# Krull dimension in set theory1112010 AMS subject classification 03E15,
03E25, 03E35.

Jindřich Zapletal
University of Florida

###### Abstract

For every number $n\geq 2$, let $\Gamma_{n}$ be the hypergraph on
$\mathbb{R}^{n}$ of arity four consisting of all non-degenerate Euclidean
rectangles. It is consistent with ZF+DC set theory that the chromatic number
of $\Gamma_{n}$ is countable while that of $\Gamma_{n+1}$ is not.

##  1 Introduction

In culmination of several years of work, Schmerl [6] classified algebraic
hypergraphs on Euclidean spaces according to how difficult it is to find
countable coloring for them. It turns out that for every such a hypergraph
$\Gamma$ there is a number $n\in\omega+1$ such that in ZFC, the assertion
“chromatic number of $\Gamma$ is countable” is equivalent to
$2^{\aleph_{0}}\leq\aleph_{n}$ (if $n\in\omega$), or it is outright provable
(if $n=\omega$). Moreover, this number $n=n(\Gamma)$ can be found
algorithmically.

Schmerl’s success raises a question: is it possible to find a similar
classification of algebraic hypergraphs with the base theory being the
choiceless ZF+DC set theory? After all, coloring an algebraic hypergraph of
any complexity requires copious amounts of the Axiom of Choice.

In this paper, I make tangible progress towards such a classification. Using
the methods of geometric set theory [4], I isolate a general approach for
separating chromatic numbers of algebraic hypergraphs which differ by
dimension, even when their dimension is high. It appears to be difficult to
state a satisfactory general theorem, so I treat a remarkable special case.
Given a number $n\geq 2$, let $\Gamma_{n}$ be the hypergraph on
$\mathbb{R}^{n}$ of arity four consisting of all non-degenerate Euclidean
rectangles. By a result of [1], in ZFC the chromatic number of every single of
these hypergraphs being countable is a statement equivalent to the Continuum
Hypothesis. In the choiceless theory ZF+DC, a more informative picture
appears.

######  Theorem 1.1.

Let $n\geq 2$ be a number. It is consistent relative to an inaccessible
cardinal that ZF+DC holds and the chromatic number of $\Gamma_{n}$ is
countable while that of $\Gamma_{n+1}$ is not.

Unlike many purely combinatorial independence proofs in ZFC, the proof of
Theorem 1.1 must consider algebraic structure of the real line, in particular
the Hilbert basis theorem and the Krull dimension of the topology of algebraic
subsets of $\mathbb{R}^{n}$. Many similar theorems can be proved on what at
the moment is a case by case basis. The method introduced in this paper is not
universally applicable. For example, the following question appears to require
a different approach.

######  Question 1.2.

Let $n\geq 2$ be a number. Writing $\Delta_{n}$ for the hypergraph of arity
four of parallelograms on $\mathbb{R}^{n}$, is it consistent with ZF+DC that
the chromatic number of $\Delta_{n}$ is countable while that of $\Delta_{n+1}$
is not?

The paper uses the set theoretic notational standard of [2]. The calculus of
balanced forcing is developed in [4]. For purposes of this paper, a rectangle
in a Euclidean space is always identified with the set of its four extreme
points. A non-degenerate rectangle is one which has four extreme points, as
opposed to degenerate rectangles which may only have one or two extreme
points. Given a set $X$, a hypergraph $\Gamma$ on $X$ is just a collection of
finite subsets of $X$, its elements are called hyperedges. A $\Gamma$-coloring
is a (possibly partial) map $c$ on $X$ such that there is no
$\Gamma$-hyperedge which is a subset of $\mathrm{dom}(c)$ and on which
$\Gamma$ is constant. The chromatic number of $\Gamma$ is countable if there
is a total $\Gamma$-coloring whose range is a subset of $\omega$. DC denotes
the Axiom of Dependent Choices.

##  2 Krull dimension

In [8], the notion of Noetherian topology is analyzed in descriptive set
theoretic context. The following definition is central.

######  Definition 2.1.

Let $X$ be a Polish $K_{\sigma}$ space and let $\mathcal{T}$ be a topology on
$X$ different from the original Polish one. Say that $\mathcal{T}$ is an
_analytic Noetherian topology_ if

  1. 1.

$\mathcal{T}$ is Noetherian. That is, there is no infinite strictly decreasing
sequence of $\mathcal{T}$-closed sets;

  2. 2.

$\mathcal{T}$ is analytic. That is, every $\mathcal{T}$-closed set is closed
in the Polish topology, and the collection of $\mathcal{T}$-closed sets is
analytic in the Effros Borel space on $X$.

The restriction to $K_{\sigma}$ spaces serves to establish the Borelness of
the union and intersection operations and the subset relation on closed sets
as in [3, Section 12.C], accommodating the necessary complexity computations
and absoluteness arguments. In this paper, I show that it is useful to
consider the usual notion of dimension of Noetherian spaces, for use in
certain choiceless independence results. Recall:

######  Definition 2.2.

Let $X$ be a set and $\mathcal{T}$ be a Noetherian topology on it. A
$\mathcal{T}$-closed set $C\subset X$ is _irreducible_ if it is not the union
of a finite collection of $\mathcal{T}$-closed proper subsets.

It is well-known and easy to prove that every $\mathcal{T}$-closed set $C$ can
be written in a unique way as a union of a finite collection of nonempty
irreducible sets, so-called irreducible components of $C$. For an analytic
Noetherian topology $\mathcal{T}$ and its closed set $C\subset X$, its
irreducibility is a coanalytic statement, therefore absolute between
transitive models of set theory.

######  Definition 2.3.

Let $X$ be a set and $\mathcal{T}$ be a Noetherian topology on it. The _Krull
dimension_ of $\mathcal{T}$ is the supremum of the lengths of finite chains of
nonempty irreducible $\mathcal{T}$-closed sets linearly ordered by strict
inclusion, minus one.

This standard definition may only yield values of dimension between $0$ and
$\omega$ inclusive. It is not difficult to see that if $\mathcal{T}$ is an
analytic Noetherian topology on a $K_{\sigma}$ Polish space $X$, then the set
of reducible $\mathcal{T}$-closed sets is analytic. The statement “the Krull
dimension of $\mathcal{T}$ is $n$” is a conjunction of a
$\mathbf{\Pi}^{1}_{2}$ and a $\mathbf{\Sigma}^{1}_{2}$-statements and as such
absolute among all forcing extensions. While there are many interesting
examples of Noetherian spaces with prescribed dimensions, this paper is
limited to the following:

######  Example 2.4.

Let $n\geq 1$ and let $\mathcal{T}$ be the topology of algebraic sets on
$\mathbb{R}^{n}$. Then $\mathcal{T}$ is analytic Noetherian and has Krull
dimension $n$. This is the main corollary of the Hilbert Basis Theorem.

The following minor variation comes very handy in this paper.

######  Example 2.5.

Let $n\geq 1$ and let $\mathcal{T}$ be the topology on
$\mathbb{R}^{n}\times\mathbb{R}$ whose closed sets are (the whole space and)
finite unions of sets of the form $f\restriction A$, where
$A\subset\mathbb{R}^{n}$ is an algebraic set distinct from $\mathbb{R}^{n}$,
and $f$ is a polynomial function from $\mathbb{R}^{n}$ to $\mathbb{R}$. Then
$\mathcal{T}$ is Noetherian of Krull dimension $n$. To see this, note that the
set of generators is closed under intersections: $(f_{0}\restriction
A_{0})\cap(f_{1}\restriction A_{1})=f_{0}\restriction(A\cap
B\cap\\{x\in\mathbb{R}^{n}\colon f_{0}(x)=f_{1}(x)\\})$ and the last set in
the intersection is algebraic. Every decreasing sequence of
$\mathcal{T}$-indecomposable sets starts possibly with
$\mathbb{R}^{n}\times\mathbb{R}$ and continuous with $f\restriction A_{i}$ for
some fixed polynomial function $f\colon\mathbb{R}^{n}\to\mathbb{R}$ and
indecomposable algebraic sets $A_{i}\subset\mathbb{R}^{n}$ distinct from the
whole space. As such, it can have length at most $n+1$, and the Krull
dimension of $\mathcal{T}$ is $n$. Note that if polynomial functions on the
whole space $\mathbb{R}^{n}$ were allowed as generators of $\mathcal{T}$, the
Krull dimension of the topology would increase to $n+1$.

The main feature connecting Krull dimension with forcing is the following
mysterious theorem.

######  Theorem 2.6.

Let $n\geq 2$ be a number and $\mathcal{T}$ be an analytic Noetherian topology
on a $K_{\sigma}$-Polish space $X$ of Krull dimension less than $n$. Let
$\langle V[G_{i}]\colon i\in n\rangle$ be a tuple of forcing extensions such
that for any sets $b_{0},b_{1}\subset n$, $V[G_{i}\colon i\in b_{0}]\cap
V[G_{i}\colon i\in b_{1}]=V[G_{i}\colon i\in b_{0}\cap b_{1}]$. For every
point $x\in X$ there is $i\in n$ such that the smallest $\mathcal{T}$-closed
set coded in $V[G_{i}]$ containing $x$ is coded in $V$.

The intersection condition on the tuples of generic extensions is satisfied
for example for a mutually generic tuple by the product forcing theorem. The
point $x\in X$ does not have to belong to any of the models mentioned. Note
that if $M$ is any generic extension and $x\in X$ is any point outside of $M$,
the Noetherian property of the topology $\mathcal{T}$ shows that there indeed
exists an inclusion-smallest $\mathcal{T}$-closed set coded in $M$ which
contains $x$ as an element. In addition, this set must be irreducible since
all of its composants are coded in $M$, and one of them must contain the point
$x$.

###### Proof.

Suppose that the models $V[G_{i}]$ for $i\in n$ and the point $x$ are given.
For each set $b\subseteq n$ write $M_{b}=V[G_{i}\colon i\in b]$ and let
$C_{b}\subset X$ be the smallest $\mathcal{T}$-closed set coded in $M_{b}$
such that $x\in C_{b}$. Note that $c\subseteq b$ implies $C_{b}\subseteq
C_{c}$; nevertheless, the status of sharp inclusion is not immediately clear.
Suppose towards a contradiction that the conclusion of the proposition fails,
i.e. $C_{0}\neq C_{\\{i\\}}$ for any $i\in n$.

For a set $b\subseteq n$, a _good sequence in $b$_ is an inclusion-increasing
sequence $\langle b_{k}\colon k\leq|b|\rangle$ of subsets of $b$ such that
$|b_{k}|=k$ and such that the sets $C_{b_{k}}$ strictly decrease with $k$. By
induction on the cardinality of the set $b\subseteq n$ I will show that there
is a good sequence in $b$. The base case $|b|=1$ follows from the initial
contradictory assumption. For the induction step, suppose that $b\subset n$ is
a set of cardinality at least two such that the statement has been verified
for all its proper subsets. Choose $i\in b$ and let $c=b\setminus\\{i\\}$. By
the induction hypothesis applied to the set $c$, there must be $j\in c$ such
that, writing $d=c\setminus\\{j\\}$, $C_{d}\neq C_{c}$ holds. Let
$e=d\cup\\{i\\}$, so $d=e\cap c$.

It cannot be the case that $C_{e}=C_{c}$. To see that, note that in such a
case, $C_{c}$ would be coded in both $M_{e}$ and $M_{c}$, and by the
intersection condition on the generic extensions, it would be coded in the
model $M_{d}$, contradicting the assumption that $C_{d}\neq C_{c}$. Thus, the
set $C_{e}\cap C_{c}$ is a proper subset either of $C_{c}$ or of $C_{e}$. If
the former case prevails, use the induction hypothesis on $c$ to find a good
sequence in $c$ and add to it $C_{b}$, which is a subset of $C_{c}\cap C_{e}$
and therefore a proper subset of $C_{c}$. Thus, a good sequence in $b$ has
been obtained. If the latter case prevails, just switch the role of $c$ and
$e$ and construct a good sequence in $b$ just the same.

After the induction is complete, consider a good sequence $\langle b_{k}\colon
k\leq n\rangle$ in the set $b=n$. The sets $C_{b_{k}}$ for $k\leq n$ form a
strictly decreasing sequence of length $n+1$ consisting of irreducible sets,
contradicting the assumption on the Krull dimension of the topology
$\mathcal{T}$. ∎

##  3 Pairs of generic extensions

The paper [8] introduces the notion of mutually Noetherian pair of generic
extensions. In this section, I develop a stratification of that concept which
takes into account the Krull dimension of the topologies concerned.

######  Definition 3.1.

Let $V[G_{0}]$, $V[G_{1}]$ be generic extensions. The _Krull dimension of
$V[G_{0}]$ over $V[G_{1}]$_, $\dim(V[G_{0}]/V[G_{1}])$, is

  1. 1.

at least $1$ always;

  2. 2.

at least $n$ (where $n\geq 2$ is a natural number) if for every $K_{\sigma}$
Polish space $X$, every basic open set $O\subset X$, and every analytic
Noetherian topology $\mathcal{T}$ on $X$ of Krull dimension less than $n$,
both in the ground model, if $C$ is a $\mathcal{T}$-closed set coded in
$V[G_{0}]$ with nonempty intersection with $O\cap V[G_{1}]$, then $C$ has
nonempty intersection with $O\cap V$;

  3. 3.

equal to $n$ if $\dim(V[G_{0}]/V[G_{1}])\geq n$ and
$\dim(V[G_{0}]/V[G_{1}])\not\geq n+1$;

  4. 4.

$\infty$ if $\dim(V[G_{0}]/V[G_{1}])\geq n$ holds for every natural number
$n$.

Noetherian topologies of dimension $0$ are uninteresting, and that is why the
counting starts at $n=1$. The case $n=2$ allows a simple characterization via
the following proposition. Higher dimensions are more difficult to understand,
and they are the bread and butter of this paper. I do not know if
$\dim(V[G_{0}]/V[G_{1}])=\dim(V[G_{1}]/V[G_{0}])$ holds in general, but this
interesting question is irrelevant for the purposes of this paper.

######  Proposition 3.2.

Let $V[G_{0}]$, $V[G_{1}]$ be generic extensions. $\dim(V[G_{0}]/V[G_{1}])\geq
2$ holds if and only if $\mathcal{P}(\omega)\cap V[G_{0}]\cap
V[G_{1}]=\mathcal{P}(\omega)\cap V$.

###### Proof.

For the left-to-right inclusion, suppose that $\dim(V[G_{0}]/V[G_{1}])\geq 2$
holds. Consider the topology $\mathcal{T}$ on $\mathcal{P}(\omega)$ whose
closed sets are the whole space, the empty set, and the finite sets. It is an
analytic Noetherian topology of Krull dimension $1$. Suppose that
$x\in\mathcal{P}(\omega)$ belongs to $V[G_{0}]\cap V[G_{1}]$. Applying the
dimension assumption to the set $C=\\{x\\}$, the conclusion is that $x\in V$
as desired.

For the right-to-left inclusion, assume that $\mathcal{P}(\omega)\cap
V[G_{0}]\cap V[G_{1}]=\mathcal{P}(\omega)\cap V$ holds. Let $X$ be a
$K_{\sigma}$ Polish space with a fixed countable basis, let $O\subset X$ be an
open set, and let $\mathcal{T}$ be an analytic Noetherian topology of Krull
dimension $1$, all coded in $V$. Let $C\in\mathcal{T}$ be a closed set coded
in $V[G_{0}]$ and $x\in V[G_{1}]$ be a point in $C\cap O$. Consider the sets
$D,D_{0},D_{1}\in\mathcal{T}$ which are the smallest $\mathcal{T}$-closed sets
containing $x$ and coded in $V,V[G_{0}],V[G_{1}]$ respectively. These sets do
exist and are indecomposable as the topology $\mathcal{T}$ is Noetherian.
Clearly, $D_{0}\subseteq D$ holds. In addition, a Shoenfield absoluteness
argument shows that $D_{1}$ is the smallest $\mathcal{T}$-closed set
containing $x$ in any model, so $D_{1}\subseteq D_{0}$ holds.

The Krull dimension assumption on the topology $\mathcal{T}$ now implies that
either $D_{0}=D$ or $D_{1}=D_{0}$ must hold. If $D_{0}=D$, then $D_{0}$ is
coded in the ground model. By a Shoenfield absoluteness argument $D_{0}\cap O$
must be nonempty in $V$ since it is nonempty in $V[G_{1}]$. Any ground model
point of $D_{0}\cap O$ verifies the $\dim(V[G_{0}]/V[G_{1}])\geq 2$ assertion.
If, on the other hand, $D_{0}=D_{1}$ holds, then the set of all basic open
sets disjoint from $D_{0}$ belongs to both $V[G_{0}]$ and $V[G_{1}]$, and
therefore to $V$ by the initial assumption. Thus, $D_{0}$ is coded in $V$, and
the proof is concluded as in the previous case. ∎

The next order of business is to show that certain common operations on
generic extensions preserve Krull dimension.

######  Proposition 3.3.

Let $V[G_{0}]$ and $V[G_{1}]$ be generic extensions and $n\geq 1$ be a number.
Suppose that $\dim(V[G_{0}]/V[G_{1}])\geq n$ holds.

  1. 1.

If $V[K_{0}]\subset V[G_{0}]$ and $V[K_{1}]\subset V[G_{1}]$ then
$\dim(V[K_{0}]/V[K_{1}])\geq n$;

  2. 2.

if $P_{0}\in V[G_{0}]$ and $P_{1}\in V[G_{1}]$ are posets and $H_{0}\subset
P_{0}$ and $H_{1}\subset P_{1}$ are filters mutually generic over
$V[G_{1},G_{0}]$, then $\dim(V[G_{0}][H_{0}]/V[G_{1}][H_{1}])\geq n$.

In particular, a mutually generic pair of extensions of $V$ has Krull
dimension $\infty$.

###### Proof.

The first item is obvious. For the second item, let $X$ be a
$K_{\sigma}$-Polish space and $\mathcal{T}$ an analytic Noetherian topology on
it of Krull dimension smaller than $n$. Suppose that $\tau_{1}\in V[G_{1}]$ is
a $P_{1}$-name for a $\mathcal{T}$-closed subset of $X$ and $\tau_{0}\in
V[G_{0}]$ is a $P_{0}$-name for an element of $X$, and $p_{1}\in P_{1}$ and
$p_{0}\in P_{0}$ are conditions such that the pair $\langle
p_{1},p_{0}\rangle$ forces in the product $P_{1}\times P_{0}$ that
$\tau_{0}\in\tau_{1}$ holds. I will find a point $x\in X\cap V$ such that
$p_{1}\Vdash\check{x}\in\tau_{1}$, proving the proposition.

Working in $V[G_{1}]$, let $M_{1}$ be a countable elementary submodel of a
large structure, and let $\mathcal{F}$ be the set of all filters on $P_{1}\cap
M_{1}$ which are generic over the model $M_{1}$ and contain the condition
$p_{1}$. Let $C=\bigcap\\{\tau_{1}/F\colon F\in\mathcal{F}\\}$; this is a
$\mathcal{T}$-closed set in the model $V[G_{1}]$. Working in $V[G_{0}]$, let
$M_{0}$ be a countable elementary submodel of a large structure, let
$F_{0}\subset P_{0}$ be a filter generic over $M_{0}$ containing the condition
$p_{0}$, and let $x_{0}=\tau_{0}/F_{0}$; this is a point in the model
$V[G_{0}]$. Observe that $x_{0}\in C$ must hold. Otherwise, there would be a
filter $F_{1}\in\mathcal{F}$ such that
$x_{0}=\tau_{0}/F_{0}\notin\tau_{1}/F_{1}$ and a basic open set $O\subset X$
in its usual Polish topology which contains $\tau_{0}/F_{0}$ and is disjoint
from $\tau_{1}/F_{1}$. By the forcing theorem applied with $P_{1}$ and
$P_{0}$, there would have to be conditions $p^{\prime}_{1}\leq p_{1}$ and
$p^{\prime}_{0}\leq p_{0}$ forcing respectively that $O\cap\tau_{1}=0$ and
$\tau_{0}\in O$. Such conditions contradict the initial assumptions on $p_{1}$
and $p_{0}$.

Now, since $\dim(V[G_{0}]/V[G_{1}])\geq n$, there must be a point $x\in X\cap
V$ such that $x\in C$ holds. I claim that $p_{1}\Vdash\check{x}\in\tau_{1}$ as
desired. If this failed, there would have to be a basic open set $O\subset X$
containing the point $x$ and a condition $p^{\prime}_{1}\leq p_{1}$ forcing
$O\cap\tau_{1}=0$. By the elementarity of the model $M_{1}$, such a condition
$p^{\prime}_{1}$ can be found in the model $M_{1}$. Let $F\in\mathcal{F}$ be a
filter containing the condition $p^{\prime}_{1}$. Observe that
$x\notin\tau_{1}/F$, contradicting the choice of the point $x$. ∎

The Krull dimension of generic extensions will be used in this paper only to
topologies associated with algebraic subsets of Euclidean spaces. The
following proposition provides their instrumental property reminiscent of
mutual genericity.

######  Proposition 3.4.

Suppose that $n\geq 1$ is a number and $V[G_{0}],V[G_{1}]$ are generic
extensions such that $\dim(V[G_{0}]/V[G_{1}])\geq n+1$. Suppose that
$f\colon\mathbb{R}^{n}\to\mathbb{R}$ is a polynomial function and
$C\subset\mathbb{R}^{n}$ is an algebraic set distinct from $\mathbb{R}^{n}$,
both coded in $V[G_{0}]$. Suppose $x\in C$ is a point in $V[G_{1}]$ such that
$f(x)\in V[G_{0}]$. Then there is an algebraic set $D\subseteq C$ and a
polynomial function $g\colon\mathbb{R}^{n}\to\mathbb{R}$, both coded in the
ground model $V$, such that $x\in D$ and $f\restriction D=g\restriction D$.

###### Proof.

Consider the analytic Noetherian topology $\mathcal{T}$ on
$\mathbb{R}^{n}\times\mathbb{R}$ generated by all restrictions of a polynomial
functions from $\mathbb{R}^{n}$ to $\mathbb{R}$ to an algebraic subset of
$\mathbb{R}^{n}$ distinct from $\mathbb{R}^{n}$. As in Example 2.5, the Krull
dimension of $\mathcal{T}$ is clearly equal to $n$. Note that $f\restriction
C$ is a $\mathcal{T}$-closed set in $V[G_{0}]$ containing the pair $\langle
x,f(x)\rangle$. By an application of the dimension assumption, there is a
sequence of pairs $\langle x_{i},y_{i}\colon i\in\omega\rangle$ converging to
$\langle x,f(x)\rangle$ such that for each $i\in\omega$, the points $x_{i}\in
C$ and $y_{i}$ are both in $V$ and $f(x_{i})=y_{i}$.

Let $e$ be a polynomial function on $\mathbb{R}^{n}$ such that
$C=\\{z\in\mathbb{R}^{n}\colon e(z)=0\\}$. Now, the polynomial functions $f,e$
are linear functions of the coefficients used in the polynomials defining
$f,e$. That is, for some $k\in\omega$ there is a polynomial function
$h\colon\mathbb{R}^{m}\times\mathbb{R}^{n}\to\mathbb{R}\times\mathbb{R}$ with
integer coefficients which is linear in the variables from $\mathbb{R}^{m}$
and such that for some $t\in\mathbb{R}^{m}\cap V[G_{0}]$, for all
$z\in\mathbb{R}^{n}$ $\langle f(z),e(z)\rangle=h(t,z)$. For each $i\in\omega$
let $F_{i}=\\{u\in\mathbb{R}^{m}\colon h(u,x_{i})=\langle y_{i},0\rangle\\}$;
this is an algebraic and even affine subset of $\mathbb{R}^{m}$. Since the
topology of algebraic subsets of $\mathbb{R}^{m}$ is Noetherian, the
intersection $F=\bigcap_{i}F_{i}$ belongs to the ground model, since for some
finite number $j\in\omega$, $\bigcap_{i\in\omega}F_{i}=\bigcap_{i\in j}F_{i}$
holds. Note that the set $F$ is nonempty, because in $V[G_{0}]$ it contains
$t$. Let $D\subset\mathbb{R}^{n}$ be the algebraic set of all points
$z\in\mathbb{R}^{n}$ such that the value of $h(u,z)$ is the same for all $u\in
F$ and has zero as its second coordinate. Note that $D$ is in the ground
model, $x_{i}\in D$ for all $i\in D$ and also $x\in D$, since all polynomial
functions are continuous. Choose a point $u\in F$ in the ground model, and let
$g$ be the polynomial function defined by $g(z)=$ the first coordinate of
$h(u,z)$. Clearly, the set $D$ and the function $g$ work as required. ∎

##  4 Examples

In order to use the concept of Krull dimension efficiently, it is necessary to
build examples of pairs of generic extensions with interesting Krull dimension
characteristics. In this section, I show how to produce such examples from
algebraic subsets of Euclidean spaces. The following abstract observation will
be useful at various points in the proofs.

######  Theorem 4.1.

Let $X_{0},X_{1}$ be Euclidean spaces and $A\subset X_{0}\times X_{1}$ be an
algebraic set, and write $\pi\colon X_{0}\times X_{1}\to X_{0}$ for the
projection function. For every nonempty relatively open set $O_{0}\subset A$
there is a nonempty relatively open set $O_{1}\subset O_{0}$ and an algebraic
set $B\subset X_{0}$ such that $\pi\restriction O_{1}$ is an open map to $B$.

###### Proof.

Fix the relatively open set $O_{0}\subset A$. To find the algebraic set $B$,
consider the collection $\mathcal{C}$ of all algebraic sets $C\subset X_{0}$
such that there is a nonempty basic relatively open set $O^{\prime}\subset
O_{0}$ such that $\pi^{\prime\prime}O^{\prime}\subset C$. Note that
$\mathcal{C}$ is nonempty, containing the whole space $X_{0}$ in particular.
By the Hilbert basis theorem, there is an inclusion-minimal element
$B\in\mathcal{C}$, with its membership witnessed by some basic relatively open
set $O^{\prime}\subset O_{0}$. Below, we need the following key feature of the
set $B$.

######  Claim 4.2.

For every algebraic set $D\subset X_{1}$, either $B\subseteq D$ holds or
$\pi^{-1}D$ is nowhere dense in $O^{\prime}$.

###### Proof.

Observe that if $\pi^{-1}D$ is somewhere dense in $O^{\prime}$, then since
$\pi^{-1}D$ is a closed set, it has to contain a nonempty relatively open
subset $O^{\prime\prime}\subseteq O^{\prime}$. By the minimal choice of $B$,
it must be the case that $D\cap B=B$ or in other words $B\subseteq D$. ∎

To find the set $O_{1}$, let $C$ be the set of those $x\in O^{\prime}$ such
that $f\restriction O^{\prime}$ is an open map to $B$ at $x$. In other words,
$C=\\{x\in O^{\prime}\colon$ for every $U\subset O^{\prime}$ basic open with
$x\in U$ there is $V\subset X_{0}$ basic open containing $\pi(x)$ such that
$V\cap B\subset\pi^{\prime\prime}U\\}$. Note that $C\subset X_{0}\times X_{1}$
is a semialgebraic set. By the quantifier elimination theorem for real closed
fields [5, Theorem 3.3.15], $C$ is a finite Boolean combination of open and
closed sets, and therefore there is a nonempty relatively open set
$O_{1}\subset O^{\prime}$ such that either $O_{1}\subset C$ or $O_{1}\cap
C=0$. The former option gives us the conclusion of the theorem. It is
therefore enough to derive a contradiction from the latter option.

Assume towards a contradiction that $O_{1}\cap C=0$. Let $\mathcal{B}$ be a
countable basis of semialgebraic relatively open subsets of $O_{1}$. For each
$U\in\mathcal{B}$, consider the set $\pi^{\prime\prime}U\subset B$. This is a
semialgebraic set; by the quantifier elimination theorem, it is a Boolean
combination of some sets of the form $\\{y\in X_{0}\colon p(y)=0\\}$ and
$\\{y\in X_{0}\colon p(y)>0\\}$ for some polynomials $p$ with real
coefficients. Let $p_{Ui}$ for $i\in i_{U}$ be a list of these polynomials.
For each $i\in i_{U}$, the algebraic set $D_{iU}=\\{y\in X_{0}\colon
p_{iU}(y)=0\\}$ is either a superset of $B$, or else its $\pi$-preimage is
nowhere dense in $O^{\prime}$ by the claim. Use the Baire category theorem to
find a point $x\in O_{1}$ which belongs to no set $\pi^{-1}D_{iU}$ where
$U\in\mathcal{B}$, $i\in i_{U}$, and $B\not\subseteq D_{iU}$.

Since the point $x$ does not belong to the set $C$, the map $\pi\restriction
O^{\prime}$ is not open to $B$ at $x$. Thus, there must be a basic relatively
open set $U\in\mathcal{B}$ such that $x\in U$ and $\pi^{\prime\prime}U$
contains no open neighborhood of $f(x)$. By the choice of the point $x$, the
point $f(x)$ belongs to none of the closed sets $D_{iU}\subset X_{0}$ except
to those for which $B\subseteq D_{iU}$. Let $V\subset X_{0}$ be an open
neighborhood of $f(x)$ such that for every $i\in i_{U}$, if $B\not\subseteq
D_{iU}$ then $V\cap D_{iU}=0$, and if $p_{iU}(f(x))\neq 0$ then $p_{iU}$ does
not change sign in $V$. The choice of the neighborhood $V$ shows that within
the set $V\cap B$, the membership in the set $\pi^{\prime\prime}U$ does not
change. Since $f(x)\in V\cap B$, it must be that $V\cap
B\subseteq\pi^{\prime\prime}U$ holds. This contradicts the choice of the open
set $U\subset A$. ∎

All generic extensions discussed in this section are Cohen generic with the
following parlance. If $X$ is a Polish space then $P_{X}$ denotes the _Cohen
forcing_ associated with $X$, the set of nonempty open subsets of $X$ ordered
by inclusion, adding a single generic point ${\dot{x}}_{\mathit{gen}}$. If $Y$
is another Polish space and $f\colon X\to Y$ is a continuous open map, then
$P_{X}\Vdash\dot{f}({\dot{x}}_{\mathit{gen}})$ is a Cohen generic point over
$V$ for the poset $P_{Y}$ by [4, Proposition 3.1.1]. An important point is
that if $n\geq 1$ is a number, then the posets $P_{X^{n}}$ and $(P_{X})^{n}$
are co-dense, therefore a $P_{X^{n}}$-generic $n$-tuple consists of mutually
$P_{X}$-generic points.

######  Definition 4.3.

Suppose that $X,Y$ are Polish spaces, $f\colon X\to Y$ is a continuous
function, and $n\geq 1$ is a number. $f^{n}\colon X^{n}\to Y^{n}$ denotes the
continuous function defined by $f^{n}(\vec{x})(i)=f(\vec{x}(i))$. $X^{n}/f$ is
the closed subset of $X^{n}$ of all points $\vec{x}\in X^{n}$ such that the
values $f(\vec{x}(i))$ are the same for all $i\in n$. In particular,
$X^{1}/f=X$.

######  Proposition 4.4.

Let $X,Y$ be Polish spaces, $f\colon X\to Y$ be a continuous open function,
and $a\subset b$ be nonempty finite sets. The projection function $\pi\colon
X^{b}/f\to X^{a}/f$ is open.

###### Proof.

Suppose that $O_{i}\subset X$ are open sets for $i\in b$, and for each $j\in
a$ let $O^{\prime}_{j}=O_{j}\cap\bigcap_{i\in b}f^{-1}f^{\prime\prime}O_{i}$.
The sets $O^{\prime}_{j}\subset O_{j}$ are open since the function $f$ is
open. It is not difficult to see that $\pi^{\prime\prime}(\prod_{i\in
b}O_{i}\cap X^{b}/f)=\prod_{j\in a}O^{\prime}_{j}\cap X^{a}/f$. ∎

Now I am ready to state and prove a theorem which connects dimension of
algebraic sets with Krull dimension of pairs of generic extensions. For
algebraic and semi-algebraic subsets of Euclidean spaces I use the tame
topological notion of dimension as developed in [7].

######  Theorem 4.5.

Let $X$ be an algebraic subset of a Euclidean space, and $f_{0}\colon X\to
Y_{0}$ and $f_{1}\colon X\to Y_{1}$ be projections of $X$ to certain sets of
coordinates. Let $n\geq 1$ be a number. If for every nonempty relatively basic
open set $U\subset X^{n}/f_{1}$, the set $(f_{0}^{n})^{\prime\prime}U\subset
Y_{0}^{n}$ has the maximal possible dimension $n\cdot\dim(Y_{0})$, then
$P_{X}\Vdash\dim(V[f_{0}({\dot{x}}_{\mathit{gen}})]/V[f_{1}({\dot{x}}_{\mathit{gen}})])\geq
n$.

###### Proof.

Let $O$ be a nonempty, relatively basic open subset of $X$, forcing the
conclusion to fail. Thinning out $O$ if necessary, I may assume that $f_{0}$
and $f_{1}$ on $O$ are open maps to some algebraic subsets $A_{0}\subseteq
Y_{0}$, $A_{1}\subseteq Y_{1}$ respectively (this is by Theorem 4.1), and
there is a Polish $K_{\sigma}$-space $Z$, a nonempty open set $W\subset Z$, an
analytic Noetherian topology $\mathcal{T}$ of Krull dimension less than $n$,
and $P_{A_{0}}$-name $\tau_{0}$ for a $\mathcal{T}$-closed subset of $Z$
containing no ground model elements of $W$, and a $P_{A_{1}}$-name $\tau_{1}$
for an element of $U$ such that
$O\Vdash\tau_{1}/f_{1}({\dot{x}}_{\mathit{gen}})\in\tau_{0}/f_{0}({\dot{x}}_{\mathit{gen}})$.

Consider the nonempty relatively open set $U_{0}=O^{n}\cap X^{n}/f_{1}$.
Observe that $X^{n}/f_{1}$ is an algebraic set. By Theorem 4.1, there is a
relatively open nonempty subset $U_{1}\subset U_{0}$ on which $f_{0}^{n}$ is
an open map to an algebraic set $A_{2}\subset Y_{0}^{n}$. Now, the algebraic
set $A_{2}$ has full dimension $n\cdot\dim(Y_{0})$ by the assumptions, so
$A_{2}=Y_{0}^{n}$ must hold. Let $\vec{x}\in U_{1}$ be a point generic over
$V$ for the poset $P_{X^{n}/f_{1}}$. By Proposition 4.4, each of the points
$\vec{x}(i)\in X$ for $i\in n$ is $P_{X}$-generic over $V$, their
$f_{1}$-images coincide and are equal to a point $y_{1}\in Y_{1}$ which is
$P_{A_{1}}$-generic over $V$, and their $f_{0}$-images
$y_{0i}=f_{0}(\vec{x}(i))$ are mutually generic elements of $Y_{0}$. Consider
the models $V[y_{0i}]$ for $i\in n$, and in each of them the
$\mathcal{T}$-closed set $C_{i}=\tau_{0}/y_{0i}$. Consider also the point
$z=\tau_{1}/y_{1}$. By Theorem 2.6, there is a number $i\in n$ and a ground
model coded $\mathcal{T}$-closed set $D\subset Z$ such that $z\in D\subseteq
C_{i}$. Now, the set $D$ contains a point in the open set $W$, namely $z$; by
a Mostowski absoluteness argument, it must contain also some other point
$z^{\prime}\in W$ in the ground model. Then $z^{\prime}\in C_{i}$ holds,
contradicting the initial assumption about the name $\tau_{0}$. ∎

The dimension demand in the assumptions of Theorem 4.5 may seem hard to
verify; the following notion will prove helpful for the ends of this paper.

######  Definition 4.6.

Let $A\subset X_{0}\times X_{1}$ be an algebraic subset of the product of two
Euclidean spaces. Say that $A$ is _narrow_ if

  1. 1.

for every nonempty relatively open set $O\subset A$ with nonempty intersection
with $A$, $\dim(O)\geq\dim(X_{0})$;

  2. 2.

there is a relatively open dense subset $C\subset A$ with all vertical
sections finite.

######  Proposition 4.7.

Suppose $A\subset X_{0}\times X_{1}$ is a narrow algebraic subset of the
product $X_{0}\times X_{1}$ of two Euclidean spaces. Let $\pi\colon A\to
X_{0}$ be the projection function. For every nonempty relatively open set
$O\subset A$, $\dim(\pi^{\prime\prime}O)=\dim(X_{0})$.

###### Proof.

Fix the set $O\subset A$. By (2) of Definition 4.6, shrinking $O$ if necessary
I may assume that $O$ has all vertical sections finite. The only way to
account for (1) of Definition 4.6 is to conclude that
$\dim(\pi^{\prime\prime}O)=\dim(X_{0})$. ∎

For all of the following examples, fix a number $n\geq 2$.

######  Example 4.8.

Let $X_{0}=(\mathbb{R}^{n})^{n}\times(\mathbb{R}^{n})^{n}$, let
$X_{1}=\mathbb{R}^{n}$, and $A=\\{\langle x_{i}\colon i\in n,y_{i}\colon i\in
n,z\rangle\in X_{0}\times X_{1}\colon$ for all $i\in n$,
$(x_{i}-z)\cdot(y_{i}-z)=0\\}$. The set $A\subset X_{0}\times X_{1}$ is
narrow.

###### Proof.

The following claim will be useful. Let $d$ denote the Euclidean metric on
$\mathbb{R}^{n}$.

######  Claim 4.9.

If $x_{i}$ for $i\in n$ are linearly independent points in $\mathbb{R}^{n}$
and $r_{i}\in\mathbb{R}$ are real numbers, then there are at most two points
$z\in\mathbb{R}^{n}$ such that $d(x_{i},z)=r_{i}$ holds for all $i\in n$.

###### Proof.

Write the equations for such points $z$: the $i$-th equation is $z\cdot
z-2x_{i}\cdot z+x_{i}\cdot x_{i}=r_{i}^{2}$. Subtract the first equation from
the others to remove the quadratic term and get a system of $n-1$ many
equations $2(x_{0}-x_{i})\cdot z=r_{i}^{2}-r_{0}^{2}+x_{0}\cdot
x_{0}-x_{i}\cdot x_{i}$ for $0<i<n$. Since the vectors of coefficients on the
left hand side are linearly independent, the set of solutions is a line. This
line intersects the hypersphere around $x_{0}$ of radius $r_{0}$ in at most
two points, and these are the only points $z\in\mathbb{R}^{n}$ such that
$d(x_{i},z)=r_{i}$ holds for all $i\in n$. ∎

To check the items of Definition 4.6, suppose that $O\subset X_{0}\times
X_{1}$ is an open set with nonempty intersection with $A$. Counting the
dimensions of $A\cap O$ yields the following. The $z$ and $x$ coordinates can
be chosen arbitrarily within certain open subsets of $\mathbb{R}^{n}$
($n+n^{2}$ many dimensions) after which the $y$-coordinates must be chosen on
the hyperplane which contains $z$ and is perpendicular to $z-y_{i}$ ($n(n-1)$
many dimensions). Together, this yields $n+n^{2}+n(n-1)=2n^{2}$ many
dimensions as desired.

Consider the set $P=\\{\langle x_{i},y_{i},z\colon i\in n\rangle\colon$ the
center points of the segments connecting $x_{i}$ and $y_{i}$ form a linearly
independent set$\\}\subset X_{0}\times X_{1}$. The set $P\subset X_{0}\times
X_{1}$ is open as the linear independence is checked by the nonzero
determinant test. The vertical sections of $P\cap A$ are finite by Claim 4.9
and Thalet’s theorem. To check that $P\cap A$ is dense in $A$, let $O\subset
X_{0}\times X_{1}$ be an open set with nonempty intersection with $A$. To find
a point in $P\cap O\cap A$, choose $z$ arbitrarily in a certain open set, and
choose the segment centers $w_{i}$ for $i\in n$ in certain open sets in a
linearly independent way, using the fact that every nonempty open set spans
the whole space $\mathbb{R}^{n}$. Then complete the construction by finding
the points $x_{i},y_{i}$ for $i\in n$. ∎

######  Example 4.10.

Let $X_{0}=(\mathbb{R}^{n})^{n}\times(\mathbb{R}^{n})^{n}$, let
$X_{1}=\mathbb{R}^{n}$, and $A=\\{\langle x_{i}\colon i\in n,y_{i}\colon i\in
n,z\rangle\in X_{0}\times X_{1}\colon$ for all $i\in n$,
$(x_{i}-y_{i})\cdot(x_{i}-z)=0\\}$. The set $A\subset X_{0}\times X_{1}$ is
narrow.

###### Proof.

There is again a preliminary claim.

######  Claim 4.11.

If $\langle x_{i}\colon i\in n,y_{i}\colon i\in n\rangle\in X_{0}$ is a point
such that the differences $x_{i}-y_{i}$ for $i\in n$ are linearly independent,
then the vertical section of $A$ above the point has exactly one element.

###### Proof.

The linear equations for the points in the vertical section have a matrix
consisting of the vectors $x_{i}-y_{i}$ for $i\in n$. ∎

To verify the items of Definition 4.6, suppose that $O\subset X_{0}\times
X_{1}$ is an open set with a nonempty intersection with $A$. To evaluate the
dimension of $O\cap A$, first choose a point $z\in X_{1}$ in a certain open
set ($n$ dimensions), then the points $y_{i}$ in certain open sets ($n^{2}$
many dimensions), and then points $x_{i}$ in the hyperplanes perpendicular to
the segments $x_{i}-y_{i}$ and passing through $y_{i}$ ($n(n-1)$ dimensions).
In total, it gives dimension $n+n^{2}+n(n-1)=n^{2}$ as desired. Now, consider
the set $P\subset X_{0}\times X_{1}$ consisting of all points $\langle
x_{i},y_{i}\colon i\in n,z\rangle$ such that $x_{i}-y_{i}$ are linearly
independent for $i\in n$. This is an open set by the nonzero determinant test
for linear independence. The claim shows that $P\cap A$ has vertical sections
of cardinality exactly one. To see that $P\cap A$ is dense in $A$, let
$O\subset X_{0}\times X_{1}$ be an open set with nonempty intersection with
$A$. To find a point in $O\cap P\cap A$, choose $z$ and $x_{i}$ for $i\in n$
in certain open neighborhoods, and then choose the points $y_{i}$ on the
hyperspheres with diameters $[x_{i},z]$, making sure that $x_{i}-y_{i}$ for
$i\in n$ are linearly independent points. This is possible since every
nonempty open subset of a hypersphere spans the whole space $\mathbb{R}^{n}$.
∎

######  Example 4.12.

Let $n\geq 2$ be a number and $X\subset(\mathbb{R}^{n})^{4}$ be the closed set
of all rectangles in $\mathbb{R}^{n}$. Let $x\in X$ be a point $P_{X}$-generic
over $V$.

  1. 1.

For every $i\in 4$, the point $x(i)\in\mathbb{R}^{n}$ is generic over $V$;

  2. 2.

for every $i\neq j$ in $4$, $V[x(i)]$ and $V[x(j)]$ are mutually generic;

  3. 3.

for any three pairwise distinct indices $i,j,k\in 4$,
$\dim(V[x_{i},x_{j}]/V[x_{k}])\geq n$.

###### Proof.

The first two items follow from the fact that the projection from $X$ to any
one or two coordinates is an open function to $\mathbb{R}^{n}$ or
$(\mathbb{R}^{n})^{2}$. This is easy and left to the reader. For the third
item, first observe that the fourth point of a rectangle is a continuous
function of the other three, and the projection from a graph of a continuous
function is an open map. Thus, the projection from $X$ to any three
coordinates is an open function to $Y$, the set of all right-angle triangles
in $(\mathbb{R}^{n})^{3}$. The dimension estimates for $P_{Y}$-generic triples
now follow from Theorem 4.5, Proposition 4.7, and Examples 4.8 and 4.10. ∎

##  5 A preservation theorem

The notion of Krull dimension of pairs of generic extension has a variation of
balance associated to it, leading to a useful preservation theorem regarding
certain balanced extensions of the choiceless Solovay model. This is explained
in the present section.

######  Definition 5.1.

A _dimension characteristic_ is a nonempty finite sequence whose entries are
positive natural numbers or the $\infty$ symbol. Dimension characteristics of
the same length are ordered by coordinatewise ordering. Let $t$ be a dimension
characteristic. A tuple $\langle V[G_{i}]\colon i\in|t|+1\rangle$ of generic
extensions has _dimension characteristic $t$_ if for every
$j\in\mathrm{dom}(t)$, $t(j)$ is the largest number such that nonempty set
$a\subset|t|+1$ of cardinality $j+1$ and every $k\in(|t|+1)\setminus a$,
$\dim(V[G_{i}\colon i\in a]/V[G_{k}])\geq t(j)$.

######  Example 5.2.

Let $n\geq 2$ be a number and let $Q_{rn}$ be the Cohen poset on the space of
all rectangles in $\mathbb{R}^{n}$. Let $\langle x_{i}\colon i\in 4\rangle$ be
a quadruple generic over $V$ for the poset $Q_{rn}$. Then each point $x_{i}$
is a Cohen-generic point of $\mathbb{R}^{n}$ and the quadruple $\langle
V[x_{i}]\colon i\in 4\rangle$ of generic extensions has dimension
characteristic $\langle\infty,n,1\rangle$. This follows from Example 4.12.

######  Definition 5.3.

Let $P$ be a Suslin poset and $t$ be a dimension characteristic.

  1. 1.

If $\bar{p}$ be a virtual condition in $P$, we say that $p$ is _$t$
-dimensionally balanced_ if for every tuple $\langle V[G_{i}]\colon
i\in|t|+1\rangle$ of generic extensions of dimension characteristic $\geq t$,
every tuple $\langle p_{i}\colon i\in j\rangle$ of conditions such that
$p_{i}\in V[G_{i}]$ and $p_{i}\leq\bar{p}$ has a common lower bound;

  2. 2.

the poset $P$ is $t$-dimensionally balanced if for every condition $p\in P$
there is a $t$-dimensionally balanced virtual condition $\bar{p}\leq p$.

######  Theorem 5.4.

Let $n\geq 2$ be a natural number. In every extension of the choiceless
Solovay model which is cofinally $\langle n,n,1\rangle$-dimensionally
balanced, every nonmeager subset of $\mathbb{R}^{n}$ contains all vertices of
a non-degenerate rectangle.

###### Proof.

Let $\kappa$ be an inaccessible cardinal. Let $P$ be a Suslin forcing which is
$\langle n,n,1\rangle$-dimensionally balanced cofinally below $\kappa$. Let
$W$ be the associated choiceless Solovay model and work in $W$. Suppose that
$p\in P$ is a condition and $\tau$ is a $P$-name for a nonmeager subset of
$\mathbb{R}^{n}$. I must find a rectangle and a strengthening of the condition
$p$ which forces the rectangle to be a subset of $\tau$. The name $\tau$ and
the condition $p$ are both definable from some parameter $z\in 2^{\omega}$ and
a parameter in the ground model $V$. Let $V[K]$ be an intermediate extension
by a poset of cardinality smaller than $\kappa$ such that $z\in V[K]$ and $P$
is $\langle n,n,1\rangle$-dimensionally balanced $V[K]$.

Work in $V[K]$. By the balance assumption, there must be a $\langle
n,n,1\rangle$-dimensionally balanced virtual condition $\bar{p}\leq p$. Write
$Q$ for the Cohen forcing associated with $\mathbb{R}^{n}$, adding a generic
point ${\dot{x}}_{\mathit{gen}}\in\mathbb{R}^{n}$. By the forcing theorem,
there must be a condition $q\in Q$, a poset $R$, and a $Q\times R$-name
$\sigma$ for a condition in $P$ stronger than $\bar{p}$ such that
$q\Vdash_{Q}R\Vdash\mathrm{Coll}(\omega,<\kappa)\Vdash\sigma\Vdash_{P}{\dot{x}}_{\mathit{gen}}\in\tau$.
Otherwise, in the model $W$, the virtual condition $\bar{p}$ would force
$\tau$ to be disjoint from the comeager set of points in $\mathbb{R}^{n}$
which are Cohen generic over the model $V[K]$, contradicting the initial
assumption on the name $\tau$.

Let $Q_{rn}$ be the Cohen poset on the space of rectangles in
$\mathbb{R}^{n}$. In the model $W$, find a rectangle $\langle x_{i}\colon i\in
4\rangle$ generic over $V[K]$ for the poset $Q_{rn}$ below the condition
$q^{4}$. Recall from Example 5.2 that each point $x_{i}$ is generic over
$V[K]$ for the poset $Q$ below the condition $q$ and the quadruple $\langle
V[G_{i}]\colon i\in 4\rangle$ has dimension characteristic
$\langle\infty,n,1\rangle$. Let $H_{i}$ for $i\in 4$ be filters on the poset
$R$ mutually generic over the model $V[x_{i}\colon i\in 4]$. By Proposition
3.3, the tuple $\langle V[x_{i}][H_{i}]\colon i\in 4\rangle$ has dimension
characteristic $\langle\infty,n,0\rangle$. Write $p_{i}=\sigma/x_{i},H_{i}\in
P$. By the forcing theorem applied in the model $V[x_{i},H_{i}]$, $p_{i}\leq
p$ is a condition forcing $x_{i}\in\tau$. By the balance assumption on the
virtual condition $\bar{p}$, the set $\\{p_{i}\colon i\in 4\\}$ has a common
lower bound in the poset $P$. That common lower bound forces each point in the
rectangle $\\{x_{i}\colon i\in 4\\}$ to belong to the set $\tau$ as required.
∎

##  6 A coloring poset

The whole development in previous sections would be worthless if no useful
posets of interesting dimension characteristics existed. One such a coloring
poset is produced in this section.

Fix a number $n\geq 2$ and let $\Gamma_{n}$ denote the hypergraph on
$\mathbb{R}^{n}$ of arity four consisting of all non-degenerate rectangles. In
this section, I define a balanced Suslin poset which adds a total
$\Gamma_{n}$-coloring. After that, I show that this poset does not add a total
$\Gamma_{n+1}$-coloring if the ground model is taken to be the choiceless
Solovay model. To start, for every subfield $F\subset\mathbb{R}$, say that an
algebraic set is visible from $F$ if there is a polynomial defining it whose
coefficients all belong to $F$. Define an equivalence relation $E_{F}$ on
$\mathbb{R}^{n}\setminus F^{n}$ by connecting points $x_{0},x_{1}$ if either
$x_{0}=x_{1}$ or there are algebraic sets $A_{0},A_{1}\subseteq\mathbb{R}^{n}$
visible from $F$ which are distinct from $\mathbb{R}^{n}$ and $x_{0}\in A_{0}$
and $x_{1}\in A_{1}$, and there are polynomial functions
$f_{0},f_{1}\colon\mathbb{R}^{n}\to\mathbb{R}^{n}$ visible from $F$ such that
$f_{0}(x_{0})=x_{1}$ and $f_{1}(x_{1})=x_{0}$. The connection between this
equivalence relation and rectangles is encapsulated in the following simple
proposition.

######  Proposition 6.1.

Let $F\subset\mathbb{R}$ be a subfield and $\\{x_{i}\colon i\in
4\\}\subset\mathbb{R}^{n}$ be a rectangle such that $x_{0},x_{1}\in F^{n}$.
Then

  1. 1.

either both points $x_{2},x_{3}$ belong to $F^{n}$ or neither of them does;

  2. 2.

if the points $x_{2},x_{3}$ do not belong to $F^{n}$, then they are
$E_{F}$-equivalent.

###### Proof.

The first item is an immediate corollary of the fact that the fourth point in
a rectangle is a linear combination of the other three. For the second item,
write $S$ for the segment connecting $x_{0}$ and $x_{1}$ and divide into two
cases. In the first case, the points $x_{0},x_{1}$ are opposite on the
rectangle $R$. In this case, write $y\in F^{n}$ for the midpoint of $S$, and
observe that both $x_{2},x_{3}$ lie on the hypersphere centered at $y$
containing $x_{0}$. Also, $x_{2},x_{3}$ can be obtained from each other by the
reflection about the point $y$. As both the hypersphere and the reflection are
visible from $F$, $x_{2}\mathrel{E_{F}}x_{3}$ follows. In the second case, the
points $x_{0},x_{1}$ are adjacent on the rectangle $R$. Here, note that
$x_{2},x_{3}$ belong to the hyperplanes perpendicular to $S$ and containing
$x_{0}$ or $x_{1}$ respectively. The points $x_{2},x_{3}$ can be obtained from
each other by the reflection about the hypeplane perpendicularly bisecting the
segment $S$. As both the hyperplanes and the reflection are visible from $F$,
$x_{2}\mathrel{E_{F}}x_{3}$ holds and the proposition follows. ∎

The appearance of a hypersphere in this proof is the only reason why
Noetherian topologies of this paper are in the algebraic category as opposed
to the much simpler affine category. Now, to define the coloring poset, fix a
Borel ideal $I$ on $\omega$ containing all singletons and such that it is not
generated by countably many sets. A particular choice of $I$ seems to be
irrelevant beyond these demands, the summable ideal will do.

######  Definition 6.2.

The poset $P_{n}$ consists of all countable functions $p$ such that there is a
countable real closed subfield $\mathrm{supp}(p)\subset\mathbb{R}$ such that
$\mathrm{dom}(p)=\mathrm{supp}(p)^{n}$ and $p$ is a $\Gamma_{n}$-coloring. The
ordering is defined by $q\leq p$ if $p\subseteq q$ and for every
$E_{\mathrm{supp}(p)}$-class $C\subset\mathrm{dom}(q)$, $q\restriction C$ is
injective and its range belongs to $I$.

######  Proposition 6.3.

$P$ is a $\sigma$-closed partial ordering.

###### Proof.

Observe that if $q\leq p$ holds then $\mathrm{dom}(q)$ is invariant under the
equivalence relation $E_{\mathrm{supp}(p)}$ and on the set
$\mathbb{R}^{n}\setminus\mathrm{dom}(q)$, $E_{\mathrm{supp}(p)}\subseteq
E_{\mathrm{supp}(q)}$ holds. The transitivity of the relation $\leq$ on $P$
follows. For the $\sigma$-closure, if $\langle p_{n}\colon n\in\omega\rangle$
is a descending chain of conditions, then $\bigcup_{n}p_{n}$ is their common
lower bound. ∎

It is important to find a precise criterion for compatibility of conditions in
the poset $P_{n}$.

######  Proposition 6.4.

Let $p_{0},p_{1}\in P_{n}$ be conditions. The following are equivalent:

  1. 1.

$p_{0},p_{1}$ are compatible;

  2. 2.

for every point $x_{0}\in\mathbb{R}^{n}$ there is a common lower bound of
$p_{0},p_{1}$ containing $x_{0}$ in its domain;

  3. 3.

    1. (a)

$p_{0}\cup p_{1}$ is a function and $\Gamma_{n}$-coloring;

    2. (b)

for every $E_{\mathrm{supp}(p_{0})}$-class $C$, $p_{1}\restriction C$ is
injective and its range belongs to $I$;

    3. (c)

same as (b) except with $p_{0}$ and $p_{1}$ interchanged.

###### Proof.

(2) implies (1), which in turn implies (3) by the definition of the ordering
on $P_{n}$. I must argue that (3) implies (2). Let $x_{0}\in\mathbb{R}^{n}$ be
an arbitrary point. Choose a countable field $F\subset\mathbb{R}$ containing
both $\mathrm{supp}(p_{0})$ and $\mathrm{supp}(p_{1})$ as subsets and all
corrdinates of the point $x_{0}$ as elements. Write
$c=F^{n}\setminus\mathrm{dom}(p_{0}\cup p_{1})$,
$E_{0}=E_{\mathrm{supp}(p_{0})}$, and $E_{1}=E_{\mathrm{supp}(p_{1})}$. Choose
an infinite set $b\subset\omega$ in the ideal $I$ which cannot be covered by
finitely many singletons and sets of the form $\mathrm{rng}(p_{1}\restriction
C)$ where $C$ is an $E_{0}$-class, or of the form
$\mathrm{rng}(p_{0}\restriction C)$ where $C$ is an $E_{1}$-class. This is
possible by items (2) and (3) and the choice of the ideal $I$. Now, let
$q\colon F^{n}\to\omega$ be a function extending $p_{0}\cup p_{1}$, such that
$q\restriction c$ is an injection, and such that for every point $x\in c$,
$q(x)\in b\setminus(p_{0}^{\prime\prime}[x]_{E_{1}}\cup
p_{1}^{\prime\prime}[x]_{E_{0}})$. This is possible by the claim and the
choice of the set $b$. I claim that $q$ is the requested common lower bound of
$p_{0}$ and $p_{1}$.

First of all, check that $q$ is a $\Gamma$-coloring. Suppose that $R$ is a
rectangle in $F^{n}=\mathrm{dom}(q)$ and argue that it is not
$q$-monochromatic. If it has all four vertices in $\mathrm{dom}(p_{0}\cup
p_{1})$ then it is not monochromatic by item (a). If it has more than one
vertex in the set $c$, then it is not monochromatic as $q\restriction c$ is an
injection. If it has exactly one vertex in $c$, call it $x$, then the three
others cannot belong to the same condition by the closure properties of
$\mathrm{dom}(p_{0})$ and $\mathrm{dom}(p_{1})$. Thus, there must be a
condition (say $p_{0}$) such that $R$ has two vertices (say $z,w$) in its
domain and the third (say $y$) is not in its domain (belongs to
$\mathrm{dom}(p_{1}\setminus p_{0})$). I will show that $x\mathrel{E_{1}}y$
holds; then $q(x)\neq q(y)$ by the choice of the coloring $q$ and $R$ is not
monochromatic. There are two cases. Suppose first that the two points $z,w$
are opposite in the rectangle $\\{z,w,y,x\\}$. Then both points $y,x$ lie on
the hypersphere of which the segment between $z,w$ is a diameter, and they can
be moved one to another by the reflection around the midpoint of this segment,
showing that $x\mathrel{E_{1}}y$. Suppose now that the points $z,w$ are
adjacent in the rectangle $\\{z,w,y,x\\}$. Then the points $y,x$ lie on the
hyperplanes perpendicular to the segment connecting $z,w$ and containing $z$
and $w$ respectively, and they can be moved to one another by the reflection
around the hyperplane perpendicularly bisecting the segment connecting $z$ and
$w$. Thus, $x\mathrel{E_{1}}y$ holds in this case too.

Finally, prove that $q\leq p_{0}$ holds; the case $q\leq p_{1}$ is symmetric.
Let $C\subset\mathrm{dom}(q)$ be an $E_{0}$-class. The choice of the coloring
$q$ implies that $q^{\prime\prime}C\subset b\cup p_{1}^{\prime\prime}C$ which
then is a set in the ideal $I$. Moreover, $q\restriction
C\cap\mathrm{dom}(p_{1})$ is an injection by item (b), and $q\restriction
C\cap c$ is an injection as well which shares no values with $q\restriction
C\cap\mathrm{dom}(p_{1})$ by the choice of $q$. In total, $q\restriction C$ is
an injection as required. ∎

######  Corollary 6.5.

$P_{n}$ is a Suslin forcing.

###### Proof.

It is immediate that the poset $P$ and its order relation are Borel.
Proposition 6.4 shows that the compatibility relation is Borel as well. ∎

######  Corollary 6.6.

$P_{n}$ forces the union of the generic filter to be a total
$\Gamma_{n}$-coloring.

###### Proof.

$P_{n}$ is a $\sigma$-closed partial order of partial $\Gamma_{n}$-colorings.
In addition, by Proposition 6.4, for each point $x_{0}\in\mathbb{R}^{n}$, the
set of conditions containing $x_{0}$ in its domain is an open dense subset of
$P_{n}$. The corollary follows by a genericity argument. ∎

######  Proposition 6.7.

In the poset $P_{n}$,

  1. 1.

for every total $\Gamma_{n}$-coloring $c$, the pair
$\langle\mathrm{Coll}(\omega,\mathbb{R}^{n}),\check{c}\rangle$ is balanced;

  2. 2.

every balanced pair is equivalent to one as in (1);

  3. 3.

distinct total $\Gamma_{n}$-colorings yield inequivalent balanced pairs.

In particular, the poset $P_{n}$ is balanced if and only if the Continuum
Hypothesis holds.

###### Proof.

For item (1), let $c\colon\mathbb{R}^{n}\to\omega$ be a total
$\Gamma_{n}$-coloring. Let $V[G_{0}],V[G_{1}]$ be mutually generic extensions
and $p_{0},p_{1}\leq c$ be conditions in $P_{n}$ in the respective models
$V[G_{0}]$ and $V[G_{1}]$; I must show that these conditions have a common
lower bound. Write $E=E_{\mathbb{R}\cap V}$, $E_{0}=E_{\mathbb{R}\cap
V[G_{0}]}$, and $E_{1}=E_{\mathbb{R}\cap V[G_{1}]}$. The following claim is
key.

######  Claim 6.8.

$E_{0}\restriction V[G_{1}]=E\restriction V[G_{1}]$ and vice versa,
$E_{1}\restriction V[G_{0}]=E\restriction V[G_{0}]$.

###### Proof.

I prove the first assertion, as the proof of the second is symmetric. The
right-to-left inclusion follows from the fact that $V\subset V[G_{0}]$ holds.
For the key left-to-right inclusion, suppose that $x,y\in V[G_{1}]\setminus V$
are $E_{0}$-related. This means that there are algebraic sets
$A,B\subset\mathbb{R}^{n}$ visible from $V[G_{0}]$, distinct from
$\mathbb{R}$, containing $x$ and $y$ respectively, and polynomial functions
$f,g\colon\mathbb{R}^{n}\to\mathbb{R}^{n}$ visible from $V[G_{0}]$ such that
$f(x)=y$ and $g(y)=x$. Now, by Proposition 3.3, $\dim(V[G_{0}]/V[G_{1}])\geq
n$ holds. By Proposition 3.4 there are algebraic sets $A^{\prime}\subseteq A$,
$B^{\prime}\subseteq B$ visible from $V[G_{0}]$ and containing $x,y$
respectively, and polynomial functions
$f^{\prime},g^{\prime}\colon\mathbb{R}^{n}\to\mathbb{R}^{n}$ visible in $V$
such that $f\restriction A^{\prime}=f^{\prime}\restriction A^{\prime}$ and
$g\restriction B^{\prime}=g^{\prime}\restriction B^{\prime}$. These objects
show that $x,y$ are $E$-related as desired. ∎

Now it is time to verify item (3) of Proposition 6.4 for $p_{0},p_{1}$. First
of all, $p_{0}\cup p_{1}$ is a function since $p_{0}\restriction
V=p_{1}\restriction V=c$ and
$\mathrm{dom}(p_{0})\cap\mathrm{dom}(p_{1})\subset V$ holds by the product
forcing theorem. To show that $p_{0}\cup p_{1}$ is a $\Gamma_{n}$-coloring,
assume that $R$ is a non-degenerate rectangle in its domain and work to show
that it is not monochromatic. There are two possible configurations. If one of
the conditions $p_{0},p_{1}$ (say $p_{0}$) contains three points of $R$, then
by the closure properties of $\mathrm{dom}(p_{0})$ it contains the fourth one
as well and $R$ is not monochromatic since $p_{0}$ is a $\Gamma_{n}$-coloring.
The only remaining option is that $p_{0}\setminus V$ contains two points (say
$x,y$) of the rectangle and $p_{1}\setminus V$ contains the remaining two (say
$z,w$). By Proposition 6.1, $z,w$ are $E_{0}$-related, and by Claim 6.8, they
are $E$-related. Then, $p_{1}(z)\neq p_{1}(w)$ follows from $p_{1}\leq c$ and
$R$ is not monochromatic. The definition of the ordering $\leq$ in Definition
6.2 is motivated exactly by this step.

Finally, we must show that $p_{0}^{\prime\prime}C$ is an injection with range
in $I$ for any $E_{0}$-class $C$ and vice versa, $p_{1}^{\prime\prime}D$ is an
injection with range in $I$ for any $E_{1}$-class $D$. The classes $C,D$ are
in fact $E$-classes by Claim 6.8, so this follows from $p_{0},p_{1}\leq c$.
Item (3) of Proposition 6.4 has been verified, which implies the compatibilty
of the conditions $p_{0},p_{1}$ and item (1) of the current proposition.

For item (2), suppose that $\langle Q,\sigma\rangle$ is a balanced pair.
Strengthening the condition $\sigma$ and the poset $Q$ if necessary, I may
assume that $Q\Vdash\mathbb{R}\cap V\subset\mathrm{dom}(\sigma)$. A balance
argument then shows that for each point $x\in\mathbb{R}^{n}$ there is a
specific number $c(x)\in\omega$ such that $Q\Vdash\sigma(\check{x})=c(x)$. It
is immediate that $c\colon\mathbb{R}^{n}\to\omega$ is a $\Gamma_{n}$-coloring.
I will show that $Q\Vdash\delta\leq\check{c}$; this will prove that the
balanced pairs $\langle Q,\tau\rangle$ and
$\langle\mathrm{Coll}(\omega,\mathbb{R}^{n}),\check{c}\rangle$ are equivalent
by [4, Proposition 5.2.6] as required. Suppose towards a contradiction that
there is a condition $q\in Q$ forcing $\sigma\not\leq\check{c}$. Let
$G_{0},G_{1}\subset Q$ be mutually generic filters containing the condition
$q$, and write $p_{0}=\sigma/G_{0}$ and $p_{1}=\sigma/G_{1}$. Since
$p_{0}\not\leq\check{q}$, the definition of the ordering in Definition 6.2
shows that there must be a $E_{\mathbb{R}}$-class $C\subset\mathbb{R}^{n}$
such that $p_{0}\restriction C$ is not an injection with range in the ideal
$I$. Since $\mathbb{R}\cap V\subset\mathrm{supp}(p_{1})$ holds,
$E_{\mathbb{R}\cap V}\subset E_{\mathrm{supp}(p_{1})}$ holds as well and $C$
is a subset of a single $E_{\mathrm{dom}(p_{1})}$-class. This shows that the
conditions $p_{0},p_{1}$ are incompatible, contradicting the initial
assumptions on $Q,\sigma$.

Item (3) is immediate. For the last sentence, if the Continuum Hypothesis
fails then there are no total $\Gamma_{n}$-colorings by the result of [1], so
no balanced virtual conditions by (2) of the present proposition. On the other
hand, assume that the Continuum Hypothesis holds and $p\in P_{n}$ is a
condition; I must produce a total $\Gamma_{n}$-coloring $c$ such that $c\leq
p$. To this end, let $\langle x_{\alpha}\colon\alpha\in\omega_{1}\rangle$ be
an enumeration of $\mathbb{R}^{n}$. By recursion on $\alpha$ construct a
descending sequence $\langle p_{\alpha}\colon\alpha\in\omega_{1}\rangle$ of
conditions such that $p_{0}=p$, $x_{\alpha}\in\mathrm{dom}(p_{\alpha+1})$, and
$p_{\alpha}=\bigcup_{\beta\in\alpha}p_{\beta}$. This is straightforward; at
the successor stage use Proposition 6.4 with the two conditions equal to
$p_{\alpha}$. In the end, the coloring $c=\bigcup_{\alpha}p_{\alpha}$
completes the proof. ∎

Finally, I must provide the dimension characteristics of the virtual balanced
conditions in the poset $P_{n}$.

######  Proposition 6.9.

In the poset $P_{n}$, every balanced virtual condition is in fact $\langle
n+1,n+1,1\rangle$-dimensionally balanced.

###### Proof.

Suppose that $c\colon\mathbb{R}^{n}\to\omega$ is a total coloring. Suppose
that $\\{V[G_{i}]\colon i\in 4\\}$ is a finite collection of generic
extensions such that for any three distinct indices $i,j,k\in 4$ it is the
case that $\dim(V[G_{i},G_{j}]/V[G_{k}])\geq n+1$ holds. Suppose that
$\\{p_{i}\colon i\in 4\\}$ is a collections of conditions in $P_{n}$ such that
$p_{i}\in V[G_{i}]$ and $p_{i}\leq c$ holds; I must produce a common lower
bound of the conditions $p_{i}$ for $i\in 4$.

Note that the dimension assumption implies that for pairwise distinct indices
$i,j,k\in 4$, $\mathbb{R}\cap V[G_{i},G_{j}]\cap V[G_{k}]=\mathbb{R}\cap V$
holds by Proposition 3.2. To set up useful notation, suppose that $i,j\in 4$
are possibly equal indices. Write $F_{ij}\subset\mathbb{R}$ for the smallest
subfield of $\mathbb{R}$ containing $\mathrm{supp}(p_{i})$ and
$\mathrm{supp}(p_{j})$ and write $E_{ij}=E_{F_{ij}}$. Also, write
$E=E_{\mathbb{R}\cap V}$. The following claim is proved exactly like Claim 6.8
using the assumption on the dimension of the quadruple $\langle V[G_{i}]\colon
i\in 4\rangle$.

######  Claim 6.10.

Let $i,j\in 4$ and let $k$ be distinct from both $i$ and $j$. Then
${E_{ij}}\restriction\mathrm{dom}(p_{k})\setminus
V=E\restriction\mathrm{dom}(p_{k})\setminus V$.

To find the requested lower bound, work in the model $V[G_{i}\colon i\in 4]$
and find a countable subfield $F\subset\mathbb{R}$ containing
$\mathrm{supp}(p_{i})$ for every $i\in 4$. Write
$d=F^{n}\setminus\mathrm{dom}(\bigcup_{i}p_{i})$. Let $b\subset\omega$ be a
set in the ideal $I$ which cannot be covered by a union of finitely many
singletons and sets of the form $p_{i}^{\prime\prime}C$ where $i\in 4$ and
$C\subset\mathrm{dom}(p_{i})\setminus V$ is an $E$-class. Note that all of the
latter sets belong to the ideal $I$ as $p_{i}\leq c$ is assumed. Let $q\colon
F^{n}\to\omega$ be any function extending $\bigcup_{i}p_{i}$ such that
$q\restriction d$ is an injection, and for each point $x\in d$, $q(x)\in b$
and $q(x)$ does not belong to any set of the form $p^{\prime\prime}_{k}C$
where $k\in 4$ and $C$ is an $E_{ij}$-class of $x$ for some $i,j\in 4$
distinct from $k$. This is possible by Claim 6.10 and the choice of the set
$b$. I will show that $q$ is the requested common lower bound of the
conditions $p_{i}$ for $i\in 4$.

The key point is showing that $q$ is in fact a $\Gamma_{n}$-coloring. Let $R$
be a rectangle in $\mathrm{dom}(q)$ and work to show that it is not
$q$-monochromatic. There are several configurations to investigate.

Case 1. Suppose that $R\subset\bigcup_{i\in 4}\mathrm{dom}(p_{i})$. Let
$a\subset 4$ be an inclusion-minimal set such that $R\subset\bigcup_{i\in
a}\mathrm{dom}(p_{i})$. Divide into subcases:

Case 1.1. $a$ is a singleton, containing some $i\in 4$. In this case, $R$ is
not monochromatic since $p_{i}$ is a $\Gamma_{n}$-coloring.

Case 1.2. $a$ contains exactly two elements $i,j\in 4$. By the closure
properties of $\mathrm{dom}(p_{i})$ and $\mathrm{dom}(p_{j})$, it is
impossible for one of them to contain three vertices of $R$, since then it
would contain the fourth one also, and Case 1.1 would be in effect. Thus,
$\mathrm{dom}(p_{i})$ contains vertices $x,y$ and $\mathrm{dom}(p_{j})$
contains vertices $z,w$, and none of these vertices can be in $V$. By
Proposition 6.1, the points $z,w$ are $E_{i}$-equivalent. By Claim 6.10, the
points are in fact $E$-equivalent. Finally, since $p_{j}\leq c$, the points
$z,w$ receive distinct $p_{j}$-color, so $R$ is not monochromatic.

Case 1.3. $a$ contains exactly three elements $i,j,k\in 4$. Then one of the
conditions (say $p_{i}$) contains two vertices $x,y\in R$, and
$\mathrm{dom}(p_{j})\setminus V$ and $\mathrm{dom}(p_{k})\setminus V$ contain
a single element $z$ and $w\in R$ respectively. By the dimension assumption
$V[G_{i},G_{j}]\cap V[G_{k}]=V$ must hold. At the same time, the point
$w\in\mathrm{dom}(p_{k})\setminus V$ is a linear combination of $x,y,z$, so it
should belong to the model $V[G_{i},G_{j}]$, contradicting the case
assumption. This case cannot occur no matter whether $R$ is monochromatic or
not.

Case 1.4. $a=4$. Let $x_{i}\in\mathrm{dom}(p_{i})$ for $i\in 4$ be the
corresponding points in the rectangle $R$; note that it must be the case that
$x_{i}\in\mathrm{dom}(p_{i})\setminus V$. Without loss assume that $x_{0}$ and
$x_{1}$ are points adjacent in $R$. Note that the point $x_{2}$ belongs to the
hyperplane $H$ perpendicular to the segment connecting $x_{0}$ and $x_{1}$
which contains either $x_{0}$ or $x_{1}$. This hyperplane belongs to the model
$V[G_{0},G_{1}]$. Since $\dim(V[G_{i},G_{j}]/V[G_{k}])\geq n+1$, by
Proposition 3.4 there must be an algebraic set $K\subset H$ in the ground
model containing the point $x_{2}$. Then, the point $x_{2}$ can be
reconstructed in $V[G_{3}]$ as the closest point on $K$ to $x_{3}$, since it
is even the closest point on $H$ to $x_{3}$. This means that $x_{2}\in
V[G_{3}]$, contradicting the case assumption. Thus, this case cannot occur no
matter whether $R$ is monochromatic or not.

Case 2. Suppose that $R$ has exactly one vertex (call it $x$) in the set $d$.
Let $a\subset 4$ be an inclusion-minimal set such that the remaining vertices
of $R$ are contained in $\bigcup_{i\in a}\mathrm{dom}(p_{i})$.

Case 2.1. $a$ is a singleton containing some index $i$. By the closure
properties of $\mathrm{dom}(p_{i})$, with the three vertices of $R$
$\mathrm{dom}(p_{i})$ would have to contain even $x$, and Case 1.1 would be in
force.

Case 2.2. $a$ contains exactly two elements $i,j\in 4$. Then one of the
conditions (say $p_{i}$) contains two elements of $R$, while
$\mathrm{dom}(p_{j})\setminus V$ contains a single vertex $y$ of $R$. Note
that $y$ is $E_{i}$-equivalent to $x$ by Proposition 6.1. By the choice of the
function $q$, $q(x)\neq q(y)$ holds and $R$ is not monochromatic.

Case 2.3. $a$ contains exactly three elements $i,j,k\in 4$. Then each of
$\mathrm{dom}(p_{i})\setminus V$, $\mathrm{dom}(p_{j})\setminus V$, and
$\mathrm{dom}(p_{k})\setminus V$ must contain exactly one vertex of $R$. Let
$y$ be the unique vertex of $R$ in $\mathrm{dom}(p_{k})\setminus V$. Then
$x,y$ are $E_{ij}$-equivalent by Proposition 6.1, $q(x)\neq q(y)$ holds and
$R$ is not monochromatic.

Case 3. Suppose finally that $R$ has more than one vertex in the set $d$. Then
$R$ is not $q$-monochromatic since $q\restriction d$ is an injection.

Finally, I need to show that for every $i\in 4$, $q\leq p_{i}$ holds. It is
clear that $p_{i}\subset q$ holds. Now, fix an $E_{i}$-class
$C\subset\mathrm{dom}(q\setminus p_{i})$ and work to show that $q\restriction
C$ is an injection with range in the ideal $I$.

First observe that there is at most one index $j\in a$ such that
$C\cap\mathrm{dom}(p_{j})\neq 0$. If there were two such indices $j,k$, with
points $x_{j}\in\mathrm{dom}(p_{j})\cap C$ and
$x_{k}\in\mathrm{dom}(p_{k})\cap C$, then use the definition of the
equivalence $E_{i}$ to show that $x_{k}\in V[G_{i},G_{j}]$, which is
impossible as $V[G_{k}]\cap V[G_{i},G_{j}]=V$. Thus, let $j$ be the unique
index (if it exists) such that $C\cap\mathrm{dom}(p_{j})\neq 0$. By Claim
6.10, $C\cap\mathrm{dom}(p_{j})$ is a single $E$-class disjoint from $V$, so
$p_{j}\restriction(C\cap\mathrm{dom}(p_{j}))$ is an injection with range in
$I$. Also, $q\restriction C\setminus\mathrm{dom}(p_{j})$ is an injection with
range in $I$, as $C\setminus\mathrm{dom}(q_{j})\subset d$ and $q\restriction
d$ is an injection with range included in the set $b\in I$. Finally, for every
point $x\in C\setminus\mathrm{dom}(p_{j})$ the color $q(x)$ was chosen exactly
so that it is different from all colors in the set
$p_{j}^{\prime\prime}(C\cap\mathrm{dom}(p_{j}))$. Thus $q\restriction C$ is an
injection with range in the ideal $I$ as desired. ∎

Finally, I am in a position to prove Theorem 1.1. Let $n\geq 2$ be a number.
Let $\kappa$ be an inaccessible cardinal, and let $W$ be the choiceless
Solovay model derived from $\kappa$. Let $G\subset P_{n}$ be a filter generic
over $W$, and consider the model $W[G]$. The poset $P_{n}$ is $\sigma$-closed,
therefore the model $W[G]$, as a $\sigma$-closed extension of a model of DC,
satisfies DC as well. By Corollary 6.6, in $W[G]$ the chromatic number of
$\Gamma_{n}$ is countable. By the conjunction of Propositions 6.7 and 6.9 and
Theorem 5.4, in the model $W[G]$, every non-meager subset of
$\mathbb{R}^{n+1}$ contains vertices of a non-degenerate rectangle, theorefore
the chromatic number of $\Gamma_{n+1}$ is uncountable. Theorem 1.1 has just
been proved.

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