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\newcommand{\UpdateDate}{Aug 25, 03}
\date{}
\title{A big symmetric planar set with small category projections}
\pagestyle{myheadings}
\markboth{K.~Ciesielski, T.~Natkaniec
\ \ \ \ \ \UpdateDate
}{A big symmetric set with small category projections
\ \UpdateDate
}



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\author{
Krzysztof Ciesielski%
\thanks{The work of the first author was partially supported by
NATO Grant PST.CLG.977652 and 2002/03 West Virginia
University Senate Research Grant.
}\\
{\footnotesize Department of Mathematics,}
{\footnotesize West Virginia University,} \\
{\footnotesize Morgantown, WV 26506-6310, USA}\\
{\footnotesize e-mail: K\_Cies@math.wvu.edu}; %\\
{\footnotesize web page: {\tt http://www.math.wvu.edu/\~{}kcies}}
\and Tomasz Natkaniec%
\thanks{Partially supported by
Gda{\'n}sk University Research Grant No. BW 5100-5-02331-2. }\\
 {\footnotesize Department of Mathematics,}
{\footnotesize Gda\'{n}sk University,} \\ {\footnotesize Wita
Stwosza 57, 80--952 Gda\'{n}sk, Poland} \\
 {\footnotesize e-mail:
mattn@math.univ.gda.pl;}}
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\begin{document}
{
\renewcommand{\thefootnote}{}
\footnotetext{AMS Subject Classification (1991): Primary {\bf 03E05}; Secondary:  03E50; 03E75.}
}
{\renewcommand{\thefootnote}{}
\footnotetext{Key words: $f$-category projections;
transfinite induction; nowhere meager sets.
Covering Property Axiom CPA; $\omega_1$-oracle.}
}

\maketitle

\begin{abstract}
We show that under appropriate set theoretic assumptions
(which follow from Martin's axiom and the continuum hypothesis)
there exists a nowhere meager set
$A\subset\mathR$ such that
\begin{itemize}
\item[(i)] for each continuous nowhere constant function
$f\colon\mathR\to\mathR$ the set
$\{c\in\mathR\colon \pi[(f+c)\cap (A\times A)]\mbox{ is not meager}\}$ is
meager, and
\item[(ii)]  the set
$\{c\in\mathR\colon (f+c)\cap (A\times A)=\emptyset\}$ is nowhere meager
for each continuous function $f\colon\mathR\to\mathR$.
\end{itemize}
The existence of such a set follows also from the principle CPA,
which holds in the iterated  perfect set model.
We also prove that the existence of a set $A$ as in (i)
cannot be proved in ZFC alone even when we restrict our attention to
homeomorphisms
of $\real$. On the other hand, for the class of real analytic functions
a Bernstein set $A$ satisfying (ii) exists in ZFC.
\end{abstract}

\section{The results}
Let $\M$ denote the class of all meager subsets of $\mathR$ and let
$\pi\colon\mathR^2\to\mathR$ be the projection into the first coordinate.
For a function $f\colon\mathR\to\mathR$ and a set $E\subset\mathR^2$
the {\em $f$-category projection}\/ of $E$ is the set
$\{c\in\mathR\colon \pi[(f+c)\cap E]\notin\M\}$. (See \cite{CG}.)
In papers  \cite{RD}, \cite{TN}, and \cite{TN1} the authors considered 
the second category sets $A\subset\mathR$ such that 
an $f$-category projection of $A\times A$ have an empty interior 
for every linear function $f(x)=ax+b$. 
Inspired by these results
Bartoszy{\'n}ski and Halbeisen~\cite{BH} recently constructed a second category
set $A\subset\mathR$ with even stronger property: 
such that for each polynomial $p$ which
is neither constant nor the identity function
the set $p\cap(A\times A)$ is finite.
In particular, if $p$ is a non-constant polynomial
then the $p$-category projection of $A\times A$ has an empty interior.
These results lead to the following,
more general question:
for which classes $\F$ of continuous functions does there exist
a ``big'' set $A\subset\real$ such that the $f$-category projection
is ``small'' for every $f\in\F$?
In particular, what happens for
the classes $\A$ of real analytic functions, $\C_0$ of all nowhere
constant functions from $\mathR$ to $\mathR$, and for the entire
class $\C$ of all continuous functions from $\mathR$ to $\mathR$?
In this note we answer these questions. We use standard
terminology as in \cite{CiBook}. We consider only real-valued
functions of one variable, unless otherwise specified. No
distinction is made between a function and its graph. For
functions $f,g$ put $[f=g]=\{ x\in\mathR\colon\; f(x)=g(x)\}$. A
set $A\subset\mathR$ is nowhere meager if $A\cap I\not\in\M$ for
each non-degenerate interval $I$. Symbol $B(x,r)$ denotes the open
ball centered at $x$ and with the radius $r$ and $\id$ stands for
the identity function. The main general theorem in a positive
direction is the following result.
\begin{thm}\label{tmain}
Let $\J\supset[\real]^{\leq\omega}$
be a translation invariant ideal on $\real$
and $B_0$ be a family of Borel sets containing
all Borel non-meager sets
such that
\begin{equation}
\label{eqMain1}
\mbox{$B\setminus\bigcup\G\neq\emptyset$ for every $B\in\B_0$
and $\G\in[\J]^{<\co}$.}
\end{equation}
If $\F\subset\C$ is such that for every $f\in\F$
\begin{equation}\label{eqMain2}
\mbox{the set
$L_f=\{y\in\mathR\colon f^{-1}(y)\notin\J\}$
belongs to $\J$}
\end{equation}
then there exists an $A\subset\mathR$ intersecting every $B\in\B_0$
such that for every $f\in\F$
\begin{itemize}
\item[{\rm (a)}] there exists a $Z\in[\real]^{<\co}$ such that
$f\cap(A\times A)\subset(A\times Z)\cup\id$.
\end{itemize}
Moreover, if\/ {\rm (\ref{eqMain2})} holds for every $f\in(\pm\F)\cup(\id-\F)$
then we can also have the following property for every $f\in\F$
\begin{itemize}
\item[{\rm (b)}] the set
$\{ c\in\mathR\colon (f+c)\cap (A\times A)=\emptyset\}$
intersects every $B\in\B_0$.
\end{itemize}
\end{thm}
We leave the proof of this theorem to the next section.
In the reminder of this section we like to discuss several
of its corollaries.
In particular, applying
Theorem~\ref{tmain} to the ideal $[\real]^{\leq\omega}$
of countable sets
and the family $\B_0$ of all uncountable Borel sets
we get the following result.
\begin{cor}\label{cor1}
Let $\F\subset\C$ be such that
the set
$L_f=\{y\in\mathR\colon |f^{-1}(y)|>\omega\}$
is empty for every $f\in\F$.
Then there exists a Bernstein
set $A\subset\mathR$ such that for every $f\in\F$
\begin{itemize}
\item[{\rm (a)}] there exists a $Z\in[\real]^{<\co}$
such that $f\cap(A\times A)\subset(A\times Z)\cup\id$.
\end{itemize}
In addition, if $L_f=\emptyset$
for every $f\in(\pm\F)\cup(\id-\F)$ then we can also have
\begin{itemize}
\item[{\rm (b)}]
$\{c\in\mathR\colon (f+c)\cap (A\times A)=\emptyset\}$ contains a Bernstein set
for every $f\in\F$.
\end{itemize}
In particular,
\begin{itemize}
\item there exists a set $A$ satisfying\/ {\rm (a)} for every countable-to-one
function $f\in\C$;
\item there exists a set $A$ satisfying\/ {\rm (a)} and\/ {\rm (b)} for every
analytic function $f\in\A$; 
moreover, if $f\neq\id$ is not constant then $|f\cap(A\times A)|<\co$.
\end{itemize}
\end{cor}
\pf Clearly the ideal $\J=[\real]^{\leq\omega}$
and the family $\B_0$ of all uncountable Borel sets
satisfy
(\ref{eqMain1}) from Theorem~\ref{tmain}.
Since we assume that (\ref{eqMain2}) is also satisfied,
the main part of the corollary follows Theorem~\ref{tmain}.
To see the additional part for analytic functions first notice that $\F=\A$
satisfies
the assumption for the part (b),
since the set $L_f=\emptyset$
is empty for every non-constant $f\in\A$.
In particular, if $f\neq\id$ then, by condition (b), the set
$f\cap(A\times A)$ is contained in
$(f-\id)^{-1}(0)\cup\bigcup_{z\in Z}f^{-1}(z)$, which is
a union of less than $\co$-many countable sets.
\qed

In the proof of the next corollary we need the following simple fact.
\begin{equation}\label{eqFact}
\mbox{If $f\in\C$ then the set $\{ c\colon\; [f+c=\id]\not\in\M\}$ is
countable.}
\end{equation}
To see it notice that for each $c\in\mathR$ the set $[f+c=\id]$ is closed.
Therefore,
if $[f+c=\id]\not\in\M$, there exists a non-empty open interval
$I_c$ with $f(x)=x-c$ for each $x\in I_c$. It is easy to observe
that $I_c\cap I_d=\emptyset$ if $c\neq d$.
\begin{cor}\label{cor1b}
There exists a Bernstein set $A\subset\real$ such that for every homeomorphism
$f\colon\real\to\real$ for all but countably many $c\in \real$
the set $\pi[(f+c)\cap (A\times A)]$ is a union of a meager set and a set
of cardinality
less than $\co$.
In particular, if $[\real]^{<\co}\subset\M$ then the set
\[
\{c\in\mathR\colon \pi[(f+c)\cap (A\times A)]\notin\M\}
\]
is countable (so meager) for every homeomorphism $f\colon\real\to\real$.
\end{cor}
\pf Let $A$ be as in Corollary~\ref{cor1}. So for every
homeomorphism $f\colon\real\to\real$ 
and $c\in\real$ there exists
a $Z\in[\real]^{<\co}$ such that $(f+c)\cap(A\times
A)\subset(A\times Z)\cup\id$. Notice that $\pi[(f+c)\cap(A\times
A)]\subset[f+c=\id]\cup\pi[(f+c)\cap(A\times Z)]$ and
$\pi[(f+c)\cap(A\times Z)]\subset f^{-1}(Z-c)\in[\real]^{<\co}$.
So there exists a $D\in[\real]^{<\co}$ such that
$\pi[(f+c)\cap(A\times A)]\subset[f+c=\id]\cup D$. The conclusion
follows from~(\ref{eqFact}). \qed

Notice that the set theoretical assumption in
Corollary~\ref{cor1b} is essential.
\begin{thm}\label{thmCons1} It is relatively consistent with the set theory ZFC
that for every nowhere meager set $A\subset\mathR$
there exists a homeomorphism $f\colon\real\to\real$
such that
$\{c\in\real\colon \pi[(f+c)\cap(A\times A)]\notin\M\}$ is  nowhere meager.
\end{thm}
\pf This follows immediately from Theorem~\ref{thConsOracle}
and Proposition~\ref{propA}
which will
be proven in Section~\ref{sec:oracle}. \qed

Applying Theorem~\ref{tmain} to the ideal $\M$
and the family $\B_0$ of non-meager Borel sets
we get also the following result.
(Recall that the set theoretic assumption
about covering by meager sets
follows form the Martin's axiom MA
and from the continuum hypothesis CH.)
\begin{cor}\label{cor2}
Assume that less than continuum many
meager sets do not cover $\real$.
Then there exists a nowhere meager set
$A\subset\mathR$ such that for every $f\in\C$
the set
$\{c\in\mathR\colon (f+c)\cap (A\times A)=\emptyset\}$
is nowhere meager in $\mathR$.
In particular, the $f$-category projection of
$A\times A$ has an empty interior.
\end{cor}
\pf The set theoretical assumption assures that
the ideal $\J=\M$, the family $\B_0$ of non-meager Borel sets,
satisfy assumptions (\ref{eqMain1}) from Theorem~\ref{tmain}.
Since, for $\J=\M$, (\ref{eqMain2}) holds for every $f\in\C$,
there is a set $A$ satisfying (b) from Theorem~\ref{tmain}.
Clearly, it has the desired properties.  \qed

In case of the class $\C_0$ of nowhere constant functions
we can have yet another corollary. In its proof we will use the following
simple fact.
\begin{equation}\label{PreimM}
f^{-1}(M)\in\M \mbox{ for every $f\in\C_0$ and $M\in\M$}.
\end{equation}
Indeed, if sets $F_n$ are closed and nowhere dense in $\real$ such that
$M\subset \bigcup_{n<\omega}F_n$ then
$f^{-1}(M)\subset\bigcup_{n<\omega}f^{-1}(F_n)$.
It is just enough to notice that $f^{-1}(F_n)$ is closed and nowhere dense
for every $f\in\C_0$.
\begin{cor}\label{cor3}
Assume that less than continuum many
meager sets do not cover $\real$
and that $[\real]^{<\co}\subset\M$.
Then there exists a nowhere meager set
$A\subset\mathR$ such that for every $f\in\C_0$
the set
$\{c\in\mathR\colon \pi[(f+c)\cap (A\times A)]\notin\M\}$
is countable.
\end{cor}
\pf Let $f\in\C_0$ and $c\in\real$. By part (a)
of Theorem~\ref{tmain} there exists a
$Z\in[\real]^{<\co}$ such that
$\pi[(f+c)\cap(A\times A)]\subset[f+c=\id]\cup\pi[(f+c)\cap(A\times Z)]$.
But, by (\ref{PreimM}),
$\pi[(f+c)\cap(A\times Z)]\subset f^{-1}(Z-c)\in\M$
since $Z-c\in[\real]^{<\co}\subset\M$.
Thus, $\pi[(f+c)\cap(A\times A)]\in\M$ as long as $[f+c=\id]\in\M$.
So, the result follows immediately from (\ref{eqFact}).
\qed

We believe that the conclusion of Corollary~\ref{cor3}
cannot be proved in ZFC, what we state as a following conjecture.
(See also the last section of the paper for some comments on it.)
\begin{conj}\label{conj1} It is relatively consistent with ZFC that
for every nowhere meager set $A\subset\mathR$
there is an $f\in\C_0$
such that $\pi[(f+c)\cap(A\times A)]\notin\M$
for every $c\in\real$.
\end{conj}
It worth to note that a set $A$ as in Corollaries~\ref{cor2}
and~\ref{cor3} can be also constructed under the Covering Property
Axiom CPA, which extracts the essence of the iterated perfect set
model.
(See \cite{CP83,CMP90,CPAbook}.)
This is of interest, since under CPA the
set theoretical assumptions of each of these corollaries are false:
CPA implies
that $\co=\omega_2$ and that $\mathR$ can be covered by $\omega_1$
meager sets.
In fact, in the theorem we will use only a simpler version of CPA
known as \psmCGame.
\begin{thm}\label{thmCPA} Assume that
\psmCGame\ holds.
Then there exists a nowhere meager set $A\subset\mathR$ of cardinality
$\omega_1<\continuum$
such that for every $f\in\C$
there exists a countable set $Z\subset A$ such that
\[
f\cap(A\times A)\subset(A\times Z)\cup\id.
\]
In particular, for every $f\in\C_0$ the set
$\{c\in\mathR\colon \pi[(f+c)\cap (A\times A)]\notin\M\}$ is countable and
the set
$\{c\in\real\colon (f+c)\cap(A\times A)=\emptyset\}$
is a complement of a set of cardinality $\omega_1<\continuum$,
so it contains a Bernstein set.
\end{thm}
\pf The main part of the theorem will be proved in Section~\ref{sec:CPA}.
To prove the additional part of the theorem fix an $f\in\C_0$.
The proof that the set 
$\{c\in\mathR\colon \pi[(f+c)\cap (A\times A)]\notin\M\}$ is countable
is exactly the same as for Corollary~\ref{cor3}.
To see that the complement $D$ of
$\{c\in\real\colon (f+c)\cap(A\times A)=\emptyset\}$
has cardinality $\omega_1$
notice that if $(f+c)\cap(A\times A)\neq\emptyset$
then there are $x,y\in A$ such that $f(x)+c=y$, that is,
$c=y-f(x)\in A - f[A]$.
So, $D$ is equal to the set $A - f[A]$ and has
cardinality~$\omega_1$. \qed

It is also good to notice that the set $A$ in the
Corollaries~\ref{cor2} and~\ref{cor3} cannot have the Baire property.
\begin{prop}\label{prop2AAA}
Suppose that $f\in\C_0$ and that $A\subset\mathR$ is
a non-meager set having the Baire property.
Then the $f$-category projection of $A\times A$ has a non-void interior.
\end{prop}
\pf
Let $G$ be a non-empty open set with $M=G\setminus A\in\M$.
Fix $x_0\in G\cap A$ and $c_0\in\mathR$ with $f(x_0)+c_0\in G\cap A$,
an $\varepsilon>0$ such that $B(f(x_0)+c_0,2\varepsilon)\subset G$, and a
$\delta>0$ such that $f[B(x_0,\delta)]\subset B(f(x_0),\varepsilon)$.
Let $c\in B(c_0,\varepsilon)$.
Since $A_0=A\cap B(x_0,\delta)\not\in\M$ and $f+c\in\C_0$,
condition (\ref{PreimM}) implies $(f+c)(A_0)\not\in\M$.
Since $(f+c)(A_0)\subset G$ and $(f+c)(A_0)\cap A\not\in\M$
point $c$ belongs to the $f$-category projection of $A\times A$.
\qed

We will finish this section with 
following result of Bartoszynski and Halbeissen, which was one of the 
departing points for this note.


\begin{thm}\label{prop2}{\rm \cite{BH}}
There exists a set $A\subset\mathR$ intersecting every perfect set
such that for each non-constant polynomial $p\neq\id$
the set $p\cap(A\times A)$ is finite.
\end{thm}

Note that Corollary~\ref{cor1} implies immediately a weaker version of
Theorem~\ref{prop2}: 
there exists a Bernstein set $A$ for which
each set $p\cap(A\times A)$ has cardinality less than $\continuum$. 
However, we see no easy way to deduce the full version of the theorem
from the results presented above. Nevertheless we like to include here
a very short proof of Theorem~\ref{prop2}, since it is
considerably simpler and completely different that the argument presented in~\cite{BH}.


\noindent
\pf First notice that if $A$ is a transcendental base of $\mathR$
(over $\mathQ$) then the set $p\cap(A\times A)$ is finite for
every polynomial $p$ which is neither constant nor the identity
function. Indeed, if $K\in[A]^{<\omega}$ is such that
$p\in\overline{\mathQ(K)}[x]$, where $\overline{\mathQ(K)}$ stands
for the algebraic closure of $\mathQ(K)$ in $\real$, then for
every $a\in A\setminus K$ we have
$p(a)\in\overline{\mathQ(K\cup\{a\})}\setminus\overline{\mathQ(K)}$,
since $A$ is algebraically independent. (See e.g., \cite[Lemma~2,
p. 99]{MK}.) So, if $p(a)\in A$ then $p(a)=a$. But this is
impossible, since $p(a)=a$ implies that $a$ is a root of a
non-zero polynomial $p-\id\in\overline{\mathQ(K)}[x]$. So
$\pi[p\cap(A\times A)]\subset K$. It is well known that there are
transcendental bases $A$ that are also Bernstein sets (in
\cite{CiBook} see corollary 7.3.6 and exercise 2 on page 126) and
any such base satisfies Proposition~\ref{prop2}. \qed

\section{Proof of Theorem~\ref{tmain}}
Let $\{\la f_{\aaa},B_{\aaa}\ra\colon\aaa<\co\}$ be
an enumeration of $\F\times\B_0$.
For each $\aaa<\co$ we will choose, by induction on $\alpha<\co$,
points $x_{\aaa}\in B_{\aaa}$ and $c_{\aaa}\in B_{\aaa}$
aiming for $A=\{ x_{\aaa}\colon\aaa<\co\}$.
We will set up the induction in such a way
that for every $\aaa<\co$ the set $Z$ satisfying (a)
for $f_{\aaa}$
will be equal to $A_{\aaa}=\{ x_{\bbb}\colon
\bbb<\aaa\}$ and, if (\ref{eqMain2}) holds for every
$f\in(\pm\F)\cup(\id-\F)$, that $(f_{\aaa}+c_{\aaa})\cap (A\times
A)=\emptyset$. So, assume that for some $\alpha<\continuum$ the
sets $\{ x_{\bbb}\colon \bbb<\aaa\}$ and $\{c_{\bbb}\colon
\bbb<\aaa\}$ are already constructed. If we need only to insure
(a), we put $c_\alpha=x_\alpha$ and choose an $x_{\aaa}\in
B_{\aaa}\setminus \bigcup_{\beta\leq\alpha}L_{f_\beta}$ such that
\[
f_\beta\cap(\{x_\alpha\}\times (A_\alpha\setminus A_\beta))=\emptyset=
f_\beta\cap(A_\alpha\times\{x_\alpha\})
\]
for every $\beta\leq\alpha$.
This is possible by (\ref{eqMain1}), since
$\{L_{f_\beta}\colon\beta\leq\alpha\}\subset\J$,
singletons $f_{\bbb}[\{a\}]\in\J$, for $\beta\leq\alpha$ and $a\in A_\alpha$,
and
$\{x\in\mathR\colon f_{\bbb}(x)=x_\gamma\}=f_{\bbb}^{-1}(x_\gamma)\in\J$
for every $\beta\leq\gamma<\alpha$ since $x_\gamma\notin L_{f_\beta}$.
It is easy to see that such a choice implies that
$f_\alpha\cap(A\times A)\subset(A\times A_\alpha)\cup\id$ for every
$\alpha<\co$.
So, in what follows we assume that (\ref{eqMain2}) holds for every
$f\in(\pm\F)\cup(\id-\F)$.
We assume also that the following inductive conditions hold
for every $\alpha<\co$.
\begin{description}
\item[(I)] $(\id-f_{\aaa})^{-1}(c_{\aaa})
\cup f_\alpha^{-1}(x_\beta-c_\alpha)\in\J$ for every $\beta<\alpha$, and
\item[(II)] $f_{\bbb}^{-1}(x_\alpha)
\cup f_{\bbb}^{-1}(x_\alpha-c_{\bbb})\in\J$
for every $\beta\leq\alpha$.
\end{description}
First choose a $c_{\aaa}\in B_{\aaa} \setminus
\left(L_{\id-f_\alpha}\cup\bigcup_{\beta<\alpha}(x_\beta+L_{-f_\alpha})\right)$
outside the set
\[
\{c\in\real\colon (f_{\aaa}+c)\cap (A_{\aaa}\times A_{\aaa})\neq\emptyset\}
=A_{\aaa}-f_{\aaa}[A_{\aaa}]\in [\mathR]^{<\co}.
\]
Such a choice is possible by
the assumption (\ref{eqMain1}), since the singletons and the sets
$L_{\id-f_\alpha}$
and $x_\beta+L_{-f_\alpha}$
belong to $\J$.
Clearly $c_\alpha\notin L_{\id-f_\alpha}$
insures that
$(\id-f_{\aaa})^{-1}(c_{\aaa})\in\J$.
Similarly, $c_\alpha\notin x_\beta+L_{-f_\alpha}$
implies that $c_\alpha-x_\beta\notin L_{-f_\alpha}$
so the set $f_\alpha^{-1}(x_\beta-c_\alpha)=(-f_\alpha)^{-1}(c_\alpha-x_\beta)$
belongs to $\J$. So, condition (I) is satisfied.
Note also that $c_\alpha\notin A_{\aaa}-f_{\aaa}[A_{\aaa}]$
implies that
\begin{description}
\item[(i)]
$(f_{\aaa}+c_{\aaa})\cap (A_{\aaa}\times A_{\aaa})=\emptyset$.
\end{description}
Next choose an
$x_{\aaa}\in B_{\aaa}\setminus
\bigcup_{\beta\leq\alpha}(L_{f_\beta}\cup(c_\beta+L_{f_\beta}))$
such that
\begin{equation}\label{eq22}
(f_\beta+c_\beta)\cap(\{x_\alpha\}\times A_\alpha)=\emptyset=
(f_\beta+c_\beta)\cap(A_\alpha\times\{x_\alpha\}),
\end{equation}
$x_\alpha\notin\bigcup_{\gamma\leq\alpha}(\id-f_\gamma)^{-1}(c_\gamma)$, and
\begin{equation}\label{eq44}
f_\beta\cap(\{x_\alpha\}\times (A_\alpha\setminus A_\beta))=\emptyset=
f_\beta\cap(A_\alpha\times\{x_\alpha\})
\end{equation}
for every $\beta\leq\alpha$.
This is possible by (\ref{eqMain1}), since
$\{L_{f_\beta}\cup(c_\beta+L_{f_\beta})\colon\beta\leq\alpha\}\subset\J$ and
by inductive assumptions (I)\&(II)
for every $a\in A_\alpha$ and $\beta\leq\gamma<\alpha$
the following sets belong to $\J$.
\begin{itemize}
\item $\{x\in\mathR\colon(f_{\bbb}+c_{\bbb})(x)=a\}=f_{\bbb}^{-1}(a-c_{\bbb})$;
\item $(f_{\bbb}+c_{\bbb})[\{a\}]$;
\item $(\id-f_\gamma)^{-1}(c_\gamma)$;
\item $\{x\in\mathR\colon f_{\bbb}(x)=x_\gamma\}=f_{\bbb}^{-1}(x_\gamma)$;
\item $f_{\bbb}[\{a\}]$.
\end{itemize}
Note that
$x_{\aaa}\notin\bigcup_{\beta\leq\alpha}L_{f_\beta}$
insures that $f_{\bbb}^{-1}(x_\alpha)\in\J$
and $x_{\aaa}\notin\bigcup_{\beta\leq\alpha}(c_\beta+L_{f_\beta})$
implies that $x_\alpha-c_\beta \notin L_{f_\beta}$
so the set $f_{\bbb}^{-1}(x_\alpha-c_{\bbb})$
is in $\J$. In particular, (II) is satisfied.
This finishes the inductive construction.

Clearly $A$ is nowhere meager, since
it meets every non-meager Borel set.
To see that (b) holds notice that by (i) we have
$(f_\beta+c_\beta)\cap (A_\beta\times A_\beta)=\emptyset$
while
$(f_\beta+c_\beta)\cap (A_{\aaa}\times A_{\aaa})=\emptyset$
for $\alpha>\beta$ is insured by the choice of $x_\alpha$
as in (\ref{eq22}) and the fact that
$(f_\beta+c_\beta)(x_\alpha)\neq x_\alpha$
since  $x_\alpha\notin(\id-f_\beta)^{-1}(c_\beta)$.
To see that (a) holds pick an $f\in\F$
and let $\beta<\co$ be such that
$f=f_\beta$. The choice of $x_\alpha$ for $\alpha>\beta$
as in (\ref{eq44}) implies that
$f\cap(A\times A)\subset(A\times A_\beta)\cup\id$.
\qed

\section{Set $A$ from CPA}\label{sec:CPA}
To formulate axiom \psmCGame\ we need a few definitions. Let
$\Cantor$ denote the Cantor set $2^\omega$. For a Polish space $X$
we use symbol $\perf(X)$ to denote the family of all subsets of
$X$ homeomorphic to $\Cantor$. A subset $C$ of a product
$\Cantor^\omega$ of the Cantor set is said to be a {\em perfect
cube}\/ if $C=\prod_{n\in\omega} C_n$, where
$C_n\in\perf(\Cantor)$ for each $n$. For a fixed Polish space $X$
let $\Fcube$ stand for the family of all continuous injections
from a perfect cube $C\subset\Cantor^\omega$ onto a set $P$ from
$\perf(X)$. We consider each function $f\in\Fcube$ from $C$ onto
$P$ as a coordinate system imposed on~$P$. We say that
$P\in\perf(X)$ is a {\em cube}\/ if we consider it with
(implicitly given) witness function $f\in\Fcube$ onto $P$, and $Q$
is a {\em subcube of a cube}\/ $P\in\perf(X)$ provided $Q=f[C]$,
where $f\in\Fcube$ is a witness function for $P$ and
$C\subset\dom(f)\subset\Cantor^\omega$ is a perfect cube.
(Here and
in what follows symbol $\dom(f)$ stands for the domain of $f$.)
We say that a family $\E\subset\perf(X)$ is {\em cube dense}\/ in
$\perf(X)$ provided every cube $P\in\perf(X)$ contains a subcube
$Q\in\E$. More formally, $\E\subset\perf(X)$ is cube dense
provided
\begin{equation}\label{eqDefFdense}
\forall f\in\Fcube\ \exists g\in\Fcube\ (g\subset f\ \&\
\range(g)\in\E).
\end{equation}
We need also a notion of a {\em constant cubes}: a family
$\Ccube(X)$ of constant ``cubes'' is defined as a family of all
constant functions from a perfect cube $C\subset\Cantor^\omega$
to $X$. We define $\Fcube^*(X)$ as
\begin{equation}\label{eqFcubeSTAR}
\Fcube^*=\Fcube\cup\Ccube.
\end{equation}
Thus, $\Fcube^*$ is the family of all continuous functions from a
perfect cube $C\subset\Cantor^\omega$ into $X$ which are either
one-to-one or constant. Now the range of every $f\in \Fcube^*$
belongs to the family $\perf^*(X)$\label{PagePerfStar} of all sets
$P$ such that either $P\in\perf(X)$ or $P$ is a singleton. The
terms ``$P\in\perf^*(X)$ is a cube'' and ``$Q$ is a subcube of a
cube $P\in\perf^*(X)$'' are defined in a natural way.
\medskip
For a Polish space $X$ consider the following game ${\rm
GAME}_{\rm cube}(X)$\label{PageGameCube} of length $\omega_1$. The
game has two players, Player~I and Player~II. At each stage
$\xi<\omega_1$ of the game Player~I can play an arbitrary cube
$P_\xi\in\perf^*(X)$ and Player~II must respond with a subcube
$Q_\xi$ of $P_\xi$. The game $\la\la P_\xi,Q_\xi\ra\colon
\xi<\omega_1\ra$ is won by Player~I provided
\[
\bigcup_{\xi<\omega_1}Q_\xi=X;
\]
otherwise the game is won by Player~II.
By a strategy for Player~II we will understand any function $S$
such that $S(\la\la P_\eta,Q_\eta\ra\colon \eta<\xi\ra,P_\xi)$ is
a subcube of $P_\xi$, where $\la\la P_\eta,Q_\eta\ra\colon
\eta<\xi\ra$ is any partial game. (We abuse here slightly the
notation, since function $S$ depends also on the implicitly given
coordinate functions $f_\eta\colon \Cantor^\omega\to P_\eta$
making each $P_\eta$ a cube.) A game $\la\la P_\xi,Q_\xi\ra\colon
\xi<\omega_1\ra$ is played according to a strategy $S$
provided $Q_\xi=S(\la\la P_\eta,Q_\eta\ra\colon
\eta<\xi\ra,P_\xi)$ for every $\xi<\omega_1$. A strategy $S$ for
Player~II is a {\em winning strategy}\/ for Player~II provided
Player~II wins any game played according to the strategy $S$.
Now we can formulate the axiom. (See~\cite{CPAbook}.)
\begin{description}
\item[{\bf \psmCGame:}] $\continuum=\omega_2$ and for any Polish space $X$
Player~II has no winning strategy in the game ${\rm GAME}_{\rm
cube}(X)$.
\end{description}
All we need to know about cube-dense families is the following
fact.
\begin{Fact}\label{fact:cube}
Let $X$ be a Polish space and let $\E\subset\perf^*(X)$ contain
all singletons. If for every $P\in\perf(X)$ and every
Borel probability measure $\mu$ on $P$ there exists a
$Q\in\perf(P)\cap\E$ such that $\mu(Q)>0$, then $\E$ is
cube-dense.
\end{Fact}
\pf This follows immediately from \cite[Claim~3.2]{CP88}. (See
also \cite[Claim~1.1.5]{CPAbook} or  \cite[Claim~2.3]{CP83}.) \qed

We will apply this fact to $X=\C$, where $\C$ is considered with
the sup norm. Notice that for every $Q\subset\C$ the set $\bigcup
Q\subset\real^2$ is a union of the graphs of all functions
belonging to $Q$, since functions are identified with their
graphs. In what follows for a set $K\subset\real^2$
and $x\in\real$ we will use the symbol
$K_x$ to denote the vertical section of $K$ above $x$, that is,
$K_x=\{y\colon\la x,y\ra\in K\}$.
Similarly, $K^x=\{x\colon\la x,y\ra\in K\}$.
For $A\in[\real]^{\leq\omega}$ let $\E(A)$ be the family
of all $Q\in\perf^*(\C)$ such that
\begin{itemize}
\item $\bigcup Q$ is nowhere dense in $\real^2$, and
\item $[\bigcup Q]_x$ is nowhere dense in $\real$ for every $x\in A$.
\end{itemize}
\begin{lem}\label{lemCPA} Family
$\E(A)$ is cube-dense for every $A\in[\real]^{\leq\omega}$.
\end{lem}
\pf Without loss of generality we can assume that $A$ is dense in
$\real$. Clearly every singleton belongs to $\E(A)$. So, let
$P\in\perf(\C)$ and let $\mu$ be a Borel
probability measure $\mu$ on $P$. By Fact~\ref{fact:cube} it is
enough to show that there exists a $Q\in\perf(P)\cap\E(A)$ such
that $\mu(Q)>0$. To see this, fix a countable base $\B$ for
$\real$ and let $\la \la a_n,J_n\ra\colon n<\omega\ra$ be an
enumeration of $A\times \B$. Notice that for every $n<\omega$
there exists a non-empty open set $U_n\subset J_n$ such that
\begin{equation}\label{eq:CPA1}
\mu(\{f\in P\colon f(a_n)\in U_n\})<2^{-(n+2)}.
\end{equation}
Indeed, if $\U_n$ is
an infinite family of non-empty pairwise disjoint open subsets of
$J_n$ then for each $U\in\U_n$ the set $\{f\in P\colon f(a_n)\in
U\}$ is open in $P$ (so $\mu$-measurable) and so condition
(\ref{eq:CPA1}) must hold for some $U\in\U_n$. Let
$W=\bigcup_{n<\omega} \{f\in\C\colon f(a_n)\in U_n\}$. It is clear
that $W$ is open and dense in $\C$. So, $Q=P\setminus W=P\setminus
\bigcup_{n<\omega} \{f\in P\colon f(a_n)\in U_n\}$ is nowhere
dense (and therefore $\bigcup Q$ is nowhere dense in $\mathR^2$)
and, by (\ref{eq:CPA1}), it has $\mu$ measure at least
$1-\sum_{n<\omega} 2^{-(n+2)}=2^{-1}>0$. It is also clear that for
every $x\in A$ the set $\bigcup\{U_n\colon a_n=x\}$ is dense open
in $\real$ and it is disjoint with $[\bigcup Q]_x$. Thus
$Q\in\E(A)$.
\qed

\begin{prop}\label{propCPA}
Assume that \psmCGame\ holds, let $X$ be a Polish space,
and let $S$ be a mapping associating to every
$\bar P\in \bigcup_{\alpha<\omega_1}(\perf^*(X))^\alpha$
a cube-dense family $\E(\bar P)\subset \perf^*(X)$.
Then there exists a sequence $\la \la P_\xi,Q_\xi\ra\colon\xi<\omega_1\ra$
such that for every $\xi<\omega_1$ we have
$Q_\xi\in\perf^*(P_\xi)\cap\E(\la P_\zeta\colon\zeta<\xi\ra)$
and $X=\bigcup_{\xi<\omega_1}Q_\xi$.
\end{prop}
\pf This follows easily 
from \psmCGame.
More precisely,
it is enough to apply \psmCGame\ to the strategy
$S^*$ such that
$S^*(\la\la P_\eta,Q_\eta\ra\colon \eta<\xi\ra,P_\xi)$
is a subcube of $P_\xi$
from $S(\la P_\eta\colon \eta<\xi\ra)$.
\qed

\noindent {\sc Proof of Theorem~\ref{thmCPA}.}
First recall that \psmCGame\ implies that the cofinality
of the ideal of meager sets is equal to $\omega_1<\continuum$,
that is,
there exists an $\M_0\in[\M]^{\omega_1}$
such that every meager set
is contained in some $M\in\M_0$.
(See e.g. \cite[sec.~4]{CP88} or \cite{CPAbook}.)
Let $\B_0$ be a countable base for $\real$ and let
$\{\la M_\xi,J_\xi\ra\colon\xi<\omega_1\}$
be an enumeration of $\M_0\times \B_0$.
By simultaneous induction on $\xi<\omega_1$, using Lemma~\ref{lemCPA},
we will define
functions $S$, $Q$,  and $k$ on $(\perf^*(\C))^\xi$
such that
\begin{itemize}
\item[(i)] $S(\la P_\zeta\colon\zeta<\xi\ra)=\E(\{a_\zeta\colon\zeta<\xi\})$,
where $a_\zeta=k(\la P_\eta\colon\eta\leq\zeta\ra)\in\real$, and
$Q_\xi=Q(\la P_\zeta\colon\zeta<\xi\ra)\in\E(\{a_\zeta\colon\zeta<\xi\})$.
\item[(ii)]  $k(\la P_\zeta\colon\zeta\leq\xi\ra)$ belongs to
$J_\xi$ and the residual set
\[
\bigcap_{\zeta\leq\xi}\{z\in\real\colon
\mbox{ $(\bigcup Q_\zeta)^z$ and $(\bigcup Q_\zeta)_z$ are nowhere dense in
$\real$}\}.
\]
\item[(iii)]  $k(\la P_\zeta\colon\zeta\leq\xi\ra)$ does not belong to the
meager set
\[
M_\xi\cup
\bigcup_{\eta\leq\xi}
\mbox{$(\{(\bigcup Q_\eta)_{a_\zeta}\colon\zeta<\xi\}\cup
\{(\bigcup Q_\eta)^{a_\zeta}\colon\eta\leq\zeta<\xi\})$}.
\]
\end{itemize}
The set as in (ii) is residual by Kuratowski-Ulam
theorem, since each set $\bigcup Q_\zeta$
is nowhere dense, as $Q_\zeta$ belongs to some $\E(A)$.
In (iii) for every $\eta\leq\zeta<\xi$ the set
$(\bigcup Q_\eta)_{a_\zeta}\cup (\bigcup Q_\eta)^{a_\zeta}$
is nowhere dense
by the choice of $a_\zeta=k(\la P_\eta\colon\eta\leq\zeta\ra)$ as in (ii).
Finally, for $\zeta<\eta$ the set $(\bigcup Q_\eta)_{a_\zeta}$
is nowhere dense since, by (i),
$Q_\eta$ belongs to
$S(\la P_\zeta\colon\zeta<\eta\ra)=\E(\{a_\zeta\colon\zeta<\eta\})$.
Now, by the axiom \psmCGame\ and Proposition~\ref{propCPA},
there exists a sequence
$\la \la P_\xi,Q_\xi,a_\xi\ra\colon\xi<\omega_1\ra$
such that $\C=\bigcup_{\xi<\omega_1}Q_\xi$
and the conditions (i)--(iii) are satisfied.
We claim that $A=\{a_\xi\colon\xi<\omega_1\}$ satisfies
Theorem~\ref{thmCPA}.
Clearly, $A$ is nowhere meager since
for every non-empty open set $U\subset\real$ and every
meager set $M$ there exists a $\xi<\omega_1$ such that
$J_\xi\subset U$ and $M\subset M_\xi$.
But then $a_\xi\in (A\cap J_\xi)\setminus M_\xi\subset(A\cap U)\setminus M$,
so $A\cap U\neq M$.
To see the main part of Theorem~\ref{thmCPA} take an $f\in\C$.
Then, there exists a $\eta<\omega_1$ such that $f\in Q_\eta$.
We claim that for $Z=\{a_\beta\colon \beta<\eta\}$ we have
$f\cap(A\times A)\subset(A\times Z)\cup\id$.
Indeed, let $\eta\leq\xi<\omega_1$
and $\zeta<\omega_1$ be such that $\zeta\neq\xi$.
We need to show that $\la a_\zeta,a_\xi\ra\notin f$.
But if $\zeta<\xi$ then, by (iii),
$a_\xi$ does not belong to $[\bigcup Q_\eta]_{a_\zeta}\ni f(a_\zeta)$,
so $\la a_\zeta,a_\xi\ra\notin f$.
Similarly, if $\xi<\zeta$ then, again by (iii),
$a_\zeta$ does not belong to $[\bigcup Q_\eta]^{a_\xi}\supset f^{-1}(a_\xi)$
and once more $\la a_\zeta,a_\xi\ra\notin f$. \qed

\section{Main consistency result}\label{sec:oracle}
The main goal of this section is to prove the following theorem.
\begin{thm}\label{thConsOracle}
It is relatively consistent with ZFC that
$\continuum=\omega_2$ and
the following two conditions
hold simultaneously.
\begin{description}
\item[(A)] For every family $\{B_\xi\colon\xi<\omega_1\}$
of pairwise disjoint nowhere meager subsets of $\real^2$
there exists an increasing homeomorphism $f\colon\real\to\real$
such that $\pi[f\cap B_\xi]$ is nowhere meager for every $\xi<\omega_1$.
\item[(B)] Every nowhere meager set $B\subset \real$
contains a nowhere meager subset of cardinality $\omega_1$.
\end{description}
\end{thm}
We use Theorem~\ref{thConsOracle}, in conjunction with the following
proposition,
to deduce Theorem~\ref{thmCons1}.
\begin{prop}\label{propA} Assume that $\continuum>\omega_1$ and that
{\rm (A)} and\/ {\rm (B)} hold.
Then for every nowhere meager set $A\subset\mathR$
there is a homeomorphism $f\colon\real\to\real$
such that
$\{c\in\real\colon \pi[(f+c)\cap(A\times A)]\notin\M\}$ is  nowhere meager.
\end{prop}
\pf By condition (B) we can assume that $|A|=\omega_1<\continuum$.
Let $B$ be a Bernstein set such that
$(b+A)\cap(b'+A)=\emptyset$ for every distinct $b,b'\in B$.
Let $\{c_\xi\colon\xi<\omega_1\}$ be a nowhere meager subset of $B$
and let $B_\xi=A\times(c_\xi+A)$ for $\xi<\omega_1$
and let $f\colon\real\to\real$ be as in (A).
Then set $\{c\in\real\colon \pi[(f+c)\cap(A\times A)]\notin\M\}$
contains a nowhere meager set $\{-c_\xi\colon\xi<\omega_1\}$
since for every $\xi<\omega_1$ we have
\[
\pi[(f-c_\xi)\cap(A\times A)]=\pi[f\cap B_\xi]\notin\M
\]
finishing the proof. \qed

The proof of Theorem~\ref{thConsOracle}
is a slight modification of the proof
of the main result
(Theorem 2) from~\cite{CS2}. Also, Theorem~\ref{thConsOracle}
easily implies
\cite[Theorem 2]{CS2}.
 We will use here terminology
and notation from \cite{CS2}. In particular, according to the
machinery used in this paper, Theorem~\ref{thConsOracle} follows
in a standard way from the following lemma.
(More precisely, condition (A) is ensured by the lemma, while
(B) and $\continuum=\omega_2$ are guaranteed by the iteration
procedure.)
\begin{lem}\label{lemMain}
For every family $\B=\{B_\xi\colon\xi<\omega_1\}$
of pairwise disjoint nowhere meager subsets of $\real^2$
and for every $\omega_1$-oracle $\M$
there exists an $\M$-cc forcing notion $Q_\B$
of cardinality $\omega_1$ such that $Q_\B$ forces
\begin{quote}
there exists an increasing homeomorphism $f\colon\real\to\real$
such that $\pi[f\cap B_\xi]$ is nowhere meager for every $\xi<\omega_1$.
\end{quote}
\end{lem}
In what follows we will present the proof of Lemma~\ref{lemMain}.
Let
\[
\Gamma=\{\lambda<\omega_1\colon \lambda \mbox{ is a limit ordinal}\}.
\]
Recall that an {\em $\omega_1$-oracle}\/
is any sequence $\M=\la
M_{\delta}\colon \delta\in\Gamma\ra$, where $M_{\delta}$ is a
countable transitive model of ZFC$^{-}$ (that is, ZFC without the
power set axiom) with a property that $\delta+1\subset
M_{\delta}$, $\delta$ is countable in $M_{\delta}$, and the set
$\{\delta\in\Gamma\colon A\cap\delta\in M_{\delta}\}$ is
stationary in $\omega_1$ for every $A\subset\omega_1$.
With each $\omega_1$-oracle $\M=\la
M_\delta\colon\delta\in\Gamma\ra$ there is associated a filter
$D_{\M}$ generated by the sets $I_\M(A)=\{\delta\in\Gamma\colon
A\cap\delta\in M_\delta\}$ for $A\subset\omega_1$. It is proved
in~\cite[Claim~1.4]{Sh:f} that $D_{\M}$ is a proper normal filter
containing every closed unbounded subset of $\Gamma$.
We will also need the following fact which, for our purposes, can
be viewed as a definition of $\M$-cc property.
 \fact{f1}{{\rm (\cite[Fact~4]{CS2})}
 Let $P$ be a
forcing notion of cardinality $\leq\omega_1$, $e\colon
P\to\omega_1$ be one-to-one, and $\M=\la
M_\delta\colon\delta\in\Gamma\ra$ be an $\omega_1$-oracle. If
there exists a $C\in D_\M$ such that for every
$\delta\in\Gamma\cap C$
\begin{quote}
$e^{-1}(E)$ is predense in $P$ for every set $E\in
M_\delta\cap\P(\delta)$, for which $e^{-1}(E)$ is predense in
$e^{-1}(\{\gamma\colon \gamma<\delta\})$,
\end{quote}
then $P$ has the $\M$-cc property. \qed}

 Let $\K$ be the family of
all sequences $\bar h=\la h_\xi\colon\xi\in\Gamma\ra$ such that
each $h_\xi$ is a function from a countable set
$D_\xi\subset\real$ onto $R_\xi\subset \real$ such that $h_\xi$ is
dense in $\real^2$ and
\[
D_\xi\cap D_\eta=R_\xi\cap R_\eta=\emptyset
\ \mbox{ for every distinct }\ \xi,\eta\in\Gamma.
\]
For each $\bar h\in\K$ we will define a forcing notion
$Q_{\bar h}$.
Forcing $Q_\B$ satisfying Lemma~\ref{lemMain}
will be chosen as $Q_{\bar h}$ for some $\bar h\in\K$.
So let $\H$ be the family of all strictly
increasing functions from finite subsets of $\real$
into $\real$ and
fix an $\bar h\in\K$. Then $Q_{\bar h}$ is defined as
\[
Q_{\bar h}=\left\{h\in\H\colon h\subset\bigcup_{\xi\in\Gamma}h_\xi
\ \&\ |h\cap h_\xi|\leq 1\mbox{ for every }\xi\in\Gamma\right\}
\]
and is ordered by the reverse inclusion.
In what follows we will use the following basic property of $Q_{\bar h}$.
\fact{fact2}{Let $\bar h=\la h_\xi\colon\xi\in\Gamma\ra\in\K$ and
$f=\bigcup  H$, where
$H$ is a $V$-generic filter over $Q_{\bar h}$.
Then $f$ is a strictly increasing function from a dense subset
$D$ of $\real$ onto a dense subset of $\real$.
In particular, $f$ can be uniquely extended to an increasing homeomorphism
$\tilde f$ of $\real$.
}
\pf Clearly $f$ is a strictly increasing function from a subset
$D$ of $\real$ onto a subset $R$ of $\real$.
Thus, it is enough to show that $D$ and $R$ are dense in $\real$.
So, let $U\neq\emptyset$ be open in $\real$
and notice that the set
\[
D=\{h\in  Q_{\bar h}\colon\dom(h)\cap U\neq\emptyset\}
\]
is dense in $Q_{\bar h}$.
Indeed, if $h_0\in  Q_{\bar h}$ is such that $\dom(h_0)\cap U=\emptyset$
then we can find $\xi\in\Gamma$ such that
$h\cap h_\xi=\emptyset$.
Since graph of $h_\xi$ is dense in $\real^2$ we can find
$\la x,y\ra\in h_\xi$ such that $x\in U$ and
$h=h_0\cup\{\la x,y\ra\}$ is strictly increasing.
Then $h\in D$ extends $h_0$.
Similarly we can proved that the set
\[
\{h\in  Q_{\bar h}\colon\range(h)\cap U\neq\emptyset\}
\]
is dense in $Q_{\bar h}$.
The rest follows from the genericity of $H$.
\qed

Now let $\B=\{B_\xi\colon\xi<\omega_1\}$ be as in the lemma
and fix an $\omega_1$-oracle
$\M=\la M_\delta\colon\delta\in\Gamma\ra$.
By Fact~\ref{fact2} in order to prove Lemma~\ref{lemMain}
it is enough to find an
$\bar h=\la h_\xi\colon\xi\in\Gamma\ra\in\K$ such that
\begin{equation}\label{eqN2}
\mbox{$Q_\B=Q_{\bar h}$ is $\M$-cc}
\end{equation}
and $Q_{\bar h}$ forces that, in $V[H]$,
\begin{equation}\label{eqN3}
\mbox{the sets
$\pi(f\cap B_\xi)$ is nowhere
meager for every $\xi<\omega_1$,}
\end{equation}
where function $f$ is defined as in Fact~\ref{fact2}.
To define $\bar h$ we will construct
a sequence
$\la \la x_\alpha,y_\alpha\ra\in \real^2\colon
\alpha<\omega_1\ra$
aiming at
$h_\xi=\{\la x_{\xi+n},y_{\xi+n}\ra\colon n<\omega\}$,
where $\xi\in\Gamma$.
So, let $\U\not\ni\emptyset$ be a standard countable basis for $\real$
and for every $\xi\in\Gamma$ let
$\la\la U^\xi_n,V^\xi_n,\zeta^\xi_n\ra\colon n<\omega\ra$
be a fixed enumeration of $\U\times\U\times\xi$.
Points $\la x_{\xi+n},y_{\xi+n}\ra$
are chosen inductively in such a way that
\begin{description}
\item{(i)} $\la x_{\xi+n},y_{\xi+n}\ra$
      is a Cohen real over
      $M_\delta[\la \la x_\alpha,y_\alpha\ra\colon \alpha<\xi+n\ra]$
      for every $\delta\leq\xi$, $\delta\in\Gamma$,
      that is, $\la x_{\xi+n},y_{\xi+n}\ra$ is outside all
      meager subsets of $\real^2$ which are
      coded in
      $M_\delta[\la \la x_\alpha,y_\alpha\ra\colon \alpha<\xi+n\ra]$;
\item{(ii)}
      $\la x_{\xi+n},y_{\xi+n}\ra\in(U^\xi_n\times V^\xi_n)\cap
B_{\zeta^\xi_n}$.
\end{description}
The choice of $\la x_{\xi+n},y_{\xi+n}\ra$
is possible since the sets $(U^\xi_n\times V^\xi_n)\cap B_{\zeta^\xi_n}$
are non-meager and
each time we need to avoid only countably many meager sets.
Condition (ii) guarantees that the graph of each of $h_\xi$
will be dense in $\real^2$.
Note also that if $\Gamma\ni\delta\leq \alpha_0<\cdots<\alpha_{k-1}$,
where $k<\omega$, then (by the product lemma in $M_\delta$)
\begin{equation}\label{PLem}
\mbox{$\la\la x_{\alpha_i},y_{\alpha_i}\ra\colon i<k\ra$
is an $M_\delta$-generic Cohen real in
$\left(\real^2\right)^k$.}
\end{equation}
For $h\in\H$ and $0<k<\omega$
let $U(h,k)$ stand for the set of all sequences
$\la \la a_i,b_i\ra\notin h\colon i<k\ra\in (\real^2)^k$
such that $h\cup\{\la a_i,b_i\ra\colon i<k\}\in\H$.
Clearly $U(h,k)$ is an open subset of $(\real^2)^k$.
In fact, it can be easily proved that
if $h=\{\la x_j,y_j\ra\colon 0<j<m\}$, where
$x_1<\cdots<x_{m-1}$, then
$\la \la a_i,b_i\ra\colon i<k\ra$ belongs to $U(h,k)$
if and only if $\{\la a_i,b_i\ra\colon i<k\}\in\H$ is disjoint with $h$ and
\begin{center}
$\{\la a_i,b_i\ra\colon i<k\}\subset \bigcup_{j< m} (x_j,x_{j+1})\times
(y_j,y_{j+1})$,
\end{center}
where $x_0=y_0=-\infty$ and $x_m=y_m=\infty$. 
In particular, if
$0<j_0<\cdots<j_{k-1}<m$ and sets $W_i\ni \la x_{j_i},y_{j_i}\ra$, $i<k$,
are
open in $\real^2$
then there are non-empty open sets $V_i\subset W_i$ such that
\begin{equation}\label{con44}
\prod_{i<k} V_i\subset U(h,k).
\end{equation}
For $\delta\in\Gamma$ let
$(Q_{\bar h})^\delta=\left\{h\in Q_{\bar h}
\colon h\subset\bigcup_{\zeta<\delta}h_\zeta\right\}$.
\fact{FACT}{Let $\delta\in\Gamma$
and let $E\in M_\delta$ be
a predense subset of $(Q_{\bar h})^\delta$. Then
for every $k<\omega$ and
$h\in (Q_{\bar h})^\delta$
the open set
\begin{equation}\label{CCC}
B_{h}^k=\bigcup\left\{
U(g,k)\colon
\mbox{ $g\in (Q_{\bar h})^\delta$ extends $h$ and some $h_0\in E$}\right\}
\end{equation}
is dense in $U(h,k)$.
}
\pf Let
$\la \la a_i,b_i\ra\colon i<k\ra\in U(h,k)\subset(\real^2)^k$
and $W$ be an open subset of $U(h,k)$
containing $\la \la a_i,b_i\ra\colon i<k\ra$.
We need to show that $W$ intersects
$U(g,k)$ for some
$g\in Q_{\bar h}$ extending $h$ and an $h_0\in E$.
Decreasing $W$, if necessary,
we can assume that it is of the form
$\prod_{i<k}W_i$.
Since $h_1=h\cup\{\la a_i,b_i\ra\colon i<k\}\in\H$, by
(\ref{con44}), there are open sets $V_i\subset W_i$ for which
$\prod_{i<k} V_i\subset U(h,k)\cap W$. In particular, for any
choice of points $\la c_i,d_i\ra\in V_i$ we have $h_1\cup\{\la
c_i,d_i\ra\colon i<k\}\in\H$. Now, since all functions $h_\xi$ are
dense in $\real^2$ we can choose points $\la c_i,d_i\ra$ from
distinct functions $h_\xi$ in such a way that $h_2=h\cup\{\la
c_i,d_i\ra\colon i<k\}\in(Q_{\bar
h})^\delta$. Now, since $E$ is
predense in $(Q_{\bar h})^\delta$, there exists a $g\in (Q_{\bar
h})^\delta$ extending
$h_2\leq h$ and some $h_0\in E$. But $\{\la c_i,d_i\ra
\colon i<k\}\subset g$ so, by (\ref{con44}), there are
non-empty open sets $U_i\subset V_i$ for which $\prod_{i<k}
U_i\subset U(g,k)\cap W$. \qed

Now we are ready to prove (\ref{eqN2}), that is, that
$Q_{\bar h}$ is $\M$-cc. So, fix a bijection
$e\colon Q_{\bar h}\to\omega_1$
and let
\[
C=\left\{\delta\in\Gamma\colon
(Q_{\bar h})^\delta=
e^{-1}(\delta)\in M_\delta\right\}.
\]
Then $C\in D_{\mathcal M}$. (See e.g. \cite[Claim~1.4(4)]{Sh:f}.)
Take a $\delta\in C$ and fix an $E\subset \delta$, $E\in M_\delta$,
for which $e^{-1}(E)$ is predense in $(Q_{\bar h})^\delta$. By
Fact~\ref{f1} it is enough to show that
\begin{equation*}\label{AAA}
\mbox{$e^{-1}(E)$ is predense in $Q_{\bar h}$.}
\end{equation*}
So, take $h_0$ from $Q_{\bar h}$,
put $h=h_0\restriction\bigcup_{\eta<\delta}D_\eta$ and
$h_1=h_0\setminus h$,
and notice that the condition
$h$
belongs to $(Q_{\bar h})^\delta$.
Assume that $h_1=\{\la x_i,y_i\ra\colon i<k\}$,
where $x_0<\cdots<x_{k-1}$.
So, $\la\la x_i,y_i\ra\colon i<k\ra\in U(h,k)$.
By Fact~\ref{FACT} the set $U(h,k)\setminus B^k_h$
is nowhere dense and belongs to $M_\delta$ (as it is defined from
$(Q_{\bar h})^\delta\in M_\delta$).
Hence, by (\ref{PLem}), $\la\la x_i,y_i\ra\colon i<k\ra$
cannot belong to this set,
so $\la\la x_i,y_i\ra\colon i<k\ra\in B^k_h$.
In particular,
there is a $g\in (Q_{\bar h})^\delta$ extending $h$ and some $h_0\in e^{-1}(E)$
such that $\la\la x_i,y_i\ra\colon i<k\ra\in U(g,k)$.
But then $g\cup h_1$ belongs to $Q_{\bar h}$ and extends $h$ and $h_0$.
This finishes the proof of~(\ref{eqN2}).
The proof of (\ref{eqN3}) is similar. So, fix a $\zeta<\omega_1$.
We will prove that $\pi(f\cap B_\zeta)$ is nowhere meager in
$\real$.
By way of contradiction assume that $\pi(f\cap B_\zeta)$ is not
nowhere meager in~$\real$. So there exists
a $U^*\in \U$ such
that $\pi(f\cap B_\zeta)$ is meager in $U^*$.
Let a condition
$h^*\in
Q_{\bar{h}}$ and $Q_{\bar h}$-names $\dot{U}_m$, for $m<\omega$,
be such that
$$
h^*\forces_{Q_{\bar{h}}}\mbox{each $\dot U_m$ is an open dense subset of
$U^*$ and }
\pi(f\cap B_\zeta)\cap\bigcap_{m<\omega}
\dot{U}_m=\emptyset.
$$
For each $m<\omega$, since $h^*$ forces that
$\dot U_m$ is an open dense subset of $U^*$,
for every subset
$U\in\U$ of $U^*$ there is a subset $V\in\U$ of $U$
and a  maximal antichain
$\la h^m_{V,k}\colon k<\kappa_V^m\rangle$ in $Q_{\bar{h}}$
such that each $h^m_{V,k}$ forces that
$V\subset\dot U_m$.
Note that each of these antichains must be  countable, since
the forcing notion $Q_{\bar h}$ is ${\mathcal M}$-cc and
therefore ccc.
Combining all these antichains we
find a $\V\subset\U$ and
a sequence
$\la h^m_{V,k}\in Q_{\bar{h}}\colon m<\omega, V\in\V, k<\kappa_V^m\ra$
such that
\begin{itemize}
\item $\kappa_V^m\leq\omega$,
\item
$h^m_{V,k}\forces_{Q_{\bar{h}}} V\subseteq \dot{U}_m$,
\item for every $m<\omega$,  $h\in Q_{\bar{h}}$ extending $h^*$,
and subset $U\in\U$ of $U^*$
there is a subset $V\in\V$ of $U$
and a $k<\kappa_V^m$ such that the conditions $h$ and $h^m_{V,k}$ are
compatible.
\end{itemize}
Note that for sufficiently large $\delta\in \Gamma$ we have
$h^m_{V,k}\in (Q_{\bar{h}})^\delta$ for all $m<\omega$, $V\in \V$,
and $k<\kappa_V^m$.
Now, by the definition of $\omega_1$-oracle, the set
$B_0$ of all $\delta\in\Gamma$ for which
\[
\la h^m_{V,k}\in Q_{\bar{h}}\colon m<\omega, V\in\V, k<\kappa_V^m\ra
\in M_\delta\ \
\mbox{ and } \ \
(Q_{\bar h})^{\delta}\in M_\delta
\]
is stationary in $\omega_1$. Thus, choose a $\delta>\zeta$ from $B_0$
and let $U,W\in\U$ be such that $U\subset U^*$ and
$h^*\cup\{\la x,y\ra\}\in\H$ for every $\la x,y\ra\in U\times W$.
Using clause (ii) of the choice of
$x_\alpha$'s we may find an $n<\omega$ such that
$\la x_{\delta+n},y_{\delta+n}\ra\in(U\times W)\cap B_\zeta$.
Then $h_0=h^*\cup\{\la x_{\delta+n},y_{\delta+n}\ra\}\in Q_{\bar{h}}$
extends $h^*$ and
$h_0\forces``x_{\delta+n}\in U^*\cap\pi(f\cap B_\zeta)$''
since $\langle x_{\delta+n},y_{\delta+n}\rangle\in f$.
We will show that
\[
h_0\forces x_{\delta+n}\in \bigcap_{m<\omega}\dot{U}_m,
\]
which contradicts the choice of $h^*$.
So, assume that this is not the case. Then there exist an $i<\omega$ and
an $h_1\in Q_{\bar h}$ stronger than $h_0$ such  that
$h_1\forces``x_{\delta+n}\notin\dot{U}_i$.''
Let us define $h=h_1\restriction\{x_\alpha\colon \alpha<\delta\}\in
(Q_{\bar h})^\delta$
and $h_1\setminus h=\{\la a_l,b_l\ra\colon l<m\}$,
where $a_0<\cdots<a_{m-1}$. Let $j<m$ be such that
$\la x_{\delta+n},y_{\delta+n}\ra=\la a_j,b_j\ra$.
Consider the set
$Z$ of all
$\la\la z_l,z'_l\ra\colon l<m\ra\in(\real^2)^m$
for which
\begin{itemize}
\item there exist $V\in \V$, $k<\kappa_V^i$, and
$g\in (Q_{\bar{h}})^\delta$ such that $z_j\in V$,
$g$ extends $h$ and $h^i_{V,k}$, and
$\la\la z_l,z'_l\ra\colon l<m\ra\in U(g,m)$.
\end{itemize}
\noindent {\bf Claim} {\em The set $Z$ belongs to the model $M_\delta$
and it is an open dense subset of
$K=\{\la\la z_l,z'_l\ra\colon l<m\ra\in U(h,m)\colon z_j\in U^* \}$.}
\medskip
\noindent {\sc Proof.} It should be clear that $Z$ is (coded) in
$M_\delta$. (Remember the choice of~$\delta$.)
It is also obvious that $Z$ is open.
To show that it is dense in  $U(h,m)$
we proceed like in the proof of Fact~\ref{FACT}.
Let $\la \la c_l,d_l\ra\colon l<m\ra\in K$
and $W$ be an open subset of $K$
containing $\la \la c_l,d_l\ra\colon l<m\ra$.
We need to show that $W\cap Z\neq\emptyset$.
As in the proof of Fact~\ref{FACT} we can find an open set
$\prod_{l<m} (V_l\times U_l)\subset K\cap W$
and points
$\la c_l,d_l\ra\in V_l\times U_l$ such that
$h_2=h\cup\{\la c_l,d_l\ra\colon l<m\}\in(Q_{\bar h})^\delta$.
Since $h_2\in (Q_{\bar h})^\delta$ extends $h^*$
and $V_j$ is an open subset of $U^*$,
there is a subset $V\in\V$ of $V_j$
and a $k<\kappa_V^i$ such that the conditions $h_2$ and $h^i_{V,k}$ are
compatible.
Let $g\in Q_{\bar{h}}$ extend $h_2$ and $h^i_{V,k}$.
Since $h_2,h^i_{V,k}\in (Q_{\bar{h}})^\delta$ we can assume that $g\in
(Q_{\bar{h}})^\delta$.
But $\{\la c_l,d_l\ra\colon l<m\}\subset g$ so,
by (\ref{con44}), there are non-empty open sets
$U'_l\subset U_l$ and $V'_l\subset V_l$
for which
$\prod_{l<m} (U'_i\times V'_i)\subset U(g,m)\cap\prod_{l<m} (V_l\times U_l)$.
Thus, $\emptyset\neq\prod_{l<m}(U'_i\times V'_i)\subset Z\cap W$. This
completes the proof of Claim.
\qed

Now, $\la\la a_l,b_l\ra\colon l<m\ra$ belongs to $K$.
Since, by Claim, $K\setminus Z\in M_\delta$ is nowhere dense,
by (\ref{PLem}) we conclude that this point does not belong to $K\setminus Z$.
So, $\la\la a_l,b_l\ra\colon l<m\ra\in Z$. But this means that there
exist
$g\in (Q_{\bar{h}})^\delta$ and a $V\in\V$ such that:
\begin{itemize}
\item $g\leq h$, $g\forces$``$V\subseteq \dot{U_i}$'', and
\item $\langle\langle a_l,b_l\rangle\colon l<m\rangle\in U(h,m)$,
and
$x_{\delta+n}=a_j\in V$.
\end{itemize}
But then $h_3=g\cup\{\la a_l,b_l\ra\colon l<m\}$
belongs to $Q_{\bar h}$ and extends both $g$ and $h_1$. So,
$h_3$
forces that $x_{\delta+n}=a_j\in V\subseteq\dot{U}_i$,
contradicting our
assumption that $h_1\forces``x_{\delta+n}\notin\dot U_i$.''
This finishes the proof of (\ref{eqN3})
and of Lemma~\ref{lemMain}.
\qed

\section{A conjecture}
We believe that Corollary~\ref{cor2}
cannot be proved in ZFC.
We believe the following is consistent with ZFC.
\begin{description}
\item[(C)] For every nowhere meager set $A\subset[0,1]$
there exists a continuous Peano-like function
$p$ from $[0,1]$ onto $[0,1]^2$
such that for every $x\in [0,1]$ the set
$p[A]\cap(\{x\}\times[0,1])$ is non-meager in $\{x\}\times[0,1]$.
\end{description}
\begin{prop}\label{propC} If (C) holds then
for every nowhere meager set $A\subset\mathR$
there is an $f\in\C_0$
such that $\pi[(f+c)\cap(A\times A)]\notin\M$
for every $c\in\real$.
\end{prop}
\pf Let $A\subset\real$ be nowhere meager,
$f$ be as in the condition (C), and let $f_0=\pi\circ p\colon[0,1]\to[0,1]$.
Since there exists a meager set
$M\subset[0,1]$ such that $p\restriction[0,1]\setminus M$
is a homeomorphism between $[0,1]\setminus M$ and $[0,1]^2\setminus p[M]$,
for every $c\in[0,1]$ we have
\[
\pi[(f_0+c)\cap(A\times A)]=
\pi[f_0\cap(A\times(A-c))]=
A\cap(f_0)^{-1}(A-c)\notin\M
\]
since
$A\cap(f_0)^{-1}(A-c)=
A\cap p^{-1}(  \pi^{-1}(A-c)  )=
A\cap p^{-1}( (A-c)\times[0,1] )$
and this last set is equal, modulo $\M$,
to $p^{-1}(p[A] \cap((A-c)\times[0,1]))$
while, by (C), the set $p[A] \cap((A-c)\times[0,1])$ is not meager
(by Kuratowski-Ulam theorem).
To get function from $\real$ to $\real$ glue countable many
shifts of $f_0$. \qed

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\end{document}
