\documentclass{rae}

\usepackage{amssymb}
\usepackage{amstex}
\usepackage{epic}
%\input{StandardMacros}

%\bysection

\received{October 26, 1994}

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

\keywords{cardinal invariants, extendable functions, 
functions with prefect road, peripherially continuous functions. }

\firstpagenumber{459}

\markboth{K. Ciesielski and I. Rec\l aw}%
{Sum of Extendable and Peripherally Continuous Functions} 

\author{Krzysztof Ciesielski, Department of Mathematics, West
Virginia University, Morgantown, WV 26506-6310, 
email: kcies@@wvnvms.wvnet.edu
\and
Ireneusz Rec\l aw\footnotemark,\addtocounter{footnote}{-1}
Department of Mathematics, University of Scranton, Scranton, PA 18510-4666,
email: ReclawI1@@jaguar.uofs.edu\\
Permanent address: Institute of Mathematics, Gda\'{n}sk University, 
Wita Stwosza~57, 80-952 Gda\'{n}sk, Poland, 
email: matir@@halina.univ.gda.pl}

\title{CARDINAL INVARIANTS CONCERNING EXTENDABLE AND PERIPHERALLY 
CONTINUOUS FUNCTIONS}

%\title{Cardinal invariants concerning extendable 
%and peripherally continuous functions.}


% IF YOUR VERSION OF LATEX DOES NOT RECOGNIZE COMMANDS \frak, \mathbb, OR
% \operatorname UNCOMMENT NEXT 3 LINES

%\def\frak{\bf}
%\def\operatorname{\rm}
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%\pagestyle{myheadings}

\renewcommand{\thefootnote}{\fnsymbol{footnote}}

\def\real{{\mathbb R}}
\def\rational{{\mathbb Q}}
\def\integer{{\mathbb Z}}
\def\continuum{{\frak c}}
\def\co{\continuum}
\def\add{{\operatorname{A}}}
\def\mul{{\operatorname M}}

\def\ext{{\operatorname{Ext}}}
\def\pr{{\operatorname{PR}}}
\def\pc{{\operatorname{PC}}}
\def\ac{{\operatorname{AC}}}
\def\conn{{\operatorname{Conn}}}
\def\darb{{\operatorname D}}

\def\cl{{\operatorname{cl}}}
\def\bd{{\operatorname{bd}}}
\def\cf{{\operatorname{cf}}}
\def\diam{{\operatorname{diam}}}


\def\rl{\real}
\def\reals{\real}

\def\ds{\bigcup}
\def\di{\bigcap}
\def\sb{\subseteq}
\def\su{\subseteq}
\newcommand{\propsub}{\subsetneqq}
\def\sm{\setminus}
\def\es{\emptyset}
\def\C{{\cal C}}
\def\F{{\cal F}}
\def\G{{\cal G}}
\def\H{{\cal H}}
\def\J{{\cal J}}
\def\T{{\cal T}}
\def\e{\varepsilon}
\def\la{\langle}
\def\ra{\rangle}
% characteristic function
     \newcommand{\charf}[1]{\mbox{\raise.48ex\hbox{$\chi$}$_{#1}$}}



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\newtheorem{definition}{Definition}
\newtheorem{remark}{Fact}[section]

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\newcommand{\cor}[2]{\begin{corollary}\label{#1}#2\end{corollary}}
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\newcommand{\defi}[2]{\begin{definition}\label{#1}#2\end{definition}}
\newcommand{\rem}[2]{\begin{remark}\label{#1}#2\end{remark}}


%End of proof marker
\def\qed{\hfill$\Box$}
%Proof
\def\pf{\noindent{\sc Proof.} }


\begin{document}

\maketitle

\addtocounter{footnote}{1}
\footnotetext{The article was written when the second
author was working at Institut fur Mathematik~II, FU~Berlin, Arnimallee 3,
14195~Berlin, Germany, under the support of Alexander von Humboldt
Foundation.}



\begin{abstract}
Let $\F$ be a family of real functions, $\F\su\real^\real$. 
In the paper we will examine the following question: 
for which families $F\su\real^\real$ 
does there exist $g\colon\real\to\real$
such that $f+g\in\F$ for all $f\in F$?
More precisely, we will study a cardinal
function $\add(\F)$ defined as 
the smallest cardinality of a family 
$F\su\real^\real$ for which there is no such $g$. 
We will prove that $\add(\ext)=\add(\pr)=\co^+$ and 
$\add(\pc)=2^{\co}$, where $\ext$, $\pr$ and $\pc$
stand for the classes of extendable functions, functions
with perfect road and peripherally continuous functions
from $\real$ into $\real$, respectively.
In particular, equation $\add(\ext)=\co^+$ implies immediately
that every real function is a sum of two extendable functions.
This solves a problem of Gibson~\cite{Gib1}. 

We will also study the multiplicative analogue $\mul(\F)$
of the function $\add(\F)$ and we prove that 
$\mul(\ext)=\mul(\pr)=2$ and 
$\add(\pc)=\co$. 

This article is a continuation of papers \cite{N2,CM,NR}
in which functions $\add(\F)$ and $\mul(\F)$ has been studied 
for the classes of almost continuous, connectivity and Darboux functions.
\end{abstract}

\section{ Introduction.}

We will use the following terminology and notation. 
Functions will be identified with their graphs. 
The family of all functions from a set $X$ into $Y$
will be denoted by $Y^X$.
Symbol $|X|$ will stand for the cardinality of a set $X$. 
The cardinality of the set $\real$ of real numbers is denoted by
$\co$.
For a cardinal number $\kappa$ we will write $\cf(\kappa)$
for the cofinality of $\kappa$.
A cardinal number $\kappa$ is regular, if $\kappa=\cf(\kappa)$.
For $A\su\real$ its characteristic function is denoted by $\charf{A}$.
In particular, $\charf\emptyset$ stands for the zero constant function. 

In his study of the class $\darb$ of Darboux functions (see definition below)
Fast \cite{fast} proved that for every family $F\su\reals^\reals$
of cardinality at most continuum there exists $g\colon\reals\to\reals$
such that $f+g$ is Darboux for every $f\in F$. 
Natkaniec \cite{N2} proved the similar result for the class $\ac$ of 
almost continuous functions and defined the following 
two cardinal invariants for every $\F\su\reals^\reals$.
\begin{eqnarray*}
\add(\F)
& = & \min\{|F|\colon F\su\reals^\reals\ \&\ \neg\exists
              g\in\reals^\reals\ \forall f\in F\ f+g\in\F\}\cup\{(2^\co)^+\}\\
& = & \min\{|F|\colon F\su\reals^\reals\ \&\ \forall
              g\in\reals^\reals\ \exists f\in F\ f+g\not\in\F\}\cup\{(2^\co)^+\}
\end{eqnarray*}
and 
\begin{eqnarray*}
\mul(\F)\!\!
& = &\! \min\{|F|\colon F\su\reals^\reals\ \&\ \neg\exists 
     g\in\reals^\reals\setminus\{\charf\emptyset\}\ 
     \forall f\in F\ f\cdot g\in\F\}\cup\{(2^\co)^+\}\\
& = &\! \min\{|F|\colon F\su\reals^\reals\ \&\ \forall 
     g\in\reals^\reals\setminus\{\charf\emptyset\}\ 
     \exists f\in F\ f\cdot g\not\in\F\}\cup\{(2^\co)^+\}.
\end{eqnarray*}
Thus, Fast and Natkaniec effectively showed that 
$\add(\darb)>\co$ and $\add(\ac)>\co$.

The extra assumption that $g\neq\charf\emptyset$ is added in the definition of
$\mul$ since otherwise for every family $\F\su\real^\real$ containing
constant zero function $\charf\emptyset$ we would have $\mul(\F)=(2^\co)^+$.

Notice the following basic properties of functions $\add$ and $\mul$.

\prop{prop1}{ Let $\F\su\G\su\reals^\reals$.
\begin{description}

\item[(1)] $\add(\F)\le\add(\G)$. 

\item[(2)] $\add(\F)\geq 2$ if $\F\neq\emptyset$.

\item[(3)] $\add(\F)\leq 2^\co$ if $\F\neq\real^\real$.

\end{description}
}

\pf (1) is obvious. To see (2) let $h\in\F$ and $F=\{f\}$ for some
$f\in\real^\real$. Then $f+g\in\F$ for $g=h-f$. 
To see (3) notice that
for $F=\real^\real$ and every $g\in\real^\real$ there is 
$f\in F$ with $f+g\not\in\F$, namely $f=h-g$, where $h\in\real^\real\setminus\F$.
\qed

%\pagebreak

\prop{prop2}{ Let $\F\su\G\su\reals^\reals$.
\begin{description}

\item[(1)] $\mul(\F)\le\mul(\G)$. 

\item[(2)] $\mul(\F)\geq 2$ if $\charf\emptyset,\charf\real\in\F$.

\item[(3)] $\mul(\F)\leq\co$ if $r\charf{\{x\}}\not\in\F$ for every
           $r,x\in\real$, $r\neq 0$.

\end{description}
}

\pf (1) is obvious. To see (2) let $F=\{f\}$ for some
$f\in\real^\real$. If there is $x\in\real$ such that $f(x)=0$ take 
$g=\charf{\{x\}}$. Otherwise, define $g(x)=1/f(x)$ for every $x\in\real$. 
Then $f\cdot g\in\{\charf\emptyset,\charf\real\}\su\F$.

To see (3) notice that
for $F=\{\charf{\{x\}}\colon x\in\real\}$ and every 
$g\in\real^\real\setminus\{\charf\emptyset\}$ there is 
$f\in F$ with $f\cdot g\not\in\F$, namely $f=\charf{\{x\}}$, where $x$ is such
that $g(x)=r\neq 0$.
\qed


\prop{prop3}{ Let $\charf\emptyset\in\F\su\reals^\reals$. Then
$\add(\F)=2$ if and only if  
$\F-\F=\{f_1-f_2\colon f_1,f_2\in\F\}\neq\real^\real$. }

\pf 
``$\Rightarrow$'' 
Assume that $\F-\F=\real^\real$. We will show that $\add(\F)>2$.
So, pick arbitrary $f_1,f_2\in\real^\real$ and put $F=\{f_1,f_2\}$.
It is enough to find $g\in\real^\real$ such that $f_1+g,f_2+g\in\F$.
But $f_1-f_2\in\real^\real=\F-\F$. So, there exist $h_1,h_2\in\F$ such that
$f_1-f_2=h_1-h_2$. Put $g=h_1-f_1=h_2-f_2$. Then 
$f_i+g=f_i+(h_i-f_i)=h_i\in\F$ for $i=1,2$.

``$\Leftarrow$'' 
By Proposition~\ref{prop1}(2) we have $\add(\F)\geq 2$. To see that
$\add(\F)\leq 2$ let $h\in \real^\real\setminus (\F-\F)$, take 
$F=\{\charf\emptyset,h\}$ and choose an arbitrary $g\in\real^\real$.
It is enough to show that $f+g\not\in\F$ for some $f\in F$. 
But if $g=\charf\emptyset+g\in\F$ and $h+g\in\F$ then
$h\in\F-g\subset\F-\F$, contradicting the choice of $h$.
\qed

\medskip

Now, let $X$ and $Y$ be topological spaces.
In what follows we will consider the following classes of
functions from $X$ into $Y$. 
(In fact, we will consider these classes mainly for
$X=Y=\real$.)

\begin{description}

\item{$\darb(X,Y)$} of {\em Darboux functions} $f\colon X\to Y$, i.e., such that 
          $f[C]$ is connected in $Y$ for every connected subset $C$ of $X$.

\item{$\conn(X,Y)$} of {\em connectivity functions} $f\colon X\to Y$, i.e., such
          that the graph of $f$ restricted to $C$ (that is $f\cap[C\times Y]$)
          is connected in $X\times Y$ for every connected subset $C$ of $X$.

\item{$\ac(X,Y)$} of {\em almost continuous functions} $f\colon X\to Y$, i.e.,
          such that every open subset $U$ of $X\times Y$ 
          containing the graph of $f$, there is a 
          continuous function $g\colon X\to Y$ with $g\subset U$. 

\item{$\ext(X,Y)$} of {\em extendable functions} $f\colon X\to Y$, i.e., such that
          there exists a connectivity function $g\colon X\times[0,1]\to Y$ 
          with $f(x)=g(x,0)$ for every $x\in X$.

\item{$\pr$} of {\em functions $f\colon\real\to\real$ with perfect road}
          ($X=Y=\real$), i.e. such that for every $x\in\real$ there
          exists a perfect set $P\su\real$ having $x$ as a bilateral limit point
          for which restriction $f|_P$ of $f$ to $P$ is continuous at $x$.

\item{$\pc(X,Y)$} of {\em peripherally continuous functions} $f\colon X\to Y$, 
          i.e., such that
          for every $x\in X$ and any pair $U\su X$ and $V\in Y$ of 
          open neighbourhoods of $x$ and $f(x)$, respectively, 
          there exists an open neighbourhood $W$ of $x$ with 
          $\cl(W)\su U$ and $f[\bd(W)]\su V$, where $\cl(W)$
          and $\bd(W)$ stand for the closure and the boundary of $W$,
          respectively.
\end{description}
We will write %simply 
$\darb$, $\conn$, $\ac$, $\ext$, and $\pc$ in place of 
$\darb(X,Y)$, $\conn(X,Y)$, $\ac(X,Y)$, $\ext(X,Y)$, and $\pc(X,Y)$
if $X=Y=\real$. 

Notice also, that function 
$f\colon\rl\to\rl$ is peripherally continuous ($f\in\pc$)
if and only if for every $x\in\rl$ there are sequences $a_{n}\nearrow x$ and
$b_{n}\searrow x$
such that $\lim_{n\to\infty} f(a_{n})=\lim_{n\to\infty} f(b_{n})=f(x)$.
In particular, if graph of $f$ is dense in $\real^2$ then
$f$ is peripherally continuous. 

For the classes of functions (from $\real$ into $\real$) defined above
we have the following proper inclusions $\subset$, marked by arrows
$\longrightarrow$. (See \cite{bhl}.)

\begin{picture}(300,100)

\thicklines
\put(10,40){\makebox(0,0){$\ext$}}
\put(80,80){\makebox(0,0){$\ac$}}
\put(160,15){\makebox(0,0){$\pr$} }
\put(160,80){\makebox(0,0){$\conn$}}
\put(240,80){\makebox(0,0){$\darb$} }
\put(320,40){\makebox(0,0){$\pc$}}

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

\end{picture}

In what follows we will also use the following theorem due to
Hagan \cite{Hagan}. 

\thm{thHagan}{ If $n\geq 2$ then $\conn(\real^n,\real)=\pc(\real^n,\real)$.
\qed}

Functions $\add$ and $\mul$ for classes $\ac$, $\conn$ and $\darb$ were
studied in \cite{N2,CM,NR}. In particular, the following is known.

\thm{thNR}{ {\rm \cite{NR}} $\mul(\ac)=\mul(\conn)=\mul(\darb)=\cf(\co)$. \qed}

\thm{thCM}{ {\rm \cite{CM}} 
$\co^+\leq\add(\ac)=\add(\conn)=\add(\darb)\leq 2^\co$,
$\cf(\add(\darb))>\continuum$ and it is pretty much all 
that can be shown in ZFC. 
More precisely, it is consistent with ZFC that $\add(\darb)$
can be equal to any regular cardinal between $\continuum^+$
and $2^{\continuum}$ and that it can be equal 
to $2^{\continuum}$ independently of the cofinality of $2^{\continuum}$. 
\qed}

The goal of this paper is to prove the following theorem.

\pagebreak

\thm{main}{
\begin{description}
\item[(1)] $\mul(\pc)=\co$.

\item[(2)] $\mul(\ext)=\mul(\pr)=2$.

\item[(3)] $\add(\pc)=2^\co$.

\item[(4)] $\add(\ext)=\add(\pr)=\co^+$.

\end{description}
}

This will be proved in the next sections. Notice only that the equation
$\add(\ext)=\co^+$ and Proposition~\ref{prop3} imply immediately 
the following corollary, which gives a positive answer to a question of Gibson
\cite{Gib1}.  (Compare also \cite{Rosen1} and \cite{Rosen2}.)

\cor{corGib}{ Every function $f\colon\real\to\real$ is a sum
of two extendable functions. \qed}


\section{Proof of Theorem~\ref{main}(1), (2) and (3).}
%$\mul(\ext)=\mul(\pr)=2$, $\mul(\pc)=\co$, $\add(\pc)=2^{\frak c}$.}

{\noindent{\sc Proof of $\mul(\ext)=\mul(\pr)=2$.}} 
The inequalities $2\leq\mul(\ext)\leq\mul(\pr)$ follow from 
Proposition~\ref{prop2}. 
To see that $\mul(\pr)\leq 2$ take $F=\{\charf{B},\charf{\real\setminus B}\}$
where $B\subset\real$ is a Bernstein set.
Then for every $g\in\real^\real\setminus\{\charf\emptyset\}$
we have $f\cdot g\neq\pr$ for some $f\in F$. To see it, take
$x\in\real$ such that $g(x)=r\neq 0$. If $x\in B$ then 
$\charf{B}\cdot g$ 
does not have a perfect road 
at $x$, since 
$(\charf{B}\cdot g)(x)=r\neq 0$ and 
$(\charf{B}\cdot g)^{-1}(0)\cap P\neq\emptyset$
for every perfect set $P\su\real$.
Similarly, $\charf{\real\setminus B}\cdot g$ 
does not have a perfect road 
at $x$ if $x\in \real\setminus B$. \qed

\medskip

{\noindent{\sc Proof of $\mul(\pc)=\co$.}} 
Inequality $\mul(\pc)\leq\co$ follows from 
Proposition~\ref{prop2}. 
So, it is enough to show that $\mul(\pc)\geq\co$.

Let $F\su\real^\real$ be a family of cardinality 
less then or equal to $\kappa$ with $\omega\leq\kappa<\co$.
We will find $g\in\real^\real\setminus{\charf\emptyset}$ such that 
$f\cdot g\in\pc$ for every $f\in F$. 

For $f\colon\real\to\real$ let $[f\ne 0]$ stand for the set 
$\{x\in\real\colon f(x)\neq 0\}$ and
define
\[
A_f=\{x\in\real\colon f(x)\ne 0\ \&\ 
[f\ne 0]\ \text{ is not bilaterally $\kappa^+$-dense at } x \},
\]
where set $S\su\real$ is said to be bilaterally $\kappa^+$-dense at $x$
if for every $\e>0$ each of the sets
$S\cap[x-\e,x]$ and $S\cap[x,x+\e]$ has cardinality at least $\kappa^+$.

Notice that $|A_f|\leq\kappa$ for every $f\colon\real\to\real$.
This is the case, since for every $x\in A_f$ there exists
a closed interval $J$ with non-empty interior such that 
$x\in J$ and $|[f\ne 0]\cap J|\leq\kappa$. 
Now, if $\J$ is the family of all maximal intervals $J$
with non-empty interior such that
$|[f\ne 0]\cap J|\leq\kappa$ then $|\J|\leq\omega$,
$A_f\su\bigcup_{J\in\J}([f\ne 0]\cap J)$ 
and $|A_f|\leq|\bigcup_{J\in\J}([f\ne 0]\cap J)|\leq\kappa$. 

Let $A=\ds_{f\in F}A_f$. Then $|A|\leq\kappa$.
Notice that the set $[f\ne 0]\sm A$ is bilaterally $\kappa^+$-dense
at $x$ for every $f\in F$ and $x$ from $[f\ne 0]\sm A$.
To define $g$ let 
$\la\la f_\alpha,q_\alpha,I_\alpha\ra\colon \alpha<\kappa\ra$ be 
the sequence of all triples with
$f_\alpha\in F$, $q_\alpha\in\rational$ and $I_\alpha$ is an open
interval with rational end points. 
Define by induction on $\alpha<\kappa$ a one-to-one sequence 
$\la x_\alpha\colon\alpha<\kappa\ra$
by choosing 
\begin{description}
\item[(i)] 
$x_\alpha\in[f_{\alpha}\ne 0]\cap(I_\alpha\sm A)\sm\{x_\beta\colon\beta<\alpha\}$
if the choice can be made, and
\item[(ii)] 
$x_\alpha\in(I_\alpha\sm A)\sm\{x_\beta\colon\beta<\alpha\}$ otherwise. 
\end{description}
Now, for $x\in\real$ we put
\[
g(x)=
\begin{cases}
\frac{q_\alpha}{f_\alpha(x)} & 
\text{ if there is $\alpha<\kappa$ with $x=x_\alpha$ and $f_\alpha(x)\neq 0$}\\
1 & 
\text{ if there is $\alpha<\kappa$ with $x=x_\alpha$ and $f_\alpha(x)=0$}\\
0 & \text{ otherwise.}
\end{cases}
\]
Then $g\neq\charf\emptyset$ and for each $q\in\rational$ and $f\in F$ 
the set $(g\cdot f)^{-1}(q)$ is
bilaterally dense at every $x$ from $[f\ne 0]\sm A$.
Moreover, $(g\cdot f)(x)=0$ outside of $[f\ne 0]\sm A$
and  $(g\cdot f)^{-1}(0)$ is bilaterally dense at every $x\in\rl$. 
So, $g\cdot f\in\pc$ for every $f\in F$.
\qed

\medskip

To prove $\add(\pc)=2^\co$ we will use the following result.

\thm{thComfNeg}{ Let $A$ and $B$ be such that $|A|=\omega$ and $|B|=\continuum$.
Then there exists a family $\C\sb A^B$ of size
$2^\co$ such that for every one-to-one sequence
$\la g_a\in\C\colon a\in A\ra$
there is $b\in B$ with $g_a(b)=a$ for every $a\in A$.}

\pf The theorem is proved in 
\cite[Corollary 3.17, p. 77]{CN} for $A=\omega$ and $B=\continuum$. 
The generalization is obvious. \qed

\medskip

From this we will conclude the following lemma.

\lem{lem1}{ If $B\su\real$ has cardinality $\co$ and
$H\su\rational^B$ is such that $|H|<2^{\co}$
then there is $g\in\rational^B$ such that $h\cap g\ne\es$
for every $h\in H$.}

\pf Let $\C$ be as in Theorem~\ref{thComfNeg} with 
$A=\rational$. 
For each $h\in H$ there only finitely many $g\in\C$
for which $h\cap g =\es$, since any countable infinite
subset of $\C$ can be enumerated as $\{g_a\in\C\colon a\in\rational\}$.
So there is $g\in\C$ such that
$h\cap g\ne\es$ for every $h\in H$. \qed

\medskip

{\noindent{\sc Proof of $\add(\pc)=2^\co$.}} 
By Proposition~\ref{prop1} to prove $\add(\pc)=2^\co$
it is enough to show that $\add(\pc)\geq 2^\co$. 
So, let $F\su\real^\real$ be such that $|F|<2^{\co}$.
We will find $g\colon\real\to\real$ such that 
$f+g\in\pc$ for every $f\in F$. 

Let $\G$ be the family of all triples $\la I,p,m\ra$ where
$I$ is a non-empty open interval with rational end points,
$p\in\rational$ and $m<\omega$.
For each $\la I,p,m\ra\in\G$ 
define a set $B_{\la I,p,m\ra}\su I$ of size $\co$ such that 
$B_{\la I,p,m\ra}\cap B_{\la J,q,n\ra}=\es$ for any distinct
$\la I,p,m\ra$ and $\la J,q,n\ra$ from $\G$. 

Next, fix $\la I,p,m\ra\in\G$ and 
for each $f\in F$ 
choose $h^f_{\la I,p,m\ra}\colon B_{\la I,p,m\ra}\to\rational$
such that
\[
|p-(f(x)+h^f_{\la I,p,m\ra}(x))|< \frac{1}{m}\ \text{ for every } 
x\in B_{\la I,p,m\ra}.
\]
Then, by Lemma~\ref{lem1} used with a set
$H_{\la I,p,m\ra}=\{h^f_{\la I,p,m\ra}\colon f\in F\}$,
there exists
$g_{\la I,p,m\ra}\colon B_{\la I,p,m\ra}\to\rational$
such that 
\[
\forall f\in F\ \exists x\in B_{\la I,p,m\ra}\ 
h^f_{\la I,p,m\ra}(x)=g_{\la I,p,m\ra}(x).
\]
In particular, if $g\colon\real\to\rational$ is a common extension
of all functions $g_{\la I,p,m\ra}$ then
for every $\la I,p,m\ra\in\G$ and every $f\in F$ there exists 
$x\in  B_{\la I,p,m\ra}\su I$ such that
\[
|p-(f(x)+g(x))|< \frac{1}{m}.
\]
So, for every $f\in F$ the graph of $f+g$ is dense in $\real^2$.
Thus, $f+g\in\pc$. \qed


\section{ Proof of Theorem~\ref{main}(4): 
$\add(\ext)=\add(\pr)=\co^+$.}

By Proposition~\ref{prop1} we have $\add(\ext)\leq\add(\pr)$.
Thus, it is enough to prove two inequalities:
$\add(\pr)\leq\co^+$ and $\add(\ext)\geq\co^+$. 

First we will prove $\add(\pr)\leq\co^+$. For this we need the 
following lemma. 

\lem{lemaddpr}{ There is a family $F\sb\rl^{\rl}$ of size $\co^{+}$
such that for every distinct $f,h\in F$, every perfect set $P$ and every
$n<\omega$ there exists an $x\in P$ with $|f(x)-h(x)|\ge n$. }

\pf The family $F=\{f_\xi\colon \xi<\co^+\}$ is constructed by 
induction using standard diagonal argument. If for some $\xi<\co^+$
the functions $\{f_\zeta\colon \zeta<\xi\}$ are already constructed, we 
construct $f_\xi$ as follows. 

Let $\la\la P_{\alpha},h_{\alpha},n_{\alpha}\ra\colon\alpha<\co\ra$
be an enumeration of all triples $\la P,h,n\ra$ where $P\su\real$ is
perfect, $h=f_\zeta$ for some $\zeta<\xi$ and $n<\omega$.
By induction on $\alpha<\co$ 
choose $x_{\alpha}\in P_{\alpha} \sm\{x_{\beta}\colon\beta < \alpha\}$ 
and define $f_\xi(x_{\alpha})=h_{\alpha}(x_{\alpha})+n_\alpha$.
Then any extension of such defined $f_\xi$ onto $\real$ will have the desired
properties. \qed

\medskip

{\noindent{\sc Proof of $\add(\pr)\leq\co^+$.}} 
Now let $F$ be a family from Lemma~\ref{lemaddpr}.
We will show that for every $g\colon\real\to\real$ 
there exists $f\in F$ such that $f+g\not\in\pr$.

By way of contradiction assume that there exists a
function $g\colon\real\to\real$  such that $f+g\in\pr$
for every $f\in F$. Then, for every 
$f\in F$ there exists a perfect set $P_f$ such that $0$ is a
bilateral limit point of $P_f$ and $(f+g)|_{P_f}$ is continuous at $0$.
Since there are $\co^{+}$-many functions in $F$ and only $\co$-many
perfect sets there are distinct $f,h\in F$ with $P_f=P_h$. Then the function
$((f+g)-(h+g))|_{P_f}=(f-h)|_{P_f}$ is continuous at $0$
contradicting the choice of family $F$. \qed

\medskip

The proof of $\add(\ext)\geq\co^+$ is based
on the following facts. 

\lem{lem11}{ For every meager subset $M$ of the real line $\rl$ there exists a
family 
$\{h_\xi\in\real^\real\colon \xi<\co\}$
of increasing homeomorphisms such
that $h_\zeta[M] \cap h_\xi[M]=\es$ for every $\zeta<\xi<\co$.}

\pf Let $\{D_\zeta:\zeta <\co\}$ be a family of pairwise disjoint 
$\co$-dense, meager $F_\sigma$-sets. 
Then by \cite[Lemma 4]{gor} there are homeomorphisms
$\{h_\zeta:\reals \to \reals: \zeta < \co\}$ such that $h_\zeta[M] \subset 
D_\zeta$.
\qed

\medskip

For $f\in\F\subset\real^\real$ we say that
a set $G\su\real$ is {\em $f$-negligible for the class $\F$} provided
$g\in\F$ for every $g\colon\real\to\real$ such that 
$g|_{\rl\sm G}=f|_{\rl\sm G}$. 
Thus, $G\su\real$ is $f$-negligible for $\F$ 
if we can modify $f$ arbitrarily
on $G$ remaining in the class~$\F$.

\thm{thConnExamp}{ There exists a connectivity function 
$f\colon\rl^2 \to \rl$
with graph dense in $\real^3$ such that some dense $G_{\delta}$
subset $G$ of $\real^2$ is $f$-negligible for the class 
$\conn(\real^2,\real)$. }

The proof of this theorem will be postponed to the end of the section.

\cor{corExtExamp}{ There exists an extendable function 
$\hat{f}\colon\rl \to \rl$
with graph dense in $\real^2$ such that some dense $G_{\delta}$
subset $\hat{G}$ of $\real$ is $\hat{f}$-negligible for the class $\ext$.}

\pf 
Let $f\colon\rl^{2}\to\rl$  and $G$ be as in Theorem~\ref{thConnExamp}.
Then there exists $y\in\rl$ with 
$G^y=\{x\colon\la x,y\ra\in G\}$ being a dense
$G_{\delta}$ subset of $\rl$. Clearly the set $\hat{G}=G^y$ and the function 
$\hat{f}\colon\rl\to\rl$ defined by 
$\hat{f}(x)=f(x,y)$ for every $x\in\real$ 
satisfy the requirements.
\qed

\medskip

The existence of a function as in Corollary~\ref{corExtExamp}
was first announced by H.~Rosen at the 10th Auburn
Miniconference in Real Analysis, April 1995. 
However, the construction presented at that time had a gap.
This gap had been removed later, as described in \cite{Rosen2}.

The construction presented 
in this paper is an independently discovered 
fix-up of the original Rosen's gap.
It is also more general than that of \cite{Rosen2},
since \cite{Rosen2} does not contain any example 
similar to that of Theorem~\ref{thConnExamp}.

Next, we will show how Lemma~\ref{lem11} and Corollary~\ref{corExtExamp}
imply $\add(\ext)\geq\co^+$.
The argument is a modification of the proof of 
Corollary~\ref{corGib}. (Compare also \cite{N11} and \cite{Rosen2}.)

\medskip

{\noindent{\sc Proof of $\add(\ext)\geq\co^+$.}} 
Let $F=\{f_\xi\in\real^\real\colon\xi<\co\}$.
We will find $g\colon\real\to\real$ such that 
$f_\xi+g\in\ext$ for every $\xi<\co$.

So, let $\hat{f}\colon\real\to\real$ and $\hat{G}\su\real$ be as in 
Corollary~\ref{corExtExamp}. Put $M=\rl\sm\hat{G}$ and take
$\{h_\xi\in\real^\real\colon \xi<\co\}$
as in Lemma~\ref{lem11}.
For $\xi<\co$ define $g$ on $h_\xi[M]$ to be 
$(\hat{f}\circ h_\xi^{-1}-f_\xi)|_{h_\xi[M]}$
and extend it onto $\real$ arbitrarily. 
To see that $f_\xi+g\in\ext$ notice that
$f_{\xi}+g=\hat{f}\circ h_{\xi}^{-1}$ on $h_{\xi}[M]$. 
But the set $\rl \sm h_{\xi}[M]=h_{\xi}[\hat{G}]$ is 
$(\hat{f}\circ h_{\xi}^{-1})$-negligible for the class $\ext$.
(See \cite{N11} for an easy proof.) 
So, each $f_{\xi}+g$ is extendable. \qed

\medskip

\begin{figure}

\begin{picture}(300,220)(-40,0)
\thicklines

\multiput(10,10)(60,0){4}{
\drawline[20](0,0)(60,0)(30,52)(0,0)}

\multiput(40,62)(60,0){3}{
\drawline[20](0,0)(60,0)(30,52)(0,0)}


\multiput(70,114)(60,0){2}{
\drawline[20](0,0)(60,0)(30,52)(0,0)}


\put(100,166){
\drawline[20](0,0)(60,0)(30,52)(0,0)}

\end{picture}
\caption{grid $S_n$\label{fig1}}
\end{figure}    

{\noindent{\sc Proof of Theorem~\ref{thConnExamp}.}} 
We will construct a peripherally continuous function 
$f\colon\rl^{2}\to\rl$ with dense $G_{\delta}$ subset $G$ of $\rl^{2}$
which is $f$-negligible for the class $\pc(\real^2,\real)$. 
It is enough since, by Theorem~\ref{thHagan},
$\conn(\real^2,\real)=\pc(\real^2,\real)$.

The construction is a modification of that from \cite{GR},
where a similar example of function from $[0,1]\times[0,1]$ onto $[0,1]$
was constructed. 
(Compare also \cite{B}.)  
The additional difficulty in our construction
is to make sure that some sequences of points in the range of $f$ (=$\real$)
have cluster points, what is obvious for all sequences in $[0,1]$.
Also, our  basic construction step will be based on a triangle, while 
construction in \cite{GR}
was based on a square. The triangles work better, since for arbitrary 
three non-collinear points
in $\rl^{3} $ there is precisely one plane
passing through them, while it is certainly false for four points. 

\pagebreak

{\bf Basic Idea}: We will construct, by induction on $n<\omega$, a
sequence $\la S_n\colon n<\omega\ra$ of triangular ``grids'' formed with
equilateral triangles of side length $1/2^{k_n}$, as in Figure~\ref{fig1}.
The grid $S_n$ will be identified with
the points on the edges of triangles forming it
and we will be assuming that $S_n\su S_{n+1}$ for all $n<\omega$.
With each grid $S_n$ we will associate a 
continuous function $f_n\colon S_n\to\real$
which is linear on each side of a triangle from $S_n$.
Moreover, each $f_{n+1}$ will be an extension of $f_n$. 
Function $f$ will be defined as an extension of $\bigcup_{n<\omega}f_n$. 

\begin{figure}

\begin{picture}(300,130)(-100,0)
\thicklines

\put(10,10){
\begin{picture}(150,150)
\drawline[20](0,0)(120,0)(60,104)(0,0)
\drawline[20](60,0)(90,52)(30,52)(60,0)
\drawline[20](45,26)(75,26)(60,52)(45,26)
\end{picture}}

\end{picture}
\caption{basic partition\label{fig2}}
\end{figure}                   

{\bf Terminology}: In what follows a {\em triangle} will be identified 
with the set of points of its interior or its  boundary. 

For a grid $S$ we say that a {\em triangle $T$ is 
from $S$} if the interior of $T$ is equal to
a component of $\real^2\setminus S$. 

For an equilateral triangle $T$, its {\em basic partition}
will be its split into seven equilateral triangles, as in Figure~\ref{fig2}. 
The central triangle $\hat{T}$ of Figure~\ref{fig2}
will be
referred as {\em the middle quarter of $T$}.
Thus, $\hat{T}\cap\bd(T)=\emptyset$
and the length of each side of $\hat{T}$ is equal to $1/4$ of the length
of a side of $T$. 

If a function $F$ is defined on the three vertices of a triangle
$T$, its {\em basic extension} is defined as
the unique function $\hat{F}\colon T\to\real$
extending $F$ whose graph is a subset of a plane. 
Notice, that $\hat{F}$ is linear on each side of the triangle $T$
and that $\hat{F}$ extends $F$ 
even if the function $F$ has already been defined on some side of $T$
as long as $F$ is linear on this side. 

{\bf Inductive construction}: 
We will define inductively three increasing sequences: 
$\la S_n\colon n<\omega\ra$
of triangular grids as in Figure~\ref{fig1}, 
$\la f_n\in\real^{S_n}\colon n<\omega\ra$ of continuous functions and
$\la k_n<\omega\colon n<\omega\ra$ of natural numbers such that
the following inductive conditions are satisfied for every $n<\omega$.

\pagebreak

\begin{description}
\item[(i)]   $f_n\colon S_n\to[-2^n,2^n]$ and it is 
             linear on each side of a triangle $T$ from $S_n$.

\item[(ii)]  The side length of each triangle from $S_{n}$ is equal to $1/2^{k_n}$.

\item[(iii)] The variation of $f_n$ on each triangle from $S_{n}$ is $\le 1/2^{n}$.

\item[(iv)]  If $n>0$ then for every triangle $T$ from $S_{n-1}$ and every dyadic    
             number $i/2^{n}\in[-2^n,2^n]$ with 
             $i\in\integer$ ($-4^{n}\leq i\leq 4^{n}$) 
             there
             is a triangle $T_i\su\hat{T}$ such that $\bd(T_i)\su S_{n}$ and
             $f_{n}(x)=i/2^{n}$ for every $x\in\bd(T_i)$. 

\item[(v)]   If $n>0$, $T$ is a triangle from $S_{n-1}$ and $T^\prime$ is a
             triangle from $S_{n}$ such that $T^\prime\su T$ and
             $T^\prime\not\su\hat{T}$ then 
             $f_{n}[\bd(T^\prime)]\su [-M,M]$, where
             $M=\max\{|f_{n-1}(x)|\colon x\in\bd(T)\}$.
\end{description}


\begin{figure}

\begin{picture}(300,220)(-40,0)
\thicklines


\multiput(10,10)(60,0){4}{
\dashline{4}(0,0)(60,0)(30,52)(0,0)}

\multiput(40,62)(60,0){3}{
\dashline{4}(0,0)(60,0)(30,52)(0,0)}


\multiput(70,114)(60,0){2}{
\dashline{4}(0,0)(60,0)(30,52)(0,0)}


\put(100,166){
\dashline{4}(0,0)(60,0)(30,52)(0,0)}

\multiput(55,36)(60,0){3}{
\drawline[20](0,0)(30,0)(15,26)(0,0)}

\multiput(85,88)(60,0){2}{
\drawline[20](0,0)(30,0)(15,26)(0,0)}

\put(115,140){
\drawline[20](0,0)(30,0)(15,26)(0,0)}


\end{picture}

\caption{some triangles of the grid of $\hat{T}$\label{fig3}}  %KC
\end{figure}                      %KC

To start the induction, take $k_0=0$, 
define grid $S_0$ as in Figure~\ref{fig1} with all sides 
of length $1=1/2^0$ and choose $f_0\colon S_0\to\real$ 
as constantly equal $0$. It is easy to see that the
conditions (i)-(v) are satisfied with such a choice.  

Next, assume that for some $n>0$ we already have 
$S_{n-1}$, $f_{n-1}$ and $k_{n-1}$ 
 satisfying (i)-(v).
We will define $S_n$, find $k_n$ and extend $f_{n-1}$ to
$f_n\colon S_n\to\real$ 
such that (i)-(v) will still hold. 

Put $F_n=f_{n-1}$. 

Step 1. Let $T$ be a triangle from $S_{n-1}$
and extend $F_n$ into each vertex of 
its middle quarter $\hat{T}$ by assigning
it the value $0$. Notice, that $F_n$ is defined on all vertices of 
the basic partition of $T$.

Partition $\hat{T}$ into a grid $S$ such that the
size of each triangle from $S$ is equal $1/2^{\hat{k}_n}$. 
The number ${\hat{k}_n}<\omega$
is chosen as a minimal number such that there are 
$2\cdot 8^n + 1$ disjoint 
triangles $\{T_i\colon i\in\integer, -4^n\leq i\leq 4^n\}$ from $S$ 
none of which intersects the boundary of $\hat{T}$. 
(See Figure~\ref{fig3}.) Notice that the value of ${\hat{k}_n}$ 
does not depend on $T$.  
On the vertices of each triangle $T_i$ define $F_n$ to
be equal $i/2^{n}$. 
On the remaining undefined vertices of $S$ 
define $F_n$ to be equal $0$.
Notice that $F_n$ is defined on all vertices of each triangle defined so far.

Step 2. Extend 
$F_n$ into $\rl^{2}$, by defining it on 
every triangle $T$ constructed so far as
the basic extension of $F_n|_{\bd(T)}$. 

Notice that if we
extend grid $S_{n-1}$ to the grid $\hat{S}_n$ 
with side length of each triangle from $\hat{S}_n$ equal 
$1/2^{\hat{k}_n}$
and put $\hat{f}_n=F_n|_{\hat{S}_n}$ then 
the triple $\la \hat{S}_n,\hat{f}_n,\hat{k}_n\ra$
satisfies conditions (i), (ii), (iv) and (v). 

Step 3. We have to modify $\hat{S}_n$, $\hat{f}_n$ and $\hat{k}_n$
to get also condition (iii), while keeping the other properties.
First notice that for every triangle $T$ from $\hat{S}_n$
and any interval $J$ inside $T$ the slope of $F_n$ on $J$
does not exceed the number
\[
\frac{\text{length of the range of $F_n$}}{\text{length of a side of $T$}}
\leq\frac{2\ 2^n}{2^{-\hat{k}_n}}=2^{n \hat{k}_n +1}.
\]
So, let $k_n\geq n\hat{k}_n+n+1$, 
in which case
\[
\frac{1}{2^{k_n}} 2^{n \hat{k}_n +1}\leq \frac{1}{2^n},
\]
let $S_n$ be a refinement of the grid $\hat{S}_n$
with triangles with side size $1/2^{k_n}$ and put 
$f_n=F_n|_{S_n}$.
It is easy to see that this gives us (iii) while preserving the other conditions. 

This finishes the inductive construction.

\medskip

Let $S=\ds_{n<\omega}S_{n}$ and define $f$ on $S$ by
$f=\ds_{n<\omega}f_{n}$.
To extend it to $\rl^{2}\sm S$
notice that
\begin{description}
\item[($\star$)] for every $x\in\rl^{2}\sm S$ there exists a number $f(x)\in\rl$
and a sequence $\la T_k\colon k<\omega\ra$
of triangles with $x$ being an interior point of each $T_k$ such that 
$\lim_{k\to\infty}\diam(T_k)\to 0$ and 
\[
f(x)=\lim_{k \to \infty}\min f[\bd(T_k)]=\lim_{k \to \infty}\max f[\bd(T_k)].
\]
\end{description}
The proof of ($\star$) will finish the construction of $f$.

To see ($\star$) fix $x\in\rl^{2}\sm S$ and let $T_{n}^{0}$ be the
triangle from $S_{n}$ such that $x$ belongs to the interior of $T_{n}^{0}$.
Let $N=\{n<\omega\colon n>0\ \&\ T_n^{0}\su\hat{T}_{n-1}^{0}\}$.
There are two cases to consider. 

Case 1. Set $N$ is infinite. Then, let $\la n_k\colon k<\omega\ra$ be a one-to-one 
enumeration of $N$ and define $T_k=\hat{T}_{n_k - 1}^0$. It is easy to see that 
this sequence satisfies ($\star$) with $f(x)=0$.

Case 2. Set $N$ is finite. Let $m<\omega$ be such that 
$T_n^{0}\not\su\hat{T}_{n-1}^{0}$ for every $n\geq m$
and let $M=\max\{|f_{m-1}(x)|\colon x\in\bd(T_{m-1}^{0})\}$.
Then, by condition (v), $f_{n}[\bd(T_n^{0})]\su [-M,M]$ for 
every $n\geq m$. So, there exists an increasing sequence
$\la n_k\geq m\colon k<\omega\ra$ such that
the limit 
$L=\lim_{k \to \infty}\max f[\bd(T_{n_k}^0)]$ exists. 
It is easy to see that the sequence 
$\la T_k\ra =\la T_{n_k}^0\ra$ satisfies ($\star$) with $f(x)=L$,
since the variation of $f$ on $\bd(T_{n_k}^0)$ tends to $0$
as $k\to\infty$.

This finishes the construction of function $f$. 
It remains to show that $f$ has the desired properties.

Clearly ($\star$) implies that $f$ is peripherally continuous at every
point $x\in\rl^{2} \sm S$.
To see that $f$ is peripherally continuous on $S$ take $x\in S$.
Then, there exists $k<\omega$ such that 
$x \in S_{n}$ for every $n\geq k$. For any such
$n$ let $\T_n$ be the set of all triangles from $S_n$ to which 
$x$ belongs. Notice that $\T_n$ has at most six elements and that
$x$ belongs to the interior of the polygon $P_n=\bigcup\T_n$. 
Hence, the variation on the boundary of $P_n$ is at most 
$6/2^n$ and the diameter of $P_n$ is at most $1/2^{n-1}$. 
So, the sequence $\la P_n\ra$ guarantees that $f$ is peripherally continuous at $x$.

To finish the proof it is enough to find a dense $G_{\delta}$ set $G$
which is 
$f$-negligible for $\pc(\real^2,\real)$. 

For any dyadic number $d$
and any $k\in\omega$ let $\F_d^k$ denote %be 
the family of all
triangles $T$ for which there exists $n\geq k$ such that
$T$ is from $S_n$ and $f_n(x)=d$ for every $x\in\bd(T)$. 
Let $G_d^k$ be the union of the interiors of all triangles $T\in\F_d^k$.
Then, by condition (iv), each set $G_d^k$ is open and dense. 
Therefore,
%Thus, 
$G=\di\{G_d^k\colon k\in\omega\ \&\ d\text{ is dyadic}\}$ 
is a dense $G_\delta$ set.
It is easy to see, that $f$ is peripherally continuous if we redefine it 
on the set $G$ in an arbitrary way.

This finishes the proof of Theorem~\ref{thConnExamp}. 

\section{Compositions of Lebesgue measurable functions.}

We can consider similar problems for compositions of functions.
For example, we know that every function is a composition of Lebesgue
measurable functions \cite{Ruz}. (See also {\sl Problem 6378},
American Mathematical Monthly {\bf 90}, 573.)
It is easy to make every function in $\real^\real$ measurable (in a sense
of definition of $\add$) using composition with just 
one function. We simply take a composition with a
constant function.
So we need to define cardinal invariants in different way.

The next definition will represent one of the way the problem can
be approached. Instead of ``making family $H$ to be in $\F$''
we will try to recover all elements of $H$
with one ``coding'' function $\hat f\in\F$ and the class $\F$ of
all codes. This leads to the following definitions. 

\[
C_{r}(\F) = \min\{|H|\colon H\su\reals^\reals\ \&\ \neg\exists
                \hat f\in\F\; \forall h\in H \exists f\in \F\ 
                f\circ \hat f = h\}\cup\{(2^\co)^+\}
\]
and
\[
C_{l}(\F) = \min\{|H|\colon H\su\reals^\reals\ \&\ \neg\exists
                \hat f\in\F\; \forall h\in H \exists f\in \F\ 
                \hat f\circ f = h\}\cup\{(2^\co)^+\}.
\]


Let $\cal L$ be the family of all Lebesgue measurable functions from
$\rl$ into $\rl$.

\thm{th:comp}{ $C_{r}({\cal L})=(2^{\co})^{+}$ and  
$C_{l}({\cal L})=\co^{+}$. }


\pf To see $C_{r}({\cal L})\geq(2^{\co})^{+}$  
we will show that the family $H=\real^\real$ of all functions
can be ``coded'' by one function $\hat f\in{\cal L}$. Simply, let 
$\hat f$ be a Borel isomorphism from $\rl$ onto Cantor ternary set $C$.
For any function $h\in\real^\real$
we define $f_h\in{\cal L}$ by putting 
$f_{h}\equiv 0$ on the
complement of $C$ and $f_{h}=h\circ \hat f^{-1}$ on $C$. 
Then $h=f_{h}\circ \hat f$.

To see that $C_{l}({\cal L})\ge \co^{+}$ let 
$H=\{h_{\xi}\colon \xi <\co\}  \sb \rl^{\rl}$ and  
let $\{C_{\xi}\colon\xi < \co\}$ be a partition of the Cantor ternary set
$C$ into perfect sets.
Then for every $\xi<\co$ take a Borel isomorphism
$f_{\xi}\colon \rl \to C_{\xi}$ and
define $\hat f(x)=(h_\xi\circ f_{\xi}^{-1})(x)$ for $x \in C_{\xi}$ and 
$\hat f(x)=0$
otherwise.
It is easy to see that $\hat f\in{\cal L}$ and
$\hat f\circ f_\xi=h_\xi$ for every $\xi<\co$ 

To prove $C_{l}({\cal L})\le \co^{+}$ take
$\{h_{\xi }\colon \xi< \co^{+}\}$ from Lemma~\ref{lemaddpr}.
Assume that there exists a sequence $\{f_{\xi}\colon \xi<\co^{+}\}$
of measurable functions and measurable function $\hat f$ such
that $h_{\xi}=\hat f\circ f_{\xi}$ for every $\xi<\co^+$. 
For each $\xi<\co^{+}$ let $P_{\xi}$  be a perfect set such that
$f_{\xi}|P_{\xi}$  is continuous. Then, by a cardinality argument,
there are $\zeta<\xi<\co^+$
such that $P_\zeta=P_\xi$ and
$f_\zeta|P_\zeta=f_\xi|P_\xi$. So,
$h_\zeta|P_\zeta=h_\xi|P_\xi$. Contradiction. \qed

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\end{document}





