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%\MathReviews{Primary:  26A15; Secondary: 03E75, 54A25.}

%\keywords{cardinal invariants; extendable, Darboux, almost continuous
%and peripherially continuous functions; functions with perfect road. }

%\coverauthor{}
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\title{A new cardinal invariant related to adding real functions}

\author{{\small Francis Jordan}%
\thanks{AMS classification numbers: Primary 26A15;  Secondary 54A25 \endgraf
Key words and phrases: cardinal functions,
extendable functions, connectivity
functions, peripherially continuous functions,
almost continuous functions,
Darboux functions. \endgraf
This paper was written under supervision of K.~Ciesielski.
The author wishes to thank him for many helpful conversations.},
\small
Department of Mathematics, University of Louisville,\\
Louisville, KY 40208\\
(fejord01@athena.louisville.edu)
}

\date{}




%% Theorems, etc.
\def\integer{{\bf Z}}
\def\natural{{\bf N}}
\def\rational{{\bf Q}}
\def\real{{\bf R}}


\newcommand{\cof}{\operatorname{cf}}
\newcommand{\add}{\operatorname{A}}
\newcommand{\cl}{\operatorname{cl}}
\newcommand{\bd}{\operatorname{bd}}
\newcommand{\lin}{\operatorname{LIN}}
%%\frak c
\newcommand{\cuum}{{\bf c}}
\newcommand{\D}{{\cal D}}
\newcommand{\F}{{\cal F}}
\newcommand{\G}{{\cal G}}
\newcommand{\pos}{{\Bbb P}}
\newcommand{\dar}{{\operatorname {Dar}}}
\newcommand{\conn}{{\operatorname {Con}}}
\newcommand{\acon}{{\operatorname {AC}}}
\newcommand{\ext}{{\operatorname {Ext}}}
\newcommand{\pr}{{\operatorname {PR}}}
\newcommand{\phc}{{\operatorname {PC}}}
\newcommand{\sz}{{\operatorname {SZ}}}
\newcommand{\diff}{{e}}
\newcommand{\same}{{d}}
\newcommand{\dom}{{D}}
\newcommand{\domain}{{\operatorname {dom}}}
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\newcommand{\comp}{{\neg}}

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





\begin{document}\maketitle
\begin{abstract}
Let $F\subseteq\real^{\real}$.  The additivity of $F$, shortly $\add(F)$, is the
minimum cardinality of a family $G\subseteq\real^{\real}$ with the property that
$h+G\subseteq F$ for no $h\in\real^{\real}$.  In this paper we consider the notion of
super-additivity which we will denote by $\add^*$.  If $F\subseteq\real^{\real}$, then
$\add^*(F)$ is the minimum cardinality of a family of functions $G$ with the property
that for any $H\subseteq\real^{\real}$ if $|H|<\add(F)$, then there is a $g\in G$
such that $g+H\subseteq F$.  We calculate the super-additivities of the families of
Darboux-like functions and their complements.
\end{abstract}
\section{Preliminaries}
In what follows we will use standard terminology and
notation as in \cite{Ci:book}.  In particular, the set of all
functions from a set $X$
into a set $Y$ will be denoted by $Y^{X}$.  Given a set $X$ and $f,g\in X^X$
we denote their composition by $f\circ g$.  The characteristic function of
a set $A\subseteq\real$ will be denoted by $\charf{A}$.
The symbol $|X|$ will denote the
cardinality of the set $X$.  The successor of a cardinal $\kappa$ will be
denoted by $\kappa^{+}$.
We denote by $[X]^{<\kappa}$, $[X]^{\kappa}$, and
$[X]^{\leq\kappa}$ the sets of all subsets of $X$ of cardinality less than $\kappa$,
equal to $\kappa$, and less than or equal to $\kappa$, respectively.
The cardinality of the real numbers $\real$
will be denoted by $\cuum$.  Given a cardinal number $\kappa$ we let $\cof(\kappa)$ denote
the cofinality of $\kappa$.  We say a cardinal $\kappa$ is regular
provided that $\cof(\kappa)=\kappa$.
For functions $f,g\in\real^{\real}$ let
$[f=g]$ denote the set $\{x\in\real\colon f(x)=g(x)\}$.  We define $[f<g]$ and
$[f\leq g]$ in a similiar way.  Functions
will be identified with their graphs.  For a set $S\subseteq X\times Y$
we let $\domain(S)=\{x\in X\colon (\exists y\in Y)(\langle x,y \rangle\in S)\}$.

We will need some cardinals which have
combinatorial descriptions.  For a cardinal $\kappa$ we define
\begin{description}
\item[ ]$\same_{\kappa}
 =\min\{|F|\colon F\subseteq \kappa^{\kappa}\ \&\ (\forall g\in\kappa^{\kappa})
(\exists f\in F) (|[f=g]|=\kappa)\}$,
\item[ ]$\diff_{\kappa}
=\min\{|F|\colon F\subseteq \kappa^{\kappa}\ \&\ (\forall g\in\kappa^{\kappa})
(\exists f\in F) (|[f=g]|<\kappa)\}$,
\item[ ]$\diff_{\kappa}^{1}=\min\{|F|\colon F\subseteq\kappa^{\kappa}\ \&\
\left(\forall G\in\left[\kappa^{\kappa}\right]^{<\diff_{\kappa}}\right)
(\exists f\in F)(\forall g\in G)(|[f=g]|=\kappa)\}$,
\item[ ]$\same_{\kappa}^{1}=\min\{|F|\colon F\subseteq\kappa^{\kappa}\ \&\
\left(\forall G\in\left[\kappa^{\kappa}\right]^{<\same_{\kappa}}\right)
(\exists f\in F)(\forall g\in G)(|[f=g]|<\kappa)\}$.
\end{description}

\section{Introduction}
The cardinal function called additivity was orginally defined
by Natkaniec \cite{NATsurvey} for families $\F\in\real^{\real}$ to
be
\[\add(\F)=\min(\{|F|\colon F\subseteq \real^{\real}\ \&\ (\forall
g\in \real^{\real}) (\exists f\in F) (f+g\notin \F)\}\cup
\{(2^{\cuum})^{+}\}).\]
This cardinal function has been studied intensively and has been
generalized to include families in $(\real^{m})^{\real^n}$ see
\cite{GNsurvey}, \cite{NATsurvey}, and \cite{KCsurvey}.  We will restrict the scope
of this paper to $\real^{\real}$ and consider a new cardinal
function which is based on the additivity function.  Before
continuing let us recall some basic facts about additivity.

\prop{prop:1}{{\rm\cite[Proposition 1]{jord1}} Let ${\cal P},\F
\subseteq \real^{\real}$.  Then,
\begin{description}
\item[(i)] $\F=\emptyset$ if and only if
$\add(\F)=1$,
\item[(ii)] $\F=\real^{\real}$ if and only if
$\add(\F)=|2^{\cuum}|^+$,
\item[(iii)] if $\F\subseteq{\cal P}$ then
$\add(\F)\leq\add({\cal P})$, and
\item[(iv)] if $\F\neq\emptyset$ then
$2=\add(\F) \text{ if and only if }\F - \F\neq\real^{\real}$.
\end{description}}
Given $\F\subseteq\real^{\real}$ the definition of additivity
implies that $\real^{\real}$ has the property
\[(*)\ \ \ \ \ \
\left(\forall G\in\left[\real^{\real}\right]^{<\add(\F)}\right)
(\exists f\in\real^{\real})(f+G\subseteq\F).\]
A natural question that arises is wether or not $\real^{\real}$ is
the only subset of $\real^{\real}$ to satisfy $(*)$.  In
particular, one might want to find the smallest cardinality  of a
family $F\subseteq\real^{\real}$ that satisfies $(*)$.  This
consideration leads to the definition of super-additivity.  If
$F\subseteq\real^{\real}$ we define the super-additvity of
$\F$ to be
\[\add^*(\F)=\min\{|F|\colon F\subseteq\real^{\real}\ \&\
\left(\forall G\in\left[\real^{\real}\right]^{<\add(\F)}\right)
(\exists f\in F)(f+G\subseteq\F)\}.\]
We list some basic facts about super-additivity

\prop{prop:2}{Let $\F,{\cal E}\subseteq\real^{\real}$.  Then,
\begin{description}
\item[(i)] $\F\in\{\emptyset,\real^{\real}\}$ if and only if
$\add^*(\F)=1$ and
\item[(ii)] if $\add(\F)=\add({\cal E})$ and
$\F\subseteq{\cal E}$ then
$\add^*(\F)\geq\add^*({\cal E})$.
\end{description}}
\proof
We show (i).  If $\F=\real^{\real}$ then, by
Proposition~\ref{prop:1}(ii),
$\add(\F)=(2^{\cuum})^+$.  Let
$G\in\left[\real^{\real}\right]^{<(2^{\cuum})^+}$.  Clearly,
$\charf{\emptyset}+G\subseteq\real^{\real}=\F$.  So,
$\add^*(\F)=1$.
If $\F=\emptyset$ then, by
Proposition~\ref{prop:1}(i),
$\add(\F)=1$.  Since $\left[\real^{\real}\right]^{<1}=\{\emptyset\}$
and $\charf{\emptyset}+\emptyset\subseteq \F$, it follows that
$\add^*(\F)=1$.  Suppose now that $\add^{*}(\F)=1$.  We show
that $\F\in\{\emptyset,\real^{\real}\}$.  Assume that
$\F\neq\emptyset$.  By Proposition~\ref{prop:1}(i), $\add(\F)>1$.
Since $\add^{*}(\F)=1$, there is an $h\in\real^{\real}$ such that
$h+g\in\F$ for any $g\in\real^{\real}$.  So,
$\F\subseteq\real^{\real}=h+\real^{\real}\subseteq\F$.  Thus,
$\F=\real^{\real}$.

We show (ii).  Let $\kappa=
\add(\F)=\add({\cal E})$.  Suppose
$F\subseteq\real^{\real}$ and $|F|<\add^*({\cal E})$.  Then, there
exists a $G\in\left[\real^{\real}\right]^{<\kappa}$ such that
$f+G$ is not contained in ${\cal E}$ for every $f\in F$.
But $\F\subseteq{\cal E}$, so $f+G$ is not contained in
$\F$ for every $f\in F$.  Thus,
$\add^*(\F)\geq\add^*({\cal E})$.\qed

Next we point out a basic relationship between additivity and
super-additivity.
\prop{prop:3}{If $\F\notin\{\real^{\real},\emptyset\}$ then
\[\max\{\add(\F),\add(\real^{\real}\setminus\F)\}
\leq \add^{*}(\F).\]}
\proof
We first show that $\add(\real^{\real}\setminus\F)
\leq\add^{*}(\F)$.
Let $F\subseteq\real^{\real}$ be a witness to the definition of
$\add^{*}(\F)$,
i.e., $|F|=\add^{*}(\F)$ and
\begin{equation*}
(\forall G\in \left[\real^{\real}\right]^{<\add(\F)}) (\exists f\in F)
(f+G\subseteq\F).
\end{equation*}
Since $\add(\F)\geq 2>1$ we see that $F$ also satisfies
\begin{equation*}
(\forall g\in\real^{\real}) (\exists f\in F) (f+g\in\F).
\end{equation*}
Since $\F=\real^{\real}\setminus(\real^{\real}\setminus\F)$
we see that
$\add^*(\F)=|F|\geq\add(\real^{\real}\setminus\F)$.

We show that $\add(\F)\leq\add^{*}(\F)$.
By way of contradiction assume
$\add(\F)>\add^{*}(\F)$.
Then there is an $F\subseteq\real$ such that
$|F|<\add(\F)$
and
\begin{equation}\label{eq:wart1}
\left(\forall G\in \left[\real^{\real}\right]^{<\add(\F)}\right)
(\exists f\in F)(f+G\subseteq\F).
\end{equation}
Since $\F\neq\real^{\real}$, for every
$f\in F$ there is a $g_f\in\real^{\real}$ such that $f+g_f\notin\F$.
Let $G=\{g_f\colon f\in F\}$.
Notice that $|G|\leq |F|<\add(\F)$.
By (\ref{eq:wart1}) there is an $f\in F$ such that $f+G\subseteq\F$.
In particular, $f+g_f\in\F$ but this contradicts the choice of $g_f$.
Thus, $\add(\F)\leq\add^{*}(\F)$.\qed

\section{ The Results}

We will primarily be concerned with calculating the super-additivities
of the following families of functions from $\real$ into $\real$ and
their complements.  Some combinatorial characterizations of these
cardinals are also given.
We give general descriptions of these families that
will work for any function from one space to another where the spaces
are assumed to have the appropriate structure.

\begin{description}
\item[$\dar$:] $f\in Y^X$ is a Darboux function if and only if $f[C]$ is connected
in $Y$ for every connected subset $C$ of $X$.

\item[$\conn$: ] $f\in Y^X$ is a connectivity function if and only if the graph of
$f$ restricted to $C$ is connected in $X\times Y$ for every connected
subset $C$ of~$X$.

\item[$\acon$: ] $f\in Y^X$ is an almost continuous function
if and only if every open set in $X\times Y$ containing $f$ also
contains some continuous function $g\in Y^X$.

\item[$\ext$:] $f\in Y^X$ is an extendable function if and only if
there is a connectivity function $g\colon X\times [0,1]\to Y$ such that
$f(x)=g(x,0)$ for every $x\in X$.

\item[$\pr$: ] $f\in \real^{\real}$ is a perfect road function
if and only if for every $x\in\real$ there is a perfect set
$P\subset\real$ such that $x$ is a bilateral limit point
of $P$ and $f|_{P}$ is continuous at $x$.

\item[$\phc$: ] $f\in Y^X$ is a peripherally continuous function if and only if
for every $x\in X$ and pair of open sets $U\subset X$ and $V\subset Y$
such that $x\in U$ and $f(x)\in V$ there is an
open neighborhood $W$ of $x$ with $\cl(W)\subset U$ and
$f[\bd(W)]\subset V$, where $\cl(W)$ and $\bd(W)$ denote the boundary
and the closure of $W$, respectively.

\item[$\sz$: ] $f\in Y^X$ is a Sierpi\'{n}ski-Zygmund function if and only if
$f|_A$ is continuous for no set $A\subseteq X$ of cardinality $\cuum$.
\end{description}
The diagrams below describe the relations between the above families in
$\real{^\real}$ except $\sz$.  The symbol $\longrightarrow$ denotes
containment.  All inclusions are proper.

\begin{picture}(300,100)

\thicklines
\put(10,40){\makebox(0,0){$\ext$}}
\put(80,80){\makebox(0,0){$\acon$}}
\put(160,15){\makebox(0,0){$\pr$} }
\put(160,80){\makebox(0,0){$\conn$}}
\put(240,80){\makebox(0,0){$\dar$} }
\put(320,40){\makebox(0,0){$\phc$}}

\put(30,50){\vector(1,1){30}}
\put(100,80){\vector(1,0){30}}
\put(190,80){\vector(1,0){30}}
\put(260,80){\vector(3,-2){40}}
\put(30,35){\vector(4,-1){100}}
\put(190,10){\vector(4,1){100}}

\end{picture}

It is clear from the
definition of super-additivity and Proposition~\ref{prop:3} that
it would useful to know the additivities of these families.
Fortunately, there is good bit that we know about these values.


\prop{prop:four}{
\begin{description}\item[ ]
\item[(i)]   {\rm (Ciesielski, Rec\l aw \cite{CR})}
             $\add(\ext)=\add(\pr)=\cuum^{+}$ and $\add(\phc)=2^{\cuum}$;
\item[(ii)]  {\rm (Ciesielski, Miller \cite{CM})}
             $\add(\dar)=\add(\conn)=\add(\acon)=\diff_{\cuum}$;
\item[(iii)] {\rm (Ciesielski, Natkaniec \cite{CN})}
             $\add(\sz)=\same_{\cuum}$;
\item[(iv)]  {\rm (Ciesielski \cite{kc}(see\cite{jord1}))}
             $\add(\comp \phc)=\omega_{1}$;
\item[(v)]    {\rm (Jordan \cite{jord1})}
             $\add(\comp \pr)=\add(\comp \ext)=2^{\cuum}$;
\item[(vi)]  {\rm (Jordan \cite{jord1})}
             If $|[\cuum]^{<\cuum}|=\cuum$ then $\add(\comp
             \sz)=\diff_{\cuum}$;
\item[(vii)] {\rm (Jordan \cite{jord1})}
             If $|[\cuum]^{<\cuum}|=\cuum$ then
             $\same_{\cuum}=\add(\comp\dar)
             =\add(\comp\conn)=\add(\comp\acon)$.\qed
\end{description}}
We first calculate the super-additivies of the families of
functions we are concerned with.
\thm{thm:1}{If $\F\in\{\ext,\pr,\phc\}$ then
$\add^*(\F)=\add^*(\comp\F)=2^{\cuum}$.}
\thm{thm:2}{$\add^*(\acon)=\add^*(\conn)=\add^*(\dar)=\diff^{1}_{\cuum}$.}
\thm{thm:3}{$\add^*(\sz)=\same_{\cuum}^{1}$.}
\thm{thm:4}{If $|[\cuum]^{<\cuum}|=\cuum$ then $\same_{\cuum}^{1}
=\add^*(\comp\dar)=\add^*(\comp\conn)=\add^*(\comp\acon)$.}
\thm{thm:5}{If $|[\cuum]^{<\cuum}|=\cuum$ then
$\diff_{\cuum}^{1}=\add^*(\comp\sz)$.}

We also have a purely combinatorial result which will allow
us to say something about the values $\same_{\cuum}^{1}$ and
$\diff_{\cuum}^{1}$.

\thm{thm:7}{If $|\cuum^{<\cuum}|=\cuum$ and $\cuum=\lambda^+$ then
$\same_{\cuum}\leq\diff_{\cuum}=
\diff_{\cuum}^{1}=\same_{\cuum}^{1}$.}

Finally, we quote two consistency results.

\prop{prop:five}{ {\rm (Ciesielski, Natkaniec \cite{CN})}
Let $\lambda\geq\kappa\geq\omega_{2}$ be cardinals such that
$\cof(\lambda)>\omega_{1}$ and $\kappa$ is regular.  Then it is relatively consistent
with ZFC+CH that $2^{\cuum}=\lambda$ and $\add(\dar)=\add(\sz)=\kappa$. \qed}

\prop{prop:six}{ {\rm (Ciesielski, Natkaniec \cite{CN})}
Let $\lambda>\omega_{2}$ be a cardinal such that
$\cof(\lambda)>\omega_{1}$.  Then it is relatively consistent with ZFC+CH that
$2^{\cuum}=\lambda$, and $\add(\sz)=\cuum^{+}<2^{\cuum}=\add(\dar)$.\qed}

Since, CH implies that $|[\cuum^{<\cuum}]|=\cuum$,
Propositions~\ref{prop:five}, \ref{prop:six} and \ref{prop:four}
together with Theorem~\ref{thm:7} imply that

\cor{cor:1}{Let $\lambda\geq\kappa\geq\omega_{2}$ be cardinals such that
$\cof(\lambda)>\omega_{1}$ and $\kappa$ is regular.  Then it is relatively consistent
with ZFC+CH that $2^{\cuum}=\lambda$ and for
$\F\in\{\acon,\conn,\dar,\sz\}$
\begin{center}
$\add(\F)=\add^{*}(\F)=\add(\comp\F)=\add^*(\comp\F)=\kappa.$\qed
\end{center}}

\cor{cor:2}{Let $\lambda>\omega_{2}$ be a cardinal such that
$\cof(\lambda)>\omega_{1}$.  Then it is relatively consistent with
ZFC+CH that
$2^{\cuum}=\lambda$, and for
$\F\in\{\dar,\conn,\acon\}$
\begin{center}
$\add(\sz)=\add(\comp\F)=\cuum^{+}<
2^{\cuum}=\add(\F)=\add^*(\F)=\add^*(\comp\F)=\add(\comp\sz)
=\add^*(\comp\sz)=\add^*(\sz).$\qed
\end{center}}

We prove Theorems~\ref{thm:1} and \ref{thm:3} at this time.
We will prove the other Theorems in later
sections since the proofs are somewhat long.

\medskip

\noindent{\sc Proof of Theorem~\ref{thm:1}.}
Let $\F\in\{\ext,\pr,\phc\}$.  By (i) and (v)
of Proposition~\ref{prop:four}, we have
$\max\{\add(\F),\add(\real^{\real}\setminus\F)\}=2^{\cuum}$.
Since $\F\notin\{\real^{\real},\emptyset\}$
Proposition~\ref{prop:3} implies that $\add^*(\F)=2^{\cuum}=
\add^*(\real^{\real}\setminus\F)$.  \qed

To begin the proof of Theorem \ref{thm:3} we quote a theorem about
$\sz$ functions which may be found in \cite{sezy}.
\prop{prop:5}{ {\rm(Sierpi\'{n}ski, Zygumund \cite{sezy})}
$f\in\real^{\real}$ is in $\sz$ if and only if $|[f=h]|<\cuum$
for every continuous function $h$ defined on a $G_{\delta}$-set
of cardinality~$\cuum$.\qed}

\medskip

\noindent{\sc Proof of Theorem~\ref{thm:3}.  }
We show that $\add^*(\sz)\leq\same_{\cuum}^{1}$.  Let $H$ stand
for the family of all functions $h\in\real^{\real}$ such that
$h|_A$ is continuous on a $G_{\delta}$-set $A$ of cardinality $\cuum$ and
equal to zero elsewhere.  Note that $|H|=\cuum$.
Pick $F\subseteq\real^{\real}$ such that $|F|=\same_{\cuum}^{1}$
and
\begin{equation}\label{eq:sz20}
\left(\forall G\in\left[\real^{\real}\right]^{<\same_{\cuum}}\right)
(\exists f\in F)(\forall g\in G)(|[f=g]|<\cuum).
\end{equation}
We claim that $F$ satisfies
\begin{equation}\label{eq:sz21}
\left(\forall G\in\left[\real^{\real}\right]^{<\add(\sz)}\right)
(\exists f\in F)(f+G\subseteq\sz).
\end{equation}
Let $G\in\left[\real^{\real}\right]^{<\add(\sz)}$ be arbitrary.
By Proposition~\ref{prop:four}(iii) we have $|G|<\same_{\cuum}$.  It
is shown in \cite{CN} that $\same_{\cuum}>\cuum$.  It follows that
$\{h-g\colon g\in G\ \&\ h\in H\}$ is a set of cardinality
strictly less than $\same_{\cuum}$.  By (\ref{eq:sz20}) there is
an $f\in F$ such $|[f=h-g]|<\cuum$ for every $g\in G$ and $h\in H$.
So, by Proposition~\ref{prop:5}, $(f+g)|_A$ is continuous for no set $A$
of cardinality~$\cuum$ for every
$g\in G$.  Thus, $F$ satisfies (\ref{eq:sz21}) and
$\add^*(\sz)\leq\same_{\cuum}^{1}$.

We show that $\same_{\cuum}^{1}\leq\add^*(\sz)$.  Pick
$F\subseteq\real^{\real}$ such that $|F|=\add^*(\sz)$ and
\begin{equation}\label{eq:sz22}
\left(\forall G\in\left[\real^{\real}\right]^{<\add(\sz)}\right)
(\exists f\in F)(\forall g\in G)(f+g\in\sz).
\end{equation}
Let $F_1=\{-f\colon f\in F\}$, notice $|F_1|=|F|$.  We show that
$F_1$ satisfies
\begin{equation}\label{eq:sz23}
\left(\forall G\in\left[\real^{\real}\right]^{<\same_{\cuum}}\right)
(\exists f\in F_1)(\forall g\in G)(|[f=g]|<\cuum).
\end{equation}
Let $G\in\left[\real^{\real}\right]^{<\same_{\cuum}}$ be arbitrary.
By Proposition~\ref{prop:four}(iii) we have $|G|<\add(\sz)$.  By
(\ref{eq:sz22}) there is an $f\in F$ such that $f+g\in\sz$ for every
$g\in G$.  In particular, $|(f+g)^{-1}(\{0\})|<\cuum$ for every
$g\in G$.  It follows that $|[-f=g]|<\cuum$ for every $g\in G$.
Thus, $F_1$ satisfies (\ref{eq:sz23}) and
$\same_{\cuum}^{1}\leq\add^*(\sz)$.\qed

\section{Proof of Theorem \ref{thm:2}}

Our first goal is to show that
$\add^*(\dar)=\add^*(\conn)=\add^*(\acon)$.  To do this we will
need to define the following family of functions.  Let
$\dar_1$ denote the collection of all $f\in\real^{\real}$ such that
$f[(a,b)]=\real$ for every $a<b$.  Clearly, $\dar_1\subseteq\dar$.

\lem{lem:20}{$\add^*(\dar)=\add^*(\dar_1)$.}
\proof
It follows from \cite[Theorem 2.4]{CM} that
$\add(\dar)=\add(\dar_1)$.  By Proposition~\ref{prop:2}(ii) we
have $\add^*(\dar)\leq\add^*(\dar_1)$.  We show that
$\add^*(\dar_1)\leq\add^*(\dar)$.  Let $F\subseteq\real^{\real}$
be such that $|F|=\add^*(\dar)$ and
\begin{equation}\label{eq:20}
\left(\forall G\in\left[\real^{\real}\right]^{<\add(\dar)}\right)
(\exists f\in F)(f+G\subseteq\dar).
\end{equation}
We show that $F$ also satisfies
\begin{equation}\label{eq:21}
\left(\forall G\in\left[\real^{\real}\right]^{<\add(\dar_1)}\right)
(\exists f\in F)(f+G\subseteq\dar_1)
\end{equation}
which will complete the proof.  Let
$G\in\left[\real^{\real}\right]^{<\add(\dar_1)}$.  Note
that $|G|<\add(\dar)$.  Put $G_1=G\cup\{
g+r\cdot\charf{\rational}\colon g\in G\ \&\ r\in\real\}$.  Since
$\add(\dar)>\cuum$ \cite{CM} we have $G_1<\add(\dar)$.
By (\ref{eq:20}) there is an $f\in F$
such that $f+G_1\subseteq\dar$.  We claim that $f+G\subseteq\dar_1$.
By way of contradiction assume there is some $g\in G$ and $a<b$
such that $(f+g)[(a,b)]\neq\real$.  Since $(f+g)[(a,b)]$ is an
interval we may assume without loss of generality that there is
some $M>0$ which is an upper bound for $(f+g)[(a,b)]$.  Let
$q\in (a,b)\cap\rational$ and $k=M-(f+g)(q)$.  Now
$f+g+(k+1)\cdot\charf{\rational}\notin\dar$, since
$(f+g+(k+1))[(a,b)\setminus\rational]$ is bounded above
by $M$ but $(f+g+(k+1)\cdot\charf{\rational})(q)=M+1$.  However,
$g+(k+1)\cdot\charf{\rational}\in G_1$ so by the choice of $f$ we
have $f+g+(k+1)\cdot\charf{\rational}\in\dar$ giving
a contradiction.  Thus, $F$ satisfies (\ref{eq:21}).\qed

We will need a result of K.Kellum \cite[Theorem 1.2]{NATsurvey}.
\prop{prop:7}{There is a family ${\cal B}$ of closed subsets
of $\real^2$ such that $|{\cal B}|=\cuum$, $\domain(B)$ is a
non-degenerate interval for every
$B\in{\cal B}$ and $f\in\acon$ if and only if
$f\cap B\neq\emptyset$ for each $B\in{\cal B}$.}
\lem{lem:21}{$\add^*(\dar)=\add^*(\conn)=\add^*(\acon)$.}
\proof
By Proposition~\ref{prop:four}(ii) and Proposition~\ref{prop:2}(ii)
we have \[\add^*(\dar)\leq\add^*(\conn)\leq\add^*(\acon).\]  So it is
enough for us to show that $\add^*(\acon)\leq\add^*(\dar)$.  By
Lemma~\ref{lem:20} there is an
$F\subseteq\real^{\real}$ be such that $|F|=\add(\dar)$ and
\begin{equation}\label{eq:22}
\left(\forall G\in\left[\real^{\real}\right]^{<\add(\dar)}\right)
(\exists f\in F)(f+G\subseteq\dar_1).
\end{equation}
We claim that $F$ also satisfies
\begin{equation}\label{eq:23}
\left(\forall G\in\left[\real^{\real}\right]^{<\add(\acon)}\right)
(\exists f\in F)(f+G\subseteq\acon)
\end{equation}
which will complete the proof.  Let ${\cal B}$ be as in
Proposition~\ref{prop:7}.  For each $B\in{\cal B}$ let
$h_B\in\real^{\real}$ be such that
$h_B|_{{\text pr}_x(B)}\subseteq B$ and zero otherwise.  Let
$G\subseteq\real^{\real}$ and $|G|<\add(\acon)=\add(\dar_1)$.
Let $G_1=\{g-h_B\colon g\in G\ \&\ B\in{\cal B}\}$.
Since $\cuum<\add(\dar)$ we have $|G_1|<\add(\dar)$.
By (\ref{eq:22}) there is an $f\in F$ such that
$f+G_1\subseteq\dar_1$.  We claim that $f+G\subseteq\acon$.  Fix
$g\in G$.  Let $B\in {\cal B}$ be arbitrary.
Since $f+(g-h_B)\in\dar_1$ there is an $r\in{\text pr}_x(B)$
such that $f+(g-h_B)(r)=0$.  So $\langle r,(f+g)(r)\rangle=
\langle r,h_B(r)\rangle\in B$.  Thus,
$f+g\in\acon$.  \qed

To complete the proof of Theorem~\ref{thm:2} it is enough to
prove the following lemma.
\lem{lem:22}{$\add^*(\dar_1)=\diff^1_{\cuum}$.}
\proof
We show that $\add^*(\dar_1)\leq\diff^1_{\cuum}$.  Let
$\{P_{\alpha}\}_{\alpha\in\cuum}$ be a partition of $\real$ such
that $|P_{\alpha}|=\cuum$ for every $\alpha\in\cuum$ and for every
non-degenerate open interval $U$ there is an $\alpha\in\cuum$ such
that $P_{\alpha}\subseteq U$.  For each $\alpha\in\cuum$ let
$\{p_{\alpha,\beta}\colon\beta\in\cuum\}$ be an injective
enumeration of $P_{\alpha}$.  Let $F\subseteq\real^{\cuum}$ be such
that $|F|=\diff^1_{\cuum}$ and
\begin{equation}\label{eq:24}
\left(\forall G\in\left[\real^{\cuum}\right]^{<\diff_{\cuum}}\right)
(\exists f\in F)(|[f=g]|=\cuum).
\end{equation}
For each $f\in F$ define $f^*\in\real^{\real}$ by
$f^*(p_{\alpha,\beta})=f(\beta)$.  Let $F^*=\{f^*\colon f\in F\}$.
Note that $|F^*|=|F|=\diff^1_{\cuum}$.  It is enough to show that
$F^*$ satisfies
\begin{equation}\label{eq:25}
\left(\forall G\in\left[\real^{\real}\right]^{<\add(\dar_1)}\right)
(\exists f^*\in F^*)(f^*+G\subseteq\dar_1).
\end{equation}
Let $G\in\left[\real^{\real}\right]^{<\add(\dar_1)}$.  By
\cite[Theorem 2.4]{CM} and Proposition~\ref{prop:four}(ii)
$\add(\dar_1)=\add(\dar)=\diff_{\cuum}$.  So $|G|<\diff_{\cuum}$.
For each $\alpha\in\cuum$, $g\in G$, and $r\in\real$ let
$g_{\alpha,r}\in\real^{\cuum}$ be defined by
$g_{\alpha,r}(\beta)=r-g(p_{\alpha,\beta})$ for each $\beta\in\cuum$.
Let $G_1=\{g_{\alpha,r}\colon r\in\real\ \&\ g\in G\}$.  Since
$\diff_{\cuum}>\cuum$ it follows that $|G_1|<\diff_{\cuum}$.
By (\ref{eq:24}) there is an $f\in F$ such that
$|[f=g_1]|=\cuum$ for every $g_1\in G_1$.  We claim that
$f^*+g\in\dar_1$ for every $g\in G$.  Fix $g\in G$, $r\in\real$,
and a non-degenerate open interval $U$.  We must show that
$r\in (f^*+g)[U]$.  By the way we defined our partition there
is an $\alpha\in\cuum$ such that $P_{\alpha}\subseteq U$.
Let $\beta\in [f=g_{\alpha,r}]$.  Then
$f^*(p_{\alpha,\beta})=r-g(p_{\alpha,\beta})$.  So,
$(f^*+g)(p_{\alpha,\beta})=r$ and
$p_{\alpha,\beta}\in P_{\alpha}\subseteq U$.  Thus, $F^*$ satisfies
(\ref{eq:25}) establishing the inequality.

We now show that $\add^*(\dar_1)\geq\diff^1_{\cuum}$.  Let
$F\subseteq\real^{\real}$ be such that $|F|=\add^*(\dar_1)$
and
\begin{equation}\label{eq:26}
\left(\forall G\in\left[\real^{\real}\right]^{<\add(\dar_1)}\right)
(\exists f\in F)(f+G\subseteq\dar_1).
\end{equation}

Let $\Theta\in\real^{\real}$
be an additive function such that
$|\Theta^{-1}(r)|=\cuum$ for every $r\in\real$.
Let $F_1=\{\Theta\circ f\colon f\in F\}$, note that $|F_1|\leq |F|$.
It is enough for us to show that $F_1$ satisfies
\begin{equation}\label{eq:27}
\left(\forall G\in\left[\real^{\real}\right]^{<\diff_{\cuum}}\right)
(\exists f_1\in F_1)(\forall g\in G)(|[f_1=g]|=\cuum).
\end{equation}
Let $G\in\left[\real^{\real}\right]^{<\diff_{\cuum}}$.
For each
$g\in G$ pick $g_1\in\real^{\real}$ such that $\Theta\circ g_1=-g$.
Let $G_1=\{g_1\colon g\in G\}$.  Notice that $|G_1|\leq |G|$.
By (\ref{eq:26})
there is an $f\in F$ such that for every $g_1\in G_1$ we have $f+g_1\in\dar_1$.
Put $f_1=\Theta\circ f$.
By our choice of $\Theta$ and the fact that $(f+g_1)[\real]=\real$, we have
\[|(f_1-g)^{-1}(0)|=|(\Theta\circ f+\Theta\circ g_1)^{-1}(0)|=
|\Theta\circ(f+g_1)^{-1}(0)|=\cuum.\]
In particular, for each $g\in G$ we have $|f_1=g|=\cuum$.
Thus, $F_1$ satisfies (\ref{eq:27}).\qed

\section{Proofs of Theorems \ref{thm:4} and \ref{thm:5}}
Our first goal will be to prove Theorem~\ref{thm:4}.  Towards this
end we introduce two more cardinals the first of which appears in
\cite{jord1}.
\[\same^*_{\cuum}
 =\min\{|F|\colon F\subseteq \cuum^{\cuum}\ \&\
(\forall G\in\left[\cuum^{\cuum}\right]^{\cuum})
(\exists f\in F)(\forall g\in G)(|[f=g]|=\cuum)\}.\]


\begin{description}
\item[ ]$\kappa_1=\min\{|F|\colon F\subseteq\left[\cuum^{\cuum}\right]^{\cuum}$
\item $\&\ \left(\forall G\in\left[\cuum^{\cuum}\right]^
{<\same^*_{\cuum}}\right)
(\exists A\in F)(\forall g\in G)(\exists f\in A)(|[f=g]|<\cuum)\}$.
\end{description}


\lem{lem:30}{If $|[\cuum]^{<\cuum}|=\cuum$ then
$\kappa_1\leq\add^*(\comp\acon)$.}
\proof
Let ${\cal B}$ be as in Proposition~\ref{prop:7}.  Enumerate ${\cal B}$
injectively by $\{B_{\alpha}\colon\alpha\in\cuum\}$.
For each $\alpha\in\cuum$ let $h_{\alpha}\in\real^{\real}$
be such that
$h_{\alpha}|_{{\text pr}_x(B_{\alpha})}\subseteq B_{\alpha}$ and
zero otherwise.  Let
$\{P_{\alpha}\}_{\alpha\in\cuum}$ be a partition of $\real$ such
that $P_{\alpha}\subseteq{\text pr}_x(B_{\alpha})$ and
$|P_{\alpha}|=\cuum$ for every $\alpha\in\cuum$.  For each
$\alpha\in\cuum$ let $\{p_{\alpha,\beta}\colon\beta\in\cuum\}$ be an injective
enumeration of $P_{\alpha}$.

Let $F\subseteq\real^{\real}$ be such that $|F|=\add^*(\comp\acon)$
and
\begin{equation}\label{eq:30}
\left(\forall G\in\left[\real^{\real}\right]^{<\add(\comp\acon)}\right)
(\exists f\in F)(f+G\subseteq\comp\acon).
\end{equation}
For each $f\in F$ and $\alpha\in\cuum$ define
$f_{\alpha}\in\real^{\cuum}$ so that
$f_{\alpha}(\beta)=(h_{\alpha}-f)(p_{\alpha,\beta})$.
For each $f\in F$ let
$A_f=\{f_{\alpha}\colon\alpha\in\cuum\}$.  Put
$F^*=\{A_f\colon f\in F\}$.  Note that $|F^*|\leq
|F|=\add^*(\comp\acon)$ and
$F^*\subseteq\left[\real^{\cuum}\right]^{\leq\cuum}$.  It is enough
for us to show that $F^*$ satisfies
\begin{equation}\label{eq:31}
\left(\forall G\in\left[\real^{\cuum}\right]^{<\same^*_{\cuum}}\right)
(\exists A\in F^*)(\forall g\in G)(\exists f\in A)(|[f=g]|<\cuum).
\end{equation}
Let $G\in\left[\real^{\cuum}\right]^{<\same^*_{\cuum}}$.
Since $|[\cuum]^{<\cuum}|=\cuum$, we have $\same_{\cuum}^*=\same_{\cuum}$ by
\cite[Corollary 12]{jord1}.  It follows from
Proposition~\ref{prop:four}(vii) that $|G|<\add(\comp\acon)$.
For every $g\in G$ define $g^*\in\real^{\real}$ so that
$g^*(p_{\alpha,\beta})=g(\beta)$.  Since
$|\{g^*\colon g\in G\}|\leq |G|<\add(\comp\acon)$, it follows
by (\ref{eq:30}) that there is an $f\in F$ such that
$f+g^*\notin\acon$ for every $g\in G$.  We show that $A_f\in F^*$
has the property that
\begin{equation}\label{eq:32}
(\forall g\in G)(\exists h\in A_f)(|[h=g]|<\cuum).
\end{equation}
Fix $g\in G$.  By
Proposition~\ref{prop:7} there is an
$\alpha\in\cuum$ such that $(f+g^*)\cap B_{\alpha}=\emptyset$.
It follows that
\begin{equation}\label{eq:33}
(f+g^*)|_{P_{\alpha}}\cap h_{\alpha}|_{P_{\alpha}}=\emptyset.
\end{equation}
Pick $f_{\alpha}\in A_f$.  By way of contradiction assume that
$f_{\alpha}(\beta)=g(\beta)$ for some $\beta\in\cuum$.  Then
$(h_{\alpha}-f)(p_{\alpha,\beta})=g(\beta)=g^*(p_{\alpha,\beta})$ but this
contradicts (\ref{eq:33}).  Thus, $[f_{\alpha}=g]=\emptyset$ so
$A_f$ satisfies (\ref{eq:32}).  Therefore, $F^*$ satisfies
(\ref{eq:31}). \qed

To continue the proof it will be useful for us to define the
following families of functions.  Let $\dar(\cuum)$ stand for
the set of all $f\in\real^{\real}$ with the property that
\[|f^{-1}(y)\cap (a,b)|=\cuum\] for all $a,b,y\in\real$ such that
$a<b$.  Let $\dar^*$ denote the set of Darboux functions $f$ which
are nowhere constant (i.e. if $a<b$ then $|f[(a,b)]|>1$).

\lem{lem:31}{$\add^*(\comp\dar)=\add^*(\comp\dar^*)$.}
\proof
By \cite[Lemma 25]{jord1} $\add(\comp\dar)=\add(\comp\dar^*)$.
It follws by Proposition~\ref{prop:2}(ii) that
$\add^*(\comp\dar^*)\leq\add^*(\comp\dar)$.

We show the other inequality.
Let $\cal{I}$ be the family of collections of mutually disjoint
non-degenerate open intervals.  Since there are continuum many
open intervals and the cardinality of any disjoint collection of
open intervals is at most $\omega$, it follows that
$|{\cal{I}}|=\cuum$.  For each $I\in{\cal{I}}$ pick
$h_{I}\in\dar(\cuum)$ such that $h_{I}(x)=0$ if $x$ is an endpoint
of any $i\in I$.  Let $k_{I}$ be defined by
$k_{I}(x)=\charf{\cup I}(x)\cdot h_{I}(x)$ for each $x\in\real$.
Let $K=\{k_{I}\colon I\in \cal{I}\}$. Note that $|K|=\cuum$.
Suppose that $F\subseteq\real^{\real}$ and $|F|<\add^*(\comp\dar)$.
Then by definition of $\add^*(\comp\dar)$ there is a
$G\subseteq\real^{\real}$ such that
$|G|<\add(\comp\dar)=\add(\comp\dar^*)$ and
\begin{equation}\label{eq:34}
(\forall f\in F)(\exists g\in G)(f+g\in\dar).
\end{equation}
Let $G_1=\{g+k\colon g\in G\ \&\ k\in K\}\cup G$.  Note that
$|G_1|<\add(\comp\dar^*)$ since $|K|=\cuum<\add(\comp\dar^*)$.
It is enough to show that $G_1$ satisfies
\begin{equation}\label{eq:35}
(\forall f\in F)(\exists g\in G)(f+g\in\dar^*).
\end{equation}
Let $f\in F$.  By (\ref{eq:34}) there is a $g\in G$ such that
$f+g\in\dar$.  If $f+g\in\dar^*$ there is nothing to do so assume
$f+g\in\dar\setminus\dar^*$.  The set
of points at which $f+g$ is constant form a
countable collection $J$ of mutally disjoint non-degenerate
open intervals such that $f+g$ is constant on each $j\in J$
and is nowhere-constant on $\real\setminus\bigcup J$.
Since $g+k_{J}\in G_1$, it is enough to show that
$(f+k_{J})+g\in\dar^{*}$.

We first show that $(f+k_{J})+g$
is nowhere-constant.  Let $x\in\real$ be arbitrary.  If
$x\in\cl\left(\bigcup J\right)$ then any open nieghborhood $U$
about $x$ contains a non-degenerate sub-interval $i$ of some
$j\in J$.  Thus,
\begin{equation}\label{equation:awaka}
((f+g)+k_{J})[U]\supseteq((f+g)+k_{J})[i]=\{r\}+k_{J}[i]=\{r\}+\real=\real,
\end{equation}
where $\{r\}=(f+g)[j]$.  So $(f+k_{J})+g$ is not constant
at $x$.  If $x\notin\cl\left(\bigcup J\right)$ then there is a
neighborhood $U\subseteq\real\setminus\cl\left(\bigcup J\right)$
of $x$ such that $k_{J}$ is equal to $0$ on $U$, and
$(f+k_{J}+g)|_{U}=(f+g)|_{U}$ which is non-constant on $U$.
So, $f+g$ is non-constant at $x$.  Thus, $(f+k_{J})+g$ is
nowhere-constant.

We now must show that $(f+k_{J})+g$ is Darboux.
Let $i\subseteq\real$ be a non-degenerate open interval.
If $i\cap j\neq\emptyset$ for some $j\in J$ then $i$ contains a
non-trival sub-interval of $j$, so, arguing as in
(\ref{equation:awaka}), $((f+k_{J})+g)[i]=\real$.
If $i\cap j=\emptyset$ for all $j\in J$ then
$((f+k_{J})+g)[i]=(f+g)[i]$.  In either case $((f+k_{J})+g)[i]$
is an interval.  Thus, $(f+k_{J})+g$ is Darboux.
So, $(f+k_{J})+g\in\dar^{*}$ and $G_1$ satisfies
(\ref{eq:35}) completing the proof.\qed

\lem{lem:32}{$\add^*(\comp\dar(\cuum))=\add^*(\comp\dar)$.}
\proof
By \cite[Lemma 27]{jord1} $\add(\comp\dar)=\add(\comp\dar(\cuum))$.
It follws by Proposition~\ref{prop:2}(ii) that
$\add^*(\comp\dar(\cuum))\leq\add^*(\comp\dar)$.

We show the other inequality.  By \cite[Lemma 26]{jord1} there is an
additive function $\Theta\in\real^{\real}$ such that
$\Theta\circ h\in\dar(\cuum)$ for every $h\in\dar^*$.  Notice
that $\Theta$ is a surjection.  Let
$F\subseteq\real^{\real}$ and $|F|<\add^*(\comp\dar)$.  For each
$f\in F$ pick $f_1\in\real^{\real}$ such that $\Theta\circ f_1=f$.
Let $F_1=\{f_1\colon f\in F\}$.  Note that $|F_1|\leq |F|<
\add^*(\comp\dar)$.  By the
definition of $\add(\comp\dar^*)$ and Lemma~\ref{lem:31} there
is a $G\subseteq\real^{\real}$ such that $|G|<\add(\comp\dar)$ and
\begin{equation}\label{eq:36}
(\forall f_1\in F_1)(\exists g\in G)(f_1+g\in\dar^*).
\end{equation}
For each $g\in G$ let $g_1=\Theta\circ g$.  Put
$G_1=\{g_1\colon g\in G\}$.  Note that $|G_1|\leq
\add(\comp\dar(\cuum))$.  We will be done if we show that $G_1$
satisfies
\begin{equation}\label{eq:37}
(\forall f\in F)(\exists g_1\in G_1)(f+g_1\in\dar(\cuum)).
\end{equation}
Let $f\in F$.  By (\ref{eq:36}) there is a $g\in G$ such that
$f_1+g\in\dar^*$.  We now have $f+g_1=(\Theta\circ f_1)+
(\Theta\circ g)=\Theta(f_1+g)\in\dar(\cuum)$.  Thus, $G_1$
satisfies (\ref{eq:37}).\qed


\lem{lem:33}{If $|[\cuum]^{<\cuum}|=\cuum$ then
$\add^*(\comp\dar)\leq\same^1_{\cuum}$.}
\proof
By Lemma~\ref{lem:32} it is enough to prove that
$\add^*(\comp\dar(\cuum))\leq\same^1_{\cuum}$.
Let $F\subseteq\real^{\real}$ be such that $|F|=\same^1_{\cuum}$
and
\begin{equation}\label{eq:38}
\left(\forall G\in\left[\real^{\real}\right]^{<\same_{\cuum}}\right)
(\exists f\in F)(\forall g\in G)(|[f=g]|<\cuum).
\end{equation}
It is enough for us to show that $F$ satisfies
\begin{equation}\label{eq:39}
\left(\forall G\in\left[\real^{\real}\right]^{<\add(\comp\dar(\cuum))}\right)
(\exists f\in F)(f+G\subseteq\comp\dar(\cuum)).
\end{equation}
Let $G\in\left[\real^{\real}\right]^{<\add(\comp\dar(\cuum))}$.
Since $|[\cuum]^{<\cuum}|=\cuum$ we have, by
\cite[Lemma 27]{jord1} and Proposition~\ref{prop:four}(vii), that
$|G|<\same_{\cuum}$.  By
(\ref{eq:38}) there is an $f\in F$ such that $|[f=-g]|<\cuum$ for
every $g\in G$.  In particular, $|(f+g)^{-1}(\{0\})|<\cuum$ for
each $g\in G$.  So $f+g\notin\dar(\cuum)$ for every $g\in G$ and so
$F$ satisfies (\ref{eq:39}).\qed


\noindent{\sc Proof of Theorem~\ref{thm:4}}
We have shown that \[\kappa_1\leq\add^*(\comp\acon)\leq
\add^*(\comp\conn)\leq
\add^*(\comp\dar)\leq\same^1_{\cuum}.\]  So it is enough to show that
$\same^1_{\cuum}\leq\kappa_1$.  Let $W=\bigcup\{\cuum^{\alpha}
\colon \alpha<\cuum\}$.  Note that $|W|=\cuum$ by
our assumption that $|[\cuum]^{<\cuum}|=\cuum$.
Let $V=\{\langle\alpha,\xi\rangle\colon\xi\leq\alpha<\cuum\}$.
Let $F\subseteq[\cuum^{V}]^{\cuum}$ be such that
$|F|=\kappa_1$ and
\begin{equation}\label{eq:310}
\left(\forall G\in\left[\cuum^{V}\right]^{<\same^*_{\cuum}}\right)
(\exists A\in F)(\forall g\in G)(\exists f\in A)(|[f=g]|<\cuum).
\end{equation}
For each $A\in F$ let $A=\{f_{\beta}\colon\beta\in\cuum\}$.
For each $A\in F$ let $f_A\in W^{\cuum}$ be such that
$f_A(\alpha)\in\cuum^{\alpha}$ and
$f_A(\alpha)(\beta)=f_{\beta}\langle\alpha,\beta\rangle$.  Let
$F^*=\{f_A\colon A\in F\}$.  Note that $|F^*|\leq |F|=\kappa_1$.
It is enough for us to show that $F^*$ satisfies
\begin{equation}\label{eq:311}
\left(\forall G\in\left[W^{\cuum}\right]^{<\same_{\cuum}}\right)
(\exists f\in F^*)(\forall g\in G)(|[f=g]|<\cuum).
\end{equation}
Let $G\subseteq W^{\cuum}$ and $|G|<\same_{\cuum}$.
For every $g\in G$ let $g_1\in\cuum^{V}$ be defined by
$g_1\langle\alpha,\beta\rangle=g(\alpha)(\beta)$ for all
$\beta\in\domain(g)$ and zero otherwise.  Let
$G_1=\{g_1\colon g\in G\}$ and notice that
$|G_1|<\same_{\cuum}\leq\same_{\cuum}^*$.  By (\ref{eq:310}) there
is an $A\in F$ such that
\begin{equation}\label{eq:312}
(\forall g_1\in G_1)(\exists f\in A)(|[f=g_1]|<\cuum).
\end{equation}
We claim that $|[f_A=g]|<\cuum$ for every $g\in G$.  Fix
$g\in G$.  There is an $f_{\beta}\in A$ such that
$|[g_1=f_{\beta}]|<\cuum$.  Thus, for all but strickly less than
$\cuum$-many $\alpha>\beta$ we have
$g(\alpha)(\beta)=g_1\langle\alpha,\beta\rangle\neq
f_{\beta}\langle\alpha,\beta\rangle=f_A(\alpha)(\beta)$.  It
follows that $|[g=f_A]|<\cuum$.  So, by (\ref{eq:312}) the claim is proved.
Therefore, $F^*$ satisfies (\ref{eq:311}).\qed

We now work to prove Theorem~\ref{thm:5}.  Towards this end
we define some other cardinals.

\[\diff_{\cuum}^*=\min\{|F|\colon F\subseteq\cuum^{\cuum}\ \&\
\left(\forall G\in\left[\cuum^{\cuum}\right]^{\cuum}\right)
(\exists f\in F)(\forall g\in G)(|[g=f]|<\cuum)\}.\]


\begin{description}
\item[ ]$\kappa_2=\min\{|F|\colon F\subseteq
\left[\cuum^{\cuum}\right]^{\cuum}$
\item[ ] $\&\ \left(\forall G\in\left[\cuum^{\cuum}\right]^{<\diff^*_{\cuum}}\right)
(\exists A\in F)(\forall g\in G)(\exists f\in A)(|[f=g]|=\cuum)\}$.
\end{description}

\lem{lem:401}{If $|[\cuum]^{<\cuum}|=\cuum$ then $\kappa_2\leq\add^*(\comp\sz)\leq\diff^1_{\cuum}$.}
\proof
Let $H$ stand
for the family of all functions $h\in\real^{\real}$ such that
$h|_A$ is continuous for some $G_{\delta}$ set $A$ of cardinality $\cuum$ and
equal to zero elsewhere.  Note that $|H|=\cuum$.

We show that $\kappa_2\leq\add^*(\comp\sz)$.  Let
$F\subseteq\real^{\real}$ be such that $|F|=\add^*(\comp\sz)$ and
\begin{equation}\label{eq:313}
\left(\forall G\in\left[\real^{\real}\right]^{<\add(\comp\sz)}\right)
(\exists f\in F)(f+G\subseteq\comp\sz).
\end{equation}
For each $f\in F$ let $A_f\in[\real^{\real}]^{\cuum}$ be defined
by $\{h-f\colon h\in H\}$.  Let $F^*=\{A_f\colon f\in F\}$.  Notice
that $|F^*|\leq |F|=\add^*(\comp\sz)$.  It is enough for us to show that
$F^*$ satisfies
\begin{equation}\label{eq:314}
\left(\forall G\in\left[\real^{\real}\right]^{<\diff^*_{\cuum}}\right)
(\exists A\in F^*)(\forall g\in G)(\exists f\in A)(|[f=g]|=\cuum).
\end{equation}
Let $G\subseteq\real^{\real}$ be such that $|G|<\diff^*_{\cuum}$.
Since $|[\cuum]^{<\cuum}|=\cuum$, we have by \cite[Corollary 13]{jord1}
$\diff_{\cuum}^*=\add(\comp\sz)$.  So
$|G|<\add(\comp\sz)$.  By (\ref{eq:313}) there is an $f\in F$
such that $f+G\subseteq\comp\sz$.  So for each $g\in G$ there
is, by Proposition~\ref{prop:5}, an $h\in H$ such that
$|[f+g=h]|=\cuum$.  It follows that for every $g\in G$ there
is a $k\in A_f$ such that $|[k=g]|=\cuum$.  Thus, $F^*$ satisfies
(\ref{eq:314}).

We now show that $\add^*(\comp\sz)\leq\diff^1_{\cuum}$.
Let $F\subseteq\real^{\real}$ be such that $|F|=\diff^1_{\cuum}$
and
\begin{equation}\label{eq:315}
\left(\forall G\in\left[\real^{\real}\right]^{<\diff_{\cuum}}\right)
(\exists f\in F)(\forall g\in G)(|[f=g]|=\cuum).
\end{equation}
It is enough for
for us to show that $F$ satisfies
\begin{equation}\label{eq:316}
\left(\forall G\in\left[\real^{\real}\right]^{<\add(\comp\sz)}\right)
(\exists f\in F)(\forall g\in G)(f+g\in\comp\sz).
\end{equation}
Let $G\subseteq\real^{\real}$ and $|G|<\add(\comp\sz)$.  Notice
that since $|[\cuum]^{<\cuum}|=\cuum$ we have, by
\cite[Corollary 13]{jord1}, $|G|<\diff_{\cuum}$.  By
(\ref{eq:315}) there is an $f\in F$ such that
$|[f=-g]|=\cuum$ for all $g\in G$.  So
$|(f+g)^{-1}(\{0\})|=\cuum$ which implies that $f+g\in\comp\sz$.
Therefore, $F$ satisfies (\ref{eq:316}).\qed

To finish the proof of Theorem \ref{thm:5} it is enough, by
Lemma~\ref{lem:401}, to prove the following lemma


\lem{lem:41}{If $|[\cuum]^{<\cuum}|=\cuum$ then
$\diff_{\cuum}^1=\kappa_2$.}
\proof
First notice that Lemma~\ref{lem:401} provides the inequality
$\kappa_2\leq\diff_{\cuum}^1$.
We show that $\diff_{\cuum}^1\leq\kappa_2$.
Let $W=\bigcup\{\cuum^{\alpha}\colon \alpha<\cuum\}$.
Note that $|W|=\cuum$ by
our assumption that $|[\cuum]^{<\cuum}|=\cuum$.
Let $V=\{\langle\alpha,\xi\rangle\colon\xi\leq\alpha<\cuum\}$.
Let $F\subseteq[W^{\cuum}]^{\cuum}$ be such that
$|F|=\kappa_2$ and
\begin{equation}\label{eq:317}
\left(\forall
G\in\left[W^{\cuum}\right]^{<\diff^*_{\cuum}}\right)
(\exists A\in F)(\forall g\in G)(\exists f\in A)(|[f=g]|=\cuum).
\end{equation}
For each $A\in F$ let $A=\{f_{\alpha}\colon\alpha\in\cuum\}$ and
define $f_A\in\cuum^V$ by $f_A\langle\alpha,\beta\rangle=
f_{\beta}(\alpha)(\beta)$ if $\beta\in\domain(f_{\beta}(\alpha))$ and
zero otherwise.  Let $F^*=\{f_A\colon A\in F\}$.  Notice that
$|F^*|\leq |F|=\kappa_2$.  It is enough for us to show that
$F^*$ satisfies
\begin{equation}\label{eq:318}
\left(\forall G\in\left[\cuum^V\right]^{<\diff_{\cuum}}\right)
(\exists f\in F)(\forall g\in G)(|[f=g]|=\cuum).
\end{equation}
Let $G\subseteq\cuum^V$ be such that $|G|<\diff_{\cuum}$.
For each $g\in G$ define $g_1\in W^{\cuum}$ by
$g_1(\alpha)(\beta)=g\langle\alpha,\beta\rangle$ where
$\domain(g_1(\alpha))=\alpha+1$.  Let $G_1=\{g_1\colon g\in G\}$ and
notice that $|G_1|\leq |G|<\diff_{\cuum}\leq\diff_{\cuum}^*$.  By (\ref{eq:317})
there is an $A\in F$ such that
\begin{equation}\label{eq:319}
(\forall g\in G_1)(\exists f\in A)(|[f=g]|=\cuum).
\end{equation}
We claim that $f_A\in F^*$ has the property that $|[f_A=g]|=\cuum$
for every $g\in G$.  Fix $g\in G$.  By (\ref{eq:319}) there is an
$\beta\in\cuum$ such that $|[g_1=f_{\beta}]|=\cuum$.  In particular,
\begin{eqnarray}
\cuum
& = & |\{\alpha>\beta\colon f_{\beta}(\alpha)=g_1(\alpha)\}| \notag \\
& \leq & |\{\alpha>\beta\colon f_{\beta}(\alpha)(\xi)=g_1(\alpha)(\xi)
\text{ for all }\xi<\alpha\}| \notag \\
& \leq & |\{\alpha>\beta\colon f_{\beta}(\alpha)(\beta)=
g_1(\alpha)(\beta)\}| \notag \\
& \leq & |\{\alpha>\beta\colon
f_A\langle\alpha,\beta\rangle=g\langle\alpha,\beta\rangle\}| \notag \\
& \leq  & |[f_A=g]|.  \notag
\end{eqnarray}
Thus, $F^*$ satisfies (\ref{eq:318}) which implies that
$\diff^1_{\cuum}\leq\kappa_2$. \qed


\section{Proof of Theorem~\ref{thm:7}}
To prove Theorem \ref{thm:7} we will need a few lemmas and
some more cardinals.


\begin{description}
\item[ ]
$\dom_{\cuum}
=\min\{|F|\colon F\subseteq \cuum^{\cuum}\ \&\
(\forall g\in\cuum^{\cuum})
(\exists f\in F) (|[f\leq g]|<\cuum)\}$,
\item[ ]
$\unb_{\cuum}
=\min\{|F|\colon F\subseteq \cuum^{\cuum}\ \&\
(\forall g\in \cuum^{\cuum})(\exists f\in F)
(|[g\leq f]|=\cuum)\}$.
\end{description}
The numbers $\unb_{\cuum}$ and $\dom_{\cuum}$ are analogs of the
bounding number
$\unb=\unb_{\omega}$ and the dominating number $\dom=\dom_{\omega}$.
We will use the following proposition from \cite[Lemma 31]{jord1} a number of times thoughout
this section.
\prop{prop:40}{If $\cuum=\lambda^{+}$ then the set
$\{\langle\alpha,\beta\rangle\in\cuum^{2}\colon\beta\leq f(\alpha)\}$
is the union of $\lambda$-many
functions in $\cuum^{\cuum}$ for every
$f\in\cuum^{\cuum}$.}
Our first goal will be to show that under the assumption of
$|[\cuum]^{<\cuum}|=\cuum$ and $\cuum=\lambda^+$ we have
$\diff_{\cuum}=\diff^1_{\cuum}=\dom_{\cuum}$.


\lem{lem:42}{If $\cuum=\lambda^+$ then
$\dom_{\cuum}=\diff^*_{\cuum}\geq\kappa_2$.}
\proof
By \cite[Lemma 33]{jord1} $\diff^*_{\cuum}=\dom_{\cuum}$ under
the assumption of $\cuum=\lambda^+$ so it is enough for us to
show that $\kappa_2\leq\dom_{\cuum}$.

Let $F\subseteq\cuum^{\cuum}$ be such that $|F|=\dom_{\cuum}$ and
\begin{equation}\label{eq:45}
(\forall g\in\cuum^{\cuum})
(\exists f\in F) (|[f\leq g]|<\cuum).
\end{equation}
For each $f\in F$ we may
find, using Proposition~\ref{prop:40}, an
$A_f\in [\cuum^{\cuum}]^{\lambda}$ such that
$\{\langle\alpha,\beta\rangle\in\cuum^{2}\colon\beta\leq
f(\alpha)\}=\bigcup A_f$.  Let $F^*=\{A_f\colon f\in F\}$.
Notice that $|F^*|\leq |F|=\dom_{\cuum}$.  It is enough for us to
show that $F^*$ satisfies
\begin{equation}\label{eq:46}
\left(\forall G\in\left[\cuum^{\cuum}\right]^{<\diff^*_{\cuum}}\right)
(\exists A\in F)(\forall g\in G)(\exists f\in A)(|[f=g]|=\cuum).
\end{equation}
Let $G\subseteq\cuum^{\cuum}$ and $|G|<\diff^*_{\cuum}=\dom_{\cuum}$.
Since $|G|<\dom_{\cuum}$ there is an $h\in\cuum^{\cuum}$ such that
$|[g\leq h]|=\cuum$ for every $g\in G$.  By (\ref{eq:45}) there is
an $f\in F$ such that $|[f\leq h]|<\cuum$ so $|[g\leq f]|=\cuum$
for every $g\in G$.  We claim that $A_f$ has the property that
\begin{equation}\label{eq:47}
(\forall g\in G)(\exists h\in A_f)(|[h=g]|=\cuum).
\end{equation}
Let $g\in G$.  By the choice of $f$ we have
$|g\cap(\bigcup A_f)|=\cuum$.  Since $|A_f|=\lambda$ it follows that
$|[g=h]|=\cuum$ for some $h\in A_f$.  Thus, $A_f$ satisfies
(\ref{eq:47}). Therefore, $F^*$ satisfies (\ref{eq:46}).  \qed

\lem{lem:43}{If $|[\cuum]^{<\cuum}|=\cuum$ and $\cuum=\lambda^+$
then $\diff_{\cuum}=\diff^1_{\cuum}=\dom_{\cuum}$.}
\proof
By \cite[Theorem 10]{jord1} and \cite[Lemma 33]{jord1}
$\diff_{\cuum}=\dom_{\cuum}$.  By Lemmas~\ref{lem:41} and
\ref{lem:42} and Theorem~\ref{thm:2} we have $\diff_{\cuum}=
\add(\dar)\leq\add^*(\dar)=\diff^1_{\cuum}\leq\dom_{\cuum}$.\qed

\lem{lem:44}{If $|[\cuum]^{<\cuum}|=\cuum$ and $\cuum=\lambda^+$
then $\diff_{\cuum}=\same^1_{\cuum}=\dom_{\cuum}$.}
\proof
By Lemma~\ref{lem:43} it is enough for us to show that
$\dom_{\cuum}=\same^1_{\cuum}$.  Since $\cuum\leq\same_{\cuum}$
it is easy to check that $\diff^*_{\cuum}\leq\same^1_{\cuum}$.
So by Lemma~\ref{lem:43} we have $\dom_{\cuum}\leq\same^1_{\cuum}$.
All we must do now is show that $\same^1_{\cuum}\leq\dom_{\cuum}$.

Let $F\subseteq\cuum^{\cuum}$ be such that $|F|=\dom_{\cuum}$ and
\begin{equation}\label{eq:48}
(\forall g\in\cuum^{\cuum})
(\exists f\in F) (|[f\leq g]|<\cuum).
\end{equation}
It is enough for us to show that $F$ satisfies
\begin{equation}\label{eq:49}
\left(\forall G\in\left[\cuum^{\cuum}\right]^{<\same_{\cuum}}\right)
(\exists f\in F)(\forall g\in G)(|[f=g]|<\cuum).
\end{equation}
Let $G\subseteq\cuum^{\cuum}$ and $|G|<\same_{\cuum}$.  By
\cite[Lemma 33]{jord1} $\same_{\cuum}=\unb_{\cuum}$.
Since $|G|<\unb_{\cuum}$ there is an $h\in\cuum^{\cuum}$ such that
$|[h\leq g]|<\cuum$ for all $g\in G$.  By (\ref{eq:48}) there is
an $f\in F$ such that $|[f\leq h]|<\cuum$.  It follows that
$|[f\leq g]|<\cuum$ for every $g\in G$.  In particular, we have
that $|[f=g]|<\cuum$ for every $g\in G$.  Thus, $F$ satisfies
(\ref{eq:49}).\qed

{\sc Proof of Theorem~\ref{thm:7}}
Lemmas~\ref{lem:43} and
\ref{lem:44} yield the equalities $\diff_{\cuum}=\diff_{\cuum}^1=\same^1_{\cuum}$.
The inequality $\same_{\cuum}\leq\same_{\cuum}^1$ is a consequence of
Proposition~\ref{prop:four}(iii) and Theorem~\ref{thm:3}. \qed

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