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\newcommand{\poset}{{\Bbb P}}
\newcommand{\proof}{\par\noindent proof:\par}
\newcommand{\propsub}{\subsetneqq}
\newcommand{\qed}{\nopagebreak\par\noindent\nopagebreak$\blacksquare$\par}
\newcommand{\ramseynull}{{r^0}}
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\newcommand{\seq}{\omega^{< \omega}}
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\newtheorem{thm}{Theorem}[section]
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\newenvironment{thmnonum}[1]{\bigskip\noindent{\bf Theorem #1.} \em}{\bigskip}
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\begin{document}
\begin{center}
  {\Huge Sacks forcing, Laver forcing, and} \\
  {\Huge Martin's Axiom}
  \footnote{in Archive for Mathematical Logic, 31(1992), 145-161.}
  \\
\end{center}



\begin{center}
  Haim Judah, Arnold W. Miller\footnote{Research partially supported by
  NSF grant 8801139.}, Saharon Shelah
\end{center}


\centerline{Abstract}
\begin{quote}
In this paper we study the question assuming MA+$\neg$CH does Sacks
forcing or Laver forcing collapse cardinals?   We show that this
question is equivalant
to question of what is the additivity of Marczewski's ideal $\szero$.
We give a proof that it is consistent that Sacks forcing
collapses cardinals.  On the other hand we show that Laver forcing
does not collapse cardinals.
\end{quote}

\section*{Introduction}

Let $\sacks$ be Sacks perfect set forcing \cite{sacks};
$p\in\sacks$ iff $p\subseteq \binseq$ is a nonempty subtree and
for every $s\in p$ there exists $t\supseteq s$ such that
$t\concat 0\in p$ and $t\concat 1\in p$.  The ordering is defined
by $p\leq q$ iff $p\subseteq q$.
Define
$[p]=\{x\in\cantorspace: \forall n\; x\res n\in p\}$.

The $\szero$ ideal of subsets of $\cantorspace$
is defined by
$X\in \szero$ iff for every $p\in\sacks$ there exists $q\leq p$
with $X\intersect [q]=\emptyset$.
Define $add(\szero)=min\{|F|: F\subseteq \szero, \Union F\notin \szero\}$.


Marczewski's ideal $\szero$, which first appeared in \cite{marc},
has been studied by a number of authors,
Aniszczyk, Frankiewicz, Plewik \cite{anisetal},
Brown \cite{brown1}\cite{brown2}\cite{brown3}, Brown, Cox \cite{browncox},
Brown, Prikry
\cite{brownprik}, Corazza \cite{cor}, Morgan \cite{morgan}, and
Pawlikowski \cite{paw}.
Aniszczyk \cite{anis} has asked if MA implies that
the ideal $\szero$ is $\continuum$-additive,
i.e., is it true that the union of fewer than continuum many $\szero$ sets
is an $\szero$ set, i.e.,
$add(\szero)=\continuum$.
It is a folklore result that assuming the proper forcing
axiom the ideal $\szero$ is $\continuum$-additive
(see Abraham \cite{AV}).   It is also an
easy exercise to show the consistency of $add(\szero)=\omega_1$ plus
the continuum is large.  This happens in the Cohen real model.

\begin{thmnonum} {\ref{my1}} (MA+$\neg$CH)
$add(\szero)$ is the minimum $\kappa$ such that for
some $p\in\sacks$ we have
$p\forces_{\sacks} cof(\continuum)=\kappa$.
\end{thmnonum}

This means that the question of the additivity of
the $\szero$ ideal is the same as the question of whether Sacks forcing
collapses cardinals.
In the proof we only use that $\continuum$ is regular and MA
holds for countable
posets.  It is well known that Sacks forcing cannot blow up the continuum.
In fact Sacks \cite{sacks} showed that  every new element $y$ of
$\cantorspace$ there is a homeomorphism (with
perfect domain and range) coded in the ground model
which maps the Sacks real to $y$.

\bigskip

Question: Is it consistent with ZFC that $\continuum=\omega_2$,
$add(\szero)=\omega_1$, and
$\sacks$ does not collapse $\continuum$?

Question (Laver): Does ${\goth d}=\aleph_1$ imply that $\sacks$
collapses $\continuum$ to $\goth d$?

\bigskip

The remainder of the section is concerned
with three other cardinals associated  with the ideal of $\szero$-sets.
These cardinals have been extensively studied for the ideals
of measure and category, see for example Judah and Shelah \cite{IS}.

$non(\szero)=min\{|X|: X\notin \szero, X\subseteq \cantorspace\}$

$cov(\szero)=min\{|F|: F\subseteq \szero, \Union F=\cantorspace\}$

$cof(\szero)=min\{|F|: F\subseteq \szero\mbox{ and }
\forall A\in \szero\;\exists B\in F\;
A\subseteq B\}$

Note that $non(\szero)=\continuum$, since any small set of reals is
in $\szero$.

In Veli\v{c}kovi\'{c}'s model (\cite{vel}) he gets MA,
$\continuum=\omega_2$, and
$cov(\szero)=\omega_1$.  The same is true in the model of
Theorem \ref{velthm}.

\begin{thmnonum}{\ref{my2}}
It is relatively consistent with ZFC that
$c=\omega_2=cov(\szero)$ and $add(\szero)=\omega_1$.
\end{thmnonum}

Fremlin noted that $cof(\szero)>\continuum$. Theorem \ref{fremlin}
slightly improves this.

\begin{thmnonum}{\ref{fremlin}}
 $cof(cof(\szero))>\continuum$
\end{thmnonum}

This argument also produces an $\szero$ set of cardinality $\continuum$
(Gurevich and Shelah \cite{shelah} see also \cite{survey}).
Pierre Matet \cite{matet} has also proved a similar result for
$cof(\ramseynull)$ where $\ramseynull$ is the ideal of Ramsey null sets.

So we get (in ZFC) that
$$\omega_1\leq add(\szero)\leq cov(\szero)\leq\continuum=
non(\szero)<cof(cof(\szero))\leq
cof(\szero)\leq 2^{\continuum}.$$

The remaining two theorems in this section are easy consistency results
to indicate that the inequality
$$\continuum<cof(cof(\szero))\leq cof(\szero)\leq 2^{\continuum}$$
is best possible.  For simplicity of notation we only consider models of CH.

\begin{thmnonum}{\ref{Cohen}}
Let $V\models$ GCH, $\kappa$ any cardinal
of cofinality greater than $\omega_1$,
and let $\poset$ be the partial order of countable functions
from $\kappa$ to $\omega_1$.  If $G$ is $\poset$-generic over
$V$, then
$V[G]\models$``$\omega_1=\continuum$ and $cof(\szero)=2^{\omega_1}=\kappa$''.
So, for example, it is consistent that
$\omega_2=cof(cof(\szero))<cof(\szero)=\aleph_{\omega_2}$.
\end{thmnonum}

\begin{thmnonum}{\ref{Cohen2}}
  Let $V$ be a model of CH and $\kappa$ be any regular
  cardinal with $\omega_1<\kappa\leq 2^{\omega_1}$.
  Then there exists a generic extension $V[G]$ (with same cardinals and
  function $2^\lambda$) in which $cof(\szero)=\kappa$.
\end{thmnonum}

%  vel.tex

In section \ref{vel} we show that it is consistent with MA+$\neg$CH that
the addivity of the $\szero$ ideal is $\omega_1$, (equivalently
Sacks forcing collapses the continuum.)

\begin{thmnonum}{\ref{velthm}}
  It is consistent with MA+$\neg$CH that the ideal of
  $\szero$ sets is not $\omega_1$-additive.
\end{thmnonum}

This result was also
obtained independently by Velickovic \cite{vel} about the same time.
Velickovic starts with a model of PFA and forces to
kill the additivity of the $\szero$ ideal and proves that his forcing
does not add any new subsets of $\omega_1$.

It is clear from the construction that
$\continuum$ can be made arbitrarily large, but for simplicity
we make it $\omega_2$.   It is not clear
that $add(\szero)$ can be made anything we want.
We assume the reader is familiar with
the usual proof of the consistency of MA (see Kunen \cite{kunen}).

A similar theorem has been proved for Silver forcing by
Steprans \cite{steprans}.  For Silver forcing
the ideal analagous to $\szero$ is the Mycielski ideal $\cite{myciel}$.

The technique we use is based on an unpublished proof
of Kunen who showed that MA+$\neg$CH does not settle the existence of
$(\omega_1^*,\continuum)$-gaps or $(\continuum^*,\continuum)$-gaps.
Kunen \cite{kunenma} also uses this argument in his proof that
it is possible for $\aleph_{\omega_1}$ to be the first cardinal for which
MA fails.   It has
also been used to prove other  statements are independent of MA+$\neg$CH,
see Abraham, Rubin, Shelah \cite{ARS}, Abraham, Todorcevic \cite{AT},
Devlin \cite{devlin}, and Steprans \cite{steprans}.

Kunen's idea can be summarized as follows.  To get the consistency
of MA+$\neg$CH+P where P is some statement, do a ccc iteration of
of small ccc posets.
At each step you either force with a ccc poset Q which preserves
the statement P or you force with a ccc poset R which destroys
the ccc-ness of Q but preserves statement P.  In either case you have
taken care of Q and preserved statement P.


% laver.tex

Section \ref{laver} is about Laver forcing $\laver$.
Conditions in  $\laver$ are
subtrees $p\subseteq\seq$ with the property that there exists
a node $s\in p$ called its root such that:
\begin{enumerate}
  \item for every $t\in p$ either $t\subseteq s$ or $s\subseteq t$,
  \item for every $t\in p$ if $s\subseteq t$, then the set
  $split(p,t)=^{def}\{n\in\omega: t\concat n\in p\}$ is infinite.
\end{enumerate}
The order relation is given by subset.   This forcing was used
by Laver \cite{laver} to prove the consistency of
the Borel conjecture.

\begin{thmnonum}{\ref{t1}}
  Assume MA.  Then $\laver$ does not collapse cardinals.
\end{thmnonum}

Analogously to the $\szero$ sets we can define the Laver null sets
$\lavernull$ to be all $X\subseteq\bairespace$ such that
for every $p\in \laver$ there exists $q\leq p$ with $q\in\laver$
such that $[q]\intersect X=\emptyset$.
It is easy to see
that Theorem \ref{t1} also shows that MA implies that
$add(\lavernull)=\continuum$.

The corresponding theorem for Mathias forcing
(\cite{happy}) is easy to prove. The fact that the additivity of the
ideal of Ramsey null sets is
greater than $\omega_1$ under MA+$\neg$CH plays a key role in
Silver's proof \cite{silver} that analytic sets are completely Ramsey.

%%%% super.tex

In section \ref{super} we consider a notion of forcing which is
half-way in between Sacks forcing $\sacks$ and Laver forcing
$\laver$.   This forcing is often called superperfect tree forcing
or rational perfect set forcing, see  Kechris \cite{superperf},
Miller \cite{ratperf}, Blass and Shelah \cite{blassshe}, Blass \cite{blass}

Let $\super$ be superperfect tree forcing which we define as follows.
For $p$ a subtree of $\seq$ define the splitting nodes of $p$:
$$\splitnode(p)=\{s\in p: \exists^\infty n\in\omega\; s\concat n\in p\}.$$
Define $p\in\super$ iff
\begin{enumerate}
  \item $p$ is a nonempty subtree of $\seq$,
  \item $\splitnodes(p)$ is dense in $p$, i.e.,
  $\forall s\in p \;\exists t\in\splitnodes(p)$ with $t\supseteq s$, and
  \item if any node in $p$ splits it is a splitting node, i.e.,
  if there exists more than one $n\in\omega$ such that $s\concat n\in p$, then there
  are infinitely many $n\in\omega$ such that $s\concat n\in p$.
\end{enumerate}

\begin{thmnonum}{\ref{t2}}
  Assume MA.  Then $\super$ does not collapse cardinals.
\end{thmnonum}

As is the case with Ramsey null and Laver null sets, MA implies
that $add(\supernull)=\continuum$, where $\supernull$ is the ideal
of all subsets $X\subseteq\bairespace$ such that
for every $p\in \super$ there exists $q\leq p$ with $q\in\super$
such that $[q]\intersect X=\emptyset$.

Its seems in general that MA can only handle those forcing which have
some kind of infinite splitting going on, e.g. Mathias, Laver, superperfect,
whereas for those whose conditions are compact, e.g. Sacks, Silver, it
is consistent with MA that they collapse the continuum.

% for.tex

In the appendix we solve a
problem posed by  M. Foreman.


\section{Sacks forcing and Marczewski's ideal}
\label{szero}

Let $\sacks$ be Sacks perfect set forcing,
$p\in\sacks$ iff $p\subseteq \binseq$ is a nonempty subtree and
for every $s\in p$ there exists $t\supseteq s$ such that
$t\concat 0\in p$ and $t\concat 1\in p$.  The ordering is defined
by $p\leq q$ iff $p\subseteq q$.
Define
$[p]=\{x\in\cantorspace: \forall n\; x\res n\in p\}$.

The $\szero$ ideal of subsets of $\cantorspace$
is defined by
$X\in \szero$ iff for every $p\in\sacks$ there exists $q\leq p$
with $X\intersect [q]=\emptyset$.

$$add(\szero)=min\{|F|: F\subseteq \szero, \Union F\notin \szero\}$$

\begin{thm} \label{my1} (MA+$\neg$CH)
$add(\szero)$ is the minimum $\kappa$ such that there exists
$p\in\sacks$ such that
$p\forces_{\sacks} cof(\continuum)=\kappa$.
\end{thm}
\proof

We will need the following lemma.

\begin{lem}
Suppose that $D\subseteq\sacks$ is open and
dense,  then there exists $A\subseteq D$ a maximal antichain with the
property that for every $p\in\sacks$ if

$[p]\subseteq \Union_{q\in A}[q]$, then there exists
$A^\prime\in [A]^{<\continuum}$
such that $[p]\subseteq \Union _{q\in A^\prime}[q]$.
\end{lem}
\proof
Let $\sacks=\{q_{\alpha}:\alpha<\continuum\}$.
Build $A=\{p_{\alpha}:\alpha<\continuum\}$ and
$\{x_\alpha:\alpha<\continuum\}\subseteq\cantorspace$
by induction on $\alpha$. At stage $\alpha$ if
$[q_{\alpha}]$ is not covered by $\Union_{\beta<\alpha} [p_{\beta}]$,
then choose $x_\alpha\in [q_{\alpha}]\setminus
\Union_{\beta<\alpha} [p_{\beta}]$ otherwise let $x_\alpha=x_0$.
If $q_{\alpha}$ is compatible with some $p_\beta$ then let
$p_\alpha=p_0$.  Otherwise
since a perfect set can be divided into perfectly many disjoint perfect
sets it is possible to find $p_\alpha\leq q_\alpha$ such
that $[p_\alpha]$ is disjoint from $\{x_\beta:\beta\leq\alpha\}$.
\qed

Now suppose $\forces cof(\continuum)=\kappa$.  We show $add(\szero)\leq\kappa$.
Let
$$\forces_{\sacks}\mbox{``}\tau:\kappa\rightarrow\continuum\mbox{ is cofinal''}.$$
For each $\alpha<\kappa$ let
$$D_{\alpha}=\{p\in\sacks: \exists\beta\; p\forces \tau(\alpha)=\beta\}.$$
Let $A_\alpha\subseteq D_\alpha$
be the antichains obtained from the lemma.  And let
$$X_{\alpha}=\cantorspace\setminus\Union_{p\in A_{\alpha}}[p].$$
Since $A_\alpha$ is a maximal antichain it is easy to see that
$X_{\alpha}$ is an $\szero$ set. Now suppose that $\continuum$ is regular and
$\kappa<\continuum$.
We claim that $X=\Union_{\alpha<\kappa} X_{\alpha}$
is not an $\szero$ set. If it is we must have $[p]\intersect X=\emptyset$
for some $p\in\sacks$.  But this implies that
$[p]\subseteq\Union_{q\in A_{\alpha}}[q]$
for each $\alpha$ so there exists $A_{\alpha}^{\prime}\subseteq A_{\alpha}$
of cardinality less than $\continuum$ such that
$[p]\subseteq\Union_{q\in A_{\alpha}^\prime}[q]$.  By the definition
of $D_{\alpha}$ this means there exists $Y_{\alpha}$ of cardinality less
than $\continuum$ such that
$$p\forces\forall\alpha<\kappa \;\tau(\alpha)\in \check{Y}_{\alpha}.$$
This contradicts the fact that $\tau$ is a cofinal map.

For the other direction suppose that $add(\szero)=\kappa$ and suppose for
contradiction that
 $\forces_{\sacks} cof(\continuum)>\kappa$. We can
assume $\kappa<\continuum$. Let
$X_{\alpha}$ be $\szero$ sets such that
$$X=\Union_{\alpha<\kappa}X_{\alpha}$$
is not $\szero$.
Working below the appropriate $p^*$ (namely, some withness to
the fact $X\notin\szero$), we have that every $p$ satisfies
$[p]\intersect X\not=\emptyset$.
Let
$D_{\alpha}=\{p\in\sacks:[p]\intersect X_{\alpha}=\emptyset\}$. It is
easy to see that $D_{\alpha}$ is open dense.
Let $A_{\alpha}\subseteq D_{\alpha}$ be any maximal antichain.
Let $\sacks=\{p_{\alpha}:\alpha<\continuum\}$ be listed without repetitions
and define an $\sacks$ name $\tau:\kappa\rightarrow\continuum$
by $\tau(\alpha)$ is the unique
$\beta$ such that $p_\beta\in G\intersect A_{\alpha}$ where $G$ is
the $\sacks$-generic filter.  Since the cofinality of $\continuum$ is
greater than $\kappa$ in the extension, we can find $\beta<\continuum$
and $p\in\sacks$ such that
$$p\forces\forall\alpha<\kappa\;\tau(\alpha)<\beta.$$
Let $H=\{p_{\alpha}:\alpha<\beta\}$.
Clearly for every $r\leq p$ and $\alpha<\kappa$
there exists $q\in A_{\alpha}\intersect H$ which is compatible with $r$.
Define for $s\in p$, $p_s=\{t\in p: s\subseteq t\mbox{ or } t\subseteq s\}$.
For a fixed $\alpha$, there are two possibilities:
\begin{enumerate}
  \item for every $q\in A_{\alpha}\intersect H$ the set $[q]\intersect [p]$ is
  nowhere dense in $[p]$, or
  \item there exists $q\in A_{\alpha}\intersect H$ such that
  for some $s\in p$ we have $p_s\leq q$.
\end{enumerate}
But (1) is impossible, since by MA we could find a perfect
$r\leq p$ such that for every $q\in A_{\alpha}\intersect H$ we have
$[r]\intersect [q]=\emptyset$.  Also (1) is impossible
for any $p_s$ in place of $p$.   Hence we can find $E_{\alpha}\subseteq p$
such that for every $s\in E_{\alpha}$ there exists $q\in A_{\alpha}$ such
that $p_s\leq q$ and $E_{\alpha}$ is a dense set of nodes in $p$, ie.
for every $t\in p$ there exists $s\supseteq t$ with $s\in E_{\alpha}$.

Consider the forcing notion:
$${\Bbb P}=\{(F,n):\emptyset\not=F\subseteq p \intersect 2^n\}$$
and ordered by $(F,n)\leq(F^\prime,n^\prime)$ iff $n\geq n^{\prime}$ and
$F^{\prime}=\{s\res n^\prime: s\in F\}$.  Since $\Bbb P$ is countable,
forcing with it is the same as Cohen real forcing. Given $G$ a
$\Bbb P$-filter let $r=\Union\{F:\exists n\; (F,n)\in G\}$.
It is easy to write down countably many dense sets which will guarantee that
$r\in\sacks$.  Also $\kappa$ many dense sets which will make it true
that for every $\alpha<\kappa$ there exists $n<\omega$ for every
$s\in r\intersect 2^n$ there exists $q\in A_{\alpha}$ with
$r_s\leq q$.  But this implies that for every $\alpha$
$$[r]\subseteq\Union_{q\in A_{\alpha}}[q]$$
and so $[r]\intersect\Union_{\alpha<\kappa}X_{\alpha} =\emptyset$.
This contradiction finishes the proof of Theorem \ref{my1}. \qed


\noindent Three other cardinal functions of an
ideal $I$ are $non$, $cov$, and $cof$:

$non(I)=min\{|X|: X\notin I, X\subseteq \cantorspace\}$

$cov(I)=min\{|F|: F\subseteq I, \Union F=\cantorspace\}$

$cof(I)=min\{|F|: F\subseteq I\mbox{ and } \forall A\in I\;\exists B\in F\;
A\subseteq B\}$


\begin{thm}\label{my2}
It is relatively consistent with ZFC that
$c=\omega_2=cov(\szero)$ and $add(\szero)=\omega_1$.
\end{thm}

\proof
The iterated Sacks forcing model is described in
Baumgartner-Laver \cite{BL}.  It is obtained by starting with a ground
model which satisfies CH and then iterating $\sacks$ $\omega_2$ times
with countable supports. The continuum ends up being $\omega_2$ and
no cardinals are collapsed.

The fact that $add(\szero)=\omega_1$ follows from half of the proof of
Theorem~\ref{my1} and
the theorem (proved in \cite{BL}) that Sacks forcing over $V_{\omega_2}$
collapses $\omega_2$.

To see
that $cov(\szero)=\omega_2$ let $V_{\alpha}$ for $\alpha\leq\omega_2$ be
the iteration sequence.  Let $\langle X_\alpha:\alpha<\omega_1\rangle\in
V_{\omega_2}$ be $\szero$ sets.  In $V_{\omega_2}$ let
$f_\alpha:\sacks\rightarrow\sacks$ be such that for every $p\in\sacks$
$f_\alpha(p)\leq p$ and $[f_\alpha(p)]\intersect X_\alpha=\emptyset$.
Since
the iteration has the $\omega_2$ chain condition by a
Lowenheim-Skolem argument it is possible to find $\gamma<\omega_2$
so that
$$\langle f_{\alpha}\res \sacks^{V_\gamma}:\alpha<\omega_1\rangle
\in V_{\gamma}.$$
But, we claim the $\gamma^{th}$ Sacks real $x_\gamma$ is not in
$\Union_{\alpha<\omega_1} X_{\alpha}$.  If it was, then for some
condition $p\in {\Bbb P}_{[\gamma,\omega_2)}$ and some
$\alpha<\omega_1$ we would have:
$$p\forces x_\gamma\in X_\alpha.$$
But letting $q=p(\gamma)\in\sacks$ and letting $r(\gamma)=f_\alpha(q)$
and $r(\beta)=p(\beta)$ for $\beta>\gamma$ we see that
$$r\forces x_\gamma\notin X_\alpha.$$
\qed

\begin{thm} \label{fremlin}
 $cof(cof(\szero))>\continuum$
\end{thm}
\proof
Let $\kappa=cof(\szero)$ and $\{Y_\alpha:\alpha<\kappa\}\subseteq \szero$ be a
cofinal family.  Let
$$\{p_{\alpha}:\alpha<\continuum\}$$
be all
perfect subsets of $\cantorspace\cross\cantorspace$,
$$\{x_{\alpha}:\alpha<\continuum\}=\cantorspace,\mbox{ and }
L_\alpha=\{(x_\alpha,y):y\in\cantorspace\}.$$
 Build $q_\alpha\subseteq
L_\alpha$ perfect such that for every $\beta<\alpha$ if $p_\beta\intersect
L_{\alpha}$ is countable, then $q_\alpha$ is disjoint from $p_\beta$.
This is easily done since any perfect set splits into $\continuum$ many
disjoint perfect sets.  Now we assume for contradiction that
$cof(\kappa)\leq\continuum$.  Let $F:\continuum\rightarrow\kappa$ have
unbounded range.   Since $\kappa=cof(\szero)$ for each
$\beta<\continuum$ the set
$$\{Y_{\alpha}\intersect q_\beta:\alpha<F(\beta)\}$$
 is not
cofinal in the $\szero$ subsets of $q_\beta$.  So there exists an $\szero$ set
$Z_\beta\subseteq q_\beta$ such that $Z_\beta$ is not covered by any
$Y_{\alpha}$ with $\alpha<F(\beta)$.   Then
$Z=\Union_{\beta<\continuum}Z_\beta$ is not covered by any $Y_\alpha$, so
it suffices to see that $Z$ is an $\szero$ set.   Let
$p\subseteq \cantorspace\cross\cantorspace$ be any perfect set.
 If $p\intersect L_\beta$ is uncountable for any $\beta$,
then let $q\subseteq p\intersect L_\beta$ be perfect such that
$q\intersect Z_{\beta}=\emptyset$, then $q\intersect Z=\emptyset$.
Otherwise $p\intersect L_\beta$ is countable for every $\beta$.
If $p=p_\alpha$, then by construction $q_{\beta}\intersect p_\alpha=\emptyset$
for every $\beta\geq\alpha$.  It follows that
$$(p\intersect Z)\subseteq(p\intersect \Union_{\beta<\alpha} L_\beta).$$
But $p\intersect (\Union_{\beta<\alpha} L_\beta)$
 is a set of cardinality less than $\continuum$ and so
there exists $q\subseteq p$ perfect such that
$q\intersect (\Union_{\beta<\alpha} L_\beta)=\emptyset$,
and hence $q\intersect Z=\emptyset$. \qed


\begin{thm}\label{Cohen}
Let $V\models$ GCH, $\kappa$ any cardinal
of cofinality greater than $\omega_1$,
and let $\poset$ be the partial order of countable functions
from $\kappa$ to $\omega_1$.  If $G$ is $\poset$-generic over
$V$, then
$V[G]\models$``$\omega_1=\continuum$ and $cof(\szero)=2^{\omega_1}=\kappa$''.
So, for example, it is consistent that
$\omega_2=cof(cof(\szero))<cof(\szero)=\aleph_{\omega_2}$.
\end{thm}
\proof
Countably closed forcing does not add any reals.
By the usual chain condition argument and decomposition as a product
forcing it is enough to see:

If $G:\omega_1\rightarrow\omega_1$ is $\omega_1^{<\omega_1}$-generic over
$V$ then in $V[G]$ there exists an $\szero$ set $X$ which is not covered
by any $\szero$ set in $V$.

By a similar construction as in the proof of Theorem \ref{fremlin} we
can  find in $V$ disjoint perfect sets $q_{\alpha}$ for
$\alpha<\omega_1=\continuum$ such that any $X\subseteq\cantorspace$
which meets each $q_{\alpha}$ in a singleton is an $\szero$-set.  Now
just use the generic function $G:\omega_1\rightarrow\omega_1$ to pick
out a single element of each $q_\alpha$.
\qed


\begin{thm}\label{Cohen2}
  Let $V$ be a model of CH and $\kappa$ be any regular
  cardinal with $\omega_1<\kappa\leq 2^{\omega_1}$,
  then there exists a generic extension $V[G]$ (with same cardinals and
  function $2^\lambda$) in which $cof(\szero)=\kappa$.
\end{thm}
\proof

To bring down $cof(\szero)$ but leave $2^\continuum$ large, force
with the following partial order $\poset$:

\noindent $(X,f)\in\poset$ iff
\begin{enumerate}
\item $X\in \szero$,
\item $f:\sacks\rightarrow\sacks$ is a countable partial function, and
\item $\forall p\in domain(f)\;$ $f(p)\leq p$ and
$[f(p)]\intersect X=\emptyset$.
\end{enumerate}
The ordering on $\poset$ is
defined by $(X^\prime,f^\prime)\leq (X,f)$ iff
$X^\prime\supseteq X$ and $f^\prime\supseteq f$.
Clearly $\poset$ is countably closed and
two elements of $\poset$ with the same $f$ are compatible,
so it is $\omega_1$-centered.
It is also true that it is well met, ie. infimums exist.

Now define $\poset_\omega^{\omega_1}$ to be those elements of
$p\in \poset^{\omega_1}$ with countable support, ie. there are at most
countably many $\alpha<\omega_1$ such that
$p(\alpha)\not=(\emptyset,\emptyset)$.  It is easy to see that
forcing with $\poset_\omega^{\omega_1}$ adds an
$\langle f_{\alpha}:\alpha<\omega_1\rangle$ such that
if
$$X_\alpha=\cantorspace\setminus\Union_{p\in\sacks}[f_\alpha(p)]$$
then each $X_{\alpha}\in \szero$ and for every $Y\in \szero\intersect V$ there
exists $\alpha<\omega_1$ such that $Y\subseteq X_{\alpha}$.
Now just like in the usual proofs of Generalized Martin's Axiom,
see \cite{GMA}, \cite{weiss}, or \cite{baum},
we iterate forcing with $\poset_\omega^{\omega_1}$ $\kappa$ many times
with countable supports.  Since cofinally often we add a function
$f:\omega_1\rightarrow\omega_1$ which is $\omega_1^{<\omega_1}$ generic,
by the argument of Theorem \ref{Cohen} we see that
$cof(\szero)\geq\kappa$.  Since the iteration satisfies the $\omega_2$
chain condition every subset of $\cantorspace$ appears at some
intermediate stage, and is covered by one of the generic
$\szero$ sets.  Hence $cof(\szero)\leq\kappa$.\qed


\section{Martin's Axiom and Marczewski's ideal}
\label{vel}

In this section we start with a model satisfying
the continuum hypothesis and by an inductive construction, we will
get a model for Martin's Axiom where the Marczewski's ideal, $\szero$,
is not $\omega_1$ additive.

\begin{thm}\label{velthm}
  It is consistent with MA+$\neg$CH that the ideal of
  $\szero$ sets is not $\omega_1$-additive.
\end{thm}

\proof

Recall that  $[T]=\{x\in\cantorspace:\forall n<\omega\; x\res n\in T\}$.

Define $(*)\langle T_{j}:j<\zeta \rangle$
where $\zeta\leq\omega_2$ and each
$T_j\in\sacks$ as follows:

\medskip

$(*)\langle T_{j}:j<\zeta \rangle$:
Given $\langle F_\alpha:\alpha<\omega_1\rangle$
pairwise disjoint finite subsets of $\zeta$
there exists $\alpha\not=\beta$ such that
$\forall i\in F_\alpha\forall j\in F_\beta\;
[T_{i}]\intersect [T_{j}]=\emptyset$.

\medskip

The Construction:   By induction on $\zeta\leq\omega_2$ we shall
\begin{enumerate}
  \item define $\langle P_i,Q_j: j<\zeta,i\leq\zeta\rangle$
      a finite support iteration of ccc forcing notions
       with as usual $P_{\zeta+1}=P_{\zeta}*Q_\zeta$,
  \item define $\langle T_{j}:j<\zeta\rangle\in V^{P_\zeta}$ such that
     $V^{P_\zeta}\models (*)\langle T_{j}:j<\zeta\rangle$,
  \item  make sure that
     $\{T_{j}:j<\omega_2\}$ is dense in
     $\sacks\intersect{V^{P_{\omega_2}}}$, and
  \item make sure that MA holds in $V^{P_{\omega_2}}$.
\end{enumerate}

In order to make MA true we list
all possible $P_{\omega_2}$ names for posets of cardinality $\omega_1$
say $R_{\zeta}$ for $\zeta<\omega_2$.
We then apply Lemma~\ref{vlemma1}
to the ground model $V^{P_\zeta}$ to get either that
forcing with ${\poset_{\zeta}*R_{\zeta}}$ satisfies
$(*)\langle T_{j}:j<\zeta\rangle$ or there exists $Q$ such that
forcing with
$P_{\zeta}*Q$ satisfies $(*)\langle T_{j}:j<\zeta\rangle$ and $R_{\zeta}$
does not have ccc.  Let $T^\prime_\zeta$ be any element
of $\sacks\intersect V^{P_\zeta}$.  However make sure that at the end
we have $\{T_\zeta^\prime:\zeta<\omega_2\}$ is dense in
$\sacks\intersect V^{P_{\omega_2}}$; because
$T_\zeta\leq T_\zeta^\prime$ (see $(**)$ below) this
ensures that $\{T_\zeta:\zeta<\omega_2\}$ is dense.
Note that since $Q_{T^\prime_\zeta}$ (defined below) is a countable poset it
cannot destroy $(*)$ nor can adding one tree destroy $(*)$.

We then let $Q_\zeta=R*Q_{T^\prime_\zeta}$ or
$Q_\zeta=Q*Q_{T^\prime_\zeta}$ which ever preserves $(*)$.
This concludes the proof of Theorem \ref{velthm}.

\qed

\begin{lem}
In $V^{P_{\omega_2}}$ the additivity of  $\szero$ ideal is $\omega_1$, in fact
$cov(\szero)=\omega_1$.
\end{lem}
\proof
Choose $\langle T_{i,j}\in\sacks:i<\omega_1, j<\omega_2\rangle$
such that for each $i,j$ there is a unique $\zeta(i,j)$ such
that $T_{i,j}\subseteq T_{\zeta(i,j)}$ and also for each $i<\omega_1$
$$A_i=\{T_{i,j}:j<\omega_2\}$$
is a maximal antichain in $\sacks$.  This is easy to do
since the family $\{T_\zeta:\zeta<\omega_2\}$ is dense in $\sacks$
and every condition in $\sacks$ has a perfect set of incompatible extensions.
If
$$X_i=\cantorspace\setminus\Union_{T\in A_i}[T]$$
then each $X_i$ is an $\szero$-set since $A_i$
is a maximal antichain.   But $\cantorspace=\Union_{i<\omega_1}X_i$,
since otherwise if $x\in \cantorspace\setminus \Union_{i<\omega_1}X_i$,
then choosing $F_{\alpha}=\{\zeta(\alpha,\beta)\}$ where
$x\in [T_{\alpha,\beta}]\subseteq [T_{\zeta(\alpha,\beta)}]$
witnesses the failure of $(*)$.
\qed

The following lemma is the key to preserving $(*)$ while at the same time
making MA true.

\begin{lem} \label{vlemma1}
  Suppose $V\models$``$(*)\langle T_{j}:j<\zeta\rangle$, $R$ is ccc'',
  and $V^R\models$``$\neg(*)\langle T_{j}:j<\zeta\rangle$''.
  Then there exists a ccc partial
  order $Q$ such that $V^Q\models$``$(*)\langle T_{j}:j<\zeta\rangle$
   and $R$ is not ccc''.
\end{lem}

\proof
Let $\langle F^\prime_\alpha:\alpha<\omega_1\rangle$
be $R$-names and $r\in R$ such that
$$r\forces_{R}\mbox{``}\langle F^\prime_\alpha :\alpha<\omega_1\rangle
\mbox{ is a counterexample to }(*)\mbox{''.}$$
Let $\langle r_{\alpha},F_\alpha:\alpha<\omega_1\rangle\in V$ be such that
each $r_\alpha \leq r$ and
$$r_\alpha\forces F^\prime_\alpha= \check{F}_\alpha.$$

For some $A\in[\omega_1]^{\omega_1}$ we have that
$\langle F_\alpha:\alpha\in A\rangle$ is a $\Delta$-system.
Note that  root of this $\Delta$-system must be empty because
if for some $\alpha\not=\beta$
$F_\alpha\intersect F_\beta\not=\emptyset$,
then $r_\alpha$ and $r_\beta$ are incompatible.  But $R$ has the ccc.

Define
$Q$ to be the set of all $q\in [A]^{<\omega_0}$ such that
if $\alpha\not=\beta\in q$, then
$$\forall i\in F_\alpha\forall j\in F_\beta\;
[T_{i}]\intersect [T_{j}]=\emptyset.$$
Order $Q$ by inclusion.  Note that if $q\in Q$ and
$\alpha,\beta\in q$ we have $r_\alpha$ and $r_\beta$ are incompatible in $R$.
Hence forcing with $Q$ adds an uncountable antichain to $R$. So it is
enough to prove the following two claims.

\bigskip
\noindent Claim 1.  $Q$ satisfies ccc.

\proof
If not let $\langle q_\alpha:\alpha<\omega_1\rangle$ be pairwise
incompatible. For some $B\in [\omega_1]^{\omega_1}$ we have
that $\langle q_\alpha:\alpha\in B\rangle$ forms a $\Delta$-system
with root $q^*$.  Now $q_\alpha\union q_\beta\in Q$ iff
$(q_\alpha\setminus q^*)\union (q_\beta\setminus q^*)\in Q$.
Therefore without loss of generality we may assume
$\langle q_\alpha:\alpha\in B\rangle$ are pairwise disjoint.
If we let $K_\alpha=\Union_{\delta\in q_\alpha}F_\delta$,
then $\langle K_\alpha:\alpha\in B\rangle$ are pairwise
disjoint. Therefore by applying $(*)$ in $V$,
there are $\alpha\not=\beta\in B$ satisfying
$$\forall i\in K_\alpha\forall j\in K_\beta\;
[T_{i}]\intersect [T_{j}]=\emptyset$$
and so $q_\alpha\union q_\beta\in Q$.
\qed

\bigskip
\noindent Claim 2. $Q$ preserves $(*)$.

\proof
Let $q\in Q$ and $\langle H^\prime_\alpha:\alpha<\omega_1\rangle$
be a $Q$-name such that
such that
$$q\forces_{Q}\mbox{``} \langle H^\prime_\alpha:\alpha<\omega_1\rangle
\mbox{ are pairwise disjoint finite subsets of }\zeta\mbox{''}.$$
Let $\langle q_\alpha,H_\alpha:\alpha<\omega_1\rangle\in V$
be such that $q_\alpha\leq q$  and
$$q_\alpha\forces\mbox{``}H^\prime_\alpha=\check{H}_\alpha\mbox{''}.$$
Then for some
$B\in [\omega_1]^{\omega_1}$ the following form  $\Delta$-systems:
$\langle q_\alpha:\alpha\in B\rangle$ and
$\langle H_\alpha:\alpha\in B\rangle$.  By ccc of
$Q$ we have that the root of $\langle H_\alpha:\alpha\in B\rangle$ must
be empty.  Let $q^*$ be the root of
the $q$'s so  $q_\alpha\intersect q_\beta=q^*$ for
$\alpha\not=\beta\in B$.   Now we define
$$K_\alpha=\Union_{\delta\in q_\alpha\setminus q^*} F_\delta\union H_\alpha.$$
Since the $H$'s and $F$'s are pairwise disjoint families it is easy to
find $C\in [B]^{\omega_1}$ such that $K_\alpha$ for $\alpha\in C$ are
pairwise disjoint.

Hence by $(*)$ in $V$ there exists
$\alpha\not=\beta\in C$ such that
$$\forall i\in K_\alpha\forall j\in K_\beta\;
[T_{i}]\intersect [T_{j}]=\emptyset.$$
But this means that $q_\alpha\union q_\beta\in Q$ and
$$q_\alpha\union q_\beta\forces_Q\mbox{``}
\forall i\in H^\prime_\alpha\forall j\in H^\prime_\beta\;
[T_i]\intersect [T_j]=\emptyset\mbox{''}.$$
Since we started with an arbitrary condition and name we have
$(*)$ holds in $V^Q$.  Hence Claim 3 is proven and
this finishes the proof of Lemma \ref{vlemma1}.
\qed

Next we show that property $(*)$ is preserved at limit stages.
Note that $(*)$ is trivially preserved at stage $\omega_2$.
The preservation of $(*)$ at stages of cofinality $\omega_1$
is more delicate and requires that we specify the details of
exactly how we pick the trees $T_{i}$.

For $T^\prime\in\sacks$ let
$Q_{T^\prime}=\{(F,n):\emptyset\not= F\subseteq T^\prime\intersect 2^n\}$

Order $Q_{T^\prime}$ by $(F,n)\leq (G,m)$ iff $n\geq m$ and
$G=\{s\res m:s\in F\}$.
Forcing with $Q_{T^\prime}$ naturally determines a perfect subtree $T$
of $T^\prime$ as follows: if $G$ is $Q_{T^\prime}$ generic, then
let
$$T=\{s:\exists (F,n)\in G\;s\in F\}.$$
In our construction $T_{j}$  is obtained by forcing with
$Q_{T^\prime}$ for some $T^\prime$.
Recall that  $T_s=\{t\in T: s\subseteq t \mbox{ or } t\subseteq s\}$.
For $T$ which is
$Q_{T^\prime}$-generic it is easy to see that
%
\begin{enumerate}
\item if $T_0\in\sacks$ is in the ground model, then for some
$n<\omega$ for every $s\in 2^n\intersect T_0\intersect T\;\;$
$T_s\subseteq T_0$, and
%
\item for any $p\in Q_{T^\prime}$ there exists
$n<\omega, t_0,t_1, q_0,q_1\leq p$ such that $t_0$ and $t_1$ are
disjoint,
$q_0\forces T\intersect 2^n=t_0$, and $q_1\forces T\intersect 2^n=t_1$.
%
\end{enumerate}
%

\noindent Our construction will satisfy the following:

\medskip
$(**)$ For every $\alpha<\omega_2\;\;T_{\alpha}$ is
$Q_{T^\prime_\alpha}$-generic over ${V^{P_\alpha}}$ for some
$T^\prime_\alpha\in \sacks\intersect{V^{P_\alpha}}$.


\begin{lem} \label{lemma1}
  Assume $(**)$ and
  suppose $F=\{\alpha_l:l<m\}\subseteq\zeta$ is finite and
  enumerated in increasing order,
  and $p\in P_\zeta$, then there exists $q\leq p$, $n<\omega$, and
  $\langle Q_{l}\subseteq 2^n:l<m\rangle$ such that
  for each $l<m$
  $$q\forces \mbox{`` }T_{\alpha_l}\intersect 2^n=\check{Q}_{l}\mbox{''}$$
  and for any $k<l<m$ if $s\in Q_l\intersect Q_k$
  then
  $$q\forces \mbox{`` }T_{\alpha_ls}\subseteq T_{\alpha_k}\mbox{''}.$$
\end{lem}
\proof
Left to the reader.
\qed

\begin{lem} \label{lemma2}
    Assume $(**)$ and
  suppose $\gamma\leq min(H)$ where $H\subseteq\zeta$ is finite
  and $p\in P_\zeta$,
  then there exists $q_0,q_1\leq p$ with $q_0\res\gamma=q_1\res\gamma$ and
  disjoint $t_0,t_1\subseteq 2^n$ for some $n<\omega$
  such that for each $\alpha\in H$ and $i=0,1$
 $$q_i\forces\mbox{``} T_\alpha\intersect 2^n\subseteq\check{t}_i\mbox{''}.$$
\end{lem}
\proof

\noindent  It is easy to show:

Suppose $P_1,P_2$ are arbitrary posets and
$T_i^\prime\in\sacks\intersect V^{P_i}$.
Given $p_1\in P_1 * Q_{T^\prime_1}$ and $p_2\in P_2 * Q_{T^\prime_2}$
there exists
$n<\omega, q_1\leq p_1,q_2\leq p_2, t_0\subseteq 2^n,t_1\subseteq 2^n$
such that $t_0\intersect t_1=\emptyset$ and for $i=0,1\;$
$q_i\forces T_i\intersect 2^n\subseteq \check{t}_i$.

The lemma follows by iteratively applying this statement to all pairs in $H$
with $V^{P_\gamma}$ as the ground model.

\qed

\begin{lem}
    Assume $(**)$ and suppose that $\zeta$ is a limit ordinal
  and for all $\alpha<\zeta$
  $$V^{P_\alpha}\models (*)\langle T_{j}:j<\alpha\rangle,$$
  then $V^{P_\zeta}\models (*)\langle T_{j}:j<\zeta\rangle$.
\end{lem}
\proof
Let
$$p\forces\mbox{``}\langle F^\prime_\alpha:\alpha<\omega_1\rangle
\mbox{ are pairwise disjoint finite subsets of }\zeta\mbox{''}.$$
We must show there exists $q\leq p$ and $\alpha\not=\beta$ such
that
$$q\forces \forall i\in F^\prime_\alpha\forall j\in F^\prime_\beta\;\;
[T_i]\intersect [T_j]=\emptyset.$$
Let $\langle p_\alpha,F_\alpha:\alpha<\omega_1\rangle\in V$ be such that
$p_\alpha\leq p$ and
$$p_\alpha\forces F_\alpha^\prime=\check{F}_\alpha.$$
For some $A\in[\omega_1]^{\omega_1}$
$\{ F_\alpha:\alpha\in A\}$ forms a $\Delta$-system.
The root of this  $\Delta$-system must be empty, since $P_\zeta$ satisfies
ccc.


\medskip
\noindent Case 1. Cofinality of $\zeta$ is $\omega$.

For some $B\in[A]^{\omega_1}$, and $\gamma<\zeta$
we have that ${F_\alpha}\subseteq \gamma$ and
$p_\alpha\in P_\gamma$ for every $\alpha\in B$.
Let $G$ be a $P_\gamma$-filter such that $C=\{\alpha\in B:p_\alpha\in G\}$
is uncountable.  ( Note that  such a $G$ must exist, else there would
exist a maximal antichain $Q$ such that for every $q\in Q$ there
exists $\alpha<\omega_1$
$q \forces C\subseteq \alpha$.  Since
$Q$ would be countable this would imply that $A$ is countable. )
 Then by applying $(*)$ in $V[G]$
there exists $\alpha\not=\beta\in C$ such that $\forall i \in F_\alpha
\forall j\in F_\beta\;
[T_i]\intersect [T_j]=\emptyset.$  Then $q\in G$ with
$q\leq p_\alpha$ and $q\leq p_\beta$
is as required.


\medskip
\noindent Case 2. Cofinality of $\zeta$ is $\omega_1$.

Apply Lemma \ref{lemma1}.  By cutting down to uncountable subset
of $A$
we can assume that $|F_\beta|=m$ and
$\langle Q_l:l< m\rangle$ are the
same for each $\beta\in A$.

By passing to an uncountable subset of $A$ we can assume that
there exists $\gamma<\zeta$ such that $F_\alpha=G_\alpha\union H_\alpha$
where for each $\alpha\in A\;$ $G_\alpha\subseteq \gamma$ and
$\langle min(H_\alpha):\alpha\in A\rangle$ is unbounded in
$\zeta$, and if $\alpha,\beta \in A$ and $\alpha<\beta$ then
$max(H_\alpha)< min(H_\beta)$.
Apply Lemma \ref{lemma2}
to each $H_\alpha$ obtaining $q_\alpha^i$ for $i=0,1$ with
$$q_\alpha=^{def}q_\alpha^0\res min(H_\alpha)=q_\alpha^1\res min(H_\alpha)$$
and disjoint
$t_0^\alpha,t_1^\alpha$. Again by passing to an uncountable subset
we may assume $t_0,t_1$ are the same for each $\alpha\in A$.
By cutting down $A$ and increasing $\gamma$ we may suppose that
$\{$support$(q_\alpha^0)\union$support$(q_\alpha^1):\alpha\in A\}$
is a $\Delta$-system whose
root is a subset of $\gamma$.  By the same argument as was used in
Case 1, we can find distinct $\alpha,\beta\in A$ such that
$q_\alpha$ and $q_\beta$ are compatible and
$$\forall i\in G_\alpha\forall j\in G_\beta\;
[T_i]\intersect [T_j]=\emptyset.$$

\medskip\noindent
Claim 1. $q_\alpha^0$ and $q_\beta^1$ are compatible.

\proof
By definition $q_\delta=q_\delta^i\res min(H_\delta)$.
Also $q_\alpha$ and $q_\beta$ are compatible elements of
$P_\gamma$.  Since the supports form a $\Delta$-system
with root contained in $\gamma$ they are compatible.
\qed

\medskip\noindent
Claim 2. $q_\alpha^0\union q_\beta^1\forces
\forall i\in F_\alpha\forall j\in F_\beta\;
[T_i]\intersect [T_j]=\emptyset.$

\proof

\medskip\noindent
Case a. $i\in G_\alpha$ and $j\in G_\beta$.

This is true by the
way we picked $\alpha$ and $\beta$.

\medskip\noindent
Case b. $i\in H_\alpha$ and $j\in H_\beta$.

$ q_\alpha^0\forces T_i\intersect 2^n\subset t_0$ and
$ q_\alpha^1\forces T_j\intersect 2^n\subset t_1$, but
$t_0\intersect t_1=\emptyset$.

\medskip\noindent
Case c. $\alpha_l=i\in H_\alpha$ and $\beta_k=j\in G_\beta$.

Let $F_\alpha=\{\alpha_l:l<m\}$ and
$F_\beta=\{\beta_l:l<m\}$, and so $k<l$.
Note that $$q_\alpha\forces T_{\alpha_l}\intersect 2^n=Q_l$$
and $$q_\beta\forces T_{\alpha_k}\intersect 2^n=Q_k.$$
If $s\in Q_l\intersect Q_k$, then
$$q_\alpha\forces T_{\alpha_ls}\subseteq T_{\alpha_k}.$$
But by Case (a) we know $$q_\alpha\union q_\beta\forces
[T_{\alpha_k}]\intersect[T_{\beta_k}]=\emptyset.$$
Consequently $q_\alpha\union q_\beta\forces
[T_{\alpha_l}]\intersect[T_{\beta_k}]=\emptyset$.

\medskip\noindent
Case d. $\alpha_l=i\in G_\alpha$ and $\beta_k=j\in H_\beta$.

Same as Case (c).

\qed




\section{Laver tree forcing}
\label{laver}

Let $\laver$ be Laver tree forcing, that is conditions are
subtrees $p\subseteq\seq$ with the property that there exists
a node $s\in p$ called its root such that:
\begin{enumerate}
  \item for every $t\in p$ either $t\subseteq s$ or $s\subseteq t$,
  \item for every $t\in p$ if $s\subseteq t$, then the set
  $split(p,t)=^{def}\{n\in\omega: t\concat n\in p\}$ is infinite.
\end{enumerate}

\begin{thm} \label{t1}
  Assume MA.  Then $\laver$ does not collapse cardinals.
\end{thm}

Given $A=\langle A_s\in\infsets:s\in\seq\rangle$
and $s\in\seq$ define $p_s(A)=p\in\laver$ to be the unique Laver tree
such that the root of $p$ is $s$ and for every $t\supseteq s$
with  $t\in p$ we have that $split(p,t)=A_t$.

\begin{lem}\label{laverlem}
  Suppose $\forces \tau\in V$ and $B=\langle B_s:s\in\seq\rangle$ where
  each $B_s\in\infsets$.  Then there exists a countable $X$ and
  $A=\langle A_s\in\infsubsets{B_s}:s\in\seq\rangle$ such that
  for every $s\in\seq$
  $$p_s(A)\forces \tau\in X.$$
\end{lem}

\proof
Laver proved that for any $p\in \laver$ there exists $q\leq p$ with
the same root and $X$ countable such that
$$q\forces \tau\in X.$$
Build $p_n\in\laver$ as follows.  At stage $n$, let $s$ be the
$n^{th}$ element of $\seq$.  If $s\in p_m$ for some $m<n$, then
do nothing.  Otherwise, take $p_n\leq p_s(B)$ and $X_n$ countable
such that $s$ is the root of $p_n$ and $p_n\forces\tau\in X_n$.
For every $t\in p_n$ such that $s\subseteq t$, let
$A_t=split(p_n,t)$.  Finally let $X=\Union_{n<\omega}X_n$.
\qed

The next lemma proves the theorem.

\begin{lem}
Suppose MA, $\kappa< \continuum$, and $p\forces \tau:\kappa\rightarrow V$,
then there exist $q\leq p$ and $\langle X_\alpha:\alpha<\kappa\rangle$
such that for every $\alpha\;$
$X_\alpha$ is countable  and
$q\forces\tau(\alpha)\in \check{X}_\alpha$.
\end{lem}
\proof
To simplify notation let $p=\seq$.

Let $Q=\{\langle A_s:s\in\seq\rangle: A_s\in\infsets\}$ and
for $A,B\in Q$ define $A\subseteq^* B$ iff for all
$s\in\seq\;$ $A_s\setminus B_s$ is finite.

Build $A_\alpha\in Q$ for $\alpha<\kappa$ such that
\begin{enumerate}
  \item $\alpha<\beta$ implies $A_\beta\subseteq^* A_\alpha$.
  \item there exists a countable set $X_\alpha$ such that
  for every $s\in\seq$
  $$p_s(A_{\alpha})\forces \tau(\alpha)\in X_{\alpha}.$$
\end{enumerate}
At stage $\alpha$ use the MA to get $A\in Q$ such that
for all $\beta<\alpha\;$
$A\subseteq^* A_{\beta}$. (This is a well known consequence
of Martin's Axiom, apply Solovay's Lemma, Kunen \cite{kunen} p.57,
to each of the
families $\{A_{\beta s}:\beta<\alpha\}$ for $s\in\seq$.)
Then use Lemma \ref{laverlem} to
get $A_{\alpha}\subseteq^* A$ as desired.

Now consider the following poset:
$${\Bbb P}=\{(T,W):T\mbox{ is a finite subtree of }\seq,
W\in [\kappa]^{<\omega}\}$$
Order ${\Bbb P}$ by $(T,W)\leq (T^\prime,W^\prime)$ iff
\begin{enumerate}
\item $W\supseteq W^\prime$,  $T\supseteq T^\prime$, and
\item $\forall n<\omega\; \forall s\in (T\setminus T^\prime)\intersect
\omega^{n+1}
\;\forall \alpha\in W^\prime
\;\;s(n)\in A_{\alpha,s}$.
\end{enumerate}
Since any two conditions with the same $T$ part are
compatible,  ${\Bbb P}$ is $\sigma$-centered, so we can apply MA to it.
For $G$ a ${\Bbb P}$-filter let
$$q=\union \{ T:\exists W\; (T,W)\in G\}.$$
If $G$ meets the dense subsets of ${\Bbb P}$ of the form
$$D_{s,n}=\{p\in{\Bbb P}: p\forces s\notin q \mbox{ or } \exists m>n\;
s\concat m\in T_p\}$$
then we will have that $q\in\laver$ with the empty sequence as its root.
For any $\alpha<\kappa$ let
$$D_{\alpha}=\{p\in{\Bbb P}: \alpha\in W_p\}.$$
Each $D_{\alpha}$ is dense.

Hence by MA we can get
$q\in\laver$ with the empty node as root such that
for every $\alpha<\kappa$ there is finite subtree
$T\subseteq q$ such that for every $t\in q\setminus T$ we
have $t\in p_{t\res (|t|-1)}(A_\alpha)$.  This implies that for
every $r\leq q$ there exists $s$ such that $p_s(A_{\alpha})$ is compatible
with $r$.
It follows that
$$q\forces \forall \alpha<\kappa\;\tau(\alpha)\in X_\alpha.$$

\qed


\section{Superperfect trees}\label{super}

Superperfect tree forcing $\super$ is defined as follows.
For $p$ a subtree of $\seq$ define the splitting nodes of $p$:
$$\splitnode(p)=\{s\in p: \exists^\infty n\in\omega\; s\concat n\in p\}.$$
Define $p\in\super$ iff
\begin{enumerate}
  \item $p$ is a nonempty subtree of $\seq$,
  \item $\splitnodes(p)$ is dense in $p$, ie
  $\forall s\in p \;\exists t\in\splitnodes(p)$ with $t\supseteq s$, and
  \item if any node in $p$ splits it is a splitting node, ie
  if there exists more than one $n\in\omega$ such that $s\concat n\in p$, then there
  are infinitely many $n\in\omega$ such that $s\concat n\in p$.
\end{enumerate}


\begin{thm} \label{t2}
  Assume MA.  Then $\super$ does not collapse cardinals.\footnote{
  Added post publication: Goldstern, Roslanowski, and Spinas found
  a mistake in this proof. The relation $\leq^*$ is not transitive.
  A correct proof is given in the paper:
   Goldstern, Martin; Johnson, Mark J.; Spinas, Otmar Towers on trees.
   Proc. Amer. Math. Soc.  122 (1994), no. 2, 557--564.}
\end{thm}
\proof
\noindent Call a sequence $\langle P_s:s\in\seq\rangle$ good iff
\begin{enumerate}
  \item each $P_s\subseteq \seq$ is infinite,
  \item $t\in P_s$ implies $s\propsub t$, and
  \item for $s\in \omega^n$ if $t,t^\prime\in P_s$ and $t\not=t^\prime$, then
  $t(n)\not=t^\prime(n)$.
\end{enumerate}
Given any good sequence $\langle P_s:s\in\seq\rangle$ we determine
$\langle p_s\in\super:s\in\seq\rangle$ as follows.
For each
$s$ let $S$ be is smallest subset of $\seq$ such that
$P_s\subseteq S$ and if $t\in S$ then $P_t\subseteq S$.
Then $p_s$ is the unique condition in
$\super$ such that $S=\splitnode(p_s)$.
In other words, $P_s$ says that $s$ is
a splitting node and the splitting nodes immediately below
$s$ are $P_s$.  Define $\langle P_s:s\in\seq\rangle\leq
\langle Q_s:s\in\seq\rangle$ iff
$p_s\subseteq q_s$ for each $s\in\seq$.  An equivalent definition
would be for each $s\in\seq$ and $t\in P_s$
there exists $k$ and $s_0,s_1,\ldots,s_k$ where
$s=s_0\subseteq s_1 \subseteq\ldots \subseteq s_k=t$
$s_{i+1}\in Q_{s_i}$ for $i=0,1,\ldots,k-1$.

\begin{lem} \label{suplem0}
Given $\tau$ such that $\forces_\super$``$\tau\in V$'', and
good $\langle P_s:s\in\seq\rangle$ there exists a good
$\langle Q_s:s\in\seq\rangle\leq \langle P_s:s\in\seq\rangle$ and
a countable set $\Sigma$ such that for every $s\in\seq$
$q_s\forces$``$\tau\in\Sigma$''.
\end{lem}
\proof
For any $p\in\super$ with smallest splitting node $s$ (ie root) there
exists $q\subseteq p$ such that $s\in\splitnodes(q)$ and
a countable $\Sigma$ such that $q\forces$``$\tau\in\Sigma$''.
Now just apply this fact repeatedly down the $s\in\seq$.
\qed

 Define
$\langle P_s:s\in\seq\rangle\leq^* \langle Q_s:s\in\seq\rangle$
iff there exists
$\langle P_s^\prime:s\in\seq\rangle\leq
\langle Q_s:s\in\seq\rangle$ such that for every $s\in\seq$
$P_s=^* P^\prime_s$ (equal mod finite).  Similarly
for $p,q\in \super$ define $p\leq^* q$ iff
there exists
$f:\splitnodes(p)\rightarrow \omega$ such
that $p_f\subseteq q$, where
$$p_f=p\setminus\{r\in\seq:\exists s\in \splitnodes(p)\exists n< f(s)\;
s\concat n\subseteq r\}.$$
We think of $p_f$ as being obtained from $p$ by pruning finitely many nodes
from beneath each splitting node of $p$.
Note that  $p_f\in\super$.

\begin{lem}\label{suplempt5}
The following are equivalent:
\begin{enumerate}
  \item $\langle P_s:s\in\seq\rangle\leq^* \langle Q_s:s\in\seq\rangle$
  \item for every $s\in \seq$, $p_s\leq^*q_s$.
\end{enumerate}
\end{lem}
\proof
Left to reader.
\qed

\begin{lem} \label{suplem1}
 (MA) Suppose $\gamma<\continuum$ and $\langle P_s^{\alpha}:s\in\seq\rangle$
 for $\alpha<\gamma$ are good and
 have the property that $\alpha>\beta$ implies
$\langle P_s^{\alpha}:s\in\seq\rangle\leq^*
\langle P_s^{\beta}:s\in\seq\rangle$.
  Then there exists a good $\langle P_s:s\in\seq\rangle$ such that
for every $\alpha<\gamma\;\;$
 $\langle P_s:s\in\seq\rangle\leq^*
\langle P_s^{\alpha}:s\in\seq\rangle$.
\end{lem}
\proof
Let $Q$ be the following poset, $(A_s:s\in F,H)\in Q$ iff
\begin{enumerate}
  \item $H\subseteq\gamma$, $F\subseteq \seq$, $A_s\subseteq\seq$ for $s\in F$
     are all finite,
  \item $t\in A_s$ implies $s\propsub t$, and
  \item $t,t^\prime\in A_s$ and $t\not=t^\prime$ implies
  $t(n)\not=t^\prime(n)$,   where $s\in \omega^n$.
\end{enumerate}
We define $(\hat{A}_s:s\in \hat{F},\hat{H})\leq (A_s:s\in F,H)$ iff
\begin{enumerate}
\item $\hat{F}\supseteq F$, $\hat{H}\supseteq H$,
 $\hat{A}_s\supseteq A_s$ for $s\in F$, and
\item for each $s\in F$ if $t\in \hat{A}_s\setminus A_s$ and
$\alpha\in H$, then $t\in\splitnodes(p_s^\alpha)$.
\end{enumerate}

Note that $Q$ is ccc, in fact $\sigma$-centered, since
$({A}_s:s\in {F},H\union \hat{H})$ extends both
$(A_s:s\in F,H)$ and $(A_s:s\in F,\hat{H})$. For any
$\alpha<\gamma$
$$\{(A_s:s\in F,H)\in Q: \alpha\in H\}$$
is dense in $Q$, since
$(A_s:s\in F,H\union \{\alpha\})\leq (A_s:s\in F,H)$.
In order to quarantee that $A_t$ grows up into an infinite set,
we need only check that the following sets are dense.
Fix $t\in\omega^n$ and $m<\omega$ and define
$$E_{t,m}=\{(A_s:s\in F,H)\in Q: t\in F, \exists r\in A_t\;\; r(n)>m\}$$
To check this let $(A_s:s\in F,H)\in Q$ and put $t$ into $F$ by
letting $A_t=\emptyset$ if neccessary.  Let $\alpha=max\{H\}$
and let $p=p_s^{\alpha}$.  Then
there exists $f:\splitnodes(p)\rightarrow\omega$ such that
for every $\beta\in H$ we have $p_f\subseteq p_s^{\beta}$.  Consequently
any  $r\in\splitnode(p_f)$ with $r\supset s$ and
$r(n)>m$ can be added to $A_t$.
Finally if $G$ is sufficiently $Q$-generic, then
$\langle P_s:s\in\seq\rangle$ defined by
$$P_t=\Union\{A_t:\exists  (A_s:s\in F,H)\in G, t\in F\}$$
is as required.
\qed

\begin{lem} \label{suplem2}
 (MA) Suppose $\kappa<\continuum$ is an uncountable regular cardinal and
 $p_\alpha\in\super$
 for $\alpha<\kappa$ and $r\in\super$
 have the property that for every $\alpha<\kappa\;$  $r\leq^* p_\alpha$.
 Then there exists $q\subseteq r$ and $\Gamma\in [\kappa]^\kappa$
 such that
 $q\subseteq p_\alpha$ for every $\alpha\in\Gamma$.
\end{lem}

\proof
For each $\alpha<\kappa$ let
$f_\alpha:\splitnode(r)\rightarrow\omega$ be such that
$r_{f_\alpha}\subseteq p_\alpha$ where as before,
$$r_f=r\setminus\{t\in\seq:\exists s\in\splitnode(f) \exists m < f(s);
s\concat m\subset t\}$$
that is $f_\alpha(s)$ tells what finite set of nodes below the
splitting node $s$ of $r$
we should prune from $r$ so as to end up with subtree of $p_\alpha$.
It is well known that MA gives us $f:\splitnode(r)\rightarrow\omega$ that
for all $\alpha<\kappa$ for all but finitely many
$s\in\splitnode(r)$ $\;\; f_\alpha(s)<f(s)$.
By changing $f$ on a finite set we can find $\Gamma\in [\kappa]^\kappa$
such that for all $\alpha\in\Gamma$ and for all $s\in\splitnodes(r)$
$f_\alpha(s)<f(s)$.  It follows from this that $q=r_f\subseteq p_\alpha$
for all $\alpha\in\Gamma$
\qed

Proof of Theorem \ref{t2}:  Suppose that forcing with $\super$ did
collapse cardinals, then there would exist regular cardinals
$\kappa<\lambda\leq\continuum$, $p\in\super$ and a name
$\tau$ such that
$$p\forces \tau:\kappa\rightarrow\lambda \mbox{ is increasing and cofinal}.$$
To simplyfy notation assume $p=\seq$.
Using Lemma \ref{suplem0} and Lemma \ref{suplem1} build a sequence of good
$\langle P_s^{\alpha}:s\in\seq\rangle$ for $\alpha<\kappa$ such that
\begin{enumerate}
  \item $\alpha>\beta$ implies
  $\langle P_s^{\alpha}:s\in\seq\rangle\leq^*
  \langle P_s^{\beta}:s\in\seq\rangle$  and
  \item for any $\alpha$ there exists a countable $\Sigma$ such
  that for every $s\in\seq$
  $p_s^\alpha\forces$``$\tau(\alpha)\in \Sigma$''.
\end{enumerate}

Use Lemma \ref{suplem1} one more time to obtain
$\langle R_s:s\in\seq\rangle$ such that
$\langle R_s:s\in\seq\rangle\leq^*
 \langle P_s^{\alpha}:s\in\seq\rangle$
for each $\alpha<\kappa$. Fix $s$ (say the empty node) and
define $r=r_s$ and $p_\alpha=p^\alpha_s$.  Since these
satisfy the hypothesis of Lemma \ref{suplem2} we
can obtain $q\subseteq r$ and $\Gamma\in[\kappa]^\kappa$
such that $q\subseteq p_\alpha$ for every $\alpha\in\Gamma$.  But now
there exists countable sets
$\{\Sigma_\alpha:\alpha\in\Gamma\}$ such that
$$q\forces\forall\alpha\in\Gamma\;\tau(\alpha)\in\Sigma_\alpha$$
which means that the range of $\tau$ cannot be cofinal in $\lambda$.
\qed


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 \begin{center}
               Addresses
 \end{center}


Haim Judah,
Department of Mathematics,  Bar-Ilan University, 52900 Ramat-Gan,
Israel, judah@bimacs.biu.AC.IL

Arnold W. Miller, Department of Mathematics, University of Wisconsin,
480 Lincoln Drive,
Madison, WI, 53706, USA, miller@math.wisc.edu

Saharon Shelah, Institute of Mathematics, Hebrew University, Jerusalem,
Israel

\section*{Appendix: Stationary subsets of $\omega_2$}

In this appendix  we solve a
problem posed by  M. Foreman:
Does forcing with $\sacks$ add a stationary subset of $\omega_2$
which does not contain a stationary subset of the ground model?

Note that if $\omega_2$ is collapsed to $\omega_1$, then
there exists a club subset of $\omega_2$ which does not
contain an unbounded subset of the ground model.
If $\continuum=\omega_1$, then every new stationary subset $S$ of $\omega_2$
contains an old stationary set.  This is because $\sacks$ has cardinality
$\omega_1$ and so some $p\in \sacks$ must force stationarily many
$\alpha\in S$.
Amoeba Sacks forcing is defined by
$\poset=\{(p,n):n<\omega, p\in\sacks\}$ where $(p,n)\leq(q,m)$ iff
$p\subseteq q$, $n\geq m$, and $p\intersect 2^m=q\intersect 2^m$.
It is proper.
If $\continuum>\omega_2$ and MA(Amoeba Sacks), then every stationary
subset of $\omega_2$ contains a ground model
stationary set.    To see this note that if
$$p\forces_{\sacks}\mbox{``}S\subseteq\omega_2 \mbox{ is stationary''}$$
then by using
Amoeba Sacks forcing we can find $q\leq p$ such that for every
$\alpha<\omega_2$ there exists $n<\omega$ such that for every
$s\in 2^n\intersect q$ either $q_s\forces \alpha\in S$ or
$q_s\forces \alpha\notin S$.  Hence there must be some $s\in q$ such that
$$\{\alpha<\omega_2: q_s\forces \alpha\in S\}$$
 is stationary.

\begin{theorem} \label{for}
  Suppose $\continuum=\omega_2$ then forcing with $\sacks$ adds
  a stationary subset of $\omega_2$ that does not contain a stationary
  set from the ground model.
\end{theorem}

\proof
Let $\sacks=\{q_\alpha:\alpha<\omega_2\}$.
Use Lemma \ref{forlem} to obtain for each $\beta<\omega_2$ an
antichain $\{p_{\beta\alpha}\leq q_{\alpha}:\alpha<\beta\}$. Let
$p_{\beta\alpha}^i\leq p_{\beta\alpha}$ for $i=0,1$ be two incompatible
extensions.

Now let $S=\{\langle p_{\beta\alpha}^1,\check{\beta}\rangle:
\alpha<\beta<\omega_2\}$.
Then the following two facts hold:
\begin{enumerate}
  \item $\forces_{\sacks} S\subseteq \check{\omega_2}$
  \item for every $p\in\sacks\;$ $\{\alpha<\omega_2: p\forces \alpha\in S$
  or $p\forces \alpha\notin S\}$ is bounded in $\omega_2$.
\end{enumerate}
Thus neither $S$ nor its complement can contain a stationary set (or
even an unbounded subset of $\omega_2$) which is in the ground model.
Since one of the two must be stationary the theorem is proved.
\qed

\begin{lemma}\label{forlem}
Suppose $\continuum=\omega_2$ and $Q=\{q_\alpha:\alpha<\omega_1\}
\subseteq \sacks$,
then there exists an antichain $\{p_\alpha:\alpha<\omega_1\}$ such
that for each $\alpha<\omega_1$  $p_\alpha\leq q_\alpha$.
\end{lemma}
\proof
Inductively construct $p_\alpha$ so that for every $\beta>\alpha$
$[p_\alpha]\intersect[q_\beta]$ is nowhere dense in
$[q_\beta]$.  This is easy to do.  At stage $\alpha$, find
$r\in \sacks$  such that
$$[r]\subseteq \left([q_\alpha]\setminus \Union_{\beta<\alpha}[p_\beta]
\right).$$
Then split $[r]$
into $\omega_2$ perfect disjoint sets.   Note that for any $\beta>\alpha$ at
most countably many of these perfect sets can fail to be nowhere
dense in $[q_\beta]$.  Let $p_\alpha$ be any of the remaining ones.
\qed


By a theorem of Balcar and Vojtas \cite{Balcar},
Lemma \ref{forlem} is true in much more generality.  All that is really
needed is: every $p\in\poset$ has $\omega_2$ incompatible extensions.
Consequently the result holds for all of the standard tree kinds of
forcing, eg Laver, Mathias, Silver, etc.


\end{document}