COLORING ISOSCELES TRIANGLES IN CHOICELESS SET THEORY

Abstract

It is consistent relative to an inaccessible cardinal that ZF+DC holds, and the hypergraph of isosceles triangles on $\mathbb {R}^2$ has countable chromatic number while the hypergraph of isosceles triangles on $\mathbb {R}^3$ has uncountable chromatic number.

1 Introduction

Let $k,n>0$ be natural numbers. An algebraic hypergraph $\Gamma $ of arity k on the Euclidean space $\mathbb {R}^n$ is a subset of $[\mathbb {R}^n]^k$ defined by a polynomial equation with real coefficients in $kn$ unknowns; its elements are called hyperedges. A $\Gamma $ -coloring is a (possibly partial) map c on $\mathbb {R}^n$ such that there is no $\Gamma $ -hyperedge on which $\Gamma $ is constant by c. If there exists a total $\Gamma $ -coloring whose range is countable, the chromatic number of $\Gamma $ is countable; otherwise, the chromatic number of $\Gamma $ is uncountable.

This paper shows one result of comparing chromatic numbers for different algebraic hypergraphs in the context of choiceless set theory ZF+DC. It contributes to one of the numerous paths that balanced forcings in Geometric Set Theory [5] can lead to. In general, one can use balanced forcings to prove independence theorems in choiceless set theory concerning various $\Sigma ^2_1$ sentences.

In the context of ZFC theory, there has been a great deal of interest from people studying the partition/coloring problems on Euclidean spaces related to graphs and hypergraphs. Let $n\geq 2$ be a natural number. Let $\Gamma _n$ be the hypergraph of isosceles triangles on $\mathbb {R}^n$ : $\{x_0,x_1,x_2\} \in \Gamma _n \Longleftrightarrow d_n(x_0,x_1) = d_n(x_1,x_2)$ where $d_n$ is the Euclidean distance on $\mathbb {R}^n$ . $\Gamma _n$ is an algebraic hypergraph arity three.

Fact 1.1 (ZFC)

[8] Let $n\geq 2$ be a natural number. The chromatic number of $\Gamma _n$ is countable.

Erdős and Kakutani conjectured that the conclusion of the theorem holds for all $n\geq 1$ , and proved the case for $n=1$ [2]. Given the Continuum Hypothesis, the theorem was proved by Kunen [4]. Without the Continuum Hypothesis, the theorem was proved by Schmerl [8]. Later, Schmerl provided a complete classification [9]. The chromatic number of any algebraic hypergraph $\Gamma $ on a Euclidean space satisfies a trichotomy: ZFC proves exactly one of the following:

• • The chromatic number is countable.

• • The chromatic number is uncountable.

• • There is a natural number $m\in \omega $ such that the chromatic number is countable if and only if $2^{\aleph _0} \leq \aleph _m$ .

In addition, there is a computer program which, for a given algebraic equation, outputs the number m for the hypergraph given by the equation.

Now in the choiceless context, we realize that there are much finer differences between the chromatic numbers of different hypergraphs. It seems to be frequently the case that the countable chromatic number of one hypergraph doesn’t imply this number for another hypergraph, except for some obvious cases; e.g., for every n, the countable chromatic number of the hypergraph of isosceles triangles on $\mathbb {R}^{n+1}$ implies the one on $\mathbb {R}^{n}$ . We notice that the dimension and the arity of a hypergraph play bigger roles for its chromatic number. The following is the main theorem of this paper.

Theorem 1.2. It is consistent relative to an inaccessible cardinal that ZF+DC holds, and ${\Gamma _2}$ has countable chromatic number while $\Gamma _3$ has uncountable chromatic number.

For this theorem, we will construct a model as a generic extension of the symmetric Solovay model. In fact, the conclusion for the model can be strengthened using the theorems of [5, Sections 9.1, 14.1, and 14.2]. We expect the theorem to hold for higher dimensions.

Conjecture 1.3. For every natural number $n \geq 2$ , it is consistent with ZF+DC that ${\Gamma _n}$ has countable chromatic number while $\Gamma _{n+1}$ has uncountable chromatic number.

There are two main difficulties. The first is to invent a balanced forcing poset which will add a countable total $\Gamma _n$ -coloring. The second difficulty is to refine the poset so that it doesn’t add any countable total $\Gamma _{n+1}$ -coloring, often by investigating the geometric, algebraic, or combinatorial differences. For example, the core distinction to be used in this paper is that $\Gamma _3$ satisfies Definition 6.13 for k at most $4$ while $\Gamma _2$ satisfies Definition 6.13 for k at most $3$ . We observe that similar distinction holds for every n and will be needed for handling the second difficulty. However, the first difficulty of getting a balanced poset increases with higher n. For example, if just adapting the poset that we use in this paper for $\mathbb {R}^3$ , the resulting poset won’t be balanced as Theorem 5.4 will fail due to higher dimension. Therefore, our approach does not generalize to separate the chromatic numbers for hypergraphs of isosceles triangles in dimension three and four. In comparison, the independence of the countable chromatic number for the hypergraph of rectangles between all dimensions has been established in [14]; the independence of the countable chromatic number between the hypergraph of rectangles and the hypergraph of equilateral triangles in all dimensions has been established in [13].

All theorems and definitions in this paper take place in ZFC set theory; let V be a model of ZFC for this paper. The notation in this paper uses the set theoretic standard of [3], and in matters of geometric set theory [5]. Section 2 introduces the basics for DC, the symmetric Solovay model W, the $\mathbb {R}$ -analytic balanced forcing, and RCF. In Section 3, we construct a suitable poset P that adds a countable coloring for $\Gamma _2$ when forced over the model W. Section 4 introduces a natural notion of amalgamation position ¹ of pairs of real closed subfields of $\mathbb {R}$ . Section 5 defines a notion of the analytic forcing with algebraic amalgamation of conditions; such forcing is cofinally balanced. Section 6 shows several open projections from some ordered hypergraphs; then we define the Cohen forcing to add a generic hyperedge for $\Gamma _3$ . Section 7 finally completes the proof of the main theorem.

2 Preliminaries

2.1 Axiom of dependent choices (DC)

Definition 2.1. The axiom of dependent choices $(DC)$ : for every nonempty set X and every entire binary relation R on X, there exists a countable sequence in X such that every adjacent pairs are related.

Theorem 2.2 [1]

Over ZF, DC is equivalent to the Baire category theorem for complete metric spaces.

DC is a fragment of AC on which the real analysis can be developed, possibly without paradoxical AC consequences such as the statement “there exists a set of reals that is not Lebesgue measurable.” The symmetric Solovay model [10] is a model of ZF+DC that we use in this paper.

2.2 Symmetric Solovay model

Definition 2.3. If X is a set, then Coll $(\omega ,X)$ is the poset of finite partial functions from $\omega $ to X ordered by reverse extension; if $\kappa $ is a cardinal, then Coll $(\omega ,< \kappa )$ is the finite support product of the posets Coll $(\omega , \alpha )$ for all $\alpha \in \kappa $ .

Any collapse poset obeys an important factoring rule.

Fact 2.4 [3, Corollary 26.11]

Let $\lambda $ be a cardinal and let P be a poset of cardinality $<\lambda $ . Suppose that $G\subset $ Coll( $\omega ,\lambda $ ) is a generic filter and in $V[G]$ , $H\subset P$ is a filter generic over $V $ . Then there is a filter $K \subset Coll(\omega , \lambda )$ generic over $V [H]$ such that $V [G] = V [H][K]$ .

Definition 2.5 [10]

Let $\kappa $ be an inaccessible cardinal and let $G\subset \mbox {Coll}(\omega , < \kappa )$ be a generic filter over V. In $V [G]$ , form HOD $(V\cup \omega ^\omega )$ , the model of all sets hereditarily definable from reals and elements of the ground model. This is the symmetric Solovay model; we denote it by W.

The theory of the symmetric Solovay model has been fulled developed, including the following one:

Fact 2.6 [10, 11]

In W, ZF+DC holds, every set has the Baire property and is Lebesgue measurable, and there is no uncountable well-ordered sequence of distinct Borel sets of bounded Borel rank.

We will make use of the following technical facts in the paper.

Fact 2.7 [3, Section 26], [10]

Let $\kappa $ be an inaccessible cardinal and let W be the symmetric Solovay extension of V associated with $\kappa $ . Then the following hold in $W:$

1. 1. Eevery set is definable from parameters in $V \cup \omega ^\omega $ .

2. 2. Whenever M is a generic extension of V by a partial order of cardinality smaller than $\kappa $ , then W is a symmetric Solovay extension of M.

3. 3. Whenever a is a set of ordinals, then $V [a]$ is a generic extension of V by a poset of cardinality smaller than $\kappa $ .

Fact 2.8 [5, Corollary 1.7.17], [10]

Suppose that $\kappa $ is an inaccessible cardinal, X is a Polish space, $\phi ( x,\bar y)$ is a formula of set theory with all free variables displayed, and $\bar z$ is a sequence of sets of the same length as $\bar y$ . The following are equivalent:

1. 1. Coll( $\omega ,< \kappa ) \Vdash \exists x \in X \ \phi (x,\bar z)$ .

2. 2. There exist a poset R of cardinality smaller than $\kappa $ and an R-name $\sigma $ for an element of X such that $R \Vdash $ Coll( $\omega , < \kappa ) \Vdash \phi (\sigma ,\bar z)$ .

2.3 $\mathbb {R}$ -analytic balanced forcing

In this paper we deal with analytic forcings, a looser notion than Suslin forcings from the book [5, Section 5.2].

Definition 2.9. An analytic forcing is a poset $\langle P,\leq \rangle $ for which there is a Polish space X so that:

1. 1. P is an analytic subset of X.

2. 2. The partial order relation $\leq $ is an analytic subset of $X^2$ .

Being different from Suslin forcings, analytic forcings don’t require the incompatibility relation $\{\langle p_0, p_1\rangle \in P^2 : \forall q \in P \ q\not \leq p_0 \vee q\not \leq p_1\}$ to be also an analytic subset of $X^2$ . Note that the compatibility relation is automatically analytic.

Definition 2.10. An $\mathbb {R}$ -analytic forcing is an analytic forcing P such that:

1. 1. Each condition $p \in P$ is a structure in some fixed relational first order language whose universe $\mathrm {supp}(p)$ is a countable real closed subfield of $\mathbb {R}$ .

2. 2. $p_1 \leq p_0$ implies $p_1 \upharpoonright \mathrm {supp}(p_0) = p_0$ .

3. 3. For every countable set $a \subset \mathbb {R}$ and every condition $p_0 \in P$ there is a condition $p_1 \leq p_0$ such that $a \subset \mathrm {supp}(p_1)$ .

4. 4. P is $\sigma $ -closed: whenever $\langle p_n : n \in \omega \rangle $ is a descending sequence of conditions then $\bigcup _n p_n \in P$ is their lower bound.

Thus, an $\mathbb {R}$ -analytic forcing adds a certain generic structure on $\mathbb {R}$ which is the union of all (countable) structures in the generic filter. By quantifier counting procedure, one can check being an $\mathbb {R}$ -analytic forcing is a $\Pi ^1_2$ statement and therefore absolute throughout all forcing extensions by the Shoenfield absoluteness [3, Theorem 25.20].

Balanced forcings are fully developed in the book [5, Section 5.2]. Intuitively speaking, a balanced condition is a condition $p_0$ such that, under some circumstance, $\forall p_1,p_2\leq p_0\ \exists p_3\leq p_1,p_2 $ holds. Rigorously, one needs the notion of balanced pairs and balanced virtual conditions ² by taking the quotient of some equivalence relation on balanced pairs as [5, Definitions 5.1.6, 5.1.7, 5.2.1, and 5.2.5].

Definition 2.11 [5, Definition 5.2.10]

Let P be an analytic forcing. P is balanced if for every condition $p \in P$ there is a balanced virtual condition below p.

Definition 2.12 [5, Section 1.7]

Let P be an analytic forcing. Let $\kappa $ be an inaccessible cardinal. P is balanced cofinally below $\kappa $ if $V_\kappa \models $ every forcing extension has a further extension in which P is balanced.

Cofinally balanced analytic forcings do not add any well-ordered sequences of elements of the symmetric Solovay model:

Theorem 2.13 [5, Theorem 9.1.1]

Let $\kappa $ be an inaccessible cardinal. Let W be the symmetric Solovay model derived from $\kappa $ . Let P be balanced cofinally below $\kappa $ . Then in any P extension of W, every well-ordered sequence of elements of W belongs to W.

As a consequence, any cofinally balanced analytic forcing has a limitation: there is no balanced analytic forcing for the hypergraph of right triangles, according to [12].

2.4 Real closed fields (RCF)

Consider the real closed field $\langle \mathbb {R}, +, \cdot , \leq \rangle $ : 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 fields (RCF) is well established in [6, Section 3.3]. We will make use of two key facts: real closed fields satisfy quantifier elimination [6, Theorem 3.3.15] and they are model complete [6, Corollary 3.3.16]: if $F\subset \mathbb {R}$ is a real closed subfield, then F is an elementary submodel of $\mathbb {R}$ .

We have a model theoretic notion of “algebraic” in the context of real closed fields.

Definition 2.14. Let $m,n>0$ be natural numbers.

• • If $x_0 \in \mathbb {R}$ is a point and $b \subset \mathbb {R}$ is a set, we say that $x_0$ is an algebraic point over b if there is a polynomial $p(x, \bar y)$ with coefficients in $\mathbb {Z}$ and a tuple $\bar y_0$ with each coordinate from $ b$ such that $p(x_0, \bar y_0) = 0$ and $\{x \in \mathbb {R} : p(x, \bar y_0) = 0\}$ is finite.

• • If $\bar x_0 \in \mathbb {R}^n$ is a point and $b\subset \bigcup _ {m'\in m}\mathbb {R}^{m'}$ is a set, we say that $\bar x_0$ is an algebraic point over b if each coordinate of $\bar x_0$ is an algebraic point in $\mathbb {R}$ over the set $\{\beta : \beta $ is a coordinate of some element in $b \} \subset \mathbb {R}$ .

• • If $A \subset \mathbb {R}^n$ is a set and $b \subset \mathbb {R}$ is a set, we say that A is an algebraic set over b if there is a polynomial $p(\bar x, \bar y)$ with coefficients in $\mathbb {Z}$ and a tuple $\bar y_0$ with each coordinate from $ b$ such that $A= \{\bar x \in \mathbb {R}^n : p(\bar x, \bar y_0) = 0\}$ .

There is a fact that bridges between algebraic points and algebraic sets.

Fact 2.15. Let $n>0$ be a natural number. Let F be a real closed field. If $A \subset F^n$ is a finite set, $b \subset F$ is a set, and A is an algebraic set over b, then every point in A is an algebraic point over b.

Proof Assume $A \subset F^n$ is a nonempty finite set, $b \subset F$ is a set, and A is an algebraic set over b. Then there is a polynomial $p(\bar x, \bar y)$ with coefficients in $\mathbb {Z}$ and a tuple $\bar y_0$ with each coordinate from $ b$ such that $A= \{\bar x \in \mathbb {R}^n : p(\bar x, \bar y_0) = 0\}$ . Fix a coordinate $i\in n$ . Then the ith projection $A|_i = \{x\in \mathbb {R}^{\{i\}} : \exists \bar x \in \mathbb {R}^{n\setminus \{i\} }\ p(x, \bar x, \bar y_0) = 0\}$ has cardinality at least one and at most $|A| < \infty $ . By quantifier elimination of the real closed field F, there is a quantifier-free formula $\phi $ such that $A|_i = \{x\in \mathbb {R}^{\{i\}} :\phi (x, \bar y_0)\}$ . We may assume that all atomic formulas appearing in $\phi $ are of the form“ $q(x, \bar y)>0$ ” or “ $q(x, \bar y)=0$ ” for some polynomial q with integer coefficients. Let $\{q_j : j \in k\}$ be a list of all polynomials appearing in $\phi $ . Let $x_i\in A|_i$ . Let $a = \{j \in k: q_j(x_i, \bar y_0) = 0\}$ . Then $\psi (x, \bar y) = \sum _{j\in a} q_j^2 (x, \bar y) $ is a polynomial with single variable x such that $\psi (x_i, \bar y_0) = 0$ where $\bar y_0$ has every coordinate from $ b$ .

Finally, $\psi (x, \bar y_0) $ cannot be the zero polynomial. Suppose it is the zero polynomial for the sake of contradiction. Then for every $ j$ , $q_j(x, \bar y_0)$ is the zero polynomial. Since $\phi (x, \bar y_0)$ is neither a tautology nor a contradiction as $0<|(A|_i) |<\infty $ , we can remove every $q_j(x, \bar y_0) = 0 $ inside the formula $\phi (x, \bar y_0)$ , so all atomic formulas appearing in $\phi (x, \bar y_0)$ are of the form“ $q(x, \bar y_0)>0$ ” for some polynomial q with integer coefficients. In any case, the set $A|_i = \{x\in \mathbb {R}^{\{i\}} :\phi (x, \bar y_0)\}$ must be either empty or infinite, which is a contradiction. Hence $\psi (x, \bar y_0) $ has finitely many roots. Hence the ith coordinate of each element in A is an algebraic point over b. Hence every point in A is an algebraic point over b.

We note that the model theoretic notion of “algebraic” agrees with the field-theoretic notion of “algebraic” for real closed fields (RCF). It is also the case for algebraic closed fields (ACF) as [6, Proposition 3.2.15]. Because quantifier elimination is satisfied in both RCF and ACF, two proofs are the same.

Fact 2.16. Let $K \models $ RCF and $A \subset K$ . Then $x $ is an algebraic point over A if and only if x is algebraic over the subfield of K generated by A.

3 Construction of the poset

Recall that $\Gamma _2$ is the hypergraph of isosceles triangles on $\mathbb {R}^2$ . In this section, we will produce a balanced $\mathbb {R}$ -analytic poset P adding a total $\Gamma _2$ -coloring. The model for the main theorem of this paper will be a P-generic extension of the symmetric Solovay model. For any set $b \subset \mathbb {R}$ , let $E^L(b)$ be the set of lines in $\mathbb {R}^2$ algebraic over b; let $E^O (b)$ be the set of circles in $\mathbb {R}^2$ algebraic over b. For any map p from a subset of $\mathbb {R}^2$ to $\omega $ and for any line $l\subset \mathbb {R}^2$ , let $s(p, l)$ be the set $\{i\in \omega : \mbox {there are points }y_0, y_1 $ in $\mathrm {dom}(p)$ symmetrical ³ with respect to $ l $ and $ p(y_0) = p(y_1) = i\}$ . Fix a Borel ideal $I $ on $\omega $ that contains all finite sets and it is not generated by countably many sets. Most ideals that people use have such property, such as the summable ideals. The particular choice of the ideal seems unimportant.

Definition 3.1. Define P to be the poset of all conditions p such that:

1. 1. $\mathrm {supp}(p)$ is a countable real closed subfield and $\mathrm {supp}(p)^2 = \mathrm {dom}(p) \subset \mathbb {R}^2$ ;

2. 2. $ p: \mathrm {dom}(p) \rightarrow \omega $ is a partial $\Gamma _2$ -coloring;

3. 3. (the symmetrical colors requirement) for each line $l \in E^L(\mathrm {supp}(p)) $ , $s(p, l) \in I$ holds.

The ordering is defined by $p_1 \leq p_0$ if:

1. 4. $\mathrm {dom}(p_1) \supset \mathrm {dom}(p_0)$ ;

2. 5. $p_1\upharpoonright {\mathrm {dom}(p_0)} = p_0$ ;

3. 6. (the symmetrical colors invariant requirement) for each $l \in \ E^L(\mathrm {supp}(p_0)) $ , $s(p_0, l) = s(p_1, l)$ holds;

4. 7. (the avoid center requirement) for each circle $e\in E^O(\mathrm {supp}(p_0))$ , for each point $ x \ \in e \cap ( \mathrm {dom}(p_1)\setminus \mathrm {dom}(p_0)), p_1(x) \neq p_0(t) $ holds where t is the center of e;

5. 8. (the algebraic points requirement) for every finite set $ a \subset \mathrm {supp}(p_1)$ , the $p_1$ -image of the set $\{x \in \mathrm {dom}(p_1)\setminus \mathrm {dom}(p_0)$ : $x $ is algebraic over $\mathrm {supp}(p_0)\cup a\} $ is in $ I$ .

First and foremost, one can check from Definition 3.1 that the underlying set and the ordering of the poset P are Borel (as each quantifier is quantifying over a countable set, by rewriting the definition if necessary). Automatically, the compatibility relation is also analytic. Therefore, the poset P is analytic. We will be checking that P is an $\mathbb {R}$ -analytic poset as Definition 2.10 for the rest of the section. For every condition $p\in P$ , we have some useful real closure facts for $\mathrm {supp}(p)$ .

Fact 3.2. Let $p\in P$ be a condition. Let $x_0,x_1,x_2$ be distinct points in $\mathrm {dom}(p) $ . Then the following hold:

1. 1. The unique line that passes through $x_0,x_1$ is in $E^L(\mathrm {supp}(p))$ .

2. 2. The unique line that is the perpendicular bisector of the line segment from $x_0$ to $x_1$ is in $E^L(\mathrm {supp}(p))$ .

3. 3. The unique circle centered at $x_0$ passes through $x_1$ is in $E^O(\mathrm {supp}(p))$ .

4. 4. If $x_0,x_1,x_2$ are not colinear, the unique circle that passes through $x_0,x_1,x_2$ is in $E^O(\mathrm {supp}(p))$ .

5. 5. For all distinct line $(s)$ or circle $(s) \ e_0,e_1\in E^L(\mathrm {supp}(p))\cup E^O(\mathrm {supp}(p))$ , $e_0\cap e_1\subset \mathrm {dom}(p)$ holds.

6. 6. For every line $e\in E^L(\mathrm {supp}(p))$ and every point $x\in \mathrm {dom}(p)\setminus e$ , the point that is symmetrical to x with respect to e is in $\mathrm {dom}(p)$ .

7. 7. For every circle $ e\in E^O(\mathrm {supp}(p))$ , the center of e is in $\mathrm {dom}(p)$ .

Proof Let us prove a few from above. Other proofs will be similar. Let $d_2$ be the Euclidean distance on $\mathbb {R}^2$ .

For (3), the unique circle centered at $x_0$ passes through $x_1$ is the set $\{y\in \mathbb {R}^2: d_2(y, x_0) = d_2(x_1, x_0)\}$ , and clearly this set is a circle algebraic over $\mathrm {supp}(p)$ since $x_0,x_1$ are distinct points in $\mathrm {dom}(p) $ ; hence this set is in $E^O(\mathrm {supp}(p))$ .

For (4), it is easy to see that $\{y: d_2(y, x_0) = d_2(y, x_1) =d_2(y, x_2) \}$ is the set $\{c\} $ for some $c\in \mathbb {R}^2$ . As a consequence of Fact 2.15, c is in $\mathrm {dom}(p)$ using the real closure property of $\mathrm {supp}(p)$ and by definition of $\mathrm {dom}(p) $ . Then the unique circle that passes through $x_0,x_1,x_2$ is just the unique circle centered at c that passes through $x_1$ , which is in $E^O(\mathrm {supp}(p))$ by (3).

(5) uses the fact that the intersection of two algebraic sets is algebraic and Fact 2.15.

For (7), fix a circle $ e\in E^O(\mathrm {supp}(p))$ . We just need to somehow find three non-colinear points on $e\cap \mathrm {dom}(p)$ , then as in the proof for (4), the center of e is in $\mathrm {dom}(p)$ . Since $\mathrm {supp}(p)$ contains all the rational points, there are densely many parallel lines algebraic over $\mathrm {supp}(p)$ . In particular, there are two lines that intersect e at at least three points that are non-colinear, and they are also in $\mathrm {dom}(p)$ by (5).

Other than the concrete proofs from above, one can also do much shorter proofs using the fact that $\mathrm {supp}(p)$ is an elementary submodel of $\mathbb {R}$ .

Fact 3.3. P is not empty.

Proof Take any set $b \subset \mathbb {R}$ a nonempty countable real closed subfield of $\mathbb {R}$ . Just take an injection $p:\mathrm {dom}(p) \rightarrow \omega $ where $\mathrm {dom}(p) = b^2 $ . It is easy to see that $p\in P$ ; therefore P is not empty.

Fact 3.4. $\langle P, \leq \rangle $ is transitive.

Proof Assume $p_2 \leq p_1 \leq p_0 \in P$ . We want to prove that $p_2 \leq p_0 $ . We will see that checking Definition 3.1(4)–(7) is not more difficult than just looking at the definitions. For Definition 3.1(4) and (5), it’s clear that $\mathrm {dom}(p_2) \supset \mathrm {dom}(p_0)$ and $p_2\upharpoonright {\mathrm {dom}(p_0)} = p_0$ . For Definition 3.1(6) the symmetrical colors invariant requirement: let $e \in E^L(\mathrm {supp}(p_0)) \subset E^L(\mathrm {supp}(p_1))$ ; then $s(p_0, e) = s(p_1, e) = s(p_2, e)$ holds because of the symmetrical colors invariant requirement for $p_2 \leq p_1$ and $p_1 \leq p_0$ .

For Definition 3.1(7) the avoid center requirement: let $e\in E^O(\mathrm {supp}(p_0))$ ; let $ x \in e \cap ( \mathrm {dom}(p_2)\setminus \mathrm {dom}(p_0))$ ; let t be the center of the circle e. Then:

• • either $ x \in e \cap ( \mathrm {dom}(p_1)\setminus \mathrm {dom}(p_0))$ : then by the avoid center requirement for $p_1\leq p_0$ , $p_2(x) = p_1(x) \neq p_0(t) $ ;

• • or $ x \in e \cap ( \mathrm {dom}(p_2)\setminus \mathrm {dom}(p_1))$ : then by the avoid center requirement for $p_2\leq p_1$ , $p_2(x) \neq p_1(t) = p_0(t) $ .

In either case, the avoid center requirement is satisfied on $x\in e$ .

For Definition 3.1(8) the algebraic points requirement: fix a finite set $ u \subset \mathrm {supp}(p_2)$ . Let v be an inclusion-maximal subset of $\mathrm {supp}(p_1)$ intersected with the algebraic closure of $\mathrm {supp}(p_0)\cup u$ , which is algebraically independent over $\mathrm {supp}(p_0)$ . Since the sets algebraically independent over $\mathrm {supp}(p_0)$ form a matroid by [7, Theorem 6.7.1], v is finite (and in fact $|v| \leq |u|$ ). Let

• • $a_1 =$ $p_2^{\prime \prime }\{x \in \mathrm {dom}(p_2)\setminus \mathrm {dom}(p_1)$ : $x $ is algebraic over $\mathrm {supp}(p_1)\cup u\}$ ;

• • $a_0 =$ $p_1^{\prime \prime }\{x \in \mathrm {dom}(p_1)\setminus \mathrm {dom}(p_0)$ : $x $ is algebraic over $\mathrm {supp}(p_0)\cup u\}.$

Then $a_1\in I$ is by the algebraic points requirement for $p_2\leq p_1$ ; $a_0 \subset p_1^{\prime \prime }\{x \in \mathrm {dom}(p_1)\setminus \mathrm {dom}(p_0)$ : $x $ is algebraic over $\mathrm {supp}(p_0)\cup v\}\in I$ is by our choice of v and by the algebraic points requirement for $p_1\leq p_0$ . Hence $p_2^{\prime \prime }\{x \in \mathrm {dom}(p_2)\setminus \mathrm {dom}(p_0)$ : $x $ is algebraic over $\mathrm {supp}(p_0)\cup u\} \subset a_0\cup a_1\in I$ .

Fact 3.5. P is $\sigma $ -closed.

Proof We will show that for any descending sequence $\{p_n\}_{n\in \omega }$ of conditions from P, $ q = \bigcup _{n\in \omega } p_n \in P$ is their lower bound. For Definition 3.1(1), $\mathbb {R} \supset \mathrm {supp}(q) = \mathrm {supp}(\bigcup _{n\in \omega } p_n) = \bigcup _{n\in \omega } \mathrm {supp}(p_n)$ is still a countable real closed subfield and $\mathrm {supp}(q)^2 = \mathrm {dom}(q) \subset \mathbb {R}^2$ .

For Definition 3.1(2), if a q-monochromatic triple were found in $\mathrm {dom}(q)$ , the triple would be found in $\mathrm {dom}(p_n)$ for some n. But because $p_n\in P$ is a ${\Gamma _2}$ -coloring, the triple is not $p_n$ -monochromatic hence not q-monochromatic, which is a contradiction.

Observe that $E^L(\mathrm {supp}(q)) = E^L(\bigcup _n\mathrm {supp}(p_n))= \bigcup _n E^L(\mathrm {supp}(p_n))$ and $E^O(\mathrm {supp}(q)) = E^O(\bigcup _n \mathrm {supp}(p_n))= \bigcup _n E^O(\mathrm {supp}(p_n))$ . For Definition 3.1(3) the symmetrical colors requirement: let $l \in E^L(\mathrm {supp}(q)) $ , then $l \in E^L(\mathrm {supp}(p_n) )$ for some n. Then $s(q, l) = \bigcup _{n< i\in \omega } s(p_i, l) = s(p_n, l) \in I $ because for every $ i> n$ , $s(p_i, l) = s(p_n, l) \in I $ by the symmetrical colors invariant requirement for $p_i \leq p_n$ .

For the rest of the proof, we will check $\forall n\: q \leq p_n $ holds. Fix a number n. Definition 3.1(4) and (5) is clear: $\mathrm {dom}(q) \supset \mathrm {dom}(p_n)$ and $q \upharpoonright {\mathrm {dom}(p_n)} = p_n$ . For Definition 3.1(6) the symmetrical colors invariant requirement: $s(q, l) = \bigcup _{n< i\in \omega } s(p_i, l) = s(p_n, l) $ because $\forall i> n\ \ s(p_i, l) = s(p_n, l)$ by the symmetrical colors invariant requirement for $p_i \leq p_n$ .

For Definition 3.1(7) the avoid center requirement: let $e\in E^O(\mathrm {supp}(p_n))$ be a circle and let t be the center of the circle e; then $t\in \mathrm {dom}(p_n) $ by Fact 3.2. Let $ x \in e \cap ( \mathrm {dom}(q)\setminus \mathrm {dom}(p_n)) $ , then there is some $m>n$ such that $ x \in e \cap ( \mathrm {dom}(p_m)\setminus \mathrm {dom}(p_n)) $ . Then $p_m(x) \neq p_m(t)$ because the avoid center requirement is satisfied for $p_m \leq p_n$ . Then $q (x) = p_m(x) \neq p_m(t) = q(t) $ .

For Definition 3.1(8) the algebraic points requirement: fix a finite set $ a \subset \mathrm {supp}(q)$ . Find a number $m>n$ such that $a\subset \mathrm {supp}(p_m)$ . Then the set $ \{x \in \mathrm {dom}(q)\setminus \mathrm {dom}(p_n)$ : $x $ is algebraic over $\mathrm {supp}(p_n)\cup a\}$ is a subset of $\mathrm {dom}(p_m)$ since $\mathrm {supp}(p_m)$ is a real closed field. Since $ p_m \leq p_n$ , the $p_m$ -image of the set $\{x \in \mathrm {dom}(q)\setminus \mathrm {dom}(p_n)$ : $x $ is algebraic over $\mathrm {supp}(p_n)\cup a\}$ equaling the q-image of the same set is in the ideal I.

Theorem 3.6. $\forall x \in \mathbb {R}^2, \forall p \in P$ , there is a condition $ q \in P$ such that $x\in \mathrm {dom} (q)$ and $ q \leq p $ hold.

In fact, this theorem is a consequence of Theorem 5.4, but we still provide this shorter and tighter proof.

Proof Given a point $x \in \mathbb {R}^2$ , a condition $p \in P$ . If $x\in \mathrm {dom}(p)$ , then there is nothing to do. So suppose $x\notin \mathrm {dom}(p)$ . Choose a countable real closed field $b\subset \mathbb {R}$ such that $\mathrm {supp}(p)\subset b $ and $x\in b^2$ . We will extend the ${\Gamma _2}$ -coloring p to some q with $\mathrm {dom}(q) = b^2$ .

Fix any set $\pi \in I $ such that $\forall l \in E^L(\mathrm {supp}(p))\: \pi \setminus s(p, l)$ is infinite. It is possible because for every line $l \in E^L(\mathrm {supp}(p))$ , $s(p, l)\in I$ by the symmetrical colors requirement for p, and by the feature of our choice of I. Fix a point $x \in b^2 \setminus \mathrm {dom}(p)$ . Note that there is at most one line or one circle $e\in E^L(\mathrm {supp}(p)) \cup E^O(\mathrm {supp}(p))$ such that $x \in e$ : using Fact 3.2(5), if there were different such $e, e' \in E^L(\mathrm {supp}(p)) \cup E^O(\mathrm {supp}(p))$ that $x\in e\cap e'$ , then x would be in $\mathrm {dom}(p)$ which is a contradiction. Let’s define the set of allowed colors for the point x, denoted by $a(x)$ :

• • If x is on some line $e\in E^L(\mathrm {supp}(p)) $ , then let $a(x) = \pi \setminus s( p, e) $ .

• • If x is on some circle $e\in E^O(\mathrm {supp}(p))$ , then let $a(x) = \pi \setminus \{p(t)\} $ where t is the center of e.

Since I is an ideal, and by our choice of $\pi \in I$ , the set $a(x) $ is infinite. Now just let $q: b^2\rightarrow \omega $ be a map such that:

• • $q\upharpoonright {\mathrm {dom}(p)} = p $ .

• • $\forall x\in b^2 \setminus \mathrm {dom}(p),\: q (x) \in a(x)$ .

• • q on $ b^2 \setminus \mathrm {dom}(p)$ is an injection.

Claim 3.7. $q \in P$ .

Proof Definition 3.1(1) is satisfied as we required $\mathrm {supp}(q) = b\subset \mathbb {R} $ to be a countable real closed subfield and $\mathrm {supp}(q)^2 = \mathrm {dom}(q) \subset \mathbb {R}^2$ .

For Definition 3.1(2), let’s check q is a ${\Gamma _2}$ -coloring for $\mathrm {dom}(q)$ . Assume there are points $ x,y,z\in \mathrm {dom}(q)$ such that $\{x,y,z\} \in {\Gamma _2}$ .

Case 1: at least two points in $\{x,y,z\}$ are from $\mathrm {dom}(q)\setminus \mathrm {dom}(p)$ . Then $q"\{x,y,z\} $ is not a singleton because we required q to be an injection on $b^2\setminus \mathrm {dom}(p) = \mathrm {dom}(q) \setminus \mathrm {dom}(p) $ .

Case 2: $x,y,z \in \mathrm {dom}(p)$ . Then because p is a ${\Gamma _2}$ -coloring on $\mathrm {dom}(p)$ , $q"\{x,y,z\} = p"\{x,y,z\} $ is not a singleton.

Without loss of generality, we have the following case left.

Case 3: $x \in \mathrm {dom}(q)\setminus \mathrm {dom}(p)$ and $y,z \in \mathrm {dom}(p)$ . We have two sub-cases depending on the pivot point of the isosceles triangle $\{x,y,z\}$ .

Case 3.1: x is the pivot point. Then x is on the perpendicular bisector l of the line segment from y to z, and because $y,z \in \mathrm {dom}(p)$ , l was in $ E^L(\mathrm {supp}(p))$ by Fact 3.2(2). But by the definition of $q(x)\in a(x)$ , $q(x) \notin s(p, l) \ni p(y) = p(z)$ so $q(x) \neq p(y) = p(z) = q(y) = q(z)$ ; hence $\{x,y,z\}$ is not monochromatic by q.

Case 3.2: y or z is the pivot point. Without loss of generality, let’s say $ y$ is the pivot point. Then $ x$ is on the circle centered at $ y$ passing through $ z$ , which was in $\mathrm {dom}(p)$ because of $ y, z \in \mathrm {dom}(p)$ and by Fact 3.2(3). But in the definition of $a(x)$ , we made sure $q( x) \neq q( y) = p( y)$ ; hence $\{x,y,z\}$ is not monochromatic by q.

For Definition 3.1(3) the symmetrical colors requirement, let $l \in E^L(\mathrm {supp}(q)) $ .

Case 1: $l \in E^L(\mathrm {supp}(p)) $ . Then $s(p, l) \in I$ since $p\in P$ ; we also have $s(q, l) = s(p, l)$ that we will check later as Definition 3.1(6); hence $s(q, l) \in I$ .

Case 2: $l \in E^L(\mathrm {supp}(q)) \setminus E^L(\mathrm {supp}(p)) $ . Then for any pair of points symmetrical with respect to l, at least one point in the pair has to be in $\mathrm {dom}(q)\setminus \mathrm {dom}(p)$ because otherwise it would be the case that $l \in E^L(\mathrm {supp}(p)) $ by Fact 3.2(2), which would lead to the previous case. Because of $q"(\mathrm {dom}(q)\setminus \mathrm {dom}(p))\in \pi $ , $s(q, l) \subset \pi \in I$ holds; thus $s(q, l) \in I$ holds.

Claim 3.8. $q\leq p$ .

Proof For Definition 3.1(4) and (5), indeed $\mathrm {dom}(q) \supset \mathrm {dom}(p)$ and $q\upharpoonright {\mathrm {dom}(p)} = p$ . For Definition 3.1(6) the symmetrical colors invariant requirement, first of all, obviously $s(p, l) \subset s(q, l)$ . We will show that $s(q, l)\subset s(p, l)$ . Let $l \in E^L(\mathrm {supp}(p)) $ . Any pair of points that are symmetrical with respect to l has to be both in $\mathrm {dom}(p)$ or both in $\mathrm {dom}(q) \setminus \mathrm {dom}(p) $ , again using Fact 3.2(6). For the pair of points that are both in $\mathrm {dom}(q) \setminus \mathrm {dom}(p) $ , they have distinct q-colors since q is an injection on $\mathrm {dom}(q)\setminus \mathrm {dom}(p)$ . Then $s(q, l)$ can’t include more colors than $s(p, l)$ does. Hence $s(p, l) = s(q, l)$ .

For Definition 3.1(7) the avoid center requirement, suppose that $x\in \mathrm {dom}(q\setminus p)$ is a point and $e\in E^C(\mathrm {dom}(p))$ is an old circle such that $x\in e$ . Since $q(x)\notin a(x)$ , it follows that $q(x)\neq p(y)$ where y is the center of e. For Definition 3.1(8) the algebraic points requirement, in fact we even have $q"(\mathrm {dom}(q) \setminus \mathrm {dom}(p)) \subset \pi \in I $ so that any subset of $q"(\mathrm {dom}(q) \setminus \mathrm {dom}(p)) $ is in the ideal I by the property of an ideal.

This completes the proof of the theorem.

As a consequence, P forces the union of the generic filter to be a total $\Gamma _2$ -coloring.

4 Fields in amalgamation position

Mutual genericity is at the heart of the balanced forcing technology. In this section, we introduce an algebraic counterpart to it which has the advantage of removing the forcing relation when showing our poset to be balanced.

Definition 4.1. Let $F_0,F_1$ be real closed subfields of $\mathbb {R}$ . We say that $F_0 , F_1$ are in amalgamation position if for every polynomial $p(\bar {x}_0,\bar {x}_1 $ ) with integer coefficients and all variables listed, and every pair of tuples $\bar r_0$ from $F_0$ and $\bar r_1$ from $F_1$ such that $p(\bar {r}_0,\bar {r}_1) = 0$ holds, and all open neighborhoods $O_0$ and $O_1$ of $\bar {r}_0,\bar {r}_1$ respectively, there are tuples $\bar {r}_0^{\prime }\in O_0,\bar {r}_1^{\prime }\in O_1$ from $F_0\cap F_1$ such that $p(\bar {r}_0^{\prime },\bar {r}_1) = p(\bar {r}_0,\bar {r}_1^{\prime }) = 0$ .

With the help of quantifier elimination for real closed fields, the amalgamation position yields seemingly stronger consequences for the two fields than stated in the definition.

Proposition 4.2. Let $F_0,F_1$ be real closed subfields of $\mathbb {R}$ in amalgamation position. Let $\phi (\bar {x}_0,\bar {x}_1)$ be a formula of real closed fields with all free variables listed. For every pair of tuples $\bar {r}_0$ from $F_0$ and $\bar {r}_1$ from $F_1$ such that $\phi (\bar {r}_0,\bar {r}_1)$ holds, there are tuples $\bar {r}_0^{\prime },\bar {r}_1^{\prime }$ from $F_0\cap F_1$ such that $\phi (\bar {r}_0^{\prime },\bar {r}_1) $ and $ \phi (\bar {r}_0,\bar {r}_1^{\prime }) $ both hold.

Proof By the quantifier elimination for real closed fields, we can assume that $\phi $ is quantifier-free. We also assume that all atomic formulas appearing in it are of the form $p(\bar x_0, \bar x_1) = 0$ or $ p(\bar x_0, \bar x_1)>0 $ for some polynomial p with integer coefficients. Let $\{p_i : i \in n\}$ be a list of all polynomials appearing in $\phi $ . Take tuples $\bar {r}_0$ from $F_0$ and $\bar {r}_1$ from $F_1$ such that $\phi (\bar {r}_0,\bar {r}_1)$ holds. Let $a = \{i \in n: p_i(\bar r_0, \bar r_1) \neq 0\}$ . Let $O_0, O_1$ be open neighborhoods of $\bar r_0, \bar r_1$ respectively such that for all $i \in a$ , the polynomials $p_i$ have constant sign on $O_0 \times O_1$ . Use the amalgamation position assumption as in Definition 4.1 to find tuples $\bar r_0^{\prime }\in O_0$ and $\bar r_1^{\prime }\in O_1$ from $F_0 \cap F_1$ such that $\sum _{i\in n\setminus a} p_i^2 (\bar r_0^{\prime }, \bar r_1) = 0 = \sum _{i\in n\setminus a} p_i^2 (\bar r_0, \bar r_1^{\prime })$ . One can check that the tuples $ \bar r_0^{\prime }, \bar r_1^{\prime }$ work as required.

Then we have a quick corollary as the following, although it is true without the amalgamation position requirement.

Corollary 4.3. Let $n>0$ be a natural number. Assume that $F_0,F_1$ are real closed subfields of $\mathbb {R}$ in amalgamation position and $A\subset \mathbb {R}^n$ is an algebraic set over both $F_0$ and $F_1$ . Then $A\subset \mathbb {R}^n$ is an algebraic set over $F_0\cap F_1$ .

Proof Let $p_0,p_1$ be polynomials such that there are some tuples $\bar y_0$ from $F_0$ , $\bar y_1$ from $F_1$ , $\bar z$ from $F_0\cap F_1$ so that $A = \{\bar x : p_0(\bar x,\bar y_0, \bar z) = 0 \} = \{\bar x : p_1(\bar x,\bar y_1, \bar z) = 0 \}$ . Let $\phi (\bar y_0^{\prime }, \bar y_1^{\prime })$ be the formula $\forall \bar x\ p_0(\bar x,\bar y_0^{\prime }, \bar z) = 0 \leftrightarrow p_1(\bar x,\bar y_1^{\prime }, \bar z) = 0 $ . Clearly, $\phi (\bar y_0, \bar y_1)$ holds. By Proposition 4.2, there is a tuple $\bar y_1^{\prime }$ from $F_0\cap F_1$ such that $\phi (\bar y_0, \bar y_1^{\prime })$ holds. Then $A = \{\bar x : p_0(\bar x,\bar y_0, \bar z) = 0 \} = \{\bar x : p_1(\bar x,\bar y_1^{\prime }, \bar z) = 0 \}$ . Thus $A\subset \mathbb {R}^n$ is an algebraic set over $F_0\cap F_1$ using the polynomial $p_1$ and parameters $\bar y_1^{\prime }, \bar z$ from $F_0\cap F_1$ .

Proposition 4.4. Let $F_0,F_1$ be real closed subfields of $\mathbb {R}$ in amalgamation position. Let $n_0,n_1>0$ be natural numbers and $B\subset \mathbb {R}^{n_0}\times \mathbb {R}^{n_1}$ be an algebraic set over $F_0 \cap F_1$ . If $a_0 \subset F^{n_0}$ and $a_1 \subset F^{n_1}$ are sets such that $a_0 \times a_1 \subset B$ , then there are algebraic sets $A_0 \subset \mathbb {R}^{n_0}$ and $A_1 \subset \mathbb {R}^{n_1}$ algebraic over $F_0\cap F_1$ such that $a_0 \subset A_0$ , $a_1 \subset A_1$ , and $A_0 \times A_1 \subset B$ .

One can use a corollary to the Hilbert basis theorem as Fact 4.5 to see that there are inclusion-smallest algebraic sets over $F_0 \cap F_1$ which contain $a_0$ and $a_1$ respectively, if there are any. The proof is to show that they in fact do work.

Fact 4.5 [6, Corollary 3.2.6]

If $X_i$ is Zariski closed for $i \in \iota $ , then there is a finite $\iota _0 \subset \iota $ such that

$$ \begin{align*} \bigcap_{i \in \iota} X_i = \bigcap_{i \in \iota_0} X_i. \end{align*} $$

Now we prove Proposition 4.4.

Proof Let $p(\bar x_0, \bar x_1)$ be a polynomial with coefficients in $F_0 \cap F_1$ whose zero locus is the set B. By Fact 4.5, there exists a finite set $b_0 \subset a_0$ such that $ \bigcap _{ \bar r_0\in a_0} \{ \bar r_1\in \mathbb {R}^{n_1}: p(\bar r_0, \bar r_1) = 0\}= \bigcap _{ \bar r_0\in b_0} \{ \bar r_1\in \mathbb {R}^{n_1}: p(\bar r_0, \bar r_1) = 0\}$ ; denote the intersection by $C_1$ ; let $C_1^{\prime } = C_1\cap (F_0 \cap F_1)^{n_1}$ . Similarly, there exists a finite set $c_1 \subset C_1^{\prime }$ such that $\bigcap _{ \bar r_1\in C_1^{\prime }} \{ \bar r_0\in \mathbb {R}^{n_1}: p(\bar r_0, \bar r_1) = 0\}= \bigcap _{ \bar r_1\in c_1} \{ \bar r_0\in \mathbb {R}^{n_1}: p(\bar r_0, \bar r_1) = 0\}$ ; denote the intersection by $A_0$ .

It is clear that $A_0 \subset \mathbb {R}^{n_0}$ is an algebraic set over $F_0 \cap F_1$ . It is also clear that $a_0 \subset A_0$ : if $\bar r_0\in a_0$ , then $ p(\bar r_0, \bar r_1) = 0$ for all $\bar r_1\in C_1$ , in particular for all $\bar r_1\in c_1$ ; by the definition of the set $A_0$ , $\bar r_0\in A_0$ .

Claim $(A_0 \cap F^{n_0}_0)\times a_1 \subset B$ . To prove this, suppose that $\bar r_0 \in A_0 \cap F^{n_0}_0$ and $\bar r_1 \in a_1$ are points such that $ p(\bar r_0, \bar r_1) \neq 0$ holds for the sake of contradiction. Let $O\subset \mathbb {R}^{n_1 } $ be an open neighborhood of $\bar r_1$ such that $\{\bar r_0\} \times O$ contains no solutions of $p = 0$ . Use the amalgamation position as in Definition 4.1 to find a point $\bar r_1^{\prime }\in O $ from $F_0 \cap F_1$ such that $\forall \bar r_0^{\prime } \in b_0\ p(\bar r_0^{\prime }, \bar r_1^{\prime }) = 0$ holds. Thus $\bar r_1^{\prime } \in C_1 $ . Adding $\bar r_1^{\prime }$ to the set $c_1$ , the resulting set $A_0$ would be smaller, excluding $\bar r_0$ from it. This contradicts the choice of the set $c_1$ .

By performing a symmetric argument, find an algebraic set $A_1 \subset \mathbb {R}^{n_1}$ over $F_0 \cap F_1$ such that $a_1 \subset A_1$ , and $(A_0 \cap F^{n_0})\times (A_1 \cap F^{n_1})\subset B$ holds. Then, the sets $A_0, A_1$ are what we wanted: $a_0 \subset A_0, a_1 \subset A_1$ hold; $A_0 \times A_1 \subset B$ holds as $(A_0 \cap (F_0 \cap F_1)^{n_0})\times (A_1 \cap (F_0 \cap F_1)^{n_1}) \subset B$ and because the real closed subfield $F_0 \cap F_1$ is an elementary submodel of $\mathbb {R}$ by the model completeness of real closed fields.

Corollary 4.6. If $F_0,F_1$ are real closed subfields of $\mathbb {R}$ in amalgamation position, $n>1$ is a natural number, and $\bar {r} \in (F_1)^n$ is a point on $a_0 \subset \mathbb {R}^n$ a line or a circle algebraic over $F_0$ , then $\bar {r}$ or $a_0$ is algebraic over $F_0 \cap F_1$ .

Proof Let us prove the case for lines. Assume $F_0, F_1$ are in amalgamation position and $\bar {r} \in (F_1)^n$ is a point on a line $a_0$ algebraic over $F_0$ . Consider the set $B \subset \mathbb {R}^{2n} \times \mathbb {R}^n$ consisting of all triples $(\bar x_0, \bar x_1, \bar x_2)$ of colinear n-tuples of reals. One can see that B is an algebraic set over $ \mathbb {Z} \subset F_0\cap F_1$ by writing down the polynomial equation defining B. Let $a_1 = \{\bar {r}\}$ . Indeed, $a_0 \times a_1 \subset B$ holds. By Proposition 4.4, there are algebraic sets $A_0 \subset \mathbb {R}^{2n}$ and $A_1 \subset \mathbb {R}^{n}$ algebraic over $F_0\cap F_1$ such that $a_0 \subset A_0$ , $a_1 \subset A_1$ , and $A_0 \times A_1 \subset B$ . We can see specifically $a_0\times A_1 \subset B$ holds. This means that $A_1$ is included in the line $a_0$ by definition of B. Recall just now $\{\bar {r}\} = a_1 \subset A_1$ . If $A_1$ has only one element, then it has to be $\bar r$ , which therefore is an algebraic point over $F_0 \cap F_1$ by Fact 2.15; if $A_1$ has at least two elements, then the line $a_0\supset A_1$ is an algebraic set over $F_0 \cap F_1$ , since $A_1$ is an algebraic set over $F_0 \cap F_1$ and two elements in $A_1$ will algebraically determine the line $a_0$ .

For the case for circles, we can instead just consider set $B \subset \mathbb {R}^{3n} \times \mathbb {R}^n$ consisting of all quadruples $(\bar x_0, \bar x_1, \bar x_2, \bar x_3 )$ of n-tuples of reals that belong to some circle. The rest of the proof will be the same.

I further strengthen Proposition 4.4 as the following.

Proposition 4.7. Let $F_0,F_1$ be real closed subfields of $\mathbb {R}$ in amalgamation position. Let $n_0,n_1>0$ be natural numbers, $f_0\subset F_0, f_1\subset F_1$ be finite sets, and $B\subset \mathbb {R}^{n_0}\times \mathbb {R}^{n_1}$ be an algebraic set over $(F_0 \cap F_1) \cup f_0\cup f_1$ . If $a_0 \subset F_0^{n_0}$ and $a_1 \subset F_1^{n_1}$ are sets such that $a_0 \times a_1 \subset B$ , then there are sets $A_0 \subset \mathbb {R}^{n_0}$ algebraic over $(F_0\cap F_1)\cup f_0$ and $A_1 \subset \mathbb {R}^{n_1}$ algebraic over $(F_0\cap F_1)\cup f_1$ such that $a_0 \subset A_0$ , $a_1 \subset A_1$ , and ${A_0 \times A_1 \subset B}$ .

Proof Fix some enumeration $\bar f_0\in \mathbb {R}^{|f_0|},\bar f_1\in \mathbb {R}^{|f_1|}$ for $f_0$ , $f_1$ respectively. Since B is algebraic over $(F_0 \cap F_1) \cup f_0\cup f_1$ , find an algebraic set $C \subset \mathbb {R}^{n_0}\times \mathbb {R}^{n_1}\times \mathbb {R}^{|f_0|}\times \mathbb {R}^{|f_1|}$ over $F_0 \cap F_1$ such that the horizontal section $C^{ \bar f_0\times \bar f_1} = ( C^{\bar f_1})^{\bar f_0}$ equals the set B. Assume $a_0 \subset F_0^{n_0}$ and $a_1 \subset F_1^{n_1}$ are sets such that $a_0 \times a_1 \subset B$ . Then $a_0 \times a_1 \times \{\bar f_0\} \times \{ \bar f_1\} \subset C$ by our choice of C, with $a_0\times \{\bar f_0\} \subset F_0^{n_0 }\times F_0^ { |f_0|} $ and $a_1\times \{\bar f_1\} \subset F_1^{n_1}\times F_1^ {|f_1|} $ . By Proposition 4.4, there are algebraic sets $A_0 \subset \mathbb {R}^{n_0}\times \mathbb {R}^{|f_0|}$ and $A_1 \subset \mathbb {R}^{n_1} \times \mathbb {R}^{|f_1|}$ over $F_0\cap F_1$ such that $a_0 \times \{\bar f_0\} \subset A_0$ , $a_1\times \{\bar f_1\} \subset A_1$ , and $A_0 \times A_1 \subset C$ . Then we get the algebraic sets $A_0^{\bar f_0} \subset \mathbb {R}^{n_0}$ over $(F_0\cap F_1)\cup f_0$ and $A_1^{\bar f_1} \subset \mathbb {R}^{n_1}$ over $(F_0\cap F_1)\cup f_1$ such that $a_0 \subset A_0^{\bar f_0}$ , $a_1 \subset A_1^{\bar f_1}$ , and $A_0^{\bar f_0}\times A_1^{\bar f_1} \subset B$ , as wanted.

Corollary 4.8. Assume that $F_0,F_1$ are real closed subfields of $\mathbb {R}$ in amalgamation position, that $ x_0 \subset F_0$ , $ x_1 \subset F_1$ are finite sets, and that ${y} \in F_1$ is a point. If y is an algebraic point over $(F_0\cap F_1) \cup x_0 \cup x_1 $ , then y is an algebraic point over ${(F_0\cap F_1) \cup x_1 }$ .

Proof Let $B = \{0\} \times \{y\}$ which is algebraic over $(F_0\cap F_1 )\cup x_0 \cup x_1 $ . Since $\{0\} \times \{y\} \subset B = \{0\} \times \{y\}$ , by Proposition 4.7, we have $A_1 = \{y\}$ algebraic over $(F_0\cap F_1) \cup x_1 $ . Therefore y is algebraic over $(F_0\cap F_1) \cup x_1 $ .

Finally, using mutual genericity and the following proposition, we can build pairs of real closed subfields of $\mathbb {R}$ that are in amalgamation position.

Proposition 4.9. Whenever $V [G_0], V [G_1]$ are mutually generic extensions, $F_0 = \mathbb {R} \cap V [G_0]$ and $F_1 = \mathbb {R} \cap V [G_1]$ are in amalgamation position.

Proof Firstly and trivially, $\mathbb {R}$ remains to be a real closed field in any generic extension. Now work in the ground model V. Let $Q_0, Q_1$ be partial orders, let $\tau _0, \tau _1$ be their respective names for strings of reals, let p be a polynomial, and let $ O$ be a basic open set of a Euclidean space such that $Q_0 \Vdash \tau _0 \in O$ and $Q_0\times Q_1 \Vdash p(\tau _0, \tau _1) = 0$ holds. Let $\langle q_0^n : n \in \omega \rangle $ be a descending sequence of conditions in $Q_0$ and let $\{ O_n \}_{n \in \omega }$ be a sequence of basic open neighborhoods such that $O = O_0, \bar O_{n+1} \subset O_n$ , $O_{n+1}$ has a metric diameter at most $2^{-n}$ , and $q_0^n \Vdash \tau _0 \in O_n$ . The intersection $\bigcap _n O_n$ contains a single point $\bar r $ . It will suffice to show that $Q_1 \Vdash p(\bar r , \tau _1) = 0$ holds.

If not, then by continuity of polynomial functions there has to be a number $n\in \omega $ and a condition $q_1 \in Q_1$ such that $q_1 \Vdash p\upharpoonright O_n\times \{\tau _1\}$ is never zero. Then, the condition $\langle q_0^n,q_1\rangle $ forces in the product that $p(\tau _0,\tau _1)\neq 0$ , contradicting the initial assumptions.

5 Forcing with algebraic amalgamation

This section bridges between an algebraic notion of amalgamation in posets and the forcing-theoretic balance.

Definition 5.1. An $\mathbb {R}$ -analytic forcing satisfies $\mathbb {R}$ -amalgamation if for any pair of $ p_0, p_1 \in P $ such that the fields $\mathrm {supp}(p_0)$ , $\mathrm {supp}(p_1) \subset \mathbb {R}$ are in amalgamation position, the following (1) implies (2):

1. 1. $p_0 \upharpoonright \mathrm {supp}(p_0)\cap \mathrm {supp}(p_1) = p_1 \upharpoonright \mathrm {supp}(p_0)\cap \mathrm {supp}(p_1)$ , and, writing $p_\cap $ for the common value, $p_\cap \in P$ , $p_0\leq p_\cap $ , $p_1\leq p_\cap $ hold.

2. 2. $ p_0, p_1$ have a common lower bound in P.

By quantifier counting procedure, one can check that $\mathbb {R}$ -amalgamation is a $\Pi ^1_2$ statement and therefore absolute throughout all forcing extensions by the Shoenfield absoluteness [3, Theorem 25.20]. The main point in this section is that $\mathbb {R}$ -amalgamation implies forcing theoretic balance as Theorem 5.3. Unfortunately, we are not able to easily provide a typical connection and classification of balanced virtual conditions as in [5, Chapters 6–8] because our poset P does not satisfy (2) $\Longrightarrow $ (1) in Definition 5.1. But this doesn’t affect the result of this paper.

Definition 5.2. Let P be an $\mathbb {R}$ -analytic forcing. A P-structure is a relational structure c that is total with universe $\mathbb {R}$ such that Coll holds.

Theorem 5.3. Let P be an $\mathbb {R}$ -analytic forcing with $\mathbb {R}$ -amalgamation. Then the following hold:

1. 1. If c is a P-structure, then the pair is balanced.

2. 2. If the Continuum Hypothesis holds, then P is balanced.

3. 3. P is balanced cofinally below $\kappa $ for every inaccessible cardinal $\kappa $ .

Proof For (1), assume c is a P-structure. Then holds by definition. Let $V [G_0], V [G_1]$ be mutually generic extensions and let $p_0, p_1$ be conditions in P in the respective models $V [G_0], V [G_1]$ such that $p_0, p_1 \leq c$ holds. Since c is total, $p_0, p_1$ are total as well, i.e., $\mathrm {supp}(p_0)= \mathbb {R}\cap V [G_0]$ and $\mathrm {supp}(p_1)= \mathbb {R}\cap V [G_1]$ hold. We must show that $p_0, p_1$ have a common lower bound in some ambient model. By Proposition 4.9, the fields $\mathrm {supp}(p_0)= \mathbb {R}\cap V [G_0]$ and $\mathrm {supp}(p_1)= \mathbb {R}\cap V [G_1]$ are in amalgamation position. By the properties of the ordering of P, $p_0 \upharpoonright V = p_1 \upharpoonright V = c $ holds. The $\mathbb {R}$ -amalgamation applied in the model $V [G_0, G_1]$ then implies that $p_0, p_1$ have a common lower bound as desired. Then by [5, Proposition 5.2.2], the pair is balanced.

For (2), let $p \in P$ be a condition and we will produce a balanced virtual condition below p. Use the Continuum Hypothesis (CH) assumption to enumerate all reals as $\{ r_\alpha \}_{\alpha \in \omega _1}$ , and by transfinite recursion on $\alpha \in \omega _1$ build conditions $p_\alpha $ so that $p_0 = p, p_{\alpha +1} \leq p_\alpha $ , $r_{\alpha } \in \mathrm {supp}(p_{\alpha +1})$ , and for limit ordinals $\alpha $ , $p_\alpha = \bigcup _{\beta \in \alpha } p_\beta $ . This is possible by Definition 2.10(3) and (4) and the result is a descending sequence of conditions in P. Let $c = \bigcup _{\alpha } p_\alpha $ . c is a relational structure on $\mathbb {R}$ and by Definition 2.10(4), Coll and . Thus c is a P-structure. By (1) of the present theorem, is a balanced pair. By [5, Theorem 5.2.8] ⁴ , the $ \equiv _b$ -class containing contains a balanced virtual condition, which is stronger than p as desired.

For (3), for every inaccessible cardinal $\kappa $ , since $V_\kappa \models $ every forcing extension has a further extension in which CH holds, P is balanced cofinally below $\kappa $ .

Now let the poset P be defined as Definition 3.1.

Theorem 5.4. P satisfies $\mathbb {R}$ -amalgamation.

Once the theorem is verified, we can conclude that if the Continuum Hypothesis holds, then P is balanced ⁵ by Theorem 5.3. Before we prove Theorem 5.4, we need a few lemmas and a definition. Fix any pair of conditions $ p_0, p_1 \in P $ with $\mathrm {supp}(p_0)$ and $\mathrm {supp}(p_1)$ being in amalgamation position. Assume $p_0 \upharpoonright \mathrm {supp}(p_0)\cap \mathrm {supp}(p_1) = p_1 \upharpoonright \mathrm {supp}(p_0)\cap \mathrm {supp}(p_1)$ . Denote $ p_\cup =p_0 \cup p_1 , p_\cap =p_0 \cap p_1 $ . Assume $p_\cap \in P$ , $p_0\leq p_\cap $ , $p_1\leq p_\cap $ hold.

Lemma 5.5. $p_\cup $ is a partial ${\Gamma _2}$ -coloring.

Proof Assume that there is an hyperedge $\{x,y,z\} \in {{\Gamma _2}}$ , and $x,y \in \mathrm {dom}(p_0)$ , $z \in \mathrm {dom}(p_1)$ without loss of generality. We will show that $\{x,y,z\} $ is not monochromatic by $p_\cup $ . There are three cases depending on which is the pivot point for the isosceles triangle.

Case 1: z is the pivot point. Let l be the perpendicular bisector for the line segment from x to y. By Proposition 4.6, $z\in \mathrm {dom}(p_\cap )$ or the line l is algebraic over $\mathrm {supp}(p_\cap )$ . If $z\in \mathrm {dom}(p_\cap )$ , then $p_\cup ^{\prime \prime }\{x,y,z\} = p_0^{\prime \prime }\{x,y,z\}$ is not monochromatic since $p_0$ is a ${\Gamma _2}$ -coloring on $\mathrm {dom}(p_0) $ . Now assume that $z\notin \mathrm {dom}(p_\cap )$ and l is algebraic over $\mathrm {supp}(p_\cap )$ . If $p_0(x) = p_0(y)$ , then $p_0(x) = p_0(y) \in s(p_0, l)$ . By Definition 3.1(6) the symmetrical colors invariant requirement, since $p_0 \leq p_\cap \geq p_1$ , $ s(p_0, l) = s( p_\cap , l) = s(p_1, l)$ , then there must have been $x', y' \in \mathrm {dom}(p_\cap )$ such that $p_0(x) = p_0(y) = p_0(x') = p_0(y')$ . Since $p_1$ is a $\Gamma _2$ -coloring, $p_1(z) \neq p_1(x') = p_1 (y') = p_0(x') = p_0(y') = p_0(x) = p_0(y)$ . Hence $\{x,y,z\} $ is not monochromatic by $p_\cup $ .

Case 2: y is the pivot point. Let O be the circle centered at y passing through x. By Proposition 4.6, $z\in \mathrm {dom}(p_\cap )$ or O is algebraic over $\mathrm {supp}(p_\cap )$ . If $z\in \mathrm {dom}(p_\cap )$ , $p_\cup ^{\prime \prime }\{x,y,z\}= p_0^{\prime \prime }\{x,y,z\}$ is not monochromatic since $p_0$ is a $\Gamma _2$ -coloring. Now assume that $z\notin \mathrm {dom}(p_\cap )$ and O is algebraic over $\mathrm {supp}(p_\cap )$ . By Definition 3.1(7) the avoid center requirement, $p_1(z) \neq p_1(y)$ . Hence $\{x,y,z\} $ is not monochromatic by $p_\cup $ .

Case 3: x is the pivot point. It’s the same argument as Case 2.

Definition 5.6. For every point $x \in \mathbb {R}^2\setminus \mathrm {dom}(p_\cup ) $ , define $d(p_0,p_1, x)$ the set of disallowed colors on x for $p_0,p_1$ :

$$ \begin{align*}d(p_0,p_1, x) = S \cup p_\cup^{\prime\prime} (T\cup \bigcup (O\setminus O^-)\cup R), \end{align*} $$

where:

1. (S) $L = \{l\in E^L(\mathrm {supp}(p_\cup )): x\in l\}$ and $S = \bigcup _{l\in L} s(p_\cup , l)$ ;

2. (O) $O = \{o\in E^O(\mathrm {supp}(p_\cup )): $ the center of o is in $\mathrm {dom}(p_\cup )$ and $x\in o\}$ , $O^- = O\cap (E^O(\mathrm {supp}(p_0))\cup E^O(\mathrm {supp}(p_1)) )$ ;

3. (T) $T = \bigcup _{o\in O^-}\{t: t \text { is the center of } o\}$ ;

4. (R) $R = \{y \in \mathrm {dom}(p_\cup ) : $ that x and y are symmetrical with respect to l for some $l \in E^L(\mathrm {supp}(p_0)) \cup E^L(\mathrm {supp}(p_1)) \}$ .

Lemma 5.7. $d(p_0,p_1, x)\in I$ .

Proof We start with the following claim.

Claim 5.8. For $i\in 2$ , if $l\in E^L(\mathrm {supp}(p_i))$ , then $s(p_\cup , l) = s(p_i, l) $ .

Proof We prove the case for $i=0$ and the proof for the other case is the same. We will rule out all the possible ways that $s(p_\cup , l)$ can have more elements than $s(p_0, l)$ . Let $x,y\in \mathrm {dom}(p_\cup )$ be a symmetrical pair of points with respect to l.

Case 1: at least one point from $\{x,y\} $ is in $\mathrm {dom}(p_0) $ . Since $l \in E^L(\mathrm {supp}(p_0)) $ and by Fact 3.2(6), $\{x,y\} \subset \mathrm {dom}(p_0)$ . So this case doesn’t witness any element in $s(p_\cup , l)\setminus s(p_0, l)$ .

Case 2: $x,y\in \mathrm {dom}(p_1) \setminus \mathrm {dom}(p_0) $ .

Then $l\in E^L(\mathrm {supp}(p_1))$ by Fact 3.2(2). Then $l{\kern-1.5pt}\in{\kern-1.5pt} E^L(\mathrm {supp}(p_0)){\kern-1.5pt}\cap{\kern-1.5pt} E^L(\mathrm {supp}(p_1)) \subset E^L(\mathrm {supp}(p_\cap ))$ by Corollary 4.3. Since $p_0, p_1 \leq p_\cap $ , $s(p_0 , l) = s( p_\cap , l) = s(p_1, l)$ . Then $ p_1(x) = p_1(y) \rightarrow ( p_1(x),p_1(y) \in s(p_1 , l)= s(p_\cap , l) = s(p_0 , l) ) \rightarrow (p_1(x),p_1(y) \notin s(p_\cup , l)\setminus s(p_0, l) )$ holds. So this case doesn’t witness any element in $s(p_\cup , l)\setminus s(p_0, l)$ .

We have ruled out all possible ways that $s(p_\cup , l)$ can have more elements than $s(p_0, l)$ . Therefore $s(p_0, l) = s(p_\cup , l)$ .

Part (S): Denote $L\cap (E^L(\mathrm {supp}(p_0))\cup E^L(\mathrm {supp}(p_1)))$ by $L^-$ . Observe that $L^-$ must be finite (in fact with cardinality two or less) using Fact 3.2(5). Let $l\in L$ . If $l\in L^-$ , then $s(p_\cup , l) = s(p_0, l) \in I$ or $s(p_\cup , l) = s(p_1, l)\in I$ by Claim 5.8. Therefore $\bigcup _{l\in L^-} s( p_\cup , l)$ is in $ I$ .

Now let’s investigate the set of lines $ L\setminus L^-$ . For every $l\in L\setminus L^-$ , let $\Xi (l) = \{(y,z)\in \mathrm {dom}(p_0)\times \mathrm {dom} (p_1) : y,z$ are symmetrical with respect to $l\}$ . In particular, we will look into the set $\Xi (l)|_1$ the projection of $\Xi (l)$ to the coordinate 1. We will show that every point in $ \bigcup _{l\in L\setminus L^-} \Xi (l) | _{1} $ is algebraic over $\mathrm {supp}(p_\cap ) \cup a$ for some finite set $a \subset \mathrm {dom}(p_1)$ .

Case S.1: the set $L\setminus L^-$ has cardinality one. Let $l\in L\setminus L^-$ . If the set $\Xi (l) $ has cardinality at most one, then $s( p_\cup , l)$ is finite (in fact of cardinality at most $1$ ), thus in the ideal I. Now suppose that $|\Xi (l)|> 1 $ holds. Fix $ (y,z)\in \Xi (l)$ . We will show that the set $\Xi (l)|_1$ consists of points algebraic over $\mathrm {supp}(p_\cap ) \cup \{z\}$ . Notice that l is the perpendicular bisector of the line segment from x to y. Let $ (y',z')\in \Xi (l)$ such that $(y,z) \neq (y',z')$ . Then $ y', z' $ are symmetrical with respect to l. Then it’s easy to see that the point $z'$ is algebraic over $\{y, z, y'\}$ , hence algebraic over $\mathrm {supp}(p_\cap ) \cup \{y, z, y'\}$ . Then by Proposition 4.8, the point $z'$ is algebraic over $\mathrm {supp}(p_\cap ) \cup \{z\}$ . Hence every point in $ \Xi (l)|_1$ is algebraic over $\mathrm {supp}(p_\cap ) \cup \{z\}$ , i.e., $\Xi (l)|_1$ consists of points algebraic over $\mathrm {supp}(p_\cap ) \cup \{z\}$ where $\{z\}\subset \mathrm {dom}(p_1)$ .

Case S.2: the set $L\setminus L^-$ has cardinality more than one. Take distinct lines $l_0,l_1 \in L\setminus L^-$ . Take pairs of points ⁶ $ (y,z)\in \Xi (l_0)$ and $ (y',z')\in \Xi (l_1)$ . Then $(y,z) \neq (y',z')$ . We see that x can be calculated from $\{y,z, y',z'\}$ using polynomials: x is the intersection of $l_0,l_1$ , the perpendicular bisectors of the line segment $(y,z)$ and the line segment $ (y',z')$ respectively. Then x is algebraic over $ \{y,z, y',z'\}$ . If possible, take another pair of points $(y",z")\in \Xi (l_2)$ distinct from $(y,z),(y',z')$ for some line $l_2 \in L\setminus L^-$ . The line $l_2$ is not necessarily distinct from $l_0,l_1$ . We will show that $z" $ is algebraic over $\mathrm {supp}(p_\cap ) \cup \{z, z'\}$ .

The points $x,y",z" $ form an isosceles triangle. Let c be the circle centered at x passing through $y"$ . Then $z"\in c$ holds. Since the point x is algebraic over $\{y,z, y',z'\}$ , the set c is algebraic over $\{y, y', y",z,z'\}$ . Then c is algebraic over $\mathrm {supp}(p_\cap ) \cup \{y, y', y"\} \cup \{z,z'\}$ . By Corollary 4.8, there is a set $A_1$ algebraic over $\mathrm {supp}(p_\cap ) \cup \{z,z'\}$ such that $c\cap \mathrm {dom}(p_1) \subset A_1\subset c$ . Using Fact 3.2(4), if there are at least three points in $A_1\cap \mathrm {dom}(p_1)$ , then c needs to be algebraic over $\mathrm {supp}(p_1)$ ; then by Fact 3.2(7), c’s center point x is in $\mathrm {dom}(p_1)$ , which is not the case. Therefore, there are at most two points (including $z"$ ) in $A_1\cap \mathrm {dom}(p_1)$ . Then in the view of the real closed field $\mathrm {supp}(p_1)$ , the set $ A_1\cap \mathrm {dom}(p_1)$ containing $z"$ is algebraic over $\mathrm {supp}(p_\cap ) \cup \{z, z'\}$ . By Fact 2.15, $z" $ is algebraic over $\mathrm {supp}(p_\cap ) \cup \{z, z'\}$ , where $\{z, z'\}\subset \mathrm {dom}(p_1)$ .

From both Case S.1 and Case S.2 above, we conclude that every point in $ \bigcup _{l\in L\setminus L^-} \Xi (l) | _{1} $ is algebraic over $\mathrm {supp}(p_\cap ) \cup a$ for some finite set $a \subset \mathrm {dom}(p_1)$ . Then the set $p_1^{\prime \prime }( \bigcup _{l\in L\setminus L^-} \Xi (l) | _{1} )$ is in I since $ p_1\leq p_\cap $ and by Definition 3.1(8). Then $S = \bigcup _{l\in L} s( p_\cup , l) =\bigcup _{l\in L^-\cup (L\setminus L^-)} s( p_\cup , l) $

$\subset \bigcup _{l\in L^-} s( p_\cup , l) \cup p_1^{\prime \prime }( \bigcup _{l\in L\setminus L^-} \Xi (l) | _{1} ) \in I$ holds. Thus $S\in I$ holds.

Part (T): Since $x\notin \mathrm {dom}(p_\cup )$ , it’s easy to see that $O^-$ must be finite (in fact with cardinality two or less) using Fact 3.2(5). Then $T = \bigcup _{o\in O^-}\{t: t$ is the center of $o\}$ is finite, then $p_\cup ^{\prime \prime } T$ is finite, hence in I.

Part (O): Observe that for every circle $o\in O\setminus O^-$ , $o\cap \mathrm {dom}(p_0)$ has cardinality at most two: otherwise, by Fact 3.2(4), $o\in O^-$ holds which is a contradiction; similarly, $o\cap \mathrm {dom}(p_1)$ has cardinality at most two. If $ O\setminus O^-$ has cardinality at most one, $p_\cup ^{\prime \prime }\bigcup (O\setminus O^-)$ has cardinality at most four, hence is in the ideal I.

Now assume $ O\setminus O^-$ has cardinality more than one. Let $o,o'\in O\setminus O^-$ be distinct circles. Then there are finite sets $a_0\subset \mathrm {dom}(p_0),a_1\subset \mathrm {dom}(p_1) $ such that $ o,o'$ are algebraic over $ a_0\cup a_1$ . We see that $ o\cap o' $ has cardinality one or two and contains the point x. Then $o\cap o' $ is algebraic over $ a_0\cup a_1$ , and by Fact 2.15, x is algebraic over $ a_0\cup a_1$ . Take a circle $o"\in O\setminus O^-$ . Then the center of $o" $ , denoted as $t"$ , is in $\mathrm {dom}(p_1)$ : if $t"\in \mathrm {dom}(p_0)$ were true, $o" \in E^O(\mathrm {supp}(p_0)) \subset O^-$ by Fact 3.2(3), as a contradiction. Assume $o"\cap \mathrm {dom}(p_0) $ is not empty. We will show that $o"\cap \mathrm {dom}(p_0)$ is a set of points algebraic over $\mathrm {supp}(p_\cap ) \cup a_0$ .

The circle $o" $ is algebraic over the set $ \{t" , x\}$ . Then $o"$ is algebraic over $ a_0\cup a_1\cup \{t"\}$ since x is algebraic over $ a_0\cup a_1$ . Then $o"$ is algebraic over $ \mathrm {supp}(p_\cap ) \cup a_0\cup a_1\cup \{t"\}$ . By Corollary 4.8, there is a set $A_0$ algebraic over $\mathrm {supp}(p_\cap ) \cup a_0$ such that $o"\cap \mathrm {dom}(p_0) \subset A_0\subset o"$ . If there are at least three points in $A_0\cap \mathrm {dom}(p_0)$ , then $t"\in \mathrm {dom}(p_0)$ , and then $o"$ becomes algebraic over $\mathrm {supp}(p_0)$ , contradicting our assumption $o"\notin O^-$ . Therefore, there are at most two points in $A_0\cap \mathrm {dom}(p_0)$ . Then using Fact 2.15, the finite set $o"\cap \mathrm {dom}(p_0)\subset A_0\cap \mathrm {dom}(p_0)$ is a set of points algebraic over $\mathrm {supp}(p_\cap ) \cup a_0$ .

By a symmetric argument, if $o"\cap \mathrm {dom}(p_1) $ is not empty, then $o"\cap \mathrm {dom}(p_1)$ is a set of points algebraic over $\mathrm {supp}(p_\cap ) \cup a_1$ . Then $ \bigcup (O\setminus O^-)\cap \mathrm {dom}(p_0) $ is still a set of points algebraic over $\mathrm {supp}(p_\cap ) \cup a_0$ ; $ \bigcup (O\setminus O^-)\cap \mathrm {dom}(p_1) $ is still a set of points algebraic over $\mathrm {supp}(p_\cap ) \cup a_1$ . Since $ p_0\leq p_\cap $ and $ p_1\leq p_\cap $ , and by Definition 3.1(8), $p_\cup ^{\prime \prime } \bigcup (O\setminus O^-) = p_0^{\prime \prime } \bigcup (O\setminus O^-) \cup p_1^{\prime \prime } \bigcup (O\setminus O^-) \in I$ holds.

Part (R): First consider $R\cap \mathrm {dom}(p_0)$ . Let $y_0\in R\cap \mathrm {dom}(p_0)$ if possible. Then x and $y_0$ are symmetrical with respect to $l_0$ for some $l_0 \in E^L(\mathrm {supp}(p_0)) \cup E^L(\mathrm {supp}(p_1)) $ . It can’t be the case $l_0\in E^L(\mathrm {supp}(p_0))$ using Fact 3.2(6). Then $l_0\in E^L(\mathrm {supp}(p_1))\setminus E^L(\mathrm {supp}(p_0))$ holds. Then the point x can be calculated from $y_0$ and $l_0$ . Take another point $y_1 \in R\cap \mathrm {dom}(p_0)$ different than $y_0$ if there is any. We will show that $y_1$ is algebraic over $\mathrm {supp}(p_\cap ) \cup \{y_0\}$ .

Since $y_1 \in R\cap \mathrm {dom}(p_0)$ , $\exists l_1 \in E^L(\mathrm {supp}(p_1))\setminus E^L(\mathrm {supp}(p_0))$ that the points x and $y_1$ are symmetrical with respect to $l_1$ . Then the point $y_1$ can be calculated from x and $l_1$ . Since x can be calculated from $y_0$ and $l_0$ , $y_1$ can be calculated from $l_0, l_1, y_0$ . Let $f \subset \mathrm {supp}(p_1)$ be a finite set over which the sets $l_0, l_1$ are algebraic. Then the point $y_1$ is algebraic over $\mathrm {supp}(p_\cap )\cup \{y_0\}\cup f$ . Then by Proposition 4.8, $y_1$ is algebraic over $\mathrm {supp}(p_\cap ) \cup \{y_0\}$ . Thus $R\cap \mathrm {dom}(p_0) $ consists of points algebraic over $\mathrm {supp}(p_\cap ) \cup \{y_0\}$ , where $\{y_0\}\subset \mathrm {dom}{(p_0)}$ . Then $p_0^{\prime \prime }(R\cap \mathrm {dom}(p_0)) \in I$ holds, since $p_0\leq p_\cap $ and by Definition 3.1(8).

Similarly, $p_1^{\prime \prime }(R\cap \mathrm {dom}(p_1))\in I$ . Then $p_\cup ^{\prime \prime }R = p_0^{\prime \prime }(R\cap \mathrm {dom}(p_0))\cup p_1^{\prime \prime }(R\cap \mathrm {dom}(p_1))\in I$ holds since the ideal I is closed under finite union.

Finally,

$$ \begin{align*} d(p_0,p_1, x) = S \cup p_\cup^{\prime\prime} (T\cup T_0\cup T_1\cup R) \end{align*} $$

because we have shown that each part in the union is in I and the ideal I is closed under finite union.

Now we prove Theorem 5.4.

Proof Fix any pair of conditions $ p_0, p_1 \in P$ with $\mathrm {supp}(p_0)$ and $\mathrm {supp}(p_1)$ being in amalgamation position. Assume $p_0 \upharpoonright \mathrm {supp}(p_0)\cap \mathrm {supp}(p_1) = p_1 \upharpoonright \mathrm {supp}(p_0)\cap \mathrm {supp}(p_1)$ . Denote $ p_\cup =p_0 \cup p_1 , p_\cap =p_0 \cap p_1 $ . Assume $p_\cap \in P$ and $p_0\leq p_\cap $ and $p_1\leq p_\cap $ holds. Then by Lemma 5.5, $p_\cup $ is a partial $\Gamma _2$ -coloring.

Let $ b \supset \mathrm {supp}(p_\cup )$ be a countable real closed subfield of $\mathbb {R}$ . We will extend $p_\cup $ to some $p_2 \in P$ with $ \mathrm {supp}(p_2) = b$ and $\mathrm {dom}(p_2) = b^2$ . Recall $d(p_0, p_1, x)$ from Definition 5.6, and by Lemma 5.7, $d(p_0,p_1, x)\in I$ . For every x, denote $d(p_0, p_1, x)$ as $d(x) $ for the rest of the proof. Fix any set $\pi \in I $ that $\forall x\in b^2 \setminus \mathrm {dom}(p_\cup ): \:\pi \setminus d( x)$ is infinite. It is possible by our choice of $\pi $ . Now just let $p_2$ be a map with domain $b^2$ such that:

• • $p_2\upharpoonright {\mathrm {dom}(p_\cup )} = p_\cup $ .

• • $\forall x\in b^2 \setminus \mathrm {dom}(p_\cup ) ,\: p_2 (x) \in \pi \setminus d(x)$ .

• • $p_2$ on $b^2 \setminus \mathrm {dom}(p_\cup )$ is an injection.

Claim 5.9. $p_2 \in P$ .

Proof Definition 3.1(1) is satisfied as we required $\mathrm {supp}(p_2) = b$ to be a countable real closed subfield of $\mathbb {R}$ and $\mathrm {supp}(p_2)^2 = \mathrm {dom}(p_2) \subset \mathbb {R}^2$ .

For Definition 3.1(2), let’s check $p_2$ is a ${\Gamma _2 }$ -coloring for $\mathrm {dom}(p_2)$ . Assume there are points $ x,y,z\in \mathrm {dom}(p_2)$ such that $\{x,y,z\} \in {\Gamma _2 }$ .

Case 1: at least two points in $\{x,y,z\}$ are from $\mathrm {dom}(p_2) \setminus (\mathrm {dom}(p_1)\cup \mathrm {dom}(p_0))$ . Then $p_2^{\prime \prime }\{x,y,z\} $ is not a singleton because $p_2$ is injective on $\mathrm {dom}(p_2) \setminus (\mathrm {dom}(p_1)\cup \mathrm {dom}(p_0))$ .

Case 2: all points $x,y,z \in \mathrm {dom}(p_\cup )$ . Because $p_\cup $ is a $\Gamma _2$ -coloring by Lemma 5.5, $\{x,y,z\}$ is not monochromatic by $p_2\supset p_\cup $ .

Case 3: precisely one point in $\{x,y,z\}$ are from $\mathrm {dom}(p_2) \setminus \mathrm {dom}(p_\cup )$ ; let x be such point, without the loss of generality.

Case 3.1: x is the pivot point. Then x is on the perpendicular bisector l of the line segment from y to z, and because $y,z \in \mathrm {dom}(p_\cup )$ and by Fact 3.2(2), the line l is in the set $ E^L(\mathrm {supp}(p_\cup ))$ . Since we made $p_2(x) \notin d(x)\supset S$ and by part (S) in Definition 5.6, it’s easy to see that $\{x,y,z\}$ is not monochromatic by $p_2$ .

Case 3.2: x is not the pivot point. Then x is on the circle $o\in E^O(\mathrm {supp}(p_\cup ))$ centered at y passing through z or centered at z passing through y. Since we made $p_2(x) \notin d(x)\supset p_\cup ^{\prime \prime } (T\cup \bigcup (O\setminus O^-))$ and by part (T) (O) in Definition 5.6, it’s easy to see that $\{x,y,z\}$ is not monochromatic by $p_2$ .

For Definition 3.1(3) the symmetrical colors requirement, let $l\in E^L(\mathrm {supp}(p_2)) $ .

Case 1: $l \in E^L(\mathrm {supp}(p_0)) $ . Then $s(p_0, l) \in I$ by Definition 3.1, and $s(p_2, l) = s(p_0, l)$ (that we will check later for Definition 3.1(6)), hence $s(p_2, l) \in I$ holds.

Case 2: $l \in E^L(\mathrm {supp}(p_1)) $ . $s(p_2, l) \in I$ holds for the same reason as Case 1.

Case 3: $l {\kern-1pt}\in{\kern-1pt} E^L(\mathrm {supp}(p_2)) \setminus (E^L(\mathrm {supp}(p_1)) {\kern-1pt}\cup{\kern-1pt} E^L(\mathrm {supp}(p_0)) )$ . For any pair of points symmetrical with respect to l, at least one of the pair’s coordinate has to be in $\mathrm {dom}(p_2)\setminus \mathrm {dom}(p_\cup )$ because otherwise it would be the case $l \in E^L(\mathrm {supp}(p_0)) \cup E^L(\mathrm {supp}(p_1)) $ by Fact 3.2(2). Then it’s easy to see $s(p_2, l) \subset p_2^{\prime \prime }( \mathrm {dom}(p_2)\setminus \mathrm {dom}(p_\cup ) ) \subset \pi \in I$ . Hence $s(p_2, l) \in I$ because the ideal I is closed under subset.

Since we just checked the definition, $p_2\in P$ holds.

Claim 5.10. $p_2 \leq p_0 $ and $p_2 \leq p_1 $ .

Proof We will prove $p_2 \leq p_0 $ , and the proof for $p_2 \leq p_1 $ is the same. The items (4) and (5) in Definition 3.1 are obviously satisfied.

For Definition 3.1(6) the symmetrical colors invariant requirement, for each $l \in E^L(\mathrm {supp}(p_0)) $ , we want to show $s(p_0, l) = s(p_2, l)$ . By Claim 5.8, $s( p_\cup , l) = s(p_0, l) $ . So just need to show that $s( p_\cup , l) = s(p_2, l) $ . Let $x,y\in \mathrm {dom}(p_2)$ be a symmetrical pair of points with respect to l.

Case 1: $x,y\in \mathrm {dom}( p_\cup )$ . This case doesn’t witness any element in $s(p_2, l)\setminus s( p_\cup , l)$ .

Case 2: one point from $\{x,y\} $ is in $\mathrm {dom}(p_0) $ . Since $l \in E^L(\mathrm {supp}(p_0)) $ and by Fact 3.2(6), $\{x,y\} \subset \mathrm {dom}(p_0)\subset \mathrm {dom}( p_\cup )$ . This leads to the previous case.

Case 3: $x,y\in \mathrm {dom}(p_2) \setminus \mathrm {dom}(p_\cup ) $ , then $p_2(x)\neq p_2(y)$ since $p_2$ is an injection on $ \mathrm {dom}(p_2) \setminus \mathrm {dom}(p_\cup )$ . So this case doesn’t witness any element in $s(p_2, l)\setminus s( p_\cup , l)$ .

Case 4: $x \in \mathrm {dom}(p_2) \setminus \mathrm {dom}(p_\cup )$ and $y\in \mathrm {dom}(p_1)$ without loss of generality. Because $ p_2(x) \notin R \ni p_2(y)$ as item (R) in Definition 5.6, $p_2(x)\neq p_2(y)$ holds. So this case doesn’t witness any element in $s(p_2, l)\setminus s( p_\cup , l)$ .

From above, we have ruled out all the possible ways that $s(p_2, l)$ can have more elements than $s( p_\cup , l)$ . Since trivially $s( p_\cup , l) \subset s(p_2, l)$ holds, $s( p_\cup , l) = s(p_2, l)$ holds.

For Definition 3.1(7) the avoid center requirement, let $x\in \mathrm {dom}(p_2)\setminus \mathrm {dom}(p_0)$ be a point, let $c\in E^O(\mathrm {supp}(p_0))$ be a circle, and let t be the center of the circle c. If $x\in \mathrm {dom}(p_2)\setminus \mathrm {dom}( p_\cup ) $ , we made $p_2(x) \notin d(x) \supset p_\cup ^{\prime \prime }T$ ; by Definition 5.6 (T), $T \ni t$ holds, then $p_2(x) \neq p_0(t)$ holds. If $x\in \mathrm {dom}(p_1) \setminus \mathrm {dom}(p_0)$ , by Case 2 in the proof of Lemma 5.5, $p_2(x) = p_1(x) \neq p_0(t)$ holds.

For Definition 3.1(8) the algebraic points requirement, take a finite set $ u \subset \mathrm {supp}(p_2)$ . We will show that the $p_2$ -image of the set $\{x \in \mathrm {dom}(p_2)\setminus \mathrm {dom}(p_0)$ : $x $ is algebraic over $\mathrm {supp}(p_0)\cup u\} $ , denoted as B, is in $ I$ . First observe that $p_2^{\prime \prime }(\mathrm {dom}(p_2)\setminus \mathrm {dom}(p_\cup )) \subset \pi \in I $ . Since ideal I is closed under union, once we have showed that the $p_2$ -image of the set $B \cap \mathrm {dom}(p_1) $ is in $I $ , we are done. To do that, let v be an inclusion-maximal subset of $\mathrm {supp}(p_1)$ intersected with the algebraic closure of $\mathrm {supp}(p_0)\cup u$ , which is algebraically independent over $\mathrm {supp}(p_0)$ . Since the sets algebraically independent over $\mathrm {supp}(p_0)$ form a matroid by [7, Theorem 6.7.1], v is finite (and in fact $|v| \leq |u|$ ). Observe that $B \cap \mathrm {dom}(p_1) = \{x \in \mathrm {dom}(p_2)\setminus \mathrm {dom}(p_0)$ : $x $ is algebraic over $\mathrm {supp}(p_0)\cup u\} \cap \mathrm {dom}(p_1) = \{x \in \mathrm {dom}(p_1)\setminus \mathrm {dom}(p_0)$ : $x $ is algebraic over $\mathrm {supp}(p_0)\cup v\} = \{x \in \mathrm {dom}(p_1)\setminus \mathrm {dom}(p_0)$ : $x $ is algebraic over $\mathrm {supp}(p_\cap )\cup v\}$ . This last equality is a consequence of amalgamation position: every $x\in \mathrm {dom}(p_1)$ algebraic over $\mathrm {supp}(p_0)\cup v$ is algebraic over $\mathrm {supp}(p_\cap )\cup v$ by Corollary 4.8, because $\mathrm {supp}(p_0)$ and $\mathrm {supp}(p_1)$ are in amalgamation position with the intersection $\mathrm {supp}(p_\cap )$ . Then we have $p_2^{\prime \prime }(B \cap \mathrm {dom}(p_1) ) =p_1^{\prime \prime } (B \cap \mathrm {dom}(p_1) ) =p_1^{\prime \prime } \{x \in \mathrm {dom}(p_1)\setminus \mathrm {dom}(p_0)$ : $x $ is algebraic over $\mathrm {supp}(p_\cap )\cup v\} \in I$ by Definition 3.1 (8) for $p_1\leq p_\cap $ .

Therefore, $p_2\leq p_0$ holds. Similarly, $p_2\leq p_1$ holds as well.

As we have found a condition $p_2\in P$ extending $p_0$ and $p_1$ , it’s the end of the proof.

6 Open projections and hyperedge forcings

This section provides some open projections and hyperedge forcings in order to obtain Proposition 6.21, a key for distinguishing ${\Gamma }_2$ and ${\Gamma }_3$ . Fix $n,m\in \omega $ . We say that G is an ordered hypergraph on $\mathbb {R}^n$ of arity m if $G \subset (\mathbb {R}^n)^m$ . For $\iota $ a list of indexes in m, for $A\subset (\mathbb {R}^n)^m$ a set, write $A|_\iota \subset (\mathbb {R}^n)^{\iota }$ to mean the projection of A into the coordinates mentioned in $\iota $ . If $ G $ is a $G_\delta $ subset of $(\mathbb {R}^n)^m$ , then $ G $ is a Polish space in the inherited topology. We want to define the Cohen forcing associated with G.

Definition 6.1. Let $ G $ be a $G_\delta $ subset of $(\mathbb {R}^n)^m$ . Equip $ G$ with its Polish topology. Let $P_G$ be the Cohen forcing associated with the space G (i.e., $P_G$ is the poset of nonempty open subsets of $ G $ ordered by reverse inclusion). We call $P_G$ the hyperedge forcing for G.

We need some basic geometry and continuity for the later open projection claims.

Definition 6.2. A similarity of $\mathbb {R}^n$ is a bijection on $\mathbb {R}^n$ that multiplies all distances by some positive real number. $\mbox {SIM}(\mathbb {R}^n)$ is the group of all similarities of $\mathbb {R}^n$ under the operation of composition.

Definition 6.3. Let $G\subset (\mathbb {R}^n)^m$ be a set. We say that G is invariant under similarities of $\mathbb {R}^n$ if for every m-tuple $g\in G$ and every $s\in \mbox {SIM}(\mathbb {R}^n)$ a similarity of $\mathbb {R}^n$ , $\bar {s}(g) \in G$ holds, where $\bar {s}$ is the bijection on $(\mathbb {R}^n)^m$ that each of its coordinate is the map s.

Fact 6.4. $\mbox {SIM}(\mathbb {R}^n)$ is a Polish group with the pointwise convergence topology, with the compatible metric

$$ \begin{align*} \delta(f,g) = \sum_{n=0}^{\infty} 2^{-n-1} \left( \frac{d(f(x_n) , g(x_n) )}{ 1 + d(f(x_n) , g(x_n) ) } + \frac{d(f^{-1}(x_n) , g^{-1}(x_n) )}{ 1 + d(f^{-1}(x_n) , g^{-1}(x_n) ) }\right), \end{align*} $$

where $\{x_n\}_{n\in \omega }$ is dense in $\mathbb {R}^n$ .

Lemma 6.5. Fix distinct points $y_0,y_1\in \mathbb {R}^n$ . Then there is an open neighborhood $U\subset (\mathbb {R}^n)^2$ of the origin and a continuous function $\Psi : U \rightarrow \mbox {SIM}(\mathbb {R}^n)$ such that each $\Psi (\delta _0,\delta _1)$ is a similarity of $\mathbb {R}^n$ that moves $y_0$ to $y_0 + \delta _0$ and $y_1$ to $ y_1+\delta _1$ .

Remark 6.6. The previous $\Psi $ being a continuous function is equivalent to “for any $y\in \mathbb {R}^n$ , the map $ (\delta _0,\delta _1) \mapsto \Psi (\delta _0,\delta _1)(y)$ is continuous from U to $\mathbb {R}^n$ .”

With the remark, let’s prove the lemma.

Proof Take an open neighborhood $U\subset (\mathbb {R}^n)^2$ of the origin such that $(y_0,y_1) + U$ is a pair of two separate open balls in $\mathbb {R}^n$ . For every $(\delta _0,\delta _1) \in U$ , let $\Psi (\delta _0,\delta _1)$ be the similarity which is a composition of the following transformations:

1. 1. shift by $\delta _0$ ;

2. 2. rotation around $y_0 + \delta _0$ by the angle from the vector $y_1 - y_0 $ to the vector $(y_1+ \delta _1) - (y_0+ \delta _0)$ ;

3. 3. stretch by the ratio $\frac {d(y_0+ \delta _0,y_1+ \delta _1) }{d(y_0,y_1)}$ from the point $y_0 + \delta _0 $ .

By working out the explicit formula, one can observe that for every $y \in \mathbb {R}^n$ , the map $(\delta _0,\delta _1) \mapsto \Psi (\delta _0,\delta _1) (y)$ is continuous. Hence $\Psi $ is continuous on U.

Proposition 6.7. Let $G\subset (\mathbb {R}^n)^m$ be $G_\delta $ and invariant under similarities of $\mathbb {R}^n$ . Then the projection of G into any two coordinates is an open map to $(\mathbb {R}^n)^2$ .

Proof For index convenience, we will only prove for the projection to the first two coordinates. The proof for any other case is identical. Denote the projection as $\mathrm {proj}: G \rightarrow G|_{0,1} $ . Take any open set $ Q\subset (\mathbb {R}^n)^m$ so that $Q\cap G\neq \emptyset $ (so $Q\cap G$ is a relative open set in the Polish space G). Take any $(y_0 ,y_1) \in \mathrm {proj}"G$ . Then $(y_0 ,y_1) $ is the projection of some $y = (y_i)_{i\in m} \in G$ . We just need to find an open set inside $\mathrm {proj}"G$ that includes $(y_0,y_1)$ .

By Lemma 6.5, take an open neighborhood $U\subset (\mathbb {R}^n)^2$ of the origin and a continuous function $\Psi : U \rightarrow \mbox {SIM}(\mathbb {R}^n)$ such that each $\Psi (\delta _0,\delta _1)$ is a similarity of $\mathbb {R}^n$ that moves $y_0$ to $y_0 + \delta _0$ and $y_1$ to $ y_1+\delta _1$ . Consider the map $ \Psi ': U\to (\mathbb {R}^n)^{m-2}$ by $ \Psi '(\delta _0,\delta _1) = (\Psi (\delta _0,\delta _1)(y_i) )_{2 \leq i\in m }$ , which is continuous too. Let $Q' = \Psi ^{\prime -1} (Q|_{2, 3, \dots , m-1 } )+ (y_0, y_1)$ , which is an open set by the continuity of $\Psi '$ , and contains $(y_0, y_1 )$ since as $\Psi ^{\prime -1} Q|_{2, 3, \dots , m-1 }$ contains the origin.

Further shrink $Q'$ : $Q" = Q'\cap \mathrm {proj}" (Q\cap G)$ . We will just show that $(y_0,y_1) \in Q" \subset \mathrm {proj}" (Q\cap G)$ . Clearly, $(y_0,y_1) \in Q"$ still. Take an arbitrary $ y' \in Q"$ , and we will show that $y' \in \mathrm {proj}" (Q\cap G)$ . The point $y'$ has the form $y ' = (y_0 + \delta _0, y_1 + \delta _1)$ , for some $\delta _0,\delta _1$ . Let $y" = (\Psi (\delta _0,\delta _1)(y_i) )_{2 \leq i\in m }$ . Then $y' \times y" \in Q\cap G$ :

• • $y' \times y" \in G$ is because $y'\times y"$ is coordinatewisely moved by a similarity of $\mathbb {R}^n$ from y and G is an ordered hypergraph on any Euclidean space invariant under similarities of $\mathbb {R}^n$ .

• • $y' \times y" \in Q$ is because $y'\in Q|_{0,1}$ and $y" \in Q|_{2,3,\dots , m-1}$ by our choice of $Q'$ and $Q"$ .

Hence $ y' = \mathrm {proj}( y' \times y" ) \in \mathrm {proj}" (Q\cap G)$ . Then $ Q" \subset \mathrm {proj}" (Q\cap G)$ . Hence $\mathrm {proj}" (Q\cap G)$ is open. Hence $\mathrm {proj}$ is an open map.

For $n \in \omega $ , let $\bar {\Gamma }_n\subset (\mathbb {R}^n)^3$ be the ordered hypergraph on $\mathbb {R}^n$ such that: $\forall \ x_0,x_1,x_2\in \mathbb {R}^n$ , $( x_0,x_1,x_2) \in \bar \Gamma _n \Longleftrightarrow (d_n(x_0,x_1) = d_n(x_1,x_2)\wedge x_0\neq x_1\neq x_2\neq x_0)$ . Recall that $\Gamma _ n\subset [\mathbb {R}^n]^3$ is the hypergraph of isosceles triangles on $\mathbb {R}^n$ .

Fact 6.8. $\forall \ x_0,x_1,x_2\in \mathbb {R}^n$ distinct points,

$$ \begin{align*}\{x_0,x_1,x_2\}\in {\Gamma_n} \Longleftrightarrow \bar{{\Gamma}}_n (x_0,x_1,x_2) \vee \bar{{\Gamma}}_n (x_1,x_0,x_2) \vee \bar{{\Gamma}}_n (x_2,x_0,x_2)\end{align*} $$

and

$$ \begin{align*}\bar{{\Gamma}}_n (x_0,x_1,x_2) \Longleftrightarrow \bar{{\Gamma}}_n (x_2,x_1,x_0).\end{align*} $$

Corollary 6.9. The projection of $\bar {\Gamma }_3$ to any two coordinates is an open map.

Proof $\bar {\Gamma }_3 \subset (\mathbb {R}^n)^m$ is $G_\delta $ and invariant under similarities of $\mathbb {R}^3$ . Then apply Proposition 6.7.

Let the ground model be V (of ZFC). According to Definition 6.1, consider the poset $P_{\bar {\Gamma }_3}$ to force over V. Let $e = ( x_0,x_1,x_2)$ be $P_{\bar {\Gamma }_3}$ -generic over V. Then $\{x_0,x_1,x_2\}$ form a ${\Gamma }_3$ -hyperedge.

Corollary 6.10. Each pair of points from $ e$ are mutually generic over V for the product forcing $P_{\mathbb {R}^3} \times P_{\mathbb {R}^3} $ .

Proof We will prove that $x_0,x_1$ are mutually generic; the proof for any other pair is the same. In V, fix an arbitrary open dense set D of conditions in $P_{ \mathbb {R}^3} \times P_{ \mathbb {R}^3} $ . Then $\bigcup D \subset \mathbb {R}^3 \times \mathbb {R}^3 $ is an open dense set. Let $\mathrm {proj}$ denote the projection of $\bar {\Gamma }_3$ to the first two coordinates. Because $\mathrm {proj}$ is a continuous open map by Corollary 6.9, $\mathrm {proj}^{-1} (\bigcup D) $ is an open dense subset of $\bar {\Gamma }_3$ .

Look in $V[e]$ . Because e is a generic tuple over V for the forcing $P_{\bar {\Gamma }_3 }$ , $e\in \mathrm {proj}^{-1} (\bigcup D)$ holds. Then $\mathrm {proj}(e) = (x_0, x_1) \in \bigcup D$ holds. Then there is some $d\in D $ that $(x_0,x_1 )\in d$ . Hence $(x_0, x_1)$ is generic over V for the product forcing $P_{ \mathbb {R}^3} \times P_{ \mathbb {R}^3} $ .

We want to define a notion of duplication on an ordered hypergraph’s coordinate, and investigate a few relevant projection maps.

Definition 6.11. Let $G\subset (\mathbb {R}^n)^m$ be an ordered hypergraph. Define $G^k\subset (\mathbb {R}^n)^{k }\times (\mathbb {R}^n)^{ m-1} = (\mathbb {R}^n)^{m + k-1} $ to be the ordered hypergraph of arity $m+k-1$ such that $ (x_{0i}, x_j)_{i\in k, j\in m-1} \in G^k $ if $ (\forall i\in k \: \ (x_{0i}, x_j)_{j \in m-1} \in G $ and $x_{0i}, i\in k \mbox { are in general position} )$ . We say that the hypergraph $G^k$ is obtained from G by duplicating its zeroth coordinate k times.

Recall that, for $d\in \omega $ , a set of points in a d-dimensional Euclidean space is in general linear position (or just general position) if no k of them lie in a $(k -2)$ -dimensional flat for $k = 2, 3, \dots , d + 1.$ In particular for $d = 3$ , four points are in general position if they are distinct, any three of them determine a plane, and any one of them is not on the plane determined by the other three. Since being in general position is an open relation, $G^k$ remains $G_\delta $ if G is $G_\delta $ . Similarly, we could define the duplication on any other coordinate too, but this is not needed for this research effort.

Proposition 6.12. Let $G\subset (\mathbb {R}^n)^m$ be $G_\delta $ . Suppose the projection of G into all but the zeroth coordinate is an open map to the range space. Consider the hypergraph $G^k$ obtained from G by duplicating its zeroth coordinate k times. Then the projection of $G^k$ to the last m many coordinates is an open map to the space G.

Proof Denote the projection of $G^k$ to the last m many coordinates as $\mathrm {proj}$ .

Suppose the projection of G into all but the zeroth coordinate, denoted as $\mathrm {proj}'$ , is an open map to the range space. Take an open set $Q = \Pi _{i\in k} Q_{0i} \times \Pi _{ j\in m-1}Q_{j} \subset (\mathbb {R}^n)^{k }\times (\mathbb {R}^n)^{ m-1}$ such that the relative open set $ Q\cap G^k$ is not empty. For each $i \in k$ , let $\mathrm {proj}_i$ be the projection of $(\mathbb {R}^n)^{k }\times (\mathbb {R}^n)^{ m-1}$ to all the coordinates in $\{ 0i, j:j\in m-1\}$ . In fact, for each $i \in k$ , $\mathrm {proj}_i^{\prime \prime }G^k$ is a copy of G. Then for each $i \in k$ , $ (\mathrm {proj}'\circ \mathrm {proj}_i)" Q$ is open to the space $\mathrm {rng}(\mathrm {proj}')$ , by the openness of $ \mathrm {proj}'$ from our assumption and the openness of $\mathrm {proj}_i$ between Euclidean spaces. We claim $\mathrm {proj}"Q =( Q_{0(k-1)}\times \bigcap _{i \in k-1} (\mathrm {proj}'\circ \mathrm {proj}_i)" Q ) \cap \mathrm {rng}(\mathrm {proj})$ . We will show both inclusions.

• ⊂ Basically just chase definitions and quantifiers. Take an arbitrary point $y' \in \mathrm {proj}"Q $ . Then $y'$ is a projection of some $y = (y_{0i},y_j)_{i\in k, j\in m-1} \in Q$ , so $y'=(y_{0(k-1)}, y_{j} )_{j\in m-1} $ . Firstly, clearly $y_{0(k-1)} \in Q_{0(k-1)}$ holds. For each $i \in k-1$ , $ (y_{j} )_{j\in m-1} = (\mathrm {proj}'\circ \mathrm {proj}_i)(y) $ holds, Then $(y_{j} )_{j\in m-1} \in \bigcap _{i \in k-1} (\mathrm {proj}'\circ \mathrm {proj}_i)" Q $ . Lastly, clearly $y' \in \mathrm {rng}(\mathrm {proj})$ holds. Hence $y'\in ( Q_{0(k-1)}\times \bigcap _{i \in k-1} (\mathrm {proj}'\circ \mathrm {proj}_i)" Q ) \cap \mathrm {rng}(\mathrm {proj})$ .

• ⊃ Take an arbitrary point $y' \times y" \in ( Q_{0(k-1)}\times \bigcap _{i \in k-1} (\mathrm {proj}'\circ \mathrm {proj}_i)" Q ) \cap \mathrm {rng}(\mathrm {proj})$ . For each $i \in k-1$ , since $y"\in (\mathrm {proj}'\circ \mathrm {proj}_i)" Q $ , there is some point $x_{0i} \in Q_{0i}$ such that $ (x_{0i} ) \times y" \in G$ . Since $ y' \times y" \in \mathrm {rng}(\mathrm {proj}) = G$ is part of our assumption, $ {(x_{0i})_{i \in k-1} }\times y' \times y" \in G^k$ by the definition of $G^k$ . Then $\mathrm {proj}( {(x_{0i})_{i \in k-1} }\times y' \times y" ) = y' \times y"$ holds. Since clearly ${(x_{0i})_{i \in k-1} }\times y' \times y" \in Q$ holds, $y' \times y" \in \mathrm {proj}"Q$ is true.

Hence we have proved $\mathrm {proj}"Q =( Q_{0(k-1)}\times \bigcap _{i \in k-1} (\mathrm {proj}'\circ \mathrm {proj}_i)" Q ) \cap \mathrm {rng}(\mathrm {proj})$ . The right-hand side of this equality, as we can see, is open in the range space $\mathrm {rng}(\mathrm {proj})=G$ , then so is the left-hand side.

Given $G\subset (\mathbb {R}^n)^m$ , $y_1, y_2, \ldots , y_{m-1} \in \mathbb {R}^n$ , for now we use an alternative notation for the horizontal section of G as $G( -,y_1, y_2, \ldots , y_{m-1} ) := \{y_0\in \mathbb {R}^n:\: (y_0, y_1, y_2, \ldots , y_{m-1} )\in G \}$ .

Definition 6.13. Let $G\subset (\mathbb {R}^n)^m$ . Let us say that G is stable against k vertices if the following holds. Whenever $x = \{x_{i}\}_{i \in k} \subset \mathbb {R}^n$ is a set of points in general position, there is an open neighborhood $U\subset (\mathbb {R}^n)^k$ of the origin and a continuous function $\Psi : U \rightarrow \mbox {SIM}(\mathbb {R}^n)$ such that for each $ u = (u_i)_{i\in k} \in U$ :

1. 1. $\Psi ( u )$ is a similarity of $\mathbb {R}^n$ .

2. 2. $\Psi (0)$ is the identity map.

3. 3. $\forall y_1, y_2, \ldots , y_{m-1} \in \mathbb {R}^n$ , if $G( -, y_1, y_2, \ldots , y_{m-1} ) \supset x$ , then:

   • • $\Psi ( u )$ moves the section $G( -, y_1, y_2, \ldots , y_{m-1} )$ to the section

     $G( -,\Psi (u)(y_1), \Psi (u)(y_2) ,\ldots , \Psi (u)(y_{m-1}) ) $ .

   • • $G( -,\Psi (u)(y_1), \Psi (u)(y_2) ,\ldots , \Psi (u)(y_{m-1}) ) \supset \{x_i + u_i\}_{i\in k}$ .

We call $\Psi $ a similarity function transferring points $x_{i}, {i \in k}$ for G.

Proposition 6.14. Let $G\subset (\mathbb {R}^n)^m$ be $G_\delta $ , and consider the hypergraph $G^k$ obtained from G by duplicating its zeroth coordinate k times. If G is stable against k vertices, then the projection of $G^k$ into the first k many coordinates is an open map to the space $(\mathbb {R}^n)^k$ .

Proof Assume that G is stable against k vertices. Denote the mentioned projection by $\mathrm {proj}$ . Take a basic open set $Q = \Pi _{i\in k} Q_{0i} \times \Pi _{j\in m-1}Q_j \subset (\mathbb {R}^n)^k\times (\mathbb {R}^n)^{m-1}$ such that the relative open set $Q\cap G^k \neq \emptyset $ . We want to show that $\mathrm {proj}"(Q\cap G^k)$ is open in $(\mathbb {R}^n)^k $ . Take an arbitrary point $x' = (x_{0i})_{i\in k} \in \mathrm {proj}"(Q\cap G^k) $ , which is the projection of some point $x = (x_{0i}, x_j)_{i\in k, j\in m-1} \in Q\cap G^k$ . We want to find an open set around $x'$ that is inside $\mathrm {proj}"(Q\cap G^k)$ .

Since $\{x_{0i}:{i\in k}\} $ is a set of points in general position by the definition of $G^k$ , we obtain the open set U and the similarity function $\Psi $ on U transferring points $x_{0i}, {i \in k}$ for G, as described in Definition 6.13. Then the map $u \mapsto (\Psi (u)(x_j) )_{ j \in m-1} $ , denoted as $\Theta $ , is continuous from U to $ (\mathbb {R}^n)^{ m - 1}$ since $\Theta $ is coordinatewisely continuous by the pointwise convergence topology of the space $\mbox {SIM}(\mathbb {R}^n)$ ; and still $ \forall u\in U \ (x' + u) \times \Theta (u) \in G^k$ holds by the properties of $\Psi $ in Definition 6.13. Let $Q' = \Theta ^{-1}(\Pi _{j\in m-1} Q_j)$ . Move and shrink $Q'$ : let $Q" = (Q'+x') \cap \Pi _{i\in k} Q_{0i}$ .

By chasing the definition and quantifiers, we can check $x'\in Q"\cap \mathrm {rng}(\mathrm {proj}) \subset \mathrm {proj}"Q$ holds:

1. (∈) Clearly $x'\in \mathrm {rng}(\mathrm {proj})$ and $x'\in \Pi _{i\in k} Q_{0i}$ hold; $x'\in x' + Q'$ holds as $0\in Q'$ which is because: by Definition 6.13(2) and the definition of $\Theta $ , $\Theta (0) = (x_j)_{j\in m-1} \in \Pi _{j\in m-1} Q_j $ holds, then $0\in Q' = \Theta ^{-1}(\Pi _{j\in m-1} Q_j)$ .

2. (⊂) Whenever $ (x'+u) \in Q"\cap \mathrm {rng}(\mathrm {proj})$ , $ (x'+u)\in \mathrm {proj}"Q$ holds since $ (x' + u) \times \Theta (u) \in G^k \cap Q$ .

It follows that $\mathrm {proj}"Q$ is open in $ \mathrm {rng}(\mathrm {proj})$ .

Let $\bar {{\Gamma }}_3^4$ be the ordered hypergraph obtained from $\bar {\Gamma }_3 $ by duplicating its zeroth coordinate four times.

Corollary 6.15. The projection of $\bar {\Gamma }_3^4$ to the first four coordinates is an open map to $(\mathbb {R}^3)^4$ .

Proof We will check Definition 6.13 is satisfied for $\bar {\Gamma }_3$ , and then just use Proposition 6.14. To check Definition 6.13 that $\bar {\Gamma }_3$ is stable against four vertices, let $x = \{x_{i}\}_{i \in 4} \subset \mathbb {R}^n$ be a set of points in general position. Then there is a unique sphere S that passes through all the points in x, with some center point c and some radius r, by elementary geometry. Take an open neighborhood $U\subset (\mathbb {R}^n)^4$ of the origin such that $\forall u = (u_i)_{i\in 4} \in U$ , $\{x_{i} + u_i:{i \in 4}\}$ still remains in general position. Let $ u = (u_i)_{i\in 4} \in U$ . Then there is a unique sphere $S'$ that passes through all the points in $\{x_{i}+u_i\}_{i \in 4}$ , with some center $c'$ and some radius $r'$ . One can explicitly calculate that $u\mapsto (c' - c, r'/r)$ is continuous to its image space. Define $\Psi (u)$ to be the similarity of $\mathbb {R}^3$ as a composition of the following maps:

• • stretch from c by $r'/r$ ;

• • shift by $c' - c$ .

As a consequence, $\Psi (u)$ is a similarity of $\mathbb {R}^3$ that moves $S $ to $S'$ . With the continuity of the map $u\mapsto (c' - c, r'/r)$ and the continuity of the transformation between the Cartesian coordinates and spherical coordinates, one can check that $\Psi $ is a continuous function with respect to the topology of its range $\mbox {SIM}(\mathbb {R}^3)$ .

For Definition 6.13 (2), it’s easy to see that $\Psi (0)$ is the identity map. Finally, let’s check Definition 6.13(3). Recall S is a unique sphere that $x\subset S$ . Let $y_1, y_2$ be such that $\bar {\Gamma }_3^4( -,y_1,y_2) = S $ ; then $y_1$ is the center of S and $y_2\in S$ . We have had S being mapped to $S'$ by $\Psi (u)$ such that $\{x_{i} + u_i:{i \in 4}\}\subset S'$ so $S'$ is uniquely determined. Observe that $S' = G( -,\Psi (u)(y_1),\Psi (u)(y_2))$ , because

• • we must have $\Psi (u)$ maps $y_1$ , the center of S, to $\Psi (u)(y_1)$ , the center of $S'$ ;

• • by $\Psi (u)"S=S'$ and $y_2\in S$ , $\Psi (u)(y_2)\in S'$ holds.

Hence we have checked the following:

• • $\Psi (u)$ moves $G( -,y_1,y_2) = S $ to $G( -,\Psi (u)(y_1),\Psi (u)(y_2)) = S' $ .

• • $\{x_{i} + u_i:{i \in 4}\}\subset G( -,\Psi (u)(y_1),\Psi (u)(y_2)) =S'$ .

Therefore, $\bar {\Gamma }_3$ is stable against four vertices. This completes the proof.

Corollary 6.16. The projection of $\bar {\Gamma }_3^4$ to the last three coordinates is an open map to the space $\bar {\Gamma }_3$ .

Proof It’s a direct consequence of Proposition 6.12.

Corollary 6.17. The projection of $\bar {\Gamma }_3^4$ to any two coordinates is an open map.

Proof We will use the fact that the composition of open maps is open. The projection to any two coordinates in the first four coordinates is an open map: by Corollary 6.15, the projection of $\bar {\Gamma }_3^4$ to the first four coordinates is an open map to the space $(\mathbb {R}^3)^4$ ; then just compose it with the projection map from $(\mathbb {R}^3)^4$ to any two coordinates.

The projection to any two coordinates in the last three coordinates is an open map: by Corollary 6.16, the projection of $\bar {\Gamma }_3^4$ to the last three coordinates is an open map to the space $\bar {\Gamma }_3 $ ; then just compose with the open map from Corollary 6.9.

The proof for the rest of the cases is the same, using the fact that $\bar {\Gamma }_3^4$ is invariant when permuting the first four coordinates.

Corollary 6.18. Let $e^4$ be generic over V for the poset $P_{\bar {\Gamma }_3 ^4} $ . Then the first four coordinate points from $ e^4$ are mutually generic over V for the product forcing $(P_{\mathbb {R}^3}) ^4$ .

Proof With the openness of each projection map in Corollary 6.15, the proof is similar to what we have done in Corollary 6.10.

Corollary 6.19. Let $e^4 = ( x_{0i}, x_1, x_{2} : i\in 4) $ be generic over V for the poset $P_{\bar {\Gamma }_3 ^4} $ . $\forall i\in 4$ , $(x_{0i}, x_1, x_{2})$ is generic over V for the poset $P_{\bar {\Gamma }_3 } $ .

Proof By Corollary 6.16, it’s easy to see that $(x_{03}, x_1, x_{2})$ is generic over V for the poset $P_{\bar {\Gamma }_3 } $ . Again using the fact that $\bar {\Gamma }_3^4$ is invariant when permuting the first four coordinates, for every $ i\in 4$ , $(x_{0i}, x_1, x_{2})$ is generic over V for the poset $P_{\bar {\Gamma }_3 } $ .

Corollary 6.20. Let $e^4$ be generic over V for the poset $P_{\bar {\Gamma }_3 ^4} $ . Then any two points from $ e^4$ are mutually generic over V for the product forcing $P_{\mathbb {R}^3} \times P_{\mathbb {R}^3} $ .

Proof Using the openness of the projection map in Corollary 6.17, the proof is similar to what we have done in Corollary 6.10.

The following is a key proposition for distinguishing ${\Gamma }_2$ and ${\Gamma }_3$ .

Proposition 6.21. There is no ${\Gamma }_2$ hyperedge in $V [e]$ with each vertex in $V [x_i]\setminus V$ respectively, where $e = ( x_0,x_1,x_2)$ is a $P_{\bar {\Gamma }_3}$ -generic hyperedge over V.

Proof First of all, for distinct $ x_i,x_j$ from e, as Corollary 6.20, $x_i$ and $x_j$ are mutually generic over V for the product forcing $P_{\mathbb {R}^3} \times P_{\mathbb {R}^3} $ ; hence $\: V[x_i] \cap V[x_j] = V$ , by basic theorem of product forcing.

Suppose towards the contradiction, in the ground model V, there is some condition $p\in P_{\bar \Gamma _3} $ forcing that there are $\Gamma _2$ -connected points $z_0 \in V[\dot {x}_ 0]\setminus V$ , $z_1 \in V[\dot {x}_ 1]\setminus V$ and $z_2 \in V[\dot {x}_ 2]\setminus V$ that are in $\mathbb {R}^2$ . Strengthen the condition p if necessary to find $P_{\mathbb {R}^3}$ -names $\tau _0,\tau _1, \tau _2$ for elements of $\mathbb {R}^2 \setminus V$ such that $p \Vdash \{\tau _0 /\dot {x}_0 ,\tau _1 / \dot {x}_ 1 , \tau _2 / \dot {x}_ 2\}\in \Gamma _2 $ . Strengthening p further, we may find disjoint basic open sets $O_0,O_1, O_2\subset \mathbb {R}^3$ such that $p=(O_0\times O_1\times O_2)\cap \bar \Gamma _3$ and p decides the pivot point for the isosceles triangle $\{\tau _0 /\dot {x}_0 ,\tau _1 / \dot {x}_ 1 , \tau _2 / \dot {x}_ 2\}$ .

Consider the hyperedge forcing $P_{\bar {\Gamma }_3^4}$ . Observe that $\bar {\Gamma }_3^4 \cap (O_0^4\times O_1\times O_2) \neq \emptyset $ holds. Obtain a generic hyperedge $e^4 = ( x_{0i}, x_1, x_{2} : i\in 4) $ in $O_0^4\times O_1\times O_2$ using the poset $P_{\bar {\Gamma }_3^4}$ . Look into the model $V[e^4]$ . We have the following observations:

1. 1. By Corollary 6.18, the points $ x_{0i} ,{i\in 4}$ are mutually Cohen generic for the product forcing $(P_{\mathbb {R}^3} )^4$ ; $V[{x}_{00},{x}_{01}] , V[{x}_{02},{x}_{03}]$ are also mutually generic extensions.

2. 2. By Corollary 6.19, for every $ i\in 4$ , $(x_{0i}, x_1, x_{2})$ is generic for the poset $P_{\bar {\Gamma }_3}$ .

3. 3. As a consequence of Corollary 6.20, the models obtained by adjoining each point from e pairwise intersect in V; using both item 1 and Corollary 6.20, $V[x_{0i}, i\in 4] $ and $V[x_1]$ are also mutually generic extensions, hence they intersect in V; the same holds for $V[x_{0i}, i\in 4] $ and $V[x_2]$ .

4. 4. By item 1, Corollary 6.20, and Proposition 4.9, the fields of reals in the models obtained by adjoining each point from e are pairwise in amalgamation position; the fields of reals $V[{x}_{00},{x}_{01}] \cap \mathbb {R} , V[{x}_{02},{x}_{03}]\cap \mathbb {R}$ are also in amalgamation position.

As a consequence of observation item 3, the points in $\{ \tau _0 / {x}_{ 0i}, \tau _1 /{x}_{1} ,\tau _2 / {x}_ 2 \} $ are pairwise distinct. With observation item 2 and by the forcing theorem applied to $p\in P_{\bar {\Gamma }_3}$ earlier, for $i\in 4$ , $ \{ \tau _0 / {x}_{ 0i}, \tau _1 /{x}_{1} ,\tau _2 / {x}_ 2 \} \in {\Gamma }_2$ holds. We have three complete cases depending on the pivot point of each isosceles triangle $ \{ \tau _0 / {x}_{ 0i}, \tau _1 /{x}_{1} ,\tau _2 / {x}_ 2 \}$ for $i\in 4 $ .

Case 1: for each $i\in 4$ , $\tau _0 / {x}_{ 0i}$ is the pivot point. Then there is a line l on which all the points $\tau _0 /{x}_{0i}, i\in 4$ are. Then $l $ is algebraic over $ \{\tau _0 / {x}_{00},\tau _0 / {x}_{01}\}\subset V[{x}_{00},{x}_{01}] \cap \mathbb {R}$ and over $\{\tau _0 / {x}_{02},\tau _0 / {x}_{03}\} \subset V[{x}_{02},{x}_{03}] \cap \mathbb {R}$ by Fact 3.2 (1). See that l is algebraic over the field $V\cap \mathbb {R}$ : l is algebraic over $ V[{x}_{00},{x}_{01}] \cap V[{x}_{02},{x}_{03}] \cap \mathbb {R}= V\cap \mathbb {R}$ by our observation item 4 and Corollary 4.3. Also see $\tau _1 /{x}_{1}, \tau _2 /{x}_{2} \in V $ : since $\tau _1 /{x}_{1},\tau _2 /{x}_{2}$ are symmetrical with respect to l, $\tau _1 /{x}_{1} \in V[x_2]$ and $\tau _2 /{x}_{2} \in V[x_1]$ hold by Fact 3.2(6); then $\tau _1 /{x}_{1} , \tau _2 /{x}_{2} \in V[x_2]\cap V[x_1] = V$ by our observation item 3. This is a contradiction to $\tau _1,\tau _2$ being forced not in V.

Case 2: $\tau _1 /{x}_{1} $ is always the pivot point. Then $\tau _1 /{x}_{1} $ is the center of the circle c on which all the points $\tau _0 /{x}_{0i}, i\in 4$ are. Then $\tau _1 /{x}_{1} \in V[x_{0i}:i\in 4] $ holds by Fact 3.2(3) and (7). Then $\tau _1 /{x}_{1} \in V[x_{0i}:i\in 4] \cap V[x_1] = V$ by our observation item 3. This is a contradiction to $\tau _1$ being forced not in V.

Case 3: $\tau _2 /{x}_{2} $ is always the pivot point. This case is handled the same as Case 2.

Therefore, our initial assumption is impossible.

7 The independence theorem

This section is devoted to the proof of Theorem 1.2.

Lemma 7.1. For each $i\in 3$ , let $V[G_i]$ be a generic extension over V such that:

1. 1. $V[G_i]$ for $i\in 3$ form a $\Delta $ -system with the root V.

2. 2. There is no isosceles triangle in $\mathbb {R}^2$ with each vertex in each $V[G_i]\setminus{\kern-1.5pt} V$ respectively.

3. 3. $P_i \in V [G_i]$ is a poset and $H_i \subset P_i$ is a filter such that $H_i, i\in 3$ are mutually generic over the model $V [G_i: i\in 3]$ .

Then:

1. 4. $V [G_i][H_i], i\in 3$ form a $\Delta $ -system with the root V.

2. 5. There is no isosceles triangle in $\mathbb {R}^2$ with each vertex in $V [G_i][H_i]\setminus V$ respectively.

Proof Let’s first show the $\Delta $ -system for (4). Fix $i\neq j$ . Let $x\in V [G_i][H_i]\cap V [G_j][H_j]$ . Then $x\in V [G_i][H_i]\cap V [G_j][H_j] \subset V [G_i, G_j][H_i]\cap V [G_i, G_j][H_j] = V[G_i,G_j] $ . Then $x \in V[G_i,G_j] \cap V [G_i][H_i] = V[G_i]$ and $x \in V[G_i,G_j] \cap V [G_j][H_j] \kern1.3pt{=}\kern1.3pt V[G_j]$ . Then $x\kern1.3pt{\in}\kern1.3pt V[G_i] \kern1.3pt{\cap}\kern1.3pt V[G_j] \kern1.3pt{=}\kern1.3pt V$ . Thus $ V [G_i][H_i]\kern1.2pt{\cap}\kern1.2pt V [G_j][H_j] \subset V$ . Thus $ V [G_i][H_i]\cap V [G_j][H_j] = V$ .

For (5), suppose towards a contradiction that for each $i\in 3$ , $p_i \in P_i$ is a condition and $\tau _i$ are $P_i$ -names in $V [G_i]$ for elements of $\mathbb {R}^2$ which do not belong to V, and $\langle p_0,p_1,p_2\rangle $ forces in $P_0 \times P_1\times P_2$ that $\{\tau _0,\tau _1, \tau _2\} \in \Gamma _2$ . Strengthen $\langle p_0,p_1,p_2\rangle $ to decide which of the points in the set $\{\tau _0,\tau _1, \tau _2\} \in \Gamma _2$ is the pivot point for the isosceles triangle. To complete the proof, we will reach a contradiction in any case as below.

Case 1: $\forall i\: p_i \Vdash \tau _i\in V[G_i]$ . Then $\forall i\: p_i \Vdash \tau _i\in V[G_i]\setminus V$ and $\langle p_0,p_1,p_2\rangle $ forces that $\{\tau _0,\tau _1, \tau _2\} \in \Gamma _2$ , which is impossible by our assumption item 2.

Case 2: $\exists i\ p_i \Vdash \tau _i\notin V[G_i]$ . Without loss of generality, $p_0 \Vdash \tau _0\notin V[G_0]$ . So $p_0 \Vdash \tau _0\in V[G_0][H_0]\setminus V[G_0]$ . Here comes a “duplication” technique again. For each $k\in 4$ , let $h_{0k}\subset P_0$ be mutually generic filters over $V[G_0] $ containing the condition $p_0$ . Let $h_{1}\subset P_1$ containing the condition $p_1$ and $h_{2}\subset P_2$ containing the condition $p_2$ be mutually generic filters over $V[G_{i}: i\in 3] [h_{0k}: k\in 4]$ . For each $k\in 4$ , let $y_{0k} = \tau _0 / h_{0k}, y_{1} = \tau _1 / h_{1}, y_{2} = \tau _2 / h_{2} \in \mathbb {R}^2$ be points. As $\langle p_0,p_1,p_2\rangle \Vdash \{\tau _0,\tau _1, \tau _2\} \in \Gamma _2$ , for each $ k\in 4, \{y_{0k}, y_1, y_2 \} \in \Gamma _2$ holds. We will reach a contradiction in two complementary cases, similar to the cases in the proof of Proposition 6.21.

Case 2.1: $p_0$ forces that $\tau _0$ is not the pivot point. Then for each k, $y_{0k}$ will not be the pivot point. Then $y_j$ is the pivot point for $j=1$ or $j = 2$ . The point $y_j$ , as the center of the circle passing through all points in $\{y_{0k}\}_{k\in 4}$ , is algebraic over $\{y_{0k}\}_{k\in 4}$ by Fact 3.2(7); hence $y_j$ is in $V [G_0][h_{0k}: k\in 4]$ . But also $y_j\in V[G_j]$ . Hence $y_j\in V[G_j]\cap V [G_0][h_{0k}: k\in 4] = V$ , contradicting our initial assumption that $\tau _j$ is forced to not belong to V.

Case 2.2: $p_0$ forces that $\tau _0$ is the pivot point. Then for each k, $y_{0k}$ is the pivot point. Then the line l, which is the perpendicular bisector of the line segment from $y_1$ to $y_2$ , passes through $y_{0k}$ for each k. Then $l $ is algebraic over $ \{y_{00}, y_{01}\}\subset V[G_0][y_{00}, y_{01}] \cap \mathbb {R}$ and over $\{y_{02}, y_{03}\} \subset V[G_0][y_{02}, y_{03}] \cap \mathbb {R}$ by Fact 3.2(1). Then l must be algebraic over $ V[G_0][y_{00}, y_{01}] \cap V[G_0][y_{02}, y_{03}] \cap \mathbb {R}= V[G_0]\cap \mathbb {R}$ by Proposition 4.9 and Corollary 4.3. Also observe $y_1 \in V[G_1],y_2 \in V[G_2] $ : since $y_1,y_2$ are symmetrical with respect to l, $y_1 \in V[G_i:i\in 3][h_2]$ and $y_2 \in V[G_i:i\in 3][h_1]$ hold by Fact 3.2(6); $V[G_i:i\in 3][h_1]\cap V[G_i:i\in 3][h_2] = V[G_i:i\in 3]$ holds; then $y_1 \in V[G_i:i\in 3] \cap V[G_1][h_1]= V[G_1],y_2 \in V[G_i:i\in 3] \cap V[G_2][h_2] = V[G_2] $ hold.

The line l algebraic over $V[G_0] \cap \mathbb {R}$ needs to contain some points in $V[G_0] $ . See that l can’t contain any point in $V[G_0]\setminus V$ , because such a point would form an isosceles triangle with $y_1,y_2$ contradicting the assumption item 2. Then l contains points with coordinates from $V\cap \mathbb {R}$ by and must be algebraic over $V\cap \mathbb {R}$ by Fact 3.2. Since $y_1,y_2$ are symmetrical with l, $y_1 \in V[G_2]$ and $y_2 \in V[G_1]$ , hence $y_1,y_2\in V[G_2] \cap V[G_1] = V$ , contradicting our initial assumption that $\tau _1,\tau _2$ are forced to not belong to V.

Now let the poset P be defined as Definition 3.1. The following gives us the ability of amalgamating three conditions in P.

Proposition 7.2. For arbitrary conditions $ p_0, p_1,p_2 \in P $ , with their common upper bound p, such that any pair of fields from $\mathrm {supp}(p_0),\mathrm {supp}(p_1) , \mathrm {supp}(p_2) \subset \mathbb {R}$ is in amalgamation position and that $\{\mathrm {supp}(p_0),\mathrm {supp}(p_1) , \mathrm {supp}(p_2)\}$ forms a $\Delta $ -system with the root $\mathrm {supp}(p_0)\cap \mathrm {supp}(p_1) \cap \mathrm {supp}(p_2) = \mathrm {supp}(p)$ , $ p_0, p_1,p_2 $ have a common lower bound in P whenever:

1. 1. $p_0 \upharpoonright \mathrm {supp}(p) = p_1 \upharpoonright \mathrm {supp}(p) = p_2 \upharpoonright \mathrm {supp}(p) $ , and $p\in P$ , $p_0\leq p$ , $p_1\leq p$ and $p_2\leq p$ hold.

2. 2. There is no isosceles triangle in $\mathbb {R}^2$ with each of its vertex in $\mathrm {dom}(p_0) \setminus \mathrm {dom}(p),\mathrm {dom}(p_1) \setminus \mathrm {dom}(p), \mathrm {dom}(p_2) \setminus \mathrm {dom}(p)$ respectively.

Proof First of all, $\forall i, j\in 3 , p_i \cup p_j$ is a $\Gamma _2$ -coloring by Lemma 5.5. Since there is no isosceles triangle in $\mathbb {R}^2$ with each of its vertex in $\mathrm {dom}(p_0) \setminus \mathrm {dom}(p),\mathrm {dom}(p_1) \setminus \mathrm {dom}(p), \mathrm {dom}(p_2) \setminus \mathrm {dom}(p)$ respectively by item 2 in our assumption, $ p_0\cup p_1\cup p_2$ is a $\Gamma _2$ -coloring.

Now we will extend the $\Gamma _2$ -coloring $ p_0\cup p_1\cup p_2$ . Let $ b \supset \mathrm {supp}(p_0) \cup \mathrm {supp}(p_1) \cup \mathrm {supp}(p_2)$ be a countable real closed subfield of $\mathbb {R}$ . We will extend $p_0 \cup p_1\cup p_2$ to some $p_3 \in P$ with $ \mathrm {supp}(p_3) = b$ and $\mathrm {dom}(p_3) = \mathrm {supp}(p_3) ^2 = b^2$ . For each point $x \in b^2 \setminus (\mathrm {dom}(p_0) \cup \mathrm {dom}(p_1)\cup \mathrm {dom}(p_2) ) $ , let $d(x) = \bigcup _{i\neq j\in 3} d(p_i,p_j,x) $ where each $d(p_i,p_j,x) $ is the set of disallowed colors on x for $p_i,p_j$ from Definition 5.6. Then $d(x)\in I$ holds as each $d(p_i,p_j,x)\in I$ by Lemma 5.7. Fix any set $\pi \in I $ so that $\forall x\in b^2 \setminus (\mathrm {dom}(p_0) \cup \mathrm {dom}(p_1)\cup \mathrm {dom}(p_2) ) : \:\pi \setminus d(x)$ is infinite, which is possible by our choice of the ideal I. Now just let $p_3$ be a map with domain $b^2$ such that:

• • $p_2\upharpoonright {\mathrm {dom}(p_0)\cup \mathrm {dom}(p_1)\cup \mathrm {dom}(p_2)} = p_0 \cup p_1\cup p_2$ .

• • $\forall x\in b^2 \setminus ( \mathrm {dom}(p_0 ) \cup \mathrm {dom}( p_1) \cup \mathrm {dom}(p_2)),\: p_3 (x) \in \pi \setminus d(x)$ .

• • $p_3$ on $b^2 \setminus ( \mathrm {dom}(p_0 ) \cup \mathrm {dom}( p_1) \cup \mathrm {dom}(p_2))$ is an injection.

Claim 7.3. $p_3 \in P$ and $p_3 \leq p_0,p_1,p_2$ hold.

Proof Similar to the proof of Claim 5.9 and Claim 5.10.

This completes the proof.

Theorem 7.4. Let $\kappa $ be an inaccessible cardinal. There is a model of ZF+DC in which ${\Gamma _2}$ has countable chromatic number while for every non-meager set $A\subset \mathbb {R}^3$ , A contains all vertices of an isosceles triangle.

Proof Let W be the symmetric Solovay model derived from $\kappa $ , and the model we need is the P extension over W, in which ${\Gamma _2}$ has countable chromatic number by our construction of P. We will show that in this model, for every non-meager set $A\subset \mathbb {R}^3$ , A contains all vertices of an isosceles triangle.

Work in W. Suppose $p\in P$ is a condition and let $\tau $ be a P-name for a nonmeager subset of $\mathbb {R}^3$ . We will find a $\Gamma _3$ -hyperedge e and a strengthening of the condition p which forces e to be a subset of $\tau $ . By the definition of W and Fact 2.7(1), the condition p as well as the name $\tau $ must be definable from ground model parameters and an additional parameter $z \in \omega ^\omega $ . Using Fact 2.7(3), let $V [K]$ be an intermediate extension obtained by a forcing of cardinality smaller than $\kappa $ with z, $p \in V[K]$ and the Continuum Hypothesis holds in $V [K]$ . Then P is balanced in $V[K]$ by Proposition 5.3(2).

Work in the model $V [K]$ . Let $\bar {p} \leq p$ be a balanced virtual condition in the poset P. Consider the Cohen poset $P_{\mathbb {R}^3}$ consisting of nonempty open subsets of ${\mathbb {R}^3}$ ordered by inclusion, with its name $ \dot {x}_{gen}$ for a generic element of the space ${\mathbb {R}^3}$ . By a standard Solovay model argument using Fact 2.8 with Fact 2.4, there must be a condition $O \in P_{\mathbb {R}^3}$ , a poset Q of cardinality smaller than $\kappa $ , and (with the convenience of product forcing) a $P_{\mathbb {R}^3} \times Q$ -name $\sigma $ for a condition in the poset P stronger than $\bar {p}$ such that

$$ \begin{align*}O \Vdash_{P_{{\mathbb{R}^3}}} Q \Vdash \mathrm{Coll} (\omega, < \kappa)\Vdash \sigma \Vdash_P \dot{x}_{gen}\in \tau. \end{align*} $$

Otherwise, in the model W, the virtual condition $\bar {p} $ would force $\tau $ to be disjoint from the comeager set of points in $\mathbb {R}^3$ which are Cohen generic over the model $V [K]$ , contradicting the initial assumption that the name $\tau $ forced to be a non-meager set.

Now recall the poset $P_{\bar \Gamma _3}$ according to Definition 6.1, and observe that $O^3$ is a condition in $P_{\bar \Gamma _3}$ . Over the ground model $V[K]$ , let e be a generic hyperedge of the poset $P_{\bar \Gamma ^3}$ that meets $O^3$ . Identify e with set of its vertices. Observe that for each vertex $x\in e$ , x is $P_{\mathbb {R}^3}$ -generic, and each pair of vertices in e is mutually generic, by Corollary 6.10. Let $H_x\subset Q$ for $x \in e$ be filters mutually generic over the model $V [K][e]$ . Work in the model $V [K][e][H_x : x \in e]$ . Use Proposition 6.21 and Lemma 7.1 to observe that:

• • The models $V [K][x][H_x]$ for $x \in e$ form a $\Delta $ -system with the root $V[K]$ .

• • There is no isosceles triangle in $\mathbb {R}^3$ with one vertex in each model $V [K][x][H_x]$ for $x \in e$ respectively.

Claim 7.5. Whenever for each vertex $x\in e$ , $p_x$ is a condition in $V [K][e][H_x]$ with $ p_x \leq \bar {p}$ , the conditions $ p_x$ for $x\in e$ have a common lower bound.

Proof Apply Proposition 7.2 after checking its assumptions.

For each $x\in e$ , write $p_x = \sigma / x$ , which is a condition in $V [K][x][H_x]$ ; by forcing theorem applied in the model $V [K][x][H_x]$ , $p_x\leq \bar {p}$ is a condition forcing $x \in \tau $ . Then by Claim 7.5, let q be a lower bound of the conditions $ p_x$ for $x\in e$ . Thus we have produced a $\Gamma _3$ -hyperedge e and a condition q stronger than p which forces all vertices of e to be in $\tau $ , as required.

Finally, Theorem 1.2 is just a corollary of Theorem 7.4. Here is the proof of Theorem 1.2.

Proof Let $\kappa $ be an inaccessible cardinal. By Theorem 7.4, there is a model of ZF+DC in which ${\Gamma _2}$ has countable chromatic number while for every non-meager set $A\subset \mathbb {R}^3$ , A contains all vertices of an isosceles triangle. Then immediately, in this model, the hypergraph $\Gamma _3$ has uncountable chromatic number: by Theorem 2.2, Baire Category Theorem holds for $\mathbb {R}^3$ due to DC; then when dividing the space $\mathbb {R}^3$ into countably many pieces, one of them has to be nonmeager; that piece contains all vertices of an isosceles triangle.