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).

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 [Reference Oxley7, 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.

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.

This completes the proof of the theorem.

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.

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 $F_0\cap F_1$ .

Fact 4.5 [Reference Marker6, 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

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

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.

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.

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.

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 $ .

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.

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 [Reference Larson and Zapletal5, 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 [Reference Larson and Zapletal5, Theorem 5.2.8]Footnote 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 $ .

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.

Proof We start with the following claim.

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 pointsFootnote 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.

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.

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

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 $\mathbb {R}^n$ .

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.

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.

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

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

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.

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 $\bar {\Gamma }_3$ .