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\controldates{20-NOV-2001,20-NOV-2001,20-NOV-2001,20-NOV-2001}
 
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\begin{document}
\title[Darboux and nowhere constant continuous functions]
{On sums of Darboux and nowhere constant continuous functions}
\author[K. Ciesielski]{Krzysztof Ciesielski}
\address{Department of Mathematics,
West Virginia University,
Morgantown, West Virginia 26506-6310}
\email{K\_Cies@math.wvu.edu}
\urladdr{http://www.math.wvu.edu/\textasciitilde{}kcies}
\thanks{The work of the second author was partially supported by
KBN Grant 2 P03A 031 14.}
\author{Janusz Pawlikowski}
\address{Department of Mathematics,
University of  Wroc\l aw,	
pl. Grunwaldzki 2/4, 50-384 Wroc\l aw, Poland -- and --	Department of 
Mathematics,
West Virginia University,
Morgantown, West Virginia 26506-6310}
\email{pawlikow@math.uni.wroc.pl}
\commby{Alan Dow}
\date{November 13, 2000 and, in revised  form, January 24, 2001}
\subjclass{Primary 26A15; Secondary  03E35}
\keywords{Darboux, nowhere constant, images of continuous functions}


\begin{abstract}
We show that the property
\begin{itemize}
\item[(P)] for every  Darboux function $g\colon{\mathbb R}\to\mathbb{R}$ there exists
a continuous nowhere constant function $f\colon{\mathbb R}\to\mathbb{R}$ 
such that $f+g$ is
Darboux
\end{itemize}
follows from the following two propositions:
\begin{itemize}
\item[(A)]
for every subset $S$ of $\mathbb{R}$ of cardinality
$\mathfrak{c}$ there  exists a uniformly continuous
function $f\colon\mathbb{R}\to[0,1]$ such that $f[S]=[0,1]$,
\item[(B)]
for an arbitrary function $h\colon\mathbb{R}\to\mathbb{R}$ whose
image $h[\mathbb{R}]$ contains a non-trivial interval
there exists an $A\subset\mathbb{R}$ of cardinality
$\mathfrak{c}$ such that the restriction $h\restriction A$ of $h$ to $A$
is uniformly continuous,
\end{itemize}
which hold in the iterated perfect set model.
\end{abstract}
\maketitle

Our set theoretic terminology is standard and follows that of~\cite{CiBook}.
In particular, functions are identified with their graphs
and $|X|$ stands for the cardinality of a set $X$.
The symbol $\continuum$ stands for $|\real|$ and  $\C(\real)$ denotes the
family of all continuous functions from $\real$ into $\real$.
A function $f\colon\real\to\real$ is {\em Darboux}\/ if
a conclusion of the intermediate value theorem holds
for $f$ or, equivalently,
when $f$ maps every interval onto an interval;
$f$ is a {\em Sierpi\'nski-Zygmund function}\/
if its restriction  $f\restriction Y$ is discontinuous for every
subset $Y$ of $\real$ of cardinality $\continuum$;
$f$ is {\em nowhere constant}\/ if it is not constant
on any non-trivial interval.



The class of Darboux functions has been studied for a long time as one of
possible generalizations of the class of continuous functions.
However, it has some peculiar properties. For
example, it is not closed under  addition.
In fact, in 1927 Lindenbaum \cite{Li}
noticed that every function $f\colon\real\to\real$ can be
written as a sum of two Darboux functions (see also \cite{SMa})
while Erd\H{o}s~\cite{Erdos} showed
that if additionally
$f$ is measurable, then both of the summands can be chosen to be measurable.
In~1990 Kirchheim and Natkaniec showed that there are
some truly bad Darboux functions by proving the following.

\begin{proposition}[Kirchheim, Natkaniec~\cite{NK}]\label{pr:NatKir} 
If
the union of less than
$\continuum$ many meager subsets of\/ $\real$ is meager (thus under CH or MA),
then there exists a Darboux function $g\colon\R\to\R$ such that $f+g$ is not
Darboux for every  continuous nowhere constant function $f\colon\R\to\R$.
\end{proposition}

The problem of whether the additional set-theoretic assumptions are necessary in
this theorem was investigated in~1992  by Komj\'ath~\cite{Km} and was settled
in~1995 by Stepr\=ans.

\begin{proposition}[Stepr\=ans~\cite{St}]\label{pr:Stepr1} 
It is consistent with ZFC that
\begin{itemize}
\item[{\rm (P)}] for every  Darboux function $g\colon\R\to\R$ there exists
a continuous nowhere
constant function $f\colon\R\to\R$ such that $f+g$ is Darboux.
\end{itemize}
\end{proposition}

In his paper Stepr\=ans showed that (P) holds in the iterated perfect set model.
His direct forcing argument is quite complicated
and difficult to follow.
We will show that (P) follows from the
principles (A) and (B) defined in the abstract.
Since the proofs that (A) and (B) hold in the
the iterated perfect set model
are much simpler than the argument given
in~\cite{St}, our argument
for (P) is essentially more accessible.

The proof that (A) holds in the iterated perfect set model
is due to Miller~\cite{Mi}.
The proof that (B) holds in this model
is due to Balcerzak, Ciesielski, and Natkaniec\linebreak \cite{BCN}
and consists of a page-long argument based on
the results from~\cite{Mi}.
In addition, properties (A) and (B) follow
quite easily from the Covering Property Axiom (CPA)
as shown
by the authors in~\cite{CPAbook}.
For more on the Darboux property see also the survey
article~\cite{KCsurv}.

Note also that condition (B) is equivalent to its following
seemingly stronger version:
\begin{itemize}
\item[(B$^*$)]
for an arbitrary $S\subset\real$ and a
function $h\colon S\to\real$ whose
image $h[S]$ contains a non-trivial interval
there exists an $A\subset S$ of cardinality
$\continuum$ such that the restriction $h\restriction A$ of $h$ to $A$
is uniformly continuous.
\end{itemize}
Clearly (B$^*$) implies (B). To see the other implication let
$h$ be  as in (B$^*$) and $f\colon\real\to\real$
be Sierpi\'nski-Zygmund. Put $h^*=f\restriction(\real\setminus S)\cup h$
and apply (B) to find $A^*\subset\real$ of cardinality
$\continuum$ such that $h^*\restriction A^*$ is uniformly continuous.
Then $A=A^*\cap S$ is as in (B$^*$).





Our main theorem is the following.

\thm{thDarb}{
Assume that {\rm (A)} and {\rm (B)} hold. Then for every Darboux
function $d\colon\real\to\real$
there exist
a complete metric $\rho$ on $\C(\real)$
and  a dense $G_\delta$
subset $G$ of $\la \C(\real),\rho\ra$ of nowhere constant functions
such that $d+g$ is Darboux for every $g\in G$.

In particular
{\rm (A)}\&{\rm (B)} implies {\rm (P)}.}



Our proof will be based on the following two lemmas.

\lem{lemDarb}{
Let $d\colon\real\to\real$ be Darboux
and $D\subset\real$ be such that $d\restriction D$ is dense in $d$.
If $g\colon\real\to\real$ is continuous and such that
\begin{equation}\label{eqD1}
[(d+g)(\alpha),(d+g)(\beta)]\subset (d+g)[[\alpha,\beta]]
\ \ \mbox{ for all \ \ $\alpha,\beta\in D$,}
\end{equation}
then $d+g$ is Darboux.
}
\begin{proof} An easy proof can be found in~\cite[lemma~4.1]{St}.
\end{proof}
\lem{lemDarb2a}{Assume that {\rm (A)} holds and for every $n<\omega$
let $A_n\in[\real]^\continuum$. Then for every $n<\omega$ there exists
a $B_n\in[A_n]^\continuum$ such that
the closures of $B_n$'s are pairwise disjoint.
}

\begin{proof} First note that (A) implies
\begin{equation}\label{eqDarb2}
\forall \la C_n\in[\real]^\continuum\colon n<\omega\ra \
\exists C\in[C_0]^\continuum \
\forall n<\omega \ |C_n\setminus\cl(C)|=\continuum.
\end{equation}

Indeed, by condition (A), we can find
a continuous function $f\colon\real\to[0,1]$
such that $|C_0\cap f^{-1}(c)|=\continuum$
for every $c\in[0,1]$.
(Just take a composition of a function from (A) with
the Peano curve followed by a projection.)
Identify $2^\omega$ with
a subset of $[0,1]$ and
by induction on $n<\omega$
choose an increasing sequence $s_n\in 2^{n+1}$ such that
$|C_n\setminus f^{-1}(\{c\in2^\omega\colon s_n\subset c\})|=\continuum.$
Put $s=\bigcup_{n<\omega}s_n$. Then $C=C_0\cap f^{-1}(s)$
satisfies (\ref{eqDarb2}).


Next note that there exists a
sequence $\la D_n\colon n<\omega\ra$ of closed subsets of
$\real$\linebreak such that
$\left|A_n\cap\left(D_n\setminus\bigcup_{i<n}D_i\right)\right|
=\continuum.$
This sequence is constructed by using\linebreak induction on $n<\omega$ with each
set $D_n$ chosen by applying (\ref{eqDarb2})
to the sequence $\la A_k\setminus \bigcup_{i<n}D_i\colon n\leq k<\omega\ra$.

Finally, since each set $D_n\setminus\bigcup_{i<n}D_i$ is an
$F_\sigma$-set, we can find closed subsets of them, $E_n$, with
$B_n=E_n\cap A_n$ having cardinality $\continuum$.
It is easy to see that the sets $B_n$ are as required.
\end{proof}

\rem{rem11}{Note that Lemma~\ref{lemDarb2a} cannot be proved in ZFC.
Indeed, if we assume that there exists a $\continuum$-Luzin
set\footnote{A set $L\subset\real$ is a
{\em $\continuum$-Luzin set\/} if $|L|=\continuum$ but
$|L\cap N|<\continuum$ for every nowhere dense subset $N$ of $\real$.
It is well known (see, e.g., \cite[sec.~2]{Mi})
and easy to see that no $\continuum$-Luzin set can be mapped
continuously onto $[0,1]$. Thus (A) implies that there is no
$\continuum$-Luzin set.}
and $\la A_n\in[\real]^\continuum\colon n<\omega\ra$
is a sequence of $\continuum$-Luzin sets such that
every non-empty open interval contains one of $A_n$'s, then
for such a sequence there are no $B_n$'s as in the lemma.
}


\lem{lemDarb3}{Assume that {\rm (A)}
and {\rm (B)} hold and $d\colon\real\to\real$
is a Darboux function. If $D$ is a countable subset of $\real$
such that
\begin{itemize}
\item[\rm ($*$)] for every $\alpha<\beta$ from $D$ for which
$d(\alpha)=d(\beta)$
             there exist $p,q\in\real$ such that $\alpha<p<q<\beta$
             and $d[(p,q)]=\{f(\alpha)\}$,
\end{itemize}
then there exists a countable family $\A\subset[\real]^\continuum$ such that
\begin{itemize}
\item[(a)] different elements of $\A$ have disjoint closures,
\item[(b)] $d\restriction A$ is uniformly continuous
           for every $A\in\A$, and
\item[(c)] for every $\alpha,\beta\in D$
           there exists an $A\in\A$
           with the property that
           $d\restriction A\subset[\alpha,\beta]\times[d(\alpha),d(\beta)]$.
\end{itemize}
}

\begin{proof} First for every $\alpha<\beta$ from $D$ we choose
$A_\alpha^\beta\in[\real]^\continuum$ such that the family
$\A_0=\{A_\alpha^\beta\colon \alpha,\beta\in D\ \&\ \alpha<\beta\}$
satisfies (b) and (c). For this fix $\alpha<\beta$ from~$D$.

If $d(\alpha)=d(\beta)$, then it is enough to put
$A_\alpha^\beta=(p,q)$, where $p$ and $q$ are from ($*$).
So, assume that $d(\alpha)\neq d(\beta)$.
Then
$I=[d(\alpha),d(\beta)]$ is a non-trivial interval,
which is a subset of $d[[\alpha,\beta]]$,
since $d$ is Darboux.
Let $S=[\alpha,\beta]\cap d^{-1}(I)$
and notice that
$h=d\restriction S$ maps $S$ onto $I$.
So, by (B$^*$),
we can find
an $A_\alpha^\beta\in[S]^\continuum$ such that
$d\restriction A_\alpha^\beta$ is uniformly continuous.

It is easy to see that $\A_0$ satisfies (b) and (c)
so it is enough to decrease its elements to
get condition (a), while assuring that they still
have cardinality~$\continuum$.
This can be done by applying  Lemma~\ref{lemDarb2a}.
\end{proof}


\begin{proof}[Proof of Theorem~\ref{thDarb}]
Take a Darboux function
$d\colon\real\to\real$
and choose a countable dense set
$D\subset\real$
satisfying condition ($*$) from Lemma~\ref{lemDarb3}
such that
$d\restriction D$ is dense in $d$.
Let
$\A\subset[\real]^\continuum$
be the family from Lemma~\ref{lemDarb3}
and let $\{A_n\colon n<\omega\}$ be an enumeration of
$\A\cup\{\{d\}\colon d\in D\setminus\bigcup\A\}$.
Let $\rho_0$ be the uniform convergence
metric on $\C(\real)$, that is,
\[
\rho_0(f,g)=\min\{1,\sup\{|f(x)-g(x)|\colon x\in\real\}\}
\]
and, for $f,g\in \C(\real)$, let
\[
\rho_1(f,g)=2^{
-\min\{n<\omega\colon f\restriction A_n\neq g\restriction A_n\}}.
\]
(If $\{n<\omega\colon f\restriction A_n\neq g\restriction A_n\}=
\emptyset$ we assume that $\rho_1(f,g)=0$.)
Then $\rho_1$ is a pseudometric on $\C(\real)$.
Consider $\C(\real)$ with the following metric $\rho$:
\[
\rho(f,g)=\max\{\rho_0(f,g),\rho_1(f,g)\}
\]
and notice that $\la \C(\real),\rho\ra$ forms a complete
metric space.
We will prove that if $G$ is the set of all nowhere constant continuous
functions
$g\colon\real\to\real$ for which $d+g$
satisfies~(\ref{eqD1}) of Lemma~\ref{lemDarb},
then $G$ contains
a dense G$_\delta$ subset of $\la \C(\real),\rho\ra$.
For this we will show that for every $\alpha<\beta$ from $D$
the following two types of sets
contain dense open subsets of $\la \C(\real),\rho\ra$:
\[
H_\alpha^\beta=\{g\in \C(\real)\colon g\ \mbox{ is not constant on }\
(\alpha,\beta)\}
\]
and
\[
G_\alpha^\beta=
\{g\in \C(\real)\colon
[(d+g)(\alpha),(d+g)(\beta)]\subset(d+g)[[\alpha,\beta]]\}.
\]
Then
\[
\bigcap\left\{H_\alpha^\beta\cap G_\alpha^\beta\colon \alpha,\beta\in D\ \&\
\alpha<\beta\right\}\subset G
\]
will contain a dense G$_\delta$ subset of  $\la \C(\real),\rho\ra$.


To see that $H_\alpha^\beta$ contains a dense open subset first note that,
by (a) and (c) of Lemma~\ref{lemDarb3}, elements of $\A$ are
nowhere dense.
Next, take an $f\in \C(\real)$, and fix an $\e>0$. Let $B(f,\e)$
be the open $\rho$-ball centered at $f$ and of radius $\e$.
We will find $g\in B(f,\e)$ and $\delta>0$ such that
$B(g,\delta)\subset B(f,\e)\cap H_\alpha^\beta$.
So take an $n<\omega$ such that $2^{-n}<\e/4$,
choose a non-empty open interval
$J\subset (\alpha,\beta)\setminus\bigcup_{i=1}^n A_i$,
and pick different $x,y\in J$.
It is easy to find $g\in \C(\real)$ such that $g(x)\neq g(y)$,
$g\restriction (\real\setminus J)=f\restriction (\real\setminus J)$,
and $\rho_0(f,g)<2^{-n}$. Then, by the choice of $J$
we also have $\rho_1(f,g)<2^{-n}$, so
$g\in B(f,\e/4)$. Now, if $\delta=\min\{|g(x)-g(y)|/4,\e/4\}$,
then $B(g,\delta)\subset B(f,\e)\cap H_\alpha^\beta$.

To see that each $G_\alpha^\beta$  contains a dense open subset
take $f\in \C(\real)$ and $\e>0$. As previously we
will find $g\in B(f,\e)$ and $\delta>0$ such that
$B(g,\delta)\subset B(f,\e)\cap G_\alpha^\beta$.
Find $\alpha=x_0<x_1<\cdots<x_m=\beta$, $x_i\in D$,
such that the variation of $f$ on each interval $[x_i,x_{i+1}]$
is less than $\e/8$. Also, since $d\restriction[x_i,x_{i+1}]$
is Darboux, we can partition each $[x_i,x_{i+1}]$ even further
to also get that $|d(x_i)-d(x_{i+1})|<\e/8$ for all $i<m$.

Pick an $n<\omega$ such that $2^{-n}<\e/8$ and
$\{x_i\colon i\leq m\}\subset\bigcup_{i=1}^n A_i$.
For every $i<m$ choose an index  $k_i>n$ such that
$d\restriction A_{k_i}\subset(x_i,x_{i+1})\times[d(x_i),d(x_{i+1})]$.
By (A)
for every $i<m$ we can also pick a uniformly continuous function
$h_i$ from $A_{k_i}$ onto
$[(d+f)(x_i),(d+f)(x_{i+1})]$.
For each $i<m$ define $h\restriction A_{k_i}=(h_i-d-f)\restriction A_{k_i}$
and note that
\[
h[A_{k_i}]\subset h_i[A_{k_i}]-d[A_{k_i}]-f[A_{k_i}]
\subset [-\e/2,\e/2]
\]
since
$[(d+f)(x_i),(d+f)(x_{i+1})]\subset[d(x_i)+f(x_i)-\e/4,d(x_i)+f(x_i)+\e/4]$,
$d[A_{k_i}]\subset[d(x_i),d(x_{i+1})]\subset[d(x_i)-\e/8,d(x_i)+\e/8]$,
and 
$f[A_{k_i}]\subset(f(x_i)-\e/8,f(x_i)+\e/8)$.
Also define $h$ as $0$ on $\bigcup_{i=1}^n A_i$
and extend it to a uniformly continuous function
from $\real$ into $[-\e/2,\e/2]$.
Put $g=f+h$ and note that $\rho(f,g)\leq\e/2$.
Also let $k=\max\{n,k_0,\ldots,k_{m-1}\}$ and $\delta\in(0,2^{-k})$.
We claim that
$B(g,\delta)\subset B(f,\e)\cap G_\alpha^\beta$.

Indeed, it is easy to see that $B(g,\delta)\subset B(f,\e)$.
To see that $B(g,\delta)\subset G_\alpha^\beta$
take a $g_0\in B(g,\delta)$. By the choice of $\delta$, $h$, and $g$
for every $i\leq m$
we have
\[
(d+g_0)(x_i)=(d+g)(x_i)=(d+f+h)(x_i)=(d+f)(x_i)
\]
and, for $A=\bigcup_{j<m}A_{k_j}$,
\[
g_0\restriction A
=g\restriction A
=(f+h)\restriction A
=(f+(h_i-d-f))\restriction A
=(h_i-d)\restriction A.
\] 
So,
\begin{eqnarray*}
\left[(d+g_0)(\alpha),(d+g_0)(\beta)\right]
& = & \left[(d+f)(x_0),(d+f)(x_m)\right]\\
& \subset & \bigcup_{j<m}\left[(d+f)(x_j),(d+f)(x_{j+1})\right]\\
& = & \bigcup_{j<m}h_j[A_{k_j}]\\
& = & \bigcup_{j<m}(d+(h_j-d))[A_{k_j}]\\
& = & \bigcup_{j<m}(d+g_0)[A_{k_j}]\\
& \subset & (d+g_0)[[\alpha,\beta]]
\end{eqnarray*}
proving that $g_0\in G_\alpha^\beta$.
This finishes the proof of the theorem.
\end{proof}


We say that a function $f\colon\R\to\R$ is a $D^1$ function
if it is differentiable with its derivative $f'(x)$ finite
at every $x\in\real$; $f$ is in the class ``$D^1$''
provided its derivative $f'(x)$ exists at every point, but it can
have an infinite value.
We say that $f$ belongs to $\C^1$ (to ``$\C^1$'')
if it belongs to $D^1$ (to ``$D^1$'', respectively)
and its derivative $f'$ is continuous.

Let us also notice that in (P) we cannot
require that function $g$ is ``$\C^1$''.
This follows from the following fact which,
for the functions from the class $\C^1$, was first noticed
by Stepr\=ans~\cite[p.~118]{St}. Since Stepr\=ans
leaves his statement without any comments concerning its proof,
we include here a missing argument.





\prop{prop:addC1}{
There exists, in ZFC, a Darboux function $d\colon\real\to\real$ such that
$d+f$ is not Darboux for every non-constant \textup{``}$\C^1$\textup{''} function $f$.}

\begin{proof} Let $\real=\{x_\xi\colon\xi<\continuum\}$
and let $\{f_\xi\colon\xi<\continuum\}$
be an enumeration of all non-constant ``$\C^1$'' functions.
Notice that for every $\xi<\continuum$ there exists
a non-empty open interval $I_\xi$ such that $f_\xi\restriction I_\xi$
is strictly monotone.
(Just take an $x\in\real$ such that $f_\xi'(x)\neq 0$,
which exists since $f_\xi$ is non-constant.
Then $I_\xi$ is chosen as a neighborhood of $x$
on which $f_\xi'$ is non-zero.)


The function $d$ that we construct will be strongly Darboux
in a sense that $d^{-1}(r)$ is dense in $\real$ for every $r\in\real$.
For such $d$ in order to show that $d+f_\xi$ is not Darboux
it is enough to show that $(d+f_\xi)^{-1}(y_\xi)$
is not dense in $\real$ for some $y_\xi\in\real$.
(See, e.g., \cite[prop.~7.2.4]{CiBook}.)

By induction we construct
a sequence $\{\la Q_\xi,d_\xi,y_\xi\ra\colon\xi<\continuum\}$
such that for every $\xi<\continuum$ we have:
\begin{itemize}
\item[(i)] Sets $\{Q_\eta\colon \eta\leq\xi\}$ are
           countable and pairwise disjoint, and
           $x_\xi\in\bigcup_{\eta\leq\xi}Q_\eta$.
\item[(ii)] $d_\xi\colon Q_\xi\to\real$
           and  $d_\xi^{-1}(x_\xi)$ is dense in $\real$.
\item[(iii)] $y_\zeta\notin(d_\eta+f_\zeta)[Q_\eta\cap I_\zeta]$
             for every $\zeta,\eta\leq\xi$.
\end{itemize}

Notice that if we have such a sequence, then, by (i),
$\real=\bigcup_{\xi<\continuum}Q_\xi$ so
$d=\bigcup_{\xi<\mu}d_\xi\colon\real\to\real$.
It is strongly Darboux by condition (ii), while, by (iii),
\pagebreak
for every $\zeta<\continuum$ the set
$(d+f_\zeta)^{-1}(z_\zeta)$ is not dense in $\real$, since it misses
$I_\zeta$. Thus, $d$ is as desired.

To make an inductive $\xi$-th step first choose
a  countable dense subset $Q^0_\xi$ of $\real$ disjoint with
\[
\bigcup_{\eta<\xi}Q_\eta \cup
\bigcup_{\zeta<\xi}\left(I_\zeta\cap f_\zeta^{-1}(y_\zeta-x_\xi)\right).
\]
This can be done since, by the choice of of the intervals $I_\zeta$'s,
each of the sets
$I_\zeta\cap f_\zeta^{-1}(y_\zeta-x_\xi)$
has at most one element.


Define $d_\xi\restriction Q^0_\xi$ as constantly equal to $x_\xi$.
This guarantees (ii), while condition (iii)
is satisfied for the part defined so far:
for every $\zeta<\xi$ and $x\in I_\zeta\cap Q_\xi^0$ we have
$d_\xi(x)+f_\zeta(x)= f_\zeta(x)+x_\xi\neq y_\zeta$,
since otherwise we would have
$x\in I_\zeta\cap f_\zeta^{-1}(y_\zeta-x_\xi)$ contradicting the choice
of $Q^0_\xi$.

If $x_\xi\in \bigcup_{\eta<\xi}Q_\eta$ we define  $Q_\xi=Q^0_\xi$.
Otherwise we put $Q_\xi=Q^0_\xi\cup\{x_\xi\}$
and if $d_\xi(x_\xi)$ is not defined yet
(i.e., if $x_\xi\notin Q^0_\xi$) we define $d_\xi(x_\xi)$
by choosing
$d_\xi(x_\xi)\notin\{y_\zeta-f_\zeta(x_\xi)\colon\zeta<\xi\}$.
This guarantees that (i) is satisfied, while (iii) is preserved.

Finally, we choose $y_\xi$ to have
\[
y_\xi\in\real\setminus\bigcup_{\eta\leq\xi} (d_\eta+f_\xi)[Q_\eta].
\]
This will guarantee that (iii) also holds for $\zeta=\xi$.
\end{proof}




Let us also note that the following question,
for the case of $D^1$ due to Stepr\=ans~\cite[Question~5.1]{St},
remains open.

\pr{pr:Stepr}{Does there exist, in ZFC,
a Darboux function $d\colon\real\to\real$ such that
$d+f$ is not Darboux for every nowhere constant ``$D^1$'' function $f$?
What if we restrict the choice of $f$ to $D^1$?
}





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\end{document}
