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% Density and \I-density continuous homeomorphisms


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\MathReviews{Primary 26A03, Secondary 28A05}
\markboth{Krzysztof Ciesielski}{Density 
and \I-density continuous homeomorphisms}
\keywords{{$\cal I$}-density continuous, density continuous, homeomorphism}

\title{Density and \I-density continuous homeomorphisms}

\author{Krzysztof Ciesielski (email:
kcies@wvnvms.bitnet), Department of Mathematics, West
Virginia University, Morgantown, WV 26506-6310
}


\begin{document}
\maketitle

\section{Preliminaries}

Let \dhomeo\ and \idhomeo\ stand for the increasing homeomorphisms that
are density and \I-density continuous and let
\phomeo\ and \iphomeo\ denote the classes of inverses of functions from
\dhomeo\ and \idhomeo, respectively; i.e., classes of increasing 
homeomorphisms that preserve density and \I-density points.
In the paper we prove that classes \dhomeo, \idhomeo\ and \iphomeo\
are closed under the addition operation. A 
similar result for the class \phomeo\ has been proved 
by Niewiarowski \cite{Niewiarowski:Density}. 
The theorem that the class \iphomeo\
is closed under the addition operation is also contained in
the paper of Aversa and Wilczy\'nski 
\cite[Theorem 4]{AversaWilczynski:Homeomorphisms}.
(See also \cite[Theorem 25]{Wilczynski:CatAn}.) However, their
proof contains an essential gap. (The gap will be discussed in
the last paragraph of the paper.) 

This paper contains also the examples showing that none of the
above theorems is correct if we admit the possibility that one of the
homeomorphism is increasing, and the second one is decreasing, even in the
case when their sum is still a homeomorphism. 

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}, 
while $B - d = \{x - d\in\reals\colon x\in B\}$
and  $dB =\{dx\in\reals\colon x\in B\}$.
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 $A\in\Lebesgue$, 
its Lebesgue measure is denoted by \measure{A}.

To define the density topology \densitytop\ and the \I-density topology
\itopology\ we need the following notions of 
density and \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  
\[
\lim_{h\to 0^+}\frac{\measure{A\cap(x-h,x+h)}}{2h}=1.
\]
The set of all density points of $A\in\Lebesgue$ we denote by \densitypts{A}.
The family of sets
\[
\densitytop=\{A\in\Lebesgue\colon A\subset\densitypts{A}\}
\]
forms a topology on \reals\ \cite{GNN:densitytop,Oxtoby:MeasCat} and is called
the {\em density topology\/} on \reals.

We say that $0$ is an {\em \I-density point\/}
of a set $A\in\BaireSets$ 
\cite[Theorem 1]{Wilczynski:CatAn} (see also \cite[Corollary 1]{PWW:RemIDens}
and \cite{PWW:CatAn})
if for every increasing sequence $\{t_k\}_{k\in\smallnats}$ of 
positive numbers diverging to infinity
there exists a subsequence $\{t_{k_i}\}_{i\in\smallnats}$ such that
\[
\lim_{i\to\infty}\charf{t_{k_i}A\cap(-1,1)} = \charf{(-1,1)} \ \mbox{\I-a.e.}
\]
It is worth noticing that the above condition is equivalent to
the fact that the set
$\liminf_{i\to\infty}t_{k_i}A =
\bigcup_{j\in\smallnats} \bigcap_{i\ge j} t_{k_i}A$
is residual in $(-1,1)$. We say that a point $a$ is an \I-density point 
of $A\in\BaireSets$ if $0$ is an \I-density point of $A-a$.
The set of all \I-density points of $A\in\Lebesgue$ we denote by
\idensitypts{A}. 
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\/} on \reals.

We also use the following notions dual to the density definitions given above.
We say that $x$ is a {\em dispersion} ({\em \I-dispersion})
point of $A$ if $x$ is a density (\I-density) point of
\complement{A}. In particular, $0$ is an \I-dispersion point of $B$ if
for every increasing sequence $\{t_k\}_{k\in\smallnats}$ of 
positive numbers diverging to infinity
there exists a subsequence $\{t_{k_i}\}_{i\in\smallnats}$ such that
\begin{equation}\label{con_idisp}
(-1,1) \cap \bigcap_{j\in\smallnats} \bigcup_{i\ge j} t_{k_i}B =
(-1,1) \cap \limsup_{i\to\infty}(t_{k_i}B) \in\I
\end{equation}
and $0$ is a dispersion point of $B$ if
\begin{equation}\label{con_disp}
\limsup_{h\to 0^+}\frac{\measure{B\cap(-h,h)}}{2h}=0.
\end{equation}

A function $f\RtoR$ is {\em density continuous} ({\em \I-density continuous})
if it is continuous with respect to the density (\I-density) topology on
the domain and the range. A homeomorphism $h\RtoR$ preserves density
(\I-density) points if $\inv{h}$ is density (\I-density) continuous.

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.

We say that a set $\bigcup_{n\in\smallnats}(a_n,b_n)$ 
is a {\em right interval set\/} if $b_{n+1} < a_n < b_n$ for
$n\in\nats$ and $\lim_{n\to\infty}a_n = 0$.

In what follows we will need the following facts.

\proposition{homeom}{Let $h\RtoR$ be an
increasing homeomorphism with the property that $h(0)=0$. Then
\begin{description}
\item[(i)]  $h$ is right density (\I-density) continuous at $0$ if,
and only if, $0$ is not a 
dispersion (\I-dispersion) point of $h(D)$
for every closed set $D\subset[0,\infty)$ such that $0$ is not a 
dispersion (\I-dispersion) point of $D$;
\item[(ii)] $h$ preserves right \I-density points at $0$ if, and only if,
for every right interval set $E$ for which $0$ is a right \I-density
point, $0$ is a right \I-density point of a right interval set $h(E)$.
\end{description}
}

\pf (i) follows easily from \cite[Theorem 3]{Bruckner:7} in density case 
and from \cite[Theorem 3]{AversaWilczynski:Homeomorphisms} in \I-density case.
For (ii) see \cite[Theorem 3]{AversaWilczynski:Homeomorphisms}. \qed

\proposition{disppts}{$0$ is a right \I-density point of a 
right interval set $E$ if, and only if, 
for every increasing sequence $\{t_k\}_{k\in\smallnats}$ 
of positive numbers diverging to infinity
and every nonempty interval $(A,B)\subset(0,1)$
there exists a nonempty subinterval $J\subset(A,B)$ and a subsequence
$\{t_{k_i}\}_{i\in\smallnats}$ such that for every $i\in\nats$ 
\[
J\subset t_{k_i}E. 
\]
}

\pf See \cite[Lemma 6.1(iii)]{CL:Examples}. \qed

\proposition{deep}{
Let $P\subset(0,1]$ be closed and nowhere dense and
let $\{d_k\}_{k\in\smallnats}$ be a sequence of positive numbers
such that $\lim_{k\to\infty}d_{k + 1}/d_k = 0$.
Then there is an open set $V\supset \bigcup_{k\in\smallnats}d_kP$
such that $0$ is an \I-dispersion point of $V$.
}

\pf See \cite[Lemma 2.4]{CL:Examples}. In fact, Proposition
\ref{deep} says that $0$ is a deep-\I-dispersion point of 
$\bigcup_{k\in\smallnats}d_kP$. \qed

\proposition{INTERdisp}{
Let $a>0$ and let
$\{d_k\}_{k\in\smallnats}$, $\{a_k\}_{k\in\smallnats}$
and $\{b_k\}_{k\in\smallnats}$ be sequences of positive
numbers such that $a<a_k<b_k$ for every $k\in\nats$ and
\[
\lim_{k\to\infty}d_k=\lim_{k\to\infty}[b_k-a_k]=0.
\]
Then, $0$ is an \I-dispersion point of 
\[
\bigcup_{k\in\smallnats}d_k(a_k,b_k).
\]
}

\pf It follows immediately from
\cite[Theorem 2]{Wilczynski:CatAn}. See also \cite[Theorem 1]{PWW:RemIDens}.
\qed

\proposition{lip}{Let $h\RtoR$ be a homeomorphism.
If $h$ and $\inv{h}$ satisfy a local Lipschitz condition then $h$ and $\inv{h}$
preserve density and \I-density points and are density and \I-density 
continuous.}

\pf For the density case see \cite[Lemma 1]{CL:SpDensCont}.
(Compare also \cite[Corollary 2]{Bruckner:7}.)
The \I-density case can be found in 
\cite[Corollary 1]{AversaWilczynski:Homeomorphisms},
\cite[Theorem 26]{Wilczynski:CatAn} or \cite[Theorem 5.8]{CL:Examples}. \qed


\section{ Density continuous homeomorphisms}

In this section we prove
that the sum of two increasing density continuous
homeomorphisms is density continuous.

\theorem{th:denscont}{If $f,g\in\dhomeo$, then $f+g\in\dhomeo$.}

\pf Let $f,g\in\dhomeo$ and let $a\in\reals$. It is enough to prove that 
$f+g$ is right density continuous at $a$, as the left-hand side 
argument is similar.
Without loss of generality we may assume that $a=f(a)=g(a)=0$.
Let $D\subset [0,\infty)$ be a closed set for which
$0$ is not a dispersion point of $D$. By Proposition \ref{homeom}(i),
it is enough to prove that
$0$ is not a dispersion point of $(f+g)(D)$.

Let $D_f=\{x\in D\colon g(x)\leq f(x)\}$ and 
$D_g=\{x\in D\colon f(x)\leq g(x)\}$. Then 
$0$ is not a dispersion point of either $D_f$ or $D_g$.
Assume that $0$ is not a dispersion point of $D_f$.
We may assume, without loss of generality, that $D=D_f$. Thus,
\[
g(x)\leq f(x)\ \mbox{ for every }\ x\in D.
\]
But $f\in\dhomeo$ and $0$ is not a dispersion point of $D$. 
So, $0$ is not a dispersion point of $f(D)\subset [0,\infty)$ and,
by (\ref{con_disp}),
there exist $\e>0$ and a decreasing sequence $\{h_n\}_{n\in\smallnats}$
of positive numbers converging to $0$ such that
\[
\limsup_{n\to\infty}\frac{\measure{f(D)\cap(0,h_n)}}{h_n}=\e.
\]
Let $h_n^\prime=\sup f(D)\cap(0,h_n]\in(0,h_n]$, 
$t_n=\inv{f}(h_n^\prime)\in D$ and define $p_n=(f+g)(t_n)\leq 2f(t_n)$.
Then, $\lim_{n\to\infty}p_n=0$ and
\begin{eqnarray}
\limsup_{n\to\infty}\frac{\measure{(f+g)(D)\cap(0,p_n)}}{p_n}
& = & 
\limsup_{n\to\infty}
\frac{\measure{(f+g)(   D\cap( 0,t_n )  )}}{p_n} \nonumber \\
& \geq & 
\limsup_{n\to\infty}
\frac{\measure{f(   D\cap( 0,t_n )  )}}{2f(t_n)} \label{cond_ineq2} \\
& = & 
\frac{1}{2} \limsup_{n\to\infty}
\frac{\measure{f(D)\cap ( 0,h_n^\prime ) }}{h_n^\prime} \nonumber \\
& = & 
\frac{1}{2} \limsup_{n\to\infty}
\frac{\measure{f(D)\cap ( 0,h_n ) }}{h_n^\prime} \nonumber \\
& \geq & 
\frac{1}{2} \limsup_{n\to\infty}
\frac{\measure{f(D)\cap ( 0,h_n ) }}{h_n} \nonumber \\
& = & 
\frac{\e}{2}>0, \nonumber
\end{eqnarray}
where the numerator part of inequality (\ref{cond_ineq2}) holds, because
$\measure{(f+g)(A)}\geq \measure{f(A)}$ for every $A\in\Lebesgue$.
Thus, by (\ref{con_disp}),
$0$ is not a dispersion point of $(f+g)(D)$. This finishes the proof of
Theorem \ref{th:denscont}. \qed

\corollary{cor1m}{If $f,g\in\dhomeo$, then $f\,g\in\dhomeo$.}

\pf By Proposition \ref{lip}, functions $\exp$ and $\ln$ are density
continuous. We show that $f\,g$ is density continuous at $a\in\reals$. 
Translating and restricting $f$ and $g$ to an open neighborhood of $a$,
if necessary, we may assume that $f$ and $g$ are positive. Then, 
$\ln f,\ln g\in\dhomeo$, as composition of density continuous increasing
homeomorphisms is a density continuous increasing homeomorphism.
Thus, by Theorem \ref{th:denscont},
density continuous is also

\medskip

%\[
\hfill $f\,g=\exp{(\ln f + \ln g)}.$ \qed
%\] 

\bigskip
\medskip

Let us also notice that in fact we proved the following result, which is
a little bit stronger that Theorem \ref{th:denscont}.

\corollary{cor1s}{Let $f$ and $g$ be increasing homeomorphisms such that
$f(a)=g(a)$ for some $a\in\reals$. If $g(x)\leq f(x)$ for every $x\geq a$ and
$f$ is right density continuous at $a$ then $f+g$ is also
right density continuous at $a$.}

\section{ \I-density continuous homeomorphisms}

The purpose of this section is to prove 
that the sum of two increasing \I-density continuous
homeomorphisms is \I-density continuous.
For this we need the following lemmas.

\lemma{subsec1}{Let $D\in\BaireSets$ be such that $0$ is not a right 
\I-dispersion point of $D$. Then there exists an increasing sequence 
$\{t_k\}_{k\in\smallnats}$ of positive numbers diverging to infinity
and a nonempty interval $(a,b)\subset(0,1)$ such that
\[ 
\left(\liminf_{k\to\infty}{t_k}D\right) \mbox{ is dense in } (a,b).
\]
}

\pf Since $0$ is not a right \I-dispersion point of $D$ then, by 
(\ref{con_idisp}), there exists an increasing sequence 
$\{s_n\}_{n\in\smallnats}$ of positive numbers diverging to infinity
such that for every its subsequence $\{s_{n_k}\}_{k\in\smallnats}$
\begin{equation}\label{condnotI}
\left(\limsup_{k\to\infty}s_{n_k}D\right)\cap(0,1)\not\in\I.
\end{equation}

Let $(p_k,q_k)\subset (0,1)$ be a sequence of all nonempty intervals with 
rational endpoints. Let us construct, by induction on $k$, sequences
$\{s_n^k\}_{n\in\smallnats}$ such that 
$\{s_n^0\}_{n\in\smallnats}=\{s_n\}_{n\in\smallnats}$ and  
$\{s_n^k\}_{n\in\smallnats}$ is a subsequence
of $\{s_n^{k-1}\}_{n\in\smallnats}$ such that 
\begin{equation}\label{condnotI2} 
\mbox{either }\ \ 
\left(\limsup_{n\to\infty}s_n^k D\right)\cap(p_k,q_k)=\emptyset
\ \ \mbox{ or }\ \ 
\left(\liminf_{n\to\infty}s_n^k D\right)\cap[p_k,q_k]\neq\emptyset.
\end{equation}
Put $t_k=s_k^k$. Then, by (\ref{condnotI}),
$\left(\limsup_{k\to\infty}t_k D\right)\cap(0,1)\not\in\I$;
i.e., there exists a nonempty interval $(a,b)\subset(0,1)$ such that
\[ 
\left(\limsup_{k\to\infty}t_k D\right) \mbox{ is dense in } (a,b).
\]
But this, together with (\ref{condnotI2}), guarantees that then also
\[ 
\left(\liminf_{k\to\infty}t_k D\right) \mbox{ is dense in } (a,b).
\]
This finishes the proof of Lemma \ref{subsec1}. \qed

\lemma{lem:44}{Let $h\RtoR$ be an increasing \I-density continuous
homeomorphism such that $h(0)=0$ and let $\{t_k\}_{k\in\smallnats}$ be an
increasing sequence of positive numbers diverging to infinity.
Then for every nontrivial interval $[a,b]\subset(0,1)$ there exists 
a nonempty interval $(c,d)\subset(a,b)$ and a subsequence 
$\{t_{k_i}\}_{i\in\smallnats}$ of $\{t_k\}_{k\in\smallnats}$ such that
the limit
\[
\lim_{i\to\infty} \frac{h(c/t_{k_i})}{h(d/t_{k_i})}
\]
exists and is positive.
}

\pf By way of contradiction assume that it cannot be done; i.e., that
\begin{equation}\label{con6}
\limsup_{k\to\infty} \frac{h(c/t_k)}{h(d/t_k)}=0
\ \mbox{ for every }\ a\leq c<d\leq b.
\end{equation}
We will show that this contradicts \I-density continuity of $h$.

So, let $\{q_k\colon k\in\nats\}$ be an enumeration of $Q=[a,b]\cap\rationals$
and for each $i\in\nats$ let $d_1,\ldots,d_i$ be an increasing enumeration
of $q_1,\ldots,q_i$. Choose $\{t_{k_i}\}_{i\in\smallnats}$
such that
\begin{equation}\label{con7}
\frac{h(b/t_{k_{i+1}})}{h(a/t_{k_i})}\leq
\frac{h(d_{j+1}/t_{k_i})}{h(d_j/t_{k_i})}\leq\frac{1}{i}
\ \mbox{ for every }\ j<i,\ i\in\nats.
\end{equation}
This can be done by (\ref{con6}). Let 
\[
U_i=\Cup_{j\leq i} 
h(d_j/t_{k_i})\left(1-\frac{1}{i},1+\frac{1}{i}\right).
\]
and put $U=\Cup_{i\in\smallnats} U_i$.
Then, by (\ref{con7}) and Proposition \ref{INTERdisp}, $0$ is 
an \I-dispersion point of $U$. But $0$ is not an \I-dispersion
point of $\inv{h}(U)$, since 
for any subsequence $\{t_m\}_{m\in\smallnats}$ of
$\{t_{k_i}\}_{i\in\smallnats}$ the open set 
$\Cup_{m\geq m_0}t_m\inv{h}(U)\supset Q$ 
is dense in $(a,b)$ for every $m_0\in\nats$, and so,
\[
(-1,1)\cap\limsup_{m\to\infty}(t_m\inv{h}(U))\not\in\I.
\]
This finishes the proof of Lemma \ref{lem:44}. \qed

\lemma{lem:conv}{Let $a<b$, $H_k\colon[a,b]\to\reals$
be a sequence of increasing homeomorphisms and let us
assume that there exists a dense subset $Q$ of $[a,b]$ containing 
$a$ and $b$ such that the limit $H(q)=\lim_{k\to\infty} H_k(q)$ exists
for every $q\in Q$. If $H(Q)$ is dense in $[H(a),H(b)]$ and
$H(x)=\inf H( Q\cap[x,\infty) )$ for every $x\in[a,b]$, 
then $H_k$ converges uniformly to $H$.}

\pf First notice that the function 
$H(q)=\lim_{k\to\infty} H_k(q)$ on $Q$
is nondecreasing, so indeed $H(q)=\inf H( Q\cap[q,\infty) )$
for every $q\in Q$.

Let us fix $\e>0$. For $x\in[a,b]$ 
choose distinct $q_1,q_2\in Q$, $q_1\leq x\leq q_2$, such that 
$q_1<x<q_2$ for $x\in(a,b)$ and
\[
|H(q_2)-H(q_1)|<\e/5.
\]
Let $N_x\in\nats$ be such that
\[
|H(q_i)-H_n(q_i)|<\e/5
\]
for every $n>N_x$ and $i=1,2$. 
Put $U_x=(q_1,q_2)$ for $x\in(a,b)$, 
$U_x=[q_1,q_2)$ for $x=a$ and $U_x=(q_1,q_2]$ for $x=b$.
Thus, $U_x$ is an open neighborhood of $x$ in $[a,b]$ and, 
for every $y\in U_x$ and $n>N_x$,
\[
H(q_1)\leq H(y)\leq H(q_2)\ \mbox{ and } \
H_n(q_1)\leq H_n(y)\leq H_n(q_2),
\]
so that
\begin{eqnarray}
|H(y)-H_n(y)| 
& \leq & |H(y)-H(q_2)| + |H(q_2)-H_n(q_2)| + |H_n(q_2)-H_n(y)| \nonumber \\
& < & |H(q_1)-H(q_2)| + \e/5 + |H_n(q_2)-H_n(q_1)| \nonumber \\
& < & \e/5 + \e/5 + \nonumber \\ 
&   & |H_n(q_2)-H(q_2)| + |H(q_2)-H(q_1)| + |H(q_1)-H_n(q_1)| \nonumber \\
& < & \e/5 + \e/5 + \e/5 + \e/5 + \e/5 =\e. \nonumber
\end{eqnarray}
Choose a finite subcover $\{U_{x_1},\ldots,U_{x_k}\}$ of 
the open cover ${\cal U}=\{U_x\}_{x\in[a,b]}$ of $[a,b]$
and put $N=\sup\{N_{x_1},\ldots,N_{x_k}\}$. Then we obtain
\[
|H(y)-H_n(y)|<\e
\]
for every $y\in[a,b]$ and $n>N$. This finishes the proof of Lemma 
\ref{lem:conv}. \qed

\lemma{lem:func}{Let $h\RtoR$ be an increasing \I-density continuous
homeomorphism such that $h(0)=0$ and let $[a,b]\subset(0,1)$ be a
nontrivial interval. If 
$\{s_k\}_{k\in\smallnats}$ and $\{t_k\}_{k\in\smallnats}$ are
increasing sequences of positive numbers diverging to infinity such
that $H_k(x)=s_kh(x/t_k)\in[0,1]$ for every $x\in[a,b]$,
then there exist a nonempty interval $(c,d)\subset(a,b)$ and a subsequence 
$\{H_{k_i}\}_{i\in\smallnats}$ of $\{H_k\}_{k\in\smallnats}$ 
such that the sequence $H_{k_i}|_{[c,d]}$
converges uniformly to a function $H\colon[c,d]\to[0,1]$. 

Moreover, if $\liminf_k H_k(a)>0$
then we can assume that the function $H$ is one-to-one.}

\pf First notice that functions $H_k$ are increasing.

Let $Q=\{q_i\colon i\in\nats\}$ 
be a dense subset of $[a,b]$ containing $a$ and $b$. 
The functions $H_k|_Q$ are elements of a compact metric space $[0,1]^Q$.
So, there exists a subsequence $\{H_{k_i}\}_{i\in\smallnats}$ of 
$\{H_k\}_{k\in\smallnats}$
that converges in $[0,1]^Q$; i.e., such
that for every $j\in\nats$ there exists $H(q_j)\in[0,1]$
with the property that
\[
\lim_{i\to\infty} H_{k_i}(q_j)=H(q_j).
\]

If $H(q_r)=0$ for some $q_r\in(a,b)$ then, by Lemma \ref{lem:conv},
interval $[c,d]=[a,q_r]$ and the function $H(x)=0$ for every $x\in[c,d]$
work. So, decreasing $[a,b]$, if necessary, we can assume that $H(a)>0$.
This is also the case when $\liminf_k H_k(a)>0$.

We prove that 
\begin{equation}\label{conP}
P=\closure{H(Q\cap[a',b'])}\subset(0,1]
\end{equation}
is not nowhere dense for every nonempty interval $(a',b')\subset(a,b)$
such that $a',b'\in Q$.
Notice that this will finish the proof, because it implies existence of
a nontrivial interval $[c,d]\subset[a,b]$, $c,d\in Q$, such that
$H(Q\cap[c,d])$ is dense in $[H(c),H(d)]$. So, Lemma \ref{lem:conv} gives
us the desired uniform convergence. Moreover, condition
(\ref{conP}) guarantees also that $H$ will be one-to-one on $[c,d]$.

By way of contradiction let us assume that condition
(\ref{conP}) fails; i.e., that $P$ is nowhere dense 
for some nonempty interval $(a',b')\subset(a,b)$ such that $a',b'\in Q$.
Choosing a subsequence, if necessary, we can assume that
\[
\lim_{i\to\infty}s_{k_{i+1}}^{-1}/s_{k_i}^{-1}=0.
\] 
Then, by Proposition \ref{deep}, there exists an open set 
$W\supset\Cup_{i\in\smallnats}\inv{s_{k_i}}P$ such that
$0$ is an \I-dispersion point of $W$. We will construct a set $V$ such that
$0$ is not an \I-dispersion point of $V$, while $h(V)\subset W$; i.e., 
$h(0)=0$ is an \I-dispersion point of $h(V)$. This will contradict the
assumption that $h$ is \I-density continuous.

So, let us choose a countable base $\{I_i\}_{i\in\smallnats}$ of $[a',b']$ and
for  every $i,j\in\nats$, $j\leq i$, choose 
$q_{i,j},q_{i,j}^\prime\in Q$ such that $q_{i,j}<q_{i,j}^\prime$,
$[q_{i,j},q_{i,j}^\prime]\subset I_j$, and
\begin{equation}\label{conWA}
H([q_{i,j},q_{i,j}^\prime])\subset s_{k_i}W.
\end{equation}
This can be done, since $P\subset s_{k_i}W$, so the distance $d_i$ between
$P$ and the complement of $s_{k_i}W$ is positive and any interval
$[q_{i,j},q_{i,j}^\prime]$ for which 
$H(q_{i,j}^\prime)-H(q_{i,j})<d_i$ satisfies condition (\ref{conWA}).
In addition, choosing subsequence of $\{k_i\}_{i\in\smallnats}$, 
if necessary, we can also assume that for every $i,j\in\nats$, $j\leq i$,
$H_{k_i}(q_{i,j})$ and $H_{k_i}(q_{i,j}^\prime)$ are closer to 
$H(q_{i,j})$ then $d_i$. This means that
\begin{equation}\label{conW}
H_{k_i}((q_{i,j},q_{i,j}^\prime))\subset s_{k_i}W
\ \mbox{ for every }i,j\in\nats,\ j\leq i. 
\end{equation}
Let $V_i=\Cup_{j\leq i} (q_{i,j},q_{i,j}^\prime)$ and 
\[
V=\Cup_{i\in\smallnats}\frac{1}{t_{k_i}}V_i.
\] 
Then, by (\ref{conW}),
\[
h\left(\frac{1}{t_{k_i}}V_i\right)=
\frac{1}{s_{k_i}}\left[s_{k_i}h\left(\frac{1}{t_{k_i}}V_i\right)\right]=
\frac{1}{s_{k_i}}H_{k_i}(V_i)
\subset 
\frac{1}{s_{k_i}}\left[s_{k_i}W\right]=W
\]
for every $i\in\nats$ and, indeed, $h(V)\subset W$.

On the other hand, 
$0$ is not \I-dispersion point of $V$, since 
for any subsequence $\{t_{k_{i_p}}\}_{p\in\nats}$ of
$\{t_{k_i}\}_{i\in\nats}$ the set
$\Cup_{p\geq p_0}t_{k_{i_p}}V\supset \Cup_{p\geq p_0}V_{i_p}$
is open and dense in $(a',b')$ for every $p_0\in\nats$, and so,
\[
(-1,1)\cap\limsup_{p\to\infty}(t_{k_{i_p}}V)\not\in\I.
\]
This finishes the proof of Lemma \ref{lem:func}. \qed

\theorem{th1}{If $f,g\in\idhomeo$, then $f+g\in\idhomeo$.}

\pf Let $f,g\in\idhomeo$  and let $a\in\reals$. It is enough to prove that 
$f+g$ is right \I-density continuous at $a$.
Without loss of generality we may assume that $a=f(a)=g(a)=0$.
Let $D_0\subset [0,\infty)$ be a closed set for which
$0$ is not an \I-dispersion point of $D_0$. By Proposition \ref{homeom}(i),
it is enough to prove that
$0$ is not an \I-dispersion point of $(f+g)(D_0)$.

Let $D_f=\{x\in D_0\colon g(x)\leq f(x)\}$ and 
$D_g=\{x\in D_0\colon f(x)\leq g(x)\}$. Then 
$0$ is not an \I-dispersion point of either $D_f$ or $D_g$.
Assume that $0$ is not an \I-dispersion point of $D_f$.
We may assume, without loss of generality, that $D_0=D_f$; i.e., that
\begin{equation}\label{condDf} 
g(x)\leq f(x)\ \mbox{ for every }\ x\in D_0.
\end{equation}

Let $D$ be the interior of $D_0$. Since $0$ is an \I-dispersion point of
the closed nowhere dense set $D_0\setminus D$, 
$0$ is not an \I-dispersion point of $D$. 
Then, by Lemma \ref{subsec1}, there is 
an increasing sequence $\{t_k\}_{k\in\smallnats}$ of positive numbers
diverging to infinity and a nontrivial interval $[a,b]\subset(0,1)$ 
such that 
\begin{equation}\label{conQ}
Q=\liminf_{k\to\infty}t_kD\cap(a,b)\ \mbox{ is dense in } (a,b).
\end{equation}
We may also easily assume that $b\in\liminf_{k\to\infty}t_kD$; i.e., that
$b/t_k\in D\subset D_0$ for almost all $k\in\nats$. This, together with
(\ref{condDf}), implies that
\begin{equation}\label{conQA}
g(b/t_k)\leq f(b/t_k)\
\mbox{ for almost all } k\in\nats.
\end{equation}

Now, by Lemma \ref{lem:44} used for the function $f$, the sequence
$\{t_k\}_{k\in\smallnats}$ and the interval $[a,b]$,
we may find a subsequence 
$\{t_{k_i}\}_{i\in\smallnats}$ of $\{t_k\}_{k\in\smallnats}$,
and a nonempty interval $(c,d)\subset(a,b)$ such that
$\lim_{i\to\infty} f(c/t_{k_i})/f(d/t_{k_i})>0$.
Without loss of generality we may assume that
$\{t_{k_i}\}_{i\in\smallnats}=\{t_k\}_{k\in\smallnats}$ and $[c,d]=[a,b]$;
i.e., that
\begin{equation}\label{conQQ}
\lim_{k\to\infty} \frac{f(a/t_k)}{f(b/t_k)}>0.
\end{equation}

\pagebreak

Let $s_k=1/(f+g)(b/t_k)$, $F_k(x)=s_k f(x/t_k)$ and 
$G_k(x)=s_k g(x/t_k)$ for $x\in[0,1]$. Then, 
\begin{equation}\label{condB}
(F_k+G_k)(x)=s_k(f+g)(x/t_k)\in[0,1]\ \mbox{ for }\ x\in[a,b],\ k\in\nats,
\end{equation} 
$s_k (f+g)(b/t_k)=(F_k+G_k)(b)=1$ and, by conditions (\ref{conQA})
and (\ref{conQQ}),
\begin{eqnarray}
\liminf_{k\to\infty} F_k(a) & = & 
                          \liminf_{k\to\infty} s_k f(a/t_k) \nonumber \\
& = & \liminf_{k\to\infty} \frac{f(a/t_k)}{(f+g)(b/t_k)} \label{conQQQ} \\
& \geq & \liminf_{k\to\infty} \frac{f(a/t_k)}{2f(b/t_k)} \nonumber \\
& > & 0. \nonumber 
\end{eqnarray}

By Lemma \ref{lem:func} used twice, we can find a nonempty interval
$(c,d)\subset(a,b)$ and a sequence $\{k_i\}_{i\in\smallnats}$ of natural numbers
such that $\{F_{k_i}|_{[c,d]}\}_{i\in\smallnats}$ converges uniformly to some
$F$ and $\{G_{k_i}|_{[c,d]}\}_{i\in\smallnats}$ converges uniformly to a
function $G$. Moreover, by (\ref{conQQQ}), we can also assume that
$F$ and $F+G$ are increasing homeomorphisms on $[c,d]$.
Without loss of generality we may assume that $[c,d]=[a,b]$.

Let $(A,B)=((F+G)(a),(F+G)(b))\subset(0,1]$.
By (\ref{conQ}), the set $(F+G)(Q)$ is dense in $(A,B)$. 
But if $q\in Q$ then, by (\ref{conQ}),
$q/t_k\in D$ for almost all $k\in\nats$. So, for every sequence 
$\{k_i\}_{i\in\smallnats}$ of natural numbers and every $j\in\nats$,
\[
(F+G)(q)       %=\lim_{i\to\infty}(F_{k_i}+G_{k_i})(q)
=\lim_{i\to\infty}s_{k_i}(f+g)(q/t_{k_i})\in 
{\rm cl}\left(\bigcup_{i\ge j} s_{k_i}(f+g)(D)\right)
\]
which implies that the set $\bigcup_{i\ge j} s_{k_i}(f+g)(D)$ is dense in 
$(A,B)$. Thus, the $G_{\delta}$ set
\[
(-1,1)\cap\limsup_{i\to\infty}s_{k_i}(f+g)(D)=
(-1,1)\cap\bigcap_{j\in\smallnats} \bigcup_{i\ge j} s_{k_i}(f+g)(D)\not\in\I;
\]
i.e.,
$0$ is not an \I-dispersion point of $(f+g)(D)$.
This finishes the proof of Theorem \ref{th1}. \qed

\corollary{cor2m}{If $f,g\in\idhomeo$, then $f\,g\in\idhomeo$.}

\pf Same as for Corollary \ref{cor1m}. \qed

\bigskip

Let us also notice that the previous proof works also for
the following result, that is a little bit stronger that Theorem
\ref{th1}.

\corollary{cor2s}{Let $f$ and $g$ be increasing homeomorphisms such that
$f(a)=g(a)$ for some $a\in\reals$. If $g(x)\leq f(x)$ for every $x\geq a$ and
$f$ is right \I-density continuous at $a$ then $f+g$ is also
right \I-density continuous at $a$.}

\section{ Homeomorphisms that preserve \I-density points}

In this section we prove the theorem that $f+g$
preserves \I-density points provided $f$ and $g$ are 
increasing homeomorphisms preserving \I-density points.
For this we need the following lemma analogous to Lemma \ref{lem:func}.

\lemma{lem:func2}{Let $h\RtoR$ be an increasing 
homeomorphism preserving \I-density points such that $h(0)=0$ and 
let $\{s_k\}_{k\in\smallnats}$ and $\{t_k\}_{k\in\smallnats}$ be the
increasing sequences of positive numbers diverging to infinity such
that $H_k(x)=s_kh(x/t_k)\in[0,1]$ for every $x\in[0,1]$.
Then for every nontrivial interval $[a,b]\subset(0,1)$ there exists 
a nonempty interval $(c,d)\subset(a,b)$ and a subsequence 
$\{H_{k_i}\}_{i\in\smallnats}$ of $\{H_k\}_{k\in\smallnats}$ 
such that the sequence $H_{k_i}|_{[c,d]}$
converges uniformly to a function $H\colon[c,d]\to[0,1]$. 

Moreover, if $\limsup_k(H_k(b)-H_k(a))>0$
then we can assume that the function $H$ is one-to-one.}

\pf Let $Q=\{q_i\colon i\in\nats\}$ 
be a dense subset of $[a,b]$ containing $a$ and $b$. 
Functions $H_k|_Q$ are elements of a compact metric space $[0,1]^Q$.
So, there exists an increasing sequence $\{k_i\}_{i\in\smallnats}$ of 
natural numbers such that $H_{k_i}|_Q$ converges in $[0,1]^Q$; i.e.,
that for every $j\in\nats$ there exists $H(q_j)\in[0,1]$ such that
\[
\lim_{i\to\infty} H_{k_i}(q_j)=H(q_j).
\]
Moreover, if $\limsup_k(H_k(b)-H_k(a))>0$ then we can also assume that
\[
H(a)<H(b).
\] 

If $H(a)=H(b)$ then, by Lemma \ref{lem:conv},
interval $[c,d]=[a,b]$ and the function $H=H(a)\charf{[c,d]}$
work. So, we can assume that $H(a)<H(b)$.

By Lemma \ref{lem:conv} in order to prove the first part of Lemma 
\ref{lem:func2} it is 
enough to show that $H(Q)$ is dense in $[H(a),H(b)]\subset[0,1]$.
So, by way of contradiction, assume that 
$H(Q)$ is not dense in $[H(a),H(b)]$. Then, there exists
a nonempty interval $(A,B)\subset[H(a),H(b)]$ such that
$H(Q)\cap[A,B]=\emptyset$, and we can find $a_i,b_i\in Q$, $0<b_i-a_i<1/i$,
such that $H(a_i)<A<B<H(b_i)$ for every $i\in\nats$. Now, taking subsequence of
$\{k_i\}_{i\in\smallnats}$, if necessary, we can conclude that
\[
s_{k_i}h(a_i/t_{k_i})=H_{k_i}(a_i)<A<B<H_{k_i}(b_i)=s_{k_i}h(b_i/t_{k_i})
\]
for every $i\in\nats$. 

Let $U=\bigcup_{i\in\smallnats}t_{k_i}^{-1}(a_i,b_i)$. Then, by Proposition
\ref{INTERdisp}, $0$ is an \I-dispersion point of $U$. But,
\[
[A,B]\subset \left(H_{k_i}(a_i),H_{k_i}(b_i)\right)=
s_{k_i} h\left(t_{k_i}^{-1}(a_i,b_i)\right)
\subset s_{k_i} h(U)
\]
for every $i\in\nats$. So, by (\ref{con_idisp}), $0$ is not \I-dispersion
point of  $h(U)$. This contradicts the assumption that $h$ preserves
\I-density points.

To prove the additional condition let us assume, by way of contradiction, 
that $H$ is not one-to-one on any nonempty interval $(c,d)\subset(a,b)$.
Then, the set 
\[
U=\bigcup\{(c,d)\subset(a,b)\colon H(c)=H(d)\}
\]
is dense in $(a,b)$ and the set $H(U)$ is countable.
In particular, the set $P=[a,b]\setminus U$ is nowhere dense in $[a,b]$, 
while $H(P)$ is dense in $[H(a),H(b)]$.
We will show that this implies that $h$
does not preserve right \I-density at $0$.

Choosing a subsequence of $\{k_i\}_{i\in\smallnats}$, 
if necessary, we may assume that
\[
\lim_{i\to\infty}\inv{ t_{k_{i+1}} }/\inv{ t_{k_i} }=0.
\]
Then, by Proposition \ref{deep}, there exists
an open set $V\supset \Cup_{i\in\smallnats}\inv{t_{k_i}}P$ 
such that $0$ is an \I-dispersion point of $V$. We will show that $0$ is not 
an \I-dispersion point of the open set $h(V)$. 

So, let $\{k_p\}_{p\in\smallnats}$ be an arbitrary subsequence of 
$\{k_i\}_{i\in\smallnats}$. Then, for every $x\in P$,
\[
H(x)=\lim_{p\to\infty}H_{k_p}(x)
=\lim_{p\to\infty}s_{k_p}h(x/t_{k_p})\in 
\mbox{cl}\left(\bigcup_{r\ge p} s_{k_r}h(V)\right)
\]
which implies that the set $\bigcup_{r\ge p} s_{k_r}h(V)$ is dense in 
$[H(a),H(b)]$ for every $p\in\nats$.
Thus, the $G_{\delta}$ set
\[
(0,1)\cap\limsup_{p\to\infty} s_{k_p}h(V)=
(0,1)\cap\bigcap_{r\in\smallnats} \bigcup_{p\ge r}  s_{k_p}h(V)
\not\in\I,
\]
because it is dense in $(H(a),H(b))\neq\emptyset$.
Now, by (\ref{con_idisp}), $0$ is not an \I-dispersion point of $h(V)$.
This finishes the proof of Lemma \ref{lem:func2}. \qed

\theorem{th2}{If $f,g\in\iphomeo$, then $f+g\in\iphomeo$.}

\pf Let $f,g\in\iphomeo$  and let $a\in\reals$. It is enough to prove that 
$f+g$ preserves right \I-density at $a$.
Without loss of generality we may assume that $a=f(a)=g(a)=0$.
Let $E$ be a right interval set such that
$0$ is a right \I-density point of $E$, let 
$\{s_k\}_{k\in\smallnats}$ be an increasing sequence of positive numbers
diverging to infinity and let $0<A<B<1$.
By Propositions \ref{homeom}(ii) and \ref{disppts},
it is enough to prove that there exist
a subsequence $\{s_{k_i}\}_{i\in\smallnats}$ of $\{s_k\}_{k\in\smallnats}$
and a nonempty open interval $J\subset(A,B)$ such that
\[ 
J\subset s_{k_i}(f+g)(E)
\]
for every $i\in\nats$. 

Let us define 
\[
t_k=1/\inv{(f+g)}(B/s_k),\ \ \ a_k=1/\inv{(f+g)}(A/s_k),
\]
\[
F_k(x)=s_k f(x/t_k),\ \  G_k(x)=s_k g(x/t_k)
\]
and
\[
H_k(x)=(F_k+G_k)(x)=s_k(f+g)(x/t_k).
\]
Then, $t_k<a_k$,
$A=s_k(f+g)(1/a_k)$ and $B=s_k(f+g)(1/t_k)$. In particular,
\[
H_k\left(\left[\frac{t_k}{a_k},1\right]\right)=
s_k(f+g)\left(\frac{1}{t_k}\left[\frac{t_k}{a_k},1\right]\right)=[A,B].
\]

Let $\{k_i\}_{i\in\smallnats}$ be a sequence of natural numbers such that
the following limits exist
\[
a=\lim_{i\to\infty}\frac{t_{k_i}}{a_{k_i}}\in[0,1],
\]
\[
F(a)=\lim_{i\to\infty}F_{k_i}(a),\ \ \ 
G(a)=\lim_{i\to\infty}G_{k_i}(a).
\]
We will show that
\begin{equation}\label{condBB}
(F+G)(a)=A.
\end{equation} 

By way of contradiction, let us assume that it is not the case.
We will assume that 
\[
s_{k_i}(f+g)(1/a_{k_i})=A<(F+G)(a)=\lim_{i\to\infty}s_{k_i}(f+g)(a/t_{k_i}).
\]
The other inequality is similar. Let $A<C<(F+G)(a)$. Then, 
\[
s_{k_i}(f+g)(1/a_{k_i})=A<C<s_{k_i}(f+g)(a/t_{k_i})
\]
for almost all $i\in\nats$. Assume that it is true for all $i\in\nats$.
Then, 
\[
\frac{f(1/a_{k_i})+g(1/a_{k_i})}{f(a/t_{k_i})+g(a/t_{k_i})}=
\frac{s_{k_i}(f+g)(1/a_{k_i})}{s_{k_i}(f+g)(a/t_{k_i})}<\frac{A}{C}<1.
\]
Hence, for every $i\in\nats$, either
$\frac{f(1/a_{k_i})}{f(a/t_{k_i}))}\leq\frac{A}{C}$ or 
$\frac{g(1/a_{k_i})}{g(a/t_{k_i}))}\leq\frac{A}{C}$. Without loss of 
generality, passing to a subsequence, if necessary, we can assume that
for all $n\in\nats$
\[
\frac{ f\left( \frac{1}{t_{k_i}} \frac{t_{k_i}}{a_{k_i}} \right)}
     { f\left( \frac{1}{t_{k_i}}       a                 \right)}=
\frac{f(1/a_{k_i})}{f(a/t_{k_i})}\leq\frac{A}{C}<1.
\]
Let $u_{k_i}=f\left( \frac{1}{t_{k_i}} a  \right)$. Then
\[
u_{k_i}^{-1}f\left( \frac{1}{t_{k_i}} \frac{t_{k_i}}{a_{k_i}} \right)
\leq\frac{A}{C}<1=u_{k_i}^{-1}f\left( \frac{1}{t_{k_i}} a  \right);
\]
i.e., 
\begin{equation}\label{condXXX}
\left(\frac{A}{C},1\right)\subset
u_{k_i}^{-1}f\left( \frac{1}{t_{k_i}} 
                               \left(\frac{t_{k_i}}{a_{k_i}},a\right)  \right)
\end{equation} 
for every $i\in\nats$. But, choosing subsequence, if necessary, we can
assume that $\lim_{i\to\infty}t_{k_{i+1}}^{-1}/t_{k_i}^{-1}=0$ and hence,
by Proposition \ref{INTERdisp}, $0$ is an \I-dispersion point of 
\[
D=\bigcup_{i\in\smallnats}\frac{1}{t_{k_i}} 
                               \left(\frac{t_{k_i}}{a_{k_i}},a\right).
\]
On the other hand, by (\ref{condXXX}),
\[
\left(\frac{A}{C},1\right)\subset u_{k_i}^{-1}f(D)
\] 
for every $i\in\nats$; i.e., $0$ is not an \I-dispersion point of $f(D)$. 
This contradicts the assumption that $f$ preserves 
\I-density points. Condition (\ref{condBB}) is proved.

\bigskip

Now notice that condition (\ref{condBB}) implies, in particular, that $a<1$,
since 
$\lim_{i\to\infty}(F_{k_i}+G_{k_i})(1)=B>A=\lim_{i\to\infty}(F_{k_i}+G_{k_i})(a)$.

Using Lemma \ref{lem:func2} twice for functions $F_{k_i}$ and 
$G_{k_i}$, passing to a subsequence, if necessary, we can find
a nontrivial interval $[c,d]\subset (a,1)$ such that 
$\{F_{k_i}|_{[c,d]}\}_{i\in\smallnats}$ converges uniformly to some 
function $F$ and that
$\{G_{k_i}|_{[c,d]}\}_{i\in\smallnats}$ converges uniformly to a function $G$.
Let us notice also that condition 
(\ref{condBB}) implies that either
$\limsup_{k\to\infty}(F_k(1)-F_k(a))>0$ or 
$\limsup_{k\to\infty}(G_k(1)-G_k(a))>0$, since
$\limsup_{k\to\infty}(F_k(1)-F_k(a))+(G_k(1)-G_k(a))=H(1)-H(a)=B-A>0.$
Thus, we can also assume that the function $H=F+G$ is a homeomorphism on $[c,d]$.

By Proposition \ref{disppts},
choosing a subsequence of $\{k_i\}_{i\in\smallnats}$ and a subinterval of $(c,d)$,
if necessary, we may also assume that
\[
(c,d)\subset t_{k_i} E\ \ \mbox{ for every } i\in\nats,
\]
which implies that 
\begin{eqnarray}
( (F_{k_i}+G_{k_i})(c) , (F_{k_i}+G_{k_i})(d) ) & = &
(F_{k_i}+G_{k_i})((c,d)) \nonumber \\
& \subset & (F_{k_i}+G_{k_i})(t_{k_i} E) \nonumber \\
& = & s_{k_i} (f+g)\left(\frac{1}{t_{k_i}} (t_{k_i}E)\right) \nonumber \\
& = & s_{k_i} (f+g)(E). \nonumber
\end{eqnarray}
Now, if $c<c'<d'<d$ then 
\[
A\leq (F+G)(c)<(F+G)(c')<(F+G)(d')<(F+G)(d)\leq B
\]
so, $J=((F+G)(c'),(F+G)(d'))\subset(A,B)$ and
\[
J\subset 
((F_{k_i}+G_{k_i})(c),(F_{k_i}+G_{k_i})(d))\subset s_{k_i} (f+g)(E)
\]
for $i's$ large enough,
since $\{F_{k_i}+G_{k_i}\}$ converges to $F+G$. Thus, we may assume that 
\[
J\subset s_{k_i} (f+g)(E)\cap (A,B)
\]
for every $i\in\nats$. 
This finishes the proof of Theorem \ref{th2}. \qed

\corollary{cor3m}{If $f,g\in\iphomeo$, then $f\,g\in\iphomeo$.}

\pf Same as for Corollary \ref{cor1m}. \qed

\section{Discussion and examples}

We start this section with an explicit statement of the density analog
of Theorem \ref{th2}, that has been proved by Niewiarowski 
\cite{Niewiarowski:Density}.

\theorem{th3}{If $f,g\in\phomeo$, then $f+g\in\phomeo$.}

We are going to present examples showing that none of the Theorems
\ref{th:denscont}, \ref{th1}, \ref{th2} or \ref{th3} remains valid
if we admit that one of the homeomorphisms is decreasing, even
when their sum is an increasing homeomorphism. Moreover, the decreasing function
can be defined by $g(x)=-x$.

\example{ex1}{There exists 
$h\in\idhomeo\cap\iphomeo\cap\phomeo\cap\cinfinity$, $h\RtoR$, such that
$f\RtoR$, $f(x)=h(x)-x$, is strictly increasing but
$f\not\in\idhomeo\cup\iphomeo\cup\phomeo$.}

\pf Let
\[
f(x)= \left\{
\begin{array}{ll}
e^{-x^{-2}} & x>0 \\
0 & x=0 \\
-e^{-x^{-2}} & x<0 \\
\end{array}\right.
\]
It is known that $f$ is \cinfinity, which is not \I-density continuous
and does not preserve \I-density points \cite[Example 10]{CL:Analytic}.
(See also \cite[Example 5.7]{CL:Examples}.)
It also follows easily from Bruckner \cite[Theorem 7]{Bruckner:7} that
$f$ does not preserve density points. 

Define $h(x)=f(x)+x$. Then $h$ is an increasing, \cinfinity\ 
homeomorphism such that $h$ and $\inv{h}$ satisfy a local Lipschitz
conditions. Thus, by Proposition \ref{lip},
$h$ and $\inv{h}$ are density and \I-density continuous.
\qed

\example{ex2}{There exists $h\in\dhomeo\cap\cinfinity$, $h\RtoR$, such that
$f\RtoR$, $f(x)=h(x)-x$, is strictly increasing but $f\not\in\dhomeo$.}

\pf In \cite[Example 1]{CL:SpDensCont} it is constructed a nondecreasing
function $f\in\cinfinity$ which is not density continuous. In fact, $f$ is
strictly increasing except for the intervals forming some right interval set.
It is not difficult to modify this function to be strictly increasing
\cinfinity\ and not density continuous. Then function $h(x)=f(x)+x$ works. \qed

\bigskip

Let us finish this paper with the remark that 
Theorem \ref{th2} is also contained in 
Aversa and Wilczy\'nski 
\cite[Theorem 4]{AversaWilczynski:Homeomorphisms}.%
\footnote{Theorem 4 of \cite{AversaWilczynski:Homeomorphisms}
is not stated explicitly for the increasing homeomorphisms. 
This additional assumption, however, is contained in 
the preliminaries of the paper.}
However, their proof, considerably
shorter that the one presented in this paper, 
contains an essential gap. In their proof
Aversa and Wilczy\'nski 
show that for every homeomorphisms $f$ 
and $g$ preserving \I-density points,
any sequence $\{n_k\}$ of natural numbers and 
any open set $A$ for which $0$ is an \I-dispersion point and a nonempty
interval $J\subset(0,1)$ there exists a nonempty interval $I\subset(0,1)$ 
and subsequence $\{n_{k_i}\}$ such that $f(I)\cup g(I)\subset J$, and
\[
f(I)\cap n_{k_i}f(A)=\emptyset=g(I)\cap n_{k_i}g(A).
\]
From this they conclude that $(f+g)(I)\cap n_{k_i}(f+g)(A)=\emptyset$.
This evidently might be false. To see this you can take, for example,
$f(x)=x^3$, $g(x)=x$, $n_{k_i}=8$, $I=(1/3,1/2)$ and 
$A=\left( [1/24,1/16]\cup[1/6,1/4] \right)^c$.

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\end{document}
