\documentstyle[12pt]{article} \setlength{\topmargin}{.1in} \setlength{\textheight}{8in} \setlength{\textwidth}{5.8in} \setlength{\evensidemargin}{.4in} \setlength{\oddsidemargin}{.4in} \def\mathR{\mbox{$\bf \rm I\makebox [1.2ex] [r] {R}$}} \newcommand{\subRe}{\mbox{\scriptsize $\bf \rm I\makebox [1.2ex] [r] {R}$}} \newcommand{\restr}{{\hbox{$\,|\grave{}\,$}}} \begin{document} \def\intersect{\cap} \def\Union{\bigcup} \def\union{\cup} \def\rmiff{\mbox{ iff }} \def\rmand{\mbox{ and }} %\def\dom{{\rm dom}} \def\blocking{{\cal B}} \def\famil{{\cal G}} \def\proof{{\sc Proof. }} %\def\reals{{\Bbb R}} %\def\rationals{{\Bbb Q}} \def\reals{{\mathR}} \def\real{{\mathR}} \def\rationals{{\mathQ}} \def\realfunc{{\reals^{\subRe}}} \def\almost{{\cal A}} \def\aa{{{\rm A}(\almost)}} \def\darboux{{\cal D}} \def\ad{{{\rm A}(\darboux)}} \def\A{{\cal H}} \def\N{{\cal N}} \def\D{{\cal D}} \def\E{{\cal E}} \def\proj{{{\rm pr}_x}} \def\goth{\frak} %\def\continuum{{\goth c}} \def\continuum{{\co}} %\def\poset{{\Bbb P}} %\def\posetx{{\Bbb P}^*} \def\poset{{\mathP}} \def\posetx{{\mathP}^*} \def\poS{{\Bbb S}} \def\ONE{{\Bbb I}} \def\su{\subseteq} \def\eq{{\goth e}} \def\la{\langle} \def\ra{\rangle} \def\implies{\longrightarrow} \newtheorem{Th}{Theorem}[section] \newtheorem{Pro}[Th]{Problem} \newtheorem{Ex}[Th]{Example} \newtheorem{Le}[Th]{Lemma} \newtheorem{Co}[Th]{Corollary} \newtheorem{Pn}[Th]{Proposition} \renewcommand{\labelenumi}{(\arabic{enumi})} \renewcommand{\theenumi}{(\arabic{enumi})} \newcommand{\pf}{{\noindent\sc Proof. }} \newcommand{\Qed}{\nopagebreak\mbox{}\hfill$\Box$\medskip} \newcommand{\RRR}{\mathR^{\subRe}} \newcommand{\F}{{\cal F}} \newcommand{\R}{{\cal R}} \newcommand{\C}{{\cal C}} \newcommand{\Cd}{{{\cal C}_{G_\delta}}} \newcommand{\G}{{\cal G}} \newcommand{\aaa}{{\alpha}} \newcommand{\bbb}{{\beta}} \newcommand{\ggg}{{\gamma}} \newcommand{\ddd}{{\delta}} \newcommand{\fff}{{\varphi}} \newcommand{\kkk}{{\kappa}} \newcommand{\lll}{{\lambda}} \newcommand{\ooo}{{\omega}} \newcommand{\eee}{{\varepsilon}} \newcommand{\card}{\mbox{\rm card}} \newcommand{\add}{\mbox{\rm add}} \newcommand{\cov}{\mbox{\rm cov}} \newcommand{\unif}{\mbox{\rm unif}} \newcommand{\dom}{\mbox{\rm dom}} \newcommand{\rng}{\mbox{\rm rng}} \newcommand{\id}{\mbox{\rm id}} %\chKC if you define like this, it is too big in subscripts %\newcommand{\cf}{\mbox{\rm cf}} \newcommand{\cf}{{\rm cf}} \newcommand{\tl}{\triangleleft} \newcommand{\tr}{\triangleright} \newcommand{\FF}{{\cal R}_1} \newcommand{\GG}{{\cal G}_1} \newcommand{\HH}{{\cal H}} \newcommand{\ch}[1]{\mbox{\raise.48ex\hbox{$\chi$}$_{#1}$}} \newcommand{\mathN}{\mbox{\rm I\kern-0.18em N}} \newcommand{\mathI}{\mbox{\rm \bf I}} \newcommand{\mathQ}{\mbox{\rm \bf Q}} \newcommand{\mathP}{\mbox{\rm I\kern-0.18em P}} \newcommand{\co}{{\bf c}} { \renewcommand{\thefootnote}{} \footnotetext{AMS Subject Classification (1991): Primary: 26A15; Secondary: 03E50, 03E65} } { \renewcommand{\thefootnote}{} \footnotetext{This work was partially supported by NSF Cooperative Research Grant INT-9600548.} } { \renewcommand{\thefootnote}{} \footnotetext{Key words: cardinal invariants, Sierpi{\'n}ski-Zygmund functions, Generalized Martin's Axiom, Lusin sequence of filters.} } \begin{center} {\Large\bf ALGEBRAIC PROPERTIES OF THE CLASS OF SIERPI\'{N}SKI-ZYGMUND FUNCTIONS} \end{center} {\small\noindent Krzysztof Ciesielski, Department of Mathematics, West Virginia University, Morgantown, WV 26506-6310, USA. (kcies@wvnvms.wvnet.edu)} \medskip {\small\noindent Tomasz Natkaniec, Department of Mathematics, Pedagogical University, Chodkiewicza 30, 85-064 Bydgoszcz, Poland. (wspb11@vm.cc.uni.torun.pl)} \vspace{.25in} \begin{abstract} Sums, products and compositions with Sierpi\'{n}ski-Zygmund functions are investigated. Moreover, cardinal invariants connected with those operations are defined and studied. \end{abstract} \section{Preliminaries.} Let us establish some terminology to be used. No distinction is made between a function and its graph. The family of all functions from a set $X$ into $Y$ will be denoted by $Y^X$. Symbol $\card(X)$ will stand for the cardinality of a set $X$. The cardinality of the set $\mathR$ of real numbers is denoted by $\co$. Symbol $[X]^{\kkk}$ denotes the family of all subsets $Y$ of $X$ with $\card(Y)=\kkk$. Similarly we define $[X]^{<\kkk}$ and $[X]^{\leq\kkk}$. For a cardinal number $\kappa$ we will write $\cf(\kappa)$ for the cofinality of $\kappa$. Recall that a cardinal number $\kappa$ is regular, if $\kappa=\cf(\kappa)$. For $A\subset\mathR$ its characteristic function is denoted by $\ch{A}$. If $A$ is a planar set, we denote its $x$-projection by $\dom(A)$ and $y$-projection by $\rng(A)$. For $f,g\in\RRR$ the notation $[f=g]$ means the set $\{x\in\real\colon f(x)=g(x)\}$. Likewise for $[f>g]$, $[f \neq g]$, etc. For $X\subset \real$ we say that a function $f\colon X\to\mathR$ is of {\it Sierpi\'{n}ski-Zygmund type\/} (shortly, an $SZ$ function), if its restriction $f\restr M$ is discontinuous for any set $M\subset X$ with $\card(M)=\co$ \cite{SZ}. The family of all $SZ$-functions from $\real$ to $\real$ will be denoted by $SZ$. The symbol $\C$ will stand for the family of all continuous functions $f\colon\real\to\real$, and $\Cd$ for the family of all continuous functions defined on $G_{\delta}$-sets $X\subset\mathR$ with $\card(X)=\co$. Recall also that a function $f\in\RRR$ is an $SZ$ function if and only if $\card([f=g])<\co$ for every $g\in\Cd$~\cite{SZ}. We will sometimes abuse this notation by writing $f\in SZ$ and $f\in\C$ for partial functions $f\colon X\to\real$ with $X\su\real$. The following fact can be proved by a slight modification of the original proof of Sierpi\'{n}ski and Zygmund~\cite{SZ}. \begin{Pn}\label{prop1} For every family $\{Y_x\colon x\in\real\}$ of subsets of $\real$ of cardinality $\co$ there exists an $SZ$ function $f\colon\real\to\real$ such that $f(x)\in Y_x$ for every $x\in\real$. \Qed \end{Pn} In particular, $\card(SZ)=2^{\co}$. For every cardinal $\kappa$ and a partially ordered set (shortly poset) $\mathP$ we shall consider the following statements. (See \cite {CM}. Compare also \cite{df,sur,mp,tod}.) \begin{description} \item[${\rm MA}_{\kappa}({\mathP})$:] ($\kappa$-Martin's Axiom for $\mathP$) For any family ${\cal D}$ of dense subsets of $\mathP$ with $\card({\cal D})<\kappa$ there exists a $\D$-generic filter $G$ in $\mathP$, i.e., such that $D\cap G\neq\emptyset$ for every $D\in {\cal D}$. \item[${\rm Lus}_{\kappa}(\mathP)$:] There exists a sequence $\la G_{\alpha}\colon\alpha <\kappa\ra$ of $\mathP$-filters, called a $\kappa$-Lusin sequence, such that $\card(\{\alpha<\kappa\colon G_{\alpha}\cap D=\emptyset\})<\kappa$ for every dense set $D\subset\mathP$. \end{description} \section{Sums.} \begin{Th}\label{TH1} For every family ${\cal F}\subset\RRR$ with $\card ({\cal F})\leq \co$ there exists an $h\in\RRR$ such that $h+f\in SZ$ for each $f\in{\cal F}$. \end{Th} \pf Let $\{g_{\alpha}\colon\alpha <\co\}=\Cd$, $\{x_{\alpha}\colon\alpha <\co\}=\real$, and $\{ f_{\alpha}\colon\alpha <\co\}=\cal F$. For every $\alpha<\co$ choose $h(x_{\alpha})\in\mathR\setminus \{ g_{\gamma}(x_{\alpha})-f_{\beta}(x_{\alpha})\colon \beta,\gamma\leq \alpha\}$. Such a function $h$ satisfies the following condition: \[ (\forall \beta<\co )\; (\forall \gamma<\co)\;\; [h+f_{\beta}=g_{\gamma}]\subset \{ x_{\alpha}\colon\alpha<\max(\beta,\gamma)\}, \] so $\card ((h+f_{\beta})\cap g_{\gamma})<\co$ for all $\beta,\gamma <\co$. \Qed \begin{Co} Every real function $f$ can be expressed as the sum of two $SZ$ functions. \end{Co} \pf Use Theorem~\ref{TH1} with $\F=\{0,f\}$. \Qed The following cardinal function has been defined in~\cite{AC} for $\G\subset\RRR$. (Compare also \cite{CM,CR}.) \begin{eqnarray*} a(\G) &=&\min\left(\{\card(\F)\colon \F\subset\RRR\ \&\ \neg\exists h\in\RRR\ \forall f\in \F\ h+f\in\G\}\cup\{(2^{\co})^+\}\right )\\ &=&\min \left (\{ \card(\F)\colon \F\subset\RRR\ \&\ \forall h\in\RRR\ \exists f\in \F\ h+f\not\in\G\}\cup\{(2^{\co})^+\}\right). \end{eqnarray*} Evidently, there is no $h\in\RRR$ such that $h+f\in SZ$ for all $f\in\RRR$. Therefore Theorem~\ref{TH1} yields to the following corollary. \begin{Co}\label{C} $\co < a(SZ)\leq 2^{\co}$. \Qed \end{Co} Hence, if $\co^+=2^{\co}$, then $a(SZ)=2^{\co}$. However, it is interesting whether or not anything more can be said about the cardinal $a(SZ)$. (The analogous problem for the classes $AC$ of almost continuous functions and $\D$ of Darboux functions is considered in \cite{CM}.) To address this question we need the following partially ordered sets $\la\poset,\leq\ra$ and $\la\posetx,\leq\ra$. \[ \poset= \{p\in\reals^X\colon X\su\reals\ \&\ \card(X)<\continuum\} \] i.e., $\poset$ is the set of all partial functions from $\reals$ to $\reals$ of cardinality less than $\continuum$. We put $p\leq q$ if and only if $p\supseteq q$, i.e., when $p$ extends $q$ as a partial function. \[ \posetx=\{\la p,E\ra\colon p\in\poset\ \&\ E\su\realfunc \ \&\ \card(E)<\continuum\}. \] The ordering on $\posetx$ is defined by \begin{eqnarray*} \la p,E\ra\leq \la q,F\ra & \rmiff & p\supseteq q\ \rmand \ E\supseteq F\\ & \rmand & \forall x\in\dom(p)\setminus\dom(q)\;\; \forall f\in F\; p(x)\not=f(x). \end{eqnarray*} The following theorem can be found in \cite[theorem 3.7]{CM}. \begin{Th}\label{TH:PStar} Let $\lambda\geq\kappa\geq\omega_2$ be cardinals such that $\cf(\lambda)>\omega_1$ and $\kappa$ is regular. Then it is relatively consistent with ZFC+CH that $2^{\continuum}=\lambda$ and Lus$_\kappa(\posetx)$ holds. \Qed \end{Th} We will prove the following theorem. \begin{Th}\label{TH:AAA} If $\kappa>\co$ is a regular cardinal then Lus$_\kappa(\posetx)$ implies that $a(SZ)=\kappa$. \end{Th} This and Theorem~\ref{TH:PStar} will immediately imply the following corollary. \begin{Co}\label{CO:AAA} Let $\lambda\geq\kappa\geq\omega_2$ be cardinals such that $\cf(\lambda)>\omega_1$ and $\kappa$ is regular. Then it is relatively consistent with ZFC+CH that $2^{\continuum}=\lambda$ and $a(SZ)=\kappa$. \Qed \end{Co} The proof of Theorem \ref{TH:AAA} will be split into three lemmas. \begin{Le}\label{L:AAAzero} \begin{description} \item[(i)] Lus$_\kappa(\posetx)\implies$ Lus$_\kappa(\poset)$. \item[(ii)] For any regular $\kappa$ we have Lus$_\kappa(\posetx)\implies$ MA$_\kappa(\posetx)$. \end{description} \end{Le} \pf The proof is implicitly contained in the proof of \cite[Lemma 3.6]{CM}. Let $\la G_\alpha\colon \alpha<\kappa\ra$ be a $\kappa$-Lusin sequence for $\posetx$. (i) follows from the fact that in some sense $\poset$ is `living inside' of $\posetx$. To see it, let $r\colon \reals\to\reals$ be a map with of $\card(r^{-1}(y))=\continuum$ for every $y\in\reals$. Define $\pi\colon\posetx\to\poset$ by \[ \pi(p,F)=r\circ p. \] Notice that if $\la p,E\ra\leq\la q,F\ra$ then $\pi(p,E)\leq\pi(q,F)$. This implies that $\pi[G]$ is a $\poset$-filter for any $\posetx$-filter $G$. Furthermore, we claim that if $D\su\poset$ is dense, then $\pi^{-1}(D)$ is dense in $\posetx$. To see this, let $\la p,F\ra\in\posetx$ be arbitrary. Since $D$ is dense, there exists $q\leq \pi(p,F)$ with $q\in D$. Now, find $s\in\poset$ extending $p$ such that $r\circ s= q\supseteq r\circ p$ and $s(x)\neq f(x)$ for every $x\in\dom(s)\setminus\dom(p)$ and $f\in F$. This can be done by choosing \[ s(x)\in r^{-1}(q(x))\setminus\{f(x)\colon f\in F\} \] for every $x\in\dom(q)\setminus\dom(p)$. Then, $\la s,F\ra\leq\la p,F\ra$ and $\la s,F\ra\in\pi^{-1}(q)\su\pi^{-1}(D)$. Now, $\la \pi[G_\alpha]\colon\alpha<\kappa\ra$ is a $\kappa$-Lusin sequence for $\poset$ since for every dense $D\su\poset$, \begin{eqnarray*} \{\alpha<\kappa\colon \;\pi[G_\alpha]\cap D=\emptyset\} & = & \{\alpha<\kappa\colon \;\pi[G_\alpha]\cap \pi[\pi^{-1}(D)]=\emptyset\}\\ & \su & \{\alpha<\kappa\colon \;G_\alpha\cap \pi^{-1}(D)=\emptyset\}. \end{eqnarray*} To see (ii) take a family $\D$ of dense subsets of $\posetx$ of cardinality less than $\kappa$. By the regularity of $\kkk$, there exists $\alpha<\kappa$ such that $G_\alpha$ meets every element of $\D$. \Qed \begin{Le}\label{L:AAA1} Assume that $\kkk$ is a regular cardinal and $\kkk>\co$. Then Lus$_\kappa(\poset)$ implies that $a(SZ)\leq\kappa$. \end{Le} \pf Let $\la G_\alpha\colon\alpha<\kappa\ra$ be a $\kappa$-Lusin sequence of $\poset$-filters and let \[ f_\alpha=\Union G_\alpha. \] Then $f_\alpha$ is a partial function from $\reals$ into $\reals$. Let \[ D_x=\{p\in\poset\colon \; x\in\dom(p)\}.$$ It is easy to see that each $D_x$ is dense in $\mathP$. Hence, since $\continuum<\kappa$ and $\kappa$ is regular, we may assume that each $f_\alpha$ is a total function. Now, let $\{x_\xi\colon\xi<\co\}=\reals$. For each $\xi<\co$, $g\in\Cd$, and $h\in\realfunc$ define \[ D_\xi(g,h)=\{p\in\poset\colon (\exists \eta\geq\xi)\,(x_\eta\in\dom(p)\cap\dom(g)\ \&\ (h+p)(x_\eta)=g(x_\eta))\}. \] Note that $D_\xi(g,h)$ is dense in $\mathP$, since for any $p\in\poset$ there is $\eta\geq\xi$ with $x_\eta\in\dom(g)\setminus\dom(p)$. Then $p\cup\{\la x_\eta,g(x_\eta)-h(x_\eta)\ra\}\in D_\xi(g,h)$ extends $p$. By the regularity of $\kkk$, for any $h\in\realfunc$ there exists $\alpha<\kappa$ such that $G_\alpha$ intersects every set $D_\xi(g,h)$ with $\xi<\co$ and $g\in\Cd$, and so, $\card((h+f_\alpha)\cap g)=\co.$ Thus, for every $h\in\realfunc$ there exists $\alpha<\kappa$ such that $h+f_\alpha\not\in SZ$, i.e., the family $\F=\{f_\alpha\colon\alpha<\kappa\}$ shows that $a(SZ)\leq\kappa$ as was to be shown. \Qed \begin{Le}\label{L:AAA2} If $\kappa>\co$ then MA$_\kappa(\posetx)$ implies that $a(SZ)\geq\kappa$. \end{Le} \pf Let $\F\su\RRR$ be such that $\card(\F)<\kappa$. We will find $h\in \RRR$ such that $h+f\in SZ$ for every $f\in\F$. Notice that for any $x\in\reals$ the set \[ D_x=\{\la p,E\ra\in\posetx\colon x\in\dom(p)\} \] is dense in $\posetx$. Indeed, let $\la q,F\ra$ be an arbitrary element of $\posetx$ and suppose it is not already an element of $D_x$. The set $Q=\{f(x)\colon f\in F\}$ has cardinality less than $\continuum$, so there exists $y\in\reals\setminus Q$. Let $p=q\union\{\la x,y\ra\}$. Then $\la p,F\ra\leq\la q,F\ra$ and $\la p,F\ra\in D_x$. Therefore $h=\bigcup\{p\colon\; (\exists E)\,(\la p,E\ra\in G)\}$ is a function from $\reals$ into $\reals$ for any $\posetx$-filter $G$ intersecting all sets $D_x$. Note also, that for $f\in\RRR$ the set \[ E_f=\{\la p,E\ra\in\posetx\colon f\in E\} \] is dense in $\posetx$ since $\la p,E\cup\{f\}\ra\in E_f$ extends $\la p,E\ra$. Let \[ \D=\{D_x\colon x\in\reals\}\cup\{E_{\bar{g}-f}\colon f\in\F\ \&\ g\in\Cd\}, \] where $\bar{g}\in\RRR$ extends $g\in\Cd$ by associating $0$ at all undefined places. Then, $\D$ is a family of less than $\kappa$ many dense subsets of $\posetx$. Let $G$ be a $\D$-generic filter in $\posetx$ and let $h=\bigcup\{p\colon (\exists E)(\la p,E\ra\in G)\}$. We have to show that $h+f\in SZ$ for every $f\in\F$. So, let $f\in\F$ and $g\in\Cd$. Then there exists $\la p,E\ra\in G\cap E_{\bar{g}-f}$. So, by the definition of order on $\poset$ it is easy to see that \[ \{x\in\reals\colon (f+h)(x)=g(x)\}\su \{x\in\reals\colon h(x)=\bar{g}(x)-f(x)\}\su\dom(p). \] Thus, $h+f\in SZ$ for every $f\in\F$. \Qed Application of Lemmas \ref{L:AAAzero}, \ref{L:AAA1} and \ref{L:AAA2} finishes the proof of Theorem~\ref{TH:AAA}. \Qed In \cite{CM} it has been proved that $a(\D)=a(AC)=e_\co$ and that this number has cofinality greater than continuum $\co$, where \[ e_\kappa=\min\{\card(F)\colon F\su \kappa^\kappa\ \&\ \forall h\in \kappa^\kappa\ \exists f\in F\ \card(f\cap h)<\kappa\}. \] Next, we will compare $a(SZ)$ with $a(\D)$, and give a characterization of $a(SZ)$ similar to that of $e_\co$. We will also address an issue of the cofinality of $a(SZ)$. Since for a regular $\kappa>\co$ an axiom Lus$_\kappa(\posetx)$ implies $a(\D)=\kappa$ \cite[section~3]{CM} we can conclude the following fact. \begin{Co}\label{CO:AAA2} Let $\lambda\geq\kappa\geq\omega_2$ be cardinals such that $\cf(\lambda)>\omega_1$ and $\kappa$ is regular. Then it is relatively consistent with ZFC+CH that $2^{\continuum}=\lambda$ and $a(\D)=a(SZ)=\kappa$. \Qed \end{Co} Note also the following strengthening of \cite[theorem~3.3]{CM}. \begin{Th}\label{TH:PStar2} Let $\lambda\geq\omega_2$ be a cardinal such that $\cf(\lambda)>\omega_1$. Then it is relatively consistent with ZFC+CH that $2^{\continuum}=\lambda$ and Lus$_\kappa(\poset)$ holds for every regular $\kappa>\co$, $\kappa\leq 2^\co$. \end{Th} \pf The proof is identical to that of \cite[theorem~3.3]{CM}. \Qed Now, recall also that Lus$_\kappa(\poset)$ implies $a(\D)\geq\kappa$ for every regular $\kappa>\co$ \cite{CM}. Thus, in a model of Theorem~\ref{TH:PStar2} we have $a(\D)=2^{\continuum}=\lambda$. On the other hand in this model we have Lus$_{\co^+}(\poset)$. So, by Lemma~\ref{L:AAA1} and Corollary~\ref{C}, $a(SZ)=\co^+$. In particular, we obtain the following corollary. \begin{Co}\label{CO:AAA3} Let $\lambda>\omega_2$ be a cardinal such that $\cf(\lambda)>\omega_1$. Then it is relatively consistent with ZFC+CH that $2^{\continuum}=\lambda$ is true, and $a(SZ)=\co^+<2^{\continuum}=a(\D)$. \Qed \end{Co} The following remains an open problem. \begin{Pro}\label{a} Is it consistent that $a(SZ)>a(\D)$? \end{Pro} For an infinite cardinal $\kappa$ define \[ d_\kappa=\min\{\card(F)\colon F\su \kappa^\kappa\ \&\ \forall h\in \kappa^\kappa\ \exists f\in F\ \card(f\cap h)=\kappa\}. \] Notice that $d_{\kappa}>\kappa$. \begin{Th}\label{THcomb} $a(SZ)=d_\co$. \end{Th} \pf To see that $d_\co\leq a(SZ)$ choose $\F\subset \RRR$ with $\card(\F)\kappa$. \end{Th} \pf Let $T$ be the set of all functions from some $\xi<\kappa$ into $\kappa$, i.e., $T=\bigcup_{\xi<\kappa}\kappa^\xi$. Thus, by our assumption, $\card(T)=\kappa$. Let $\la F_\xi\subset T^\kappa\colon \xi<\kappa\ra$ be an increasing sequence such that $\card(F_\xi)\co$. \Qed \end{Co} The following remains an open problem. \begin{Pro}\label{PRO1} Can $a(SZ)$ be a singular cardinal? \end{Pro} Since $a(SZ)=d_\co$ and $a(\D)=e_\co$, Problems~\ref{a} and \ref{PRO1} can be rephrased as follows. \begin{description} \item{($\bullet$)} Let $\kappa=\co$. Is it consistent that $d_\kappa>e_\kappa$? Can $d_\kappa$ be singular? \end{description} Notice that for $\kappa=\omega$ the answer for these problems is well known, since $d_\omega=non(meager)$ is the minimum cardinality of a non meager subset of $\real$, and $e_\omega=cov(meager)$ is the minimum cardinality of a family of meager subset of $\real$ whose union is equal to $\real$. (See \cite{Bar}.) Thus, for $\kappa=\omega$ the answer for both questions is positive. (Compare also \cite{Lan} for some results concerning $e_\kappa$ for $\kappa>\omega$.) \medskip Next, let ${\cal M}_a(SZ)$ denote the {\em maximal additive family\/} for the class $SZ$, i.e., \[ {\cal M}_a(SZ)=\{ f\in\RRR\colon f+h\in SZ \mbox{ for each } h\in SZ\}. \] To describe the structure of ${\cal M}_a(SZ)$ we need the following easy lemma. \begin{Le}\label{L} Let $X\subset\mathR$ and $f\colon X\to\mathR$ be an $SZ$-function. Then there exists an $SZ$-extension of $f$, i.e., an $f^{\ast}\in\RRR$ that $f^{\ast}\in SZ$ and $f^{\ast}\restr X=f$. \end{Le} \pf Obviously for each $h\colon\mathR\to\mathR$, $h\in SZ$ if and only if $h\restr X\in SZ$ and $h\restr (\mathR\setminus X)\in SZ$. Moreover, we can use the Sierpi\'{n}ski \& Zygmund's method to obtain an $SZ$-function defined on any subset of $\mathR$. Therefore it is enough to construct an $SZ$-function $g\colon \mathR\setminus X\to \mathR$ and put $f^{\ast}=f\cup g$. \Qed \begin{Th}\label{TH2} For every function $f\in\RRR$ the following conditions are equivalent \begin{description} \item[(i)] $f\in {\cal M}_a(SZ)$; \item[(ii)] for each $X\in [\mathR]^{\co}$ there exists a $Y\in [X]^{\co}$ such that $f\restr Y\in{\cal C}$. \end{description} \end{Th} \pf (ii)$\Rightarrow$(i) Suppose that $f$ satisfies the condition (ii) and $h+f\not\in SZ$ for some $h\in SZ$. Then $(h+f)\restr X\in{\cal C}$ for some set $X\in [\mathR]^{\co}$. Let $Y\in [X]^{\co}$ be a set such that $f\restr Y\in{\cal C}$. Then $h\restr Y\in{\cal C}$, in contradiction with $h\in SZ$. (i)$\Rightarrow$(ii) Suppose that $f$ does not fulfill the condition (ii). Then there exists $X\in[\mathR]^{\co}$ such that $f\restr Y\not\in {\cal C}$ for each $Y\in [X]^{\co}$, i.e., $f\restr X\in SZ$. Let $f^{\ast}\in \RRR$ be an $SZ$-extension of $f$. Then $-f^{\ast}\in SZ$ and $(f-f^{\ast})\restr X\in{\cal C}$, so $f\not\in{\cal M}_a(SZ)$. \Qed \noindent {\bf Remark.} U. Darji proved under CH that a Borel function $f$ satisfies the the condition (ii) if and only if it is countably continuous \cite[Thm.~10]{Da}. In the same way one can prove that (ii) implies the following condition: \begin{description} \item[(iii)] $f$ is the union of less than $\co$ many continuous functions; \end{description} and, assuming regularity of $\co$, that (iii) implies (ii). \pf (ii)$\implies$(iii). Let $\{g_{\alpha}\colon\alpha <\co\}=\Cd$. Suppose that $f$ is not the union of less than $\co$ many continuous functions. Then $\card(\dom(f\setminus\bigcup_{\bbb<\aaa}g_{\aaa}))=\co$ for each $\aaa<\co$. For every $\aaa<\co$ choose $x_{\aaa}\in \dom(f\setminus\bigcup_{\bbb<\aaa}g_{\aaa})\setminus \{ x_{\bbb}\colon \bbb<\aaa\}$ and set $X=\{ x_{\aaa}\colon \aaa<\co\}$. By (ii), there exists $Y\in [X]^{\co}$ such that $f\restr Y$ is continuous. Therefore $f\restr Y=g_{\aaa}\restr Y$ for some $\aaa<\co$, so $\card(f\cap g_{\aaa})=\co$, contrary to the construction of $X$. Now assume that $\co$ is a regular cardinal and $f$ satisfies (iii). Then $f=\bigcup_{\aaa<\kkk}f\restr X_{\aaa}$ for some $\kkk<\co$ and all functions $f\restr X_{\aaa}$ are continuous. Fix $X\in [\mathR]^{\co}$. By the regularity of $\co$, $\card(X\cap X_{\aaa})=\co$ for some $\aaa<\kkk$ and, for $Y=X\cap X_{\aaa}$, $f\restr Y$ is continuous. \Qed It is also worth to notice in this context that if $f\colon X\to\real$ is $SZ$ for some $X\subset\real$ then for every $Y\in[X]^{\co}$ its restriction $f\restr{Y}$ is not countably (even $\kkk<\cf(\co)$) continuous. \section{Products.} In this section we will examine for which functions $f\in\RRR$ there exists $h\in\RRR$ such that $hf\in SZ$. First note that if $\card([f=0])=\co$ then $hf\in SZ$ for no $h\colon\mathR \to\mathR$. Thus, we will restrict our attention to the family \[ \R_0=\{f\in\RRR\colon\card([f=0])<\co\}. \] \begin{Th}\label{TH3} For every family $\F\subset\R_0$ with $\card({\cal F})\leq\co$ there exists an $h\colon\mathR\to\mathR\setminus\{ 0\}$ such that $hf\in SZ$ for each $f\in {\cal F}$. \end{Th} \pf Let $\{g_{\alpha}\colon\alpha <\co\}=\Cd$, $\{x_{\alpha}\colon\alpha <\co\}=\real$, and $\{f_{\alpha}\colon\alpha <\co\}=\cal F$. For $\alpha<\co$ choose \[ h(x_{\alpha})\in \mathR \setminus \left(\{ 0\}\cup \left\{\frac{g_{\gamma}(x_{\alpha})}{f_{\beta}(x_{\alpha})}\colon \beta,\gamma\leq\alpha\ \&\ f_{\bbb}(x_{\aaa})\neq 0\right\}\right). \] Such a function $h$ satisfies the following condition: \[ (\forall\beta<\co )\; (\forall\gamma<\co)\;\; [hf_{\beta}=g_{\gamma}]\subset [f_{\beta}=0]\cup\{x_{\alpha}\colon\alpha<\max (\beta,\gamma )\}, \] so $\card((hf_{\beta})\cap g_{\gamma})<\co$ for all $\beta,\gamma<\co$. \Qed \begin{Co} For every function $f\in\RRR$ the following conditions are equivalent: \begin{description} \item[(i)] $\card([f=0])<\co$, \item[(ii)] $f$ is the product of two $SZ$ functions. \Qed \end{description} \end{Co} Let $m(SZ)$ denote the least cardinal $\kappa$ for which there exists a family $\F\subset\R_0$ such that $\card(\F)=\kkk$ and for every $h\colon\mathR\to\mathR$ there exists $f\in {\cal F}$ with $hf\not\in SZ$. (Note that this definition is different from the definition of the cardinal function $m$ defined in \cite{AC}; cf., \cite{NR}.) \begin{Th}\label{TH4} $a(SZ)=m(SZ)$. \end{Th} \pf ``$a(SZ)\leq m(SZ)$.'' Assume that ${\cal F}\subset\R_0$ is a family of functions such that $\card({\cal F})0]$ and $X_0=X\cap [f=0]$. Note that $\card(X_0)<\co$. Also, $\card(X_+)<\co$, since $\exp(h)\tilde{f}\restr{X_+}=\exp(h)f\restr{X_+}\in{\cal C}$. Similarly, $\card(X_-)<\co$, since $\exp(h)\tilde{f}\restr{X_-}=-\exp(h)f\restr{X_-}\in{\cal C}$. Thus $\card(X)<\co$ and consequently, $\exp(h)f\in SZ$. ``$m(SZ)\leq a(SZ)$.'' Now assume that ${\cal F}\subset\RRR$ is a family of functions such that $\card({\cal F})0]$ and $X_0=X\cap [h=0]$. Of course, $\card(X_0)<\co$. Moreover, $\exp(f)\tilde{h}\restr{X_+}= \exp(f)h\restr{X_+}\in{\cal C}$ and $\exp(f)\tilde{h}\restr{X_-}=-\exp(f)h\restr{X_-}\in{\cal C}$, so $\card(X_+)<\co$ and $\card(X_-)<\co$. Hence $\card(X)<\co$. Therefore $\ln(\exp(f)\tilde{h})\in SZ$, so $\ln(\tilde{h})+f\in SZ$ for each $f\in{\cal F}$. \Qed Let ${\cal M}_m(SZ)$ denote the {\em maximal multiplicative family\/} for the class $SZ$, i.e., $${\cal M}_m(SZ)=\{ f\in\RRR\colon fh\in SZ \mbox{ for each } h\in SZ\}. $$ \begin{Th}\label{TH5} For every function $f\in\RRR$ the following conditions are equivalent: \begin{description} \item[(i)] $f\in {\cal M}_m(SZ)$; \item[(ii)] $\card([f=0])<\co$ and for each $X\in [\mathR]^{\co}$ there exists a $Y\in [X]^{\co}$ such that $f\restr Y\in{\cal C}$. \end{description} \end{Th} \pf (ii)$\Rightarrow$(i) Suppose that $f$ satisfies the condition (ii) and $hf\not\in SZ$ for some $h\in SZ$. Then $hf\restr X\in{\cal C}$ for some set $X\in[\mathR]^{\co}$. Let $Y\in[X\setminus[f=0]]^{\co}$ be a set such that $f\restr Y\in{\cal C}$. Then $h\restr Y=(hf)/f\restr Y\in{\cal C}$, in contradiction with $h\in SZ$. (i)$\Rightarrow$(ii) Assume that $f\in{\cal M}_m(SZ)$. Note that $\card([f=0])<\co$. Fix $X\in [\mathR]^{\co}$ and set $X_0=X\setminus [f=0]$. Obviously, $\card(X_0)=\co$. Suppose that $f\restr Y\in{\cal C}$ for no $Y\in [X_0]^{\co}$, i.e., $f\restr {X_0}\in SZ$. Then $(1/f)\restr {X_0}\in SZ$ and there exists an $SZ$-extension $f^{\ast}\in \RRR$ of the function $(1/f)\restr {X_0}$. Then $(f^{\ast}f)\restr {X_0}\in{\cal C}$, a contradiction. Hence there exists $Y\in [X]^{\co}$ such that $f\restr Y\in{\cal C}$. \Qed %%%%%%% SZ FUNCTIONS. PART II %%%%%%%%%%%%%% \section{Compositions.} Let \[ \begin{array}{rcl} {\cal M}_{out}(SZ) & = & \{f\in\RRR\colon f\circ h\in SZ \mbox{ for each } h\in SZ\},\\ {\cal M}_{in}(SZ) & = & \{ f\in\RRR\colon h\circ f\in SZ \mbox{ for each } h\in SZ\}. \end{array} \] \begin{Th}\label{TH6} Assume that $\co$ is a regular cardinal. Then for every function $f\in\RRR$ the following conditions are equivalent: \begin{description} \item[(i)] $f\in {\cal M}_{out}(SZ)$; \item[(ii)] $\card(f^{-1}(y))<\co$ for each $y\in\mathR$, and every choice function $g\colon \rng(f)\to\mathR$, $g(y)\in f^{-1}(y)$, satisfies the following condition \begin{description} \item[($\ast$)] for each $X\in [\rng(f)]^{\co}$ there exists a $Y\in [X]^{\co}$ such that $g\restr Y\in{\cal C}$; \end{description} \item[(iii)] $f\in {\cal M}_{in}(SZ)$. \end{description} \end{Th} \pf (i)$\Rightarrow$(ii). Fix $f\in{\cal M}_{out}(SZ)$. Suppose that $\card(f^{-1}(y))=\co$ for some $y\in\mathR$. By Proposition~\ref{prop1} we can choose an $SZ$-function $g\in\RRR$ with $\rng(g)\subset f^{-1}(y)$. Then $f\circ g\in {\cal C}$, a contradiction. Suppose that there exists a choice function $g\colon\rng(f)\to\mathR$, $g(y)\in f^{-1}(y)$, without the property ($\ast$), i.e., that there exist $X\in [\rng(f)]^{\co}$ and $g\in\mathR^X$ such that $g\in SZ$ and $f\circ g=\id_X$. Let $g^{\ast}\in\RRR$ be an $SZ$-extension of $g$. Then $f\circ g^{\ast}\restr X\in{\cal C}$, so $f\circ g^{\ast}\not\in SZ$ and consequently, $f\not\in{\cal M}_{out}(SZ)$, a contradiction. (ii)$\Rightarrow$(i). Suppose that $f\circ h\not\in SZ$ for some $SZ$-function $h\in\RRR$. Then there exists $X\in [\mathR]^{\co}$ such that $f\circ h\restr X\in{\cal C}$. Note that $\card (\rng(f\circ h\restr X))=\co$. Indeed, otherwise, by regularity of $\co$, $f\circ h$ is constant on some set $X_0\in [X]^{\co}$ and because $\card(f^{-1}(y)) <\co$ for each $y$, $h$ is constant on some set $X_1\in [X_0]^{\co}$, a contradiction. Let $g\colon \rng(f)\to\mathR$, $g(y)\in f^{-1}(y)$, be a choice function such that $g(t)\in\rng(h\restr X)$ for $t\in \rng(f\circ h\restr X)$. Let $g\restr Y\in{\cal C}$ for $Y\in [\rng(f\circ h\restr X)]^{\co}$. Then $X_0=(f\circ h)^{-1}(Y)\cap X\in [X]^{\co}$ and $h\restr {X_0}=g\circ (f\circ h\restr {X_0})\in{\cal C}$, a contradiction. (iii)$\Rightarrow$(ii). Fix $f\in{\cal M}_{in}(SZ)$. Obviously, $\card(f^{-1}(y))<\co$ for every $y\in\mathR$. Suppose that $g\colon \rng(f)\to\mathR$, $g(y)\in f^{-1}(y)$, is a choice function without the property ($\ast$), i.e., that there exists $X\in [\rng(f)]^{\co}$ such that $g\restr X\in SZ$. Let $g^{\ast}\in\RRR$ be an $SZ$-extension of $g\restr X$. Then $g^{\ast}\circ f\restr {(\rng(g|X))}=\id_{\rng(g\restr X)}$. But $g$ is one-to-one. So, $\card(\rng(g\restr X))=\co$ and $g^{\ast}\circ f\not\in SZ$. A contradiction with $f\in{\cal M}_{in}(SZ)$. (ii)$\Rightarrow$(iii). Suppose that $h\circ f\not\in SZ$ for some $h\in SZ$. Then $h\circ f\restr X\in{\cal C}$ for some $X\in [\mathR]^{\co}$. Note that $\card(\rng(f\restr X))=\co$ since $\card(f^{-1}(y))<\co$ for each $y\in\mathR$ and $\co$ is regular. Let $g\colon \rng(f)\to\mathR$, $g(y)\in f^{-1}(y)$, be a choice function such that $g(y)\in X$ for $y\in \rng(f\restr X)$ and let $Y\in [\rng(f\restr X)]^{\co}$ be such that $g\restr Y\in{\cal C}$. Then $h\restr Y= (h\circ f)\circ g\restr Y\in{\cal C}$, a contradiction. \Qed Notice that in the proofs of implications (i)$\Rightarrow$(ii) and (iii)$\Rightarrow$(ii) we did not use the assumption that $\co$ is regular. Moreover, in the above proof of implication (ii)$\Rightarrow$(iii) we do not have to use the assumption of regularity of $\co$ if we additionally assume that $f$ is one-to-one. (Or even only that $\sup\{\card(f^{-1}(y))\colon y\in\real\}<\co$.) This implies the following two corollaries. \begin{Co}\label{coComp1} If $\co$ is regular then ${\cal M}_{out}(SZ)={\cal M}_{in}(SZ)$.\Qed \end{Co} \begin{Co}\label{coComp2} If a one-to-one function $f\colon\real\to\real$ satisfies condition (ii) from Theorem~\ref{TH6} then $f\in{\cal M}_{in}(SZ)$.\Qed \end{Co} %\chTN The next result, being a version of Sierpi{\'n}ski-Zygmund theorem, will be used to show that Corollary~\ref{coComp1} is false when $\co$ is singular. \begin{Th} Suppose that %\chKC a cardinal $\kkk\leq\co$ is cofinal with $\co$. $\kkk\leq\co$ is a cardinal such that $\cf(\kappa)=\cf(\co)$. Then for every $X\in [\mathR]^{\kkk}$ there exists $f\colon X\to\mathR$ such that $\card(\rng f)=\cf(\co)$ and $f\restr X_0$ is continuous for no $X_0\in [X]^{\kkk}$. \end{Th} \pf Let $\{\lambda_\xi\colon\xi<\cf(\co)\}$ and $\{\mu_\xi\colon\xi<\cf(\co)\}$ be increasing sequences of %chKC the sequence of cardinals does not have to exist ordinal numbers such that $\kkk=\bigcup_{\xi<\cf(\co)}\lambda_\xi$ and $\co=\bigcup_{\xi<\cf(\co)}\mu_\xi$ %chKC and let $X=\{x_{\xi}\colon \xi<\kkk\}$. %chKC Choose a partition $\{X_\xi\colon \xi<\cf(\co)\}$ of $X$ such that %chKC $\card(X_\xi)=\card(\lambda_\xi)$ for every $\xi<\cf(\co)$ %chKC and let $\{g_\xi\colon\xi<\co\}$ be an enumeration of $\Cd$. By induction on $\xi<\kkk$ define a sequence $\la y_\xi\in\real\colon\xi<\cf(\co)\ra$ such that for every $\xi<\kkk$ \[ y_\xi\in\real\setminus\{g_\eta(x)\colon\eta<\mu_\xi\ \&\ x\in X_\xi\}. \] Now, define $h$ by putting $h(x)=y_\xi$ for $x\in X_\xi$ and $\xi<\cf(\co)$. It is easy to see that $\rng(h)=\{y_\xi\colon \xi<\cf(\co)\}$. Also, if $g=g_\eta\in\Cd$ and $\eta<\mu_\xi$ then $[h=g]\subset\bigcup_{\zeta\leq\xi}X_\zeta$. Thus, $\card([h=g])<\kkk$ and, as in Sierpi{\'n}ski-Zygmund's proof, we conclude that $h\restr X_0$ is continuous for no $X_0\in [X]^{\kkk}$. \Qed \begin{Co}\label{CoNew} There exists an SZ function $h\colon\mathR\to\mathR$ with $\card(\rng(h))=\cf(\co)$. %chKC \Qed \end{Co} \begin{Pro} Does there exist an SZ function $h\colon\mathR\to Y$ for every $Y\in [\mathR]^{\cf(\co)}$? \end{Pro} \begin{Co}\label{coComp3} If $\co$ is singular then ${\cal M}_{in}(SZ)\not\subset{\cal M}_{out}(SZ)$. \end{Co} \pf %\chTN Let $h$ be as in Corollary \ref{CoNew}. Fix $x_0\in\rng(h)$ and define a function $f$ by putting $f(x)=x_0$ for $x\in\rng(h)$ and $f(x)=x$ otherwise. Notice that $f\in {\cal M}_{in}(SZ)$. Indeed, %chKC consider $g\in SZ$. In order to show that $g\circ f\in SZ$ by way of contradiction suppose that there is an $X\in [\mathR]^{\co}$ such that $g\circ f\restr X$ is continuous. But $\card(X\setminus\rng(h))=\co$, since $\cf(\co)<\co$. Moreover, $f(x)=x$ for every $x\in X\setminus\rng(h)$. So, $g\restr{X\setminus\rng(h)}=g\circ f\restr{X\setminus\rng(h)}$ is continuous on a set of cardinality $\co$, contradicting $g\in SZ$. On the other hand, $f\circ h$ is constant, so $f\circ h\not\in SZ$, while $h\in SZ$. Thus, $f\not\in {\cal M}_{out}(SZ)$. \Qed \begin{Pro}\label{PR:CompReg} Can inclusion ${\cal M}_{out}(SZ)\subset{\cal M}_{in}(SZ)$ be proved without the assumption that $\co$ is regular? \end{Pro} \subsection{Compositions with $SZ$ functions from the left.} \begin{Th}\label{TH7} For each $f\colon\mathR\to\mathR$ the following conditions are equivalent: \begin{description} \item[(i)] there exists $h\in SZ\cap\RRR$ such that $h\circ f\in SZ$; \item[(ii)] there exists $h\colon\mathR\to\mathR$ such that $h\circ f\in SZ$; \item[(iii)] $\card (f^{-1}(y))<\co$ for each $y\in\mathR$. \end{description} \end{Th} \pf (i)$\Rightarrow$(ii). Obvious. (ii)$\Rightarrow$(iii). Suppose that $\card (f^{-1}(y_0))=\co$ for some $y_0\in\mathR$. Then $h\circ f$ is constant on $f^{-1}(y_0)$, a contradiction. (iii)$\Rightarrow$(i). First notice that there exists $\E\su\co$ and a one-to-one enumeration $\{y_\alpha\colon \alpha\in\E\}$ of $\real$ such that \begin{description} \item{($\star$)} $\card(f^{-1}(y_\alpha))\leq\card(\alpha)$ for every $\alpha\in\E$. \end{description} To see it, let $\{y_\alpha\colon \alpha<\co\}$ be an enumeration of $\real$ with each number appearing $\co$ many times. For $y\in\real$ let $\alpha(y)=\min\{\alpha<\co\colon y_{\alpha}=y\,\&\,\card(f^{-1}(y))\leq\card(\alpha)\}$ and put $\E=\{\alpha(y)\colon y\in\real\}$. Then $\{y_\alpha\colon \alpha\in\E\}$ has the desired properties. Next, let $\{g_\xi\colon\xi<\co\}=\Cd$ and let $\{\alpha_\xi\colon\xi<\co\}$ be an increasing enumeration of $\E$. Then $\{y_{\aaa_\xi}\colon\xi<\co\}$ is a one-to-one enumeration of $\real$. For each $\xi<\co$ choose \[ h(y_{\aaa_\xi})\in\mathR\setminus \left(\{g_{\zeta}(y_{\alpha_\xi})\colon\zeta<\xi\}\cup \bigcup\left\{g_\zeta\left[f^{-1}(y_{\alpha_\xi})\right] \colon\zeta<\xi\right\}\right). \] Such a choice can be made, since the set $\bigcup\{g_\zeta[f^{-1}(y_{\alpha_\xi})]\colon\zeta<\xi\}$ is a union of $\card(\xi)<\co$ many sets, each set of cardinality $\leq\card(\alpha_\xi)<\co$. It is clear that $h\in SZ$. To verify that $h\circ f\in SZ$ fix $\zeta<\co$. Observe that $[h\circ f=g_\zeta]\su\bigcup_{\xi\leq\zeta}f^{-1}(y_{\alpha_\xi})$. Indeed, if $h\circ f(x)=g_\zeta(x)$ and $f(x)=y_{\aaa_\xi}$ some $\xi<\co$ then $h(y_{\aaa_\xi})\in g_\zeta\left[f^{-1}(y_{\aaa_\xi})\right]$. So $\xi\leq\zeta$ and $x\in\bigcup_{\xi\leq\zeta}f^{-1}(y_{\alpha_\xi})$. Thus, by ($\star$), $\card((h\circ f)\cap g_\zeta) \leq\card(\bigcup_{\xi\leq\zeta}f^{-1}(y_{\alpha_\xi})) \leq\card(\zeta) \cdot\card(\alpha_\zeta)<\co$. \Qed Theorem~\ref{TH7} justifies restriction of our attention only to the functions from a family \[ \R_1=\{f\in\RRR\colon\card (f^{-1}(y))<\co\mbox{ for every }y\in\mathR\} \] and definition \begin{eqnarray*} c_{out}(SZ) \!\! & = & \!\! \min(\{\card(\F)\colon \F\subset\R_1\ \&\ \neg\exists h\in\RRR\ \forall f\in \F\ h\circ f\in SZ\} \cup\{(2^{\co})^+\})\\ & = & \!\! \min(\{\card(\F)\colon \F\subset\R_1\ \&\ \forall h\in\RRR\ \exists f\in \F\ h\circ f\not\in SZ\} \cup\{(2^{\co})^+\}). \end{eqnarray*} Note that $SZ\su\R_1$, so $\card(\R_1)=2^\co$. %\chTN Now, we have the following analog of Theorem~\ref{TH1}. \begin{Th}\label{TH8} If $\co$ is a regular cardinal then \[ \co\omega_1$ and $\kappa$ is regular. Then it is relatively consistent with ZFC that the Continuum Hypothesis ($\continuum=\aleph_1$) is true, $2^{\continuum}=\lambda$, and $c_{out}(SZ)=\kappa$. \Qed \end{Co} \subsection{Compositions with $SZ$ functions from the right.} In this section we will examine for which functions $f\in\RRR$ there exists an $h\in\RRR$ such that $f\circ h\in SZ$. The class of all functions $f\in\RRR$ having this property will be denoted by $\R_2$. Also, as in previous sections, we will define the cardinal $c_{in}(SZ)$ analogous to $c_{out}(SZ)$ restricting our attention to the maximal family for which such a definition has a sense, i.e., to $\R_2$. Thus, we define \begin{eqnarray*} c_{in}(SZ) \! & = & \! \min(\{\card(\F)\colon \F\subset\R_2\ \&\ \neg\exists h\in\RRR\ \forall f\in \F\ f\circ h\in SZ\} \cup\{(2^{\co})^+\})\\ & = & \! \min(\{\card(\F)\colon \F\subset\R_2\ \&\ \forall h\in\RRR\ \exists f\in \F\ f\circ h\not\in SZ\} \cup\{(2^{\co})^+\}). \end{eqnarray*} The next theorem gives a characterization of the family $\R_2$ in case when $\co$ is regular. \begin{Th}\label{TH9} Assume that $\co$ is a regular cardinal. For each $f\colon\mathR\to\mathR$ the following conditions are equivalent: \begin{description} \item[(i)] there exists $h\in SZ\cap\RRR$ such that $f\circ h\in SZ$; \item[(ii)] there exists $h\colon\mathR\to\mathR$ such that $f\circ h\in SZ$; \item[(iii)] $\card (\rng(f))=\co$. \end{description} \end{Th} \pf (i)$\Rightarrow$(ii). Obvious. (ii)$\Rightarrow$(iii). Note that $\card (\rng (h))=\co$. Indeed, otherwise, by regularity of $\co$, $\card (h^{-1}(y_0))=\co$ for some $y_0\in\mathR$ and then $f\circ h$ is constant on $h^{-1}(y_0)$ for any $f$, a contradiction. Next, by way of contradiction, suppose that $\card (\rng(f))<\co$. Then, there exists a $y_0\in\mathR$ such that $\card(f^{-1}(y_0) \cap\rng (h))=\co$. Therefore, $\card ((f\circ h)^{-1}(y_0))=\co$, a contradiction. (iii)$\Rightarrow$(i). Let $\{g_{\alpha}\colon\alpha<\co\}=\Cd$, and $\{x_{\alpha}\colon\alpha <\co\}=\real$. For every $\aaa <\co$ choose $$ h(x_{\aaa})\in\mathR\setminus\Bigl( \{g_{\bbb}(x_{\aaa})\colon\, \bbb\leq\aaa \} \cup \bigcup_{\bbb\leq\aaa}f^{-1} (g_{\bbb}(x_{\aaa}) )\Bigl). $$ The choice can be made since, by (iii), $\mathR\setminus (\{g_{\bbb}(x_{\aaa})\colon\,\bbb\leq\aaa \} \cup \bigcup_{\bbb\leq\aaa}f^{-1}(g_{\bbb}(x_{\aaa})))$ is not empty. Obviously, $h\in SZ$. It is enough to verify that $f\circ h\in SZ$. So, fix $\aaa <\co$. Then $\{ x\colon\, f\circ h(x)=g_{\aaa}(x)\}= \{x\colon\, h(x)\in f^{-1}(g_{\aaa}(x))\}\subset \{x_{\bbb}\colon\, \bbb<\aaa\}$, and so $\card ((f\circ h)\cap g_{\aaa})<\co$. \Qed Note that we did not use the regularity assumption in implications (iii)$\Rightarrow$(i) and (i)$\Rightarrow$(ii). In particular, if \[ \R_2^\star=\{f\in\RRR\colon\card(\rng(f))=\co\} \] then \begin{Co}\label{CorRight1} $\R_2^\star\subset\R_2$. \Qed \end{Co} We have also. \begin{Co}\label{CorRight2} If $\co$ is a regular cardinal then $\R_1\subset\R_2=\R_2^\star$. \Qed \end{Co} \begin{Ex}\label{Ex2} There exist functions $f_0,f_1\in\R_2^\star$ such that for every $h\colon\mathR\to\mathR$ either $f_0\circ h\not\in SZ$ or $f_1\circ h\not\in SZ$. \end{Ex} \pf Indeed, decompose the real line onto two sets $A_0$ and $A_1$ such that $\card(A_i)=\co$ for $i<2$, and define a function $f_i$ such that $f_i(A_i)=0$ and $f_i\restr {A_{1-i}}$ is one-to-one. Fix an $h\colon\mathR\to\mathR$. Since $\real=h^{-1}(\real)=h^{-1}(A_0)\cup h^{-1}(A_1)$ there exists $i<2$ such that $\card(h^{-1}(A_i))=\co$. Then $\card((f_i\circ h)^{-1}(0))=\card(h^{-1}(A_i))=\co$, so $f_i\circ h\not\in SZ$. \Qed \begin{Co}\label{C2} $c_{in}(SZ)=2$.\Qed \end{Co} %%%%%%% SZ FUNCTIONS. PART III %%%%%%%%%%%%%% \subsection{Coding functions by $SZ$ functions.} In the previous sections we examined when for a given function $f\in\RRR$ there exist two $SZ$ functions $g,h\in\RRR$ such that $f\circ h=g$ or $h\circ f=g$. In this section we will ask for which $f\in\RRR$ there exist $SZ$ functions $g,h\in\RRR$ such that $f=g\circ h$ or $f=h\circ g$, i.e., that $f$ is coded by two $SZ$ functions. Note that even when for some $f$ the first set of questions have a positive answer with $h$ being one-to-one, this does not imply the positive answer for the second set of questions, since the inverse of an $SZ$ function does not have to be $SZ$. In fact, it is consistent with ZFC that no $SZ$ function $h\colon\real\to\real$ has an $SZ$ inverse. This happens in the iterated perfect set model, where there is no $SZ$ function from $\real$ onto $\real$~\cite{BCN}. (If $h^{-1}$ is $SZ$ then it is onto $\real$ and any of its $SZ$ extension is an $SZ$ function from $\real$ onto $\real$.) The same example also shows, that the set of questions we consider in this section cannot have a positive answer in ZFC for any function from $\real$ onto $\real$, even for the identity function. Thus, we will work here with the additional set theoretical assumptions. We will start with the following lemmas. \begin{Le}\label{NL1} Assume that $\co$ is a regular cardinal. Then the class $\FF$ is closed under the compositions of functions. \end{Le} \pf Suppose that $f=f_2\circ f_1$, $f_1,f_2\in \FF$ and $\card (f^{-1}(y_0))=\co$ for $y_0\in\mathR$. Then $f$ is constant on the set $X=f^{-1}(y_0)=\bigcup\{ (f_1)^{-1}(t)\colon\, t\in (f_2)^{-1}(y_0)\}$, so either $f_1$ or $f_2$ is constant on a set of cardinality $\co$, a contradiction. \Qed Note that if $\co$ is a singular cardinal then %\chTN by Example~\ref{ExComp1}, the conclusion of Lemma~\ref{NL1} is false. \begin{Pn} If $\co$ is a singular cardinal then every function from $\mathR$ into $\mathR$ is a composition of two functions from the class $\R_1$. \end{Pn} \pf Suppose that $\mathR=\{ x_{\aaa}\colon\aaa<\co\}$, $\kkk=\cf(\co)<\co$ and $\la\lll_{\aaa}\colon\aaa<\kkk\ra$ is an increasing sequence of cardinals such that $\co=\bigcup_{\aaa<\kkk} \lll_{\aaa}$. Fix $f\in\RRR$. For every $\aaa<\co$ let $X_{\aaa}=f^{-1}(x_{\aaa})$ and let $X_{\aaa}=\bigcup_{\bbb<\kkk}X_{\aaa,\bbb}$ be a partition such that $\card(X_{\aaa,\bbb})\leq\lll_{\bbb}$ for every $\bbb<\kkk$. Choose a sequence $\la Y_{\aaa}\colon\aaa<\co\ra$ of pairwise disjoint sets of reals, each of cardinality equal to $\kkk$; $Y_{\aaa}=\{ y_{\aaa,\bbb} \colon\bbb<\kkk\}$ and define $f_{1}(x)=y_{\aaa,\bbb}$ for $x\in X_{\aaa,\bbb}$ and $\hat{f}_2(y_{\aaa,\bbb})=x_{\aaa}$ for $\aaa<\co$, $\bbb<\kkk$. Let $f_2\in\R_1$ be any extension of $\hat{f}_2$. Then $f=f_2\circ f_1$. \Qed \begin{Le}\label{L1} Assume $f\in\FF$. Then $f\in SZ$ if and only if $\card (f\cap g)<\co$ for each continuous nowhere constant function $g$ defined on a $G_{\delta}$-set. \end{Le} \pf The implication ``$\Rightarrow$'' is obvious. To prove ``$\Leftarrow$'' assume that $g$ is a continuous function defined on a $G_{\delta}$-set $G$. Let $\la G_n\ra_{n<\omega}$ be a sequence of all maximal intervals in $G$ (i.e., non-empty sets of the form $G\cap (a,b)$, for $a\omega$. \Qed The next theorem tells us that for every sequence $\la f_{\aaa}\colon\aaa <\co\ra$ of $\R_1$ functions there exists a sequence $\la f^{\tr}_{\aaa}\colon\aaa <\co\ra$ of their $SZ$ codes and an $\circ$-decoder function $h\in SZ$ such that every $f_{\aaa}$ can be ``right $\circ$-decoded'' by $h$ from $f^{\tr}_{\aaa}$. \begin{Th}\label{TH10} Assume that the real line is not a union of less than $\co$ many meager sets. Then for every family $\{f_{\aaa}\colon\aaa <\co\}\subset\FF$ there is a family $\{f^{\tr}_{\aaa}\colon\, \aaa <\co\}$ of $SZ$-functions and a ``decoding'' function $h\in SZ$ and such that $f^{\tr}_{\aaa}\circ h=f_{\aaa}$ for each $\aaa <\co$. \end{Th} \pf Let $\C_n=\{g_{\alpha}\colon\alpha <\co\}$ be an enumeration of all nowhere constant $g\in\Cd$ and let $\{x_{\alpha}\colon\alpha <\co\}=\real$. For every $\aaa <\co$ choose \[ h(x_{\aaa})\in\mathR\setminus \Bigl(\{g_{\bbb}(x_{\aaa})\colon\bbb\leq\aaa\} \cup\{h(x_{\bbb})\colon\bbb<\aaa\} \cup\bigcup\{g_{\bbb}^{-1}(f_{\nu}(x_{\aaa}))\colon \bbb,\nu\leq\aaa\}\Bigl). \] Note that the choice can be made since every set $g_{\bbb}^{-1}(f_{\nu}(x_{\aaa}))$ is meager and $\real$ is not a union of less than $\co$ many meager sets. It is easy to observe that the function $h$ is one-to-one and so, $h\in\R_1$. Also, by our choice, $\card([h=g])<\co$ for every $g\in\C_n$. So, by Lemma~\ref{L1}, $h\in SZ$. Now for $\nu<\co$ define $f_{\nu}^{\tr}$. Put $f_{\nu}^{\tr}(h(x_{\aaa}))=f_{\nu}(x_{\aaa})$ for every $\aaa<\co$ and for $x\not\in\rng (h)$ define $f_{\nu}^{\tr}(x)=h(x)$. Clearly $f_{\nu}=f_{\nu}^{\tr}\circ h$ for every $\nu<\co$. To see that $f_{\nu}^{\tr}\in SZ$ first notice that $f_{\nu}^{\tr}\in\R_1$, since for every $y\in\real$ the set \begin{eqnarray*} (f_{\nu}^{\tr})^{-1}(y) & = & \{h(x)\colon f_{\nu}^{\tr}(h(x))=y)\} \cup \{z\in\real\setminus\rng(h)\colon f_{\nu}^{\tr}(z)=y\}\\ & \subset & h[f_{\nu}^{-1}(y)] \cup h^{-1}(y) \end{eqnarray*} has cardinality less than $\co$ as $h,f_{\nu}\in\R_1$. Moreover, for every $\beta<\co$ \begin{eqnarray*} [f_{\nu}^{\tr}=g_\beta] & = & \{h(x)\colon f_{\nu}^{\tr}(h(x))=g_\beta(h(x))\} \cup \{z\in\real\setminus\rng(h)\colon f_{\nu}^{\tr}(z)=g_\beta(z)\}\\ & = & h[\{x\colon f_{\nu}(x)=g_\beta(h(x))\}] \cup \{z\in\real\setminus\rng(h)\colon h(z)=g_\beta(z)\}\\ & = & h[\{x\colon h(x)\in g_\beta^{-1}(f_{\nu}(x))\}] \cup ([h=g_\beta]\setminus\rng(h))\\ & \subset & h[\{x_\alpha\colon \alpha<\max\{\beta,\nu\}\}] \cup [h=g_\beta]. \end{eqnarray*} Thus, $\card([f_{\nu}^{\tr}=g])<\co$ for every $g\in\C_n$ and, by Lemma~\ref{L1}, $f_{\nu}^{\tr}\in SZ$. \Qed Lemma~\ref{NL1} together with Theorem~\ref{TH10} yield to the following result: \begin{Co}\label{COO} Assume that the real line is not a union of less than $\co$ many meager sets and that $\co$ is a regular cardinal. For every $f\colon \mathR\to\mathR$ the following conditions are equivalent: \begin{description} \item[(i)] there exist $h, f^{\tr}\in SZ$ such that $f=f^{\tr}\circ h$; \item[(ii)] $f\in\FF$. \Qed \end{description} \end{Co} Note that Theorem~\ref{TH10} cannot be proved in ZFC since, as mentioned above, there exists a model $V$ of ZFC in which no real function onto $\real$ (including the identity function) is a composition of two $SZ$ functions. Nevertheless, we have the following example. \begin{Ex}\label{Ex3} There exists an $SZ$ function $h\colon\real\to\real$ such that its $n$-th composition $h^n$ is $SZ$ for every $n>0$. \end{Ex} \pf Let $\{g_{\alpha}\colon\alpha <\co\}=\Cd$ and $\{x_{\alpha}\colon\alpha <\co\}=\mathR$. For every $\ggg<\co$ choose $$ h(x_{\ggg})\in\mathR\setminus\Bigl (\{ g_{\bbb}(x_{\aaa})\colon \aaa,\bbb\leq\ggg\}\cup\{ x_{\aaa}\colon\aaa\leq\ggg\}\Bigl ). $$ Observe that $h\in SZ$. We shall verify that $h^n\in SZ$ for $n>1$. Suppose that $g_{\bbb}(x_{\aaa})=h^n(x_{\aaa})$. Let $x_{\ggg}=h^{n-1}(x_{\aaa})$. Note that $\ggg>\aaa$ and $g_{\bbb}(x_{\aaa})=h(x_{\ggg})$, so $\ggg<\bbb$. Therefore $\{x\colon h^n(x)=g_{\bbb}(x)\}\subset\{x_{\aaa}\colon\aaa<\bbb\}$, so $\card(h^n\cap g_{\bbb})<\co$. \Qed Now, we consider the following cardinals. (See~\cite{CR}.) \begin{eqnarray*} c_{r}(SZ) &= & \min\{\card(\F)\colon\F\subset\R_1\ \&\ \neg\exists h\in SZ\ \forall f\in \F\ \exists f^{\tr}\in SZ\ f=f^{\tr}\circ h\}\\ &= & \min\{\card(\F)\colon \F\subset\R_1\ \&\ \forall h\in SZ\ \exists f\in \F\ \forall f^{\tr}\in SZ\ f\neq f^{\tr}\circ h\} \end{eqnarray*} and \begin{eqnarray*} c_{l}(SZ) &= & \min\{\card(\F)\colon \F\subset\R_1\ \&\ \neg\exists h\in SZ\ \forall f\in \F\ \exists f^{\tl}\in SZ\ f=h\circ f^{\tl}\}\\ &= & \min\{\card(\F)\colon \F\subset\R_1\ \&\ \forall h\in SZ\ \exists f\in \F\ \forall f^{\tl}\in SZ\ f\neq h\circ f^{\tl}\}. \end{eqnarray*} (We will assign the value $(2^\co)^+$ in case when the minimum is run over the empty set.) Note, that by the remark above in the iterated perfect set model the following corollary holds. \begin{Co}\label{CorIndep} It is consistent with ZFC that $\co=\omega_2$ and $c_r(SZ)=c_l(SZ)=1$.\Qed \end{Co} \begin{Th}\label{TH11} Assume that the real line is not a union of less than $\co$ many meager sets and that $\co$ is a regular cardinal. Then $$\co < c_r(SZ)\leq 2^{\co}.$$ \end{Th} \pf The inequality $\co \omega_1$ and $\kappa$ is regular. Then it is relatively consistent with ZFC+CH that $2^{\continuum}=\lambda$ and Lus$_\kappa(\poset^{\tr})$ holds. \Qed \end{Th} We will prove the following theorem. \begin{Th}\label{TH:CR1} If $\co=\omega_1$ and $\kappa>\co$ is a regular cardinal then Lus$_\kappa(\poset^{\tr})$ implies that $c_r(SZ)=\kappa$. \end{Th} This and Theorem~\ref{TH:CR} will immediately imply the following corollary. \begin{Co}\label{CO:CR} Let $\lambda\geq\kappa\geq\omega_2$ be cardinals such that $\cf(\lambda)>\omega_1$ and $\kappa$ is regular. Then it is relatively consistent with ZFC+CH that $2^{\continuum}=\lambda$ and $c_r(SZ)=\kappa$. \Qed \end{Co} The proof of Theorem \ref{TH:CR1} will be split into three lemmas. \begin{Le}\label{L:CR1} \begin{description} \item[(i)] Assume that a union of less than continuum many meager sets is meager again. Then Lus$_\kappa(\poset^{\tr})\implies$ Lus$_\kappa(\poset)$. \item[(ii)] For any regular $\kappa$ we have Lus$_\kappa(\poset^{\tr})\implies$ MA$_\kappa(\poset^{\tr})$. \end{description} \end{Le} \pf The proof is similar to the proof of Lemma~\ref{L:AAAzero}. The only modification is that in the proof of (i) we must replace the condition ``$\card(r^{-1}(y))=\co$ for every $y\in\mathR$'' by ``for every $y\in\mathR$ the level set $r^{-1}(y)$ is not meager'' and that we choose $s(x)\in r^{-1}(q(x))\setminus \bigcup\{g^{-1}(f(x))\colon f\in E\ \$\ g\in G\}$. \Qed \begin{Le}\label{L:CR2} Assume that $\co$ and $\kkk$ are regular cardinals and $\kkk>\co$. Then Lus$_\kappa(\poset)$ implies that $c_r(SZ)\leq\kappa$. \end{Le} \pf Let $\la G_\alpha\colon\alpha<\kappa\ra$ be a $\kappa$-Lusin sequence of $\poset$-filters and define \[ g_\alpha=\Union G_\alpha. \] Then similarly as in the proof of Lemma~\ref{L:AAA1} we can assume that each $g_\alpha$ is a total function from $\reals$ into $\reals$. Let $\{x_{\xi}\colon \xi<\co\}$ be an enumeration of $\mathR$. For every $\aaa<\kkk$ put $$ X_{\aaa}=\{ x_{\xi}\colon g_{\aaa}(x_{\xi})\neq g_{\aaa}(x_{\eta})\ \mbox{ for every } \ \eta<\xi\} $$ and let $f_{\aaa}\in\FF$ be an extension of $g_{\aaa}\restr X_{\aaa}$. We will show that for an arbitrary $h\in\RRR$ there is an $\aaa<\kkk$ such that $f_{\aaa}=f^{\tr}_{\aaa}\circ h$ for no $f^{\tr}_{\aaa}\in SZ$. If $h\not\in\FF$ then $f^{\tr}_{\aaa}\circ h\not\in\FF$ for each $f^{\tr}_{\aaa}\in\RRR$ and, since $f_{\aaa}\in\FF$, $f_{\aaa}\neq f_{\aaa}^{\tr}\circ h$. So, assume that $h\in\FF$. Then $\card(\rng(h))=\co$, because $\co$ is a regular cardinal. For $\xi<\co$ let $D_\xi$ be the set of all $p\in\poset$ such that \[ \exists\ggg\geq\xi\ \ \Bigl[\Bigl(\forall\aaa\leq\ggg\Bigl)\Bigl(x_{\aaa}\in\dom(p)\Bigl)\ \&\ \Bigl(\forall\aaa<\ggg\Bigl) \Bigl(p(x_{\aaa})\neq p(x_{\ggg})\Bigl) \ \&\ p(x_{\ggg})=h(x_{\ggg})\Bigl] \] and observe that every $D_\xi$ is dense in $\poset$. Indeed, for every $p\in\poset$ there is $\ggg\geq \xi$ with $x_{\ggg}\not\in\dom(p)$ and $h(x_{\ggg})\not\in\rng(p)$. Choose $y\neq h(x_{\ggg})$ and set $q=p\cup\{ (x_{\ggg},h(x_{\ggg}))\}\cup\{ (x_{\eta},y)\colon \ \eta<\ggg \ \&\ x_{\eta}\not\in\dom(p)\}$. Then $q\in D_{\xi}$ and $q\leq p$. By the regularity of $\kkk$, there exists $\alpha<\kappa$ such that $G_\alpha$ intersects every set $D_\xi$ with $\xi<\co$. Note that this implies that $\card(X_{\aaa})=\co$. Now, suppose that $f_{\aaa}=f^{\tr}_{\aaa}\circ h$. We will show that $f^{\tr}_{\aaa}\not\in SZ$. To see it note first that if $Y_{\aaa}=\{x\in X_{\aaa}\colon f_{\aaa}(x)=h(x)\}$, then $\card(Y_{\aaa})=\co$, since $G_\alpha$ intersects every set $D_\xi$. So, $h\in\FF$ and the regularity of $\co$ imply that $\card(\rng(h\restr Y_{\aaa}))=\co$. Finally, observe that $f^{\tr}_{\aaa}(h(x))=h(x)$ when $f_{\aaa}(x)=h(x)$, so $\rng(h\restr Y_{\aaa})\subset[f^{\tr}_{\aaa}=\id]$. Therefore $\card(f^{\tr}_{\aaa} \cap\id)=\co$ and consequently, $f^{\tr}_{\aaa}\not\in SZ$. \Qed \begin{Le}\label{L:CR3} If $\kappa>\co=\omega_1$ then MA$_\kappa(\poset^{\tr})$ implies that $c_r(SZ)\geq\kappa$. \end{Le} \pf Let $\F\su\FF$ be such that $\card(\F)<\kappa$. We shall find $h\in SZ$ such that for every $f\in\F$ there exists $f^{\tr}\in SZ$ with $f=f^{\tr}\circ h$. Observe that for any $x\in\mathR$ the set $$ D_x=\{\la p,E,H\ra\in\poset^{\tr}\colon x\in\dom(p)\} $$ is dense in $\poset^{\tr}$. Indeed, for $\la q,E,H\ra\in\poset^{\tr}\setminus D_x$ choose $y\in\mathR\setminus\bigcup\{ g^{-1}(f(x))\colon g\in H\ \&\ f\in E\}$. The choice is possible since, by CH, the set $\bigcup\{g^{-1}(f(x))\colon\ g\in H\ \&\ f\in E\}$ is meager as a countable union of nowhere dense sets. Put $p=q\cup\{\la x,y\ra\}$. Then $\la p,E,H\ra\leq\la q,E,H\ra$ and $\la p,E,H\ra\in D_x$. Note also that for any $f\in\RRR$ and $g\in\C_n$ the set $$ E_{f,g}=\{ \la p,E,H\ra\colon\ f\in E \ \&\ g\in H\} $$ is dense in $\poset^{\tr}$, because $\la p,E\cup\{f\},H\cup\{g\}\ra$ extends $\la p,E,H\ra$. Let $$ \D=\{D_x\colon x\in\mathR\} \cup\{E_{\bar{g},{\rm id}}\colon g\in\Cd\} \cup\{E_{f,k}\colon f\in\F\ \&\ k\in\C_n\}, $$ where $\bar{g}\in\RRR$ extends $g\in\Cd$ by associating $0$ at all undefined places. Then $\D$ is a family of less than $\kkk$ many dense subsets of $\poset^{\tr}$. Let $G$ be a $\D$-generic filter in $\poset^{\tr}$ and let $$ \hat{h}=\bigcup\{p\colon\ \exists E\subset\RRR\ \exists H\subset\C_n\ \la p,E,H\ra\in G\}. $$ Since $G\cap D_x\neq\emptyset$ for every $x\in\mathR$, $\hat{h}$ is a total function from $\mathR$ into $\mathR$. Observe that $\hat{h}\in SZ$. Indeed, fix $g\in\Cd$ and $\la p,E,H\ra \in E_{\bar{g},{\rm id}}\cap G$. Then $\{x\colon \hat{h}(x)=g(x)\}\subset \{ x \colon\ \hat{h}(x)=\bar{g}(x)\}\subset\dom(p)$, so $\card(\hat{h}\cap g)<\co$. To define $h$ note that by CH all level sets of $\hat{h}$ are countable. In particular, the set $\hat{h}^{-1}(\rationals)$ is also countable. For every $y\in\rng(h)\cap\N$ let $\hat{h}^{-1}(y)=\{ x_{y,n}\colon n<\omega\}$. Choose a one-to-one sequence $\la s_n\colon n<\omega\ra$ of irrationals and define a function $h^{\ast}\colon\mathR\setminus h^{-1}(\rationals) \to\N$ by $h^{\ast}(x_{y,n})=\la s_n,y\ra$, where we identify $\N=\mathR\setminus\rationals$ with $\N\times\N$ via natural homeomorphism. Note that $h^{\ast}$ is one-to-one. Let $h\colon\mathR\to\mathR$ be a one-to-one extension of $h^{\ast}$. Then $h\in SZ$. Indeed, suppose that $h\restr X\in\C$ for some $X\in [\mathR]^{\co}$. Then $X_0=X\setminus\hat{h}^{-1}(\rationals)\in[\mathR]^{\co}$ and $h^{\ast}\restr X_0=h\restr X_0\in\C$, so $\hat{h}\restr X_0 ={\rm pr}_y\circ h^{\ast}\restr X_0\in\C$, contrary to $\hat{h}\in SZ$. Now, for arbitrary $f\in\F$ define $\tilde{f}\colon\rng(h)\to\mathR$ by $\tilde{f}(y)=f(x)$ for $x=h^{-1}(y)$. We shall verify that $\tilde{f}\in SZ$. First, note that$\tilde{f}\in\FF$, because $f\in\FF$. So, by Lemma~\ref{L1}, it is enough to verify that $\card(\tilde{f}\cap g)<\co$ for every $g\in \C_n$. So, fix $g\in\C_n$ and suppose that $X=[\tilde{f}=g]\in [\rng(h)]^{\co}$. Then there exists $X_0\in [\rng(h^{\ast})]^{\co}$ such that $X_0\subset [\tilde{f}=g]$. Therefore there are $n<\omega$ and $Z\in [\rng(\hat{h})\cap\N]^{\co}$ such that $\{s_n\}\times Z\subset X_0$, so $Z\subset\{\hat{h}(x)\colon\la s_n,\hat{h}(x)\ra\in g^{-1}(f(x))\}$. Let $\varphi\colon\N\to\{s_n\}\times\N$ be a function defined by: $\varphi(y)=\la s_n,y\ra$. Then $\varphi$ is a homeomorphism, so $k=g\circ\varphi\restr \varphi^{-1}(\dom(g))\in\C_n$. Let $\la p,E,H\ra\in G\cap E_{f,k}$. Then $Y=\{ x\colon \hat{h}(x)\in k^{-1}(f(x))\}\subset\dom(p)$, so $\card(Y)<\co$. But $Z\subset \hat{h}(Y)$, contrary to $\card(Z)=\co$. Finally, let $f^{\tr}\colon\mathR\to\mathR$ be an SZ-extension of $\tilde{f}$. Then $f=f^{\tr}\circ h$. \Qed \begin{Th}\label{TH12a} Assume that the real line is not a union of less than $\co$ many meager sets and that $\co$ is a regular cardinal. Then $$c_l(SZ)>\co.$$ \end{Th} \pf To see it take $\{f_\beta\colon\beta<\co\}\subset\FF$. We will construct an $h\in SZ$ and a family $\{f^{\tl}_\beta\colon\beta<\co\}$ of $SZ$-functions such that $f_\beta=h\circ f^{\tl}_\beta$ for each $\beta<\co$. Let $\C_n=\{g_{\alpha}\colon\alpha <\co\}$ be an enumeration of all nowhere constant $g\in\Cd$ and $\{z_\xi\colon\xi<\co\}$ be a one-to-one enumeration of $Z=\bigcup_{\xi<\co}\rng(f_\xi)$. Define inductively a sequence $\la y_\xi\colon\xi<\co\ra$ by choosing for every $\xi<\co$ \[ y_\xi\in\mathR\setminus \Bigl(\{y_\zeta\colon\zeta<\xi\} \cup\bigcup\{g_\alpha^{-1}(z_\xi)\colon\alpha\leq\xi\}\cup \bigcup\{g_\alpha[f_\beta^{-1}(z_\xi)]\colon\alpha,\beta\leq\xi\} \Bigl). \] The choice can be made, since the exceptional set is a union of less than $\co$ many meager sets. Now, let $Y=\{y_\xi\colon\xi<\co\}$, and define $h\colon Y\to\real$ by putting \[ h(y_\xi)=z_\xi \] for every $\xi<\co$. Moreover, for every $\beta<\co$ define $f^{\tl}_{\bbb}\colon\real\to\real$ by a formula \[ f^{\tl}_{\bbb}(x)= y_\xi\ \mbox{ iff }\ x\in f_\beta^{-1}(z_\xi). \] Note that $f^{\tl}_{\bbb}$ is defined on $\real$ since $\rng(f_\beta)\subset Z$. Also, $f_\beta=h\circ f^{\tl}_\beta$ for every $\beta<\co$, since for every $x\in\real$ there exists $\xi<\co$ such that $f_\beta(x)=z_\xi$, and $f_\beta(x)=z_\xi=h(y_\xi)=h(f^{\tl}_\beta(x))$, as $x\in f_\beta^{-1}(z_\xi)$. To see that $h\in SZ$ note first that $h$ is one-to-one, so $h\in\R_1$. Thus, by Lemma~\ref{L1}, it is enough to show that $\card([h=g_\alpha])<\co$ for every $\alpha<\co$. But if $g_\alpha(y_\xi)=h(y_\xi)=z_\xi$ then $y_\xi\in g_\alpha^{-1}(z_\xi)$ and, by the choice of $y_\xi$, $\alpha>\xi$. So, $[h=g_\alpha]\su\{y_\xi\colon\xi<\alpha\}$ has cardinality less than $\co$, and $h\in SZ$. Next fix $\beta<\co$ and notice that $f^{\tl}_\beta\in\R_1$. To see that $f^{\tl}_\beta\in SZ$ fix $\alpha<\co$. We will show that $\card([f^{\tl}_\beta=g_\alpha])<\co$. So, let $g_\alpha(x)=f^{\tl}_\beta(x)=y_\xi$. Then $x\in f_\beta^{-1}(z_\xi)$ and $y_\xi=g_\alpha(x)\in g_\alpha[f_\beta^{-1}(z_\xi)]$. So, by the choice of $y_\xi$, $\alpha>\xi$ or $\beta>\xi$. In particular, $[f^{\tl}_\beta=g_\alpha]\su\{y_\xi\colon\xi< \max(\alpha,\beta)\}$ has cardinality less than $\co$. \Qed \begin{Pro}\label{PRO4} Can it be proved in ZFC that $c_l(SZ)\leq 2^{\co}$? What about under CH? \end{Pro} \section{Final remarks.} Proofs of the following statements are left to the reader. \begin{enumerate} \item Every function $f\in\RRR$ is the uniform limit of a sequence of $SZ$-functions. \item Assuming $\cf(\co)=\omega_1$, every function $f\in\RRR$ is the transfinite limit of a sequence of $SZ$-functions (cf.,~\cite{Si}). \item Assuming $\co$ is a regular cardinal, the discrete limits of sequences of $SZ$-functions are in the class $SZ$ (cf.,~\cite{CL}). \item If $f,g\in SZ$, then $\max(f,g)\in SZ$ and $\min(f,g)\in SZ$ (hence the family $SZ$ forms a lattice of functions). \end{enumerate} \begin{thebibliography}{22} \bibitem{BCN} M.~Balcerzak, K.~Ciesielski and T.~Natkaniec, {\it Sierpi{\'n}ski-Zygmund functions that are Darboux, almost continuous or have a perfect road}, Archive for Math. Logic, to appear. \bibitem{Bar} T. Bartoszy{\'n}ski, {\it Combinatorial aspects of measure and category}, Fund. Math. {\bf 127} (1987), 225--239. \bibitem{CM} K.~Ciesielski and A.~W.~Miller, {\it Cardinal invariants concerning functions, whose sum is almost continuous\/}, Real Anal. Exchange {\bf 20} (1994--95), 657--673. \bibitem{CR} K.~Ciesielski and I.~Rec{\l}aw, {\it Cardinal invariants concerning extendable and peripherally continuous functions}, Real Anal. Exchange {\bf 21} (1995-96), 459--472. \bibitem{CL} A.~Csasz\'{a}r and M.~Laczkovich, {\it Discrete and equal convergence}, Studia Sci. Math. Hungar. {\bf 10} (1975), 463--472. \bibitem{Da} U.~Darji, {\it Decomposition of functions}, Invited talk during the 19th Summer Symposium in Real Analysis, Real Anal. Exchange {\bf 21(1)} (1995--96), 19--25. \bibitem{df} E.~van Douwen and W.~Fleissner, {\it Definable forcing axiom: an alternative to Martin's axiom}, Topology Appl.~{\bf 35} (1990), 277--289. \bibitem{Lan} A. Landver, {\it Baire numbers, uncountable Cohen sets and perfect-set forcing}, Journal of Symbolic Logic {\bf 57} (1992), 1086--1107. \bibitem{sur} A.~Miller, {\it Special sets of reals}, in {\bf Set Theory of the Reals} (edited by Haim Judah), Israel Mathematical Conference Proceedings, {\bf 6} (1993), 415--432, Amer. Math. Soc. \bibitem{mp} A.~Miller, K.~Prikry, {\it When the continuum has cofinality $\omega_1$}, Pacific J. Math. {\bf 115} (1984), 399--407. \bibitem{AC} T.~Natkaniec, {\it Almost continuity}, Real Anal. Exchange {\bf 17}, (1991--92), 462--520. \bibitem{SU} T.~Natkaniec, {\it New cardinal invariants in real analysis}, %\chTN Bull. Polish Acad. Sci.~{\bf 44}, 1996, 251--256. \bibitem{NR} T.~Natkaniec and I.~Rec{\l}aw, {\it Cardinal invariants concerning functions whose product is almost continuous}, Real Anal. Exchange {\bf 20} (1994--95), 281--285. \bibitem{Si} W.~Sierpi\'{n}ski, {\it Sur les suites transfinies convergentes de fonctions de Baire}, Fund. Math. {\bf 1} (1920), 132--141. \bibitem{SZ} W.~Sierpi\'{n}ski, A.~Zygmund, {\it Sur une fonction qui est discontinue sur tout ensemble de puissance du continu}, Fund. Math. {\bf 4} (1923), 316--318. \bibitem{tod} S.~Todorcevic, {\it Remarks on Martin's axiom and the continuum hypothesis}, Canad. J. Math. {\bf 43} (1991), 832--851. \end{thebibliography} \end{document}