# Noetherian spaces in choiceless set theory1112020 AMS subject classification
03E25, 03E35.

Jindřich Zapletal
University of Florida
Institute of Mathematics, Czech Academy of Sciences

###### Abstract

I prove several independence results in the choiceless ZF+DC theory which
separate algebraic and non-algebraic consequences of the axiom of choice.

##  1 Introduction

Geometric set theory [5] was developed in part to produce consistency results
in choiceless ZF+DC set theory regarding various $\Sigma^{2}_{1}$ sentences.
In this paper, I produce several such consistency results which separate
$\Sigma^{2}_{1}$ sentences dealing with algebraic topics from those which do
not. While the distinction may seem vague, the techniques of the paper show
that there is in fact a clearly visible fracture line.

I isolate the notion of Noetherian balanced forcing and show that in the
generic extensions of the choiceless Solovay model by such forcings, numerous
$\Sigma^{2}_{1}$ consequences of the axiom of choice fail. The following
theorem, stated using the conventions of geometric set theory [5, Convention
1.7.18], provides a somewhat representative list.

######  Theorem 1.1.

In cofinally Noetherian balanced extensions of the choiceless Solovay model,

  1. 1.

if $A_{0},A_{1}\subset 3^{\omega}$ are non-meager sets, then there are points
$x_{0}\in A_{0}$ and $x_{1}\in A_{1}$ such that the set $\\{n\in\omega\colon
x_{0}(n)=x_{1}(n)\\}$ is finite;

  2. 2.

if $\Gamma\curvearrowright X$ is a turbulent Polish group action with all
orbits dense and meager, and $A_{0},A_{1}\subset X$ are non-meager sets then
there are points $x_{0}\in A_{0}$ and $x_{1}\in A_{1}$ which are orbit
equivalent;

  3. 3.

(if the extension is $\sigma$-closed) outer Lebesgue measure is continuous in
arbitrary increasing unions.

The theorem implies in particular that there is no non-principal ultrafilter
on natural numbers in the extensions. Such an ultrafilter $U$ could be used to
divide sequences in $3^{\omega}$ into three non-meager sets according to their
$U$-prevalent value, violating item (1). It also implies that if
$\Gamma\curvearrowright X$ is a turbulent Polish group action with an orbit
equivalence relation $E$, then every $E$-invariant function $f$ from $X$ to
another Polish space $Y$ stabilizes on a co-meager set in the extensions. The
reason is that for each basic open set $O\subset Y$, the set $\\{x\in X\colon
f(x)\in O\\}$ is $E$-invariant, and therefore meager or co-meager by (2).

It turns out that many posets adding objects to the choiceless Solovay model
which are related to algebraic applications of the Axiom of Choice are
Noetherian balanced. The following theorem provides a good list.

######  Theorem 1.2.

The following Suslin posets are $\sigma$-closed and Noetherian balanced:

  1. 1.

a poset adding a Hamel basis, for a Polish field $X$ over a countable subfield
$F$ [5, Example 6.3.6];

  2. 2.

a poset adding a transcendence basis, for a Polish field $X$ over a countable
subfield $F$ [5, Example 6.3.10] ;

  3. 3.

a poset coloring a given closed Noetherian graph without an uncountable clique
[8];

  4. 4.

a poset coloring $\sigma$-algebraic graphs without an uncountable clique [7];

  5. 5.

a poset coloring the hypergraph of isosceles triangles in $\mathbb{R}^{2}$
[10];

  6. 6.

a poset coloring a given redundant $\sigma$-algebraic hypergraph [9].

There are attendant consistency results too numerous to list separately. For
example, it is consistent relative to an inaccessible cardinal that ZF+DC
holds, there is a transcendence basis for $\mathbb{R}$ over $\mathbb{Q}$,
there is no nonprincipal ultrafilter on $\omega$, and outer Lebesgue measure
is continuous under arbitrary increasing unions. Since the class of Noetherian
balanced posets is closed under countable product, the consistency results can
be even combined.

Section 2 discusses the notion of a subbasis for a Noetherian topology and
that of a mutually Noetherian pair of generic extensions of a model of ZFC.
This is an instrumental weakening of mutual genericity. In Section 3 I produce
several interesting mutually Noetherian pairs of generic extensions, notably
one induced by a turbulent action of a Polish group. Section 4 defines the
notion of a Noetherian balanced Suslin forcing–this is a forcing in which
conditions can be successfully amalgamated across mutually Noetherian pairs of
generic extensions. There are several attendant preservation theorems for
Noetherian balanced Suslin forcings, resulting in the proof of various items
of Theorem 1.1. Section 5 lists some Noetherian balanced Suslin forcings,
proving Theorem 1.2. Section 6 finally shows that the conclusions of Theorem
1.1 can be violated in suitable balanced, not Noetherian balanced, extensions
of the Solovay model.

The paper uses the notation standard of [3]. In matters pertaining to
geometric set theory it follows the terminology and notation of [5]. DC
denotes the Axiom of Dependent Choices. The inaccessible cardinal in the
assumptions of Theorems 1.1 and 1.2 is used to start the method of balanced
forcing as in [5]; I do not know if it is necessary.

Thanks are due to Chris Lambie-Hanson for suggesting the terminology of
Definition 2.1. The author was partially supported by grant EXPRO 20-31529X of
GA ČR.

##  2 Noetherian pairs of generic extensions

The technology of this paper rests on the treatment of Noetherian topologies
from a descriptive set theoretic point of view. After plenty of hesitation, I
have decided in the favor of the following presentation.

######  Definition 2.1.

Let $X,Y$ be Polish spaces and $C\subset Y\times X$ be a closed set. The set
$C$ is a _Noetherian subbasis_ if there is no sequence $\langle b_{n}\colon
n\in\omega\rangle$ of finite subsets of $Y$ such that the sets
$D_{n}=\bigcap_{y\in b_{n}}C_{y}$ for $n\in\omega$ strictly decrease with
respect to inclusion.

It is not difficult to see that being a Noetherian subbasis is a coanalytic
property of the set $C$, therefore transitive among well-founded models of set
theory. The terminology is justified by the following routine proposition.

######  Proposition 2.2.

Let $C\subset Y\times X$ be a Noetherian subbasis. Then the smallest topology
on $X$ containing all complements of vertical sections of $C$ is Noetherian.

###### Proof.

Close the collection of vertical sections of $C$ first under finite
intersections, then under finite unions, obtaining a collection $\mathcal{A}$.

######  Claim 2.3.

$\mathcal{A}$ contains no infinite sequences strictly decreasing with respect
to inclusion.

###### Proof.

Suppose towards a contradiction that $\langle A_{n}\colon n\in\omega\rangle$
is such a sequence. The definitory property of a Noetherian subbasis shows
that there is a finite set $b\subset Y$ such that, writing
$C_{b}=\bigcap_{y\in b}C_{y}$, the sequence of intersections $\langle
A_{n}\cap C_{b}\colon n\in\omega\rangle$ does not stabilize and the set
$C_{b}$ is inclusion-minimal such. There must be $n\in\omega$ such that
$A_{n}\cap C_{b}$ is a proper subset of $C_{b}$. Find a finite collection
$\\{c_{i}\colon i\in m\\}$ of finite subsets of $Y$ such that
$A_{n}=\bigcup_{i}B_{i}$ where $B_{i}=\bigcap_{y\in c_{i}}C_{y}$. It is clear
that there must be $i\in m$ such that the sequence of intersections $\langle
A_{k}\cap C_{b}\cap B_{i}\colon k\in\omega\rangle$ does not stabilize,
contradicting the minimal choice of the set $C_{b}$. ∎

######  Claim 2.4.

$\mathcal{A}$ is closed under finite unions and arbitrary intersections.

###### Proof.

$\mathcal{A}$ is closed under finite unions and finite intersections by its
definition. Now, if $a\subset\mathcal{A}$ is any set, then by the previous
claim there is a finite subset $b\subset a$ such that $\bigcap a=\bigcap b$.
Since $\bigcap b\in\mathcal{A}$, it is also the case that $\bigcap
a\in\mathcal{A}$ as desired. ∎

It is now clear from Claim 2.4 that the collection of the complements of sets
in $\mathcal{A}$ is a topology, it is the smallest topology containing all
complements of vertical sections of the set $C$, and that it is Noetherian by
Claim 2.3. ∎

The following observation will be key in numerous contexts.

######  Proposition 2.5.

Let $X,Y$ be Polish spaces and $C\subset Y\times X$ be a Noetherian subbasis.
If $a\subset Y$ is any set, then there is a finite set $b\subset a$ such that
$\bigcap_{y\in a}C_{y}=\bigcap_{y\in b}C_{y}$.

###### Proof.

If this failed for some set $a\subset Y$, it would be possible to build finite
sets $b_{n}\subset a$ for $n\in\omega$ such that $b_{n}\subset b_{n+1}$ and
$\bigcap_{y\in b_{n}}C_{y}\neq\bigcap_{y\in b_{n+1}}C_{y}$. The sets $b_{n}$
for $n\in\omega$ would then violate the Noetherian property of the set $C$. ∎

In the way of an illustration, the following example of a Noetherian subbasis
is ubiquitous in algebra.

######  Example 2.6.

Let $n\in\omega$ be a number and let $X$ be a Polish field. Let $Y$ be a
Polish space indexing all polynomials with $n$ free variables and coefficients
in $X$. Then $C\subset Y\times X^{n}$, defined by $\langle y,\bar{x}\rangle\in
C$ if and only if $y(\bar{x})=0$ is a Noetherian subbasis. The nonexistence of
decreasing sequences of finite intersections of vertical sections of the set
$C$ is verified via the Hilbert Basis Theorem [2].

The following definition is the key tool for connecting Noetherian topologies
with geometric set theory.

######  Definition 2.7.

Let $X,Y$ be Polish spaces and $C\subset Y\times X$ be a Noetherian subbasis.
Let $M$ be a transitive model of set theory coding $X,Y$, and $C$. Let
$A\subset X$ be a set. The symbol $C(M,A)$ denotes the set
$\bigcap\\{C_{y}\colon y\in Y\cap M$ and $A\subseteq C_{y}\\}$. If the set in
the scope of the $\bigcap$ sign is empty, then set $C(M,A)=X$.

The most common situation for applying this definition is that neither the set
$A$ nor any of its elements belong to the model $M$. Note that the set
$C(M,A)$ is coded in the model $M$ as the intersection defining it can be
obtained as an intersection $\bigcap_{y\in b}C_{y}$ for a suitable finite set
$b\subset Y\cap M$ by Proposition 2.5.

######  Definition 2.8.

Let $V[G_{0}]$ and $V[G_{1}]$ be generic extensions of $V$ inside an ambient
generic extension. Say that $V[G_{1}]$ is a _Noetherian to_ $V[G_{0}]$ (over
$V$) if for every Noetherian subbasis $C\subset Y\times X$ coded in $V$ and
for every set $A\subset X$ in $V[G_{1}]$, $C(V[G_{0}],A)=C(V,A)$. Say that the
extensions $V[G_{0}],V[G_{1}]$ are _mutually Noetherian_ if each is Noetherian
to the other. Say that $V[G_{1}]$ is _$K_{\sigma}$ -Noetherian to_ $V[G_{0}]$
if the above equation holds for all $K_{\sigma}$ Noetherian subbases coded in
$V$, and similarly for the mutual $K_{\sigma}$-Noetherian relation.

It is useful to introduce special terminology for $K_{\sigma}$ Noetherian
subbases, since most natural examples are $K_{\sigma}$ and $K_{\sigma}$
subbases are easier to deal with in several respects which appear elsewhere
[7]. Similar notions of perpendicularity always have a friendly relationship
with product forcing, as recorded in the following proposition.

######  Proposition 2.9.

Let $n\geq 1$ be a number. Let $V[G_{0}],V[G_{1}]$ be generic extensions and
$V[G_{1}]$ is Noetherian to $V[G_{0}]$. Suppose that $P_{0}\in V[G_{0}]$ and
$P_{1}\in V[G_{1}]$ be posets and $H_{0}\subset P_{0}$ and $H_{1}\subset
P_{1}$ be filters mutually generic over $V[G_{0},G_{1}]$. Then
$V[G_{1}][H_{1}]$ is Noetherian to $V[G_{0}][H_{0}]$. Similarly for
$K_{\sigma}$-Noetherian.

###### Proof.

Work in the model $V[G_{0},G_{1}]$ and consider the poset $P_{0}\times P_{1}$.
Let $X,Y$ be Polish spaces and $C\subset Y\times X$ be a Noetherian subbasis
coded in $V$. Let $p_{0}\in P_{0}$ and $p_{1}\in P_{1}$ be conditions and let
$\tau_{0},\tau_{1}$ be respective $P_{0},P_{1}$-names in the models
$V[G_{0}],V[G_{1}]$ such that $p_{0}\Vdash\tau_{0}\subset X$, $\tau_{1}\subset
Y$ is a finite set, and $\langle
p_{0},p_{1}\rangle\Vdash\bigcap_{y\in\tau_{0}}C_{y}=C(V[G_{0}][H_{0}],\tau_{1})$;
I must produce a ground model coded closed set $D\subset X$ such that
$p_{0}\Vdash\bigcap_{y\in\tau_{0}}C_{y}=D$.

Working in $V[G_{1}]$, form the closed set $A\subset X$ as
$A=X\setminus\bigcup\\{O\colon O\subset X$ is open and $p_{1}\Vdash
O\cap\tau_{1}=0$. By the initial assumptions on the models $V[G_{0}]$ and
$V[G_{1}]$, $C(V[G_{0}],A)=C(V,A)$ holds; write $D$ for the common value.
Observe that $p_{1}\Vdash\tau_{1}\subset D$. It will be enough to show that
$p_{0}\Vdash\bigcap_{y\in\tau_{0}}C_{y}=D$.

Since $p_{1}\Vdash\tau_{1}\subset D$, the only way how the equality can fail
is that there is a condition $p^{\prime}_{0}\leq p_{0}$ forcing $\tau_{0}$ to
be a proper subset of $D$. Working in $V[G_{0}]$ under this assumption, let
$M_{0}$ be a countable elementary submodel of some large structure containing
$\tau_{0},D$, and $p^{\prime}_{0}$. Let $h_{0}\subset P_{0}\cap M_{0}$ be a
filter generic over the model $M_{0}$ and let $b=\tau_{0}/h_{0}$. The set
$E=\bigcap_{y\in b}C_{y}$ is a proper subset of $D$, so $A\subseteq E$ fails.
Thus, there must be a basic open set $O\subset X$ disjoint from $E$ which
contains some element of the set $A$. By the definitions, this means that
there is a condition $p^{\prime\prime}_{0}\leq p^{\prime}_{0}$ in the filter
$h_{0}$ which forces $\tau_{0}\cap O=0$, and a condition $p^{\prime}_{1}\leq
p_{1}$ which forces $\tau_{1}\cap O\neq 0$. This contradicts the initial
assumptions on the conditions $p_{0},p_{1}$. ∎

######  Corollary 2.10.

Mutually generic extensions are mutually Noetherian.

The following more elaborate product forcing feature will come handy as well.

######  Proposition 2.11.

Suppose that $V[G_{0},G_{1}]$ is a generic extension. Suppose that
$V[G_{0},G_{1}][K_{0}]$ and $V[G_{0},G_{1}][K_{1}]$ are mutually generic
extensions of $V[G_{0},G_{1}]$ and $H_{0}\in V[G_{0},G_{1}][K_{0}]$ and
$H_{1}\in V[G_{0},G_{1}][K_{1}]$ are filters such that

  1. 1.

$G_{0}\in V[H_{0}]$ and $G_{1}\in V[H_{1}]$;

  2. 2.

$V[G_{0},G_{1}]$ is Noetherian to $V[H_{0}]$ over $V[G_{0}]$;

  3. 3.

$V[G_{1}]$ is Noetherian to $V[H_{0}]$ over $V$.

Then $V[H_{1}]$ is Noetherian to $V[H_{0}]$ over $V$. Similarly for
$K_{\sigma}$-Noetherian.

###### Proof.

Let $X,Y$ be Polish spaces, $C\subset Y\times X$ a Noetherian basis coded in
$V$, and suppose that $A\subset X$ is a set in $V[H_{1}]$. To see that
$C(V[H_{0}],A)=C(V,A)$, consider the set $B=C(V[G_{0},G_{1}],A)$. By the
mutual genericity of $K_{0},K_{1}$ and Proposition 2.9,
$C(V[G_{0},G_{1},K_{0}],A)=B$. Now, observe that equality
$C(V[H_{0}],A)=C(V[H_{0}],B)$ must hold: the left-to-right inclusion follows
from $A\subseteq B$, and the right-to-left inclusion follows the fact that
$B=C(V[G_{0},G_{1},K_{0}],A)\subseteq C(V[H_{0}],A)$. By (2),
$C(V[H_{0}],B)=C(V[G_{0},B])$ holds. Observe that
$C(V[G_{0}],A)=C(V[G_{0}],B)$ must hold: the left-to-right inclusion follows
from $A\subseteq B$ again, and the right-to-left follows from the fact that
$B=C(V[G_{0},G_{1}],A)\subseteq C[V[G_{0}],A)$. By (3), $C(V[G_{0}],A)=C(V,A)$
holds. In conclusion, $C(V[H_{0}],A)=C(V,A)$ holds as required. ∎

The following proposition collects several useful features of mutually
Noetherian extensions.

######  Proposition 2.12.

Let $V[G_{0}],V[G_{1}]$ be mutually Noetherian extensions. Then

  1. 1.

$2^{\omega}\cap V[G_{0}]\cap V[G_{1}]=2^{\omega}\cap V$;

  2. 2.

let $X$ be a Polish field, $p(\bar{v}_{0},\bar{v}_{1})$ a polynomial with all
parameters in $V$ and all free variables listed. Let $\bar{x}_{0}\in V[G_{0}]$
and $\bar{x}_{1}\in V[G_{1}]$ be tuples such that $X\models
p(\bar{x}_{0},\bar{x}_{1})=0$. Then there is a tuple $\bar{x}_{1}^{\prime}\in
V$ such that $X\models p(\bar{x}_{0},\bar{x}^{\prime}_{1})=0$;

  3. 3.

let $\phi(\bar{v}_{0},\bar{v}_{1})$ be a formula of the language of real
closed fields with real parameters in $V$, with all free variables listed. Let
$V[G_{0}],V[G_{1}]$ be mutually Noetherian extensions and $\bar{x}_{0}\in
V[G_{0}]$ and $\bar{x}_{1}\in V[G_{1}]$ be tuples of reals such that
$\mathbb{R}\models\phi(\bar{x}_{0},\bar{x}_{1})$ holds. Then there is a tuple
$\bar{x}^{\prime}_{1}\in V$ of reals such that
$\mathbb{R}\models\phi(\bar{x}_{0},\bar{x}^{\prime}_{1})$ holds.

###### Proof.

For (1), let $X=Y=2^{\omega}$ and let $C\subset Y\times X$ be the diagonal;
this is a Noetherian basis. Now, suppose that $x\in X\cap V[G_{0}]\cap
V[G_{1}]$ is a point. By the Noetherian assumption,
$C(V[G_{0}],\\{x\\})=C(V,\\{x\\})=\\{x\\}$. Since the set $C(V,\\{x\\})$ is
coded in $V$, this is possible only if $x\in V$.

For (2), let $n_{0}=|\bar{v}_{0}|$ and $n_{1}=|\bar{v}_{1}|$. Consider the
closed set $C\subset X^{n_{0}}\times X^{n_{1}}$ consisting exactly the pairs
$\langle\bar{y}_{0},\bar{y}_{1}\rangle$ if $p(\bar{y}_{0},\bar{y}_{1})=0$;
this is a Noetherian subbasis by the Hilbert Basis Theorem. The Noetherian
assumption shows that $C(V[G_{0}],\\{\bar{x}_{1}\\})=C(V,\\{\bar{x}_{1}\\})$.
Any tuple $x^{\prime}_{1}$ in this set in the ground model will work as in
(2), since $C(V[G_{0}],\bar{x}_{1})\subseteq\\{\bar{y}_{1}\colon
p(\bar{x}_{0},\bar{y}_{1})=0\\}$.

For (3), let $n_{0}=|\bar{v}_{0}|$ and $n_{1}=|\bar{v}_{1}|$. Use the
quantifier elimination theorem for real closed fields [6, Theorem 3.3.15] to
assume that $\phi$ is quantifier-free. Then $\phi$ is a boolean combination of
statements of the form $p(\bar{v}_{0},\bar{v}_{1})>0$ and
$p(\bar{v}_{0},\bar{v}_{1})=0$ for some polynomials $p$ with coefficients in
$V$. Let $p_{i}$ for $i\in j$ be a list of all polynomials used in this
boolean combination. Let $a\subset j$ be the set of all indices $i$ such that
$p_{i}(\bar{x}_{0},\bar{x}_{1})=0$ and let
$q(\bar{v}_{0},\bar{v}_{1})=\Sigma_{i\in
a}p_{i}(\bar{v}_{0},\bar{v}_{1})^{2}$. Consider the set
$C\subset\mathbb{R}^{n_{0}}\times\mathbb{R}^{n_{1}}$ consisting of all pairs
$\langle\bar{y}_{0},\bar{y}_{1}\rangle$ such that
$q(\bar{y}_{0},\bar{y}_{1})=0$. $C$ is a Noetherian subbasis by the Hilbert
Basis Theorem, and by the Noetherian assumption on the models
$V[G_{0}],V[G_{1}]$, $C(V[G_{0}],\bar{x}_{1})=C(V,\bar{x}_{1})$. Let
$O\subset\mathbb{R}^{n_{1}}$ be a rational open box containing $\bar{x}_{0}$
such that for each $i\in j\setminus a$, the values
$p_{i}(\bar{x}_{0},\bar{y}_{1})$ have the same sign for all $\bar{y}_{1}\in
O$. By a Mostowski absoluteness argument, there must be a point
$\bar{x}^{\prime}_{1}\in O\cap C(V,\bar{x}_{1})$ in $V$ since there is such a
point $\bar{x}_{1}$ in $V[G_{1}]$. The point $\bar{x}^{\prime}_{1}$ works as
required. ∎

It may seem difficult to verify that given two generic extensions are mutually
Noetherian. In this paper, this is always done using the following
_duplication criterion_.

######  Definition 2.13.

Suppose that $P$ is a poset, $\tau_{0},\tau_{1}$ are $P$-names for subsets of
the ground model. Say that $\tau_{0}$ is _duplicable over_ $\tau_{1}$ in $P$
if for every condition $p\in P$ there is a generic extension $V[K]$ and in it,
a sequence $\langle H_{\alpha}\colon\alpha\in\kappa\rangle$ such that

  1. 1.

$\kappa$ is an uncountable ordinal in $V[K]$;

  2. 2.

each $H_{\alpha}\subset P$ is a filter generic over the ground model
containing the condition $p$;

  3. 3.

$\tau_{1}/H_{\alpha}$ is the same for all $\alpha\in\kappa$;

  4. 4.

for disjoint finite sets $a,b\subset\kappa$,
$V[\tau_{0}/H_{\alpha}\colon\alpha\in a]\cap
V[\tau_{0}/H_{\alpha}\colon\alpha\in b]=V$.

######  Proposition 2.14.

Suppose that $P$ is a poset, $\tau_{0},\tau_{1}$ are $P$-names for subsets of
the ground model, and $\tau_{0}$ is duplicable over $\tau_{1}$. Then $P$
forces $V[\tau_{0}]$ to be Noetherian to $V[\tau_{1}]$ over $V$.

###### Proof.

Suppose towards a contradiction that the conclusion fails. Then there must be
a condition $p\in P$, Polish spaces $X,Y$, a Noetherian subbasis $C\subset
Y\times X$ coded in $V$, and a name $\sigma$ for a subset of $X$ in the model
$V[\tau_{1}]$ such that $p$ forces $C(V[\tau_{0}],\sigma)$ to be strictly
smaller than $C(V,\sigma)$. Basic forcing theory shows that $V[\tau_{1}]$ is a
generic extension of the ground model, and that we may assume that there is a
poset $Q_{1}$ such that $p\Vdash\tau_{1}\subset\check{Q}_{1}$ is a filter
generic over the ground model. We also may assume that $\sigma$ is in fact a
$Q_{1}$-name.

Move to a generic extension $V[K]$ in which a sequence $\langle
H_{\alpha}\colon\alpha\in\kappa\rangle$ satisfies the items of Definition
2.13. Write $A\subset X$ for the common value of
$\sigma/(\tau_{1}/H_{\alpha})$ for all ordinals $\alpha\in\kappa$ and write
$C_{\alpha}=C(V[H_{\alpha}],A)$. For each ordinal $\beta\in\omega_{1}$ let
$D_{\beta}=\bigcap_{\alpha\in\omega_{1}\setminus\beta}C_{\beta}$. The sequence
$\langle D_{\beta}\colon\beta\in\omega_{1}\rangle$ is an inclusion-increasing
uncountable sequence of closed subsets of $X$, and as such it has to stabilize
at some ordinal $\beta_{0}$. Write $D\subset X$ for the stable value. Use the
Noetherian property of the subbasis $C$ to find a finite set
$b_{0}\subset\omega_{1}\setminus\beta_{0}$ such that $D=\bigcap_{\alpha\in
b_{0}}C_{\alpha}$. Let $\beta_{1}\in\omega_{1}$ be an ordinal larger than
$\max(b_{0})$ and find a finite set $b_{1}\subset\omega_{1}\setminus\beta_{1}$
such that $D=\bigcap_{\alpha\in b_{1}}C_{\alpha}$. By the intersection
assumption on the sequence of the generic extensions,
$V[G_{\alpha}\colon\alpha\in b_{0}]\cap
V[G_{\alpha}\colon\alpha\in\beta_{1}]=V$ must hold. Thus, the set $D$ is in
fact coded in $V$. Now, for every ordinal
$\alpha\in\omega_{1}\setminus\beta_{0}$ it is the case that $A\subseteq
D\subseteq C_{\alpha}$. The definition of the set $C_{\alpha}$ then shows that
$D=C_{\alpha}$ and the proposition follows. ∎

##  3 Examples I

This section contains several example of mutually duplicable names in forcing,
which by Proposition 2.14 always lead to mutually Noetherian pairs of
extensions. First, a warm-up example.

######  Example 3.1.

If $Q_{0},Q_{1}$ are posets with the respective names $\tau_{0},\tau_{1}$ for
their generic filters and $P=Q_{0}\times Q_{1}$, then $\tau_{0},\tau_{1}$ are
mutually duplicable in $P$. To see that $\tau_{0}$ is duplicable over
$\tau_{1}$, let $\kappa$ be the successor of the maximum of $|Q_{0}|$ and
$|Q_{1}|$. For every condition $p=\langle q_{0},q_{1}\rangle\in P$, consider
the product forcing of $\kappa$-many copies of $Q_{0}\restriction q_{0}$ and a
single copy of $Q_{1}\restriction q_{1}$, yielding filters
$G_{0}^{\alpha}\subset Q_{0}$ and $G_{1}\subset Q_{1}$. Let
$H_{\alpha}=G_{0}^{\alpha}\times H_{1}$ and observe that for finite disjoint
sets $a,b\subset\kappa$, the intersection $V[G_{0}^{\alpha}\colon\alpha\in
a]\cap V[G_{0}^{\alpha}\colon\alpha\in b]$ is equal to $V$ by the product
forcing theorem.

The following two classes of examples deal with Cohen forcing. Recall that if
$X$ is a Polish space then $P_{X}$ is the poset of nonempty open subsets of
$X$ ordered by inclusion; it adds a single _Cohen generic_ point
${\dot{x}}_{\mathit{gen}}$ which belongs to all sets in the generic filter. If
$f\colon X\to Y$ is a continuous open map from $X$ to $Y$ then $P_{X}$ forces
$\dot{f}({\dot{x}}_{\mathit{gen}})$ to be $P_{Y}$ generic over the ground
model by [5, Proposition 3.1.1].

The first Cohen forcing example deals with product graphs. If $X_{n}$ for
$n\in\omega$ are countable sets and $\Gamma_{n}$ are graphs on each, then the
product $\prod_{n}\Gamma_{n}$ is the graph on $X=\prod_{n}X_{n}$ connecting
sequences $x_{0},x_{1}$ if for every $n\in\omega$,
$x_{0}(n)\mathrel{\Gamma}_{n}x_{1}(n)$ holds.

######  Example 3.2.

Let $\Gamma_{n}$ be graphs on nonempty countable sets $X_{n}$, each of which
is connected and not bipartite. Let $X=\prod_{n}X_{n}$ and
$\Gamma=\prod_{n}X_{n}$. The poset $P_{\Gamma}$ adds a generic pair
$\langle\tau_{0},\tau_{1}\rangle\in\Gamma$. The names $\tau_{0},\tau_{1}$ are
mutually duplicable.

Note that the assumptions on the graphs $\Gamma_{n}$ are best possible. If
they are not connected, then the sequence of the $\Gamma_{n}$-components of
$\tau_{0}(n)$ belongs to both $V[\tau_{0}]$ and $V[\tau_{1}]$ and not to $V$.
If the graphs $\Gamma_{n}$ are bipartite as witnessed by partitions
$X_{n}=A_{n0}\cup A_{n1}$ for each $n\in\omega$, then the binary sequence
$y\in 2^{\omega}$ defined by $y(n)=i$ if $\tau_{0}(n)\in A_{ni}$ belongs to
$V[\tau_{0}]\cap V[\tau_{1}]$ (the same definition applied to $\tau_{1}$
yields the sequence $1-y$) but not to $V$. In both cases, the conclusion shows
that the models $V[\tau_{0}]$ and $V[\tau_{1}]$ are forced not to be mutually
Noetherian by Proposition 2.12(1).

###### Proof.

Let $P$ be the combinatorial presentation of the poset $P_{X}$: its conditions
are finite sequences $p$ such that $\forall n\in\mathrm{dom}(p)$, $p(n)\in
X_{n}$, ordered by reverse inclusion. The following technical claim uses the
full strength of initial assumptions on the graphs:

######  Claim 3.3.

Let $p\in P$ and let $\sigma$ be a $P$-name for a set of ordinals not in the
ground model. Then there is an ordinal $\alpha$ and conditions
$p_{0},p_{1}\leq p$ such that

  1. 1.

$\mathrm{dom}(p_{0})=\mathrm{dom}(p_{1})$, there is a unique
$n\in\mathrm{dom}(p_{0})$ such that $p_{0}(n)\neq p_{1}(n)$, and there is
$x\in X_{n}$ $\Gamma_{n}$ related to both $p_{0}(n)$ and $p_{1}(n)$;

  2. 2.

$p_{0}\Vdash\check{\alpha}\in\sigma$ and
$p_{1}\Vdash\check{\alpha}\notin\sigma$.

###### Proof.

First, since $\sigma$ is forced not to be in the ground model, there must be
an ordinal $\alpha$ and conditions $p_{0}^{0},p_{0}^{1}\leq p$ with the same
domain $m\in\omega$ such that $p_{0}^{0}\Vdash\check{\alpha}\in\sigma$ and
$p_{0}^{1}\Vdash\check{\alpha}\notin\sigma$. The next move is to find such two
conditions with the same domain and so that they differ in exactly one entry.

For this, for each condition $q\leq p_{0}^{0}$ and every number $k\leq m$,
write $q_{n}$ for the condition obtained from $q$ by rewriting the entry
$q(k)$ by $p_{0}^{1}(k)$ for every $k\in n$. Strengthening repeatedly the
condition $p_{0}^{0}$, find $q\leq p_{0}^{0}$ such that for every number
$n\leq m$ the condition $q_{n}$ decides the statement
$\check{\alpha}\in\sigma$. Since $q_{0}\Vdash\check{\alpha}\in\sigma$ and
$q_{m}\Vdash\check{\alpha}\notin\sigma$, there must be a number $n\in m$ such
that $q_{n}\Vdash\check{\alpha}\in\sigma$ and
$q_{n+1}\Vdash\check{\alpha}\notin\sigma$. The conditions $q_{n}$ and
$q_{n+1}$ are as required.

Thus, there must be an ordinal $\alpha$, a number $n\in\omega$, and conditions
$p_{0}^{0},p_{0}^{1}\leq p$ with the same domain $m\in\omega$ such that
$p_{0}^{0}\Vdash\check{\alpha}\in\sigma$ and
$p_{0}^{1}\Vdash\check{\alpha}\notin\sigma$ and $p_{0}^{0},p_{0}^{1}$ differ
only at entry $n$. Now, let $\\{x_{i}\colon i\in\omega\\}$ be an enumeration
of $X_{n}$ with possible repetitions. By recursion on $i\in\omega$ build
conditions $q_{i}\in P$ such that

  * •

$p_{0}^{0}=q_{0}\geq q_{1}\geq\dots$;

  * •

writing $q^{\prime}_{i}$ for the condition obtained from $q_{i}$ by replacing
its $n$-th entry with $x_{i}$, $q^{\prime}_{i}$ decides the membership of
$\alpha$ in $\sigma$.

Let $a=\\{x_{i}\colon q^{\prime}_{i}\Vdash\check{\alpha}\in\sigma\\}$ and
$b=\\{x_{i}\colon q^{\prime}_{i}\Vdash\check{\alpha}\notin\sigma\\}$. This is
a partition of $X_{n}$ into two nonempty sets by the choice of the condition
$p_{0}^{0}$. Since the graph $\Gamma_{n}$ is not bipartite, one of them (say
$a$) must contain an edge. Let $c\subseteq a$ be a maximal
$\Gamma_{n}$-connected component of $a$ containing more than one element.
Since the graph $\Gamma_{n}$ is connected, there must be a point $x_{j}$ which
a $\Gamma_{n}$-neighbor of $c$, not in $c$ itself. Then $x_{j}\in b$ and there
must be elements $x_{k},x_{l}\in c$ such that
$x_{l}\mathrel{\Gamma}_{n}x_{k}\mathrel{\Gamma}_{n}x_{j}$.

Let $i=\max\\{j,l\\}$ and consider the conditions $p_{0},p_{1}$ obtained from
$q_{i}$ by replacing its $n$-th entry with $x_{l}$ and $x_{j}$ respectively.
The conditions $p_{0},p_{1}$ and the ordinal $\alpha$ are as required. ∎

Now, let $a$ be a finite set, and consider the posets $Q_{a}$ and $R_{a}$.
$Q_{a}$ is a set of tuples $q$ such that for some $m=m_{q}$, $q\colon a\to P$
is a function and for every $i\in a$, $\mathrm{dom}(q(i))=m$, and for every
$n\in m$ there is a point $x\in X_{n}$ which is $\Gamma_{n}$ related to all
points $q(i)(n)$ for all $i\in a$. The ordering is coordinatewise inclusion.
The poset $Q_{a}$ adds a generic point in $X^{a}$ which is the coordinatewise
union of the conditions in the generic filter. $R_{a}$ is the collection of
tuples $r=\langle p_{r},q_{r}\rangle$ where $p_{r}\in P$ is a tuple with
domain $m_{r}$, $q_{r}\in Q_{a}$ is a condition with $m_{q_{r}}=m_{r}$, and
for each $n\in m_{r}$ and each $i\in a$,
$p_{r}(n)\mathrel{\Gamma}_{n}q_{r}(i)(n)$ holds. The ordering in $r$ is
coordinatewise again. The poset $R_{a}$ adds points $\tau\in X$ and
$\sigma_{a}\in X^{a}$ which are the union of first and coordinatewise union of
second conditions in the generic filters respectively.

######  Claim 3.4.

$R_{a}$ forces the following:

  1. 1.

$\tau$ is $P_{X}$-generic over the ground model and $\sigma_{a}$ is
$Q_{a}$-generic over the ground model;

  2. 2.

if $b\subset a$ then $\langle\tau,\sigma_{a}\restriction b\rangle$ is
$R_{b}$-generic over the ground model;

  3. 3.

$V[\tau]\cap V[\sigma_{a}]=V$.

###### Proof.

The first two items are elementary density arguments. For the last item,
suppose that $r\in R$ is a condition, $\eta$ is a $Q_{a}$-name for a new set
of ordinals and $\theta$ is a $P_{X}$-name for a new set of ordinals. I must
find a condition $s\leq r$ and an ordinal $\alpha$ such that
$s\Vdash\check{\alpha}\in(\eta/\sigma_{a})\Delta(\theta/\tau)$.

To this end, first find an ordinal $\alpha$ and conditions $p_{0},p_{1}\leq
p_{r}$ in the poset $P$ as in Claim 3.3. Let $n\in\omega$ be the unique number
such that $p_{0}(n)\neq p_{1}(n)$, and let $x\in X_{n}$ be a point which is
$\Gamma_{n}$-related to both $p_{0}(n)$ and $p_{1}(n)$. Let $q^{\prime}\leq
q_{r}$ be a condition in $Q_{a}$ such that for each $i\in a$ the domain of
$q^{\prime}(i)$ is equal to that of $p_{0}$, for each
$m\in\mathrm{dom}(q^{\prime}(i))$ the point $q^{\prime}_{i}(m)$ is
$\Gamma_{m}$-related to $p_{0}(m)$, and $q^{\prime}_{i}(n)=x$. Let
$q^{\prime\prime}\leq q^{\prime}$ be a condition in $Q_{a}$ deciding the
membership of $\alpha$ in $\eta$; for definiteness, assume that
$q^{\prime\prime}\Vdash\check{\alpha}\in\eta$. It is not difficult to see that
there is a condition $s\leq r$ such that $p_{s}\leq p_{1}$ and
$q_{s}=q^{\prime\prime}$. Such a condition forces
$\check{\alpha}\in\eta/\sigma_{a}$ and $\check{\alpha}\notin\theta/\tau$ as
desired. ∎

Finally, the ground is ready for a quick proof of the example. Let $S$ be the
combinatorial presentation of the poset $P_{\Gamma}$: it consists of pairs
$s=\langle u_{s},t_{s}\rangle$ where $u_{s},t_{s}\in P$ have the same domain
and for every $n\in\mathrm{dom}(u_{s})$,
$t_{s}(n)\mathrel{\Gamma}_{n}u_{s}(n)$. The ordering is that of reverse
coordinatewise inclusion. Now, suppose that $s\in S$ is a condition. Let
$x_{0}\in X$ be $P$-generic over the ground model, extending $u_{q}$. In the
model $V[x_{0}]$, consider the poset $T$ of all conditions $t\in P$ such that
$t_{q}\subset t$ and for all $n\in\mathrm{dom}(t)$,
$t(n)\mathrel{\Gamma}_{n}x_{0}(n)$. The ordering is that of inclusion. Force
with the finite support product of $\omega_{1}$ many copies of $T$, resulting
in points $x_{1\alpha}\in X$ for $\alpha\in\omega_{1}$.

A brief density argument shows that each pair $\langle
x_{0},x_{1\alpha}\rangle$ is $S$-generic over the ground model, and for every
finite set $a\subset\omega_{1}$, the pair $\langle x_{0},\langle
x_{1\alpha}\colon\alpha\in a\rangle\rangle$ is $R_{a}$-generic over the ground
model. Let $a,b\subset\omega_{1}$ be disjoint finite sets. By a mutual
genericity argument, $V[x_{0}][x_{1\alpha}\colon\alpha\in a]\cap
V[x_{0}][x_{1\alpha}\colon\alpha\in b]\subseteq V[x_{0}]$ must hold. Thus,
$V[x_{1\alpha}\colon\alpha\in a]\cap V[x_{1\alpha}\colon\alpha\in b]\subseteq
V[x_{0}]$. By Claim 3.4(3), $V[x_{1\alpha}\colon\alpha\in a]\cap
V[x_{1\alpha}\colon\alpha\in b]=V$ as desired. ∎

######  Corollary 3.5.

Let $\Gamma_{n}$ be graphs on nonempty countable sets $X_{n}$, each of which
is connected and not bipartite. Let $X=\prod_{n}X_{n}$ and
$\Gamma=\prod_{n}X_{n}$. The poset $P_{\Gamma}$ adds a generic pair
$\langle\tau_{0},\tau_{1}\rangle\in\Gamma$. It forces $V[\tau_{0}]$,
$V[\tau_{1}]$ to be mutually Noetherian extensions.

The second Cohen forcing example deals with turbulent actions of Polish groups
as outlined in [4, Section 13.1].

######  Example 3.6.

Let $\Gamma$ be a Polish group acting turbulently on a Polish space $X$ with
all orbits dense and meager. Let $P=P_{\Gamma}\times P_{X}$, let
$\dot{\gamma}_{\mathrm{gen}}$ be the $P$-name for the $P_{\Gamma}$-generic
point, $\tau_{0}$ the $P$-name for the $P_{X}$-generic point, and
$\tau_{1}=\dot{\gamma}_{\mathrm{gen}}\cdot\tau_{0}$. Then $\tau_{0},\tau_{1}$
are mutually duplicable names in $P$.

###### Proof.

By a symmetry argument, it is enough to show that $\tau_{1}$ is duplicable
over $\tau_{1}$. Suppose that $p=\langle U,O\rangle$ is a condition in the
poset $P$, where $U\subset\Gamma$ and $O\subset X$ are nonempty open sets. Let
$\kappa=\omega_{1}$, let $x\in O$ be a point $P_{X}$-generic over $V$. Force
with a finite support product of $\kappa$-many copies of the poset
$P_{\Gamma}\restriction U$ to obtain a sequence $\langle
g_{\alpha}\colon\alpha\in\kappa\rangle$ of points in $U$ which are in finite
tuples mutually $P_{\Gamma}$-generic over $V[x]$; write
$x_{\alpha}=g_{\alpha}\cdot x$. Each of the filters $H_{\alpha}\subset P$ for
$\alpha\in\kappa$ given by the pair $\langle g_{\alpha},x\rangle$ is
$P$-generic over $V$; I will show that the sequence $\langle
H_{\alpha}\colon\alpha\in\kappa\rangle$ witnesses the duplicability of
$\tau_{1}$ over $\tau_{0}$.

######  Claim 3.7.

Let $a\subset\kappa$ be a finite set. Then $V[x_{\alpha}\colon\alpha\in a]\cap
V[x]=V$.

###### Proof.

Without loss, assume that the set $a$ is nonempty, and write $\beta=\min(a)$.
The point $g_{\beta}^{-1}$ is $P_{\Gamma}$-generic over $V[x_{\beta}]$, and
the tuple $t=\langle g_{\alpha}\cdot g_{\beta}^{-1}\colon\beta\in
a\setminus\\{\beta\\}\rangle$ is generic over $V[x_{\beta}][g_{\beta}^{-1}]$
for a product of the posets $P_{\Gamma}$. Use the product forcing theorem to
conclude that $V[x_{\beta}][g_{\beta}^{-1}]\cap V[x_{\beta}][s]=V[x_{\beta}]$
and therefore $V[x]\cap V[x_{\alpha}\colon\alpha\in a]\subseteq V[x_{\beta}]$.
However, $V[x]\cap V[x_{\beta}]=V$ holds by the turbulence assumption and [5,
Theorem 3.2.]. It follows that $V[x]\cap V[x_{\alpha}\colon\alpha\in a]=V$ as
desired. ∎

######  Claim 3.8.

If $a,b\subset\kappa$ are disjoint finite sets then
$V[x_{\alpha}\colon\alpha\in a]\cap V[x_{\alpha}\colon\alpha\in b]=V$.

###### Proof.

By the product forcing theorem, $V[x][g_{\alpha}\colon\alpha\in a]\cap
V[x][g_{\alpha}\colon\alpha\in b]=V[x]$ holds, and therefore
$V[x_{\alpha}\colon\alpha\in a]\cap V[x_{\alpha}\colon\alpha\in b]\subseteq
V[x]$ holds. By the previous claim, $V[x_{\alpha}\colon\alpha\in a]\cap
V[x]=V$ holds, and in consequence $V[x_{\alpha}\colon\alpha\in a]\cap
V[x_{\alpha}\colon\alpha\in b]=V$ as desired. ∎

The duplicability follows. ∎

######  Corollary 3.9.

The poset $P$ of Example 3.6 forces $V[\tau_{0}]$, $V[\tau_{1}]$ to be
mutually Noetherian extensions.

The following example uses the notion of Suslin forcing which occurs several
times in this paper.

######  Definition 3.10.

[1, Definition 3.6.1] A forcing $Q$ is _Suslin_ if there s an _ambient_ Polish
space $X$ in which the conditions of $Q$ form an analytic set, and the
ordering and incompatibility relations on $Q$ are analytic relations on $X$.

The following fact regarding Suslin forcing will be used repeatedly throughout
the paper.

######  Fact 3.11.

Let $Q$ be a c.c.c. Suslin forcing and let $V[G]$ be a forcing extension. Then

  1. 1.

[1, Theorem 3.6.6] the reinterpretation $Q^{V[G]}$ is a c.c.c. Suslin forcing
in $V[G]$;

  2. 2.

[1, Corollary 3.6.5] if $H\subset Q^{V[G]}$ is a filter generic over $V[G]$,
then $H\cap V\subset Q^{V}$ is a filter generic over $V$.

######  Example 3.12.

Let $Q_{0}$ be an arbitrary forcing and $Q_{1}$ be a Suslin c.c.c. forcing.
Let $P$ be the iteration $Q_{0}*\dot{Q}_{1}$ where the definition of $Q_{1}$
is reinterpreted in the $Q_{0}$-forcing extension. Let $\tau_{0}$ be the
$P$-name for the filter on the first iterand and $\tau_{1}$ be the $P$-name
for the intersection of the filter on the second iterand with the ground
model. Then $\tau_{0},\tau_{1}$ are mutually duplicable names in $P$.

Note that by Fact 3.11(2) $\tau_{1}$ is forced to be a filter on $Q_{1}$
generic over $V$.

###### Proof.

To show that $\tau_{1}$ is duplicable over $\tau_{0}$, let $p=\langle
q_{0},\dot{q}_{1}\rangle$ be a condition in the poset $P$. Let $\kappa$ be a
regular cardinal larger than $|Q_{0}|$. Let $G_{0}\subset Q_{0}$ be a filter
generic over $V$ containing the condition $q_{0}$, and force with the finite
support product of $\kappa$-many copies of the forcing
$Q_{1}\restriction\dot{q}_{1}/G_{0}$ to obtain filters $G_{1\alpha}\subset
Q_{1}$ for $\alpha\in\kappa$. I claim that the filters
$H_{\alpha}=G_{0}*G_{1\alpha}\subset P$ for $\alpha\in\kappa$ witness the
duplicability of $\tau_{1}$ over $\tau_{0}$. To see this, for each
$\alpha\in\kappa$ write $K_{\alpha}=G_{1\alpha}\cap V=\tau_{1}/H_{\alpha}$.
The following claim completes the proof by the product forcing theorem.

######  Claim 3.13.

If $a\subset\kappa$ is a finite set, then $\langle K_{\alpha}\colon\alpha\in
a\rangle$ are filters on $Q_{1}$ which are mutually generic over $V$.

###### Proof.

Consider the poset $R$ which is the product of $a$-many copies of $Q_{1}$. It
is easy to check that $R$ is Suslin, and by Fact 3.11(1), $R$ is c.c.c. The
filters $\langle G_{1\alpha}\colon\alpha\in a\rangle$ form an $R$-generic
sequence over $V[G_{0}]$. By Fact 3.11(2), the sequence $\langle
K_{\alpha}\colon\alpha\in a\rangle$ is an $R$-generic sequence over $V$ as
desired. ∎

To show that $\tau_{0}$ is duplicable over $\tau_{1}$, let $p=\langle
q_{0},\dot{q}_{1}\rangle$ be a condition in the poset $P$. Let $\kappa$ be a
regular cardinal larger than $|Q_{0}|$. Let $s=\langle
G_{0\alpha}\colon\alpha\in\kappa\rangle$ be a mutually generic sequence of
filters on $Q_{0}\restriction q_{0}$. In the model $V[s]$, consider the
(reinterpretation of the) poset $Q_{1}$ and the conditions
$r_{\alpha}=\dot{q}_{1}/G_{0\alpha}$ in it for $\alpha\in\kappa$. Since
$Q_{1}$ is c.c.c. in $V[s]$ by Fact 3.11(1) there must be a condition $r\in
Q_{1}$ which forces that the set $\\{\alpha\in\kappa\colon r_{\alpha}$ belongs
to the generic filter$\\}$ is cofinal in $\kappa$. Let $G_{1}\subset Q_{1}$ be
a filter generic over $V[s]$, containing the condition $r$. By Fact 3.11(2),
for each ordinal $\alpha\in\kappa$, $G_{1\alpha}=G_{1}\cap V[G_{0\alpha}]$ is
a filter on $Q_{1}$ generic over $V[G_{0\alpha}]$. I claim that the filters
$H_{\alpha}=G_{0\alpha}*G_{1\alpha}$ for $\alpha\in\kappa$ such that
$r_{\alpha}\in G_{1}$ witness the duplicability of $\tau_{0}$ over $\tau_{1}$.
This is an immediate corollary of the product forcing theorem applied to the
models $V[G_{0\alpha}]$ for $\alpha\in\kappa$. ∎

######  Corollary 3.14.

Let $\mu$ be a Borel probability measure on a Polish space $X$, and let
$x_{0},x_{1}\in X$ be mutually $\mu$-random elements of $X$. Then the models
$V[x_{0}]$, $V[x_{1}]$ are mutually Noetherian.

##  4 Preservation theorems

Any notion of perpendicularity similar to Definition 2.7 comes with a natural
notion of balance for analytic forcings.

######  Definition 4.1.

Let $P$ be an analytic forcing.

  1. 1.

A pair $\langle Q,\sigma\rangle$ is _Noetherian balanced_ if $Q\Vdash\sigma\in
P$ and for any pair $V[G_{0}]$, $V[G_{1}]$ of mutually Noetherian extensions
of the ground model, every pair $H_{0},H_{1}\subset Q$ of filters generic over
$V$ and for every pair $p_{0}\in V[G_{0}]$, $p_{1}\in V[G_{1}]$ of conditions
stronger than $\sigma/H_{0},\sigma/H_{1}$ respectively and belonging to the
respective models $V[G_{0}],V[G_{1}]$, the conditions $p_{0},p_{1}\in P$ have
a common lower bound.

  2. 2.

$P$ is _Noetherian balanced_ if for every condition $p\in P$ there is a
Noetherian balanced pair $\langle Q,\sigma\rangle$ such that
$Q\Vdash\sigma\leq\check{p}$.

Similar definitions apply for $K_{\sigma}$-Noetherian.

Note that if a poset is $K_{\sigma}$-Noetherian balanced, then it is
Noetherian balanced, since the key quantification over pairs of generic
extensions includes more of them in case of mutually $K_{\sigma}$-Noetherian
pairs.

The supply of mutually Noetherian pairs of extensions provided in the previous
section now makes it possible to prove several preservation theorems. They are
stated using the parlance of [5, Convention 1.7.18]. Thus, given an
inaccessible cardinal $\kappa$, a Suslin poset $P$ is Noetherian balanced
cofinally below $\kappa$ if for every generic extension $V[K_{0}]$ generated
by poset of cardinality smaller than $\kappa$ there is a larger generic
extension $V[K_{1}]$ generated by a poset of cardinality smaller than $\kappa$
such that $V_{\kappa}[K_{1}]\models P$ is Noetherian balanced.

######  Theorem 4.2.

In cofinally Noetherian balanced extensions of the choiceless Solovay model,
if $A_{0},A_{1}\in 3^{\omega}$ are non-meager sets then there are points
$y_{0}\in A_{0}$ and $y_{1}\in A_{1}$ such that the set $\\{i\in\omega\colon
y_{0}(i)=y_{1}(i)\\}$ is finite.

This proves Theorem 1.1(1). I do not know if the conclusion holds also for
non-null sets for the usual Borel probability measure on $3^{\omega}$.

###### Proof.

Let $\kappa$ be an inaccessible cardinal. Let $P$ be a Suslin forcing which is
Noetherian balanced cofinally below $\kappa$. Let $W$ be the choiceless
Solovay model derived from $\kappa$. Work in $W$. Let $p\in P$ be a condition
and let $\tau_{0},\tau_{1}$ be $P$-names such that
$p\Vdash\tau_{0},\tau_{1}\subset 3^{\omega}$ are non-meager sets. I have to
find two points $y_{0},y_{1}$ such that the set $\\{i\in\omega\colon
y_{0}(i)=y_{1}(i)\\}$ is finite and a condition stronger than $p$ which forces
them to $\tau_{0},\tau_{1}$ respectively.

Both $p,\tau_{0},\tau_{1}$ are definable from some elements of the ground
model and an additional parameter $z\in 2^{\omega}$. Let $V[K]$ be an
intermediate extension obtained by a partial order of cardinality less than
$\kappa$ such that $z\in V[K]$, and such that $V[K]\models P$ is Noetherian
balanced. Work in $V[K]$. Let $\langle Q,\sigma\rangle$ be a Noetherian
balanced pair such that $Q\Vdash\sigma\leq p$. Let $R$ be the Cohen poset of
nonempty open subsets of $3^{\omega}$ ordered by inclusion, adding a Cohen
generic point $\dot{y}$.

For each $i\in 2$, there has to be a poset $S_{i}$ of cardinality smaller than
$\kappa$, a $Q\times R\times S_{i}$-name $\eta_{i}$ for a condition in $P$
stronger than $\sigma$, and conditions $q_{i}\in Q$, $r_{i}\in R$ and
$s_{i}\in S_{i}$ which force in the product $Q\times R\times S_{i}$ the
following. In the $\mathrm{Coll}(\omega,<\kappa)$-extension, in the poset $P$,
$\eta_{i}\Vdash\dot{y}\in\tau_{i}$ holds. Otherwise, in the model $W$, if
$H\subset Q$ is a filter generic over $V[K]$ then the condition $\sigma/H$
would force in $P$ that the comeager set of points $R$-generic over $V[K][H]$
to be disjoint from $\tau_{i}$, contradicting the initial assumptions on $p$
and $\tau_{i}$.

In the model $W$, use Corollary 3.5 to produce points $x_{0},x_{1}\in
3^{\omega}$ which are are separately $R$-generic over $V[K]$, such that for
all $i\in\omega$ $x_{0}(i)\neq x_{1}(i)$, and such that the models
$V[K][x(0)]$, $V[K][x_{1}]$ are mutually Noetherian. Let $y_{0}\in 3^{\omega}$
be a finite modification of $x_{0}$ which belongs to $r_{0}$ and let $y_{1}\in
3^{\omega}$ be a finite modification of $x_{1}$ which belongs to $r_{1}$. Let
$H_{0}\subset Q\times S_{0}$, $H_{1}\subset Q\times S_{1}$ be filters mutually
generic over $V[K][x_{0}][x_{1}]$ meeting the respective conditions
$q_{0},q_{1}\in Q$ and $s_{0}\in S_{0}$ and $s_{1}\in S_{1}$. By Proposition
2.9, the models $V[K][y_{0}][H_{0}]$ and $V[K][y_{1}][H_{1}]$ are mutually
Noetherian extensions of $V[K]$. Let $p_{0}=\sigma/y_{0},H_{0}$ and
$p_{1}=\sigma/y_{1},H_{1}$. These are conditions in $P$ in the respective
models stronger than $\sigma/H_{0}$ and $\sigma/H_{1}$ respectively. By the
balance assumption on the pair $\langle Q,\sigma\rangle$, the conditions
$p_{0},p_{1}$ are compatible. By the forcing theorem applied in the respective
models $V[K][y_{0}][H_{0}]$ and $V[K][y_{1}][H_{1}]$, the common lower bound
of these two conditions forces $\check{y}_{0}\in\tau_{0}$ and
$\check{y}_{1}\in\tau_{1}$ as required. ∎

######  Corollary 4.3.

In cofinally Noetherian balanced extensions of the choiceless Solovay model,
there are no non-principal ultrafilters on $\omega$.

###### Proof.

If $U$ is a non-principal ultrafilter on $\omega$, then the map $c\colon
3^{\omega}\to 3$ defined by $c(x)=i$ if $\\{n\in\omega\colon c(n)=i\\}\in U$
partitions of $3^{\omega}$ into three pieces neither of which contains points
$y_{0},y_{1}$ such that the set $\\{i\in\omega\colon y_{0}(i)=y_{1}(i)\\}$ is
finite. One of these pieces must be non-meager. Theorem 4.2 concludes the
proof. ∎

######  Theorem 4.4.

Let $\Gamma\curvearrowright X$ be a turbulent action of a Polish group with
all orbits dense and meager. In cofinally Noetherian balanced extensions of
the choiceless Solovay model, if $A_{0},A_{1}\subset X$ are non-meager sets,
then there are points $x_{0}\in A_{0}$, $x_{1}\in A_{1}$, and
$\Gamma\in\Gamma$ such that $\gamma\cdot x_{0}=x_{1}$.

This proves Theorem 1.1(2).

###### Proof.

Let $\kappa$ be an inaccessible cardinal. Let $P$ be a Suslin forcing which is
Noetherian balanced cofinally below $\kappa$. Let $W$ be the symmetric Solovay
model derived from $\kappa$. Work in $W$. Let $p\in P$ be a condition and let
$\tau_{0},\tau_{1}$ be $P$-names such that $p\Vdash\tau_{0},\tau_{1}\subset X$
are both non-meager sets. I have to find an element $\gamma\in\Gamma$ and
points $x_{0},x_{1}\in X$ such that $\gamma\cdot x_{0}=x_{1}$, and a condition
stronger than $p$ which forces $\check{x}_{0}\in\tau_{0}$ and
$\check{x}_{1}\in\tau_{1}$.

Both $p,\tau$ are definable from some elements of the ground model and an
additional parameter $z\in 2^{\omega}$. Let $V[K]$ be an intermediate
extension obtained by a partial order of cardinality less than $\kappa$ such
that $z\in V[K]$, and such that $V[K]\models P$ is Noetherian balanced. Work
in $V[K]$. Let $\langle Q,\sigma\rangle$ be a Noetherian balanced pair such
that $Q\Vdash\sigma\leq\check{p}$. Let $R$ be the Cohen poset of all nonempty
open subsets of $X$, adding a point ${\dot{x}}_{\mathit{gen}}\in X$. For each
$i\in 2$ there must be a poset $S_{i}$ of cardinality smaller than $\kappa$, a
$Q\times R\times S_{i}$-name $\eta_{i}$ for a condition in $P$ stronger than
$\sigma$, and conditions $q_{i}\in Q$, $r_{i}\in R$ and $s_{i}\in S_{i}$
forcing in the product $Q\times R\times S_{i}$ the following. In the
$\mathrm{Coll}(\omega,<\kappa)$-extension, in the poset $P$,
$\eta_{i}\Vdash{\dot{x}}_{\mathit{gen}}\in\tau_{i}$. Otherwise, in the model
$W$, if $H\subset Q$ is a filter generic over the model $V[K]$, the condition
$\sigma/H\in P$ would force $\tau_{i}$ to be disjoint from the co-meager set
of elements of $X$ Cohen-generic over the model $V[K][H]$. This would
contradict the initial assumptions on the name $\tau$.

Now, since the group $\Gamma$ acts on $X$ continuously and with dense orbits,
thinning out the open set $r_{0}\subset X$ if necessary, one can find an open
set $U\subset\Gamma$ such that $U\cdot r_{0}\subset r_{1}$. In the model $W$,
find points $\gamma\in U$ and $x_{0}\in r_{0}$ which are $P_{\Gamma}\times
R$-generic over $V[K]$, and let $x_{1}=\gamma\cdot x_{0}$. By the turbulence
assumption and Example 3.6, the models $V[K][x_{0}]$ and $V[K][x_{1}]$ are
mutually Noetherian extensions of $V[K]$. Let $H_{0}\subset Q\times S_{0}$ and
$H_{1}\subset Q\times S_{1}$ be filters mutually generic over
$V[K][x_{0}][x_{1}]$ and containing the respective conditions
$q_{0},s_{0},q_{1},s_{1}$. By Proposition 2.9, the models $V[K][x_{0}][H_{0}]$
and $V[K][x_{1}][H_{1}]$ are mutually Noetherian as well.

Let $p_{0}=\eta_{0}/x_{0},H_{0}$ and $p_{1}=\eta_{1}/x_{1},H_{1}$. These are
conditions in $P$ in mutually Noetherian extensions extending the conditions
$\sigma/H_{0}$ and $\sigma/H_{1}$. By the balance assumption on the pair
$\langle Q,\sigma\rangle$, the conditions $p_{0},p_{1}$ have a common lower
bound in $P$. In the model $W$, the common lower bound forces both
$\check{x}_{0}\in\tau$ and $\check{x}_{1}\notin\tau$, while it is also the
case that $\gamma\cdot x_{0}=x_{1}$. This completes the proof of the theorem.
∎

For the following corollary, recall that a tournament on a set $X$ is a an
orianted graph on $X$ which for any two elements of $X$ contains exactly one
ordered pair consisting of the two.

######  Corollary 4.5.

Let $E$ be an orbit equivalence relation of a turbulent action of a Polish
group on a Polish space $Y$. In cofinally Noetherian balanced extensions of
the choiceless Solovay model, there is no tournament on the quotient space
$Y/E$.

###### Proof.

Suppose towards a contradiction that $T$ is a tournament on the $Y/E$-space.
Let $\Gamma$ be a Polish group turbulently acting on $Y$, inducing the orbit
equivalence relation $E$. The product action of $\Gamma\times\Gamma$ on
$Y\times Y$ is easily checked to be turbulent as well, inducing the
equivalence relation $E\times E$. The set $B=\\{\langle y_{0},y_{1}\rangle\in
Y\times Y\colon\langle[y_{0}]_{E},[y_{1}]_{E}\rangle\in T\\}$ is $E\times
E$-invariant, and therefore meager or co-meager by Theorem 4.4. For
definiteness, assume that $B$ is co-meager. Then, the set
$B^{\perp}=\\{\langle y_{0},y_{1}\rangle\in Y\times Y\colon\langle
y_{1},y_{0}\rangle\in B\\}$ is co-meager as well. Let $\langle
y_{0},y_{1}\rangle$ be a pair in the intersection of the two co-meager sets
$B$ and $B^{\perp}$. It follows that both
$\langle[y_{0}]_{E},[y_{1}]_{E}\rangle$ and
$\langle[y_{1}]_{E},y_{0}]_{E}\rangle$ should be in the tournament $T$, an
impossibility. ∎

######  Theorem 4.6.

In cofinally Noetherian balanced $\sigma$-closed extensions of the choiceless
Solovay model, outer Lebesgue measure on $[0,1]$ is continuous in arbitrary
increasing unions.

That is to say, writing $X=[0,1]$ and $\lambda$ for the Lebesgue measure on
$X$ and $\lambda^{*}$ for the outer Lebesgue measure, if
$A\subset\mathcal{P}(X)$ is a collection of sets linearly ordered by
inclusion, then $\lambda^{*}(\bigcup A)=\sup_{B\in A}\lambda^{*}(B)$. This
proves Theorem 1.1(3). One should note that in most Noetherian balanced
extensions, there are non-measurable sets; the conclusion of the theorem does
not rule these out.

###### Proof.

Let $\kappa$ be an inaccessible cardinal. Let $P$ be a $\sigma$-closed Suslin
forcing which is Noetherian balanced cofinally below $\kappa$. Let $W$ be the
choiceless Solovay model derived from $\kappa$. Work in $W$.

Suppose towards a contradiction that $p\in P$ is a condition, $\varepsilon$ is
a real number, $\tau$ is a $P$-name, and the condition $p$ forces $\tau$ to be
a collection of subsets of $X$ which is linearly ordered by inclusion, each
element of $\tau$ has outer Lebesgue measure smaller than $\varepsilon$, and
$\lambda^{*}(\bigcup\tau)>\varepsilon$.

######  Claim 4.7.

There is an open set $O\subset X$ such that $\lambda(O)<\varepsilon$ and a
condition stronger than $p$ forcing $\forall a\in\tau\ \forall U(\lambda(O\cap
U)<\lambda(U)/4\to\lambda^{*}(a\cap U)<\lambda(U)/4$.

###### Proof.

Using DC in $W$, find a descending sequence $\langle p_{n}\colon
n\in\omega\rangle$ starting with $p=p_{0}$ and a sequence
$\langle\dot{a}_{n}\colon n\in\omega\rangle$ such that for every $n\in\omega$,
$p_{n+1}$ forces that $\dot{a}_{n}\in\tau$ and if there is an $a\in\tau$ such
that $\lambda^{*}(a\cap U_{n})\geq 1/4$ then $\dot{a}_{n}$ is such. Use the
$\sigma$-closure assumption on the poset $P$ to find a lower bound
$p_{\omega}$ of the sequence.

Now, $p_{\omega}\Vdash\\{\dot{a}_{n}\colon n\in\omega\\}$ is not a cofinal
subset of $\tau$. The reason is that $\lambda^{*}$ is continuous in increasing
unions of countable length. Therefore, strengthening $p_{\omega}$ if
necessary, it is possible to find a name $\dot{a}_{\omega}$ and an open set
$O\subset X$ such that $\lambda(O)<\varepsilon$ and $p_{\omega}$ forces
$\dot{a}_{\omega}\in\tau$, $\bigcup_{n}\dot{a}_{n}\subset\dot{a}_{\omega}$,
and $\dot{a}_{\omega}\subseteq O$. The condition $p_{\omega}$ and the set
$O\subset X$ are as required. ∎

Let $O\subset X$ be an open set as in the claim, and strengthen the condition
$p$ as to force the conclusion of the claim. Now, the name $\tau$ is definable
from a parameter $z\in 2^{\omega}$ and additional parameters from the ground
model. Let $V[K]$ be an intermediate extension containing $z$, $p$, and $O$
and such that $P$ is balanced in $V[K]$. In the model $V[K]$, let $\langle
Q,\sigma\rangle$ be a Noetherian balanced pair in the poset $P$ such that
$Q\Vdash\sigma\leq p$.

The following considerations take place in the model $W$, using the fact that
every set is Lebesgue measurable in $W$. Let $B=\\{x\in X\colon\exists
G\subset Q$ generic over $V[K]\ \exists p\leq\sigma/G\ p\Vdash
x\in\bigcup\tau\\}$. Note that $\lambda(B)\geq\varepsilon$ since whenever
$G\subset Q$ is generic over $V[K]$, then the condition $\sigma/G$ forces
$\bigcup\tau$ to be a subset of $B$. Thus, $\lambda(B\setminus O)>0$ and by
the Lebesgue density theorem, there must be a basic open set $U\subset X$ such
that $\lambda(B\setminus O\cap U)>3\lambda(U)/4$. Note that $p$ forces that
every element of $\tau$ has relative outer measure in $U$ smaller than $1/4$
by the choice of the open set $O$.

Let $C_{0}=\\{\langle x_{0},x_{1}\rangle\in U\times U\colon\exists
G_{0}\subset Q$ generic over $V[K]\ \exists p_{0}\leq\sigma/G_{0}\ \exists
O_{0}\subset X$ of relative open measure in $U$ smaller than 1/4 such that
$x_{1}\notin O_{0}$ is random over $V[x_{0}][G_{0},p_{0},O_{0}]$ and
$p_{0}\Vdash x_{0}$ belongs to some element of $\tau$ which is a subset of
$O_{0}$.

######  Claim 4.8.

The relative product measure of $C_{0}$ in $U\times U$ is greater than $1/2$.

###### Proof.

Use the Fubini theorem. Let $x_{0}\in B\cap U\setminus O$ be a point random
over $V[K]$–there are relative measure $>3/4$ of such points in the set $U$.
Let $G_{0},p_{0}$ witness the membership of $x_{0}$ in $B$, and strengthen
$p_{0}$ if necessary to find an open set $O_{0}\subset X$ of relative measure
in $U$ smaller than $1/4$ such that $p_{0}\Vdash\check{x}_{0}$ belongs to some
element of $\tau$ which is a subset of $O_{0}$. Now, any point $x_{1}\in U$
which is random over $V[K][x_{0},G_{0},p_{0},O_{0}]$ and does not belong to
$O_{0}$ (and there is more than relative measure $3/4$ of these) belongs to
the vertical section of $C_{x_{0}}$. The relative product measure of $C$ is
thus greater than $3/4\cdot 3/4=1/2$ as desired. ∎

Let $D=\\{\langle x_{0},x_{1}\rangle\in U\times U\colon\langle
x_{0},x_{1}\rangle\in C$ and $\langle x_{1},x_{0}\rangle\in C\\}$. By the
claim, the set $D$ has positive product measure. Working in the model $W$, let
$\langle x_{0},x_{1}\rangle\in D$ be a pair random over $V[K]$. In
$V[K][x_{0},x_{1}]$, there must be a poset $R_{0}$ of cardinality smaller than
$\kappa$ adding witnesses to the statement that $\langle x_{0},x_{1}\in
C_{0}$. There also must be a poset $R_{1}$ of cardinality smaller than
$\kappa$ adding witnesses to the statement that $\langle x_{1},x_{0}\rangle\in
C$. Let $H_{0}\subset R_{0}$ and $H_{1}\subset R_{1}$ be filters mutually
generic over $V[K][x_{0},x_{1}]$. Let $G_{0},p_{0},O_{0}$ be the witnesses to
$\langle x_{0},x_{1}\rangle$ added by $H_{0}$. Let $G_{1},p_{1},O_{1}$ be the
witnesses to $\langle x_{1},x_{0}\rangle$ added by $H_{1}$.

Note that $V[K][x_{0},G_{0},p_{0},O_{0}]$ and $V[K][x_{0},x_{1}]$ are mutually
Noetherian over $V[K][x_{0}]$, and $V[K][x_{0},G_{0},p_{0},O_{0}]$ and
$V[K][x_{1}]$ are mutually Noetherian over $V[K]$ by Example 3.12 applied in
the model $V[K]$ and $V[K][x_{0}]$ respectively to the Suslin c.c.c. poset of
Borel non-null sets ordered by inclusion. The same assertions are true with
$0,1$ interchanged. Proposition 2.11 shows that the models
$V[K][x_{0},G_{0},p_{0},O_{0}]$ and $V[K][x_{1},G_{1},p_{1},O_{1}]$ are
mutually Noetherian.

The balance assumption on the pair $\langle Q,\sigma\rangle$ now shows that
the conditions $p_{0},p_{1}$ are compatible. Their common lower bound forces
that there are two elements of $\tau$, one of them containing $x_{0}$ but not
$x_{1}$, the other containing $x_{1}$ but not $x_{0}$. In other words, $\tau$
is not linearly ordered by inclusion, a contradiction. ∎

######  Corollary 4.9.

In cofinally Noetherian balanced extensions of the choiceless Solovay model,
the Lebesgue null ideal is closed under well-ordered unions.

###### Proof.

Let $\langle B_{\alpha}\colon\alpha\in\kappa\rangle$ be a well-ordered
sequence of Lebesgue null sets. By induction on $\alpha\leq\kappa$ argue that
the union of the sets on the sequence up to $\alpha$ is null, using the
continuity of outer Lebesgue measure under increasing unions at every stage. ∎

##  5 Examples II

In this section, I produce several classes of Noetherian balanced Suslin
forcings. This will verify all items of Theorem 1.2.

######  Example 5.1.

Let $X$ be a Polish field and $F$ be a countable subfield. Let $P$ be the
partial order of countable subsets of $X$ which are linearly over $F$. The
ordering is reverse inclusion [5, Example 6.3.6]. Then $P$ is $\sigma$-closed,
Suslin and Noetherian balanced.

###### Proof.

If $p\subset X$ is a Hamel basis of $X$ over $F$, $V[G_{0}]$ and $V[G_{1}]$
are generic extensions such that $V[G_{0}]\cap V[G_{1}]\cap 2^{\omega}=V\cap
2^{\omega}$ and $p_{0}\in V[G_{0}]$ and $p_{1}\in V[G_{1}]$ are conditions
containing $p$ as a subset, then $p_{0},p_{1}$ are compatible, as the
inspection of the proof of [5, Theorem 6.3.4] reveals. At the same time, if
$V[G_{0}]$ and $V[G_{1}]$ are mutually Noetherian extensions, then
$V[G_{0}]\cap V[G_{1}]\cap 2^{\omega}=V\cap 2^{\omega}$ holds by Proposition
2.12(1). ∎

It is clear that this argument extends to cover most posets which are placid
in the sense of [5, Definition 9.3.1], such as the posets adding a maximal
acyclic subset to a given Borel graph [5, Example 6.3.7], or posets adding a
selector to a given countable Borel equivalence relation.

######  Example 5.2.

Let $X$ be a Polish field and $F$ be a countable subfield. Let $P$ be the
partial order of countable algebraically free over $F$ subsets of $X$, ordered
by reverse inclusion [5, Example 6.3.10]. Then $P$ is $\sigma$-closed, Suslin
and Noetherian balanced.

###### Proof.

Let $p$ be a transcendence basis for $X$ over $F$; it will be enough to show
that $\langle\mathrm{Coll}(\omega,X),\check{p}\rangle$ is a Noetherian
balanced pair. To this end, suppose that $V[G_{0}]$, $V[G_{1}]$ are mutually
Noetherian extensions, containing respective algebraically free sets
$p_{0},p_{1}$ containing $p$; it is necessary to show that $p_{0}\cup p_{1}$
is algebraically free. Towards a contradiction, suppose that this fails and
let $\phi(\bar{v}_{0},\bar{v}_{1})$ be a nonzero polynomial with coefficients
in $F$, and $\bar{x}_{0}\in p_{0}$ and $\bar{x}_{1}\in p_{1}\setminus p_{0}$
are tuples such that $\phi(\bar{x}_{0},\bar{x}_{1})=0$. By Proposition
2.12(3), there is a tuple $\bar{x}_{0}^{\prime}$ in $V$ such that
$\phi(\bar{x}_{0}^{\prime},\bar{x}_{1})=0$ holds. Since $p$ is a transcendence
basis for $X$ over $F$ in $V$, all elements of $\bar{x}^{\prime}_{0}$ are
algebraic over $p$. In conclusion, the tuple $\bar{x}_{1}$ satisfies a
nontrivial algebraic identity over $p$, violating the assumption that $p_{1}$
is a set algebraically free over $F$. ∎

######  Example 5.3.

Let $\Gamma$ be a closed Noetherian graph on a Polish space $X$ with no
uncountable cliques. The coloring poset of [8] is $\sigma$-closed, Suslin and
Noetherian balanced.

###### Proof.

A closed graph $\Gamma$ on a Polish space $X$ is Noetherian if the set
$C\subset X\times X$ consisting of all pairs $\langle x_{0},x_{1}\rangle$ such
that $x_{0}=x_{1}$ or $x_{0}\mathcal{\Gamma}x_{1}$ is a Noetherian subbasis.
An inspection of the proof of balance in [8] shows that the only feature of
mutually generic extensions $V[G_{0}],V[G_{1}]$ used is that for every pair of
$\Gamma$-connected points $x_{0}\in V[G_{0}]$ and $x_{1}\in V[G_{1}]$, there
are points $x_{1}^{\prime}\in V$ $\Gamma$-connected to $x_{0}$ and arbitrarily
close to $x_{1}$. This holds already when the models $V[G_{0}],V[G_{1}]$ are
mutually Noetherian. To see this, consider the set
$C(V[G_{0}],x_{1})=C(V,x_{1})$. All of its points are $\Gamma$-connected (or
equal) to $x_{0}$. If $O\subset X$ is any basic open neighborhood of $x_{1}$,
a Mostowski absoluteness argument shows that there must be a point
$x^{\prime}_{1}\in C(V,x_{1})\cap O$ in $V$ as there is such a point, namely
$x_{1}$, in $V[G_{1}]$. The point $x^{\prime}_{1}$ is as desired. ∎

######  Example 5.4.

Let $\Gamma$ be a $\sigma$-algebraic graph on a Euclidean space $X$ without
uncountable cliques. The coloring poset $P$ of [7] is $\sigma$-closed, Suslin
and Noetherian balanced.

######  Example 5.5.

Let $P$ be the coloring poset for the hypergraph of isosceles triangles in
$\mathbb{R}^{2}$ isolated in [10]. The poset is $\sigma$-closed and Noetherian
balanced.

######  Example 5.6.

Let $\Gamma$ be a $\sigma$-algebraic redundant hypergraph on a Euclidean space
$X$. The coloring poset $P$ for $\Gamma$ in [9] is $\sigma$-closed and
Noetherian balanced.

###### Proof.

An inspection of the proof of balance in the above three examples shows that
the only property of generic extensions $V[G_{0}],V[G_{1}]$ used is the coheir
property of Proposition 2.12(3). ∎

##  6 Examples III

As a final note in this paper, it is interesting to see how the conclusions of
Theorem 1.1 can be violated by balanced posets which are not Noetherian
balanced. The conclusions of the first two items of Theorem 1.1 are easiest to
violate using the following theorem.

######  Theorem 6.1.

Let $\Gamma$ be a Borel meager graph on a Polish space $X$. In a balanced
extension of the Solovay model, there are non-meager sets $A_{0},A_{1}\subset
X$ such that $(A_{0}\times A_{1})\cap\Gamma=0$.

###### Proof.

Without loss, assume that $\Gamma$ is $F_{\sigma}$. Removing the meager set of
all points $x\in X$ such that the set $\\{y\in X\colon x\mathrel{\Gamma}y\\}$
is non-meager from the space $X$ if necessary, we may assume that for the set
$\\{y\in X\colon x\mathrel{\Gamma}y\\}$ is meager for every $x\in X$.

Let $P$ be the partial order of all triples $p=\langle
a_{p},b_{p},c_{p}\rangle$ such that $a_{p},b_{p},c_{p}\subset X$ are all
countable sets, $a_{p}\cap b_{p}=0$, and $(a_{p}\times b_{p})\cap\Gamma=0$.
The ordering is defined by $q\leq p$ if $a_{p}\subseteq a_{q}$,
$b_{p}\subseteq b_{q}$, $c_{p}\subseteq c_{q}$, and no element of $c_{p}$ is
$\Gamma$-connected to any element of $(a_{q}\setminus
a_{p})\cup(b_{q}\setminus b_{p})$. It is not difficult to check that $P$ is a
$\sigma$-closed Suslin poset.

######  Claim 6.2.

$P$ is a balanced forcing.

###### Proof.

Let $p\in P$ be a condition. It will be enough to show that the pair
$\langle\mathrm{Coll}(\omega,X),\langle\check{a}_{p},\check{b}_{p},(X\cap
V)\rangle\rangle$ is balanced in the poset $P$.

Indeed, let $V[G_{0}]$ and $V[G_{1}]$ be mutually generic extensions
containing conditions $p_{0}=\langle a_{0},b_{0},c_{0}\rangle\in V[G_{0}]$ and
$p_{1}=\langle a_{1},b_{1},c_{1}\rangle\in V[G_{1}]$ respectively, both
stronger than $\langle a_{p},b_{p},(X\cap V)\rangle$. To show that $\langle
a_{0}\cup a_{1},b_{0}\cup b_{1},c_{0}\cup c_{1}\rangle$ is a common lower
bound of the conditions $p_{0},p_{1}$, It is only necessary to observe that no
element in $a_{0}\setminus a_{p}$ is $\Gamma$-connected with any element of
$b_{1}\cup c_{1}$; the other verification items are proved in a symmetric way.

Thus, towards a contradiction assume that $x_{0}\in a_{0}\setminus a_{p}$ and
$x_{1}\in b_{1}\cup c_{1}$ are $\Gamma$-connected. Find a closed set
$C\subset\Gamma$ such that $\langle x_{0},x_{1}\rangle\in C$. Let $e_{0}$ be
the set of basic open sets $O\subset X$ such that $\\{x_{0}\\}\times
O\cap\Gamma=0$; let $e_{1}$ be the set of basic open sets $O\subset X$ such
that $x_{1}\in O_{1}$. The sets $e_{0}\in V[G_{0}]$ and $e_{1}\in V[G_{1}]$
are disjoint. By a mutual genericity argument [5, Proposition 1.7.9], there
are disjoint ground model sets $f_{0}\supset e_{0}$ and $f_{1}\supset e_{1}$.
By a Mostowski absoluteness argument between $V[G_{1}]$ and $V$ there is a
point $x^{\prime}_{1}\in V$ such that every basic open neighborhood of it
belongs to $f_{1}$–since the point $x_{1}\in V[G]$ is such. It follows that
$\langle x_{0},x^{\prime}_{1}\rangle\in\Gamma$, and this contradicts the
assumption that $p_{0}\leq\langle a_{p},b_{p},X\cap V\rangle$. ∎

If $G\subset P$ is a filter generic over the choiceless Solovay model (or
indeed any other model), let $A=\\{a_{p}\colon p\in G\\}$ and
$B=\\{b_{p}\colon p\in G\\}$. It is clear from the definition of the poset $P$
that $(A\times B)\cap\Gamma=0$. To complete the proof of the theorem, it is
enough to argue that the sets $A,B\subset X$ are non-meager.

For this, return to the ground model, let $p\in P$ be a condition, and let
$C_{n}\subset X$ be closed nowhere dense sets for $n\in\omega$. By a density
argument, it is only necessary to find a condition $q\leq p$ such that $a_{q}$
contains a point in $X\setminus\bigcup_{n}C_{n}$. The arguument for $b_{q}$ is
the same. To produce the condition $q$, note that the set $D=\\{y\in
X\colon\exists z\in b_{p}\cup c_{p}\ z\mathrel{\Gamma}y\\}$ is meager by the
initial assumptions on the graph $\Gamma$. Just pick a point $x\in X$ which
does not belong to the meager set $\bigcup_{n}C_{n}\cup D\cup b_{p}\\}$, let
$q=\langle a_{p}\cup\\{x\\},b_{p},c_{p}\rangle$, and note that $q\leq p$
holds. The proof is complete. ∎

The conclusion of Theorem 1.1(3) is more difficult to violate.

######  Theorem 6.3.

Let $X$ be a Polish space and $I$ a suitably definable ideal on $X$ such that
every countable subset of $X$ is a subset of a $G_{\delta}$ set in $I$. In a
balanced extension of the Solovay model, $X$ is the union of a collection of
$I$-small sets, linearly ordered by inclusion.

In particular, it is possible to exhaust the unit interval $[0,1]$ as a union
of Lebesgue null sets linearly ordered with respect to inclusion.

###### Proof.

The partial order $P$ consists of conditions
$p=\langle\preceq_{p},b_{p}\rangle$ such that for some countable set
$\mathrm{supp}(p)\subset X$, $\preceq_{p}$ is a linear preorder on
$\mathrm{supp}(p)$ and $b_{p}$ is a countable set of pairs $\langle
x,O\rangle$ where $x\in\mathrm{supp}(p)$ and $O\subset X$ is an open set. In
addition, if $y\preceq_{p}x$ are points in $\mathrm{supp}(p)$ and $\langle
x,O\rangle\in b_{q}$ then $y\in O$. The ordering is defined by $q\leq p$ if
$\mathrm{supp}(p)\subseteq\mathrm{supp}(q)$,
$\preceq_{p}=\preceq_{p}\restriction\mathrm{supp}(p)$, $b_{p}\subseteq b_{q}$,
and for every $y\in\mathrm{supp}(q)$, if $x$ is $\preceq_{q}$-smallest element
of $\mathrm{supp}(p)$ such that $y\preceq_{q}x$, then $x\preceq_{q}y$.

To see the idea behind the definition, suppose that $G\subset P$ is a generic
filter and let $\preceq=\bigcup_{p\in G}\preceq_{p}$. An elementary density
argument shows that $\preceq$ is a linear preorder on all of $X$. The
assumptions on the ideal $I$ imply that for every element $x\in X$, the
initial segment of $\preceq$ determined by $x$ belongs to the ideal $I$: if
$p\in P$ is a condition containing $x$ in its support, there is a $G_{\delta}$
set $\bigcap_{n}O_{n}\in I$ containing all elements of $\mathrm{supp}(p)$, and
then the condition $q\leq p$ obtained from adding the pairs $\langle
x,O_{n}\rangle$ for $n\in\omega$ into $b_{q}$ ensures that the initial segment
determined by $x$ is a subset of $\bigcap_{n}O_{n}$.

It is not difficult to check that the poset $P$ is analytic and
$\sigma$-closed. The balance is the only hard part. Let $p\in P$ be a
condition. Adding some ordered pairs to $b_{p}$ if necessary I may assume that
if $x,y\in\mathrm{supp}(p)$ then

  1. 1.

if $y\preceq_{p}x$ and $\langle x,O\rangle\in b_{p}$ then $\langle
y,O\rangle\in b_{p}$;

  2. 2.

if $x\not\preceq_{p}y$ then $\langle y,2^{\omega}\setminus\\{x\\}\rangle\in
b_{p}$.

For each $y\in X$ let $a(y)=\\{x\in\mathrm{supp}(p)\colon y\in O$ for every
$O$ with $\langle x,O\rangle\in b_{p}\\}$. Note that by for each
$x\in\mathrm{supp}(p)$, $a(x)=[x,\infty)$ by (2) above. Define a relation
$\preceq$ on $2^{\omega}$ by $y_{0}\preceq y_{1}$ if $a_{y_{1}}\subseteq
a_{y_{0}}$. Let also $b$ be the set of all pairs $\langle y,O\rangle$ such
that for some $x\in a(y)$, $\langle x,O\rangle\in b_{p}$.

######  Claim 6.4.

$\mathrm{Coll}(\omega,X)$ forces $\langle\preceq,b\rangle$ is a condition in
$P$ stronger than $p$.

###### Proof.

I first show that $\langle\preceq,b\rangle$ is forced to be a condition in the
poset $P$. First of all, $\preceq$ is clearly a preorder. It is linear: by (1)
above each set $a(y)$ is a cofinal segment of $\preceq_{p}$, and such sets are
linearly ordered by inclusion. Now, suppose that $y\preceq x$ and $\langle
x,O\rangle\in b$. By the definitions, $\langle z,O\rangle\in b_{p}$ for some
$z\in a(x)$. Since $a(x)\subseteq a(y)$, $z\in a(y)$ holds, so $y\in O$ holds
as desired.

To show that $\langle\preceq,b\rangle$ is forced to be stronger than $p$,
first note that $\preceq\restriction\mathrm{supp}(p)=\preceq_{p}$ holds. To
see that, let $x,y\in\mathrm{supp}(p)$ be any points. If $y\preceq_{p}x$ then
$a(x)=[x,\infty)\subseteq a(y)=[y,\infty)$. If $y<x$ are points in
$\mathrm{supp}(p)$ then $x\not\preceq y$ by item (2) above. Finally, note that
if $x$ is the smallest element of $\mathrm{supp}(p)$ such that $y\preceq x$,
then $x\preceq y$. To see that, note that the assumptions imply that
$a(y)=[x,\infty)$. The definition of $\preceq$ implies that $x\preceq y$. ∎

For future reference, note that the set $\mathrm{supp}(p)$ is dense in the
preorder $\preceq$ by the definitions. Also, for each $y\in X$, let
$B_{y}=\bigcap\\{O\colon\langle y,O\rangle\in b\\}$ and observe that
$B_{y}=\\{x\in X\colon x\preceq y\\}$. This feature, that the initial segments
of $\preceq$ which include their supremum are the largest possible
$G_{\delta}$ sets, is the whole purpose of the construction of the preorder
$\preceq$.

Now, I must show that
$\langle\mathrm{Coll}(\omega,X),\langle\preceq,\check{b}\rangle\rangle$ is a
balanced pair. Suppose that $V[G_{0}],V[G_{1}]$ are mutually generic
extensions and $p_{0}=\langle\preceq_{0},b_{0}\rangle$ and
$p_{1}=\langle\preceq_{1},b_{1}\rangle$ be any conditions in these respective
models stronger than $\langle\preceq,b\rangle$. I must find a common lower
bound of conditions $p_{0},p_{1}$. The common lower bound
$p_{01}=\langle\preceq_{01},b_{01}\rangle$ will have
$\mathrm{supp}(p_{01})=\mathrm{supp}(p_{0})\cup\mathrm{supp}(p_{1})$ and
$b_{01}=b_{0}\cup b_{1}$. The definition of the preorder is the more difficult
point.

I will have $\preceq_{0}\cup\preceq_{1}\subset\preceq_{01}$, and, naturally,
if $x\preceq_{0}y\preceq_{1}z$ for $x\in\mathrm{supp}(p_{0})$, $y\in V$ and
$z\in\mathrm{supp}(p_{1})$, then $x\preceq_{01}yz$. Finally suppose that
$X\cap V=c\cup d$ is a cut in the preorder $\preceq$ and containing some
elements of $\mathrm{supp}(p_{0})$ (gathered in the set denoted by $e_{0}$)
and some lements of $\mathrm{supp}(p_{1})$ (gathered in a set $e_{1}$). Note
that this means that $d$ has no smallest element. I must show how to linearly
preorder $e_{0}\cup e_{1}$ in $\preceq_{01}$.

Let $f_{0},f_{1}$ be the respective longest prewellordered initial segments of
$\langle e_{0},\preceq_{0}\rangle$ and $\langle e_{1},\preceq_{1}\rangle$
inside the cut $c\cup d$. For definiteness assume that $f_{1}$ is not shorter
than $f_{0}$. Then, for $x_{0}\in e_{0}$ and $x_{1}\in e_{1}$, put $\langle
x_{0},x_{1}\rangle$ into $\preceq_{01}$ if $x_{0}\in f_{0}$ and $x_{1}$ is
either in $e_{1}\setminus f_{1}$ or $x_{1}$ is in $f_{1}$ at an ordinal rank
at least that of the rank of $x_{0}$ in $f_{0}$. Otherwise, put $\langle
x_{1},x_{0}\rangle$ into $\preceq_{01}$.

######  Claim 6.5.

$p_{01}\in P$.

###### Proof.

Suppose that $x_{0},x_{1}\in\mathrm{supp}(p_{01})$, $x_{0}\preceq_{01}x_{1}$
and $\langle x_{1},O\rangle\in b_{01}$ is a pair. I must show that $x_{0}\in
O$ holds. For definiteness assume that $x_{0}\in\mathrm{supp}(p_{0})$ and
$x_{1}\in\mathrm{supp}(p_{1})$ and $\langle x_{1},O\rangle\in b_{1}$ all hold.

Case 1. There is $y\in V$ such that $x_{0}\preceq_{0}y_{0}\preceq_{1}x_{1}$.
In such a case, in the ground model consider the set $B_{y}$. Since $p_{0}\in
P$, it must be the case that $x_{0}\in B_{y}$. Since $p_{1}\in P$, it must be
the case that $B_{y}\cap V\subset O$, and by a mutual genericity argument
$B_{y}\cap V[G_{0}]\subset O$. It follows that $x_{0}\in O$ holds as desired.

Case 2. Not case 1. Then $x_{0},x_{1}$ are in the same $\preceq$-cut $X\cap
V=c\cup d$. Since $c,d\in V[G_{0}]\cap V[G_{1}]$, a mutual genericity argument
shows that $c,d\in V$. Since $p_{0},p_{1}\leq p$, it must be the case that the
set $d$ has no smallest element. Work in $V$. Note that the set
$\mathrm{supp}(p)$ is coinitial in $d$ and that $c=\bigcap\\{B_{y}\colon
y\in\mathrm{supp}(p)\cap d\\}$ is a $G_{\delta}$-set. Since $p_{0}\in P$, it
must be the case that $x_{0}\in c$. Since $p_{1}\in P$, it must be the case
that $c\cap V\subset O$, and by a mutual genericity argument $c\cap
V[G_{0}]\subset O$. It follows that $x_{0}\in O$ holds as desired. ∎

It is now not difficult to show that $p_{01}$ is a common lower bound of
$p_{0}$ and $p_{1}$. To show for example that $p_{01}\leq p_{0}$, suppose that
$x\in\mathrm{supp}(p_{01})$ and suppose that $y$ is the smallest element of
$\mathrm{supp}(p_{0})$ such that $x\preceq_{01}y$; it must be proved that
$y\preceq_{01}x$ holds. If $x\in\mathrm{supp}(p_{0})$ then this is clear from
the fact that $\preceq_{0}$ is a preorder. Suppose then that
$x\in\mathrm{supp}(p_{1})$ holds. If $y$ is $\preceq_{0}$-equivalent to some
element in $V$, then $y\preceq_{01}x$ follows from
$p_{1}\leq\langle\preceq,b\rangle$. Suppose then that $y$ is not
$\preceq_{0}$-equivalent to any point in $V$. Let $X\cap V=c\cup d$ be the cut
in $\preceq$ into which both $x,y$ fall. It cannot be the case that $y$ falls
into the illfounded part of $\mathrm{supp}(p_{0})$ inside this cut, since that
illfounded part is convex in $\preceq_{01}$ and does not have a smallest
element. Thus, $y$ falls into the longest pre-well-ordered initial segment of
this cut in $\mathrm{supp}(p_{0})$. Since $x\preceq_{01}y$ holds, it must be
the case that $x$ falls into the pre-well-ordered segment of this cut in
$\mathrm{supp}(p_{1})$ and the ordinal rank of $x$ and $y$ in these two must
be the same. Again, $y\preceq_{01}x$ holds as desired. ∎

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