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\title{Compositions of Sierpi\'{n}ski-Zygmund functions from the left}

%\MathReviews{Primary:  26A15; Secondary: 03E75, 54A25.}
%\keywords{cardinal functions; extendable, Darboux, almost continuous
%and peripherally continuous functions; functions with perfect road. }


\author{{\small Francis Jordan}%
\thanks{AMS classification numbers: Primary 26A15;  Secondary 54A25
\endgraf  Key words and phrases: Sierpi\'{n}ski-Zygmund functions, cardinal functions.
\endgraf The author wishes to thank K. Ciesielski, M. Morayne, and the referees for their comments. \endgraf},
\small  Department of Mathematics, University of Louisville,\\
Louisville, KY 40292}

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\newcommand{\cof}{\operatorname{cf}}
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\newcommand{\meager}{{\cal M}}
\newcommand{\cuum}{{\frak c}}
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\newcommand{\G}{{\cal G}}
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\newcommand{\conn}{{\operatorname {Con}}}
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\begin{document}\maketitle
\begin{abstract}
A cardinal related to compositions of Sierpi\'{n}ski-Zygmund functions will be
considered. We answer a question of K. Ciesielski and T. Natkaniec.
\end{abstract}
\section{Preliminaries}

In what follows we will use the following terminology and notation.  Any
notation not specifically defined can be found in \cite{cibook}.  Functions will be
identified with their graphs.  The set of all functions from a set $X$ into a set $Y$
will be denoted by
$Y^{X}$.  Given sets
$X,Y,W$ and $f\in W^Y$ and $g\in Y^X$ we denote their composition by $f\circ g$.
We will denote the collection of meager sets in $\real$ by $\meager$.

The symbol $|X|$ will denote the
cardinality of the set $X$.  The successor of a cardinal $\kappa$ will be
denoted by $\kappa^{+}$.
For a successor cardinal $\kappa=\lambda^{+}$ its predecessor will be denoted by
$\pred(\kappa)=\lambda$.   We denote by $[X]^{<\kappa}$, $[X]^{\kappa}$, and
$[X]^{\leq\kappa}$ the sets of all subsets of $X$ of cardinality less than $\kappa$,
equal to $\kappa$, and less than or equal to $\kappa$, respectively.

The cardinality of the real numbers $\real$
will be denoted by $\cuum$.  We let $\irr$ stand for the irrational numbers.  Given a
cardinal number
$\kappa$ we let
$\cof(\kappa)$ denote the cofinality of $\kappa$.  We say a cardinal $\kappa$ is regular
provided that $\cof(\kappa)=\kappa$.
For functions $f,g\in Y^{X}$ let
$[f=g]$ denote the set $\{x\in X\colon f(x)=g(x)\}$.  We define $[f<g]$ and
$[f\leq g]$ in a similar way when $\leq$ and $<$ are defined for $Y$.



We also will consider the following cardinal invariant related to $\cuum$.
\begin{description}
\item[ ]
$\bd_{\cuum}
=\min\{|F|\colon F\subseteq \cuum^{\cuum}\ \&\
(\forall g \in\cuum^{\cuum})(\exists f\in F)(|[f\geq g]|=\cuum\}$.
\end{description}
The number $\bd_{\cuum}$ is defined similarly to the bounding
number $\bd_{\omega}$ which has been  heavily studied, e.g.
\cite{BJ}.  We mention some well known facts about $\bd_{\cuum}$.
If $\cuum$ is regular, then $\cuum<\bd_{\cuum}\leq 2^{\cuum}$. If
$\cuum$ is a successor, we have $\bd_{\cuum}=\same_{\cuum}$ (see
\cite{jord1} for definition and proof).  If $\cuum=\omega_1$, then
$\bd_{\cuum}$ may be any regular cardinal between $\cuum$ and
$2^{\cuum}$ (Cor. 2.6. of \cite{CN}).


\section{Introduction}
We say $f\in \real^{\real}$ is a Sierpi\'{n}ski-Zygmund function provided that
the restriction $f|_X$ is not continuous for any set $X\subseteq\real$ of
cardinality $\cuum$.  We let ${\cal C}_{G_{\delta}}$ denote the real valued continuous
functions which are defined on some $G_{\delta}$-set of $\real$.
Recall the following result of W.~Sierpi\'{n}ski and A.~Zygmund \cite{sezy}:
\prop{prop:50}{$f\in\real^{\real}$ is in $\sz$ if and only if $|[f=h]|<\cuum$
for every continuous function $h$ defined on a $G_{\delta}$ set of
cardinality~$\cuum$.}
We will use the following statement which follows from the fact that meager sets form
an ideal which has the countable chain condition with respect to closed sets.
\prop{prop:1}{If $A\subseteq\real$ and $f\colon A\to\real$ is continuous, then
there are at most countably many $r\in\real$ such that
$f^{-1}(r)\notin\meager$.}


We will be interested in resolving some problems in
\cite{CN} about Sierpi\'{n}ski-Zygmund functions.  To state the problems we will need some
definitions.  For sets $X$ and $Y$ define ${\cal
R}=\{f\in \real^{\real}\colon |f^{-1}(y)|<\cuum \text{ for every
}y\in\real\}$.  In \cite[Cor. 4.26]{CN} it is shown that if $\real$ is not the union of
less than $\cuum$-many meager sets and $\cuum$ is regular, then $f\in {\cal R}$ if and only if there exist
$h,g\in\sz$ such that
$h\circ g=f$.

The following cardinal is defined in~\cite{CN}
\[
\cmp_{l}(\sz)\!=\!\min
\{|F|\colon F\subseteq {\cal R}\ \&\
(\forall h\in\sz)(\exists f\in F)(\forall g\in\sz)(f\neq h\circ
g)\}.
\]
If the set we take the minimum of in the definition is empty, then we let
$\cmp_{l}(\sz)=(2^{\cuum})^+$.


The following proposition is established in \cite{CN}:

\prop{prop:10}{If $\cuum$ is a regular cardinal and $\real$ is not the union of less
$\cuum$-many meager sets, then $\cuum<\cmp_{l}(\sz)$.}

The following problem about the cardinal
$\cmp_{l}(\sz)$ is posed in \cite{CN}.

\prob{prob:1}{Can it be shown that $\cmp_{l}(\sz)\leq 2^{\cuum}$ in ZFC?
What if $\cuum=\omega_1$?}

We answer this question with two theorems.

\thm{thm:1}{If $\cuum=\omega_1$, then $\cmp_{l}(\sz)=\bd_{\cuum}\leq 2^{\cuum}$.}

\thm{thm:2}{If $\cuum$ is a regular limit cardinal and $\real$ is
not the union of less than $\cuum$-many meager sets, then there
exists an $h\in\sz$ such that \[{\cal R}=\{h\circ m\colon
m\in\sz\}.\] In particular, $\cmp_{l}(\sz)=(2^{\cuum})^{+}$.}

In the case when $\cuum$ is singular we can get an even stronger conclusion.

\thm{thm:2a}{If $\cuum$ is a singular cardinal and $\real$ is not
the union of less than $\cuum$-many meager sets, then there is an
$h\in\sz$ such that
\[\real^{\real}=\{h\circ m\colon m\in\sz\}.\]}


\section{Proof of Theorem~\ref{thm:1}}
We first prove a lemma.

\lem{lem:1}{There is a countable collection $\{j_{n}\colon
n\in\omega\}$ of continuous functions $j_n\colon\irr\to\real$ such
that for any countable set $A\subseteq\real$ there is an
$x_A\in\irr$ such that $A=\{j_n(x_A)\colon n\in\omega\}$.} \proof
Since $\real^{\omega}$ is a separable complete metric space, there
is a continuous onto function $J\colon\irr\to\real^{\omega}$ (Thm.
7.9 of \cite{Kech}). For each $n\in\omega$ let
$\pi_{n}\colon\real^{\omega}\to\real$ be the projection onto the
$n^{th}$ coordinate.  It is easily verified that $\{\pi_{n}\circ
J\colon n\in\omega\}$ has the desired property, since every
countable set in $\real$ can be thought of as sequence of real
numbers.\qed

{\sc Proof of Theorem~\ref{thm:1}}

We show that $\cmp_l(\sz)\leq\bd_{\omega_1}$.
It is enough for us to find $F\in [{\cal R}]^{\leq\bd_{\omega_1}}$
such that
\begin{equation}\label{eq:good0}
(\forall h\in\sz)(\exists f\in F)(\forall g\in\sz)(f\neq h\circ g).
\end{equation}
Let $\{r_{\alpha}\colon\alpha\in\omega_1\}$ be a well ordering of $\real$.  Let
$K\in [\omega_1^{\omega_1}]^{\bd_{\omega_1}}$ be a family of injections which witness the
definition of $\bd_{\omega_1}$, i.e.,
\[(\forall f\in\omega_1^{\omega_1})(\exists k\in K)(|[f\leq k]|=\omega_1).\]

Let $\{j_n\colon n\in\omega\}$ be
the collection of functions from Lemma~\ref{lem:1}.    For each $\alpha\in\omega_1$
pick $x_{\alpha}\in\real$ such that
$\{r_{\xi}\colon\xi\leq\alpha\}=\{j_n(x_{\alpha})\colon n\in\omega\}$.  For
each $k\in K$ define $f_{k}\in {\cal R}$ so that
$f_k(x_{k(\alpha)})=r_{\alpha}$ for every $\alpha\in\omega_1$ and arbitrarily
elsewhere.  Let $F=\{f_k\colon k\in K\}$ and notice that $|F|\leq\bd_{\omega_1}$.

We show that $F$ satisfies (\ref{eq:good0}).  Let $h\in\sz$.  Since $h\in\sz$,
we may define  a function $\rho_h\colon\omega_1\to\omega_1$ such that for every
$\alpha\in\omega_1$ we have
$h^{-1}(r_{\alpha})\subseteq\{r_{\xi}\colon\xi\in\rho_h(\alpha)\}$.  Pick $k\in K$ such
that
$|[\rho_h\leq k]|=\omega_1$.

We show that $f_k$ has the property that $(\forall g\in\sz)(f_k\neq h\circ g)$.
Suppose $g\in\real^{\real}$ is such that
$f_k=h\circ g$.
For every $\alpha\in\omega_1$ we must have
\begin{equation}\label{eq:stv1}
g(x_{k(\alpha)})\in h^{-1}(f_k(x_{k(\alpha)}))=h^{-1}(r_{\alpha})
\subseteq\{r_{\xi}\colon\xi\leq\rho_h(\alpha)\}.
\end{equation}
We also have that there are $\omega_1$-many values of $\alpha$
such that
\begin{equation}\label{eq:stv2}
\{r_{\xi}\colon\xi\leq\rho_h(\alpha)\}\subseteq\{r_{\xi}\colon\xi\leq
k(\alpha)\}=\{j_n(x_{k(\alpha)})\colon n\in\omega\}.
\end{equation}
It follows from (\ref{eq:stv1}) and (\ref{eq:stv2}) that for $\omega_1$-many
$\alpha\in\omega_1$ there is an
$n_{\alpha}\in\omega$  such that
$g(x_{k(\alpha)})=j_{n_{\alpha}}(x_{k(\alpha)})$.  So,
$|[g=j_n]|=\omega_1$ for some $n\in\omega$.   Therefore,
$g\notin\sz$ so (\ref{eq:good0}) holds.

\medskip

We now show that $\cmp_{l}(\sz)\geq\bd_{\omega_1}$.  Let
$F\in[{\cal R}]^{<\bd_{\omega_1}}$.  Let
$\{g_{\beta}\colon\beta\in\omega_1\}$ be an enumeration of ${\cal
C}_{G_{\delta}}$ and $\{r_{\beta}\colon\beta\in\omega_1\}$ be an
enumeration of $\real$.  For the remainder of the proof we write
$r\leq^* s$ if and only if $r$ appears before $s$ in the
enumeration of $\real$.  Since $\omega_1$ is regular, for any
$f\in F$ and $\alpha\in\omega_1$ the set
\[\bigcup_{\xi\leq\alpha}g_{\xi}[f^{-1}(r_{\alpha})]\]
is $\leq^*$-bounded above by some $r^{f,\alpha}\in\real$.
For each
$f\in F$ let $f^*(r_{\alpha})=r^{f,\alpha}$ for each $\alpha\in\omega_1$.  Since
$|F|<\bd_{\omega_1}$, there is a $m\in\real^{\real}$ such that for every $f\in F$ we
have $|[m\leq^* f^*]|<\omega_1$.

Define a sequence $\{y_{\alpha}\}_{\alpha\in\omega_1}$ inductively
so that for every $\alpha\in\omega_1$
\[y_{\alpha}\in\real\setminus(K_1\cup K_2)\] where
\[K_1=\{y_{\xi}\colon\xi\in\alpha\}\cup\bigcup\left\{
g_{\xi}^{-1}(r_{\alpha})\colon\xi\leq\alpha\ \&\
g_{\xi}^{-1}(r_{\alpha}) \in\meager\right\}\] and
\[K_2=\bigcup\left\{g_{\xi}[f^{-1}(r_{\alpha})]\colon
f^*(r_{\alpha})\leq^* m(r_{\alpha})\ \&\ \xi\leq\alpha\right\}.\]
We show why such choices are possible. Clearly the set $K_1$ is
the union of less than $\omega_1$-many meager sets.  To see that
$|K_2|<\omega_1$, notice that for each $f\in F$ such that
$f^*(r_{\alpha})\leq^* m(r_{\alpha})$ we have
\[\bigcup_{\xi\leq\alpha}g_{\xi}[f^{-1}(r_{\alpha})]\subseteq
\{r\colon r\leq^* r^{f,\alpha}\}\subseteq \{r\colon r\leq^*
m(r_{\alpha})\}.\] Since both of the sets to be avoided are unions
of less than $\omega_1$-many meager sets we may make the desired
choice.

Let $Y=\{y_{\alpha}\colon\alpha\in\omega_1\}$.  Now define
$h^*\in\real^{Y}$ so that $h^*(y_{\alpha})=r_{\alpha}$.  We claim
that $h^*\in\sz$. Fix $\beta\in\omega_1$ and suppose
$h^*(y_{\alpha})=g_{\beta}(y_{\alpha})$.  Then
$g_{\beta}(y_{\alpha})=h^*(y_{\alpha})=r_{\alpha}$, so
$y_{\alpha}\in g_{\beta}^{-1}(r_{\alpha})$.  By our choice of
$y_{\alpha}$, we have $\alpha<\beta$ or
$g_{\beta}^{-1}(r_{\alpha})\notin\meager$. By
Proposition~\ref{prop:1},
$g_{\beta}^{-1}(r_{\alpha})\notin\meager$ for at most countably
many $r\in\real$, so we have $|[h^*=g_{\beta}]|<\omega_1$.  Thus,
$h^*\in\sz$.  Let $h\in\sz$ be an extension of $h^*$ to a function
defined on all of $\real$.

For each $f\in F$ define $e_f\in\real^{\real}$ so that
\[e_f(r)=y_{\alpha}\text{ if and only if } r\in f^{-1}(r_{\alpha}).\]
Notice that $f=h\circ e_f$ for every $f\in F$, since for every
$r\in\real$ there is an $\alpha$ such that $r\in f^{-1}(r_{\alpha})$ but then
$f(r)=r_{\alpha}= h(y_{\alpha})=h(e_f(r))$.  Since $f\in{\cal R}$, we also
have that $e_f\in{\cal R}$.

We now show that $e_f\in\sz$ for each $f\in F$.  Fix $\beta\in\omega_1$ and suppose
$e_f(r_{\alpha})=g_{\beta}(r_{\alpha})$.  Let $y_{\xi}=e_{f}(r_{\alpha})$.  By
definition of $e_f$ we have $r_{\alpha}\in f^{-1}(r_{\xi})$.  Now
$y_{\xi}=g_{\beta}(r_{\alpha})\in g_{\beta}[f^{-1}(r_{\xi})]$.  So,
either $\xi<\beta$ or $m(r_{\xi})\leq^* f(r_{\xi})$.  It follows that
$M=\{y_{\xi}\colon (\exists\alpha\in\omega_1)(
e_f(r_{\alpha})=g_{\beta}(r_{\alpha})=y_{\xi})\}$ has cardinality less than
$\omega_1$.  Since $e_f\in{\cal R}$ and $\omega_1$ is regular, we have
$|e_f^{-1}(M)|<\omega_1$.  To finish the proof notice that $[e_f=g_{\beta}]\subseteq
e_f^{-1}(M)$.  Thus,
$e_f\in\sz$.\qed



\section{ Proof of Theorem~\ref{thm:2}}

For the remainder of this section we assume that $\cuum$ is a
regular limit cardinal. Let $\{\lambda_{\beta}\}_{\beta\in\cuum}$
be an increasing sequence of infinite successor cardinals such
that $\bigcup_{\beta\in\cuum}\lambda_{\beta}=\cuum$.  Let
$\{r_{\beta}\colon\beta\in\cuum\}$ be an enumeration of $\real$
and $\{g_{\beta}\colon\beta\in\cuum\}$ be an enumeration of ${\cal
C}_{G_{\delta}}$.

\lem{lem:2}{If $\real$ is not the union of less than $\cuum$-many
meager sets, then there is an onto function $h\in\sz$ such that
for every $\alpha\in\cuum$  we have
$|h^{-1}(r_{\alpha})|=|\lambda_{\alpha}|$.} \proof Define a
sequence $\{W_{\alpha}\}_{\alpha\in\cuum}$ so that
$|W_{\alpha}|=|\lambda_{\alpha}|$  and
\[W_{\alpha}\subseteq\real
\setminus\left(\bigcup_{\xi\in\alpha}W_{\xi}\cup\bigcup\left\{
g_{\xi}^{-1}(r_{\alpha})\colon\xi\leq\alpha\ \&\
g_{\xi}^{-1}(r_{\alpha}) \in\meager\right\}\right).\]  To see that
such choices can be made notice that the set of forbidden values
is a union of less than $\cuum$-many meager sets.  Hence,
\[\real\setminus\left(\bigcup_{\xi\in\alpha}W_{\xi}\cup\bigcup\left\{
g_{\xi}^{-1}(r_{\alpha})\colon\xi\leq\alpha\ \&\
g_{\xi}^{-1}(r_{\alpha}) \notin\meager\right\}\right)\] must have
cardinality $\cuum$.  Otherwise, $\real$ could be covered less
than $\cuum$-many meager sets.

Define $h_1\colon\bigcup_{\alpha\in\cuum}W_{\alpha}\to\real$ so
that $h_1[W_{\alpha}]=\{r_{\alpha}\}$ for all $\beta\in\cuum$. We
show that $h_1\in\sz$.  Fix $\xi\in\cuum$.  Suppose that
$h_1(r)=g_{\xi}(r)$.  There is a $\alpha\in\cuum$ such that $r\in
W_{\alpha}$.   By definition of $h_1$, we have that
$g_{\xi}(r)=h_1(r)=r_{\alpha}$.  So, $r\in
g_{\xi}^{-1}(r_{\alpha})$.  By our choice of $W_{\alpha}$, we have
that $\alpha<\xi$ or $g_{\xi}^{-1}(r_{\alpha})\notin\meager$.  So,
$r\in\bigcup\{W_{\alpha}\colon\alpha<\xi\}\cup\{r\colon
g_{\xi}^{-1}(r)\notin\meager\}$. By Proposition~\ref{prop:1},
$g_{\beta}^{-1}(r_{\alpha})\notin\meager$ for at most countably
many $r\in\real$, so we have $|[h_1=g_{\xi}]|<\cuum$.   So,
$h_1\in\sz$. Clearly, $|h_1^{-1}(r_{\beta})|=\lambda_{\beta}$ for
every $\beta\in\cuum$. Now extend $h_1$ to a function $h\in\sz$
defined on all of $\real$ which is an injection outside the domain
of $h_1$.  Now $h\in\sz$, $h$ is onto, and
$|h^{-1}(r_{\beta})|=\lambda_{\beta}$ for every $\beta\in\cuum$.
\qed

{\sc Proof of Theorem~\ref{thm:2}}

Let $h$ be as in Lemma~\ref{lem:2}.  Let $f\in {\cal R}$ be
arbitrary.  We will be done if we show that there is a $m\in\sz$
such that $f=h\circ m$. Define $k\colon\cuum\to\cuum$ by
$k(\beta)=\pred(|h^{-1}(f(r_{\beta}))|)$.

We claim that $|\{\xi\colon k(\xi)\leq\lambda_{\rho}\}|<\cuum$ for
every $\rho\in\cuum$.  Let $\xi\in\{\xi\colon
k(\xi)\leq\lambda_{\rho}\}$.  By definition of $k$, we have
$|h^{-1}(f(r_{\xi}))|\leq\lambda_{\rho}^{+}\leq\lambda_{\rho+1}$.
By the way we defined $h$, we  must have that
$f(r_{\xi})\in\{r_{\beta}\colon\beta\leq\rho+1\}$.  So,
$r_{\xi}\in f^{-1}(\{r_{\beta}\colon\beta\leq\rho+1\})$.  By
regularity and the fact that $f\in{\cal R}$, we have that
$|f^{-1}(\{r_{\beta}\colon\beta\leq\rho+1\})|<\cuum$.  Thus, we
have the claim.

Define $m\in\real^{\real}$
so that for every $\beta\in\cuum$ we have
\[m(r_{\beta})\in h^{-1}(f(r_{\beta}))\setminus
\{g_{\rho}(r_{\beta})\colon\lambda_{\rho}<k(\beta)\}.\]
Notice such choices can be made since $|h^{-1}(f(r_{\beta}))|>k(\beta)$.
Clearly,
$f=h\circ m$.  We show that $m\in\sz$.  Let $\rho\in\cuum$ and suppose
$m(r)=g_{\rho}(r)$.  There is a $\beta\in\cuum$ such that $r=r_{\beta}$.  By
our choice of $m$ we must have that $k(\beta)\leq\lambda_{\rho}$.  So,
$\beta\in\{\xi\colon k(\xi)\leq\lambda_{\rho}\}$.  Since $|\{\xi\colon
k(\xi)\leq\lambda_{\rho}\}|<\cuum$, we have that $|m\cap g_{\rho}|<\cuum$.   So,
$m\in\sz$.\qed

\section {Proof of Theorem~\ref{thm:2a}}

\lem{lem:2d}{If $\{A_{\alpha}\colon\alpha\in\cuum\}$ is sequence
of sets such that $A_{\alpha}\in [\cuum]^{<\cuum}$, then there is
a bijection $k\in\cuum^{\cuum}$ such that
\[|\bigcup_{\xi\in\alpha}A_{k(\xi)}|<\cuum\]
for every $\alpha\in\cuum$.} \proof Define $f\in\cuum^{\cuum}$ by
$f(\alpha)=|A_{\alpha}|$.   There is an injection
$g\in\cuum^{\cuum}$ such that $f(\alpha)<g(\alpha)$  for all
$\alpha\in\cuum$. Since $|g[\cuum]|=\cuum$, there is an increasing
onto function $h\colon g[\cuum]\to\cuum$.  Let $k=g^{-1}\circ
h^{-1}$.  Since $k$ is a bijection, we will be done if we show
that
\begin{equation}\label{eq:lamp}
|\bigcup_{\alpha<\gamma}A_{k(\alpha)}|<\cuum
\end{equation}
for all $\gamma\in\cuum$.  Let $\gamma\in\cuum$.  Since $h$ is
increasing and $g$ has unbounded range, there is a
$\beta_{\gamma}\in\cuum$ such that $h^{-1}(\xi)\leq
h^{-1}(\gamma)\leq g(\beta_{\gamma})$ for all $\xi\in\gamma$.
Consider $\alpha\in\gamma$.  Notice that we have
$k(\alpha)=g^{-1}(h^{-1}(\alpha))\neq\zeta$ for any
$\zeta\in\cuum$ such that $g(\zeta)>g(\beta_{\gamma})$, since
otherwise we would have
\[g(\beta_{\gamma})\geq
h^{-1}(\alpha)=g(g^{-1}(h^{-1}(\alpha)))=g(\zeta)\] a
contradiction.  Now,
\[|A_{k(\alpha)}|=f(k(\alpha))<g(k(\alpha))\leq g(\beta_{\gamma}).\]  Thus,
$|\bigcup_{\alpha<\gamma}A_{k(\alpha)}|\leq\max\{|\gamma|,|g(\beta_{\gamma})|\}<\cuum$
so we have (\ref{eq:lamp}).  \qed

\lem{lem:sing}{If ${\cal A}\in\left[[\cuum]^{<\cuum}\right]^{\cuum}$ then there
is a $B\in[\cuum]^{\cof(\cuum)}$ such that $|B\cap A|<\cof(\cuum)$ for all $A\in {\cal
A}$.}
\proof
Fix an enumeration $\{A_{\alpha}\colon\alpha\in\cuum\}$ of ${\cal A}$.  By
Lemma~\ref{lem:2d}, there is a bijection $k\colon\cuum\to\cuum$ such that
\begin{equation}\label{eq:200}
|\bigcup_{\alpha<\gamma}A_{k(\alpha)}|<\cuum
\end{equation}
for all $\gamma\in\cuum$.

We now define the set $B$.  Let
$\langle\alpha_{\xi}\rangle_{\xi\in\cof(\cuum)}$ be a cofinal
increasing sequence of ordinals.  For each $\xi\in\cof(\cuum)$
pick
$b_{\xi}\in\cuum\setminus(\bigcup_{\beta\leq\alpha_{\xi}}A_{k(\beta)}\cup\{b_{\beta}\colon\beta<\xi\})$.
Notice the choice can be made by (\ref{eq:200}).  Let
$B=\{b_{\xi}\colon\xi\in\cof(\cuum)\}$.  Clearly,
$|B|=\cof(\cuum)$.  We show that $B$ has the other desired
property. Pick $A\in{\cal A}$.  There is a $\gamma\in\cuum$ such
that $A=A_{k(\gamma)}$.  Let $\xi\in\cof(\cuum)$ be such that
$\gamma\leq\alpha_{\xi}$.  If $b_{\zeta}\in A_{k(\gamma)}\cap B$,
then, by the way we defined $B$, we must have $\zeta<\xi$.  Thus,
$|B\cap A|<\cof(\cuum)$. \qed

{\sc Proof of Theorem~\ref{thm:2a}}

Let $\{r_{\beta}\colon\beta\in\cuum\}$ be an enumeration of
$\real$ and $\{g_{\beta}\colon\beta\in\cuum\}$ be an enumeration
of ${\cal C}_{G_{\delta}}$.

We first define $h\in\sz$. Let ${\cal
A}=\{\{g_{\xi}(r_{\alpha})\colon\xi\leq\alpha\}\colon\alpha\in\cuum\}
\subseteq[\cuum]^{<\cuum}$. Define a sequence
$\{W_{\alpha}\}_{\alpha\in\cuum}$ so that we have
$|W_{\alpha}|=\cof(\cuum)$, and
\begin{equation}\label{eq:1a1}
|W_{\alpha}\cap A|<\cof(\cuum) \text{ for every } A\in{\cal A},
\end{equation}
and
\begin{equation}\label{eq:2a2}
W_{\alpha}\subseteq\real\setminus\left(\bigcup\{g^{-1}_{\xi}(r_{\alpha})
\colon\xi\leq\alpha\ \&\ g^{-1}_{\xi}(r_{\alpha})\in\meager\}\cup
\bigcup_{\zeta<\alpha}W_{\zeta}\right)
\end{equation}
for every $\alpha\in\cuum$. To see that such choices can be made
notice that the set of forbidden values is a union of less than
$\cuum$-many meager sets.  Hence, the set
\[M=\real\setminus\left(\bigcup_{\xi\in\alpha}W_{\xi}\cup\bigcup\left\{
g_{\xi}^{-1}(r_{\alpha})\colon\xi\leq\alpha\ \&\
g_{\xi}^{-1}(r_{\alpha})\in\meager\right\}\right)\] must have
cardinality $\cuum$.  Otherwise, $\real$ could be covered less
than $\cuum$-many meager sets.  By Lemma~\ref{lem:sing}, there is
a $W_{\beta}\in [M]^{\cof(\cuum)}$ with the property that
$|W_{\beta}\cap(A\cap M)|<\cof(\cuum)$ for every $A\in{\cal A}$.
Now $W_{\beta}$ satisfies (\ref{eq:1a1}) and (\ref{eq:2a2}).

Define $h_1\colon\bigcup_{\alpha\in\cuum}W_{\alpha}\to\real$ so
that $h_1[W_{\alpha}]=\{r_{\alpha}\}$ for all $\alpha\in\cuum$. We
show that $h_1\in\sz$.  Fix $\xi\in\cuum$.  Suppose that
$h_1(r)=g_{\rho}(r)$.  There is a $\alpha\in\cuum$ such that $r\in
W_{\alpha}$.   By definition of $h_1$, we have that
$g_{\xi}(r)=h_1(r)=r_{\alpha}$.  So, $r\in
g_{\xi}^{-1}(r_{\alpha})$.  By (\ref{eq:2a2}), we have that
$\alpha<\xi$ or $g_{\xi}^{-1}(r_{\alpha})\notin\meager$.  So,
$r\in\bigcup\{W_{\alpha}\colon\alpha<\xi\ \text{or}\
g_{\xi}^{-1}(r_{\alpha})\notin\meager\}$.  By
Proposition~\ref{prop:1} and the fact that
$|W_{\alpha}|=\cof(\cuum)$ for all $\alpha\in\cuum$, we have
$|\bigcup\{W_{\alpha}\colon\alpha<\xi\ \text{or}\
g_{\xi}^{-1}(r_{\alpha})\notin\meager\}|<\cuum$.  So, $h_1\in\sz$.
Now extend $h_1$ to a function $h\in\sz$ defined on $\real$.

Let $f\in\real^{\real}$ be arbitrary.  We will be done if we show
that there is a $m\in\sz$ such that $f=h\circ m$.  Define
$m\in\real^{\real}$ so that for every $\alpha\in\cuum$ we have
\[m(r_{\alpha})\in h^{-1}(f(r_{\alpha}))\setminus
\{g_{\xi}(r_{\alpha})\colon\alpha\leq\xi\}.\] Notice such choices
can be made by (\ref{eq:1a1}) and the fact that $W_{\xi}\subseteq
h^{-1}(f(r_{\alpha}))$ for some $\xi\in\cuum$. Clearly, $f=h\circ
m$. We show that $m\in\sz$.  Let $\rho\in\cuum$ and suppose
$m(r)=g_{\rho}(r)$.  There is a $\beta\in\cuum$ such that
$r=r_{\beta}$.  By our choice of $m$ we must have that
$\beta<\rho$.  So, we have that $|m\cap g_{\rho}|<\cuum$.   So,
$m\in\sz$.\qed



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\end{thebibliography}

\end{document}
