Fact 2.2. Let $n\geq 1$ be a number.

1. 1. (Hilbert basis theorem) There is no infinite sequence of algebraic sets which is strictly decreasing with respect to inclusion;

2. 2. [Reference Bochnak, Coste and Roy1, Theorem 2.8.3] every algebraic set $A\subseteq \mathbb {R}^d$ has a unique decomposition into irreducible components: $A=\bigcup \{A_i\colon i\in j\}$ where $j\geq 1$ is a number and each sets $A_i$ is algebraic, irreducible, and not included in the union of the others;

3. 3. [Reference Zapletal10, Claim 4.9] if $A, B\subseteq \mathbb {R}^d$ are algebraic sets and A is irreducible, then either $A\subseteq B$ or $A\cap B$ is nowhere dense in the set of nonsingular points of A.

Proof. (1) follows from the fact that the components of an algebraic set are definable from the algebraic set. To fill in the details, let $A\subset \mathbb {R}^d$ be an algebraic set. Let $p_i$ for $i\in j$ be polynomials with real coefficients whose zero sets are the components of the set A. Using dummy zero coefficients if necessary, we may assume that there is a number m such that the sequence of coefficients of $p_i$ is an element of $\mathbb {R}^m$ for every $i\in j$ . For each $i\in j$ let $O_i\subset \mathbb {R}^m$ be a basic open set containing the tuple of coefficients of $p_i$ , so that the sets $O_i$ for $i\in j$ are pairwise disjoint. Now, the zero set of (say) the polynomial $p_0$ is defined as the set of all $x\in A$ such that there is a sequence $\langle q_i\colon i\in j\rangle $ of polynomials whose coefficients come from the set $O_i$ respectively, each of whose zero sets is a proper subset of A, and such that $q_0(x)=0$ . This follows from the uniqueness of components of algebraic sets.

(2) follows from the fact that the set of singular points of an irreducible algebraic set A is an algebraic, proper subset of it definable from A. The algebraicity and proper inclusion is proved in [Reference Bochnak, Coste and Roy1, Proposition 3.3.14]. For the definability, it is enough to restate the definitions. Namely, for every polynomial p such that $A=\{y\in \mathbb {R}^d\colon p(y)=0\}$ , the Zariski tangent space to A at y is the set of all vectors perpendicular to the vector of partial derivatives of p at y. The tangent space does not depend on the choice of p. The point y is then nonsingular in A if this space has maximal dimension possible, namely $\dim (A)$ . This can be mechanically restated by a rather long formula in the language of real closed fields.

(3) follows from the fact that a real closed subfield is an elementary submodel of $\mathbb {R}$ . If $D(F, x)=\{x\}$ then each coordinate of x is definable from parameters in F and therefore an element of F.

Proof. Let $v\in \mathbb {R}^m$ be a tuple enumerating all elements of a. For each $x\in F_1^n$ , let $p_x$ be a polynomial with $n+m$ free variables and coefficients in the field $F_0$ such that $p_x(x, v)=0$ and the set $A_x=\{u\in \mathbb {R}^n\colon p_x(u, v)=0\}$ is as small as possible. Such a polynomial exists by the Hilbert basis theorem. Let $B_x=\{w\in \mathbb {R}^m\colon p_x(x, w)=0\}$ ; this is an algebraic set containing v as an element. Let $C=\bigcap \{B_x\colon x\in F_1^n\}$ ; this is an algebraic set containing v as an element. Let $c\subset F_1^n$ be a finite set such that $C=\bigcap \{B_x\colon x\in c\}$ . Let $b\subset F_1$ be a finite set such that all coordinates of points in c can be found in the set b. I claim that the set b works.

To show this, suppose that $x\in F_1^n$ is an arbitrary point, and work to show that $D(F_0\cup b, x)\subseteq D(F_0\cup a, x)$ holds. Use quantifier elimination for real closed fields to define the set $D(F_0\cup a, x)$ by a quantifier-free formula with parameters in $F_0\cup a$ . There is a finite list of polynomials $q_i\colon i\in j$ , each having parameters in $F_0$ and $n+m$ free variables such that the quantifier-free formula is a Boolean combination of formulas of the form $q_i(u, v)\geq 0$ . Let $d=\{i\in j\colon q_i(x, v)=0\}$ , let $D_x=\{u\in \mathbb {R}^n\colon q_i(u, v)=0$ for all $i\in n\}$ , and let $O\subset \mathbb {R}^n$ be a basic open neighborhood of x in which the polynomials $q_i(u, v)$ for $i\in j\setminus d$ do not change sign. Clearly, $A_x\subseteq D_x$ holds by the minimal choice of the polynomial $p_x$ , and $D_x\cap O\subset D(F_0\cup a, x)$ holds by definitions.

Consider the set $E_x=\{u\in \mathbb {R}^d\colon \forall w\in C\ p_x(u, w)=0\}$ . This is an algebraic set as well as a $F_0\cup b$ -semi-algebraic set containing x as an element. By its definition, $E_x\subseteq A_x$ holds. Thus, $E_x\cap O\subset D(F_0\cup a, x)$ must hold as well. Finally, since $D(F_0\cup b, x)\subseteq E_x$ holds by the definition of $D(F_0\cup b, x)$ , it must be the case that $D(F_0\cup b, x)\cap O\subseteq D(F_0\cup a, x)$ .

Now, consider the disjunction in Fact 2.2(3) applied to $A=D(F_0\cup b, x)$ and $B=D(F_0\cup a, x)$ . It is impossible for the set $A\cap B$ to be nowhere dense in the set of nonsingular points of A, because $x\in O$ is a nonsingular point of A by Proposition 2.5 and B contains the whole neighborhood $A\cap O$ of x in A. Thus, it must be the case that $A\subseteq B$ holds as desired.

Proof. Let $v\in \mathbb {R}^m$ be a finite sequence enumerating all elements of a. It follows directly from the definitions that $D((\mathbb {R}\cap V)\cup a, x)$ is just the vth section of the set $D(\mathbb {R}\cap V, v^\smallfrown x)$ , and similarly for $D((\mathbb {R}\cap V[G_0])\cup a, x)$ . However, the sets $D(\mathbb {R}\cap V, v^\smallfrown x)$ and $D(\mathbb {R}\cap V[G_0], v^\smallfrown x)$ are equal by [Reference Zapletal10, Corollary 2.7] applied to the Noetherian topology of algebraic sets.

Proof. For (1), recall that x is algebraic over A if and only if $D(A, x)=\{x\}$ by Proposition 2.5(3). Now, if $D(A, x)=\{x\}$ then $\varepsilon (A, x)=0$ by the definitions. If $D(A, x)\neq \{x\}$ then pick a point $y\in D(A, x)$ distinct from x. If for every $\varepsilon>0$ there is a set $b_\varepsilon $ consisting of points closer to x than $\varepsilon $ such that $b_\varepsilon \cup \{y\}\in \Gamma $ , then $\{x, y\}$ belongs to the closure of $\Gamma $ in the space $X^{\leq 3}$ , contradicting the assumption of $2$ -irreflexivity on $\Gamma $ . Thus, $\varepsilon (A, x)>0$ holds as desired.

(2) follows directly from the definitions, noting that $D(B, x)\subseteq D(A, x)$ . For (3), suppose toward a contradiction that $b\subset F^d$ is a set consisting of points closer than $\varepsilon (F, x)$ to x such that $b\cup \{x\}\in \Gamma $ . Let $B=\{y\in X\colon b\cup \{y\}\in \Gamma \}\cup b$ . This is an F-algebraic set containing x, and therefore $D(F, x)\subseteq B$ holds. It follows that for every point $y\in D(F, x)\setminus (b\cup \{x\})$ , $b\cup \{y\}\in \Gamma $ holds, and by the definitions, the set b cannot consist of points which are closer than $\varepsilon (F, x)$ to x.

Proof. Let $r\leq q\leq p$ be conditions in the coloring poset P; I must show that ${r\leq p}$ holds. For Definition 3.4(1), since $p\subseteq q$ and $q\subseteq r$ holds, $p\subseteq r$ certainly follows. For Definition 3.4(2), suppose that $x\in \mathrm {dom}(r)\setminus \mathrm {dom}(p)$ is a point. If $x\in \mathrm {dom}(q)$ , then Definition 3.4(2) applied to $q\leq p$ implies that the diameter of $r(x)=q(x)$ is smaller than $\varepsilon (\mathrm {supp}(p), x)$ . If $x\in \mathrm {dom}(r)\setminus \mathrm {dom}(q)$ , then Definition 3.4(2) applied to $r\leq q$ implies that $\mathrm {diam}(r(x))<\varepsilon (\mathrm {supp}(q), x)\leq \varepsilon (\mathrm {supp}(p), x)$ where the second inequality follows from Proposition 3.3(2).

The key point is the verification of Definition 3.4(3). Suppose that $a\subset \mathrm {supp}(r)$ is a finite set. By Proposition 2.6, find a finite set $b\subset \mathrm {supp}(q)$ such that for every point $x\in \mathrm {dom}(q)$ , $D(\mathrm {supp}(p)\cup b, x)\subseteq D(\mathrm {supp}(p)\cup a, x)$ . It is then clear that the set $C(p, r, a)$ is a subset of $C(p, q, b)\cup C(q, r, a)$ . Now, $q^*C(p, q, b)\in I$ and $r^*C(q, r, a)\in I$ holds by Definition 3.4(3) applied to $q\leq p$ and to $r\leq q$ . Thus, $r^*C(p,r,a)\in I$ and the proof is complete.

Proof. Fix $p\in P$ and $x\in \mathbb {R}^d$ . Let $G\subset \mathbb {R}$ be any countable real closed subfield of $\mathbb {R}$ containing $\mathrm {supp}(p)$ as a subset and coordinates of the point x as elements. Choose an infinite set b of positive rationals converging to zero. Let q be a function with domain $G^d$ which extends p such that for every $y\in G^d\setminus \mathrm {dom}(p)$ $q(y)$ is a basic open neighborhood of y of diameter which is in the set b and smaller than $\varepsilon (\mathrm {supp}(p), y)$ and such that $q\restriction G^d\setminus \mathrm {dom}(p)$ is an injection. It will be enough to show that $q\in P$ is a condition stronger than p.

To show that q is a coloring, suppose that e is a hyperedge in $\mathrm {dom}(q)$ , and work to show that e is not monochromatic. If all its vertices belong to p, then this follows from p being a coloring. If more than one of its vertices belongs to $\mathrm {dom}(q\setminus p)$ then this follows from q being an injection on this set. Finally, in the event that e has exactly one vertex y in the set $\mathrm {dom}(q\setminus p)$ , apply Proposition 3.3(3) to show that e is not monochromatic in this case either.

Now, to show that $q\leq p$ , it is clear that $p\subseteq q$ holds. For every point $y\in \mathrm {dom}(q\setminus p)$ , the value $q(y)$ is a basic open neighborhood whose diameter is smaller than $\varepsilon (\mathrm {supp}(y), y)$ . This verifies Definition 3.4(2) for p and q. Finally, for any finite set $a\subset \mathrm {supp}(q)$ the set $q^*C(p, q, a)$ belongs to the ideal I, because it is a subset of $q^*\mathrm {dom}(q\setminus p)$ which is in turn a subset of b. This verifies Definition 3.4(3) for p and q and completes the proof.

Proof. If $\langle p_i\colon i\in \omega \rangle $ is a descending sequence of conditions then $p=\bigcup _ip_i$ is their common lower bound. To see this, first note that $\mathrm {supp}(p)=\bigcup _i\mathrm {supp}(p_i)$ is a real closed subfield of $\mathbb {R}$ , $\mathrm {dom}(p)=\mathrm {supp}(p)^d$ , and p is a function and a coloring, therefore a condition in the poset P. Now, fix $i\in \omega $ and work to show that $p\leq p_i$ holds. Items (1) and (2) of Definition 3.4 are immediate. For item (3), if $a\subset \mathrm {supp}(p)$ is a finite set, then there is $j\geq i$ such that $a\subset \mathrm {supp}(p_j)$ , and then $C(p_i, p, a)=C(p_i, p_j, a)$ and (3) follows from the same item of the definition applied to $p_j\leq p_i$ .

Proof. Use the CH assumption to find an enumeration $\langle r_\alpha \colon \alpha \in \omega _1\rangle $ of all reals. By transfinite recursion on the countable ordinal $\alpha $ build conditions $p_\alpha \in P$ so that $p_0=p$ , $\alpha \in \beta $ implies $p_\beta \leq p_\alpha $ , and $r_\alpha \in \mathrm {supp}(p_{\alpha +1})$ . This is easy to do using Proposition 3.6 in successor stages, and Proposition 3.7 in limit stages. Finally, let $\bar p=\bigcup _\alpha p_\alpha $ . This is a total $\Gamma $ -coloring. The proof of Proposition 3.7 in the $\mathrm {Coll}(\omega , \mathbb {R})$ -extension shows that $\mathrm {Coll}(\omega , \mathbb {R})\Vdash \bar p$ is a common lower bound of all $p_\alpha $ for $\alpha \in \omega _1^V$ as desired.

Proof. The first observation must be the following.

It is important to understand that the work does not end here. While the function $\bigcup _{i\in k}p_i$ is a coloring, its domain is not in the required form for a condition in the coloring poset; it is necessary to extend its domain. Toward this end, I need to analyze the situation closer.

Now, let F be any countable real closed subfield of $\mathbb {R}$ such that $\bigcup _{i\in k}\mathrm {supp}(p_i)\subset F$ . Let $G=F^d\setminus \bigcup _{i\in k}\mathrm {dom}(p_i)$ . Let $b\subset \mathbb {Q}^+$ be an infinite sequence converging to zero which has finite intersection with every set $b_{ij}(x)$ for distinct indices $i, j\in k$ and points $x\in G$ . It is not difficult to find a map q on $F^d$ extending $\bigcup _{i\in k}p_i$ , such that $q\restriction G$ is an injection and such that for every $x\in F^n\setminus \bigcup _{i\in k}\mathrm {dom}(p_i)$ , and all indices $i, j\in k$ , $\mathrm {diam}(q(x))<\varepsilon (\mathrm {supp}(p_i), x)$ and $\mathrm {diam}(q(x))\in b\setminus b_{ij}$ both hold. This is easy to do as the set G countable, and for each point $x\in G$ the last two demands still leave infinitely many options for the value $q(x)$ . It will be enough to show that q is the desired common lower bound of the conditions $p_i$ for $i\in k$ .

First of all, q is a function and $\bigcup _ip_i\subset q$ by the definitions. To see that q is a coloring, let $e=\{x_0, x_1, x_2\}$ be a $\Gamma $ -hyperedge in $\mathrm {dom}(q)$ ; I must show that it is not monochromatic. There are several possible configurations.

Case 1. $e\subset \mathrm {dom}(\bigcup _ip_i)$ . This configuration is resolved in Claim 3.10.

Case 2. If e contains exactly two vertices in $\mathrm {dom}(\bigcup _ip_i)$ and there is $i\in k$ such that these two vertices, say $x_1, x_2$ , both belong to $\mathrm {dom}(p_i)$ . Then the hyperedge is not monochromatic by Proposition 3.3(3) and $\mathrm {diam}(q(x))<\varepsilon (\mathrm {supp}(p_i), x)$ .

Case 3. If e contains exactly two vertices in $\mathrm {dom}(\bigcup _ip_i)$ and there are distinct indices $i, j\in k$ such that (say) $x_1\in \mathrm {dom}(p_i\setminus \bar p)$ and $x_2\in \mathrm {dom}(p_j\setminus \bar p)$ then either $\{x_0, x_2\}\not \subset p_i(x_1)$ or $p_i(x_1)$ has diameter in $b_{ij}$ while $q(x_0)$ does not, by the choice of the set $b_{ij}(x)$ . This means that the hyperedge e is not monochromatic.

Case 4. If e contains two or three vertices in the set G then e is not monochromatic since the function $q\restriction G$ is an injection.

Finally, I have to show that for all $i\in k$ , $q\leq p_i$ holds. This is in fact another small ordeal. Fix $i\in k$ ; I must verify items (2) and (3) of Definition 3.4. Start with item (2). Let $x\in F^d$ be a point not in $\mathrm {dom}(p_i)$ . There are two cases.

Case 1. There is $j\neq i$ such that $x\in \mathrm {dom}(p_j)$ . In this case, use Proposition 2.7 and the mutual genericity assumption to see that $D(\mathbb {R}\cap V, x)=D(\mathrm {supp}(p_i), x)$ and then apply item (2) of Definition 3.4 of $p_j\leq \bar p$ to see that $q(x)=p_j(x)$ has diameter smaller than $\varepsilon (\mathrm {supp}(p_i), x)$ .

Case 2. $x\notin \bigcup _j\mathrm {dom}(p_j)$ ; in this case, $\mathrm {diam}(q(x))<\varepsilon (\mathrm {supp}(p_i), x)$ by the choice of the function q.

Now, to verify item (3), let $a\subset \mathbb {R}$ be an arbitrary finite set. For every index $j\neq i$ , use Proposition 2.6 to find a finite set $b_j\subset \mathrm {supp}(p_j)$ such that for every point $x\in \mathrm {dom}(p_j)$ , $D(\mathrm {supp}(p_i)\cup b_j, x)\subseteq D(\mathrm {supp}(p_i)\cup a, x)$ . Use Proposition 2.7 and the mutual genericity assumption to conclude that for every point $x\in \mathrm {dom}(p_j)$ , $D((\mathbb {R}\cap V)\cup b_j, x)\subseteq D(\mathrm {supp}(p_i)\cup a, x)$ . Conclude that $C(p_i, p_j, a)\cap \mathrm {dom}(p_j)\subseteq C(\bar p, p_j, b_j)$ . Now, it is clear that $C(p_i, q, a)\subseteq G\cup \bigcup _{j\neq i}C(p_i, q, a)\cap \mathrm {dom}(p_j)\subseteq G\cup \bigcup _{j\neq i}C(\bar p, p_j, b_j)$ . It follows that the set $q^*C(p_i, q, a)$ belongs to the ideal I: $q^*G\in I$ holds by the construction of the map q, and for every $j\in i$ $q^*C(p_i, q, a)\cap \mathrm {dom}(p_j)\subseteq q^*C(\bar p, p_j, b_j)\in I$ by Definition 3.4(3) applied to $p_j\leq \bar p$ . The proof is complete.

Proof. Let $p\in P$ be any condition. Proposition 3.8 provides a total coloring $\bar p$ such that $\mathrm {Coll}(\omega , \mathbb {R})\Vdash \bar p\leq p$ . Proposition 3.9 then shows that $\langle \mathrm {Coll}(\omega , \mathbb {R}), \bar p\rangle $ is a (4,3)-balanced pair below p as required by the definition of balance in [Reference Larson and Zapletal4, Definition 13.1.1].

Proof. The proof is literally identical to the proof of Proposition 3.9 except for the configuration in Case 3 of the proof of Claim 3.10. To argue for that in the special case of equilateral triangles from the weaker assumptions on the tuple of generic extensions, proceed as follows.

Suppose that $\{x_0, x_1, x_2\}$ is an equilateral triangle in the domain of $\bigcup _{i\in k}p_i$ and there are pairwise distinct numbers $i_0, i_1, i_2\in j$ such that $x_k\in \mathrm {dom}(p_{i_j}\setminus p)$ for all $j\in 3$ ; re-numbering if necessary, we may assume that $i_j=j$ for all $j\in 3$ . Let $\delta>0$ be the common length of the edges of the equilateral triangle $\{x_0, x_1, x_2\}$ . Since $\delta \in V[G_0, G_1]\cap V[G_1, G_2]\cap V[G_0, G_2]$ , the amalgamation assumption (2) implies that $\delta \in V$ holds. Now, the pairwise genericity assumption (1) and Proposition 2.7 show that $D(V\cap \mathbb {R}, x_1)$ is a subset of both spheres of radius $\delta $ centered at $x_0$ and $x_2$ , as these spheres are algebraic sets defined from parameters in $V[x_0]$ and $V[x_2]$ respectively and contain $x_1$ . It follows that for every point $y\in D(V\cap \mathbb {R}, x_1)$ , $\{x_0, y, x_2\}\in \Gamma $ holds. Now, Definition 3.4(2) applied to $p_1\leq p$ shows that the diameter of $p_1(x)$ is smaller than $\varepsilon (V\cap \mathbb {R}, x_1)$ . The definition of $\varepsilon (V\cap \mathbb {R}, x_1)$ implies that $\{x_0, x_2\}\not \subset p_1(x_1)$ ; therefore, the triangle $\{x_0, x_1, x_2\}$ cannot be monochromatic.

Proof. The proof is a rather mechanical application of the calculus of generic points of semi-algebraic sets. For completeness, I recall the basic tenets of this calculus. If X is a Polish space, a Cohen-generic element of X over a transitive model V of ZFC is a point $x\in X$ which belongs to all dense open subsets of X coded in V. It is obtained by forcing with the Cohen poset $P_X$ of all nonempty open subsets of X ordered by inclusion; its name for the generic point of X will be denoted by $\dot x$ . The following is proved by an immediate density argument.

If $m\geq 1$ is a number, then every semi-algebraic set $C\subset \mathbb {R}^m$ is a Boolean combination of closed sets by the quantifier elimination theorem, therefore it is $G_\delta $ in $\mathbb {R}^m$ and Polish in the inherited topology. In this sense I speak of Cohen-generic elements of C. The poset $P_C$ is realized as the poset of all basic open sets $O\subset \mathbb {R}^m$ with nonempty intersection with C, ordered by inclusion.

Let $C\subset (\mathbb {R}^2)^3$ be the semi-algebraic set of all triples $\langle u_0, u_1, u_2\rangle $ consisting of pairwise distinct points such that the Euclidean distance from $u_0$ to $u_1$ or $u_2$ is the same. Denote the $P_C$ -names for points in the generic triple by $\dot x_0, \dot x_1, \dot x_2$ .

The following computation is a little more involved. The proof is a great example of a powerful method I call the duplication technique.

A symmetrical argument shows the following.

Now, Claims 4.4 and 4.5 show that the intersection of the models $V[\dot x_0, \dot x_1]$ , $V[\dot x_1, \dot x_2]$ and $V[\dot x_0, \dot x_2]$ is forced to be equal to the intersection of $V[\dot x_0]$ and $V[\dot x_1]$ . Claim 4.3 shows that these two extensions are mutually generic extensions of the ground model, showing that their intersection is V and completing the proof of the proposition.

Claim 4.6. The models $V[K][x_i][H_i]$ for $i\in 3$ are pairwise mutually generic extensions of $V[K]$ . In addition, $\bigcap _{a\in [3]^2}V[K][x_i\colon i\in a][H_i\colon i\in a]=V[K]$ .

Proof. The first sentence is immediate. For the second sentence, I will show that $V[K][x_0, x_1][H_0]\cap V[K][x_0, x_2][H_0]\cap V[K][x_1, x_2][H_0]=V[K]$ ; the claim is then obtained by repeating this argument three times.

Suppose that b is a set of ordinals in the intersection of the three models. Since $H_0$ is generic over $V[K][x_0, x_1, x_2]$ and $b\in V[K][x_1, x_2]$ holds, a genericity argument with H in the models $V[K][x_0, x_1]$ and $V[K][x_0, x_2]$ shows that b must be in both of these models. It follows that $b\in \bigcap _{a\in [3]^2}V[K][x_i\colon i\in a]$ and therefore in $V[K]$ as desired.