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\title{Cardinal numbers connected with adding Darboux-like
functions}

%\MathReviews{Primary:  26A15; Secondary: 03E75, 54A25.}
%\keywords{cardinal functions; extendable, Darboux, almost continuous
%and peripherially continuous functions; functions with perfect road. }


\author{DISSERTATION\\
Submitted to the College of Arts and Science\\
of\\
West Virginia University\\
In Partial Fulfillment of the Requirements for\\
The Degree of Doctor of Philosophy\\
by\\
Francis Edmund Jordan\\ \vspace{.5in}$1998$}
\date{1997}
%\address{418 Armstrong Hall, Morgantown, WV 
%\quad{\tt FJordan@math.wvu.edu} }

\date{}


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


\newcommand{\cof}{\operatorname{cf}}
\newcommand{\cl}{\operatorname{cl}}
\newcommand{\bd}{\operatorname{bd}}
\newcommand{\interior}{\operatorname{int}}
\newcommand{\F}{{\mathcal F}}
\newcommand{\G}{{\mathcal G}}
\newcommand{\baire}{{\mathcal B}}
\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{\swiat}{{\operatorname {Sw}}}
\newcommand{\quasi}{{\operatorname {Qc}}}
\newcommand{\cont}{{\operatorname {cont}}}
\newcommand{\domain}{{\operatorname{dom}}}
\newcommand{\osc}{{\operatorname{osc}}}
\newcommand{\acc}{{\operatorname{SC}}}
\newcommand{\wacc}{{\operatorname{C}}}
\newcommand{\norm}{{\operatorname{norm}}}
\newcommand{\supp}{{\operatorname{supp}}}
\newcommand{\cliq}{{\operatorname{cliq}}}
\newcommand{\cnwd}{{\mathcal K}}
\newcommand{\meager}{{\mathcal M}}
\newcommand{\dense}{{\mathcal G}}
\newcommand{\cuum}{{\frak c}}
\newcommand{\compact}{{\mathcal P}}
\newcommand{\add}{{\operatorname{A}}}
\newcommand{\lin}{\operatorname{LIN}}
\newcommand{\sz}{{\operatorname {SZ}}}
\newcommand{\rep}{{\operatorname {rep}}}
\newcommand{\diff}{{e}}
\newcommand{\same}{{d}}
\newcommand{\dom}{{D}}
\newcommand{\unb}{{b}}
\newcommand{\comp}{{\neg}}
\newcommand{\floor}[1]{{\lfloor #1 \rfloor}}


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\newtheorem{theorem}{Theorem}[section]
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\newcommand{\thm}[2]{\begin{theorem}\label{#1}#2\end{theorem}}
\newcommand{\cor}[2]{\begin{corollary}\label{#1}#2\end{corollary}}
\newcommand{\prop}[2]{\begin{proposition}\label{#1}#2\end{proposition}}
\newcommand{\lem}[2]{\begin{lemma}\label{#1}#2\end{lemma}}
\newcommand{\prob}[2]{\begin{problem}\label{#1}#2\end{problem}}
\newcommand{\ex}[2]{\begin{example}\label{#1}#2\end{example}}

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




\begin{document}
\pagenumbering{roman}
\maketitle
\pagestyle{headings}
\tableofcontents
\listoffigures

\newpage

\pagenumbering{arabic}
\pagestyle{plain}

\chapter{Preliminaries}

\section{Introduction}\label{intro}
We say a class of functions is Darboux-like if it generalizes 
some notion of continuity.  An old problem in real analysis is 
the characterization of functions which are derivatives.  In \cite{DAR} 
G. Darboux showed that derivatives of real valued functions defined
on $\real$ possess the intermediate value property, that is, if
$a<b$ and $c$ is between $f(a)$ and $f(b)$, then there is an $x\in
(a,b)$ such that $f(x)=c$.  Today the Darboux functions from 
$\real$ to $\real$ are defined to be precisely those functions which satisfy the 
intermediate value property.  Another well known class of
Darboux-like functions is the almost continuous functions which
were introduced in 1959 by J. Stallings \cite{STAL} in connection
with fixed point theorems in topology.  Below, we define the family of almost
continuous functions as well as many other Darboux-like families. 
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 topological structure.  To find out more about these 
families see the survey articles by Gibson and Natkaniec 
\cite{GNsurvey}, Natkaniec 
\cite{NATsurvey}, and Ciesielski \cite{CiSurvey}.  For notation see the next section.  


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

\item[$\conn(X,Y)$:] $f\in Y^X$ is a {\em 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[$\acon(X,Y)$:] $f\in Y^X$ is an {\em 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[$\ext(X,Y)$:] $f\in Y^X$ is an {\em 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 {\em 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 $f|_{P}$, that is, the restriction of $f$ to $P$, 
is continuous at $x$.

\item[$\phc(X,Y)$:] $f\in Y^X$ is a {\em 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.

\newpage

\item[$\quasi(X,Y)$:] $f\in Y^X$ is a {\em quasi-continuous} function  if
and only if  at each point $p\in X$ the following conditon holds: 
for every open set $U\subseteq X$ with 
$p\in U$ and open set $V\subseteq Y$ with $f(p)\in V$ there exists  a
non-empty open set $W\subseteq U$ such that $f[W]\subseteq V$.

\item[$\swiat$:] $f\in \real^{\real}$ is a {\em strong {\'S}wi{\c
a}tkowski} function if and only if for every $a<b$ and for every $y$ strictly
between $f(a)$ and $f(b)$ there is an $x\in (a,b)$ such that 
$f(x)=y$ and $x$ is a point of continuity for $f$.
\end{description}

A family of functions which stands in opposition to the
Darboux-like functions is the family of
{\em Sierpi\'{n}ski-Zygmund} functions.  We say $f\in Y^X$ is a
Sierpi\'{n}ski-Zygmund function ($f\in\sz(X,Y)$) if and only if $f$ 
is continuous on no set of cardinality $|X|$.  It is interesting 
to note that under certain set-theoretical assumptions 
$\sz(\real,\real)\cap\acon(\real,\real)\cap\pr(\real,\real)
\neq\emptyset$, while on the other hand there is a 
model of set theory ZFC such that 
$\acon(\real,\real)\cap\sz(\real,\real)=\emptyset$ \cite{BCN}.  When the context 
is clear we will often drop the symbols $X$ and $Y$, for example, if we are talking 
about functions from $\real$ to $\real$ we will write $\dar$ instead of 
$\dar(\real,\real)$.

The diagram below describes the relations between some of the above families in
$\real{^\real}$.  The symbol $\longrightarrow$ denotes
containment.  All inclusions are proper.  See Brown, Humke, and 
Laczkovich \cite{BHL} for all inclusions except 
$\ext\subseteq\comp\sz$ which follows directly from 
the work of Rosen, Gibson, and Roush \cite{RGR}.  The characteristic 
function of a point shows that the inclusion $\ext\subseteq\comp\sz$ 
is proper.


\begin{figure}[h]
\begin{picture}(300,100)

\thicklines
%\put(10,40){\makebox(0,0){$n=1$:}}
\put(60,40){\makebox(0,0){$\ext$}}
\put(110,80){\makebox(0,0){$\comp\sz$}}
\put(120,40){\makebox(0,0){$\acon$}}
\put(185,10){\makebox(0,0){$\pr$} }
\put(180,40){\makebox(0,0){$\conn$}}
\put(240,40){\makebox(0,0){$\dar$} }
\put(310,40){\makebox(0,0){$\phc$}}

\put(70,45){\vector(1,1){30}}
\put(70,40){\vector(1,0){30}}
\put(130,40){\vector(1,0){30}}
\put(190,40){\vector(1,0){30}}
\put(260,40){\vector(1,0){40}}
\put(70,35){\vector(4,-1){100}}
\put(200,10){\vector(4,1){100}}

\end{picture}
\caption{Containments for $\real^{\real}$.}
\label{fig:1}
\end{figure}

\pagebreak%\newpage

The above diagram changes significantly for $(\real)^{\real^n}$ when 
$n>1$.  The inclusion $\conn\subseteq\phc$ is due to Hamilton \cite{HAM} and Stallings 
\cite{STAL} and the inclusion $\phc\subseteq\conn$ is due to Hagan \cite{HAG}.  
The inclusion $\conn\subseteq\acon\cap\dar$ is proved in \cite{STAL}, 
and a Baire class one example showing the inclusion is proper can 
be found in \cite[example 1]{RGR}.  The 
relation of $\acon$ and $\dar$ in the diagram above is also 
established in Baire class one.  For $\dar\setminus\acon$ see 
\cite{NATsurvey} or for the Baire class one example see 
\cite{NATsurvey1}.  For $\acon\setminus\dar$ see \cite{NATsurvey}.  
Inclusions $\acon\subseteq\comp\sz$ and $\dar\subseteq\comp\sz$ are 
shown in the present work, the characteristic function of a point 
shows that both inclusions are proper.  

\begin{figure}[h]
\begin{picture}(300,100)

\thicklines
%\put(20,40){\makebox(0,0){$n>1$:}}
\put(100,40){\makebox(0,0){$\ext=\conn=\phc$}}
\put(205,40){\makebox(0,0){$\dar\cap\acon$}}
\put(270,75){\makebox(0,0){$\dar$}}
\put(270,10){\makebox(0,0){$\acon$}}
\put(325,40){\makebox(0,0){$\comp\sz$}}
\put(150,40){\vector(1,0){30}}
\put(230,40){\vector(1,1){30}}
\put(230,40){\vector(1,-1){30}}
\put(280,70){\vector(1,-1){30}}
\put(280,10){\vector(1,1){30}}
%\put(190,10){\vector(4,1){100}}

\end{picture}
\caption{Containments for $\real^{\real^n}$ when $n>1$.}
\label{fig:2}
\end{figure}

The contrast between the situation when we consider $\real^{\real}$ as 
opposed to $(\real)^{\real^n}$ where $n>1$ is further magnified 
when one realizes that the second diagram remains the same when we 
consider only the Baire class one functions in $\real^{\real^n}$ but, 
that the diagram 
for the Baire class one functions Figure~\ref{fig:3} where $n=1$ is different 
from both Figure~\ref{fig:1} and Figure~\ref{fig:2}.  
The containment of $\swiat\subseteq\quasi\cap\dar$ is straight 
forward and left without proof.  
To see that the containment is proper consider the following 
example \cite[p.10]{AM}:
\begin{equation*} g(x)=
\begin{cases} 1+x+\text{sin}(1/x)& \text{if $x>0$;}\\
-1+x+\text{sin}(1/x)& \text{if $x<0$}\\ 0& \text{if $x=0$.}
\end{cases}
\end{equation*}
That the containment $\dar\cap\quasi\subseteq\quasi$ is proper is 
shown by $\charf{(0,1)}$.  That the containment 
$\dar\cap\quasi\subseteq\dar$ 
is proper follows from the existance of a Darboux Baire class one 
function which is non-constant and identically zero outside the 
Cantor set \cite{BL}.   
The equalities 
$\dar=\F$ in Baire class one for $\F=\ext,\acon,\pr,\conn,\phc$ are
shown in \cite{BHL}, 
\cite{JB}, \cite{MAX}, \cite{KKWS}, and \cite{JY}, respectively.   

 
\begin{figure}[h]
\begin{picture}(300,100)

\thicklines
%\put(40,60){\makebox(0,0){$n=1$ Baire class one:}}
\put(250,40){\makebox(0,0){$\ext=\acon=\conn=\dar=\phc=\pr$}}
\put(30,40){\makebox(0,0){$\swiat$}}
\put(95,40){\makebox(0,0){$\dar\cap\quasi$}}
\put(170,70){\makebox(0,0){$\quasi$}}
\put(130,40){\vector(1,0){30}}
\put(40,40){\vector(1,0){30}}
\put(130,40){\vector(1,1){30}}
%\put(200,40){\vector(1,0){30}}
%\put(190,10){\vector(4,1){100}}

\end{picture}
\caption{Containments for Baire class 1 in $\real^{\real}$.}
\label{fig:3}
\end{figure}

While the Darboux-like families all capture some aspect of continuity 
they can be quite pathological.  This fact is illustrated by the following 
result.  
\noindent\prop{prop:1}{Every function in $\real^{\real^n}$ is the sum of 
$n+1$ functions from $\ext(\real^n,\real)$.  Every Baire class one 
function in $\real^{\real}$ is the sum of two Baire class one 
functions in $\swiat(\real,\real)$.}
\proof
The first statement was proven independently for the case $n=1$ by 
Ciesielski and Rec{\l}aw \cite{CR} and Rosen \cite{ROS}.  The result 
for $n>1$ is due to Ciesielski and Wojciechowski~\cite[Theorem 2.1]{CW}. 
The second result was shown by Maliszewski \cite{AM}.\qed

The first result of this kind for the general 
case was that 
every function in $\real^{\real}$ is the sum of two Darboux 
functions.  
It was stated without proof by A.~ Lindenbaum \cite{LIND}, and 
a proof was provided by W. ~Sierpi{\'n}ski \cite{SIER1}.  
Other papers leading up to Proposition~\ref{prop:1} in the 
general case are \cite{FAS} and \cite{KEL1}.  Papers associated with 
the Baire class one case are \cite{BCK}, \cite{CED1}, \cite{ERD}, \cite{KEL1}, 
\cite{AM1}, \cite{MARC1}, \cite{MINSK1}, and \cite{PUPU}.  

A result that ultimately led to the formulation of the cardinal 
functions dealt with in this work is due to Fast \cite{FAS} which 
states that if $F\subseteq\real^{\real}$ and $|F|\leq\cuum$, then 
there is a $g\in\real^{\real}$ such that 
$g+F=\{g+f\colon g\in G\}\subseteq\dar$.  Later, Kellum \cite{KEL1} 
proved the analogous result for $\acon(\real,\real)$.  In 
\cite{NATsurvey} and \cite{NAT2} the notion of {\em additivity} is 
defined for families of functions in $\real^{\real}$.  For 
$\F\subseteq\real^{\real}$ the additivity of $\F$ is defined to be 
\begin{equation*}
\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})^{+}\}).
\end{equation*}
We refer to the cardinal $\add(\F)$ as the additivity of $\F$.  In 
this language the results of Fast and Kellum mentioned above become 
$\cuum<\add(\dar)$ and $\cuum<\add(\acon)$, respectively.  The exact 
additivities of all the Darboux-like families in $\real^{\real}$ 
have been calculated, 
and these results will be stated in Chapter~\ref{first_chapter}.  In the present work 
we will consider the additivities of the complements of the 
Darboux-like families and develop variations of additivity which 
will allow us to extend the known results in significant ways. 

\section{Notation}
Throughout 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 that a cardinal $\kappa$ is regular
provided that $\cof(\kappa)=\kappa$, otherwise we say $\kappa$ is singular.  
The symbol $\oplus$ will stand for
cardinal addition.   Recall that $\kappa\oplus\lambda=\kappa+\lambda$ if
$\kappa$ and
$\lambda$ are finite cardinals and that
$\kappa\oplus\lambda=\max\{\kappa,\lambda\}$ if either one of
$\kappa$ or $\lambda$ is an infinite cardinal.  For $\alpha\in\real$ we 
let $\floor{\alpha}$ denote the greatest integer less than or equal 
to $\alpha$.
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.  If $S\subseteq X\times Y$,
then the set $\{x\in X\colon\langle x,y 
\rangle\in S \text{ for some $y \in Y$}\}$ is denoted by $\domain(S)$ .  In particular, if 
$f$ is a function, then $\domain(f)$ is just the domain of $f$.  

We 
denote the topological closure, interior, and boundary of a set $A$ 
by $\cl(A)$, $\interior(A)$, and $\bd(A)$, respectively.  
We say that a subset $A$ of $\real$ is 
$\cuum$-dense provided that $|A\cap(a,b)|=\cuum$ for all 
$a,b\in\real$ such that $a<b$.  Given a topological space
$X$ and a natural  number $n$, we let $X^{n}$ stand for the product of the space
$X$ with itself $n$-times with the usual topology.  A topological 
space $X$ is said to be perfect if $X$ is compact and has no 
isolated points.  We say a 
$B\subseteq\real^n$ is a Bernstein set provided that $B$ and 
$\real^n\setminus B$ both have non-empty intersection with every 
perfect set $P\subseteq\real^n$.  
Recall that 
a function is said to be in Baire class one provided that it is 
the pointwise limit of continuous functions.  We denote the family 
of Baire class one functions from $X$ into $Y$ by $\baire_{1}(X,Y)$. 

When the 
appropriate algebraic structure exists and $G$ is a set, we will often 
write $f+G$ to denote $\{f+g\colon g\in G\}$, 
$-G$ to denote $\{-g\colon g\in G\}$, $G-G$ to denote $\{g-h\colon g,h\in G\}$; 
and for positive integers $n$, 
the set $\{g_{0}+\cdots+g_{n-1}\colon g_i\in G\}$ will be denoted by 
$nG$. 

%%%CHAPTER ONE
\chapter{Additivity in $\real^{\real}$}\label{first_chapter}
\section{Introduction}
Before we begin an investigation of the additivities of the Darboux-like 
and $\sz$ functions, we should say a word about why such cardinals 
are of interest.  In particular, we will look at some easy 
interpretations of the meaning of the cardinals associated with 
additivity.
  
Towards this end let us first give a more general 
defintion of additivity which will serve us well later.  In what follows 
we will assume that $X$ is a non-empty set.  Let $Y$ be an 
Abelian group, that is, $x+y=y+x$ for all $x,y\in Y$, with identity 
zero. We denote the function in $Y^X$ with range $\{0\}$ by $\theta$.  
The obvious definition for additivity in this situation for families 
$\F\subseteq Y^X$ is  
\begin{equation*}
\add(\F)=\min\left(\{|F|\colon F\subseteq Y^{X}\ \&\ (\forall  
g\in Y^{X}) (\exists f\in F) (f+g\notin \F)\}\cup \{(|Y^X|)^{+}\}\right).
\end{equation*}
We refer to the cardinal $\add(\F)$ as the additivity of
$\F$ in $Y^X$.  
Below, we list some basic facts about the additivity function 
which are stated in a less general way in \cite[Proposition~1]{jord1} 
which is itself a modification of \cite[Proposition~1.1]{CR}.  
\noindent\prop{prop1:1}{Let $\G,\F\subseteq Y^X$ and $|Y|>1$.  Then,
\begin{description}
\item[(i)] $\add(\F) = 1 \text{ if and only if } \F=\emptyset$; 
\item[(ii)] $\add(\F)\leq |Y^X| \text{ if and only if } \F\neq Y^X$; 
\item[(iii)] if $\F\subseteq\G$, then $\add(\F)\leq\add(\G)$; 
\item[(iv)] $2<\add(\F) \text{ if and only if } \F - \F = Y^X$.  
\end{description}} 
\proof 
We show (i).  If $\F=\emptyset$, then $\theta$ would 
have the property that $\theta+g\notin\F$ for all $g\in Y^X$.  
Thus, $\add(\F)=1$.  If $\F\neq\emptyset$, then there is some $f\in\F$.  
For any $g\in Y^X$ we have $(f-g)+g=f\in\F$ so $\add(\F)>1$.

We show (ii).  Suppose $\add(\F)<|Y^X|^+$.  
Then, there 
is an $F\subseteq Y^X$ with the property that for every 
$g\in Y^X$ there is some $f\in F$ such that $f+g\notin\F$.  
In particular, there is some 
$f\in F$ such that $f=f+\theta\notin\F$.  Thus, $\F\neq X^Y$.  
Suppose that $\add(\F)=|Y^X|^+$.  Since $|Y^X|<|Y^X|^+$ there 
is a $g\in Y^X$ such that $g+Y^X\subseteq\F$.  But, $Y^X$ is a group, 
so $g+Y^X=Y^X$.  Thus, $\F=Y^X$.   

We show (iii).  Suppose $|F|<\add(\F)$. Then, there is a $g\in Y^X$ such 
that $g+F\subseteq\F$.  Since $\F\subseteq\G$, we also have that 
$g+F\subseteq\G$.  Thus, $\add(\F)\leq\add(\G)$.  

We show (iv).  Suppose that $\F-\F=Y^X$.  
We show $\add(\F)>2$.  Let 
$f_{1},f_{2}\in Y^X$ be arbitrary and $F=\{f_{1},f_{2}\}$.  
We find  a $g\in Y^X$ such that $g+f_{1},g+f_{2}\in\F$.  
Since $\F-\F=Y^X$, 
there exists $h_{1},h_{2}\in\F$ such that 
$f_{1}-f_{2}=h_{1}-h_{2}$.  Let 
$g=h_{1}-f_{1}=h_{2}-f_{2}$.  Then 
$f_{i}+g=f_{i}+(h_{i}-f_{i})=h_{i}\in\F$ for $i=1,2$.  Thus, 
$\add(\F)>2$.
To see the other implication, suppose that $\F-\F\neq Y^X$.  
Pick $h\in Y^X\setminus(\F-\F)$ and put 
$F=\{\theta,h\}$.  
Let $g\in Y^X$ be arbitrary.  It is enough to show that 
$g+f\notin\F$  for some $f\in F$.  However, if 
$g=\theta+g\in\F$ and $h+g\in\F$ 
then $h\in\F-g\subseteq\F-\F$, contradicting our choice of $h$.  
Thus, $\add(\F)\leq 2$.\qed 

We prove a proposition which yields an interpretation of $\add$ 
in terms of covering type statements.

\noindent\prop{prop1:20}{Let $\F\subseteq Y^X$ and $\add(Y^X\setminus\F)\neq |Y^X|^+$.  
The additivity of $Y^X\setminus\F$ in 
$Y^X$ is the smallest cardinality $\kappa$ of a collection 
$F\subseteq Y^X$ with the property that 
\begin{equation}\label{eq1:1}
\bigcup\{f+\F\colon f\in F\}=Y^X.
\end{equation}}
\proof
We show that $\kappa\leq\add(Y^X\setminus\F)$.   
Suppose $F_1\subseteq Y^X$ is a witness to the definition of 
$\add(Y^X\setminus\F)$, i.e., $|F_1|=\add(Y^X\setminus\F)$ and 
\begin{equation}\label{eq1:2}
(\forall g\in Y^{X}) (\exists f\in F_1) (f+g\notin Y^X\setminus\F).
\end{equation}
We claim that $\bigcup\{-f+\F\colon f\in F_1\}=Y^X$.  To see this let 
$g\in Y^X$.  By (\ref{eq1:2}) there is an $f\in F_1$ such that 
$f+g\notin Y^X\setminus\F$, which is to say $f+g\in\F$.  So, $g\in -f+\F$.  
Since $g$ was arbitrary, it follows that $F=\{-f\colon f\in F_1\}$ 
satisfies (\ref{eq1:1}).  Thus, $\kappa\leq\add(Y^X\setminus\F)$.  

We show that $\add(Y^X\setminus\F)\leq\kappa$.  Let $F\subseteq Y^X$ be a set 
of cardinality $\kappa$ which satisfies (\ref{eq1:1}).  Since 
the set $F_1=\{-f\colon f\in F\}$ also has cardinality $\kappa$, 
it is enough for us to show that $F_1$ satisfies
\begin{equation}\label{eq1:3}
(\forall g\in Y^{X}) (\exists f\in F_1) (f+g\notin Y^X\setminus\F).
\end{equation}
Let $g\in Y^X$.  Since $F$ satisfies (\ref{eq1:1}), there is an 
$f\in F$ such that $g\in f+\F$.  So, $-f+g\notin Y^X\setminus\F$.  Thus, 
$F_1$ satisfies (\ref{eq1:3}).\qed 

So, for example, the cardinal $\add(\comp\dar(\real,\real))$ 
is the minimum cardinality of a collection 
$G\subseteq\real^{\real}$ such that 
$\real^{\real}=\{g+\dar\colon g\in G\}$.  This is just another way 
of saying that $\real^{\real}$ maybe covered by $\add(\comp\F)$ 
many translations of $\dar$ and not less.  

Throughout the remainder of this chapter we will only consider 
families of functions contained in $\real^{\real}$.  In particular, 
we will always write, for example, $\dar$ instead of 
$\dar(\real,\real)$ and $\comp\F$ will always mean 
$\real^{\real}\setminus\F$.  Before we continue our discussion, 
let us notice that the relationship between the additivity of a 
family $\F\subseteq\real^{\real}$ and the additivity of $\comp\F$ 
is by no means clear.  This is shown by the following two 
propositions which may be found in \cite{jord1}.  

\noindent\prop{prop1:2}{Let $\omega\leq\lambda\leq 2^{\cuum}$.  There is an 
$\F\subseteq\real^{\real}$ such that $\add(\F)=2$ and 
$\add(\comp\F)=\lambda$.} 
\proof 
Let $H$ be an additive subgroup of $\real^{\real}$ such that 
$|H|=\lambda$ and  the set $\{g+H\colon 
g\in\real^{\real}\}=\{H_{\alpha}\colon\alpha\in 2^{\cuum}\}$ 
has cardinality $2^{\cuum}$.  Let $\F$ be a selector of 
$\{H_{\alpha}\colon\alpha\in 2^{\cuum}\}$.  

We prove that $\add(\comp\F)=\lambda$.  We first show that 
$\add(\comp\F)\leq\lambda$.  Since $|H|=\lambda$, it is enough to 
show that 
\begin{equation}\label{eq1:5} 
(\forall g\in\real^{\real})(\exists h\in H)(g+h\in\F). 
\end{equation} 
Let $g\in\real^{\real}$ be arbitrary.  
Clearly, $(g+H)\cap\F\neq\emptyset$.  So, there exists an $h\in H$ 
such that 
$g+h\in\F$.  Thus, (\ref{eq1:5}) holds and 
$\add(\comp\F)\leq |H|=\lambda$.  
Next we prove the other inequality.  
Let $F\subseteq\real^{\real}$ be such that $|F|=\kappa<\lambda$.  We 
find a 
$g\in\real^{\real}$ such that $f+g\in\comp\F$ for all $f\in F$.  
We enumerate $F$ by $\{f_{\xi}\colon\xi<\kappa\}$.  Each $f_{\xi}$ lies 
in some translation $H_{\alpha_{\xi}}$ of $H$.  Since $\F$ is a 
selector of 
$\{H_{\alpha}\colon\alpha\in 2^{\cuum}\}$, 
\begin{equation}\label{eq1:6} 
\text{ there is a unique } 
h_{\xi}\in H \text{ such that }f_{\xi}+h_{\xi}\in 
H_{\alpha_{\xi}}\cap\F. 
\end{equation}
Since $|F|=\kappa<\lambda$, there is an $h\in 
H\setminus\{h_{\xi}\colon\xi\in\kappa\}$.  Thus,  by the uniqueness 
part of (\ref{eq1:6}), 
$h+f_{\xi}\in (H_{\alpha_{\xi}}\setminus\F)\subseteq\comp\F$ for 
every $\xi<\kappa$.  So, letting $g=h$, we have 
$\lambda\leq\add(\comp\F)$.  

We show that $\add(\F)=2$.  Since $\F\neq\emptyset$, it is enough to 
show that $\add(\F)\leq 2$.  We pick 
$h\in H\setminus\{\charf{\emptyset}\}$.  Let 
$F=\{\charf{\emptyset},h\}$ and 
$g\in\real^{\real}$ be arbitrary.  
It is enough to show that $g+f\notin\F$ for some $f\in\F$.  
If $g\notin\F$, then $g+\charf{\emptyset}\notin\F$.  
On the other hand, if  $g\in\F$, then $h+g\notin\F$ since 
$h\neq\charf{\emptyset}$ and 
$\F\cap(g+H)=\{g\}$.  
So, for any $g\in\real^{\real}$ either $h+g\in\comp\F$ or 
$g+\charf{\emptyset}\in\comp\F$.  Thus, $\add(\F)\leq 2$.\qed 



An immediate corollary of Proposition~\ref{prop1:2} is the 
following fact.  
\noindent\cor{cor1:1}{There exists families $\G,\F\subseteq\real^{\real}$
such that 
$\add(\F)=\add(\G)$ and $\add(\comp\F)\neq\add(\comp\G)$.\qed }

Using a different argument, we may extend Proposition~\ref{prop1:2} 
to include  all finite cardinals.  
\noindent\prop{prop1:3}{For every $2\leq n<\omega$ 
there exists an $\F\subseteq\real^{\real}$ such that $\add(\F)=n$ and 
$\add(\comp\F)=2$.}
\proof 
Let $S$ be the family of constant functions from $\real$ into 
$\integer$ and put 
$A_{k}=\{f\in S\colon f(0)=k\ {\rm modulo}\ n\}$ 
for every $0\leq k \leq n-1$.  Let $T$ be a selector of 
$\{h+S\colon h\in\real^{\real}\}$.  Put 
$\F=\bigcup\{T+A_{k}\colon 1\leq k \leq n-1\}$ and note that 
$\comp\F=T+A_{0}$.  

We prove that $\add(\F)=n$.  We first show that $\add(\F)\geq n$.  
Let 
$F=\{f_{0},\ldots,f_{n-2}\}\subseteq\real^{\real}$ be arbitrary.  We 
find  a $g$ such that $g+F\subseteq\F$.  For every $0\leq t \leq 
n-2$ there is 
an $s_{t}\in S$ such that $f_{t}\in s_{t}+T$.  Since 
$|\{A_{0},\ldots,A_{n-1}\}|=n$ and $|\{s_{0},\ldots,s_{n-2}\}|\leq n-1$, 
there is a $k^{*}\in\{0,\ldots,n-1\}$ such that 
$s_{t}\notin A_{k^{*}}$ for every $0\leq t \leq n-2$.  
Let $s\in A_{k^{*}}$.  Then, $-s+s_{t}\notin A_{0}$ for all $0\leq 
t\leq n-2$.  Thus, putting $g=-s$, we have $g+f_{t}\in 
(-s+s_{t})+T\subseteq\F$ for every 
$f_{t}\in F$.  So, $\add(\F)\geq n$.  
We now show that 
$\add(\F)\leq n$.  Let $F=\{f_{0},\ldots,f_{n-1}\}$ where $f_{k}\in 
A_{k}$ for all 
$0\leq k\leq n-1$.  It is enough to show that 
\begin{equation}\label{eq1:7}
\left(\forall g\in\real^{\real}\right)(\exists f\in F)(g+f\notin\F).
\end{equation} 
Let $g\in\real^{\real}$ be arbitrary.  Since $F\subseteq S$, we have 
$g+F\subseteq t+S$ for some $t\in T$.  By way of
contradiction, assume that 
$g+F\subseteq\F$.  Since $|g+F|=n$ and 
$|\{A_{0},\ldots,A_{n-1}\}|=n$ and 
$(g+F)\cap (t+A_{0})=\emptyset$, there exist $0\leq i<j\leq n-1$  
such that 
$g+f_{i},g+f_{j}\in t+A_{k}$ for some $1\leq k\leq n-1$.  So, there 
exist 
$s_{i},s_{j}\in A_{k}$ such that $g+f_{i}=s_{i}+t$ and 
$g+f_{j}=s_{j}+t$.  Then, 
$f_{i}-f_{j}=(g+f_{i})-(g+f_{j})=(s_{i}+t)-(s_{j}+t)=s_{i}-s_{j}\in 
A_{0}$.  It follows that there is some $1\leq k\leq n-1$ such that 
$f_{i},f_{j}\in A_{k}$, which contradicts our choice of $F$.  Thus, 
$F$  satisfies (\ref{eq1:7}) and $\add(\F)\leq n$. 
  
We show that $\add(\comp\F)=2$.  Since $\comp\F\neq\emptyset$ 
by Proposition~\ref{prop1:1}(i) it is enough 
to show that $\add(\comp\F)\leq 2$.  
Let $h_{0}=\charf{\emptyset}\in A_{0}$ and $h_{1}\in A_{1}$.  We 
show that 
$\{h_{0},h_{1}\}$ witnesses $\add(\comp\F)\leq 2$, i.e, 
\begin{equation}\label{eq1:8} 
g+h_{0}\in\F\text{ or }g+h_{1}\in\F\text{ for all }g\in\real^{\real}.
\end{equation}
Let $g\in\real^{\real}$.  Since $h_{0},h_{1}\in S$, 
there exists  a $t\in T$ such that $g+h_{0},g+h_{1}\in(t+S)$.  In 
particular, there exist  
$s_{0},s_{1}\in S$ with the property that $g+h_{0}=t+s_{0}$ and 
$g+h_{1}=t+s_{1}$.  Since 
$s_{1}-s_{0}=(g+h_{1})-(g+h_{0})=h_{1}\in A_{1}$, one of $s_{0}$ or  
$s_{1}$  is not in $A_{0}$.  Thus, $\{h_{0},h_{1}\}$ 
satisfies~(\ref{eq1:8}).  So $\add(\F)\leq 2$. \qed

\section{The Results}
To state the main results of this chapter we will need some 
cardinal numbers which have the following combinatorial descriptions.  
For a fixed infinite cardinal $\kappa$ we define four cardinals.   
\begin{flushleft}
$\same_{\kappa} 
 =\min\{|F|\colon F\subseteq \kappa^{\kappa}\ \&\ (\forall 
g\in\kappa^{\kappa}) (\exists f\in F) (|[f=g]|=\kappa)\}.$
\end{flushleft}

\medskip
 
\begin{flushleft}
$\same_{\kappa}^{*} =\min\{|F|\colon F\subseteq \kappa^{\kappa}\ \&\ 
(\forall G \in [\kappa^{\kappa}]^{\kappa})(\exists f\in F)(\forall 
g\in G) (|[f=g]|=\kappa\}.$
\end{flushleft}

\medskip

\begin{flushleft}
$\diff_{\kappa} =\min\{|F|\colon F\subseteq \kappa^{\kappa}\ \&\  
(\forall g\in\kappa^{\kappa}) (\exists f\in F) (|[f=g]| <\kappa)\}.$
\end{flushleft}

\medskip
 
\begin{flushleft}
$\diff_{\kappa}^{*} =\min\{|F|\colon F\subseteq \kappa^{\kappa}\ \&\ 
(\forall G \in [\kappa^{\kappa}]^{\kappa})(\exists f \in F)(\forall 
g\in G) (|[f=g]|<\kappa)\}$.
\end{flushleft}

\medskip 

Note that $\same_{\kappa}\leq\same_{\kappa}^{*}$ and
$\diff_{\kappa}\leq\diff_{\kappa}^{*}$.  For 
$\kappa=\omega$ 
these cardinals have been intensively studied in conjunction with certain 
cardinal invariants of the ideals of measure zero and meager sets in 
$\real$; see \cite{BAJU}.  It is easy to prove that 
$\cuum<\diff_{\cuum}\leq 2^{\cuum}$ and 
$\cuum<\same_{\cuum}\leq 2^{\cuum}$.  In \cite{CM} and \cite{CN} 
it is shown that this 
is about all that can be said about the values of $\same_{\cuum}$ and 
$\diff_{\cuum}$ in ZFC.

We will now discuss what may be said about the additivities of the 
families above and their respective complements in $\real^{\real}$.  
The additivities of the families $\ext$, $\pr$, $\phc$, 
$\dar$, $\conn$, $\acon$, $\comp\phc$, and $\sz$ are known.  

\noindent\prop{prop1:4}{ \mbox{ }
\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{jord1})} 
$\add(\comp\phc)=\omega_{1}$.\qed 
\end{description}} 
First, we calculate the numbers $\add(\comp\ext)$, $\add(\comp\pr)$ 
as stated in the theorem below.  
\noindent\thm{thm1:1}{$\add(\comp \pr)=\add(\comp \ext)=2^{\cuum}$.}
Next, we prove two theorems relating the additivies of $\comp\dar$, 
$\comp\conn$, 
$\comp\acon$, and $\comp\sz$ to 
some of the combinatioral cardinals listed above.  
\noindent\thm{thm1:2}{$\add(\sz)=\same_{\cuum}\leq\add(\comp\dar) 
\leq\add(\comp\conn)\leq\add(\comp\acon)\leq\same_{\cuum}^{*}$.}
\noindent\thm{thm1:3}{$\add(\acon)=\add(\conn)=\add(\dar)= 
\diff_{\cuum}\leq\add(\comp\sz)\leq\diff_{\cuum}^{*}$.}

We then have two purely combinatorial theorems.  The first will 
allow us, under certain 
assumptions, to replace the inequalities of Theorems~\ref{thm1:2} and 
\ref{thm1:3} with equalities.  The second will allow us, under certain 
assumptions, to relate the 
additivies of some of these families with respect to 
complementation, e.g., 
$\add(\comp\dar)\leq\add(\dar)$.  
The first part of Theorem~\ref{thm1:4} is due to 
Ciesielski~\cite{jord1}.  

\noindent\thm{thm1:4}{If $|[\cuum]^{<\cuum}|=\cuum$ then 
$\same_{\cuum}=\same_{\cuum}^{*}$ and 
$\diff_{\cuum}=\diff_{\cuum}^{*}$.}  

\noindent\thm{thm1:5}{If $|[\cuum]^{<\cuum}|=\cuum$ and 
$\cuum=\lambda^{+}$ then 
$\same_{\cuum}\leq\diff_{\cuum}$.}  

A result of S.~Shelah gives a simple condition for when the opposite 
inequality holds.  
\noindent\prop{prop1:5}{ {\rm (Shelah) \cite[Proposition~2.2]{SHEL1}} If 
$\cuum$ is a singular cardinal, then $\diff_{\cuum}=\diff^*_{\cuum}=\cuum^+\leq\same_{\cuum}$.\qed} 

Theorem~\ref{thm1:2} and the first part of Theorem~\ref{thm1:4}
yield 
\noindent\cor{cor1:2}{If $|[\cuum]^{<\cuum}|=\cuum$, then 
\begin{center} 
\hfill $\add(\sz)=\same_{\cuum}=\add(\comp\dar)= 
\add(\comp\conn)=\add(\comp\acon))=\same_{\cuum}^{*}.$\qed 
\end{center}
} 

Theorem~\ref{thm1:3} and the second part of 
Theorem~\ref{thm1:4} imply 

\noindent\cor{cor1:3}{If $|[\cuum]^{<\cuum}|=\cuum$, then 
\begin{center} 
\hfill$\add(\dar)=\add(\conn)=\add(\acon)=\diff_{\cuum}=
\add(\comp\sz)=\diff_{\cuum}^{*}$.\qed 
\end{center}}

From Theorem~\ref{thm1:5} and Corollaries~\ref{cor1:2} and 
\ref{cor1:3}  we can also conclude that 

\noindent\cor{cor1:4}{If $|[\cuum]^{<\cuum}|=\cuum$ and 
$\cuum=\lambda^{+}$, then 
\begin{center} 
\hfill$\add(\sz)=\add(\comp\dar)=\add(\comp\acon)=\same_{\cuum}\leq 
\diff_{\cuum}=\add(\dar)=\add(\acon)=\add(\comp\sz).$\qed 
\end{center}}

As a corollary of Proposition~\ref{prop1:5} and Theorem~\ref{thm1:3} 
we have 
\noindent\cor{cor1:5}{If $\cuum$ is a singular cardinal, then 
\begin{center} 
\hfill$\add(\dar)=\add(\acon)=\add(\comp\sz)=\cuum^+
\leq\same_{\cuum}=\add(\sz)=\add(\comp\dar)=\add(\comp\acon) .$\qed 
\end{center}}

The importance of the assumptions in Propostion~\ref{prop1:5} and 
Theorems~\ref{thm1:4} and \ref{thm1:5} is not clear.  In particular, 
the following problems are still open although some work on the first 
has been done \cite{SHEL1}.

\noindent\prob{prob:one}{Is it consistent with ZFC that either $\same_{\cuum}<\same_{\cuum}^{*}$ or 
$\diff_{\cuum}<\diff_{\cuum}^{*}$?}

\noindent\prob{prob:two}{Is it consistent with ZFC that $\cuum$ is regular and 
$\diff_{\cuum}<\same_{\cuum}$.}  

Finally, we quote three consistency results.

\noindent\prop{prop1:6}{ {\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} 

\noindent\prop{prop1:7}{ {\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} 

\noindent\prop{prop1:8}{ {\rm (Shelah) \cite[Conclusion~1.9 and 
Proposition~2.2]{SHEL1}} If ZFC + (existence of certain measurable cardinals) 
is consistent, then it is consistent with ZFC that $\same_{\cuum}$ is 
singular and 
\begin{center}
\hfill $\cuum^+=\diff_{\cuum}=\diff^*_{\cuum}<\same_{\cuum}=2^{\cuum}$.\qed
\end{center}}

Since, CH implies that $|[\cuum]^{<\cuum}|=\cuum$, 
Propositions~\ref{prop1:6} and \ref{prop1:7} together 
with Corollaries \ref{cor1:2} and \ref{cor1:3} imply immediately 
the following two corollaries: 

\noindent\cor{cor1:6}{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 
\begin{center} 
\hfill$\add(\acon)=\add(\dar)=\add(\sz)=\add(\comp(\dar)=\add(\comp\acon)=\kappa.$\qed 
\end{center}}

\noindent\cor{cor1:7}{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 
\begin{center} 
\hfill$\add(\comp\dar)=\add(\comp\acon)=\add(\sz)=\cuum^{+}<  
2^{\cuum}=\add(\dar)=\add(\acon)=\add(\comp\sz).$\qed 
\end{center}} 

Proposition~\ref{prop1:8} together with Theorems \ref{thm1:2} and 
\ref{thm1:3} imply immediately the following corollary.  

\noindent\cor{cor1:8}{If ZFC+(existence of certain 
measurable cardinals) is consistent, then it is consistent with ZFC 
that $\same_{\cuum}$ is singular and 
\begin{center} 
\hfill$\add(\dar)=\add(\acon)=\add(\comp\sz)=\cuum^{+}<  
2^{\cuum}=\add(\comp\dar)=\add(\comp\acon)=\add(\sz).$\qed 
\end{center}} 


\section{Proof of Theorem~\ref{thm1:1}}

To prove Theorem~\ref{thm1:1} we will use the following lemma which 
can be found in \cite{CR}.  

\noindent\prop{prop1:9}{If $B\subset\real$ has cardinality $\cuum$, 
$H\subset\rational^{B}$, and $|H|<2^{\cuum}$, then there is a 
$g\in\rational^{B}$ such that 
$h\bigcap g \neq\emptyset$ for every $h\in H$. \qed} 

It may be of interest to notice that Proposition~\ref{prop1:9} is used in 
\cite{CR} to prove that $\add(\phc)=2^{\cuum}$.  

\noindent\lem{lem1:2}{Let $A\subset\real$, $|A|=\cuum$, and 
\[
\kappa=\min\left\{|F|\colon F\subset \real^{A}\&\ 
(\forall g\in \real^{A}) (\exists f\in F)
(g+f\in \real^{A} \text{ is bounded})\right\}.
\]  
Then, $\kappa=2^{\cuum}$.}
\proof 
It is enough to show that $2^{\cuum}\leq\kappa$.  Let 
$F\in[\real^{A}]^{<2^{\cuum}}$.  We will find $g\in\real^{A}$ such 
that $f+g$ is not bounded for every $f\in F$.  
Let $\{B_{n}\colon n\in\omega\}$ be a partition of $A$ such that 
$|B_{n}|=\cuum$ for all $n\in\omega$.  
Fix $n\in\omega$.  
For each $f\in F$ choose $h^{f}_{n}\colon B_{n}\rightarrow\rational$ 
such that 
\begin{equation}\label{eq1:9} 
f(x)+h^{f}_{n}(x)>n \text{ for every } x\in B_{n}.  
\end{equation} 
By Proposition~\ref{prop1:9} used with the sets $B_{n}$ and 
$\{h^{f}_{n}\colon f\in F\}$, there is a $g_{n}\colon 
B_{n}\rightarrow\rational$ such that 
\begin{equation}\label{eq1:10}
(\forall f\in F) (\exists x\in B_{n})(h^{f}_{n}(x)=g_{n}(x)). 
\end{equation} 

Let $g=\bigcup\{g_{n}\colon n\in\omega\}$.  Then, by 
(\ref{eq1:9}) and (\ref{eq1:10}), 
for every $n\in\omega$ and $f\in F$ 
there exists $x\in B_{n}\subset A$ such that 
$f(x)+g_{n}(x)=f(x)+g(x)>n$.  So, $f+g$ is unbounded for every $f\in 
F$.  Thus, $2^{\cuum}\leq\kappa$ 
and the proof is complete.  \qed 

\medskip


{\sc Proof of Theorem~\ref{thm1:1}.} 
Notice that, by Proposition~\ref{prop1:1}(ii) and (iii), to prove 
Theorem~\ref{thm1:1} it  is enough to show that $\add(\comp 
\pr)\geq 2^{\cuum}$ since 
$\comp\pr\subseteq\comp\ext$ and $\comp\ext\neq\real^{\real}$.  

Let $F\subset\real^{\real}$ and $|F|< 2^{\cuum}$.  
Choose a partition $\{B_{\alpha}\colon \alpha\in \cuum \}$ of 
$\real$ into disjoint  Bernstein sets; let $\{P_{\alpha}\colon 
\alpha<\cuum\}$ be an enumeration of the 
perfect sets in $\real$; and put $I_{\alpha}=P_{\alpha}\cap 
B_{\alpha}$.  Note that $|I_{\alpha}|=\cuum$ and $I_{\alpha}\cap 
I_{\beta}=\emptyset$ for $\alpha<\beta<\cuum$.  
Fix $\alpha <\cuum$ and 
let $F_{\alpha}=\{f|_{I_{\alpha}}\colon f\in F\}$.  By 
Lemma~\ref{lem1:2}, there is some $g_{\alpha}\colon 
I_{\alpha}\rightarrow \real$ such that 
$g_{\alpha}+f|_{I_{\alpha}}$ is not bounded for every 
$f|_{I_{\alpha}} \in F_{\alpha}$.  Let $g\in\real^{\real}$ extend 
$\bigcup\{g_{\alpha}\colon \alpha < \cuum \}$.  Clearly, $f+g$ is not 
bounded on any  perfect set for every $f\in F$.  In particular, 
$f+g$ is nowhere continuous on every perfect set, 
so $f+g\notin \pr$ for every $f\in F$.  Therefore, 
$\add(\comp \pr)=2^{\cuum}$.\qed



\section{Proof of Theorem~\ref{thm1:2}}
In this section we prove Theorem \ref{thm1:2}.  By containment, we 
immediately  have 
$\add(\comp\dar)\leq\add(\comp\conn)\leq\add(\comp\acon)$.  Thus, 
it is enough to prove the inequalities 
$\same_{\cuum}\leq\add(\comp\dar)$ and 
$\add(\comp\acon)\leq\same_{\cuum}^{*}$.  
For the proof of $\same_{\cuum}\leq\add(\comp\dar)$, we need the 
following definitions.  For $A\subseteq\real$, let $\lin(A)$ denote 
the linear subspace of $\real$ over $\rational$ spanned by $A$.  
Given a linear subspace $S$ of $\real$ over 
$\rational$, we call any set of the form $t+S$ with $t\in\real$, 
a translation of $S$.  
Let $\dar^{*}\subseteq\real^{\real}$ stand for the family of 
nowhere-constant  Darboux functions and $\dar(\cuum)$ for the family 
of functions $f\in\real^{\real}$ such that
\[|f^{-1}(y)\cap(a,b)|=\cuum\] 
for all $a,b,y\in\real^{\real}$ with $a<b$.  
We will need a number of lemmas to prove the Theorem.  

\noindent\lem{lem1:skid}{Let $H$ be a linear base of $\real$ over $\rational$ 
and $H_{1}\subseteq H$ be such that $|H\setminus H_{1}|=\cuum$.  
Then, $\lin(S\cup H_{1})\neq\real$ for any 
linear subspace $S$ such that $|S|<\cuum$.}
\proof 
Let $H_{0}=H\setminus H_{1}$ and $N=\lin(H_{1})$.  By the linear 
independence of $H$, we have that 
$\{h+N\colon h\in H_{0}\}$ is a collection of pairwise disjoint 
translations of $N$.  In particular, $N$ has $\cuum$-many pairwise 
disjoint translations.  Since $|S|<\cuum$, 
there is a translation $T$ of $N$ such that $(s+N)\cap T=\emptyset$ 
for every $s\in S$.  Thus, $T\cap(\bigcup\{s+N\colon s\in 
S\})=\emptyset$.   Since 
$\bigcup\{s+N\colon s\in S\}=\lin(S\cup H_{1})$, we have $\lin(S\cup 
H_{1})\neq\real$.\qed 


\noindent\lem{lem1:row}{Let $N$ be a linear subspace of $\real$ over 
$\rational$ with $|N|=\cuum$.  
Then, for every non-degenerate interval $I\subseteq\real$, we 
have $|I\cap N|=\cuum$.}  
\proof 
Let $H$ be a linear base for $N$ over $\rational$.  Note that
$|H|=\cuum$.  Since $H$ is linearly independent, 
$\{h\cdot(\rational\setminus\{0\})\colon h\in H\}$ is a collection 
of pairwise disjoint dense subsets of $\real$.  Since 
$\bigcup\{h\cdot(\rational\setminus\{0\})\colon h\in H\}\subseteq 
N$,  it follows that $|I\cap N|=\cuum$ 
for all non-degenerate intervals $I\subseteq\real$.\qed



\noindent\lem{lem1:five}{$\add(\comp\dar)>\cuum$.} 
\proof
Let $F=\{f_{\alpha}\in\real^{\real}\colon\alpha<\cuum\}$ 
be an arbitrary family of functions of cardinality $\cuum$.  
It is enough to find a $g\in \real^{\real}$ such that 
$f+g\notin\dar$ for all $f\in F$.  

We construct $g$.  Let $H$ be a linear base for $\real$ over $\rational$ and 
$H_{1}\subseteq H$  be such that $|H_{1}|=|H\setminus H_{1}|=\cuum$.  
Let $N=\lin(H_{1})$ and 
$P_{\alpha,x}=\{f_{\beta}(x)\colon\beta\leq\alpha\}$ for 
every $\alpha<\cuum$ and $x\in\real$.  Let 
$\{x_{\alpha}\colon\alpha<\cuum\}$ be an enumeration  of $\real$.  
Define $g\in\real^{\real}$ so that 
\begin{equation}\label{eq1:11} 
g(x_{\alpha})\in\real\setminus\lin(N_{\alpha,x_{\alpha}}\cup N) 
\end{equation} 
where 
$N_{\alpha,x_{\alpha}}=\lin(P_{\alpha,x_{\alpha}}\cup\{(f_{\alpha}-g)(x_{0})\})$.  
We may make such a choice since $H_{1}$ and 
$S=N_{\alpha,x_{\alpha}}$ satisfy the hypothesis of 
Lemma~\ref{lem1:skid}.  
We show $g$ is the desired function.  Fix $\alpha<\cuum$.  If 
$\alpha\leq\beta<\cuum$ then, by (\ref{eq1:11}), 
$g(x_{\beta})\notin N_{\beta,x_{\beta}}$.  Since 
$f_{\alpha}(x_{\beta})\in N_{\beta,x_{\beta}}$ and 
$g(x_{\beta})\notin \lin(N_{\beta,x_{\beta}}\cup N)$, we have 
$f_{\alpha}(x_{\beta})+g(x_{\beta})\notin 
\lin(N_{\beta,x_{\beta}}\cup N)$.  In particular, 
$f_{\alpha}(x_{\beta})+g(x_{\beta})\notin N$ for all 
$\beta>\alpha$.  Thus, 
$|(f_{\alpha}+g)[\real]\bigcap N|<\cuum$.  Since $|N|=\cuum$, by
Lemma~\ref{lem1:row}, 
$|I\cap N|=\cuum$ for every non-degenerate interval.  It 
follows that the range of $f_{\alpha}+g$ contains no non-degenerate 
interval.  But by (\ref{eq1:11}), $f_{\alpha}+g$ takes at 
least two values and so, it is not constant.  Therefore, 
$g+f_{\alpha}\notin\dar$, which completes the proof since 
$\alpha<\cuum$ was arbitrary.  \qed

\noindent\lem{lem1:six}{$\add(\comp \dar)=\add(\comp \dar^{*})$.} 
\proof 
Since $\comp\dar\subseteq\comp\dar^{*}$, the inequality 
$\add(\comp\dar)\leq\add(\comp\dar^{*})$ is obvious.  We show the 
other inequality.  

Let $F\subseteq\real^{\real}$ be a family of cardinality 
$\kappa=\add(\comp\dar)$ that witnesses the definition of 
$\add(\comp\dar)$, i.e., 
\begin{equation}\label{eq1:12} 
(\forall g\in\real^{\real})(\exists f\in F)(f+g\in\dar). 
\end{equation} 
It is enough to find a family $F^{*}\subseteq\real^{\real}$ 
of cardinality $\kappa$ such that 
\begin{equation}\label{eq1:13} 
(\forall g\in\real^{\real})(\exists f\in F^{*})(f+g\in\dar^{*}).  
\end{equation} 

Let $\mathcal{I}$ 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 
$|{\mathcal{I}}|=\cuum$.  For each $I\in{\mathcal{I}}$ 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 
$k_{I}(x)=\charf{\cup I}(x)h_{I}(x)$ for each $x\in\real$.  Let 
$K=\{k_{I}\colon I\in \mathcal{I}\}$. Note that $|K|=\cuum$.  
Define $F^{*}=\{f+h\colon f\in F\ \&\ h\in K \}$.  
Since $|K|=\cuum$ and by Lemma~\ref{lem1:five}, $\cuum<\kappa$, 
we have $|F^{*}|=\kappa$.  So it is enough to show that 
$F^{*}$ satisfies (\ref{eq1:13}).  

Let $g\in\real^{\real}$ be 
arbitrary.  By (\ref{eq1:12}), there is an $f\in F$ 
such that 
$f+g\in\dar$.  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 $(f+k_{J})\in F^{*}$, 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{eq1:14}
\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
((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 locally 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 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{eq1:14}), $((f+k_{J})+g)[i]=\real$.  
If $i\cap j=\emptyset$ for all $j\in J$, then we have 
$((f+k_{J})+g)[i]=(f+g)[i]$.  In either case, $((f+k_{J})+g)[i]$ is 
an interval.  Thus, $(f+k_{J})+g$ is Darboux.  
So, $(f+k_{J})+g\in\dar^{*}$.  Thus, $F^{*}$ satisfies 
(\ref{eq1:13}), completing the proof.  
\qed



\noindent\lem{lem1:seven}{There is an additive function $\Theta\in 
\real^{\real}$ such  that $\Theta\circ f\in\dar(\cuum)$ for every 
$f\in\dar^{*}$.} 
\proof 
Let $\Theta\in\dar(\cuum)$ be an additive function.  
To construct such a function, pick a linear base 
$H$ for 
$\real$ over $\rational$.  Partition $H$ into two sets $H_{1}$ and 
$H_{2}$  of size $\cuum$ and define $f\colon H\to\real$ so that 
$f[H_{1}]={0}$ and 
$f[H_{2}]=\real$.  The unique additive extension $\Theta$ of $f$   
belongs to $\dar(\cuum)$.  

We show that $\Theta$ has the desired property.  Let $f\in\dar^{*}$ 
be arbitrary.  For any $a<b$ and $y\in\real$, we have 
\begin{eqnarray} 
|(a,b)\cap(\Theta\circ f)^{-1}(y)| 
& = & |\{x\in(a,b)\colon\Theta(f(x))=y\}| \notag \\ 
& \geq & |\{z\in f[(a,b)]\colon\Theta(z)=y\}| \notag \\ 
& = & \cuum.  \notag 
\end{eqnarray} 
The last equality follows from the fact that 
$f[(a,b)]$ is a non-degenerate interval since $f\in\dar^{*}$ and 
that $\Theta\in\dar(\cuum)$.  \qed



\noindent\lem{lem1:eight}{$\add(\comp\dar)=\add(\comp\dar(\cuum))$.} 
\proof 
Since $\comp\dar\subseteq\comp\dar(\cuum)$, we have 
$\add(\comp\dar)\leq\add(\comp\dar(\cuum))$.  We show the other 
inequality.  

Let $F\subseteq \real^{\real}$ have cardinality 
$\kappa=\add(\comp\dar)$.  By Lemma~\ref{lem1:six}, we may assume 
that $F$ witnesses the definition 
of $\add(\comp\dar^{*})$, i.e., 
\begin{equation}\label{eq1:15} 
(\forall g\in \real^{\real})(\exists f\in F)(f+g\in\dar^{*}).  
\end{equation} 
It is enough to find a family $F^{*}$ of cardinality at most 
$\kappa$ such that 
$F^{*}$ witnesses the definition of $\add(\comp\dar(\cuum))$, i.e., 
\begin{equation}\label{eq1:16} 
(\forall g\in \real^{\real})(\exists f^{*}\in 
F^{*})(f^*+g\in\dar(\cuum)).  
\end{equation} 

Let $F^{*}=\{\Theta\circ f\colon f\in F\}$ where $\Theta$ is the 
additive function  from Lemma~\ref{lem1:seven}.  Clearly, 
$|F^{*}|\leq\kappa$.  It is enough to show that 
$F^{*}$ satisfies (\ref{eq1:16}).  Let $g\in\real^{\real}$ be 
arbitrary.  Since $\Theta$ is a surjection, we may pick 
$g_{1}\in\real^{\real}$ so that $\Theta\circ g_1=g$.  
By (\ref{eq1:15}) there is an $f\in F$ 
such that $f+g_{1}\in\dar^{*}$.  By our choice of 
$\Theta$, we have 
\[
(\Theta\circ f) + g = \Theta\circ f + \Theta\circ g_{1} =
\Theta\circ(f+g_{1})\in\dar(\cuum).
\] 
Thus, $F^{*}$ satisfies (\ref{eq1:16}), completing the proof.  
\qed

\medskip

{\sc Proof of $\same_{\cuum}\leq\add(\comp\dar)$.}
Since $\add(\comp\dar)=\add(\comp\dar(\cuum))$, it is enough for us 
to show that 
$\same_{\cuum}\leq\add(\comp\dar(\cuum))$.  Let 
$\kappa=\same_{\cuum}$ and 
$F\subseteq\real$ be such that $|F|<\kappa$.  By definition of 
$\same_{\cuum}$, there is a $g\in\real^{\real}$ such that 
$|[f=g]|<\cuum$  for all $f\in F$.  Let $g_{1}=-g$.  
Now, $|(f+g_{1})^{-1}(0)|<\cuum$ for all $f\in F$.  Thus, 
$f+g_{1}\in\comp\dar(\cuum)$ for all $f\in F$.  
Since $F$ was an arbitrary set of cardinality less than $\kappa$, we 
have 
$\kappa<\add(\comp\dar(\cuum))$.  It follows that 
$\same_{\cuum}\leq\add(\comp\dar(\cuum))$.  \qed

\medskip

We now prepare to prove that $\add(\comp\acon)\leq\same_{\cuum}^{*}$.

\noindent\lem{lem1:ten}{Let 
$\{P_{\alpha}\colon\alpha<\cuum\}$ be a partition of $\cuum$ 
into $\cuum$-many sets of cardinality $\cuum$ and let 
\begin{eqnarray*} 
\kappa 
& =  & \min\{|F|\colon F\subseteq\cuum^{\cuum}\\ 
& \& & (\forall G\in[\cuum^{\cuum}]^{\cuum}) 
(\exists f\in F)(\forall g\in G)(\forall\alpha<\cuum) 
(|P_{\alpha}\cap[f=g]|=\cuum)\}.  
\end{eqnarray*} 
Then, $\kappa=\same_{\cuum}^{*}$.} 
\proof 
We first show that $\kappa\leq\same_{\cuum}^{*}$.  
Let $F\subseteq\cuum^{\cuum}$ witness 
the definition of $\same_{\cuum}^{*}$, i.e., $|F|=\same_{\cuum}^{*}$ 
and 
\begin{equation}\label{eq1:17} 
(\forall G\in[\cuum^{\cuum}]^{\cuum})(\exists f\in F)(\forall g\in 
G)(|[f=g]|=\cuum). \end{equation} 
It is enough to find a family $F^{*}\subseteq\cuum^{\cuum}$ such that 
$|F^{*}|=|F|$ and 
\begin{equation}\label{eq1:18} 
(\forall G\in[\cuum^{\cuum}]^{\cuum})(\exists f^{*}\in 
F^{*})(\forall g\in G) 
(\forall\alpha<\cuum)(|P_{\alpha}\cap[f^{*}=g]|=\cuum).  
\end{equation} 

For every $\alpha<\cuum$ enumerate $P_{\alpha}$ as 
$\{x_{\langle\alpha,\beta\rangle}\colon\beta<\cuum\}$.  
For each $f\in F$ let $f^{*}\in\cuum^{\cuum}$ be given by 
$f^{*}(x_{\langle\alpha,\beta\rangle})=f(\beta)$ for all 
$\langle\alpha,\beta\rangle\in\cuum^{2}$.  
Clearly, $|F^{*}|=|F|$.  We show that $F^{*}$ satisfies 
(\ref{eq1:18}).  
Let $G\in[\cuum^{\cuum}]^{\cuum}$ be arbitrary.  Enumerate $G$ by 
$\{g_{\xi}\colon\xi<\cuum\}$.  For each 
$\langle\alpha,\xi\rangle\in\cuum^{2}$ define 
$g_{\langle\alpha,\xi\rangle}\in\cuum^{\cuum}$ so that 
$g_{\langle\alpha,\xi\rangle}(\beta)=g_{\xi}
(x_{\langle\alpha,\beta\rangle})$ for all 
$\beta<\cuum$.  By (\ref{eq1:17}), there exists an 
$f\in F$ such that $|[g_{\langle\alpha,\xi\rangle}=f]|=\cuum$ for all 
$\langle\alpha,\xi\rangle\in\cuum^{2}$.  
Fix $\langle\alpha,\xi\rangle\in\cuum^{2}$.  Then, 
$|[g_{\langle\alpha,\xi\rangle}=f]|=\cuum$ 
and for every $\beta\in[g_{\langle\alpha,\xi\rangle}=f]$ 
\[
f^{*}(x_{\langle\alpha,\beta\rangle})=f(\beta)=
g_{\langle\alpha,\xi\rangle}(\beta)=g_{\xi}(x_{\langle\alpha,\beta\rangle}).
\]
Thus, $|P_{\alpha}\cap[f^{*}=g_{\xi}]|=\cuum$.  Since 
$\langle\alpha,\xi\rangle\in\cuum^{2}$ was 
arbitrary, we conclude that $F^{*}$ satisfies (\ref{eq1:18}).  
So $\kappa\leq\same_{\cuum}^{*}$.  

We now show the other inequality.  Let $F\subseteq\cuum^{\cuum}$ 
witness the definition  of $\kappa$, i.e., $F$ satisfies 
(\ref{eq1:18}).  It is enough to show that $F$ 
satisfies (\ref{eq1:17}).  Let $G\in[\cuum^{\cuum}]^{\cuum}$ 
be arbitrary.  Let $\{g_{\alpha}\colon\alpha<\cuum\}$ be an 
enumeration of $G$.  Define $g\in\real^{\real}$ so that 
$g|_{P_{\alpha}}=g_{\alpha}|_{P_{\alpha}}$.  
Pick $f\in F$ so that $|P_{\alpha}\bigcap[f=g]|=\cuum$ for every 
$\alpha<\cuum$.   Clearly, 
we have $|[f=g_{\alpha}]|=\cuum$ for every $g_{\alpha}\in G$.  Thus, $F$ satisfies 
(\ref{eq1:17}).  The proof is complete.\qed 

We will also need the following fact about almost continuous functions which may be 
found in \cite{NATsurvey}: 


\noindent\prop{prop1:10}{ {\rm (Kellum \cite{NATsurvey})} 
If $f\in\real^{\real}$ has non-empty intersection with every 
upper semi-continuous function 
$u$ defined on a non-degenerate interval, then $f\in\acon$.\qed} 

\medskip
{\sc Proof of $\add(\comp\acon)\leq\same_{\cuum}^{*}$.} 
Let $\kappa=\same_{\cuum}^{*}$ and $\{P_{\alpha}\colon\alpha<\cuum\}$ 
be a partition of $\real$ into $\cuum$-many sets of cardinality 
$\cuum$ so that for every open interval $I$ there is an 
$\alpha<\cuum$ such that $P_{\alpha}\subseteq I$.  
By Lemma~\ref{lem1:ten} there is an 
$F\in\left[\real^{\real}\right]^{\kappa}$ such that 
\begin{equation}\label{eq1:19} 
\left(\forall G\in\left[\real^{\real}\right]^{\cuum}\right)(\exists 
f\in F)(\forall g\in G)(\forall\alpha<\cuum) 
(|P_{\alpha}\cap[f=g]|=\cuum).  
\end{equation} 
It is enough to show that 
\begin{equation}\label{eq1:20} 
\left(\forall g\in\real^{\real}\right)(\exists f\in F)(f+g\in\acon).  
\end{equation}
Let $U$ denote the family of upper semi-contiuous functions in 
$\real^{\real}$ defined  on a non-degenerate interval.  We extend 
the domain of 
any partial function $u\in U$ to all of $\real$ in an abitrary 
manner.  Note that $|U|=\cuum$.  
Let $g\in\real^{\real}$ be arbitrary and put $G=\{u-g\colon u\in 
U\}$.  By (\ref{eq1:19}), there is an $f\in F$ such that for 
every $\alpha\in\cuum$ and 
$u\in U$ we have $|P_{\alpha}\bigcap[f=u-g]|=\cuum$.  
So, $[f+g=u]$ is dense for each $u\in U$ by our choice of 
$\{P_{\alpha}\colon\alpha<\cuum\}$.  Thus, by 
Proposition~\ref{prop1:10}, 
$f+g\in\acon$.  Therefore, $F$ satisfies (\ref{eq1:20}) which 
completes the  proof. \qed 

\medskip

{\sc Proof of Theorem \ref{thm1:2}.} 
We have proved $\same_{\cuum}\leq\add(\comp\dar)$ and 
$\add(\comp\acon)\leq\same_{\cuum}^{*}$  while, by containment, we 
have $\add(\comp\dar)\leq\add(\comp\conn)\leq\add(\comp\acon)$.  
\qed 


\section{Proof of Theorem~\ref{thm1:3}} 
To prove Theorem \ref{thm1:3} we quote a theorem about
$\sz$ functions which may be found in \cite{sezy}.  
\noindent\prop{prop1:11}{ {\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 defined on a $G_{\delta}$ set of 
cardinality~$\cuum$.\qed} 

\medskip

{\sc Proof of Theorem \ref{thm1:3}.} We first show that 
$\add(\comp\sz)\leq\diff_{\cuum}^{*}$. Let $H$ stand for the family 
of all functions $h\in\real^{\real}$ such that 
$h$ is continuous on a $G_{\delta}$ set of cardinality $\cuum$ and 
equal to zero elsewhere. Note that $|H|=\cuum$.  
Let $F\subseteq\real^{\real}$ witness the definition of 
$\diff_{\cuum}^{*}$, i.e., 
$|F|=\diff^{*}_{\cuum}$ and 

\begin{equation}\label{eq1:21} 
\left(\forall G\in\left[\real^{\real}\right]^{\cuum}\right)(\exists 
f\in F)(\forall g\in G)(|[f=g]|<\cuum).  
\end{equation} 
It is enough to show that $F$ satisfies 
\begin{equation}\label{eq1:22} 
(\forall g\in G)(\exists f\in F)(f+g\in\sz).
\end{equation}
Let $g\in\real^{\real}$ be arbitrary.  Let $G=\{h-g\colon h\in 
H\}$.  By  (\ref{eq1:21}), there exists an $f\in F$ such 
that
$|[f=h-g]|<\cuum$ for each $h\in H$.  So, $|[f+g=h]|<\cuum$ for each 
$h\in H$.  Thus, by Proposition~\ref{prop1:11}, $f+g\in\sz$ and 
$F$ satisfies (\ref{eq1:22}).  
So, $\add(\comp\sz)\leq\diff_{\cuum}^{*}$.

Next, we show that $\diff_{\cuum}\leq\add(\comp\sz)$.  Let 
$F\subseteq\real^{\real}$ witness the  definition of 
$\add(\comp\sz)$, i.e., $|F|=\add(\comp\sz)$ and $F$ satisfies 
(\ref{eq1:22}).  It is enough to show that $F$ satisfies 
\begin{equation}\label{eq1:23} 
(\forall g\in\real^{\real})(\exists f\in F)(|[f=g]<\cuum).  
\end{equation} 
Let $g\in\real^{\real}$ be arbitrary.  By (\ref{eq1:22}) 
there is an $f\in F$  such that $-g+f\in\sz$.  Then 
$|[-g+f=\charf{\emptyset}]|<\cuum$, so $|[f=g]|<\cuum$.  
Thus, $F$ satisfies (\ref{eq1:23}).\qed

\section{Proof of Theorems~\ref{thm1:4} and \ref{thm1:5}}

{\sc Proof of Theorem \ref{thm1:4}.} 
Let $W=\bigcup\{\cuum^{\alpha}\colon \alpha<\cuum\}$. Note that 
$|W|=\cuum$ by  our assumption that $|[\cuum]^{<\cuum}|=\cuum$.  
Let $V=\{\langle\alpha,\xi\rangle\colon\xi\leq\alpha<\cuum\}$.  We 
will  use these sets throughout the proofs of 
$\same_{\cuum}=\same_{\cuum}^{*}$ and 
$\diff_{\cuum}=\diff_{\cuum}^{*}$.  

We prove $\same_{\cuum}=\same_{\cuum}^{*}$.  
Since the inequality $\same_{\cuum}\leq\same_{\cuum}^{*}$ is 
obvious, we  prove only that $\same_{\cuum}^{*}\leq\same_{\cuum}$.  
Let $F\subseteq W^{\cuum}$ witness the definition of 
$\same_{\cuum}$, i.e., $|F|=\same_{\cuum}$ and 
\begin{equation}\label{eq1:24} 
(\forall g\in W^{\cuum})(\exists f\in F)(|[f=g]|=\cuum). 
\end{equation} 
It is enough to find a family $F^{*}\subseteq\cuum^{V}$ 
of cardinality less than or equal to $|F|$ witnessing the defintion 
of $\same_{\cuum}^{*}$, i.e., 
\begin{equation}\label{eq1:25} 
(\forall G\in [\cuum^{V}]^{\cuum})(\exists f\in F^{*})(\forall g\in 
G)  (|[f=g]|=\cuum).  
\end{equation}

For each $f\in F$ let $f^{*}\in \cuum^{V}$ be such that 
$f^{*}(\alpha,\beta)=f(\alpha)(\beta)$ if $\beta \in\domain(f(\alpha))$  and zero otherwise.  Let $F^{*}=\{f^{*}\colon f\in 
F\}$, note $|F^{*}|\leq |F|$.  
We show that $F^{*}$ satisfies (\ref{eq1:25}) which will 
complete  the proof.  Let $G\in [\cuum^{V}]^{\cuum}$ be arbitrary 
and let $\{g_{\alpha}\colon\alpha<\cuum\}$ be an enumeration of $G$.  
Define $h \in W^{\cuum}$ so that 
\begin{equation}\label{eq1:26} 
h(\alpha)\in \cuum^{\alpha} \text{ and } h(\alpha)(\beta)= 
g_{\beta}(\alpha,\beta) \text{ for all }\beta<\alpha. 
\end{equation} 
By (\ref{eq1:24}), there exists an $f\in F$ such that 
$|[f=h]|=\cuum$.  We will now show that $|[f^{*}=g_{\alpha}]|=\cuum$ for all 
$\alpha<\cuum$.  Fix 
$\alpha<\cuum$.  Since $|[f=h]|=\cuum$, the relations 
\begin{equation}\label{eq1:27}
\alpha<\beta<\cuum \text{ and } f(\beta)=h(\beta) 
\end{equation} 
are satisfied by $\cuum$-many $\beta<\cuum$.  Consider $\beta$ 
satisfing  (\ref{eq1:27}). 
By (\ref{eq1:26}), 
$h(\beta)\in\cuum^{\beta}$. So, 
\[
f^{*}(\beta,\alpha)=f(\beta)(\alpha)=h(\beta)(\alpha)=g_{\alpha}(\beta,\alpha).
\notag 
\] 
We conclude that $|[f^{*}=g_{\alpha}|=\cuum$.  Thus, $F^{*}$ 
satisfies (\ref{eq1:25}), completing the proof of 
$\same_{\cuum}=\same_{\cuum}^{*}$.  

We prove $\diff_{\cuum}=\diff_{\cuum}^{*}$.  
The inequality $\diff_{\cuum}\leq\diff_{\cuum}^{*}$ is obvious.  To 
prove 
$\diff_{\cuum}^{*}\leq\diff_{\cuum}$ 
let $F\subseteq \cuum^{V}$ witness the definition 
of $\diff_{\cuum}$, i.e., $|F|=\diff_{\cuum}$ and 
\begin{equation}\label{eq1:28} 
(\forall g\in \cuum^{V})(\exists f\in F)(|[f=g]|<\cuum).  
\end{equation} 
It is enough to find a family $F^{*}\subset W^{\cuum}$ of 
cardinality less  than or equal to $|F|$ witnessing 
$\diff_{\cuum}^{*}$, i.e., 
\begin{equation}\label{eq1:29} 
(\forall G\in [W^{\cuum}]^{\cuum})(\exists f^{*}\in F^{*}) 
(\forall g\in G)(|[f^{*}=g]|<\cuum).  
\end{equation} 

For each $f\in F$ let $f^{*}\in W^{\cuum}$ be such that 
$f^{*}(\alpha)\in\cuum^{\alpha}$  for every $\alpha<\cuum$ and let 
$f^{*}(\alpha)(\beta)=f(\alpha,\beta)$ for every $\beta<\alpha$.  
Let $F^{*}=\{f^{*}\colon f\in F\}$ and note that $|F^{*}|\leq |F|$.  
We show that $F^{*}$ satisfies (\ref{eq1:29}).  Let 
$G\in[W^{\cuum}]^{\cuum}$, and let $\{g_{\alpha}\colon\alpha<\cuum\}$ 
enumerate 
$G$.  We find an $f^{*}\in F^{*}$ such that 
\begin{equation}\label{eq1:30} 
|[f^{*}=g_{\alpha}]|<\cuum\text{ for every }\alpha<\cuum.  
\end{equation} 
Define $g\in \cuum^{V}$ by 
$g(\beta,\alpha)=g_{\alpha}(\beta)(\alpha)$ if 
$\alpha\in{\rm dom}(g_{\alpha}(\beta))$ and zero otherwise.  
By (\ref{eq1:28}) there is some $f\in F$ such that 
$|f\cap g|<\cuum$.  We show $f^{*}$ satisfies 
(\ref{eq1:30}),  which will complete the proof.  Fix 
$\alpha<\cuum$.  For any 
$\alpha<\beta<\cuum$, if $f^{*}(\beta)=g_{\alpha}(\beta)$, then  
$f^{*}(\beta)(\gamma)=f(\beta,\gamma)=g_{\alpha}(\beta)(\gamma)$ for 
every $\gamma<\beta$.  In particular, 
$f(\beta,\alpha)=g_{\alpha}(\beta)(\alpha)$ so 
$f(\beta,\alpha)=g(\beta,\alpha)$.  Since $|[f=g]|<\cuum$, there 
must be less  than $\cuum$ many $\beta>\alpha$ such that 
$f^{*}(\beta)=g_{\alpha}(\beta)$.  
Thus, $|[f^{*}=g_{\alpha}]|<\cuum$.  Since $\alpha<\cuum$ was 
arbitrary, 
$f^{*}$ satisfies (\ref{eq1:30}) completing the proof.  
\qed

\medskip

To prove Theorem \ref{thm1:5} we will need three more lemmas and 
will define two more cardinals:
\begin{description} 
\item[ ] 
$\dom_{\kappa}
=\min\{|F|\colon F\subseteq \kappa^{\kappa}\ \&\ (\forall 
g\in\kappa^{\kappa})  (\exists f\in F) (|[f\leq g]|<\kappa)\}$,
\item[ ] 
$\unb_{\kappa}
=\min\{|F|\colon F\subseteq \kappa^{\kappa}\ \&\
(\forall g\in \kappa^{\kappa})(\exists f\in F) (|[g\leq
f]|=\kappa)\}$. 
\end{description} 
The numbers $\unb_{\kappa}$ and $\dom_{\kappa}$ are analogs of the 
bounding number 
$\unb=\unb_{\omega}$ and the dominating number $\dom=\dom_{\omega}$; 
see \cite{BAJU}.  

\noindent\lem{lem1:fourteen}{If $\cuum=\lambda^{+}$, then for every 
$f\in\cuum^{\cuum}$ the set 
$\{\langle\alpha,\beta\rangle\in\cuum^{2}\colon\beta\leq 
f(\alpha)\}$ is the union of $\lambda$-many  functions in 
$\cuum^{\cuum}$.} 
\proof 
Let $f\in\cuum^{\cuum}$ be arbitrary.  We find a family 
$\{f_{\zeta}\in\cuum^{\cuum}\colon\zeta<\lambda\}$ such that 
\begin{equation}\label{eq1:31}
\{\langle\alpha,\beta\rangle\in\cuum^{2}\colon\beta\leq
f(\alpha)\}=\bigcup\{f_{\zeta}\colon\zeta<\lambda\}. 
\end{equation} 
For each $\alpha<\cuum$, let  
$\{\gamma_{\langle\alpha,\zeta\rangle}\colon\zeta<\lambda\}$ be an 
enumeration of $f(\alpha)+1$.  For each $\zeta<\lambda$ define  
$f_{\zeta}\in\cuum^{\cuum}$ so that 
$f_{\zeta}(\alpha)=\gamma_{\langle\alpha,\zeta\rangle}$.  We show 
that $\{f_{\zeta}\colon\zeta<\lambda\}$ satisfies 
(\ref{eq1:31}).  Fix 
$\alpha<\cuum$.  Then, for any $\beta<\cuum$ we have 
\[
\beta\leq 
f(\alpha)\Leftrightarrow(\exists\zeta<\lambda)
(\gamma_{\langle\alpha,\zeta\rangle}=\beta) 
\Leftrightarrow(\exists\zeta<\lambda)(f_{\zeta}(\alpha)=\beta).  
\]
Thus, (\ref{eq1:31}) is satisfied.\qed  



\noindent\lem{lem1:rats}{If $\cof(\cuum)=\cuum$, then $\unb_{\cuum}>\cuum$.}

\proof 
Let $F\in[\cuum^{\cuum}]^{\cuum}$ be arbitrary.  It is enough to
find $g\in\cuum^{\cuum}$  such that $|[g\leq f]|<\cuum$ for every
$f\in F$.  Let $\{f_{\alpha}\colon\alpha\in\cuum\}$
be an enumeration of $F$.  Define $g\in\cuum^{\cuum}$ so that
$g(\alpha)>f_{\beta}(\alpha)$ for all $\beta\leq\alpha$.  We may
make such a choice  since $\cof(\cuum)=\cuum$.
To see that $g$ is as desired,
let $\alpha<\cuum$ be arbitrary.  Then $g(\beta)>f_{\alpha}(\beta)$ for all
$\beta>\alpha$. \qed

The first part of the next lemma is due to Ciesielski~\cite{jord1}.  
\noindent\lem{lem1:thirteen}{If $\cuum=\lambda^{+}$, then 
$\unb_{\cuum}=\same_{\cuum}$ 
and $\dom_{\cuum}=\diff_{\cuum}^{*}$.  In particular, 
$\same_{\cuum}\leq\diff_{\cuum}^*$.} 
\proof 
We prove $\unb_{\cuum}=\same_{\cuum}$.  
To show that $\same_{\cuum}\leq\unb_{\cuum}$, 
let $F\subseteq\cuum^{\cuum}$ witness the definition of 
$\unb_{\cuum}$, i.e., 
$|F|=\unb_{\cuum}$ and 
\begin{equation}\label{eq1:32} 
(\forall g\in\cuum^{\cuum})(\exists f\in F)(|[g\leq f]|=\cuum).  
\end{equation}
It is enough to find $F^{*}\subseteq\cuum^{\cuum}$ such that 
$|F|=|F^{*}|$ and 
\begin{equation}\label{eq1:33} 
(\forall g\in\cuum^{\cuum})(\exists f^{*}\in F^{*})(|[f=g]|=\cuum).  
\end{equation}
For each $f\in F$, let $\{f_{\zeta}\colon\zeta<\lambda\}$ be as in 
Lemma~\ref{lem1:fourteen} and put 
\[F^{*}=\bigcup\{\{f_{\zeta}\colon\zeta<\lambda\}\colon f\in F\}.\]  
By Lemma~\ref{lem1:rats}, $|F|=|F^{*}|$.
We now work to show that $F^{*}$ satisfies (\ref{eq1:33}).  Let 
$g\in\cuum^{\cuum}$ be arbitrary.  By (\ref{eq1:32}), there 
is an $f\in F$ such that $|[g\leq f]|=\cuum$. 
Since $|g\bigcap\{\langle\alpha,\beta\rangle\colon\beta\leq 
f(\alpha)\}|=\cuum$ and 
$\{\langle\alpha,\beta\rangle\colon\beta\leq 
f(\alpha)\}=\bigcup\{f_{\zeta}\colon\zeta<\lambda\}$, there is some 
$\zeta<\lambda$ such that $|[f_{\zeta}=g]|=|f_{\zeta}\cap g|=\cuum$.  
Thus, $F^{*}$ satisfies (\ref{eq1:33}).  So, 
$\same_{\cuum}\leq\unb_{\cuum}$.  To see the other inequality notice 
that any family $F$ satisfying (\ref{eq1:33}) 
also satisfies (\ref{eq1:32}).  

We prove $\dom_{\cuum}=\diff_{\cuum}^{*}$.  
To show that $\dom_{\cuum}\leq\diff_{\cuum}^{*}$, let 
$F\subseteq\cuum^{\cuum}$ witness the definition of 
$\diff_{\cuum}^{*}$, i.e., 
$|F|=\diff_{\cuum}^{*}$ and 
\begin{equation}\label{eq1:34} 
(\forall G\in[\cuum^{\cuum}]^{\cuum})(\exists f\in F)(\forall g\in 
G)(|[f=g]|<\cuum).  
\end{equation} 
It is enough to find $F^{*}\subseteq\cuum^{\cuum}$ such that 
$|F|=|F^{*}|$ and 
\begin{equation}\label{eq1:35} 
(\forall g\in\cuum^{\cuum})(\exists f\in F)(|[f\leq g]|<\cuum).  
\end{equation} 
We show that $F$ itself satisfies (\ref{eq1:35}).  
Let $g\in\cuum^{\cuum}$ be arbitrary.  Let 
$\{g_{\zeta}\colon\zeta<\lambda\}$ be as in 
Lemma~\ref{lem1:fourteen}.  By (\ref{eq1:34}) there exists an 
$f\in F$ such that 
$|[f=g_{\zeta}]|<\cuum$ for every $\zeta<\lambda$.  Since 
$\cof(\cuum)=\cuum$ and 
$\{\langle\alpha,\beta\rangle\colon\beta\leq 
g(\alpha)\}=\bigcup\{g_{\zeta}\colon\zeta<\lambda\}$, we have 
$|[f\leq g]|<\cuum$.  Thus, $F$ satisfies (\ref{eq1:35}).  
So, $\dom_{\cuum}\leq\diff_{\cuum}^{*}$.  

We show that $\diff_{\cuum}^{*}\leq\dom_{\cuum}$.  
It is enough to prove that any family $F$ satifying 
(\ref{eq1:35}) also satisfies  (\ref{eq1:34}).  
Suppose $F$ satifies (\ref{eq1:35}).  
Let $G\in[\cuum^{\cuum}]^{\cuum}$ be arbitrary and 
$\{g_{\alpha}\colon \alpha<\cuum\}$ be an  enumeration of $G$.  By 
Lemma~\ref{lem1:rats} there is an $h\in\cuum^{\cuum}$ such that 
$|[h\leq g_{\alpha}]|<\cuum$ for each $\alpha<\cuum$.  
By (\ref{eq1:35}) there exists an $f\in F$ such that 
$|[f\leq h]|<\cuum$.  It follows that $|[f\leq g_{\alpha}]|<\cuum$ 
for all $\alpha<\cuum$.  In particular, 
$|[f=g]|<\cuum$ for every $g\in G$, so $F$ satisfies 
(\ref{eq1:34}).  Thus, $\diff_{\cuum}^{*}\leq\dom_{\cuum}$.\qed 

\medskip

{\sc Proof of Theorem \ref{thm1:5}.}
Clearly, $\unb_{\cuum}\leq\dom_{\cuum}$.  By Lemma~\ref{lem1:thirteen}, 
we have 
$\same_{\cuum}\leq\diff_{\cuum}^{*}$.  By the second part of Theorem 
\ref{thm1:4}, 
$\diff_{\cuum}^{*}=\diff_{\cuum}$.  Thus, 
$\same_{\cuum}\leq\diff_{\cuum}$.  
\qed

%%%CHAPTER TWO
\chapter{Generalizing additivity for higher dimensions}\label{second_chapter}

%\setcounter{figure}{3}
\section{Introduction}
The results of Chapter~\ref{first_chapter} may lead us to ask to what 
extent those results may be extended to the case of
$\real^{\real^n}$ and $(\real^m)^{\real^n}$.  The fact that the 
containment relations between the Darboux-like families change so 
dramatically in the two cases suggests that the problem is
nontrivial.  But, with our present definition of additivity
the problem is trivial.  However, the
present definition of additivity is actually inappropriate for
the problem at hand, not just because it yields trivial results,
but because it no longer captures the ideas it reflects in the
results for $\real^{\real}$.  This idea is recaptured in the
definition of $(n,k)$-additivity.  With this definition we can
extend our results from $\real^{\real}$ with great success.   

We now discuss the notion of
$(n,k)$-additivity.   In what follows we will assume that $X$ is
a non-empty set and $Y$ is an Abelian group, that is, for all $x,y\in Y$ 
we have $x+y=y+x$, with identity $0$.  We will let 
$\theta$ stand for the function in $Y^X$ with range $\{0\}$. 

We define the repeatability of $\F\subseteq Y^X$ to be 
\begin{equation*}
\rep(\F)=\min(\{n\in\omega\colon n\F=Y^X\}\cup\{\omega\}).
\end{equation*} 
The repeatablities of the Darboux-like families are of particular 
interest to us becuase of the following result.
\noindent\prop{prop2:rep}
{\rm{\cite{CW}} $\rep(\F(\real^n,\real))=n+1$ for  
$\F\in\{\ext,\conn,\phc\}$ and $n>0$.}
We now mention an easy observation about repeatablity that will 
often be used without reference.
\noindent\prop{prop2:rep1}{If $\rep(\F)=n$ then $m\F=Y^X$ for all $m\geq n$.}
\proof
If $\rep(\F)=n$ then $n\F=Y^X$.  Let $k>0$ be such that $m=n+k$.  
Let $g\in Y^X$.  Pick $f\in k\F$.  Since $Y^X$ is a group, there is 
an $h\in Y^X=n\F$ such that $f+h=g$.  Thus, $g\in m\F$.\qed 

For the convenience of the discussion we restate
Proposition~\ref{prop1:1}.  
\noindent\prop{prop2:one}{Let $\G,\F\subseteq Y^X$ and $|Y|>1$.  Then,
\begin{description}
\item[(i)] $\add(\F) = 1 \text{ if and only if } \F=\emptyset$;
\item[(ii)] $\add(\F)\leq |Y^X| \text{ if and only if } \F\neq Y^X$;
\item[(iii)] if $\F\subset\G$, then $\add(\F)\leq\add(\G)$;  
\item[(iv)]$2<\add(\F) \text{ if and only if } \F - \F = Y^X$.
\end{description}}

When $\F\subseteq Y^X$ has the property that $\F=-\F$, 
the statements (ii) and 
(iv) of Proposition~\ref{prop2:one} imply, using the notation of repeatablity, that 
\[(*)\ \ \ \ \ 2<\add(\F)\leq |Y^X|\text{ if and only if } \rep(\F)=2.\]
One implication of ($*$) is that for reasonable families $\F$ of functions the 
additivity is a generalization of repeatability when $\rep(\F)=2$.  
One might also notice that under certain conditions we have a nice restatement
of the definition of $\add$ in terms of coding functions as sums of
functions in $\F$.  
\noindent\prop{prop2:will}{Let $\F=-\F$ and $\add(\F)\geq\omega$.  Then,
$\add(\F)$ is the largest cardinal $\kappa$ such that for any 
$F\subseteq Y^X$
if $|F|<\kappa$, then there exists a $g\in\F$ such that 
\begin{equation}\label{eq2:last}
(\forall f\in F)(\exists g_f\in\F)(g+g_f=f).
\end{equation}}
\proof
Let $F\subseteq Y^X$ and $|F|<\add(\F)$.  We show that there is a 
$g\in\F$ which 
satisfies (\ref{eq2:last}).  Since $\add(\F)\geq\omega$, we may
assume the zero function $\theta$ is an element of $F$.  There is
a $g'\in Y^X$ such that
$g'+F\subseteq\F$.  Notice that since $\theta\in F$, we actually have $g'\in\F$. 
For each $f\in F$ let $g_f=g'+f\in\F$.  Since $g'\in\F$ and $-\F=\F$, we have
$-g'\in\F$.  Now $g=-g'$ satisfies (\ref{eq2:last}).  

Now assume $\kappa$ is
a cardinal that is larger than $\add(\F)$.  We will find a family $F\subseteq
Y^X$ such that $|F|<\kappa$, and there is no $g\in\F$ which satisfies
(\ref{eq2:last}).  Let $F\subseteq Y^X$ witness the definition of
$\add(\F)$, i.e., $|F|=\add(\F)<\kappa$ and 
\[(\forall g\in Y^{X}) (\exists f\in F) (f+g\notin \F).\]
By way of contradiction, assume that there is some $g\in\F$ satisfying
(\ref{eq2:last}).  
Then $f+(-g)=g_f\in\F$ for each $f\in F$ which contradicts our choice of
$F$.\qed

When 
the repeatability of a family of functions is greater than 2, the additivity 
function tells us at most that the repeatability is not 2.   
Since we will be considering 
families of functions with repeatabilities that may be larger than 2, 
we want to define a cardinal function which 
will be of use for these families.  We would like this cardinal function 
(actually, it will be a collection of cardinal functions) to satisfy some statements 
similar to ($*$) and Proposition~\ref{prop2:will}.  It would also be appealing if
we could have a statement like ($*$) that held for all $\F\subseteq Y^X$, not
just those  where $\F=-\F$.  We might also want to remove some of the
hypothesises of Proposition~\ref{prop2:will} as well.  Removal of the need for the
hypothesises $\F=-\F$ and
$\add(\F)\geq\omega$ is fairly simple.  We define
\begin{equation*}
\add^{\circ}(\F)=\!\!\min\left(\left\{|F|\colon F\subseteq Y^{X}\ \!\!\&\ \!(\forall  
g\in \F) (\exists f\in F) (f-g\notin \F)\right\}\!\cup\! \left\{\left(|Y^X|\right)^{+}\right\}\right)\!\!.
\end{equation*}

We must now ask how closely $\add^{\circ}$ corresponds to $\add$ as a function.  
The following proposition shows that the correspondence is very close for
reasonable families of functions.  

\noindent\prop{prop2:3}{If $\F\subseteq Y^X$ and $\F=-\F$, 
then $\add(\F)=1\oplus\add^{\circ}(\F)$.  In particular if
$\add(\F)\geq\omega$ then
$\add^{\circ}(\F)=\add(\F)$.}
\proof
The inequality 
\begin{equation}\label{eq2:fa}
\add^{\circ}(\F)\leq\add(\F)
\end{equation}
is immediate from the definitions.

We show that $\add(\F)\leq 1\oplus\add^{\circ}(\F)$.  
Let $F\subseteq Y^X$ be such that $|F|=\add^{\circ}(\F)$ and
\begin{equation*}
(\forall g\in \F) (\exists f\in F) (f+g\notin \F).
\end{equation*}
Put $F_1=F\cup\{\theta\}$ and notice that 
\begin{equation*}
(\forall g\in Y^X) (\exists f\in F_1) (f+g\notin \F).
\end{equation*}
Thus, 
\begin{equation}\label{eq2:fb}
\add(\F)\leq 1\oplus\add^{\circ}(\F).  
\end{equation}

For the other inequality, we consider two cases.  
First assume $\add(\F)\geq\omega$.  Then, by (\ref{eq2:fb}), 
$\add^{\circ}(\F)$ is infinite, so $1\oplus\add^{\circ}(\F)=\add^{\circ}(\F)$.  
Now by
(\ref{eq2:fa}) we have $1\oplus\add^{\circ}(\F)=\add^{\circ}(\F)\leq\add(\F)$.
Next assume $\add(\F)<\omega$.  Note that (\ref{eq2:fa}) 
implies that $\add^{\circ}(\F)$ is finite.
We will show that $1\oplus\add^{\circ}(\F)\leq\add(\F)$.  Assume that
$\add^{\circ}(\F)=0$.  Since the empty set $\emptyset$ is the only subset of $Y^X$ with
cardinality zero we have, by the definition of $\add^{\circ}(\F)$,
\begin{equation*}
(\forall g\in \F) (\exists f\in\emptyset) (f-g\notin \F).
\end{equation*}
The only way the above statement may be true is if 
$\F=\emptyset$.  By Proposition~\ref{prop2:one}, 
$\add(\emptyset)=1$.  So, in this case $1+\add^{\circ}(\F)=\add(\F)$.  So we may assume
that $\add^{\circ}(\F)>0$.  Take
$F\subseteq Y^X$ with $|F|<1\oplus\add^{\circ}(\F)$.  Fix $f\in F$ and put 
$F_1=(-f+F)\setminus\{\theta\}$.  We have
$|F_1|<\add^{\circ}(\F)$. So, there is a $g\in\F$ such that
$-g+F_1\subseteq\F$.   Since $\F=-\F$, we have
$-g+(F_1\cup\{\theta\})\subseteq\F$.  Now,
$(-g-f)+F=-g+(F_1\cup\{\theta\})\subseteq\F$. 
Thus,
$1\oplus\add^{\circ}(\F)\leq\add(\F)$. 
\qed

Now the $\add^{\circ}$ analog of ($*$) is that for any $\F\subseteq Y^X$ 
\[(*_1)\ \ \ \ \  1<\add^{\circ}(\F)\leq |Y^X|\text{ if and only if } \rep(\F)=2.\]
The version of Proposition~\ref{prop2:will} for $\add^{\circ}$ is more simple to state 
and prove.
\noindent\prop{prop2:was}{$\add^{\circ}(\F)$ is the largest cardinal $\kappa$ such that for 
any $F\subseteq Y^X$ if $|F|<\kappa$, then there
exists a $g\in\F$ such that 
\begin{equation}\label{eq2:tour}
(\forall f\in F)(\exists g_f\in\F)(g+g_f=f).
\end{equation}}
\proof
Let $F\subseteq Y^X$ and $|F|<\add^{\circ}(\F)$.  We show that there is a $g\in\F$ 
which satisfies (\ref{eq2:tour}).  Clearly there is a $g\in\F$ such that
$-g+F\subseteq\F$.  For each $f\in F$ let $g_f=-g+f\in\F$.  Then $g$
satisfies (\ref{eq2:tour}).

Now assume $\kappa$ is a cardinal that is
larger than $\add^{\circ}(\F)$.  We find a family $F\subseteq Y^X$ such that
$|F|<\kappa$, and there is no $g\in\F$ which satisfies (\ref{eq2:tour}).  
Let
$F\subseteq Y^X$ witness the definition of $\add^{\circ}(\F)$, i.e.,
$|F|=\add^{\circ}(\F)<\kappa$ and 
\[(\forall g\in\F) (\exists f\in F) (f-g\notin \F).\]
By way of contradiction, assume that there is some $g\in\F$ satisfying
\begin{equation*}
(\forall f\in F)(\exists g_f\in\F)(g+g_f=f).
\end{equation*}
Then, $f-g=g_f\in\F$ for each $f\in F$, which contradicts our choice of
$F$.\qed

Now our goal will be to construct a family of cardinal functions that will
allow us to generalize $(*_1)$ and Proposition~\ref{prop2:was} to include
families with repeatablities greater than $2$.   Let $n,k\in\omega$ be such that
$k<n$, and let $\F\subseteq Y^X$.  We define the $(n,k)$-additivity of 
$\F$ to be
\begin{equation*}
\add_{n,k}(\F)=
\min(\{|F|\colon F\subseteq Y^X\ \&\ \Psi_{n,k}(F,\F) \text{
holds}\}\cup\{|Y^X|^+\})
\end{equation*}
where $\Psi_{n,k}(F,\F)$ denotes the statement
\begin{equation*}
(\forall g\in (n-k)\F) 
(\exists f\in F)(f-g\notin (k+1)\F).
\end{equation*}
Notice that $\add^{\circ}=\add_{1,0}$.  We can now restate
Proposition~\ref{prop2:3}  in this language.
\noindent\prop{prop2:again}{If $\F\subseteq Y^X$ and $\F=-\F$, 
then $\add(\F)=1\oplus\add_{1,0}(\F)$.  In particular, if
$\add(\F)\geq\omega$, then
$\add^{\circ}(\F)=\add_{1,0}(\F)=\add(\F)$ as in Proposition~\ref{prop2:3}.}

We see
that Proposition~\ref{prop2:was} is generalized as follows: 
\noindent\prop{prop2:was1}{$\add_{n,k}(\F)$ is the largest cardinal 
$\kappa$ such that for any $F\subseteq Y^X$ if $|F|<\kappa$, then 
there exist $g^1,\ldots,g^{n-k}\in\F$ such that
\[(\forall f\in F)(\exists g^1_f,\ldots,g^{k+1}_f\in\F)(g^1+\ldots+g^{n-k}
+g^1_f+\ldots+g^{k+1}_f=f).\]}
We leave the proof to
the reader since it is of the same form as the proof of
Proposition~\ref{prop2:was}.  

We state an expanded version of Proposition~\ref{prop2:one} for 
$(n,k)$-additivity and include some other facts.

\pagebreak 

\noindent\prop{prop2:two}{Let $\F,\G\subseteq Y^X$ and $n\in\omega$.  Then
\begin{description}
\item[(i)] if $k<n$, then $\add_{n,k}(\F)=0 \text{ if and only if
}\F=\emptyset$;
\item[(ii)] if $k<n$ and $\F\subseteq\G$, then 
$\add_{n,k}(\F)\leq\add_{n,k}(\G)$; 
\item[(iii)] if $i<k<n$, then $\add_{n,i}(\F)\leq\add_{n,k}(\F)$;
\item[(iv)] if $k<n$, then $\add_{n,k}(\F)\leq |Y^X| \text{ if and only if } 
\rep(\F)> k+1$;
\item[(v)] if $k<n$, then $1<\add_{n,k}(\F)\text{ if and only if } 
\rep(\F)\leq n+1$;
\item[(vi)] $1<\add_{n,k}(\F)\leq |Y^X| \text{ for all }k<n\text{ if and only
if } 
\rep(\F)=n+1$;
\item[(vii)] if $\rep(\F)\geq n+1$ and $\F=-\F$, then $\add_{n,k}(\F)\leq 2$
for all $k<\floor{n/2}$;
\item[(viii)] if $k<n$, then $\add_{n,k}(\F)\leq\add((k+1)\F)$.
\end{description}}
\proof
Items (i) and (ii) are straight forward and will be left without proof.  

We prove (iii).  Let $F\subseteq Y^X$ and $|F|<\add_{n,i}(\F)$.  By
Proposition~\ref{prop2:was1} there exist $g^1,\ldots,g^{n-i}\in\F$ such that
\begin{equation}\label{eq2:slap}
(\forall f\in F)(\exists g^1_f,\cdots,g^{i+1}_f\in\F)
(g^1+\cdots+g^{n-i}+g^1_f+\cdots+g^{i+1}_f=f).
\end{equation}  
Let $m=k-i$.  By
(\ref{eq2:slap}), we have for each $f\in F$ 
\begin{equation}\label{eq2:smick}
g^1_f+\cdots+g^{i+1}_f+g^1+\cdots+g^m=f-(g^{m+1}+\cdots+g^{n-i}).
\end{equation}
Note that $n-i-(m+1)+1=n-k$; so $g=g^{m+1}+\cdots+g^{n-i}\in (n-k)\F$.  Notice
also that $i+1+m=k+1$; so $g^1_f+\cdots+g^{i+1}_f+g^1+\cdots+g^m\in (k+1)\F$. 
So (\ref{eq2:smick}) implies that there is a $g\in (n-k)\F$ such that
$-g+F\subseteq (k+1)\F$.  Thus, $\add_{n,i}(\F)\leq\add_{n,k}(\F)$.

We prove (iv).  Suppose $\add_{n,k}(\F)=|Y^X|^+$.  Then there is a 
$g\in(n-k)\F$ such that $(-g)+(Y^X)\subseteq (k+1)\F\subseteq Y^X$.  Thus, 
$(k+1)\F=Y^X$, hence $\rep(\F)\leq k+1$.  To see the other implication, assume 
that $\rep(\F)\leq k+1$ which is to say $(k+1)\F=Y^X$.  
Then $g+(k+1)\F=Y^X$ for any $g\in Y^X$.  In particular, we may pick $g$ to be 
in $(n-k)\F$.  Thus, $(-g)+(Y^X)\subseteq (k+1)\F$ for some $g\in (n-k)\F$.  So, 
$\add_{n,k}(\F)=|Y^X|^+$.

We prove (v). Suppose $\add_{n,k}(\F)\geq 2$.  Then, for any $f\in Y^X$ 
there is a $g\in (n-k)\F$ such that $f-g\in (k+1)\F$, which is to say 
$f\in(n+1)\F$.  Thus, $(n+1)\F=Y^X$, and so $\rep(\F)\leq n+1$.  
Now, assume $\add_{n,k}(\F)\leq 1$ and let 
$G=\{g\}$ witness the definition of $\add_{n,k}(\F)$, i.e., 
$f-g\notin (k+1)\F$ for every $g\in(n-k)\F$.  Clearly, $f\notin (n+1)\F$, 
so $(n+1)\F\neq Y^X$, which is to say $\rep(\F)>n+1$.   

Item (vi) is a direct consequence of (iv) and (v).  

We now prove (vii).  
Let $k<\floor{n/2}$. 
Since $\rep(\F)\geq n+1$, there is a function $h\colon X\to Y$
that is not an element of $(2k+2)\F$.  By way of contradiction, assume
$\add_{n,k}(\F)>2$.  Pick $f_1,f_2\in Y^X$ such that $f_1-f_2=h$.  
Since
$\add_{n,k}(\F)>2$, we can find a $g\in (n-k)\F$ such that
$\{f_1-g,f_2-g\}\subseteq(k+1)\F$.  
Note that $-(f_2-g)\in (k+1)\F$ since
$\F=-\F$.  Thus,
\[h=f_1-f_2=(f_1-g)-(f_2-g)\in(2k+2)\F\] 
which contradicts our choice
of $h$.

We prove (viii).  Let $F\subseteq Y^X$ and $|F|<\add_{n,k}(\F)$.  There
is a $g\in (n-k)\F$ such that $-g+F\subseteq (k+1)\F$.  Thus,
$\add_{n,k}(\F)\leq\add((k+1)\F)$. \qed

Notice that (vi) of
Proposition~\ref{prop2:two} is a good generalization of $(*_1)$. 


\section{The Results}  
We intend to prove six theorems.  The first four deal with the
$(n,k)$-additivities of some of the Darboux-like families of 
functions in $(\real^m)^{\real^n}$.  The
remaining theorems deal with some containments 
between the Darboux-like families of functions as well as the family 
$\sz$.  Recall that 
$\ext(\real^n,\real)=\phc(\real^n,\real)=\conn(\real^n,\real)$ for 
$n>1$.

\noindent\thm{thm2:two}{Let $n>0$. Then,
\begin{equation}\label{eq2:zhz}
\add_{n,n-1}(\ext(\real^n,\real))=\add(n\ext(\real^n,\real))=\cuum^+.
\end{equation}
Moreover, when $n>1$ we may replace $\ext$ in (\ref{eq2:zhz}) by 
either $\conn$ or $\phc$.}

\noindent\thm{thm2:one}{Let $n,m\geq 1$.  Then, 
\[\add_{n,n-1}(\dar(\real^n,\real^m))=\add(n\dar(\real^n,\real^m))=\diff_{\cuum}.\]}

\noindent\thm{thm2:onea}{Let $n,m\geq 1$.  Then, 
$\add(\acon(\real^n,\real^m))=\diff_{\cuum}$.}

\noindent\thm{thm2:three}{Let $n,m\geq 1$. Then,
\begin{equation}\notag
\add_{n,j}(\dar(\real^n,\real^m))=\add((j+1)\dar(\real^n,\real^m))=\cuum^+
\end{equation}
for $n-1>j\geq \floor{n/2}$.}

Before stating the other theorems, we consider some
implications of Theorems~\ref{thm2:two}, \ref{thm2:one}, 
and \ref{thm2:onea}.  Using Propositions \ref{prop2:3} and
\ref{prop2:two}(vi), we see that Theorems \ref{thm2:two}, 
\ref{thm2:one}, and \ref{thm2:onea} generalize the following results:
\noindent\prop{prop2:foura}{\ 
\begin{description}
\item[(i)] \rm{\cite{CR}} $\add(\ext(\real,\real))=\cuum^{+}$.
\item[(ii)] \rm{\cite{CW}} $\rep(\F(\real^n,\real))=n+1$ for  
$\F\in\{\ext,\conn,\phc\}$ and $n>0$. 
\item[(iii)] \rm{\cite{CM}}
$\add(\dar(\real,\real))=\add(\acon(\real,\real))=\diff_{\cuum}$. 
\end{description}}

Theorem~\ref{thm2:one} has as a corollary, using Proposition~\ref{prop2:two}(v),
the answer to a question of Ciesielski and Wojociechowski \cite{CW}.
\noindent\cor{cor2:9}{Let $n,m\geq 1$.  Then $\rep(\dar(\real^n,\real^m))=n+1$.} 
Using (iv), (vi), and (vii) of Proposition~\ref{prop2:two} and
Theorems \ref{thm2:two} and \ref{thm2:one} we have the following
corollaries:  
\noindent\cor{cor2:20}{If $k<\floor{n/2}$, then $\add_{n,k}(\F(\real^n,\real))=2$; and 
if $m>k\geq n$, then
$\add_{m,k}(\F(\real^n,\real))=(2^{\cuum})^+$ for 
$\F\in\{\ext,\conn,\phc\}$.}
\noindent\cor{cor2:21}{Let $n,m>0$.  If $k<\floor{n/2}$, then $\add_{n,k}(\dar(\real^n,\real^m))=2$;
and if $p>k\geq n$ then,
$\add_{p,k}(\dar(\real^n,\real^m))=(2^{\cuum})^+$.}
The above results for $n>1$ and $m\geq 1$ are summarized in the figuires below.

\bigskip
\begin{figure}[h]\label{fig:4}
\begin{center}
\begin{tabular}{|l|l|l|l|}\hline
    & $k<p<n$  & $\ \ \ \ k<n=p$ &$n\leq k< p$ \\ \cline{1-4}
$\F=\ext(\real^n,\real)$   &\ \ \ \ \ \ $1$ &see table below &\ \ \ \
$\left(2^{\cuum}\right)^+$ \\ \cline{1-4}
$\F=\dar(\real^n,\real^m)$   &\ \ \ \ \ \ $1$ &see table below &\ \ \ \ 
$\left(2^{\cuum}\right)^+$  \\ \cline{1-4}
\end{tabular}
\caption{Values of $\add_{p,k}(\F)$.}
\end{center}
\end{figure}

\medskip

\begin{figure}[h]\label{fig:5}
\begin{center}
\begin{tabular}{|l|l|l|l|l|}\hline
    & $k < \floor{n/2}$   & $\floor{n/2}\leq k\leq n-2$  & $k=n-1$ \\
\cline{1-4}
$\F=\ext(\real^n,\real)$   &\ \ \ \ \ \ $2$   &\ \ \ \ \ \ \ \ \ \ \ ?     &\ \ \
\ \ $\cuum^+$ 
\\ \cline{1-4}
$\F=\dar(\real^n,\real^m)$   &\ \ \ \ \ \ $2$   &\ \ \ \ \ \ \ \ \ \ \
$\cuum^+$     &\ \ \ \ \ $\diff_{\cuum}$ 
  \\ \cline{1-4}
\end{tabular}
\caption{Values of $\add_{n,k}(\F)$.}
\end{center}
\end{figure}

\bigskip
The first table gives the values of $\add_{p,k}(\F)$; the second table
gives the values of $\add_{n,k}$.  We now state the remaining theorems.  
\noindent\thm{thm2:four}{$\acon(\real^n,\real^m)\subsetneq n\dar(\real^n,\real^m)$.  
Moreover, there is an almost continuous function $f$ such that $f\notin
 (n-1)\dar(\real^n,\real^m)$.}

\noindent\thm{thm2:five}{$n\dar(\real^n,\real)\cap\sz(\real^n,\real)=\emptyset$ for
$n>1$.}

An immediate corollary of Theorems \ref{thm2:four} and \ref{thm2:five} is 

\noindent\cor{cor2:10}{If $n>1$,
then $\acon(\real^n,\real)\cap\sz(\real^n,\real)=\emptyset$.}

In \cite{BCN} it is shown that the conclusions of 
Theorem~\ref{thm2:five} and Corollary~\ref{cor2:10} are independent of ZFC when
$n=1$.

Corollary~\ref{cor2:9} and Theorem~\ref{thm2:four} might lead one to conjecture 
that every function from $\real^n$ into $\real$ is the sum of an almost
continuous function and a Darboux function.  
The following example which is constructed in Section \ref{sec:ex1} shows
that this is very much not the case when $n>1$. 

\noindent\ex{ex:1}{There exist a Baire class 1 function 
$f\colon\real^n\to\real$ with the property that 
$f\notin n\dar(\real^n,\real)$.  Moreover, for $n>1$ we have that 
$f$ is not the sum of an
almost continuous function and $n-1$ Darboux functions.}

\pagebreak

\section{Proofs of Theorems \ref{thm2:one} and \ref{thm2:five}.}
 
\noindent\lem{lem2:a}{Let $n,m\geq 1$, $f\colon\real^n\to\real^m$ be a function and 
$A,B\subseteq\real$ be a partition of $\real$ into two disjoint $\cuum$-dense 
sets.  If $\pi:\real^n\to\real$ denotes the projection of $\real^n$ onto a fixed
coordinate and
\begin{description}
\item[(i)] $f|_{\pi^{-1}(y)}\in\dar(\pi^{-1}(y),\real^m)$ for every $y\in B$,
\item[(ii)] $f|_{\pi^{-1}(y)}$ is constant for every $y\in A$, and
\item[(iii)] $\{y\in A\colon f[
\pi^{-1}(y)]=\{r\}\}$ is dense in $\real$ 
for every $r\in\real^m$,
\end{description}
then $f\in\dar(\real^n,\real^m)$.}
\proof
Suppose $C\subseteq\real^n$ is connected.  We show that $f[C]$ is a connected 
subset of $\real^m$.  There are two possible cases that may occur.  
The first case is when there is a $y\in\real$ such that $C\subseteq\pi^{-1}(y)$.  
In this case, one of (i) or (ii) applies to show that $f[C]$ is a connected set.  
The other case is when there exist distinct $y_1<y_2\in\real$ such that 
$C\cap\pi^{-1}(y_1)\neq\emptyset\neq C\cap\pi^{-1}(y_1)$.  Since $C$ is connected, 
we have $C\cap\pi^{-1}(y)\neq\emptyset$ for all $y\in [y_1,y_2]$.  
So, by (iii), $f[C]=\real^m$, which is connected.\qed



We now prove one of the main inequalities of Theorem~\ref{thm2:one}.  
\noindent\lem{lem2:ins}{$\add_{n,n-1}(\dar(\real^n,\real^m))\geq\diff_{\cuum}$.}
\proof
We proceed by induction on $n$. The inequality is proven in \cite{CM} for 
the case $\dar(\real,\real)$, and the methods used in \cite{CM} can clearly 
be used to establish the inequality for $\dar(\real,\real^m)$ when $m>1$.  So 
we may assume that $n>1$ and 
$\add_{n-1,n-2}(\dar(\real^{n-1},\real^m))\geq\diff_{\cuum}$.  
Let $F\subseteq(\real^m)^{\real^n}$ be an arbitrary collection of functions such 
that $|F|<\diff_{\cuum}$.  We must find a $g\in\dar(\real^n,\real^m)$ such that
$-g+F\subseteq n\dar(\real^n,\real^m)$. Let $\{A_k\}_{k\in n+1}$ be a partition
of $\real$ into $n+1$ $\cuum$-dense sets.  Define 
$h\colon\real\to\real^m$ so that for each $p\in\real^m$ and $k\in n+1$  
\begin{equation*}
\{y\in A_k\colon h(y)=p\} \text{ is dense in $\real$}.
\end{equation*}
Let $\pi\colon\real^n\to\real$ denote the projection of $\real^n$ onto a fixed 
coordinate.  Note that $\pi^{-1}(y)$ is homeomorphic to
$\real^{n-1}$ for every
$y\in\real$.  For each $y\in\real$ let $F^y=\{f|_{\pi^{-1}(y)}-h(y)\colon f\in
F\}$.  Since $|F^y|\leq |F|<\diff_{\cuum}$, we may apply the inductive hypothesis
to find for each $l\in n$ and $y\in A_l$ a Darboux function
$g^{y}\colon\pi^{-1}(y)\to\real^m$ such that
$-g^y+F_y\subseteq(n-1)\dar(\pi^{-1}(y),\real^m)$.  So, for every $f\in F$ and
$y\in A_l$ there exist 
$\{g^{f,y}_{k}\in\dar(\pi^{-1}(y),\real^m)\colon k\in n\setminus\{l\}\}$ such
that 
\begin{equation}\label{eq2:gtrav}
g^{y}+\sum_{\{k\in n\colon k\neq l\}} g^{f,y}_{k}=f|_{\pi^{-1}(y)}-h(y).
\end{equation}
Define $g\colon\real^n\to\real^m$ by 
\begin{equation*}
g(p)= 
\begin{cases}
h(y)& \text{ if $\pi(p)=y\in A_{n}$, }\\ 
g^{y}(p)& \text{ if $\pi(p)=y\notin A_{n}$.}
\end{cases}
\end{equation*}
We claim $g$ is as desired.  
To see that $g$ is Darboux let $A=A_{n}$, 
$B=\real\setminus A_n$, 
and apply Lemma \ref{lem2:a}.  By inductive hypothesis, 
$\add_{n-1,n-2}(\dar(\pi^{-1}(y),\real^m)
\geq\diff_{\cuum}$ for each $y\in A_{n}$.  By
Proposition~\ref{prop2:two}(v), $\rep(\dar(\pi^{-1}(y),\real^m))\leq n$. 
Thus, for each $y\in A_{n}$ and $f\in F$, we may find
$\{g^{f,y}_k\in\dar(\pi^{-1}(y),\real^m)\colon k\in n\}$ such that 
\begin{equation}\label{eq2:trav1}
\sum_{k=0}^{n-1} g^{f,y}_k=f|_{\pi^{-1}(y)}-h(y).
\end{equation}
For each $f\in F$ and $k\in n$ define $g^f_k\colon\real^n\to\real^m$ so that 
\begin{equation}\label{howl}
g_k^f(p)= 
\begin{cases}
h(y)& \text{ if $\pi(p)=y\in A_{k}$,}\\
g^{f,y}_k& \text{ if $\pi(p)=y\notin A_{k}$.}
\end{cases}
\end{equation}
Note that $g^f_k\in\dar(\real^n,\real^m)$ for each $f\in F$ and $k\in n$. 
To see it, take $A=A_{k}$, $B=\real\setminus A_{k}$ and apply Lemma
\ref{lem2:a}.   We now show that $-g+F\subseteq n\dar(\real^n,\real^m)$.  More
precisely we show that for each $f\in F$
\begin{equation}\label{eq2:40b}
g+\sum_{k=0}^{n-1} g^f_k=f.
\end{equation}
Let $p\in\real^n$.  We must consider two cases.
First, assume there is an $l\in n$ and a $y\in A_l$ such that 
$p\in\pi^{-1}(y)$. 
Then,   
\begin{eqnarray}
g(p)+\sum_{k=0}^{n-1} g^f_k(p)
& = & g^{y}(p)+\left(\sum_{\{k\in n\colon k\neq l\}}g^{f,y}_k(p)
+g^f_l(p)\right)\notag
\\ & = & \left(g^y(p)+\sum_{\{k\in n\colon
k\neq l\}}g^{f,y}_k(p)\right)+g^f_l(p)\notag\\ 
& = & (f(p)-h(y))+h(y)\label{xues}\\  
& = & f(p), \notag
\end{eqnarray}
where (\ref{xues}) follows from (\ref{eq2:gtrav}) and
(\ref{howl}).   We now consider the case in which there exists a $y\in
A_{n}$ such that
$p\in\pi^{-1}(y)$.  Then, 
\begin{eqnarray}
g(p)+\sum_{k=0}^{n-1} g_k^f(p)
& = & h(y)+\sum_{k=0}^{n-1} g^{f,y}_k(p)\notag \\
& = & h(y)+(f(p)-h(y)) \label{gor} \\
& = & f(p), \notag
\end{eqnarray}
where (\ref{gor}) follows from (\ref{eq2:trav1}). 
So, (\ref{eq2:40b}) holds completing the inductive step.\qed 



Our next goal is to show that $\add(n\dar(\real^n,\real^m))\leq\diff_{\cuum}$.  
Since Proposition \ref{prop2:two}(viii) implies that
$\add_{n,n-1}(\dar(\real^n,\real^m))\leq\add(n\dar(\real^n,\real^m))$, this
will complete the proof.  We will need some results and definitions from
dimension theory, all of which  may be found in \cite{dim}.  


The dimension of a topological space is defined as follows.  
\begin{description}
\item[(i)] $\dim X=-1$ if and only if $X=\emptyset$.
\item[(ii)] $\dim X\leq n$ if for any $p\in X$ and any open neighborhood $W$ of $p$ 
there exists an open neighborhood $U\subseteq W$ of $p$ such that 
$\dim\bd_X(U)\leq n-1$.
\item[(iii)] $\dim X=n$ if $\dim X\leq n$ and it is not true that $\dim X\leq n-1$.
\end {description}
An n-dimensional Cantor manifold ($n \geq 1$) is a compact n-dimensional space 
that 
cannot be disconnected by a subset of dimension less than $n-1$.  We will 
need three facts about n-dimensional Cantor manifolds.  We say $X$ is a
{\em continuum} if 
$X$ is a compact, connected, nonempty metric space.  
We say a continuum $X$ is {\em degenerate} if $|X|=1$; otherwise, we say $X$ is
{\em non-degenerate}.

\noindent\prop{prop2:three}{Every n-dimensional Cantor manifold is a
continuum.}

\noindent\prop{prop2:four}{If $X$ is a compact n-dimensional space, then $X$ contains an
n-dimensional Cantor manifold.}

\noindent\prop{prop2:five}{$[0,1]^n$ is an $n$-dimensional Cantor manifold for all $n>0$.}


To establish the inequality $\add(n\dar(\real^n,\real^m))\leq\diff_{\cuum}$, we
must first prove some lemmas.

\noindent\lem{lem2:c}{Suppose $n>1$ and M is an n-dimensional Cantor manifold.  
If $B\subseteq M$ disconnects $M$, then 
there is an $(n-1)$-dimensional Cantor manifold contained in $B$ .}
\proof
Since $M$ is a continuum, there is a compact set $C\subseteq B$ that disconnects 
$M$.  By the definition of Cantor manifold, the dimension of $C$ is at least
$n-1$.   So, by Proposition~\ref{prop2:four}, there is an $(n-1)$-dimensional
Cantor manifold $N$  such that $N\subseteq C\subseteq B$.\qed 

\noindent\lem{lem2:d}{Assume that $n>1$, $M$ is an n-dimensional Cantor manifold, and 
$f\colon M\to\real$ is Darboux.  There is a collection $\{N_{\alpha}\}_{\alpha\in\cuum}$ 
of pairwise disjoint $(n-1)$-dimensional Cantor manifolds such that 
$f|_{N_{\alpha}}$ is constant for each $\alpha\in\cuum$.}
\proof
We first assume $f$ is not constant.  Let $r$ be in the interior of $f[M]$. 
Since $f[M]\setminus\{r\}$ is  not connected and $f$ is Darboux, it follows that
$M\setminus f^{-1}(\{r\})$ is not  connected.  By Lemma~\ref{lem2:c} there is a 
$(n-1)$-dimensional Cantor manifold $N$ such that $N\subseteq f^{-1}(\{r\})$. 
Thus, $f|_N$ is constant.  Since we may do this for each $r$ in the interior
of $f[M]$ and  preimages of distinct points  are disjoint, 
we can find the desired collection of continua.

If $f$ is a constant function, the lemma reduces to the question of whether
there exist $\cuum$-many pairwise disjoint $(n-1)$-dimensional Cantor
manifolds contained in $M$.  By the first part of the proof of this lemma, it
is enough to show there is a non-constant Darboux function $h\colon
M\to\real$.  Fix $x_0\in M$ and let $h\colon M\to\real$ be the 
continuous function defined by 
$h(x)=\rm{dist}(x,x_0)$.  Clearly, $h$ is non-constant and Darboux.   
\qed


\noindent\lem{lem2:b}{Suppose that $n>1$ and $M$ is a $n$-dimensional Cantor manifold.  If 
$n\geq k\geq 1$ and $f\in k\dar(M,\real)$, then there is a 
collection of pairwise disjoint non-degenerate continua $\{C_{\alpha}\}_{\alpha\in\cuum}$ 
such that $f|_{C_{\alpha}}$ is Darboux for all $\alpha\in\cuum$.  
Moreover, if $n>k$ then we may assume 
$f|_{C_{\alpha}}$ is constant for all $\alpha\in\cuum$.}
\proof
Let $M$ be an $n$-dimensional Cantor manifold and 
$f\in k\dar(M,\real)$.  We proceed by induction on $n$.  First, we establish
the  Lemma for $n=2$.
When $n=2$ and $k=1$, the result is immediate by Lemma \ref{lem2:d}.  
When $n=2$ and $k=2$, then $f=g_1+g_2$ where $g_1,g_2\in\dar(M,\real)$.  By 
Lemma \ref{lem2:d}, there is a collection of pairwise disjoint 
$1$-dimensional Cantor manifolds 
$\{C_{\alpha}\}_{\alpha\in\cuum}$ such that 
$g_1|_{C_{\alpha}}$ is constant for every $\alpha\in\cuum$.  
It follows that $f|_{C_{\alpha}}$ is Darboux for every $\alpha\in\cuum$.  
This completes the proof 
of the lemma for $n=2$ since each $C_{\alpha}$ is a continuum.

So, suppose the lemma holds for all $m$ less than $n$.  We show it also holds for
$n$.  If $k=1$, the result is immediate by Lemma
\ref{lem2:d}.  So, we  may assume $k>1$.
Let $f=g_1+\cdots+g_k$ where $g_1,\ldots,g_k\in\dar(M,\real)$.  By Lemma~
\ref{lem2:d} there is an $(n-1)$-dimensional Cantor manifold $N$ such that 
$g_1|_N$ is a constant function.  Thus, $f|_N\in (k-1)\dar(N,\real)$.  By
inductive hypothesis there  is a collection of pairwise disjoint continua, 
$\{C_{\alpha}\}_{\alpha\in\cuum}$, such that 
$C_{\alpha}\subseteq N\subseteq M$ and 
$f|_{C_{\alpha}}$ is Darboux for every $\alpha\in\cuum$.  Finally, if 
$n>k$ then $n-1>k-1$.  So we may assume by inductive hypothesis that 
$f|_{C_{\alpha}}$ is constant for every $\alpha\in\cuum$ 
since $f|_{N}$ is constant.  
So, the lemma holds for all $n\geq 2$.\qed

\noindent\lem{lem2:e}{Let $n>1$,  
$n\geq k\geq 1$, and $f\in k\dar(\real^n,\real)$.  
Then, there is a collection of pairwise disjoint non-degenerate continua
$\{C_{\alpha}\}_{\alpha\in\cuum}$   such that $f|_{C_{\alpha}}$ is Darboux for each
$\alpha\in\cuum$.   Moreover, if $n>k$, then we may assume 
$f|_{C_{\alpha}}$ is constant for every $\alpha\in\cuum$.}
\proof
If $f$ is Darboux then its restriction to $[0,1]^n$ is also 
Darboux.  By Proposition~\ref{prop2:five}, $[0,1]^n$ is an $n$-dimensional Cantor 
manifold.  The lemma now follows from Lemma \ref{lem2:b}.\qed

Lemma~\ref{lem2:e} has an easy consequence.

\noindent\lem{lem2:f}{Let $n>1$,  
$n\geq k\geq 1$, and $f\in k\dar(\real^n,\real)$.  Then either $f$ is constant 
on some non-degenerate continuum or there is a rational number $q$ 
such that $|f^{-1}(\{q\})|=\cuum$.}
\proof
By Lemma \ref{lem2:e} there are pairwise disjoint non-degenerate 
continua $\{C_{\alpha}\}_{\alpha\in\cuum}$ such that $f|_{C_{\alpha}}$ is 
Darboux for every $\alpha \in\cuum$.  So, if $f$ is constant on no non-degenerate 
continuum, $f[C_{\alpha}]$ must be a 
non-degenerate interval for each $\alpha\in\cuum$ and must contain a rational 
number.   It follows that $|f^{-1}(q)|=\cuum$ for some $q\in\rational$.\qed

We now prove Theorem~\ref{thm2:five}.

\bigskip
\noindent{\sc{Proof of Theorem~\ref{thm2:five}.}}
By Lemma \ref{lem2:f}, if $f\in n\dar(\real^n,\real)$, then $f$ is continuous on 
a set of cardinality continuum.  In particular, $f$ is not a
Sierpi\'{n}ski-Zygmund function.\qed

To get the upper bound of Theorem~\ref{thm2:one}, we will 
need the following combinatorial lemma which concerns the
cardinal 
$\diff_\cuum$.  

\noindent\lem{lem2:h}{$\diff_\cuum=\kappa$ where 
\begin{equation}\notag
\kappa=\min\{|F|\colon F\subseteq \cuum^{\cuum} \&\ (\forall G\in
[\cuum^\cuum]^\omega) (\exists f\in F)(\forall g\in G)(|[f=g]|<\cuum)\}.
\end{equation}}
\proof
We show that $\kappa\leq\diff_\cuum$.  
Let $V=\cuum\times\omega$ and $W=\cuum^\omega$.  
Take $F\subseteq \cuum^V$, witnessing the definition of $\diff_{\cuum}$, i.e., 
$|F|=\diff_{\cuum}$ and 
\begin{equation}\label{eq2:31}
(\forall g\in \cuum^V)(\exists f\in F)(|[f=g]|<\cuum).
\end{equation}
It is enough to construct a family $F^{*}\subseteq W^{\cuum}$ such that
$|F^*|\leq |F|$ and 
\begin{equation}\label{eq2:32}
(\forall G\in [W^\cuum]^\omega)(\exists f\in F^*)(\forall g\in G)(|[f=g]|<\cuum).
\end{equation}
For every $f\in F$ let $f^*\in W^{\cuum}$ be defined so that $f^{*}(\alpha)(n)=f(\alpha,n)$ 
for every $\alpha\in\cuum$ and $n\in\omega$.  
Let $F^*=\{f^*\colon f\in F\}$ and note that $|F^*|\leq |F|$.  We show that $F^*$
satisfies 
(\ref{eq2:32}).  Let $G\in [W^{\cuum}]^{\omega}$ and enumerate $G$ by 
$G=\{g_n \colon n\in\omega\}$.  Define $g'\in \cuum^V$ so that 
$g'(\alpha,n)=g_n(\alpha)(n)$ for $\alpha\in\cuum$ and $n\in\omega$.   
By (\ref{eq2:31}), there is an $f\in F$ such that $|[f=g'|<\cuum$.  
We show that $|[f^*=g_n]|<\cuum$ for every $n\in\omega$.  By way of contradiction, 
assume that $|[f^*=g_n]|=\cuum$ for some $n\in\omega$.  Then, 
$f^*(\alpha)(k)=g_n(\alpha)(k)$ for every $k\in\omega$ and $\alpha\in [f^*=g_n]$.  
In particular, we have 
$f^*(\alpha)(n)=g_n(\alpha)(n)$ for each $\alpha\in [f^*=g_n]$.  
So, $f(\alpha,n)=g'(\alpha,n)$ for every $\alpha\in [f^*=g_n]$.  
Since $|[f^*=g_n]|=\cuum$ we have $|[f=g']|=\cuum$ , contradicting 
our choice of $f$. 
So $F^*$ satisfies (\ref{eq2:32}) and
$\kappa\leq\diff_\cuum$.   The other inequality is trivial.\qed

We may now prove the remaining inequality for $m=1$.

\noindent\lem{lem2:i}{If $n>1$, then $\add(n\dar(\real^n,\real))\leq\diff_{\cuum}$.}
\proof
By Lemma \ref{lem2:h} there is a family $F\subseteq (\real)^{\real^n}$ such that 
$|F|=\diff_{\cuum}$ and 
\begin{equation}\label{eq2:33}
\left(\forall H\in \left[(\real)^{\real^n}\right]^\omega\right)\left(\exists f\in
F\right)\left(\forall h\in H\right)\left(|[f=h]|<\cuum\right).
\end{equation}
It is enough to show that $F$ satisfies 
\begin{equation}\label{eq2:34}
\left(\forall g\in \left(\real\right)^{\real^n}\right)\left(\exists f\in
F\right)\left(f+g\notin n\dar\left(\real^n,\real\right)\right).
\end{equation}
So let $g\in (\real)^{\real^n}$ be arbitrary.  To find the appropriate element of 
$F$ we must first define some other functions. 

Let $\{r_{\alpha}\}_{\alpha\in\cuum}$ be an enumeration of $\real$ and 
$\{B_{\alpha}\}_{\alpha\in\cuum}$ be a partition of $\real^n$ into Bernstein sets.  
Define $k\colon\real^n\to\real$ so that $k^{-1}(\{r_{\alpha}\})=B_{\alpha}$ for 
each $\alpha\in\cuum$.  Also, for each $q\in\rational$ let
$k_q\colon\real^n\to\real$ be such that $k_q[\real^n]=\{q\}$.  

Now, let $H=\{k_q-g\colon q\in\rational\}\cup\{k-g\}$.  By (\ref{eq2:33})
there is an 
$f\in F$ such that $|[f=h]|<\cuum$ for all $h\in H$.  We now show that 
$f+g\notin n\dar(\real^n,\real)$ using Lemma \ref{lem2:f}.  We first claim that $f+g$ is constant on 
no non-degenerate continuum in $C\in\real^n$.  By way of contradiction, assume there is a 
non-degenerate continuum $C\in\real^n$ such that $(f+g)|_{C}$ is constant.  Then, there is 
an $\alpha\in\cuum$ such that $(f+g)[C]=\{r_{\alpha}\}$.  Since $B_{\alpha}$ 
is a Bernstein set, $|C\cap B_{\alpha}|=\cuum$.  But 
$f(x)=r_{\alpha}-g(x)=(k-g)(x)$ for each $x\in C\cap B_\alpha$, which contradicts our 
choice of $f$.  So the claim is established.  Next, we claim that
$|(f+g)^{-1}(\{q\})|<\cuum$  for each $q\in\rational$.  This follows from our
choosing $f$ so that 
$|[f=k_q-g]|<\cuum$ 
for each $q\in\rational$.   Thus, $(f+g)$ is constant on no 
non-degenerate continuum and $|(f+g)^{-1}(q)|<\cuum$ for every $q\in\rational$. 
So, by  Lemma~\ref{lem2:f}, $f+g\notin n\dar(\real^n,\real)$.  Therefore, $F$
satisfies (\ref{eq2:34}), which completes the proof.\qed  




Finally, we generalize Lemma \ref{lem2:i} to include more general range spaces.

\noindent\lem{lem2:j}{If $n,m \geq 1$, then $\add(n\dar(\real^n,\real^m))\leq\diff_{\cuum}$.}
\proof
By Lemma \ref{lem2:i} there is an $F\subseteq\real^{\real^n}$ such that
$|F|=\diff_{\cuum}$ and $F$ satisfies the definition of
$\add(n\dar(\real^n,\real))$, i.e.,  
\begin{equation}\label{eq2:35}
\left(\forall g\in (\real)^{\real^n}\right)(\exists f\in F)(f+g\notin
n\dar(\real^n,\real)).
\end{equation}
Let $\pi\colon\real^m\to\real$ be the projection function of $\real^m$ onto some fixed 
coordinate.  Since $\pi$ is onto, for every $f\in F$ there
is an
$f^*\colon\real^n\to\real^m$ such that $f=\pi\circ f^*$.  Let $F^*=\{f^*\colon
f\in F\}$ and note that $|F^*|\leq|F|=\diff_{\cuum}$.  We will be done if we
show that $F^*$  satisfies
\begin{equation}\label{eq2:vill}
\left(\forall g\in (\real^m)^{\real^n}\right)(\exists f^{*}\in
F^{*})(f^{*}+g\notin n\dar(\real^n,\real^m)).
\end{equation}
So let $g\in (\real^m)^{\real^n}$ be arbitrary and put $g_1=\pi\circ g$.  By
(\ref{eq2:35}), there is an $f\in F$ such that $(f+g_1)\notin
n\dar(\real^n,\real)$.   We claim that $(f^*+g)\notin n\dar(\real^n,\real^m)$, 
which will complete the proof.   Assume that $(f^*+g)\in
n\dar(\real^n,\real^m)$. Then there  exist
$\{d_j\in\dar(\real^n,\real^m)\}_{j=1}^{n}$ such that
\begin{equation}\notag
f^*+g=d_1+\ldots+d_{n}. 
\end{equation}
Since the projection onto a coordinate is additive and continuous, 
we now have 
\begin{eqnarray*}
f+g_1 
& = & (\pi\circ f^*)+(\pi\circ g) \notag \\
& = & \pi\circ (f^*+g) \notag \\
& = &(\pi\circ d_1)+\ldots+(\pi\circ d_{n})\in n\dar(\real^n,\real), \notag
\end{eqnarray*}
which contradicts our choice of $f$.  Thus, $F^*$ satisfies
(\ref{eq2:vill}).\qed

\section{Example \ref{ex:1}}\label{sec:ex1}
In this section, we construct the 
Baire class 1 function of Example \ref{ex:1} and use Lemmas \ref{lem2:e} and
\ref{lem2:f} to show it has the required properties.  We denote the
linear span of a collection of  real numbers $A$ over $\rational$ by
$\lin_{\rational}(A)$.  Recall that if
$A$ is countable, then so is $\lin_{\rational}(A)$.    
\noindent\lem{lem2:b1}{Let $n\geq1$.  There is a Baire class 1 function $f\colon\real^n\to\real$ 
such that $|f[\real^n]|=\omega$ and $f[C]$ is an interval for no non-trivial
connected subset $C$ of
$\real^n$.}
\proof 
Let $A=\{\alpha_{i}\colon i\in n\}$ be a collection of distinct real numbers which are
linearly independent over the rationals.  Let $\{q_{k}\}_{k=1}^{\infty}$ be an enumeration of
$\rational$.  For each $i\in n$, define $g_{i}\colon\real\to\real$ by 
\begin{equation*}
g_i(x)=
\begin{cases}
\alpha_{i}/k& \text{if $x=q_{k}$}\\
0& \text{if $x\notin\rational$.}
\end{cases}
\end{equation*}
Notice that each $g_i$ is in Baire class 1.  Define $f\colon\real^n\to\real$ by 
\[f(\langle x_1,\ldots ,x_n\rangle)=\sum_{i=1}^n g_{i}(x_i).\]
It is easy to check that $f$ is in Baire class 1.  We show that $f$ has the desired 
properties.  We first notice that $f[\real^n]\subseteq\lin_{\rational}(A)$ so 
$|f[\real^n]|=\omega$. We now check the other property.
Let $C\subseteq\real^n$ be a non-trivial connected set.  Since $C$ is non-trivial, 
there is some $1\leq i\leq n$ such that the projection $\pi_i[C]$ of $C$ onto the 
$i^{\text{th}}$ coordinate is an interval with non-empty
interior.  Let 
$t\in \pi_{i}[C]\cap\rational$, $s\in\pi_{i}[C]\setminus\rational$, and
$p,q\in C$  be such that $\pi_i(q)=t$ and $\pi_i(p)=s$.  By definition of $f$ we
have 
$f(q)\notin\lin_{\rational}(A\setminus\{\alpha_i\})$ and 
$f(p)\in\lin_{\rational}(A\setminus\{\alpha_i\})$.  In particular, $f(q)\neq
f(p)$ so $f|_C$ is not constant.  Since $f|_C$ is not constant and
$|f[\real^n]|=\omega$, it is clear that 
$f[C]$ is not an interval.\qed

It is immediate by Lemma \ref{lem2:e} that $f\notin n\dar(\real^n,\real)$ for all
$n\geq 1$.  We now show that $f$ is not the sum of an almost continuous
function, $h$, and $n-1$ Darboux functions when $n>1$.  By Lemma \ref{lem2:e}, for
any
$g\in (n-1)\dar(\real^n,\real)$ there is a non-degenerate continuum $C$ upon which
$g$ is constant.  Thus, $(h+g)|_{C}$ would be
almost continuous (Restrictions of almost continuous functions to closed sets
are almost continuous \cite{STAL}.); in which case $(h+g)[C]$ is an interval
\cite[Theorem~1.7]{NATsurvey}.  Therefore, $h+g\neq f$. 


\section{Proof of Theorem~\ref{thm2:four}}
We first show that the containment $\acon(\real^n,\real^m)\subseteq
n\dar(\real^n,\real^m)$ holds.
\noindent\lem{lem2:hap}{$\acon(\real^n,\real^m)\subseteq n\dar(\real^n,\real^m)$.}
\proof 
The Lemma is known for the case $n=1$; see \cite{BHL}.  We proceed by
induction on $n$.  Assume $n-1\geq 1$ and   
$\acon(\real^{n-1},\real^m)\subseteq(n-1)\dar(\real^{n-1},\real^m)$.   
Let $g\in\acon(\real^n,\real^m)$.  We show $g\in n\dar(\real^n,\real^m)$.

Let $\{A_k\}_{k\in n}$ be a partition of $\real$ into $n$ $\cuum$-dense sets.  
Define 
$h\colon\real\to\real^m$ so that for each $p\in\real^m$ and $k\in n$  
\begin{equation*}
\{y\in A_k\colon h(y)=p\} \text{ is dense in $\real$}.
\end{equation*}
Let $\pi\colon\real^n\to\real$ denote the projection of $\real^n$
onto a fixed coordinate. Since restrictions of almost
continuous functions to closed sets are almost continuous
\cite{STAL}, 
$g|_{\pi^{-1}(y)}$ is almost continuous for every $y\in\real$.  Note that $\pi^{-1}(r)$ is homeomorphic to $\real^{n-1}$ for every $r\in\real$.
By inductive hypothesis, for each $l\in n$ and $y\in A_l$ we may find Darboux
functions 
$\{g_{y_k}\colon\pi^{-1}(y)\to\real^m\colon k\in n\setminus\{l\}\}$ such that 
\begin{equation*}
\sum_{\{k\in n-1\colon k\neq l\}}  g^y_k=g|_{\pi^{-1}(y)}-h(y).
\end{equation*}
Now for each $k\in n$, define $g_{k}\colon\real^n\to\real^m$ by 
\begin{equation}\label{eq2:39}
g_k(p)= 
\begin{cases}
h(y)& \text{ if $\pi(p)=y\in A_{k}$, }\\ 
g^y_k(p)& \text{ if $\pi(p)=y\notin A_{k}$.}
\end{cases}
\end{equation}
We claim that the functions of (\ref{eq2:39}) are as desired.  Let $k\in
n$.  To see that $g_k$ is Darboux put $A=A_k$ and  
$B=\cup\{A_i\colon i\in n \text{ and } i\neq k\}$, 
then apply Lemma~\ref{lem2:a}.  We now show that
\begin{equation}\label{eq2:40}
\sum_{k=0}^{n-1} g_k=g.
\end{equation}
If $p\in\real^n$, then there is an $l\in n$ such that $y\in A_l$, 
$p\in\pi^{-1}(y)$, and   
\begin{eqnarray*}
\sum_{k=0}^{n-1} g_k(p)
& = & h(y)+\sum_{\{k\in n\colon k\neq l\}}g_{y_k}(p) \notag \\
& = & h(y)+(g(p)-h(y)) \notag \\
& = & g(p). \notag
\end{eqnarray*}
So (\ref{eq2:40}) holds, completing the inductive step.    
\qed  

We now show that the containment of Lemma \ref{lem2:hap} is proper. In fact, we
show more.  We note that the Lemma below has been shown for the case $n=2$ in
\cite[Example 1.6]{NATsurvey} and for $n=1$ in \cite{BHL}.   
\noindent\lem{lem2:hap1}{
$\dar(\real^n,\real^m)\setminus\acon(\real^n,\real^m)\neq\emptyset$.}
\proof 
Since $\real$
can be embedded in $\real^m$ for any $m\geq 1$, it is enough to show
that 
\begin{equation}\label{eq2:rrr}
\dar(\real^n,\real)\setminus\acon(\real^n,\real)\neq\emptyset.
\end{equation}
Take $f\in\dar(\real,\real)\setminus\acon(\real,\real)$, $n>1$ and let 
$\pi\colon\real^n\to\real$ be the projection of $\real^n$ onto the first
coordinate, i.e., $\pi\langle r_0,\ldots ,r_{n-1}
\rangle=r_0$.  Define
$g\colon\real^n\to\real$ by $g=f\circ\pi$.  Since
$f$ is Darboux and $\pi$ is continuous, $g\in\dar(\real^n,\real)$.  We show
that $g\notin\acon(\real^n,\real)$.  Consider the set 
$S=\{p\in\real^n\colon p=\langle r,0,\ldots,0\rangle\ \&\, r\in\real\}$.  Since
$g|_S$ is an exact copy of $f$, it follows that
$g|_S\notin\acon(\real^n,\real)$.  However, restrictions of almost continuous
functions to closed sets are almost continuous \cite{STAL}, so we must conclude
that
$g$ is not almost continuous.\qed

We now prove the last part of Theorem~\ref{thm2:four}.  We
will need the following fact which may be found in
\cite{NATsurvey}.  

\noindent\prop{prop2:a1}{Let $n,m\in\omega\setminus\{0\}$.  There exists a
family $\mathcal{B}$ of closed sets in 
$\real^n\times\real^m$, a blocking family, with the following properties:
\begin{description}
\item[(i)] $f\in\acon(\real^n,\real^m)$ if and only if $f\cap B\neq\emptyset$ for each
$B\in\mathcal{B}$ and   
\item[(ii)] for every $B\in\mathcal{B}$ the projection of $B$ onto $\real^n$ is a non-degenerate connected set.
\end{description}}

\noindent\lem{lem2:jaba}{If $n\geq 2$ and $m \geq 1$, then there is an almost continuous 
function $f$ such that $f\notin (n-1)\dar(\real^n,\real^m)$.}
\proof
Let $\pi\colon\real^m\to\real$ be the projection function of $\real^m$ onto some fixed 
coordinate.  Since $\pi$ is additive and continuous, if
$f\in (n-1)\dar(\real^n,\real^m)$, then $(\pi\circ f)\in
(n-1)\dar(\real^n,\real)$.   It then follows from Lemma \ref{lem2:e} that
$(\pi\circ f)$ is  constant on some non-degenerate continuum in $\real^n$.  So,
if we show there is a $g\in\acon(\real^n,\real^m)$ such that $g_1=\pi\circ g$
is constant on no non-degenerate continuum in $\real^n$, we will be done.  We
now construct $g$.  Let 
$\{P_{\langle\alpha,i\rangle}\}_{\langle\alpha,i\rangle\in\cuum\times 2}$ be a 
partition of $\real^n$ into Bernstein sets.  Let
$\{B_{\alpha}\}_{\alpha\in\cuum}$ be an enumeration of the elements of the blocking family of
Proposition~\ref{prop2:a1}, and for each $\alpha\in\cuum$ let 
$B_{\alpha}^*$ denote the
projection of $B_{\alpha}$ onto $\real^n$.  Take 
an enumeration $\{r_{\alpha}\}_{\alpha\in\cuum}$ of $\real^n$.  By
Proposition~\ref{prop2:a1},
$|B_{\alpha}^*|=\cuum$ and $B^*_{\alpha}$ is an $F_{\sigma}$-set. It follows that
$|P_{\langle\alpha,i\rangle}\cap B_{\alpha}^*|=\cuum$ for every 
$\langle\alpha,i\rangle\in\cuum\times 2$.  
For each $\alpha\in\cuum$, define $h^{\alpha}\colon\real^n\to\real^m$ so that
$h^{\alpha}|_{P_{\langle\alpha,0\rangle}\cap B^*_{\alpha}}\subseteq
B_{\alpha}$, $h^{\alpha}[P_{\langle\alpha,1\rangle}\cap
B^*_{\alpha}]=\{r_{\alpha}\}$, and let
$h^{\alpha}$ be arbitrary elsewhere.  Let $g\colon\real^n\to\real^m$ be defined
by
\begin{equation*}
g(x)=
\begin{cases}
h^{\alpha}(x)& \text{if $x\in B^*_{\alpha}$}\\
0& \text{otherwise.}
\end{cases}
\end{equation*}
We claim $g$ is as desired.   Since
$g|_{P_{\alpha,0}\cap B^*_{\alpha}}=h^{\alpha}|_{P_{\alpha,0}\cap
B^*_{\alpha}}\subseteq B_{\alpha}$ for every $\alpha\in\cuum$,
$g\in\acon(\real^n,\real^m)$.  We now show that
$g_1=\pi\circ g$ is constant on no continuum.  Since Bernstein sets intersect
all perfect sets, they intersect all non-degenerate continua.  It follows from
the way we defined $g$ that if $C\subseteq\real^n$ is a non-degenerate continuum,
then $(\pi\circ g)[C]=\pi[\real^m]=\real$.  Thus, $g_1$ is constant on no
non-degenerate subcontinuum of $\real^n$.\qed


\section{Proof of Theorem~\ref{thm2:onea}}
By Theorem~\ref{thm2:four} $\acon(\real^n,\real^m)\subseteq
n\dar(\real^n,\real^m)$.  By Proposition~\ref{prop1:1}(iii), we have, using Theorem~\ref{thm2:one}, 
$\add(\acon(\real^n,\real^m))\leq\add(n\dar(\real^n,\real^m))=\diff_{\cuum}$.  
So we only need to prove that $\add(\acon(\real^n,\real^m))\geq\diff_{\cuum}$.  
To see this, 
let $F\subseteq (\real^m)^{\real^n}$ and $|F|<\diff_{\cuum}$.  
We must find a $g\colon\real^n\to\real^m$ such that 
$g+F\subseteq\acon(\real^n,\real^m)$.  Let 
$\{P_{\alpha}\}_{\alpha\in\cuum}$ be a partition of $\real^n$ into 
Bernstein sets, 
$\{B_{\alpha}\}_{\alpha\in\cuum}$ be an enumeration of the elements 
of the blocking family of
Proposition~\ref{prop2:a1}; and for each $\alpha\in\cuum$ let $B_{\alpha}^*$ 
denote the
projection of $B_{\alpha}$ onto $\real^n$.  By Proposition~\ref{prop2:a1}
$|B_{\alpha}^*|=\cuum$ and $B_{\alpha}^*$ is an $F_{\sigma}$-set. 
It follows that $|P_{\alpha}\cap B_{\alpha}^*|=\cuum$ for each 
$\alpha\in\cuum$.  For each $\alpha\in\cuum$, define 
$h^{\alpha}\colon\real^n\to\real^m$ so that $h^{\alpha}|_{P_{\alpha}\cap
B^*_{\alpha}}\subseteq B_{\alpha}$ and let $h^{\alpha}$ be arbitrary elsewhere. 
Put
\[F^*=\{h^{\alpha}-f\colon f\in F \text{ and } \alpha\in\cuum\}.\]  Since
$\cuum<\diff_{\cuum}$, it follows that $|F^*|<\diff_{\cuum}$.  So, for
every $\alpha\in\cuum$ there is a function 
$g_{\alpha}\colon (P_{\alpha}\cap B^*_{\alpha})\to\real^m$ such that 
$|\{x\in P_{\alpha}\cap B^*_{\alpha}\colon g_{\alpha}(x)=(h^{\alpha}-f)(x)\}|=\cuum$.  Let $g\colon\real^n\to\real^m$ be defined by
\begin{equation*}
g(x)=
\begin{cases}
g_{\alpha}(x)& \text{if $x\in P_{\alpha}\cap B^*_{\alpha}$}\\
0& \text{otherwise.}
\end{cases}
\end{equation*}
We claim $g$ is as desired.  Let $f\in F$ and $B\in\mathcal{B}$.  There is an 
$\alpha\in\cuum$ such that $B_{\alpha}=B$.  By the way we defined $g$, 
we have
\begin{eqnarray*}
\left|(f+g)|_{P_{\alpha}\cap
B^*_{\alpha}}=h^{\alpha}|_{P_{\alpha}\cap B^*_{\alpha}}\right|
& = &\left|f|_{P_{\alpha}\cap
B^*_{\alpha}}+g_{\alpha}=h^{\alpha}|_{P_{\alpha}\cap B^*_{\alpha}}\right|\\  & =
& \left|g_{\alpha}=(h^{\alpha}-f)|_{P_{\alpha}\cap B^*_{\alpha}}\right|\\  & = &
\cuum.
\end{eqnarray*}  Thus, 
$|(g+f)|_{P_{\alpha}\cap B^*_{\alpha}}=h^{\alpha}|=\cuum$.  In particular,
$(f+g)\cap B_{\alpha}\neq\emptyset$.  Since $B$ was arbitrary, we conclude that
$f+g\in\acon(\real^n,\real^m)$.\qed



\section{Proof of Theorem~\ref{thm2:two} }
Our first goal is to show that $\add_{n,n-1}(\conn(\real^n,\real))\leq\cuum^+$ for $n>1$.  
The following proposition can be found in \cite[Lemma 3.1]{CR} where it is
stated for 
$\real^{\real}$.  The proof is essentially the same for $(\real)^{\real^n}$.  

\noindent\prop{prop2:9}{There is a family $F\subseteq (\real)^{\real^n}$ of
cardinality $\cuum^+$ such that for  every distinct $f,h\in F$, every perfect set
$P\subseteq\real^n$, and every
$k<\omega$ there exists an $x\in P$ such that $|f(x)-h(x)|\geq k$.}   

The next proposition can be found in \cite[Proposition~2.8]{CW}.  

\noindent\prop{prop2:10}{If $n>1$ and $g\in n\conn(\real^n,\real)$, then there exists a perfect set 
$P$ such that $g|_{P}$ is continuous.} 

\noindent\lem{lem2:m}{If $n>1$, then $\add_{n,n-1}(\conn(\real^n,\real))\leq\cuum^+$.}
\proof
It is enough to show that $\add(n\conn(\real^n,\real))\leq\cuum^+$ since,
by Proposition~\ref{prop2:two} (viii), 
$\add_{n,n-1}(\conn(\real^n,\real))\leq\add(n\conn(\real^n,\real))$. Let
$F\subseteq (\real)^{\real^n}$ be as in Proposition~\ref{prop2:9}.  We claim
that $g+F\subseteq
n\conn(\real^n,\real)$ for no $g\colon\real^n\to\real$.   By way of contradiction, assume that such a
$g$ exists.  By Proposition~\ref{prop2:10}, for every $f\in F$ there is a
perfect set $P_{f}$ such that the restriction of $g+f$  to $P_{f}$ is
continuous.  Since
$|F|=\cuum^+$ and there are only
$\cuum$-many perfect sets, there are $f,h\in F$ such that $P_{f}=P_{h}$.  It
follows that $f-h=(g+f)-(g+h)$  is continuous on $P_{f}$, which contradicts our
choice of $F$.\qed

 
For what follows we will use the notation of \cite[sec. 6]{CW}.     
For sets $\{ A_i\colon i\in n\}$ and $\{B_i\colon i\in n\}$
and for $f\colon n\to 2$ we let 
\begin{equation}\notag
A_i\vee_f B_i=
\begin{cases}
A_i& \text{if } f(i)=0 \\
B_i& \text{if } f(i)=1.
\end{cases}  
\end{equation}
If $j\in n$ and $C$ is a set, we define
\begin{equation}\notag
A_i\vee_f B_i\vee_j C=
\begin{cases}
C& \text{if $i=j$}\\
A_i& \text{if $i\neq j$ and $f(i)=0$} \\
B_i& \text{if $i\neq j$ and  $f(i)=1$.}
\end{cases}  
\end{equation}
We call $M\subseteq\real$ a thick meager set provided that $M$ is dense and is a 
countable  union of nowhere dense perfect sets.  Notice that any thick meager
set is 
$\cuum$-dense in $\real$.  
We will also need the following four propositions from \cite{CW}. 

\noindent\prop{prop2:29}{{\rm(\cite[Lemma 4.1]{CW})}If $G$ is a dense 
$G_{\delta}$-set in
$\real^n$, then for each $i\in n$ there is a countable dense set 
$B_i\subseteq\real$ and a thick meager set $Y_i\subseteq\real$ such that 
$B_i\cap Y_i=\emptyset$ and 
\[\prod_{i=0}^{n-1}(B_i\cup Y_i)\subseteq G.\]}

\noindent\prop{prop2:30}{{\rm(\cite[Proposition 2.3]{CW})}Let $n\geq 1$.  There is a dense
$G_{\delta}$-set
$G\subseteq\real^n$ and  a function $f\colon\real^n\to\real$ such that for any
$g\colon\real^n\to\real$ if 
$g(x)=f(x)$ for every $x\notin G$, then $g\in\conn(\real^n,\real)$.}  
Since $\conn(\real^n,\real)=\ext(\real^n,\real)$ for $n>1$, 
we have an obvious corollary of Proposition~\ref{prop2:30}. 

\noindent\cor{cor2:49}{Let $n\geq 1$.  There is a dense $G_{\delta}$-set
$G\subseteq\real^n$ and  a function $f\colon\real^n\to\real$ such that for any
$g\colon\real^n\to\real$ if 
$g(x)=f(x)$ for every $x\notin G$, then $g\in\ext(\real^n,\real)$.} 
\proof
When $n>1$ the proposition follows from Proposition~\ref{prop2:30} and the fact that 
$\conn(\real^n,\real)=\ext(\real^n,\real)$.  When $n=1$ then the 
proposition reduces to \cite[Corollary 3.4]{CR}.\qed

\noindent\prop{prop2:32}{{\rm(\cite[Lemma 4.3]{CW})}
Let $n>0$ and $G\subseteq\real^n$ be a $G_{\delta}$-set.  If 
$f\colon n\to 2$ is a function, $i\in n$, and 
$\langle b_0,\ldots,b_{n-1}\rangle\in\real^n$, then the set 
\[
\left\{x\in\real\colon\prod_{j=0}^{n-1} 
(\{b_j\}\vee_f \real\vee_i \{x\})
\subseteq  G\right\}
\] 
is a $G_{\delta}$-subset of $\real$.
}

The next proposition is a more detailed statement of
\cite[Proposition~2.4]{CW}. 
\noindent\prop{prop2:33}{Let $G$ be a dense $G_{\delta}$-subset of 
$\real^n$.  Then there exist countable dense sets
$\{B_i\}_{i=0}^{n-1}$ of $\real$ and homeomorphisms
$h_1,\ldots,h_n\colon\real^n\to\real^n$ such that
\begin{equation*} 
\prod_{i=0}^{n-1}(B_i\vee_{f}\real)\subseteq G\cup\left(\bigcup_{i=1}^{k}
h_i(G)\right)
\end{equation*}
for each $f\in 2^n$ with $|f^{-1}(1)|=k$.  In particular, if $k=n$ then, 
\begin{equation*} 
\real^n=G\cup\left(\bigcup_{i=1}^{n} h_i(G)\right).
\end{equation*}}
\proof
The statement of the proposition follows directly from consideration of the 
inductive step of the proof of Proposition~2.4 in \cite{CW}.  \qed

\noindent\lem{lem2:51}{Let $n\geq 1$ and $G\subseteq\real^n$ be a dense $G_{\delta}$. 
There exist homeomorphisms $h_1,\ldots, h_{n-1}\colon\real^n\to\real^n$ and
meager subsets $\{M_i\}_{i=0}^{n-1}$ of $\real$ such that 
\begin{equation}\label{eq2:a}
\real^n\setminus \left(G\cup \left(\bigcup_{j=1}^{n-1}
h_j[G]\right)\right)\subseteq\prod_{i=0}^{n-1}M_i.
\end{equation} } 
\proof
Let $\{B_i\}_{i=0}^{n-1}$ be the countable dense subsets of $\real$ from 
Proposition~\ref{prop2:33} and 
$h_{1},\ldots ,h_{n-1}\colon\real^n\to\real^n$ be the first $n-1$ 
homeomorphisms of Proposition~\ref{prop2:33}.  Then  
\begin{equation}\label{eq2:b}
\prod_{j=0}^{n-1}(B_j\vee_{f}\real)\subseteq
G\cup\left(\bigcup_{i=1}^{n-1} h_i(G)\right)
\end{equation}
for each $f\in 2^n$ such that $|f^{-1}(1)|=n-1$.  Let $i\in n$ and 
$f_i\colon n\to 2$ be such that $f_i^{-1}(0)=\{i\}$.  
By Proposition~\ref{prop2:32}, for each $b=\langle b_0,\ldots,
b_{n-1}\rangle\in B_0\times\cdots\times B_{n-1}$ there is a $G_{\delta}$-subset
$K_i^b$ of $\real$ such that
\[\prod_{j=0}^{n-1}(\{b_j\}\vee_{f_i}\real\vee_i K_i^b)\subseteq
G\cup\left(\bigcup_{j=1}^{n-1}h_j[G]\right).\]   By (\ref{eq2:b}), we know
that
$B_i\subseteq K_i^b$; so, $K_i^b$ is a dense
$G_{\delta}$-subset of $\real$.  So the set 
\[K_i=\bigcap\{K_i^b\colon b\in B_0\times\cdots\times B_{n-1}\}\]
is a dense $G_{\delta}$-subset of $\real$, satisfying 
\[\prod_{j=0}^{n-1}(B_j\vee_{f_i}\real\vee_i K_i)\subseteq
G\cup\left(\bigcup_{j=1}^{n-1}h_j[G]\right).\]   Since
$f_{i}[n\setminus\{i\}]=\{1\}$, we have 
\[\prod_{j=0}^{n-1}(K_j\vee_{f_i}\real)=\prod_{j=0}^{n-1}(B_j\vee_{f_i}\real\vee_i
K_i)\subseteq G\cup\left(\bigcup_{j=1}^{n-1}h_j[G]\right).\]   Letting
$M_i=\real\setminus K_i$ for each $i\in n$, we have 
\begin{equation*}
\real^n\setminus \left(G\cup \left(\bigcup_{j=1}^{n-1}h_j[G]\right)\right)
\subseteq\real^n\setminus\bigcup_{i=0}^{n-1}\left(\prod_{j=0}^{n-1}(K_j\vee_{f_i}\real)\right)
=\prod_{i=0}^{n-1}M_i.
\end{equation*}
Thus, (\ref{eq2:a}) holds. \qed

\noindent\lem{lem2:53}{Let $\{M_i\}_{i=0}^{n-1}$ be meager subsets of $\real$ and 
$M=\prod_{i=0}^{n-1}M_i\subseteq\real^n$.  For any 
dense $G_{\delta}$-set $G\subseteq\real^n$ there exist homeomorphisms 
$\{h_{\xi}\colon\real^n\to\real^n\}_{\xi\in\cuum}$ such that 
\begin{description}
\item[(i)] $h_{\xi}[M]\cap h_{\zeta}[M]=\emptyset$ for all $\zeta<\xi<\cuum$
and
\item[(ii)] $h_{\xi}[M]\subseteq G$.
\end{description}}
\proof
By Proposition~\ref{prop2:29} there exist thick meager subsets 
$\{N_i\}_{i=0}^{n-1}$ of $\real$ such that 
\begin{equation}\label{eq2:68}
\prod_{i=0}^{n-1} N_i\subseteq G.
\end{equation}  
Let $\{N_{i,\xi}\}_{\xi\in\cuum}$ be a partition of 
$N_i$ into thick meager sets.  By \cite[Lemma 3.2]{WG}, for each $i\in n$ and 
$\xi\in\cuum$ there exists a
homeomorphism $h_{i,\xi}\colon\real\to\real$ such that 
\begin{equation}\label{eq2:69}
h_{\xi}[M_i]\subseteq N_{i,\xi}.
\end{equation}  
For each $\xi\in\cuum$ let 
$h_{\xi}=h_{0,\xi}\times\cdots\times h_{n-1,\xi}\colon\real^n\to\real^n$.  The 
homeomorphisms $\{h_{\xi}\}_{\xi\in\cuum}$ satisfy (i) and (ii).   Indeed,
$\{N_{i,\xi}\}_{\xi\in\cuum}$ partitions $N_i$ for each $i\in n$ so (i) follows
from (\ref{eq2:69}).  Using (\ref{eq2:68}) and (\ref{eq2:69}), we
conclude (ii).\qed 


\noindent\lem{lem2:54}{Let $n\geq 1$.  There exist a meager subset $M$ of $\real^n$, 
meager subsets $\{M_i\colon i\in n\}$ of $\real$, and a function 
$f\colon\real^n\to\real$ such that 
\[M=\prod_{i=0}^{n-1} M_i\]
and for any $g\colon\real^n\to\real$, if $g|_M=f|_M$, then 
$g\in n\ext(\real^n,\real)$.}
\proof
By Corollary~\ref{cor2:49} there is an extendable function
$l\colon\real^n\to\real$  and a dense $G_{\delta}$-set $G\subseteq\real^n$   
such that for any function $k\colon\real^n\to\real$ if 
$k(x)=l(x)$ for every $x\notin G$, then $k\in\ext(\real^n,\real)$.  
Let $h_{0}\colon\real^n\to\real^n$
be the identity homeomorphism. Pick $h_1,\ldots, h_{n-1}\colon\real^n\to\real^n$ 
and $\{M_i\}_{i=0}^{k-1}$, as in Lemma~\ref{lem2:51} for $G$.  Put 
\[M=\prod_{i=0}^{n-1} M_i\]  
and let $f\colon\real^n\to\real$ be defined by 
\begin{equation*}
f(x)=
\begin{cases}
\sum_{i=0}^{n-1} (l\circ h_{i}^{-1})(x)& \text{ if $x\in M$}\\
0& \text{ otherwise.}
\end{cases}
\end{equation*}
We show $f$ is as desired.  Let $g\colon\real^n\to\real$ be such that $g|_M=f|_M$.  
We show $g\in n\ext(\real^n,\real)$.  Let $G_{0}=G$, and for each $0<i\leq n$ let 
\[G_i=h_i[G]\setminus\left( \bigcup_{j=0}^{i-1} h_j[G]\right).\]
Note that $\{G_i\colon i\in n\}$ is a collection of pairwise disjoint sets such that \[\real^n\setminus M\subseteq \bigcup_{i=0}^{n-1} G_i.\]
For each $i\in n$ we define $g_i\colon\real^n\to\real$ by 
\begin{equation*}
g_i(x)=
\begin{cases}
g(x)-\sum_{\{j\in n\colon j\neq i\}}(l\circ h^{-1}_j)(x)& \text{ if $x\in G_i$}\\
(l\circ h_{i}^{-1})(x)& \text{ if $x\notin G_i$.}
\end{cases}
\end{equation*}
Now $g=g_0+\cdots +g_{n-1}$.  We show that each $g_i$ is in 
$\ext(\real^n,\real)$.  Fix an $i\in n$.  If 
$x\in\real^n\setminus G$, we have 
$g_i(h_i(x))=(l\circ h_{i}^{-1})(h_{i}(x))=l(x)$ since $h_i(x)\notin G_i$.  Thus,
$(g_i\circ h_i)\in\ext(\real^n,\real)$ by our choice of $l$.   Since the
composition of a extendable function and a homeomorphism is extendable
\cite[Lemma 1]{what}, it then follows that 
$g_i=((g_i\circ h_i)\circ h^{-1}_i)\in\ext(\real^n,\real)$.\qed


\noindent\lem{lem2:55}{$\add_{n,n-1}(\ext(\real^n,\real))\geq\cuum^+$.}
\proof
Let $\{g_{\xi}\}_{\xi\in\cuum}$ be a collection of functions from $\real^n$ into 
$\real$.  Suppose $M$, $f\colon\real^n\to\real$ are as in
Lemma \ref{lem2:54}.  Take 
an extendable function $f^*\colon\real^n\to\real$ and 
a dense $G_{\delta}$-set $G\subseteq\real^n$, as in Corollary~\ref{cor2:49}. 
By Lemma \ref{lem2:53} there exist homeomorphisms
$\{h_{\xi}\colon\real^n\to\real^n\}_{\xi\in\cuum}$ such that
\begin{description}
\item[(i)] $h_{\xi}[M]\cap h_{\zeta}[M]=\emptyset$ for all $\zeta<\xi<\cuum$.
\item[(ii)] $h_{\xi}[M]\subseteq G$ for all $\xi\in\cuum$.
\end{description}
Define $l\colon\real^n\to\real$ so that 
\begin{equation*}
l(x)=
\begin{cases}
g_{\xi}(x)-(f\circ h^{-1}_{\xi})(x)& \text{if $x\in h_{\xi}[M]$ for some
$\xi\in\cuum$}\\ f^*(x)& \text{ otherwise.} 
\end{cases}
\end{equation*}
Note that $l\in\ext(\real^n,\real)$ by our choice of $f^*$ and $G$.  We will be done 
if we show that $g_{\xi}-l\in n\ext(\real^n,\real)$ for each $\xi\in\cuum$.  So fix 
$\xi\in\cuum$.  If $x\in M$, then
$(g_{\xi}-l)(h_{\xi}(x))=g_{\xi}(h_{\xi}(x))-(g_{\xi}(h_{\xi}(x))-(f\circ
h^{-1}_{\xi})(h_{\xi}(x)))=(f\circ h^{-1}_{\xi})(h_{\xi}(x))=f(x)$.  
So for each $x\in M$ we have
$(g_{\xi}-l)(h_{\xi}(x))=f(x)$.  By our choice  of $M$ and $f$ it follows
that there exist
$f_{0,\xi},\ldots,f_{n-1,\xi}\in\ext(\real^n,\real)$ such that 
\begin{equation*}
(g_{\xi}-l)\circ h_{\xi}=\sum_{i=0}^{n-1}f_{i,\xi}.
\end{equation*} 
So, 
\begin{equation*}
(g_{\xi}-l)=\left(\sum_{i=0}^{n-1}f_{i,\xi}\right)\circ
h^{-1}_{\xi}=\sum_{i=0}^{n-1}(f_{i,\xi}\circ h^{-1}_{\xi}).
\end{equation*} 
Since the composition of an extendable function and a homeomorphism 
is extendable, $(f_{i,\xi}\circ h^{-1}_{\xi})\in\ext(\real^n,\real)$ for each 
$i\in n$.  Thus, $(g_{\xi}-l)\in n\ext(\real^n,\real)$. The proof is
complete.\qed

\section{Proof of Theorem~\ref{thm2:three}}
In this section we will use the notation of the section above.  
Additionally for $k\leq n$, we make the definition $F_{k}^n=\{f\in 2^n\colon |f^{-1}(0)|=k\}$.
We first show that 
$\add_{n,\floor {n/2}}(\dar(\real^n,\real^m))\geq\cuum^+$.  

\noindent\lem{lem2:n}{Let $n\geq k\geq 1$, $\{M_i\subseteq\real\colon i\in n\}$ be a
collection  of thick meager sets and
\[M=\bigcup_{f\in F_{k}^n}\prod_{i=0}^{n-1}(M_i\vee_f \real).\]
Then there is a function $f\colon\real^n\to\real^m$ such that for any  
$g\colon\real^n\to\real^m$ if $g|_M=f|_M$, then $g\in k\dar(\real^n,\real^m)$.}
\proof
Throughout this proof we assume that $\real^n$ is written in the form
$\real^n=\{\langle r_0,\ldots, r_{n-1}\rangle\colon r_j\in\real\text{ for each
}0\leq i\leq n-1\}$.  We do an induction on
$n$.  When n=1 the lemma is easily verified.  So, let 
$n\geq k\geq 1$ and assume the lemma holds for $n-1\geq 1$.  We now show that 
the lemma holds for $n$.  
Let $\pi\colon\real^n\to\real$ be the projection of 
$\real^n$ onto the last coordinate; and note that
$\pi^{-1}(r)$  is homeomorphic to $\real^{n-1}$ for each 
$r\in\real$.  For every $r\in\real$, put 
$M^r=M\cap \pi^{-1}(r)$.  Let $\{B_i\}_{i=0}^{k-1}$ be a partition of $M_{n-1}$
into $\cuum$-dense sets and $d\colon\real\to\real$ be a function such
that for each $y\in\real^m$ and $i\in k$  
\begin{equation*}
\{x\in B_i\colon d(x)=y\} \text{ is dense in $\real$}.
\end{equation*}

We claim that for each $r\in M_{n-1}$ we have 
\[M^r=\left(\bigcup_{f\in F_{k-1}^{n-1}}\prod_{i=0}^{n-2}(M_i\vee_f
\real)\right)\times\{r\}.\]  
Notice that $\langle x_0,\ldots, x_{n-1}\rangle\in
M\cap\pi^{-1}(r)$ provided that 
$|\{i\in n\colon x_i\in M_i\}|\geq k$ and $x_{n-1}=r\in M_{n-1}$.  It
follows that $|\{i\in n-1\colon x_i\in M_i\}|\geq k-1$. So, $x\in M^r$ if
and only if $x$ has the form $x_{n-1}=r$ and $|\{i\in n-1\colon x_i\in
M_i\}|\geq k-1$, which is to say 
\begin{equation}\label{eq2:bee}
x\in\left(\bigcup_{f\in F_{k-1}^{n-1}}\prod_{i=0}^{n-2}(M_i\vee_f
\real)\right)\times\{r\}.
\end{equation}
So if
$k>1$, we may use the inductive hypothesis to find for each $r\in M_{n-1}$ a
function 
$f^r\colon \pi^{-1}(r)\to\real^m$ such that for every $g\colon
\pi^{-1}(r)\to\real^m$ if $g|_{M^r}=f^r|_{M^r}$, then $g\in
(k-1)\dar(\pi^{-1}(r),\real^m)$.  When $k=1$, then $M^r=\pi^{-1}(r)$, and we define
$f^r\colon\pi^{-1}(r)\to\real^m$ so that $f^r[\pi^{-1}(r)]=\{d(r)\}$.

When $r\in\real\setminus M_{n-1}$, we must again consider two cases.  
If $k<n$, then
\begin{equation}\label{eq2:bee1}
M^r=\left(\bigcup_{f\in F_{k}^{n-1}}\prod_{i=0}^{n-1}(M_i\vee_f
\real)\right)\times\{r\}.
\end{equation}
So, by the inductive hypothesis, there is a function 
$f^r\colon \pi^{-1}(r)\to\real^m$ such that for every $g\colon
\pi^{-1}(r)\to\real^m$ if $g|_{M_r}=f^r|_{M_r}$, then $g\in
k\dar(\pi^{-1}(r),\real^m)$.  In the case when $n=k$, we have $M_r=\emptyset$.  
It follows from Corollary~\ref{cor2:9} that $g\in k\dar(\pi^{-1}(r),\real^m)$ for every
$g\colon\pi^{-1}(r)\to\real^m$. So, in this case, we may define $f^r\colon\pi^{-1}(r)\to\real^m$
as we please.

Let $f=\bigcup_{r\in\real}f^{r}\colon\real^n\to\real^m$.  
We show $f$ is as desired.  Let $g\colon\real^n\to\real^m$ be a function 
such that $g|_M=f|_M$.  We must show that $g\in k\dar(\real^n,\real^m)$.  For
each $r\in\real$, let $g^r=g|_{\pi^{-1}(r)}$.   For $i\in k$ and $r\in
B_i\subseteq M_{n-1}$, there exist
$\{g^r_{j}\in\dar(\pi^{-1}(r),\real^m)\colon j\in k\setminus\{i\}\}$  such that 
\begin{equation}\notag
\sum_{j\in k\setminus\{i\}} g^r_{j}=g^r 
\end{equation}
since $g^r|_{M^r}=f^r|_{M^r}$.
The above sum does not make sense when $k=1$.  
In this case, (\ref{eq2:bee}) implies that $M^r=\pi^{-1}(r)$, so 
$g^r=f^r\in\dar(\pi^{-1}(r),\real^m)$.

If $r\in\real\setminus M_{n-1}$ there exist $g^r_{0},\ldots
,g^r_{k-1}\in\dar(\pi^{-1}(r),\real^m)$  such that 
\begin{equation}\notag
g^r_0+\cdots +g^r_{k-1}=g^r 
\end{equation}
since $g^r|_{M^r}=f^r|_{M^r}$ or $k=n$.

For each $i\in k$, let 
$g_i\colon\real^n\to\real^m$ be defined by,
\begin{equation*}
g_i(x)=
\begin{cases}
d(r)& \text{if $x\in \pi^{-1}(r)$ and $r\in B_{i}$}\\
g^r_i(x)-(d(r)/(k-1))& \text{if $x\in \pi^{-1}(r)$ and $r\in M_{n-1}\setminus
B_i$}\\ g^r_i(x)& \text{if $x\in \pi^{-1}(r)$ and $r\notin M_{n-1}$}.
\end{cases}
\end{equation*}
Notice there is no problem with division by zero when $k=1$ since in this case
$M_{n-1}=B_0$.  We claim that $g_i\in\dar(\real^n,\real^m)$ for each $i\in
k$.  To see this let $A=B_i$ and $B=\real\setminus B_i$ and apply Lemma
\ref{lem2:a}.   It is easily checked that, 
\begin{equation}\notag
\sum_{i=0}^{k-1} g_i=g.
\end{equation}
So $f\colon\real^n\to\real^m$ is as desired, 
completing the inductive step.\qed

We may now prove one of the inequalities of Theorem~\ref{thm2:three}
\noindent\lem{lem2:q}{If $n\geq 1$, then
$\add_{n,\floor{n/2}}(\dar(\real^n,\real^m))\geq\cuum^+$.}
\proof
Let $\{g_{\xi}\colon\xi\in\cuum\}$ be a collection of functions from $\real^n$ 
into $\real^m$.  We must find 
$f\in(n-\floor{n/2})\dar(\real^n,\real^m)$ such that  
$g_{\xi}-f\in (\floor{n/2}+1)\dar(\real^n,\real^m)$ for every 
$\xi\in\cuum$.  

Let $\{M_{i,\xi}\subseteq\real\colon i\in n \text{ and } \xi\in\cuum\}$ and  
$\{K_i\subseteq\real\colon i\in n\}$ be thick meager sets such that for every 
$i\in n$
\begin{description}\label{eq2:76}
\item[(i)] $M_{i,\xi}\cap M_{i,\zeta}=\emptyset$ for $\xi<\zeta<\cuum$ and 
\item[(ii)] $K_i\cap (\bigcup_{\xi\in\cuum}M_{i,\xi})=\emptyset$.
\end{description}
For every $\xi\in\cuum$, let
\begin{equation}\label{eq2:77}
M_{\xi}=\bigcup_{f\in F^n_{\floor{n/2}+1}}\prod_{i=0}^{n-1}(M_{i,\xi}\vee_f\real).
\end{equation}
We let
\begin{equation}\label{eq2:78}
K=\bigcup_{f\in F^n_{n-\floor {n/2}}}\prod_{i=0}^{n-1}(K_i\vee_f\real).
\end{equation}
We now claim that
\begin{description}
\item[(a)] $M_{\xi}\cap M_{\zeta}=\emptyset$ for $\zeta<\xi<\cuum$ and
\item[(b)] $K\cap(\bigcup_{\xi\in\cuum}M_{\xi})=\emptyset$.
\end{description}
We show (b).  Fix $\xi\in\cuum$.  Let \[x=\langle x_0,\ldots ,x_{n-1}\rangle\in M_{\xi}
\text{ and }
y=\langle y_0,\ldots ,y_{n-1}\rangle\in K.\]  We claim 
that $x\neq y$.  By (\ref{eq2:77}) we have 
\[|\{i\in n\colon x_i\in M_{i,\xi}\}|\geq\floor{n/2}+1.\]   
By (\ref{eq2:78}) we have 
\[|\{i\in n\colon y_i\in K_{i}\}|\geq n-\floor{n/2}.\]
Since $n-\floor{n/2}+\floor{n/2}+1=n+1$,
by the Pigeonhole Principle there is an $i\in n$ such that $x_i\in M_{i,\xi}$
and 
$y_i\in K_{i}$.  So, by (ii), $x_i\neq y_i$ which implies that 
$x\neq y$.  A similar argument together with the fact that
$2(\floor{n/2}+1)\geq n+1$ shows that (a) holds.     

Using Lemma \ref{lem2:n}, we may find for each $\xi\in\cuum$ a function 
$f_{\xi}\colon\real^n\to\real^m$ such that for any $g\colon\real^n\to\real^m$ if 
$g|_{M_{\xi}}=f|_{M_{\xi}}$, then $g\in (\floor{n/2}+1)\dar(\real^n,\real^m)$.  
Again applying Lemma \ref{lem2:n}, we may find a function 
$f^*\colon\real^n\to\real$ such for any $g\colon\real^n\to\real^m$ if 
$g|_{K}=f^*|_{K}$ then, $g\in (n-\floor{n/2})\dar(\real^n,\real^m)$.  
Define $f\colon\real^n\to\real^m$ by 
\begin{equation}\notag
f(x)=\begin{cases}
f^*(x)& \text{if $x\notin\bigcup_{\xi\in\cuum}M_{\xi}$}\\
g_{\xi}(x)-f_{\xi}(x)& \text{if $x\in M_{\xi}$}.
\end{cases}
\end{equation}
We claim that $f$ has the desired property.  Since $f|_{K}=f^*|_K$, it follows 
from (b) that 
$f\in (n-\floor{n/2})\dar(\real^n,\real^m)$.  
Now, we only need to show that for each $\xi\in\cuum$ we have
$g_{\xi}-f\in(\floor{n/2}+1)\dar(\real^n,\real^m)$.  Fix 
$\xi\in\cuum$.  Let $x\in M_{\xi}$.  Then, 
$(g_{\xi}-f)(x)=g_{\xi}(x)-(g_{\xi}(x)-f_{\xi}(x))=f_{\xi}(x)$.  Since 
$(g_{\xi}-f)|_{M_{\xi}}=f_{\xi}|_{M_{\xi}}$, we have 
$g_{\xi}-f\in(\floor{n/2}+1)\dar(\real^n,\real^m)$.\qed

To complete the proof it is enough to show that 
$\add((n-1)\dar(\real^n,\real^m))\leq\cuum^{+}$ for $n>2$.  To see this,
notice that by (iii) of Proposition~\ref{prop2:two} we have
\[\add_{n,\floor{n/2}}(\dar(\real^n,\real^m))\leq
\add_{n,n-2}(\dar(\real^n,\real^m)).\]
By Proposition~\ref{prop2:two}(viii), we have 
\[\add_{n,n-2}(\dar(\real^n,\real^m))
\leq\add((n-1)\dar(\real^n,\real^m)).\]  Finally, notice that
$\floor{n/2}<n-1$ only if $n>2$.   

\noindent\lem{lem2:r}{If $n>2$, then
$\add((n-1)\dar(\real^n,\real))\leq\cuum^+$.}
\proof
Let $F\subseteq (\real)^{\real^n}$ be as in Proposition~\ref{prop2:9}.  We claim
that  there is no $g\colon\real^n\to\real$ such that 
$g+F\subseteq (n-1)\dar(\real^n,\real)$.  
By way of contradiction assume that there is such a $g$.  
Since $n-1<n$, Lemma~\ref{lem2:f} implies that for every $f\in F$ there is  
a perfect set $P_{f}$ such that the restriction of $g+f$ 
to $P_{f}$ is continuous.  Since $|F|=\cuum^+$ and there are only $\cuum$-many perfect 
sets, there exist $f,h\in F$ such that $P_{f}=P_{h}$.  It follows that $f-h=(g+f)-(g+h)$ 
is continuous on $P_{f}$, which contradicts our choice of $F$.\qed

We now work to generalize the range space of Lemma~\ref{lem2:r} to 
$\real^m$.

\noindent\lem{lem2:s}{If $n,m \geq 1$ and $n>2$, then
$\add((n-1)\dar(\real^n,\real^m))\leq\cuum^+$.}
\proof
Let $F\subseteq\real^{\real^n}$ be such that $|F|=\cuum^+$, and $F$ 
satisfies the definition of 
$\add((\floor{n/2}+1)\dar(\real^n,\real))$, i.e.,  
\begin{equation}\label{eq2:35a}
\left(\forall g\in (\real)^{\real^n}\right)(\exists f\in
F)(f+g\notin (n-1)\dar(\real^n,\real)).
\end{equation}
Let $\pi\colon\real^m\to\real$ be the projection function of $\real^m$ onto some fixed 
coordinate.  Since $\pi$ is onto, for every $f\in
F$ there is an
$f^*\colon\real^n\to\real^m$  such that $f=\pi\circ f^*$.  Let $F^*=\{f^*\colon
f\in F\}$ and note that $|F^*|\leq|F|=\cuum^+$.  We will be done if we show that
$F^*$  satisfies
\begin{equation}\label{eq2:spoon}
(\forall g\in (\real^n)^{\real^m})(\exists f^{*}\in F^{*})(f^{*}+g\notin
(n-1)\dar(\real^n,\real^m)).
\end{equation}
So let $g\in (\real^m)^{\real^n}$ be arbitrary.  Take $g_1=\pi\circ g$.  By
(\ref{eq2:35a})  there is an $f\in F$ such that $(f+g_1)\notin
(n-1)\dar(\real^n,\real)$.   The proof will be complete if we can 
show that 
$(f^*+g)\notin (n-1)\dar(\real^n,\real^m)$.   By
way of contradiction, we assume that $(f^*+g)\in (n-1)\dar(\real^n,\real^m)$.  Then,
there exist
$\{d_j\in\dar(\real^n,\real^m)\}_{j=1}^{n-1}$ such that 
\begin{equation}\notag
f^*+g=d_1+\ldots+d_{n-1}. 
\end{equation}
Since the projection onto a coordinate is additive, we now
have 
\begin{eqnarray*}
f+g_1 
& = & (\pi\circ f^*)+(\pi\circ g) \notag \\
& = & \pi\circ (f^*+g) \notag \\
& = &(\pi\circ d_1)+\ldots+(\pi\circ d_{n-1})\in
(n-1)\dar(\real^n,\real),
\notag
\end{eqnarray*}
which contradicts our choice of $f$.  So $F^*$ satisfies
(\ref{eq2:spoon}).\qed

%%%%%%%%%CHAPTER3
\chapter{Additivity for nice functions}\label{third_chapter}
\section{Introduction}  
Thus far we have been discussing propreties of general functions in 
$(\real^m)^{\real^n}$.  It is 
reasonable to ask what the analogs of the results in Chapter~\ref{first_chapter} are 
for the Darboux-like families if we restict ourselves to
considering only nice functions in $\real^{\real}$ such as 
those in Baire class one.  For this task we will need some extra
notation which will be used only in this chapter.  We should also
mention that all work in this chapter will be done in
$\real^{\real}$.   So, as in Chapter~\ref{first_chapter}, 
we will always write, for example, $\dar$ when we mean
$\dar(\real,\real)$.

A subset $M$ of $\real$ is said to be meager if it is contained in 
the countable union of closed nowhere dense sets.  A subset $G$ of 
$\real$ is said to be co-meager provided that $\real\setminus G$ is 
meager.  
Let $\cnwd$, $\meager$, and
$\dense$ denote the closed nowhere dense, meager, and co-meager subsets
of 
$\real$, respectively.   For a function $f\colon\real\to\real$, we let
$\cont(f)$ denote  the set of points at which $f$ is continuous.   For a
function $f\colon\real\to\real$ and  point $x\in\real$, we define the 
oscillation of $f$ at $x$ to be 
\begin{equation*}
\osc(f,x)=\lim_{\delta\to 0}\sup\{|f(x)-f(t)|\colon t\in
(x-\delta,t+\delta)\}.
\end{equation*} Note that $f$ is continuous at $x$ if and only if
$\osc(f,x)=0$. For $n\in\omega$, we let $\osc_n(f)$ denote the set 
$\{x\in\real\colon\osc(f,x)\geq 1/2^n\}$.  We note the following  fact
about oscillation.
\noindent\prop{prop3:osc}{$\cl(\osc_{n}(f))\subseteq \osc_{n+1}(f)$  for any
$f\in\real^{\real}$ and $n\in\omega$.}
\proof Let $x\in\cl(\osc_{n}(f))$.   By definition
$\osc_{n}(f)\subseteq\osc_{n+1}(f)$, so we may  assume that
$x\notin\osc_{n}(f)$.  Let 
$\{x_k\}_{k=1}^{\infty}$ be a sequence of points in $\osc_n(f)$ such that 
$\lim_{k\to\infty}x_k=x$.  For each $k\in\omega\setminus\{0\}$ we may
find a point 
$w_k$ that satisfies $|w_k-x_k|<1/k$ and $|f(x_k)-f(w_k)|\geq
1/2^n-1/k$.  Since 
$x\notin\osc_{n}(f)\supseteq\osc_{0}(f)$, it follows that for  sufficently
large values of $k$ we have $|f(x)-f(x_k)|<1$ and 
$|f(x)-f(w_k)|<1$.   Thus, taking subsequences if necessary, we may
assume that there exist numbers 
$r$ and $s$ such that $\lim_{k\to\infty}f(x_k)=r$ and 
$\lim_{k\to\infty}f(w_k)=s$.  Clearly, $|r-s|\geq 1/2^n$, so either 
$|f(x)-r|\geq 1/2^{n+1}$ or $|f(x)-s|\geq 1/2^{n+1}$.  In either case we 
must conclude that $\osc(f,x)\geq 1/2^{n+1}$.   So,
$x\in\osc_{n+1}(f)$.\qed

Given $f\in\real^{\real}$, $x\in\real$, and $S\subseteq\real$, we  define
the right cluster set of 
$f$ with respect to $S$ at $x$ by 
\begin{equation*}
\wacc^{+}_{S}(f,x)=\bigcap_{n=1}^{\infty}(\cl(f[(x,x+1/n)\cap S])).   
\end{equation*} We define the left cluster set of $f$ with respect to $S$
at $x$, denoted   by $\wacc^{-}_{S}(f,x)$, in a similiar way.  Finally,
we let 
$\wacc_{S}(f,x)=\wacc_{S}^{-}(f,x)\cap\wacc_{S}^{+}(f,x)$ denote  the
bilateral cluster set of $f$ with respect to $S$ at $x$.   We define the
strong right cluster set of $f$  with respect to $S$ at $x$ by 
\begin{equation*}
\acc^{+}_{S}(f,x)=\bigcap_{n=1}^{\infty}(f[(x,x+1/n)\cap S]).   
\end{equation*} The strong left cluster set of $f$ with respect to $S$ at 
$x$ is defined in a similiar manner and denoted by 
$\acc^{-}_{S}(f,x)$.  We call $\acc_{S}^{-}(f,x)\cap\acc_{S}^{+}(f,x)$
the  strong bilateral cluster set of $f$ with respect to $S$ at
$x\in\real$  and denote it by $\acc_{S}(f,x)$.   If
$f\colon\real\to\real$ is a function, we let $\norm(f)=\sup\{|f(x)|\colon
x\in\real\}$.  

We say $f\colon\real\to\real$ is a {\em cliquish}
function provided  that $\cont(f)$ is a co-meager subset of $\real$.   We
denote the family of all cliquish functions by $\cliq$.   We note the
following facts about Baire class one and cliqish functions. 
\noindent\prop{prop3:aaa}{
\begin{description}
\item[(1)] $\baire_1\subseteq\cliq$,
\item[(2)] $\osc_{n}(f)$ is nowhere dense for any 
$f\in\cliq$ and $n\in\omega$, and
\item[(3)] $\baire_1$ and $\cliq$ are additive groups.
\end{description}}
\proof The containment $\baire_1\subseteq\cliq$ follows from
\cite[p.394]{KUR}.   To see (2) notice that, by
Proposition~\ref{prop3:osc}, 
$\cl(\osc_{n}(f))\subseteq\osc_{n+1}(f)$.  Since $f\in\cliq$,  we know
that $\osc_{n+1}(f)$ is meager.  Thus, $\cl(\osc_{n}(f))$  contains no
non-trivial interval.  To see that the cliquish functions  form an
additive group, it is enough to notice that if $f,g\in\cliq$, then
$\cont(f+g)\supseteq\cont(f)\cap\cont(g)$  and 
$\cont(f)\cap\cont(g)$ must be a co-meager set by the Baire Category 
theorem.  The Baire class one functions form an  additive group because
the sum of the limits of  two pointwise convergent sequences of functions
is equal to the  pointwise limit of the sums of the terms of the
sequences.\qed
  
We collect some facts concerning the relations between the
Darboux-like families in the setting of $\baire_1$ and $\cliq$.

\noindent\prop{prop3:2}{The following containments and equalities hold and all 
containments mentioned are proper.
\begin{description}
\item[(i)] If $\F\in\{\ext,\acon,\pr,\conn,\phc\}$, then
$\dar\cap\baire_1=\F\cap\baire_1$.
\item[(ii)] $\swiat\subseteq\dar\cap\quasi$. 
\item[(iii)] $\quasi\subseteq\cliq$.
\item[(iv)] $\quasi\cap\pr=\quasi\cap\phc$.  
\item[(v)] $\pr\cup\dar\subseteq\phc$.
\end{description}}
\proof  
For (i) and (ii) see Figure~\ref{fig:3}.  For (v) see Figure~\ref{fig:1}.
Item (iii) can be found in \cite{MAR}.  The equality 
$\quasi\cap\pr=\quasi\cap\phc$ is shown in
\cite{GR}. \qed 
 
In \cite[p.17]{AM}, A.~Maliszewski proved the following.
\noindent\prop{prop3:4}{Let $H$ be a finite family of cliquish functions.   There
exists an $f\in\swiat\cap\baire_1$ such that 
$f+H\subseteq\swiat$ and $\bigcap\{\cont(g)\in H\}\subseteq\cont(f)$.  
Moreover, given $\epsilon>0$, we may assume that 
$\norm(f)\leq\sup\{\osc(g,x)\colon g\in H\ \&\ x\in\real\}+\epsilon$.}

It was also shown in \cite[p.19]{AM} that the above proposition could 
not be improved to include infinite families of functions.  
\noindent\prop{prop3:45}{For any function $g$, if $\cont(g)\neq\emptyset$, then 
there is a $q\in\rational$ such that 
$g+\charf{\{q\}}\notin\dar\cup\quasi$.}

The two propositions above may sound like statements about
additivity and, in fact, if we make a resticted definition of 
additivity, we can actually formulate their statements in such terms.  
The definition of additivity for $\real^{\real}$ may be restricted to
smaller classes of functions in the following way.  Let
$\mathcal{H},\F\subseteq\real^{\real}$.  The additivity of $\F$ in 
${\mathcal H}$,  denoted by $\add_{{\mathcal H}}(\F)$, is defined to be 
\begin{equation*}
\add_{{\mathcal H}}(\F)=\min\left(\{|F|\colon F\subseteq {\mathcal H}\ \&\
\left(\forall   g\in{\mathcal H}\right) (\exists f\in F) (f+g\notin\F)\}\cup
\{|{\mathcal H}|^{+}\}\right).
\end{equation*} 
The following proposition is an adaptation of 
Proposition~\ref{prop1:1} which dealt with the case 
when ${\mathcal H}=Y^X$ and $Y$ was an Abelian group with $|Y|>1$.  

\noindent\prop{prop3:667}{Let ${\mathcal P},\F,{\mathcal H}\subseteq \real^{\real}$
and 
$\charf{\emptyset}\in{\mathcal H}$.  Then,
\begin{description}
\item[(i)] if $\F\cap{\mathcal H}=\emptyset$, then $\add_{{\mathcal H}}(\F)=1$;
\item[(ii)]  if $\F\cap{\mathcal H}={\mathcal H}$, then $\add_{{\mathcal
H}}(\F)=|{\mathcal H}|^+$;
\item[(iii)] if $\F\subseteq{\mathcal P}$, then $\add_{{\mathcal
H}}(\F)\leq\add_{{\mathcal H}}({\mathcal P})$; and 
\item[(iv)] if ${\mathcal H}$ is an additive group with more than  one
element and $\F\cap {\mathcal H}\neq\emptyset$, then $2=\add_{{\mathcal H}}(\F) 
\text{ if and only if } (\F\cap{\mathcal H}) - (\F\cap{\mathcal H})\neq{\mathcal H}$;
\item[(v)] if $2<\add_{{\mathcal H}}(\F) 
\text{, then } (\F-\F)\cap{\mathcal H}={\mathcal H}$.
\end{description}}
\proof We show (i).  If $\F\cap{\mathcal H}=\emptyset$, then
$\charf{\emptyset}$ has the property that
$\charf{\emptyset}+h\notin\F$ for all $h\in {\mathcal H}$.   Thus,
$\add_{{\mathcal H}}(\F)=1$.

We show (ii).  Suppose $\add_{{\mathcal H}}(\F)<|{\mathcal H}|^+$.  Then there 
is an $F\subseteq {\mathcal H}$ with the property that for every $h\in {\mathcal
H}$ there is  some $f\in F$ such that $f+h\notin\F$.  In particular,
there is some 
$f\in F\subseteq H$ such that $f=f+\charf{\emptyset}\notin\F$.  

Since the proof of (iii) differs only slightly from the proof of its
equivalent  statement in Proposition~\ref{prop1:1}, we
exclude the proof.  

We show (iv).  Suppose that $(\F\cap{\mathcal H}) - (\F\cap{\mathcal H})={\mathcal
H}$.   We consider the two possible cases.  Let 
$S=\{|F|\colon F\subseteq {\mathcal H}\ \&\ (\forall g\in{\mathcal H})  (\exists
f\in F) (f+g\notin\F)\}$.  First assume $S=\emptyset$.   Then
$\add_{{\mathcal H}}(\F)=|{\mathcal H}|^+>2$.  Now assume that there exists an 
$F\in S$.  We show that $|F|>2$.   By way of contradiction, assume that
$F=\{f_1,f_2\}\in[{\mathcal H}]^{<3}$.   Since $(\F\cap{\mathcal H}) -
(\F\cap{\mathcal H})={\mathcal H}$, there exist 
$h_1,h_2\in\F\cap{\mathcal H}$ such that $f_1-f_2=h_1-h_2$.  Let 
$g=f_1-h_1=f_2-h_2$.  Notice that $g\in{\mathcal H}$ since ${\mathcal H}$ is a
group.   For $i\in\{1,2\}$, we have $f_i+g=f_i+(h_i-f_i)=h_i\in\F$.  Thus,
there is  a $g\in{\mathcal H}$ such that $g+F\subseteq\F$, which contradicts
our choice of 
$F$.  Hence, $\add_{{\mathcal H}}(\F)>2$.  To see the other implication
suppose that $(\F\cap{\mathcal H}) - (\F\cap{\mathcal H})\neq{\mathcal H}$.  Since
${\mathcal H}$  is a group and $\F\cap{\mathcal H}\neq\emptyset$ it is easy to
see that 
$\add_{\mathcal H}(\F)\geq 2$.  So we show that $\add_{\mathcal H}(\F)\leq 2$. 
Pick 
$h\in{\mathcal H}\setminus((\F\cap{\mathcal H}) - (\F\cap{\mathcal H}))$ and put 
$F=\{h,\charf{\emptyset}\}$.  Let $g\in{\mathcal H}$ be arbitrary.  It is
enough to  show that $f+g\notin\F$ for some $f\in F$.  However, if 
$g=\charf{\emptyset}+g\in\F$ and $h+g\in\F$ then, since ${\mathcal H}$ is a
group,  we have $h\in(F\cap{\mathcal H})-g\subseteq (\F\cap{\mathcal H}) -
(\F\cap{\mathcal H})$,  contradicting the choice of $h$.  

We prove (v). Suppose that $(\F-\F)\cap{\mathcal H}\neq{\mathcal H}$.  Pick 
$h\in {\mathcal H}\setminus(\F-\F)$ and let 
$F=\{\charf{\emptyset},h\}$.  By way of contradiction, assume there  is
some $g\in{\mathcal H}$ such that $g+F\subseteq\F$.  Then, 
$g=g+\charf{\emptyset}\in\F$, so 
$h\in\F-g\subseteq (\F-\F)\cap{\mathcal H}$ which contradicts the  choice of
$h$.  Thus, $\add_{{\mathcal H}}(\F)\leq 2$.\qed

Like additivity, the restricted version of additivity has 
interpretation in terms of coverings of one family of functions by
another.  We state this relationship in the following proposition: 
 
\noindent\prop{prop3:47}{Let $\F,{\mathcal H}\subseteq\real^{\real}$ and 
$\add_{{\mathcal H}}(\F)<|{\mathcal H}|^+$.   Then $\add_{{\mathcal H}}(\F)$ is the
minimum cardinality of a set 
$F\subseteq{\mathcal H}$ such that ${\mathcal H}\cap \bigcap\{-f+\F\colon f\in
F\}=\emptyset$; or, equivalently, that ${\mathcal
H}\subseteq\bigcup\{-f+(\real^{\real}\setminus\F)\colon  f\in F\}$.}
\proof  Since $\add_{{\mathcal H}}(\F)<|{\mathcal H}|^+$, there is an 
$F\subseteq{\mathcal H}$ that witnesses the main part of the definition of 
$\add_{\mathcal H}(\F)$, i.e., $|F|=\add_{{\mathcal H}}(\F)$ and 
\begin{equation}\label{eq3:wart} (\forall h\in{\mathcal H}) (\exists f\in F)
(h+f\notin\F).
\end{equation} We claim that ${\mathcal H}\cap\bigcap\{-f+\F\colon f\in
F\}=\emptyset$.   To see this, assume there is some 
$h\in{\mathcal H}\cap\bigcap\{-f+\F\colon f\in F\}$.   Such an $h$ would
have the property that $h+f\in\F$ for each 
$f\in F$.  But this would  contradict (\ref{eq3:wart}) since $h\in{\mathcal
H}$.   Thus, \[{\mathcal H}\cap\bigcap\{-f+\F\colon f\in F\}=\emptyset\]  for
some $F$ of cardinality $\add_{{\mathcal H}}(\F)$.  

Now assume $F\subseteq{\mathcal H}$ and $|F|<\add_{{\mathcal H}}(\F)$.   There is
an $h\in{\mathcal H}$ such that $h+F\subseteq\F$.   So, $h\in -f+\F$ for each
$f\in F$.  Thus, 
${\mathcal H}\cap\bigcap\{-f+\F\colon g\in F\}\neq\emptyset$ for any 
$F$ such that $|F|<\add_{{\mathcal H}}(\F)$, which completes the  proof.\qed

Using the language of additivity, we will now state some corollaries  to
Propositions \ref{prop3:4} and \ref{prop3:45}.  
\noindent\cor{cor3:1}{If
$\F\in\{\swiat,\dar\cap\quasi,\dar,\quasi,\dar\cup\quasi\}$, then 
\[\add_{\cliq}(\F)=\add_{\baire_1}(\F)=\omega.\]}
\proof By containment, we have that 
$\add_{\cliq}(\swiat)\leq
\add_{\cliq}(\F)\leq\add_{\cliq}(\quasi\cup\dar)$ for 
$\F\in\{\dar,\quasi,\swiat,\dar\cap\quasi,\dar\cup\quasi\}$.  Since 
$\baire_1\subseteq\cliq$, it follows from Proposition~\ref{prop3:4}  that
$\omega\leq\add_{\cliq}(\swiat)$.  Using  Proposition~\ref{prop3:45}, the
fact that any cliquish function must have at least one point of continuity, and
noting that the  characteristic function of a point is cliquish, we have 
$\add_{\cliq}(\dar\cup\quasi)\leq\omega$.  It now follows that 
$\omega=\add_{\cliq}(\F)$ for all $\F\in\{\dar,\quasi,\swiat,
\dar\cap\quasi,\dar\cup\quasi\}$.   A similiar argument yields the
equalities for $\add_{\baire_1}$.\qed 

Corollary~\ref{cor3:1} together with Proposition~\ref{prop3:47} implies 
that the minimum number of translations of $\baire_1\setminus\dar$ by 
Baire class one functions required to cover Baire class one is $\omega$.

To get a statement similiar to that of  Corollary~\ref{cor3:1} for
quasi-continuous functions, we need to make a minor modification of 
Proposition~\ref{prop3:45}.  In particular, the characteristic  functions
of singletons are not quasi-continuous so  Proposition~\ref{prop3:45} will
not be useful to us  in the quasi-continuous case.  So, we will consider
characteristic functions of open intervals, which are quasi-continuous.  
It should be pointed out that the proof here is  essentially identical to
the one which appears in \cite{AM} for  Proposition~\ref{prop3:45}.  
\noindent\prop{prop3:46}{For any function $g$, if $\cont(g)\neq\emptyset$, then
there is a $q\in\rational$ and an $n\in\omega$ such that 
$g+\charf{(q-1/n,q+1/n)}\notin\dar$.}
\proof Suppose $g$ is a function such that $\cont(g)\neq\emptyset$.   Let
$x\in\cont(g)$.  There is a $\delta>0$ such that 
$g[(x-\delta,x+\delta)]\subseteq (g(x)-1/3,g(x)+1/3)$.   Pick
$q\in\rational$ and $n\in\omega$ such that 
$[q-1/n,q+1/n]\subseteq (x-\delta,x+\delta)$.   For 
$w\in (x-\delta,x+\delta)\cap (q-1/n,q+1/n)$, we have
\[(g+\charf{(q-1/n,q+1/n)})(w)>1+g(x)-1/3=g(x)+2/3.\]   For points 
$z\in (x-\delta,x+\delta)\setminus [q-1/n,q+1/n]$, we have 
\[(g+\charf{(q-1/n,q+1/n)})(z)<g(x)+1/3.\]  Thus, 
$(g+\charf{(q-1/n,q+1/n)})[(x-\delta,x+\delta)]$ is not an interval.  
Therefore, we must conclude that $g+\charf{(q-1/n,q+1/n)}$ is not Darboux.\qed

\noindent\cor{cor3:2}{$\add_{\baire_1\cap\quasi}(\swiat)=
\add_{\baire_1\cap\quasi}(\dar)=
\add_{\quasi}(\dar)=\add_{\quasi}(\swiat)=\omega$.}
\proof We first show that
$\add_{\quasi}(\swiat)=\add_{\quasi}(\dar)=\omega$.   Since
$\swiat\subseteq\quasi\subseteq\cliq$, it follows from  
Proposition~\ref{prop3:4} that $\omega\leq\add_{\quasi}(\swiat)$.   The
countable family of functions constructed in  Proposition~\ref{prop3:46}
were all quasi-continuous and any quasi-continuous function has a point
of continuity; so, we have 
$\add_{\quasi}(\dar)\leq\omega$.  Finally, we have
$\add_{\quasi}(\swiat)\leq\add_{\quasi}(\dar)$.   Thus,
$\add_{\quasi}(\swiat)=\add_{\quasi}(\dar)=\omega$.   A similiar argument
will give the equalities for 
$\add_{\quasi\cap\baire_1}$.\qed  

Corollary~\ref{cor3:1} tells us that for any finite collection $G$  of
Baire one functions there is an $f\in\baire_1$ such that 
$f+G\subseteq\dar$.  One may ask what the minimal cardinality of a 
family of Baire one functions $F$ such that for every finite family of
functions $G$ there is an $f\in F$ such that $f+G\subseteq\dar$ is.   This
question leads us to define a new cardinal function which,  like
additivity, has meaning in more general settings (see Chapter~\ref{fourth_chapter}) and
also has a nice interpetation in terms of coverings.  If 
${\mathcal H},\F\subseteq\real^{\real}$,  then
the {\it super-additivity} of $\F$ in ${\mathcal H}$ is defined to be 
\begin{equation*}
\add^{*}_{{\mathcal H}}(\F)=\min\left\{|F|\colon F\subseteq{\mathcal H} 
\ \&\ \left(\forall G\in [{\mathcal H}]^{<\add_{{\mathcal H}}(\F)}\right)
(\exists f\in F) (f+G\subseteq\F)\right\}
\end{equation*}
This notion of super-additivity was developed by myself and T.~Natkaniec 
during his visit to West Virginia University in October of 1997.

We now state and prove some basic facts about super-additivity.  
\noindent\prop{prop3:sup}{Let $\F,{\mathcal E},{\mathcal H}\subseteq\real^{\real}$ and 
$\charf{\emptyset}\in {\mathcal H}$.  Then 
\begin{description}
\item[(i)] if $\F\cap{\mathcal H}={\mathcal H}$, then $\add^*_{{\mathcal H}}(\F)=1$;
\item[(ii)] if $\F\cap{\mathcal H}=\emptyset$, then 
$\add_{{\mathcal H}}^*(\F)=1$;
\item[(iii)] if $\add_{{\mathcal H}}(\F)=\add_{{\mathcal H}}({\mathcal E})$ and 
$\F\subseteq{\mathcal E}$ then, 
$\add^*_{{\mathcal H}}(\F)\geq\add^*_{{\mathcal H}}({\mathcal E})$.
\end{description}}
\proof We show (i).  If ${\mathcal H}=\F\cap{\mathcal H}$ then, by 
Proposition~\ref{prop3:667}(ii), 
$\add_{{\mathcal H}}(\F)=|{\mathcal H}|^+$.  Let 
$G\in[{\mathcal H}]^{<|{\mathcal H}|^+}$.  Clearly, 
$\charf{\emptyset}+G\subseteq{\mathcal H}=\F\cap{\mathcal H}$.  So, 
$\add_{{\mathcal H}}^*(\F)=1$.  

We show (ii).  If $\F\cap{\mathcal H}=\emptyset$ then, by 
Proposition~\ref{prop3:667}(i), 
$\add_{{\mathcal H}}(\F)=1$.  Since $[{\mathcal H}]^{<1}=\{\emptyset\}$ and 
$\charf{\emptyset}+\emptyset\subseteq \F$, it follows that 
$\add_{{\mathcal H}}^*(\F)=1$.  

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

The next proposition brings out the covering interpetation  of this
cardinal function. (Compare with Proposition~\ref{prop1:2}.)  
\noindent\prop{prop3:52}{Let $\charf{\emptyset}\in{\mathcal H}\subseteq\real^{\real}$ 
and 
$\F\subseteq\real^{\real}$ be such that ${\mathcal H}\cap\F\neq{\mathcal H}$.  
Then, $\add_{{\mathcal H}}(\F)$ is the  minimum cardinality of a family
$F\subseteq{\mathcal H}$  such that 
\begin{equation}\label{eq3:cov} 
[{\mathcal H}]^{<\add_{{\mathcal H}}(\F)}\subseteq
\bigcup_{f\in F}[-f+\F]^{<\add_{{\mathcal H}}(\F)}.
\end{equation}}
\proof Let $F\subseteq{\mathcal H}$ be such that 
$|F|=\add_{{\mathcal H}}^*(\F)$ and 
\begin{equation*}
\left(\forall   G\in [{\mathcal H}]^{<\add_{{\mathcal H}}(\F)}\right) (\exists
f\in F)  (f+G\subseteq\F).
\end{equation*} We show $F$ satisfies (\ref{eq3:cov}).  Suppose 
$G\in[{\mathcal H}]^{<\add_{{\mathcal H}}(\F)}$.  There is an $f\in F$ such  that
$f+G\subseteq\F$.  This means that 
$G\subseteq-f+\F$.  Thus, 
$G\in [-f+\F]^{<\add_{{\mathcal H}}(\F)}$.   So, $F$ satisfies (\ref{eq3:cov}).

Suppose now that $F\subseteq{\mathcal H}$ satisfies (\ref{eq3:cov}).   Let
$G\in [{\mathcal H}]^{<\add_{\mathcal H}(\F)}$.  By (\ref{eq3:cov}),  there is an
$f\in F$ such that 
$G\in [-f+\F]^{<\add_{{\mathcal H}}(\F)}$.  So, 
$f+G\subseteq\F$.  Thus, $|F|\geq\add^*_{{\mathcal H}}(\F)$.\qed

A basic relationship between additivity and 
super-additivity observed by myself and T.~Natkaniec is stated 
in the next proposition.  
\noindent\prop{prop3:51}{If $\F,{\mathcal H}\subseteq\real^{\real}$ and 
$2\leq\add_{\mathcal H}(\F)\leq |{\mathcal H}|$, then 
\[\max\{\add_{{\mathcal H}}(\F),\add_{{\mathcal H}}(\real^{\real}\setminus\F)\}
\leq \add^{*}_{{\mathcal H}}(\F).\]}
\proof We first show that $\add_{{\mathcal H}}(\real^{\real}\setminus\F)
\leq\add^{*}_{{\mathcal H}}(\F)$.   Let $F\subseteq{\mathcal H}$ be a witness to
the definition of $\add^{*}_{{\mathcal H}}(\F)$,  i.e., $|F|=\add^{*}_{{\mathcal
H}}(\F)$ and 
\begin{equation*}
\left(\forall G\in [{\mathcal H}]^{<\add_{{\mathcal H}}(\F)}\right) (\exists f\in
F)  (f+G\subseteq\F).
\end{equation*} Since $\add_{{\mathcal H}}(\F)\geq 2>1$, we see that $F$ also
satisfies 
\begin{equation*} (\forall g\in{\mathcal H}) (\exists f\in F) (f+g\in\F).
\end{equation*} Since
$\F=\real^{\real}\setminus(\real^{\real}\setminus\F)$,  we see that 
$\add_{\mathcal H}^*(\F)=|F|\geq\add_{{\mathcal H}}(\real^{\real}\setminus\F)$.  

We show $\add_{{\mathcal H}}(\F)\leq\add^{*}_{{\mathcal H}}(\F)$.   By way
of contradiction, assume that we have 
$\add_{{\mathcal H}}(\F)>\add^{*}_{{\mathcal H}}(\F)$.   Then there is an
$F\subseteq{\mathcal H}$ such that 
$|F|<\add_{{\mathcal H}}(\F)$  and 
\begin{equation}\label{eq3:wart1}
\left(\forall G\in [{\mathcal H}]^{<\add_{{\mathcal H}}(\F)}\right) (\exists f\in
F)  (f+G\subseteq\F).
\end{equation}   We claim that for each 
$f\in F$ there is a $g_f\in{\mathcal H}$ such that $f+g_f\notin\F$.  
Otherwise, there would be some $f\in F \subseteq{\mathcal H}$ such  that
$f+{\mathcal H}\subseteq\F$, which would imply that 
$\add_{{\mathcal H}}(\F)=|{\mathcal H}|^+$, contradicting the  assumption that
$\add_{\mathcal H}(\F)\leq |{\mathcal H}|$.   Let $G=\{g_f\colon f\in F\}$.  
Notice that $|G|\leq |F|<\add_{{\mathcal H}}(\F)$.   By (\ref{eq3: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,
$\add_{{\mathcal H}}(\F)\leq\add^{*}_{{\mathcal H}}(\F)$.\qed

\section{The Results}
To state the theorems dealing with the super-additivities of the 
families under consideration, we must first make a definition.  Let
$\F$  be a collection of subsets of $\real$.  We define the {\em
cofinality of $\F$} to be
\[\cof(\F)=\min\{|F|\colon F\subseteq\F\ 
\&\ (\forall M\in \F)(\exists N\in F)(M\subseteq N)\}.\] This cardinal
has been studied intensively for the case  when $\F$ is some ideal of
subsets of $\real$ \cite{BAJU}.   We now state the main theorems of this
chapter and then discuss  their interpetations in terms of super-additivity
and additivity.  

\noindent\thm{thm3:1}{There exists a family $F\subseteq\swiat\cap\baire_1$  such
that 
$|F|=\cof(\meager)$ and for any $H\in [\baire_{1}]^{<\omega}$  there is
an $f\in F$ such that $f+H\subseteq\dar\cap\quasi$.   Moreover, given
$\epsilon>0$, we may assume that 
$\norm(f)\leq\sup\{\osc(g,x)\colon g\in H\ \&\ x\in\real\}+\epsilon$.} 

\noindent\thm{thm3:12}{There exists a family $F\subseteq\swiat\cap\baire_1$  such
that 
$|F|=\cof(\meager)$ and for any $H\in [\cliq]^{<\omega}$ there is an
$f\in F$ such that $f+H\subseteq\pr\cap\quasi$.  Moreover, given 
$\epsilon>0$ we may assume that 
$\norm(f)\leq\sup\{\osc(g,x)\colon g\in H\ \&\ x\in\real\}+\epsilon$.} 

The next theorem will be used to show that $\cof(\meager)$ is  actually
the smallest cardinal for which either Theorem~\ref{thm3:1} or 
Theorem~\ref{thm3:12} remains true.

\noindent\thm{thm3:3}{There is a family $F\subseteq\baire_1$ such that 
$|F|=\cof(\meager)$ and 
\begin{equation}\label{eq3:heav}
\left(\forall G\in [\cliq]^{<\cof(\meager)}\right)(\exists f\in F)
(f+G\subseteq (\real^{\real}\setminus(\quasi\cup\pr\cup\dar))).
\end{equation}} We also have a version of Theorem~\ref{thm3:3} which will 
allow us to say some things about Darboux quasi-continuous functions.

\noindent\thm{thm3:31}{There is a family $F\subseteq\baire_1\cap\quasi$  such that
$|F|=\cof(\meager)$ and 
\begin{equation}\label{eq3:heav1}
\left(\forall G\in [\cliq]^{<\cof(\meager)}\right)(\exists f\in F)
(f+G\subseteq (\real^{\real}\setminus\dar)).
\end{equation}}

The above theorems yield a large number of statements about the  additivities
and super-additivities of some of the families which  we are studying. 

\noindent\cor{cor3:10}{ \ 
\begin{description}
\item[(i)]  For $\F\in\{\dar,\quasi,\dar\cap\quasi,\dar\cup\quasi\}$,  we
have \[\add_{\baire_1}(\real^{\real}\setminus\F)=
\add^*_{\baire_1}(\real^{\real}\setminus\F)=\add^*_{\baire_1}(\F)
=\cof(\meager).\]
\item[(ii)] For $\F\in\{\pr,\quasi,\pr\cap\quasi,\pr\cup\quasi\}$,  we
have \[\add_{\cliq}(\real^{\real}\setminus\F)=
\add^*_{\cliq}(\real^{\real}\setminus\F)=\add^*_{\cliq}(\F)=\cof(\meager).\]
\item[(iii)] $\add_{\quasi\cap\baire_1}(\real^{\real}\setminus\dar)=
\add^*_{\quasi\cap\baire_1}(\real^{\real}\setminus\dar)=
\add^*_{\quasi\cap\baire_1}(\dar)=\cof(\meager)$.
\end{description}}
\proof We show (i).  By containment, we have, using Corllary~\ref{cor3:1}
and  Proposition~\ref{prop3:sup}(iii),
\begin{equation}\label{eq3:wwt} 
\add^*_{\baire_1}(\dar\cap\quasi)\geq
\add^*_{\baire_1}(\F)\geq\add^*_{\baire_1}(\quasi\cup\dar)
\end{equation}  for $\F\in\{\dar,\quasi\}$.  Since 
$\add_{\baire_1}(\dar\cap\quasi)=\omega$, Theorem~\ref{thm3:1} implies
that 
\begin{equation}\label{eq3:wt1}
\cof(\meager)\geq\add_{\baire_1}^*(\dar\cap\quasi)
\end{equation}   
since, letting $F$ be as in Theorem~\ref{thm3:1}, if
$G\in[\baire_1]^{<\omega}$,  then there is an $f\in F\subseteq\baire_1$
such that 
$f+G\subseteq\dar\cap\quasi$.   By (\ref{eq3:wwt}) and (\ref{eq3:wt1})
together with Proposition~\ref{prop3:51}, we  have 
\begin{equation}\label{eq3:wwtt:1}
\cof(\meager)\geq\add^*_{\baire_1}(\quasi\cup\dar)\geq
\add_{\baire_1}(\real^{\real}\setminus(\dar\cup\quasi)).
\end{equation} 
We claim that  Theorem~\ref{thm3:3} implies that 
%%%%%%%
\begin{equation}\label{eq3:wt2}
\add_{\baire_1}(\real^{\real}\setminus(\dar\cup\quasi))\geq\cof(\meager).  
\end{equation} 
To see the inequality, let $F$ be as in
Theorem~\ref{thm3:3}.  Let 
$G\in[\baire_1]^{<\cof(\meager)}$.  Since $\baire_1\subseteq\cliq$, there  
is an $f\in F\subseteq\baire_1$ such that 
$f+G\subseteq\real^{\real}\setminus (\dar\cup\quasi)$.  So the
inequality  holds. By (\ref{eq3:wwt}), (\ref{eq3:wt1}), (\ref{eq3:wwtt:1}),
and (\ref{eq3:wt2}) we have 
\[\add^*_{\baire_1}(\F)=\add_{\baire_1}(\real^{\real}\setminus(\dar\cup\quasi))
=\cof(\meager)\]  for
$F\in\{\dar,\quasi,\dar\cap\quasi,\dar\cup\quasi\}$.   Using
Proposition~\ref{prop3:667}(iii), we have 
\begin{equation}\notag
\add_{\baire_1}(\real^{\real}\setminus(\dar\cap\quasi))\geq
\add_{\baire_1}(\real^{\real}\setminus\F)\geq
\add_{\baire_1}(\real^{\real}\setminus(\dar\cup\quasi))=
\cof(\meager)  
\end{equation}   for $\F\in\{\dar,\quasi\}$.  Theorem~\ref{thm3:1} implies
that 
$\add_{\baire_1}(\real^{\real}\setminus(\dar\cap\quasi))\leq\cof(\meager)$.  
Thus, 
\begin{equation}\label{eq3:wt3}
\add_{\baire_1}(\real^{\real}\setminus\F)=\cof(\meager)
\end{equation} for $\F\in\{\dar,\quasi,\dar\cap\quasi,\dar\cup\quasi\}$. 
Using  (\ref{eq3:wt3}) and Proposition~\ref{prop3:sup}(iii), we have, by 
containments, 
\begin{equation}\label{eq3:wt4}
\add^*_{\baire_1}(\real^{\real}\setminus(\dar\cap\quasi))\leq
\add^*_{\baire_1}(\real^{\real}\setminus\F)\leq
\add^*_{\baire_1}(\real^{\real}\setminus(\dar\cup\quasi))
\end{equation} for $\F\in\{\dar,\quasi\}$.   Since
$\add_{\baire_1}(\real^{\real}\setminus(\dar\cup\quasi))=\cof(\meager)$, 
Theorem~\ref{thm3:3} implies that 
\begin{equation}\label{eq3:wt5}
\add^*_{\baire_1}(\real^{\real}\setminus(\dar\cup\quasi))\leq\cof(\meager).
\end{equation}  
By Proposition~\ref{prop3:51} and (\ref{eq3:wt3}), 
$\add^*_{\baire_1}(\real^{\real}\setminus(\dar\cap\quasi))\geq
\add_{\baire_1}(\real^{\real}\setminus(\dar\cap\quasi))=\cof(\meager)$, 
so we have 
\begin{equation}\label{eq3:wt6}
\cof(\meager)\leq\add^*_{\baire_1}(\real^{\real}\setminus(\dar\cap\quasi)).
\end{equation} Thus, by (\ref{eq3:wt5}), (\ref{eq3:wt4}), and
(\ref{eq3:wt6}),  it follows that
$\add^*_{\baire_1}(\real^{\real}\setminus\F)=\cof(\meager)$  for every 
$F\in\{\dar,\quasi,\dar\cap\quasi,\dar\cup\quasi\}$, which completes the
proof  of (i).

The proof of (ii) follows that of (i) except Theorem~\ref{thm3:12} is 
used instead of Theorem~\ref{thm3:1}, and $\dar$ is replaced by 
$\pr$.  One must also use the fact, see Proposition~\ref{prop3:2}, that 
$\dar\cap\quasi\subseteq\phc\cap\quasi=\pr\cap\quasi$.

We now prove (iii).  Since $\swiat\subseteq\quasi\subseteq\cliq$  and
$\add_{\quasi\cap\baire_1}(\dar)=\omega$,  Theorem~\ref{thm3:1} implies
that 
$\add_{\quasi\cap\baire_1}^*(\dar)\leq\cof(\meager)$.  By 
Proposition~\ref{prop3:51}, 
\begin{equation}\label{eq3:wt7}
\add_{\quasi\cap\baire_1}(\real^{\real}\setminus\dar)
\leq\add_{\quasi\cap\baire_1}^*(\dar)\leq\cof(\meager).
\end{equation} 
Theorem~\ref{thm3:31} implies that 
$\add_{\quasi\cap\baire_1}(\real^{\real}\setminus\dar)\geq\cof(\meager)$;
so, by  (\ref{eq3:wt7}),
\[\add_{\quasi\cap\baire_1}(\real^{\real}\setminus\dar)
=\add_{\quasi\cap\baire_1}^*(\dar)=\cof(\meager).\]  Since 
$\add_{\quasi\cap\baire_1}(\real^{\real}\setminus\dar)=\cof(\meager)$, 
Theorem~\ref{thm3:31} implies that 
\[\add_{\quasi\cap\baire_1}^*(\real^{\real}\setminus\dar)\leq\cof(\meager),\]  
and Proposition~\ref{prop3:51} implies that
\[\add_{\quasi\cap\baire_1}^*(\real^{\real}\setminus\dar)\geq
\add_{\quasi\cap\baire_1}(\real^{\real}\setminus\dar)=\cof(\meager).\]  
Thus, 
$\add_{\quasi\cap\baire_1}^*(\real^{\real}\setminus\dar)=\cof(\meager)$ 
which completes the proof of (iii).\qed

The fact that $\add_{\baire_1}(\real^{\real}\setminus\dar)=\cof(\meager)$, 
together with Proposition~\ref{prop3:47}, implies  that the minimum number
of translations of $\baire_1\cap\dar$ by  Baire class one functions
required to cover Baire class one is $\cof(\meager)$.


We make heavy use of the following proposition thoughout the remainder of
this chapter.
\noindent\prop{prop3:frem}{{\rm \cite[Theorem 3B]{FREM}}
$\cof(\meager)=\cof(\cnwd)$.}


\section{Proof of Theorems~\ref{thm3:3} and \ref{thm3:31}}

{\sc Proof of Theorem~\ref{thm3:3}}.   By Proposition~\ref{prop3:frem},
there exists 
$\cnwd_0\subseteq\cnwd$ such that $|\cnwd_0|=
\cof(\meager)$ and $\cnwd_0$ satisfies
\begin{equation}\label{eq3:he1} 
(\forall M\in\cnwd)(\exists K\in\cnwd_0)(M\subseteq K).
\end{equation} 
For each $K\in\cnwd_0$, let $f_K=4\cdot\charf{K}$, notice $f_K\in\baire_1$.  
Put $F=\{f_K\colon K\in\cnwd_0\}$.  Since
$|F|=\cof(\meager)$, it is enough for us to show that $F$ satisfies
(\ref{eq3:heav}).   Let $G\subseteq\cliq$ and $|G|<\cof(\meager)$.  We
find an 
$f\in F$ such that $f+g\notin\quasi\cup\pr\cup\dar$ for every $g\in G$.  
For each $g\in G$ let $A_g=\cl(\osc_1(g))$.  Note that by
Proposition~\ref{prop3:aaa}(2), we  have
$A_g\in\cnwd$.   Since $|\{A_g\colon g\in
G\}|\leq |G|<\cof(\meager)$ and 
$\cof(\meager)=\cof(\cnwd)$, there is a
$M\in\cnwd$ such that $M\setminus A_g\neq\emptyset$ for every 
$g\in G$.  By (\ref{eq3:he1}), there is a $K\in\cnwd$ such that 
$M\subseteq K$.   Notice that, since $M\subseteq K$, we have 
$K\setminus A_g\neq\emptyset$ for every $g\in G$.   We claim that
$f_K+g\notin\quasi\cup\pr\cup\dar$ for each $g\in G$.   Fix $g\in G$.  

We first show that $f_K+g\notin\pr\cup\dar$.   By
Proposition~\ref{prop3:2}(v), it is enough for us  to show that
$f_K+g\notin\phc$. Since $K\setminus A_g\neq\emptyset$, there is an open
interval 
$U$ such that $U\cap A_g=\emptyset$ and $U\cap K\neq\emptyset$.   Since
$K\in\cnwd$, there is an $x_0\in K\cap U$ that is not a  bilateral  limit
point of $K$.  Without loss of generality, we may assume that  there is a
$\delta>0$ such that $J=(x_0,x_0+\delta)\subseteq U$ and 
$J\cap K=\emptyset$.  Moreover, since $x_0\notin A_g$, we may assume that 
$g[J]\subseteq (g(x_0)-1,g(x_0)+1)$.  Since $J\cap K=\emptyset$  implies
$f_K[J]=\{0\}$, we  have $(f_K+g)[J]=g[J]\subseteq (g(x_0)-1,g(x_0)+1)$. 
On the  other hand, $x_0\in K$ so $(f_K+g)(x_0)\geq
4+g(x_0)-1>g(x_0)+1$.   Thus, $f_K+g\notin\phc$.
  
We now show that $f_K+g\notin\quasi$.  Notice that 
\[(f_K+g)(x_0)\in (g(x_0)+3,g(x_0)+5).\]   It is enough for us to show
that there is no non-empty open set 
$W\subseteq U$ such that
$(f_K+g)[W]\subseteq (g(x_0)+3,g(x_0)+5)$.  For any non-empty open set 
$W\subseteq U$, there is some 
$x\in W\setminus K$.  Since $x\notin K$, we have
\[(f_K+g)(x)=0+g(x)\in (g(x_0)-1,g(x_0)+1).\]   So, $(f_K+g)[W]$ is not a
subset of $(g(x_0)+3,g(x_0)+5)$ for any non-empty open $W\subseteq U$.   Thus,
$f_K+g\notin\quasi$.\qed


To prove Theorem~\ref{thm3:31}, it will be helpful to have some definitions
and  lemmas.  An open set $U\subseteq\real$ is said to be regular
provided that 
$\interior(\cl(U))=U$. 

\noindent\lem{lem3:nice1}{The charcteristic function of an open set $U$  is
quasi-continuous if and only if $U$ is regular.}
\proof 
Assume $U$ is regular.  We show that $\charf{U}$ is
quasi-continuous.   If $x\in U$, then $x$ is clearly a point of continuity
and thus a point of quasi-continuity for $\charf{U}$.  So assume $x\notin U$.  Let 
$W$ be an open set such that $x\in W$.  By regularity, we have that
$x\notin\interior(\cl(U))$.   So, $W\setminus\cl(U)\neq\emptyset$ and
open.   It follows that, $\charf{U}(x)=0$ and
$\charf{U}[W\setminus\cl(U)]=\{0\}$; so, $\charf{U}$  is quasi-continuous.

Assume $\charf{U}$ is quasi-continuous.   We show that $U$ is regular. 
Let $x\in\real\setminus U$.   Since $\charf{U}(x)=0$, quasi-continuity
implies that for  every open neighborhood $W$ of $x$ there is an open set
$\emptyset\neq V\subseteq W$ such that $\charf{U}[V]=\{0\}$. Since
$V\subseteq (W\setminus U)$ is non-empty and open, it follows that
$W\setminus\cl(U)\neq\emptyset$.  So, 
$x\notin\interior(\cl(U))$.  Thus, $\interior(\cl(U))\subseteq U$.  The
containment 
$U\subseteq\interior\cl(U)$ holds for any open set.  Thus, $U$  is
regular.\qed   

\noindent\lem{lem3:nice2}{Let $E\subseteq\real$ be a closed set which is nowhere
dense. Then, there is a regular open set $U$ such that
$E\subseteq\bd(U)$.}
\proof 
We may write $\real\setminus E$ as the union of countably  many
mutually disjoint intervals $\{(a_n,b_n)\colon n\in\omega\}$ such  that
the finite endpoints of each interval lie in $E$.   For each $n\in\omega$,
let $\langle x_{n,k}\rangle_{k\in\omega}$ and 
$\langle y_{n,k}\rangle_{k\in\omega}$ be sequences of points in 
$(a_n,b_n)\cap G$ such that $x_{n,0}=y_{n,0}$, $x_{n,k}\searrow a_n$, 
and $y_{n,k}\nearrow b_n$.  Let 
\begin{equation*} 
U=\bigcup\{(x_{n,2k+1},x_{n,2k})\colon n,k\in\omega\}
\cup\bigcup\{(y_{n,2k},y_{n,2k+1})\colon n,k\in\omega\}.
\end{equation*} 
It is now easily checked that $U$ has the desired
properties.\qed

\noindent{\sc Proof of Theorem~\ref{thm3:31}.}   By
Proposition~\ref{prop3:frem} there is a $\cnwd_0\subseteq\cnwd$ such that 
$|\cnwd_0|=\cof(\meager)$ and 
$\cnwd_0$ satisfies
\begin{equation}\label{eq3:he2} (\forall M\in\cnwd)(\exists
K\in\cnwd_0)(M\subseteq K).
\end{equation} By Lemma~\ref{lem3:nice2}, we may find for each
$K\in\cnwd_0$ a regular open  set $U_K$ such that $K\subseteq \bd(U_K)$.  
For each $K\in\cnwd_0$, let $f_K=4\cdot\charf{U_K}$.   By
Lemma~\ref{lem3:nice1}, $f_K$ is quasi-continuous.  Moreover, $f_K$  is in
Baire class 1.  Put $F=\{f_K\colon K\in\cnwd_0\}$.  Since
$|F|\leq\cof(\meager)$, it  is enough for us to show that $F$ satisfies
(\ref{eq3:heav1}).   Let $G\subseteq\cliq$ and $|G|<\cof(\meager)$.  We
must find an 
$f\in F$ such that $f+g\notin\dar$ for every $g\in G$.   For each $g\in
G$, let $A_g=\cl(\osc_1(g))$.  By Proposition~\ref{prop3:aaa}(2), 
we have $A_g\in\cnwd$.  Since $|\{A_g\colon g\in
G\}|\leq |G|<\cof(\meager)=\cof(\cnwd)$,  there is an 
$M\in\cnwd$ such that $M\setminus A_g\neq\emptyset$ for every 
$g\in G$.  By (\ref{eq3:he2}), there is a $K\in\cnwd_0$ such that 
$M\subseteq K$.   Notice that $K\setminus A_g\neq\emptyset$ for every
$g\in G$.   We claim that $f_K+g\notin\dar$ for each $g\in G$.  Fix $g\in
G$.   Let $x_0\in K\setminus A_g$.  Since $x_0\notin A_g$, there is an 
open interval 
$V$ such that $x_0\in V$ and $|g(x_0)-g(x)|<1$ for all $x\in V$.  
Clearly, $g[V]\subseteq (g(x_0)-1,g(x_0)+1)$.  Note that
$\sup((f_K+g)[V])\geq 4+g(x_0)-1=g(x_0)+3$ since $x_0\in\cl(U_K)$.  
Since $x_0\notin U_K$, we also have $\inf((f_K+g)[V])\leq g(x_0)+1$.  
We claim that $g(x_0)+2\notin (f_K+g)[V]$ which will show that 
$f_K+g\notin\dar$.   To see this, notice that if $x\in U_K\cap V$, then
$(f_K+g)(x)=g(x)+4\leq(g(x_0)-1)+4=g(x_0)+3$; and if 
$x\in V\setminus U_K$, then 
$(f_{K}+g)(x)=0+g(x)<g(x_0)+1$.  Thus, $f_K+g\notin\dar$. \qed


\section{Proof of Theorems~\ref{thm3:1} and \ref{thm3:12}} To prove
Theorems~\ref{thm3:1} and \ref{thm3:12} we must have an alternative
description of $\cof(\meager)$.  To this end, we introduce some new
notation.  For $S\subseteq\real$, we  define the bilateral closure of $S$
to be 
$\cl^*(S)=\{x\in\real\colon x\text{ is a bilateral limit point of S}\}$.  
For $F_1,F_2\in\cnwd$ and 
$G\subseteq\real$, we write
$F_1\prec_{G} F_2$ provided that $F_1\subseteq\cl^*(F_2\cap G)$.   If
$G\subseteq\real$ and 
$\Theta,\Psi\in \cnwd^{\omega}$, we write 
$\Theta\prec^*_{G}\Psi$ provided that $\Theta(n)\prec_{G}\Psi(n)$  for
every $n\in\omega$.  In what follows we will  call $K\in\cnwd$ a {\em
Cantor set} provided that $K$ is dense in itself.


\noindent\lem{lem3:12}{Let $E\subseteq\real$ be a closed set nowhere dense and
$G\subseteq\real$ be a dense $G_{\delta}$-set.   Then there is a Cantor
set $N$ such that $N\setminus G\subseteq E$ and 
$E\prec_G N$.}
\proof We may write $\real\setminus E$ as the union of countably  many
mutually disjoint intervals $\{(a_n,b_n)\colon n\in\omega\}$ such  that
the finite endpoints of each interval lie in $E$.   For each $n\in\omega$,
let $\langle P_{n,k}\rangle_{k\in\omega}$ and 
$\langle Q_{n,k}\rangle_{k\in\omega}$ be sequences of perfect sets in 
$(a_n,b_n)\cap G$ with the properties that
$\lim_{k\to\infty}sup\{|x-a_n|\colon  x\in P_{n,k}\}=0$ and
$\lim_{k\to\infty}sup\{|x-b_n|\colon  x\in Q_{n,k}\}=0$.  Let
\[N=\cl\left[\left(\bigcup\{P_{n,k}\colon n,k\in\omega\}\right)
\cup\left(\bigcup\{Q_{n,k}\colon n,k\in\omega\}\right)\right].\]   It is easily
checked that $N$ has the desired properties.\qed

We are now ready to give our description of $\cof(\meager)$.   
\noindent\lem{lem3:4}{$\cof(\meager)=\kappa$ where 
\[\kappa=\min\{|F|\colon F\subseteq \cnwd^{\omega}\ \&\  (\forall
G\in\dense)(\forall\Theta\in \cnwd^{\omega}) (\exists\Psi\in
F)(\Theta\prec^*_{G}\Psi)\}.\]}
\proof 
We first show that $\cof(\meager)\leq\kappa$.   Let $F\subseteq
\cnwd^{\omega}$ be such that $|F|=\kappa$ and 
\begin{equation*} (\forall G\in\dense)(\forall\Theta\in (\cnwd)^{\omega})
(\exists\Psi\in F)(\Theta\prec^*_{G}\Psi).
\end{equation*} 
Let $\cnwd_1=\{\Psi(0)\colon\Psi\in F\}$.  Note that
$|F|\leq\kappa$.   Let $K\in\cnwd$.  Define $\Theta\in \cnwd^{\omega}$ so
that 
$\Theta[\omega]=\{K\}$.  There is a $\Psi\in F$ such that 
$\Theta\prec^*_{\real}\Psi$.  In particular,
$K=\Theta(0)\subseteq\Psi(0)$  since $\Psi(0)$ is closed and
$\Theta(0)\prec_{\real}\Psi(0)$.  So, for every $K\in\cnwd$ 
there is a
$K_1\in\cnwd_1$ such that $K\subseteq K_1$.   Thus,
$\cof(\cnwd)\leq\kappa$.  Hence, by Proposition~\ref{prop3:frem}, 
$\cof(\meager)\leq\kappa$.  

We now work for the other inequality.   Let $\cnwd^*$ and $\dense^*$
stand for the collections of subsets in 
$(0,1)$ which are closed nowhere dense in $(0,1)$ and co-meager  in
$(0,1)$, respectively.  Let 
\[\kappa_1=\min\{|F|\colon F\subseteq (\cnwd^*)^{\omega}\ \&\  (\forall
G\in\dense^*)(\forall\Theta\in (\cnwd^*)^{\omega}) (\exists\Psi\in
F)(\Theta\prec^*_{G}\Psi)\}.\]    Notice that, by homeomorphism, 
$\kappa_1=\kappa$.   Recalling the defintion of $\cof(\meager)$ and
considering complements,  there is a collection $D\subseteq\dense$  such
that $|D|=\cof(\meager)$ and 
\begin{equation}\label{e4} (\forall G\in\dense)(\exists H\in
D)(H\subseteq G).
\end{equation} By Proposition~\ref{prop3:frem}, there is a family 
$F\subseteq\cnwd$ such that $|F|=\cof(\meager)$ and 
\begin{equation}\label{e5a} (\forall M\in \cnwd)(\exists N\in
F)(M\subseteq N).
\end{equation} For each $M\in F$ and $H\in D$, define 
$\Psi_M^H\subseteq (\cnwd^*)^{\omega}$ as follows.  By Lemma \ref{lem3:12}
there is a $K_M^H\in\cnwd$  such that
$M\prec_{H} K_M^H$.  We define for each $n\in\omega$
\begin{equation}\label{eq3:wart3}
\Psi_M^H(n)=(K_M^H\cap(n,n+1))-n.
\end{equation}    Let
$F^*=\{\Psi_M^H\colon M\in F\text{ and }H\in D\}$.  Since
$|D|=\cof(\meager)=|F|$, it follows that 
$|F^*|\leq\cof(\meager)$.   We will be done if we show that $F^*$
satisfies
\begin{equation}\label{e5} (\forall G\in\dense^*)(\forall\Theta\in
(\cnwd^*)^{\omega})(\exists\Psi\in F^*)(\Theta\prec^{*}_{G}\Psi).
\end{equation} So let $G\in\dense^*$ and $\Theta\in (\cnwd^*)^{\omega}$.  
We find a
$\Psi\in F^*$ such that $\Theta\prec^*_{G}\Psi$.  Let
$G_1=\left(\bigcup_{n\in\omega}(n+G)\right)\cup (-\infty,0)$; notice that 
$G_1\in\dense$.  Define 
$N=\cl\left(\bigcup_{n\in\omega}(n+\Theta(n))\right)$ and notice that
$N\in\cnwd$.  Pick $H\in D$ and $M\in F$ so that 
$H\subseteq G_1$ and $N\subseteq M$.  We claim that
$\Theta\prec_{G}^*\Psi_M^H$.   Let $K_M^H$ be as in the definition of
$\Psi_M^H$; in particular, recall that 
$M\prec_{H} K_M^H$.  Since $H\subseteq G_1$, we have $M\prec_{G_1}
K_M^H$.   Let $k\in\omega$  be arbitrary and 
$p\in\Theta(k)$.  Then
$p+k\in\Theta(k)+k\subseteq N\cap (k,k+1)$.  Since $N\subseteq M$ and 
$M\prec_{G_1} K_M^H$, we have 
$\{p+k\}\prec_{G_1} K_M^H\cap (k,k+1)$.  But then, 
$\{p\}\prec_{G} (K_M^H\cap (k,k+1))-k=\Psi_M^H(k)$.   So,
$\Theta(k)\prec_G\Psi^{H}_{M}(k)$.  Since
$k$ was arbitrary, we have $\Theta\prec_{G}^*\Psi_M^K$.  Thus,
(\ref{e5}) is satisfied.\qed

We now work to construct a family $F\subseteq\swiat\cap\baire_1$  of
cardinality $\cof(\meager)$ that will work for both  Theorems~\ref{thm3:1}
and \ref{thm3:12}.  
\noindent\lem{lem3:2b}{Let $\rho\in (0,+\infty]$, $K,M\in\cnwd$, and 
$F$ be a Cantor set such that $M\subseteq\cl^*(F\setminus M)$ and 
$F\cap K\subseteq M$.  Given an $f\in\swiat\cap\baire_1$ with 
\begin{description}
\item[($p_1$)] $\real\setminus M\subseteq\cont(f)$ and 
\item[($p_2$)] if $x\in M\setminus\cont(f)$ then $\acc_{K\setminus M}
(f,x)\supseteq(f(x)-\rho,f(x)+\rho)$, then
\end{description} one may find a $g\in\baire_1$ such that
$g+f\in\swiat\cap\baire_1$ and 
\begin{description}
\item[($a$)] $\real\setminus M=\cont(g+f)$
\item[($b$)] $g[M\cup K]=\{0\}$
\item[($c$)] $\acc_{(F\cup K)\setminus M}
((g+f),x)\supseteq((f+g)(x)-\rho,(g+f)(x)+\rho)$  for each $x\in M$
\item[($d$)] $\norm(g)\leq 2\rho$.
\end{description} Moreover, if $f=\charf{\emptyset}$, we may improve ($d$)
to 
$\norm(g)\leq\rho$.}
\proof We may write $\real\setminus M$ as the union of countably  many
mutually disjoint intervals $\{(a_n,b_n)\colon n\in\omega\}$ such  that
the finite endpoints of each interval lie in $M$.   Let $I$ be such an
open interval in the complement of $M$.   We show how to define 
$g$ on $I$.   First assume that $I$ has finite endpoints $a<b$.   We show
how to define $g$ on $(a,(b+a)/2]$.  If $a\notin\cont(f)$, then  let
$g[(a,(b+a)/2]=\{0\}$.  Suppose that $a\in\cont(f)$.   Since $F$ is a 
Cantor set $f|_{I\cup\{a\}}$ is continuous, and $a\in M$ we may find a
sequence of mutally disjoint Cantor sets $\{F_n\}_{n\in\omega}$ such
that 
\begin{description}
\item[(1)] $\bigcup_{n\in\omega}F_n\subseteq (a,(b+a)/2)\cap F$, 
\item[(2)] $f|_{F_n}$ is continuous for each $n\in\omega$,  
\item[(3)] $\lim_{n\to\infty}\sup\{|x-a|\colon x\in F_n\}=0$, and 
\item[(4)] $\lim_{n\to\infty}\sup\{|f(a)-f(x)|\colon x\in F_n\}=0$.   
\end{description}
We must
consider two cases.  First, assume that $\rho$ is finite.   For every
$n\in\omega$ large enough, we have 
$\sup\{|f(a)-f(x)|\colon x\in F_n\}<\rho$.  For all such $n$ there is a
continuous function 
$g_n\colon F_n\to\real$ such that
$(g_n+f)[F_n]=[f(a)-\rho,f(a)+\rho]$ and 
$\norm(g_n)\leq 2\rho$.  For small values of $n$, we define $g_n$ so  that
$g_n[F_n]=\{0\}$. Notice that if $f=\charf{\emptyset}$, we may for every 
$n\in\omega$ pick a $g_n$ such that 
$(g_n+f)[F_n]=[-\rho,\rho]$ and $\norm(g_n)\leq\rho$.   By the Tietze 
extension theorem \cite[p.127]{KUR}, there is a continuous function 
$g\colon (a,(b+a)/2]\to\real$  such that
$g|_{F_n}=g_n$ for every $n\in\omega$, 
$g[K\cup\{(b+a)/2\}]=\{0\}$, and $\norm(g)\leq 2\rho$  (or
$\norm(g)\leq\rho$).   In the case when $\rho=+\infty$, we define $g$ on 
$(a,(b+a)/2]$ as we did when $\rho$ was finite, except we define 
$g_n\colon F_n\to\real$ so that $(g_n+f)[F_n]=[f(a)-n-1,f(a)+n+1]$  for
every $n\in\omega$.   We may do the symmetrical construction to define
$g$ on $[(b+a)/2,b)$.    This completes the construction of $g$  for when
$I$ has two finite endpoints.   

If $I$ has one finite endpoint, $p$ repeat
the construction above for $p$.   For example if $I=(p,+\infty)$, the
desired continuous function would be of the form $g$, above, for points
close to $p$ and constantly equal to zero otherwise.   We do one of the
constructions mentioned above for each interval of the complement of $M$
and let 
$g|_M=\{0\}$ to complete the construction of
$g$.

We now show that $g$ has the desired properties.   First, notice that
($b$) and ($d$) (or $\norm(g)\leq\rho$)  are immediate from the 
construction.  To show $g\in\baire_1$, it is enough to show that 
$g^{-1}((r,+\infty))$ and $g^{-1}((-\infty,r))$ are $F_{\sigma}$-sets 
for every $r\in\real$ \cite[p.373]{KUR}. Since $\real\setminus
M\subseteq\cont(g)$, it follows that 
$g^{-1}((r,+\infty))\cap(\real\setminus M)$ is open and thus an 
$F_{\sigma}$-set.  By ($b$), if $r<0$, then 
$g^{-1}((r,+\infty))=(g^{-1}((r,+\infty))\cap(\real\setminus M))\cup M$,
which is an $F_{\sigma}$-set.  By ($b$), if $r\geq 0$, then 
$g^{-1}(r,+\infty))=g^{-1}((r,+\infty))\cap(\real\setminus M)$, 
which is an 
$F_{\sigma}$-set.  The argument that $g^{-1}((-\infty,r))$ is an
$F_{\sigma}$-set is similar.   We show that ($c$) holds.  Let $x\in M$. 
If $x\in M\setminus\cont(f)$, then  we are done by ($b$) and ($p_2$).  So,
we may now assume that $x\in\cont(f)\cap M$.   To complete the proof of
($c$), we only show that 
\begin{equation}\label{eq3:tew}
\acc^{+}_{F\setminus M}(g+f,x)\supseteq((g+f)(a)-\rho,(g+f)(a)+\rho)
\end{equation}
since the containment  
$\acc^{-}_{F\setminus M}\supseteq((g+f)(a)-\rho,(g+f)(a)+\rho)$ follows by a 
similar argument and $F\setminus M\subseteq (K\cup F)\setminus M$.   It
is clear from the construction and (b) that if $a\in M$  is the left hand
end-point of  some complementary interval, then
\begin{equation}\label{eq3:zsa} 
\acc^{+}_{F\setminus M}(g+f,a)\supseteq((g+f)(a)-\rho,(g+f)(a)+\rho).
\end{equation}  If $x$ is the left-hand endpoint of some complementary
interval of $M$, then we are done by (\ref{eq3:zsa}).   So, assume that $x$
is not the left-hand endpoint of some  complementary interval.   Then we
may find a sequence $\{x_n\}_{n\in\omega}$ of left-hand endpoints of
complementary intervals with the property that 
$\lim_{n\in\omega}x_n=x$.  Since $x\in\cont(f)$, we may also assume that 
$\lim_{n\in\omega}f(x_n)=f(x)$.  It now follows by (\ref{eq3:zsa}) and
the fact that $g[M]=\{0\}$ that \[\acc^{+}_{F\setminus
M}(g+f,x)\supseteq((g+f)(x)-\rho,(g+f)(x)+\rho).\]   

Thus, (\ref{eq3:tew})
holds, and ($c$) is established.    We show that ($a$) holds.  It is clear
from the construction  that $\real\setminus M\subseteq\cont(g)$.  So by,
($p_1$) we have 
$\real\setminus M\subseteq\cont(f+g)$.  By ($c$), we have 
$\cont(f+g)\subseteq\real\setminus M$.  Thus, ($a$) holds.  

It remains to be shown that $g+f\in\baire_1\cap\swiat$.   We first show 
that $g$ is peripherally continuous.  Let $x\in\real$.   If $x\notin M$,
then by ($a$) and ($p_1$), $x$ is a point of continuity for 
$g+f$.  Thus, $x$ is a point of  peripheral continuity for $g+f$.  If
$x\in M$, then $x$ is a point of  peripheral continuity for $g+f$ by 
($c$).  Thus, $g+f$ is  peripherally continuous and so, by 
Proposition~\ref{prop3:2}(i), Darboux.  We now work to show that 
$g+f\in\baire_1\cap\swiat$.  Let $[a,b]$ be an arbitrary  non-trivial
interval.  If $c$ is some number strictly between 
$(g+f)(a)$ and $(g+f)(b)$, then since $g+f$ is Darboux, there is some 
$x\in (a,b)$ such that $(g+f)(x)=c$.  If $x$ is a point of continuity,
then we are done.  If $x$ is not a point of continuity, then $x\in M$; so,  
$\acc_{(F\cup K)\setminus
M}(g+f,x)\supseteq((g+f)(x)-\rho,(g+f)(x)+\rho)$.   It follows that there
is some 
$p\in ((F\cup K)\setminus M)\cap(a,b)$  such that $(g+f)(p)=c$, but
$((F\cup K)\setminus M)\subseteq\cont(f+g)$ by  ($a$).   Therefore, 
$g+f\in\swiat\cap\baire_1$. \qed


\noindent\lem{lem3:3}{Let $\langle E_n \rangle_{n\in\omega}$ be a  sequence of
nowhere dense closed subsets of $\real$ and for each 
$n\in\omega$ put $E^*_n=E_n\setminus\bigcup_{k<n}E_k$.   Given $\tau\in
(0,+\infty]$ and $0<\epsilon<\tau$, there is a $f\in\swiat\cap\baire_1$ such that
\begin{description}
\item[(i)] $\cont(f)=\real\setminus\bigcup_{n\in\omega}E_n$;
\item[(ii)] $f|_{E^*_n}$ is continuous for each $n\in\omega$ and
$f[E_0]=\{0\}$; 
\item[(iii)] if $x\in E_0$, then 
$\acc_{\cont(f)}(f,x)\supseteq (-\tau,+\tau)$; 
\item[(iv)] if $x\in E_{n}$ and $n>0$, then 
$\acc_{\cont(f)}(f,x)\supseteq  (f(x)-\epsilon/2^n,f(x)+\epsilon/2^n)$; and
\item[(v)] $\norm(f)<\tau+2\epsilon$.
\end{description}}
\proof Let $E=\bigcup_{n\in\omega}E_n$ and 
$G=\real\setminus E$.  Clearly, $G$ is a dense 
$G_{\delta}$-set. We now construct a sequence $\langle
f_n\in\baire_{1}\cap\swiat\rangle_{n\in\omega}$  such that
$f=\lim f_n$ will have the desired properties.  Since $E_0$  is nowhere
dense, we may, by Lemma~\ref{lem3:12}, find a Cantor set $F^{*}_0$  such that
$E_0\subseteq\cl^*(F^{*}_0\cap G)$.  By Lemma~\ref{lem3:2b} used with $M=E_0$,
$K=\emptyset$, $F=F^{*}_0$, $\rho=\tau$ and 
$f=\charf{\emptyset}$, we may construct a 
$g_0\in\swiat\cap\baire_1$ such that
\begin{description}
\item[($a_0$)] $\real\setminus E_0=\cont(g_0)$,
\item[($b_0$)] $g_0[E_0]=\{0\}$,
\item[($c_0$)] $\acc_{F^{*}_0\cap G}(g_0,x)\supseteq(-\tau,+\tau)$  for each
$x\in E_0$, and 
\item[($d_0$)] $\norm(g_0)\leq\tau$.
\end{description} 
For the inductive step, assume we have constructed  a
sequence $\{f_k\in\swiat\cap\baire_1\}_{k=0}^{n}$ and a sequence
$\{F_k\}_{k=0}^n$ of  Cantor sets such that $f_0=g_0$ and $F_0=F^{*}_0$; and  for $0<k\leq
n$ we have
\begin{description}
\item[($a_k$)] $\cont(f_k)=\real\setminus\bigcup_{i\leq k} E_i$,
\item[($b_k$)] $f_k|_{E_i\cup F_i}=f_i|_{E_i\cup F_i}$  for all $i<k$,
\item[($c_k$)] $\acc_{F_k\cap G}(f_k,x)\supseteq
(f_k(x)-\epsilon/2^k,f_k(x)+\epsilon/2^k)$ for every 
$x\in\bigcup_{l\leq k} E_l$,
\item[($d_k$)] $\norm(f_k-f_{k-1})\leq\epsilon/2^{k-1}$.
\end{description} We find a Cantor set $F_{n+1}$ and 
$f_{n+1}\in\swiat\cap\baire$ so that ($a_{n+1}$), ($b_{n+1}$), 
($c_{n+1}$), and ($d_{n+1}$) are all satisfied.  Since 
$G_1=\real\setminus\left((\bigcup_{k\leq n}F_k)\cup (\bigcup_{k\leq
n+1}E_{k})\right)$  is a dense $G_{\delta}$-set, there is, by
Lemma~\ref{lem3:12}, a Cantor set $F^{*}_{n+1}$  such that 
\begin{equation}\label{eq3:kle}
\bigcup_{k\leq n+1}E_k\subseteq\cl^*(F^{*}_{n+1}\cap G_1) 
\text{ and }F^{*}_{n+1}\setminus\bigcup_{k\leq n+1}E_k\subseteq G_1.  
\end{equation} 
We claim that $f_n$ with $M=(\bigcup_{k\leq n+1}E_{k})$, 
$K=(\bigcup_{k\leq n}F_k)$, and $F=F^{*}_{n+1}$ satisfy the conditions of 
Lemma~\ref{lem3:2b} with 
$\rho=\epsilon/2^{n+1}$.  To  see ($p_1$) and ($p_2$) are satisfied, notice
that $f_n$ satisfies ($a_n$)  and ($c_n$), respectively.  Also notice
that, by (\ref{eq3:kle}), we have 
$M\subseteq \cl^*(F\setminus M)$ and $F\cap K\subseteq M$.   Applying 
Lemma~\ref{lem3:2b} and letting 
$S=(F_{n+1}^{*}\cup\bigcup_{k\leq n}F_k)\setminus\bigcup_{k\leq n+1}E_k$, 
we may find a $g\in\baire_1$ such that $g+f_n\in\swiat\cap\baire_1$ and 
\begin{description}
\item[($a_{n+1}^{*}$)] $\real\setminus\bigcup_{k\leq n+1}
E_{k}=\cont(f_n+g)$,
\item[($b_{n+1}^{*}$)] $g[ (\bigcup_{k\leq n}F_k)\cup(\bigcup_{k\leq
n+1}E_{k})]=\{0\}$,
\item[($c_{n+1}^{*}$)] $\acc_{S}(g+f_n,x)
\supseteq((g+f_n)(x)-\epsilon/2^{n+1},(g+f_n)(x)+\epsilon/2^{n+1})$ for
each 
$x\in\bigcup_{k\leq n+1}E_{k}$, and 
\item[($d_{n+1}^{*}$)] $\norm(g)\leq\epsilon/2^{n}$.
\end{description} 

We claim that $f_{n+1}=f_n+g$ and 
$F_{n+1}=F^{*}_{n+1}\cup(\bigcup_{k\leq n} F_k)$  are as desired.  Since 
$f_{n+1}-f_n=g$ and $\norm(g)\leq\epsilon/2^{n}$, we have  ($d_{n+1}$). 
We show that ($b_{n+1}$) is satisfied.   If $k<n+1$, then, by 
($b_{n+1}^{*}$), $f_n|_{F_k\cup E_k}=f_{n+1}|_{F_k\cup E_k}$.  So by 
($b_{n}$), we have $f_k|_{F_k\cup E_k}=f_{n+1}|_{F_k\cup E_k}$.  So 
($b_{n+1}$) holds.  We now work to establish ($c_{n+1}$).  Let 
$x\in\bigcup_{k\leq n+1} E_{k}$.  We consider two cases.  First, assume 
$x\notin\cont(f_n)$.  By ($a_n$), there is a $k\leq n$ such that 
$x\in E_k$.  By ($b_{n+1}$), we have 
$f_k|_{F_k\cup E_k}=f_{n+1}|_{F_k\cup E_k}$.  This together with  ($c_k$)
and $F_k\subseteq F_{n+1}$ implies that 
$\acc_{F_{n+1}\cap G} (f_{n+1},x)\supseteq (f_{n+1}(x)-\epsilon/2^k,
f_{n+1}(x)+\epsilon/2^k)$ if $k>0$; and, when $k=0$, we have 
$\acc_{F_{n+1}\cap G} (f_{n+1},x)\supseteq
(f_{n+1}(x)-\tau,f_{n+1}(x)+\tau)$.  In  both cases we have 
$\acc_{F_{n+1}\cap G} (f_{n+1},x)\supseteq (f_{n+1}(x)-\epsilon/2^{n+1},
f_{n+1}(x)+\epsilon/2^{n+1})$ since $\epsilon\leq\tau$.  We now
consider the case when 
$x\in\cont(f_n)$.  By ($a_n$), we know that $x\in E_{n+1}$ so, by 
($c_{n+1}^{*}$), $\acc_{S}(f_{n+1},x)
\supseteq(f_{n+1}(x)-\epsilon/2^{n+1},f_{n+1}(x)+\epsilon/2^{n+1})$.  
Since $S\subseteq G_1\cap F_{n+1}\subseteq G\cap F_{n+1}$, we have 
($c_{n+1}$).  Finally, notice that ($a_{n+1}$) follows immediately from
($a_{n+1}^*$).   This completes the inductive step.  We now let
$f=\lim_{n\to\infty} f_n$.  

We show that $f$ satisfies (i)-(v).  We first  note that $f$ is the
uniform limit of 
$\langle f_n \rangle_{n\in\omega}$ by ($d_n$).  Notice that ($d_n$)
together with ($a_0$) also yields (v).   We show (i).  If $x\notin E$
then, for every 
$n\in\omega$ we have, by ($a_{n}$), $x\in\cont(f_n)$.   So by uniformity
of the convergence, $x\in\cont(f)$.  Thus, 
$(\real\setminus E)\subseteq\cont(f)$.   Let $x\in E$.  There is an
$n\in\omega$ such that $x\in E_n$.   By $(c_n)$, $f_n|_{F_n}$ is
discontinuous at $x$.  Since $(b_m)$  holds for all $m>n$, we have that
$f|_{F_n}$ is discontinuous at $x$.   Thus, $x\notin\cont(f)$.  So
$\cont(f)\subseteq (\real\setminus E)$,  which establishes (i).   To see
that (ii) holds let $n\in\omega$.  If $n=0$, then 
$f_0|_{E_0}$ is constant by ($b_0$).  By ($b_m$) for $m>0$, we  have
$f|_{E_0}=f_0|_{E_0}$; so, $f|_{E_0}$ is continuous.  If $n>0$  then, by
($a_n$), $E_{n+1}^*\subseteq\cont(f_n)$; so, 
$f_n|_{E^*_{n+1}}$ is continuous.  By ($b_m$) for $m>n$, we  have
$f|_{E_{n+1}^*}=f_n|_{E_{n+1}^*}$; so, $f|_{E^*_{n+1}}$ is continuous.  
Thus, (ii) holds. We show (iii) and (iv) hold.  Fix
$n\in\omega\setminus\{0\}$ and let $x\in E_n$.   We have, by ($c_n$) of the
construction, that $\acc_{F_{n}\cap G}(f_n,x)\supseteq
(f_{n}(x)-\epsilon/2^n,f_{n}(x)+\epsilon/2^n)$.  It follows, by ($b_m$)
for $m>n$, that
$\acc_{F_{n}\cap G}(f,x)\supseteq(f(x)-\epsilon/2^n,f(x)+\epsilon/2^n)$.  
Using (i), we have $G=\cont(f)$ so 
$\acc_{\cont(f)}(f,x)\supseteq(f(x)-\epsilon/2^n,f(x)+\epsilon/2^n)$.  
Thus, (iv) holds.  A similar argument shows that (iii) holds.   We must
now show that $f\in\swiat$.  Note that since the uniform limit of
Darboux Baire 1 functions is again Darboux Baire 1 
\cite[p.72]{BCW},
$f\in\dar\cap\baire_1$.  Let $a<b$ and assume without loss of generality
that $f(a)<f(b)$.   Let $f(a)<w<f(b)$ be arbitrary.  Since $f\in\dar$,
there is a $z\in (a,b)$ such that $f(z)=w$.  If $z\in\cont(f)$  then we
are done.  So, assume that 
$z\notin\cont(f)$.  Since $\real\setminus\cont(f)=E$, there is some 
$n\in\omega$ such that $z\in E_n$.  By (iii) or (iv), we may find an
$x\in\cont(f)\cap (a,b)$ such that $f(x)=w$.  Thus, 
$f\in\swiat\cap\baire_1$.
\qed

To simplify the proof of Theorems~\ref{thm3:1} and \ref{thm3:12}, we
introduce another lemma.
\noindent\lem{lem3:suff}{Let $f\in\real^{\real}$.  
\begin{description}
\item[(a)] If $f\in\baire_1$ and $f(x)\in\wacc_{\cont(f)}(f,x)$  for
every $x\in\real$, then $f\in\quasi\cap\dar$.
\item[(b)] If $f\in\cliq$ and $f(x)\in\wacc_{\cont(f)}(f,x)$  for every
$x\in\real$, then $f\in\quasi\cap\pr$.
\item[(c)] If $g\in\real^{\real}$, $x\in\cont(g)$, and 
$\cont(f)\subseteq\cont(g)$, then 
\[g(x)+\wacc_{\cont(f)}(f,x)\subseteq\wacc_{\cont(f+g)}(f+g,x).\]
\item[(d)] If $x\in\cont(f)$, then $x$ is a point of quasi-continuity  and
peripheral continuity of $f$.
\end{description}}
\proof We show (a) and (b).  By (i) and (iv) of Proposition~\ref{prop3:2},
it enough for us to show that if $f(x)\in\wacc_{\cont(f)}(f,x)$ for
every $x\in\real$, then $x\in\quasi\cap\phc$.  Let $x\in\real$.   Since
$f(x)\in\wacc_{\cont(f)}(f,x)$, there is a bilateral sequence 
$\{x_k\}_{k\in\omega}$ such that $x_k\in\cont(f)$, 
$\lim_{k\to\infty}x_k=x$, and $\lim_{k\to\infty}f(x_k)=f(x)$.   Clearly,
$x$ is a point of peripheral continuity.  We show that 
$x$ is a point of quasi-continuity.  Let $U$ and $V$ be open sets  such
that $x\in U$ and $f(x)\in V$.  Since 
$\lim_{k\to\infty}x_k=x$ and $\lim_{k\to\infty}f(x_k)=f(x)$, there  is a
$k$ such that $f(x_k)\in V$ and $x_k\in U$.  Since 
$x_k\in\cont(f)$, there is an open set $W$ such that 
$x_k\in W\subseteq U$ and $f[W]\subseteq V$.  Thus, $x$ is a point  of
quasi-continuity.   

We show (c).  Let $r\in g(x)+\wacc_{\cont(f)}(f,x)$.  Then, there is  a
bilateral sequence $\{x_k\}_{k\in\omega}$ such that $x_k\in\cont(f)$, 
$\lim_{k\to\infty}x_k=x$, and $\lim_{k\to\infty}f(x_k)=r-g(x)$.  Since 
$x\in\cont(g)$, we have $\lim_{k\to\infty}g(x_k)=g(x)$.  Now 
$\lim_{k\to\infty}(f+g)(x_k)=r$ and 
$x_k\in\cont(f)\subseteq\cont(g)$ for every $k\in\omega$.  Thus, 
$r\in\wacc_{\cont(f+g)}(f+g,x)$.  

We leave (d) without proof as it follows immediately from the 
definitions.  \qed



\bigskip

\noindent {\sc Proof of Theorem \ref{thm3:1}.}   By Lemma \ref{lem3:4} 
there exists $F^*\subseteq \cnwd^{\omega}$  such that
$|F^*|=\cof(\meager)$ and
\begin{equation}\label{e7} (\forall G\in\dense)(\forall\Theta\in
\cnwd^{\omega}) (\exists\Psi\in F^*)(\Theta\prec_{G}\Psi).
\end{equation}
For each $\Psi\in F^*$, $q\in\rational^+\cup\{\infty\}$,
and 
$n\in\omega$, define 
$f_{\Psi}^{q,n}\in\swiat\cap\baire_1$ to be the function  constructed in
Lemma \ref{lem3:3} with
$E_k=\Psi(k)$, $\tau=q$, and $\epsilon=\min\{\tau,2^{-n}\}$.   Let
$F=\{f^{q,n}_{\Psi}\colon q\in\rational^+\cup\{\infty\}\ \&\ n\in\omega\
\&\ \Psi\in F^*\}$; since
$\cof(\meager)>\omega$, we see that 
$|F|=\cof(\meager)=|F^*|$.  

We show that $F$ is as desired.  That is to say, we show that for  every
$\epsilon>0$ and $H\in[\baire_1]^{<\omega}$ there is an $f\in F$  with
the property that 
$f+H\subseteq\dar\cap\quasi$ and 
$\norm(f)\leq\max(\{\osc(g,x)\colon x\in\real\ \&\ g\in H\})+\epsilon$.
Let $H\in[\baire_1]^{<\omega}$ and $\epsilon>0$.  If 
$\max\{\osc(g,x)\colon x\in\real\ \&\ g\in H\}=+\infty$, then  let
$\tau=+\infty$.  In the other case, we pick
$\tau\in\rational\cap (0,+\infty)$ such that 
\begin{equation}\label{eq3:afinal}
0<\tau-\max\{\osc(g,x)\colon x\in\real\ \&\ g\in H\}<2^{-m}
\end{equation} 
where
$m\in\omega$ is such that $2^{-m}<\min\{\tau,\epsilon/3\}$.   Define $\Theta_H\in
\cnwd^{\omega}$ so that 
\[\Theta_H(n)=\bigcup_{g\in H}\cl(\osc_{n+m+2}(g))\] for each
$n\in\omega$.  There is a $\Psi\in F^*$ such that $\Theta_H\prec^*_G\Psi$
where
$G\!=\!\bigcap_{g\in H}\cont(g)$.  We claim that 
$f=f_{\Psi}^{\tau,m}\in F$ is the desired function.  It follows from (v)
of Lemma~\ref{lem3:3} and  the choice of $m$ that 
$\norm(f)\leq\tau+2\epsilon/3$.  By our choice of $\tau$ and $m$, we have 
\[\tau+2\epsilon/3 <\max\{\osc(g,x)\colon x\in\real\ \&\ g\in
H\}+\epsilon/3+2\epsilon/3.\]  Thus, $\norm(f)<\max\{\osc(g,x)\colon
x\in\real\text{ and } g\in H\}+\epsilon$.   Fix $g\in H$.   We show that
$f+g\in\dar\cap\quasi$.  By Lemma~\ref{lem3:suff}(a), we only have to show
that for every $x\in\real$
\begin{equation}\label{eq3:nuff}  (f+g)(x)\in\wacc_{\cont(f+g)}(f+g,x).
\end{equation}  
 Before proving (\ref{eq3:nuff}) we note that
\begin{equation}\label{eq3:300}
\cont(f)\subseteq\cont(g)
\end{equation} 
since by (i) of Lemma~\ref{lem3:3},  
$\cont(f)=\real\setminus\bigcup_{i\in\omega}
\Psi(i)\subseteq
\real\setminus\bigcup_{i\in\omega}\Theta_{H}(i)\subseteq\cont(g)$.  

Let $x\in\real$.  If $x\in\cont(g)$, then, by (\ref{eq3:300}), (c) of
Lemma~\ref{lem3:suff}, and the fact that 
$f(x)\in\wacc_{\cont(f)}(f,x)$ (Lemma~\ref{lem3:3}(iii) and (iv)) 
we have  
\[(f+g)(x)\in\wacc_{\cont(f+g)}(f+g,x).\]   
If $x\notin\cont(g)$, then, by
(\ref{eq3:300}), 
$x\notin\cont(f)$.  By (i) of Lemma~\ref{lem3:3}, there is a minimal number 
$n\in\omega$ such that
$x\in\Psi(n)$.  It is also true, since $\Theta_{H}(k)\subseteq\Psi(k)$ 
for every $k\in\omega$, that $n$ is less than or equal to the minimal 
number $p$ such that 
$x\in\Theta_H(p)$.   Since $p$ is minimal and $n\leq p$, we have in 
the case when $n>0$ that
$x\notin\Theta_H(n-1)\supseteq\osc_{n+m+1}(g)$.   Thus, there is a 
$\delta>0$ such that
\begin{equation}\label{eq3:100} |g(w)-g(x)|<2^{-m-n-1} \text{ for every }
w\in (x-\delta,x+\delta)
\end{equation} 
if $n>0$.  If $n=0$ then by (\ref{eq3:afinal}). 
\begin{equation}\label{eq3:200} |g(w)-g(x)|<\tau \text{ for every } w\in
(-\delta+x,\delta+x)
\end{equation} 
if $n=0$. Since $x\in\Psi(n)\setminus\bigcup_{k\leq
n-1}\Psi(k)$ and 
$\bigcup_{k\leq n-1}\Psi(k)$ is closed it follows, by
Lemma~\ref{lem3:3}(ii), that
\begin{equation}\label{eq3:101} f|_{\Psi(n)} \text{ is continuous at $x$.}
\end{equation}

We consider the case when $n=0$.  By Lemma~\ref{lem3:3}(ii), 
$f[\Psi(0)]=\{0\}$.  Since $\Theta_H(0)\prec_G\Psi(0)$, we may find a 
bilateral sequence
$\{x_k\}_{k\in\omega}$ in
$\cont(g)\cap\Psi(0)$ such that $\lim_{k\to\infty} x_k=x$.   By (iii) of
Lemma~\ref{lem3:3}, 
$\acc_{\cont(f)}(f,x_k)\supseteq(-\tau,+\tau)$.  Since
$\cont(f)\subseteq\cont(g)$ and 
$x_k\in\cont(g)$ for each $k$, it follows from (c) of 
Lemma~\ref{lem3:suff} that 
\begin{equation}\label{eq3:400}
\wacc_{\cont(f+g)}(f+g,x_k)\supseteq  [g(x_k)-\tau,g(x_k)+\tau]
\end{equation} for all $k\in\omega$.   If $\tau=\infty$, then, by
(\ref{eq3:400}), we have 
$\wacc_{\cont(f+g)}(f+g,x)\supseteq (-\infty,+\infty)$; and so 
$(f+g)(x)\in\wacc_{\cont(f+g)}(f+g,x)$.   If $\tau<\infty$, then we may
assume, taking a subsequence if necessary, that there is an $r\in\real$
such  that $\lim_{k\to\infty}(f+g)(x_k)=\lim_{k\to\infty}g(x_k)=r$.   So,
by (\ref{eq3:400}),  
\[\wacc_{\cont(f+g)}(f+g,x)\supseteq [r-\tau,r+\tau].\]  
By
(\ref{eq3:200}), $|(f+g)(x)-r|\leq\tau$.  Thus, 
$(f+g)(x)\in\wacc_{\cont(f+g)}(f+g,x)$ when $n=0$.
  
We now consider the case when $n>0$.     Since
$\Theta_H(n)\prec_G\Psi(n)$, we may find a bilateral sequence
$\{x_k\}_{k\in\omega}$ in
$\cont(g)\cap\Psi(n)$ such that $\lim x_k=x$.  Taking a  subsequence if
necessary, we may assume, since $x\notin\Theta_H(0)$, that there is a number 
$r$ such that $\lim (f+g)(x_k)=r$.  By (iv) of  Lemma~\ref{lem3:3}, 
\[\acc_{\cont(f)}(f,x_k)\supseteq(-2^{-n-m}+f(x_k),2^{-n-m}+f(x_k)).\]   
Since $\cont(f)\subseteq\cont(g)$ and 
$x_k\in\cont(g)$ for each $k$ it follows from (c) of 
Lemma~\ref{lem3:suff} that 
\begin{equation}\label{eq3:500}
\wacc_{\cont(f+g)}(f+g,x_k)\supseteq 
[-2^{-n-m}+(f+g)(x_k),2^{-n-m}+(f+g)(x_k)]
\end{equation}  for all $k\in\omega$. By (\ref{eq3:500}) and
$\lim_{k\to\infty}(f+g)(x_k)=r$, we have 
\begin{equation}\label{eq3:102}
\wacc_{\cont(f+g)}(f+g,x)\supseteq  [-2^{-n-m}+r,2^{n-m}+r].
\end{equation} Now, by (\ref{eq3:100}) and (\ref{eq3:101}), 
$\lim_{k\to\infty}f(x_k)=f(x)$ and $|g(x)-g(x_k)|<2^{-n-m-1}$ for all
$k$ large enough.  It follows that
\begin{equation}\label{eq3:103}
|(f+g)(x)-r|\leq|\lim_{k\to\infty}(f(x)-f(x_k)+g(x)-g(x_k))|\leq
2^{-n-m-1}.
\end{equation} 
By (\ref{eq3:103}) and (\ref{eq3:102}), we have that
$(f+g)(x)\in\wacc_{\cont(f+g)}(f+g,x)$ when $n>0$.   Thus,
(\ref{eq3:nuff}) is established, and so 
$f+g\in\quasi\cap\dar$.\qed  

\bigskip

\noindent{\sc Proof of Theorem~\ref{thm3:12}}.   Notice that in the proof
of Theorem~\ref{thm3:1} the only place we needed to use the fact that
$f+H\subseteq\baire_1$ was when  we used (a) of Lemma~\ref{lem3:suff} to
say that  (\ref{eq3:nuff}) implied $f+g\subseteq\dar\cap\quasi$.  So, if we 
had just wanted to  prove that $f+H\subseteq\quasi\cap\pr$, we could have
just assumed that 
$H\subseteq\cliq$ and used (b) of Lemma~\ref{lem3:suff} to  say that
$f+g\in\quasi\cap\pr$.\qed

%%%%%CHAPTER4
\chapter{Super-additivity in the general case}\label{fourth_chapter}
\section{Introduction}
Thus far we have only considered the 
super-additivities of the Darboux-like families of functions in the 
realm of Baire class one or cliquish functions in $\real^{\real}$.  
We now come back to the general case of considering all functions in 
$\real^{\real}$ and consider the super-additivities of the 
Darboux-like families in that light.  
So again we restrict the scope 
of this chapter to $\real^{\real}$ and keep the convention of 
Chapters~\ref{first_chapter} of writing, for example, $\dar$ when we mean 
$\dar(\real,\real)$ and $\comp\dar$ instead of $\real^{\real}\setminus\dar$.


Given $\F\subseteq\real^{\real}$, the definition of the additivity 
of $\F$ in $\real^{\real}$ implies that  
\[(*)\ \ \ \ \ \   
\real^{\real}\in\left\{F\subseteq\real^{\real}\colon
\left(\forall G\in\left[\real^{\real}\right]^{<\add(\F)}\right)
(\exists f\in\real^{\real})(f+G\subseteq\F)\right\}.\]
A natural question that arises is whether or not $\real^{\real}$ is 
the only subset of $\real^{\real}$ which is an element of the set 
in $(*)$.  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 for the case 
$\real^{\real}$. (Compare with Propositon~\ref{prop3:sup}.)  

\noindent\prop{prop4:2}{Let $\F,{\mathcal E}\subseteq\real^{\real}$.  Then, 
\begin{description}
\item[(i)] $\F\in\{\emptyset,\real^{\real}\}$ if and only if 
$\add^*(\F)=1$ and
\item[(ii)] if $\add(\F)=\add({\mathcal E})$ and 
$\F\subseteq{\mathcal E}$, then 
$\add^*(\F)\geq\add^*({\mathcal E})$.
\end{description}}
\proof
We show (i).  If $\F=\real^{\real}$, then, by 
Proposition~\ref{prop1:1}(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{prop1:1}(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{prop1:1}(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, 
$\F=\real^{\real}$.

Item (ii) is just item (iii) of Propsition~\ref{prop3:sup}.  \qed

Since $\real$ is a group, 
Propositions~\ref{prop1:1} and \ref{prop3:51} give us the following 
proposition.  
\noindent\prop{prop4:3}{If $\F\notin\{\real^{\real},\emptyset\}$, then 
\[\max\{\add(\F),\add(\real^{\real}\setminus\F)\}
\leq \add^{*}(\F).\]}

\section{ The Results}

To state our Theorems we will need some cardinals which have 
combinatorial descriptions.  For a cardinal $\kappa$, we define  

\[\diff_{\kappa}^{1}=\min\left\{|F|\colon F\subseteq\kappa^{\kappa}\ \&\ 
\left(\forall G\in\left[\kappa^{\kappa}\right]^{<\diff_{\kappa}}\right)
(\exists f\in F)(\forall g\in G)([f=g]\neq\emptyset)\right\},\]
\[\same_{\kappa}^{1}=\min\left\{|F|\colon F\subseteq\kappa^{\kappa}\ \&\ 
\left(\forall G\in\left[\kappa^{\kappa}\right]^{<\same_{\kappa}}\right)
(\exists f\in F)(\forall g\in G)(|[f=g]|<\kappa)\right\}.\]


We first calculate the super-additivities of the families of 
functions that concern us.
\noindent\thm{thm4:1}{If $\F\in\{\ext,\pr,\phc\}$, then 
$\add^*(\F)=\add^*(\comp\F)=2^{\cuum}$.}
\noindent\thm{thm4:2}{$\add^*(\acon)=\add^*(\conn)=\add^*(\dar)=\diff^{1}_{\cuum}$.}
\noindent\thm{thm4:3}{$\add^*(\sz)=\same_{\cuum}^{1}$.}
\noindent\thm{thm4:4}{If $|[\cuum]^{<\cuum}|=\cuum$, then $\same_{\cuum}^{1}
=\add^*(\comp\dar)=\add^*(\comp\conn)=\add^*(\comp\acon)$.}
\noindent\thm{thm4:5}{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 
$\diff_{\cuum}^{1}$.  
   
\noindent\thm{thm4:7}{If $|[\cuum]^{<\cuum}|=\cuum$ and $\cuum=\lambda^+$, then 
$\same_{\cuum}\leq\diff_{\cuum}=
\diff_{\cuum}^{1}=\same_{\cuum}^{1}$.}

Proposition~\ref{prop1:5} will allow us to make a statement similar 
to that of Theorem~\ref{thm4:7} for the case when $\cuum$ is 
not regular.  
\noindent\cor{cor4:0}{If $\cuum$ is not regular, then 
$\diff_{\cuum}=\diff^*_{\cuum}=\cuum^+\leq\same_{\cuum}\leq
\min\{\diff^1_{\cuum},\same^1_{\cuum}\}$.}
\proof 
The inequality $\cuum^+\leq\same_{\cuum}$ is obvious since we know 
$\same_{\cuum}>\cuum$.  By Propositions~\ref{prop4:3} and 
Theorems~\ref{thm4:2} and \ref{thm1:2},  
$\diff^1_{\cuum}=\add^*(\dar)\geq
\add(\comp\dar)\geq\same_{\cuum}$.  By Proposition~\ref{prop4:3} 
and \ref{prop1:3} and Theorem~\ref{thm4:3}, 
$\same^1_{\cuum}=\add^*(\sz)\geq\add(\sz)=\same_{\cuum}$. \qed
 
Since CH implies that $|[\cuum]^{<\cuum}|=\cuum$,
Propositions~\ref{prop1:6}, \ref{prop1:7} and 
Corollaries~\ref{cor1:2} and \ref{cor1:3} 
together with Theorem~\ref{thm4:7} yeild two corollaries. 

\noindent\cor{cor4:1}{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}
\hfill$\add(\F)=\add^{*}(\F)=\add(\comp\F)=\add^*(\comp\F)=\kappa$.\QED
\end{center}}

\noindent\cor{cor4:2}{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 
$\F\in\{\dar,\conn,\acon\}$
\[
\add(\sz)=\add(\comp\F)=\cuum^{+}
\]
which is strictly less than
\begin{center}
\hfill$\add(\F)=\add^*(\F)=\add^*(\comp\F)=\add(\comp\sz)
=\add^*(\comp\sz)=\add^*(\sz)=2^{\cuum}$.\QED
\end{center}
}

We prove Theorems~\ref{thm4:1} and \ref{thm4:3} at this time.  
We will prove the other theorems in later 
sections since their proofs are somewhat long.

\medskip

\noindent{\sc Proof of Theorem~\ref{thm4:1}.}  
Let $\F\in\{\ext,\pr,\phc\}$.  By  
Proposition~\ref{prop1:4}(i) and Theorem~\ref{thm1:1}, we have 
$\max\{\add(\F),\add(\real^{\real}\setminus\F)\}=2^{\cuum}$.  
Since $\F\notin\{\real^{\real},\emptyset\}$, 
Proposition~\ref{prop4:3} implies that $\add^*(\F)=2^{\cuum}=
\add^*(\real^{\real}\setminus\F)$.  \qed

\medskip

\noindent{\sc Proof of Theorem~\ref{thm4:3}.  }
We show that $\add^*(\sz)\leq\same_{\cuum}^{1}$.  Let $H$ stand 
for the family of all functions $h\in\real^{\real}$ such that
$h$ is continuous on a $G_{\delta}$ set 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{eq4: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{eq4: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 Corollary~\ref{cor1:2} we have $|G|<\same_{\cuum}$.  Since 
$\same_{\cuum}>\cuum$, it follows that 
$\{h-g\colon g\in G\ \&\ h\in H\}$ is a set of cardinality 
strictly less than $\same_{\cuum}$.  By (\ref{eq4:sz20}), there is 
an $f\in F$ such that for every $g\in G$ and $h\in H$, 
we have $|[f=h-g]|<\cuum$.  
So, by Proposition~\ref{prop1:11}, $f+g$ is continuous on no set of cardinality~$\cuum$ for every 
$g\in G$.  Thus, $F$ satisfies (\ref{eq4: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{eq4:sz22}
\left(\forall G\in\left[\real^{\real}\right]^{<\add(\sz)}\right)
(\exists f\in F)(f+G\subseteq\sz).
\end{equation}
Let $F_1=\{-f\colon f\in F\}$; notice $|F_1|=|F|=\add^*(\sz)$.  
It is enough to show that $F_1$ satisfies 
\begin{equation}\label{eq4:sz23}
\left(\forall G\in\left[\real^{\real}\right]^{<\same_{\cuum}}\right)
(\exists f\in F_1)(|[f=g]|<\cuum).
\end{equation}
Let $G\in\left[\real^{\real}\right]^{<\same_{\cuum}}$ be arbitrary.  
By Proposition~\ref{prop1:4}(iii), we have $|G|<\add(\sz)$.  By 
(\ref{eq4:sz22}), there is an $f\in F$ such that $f+g\in\sz$ for every 
$g\in G$.  In particular, we have $|(f+g)^{-1}(\{0\})|<\cuum$ for every 
$g\in G$.  It follows that, $|[-f=g]|<\cuum$ for every $g\in G$.  
Thus, $F_1$ satisfies (\ref{eq4:sz23}) and 
$\same_{\cuum}^{1}\leq\add^*(\sz)$.\qed

\section{Proof of Theorem \ref{thm4:2}}

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, $\dar_1\subseteq\dar$.  

\noindent\lem{lem4:20}{$\add^*(\dar)=\add^*(\dar_1)$.}
\proof
It follows from \cite[Theorem 2.4]{CM} that 
$\add(\dar)=\add(\dar_1)$.  By Proposition~\ref{prop4:2}(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{eq4:20}
\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{eq4:21}
\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{eq4:20}) 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{eq4:21}).\qed

\noindent\lem{lem4:21}{$\add^*(\dar)=\add^*(\conn)=\add^*(\acon)$.}
\proof 
By Propositions~\ref{prop1:4}(ii) and \ref{prop4:2}(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{lem4:20}, there is an  
$F\subseteq\real^{\real}$ such that $|F|=\add^*(\dar)$ and 
\begin{equation}\label{eq4:22}
\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{eq4:23}
\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 ${\mathcal B}$ be as in 
Proposition~\ref{prop2:a1}.  For each $B\in{\mathcal B}$ let 
$h_B\in\real^{\real}$ be such that 
$h_B|_{\domain(B)}\subseteq B$ and zero otherwise.  Let 
$G\subseteq\real^{\real}$ and $|G|<\add(\acon)=\add(\dar)$.  
Let $G_1=\{g-h_B\colon g\in G\ \&\ B\in{\mathcal B}\}$.  
Since $\cuum<\add(\dar)$, we have $|G_1|<\add(\dar)$.  
By (\ref{eq4:22}) 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 $B\in {\mathcal B}$ be arbitrary.  
Since $f+(g-h_B)\in\dar_1$ there is an $r\in\domain(B)$ 
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{thm4:2}, it is enough to 
prove the following lemma.  
\noindent\lem{lem4:22}{$\add^*(\dar_1)=\diff^1_{\cuum}$.}
\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 
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{eq4:24}
\left(\forall G\in\left[\real^{\cuum}\right]^{<\diff_{\cuum}}\right)
(\exists f\in F)([f=g]\neq\emptyset).
\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{eq4:25}
\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{prop1:4}(ii), it follows that 
$\add(\dar_1)=\add(\dar)=\diff_{\cuum}$.  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 
\[G_1=\{g_{\alpha,r}\colon r\in\real\ \&\ g\in G\ \&\ \alpha\in\cuum\}.\]  Since 
$\diff_{\cuum}>\cuum$, it follows that $|G_1|<\diff_{\cuum}$.  
By (\ref{eq4:24}), there is an $f\in F$ such that 
$[f=g_1]\neq\emptyset$ 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{eq4:25}), 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{eq4:26}
\left(\forall G\in\left[\real^{\real}\right]^{<\add(\dar_1)}\right)
(\exists f\in F)(f+G\subseteq\dar_1).
\end{equation}
It is enough for us to show that $F$ also satisfies 
\begin{equation}\label{eq4:27}
\left(\forall G\in\left[\real^{\real}\right]^{<\diff_{\cuum}}\right)
(\exists f\in F)([f=g]\neq\emptyset).
\end{equation}
Let $G\in\left[\real^{\real}\right]^{<\diff_{\cuum}}$.  Clearly, 
$|-G|=|\{-g\colon g\in G\}|=|G|<\diff_{\cuum}$.  By (\ref{eq4:26}), 
there is an $f\in F$ such that $f+(-G)\subseteq\dar_1$.  In 
particular, there is for each $g\in G$ an $r\in\real$ such 
that $(f-g)(r)=0$.  So $[f=g]\neq\emptyset$ for all $g\in G$.  
Thus, $F$ satisfies (\ref{eq4:27}).\qed

\section{Proofs of Theorems \ref{thm4:4} and \ref{thm4:5}}
Before we begin we note that if $|[\cuum]^{<\cuum}|=\cuum$ then 
$\cuum$ is regular \cite{Ci:book}.  
Our first goal will be to prove Theorem~\ref{thm4:4}.  
To do this we will need to define another cardinal,
\begin{eqnarray}
\kappa_1 & = & \min\{|F|\colon F\subseteq 
\left[\cuum^{\cuum}\right]^{\cuum} \notag \\
& \& & \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)
\}.\notag
\end{eqnarray}

\noindent\lem{lem4:30}{If $|[\cuum]^{<\cuum}|=\cuum$, then 
$\kappa_1\leq\add^*(\comp\acon)$.}
\proof
Let ${\mathcal B}$ be as in Proposition~\ref{prop2:a1}.  Enumerate ${\mathcal B}$ 
by $\{B_{\alpha}\colon\alpha\in\cuum\}$.  
For each $\alpha\in\cuum$, let $h_{\alpha}\in\real^{\real}$ 
be such that 
$h_{\alpha}|_{\domain(B_{\alpha})}\subseteq B_{\alpha}$ and 
zero otherwise.  Let 
$\{P_{\alpha}\}_{\alpha\in\cuum}$ be a partition of $\real$ such 
that $P_{\alpha}\subseteq\domain(B_{\alpha})$ 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{eq4:30}
\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{eq4:31}
\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, by 
Corollary~\ref{cor1:2}, that $\add(\comp\acon)=\same^*_{\cuum}$; so, 
$|G|<\add(\comp\acon)$.  
For every $g\in G$ define $g^*\in\real^{\real}$ so that for any 
$\alpha,\beta\in\cuum$ we have 
$g^*(p_{\alpha,\beta})=g(\beta)$.  Since 
$|\{g^*\colon g\in G\}|\leq |G|<\add(\comp\acon)$, it follows 
by (\ref{eq4:30}) 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{eq4:32}
(\forall g\in G)(\exists f\in A)([f=g]=\emptyset).
\end{equation}
Fix $g\in G$.  By 
Proposition~\ref{prop2:a1}, there is an 
$\alpha\in\cuum$ such that $(f+g^*)\cap B_{\alpha}=\emptyset$.  
It follows that
\begin{equation}\label{eq4:33} 
(f+g^*)|_{P_{\alpha}}\cap h_{\alpha}|_{P_{\alpha}}=\emptyset.  
\end{equation}

We will show that $f_{\alpha}\in A_f$ is such that 
$[f_{\alpha}=g]=\emptyset$.  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{eq4:33}) since $p_{\alpha,\beta}\in B_{\alpha}\subseteq 
P_{\alpha}$.  
Thus, $[f_{\alpha}=g]=\emptyset$ so 
$A_f$ satisfies (\ref{eq4:32}).  Therefore, $F^*$ satisfies 
(\ref{eq4:31}). \qed

To continue the proof it will be useful for us to recall 
some of the families of functions from Chapter~\ref{first_chapter}.  
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 $|f[(a,b)]|>1$.

\noindent\lem{lem4:31}{$\add^*(\comp\dar)=\add^*(\comp\dar^*)$.}
\proof 
By Lemma~\ref{lem1:six} we have $\add(\comp\dar)=\add(\comp\dar^*)$.  
It follows by Proposition~\ref{prop4:2}(ii) that 
$\add^*(\comp\dar^*)\leq\add^*(\comp\dar)$.  

We show the other inequality. 
Let $\mathcal{I}$ 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 
$|{\mathcal{I}}|=\cuum$.  For each $I\in{\mathcal{I}}$ pick 
$h_{I}\in\dar(\cuum)$ such that $h_{I}(x)=0$ if $x$ is an end-point 
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 $K=\{k_{I}\colon I\in \mathcal{I}\}$. 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{eq4:34}
(\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{eq4:35}
(\forall f\in F)(\exists g\in G_1)(f+g\in\dar^*).
\end{equation}
Let $f\in F$.  By (\ref{eq4:34}), 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 is 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 neighborhood $U$ 
about $x$ contains a non-degenerate sub-interval $i$ of some 
$j\in J$.  Thus,
\begin{equation}\label{eq4: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 nowhere 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 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{eq4: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{eq4:35}), completing the proof.\qed
 
\noindent\lem{lem4:32}{$\add^*(\comp\dar(\cuum))=\add^*(\comp\dar)$.}
\proof
By Lemma~\ref{lem1:eight}, we have $\add(\comp\dar)=\add(\comp\dar(\cuum))$.  
It follws by Proposition~\ref{prop4:2}(ii) that 
$\add^*(\comp\dar(\cuum))\leq\add^*(\comp\dar)$.  

By Lemma~\ref{lem4:31}, it is enough for us to show that 
$\add^*(\comp\dar^*)\leq\add^*(\comp\dar(\cuum))$.  By Lemma~\ref{lem1:seven} 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.  Assume that 
$F\subseteq\real^{\real}$ and $|F|<\add^*(\comp\dar^*)$.  Since $\Theta$ 
is a surjection, we may for each 
$f\in F$ pick $f_1\in\real^{\real}$ so 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 $\add(\comp\dar^*)$, there 
is a $G\subseteq\real^{\real}$ such that $|G|<\add(\comp\dar^*)=
\add(\comp\dar)$ and 
\begin{equation}\label{eq4:36}
(\forall f_1\in F_1)(\exists g\in G)(f_1+g\in\dar^*).
\end{equation}
Define $g_1=\Theta\circ g$ for each $g\in G$.  Define $G_1$ to 
be the set $\{g_1\colon g\in G\}$.  Notice that $|G_1|<\add(\comp\dar)\leq
\add(\comp\dar(\cuum))$.  We will be done if we show that $G_1$ 
satisfies
\begin{equation}\label{eq4:37}
(\forall f\in F)(\exists g_1\in G_1)(f+g_1\in\dar(\cuum)).
\end{equation}
Let $f\in F$.  By (\ref{eq4:36}), 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{eq4:37}).\qed


\noindent\lem{lem4:33}{If $|[\cuum]^{<\cuum}|=\cuum$, then 
$\add^*(\comp\dar)\leq\same^1_{\cuum}$.}
\proof
By Lemma~\ref{lem4:32} 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{eq4:38} 
\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{eq4:39}
\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 
Lemma~\ref{lem1:eight} and Corollary~\ref{cor1:2}, that 
$|G|<\same_{\cuum}$.  By 
(\ref{eq4:38}), there exists 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{eq4:39}).\qed


\noindent{\sc Proof of Theorem~\ref{thm4:4}.}
We have shown that if we assume $|[\cuum]^{<\cuum}|=\cuum$ then
\[\kappa_1\leq\add^*(\comp\acon)\leq
\add^*(\comp\conn)\leq
\add^*(\comp\dar)\leq\same^1_{\cuum}.\]  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 $V=\{\langle\alpha,\xi\rangle\colon\xi<\alpha<\cuum\}$.  
Let $F\subseteq[\cuum^{V}]^{\cuum}$ be such that 
$|F|=\kappa_1$ and
\begin{equation}\label{eq4:310} 
\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 $A=\{f_{\beta}\colon\beta\in\cuum\}$ 
and 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{eq4:311}
\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 
$\beta\in\domain(g(\alpha))\cap\alpha$, and define 
$g(\langle\alpha,\beta\rangle)$ to be zero otherwise.  Define 
$G_1$ to be the set $\{g_1\colon g\in G\}$ and notice that 
$|G_1|<\same_{\cuum}\leq\same^*_{\cuum}$.  By (\ref{eq4:310}) there 
is an $A\in F$ such that 
\begin{equation}\notag
(\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 
$[g_1=f_{\beta}]<\cuum$.  So there is, by regularity of $\cuum$, a 
$\zeta\in\cuum$ such that 
for any $\xi\in\cuum$ and $\alpha\geq\zeta$, 
we have that $g_1(\langle\alpha,\xi\rangle)\neq f_{\beta}(\langle\alpha,
\xi\rangle)$.  Hence for all $\alpha>\max\{\zeta,\beta\}$, 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, the claim is proved.  
Therefore, $F^*$ satisfies (\ref{eq4:311}).\qed

We now work to prove Theorem~\ref{thm4:5}.  Towards this end we must 
define another cardinal,  
\begin{eqnarray}
\kappa_2 & = & \min\{|F|\colon F\subseteq 
\left[\cuum^{\cuum}\right]^{\cuum} \notag \\
& \&\ & \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)\}.\notag
\end{eqnarray}

\noindent\lem{lem4:401}{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 
$h$ is continuous on a $G_{\delta}$ set 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{eq4:313}
\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 $A_f\in[\cuum^{\cuum}]^{\cuum}$ be the set 
$\{h-f\colon h\in H\}$.  Let $F^*=\{A_f\colon f\in F\}$.  Notice 
that $|F^*|\leq |F|=\add^*(\sz)$.  It is enough for us to show that 
$F^*$ satisfies 
\begin{equation}\label{eq4:314}
\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 $|G|<\diff^*_{\cuum}$.  
So by Corollary~\ref{cor1:3}, $|G|<\diff^*_{\cuum}=\add(\comp\sz)$.  By (\ref{eq4:313}), there is an $f\in F$ 
such that $f+G\subseteq\comp\sz$.  So, for each $g\in G$ there 
is, by Proposition~\ref{prop1:11}, 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{eq4:314}).

We now show that $\add^*(\comp\sz)\leq\diff^1_{\cuum}$.  
Let $F\subseteq\real^{\cuum}$ be such that $|F|=\diff^1_{\cuum}$ 
and
\begin{equation}\label{eq4:315}
\left(\forall G\in\left[\real^{\cuum}\right]^{<\diff_{\cuum}}\right)
(\exists f\in F)(\forall g\in G)([f=g]\neq\emptyset).
\end{equation}
Let $\{S_{\alpha}\colon\alpha\in\cuum\}$ be 
a partition of $\real$ into sets of size $\cuum$.  
For each $\alpha\in\cuum$ let 
$\{s^{\xi}_{\alpha}\colon\xi\in\cuum\}$ be an injective 
enumeration of $S_{\alpha}$.  For each $f\in F$ let 
$f^*\in\real^{\real}$ be defined 
so that $f^*(s^{\xi}_{\alpha})=f(\xi)$ for every 
$\alpha,\xi\in\cuum$.  Let 
$F^*=\{f^*\colon f\in F\}$.  Notice that 
we have $|F^*|\leq |F|=\diff^1_{\cuum}$.  It is enough 
for us to show that $F^*$ satisfies
\begin{equation}\label{eq4:316}
\left(\forall G\in\left[\real^{\real}\right]^{<\add(\comp\sz)}\right)
(\exists f^*\in F^*)(f^*+G\subseteq\comp\sz).
\end{equation}
Let $G\subseteq\real^{\real}$ and $|G|<\add(\comp\sz)$.  Notice 
that since $|[\cuum]^{<\cuum}|=\cuum$ we have, by 
Corollary~\ref{cor1:3}, $|G|<\diff_{\cuum}$.  For 
each $g\in G$ 
and $\alpha\in\cuum$ let $g_{\alpha}\in\real^{\cuum}$ be defined 
so that $g_{\alpha}(\xi)=-g(s^{\xi}_{\alpha})$ for every 
$\xi\in\cuum$.  Let $G_1=\{g_{\alpha}\colon
\alpha\in\cuum\ \&\ g\in G\}$.  Since $\cuum<\diff_{\cuum}$, we 
have $|G_1|\leq\max\{\cuum,|G|\}<\diff_{\cuum}$.  By 
(\ref{eq4:315}) there is an $f\in F$ such that 
$[f=g_{\alpha}]\neq\emptyset$ for all $g\in G$ and 
$\alpha\in\cuum$.  We claim that $f^*+G\subseteq\comp\sz$.  
Let $g\in G$ and $\alpha\in\cuum$.  Notice that for any $\beta\in\cuum$ 
we have 
$s_{\alpha}^{\beta}\in[f^*|_{S_{\alpha}}=-g|_{S_{\alpha}}]$ 
if and only if $\beta\in[f=g_{\alpha}]$.  By the choice of $f$ we have 
$[f=g_{\alpha}]\neq\emptyset$ so 
$[f^*|_{S_{\alpha}}=-g|_{S_{\alpha}}]\neq\emptyset$.  
Since $\{S_{\alpha}\}_{\alpha\in\cuum}$ is a mutally disjoint collection 
of sets it follows that $|[f^*=-g]|=\cuum$.  So 
$|(f^*+g)^{-1}(\{0\})|=\cuum$, which implies that $f^*+g\in\comp\sz$.  
Therefore, $F^*$ satisfies (\ref{eq4:316}).\qed

To finish the proof of Theorem \ref{thm4:5} it is enough, by 
Lemma~\ref{lem4:401}, to prove the following lemma.


\noindent\lem{lem4:41}{If $|[\cuum]^{<\cuum}|=\cuum$, then 
$\diff_{\cuum}^1=\kappa_2$.}
\proof
First notice that Lemma~\ref{lem4:401} 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 $V=\{\langle\alpha,\xi\rangle\colon\xi<\alpha<\cuum\}$.  
Let $F\subseteq[W^{\cuum}]^{\cuum}$ be such that 
$|F|=\kappa_2$ and 
\begin{equation}\label{eq4:317}
\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{eq4:318}
\left(\forall G\in\left[\cuum^V\right]^{<\diff_{\cuum}}\right)
(\exists f\in F)(\forall g\in G)([f=g]\neq\emptyset).
\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 
$\domain(g_1(\alpha))=\alpha$.  Let $G_1=\{g_1\colon g\in G\}$ and 
notice that $|G_1|\leq |G|<\diff_{\cuum}$.  By (\ref{eq4:317}) 
and the fact that $\diff^*_{\cuum}=\diff_{\cuum}$ under 
$|[\cuum]^{<\cuum}|=\cuum$, there is an $A\in F$ such that 
\begin{equation}\label{eq4:319} 
(\forall g\in G)(\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 a $g\in G$.  By (\ref{eq4:319}), there is an 
$\beta\in\cuum$ such that $f_{\beta}\in A$ and 
$|[g_1=f_{\beta}]|=\cuum$.  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\in\alpha\cap\domain(f_{\beta}(\alpha))\}| \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{eq4:318}) which implies that 
$\diff^1_{\cuum}\leq\kappa_2$. \qed 


\section{Proof of Theorem~\ref{thm4:7}}

Our first goal will be to show that under the assumption of
$|[\cuum]^{<\cuum}|=\cuum$ and $\cuum=\lambda^+$, we have 
$\diff_{\cuum}=\diff^1_{\cuum}=\dom_{\cuum}$.  

\noindent\lem{lem4:42}{If $\cuum=\lambda^+$ then 
$\dom_{\cuum}=\diff^*_{\cuum}\geq\kappa_2$.}
\proof 
By Lemma~\ref{lem1:thirteen}, $\diff^*_{\cuum}=\dom_{\cuum}$ 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{eq4:45}
(\forall g\in\cuum^{\cuum})
(\exists f\in F) (|[f\leq g]|<\cuum).
\end{equation}
For each $f\in F$ we may 
find, using Lemma~\ref{lem1:fourteen}, 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{eq4:46} 
\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{eq4:45}), 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{eq4:47}
(\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{eq4:47}). Therefore, $F^*$ satisfies (\ref{eq4:46}).  \qed
 
\noindent\lem{lem4:43}{If $|[\cuum]^{<\cuum}|=\cuum$ and $\cuum=\lambda^+$, 
then $\diff_{\cuum}=\diff^*_{\cuum}=\diff^1_{\cuum}=\dom_{\cuum}$.}
\proof
By Theorem~\ref{thm1:4} and Lemma~\ref{lem1:thirteen}, 
$\diff_{\cuum}=\diff^*_{\cuum}=\dom_{\cuum}$.  By 
Lemmas~\ref{lem4:41} and \ref{lem4:42}, we have 
$\diff^1_{\cuum}=\kappa_2\leq\dom_{\cuum}$.  So to complete the 
proof it is enough for us to show that 
$\diff^1_{\cuum}\geq\diff_{\cuum}$.  However, using 
Propositions~\ref{prop1:4}(ii) and \ref{prop4:3} 
and Theorem~\ref{thm4:2}, we have $\diff_{\cuum}=
\add(\dar)\leq\add^*(\dar)=\diff^1_{\cuum}$.\qed

\noindent\lem{lem4:44}{If $\cuum=\lambda^+$, 
then $\same^1_{\cuum}=\dom_{\cuum}$.}
\proof
Since $\cuum<\same_{\cuum}$, 
it is easy to check that $\diff^*_{\cuum}\leq\same^1_{\cuum}$.  
So, by Lemma~\ref{lem1:thirteen}, we have 
$\dom_{\cuum}=\diff^*_{\cuum}\leq\same^1_{\cuum}$.  
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{eq4:48}
(\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{eq4:49}
\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 
Lemma~\ref{lem1:thirteen}, $\same_{\cuum}=\unb_{\cuum}$.  
Since $|G|<\unb_{\cuum}$, there is an $h\in\cuum^{\cuum}$ such that
$|[h\leq g]|<\cuum$ for all $g\in G$.  By (\ref{eq4:48}), 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{eq4:49}).\qed

{\sc Proof of Theorem~\ref{thm4:7}}  
Since, by Lemma~\ref{lem1:thirteen}, $\same_{\cuum}=\unb_{\cuum}$ and 
$\unb_{\cuum}\leq\dom_{\cuum}$, Lemmas~\ref{lem4:43} and 
\ref{lem4:44} yield the theorem.\qed


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