% LaTeX2e
% A. Miller   Sept 2007
\documentclass[12pt]{article}
\usepackage{amssymb}

\def\om{\omega}
\def\de{\delta}
\def\la{\langle}
\def\ra{\rangle}
\def\proof{\par\noindent Proof\par\noindent}
\def\qed{\par\noindent QED\par\bigskip}
\def\res{\upharpoonright}
\def\rmiff{\mbox{ iff }}
\def\su{\subseteq}
\def\fn(#1){{\rm Fn(#1\times\om,2)}}
\def\cube{(2^\om)^\om}

\newtheorem{theorem}{Theorem}
\newtheorem{lemma}[theorem]{Lemma}

\begin{document}

\begin{flushright}
A.Miller \\
Sept 27, 2007
\end{flushright}


\begin{theorem}
Assume CH.  Then there exists a Luzin set $X\su \cube$, a Sierpinski 
set $Y\su 2^\om$, and a 
Borel function $f:\cube\times 2^\om \to 2^\om$ such that
$f(X\times Y)=2^\om$.
\end{theorem}
\proof

For $x\in 2^\om$ define
$$G_x=\{u\in 2^\om : \exists^\infty n \;\;\;\; u\res [n,2n)=x\res[n,2n)\}.$$
Note that $G_x$ is a measure zero comeagre $G_\de$ set
for any $x\in 2^\om$.  Also if $x=^*y$ (equal mod finite), then
$G_x=G_y$.  The function $f$ is defined by:
$$f(\la x_n:n<\om\ra,y)=z \rmiff (\forall n\;\; z(n)=1 \rmiff y\in G_{x_n}).$$


As in Kunen's set theory book, define
$\fn(\om)$ to be the partial order of finite partial
functions from $\om\times\om$ into $2$.

\begin{lemma}
Suppose $M$ is a countable transitive model of a sufficiently large 
finite fragement of ZFC and $z\in 2^\om$ is arbitrary.  Then there exists
$x=\la x_n:n<\om\ra$ which is $\fn(\om)$-generic over $M$ and
$y$ which is random over $M$ such that $f(x,y)=z$.
\end{lemma}
\proof
Let $u=\la u_n:n<\om\ra$ be $\fn(\om)$-generic over $M$. Let
$H$ be measure amoeba generic over $M[u]$.  Since $H$ makes the union
of all measure zero sets coded in $M[u]$ measure zero, there exist
in $M[u,H]$ a perfect tree
$T\su 2^{<\om}$ such that the set
of infinite branchs of $T$, $[T]$, is disjoint from 
every measure zero set coded
in $M[u]$. Note that:
\begin{itemize}
\item $[T]\cap G_{u_n}=\emptyset$ for every $n$, and
\item every $y\in [T]$ is random over $M$.
\end{itemize}

Let $v=\la v_n:n<\om\ra$ be $\fn(\om)$-generic 
over $M[u,H]$.  An easy density argument shows that 
for every $n$ the set $G_{v_n}$ is dense in $[T]$ and hence
comeager. 

Define:
$$w_n=\left\{
\begin{array}{ll}
u_n & \mbox{ if } z(n)=0\\
v_n & \mbox{ if } z(n)=1.\\
\end{array}\right.$$
It may be that $w$ is not $\fn(\om)$-generic over $M$.  However,
it is easy to see by the usual facts of iterated forcing that for
every $N<\om$ 
$\la w_n:n<N\ra$ is $\fn(N)$-generic over $M$.  According to 
a lemma of Harvey Friedman\footnote{
Friedman, Harvey;
Large models of countable height.
Trans. Amer. Math. Soc. 201 (1975), 227--239.  Lemma 3.
}, there exist
$x=\la x_n:n<\om\ra$ which is $\fn(\om)$-generic over $M$ and
$x_n=^*w_n$ for every $n$.

We can choose
$$y\in [T]\cap \bigcap\{G_{x_n}:z(n)=1\}$$ 
because these sets are all comeager in $[T]$.
And hence, $f(x,y)=z$.
\qed

From the Lemma and CH it is easy to construct the sets $X$ and $Y$ as
required.

\qed

\end{document}