% Corrected according to the referee suggestions. 6/27/97;
% Corrected again by TN 7/17/97 when Rosen found error
% Corrected again by TN 7/25/97 when Rosen found another small error
% Corrected once more by KC 9/15/97 

\documentclass{rae}\usepackage{amsfonts}
\usepackage{amssymb}\usepackage{latexsym}
\usepackage{amsmath}

\newtheorem{Th}{Theorem}[section]
\newtheorem{Le}{Lemma}[section]
\newtheorem{Co}{Corollary}[section]
\newtheorem{Po}{Proposition}[section]
\newtheorem{Ro}{Remark}[section]
\newtheorem{Pro}{Problem}
\newcommand\mathN{{\mathbb N}}
\newcommand\mathI{{\mathbb I}}
\newcommand\mathR{{\mathbb R}}
\newcommand\mathQ{{\mathbb Q}}
\newcommand\real{\mathR}
\newcommand{\RRR}{\mathR^{\mathR}}
\newcommand{\restr}{{\restriction}}
\newcommand{\co}{{\frak c}}
\newcommand{\Qed}{\unskip\nolinebreak\quad\hfill$\Box\;\;$\medskip}

\def\PO{{\cal P}(\mathR)}
\newcommand{\F}{{\cal F}}
\def\P{{\cal P}}
\newcommand{\A}{{\cal A}}
\newcommand{\B}{{\cal B}}
\newcommand{\C}{{\cal C}}
\newcommand{\K}{{\cal K}}
\newcommand{\CAB}{{\C_{\A,\B}}}
\newcommand{\DAB}{{\C^{-1}_{\A,\B}}}
\newcommand{\AC}{{\rm AC}}
\newcommand{\Const}{{\rm Const}}
\newcommand{\G}{{\cal G}}
\newcommand{\CC}{{\rm C}}
\newcommand{\PR}{{\rm PR}}
\newcommand{\PC}{{\rm PC}}
\newcommand{\D}{{\rm D}}
\newcommand{\Conn}{{\rm Conn}}
\newcommand{\CIVP}{{\rm CIVP}}
\newcommand{\SCIVP}{{\rm SCIVP}}
\newcommand{\WCIVP}{{\rm WCIVP}}
\newcommand{\DIVP}{{\rm DIVP}}
\newcommand{\QU}{{\rm QU}}
\newcommand{\Ext}{{\rm Ext}}
\newcommand{\Q}{{\rm Q}}
\newcommand{\SZ}{{\rm SZ}}

\newcommand\cf{{\rm cf}}
\newcommand\id{{\rm id}}
\newcommand\cl{{\rm cl}\,}
\newcommand\Cl{{\rm cl}\,}
\newcommand\Int{{\rm int}\,}
\newcommand{\pf}{{\noindent\sc Proof. }}
\newcommand{\charf}[1]{\mbox{\raise.48ex\hbox{$\chi$}$_{#1}$}}


\author{
Krzysztof Ciesielski,%
\thanks{This work was partially 
supported by NSF Cooperative Reasearch Grant INT-9600548 with its Polish part
being  financed by Polish Academy of Science PAN.}\ \
Department of Mathematics, West Virginia University,
Morgantown, WV 26506-6310, USA
(KCies@wvnvms.wvnet.edu)\\
Tomasz Natkaniec, Department of Mathematics, Gda{\'n}sk
 University, Wita Stwosza 57, 80-952 Gda{\'n}sk, Poland
(mattn@ksinet.univ.gda.pl)}

 \title{Darboux like functions that are characterizable by images,
preimages and associated sets%
\thanks{The first author is a Contributing Editor of the 
{\em Real Analysis Exchange}. This paper was managed by one of the 
other editors.}}
 \date{}

\MathReviews{Primary 26A15; Secondary 54C30.}
\keywords{Darboux functions, extendable functions, almost
continuous functions, connectivity functions, functions with
 perfect road, peripherally continuous functions, DIVP-functions,
CIVP-functions, SCIVP-functions, WCIVP-functions, Sierpi{\'n}ski-
Zygmund functions, associated sets}

 \markboth{K. Ciesielski and T. Natkaniec}{Darboux like functions}

\begin{document}
\maketitle
\begin{abstract}
For $\A,\B\subset\P(\mathR)$ let
$\CAB=\{ f\in\RRR\colon(\forall A\in\A)\,(f(A)\in\B)\}$ and
$\DAB=\{ f\in\RRR\colon(\forall B\in\B)\,(f^{-1}(B)\in\A)\}$.
A family $\F$ of real
functions is characterizable by images (preimages) of sets if
$\F=\CAB$ ($\F=\DAB$, respectively) for some $\A,\B\subset\P(\mathR)$.
We study which of classes of Darboux like functions can be
characterized in this way. Moreover, we prove that the class of all
Sierpi{\'n}ski-Zygmund functions can be characterized by neither
images nor preimages of sets.
\end{abstract}

\section{The definitions and preliminary results}

Our terminology is standard
and follows \cite{CiBook}. We consider only real-valued
 functions of one real variable. No distinction is made between
 a function and its graph. By $\mathR$ and $\mathI$ we denote the set
of all reals and the interval $[0,1]$, respectively.
 The family of all subsets of a set $X$ is
denoted by $\P(X)$. The family of all functions from a
 set $X$ into $Y$ is denoted by $Y^X$.
 By $\CC$ and Const we denote the families of all continuous
 functions and all constant functions. The symbol $|X|$
 stands for the cardinality of a set $X$. The cardinality of
 $\mathR$ is denoted by $\co$. For the cardinal number $\kappa$ we
write
$[X]^{\kappa}$ to denote the family of all subsets $Y$
 of $X$ with $|Y|=\kappa$. In particular, $[X]^1$ stands for the
 family of all singletons in $X$ and $[X]^2$ for the family of
 all doubletons in $X$.  By a Cantor set we mean any non-empty
perfect nowhere dense
 subset of $\mathR$. Moreover, we say that a set $A\subset\mathR$
 is Cantor dense in a set $X\subset\mathR$, if $A\cap J$ contains a
 Cantor set whenever $J$ is a non-empty open interval $J$ with $J\cap
X\neq\emptyset$. By $(a,b)$ we denote an open interval with
end-points $a$ and $b$, i.e., the set of all $x\in\mathR$ such that
$\min\{a,b\}<x<\max\{a,b\}$.

For the families $\A,\B\subset\P(\mathR)$ we define
\begin{align*}
%\begin{eqnarray*}
\CAB &=\{ f\in\RRR\colon(\forall A\in\A)\, (f(A)\in\B)\},\\
\intertext{and}
\DAB &=\{ f\in\RRR\colon(\forall B\in\B)\,(f^{-1}(B)\in\A)\}.
\end{align*}
%\end{eqnarray*}
%
%\begin{eqnarray*}
%\CAB &= &\{ f\in\RRR\colon(\forall A\in\A)\, (f(A)\in\B)\},\\
%\DAB &= &\{ f\in\RRR\colon(\forall B\in\B)\,(f^{-1}(B)\in\A)\}.
%\end{eqnarray*}
%
Also, for a family $\F$ of real functions we will consider the
following properties.
\begin{itemize}
\item
 We say that $\F$ is {\em characterizable by images}\/ of sets
 when $\F=\CAB$ for
 some $\A,\B\subset\P(\mathR)$.
\item
Family $\F$ is {\em characterizable by preimages}\/ of sets if
$\F=\DAB$ for some $\A,\B\subset\P(\mathR)$.
\item
$\F$ is {\em topologized}\/ if $\F=\DAB$ for some
\underline{topologies} $\A,\B$ on $\mathR$; while,
 \item
  $\F$ is {\em characterizable by associated sets}\/ if there
exists an $\A\subset\P(\mathR)$ such that
\begin{quote}
$f\in\F$ if and only if for every
$\alpha\in\mathR$, the ``associated'' sets
$E^{\alpha}(f)=\{ x\colon f(x)<\alpha\}$ and
$E_{\alpha}(f)=\{ x\colon f(x)>\alpha\}$ belong to $\A$.
\end{quote}
\end{itemize}


Clearly the class $\CC$ can be defined by preimages of
open sets, so it can be topologized and characterized
by associated sets.
%It is well known that the class of all continuous
%functions can be defined by preimages of sets.
On the other hand, this class cannot be
characterized by images of sets \cite{Ve,CDW}. Nevertheless, some
 classes of functions, often considered in real analysis, have
 such characterizations. For example, the family $\D$ of all
 Darboux functions can be defined as the class of functions which map
connected sets onto connected sets.
  We will study which of other
 classes of Darboux like functions can be characterized by
 images of sets. We consider also the analogous problem: which of
classes of Darboux like functions can be characterized by preimages
of sets.
Note that the problem of characterization of $\F$ by preimages is
strongly connected with the problem of characterization of $\F$ by
associated
sets. In fact, if $\F\subset\RRR$ is characterizable by associated
sets then it is
also characterizable by preimages. On the other hand, there exist
families of functions that are characterizable by preimages but not
by associated sets. (For example, the class of all quasi-continuous
functions has this property~\cite{EL}. Also, under GCH the class of
all derivatives can be characterized by preimages \cite{KC:Der},
while it is  not characterizable by associated sets \cite{AB1}.)
Note also that those problems are connected with the problem of
topologizing $\F$ that was studied recently in several papers. (See,
e.g., \cite{KC}.)

 Following Gibson and Natkaniec~\cite{GN}, by ``Darboux
 like'' functions we understand the following classes of functions
 (from $\real$ into $\real$, unless otherwise specified).
 \begin{description}
 \item[$\D$]
 --- the family of {\it Darboux\/} functions, i.e., such that
 map connected sets onto connected sets.
 \item[$\AC$]
 --- the class of {\it almost continuous\/} functions in the
 sense of Stallings, i.e., such that every open neighborhood of
 $f$ in $\mathR \times \mathR$ contains a continuous function
 from $\mathR$ into $\mathR$.
 \item[$\Conn(X)$]
 --- the class of {\it connectivity\/} functions
 from a topological space $X$ into $\real$, i.e., functions
 $f\colon X\to\mathR$ such
 that the restriction $f\restr C$ is a connected subset of
 $X\times\mathR$ whenever $C$ is a connected subset
 of~$\mathR$. We will write $\Conn$ for $\Conn(\real)$.
\item[$\Ext$]
 --- the family of {\it extendable\/} functions, i.e.,
 functions $f\colon\real\to\real$ for which
 that there exists a connectivity function
 $F\colon \mathR \times \mathI \to \mathR$
 with the property that $F(x,0) = f(x)$ for every $x\in \mathR$.
\item[$\PR$]
 --- the class of functions with {\it perfect road\/}, i.e.,
 such that for every $x\in\mathR$ there exists a perfect set
 $P\subset\mathR$ having $x$ as a bilateral limit point for which
 the restriction $f\restr P$ of $f$ to $P$ is continuous at $x$.
 \item[$\PC$]
 --- the class of {\it peripherally continuous\/} functions,
 i.e., functions $f\colon\real\to\real$ which satisfy
 the {\it Young condition\/} at every $x\in\mathR$,
 that is, such that there are monotone sequences
 $a_{n}\nearrow x$ and $b_{n}\searrow x$ with
 the property that
 $\lim_{n\to\infty} f(a_{n})=\lim_{n\to\infty} f(b_{n})=f(x)$.
\item[$\CIVP$]
 --- the family of functions $f$ having the {\it Cantor
 intermediate value property\/},
i.e., such that for every $x,y\in
 \mathR$ and for each Cantor set $K$ between $f(x)$ and $f(y)$
 there is a Cantor set $C$ between $x$ and $y$ such that
 $f(C)\subset K$.
\item[$\SCIVP$]
 --- the family of functions $f$ having the {\it strong Cantor
 intermediate value property\/},
i.e., such that for every $x,y\in
 \mathR$ and for each Cantor set $K$ between $f(x)$ and $f(y)$
 there is a Cantor set $C$ between $x$ and $y$ such that
 $f(C)\subset K$ and $f\restr C$ is continuous.
\item[$\WCIVP$]
 --- the family of functions $f$ having the {\it weakly Cantor
 intermediate value property\/},
i.e., such that for every $x,y\in
 \mathR$ with $f(x)\neq f(y)$ there is a Cantor set $C$ between
 $x$ and $y$ such that $f(C)\subset (f(x),f(y))$.
 \end{description}

An excellent description of the properties of these families is
 presented in a survey paper of Gibson and Natkaniec~\cite{GN}. In particular,
 the following inclusions $\subset$, denoted by $\longrightarrow$,
 hold.

%\bigskip
   \begin{picture}(0,90)
 \put(20,55){\makebox(0,0){$\CC$}}
 \put(35,55){\vector(1,0){20}}
  \put(75,55){\makebox(0,0){$\Ext$}}
 \put(88,60){\vector(2,1){18}}
 \put(120,70){\makebox(0,0){${\AC}$}}
 \put(132,70){\vector(1,0){30}}
 \put(180,70){\makebox(0,0){$\Conn$}}
 \put(195,70){\vector(1,0){20}}
 \put(224,70){\makebox(0,0){${\D}$}}
 \put(233,65){\vector(2,-1){18}}
 \put(270,55){\makebox(0,0){${\PC}$}}
 \put(127,40){\makebox(0,0){${\SCIVP}$}}
  \put(180,40){\makebox(0,0){${\CIVP}$}}
   \put(225,40){\makebox(0,0){${\PR}$}}
\put(145,40){\vector(1,0){20}}
 \put(195,40){\vector(1,0){20}}
 \put(88,52){\vector(2,-1){18}}
 \put(233,43){\vector(2,1){18}}
 \put(195,38){\vector(1,-1){20}}
   \put(235,15){\makebox(0,0){${\WCIVP}$}}
\end{picture}
\begin{center} Chart~1 \end{center}

Recall
also that, generally, all those classes are different. However
 in the first class of Baire ${\rm B}_1$,all of them,
 except for $\CC$ and $\WCIVP$,
are equal~\cite{BHL}. Recall
 also that for the functions from $\mathR^2$ to $\mathR$ the
 notions of peripherally continuous and of connectivity are
 equivalent~\cite{Ha}.

The following remarks are proved in \cite[Fact~1.2]{CDW}.
\begin{Ro}
Assume that $\F=\CAB$ and $\F\neq \RRR$. Then
\begin{description}
 \item[(1)]
 if $\Const\subset\F$ then $[\mathR]^1\subset\B$;
 \item[(2)]
 if the identity function $\id$ belongs to $\F$ then $\A\subset\B$;
 \item[(3)]
 $\F=\C_{\A,\A^{\ast}}$, where
 $\A^{\ast}=\{ f(A)\colon f\in\F\ \&\  A\in\A\}$;
 \item[(4)]
 if $[\mathR]^1\subset\B$ and $B\in\B\cap [\mathR]^2$ then
 $B^{\mathR}\subset\F$.
 \Qed
\end{description}
 \end{Ro}

\begin{Co}\label{C1}
 Assume that $\F$ satisfies the following conditions:
 \begin{description}
 \item[(1)]
 $\Const\subset\F$;
 \item[(2)]
 for every distinct $a,b\in\mathR$ there exists $f\in\F$
 with $f(\mathR)=\{ f(a),f(b)\}\in [\mathR]^2$;
 \item[(3)]
 there exists $Z\subset\mathR$ such that any distinct $a,b\in\mathR$
the  ``characteristic'' function
 $$\varphi^Z_{a,b}=\left\{
 \begin{array}{ll}
 a &\mbox{if $x\in Z$,}\\
 b &\mbox{if $x\not\in Z$}
 \end{array}\right.$$
does not belong to $\F$.
 \end{description}
 Then $\F$ cannot be characterized by images of sets.
 \Qed
 \end{Co}

In particular, none of the following classes of functions can be
 characterized by images of sets. (See also \cite[Section 4]{CDW}.)
 \begin{itemize}
 \item
 The class of all Lebesgue measurable functions;
 \item
 the class of all functions having the Baire property;
 \item
 the class of all Borel functions;
 \item
 the class of all quasi-continuous functions;
 \item
 the class of all cliquish functions.
 \end{itemize}
(For more on quasi-continuous and cliquish functions
 see~\cite{SK} and~\cite{HT}, respectively.)

The next theorem shows that there is
a class $\F$ of functions with Baire property such that
$\F$ contains all continuous functions and
it can be characterized by images of sets.
This stays in contrast with a theorem
of Ciesielski, Dikranjan and Watson from~\cite{CDW}
in which the authors show that
every class $\F$ of real functions
which contains all continuous functions and
can be characterized by images of sets
must contain a non-measurable function.

Let
\[
 % chTN wszedzie dalej poprawiam \mathcal{D} na {\mathcal % D} -
 % u mnie to zle wygladalo
{\mathcal D}_0=\{D\cap I\colon D\
\mbox{ is dense in $\mathR$ and $I\neq\emptyset$ is an interval}\}.
\]
We say that $f\colon\mathR\to\mathR$ has a
{\em Dense Intermediate Value Property (\/$\DIVP$)}
if $f[A]\in{\mathcal D}_0$ for every $A\in{\mathcal D}_0$.
Clearly every continuous function is $\DIVP$.

\begin{Th}\label{th:Baire}
If $f$ is $\DIVP$ then $f$ is continuous on a dense set.
In particular $f$ has the Baire property.
\end{Th}
\pf
 Let $C(f)$ be the set of points of continuity of
$f$. So, $C(f)$ is a G$_\delta$ set.
By way of contradiction assume that $C(f)$ is not dense.
Then there exists a non-empty open interval
$U$ such that $f$ is discontinuous at every
point of $U$.

For every $x\in U$ let $n_x\in\{1,2,3,\ldots\}$
be the smallest number $n$ such that the oscillation of $f$ at
$x$ is greater than $1/n$.
Then, by Baire Category Theorem, there exists
$n$ such that the set
\[
S_n=\{x\in U\colon n_x=n\}
\]
is of second category. Once again by Baire Category Theorem
we can find a rational number $q$ such that the set
\[
T_0=\{x\in S_n\colon |f(x)-q|<1/3n\}
\]
is of second category.
Let $W$ be a non-empty open interval such that the set
$T=T_0\cap W$ is dense in $W$.
For every $x\in T$ choose $z_x\in W$ such that
$|f(x)-f(z_x)|>1/n$. This can be done
by the oscillation requirement.
Let $T'=\{z_x\colon x\in T\}$ and $A=T\cup T'$.
Then $A\in{\mathcal D}_0$. However, $f[A]\not\in{\mathcal D}_0$, since
$f[A]=f[T]\cup f[T]$ and
\[
f[T]\subset\{y\colon |y-q|<1/3n\}
\]
while
\[
f[T']\subset\{y\colon |y-q|>2/3n\}.
\]
Consequently, $f\not\in\DIVP$.
\Qed

It is worth to notice that the class $\DIVP$ can be defined in a
``natural'' way as the uniform limit of the family of
quasi-continuous Darboux functions.
This class is usually denoted  $\QU$.
It has been studied in~\cite{QU} and, recently, in~\cite{AM}. Recall
that $f\in\QU$ if and only if $f$ satisfies the following conditions:
\begin{itemize}
\item
$f$ is quasi-continuous ($f\in\Q$), i.e.,  $f\restr C(f)$ is dense in
$f$;
\item
$f$ belongs to the class ${\cal U}_0$ (see \cite{BCW}), i.e., for all
$a<b$ the set $f[(a,b)]$ is dense in $(f(a),f(b))$.
\end{itemize}

\begin{Th}\label{th:characDIVP}
$\DIVP=\QU$.
\end{Th}
\pf
$\DIVP\subset\QU$. It is clear that $\DIVP\subset{\cal U}_0$,
thus it is enough to verify that $\DIVP\subset\Q$. Suppose that
$f\in\DIVP$ is not quasi-continuous, i.e., $f(x_0)\not\in\Cl(f\restr
C(f))$ for some $x_0\in\mathR$.
Thus there exist open intervals $U$ containing $x_0$ and $V$
containing
$f(x_0)$  such that $f(x)\not\in V$ for each $x\in C(f)\cap U$. Since
$f\in\DIVP$, the set $A=(C(f)\cap U)\cup\{ x_0\}$ is dense in $U$, so
$A\in{\mathcal D}_0$. On the other hand, $f[A]\not\in{\mathcal D}_0$, a
contradiction.

$\QU\subset\DIVP$. Fix $f\in\QU$ and $A\in{\mathcal D}_0$.
Then $A$ is dense in
$[a,b]$, where $a=\inf(A)$ and $b=\sup(A)$. We will prove that
$f[A]$ is dense in $[c,d]$, for $c=\inf(f[a,b])$, $d=\sup(f[a,b])$.
Fix $y\in (c,d)$ and a neighborhood $V$ of $y$. Since $f\in{\cal
U}_0$, there is $x\in (a,b)$ such that $f(x)\in V$ (cf., \cite{BCW}).
Because $f\in\Q$, we can assume that $x\in C(f)$. Thus there exists a
neighborhood $U$ of $x$ such that $U\subset (a,b)$ and $f[U]\subset
V$. Since $A$ is dense in $U$, there is $x_0\in A\cap U$. Therefore,
$f[A]\cap V\neq\emptyset$, thus $f[A]$ is dense in $[c,d]$.
\Qed

\begin{Co}
The relation of the class $\DIVP$ to the classes
from Chart~1 is as follows.
\begin{description}
\item[(1)]
${\CC}\subset\DIVP\subset\PR\cap\WCIVP$, and the inclusions
are proper.
\end{description}
Moreover, these are the only inclusions between
the class $\DIVP$ to the classes
from Chart~1, i.e.,
\begin{description}
\item[(2)]
$\Ext\not\subset\DIVP$;
\item[(3)]
$\DIVP\not\subset\CIVP$; and,
\item[(4)]
$\DIVP\not\subset\D$.
\end{description}
\end{Co}
\pf
(1). The proper inclusion ${\rm C}\subset\DIVP$ is obvious.
To prove the other inclusion assume that $f\in\DIVP$.

To prove $\DIVP\subset\WCIVP$,
take $a<b$ with $f(a)\neq f(b)$.
Since $C(f)$ is dense in $\real$, by the definition of
the class $\DIVP$ we can choose
$x_0\in (a,b)\cap C(f)$ such that $f(x_0)\in (f(a),f(b))$.
Then,
by the continuity of $f$ at $x_0$,
there exists a Cantor set $C\subset (a,b)$ such that
$f[C]\subset(f(a),f(b))$. Thus $f$ has $\WCIVP$.

Next, to show that $\DIVP\subset\PR$
fix an $x\in\mathR$. Because $f\in\QU$, there exists a sequence
$\{x_n\}_{n=0}^{\infty}$ of points
at which $f$ is continuous such that
$\{x_{2n}\}_{n=0}^{\infty}$ is increasing to $x$,
$\{x_{2n+1}\}_{n=0}^{\infty}$ is decreasing to $x$ and
$\lim_{n\to\infty}f(x_n)=f(x)$.
(This follows easily from the definition of the class $\DIVP$
and the fact that $C(f)$ is dense in $\real$,
by the argument similar to that of the previous paragraph.
But see also~\cite[Lemma~2]{QU}.)
Now, as in the previous paragraph, for each
$n\in\mathN$ we can choose a perfect set
$C_n$ such that
\begin{itemize}
\item
$x_n$ is a bilateral limit point of $C_n$;
\item
$f[C_n]\subset (f(x_n)-1/n, f(x_n)+1/n)$;
\item
$C=\bigcup_nC_n\cup \{ x\}$ is a perfect set.
\end{itemize}
Then $f\restr C$ is continuous at $x$, so $C$ is a perfect road of $f$
at $x$.

The fact that the inclusion
$\DIVP\subset\PR\cap\WCIVP$ is proper
follows from Theorem~\ref{th:Baire}, since
there are functions in $\Ext\subset\PR\cap\WCIVP$
without the Baire property. (In fact, every real
function $f\colon\real\to\real$ is a sum of two extendable
functions~\cite{CR,HR3},
which clearly implies that there are many extendable functions
without the Baire property.)

(2). $\Ext\setminus\DIVP\neq\emptyset$, since
$C(f)\neq\emptyset$ for every $f\in\DIVP$ and there are
$f\in\Ext$ with $C(f)=\emptyset$~\cite{JB,GR1,HR2,CR}.
(Also, every $f\in\DIVP$ is Baire, while it is not the case
for the functions from $\Ext$.)

(3). To see that $\DIVP\not\subset\CIVP$ let $C$
be the Cantor ternary set and ${\cal J}_n$ be the union of
all components of $\mathI\setminus C$ with the length $3^{-n-1}$.
Choose an enumeration
$\{q_n\colon n\in\mathN\}$ of $\mathQ$ and define
$f\colon\real\to\real$ by putting
$f(x)=q_n$ for $x\in{\cal J}_n$, $n\in\mathN$, and $f(x)=0$ otherwise.
Then $f\in\DIVP$ and $f[\real]=\mathQ$.
So, $f[\real]\not\subset K$ for any Cantor set
$K\subset\real\setminus\mathQ$, and $f\not\in\CIVP$.

(4). To see that $\DIVP\not\subset\D$ let $\{X,A,B\}$ be a partition
of $\real$ onto $\co$-dense sets. Define
$f\colon\real\to\real$ such that $f[A]=\{0\}$, $f[B]=\{1\}$, and
$f[(a,b)\cap X]=\real$ for every $a<b$.
Then $f\in\D$ and $f\notin\DIVP$, since
$f[A\cup B]=\{0,1\}\notin{\mathcal D}_0$ while
$A\cup B\in{\mathcal D}_0$.
\Qed

%\medskip
\begin{Th}
The class $\DIVP$ cannot be characterized by preimages.
\end{Th}
\pf\!\!\!\footnote{The authors would like to thank Professor Havrey
 Rosen for pointing out a mistake in the first version of this
 proof.}
 By way of contradiction
suppose that there exist $\A,\B\in\P(\real)$ such that
$\DIVP=\DAB$.
We may assume that
$\A=\{ f^{-1}(B)\colon\, f\in\DIVP, B\in\B\}$
and
$\B\not\subset\{\emptyset, \mathR\}$.
So, fix $B\in\B\setminus\{\emptyset,\mathR\}$.
Let $(d_n)_n$ be a sequence of reals such that
 \begin{itemize}
\item
the set $D=\{ d_n\colon n=0,1,\ldots\}$ is dense;
\item
if $n$ is even then $d_n\in B$;
 \item
 if $n$ is odd then $d_n\in \mathR\setminus B$.
 \end{itemize}

Let $C$ be the Cantor ternary set and let $J_n$ be the union of
 closures of all
components of $\mathI\setminus C$ with the length $3^{-n-1}$. Now,
define
$A_0=\real\setminus\bigcup_{n=0}^{\infty}J_{2n+1}$
and
$A_1=\bigcup_{n=0}^{\infty}J_{2n+1}$. Note that $A_0\cup A_1=\mathR$
and $A_0\cap A_1=\emptyset$.
Put
$$\begin{array}{ll}
f_0(x)=\left\{
\begin{array}{ll}
d_0 &\mbox{for $x\not\in\bigcup_{n=0}^{\infty}J_n$}\\
d_n &\mbox{for $x\in J_n$}
\end{array}\right.
&
\!\!\text{and \ \  }
f_1(x)=\left\{
\begin{array}{ll}
d_1 &\mbox{for $x\not\in\bigcup_{n=0}^{\infty}J_n$}\\
d_{n+1} &\mbox{for $x\in J_n$}
\end{array}\right.
\end{array}
$$
It is easy to observe that $f_0,f_1\in\DIVP$ and $A_0=f_0^{-1}(B)$,
$A_1=f_1^{-1}(B)$, so $A_0, A_1\in\A$.
Moreover, $\{\mathR,\emptyset\}\subset\A$, because all constant
functions are in $\DIVP$. Now, define $h\in\RRR$ by $h(x)=i$ for
$x\in A_i$, $i=0,1$. Then $h\in\DAB\setminus\DIVP$.
\Qed

\section{Classes of functions from Chart~1}
In the next part of this paper we  will use the following
 lemma.
 \begin{Le}\label{L1}
 If ${\rm DB}_1\subset\CAB$ and $\A$ contains a non-degenerate
interval
 then every interval belongs to $\B$.
 \end{Le}
 \pf
 It is well-known (and easy to verify) that for all intervals
 $I$ and $J$, if $|I|=|J|$ then there exists a Darboux, Baire one
 function $f\in\RRR$ such that $f(I)=J$.
 \Qed

\begin{Th}\label{CAB}
 The following classes of Darboux like functions can be
 characterized by images of sets:
$$
 \begin{array}{|c|c|c|c|c|c|c|c|c|}
 \hline
 \Ext & \AC & \Conn & \D & \PC & \SCIVP & \CIVP & \WCIVP & \PR \\
 \hline
 - & - & - & + & - & - & + & + & - \\
 \hline
\end{array}
$$
In this table the symbol ``$+$''  (``$-$'') means that
the given class
can (respectively, cannot) be characterized by images.
\end{Th}
\pf
 We will use sets $A\subset\mathR$ that have the following
 properties:
 \begin{description}
 \item[$(C_1)$]
$A$ is an interval, i.e., if $a,b\in A$ then $(a,b)\subset A$;
 \item[$(C_2)$]
  for every $a,b\in A$ if $C\subset (a,b)$ is a Cantor set then
 $C\cap A\neq\emptyset$;
\item[$(C_3)$]
  for every $a,b\in A$ with $a<b$ and for every $F_{\sigma}$ set
 $E\subset (a,b)$ if $E$ is Cantor dense in $(a,b)$
 then $E\cap A\neq\emptyset$;
 \item[$(C_4)$]
 $A$ is ordinarily dense in itself, i.e., if $a,b\in A$ and $a<b$,
 then there is $c\in A$ with $a<c<b$.
 \end{description}
 Note that for each $i=1,2,3$ the property $(C_i)$ implies
 $(C_{i+1})$.

\bigskip
 {\bf 1.} $\D=\CAB$, where $\A=\B$ is the family of all intervals
 in $\mathR$ (i.e., all sets that satisfy the condition $(C_1)$).
 Thus $\D$ is characterizable by images of sets.

\bigskip
{\bf 2. }Neither $\PR$ nor $\PC$ can be characterized by images
 of sets.

%\noindent
Indeed, let $(Z,\mathR\setminus Z)$ be a partition of
 $\mathR$ onto sets that are Cantor dense in $\mathR$. Then for each
 $a,b\in\mathR$ with $a\neq
 b$, the function $\varphi^Z_{a,b}$ belongs to $\PR$, thus also
 to $\PC$. On the other hand, characteristic function of no
 singleton belongs to $\PC$ and therefore to $\PR$. Hence
 by Corollary~\ref{C1}, the classes $\PC$ and $\PR$ are not
 characterizable by images of sets.

\bigskip
{\bf 3.} None of the classes $\Ext$, $\AC$, and $\Conn$
can be characterized by images of sets.

%\noindent
Indeed, suppose that $\Ext\subset\F\subset\D$ and $\F=\CAB$.
We can assume that
$\B=\{ f[A]\colon A\in\A \ \&\  f\in\F\}$.
We will show that $\F=\D$.
 Note that $[\mathR]^1\subset\B$, $[\mathR]^1\neq\B$ and
$\A\subset\B$.

\medskip

\noindent
{\bf Claim. }Every $A\in\A$ has the property $C_3$.
\medskip

 Indeed, suppose that there are $a,b\in A$ and an $F_{\sigma}$ set
$E\in (a,b)$ that is Cantor dense in $(a,b)$ and $E\cap
 A=\emptyset$. We will construct an extendable function $f\colon
 \mathR\to\mathI$ such that $f[\mathR\setminus E]=\{ 0,1\}$.

Let $g\colon \mathI\to \mathI$ be an extendable function
 whose graph is dense in $\mathI^2$.
 (See \cite{JB,GR1}. Compare also \cite{CR} and \cite{HR2}.)
 Then there exists a Cantor
 dense $F_{\sigma}$ set $F\subset (0,1)$ such that $\mathI\setminus F$
 is $g$-negligible~\cite{HR}. (This means that every function
 $\tilde{g}\colon \mathI\to \mathI$ with
 $\tilde{g} \restr F=g\restr F$ is
 still extendable). Let $h\colon [a,b]\to \mathI$ be a homeomorphism
 such that $h[E]=F$. (See \cite[Lemma 3]{Go}.) Then
 the composition
 $g\circ h\colon[a,b]\to \mathI$
 is  an extendable function
 and the set
 $[a,b]\setminus E$ is $(g\circ h)$-negligible~\cite{TN}. Thus
 $f_0\colon [a,b]\to \mathI$ defined by
 $$f_0(x)=\left\{
\begin{array}{ll}
 0 & \mbox{if $x=a$,}\\
 g\circ h(x) & \mbox{if $x\in E$,}\\
 1 & \mbox{otherwise}
 \end{array}\right.
 $$
is an extendable function.
 Let $f\colon\mathR\to \mathI$ be the extension of $f_0$ such that
 $f(x)=0$ for $x<a$ and $f(x)=1$ for $x>b$. Observe that
 $f\in\Ext$. Indeed, according to \cite{GR}, there exists a
 peripherally continuous function
 $F_0\colon [a,b]\times \mathI\to \mathI$
 such that $F_0\restr ([a,b]\times\{0\})=f_0$.
%ckKC
 Moreover, we can assume that
 $F_0\restr(\{a,b\}\times \mathI)$ is continuous.
 (Actually, $F_0$ can be constant on intervals
 $\{ a\}\times \mathI$ and $\{ b\}\times\mathI$.
%While this is not mentioned
%explicite in \cite{GR}, but the proof is the same.)
 While this is not mentioned explicitly in \cite{GR}, it can be
 achieved by a minimal modification%
 \footnote{
 Let $\cal J$ be a family of peripheral intervals as defined in \cite{GR}
 and let ${\cal J}_0$ be the set of all
 $\langle I,J\rangle\in {\cal J}$ such that if $0\in I$ ($1\in I$)
 then $f(0)\in J$ ($f(1)\in J$). Then
 ${\cal J}_0$ is also a family of peripheral intervals.
 Now, in the definition of $g$ (described in \cite{GR})
 we can additionally assume that
 $g(\{ 0\}\times I)=\{f(0)\}$ and $g(\{ 1\}\times I)=\{ f(1)\}$.
 Then $F_0=g$ has the desired properties. }
 of the proof presented there.)
 Then $F\colon\mathR\times \mathI\to \mathI$
 defined by
 $$F(x,y)=\left\{
 \begin{array}{ll}
 F_0(x,y) &\mbox{if $x\in [a,b]$,}\\
 F_0(a,y) &\mbox{if $x<a$,}\\
 F_0(b,y) &\mbox{if $x>b$,}
 \end{array}\right.$$
 also is peripherally continuous, so
 $f=F\restr\mathR\times\{ 0\}$ is extendable.

 Let $f$ be such a function. Then $f[A]=\{ 0,1\}$, so $\CAB$
 contains characteristic functions of all subsets of $\mathR$,
 contrary to $\CAB=\F\subset\D$.
 The Claim has been proved.

 \smallskip

Now we will prove that every $A\in\A$ is an interval.
Indeed, suppose that there is $B\in\A$ that is not an interval.
%Since $\A\subset\B$, $B\in\B$.
Then $B\in\B$, since $\A\subset\B$.
Let $f\colon\mathR\to B$ be a surjection
such that for every $y\in B$ the
level set $f^{-1}(y)$ is Cantor dense in
$\mathR$. Then $f[A]=B$ for $A\in\A\setminus [\mathR]^1$, so
$f\in\CAB$. On the
 other hand, $f[\mathR]$ is not connected, thus $f\not\in\D$,
 contrary to $\CAB=\F\subset\D$.

Since $\A\neq [\mathR]^1$, there is $A\in\A$ that is not a
non-degenerate interval.
Thus, by Lemma~\ref{L1}, every interval $I$
belongs to $\B$ and consequently, $\CAB=\D$.

\bigskip
{\bf 4. }The class $\CIVP$ is characterizable by images of sets.

%\noindent
 Let $\A$ be the family of all sets $A\subset\mathR$ that satisfy
 the condition $(C_2)$ and let $\B=\A$. We will prove that
 $\CIVP=\CAB$. Fix $f\in\CIVP$,
 $A\in\A$, $a,b\in A$ and a Cantor set $C\subset (f(a),f(b))$.
 Then there exists a Cantor set $K\subset (a,b)$ such that
 $f[K]\subset C$. Then  $C\cap f[A]\neq\emptyset$, since
 $K\cap A\neq\emptyset$, and
 so $f[A]\in\B$. Hence $\CIVP\subset\CAB$. Thus, $f\in\CAB$
 proving $\CIVP\subset\CAB$.

 Now fix $f\in\CAB$
 and by way of contradiction suppose that $f\not\in\CIVP$. So there
 exist
 $a,b\in\mathR$ and a Cantor set $C\subset (f(a),f(b))$ such that
 $f[K]\subset C$ for no Cantor set $K\subset (a,b)$. Thus
 $A=[a,b]\setminus f^{-1}(C)\in\A$ and $f[A]\not\in\B$, contrary to
$f\in\CAB$.

\bigskip
 {\bf 5. }The class $\SCIVP$ cannot be characterized by images of
 sets.

%\noindent
 Suppose that $\SCIVP=\CAB$, where
 $\B=\{ f[A]\colon f\in\SCIVP\, \&\, A\in\A\}$.
 Observe that $[\mathR]^1\subset\B$,
 $[\mathR]^1\neq \B$,
 and moreover, if $A\in\A$ then $A$
 satisfies the condition $(C_2)$. Indeed, suppose that
 there exist $A\in\A$, $a,b\in A$ and a Cantor set
 $C\subset (a,b)$ such that $C\cap A=\emptyset$.
 Decompose $C$ into $\co$ many
 sets $\{ C_{\alpha}\colon \alpha<\co\}$, where each
 $C_{\alpha}$ is Cantor dense in $C$. Let
 $\mathR=\{ r_{\alpha}\colon \alpha<\co\}$ and put
$$f(x)=\left\{
 \begin{array}{ll}
 0 &\mbox{if $x$ and $a$ belong to the same component of
 $\mathR\setminus C$,}\\
 r_{\alpha} &\mbox{if $x\in C_{\alpha}$, $\alpha<\co$,}\\
 1 &\mbox{otherwise.}
 \end{array}
 \right.$$
 Then $f\in\SCIVP$ and $f[A]=\{ 0,1\}\in\B$, thus $\CAB$
 contains characteristic functions of all subsets of $\mathR$,
 contrary to $\CAB=\SCIVP$.

 Since $\SCIVP\subset\CIVP$, by case {\bf 4} each $B\in\B$ has the
 property $(C_2)$. On the other hand, each $B\subset\mathR$ that
 satisfies $(C_2)$ belongs to $\B$. Indeed, fix such a $B$. Let
 $f\colon\mathR\to B$ be a function such that for each $y\in B$
 the level set $f^{-1}(y)$ is Cantor dense in $\mathR$. Then
 $f\in\SCIVP$ and $f[A]=B$ for each $A\in\A\setminus [\mathR]^1$.
 Thus $B\in\B$. Consequently, $\CIVP\subset\CAB$, a
 contradiction.

\bigskip
{\bf 6. }The class $\WCIVP$ is characterizable by images of sets.

%\noindent
 Let $\A$ be the family of all sets that satisfy the condition
 $(C_2)$ and let $\B$ be the family of all $B\subset\mathR$
 that satisfy the statement $(C_4)$.
 We will verify that $\WCIVP=\CAB$. The inclusion
 $\WCIVP\subset\CAB$ is obvious. Now assume that
 $f\not\in\WCIVP$. Then there are $a,b\in\mathR$ such that $a<b$
 and $f[C]\not\subset(f(a),f(b))$ for each Cantor set $C\subset
 (a,b)$. Put $A=[a,b]\setminus f^{-1}(f(a),f(b))$. Then $A\in\A$
 and $f[A]\not\in\B$, thus $f\not\in\CAB$. Hence
 $\CAB\subset\WCIVP$, and consequently, we have the equality
 $\WCIVP=\CAB$.
 \Qed

%\bigskip
Now we will consider the problem which of Darboux like functions
from Chart~1 can
be characterized by preimages or by associated sets. The question
whether the class $\F$ is characterizable by associated sets have
been studied  for the following classes of Darboux like functions:
$\D$~\cite{AB}, $\Conn$~\cite{CT}, $\AC$~\cite{KK} and
$\Ext$~\cite{HR3}.
Recall that none of those classes can be characterized by
associated sets. The next theorem generalizes these results.

\begin{Th}\label{DAB}
 The following classes of Darboux like functions can be
 characterized by preimages:
$$
 \begin{array}{|c|c|c|c|c|c|c|c|c|}
 \hline
 \Ext & \AC & \Conn & \D & \PC & \SCIVP & \CIVP & \WCIVP & \PR \\
 \hline
 - & - & - & - & + & - & - & - & + \\
 \hline
\end{array}
$$
In this table the symbol ``$+$''  (``$-$'') means that the given class
can (respectively, cannot) be characterized by preimages.
\end{Th}
\pf
The argument will be split into three cases.

{\bf 1.} None of the classes: $\Ext$, $\AC$, $\Conn$, $\D$,
$\SCIVP$, $\CIVP$, $\WCIVP$ can be characterized by preimages of sets.

%\noindent
Assume that $\Ext\subset\F=\DAB$
for some $\A,\B\subset\P(\mathR)$. We
will prove that $\F\setminus(\D\cup\WCIVP)\neq\emptyset$. We can
assume that:
\begin{itemize}
\item
$\A=\{ f^{-1}(B)\colon f\in\F\ \&\  B\in\B\}$;
\item
$\B\neq\emptyset$ and $\{\emptyset,\mathR\}\cap\B=\emptyset$;
\item
$\{\emptyset,\mathR\}\subset\A$.
\end{itemize}
Fix $B\in\B$, $y_0\in\B$ and $y_1\not\in\B$. Let $f\in\Ext$ be dense
in $\mathR^2$.
(See \cite{HR2} or \cite{CR} for
examples of such functions.) By a result from~\cite{HR}, there
exists an
$F_{\sigma}$ meager set $C\subset\mathR$ such that
$\mathR\setminus C$ is $f$-negligible.
As in \cite{TN} (and similarly to the argument in the proof of
Claim from case {\bf 3} of Theorem~\ref{CAB}),
we can construct  $f_1\in\Ext$
 and an $F_{\sigma}$ meager set $D\subset \mathR$ such that $f_1$ is
dense in $\mathR^2$, $D$ is $f_1$-negligible, and $C\cap D=\emptyset$.
Set $C_0=C\cap f^{-1}(B)$, $C_1=C\setminus C_0$,
$D_0=D\cap f^{-1}(B)$, $D_1=D\setminus D_0$,
$$\begin{array}{ll}
g(x)=\left\{
\begin{array}{ll}
f(x) &\mbox{for $x\in C$}\\
y_0 &\mbox{for $x\in D_1$}\\
y_1 &\mbox{otherwise}
\end{array}\right.
&
\text{ and \ \ \ \ }
g_1(x)=\left\{
\begin{array}{ll}
f_1(x) &\mbox{for $x\in D$}\\
y_1 &\mbox{for $x\in C_0$}\\
y_0 &\mbox{otherwise}
\end{array}\right.
\end{array}
$$
Then $g,g_1\in\Ext$, thus $E=C_0\cup D_1=g^{-1}(B)$ and
$F=\mathR\setminus (C_0\cup D_1)=g_1^{-1}(B)$ belong to $\A$. Note
that $E$ and $F$ are dense in $\mathR$, $E\cup F=\mathR$ and
$E\cap F=\emptyset$.
Let $h\in\RRR$ be the characteristic function of $E$.
Then $h\in\DAB\setminus (\D\cup\WCIVP)$.

\medskip
{\bf 2.} The class $\PC$ can be characterized by preimages.

Indeed,
$\PC=\DAB$, where $\A$ is the family of all bilaterally dense in
itself subsets of $\mathR$ and $\B$ is the family of open intervals.

\medskip
{\bf 3.} The class $\PR$ can be characterized by preimages.

Indeed,
let $\A$ be the family of all bilaterally Cantor dense
in itself subsets of $\mathR$ and $\B$ be the family of open
intervals. Then $\PR=\DAB$.

(Notice that the last two equalities follow also from the fact that
the classes $\PC$ and $\PR$ can be defined in terms of the continuity
with respect to systems of paths. See~\cite{KB}.)
\Qed

\begin{Co}
None of the classes of Darboux like functions
from Chart~1 can be topologized.
\end{Co}
\pf
 By Theorem~\ref{DAB} we must consider only two classes: $\PC$ and
$\PR$. Assume that $\A$ and $\B$ are topologies on $\mathR$ and
$\PR\subset\F=\DAB$. We will prove that $\F=\RRR$. This is obvious
if $\B=\{\emptyset,\mathR\}$.
Thus suppose that there exists $B\in\B\setminus\{\emptyset,\mathR\}$
and fix $y_0\in B$, $y_1\not\in B$. We will prove that
$[\mathR]^1\subset\A$, so
$\A=\P(\mathR)$ since $\A$ is a topology.
For an $x_0\in\mathR$ divide the
set $\mathR\setminus\{ x_0\}$ onto two sets $C_0$ and $C_1$, each
Cantor dense in $\mathR$. Put
$$\begin{array}{ll}
f_0(x)=\left\{
\begin{array}{ll}
y_0 &\mbox{for $x\in \{x_0\}\cup C_0$}\\
y_1 &\mbox{for $x\in C_1$}
\end{array}\right.
&
\!\!\!\text{and \ \ }
f_1(x)=\left\{
\begin{array}{ll}
y_0 &\mbox{for $x\in \{x_0\}\cup C_1$}\!\!\!\\
y_1 &\mbox{for $x\in C_0$}\!\!\!
\end{array}\right.
\end{array}
$$
Then $f_0, f_1\in\DAB$, since $f_0, f_1\in\PR$. Thus
$\{x_0\}=f_0^{-1}(B)\cap f_1^{-1}(B)\in\A$.
It follows that $\A=\P(\mathR)$.
\Qed

\begin{Co}
None of the classes of Darboux like functions
from Chart~1 can be defined by
associated sets.
\end{Co}
\pf
By Theorem~\ref{DAB} we must consider only two classes: $\PC$ and
$\PR$. Assume that $\PR\subset\F$ and $\F$  can be characterized by
associated sets. We will prove that $\F\setminus\PC\neq\emptyset$.

Let $\A$ denote the family of all associated sets of $\F$.
Divide the set $\mathR\setminus\{0\}$ onto two sets $C$ and $D$,
each Cantor dense in $\mathR$.
Since the characteristic functions
$\charf{C},\charf{D}\in\PR\subset\F$,
the sets $C$, $\mathR\setminus C$, $D$, and $\mathR\setminus D$
belong to $\A$.
Then $f=\charf{C}-\charf{D}\in\F\setminus\PC$, with $0$ being
a point in which $f$ is not peripherally continuous.
\Qed

\section{The class of Sierpi{\'n}ski-Zygmund functions}

In this section we consider the problem whether the class of all
Sierpi{\'n}ski-Zygmund functions can be characterized either by
images or by preimages.
Recall that for $X\subset\mathR$
the class $\SZ(X)$ of
{\em Sierpi\'nski-Zygmund functions}\/ is the class of all
functions $f\colon X\to\mathR$ whose restrictions $f\restriction Y$
are discontinuous for all subsets $Y$ of $X$ of cardinality continuum.
We will write $\SZ$ for $\SZ(\mathR)$.

\begin{Th}\label{SZ}
The class $\SZ$ can be characterized neither by
images nor preimages of sets.
\end{Th}


\pf
First, by way of contradiction assume that
$\SZ=\C^{-1}_{\A,\B}$ for some $\A,\B\subset\real$.

Note that $\B\not\subset\{\emptyset,\real\}$, since otherwise
either $\SZ=\C^{-1}_{\A,\B}=\real^\real$ (if $\B\subset\A$)
or $\SZ=\C^{-1}_{\A,\B}=\emptyset$ (if $\B\not\subset\A$), a contradiction.
So, let $B_0\in\B\setminus\{\emptyset,\real\}$ and pick
$x\in B_0$.

If every non-empty $B\in\B$ has cardinality $<\co$ then $\A$ contains
every subset $A$ of cardinality $<\co$.
(Since $A=f^{-1}(B_0)\in\A$, where $f\in\SZ$ is such that
$f[A]=\{x\}$  and $f[\mathR\setminus A]\subset\mathR\setminus B_0$.)
Then the identity is in $\DAB$, a contradiction.
So, $\B$ contains a set $B$ of cardinality $\co$.
Thus,
\[
\mathR\in\A,
\]
since $\mathR=f^{-1}(B)\in\A$, where $f\in\SZ$ is such that
$f[\mathR]\subset B$. Notice also that
\[
\emptyset\in\A.
\]
Indeed, if there is $B\in\B$ such that $|\real\setminus B|=\co$
then $\emptyset=f^{-1}(B)\in\A$, where $f\in\SZ$ is such that
$f[\mathR]\subset\real\setminus B$.
So, by way of contradiction assume that
$|\real\setminus B|<\co$ for every $B\in\B$.
Then $\A$ contains every set $A\subset\real$ with
$|\real\setminus A|<\co$ since $A=f^{-1}(B_0)\in\A$,
where $f\in\SZ$ is such that
$f[A]\subset B_0$ and $f[\real\setminus A]\subset\real\setminus B_0$.
But then the identity is in $\DAB$, a contradiction.

So, $\emptyset,\real\in\A$, implying that
$\SZ=\DAB$ contains all constants, a contradiction.

\medskip

Next, by way of contradiction assume that
$\SZ=\CAB$ for some families $\A,\B\subset\PO$.
Clearly we can assume that
$\emptyset\notin\A$ and that $\A\neq\emptyset$.

First note that $\A\subset [\mathR]^{\co}$. Indeed,
suppose that
$A\in\A\cap [\mathR]^{<\co}$. Since there exist $\SZ$
functions that are constant on $A$,
thus $\B$ contains a singleton and consequently,  $\CAB$ contains a
constant function, a contradiction.

So, take $A_0\in\A$ of cardinality $\co$ and let
$f\in\SZ$ be one-to-one. Then $B=f[A_0]\in\B$ has cardinality
$\co$. Note that $[B]^\co\subset\B$.
Indeed, if $C\in[B]^\co$ let $X=f^{-1}(C)$
and let $g\in\SZ$ be such that $g\restr X=f\restr X$ and
$g[\mathR\setminus X]\subset C$. Then $C=g[A_0]\in\B$.

Now, pick one-to-one $g\in\SZ$ with $g[\mathR]\subset B$
and let $h\colon\mathR\to\mathR$ be such that
$h\restr (\mathR\setminus B)=g\restr (\mathR\setminus B)$
and $h(x)=x$ for every $x\in B$. Then clearly $h\notin\SZ$.
However, $h\in\CAB$ since $h[A]\in\B$ for every $A\in\A$.
Indeed, because  $|A|=\co$, we clearly have
$h[A]\in[B]^\co\subset\B$. This finishes the proof.
\Qed

\medskip
Now, recall the following theorem.

\begin{Th}
 {\rm (Balcerzak, Ciesielski, Natkaniec \cite{BCN})}
\begin{description}
\item[(a)] If\/ $\mathR$ is not a union of less
           than continuum many of its meager subsets
           (thus under CH and MA)
           then there exists an $f\in\SZ\cap\D$.
\item[(b)] There is a model of ZFC in which every
           Darboux function $f\colon\mathR\to\mathR$
           is continuous on some set of cardinality continuum.
           In particular, in this model we have
           $\SZ\cap\D=\emptyset$.
\end{description}
\end{Th}

Note also that if $\SZ\cap\D=\emptyset$, which is consistent with
ZFC, then
\begin{description}
\item[(1)]
 $\SZ\cap\D$ can be characterized by images and by preimages;
\item[(2)]
$\D\setminus\SZ$ can be characterized by images, but cannot be
characterized by preimages;
\item[(3)]
$\SZ\setminus\D$ can be characterized by neither images no preimages.
\end{description}

On the other hand,  the statement (3)  can be proved in ZFC, by an
easy modification of the proof of Theorem \ref{SZ}.
%\chKC
Moreover, since $\Ext\subset\D\setminus\SZ$
\begin{itemize}
\item the argument from Theorem~\ref{DAB} shows in ZFC that
      $\D\setminus\SZ$ cannot be characterized by preimages; and,
\item the argument from Theorem~\ref{CAB} shows that
      $\D\setminus\SZ$ cannot be characterized by preimages
      as long as $\D\setminus\SZ\neq\D$;
      in particular, the statement
      \begin{quote}
      the class $\D\setminus \SZ$ can be characterized by images
      \end{quote}
      is equivalent to the equation $\D\setminus\SZ=\D$ and so, it
      cannot be proved in~ZFC.
\end{itemize}


\begin{Pro}
 Can the class $\SZ\cap\D$ be characterized by images or preimages if
$\SZ\cap\D\neq\emptyset$?
%In particular, what happens under CH?
\end{Pro}
Under CH this problem is probably not very difficult.
The interesting part is, whether in ZFC alone
the assumption $\SZ\cap\D\neq\emptyset$
decides whether the class
$\SZ\cap\D$ can be characterized by images or preimages.

\begin{thebibliography}{33}
\bibitem{BCN} M.~Balcerzak, K.~Ciesielski, T.~Natkaniec,
{\it Sierpi\'nski--Zygmund functions that are
 Darboux, almost continuous, or have a perfect road},
Arch. Math. Logic, to appear. (Preprint$^\star$ available.%
\footnote{Preprints marked by $^\star$ are available
in electronic form. They can be accessed
from K.~Ciesielski web page:
http://www.math.wvu.edu/homepages/kcies/})


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\end{thebibliography}

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