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% A product of $\ga$-sets which is not Menger.
% A. Miller
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\def\be{\beta}
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\def\lm{\lambda}
\def\om{\omega}
\def\poset{{\mathbb P}}
\def\proof{\par\noindent Proof\par\noindent}
\def\qed{\par\noindent QED\par\bigskip}
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\def\star{0}
\def\su{{\;\subseteq\;}}
\def\uu{{\mathcal U}}

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

\begin{document}

\begin{center}
A product of $\ga$-sets which is not Menger.
\end{center}

\begin{flushright}
A. Miller \\
Dec 2009
\end{flushright}


\bigskip


{\bf Theorem}. Assume CH. Then there exists 
$\ga$-sets $A_0,A_1\su 2^\om$ such that
$A_0\times A_1$ is not Menger.

\bigskip\bigskip

We use perfect sets determined by Silver forcing
(see Grigorieff \cite{grig}) and construct an
Aronszajn-tree of such perfect sets in the style of 
Todorcevic (see Galvin-Miller \cite{gal}).

Define $p\in\poset$ iff $p:D\to 2$ where
$D\su\om$ is co-infinite, i.e., $\comp(D)=\om\sm D$
is infinite.  For $p\in\poset$ define
$$[p]=\{x\in 2^\om\;:\; p\su x\}.$$

Define $p\leq q$ iff\footnote{We follow the convention that $p\leq q$ means that $p$ is
stronger than $q$ or equivalently that $p$ extends $q$.} $p\supseteq q$ 
or equivalently $[p]\su [q]$.

For $n<\om$ define $p\leq_n q$ iff $p\leq q$ and the first
$n$ elements of $\comp(D_p)$ are the same as the first
$n$ elements of $\comp(D_q)$.
The fusion property is important:
\begin{quote}
{\bf Fusion Lemma}. Suppose $(p_n\in\poset:n<\om)$ has the
property that $p_{n+1}\leq_n p_n$ for every $n<\om$.  Then
the fusion $q=\bigcup_{n<\om}p_n$ is in $\poset$ and
$q\leq_n p_n$ for every $n$.
\end{quote}

Define
$$Q^0(p)=\{x\in [p]: \forall^\infty n\in\comp(D_p)\;\; x(n)=0\} $$
and
$$ Q^1(p)=\{x\in [p]: \forall^\infty n\in\comp(D_p)\;\; x(n)=1\}.$$
Each of these are countable dense
subsets of $[p]$.

Define $q\leq_n^\star p$ iff $q\leq_n p$ and $q$ restricted to
$D_q\sm D_p$ is identically zero.

\begin{lemma}\label{lemone}
Given $\uu$ an $\om$-cover of $Q^0(p)$ and $n<\om$
there exist $U\in\uu$ and $q\leq_n^\star p$ such that $[q]\su U$.
\end{lemma}
\proof
To see why this is true let $F$ be the first $n$ elements
of $\comp(D_p)$.  For each $s\in 2^F$ let $x_s\in Q^0(p)$ such
that $x_s\res F=s$ and $x_s(n)=0$ for every $n\in \comp(D_p)\sm F$.
Take $U\in\uu$ with $\{x_s:s\in 2^F\}\su U$.  Since
$U$ is open there exists $N<\om$ with 
$[x_s\res N]\su U$ for all $s\in 2^F$.  Define $q\leq_n^\star p$ 
by
$$q=p\cup\{\langle k,0\rangle: k<N \rmand k\in (\comp(D_p)\sm F)\}$$

\qed



\begin{lemma}
Given $p_n\in\poset$ and $k_n<\om$ for $n<N$ and an $\om$-cover
$\uu$ of $\bigcup_{n<N}Q^0(p_n)$ there exists $U\in\uu$ and
$(q_n\leq_{k_n}^\star p_n : n<N)$ such that
$$\bigcup_{n<N} [q_n]\su U.$$
\end{lemma}
\proof
Let $F_n$ be the first $k_n$ elements of $\comp(D_{p_n})$.
For $s\in 2^{F_n}$ define $x_s^n\in Q^0(p_n)$ as in the 
proof of Lemma \ref{lemone}.
Let $H\su \bigcup_{n<N}Q^0(p_n)$ be a finite set containing
all such $x_s^n$. Choose $U\in\uu$ with $H\su U$ and determine the $q_n$
for $n<N$ as in Lemma \ref{lemone}.
\qed

Remark. Note that if $q\leq_k^\star p$ then $Q^0(q)\su Q^0(p)$ and
hence any $\om$-cover of  $Q^0(p)$ is still an $\om$-cover of $Q^0(q)$.
In these two lemmas, the $q$ we obtain are also equal mod finite
to the $p$, which also implies this.

\begin{lemma}\label{gamcover}
Given $(p_n,k_n:n\in\om)$ elements of $\poset\times\om$ and
$(\uu_n:n<\om)$ which are $\om$-covers of $Q=\bigcup_{n<\om}Q^0(p_n)$
there exists $(U_m\in\uu_m:m<\om)$ and 
$(q_n\leq_{k_n}p_n : n\in\om)$ such that 
$$\forall n<\om  \;\;  
\forall m\geq n \;\;\;[q_n]\su U_m.$$
\end{lemma}

\proof
Construct
$(q_n^m:n,m<\om)$ and $(U_n\in\uu_n:n<\om)$ inductively.
Given $(q_n^m:n<\om)$ 
and $(U_n: n<m)$, we construct $q_n^{m+1}$
and $U_m\in\uu_m$ so that

\begin{enumerate}
\item $q_n^{m+1}=p_n$ for $n\geq m+1$,
\item $q_n^{m+1}\leq_{k_n+m}^\star q_n^{m}$ for $n\leq m$, and
\item $[q_n^{m+1}]\su U_m$ for $n\leq m$.
\end{enumerate}
Then we let $q_n=\bigcup_{m>n}q_n^m$ be the fusion.  We
have that $q_n\leq_{k_n}q_n^n=p_n$ and $[q_n]\su U_m$ whenever
$m\geq n$.
\qed

Remark. Obviously, the analogue of this Lemma for $Q^1$ is
also true.

\begin{lemma}\label{disj}
Suppose $(p_n,k_n:n\in\om)$ are elements of $\poset\times\om$.
Then there exists $(q_n\leq_{k_n}p_n:n\in\om)$ such that for
$n\neq m$, 
$q_n$ and $q_m$ are strongly disjoint, i.e., there
are infinitely many $k\in (D_{q_n}\cap D_{q_m})$ with
$q_n(k)\neq q_m(k)$.
\end{lemma}
\proof
Given $p_1,p_2$ and $n$ it is easy to find $q_1\leq_n p_1$ and
$q_2\leq_n p_2$ which are strongly disjoint.  A fusion
argument produces a sequence $(q_n:n<\om)$ where all pairs
have been considered and made strongly disjoint.
\qed


Now we construct the Aronszajn tree of Silver conditions.
Let $\uu_\al$ for $\al<\om_1$ list all $\om$ sequences of
countable families of open subsets of $2^\om$.  Make sure
that each such sequence occurs as a $\uu_\al$ for
both $\al$ even and $\al$-odd.

We can construct a tree $T\su \om^{<\om_1}$ and
$(p_s\in\poset\;:\;s\in T)$ which has the following
properties.

\begin{enumerate}
\item $T\su \om^{<\om_1}$ is a subtree, i.e.,
$s\su t\in T$ implies $s\in T$.
\item $T_\al=T\cap \om^\al$ is countable for each $\al<\om_1$.
\item $s\su t\in T$ implies $p_t\leq p_s$.
\item If $s,t\in T$ are incomparable, then $p_s$ and
$p_t$ are strongly disjoint (as in Lemma \ref{disj}).
\item For any $\al<\be<\om_1$ and any $s\in T_\al$ and
$n<\om$ there exists $t\in T_\be$ with $p_t\leq_n p_s$.
\item Define
$$Q^0_\al=\bigcup\{Q^0(p_t):t\in T_{\leq\al}\}
\rmand 
Q^1_\al=\bigcup\{Q^1(p_t):t\in T_{\leq\al}\}.
$$
\begin{enumerate}
\item
For $\al$ an even ordinal, if $\uu_\al=(\uu_n^\al:n<\om)$ is
a sequence of $\om$-covers of $Q^0_\al$ then
there exists a sequence $(U_n\in\uu_n^\al:n<\om)$ which is a
$\ga$-cover of 
$$Q^0_\al\cup\bigcup\{[p_s]: s\in T_{\al+1}\}.$$
\item For $\al$ odd, the analogous statement
is true but with $Q^1_\al$ in place of $Q^0_\al$.
\end{enumerate}
\item Let $\dd$ be the family of $f\in\om^\om$ such that
for some $s\in T$ the function $f:\om\to \comp(D_{p_s})$
is the unique order preserving bijection.  Then $\dd$ is a
dominating family, i.e., for all $g\in\om^\om$ there
exists $f\in\dd$ such that $g(n)<f(n)$ for all $n<\om$.
\end{enumerate}

To construct $T_\lm$ and $p_s$ for $s\in T_\lm$ where
$\lm$ is a countable limit ordinal, proceed as follows.
For any $s\in T_{<\lm}$ and $N<\om$ 
choose a strictly increasing sequence $\lm_n$ cofinal in $\lm$ with 
$s=s_0\in T_{\lm_0}$.  By inductive hypothesis we can find
$s_n\in T_{\lm_n}$ with $p_{s_{n+1}}\leq_{N+n} p_{s_n}$.
Take $p_t$ for $t=\bigcup_{n<\om}s_n$ to be the fusion of
this sequence.   Repeat countably many times, to take care
of all $s\in T_{<\lm}$ and $N<\om$.

At successor stages for $\al$ even, check to see if $\uu_\al$ is an
$\om$-sequence of $\om$-covers of $Q^0_\al$.  If it is not, 
we need never worry about
it since the set we are building will contain $Q^0_\al$.
If it is, let $\{x_n:n<\om\}=Q^0_\al$ and let
$$\uu_n=\{U\in\uu_n^\al\;:\; \{x_i:i<n\}\su U\}.$$
Let $(p_n,k_n:n<\om)$ list all elements of 
$$\{p_s:s\in T_\al\}\times \om$$
with infinite repetitions and apply Lemma \ref{gamcover}
followed by Lemma \ref{disj}.  From the resulting
sequence we may find for each $s\in T_\al$ and $n<\om$ a
distinct condition $q\leq_n p_s$ which we label $p_{s\concat(n)}$.
Obtaining the last condition (7), is easy
(in fact, it seems hard to avoid), and is left to the reader.

\bigskip
The $\ga$ sets are the sets:
$$A_0=\bigcup_{s\in T} Q^0(p_s)
\;\;\;\;\;\;\rmand\;\;\;\;\;\;
A_1=\bigcup_{s\in T} Q^1(p_s).
$$
For $x\in A_0$ and $y\in A_1$ we note that
there are infinitely many $n$ with $x(n)\neq y(n)$.  To see
this note that if $x\in Q^0(p_s)$ and $y\in Q^1(p_t)$,
and $s$ and $t$ are incomparable, then $p_s$ and
$p_t$ are strongly disjoint.  On the other hand, if they
are comparable, for example, $s\su t$, then
since $p_t\leq p_s$, we have that $\comp(D_{p_t})\su \comp(D_{p_s})$.
So for all but finitely many $n\in \comp(D_{p_t})$ we will have
that $y(n)=1$ and $x(n)=0$.

Condition (7) gives us a continuous map from $A_0\times A_1$ onto
a dominating family $\dd\su\om^\om$.  Namely, if
$x_0\in Q^0(p_s)$ is identically zero on $\comp(D_{p_s})$
and  $x_1\in Q^1(p_s)$ identically one on $\comp(D_{p_s})$,
then $\comp(D_{p_s})=\{n:x_0(n)\neq x_1(n)\}$.
So the continuous map is 
$\phi:A_0\times A_1\to \om^\om$ 
defined by $\phi(x,y)=f$ where $f$ is the unique order preserving
map from $\om$ to the infinite set $\{n:x(n)\neq y(n)\}$.

\qed

Remark. Our result also shows that assuming CH there are
Borel-cover $\ga$-sets whose product is not Menger.
To see this note the following:


\begin{lemma}\label{borel}
Suppose $p\in\poset$, $n<\om$, and $B\su 2^\om$ a
Borel set.  Then there exists
$q\leq_n p$ such that $[q]\cap B$ is clopen
in $[q]$.
\end{lemma}
\proof

Let $F$ be the first $n$ elements of $\comp(D_p)$
and let $\phi:\om\to (\comp(D_p)\sm F)$ be a bijection.
For $X\su\om$ let $\psi_X:(\comp(D_p)\sm F)\to 2$ be
the restriction of the characteristic function of $\phi(X)$.
For each $s\in 2^F$ define 
$$C_s=\{X\in [\om]^\om\;:\; (p\cup s\cup \psi_X)\in B\}.$$
Since these are Borel sets
by the Galvin-Prikry Theorem \cite{gp} there exists
$H\in[\om]^\om$ such that for each $s\in 2^F$ either
$[H]^\om\su C_s$ or $[H]^\om\cap C_s=\emptyset$.
Let $H_1\su H$ be infinite such that $H\sm H_1$ is also infinite.
Let
$$q=p\cup (\phi(\comp(H))\times\{0\})\cup (\phi(H_1)\times\{1\}).$$
Note that $\comp(D_q)=F\cup \phi(H\sm H_1)$.
We claim that given any $x,y\in [q]$
if $x\res F= y \res F = s$, then $x\in B$ iff $y\in B$.
Letting $H_x=\phi^{-1}(x^{-1}(1))$ we see that
$H_1\su H_x\su H$ and so $H_x$ is an infinite subset of $H$.
Similarly for $H_y$.  By choice of $H$ we have that
$H_x\in C_s$ iff $H_y\in C_s$ and so the claim follows.

\qed

Modify our construction in the open cover case as follows.
Let $B_\be$ for $\be<\om_1$ list all Borel sets.  At
the end of each stage $\be$ apply Lemma \ref{borel} accross
the level to make sure that $[p_s]\cap B_\be$ is
relatively clopen for each $s\in T_{\be+1}$.

Let $\bb_\al=(\bb_\al^n:n<\om)$ for $\al<\om_1$ all $\om$ sequences of
countable families of Borel sets.
We may assume that each
element of $\bigcup \bb_\al$ has already been listed
as a $B_\be$ for some $\be<\al$.

At successor stages for $\al$ even, check to see
if $\bb_\al=(\bb_n^\al:n<\om)$ is an
$\om$-sequence of countable Borel $\om$-covers of $Q^0_\al$.
If it is not, we need never worry about
it since the set we are building will contain $Q^0_\al$.
If it is, let $\{x_n:n<\om\}=Q^0_\al$ and let
$$\bb_n=\{U\in\bb_n^\al\;:\; \{x_i:i<n\}\su U\}.$$
Since the elements of each $\bb_n$ are relatively open
in the $[p_s]$ the rest of the argument is the same as
in the open case.


\begin{thebibliography}{99}

\bibitem{gp}
Galvin, Fred; Prikry, Karel;
Borel sets and Ramsey's theorem.
J. Symbolic Logic 38 (1973), 193--198.

\bibitem{gal}
Galvin, Fred; Miller, Arnold W.;
$\gamma $-sets and other singular sets of real numbers.
Topology Appl. 17 (1984), no. 2, 145--155.

\bibitem{grig} Grigorieff, Serge;
Combinatorics on ideals and forcing.
Ann. Math. Logic 3 (1971), no. 4, 363--394.

\end{thebibliography}

\end{document}
