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% Compositions of two additive almost continuous functions
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  \author{Dariusz Banaszewski,
  Mathematics Department, Pedagogical University,
  Chodkiewicza 30, 85--064 Bydgoszcz, Poland.\\
  Krzysztof Ciesielski,
  Department of Mathematics, West Virginia University,
  Morgantown, WV 26506--6310.
 }
\title{ Compositions of two additive almost continuous functions}
\date{}

\MathReviews{Primary: 26A15; Secondary: 26A51.}
\keywords{Darboux function, connectivity function, almost
continuous function, additive function, composition of functions.}
\markboth{Dariusz Banaszewski and Krzysztof Ciesielski }{Compositions
of two additive almost continuous functions}

\begin{document}
\maketitle

 \begin{abstract}
In the paper we prove that an additive Darboux function
$f\colon \mathR\to \mathR$ can be expressed as a
composition of two additive almost continuous (connectivity) 
functions if and only if 
either $f$ is almost continuous (connectivity)  function 
or $\dim(\ker(f))\neq 1$. We also show that for every cardinal 
number $\lambda\leq \co$ there exists an 
additive almost continuous functions with $\dim(\ker(f))=\lambda$.
A question whether every Darboux function
$f\colon\mathR\to\mathR$ can be expressed as a
composition of two almost continuous functions 
(see \cite{TN} or \cite{GMN}) remains open. 
 \end{abstract}
 
 \section{Definitions and notation} 

Our terminology and notation is standard. 
In particular, functions will be identified with
their graphs, and for a subset $A$ of $\mathR\times\mathR$
(possibly, but not necessarily, a graph of a function) 
we will write $\dom(A)$ and $\rng(A)$ to denote 
the $x$-projection 
(the domain) and the $y$-projection (the range) of $A$, respectively. 
The cardinality of a set $A$ will be denoted by 
$\card(A)$.
Cardinals will be identified with the initial ordinals.
The cardinality of the set $\real$ of real numbers, 
the continuum, will be denoted by $2^{\omega }$.

Throughout the paper we will consider $\mathR$ as a linear space over
the field $\mathQ$ of rational numbers.
A linear basis of this space will be referred to as a
{\em Hamel basis}. It is evident that the cardinality
of every Hamel basis is equal to $2^{\omega }$.

For 
an arbitrary set $A\subset \mathR$ the symbol 
$L(A)$ will denote the linear subspace of $\mathR$ over $\rational$ 
spanned by $A$, i.e., the set of all
finite linear combinations of elements of $A$
with coefficients from $\mathQ$. 
Similarly for an arbitrary planar set
$A\subset \mathR\times \mathR$
we define the set $L_{2}(A)$.
Also, for $A\subset\mathR$ and $x\in \mathR $
we write $x+A$ for $\{x+a\colon a\in A\}$.

Now, let $L \neq \emptyset$ be a linear subspace of $\mathR$
over $\mathQ $.
A function $f\colon L\to\mathR$ is said to be {\em additive} if it
satisfies the Cauchy's equation $f(x+y)= f(x)+f(y)$ for every 
$x,y\in L$. (See \cite{C} or \cite[p.~120]{4}.) 
The class of all additive functions from $\real$ to $\real$ 
will be denoted by ${\cal A}dd$.
Recall that if $H\subset\real$ is a Hamel basis then
every function $f_0\colon H\to\real$ can be uniquely
extended to the additive function $f\colon\real\to\real$.
In fact, $f=L_2(f_0)$. 

For $f\in{\cal A}dd$ its kernel $\ker(f)$ is defined as $f^{-1}(0)$.
Clearly $\ker(f)$ is a linear subspace of $\real$. Thus, 
$\dim(\ker(f))$ denotes the (linear) dimension of $\ker(f)$
over $\rational$. 

A function $f\colon\mathR\longrightarrow \mathR $
is a {\em Darboux function} if it has the intermediate value property, i.e.,
whenever for every $x_{1},x_{2}\in\real$, $x_{1}< x_{2}$, and
every point $c$ between $f(x_{1})$ and $f(x_{2})$
there exists $x \in [x_{1},x_{2}]$
such that $f(x)=c$. The family of all Darboux 
functions from $\real$ to $\real$ will be denoted by ${\cal D}$.

A function $f:\mathR\to\mathR$ is said to be {\em almost
continuous\/} in the sense of Stallings if each open set (in $\real^2$)
containing $f$ contains also a (graph of) continuous function
$g:\mathR\to\mathR $~\cite{S}.
The class of all almost continuous functions from
$\mathR$ into $\mathR$
will be denoted by ${\cal AC}$.

A closed set $K\subset\mathR\times\mathR $
is said to be a {\em blocking set\/}
for a function $f\colon \mathR \to \mathR$ provided 
$f\cap K=\emptyset $ while 
$g\cap K\ne \emptyset$ for every continuous function
$g\colon\mathR\to\mathR$. A blocking set $K\subset\mathR\times\mathR$
for $f$ is {\em irreducible} if no
proper subset of $K$  is a blocking set for $f$ \cite{3}.

It is known that $f$ is almost continuous
if and only if it has no blocking set.
Moreover, if $f$ is not almost continuous then
there is an irreducible blocking set
$K$ for $f$, and the $x$-projection
of $K$ is a non--degenerate connected set \cite{3}.
Thus, if $f\colon\real\to\real$ intersects all
closed sets $K\subset\mathR^2$ with the domain being a
non-degenerate interval, then it is almost continuous
(cf.~\cite{KG}).

A function $f\colon \mathR \to \mathR$
is a {\it connectivity function\/}
if its graph is connected (in $\real^2$). 
We will use a symbol ${\cal C}onn$
to denote the class of all connectivity functions $f\colon\mathR\to\mathR$.
The class of all continuous functions $f\colon\mathR\to\mathR$
will be denoted by $\cal C$.
We have the following chain of proper inclusions \cite{S}:
\[
{\cal C}\subset {\cal AC}\subset{\cal C}onn \subset {\cal D}.
\]

It is well--known that the composition of
two Darboux functions is a Darboux function again.
The problem of characterization of these Darboux
functions which can be expressed as a composition
of two almost continuous functions was considered
in \cite{TN}. (See also \cite{GMN}.)
In this paper we will consider the analogous
problem in the class of additive functions.

\section{Main theorem} 

Let $\B$ be the family of all closed sets
$B\subset\mathR\times\mathR$ such that $\dom(B)$ 
is a non-degenerate interval and either
\begin{description}
\item{(A)} $B= \mathR\times\{y\}$; or,
\item{(B)} $B^y= \{x\in\mathR\colon \la x,y\ra\in B\}$
           is nowhere dense for each $y\in\mathR$.
\end{description}
We will use this family throughout the paper. 

In what follows we will use repeatedly the following lemma. 

\begin{lem}\label{llee1}
Let $f\in\Add$ be such that $\ker(f)\neq\{0\}$.
If $f\cap B\neq\emptyset$ for every $B\in\B$ then $f\in\AC$.
\end{lem}

 \pf
 Fix an arbitrary closed set $K\subset\mathR^2$ such that $\dom(K)$ is a
non-degenerate interval. It is enough to show that
$f\cap K\neq\emptyset$.
  
If $K^y$ is nowhere dense for each $y\in\mathR$ then $K\in\B$
and $f\cap K\neq\emptyset$. So, assume otherwise. 

Then there is $y\in\mathR$ such that $K^y$
contains a non-degenerate interval $I$.
But $\mathR\times\{y\}\in\B$ so $f\cap(\mathR\times\{y\})\neq\emptyset$.
In particular, there exists $x\in\real$ such that $f(x)=y$.
Also, $\ker(f)$ is dense, since  $\ker(f)\neq\{0\}$,
and so $f^{-1}(y)$ contains a dense set $x+\ker(f)$.
Thus $f^{-1}(y)\cap I\supset (x+\ker(f))\cap I\neq\emptyset$
and $\emptyset\neq f\cap (I\times\{y\})\subset f\cap K$.
\Qed

The next theorem constitutes one direction of
our main characterization theorem. 

\begin{thm}\label{tw1}
Let $f\in {\cal D}\cap {\cal A}dd$
be such that $\dim(\ker (f))\neq 1.$
Then $f$ is a composition of two additive
almost continuous functions.
\end{thm}

\pf
Fix $f\in{\cal D}\cap{\cal A}dd$ with
$\dim(\ker (f))\neq 1$. 
If $\dim(\ker (f))= 0$ then $f$ is continuous 
(see \cite{DG}) and $f= f\circ id$. Similarly, if $f\equiv 0$
then $f= f\circ id$. 
Hence we can assume that $\dim(\ker (f))\geq 2$ and $f\not \equiv 0$.
  
Let $\{K_{\alpha }\colon \alpha <\co\}$ be an enumeration 
of the family $\B$ such that $K_{0}= \mathR\times \{0\}$
and let $\{b_{\alpha }\colon \alpha <\co\}$ be an enumeration
of a fixed Hamel basis with $b_{0}\in\ker (f)$.

We construct, by induction on $\alpha<\co$, the sequences
$\la g_\alpha\colon\alpha<\co\ra$ and 
$\la h_\alpha\colon\alpha<\co\ra$ of 
additive functions from subsets of $\real$ into $\real$
maintaining the following inductive properties
for every $\alpha<\co$.
\begin{enumerate}
\item[(i)] $g_{\beta }\subset g_{\alpha }$ 
          and $h_{\beta }\subset h_{\alpha }$ for every $\beta <\alpha $;

\item[(ii)] $\card(\dom(g_{\alpha}))\leq\max(\omega,\alpha)$, and
             $\card(\dom(h_{\alpha}))\leq\max(\omega,\alpha$);

\item[(iii)] $\rng(g_{\alpha })=\dom(h_{\alpha })$ and
             $h_{\alpha }\circ g_{\alpha }= f|\dom(g_{\alpha })$;


\item[(iv)]  $g_{\alpha}\cap K_{\alpha}\neq\emptyset$ and 
             $h_{\alpha}\cap K_{\alpha}\neq\emptyset$;

\item[(v)] $b_{\alpha }\in \dom(g_{\alpha })$.
\end{enumerate}

To make an inductive step 
assume that for some $\alpha<\co$ the functions
$g_\beta$ and $h_\beta$ satisfying conditions (i)--(v) have 
already been constructed for every $\beta<\alpha$.

If $\alpha=0$ choose $s_{0}\in\ker(f)\setminus L(\{b_{0}\})$.
Such a choice is possible, since $\dim(\ker(f))\geq 2$.
Put $g_{0}= L_{2}(\{\la b_{0},0\ra,\la s_{0},s_{0}\ra\})$ and
$h_{0}= L_{2}(\{\la s_{0},0\ra\})$.
It is easy to see that $g_{0}$ and $h_{0}$ fulfill
the conditions (i)--(v).

So, assume that $\alpha>0$ and put
\[\overline{g}_{\alpha }= 
  \bigcup_{\beta < \alpha }g_{\beta },\ \text{ \ \ and \ \ }
  \overline{h}_{\alpha }= 
  \bigcup _{\beta < \alpha }h_{\beta }.
\]
Clearly functions $\overline{g}_{\alpha }$ and 
$\overline{g}_{\alpha}$ satisfy the conditions (i)-(iii).
We will find $x_\alpha,y_\alpha,s_\alpha,v_\alpha,c_\alpha\in\real$
such that
\begin{enumerate}
\item[(a)] $\la x_\alpha,y_\alpha\ra\in K_\alpha$;
\item[(b)] $\la v_\alpha,f(s_\alpha)\ra\in K_\alpha$;
\item[(c)] $g_\alpha=L_2(\overline{g}_{\alpha}\cup
\{\la x_\alpha,y_\alpha\ra,\la b_\alpha,c_\alpha\ra,\la s_\alpha,v_\alpha\ra\})$
and\\
$h_\alpha=L_2(\overline{h}_{\alpha}\cup
\{\la y_\alpha,f(x_\alpha)\ra,
\la c_\alpha,f(b_\alpha)\ra,\la v_\alpha,f(s_\alpha)\ra\})$
remain the functions. 
\end{enumerate}
It is easy to see that such $g_\alpha$ and $h_\alpha$ will satisfy
the conditions (i)--(v).

As a first step we will construct $x_\alpha$ and $y_\alpha$.
If $K_{\alpha }\cap\overline{g}_{\alpha}\neq\emptyset$
we simply choose 
$\la x_\alpha,y_\alpha\ra\in K_\alpha\cap\overline{g}_{\alpha}$.
So, assume that $K_{\alpha }\cap\overline{g}_{\alpha}=\emptyset$.
In this case we will find
$\la x_\alpha,y_\alpha\ra\in K_\alpha$ such that
\begin{equation}\label{con1}
x_{\alpha}\notin\dom(\overline{g}_{\alpha }),\ \text{ \ \ and \ \ }
y_{\alpha}\notin\dom(\overline{h}_{\alpha })=\rng(\overline{g}_{\alpha }).
\end{equation}
Such a restriction is necessary to guarantee the condition (c). 

Let $X_\alpha=\dom(\overline{g}_{\alpha})$, and
$Y_\alpha=\dom(\overline{h}_{\alpha })=\rng(\overline{g}_{\alpha })$.
Then $\card(X_\alpha)<\co$ and $\card(Y_\alpha)<\co$. 

If $K_\alpha$ was chosen according to the part (A) of the definition
of $\B$ then $K_\alpha=\mathR\times\{y\}$ for some $y\in\mathR$.
Hence $y\not\in Y_\alpha$, since 
$K_{\alpha }\cap\overline{g}_{\alpha}=\emptyset$. Put $y_\alpha=y$ and 
choose $x_\alpha\not\in X_\alpha$.
Then $\la x_\alpha,y_\alpha\ra\in K_\alpha$ and the condition
(\ref{con1}) is satisfied.

So, assume that $K_\alpha$ was chosen according to the part (B) 
of the definition of $\B$, i.e., that 
$K_\alpha^y$ is nowhere dense for every $y\in\mathR$.
To deal with this case recall the following fact. 
(See \cite[Th. 29.19, p.~231]{keh}.)
\begin{quote}
  For every closed set $K\subset\mathR^2$ the set
  \[
  Z(K)=\{y\in\mathR\colon K^y \mbox{ contains a
                                    non-empty perfect set}\}
  \]
  is either countable or is of power continuum.
\end{quote}
This leads us to the two natural subcases.

\begin{enumerate}
  \item[$\bullet$] $\card(Z(K_\alpha))=\co$. Then
   $\card(Z(K_\alpha)\setminus Y_\alpha)= \co$ and we can choose
   $y_\alpha\in Z(K_\alpha)\setminus Y_\alpha$. Moreover, 
   $\card(K_\alpha^{y_\alpha})= \co$, and so we can pick
   $x_\alpha\in K_\alpha^{y_\alpha}\setminus X_\alpha$.
   Then $\la x_\alpha,y_\alpha\ra\in K_\alpha$ satisfies (\ref{con1}). 
\item[$\bullet$] $\card(Z(K_\alpha))\leq\omega$. Then the set 
    \[
      E_\alpha=\dom(K_\alpha)\setminus
     \bigcup\left\{K_\alpha^y\colon {y\in Z(K_\alpha)}\right\}
    \]
   is a residual subset of the interval $\dom(K_\alpha)$
   since each set $K_\alpha^y$ is nowhere dense. In particular,
   $\card(E_\alpha)=\co$. Moreover, $K_\alpha^y$ is countable for 
   every $y\in \real\setminus Z(K_\alpha)$. So the set
   \[
     E_\alpha^1= E_\alpha\setminus\left(X_\alpha\cup
     \bigcup\left\{K_\alpha^y\colon{y\in Y_\alpha\setminus
                                     Z(K_\alpha)}\right\}\right)
   \]
   has cardinality $\co$. 
   Choose 
   $x_\alpha\in E_\alpha^1\subset\dom(K_\alpha)\setminus
   \left(X_\alpha\cup\bigcup_{y\in Y_\alpha}K_\alpha^y\right)$ and
   $y_\alpha\in\mathR$
   with $\la x_\alpha ,y_\alpha\ra\in K_\alpha$.
   Then $y_\alpha\not\in Y_\alpha$ and (\ref{con1}) is satisfied. 
 \end{enumerate}
This finishes the construction of $x_\alpha$ and $y_\alpha$. 

To construct $s_\alpha$ and $v_\alpha$ first note that
by (\ref{con1}), 
\[
\underline{g}_{\alpha}=
L_{2}(\overline{g}_{\alpha }\cup\{\la x_{\alpha},y_{\alpha}\ra\}),
\ \text{ \ \ and \ \ }
\underline{h}_{\alpha}= 
L_{2}(\overline{h}_{\alpha }\cup\{(y_{\alpha },f(x_{\alpha }))\})
\]
are the additive functions. 
If $K_{\alpha }\cap\underline{h}_{\alpha}\neq\emptyset$
we choose 
$\la v_\alpha,w_\alpha\ra\in K_\alpha\cap\underline{h}_{\alpha}$
and take $s_\alpha$ such that $\underline{g}_{\alpha}(s_\alpha)=v_\alpha$.
Such an $s_\alpha$ exists since
$\dom(\underline{h}_{\alpha})=\rng(\underline{g}_{\alpha })$.
Then 
$w_\alpha=\underline{h}_{\alpha}(v_\alpha)=
\underline{h}_{\alpha}(\underline{g}_{\alpha }(s_\alpha))=f(s_\alpha)$, 
so the condition (b) is satisfied. 

So, assume that $K_{\alpha }\cap\underline{h}_{\alpha}=\emptyset$.
Then, as in the construction of $x_\alpha$ and $y_\alpha$,
we can find $\la v_{\alpha },w_\alpha\ra\in K_\alpha$ such that
\begin{equation}\label{eq2}
v_{\alpha}\notin\dom(\underline{h}_{\alpha })=\rng(\underline{g}_{\alpha }),
\ \text{ \ \ and \ \ }
w_{\alpha}\notin\rng(\underline{h}_{\alpha }).
\end{equation}
Now, note that $\rng(f)= \mathR$, since $f$
is a non-zero additive Darboux function.
Choose $s_{\alpha }\in f^{-1}(w_{\alpha })$ and notice that
$s_{\alpha }\not \in \dom(\underline{g}_{\alpha })$
since otherwise 
$w_\alpha=f(s_\alpha)=
\underline{h}_{\alpha}(\underline{g}_{\alpha}(s_\alpha))=
\underline{h}_{\alpha}(v_\alpha)\in\rng(\underline{h}_{\alpha })$,
contradicting (\ref{eq2}). Thus, $\la v_\alpha,f(s_\alpha)\ra\in K_\alpha$,
as required in (b). 

Finally, to choose $c_\alpha$ note that 
\[
G_{\alpha}=
L_{2}(\underline{g}_{\alpha }\cup\{\la s_{\alpha},v_{\alpha}\ra\}),
\ \text{ \ \ and \ \ }
H_{\alpha}= 
L_{2}(\underline{h}_{\alpha }\cup\{\la v_{\alpha },f(s_{\alpha })\ra\})
\]
are the additive functions. If $b_\alpha\in\dom(G_\alpha)$
we put $c_\alpha=G_\alpha(b_\alpha)$. Otherwise we choose
$c_\alpha\in\real\setminus\dom(H_\alpha)$. 

It is easy to see that 
$x_\alpha$, $y_\alpha$, $s_\alpha$, $v_\alpha$, and $c_\alpha$
chosen above satisfy (a)--(c). This finishes the inductive
construction. 

Having constructed functions $g_\alpha$ and $h_\alpha$
let 
\[
g=\bigcup_{\alpha<\co}g_\alpha,\ \text{ \ \ and \ \ }
h^0=\bigcup_{\alpha<\co}h_\alpha.
\]
It is easy to see that $g$ and $h^0$ are additive functions
such that $\dom(g)=\real$ (by (v)) and that $f=h^0\circ g$.
Now, if $h\colon\real\to\real$ is any additive extension
of $h^0$ then, by (iv), $g$ and $h$ are almost continuous,
while we still have $f=h\circ g$. This finishes the proof. 
\Qed

Next we will prove the converse of Theorem~\ref{tw1}. 
For this we will need the following simple and well known fact.

\begin{lem}\label{kerf+g}
If $g,h\in {\cal A}dd$ and $g$ is a surjection then
\[
\dim(\ker (h\circ g))= \dim(\ker (h))+\dim(\ker (g)).
\]
\end{lem}

\pf Let $G,H$ be linearly independent sets such that
$L(G)= \ker (g)$ and $L(H)= \ker (h)$.
For every $w\in H$ choose $s_{w}\in g^{-1}(w)$ and
notice that $F= G\cup\{s_{w}\colon w\in H\}$ is linearly independent.

Indeed, suppose that
\begin{equation}\label{rf+g}
x=\sum _{i= 1}^{n} q_{i}v_{i}+\sum _{j= 1}^{k} p_{j}s_{w_{j}}= 0
\end{equation}
for some $n,k\in \mathN$, $q_{i},p_{j}\in \mathQ$,
$v_{i}\in G$, and $w_{j}\in H$, where $i= 1,\ldots n$,
and $j= 1,\ldots k$. Then
\[
g(x)= \sum _{i= 1}^{n} q_{i}g(v_{i})
            +\sum _{j= 1}^{k} p_{j}g(s_{w_{j}})= 
\sum _{j= 1}^{k} p_{j}g(s_{w_{j}})= 
  \sum _{j= 1}^{k} p_{j}w_{j}= 0
 \]
which shows that $p_{j}= 0$ for $j= 1,\ldots ,k$.
Hence, by (\ref{rf+g}),  $\sum _{i=1}^{n}
q_{i}v_{i}=0$, which implies that $q_{i}=0$ for $i=1,\ldots ,n$.

It is easy to see that $L(F)=\ker(h\circ g)$ and consequently,
\[
\dim(\ker (g))+\dim(\ker (h))=\card(G)+\card(H)=\card(F)=
\dim(\ker (h\circ g)).
\]
This finishes the proof. \Qed

With this lemma in hand we are ready for the next theorem.

\begin{thm}\label{tw2}
Assume $f\in {\cal A}dd $ and $\dim(\ker (f))=1$.
\begin{description}
\item[(I)]   If $f\notin {\cal AC}$ then $f=h\circ g$ 
             for no $h,g\in {\cal A}dd\cap {\cal AC}$.
\item[(II)]  If $f\notin {\cal C}onn$ then $f=h\circ g$ 
             for no $h,g\in {\cal A}dd\cap {\cal C}onn$.
\end{description}
\end{thm}

\pf Fix $f\in {\cal A}dd\cap\D$ such that
$\dim(\ker(f))=1$ and suppose that there exist
$g,h\in {\cal A}dd\cap\D$ with $f=h\circ g$.
Then, $g$ is surjection, since $g\not\equiv 0$. 

By Lemma~\ref{kerf+g}, either
$\dim(\ker (g))=0$ or $\dim(\ker (h))=0$.
Consequently, either $g$ or $h$ is a Darboux injection, so it is 
equal to a linear homeomorphism 
$L(x)=ax$. (Any other additive function has a dense graph,
so it cannot be Darboux and one-to-one at the same time.) 
Since the classes ${\cal AC}$ and ${\cal C}onn$
are closed under composition with homeomorphisms
(cf, e.g., \cite{NA}) we conclude that 
$f\in{\cal AC}$ ($f\in{\cal C}onn$)
if and only if
$g,h\in{\cal AC}$ ($g,h\in{\cal C}onn$). 
\Qed

Theorems~\ref{tw1} and \ref{tw2} give us as a corollary 
the main characterization. (Since ${\cal AC}\subset{\cal C}onn$.) 


\begin{Co}\label{coMain}
Let $f\colon\real\to\real$ be an additive Darboux function. Then
\begin{description}
\item[(I)] $f$ is a composition of two additive almost continuous functions
           if and only if either $f$ is almost continuous 
                          or $\dim(\ker (f))\neq 1$;
\item[(II)] $f$ is a composition of two additive connectivity functions
           if and only if either $f$ is a connectivity function 
                          or $\dim(\ker (f))\neq 1$.
\end{description}
\end{Co}

\section{Final remarks} 

Although Corollary~\ref{coMain} gives a full characterization 
of additive Darboux functions which can be expressed as 
a composition of two additive almost continuous (or connectivity)
functions it still does not exclude the possibility that every 
additive Darboux function can be expressed as a such composition. 
To conclude this, we need also the following example. 

\begin{ex}\label{D-Conn}
  There exists a function $f\colon\real\to\real$ such that 
  $\dim(\ker (f))=1$ and
  $f\in {\cal A}dd \cap {\cal D} \setminus {\cal C}onn$.
 \end{ex}

\pf Let $H$ be a Hamel basis and  $H_{0}$ be a proper subset 
of $H$ be with $\card(H_{0})=2^{\omega }$.
Choose $h_{0}\in H_{0}$, fix a bijection
$\varphi\colon H_{0}\setminus \{h_{0}\} \longrightarrow H_{0}$
and define
$\overline{f}\colon H\to\real$ as follows.
\[ 
\overline{f}(h)=
   \left\{
    \begin{array}{lll}
     0 & \mbox{ for } & h=h_{0}\\
     \varphi (h) & \mbox{ for } & h\in H_{0}\setminus \{h_{0}\} \\
     h & \mbox{ for } & h\in H\setminus H_{0}\\
    \end{array}
   \right. 
\]
Let $f$ be the additive extension of $\overline{f}$.

It is easy to observe that
\begin{equation}\label{qqq}
  \overline{f}(h)\in h+L(H_{0})\; \mbox{ for } \; h\in H,
 \end{equation}
and therefore
 \begin{equation}\label{qq2}
  f(x)\in x+L(H_{0}) \; \mbox{ for every } \; x\in \mathR.
\end{equation}

It is obvious that $\ker(f)=L(\{h_0\})$. Also $\rng(f)=\mathR$,
since $\rng(\overline{f})=H$.
Thus $f^{-1}(y)\neq \emptyset$ for every $y\in \mathR$.
Moreover, since all level sets are congruent under translations
and $\ker (f)$ is dense~\cite{4}, $f^{-1}(y)$ is dense
for every $y\in\real$. Hence, the graph of $f$ is dense in $\real^2$ and 
$f[J]=\mathR$ for every interval $J\subset \mathR$.
In particular, $f\in {\cal D}$.
Moreover, by (\ref{qq2}),
\[
f\subset \bigcup _{b\in L(H_{0})}\{\la x,x+b\ra\colon x\in \mathR\}
\]
and consequently, the line $y=x+c$
separates the graph of $f$ for every number $c\in\real\setminus L(H_{0})$.
So, $f$ is not a connectivity function. \Qed

\begin{Co}
There exists an additive Darboux function $f\colon\real\to\real$ 
such that $f=h\circ g$ for no $f,g\in {\cal A}dd\cap {\cal C}onn$.
\end{Co}

Our last theorem is a variation of the example above. 
For its proof we will need one more easy lemma. 

\begin{lem}\label{lker}
Let $f$ be an additive function and $F=L_2(f\cup\{\la u,v\ra\})$
where $u\notin \dom(f)$ and $v\notin \rng(f)$. Then $\ker(F)=\ker(f)$.
\end{lem}
 
\pf Obviously $\ker (f)\subset \ker (F)$. To prove that
$\ker (F)\subset \ker (f)$, fix an arbitrary $x\in \ker (F)$.
Then
\[
x=q_{0}u+q_{1}w \  \mbox{ where }\ q_{0},q_{1}\in\mathQ \ 
\mbox{ and }\  w\in\dom(f).
\]
Since $F(x)=q_{0}v+q_{1}f(w)=0$, $q_{0}v=-q_{1}f(w)$.
Because $v\not \in \rng(f)$, $q_{0}=0$ and consequently,
$x\in \dom(f)$. Which shows that $x\in \ker (f)$. \Qed

\begin{thm}\label{teo3}
For every cardinal number $\lambda \leq 2^{\omega }$
there exists an additive almost continuous 
function $f\colon\real\to\real$ such that $\dim(\ker(f))=\lambda $.
\end{thm}

\pf Since function $f\equiv 0$ is almost continuous and 
$\dim(\ker(f))=\co$ for such $f$, 
we can assume that $\lambda<2^\omega$.
If $\lambda=0$ then the identity function $id$ has required 
properties and so we may also assume that $\lambda>0$.

Now, let $H\subset \mathR $ be a Hamel basis and $H_{0}\subset H$
be such that $\card(H_{0})=\lambda$. Also, let
$\{b_\alpha\colon \alpha <2^{\omega }\}=H\setminus H_{0}$ and
choose an enumeration $\{K_{\alpha }\colon \alpha <2^\omega\}$ 
of the family $\B$ of blocking sets from Lemma \ref{llee1}, with
$K_{0}=\mathR\times\{0\}$.
The construction will be a slight modification of that 
in the proof of Theorem~\ref{tw1}. 

By transfinite induction construct a sequence 
$\la f_\alpha\colon\alpha <2^{\omega}\ra$ of additive partial
functions from $\real$ into $\real$ such 
the that following inductive conditions are satisfied 
for every $\alpha<\co$. 
\begin{enumerate}
\item[(i)] $f_{\beta }\subset f_{\alpha}$ for every $\beta <\alpha$;
\item[(ii)] $f_{\alpha }\cap K_{\alpha }\neq \emptyset $;
\item[(iii)] $b_\alpha\in\dom(f_{\alpha })$ and 
             $\card(f_{\alpha })\leq\max(\lambda,\omega,\alpha )$;
\item[(iv)] $\ker(f_{\alpha })=L(H_{0})$.
\end{enumerate}
We start the induction with putting 
$f_{0}=L_{2}((H_{0}\times \{0\})\cup\{\la b_0,1\ra\})$.
It is obvious that $f_{0}$ fulfills the conditions (i)--(iv).

To make an inductive step, fix 
$\alpha<2^{\omega }$, $\alpha>0$,  and assume that we have 
already chosen functions $f_\beta$ for $\beta<\alpha$ which 
satisfy conditions (i)--(iv).

If $b_\alpha\in\dom(\bigcup_{\beta < \alpha }f_{\beta })$,
we put $\overline{f}_{\alpha }=\bigcup_{\beta<\alpha }f_{\beta }$.
Otherwise, by (iii),
$\card(\rng(\bigcup _{\beta < \alpha }f_{\beta }))<2^{\omega }$
and we can choose 
$c_{\alpha }\in\mathR\setminus\rng(\bigcup_{\beta <\alpha }f_{\beta })$.
Put
\[
\overline{f}_{\alpha }=L_{2}\left(\{\la b_\alpha ,c_{\alpha }\ra\}\cup
\bigcup _{\beta < \alpha }f_{\beta }\right).
\]
Clearly $\overline{f}_{\alpha }$ satisfies (i), (iii) and (iv).
Also, if $K_{\alpha}\cap\overline{f}_{\alpha }\neq\emptyset$ then
$f_{\alpha}=\overline{f}_{\alpha}$ satisfies (ii) as well and 
the construction is completed. 

So, assume that $K_{\alpha}\cap\overline{f}_{\alpha}=\emptyset$ and
let $X_\alpha=\dom(\overline{f}_\alpha)$, and 
$Y_\alpha=\rng(\overline{f}_\alpha)$.
From (iii) we have that 
$\card(Y_\alpha)\leq\card(X_\alpha)\leq\max(\omega,\alpha,\lambda)<\co$.
We will choose $\la x_\alpha,y_\alpha\ra\in K_\alpha$ such that
\begin{equation}\label{conLast}
x_\alpha\notin X_\alpha\ \ \ \text{ and }\ \ \ 
y_\alpha\notin Y_\alpha
\end{equation}
and define 
$f_\alpha=L_2(\overline{f}_{\alpha}\cup\{\la x_\alpha,y_\alpha\ra\})$.
This will finish the construction
since then, by Lemma~\ref{lker},
$\ker(f_{\alpha })=\ker(\overline{f}_{\alpha })=L(H_{0})$.


Now, if $K_\alpha=\mathR\times\{y\}$ for some $y\in\mathR$
then $y_\alpha=y\not\in Y_\alpha$, since 
$K_\alpha\cap\overline{f}_\alpha=\emptyset$.
Choose an arbitrary $x_\alpha\in\mathR\setminus X_\alpha$.
Then $\la x_\alpha,y_\alpha\ra$ satisfies (\ref{conLast}). 

So, assume that $K_\alpha=\mathR\times\{y\}$ for no $y\in\mathR$.
Then $K_\alpha^y$ is nowhere dense for every  $y\in\mathR$.
Since $Z(K_\alpha)=\{y\in\mathR\colon K_\alpha^y
\mbox{ contains non-empty perfect set}\}$
is either countable or has cardinality continuum thus we have the following
two cases to consider.
\begin{enumerate}
\item[$\bullet$] $\card(Z(K_\alpha))=2^\omega$.
Choose $y_\alpha\in Z(K_\alpha)\setminus Y_\alpha$. Then
$\card(K_\alpha^{y_\alpha})=2^\omega$ and we may
choose $x_\alpha\in K_\alpha^{y_\alpha}\setminus X_\alpha$.
Clearly $\la x_\alpha,y_\alpha\ra$ satisfies (\ref{conLast}). 

\item[$\bullet$] $\card(Z(K_\alpha))\leq\omega$.
Then the set
\[
E_\alpha=\dom(K_\alpha)\setminus
    \bigcup\{K_\alpha^y\colon{y\in Z(K_\alpha)}\}
\]
is residual in $\dom(K_\alpha)$ and the set
\[
E_\alpha^1=E_\alpha\setminus\left(X_\alpha\cup
\bigcup\{K_\alpha^y\colon{y\in Y_\alpha\setminus Z(K_\alpha)}\}\right)
\]
has cardinality continuum.
Choose $x_\alpha\in E_\alpha^1$ and $y_\alpha\in\mathR$
such that $\la x_\alpha,y_\alpha\ra\in K_\alpha$.
Then $\la x_\alpha,y_\alpha\ra$ satisfies (\ref{conLast}) as well. 
\end{enumerate}
This finishes the inductive construction. 

Now, put
\[
f=\bigcup _{\alpha<\co}f_{\alpha }.
\]
It is easy to see that $f$ has the desired properties. \Qed


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