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\markright{Various continuities}
 
\marginlabels{n}
 
\title{
Various continuities with the density, \idens\ and ordinary topologies
on {\amssymbig R}}

\date{}

\begin{document}


\author{Krzysztof Ciesielski, Department of Mathematics, West Virginia
University, Morgantown, WV 26506\footnote{Partially supported by a West
Virginia University research grant.} \and
Lee Larson, Department of Mathematics,
University of Louisville, Louisville, KY 40292\footnote{This author was given
support by a University of Louisville College of Arts and Sciences grant.}}

\maketitle 


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

This paper is concerned with four topologies on the
real line: the ordinary topology, density topology, \idens\ topology and
deep-\idens\ topology. Using different topologies on the domain and range,
these four topologies determine sixteen different ways a real function can be
continuous. These continuities are examined and their relationships are
determined. In addition, they are compared with the classes of analytic,
\cinfinity, Baire*1, and Baire~1 functions.

The notation used throughout this paper is standard. In particular,
\reals\ stands for the set of real numbers and $\nats=\{1,2,3,\ldots\}$.
The symmetric difference of the sets $A$ and $B$ is denoted by $A\triangle B$
and the complement of $A\subset\reals$ by \complement{A}. 
The symbols \Lebesgue\ and
\BaireSets\  stand for the families of subsets of \reals\ 
which are 
Lebesgue measurable and have the Baire property,
respectively. \N\ and \I\ denote the ideals of Lebesgue measure zero and 
first category subsets of \reals.
If $A\in\Lebesgue$, then
its Lebesgue measure is denoted by \measure{A}.

It is easy to see that for any set
$B\in\BaireSets$ there 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 symbol \dist{A}{B}\ stands for the Euclidean distance between the sets
$A$ and $B$; i.e., $\dist{A}{B} = \inf\{|x - y|: x\in A, y\in B\}$.
For $a\in\reals$ and $E\subset\reals$, define
$E - a = \{x - a\in\reals\colon x\in E\}$ and 
$aE =\{ax\in\reals\colon x\in E\}$.

The symbols \dbone, $\cal C$, \cinfinity\ and $\cal A$ stand for the classes of
Darboux Baire one,
ordinary continuous, infinitely many times differentiable and
analytic functions from \reals\ to \reals, respectively.
\dbstar\ denotes
functions in the Darboux Baire*1 class; i.e., the class of all 
Darboux 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}. 

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 $a < b_{n+1} < a_n < b_n$ for
$n\in\nats$ and $\lim_{n\to\infty}{a_n} = a$. In the case when $a=0$ we simply
call it a {\em a right interval set}. Similarly we define 
a {\em left interval set}
of a point $a\in\reals$. We say that $E$ is an {\em interval set\/} 
if it is the union
of a right interval set and a left interval set.


\section{\hskip -1em{.} Topologies}

The ordinary topology on \reals\ is denoted by \ordinarytop.

Next, the density topology, \idens\ topology and deep \idens\ topology
are defined.
For this, are needed the notions 
of a density point, \idens\ point and deep-\idens\ point.
(See \cite{Oxtoby:MeasCat,Wilczynski:CatAn}. For a discussion of these ideas
read \cite{CL:Refinements}.)

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 by \densitypts{A}.
It is a consequence of the Lebesgue density theorem that for a measurable set
$A$ almost every point of $A$ is a density point of $A$. 
Therefore, whenever $A$ is measurable, then so is \densitypts{A}\ and
$\measure{A\triangle\densitypts{A}}=0$
\cite[p. 107]{WheedenZygmund:MeasureandIntegral}.
It is also easy to see that
\begin{equation}\label{con_dens}
\densitypts{A}=\densitypts{B}\,\,\,
\mbox{ for every $A,B\in\Lebesgue$ such that }\,\,\,  A\triangle B\in\N. 
\end{equation}
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\/}.

If $A\in\BaireSets$, then $0$ is
an {\em \idens\ point\/} of $A$ if
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 $a$ is an 
\idens\ point of $A\in\BaireSets$ if $0$ is an \idens\ point of $A-a$.
The set of all \idens\ points of $A\in\BaireSets$ is denoted by 
\idensitypts{A}. It is not difficult to see that 
$A\triangle\idensitypts{A}\in\I$ for every $A\in\BaireSets$
\cite[Theorem 3]{Wilczynski:CatAn} and that 
\begin{equation}\label{con_idens}
\idensitypts{A}=\idensitypts{B}\,\,\,
\mbox{ for every $A,B\in\BaireSets$ such that }\,\,\,  A\triangle B\in\I. 
\end{equation}
The family of sets
\[
\itopology=\{A\in\BaireSets\colon A\subset\idensitypts{A}\}
\]
forms a topology on \reals\ \cite{PWW:CatAn,Wilczynski:CatAn} called
the {\em \idens\ topology\/}.

Finally, 
a point $a\in\reals$ is a {\em deep-\idens\ point\/}
\cite{Wilczynski:CatAn} of an $A\in\BaireSets$ if there exists a
closed set $F\subset A\cup \{a\}$ such that $a$ is an \idens\
point of $F$.
The set of all deep-\idens\ points of $A\in\BaireSets$ is denoted by 
\deepidensitypts{A}.
The family of sets
\[
\deepitopology=\{A\in\BaireSets\colon A\subset\deepidensitypts{A}\}
\]
forms a topology on \reals\ 
called the {\em deep-\idens\ topology\/} 
\cite{Lazarow:Coarsest,Wilczynski:CatAn}.

A point
$x$ is a {\em dispersion} ({\em \I-dispersion}, {\em deep-\I-dispersion})
point of $A$ if $x$ is a density (\idens\, deep-\idens\ ) point of
\complement{A}. In particular, $0$ is an \I-dispersion point of $A$ if
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}
and $x$ is a deep-\I-dispersion point of $A$ if there exists an open
$U\supset (A\setminus\{x\})$ such that $x$ is an \I-dispersion point of $U$.

\bigskip

We also need the following fact that easily follows
from either of \cite[Lemma 2.4]{CL:Baire},
\cite[Theorem 1]{PWW:RemIDens} or \cite[Theorem 2]{Wilczynski:CatAn}.

\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, $E^c\in\deepitopology$.
}
 
The next theorem summarizes the topological properties of the 
topologies defined above.

\theorem{topolprop}{The topologies \densitytop, \itopology\ and \deepitopology\
on \reals\ have the following properties:
\begin{description}
\item[(i)] $\ordinarytop\subset\densitytop$ 
and $\ordinarytop\subset\deepitopology\subset\itopology$;
in particular, \densitytop, \itopology\ and \deepitopology\ are Hausdorff;
\item[(ii)] a subset $C$ of \reals\ is closed and nowhere dense with respect
to \densitytop\ if, and only if, $C\in\N$;
a subset $C$ of \reals\ is closed and nowhere dense with respect
to \itopology\ if, and only if, $C\in\I$;
\item[(iii)] \densitytop\ and \deepitopology\ are 
completely regular; \itopology\ is not regular;
\item[(iv)] \densitytop\ and \itopology\ are not separable;
on the other hand, \deepitopology\ is separable; moreover, every set $D$ which
is dense with respect to \ordinarytop\ is also dense with respect to
\deepitopology;
\item[(v)] every closed subinterval of \reals\ is connected in the
topologies \densitytop, \itopology\ and \deepitopology;
\item[(vi)] a set $A$ is compact with respect to any
of the topologies
\densitytop, \itopology\ and \deepitopology\ if, and only if, it
is finite.
\end{description}
}

Proof. (i) is obvious from the definitions. (ii) follows immediately from
(\ref{con_dens}) and (\ref{con_idens}).
The first part of (iii) can be found in \cite{Zachorski}
or \cite{GNN:densitytop} for the density topology and in \cite{Lazarow:Coarsest}
or \cite{Wilczynski:CatAn} for the deep-\idens\ topology.
The second part follows from the
fact that any set $D$ which is dense with respect to \ordinarytop\ cannot
be separated in $\itopology$ from any point
$x\in\reals\setminus D$ \cite[Theorem 5]{PWW:CatAn}.
The above, together with the regularity of \deepitopology, 
also implies the second part
of (iv), while its first part follows immediately from (ii). 
A proof of (v) can be found in \cite{GW:AppContTrans} in the 
case of the density
topology and in \cite[Corollary 2]{PWW:CatAn} for the \idens\ topology.
The deep-\idens\ topology case follows from that of \idens\ topology.
To argue condition (vi) we need to show that an arbitrary infinite set
is not compact with respect to any of \densitytop, \itopology\ or 
\deepitopology. So, let $A$ be an infinite set and let us choose 
a set $S=\{c_n\colon n\in\nats\}\subset A$ such that
sequence $\{c_n\}_{n\in\smallnats}$ is monotone.
By (ii), $S$ is closed with respect to \densitytop\ and \itopology.
It is also closed in \deepitopology\, if $\{c_n\}_{n\in\smallnats}$ is unbounded.
If $\{c_n\}_{n\in\smallnats}$ is bounded, say by $c$, and if
we additionally assume that
\[
\lim_{n\to\infty}\frac{c_{n+1}-c}{c_n-c} = 0,
\]
then, using Proposition \ref{sequence}, we can also show that
$S$ is closed in \deepitopology. Now, the family of sets
$W_m=\reals\setminus\{c_n\colon n\geq m\}\in\densitytop\cap\deepitopology$ 
forms an infinite open cover of $A$ without a finite subcover.

\bigskip

In the sequel the following two lemmas are needed.

\lemma{lem:denseseq}{
There exists a decreasing sequence $S=\{s_n\}_{n\in\smallnats}$ converging to $0$
such that $0$ is not an \I-dispersion point of any set $B\in\itopology$
with the property that 
$B\cap[s_{n+1},s_n]\neq\emptyset$ for every $n\in\nats$.
In particular, if $B\supset S$ then $0\not\in\idensitypts{\complement{B}}$.
}
 
Proof. Let
\[
D_n=\{\frac{i}{2^n}\colon i=1,2,\ldots,2^n\}\subset (0,1]
\]
and let
\[
S=\bigcup_{n\in\smallnats} \frac{1}{n} D_n.
\]
Notice that $S\subset (0,1]$ and that $S\setminus (0,\frac{1}{n}]$ is finite
for every $n\in\nats$. Thus, S can be enumerated as a 
decreasing sequence converging to $0$.

Now, let $B$ satisfy our assumptions. Then, $\tilde B$ also satisfies them
and we may assume that $B=\tilde B$. Moreover, 
\[
nB\cap \left[\frac{i}{2^n},\frac{i+1}{2^n}\right]\neq\emptyset
\]
for every $i,n\in\nats$, $i<2^n$. 
In particular,
for every increasing sequence $\{n_{m_p}\}_{p\in\smallnats}$ of positive
integers, the set $\Cup_{p\in\smallnats} n_{m_p} B$ 
is dense and open. 
Hence, $\limsup_{p\in\smallnats} n_{m_p} B$ 
contains a dense \Gdelta\ subset of $(0,1)$.
Thus, by (\ref{con_idisp}), $0$ cannot be an \idisp\ point of $B$.

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

Proof. According to
Zaj\'{\i}\v{c}ek \cite[Theorem (5)]{Zajicek:AlternativeDefinition},
it is enough to prove that
for every $c\in(0,1)$ there exist $\e>0$ and $\d>0$ 
such for any $x\in(0,\d)$ there exists a closed interval 
$I\subset\complement{Q}\cap (cx,x)$ such that $\measure{I}\geq x\e$.

Let $c\in(0,1)$, $p=\min P$ and let $\d>0$ be such that
\begin{equation}\label{eq:ddef}
\frac{b_{n+1}}{b_n}<pc\ \mbox{ for every }\ 
n\in\nats\ \mbox{ for which }\ pb_n\leq \d.
\end{equation}

Moreover, let $\e_0>0$ be a 
number such that for every interval $K\subset(0,1)$ of length 
$\geq p(1-c)/2$ there exists a closed interval 
$J\subset K\setminus P$ of length $\geq \e_0$. 
Such a number can be found, by partitioning
$(0,1)$ into intervals $J_1,\ldots,J_k$, of length 
$< p(1-c)/4$ and defining 
\[ 
\e_0<\min\{\sup\{b-a\colon [a,b]\subset J_i\setminus P\}\colon 1\leq i\leq k\}.
\]
Put $\e=\min\{\e_0/2,(1-c)/2\}$.

Now, let $x\in(0,\d)$ and define
\[
m=\min\{n\colon pb_n\leq x\}.
\]
Then, by (\ref{eq:ddef}), $b_{m+1}<pb_mc\leq xc$. In particular,
\[
\complement{Q}\cap (cx,x)=(cx,x)\setminus b_m P.
\]
Let $a$ be the middle point of $(cx,x)$. 
If $pb_m>a$ then $I=[xc,a]$ works, as $\measure{I}=x(1-c)/2\geq x\e$.
Similarly, if $b_m<a$ then it suffices to set $I=[a,x]$.

So, let us assume that $pb_m\leq a\leq b_m$.
Then $x\leq 2a\leq 2b_m$ and
$pb_m<x=\frac{2}{1-c}(x-a)$, as $2(x-a)=(x-xc)$.
In particular, 
$(x-a)/b_m>p(1-c)/2$. 
Thus, by the definition of $\e_0$, there
exists a closed interval $J\subset \frac{1}{b_m}(a,x)\setminus P$ of the length
$\geq \e_0$. Hence, 
\[
I=b_mJ\subset(a,x)\setminus b_mP\subset(cx,x)\setminus b_m P
=\complement{Q}\cap(cx,x)
\]
has length $\geq \e_0 b_m=2\e_0 b_m/2\geq x\e$.
This finishes the proof.

\bigskip

This section is concluded with the fact that Theorem \ref{topolprop}(i)
lists all possible inclusions between the topologies
\ordinarytop, \densitytop, \itopology\ and \deepitopology. 

\theorem{th:densvsidens}{If ${\cal P}(\reals)$ stands for 
the discrete topology on \reals\ then
\[
\begin{array}{ccccccc}
\ordinarytop\cap\densitytop & \subset & \deepitopology\cap\densitytop &
\subset & \itopology\cap\densitytop & \subset & \densitytop \nonumber \\

\| & & \cap & & \cap & & \cap \nonumber \\

\ordinarytop & \subset & \deepitopology &
\subset &   \itopology & \subset & {\cal P}(\reals) \nonumber
\end{array}
\]
Moreover, all the inclusions are proper.
}
 
Proof. All the inclusions follow immediately from Theorem \ref{topolprop}(i).

To show that the horizontal inclusions are proper, it is enough to argue for the
inclusions in the first row. Thus, it is enough to prove that 
$\deepitopology\cap\densitytop\not\subset\ordinarytop$,  
$\itopology\cap\densitytop\not\subset\deepitopology$  and  
$\densitytop\not\subset\itopology$.
To show that the vertical inclusions are proper we will show that
$\deepitopology\not\subset\densitytop$.

$\deepitopology\cap\densitytop\not\subset\ordinarytop$. Let 
$E = \bigcup_{n\in\smallnats}[a_n,b_n]$ be as in Proposition \ref{sequence}.
It is easy to check that $0$ is also a dispersion point of $E$. 
So, $\complement{E}\in\deepitopology\cap\densitytop$ while, evidently,
$\complement{E}\not\in\ordinarytop$.

$\itopology\cap\densitytop\not\subset\deepitopology$.  
Let  $E=\reals\setminus\rationals$. By Theorem \ref{topolprop}(ii)
\rationals\ is closed in 
\itopology\ and \densitytop\ and so $E\in\itopology\cap\densitytop$. But,
\rationals\ is also dense in \deepitopology. Hence, $E\not\in\deepitopology$.

$\densitytop\not\subset\itopology$.  Let $C$ be a nowhere dense set of positive
Lebesgue measure. By the Lebesgue density theorem,
$D=C\cap\densitypts{C}\in\densitytop$ and $D\neq\emptyset$. Moreover, by Theorem
\ref{topolprop}(ii), $D$ is  nowhere dense in \itopology, so
$D\not\in\itopology$.

$\deepitopology\not\subset\densitytop$. Let $C\subset [\frac{1}{2},1]$
be a closed nowhere dense set with positive Lebesgue measure and 
let $\{b_n\}_{n\in\smallnats}$ be a decreasing sequence of positive numbers
such that $\lim_{n\to\infty}b_{n + 1}/b_n = 0$. Then, 
by Lemma \ref{lem:deepidisp}, $0$ is a deep-\I-dispersion point of
$E=\bigcup_{n\in\smallnats}b_nC$ and $E^c\in\deepitopology$.
On the other hand, for every $n\in\nats$,
\[
\frac{m\left(E\cap (0,b_n)\right)}{b_n} =
m\left(\inv{b_n} E\cap (0,1)\right) \geq m(C) >0.
\]
Thus, $0$ is not a dispersion point of $E$ and $E^c\not\in\densitytop$.


\section{\hskip -1em{.} Classes of functions}

For $\J,\K\in\{\N,\I,\D,\O\}$ let
$C_{\cal JK}$ stand for the family of all continuous functions
$f\colon (\reals,\T_{\cal J})\to(\reals,\T_{\cal K})$. It is easy to see
that we have sixteen different classes:
\[
\begin{array}{ccccccl}
C_{\cal OO} & \,\,\, & C_{\cal DO} & & C_{\cal IO} & & C_{\cal NO} \nonumber \\
C_{\cal OD} & & C_{\cal DD} & \,\,\, & C_{\cal ID} & & C_{\cal ND} \nonumber \\
C_{\cal OI} & & C_{\cal DI} & & C_{\cal II} & \,\,\, & C_{\cal NI} \nonumber \\
C_{\cal ON} & & C_{\cal DN} & & C_{\cal IN} & & C_{\cal NN}. \nonumber \\
\end{array}
\]
Some of these classes have the following names associated with them:
\begin{description}
\item{$C_{\cal OO}=\cal C$} -- the class of (ordinary) continuous functions; 
\item{$C_{\cal NN}$} -- the class of density continuous functions; 
\item{$C_{\cal II}$} -- the class of \idens\ continuous functions; 
\item{$C_{\cal DD}$} -- the class of deep-\idens\ continuous functions; 
\item{$C_{\cal NO}$} -- the class of approximately continuous functions; 
\item{$C_{\cal IO}$} -- the class of \I-approximately continuous functions.
\end{description} 
We start our investigation by showing that the other classes 
coincide with them or with the class of all constant functions.

\theorem{const}{If \const\ stands for the class of all constant functions then
\[
C_{\cal OD}=C_{\cal OI}=C_{\cal ON}=C_{\cal DI}=C_{\cal DN}=C_{\cal IN}=
C_{\cal ND}=C_{\cal NI}=\const.
\]
}

Proof. It is obvious that the constant functions are members of all these
classes, as they are universally continuous. Thus, we need only show that
the functions from the classes listed above $C_{\cal JK}$ are constant.

By way of contradiction, let us assume that there is a nonconstant function
$f$ in some class $C_{\cal JK}$ from the theorem.

Case 1$^{\rm o}$. $C_{\cal JK}\in\{C_{\cal OD},C_{\cal OI},C_{\cal ON}\}$. 
By our assumption there are $a<b$ such that $f([a,b])$ has more than one
point. But, by Theorem \ref{topolprop}(vi),
$f([a,b])$ is finite, because it is the continuous image of compact set.
This means that $f([a,b])$ is disconnected, 
which is impossible, because the
interval $[a,b]$ is connected in \ordinarytop.

Case 2$^{\rm o}$. $C_{\cal JK}\in\{C_{\cal DI},C_{\cal DN}\}$.
Let $D$ be a countable dense subset of $(\reals,\ordinarytop)$ such that
$A=f(D)$ has more than one point. By Theorem \ref{topolprop}(ii),
$A$ is closed and discrete in \itopology\ and \densitytop. We will show that
$B=\inv{f}(A)$ is not closed in \deepitopology.
First notice that $A$, 
as a discrete set having more than one element, is disconnected
and thus, $B$ is disconnected as the continuous preimage of disconnected set.
So, by Theorem \ref{topolprop}(v), $B\neq\reals$. But, $B\supset D$ is
dense in \deepitopology, by Theorem \ref{topolprop}(iv). Hence,
$B$ is not closed in \deepitopology. 

Case 3$^{\rm o}$. $C_{\cal JK}=C_{\cal IN}$. As in the previous case, let
$D$ be a countable dense subset of $(\reals,\ordinarytop)$ such that
$A=f(D)$ has more than one point and let 
$a\not\in B=\inv{f}(A)$. Without any loss of generality we may assume
that $a=0=f(a)$. Let $S=\{s_n\}_{n\in\smallnats}$ be as in Lemma 
\ref{lem:denseseq} and, for every $n\in\nats$, let $d_n\in B\cap[s_{n+1},s_n]$.
Now, choose $a_n<f(d_n)<b_n$ such that $0$ is a dispersion point of
$E=\Cup_{n\in\smallnats} [a_n,b_n]$. 
Let $U\in\densitytop$ be such that $0\in U\subset\complement{E}$.
But then, by  Lemma 
\ref{lem:denseseq}, $0$ is not an \I-dispersion point of 
$\inv{f}(E)\subset\inv{f}(\complement{U})$; i.e.,
$\inv{f}(U)\not\in\itopology$.

Case 4$^{\rm o}$. $C_{\cal JK}\in\{C_{\cal ND},C_{\cal NI}\}$. As 
$C_{\cal NI}\subset C_{\cal ND}$ we may assume that 
$C_{\cal JK}=C_{\cal ND}$. This case can be found in \cite{KC:Cndfunctions}.

\bigskip

Theorem \ref{const} shows that the classes listed above are trivial.
Thus, we need no longer consider them. Similarly, we can ignore 
the class \deepiappr\ because of the theorem below 
\cite[Theorem 2]{Lazarow:Coarsest}.

\theorem{iapp}{ $\iappr=\deepiappr$.}

To argue the next theorem we use the following easy lemma. (See
\cite[Theorems 14.12 and  8.10]{Willard:GenTop}.)

\lemma{lem:will}{
Let $X$ be a completely regular topological space and $Y$ any topological
space. The following are equivalent:
\begin{description}
\item[(i)]  the function $f\colon Y\to X$ is continuous;
\item[(ii)] the function $g\circ f\colon Y\to [0,1]$ is continuous for every
continuous $g\colon X\to [0,1]$.
\end{description}
}
 
\theorem{deepidens}{ $\deepidenscont=C_{\cal ID}$.}

Proof. Let 
$f\in C_{\cal ID}$. We have to prove that $f$ is continuous as
a function from $(\reals,\deepitopology)$ to $(\reals,\deepitopology)$.
But, evidently, $g\circ f\in\iappr=\deepiappr$ for every 
$g\in\deepiappr$. Hence, by Lemma \ref{lem:will}, $f\in\deepidenscont$.

\bigskip

By the previous theorem we
need no longer consider the class $C_{\cal ID}$ in our discussion.
The other classes will be discussed in the following sections.


\section{\hskip -1em{.} Main Theorem}

We start this section with the following list of inclusions.

\theorem{thm:incl}{ The following relations hold.
\begin{description}
\item[(i)]  $\cal A\subset\denscont\subset\appr\subset\dbone$;
\item[(ii)] $\denscont\subset\dbstar$;
\item[(iii)] $\cal C\subset\appr$;
\item[(iv)]
$\cal A\subset\idenscont\subset\deepidenscont\subset\iappr\subset\dbone$;
\item[(v)] $\deepidenscont\subset\dbstar$;
\item[(vi)] $\cal C\subset\iappr$;
\item[(vii)] 
$\cal A\subset\cinfinity\subset\cal C\subset\dbstar\subset\dbone$.
\end{description}
}

Proof. (i). The inclusions $\cal A\subset\denscont$ and
$\appr\subset\dbone$ can be found in \cite[Corollary 3]{CL:SpDensCont}
and \cite[Theorem 5.5(b)]{Bruckner:DiffReal}, respectively. 
$\denscont\subset\appr$ follows from Theorem \ref{topolprop}(i).

(ii) can be found in \cite[Theorem 3]{CLO:DensContCont}.

(iii) follows immediately from Theorem \ref{topolprop}(i).

(iv). The inclusions $\cal A\subset\idenscont$ and
$\iappr\subset\dbone$ can be found in \cite{CL:Analytic}
and \cite[Theorem 3.2]{CL:Baire}, \cite[Theorem 8]{PWW:CatAn}, respectively. 
$\deepidenscont\subset\iappr$ follows from Theorems \ref{topolprop}(i) and
\ref{deepidens}.
To argue the inclusion $\idenscont\subset\deepidenscont$ let $f\in\idenscont$.
Then, evidently, $g\circ f\in\iappr=\deepiappr$ for every 
$g\in\deepiappr=\iappr$. Hence, by Lemma \ref{lem:will}, 
$f\in\deepidenscont$.

(v) can be found in \cite[Theorem 4.2]{CL:Baire}.

(vi) follows immediately from Theorem \ref{topolprop}(i).

(vii) is well known.

\bigskip

The inclusions between the discussed classes are summarized in
the following charts.
We use the convention
that, for the classes of functions $\cal X$ and $\cal Y$, the symbol ${\cal
X\,Y}$  stands for the class $\cal X\cap\cal Y$.

\theorem{thm:idensVSdens}{The following inclusions hold
\[
\begin{array}{ccccccc}

            
&         &  
&         & \iappr 
& \subset &                      \dbone                   \nonumber \\

  & &   & & \cup & & \cup                                 \nonumber \\

            \idenscont        
& \subset & \deepidenscont   
& \subset & \iappr           \,\,\dbstar 
& \subset &                      \dbstar                  \nonumber \\

\cup & & \cup & & \cup & &                                \nonumber \\

            \idenscont        \,\,\cal C
& \subset & \deepidenscont    \,\,\cal C
& \subset &                       \cal C
&         &                                               \nonumber \\

\cup & & \cup & & \cup & &                                \nonumber \\

            \idenscont        \,\,\cinfinity
& \subset & \deepidenscont    \,\,\cinfinity
& \subset &                       \cinfinity
&         &                                               \nonumber \\

\cup & &      & &      & &                                \nonumber \\

            \cal A
&   &       
&   &       
&   &                                                     \nonumber \\


\end{array}
\]
and the relativization of the above chart to (i.e., intersecting
each of the element of the chart by) \appr\ and \denscont\ gives respectively
\[
\begin{array}{ccccccc}

            
&         & 
&         & \iappr                           \,\,\appr
& \subset &                                      \appr    \nonumber \\

   & &    & & \cup & & \cup                               \nonumber \\

            \idenscont                       \,\,\appr
& \subset & \deepidenscont                   \,\,\appr
& \subset & \iappr           \,\,\dbstar     \,\,\appr 
& \subset &                      \dbstar     \,\,\appr    \nonumber \\

\cup & & \cup & & \cup & &                                \nonumber \\

            \idenscont        \,\,\cal C
& \subset & \deepidenscont    \,\,\cal C
& \subset &                       \cal C
&         &                                               \nonumber \\

\cup & & \cup & & \cup & &                                \nonumber \\

            \idenscont        \,\,\cinfinity
& \subset & \deepidenscont    \,\,\cinfinity
& \subset &                       \cinfinity
&         &                                               \nonumber \\

\cup & &      & &      & &                                \nonumber \\

            \cal A 
&   &       
&   &       
&   &                                                     \nonumber \\


\end{array}
\]
and
\[
\begin{array}{ccccccc}

            
&         & 
&         & \iappr                           \,\,\denscont
& \subset &                                  \,\,\denscont    \nonumber \\

   & &    & & \| & & \|                                       \nonumber \\

            \idenscont                       \,\,\denscont
& \subset & \deepidenscont                   \,\,\denscont
& \subset & \iappr                           \,\,\denscont 
& \subset &                                      \denscont    \nonumber \\

\cup & & \cup & & \cup & &                                    \nonumber \\

            \idenscont        \,\,\cal C     \,\,\denscont
& \subset & \deepidenscont    \,\,\cal C     \,\,\denscont
& \subset &                       \cal C     \,\,\denscont
&         &                                                   \nonumber \\

\cup & & \cup & & \cup & &                                    \nonumber \\

            \idenscont        \,\,\cinfinity \,\,\denscont
& \subset & \deepidenscont    \,\,\cinfinity \,\,\denscont
& \subset &                       \cinfinity \,\,\denscont
&         &                                                   \nonumber \\

\cup & &      & &      & &                                    \nonumber \\

            \cal A
&   &       
&   &       
&   &                                                         \nonumber \\


\end{array}
\]
Moreover, the inclusions in the charts and the nontrivial inclusions
between corresponding parts of the charts, i.e.,
$\cal X\,\,\appr\subset\cal X$ and
$\cal Y\,\,\denscont\subset\cal Y\,\,\appr$ for
\[
\cal X\in\{
\idenscont,
\deepidenscont,
\iappr            \,\,\dbstar,
\dbstar,
\iappr,
\dbone
\}
\]
and
\[
\cal Y\in\{
\idenscont        \,\,\cinfinity,
\deepidenscont    \,\,\cinfinity,
\cinfinity,
\idenscont        \,\,\cal C,
\deepidenscont    \,\,\cal C,
\cal C,        
\idenscont,
\deepidenscont,
\iappr            \,\,\dbstar,
\dbstar,
\},
\]
are proper.
}

Proof. The fact that the second and third charts are appropriate relativizations
follows from Theorem \ref{thm:incl}. Theorem \ref{thm:incl} also implies
all the inclusions.

To prove 
that
the inclusions in the main charts are proper it is enough to
reduce our task to the following inclusions. 
To argue for the inclusions between first and second rows it is enough to
show that 
  $\iappr\,\,\appr\not\subset\dbstar$.
For the other inclusions it is enough to consider 
the inclusions in the third chart.
The vertical containments are 
proper because
of
  $\idenscont\,\,\cinfinity\,\,\denscont\not\subset\cal A$,
  $\idenscont\,\,\cal C\,\,\denscont\not\subset\cinfinity$ and
  $\idenscont\,\,\denscont\not\subset\cal C$.
To prove that the horizontal inclusions are proper it is enough to show that
  $\denscont\not\subset\iappr$,
  $\cinfinity\,\,\denscont\not\subset\deepidenscont$ and
  $\deepidenscont\,\,\cinfinity\,\,\denscont\not\subset\idenscont$.
To argue the additional part it is enough to show that
  $\idenscont\not\subset\appr$ and 
  $\idenscont\,\,\cinfinity\not\subset\denscont$. 

\medskip

  $\iappr\,\,\appr\not\subset\dbstar$. See Example \ref{notbstar}.

  $\idenscont\,\,\cinfinity\,\,\denscont\not\subset\cal A$. 
  See Example \ref{notanal}.

  $\idenscont\,\,\cal C\,\,\denscont\not\subset\cinfinity$. 
  See Example \ref{contnotcinf}.

  $\idenscont\,\,\denscont\not\subset\cal C$. See Example \ref{notcont}.

  $\denscont\not\subset\iappr$. See Example \ref{notiappr}.

  $\cinfinity\,\,\denscont\not\subset\deepidenscont$. 
  See Example \ref{cinfnotid}.

  $\deepidenscont\,\,\cinfinity\,\,\denscont\not\subset\idenscont$.
  See Example \ref{deepInotI}.

  $\idenscont\not\subset\appr$. See Example \ref{notappr}.
 
  $\idenscont\,\,\cinfinity\not\subset\denscont$. 
  See Example \ref{cinfnotdendcont}.

The examples needed for this proof are constructed in the next sections.


\section{\hskip -1em{.} Easy examples}

It is easy to see that linear functions are both density and \idens\
continuous. (This also follows immediately from \cite[Lemma
1]{CL:SpDensCont} and Theorem \ref{thm:lipschitz}.)
In particular, we can notice that

\proposition{plinear}{ If $f$ is piecewise linear, then 
$f\in\denscont\,\,\idenscont$.}

This gives us immediately

\example{contnotcinf}{There is an 
$f\in\idenscont\,\,\cal C\,\,\denscont\setminus\cinfinity$.}

Proof. The function $f(x)=2x+|x|$ satisfies the conditions.

\bigskip

It is also easy to construct the next example.

\example{notcont}{There is an 
$f\in\idenscont\,\,\denscont\setminus\cal C$.}

Proof. Let $E = \bigcup_{n\in\nats}[a_n,b_n]$ be a right interval set as in
Proposition \ref{sequence}. Then, $0$ is an \I-dispersion point of $E$.
It is also easy to see that $0$ is a dispersion point of $E$. In particular,
$\complement{E}\in\itopology\cap\densitytop$.
Define $f\colon\reals\to[0,1/2]$ by
\begin{equation}\label{def_notcont}
f(x)=\left\{
\begin{array}{ll}
0 & \mbox{if $x\not\in E$} \\
\frac{ \dist{\{x\}} {\textstyle (a_n,b_n)^c} }{b_n-a_n} & \mbox{for $x\in [a_n,b_n]$}.
\end{array}
\right.
\end{equation}
 
By Proposition \ref{plinear}, $f$ is density and \idens\ continuous at every
point $x\neq 0$. Also, $f$ is density and \idens\ continuous at $0$
because it is constant on $\complement{E}\in\itopology\cap\densitytop$, 
while $0\in\complement{E}$. Finally, $f$ is
not continuous because the value of $f$ equals $\frac{1}{2}$ at the
center of every $(a_n,b_n)$, while $f(0)=0$.
 
\bigskip

The next two examples are done in a manner very similar to the previous ones.

\example{notiappr}{There is an $f\in\denscont\setminus\iappr$.}

Proof. Let $S=\{s_n\}_{n\in\smallnats}$ be as in Lemma \ref{lem:denseseq}
and let $E = \bigcup_{n\in\smallnats}[a_n,b_n]$ be a right interval set 
such that $s_n$ is the center of each interval $[a_n,b_n]$ and 
$0$ is a dispersion point of $E$. (For example, taking $b_n-a_n=s_n/(n!)$.)
Define $f$ as in (\ref{def_notcont}). Then, as in Example \ref{notcont},
$f$ is density continuous. But $f$ is not \I-approximately continuous at $0$,
as 
$f(0)=0$ and $0$ is not an
\I-dispersion point of the open set 
$B=\inv{f}((1/4,\infty))\subset\complement{\inv{f}((-\infty,1/4))}$ containing
the set $S$.

\example{notappr}{There is an $f\in\idenscont\setminus\appr$.}

Proof. 
Let $C\subset [\frac{1}{2},1]$
be a closed nowhere dense set with positive Lebesgue measure and 
let $\{d_n\}_{n\in\smallnats}$ be a decreasing sequence of positive numbers
such that $\lim_{n\to\infty}d_{n + 1}/d_n = 0$. Then, 
by Lemma \ref{lem:deepidisp}, $0$ is a deep-\I-dispersion point of
$D=\bigcup_{n\in\smallnats}d_nC$. Thus, there exists an open set $U\supset D$
such that $0$ is an \I-dispersion point of $U$. Decreasing $U$, if necessary, 
we may assume that,
for some sequence $\{\e_n\}_{n\in\smallnats}$ of positive numbers, 
$U=\Cup_{n\in\smallnats}(a_n+\e_n,b_n-\e_n)$ and
$0$ is a deep-\I-dispersion point of a right interval set
$E=\Cup_{n\in\smallnats}[a_n,b_n]$. Then, 
$\complement{E}\in\itopology$ and, 
as in Theorem \ref{th:densvsidens}($\deepitopology\not\subset\densitytop$),
$0$ is not a dispersion point of $U$.
Define $f\colon\reals\to\reals$ by
\[
f(x)=\left\{
\begin{array}{ll}
0 & \mbox{if $x\not\in E$} \\
\frac{ \dist{\{x\}} {(a_n,b_n)^c} }{\e_n} & \mbox{for $x\in (a_n,b_n)$}.
\end{array}
\right.
\]
 
Then, as in Example \ref{notcont}, $f$ is \idens\ continuous. 
But $f$ is not approximately continuous at $0$,
as 
$f(0)=0$ and $0$ is not a
dispersion point of the set 
$\inv{f}([1,\infty))\supset U$.

\bigskip

For the next example we use the following theorem \cite{CL:Analytic}.

\theorem{th:Cinfinityex}{ Let $f\in\cal C^{\infty}$ be such that for every
$n\geq 0$
\[
f^{(n)}(0)=0 \,\,\, \mbox{ and } \,\,\, f^{(n)}( (0,\varepsilon_n)) \subset
(0,\infty) \,\,\, \mbox{ for some } \,\, \varepsilon_n > 0.
\]
Then $f$ is not deep-\idens\ continuous.
}

\example{cinfnotid}{There is an increasing homeomorphism 
$f\in\cinfinity\,\,\denscont\setminus\idenscont$.}

Proof. Define
\begin{equation}\label{def_convnotid}
f(x)= \left\{
\begin{array}{ll}
e^{-x^{-2}} & x>0 \\
0 & x=0 \\
-e^{-x^{-2}} & x<0 \\
\end{array}\right.
\end{equation}
It is known that $f\in\cal C^{\infty}$ and that $f^{(n)}(0)=0$ for all
$n\in\nats$. It is also easy to see that, for every $n\in\nats$, there exists
$\varepsilon_n > 0$ such that $f^{(n)}(x)>0$ for every $x\in (0,\varepsilon_n)$.
Thus, by Theorem \ref{th:Cinfinityex}, $f\not\in\idenscont$.
It is also known \cite[Theorem 1]{CL:SpDensCont} that functions convex
on open intervals are density continuous. This easily implies that
$f$ is unilaterally density continuous at every point and so $f\in\denscont$.

\bigskip

Aversa and Wilczy\'nski \cite{AversaWilczynski:Homeomorphisms} study 
the homeomorphisms which preserve \idens\ points. In our terminology,  
the homeomorphism $h$ preserves \idens\ points when
$\inv{h}\in\idenscont$. In particular, they proved the following theorem 
\cite[Corollary 1]{AversaWilczynski:Homeomorphisms}, which also follows 
easily from Theorem \ref{thm:genlipsch}.

\theorem{thm:lipschitz}{If $h$ is a homeomorphism such that $h$ and $h^{-1}$
fulfill a local Lipschitz condition, then $h$ and $h^{-1}$ are \idens\/
continuous.}

Now we are ready to present the next example.

\example{notanal}{There is an increasing homeomorphism 
$h\in\idenscont\,\,\cinfinity\,\,\denscont\setminus\cal A$.}

Proof. Let $h(x)=x+f(x)$, where $f$ is from (\ref{def_convnotid}).
The function $f$ is \cinfinity\ with a pole, so $f\in\cinfinity\setminus\cal A$
and also  $h\in\cinfinity\setminus\cal A$. But, evidently,
$h$ is a homeomorphism such that $h$ and $h^{-1}$
fulfill a local Lipschitz condition. So, by Theorem \ref{thm:lipschitz},
$h\in\idenscont$ and, by its density analog \cite[Lemma 1]{CL:SpDensCont}, 
$h\in\denscont$.

\bigskip

For the last example of this section we need the following obvious fact.

\proposition{pr:uniformconv}{ If $\{g_n\}$ is a sequence of \I-approximately
(approximately) continuous functions converging uniformly to $g$ then 
the function $g$ is
\I-approximately (approximately) continuous.}

Proof. This is well known for approximately continuous functions 
\cite[Theorem 5.7]{Bruckner:DiffReal}. 
The same easy proof also works for the \I-approximately continuous case.

\bigskip

\example{notbstar}{There is a $g\in\iappr\,\,\appr\setminus\dbstar$.}

Proof. Let $\{q_n\colon n\in\nats\}$ be an enumeration of \rationals. Let $f$
be as in (\ref{def_notcont}) and put
\[
h_n(x)=\frac{2}{3^n}f(x-q_n).
\]
Then it is easy to see that $h_n(\reals)=[0,1/3^n]$ and that $h_n$ is piecewise
linear at points $\neq q_n$ while $\osc(h_n,q_n)=1/3^n$.
Define 
$g_n=\Sigma_{i\leq n}h_i$ and let $g$ be the limit of $\{g_n\}$.
Notice that 
\[
|g-g_n|\leq \frac{1}{2}\frac{1}{3^n}
\]
so, $\{g_n\}$ converges to $g$
uniformly. The same argument as in Example \ref{notcont} shows that the
functions $g_n\in\iappr\,\,\appr$. Thus, by Proposition \ref{pr:uniformconv},
$g\in\iappr\,\,\appr$. Also notice that, for every $n\in\nats$, 
\[
\osc(h,q_n)=\osc(h_n+(g-g_n),q_n)
\geq\osc(h_n,q_n)-\sup_{x\in\smallreals}(g(x)-g_n(x)) \geq 
\frac{1}{3^n}-\frac{1}{3^{n+1}}>0
\]
so, $g$ is discontinuous at every rational number. But, 
Baire*1 functions are continuous on a dense open set. Thus,
$g\not\in\dbstar$. 


\section{\hskip -1em{.} \cinfinity\ deep-\idens\ continuous examples}

We start this section with the following technical lemma. In what follows
\nwdense\ stands for the ideal of nowhere dense subsets of \reals.

\lemma{lem:longlist}{
Let $B\in\BaireSets$. The condition that $0$ is an \idisp\/ point of $B$
is equivalent to any of the following:
 
\begin{description}

\item[(i)] for every increasing sequence $\{n_k\}_{k\in\smallnats}$ 
of natural numbers there exists a
subsequence $\{n_{k_p}\}_{p\in\smallnats}$ such that \[ 
\limsup_{p\to\infty}\left({n_{k_p}}\tilde B\right)\cap(-1,1)\in\I; \]
 
\item[(ii)] for every increasing sequence $\{n_k\}_{k\in\smallnats}$ 
of natural numbers there exists a
subsequence $\{n_{k_p}\}_{p\in\smallnats}$ such that \[
\limsup_{p\to\infty} \left(n_{k_p}\tilde{B}\right)\cap(-1,1)\in\nwdense; \]
 
\item[(iii)] for every increasing sequence $\{n_k\}_{k\in\smallnats}$ 
of natural numbers 
and every nonempty interval $(a,b)\subset(-1,1)$
there exists a nonempty subinterval $(c,d)\subset(a,b)$ and a subsequence
$\{n_{k_p}\}_{p\in\smallnats}$ such that for every $p\in\nats$ 
\[
(c,d)\cap n_{k_p}\tilde B = \emptyset; 
\]

\item[(iv)]  for every sequence $\{t_n\}_{n\in\smallnats}$ of
positive numbers diverging to 
infinity
and every nonempty interval $(a,b)\subset(-1,1)$
there exists a nonempty subinterval $(c,d)\subset(a,b)$ and a subsequence
$\{t_{n_k}\}_{k\in\smallnats}$ such that for every $k\in\nats$ 
\[
(c,d)\cap t_{n_{k}}\tilde B = \emptyset; 
\]
 
\item[(v)]  for every sequence $\{t_n\}_{n\in\smallnats}$ of
positive numbers diverging to  
infinity 
there exists a subsequence $\{t_{n_k}\}_{k\in\smallnats}$ 
such that for every nonempty interval
$(a,b)\subset(-1,1)$ there exists a nonempty subinterval $(c,d)\subset(a,b)$
and $k_0\in\nats$ such that for all $k\ge k_0$ 
\[ (c,d)\cap t_{n_{k}}\tilde B = \emptyset. \]
 
\end{description}
 }
 
Proof. By (\ref{con_idens}) and (\ref{con_idisp}) condition (i) is 
equivalent to the fact that $0$ is an \idisp\ point of $B$.  
To finish the proof, it will be shown that
(i) through (iv) each imply the next item and that (v)
implies (i).
 
(i)$\Rightarrow$(ii). 
Note that $\limsup_{p\to\infty}(n_{k_p}\tilde B)\cap(-1,1)$
is a \Gdelta\ set, and a \Gdelta\ set belongs to \I\ if, and only if,
it belongs to \nwdense.

(ii)$\Rightarrow$(iii).
The sets 
\begin{equation}\label{decreasingsets}
\{ \Cup_{p\ge m}n_{k_p}\tilde{B}: m\in\nats\}
\end{equation} 
are open and their intersection is nowhere dense.  If
each of these open sets were dense in $(a,b)$, then their intersection
would be a dense \Gdelta\ subset of $(a,b)$.  Since this contradicts
(ii), there must exist an $m_0\in\nats$ and $(c,d)\subset (a,b)$ such
that $(c,d)\cap \Cup_{p\ge m_0}n_{k_p}\tilde{B}=\emptyset$.
(iii) follows for the sequence $\{n_{k_p}\}_{p\geq m_0}$.
  
(iii)$\Rightarrow$(iv).
There is no generality lost with the assumptions that $t_1\ge 1$, that
$t_{n+1}-t_n\ge 1$ for all $n$ and $(a,b)\subset [a,b]\subset (0,1)$.  
Let $p_n=\lfloor t_n\rfloor$, where $\lfloor x\rfloor$ stands for the 
greatest integer function.  
The sequence of integers $\{p_n\}$ satisfies 
the conditions of (iii), so there must exist a subsequence 
$\{p_{n_k}\}$ of $\{p_n\}$ and a nonempty $(c,d)\subset(a,b)$
such that $p_{n_k}\tilde{B}\cap(c,d)=\emptyset$ for all $k\in\nats$.

Since $t_{n_k}/p_{n_k}\to 1$ as $k\to\infty$, there exists a $k_0$ such that
\[
1\le\frac{t_{n_k}}{p_{n_k}}<1+\frac{d-c}{3c},
\ \ \mbox{ for all } \ \  k\ge k_0. 
\]
Let $J=(c+(d-c)/3,d)$ and $k\ge k_0$.
Then
\[
\emptyset = \frac{t_{n_k}}{p_{n_k}}\left(
(c,d)\cap p_{n_k}\tilde{B}\right) =
\left(\frac{t_{n_k}c}{p_{n_k}},\frac{t_{n_k}d}{p_{n_k}}\right)
\cap t_{n_k}\tilde{B}
\supset J\cap t_{n_k}\tilde{B}.
\]
Part (iv) follows easily from this.
 
(iv)$\Rightarrow$(v). 
Without loss of generality we may assume that
$a$ and $b$ are rational. Let $\{(a_s,b_s)\}_{s\in\smallnats}$ be an 
enumeration
of all such intervals, and let us choose a sequence $\{t_n\}$.
The idea of the proof is an application of (iv) infinitely many
times, and diagonalization.
 
Let $m_k^0 = k$ for $k\in\nats$. We will, by induction on $s\in\nats$,
construct sequences $\{m_k^s\}_{k\in\smallnats}$ such that, 
for every $s\in\nats$,
$\{m_k^s\}_{k\in\smallnats}$ will be a subsequence of
$\{m_k^{s-1}\}_{k\in\smallnats}$, and there will be a nonempty 
interval $(c,d)$
(possibly dependent on $s$) contained in $(a_s,b_s)$ such that for all
$k\in\nats$
\[
(c,d)\cap(t_{m_k^s}\tilde B) = \emptyset.
\]
The construction is facilitated by (iv). Now, define 
$t_{n_k} = t_{m_k^k}$. 
Then,
for every $k_0\in\nats$, there exists 
$(c,d)\subset(a_{k_0},b_{k_0})$
such that for every $k \geq k_0$
\[
(c,d)\cap(t_{n_k}\tilde B) = \emptyset.
\]
This proves (v).

(v)$\Rightarrow$(i).
Let $\{n_k\}_{k\in\smallnats}$ be an increasing sequence of natural
numbers and let 
$\{(a_p,b_p)\}_{p\in\smallnats}$ be an
enumeration of all nonempty subintervals of $(-1,1)$ with rational
endpoints. Using (v), there 
is a subsequence $\{n_k^1\}$ of $\{n_k\}$ and a nonempty open
interval $U_1\subset (a_1,b_1)$ such that
\[
U_1\cap n_k^1\tilde B=\emptyset,\ \ \mbox{ for all } \ \  k\in\nats.
\]
Continuing inductively, for each $p>1$ we can choose a subsequence 
$\{n_k^p\}$ of $\{n_k^{p-1}\}$
and a nonempty interval $U_p\subset (a_p,b_p)$ such that
\begin{equation}\label{eq:emptyIntersection}
U_p\cap n_k^p\tilde B=\emptyset,\ \ \mbox{ for all } \ \  k\in\nats.
\end{equation}
If $n_{k_p}=n_p^p$ then (\ref{eq:emptyIntersection}) implies
\[
\Cup_{k\in\smallnats}U_k\cap\limsup_{p\to\infty} n_{k_p}\tilde{B}=\emptyset.
\]
Since $\Cup_{k\in\smallnats}U_k$ is a dense open subset of $(-1,1)$, this
implies (i).

This finishes the proof of Lemma \ref{lem:longlist}.

\bigskip 
Notice that Lemma \ref{lem:longlist} can be viewed as a version of
\cite[Lemma 1]{PWW:RemIDens}. (Compare also \cite[Corollary 1]{PWW:RemIDens}.)
In particular, an alternative prove of Lemma \ref{lem:longlist} is as
follows. Implication 
``$0$ is an \I-dispersion point of $B$''$\Rightarrow$(v) is implicitely
contained in the proof of \cite[Lemma 1]{PWW:RemIDens}. Since the implications
(v)$\Rightarrow$(iv)$\Rightarrow$(iii) and
(ii)$\Rightarrow$(i)$\Rightarrow$``$0$ is an \I-dispersion point of $B$''
are obvious, it is enough to show (iii)$\Rightarrow$(ii). But it is included
in this paper as a proof of (v)$\Rightarrow$(i).

\bigskip
In what follows we also need the following theorem. 
Notice that Theorem \ref{thm:lipschitz} that can be obtained from the
following theorem by taking an arbitrary set $E$ and $u_k=1$.

\theorem{thm:genlipsch}{Let $0$ be a right \idens\ point of a right
interval set $E=\Cup_{k\in\smallnats} (a_k,b_k)\subset[0,1]$ and let 
$h\colon[-1,1]\to[-1,1]$ be such that the restricted function
$h|_{(a_k,b_k)}$ is a homeomorphism for every $k\in\nats$. Moreover,
let us assume that there  % I still think, it should be plural
exist a nondecreasing sequence 
$\{u_k\}_{k\in\smallnats}$ of positive numbers and constants
$K,L>0$ such that for every $k\in\nats$ the functions
\[
h_k=u_k h|_{\{0\}\cup(a_k,b_k)}\colon \{0\}\cup(a_k,b_k)\to [-K,K]
\quad\mbox{and}\quad 
\left(h_k|_{(a_k,b_k)}\right)^{-1}
\] 
satisfy a Lipschitz condition with constant $L$. Then $h$
is right \idens\ continuous at $0$.
}

Before proving the theorem let us notice that it implies
the following corollary. It can be considered as an \idens\ analog of the
theorem that convex functions defined on open intervals are density continuous
\cite[Theorem 1]{CL:SpDensCont}. Notice also that an easy modification of the
Example \ref{cinfnotid} shows that not all convex functions defined on open
intervals are \idens\ continuous.

\corollary{cor:genlinear}{Let $0$ be a right \idens\ point of a right
interval set $\Cup_{k\in\smallnats} (a_k,b_k)$ and let 
$h\colon[-1,1]\to[-1,1]$ be a convex function such that 
the restricted functions
$h|_{(a_k,b_k)}$ are linear for every $k\in\nats$.
Then $h$ is right \idens\ continuous at $0$.
}

Proof. Let $c_k=h^{\prime}((a_k+b_k)/2)$ for $k\in\nats$.
Then, by the convexity of $h$, the sequence $\{c_k\}$ is nonincreasing. 
Let, $c=\inf_{k\in\smallnats}c_k$. Notice that $c\neq -\infty$, as 
$c=-\infty$
would contradict the
convexity of $h$ on $(-1,1)$. 
If $c\neq 0$ then $h$ and $\inv{h}$ both satisfy a Lipschitz
condition on a right neighbourhood of  
$0$. By Theorem
\ref{thm:lipschitz}, this implies  that $h$ is \idens\ continuous at $0$.
(See the remark before  
Theorem \ref{thm:genlipsch}.) If $c=0$, we may also assume that $h(0)=0$. 
There are two cases to consider. If $P=\{k:c_k=0\}$ is infinite, then
  $h$ must be linear on a right-hand
neighborhood of $0$ since $h$ is convex on $(0,1)$. 
Otherwise, assume $P$ is finite. In this case there is no generality lost in
assuming that $P=\emptyset$. Evidently, the
sequence $u_k=c_k^{-1}$ is nondecreasing and the
derivative of $u_kh$ at points from $(a_k,b_k)$ equals $1$.
Moreover, the convexity of $h$ implies that 
$h_k=u_k h|_{\{0\}\cup(a_k,b_k)}\colon \{0\}\cup(a_k,b_k)\to [0,1]$.
An application of Theorem \ref{thm:genlipsch} finishes the proof.

\bigskip

We also need the following easy lemma.

\lemma{nest}{ Let 
$\varepsilon, M >0$ and let $\{(a_n,b_n)\}_{n\in\nats}$
be a sequence of subintervals of 
$[-M,M]$ such that $b_n-a_n>\varepsilon$ for
every $n\in\nats$. Then there exists a nonempty interval $(a,b)\subset
[-M,M]$ and an increasing sequence $\{n_k\}_{k\in\nats}$
of positive numbers such that
$(a,b)\subset (a_{n_k},b_{n_k})$ for every $k\in\nats$. }
 
Proof. Let $c_n$ be the center of $(a_n,b_n)$ for every $n\in\nats$ 
and let 
$-M=d_0<d_1<\ldots<d_m=M$ be such that $d_i-d_{i-1}<\e/2$
for every $i=1,2,\ldots,m$. By the Pigeon Hole Principle, there exists
$i\leq m$ and a sequence $\{n_k\}_{k\in\nats}$ such that
$d_i\in (a_{n_k},c_{n_k})$ for every $k\in\nats$. Then the sequence
$\{n_k\}_{k\in\smallnats}$ and the interval
$(a,b)=(d_i,d_i+\frac{\varepsilon}{2})$ suffice.

\bigskip

Proof of Theorem \ref{thm:genlipsch}. Let $h$ be as in the 
assumptions. Replacing $h(x)$ by $(h(x)-h(0))/K$, if necessary, 
we may assume that $h(0)=0$ and $K=1$. 

For $k,n\in\nats$, let 
\[
h_{k,n}(x)=n\, h_k\left(\frac{1}{n}x\right)=
n\, u_k\, h\left(\frac{1}{n}x\right) \mbox{ with domain }
\{0\}\cup(na_k,nb_k).
\]
The functions $h_{k,n}$ and 
$\left(h_{k,n}|_{(na_k,nb_k)}\right)^{-1}$ also 
satisfy a Lipschitz condition with $L$ as the constant.

To prove that $h$ is right \idens\ continuous at $0$, let $B\in\BaireSets$,
$0\not\in B$, be such that $0$ is an \I-dispersion point of $B$.
It is enough to prove that $0$ is a right \I-dispersion point of
$D=h^{-1}(B)\cap E$. Notice that
$\tilde D=h^{-1}(\tilde B)\cap E$. Thus, we may assume from the beginning
that both $B$ and $D$ are regular open sets.

It will be shown that $0$ is an \I-dispersion point of 
$h^{-1}(B)\cap E$ by using Lemma \ref{lem:longlist}(iii).

So, let $\{n_k\}_{k\in\smallnats}$ be an
increasing sequence of integers and let $(a,b)\subset [-1,1]$ be a nonempty
interval. We must find a nonempty interval $(c,d)\subset (a,b)$ and a
subsequence $\{n_{k_p}\}_{p\in\smallnats}$
of $\{n_k\}_{k\in\smallnats}$ such that for every $p\in\nats$ 
\begin{equation}\label{condC}
(c,d)\cap n_{k_p} D = \emptyset.
\end{equation}

First, it may be assumed that $(a,b)\subset(0,1)$, else (\ref{condC}) is
trivially satisfied. Since 
$(0,\infty)\setminus\closure{E}$ 
is a regular open set having $0$ as 
an \idisp\/ point, Lemma \ref{lem:longlist}(iii) implies the
existence of a nonempty 
open interval $I\subset(a,b)$ and a subsequence 
$\{n'_k\}$ of $\{n_k\}$
such that $I\subset n'_kE$ for all $k$. There is no generality lost with the
assumptions that $I=(a,b)$ and $n'_k=n_k$ for all $k$. Thus, with these
assumptions, there exists a nondecreasing sequence of integers $m_k$ such that
$(a,b)\subset n_k(a_{m_k},b_{m_k})$ and $(a,b)$ is in the domain of
$h_{m_k,n_k}$ for every $k\in\nats$.

But the functions 
$h_{m_k,n_k}$ and 
$\left(h_{m_k,n_k}|_{(n_ka_{m_k},n_kb_{m_k})}\right)^{-1}$ satisfy 
a Lipschitz condition with constant $L$. So, the intervals 
\[
h_{m_k,n_k}((a,b))\subset h_{m_k,n_k}(\{0\}\cup(a,b))\subset[-L,L]
\] 
have length at least
$(b-a)/L$ and, using Lemma \ref{nest}, we may 
also assume, choosing a subsequence, if necessary, that for some
nonempty interval $(a^\prime,b^\prime)\subset[-L,L]$ and every $k\in\nats$
\[
(a^\prime,b^\prime)\subset h_{m_k,n_k}((a,b)).
\]

Now, by Lemma \ref{lem:longlist}(iv) used with $B$, the divergent sequence
$\{u_{m_k} n_k\}_{k\in\smallnats}$ and $(a^\prime,b^\prime)$,
we may assume, passing to a subsequence, if necessary, that for some
nonempty interval 
$(c^\prime,d^\prime)\subset(a^\prime,b^\prime)$ and every $k\in\nats$
\[
(c^\prime,d^\prime)\cap u_{m_k} n_k B = \emptyset,
\]
which is equivalent to
\[
n_k h^{-1}\left(\frac{1}{u_{m_k} n_k}(c^\prime,d^\prime)\right)
\cap n_k h^{-1}\left(B\right) = \emptyset.
\]
For $x$ in the domain of $h_{m_k,n_k}^{-1}$, we have
$n_k h^{-1}\left(\frac{1}{u_{m_k} n_k}x\right) = h_{m_k,n_k}^{-1}(x)$. Thus,
the above is equivalent to
\[
h_{m_k,n_k}^{-1}\left((c^\prime,d^\prime)\right)
\cap n_k h^{-1}\left(B\right) = \emptyset.
\]

But,
\[
h_{m_k,n_k}^{-1}\left((c^\prime,d^\prime)\right)
\subset h_{m_k,n_k}^{-1}\left((a^\prime,b^\prime)\right)\subset (a,b)
\]
for every $k\in\nats$.
Using Lemma \ref{nest} once more and the fact that 
the functions $h_{m_k,n_k}$
satisfy a Lipschitz condition with the same constant $L$, we may 
choose an increasing sequence $\{k_p\}_{p\in\nats}$
of natural numbers and a nonempty interval $(c,d)$
such that for every $p\in\nats$  
\[
(c,d)\subset h_{m_{k_p},n_{k_p}}^{-1}\left((c^\prime,d^\prime)\right)
\subset (a,b).
\]
The last two conditions easily imply (\ref{condC}). 
The proof of Theorem \ref{thm:genlipsch} is finished.

\bigskip

The next two examples
are technical and will be used to construct other examples.

\example{ex:idensewithpole}{Let $\{c_n\}_{n\in\smallnats}$ be a sequence 
of positive numbers. For every right
interval set $E=\Cup_{n\in\smallnats} [a_n,b_n]\subset[0,1]$
for which $0$ is a right \idens\ point there exists 
an \idens\ 
and density continuous, \cinfinity, convex increasing homeomorphism 
$h\colon[0,b_1]\to[0,\infty)$ such that
\begin{description}
\item[(1)]  $h^{(n)}(0)=0$ for every $n\geq 0$; and,
\item[(2)]  for every $n\in\nats$ there exists a positive number $d_n\leq c_n$
            such that $h^\prime(x)=d_n$ for every $x\in(a_n,b_n)$.
\end{description}
}

Proof. Let 
$D=\Cup_{n\in\smallnats} (l_n,r_n)=(0,b_1]\setminus E$
and let $f$ be a nonnegative \cinfinity\ function
such that $f(x)=0$ for 
$x\in\complement{D}$ while $f(x)>0$ for $x\in D$.
To choose such a function it is enough to define $f$ on $(l_n,r_n)$ by 
\[
f(x)=u_n e^{-(x-l_n)^{-2}-(x-r_n)^{-2}},
\] 
which is known to be \cinfinity\ with poles at $l_n$ and $r_n$, 
and in which the constants $u_n\leq c_n$ are chosen in such a way that 
$f^{(i)}(x)\leq\frac{1}{n}$ 
for every $i\leq n$ and $x\in[l_n,r_n]$.

Let $g(x)=\int_0^x f(y)\,dy$ and $h(x)=\int_0^x g(y)\,dy$.

Evidently $h$ is a \cinfinity\ increasing and convex homeomorphism
satisfying (1). It is also easy to see that (2) holds, as 
$f(x)\leq u_n\leq c_n$ for all $x\in[0,b_n]\subset[0,1]$.
Moreover, by Theorem \ref{thm:lipschitz},
$h$ is \idens\ continuous at every point $\neq 0$ and, 
by Corollary \ref{cor:genlinear}, $h$ is right \idens\ continuous at $0$.

The function $h$ is density continuous, since it can be extended to a convex
function on an open interval \cite[Theorem 1]{CL:SpDensCont}.

\example{ex:filings}{Let $k<l<m<a<b<c$. There exists 
an \idens\ continuous, \cinfinity\ function
$g\colon[k,c]\to\reals$ constant on $[m,a]$ and
such that $g(x)=x$ for $x\in [k,l]\cup[b,c]$.
}

Proof. Without loss of generality we may assume that $m<0<a$.
It suffices to 
define $g$ on $[0,c]$, because defining $g$ on $[k,0]$ merely involves
homothetically altering the odd extension of $g$.

Let $a_0, b_0$ be such that $a<a_0<b_0<b$ and choose
$h\colon[0,a_0-a]\to[0,\infty)$ as in Example \ref{ex:idensewithpole}
in such a way that $h(a_0-a)<b_0$. Define $g_0\colon[0,c]\to\reals$ by
putting $g_0(x)=0$ for $x\in [0,a]$, $g_0(x)=x$ for $x\in[b_0,c]$,
$g_0(x)=h(x-a)$ for $x\in [a,a_0]$ and extend $g_0$ on $[a_0, b_0]$
in a continuous manner as a linear function.

Evidently,
$g_0$ is \idens\ continuous. It is also \cinfinity\ at all
points with the exceptions of $a_0$ and $b_0$. To obtain the desired function
$g$ we will modify $g_0$ on $[a,b]$ by ``rounding its corners'' at the points
$a_0$ and $b_0$. But notice that $g_0$ is increasing and linear on some
left and right hand neighborhoods of $a_0$ and $b_0$. Thus, using an 
appropriate homothetic transformation of a function $L(x)+rh(x)$ where
$L(x)$ is a linear function, $r\in\reals$ and $h$ is from Example 
\ref{ex:idensewithpole} (or even
$h(x)=\int_0^x\int_0^y\exp\left(-z^{-2}-(z-1)^{-2}\right)\,dz\,dy$)
we can easily modify
$g_0$ to be a \cinfinity\ function $g$  while keeping
the property that $g^\prime(x)>0$ on the modification set. Thus, $g$ is 
\idens\ continuous.

\example{deepInotI}{There is an 
$f\in\deepidenscont\,\,\cinfinity\,\,\denscont\setminus\idenscont$.}

Proof. Let $g\colon[0,1]\to[0,1]$ be as in Example \ref{ex:filings}, 
constant on an interval containing $1/2$ and
choose constants $c_k\leq\frac{1}{k}$ such that
$c_kg^{(i)}(x)\leq\frac{1}{k}$ for every $x\in [0,1]$ and $i\leq k$. 
Next, choose a right interval set
$E=\Cup_{k\in\smallnats} [p_k,q_k]\subset[0,1]$ for which $0$ is a
right \idens\ point and let 
$h$ and $\{d_k\}$ be choosen as in Example
\ref{ex:idensewithpole} for the sequence $\{c_k\}$ and the set $E$. 
Extend $h$ onto \reals\ by putting $h(x)=0$ for $x<0$ and as a linear
function on $[q_1,\infty)$ by the same formula as on $[p_1,q_1]$. It is easy
to see that 
\[
h\in\deepidenscont\,\,\cinfinity\,\,\denscont.
\]

Notice that for every $t>0$ the function $g_{k,t}\colon [0,t]\to[0,t]$
defined by 
$g_{k,t}(x)=t\,d_kg(x/t)$ has the properties that
\[
g_{k,t}^{(i)}(x)\leq\frac{1}{k}\mbox{ for every }x\in [0,t]
\mbox{ and }i\leq k
\]
and 
\[
g_{k,t}(x)=d_k x\mbox{ for every }x\in t\left( [0,l]\cup[b,1]\right).
\]

Moreover, let $S=\{s_n\}_{n\in\nats}$ be
from Lemma \ref{lem:denseseq}. Without loss of generality we may assume 
that $S\subset\interior{E}$, as $0$ is a right deep-\idens\ point of $E$.
Notice, that each component of $E$ contains only finitely many points from
$S$. 
Denote the component of $E$ which contains $s_n$ by $[p_{k(n)},q_{k(n)}]$.

We need also to choose nonempty pairwise disjoint intervals
$(a_n,b_n)\subset [a_n,b_n]\subset(p_{k(n)},q_{k(n)})$, 
centered at $s_n$ such that $0$ is a dispersion point of 
$P=\UoverN(a_n,b_n)$ and
\begin{equation}\label{condsmall}
\lim_{n\to\infty} \frac{h(b_n)-h(a_n)}{h(a_n)}=0.
\end{equation}

Define $h_n(x)=g_{k(n),b_n-a_n}(x-a_n)$ for $x\in(a_n,b_n)$, with the interval 
in $(a_n,b_n)$ on which $h_n$ is constant denoted by $I_n$. 
Notice that $s_n\in I_n$. Define $f$ by putting $f|_{P^c}=h|_{P^c}$
and $f(x)=h_n(x) + h(a_n)$ for $x\in(a_n,b_n)$. 

Obviously, $f$ is \cinfinity, \idens\ continuous
and density continuous at any point $\neq 0$. 
It is also clear, by the choice of the constants $c_k$, that
$f$ is infinitely differentiable at $0$. 

To see that $f$ is density continuous at $0$ it is enough to notice that
$f|_{(-\infty,0]\cup\interior{E}}=h|_{(-\infty,0]\cup\interior{E}}$, 
while $h\in\denscont$ and $(-\infty,0]\cup\interior{E}\in\densitytop$.

It is also easy to see that $f\not\in\idenscont$, as $0$ is an \I-dispersion
point of the countable set
$f(S)$, while $0$ is not an \I-dispersion point of 
$f^{-1}(f(S))\supset \Cup_{n\in\nats}I_n\supset S$.

It remains to prove that $f$ is right deep-\idens\ continuous at $0$,
as the left hand case is obvious.
So, let $0$ be an \I-dispersion point of a right interval set 
$B=\Cup_{n\in\nats} (v_n,w_n)$.
It is enough to prove that $0$ is an \I-dispersion point of $\inv{f}(B)$, as 
the sets $E\cup\{0\}\in\itopology$, where $E$ is an open interval set, form
a basis of \deepitopology\ at $0$ \cite{Lazarow:Coarsest}.

We claim that the lengths of some of the intervals $(v_n,w_n)$
can be increased slightly to satisfy the condition
\begin{equation}\label{con:inclus}
(h(a_k),h(b_k))\cap(v_n,w_n)=\emptyset\,\,
\mbox{ or }\,\,
(h(a_k),h(b_k))\subset(v_n,w_n),
\end{equation}
while still preserving the property that $0$ is an \I-dispersion point of $B$.

To see this, define $B^\prime=
B\cup\Cup\{(h(a_k),h(b_k))\colon (h(a_k),h(b_k))\cap B\neq\emptyset\}$.
Evidently, $B^\prime$ satisfies (\ref{con:inclus}). We have to prove only
that $0$ is an \I-dispersion point of $B^\prime$.

So, let $(a,b)\subset(0,1)$ and $\{n_k\}_{k\in\smallnats}$ be an
increasing sequence of natural numbers.
Using Lemma 
\ref{lem:longlist}(iii), we can find an increasing subsequence 
$\{n_{k_p}\}_{p\in\smallnats}$ and a nonempty interval $(c,d)\subset(a,b)$
such that
\[
(c,d)\cap n_{k_p} B=\emptyset\,\,\mbox{ for all }\,\,p\in\nats.
\]
According to (\ref{condsmall}), there is an $p_0\in\nats$ such that
\[
\frac
{h(b_{n_{k_p}})-h(a_{n_{k_p}})}
{h(a_{n_{k_p}})}<\frac{d-c}{3}
\,\,\,\mbox{ for all }\,\,\, p\geq p_0.
\]
But, then it is clear that
\[
(c+\frac{d-c}{3},d-\frac{d-c}{3})\cap n_{k_p} B^\prime=
\emptyset\,\,\mbox{ for all }\,\,p\geq p_0,
\]
so, by Lemma \ref{lem:longlist}(iii), 
$0$ is an \I-dispersion point of $B^\prime$.

But (\ref{con:inclus}) implies that $h^{-1}(B)=f^{-1}(B)$, so $0$ is 
a deep-\I-dispersion point of $f^{-1}(B)$, as $h$ is deep-\idens\ continuous.

\example{cinfnotdendcont}{There is an 
$f\in\idenscont\,\,\cinfinity\setminus\denscont$.}

Proof. We will proceed in a method similar to Example
\ref{deepInotI}.

Let $g\colon[0,1]\to[0,1]$ be as in Example \ref{ex:filings} such that
$g$ is constant on $[1/3,2/3]$.
Choose constants $c_k\leq\frac{1}{k}$ such that
$c_kg^{(i)}(x)\leq\frac{1}{k}$ for every $x\in [0,1]$ and $i\leq k$. 
Next, choose a right interval set
$E=\Cup_{k\in\smallnats} [p_k,q_k]\subset[0,1]$
for which $0$ is a right density and deep-\idens\ point. Let 
$h$ and $\{d_k\}$ be choosen as in
Example \ref{ex:idensewithpole} for the sequence $\{c_k\}$ and set $E$.
Extend $h$ onto \reals\ by putting $h(x)=0$ for $x<0$ and as a linear
function on $[q_1,\infty)$ by the same formula as on $[p_1,q_1]$. It is easy
to see that 
\[
h\in\idenscont\,\,\cinfinity\,\,\denscont.
\]

Notice that for every $t>0$ the function $g_{k,t}\colon [0,t]\to[0,t]$
defined by 
$g_{k,t}(x)=t\,d_kg(x/t)$ has the properties that
\begin{equation}\label{con:gfunct1}
g_{k,t}^{(i)}(x)\leq\frac{1}{k}\mbox{ for every }x\in [0,t]
\mbox{ and }i\leq k
\end{equation}
and 
\begin{equation}\label{con:gfunct2}
g_{k,t}(x)=d_k x\mbox{ for every }x\in t\left( [0,l]\cup[b,1]\right).
\end{equation}

Moreover,
let $D=\UoverN [a_n,b_n]$ be a right interval set such that 
$D^c\in\deepitopology\setminus\densitytop$. (See Theorem \ref{th:densvsidens}.)
Without loss of generality we may assume 
that $D\subset\interior{E}$, as $0$ is a right density and
deep-\idens\ point of $E$. 
Notice, that each component of $E$ contains only finitely many intervals
$[a_n,b_n]$. 
Denote the component of $E$ which contains $[a_n,b_n]$ by $[p_{k(n)},q_{k(n)}]$.

Let $h_n=g_{k(n),b_n-a_n}$, with the interval in $(0,b_n-a_n)$ on which
$h_n$ is constant denoted by $I_n$. Define $f$ by putting 
$f|_{\left(\UoverN(a_n,b_n)\right)^c}=h|_{\left(\UoverN(a_n,b_n)\right)^c}$
and $f(x)=h_n(x-a_n) + h(a_n)$ for $x\in[a_n,b_n]$. 

Obviously $f$ is \cinfinity, \idens\ continuous
and density continuous at any point $\neq 0$. 
It is also clear, by the choice of the constants $c_k$, that
$f$ is infinitely differentiable at $0$. 
Also, $f$ is \idens\ continuous at $0$ as
$f|_{D^c}=h|_{D^c}$, 
while 
$h\in\idenscont$ and $D^c\in\itopology$.

Finally, $f$ is not density continuous at $0$, as $0$ is a dispersion point of
the countable set $f(\UoverN I_n)$, while $0$ is not a dispersion point
of a set $f^{-1}(f(\UoverN I_n))=\UoverN I_n$. 
This last statement is correct, as $\UoverN I_n$ is obtained by taking 
the middle third from every component of $D=\UoverN [a_n,b_n]$.

\bigskip

\bibliographystyle{plain}
%\bibliography{examples}
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\end{thebibliography}

 
\end{document}
