .Building pairs of real closed subfields of R appears to be a difficult task. Inthis paper, we will use only one way of doing it, using mutual genericity andthe following proposition. Other, more sophisticated approaches will appear inforthcoming work. Proposition 2.6.
Let V [ G ] , V [ G ] be mutually generic extensions. Then F = R ∩ V [ G ] and F = R ∩ V [ G ] are in amalgamation position.Proof. Work in the ground model. Let Q , Q be partial orders, τ , τ be theirrespective names for strings of reals, and let O be a basic open neighborhood ofa Euclidean space such that Q (cid:13) τ ∈ O and Q × Q (cid:13) p ( τ , τ ) = 0 holds. Let h q n : n ∈ ω i be a descending sequence of conditions in Q and O n be a sequenceof basic open neighborhoods such that O = O , ¯ O n +1 ⊂ O n , O n +1 has a metricdiameter at most 2 − n , and q n (cid:13) τ ∈ O n . The intersection T n O n contains asingle point ¯ r . It will be enough to show that Q (cid:13) p (¯ r, τ ) = 0 holds.If not, then by continuity of the polynomial functions there has to be anumber n ∈ ω and a condition q ∈ Q such that q (cid:13) p ↾ O n × { τ } is neverzero. Then, the condition h q n , q i forces in the product that p ( τ , τ ) = 0,contradicting the initial assumptions. This section isolates an algebraic notion of amalgamation in posets which impliesforcing-theoretic balance and which can be flexed to yield more sophisticatedversions of balance.
Definition 3.1. An R -Suslin forcing is a Suslin forcing P such that:1. each condition p ∈ P is a structure in some fixed relational first orderlanguage whose universe supp( p ) is a countable real closed subfield of R ;2. p ≤ p implies p ↾ supp( p ) = p ;3. for every countable set a ⊂ R and every condition p ∈ P there is acondition p ≤ p such that a ⊂ supp( p );4. whenever h p n : n ∈ ω i is a descending sequence of conditions then S n p n ∈ P is their lower bound.Thus, R -Suslin forcing adds a certain generic structure on R which is the unionof all the (countable) structures in the generic filter. It is not difficult to seethat being an R -Suslin forcing is a Π property of the real coding the forc-ing and therefore absolute throughout all forcing extensions by the Shoenfieldabsoluteness [3, Theorem 25.20]. 6 xample 3.2. Consider the graph Γ on R connecting two points if they havea rational Euclidean distance. Consider the poset P of all conditions p suchthat for some countable real closed subfield supp( p ) ⊂ R , p is a Γ-coloring withdomain (supp( p )) and range included in ω . The ordering is that of extension.This is a R -Suslin forcing. Each condition p ∈ P can be viewed as a system { R n : n ∈ ω } of binary relations on supp( p ) such that { r , r } ∈ R n if and onlyif p ( r , r ) = n . Items (2, 3, 4) are easily checked.The way it is stated above above, the definition of R -Suslin forcing has littlecontent, and it comes to life only with the following amalgamation concept. Definition 3.3. An R -Suslin forcing P has R -amalgamation if for any pair p , p of conditions such that the fields supp( p ) , supp( p ) ⊂ R are in amalga-mation position, the following are equivalent:1. p ↾ supp( p ) ∩ supp( p ) = p ↾ supp( p ) ∩ supp( p ), and, writing p forthe common value, p ∈ P and p ≤ p and p ≤ p holds;2. p , p have a common lower bound in P . Example 3.4.
The coloring poset P in Example 3.2 has R -amalgamation. Tosee this, let p , p ∈ P be two conditions such that supp( p ) , supp( p ) are inamalgamation position. Let F = supp( p ) ∩ supp( p ); this is a countable realclosed subfield of R . If p ↾ F = p ↾ F then the conditions cannot becompatible. If p ↾ F = p ↾ F , then the common value, denoted by p ,belongs to P and p , p ≤ p holds by the definitions. In such a case, it is notdifficult to construct a common lower bound of p , p . The main point is that p ∪ p is a coloring for the rational distance graph. To see this, if x ∈ supp( p ) and x ∈ supp( p ) are points at rational Euclidean distance of each other. Ifone of them, say x , belongs to F , then the have a different color since p is acoloring. It cannot occur though that none of the two belongs to F . To see this,assume that x / ∈ F and note that d ( x , x ) = q is a formula of the languageof real closed fields for a rational number q , and by the amalgamation positionassumption there would have to be two (in fact, infinitely many) distinct points x ′ , x ′′ ∈ F such that d ( x ′ , x ) = d ( x ′′ , x ) = q . However, in the plane there areonly two points x satisfying the latter system of equations and so x ∈ F .A routine quantifier counting procedure shows that R -amalgamation is a Π statement and therefore absolute throughout all forcing extensions by the Shoen-field absoluteness [3, Theorem 25.20]. The whole point of the development inthis section is that R -amalgamation implies forcing theoretic balance. The fol-lowing theorem provides the connection and the classification of balanced virtualconditions as in [4, Chapters 6, 7, 8]. Definition 3.5.
Let P be an R -Suslin forcing. A P -structure is a relationalstructure c with universe R such that Coll( ω, R ) (cid:13) ˇ c ∈ P . Theorem 3.6.
Suppose that P is an R -Suslin forcing with R -amalgamation. . if c is a P -structure, then the pair h Coll( ω, R ) , ˇ c i is balanced;2. if h Q, τ i is a balanced pair, then there is a P -structure such that the pairs h Q, τ i and h Coll( ω, R ) , ˇ c i are equivalent;3. distinct P -structures provide inequivalent balanced pairs.In particular, if the Continuum Hypothesis holds then P is balanced.Proof. For (1), let V [ G ] , V [ G ] be mutually generic extensions and p , p leqc be conditions in P in the respective models V [ G ] , V [ G ]. We must showthat p , p have a common lower bound. By Proposition 2.6, the fields R ∩ V [ G ]and R ∩ V [ G ] are in amalgamation position and then so are the fields supp( p )and supp( p ). By the properties of the ordering P , p ↾ V = p ↾ V = c holds.The R -amalgamation applied in the model V [ G , G ] then implies that p , p have a common lower bound as desired.For (2), strengthening the name τ if necessary, using Definition 3.1(3) in the Q -extension, we may assume that Q (cid:13) R ∩ V ⊂ supp( τ ). Further, by a balanceargument, there is a relational structure c on R such that Q (cid:13) τ ↾ ( R ∩ V ) = ˇ c .We will show that c is a P -structure and Q (cid:13) τ ≤ ˇ c , which will prove item(2) by [4, Proposition 5.2.6]. Towards this end, let G , G ⊂ Q be mutuallygeneric filters and let p = τ /G and p = τ /G . By the balance of the pair h Q, τ i , the conditions p , p have a common lower bound in P . At the sametime, Proposition 2.6 shows that the fields R ∩ V [ G ] and R ∩ V [ G ] are inamalgamation position, and consequently so are the fields supp( p ), supp( p ).The R -amalgamation applied in V [ G , G ] then shows that c ∈ P and p , p ≤ c as desired.(3) is immediate by Definition 3.1(2). For the last sentence, let p ∈ P be acondition and work to produce a balanced virtual condition below p . Use theContinuum Hypothesis assumption to find an enumeration h r α : α ∈ ω i of allreals, and by transfinite recursion on α ∈ ω build conditions p α so that p = p , p α +1 ≤ p α , r α ∈ supp( p α +1 ), and for limit ordinals α , p α = S β ∈ α p β . This ispossible by Definition 3.1(3) and (4) and the result is a descending sequenceof conditions in P . Let c = S α p α . This is a relational structure on R andby Definition 3.1(4), Coll( ω, R ) (cid:13) ˇ c ∈ P and ˇ c ≤ ˇ p . By (1) of the presenttheorem, the P -structure c yields a balanced virtual condition stronger than p as desired.The R -amalgamation is a flexible concept which allows for many variations suchas how many variables are allowed in the polynomials or how many fields appearin the amalgamation scheme. In this paper, we will use only one variationdescribed in the following definitions and theorems. Definition 3.7.
Let n ≥ { F i : i ∈ n } be real closedsubfields of R . We say that the fields are in amalgamation position if1. the field F i ∩ F j is the same for any choice of distinct indices i, j ∈ n ;8. whenever p (¯ x i : i ∈ n ) is a polynomial with integer coefficients and ¯ r i aretuples in the respective fields F i for i ∈ n such that p (¯ r i : i ∈ n ) = 0 holds,and whenever j ∈ n is an index and O is an open neighborhood of ¯ r j , thenthere is a tuple ¯ r ′ j ∈ O in T i F i such that p (¯ r i : i = j, ¯ r ′ j ) = 0 holds.The following is proved just like Proposition 2.6. Proposition 3.8. If n ≥ is a natural number and V [ G i ] for i ∈ n are mu-tually generic extensions of V , then the fields F i = R ∩ V [ G i ] for i ∈ n are inamalgamation position. Definition 3.9.
Let 2 ≤ n ≤ m be natural numbers. Let P be an R -Suslinforcing. We say that P has m, n - R -amalgamation if, whenever { F i : i ∈ m } and { p i : i ∈ m } are such that1. F i ’s are countable real closed subfields of R , their pairwise intersectionsare all the same, equal to some F , and any n of them are in amalgamationposition;2. p i ’s are conditions in P such that supp( p i ) = F i holds for all i ∈ m , p = p i ↾ F is a condition in P which does not depend on i and for all i ∈ m p i ≤ p holds,then the set { p i : i ∈ m } has a common lower bound in P .The m, n -amalgamation is a forcing free criterion for m, n -balance, a propertyused in [4, Chapter 13] for many independence results. Theorem 3.10. (CH) Let ≤ n ≤ m be natural numbers. Suppose that P isa R -Suslin forcing with R -amalgamation and m, n - R -amalgamation. Then theposet P is m, n -balanced. It appears that there is no canonical candidate for a σ -closed Suslin poset addinga coloring to a given redundant algebraic hypergraph. The following definitionworks in general, but in many specific cases it does not provide the mildestcoloring forcing possible. Definition 4.1.
Let n ≥ , k ≥ n on the Euclidean space of dimension k . The coloring poset P consists of all conditions p such that there is an algebraicallyclosed countable subfield supp( p ) ⊂ R for which p : supp( p ) k → ω × ω is a Γ-coloring. For a condition p ∈ P , an algebraically closed subfield d ⊂ R and afinite set u ⊂ R , define A ( p, d, u ) to be the set of all nonempty monochromaticsets a ⊂ dom( p ) \ d k of cardinality smaller than n − b ⊂ d k and a point x ∈ R k algebraic over d ∪ u such that a ∪ b ∪ { x } ∈ Γ. Theordering on P is defined by p ≤ p if 9. p ⊂ p ;2. for every set a ⊂ dom( p ) \ dom( p ) of cardinality at least two, if there is b ⊂ dom( p ) such that a ∪ b ∈ Γ, then a is not p -monochromatic;3. for every finite set u ⊂ supp( p ), the set p ′′ S A ( p , supp( p ) , u ) ⊂ ω × ω has all vertical sections finite.The choice of ω × ω as the target set of the coloring and the choice of the idealof sets with all vertical sections finite are a matter of convenience. However,some type of restriction as in (3) is necessary for the amalgamation argumentsin Claim 4.3 and later, and it is exactly this item which makes the discovery andanalysis of the coloring poset difficult. Elimination of quantifiers shows that inthe real closed field R , every countable set definable from a tuple of parametersis in fact finite and consists of points algebraic in that tuple. Therefore, if p ∈ P is a condition, a ⊂ dom( p ) is a set of cardinality n −
1, and x ∈ R k is a pointsuch that a ∪ { x } ∈ Γ, then x ∈ dom( p ) must hold. This closure property ofconditions in the coloring poset will be used below repeatedly without furtherjustification. Theorem 4.2.
Let Γ be a redundant algebraic hypergraph on the Euclideanspace R k of arity n ≥ .1. The coloring poset P is an R -Suslin forcing;2. P has R -amalgamation;3. for every number m ≥ n , P has m, n - R -amalgamation.Proof. Let n ∈ ω be the arity of the hypergraph. First argue that ≤ is atransitive relation. Let p ≤ p ≤ p be conditions in P , and work to prove that p ≤ p holds. It is clear that p ⊂ p holds. To verify (2) of Definition 4.1,suppose that a ⊂ dom( p ) \ dom( p ) is a set of cardinality at least 2 suchthat there is b ⊂ dom( p ) with a ∪ b ∈ Γ; we need to show that a is not p -monochromatic. If a ⊂ dom( p ) then we are done by Definition 4.1(2) of p ≤ p . By the Γ-closure of dom( p ), it cannot be the case that | a \ dom( p ) | = 1. If | a \ dom( p ) | ≥ p ≤ p . This coversall possible cases and Definition 4.1(2) follows.For Definition 4.1(3), let u ⊂ supp( p ) be a finite set; we need to show that p ′′ S A ( p , supp( p ) , u ) has all vertical sections finite. Let v be an inclusion-maximal subset of supp( p ) intersected with the algebraic closure of supp( p ) ∪ u ,which is algebraically independent over supp( p ). Since the sets algebraicallyindependent over supp( p ) form a matroid, v is finite and in fact | v | ≤ | u | . Wewill show that p ′′ S A ( p , supp( p ) , u ) is a subset of p ′′ S A ( p , supp( p ) , v ) ∪ p ′′ S A ( p , supp( p ) , u ). This will complete the proof as the latter two sets haveall vertical sections finite by Definition 4.1(3) applied to p ≤ p and p ≤ p respectively.Let a ∈ A ( p , supp( p ) , u ) be a set of cardinality smaller than n − A ( p , supp( p ) , u ) is witnessed by some point x . If a ⊂ dom( p )10hen by the real closure of supp( p ) x ∈ dom( p ) must hold, by the max-imality of the set v all coordinates of x must belong to the algebraic clo-sure of supp( p ) ∪ v , and so a ∈ A ( p , supp( p ) , v ). If, on the other hand, a dom( p ), then a \ dom( p ) is a nonempty monochromatic set which belongsto A ( p , supp( p ) , u ) and contributes the same color to A ( p , supp( p ) , u ) as a does to A ( p , supp( p ) , u ). This covers all possible cases and Definition 4.1(3)follows.Now we check that the poset P is σ -closed. Let h p m : m ∈ ω i be a descendingsequence of conditions in P ; we will show that p ω = S m p m is their commonlower bound. To see this, fix a number m ∈ ω and argue for p ω ≤ p m . It isimmediately clear that p m ⊂ p ω . To verify Definition 4.1(2), let a ⊂ dom( p ω ) \ dom( p m ) be a set of cardinality at least 2 such that there is b ⊂ dom( p m ) with a ∪ b ∈ Γ. Since the set a is finite, there must be m ′ > m such that a ⊂ dom( p m ′ ).Definition 4.1(2) for p ω ≤ p m then follows from Definition 4.1(2) for p m ′ ≤ p m .To verify Definition 4.1(3), let u ⊂ supp( p ω ) be a finite set. Find a number m ′ > m such that u ⊂ supp( p m ′ ); we will show that A ( p ω , supp( p m ) , u ) = A ( p m ′ , supp( p m ) , u ) which will finish the proof. Indeed, let a ⊂ dom( p ω ) \ dom( p m ) be a set whose membership in A ( p ω , supp( p m ) , u ) is witnessed by some b ⊂ dom( p m ) and x . The coordinates of the point x is in the algebraic closureof supp( p m ) ∪ u and therefore x ∈ dom( p m ′ ) holds. Now, if a ⊂ dom( p m ′ ) then a ∈ A ( p m ′ , supp( p m ) , u ) as witnessed by the same b, x . If a \ dom( p m ′ ) is anonempty set, it must contain more than one element by the real closure ofsupp( p m ′ ). Let m ′′ > m ′ be a number large enough such that a ⊂ dom( p m ′′ ).Applying Definition 4.1(2) to p m ′′ ≤ p m ′ and the set a \ dom( p m ′ ), we concludethat the set a cannot be monochromatic. This concludes the proof of the σ -closure.Now we turn to the complexity calculation. Let p , p be conditions. Write d ( p , p ) ⊂ R for the real closure of the set supp( p ) ∪ supp( p ). Claim 4.3. p , p have a common lower bound in P if and only if the followingitems hold in conjunction:1. p ∪ p is a function which is a Γ -coloring;2. there are no sets a ⊂ dom( p ) and a ⊂ dom( p ) \ dom( p ) which arerespectively p - and p -monochromatic and such that a ∪ a ∈ Γ andsimilarly with subscripts , interchanged;3. for each finite set u ⊂ d ( p , p ) , the images p ′′ S A ( p , supp( p ) , u ) as wellas p ′′ S A ( p , supp( p ) , u ) ⊂ ω × ω have all vertical sections finite.Proof. Any common lower bound q ∈ P of p , p must satisfy d ( p , p ) k ⊂ supp( q ). It then follows immediately from the definition of the ordering P thatfailure of any of the three items would result in an impossibility to find such acommon lower bound.Now suppose that the three items hold; we will construct a common lowerbound q of p , p which satisfies supp( q ) = d ( p , p ). Write e = d ( p , p ) k \ p ) ∪ dom( p )). Fix a bijection π : e → ω . Consider the map q : d ( p , p ) k → ω × ω satisfying p ∪ p ⊂ q and for each point x ∈ e , q ( x ) = h π ( x ) , l i for somenumber l ∈ ω which does not belong to the finite π ( x )-th sections of the sets p ′′ S A ( p , supp( p ) , x ) and p ′′ S A ( p , supp( p ) , x ). We claim that q ∈ P is acommon lower bound of p and p .To show that q ∈ P , it is just necessary to argue that there is no hyperedge a ⊂ dom( q ) in Γ is q -monochromatic. To see that, let a ⊂ dom( q ) be a hyper-edge. It cannot be the case that a ∩ e = 0 by item (2). If | a ∩ e ) | > a is not monochromatic as q ↾ e is an injective function. Finally, if a ∩ e is asingleton containing some point x ∈ e , then by the closure properties of dom( p )and dom( p ), both sets a ∩ p and a ∩ p must be nonempty and therefore ofcardinality smaller than n − a cannot be monochromatic by the choiceof the color q ( x ). Since this covers all possible cases, we have proved q ∈ P .Finally, we show that q ≤ p ; the proof of q ≤ p is symmetric. Clearly, p ⊂ q . To verify Definition 4.1(2), suppose that a ⊂ dom( q ) \ dom( p ) is anonempty set such that there is b ⊂ dom( p ) such that a ∪ b is monochromatic;we must show that a is not monochromatic. If a ⊂ dom( p ) then this followsfrom item (2) of this claim. If | a \ dom( p ) | > q ↾ e is an injective function. If a \ dom( p ) is a singleton containing some x ∈ e , then a ∩ dom( p ) must be a nonempty set by the closure of dom( p ).Now, if a ∩ dom( p ) is not monochromatic, we are done. If it is monochromatic,then it appears in the set B (0 , x ) and so its p -color is distinct from q ( x ) by thechoice of q ( x ), and a is not monochromatic either. Since this covers all possiblecases, we have verified Definition 4.1(2).To verify Definition 4.1(3), suppose that u ⊂ d ( p , p ) is a finite set. The setsin A ( q, supp( p ) , u ) can be divided into two groups: those which intersect theset e , and those which are subsets of dom( p ). The first group contributes onlyone element to each vertical section of the set q ′′ S A ( q, supp( p ) , u ) ⊂ ω × ω as q ′′ e ⊂ ω × ω is a function. The second group is a subset of A ( p , supp( p ) , u )and therefore contributes only finitely many elements to each vertical section ofthe set q ′′ S A ( q, supp( p ) , u ) ⊂ ω × ω by item (3) of the present claim. Thiscompletes the proof of q ≤ p and the proof of the claim.It now follows immediately that the incompatibility relation in the poset P isBorel and so P is a Suslin partial order.For the density item of Definition 3.1, let p ∈ P be a condition and d ⊂ R k be a countable set. Enlarging the set d if necessary, we may assume that d = e k for some real closed field e ⊂ R . Now, let π : d \ dom( p ) → ω be an injectionand let p = p ∪ {h x, h π ( x ) , ii : x ∈ d \ dom( p ) } . It is not difficult to see that p ∈ P is a condition and p ≤ p holds.For (2) of the theorem, let p , p ∈ P be conditions such that the fieldssupp( p ) , supp( p ) are in amalgamation position, and write F = supp( p ) ∩ supp( p ). We must verify the equivalence in Definition 3.3. For the implication(2) → (1) of that definition, it is clear that if p ↾ F = p ↾ F then p , p areincompatible. If p ↾ F = p ↾ F , then the common value p belongs to the poset P . If, say, p p , then the same reason for the failure of the inequality p ≤ p p and p .The implication (1) → (2) of Definition 3.3 is the heart of the matter. Supposethat p ↾ F = p ↾ F , denote the common value by p , assume that p ≤ p and p ≤ p holds, and work to verify the assumptions of Claim 4.3 to find a commonlower bound of p , p .To verify (1) of Claim 4.3, it is clear that p ∪ p is a function. To checkthat p ∪ p is a Γ-coloring, suppose that a ⊂ dom( p ∪ p ) is a set in Γ andlet a = a ∩ dom( p ) and a = a ∩ dom( p ) \ V ; so a = a ∪ a . We breakinto cases according to the cardinality of a . If a = 0 then a = a cannot bemonochromatic as p is a Γ-coloring. By the closure properties of dom( p ) itcannot be the case that | a | = 1. Assume now that | a | ≥
2. By Proposition 2.2there must be a tuple a ′ ⊂ F such that a ′ ∪ a ∈ Γ. Then, since p ≤ p holds,it must be the case that a is not monochromatic in p and consequently a isnot monochromatic in p ∪ p .The same argument verifies (2) of Claim 4.3. Verification of (3) of Claim 4.3is the key to the whole construction in this paper. Let u ⊂ R be a finite setand write A = A ( p , supp( p ) , u ); I must show that p ′′ S A ⊂ ω × ω has allvertical sections finite. This will be done by finding a finite set v ⊂ supp( p )such that A ⊂ A ( p , F, v ) and using the assumption p ≤ p . To find the finiteset v , for each a ∈ A find a point g ( a ) ∈ R k such that g ( a ) is in the algebraicclosure of supp( p ) ∪ u and such that there is some h ( a ) ⊂ dom( p ) such that a ∪ h ( a ) ∪ { g ( a ) } ∈ Γ. By a transcendental dimension argument, there is a finiteset A ′ ⊂ A (even | A ′ | ≤ | u | ) such that each point g ( a ) for a ∈ A is algebraicover dom( p ) and the points g ( a ′ ) for a ′ ∈ A ′ . Let v ⊂ supp( p ) be the set ofall reals used in points in A ′ ; I claim that the set v works as required.To see this, suppose that a ∈ A is an arbitrary set. Note that there is afinite set w ⊂ supp( p ) such that g ( a ) is algebraic in v ∪ w : namely, one canpick the set w so that it contains the coordinates of the points h ( a ′ ) for a ′ ∈ A ′ and so that g ( a ) is in the algebraic closure of w and the points g ( a ′ ) for a ′ ∈ A ′ .Then, each point g ( a ′ ) for a ′ ∈ A ′ is algebraic in a ′ and h ( a ′ ) by the redundanceproperty of Γ, so algebraic in v ∪ w . The point g ( a ) is algebraic in w and thepoints g ( a ′ ) for a ′ ∈ A ′ , so it is algebraic in v ∪ w as well.Now, fix an algebraic definition ψ of g ( a ) from v and w . Consider the formula φ ( a, v, ¯ x, ¯ y ) saying that the point defined by ψ from v and ¯ y completes the set a ∪ ¯ x into a Γ-edge. This formula (with parameters a, v in supp( p )) has asolution in supp( p ), namely h ( a ) and w . By the amalgamation position andProposition 2.2, it has a solution in F . This means that a ∈ A ( p , F, v ) holdsas desired and (3) of Claim 4.3 follows.The proof of m, n - R -amalgamation is the same. The only extra observationnecessary is that given conditions { p i : i ∈ m } such that the pairwise intersec-tions of their supports are the same, equal to some real closed field F , thenevery Γ-hyperedge has nonempty intersection with at most n many elements ofsupp( p i ) k \ F k by the arity assumption on Γ. Corollary 4.4. (CH) Let Γ be a redundant algebraic hypergraph on R k of arity n . The coloring poset P is balanced and in fact m, n -balanced for all m ≥ n . roof. This is just a conjunction of Theorem 3.6, Theorem 3.10, and Theo-rem 4.2.The Continuum Hypothesis assumption is necessary to some extent. For exam-ple, the rectangle hypergraph Γ is algebraic and redundant, and existence ofa total Γ -coloring is equivalent to the Continuum Hypothesis by [2]. It wouldbe interesting though to weaken the Continuum Hypothesis assumption to theweakest possible, namely to the existence of a total Γ-coloring. We do not knowif this is possible, even though certain coloring posets do have this property.The following corollary proves Theorem 1.2.
Corollary 4.5.
Let Γ be a redundant algebraic hypergraph on R k of arity n .1. Let P be the coloring poset for Γ . Then in the P -extension of the symmet-ric Solovay model, there are no uncountable sequences of pairwise distinctBorel sets of bounded complexity and no discontinuous homomorphismsbetween Polish groups.2. It is consistent relative to an inaccessible cardinal that ZF+DC holds, Γ has countable chromatic number, and there are no uncountable sequencesof pairwise distinct Borel sets of bounded complexity and no discontinuoushomomorphisms between Polish groups.Proof. Uncountable sequences of pairwise distinct Borel sets of bounded com-plexity are ruled out by the balance of P and [4, Corollary 9.1.2]. Discontin-uous homomorphisms are ruled out by the m, n -balance of P and [4, Theorem13.2.1]. This section is devoted to the proof of Theorem 1.4. In principle, the outline ofproof is straightforward. We sharpen Definition 4.1 to develop an R -Suslin forc-ing P adding a total Γ -coloring, which in addition has 3 , R -amalgamation–this occurs in Definition 5.1. After that, we show that in 3 , △ and Γ remain uncountable–Theorem 5.9. Checking all the amalgamation details is painstaking work; asanother aggravating circumstance, there is apparently no canonical choice forthe poset P . This turns a proof with a straighforward outline into a rather longslog.It will be useful to introduce some elementary geometric parlance for thefollowing definitions and arguments. Given two distinct parallel lines l, l ′ in R and points x ∈ l and x ′ ∈ l ′ , we say that x, x ′ are antipodal if the segmentconnecting them is perpendicular to both l, l ′ . Two points x, x ′ ∈ R on a circleare antipodal if the segment connecting them passes through the center of thecircle. A circle is visible in a set a ⊂ R if a contains its center and one of itspoints; a line is visible in a if a contains two of its points.14 efinition 5.1. The rectangle coloring poset P consists of partial functions p : R → ω × ω which are Γ -colorings and such that for some relatively al-gebraically closed countable subfield supp( p ) ⊂ R , dom( p ) = supp( p ) . Theordering is defined by p ≤ p if1. p ⊂ p ;2. for any pair of parallel lines visible in p and any pair of antipodal points x, y on these lines in dom( p \ p ), p ( x ) = p ( y );3. for any circle visible in p and any pair of antipodal points x, y on thiscircle in dom( p \ p ), p ( x ) = p ( y );4. for every finite set u ⊂ supp( p ), the p -image of the set { x ∈ dom( p \ p ) : x is algebraic over supp( p ) ∪ u } has all vertical sections finite. Theorem 5.2.
Let P be the rectangle coloring poset.1. P is an R -Suslin forcing;2. P has R -amalgamation;3. P has , - R -amalgamation.Proof. An important point is the transitivity of the relation ≤ . Suppose that p ≤ p ≤ p holds and argue that p ≤ p must follow. Clearly, p ⊂ p .For Definition 5.1(2), let l, l ′ be parallel lines visible from p and let x, x ′ ∈ dom( p \ p ) be a pair of antipodal points on them. By the closure properties ofdom( p ), it must be the case that either x, x ′ ∈ dom( p ) or x, x ′ ∈ dom( p \ p ).In the former case, p ( x ) = p ( x ′ ) follows from Definition 5.1(2) applied with p ≤ p , in the latter case p ( x ) = p ( x ′ ) follows from Definition 5.1(2) appliedwith p ≤ p . (3) of Definition 5.1 for p ≤ p is verified in the same way. Forthe verification of (4), suppose that a ⊂ supp( p ) is a finite set and let d ⊂ R be the algebraic closure of supp( p ) ∪ a . Let b ⊂ supp( p ) be a transcendencebasis of d = d ∩ supp( p ) over supp( p ); so b is finite and even | b | ≤ | a | holds.Then Definition 5.1(4) applied with p ≤ p shows that p ′′ ( d \ dom( p )) ⊂ ω × ω has all vertical sections finite. Moreover, Definition 5.1(4) applied with p ≤ p shows that p ′′ ( d \ dom( p )) ⊂ ω × ω has all vertical sections finite. In total,the set p ′′ ( d \ dom( p )) has all vertical sections finite as desired.For the σ -closure, suppose that h p n : n ∈ ω i is a descending sequence of con-ditions in P ; it will be enough to argue that p ω = S n p n is a common lowerbound of the sequence. It is clear that p ω is a Γ -coloring with an algebraicallyclosed domain and so p ω ∈ P holds. Let n ∈ ω be an arbitrary number; weneed to show that p ω ≤ p n holds. Clearly p n ⊂ p ω holds. To argue for Defini-tion 5.1(2), suppose that l, l ′ are two lines visible in p n and x, x ′ ∈ dom( p ω \ p n )are antipodal points on them. By the closure properties of the domains of theconditions p m for m ≥ n , there has to be a number m ≥ n such that both x, x ′ belong to dom( p m +1 \ p m ), and then the colors p m +1 ( x ) = p ω ( x ) and15 m +1 ( x ′ ) = p ω ( x ′ ) must be distinct by Definition 5.1(2) applied to p m +1 ≤ p m .(3) is verified in the same way. Finally, for Definition 5.1(4), let u ⊂ supp( p ω )be a finite set. There must be a number m ≥ n such that u ⊂ supp( p m ) andthen the algebraic closure of supp( p n ) ∪ u is already a subset of supp( p m +1 ).An application of (4) of Definition 5.1 to p m +1 ≤ p m concludes the proof.For the Suslinity of P , it is not difficult to see that both P and ≤ are Borelsets in a suitable Polish space. The challenge is proving that the incompatibilityrelation is Borel as well. To this aim, we prove a claim useful later. For condi-tions p , p write d ( p , p ) ⊂ R for the algebraic closure of supp( p ) ∪ supp( p )in R . Claim 5.3.
Let p , p ∈ P be any conditions. Then p , p are compatible justin case the conjunction of the following items occurs:1. p ∪ p is a function;2. for any pair of parallel lines visible in p and any pair of antipodal points x, y on these lines in dom( p \ p ) , p ( x ) = p ( y ) ;3. for any circle visible in p and any pair of antipodal points x, y on thiscircle in dom( p \ p ) , p ( x ) = p ( y ) ;4. for every finite set u ⊂ d ( p , p ) , the p -image of the set { x ∈ dom( p \ p ) : x is algebraic over dom( p ) ∪ u } has all vertical sections finite;5. items (2-4) with the subscripts , interchanged. It is clear that the claim shows that compatibility (and therefore incompat-ibility) of conditions in the poset P is a Borel statement. Proof.
Fix conditions p , p ∈ P . Any putative lower bound q of p , p musthave d ( p , p ) ⊂ supp( q ), and so the failure of any of the above items impliesthat such a common lower bound does not exist. Suppose now that the itemsall hold and work to find a lower bound q with supp( q ) = d ( p , p ).For each point x ∈ d ( p , p ) \ dom( p ∪ p ) consider sets A ( x ) ⊂ dom( p \ p ) and A ( x ) ⊂ dom( p \ p ) defined in the following way. A ( x ) = { x ∈ dom( p \ p ) : some points x , x ′ ∈ dom( p ) form a rectangle with x, x } and A ( x ) = { x ∈ dom( p \ p ) : some points x , x ′ ∈ dom( p ) form a rectanglewith x, x } . Observe that every element of A is algebraic over supp( p ) andany other element of A : if x , y ∈ A are distinct points and their membershipin A is witnessed by some points x , x ′ and y , y ′ ∈ dom( p ), then x can bediscovered from x , x , x ′ and then y can be discovered from x, y , y ′ . In thesame way, every element of A is algebraic over supp( p ) and any other elementof A .Now we are ready to define the lower bound q ≤ p , p . We set p ∪ p ⊂ q .To define q on d ( p , p ) \ dom( p ∪ p ), let h x n : n ∈ ω be an enumeration of thelatter set and for each n ∈ ω let q ( x n ) = h n, m i for some number m ∈ ω whichdoes not belong to the n -th vertical section of the sets p ′′ A or p ′′ A . Note that16hese two sets have all vertical sections finite by the previous paragraph anditem (4) of the claim. We claim that q is the desired common lower bound of p , p .To prove this, the most important thing is to verify that q is in fact a Γ -coloring. To this aim, let a ⊂ dom( q ) be a set of size four forming a rectangle.The proof that the rectangle is not monochromatic breaks into cases. Case 1.
If more than one point in a belongs to dom( q ) \ dom( p ∪ p ) then therectangle is not monochromatic as q is injective on the latter set. Case 2. If a ⊂ dom( p ∪ p ), we split into subcases: Case 2a.
For one of the conditions p , p (say p for definiteness) it is thecase that dom( p ) contains three elements of a . In such a case, in fact a ⊂ dom( p holds by the closure properties of dom( p ), and the rectangle is notmonochromatic since p is a Γ -coloring. Case 2b.
If Case 2a fails then a = a ∪ a where a = a ∩ dom( p \ p ) and a = a ∩ dom( p \ p ) are both sets of cardinality two. We break into twosubcases again. Case 2ba.
The set a (and also the set a ) consists of neighboring points of therectangle. In such a case, the two lines passing through the respective pointsof a and perpendicular to the segment connecting the two elements of a arevisible in p and the two points in a are antipodal on these two lines. It followsfrom item (2) that the colors assigned to the points in a are distinct and therectangle is not monochromatic. Case 2bb.
The set a (and also the set a ) consists of opposite points in therectangle. In such a case, the circle containing both points in a and center inthe middle between them is visible in p . The two points in a are antipodal onthis circle. It follows from item (3) that the colors assigned to the points in a are distinct and the rectangle is not monochromatic. Case 3.
The set a contains exactly one element of dom( q ) \ dom( p ∪ p );call this point x . Now, the remaining points of a cannot be all in dom( p ) orall in dom( p ) since even x would then be there by the closure properties ofdom( p ) and dom( p ). It follows that there is one of the conditions p , p (say p ) such that dom( p ) contains two elements of a while dom( p \ p ) containsthe remaining point of a , call it y . But then, y ∈ A ( x ) and so q ( x ) = p ( x )by the choice of the color of q ( x ). Thus, the rectangle is not monochromatic inthis case either.The long break into cases has succeeded in showing that q ∈ P holds. Nowwe have to show that q ≤ p , p holds; by symmetry, it is enough to argue for q ≤ p . Clearly p ⊂ q holds. To verify Definition 5.1(2), suppose that l, l ′ arelines visible in p and x, x ′ ∈ dom( q \ p ) are antipodal points on them. Thereis again a split into cases. If both points x, x ′ belong to dom( p ), then item (2)of the claim shows that the colors p ( x ) = q ( x ) and p ( x ′ ) = q ( x ′ ) are distinct.If both points x, x ′ belong to dom( q \ p ) then their colors are distinct since q is injective outside of dom( p ∪ p ). If one of the points, say x , is outsideof dom( p ) while x ′ is in, then observe that x ′ ∈ A ( x ); thus, their colors aredistinct again by the choice of q ( x ). Definition 5.1(3) is verified in the same way.To check Definition 5.1(4), use item (4) of the present claim, together with the17act that the range of q \ ( p ∪ p ) has all vertical sections of cardinality one.For the density statement, if p ∈ P is a condition and d ⊂ R is a countableset, first find a countable algebraically closed subfield e ⊂ R such that dom( p ) ∪ d ⊂ e and then find a map q : e → ω × ω such that p ⊂ q , q \ p is an injection,and the range of q \ p has all vertical sections finite. It is not difficult to checkthat q ≤ p is a condition in the poset P ; clearly, d ⊂ dom( q ).For the R -amalgamation (item (2) of the theorem), suppose that p , p ∈ P are conditions such that the fields supp( p ) , supp( p ) are in amalgamationposition; write F = supp( p ) ∩ supp( p ). We need to check the equivalence inDefinition 3.3. For the (2) → (1) direction, note that if p ↾ F = p ↾ F , thenthe conditions p , p are certainly incompatible. Suppose that p ↾ F = p ↾ F holds, with the common value denoted by p . If, say, p ≤ p fails, then the reasonfor its failure also shows that p , p cannot have a common lower bound.The opposite direction is the heart of the matter. Suppose that p ↾ F = p ↾ F with the common value denoted by p , and suppose that p , p ≤ p holds;we must find a common lower bound for p , p . To do this, we check items (1-5)of Claim 5.3 one by one. Clearly, p ∪ p is a function. For item (2), let l, l ′ belines visible in p and x ′ , x be antipodal points in dom( p ). By Proposition 2.4,the lines l, l ′ must be in fact visible from supp( p ); then p ( x ) = p ( x ′ ) followsfrom p ≤ p . Item (3) is verified in a similar way. For item (4), suppose that u ⊂ R is a finite set. By Proposition 2.3, there is a finite set v ⊂ supp( p ) suchthat the intersection of the algebraic closure of supp( p ) ∪ u with supp( p ) isequal to the algebraic closure of supp( p ) ∪ v . Item (4) of Claim 5.3 for u thenfollows from item (4) of Definition 5.1 applied to p ≤ p . The verification ofitem (5) of Claim 5.3 is symmetric.For the 3 , R -amalgamation, suppose that p , p , p ∈ P are conditionssuch that the fields supp( p ), supp( p ), supp( p ) are pairwise in amalgamationposition and their pairwise intersections are all the same, equal to some field F . Assume that p i ↾ F is the same for all i ∈
3, and denote the common valueby p ∈ P . Assume that p i ≤ p holds for all i ∈ x ∈ R \ (dom( p ) ∪ dom( p ) ∪ dom( p )) be an arbitrary point. Let i = j be distinct elements of 3, and let A ( x, i, j ) = { y ∈ dom( p i ) \ F : for some z ∈ dom( p j ) the points x, y, z make a right triangle with the right angle at y } . Claim 5.4.
There is a finite set a ( x, i, j ) ⊂ A ( x, i, j ) such that all points of A ( x, i, j ) are algebraic over F ∪ a ( x, i, j ) .Proof. For each point y ∈ A ( x, i, j ) select a witness point z ( y ) ∈ dom( p j )such that the points x, y, z ( y ) make a right angle at y . Let A = { y ∈ A ( x, i, j ) : z ( y ) ∈ V } and A = { y ∈ A ( x, i, j ) : z ( y ) / ∈ V } , and prove theconclusion for A , A separately.To deal with A , note that the line connecting y ∈ A with x is visible in p i ,as it is just the line through y and perpendicular to the segment connecting y and z ( y ). These lines must be the same for all y ∈ A , since otherwise the point x as their intersection would be an element of dom( p i ). But then, whenever18 , y ∈ A are distinct points then they are mutually algebraic over F : y is apoint such that the broken line connecting z ( y ) , y , y , z ( y ) in this order hasright angles at both y and y . To deal with A , we break into cases. Case 1.
For all points y ∈ A , the line connecting y to x is the same. Here, thetreatment is similar to that of A : whenever y , y ∈ A are distinct points thenthey are mutually algebraic over supp( p j ): y is a point such that the brokenline connecting z ( y ) , y , y , z ( y ) in this order has right angles at both y and y . By the amalgamation position of supp( p i ) and supp( p j ) and Proposition 2.2, y , y are mutually algebraic over F . Case 2.
If the assumption of Case 1 fails, pick two points y , y ∈ A such thatthe lines connecting them to x are distinct. We will show that any other point y ∈ A is algebraic over F ∪ { y , y } , proving the claim in this case as well. Tothis end, note that the point x is algebraic over z ( y ) , z ( y ) and y and y as theunique intersection of the lines passing through y or y and perpendicular tothe segments connecting y , y with z ( y ) , z ( y ) respectively. It follows that thecircle C having the segment connecting x and z ( y ) as a diameter is algebraicover z ( y ) , z ( y ) , z ( y ) and y , y . By Thalet’s theorem, the circle C passesthrough y .By Proposition 2.2, there are circles algebraic over F and y , y which passthrough y and arbitrarily close to x and z ( y ). If there is more than one sucha circle, then y is algebraic over F and y , y , since the intersection of twodistinct circles is a finite set. Thus, it will be enough to derive a contradictionfrom the assumption that there is only one such circle. In such a case, this circlemust be equal to C . Now, since C is visible in supp( p i ) and contains the point z ( y ) ∈ V [ G j ] \ V , it must be the case by Proposition 2.2 that the circle C is infact visible in F . However, then supp( p j ) must hold as x is a point antipodalto z ( y ) on C . This contradicts the original assumption on the point x .Now, let d ⊂ R be any countable real closed subfield of R containing supp( p ) ∪ supp( p ) ∪ supp( p ). We will find a lower bound q of p , p , p with supp( q ) = d .We set p ∪ p ∪ p ⊂ q . To find the remaining values of q , set e = d \ (dom( p ) ∪ dom( p ) ∪ dom( p )) and find a bijection π : e → ω . For each x ∈ e define q ( x ) ∈ ω × ω to be some point h π ( x ) , m i such that m is greater than anynumber in the π ( x )-th section of the set S ij p ′′ i ( A ( x, i, j ) \ F ). Note that thelatter images have all vertical sections finite by the claim and the assumptionthat p i ≤ c . We claim that q is the required common lower bound.The key thing is to check that q is in fact a Γ -coloring. Suppose that z i for i ∈ q which form a rectangle. There is aconsideration of numerous subcases. Case 1.
If more than two of z i ’s fall out of the set dom( p ) ∪ dom( p ) ∪ dom( p ),then the rectangle is not monochromatic as q is injective on the remainder ofits domain. Case 2.
If all of the z i ’s belong to dom( p ) ∪ dom( p ) ∪ dom( p ), then two ofthem (say z , z ) belong to one of the conditions (say p ). Case 2a.
If either of the remaining points belongs to dom( p ) then the last onedoes too and the rectangle is not monochromatic as p is a Γ -coloring. Note19hat the failure of this case implies that no remaining points belong to V . Case 2b.
Suppose that Case 2a fails, and z , z are opposing vertices in therectangle. Then, by Thalet’s theorem, z belongs to the circle C whose antipodalpoints are z , z . By the pairwise mutual genericity and Proposition 2.4, thecircle C is visible from the ground model. The points z , z are antipodal on C and so belong to the same condition, say p . It follows that p ( z ) = p ( z )must hold by Definition 5.1(3) applied to C with p ≤ c . Thus, the rectangle isnot monochromatic in this case. Case 2c.
Suppose that Case 2a fails, and z , z are neighboring vertices in therectangle. Then, z and z belong to the lines perpendicular to the segmentconnecting z and z and passing through z or z respectively. By the pairwisemutual genericity and Proposition 2.4, these lines are visible from the groundmodel. The points z , z are antipodal on these lines and so belong to the samecondition, say p . It follows that p ( z ) = p ( z ) must hold by Definition 5.1(2)applied to the two lines with p ≤ c . Thus, the rectangle is not monochromaticin this case. Case 3.
There is exactly one point (say z ) which does not belong to dom( p ) ∪ dom( p ) ∪ dom( p ). For definiteness assume that z , z , z enumerate the remain-ing vertices of the rectangle counterclockwise starting from z . Case 3a.
Suppose that z , z do not belong to the same condition. Assumingthat z ∈ dom( p ) and z ∈ dom( p ) for definiteness, we see that z / ∈ V , z ∈ A ( z , ,
1) and therefore q ( z ) = p ( z ) holds by the choice of the color q ( z ). The rectangle is not monochromatic in this case either. Case 3b.
Suppose that Case 3a fails; then z cannot belong to the condition z , z do, since then z would have to belong to it too, which is not the case.Assuming that z , z ∈ dom( p ) and z ∈ dom( p ) for definiteness, we see that z / ∈ V , z ∈ A ( z , ,
1) and therefore q ( z ) = p ( z ) holds by the choice of thecolor q ( z ). The monochromaticity of the rectangle fails in this case as well.The long discussion of the various configurations has ended in the conclusionthat q ∈ P holds. Now we need to show that q is a common lower bound ofthe conditions p , p , p ; by symmetry, it is enough to show q ≤ p . Clearly, p ⊂ q holds. To verify Definition 5.1(2), suppose that l , l are two lines visiblein p and x , x ∈ dom( q \ p ) are antipodal points on them and argue for q ( x ) = q ( x ). If both x , x belong to e then q ( x ) = q ( x ) holds as q ↾ e isan injection. If exactly one of the points x , x (say x ) belongs to e , then x is algebraic over l , l , x and so q ( x ) = q ( x ) by the choice of q ( x ). Finally,if none of the points x , x belongs to e , then the lines l , l are visible in F by Proposition 2.4, so x , x belong to the same condition (say p ) and then q ( x ) = p ( x ) = p ( x ) = q ( x ) follows from p ≤ c . Definition 5.1(3) isverified in the same way.To verify Definition 5.1(4), let u ⊂ d be a finite set and argue that the q -image of the set B = { x ∈ dom( q \ p ) : x is algebraic over supp( p ) ∪ a } has allvertical sections finite. To this end, let B = B ∪ B ∪ B where B = e ∩ B , B = dom( p ) ∩ B and B = dom( p ) ∩ B . Let v ⊂ B be a maximal subsetalgebraically independent over supp( p ) and let v ⊂ B be a maximal subset al-gebraically independent over supp( p ); thus, | v | , | v | ≤ | u | . Use Proposition 2.220o observe that all elements of B are algebraic over F and v , and all elementsof B are algebraic over F and v . It follows then from p ≤ p that the set p ′′ B has all vertical sections finite; similarly, p ≤ p implies that p ′′ B has allvertical sections finite. Also, q ′′ B ⊂ q ′′ e has vertical sections of cardinality atmost one. In total, the set q ′′ B = q ′′ B ∪ q ′′ B ∪ q ′′ B has all vertical sectionsfinite as required. This completes the proof of the theorem.It follows now from Theorem 3.10 that under CH, the rectangle coloring posetis 3 , , Definition 5.5.
Let m ≤ n be natural numbers greater than zero and letΓ ⊂ [ X ] n be a closed hypergraph of arity n . We say that Γ has m degrees offreedom if1. every nonempty open subset of X contains a hyperedge;2. conflating Γ with the subset of X n of all enumerations of hyperedges ofΓ, the projection map from Γ to X m is continuous and open.The verification of the following examples is trivial and left to the reader. Example 5.6.
The equilateral triangle hypergraph Γ △ on R has two degreesof freedom. Example 5.7.
The rectangle hypergraph Γ on R has two degrees of freedom. Example 5.8.
The parallelogram hypergraph Γ on R has three degrees offreedom.The following theorem identifies a class of generic extensions in which hyper-graphs with many degrees of freedom remain uncountably chromatic. Theorem 5.9.
Let Γ be a closed hypergraph on a Polish space X of arity n ≥ and m degrees of freedom where m = 2 ⌊ n/ ⌋ + ( n mod 3) . In cofinally , -balanced extensions of the symmetric Solovay model, the chromatic number of Γ is uncountable.Proof. Let κ be an inaccessible cardinal. Let P be a Suslin forcing which is3 , κ . Let W be a symmetric Solovay model derivedfrom κ , and work in W . Suppose that p ∈ P is a condition and τ is a P -namesuch that p (cid:13) τ : X → ω is a function. We will produce a Γ-hyperedge e ∈ Γ anda condition stronger than p which forces τ ↾ e to be constant. The condition p as well as the name τ must be definable from ground model parameters and anadditional parameter z ∈ ω . Let V [ K ] be an intermediate extension obtainedby a forcing of cardinality smaller than κ such that z, p ∈ V [ K ] and P is 3 , V [ K ]. Work in the model V [ K ].Let ¯ p ≤ p be a 3 , P . Consider theCohen poset P X consisting of nonempty open subsets of X ordered by inclusion,21ith its name ˙ x gen for a generic element of the space X . By a standard Solovaymodel argument, there must be a condition O ∈ P X , a natural number k ∈ ω , aposet Q of cardinality smaller than κ , and a P X × Q -name σ for a condition in theposet P stronger than ¯ p such that O (cid:13) P X Q (cid:13) Coll( ω, < κ ) (cid:13) σ (cid:13) P τ ( ˙ x gen ) = ˇ k .Consider the poset P Γ of nonempty open subsets of Γ ordered by inclusion.The set O ′ ⊂ Γ of all hyperedges which are subsets of O is open and nonemptyby Definition 5.5(1) and therefore a condition in the poset P Γ . Let e be ahyperedge generic over V [ K ] which is a subset of O ′ . By Definition 5.5(2) and[4, Proposition 3.1.1], every m -tuple of elements of e is generic for the m -foldproduct of the poset P X . Let H x ⊂ Q for x ∈ e be filters mutually generic overthe model V [ K ][ e ]. Work in the model V [ K ][ e ][ H x : x ∈ e ].By the arithmetical assumption on m and n , one can find a partition e = e ∪ e ∪ e such that for any choice of indices i, j ∈ | e i ∪ e j | ≤ m . Consider themodels M i = V [ K ][ e i ][ H x : x ∈ e i ] for i ∈
3; these models are pairwise mutuallygeneric extensions of V [ K ]. In each model M i , for each point x ∈ e i consider themodel N x = V [ K ][ x ][ H x ] and the condition p x = σ/x, H x ≤ ¯ p in the model N x .The models N x for x ∈ e i are mutually generic, and by the balance assumptionon the virtual condition ¯ p in V [ K ], the conditions p x for x ∈ e i have a commonlower bound p i ∈ M i . By the 3 , p in V [ K ], the conditions p i for i ∈ W , by the forcing theorem applied in each of the models N x , it must be the case that the common lower bound of the conditions p i for i ∈ P that τ ↾ ˇ e is constant, with the constant value equal to k . Thiscompletes the proof of the theorem.Finally we are in a position to prove Theorem 1.4. The hypergraph Γ △ hasarity three and two degrees of freedom, the hypergraph Γ has arity four andthree degrees of freedom. In both cases, Theorem 5.9 says that in cofinally3 , P is cofinally 3 , P -extension of the symmetricSolovay model, Γ has countable chromatic number, while both Γ △ and Γhave uncountable chromatic number as desired. In this section we prove Theorem 1.6. The simple algebraic hypergraph Γ on R of arity three consisting of those sets a whose projection to both coordinateaxes has cardinality two has the following property. In ZF, if the chromaticnumber of Γ is countable, then there is a countable-to-one map from R to ω .For the proof, argue in ZF. Let c : R → ω be the Γ coloring. For x ∈ R write M x for the model of all sets hereditarily ordinally definable from x and c . Notethat c ↾ M x ∈ M x holds. Case 1.
There is a real x such that R ∩ M x is uncountable. In this case, weshow that R ⊂ M x and M x | =CH, which will prove the theorem.22o show that R ⊂ M x holds, suppose towards contradiction that it doesnot, and pick z ∈ R \ M x . By a counting argument, there are distinct points y , y ∈ R ∩ M x such that c ( y , z ) = c ( y , z ), the common value being some n ∈ ω . Then z is not the unique point such that c ( y , z ) = n –otherwise itwould be definable from c and y , and therefore from c and x , contradictingthe choice of z . Let u ∈ R be a point different from y such that c ( y , u ) = n .Then {h y , z i , h y , z i , h y , u i} is a monochromatic Γ -hyperedge of color n , acontradiction.To show that M x | =CH, suppose towards a contradiction that it fails. Workin M x ; observe that it is a model of AC. Let N be an elementary submodelof some large structure containing c ↾ M x such that N has cardinality ℵ ; let x ∈ X \ N . Let N be an elementary submodel of a large structure containing c ↾ M x , x , and N , such that N is countable. Let x ∈ R ∩ N \ N . Let n = c ( x , x ). By the elementarity of N , there must be u ∈ N such that c ( u, x ) = n . By the elementarity of N , there must be v ∈ N ∩ N such that c ( x , v ). Note that u = x and v = x holds. Clearly, {h x , x i , h u, x i , h x , v i} is a monochromatic Γ -hyperedge of color n , a contradiction. Case 2.
Case 1 fails. Let π : R → ω be the map defined by π ( x ) = ω M x . Thecase assumption shows that the range of this map is indeed a subset of ω . Wewill show that π is in fact countable-to-one. Suppose towards contradiction thatit is not, and let α ∈ ω be an ordinal such that the set { x ∈ R : π ( x ) = α } isuncountable. By the case assumption, there have to be points x , x in this setsuch that x / ∈ M x . We will reach a contradiction by a split into cases.Suppose first that x / ∈ M x . Let L be the line in R consisting of pointswhose 0-th coordinate is equal to x and let L be the line in R consisting ofpoints whose 1-st coordinate is equal to x . Let n = c ( x , x ). Then h x , x i is not the only point on L which gets color n –otherwise x would be definablefrom x . Let h x , x i ∈ L be a different point which gets color n . By the sameargument, h x , x i is not the only point on L which gets color n –otherwise x would be definable from x . Let h x , x i ∈ L be a different point which getscolor n . Then {h x , x i , h x , x i , h x , x i} is a monochromatic Γ -hyperedge ofcolor n . A contradiction.Assume now that x ∈ M x . The set R ∩ M x then belongs to M x andmust be uncountable there because the two models have the same ω . By acounting argument in M x , there must be distinct points y , y ∈ R ∩ M x suchthat h y , x i and h y , x i get the same c -color, say n . Now, x cannot be theonly point such that h y , x i gets the color n –otherwise x would be definablefrom y and then also from x . So, pick a point z ∈ R such that c ( y , z ) = n andnote that the set {h y , x i , h y , x i , h y , z i} is a c -monochromatic Γ -hyperedgeof color n . This is a final contradiction proving Theorem 1.6. References [1] Jack Ceder. Finite subsets and countable decompositions of Euclideanspaces.
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