% Baire classification of \I-density and \I-approximately
% continuous functions.

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     \newcommand{\iacont}{\mbox{$\C_{\cal IO}$}}
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     \newcommand{\deepiacont}{\mbox{$\C_{\cal DO}$}}
%I-dens cont funct
     \newcommand{\idcont}{\mbox{$\C_{\cal II}$}}
%deep I-dens cont funct
     \newcommand{\deepidcont}{\mbox{$\C_{\cal DD}$}}
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     \newcommand{\idensitypts}[1]{\mbox{$\Phi_{\cal I}(#1)$}}
     \newcommand{\deepidensitypts}[1]{\mbox{$\Phi_{\cal D}(#1)$}}
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\newcommand{\e}{\mbox{$\varepsilon$}}
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\title {Baire classification of \I-density and \I-approximately continuous
functions.}
\date{}
\author{}

\begin{document}
\maketitle

{\small\noindent Krzysztof Ciesielski, Department of Mathematics, West
Virginia University, Morgantown, WV 26506
      \footnote
      { This author was partially supported by
        West Virginia University Senate Research Grant.
      }

\noindent Lee Larson, Department of Mathematics, University of
Louisville, Louisville, KY 40292\footnote{This author was partially supported
by
a University of Louisville Arts and Sciences Research grant.
\par\vskip 5pt
\begin{flushleft}
Key Words: \idens\ topology, fine topologies, Baire classification\\
AMS Subject Classification: 26A03, 26A21
\end{flushleft}
}}

\begin{abstract}
Let \otop\ be the ordinary topology, \itop\ be the \idens\ topology and
\deepitop\ be the deep-\idens\ topology on the real numbers, \reals. Any
continuous function $f:(\reals,\itop)\to(\reals,\otop)$ is a Darboux function
of the first Baire class. Any unilaterally continuous function
$f:(\reals,\deepitop)\to(\reals,\deepitop)$ is a Darboux function of the
Baire*1 class.
\end{abstract}

\section{$\!\!\!\!\!\!\!${\bf.} Introduction}

The terminology and notation we use is standard. In particular,
\reals\ denotes the set of real numbers and $\nats=\{1,2,3,\ldots\}$. 
For $A,B\subset\reals$, $\complement{A}$ stands for complement
of $A$, $A\triangle B$
for the symmetric difference of $A$ and $B$ and $\dist{A}{B}$ for the Euclidean
distance between $A$ and $B$.
The natural topology on \reals\ is denoted by \otop. Topological terms in which
we do not specifically state the topology concern the natural topology.
For example, $\interior{A}$ and \closure{A}\
stand for the interior and closure of $A$ with respect to \otop, respectively.
The oscilation of a function $f$ at point $x$ will be denoted by $\osc(f,x)$. 

W. Wilczy{\'n}ski \cite{Wilczynski:GenDense} introduced a topology on \reals\
which has many properties in common with the ordinary density topology,
except that it is based upon category instead of measure. This topology, called
the \idens\ topology, is defined as follows.

Let \I\ stand for the ideal of first category subsets of \reals. 
A Boolean function $P$ defined on \reals\ is said to be true {\em\I-almost
everywhere} (\I-a.e.), if 
\[
\{ x: P(x)\mbox{ is false}\}\in\I.
\]
A sequence of
functions $f_n\RtoR$ {\em converges (\I)} to a function $f$, if for each
increasing sequence $\{n_j\}$
of natural numbers, there is a further subsequence $\{n_{j_k}\}$
such that $f_{n_{j_k}}$ converges pointwise to $f$, \I-a.e. Notice that this
definition mimics the standard definition of convergence in measure, or
stochastic convergence
\cite{Bauer:ProbMeasure}, with the exception that the ideal \N\
of Lebesgue null sets is replaced by \I.

If $A\subset\reals$, then a point $a\in\reals$ is an {\em \idens\ point} of $A$ if
\charf{n(A-a)\cap (-1,1)} converges (\I) to \charf{(-1,1)}, where \charf{S} is
the characteristic function of the set $S$. The set of all \idens\ points of
the set $A$ is written \idensitypts{A}.
It is obvious from the above definition that
\begin{equation}\label{eq:idenspts}
\idensitypts{A}=\idensitypts{B}\,\,\,\mbox{ whenever } A\triangle B\in\I.
\end{equation}
A point $a$ is an {\em\idisp\ point} of $A$, if
$a\in\idensitypts{\complement{A}}$. 

Closely related to the notion of an \idens\ point is that of a {\em deep-\idens\
point}. An \idens\ point $a$ of the set $A$ is a deep-\idens\ point of $A$, 
if there is a
closed set $F\subset \interior{A}\cup\{a\}$ such that $a\in\idensitypts{F}$. 
The set
of all deep-\idens\ points of $A$ is denoted by \deepidensitypts{A}. 
From the definitions, it
is obvious that for any set $A\subset\reals$, 
\begin{equation}\label{eq:deepinregular}
\deepidensitypts{A}\subset\idensitypts{A}.
\end{equation}

Denote the Baire subsets of \reals\ by \BaireSets. A standard definition of
the Baire sets is the following \cite{Kuratowski:Topology1}:
\[
\BaireSets =\{ G\triangle I:G\in\otop\mbox{ and }I\in\I\}.
\]
From this definition, it is not hard to show that associated with 
every $B\in\BaireSets$ is a unique open set $\tilde{B}$ such that
$\tilde{B}=\interior{\closure{\tilde{B}}}$ and $B=\tilde{B}\triangle I$ 
for some $I\in\I$. Any open set $G$ for which 
$G=\interior{\closure{G}}$ is called a {\em regular} open set.
In a sense, $\tilde{B}$ is the largest open set such that $B$ can
be written as $\tilde{B}\triangle I$ for some $I\in\I$.


The {\em\idens\ topology}
\cite{Wilczynski:GenDense,Wilczynski:CatAn} is defined as
\[
\itop = \{ A\in\BaireSets : A\subset\idensitypts{A}\}.
\]
Similarly, the {\em deep-\idens\ topology} 
\cite{Lazarow:Coarsest,Wilczynski:CatAn} is defined as
\[
\deepitop = \{ A\in\BaireSets : A\subset\deepidensitypts{A}\}.
\]
It is clear from (\ref{eq:deepinregular}) and the
definitions of \itop\ and \deepitop\ that 
\begin{equation}\label{eq:ToinTi}
\otop\subset\deepitop\subset\itop.
\end{equation}
It is known that these containments are proper.
(The first inclusion is proper by Lemma \ref{lem:intervalset}
\cite[Theorem 2]{Wilczynski:CatAn}. The second inclusion is proper because
\deepitop\ is completely regular \cite{Lazarow:Coarsest} while \itop\
is not \cite[Theorem 5]{PWW:RemIDens}.)

Using the three topologies, \otop, \itop\ and \deepitop, there are nine different
definitions of continuity possible. For ${\cal X,Y}\in\{{\cal I,D,O}\}$, let
\[
{\cal C}_{\cal XY}=\{ f:(\reals,\T_{\cal X})\to(\reals,\T_{\cal Y}):
f\mbox{ is continuous}\}.
\]
For example, \ocont\ is the set of all functions $f\RtoR$ continuous in the 
ordinary sense. It is not hard to show that $\C_{\cal OI}=\C_{OD}$ consists 
exactly of the constant functions \cite{CL:Examples}. 
It is also known that $\C_{\cal DO}=\C_{IO}$, and these are called the
{\em \iapp\ continuous functions} \cite{Lazarow:Coarsest}. The remaining classes 
with which this paper is concerned are the {\em\idens\ continuous} and 
{\em deep-\idens\ continuous} functions, \idcont\ and \deepidcont, respectively.

It is known that the following relationships hold
\begin{equation}\label{eq:toprels}
\ocont\subset\iacont\quad\mbox{and}\quad\idcont\subset\deepidcont\subset
\deepiacont=\iacont.
\end{equation}
Moreover, the inclusions in (\ref{eq:toprels}) are proper 
and these are the only inclusions between those classes \cite{CL:Examples}.

All the continuity and density definitions given above can be restated in
more-or-less obvious ways in one-sided versions. For technical reasons it is
often more convenient to work with one-sided density or continuity. For
example, to show that a point $a$ is an \idens\ point of a set $A$, it is often
easier to establish that it is both a left and right \idens\ point. Such
simple technical extensions to the definitions will be used without further
comment.

Let \Darboux\ stand for the functions $f\RtoR$ with the Darboux (intermediate
value) property, \bone\ the functions of the first Baire class and \bstar\ the
functions in the Baire*1 class; i.e., the class of all 
functions $f\RtoR$ with the property that for every
perfect set $P$ there is its nonempty portion $Q=P\cap(a,b)$ such that $f$
restricted to $Q$ is continuous \cite{RO:Baire*1}. 

In Section \ref{sec:CioDB1} it is shown
that the one-sided \idens\ continuous functions are in $\Darboux\cap\bone$. Section
\ref{sec:CddDB*} has as its main purpose a proof that
$\deepidcont\subset\Darboux\cap\bstar$. But first, the next section is devoted to
presenting some technical lemmas which are needed in the later sections.

%%%%%
%%%%%
%%%%%

\section{$\!\!\!\!\!\!\!${\bf.} Some Technical Lemmas}\label{sec:tech}

In this section some technical lemmas are presented which are used in the later
sections. The first of these is a restatement of the definition of \idens\ 
points \cite{Wilczynski:CatAn}.

\lemma{lem:obvious1}{$x\in\idensitypts{A}$ if, and only
if, for every increasing sequence $\{n_m\}$ of natural numbers
there is a subsequence $\{n_{m_p}\}$ such that
\[
(-1,1) \cap \left(\bigcup_{q\in\smallnats} \bigcap_{p\ge q} n_{m_p}(A-x)\right)^c
=
(-1,1) \cap \left(\liminf_{p\to\infty}\left(n_{m_p}(A-x)\right)\right)^c\in\I.
\]
}
 
The next lemma is a dual version of Lemma \ref{lem:obvious1}.
 
\lemma{lem:obvious2}{Let $B\subset\reals$ and $x\in\reals$. 
The following are equivalent:
\begin{description}
\item[(i)]   $x$ is a \idisp\ point of $B$;
\item[(ii)]  for every increasing sequence $\{n_m\}$ of natural
numbers there exists a subsequence $\{n_{m_p}\}$ such that
\[
\lim_{p\to\infty}\charf{\left(n_{m_p}(B-x)\right)\cap(-1,1)} = 0,
\,\,\,\, \I-\mbox{a.e.};
\]
\item[(iii)] for every increasing sequence $\{n_m\}$ of natural
numbers there exists a subsequence $\{n_{m_p}\}$ such that
\[
(-1,1) \cap \bigcap_{q\in\smallnats} \bigcup_{p\ge q} n_{m_p}(B-x) =
(-1,1) \cap \limsup_{p\to\infty}\left(n_{m_p}(B-x)\right) \in\I.
\]
\end{description}
}


In the definition of \idens\ given above, the divergent sequence of ``multipliers'',
$\{n_k\}$, is limited to the positive integers, rather than arbitrary sequences 
of positive numbers diverging to infinity. This restriction is removed by the 
following theorem.

\theorem{thm:list}{
Let $B\in\BaireSets$ and $x\in\reals$. 
The following statements are equivalent to each other.
\begin{description}
\item[(i)]   $x$ is an \idisp\ point of $B$;
\item[(ii)]  for every divergent increasing sequence $\{t_n\}$ of positive
             numbers there exists a
             subsequence $\{t_{n_{k}}\}$ such that 
\[
\limsup_{k\to\infty}\left(t_{n_{k}}(\tilde B-x)\right)\cap(-1,1)\in\I; 
\] 
\item[(iii)] for every divergent increasing sequence $\{t_n\}$ of positive
             numbers there exists a
             subsequence $\{t_{n_{k}}\}$ such that 
\[
\limsup_{k\to\infty}\left(t_{n_{k}}(B-x)\right)\cap(-1,1)\in\I. 
\]
\end{description}
}

Proof. The equivalence of 
(iii) and (i) is proved by W. Poreda, E. Wagner-Bojakowska and
W. Wilczy\'{n}ski \cite[Corollary 1]{PWW:RemIDens}. Thus, (ii) is equivalent to
the fact that $x$ is an \idisp\ point of $\tilde B$. But the
last fact is equivalent to (i) by (\ref{eq:idenspts}).

\bigskip

The next lemma and its corollary provide a tool for constructing sets which are in
\itop\ and \deepitop. They are similar to theorems originally proved by W. Poreda,
E. Wagner-Bojakowska and W. Wilczy{\'n}ski \cite[Theorem 1]{PWW:RemIDens}
\cite[Theorem 2]{Wilczynski:CatAn}. To state them we need first the
following definition.

We say that any of the sets $\bigcup_{n\in\smallnats}(a_n,b_n)$ or
$\bigcup_{n\in\smallnats}[a_n,b_n]$ is a {\em right interval set of a point\/}
$a\in\reals$ if $a_{n+1} < b_{n+1} < a_n < b_n$ for
$n\in\nats$ and $\lim_{n\to\infty}{a_n} = a$. In the case when $a=0$ we simply
say that it is {\em a right interval set}.

\lemma{lem:intervalset}{
If $E = \bigcup_{n\in\smallnats}[a_n,b_n]$ is a right interval set such that
\begin{description}
\item[(i)] $\lim_{n\to\infty}{(b_n - a_n)/a_n} = 0$; and
\item[(ii)] $\lim_{n\to\infty}b_{n + 1}/a_n = 0,$
\end{description}
then $0$ is an \idisp\ point of $E$. In particular,
$\complement{E}\in\deepitop$.
}

Proof. It follows immediatelly from the proof of 
\cite[Theorem2]{Wilczynski:CatAn}.

\bigskip

From the lemma above we obtain easily the following corollary.
(Compare also, \cite[Lemma 3]{AversaWilczynski:Homeomorphisms}.)

\corollary{lem:awsequence}{
If $\bigcup_{n\in\smallnats}[a_n,b_n]$ is a right interval set with
\[
\lim_{n\to\infty}\frac{(b_n - a_n)}{b_n} = 0,
\]
then there exists an increasing sequence $\{n_m\}_{m\in\smallnats}$ of
natural numbers such that $0$ is an \idisp\ point of
\[
\bigcup_{m\in\smallnats}[a_{n_m},b_{n_m}].
\]
}

Finally, the following example provides a way to construct functions which are
well-behaved in the ordinary sense, but are not well-behaved with regard to
density continuity.

\example{ex:notinCdd}{ There is a monotone continuous function 
$f\RtoR$ which is not in \idcont\ or \deepidcont.}

Proof.
To construct such a function, let $E=\bigcup_{n\in\smallnats}[a_n,b_n]$
be as in Lemma \ref{lem:intervalset} and let
$D=\bigcup_{n\in\smallnats}[c_n,d_n]$, where $[c_n,d_n]=[b_{n+1},a_n]$
for each $n\in\nats$. Then, $\complement{D}\notin\itop$. It
is also easy to see, that by decreasing the intervals $[a_n,b_n]$, if necessary,
that $\complement{E}\in\deepitop$.

Define the function $f$ by letting $f(x)=0$
whenever $x\le 0$, $f(c_n)=a_n$ and $f(d_n)=f(b_n)$ for all $n$. Make $f$ piecewise
linear between the points on which it has already been defined. 
Then $\inv{f}(\complement{E})=\complement{D}$. So, 
$f\notin\idcont\cup\deepidcont$.

%%%%%
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%%%%%

\section{$\!\!\!\!\!\!\!${\bf.}
\I-approximately Continuous Functions}\label{sec:CioDB1}

In this section it is shown that $\iacont\subset\Darboux\cap\bone$.
The following lemma is used in the proof.

\lemma{lem:openDense}{
If $f$ is right \iapp\ continuous at each of its points, 
$a\in\reals$ and $A=\{x:f(x)>a \}$, then \interior{A}\/ 
is dense in $A$.
}

Proof.
It may be supposed without loss of generality that $a=0$.
Let $A_n=\{x: f(x)\ge 1/n\}\in\BaireSets$ and let $A=\UoverN{A_n}$.  
If $x\in\tilde A_n$, then, by Lemma \ref{lem:obvious1},
$x$ is an \idens\/ point of $A_n$, as $A_n\triangle\tilde A_n\in\I$.
The definition of right \idens\/ continuity shows that $f(x)\ge 1/n$.
It follows from this that 
$\tilde A_n\subset A_n$.
Therefore, $G=\UoverN{\tilde A_n}\subset\UoverN{A_n}=A$.
To see that $G$ is dense in $A$, let $x\in A$ and choose $n\in\nats$ such that
$1/n < f(x)$.
Then $x$ is a right 
\idens\/ point of $\{w: f(w) > 1/n\}$ and it is apparent that $x$
must be a limit point of $\tilde A_n$.  
From this, it follows that $G$ is dense in $A$.

\theorem{thm:B1Darboux}
{Every right \I-approximately continuous function is of the
first Baire class.}
 
Proof. Poreda,
Wagner-Bojakowska and Wilczy\'{n}ski \cite{PWW:CatAn} 
proved a slightly weaker version of this theorem
for two-sided
\I-approximate continuity.   The following alternative proof is presented here
because it is somewhat shorter and the result is a little sharper.



Let $f$ be right \idens\/ continuous on \reals.
It suffices to show that $\{x: f(x)\ge 0\}$ is a \Gdelta\/ set.  To do
this, for each $p\in\nats$, let $U_p=\{x\colon f(x)>-1/p \}$ 
and, for $p,q,r,k\in\nats$, define
\begin{equation}\label{eq:Akdef}
A(p,q,r,k)=\left\{x\in\reals\colon
\left(\frac{k-1}{q},\frac{k}{q}\right)
\cap r\left(U_p - x\right)\ne\emptyset\right\}
\end{equation}
and
\begin{equation}\label{eq:Adef}
A(p,q,r)=\Cap_{k=1}^q A(p,q,r,k).
\end{equation}
It is easy to see that each $A(p,q,r)$ is an open set.

Next, define
\begin{equation}\label{eq:iiu}
     U=\Cap_{p\in\smallnats}\Cap_{q\in\smallnats}\Cup_{r\ge q}A(p,q,r).
\end{equation}
It is clear that $U$ is a \Gdelta\/ set.
It will be shown that $U=\{x\colon f(x)\ge 0\}$.

To show that $U\subset\{x\colon f(x)\ge 0\}$, fix
$p\in\nats$ and let 
\[
V_p=\Cap_{q\in\smallnats}\Cup_{r\ge q}A(p,q,r).
\]

Suppose that $x\in V_p$.
For each $q\in\nats$ there is an $r_q\in\nats$, $r_q\ge q$,
such that when $1\le k\le q$,
then
\[
\left( \frac{k-1}{q},\frac{k}{q}\right)\cap r_q\left(U_p-x\right) \ne\emptyset.
\]



From this and Lemma \ref{lem:openDense} it is apparent that
\begin{equation}\label{eq:blowup}
\left(\frac{k-1}{q},\frac{k}{q}\right)\cap r_q\left(\interior{U_p}-x\right)
\ne\emptyset
\,\,\,\, \mbox{ for } k=1,2,\ldots,q.
\end{equation}

Let $\{r_{q_i}\}_{i\in\smallnats}$ be an increasing subsequence of 
$\{r_q\}_{p\in\smallnats}$ and put $n_i=r_{q_i}$ for $i\in\nats$.
From (\ref{eq:blowup}) it follows that 
$\Cup_{j\in\smallnats}n_{i_j}\,\interior{U_p}$ is a dense open subset of $(0,1)$
for every subsequence $\{n_{i_j}\}_{j\in\smallnats}$ of 
$\{n_i\}_{i\in\smallnats}$. Therefore, 
\[
\limsup_{j\to\infty}n_{i_j}U_p\cap(0,1)
\]
is a residual subset of $(0,1)$.
It follows, by Lemma \ref{lem:obvious2}, 
that $x$ is not a right \I-dispersion of 
$U_p$ and the right \idens\/ continuity of $f$ shows that 
$x\in \{x\colon f(x)\ge -1/p\}$.
Thus, $V_p\subset\{x\colon f(x)\ge -1/p\}$ and
\[
U=\Cap_{p\in\smallnats}V_p\subset\{x\colon f(x)\ge 0\}.
\]

To show that $\{x\colon f(x)\ge 0\}\subset U$ let us fix $x$ such that 
$f(x)\ge 0$ and $p,q\in\nats$.
It must be shown that there is an $r\in\nats$, $r\ge q$, such that $x\in
A(p,q,r)$. If not, for every $r\ge q$ there must be an integer $k_r$, with $1\le
k_r\le q$, such that 
\[
\left(\frac{k_r-1}{q},\frac{k_r}{q}\right)\cap r\left(U_p-x\right)=\emptyset.
\]
There must exist an increasing sequence of natural numbers
$r_i$ such that $k_{r_i}=k$ for some $1\le k\le q$.
This gives
\[
\left(\frac{k-1}{q},\frac{k}{q}\right)\cap r_i\left(U_p-x\right)=\emptyset
\]
for all $i$ so that for any subsequence $\{r_{i_j}\}$ of $\{r_i\}$
\[
\liminf_{j\to\infty}r_{i_j}(U_p-x)\cap\left(\frac{k-1}{q},\frac{k}{q}\right)
=\emptyset.
\]
Therefore, $x$ is not a point of right \I-density of $U_p$.  But, this is
impossible because 
$f(x)\ge 0$
and $f$ is right \I-approximately continuous at $x$.

Therefore $U\supset\{x:f(x)\ge 0\}$ and consequently 
$U=\{ x: f(x)\ge 0\}$, which finishes the
proof of the theorem.

\medskip

The following corollary is immediate.

\corollary{cor:DenseCont}{If $f\in\iacont$, then $f$ is continuous in the ordinary
sense on a dense \Gdelta\ subset of \reals.}

\corollary{cor:Darboux}{
$\iacont\subset\Darboux\cap\bone$.
}

Proof.
Since sets which are open in the \idens\ topology must be bilaterally $c$-dense
in themselves,
this is an immediate consequence of the preceding theorem and Young's criterion.
(See Bruckner \cite{Bruckner:DiffReal}.)

%%%%%
%%%%%
%%%%%

\section{$\!\!\!\!\!\!\!${\bf.}
\idens\ Continuous Functions}\label{sec:CddDB*}

The goal of this section is to prove that $\deepidcont\subset\Darboux\cap\bstar$.
To do this, the following definition and lemma are needed 
\cite[Lemma 29.1]{Jech:SetTheory}.

A {\em partition} of a set $E$ is a pairwise disjoint family $\Pi =
\{E_i\colon i\in\Lambda\}$ such that $\bigcup_{i\in\Lambda}E_i = E$. Note that
any partition $\Pi$ can be associated with a function $F\colon
E\to\Lambda$ such that $F(x) = F(y)$ if, and only if, $x$ and $y$ belong to
the same $E_i\in\Pi$. Conversely, any function $F\colon E\to\Lambda$
determines a partition of $E$.
 
For a set $A$ and $n\in\nats$ define
\[
[A]^n = \left\{ B\subset A\colon \mbox{ card}(B) = n\right\}.
\]
If $\Pi = \{ E_i\colon i\in\Lambda\}$ is a partition of $[A]^n$, then a set
$H\subset A$ is {\em homogeneous} for the partition $\Pi$ if, for some
$i\in\Lambda$, $[H]^n\subset E_i$. That is, all $n$-element subsets of $H$ are
in the same piece of the partition $\Pi$.
 
\lemma{lemma37}{{\bf (Ramsey's Theorem)} If $n,k\in\nats$, then every 
finite partition $\Pi = \{E_1, E_2, \ldots, E_k\}$ of $[\nats]^n$ has an
infinite homogeneous set.
In other words, for every $F\colon[\nats]^n\to\{1,2,\ldots,k\}$ there
exists an infinite $H\subset\nats$ such that $F$ is constant on
$[H]^n$.
} 

\theorem{thm:CddinBstar}{$\deepidcont\subset\Darboux\cap\bstar$.}
 
Proof. Assume to the contrary that for some perfect set $P$ the set
 
\[
Z = \{x\in P\colon\fP \mbox{ is not continuous at }x\}
\]
is dense in $P$.
 
We will construct sequences: $\{x_n\}_{n\in\smallnats}$ of points of $P$,
$\{(a_n,b_n)\}_{n\in\smallnats}$ of open intervals,
$\{J_n\}_{n\in\smallnats}$ of compact intervals,
and $\{I_n\}_{n\in\smallnats}$ of open intervals having the
same midpoint as the corresponding $J_n$, and contained in that corresponding
$J_n$. The construction is inductive, and aimed at having all the objects
obtained satisfy the conditions (a) through (f) listed below.
 
For the reminder of this proof let $\widetilde{f^{-1}}(A)$ stands for
$\tilde B$, where $B=f^{-1}(A)$.

Start by choosing $x_0\in Z$, $(a_0,b_0) = (x_0 - 1, x_0 + 1)$ and
$I_0 = J_0 = \emptyset$. 
Assume that for all $n\in\nats$ and all 
$i\in\nats$, $i\le n$, it holds that:
 
\begin{description}
 
\item[(a)] $f(x_i)\in I_i$;
 
\item[(b)] $J_{i - 1}\cap J_i = \emptyset$ and, for $i>2$,
\[
\vert J_i\vert \le \frac{1}{3}
\min \{\mbox{dist}(J_k,J_{k+1})\colon k\in\nats, k < i - 1\};
\]
 
\item[(c)] $\vert J_i\vert < \osc(\fP,x_i)$ and $0<\vert I_i\vert <2^{-i}\vert
J_i\vert$;
 
\item[(d)] $x_i\in(a_i,b_i)\cap Z \subset [a_i,b_i]\subset(a_{i-1},b_{i-1})$
and $\vert b_i - a_i\vert < 2^{-i}$;
 
\item[(e)] for every $k\in\nats$, $2^i \le k < 4^i$,
 
\[
\left(\frac{1}{b_i - x_i}\left(\widetilde{f^{-1}}(I_i) - x_i\right)\right)
\cap
\left(\frac{k}{4^i}, \frac{k+1}{4^i}\right)\ne\emptyset;
\]
 
\item[(f)] for every $x\in[a_i,b_i]$ and every
$k\in\nats$, $2^{i-1} \le k < 4^{i-1}$,
 
\[
\left(\frac{1}{b_{i-1} - x}\left(\widetilde{f^{-1}}
(I_{i-1}) - x\right)\right)
\cap
\left(\frac{k}{4^{i-1}}, \frac{k+1}{4^{i-1}}\right)\ne\emptyset.
\]
\end{description}
 
Let us present the inductive construction. Assume it is done for some $n\ge
0$. We will show the next step. 
Start with condition (f). If $n + 1 = 1$,
(f) is void and can be ignored by defining $U=\reals$. 
Otherwise, by (e), the set
\[
U_k = \left\{x\colon
\left(\frac{1}{b_n - x}
\left(\left(\widetilde{f^{-1}}(I_n) \right) -x\right)
\right)\cap\left(\frac{k}{4^n},\frac{k +1}{4^n}\right)
\ne\emptyset\right\}
\]
contains $x_n$ for every $k\in\nats$, $2^n\le k < 4^n$. 
It is also not difficult to see that the sets $U_k$ are open. Therefore
\[
U = \bigcap_{2^n\le k < 4^n}U_k
\]
is also open and contains $x_n$. 
It is easy to see that condition (f) is satisfied for $x\in U$.

Now, find
\[
y\in P\cap\inv{f}\left(J_n^c\right)\cap\left(\left(a_n,b_n\right)\cap U\right).
\]
The existence of such a $y$ is guaranteed because $U$ is open, $x_n\in U$
and (c). If $y\in Z$, let $x_{n + 1} = y$. Otherwise \fP\ is
continuous at $y$. In this case, the fact that $Z$ is dense in $P$ 
and $U$ is open guarantees the existence of
\[
x_{n + 1}\in
P\cap\inv{f}\left(J_n^c\right)\cap\left(\left(a_n,b_n\right)\cap U\right)
\cap Z.
\]
Since $f(x_{n + 1})\notin J_n$ and $x_{n + 1}\in Z$, there exists a small
interval $J_{n + 1}$ centered at $f(x_{n + 1})$ satisfying conditions (b)
and (c). Choosing $I_{n + 1}$ centered at $f(x_{n + 1})$ of length
\[
\frac{\vert J_{n + 1}\vert}{2^{n + 2}}
\]
guarantees (a), (b), and (c).
 
Defining $\left(a_{n + 1}^\prime,b_{n + 1}^\prime\right)$ to be centered at
$x_{n + 1}$ and such that
\[
\left[a_{n + 1}^\prime,b_{n + 1}^\prime\right]\subset\left(a_n,b_n\right)\cap
U\,\,\mbox{ and }\,\, b_{n + 1}^\prime - a_{n + 1}^\prime < \frac{1}{2^{n + 1}}
\]
guarantees (d) and (f) for the interval 
$\left[a_{n + 1}^\prime,b_{n + 1}^\prime\right]$. 
However, it still must be shown that condition
(e) is satisfied. This is done by choosing interval
$(a_{n+1},b_{n+1})\subset(a_{n+1}^\prime,b_{n+1}^\prime)$.
 
Note that $x_{n + 1}$ is an \I-density point of 
$\inv{f}\left(I_{n + 1}\right)$.
Therefore, by Lemma \ref{lem:obvious1}, there exists an increasing 
sequence $\{n_i\}_{i\in\nats}$ of natural numbers such that the set
\[
S = \liminf_{i\to\infty}\left(n_i\left(
\widetilde{f^{-1}}(I_{n + 1}) - x _{n + 1}
\right)\right)\cap\left(-1,1\right)
\]
is residual in $(-1,1)$. Define
\[
W_i = n_i \left(\widetilde{f^{-1}}(I_{n + 1}) - x_{n + 1}\right).
\]
The set
\[
\bigcup_{r = 1}^{+\infty}\bigcap_{i\ge r} W_i
\]
is residual in $(-1,1)$. In particular, for every $k\in\nats$,
$2^{n + 1}\le k < 4^{n + 1}$,
\[
\left(\bigcup_{r = 1}^{+\infty}\bigcap_{i\ge r} W_i\right)\cap\left(
\frac{k}{4^{n + 1}}, \frac{k + 1}{4^{n + 1}}\right) \ne\emptyset.
\]
The sequence
$\left\{\bigcap_{i\ge r} W_i\right\}_{r\in\nats}$
is increasing. Thus, there is an $r_0\in\nats$ such that
\[
W_i\cap\left(\frac{k}{4^{n + 1}},\frac{k + 1}{4^{n + 1}}\right)\ne\emptyset
\]
for every $i\ge r_0$ and $k\in\nats$, $2^{n + 1}\le k < 4^{n + 1}$.
But
\[
W_i = n_i \left(\widetilde{f^{-1}}(I_{n + 1}) - x_{n + 1}\right) =
\frac{1}{x_{n + 1} + \frac{1}{n_i} - x_{n + 1}}
\left(\widetilde{f^{-1}}(I_{n +1}) - x_{n + 1}\right).
\]
Define $(a_{n + 1}, b_{n + 1})$ as
\[
\left(x_{n + 1} - \frac{1}{n_i}, x_{n + 1} + \frac{1}{n_i}\right),
\]
where $i\ge r_0$ and $[a_{n + 1}, b_{n + 1}]\subset[a_{n + 1}^\prime, 
b_{n + 1}^\prime]$. The desired condition (e) is satisfied.
This ends the inductive construction.
 
\medskip

It will now be shown how the conclusion of the theorem follows from the
construction.
 
Let
\[
\{x\} = \bigcap_{n\in\smallnats}[a_n,b_n] =
\bigcap_{n\in\smallnats}\left([a_n,b_n]\cap Z\right).
\]
We will show that $f$ is not deep-\I-density continuous at $x$. To be more
specific, we will find a sequence $\{n_i\}_{i\in\smallnats}$ such that
 
\begin{description}
\item[(1)] $f(x)$ is a deep-\I-dispersion 
point of $\bigcup_{i\in\smallnats}I_{n_i}$, and
\item[(2)] $x$ is not a deep-\I-dispersion 
point of $\inv{f}\left(\bigcup_{i\in\smallnats}I_{n_i}\right)$.
\end{description}
 
We will first show $x$ is not an \I-dispersion point of
$\inv{f}\left(\bigcup_{i\in\smallnats}I_{n_i}\right)$ for every sequence
$\{n_i\}_{i\in\smallnats}$.
 
Let $\{n_i\}_{i\in\smallnats}$ be any increasing sequence of natural numbers.
By the definition of $x$, condition (f) implies
\[
\left(t_n\left(\widetilde{f^{-1}}(I_n) - x\right)\right) \cap
\left(\frac{k}{4^n},\frac{k + 1}{4^n}\right)\ne\emptyset
\]
for every $k\in\nats$, $2^n\le k < 4^n$, where $t_n$ is defined as
$1/(b_n - x)$. Note that sequence $\{t_n\}_{n\in\smallnats}$ is increasing 
and diverging to $\infty$. Thus, the open set
$U_n = t_n\left(\widetilde{f^{-1}}(I_n) - x \right)$
intersects every interval
$\left(\frac{k}{4^n},\frac{k+1}{4^n}\right)\subset\left[\frac{1}{2},1\right].$
This implies that for every increasing sequence $\{n_i\}_{i\in\smallnats}$
of natural numbers and for every $s\in\nats$ the set 
$\bigcup_{i\ge s}U_{n_i}$
is dense in $\left[\frac{1}{2}, 1\right]$. Hence,
\[
\limsup_{j\to\infty}t_{n_{i_j}}
\left( 
\widetilde{f^{-1}}\left(\bigcup_{i\in\smallnats}{I_{n_{i_j}}}\right) -x\right)
\supset \limsup_{j\to\infty}U_{n_{i_j}} \not\in \I
\]
for every subsequence $\{n_{i_j}\}_{j\in\smallnats}$
of $\{n_i\}_{i\in\smallnats}$. Thus, by Theorem
\ref{thm:list}(ii), $x$ is not an \I-dispersion point of
$\inv{f}\left(\bigcup_{i\in\smallnats}I_{n_i}\right)$.
 
Let us now turn to the proof of condition {\bf (1)}. 
We must find an
increasing sequence $\{n_i\}_{i\in\smallnats}$ of natural numbers such that
$f(x)$ is a deep-\I-dispersion point of
$\bigcup_{i\in\smallnats}I_{n_i}$. The set
$\bigcup_{i\in\smallnats}I_{n_i}$ is open. Therefore,
it suffices to find a sequence
$\{n_i\}_{i\in\smallnats}$ such
that $f(x)$ is an \I-dispersion point of
$\bigcup_{i\in\smallnats}I_{n_i}$. For the
sake of simplicity, let us assume that $f(x) = 0$.
 
\medskip
 
There are two cases to consider.
 
\smallskip
 
Case 1${}^o$. There exists an increasing sequence 
$\{n_i\}_{i\in\smallnats}$ of
natural numbers such that the $J_{n_i}$ are pairwise disjoint.
 
By taking a subsequence of $\{n_i\}_{i\in\smallnats}$, 
if necessary, it may be assumed that
\[
\bigcup_{i\in\smallnats}J_{n_i}
\]
is either a right or left  
interval set. For simplicity, assume it is a right interval set.
 
Let $J_{n_i} = [c_i,d_i]$ and $I_{n_i} = (\alpha_i,\beta_i)$. Then
\[
f(x) = 0 < d_{i - 1} < c_i < \alpha_i < \beta_i < d_i
\]
for all $i$. Condition (c) states that
\[
\frac{\beta_i - \alpha_i}{d_i - c_i} = 
\frac{\vert I_{n_i}\vert}{\vert J_{n_i}\vert} <
\frac{1}{2^{n_i}}.
\]
Let $z_n$ be the common center of $I_n$ and $J_n$, for $n\ge 0$. Then
\[
\lim_{i\to\infty}\frac{\beta_i - \alpha_i}{\beta_i}\le\lim_{i\to\infty}
\frac{\beta_i - \alpha_i}{z_{n_i}}
\le\lim_{i\to\infty}\frac{\beta_i - \alpha_i}{z_{n_i} - c_i} =
2\lim_{i\to\infty}\frac{\beta_i - \alpha_i}{d_i - c_i} = 0.
\]
By Corollary \ref{lem:awsequence} choose a subsequence of 
$\{n_i\}_{i\in\smallnats}$
with the desired properties.
 
\smallskip
 
Case 2${}^o$. There is no pairwise disjoint subsequence
$\{J_{n_i}\}_{i\in\smallnats}$ of the sequence $\{J_n\}_{n\in\smallnats}$.
 
\smallskip
 
Let us first consider the subsequence $\{J_{2n + 1}\}_{n\in\smallnats}$,
indexed by the odd numbers, of the sequence
$\{J_n\}_{n\in\smallnats}$. Define a partition
function $F\colon[\nats]^2\to\{0,1\}$ by
\[
F\left(\{n,m\}\right) = 1\,\,\,
\mbox{ if, and only if, }\,\,\,
J_{2n+1}\cap J_{2m+1} \ne\emptyset.
\]
By Lemma \ref{lemma37} (Ramsey's Theorem) there exists an infinite homogeneous
subset $\{n_i\}_{i\in\smallnats}$ of \nats;
i.e., the sequence $\{n_i\}_{i\in\smallnats}$ of
natural numbers such that for some $k\in\{0,1\}$,
$F\left(\{n_i,n_j\}\right) = k$
for all positive integers $i\ne j$. But $k = 0$ would contradict the
definition of the case 2${}^o$, which is currently considered. Thus $k = 1$;
i.e.,
\begin{equation}\label{oddsequence}
J_{2n_i + 1}\cap J_{2n_j + 1}\ne\emptyset
\end{equation}
for all nonnegative integers $i\ne j$.
 
Now let us repeat the Ramsey-type argument, which was used above, for the
even-numbered counterparts of $\{J_{2n_i + 1}\}_{i\in\smallnats}$.
Define $G\colon[\nats]^2\to\{0,1\}$ by
\[
G\left(\{i,j\}\right) = 1\,\,\,
\mbox{ if, and only if, }\,\,\,
J_{2n_i}\cap J_{2n_j}\ne\emptyset.
\]
By Lemma \ref{lemma37} (Ramsey's Theorem) there 
exists a subsequence
$\{n_{i_s}\}_{s\in\smallnats}$ of $\{n_i\}_{i\in\smallnats}$ such that
\begin{equation}\label{evennumbered}
J_{2n_{i_s}}\cap J_{2n_{i_t}}\ne\emptyset
\end{equation}
for all nonnegative integers $s\ne t$, while condition (\ref{oddsequence})
is still preserved, or more precisely
\begin{equation}\label{oddnumbered}
J_{2n_{i_s} + 1}\cap J_{2n_{i_t} + 1}\ne\emptyset
\end{equation}
for $s\ne t$. Define 
$\varepsilon = \mbox{dist}\left(J_{2n_{i_0}}, J_{2n_{i_0} + 1}\right)$.
By (b), $\varepsilon > 0$. Moreover, by (b), 
(\ref{evennumbered}) and (\ref{oddnumbered})
\[
B_0 = \bigcup_{s\in\smallnats}J_{2n_{i_s}}\subset
\left\{x\colon\mbox{dist}\left(x,J_{2n_{i_0}}\right) <
\frac{\varepsilon}{3}\right\}
\]
and
\[
B_1 = \bigcup_{s\in\smallnats}J_{2n_{i_s}+1}\subset
\left\{x\colon\mbox{dist}\left(x,J_{2n_{i_0}+1}\right) <
\frac{\varepsilon}{3}\right\}.
\]
Hence
\[
\mbox{dist}(B_0,B_1)\ge\frac{\varepsilon}{3} > 0.
\]
Note that
\[
S_0 = \bigcup_{s\ge 0}I_{2n_{i_s}}\subset B_0
\]
and
\[
S_1\ = \bigcup_{s\ge 0}I_{2n_{i_s} + 1}\subset B_1.
\]
Thus $\mbox{dist}(S_0,S_1)>0$, which implies that either
\[
\mbox{dist}(f(x),
S_0) > 0
\]
or
\[
\mbox{dist}(f(x),S_1) > 0.
\]
This clearly means that $f(x)$ is an
\I-dispersion point of either $S_0$ or $S_1$.
 
This finishes the proof of Theorem \ref{thm:CddinBstar}.

\medskip

Since every function in \bstar\ is continuous in the ordinary sense on a dense open
set, the following corollary is obvious.

\corollary{cor:denseopen}{If $f$ belongs to \deepidcont\ or \idcont, then there is
a dense \otop-open set $G$ such that $f|_G$ is \otop-continuous.}


\theorem{theorem24}{The spaces \deepidcont\ and \idcont\,
equipped with the
topology of uniform convergence, are of the first category in themselves.}

Proof. This will only be proved for the class \deepidcont\ as the other case
is essentially the same.

Let $\{I_n\}_{n\in\nats}$ be the sequence of all 
open intervals with rational endpoints and let $C_n$ be the family of all
deep \idens\ continuous functions that are continuous on $I_n$ in the 
ordinary sense. By Theorem \ref{thm:CddinBstar},
$\deepidcont = \UoverN C_n$. Also, it is evident that the sets $C_n$ are
closed in the topology of uniform convergence. Finally, for any function
$g\in C_n$ and any 
of its neighborhoods $U$ it is easy to slightly modify the
function $f$ from Example \ref{ex:notinCdd} 
in such a way that
\[
f\in U\cap(C_n\setminus\deepidcont).
\]
Thus, the sets $C_n$ are nowhere dense.

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\end{document}