% Refinements of the density and \I-density topologies

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\title[Density and \I-density topologies]
{Refinements of the density and \I-density topologies}

%author info************************************************
%% First author
\author{Krzysztof Ciesielski}
\address[K. Ciesielski]{Department of Mathematics\\ West Virginia University\\
Morgantown, West Virginia 26506}
\thanks{The first author was partially supported by a West Virginia University
Senate research grant.}

%% Second author
\author{Lee Larson}
\address[L. Larson]{Department of Mathematics\\ University of Louisville\\
Louisville, Kentucky 40292}
\thanks{The second author was aided by a grant from the University
of Louisville College of Arts and Sciences.}

 

%*************************************************************

\subjclass{26A03, 28A05}
\keywords{density topology, \I-density topology, fine topologies}

\date{}

\begin{document}

\maketitle

\begin{abstract}
Given an arbitrary ideal \J\ on the real numbers, two topologies are defined
which are both finer than the ordinary topology. There are nonmeasurable,
non-Baire sets which are open in all of these topologies, independent of \J.
This shows why the restriction to Baire sets is necessary in the usual
definition of the \idens\ topology. It appears to be difficult to find such
restrictions in the case of an arbitrary ideal. 
\end{abstract}

 
In studying real functions it is often helpful to endow the real numbers
\reals\ with a topology finer than the natural one, denoted by \ordinarytop,
arising from its order.
The most common examples of such topologies are the density topology
\densitytop\ and its category analog, the \I-density topology \itopology. 

To recall the definitions of these topologies we need the following notions. 
A point $x\in\reals$ is said to be 
a {\em density point\/} of $A\subset\reals$ if
\begin{equation}\label{def_n}
\lim_{n\to\infty}\frac{\measure{A\cap(x-\frac{1}{n},x+\frac{1}{n})}}{2/n}=1,
\end{equation}
or, equivalently,
\begin{equation}\label{def_h}
\lim_{h\to0^+}\frac{\measure{A\cap(x-h,x+h)}}{2h}=1,
\end{equation}
where \measure{A}\ denotes the inner Lebesgue measure of $A$.

We let \densitypts{A}\ be the set of all density points of $A\subset\reals$.
The density topology is defined by
\[
\densitytop=\{ A\subset\reals:A\subset\densitypts{A}\}.
\]
It is a well-known theorem
that $\densitytop\subset\Lebesgue$, where
\Lebesgue\ stands for the family of all Lebesgue measurable sets.

To motivate the definition of the \I-density topology \itopology, 
the following reformulation of the definition of a density
point is usually used \cite{PWW:CatAn,Wilczynski:CatAn}.

For a measurable set $A$ the following conditions are equivalent:

\begin{description}

\item[(A)] $0$ is a density point of $A$;

\item[(B)] $\lim_{n\to\infty}\frac{\measure{A\cap(-1/n,1/n)}}{2/n} = 1$;

\item[(C)] $\lim_{n\to\infty}\measure{nA\cap(-1,1)} = 2$;

\item[(D)] $\charf{nA\cap(-1,1)}$ converges to $\charf{(-1,1)}$ in measure; and,

\item[(E)] for every increasing sequence $\{n_m\}_{m\in\nats}$ of natural
numbers there exists a subsequence $\{n_{m_p}\}_{p\in\nats}$ such that
\[
\lim_{p\to\infty}\charf{n_{m_p}A\cap(-1,1)} = \charf{(-1,1)}
\ \mbox{a.e.}
\]

\end{description}

The equivalence of (D) and (E) follows from a well-known theorem of Riesz
concerning convergence in measure (or stochastic convergence)
\cite[Theorem 2.11.6]{Bauer:ProbMeasure}.

Using the translation invariance of Lebesgue measure, the same sequence
of equivalences can be rewritten for any density point of $A$.
The significance of the equivalence of (E) and (A) is that
it clearly shows the
measure function itself is not vital to the definition of a density point. What
are needed is the $\sigma$-algebra \Lebesgue\ of measurable sets and the ideal 
\N\ of measure zero sets.

So, let us make an attempt to define a \J-density topology for an arbitrary 
ideal \J\ of subsets of \reals. We start with the following definitions.

A proposition is said to be
{\em true \J-a.e.} if the set of points at which it does not hold is a
member of \J; more formally, a Boolean function $P$ is true \J-a.e. if,
and only if,
\[
\{ x:\neg P(x) \}\in\J.
\]

If $f_n$ is a sequence of real-valued functions, we say that $f_n$
{\em converges (\J)\/} to a function $f$ if for every subsequence
$f_{n_p}$ of $f_n$, there exists a further subsequence $f_{n_{p_q}}$
such that  $f_{n_{p_q}}$ converges pointwise to $f$, \J-a.e.

Finally, a point $a$ is a \J-{\em density point\/} of a set $S\subset\reals$
if \charf{n(S-a)\cap(-1,1)}\ converges (\J) to \charf{(-1,1)}.
Let \jdensitypts{J}{S}\ denote the set of all \J-density points of the
set $S$.

In this way we have defined a \J-density which is analogous to ordinary
Lebesgue density.  In fact, when the ideal \J\ is taken to be the ideal
of Lebesgue null sets \N, and the set $S$ is measurable,
then the equivalence of (A) with (E) shows
that the \J-density points are precisely the
points of ordinary Lebesgue density. To continue the analogy, let
\[
\phitopology{J}=\{ S\subset\reals: S\subset\jdensitypts{J}{S} \}.
\]
It is quite easy to see that the family \phitopology{J}\ is a topology on 
\reals\ and that in the case $\J=\N$, it contains the ordinary density
topology; i.e., that $\densitytop\subset\phitopology{N}$.

At this point, one might suspect that the analogy will continue and that
the properties of \phitopology{J}\ could be developed very naturally along
the same lines as the Lebesgue density topology. However, in the usual
definition of the \I-density topology, \I\ being the ideal of first category
sets, it is additionally assumed that the sets in \itopology\ have the Baire 
property; i.e., the \I-density topology is defined by
\[
\itopology =\phitopology{I} \cap \BaireSets
= \{ S\in\BaireSets: S\subset\idensitypts{S} \},
\]
where \BaireSets\ stands for the family of all subsets of \reals\ with
the Baire property.

It is natural to ask whether this additional assumption in the definition is
essential. The following theorem shows this is really the case.

\theorem{Ex:UnivSet}{
There exists a nonmeasurable set $A\subset\reals$,
which does not have the Baire property, such that
$\lim_{n\to\infty}\charf{n(A - a)\cap(-1,1)} = \charf{(-1,1)}$
for every $a\in A$.
}

Proof. Let $B$ be a Hamel basis
which is also a Bernstein set; i.e., a linear basis
of $\reals$ over $\rationals$, such that $B$ intersects
every nonempty perfect set\footnote
{To construct such a basis 
$B=\{b_\zeta\colon \zeta < c\}$ it is enough
to define it by tranfinate induction on $\zeta$. 
We can simply choose $b_\zeta$
from $P_\zeta\setminus\rationals(\{b_\xi\colon \xi < \zeta \})$, where
$\{P_\zeta\colon \zeta < c\}$ is a fixed enumeration of the nonempty perfect
subsets of \reals\ and $\rationals(\{b_\xi\colon \xi < \zeta \})$ stands for a
subfield of \reals\ generated by $\{b_\xi\colon \xi < \zeta\}$.
Compare also \cite{Morgan}. }. 
For $x\in\reals$ let 
$\rho^\prime(x) = \sumi{n}{|\alpha_i|}$ where $x = 
\alpha_1 b_1 + \alpha_2 b_2 + \ldots + \alpha_n b_n$ is a representation
of $x$ in the base $B$. Define
\[
A = \{x\in\reals\colon \rho^\prime(x) < s\},
\]
where $s\in (1/2,1)$.
Obviously
\[
\frac{1}{2} B\subset A,
\]
so that $A$ intersects every nonempty perfect set. Also,
$B\subset\reals\setminus A$, and thus the complement of $A$ also intersects
every nonempty perfect set. This proves that $A$ is neither Lebesgue
measurable, nor does it have the Baire property.

Let $a\in A$. We will prove that
\[
\lim_{n\to\infty}\charf{n(A - a)\cap(-1,1)} = \charf{(-1,1)}
\]
everywhere.

Let $x\in(-1,1)$. There is a natural number $n_0$ such that
\[
\frac{1}{n} x + a \in A
,\]
for all $n\ge n_0$, $n\in\nats$. In fact, for $x = \beta_1 b_1 + \beta_2 b_2 +
\ldots + \beta_k b_k$ and $a = \alpha_1 b_1 + \alpha_2 b_2 + \ldots + \alpha_k
b_k$ it suffices to choose $n_0$ such that
\[
\sum_{i = 1}^{k}\vert\alpha_i\vert + \frac{1}{n_0}\sum_{i =
1}^{k}\vert\beta_i\vert < s.
\]
Then, for every $n\ge n_0$,
\[
x = n\left(\left(\frac{1}{n} x + a\right) - a\right)\in n(A - a).
\]
This ends the proof.

\bigskip

Let $\O = \{\emptyset\}$. As an immediate corollary we obtain

\corollary{cor1}{ $\phitopology{O}\not\subset\densitytop\cap\itopology$.
In particular, $\phitopology{N}\not\subset\densitytop$ and 
$\phitopology{I}\not\subset\itopology$.
}

It is also easy to see that if $s$ is choosen to be irrational 
in the proof of the previous theorem
then the complement $A^c$ of $A$ satisfies 
a condition similar to $A$. Thus, although the topologies 
\densitytop\ and \itopology\ are connected 
\cite{GW:AppContTrans,Wilczynski:CatAn}, this is not the case 
with \phitopology{N}\ and \phitopology{I}, as stated below.

\corollary{cor:psiex}{Let \J\ be an arbitrary ideal. Then there
exists a nonmeasurable set $A\subset\reals$,
which does not have the Baire property, such that
$A,A^c\in\phitopology{O}\subset\phitopology{J}$. In particular,
\phitopology{J} is disconnected.}

The set $A$ in this corollary is a ``universal'' clopen set in all
the topologies $\phitopology{J}$. In particular, 
$\densitytop\neq\phitopology{N}$.
So the logical question is, why is
it avoided in the ordinary density topology?  A careful reading of the
equivalences (A) and (E) shows that those equivalences are
only valid when the set is assumed to be Lebesgue measurable. Thus, the
ordinary density topology is all sets $S$ such that
$S\subset\jdensitypts{N}{S}$ {\em and} $S$ is measurable, which excludes the
set defined in Theorem \ref{Ex:UnivSet}.
As we mentioned before,
this restriction to measurable sets is a consequence of the normal
definition of the density topology, contained in (\ref{def_n}), but it is lost
with the more general approach, based on ideals.

The above arguments also justify the definition of the \I-density topology as
$\itopology=\phitopology{I}\cap\BaireSets$. In addition, by letting
$f=\charf{A}$, where $A$ is from Corollary \ref{cor:psiex}, the following
corollary is easily seen.

\corollary{cor:fromReferee}{If $\J\in\{\I,\N\}$, then family of all
continuous 
$f:(\reals,\phitopology{\J})\to (\reals,\ordinarytop)$
is strictly larger than both
$f:(\reals,{\cal T}_{\cal J})\to (\reals,\ordinarytop)$.
}

Perhaps in the equivalences (A) through (E) we should replace 
the definition (\ref{def_n}) by (\ref{def_h}) in condition (B)? If we do, we
obtain the following conditions which are
equivalent for every measurable set $A$.

\begin{description}

\item[(A)] $0$ is a density point of $A$;

\item[(B$^\prime$)] for every sequence
$\{t_n\}_{n\in\nats}$
of positive numbers diverging to $\infty$
\[
\lim_{n\to\infty}\frac{\measure{A\cap(-1/t_n,1/t_n)}}{2/t_n} = 1;
\]

\item[(C$^\prime$)] for every sequence
$\{t_n\}_{n\in\nats}$
of positive numbers diverging to $\infty$
\[
\lim_{n\to\infty}\measure{t_nA\cap(-1,1)} = 2;
\]

\item[(D$^\prime$)] for every sequence
$\{t_n\}_{n\in\nats}$
of positive numbers diverging to $\infty$
\[
\charf{t_nA\cap(-1,1)} \,\,\,\mbox{ converges to }\,\,\, \charf{(-1,1)}
 \,\,\,\mbox{ in measure; and, }
\]

\item[(E$^\prime$)] for every sequence
$\{t_n\}_{n\in\nats}$
of positive numbers diverging to $\infty$
there exists a subsequence $\{t_{n_p}\}_{p\in\nats}$ such that
\[
\lim_{p\to\infty}\charf{t_{n_p}A\cap(-1,1)} = \charf{(-1,1)}
\ \mbox{a.e.}
\]

\end{description}

To follow the path already familiar from the previous definitions, let
$\{t_n\}$ be any sequence of positive numbers diverging to $\infty$ and 
for any $S\subset\reals$, define $\Psi_{\cal J}(S,\{t_n\})$
to be the set of all $a\in S$
such that $\charf{t_n(S-a)\cap (-1,1)}$
converges (\J) to \charf{(-1,1)}. Then set
\[
\Psi_{\cal J}(S)=\Cap_{\{t_n\}}\Psi_{\cal J}(S,\{t_n\}),
\]
where the intersection is over all sequences $\{t_n\}$ of positive numbers
diverging to $\infty$. It is not hard to show that
\[
\psitopology{J}=\{ S\subset\reals: S\subset\Psi_{\cal J}(S) \}\subset
\phitopology{J}
\]
is a topology on \reals. This is termed an {\em abstract \J-density
topology.}

It follows, by the equivalence of (E), (E$^\prime$) and (A), that 
$\densitytop=\phitopology{N}\cap\Lebesgue=\psitopology{N}\cap\Lebesgue$.
It is also known that
$\itopology=\phitopology{I}\cap\BaireSets=\psitopology{I}\cap\BaireSets$
\cite{Wilczynski:RemDense,Wilczynski:CatAn}.
But we may hope that at least $\psitopology{N}=\densitytop$
or $\psitopology{I}=\itopology$. However, as the following theorem shows,
this is not the case.

\theorem{ex:generalt}{
There exists a nonmeasurable set $A\subset\reals$,
which does not have the Baire property, such that
$A\subset\Psi_{{\cal I}_c}(A)$,
where $\I_{c}$ is the ideal of all sets of cardinality less than the continuum
$c$. }

Proof.
Let $B = \{b_{\zeta + 1}\colon \zeta < c\}$ be a transcendental base
of $\reals$ over $\rationals$ such that $B$ intersects every perfect
set\footnote
{For the definition of transcendental base see, e.g., \cite{Lang}.
The additional requirements are obtained in the same way as described in
the footnote for Theorem \ref{Ex:UnivSet}.}. 
For $\xi < c$ let $K_\xi$ be the algebraic closure of $\{b_\zeta\colon
\zeta < \xi\}$ in $\reals$ and let $B^{\xi}$ be a linear base of $K_{\xi + 1}$ 
over $K_\xi$ containing $1$ and $b_{\xi}$. 
For $a\in K^{\xi + 1}\setminus K ^\xi$, let 
$\rho^{\prime\prime}(a) = \vert \alpha_0\vert$, where $a = \alpha_0
b_{\xi} + \alpha_1 b_1 + \ldots + \alpha_k b_k$ is a representation  of
$a$ in the base $B^{\xi}$.

Define $A = \{a\in\reals\colon
\rho^{\prime\prime}(a) < 1\}$. As in Theorem \ref{Ex:UnivSet},
\[
\frac{1}{2}B\subset A\hbox{  and  }B\subset\reals\setminus A,
\]
so that $A$ is neither measurable, nor does it have the Baire property. Note
that if $x\in K^\xi$ and $y\notin K^\xi$ then 
$\rho^{\prime\prime}(x + y) = \rho^{\prime\prime}(y)$. Let
$a\in A$ and let $\{t_n\}_{n\in\nats}$ be an increasing sequence diverging to
infinity. Let $\xi < c$ be such that $a, t_n\in K^\xi$ for every $n\in\nats$.
We have $\mbox{card}(K^\xi) < c$. It suffices to show that for every
$x\in(-1,1)\setminus K^\xi$
\[
\lim_{n\to\infty}\charf{t_n(A - a)\cap(-1,1)}(x) = \charf{(-1,1)}(x) = 1.
\]
It is enough to show that $x\in t_n(A - a)$ for all but finitely many $n$. This
is equivalent to the fact that
\[
\frac{x}{t_n} + a \in A.
\]
But,
\[
\frac{1}{t_n}\in K_\xi,
\]
so that
\[
\frac{x}{t_n}\notin K_\xi,
\]
and $a\in K_\xi$. Therefore
\[
\rho^{\prime\prime}\left(\frac{x}{t_n} + a\right) = 
\rho^{\prime\prime}\left(\frac{x}{t_n}\right) =
\frac{\rho^{\prime\prime}(x)}{\vert t_n\vert}.
\]
If we choose $n\in\nats$ such that
\[
\frac{1}{\vert t_n\vert} < 1,
\]
then $x\in t_n(A - a)$. This finishes the proof.

\bigskip

In particular, the previous theorem implies

\corollary{corollary:CHandMA1}{
If the Continuum Hypothesis, or Martin's Axiom, holds, then there is a
nonmeasurable set $A$ without the Baire property such that
$A\subset\Psi_{\cal I}(A)\cap\Psi_{\cal N}(A)$; i.e.,
$A\in\psitopology{N}\cap\psitopology{I}\setminus(\Lebesgue\cup\BaireSets)$.
}

Proof. If the Continuum Hypothesis holds, then
\[
\lessthanc = \left\{B: card(B)\leq{\aleph}_0\right\}
\subset {\cal I}\cap\N.
\]
Martin's Axiom also implies $\lessthanc\subset\I\cap\N$. (See \cite{Kunen}.)

\bigskip

The previous examples show that the topology \psitopology{J}\ does not 
behave as nicely as the ordinary density topology even when $\J=\N$.
Thus, to obtain a good analogy, we must refine the definition somewhat. 
However, the topologies \psitopology{J}\ perhaps deserve closer scrutiny. 
For example, the following 
open problems seem to be interesting.

\problem{problem:regularity}{
Are the topologies \phitopology{I}, \phitopology{N},
\psitopology{I}\ and \psitopology{N}\ on
\reals\ regular? completely regular? normal?}

\problem{problem:disconnected}{
Are the topologies \psitopology{I}\ and \psitopology{N}\ disconnected?}

\problem{problem:ZFC}{
Can we prove, without additional set theoretical assumptions, that
$\psitopology{I}\neq\itopology$ and $\psitopology{N}\neq\densitytop$?}

The discussion above shows clearly that in order to define a ``reasonable''
\J-density topology $\T_{\cal J}$ we should define it as
\[
\T_{\cal J}=\psitopology{J}\cap\F_{\cal J}
\]
for some family $\F_{\cal J}$ of subsets of \reals. However, there is 
an evident problem with this definition. How do we correctly  choose
the family $\F_{\cal J}$ for the given ideal \J? Our choice should 
at least guarantee that the family
\begin{equation}\label{cond:topol}
\T_{\cal J}=\psitopology{J}\cap\F_{\cal J}\,\,\,\mbox{ is a topology on \reals.}
\end{equation}
It would be also very desirable to have the equation
\begin{equation}\label{cond:equation}
\T_{\cal J}=\phitopology{J}\cap\F_{\cal J}.
\end{equation}

In the remaining part of this note we will argue that
for a general ideal \J, finding the natural and
nontrivial family $\F_{\cal J}$ satisfying conditions (\ref{cond:topol}) and
(\ref{cond:equation}) could be very difficult, if not impossible.

The easiest example of these difficulties comes from the ordinary topology
\ordinarytop. 
In particular, it is clear from the next proposition that
we cannot even allow the family $\F_{\cal O}$ to contain all
complements of converging sequences and simultaneously have condition 
(\ref{cond:equation}).

\proposition{naturaltop}{There is a decreasing sequence $S$
converging to $0$ such that $S^c\in\phitopology{O}$. 
Moreover, $\ordinarytop=\psitopology{O}$. 
In particular,
\[
\phitopology{O}\cap\Fsigma\cap\Gdelta\not\subset
\ordinarytop=\psitopology{O}.
\]
}

Proof. The inclusion $\ordinarytop\subset\psitopology{O}$ is obvious. To see the
reverse containement let us take $A\not\in\ordinarytop$. Then, there exists
a point $a\in A$ and a monotone sequence $\{p_n\}\subset\complement{A}$
converging to $a$. Translating the set $A$, if necessary, we may assume that
$a=0$. Then, it is easy to see that $0\not\in\Psi_{\cal O}(A,\{|p_n^{-1}|/2\})$;
i.e., $A\not\in\psitopology{O}$.

For the second part it is enough to choose a
decreasing sequence $S$ converging to $0$ such that the set $S$ is linearly
independent over \rationals. Then 
\charf{nS^c\cap(-1,1)}\ converges everywhere to \charf{(-1,1)},
because for every point $x\in(-1,1)$ there is at most one
$n$ such that $x\in nS^c$. Hence, $S^c\in\phitopology{O}$. 

\bigskip

To see how bad things can be, consider the ideal $\I_\omega$
of countable subsets of \reals. Then we have

\theorem{countebleideal}{There exist nonempty perfect 
sets $P$ and $C$ such that
$\{0\}\cup P^c\in\phitopology{I_\omega}\setminus\psitopology{I_\omega}$ and
$Z^c\in\psitopology{I_\omega}$ for every set $Z\subset C$. In particular,
\[
\phitopology{I_\omega}\cap\Fsigma\cap\Gdelta \not\subset
\psitopology{I_\omega}\,\,\,\mbox{ and }\,\,\,
\psitopology{I_\omega}\not\subset\BaireSets.
\]
}

Proof. We start the proof with the construction of the set $C$.
Define by induction on $n$ a decreasing
sequence of closed sets $\{C_n\}_{n\in\nats}$ such that each $C_n$ is formed
from $2^n$ pairwise disjoint closed subintervals of $[0,1]$ of equal length.
Put $C_0=[0,1]$. To form $C_{n+1}$ from $C_n$ we remove from every component
interval $[a,b]$ of $C_n$ the open interval $(c,d)$ with the same center
and such that $\frac{d-c}{b-a}=\frac{(n+2)! - 1}{(n+2)!}$. Put
$C=\bigcap_{n\in\nats} C_n$.
 
It is not difficult to check that for every $x\in C$ there an interval set
$E=\UoverN [a_n,b_n] \cup \UoverN [c_n,d_m]\supset C$ such that
$a_n<b_n<a_{n+1} < x < d_{n+1}<c_n<d_n$ for every $n\in\nats$,
\[
\lim_{n\to\infty}{(b_n - a_n)/(x-b_n)}=
\lim_{n\to\infty}{(d_n - c_n)/(c_n-x)}=0 
\]
and
\[
\lim_{n\to\infty}(x-a_{n + 1})/(x-b_n)=
\lim_{n\to\infty}(d_{n + 1}-x)/(c_n-x)=0. 
\]
But, it is a consequence of the proof of 
\cite[Theorem 2]{Wilczynski:CatAn} that 
$x\in\Psi_{{\cal I}_\omega}(E^c)\subset\Psi_{{\cal I}_\omega}(C^c)$.
In particular, $\{x\}\cup C^c\in\psitopology{I_\omega}$ for every
$x\in\reals$. This implies that
$Z^c\in\psitopology{I_\omega}$ for every set $Z\subset C$.

For the construction of the set $P$ we use a nonempty perfect set 
$T\subset [\frac{1}{2},1]$ to be constructed later.

Put $t_n = e^n$ and define 
\[
P=\{0\}\cup \UoverN t_n^{-1} T.
\]
Evidently, $P$ is perfect. Also,
$\limsup_{n\to\infty}\charf{t_nP^c\cap(-1,1)}\leq\charf{(-1,1)\setminus T}$,
as $T\subset t_nP$ for all $n\in\nats$.
So, $0\not\in\Psi_{{\cal I}_\omega}(P^c,\{t_n\})$; i.e.,
$\{0\}\cup P^c\not\in\psitopology{I_\omega}$.

To prove that $\{0\}\cup P^c\in\phitopology{I_\omega}$ let us notice that
it suffices to have the following condition:
\begin{equation}\label{condx}
k\,t_n^{-1} T\cap l\,t_m^{-1} T = \emptyset 
\,\,\,\mbox{ for all }\,\,\, k,l,m,n\in\nats, n\neq m.
\end{equation}
This implies that $\{0\}\cup P^c\in\phitopology{I_\omega}$, because
for every $x\in (-1,1)$, $x\neq 0$, there is at most one number $n$ such 
that $x\in\Cup_{k\in\nats} k\,t_n^{-1} T$. So, $x$ belongs to at most 
finite number of sets $k\,P=\{0\}\cup \UoverN k\,t_n^{-1} T$, as for every
$n\in\nats$ there are at most finitely many $k$ for which $x\in k\,t_n^{-1} T$.

To finish the proof we must find a perfect set
$T\subset [\frac{1}{2},1]$ satisfying (\ref{condx}).
We construct $T$ as an intersection $T=\bigcap_{i\in\nats} T_i$,
where the sets $T_i$ are defined by induction on $i$. 
Put $T_0=[\frac{1}{2},1]$.
Every set $T_i\subset T_{i-1}$, $i>0$, is formed
by $2^i$ pairwise disjoint closed intervals of equal length
such that the set $A_i$ of their 
endpoints contains only rational numbers. We construct the
$T_i$ from $T_{i-1}$ by removing from every component interval 
$[a,b]$ of $T_{i-1}$ an open interval $(c,d)$, $a<c<d<b$, with the same center
in such a way that
\[ 
k\,t_n^{-1} T_i\cap l\,t_m T_i^{-1} = \emptyset 
\,\,\,\mbox{ for all }\,\,\, k,l,m,n<i, n\neq m.
\]
But, the condition above is equivalent to the condition
\begin{equation}\label{condy}
T_i\cap \frac{k}{l}e^n T_i = \emptyset 
\,\,\,\mbox{ for all }\,\,\, k,l,n<i, n>0.
\end{equation}
Let $B_{i-1}$ be the union of the sets $\frac{k}{l}e^n A_{i-1}$ for 
$k,l,n<i$, $n>0$. Then $B_{i-1}\cap\rationals=\emptyset$, as 
$A_{i-1}\subset\rationals$ and $e^n$ is irrational for every $n\in\nats$, $n>0$.
Hence, $\varepsilon = \dist{A_{i-1}}{B_{i-1}}>0$. Let $\d >0$
be a number such that $\frac{k}{l}e^n\,\d <\frac{\varepsilon}{2}$ 
for all $k,l,n<i$ and let
us choose an interval $(c,d)\subset [a,b]$ in such a way that
$b-d=c-a<\d$. It is easy to see that this choice guarantees satisfaction of 
condition (\ref{condy}). This finishes the proof of 
Theorem \ref{countebleideal}.

\bigskip

We finish this paper with the following problem.

\problem{problem:lebesgue}{Is $\psitopology{I_\omega}\not\subset\Lebesgue$?}


\ifx\undefined\bysame
\newcommand{\bysame}{\leavevmode\hbox to3em{\hrulefill}\,}
\fi

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\end{document}