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\markright{Density-to-deep-\I-density continuous functions}
 
\marginlabels{n}
 
\title{Density-to-deep-\I-density continuous functions}
\author{Krzysztof Ciesielski\footnote{Received support from a West Virginia
University Senate research grant.}, Department of Mathematics, West Virginia
University, Morgantown, WV 26506}
\date{}

\begin{document}

\maketitle 
 
\section{\hskip -1em{.} Preliminaries}

The class of real functions continuous
with respect to the deep-\I-density topology on the range and the density
topology on the domain coincides with the class of all constant functions.
This determines of the last class from the sixteen classes
of continuous functions $C(T_1,T_2)=\{f\colon(\reals,T_1)\to(\reals,T_2)\}$,
where $T_i$ stands for ordinary, density, \I-density or deep-\I-density
topology \cite{CL:Examples}.
 
The notation used throughout this paper is standard. In particular,
\reals\ stands for the set of real numbers and $\nats=\{1,2,3,\ldots\}$.
For $A,B\subset\reals$ and $d\in\reals$ the
complement of $A$ is denoted by \complement{A}, 
the Euclidean distance between $A$ and $B$ by \dist{A}{B}; i.e.,
$\dist{A}{B} = \inf\{|x - y|: x\in A, y\in B\}$ and we define
$B - d = \{x - d\in\reals\colon x\in B\}$ and 
$dB =\{dx\in\reals\colon x\in B\}$.
The families of Lebesgue measurable subsets of \reals\
and of subsets of \reals\ with Baire property are denoted by
\Lebesgue\ and \BaireSets, while
\N\ and \I\ stand for the ideals of Lebesgue measure zero and 
of first category subsets of \reals.
If $A\in\Lebesgue$, we denote its Lebesgue measure by \measure{A}.

To define the density topology \densitytop\ and the deep-\I-density
topology \deepitopology\ we need the following notions of 
density and deep-\I-density points
\cite{Oxtoby:MeasCat,Wilczynski:CatAn}.

Let $A\in\Lebesgue$. A number $x$, not necessarily in $A$, is a 
{\em density point\/} of $A$ if  
\begin{equation}\label{def_dens}
\lim_{h\to 0^+}\frac{\measure{A\cap(x-h,x+h)}}{2h}=1.
\end{equation}
The set of all density points of $A\in\Lebesgue$ is denoted as
\densitypts{A}. 
The family of sets
\[
\densitytop=\{A\in\Lebesgue\colon A\subset\densitypts{A}\}
\]
forms a topology on \reals\ \cite{Oxtoby:MeasCat,GW:AppContTrans} called
the {\em density topology\/}.

We say that $0$ is a {\em deep-\I-density point\/}
of a set $B\in\BaireSets$ \cite{Wilczynski:CatAn}
if there exists a closed set $A\subset B\cup \{0\}$ such that
for every increasing sequence $\{n_m\}_{m\in\smallnats}$ of natural
numbers there exists a subsequence $\{n_{m_p}\}_{p\in\smallnats}$ such that
\begin{equation}\label{def_idens1}
\lim_{p\to\infty}\charf{n_{m_p}A\cap(-1,1)} = \charf{(-1,1)} \ \mbox{\I-a.e.}
\end{equation}
It is worth noticing that condition (\ref{def_idens1}) is equivalent to
the fact that the set
\begin{equation}\label{def_idens2}
\liminf_{p\to\infty}n_{m_p}A =
\bigcup_{q\in\smallnats} \bigcap_{p\ge q} n_{m_p}A
\end{equation}
is residual in $(-1,1)$. We say that a point $b$ is a deep-\I-density point 
of $B\in\BaireSets$ if $0$ is a deep-\I-density point of $B-b$.
The set of all deep-\I-density points of $B\in\BaireSets$ is denoted as
\deepidensitypts{B}. 
The family of sets
\[
\deepitopology=\{B\in\BaireSets\colon B\subset\deepidensitypts{B}\}
\]
forms a topology on \reals\ 
called the {\em deep-\idens\ topology\/} 
\cite{Lazarow:Coarsest,Wilczynski:CatAn}.

We will use also the following dual versions of the density points. We say
that $x$ is a {\em dispersion} ({\em deep-\I-dispersion})
point of $A$ if $x$ is a density (deep-\I-density) point of
\complement{A}. In particular, $0$ is a deep-\I-dispersion point of $A$ if
there exists an open $B\supset (A\setminus\{0\})$ such that
for every increasing sequence $\{n_m\}_{m\in\smallnats}$ of natural
numbers there exists a subsequence $\{n_{m_p}\}_{p\in\smallnats}$ such that
\begin{equation}\label{con_idisp}
(-1,1) \cap \bigcap_{q\in\smallnats} \bigcup_{p\ge q} n_{m_p}B =
(-1,1) \cap \limsup_{p\to\infty}(n_{m_p}B) \in\I.
\end{equation}

The symbols \const,
$\cal C$ and $C_{\cal ND}$ stand for the classes of real functions that are
constant, ordinary continuous and continuous with the
density topology on the domain and deep-\I-density topology on the range,
respectively. \bstar\ denotes 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}. 

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}.

We need also the following two propositions. The first one can be found in
\cite[Lemma 2.4]{CL:Baire}. (Compare also \cite[Theorem 1]{PWW:RemIDens} and 
\cite[Theorem 2]{Wilczynski:CatAn}.) The second in 
\cite[Lemma 2.4]{CL:Examples}.

\proposition{sequence}{
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 a 
deep-\I-dispersion point of $E$. In particular, $E^c\in\deepitopology$.}

\proposition{deepidisp}{
Let $C\subset(0,1]$ be a closed nowhere dense set and
let $\{b_n\}_{n\in\smallnats}$ be a decreasing sequence of positive numbers
such that the limit $\lim_{n\to\infty}b_{n + 1}/b_n = 0$.
Then $0$ is a deep-\I-dispersion point of the set
\[
E=\bigcup_{n\in\smallnats}b_nC.
\] 
In particular, $E^c\in\deepitopology$.
}

\section{\hskip -1em{.} Continuous functions}

We will start this section with the following lemma. 

\lemma{lem:intervalsets}{Let $f\RtoR$ be a measurable function such that $f(0)=0$
and let $c\in(0,1)$. If $E=\Cup_{n\in\smallnats}[a_n,b_n]$ is a right interval
set  such that, for every $n\in\nats$, $a_n\geq c\,b_n$ and 
$\{d_n\}_{n\in\smallnats}$ is a sequence from $(0,1)$ such that
\begin{equation}\label{assumption1}
\measure{\inv{f}([a_n,b_n])\cap(0,d_n)}\geq d_n/2,
\end{equation}
then $f\not\in C_{\cal ND}$.
}

Proof. The sets $\inv{f}([a_n,b_n])\cap (0,1)$ are pairwise disjoint
and bounded. Therefore 
$\lim_{n\to\infty}\measure{\Cup_{k\ge n}\inv{f}([a_k,b_k])\cap(0,1)}=0$.
From this and (\ref{assumption1}) it follows that $\lim_{n\to\infty}d_n=0$.

Taking a subsequence, if necessary, we can assume that
\[
\lim_{n\to\infty}b_{n + 1}/b_n = 0.
\] 
We may also assume that 
$a_n= c\,b_n$, because decreasing $a_n$ does not change 
(\ref{assumption1}).

There are two cases to consider:

\begin{description}
 
\item[(1)] there is a point $x\in(c,1)$ and an $\e>0$ such that
such that for every nontrivial interval
$I\subset(c,1)$ containing $x$ and for every $k\in\nats$ there is
$n_k\geq k$ with $\measure{\inv{f}(b_{n_k}I)\cap(0,d_{n_k})}\geq \e d_{n_k}$; and
 
\item[(2)] for every $x\in(c,1)$ and every $\e>0$ there exists  an
interval $I\subset(c,1)$ containing $x$ and an $k\in\nats$ such that
$\measure{\inv{f}(b_n I)\cap(0,d_n)}< \e d_n$ for every $n\geq k$.

\end{description}

Case (1). Let $x$ and $\e$ be as in the assumption. 
Put $n_0=0$ and, by induction on $k\in\nats$, define
a closed interval $I_k\subset(x-1/k,x+1/k)\cap(c,1)$
and an $n_k> n_{k-1}$ such that
\begin{equation}\label{cond:big1}
\measure{\inv{f}(b_{n_k}I_k)\cap(0,d_{n_k})}\geq \e d_{n_k}.
\end{equation}
Then, by Proposition \ref{sequence}, 
$0$ is a deep-\I-dispersion point of the interval set
$D=\Cup_{n\in\smallnats} b_{n_k}I_k$, while $d(\inv{f}(D),0)\neq 0$,
because 
\[
\liminf_{k\to\infty}
\frac{\measure{\inv{f}(D)\cap(0,d_{n_k})}}{d_{n_k}}\geq \e>0.
\]
Hence, $\complement{D}\in\deepitopology$ and 
$\inv{f}(\complement{D})\not\in\densitytop$;
i.e., $f\not\in C_{\cal ND}$.

Case (2). Let $\{q_k\colon k\in\nats\}$ be an enumeration of the
rational numbers in $(c,1)$ and let $\delta\in(0,1/2)$. 
Put $n_0=0$ and, by induction on $k\in\nats$, define an open
interval $I_k$ containing $q_k$ and a number $n_k>n_{k-1}$ such that
\[
\measure{\inv{f}(b_{n_k}I_{n_k})\cap(0,d_{n_k})}<\frac{\delta}{2^{n_k}}d_{n_k}.
\]
Let $C=[c,1]\setminus \Cup_{k\in\smallnats} I_{n_k}$. Then, $C$ is closed 
and nowhere dense. By Proposition \ref{deepidisp}, $0$ is a
deep-\I-dispersion point of 
$E=\bigcup_{k\in\smallnats}b_{n_k}C$. So, $E^c\in\deepitopology$.
On the other hand, $0$ is not a dispersion point of $\inv{f}(E)$, as 
\begin{eqnarray*}
\liminf_{k\to\infty}
\frac{\measure{\inv{f}(E)\cap(0,d_{n_k})}}{d_{n_k}}
& \geq &
\liminf_{k\to\infty}
\frac{\measure{\inv{f}(b_{n_k}C)\cap(0,d_{n_k})}}{d_{n_k}}\\
&\geq &
\frac{1}{2}-\sum_{k=1}^{\infty}\frac{\delta}{2^{n_k}}
\geq\frac{1}{2}-\delta>0.
\end{eqnarray*}
Thus, $\inv{f}(\complement{E})\not\in\densitytop$ and $f\not\in C_{\cal ND}$.

\bigskip

Now we are ready for the main result of this section.

\lemma{continuous}{$\cal C\cap C_{\cal ND}=\const$.}

Proof. Evidently, $\const\subset\cal C\cap C_{\cal ND}$. To prove the
opposite inclusion, let $f\in\cal C\setminus\const$. We
will show that $f\not\in C_{\cal ND}$ by using Lemma \ref{lem:intervalsets}.
Let $a<b$ be such that $f(a)\neq f(b)$. We may assume that 
$f(a) < f(b)$ and, 
by the continuity of $f$, that $f((a,b))=(f(a),f(b))$.
We may also assume,
modifying of $f$ in a linear way, if necessary, that $f(a)=a=-1$ and $f(b)=b=1$.
Then, we obtain $f(-1)=-1$, $f(1)=1$ and $f((-1,1))=(-1,1)$.

We construct, by induction on $n\in\nats$, the sequences:
$\{a_n\}$, $\{b_n\}$, $\{c_n\}$ and $\{d_n\}$ of real numbers and 
sequences $\{I_n\}$ and $\{J_n\}$ of intervals.
We start by putting $a_0=c_0=-1$, $b_0=d_0=1$ and $I_0=[-1,1]$.
Then we procede inductively to obtain the following conditions:

\begin{description}
 
\item[(a)] $I_n=[a_n,b_n]$;

\item[(b)] $f(c_n)=a_n$ and $f(d_n)=b_n$;

\item[(c)] $f((c_n,d_n))=(a_n,b_n)$;

\item[(d)] 
$I_n\in\{[a_{n-1},(a_{n-1}+b_{n-1})/2],[(a_{n-1}+b_{n-1})/2,b_{n-1}]\}$;

\item[(e)] $J_n=\closure{I_{n-1}\setminus I_n}$;

\item[(f)] $\measure{\inv{f}(J_n)\cap[c_{n-1},d_{n-1}]}\geq(d_{n-1}-c_{n-1})/2$.

\end{description}

The inductive step is self-explanatory. First, select $I_n$ as
in (d) to satisfy (f). If $I_n=[a_{n-1},(a_{n-1}+b_{n-1})/2]$ then we put
$c_n=c_{n-1}$ and 
$d_n=\min\{x\in[c_{n-1},d_{n-1}]\colon f(x)=(a_{n-1}+b_{n-1})/2\}$. In the other
case, proceed similarly.

Let $x\in\Cap_{n\in\smallnats}[c_n,d_n]$. 
Then, $f(x)\in\Cap_{n\in\smallnats}I_n$.
We may assume, translating $f$, if necessary, that $x=0=f(x)$. 

Evidently, $\measure{I_n}=2\measure{I_{n+1}}$. A simple argument shows that
for every $n\in\nats$ either $\dist{J_n}{I_j}\geq\measure{J_n}/4$
or $\dist{J_{n+1}}{I_j}\geq\measure{J_{n+1}}/4$ for all $j\geq n+2$.
This allows us to choose a subsequence $\{n_k\}$ such that
\begin{equation}\label{cond:dist} 
\dist{J_{n_k}}{0}\geq\measure{J_{n_k}}/4
\end{equation} 
for all $k\in\nats$. It is easy to assume that
a subsequence $\{n_k\}$ is choosen in such a way that the intervals
$\{J_{n_k}\}$ are monotone and on one side of $0$.
For simplicity we assume that 
$E=\Cup_{k\in\smallnats}J_{n_k}$ is a right interval set.
Then,  condition (\ref{cond:dist}) implies that  
$\min{J_{n_k}}\geq(1/5)\max{J_{n_k}}$. Thus,
the first part of the assumptions from  
Lemma \ref{lem:intervalsets} is satisfied
for the set $E$ with $c=1/5$. 
To finish the proof we will show that the second part is satisfied
as well. 

First notice that for infinitely many $k$ we have either 
\begin{equation}\label{cond:ineq} 
\frac{\measure{\inv{f}(J_{n_k})\cap [0,d_{{n_k}-1}]}}{d_{{n_k}-1}}\geq
\frac{\measure{\inv{f}(J_{n_k})\cap [c_{{n_k}-1},0]}}{-c_{{n_k}-1}}
\end{equation} 
or the converse inequality (where, $0/0$ is considered to be $0$.)
Without loss of the generality we may assume that (\ref{cond:ineq}) holds
for every $k$. But this, together with (f), implies that
\[
\measure{ \inv{f}(J_{n_k})\cap [0,d_{{n_k}-1}] }\geq  d_{{n_k}-1}/2.
\]
Thus, the assumptions of Lemma \ref{lem:intervalsets} are satisfied and 
Lemma \ref{continuous} is proved.

\section{\hskip -1em{.} General case}

For the next step, 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
$[H]^n$.
} 

The next lemma combines the proofs of the theorems that density 
continuous functions and 
deep-\I-density continuous functions are in \bstar\ class 
\cite[Theorem 3]{CLO:DensContCont}, \cite[Theorem 4.2]{CL:Baire}.

\lemma{CndinBstar}{$C_{\cal ND}\subset\bstar$.}
 
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.
 
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 
$i\in\nats$, $1\le i\le n$, it holds that:
 
\begin{description}
 
\item[(a)] $f(x_i)\in I_i\subset J_i$;
 
\item[(b)] $J_{i - 1}\cap J_i = \emptyset$ and, for $i>2$,
\[
\measure{J_i}\leq\frac{1}{3}\min\{\dist{J_k}{J_{k+1}}\colon k\in\nats, k<i-1\}; 
\]
 
\item[(c)] $\measure{J_i} < \osc(\fP,x_i)$ and 
$0<\measure{I_i}<2^{-i}\measure{J_i}$;
 
\item[(d)] $x_i\in(a_i,b_i)\cap Z \subset [a_i,b_i]\subset(a_{i-1},b_{i-1})$;
 
\item[(e)] $(b_i - a_i) < 2^{-i}$; and,

\item[(f)] $\measure{f^{-1}(I_i)\cap (a_i,b_i)}>(1-2^{-i})(b_i-a_i)$.

\end{description}

To continue with the inductive step, note that by (c) and (d),
we are able to choose
\[
y\in P\cap f^{-1}(J_n^c)\cap (a_n,b_n).
\]
If $y\in Z$, then let $x_{n+1}=y$. Otherwise, $f|_P$ is continuous
at $y$. In this case, the fact that $Z$ is dense in $P$ guarantees
the existence of
\[
x_{n+1}\in  P\cap f^{-1}(J_n^c)\cap (a_n,b_n)\cap Z.
\]

Because $J_n$ is closed and $x_{n+1}\in Z$, there is a
closed interval $J_{n+1}$ centered at $f(x_{n+1})$ such that
$J_{n+1}\cap J_n=\emptyset$, $0<\measure{J_{n+1}}<\osc(\fP,x_{n+1})$
and, for $i>2$,
\[
\measure{J_i}\leq\frac{1}{3}\min\{\dist{J_k}{J_{k+1}}\colon k\in\nats, k<i-1\}.
\]
Setting $I_{n+1}$ to be the closed interval centered at $f(x_{n+1})$
with length equal 
$\measure{J_{n+1}}/2^{n+1}$, it follows that (a), (b) and (c)
are true with $i=n+1$. Next, use the
approximate continuity of $f$ at $x_{n+1}$ to find an interval
$(a_{n+1},b_{n+1})\subset (a_n,b_n)$ containing $x_{n+1}$ such that
(d), (e) and (f) are satisfied. The induction is complete.

Let
\[
\{x\} = \bigcap_{n\in\smallnats}[a_n,b_n].
\]
We show that there is an increasing sequence $\{n_i\}_{i\in\smallnats}$ 
of natural numbers 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 dispersion 
point of $\inv{f}\left(\bigcup_{i\in\smallnats}I_{n_i}\right)$.
\end{description}
This implies that $f\not\in C_{\cal ND}$.
 
First notice that $x$ is not a dispersion point of
$\inv{f}\left(\bigcup_{i\in\smallnats}I_{n_i}\right)$ for every sequence
$\{n_i\}_{i\in\smallnats}$ as, by condition (f),
\[
\lim_{i\to\infty}\frac{\measure{f^{-1}(I_{n_i})\cap (a_{n_i},b_{n_i})}}
{\measure{(a_{n_i},b_{n_i})}} =1.
\]

To find an
increasing sequence $\{n_i\}_{i\in\smallnats}$ of natural numbers such
that condition (2) is satisfied 
we will consider two cases.
 
\medskip

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
\[
\frac{\beta_i - \alpha_i}{d_i - c_i} = 
\frac{\measure{I_{n_i}}}{\measure{J_{n_i}}} <
\frac{1}{2^{n_i}}.
\]
Let $z_n$ be the common center of $I_n$ and $J_n$, for $n\ge 0$. Then
\[
\lim_{i\to\infty}\frac{\beta_i - \alpha_i}{\beta_i}\le\lim_{i\to\infty}
\frac{\beta_i - \alpha_i}{z_{n_i}}
\le\lim_{i\to\infty}\frac{\beta_i - \alpha_i}{z_{n_i} - c_i} =
2\lim_{i\to\infty}\frac{\beta_i - \alpha_i}{d_i - c_i} = 0.
\]
The above allows us to choose a subsequence of 
$\{n_i\}_{i\in\smallnats}$ satisfying the assumptions of Proposition
\ref{sequence}.
 
\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., a 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 Lemma \ref{CndinBstar}.

\bigskip

Now, we are ready to prove our main theorem.

\theorem{main}{$C_{\cal ND}=\const$.}

Proof. Evidently, $\const\subset C_{\cal ND}$. 
To prove the other inclusion, let $f\in C_{\cal ND}$. By Lemma \ref{CndinBstar}
the set 
\[
U=\interior{\{x\in\reals\colon f \mbox{ is continuous at } x\}}
\]
is dense. Notice that $\complement{U}$ does not have any isolated points,
because the approximately continuous function $f$ has the Darboux
property. Thus, the set $P=\complement{U}$ is perfect. We prove that
$P=\emptyset$. By way of contradictions let us assume that $P\neq\emptyset$ and
let $\{(a_n,b_n)\colon n\in\nats\}$ be an enumeration of all components of $U$.
Notice that, by Lemma \ref{continuous}, $f$ is constant on any interval
$(a_n,b_n)$ and, by the Darboux property, also on $[a_n,b_n]$.

Now, let us use Lemma \ref{CndinBstar} for $f$ and $P$. Then, there is a
nonempty portion $Q=P\cap(c,d)$ on which $f$ is continuous. 
The set $P$ is nowhere dense, so there exists an $n$ such that
$(c,d)\cap(a_n,b_n)\neq\emptyset$. Then, $(c,d)\cup(a_n,b_n)$ is an interval
properly containing $(a_n,b_n)$. We will obtain a contradiction
with  the assumption that $(a_n,b_n)$ is a component of $U$ by showing that $f$
is continuous on $J=(c,d)\cup(a_n,b_n)$. So, let $x\in J$.
If $x\in U$, then evidently $f$ is continuous at $x$.
If $x\in P$, then choose a sequence 
$\{x_i\}_{i\in\smallnats}$ converging to $x$ and define
\[
y_i=\left\{
\begin{array}{ll}
x_i & \mbox{if }\  x_i\in P \\
a_n & \mbox{for }  x_i\in (a_n,b_n).
\end{array}
\right.
\]
Then, $y_i\in P$, $f(x_i)=f(y_i)$ for $i\in\nats$ and $\lim_{i\to\infty}y_i=x$.
Moreover,
\[
\lim_{i\to\infty}f(x_i)=\lim_{i\to\infty}f(y_i)=f(x)
\]
as $\fP$ is continuous at $x$. Hence, $f$ is continuous at $x$.

This finishes the proof of Theorem \ref{main}.

\bibliographystyle{plain}
%\bibliography{Cnd}

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\end{document}
