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% Ultrafilters with property (s)
% A. Miller Oct 2002 last revised Jan 15, 2004
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\begin{document}
\begin{center}
{\large Ultrafilters with property (s)}
\end{center}
\begin{flushright}
Arnold W. Miller\footnote{
Thanks to the Fields Institute, Toronto for their support
during the time these results were proved and to Juris Steprans
for helpful conversations and thanks to Boise State University
for support during the time this paper
was written.
\par Mathematics Subject Classification 2000: 03E35; 03E17; 03E50
}
\end{flushright}
\begin{center}
Abstract
\end{center}
\begin{quote}
A set $X\subseteq 2^\omega$ has property (s) (Marczewski (Szpilrajn)) iff
for every perfect set $P\subseteq 2^\omega$ there exists a perfect set
$Q\subseteq P$ such that $Q\subseteq X$ or $Q\cap X=\emptyset$.
Suppose $\uu$ is a nonprincipal ultrafilter on $\omega$.
It is not difficult to see that if $\uu$ is preserved by Sacks forcing,
i.e., it generates an ultrafilter in the generic extension after forcing
with the partial order of perfect sets, then $\uu$ has property (s) in
the ground model. It is known that selective ultrafilters or even
P-points are preserved by Sacks forcing. On the other hand (answering
a question raised by Hrusak) we show that assuming CH (or more generally
MA) there exists an ultrafilter $\uu$ with property (s) such that
$\uu$ does not generate an ultrafilter in any extension which adds a new
subset of $\omega$.
\end{quote}
It is a well known classical result due to Sierpinski (see \cite{bj})
that a nonprincipal ultrafilter $\uu$ on $\omega$ when considered as a
subset of $P(\omega)=2^\omega$ cannot have the property of Baire or
be Lebesgue measurable. Here we identify
$2^\omega$ and $P(\omega)$ by identifying a subset of $\omega$ with
its characteristic function.
Another very weak regularity property
is property (s) of Marczewski (see Miller \cite{survey}). A set
of reals $X\subseteq 2^\omega$ has property (s) iff for every perfect
set $P$ there exists a perfect subset $Q\subseteq P$ such that
either $Q\subseteq X$ or $Q\cap X=\emptyset$. Here by perfect we mean
homeomorphic to $2^\omega$.
It is natural to ask:
\bigskip\noindent
{\bf Question.} (Steprans) Can a nonprincipal ultrafilter
$\uu$ have property (s)?
\bigskip
If $\uu$ is an ultrafilter in a model of set theory $V$ and
$W\supseteq V$ is another model of set theory then we say
$\uu$ generates an ultrafilter in $W$ if for every
$z\in P(\omega)\cap W$ there exists $x\in\uu$ with
$x\subseteq z$ or $x\cap z=\emptyset$. This means that the
filter generated by $\uu$ (i.e. closing under supersets)
is an ultrafilter in $W$.
We begin with the following result:
\begin{theorem}\label{one}
For $\uu$ a nonprincipal ultrafilter on $\omega$ in $V$ the following are
equivalent:
\begin{enumerate}
\item For some Sack's generic real $x$ over $V$
$$V[x]\models \uu \mbox{ generates an ultrafilter. }$$
\item In $V$, for every perfect set $P\subseteq P(\omega)$ there
exists a perfect set
$Q\subseteq P$
and a $z\in \uu$ such that either $\forall x\in Q\;\; z\subseteq x$
or $\forall x\in Q\;\; z \cap x=\emptyset$.
\item For some extension $W\supseteq V$ with a new subset of $\omega$
$$W\models \uu \mbox{ generates an ultrafilter. }$$
\end{enumerate}
\end{theorem}
\proof
To see that $(3)\to (2)$, let $P$ be any perfect set coded in $V$. Since $W$
contains a new subset of $\omega$ there exists $x\in (P\cap W)\setminus V$.
Since $\uu$ generates an ultrafilter in $W$ there exists $z\in \uu$ so that
either $z\subseteq x$ or $z\cap x=\emptyset$. Suppose the first happens. In
$V$ consider the set
$$Q=\{y\in P: z\subseteq y\}$$
Note that the new real $x$ is in the closed set $Q$. It follows that $Q$ must
be an uncountable closed set and so it contains a perfect subset. The other
case is exactly the same.
One way to see that $Q$ must be uncountable is to note that if (in $V$)
$Q=\{x_n:n<\omega\}$, then the $\pioneone$ sentence
$$\forall x\in 2^\omega (x\in Q\rmiff \exists n<\omega \;x=x_n)$$
would be true in $V$ and since $\pioneone$
sentences are absolute (Mostowski absoluteness, see \cite{kechris}) true in
$W$. Another way to prove it is to do the standard derivative Cantor argument
to the closed set $Q$ removing isolated points and
iterating thru the transfinite and noting that each real removed is in $V$,
while the new
real is never removed, and hence the kernel of $Q$ is perfect. See Solovay
\cite{solovay}, for a similar proof of Mansfield's theorem that a (lightface)
$\pionetwo$ set with a nonconstructible element contains a perfect set.
Now we see that $(2)\to (1)$.
A basic property of Sack's forcing is that for every $y\in 2^\omega \cap M[x]$
in a Sacks extension is either in $M$ or is itself Sacks generic over $M$ (see
Sacks \cite{sacks}). Hence we need only show that if $y\subseteq \omega$ is
Sacks generic over $M$, then there exists $z\in \uu$ with $z\subseteq y$
or $z\cap y=\emptyset$. Recall also that the Sacks
real $y$ satisfies that the generic filter $G$ is
exactly the set of all perfect sets $Q$ coded in $V$ with $y\in Q$.
Condition (2) says that the set of such $Q$ are dense and hence there exists
$Q$ in the generic filter determined by $y$ and $z\in \uu$ such that either
$z\subseteq u$ for every $u\in Q$ or either $z\cap u=\emptyset$ for every
$u\in Q$. But this means that either $z\subseteq y$ or $z\cap y=\emptyset$.
$(1)\to (3)$ is obvious.
\qed
\bigskip
Remark.
The above proof also shows that if an ultrafilter is preserved
in one Sacks extension, then it is preserved in all Sacks extensions.
\bigskip
Remark. In Baumgartner and Laver \cite{bl} it is shown that selective
ultrafilters are preserved by Sacks forcing. In Miller \cite{super}
it is shown that $P$-points are preserved by superperfect set forcing
(and hence by Sacks forcing also).
\bigskip
We say that an ultrafilter $\uu$ is preserved by Sacks forcing iff
for some (equivalently all) Sacks generic real $x$ that
$\uu$ generates an ultrafilter in $V[x]$. Recall that $\uu\times \vv$
is the ultrafilter on $\omega\times\omega$ defined by
$$A\in \uu\times \vv \rmiff \{n:\{m:(n,m)\in A\}\in\uu\}\in\vv$$
If $\uu$ and $\vv$ are nonprinciple ultrafilters, then $\uu\times\vv$
is not a P-point. Also recall that $\uu\leq_{RK}\vv$ (Rudin-Keisler)
iff there exists $f\in\omega^\omega$ such that for every $X\subseteq \omega$
$$X\in\uu \rmiff f^{-1}(X)\in \vv$$
\bigskip
\begin{prop}\label{two}
If $\uu$ and $\vv$ are preserved by Sacks forcing, then so is $\uu\times\vv$.
If $\uu\leq_{RK}\vv$ and $\vv$ is preserved by Sacks forcing, then so
is $\uu$.
\end{prop}
\proof
Suppose $A\subseteq \omega\times\omega$ and $A\in V[x]$. For
each $n<\omega$ let $A_n=\{m:(n,m)\in A\}$. Since $\uu$ is preserved
there exists $B_n\in\uu$ with $B_n\subseteq A_n$ or $B_n\cap A_n=\emptyset$.
By the preservation of $\vv$ there exists $C\in\vv$ such that either
$B_n\subseteq A_n$ for all $n\in C$ or
$B_n\cap A_n=\emptyset$ for all $n\in C$.
By the Sacks property there exists
$(\bb_n\in [\uu]^{2^n}:n<\omega)\in V$ such that $B_n\in\bb_n$
for every $n$. Let $B_n^0=\cap \bb_n$. Then
$$\cup_{n\in C}\{n\}\times B_n^0\subseteq A \rmor
\cup_{n\in C}\{n\}\times B_n^0\cap A=\emptyset
$$
Suppose $\uu\leq_{RK}\vv$ via $f$. If $A\subseteq \omega$, then
since $\vv$ is preserved, there exists $B\in\vv$ such that
either $B\subseteq f^{-1}(A)$ or $B\subseteq f^{-1}(\overline{A})$
where $\overline{A}$ is the compliment of $A$.
But then $f(B)\subseteq A$ or $f(B)\subseteq \overline{A}$
and since $f(B)\in\uu$ we are done.
\qed
\bigskip Remark. The Rudin-Keisler result is generally true, but the
product result depends on the bounding property. For example, if $\uu$ is a
P-point, then $\uu$ is preserved in the superperfect extension, but
$\uu\times\uu$ is not.
\bigskip
It is clear that property (2) of Theorem \ref{one}
implies that any ultrafilter which is
preserved by Sacks forcing has property (s). But what about the converse?
The main result of this paper is that the reverse implication is false.
This answers a question raised by Hrusak.
\begin{theorem}
Suppose the CH is true or even just that the real line cannot be
covered by fewer than continuum many meager sets.
Then there exists an ultrafilter $\uu$ on
$\omega$ which has property (s) but is not preserved by Sacks forcing.
\end{theorem}
\proof
We give the proof in the case of the continuum hypothesis and indicate
how to do it under the more general hypothesis.
Let $\ii\subseteq\infsub$ be an independent perfect family.
Independent means that for every $m,n$ and distinct
$x_1,\ldots,x_m,y_1,\ldots,y_n\in\ii$ the set
$$x_1\cap\ldots\cap x_m\cap \overline{y}_1
\cap \ldots\cap \overline{y}_n\mbox{ is infinite}$$
where $\overline{y}$ means the complement of $y$ in $\omega$.
The usual construction (probably due to Hausdorff)
is the following.
Let
$Q=\{(F,s)\;:\; F\subseteq P(s),\; s\in [\omega]^{<\omega}\}$.
Given any $A\subseteq \omega$ define
$x_A=\{(F,s)\in Q\;:\; A\cap x\in F\}$,
then $\{x_A\;:\;A\subseteq \omega\}$ is a perfect independent family
in the Cantor space $P(Q)$.
We claim that the following family
$$\ii\cup\{\overline{z}:\;\exists^\infty \; x\in \ii\;\; z\subseteq^*x\}$$
has the finite intersection property. ($\exists^\infty$ means
there exists infinitely many). To see this suppose that we are given
$x_1,\ldots,x_m\in \ii$ and $z_1,\ldots, z_n$ such that
$\exists^\infty \; x\in \ii\;\; z_i\subseteq^*x$ for each $i$. Then
we can choose $y_i\in \ii$ distinct from each other and the $x's$ so
that each $z_i\subseteq^* y_i$. But since $\overline{y}_i\subseteq
\overline{z}_i$ we have that
$$x_1\cap\ldots\cap x_m\cap \overline{y}_1 \cap \ldots\cap \overline{y}_n
\subseteq^* x_1\cap\ldots\cap x_m\cap \overline{z}_1 \cap \ldots\cap
\overline{z}_n $$
By independence the set on the left is infinite and hence so
is the set on the right.
Thus this family has the finite intersection property.
Now let
$\ff_0$ be the filter generated by
$\ii\cup\{\overline{z}:\exists^\infty \; x\in \ii\;\; z\subseteq^*x\}$.
Note that if $\uu\supseteq\ff_0$ is any ultrafilter then it cannot be
preserved by Sacks forcing. This is because $\ii$ is a perfect subset
of $\uu$, however there is no $z\in \uu$ with $z\subseteq x$ for
all $x\in \ii$ or even infinitely many $x\in\ii$
or else $\overline{z}\in\ff_0\subseteq\uu$, hence Theorem \ref{one} (2)
fails.
Note that since $\ii$ was perfect the filter $\ff_0$ is a $\sig$
subset of $P(\omega)$.
\begin{lemma}
Suppose that $P\subseteq\infsub$ is a perfect set and
$\ff$ is a $\sig$ filter extending the cofinite filter
on $\omega$. Then there exists a perfect $Q\subseteq P$ such that
either
\begin{enumerate}
\item $\ff\cup Q$ has the finite intersection property or
\item there exists $z\subseteq \omega$ so that $\ff\cup\{z\}$ has
the finite intersection property and for every $x\in Q$ we have that
there exists $y\in\ff$ with $x\cap y\cap z=\emptyset$.
\end{enumerate}
\end{lemma}
\proof
The strategy is try to do a fusion argument to get case (1). If it
ever fails, then stop and get case (2).
\bigskip {\bf Claim}. Suppose $(Q_i:im\}$$
$$D_m^*=\{p\in\pp\;:\; \forall i