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\author{
Krzysztof Ciesielski%
\thanks{Work partially supported by the NATO 
Collaborative Research Grant CRG~950347
and 1996/97 West Virginia
University Senate Research Grant.
\endgraf 
Papers authored or co-authored by a Contributing Editor are managed 
by a Managing Editor or one of the other Contributing Editors.},
Department of Mathematics, West Virginia University, \\
Morgantown, WV 26506-6310, USA
(KCies@wvnvms.wvnet.edu) }

\title{Characterizing derivatives by preimages of sets}
\date{}
\MathReviews{Primary 26A24. Secondary 03E35.}
\keywords{derivatives, preimages of sets.}

\markboth{Krzysztof Ciesielski}{Characterizing 
derivatives by preimages of sets}



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\begin{document}

\maketitle
\begin{abstract}

In this note we will show that many classes $\F$ of real functions
$f\colon\real\to\real$ can be characterized by 
preimages
of sets in a sense that there exist the families $\A$ and $\D$ 
of subsets of $\real$ such that $\F=\C(\D,\A)$, where
$\C(\D,\A)=\{f\in\real^\real\colon f^{-1}(A)\in \D\ \text{ for every }
A\in\A\}.$
In particular, we will show that there exists a Bernstein 
$B\subset \real$ such that the 
family $\Delta$ of all derivatives 
can be represented as $\Delta=\C(\D,\A)$, where 
$\A=\bigcup_{c\in\real}\{(-\infty,c),(c,\infty),B+c\}$
and $\D=\{g^{-1}(A)\colon A\in\A\ \&\ g\in\Delta\}$. 


\end{abstract}

\section{Introduction}

Our terminology is standard and follows \cite{Ci:book}. 
By $\real$ and $\rational$ we denote the set
of all real and rational numbers, respectively.
Symbol $\P(X)$ will stand for the family of all subsets of $X$. 
The family of all functions from a
set $X$ into $Y$ is denoted by $Y^X$. In particular, 
$\real^\real$ will stand for the set of all functions
$f\colon\real\to\real$. 
For a set $S\subset\real$ the symbol $S^c$ will denote the complement of
$S$, i.e., $S^c=\real\setminus S$. 
We will write $\bor$ for the family of all Borel functions
$f\colon\real\to\real$ and $\B$ for 
the family of all Borel subsets of $\real$. 
The ordinal numbers will be identified with the sets of 
all their predecessors, and cardinals with the initial ordinals. 
The cardinality of a set $X$ will be denoted by $|X|$. 
The cardinality of
$\real$ is denoted by $\co$ and referred as {\em continuum}. 


The problem of characterizing the real functions $f\colon\real\to\real$
that are derivatives of some function $F\colon\real\to\real$
preoccupied many authors for most of this century. 
The development led, for example, to a characterization of
associated sets (i.e., sets of the form 
$\{x\in\real\colon f(x)<b\}$) for the derivatives
(\cite{Za,Pr}) and many other results in this direction
(\cite{Br,BL}). However, it is known already from
1936 paper \cite{Ma} of Mazurkiewicz that a ``simple'' characterization
of derivatives might not exist, since the set
of all differentiable functions is a true co-analytic set.
Also, Freiling in his recent article~\cite{Freil} 
gives a convincing argument that any nice structural
characterization of derivatives is circular in a sense 
that it allows us to solve, to some extend, the problem of finding
the primitive of a derivative. 

The main goal of this article is to show that 
many classes $\F$ of real functions, including the family 
$\Delta$ of all derivatives, 
can be characterized by means of preimages of some sets
similarly as the class of all continuous functions, that is
as a family of the form
\[
\C(\D,\A)=\{f\in\real^\real\colon f^{-1}(A)\in \D\ \text{ for every }
A\in\A\},
\]
where $A$ is a family of subsets of $\real$ and 
$\D=\{f^{-1}(A)\colon f\in\F\ \&\ a\in\A\}$. 

The general theorem in this direction we prove here is the following. 

\thm{th:mainGen}{ Let $\F,\R\subset\real^\real$ be 
such that $|\R|\leq\co^+$, $|\F|\leq\co$, 
$\F$ contains all constant functions, and
$|g[\real]|=\continuum$ for any non-constant function $g$
which is a difference of two functions from $\F$.
Then there exists a family
$\A\subset\P(\real)$ of cardinality less than or equal to 
$|\R|$ such that
\[
\F\cap\R=\R\cap\C(\D,\A)
\]
where $\D=\{f^{-1}(A)\colon f\in\F\ \&\ A\in\A\}$.
}

Using this theorem to $\F=\Delta$ and $\R$ equal to $\bor$ and $\real^\real$
we obtain immediately the following two corollaries. 


\cor{corDouble}{There exists a family $\A\subset\P(\real)$ such that 
$|\A|\leq\co$ and
\[
\Delta=\bor\cap\C(\D,\A),
\]
where $\D=\{f^{-1}(A)\colon f\in\Delta\ \&\ A\in\A\}$.}


\cor{corGCH}{If the Generalized Continuum Hypothesis holds
(more specifically, if $2^\continuum=\continuum^+$) then there 
exists a family $\A\subset\P(\real)$ such that 
\[
\Delta=\C(\D,\A),
\]
where $\D=\{f^{-1}(A)\colon f\in\Delta\ \&\ A\in\A\}$.}

Certainly, we can obtain the similar corollaries for a wide
variety of classes $\F$. Moreover, specifically for the class $\Delta$
the following stronger characterization will be proved,
where $\D\B_1$ stands for the class of Darboux Baire one functions. 
Recall also that $B\subset\real$ is {\em Bernstein\/} if $B$ 
and its complement intersect every perfect subset of $\real$.

\thm{thZFCdel}{There exists a Bernstein set $B\subset\real$ such that 
\[
\Delta=\D\B_1\cap\C(\D_0,\{B+c\colon c\in\real\})=\C(\D,\A),
\]
where $\A=\bigcup_{c\in\real}\{(-\infty,c),(c,\infty),B+c\}$,
$\D_0=\{f^{-1}(B+c)\colon f\in\Delta\ \&\ c\in\real\}$,
and $\D=\{f^{-1}(A)\colon f\in\Delta\ \&\ A\in\A\}$.
}

Note that 
Theorem~\ref{thZFCdel} and Corollary~\ref{corGCH} (under the assumption
$2^\continuum=\continuum^+$) generalize
the following theorem of Preiss and Tartaglia~\cite{PT},
which was a motivation for this paper. 

\prop{prop1}{ For every subset $E$ of $\real$ there exists
a family $\D_E$ (equal to $\{f^{-1}(E)\colon f\in\Delta\}$)
such that $\Delta$ is equal to
\[
\C(\{\D_E\}_{E\in\P(\real)},\P(\real))=
\{f\colon\real\to\real\colon  f^{-1}(E)\in\D_E\ 
\text{ for every } E\in\P(\real)\}.
\]
}

The obvious disadvantage of the characterization of
$\Delta$ as in Theorem~\ref{thZFCdel} (and 
Corollaries~\ref{corDouble} and \ref{corGCH}) 
is its circular character: 
the family $\D$ is defined with the use of $\Delta$
as a kind of weak ``topology''
for the family $\Delta$ generated by a ``topology'' $\A$.
However, by an argument of Freiling~\cite{Freil},
any characterization of $\Delta$ will be, to some extend, circular.

Another disadvantage of the characterization from Theorem~\ref{thZFCdel}
is that it uses
a Bernstein set which is highly unconstructive. (It is
is non-measurable, does not have a Baire property, and 
its existence cannot be proved without the Axiom of Choice.
In fact, even the Dependent Choice Axiom, which is 
a part of the Axiom of Choice that 
implies the classical induction
theorem, is not sufficient for deducing its existence.)
It would be nicer to have a similar characterization with
$\A$ being a subfamily of Borel sets. However, the existence of 
such a characterization is not clear at this point.

Despite all of these reservations, 
the characterization from Theorem~\ref{thZFCdel}
really says something. If a 
Darboux Baire one function $f\colon\real\to\real$
fails to be a derivative, then it is prevented from being so 
solely because of the form of its preimage 
$f^{-1}(B+c)$ of a translation of a single set $B$. 


Notice also that although in the characterizations
\[
\Delta=\C(\D,\A)
\]
the family $\D$ is a weak ``topology''
for a family $\Delta$ generated by a ``topology'' $\A$, 
the family $\A$ cannot be a topology.  
This follows from the next proposition, which has been proved
by  the author~\cite[Corollary~3]{C} and, independently,
by Tartaglia~\cite{T}.%
\footnote{The information on Tartaglia's comes from~\cite{PT}
since the preprint~\cite{T} is not available to the author.}

\prop{propTopolo}{There are no topologies $\tau_0$ and $\tau$ on $\real$
with the property that  
$\Delta=\C(\tau_0,\tau)$.}

Also, if $\F=\C(\B,\A)$ for some families $\A$ and $\B$ of subsets of $\real$
then we have also $\F=\C(\D,\A)$, where 
$\D=\{f^{-1}(A)\colon A\in\A\ \text{ and }\ f\in\F\}$, since 
$\F\subset\C(\D,\A)\subset\C(\B,\A)=\F$. Thus, 
the form of the family $\D$ in 
the above characterizations 
is, in a sense, forced on us. Also
\begin{equation}\label{eq:subs}\text{
if $\Delta=\C(\D,\A)$ or $\Delta=\bor\cap\C(\D,\A)$
then $\A\subset\D$
}\end{equation}
since the identity function belongs to $\Delta$. 
However in Theorem~\ref{thZFCdel} and Corollary~\ref{corGCH}
we cannot have $\A=\D$, since the class $\C(\A,\A)$ is
closed under the composition, 
while there exist a derivative $f$ and a homeomorphism $h$ (which is also a
derivative) such that $h\circ f$ is not a derivative.
(See e.g.~\cite{Br}.)

Notice also that in the characterization  $\Delta=\C(\D,\A)$
neither $\A$ nor $\D$ can be an algebra, as follows from the following fact. 

\prop{propAlg}{If $\Delta=\C(\D,\A)$ for some families 
$\D$ and $\A$ of subsets of $\real$ then neither $\A$ nor $\D$ 
contain simultaneously a non-empty proper subset 
$S$ of $\real$ and its complement $S^c$.

In particular, neither $\A$ nor $\D$ is an algebra.}

\proof First note that 
$\A\not\subset\{\emptyset,\real\}$, since 
this and the inclusion $\A\subset\D$ would imply that 
$C(\D,\A)$ consists of all real functions, 
contradicting  $\Delta=\C(\D,\A)$.
In particular, $\real\in\D$, since $\Delta$ contains all constant functions.

Now, if $S,S^c\in\D$ then 
the characteristic function $\charf{S}$
of $S$ belongs to $\C(\D,\A)=\Delta$. 
So $S\in\{\emptyset,\real\}$, since otherwise $\charf{S}$
does not belong to $\Delta$. 
(The derivatives satisfy the Intermediate Value Theorem \cite{Br}.)
This implies the main part of the proposition, as $\A\subset\D$. 

The additional part follows immediately from 
the first part, inclusion \mbox{$\A\subset\D$}, and the fact that 
$\A\not\subset\{\emptyset,\real\}$. \qed

\medskip

Propositions~\ref{propTopolo} and \ref{propAlg} and 
the condition (\ref{eq:subs})
show, in particular, that we cannot expect to improve 
in any essential way 
the structure of the families $\D$ and $\A$ in $\Delta=\C(\D,\A)$. 

It is also worth to mention that the class $\Delta$
cannot be characterized by images of sets 
in a way similar to $\Delta=\C(\D,\A)$
in a sense that 
\[
\Delta\neq
\{f\in\real^\real\colon f[A]\in\B\ \text{ for every } A\in\A\}
\]
for any families $\A$ and $\B$ of subsets $\real$. 
This follows immediately from 
the following theorem~\cite[Thm. 4.1]{CDW}.
(The proof of the theorem is a modification of 
a theorem of Velleman from~\cite{V}.
Compare also \cite{CN}.)


\thm{thCDW}{If $\A$ and $\B$ are the families of subsets of $\real$ 
with the property
that 
$\C_{\A,\B}=\{f\in\real^\real\colon f[A]\in\B\ \text{ for every } A\in\A\}$ 
contains all continuous functions 
then there is a non-measurable function $f\in\C_{\A,\B}$.}






\section{Proof of Theorem \ref{thZFCdel}}

The results presented in this section are a modification of
an argument sent to the author by an anonymous referee
of a previous version of the paper which consisted mainly of the results
presented in the next section. 

The proof of the theorem presented below will be based on the following two
lemmas. Recall that a set $T\subset\real^n$ is analytic, if it is a continuous
image of a Borel subset of $\real^m$.

\lem{A1}{There exists a Bernstein set $B$ such that for every analytic
set $A\subset\real^2$
\begin{description}
\item{{\rm (A)}} if $T\cap(B\times B)=\emptyset$ then 
                 $T\subset (C\times\real)\cup(\real\times C)$
                 for some countable set $C\subset B^c$;
\item{{\rm (B)}} if $T\cap(B\times B^c)=\emptyset$ then 
                 $T\setminus\{\la x,x\ra\colon x\in\real\}
                 \subset (C\times\real)\cup(\real\times D)$
                 for some countable sets $C\subset B^c$
                 and $D\subset B$.
\item{{\rm (C)}} if $T\cap(B^c\times B)=\emptyset$ then 
                 $T\setminus\{\la x,x\ra\colon x\in\real\}
                 \subset (C\times\real)\cup(\real\times D)$
                 for some countable sets $C\subset B$
                 and $D\subset B^c$.
\end{description}
}

\proof Let $\{A_0,A_1,A_2\}$ be a partition of $\co$ onto the sets
of cardinality $\co$ and for $i<3$ let 
$\la T_\xi\colon \xi\in A_i\ra$ be an enumeration of all 
analytic subsets of $\real^2$. 
By transfinite 
induction on $\xi<\co$ we will choose disjoint four-element sets 
$D_\xi=\{a_\xi,b_\xi,c_\xi,d_\xi\}$
aiming for $B=\bigcup_{\xi<\co}\{a_\xi,b_\xi\}$
(thus, also for $B^c\supset\bigcup_{\xi<\co}\{c_\xi,d_\xi\}$).
The construction is done maintaining the following conditions
for every $\xi<\co$, where 
$D_\xi=\bigcup_{\zeta<\xi}\{a_\zeta,b_\zeta\}$
and $C_\xi=\bigcup_{\zeta<\xi}\{c_\zeta,d_\zeta\}$.
\begin{description}
\item{Choose different $a,b,c,d\in\real\setminus(C_\xi\cup D_\xi)$.}

\item[For $\xi\in A_0$:] 
       If $T_\xi\subset(C_\xi\times\real)\cup(\real\times C_\xi)$
       put $a_\xi=a$, $b_\xi=b$, $c_\xi=c$, and $d_\xi=d$. Otherwise choose 
  $\la z,d_\xi\ra\in T_\xi\setminus(C_\xi\times\real)\cup(\real\times C_\xi)$.
       If $z\neq d_\xi$ we put $c_\xi=z$ and choose different
       $a_\xi,b_\xi\in\{a,b,c,d\}\setminus\{c_\xi,d_\xi\}$.
       Otherwise we choose different
       $a_\xi,b_\xi,c_\xi\in\{a,b,c,d\}\setminus\{d_\xi\}$.


\item[For $\xi\in A_1$:] 
       If $T_\xi\setminus\{\la x,x\ra\colon x\in\real\}\subset
       (C_\xi\times\real)\cup(\real\times D_\xi)$
       put $a_\xi=a$, $b_\xi=b$, $c_\xi=c$, and $d_\xi=d$. Otherwise choose 
$\la c_\xi,a_\xi\ra\in T_\xi\setminus(C_\xi\times\real)\cup(\real\times D_\xi)$
       with $a_\xi\neq c_\xi$. Then choose different
       $b_\xi,d_\xi\in\{a,b,c,d\}\setminus\{a_\xi,c_\xi\}$.

\item[For $\xi\in A_2$:] 
       If $T_\xi\setminus\{\la x,x\ra\colon x\in\real\}\subset
       (D_\xi\times\real)\cup(\real\times C_\xi)$
       put $a_\xi=a$, $b_\xi=b$, $c_\xi=c$, and $d_\xi=d$. Otherwise choose 
$\la a_\xi,c_\xi\ra\in T_\xi\setminus(D_\xi\times\real)\cup(\real\times C_\xi)$
       with $a_\xi\neq c_\xi$. Then choose different
       $b_\xi,d_\xi\in\{a,b,c,d\}\setminus\{a_\xi,c_\xi\}$.
\end{description}
This finishes the construction. 

The construction gives us immediately (A)-(C) with sets $C$ and $D$ having
cardinality less than continuum. But this implies that 
the appropriate analytic set is covered by 
less than $\continuum$ many
horizontal and vertical lines, and hence it is covered by countably many
of these lines~\cite{vEKM}.  

To see that $B$ is Bernstein, take an arbitrary perfect set $P\subset\real$
and notice that $P\times\real$ must be intersected by
$B\times B$ and $B^c\times B$. \qed

\medskip

In what follows we will use the following notation. 
We will write $\J$ for the family of  
all intervals in the form $(-\infty,c)$ 
and $(c,\infty)$ with $c\in\real$, and $M_0$ for the family of all
$F_\sigma$ subsets $E$ of $\real$ such that 
every point of $E$ is a point of bilateral accumulation of $E$. 
Recall also (see e.g. \cite[page 62]{Br})
that $\C(M_0,\J)$ is equal to the family $\D\B_1$ of Darboux
Baire class one functions, and that $\Delta\subset\D\B_1$. 

The following lemma has been proved by 
Preiss and Tartaglia~\cite[Lemma~2]{PT}. (The lemmas 
in the paper is stated there only 
for the family $\Delta$. However, at the end of the paper
the authors remark that it is true for a lot wider
classes of functions, including the case presented below.)


\lem{lem:PTnew}{Let $\F\subset\real^\real$ be such that it contains 
all constant functions and that $|g[\real]|=\co$ for every non-constant
$g$  which is a difference of two functions from $\F$. 
Then for every $h\in\real^\real$ there exists at most one non-constant 
$f\in\F$ such that for some $Z\subset\real$,
$|Z|<\continuum$,
\begin{center}
$f(x)=h(x)$ for every $x\in\real$ such that 
$\{f(x),h(x)\}\not\subset Z$.
\end{center}
} 

\thm{ZFCdelta}{Assume that $\F\subset\D\B_1$  
contains all constant functions, is closed under 
constant addition, and that any non-constant $g$ which is
a difference of two functions from $\F$ has uncountable range. 
Then
\[
\F=\C(M_0,\J)\cap\C(\D_0,\{B+c\colon c\in\real\})=\C(\D,\A),
\]
where $B\subset\real$ is a Bernstein set from Lemma~\ref{A1}, 
$\A=\J\cup\{B+c\colon c\in\real\}$,
$\D_0=\{f^{-1}(B+c)\colon f\in\F\ \&\ c\in\real\}$,
and $\D=\{f^{-1}(A)\colon f\in\F\ \&\ A\in\A\}$.
}

\proof Note that $D_0=\{f^{-1}(B)\colon f\in\F\}$,
since $f^{-1}(B+c)=(f-c)^{-1}(B)$ and $\F$ is closed under 
constant addition. Also 
$\D\subset M_0\cup\D_0$. 



Clearly $\F\subset\C(\D,\A)$. We will show that 
\[
\C(\D,\A)\subset\C(M_0,\J)\cap\C(\D_0,\{B+c\colon c\in\real\})
\subset\F.
\]

To argue for  the first inclusion
fix an $h\in \C(\D,\A)$. Our first goal will be to show that
$h\in\bor$. For this is enough to prove that
\begin{equation}\label{eqZ1}
\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\text{for every $c\in\real$ there is $E\in\B$ with
$h^{-1}((-\infty,c))\subset E\subset h^{-1}((-\infty,c])$.}
\!\!\!\!\!\!\!
\end{equation}

To see (\ref{eqZ1}) fix $c\in\real$. If either 
$h^{-1}((-\infty,c))\in\B$ or $h^{-1}((-\infty,c])\in\B$
then (\ref{eqZ1}) clearly holds. So, we can assume that 
it is not the case. Then
$h^{-1}((-\infty,c)),h^{-1}((c,\infty))\in\D_0$.
Therefore there exist $f,g\in\F$ such that 
$h^{-1}((-\infty,c))=f^{-1}(B)$
and 
$h^{-1}((c,\infty))=g^{-1}(B)$. But then
\[
T=\{\la f(x),g(x)\ra\colon x\in\real\}
\]
does not intersect $B\times B$, since 
$f(x)\in B$ implies $h(x)\in(-\infty,c)$, and
$g(x)\in B$ implies $h(x)\in(c,\infty)$.
However, $T$ is analytic as an image of $\real$ under a 
Borel function $\la f,g\ra$. 
So, by Lemma~\ref{A1}(A), 
there exists a countable set $C\subset B^c$
such that
$T\subset(C\times\real)\cup(\real\times C)$. But then
\[
h^{-1}((-\infty,c))=f^{-1}(B)
\subset g^{-1}(C) \subset g^{-1}(B^c) 
=h^{-1}((-\infty,c]),
\]
where the first inclusion follows from the fact that
$x\in f^{-1}(B)$ implies that
$f(x)\in B$, so $f(x)\notin C$, and $g(x)\in C$.
Therefore (\ref{eqZ1}) is satisfied by a Borel set
$E=g^{-1}(C)$.

To show that $h\in\C(M_0,\J)=\D\B_1$ notice first that 
\begin{equation}\label{eqZ2}
\text{$g^{-1}(B)\notin\B$ for every non-constant $g\in\D\B_1$.}
\end{equation}
Indeed, if $g^{-1}(B)\in\B$ then
$g[g^{-1}(B)]$ is an analytic subset of a Bernstein set $B$, so it is
countable. Similarly, $g[g^{-1}(B^c)]\subset B^c$ is analytic, thus 
countable. Therefore 
\[
g[g^{-1}(B)]\cup g[g^{-1}(B^c)]=
g[g^{-1}(B\cup B^c)]=
g[\real]
\]
is countable as well. So $g$, being Darboux, must be constant.

Now, to prove that $h\in\D\B_1=\C(M_0,\J)$ fix $J\in\J$.
We have to show that
\[
h^{-1}(J)\in M_0.
\]
Indeed, we know that $h^{-1}(J)\in\D\subset M_0\cup\D_0$. 
If $h^{-1}(J)\in M_0$ there is nothing to prove. But if 
$h^{-1}(J)\in\D_0$ then $h^{-1}(J)=g^{-1}(B)$ for some $g\in\F$.
In particular $g^{-1}(B)\in\B$, since $h$ is Borel. So, by (\ref{eqZ2}), 
$g$ is constant. Therefore, 
$h^{-1}(J)=g^{-1}(B)\in\{\emptyset,\real\}\subset M_0$.
 
Next we will show that $h\in\C(\D_0,\{B+c\colon c\in\real\})$.
Indeed, it is obvious if $h$ is constant. 
So assume that $h$ is not constant. Then, by (\ref{eqZ2}),
\begin{equation}\label{eqZ3}
h^{-1}(B+c)=(h-c)^{-1}(B)\in\D\setminus\B\subset\D_0
\ \ \ \text{ for every }\ \ \ c\in\real,
\end{equation}
since $g=h-c$ is a Darboux non-constant function. The proof of the inclusion
$\C(\D,\A)\subset\C(M_0,\J)\cap\C(\D_0,\{B+c\colon c\in\real\})$
has been completed.  

To show that 
$\C(M_0,\J)\cap\C(\D_0,\{B+c\colon c\in\real\})\subset\F$
fix an arbitrary $h\in \C(M_0,\J)\cap\C(\D_0,\{B+c\colon c\in\real\})$. 
We will show that $h\in\F$. 

Clearly $h^{-1}(B+c)\in\D_0$ for every $c\in\real$,
since $h\in\C(\D_0,\{B+c\colon c\in\real\})$.
In particular, for every
$c\in\real$ there exists $f_c\in\F$ such that 
\begin{equation}\label{eqZ4}
h^{-1}(B+c)=f_c^{-1}(B).
\end{equation}
We claim that $h=f_0$, which will finish the proof, since $f_0\in\F$. 



To see this, first we note that
\[
f_c=f_0-c\ \ \text{ for every }c\in\real.
\]
Indeed, $(h-c)^{-1}(B)=f_c^{-1}(B)$, so
$U=\{\la h(x)-c, f_c(x)\ra\colon h(x)-c\ne f_c(x)\}$ 
is an analytic subset of
$[(B\times B)\cup (B^c\times B^c)]\setminus\{\la x,x\ra\colon x\in\real\}$.
Thus, by parts (B) and (C) of Lemma~\ref{A1},
there exist countable sets 
$C\subset B^c$, $D\subset B$,
$C_1\subset B$, and $D_1\subset B^c$ such that 
$U$ is a subset of a countable set
\[
[(C\cup\real)\cup(\real\times D)]\cap
[(C_1\cup\real)\cup(\real\times D_1)]=(C\times D_1)\cup(C_1\times D).
\]
Thus the set
\[
U+\la c,c\ra=\{\la h(x),(f_c+c)(x)\ra\colon h(x)\ne f_c(x)+c\}
\]
is countable too. Similarly we show that the set
$\{\la h(x),f_0(x)\ra\colon h(x)\ne f_0(x)\}$
is countable. Thus, by Lemma~\ref{lem:PTnew}, $f_0=f_c+c$. 

Now, to prove that $h=f_0$ assume, by way of contradiction,
that there exists an $x\in\real$ such that $h(x)\neq f_0(x)$.
Then $b=f_0(x)- h(x)\ne 0$.
Applying Lemma~\ref{A1}(B) to $T=\{\la y,y+b\ra\colon y\in\real\}$
we can find $y\in\real$ with the property that 
$\la y,y+b\ra\in B\times B^c$.
But then $x\notin f_0^{-1}(B+h(x)-y)=f_{h(x)-y}^{-1}(B)$
while $x\in h^{-1}(B+h(x)-y)$, contradicting (\ref{eqZ4}).
\qed

\medskip

Since for $\F=\Delta$ the assumptions of Theorem~\ref{ZFCdelta}
are clearly satisfied, 
Theorem~\ref{thZFCdel} can be easily deduced. 



\section{Proof of Theorem \ref{th:mainGen}}


Clearly, we can assume that $\F$ contains non-constant functions, since
otherwise $\F$ is equal to the class of all constant functions, 
and for such $\F$ the theorem is obvious. 
Also, independently of the choice of the family $\A$, we
will have $\F\subset\C(\D,\A)$, by the definition of family $\D$.
So, we do not have to worry about the inclusion~$\subset$.

To have the converse inclusion we have to find $\A$ such that
no function $h\in\R\setminus\F$ belongs to $\C(\D,\A)$.
This will be done by choosing for every  $h\in\R\setminus\F$
a non-empty ``witness set'' $A_h\subset\real$ such that 
\begin{equation}\label{con:goal}
h^{-1}(A_h)\notin\D,
\end{equation}
where $\A=\{A_h\colon h\in \R\setminus\F\}$.
Evidently such an $\A$ will have all desired properties.




The construction of sets $A_h$ will be done by transfinite induction.
More precisely, let $\kappa=|\R|\leq\continuum^+$ and let
$\{h_\alpha\colon\alpha<\kappa\}$ be an enumeration of
$\R\setminus\F$. The induction will be 
on the length $\kappa$ and at stage
$\alpha<\kappa$ we will choose a set $A_\alpha$ playing the  role
of $A_{h_\alpha}$, that is, satisfying (\ref{con:goal}).

A technical problem with choosing at stage $\alpha$ an $A_\alpha$
satisfying (\ref{con:goal}) is that at this moment we do not know yet
the entire set $\D$, which will be equal
to $\{f^{-1}(A_\beta)\colon f\in\F\ \text{ and }\ \beta<\kappa\}$.
The best we can do at this point is to choose 
$A_\alpha$ with $h_\alpha^{-1}(A_\alpha)\notin
\{f^{-1}(A_\beta)\colon f\in\F\ \&\ \beta<\alpha\}$, i.e., 
such that
\begin{description}
\item[(I$_\alpha$)] $h_\alpha^{-1}(A_\alpha)\neq f^{-1}(A_\beta)$
                    for every $f\in\F$ and $\beta<\alpha$,
\end{description}
and with 
$h_\alpha^{-1}(A_\alpha)\notin
\{f^{-1}(A_\alpha)\colon f\in\F\}$, that is, 
\begin{description}
\item[(II$_\alpha$)] $h_\alpha^{-1}(A_\alpha)\neq f^{-1}(A_\alpha)$
                    for every $f\in\F$.
\end{description}
In order to have $h_\alpha^{-1}(A_\alpha)\notin
\{f^{-1}(A_\beta)\colon f\in\F\ \&\ \alpha<\beta<\kappa\}$
we will make sure that $h_\alpha^{-1}(A_\alpha)\neq f^{-1}(A_\beta)$
for every $f\in\F$ and $\alpha<\beta<\kappa$.
Thus, we will be choosing ``future'' $A_\beta$'s ($\alpha<\beta<\kappa$)
to satisfy this requirements. But, by 
interchanging $\beta$ with $\alpha$, it is the same 
as choosing at each step $\alpha$ the set $A_\alpha$ 
satisfying
\begin{description}
\item[(III$_\alpha$)] $f^{-1}(A_\alpha)\neq h_\beta^{-1}(A_\beta)$
                    for every $f\in\F$ and $\beta<\alpha$.
\end{description}
Thus, the choice of $A_\alpha$ satisfying 
conditions (I$_\alpha$)-(III$_\alpha$) will result in ensuring
(\ref{con:goal}) to be satisfied. 

In addition to these conditions, however, 
in order to make sure that the construction can continue to completion 
we still have to make sure that 
the conclusion of Proposition~\ref{propAlg} is satisfied by 
$\D$. To have this, our induction will be aimed to 
satisfy the following inductive condition 
\begin{description}
\item[($\triangle_\alpha$)] 
$\{A,\real\setminus A\}\not\subset
\{f^{-1}(A_\alpha)\colon f\in\F\ \&\ \beta<\alpha\}$
for every $A\in\P(\real)\setminus\{\emptyset,\real\}$.
\end{description}
In order to preserve this condition while choosing $A_\alpha$
we will ensure that
\begin{description}
\item[(IV$_\alpha$)] 
$f^{-1}(A_\alpha)\neq\real\setminus g^{-1}(A_\beta)$
for every non-constant $f,g\in\F$ and $\beta<\alpha$
\end{description}
and
\begin{description}
\item[(V$_\alpha$)] 
$f^{-1}(A_\alpha)\neq \real\setminus g^{-1}(A_\alpha)$
for every distinct non-constant $f,g\in\F$.
\end{description}
At this point of the proof the reader should be convinced that
choosing $A_\alpha$ satisfying (I$_\alpha$)--(V$_\alpha$)
and ($\triangle_{\alpha+1}$)
will finish the proof, as long as we have already
chosen a sequence $\la A_\beta\colon \beta<\alpha\ra$ 
of non-empty subsets of $\real$ satisfying ($\triangle_\alpha$)
and (I$_\beta$)--(V$_\beta$) for all $\beta<\alpha$.






The construction of $A_\alpha$ satisfying (I$_\alpha$)--(V$_\alpha$)
and ($\triangle_{\alpha+1}$)
will be done by yet another transfinite induction argument. 
For this, let 
$\{\la f_\xi,g_\xi,\beta_\xi\ra\colon 0<\xi<\continuum\}$
be an enumeration of the set 
$\F\times\F\times\{\beta\colon\beta<\alpha\}$.
This can be found, since $|\F|\leq\co$ 
and $\alpha<\kappa\leq\continuum^+$. 
We will construct inductively increasing sequences 
$\la Y_\xi\subset\real\colon\xi<\continuum\ra$
and
$\la Z_\xi\subset\real\colon\xi<\continuum\ra$
with sets $Y_0$ and $Z_0$ having cardinality less than $\continuum$ and 
such that 
\begin{description}
\item[($\star_\xi$)] $Y_\xi\cap Z_\xi=\emptyset$
for every $\xi<\continuum$, and 
\item[($\star\star_\xi$)] the sets 
$Y_\xi\setminus \bigcup_{\zeta<\xi}Y_\zeta$ and 
$Z_\xi\setminus \bigcup_{\zeta<\xi}Z_\zeta$
are finite for every $0<\xi<\continuum$. 
\end{description}
The construction is aimed to ensure that
$A_\alpha=\bigcup_{\xi<\continuum}Y_\xi$
satisfies (I$_\alpha$)--(V$_\alpha$)
and that ($\triangle_{\alpha+1}$) holds.

Note that the condition ($\star\star_\xi$) together with the cardinality 
assumption
on $Y_0$ and $Z_0$ guarantee that all sets
$Y_\xi$ and $Z_\xi$ will have cardinalities less than~$\continuum$. 

To construct $Y_0$ and $Z_0$ let $f_0\in\F$ 
be the unique non-constant 
function from Lemma~\ref{lem:PTnew} for $h=h_\alpha$, if it exists,
and arbitrary non-constant function from $\F$ otherwise. 



If $|h_\alpha[\real]|=\continuum$ choose $x_0\in\real$ such that
$h_\alpha(x_0)\neq f_0(x_0)$ and define
$Y_0=\{h_\alpha(x_0)\}$ and $Z_0=\{f_0(x_0),z\}$,
where $z\in h_\alpha[\real]\setminus\{h_\alpha(x_0)\}$.
Notice that this implies that $A_\alpha$ will 
be non-empty and will have the following properties:
\begin{description}
\item[(R$_\alpha$)] $h_\alpha^{-1}(A_\alpha)\neq\real$, and
\item[(ii$_0$)] $h_\alpha^{-1}(A_\alpha)\neq f_0^{-1}(A_\alpha)$,
\end{description}
as $x_0\in h_\alpha^{-1}(A_\alpha)\setminus f_0^{-1}(A_\alpha)$.

If $|h_\alpha[\real]|<\continuum$ let $\bar Y_0$ and $\bar Z_0$
be non-empty sets forming a partition of $h_\alpha[\real]$.
Thus, $h_\alpha^{-1}(\bar Y_0)$ and $h_\alpha^{-1}(\bar Z_0)$
are non-empty, disjoint sets forming a partition of $\real$.
But, by condition ($\triangle_\alpha$),
at least one of these sets does not belong to
$\{f^{-1}(A_\beta)\colon f\in\F\ \&\ \beta<\alpha\}$.
Without loss of generality we can assume that
\begin{equation}\label{eq6}
h_\alpha^{-1}(\bar Y_0)\notin
\{f^{-1}(A_\beta)\colon f\in\F\ \&\ \beta<\alpha\}.
\end{equation}
Next choose $x_0\in\real$ such that 
$f_0(x_0)\not\in \bar Y_0\cup \bar Z_0=h_\alpha[\real]$.
This can be done since $|f_0[\real]|=\continuum$,
as $f_0$ is non-constant and is a difference
of two functions from $\F$. 
If $h_\alpha(x_0)\in \bar Y_0$ we put
$Y_0=\bar Y_0$ and $Z_0=\bar Z_0\cup\{f_0(x_0)\}$.
Otherwise we put
$Y_0=\bar Y_0\cup\{f_0(x_0)\}$ and $Z_0=\bar Z_0$.

Notice that the condition (R$_\alpha$) is guaranteed and that
$x_0$ distinguishes between 
$h_\alpha^{-1}(A_\alpha)$ and $f_0^{-1}(A_\alpha)$
implying (ii$_0$).
Also $h_\alpha^{-1}(A_\alpha)=h_\alpha^{-1}(\bar Y_0)$.
So, by (\ref{eq6}), (I$_\alpha$) holds. 

To proceed farther assume that for some 
ordinal $0<\xi<\continuum$
the sequences $\la Y_\zeta\colon\zeta<\xi\ra$
and $\la Z_\zeta\colon\zeta<\xi\ra$
have been already constructed. So
$Y^0=\bigcup_{\zeta<\xi}Y_\zeta$ and $Z^0=\bigcup_{\zeta<\xi}Z_\zeta$
are disjoint and have cardinalities less than $\continuum$.  
Sets $Y_\xi$ and $Z_\xi$ will be disjoint finite extensions
of $Y^0$ and $Z^0$, respectively, and will imply 
the following properties:
\begin{description}
\item[(i$_\xi$)] 
        $h_\alpha^{-1}(A_\alpha)\neq f_\xi^{-1}(A_{\beta_\xi})$;

\item[(ii$_\xi$)] 
        $h_\alpha^{-1}(A_\alpha)\neq f_\xi^{-1}(A_\alpha)$;

\item[(iii$_\xi$)] 
        $f_\xi^{-1}(A_\alpha)\neq h_{\beta_\xi}^{-1}(A_{\beta_\xi})$;

\item[(iv$_\xi$)] 
$f_\xi^{-1}(A_\alpha)\neq\real\setminus g_\xi^{-1}(A_{\beta_\xi})$
provided $f_\xi$ and $g_\xi$ are non-constant; and, 

\item[(v$_\xi$)] 
$f_\xi^{-1}(A_\alpha)\neq \real\setminus g_\xi^{-1}(A_\alpha)$
provided $f_\xi$ and $g_\xi$ are non-constant.
\end{description}
This will finish the proof, since 
our choice of 
triples $\la f_\xi,g_\xi,\beta_\xi\ra$
guarantee that all conditions (i$_\xi$) imply (I$_\alpha$),
all conditions (ii$_\xi$) imply (II$_\alpha$), and 
similarly for conditions (III$_\alpha$)--(V$_\alpha$).

To fulfill these requirements we will construct
increasing disjoint sequences $\la Y^i\colon i=0,\ldots,5\ra$ and
$\la Z^i\colon i=0,\ldots,5\ra$, at each step taking care 
of one of the above conditions ensuring that 
$Y_\xi=Y^5$ and $Z_\xi=Z^5$ will have the desired properties. 

Step (i). If $|h_\alpha[\real]|<\continuum$ then the choice of
$Y_0\subset Y^0$ and $Z_0\subset Z^0$ guarantee already (I$_\alpha$) and 
so, also (i$_\xi$). Then we can put $Y^1=Y^0$ and $Z^1=Z^0$.

Otherwise, choose $x_1\in\real$ such that 
$h_\alpha(x_1)\notin Y^0\cup Z^0$. 
If $x_1\notin f_\xi^{-1}(A_{\beta_\xi})$ 
put $Y^1=Y^0\cup\{h_\alpha(x_1)\}$ and $Z^1=Z^0$. 
Otherwise put $Y^1=Y^0$ and $Z^1=Z^0\cup\{h_\alpha(x_1)\}$.
It is easy to see that this guarantees (i$_\xi$),
with $x_1$ distinguishing between $h_\alpha^{-1}(A_\alpha)$ and 
$f_\xi^{-1}(A_{\beta_\xi})$.

Step (ii). If $f_\xi=f_0$ or $f_\xi$ is constant 
then (ii$_\xi$) is already 
implied either by (ii$_0$) or by (R$_\alpha$)
and we can define $Y^2=Y^1$ and $Z^2=Z^1$.

Otherwise, by Lemma~\ref{lem:PTnew} and the choice of $f_0$, 
there is an 
$x_2\in\real$ such that 
$h_\alpha(x_2)\neq f_\xi(x_2)$ and 
$\{h_\alpha(x_2),f_\xi(x_2)\}\not\subset Y^1 \cup Z^1$.
This ensures
that one can write $\{h_\alpha(x_2),f_\xi(x_2)\}$ as $\{y,z\}$ with $y
\notin Z^1$ and $z \notin Y^1$.  Then one can let $Y^2 = Y^1 \cup \{y\}$
and $Z^2 = Z^1 \cup \{z\}$.
Then $x_2$ distinguishes between 
$h_\alpha^{-1}(A_\alpha)$ and $f_\xi^{-1}(A_\alpha)$, implying 
(ii$_\xi$).

Step (iii). If $f_\xi$ is constant then $Y^3=Y^2$ and 
$Z^3=Z^2$ imply (iii$_\xi$) by (II$_{\beta_\xi}$). 

Otherwise, there exists $x_3\in\real$ such that 
$f_\xi(x_3)\in f_\xi[\real]\setminus(Y^2\cup Z^2)$. 
If $x_3\notin h_{\beta_\xi}^{-1}(A_{\beta_\xi})$ 
put $Y^3=Y^2\cup\{f_\xi(x_3)\}$ and $Z^3=Z^2$.
Otherwise define
$Y^3=Y^2$ and $Z^3=Z^2\cup\{f_\xi(x_3)\}$. 
Then $x_3$ distinguishes between 
$h_{\beta_\xi}^{-1}(A_{\beta_\xi})$ and $f_\xi^{-1}(A_\alpha)$, 
implying (iii$_\xi$).

Step (iv). If $f_\xi$  is constant 
then $Y^4=Y^3$ and  $Z^4=Z^3$ imply (iv$_\xi$). 

Otherwise, there exists $x_4\in\real$ such that 
$f_\xi(x_4)\in f_\xi[\real]\setminus(Y^3\cup Z^3)$. 
If $x_4\notin\real\setminus g_\xi^{-1}(A_{\beta_\xi})$
put $Y^4=Y^3\cup\{f_\xi(x_4)\}$ and $Z^4=Z^3$.
Otherwise define
$Y^4=Y^3$ and $Z^4=Z^3\cup\{f_\xi(x_4)\}$. 
Then $x_4$ distinguishes between 
$\real\setminus g_\xi^{-1}(A_{\beta_\xi})$ and $f_\xi^{-1}(A_\alpha)$, 
implying (iv$_\xi$).

Step (v). If $f_\xi$  is constant 
then $Y^5=Y^4$ and  $Z^5=Z^4$ imply (v$_\xi$). 

So, assume that $f_\xi$ is not constant.
Then there exists $x_5\in\real$ such that 
$f_\xi(x_5)\in f_\xi[\real]\setminus(Y^4\cup Z^4)$. 
If $g_\xi(x_5)\in Y^4$ 
put $Y^5=Y^4\cup\{f_\xi(x_5)\}$ and $Z^5=Z^4$.
This implies that 
$x_5\in g_\xi^{-1}(A_\alpha)\cap f_\xi^{-1}(A_\alpha)$,
so (v$_\xi$) holds.

If $g_\xi(x_5)\notin Y^4$ define
$Y^5=Y^4$ and $Z^5=Z^4\cup\{f_\xi(x_5),g_\xi(x_5)\}$.
Then $x_5\in
(\real\setminus g_\xi^{-1}(A_\alpha))\setminus f_\xi^{-1}(A_\alpha)$
again implying (v$_\xi$).

Since the construction clearly preserves ($\star_\alpha$)
the construction and the proof are completed.
\qed

\medskip

The following problems seem to be interesting.

\pr{pr1}{Can the family $\A$ in either Theorem~\ref{thZFCdel}
or Corollaries~\ref{corDouble}
or \ref{corGCH} consist of any kind of regular sets like
Lebesgue measurable, Borel, or sets with Baire property?}

\pr{pr2}{Can Corollary~\ref{corGCH} be proved without extra set theoretic
assumptions?}



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\end{document}