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\def\continuum{{\mathfrak c}}
\def\co{\continuum}

% characteristic function
     \newcommand{\charf}[1]{\mbox{\raise.48ex\hbox{$\chi$}$_{#1}$}}

\def\proof{\noindent{\sc Proof. }}
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\newcommand{\bd}{{\rm bd}}
%\newcommand{\mathR}{\/\mbox{\rm I\kern -0.18em{R}}}
\newcommand{\mathR}{{\mathbb R}}
\newcommand{\real}{\mathR}
\newcommand{\rational}{{\mathbb Q}}
%\newcommand{\mathI}{\/\mbox{\rm I\kern -0.18em{I}}}
\newcommand{\mathI}{{\mathbb I}}
%\newcommand{\mathN}{\/\mbox{\rm I\kern -0.18em{N}}}
\def\lin{{\operatorname{LIN}_\rational}}
\def\proj{{\operatorname{pr}}}
\def\supp{{\operatorname{supp}}}

\newcommand{\F}{{\cal F}}
\newcommand{\e}{\varepsilon}
\newcommand{\J}{{\cal J}}
\newcommand{\la}{\langle}
\newcommand{\ra}{\rangle}
\def\P{{\cal P}}
\newcommand{\A}{{\cal A}}
\newcommand{\B}{{\cal B}}
\newcommand{\C}{{\cal C}}
\newcommand{\K}{{\cal K}}
\newcommand{\X}{{\cal X}}
\newcommand{\Ma}{{\cal M}_{a}}
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\newcommand{\Const}{{\rm Const}}
\newcommand{\G}{{\cal G}}
\newcommand{\I}{{\cal I}}
\newcommand{\CC}{{\rm C}}
\newcommand{\PC}{{\rm PC}}
\newcommand{\D}{{\rm D}}
\newcommand{\M}{{\rm M}}
\newcommand{\AC}{{\rm AC}}
\newcommand{\Conn}{{\rm Conn}}
\newcommand{\Add}{{\rm Add}}
\newcommand{\darb}{{\rm D}}
\newcommand{\EE}{{\cal E}}
\newcommand{\Ext}{{\rm Ext}}

\newcommand{\PR}{{\rm PR}}
\newcommand{\CIVP}{{\rm CIVP}}
\newcommand{\SCIVP}{{\rm SCIVP}}
\newcommand{\WCIVP}{{\rm WCIVP}}
\newcommand{\DIVP}{{\rm DIVP}}
\newcommand{\bor}{{{\cal B}{\rm or}}}




%\newcommand{\E}{{\rm E}}
\newcommand{\cl}{{\rm cl}}

\newcommand{\dist}{{\rm dist}}
\newcommand{\diam}{{\rm diam}}


\title{Darboux-like functions within the classes of Baire one, Baire two,
and additive functions}

\author{
Krzysztof Ciesielski%
\thanks{
1991 {\it Mathematics Subject Classification}.
Primary 26A15, 54C30; secondary 26A21, 54C08
\endgraf %\newline 
\hspace*{4pt}
{\it Key words and phrases}. Extendable, almost continuous, connectivity, additive functions.
\endgraf %\newline 
\hspace*{4pt}
This work was partially supported by NSF Cooperative
Research Grant INT-9600548 with its Polish part being
financed by Polish Academy of Science PAN. The results were obtained
during a visit of the second author at West Virginia University. 
}\\
{\footnotesize Department of Mathematics}\\
{\footnotesize West Virginia University}\\
{\footnotesize Morgantown, WV 26506-6310}\\
{\footnotesize USA}\\
{\footnotesize KCies@wvnvms.wvnet.edu}
%Department of Mathematics, West
%Virginia University, Morgantown, WV 26506-6310, USA (kcies@wvnvms.wvnet.edu)
\and
Jan Jastrz{\c{e}}bski$^*$\\
{\footnotesize Department of Mathematics}\\
{\footnotesize Gda{\'n}sk University}\\
{\footnotesize Wita Stwosza 57}\\
{\footnotesize 80-952 Gda{\'n}sk, Poland}\\
{\footnotesize jjas@ksinet.univ.gda.pl}}




\date{}



\begin{document}




\maketitle

\begin{abstract}
In the paper we present an exhaustive discussion of 
the relations between Darboux-like
functions within the classes of Baire one, Baire two, 
Borel, and additive functions
from $\real^n$ into $\real$.
In particular we 
construct an additive extendable discontinuous function
$f\colon\real\to\real$, answering a question of Gibson and Natkaniec \cite[p.~499]{GN}, 
and show that there is no similar function
from $\real^2$ into $\real$. 
We also describe a Baire class two
almost continuous function $f\colon\real\to\real$ which is not extendable.
This gives a negative  answer to a problem of Brown, Humke, and Laczkovich
\cite[prob.~1]{BHL}. (See also \cite[prob.~3.21]{GN}.)
%Finally, assuming the continuum hypothesis, we will construct
%an additive connectivity function $f\colon\real\to\real$
%which is not almost continuous. This gives a partial answer to another 
%question of Gibson and Natkaniec \cite[prob.~5.5]{GN}.

\end{abstract}

\section{Introduction}

The study of different generalizations of continuity of functions
from $\real^n$ into $\real$ has a long history. 
In this paper we will be interested in the functions with some of these 
generalized continuities that are known under the common name of 
Darboux-like functions. The readers unfamiliar with their definitions 
can find them in the next section.

The basic relations between these classes, for the functions from $\real$ to
$\real$, are given in the following chart, in which arrows
$\longrightarrow$ denote strict inclusions. 
Moreover, all other possible ``natural intersection'' inclusions 
(in a form of $\AC\cap\CIVP\subseteq\Conn\cap\CIVP$)
obtained from different classes of this chart remain strict. 

\hspace{1.5pc}
\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}
The inclusions $\CC\subset\Ext$, $\Conn\subset\darb\subset\PC$,
$\SCIVP\subset\CIVP\subset\WCIVP$, and $\PR\subset\PC$
are obvious from the definitions. The inclusions 
$\Ext\subset\AC\subset\Conn$ 
were proved by Stallings~\cite{JS}.
The inclusion $\CIVP\subset\PR$ was stated without the proof in 
\cite{GRou}. The proof can be found in \cite[thm.~3.8]{GN}.
The inclusion $\Ext\subset\SCIVP$ was proved 
by Rosen, Gibson, and Roush in~\cite{RGR}.
An excellent discussion of this chart can be found in a recent survey 
by Gibson and Natkaniec~\cite[sec.~3]{GN}. 
The examples concerning the properness of all intersection inclusions 
can be also found, in stronger versions, in Theorems~\ref{thA}, \ref{thB}, and
\ref{thE} stated below. 
Also, function $F$ from Corollary~\ref{cor3} is the first simple ZFC
example of almost continuous SCIVP function which is not extendable. 


For the functions from $\real^n$ into $\real$ with $n>1$ the classes 
from the lower part of Chart~1 are not defined. The relations
between the classes in the upper part of the chart 
change to the following. 

\begin{picture}(0,90)
 \put(0,55){\makebox(0,0){$\CC(\real^n)$}}
 \put(19,55){\vector(1,0){15}}
 \put(120,55){\makebox(0,0){$\Ext(\real^n)=\Conn(\real^n)=\PC(\real^n)$}}
   \put(205,55){\vector(1,0){15}}
   \put(265,55){\makebox(0,0){$\AC(\real^n)\cap\D(\real^n)$}}
 \put(310,60){\vector(2,1){18}}
 \put(352,70){\makebox(0,0){${\AC(\real^n)}$}}
 \put(350,40){\makebox(0,0){${\D(\real^n)}$}}
 \put(310,52){\vector(2,-1){18}}
\end{picture}
\vspace{-2pc}\begin{center} Chart~2 \end{center}
The inclusions $\CC(\real^n)\subset\Ext(\real^n)\subset\Conn(\real^n)$
are obvious from the definitions. 
The inclusion  $\Conn(\real^n)\subset\Ext(\real^n)$
was recently proved by Ciesielski, Natkaniec, and Wojciechowski~\cite{CNW}. 
The containment  
$\Conn(\real^n)\subset\PC(\real^n)$ was proved by    
Hamilton~\cite{OH} and                  
Stallings~\cite{JS}, and the inclusion 
$\PC(\real^n)\subset\Conn(\real^n)$ by Hagan~\cite{H}.
(See also Whyburn~\cite{Why} and \cite[thm,~8.1]{GN}.)
The relation %inclusion 
$\Conn(\real^n)\subset\AC(\real^n)\cap\darb(\real^n)$
was proved by Stallings~\cite{JS}.
The examples concerning the properness of the inclusions 
can be found, in the Baire class one, in Theorem~\ref{thC}.

The main goal of this paper is to discuss these two charts when
we restrict the function in all these classes to
the following four classes of functions:
Baire one $B_1$, Baire two $B_2$, Borel $\bor$, and additive functions $\Add$.
Notice that an intersection of any two of these classes is trivial, since
$B_1\subset B_2\subset\bor$ and 
$\Add\cap\bor\subset\CC$. 

\thm{thA}{For the Baire one functions $B_1$ from $\real$ to $\real$ the following
holds.
\[
\CC\subsetneq\Ext=\AC=\Conn=\darb=\PC=\SCIVP=\CIVP=\PR\subsetneq\WCIVP.
\]
}

\proof The proof of the equation $\Ext\cap B_1=\PC\cap B_1$ can be found in~\cite{BHL}.
This equation and Chart~1 imply all the other equations. 
The properness of the inclusions is justified as follows.
\begin{description}
\item{$B_1\cap\darb\setminus\CC\neq\emptyset$ --} It is witnessed by the function 
$f_0\colon\real\to\real$ defined by 
$f_0(x)=\sin(1/x)$ for $x\neq 0$ and  $f_0(0)=0$.
(See \cite[example~1.1]{N1}.) 
\item{$B_1\cap\WCIVP\setminus\PC\neq\emptyset$ --} It is witnessed by the function 
$g\colon\real\to\real$ defined by 
$g(x)=x^2$ for $x\neq 0$ and  $g(0)=-1$.\qed
\end{description}

\thm{thB}{The classes from Chart~1 restricted to either 
Baire two functions $B_2$ or Borel functions 
from $\real$ to $\real$ leads to the following chart.
Moreover, all possible ``natural intersection'' inclusions 
obtained from different classes of this chart remain strict. 


\begin{picture}(0,90)
 \put(130,55){\makebox(0,0){$\CC%\longrightarrow
 \longrightarrow\Ext\longrightarrow\AC\longrightarrow
 \Conn\longrightarrow\darb\longrightarrow\SCIVP=\CIVP$}}
 \put(280,60){\vector(2,1){18}}
 \put(332,70){\makebox(0,0){$\PR\longrightarrow\PC$}}
 \put(328,40){\makebox(0,0){${\WCIVP}$}}
 \put(280,52){\vector(2,-1){18}}
\end{picture}\vspace{-2pc}
}

\proof To see that the inclusion $\darb\subseteq\SCIVP$ holds in the class of Borel
functions let $x<y$ and $K$ be a perfect set between 
$f(x)$ and $f(y)$. Since $f$ is Darboux, $f^{-1}(K)\cap(x,y)$ is an uncountable
Borel set. Thus, it contains a perfect subset $C_0$. Moreover, we can find
a perfect set $C\subset C_0$ for which $f\restriction C$ is continuous.
Similarly we can argue that the inclusion $\CIVP\subseteq\SCIVP$ holds
in the class of Borel functions.

The properness of the inclusions is justified as follows.
\begin{description}
\item{$B_2\cap\Ext\setminus\CC\neq\emptyset$ --} See Theorem~\ref{thA}.
\item{$B_2\cap\AC\setminus\Ext\neq\emptyset$ --} See Corollary~\ref{cor3}.
\item{$B_2\cap\Conn\setminus\AC\neq\emptyset$ --} See Brown~\cite{Br} or
      Jastrz\c{e}bski~\cite{Jas}.
\item{$B_2\cap\darb\setminus\Conn\neq\emptyset$ --} See Brown~\cite{Br}.
\item{$B_2\cap\SCIVP\setminus\darb\neq\emptyset$ --} See Example~\ref{B2SCIVPnotD}.

\item{$B_2\cap\WCIVP\cap\PR\setminus\CIVP\neq\emptyset$ --} Let $\{F_q\colon q\in\rational\}$
be a family of pairwise disjoint $\co$-dense $F_\sigma$ sets.
Then 
$f=\sum_{q\in\rational\cap(0,1)}q\ \charf{F_q}$ is as desired. 

\item{$B_2\cap\WCIVP\cap\PC\setminus\PR\neq\emptyset$ --} It is witnessed by 
$g=f+2\ \charf{D}$, where $f$ is as above and $D$ is a countable dense subset of $F_2$.
      
\item{$B_2\cap\WCIVP\setminus\PC\neq\emptyset$ --} See Theorem~\ref{thA}.\qed
\end{description}


Restricting functions from Chart~2 to Baire one functions 
has a lot simpler solution.

\thm{thC}{For the Baire one functions $B_1(\real^n)$ from $\real^n$ to $\real$,
$n>1$, Chart~2 remains unchanged.}

\proof 
The properness of all the inclusions, as well
as of their other possible combinations, is justified by the following facts.
\begin{description}
\item{$B_1(\real^n)\cap\Conn(\real^n)\setminus\CC(\real^n)\neq\emptyset$ --} 
It is witnessed by a function $f\colon\real^2\to\real$ given by
$f(x,y)=\sin(1/(x^2+y^2))$ for $\la x,y\ra\neq \la 0,0\ra$ and $f(0,0)=0$.
It is in $\PC(\real^2)$ straight from the definition. 

\item{$B_1(\real^n)\cap\AC(\real^n)\cap
\darb(\real^n)\setminus\Conn(\real^n)\neq\emptyset$ --} 
Rosen, Gibson, Roush~\cite[example~1]{RGR} proved that
a function $f\colon[-1,1]\times[0,1]\to[-1,1]$ given by
$f(x,y)=\sin(1/y)$ for $y>0$ and $f(x,0)=x$ is 
Baire one, almost continuous, Darboux, but not connectivity. It is easy to extend it
to a finite support function on $\real^2$ with the same properties. 

\item{$B_1(\real^n)\cap\darb(\real^n)\setminus\AC(\real^n)\neq\emptyset$ --} 
It is justified by the function $F(x,y)=f_0(x)$, where $f_0$ is a function from
Theorem~\ref{thA}. 
(See Natkaniec~\cite[example~1.7]{N1} or \cite[example~1.1.9]{N1hab}.)

\item{$B_1(\real^n)\cap\AC(\real^n)\setminus\darb(\real^n)\neq\emptyset$ --} 
Let $f(x)=\sin(1/x)$ for $x\neq 0$ 
and  $f(0)=1$, and let $F\colon[-1,1]^2\to[-1,1]$ be given by 
the formula $F(x,y)=y\ f(x)$. 
It was proved by Natkaniec \cite[example~1.1.10]{N1hab} 
that $F$ is Baire one, almost continuous, and not Darboux. 
It is easy to extend $F$ to a function $\bar F\colon\real^2\to[0,1]$ 
with compact support while preserving these properties.
\qed

\end{description}

The study of classes of additive functions from Charts~1 and~2 
were initiated by D.~Banaszewski~\cite{Ban}. (See also \cite[sec.~5]{GN}.)
In this direction we have the following results.


\thm{thD}{For the additive functions $\Add(\real^n)$ from $\real^n$ to $\real$,
$n>1$, Chart~2 changes as follows:

\hspace{6pt}\begin{picture}(0,90)
 \put(140,55){\makebox(0,0){$\CC(\real^n)=\Ext(\real^n)=\Conn(\real^n)=\PC(\real^n)
 =\AC(\real^n)\cap\D(\real^n)$}}
 \put(300,60){\vector(2,1){18}}
 \put(345,70){\makebox(0,0){${\AC(\real^n)}$}}
 \put(342,40){\makebox(0,0){${\D(\real^n)}$}}
 \put(300,52){\vector(2,-1){18}}
\end{picture}
\vspace{-2pc}
}

\proof The inclusion $\Add(\real^n)\cap\AC(\real^n)\cap\D(\real^n)\subset\CC(\real^n)$
is proved in Theorem~\ref{thm2}. 
The properness of the inclusions is justified by the following facts.
\begin{description}
\item{$\Add(\real^n)\cap\AC(\real^n)\setminus\darb(\real^n)\neq\emptyset$ --} 
See Example~\ref{ACnotD}.
\item{$\Add(\real^n)\cap\darb(\real^n)\setminus\AC(\real^n)\neq\emptyset$ --} See
Example~\ref{DnotAC}.\qed
\end{description}


\thm{thE}{For the additive functions $\Add$ from $\real$ to $\real$
we have the equation $\PR=\WCIVP$. 
The other inclusions of Chart~1 remain unchanged, 
except possibly for the inclusion $\AC\subset\Conn$. Thus, we have  

\hspace{1.5pc}
\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}}
\end{picture}
\vspace{-2pc}
%\begin{center} Chart~4 \end{center}

\noindent 
Moreover all possible ``natural intersection'' inclusions
obtained from the different classes of this chart 
and not involving $\AC\subset\Conn$ remain strict. 
The inclusion $\Add\cap\CIVP\cap\AC\subset\Conn$
is strict if union of less than continuum
many meager subsets of $\real$ is meager in $\real$. 
}

\proof The properness of all the inclusions is justified by the following facts.
\begin{description}
\item{$\Add\cap\Ext\setminus\CC\neq\emptyset$ --} See Corollary~\ref{cor1}.

\item{$\Add\cap\SCIVP\cap\AC\setminus\Ext\neq\emptyset$ --} See 
Ciesielski, Ros{\l}anowski~\cite{CRo}. 
Compare also Ciesielski~\cite[thm.~3.1]{SomeDarbF}.

\item{$\Add\cap\SCIVP\cap\darb\setminus\Conn\neq\emptyset$ --} 
See Example~\ref{SCIVPandDnotCon}.
(An example of a function from 
$\Add\cap\darb\setminus\Conn$ was earlier given in~\cite{Ban}.)

\item{$\Add\cap\PC\setminus\darb\neq\emptyset$ --} 
See Example~\ref{SCIVPandPCnotD}.
(An example of a function from 
$\Add\cap\CIVP\setminus\darb$ was earlier given in~\cite{Ban}.)

\item{$\Add\cap\AC\cap\CIVP\setminus\SCIVP\neq\emptyset$ --} 
See Ciesielski~\cite[thm.~4.1]{SomeDarbF}.

\item{$\Add\cap\AC\cap\PR\setminus\CIVP\neq\emptyset$ --} See Example~\ref{ACandPRnotCIVP}.

\item{$\Add\cap\AC\setminus\PR\neq\emptyset$ --} See D.~Banaszewski~\cite{Ban}.

\item{$\Add\cap\CIVP\cap\Conn\setminus\AC$ --} Such a function,
under the assumption that union of less than continuum
many meager subsets of $\real$ is meager, has been constructed by 
Ciesielski and Ros{\l}anowski~\cite{CRo}. 
(Example for $\Add\cap\Conn\setminus\AC$ requires only that
union of less than continuum
many meager subsets of $\real$ does not cover $\real$.)
\qed
\end{description}

The following questions, which are variants of D.~Banaszewski's question 
\cite[Question~5.5]{GN}, remain open. 

\pr{prob1}{
\begin{description} 
\item[(1)] Does there exist a ZFC example of 
an additive connectivity function $f\colon\real\to\real$
which is not almost continuous?
\item[(2)] Does there exist 
an additive connectivity function $f\colon\real\to\real$
with $\SCIVP$ property 
which is not almost continuous?
\end{description}
}




\section{Definitions and notation}

Our terminology is standard and follows \cite{Ci:book}. 
We consider only real-valued
functions of one or more 
real variables. No distinction is made between
a function and its graph. 
A restriction of a function $f\colon X\to Y$ to a set $A\subset X$
is denoted by $f\restriction A$. Symbol $\charf{A}$ will be used
for a characteristic function of a subset $A$ of a fixed space $X$. 
By $\real$ and $\rational$ we denote the set
of all real and rational numbers, respectively.
We will consider $\real^n$ as linear spaces over $\rational$.
In particular, for $X\subset\real^n$
we will use the symbol $\lin(X)$ to denote the smallest
linear subspace of $\real^n$ over $\rational$
that contains $X$.
Recall also that if $D\subset\real^n$ is linearly independent
over $\rational$ and $f\colon D\to\real$ then
$F=\lin(f)\subset\real^{n+1}$ is an additive function (see definition below) 
from $\lin(D)$ into $\real$ which extends $f$. 
Any linear basis of $\real$ over $\rational$ will be referred
as a {\em Hamel basis}. 

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 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 $a<b$. The closure of a set $A\subset\real^n$
is denoted by $\cl(A)$, its boundary by $\bd(A)$, and its diameter by $\diam(A)$.  



We will use also the following terminology~\cite{GN}.
For $X\subseteq\real^n$
a function $f\colon X\to\real$ is:
\begin{itemize}
\item {\em additive\/} 
if $X$ is closed under the addition and $f(x+y)=f(x)+f(y)$ for every $x,y\in X$;

\item {\em Darboux\/} if $f[K]$ is a connected subset of $\real$ (i.e.,
an interval) for every connected subset $K$ of $X$;

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

\item 
 {\it connectivity\/} function if the graph of $f\restriction Z$
 is connected in $Z\times\real$ for any connected subset $Z$ of~$X$;

\item 
 {\it extendability\/} function provided 
 there exists a connectivity function $F\colon X\times[0,1]\to\real$
 such that $f(x)=F(x,0)$ for every $x\in X$;
 
 \item
{\it peripherally continuous\/} if for
every $x\in X$ and for all pairs of open sets $U$ and $V$ containing
$x$ and $f(x)$, respectively, there exists an open subset $W$ of $U$ 
such that $x \in W$ and $f[\bd (W)]\subset V$.
\end{itemize}

The classes of these functions are denoted by $\Add(X)$, $\D(X)$,
$\AC(X)$, $\Conn(X)$, 
$\Ext(X)$, and $\PC(X)$, respectively. The class of continuous functions
from $X$ into $\real$ is denoted by $\CC(X)$. 
We will drop the index $X$ if $X=\real$. 
 
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}.) Similarly, if 
the graph of $f\colon\real\to\real$
intersects every compact connected subset $B$ of $\real^2$ 
with $|\proj(B)|>1$ then $f$ is connectivity. 
This follows from the following well known fact.

\fac{fact1}{If $S\subset\real^2$ disconnects $\real^2$ then it contains 
a nontrivial compact connected subset.}


A function $f\colon\real\to\real$ has: 
\begin{itemize}
\item
{\it Cantor intermediate value property}
if for every $x,y\in\mathR$ and for each perfect set $K$ between $f(x)$ and $f(y)$
there is a perfect set $C$ between $x$ and $y$ such that $f[C]\subset K$;

\item 
{\it strong Cantor intermediate value property\/} if
for every $x,y\in\mathR$ and for each perfect set $K$ between $f(x)$ and $f(y)$
 there is a perfect set $C$ between $x$ and $y$ such that
 $f[C]\subset K$ and $f\restriction C$ is continuous;

\item
{\it weak Cantor intermediate value property} if for every $x,y\in\real$
with $f(x)<f(y)$ there exists a perfect set $C$ between
$x$ and $y$ such that $f[C]\subset (f(x),f(y))$;

\item
{\it perfect road} if for every $x\in\mathR$ there exists a perfect set
$P\subset\mathR$ having $x$ as a bilateral (i.e., two sided)
limit point for which
$f\restriction P$ is continuous at $x$.

\end{itemize}
The above classes of these functions are denoted by $\CIVP$, $\SCIVP$, $\WCIVP$, 
and $\PR$, respectively.


\section{Almost continuous Baire two class function which is not extendable}


The main example described in this section 
answers problem~2 and the main part 
of problem~1 from \cite{BHL}, as well as problem 3.21 from~\cite{GN}.

Let $C\subset [0,1]$ be the ternary Cantor set and let ${\cal J}$ 
be the family of all component intervals of $[0,1] \setminus C$. We put
\[
{\cal J}_0 =\{ J \in {\cal J} \colon
\mbox{ the length of } J \mbox{ is } 3^{-n}
\mbox{ with }   n < {\omega} \mbox{ even} \}
\]
and
\[
{\cal J}_1 =\{ J \in {\cal J} \colon
\mbox{ the length of } J \mbox{ is } 3^{-n}
\mbox{ with }   n < {\omega} \mbox{ odd} \}.
\]
Let $\{(a_n,b_n)\colon n<\omega\}$ and $\{(c_n,d_n)\colon n<\omega\}$
be the enumerations of $\J_0$ and $\J_1$, respectively. 
Define function $f \colon [0,1] \to [0,1]$ in the following way:
\begin{itemize}
\item for every $n<\omega$ we put $f(a_n)=f(d_n)=0$, $f(b_n)=f(c_n)=1$,
and extend it linearly on $[a_n, b_n]$ and $[c_n, d_n]$;
\item for all other $x$'s we put $f(x)=0$. 
\end{itemize}

\thm{thm3}{
The function $f$ is almost continuous, Baire class two, but not extendable.}

\proof We will start with showing that $f$ is not extendable.
By way of contradiction, assume that $f$ can be extendable, that is, that there is 
a connectivity function
$F \colon [0,1]^2 \to [0,1]$ with $F(x,0)=f(x)$ for all $x \in [0,1]$.
Thus $F$ is peripherally continuous. We will deduce from this that
there exists a perfect set $P\subset C\times\{0\}$ on which $F$ is constantly equal to $1$,
which evidently contradicts our definition of~$f$. 

We will define the families: 
$\la p_s\in C\colon s\in 2^{<\omega}\ra$,
$\la B_s\subset [0,1]^2\colon s\in 2^{<\omega}\ra$, and
$\la U_s\subset [0,1]^2\colon s\in 2^{<\omega}\ra$
such that the following conditions hold for every 
$s\in 2^n$ and different $t,t'\in 2^{n+1}$ extending $s$.
\begin{description}
\item{(i)}   $\la p_s,0\ra\in U_s$, $f(p_s)=1$, and $U_s$ is open with 
             a diameter at most $2^{-n}$.
\item{(ii)}  $B_s$ is closed, connected, and $F[B_s]\subset[1-2^{-n},1]$.
\item{(iii)} $\cl(U_t)\cup B_t\subset U_s$, $\la p_s,0\ra\notin U_t$, and
             $B_t\cap B_s\neq\emptyset$.
\item{(iv)}  $\cl(U_t)\cap\cl(U_{t'})=\emptyset$.
\end{description}

For $s=\emptyset\in 2^0$ we put $U_s=B_s=[0,1]^2$
and choose an arbitrary $p_s$ with $f(p_s)=1$.
Farther, the construction goes by the induction on the length $n$ of $s\in 2^{<\omega}$.
Thus, assume that for some $s\in 2^n$ the sets $B_s$, $U_s$, 
and the point $p_s$ are already chosen. Let $t$ and $t'$ be different 
sequences from $2^{n+1}$ extending $s$. 
To choose $B_t$, $p_t$, and $U_t$ we proceed as follows.

Note that $p_s$ is an endpoint of some $J\in\J$ since $f(p_s)=1$. 
If $p_s$ is a left endpoint of $J$ we put $I=[0,p_s)$. Otherwise we put $I=(p_s,1]$. 
Choose an $\e>0$ less than the diameters of $I$ and $B_s$ and 
such that 
\begin{description}
\item{($*$)} for every $J\in\J$ with $J\subset I$ if the 
distance from $J$ to $p_s$ is less than $\e$ then $J\times\{0\}\subset U_s$. 
%$z\in I$ with $|z-p_s|<\e$ and for every $p\in[0,1]$ such that 
%either $p=z$ or $p,z\in\cl(J)$ for some $J\in\J$ we have $\la p,0\ra\in U_s$. 
\end{description}
Let $W_s\subset[0,1]^2$ be an open 
neighborhood of 
$\la p_s,0\ra$ with diameter less than $\e$
and such that $F[\bd(W_s)]\subset[1-2^{-n},1]$. It  exists since $F$ is peripherally
continuous at $\la p_s,0\ra$ and $F(p_s,0)=f(p_s)=1$.
Without loss of generality we can assume that $\bd(W_s)$ is connected. 
(Replacing $W_s$ by its component, if necessary, we can assume that $W_s$ is connected. 
Then we can increase $W_s$ to $[0,1]^2\setminus V_s$, 
where $V_s$ is an ``unbounded'' component of $[0,1]^2\setminus W_s$,
that is the one which contains the boundary of $[0,1]^2$.
This decreases $\bd(W_s)$ and  makes it connected.) 
Note that the choice of $\e$ guarantees that 
$\bd(W_s)\cap B_s\neq\emptyset\neq\bd(W_s)\cap (I\times\{0\})$ 
since $B_s$ and $I\times\{0\}$ are connected and
$\bd(W_s)$ disconnects $[0,1]^2$. Let $z\in I$ be such that 
$\la z,0\ra\in \bd(W_s)\cap (I\times\{0\})$.
Since $F(z,0)\in F[\bd(W_s)]\subset[1-2^{-n},1]\subset(0,1]$ there exists a $J\in\J$
such that $z\in\cl(J)$. Let $p_t$ be the endpoint of $\cl(J)$ for which $f(p_t)=1$.
By ($*$) we have $\la p_t,0\ra\in U_s$.
The set $B_t$ is defined as a union of $\bd(W_s)$ and 
a closed segment joining $\la z,0\ra$ and $\la p_t,0\ra$.
The open neighborhood $U_t$ of $p_t$ is chosen such that 
its diameter is at most $2^{-(n+1)}$ and that 
$p_s\notin\cl(U_t)\subset U_s$. 
It is easy that conditions (i), (ii), and (iii) are satisfied. 

To choose $B_{t'}$, $p_{t'}$, and $U_{t'}$ we replace $U_s$ with
$U^\prime_s=U_s\setminus \cl(U_t)$ and repeat the process described above. 
This finishes the inductive construction. 

Now, to finish the argument take an arbitrary $s\in 2^\omega$  and note 
that by (i) and (iii) the limit $\lim_{n\to\infty}\la p_{s\restriction n},0\ra$ exists
and is equal to a point $p_s$ which belongs to $C\times\{0\}$. 
Also, by (ii), the set $B_s=\bigcup_{0<n<\omega}B_{s\restriction n}$ is connected
and $p_s$ is its only accumulation point.
Since $P=\{p_s\colon s\in 2^\omega\}$ is evidently equal to a perfect set
$\bigcap_{n<\omega}\bigcup\{\cl(U_t)\colon t\in 2^n\}$
it is enough to prove that $F(p_s)=1$ for every $s\in 2^\omega$. 
But if $F(p_s)\neq 1$ then $F(p_s)=0$, since $p_s\in C\times\{0\}$.
Take $\e>0$ is less then the diameter of $B_s$ and let $U$ be an open neighborhood
of $p_s$ of diameter less than $\e$ and such that $F[\bd(U)]\subset[0,1/2)$.
Then for a point $w\in B_s\cap\bd(U)$ we have $F(w)\in[0,1/2)\cap[1/2,1]$,
a contradiction. This finishes the proof that $f$ is not extendable. 


Next we will show that $f$ is almost continuous. Let $G$ be an open set contained in $[0,1]^2$
containing the graph of $f$. For every $x\in[0,1]$ there exists an interval $(a_x,b_x)$
such that
\begin{itemize}
\item $x\in(a_x, b_x)$,
\item $f(a_x)=f(b_x)=0$, and
\item there is a continuous function $g_x \colon [0,1] \to \mathR$ 
  with $g_x \restriction{[a_x,b_x]} \subset G$ and such that 
  $g_x(t)=0$ for $t \notin (a_x, b_x)$.
\end{itemize}

Indeed, if $f(x)=0$ then it is easy to find $(a_x, b_x)$ for which $g_x\equiv 0$ works. 
If $f(x)\neq 0$ then $x\in J$ for some $J\in\J$, say $J=(A,B)$. 
Assume that $f$ is increasing on $J$, the other case is similar.
Then $f(B)=1$. Find an interval $J'=(C,D)\in\J$ on which function $f$ is decreasing and
such that $[B,C]\times\{1\}\subset G$. 
Put $a_x=A$, $b_x=D$, define $g_x(B)=g_x(C)=1$, $g_x(0)=g_x(a_x)=g_x(b_x)=g_x(1)=0$, 
and extend $g_x$ in a linear way on each interval with these endpoints.
Thus $g_x$ has a hat shape.

Now choose a finite subcover $\{(a_{x_i},b_{x_i})\colon i<n\}$, with $n< \omega$, of the cover
$\{(a_x, b_x)\colon x \in [0,1]\}$ of the interval $[0,1]$. Then the function
$$g(x)= \max \{ g_{x_i}(x) : i<n\}$$
is continuous and $g \subset G$. This ends the proof that $f$ is almost continuous.

To see that $f$ is of Baire class two it is enough to notice that
the preimage of every open (and closed) set $U$ 
is a countable union of F$_\sigma$ (closed) sets $f^{-1}(U)\cap\cl(J)$ with $J\in\J$
and, possible, of a G$_\delta$ set $C\setminus \bigcup\{\cl(J)\colon J\in\J\}$.
\qed

In \cite[prob.~1]{BHL} the authors asked also whether the function as above can be
in the class $J_1$ of all functions (from $\real$ or $[0,1]$ into $\real$) that are 
pointwise limits of functions which have only discontinuities of the first class, 
that is, these functions for which both one sided limits exist at each point. 
Clearly $B_1\subset J_1\subset B_2$. Our function gives also an answer to this
question.

\thm{thm3a}{
There exists a $J_1$ function $f\colon[0,1]\to[0,1]$ which is almost continuous but not
extendable.}

\proof Let $f$ be a function from Theorem~\ref{thm3} and let 
$\charf{D}$ stand for a characteristic function of $D\subset[0,1]$.
If $\{J_n\colon n<\omega\}$ is an enumeration of $\{\cl(J)\colon J\in\J\}$
and $D_n=\bigcup_{i<n}J_i$ then $f$ is a pointwise limit of
functions $f\ \charf{D_n}$. Thus, $f$ is in $J_1$. \qed 

\cor{cor3}{
There exists a $J_1$ function $F\colon\real\to\real$ which is almost continuous but not
extendable.}

\proof Extend the function $f$ from Theorem~\ref{thm3a} to $F$ by putting
$F(x)=0$ for all $x\in\real\setminus[0,1]$. \qed

The main core of the proof that the function $f$ from Theorem~\ref{thm3}
is not extendable is that the set $f^{-1}(1)$ is countable. 
The next proposition shows that it is essentially the only obstacle
for $f$ to be extendable, in a sense that we can redefine $f$ on a subset
of $C$ to get an extendable function. 

\prop{prop3}{If $f$ is from Theorem~\ref{thm3} then there
exists a meager $F_\sigma$ subset $B$ of $C_0=C\setminus\bigcup\{\cl(J)\colon J\in\J\}$
such that $f_0=f+\charf{B}$ is extendable.}

\proof Gibson and Roush \cite{GR} proved that a function $g\colon[0,1]\to[0,1]$
is extendable if and only if there exists a sequence
$\la \la I_n,J_n\ra\colon n<\omega\ra$ of pairs of open intervals,
called a {\em PI family for $g$},
such that
\begin{description}
\item{(a)} $\lim_{n\to\infty}\diam(I_n)=0$,
\item{(b)} $g[\bd(I_n)]\subset J_n$ for every $n<\omega$,
\item{(c)} for every $x\in[0,1]$ and $\e>0$ there exists an $n<\omega$ such that
   $x\in I_n$, $g(x)\in J_n$, and $\max\{\diam(I_n),\diam(J_n)\}<\e$,
\item{(d)} for every $n<m<\omega$ if the sets $I_n\cap I_m$, $I_n\setminus I_m$, and
   $I_m\setminus I_n$ are nonempty then $J_n\cap J_m\neq\emptyset$.
\end{description}
It was also noticed in \cite{CNW} that in (c) it is enough to 
consider only discontinuity points $x$ of $g$. 
In what follows we will construct a PI family for our future $f_0$.

Let $T$ be the set of all endpoints of all intervals $J\in\J$ 
and let $\{t_n\colon n<\omega\}$ be an enumeration of $T$.
We construct by induction on $n<\omega$ the 
finite families $F_n$ of triples $\la t,I,J\ra$ such that 
\begin{description}
\item{(1)} $I$ and $J$ are open intervals in $[0,1]$, $t\in T\cap I$, and $f(t)\in J$;
\item{(2)} $f[\bd(I)]\subset(0,1)\cap J$ and $\diam(J)\leq 2^{-n}$;
\item{(3)} $I\cap I'=\emptyset$ for different $\la t,I,J\ra,\la t',I',J'\ra\in F_n$;
\item{(4)} $C\subset\bigcup\{I\colon \la t,I,J\ra\in F_n\}$ and 
           $\{t_i\colon i<n\}\subset\{t\colon \la t,I,J\ra\in F_n\}$;
\item{(5)} if $n>0$ then for every $\la t,I,J\ra\in F_n$ there exists 
           $\la t_0,I_0,J_0\ra\in F_{n-1}$ such that $I\subset I_0$.
\end{description}
The induction can be started with $F_0=\{\la t_0,(0,1),(0,1)\ra\}$, 
and can be easily carried through since $C$ is compact zero dimensional
and the sets $f^{-1}(1)$ and $T\cap f^{-1}(0)$ are dense in $C$.

Now, let $\{\la t_n,I_n,J_n\ra\colon n<\omega\}$ be an enumeration of $\bigcup_{n<\omega}F_n$.
We claim that $\la \la I_n,J_n\ra\colon n<\omega\ra$ is a PI family for 
$f_0=f+\charf{B}$ for an appropriately chosen set $B$.

Clearly (3) and (5) imply that condition (d) is satisfied in void. Condition (a) 
can be deduced from the density of $T$ in $C$ and (1--3). 
(b) for $g=f_0$ is implied by (2) and the fact that $f_0(x)=f(x)$ for 
$x\in[0,1]\setminus C$.
Similarly (c) for the points $x\in T$ is implied by (4) and 
$f_0\restriction T=f\restriction T$. To finish the proof it is enough to
show that (c) holds for points $x\in C_0$ for an appropriate choice of $B$. 
But every $x\in C_0$ the set $S_x=\{k<\omega\colon x\in I_k\}$ is infinite.
Let $B$ be the set of all those points $x\in C_0$ for which the set 
$\{k\in S_x\colon f(t_k)=0\}$ is finite. 
Then for every $x\in C_0$ the set $\{k\in S_x\colon f_0(x)\in J_k\}$
is infinite proving (b).

The fact that $B$ is a meager $F_\sigma$ subset of $C_0$ is left as an exercise. \qed

\ex{B2SCIVPnotD}{There exists an $f\in B_2\cap\SCIVP\setminus\darb$.
}

\proof Define $g$ from $[-1,1]$ onto $(0,1]$ by $g(x)=(x^2-1)\sin^2(1/x)+1$
for $x\neq 0$ and $f(0)=1$. Note that $g(-1)=g(1)=1$ and that $g$ is $\SCIVP$.
For each component $J$ of $[0,1]\setminus C$ of length $1/3^{-n}$ define 
$f\restriction\cl(J)$ as $(-1)^n g\circ h_J$, where $h_J$ is an increasing
linear function with $h_J[\cl(J)]=[-1,1]$. For all other points $x$
we put $f(x)=1$. 
Note that $f[\real]=[-1,0)\cup(0,1]$ and that preimage of every
open set is a union of a $G_\delta$ and an $F_\sigma$ set. So $f$ is
Baire two and not Darboux. To see that it is $\SCIVP$
take $x<y$ and a perfect set $K$ between $f(x)$ and $f(y)$.
We have to find a perfect set $P\subset(x,y)$ on which $f$ is continuous
and with $f[P]\subset K$.
If both $x$ and $y$ belong to the closure of the same component of $C$ then 
the existence of $K$ follows from $\SCIVP$ property of $g$. 
Otherwise there exist two components $J_0$ and $J_1$ of $C$ 
between $x$ and $y$ with $f[J_0\cup J_1]=[-1,0)\cup(0,1]$.
This we can choose appropriate $K\subset J_0\cup J_1$. \qed




\section{Additive extendable discontinuous function}

Let $h\in\Ext(\real)$. 
We say that a set $G\subset\real$  
is {\em $h$-negligible for a function $h$\/}
provided $f \in\Ext$ for every 
function $f\colon\mathR\to\mathR$ for which $f=h$ on a set $\real\setminus G$.

\lem{lem1}{{\rm (Rosen~\cite{Rosen2}, Ciesielski, Rec{\l}aw~\cite{CR})}
There exists an extendable function $h\colon\real\to\real$ 
such that some dense $G_{\delta}$-set $G\subset \mathR$
is $h$-negligible.
}

\lem{lem2}{ 
If $g \colon \mathR \to \mathR$ is homeomorphism,
$h\in\Ext$, and $G$ is $h$-negligible
then $h \circ g^{-1}\in\Ext$ and $g[G]$ is ($h \circ g^{-1}$)-negligible.
}

\proof This is a simple corollary from \cite[Lemma 2.2]{JN}. 
(See also \cite{N}.)\Qed

\prop{prop1}{
For every $\co$-dense meager $F_{\sigma}$-set $F\subset\real$ 
there exists an extendable function $f\colon\real\to\real$  such that 
$\real\setminus F$ is $f$-negligible. }

\proof  Let $h$ and $G$ be as in Lemma~\ref{lem1}. By Lemma~4 of
\cite{G} there exists a homeomorphism $g \colon \mathR \to \mathR$
such that $g[\real\setminus G] \subset F$. 
Then $\real\setminus F\subset g[G]$ so, by Lemma~\ref{lem2}, function 
$f=h \circ g^{-1}$ is extendable and the set $\real\setminus F$ 
is $f$-negligible.\Qed

\cor{cor1}{
There exists an additive extendable function $g\colon\real\to\real$
with a dense graph. In particular $g\in\Add\cap\Ext\setminus\CC$. 
}

\proof Let $F$ be $\co$-dense meager $F_{\sigma}$-set which is linearly independent
over $\rational$. Such a set can be easily constructed 
from a linearly independent perfect set, which 
description can be found in \cite[thm.~2, Ch.~XI sec.~7]{Kucz}.
By Proposition~\ref{prop1} there exists an $f \in \Ext$ such that 
$\real\setminus F$ is $f$-negligible.
In  particular $f \restriction F$ 
must be discontinuous. Let $g$ be a linear extension
of $f \restriction F$. Then $g\in\Ext$ since $\real\setminus F$ is $f$-negligible.
Clearly $g$ is additive and discontinuous, so it has a dense graph. 
\Qed

Next we prove that $\Add(\real^n)\cap\AC(\real^n)\cap\darb(\real^n)\subset\CC(\real^n)$ for
$n>1$. Its proof will be based on the following two propositions, 
the first of which was proved by Lipi\'{n}ski~\cite{Lip}.
(See also Maliszewski and Natkaniec~\cite{MN}.) 
This fact was noticed independently by the authors of this paper
and our proof is enclosed below. 


\prop{propA}{If $f\colon\real\to\real$ is discontinuous then 
$F\colon\real^2\to\real$ given by $F(x,y)=f(x)$ is not almost continuous. 
}

\proof Assume that $f\colon\real\to\real$ is discontinuous at some point $x$.
Taking a translation of a graph of $f$, if necessary, we can assume that
$x=0$ and $f(0)=0$. So, there exists a sequence $\{x_n\}_{n<\omega}$ converging to $0$
such that $\lim_{n\to\infty}f(x_n)=L_0\neq 0$. Multiplying $f$ by $3/L_0$, if necessary,
we can assume that $L_0=3$. We will also assume that $f(x_n)>2$ for all $n<\omega$.

Consider the closed set 
\[
A=\{\la x,y,z\ra\colon x=x_n\mbox{ for some } n<\omega, 
y\in\real, \mbox{ and }z\leq 2\}\cup(\{0\}\times\real^2)
\]
and let $G=\real^3\setminus A$. Then $G$ is an open set containing
all the graph of $F$ except of a line $L=\{0\}\times\real\times\{0\}$. 
Let $H=\{\la x,y,z\ra\in\real^3\colon x^2+z^2<e^{-y^2}\}$.
Then $H$ is an open and contains $L$, so $U=G\cup H$ contains $F$.
We will show that $U$ does not contain a graph of a continuous function.

By way of contradiction assume that there exists a continuous $g\colon\real^2\to\real$
with $g\subset U$. Then $\la 0,0,g(0,0)\ra\in H$. In particular, $g(0,0)\in(-1,1)$.
So, by the continuity of $g$, there exists an $n<\omega$ such that 
$g(x_n,0)\in(-1,1)$. We claim that $g\restriction(\{x_n\}\times\real)$
is discontinuous. 
Indeed, notice that 
\[
(\{x_n\}\times\real^2)\cap U\subset\{x_n\}\times(G_0\cup H_0),
\]
where $G_0=\real\times(2,\infty)$, $H_0=(-b,b)\times(-\infty,1)$, and 
$b>0$ is such that $e^{-(|x_n|+b)^2}=1$. Moreover, 
$\la x_n,0,g(x_n,0)\ra\in \{x_n\}\times H_0$. But clearly there is no
continuous function on $\real$  whose graph is contained in $G_0\cup H_0$
and intersects $H_0$. This contradiction finishes the proof. \qed

Notice that Proposition~\ref{propA} stay in contrast with the following 
fact.

\fac{factContr}{ {\rm (Natkaniec~\cite[cor.~4.2(1)]{N1})}
If $f\colon\real\to\real$ is almost continuous and 
$Y$ is a compact topological space then 
$F\colon\real\times Y\to\real$ given by $F(x,y)=f(x)$ is almost continuous.}



\prop{propB}{If $n>1$ and $F\in\Add(\real^n)\cap\darb(\real^n)$ then
$F^{-1}(0)$ contains a straight line. 
}

\proof If $F$ is constantly equal to $0$ then there is nothing to prove.
So, assume that this is not the case. Then $f[\real^2]=\real$.
In particular $V=F^{-1}(0)$ disconnects $\real^2$. (Otherwise its complement 
$\real^2\setminus V$ would be connected, while 
$F[\real^2\setminus V]=\real\setminus\{0\}$ would not, contradicting Darboux
property.) Thus, by Fact~\ref{fact1}, $V$ 
contains  a nontrivial compact connected subset $K$. 
Pick $a,b\in K$ with $\diam(\{a,b\})=\diam(K)$. 
Since every rotation $r$ is a linear homeomorphism, 
replacing $F$ with $F\circ r$ for an appropriate $r$ 
if necessary, we can assume that $a$ and $b$ are on the same vertical line. 
If $\proj(K)$ is a singleton, then 
$K$ contains a straight line segment connecting $a$ with $b$. 
This, and the fact that $V$ is linear over $\rational$, 
easily imply that $V$ contains a straight line. 
So, assume that $\proj(K)=[x_0,x_1]$ for some $x_0<x_1$. 
We claim that this implies that 
\begin{description}
\item{($*$)} there exists a bounded open set $U\subset\real^2$ with $\bd(U)\subset V$.
\end{description}

To see it take $c,d\in K$ with $\proj(c)=x_0$ and $\proj(d)=x_1$,
and let $[y_0,y_1]$ be a projection of $K$ to the second coordinate. 
Let $P=[x_0,x_1]\times[y_0,y_1]$. Notice that 
$K\subset P$ and that $a$ and $b$ lie on the opposite horizontal sides
of $P$, and $c$ and $d$ on opposite vertical sides of it. 
Note also that vectors $v=b-a$ and $w=d-c$ belong to $V$. 
Now consider the parallelogram-like set 
\[
B=\bigcup_{m=0}^4[m v+(K\cup (4w+K))\cup m w+(K\cup (4v+K))]\subset V.
\]
(The ``sides'' are formed from translated ``roads'' from $a$ to $b$ and
from $c$ to $d$.)
Note that the interior $U_0$ of $2v+w+P$ is disjoint with $B$ and that
$B$ separates it from the infinity. Then the component $U$ of $\real^2\setminus B$
containing $U_0$ satisfies~($*$). 

Now, as in \cite[lem.~5.4]{CW} 
we can find a nonempty bounded connected open set $W\subset\real^2$
with connected boundary $\bd(W)\subset B\subset V$. 
Take an $\e>0$ such that some open disk of radius $\e$
is contained in $B$. Then, if 
$B(\e)=\{v\in\real^2\colon\mbox{ length of $v$ is less than }\e\}$
then for every $v\in B(\e)$ we have $\bd(W)\cap(v+\bd(W))\neq\emptyset$.
So, $v\in\bd(W)-\bd(W)\subset V$. Therefore, $B(\e)\subset V$, and so
$V=\real^2$ contradicting our assumption that $F$ is non-zero. \qed






\thm{thm2}{$\Add(\real^n)\cap\AC(\real^n)\cap\darb(\real^n)\subset\CC(\real^n)$ for $n>1$.
}

\proof Let $F\in \Add(\real^n)\cap\AC(\real^n)\cap\darb(\real^n)$.
We will show that it is continuous. 

Since it is enough to prove continuity of 
any restriction of $F$ to a plane containing the origin, we can assume that $n=2$. 
Now, by Proposition \ref{propB}, $f^{-1}(0)$ contains a line $L_0$. 
Since $f^{-1}(0)$ is close under addition, it contains also 
a parallel line $L$ which contains the origin.
Since all the classes under consideration are closed under
inner composition with a rotation, we can assume that $L$ is a vertical line. 
But this means that there exists additive $f\colon\real\to\real$ such that 
$F(x,y)=f(x)$ for every $x,y\in\real$. 
So, by Proposition~\ref{propA}, function $f$ is continuous. 
Thus $F\in\CC(\real^2)$. \qed


\ex{ACnotD}{There exists an
$f\in\Add(\real^2)\cap\AC(\real^2)\setminus\darb(\real^2)$.}

\proof Let $\B$ be a family of all closed subsets $B$ of $\real^2\times\real$
for which the projection $\proj(B)$ onto first coordinate $\real^2$
has cardinality $\co$. (Thus, it contains
a perfect set.) It is known that if $f\colon\real^2\to\real$ intersects
every element of $\B$ then $f\in \AC(\real^2)$. (See \cite[prop.~1.2]{N1}.)
Let $\{F_\xi\colon\xi<\co\}$ be an enumeration of $\B$. By induction on $\xi<\co$ 
define a sequence $\la \la a_\xi,b_\xi,y_\xi\ra\colon \xi<\co\ra$ such that
for every $\xi<\co$ the following conditions hold.
\begin{description}
\item{(i)} $a_\xi\in\proj(B_\xi)\setminus
\lin(\{a_\zeta\colon \zeta<\xi\}\cup\{b_\zeta\colon \zeta<\xi\})$.
\item{(ii)} $b_\xi\in\proj(B_\xi)\setminus
\lin(\{a_\zeta\colon \zeta\leq\xi\}\cup\{b_\zeta\colon \zeta<\xi\})$.
\item{(iii)} $\la a_\xi,y_\xi\ra\in B_\xi$.
\end{description}

Let $A=\{a_\xi\colon\xi<\co\}$, $B=\{b_\xi\colon\xi<\co\}$, and 
$H$ be a Hamel base containing $A\cup B$. 
Define $f_0\colon H\to\real$ by putting $f_0(a_\xi)=y_\xi$ and $f_0(x)=1$
for $x\in H\setminus A$. Let $f\colon\real^2\to\real$ be a linear
extension of $f_0$. Then $f$ is additive and almost continuous, since 
it intersects every $F\in\F$. However, it is not Darboux, since 
the set $D=B\cup\{0\}$ is connected, as it intersects every perfect subset of
$\real^2$, while $f[D]=\{0,1\}$ is not connected. \qed

\ex{DnotAC}{There exists an
$f\in\Add(\real^2)\cap\darb(\real^2)\setminus\AC(\real^2)$.}

\proof 
Take an additive Darboux function $g\colon\real\to\real$
with the property that
\begin{description}
\item{($*$)} $g[(a,b)]=\real$ for every $a<b$. 
\end{description}
A function $f$ from Example~\ref{SCIVPandDnotCon} has the property. 
Define $f\colon\real^2\to\real$ by $f(x,y)=g(x)$. Clearly $f$ is additive
and discontinuous. 
To see that $f$ is Darboux take a non-empty connected set $D\subset\real^2$.
If $J=\proj(D)$ is a singleton then 
$f[D]=g[\proj(D)]$ is a singleton, so it is connected. 
Otherwise $\proj(D)$ contains an interval $(a,b)\neq\emptyset$ and, by ($*$),
$f[D]=g[\proj(D)]\supset g[(a,b)]=\real$. 
Now, by Theorem~\ref{thm2}, $f$ is not almost continuous. 
\qed

As noted above, function $f$ from Example~\ref{DnotAC} cannot be almost continuous.
It is interesting however, that if function $g$ used to define $f$ is almost continuous,
which can be easily constructed, then $f\restriction \real\times[-k,k]$
is almost continuous for every $k>0$. This follows from Fact~\ref{factContr}.


\section{Some missing examples of 
additive Darboux-like functions on $\real$}

\ex{ACandPRnotCIVP}{There exists an
$f\in\Add\cap\AC\cap\PR\setminus\CIVP$. Moreover, $f[K]$ is not nowhere dense 
for every perfect set $K\subset\real$. }

\proof Let $\P$ be a family of pairwise disjoint perfect sets such that 
the set $\bigcup \P$ is linearly independent and 
$|\{P\in\P\colon P\subset (a,b)\}|=\co$ for every $a<b$. Such a family can be easily 
constructed from a linearly independent perfect set. 
(See e.g. \cite[thm.~2, Ch.~XI sec.~7]{Kucz}.)
Let $\J$ be a family of all non-empty open intervals and let
$\{\la I_\xi,J_\xi\ra\colon \xi<\co\}$ be an enumeration of $\J\times\J$. 
By an easy induction we can find a one-to-one sequence
$\{P_\xi\in\P\colon \xi<\co\}$ such that $P_\xi\subset I_\xi$ for every $\xi<\co$. 
Let $H\subset\real$ be a Hamel basis containing $\bigcup\P$
and for each $h\in H$ let $J_h=J_\xi$ if $h\in P_\xi$, and $J_h=(0,1)$ otherwise.
Our $f$ will be a linear extension of some function $f_0\colon H\to\real$
such that 
\begin{description}
\item{($*$)} $f_0(h)\in J_h$ for every $h\in H$. 
\end{description}
It is easy to see that any
such an $f$ will have a perfect road. 
To make sure that $f$ has the additional property, which make it not $\CIVP$,
we will need to make some more work.

Let $\F$ be the family of all perfect nowhere dense subsets of $\real$ 
and let $\{\la K_\xi,S_\xi\ra\colon \xi<\co\}$ be an enumeration of $\F\times\F$.
We will make sure for every $\xi<\co$ there exists an $x_\xi\in K_\xi$
such that $f(x_\xi)\notin S_\xi$. Clearly such a function will have all
the desired property. 
%So, let $\{h_\xi\colon \xi<\co\}$ be an enumeration of $H$. 
For this we will construct a sequences $\la H_\xi\colon \xi<\co\ra$ 
of pairwise disjoint finite subsets of $H$ and 
$\la g_\xi\colon \xi<\co\ra$ of functions from $H_\xi$ into $\real$ 
such that for every $\xi<\co$
\begin{description}
\item{(i)}  $g_\xi(h)\in J_h$ for every $h\in H_\xi$, 
\item{(ii)}  there exists an 
$x_\xi\in K_\xi\cap \lin\left(\bigcup\{H_\zeta\colon\zeta\leq\xi\}\right)$
such that $G_\xi(x_\xi)\notin S_\xi$, where 
$G_\xi=\lin\left(\bigcup\{g_\zeta\colon\zeta\leq\xi\}\right)$.
\end{description}

Now, the inductive choice of $H_\xi$ and $g_\xi$ is quite simple. 
We choose an $x_\xi \in K_\xi\setminus\lin\left(\bigcup\{H_\zeta\colon\zeta<\xi\}\right)$
and represent $x_\xi$ as $z+q_1 h_1+\cdots +q_n h_n$ where 
$z\in \lin\left(\bigcup\{H_\zeta\colon\zeta<\xi\}\right)$,  
$h_1,\ldots,h_n\in H\setminus\bigcup\{H_\zeta\colon\zeta<\xi\}$, and
$q_1,\ldots,q_n\in\rational$. 
We put $H_\xi=\{h_1,\ldots, h_n\}$ and define $g_\xi$
such that (i) and (ii) are satisfied. This can be done, since
we have an open interval of possible choices for each value of
$g_\xi(h_i)$, while we have to omit only a nowhere dense set $S_\xi$ 
for the value of $G_\xi(x_\xi)=G_\xi(z)+q_1 g_\xi(h_1)+\cdots +q_n g_\xi(h_n)$.
Now, if $f_0\colon H\to\real$ is any extension of $\bigcup_{\xi<\co}g_\xi$
which satisfies ($*$) then a linear extension $f$ of $f_0$
has all the desired properties. 
\qed

\ex{SCIVPandPCnotD}{There exists an
$f\in\Add\cap\SCIVP\cap\PC\setminus\darb$. }

\proof Let $H_0$ be a Hamel basis such that $|H_0 \cap K|=\co$ for every perfect set
$K \subset\real$. (See \cite[cor.~7.3.7]{Ci:book}). Let $h_0 \in H_0$ and 
$V=\lin(H_0\setminus \{h_0\})$. 
Then $V$ is a proper linear subspace of $\real$ with $|V \cap K|=\co$
for every perfect $K \subset \real$.

As in the example above, let $\P$ be a family of pairwise disjoint perfect sets such that 
the set $\bigcup \P$ is linearly independent and 
$|\{P\in\P\colon P\subset (a,b)\}|=\co$ for every $a<b$. 
Let $\J$ be a family of all non-empty open intervals and let
$\{\la I_\xi,v_\xi\ra\colon \xi<\co\}$ be an enumeration of $\J\times V$. 
Find a one-to-one sequence
$\{P_\xi\in\P\colon \xi<\co\}$ such that $P_\xi\subset I_\xi$ for every $\xi<\co$ and  
let $H\subset\real$ be a Hamel basis containing $\bigcup\P$.

Define $f_0\colon H\to V$ by putting $f_0(h)=v_\xi$ for $h\in P_\xi$ and
choose an arbitrary $f_0(h)\in V$ for all other $h\in H$. 
Let $f\colon\real\to\real$ be a linear extension of $f_0$. 
Then $f$ is additive and non-zero, so $f$ has a dense graph. Thus $f\in\PC$. 
Also $f[\real]=V$ implying $f\notin \darb$. 
It is $\SCIVP$ since for every perfect set $K\subset\real$ and $a<b$ 
there exist $v\in V\cap K$ and $\xi<\co$ such that $\la I_\xi,v_\xi\ra=\la(a,b),v\ra$.
So $f[P_\xi]=\{v_\xi\}=\{v\}\subset K$ and $f\restriction P_\xi$ is continuous.
\qed

\ex{SCIVPandDnotCon}{
There exists an
$f\in\Add\cap\SCIVP\cap\darb\setminus\Conn$.}

\proof The construction is very similar to that for function $g$
from Example~\ref{DnotAC}. 
Let $L=\{\la x,x+1\ra \colon x\in \real\}$.
As above choose a family $\P$ of pairwise disjoint perfect sets such that 
the set $\bigcup \P$ is linearly independent and 
$|\{P\in\P\colon P\subset (a,b)\}|=\co$ for every $a<b$.
Let $\{A,B\}$ be a partition of continuum $\co$ with $|A|=|B|=\co$. 
Put $\J=\{(a,b)\colon a<b\}$ and let $\{\la I_\xi,r_\xi\ra\colon \xi\in A\}$
be an enumeration of $\J\times\real$.
Also, find a one-to-one sequence
$\{P_\xi\in\P\colon \xi<\co\}$ such that $P_\xi\subset I_\xi$ for every $\xi<\co$.
Finally, let 
$H\subset\real$ be a Hamel basis containing $\bigcup\P$ 
and let $H=\{h_\xi\colon x\in B\}$.


By induction on $\xi<\co$ construct 
a sequence $\la \la K_\xi,y_\xi\ra\colon \xi<\co\ra$ such that
for every $\xi<\co$ the following conditions hold.
\begin{description}
\item{(i)} $K_\xi\cap\lin(\bigcup\{K_\zeta\colon \zeta<\xi\})=\emptyset$.
\item{(ii)} $L\cap\lin(\bigcup\{K_\zeta\times\{y_\zeta\}\colon \zeta\leq\xi\})=\emptyset$.
\item{(iii)} If $\xi\in A$ then 
$K_\xi$ is a perfect subset of $P_\xi$ and $y_\xi=r_\xi$. 
\item{(iv)} If $\xi\in B$ then $|K_\xi|\leq 1$ and 
$h_\xi\in\lin(\bigcup\{K_\zeta\colon \zeta\leq\xi\})$. 
\end{description}

To find such a sequence assume that a sequence
$\la \la a_\zeta,y_\zeta\ra\colon \zeta<\xi\ra$
is already constructed for some $\xi<\co$. 
If $\xi\in A$ put $y_\xi=r_\xi$ and look at the set 
$Z=L\cap
\lin(\bigcup\{K_\zeta\times\{y_\zeta\}\colon \zeta\leq\xi\}\cup(P_\xi\times\{y_\xi\})$.
Notice that $Z$  has cardinality less than $\co$
since $Z\subset\real\times\lin(\bigcup\{y_\zeta\colon \zeta\leq\xi\})$
and $L$ intersects every horizontal line at exactly one point.
Let $T=\proj(Z)$ and notice that by the inductive assumption of (ii)
we have $T\cap\lin(\bigcup\{K_\zeta\colon \zeta<\xi\})=\emptyset$.
For every $t\in T$ choose an $h_t\in H\setminus \bigcup\{K_\zeta\colon \zeta<\xi\}$
with $t\notin \lin(H\setminus\{h_t\})$. Then the set 
$S=\{h_t\colon t\in T\}\cup\bigcup\{K_\zeta\colon \zeta<\xi\ \&\ \zeta\in B\}$
has cardinality less than $\co$. Choose a perfect set $K_\xi\subset P_\xi\setminus S$.
Then $K_\xi$ satisfies (i)--(iii). 

Next assume that $\xi\in B$ and let $V=\lin(\bigcup\{K_\zeta\colon \zeta<\xi\})$. If 
$h_\xi\in V$
we put $K_\xi=\emptyset$ and choose $y_\xi$ arbitrarily. 
So assume that $h_\xi\notin V$. We put $K_\xi=\{h_\xi\}$.
This guarantees (i) and (iv). 
To get (ii) we have to find $y_\xi$ such that the set 
$\lin(V\cup\{\la h_\xi,y_\xi\ra\})=
\{\la x,y\ra+q\la h_\xi,y_\xi\ra\colon \la x,y\ra\in V\ \&\ q\in\rational\}$
is disjoint with $L$. Thus, we must have
$y+q y_\xi\neq x+q h_\xi + 1$, that is,
$y_\xi\neq p x + h_\xi - p y  + p$ 
for every $\la x,y\ra\in V$ and $q=p^{-1}\in\rational\setminus\{0\}$.
Therefore it is enough to choose
\[
y_\xi\notin\lin\left(\bigcup\{K_\zeta\colon \zeta\leq\xi\}\cup
\{y_\zeta\colon \zeta<\xi\}\cup\{1\}
\right)
\]
which can be done, since %the space 
$\lin\left(\bigcup\{K_\zeta\colon \zeta\leq\xi\}\right)$
has co-dimension $\co$. 
This finishes the inductive construction.

Now put $f=\lin(\bigcup\{K_\xi\times\{y_\xi\}\colon \xi<\co\}$.
Then $f\colon\real\to\real$ is additive and misses $L$, so it is not
connectivity. It is Darboux and $\SCIVP$ since for every
$a,b,y\in\real$, $a<b$, 
there exists $\xi<\co$ such that $\la I_\xi,y_\xi\ra=\la(a,b),y\ra$.
So $K_\xi\subset P_\xi\subset I_\xi=(a,b)$,
$f[K_\xi]=\{y_\xi\}$, and $f\restriction K_\xi$ is continuous.\qed



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