# Mathias and Silver forcing parametrized by density

Giorgio Laguzzi, Heike Mildenberger, Brendan Stuber-Rousselle

(Date: Feb. 9, 2021)

###### Abstract.

We define and investigate versions of Silver and Mathias forcing with respect
to lower and upper density. We focus on properness, Axiom A, chain conditions,
preservation of cardinals and adding Cohen reals. We find rough forcings that
collapse $2^{\omega}$ to $\omega$, while others are surprisingly gentle. We
also study connections between regularity properties induced by these
parametrized forcing notions and the Baire property.

##  1\. Introduction

Forcings consisting of tree conditions are the _dramatis personae_ extensively
studied in descriptive set theory and set theory of the reals. The original
interest on this type of forcings was mainly due to their applications in
resolving problems of independence risen from the study of cardinal
characteristics, infinitary combinatorics and regularity properties. Through
the years, the study of the combinatorial properties of tree-forcings has been
intensified and has been interesting on its own.

From the forcing point of view, trees can be thought as conditions whose stem
decides a finite fragment of the generic real, and the rest of the tree above
describes the possible paths of the generic real. In this sense, Cohen forcing
can be understood as the simplest tree-forcing notion, whose conditions decide
a finite fragment of the generic and then leave all paths extending the stem
possible. In general, the fatter the tree, the more freedom the path of the
generic real; the slimmer the tree, the more restrictive the conditions on the
path of the generic real. Under this point of view a Sacks tree can be seen as
the other extreme, since in this case the tree can be shrunk as much as
desired, and the only requirement is to keep perfectness.

Many other tree-forcings in between have been introduced and extensively
studied, among them Silver and Mathias trees111The nodes of the Mathias (or
Silver) tree $p$ are all $t\in 2^{<\omega}$ that are initial segments of one
of the possible generic branches that are compatible with a Mathias condition
$(s,A)$ (or a Silver condition $f_{p}$).. In this paper we focus on studying
some variants where the set of nodes above the stem are governed by
restrictions imposed on the density of the set of splitting nodes. Imposing
fatness or slimness conditions though, may result in a non-proper forcing.

Each condition in Mathias forcing and in Silver forcing, when conceived as a
tree, comes with an infinite set $A_{p}$ such that any node $t$ of $p$ whose
length is in $A_{p}$ is a splitting node. Also other tree forcings, like Sacks
or Miller, can be thinned out to the subsets of those conditions who come with
such a set $A_{p}$ of overall splitting levels. In Bukovský-Namba forcing the
conditions with uniform splitting levels are even dense, see e.g. [4, Theorem
2.2].

We let the $A_{p}$ range over prescribed families that are sets of positive
lower or upper density in $\omega$. There is some work on forcings of this
type. Grigorieff [9] parametrised Silver forcing by having the $A_{p}$ range
over a $P$-point and Halbeisen [10, Ch 24] generalised this to $P$-families.
Mathias forcing with a Ramsey ultrafilter [17] is a versatile notion of
forcing. Farah and Zapletal [8, Ch. 9] used the coideal of the density zero
ideal as a reservoir for the infinite component of Mathias conditions.

In Sections 2 to 5 we investigate whether the variants we introduce of Silver
and of Mathias forcing are proper at all. We show that Mathias with lower
density $\geq\varepsilon$ is a disjoint union of $\sigma$-centered forcings
(see Proposition 3.6) and that Silver forcing with positive lower density
collapses $2^{\omega}$ to $\omega$ (see Theorem 5.1). We also show that the
lower density is far from the notion of a measure (see Proposition 5.3). This
result fits in the framework presented in [15].

In Section 6 and 7 we are concerned with regularity properties of these
forcings. Regularity properties of tree-forcings by themselves are a well
studied field in mathematics. The following table illustrates which of the
most popular regularity property of a subset of $2^{\omega}$ correspond to
which tree-forcing $\mathbb{P}$.

Tree-forcing $\mathbb{P}$ |  $\mathbb{P}$-measurable
---|---
Cohen $\mathbb{C}$ | Baire property
Random $\mathbb{B}$ | Lebesgue measurable
Silver $\mathbb{V}$ | completely doughnut [18]
Sacks $\mathbb{S}$ | Marczewski set [22]
Mathias $\mathbb{MA}$ | completely Ramsey

In many cases adding a Cohen real is enough to establish a dependence between
Baire property and the measurability given by the tree-forcing. This was made
explicit in [14, Proposition 3.1.]. Proposition 7.1 is an improvement of this
result.

In Section 8 we construct a model in which all $\text{On}^{\omega}$-definable
sets are $\mathbb{V}^{+}_{\varepsilon}$-measurable but
$\Sigma^{1}_{2}(\mathbb{C})$ fails. In particular
$\Sigma^{1}_{2}(\mathbb{V}^{+}_{\varepsilon})\Rightarrow\Sigma^{1}_{2}(\mathbb{C})$
fails.

In Section 9 we return to a question we asked in [13], namely: Does the
splitting forcing $\mathbb{SP}$ have the Sacks property? We give a partial
negative answer in Theorem 9.13. Independently, Jonathan Schilhan found a
complete negative answer to our question [21].

In the remainder of this introduction, we set up our notation.

######  Definition 1.1.

Let $X$ be non-empty.

  1. (a)

We let $X^{<\omega}=\\{s\>:\>(\exists n<\omega)(s\colon n\to X)\\}$. The set
$X^{<\omega}$ is partially ordered by the initial segment relation
$\trianglelefteq$, namely $s\trianglelefteq t$ if
$s=t{\upharpoonright}\operatorname{dom}(s)$. We use $\triangleleft$ for the
strict relation.

  2. (b)

A set $p\subseteq X^{<\omega}$ is called a _tree_ if it is closed under
initial segments, i.e. $(t\in p\wedge s\trianglelefteq t)\rightarrow s\in p$.
The elements of $p$ are called _nodes_.

  3. (c)

A node is called a _splitting node_ if it has at least two immediate
successors in $p$. We write $\operatorname{Split}(p)$ for the set of splitting
nodes of $p$.

  4. (d)

A tree p is called _perfect_ if for every $s\in p$ there is a splitting node
$t\trianglerighteq s$.

  5. (e)

For $t\in p$, we write $\operatorname{splsuc}(t)$ for the shortest splitting
node extending $t$. When $t$ is splitting, then $\operatorname{splsuc}(t)=t$.

  6. (f)

The _stem of $p$_, short $\operatorname{stem}(p)$, is the
$\trianglelefteq$-least splitting node of $p$ if it exists.

  7. (g)

For $n<\omega$ we let $\operatorname{Split}_{n}(p)$ consist of all splitting
nodes $t$ in $p$ such that there are exactly $n$ splitting nodes preceding $t$
i.e., $\operatorname{stem}(p)=t_{0}\triangleleft
t_{1}\triangleleft\dots\triangleleft t_{n-1}\triangleleft t$, in particular
$\operatorname{Split}_{0}(p)=\\{\operatorname{stem}(p)\\}$. Analogously, we
define $\operatorname{Split}_{\leq n}(p)$.

  8. (h)

For $n<\omega$, let $\operatorname{Lev}_{n}(p):=\\{t\in p\>:\>|t|=n\\}$.

  9. (i)

For $t\in p$ we let $p{\upharpoonright}t=\\{s\in p\>:\>s\trianglelefteq t\vee
t\triangleleft s\\}$.

  10. (j)

Let $p\subseteq X^{<\omega}$ be a tree such that for any $t\in p$ and any $n$
there is $s\in p$, $t\trianglelefteq s$ such that $|s|>n$. The body or rump of
a tree $p$, short $[p]$, is the set $\\{f\in X^{\omega}\>:\>(\forall
n)(f{\upharpoonright}n\in p)\\}$.

######  Definition 1.2.

A partial ordering $(\mathbb{P},\leq)$ is called a _tree-forcing_ , if there
is a non-empty set $X$ such that

  1. -

All conditions $p\in\mathbb{P}$ are perfect trees on $X^{<\omega}$.

  2. -

For all $p\in\mathbb{P}$ and $t\in p$ the restriction $p{\upharpoonright}t$ is
again a condition in $\mathbb{P}$.

  3. -

The partial order is the inclusion i.e., $q\leq p$ iff $q\subseteq p$.

We are interested in tree-forcings $\mathbb{P}$ defined over $2^{<\omega}$ or
$\omega^{<\omega}$ with the following additional property.

######  Definition 1.3.

Let $p$ be a perfect tree defined over $2^{<\omega}$ or $\omega^{<\omega}$.
The tree $p$ is called _uniformly splitting_ , if there is an infinite set
$A_{p}\in[\omega]^{\omega}$ such that

$\forall s\in p(|s|\in A_{p}\Leftrightarrow s\in\operatorname{Split}(p)).$

A tree-forcing $\mathbb{P}$ is called _uniformly splitting tree-forcing_ , if
all conditions $p\in\mathbb{P}$ are uniformly splitting.

In this paper we study two well-known examples of uniformly splitting tree
forcings: The Mathias forcing $\mathbb{MA}$ and the Silver forcing
$\mathbb{V}$.

######  Example 1.4.

  1. (1)

The Mathias forcing is given by conditions $(s,A)\in\mathbb{MA}$ if
$s\in[\omega]^{<\omega}$ and $A\in[\omega]^{\omega}$ and $\max(s)<\min(A)$.
$(t,B)\leq(s,A)$ if $t\setminus s\subseteq A$, $t\supseteq s$, and $B\subseteq
A$. Now a Mathias tree $p=p(s,A)$ is given by $p\subseteq 2^{<\omega}$ and
$t\in p$ if $t$ is the characteristic function of $s\cup s^{\prime}$ for some
finite $s^{\prime}\subseteq A$. Of course, now $A_{p}=A$, and stronger
conditions correspond to subtrees. All trees are perfect and restrictions
$p{\upharpoonright}t$ are again conditions.

  2. (2)

A condition $f$ is a Silver condition if there is an infinite set $A$ such
that $f:\omega\setminus A\rightarrow 2$. So we get a Silver tree $p\subseteq
2^{<\omega}$ by letting

$t\in p\leftrightarrow(\forall n\in|t|\setminus A)(t(n)=f(n)).$

Then $A_{p}=A$. Stronger conditions are extensions and again correspond to
subtrees.

######  Definition 1.5.

For a set $A\subseteq\omega$ we define the upper density $d^{+}(A)$ and lower
density $d^{-}(A)$ of $A$ via:

  1. (1)

$d^{+}(A):=\limsup\limits_{n\rightarrow\infty}\frac{|A\cap n|}{n}$,

  2. (2)

$d^{-}(A):=\liminf\limits_{n\rightarrow\infty}\frac{|A\cap n|}{n}.$

######  Definition 1.6.

Let $\mathbb{P}$ be a uniformly splitting tree-forcing defined over
$2^{<\omega}$ or $\omega^{<\omega}$. For $\varepsilon\in(0,1]$ we define two
subforcings $\mathbb{P}^{+}_{\varepsilon}$ and $\mathbb{P}^{-}_{\varepsilon}$,
and we define the upper and lower positive density versions:

  1. (1)

$p\in\mathbb{P}^{+}_{\varepsilon}$ if $p\in\mathbb{P}$ and $A_{p}$ has upper
density $\geq\varepsilon$.

  2. (2)

$p\in\mathbb{P}^{-}_{\varepsilon}$ if $p\in\mathbb{P}$ and $A_{p}$ has lower
density $\geq\varepsilon$.

  3. (3)

$p\in\mathbb{P}^{+}$ if $p\in\mathbb{P}$ and $A_{p}$ has upper density $>0$.

  4. (4)

$p\in\mathbb{P}^{-}$ if $p\in\mathbb{P}$ and $A_{p}$ has lower density $>0$.

In all forcing orders, a condition $q$ is stronger than $p$ iff $q$ is a
subset of $p$.

We focus our attention on $\mathbb{MA}^{+}_{\varepsilon}$,
$\mathbb{MA}^{-}_{\varepsilon}$, $\mathbb{MA}^{+}$, $\mathbb{MA}^{-}$ and the
same for Silver. We order our investigation now in pairs, according to the
density requirement. Some steps work also for general $\mathbb{P}$.

##  2\. Upper density $\geq\varepsilon$

######  Definition 2.1.

A notion of forcing $(\mathbb{P},\leq)$ has _Axiom A_ if there are partial
order relations $\langle\leq_{n}\>:\>n<\omega\rangle$ such that

  * (a)

$q\leq_{n+1}p$ implies $q\leq_{n}p$ , $q\leq_{0}p$ implies $q\leq p$,

  * (b)

If $\langle p_{n}\>:\>n<\omega\rangle$ is a fusion sequence, i.e., a sequence
such that for any $n$, $p_{n+1}\leq_{n}p_{n}$, then there is a lower bound
$p\in\mathbb{P}$, $p\leq_{n}p_{n}$.

  * (c)

For any maximal antichain $A$ in $\mathbb{P}$ and and $n\in\omega$ and any
$p\in\mathbb{P}$ there is $q\leq_{n}p$ such that only countably many elements
of $A$ are compatible with $q$. Equivalently, for any open dense set $D$ and
any $n$, $p$, there is a countable set $E_{p}$ of conditions in $D$ and
$q\leq_{n}p$ such that $E_{p}$ is predense below $q$.

A notion of forcing $(\mathbb{P},\leq)$ has _strong Axiom A_ if the set of
compatible elements in (c) is even finite.

Axiom A entails properness and strong Axiom A implies
${}^{\omega}\omega$-bounding (see, e.g., [20, Theorem 2.1.4, Cor 2.1.12]).

######  Remark 2.2.

Let $p\in\mathbb{P^{+}_{\varepsilon}}$. We define an increasing sequence
$\langle k^{p}_{n}\in\omega\>:\>n\in\omega\rangle$ as follows:

$\displaystyle n=0:\text{ We put }k^{p}_{0}:=\min(A_{p}),$ $\displaystyle
n>0:\text{ We let
}k^{p}_{n}:=\min\\{k\in\omega\>:\>(k>k^{p}_{n-1}\wedge|A_{p}\cap k|\geq
k(\varepsilon-2^{-n}))\\}.$

Such a sequence has the following property for $n\in\omega$:

$\frac{|A_{p}\cap k^{p}_{n}|}{k^{p}_{n}}\geq\varepsilon-2^{-n},$

and therefore witnesses $d^{+}(A_{p})\geq\varepsilon$.

We use the sequences $\langle k^{p}_{n}\in\omega\>:\>n\in\omega\rangle$ to
define a _stronger- $n$_-relation $\leq_{n}$ on
$\mathbb{P^{+}_{\varepsilon}}$.

######  Definition 2.3.

We define a decreasing sequence of partial order relations
$\langle\leq_{n}\>:\>n\in\omega\rangle$ on $\mathbb{P}^{+}_{\varepsilon}$ as
follows:

$q\leq_{n}p\Leftrightarrow q\leq p\wedge k^{q}_{n}=k^{p}_{n}\wedge A_{q}\cap
k^{q}_{n}=A_{p}\cap k^{p}_{n}.$

Observe that given two conditions satisfying $q\leq_{n}p$ we must have
$k^{q}_{i}=k^{p}_{i}$ for $i\leq n$.

######  Fact 2.4.

Let $\mathbb{P}\in\\{\mathbb{MA},\mathbb{V}\\}$. Let $\langle
q_{n}\>:\>n\in\omega\rangle$ be a fusion sequence in
$\mathbb{P}^{+}_{\varepsilon}$. Then, fusions exist. Especially, $q=\bigcap
q_{n}$ is a condition in $\mathbb{P}^{+}_{\varepsilon}$.

###  2.1. Silver forcing with upper density $\varepsilon>0$

We quickly establish that the Silver forcing with positive upper density
$\varepsilon$ has strong Axiom A and thus is a proper forcing that does not
add unbounded reals. The proof is a straightforward generalization of the
standard case.

######  Lemma 2.5.

The forcing $\mathbb{V}^{+}_{\varepsilon}$ has strong Axiom A.

###### Proof.

We take the partial order relations $\leq_{n}$ as defined in Definition 2.3.
It is easy to see that the requirements (a) and (b) from Definition 2.1 are
fulfilled. We make sure that the strong version of (c) can be fulfilled as
well. So, fix an open dense set $D\subseteq\mathbb{V}^{+}_{\varepsilon}$,
$n\in\omega$ and a condition $p\in\mathbb{V}^{+}_{\varepsilon}$. We have to
find a finite set $E_{p}\subseteq D$ and a condition $q\leq_{n}p$ such that
$E_{p}$ is predense below $q$.
To this end, let $\langle k^{p}_{m}\>:\>m<\omega\rangle$ be the sequence
associated to the condition $p$ as given in Remark 2.2. Let $k:=|A_{p}\cap
k_{n}^{p}|$ denote the number of splitting levels up to $k_{n}^{p}$ and
$\\{t_{i}\in\operatorname{Split}(p)\>:\>i<2^{k}\\}$ enumerate all splitting
nodes of length exactly $k_{n}^{p}$. We construct a decreasing sequence
$p=:q_{0}\geq q_{1}\geq\dots\geq q_{2^{k}}=:q$ such that
$q_{i+1}{\upharpoonright}t_{i}\in D,i<2^{k}$. Fix $i<2^{k}$. Now, pick
$p_{i}\leq q_{i}{\upharpoonright}t_{i}$ in $D$ and copy $p_{i}$ into $q_{i}$
above level $k_{n}^{p}$, more precisely:

$q_{i+1}:=\\{s\in q_{i}\>:\>(|s|\leq k_{n}^{p}\vee(|s|>k_{n}^{p}\wedge\exists
t\in p_{i}(k_{n}^{p}\leq m<|s|\rightarrow s(m)=t(m))))\\}.$

It is clear that the resulting tree $q$ satisfies $q{\upharpoonright}t\in D$,
whenever $t\in q$ is of length at least $k_{n}^{p}$. Moreover, since we have
pruned the tree $q$ only above $k_{n}^{p}$ we also made sure that

$k^{p}_{n}=k^{q}_{n}\wedge A_{p}\cap k^{p}_{n}=A_{q}\cap k^{p}_{n},$

especially $q\leq_{n}p$. For the finite set $E_{p}$ we can simply put
$E_{p}:=\\{q{\upharpoonright}t\>:\>t\in\operatorname{Lev}_{k^{p}_{n}}(q)\\}$.
This completes the proof. ∎

######  Corollary 2.6.

The forcing $\mathbb{V}^{+}_{\varepsilon}$ is proper and
$\omega^{\omega}$-bounding for each $\varepsilon\in(0,1]$.

###  2.2. Mathias forcing with upper density $\varepsilon>0$

In the following, we investigate the differences between
$\mathbb{MA}^{+}_{\varepsilon}$ and the classical Mathias forcing
$\mathbb{MA}$ and Mathias forcing $\mathbb{MA}\mathcal{(}\mathcal{F})$ with
respect to a filter $\mathcal{F}$. The decisive difference to the classical
forcing is that $\mathbb{MA}^{+}_{\varepsilon}$ adds Cohens. From
$\mathbb{MA}(\mathcal{F})$ it already differs by the fact that the set
$\\{A\subseteq\omega\>:\>d^{+}(A)\geq\varepsilon\\}$ is not closed under
finite intersection and thus not a filter. Note that this does not yet imply
that the two forcing are not forcing equivalent.

######  Definition 2.7.

Let $\mathcal{F}$ be a filter over $\omega$. The partial order
$\mathbb{MA}(\mathcal{F})$ consists of all $p\in\mathbb{MA}$ such that the set
of splitting levels $A_{p}$ is an element of the filter $\mathcal{F}$.
$\mathbb{MA}(\mathcal{F})$ is ordered by inclusion.

Remember that a filter $\mathcal{F}$ over $\omega$ is called _Canjar_ , if the
corresponding Mathias forcing $\mathbb{MA}(\mathcal{F})$ does not add
dominating reals. See for instance [5], where Michael Canjar constructed an
ultrafilter $\mathcal{U}$ under the assumption $\mathfrak{d}=\mathfrak{c}$
such that $\mathbb{MA}(\mathcal{U})$ does not add dominating reals. The
following lemma shows that $\mathbb{MA}^{+}_{\varepsilon}$ cannot be
equivalent to $\mathbb{MA}(\mathcal{F})$, for $\mathcal{F}$ Canjar.

######  Lemma 2.8.

$\mathbb{MA}^{+}_{\varepsilon}$ adds dominating reals.

###### Proof.

Let $G$ be $\mathbb{MA}^{+}_{\varepsilon}$-generic over $V$ and let
$x_{G}=\bigcup\\{s\>:\>\exists A((s,A)\in G)\\}$ denote the the
$\mathbb{MA}^{+}_{\varepsilon}$-generic real. Observe that $x_{G}$ has upper
density $\varepsilon$ in the generic extension $V[G]$. Therefore, the
following function is well defined in $V[G]$:

${f}(n):=\min\left\\{k\>:\>\frac{|x_{G}\cap
k|}{k}\geq\varepsilon-2^{-n}\right\\}.$

Remember that in Remark 2.2 we assigned to each condition
$p\in\mathbb{MA}^{+}_{\varepsilon}$ a sequence $\langle
k^{p}_{n}\in\omega\>:\>n<\omega\rangle$. That $f$ is dominating now follows
from the following two facts:

  1. (i)

$\forall p\in\mathbb{MA}^{+}_{\varepsilon},p\Vdash\forall n<\omega(f(n)\geq
k^{p}_{n})$,

  2. (ii)

$\forall g\in\omega^{\omega}\forall p\in\mathbb{MA}^{+}_{\varepsilon}\exists
q\leq p,q\Vdash\forall^{\infty}n(k^{q}_{n}\geq g(n))$.

∎

There is a more general reason why $\mathbb{MA}^{+}_{\varepsilon}$ and
$\mathbb{MA}(\mathcal{F})$ for any filter $\mathcal{F}$ cannot be forcing
equivalent. Since $\mathcal{F}$ is a filter, any two conditions
$p,q\in\mathbb{MA}(\mathcal{F})$ with the same stem are compatible and in
particular $\mathbb{MA}^{+}_{\varepsilon}$ is $\sigma$-centered.

######  Lemma 2.9.

$\mathbb{MA}^{+}_{\varepsilon}$ does not satisfy the countable chain
condition.

###### Proof.

Without loss of generality let $\varepsilon=1$. We construct an injective
function $f:2^{\omega}\rightarrow[\omega]^{\omega}$ such that each $x\in
2^{\omega}$ with $|x^{-1}(1)|=\omega$ get mapped to a set with upper density
$1$. To this end we define $f$ together with an auxiliary function
$g:\omega\rightarrow\omega$

$\displaystyle g(0)$ $\displaystyle:=0$ $\displaystyle g(i+1)$
$\displaystyle:=2^{g(i)}$ $\displaystyle n\in f(x)$
$\displaystyle:\Leftrightarrow\text{ there is }i\in\omega,x(i)=1\text{ and
}n\in[g(i),g(i+1))$

Now, choose an almost disjoint family
$\\{A_{\alpha}\in[\omega]^{\omega}\>:\>\alpha<\mathfrak{c}\\}$ of size
continuum and let $x_{\alpha}$ denote the corresponding characteristic
function i.e., $x_{\alpha}(i)=1$ iff $i\in A_{\alpha}$. Then,
$\\{f(x_{\alpha})\>:\>\alpha<\mathfrak{c}\\}$ is an almost disjoint family
consisting of sets of upper density $1$. In particular, the conditions
$\\{(\langle\rangle,f(x_{\alpha}))\>:\>\alpha<\mathfrak{c}\\}$ form an
antichain in $\mathbb{MA}^{+}_{1}$ of size the continuum. ∎

Now we turn our attention to a comparison with the classical Mathias forcing
$\mathbb{MA}$. Usually the first step in showing that a given forcing
satisfies _pure decision_ (compare [1, Lemma 7.4.5]) is to show that the
forcing satisfies _quasi pure decision_ (compare [1, Lemma 7.4.6]). We will
see that although $\mathbb{MA}^{+}_{\varepsilon}$ fails to satisfy pure
decision it still satisfies quasi pure decision.

######  Lemma 2.10.

$\mathbb{MA}^{+}_{\varepsilon}$ satisfies quasi pure decision i.e., given a
condition $p=(s,A)\in\mathbb{MA}^{+}_{\varepsilon}$ and an open dense set
$D\subseteq\mathbb{MA}^{+}_{\varepsilon}$ there is $B\subseteq A$ such that
the following holds:

$\text{If there is }(t,C)\leq(s,A)\text{ and }(t,C)\in D,\text{ then
}(t,B\setminus(\max(t)+1))\in D.$

###### Proof.

The proof is a straightforward generalization of [1, Lemma 7.4.5.]. To make
sure that the final set $B$ has upper density $\geq\varepsilon$ we have to use
finite sets $b_{n}$ instead of singletons.
We construct $B=\bigcup\\{b_{n}\in[\omega]^{<\omega}\>:\>n\in\omega\\}$
together with a decreasing sequence $\\{B_{n}\subseteq A\>:\>n\in\omega\\}$.
We start with $B_{0}=A$ and $b_{0}=\min(A)$. So, assume we have constructed
all sets up to $B_{n}$ and $b_{n}$. Let $\\{t_{0},\dots,t_{k-1}\\}$ enumerate
all subsets of $\bigcup\\{b_{0},\dots,b_{n}\\}$. We construct $B_{n+1}$ as a
decreasing sequence $B_{n}=:B^{0}_{n+1}\supseteq\dots\supseteq
B^{k}_{n+1}=:B_{n+1}$. Let $j<k$ be given.

  1. Case 1:

There exists $C\subseteq B^{j}_{n+1}\setminus(\max(t_{j})+1)$ such that
$(s\cup t_{j},C)\in D$. Then put $B^{j+1}_{n+1}:=C$.

  2. Case 2:

Otherwise put $B^{j+1}_{n+1}:=B^{j}_{n+1}$.

Finally put $B_{n+1}:=B^{k}_{n+1}$. Since $B_{n+1}$ has upper density
$\geq\varepsilon$ we can find $k_{n+1}\in\omega$ such that

$\frac{|(\bigcup_{j\leq n}b_{j}\cup B_{n+1})\cap
k_{n+1}|}{k_{n+1}}\geq\varepsilon-2^{-n-1}$

and set $b_{n+1}:=B_{n+1}\cap k_{n+1}$. ∎

######  Corollary 2.11.

The forcing $\mathbb{MA}^{+}_{\varepsilon}$ has Axiom A.

###### Proof.

We take the partial order relations $\leq_{n}$ as defined in Definition 2.3
and also recall the sequences $\langle k^{p}_{n}\>:\>n<\omega\rangle$ from
Remark 2.2. The crucial part is to make sure that the requirement (c) from
Axiom A (Definition 2.1) is fulfilled. So, fix an open dense set
$D\subseteq\mathbb{MA}^{+}_{\varepsilon},n<\omega$ and a condition
$p=(s,A_{p})\in\mathbb{MA}^{+}_{\varepsilon}$. Let $N$ be big enough and
$\\{t_{i}\in[\omega]^{<\omega}\>:\>i<N\\}$ enumerate all subsets of $A_{p}\cap
k^{p}_{n}$. We define a decreasing sequence $A_{p}\supseteq B_{0}\supseteq
B_{1}\supseteq,\dots,\supseteq B_{N}$ such that for $i<N$:

$\displaystyle\forall(t,C)\in D\big{(}(t,C)\leq(s\cup
t_{i},B_{i})\rightarrow(t,B_{i+1}\setminus(\max(t)+1))\in D\big{)}.$

We start with $B_{0}:=A_{p}\setminus(\max(t_{0})+1)$. To get the sets $B_{i}$
for $i>0$, simply apply Lemma 2.10 to the condition $(s\cup t_{i-1},B_{i-1})$
and the open dense set $D$.
Finally, we set $B:=(A\cap k^{p}_{n})\cup B_{N}$ and
$E_{p}:=\\{(t,B\setminus(\max(t)+1))\>:\>\exists
C\in[\omega]^{\omega}((t,C)\leq(s,B)\wedge(t,C)\in D)\\}$. Then,
$(s,B)\leq_{n}(s,A)$ and $E_{p}$ is a countable predense set below $(s,B)$. ∎

######  Proposition 2.12.

$\mathbb{MA}^{+}_{\varepsilon}$ adds a Cohen real.

###### Proof.

Take $N\in\omega$ such that $1/N<\varepsilon$. Divide $\omega$ into $N+1$
disjoint sets $a_{i}\subseteq\omega,i<N+1$ of density $1/(N+1)$ i.e.,
$d^{+}(a_{i})=d^{-}(a_{i})=1/(N+1),i<(N+1).$ Then each set $A\subseteq\omega$
with upper density $\geq 1/N$ cannot be completely contained in a single set
$a_{i}$. Furthermore it must intersect at least two of them infinitely often.
Let $x_{G}$ be the canonical name for the
$\mathbb{MA}^{+}_{\varepsilon}$-generic real and $\langle
n_{k}\>:\>k\in\omega\rangle$ enumerate all integers $n$ such that
$x_{G}(n)=1$. We define a $\mathbb{MA}^{+}_{\varepsilon}$-name $\dot{c}$ via:

$\dot{c}(k):=\begin{cases}0,&\exists i<(N+1)\\{n_{2k},n_{2k+1}\\}\subseteq
a_{i}\\\ 1,&\text{ else.}\end{cases}$

We claim that $\dot{c}$ is Cohen. For this purpose, fix $t\in
2^{<\omega},(s,A)\in\mathbb{MA}^{+}_{\varepsilon}$. W.l.o.g. we can assume
that $|s^{-1}(\\{1\\})|$ is even. Let $r\in 2^{<\omega}$ maximal such that
$(s,A)\Vdash r\trianglelefteq\dot{c}$. We have to find
$(s^{\prime},A^{\prime})\leq(s,A)$ with the property
$(s^{\prime},A^{\prime})\Vdash r^{\smallfrown}t\trianglelefteq\dot{c}$. We
construct $(s^{\prime},A^{\prime})$ as a decreasing sequence
$(s,A)=:(s_{0},A_{0})\geq(s_{1},A_{1})\geq\dots\geq(s_{|t|},A_{|t|})=(s^{\prime},A^{\prime})$
such that $|{s_{i}}^{-1}(\\{1\\})|+2=|{s_{i+1}}^{-1}(\\{1\\})|,i<|t|$ and
$(s_{i},A_{i})\Vdash
r^{\smallfrown}t{\upharpoonright}i\trianglelefteq\dot{c}$. We only carry out
the first step of the construction. Take $i<(N+1)$ such that $A_{0}\cap a_{i}$
is infinite. Put $m:=\min(A_{0}\cap a_{i})$. There are two cases:

$\displaystyle t(0)=0:$ $\displaystyle\text{ Define }M:=\min((A_{0}\cap
a_{i})\setminus(m+1))\text{ and put
}s_{1}:={s_{0}}^{\smallfrown}m^{\smallfrown}M,$ $\displaystyle
A_{1}:=A_{0}\setminus(M+1).\text{ Then }(s_{1},A_{1})\Vdash
r^{\smallfrown}t(0)\trianglelefteq\dot{c}.$ $\displaystyle t(0)=1:$
$\displaystyle\text{ Define }M:=\min(A_{0}\setminus((m+1)\cup a_{i}))\text{
and put }s_{1}:={s_{0}}^{\smallfrown}m^{\smallfrown}M,$ $\displaystyle
A_{1}:=A_{0}\setminus(M+1).\text{ Then }(s_{1},A_{1})\Vdash
r^{\smallfrown}t(0)\trianglelefteq\dot{c}.$

The rest of the construction is carried out analogously.

∎

Thus, in contrast to $\mathbb{MA}$ we get.

######  Corollary 2.13.

$\mathbb{MA}^{+}_{\varepsilon}$ does not satisfy pure decision.

##  3\. Lower density $\geq\varepsilon$

In this section we investigate the Mathias forcing with lower density
$\geq\varepsilon$. The first observation is.

######  Observation 3.1.

The family of sets $\mathcal{F}_{1}:=\\{A\subseteq\omega\>:\>d^{-}(A)=1\\}$ is
a filter.

So, $\mathbb{MA}^{-}_{1}$ is in fact equivalent to Mathias forcing
$\mathbb{MA}(\mathcal{F}_{1})$ with respect to the filter $\mathcal{F}_{1}$.

Thus, we get:

######  Fact 3.2.

$\mathbb{MA}^{-}_{1}$ satisfies the countable chain condition.

Now, we will see that the forcing $\mathbb{MA}^{-}_{\varepsilon}$ is a
disjoint union of $\sigma$-centred forcings.

######  Definition 3.3.

Let $A,B\subseteq\omega$. The lower density of $B$ with respect to $A$ is
defined by:

$d_{A}^{-}(B):=\liminf_{n\rightarrow\infty}\frac{|A\cap B\cap n|}{|A\cap n|}.$

######  Lemma 3.4.

Let $B\subseteq A\subseteq\omega,d^{-}(A)>0$ and $d_{A}^{-}(B)<1$.
Then, $d^{-}(B)<d^{-}(A).$

###### Proof.

Let $B\subseteq A\subseteq\omega$ be as in the lemma. Then, $d_{A}^{-}(B)<1$
implies

$(\exists n_{0}<\omega)(\forall k<\omega)(\exists m\geq k)\left(\frac{|B\cap
m|}{|A\cap m|}<1-2^{-n_{0}}\right).$

So, we in fact get

$(\exists^{\infty}m)\left(|B\cap m|<(1-2^{-n_{0}})\cdot|A\cap m|\right).$

∎

######  Corollary 3.5.

If $d^{-}(A)=\varepsilon$, then ${\mathcal{F}}(A):=\\{B\subseteq
A\>:\>d^{-}(B)=\varepsilon\\}$ is a filter.

######  Proposition 3.6.

Let $\varepsilon\in(0,1]$.

The forcing notion $\mathbb{MA}^{-}_{\varepsilon}$ is equivalent to a disjoint
union of $\sigma$-centred forcings.

###### Proof.

First, note that
$\mathcal{D}:=\\{p=(s,A_{p})\in\mathbb{MA}^{-}_{\varepsilon}\>:\>d^{-}(A_{p})=\varepsilon\\}$
is an open dense subset of $\mathbb{MA}^{-}_{\varepsilon}$. Fix a maximal
antichain $\mathcal{A}\subseteq\mathcal{D}$. For each $p\in\mathcal{A}$ the
restriction of $\mathbb{MA}^{-}_{\varepsilon}$ to $p$ is denoted by
$\mathbb{MA}^{-}_{\varepsilon}{\upharpoonright}p=\\{q\in\mathbb{MA}^{-}_{\varepsilon}\>:\>q\leq
p\\}$. Then, each $\mathbb{MA}^{-}_{\varepsilon}{\upharpoonright}p$ has the
countable chain condition. Indeed, fix $p=(s,A_{p})\in\mathcal{A}$. By
Corollary 3.5 we know that the set $\\{B\subseteq
A_{p}\>:\>d^{-}(B)=\varepsilon\\}$ is closed under finite intersections. Thus,
any two conditions $q_{0},q_{1}\leq p$ which have the same stem, are
compatible.

Finally, we can set
$\mathbb{MA}^{-}_{\varepsilon}=\bigcup_{p\in\mathcal{A}}\mathbb{MA}^{-}_{\varepsilon}{\upharpoonright}p.$
∎

######  Corollary 3.7.

$\mathbb{MA}^{-}_{\varepsilon}$ is proper for each $\varepsilon\in(0,1)$.

Next, we show that $\mathbb{MA}^{-}_{\varepsilon}$ has antichains of size the
continuum, whenever $\varepsilon<1$.

######  Lemma 3.8.

Let $\varepsilon\in(0,1)$. There is a family of sets $\\{A_{f}\>:\>f\in I\\}$
such that

  1. (1)

$I\subseteq 2^{\omega}$ has size continuum,

  2. (2)

$d^{-}(A_{f})\geq\varepsilon,$ for all $f\in I$,

  3. (3)

$d^{-}(A_{f}\cap A_{g})<\varepsilon$, for all $f\neq g$ in $I$.

###### Proof.

Let $\varepsilon\in(0,1)$. First, we fix a suitable set of indices $I\subseteq
2^{\omega}$. To this end, let $I$ be any family of functions of size continuum
with the property

(3.1) $\displaystyle\forall f,g\in I(f\neq
g\rightarrow\exists^{\infty}i,j(f(i)=1>g(i)\wedge f(j)=0<g(j))).$

Next, we define for $n\in\omega$ the interval
$I_{n}:=[2^{n+1},2^{n+2})\subseteq\omega$. Then, all intervals are pairwise
disjoint and each interval $I_{n}$ has size $2^{n+1}$. Now, we define for each
$f\in I$ a function $F_{f}\in\omega^{\omega}$ via

$F_{f}(n):=\sum_{i=0}^{n}f(i)\cdot 2^{n-i}.$

Observe, that the functions $F_{f}$ have the property that for
$n\in\omega(F_{f}(n)\in[0,2^{n+1}))$, in other words there are $2^{n+1}$
possible values for $F_{f}(n)$.
We define the sets $A_{f}$ of lower density $\geq\varepsilon$ recursively over
$n$ as a disjoint union $A_{f}=\bigcup_{n}A^{n}_{f}$, such that
$A^{n}_{f}\subseteq I_{n}$. To this end, fix $n$ and let $k$ be the closest
natural number $\leq\varepsilon\cdot 2^{n+1}$, so
$k:=\max\\{j\in\omega\>:\>j\leq\varepsilon\cdot 2^{n+1}\\}$. We differentiate
two cases:
If $2^{n+1}+F_{f}(n)+k-1<2^{n+2}:$ We set

$A^{n}_{f}:=\\{2^{n+1}+F_{f}(n),2^{n+1}+F_{f}(n)+1,\dots,2^{n+1}+F_{f}(n)+k-1\\}$

Otherwise, there is $j\leq k-1$ such that $2^{n+1}+F_{f}(n)+j=2^{n+2}-1$ and
we can set

$A^{n}_{f}:=\\{2^{n+1},\dots,2^{n+1}+k-1-j\\}\cup\\{2^{n+1}+F_{f}(n),\dots,2^{n+1}+F_{f}(n)+j\\}$

Finally we set $A_{f}:=\bigcup A^{n}_{f}$. It follows directly from the
definition that each set $A_{f}$ has density $\varepsilon$ and for
$n\in\omega$ we get $(A_{f}\cap I_{n}=A^{n}_{f})$.
It is left to show that for $f\neq g$ the set $A_{f}\cap A_{g}$ has lower
density $<\varepsilon$. So, fix $f,g\in I$ and let $i_{0}$ be minimal such
that $f(i_{0})\neq g(i_{0})$. W.l.o.g assume $f(i_{0})=1$ and $g(i_{0})=0$, so
in particular $F_{f}(n)>F_{g}(n)$ for all $n\geq i_{0}$. By property (3.1) of
$I$ there is $i_{1}>i_{0}$ such that $f(i_{1})=1$ and $g(i_{1})=0$. Then, for
all $n>i_{1}$ we have

$|F_{f}(n)-F_{g}(n)|\geq
2^{n-i_{0}}+2^{n-i_{1}}-(\sum_{i>i_{0}}^{i_{1}-1}2^{n-i}+2^{n-i_{1}}-1)>2^{n-i_{1}+1}.$

######  Claim 3.9.

There is a strictly positive constant $c\in(0,1)$ such that for all
$n\in\omega$

$\frac{|A_{f}\cap A_{g}\cap I_{n}|}{|I_{n}|}\leq\varepsilon-c.$

Clearly, the claim above implies $d^{-}(A_{f}\cap A_{g})<\varepsilon$. So, fix
$n\in\omega$. Since we are only interested in the size of $A_{f}\cap A_{g}\cap
I_{n}$, we might assume, by a transformation argument, that $g(i)=0,i<n$ and
$f(0)=0$. Thus, $A^{n}_{g}$ consists of the first $k$ elements of $I_{n}$
i.e., $\\{2^{n+1},\dots,2^{n+1}+k-1\\}$ and
$F_{f}(n)-F_{g}(n)=F_{f}(n)<2^{n}$. Again, we have to differentiate two cases:
If $F_{f}(n)+2^{n+1}+k-1<2^{n+2}$: Then,

$\frac{|A^{n}_{f}\cap A^{n}_{g}\cap
I_{n}|}{2^{n+1}}\leq\frac{k-2^{n-i_{1}+1}}{2^{n+1}}\leq\varepsilon-\frac{1}{2^{i_{1}}}.$

Otherwise $F_{f}(n)+2^{n+1}+k-1\geq 2^{n+2}$. In this case we must have
$\varepsilon>1/2$ and we get:

$\frac{|A^{n}_{f}\cap A^{n}_{g}\cap
I_{n}|}{2^{n+1}}\leq\frac{2^{n+1}-2(2^{n+1}-k)}{2^{n+1}}\leq
2\cdot\varepsilon-1.$

We can set $c:=\min\\{2^{-i_{1}},(1-\varepsilon)\\}$. ∎

######  Corollary 3.10.

$\mathbb{MA}^{-}_{\varepsilon}$ has antichains of size continuum for
$\varepsilon\in(0,1)$.

###### Proof.

Take the family of sets $\\{A_{f}\>:\>f\in I\\}$ as in the lemma above. Then,
$\\{(\langle\rangle,A_{f})\>:\>f\in
I\\}\subseteq\mathbb{MA}^{-}_{\varepsilon}$ is an antichain of size continuum.
∎

######  Corollary 3.11.

There is no filter $\mathcal{F}$ such that $\mathbb{MA}(\mathcal{F})$ and
$\mathbb{MA}^{-}_{\varepsilon}$ are forcing equivalent.

###### Proof.

This follows from the previous corollary, together with the fact, that
$\mathbb{MA}(\mathcal{F})$ is a $\sigma$-centered forcing for each filter. ∎

Analog to the upper density case $\mathbb{MA}^{+}_{\varepsilon}$ we get that
$\mathbb{MA}^{-}_{\varepsilon}$ adds Cohen reals.

######  Proposition 3.12.

$\mathbb{MA}^{-}_{\varepsilon}$ adds Cohen reals.

###### Proof.

We can repeat the proof of Proposition 2.12. Again we divide $\omega$ into
$N+1$ disjoint sets $a_{i}\subseteq\omega,i<N+1$ of density $1/(N+1)$, where
$N$ is such that $1/N<\varepsilon$. Now to carry out the rest of the
construction it enough to see that any set of lower density $\varepsilon$
intersects at least two of the sets $a_{i}$ infinitely often. ∎

Let $x_{G}$ be the generic real added by $\mathbb{MA}^{-}_{\varepsilon}$.
Then, $x_{G}$ has lower density $0$ and upper density $\varepsilon$ in the
generic extension. So one might try to use the same recipe as in Lemma 2.8,
where it was proven that $\mathbb{MA}^{+}_{\varepsilon}$ adds dominating
reals. However, condition (ii) from the proof is not satisfied and the
following question remains open.

######  Question 3.13.

Does $\mathbb{MA}^{-}_{\varepsilon}$ add dominating reals?

Note, a positive answer to this question for $\varepsilon=1$ would also be a
positive answer to [19][Question 38] from Raghavan.

##  4\. Upper density $>0$

In this section we investigate $\mathbb{MA}^{+}$ i.e., the forcing consisting
of Mathias conditions $p\subseteq 2^{<\omega}$ such that the corresponding set
of splitting levels $A_{p}$ has strictly positive upper density.

######  Definition 4.1.

The density zero ideal $\mathcal{Z}$ is defined by:

$\mathcal{Z}:=\\{A\subseteq\omega\>:\>d^{+}(A)=0)\\},$

and the corresponding coideal is denoted by
$\mathcal{Z}^{+}=\\{A\subseteq\>:\>d^{+}(A)>0\\}$.

Observe that for $p\in\mathbb{MA}$ we have $p\in\mathbb{MA}^{+}$ iff
$A_{p}\in\mathcal{Z}^{+}$.
We show that the forcing $\mathbb{MA}^{+}$ is proper. In order to do this, we
need the following Lemma (compare [8, Lemma 9.6.]).

######  Lemma 4.2.

$\mathbb{MA}^{+}$ is forcing equivalent to the two step iteration of
$\mathcal{P}(\omega)/\mathcal{Z}^{+}*\mathbb{MA}(\dot{\mathcal{F}})$, where
$\dot{\mathcal{F}}$ is a $\mathcal{P}(\omega)/\mathcal{Z}^{+}$-generic filter.

For sake of completeness we sketch a proof.

###### Proof.

We define a map $i:\mathbb{MA}^{+}\longrightarrow
P(\omega)/\mathcal{Z}^{+}*\mathbb{MA}(\dot{\mathcal{F}})$ via
$i((s,A)):=([A]_{\mathcal{Z}^{+}},(s,A))$, where
$[A]_{\mathcal{Z}^{+}}=\\{B\subseteq\omega\>:\>A\Delta B\in\mathcal{Z}\\}$ is
the corresponding equivalence class. It is not hard to see that $i$ preserves
being stronger and being incompatible. We show that the range of $i$ is dense.
To this end, fix a condition $([A]_{\mathcal{Z}^{+}},(s,\dot{C}))$. Then
$\dot{C}$ is a $P(\omega)/\mathcal{Z}^{+}$-name for an infinite subset of
$\omega$ such that $[A]_{\mathcal{Z}^{+}}\Vdash\dot{C}\in\dot{\mathcal{F}}$.
This implies $[A]_{\mathcal{Z}^{+}}\Vdash A\setminus\dot{C}\in\mathcal{Z}$. So
we get
$i((s,A))=([A]_{\mathcal{Z}^{+}},(s,A))\leq([A]_{\mathcal{Z}^{+}},(s,\dot{C})).$
∎

######  Corollary 4.3.

$\mathbb{MA}^{+}$ is proper.

###### Proof.

Let $\mathsf{Fin}$ denote the ideal of finite subsets of $\omega$. In [7,
Theorem 1.3.] Farah proved that $P(\omega)/\mathcal{Z}^{+}$ is forcing
equivalent to the two step iteration of $\mathcal{P}(\omega)/\mathsf{Fin}$ and
a measure algebra of Maharam character $\mathfrak{c}$ and therefore is proper.
So, $\mathbb{MA}^{+}$ is a finite iteration of proper forcings. ∎

######  Theorem 4.4.

$\mathbb{MA}^{+}$ adds Cohen reals.

The Theorem follows from [8, Lemma 9.8.] and the fact that the coideal
$\mathcal{Z}^{+}$ is not semiselective (compare [6, Definition 2.1.]).
However, since we will make use of the explicit construction of the Cohen
real, we also give a proof. We freely identify sequences $x\in 2^{\leq\omega}$
with their corresponding sets of natural numbers $x\in[\omega]^{\leq\omega}$
via $x\mapsto\\{n\>:\>x(n)=1\\}$.

###### Proof.

We define maximal antichains $\mathcal{A}_{n}$ in
$(\mathcal{Z}^{+},\subseteq^{*})$ as follows: For $n\in\omega$ let
$A^{i}_{n}:=\\{k\in\omega\>:\>k=i\mod 2^{(n+1)}\\}$ and put
$\mathcal{A}_{n}:=\\{A^{i}_{n}\>:\>i<2^{(n+1)}\\}$, e.g. $\mathcal{A}_{0}$
consists of the even and odd numbers. Let $x_{G}\in 2^{\omega}$ be a
$\mathbb{MA}^{+}$-generic real. Then, in the generic extension $V[x_{G}]$,
there is for each $n\in\omega$ exactly one $i_{n}<2^{(n+1)}$ such that
$x_{G}\subseteq^{*}A^{i_{n}}_{n}$. To simplify notations we denote with
$A_{n}$ this unique $A^{i_{n}}_{n}$. We define two sequences $\langle
n_{i}\>:\>i\in\omega\rangle$ and $\langle
m_{i}\>:\>i\in\omega\setminus\\{0\\}\rangle$ as follows: We start with
$n_{0}:=\min(x_{G})$. When $n_{i}$ is known, we put
$m_{i+1}:=\min(x_{G}\setminus(n_{i}+1))$. To define $n_{i+1}$ we differentiate
two cases: If $x_{G}\setminus(m_{i+1}+1)\subseteq A_{m_{i+1}}$ we put
$n_{i+1}:=m_{i+1}$. Otherwise we put
$n_{i+1}:=\min(x_{G}\setminus((m_{i+1}+1)\cup A_{m_{i+1}})).$ Observe that by
definition we have $m_{i}\leq n_{i}$. We put

$c(i):=\begin{cases}0,&\text{ if }x_{G}\setminus(m_{i+1}+1)\subseteq
A_{m_{i+1}}\\\ 1,&\text{ else.}\end{cases}$

We show that $c$ is Cohen. For this purpose, fix a condition
$(s,A)\in\mathbb{MA}^{+}$. By density we can assume that there is $i<\omega$
such that $m_{i},n_{i}$ and $c{\upharpoonright}i$ are decided by $(s,A)$ but
none of the values of $m_{i+1},n_{i+1}$ nor $c(i)$. We must have
$\max(s)=n_{i}$. Fix $j\in 2$. It is enough to find $(t,B)\leq(s,A)$ such that
$(t,B)\Vdash c(i)=j$ and in addition $(t,B)$ does not decide $m_{i+2},n_{i+2}$
or $c(i+1)$. Depending on the value of $j$ we distinguish two cases:

  1. $j=0$:

Let $m:=\min(A),t:=s\cup\\{m\\}$ and $B:=A_{m}\cap A\setminus(m+1)$.

  2. $j=1$:

Find $m\in A$ such that $A\not\subseteq A_{m}$ (such an $m$ always exists
since $A$ has positive upper density and $A_{m}$ has density $2^{-(m+1)}$).
Next, pick $n\in A\setminus A_{m}$ and put $t:=s\cup\\{m\\}\cup\\{n\\}$.
Finally, define $B:=A\setminus(n+1)$.

In both cases we have $\max(t)=n_{i+1},(t,B)$ decides the value of $c(i)$ to
be $j$ but neither does it decide $m_{i+2},n_{i+2}$ and nor $c(i+1)$. ∎

Next, we show that $\mathbb{MA}^{+}$ adds dominating reals.

######  Theorem 4.5.

$\mathbb{MA}^{+}$ adds dominating reals.

###### Proof.

Let $\mathcal{Z}^{*}=\\{A\subseteq\omega\>:\>\omega\setminus
A\in\mathcal{Z}\\}$ be the dual filter of the density zero ideal. In [11,
Corollary 3] Hrušak and Minami showed that $\mathbb{MA}(\mathcal{F})$ adds
dominating reals, whenever $\mathcal{F}$ is a filter extending
$\mathcal{Z}^{*}$. We use Lemma 4.2. Let $\dot{\mathcal{F}}$ denote the
$P(\omega)/\mathcal{Z}^{+}$-generic filter. Then, in the generic extension
$V[\dot{\mathcal{F}}]$ the filter $\dot{\mathcal{F}}$ extends
$(\mathcal{Z}^{*})^{V}$ and therefore $\mathbb{MA}(\dot{\mathcal{F}})$ adds
dominating reals over $V$. Specifically, $\mathbb{MA}^{+}$ adds dominating
reals over $V$. ∎

##  5\. Positive lower density

In this section we show that $\mathbb{V}^{-}$ collapses the continuum to
$\omega$. We also construct an uncountable antichain in the partial order
consisting of sets with strictly positive lower density ordered by inclusion.

######  Theorem 5.1.

The forcing $\mathbb{V}^{-}$ collapses the continuum to $\omega$.

We first prove a lemma:

######  Lemma 5.2.

There is a strictly increasing function
$\ell\colon\omega\times\omega\times\omega\cup\\{-1\\}\to\omega$, a small
natural number $r$ and a $\mathbb{V}^{-}$-name $\dot{F}$ for a function
$F\colon\omega\times\omega\times\omega\to 2$ in $V[G]$ with the following
property:

(5.1) $\begin{split}&(\forall\varrho\in 2^{\omega})(\forall
p\in\mathbb{V}^{-})(\forall k\in\omega\setminus\\{0\\})(\forall
k_{0}\in\omega)\\\ &\Bigl{(}\bigl{(}d^{-}(A_{p})\geq\frac{2}{k}\wedge\forall
n\geq
k_{0}\frac{|A_{p}\cap[\ell(k,k_{0},n),\ell(k,k_{0},n+1))|}{\ell(k,k_{0},n+1)-\ell(k,k_{0},n)}\geq\frac{1}{k}\bigr{)}\rightarrow\\\
&(\exists q_{\varrho}\leq
p)\bigl{(}d^{-}(A_{q_{\varrho}})\geq\frac{1}{rk}\wedge
q_{\varrho}\Vdash(\forall
n\in\omega)\dot{F}(k,k_{0},n)=\varrho(n)\bigr{)}\Bigr{)}\end{split}$

###### Proof.

$r$, $z_{1}$, $z_{2}$ and $m$ will be arranged later. The function $\ell$ is
defined recursively by

$\ell(k,k_{0},n)=\begin{cases}0,&\text{ if }n=-1,\\\ z_{1}k(k_{0}+1),&\text{
if }n=0,\\\ mk\cdot\ell(k,k_{0},n-1),&\text{ if }n>0.\end{cases}$

Although the function $\ell$ in fact depends on three variables it makes sense
to abbreviate it when the values $k$ and $k_{0}$ are clear from the context.
In such a situation we also write $\ell_{n}$ instead of $\ell(k,k_{0},n)$.
We let $x_{G}$ be the generic branch and define

$f(k,k_{0},n)=\mbox{ the closest natural number to
}\frac{|x_{G}^{-1}[\\{1\\}]\cap\ell_{n}|}{\ell_{n}}\cdot\frac{z_{1}k}{z_{2}}.$

We define in $V[G]$ the following function:

$F(k,k_{0},n)=\begin{cases}0,&\mbox{ if }f(k,k_{0},n)\mbox{ is even;}\\\
1,&\mbox{ else.}\end{cases}$

So, assume that we are given a condition $p\in\mathbb{V}^{-}$ that meets the
premise of the implication (5.1) for $k$ and $k_{0}$ i.e.,
$d^{-}(A_{p})\geq\frac{2}{k}$ and $(\forall n\geq
k_{0})\frac{|A_{p}\cap[\ell_{n},\ell_{n+1})|}{\ell_{n+1}-\ell_{n}}\geq\frac{1}{k}$.
Let $\varrho\in 2^{\omega}$ be given. Recall any Silver condition $p$ is
uniquely described by a function $f_{p}\colon\omega\setminus A_{p}\to 2$. By
induction on $n<\omega$ we define an increasing sequence of partial functions
$f_{p}=f_{q_{-1}}\subseteq f_{q_{0}}\subseteq f_{q_{1}}\subseteq\dots$ such
that for all $n\geq 0$

  1. (1)

$f_{q_{n}}{\upharpoonright}\ell_{n-1}=f_{q_{n-1}}{\upharpoonright}\ell_{n-1}$,

  2. (2)

$f_{q_{n}}{\upharpoonright}[\ell_{n-1},\ell_{n})\supseteq
f_{q_{n-1}}{\upharpoonright}[\ell_{n-1},\ell_{n}),$

  3. (3)

$f_{q_{n}}{\upharpoonright}[\ell_{n},\infty)=f_{p}{\upharpoonright}[\ell_{n},\infty),$

From the three conditions above it already follows that each corresponding
tree $q_{n}$ will be a member of $\mathbb{V}^{-}$. Additionally, we make sure
that the following holds as well for $n\geq 0$:

  1. (4)

$\frac{|A_{q_{n}}\cap[\ell_{n-1},\ell_{n})|}{\ell_{n}-\ell_{n-1}}\geq\frac{1}{rk}$

  2. (5)

$q_{n}\Vdash\forall m\leq nF(k,k_{0},m)=\varrho(m).$

The conditions $(1)-(4)$ together make sure that the decreasing sequence
$\langle q_{n}\>:\>n<\omega\rangle$ has a lower bound in $\mathbb{V}^{-}$,
namely $q:=\bigcap_{n}q_{n}$. Condition $(5)$ ensures that $q\Vdash\forall
n\in\omega F(k,k_{0},n)=\varrho(n)$. Thus, we can set $q_{\varrho}:=q$ and are
done.

Now for the step from $n-1$ to $n$: We have

$\frac{|A_{p}\cap[\ell_{n-1},\ell_{n})|}{\ell_{n}-\ell_{n-1}}\geq\frac{1}{k}.$

This means we can add at most
$\frac{(r-1)}{r}\cdot\frac{1}{k}\cdot(\ell_{n}-\ell_{n-1})$ elements of
$[\ell_{n-1},\ell_{n})\cap A_{p}$ to $\operatorname{dom}(f_{q_{n}})$ and still
make sure that condition $(4)$ is met. We will later choose which of them are
mapped to 0 and which are mapped to 1 by $f_{q_{n}}$.

We define two approximations to $f$ and $F$ respectively, which do not depend
on the generic element $x_{G}$. We let

$f(k,k_{0},n,p)=\mbox{ the closest natural number to
}\frac{|f_{p}^{-1}[\\{1\\}]\cap\ell_{n}|}{\ell_{n}}\cdot\frac{z_{1}k}{z_{2}},$

and

$F(k,k_{0},n,p)=\begin{cases}0,&\mbox{ if }f(k,k_{0},n,p)\mbox{ is even;}\\\
1,&\mbox{ else.}\end{cases}$

Now at most $\frac{z_{2}}{z_{1}}\cdot\frac{\ell_{n}}{k}$ many new values are
needed to change $f(k_{0},k,n,p)$ to $f(k_{0},k,n,q)$ such that the quotient

$\frac{|f_{p}^{-1}[\\{1\\}]\cap\ell_{n}|}{\ell_{n}}\cdot\frac{z_{1}k}{z_{2}}\in\omega$

and such that $F(k_{0},k,n,q)$ coincides with $\varrho(n)$. However, we have
to be careful not to contradict condition $(4)$. Especially, the following
must hold:

$\frac{z_{2}}{z_{1}}\cdot\frac{\ell_{n}}{k}\leq\frac{(r-1)}{r}\cdot\frac{1}{k}\cdot(\ell_{n}-\ell_{n-1})=\frac{(r-1)}{r}\cdot\frac{1}{k}\cdot\frac{(mk-1)}{mk}\cdot\ell_{n}$

On the other hand, we need to ensure that $q_{n}$ decides $F(k,k_{0},n)$ to be
$F(k_{0},k,n,q_{n})$. By construction and in particular condition $(4)$, we
get that the amount of digits that are in
$\ell_{n}\setminus\operatorname{dom}(f_{q_{n}})$ is $\frac{\ell_{n}}{rk}$.
Hence we need

$\frac{z_{2}}{z_{1}}\cdot\frac{\ell_{n}}{k}>2\cdot\frac{\ell_{n}}{rk}$

Both inequalities are true if we have:

$2\cdot\frac{1}{r}<\frac{z_{2}}{z_{1}}\cdot\frac{\ell_{n}}{k}\leq\frac{(r-1)}{r}\cdot\frac{(mk-1)}{mk}.$

We can take, e.g., $r=m=4$, $z_{1}=3$ and $z_{2}=2$ and get
$\frac{1}{2}<\frac{2}{3}\leq\frac{3}{4}\cdot\frac{15}{16}$. ∎

###### Proof of the Theorem.

Now it is easy to see that

$\Vdash_{\mathbb{V}^{-}}(\forall\varrho\in V\cap 2^{\omega})(\exists
k>0)(\exists k_{0})(\forall nF(k,k_{0},n)=\varrho(n)).$

Simply fix $p$ and $\varrho$. Compute $k$ and $k_{0}$ for $p$ such that the
prerequirement of the implication (5.1) are fulfilled. This is always possible
since $d^{-}(A_{p})>0$. Then construct $q_{\varrho}\leq p$ as in the lemma
above.

Hence, $\Vdash_{\mathbb{V}^{-}}(k,k_{0})\mapsto F(k,k_{0},\cdot)$ is a
surjection from $\omega\times\omega$ onto $2^{\omega}\cap V$.

∎

###  5.1. $\mathbb{MA}^{-}$ has large antichains

The following lemma establishes that below any condition $p\in\mathbb{MA}^{-}$
there is an antichain of size continuum.

######  Proposition 5.3.

There is a family of sets $\\{A_{f}\subseteq\omega\>:\>f\in I\\}$ such that

  1. (1)

$I\subseteq 3^{\omega}$ has size continuum,

  2. (2)

$d^{-}(A_{f})\geq 1/2$, for all $f\in I$.

  3. (3)

$d^{-}(A_{f}\cap A_{g})=0,$ for all $f\neq g$ in $I$.

###### Proof.

First, we fix a suitable set of indices $I\subseteq 3^{\omega}$. Therefore
pick any family of functions $I$ of size continuum with the property, that for
any two different functions $f,g\in I$ there are infinitely many $n\in\omega$
$(f(n)\neq g(n))$. In order to define the sets $A_{f}$ we will define three
auxiliary sets $B_{0},B_{1},B_{2}\subseteq\omega$ and a sequence $\langle
k^{n}_{j}\>:\>n\in\omega,j\in 3\cup\\{-1\\}\rangle$ such that

  1. i)

$k^{n}_{-1}<k^{n}_{0}<k^{n}_{1}<k^{n}_{2}=k^{n+1}_{-1}$, for $n<\omega$,

  2. ii)

$\frac{k^{n}_{j-1}}{k^{n}_{j}}<2^{-n}$, for $j<3,n<\omega$,

  3. iii)

$d^{-}(B_{i}\cup B_{j})\geq 1/2$, for $i\neq j$,

  4. iv)

$d^{-}(B_{i})=0$, for $i<3$.

We construct the sets $B_{i}$ recursively over $n\in\omega$ as a disjoint
union of sets $\\{B^{n}_{i}\\}_{n}$ such that each set $B^{n}_{i}$ is a subset
of $[k^{n}_{-1},k^{n}_{2})$. Start by defining $B^{0}_{i}:=\emptyset,i<3$ and
$k_{j}^{0}:=j+1,j\in 3\cup\\{-1\\}$. Now assume we have constructed
$B^{m}_{i}$ and $k^{m}_{j}$ for $m<n,i<3$ and $j\in 3\cup\\{-1\\}$. We start
with $k^{n}_{j}$. Let $k^{n}_{-1}:=k^{n-1}_{2}$ and choose $k^{n}_{j},j<3$ big
enough such that conditions $i)$ and $ii)$ are fulfilled and additionally the
difference $k^{n}_{j}-k^{n}_{j-1}$ is an even natural number for each $j<3$.
We perform three construction steps to define the sets $B^{n}_{i}$:

  1. a)

Divide the interval $[k^{n}_{-1},k^{n}_{0})$ evenly between the two sets
$B^{n}_{1}$ and $B^{n}_{2}$ and avoid the set $B^{n}_{0}$ entirely i.e.,

$\displaystyle[k^{n}_{-1},k^{n}_{0})\cap B^{n}_{0}$
$\displaystyle:=\emptyset,$ $\displaystyle[k^{n}_{-1},k^{n}_{0})\cap
B^{n}_{1}$ $\displaystyle:=\\{k^{n}_{-1},k^{n}_{-1}+2,\dots,k^{n}_{0}-2\\},$
$\displaystyle[k^{n}_{-1},k^{n}_{0})\cap B^{n}_{2}$
$\displaystyle:=\\{k^{n}_{-1}+1,k^{n}_{-1}+3,\dots,k^{n}_{0}-1\\}.$

  2. b)

Divide the interval $[k^{n}_{0},k^{n}_{1})$ evenly between the two sets
$B^{n}_{0}$ and $B^{n}_{2}$ and avoid the set $B^{n}_{1}$ entirely i.e.,

$\displaystyle[k^{n}_{0},k^{n}_{1})\cap B^{n}_{1}$ $\displaystyle:=\emptyset,$
$\displaystyle[k^{n}_{0},k^{n}_{1})\cap B^{n}_{0}$
$\displaystyle:=\\{k^{n}_{0},k^{n}_{0}+2,\dots,k^{n}_{1}-2\\},$
$\displaystyle[k^{n}_{0},k^{n}_{1})\cap B^{n}_{2}$
$\displaystyle:=\\{k^{n}_{0}+1,k^{n}_{0}+3,\dots,k^{n}_{1}-1\\}.$

  3. c)

Divide the interval $[k^{n}_{1},k^{n}_{2})$ evenly between the two sets
$B^{n}_{0}$ and $B^{n}_{1}$ and avoid the set $B^{n}_{2}$ entirely i.e.,

$\displaystyle[k^{n}_{1},k^{n}_{2})\cap B^{n}_{2}$ $\displaystyle:=\emptyset,$
$\displaystyle[k^{n}_{1},k^{n}_{2})\cap B^{n}_{0}$
$\displaystyle:=\\{k^{n}_{1},k^{n}_{1}+2,\dots,k^{n}_{2}-2\\},$
$\displaystyle[k^{n}_{1},k^{n}_{2})\cap B^{n}_{1}$
$\displaystyle:=\\{k^{n}_{1}+1,k^{n}_{1}+3,\dots,k^{n}_{2}-1\\}.$

This completes the construction of the sets $B^{n}_{i}$ and we can put
$B_{i}:=\bigcup_{n}B^{n}_{i}$. We check that conditions iii) and iv) are
fulfilled. Let $i\neq j$ be given. By construction steps $a)-c)$, we know that
$B_{i}\cup B_{j}$ selects at least each second natural number of each interval
$[k^{n}_{-1},k^{n}_{2})$. Since the intervals $[k^{n}_{-1},k^{n}_{2})$
partition $\omega$ we get iii). Condition iv) follows from

$\frac{|B_{i}\cap
k^{n}_{i}|}{k^{n}_{i}}\leq\frac{k^{n}_{i-1}}{k^{n}_{i}}<2^{-n}.$

Now, we are in a position to define the sets $A_{f},f\in I$. For $n\in\omega$
and $i\in 3$, we let $A^{n}_{i}:=B^{n}_{i_{0}}\cup B^{n}_{i_{1}}$, where
$i_{0}$ and $i_{1}$ are chosen such that $\\{i,i_{0},i_{1}\\}=3$. Then, we set
$A_{f}:=\bigcup_{n}A^{n}_{f(n)}$. We have to verify that the sets $A_{f}$
satisfy conditions $(2)$ and $(3)$. That each set $A_{f}$ has lower density
$\geq 1/2$ follows directly from condition iii) for the sets $B_{i}$. So let
$f,g\in I$ be two different functions and take $n$ such that $f(n)\neq g(n)$.
W.l.o.g. assume $f(n)=0$ and $g(n)=1$. Then, from construction step c) it
follows that $B_{0}\cap B_{2}\cap[k^{n}_{1},k^{n}_{2})=\emptyset$ and thus

$\frac{|A_{f}\cap A_{g}\cap
k^{n}_{2}|}{k^{n}_{2}}\leq\frac{k^{n}_{1}}{k^{n}_{2}}<2^{-n}.$

Since $f$ and $g$ differ on infinitely many $n$ we get $d^{-}(A_{f}\cap
A_{g})=0$. ∎

The results of the previous sections are summarized in the table below.

$\mathbb{P}$ | $\mathbb{MA}^{+}_{\varepsilon}$ | $\mathbb{V}^{+}_{\varepsilon}$ | $\mathbb{MA}^{-}_{1}$ | $\mathbb{MA}^{-}_{\varepsilon}$ | $\mathbb{V}^{-}_{\varepsilon}$ | $\mathbb{MA}^{+}$ | $\mathbb{V}^{+}$ | $\mathbb{MA}^{-}$ | $\mathbb{V}^{-}$
---|---|---|---|---|---|---|---|---|---
proper | ✓ | ✓ | ✓ | ✓ |  | ✓ |  |  | ✗
c.c.c | ✗ | ✗ | ✓ | ✗ | ✗ | ✗ | ✗ | ✗ | ✗
Cohen | ✓ | ✗ | ✓ | ✓ |  | ✓ |  | ✓ |
dominating | ✓ | ✗ |  |  |  | ✓ |  |  |

##  6\. Measurability

In this section we compare different notions of measurability. We first
establish some notations.

######  Definition 6.1.

Let $\mathcal{X}$ be a non-empty set and $\mathbb{P}$ be a tree-forcing
defined over $\mathcal{X}^{\omega}$.

  * (1)

A subset $X\subseteq\mathcal{X}^{\omega}$ is called _$\mathbb{P}$ -nowhere
dense,_ if

$(\forall p\in\mathbb{P})(\exists q\leq p)([q]\cap X=\emptyset).$

We denote the ideal of $\mathbb{P}$-nowhere dense sets with
$\mathcal{N}_{\mathbb{P}}$.

  * (2)

A subset $X\subseteq\mathcal{X}^{\omega}$ is called _$\mathbb{P}$ -meager_ if
it is included in a countable union of $\mathbb{P}$-nowhere dense sets. We
denote the $\sigma$-ideal of $\mathbb{P}$-meager sets with
${\mathcal{I}}_{\mathbb{P}}$.

  * (3)

A subset $X\subseteq\mathcal{X}^{\omega}$ is called _ $\mathbb{P}$-measurable_
if

$(\forall p\in\mathbb{P})(\exists q\leq p)([q]\setminus
X\in\mathcal{I}_{\mathbb{P}}\vee[q]\cap X\in\mathcal{I}_{\mathbb{P}}).$

  * (4)

A family $\Gamma\subseteq\mathcal{P}(\mathcal{X}^{\omega})$ is called _well-
sorted_ if it is closed under continuous pre-images. We abbreviate the
sentence “every set in $\Gamma$ is $\mathbb{P}$-measurable” by
$\Gamma(\mathbb{P})$.

We make two useful observations concerning the measurability of a set $X$.

######  Observation 6.2.

Let $\mathbb{P}$ be a tree-forcing and $X\subseteq\mathcal{X}^{\omega}$ a set
of reals.

  1. (1)

If $H$ is $\mathbb{P}$-comeager, then $X$ is $\mathbb{P}$-measurable if and
only if $H\cap X$ is $\mathbb{P}$-measurable.

  2. (2)

If $\mathcal{I}_{\mathbb{P}}=\mathcal{N}_{\mathbb{P}}$, then $X$ is
$\mathbb{P}$-measurable if and only if

$(\forall p\in\mathbb{P})(\exists q\leq p)([q]\subseteq X\vee[q]\cap
X=\emptyset).$

For Sacks, Laver, Miller, Silver and Mathias forcing we have
$\mathcal{N}_{\mathbb{P}}=\mathcal{I}_{\mathbb{P}}$. In case of the Silver
forcing the usual proof for
$\mathcal{I}_{\mathbb{V}}=\mathcal{N}_{\mathbb{V}}$ also works for
$\mathbb{V}^{+}_{\varepsilon}$, since it only makes use of fusion sequences.
But it is unclear whether we can expect the same for the other versions of
Silver forcing.
In case of the Mathias forcing however, we don’t have an equality in none of
the four versions of the forcing.

######  Theorem 6.3.

  1. (1)

$\mathcal{I}_{\mathbb{V}^{+}_{\varepsilon}}=\mathcal{N}_{\mathbb{V}^{+}_{\varepsilon}}$,

  2. (2)

$\mathcal{I}_{\mathbb{P}}\neq\mathcal{N}_{\mathbb{P}}$ for
$\mathbb{P}\in\\{\mathbb{MA}^{+}_{\varepsilon},\mathbb{MA}^{-}_{\varepsilon},\mathbb{MA}^{+},\mathbb{MA}^{-}\\}$.

###### Proof.

(1) As we mentioned above, the first part of the Theorem is a straightforward
generalization of the proof for the usual Silver forcing. The only difference
being that one has to use the partial orderings defined as in Definition 2.3
to ensure that fusions exist in $\mathbb{V}^{+}_{\varepsilon}.$
(2) We divide the proof into cases.

  1. $(\geq\varepsilon)$

Let $\varepsilon\in(0,1]$ be given. The proof is closely intertwined with the
fact that the forcings $\mathbb{MA}^{+}_{\varepsilon}$ and
$\mathbb{MA}^{-}_{\varepsilon}$ add Cohen reals. We quickly recall the
construction of the Cohen real from Lemma 2.12 and define a function $\varphi$
such that the image of the generic real is a Cohen real. Fix
$\varepsilon>0,N\in\omega$ such that $1/N<\varepsilon$ and a partition
$\\{a_{i}\\}_{i<N+1}$ of $\omega$ such that each set $a_{i}$ has density
$1/(N+1)$. Now let $H:=\\{x\in 2^{\omega}\>:\>\exists^{\infty}i(x(i)=1)\\}$.
For $x\in H$ let $\langle n_{k}^{x}\>:\>k<\omega\rangle$ enumerate all
integers $n$ such that $x(n)=1$. We define a function $\varphi:H\rightarrow
2^{\omega}$ as follows:

$\varphi(x)(k):=\begin{cases}0,&\exists
i<(N+1)\\{n_{2k}^{x},n_{2k+1}^{x}\\}\subseteq a_{i}\\\ 1,&\text{
else.}\end{cases}$

For $n\in\omega$ let $M_{n}:=\\{x\in H\>:\>\forall k\geq
n(\varphi(x)(k)=0)\\}$. It is not hard to see that for
$\mathbb{P}\in\\{\mathbb{MA}^{+}_{\varepsilon},\mathbb{MA}^{-}_{\varepsilon}\\}$,
and each $n$ we have $M_{n}\in\mathcal{N}_{\mathbb{P}}$, but
$M:=\bigcup_{n}M_{n}\not\in\mathcal{N}_{\mathbb{P}}$.

  2. $(>0)$

In this case the meager set which is not nowhere dense lies directly at hand.

######  Claim 6.4.

The set $\mathcal{Z}^{+}$ is $\mathbb{P}$-meager but not $\mathbb{P}$-nowhere
dense, for $\mathbb{P}\in\\{\mathbb{MA}^{+},\mathbb{MA}^{-}\\}$.

The proof works for both forcings analogously. We only check it for
$\mathbb{MA}^{+}$ explicitly. For $n\in\omega$ we define the sets
$N_{n}:=\\{x\in 2^{\omega}\>:\>d^{+}(x)\geq 1/n\\}$. Let $n\in\omega$ and
$p\in\mathbb{MA}^{+}$ be fixed. We can easily find $q\leq p$ such that
$d^{+}(A_{q})<1/n$. Such a condition $q$ satisfies the property $\forall
x\in[q](d^{+}(x)<1/n)$ and in particular $[q]\cap N_{n}=\emptyset.$ This
proves that each set $N_{n}$ is $\mathbb{MA}^{+}$-nowhere dense and so
$\mathcal{Z}^{+}=\bigcup_{n}N_{n}$ is $\mathbb{MA}^{+}$-meager. To see that
$\mathcal{Z}^{+}$ cannot be $\mathbb{MA}^{+}$-nowhere dense it is enough to
check that each condition $p=(s,A_{p})\in\mathbb{MA}^{+}$ contains a branch
$x$ of positive upper density. Clearly, the rightmost branch (i.e.
$x(i)=1\Leftrightarrow i\in s\cup A_{p}$) fulfills this requirement.

∎

Since subsets of $\mathbb{P}$-meager sets are $\mathbb{P}$-meager as well,
$\mathbb{P}$-meager and $\mathbb{P}$-comeager sets are $\mathbb{P}$-measurable
we get the following.

######  Corollary 6.5.

The sets $\mathcal{Z},\mathcal{Z}^{*}$ and $\mathcal{Z}^{+}$ are
$\mathbb{P}$-measurable, $\mathbb{P}\in\\{\mathbb{MA}^{+},\mathbb{MA}^{-}\\}$.

Our next goal is to compare different notions of measurability. Let
$\mathbb{P}$ be any tree-forcing. The statement

$``\Gamma(\mathbb{P})\Rightarrow\Gamma(\mathbb{C})\text{, for each well-sorted
family }\Gamma"$

is true for various tree-forcings adding Cohen reals e.g. Hechler forcing
$\mathbb{D}$, Eventually different forcing $\mathbb{E}$, a Silver like version
of Mathias forcing $\mathbb{T}$ introduced in [14, Definition 2.1.] and the
full-splitting Miller forcing $\mathbb{FM}$ [12, Definition 1.1.]. In fact, it
appears reasonable enough to ask.

######  Question 6.6.

Does each tree-forcing $\mathbb{P}$ adding a Cohen real, necessarily satisfy
$\Gamma(\mathbb{P})\Rightarrow\Gamma(\mathbb{C})$, for each well sorted family
$\Gamma$?

A partial answer to this question was given in [14, Proposition 3.1.]. We
restate the proposition in a slightly modified version, that will allow us to
generalize the result later on.

######  Proposition 6.7.

Let $\mathcal{X}$ be a set of size $\leq\omega$ and $\mathbb{P}$ be tree-
forcing defined on $\mathcal{X}^{<\omega}$. Equip $\mathcal{X}$ with the
discrete topology and $\mathcal{X}^{\omega}$ with the product topology. Let
$\varphi^{*}:\mathcal{X}^{<\omega}\rightarrow 2^{<\omega}$ and
$\varphi:\mathcal{X}^{\omega}\rightarrow 2^{\omega}$ be two mappings that
satisfy the following conditions:

  1. (1)

$\varphi^{*}$ is order preserving and
$\varphi^{*}(\langle\rangle)=\langle\rangle$,

  2. (2)

$\varphi$ is continuous,

  3. (3)

$(\forall p\in\mathbb{P})\varphi[p]$ is open dense in
$[\varphi^{*}(\operatorname{stem}(p))]$,

  4. (4)

$(\forall p\in\mathbb{P})\forall
t\trianglerighteq\varphi^{*}(\operatorname{stem}(p))\exists p^{\prime}\leq
p(\varphi^{*}(\operatorname{stem}(p^{\prime}))\trianglerighteq t).$

Then $\Gamma(\mathbb{P})\Rightarrow\Gamma(\mathbb{C})$, for each well-sorted
family.

To understand why Proposition 6.7 is a partial answer to Question 6.6, it
should be noted that given $\mathbb{P},\varphi$ as in the proposition and
$x_{G}$ a $\mathbb{P}$-generic real, we have that $\varphi(x_{G})$ is Cohen.

######  Theorem 6.8.

$\Gamma(\mathbb{MA}^{+}_{\varepsilon})\Rightarrow\Gamma(\mathbb{C})$, for each
well sorted family $\Gamma$.

###### Proof.

Our aim is to apply Proposition 6.7. To this end let $H^{*}:=\\{s\in
2^{<\omega}\>:\>|s^{-1}(\\{1\\})|\text{ is even }\\}$ denote the set of all
finite sequences with an even number of $1^{\prime}s$. For $s\in H^{*}$ let
$\langle n_{k}\>:\>k<|s^{-1}(\\{1\\})|\rangle$ enumerate $s^{-1}(\\{1\\})$.
Observe that $H^{*}$ is dense in $2^{<\omega}$. Now fix $N\in\omega$ and a
partition $\\{a_{i}\\}_{i<(N+1)}$ of $\omega$ such that $\varepsilon>1/(N+1)$
and such that each set $a_{i}$ has density $1/(N+1)$. We define the function
$\varphi^{*}$ as expected. Let $k\in\omega$ and $s\in H^{*}$ be such that
$2k<|s^{-1}(\\{1\\})|$. We define:

$\varphi^{*}(s)(k):=\begin{cases}0,&\exists
i<(N+1)(\\{n_{2k},n_{2k+1}\\}\subseteq a_{i})\\\ 1,&\text{ else.}\end{cases}$

Let $H$ and $\varphi:H\rightarrow 2^{<\omega}$ be defined as in the proof of
Theorem 6.3 (2). Now, given $x\in H$ let $\langle
n_{k}^{x}\>:\>k<\omega\rangle$ enumerate $x^{-1}\\{1\\}$. Then,
$\varphi^{*}(x{\upharpoonright}n_{2k}^{x})$ is defined for each $k$ and
$\varphi(x)=\bigcup_{k}\varphi^{*}(x{\upharpoonright}n_{2k}^{x})$.
Note that the same argument used in the proof to show that
$\mathbb{MA}^{+}_{\varepsilon}$ adds Cohen reals, gives us
$\varphi[p]=[\varphi^{*}(\operatorname{stem}(p))]$. Especially, condition (3)
from the proposition is fulfilled. It is straightforward to check that the
other three conditions are satisfied as well and since the set $H$ is
$\mathbb{MA}^{+}_{\varepsilon}$-comeager, we can apply the proposition. ∎

##  7\.  $\mathbb{MA}^{+}$-measurability

In this section we examine $\mathbb{MA}^{+}$-measurability and give a
generalization of Proposition 6.7.
By Theorem 4.4 we know that $\mathbb{MA}^{+}$ adds Cohen reals. So, it seems
reasonable enough to try the same method used in Section 6 to prove
$\Gamma(\mathbb{MA}^{+}_{\varepsilon})\Rightarrow\Gamma(\mathbb{C})$, with the
forcing $\mathbb{MA}^{+}$. However, in doing so one encounters the following
problem. The coding from Theorem 4.4 used to generate the Cohen real, uses
information of the _whole_ condition and does not depend solely on the stem.
To make it clear what we mean, we quickly explain how using the proof of
Theorem 4.4 one gets a coding function defined on a $\mathbb{MA}^{+}$-comeager
set.

###  7.1. Construction of the coding function $\varphi$

Recall that we defined in the proof of Theorem 4.4 families
$\mathcal{A}_{n}:=\\{A^{i}_{n}\>:\>i<2^{(n+1)}\\}$, where
$A^{i}_{n}:=\\{k\in\omega\>:\>k=i\mod 2^{(n+1)}\\}$. Each $\mathcal{A}_{n}$ is
a maximal antichain in $(\mathcal{Z}^{+},\subseteq^{*})$ and each set
$A^{i}_{n}$ has density $2^{-(n+1)}$. Now we define
$D_{n}:=\\{A\in[\omega]^{\omega}\>:\>\exists
i<2^{-(n+1)}(A\subseteq^{*}A^{i}_{n})\\}$. Then for each $n\in\omega$ and
$(s,A)\in\mathbb{MA}(\mathcal{Z}^{+})$ there is $B\in D_{n}$ such that
$(s,B)\leq(s,A)$. This already implies that the set $D:=\bigcap D_{n}$ is
$\mathbb{MA}(\mathcal{Z}^{+})$-comeager and thus it is enough to find a
suitable coding function $\varphi$ defined on $D$ instead of $2^{\omega}$.
First, note that for $x\in D$ we also must have the property $(\forall
n\exists!i_{n}(x\subseteq^{*}A^{i_{n}}_{n}))$ and we abbreviate
$A^{i_{n}}_{n}$ with $A_{n}$. So, we can define for $x\in D$ two sequences
$\langle n_{i}\>:\>i<\omega\rangle$ and $\langle
m_{i}\>:\>i<\omega\setminus\\{0\\}\rangle$ as in the proof of Theorem 4.4.
Finally, $\varphi:D\rightarrow 2^{\omega}$ is defined as

$\varphi(x)(i):=\begin{cases}0,&\text{ if }x\setminus(m_{i+1}+1)\subseteq
A_{m_{i+1}}\\\ 1,&\text{ else. }\end{cases}$

This definition of the coding function $\varphi$ seems promising. However,
when one takes a close look at possible candidates for a corresponding
$\varphi^{*}$, it becomes clear that $\varphi^{*}$ cannot depend solely on the
stem of a condition. Especially, we cannot apply Proposition 6.7 as it is
stated in Section 6. We have to find a generalization like the following.

######  Proposition 7.1.

Let $\mathcal{X},\mathcal{Y}$ be sets of size $\leq\omega$,
$\mathbb{P},\mathbb{Q}$ be tree-forcings defined on
$\mathcal{X}^{<\omega},\mathcal{Y}^{<\omega}$ respectively. Equip
$\mathcal{X},\mathcal{Y}$ with the discrete topology and
$\mathcal{X}^{\omega},\mathcal{Y}^{\omega}$ with the product topology. Let
$\varphi^{*}:\mathbb{P}\rightarrow\mathbb{Q}$ and
$\varphi:\mathcal{X}^{\omega}\rightarrow\mathcal{Y}^{\omega}$ be two mappings
that satisfy the following conditions:

  1. (1)

$\varphi^{*}$ is order preserving and
$\varphi^{*}(1_{\mathbb{P}})=1_{\mathbb{Q}}$,

  2. (2)

$\varphi$ is continuous,

  3. (3)

$\forall p\in\mathbb{P}\varphi[p]$ is $\mathbb{Q}$-open dense in
$[\varphi^{*}(p)]$,

  4. (4)

$\forall p\in\mathbb{P}\forall q\leq\varphi^{*}(p)\exists p^{\prime}\leq
p(\varphi^{*}(p^{\prime})\leq q).$

Then $\Gamma(\mathbb{P})\Rightarrow\Gamma(\mathbb{Q})$, for each well-sorted
family $\Gamma$.

The key difference is that $\varphi^{*}$ is a map from $\mathbb{P}$ to
$\mathbb{Q}$, instead of being defined for finite sequences.
Before we turn to the proof of the proposition we investigate further if we
might apply it to $\mathbb{MA}^{+}$ and $\mathbb{C}$. To this end we already
have defined a $\mathbb{MA}^{+}$-comeager set $D$ and a coding function
$\varphi:D\rightarrow 2^{\omega}$ satisfying $\varphi(x_{G})$ is Cohen, where
$x_{G}$ is $\mathbb{MA}^{+}$-generic. We want to define
$\varphi^{*}:\mathbb{MA}^{+}\rightarrow\mathbb{C}$. Let $\dot{x}_{G}$ be the
canonical name for the $\mathbb{MA}^{+}$-generic real. For
$p\in\mathbb{MA}^{+}$ let $r_{p}\in 2^{<\omega}$ be maximal such that
$p\Vdash\varphi(\dot{x}_{G})\trianglerighteq r_{p}$ and put
$\varphi^{*}(p):=r_{p}$. Observe that we have the following properties:

  * •

$\varphi^{*}$ is order preserving and
$\varphi^{*}((\langle\rangle,\omega))=\langle\rangle$,

  * •

$\varphi$ is not continuous,

  * •

It follows from the proof that $\varphi(\dot{x}_{G})$ is Cohen, that
conditions (3) and (4) of Proposition 7.1 are satisfied.

This shows that we almost get
$\Gamma(\mathbb{MA}^{+})\Rightarrow\Gamma(\mathbb{C})$, for any well-sorted
family $\Gamma$. In fact, the only time we need $\varphi$ to be continuous is
to ensure that the pre-image of a regular set $Y\in\Gamma$ is again regular.
So, if we change the requirement of $\Gamma$ of being well-sorted and instead
assume that the family of sets $\Gamma$ is closed under pre-images of the
$\varphi$ constructed above we get:

######  Corollary 7.2.

Let $\varphi$ be defined as in the beginning of this section and $\Gamma$ be a
family of sets closed under pre-images of $\varphi$. Then
$\Gamma(\mathbb{MA}^{+})\Rightarrow\Gamma(\mathbb{C})$ holds.

Now we prove Proposition 7.1. The key step is the following lemma.

######  Lemma 7.3.

Let $\mathbb{P},\mathbb{Q},\varphi,\varphi^{*}$ be as in the Proposition and
$Y\subseteq\mathcal{Y}^{\omega}$. Define $X:=\varphi^{-1}[Y]$. Assume there is
$p\in\mathbb{P}$ such that $X\cap[p]$ is $\mathbb{P}$-comeager in $[p]$. Then
$Y\cap[\varphi^{*}(p)]$ is $\mathbb{Q}$-comeager in $[\varphi^{*}(p)]$.

###### Proof.

We are assuming $X\cap[p]$ is $\mathbb{P}$-comeager, for some
$p\in\mathbb{P}$. This implies that there is a collection
$\\{A_{n}\>:\>n<\omega\wedge A_{n}\text{ is }\mathbb{P}\text{-open dense in
}[p]\\}$ such that $\bigcap_{n}A_{n}\subseteq[p]\cap X$. W.l.o.g. assume
$A_{n}\supseteq A_{n+1}$, for all $n$. Let $q=\varphi^{*}(p)$. We want to show
that $\varphi[X]\cap[q]=Y\cap[q]$ is $\mathbb{Q}$-comeager in $[q]$ i.e., we
want to find $\\{B_{n}\>:\>n<\omega\\}$ $\mathbb{Q}$-open dense sets in $[q]$
such that $\bigcap_{n}B_{n}\subseteq Y\cap[q]$. Given
$\sigma\in\mathfrak{c}^{<\omega}$ we recursively define on the length of
$\sigma$ a set
$\\{p_{\sigma}\>:\>\sigma\in\mathfrak{c}^{<\omega}\\}\subseteq\mathbb{P}$ with
the following properties:

1.:

$p_{\langle\rangle}=p$,

2.:

$\forall\sigma\in\mathfrak{c}^{<\omega}\;\bigcup_{i}[\varphi^{*}(p_{\sigma^{\smallfrown}i})]$
is $\mathbb{Q}$-open dense in $[\varphi^{*}(p_{\sigma})]$,

3.:

$\forall\sigma\in\mathfrak{c}^{<\omega}\forall
i\in\omega\;([p_{\sigma^{\smallfrown}i}]\subseteq\bigcap_{k\leq|\sigma|}A_{k}\wedge
p_{\sigma^{\smallfrown}i}\leq p_{\sigma})$.

Assume we are at step $n$. Fix $\sigma\in\mathfrak{c}^{n}$ arbitrarily and
then put $q_{\sigma}=\varphi^{*}(p_{\sigma})$. We first make sure that $2.$
holds. For this purpose, let $\\{q_{i}\>:\>i<\mathfrak{c}\\}$ enumerate all
conditions in $\mathbb{Q}$ below $q_{\sigma}$. By condition (4) from
Proposition 7.1 we can find $p_{i}\leq p_{\sigma}$ such that
$\varphi^{*}(p_{i})\leq q_{i}$. Since each $A_{k}$ is $\mathbb{P}$-open dense
in $[p]$ we can find for each $i<\mathfrak{c}$ an extension
$p_{\sigma^{\smallfrown}i}\leq p_{i}$ such that
$[p_{\sigma^{\smallfrown}i}]\subseteq\bigcap_{k\leq n}A_{k}$. This ensures
that also $3.$ holds. Finally, we put
$B_{n}:=\bigcup\\{\varphi[[p_{\sigma}]]\>:\>\sigma\in\mathfrak{c}^{n}\\}$.
We have to check that each set $B_{n}$ is $\mathbb{Q}$-open dense in $[q]$ and
$\bigcap_{n}B_{n}\subseteq Y\cap[q]$. So fix $n\in\omega$ and $q^{\prime}\leq
q=\varphi^{*}(p)$. In the first construction step this $q^{\prime}$ was
enumerated, say by $i<\mathfrak{c}$ so $q^{\prime}=q_{i}$ and
$\varphi^{*}(p_{\langle i\rangle})\leq q^{\prime}$. Especially,
$\varphi^{*}(p_{\sigma})\leq q^{\prime}$, whenever
$\sigma\in\mathfrak{c}^{n},\sigma(0)=i$. By condition $(3)$ from the
Proposition $\varphi[[p_{\sigma}]]$ is $\mathbb{Q}$-open dense in
$[\varphi^{*}(p_{\sigma})]$. This proves that $B_{n}$ is $\mathbb{Q}$-open
dense in $[q].$
By construction of $B_{n+1}$ we know $B_{n}\subseteq\varphi[\bigcap_{k\leq
n+1}A_{k}]$ and hence

$\bigcap B_{n}\subseteq\varphi[\bigcap_{n}A_{n}]\subseteq\varphi[[p]\cap
X]\subseteq\varphi[p]\cap Y\subseteq[q]\cap Y.$

∎

###### Proof of the proposition.

Let $Y\in\Gamma$ be given and put $X:=\varphi^{-1}[Y]$. Then also
$X\in\Gamma$, since $\Gamma$ is well-sorted and $\varphi$ is continuous. We
now use the lemma to show that for every $q\in\mathbb{Q}$ there exists
$q^{\prime}\leq q$ such that $Y\cap[q^{\prime}]$ is $\mathbb{Q}$-meager or
$Y\cap[q^{\prime}]$ is $\mathbb{Q}$-comeager.

Observe that by conditions $(1)$ and $(4)$ we get $\varphi^{*}[\mathbb{P}]$ is
dense in $\mathbb{Q}$. Now fix $q\in\mathbb{Q}$ arbitrarily and pick
$p\in\mathbb{P}$ such that $\varphi^{*}(p)\leq q$. By assumption $X$ is
$\mathbb{P}$-measurable, and so:

  * •

in case there exists $p^{\prime}\leq p$ such that $X\cap[p^{\prime}]$ is
$\mathbb{P}$-comeager; put $q^{\prime}:=\varphi^{*}(p^{\prime})$. By the lemma
above, $Y\cap[q^{\prime}]$ is $\mathbb{Q}$-comeager in $[q^{\prime}]$;

  * •

in case there exists $p^{\prime}\leq p$ such that $X\cap[p^{\prime}]$ is
$\mathbb{P}$-meager, then apply the lemma above to the complement of $X$, in
order to get $Y\cap[q^{\prime}]$ be meager in $[q^{\prime}]$, with
$q^{\prime}:=\varphi^{*}(p^{\prime})$.

∎

######  Corollary 7.4.

Let $\mathbb{P},\mathbb{Q},\varphi^{*}$ and $\varphi$ be as in proposition.
Then $\varphi(x_{G})$ is $\mathbb{Q}$-generic, where $x_{G}$ is a
$\mathbb{P}$-generic real.

###### Proof of the Corollary.

Fix an $\mathbb{Q}$-open dense set $D\subseteq\mathbb{Q}$. We want to show
that the conditions $p\in\mathbb{P}$ such that $p\Vdash\exists q\in
D(\varphi(x_{G})\in[q])$ is dense in $\mathbb{P}$. To this end, fix a
condition $p\in\mathbb{P}$. Then, since $D$ is dense in $\mathbb{Q}$ there is
$q^{\prime}\in D$ below $\varphi^{*}(p)$. By condition $(4)$ of Proposition
7.1 there is $p^{\prime}\leq p$ such that $\varphi^{*}(p^{\prime})\leq
q^{\prime}$. By condition $(3)$ we know that each condition $r\in\mathbb{P}$
forces $\varphi(x_{G})$ into $[\varphi^{*}(r)]$ and hence
$p^{\prime}\Vdash\varphi(x_{G})\in[\varphi^{*}(p^{\prime})]\subseteq[q^{\prime}]$.

∎

In light of this one might also generalize Question 6.6 to the following.

######  Question 7.5.

Let $\mathbb{P},\mathbb{Q}$ be two tree-forcings and assume $\mathbb{P}$ adds
a $\mathbb{Q}$-generic. Does $\Gamma(\mathbb{P})\Rightarrow\Gamma(\mathbb{Q})$
hold, for each well-sorted family $\Gamma$?

##  8\. A model for
$\Sigma^{1}_{2}(\mathbb{V}^{+}_{\varepsilon})\wedge\neg\Sigma^{1}_{2}(\mathbb{C})$

We construct a model in which the implication
$\Gamma(\mathbb{V}^{+}_{\varepsilon})\Rightarrow\Gamma(\mathbb{C})$ fails for
$\Gamma=\Sigma^{1}_{2}$.
We recall that the shortest splitting node extending $s\in 2^{<\omega}$ is
denoted by $\operatorname{splsuc}(s)$ (see Definition 1.1 (e)).

######  Lemma 8.1.

Let $\varepsilon\in(0,1]$ and $p\in\mathbb{V}^{+}_{\varepsilon}$. Let
$\bar{\varphi}:\operatorname{Split}(p)\rightarrow 2^{<\omega}$ such that
$\bar{\varphi}(\operatorname{stem}(p)):=\langle\rangle$ and for every
$t\in\operatorname{Split}(p)$ and $j\in\\{0,1\\}$,

$\bar{\varphi}(\operatorname{splsuc}(t^{\smallfrown}\langle
j\rangle)):=\bar{\varphi}(t)^{\smallfrown}\langle j\rangle.$

Let $\varphi:[p]\rightarrow 2^{\omega}$ be the expansion of $\bar{\varphi}$,
i.e. for every $x\in[p]$,
$\varphi(x):=\bigcup_{n\in\omega}\bar{\varphi}(t_{n})$, where $\langle
t_{n}\>:\>n\in\omega\rangle$ is a $\trianglelefteq$-increasing sequence of
splitting nodes in $p$ such that $x=\bigcup_{n\in\omega}t_{n}$.

If $c$ is Cohen generic over $V$, then

$V[c]\models\exists p^{\prime}\in\mathbb{V}^{+}_{\varepsilon}\land
p^{\prime}\subseteq p\land\forall x\in[p^{\prime}](\varphi(x)\text{ is Cohen
over }V).$

###### Proof.

For a finite tree $T\subseteq 2^{<\omega}$ we define the set of terminal nodes
$\operatorname{Term}(T):=\\{s\in T\>:\>\lnot\exists t\in T(s\triangleleft
t)\\}$. Consider the following forcing $\mathbb{P}$ consisting of finite trees
$T\subseteq 2^{<\omega}$ such that for all $s,t\in T$ the following holds:

  1. (1)

If $s,t\in\operatorname{Term}(T)$, then $|s|=|t|$.

  2. (2)

If $s,t\not\in\operatorname{Term}(T)$ and $|s|=|t|$, then $s^{\smallfrown}i\in
T$ iff $t^{\smallfrown}i\in T$, $i\in 2$.

The partial order $\mathbb{P}$ is ordered by end-extension: $T^{\prime}\leq T$
iff $T^{\prime}\supseteq T$ and $\forall t\in T^{\prime}\setminus T\exists
s\in\operatorname{Term}(T)(s\trianglelefteq t)$.

Note $\mathbb{P}$ is countable and non-trivial, thus it is equivalent to Cohen
forcing $\mathbb{C}$. Let $p_{G}:=\bigcup G$, where $G$ is
$\mathbb{P}$-generic over $V$. We claim that
$p^{\prime}:=\bar{\varphi}^{-1}"p_{G}$ satisfies the required properties. It
is left to show that:

  1. (1)

for every $x\in[p^{\prime}]$ one has $\varphi(x)$ is Cohen generic, i.e.,
every $y\in[p_{G}]$ is Cohen generic;

  2. (2)

$p^{\prime}\in\mathbb{V}_{\varepsilon}^{+}$.

For proving (1), let $D$ be an open dense subset of $\mathbb{C}$ and
$T\in\mathbb{P}$. It is enough to find $T^{\prime}\leq T$ such that every
$t\in\operatorname{Term}(T^{\prime})$ is a member of $D$.

Let $\\{t_{j}:j<N\\}$ enumerate all terminal nodes in $T$ and pick $r_{N}$, so
that for every $j<N$, $(t_{j}{}^{\smallfrown}r_{N}\in D)$. Then put
$T^{\prime}:=\\{t\in 2^{<\omega}\>:\>\exists
t_{j}\in\operatorname{Term}(T)(t\trianglelefteq{t_{j}}^{\smallfrown}r_{N}\\}$.
Hence $T^{\prime}\leq T$ and $T^{\prime}\Vdash\forall y\in[p_{G}]\exists
t\in(T^{\prime}\cap D)(t\triangleleft y)$. Hence we have proven that

$\Vdash_{\mathbb{C}}\forall y\in[p_{G}]\exists t\in D(t\triangleleft y),$

which means every $y\in[p_{G}]$ is Cohen generic over $V$.

For proving (2), one has to verify that the resulting set of splitting levels
$A_{p^{\prime}}$ has upper density $\geq\varepsilon$. It is easy to see that
given any condition $T\in\mathbb{P}$ and $n\in\omega$ one can always find an
end-extension $T_{n}\leq T$ such that

$T_{n}\Vdash\exists k<\omega\left(\frac{|A_{p^{\prime}}\cap
k|}{k}\geq\varepsilon-2^{-n}\right).$

∎

######  Proposition 8.2.

Let $\mathbb{C}_{\omega_{1}}$ be an $\omega_{1}$-product with finite support
and let $G$ be $\mathbb{C}_{\omega_{1}}$-generic over the constructible
universe $L$. Then, for every $\varepsilon\in(0,1]$

$L(\mathbb{R})^{L[G]}\models\text{``All }\text{On}^{\omega}\text{-definable
sets are
}(\mathbb{V}^{+}_{\varepsilon})\text{-measurable''}\land\neg\Sigma^{1}_{2}(\mathbb{C}).$

###### Proof.

The argument is the same as in the proof of [2, Proposition 3.7]. Fix
$\varepsilon\in(0,1]$. Let for $\alpha\leq\omega_{1}$ $\mathbb{C}_{\alpha}$
denote the forcing adding $\alpha$ Cohen reals. Let $G$ be
$\mathbb{C}_{\omega_{1}}$-generic over $L$.

Let $X$ be an $\text{On}^{\omega}$-definable set of reals, i.e. $X:=\\{x\in
2^{\omega}:\psi(x,v)\\}$ for a formula $\psi$ with a parameter
$v\in\text{On}^{\omega}$, and let $p\in\mathbb{V}^{+}_{\varepsilon}$. We aim
to find $q\leq p$ such that $[q]\subseteq X$ or $[q]\cap X=\emptyset$.

We can find $\alpha<\omega_{1}$ such that $v,p\in
L[G{\upharpoonright}\alpha]$. Let $\varphi:[p]\rightarrow 2^{\omega}$ be as in
Lemma 8.1. Let $c=G(\alpha)$ be the next Cohen real and write $\mathbb{C}$ for
the $\alpha$-component of $\mathbb{C}_{\omega_{1}}$. We let

$b_{0}=\big{\llbracket}\llbracket(\psi(\varphi^{-1}(c),v))\rrbracket_{\mathbb{C}_{\alpha}}=\mathbf{0}\big{\rrbracket}_{\mathbb{C}}\quad\text{and}\quad
b_{1}=\big{\llbracket}\llbracket\psi(\varphi^{-1}(c),v))\rrbracket_{\mathbb{C}_{\alpha}}=\mathbf{1}\big{\rrbracket}_{\mathbb{C}}.$

Then, by $\mathbb{C}$-homogeneity, $b_{0}\land b_{1}=\mathbf{0}$ and
$b_{0}\vee b_{1}=\mathbf{1}$. Hence, by applying Lemma 8.1, one can then find
$q\leq p$ such that $q\subseteq b_{0}$ or $q\subseteq b_{1}$ and for every
$x\in[q]$, $\varphi(x)$ is Cohen over $L[G{\upharpoonright}\alpha]$. We claim
that $q$ satisfies the required property.

  * •

Case $q\subseteq b_{1}$: note for every $x\in[q]$, $\varphi(x)$ is Cohen over
$L[G{\upharpoonright}\alpha]$, and so
$L[G{\upharpoonright}\alpha][\varphi(x)]\models\llbracket\psi(\varphi^{-1}(\varphi(x)),v)\rrbracket_{\mathbb{C}_{\alpha+1}}$.
Hence $L[G]\models\forall x\in[q](\psi(x,v))$, which means
$L[G]\models[q]\subseteq X$.

  * •

Case $q\subseteq b_{0}$: we argue analogously and get $L[G]\models\forall
x\in[q](\neg\psi(x,v))$, which means $L[G]\models[q]\cap X=\emptyset$.

Moreover in $L[G]$ it is well-known that $\mathbf{\Sigma}^{1}_{2}(\mathbb{C})$
fails (see [3, Theorem 5.8] and [1, 6.5.3, p. 313]). Hence in $L[G]$ all
$\text{On}^{\omega}$-definable sets are
$\mathbb{V}^{+}_{\varepsilon}$-measurable, but there is a
$\mathbf{\Sigma}^{1}_{2}$ set not satisying the Baire property. As a
consequence, in particular we obtain
$L(\mathbb{R})^{L[G]}\models\Sigma^{1}_{2}(\mathbb{V}^{+}_{\varepsilon})\land\neg\Sigma^{1}_{2}(\mathbb{C})$.
∎

We conclude this section by summarizing our results from Sections 6,7 and 8.

$\mathbb{P}$ | $\mathbb{MA}^{+}_{\varepsilon}$ | $\mathbb{MA}^{+}$ | $\mathbb{V}^{+}_{\varepsilon}$
---|---|---|---
$\mathcal{I}_{\mathbb{P}}=\mathcal{N}_{\mathbb{P}}$ | ✗ | ✗ | ✓
$\Gamma(\mathbb{P})\Rightarrow\Gamma(\mathbb{C})$ | for all $\Gamma$ | $\Gamma=\mathcal{P}(\omega)$ | $\Sigma^{1}_{2}(\mathbb{V}^{+}_{\varepsilon})\not\Rightarrow\Sigma^{1}_{2}(\mathbb{C})$

##  9\. $\mathbb{SP}$ does not have the Sacks property

In this section we return to one of our questions from [13]. First, we recall
the definition of splitting tree.

######  Definition 9.1.

A tree $p\subseteq 2^{<\omega}$ is called _splitting tree_ , short
$p\in\mathbb{SP}$, if for every $t\in p$ there is $k\in\omega$ such that for
every $n\geq k$ and every $i\in\\{0,1\\}$ there is $t^{\prime}\in p$,
$t\trianglelefteq t^{\prime}$ such that $t^{\prime}(n)=i$. We denote the
smallest such $k$ by $K_{p}(t)$. The set $\mathbb{SP}$ is partially ordered by
$q\leq_{\mathbb{SP}}p$ iff $q\subseteq p$.

On $\mathbb{SP}$ we can define a stronger $n$ relation.

######  Definition 9.2.

For two conditions $p,q\in\mathbb{SP}$ we let $q\leq_{n}p$ if:

$q\leq p\wedge\operatorname{Split}_{\leq n}(p)=\operatorname{Split}_{\leq
n}(q)\wedge\forall t\in\operatorname{Split}_{\leq n}(p)(K_{p}(t)=K_{q}(t)).$

Assume we are given a condition $p\in\mathbb{SP}$ and a node $t\in p$. Then,
for each $q\leq_{0}p{\upharpoonright}t$ and $n>K_{p}(t)$ there are two nodes
$t_{0},t_{1}\in\operatorname{Lev}_{n}(q)$ such that $t_{i}(n-1)=i,(i\in 2)$.
We make use of this observation in the following definition.

######  Definition 9.3.

Let $p\in\mathbb{SP}$. For each $t\in p$ and each $n>K_{p}(t)$ we define

(9.1)
$K(p,t,n)=\min\\{|\operatorname{Lev}_{n}(q)|\>:\>q\leq_{0}p{\upharpoonright}t)\\}$

The key observation is

######  Lemma 9.4.

For $q\leq p$, $t\in q$, and $n>K_{q}(t)=K_{p}(t)$ we have

(9.2) $K(q,t,n)\geq K(p,t,n).$

###### Proof.

Fix $r\leq_{0}q{\upharpoonright}t$ such that
$|\operatorname{Lev}_{n}(r)|=K(q,t,n)$. Then, also
$r\leq_{0}p{\upharpoonright}t$ since $K_{q}(t)=K_{p}(t)$ and thus
$|\operatorname{Lev}_{n}(r)|\geq K(p,t,n)$. ∎

However, the observed property is not yet strong enough, since it needs the
proviso $K_{q}(t)=K_{p}(t)$.

######  Theorem 9.5.

There is $p\in\mathbb{SP}$ such that for any $q\leq p$, $t\in q$,

(9.3) $\begin{split}&(\forall^{\infty}\ell)(\forall
t\in\operatorname{Lev}_{\ell}(q))(|t^{-1}\\{1\\}|<\log_{2}(\ell)^{2})\end{split}$

The proof of the theorem consists of the next three lemmata.

######  Definition 9.6.

To improve readability we fix the following sequence $\langle
l_{n}\>:\>n<\omega\rangle$ given by

$l_{n}=2^{n}-1.$

######  Lemma 9.7.

There is a perfect tree $T_{0}$ such that

  * (1)

$K_{T_{0}}(\emptyset)=0$,

  * (2)

$(\forall\ell)(|\operatorname{Lev}_{\ell}(T_{0})|=\ell+1)$,

  * (3)

$(\forall n)(\forall\ell\in[\ell_{n},\ell_{n+1})(\forall
t\in\operatorname{Lev}_{\ell}(T_{0}))|t^{-1}\\{1\\}|\leq n$,

  * (4)

$(\forall t\in T_{0})(\forall
n\in\omega)(t{}^{\smallfrown}\langle\underbrace{0,\dots,0}_{n}\rangle\in
T_{0})$.

###### Proof.

We let $\operatorname{Lev}_{1}(T_{0})=\\{\langle 0\rangle,\langle
1\rangle\\}$. We go by induction on $n$. Suppose
$\operatorname{Lev}_{\ell_{n}}(T_{0})$ is $\leq_{\rm lex}$\- increasingly
enumerated by $s_{\ell_{n},0},\dots,s_{\ell_{n},\ell_{n}}$. Now for $1\leq
j<\ell_{n}$ we let $\operatorname{Lev}_{\ell_{n}+j}(T_{0})$ be increasingly
enumerated by $s_{\ell_{n}+j,i}$, $i\leq\ell_{n}+j+1$, as follows

$\displaystyle s_{\ell_{n}+j,i}$ $\displaystyle=$
$\displaystyle{s_{\ell_{n}+j-1,i}}{}^{\smallfrown}0,\mbox{ for }i<2(j-1),$
$\displaystyle s_{\ell_{n}+j,2j-2}$ $\displaystyle=$ $\displaystyle
s_{\ell_{n}+j-1,2j-2}{}^{\smallfrown}0,$ $\displaystyle s_{\ell_{n}+j,2j-1}$
$\displaystyle=$ $\displaystyle s_{\ell_{n}+j-1,2j-2}{}^{\smallfrown}1,$
$\displaystyle s_{\ell_{n}+j,i+1}$ $\displaystyle=$ $\displaystyle
s_{\ell_{n}+j-1,i}{}^{\smallfrown}0,\mbox{ for }i\in[2j-1,\ell_{n}+j).$

Then each $t\in\operatorname{Lev}_{\ell_{n+1}}(T_{0})$ has exactly one
$i\in[\ell_{n},\ell_{n+1})$ such that $t(i)=1$. Items (1), (2) and (4) follow
immediately from the construction. ∎

In many cases a picture of the object helps to understand the cumbersome
indices. Here is the specific construction of $T_{0}$ up to level $l_{3}=7$:

$\emptyset$

######  Definition 9.8.

For $T\subseteq 2^{<\omega}$ and $s\in 2^{<\omega}$ we let
$s{}^{\smallfrown}T=\\{s{}^{\smallfrown}t\>:\>t\in T\\}$.

######  Definition 9.9.

By induction on $m$ we define $T_{m}$. $T_{0}$ is as in Lemma 9.7. Now suppose
that $T_{m}$ is defined. For each $t\in\operatorname{Split}(T_{m})\setminus
T_{m-1}$ (for $m=0$, we let $T_{-1}=\\{\emptyset\\}$) we choose the minimal
$n$ such that $|t|\in[\ell_{n},\ell_{n+1})$. Now we let
$f(t)=t{}^{\smallfrown}\langle 0,0,0,\dots,0\rangle$ such that the string of
zeros is so long that $|f(t)|=\ell_{n+1}$. Note that $f(t)\in T_{m}$. Now we
let

$T_{m+1}=\bigcup\\{f(t){}^{\smallfrown}T_{0}\>:\>t\in\operatorname{Split}(T_{m})\setminus
T_{m-1}\\}.$

Note that the $T_{0}$ is on purpose. We let
$p=\bigcup\\{T_{m}\>:\>m<\omega\\}$.

######  Lemma 9.10.

Let $p$ be as in Definition 9.9. Then $p\in\mathbb{SP}$.

###### Proof.

Let $t\in\operatorname{Split}(p)$ and let $m$ be minimal such that $t\in
T_{m}$. If $m=-1$, then $t=\operatorname{stem}(T_{0})=\emptyset$ and already
$K_{T_{0}}(t)=0$. Now assume that $m\geq 0$. Then $f(t)\in T_{m}$ and $f(t)$
is the stem of $f(t){}^{\smallfrown}T_{0}\subseteq T_{m+1}$. So
$K_{T_{m+1}}(t)=|f(t)|$. In particular $K_{p}(t)$ is defined and at most
$|f(t)|$. ∎

######  Lemma 9.11.

Let $p$ be as in the definition and $\ell\in\omega$. For any
$t\in\operatorname{Lev}_{\ell}(p)$, there are at most $\log_{2}(\ell)^{2}$
elements $i<|t|$ such that $t(i)=1$.

###### Proof.

For each $t\in p$, $|t|\leq\ell_{n+1}$, there is some $j\leq
n\leq\log_{2}(|t|)$ and are $t_{i}\in T_{i}$, $i\leq j$, such that

$t_{0}\triangleleft f(t_{0})\trianglelefteq t_{1}\triangleleft
f(t_{1})\trianglelefteq\dots\trianglelefteq t_{j}=t$

and for $i\leq j$ there are $n_{i}\leq n$ such that $n_{i}<n_{i+1}$ for $0\leq
i\leq j$ and and $|t_{i}|\in[\ell_{n_{i}},\ell_{n_{i}+1})$. We let
$t_{-1}=\emptyset$. Now by the properties of $T_{0}$, the number of
$k\in[|t_{i-1}|,|t_{i}|)$ such that $t_{i}(k)=1$ is bounded by $n_{i+1}\leq
n+1\leq\log_{2}(\ell)$. Since also $j\leq\log_{2}(\ell)$, we have

$|t^{-1}\\{1\\}|\leq\log_{2}(\ell)^{2}.$

∎

Now we turn our attention to the Sacks property and the question whether
$\mathbb{SP}$ satisfies this property.

######  Definition 9.12.

  1. (1)

$\langle S_{n}\>:\>n<\omega\rangle$ is called an _$f$ -slalom_ if
$S_{n}\subseteq[\omega]^{f(n)}$.

  2. (2)

A forcing $\mathbb{P}$ has _the Sacks property_ if for any
$f\colon\omega\to\omega\setminus\\{\emptyset\\}$ such that
$\lim_{n}f(n)=\infty$ and any $\mathbb{P}$-name $\tau$ for a real and any
condition $p$ there is an $f$-slalom $\langle S_{n}\>:\>n<\omega\rangle$ and
there is $q\leq p$ such that

$q\Vdash(\forall^{\infty}n)(\tau(n)\in S_{n}).$

######  Theorem 9.13.

At least under a condition $p$ as in Theorem 9.5 the forcing $\mathbb{SP}$
does not satisfy the Sacks property.

The proof consists of the following two lemmata.

######  Lemma 9.14.

Let $p$ be as in the Theorem 9.5. Then

$(\forall q\leq p)(\forall
t\in\operatorname{Split}(q))(\forall^{\infty}n)\Bigl{(}K(q,t,\ell_{n})\geq\frac{\ell_{n}}{2\cdot\log_{2}(\ell_{n})^{2}}\Bigr{)}.$

###### Proof.

Suppose not and fix a $q\leq p$ and a node $t\in\operatorname{Split}(q)$ such
that $K_{q}(t)=k$ and

$(\exists^{\infty}n)K(q,t,\ell_{n})<\frac{\ell_{n}}{2\cdot\log_{2}(\ell_{n})^{2}}=:M_{n}.$

We fix such an $n$ such that $\ell_{n}>2k$. We fix a slimmest $K_{q}(t)$
witnessing tree
$W\subseteq\operatorname{Lev}_{\leq\ell_{n}}(q{\upharpoonright}t)$ of width
$K(q,t,\ell_{n})<M_{n}$ at level $\ell_{n}$. That means
$|\operatorname{Lev}_{\ell_{n}}(W)|=K(q,t,\ell_{n})$ and on each height
$i\in[k,\ell_{n})$ there is at least one $s\in W$ such that $s(i)=1$. Hence
there is $s\in\operatorname{Lev}_{\ell_{n}}(W)$ such that $s$ has at least
$\frac{1}{M_{n}}\cdot(\ell_{n}-k)$ many $i\in[k,\ell_{n})$ such that $s(i)=1$.
We pick such an $s\in W\cap\operatorname{Lev}_{\ell_{n}}(q)$. However now $s$
shows

$\frac{1}{M_{n}}\cdot(\ell_{n}-k)>\frac{\ell_{n}}{2\cdot
M_{n}}=\log_{2}(\ell_{n})^{2}.$

This contradicts the property (9.3) that says that there are at most
$\log_{2}(\ell_{n})^{2}$ many $i<|s|$ such that $s(i)=1$. ∎

Now we apply the next lemma with

$g(n)=\frac{n}{2\cdot\log_{2}(n)^{2}}.$

######  Lemma 9.15.

Let $g\colon\omega\to\omega$ be such that $\lim_{n\to\infty}g(n)=\infty$. Let
$p$ be such that $(\forall q\leq p)(\forall
t\in\operatorname{Split}(q))(\forall^{\infty}n)(K(q,t,\ell_{n})\geq
g(\ell_{n}))$. For any function
$f\colon\omega\to\omega\setminus\\{\emptyset\\}$ there is a $\mathbb{SP}$-name
$\tau$ such that for any $f$-slalom $\langle S_{n}\>:\>n<\omega\rangle\in V$

$p\Vdash_{\mathbb{SP}}(\exists^{\infty}n)(\tau(n)\not\in S_{n}).$

###### Proof.

Fix $f\in\omega^{\omega}\cap V$. We give a name
$\dot{h_{f}}\in\omega^{\omega}\cap V^{\mathbb{SP}}$ so that for every slalom
$S\in([\omega]^{<\omega})^{\omega}\cap V$, with $|S(n)|\leq f(n)$, we have
that $p$ forces that $\dot{h_{f}}$ is not captured by $S$, i.e.,

$p\Vdash(\exists n\in\omega)\dot{h_{f}}(n)\notin S(n).$

We let $\dot{x}$ be the name for the $\mathbb{SP}$-generic real, i.e. the
union $\bigcup\\{\operatorname{stem}(p)\>:\>p\in G\\}$.

Let $\operatorname{code}\colon 2^{<\omega}\to\omega$ be an injective function.

We fix an increasing subsequence $\langle
m_{f,i}\>:\>i<\omega\rangle\subseteq\langle\ell_{n}\>:\>n<\omega\rangle$ such
that

$(\forall i)(g(m_{f,i})>f(i)).$

We define

$\tau=\dot{h_{f}}:=\langle\operatorname{code}(\dot{x}{\upharpoonright}m_{f,n})\>:\>n\in\omega\rangle.$

We aim to show that any $q\leq p$ forces that $\dot{h_{f}}$ cannot be captured
by any $f$-slalom in the ground model. So fix an $f$-slalom $S\in V$ and
$q\leq p$.

Let $t=\operatorname{stem}(q)$. It is enough to find $n\in\omega$, $r\leq q$
such that $r\Vdash\dot{h_{f}}(n)\notin S(n)$. Pick $n_{0}\in\omega$ such that
for any $n^{\prime}\geq n_{0}$

$K(q,t,{m_{f,n^{\prime}}})\geq g(m_{f,n^{\prime}})>f(n^{\prime}).$

Hence $|\operatorname{Lev}_{m_{f,n_{0}}}(q{\upharpoonright}t)|\geq
g({m_{f,n_{0}}})>f(n_{0})$. Let $\\{t_{k}\in
q{\upharpoonright}t\>:\>|t|=m_{f,n_{0}},k<g(m_{f,n_{0}})\\}$ enumerate the
first $g(m_{f,n_{0}})$ nodes of level ${m_{f,n_{0}}}$ of
$q{\upharpoonright}t$. Then

$q{\upharpoonright}t_{k}\Vdash\dot{h_{f}}(n_{0})=\operatorname{code}(\dot{x}{\upharpoonright}m_{f,n_{0}})=\operatorname{code}(t_{k}),$

Since $|S(n_{0})|\leq f(n_{0})<g(m_{f,n_{0}})$ there is a $t_{k}$ such that

$q{\upharpoonright}t_{k}\Vdash\operatorname{code}(\dot{x}{\upharpoonright}m_{f,n_{0}})=\operatorname{code}(t_{k})=\dot{h_{f}}(n_{0})\notin
S(n_{0}).$

So, $r:=q{\upharpoonright}t_{k}$ is the condition with the desired property. ∎

As we already mentioned, meanwhile Jonathan Schilhan found a complete negative
answer to our question [21].

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