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\title{Some additive Darboux-like functions}
 
\author{
Krzysztof Ciesielski%
\thanks{AMS classification
numbers: Primary  26A15, 26A30; % ???????. \endgraf
Secondary  03E50. \newline %\endgraf
Key words and phrases: additive, Darboux, almost continuous, 
extendability functions.
\newline
This work was partially supported by 
NSF Cooperative Research Grant INT-9600548 and KBN Grant .... .}
\\
{\footnotesize
Department of Mathematics, West Virginia University,} \\
{\footnotesize  Morgantown, WV 26506-6310} \\
{\footnotesize  KCies@@wvnvms.wvnet.edu}
}
\date{}
 
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\begin{document}
 
\maketitle

\begin{abstract}

In this note we will construct several additive
Darboux-like functions $f\colon\real\to\real$
answering some problems from (an earlier version of)~\cite{GNsurv}. 
In particular,
in Section~\ref{sec1} we will construct, under different 
additional set theoretical assumptions, 
additive almost continuous (in sense of Stallings)
functions $f\colon\real\to\real$ whose graph is either meager or null
in the plane. 
In Section~\ref{sec2} we will construct, under different 
additional set theoretical assumptions, a function $f\colon\real\to\real$
which is additive, almost continuous, has the 
strong Cantor 
intermediate value property, but is not extendable.
In Section~\ref{sec3} we will construct
an additive almost continuous function $f\colon\real\to\real$
which has the Cantor intermediate value property
but is discontinuous on any perfect set. In particular, such an $f$
does not have the strong Cantor intermediate value property.

\end{abstract}

\section{Preliminaries}\label{sec0}

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 $\real$ and $\rational$ we denote the set
of all real and rational numbers, respectively.
We will consider $\real$ and $\real^2$ as linear spaces over $\rational$.
In particular, for a subset $X$ of either $\real$ or $\real^2$
we will use the symbol $\lin(X)$ to denote the smallest
linear subspace (of $\real$ or $\real^2$) over $\rational$.
Recall also that if $D\subset\real$ is linearly independent
over $\rational$ and $f\colon D\to\real$ then
$F=\lin(f)\subset\real^2$ is an additive function (see definition below) 
from $\lin(D)$ into $\real$. 
Any linear basis of $\real$ over $\rational$ will be referred
as a {\em Hamel basis}. By a Cantor set we mean any nonempty
perfect nowhere dense subset of $\real$.

The ordinal numbers will be identified with the sets of 
all their predecessors, and cardinals with the initial ordinals. In particular
$2=\{0,1\}$, and the first infinite ordinal $\omega$ number is
is equal to the set of all natural numbers $\{0,1,2,\ldots\}$. 
The family of all functions from a
set $X$ into $Y$ is denoted by $Y^X$. In particular, 
$2^n$ will stand for the set of all sequences
$s\colon \{0,1,2,\ldots,n-1\}\to\{0,1\}$, while
$2^{<\omega}=\bigcup_{n<\omega}2^n$ is the set of all finite sequences 
into $2$. 
The symbol $|X|$ stands for the cardinality of a set $X$. The cardinality of
$\real$, is denoted by $\co$ and referred as {\em continuum}. 
A set $S\subset\real$ is said to be {\em $\co$-dense\/}
if $|S\cap(a,b)|=\co$ for every $a<b$. 



We will use also the following terminology~\cite{GNsurv}.
A function $f\colon\real\to\real$ is:
\begin{itemize}
\item {\em additive\/} 
if $f(x+y)=f(x)+f(y)$ for every $x,y\in\real$;

\item {\em almost continuous\/} (in sense of Stallings)  
if each open subset of $\real\times\real$ containing the graph of
$f$ contains also a continuous function from $\real$ to $\real$~\cite{Stal};

\item has the 
 {\it Cantor intermediate value property\/} if for every 
 $x,y\in\real$ 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
  has the 
 {\it strong Cantor intermediate value property\/} if for every 
 $x,y\in\real$ 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 $f\restriction C$ is continuous;

\item is an 
 {\it extendability\/} function if 
 there exists a connectivity function $F\colon\real\times[0,1]\to\real$
 such that $f(x)=F(x,0)$ for every $x\in\real$, where
 
\item for a topological space $X$ a function $f\colon X\to\real$ is a
 {\it connectivity\/} function if the graph of $f\restriction Z$
 is connected in $Z\times\real$ for any connected subset $Z$ of~$X$.
\end{itemize}
Recall also that if the graph of $f\colon\real\to\real$
intersects every closed subset $B$ of $\real^2$ 
which projection $\proj(B)$ onto the $x$-axis
has nonempty interior then $f$ is almost continuous.
(See e.g.~\cite{N1}.)

\section{An additive discontinuous almost continuous function with 
a small graph}\label{sec1}


In this section we will show that the continuum hypothesis implies
the existence of an additive almost continuous 
function $f\colon\real\to\real$ whose graph is first category
(or null) in the plane. 
This answers a question of Grande \cite{ZG}. 
(See also \cite{GMN} and \cite[Question 5.2]{GNsurv}.)
The author likes here to thank Udayan B. Darji
for very helpful conversations on the subject.





\thm{thMain1}{For $i=1,2$ let $S_i\subset\real$ be such that 
$q\cdot S_i\subset S_i$ for every $q\in\rational$ and that
the set
\[
\bigcap_{r\in T}(r+S_i)
\]
is $\continuum$-dense for any subset $T$ of $\real$
of cardinality less than continuum. 
Then there exists an additive discontinuous almost continuous function
$f\colon\real\to\real$ 
such that $f\subset(S_1\times\real)\cup(\real\times S_2)$.
}



Before we prove this theorem we like to notice the following corollary.

\cor{cor1}{\text{ }
\begin{itemize}
\item[{\rm (1)}] If $\real$ is not a union of less than continuum meager sets
                 then there exists an additive discontinuous almost continuous
                 function $f\colon\real\to\real$ with the graph of measure zero. 

\item[{\rm (2)}] If $\real$ is not a union of less than continuum sets
                 of measure zero then there exists an additive discontinuous 
                 almost continuous
                 function $f\colon\real\to\real$ with a meager graph.
\end{itemize}
}

\proof (1) Let $S$ be a dense G$_\delta$ subset of $\real$ of measure zero
and put 
$S_1=S_2=\bigcup_{q\in\rational}q\cdot S$.
Then the sets $S_1$ and $S_2$ satisfy the assumptions
of Theorem~\ref{thMain1}, while the set 
$(S_1\times\real)\cup(\real\times S_2)$ has measure zero. 

(2) Replace $S$ with a meager set of full measure. \qed

\medskip

{\sc \noindent Proof of Theorem~\ref{thMain1}. }
Let $S=(S_1\times\real)\cup(\real\times S_2)$, and 
$\{A,C\}$ be a partition of $\continuum$ with each set having cardinality
$\continuum$. Let $\{B_\xi\colon \xi\in A\}$ be an enumeration of all
closed subsets $B$ of $\real^2$ with $\proj(B)$ 
having nonempty interior,
and %let 
$\{r_\xi\colon \xi\in C\}$ be an enumeration of $\real$. 
By induction on $\xi<\continuum$ we will choose
a sequence $\la\la x_\xi,y_\xi\ra\colon \xi<\continuum\ra$ 
such that the following inductive assumptions are satisfied
for every $\xi<\continuum$. 
\begin{itemize}
\item[{\rm (i)}]    $x_\xi\notin\lin(\{x_\zeta\colon\zeta<\xi\})$. 

\item[{\rm (ii)}]   $f_\xi=\lin(\{\la x_\zeta,y_\zeta\ra\colon\zeta\leq\xi\})
                    \subset S$.

\item[{\rm (iii)}]  If $\xi\in A$ then $\la x_\xi,y_\xi\ra\in B_\xi$.

\item[{\rm (iv)}]   If $\xi\in C$ then
                    $r_\xi\in\lin(\{x_\zeta\colon\zeta\leq\xi\})$.
\end{itemize}
Note first if we have such a sequence then, by (i) and (iv) the set
$\{x_\xi\colon \xi<\continuum\}$ is a Hamel basis.
Thus $f=\lin(\{\la x_\xi,y_\xi\ra\colon\xi<\continuum\})$
is an additive function from $\real$ into $\real$ for which, by (ii), 
$f\subset S=(S_1\times\real)\cup(\real\times S_2)$.
Moreover, by (iii), $f$ is almost continuous and has a dense graph in $\real^2$. 

To construct a sequence as described above 
assume that for some $\xi<\continuum$ the sequence
$\la\la x_\zeta,y_\zeta\ra\colon \zeta<\xi\ra$
satisfying (i)--(iv) is already constructed. Then,
by the inductive hypothesis, 
\[%\begin{equation}\label{eq1}
g_\xi=\lin(\{\la x_\zeta,y_\zeta\ra\colon\zeta<\xi\})
=\bigcup_{\zeta<\xi}f_\zeta\subset S.
\]%\end{equation}
Let $D_\xi$ be the domain of $g_\xi$. 
The difficulty in choosing $\la x_\xi,y_\xi\ra$ is to make sure that
\[
f_\xi=\{\la x,y\ra+q\cdot\la x_\xi,y_\xi\ra\colon 
\la x,y\ra\in g_\xi\ \&\ q\in\rational\}
%\lin(g_\xi\cup\{q\cdot\la x_\xi,y_\xi\ra\colon q\in\rational\})
\subset S
\]
which is equivalent to the choice of $\la x_\xi,y_\xi\ra$
from the set
\[
\bigcap_{\la x,y\ra\in g_\xi}\left[
\la x,y\ra+
S\right)]\supset
\left[\left(\bigcap_{x\in D_\xi}(x+S_1)\right)\times\real\right]\cup
\left[\real\times\left(\bigcap_{x\in D_\xi}(g_\xi(x)+S_2)\right)\right].
\]
(Note that $S$ is closed under rational multiplication.)

Assume first that $\xi\in C$. If $r_\xi\notin D_\xi$
put $x_\xi=r_\xi$. Otherwise pick an arbitrary $x_\xi\in\real\setminus D_\xi$.
This will guarantee (i) and (iv). In order to have (ii) 
choose $y_\xi$ from $\bigcap_{x\in D_\xi}(g_\xi(x)+S_2)$,
which is nonempty by the assumption from the theorem
since $|D_\xi|\leq |\xi|+\omega<\co$.
Then $\la x_\xi,y_\xi\ra\in 
\real\times\left(\bigcap_{x\in D_\xi}(g_\xi(x)+S_2)\right)$
implying (ii).

To finish the proof, assume that $\xi\in A$. The set 
$T=\bigcap_{x\in D_\xi}(x+S_1)$ is $\continuum$-dense
so we can choose 
$x_\xi\in T\cap \proj(B_\xi)\setminus D_\xi$.
Take $y_\xi$ such that $\la x_\xi,y_\xi\ra\in B_\xi$.
Then (i), (ii), and (iii) are satisfied.
\qed

\medskip

To state the last corollary of this section we need the following lemma,
that seems to have an interest of its own.

\lem{lemMain1}{There exists a meager set $S\subset\real$ of measure zero 
with the properties that %such that 
$p+q\cdot S\subset S$ for every $p,q\in\rational$, and %that
the set
\[
\bigcap_{i<\omega}(r_i+S)
\]
contains a perfect set for every sequence $\la r_i\in\real\colon i<\omega\ra$. 
}

\proof 
For $1<k<\omega$ and 
a sequence $\la s_n\subset n\colon k\leq n<\omega\ra$ of nonempty sets let
\[
T(\la s_n\ra)=
\left\{\sum_{n=2}^\infty\frac{i_n}{n!}\colon\forall k\leq n<\omega\; (i_n\in s_n)\right\}.
\]
Note that $T(\la s_n\ra)$ is a nonempty closed subset of $[0,1]$.
It is nowhere dense, unless $s_n=n$ for all but finitely many $n$.
Moreover, if there exists $k\leq N<\omega$ such that
$s_n=n$ for all $n>N$ then
\[
m(T(\la s_n\ra))=\prod_{n=k}^N\frac{|s_n|}{n}.
\]
Also if $c_n=n-1$ for
$k\leq n<\omega$ then we denote the set $T(\la c_n\ra)$
by $T^k$. It is easy to see that
\[%\begin{equation}\label{eqAA}
m(T^k)=\prod_{n=k}^\infty\frac{n-1}{n}=0\ \text{ for every } k.
\]%\end{equation}
Define 
\[
S=\bigcup\left\{p + q T^k\colon p,q\in\rational\ \&\ 1<k<\omega\right\}.
\]
Then $S$ is meager, has measure zero, and is closed
under $\rational$ addition and multiplication. 
To finish the proof, choose a sequence $\la r_i\in\real\colon i<\omega\ra$.
It is enough to prove that $\bigcap_{3<i<\omega}(r_i+S)$ contains a perfect set. 
To prove this notice first that for every
$r\in\real$ and every $1<k<\omega$ there exists 
a sequence $\la s_n\subset n\colon k\leq n<\omega\ra$ 
with each $|s_n|\geq n-2$ and such that
\[
T(\la s_n\ra)\subset 
r+\bigcup\left\{p + T^k\colon p\in\rational\right\}\subset r+S.
\]
This follows from the fact that if 
$x,y\in T^k$ have the same ``$m$-th digit'' $i_m$ in the representation
$\sum_{n=2}^\infty\frac{i_n}{n!}$, then the ``$m$-th digits''
of $r+x$ and $r+y$ can differ by at most $1$ modulo $m$. 
(To see it, assume that $r$ is of the form
$p+\sum_{n=2}^\infty\frac{j_n}{n!}$ with $p$ being an integer. Then 
$x+p+\sum_{n=2}^m\frac{j_n}{n!}$ and 
$y+p+\sum_{n=2}^m\frac{j_n}{n!}$ have the same ``$m$-th digit,''
while by adding to any number the reminder 
$\sum_{n=m+1}^\infty\frac{j_n}{n!}$ of $r$,
we increase the ``$m$-th digit'' by either $0$ or $1$ modulo $m$.)

Now, for each $3<i<\omega$ choose 
a sequence $\la s^i_n\subset n\colon 2i\leq n<\omega\ra$
with $|s^i_n|\geq n-2$ for every $n\geq 2$ 
for which 
\[
T(\la s^{i}_n\ra)\subset
r_i+\bigcup\left\{p + T^{2i}\colon p\in\rational\right\}\subset r_i+S.
\]
For every $8\leq n<\omega$ let 
$s_n=\bigcap_{8\leq 2i\leq n} s^i_n\subset n$. Then each $s_n$ has at least two
elements and 
%
%Then there exists a sequence $\la s_n\subset n\colon 8\leq n<\omega\ra$
%with each $s_n$ having at least two elements and such that
\[
%T(\la s_n\ra)\subset\bigcap_{3<i<\omega}(r_i+S).
T(\la s_n\ra)\subset\bigcap_{3<i<\omega}T(\la s^{i}_n\ra)\subset
\bigcap_{3<i<\omega}(r_i+S).
\]
This finishes the proof. \qed


\cor{cor12}{If the continuum hypothesis holds then there exists an additive
discontinuous almost continuous
function $f\colon\real\to\real$ with the graph 
which is simultaneously meager and of measure zero.
}

\proof Apply Theorem~\ref{thMain1} to $S_1=S_2=S$, where
$S$ is from Lemma~\ref{lemMain1}. \qed


\pr{pr11}{Is it possible to find in ZFC 
an example of additive discontinuous almost continuous
function $f\colon\real\to\real$ with small graph
(in sense of measure, category, or both)?}




\section{An additive almost continuous nonextendable function with 
the strong Cantor intermediate value property}\label{sec2}

In~\cite{Ro} H.~Rosen proved that under the continuum hypothesis there
exists an almost continuous nonextendable function $f\colon[0,1]\to[0,1]$
with the strong Cantor intermediate value property.
He asked also whether there exists such a function
$f\colon\real\to\real$ which is additive. 
The goal of this section is to show that, under 
the continuum hypothesis and Martin's axiom, the answer to this question 
is positive. The construction presented below
is a refinement of that from~\cite{Ro}. 

\thm{thMain2}{If union of less than continuum many meager sets is meager then 
there exists an additive nonextendable almost continuous function
$f\colon\real\to\real$ with the strong Cantor intermediate value property. 
}

The construction will be based on the following two facts.

\lem{lemBasis}{Every perfect set $P_0\subset\real$ has a perfect
subset $P\subset P_0$ which is linearly independent over $\rational$.}

\proof This can be proved by a minor modification of 
the proof presented in \cite[thm.~2, Ch.~XI sec.~7]{Kucz}
that there exists a perfect subset of $\real$ which is
linearly independent over $\rational$. \qed

\lem{lemPerf}{Let $V$ be a linear subspace of $\real$ over $\rational$
such that $V$ is meager in $\real$. Then for every 
perfect set $P_0\subset\real$ there exists a perfect
subset $P\subset P_0$ such that $\lin(V\cup P)$ is meager.}

\proof First notice that 
\begin{description}
\item{($\star)$}
for every compact nowhere dense sets 
$F$ and $P$ there exists perfect $P^\prime\subset P$ such that
$F+P^\prime$ is nowhere dense. 
\end{description}

To see it, let $\la I_n\colon n<\omega\ra$ be an enumeration
of all nonempty open intervals with rational endpoints. 
Construct a family $\{P_s\colon s\in 2^{<\omega}\}$
of perfect subsets of $P$ by induction on length of a sequence $s$.
We start with putting $P_\emptyset=P$.
Next, if for some $n<\omega$ the sets $\{P_s\colon s\in 2^n\}$
are already chosen, for every $s\in 2^n$ pick 
disjoint perfect subsets 
$P_{s^\frown 0}$ and $P_{s^\frown 1}$ of $P_s$ such that 
\[
I_n\not\subset F+\bigcup\left\{P_t\colon t\in 2^{n+1}\right\}
\]
Such a choice is possible, since $F$ is nowhere dense. 
Now, if $P^n=\bigcup\{P_s\colon s\in 2^s\}$ and
$P^\prime=\bigcap_{n<\omega}P^n$, then $P^\prime$ 
satisfies ($\star$). 

To prove the main part of the lemma,
let $\{V^n\colon n<\omega\}$
be an increasing sequence of compact perfect nowhere dense sets
such that $V\subset\bigcup_{n<\omega}V^n$, and 
let $\la q_n\colon n<\omega\ra$ be a sequence
of the rational numbers, with each number appearing
infinitely many times. 
By a similar tree induction as above, using property ($\star$) 
at the inductive step, 
construct a family $\{P_s\colon s\in 2^{<\omega}\}$
of perfect subsets of $P_\emptyset=P_0$ 
such that $P_{s^\frown 0}$ and $P_{s^\frown 1}$
are disjoint perfect subsets of $P_s$
and if $P^n=\bigcup\{P_s\colon s\in 2^n\}$ then 
$V_n+q_0\cdot P^0+\cdots +q_n\cdot P^n$ is nowhere dense.
Then $P=\bigcap_{n<\omega}P^n$ is perfect 
and  
$\lin(V\cup P)\subset\bigcup_{n<\omega}(V_n+q_0\cdot P^0+\cdots +q_n\cdot P^n)$ is
meager. 
\qed

\medskip

{\sc \noindent Proof of Theorem~\ref{thMain2}. }
Let $\{B_\xi\colon \xi<\continuum\}$ be an enumeration of all
closed subsets $B$ of $\real^2$ with $\proj(B)$ 
having nonempty interior,
and $\{G_\xi\colon \xi<\continuum\}$
be an enumeration of all dense G$_\delta$ 
subsets of $\real$.
By induction on $\xi<\continuum$ construct
a sequence of functions 
$\la f_\xi\colon D_\xi\to\real\colon \xi<\continuum\ra$ 
such that the following inductive assumptions are satisfied
for every $\xi<\continuum$. 
\begin{itemize}
\item[{\rm (i)}] $D_\xi\subset\real$ is perfect, linearly independent over
                 $\rational$, and
\[
D_\xi\subset\left(\bigcap_{\zeta\leq\xi}G_\zeta\right)\setminus 
\lin\left(\bigcup_{\zeta<\xi}D_\zeta\right).
\]

\item[{\rm (ii)}]   $f_\xi\subset B_\xi$ and $f_\xi$ is continuous.

\item[{\rm (iii)}]  If $F_\xi=\lin\left(\bigcup_{\zeta\leq\xi}f_\zeta\right)$
                    then the domain and the range of $F_\xi$ 
                    are meager subsets of $\real$. 
\end{itemize}
To construct such a sequence 
assume that for some $\xi<\continuum$ a sequence
$\la f_\zeta\colon \zeta<\xi\ra$
satisfying (i)--(iii) is already constructed. Then,
by the inductive hypothesis (iii), 
the set $\lin\left(\bigcup_{\zeta<\xi}D_\zeta\right)$
is a union of less than continuum many meager sets, so
it is meager. Thus
$E=\left(\bigcap_{\zeta\leq\xi}G_\zeta\right)\setminus 
\lin\left(\bigcup_{\zeta<\xi}D_\zeta\right)$ contains a dense 
G$_\delta$ subset. Now, by Baire category theorem,
we can find an $M\in\real$ is such that
$\proj\left(B_\xi\cap(\real\times(-\infty,M])\right)$
has a nonempty interior. Fix such an $M$ and  
pick a perfect set 
$P_0\subset E\cap\proj\left(B_\xi\cap(\real\times(-\infty,M])\right)$.
Then, by Lemma~\ref{lemBasis}, we can find a perfect set $P_1\subset P_0$
which is linearly independent over $\rational$. 
Our goal is to choose $D_\xi$ as a subset of $P_1$.
Note that any such $D_\xi$ will satisfy (i). 

Next, define function $g$ on $P_1$ by putting
$g(x)=\max\{y\leq M\colon \la x,y\ra\in B_\xi\}$
for every $x\in P_1$. Note that $g$ is upper semicontinuous, 
so Baire class one. In particular, there exists a perfect
set $D\subset P_1$ such that $g\restriction D$ is continuous.
We will define $f_\xi$ as $g\restriction D_\xi$ for some 
perfect $D_\xi\subset D$. Notice, that 
this will guarantee (ii).

To choose appropriate $D_\xi\subset D$ first
recall that 
the set $V=\lin\left(\bigcup_{\zeta<\xi}D_\zeta\right)$
is meager. So, by  Lemma~\ref{lemPerf},
we can choose a perfect set 
$D^\prime\subset D$ such that 
$\lin(V\cup D^\prime)$ is meager. 
Since our goal is to choose $D_\xi\subset D^\prime$
this will guarantee that 
the domain $\dom(F_\xi)=\lin(V\cup D_\xi)$ is meager as well. 
Now, put $P=g[D^\prime]$ and 
$W=\bigcup_{\zeta<\xi}\range(F_\zeta)$.
Then, $P$ is compact and, by the inductive hypothesis,
$W$ is meager. If there is $y\in P$ such that $D^\prime\cap g^{-1}(y)$
is uncountable, choose $D_\xi$ as a perfect subset
of $D^\prime\cap g^{-1}(y)$. Then 
$\range(F_\xi)=\lin(W\cup\{y\})$ is clearly meager.
So assume that $D^\prime\cap g^{-1}(y)$ is countable
for every $y\in P$. Then $P$ is uncountable, and 
it contains a perfect subset $P^\prime$.
So, by  Lemma~\ref{lemPerf}, there exists
perfect set $P^{\prime\prime}\subset P^\prime$
such that $\lin(W\cup P^{\prime\prime})$ is meager.
The construction is finished by choosing 
$D_\xi$ as a perfect subset of $D^\prime\cap g^{-1}(P^{\prime\prime})$.

To construct function $f$ let $H$ be a Hamel basis extending 
$\bigcup_{\xi<\continuum} D_\xi$
and let $D_\continuum=H\setminus \bigcup_{\xi<\continuum} D_\xi$.
Define $f_\continuum(x)=0$ for every $x\in D_\continuum$ and
$f=\lin\left(\bigcup_{\xi\leq\continuum} f_\xi\right)$.
Then $f$ is an additive function from $\real$ to $\real$ and it extends
each $f_\xi$. In particular, $f$ intersects each $B_\xi$, so 
it is almost continuous and has a dense graph. 

The rest of the argument goes pretty much as in~\cite{Ro}.
So see that $f$ 
has the strong Cantor intermediate value property
choose $x<y$ with $f(x)\neq f(y)$, and 
a Cantor set $K$ between $f(x)$ and $f(y)$.
Choose non-decreasing Cantor like continuous function
$h$ from $\real$ onto $[x,y]$ which is constant on
every component interval of $\real\setminus K$. 
Then there is a $\xi<\continuum$ such that
$B_\xi=h^{-1}=\{\la h(z),z\ra\colon z\in\real\}$.
So, $f_\xi\subset f\cap B_\xi$.
Now, if $A=h[\real\setminus K]$ then $A$ is countable 
and for any perfect subset $P$ of $D_\xi\setminus A$ 
the restriction $f\restriction P$ is continuous and $f[P]\subset K$.

To see that $f$ is not extendable, recall that 
for every extendable function $f\colon\real\to\real$
with the dense graph there exists
a dense G$_\delta$ set $G$ such that 
any modification of $f$ on the set $G$ is still an extendable function.
(See \cite{Ro1}, where it is proved for functions
from $[0,1]$ to  $[0,1]$. The generalization to functions 
from $\real$ to $\real$ is obvious.)
So, by way of contradiction, assume that $f$ is extendable
and let $G$ be a dense G$_\delta$ set mentioned above.
Then there exists a $\xi<\continuum$ such that $G_\xi=G$.
Modify $f$ to a function $h$ by assigning 
value $0$ to every $x\in G_\xi$. 
Then 
\[
h[\real]\subset 
f\left[\lin\left(D_\continuum\cup\bigcup_{\zeta<\xi}D_\zeta\right)\right]
\subset \range(F_\xi)
\]
is meager. Since $h$ is Darboux, as an extendable function, this implies
that $h$ is constant. But then,
any modification of $h$ on a point from $G$ gives
an extendable non-Darboux function, which is impossible.
This contradiction
finishes the proof. \qed


\section{An additive almost continuous function with 
the Cantor intermediate value property
which is discontinuous on any perfect set}\label{sec3}

In this section we will construct in ZFC an additive almost continuous function
$f\colon\real\to\real$
with the Cantor intermediate value property
which is discontinuous on any perfect set.
In particular, such a function does not have a strong 
Cantor intermediate value property. 
A similar example has been constructed by 
K.~Banaszewski and T.~Natkaniec \cite{KBanNat}:
they constructed an almost continuous function
$f\colon\real\to\real$
with the Cantor intermediate value property
which is of Sierpi\'nski--Zygmund type, i.e., is 
discontinuous on any set of cardinality continuum.
However, they had to use an additional set theoretical assumption
in their construction 
($\real$ is not a union of less than continuum many meager sets)
which is necessary, since there is a model
of ZFC with no
Darboux (so almost continuous) 
Sierpi\'nski--Zygmund function~\cite{BCN}.
The constructed example answers Question 3.11 
from~\cite{GNsurv}.

\thm{thMain3}{There exists an additive almost continuous function 
$f\colon\real\to\real$
which has the Cantor 
intermediate value property, but is not continuous on any perfect set.
In particular, $f$ does not have the strong 
Cantor intermediate value property.
}

The proof of the theorem is based on the following lemma.

\lem{lem31}{There exists a Hamel basis $H$ which can be partitioned 
onto the sets $\{P_\xi\colon \xi\leq\continuum\}$
with the following properties.
\begin{itemize}
\item[{\rm (1)}] 
For every $\xi<\continuum$ the set $P_\xi$ is perfect.

\item[{\rm (2)}] 
Every nonempty interval contains continuum many
sets $P_\xi$ and continuum many points from $P_\continuum$. 
\end{itemize}
}

\proof Let $P$ be a perfect set which is linearly independent over $\rational$.
(See Lemma~\ref{lemBasis}.) Let $K$ be a proper perfect subset of
$P$ and $\{x_\xi\colon \xi\leq\continuum\}$ be an enumeration of
$P\setminus K$. Then there is a sequence
$\la \la p_\xi,q_\xi\ra\in(\rational\setminus\{0\})^2\colon\xi<\co\ra$
such that the sets
$P_\xi=p_\xi x_\xi + q_\xi K$ satisfy the first
part of (2). They also clearly satisfy (1).
Now, it is easy to extend $\bigcup_{\xi<\co}P_\xi$
to a Hamel basis $H$ such that $P_\co=H\setminus \bigcup_{\xi<\co}P_\xi$
is a $\co$-dense. \qed

\medskip


{\sc \noindent Proof of Theorem~\ref{thMain3}.}
Let $\la\la I_\alpha,K_\alpha\ra\colon\alpha<\co\ra$ be a list of all
pairs $\la I,K\ra$ such that $I$ is a nonempty open interval and
$K$ is a perfect set. By (2) of Lemma~\ref{lem31}
we can reenumerate sets $P_\xi$ to have
$P_\alpha\subset I_\alpha$ for every $\alpha<\co$.
We will construct function $f$ to have 
$f[P_\alpha]\subset K_\alpha$. This will guarantee
the Cantor intermediate property of $f$.
Next, let 
$\{B_\xi\colon \xi<\continuum\}$ be an enumeration of all
closed subsets $B$ of $\real^2$ with $\proj(B)$ 
having nonempty interior,
$\{x_\xi\colon \xi<\continuum\}$
be an enumeration of the Hamel basis $H$ from Lemma~\ref{lem31},
and $\{C_\xi\colon \xi<\continuum\}$
be an enumeration of all perfect sets $C$ in $\real$ such that
$C$ is linearly independent over $\rational$. 
By induction on $\xi<\continuum$ construct
a sequence of functions 
$\la f_\xi\colon D_\xi\to\real\colon \xi<\continuum\ra$ 
such that the following inductive assumptions are satisfied
for every $\xi<\continuum$. 

\begin{itemize}
\item[{\rm (i)}]    $\{D_\zeta\colon \zeta\leq\xi\}$  are countable pairwise
                    disjoint subsets of $H$.

\item[{\rm (ii)}]   If $x\in D_\xi\cap P_\alpha$ for some $\alpha<\co$
                    then $f_\xi(x)\in K_\alpha$.

\item[{\rm (iii)}]  There exists $z\in D_\xi$ such that $\la z,f_\xi(z)\ra\in B_\xi$.

\item[{\rm (iv)}]   If $F_\xi=\lin\left(\bigcup_{\zeta\leq\xi}f_\zeta\right)$
                    then $x_\xi\in\dom(F_\xi)$ and $F_\xi\restriction C_\xi$ 
                    is discontinuous. 
\end{itemize}
To construct such a sequence 
assume that for some $\xi<\continuum$ a sequence
$\la f_\zeta\colon \zeta<\xi\ra$
satisfying (i)--(iv) is already constructed. 
Let $V=\lin\left(\bigcup_{\zeta<\xi}D_\zeta\right)$, choose a 
perfect subset $Z\subset C_\xi \setminus V$, and a 
countable dense subset $D$ of $Z$. 
Also, let $A=\bigcup_{z\in D}\supp(z)$, where 
$\supp(z)$ is the support of $z$, i.e., the smallest set $S\subset H$ for
which $z\in\lin(S)$. Then $A$ is countable, so we can choose 
$z_\omega\in Z\setminus\lin(A)$ and 
a sequence $\la z_n\in D\colon n<\omega\ra$ 
converging to $z_\omega$. Then 
$\{z_n\colon n\leq\omega\}\subset C_\xi\setminus V$.
Moreover, if $H_\eta=\supp(z_\eta)$ for $\eta\leq\omega$
then there exists 
$y\in H_\omega\setminus(V\cup\bigcup\{H_n\colon n<\omega\})$.
Choose $z\in \proj(B_\xi)\cap P_\co\setminus(V\cup\bigcup\{H_n\colon
n\leq\omega\})$ and define 
\[
D_\xi=\left(\{x_\xi,z\}\cup\bigcup\{H_n\colon n\leq\omega\}\right)\setminus V.
\]
Function $f_\xi$ is defined on $D_\xi$ as follows. 
For $x\in D_\xi\setminus\{y,x_\xi,z\}$ we define $f_\xi(x)$
arbitrarily, making only sure that condition (ii) is satisfied. 
By now, $F_\xi$ is already defined on 
$\bigcup\{H_n\colon n\leq\omega\}\setminus\{y\}$.
Thus, the sequence $\la F_\xi(z_n)\colon n<\omega\ra$
is already determined. If it does not converge, define $f_\xi$ on $y$
arbitrarily, making sure that condition (ii) is satisfied. 
If it converges to a limit $L$, define
$f_\xi(y)$ to force $F_\xi(z_\omega)\neq L$.
This can be done even if $y\in P_\alpha$ for some $\alpha<\co$
since we still have many choices (all elements from $K_\alpha$)
for the value of $f_\xi(y)$.
Notice that such a choice implies that 
$F_\xi\restriction\{z_n\colon n\leq\omega\}$ is discontinuous,
while $\{z_n\colon n\leq\omega\}\subset C_\xi$.
Thus (iv) is satisfied. We finish the 
construction by choosing $f_\xi(z)$ such that (iii) is satisfied,
and $f_\xi(x_\xi)$ in arbitrary way (subject to condition (ii))
if $x_\xi\in D_\xi$ and $f_\xi(x_\xi)$ is not defined so far. 
The construction is completed.

Now, define $f=\lin\left(\bigcup_{\xi<\co}f_\xi\right)$.
Then $f\colon\real\to\real$ is additive, since $\bigcup_{\xi<\co}D_\xi=H$.
Clearly, by (iv), $f$ is discontinuous on any perfect set since,
by Lemma~\ref{lemBasis}, every perfect set contains some $C_\xi$.
Also, (iii) guarantees that $f$ is almost continuous, while
(ii) guarantees that $f$ has the Cantor intermediate value property. \qed



 











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\end{document}

