# The Slicing Axioms

Ziemowit Kostana111Research of Z. Kostana was supported by the GAČR project
EXPRO 20-31529X and RVO: 67985840., Saharon Shelah222Research of second author
partially supported by Israel Science Foundation (ISF) grant no: 1838/19.
Paper number 1210 on the publication list. 333Research of both authors
partially supported by NSF grant no: DMS 1833363.

###### Abstract

We introduce a family of axioms $\operatorname{Slice}_{\kappa}$, which claim
the existence of nontrivial decompositions of the form

$2^{<\kappa}=\bigcup_{\alpha<\kappa}2^{<\kappa}\cap M_{\alpha},$

where $\\{M_{\alpha}|\;\alpha<\kappa\\}$ is a sequence of transitive models of
set theory. We study compatibility of these axioms with versions of Martin’s
Axiom, and in particular show that $\operatorname{Slice}_{\omega_{1}}$ is
compatible only with some very weak form of $MA$.

Keywords: Martin’s Axiom, Suslin forcing, transitive models
MSC classification: 03E50 03E17 03E35

##  1 Introduction

###  1.1 How "compact" is the real line?

We introduce and study a family of axioms $\operatorname{Slice}_{\kappa}$ for
cardinal numbers $\kappa$. The axiom $\operatorname{Slice}_{\kappa}$ basically
claims that there exists an increasing sequence
$\\{M_{\alpha}|\;\alpha<\kappa\\}$ of transitive models of $ZFC$, which
decomposes $2^{<\kappa}$ into an increasing union

$2^{<\kappa}=\bigcup_{\alpha<\kappa}2^{<\kappa}\cap M_{\alpha}.$

Our initial motivation was to find a single model of Martin’s Axiom, which
doesn’t satisfy typical consequences of $PFA$. This was in turn motivated by
the following intuition:

If the universe is sufficiently complete, in the sense that it has many
generic filters, then any transitive submodel containing enough reals, contain
all the reals.

This intuition is supported for example by the following result:

######  Theorem 1 (Thm. 8.6, [10]).

If $MM$ holds, then any inner model with correct $\omega_{2}$ contains all
reals.

The conclusion is quite strong, so it makes sense to ask what is left if we
weaken $MM$ to $MA_{\omega_{1}}$. This motivated us to formulate the axiom
$\operatorname{Slice}_{\omega_{1}}$, which turned out to be inconsistent with
$MA_{\omega_{1}}$. The main results of this paper are the following

######  Theorem 1 (Thm. 6).

$\operatorname{Slice}_{\omega_{1}}\implies\lnot\operatorname{MA}_{\omega_{1}}(\text{$\sigma$-centred})$.

######  Theorem 2 (Thm. 11).

If $\kappa$ is a regular cardinal such that $\kappa^{\omega}=\kappa$, then the
following theory is consistent

$ZFC+MA(\text{Suslin})+\operatorname{Slice}_{\omega_{1}}+``2^{\omega}=\kappa".$

######  Theorem 3 (Thm. 18).

Assume that $\omega<\kappa\leq\theta$ are regular cardinals, such that
$\theta^{<\kappa}=\theta$. Then the following theory is consistent

$ZFC+MA_{<\kappa}+\operatorname{Slice}_{\kappa}+``2^{\omega}=\theta".$

The first of these results provides another argument in favor of the informal
claim from the beginning. The class of Suslin forcings is a class of c.c.c.
forcings, which admit simple (analytic) definitions (see Definition 10). This
class is more extensively described in [2]. Martin’s Axiom for this class is a
considerable weakening of the full $MA$.

######  Theorem 4 ([8]).

$MA(\text{Suslin})$ implies each of the following:

  1. 1.

$\operatorname{Add}(\mathcal{N})=2^{\omega}$,

  2. 2.

$\operatorname{Add}(\mathcal{SN})=2^{\omega}$,

  3. 3.

$2^{\omega}$ is regular,

  4. 4.

each MAD family of subsets of $\omega$ has size $2^{\omega}$.

It follows from 1. that all cardinal characteristics in the Cichoń’s diagram
have value $2^{\omega}$. $\mathcal{SN}$ stands for the class of strong measure
zero sets.

######  Theorem 5 ([8]).

$MA(\text{Suslin})$ does not imply any of the following:

  1. 1.

$\mathfrak{t}=2^{\omega}$,

  2. 2.

$\mathfrak{s}=2^{\omega}$,

  3. 3.

$\forall{\kappa<2^{\omega}}\quad 2^{\kappa}=2^{\omega}$,

  4. 4.

there is no Suslin tree.

For an elaborated discussion of cardinal invariants of the continuum we refer
the reader to [4].

###  1.2 Preliminaries

All non-standard notions are introduced in the subsequent sections. By _reals_
we denote elements of the sets $\omega^{\omega}$, $2^{\omega}$, or seldom
$\mathbb{R}$. We take the liberty to freely identify Borel functions with
their Borel codes, so whenever we claim that

$f\in M,$

for some Borel function $f\subseteq 2^{\omega}\times 2^{\omega}$, and
$M\models ZFC$, it should be understood that it is the Borel code of $f$ that
belongs to $M$ (so we don’t bother if, for instance,
$\operatorname{dom}{f}\not\subseteq M$).
When we write
$\mathbb{P}=\\{\mathbb{P}_{\alpha}\ast\dot{\mathbb{Q}}_{\alpha}|\;\alpha<\theta\\}$
for a finite-support iteration of forcings, we sometimes denote by
$\mathbb{P}$ the final step of the iteration, that is
$\mathbb{P}=\mathbb{P}_{\theta}$. When dealing with infinite iterations, we
assume that $\mathbb{P}_{0}$ is the trivial forcing. A function
$i:\mathbb{P}_{0}\hookrightarrow\mathbb{P}_{1}$ is a _complete embedding_ if
the following assertions hold:

  1. 1.

$\forall{p,q\in\mathbb{P}_{0}}\quad p_{0}\leq p_{1}\implies i(p_{0})\leq
i(p_{1})$,

  2. 2.

$\forall{p,q\in\mathbb{P}_{0}}\quad p_{0}\bot p_{1}\implies i(p_{0})\bot
i(p_{1})$,

  3. 3.

If $\mathcal{A}\subseteq\mathbb{P}_{0}$ is a maximal antichain, then
$i[\mathcal{A}]\subseteq\mathbb{P}_{1}$ is a maximal antichain.

We write $\mathbb{P}_{0}\lessdot\mathbb{P}_{1}$ if
$\mathbb{P}_{0}\subseteq\mathbb{P}_{1}$ and the inclusion is a complete
embedding. The importance of complete embeddings comes from the closely
related notion of _quotient forcing_. If
$\mathbb{P}_{0}\lessdot\mathbb{P}_{1}$, and $G\subseteq\mathbb{P}_{0}$ is a
filter generic over $V$, then $V[G]$ contains the quotient forcing
$\mathbb{P}_{1}/G\subseteq\mathbb{P}_{1}$, consisting of all conditions
$p\in\mathbb{P}_{1}$ that are compatible with every element of $G$. The
crucial property of this notion is that forcing with $\mathbb{P}_{1}$ is
equivalent to forcing with a two-step iteration:
$\mathbb{P}_{0}\ast\mathbb{P}_{1}/\dot{G}$ ([9], p.244).

We will be frequently using the following two observations.

######  Proposition 2.

If $V$ is a countable transitive model of $ZFC$,
$\mathbb{P}_{0},\mathbb{P}_{1}\in V$, and
$\mathbb{P}_{0}\subseteq\mathbb{P}_{1}$ is an inclusion of partial orders,
then the following conditions are equivalent:

  1. 1.

$\mathbb{P}_{0}\lessdot\mathbb{P}_{1}$,

  2. 2.

If a filter $G\subseteq\mathbb{P}_{1}$ is $\mathbb{P}_{1}$-generic over $V$
then $G\cap\mathbb{P}_{0}$ is $\mathbb{P}_{0}$-generic over $V$.

######  Proposition 3.

Let $\mathbb{P}\lessdot\mathbb{S}$ be any forcing notions, and
$p\in\mathbb{P}$. Let $\dot{q}$, $\dot{r}$ be $\mathbb{P}$-names, and finally
let $\sigma(-,-)$ be a formula with parameters in the ground model, which is
also absolute between transitive models of set theory (for example a
$\Sigma^{1}_{1}$ formula in the language of arithmetic, or a bounded formula
in the language of set theory). Then

$p\Vdash_{\mathbb{S}}\sigma(\dot{q},\dot{r})\iff
p\Vdash_{\mathbb{P}}\sigma(\dot{q},\dot{r}).$

###### Proof.

In the direction from left to right, if $p\in G\subseteq\mathbb{P}$ is generic
over the ground model $V$, then we can extend $G$ to a generic filter
$G^{\prime}\subseteq\mathbb{S}$. Notice that $\dot{q}[G]=\dot{q}[G^{\prime}]$,
and $\dot{r}[G]=\dot{r}[G^{\prime}]$. By the absoluteness of $\sigma(-,-)$ we
have

$V[G]\models\sigma(\dot{q}[G],\dot{r}[G])\iff
V[G^{\prime}]\models\sigma(\dot{q}[G^{\prime}],\dot{r}[G^{\prime}]).$

In the other direction, we proceed in a similar way, using the fact that for
any generic filter $G\subseteq\mathbb{S}$, the intersection $G\cap\mathbb{P}$
is $\mathbb{P}$-generic. ∎

##  2 The Slicing Axioms

######  Definition 4.

Let $\kappa$ be any uncountable cardinal. We will say that
$\operatorname{Slice}_{\kappa}$ holds if there exists a sequence of transitive
classes (not necessarily proper) $\\{M_{\alpha}|\;\alpha<\kappa\\}$, such that
the following conditions are satisfied

  * •

$\forall\;{\alpha<\kappa}\quad M_{\alpha}\models ZFC$,

  * •

$\forall\;{\alpha<\kappa}\quad\omega_{1}^{M_{\alpha}}=\omega_{1}$,

  * •

$\forall\;{\delta\in[\omega,\kappa)}\quad
2^{\delta}=\displaystyle{\bigcup_{\alpha<\kappa}2^{\delta}\cap M_{\alpha}}$,

  * •

$\forall\;{\alpha<\beta<\kappa}\;\forall\;{\delta\in[\omega,\kappa)}\quad
2^{\delta}\cap M_{\alpha}\subsetneq 2^{\delta}\cap M_{\beta}$.

We will say that the sequence $\\{M_{\alpha}|\;\alpha<\kappa\\}$ _preserves
cardinals_ if $\kappa\in M_{0}$ and for each cardinal $\lambda\in M_{0}$ and
each $\alpha<\kappa$, $\lambda^{M_{\alpha}}$ is a cardinal.

Let us observe that the axiom $\operatorname{Slice}_{\kappa}$ is outright
false for any singular $\kappa$.

######  Proposition 5.

If $\operatorname{Slice}_{\kappa}$ is true, than $\kappa$ is regular.

###### Proof.

If $\delta=\operatorname{cof}{\kappa}<\kappa$, then
$\operatorname{Slice}_{\kappa}$ gives us a nontrivial decomposition of the
form

$2^{\delta}=\bigcup_{\alpha<\kappa}2^{\delta}\cap M_{\alpha},$

and after passing to a cofinal sequence, also

$2^{\delta}=\bigcup_{\alpha<\delta}2^{\delta}\cap M_{\alpha}.$

For each $\alpha<\delta$ we pick $s_{\alpha}\in 2^{\delta}\setminus
M_{\alpha}$. Using a definable bijection between $\delta$ and
$\delta\times\delta$, we see that the sequence $(s_{\alpha})_{\alpha<\delta}$
must belong so some model $M_{\alpha}$. But this contradicts the choice of
$s_{\alpha}$. ∎

The most important of the slicing axioms is perhaps
$\operatorname{Slice}_{\omega_{1}}$, since it claims that the real line can be
decomposed into an increasing union of $\omega_{1}$ many sets, which belong to
bigger and bigger models. The fact that $MA_{\omega_{1}}$ is inconsistent with
$\operatorname{Slice}_{\omega_{1}}$ shows, that the Martin’s Axiom on
$\omega_{1}$ imposes certain compactness on the real line.

##  3 Slicing the real line

We begin with showing that Martin’s Axiom on $\omega_{1}$ is not compatible
with $\operatorname{Slice}_{\omega_{1}}$.

######  Theorem 6.

$\operatorname{Slice}_{\omega_{1}}\implies\lnot\operatorname{MA}_{\omega_{1}}(\text{$\sigma$-centred})$.

In the proof we will utilize the known result from [6]. Recall that a set
$A\subseteq\mathbb{R}$ is a _$Q$ -set_, if each subset of $A$ is a relative
$F_{\sigma}$, i.e. for each $B\subseteq A$ there exists an $F_{\sigma}$ subset
$F\subseteq\mathbb{R}$ such that $A\cap F=B$.

######  Theorem 7 ([6]).

$\operatorname{MA}_{\omega_{1}}(\sigma\text{-centred})$ implies that each set
of cardinality $\omega_{1}$ is a $Q$-set.

###### Proof of Theorem 6.

Assume that $\operatorname{MA}_{\omega_{1}}$ holds, and
$(M_{\alpha})_{\alpha<\omega_{1}}$ is a sequence of models witnessing
$\operatorname{Slice}_{\omega_{1}}$. $M_{0}\models``\text{$2^{\omega}$ is
uncountable}"$, so there exists a sequence of pairwise distinct reals
$X=\\{x_{\alpha}|\;\alpha<\omega_{1}\\}\in M_{0}$ (note that this sequence is
_really_ of the length $\omega_{1}$). Let $f:\omega_{1}\hookrightarrow
2^{\omega}$ be a function such that $\forall\alpha<\omega_{1}\;f(\alpha)\notin
M_{\alpha}$. We will obtain a contradiction, by showing that there exists some
$\eta<\omega_{1}$, for which $\operatorname{rg}(f)\subseteq M_{\eta}$.

For every natural number $m$, let
$A_{m}=\\{x_{\alpha}|\;f(\alpha)(m)=1\\}=X\cap F_{m}$, where $F_{m}$ is an
$F_{\sigma}$ subset of reals. Since the sequence $(F_{m})_{m<\omega}$ can be
coded by a real, clearly it belongs to some model $M_{\eta}$. It is enough to
show that using this sequence and $X$ we can give a definition of
$\operatorname{rg}(f)$. But

$\operatorname{rg}(f)=\\{x\in 2^{\omega}|\;\exists\alpha<\omega_{1}\;\forall
m<\omega\quad x_{\alpha}\in F_{m}\iff x(m)=1\\}.$

∎

It is compatible with any value of $2^{\omega}$ that
$\operatorname{Slice}_{\omega_{1}}$ holds and is witnessed by a cardinal
preserving sequence.

######  Proposition 8.

Let $\mathbb{P}$ be any finite-support product of c.c.c. forcings adding
reals, of the length at least $\omega_{1}$. Then
$\mathbb{P}\Vdash\operatorname{Slice}_{\omega_{1}}$, and the corresponding
sequence of models is cardinal preserving.

###### Proof.

Let us consider a finite-support product of c.c.c. forcings

$\mathbb{P}=\prod_{i\in I}\mathbb{P}_{i},$

where each $\mathbb{P}_{i}$ adds some real number, and $|I|\geq\omega_{1}$. We
can decompose $I$ into a strictly increasing union
$\displaystyle{I=\bigcup_{\gamma<\omega_{1}}I_{\gamma}}$. For each
$\alpha<\omega_{1}$ the product $\displaystyle{\prod_{i\in
I_{\alpha}}\mathbb{P}_{i}}$ can be identified with a complete suborder of
$\mathbb{P}$.

If $G\subseteq\mathbb{P}$ is generic over some model $V$, then
$\operatorname{Slice}_{\omega_{1}}$ is witnessed by the sequence

$M_{\alpha}=V[G\cap\prod_{i\in I_{\alpha}}\mathbb{P}_{i}].$

∎

Recall that a set of reals is called _$\omega_{1}$ -dense_, if each nonempty
open interval in this set has size $\omega_{1}$. The following was proved by
Baumgartner in [3]. The following was proved by Baumgartner in [3].

######  Theorem 6 ([3]).

It is consistent with $MA_{\omega_{1}}$, that all $\omega_{1}$-dense subsets
of reals are order-isomorphic. In particular, each $\omega_{1}$-dense set of
reals has a non-trivial order-automorphism.

The natural question whether this assertion follows from $MA_{\omega_{1}}$ was
resolved by Avraham and the second author in [1].

######  Theorem 7 ([1]).

It is consistent with $MA_{\omega_{1}}$, that there exists a rigid
$\omega_{1}$-dense real order type.

This is also an easy consequence of $\operatorname{Slice}_{\omega_{1}}$.

######  Theorem 9.

$\operatorname{Slice}_{\omega_{1}}$ implies that there is an
$\omega_{1}$-dense rigid subset of the real line.

###### Proof.

Let $(M_{\alpha})_{\alpha<\omega_{1}}$ be a sequence witnessing
$\operatorname{Slice}_{\omega_{1}}$. For each $\alpha$, we choose

$x_{\alpha}\in\mathbb{R}\cap(M_{\alpha}\setminus\bigcup_{\beta<\alpha}M_{\beta}).$

We can easily arrange the construction, so that we hit each open interval
$\omega_{1}$-many times. The set $X=\\{x_{\alpha}|\;\alpha<\omega_{1}\\}$ is
$\omega_{1}$-dense, and it remains to prove, that it is also rigid. Suppose
that $f:X\rightarrow X$ is an order isomorphism. $f$ extends uniquely to a
continuous function $f^{\prime}:\mathbb{R}\rightarrow\mathbb{R}$, and each
such function can be coded by a real number. Therefore there is some
$\eta<\omega_{1}$, such that $f^{\prime}\in M_{\eta}$. Now, for any
$\xi>\eta$, it is not possible that $f(x_{\eta})=x_{\xi}$, because it would
mean $x_{\xi}\in M_{\eta}$, contrary to the choice of $x_{\xi}$. But,
likewise, it is not possible that $f^{-1}(x_{\eta})=x_{\xi}$. The conclusion
is that for all $\xi>\eta$, $f(x_{\xi})=x_{\xi}$. But this means that $f$ is
identity on a dense set, and therefore everywhere. ∎

##  4 Slicing the real line while preserving MA(Suslin)

We are going to show that $\operatorname{Slice}_{\omega_{1}}$ is consistent
with a version of Martin’s Axiom which takes into account only partial orders
representable as analytic sets (see [2], Ch. 3.6, or [8]).

######  Definition 10.

A partial order $(\mathbb{P},\leq)$ has a _Suslin definition_ if
$\mathbb{P}\in\Sigma^{1}_{1}(\omega^{\omega})$, and both ordering and
incompatibility relations in $\mathbb{P}$ are analytic relations on
$\omega^{\omega}$. $\mathbb{P}$ is _Suslin_ if it has a Suslin definition and
is c.c.c.

The following is the main result of this Section.

######  Theorem 11.

If $\kappa$ is a regular cardinal such that $\kappa^{\omega}=\kappa$, then the
following theory is consistent

$ZFC+MA(\text{Suslin})+\operatorname{Slice}_{\omega_{1}}+``2^{\omega}=\kappa".$

Let $\psi(-,-,-,-)$ be a universal analytic formula, i.e. a $\Sigma_{1}^{1}$
formula with the property that for each analytic set
$P\subseteq\omega^{\omega}\times\omega^{\omega}\times\omega^{\omega}$ there
exists $r\in\omega^{\omega}$ such that

$P=\\{x\in\omega^{\omega}\times\omega^{\omega}\times\omega^{\omega}|\;\psi(x,r)\\}.$

We want to use $\psi$ to add generic filters to all possible Suslin forcings.
We will say that $\psi(-,-,-,\dot{r}_{\alpha})$ _defines_
$\dot{\mathbb{Q}}_{\alpha}$ if $\dot{r}_{\alpha}$ is a
$\mathbb{P}_{\alpha}$-name for a real and $\mathbb{P}_{\alpha}$ forces each of
the following

$\dot{\mathbb{Q}}_{\alpha}\text{ is a separative partial order with the
greatest element }0,$ $\psi(x,1,1,\dot{r}_{\alpha})\iff
x\in\dot{\mathbb{Q}}_{\alpha},$ $\psi(x,y,2,\dot{r}_{\alpha})\iff
x\leq_{\dot{\mathbb{Q}}_{\alpha}}y,$ $\psi(x,y,3,\dot{r}_{\alpha})\iff
x\bot_{\dot{\mathbb{Q}}_{\alpha}}y.$

We will write $\psi^{\in}(x,z)$ for $\psi(x,1,1,z)$, $\psi^{\bot}(x,y,z)$ for
$\psi(x,y,3,z)$, and $\psi^{\leq}(x,y,z)$ for $\psi(x,y,2,z)$.

We are going to iterate all Suslin forcings, each of them cofinally many
times. More precisely, we define by induction a finite-support iteration
$\\{\mathbb{P}_{\alpha}\ast\dot{\mathbb{Q}}_{\alpha}|\;\alpha<\kappa\\}$:

  * •

$\mathbb{P}_{0}=\\{0\\}$,

  * •

$\mathbb{P}_{\alpha}\Vdash``\dot{\mathbb{Q}}_{\alpha}=\\{x\in\omega^{\omega}|\;\psi(x,\dot{r}_{\alpha})\\}\text{
if this formula defines a Suslin forcing;\; \\\ else
}\dot{\mathbb{Q}}=\\{0\\}"$,

The variable $\dot{r}_{\alpha}$ ranges over all reals, and all possible names
for reals, each of them cofinally many times. In order to iterate through all
possible parameters using a suitable bookkeeping, we introduce the class of
_simple_ conditions, following [2].

######  Definition 12.

By induction on $\alpha$ we define _simple_ conditions in
$\mathbb{P}_{\alpha}$.

  * •

$\alpha=0.$ $\mathbb{P}_{0}=\\{0\\}$, and we declare $0$ to be simple.

  * •

$\alpha+1.$ $(p,\dot{q})\in\mathbb{P}_{\alpha+1}$ is simple if
$p\in\mathbb{P}_{\alpha}$ is simple and

$\dot{q}=\\{(m,n,p^{m}_{n})|\;m,n<\omega,\,p^{m}_{n}\in\mathbb{P}_{\alpha}\\},$

where each $p^{m}_{n}$ is a simple condition in $\mathbb{P}_{\alpha}$. (for
each $m\in\omega$, the set $\\{p^{m}_{n}|\;n<\omega\\}$ is a maximal antichain
deciding $\dot{q}(m)$, i.e. $p^{m}_{n}\Vdash\dot{q}(m)=n$)

  * •

$\lim{\alpha}.$ $p\in\mathbb{P}_{\alpha}$ is simple if for each
$\beta<\alpha$, $p\restriction\beta\in\mathbb{P}_{\beta}$ is simple.

It is straightforward to check by induction, that the set of simple
$\mathbb{P}_{\alpha}$-conditions is dense in $\mathbb{P}_{\alpha}$, and that
each $\mathbb{P}_{\alpha}$ has at most $\kappa$ many names for reals (if we
restrict to names with simple conditions). We declare the forcings
$\mathbb{P}_{\alpha}$ to consist only of simple conditions, so formally we
write

$\mathbb{P}_{\alpha+1}=\\{(p,\dot{q})\in\mathbb{P}_{\alpha}\ast\dot{\mathbb{Q}}_{\alpha}|\;(p,\dot{q})\text{
is simple}\\}.$

######  Proposition 13.

If $2^{\omega}\leq\kappa$ is an uncountable regular cardinal such that
$\kappa^{\omega}=\kappa$, then

$\mathbb{P}_{\kappa}\Vdash MA(\text{Suslin})+``2^{\omega}=\kappa".$

###### Proof.

Let us denote by $W_{\alpha}$ the corresponding extensions of $V$ by
$\mathbb{P}_{\alpha}$. Let $(S,\leq)$ be a Suslin forcing in $W_{\kappa}$.
Assume $S$ is defined by the formula $\psi(-,r)$. We fix a family
$\\{A_{\gamma}|\;\gamma<\lambda\\}$ of maximal antichains in $S$, where
$\lambda<\kappa$. Of course the formula $\psi(-,r)$ defines different sets in
different models of set theory, so following the common custom we will denote
by $S^{N}$ the interpretation of $S$ in the model $N$, i.e.

$S^{N}=\\{x\in\omega^{\omega}\cap N|\;N\models\psi^{\in}(x,r)\\}.$

Notice that the family $\\{A_{\gamma}|\;\gamma<\lambda\\}$ is a function from
$\lambda$ to $[\omega^{\omega}]^{\omega}$, and so is added in some
intermediate step if the iteration. Let us fix an ordinal $\delta<\kappa$ such
that $\\{A_{\gamma}|\;\gamma<\lambda\\}\in W_{\delta}$, and
$\mathbb{P}_{\delta}\Vdash\dot{r}_{\delta}=r$. Now
$\mathbb{P}_{\delta}\Vdash\dot{\mathbb{Q}}_{\delta}=S^{W_{\delta}}$, so
$W_{\delta+1}$ contains a filter $G_{0}\subseteq S^{W_{\delta}}$ intersecting
all $A_{\gamma}$’s (note that by absoluteness of $\psi^{\in}$ and
$\psi^{\bot}$, the sets $A_{\gamma}$ are maximal antichains in
$S^{W^{\delta}}$). The filter generated by $G_{0}$ in $S^{W_{\kappa}}$ is the
required generic filter. ∎

If $N$ is a transitive class containing $\kappa$, we can define by induction
the relativized iteration
$\mathbb{P}^{N}_{\kappa}\subseteq\mathbb{P}_{\kappa}$, taking into account
only names from $N$.

  * •

$\mathbb{P}^{N}_{0}=\\{0\\}$,

  * •

$\mathbb{P}^{N}_{\alpha}\Vdash``\dot{\mathbb{Q}}^{N}_{\alpha}=\\{x\in\omega^{\omega}|\;\psi^{\in}(x,\dot{r}_{\alpha})\\}\text{
if this formula defines a Suslin forcing,}\\\ \text{ $\dot{r}_{\alpha}\in N$,
and $\dot{r}_{\alpha}$ is a $\mathbb{P}_{\alpha}^{N}$-name; else
}\dot{\mathbb{Q}}_{\alpha}^{N}=\\{0\\}"$,

  * •

$\mathbb{P}^{N}_{\alpha+1}=\mathbb{P}^{N}_{\alpha}\ast\dot{\mathbb{Q}}^{N}_{\alpha}$.

We take direct limits in the limit step, so that $\mathbb{P}_{\alpha}^{N}$ is
really a subset of $\mathbb{P}_{\alpha}$. Note, that we do not define names
$\dot{r}_{\alpha}$ inductively along the way, since they have already been
defined in the construction of $\mathbb{P}_{\kappa}$, which we take as
granted. This construction is inspired by the lemmas 1.4 and 1.5 from [8], and
conceptually is very similar. In order for it to work as desired, we prove by
induction some properties of $\mathbb{P}_{\alpha}^{N}$.

######  Theorem 14.

If $N$ is a transitive class containing $\kappa$, then for all
$\alpha\leq\kappa$

$\mathbb{P}^{N}_{\alpha}\lessdot\mathbb{P}_{\alpha}.$

Specifically:

  1. 1.

If $p_{0}\bot p_{1}$ in $\mathbb{P}_{\alpha}^{N}$, then $p_{0}\bot p_{1}$ in
$\mathbb{P}_{\alpha}$.

  2. 2.

If $p_{0}\leq p_{1}$ in $\mathbb{P}_{\alpha}^{N}$, then $p_{0}\leq p_{1}$ in
$\mathbb{P}_{\alpha}$.

  3. 3.

If $\mathcal{A}\subseteq\mathbb{P}^{N}_{\alpha}$ is a maximal antichain, then
$\mathcal{A}$ is maximal in $\mathbb{P}_{\alpha}$.

###### Proof.

$\text{We proceed by induction on }\alpha.$

1.

  * •

$\alpha=0$. Clear.

  * •

$\alpha+1$. We can assume that $\dot{\mathbb{Q}}^{N}_{\alpha}$ is defined by
the formula $\psi(-,\dot{r}_{\alpha})$, for otherwise
$\mathbb{P}^{N}_{\alpha+1}=\mathbb{P}^{N}_{\alpha}$, and we are done by the
induction hypothesis. Fix two incompatible conditions
$p_{0},p_{1}\in\mathbb{P}_{\alpha+1}^{N}$. Then
$p_{0}=(p^{\prime}_{0},\dot{q}_{0})$, $p_{1}=(p^{\prime}_{1},\dot{q}_{1})$,
where $p^{\prime}_{0},p^{\prime}_{1}\in\mathbb{P}_{\alpha}^{N}$, and

$p^{\prime}_{0}\Vdash\psi^{\in}(\dot{q}_{0},\dot{r}_{\alpha}),$
$p^{\prime}_{1}\Vdash\psi^{\in}(\dot{q}_{1},\dot{r}_{\alpha}).$

The forcing relation used above is a relation from $\mathbb{P}^{N}_{\alpha}$,
however since $\dot{r}_{\alpha}$, $\dot{q}_{0}$ and $\dot{q}_{1}$ are
$\mathbb{P}^{N}_{\alpha}$-names, this is the same relation as coming from
$\mathbb{P}_{\alpha}$ (recall Proposition 3). We aim to show that $p_{0}\bot
p_{1}$ in $\mathbb{P}_{\alpha+1}$.

If $p^{\prime}_{0}\bot p^{\prime}_{1}$ in $\mathbb{P}_{\alpha}$, then clearly
$p_{0}\bot p_{1}$ in $\mathbb{P}_{\alpha+1}$, so assume otherwise, and fix
$p\leq p^{\prime}_{0},p^{\prime}_{1}$ (in $\mathbb{P}_{\alpha}$). Let $p\in
G\subseteq\mathbb{P}_{\alpha}$ be a filter generic over $V$. Conditions
$p_{0}$ and $p_{1}$ were incompatible in $\mathbb{P}^{N}_{\alpha+1}$ and, by
the induction hypothesis,
$G\cap\mathbb{P}^{N}_{\alpha}\subseteq\mathbb{P}^{N}_{\alpha}$ is generic over
$V$, therefore

$V[G\cap\mathbb{P}^{N}_{\alpha}]\models\psi^{\bot}(\dot{q}_{0}[G],\dot{q}_{1}[G],\dot{r}_{\alpha}[G]).$

By absoluteness

$V[G]\models\psi^{\bot}(\dot{q}_{0}[G],\dot{q}_{1}[G],\dot{r}_{\alpha}[G]).$

Since $p$ was arbitrary, it follows that $p_{0}\bot p_{1}$ in
$\mathbb{P}_{\alpha+1}$.

  * •

$\lim{\alpha}$. Follows from the induction hypothesis, since conditions have
finite supports.

2.

  * •

$\alpha=0$. Clear.

  * •

$\alpha+1$. Again, we can assume that $\dot{\mathbb{Q}}^{N}_{\alpha}$ is
defined by the formula $\psi(-,\dot{r}_{\alpha})$. Fix two conditions
$p_{0}\leq p_{1}\in\mathbb{P}_{\alpha+1}^{N}$. Then
$p_{0}=(p^{\prime}_{0},\dot{q}_{0})$, $p_{1}=(p^{\prime}_{1},\dot{q}_{1})$,
where $p^{\prime}_{0},p^{\prime}_{1}\in\mathbb{P}_{\alpha}^{N}$, and

$p^{\prime}_{0}\Vdash\psi^{\in}(\dot{q}_{0},\dot{r}_{\alpha}),$
$p^{\prime}_{1}\Vdash\psi^{\in}(\dot{q}_{1},\dot{r}_{\alpha}).$

By the induction hypothesis $p_{0}^{\prime}\leq p_{1}^{\prime}$ in
$\mathbb{P}_{\alpha}$. Moreover $\dot{r}_{\alpha}$, $\dot{q}_{0}$ and
$\dot{q}_{1}$ are $\mathbb{P}^{N}_{\alpha}$-names, so – in the light of
Proposition 3 – the forcing relation

$p_{0}^{\prime}\Vdash\dot{q}_{0}\leq\dot{q}_{1}$

holds in $\mathbb{P}^{N}_{\alpha}$ as well as in $\mathbb{P}_{\alpha}$.

  * •

$\lim{\alpha}$. Follows from the induction hypothesis, since conditions have
finite supports.

3.

  * •

$\alpha=0$. Clear.

  * •

$\lim{\alpha}$. Let $\mathcal{A}$ be a maximal antichain in
$\mathbb{P}^{N}_{\alpha}$, and $\overline{p}\in\mathcal{A}$. Given that
$\mathbb{P}_{\alpha}^{N}$ is a finite-support iteration of c.c.c. forcings, it
satisfies the countable chains condition, therefore we can assume that
$\mathcal{A}=\\{\overline{p}_{n}|\;n<\omega\\}$. There is some $\gamma<\alpha$
such that $\overline{p}\in\mathbb{P}_{\gamma}$.
$\\{\overline{p}_{n}\restriction\gamma|\;n<\omega\\}$ might not be an
antichain in $\mathbb{P}^{N}_{\gamma}$, however each condition in
$\mathbb{P}^{N}_{\gamma}$ is compatible with some $p_{n}\restriction\gamma$.
We can refine $\\{\overline{p}_{n}\restriction\gamma|\;n<\omega\\}$ to an
antichain in $\mathbb{P}^{N}_{\gamma}$, and this antichain will remain maximal
in $\mathbb{P}_{\gamma}$ by the induction hypothesis. Therefore
$\\{\overline{p}_{n}\restriction\gamma|\;n<\omega\\}$ meets every condition in
$\mathbb{P}_{\gamma}$, and in particular some
$\overline{p}_{n}\restriction\gamma$ is compatible with $\overline{p}$ in
$\mathbb{P}_{\gamma}$. But then $\overline{p}_{n}$ is compatible with
$\overline{p}$ in $\mathbb{P}_{\alpha}$.

  * •

$\alpha+1$. In the light of Proposition 2, it is sufficient to show that for
any $G\subseteq\mathbb{P}_{\alpha+1}$ generic over $V$,
$G\cap\mathbb{P}^{N}_{\alpha+1}$ is also generic over $V$.

######  Lemma 15.

If $G\subseteq\mathbb{P}_{\alpha}$ is generic over $V$, and
$H\subseteq\dot{\mathbb{Q}}_{\alpha}[G]$ is generic over $V[G]$, then
$H\cap\dot{\mathbb{Q}}^{N}_{\alpha}[G]\subseteq\dot{\mathbb{Q}}^{N}_{\alpha}[G]$
is generic over $V[G\cap\mathbb{P}_{\alpha}^{N}]$.

Why is this sufficient? Let
$\overline{G}\subseteq\mathbb{P}_{\alpha}\ast\dot{\mathbb{Q}}_{\alpha}$ be a
filter generic over $V$. Recalling the notation from [9],

$\overline{G}=G\ast H=\\{(p,\dot{q})|\;p\in G,\;\dot{q}[G]\in H\\},$

where

$G=\\{p\in\mathbb{P}_{\alpha}|\;\exists\dot{q}\;\in\dot{\mathbb{Q}}\quad(p,\dot{q})\in\overline{G}\\},$

and

$H=\\{\dot{q}[G]|\;\exists p\in G\quad(p,\dot{q})\;\in\overline{G}\\}.$

It is known that for any iteration $\mathbb{P}\ast\dot{\mathbb{Q}}$, if
$G\subseteq\mathbb{P}$ is generic over $V$ and $H\subseteq\dot{\mathbb{Q}}[G]$
is generic over $V[G]$, then $G\ast H$ is generic for
$\mathbb{P}\ast\dot{\mathbb{Q}}$ over $V$ (for details consult for example
[9], Section 5, Chapter VIII). Let $G^{\prime}=G\cap\mathbb{P}^{N}_{\alpha}$.
It is generic for $\mathbb{P}^{N}_{\alpha}$ over $V$ by the induction
hypothesis. Now for filters $G$ and $H$ defined above

$(G\ast
H)\cap(\mathbb{P}^{N}_{\alpha}\ast\dot{\mathbb{Q}}^{N}_{\alpha})=\\{(p,\dot{q})|\;p\in
G^{\prime},\;\dot{q}[G]\in H,\;\dot{q}\in\dot{\mathbb{Q}}^{N}_{\alpha}\\}=$
$\\{(p,\dot{q})\in\mathbb{P}^{N}_{\alpha}\ast\dot{\mathbb{Q}}^{N}_{\alpha}|\;p\in
G^{\prime},\;\dot{q}[G^{\prime}]\in
H\\}=G^{\prime}\ast(H\cap\dot{\mathbb{Q}}^{N}_{\alpha}[G^{\prime}]).$

But if the conclusion of Lemma 15 holds, this is a
$\mathbb{P}^{N}_{\alpha}\ast\dot{\mathbb{Q}}^{N}_{\alpha}$-generic filter over
$V$.

We turn to the proof of Lemma 15.

###### Proof.

Fix a maximal antichain
$\mathcal{A}\subseteq\dot{\mathbb{Q}}^{N}_{\alpha}[G]=\dot{\mathbb{Q}}^{N}_{\alpha}[G^{\prime}]$,
belonging to $V[G^{\prime}]$. As $\mathcal{A}$ is a countable set of reals, it
can be coded using a single real $z\in\omega^{\omega}$. Recall that
$\dot{\mathbb{Q}}^{N}_{\alpha}[G^{\prime}]$ is defined in $V[G^{\prime}]$ by
the formula $\psi$ with the parameter
$\dot{r}_{\alpha}[G^{\prime}]=\dot{r}_{\alpha}[G]$. It is standard to check,
that the following claim can be written as a $\Pi^{1}_{1}$ formula.

$\phi(x,y)=``x$ is a real coding a maximal antichain in the partial ordering
defined by the formula $\psi(-,-,-,y)"$.

Now

$V[G^{\prime}]\models\phi(z,\dot{r}_{\alpha}[G^{\prime}]),$

and so by absoluteness

$V[G]\models\phi(z,\dot{r}_{\alpha}[G]).$

But $\psi(-,\dot{r}_{\alpha}[G])$ is the formula defining
$\dot{\mathbb{Q}}_{\alpha}[G]$ in $V[G]$. Therefore $\mathcal{A}$ remains
maximal in $\dot{\mathbb{Q}}_{\alpha}[G]$, and conclusion of the Lemma easily
follows. ∎

This concludes the proof.

∎

Let us note that even if $N$ is an inner model of ZFC, usually
$\mathbb{P}^{N}_{\kappa}\notin N$. Definition of $\mathbb{P}^{N}_{\kappa}$
makes use of a list of $\mathbb{P}^{N}_{\alpha}$-names, for all
$\alpha<\kappa$, and although _some_ such enumeration belongs to $N$ (as it is
a model of choice), this particular might not. In what sense is
$\mathbb{P}^{N}_{\kappa}$ a _relativized_ version of $\mathbb{P}_{\kappa}$, is
explained by the next lemma.

######  Lemma 16.

For each $\alpha\leq\kappa$, if $p\in\mathbb{P}_{\alpha}$ is simple then $p$
is definable (in the language of set theory) with a parameter from
$\kappa^{\omega}$.

###### Proof.

$\text{We proceed by induction on }\alpha.$

  * •

$\alpha=0.$ Clear, since each real is definable with a real parameter.

  * •

$\alpha+1$. Let $r=(p,\dot{q})$ be simple. We can write

$\dot{q}=\\{(m,n,p^{m}_{n})|\;m,n\in\omega,\,p^{m}_{n}\in\mathbb{P}_{\alpha}\\},$

where each $p^{m}_{n}$ is simple. By the induction hypothesis each $p^{m}_{n}$
is definable with a parameter from $\kappa^{\omega}$, and so is $p$. Clearly
$r$ can be defined from them, and so $r$ is definable with countably many
parameters from $\kappa^{\omega}$. We can easily code them as a single
parameter.

  * •

$\lim{\alpha}$. Fix $r\in\mathbb{P}_{\alpha}$. $r$ has finite support, so
there exists $\beta<\alpha$ containing the support of $r$. By the induction
hypothesis $p\restriction\beta$ is definable with a parameter from
$\kappa^{\omega}$, and $p$ is definable with parameters $p\restriction\beta$,
$\beta$, and $\alpha$.

∎

From this point, we fix a list $\\{\sigma_{\alpha}|\;\alpha<\kappa\\}$ of
sequences from $\kappa^{\omega}$, such that each name $\dot{r}_{\alpha}$ is
definable from $\sigma_{\alpha}$.

######  Lemma 17.

Let $N$ be any transitive model of $ZFC$ containing $\kappa$, and let $M\prec
H((2^{\kappa})^{+})$ be a countable elementary submodel such that
$\\{\sigma_{\alpha}|\;\alpha<\kappa\\}\in M$, and
$M\cap\kappa^{\omega}\subseteq N$. Then for any $\alpha\leq\kappa$,
$\mathbb{P}_{\alpha}\cap M\subseteq\mathbb{P}_{\alpha}^{N}$.

###### Proof.

We proceed by induction.

  * •

$\alpha=0$. Clear.

  * •

$\lim{\alpha}.$ Fix $r\in\mathbb{P}_{\alpha}\cap M$. By the elementarity of
$M$, there exists $\gamma\in\alpha\cap M$ that contains the support of $r$.
From the induction hypothesis it follows that
$r\restriction\gamma\in\mathbb{P}_{\gamma}^{N}$. It is routine to verify by
induction that for all $\gamma\leq\delta\leq\alpha$,
$r\restriction\delta\in\mathbb{P}_{\delta}^{N}$.

  * •

$\alpha+1.$ Fix $r=(p,\dot{q})\in
M\cap(\mathbb{P}_{\alpha}\ast\dot{\mathbb{Q}}_{\alpha})$. Without loss of
generality we can assume that $\dot{\mathbb{Q}}_{\alpha}$ is defined by the
formula $\psi(-,\dot{r}_{\alpha})$. We need to check that $\dot{q}$ is a
$\mathbb{P}_{\alpha}^{N}$-name for an element of
$\dot{\mathbb{Q}}^{N}_{\alpha}$, which means that in particular
$\dot{\mathbb{Q}}_{\alpha}^{N}$ needs to be defined by the same formula as
$\dot{\mathbb{Q}}_{\alpha}$. In summary, our task is to verify the following
three claims:

    * –

$\dot{q}$ is a (simple) $\mathbb{P}^{N}_{\alpha}$-name,

    * –

$\dot{r}_{\alpha}$ is a (simple) $\mathbb{P}^{N}_{\alpha}$-name,

    * –

$\dot{r}_{\alpha}\in N$.

The condition $\dot{q}$ is of the form
$\dot{q}=\\{(m,n,p^{m}_{n})|\;m,n<\omega,\;p^{m}_{n}\in\mathbb{P}_{\alpha}\\}$.
Given that all conditions $p^{m}_{n}$ belong to $M$, they also belong to
$\mathbb{P}_{\alpha}^{N}$ be the induction hypothesis. For the same reason
$\dot{r}_{\alpha}$ is a $\mathbb{P}_{\alpha}^{N}$-name, once we show that
$\dot{r}_{\alpha}\in M$. But the model $M$ contains the list
$\\{\sigma_{\alpha}|\;\alpha<\kappa\\}$ and the ordinal $\alpha$. Therefore
$\sigma_{\alpha}\in M$, and $\dot{r}_{\alpha}\in M$. Finally, given that
$\sigma_{\alpha}\in\kappa^{\omega}\cap M\subseteq N$, we conclude that
$\dot{r}_{\alpha}\in N$.

∎

###### Proof of Theorem 11.

We start with a model
$V\models\operatorname{Slice}_{\omega_{1}}+``2^{\omega}=\kappa"$, and we
assume moreover that the sequence $\\{M_{\alpha}|\;\alpha<\omega_{1}\\}$
witnessing $\operatorname{Slice}_{\omega_{1}}$ satisfies the following
stronger property:

$\kappa^{\omega}=\displaystyle{\bigcup_{\alpha<\omega_{1}}\kappa^{\omega}\cap
M_{\alpha}}.$

Such a model is easy to get, for example by adding $\kappa$ many Cohen reals
to a model of $CH$ using the finite support-iteration, and proceeding like in
the proof of Proposition 8.

Let
$\mathbb{P}=\\{\mathbb{P}_{\alpha}\ast\dot{\mathbb{Q}}_{\alpha}|\;\alpha<\kappa\\}$
be the iteration described above, which forces

$MA(\text{Suslin})+``2^{\omega}=\kappa".$

We claim that if $G\subseteq\mathbb{P}$ is generic over $V$, then the sequence
$V[G\cap\mathbb{P}^{M_{\alpha}}]$ witnesses
$\operatorname{Slice}_{\omega_{1}}$ in $V[G]$. For this we need to show two
things

  1. 1.

If $r\in\omega^{\omega}\cap V[G]$, then $r\in V[G\cap\mathbb{P}^{M_{\alpha}}]$
for some $\alpha<\kappa$.

  2. 2.

None of the models $V[G\cap\mathbb{P}^{M_{\alpha}}]$ contains all reals in
$V[G]$.

Concerning $1.$ assume that
$\mathbb{P}_{\kappa}\Vdash\dot{r}\in\omega^{\omega}$. We can assume that

$\dot{r}=\\{(m,n,p_{n}^{m})|\;m,n<\omega\\},$

and all conditions $p_{n}^{m}$ are simple. By Lemma 16 each condition
$p_{n}^{m}$ is definable with a parameter from $\kappa^{\omega}$. It follows,
that $\dot{r}$ is definable with a parameter from $\kappa^{\omega}$, and by
our assumption this parameter belongs to some model $M_{\alpha}$. Therefore
$\dot{r}$ is a $\mathbb{P}^{M_{\alpha}}$-name, and so

$\dot{r}[G]=\dot{r}[G\cap\mathbb{P}^{M_{\alpha}}]\in
V[G\cap\mathbb{P}^{M_{\alpha}}].$

Concerning $2.$ fix a real $r\in\omega^{\omega}\setminus M_{\alpha}$. There
exists a representation of the Cohen forcing as a Borel subset of
$\omega^{\omega}$, from which the real $r$ is definable. For concreteness, let
us put

$\mathbb{C}_{r}=\omega^{<\omega}\cup\\{r\\}\subseteq\omega^{\omega},$

where $\omega^{<\omega}$ is identified with the set of sequences from
$\omega^{\omega}$ that are eventually equal zero. We order $\omega^{<\omega}$
by the end-extension and we declare that

$\forall{s\in\omega^{<\omega}}\;s\bot r.$

Since $\mathbb{C}_{r}$ is clearly Suslin, there exists a real $r^{\prime}$
such that

$\mathbb{C}=\\{x\in\omega^{\omega}|\;\psi^{\in}(x,r^{\prime})\\}.$

We claim that $r^{\prime}\notin M_{\alpha}$. Suppose otherwise. Let
$\sigma(x)$ stand for

$\psi^{\in}(x)\wedge x\notin\omega^{<\omega}.$

Note that

$V\models\exists\;x\in\omega^{\omega}\;\sigma(x),$

and so by absoluteness the same holds in $M_{\alpha}$. Fix
$r^{\prime\prime}\in\omega^{\omega}\cap M_{\alpha}$, such that

$M_{\alpha}\models\sigma(r^{\prime\prime}).$

Again $V\models\sigma(r^{\prime\prime})$ by absoluteness. But this shows that
$r=r^{\prime\prime}$, and therefore $r\in M_{\alpha}$, contradicting the
choice of $r$.

Let us fix $\gamma<\kappa$ such that
$\mathbb{P}_{\gamma}\Vdash\dot{r}_{\gamma}=r^{\prime}$. It follows that

$\mathbb{P}^{M_{\alpha}}_{\gamma}\Vdash\dot{\mathbb{Q}}^{M_{\alpha}}_{\gamma}=\\{0\\},$

and

$\mathbb{P}_{\gamma}\Vdash\dot{\mathbb{Q}}_{\gamma}=\mathbb{C}_{r}.$

The mapping from $\mathbb{C}_{r}$ into $\mathbb{P}_{\gamma+1}$ given by the
formula

$x\mapsto(1_{\mathbb{P}_{\gamma}},x)$

is a complete embedding, and it extends to a complete embedding

$f:\mathbb{C}_{r}\hookrightarrow\mathbb{P}_{\kappa}.$

Let $\dot{H}$ be a $\mathbb{P}_{\kappa}^{M_{\alpha}}$-name for a generic
filter. In the generic extension $V[H]$, the function $f$ remains a complete
embedding of a Cohen forcing into $\mathbb{P}_{\kappa}$, and by the choice of
$\gamma$, $f$ is in fact a a complete embedding into $\mathbb{P}_{\kappa}/H$.
In conclusion, forcing with $\mathbb{P}_{\kappa}$ adds a Cohen real over
$V[G\cap\mathbb{P}_{\kappa}^{M_{\alpha}}]$. ∎

###  4.1 Computation of cardinal invariants

For any given group $G$, one can study an associated cardinal invariant
$\operatorname{c(Sym(G))}$ that stands for the minimal cardinality $\kappa$,
for which $G$ can be represented as a union of a chain of the length $\kappa$,
consisting of proper subgroups of $G$. A substantial amount of literature is
devoted to study this cardinal invariant for symmetric groups of infinite sets
(for example [11], [14], [13], [7]). It is known that

$\mathfrak{g}\leq\operatorname{c(Sym(\omega))}\leq\mathfrak{d},$

where $\mathfrak{g}$ is the _groupwise density number_. The first inequality
was proved by Brendle and Losada [5], and the second by Sharp and Thomas [12].

It is easy to observe that
$\operatorname{Slice}_{\omega_{1}}\implies``\operatorname{c(Sym(\omega))}=\omega_{1}"$:
if $\operatorname{Slice}_{\omega_{1}}$ is witnessed by a sequence
$(M_{\alpha})_{\alpha<\omega_{1}}$, then the equality
$\operatorname{c(Sym(\omega))}=\omega_{1}$ is witnessed by the sequence of
groups $(M_{\alpha}\cap\operatorname{Sym}(\omega))_{\alpha<\omega_{1}}$. As a
matter of fact, this observation shows that Theorem 11 generalizes Lemma 2.6
from [15], which claims that the equality
$\operatorname{c(Sym(\omega))}=\omega_{1}$ is preserved under finite-support
iterations of Suslin forcings.

Together with some well-known results (see [4]), we have the following series
of inequalities:

$\operatorname{c(Sym(\omega))}\geq\mathfrak{g}\geq\mathfrak{h}\geq\mathfrak{t}\geq\mathfrak{m}\geq\omega_{1}.$

It follows that in our model all these invariants are equal $\omega_{1}$.
Together with the fact that $MA(\text{Suslin})$ decides all cardinal
characteristics from the Cichoń’s diagram to be equal $2^{\omega}$, we have
computed all of the classical cardinal invariants of the continuum, except
$\mathfrak{s}$. But the proof of Theorem 11 shows that
$\operatorname{Slice}_{\omega_{1}}$ is witnessed by some sequence
$(M_{\alpha})_{\alpha<\omega_{1}}$, for which there exists a Cohen real over
each of the models. Given that a Cohen real splits all reals from the ground
model, we conclude that $\mathfrak{s}=\omega_{1}$.

##  5 Slicing $2^{<\kappa}$

Although $MA_{\omega_{1}}$ is inconsistent with
$\operatorname{Slice}_{\omega_{1}}$, it is consistent with
$\operatorname{Slice}_{\kappa}$ for any $\kappa>\omega_{1}$. The idea of the
proof is very much like that of Theorem 11, and actually even simpler, because
we iterate forcings directly, instead of coding them as analytic sets.

######  Theorem 18.

Assume that $\omega<\kappa\leq\theta$ are regular cardinals, such that
$\theta^{<\kappa}=\theta$. Then the following theory is consistent for any
cardinal $\lambda<\kappa$:

$ZFC+MA_{\lambda}+\operatorname{Slice}_{\kappa}+``2^{\omega}=\theta".$

We are going to apply a finite-support iteration of the form

$\mathbb{P}=\\{\mathbb{P}_{\alpha}\ast\dot{\mathbb{Q}}_{\alpha}|\;\alpha<\theta\\},$

where for each $\alpha<\theta$

$\mathbb{P}_{\alpha}\Vdash\dot{\mathbb{Q}}_{\alpha}=(\lambda,\dot{\leq}_{\alpha}).$

We also assume that $0\in\lambda$ is always the largest element in
$\dot{\mathbb{Q}}_{\alpha}$. We want to arrange the iteration so that each
c.c.c. partial order of size $\lambda$ will appear cofinally many times (see
[9], p. 278), and for this reason, we will be considering only names of the
form

$\dot{\leq}_{\alpha}=\\{(\phi(\beta),p_{i}^{\beta})|\;i<\omega,\beta<\lambda\\},$

where $\phi:\lambda\rightarrow\lambda\times\lambda$ is a fixed bijection,
definable from $\lambda$. A standard induction shows that for any
$\alpha\leq\theta$ there exists at most $\theta$-many such names, and
$|\mathbb{P}_{\alpha}|\leq\theta$. Using an appropriate bookkeeping we can
include all c.c.c. partial orders of size $\lambda$ in our iteration, and
therefore we obtain:

######  Theorem 19.

$\mathbb{P}_{\theta}\Vdash MA_{\lambda}+``2^{\omega}=\theta".$

######  Definition 20.

By induction on $\alpha$, we define the class of _simple_
$\mathbb{P}_{\alpha}$-conditions.

  * •

$\alpha=0.$ $\mathbb{P}_{0}=\\{0\\}$, and we declare $0$ to be simple.

  * •

$\alpha+1.$ $(p,\dot{q})\in\mathbb{P}_{\alpha+1}$ is simple if
$p\in\mathbb{P}_{\alpha}$ is simple and
$\dot{q}=\\{(\gamma_{n},p_{n})|\;n<\omega\\}$, where conditions $p_{n}$ are
simple.

  * •

$\lim{\alpha}.$ $p\in\mathbb{P}_{\alpha}$ is simple if for each
$\beta<\alpha$, $p\restriction\beta\in\mathbb{P}_{\beta}$ is simple.

Like in the previous section, it is easy to check that the set of simple
conditions is always dense.

######  Lemma 21.

For each $\alpha\leq\theta$, if $p\in\mathbb{P}_{\alpha}$ is simple then $p$
is definable (in the language of set theory) with a parameter from
$\theta^{\omega}$.

###### Proof.

  * •

$\alpha=0.$ Clear.

  * •

$\alpha+1$. Let $r=(p,\dot{q})$ be simple. We can write
$\dot{q}=\\{(\gamma_{n},p_{n})|\;n<\omega\\}$, where conditions $p_{n}$ are
simple. By the induction hypothesis each $p_{n}$ is definable with a parameter
from $\kappa^{\omega}$, and so is $p$. Clearly $r$ can be defined from them,
and so $r$ is definable with countably many parameters, which we can code as
one.

  * •

$\lim{\alpha}$. Fix $r\in\mathbb{P}_{\alpha}$. $r$ has finite support, so
there exists $\beta<\alpha$ containing the support of $r$. By the induction
hypothesis $p\restriction\beta$ is definable with a parameter from
$\kappa^{\omega}$, and $p$ is definable with parameters $p\restriction\beta$,
$\beta$, and $\alpha$.

∎

An immediate consequence is that each of the names $\dot{\leq}_{\alpha}$ is
definable with some parameter $\sigma_{\alpha}\in\theta^{\lambda}$. Like in
the previous Section, we can define by induction the relativized forcings
$\mathbb{P}^{N}_{\kappa}\subseteq\mathbb{P}_{\kappa}$, taking into account
only names from $N$.

  * •

$\mathbb{P}^{N}_{0}=\\{0\\}$,

  * •

Assume $\mathbb{P}^{N}_{\alpha}$ is defined. We define a
$\mathbb{P}^{N}_{\alpha}$-name $\dot{\mathbb{Q}}^{N}_{\alpha}$ as follows

    * –

$\dot{\mathbb{Q}}^{N}_{\alpha}=\dot{\mathbb{Q}}_{\alpha}$ if
$\dot{\mathbb{Q}}_{\alpha}\in N$, and $\dot{\mathbb{Q}}_{\alpha}$ is a
$\mathbb{P}^{N}_{\alpha}$-name,

    * –

$\dot{\mathbb{Q}}^{N}_{\alpha}=\\{0\\}$ otherwise.

  * •

$\mathbb{P}^{N}_{\alpha+1}=\mathbb{P}^{N}_{\alpha}\ast\dot{\mathbb{Q}}^{N}_{\alpha}$.

In limit steps we take direct limits, so
$\mathbb{P}^{N}_{\kappa}\subseteq\mathbb{P}_{\kappa}$. Repeating the proof of
Lemma 17, we obtain

######  Lemma 22.

Let $N$ be a transitive model of $ZFC$, containing $\theta$. Let $M\prec
H((2^{\theta})^{+})$ be an elementary submodel of size $\lambda$, such that
$\lambda+1\subseteq M$, and $\\{\sigma_{\alpha}|\;\alpha<\theta\\}\in M$ (see
the remark after Lemma 21). Assume moreover that $\theta^{\lambda}\cap
M\subseteq N$. Then for each $\alpha\leq\theta$, $\mathbb{P}_{\alpha}\cap
M\subseteq\mathbb{P}_{\alpha}^{N}$.

###### Proof.

We proceed by induction.

  * •

$\alpha=0$. Clear.

  * •

$\lim{\alpha}.$ Fix $r\in\mathbb{P}_{\alpha}\cap M$. By the elementarity of
$M$, there exists $\gamma\in\alpha\cap M$ that contains the support of $r$.
From the induction hypothesis it follows that
$r\restriction\gamma\in\mathbb{P}_{\gamma}^{N}$. It is routine to verify by
induction that for all $\gamma\leq\delta\leq\alpha$,
$r\restriction\delta\in\mathbb{P}_{\delta}^{N}$.

  * •

$\alpha+1.$ Fix $r=(p,\dot{q})\in
M\cap(\mathbb{P}_{\alpha}\ast\dot{\mathbb{Q}}_{\alpha})$.
$p\in\mathbb{P}_{\alpha}^{N}$ be the induction hypothesis. The name $\dot{q}$
is of the form $\dot{q}=\\{(\gamma_{n},p_{n})||;n<\omega\\}$, and for each
$n<\omega$, $p_{n}\in\mathbb{P}_{\alpha}\cap
M\subseteq\mathbb{P}_{\alpha}^{N}$. This shows that $\dot{q}$ is a
$\mathbb{P}_{\alpha}^{N}$-name. It remains to show that
$\mathbb{P}_{\alpha}^{N}\Vdash\dot{\mathbb{Q}}_{\alpha}^{N}=\dot{\mathbb{Q}}_{\alpha}$,
and this in turn reduces to showing that $\dot{\mathbb{Q}}_{\alpha}^{N}$ is a
$\mathbb{P}_{\alpha}^{N}$-name belonging to $N$. To see this, let us note that
given $r\in M$, also $\alpha\in M$, and so
$\sigma_{\alpha}\in\theta^{\lambda}\cap M\subseteq N$. It follows that
$\dot{\mathbb{Q}}_{\alpha}\in M\cap N$. Recall, that $\dot{\mathbb{Q}}$ is a
$\mathbb{P}_{\alpha}$-name for a partial ordering of the form

$\dot{\leq}_{\alpha}=\\{(\phi(\beta),p^{i}_{\beta})|\;\beta<\lambda,i<\omega\\}.$

Given that $\lambda+1\subseteq M$, we conclude that each of the conditions
$p^{i}_{\beta}$ belongs to $M$, and by the induction hypothesis, also to
$\mathbb{P}_{\alpha}^{N}$. This shows that $\dot{\leq}_{\alpha}$, and
effectively also $\dot{\mathbb{Q}}_{\alpha}$, are
$\mathbb{P}_{\alpha}^{N}$-names, and concludes the proof.

∎

######  Lemma 23.

If $N$ is a transitive class, then for all $\alpha\leq\kappa$

$\mathbb{P}^{N}_{\alpha}\lessdot\mathbb{P}_{\alpha}.$

Specifically:

  1. 1.

If $p_{0}\bot p_{1}$ in $\mathbb{P}_{\alpha}^{N}$, then $p_{0}\bot p_{1}$ in
$\mathbb{P}_{\alpha}$.

  2. 2.

If $p_{0}\leq p_{1}$ in $\mathbb{P}_{\alpha}^{N}$, then $p_{0}\leq p_{1}$ in
$\mathbb{P}_{\alpha}$.

  3. 3.

If $\mathcal{A}\subseteq\mathbb{P}^{N}_{\alpha}$ is a maximal antichain, then
$\mathcal{A}$ is maximal in $\mathbb{P}_{\alpha}$.

###### Proof.

We proceed by induction on $\alpha$.

1.

  * •

$\alpha=0.$ Clear.

  * •

$\alpha+1.$ Assume $(p_{0},\dot{q}_{0})\bot(p_{1},\dot{q}_{1})$ in
$\mathbb{P}^{N}_{\alpha+1}$. If $p_{0}\bot p_{1}$ in
$\mathbb{P}_{\alpha}^{N}$, then by the induction hypothesis $p_{0}\bot p_{1}$
in $\mathbb{P}_{\alpha}$ and we are done. Suppose otherwise, and fix a
condition $p\leq p_{0},p_{1}$ from $\mathbb{P}_{\alpha}$. Let
$G\subseteq\mathbb{P}_{\alpha}$ be any filter generic over $V$, containing
$p$. $p_{0},p_{1}\in G\cap\mathbb{P}_{\alpha}^{N}$, so

$\dot{q}_{0}[G\cap\mathbb{P}_{\alpha}^{N}]\bot\dot{q}_{1}[G\cap\mathbb{P}_{\alpha}^{N}]$

in model $V[G\cap\mathbb{P}_{\alpha}^{N}]$, and in $V[G]$ as well. Since $p$
and $G$ were arbitrary, it follows that
$(p_{0},\dot{q}_{0})\bot(p_{1},\dot{q}_{1})$ in $\mathbb{P}_{\alpha+1}$.

  * •

$\lim{\alpha}.$ Follows from the induction hypothesis, since the supports are
finite.

2.

  * •

$\alpha=0.$ Clear.

  * •

$\alpha+1.$ Assume $(p_{0},\dot{q}_{0})\leq(p_{1},\dot{q}_{1})$ in
$\mathbb{P}^{N}_{\alpha+1}$. From the induction hypothesis we know, that
$p_{0}\leq p_{1}$ in $\mathbb{P}_{\alpha}$, and
$p_{0}\Vdash\dot{q}_{0}\leq\dot{q}_{1}$ in the sense of
$\mathbb{P}_{\alpha}^{N}$. We must show that the assertion

$p_{0}\Vdash\dot{q}_{0}\leq\dot{q}_{1}$

holds also in the sense of $\mathbb{P}_{\alpha}$. If
$\dot{\mathbb{Q}}^{N}_{\alpha}=\\{0\\}$ it is trivial. Otherwise
$\dot{\mathbb{Q}}^{N}_{\alpha}=\dot{\mathbb{Q}}_{\alpha}$. In that case
$\dot{q}_{0}$ and $\dot{q}_{1}$ are $\mathbb{P}_{\alpha}^{N}$-names, and the
$\Vdash$ relation for them is the same in $\mathbb{P}_{\alpha}^{N}$ as in
$\mathbb{P}_{\alpha}$.

  * •

$\lim{\alpha}.$ Follows from the induction hypothesis, since the supports are
finite.

3.

  * •

$\alpha=0.$ Clear.

  * •

$\alpha+1.$ The proof is exactly the same, as in the paragraph after Lemma 15,
so we need to prove the conclusion of Lemma 15 in the current setting. But
this is trivial, once we recall that

$\mathbb{P}^{N}_{\alpha}\Vdash\dot{\mathbb{Q}}^{N}_{\alpha}=\\{0\\},$

or

$\mathbb{P}^{N}_{\alpha}\Vdash\dot{\mathbb{Q}}^{N}_{\alpha}=\dot{\mathbb{Q}}_{\alpha}.$

  * •

$\lim{\alpha}$. Let $\\{\overline{p}_{n}|\;n<\omega\\}$ be a maximal antichain
in $\mathbb{P}^{N}_{\alpha}$, and $\overline{p}\in\mathbb{P}_{\alpha}$. There
is some $\gamma<\alpha$ such that $\overline{p}\in\mathbb{P}_{\gamma}$.
$\\{\overline{p}_{n}\restriction\gamma|\;n<\omega\\}$ might not be an
antichain in $\mathbb{P}^{N}_{\gamma}$, however each condition in
$\mathbb{P}^{N}_{\gamma}$ is compatible with some $p_{n}\restriction\gamma$.
We can refine $\\{\overline{p}_{n}\restriction\gamma|\;n<\omega\\}$ to an
antichain in $\mathbb{P}^{N}_{\gamma}$, and this antichain will remain maximal
in $\mathbb{P}_{\gamma}$ by the induction hypothesis. Therefore
$\\{\overline{p}_{n}\restriction\gamma|\;n<\omega\\}$ meets every condition in
$\mathbb{P}_{\gamma}$, and in particular some
$\overline{p}_{n}\restriction\gamma$ is compatible with $\overline{p}$ in
$\mathbb{P}_{\gamma}$. But then $\overline{p}_{n}$ is compatible with
$\overline{p}$ in $\mathbb{P}_{\alpha}$.

∎

###### Proof of Theorem 18.

Let

$V\models ZFC+GCH+\operatorname{Slice}_{\kappa}$

and let $\mathbb{P}$ be the forcing defined in the beginning of the Section.
Suppose that $\\{M_{\alpha}|\;\alpha<\kappa\\}$ witnesses
$\operatorname{Slice}_{\kappa}$ in $V$, and $G\subseteq\mathbb{P}$ is generic
over $V$. We aim to shows that the sequence $V[G\cap\mathbb{P}^{M_{\alpha}}]$
witnesses $\operatorname{Slice}_{\kappa}$ in $V[G]$. For this we need to show
two things

  1. 1.

If $F\in 2^{<\kappa}\cap V[G]$, then $F\in V[G\cap\mathbb{P}^{M_{\alpha}}]$
for some $\alpha<\kappa$.

  2. 2.

None of the models $V[G\cap\mathbb{P}^{M_{\alpha}}]$ contains all sequences
from $2^{<\kappa}\cap V[G]$ (here we use the regularity of $\kappa$, so we can
later pass to a strictly increasing subsequence).

Concerning $1.$ assume that $\mathbb{P}_{\kappa}\Vdash\dot{F}\in 2^{\delta}$,
for some ordinal $\delta<\kappa$. We can assume that
$\dot{F}=\\{(\alpha,\alpha_{n},p_{n}^{\alpha})|\;\alpha<\delta,\;n<\omega\\}$,
and all conditions $p_{n}^{\alpha}$ are simple. By Lemma 21 each condition
$p_{n}^{\alpha}$ is definable with a parameter
$E_{n}^{\alpha}\in\kappa^{\omega}$. The set
$\\{E^{\alpha}_{n}|\;\alpha<\delta,n<\omega\\}$ is definable from a binary
sequence of the length $<\kappa$, so it belongs to some model $M_{\alpha}$,
and so $\dot{F}\in M_{\alpha}$. Therefore $\dot{F}$ is a
$\mathbb{P}_{\kappa}^{M_{\alpha}}$-name, and it follows that

$\dot{F}[G]=\dot{F}[G\cap\mathbb{P}^{M_{\alpha}}]\in
V[G\cap\mathbb{P}^{M_{\alpha}}].$

Concerning $2.$ fix a sequence $F\in 2^{\delta}\setminus M_{\alpha}$,
$\delta<\kappa$. Let $\mathbb{C}_{F}$ be any representation of the Cohen
forcing, from which the sequence $F$ is definable, and
$|\mathbb{C}_{F}|<\kappa$. This of course leaves plenty of space for what
specifically $\mathbb{C}_{F}$ might be, but for the sake of concreteness let
us define $\mathbb{C}_{F}=(\delta\cdot\omega,\leq_{F})$, where $\leq_{F}$ is
the transitive closure of the union of the following three relations:

  1. 1.

$\leq_{F}\restriction\omega\times\omega$ is isomorphic to the countable
atomless Boolean algebra,

  2. 2.

$\forall\;\alpha<\delta\quad\alpha\cdot\omega<_{F}\alpha\cdot\omega+1\iff
F(\alpha)=1$,

  3. 3.

$\forall\;\alpha<\delta\quad\alpha\cdot\omega>_{F}\alpha\cdot\omega+1\iff
F(\alpha)=0$.

We pick $\gamma<\kappa$ for which

$\mathbb{P}_{\gamma}\Vdash\dot{\mathbb{Q}}_{\gamma}=\mathbb{C}_{F}.$

In this case, we also have

$\mathbb{P}^{M_{\alpha}}_{\gamma}\Vdash\dot{\mathbb{Q}}^{M_{\alpha}}_{\gamma}=\\{0\\},$

since $F\notin M_{\alpha}$.

The mapping from $\mathbb{C}_{F}$ into $\mathbb{P}_{\gamma+1}$ given by the
formula

$x\mapsto(1_{\mathbb{P}_{\gamma}},x)$

is a complete embedding, and it extends to a complete embedding

$f:\mathbb{C}_{F}\hookrightarrow\mathbb{P}_{\kappa}.$

Let $\dot{H}$ be a $\mathbb{P}_{\kappa}^{M_{\alpha}}$-name for a generic
filter. In the generic extension $V[H]$, the function $f$ remains a complete
embedding of a Cohen forcing into $\mathbb{P}_{\kappa}$, and by the choice of
$\gamma$, $f$ is in fact a complete embedding into $\mathbb{P}_{\kappa}/H$. In
conclusion, forcing with $\mathbb{P}_{\kappa}$ adds a Cohen real over
$V[G\cap\mathbb{P}_{\kappa}^{M_{\alpha}}]$. ∎

######  Corollary 24.

The following theories are consistent

$ZFC+MA_{\omega_{1}}+\operatorname{Slice}_{\omega_{2}}+``2^{\omega}=\omega_{2}",$
$ZFC+MA_{\omega_{1}}+\operatorname{Slice}_{\omega_{2}}+``2^{\omega}=\omega_{3}",$
$ZFC+MA_{\omega_{2}}+\operatorname{Slice}_{\omega_{3}}+``2^{\omega}=\omega_{29}".$

##  6 Final comments

It is easy to see that all sequences of models witnessing
$\operatorname{Slice}_{\kappa}$ that we built, are cardinal preserving.
Moreover, we proved that $MA_{\omega_{1}}$ and
$\operatorname{Slice}_{\omega_{1}}$ are not compatible. It looks reasonable to
expect that for any regular cardinal $\kappa$

$MA_{\kappa}\implies\lnot\operatorname{Slice}_{\kappa}.$

I would like to thank prof. Piotr Zakrzewski for giving the idea of looking at
the cardinal invariant $\operatorname{c(Sym(\omega))}$.

## References

  * [1]
  * [1] U. Avraham, S. Shelah, Martin’s Axiom does not imply that every two $\aleph_{1}$-dense sets of reals are isomorphic, Israel Journal of Mathematics, vol. 38, Nos. 1-2 (1981)
  * [2] T. Bartoszyński, H. Judah, Set Theory: On The Structure of The Real Line, A.K. Peters Ltd. 1995
  * [3] J.E. Baumgartner, All $\aleph_{1}$-dense sets of reals can be isomorphic, Fund. Math. 79 (1973), 101-106
  * [4] A. Blass, Combinatorial cardinal characteristics of the continuum, Handbook of set theory. Springer, Dordrecht, 2010, 395-489
  * [5] J. Brendle, M. Losada, The cofinality of the infinite symmetric group and groupwise density, The Journal of Symbolic Logic, vol. 68, No.4 (2003): 1354-1361
  * [6] W. G. Fleissner, A.W. Miller, On $Q$-sets. Proc. Am. Math. Soc. vol. 78, n. 2 (1980)
  * [7] S. D. Friedman, L. Zdomskyy, Measurable cardinals and the cofinality of the symmetric group, Fundamenta Mathematicae 207.2 (2010): 101-122
  * [8] J. Ihoda, S. Shelah, Souslin Forcing, The Journal of Symbolic Logic, vol. 53, No. 4 (1988)
  * [9] K. Kunen, Set Theory. Introduction to Independence Proofs, 1980 Elsevier Science B.V.
  * [10] J. T. Moore, What makes the continuum $\aleph_{2}$, Foundations of mathematics, essays in honor of W. Hugh Woodin’s 60th birthday, Harvard University (Caicedo et al., eds.) (2017), 259-287
  * [11] H. D. Macpherson, P. M. Neumann, Subgroups of infinite symmetric groups., Journal of the London Mathematical Society 2.1 (1990): 64-84
  * [12] J. D. Sharp, S. Thomas, Unbounded families and the cofinality of the infinite symmetric group, Archive for Mathematical Logic 34.1 (1995): 33-45
  * [13] S. Shelah, S. Thomas, The cofinality spectrum of the infinite symmetric group, The Journal of Symbolic Logic, vol. 62, No.3 (1997): 902-916
  * [14] S. Thomas, Groupwise density and the cofinality of the infinite symmetric group, Archive for Mathematical Logic 37.7 (1998): 483-493.
  * [15] Y. Zhang, Permutation groups and covering properties, Journal of the London Mathematical Society 63.1 (2001): 1-15