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\markboth{Muthuvel and Miller}{EVERYWHERE OF SECOND CATEGORY SETS}
\begin{document}
\noindent Kandasamy Muthuvel, University of Wisconsin, Department of
Mathematics, 800 Algoma Blvd., Oshkosh, WI 54901, muthuvel@uwosh.edu
\bigskip
\noindent Arnold W. Miller, University of Wisconsin, Department
of Mathematics, Van Vleck Hall, 480 Lincoln Drive, Madison, WI 53706,
miller@math.wisc.edu
\begin{center}
\large EVERYWHERE OF SECOND CATEGORY SETS
\end{center}
\footnotetext[1]{Key Words: everywhere of second category,
arithmetic progression, set having distinct distances}
\footnotetext[2]{Mathematical Reviews subject classification:
26A03, 26A15, 04A15, 04A20}
\footnotetext[3]{Real Analysis Exchange, 24(1999), 607-614, received
by the editors May 28, 1998}
\begin{abstract}
The main result of this paper states the following. For each
natural number i, let $G_i$ be a proper additive subgroup of the reals,
$A_i$ a set that contains no arithmetic progression of length three, $H_i$
a basis for the vector space $\R$ over the field of rationals, and
$E^{+}(H_i)$ the set of all finite linear combinations from the elements
of $H_i$ with nonnegative rational coefficients. Then the complement of a
finite union of sets $G_i\cup A_i\cup E^{+}(H_i)$ is everywhere of
second category. We also prove that the complement of a union of fewer
than continuum many translates of sets that have distinct distances is
everywhere of second category.
\end{abstract}
\section{Introduction}
P. Erd\"os and S. Kakutani [3] showed that the continuum hypothesis is
equivalent to the statement that the set $\R$ of real numbers can be
partitioned into countably many pieces such that each piece has distinct
distances. (That is, if $x, y, z, w$ are real numbers with
$\mid x-y\mid =\mid z-w\mid ,$ then $ \{x, y\}=\{z, w\}$.) Under the
assumption of the continuum hypothesis, K. Kunen [6] generalized
that, for each positive integer $n$, $\R^n$ can be partitioned into
countably many pieces such that each piece has distinct distances. It is
interesting to note that, by using the fact that the set $\R$ can
be partitioned into countably many sets such that each set contains no
arithmetic progression of length three [1, Thm 1.1], K. Ciesielski [2, Thm
3] showed that there exists a uniformly anti-Schwartz function from the
reals $\R$ to the natural numbers $\N$.
It is natural to look at the thickness of the complement of a finite union
of the above mentioned sets. Suppose $H$ is a basis for the
vector space $\R$ over the field of rationals, $P$
is a subset of $\R$ that has distinct distances, $A$
is a subset of $\R$ that has no arithmetic progression of
length three, $G$ is a proper
additive subgroup of $\R$, and $E^{+}(H)$
is the set of all finite linear combinations from the
elements of $H$ with nonnegative rational coefficients. Note
that any $H$ set is a $P$ set and any $P$
set is an $A$ set.
In this paper we prove that the complement of a finite union of sets of
the form $A\cup G\cup E^{+}(H)$ is
everywhere of second category. This is a generalization of a classical
result of Sierpi\'nski ``the complement of a Hamel basis is everywhere of
second category'' and a result in [9] ``the complement of a finite union of
Hamel bases is everywhere of second category.'' We also prove
that the complement of a union of fewer than continuum many translates of
sets of the form $P+G$ is everywhere of second
category and this is a generalization of Theorem 7 in [7] that the
complement of a union of fewer than continuum many translates of Hamel bases
is everywhere of second category.
It is interesting to note that there is an additive subgroup, namely, the
group generated by the set $H,$ which is everywhere of second category. The
proof of this follows from the proof of Theorem 1 in [4] with minor suitable
modification. Under the assumption of the continuum hypothesis,
$E^{+}(H)$ is a Lusin set for some set $H$
(see [4]). Under the assumption of Martin's axiom,
$E^{+}(H)$ is a meager set for some set $H$ (see [8]).
\section{Notation}
$\R$ denotes the set of all real
numbers, $\N$ is the set of all natural numbers and $\Q$
is the set of all rational numbers. Let $A$
and $B$ be subsets of $\R$. The symbols $A-B$ and $AB$ stand
for the sets $\{x-y:x\in A$ and $y\in B\}$ and $\{xy:x\in A$ and $y\in B\}$,
respectively. For $r\in \R, A+r=\{x+r:x\in A\}$. The notation
$A\backslash B$ stands for the set-theoretic difference of sets $A$ and
$B$. All sets considered in this paper are subsets of reals.
\section{Results}
\begin{theorem}
For each $i\in \N$, let $G_i$ be a proper additive subgroup
of the reals, $A_i$ a set that contains no arithmetic progression of
length three, $H_i$ a basis for the vector space $\R$ over the field of
rationals, and $E^{+}(H_i)$ the set of all finite linear combinations from
the elements of $H_i$ with nonnegative rational coefficients. Then the
complement of a finite union of sets $G_i\cup A_i\cup E^{+}(H_i$) is
everywhere of second category.
\end{theorem}
First, we prove the following lemmas.
\begin{lemma}
Let $I$ be a nonempty
open interval and let $F$ be a meager subset of $\R$.
Then the set $\N (I\backslash F)$ contains a
translated copy of every countable bounded subset of $\R$.
\end{lemma}
\pf Let $B$ be a
countable bounded subset of $\R$. Then for some nonempty open interval
$J$, $J+B\subseteq mI$ for sufficiently
large $m\in \N $. Since $\N F-B$ is meager,
$J$ is not contained in the set $\N F-B$. Let $j$ be an
element of $J\backslash (\N F-B)$. Then $j+B\subseteq (\N I)\backslash
(\N F)\subseteq \N (I\backslash F)$.\qed
\begin{lemma}
Let $\kappa$ be an infinite
cardinal number smaller than the cardinality of the continuum and let $B$ be
a subset of $\R$ of size $\kappa$. If a subset $C$ of
$\R$ contains a translated copy of every bounded subset of $\R$ of size
$\kappa$, then so does the set $C\backslash (E^{+}(H)+B).$ (Note that
$E^{+}(H)$ is defined in Theorem 1.)
\end{lemma}
\pf Assume that the conclusion of the lemma is
false. Then, for some bounded subset $D$ of $\R$ of size $\kappa $,
the set
$(D+r)\cap (${\it $E^{+}(H)+B)$} is nonempty for every $r$ in
$\{t\in \R:D+t\subseteq C\}$. For simplicity denote $\{t\in
\R:D+t\subseteq C\}$ by $Tr(D, C)$.
\medskip
\noindent {\bf (1)} Hence $Tr(D, C)\subseteq E^{+}(H)+B-D$.
\medskip
Each nonzero real number can be written uniquely as a finite
linear combination of elements of $H$ with nonzero rational
coefficients. For each nonzero $t\in B-D$, let $\supp(t)=\{h\in
H:q_h\neq 0\}$, where $t=\sum\limits_{h\in H}q_hh$. Let $X$ be a
bounded subset of $ -(H\backslash \bigcup\limits_{t\in B-D}\supp(t))$
of size $\kappa $. Since $\mid D+X\mid =\kappa$, by the
definition of $C$, we have $D+X+r\subseteq C$ for some $r $ in $\R$ and
hence $X+r\subseteq Tr(D, C)$. (In fact $Tr(D, C)$ contains a translated
copy of every bounded subset of $\R$ of size $\kappa $.)
According to (1),
\medskip
\noindent {\bf (2)}
$X+r\subseteq ${\it $E^{+}(H)+B-D$} for some $r$ in $\R$.
\medskip
Since $\supp(r)$ is finite and $X$ is an infinite subset of $-H$, there
exists an element $h_1$ in $H$ such that $-h_1\in X$ and
$h_1\notin \supp(r).$
It follows from the definition of $X$ that
$h_1\notin
\supp(t)$ for all $t\in (B-D)$.
By (2),
$$-h_1+r\in E^{+}(H)+B-D,$$
which is impossible, because every element of
$E^{+}(H)$ is a finite linear combination from the elements of $H$ with
nonnegative rational coefficients and $h_1\notin
\supp(r)\cup
\bigcup\limits_{t\in (B-D)}\supp(t)$.\qed
\begin{lemma}
If a subset $C$ of
$\R$ contains a translated copy of every finite subset
of $\R$, so does $C\backslash G$, where $G$ is any
proper additive subgroup of $\R$.
\end{lemma}
\pf Let $X$ be a finite subset of $\R$
and let
$$Tr(X, C)=\{r\in\R:X+r\subseteq C\}.$$
It is
easy to see that $Tr(X, C)-Tr(X, C)=\R$. If
$(X+r)\cap G$ is nonempty for every element $r\in
Tr(X, C)$, then
$$\R=Tr(X, C)-Tr(X, C)\subseteq (G-X)-(G-X)=G-X+X,$$
which contradicts the fact that the index of any proper
additive subgroup of $\R$ is infinite. (To see this fact, assume that
$\R=G+M$, where $G$ is a proper subgroup of $\R$ and $M$ is a finite
subset of $\R$.
Then for each $m\in M, \exists n_m\in \N$ such that $n_mm\in G$. Let $n$ be
the least common multiple of the integers $n_m$. Then
$\R=n\R=nG+nM\subseteq G+nM=G,$ which is a contradiction.)
Thus $X+t\subseteq C\backslash G$
for some $t$ in $Tr(X, C)$.\qed
\begin{lemma}
Let $I$
be a nonempty open interval and let $F$ be a meager subset of
$\R$. Given a positive integer $n$, let $M$
be a finite subset of $\N$ so that if $M$
is partitioned into $n$ classes, then at least one class contains
an arithmetic progression of length three. (Existence of such a set $M$
follows from Van der Waerden's Theorem, see [5, p.28].) Then
there exists a nonzero rational number $d$ and a nonempty open interval $J$
such that $dM+j\subseteq I\backslash F$ for all $j$ in $J$
except for a meager subset of $J$.
\end{lemma}
\pf Let $a$ and $b$
be two distinct elements of $ I$. Then there exist a nonzero rational
number $d$ and a real number $r$ such that
$dM+r\subseteq (\frac a2, \frac b2)$. Let
$F_1=(\frac a2, \frac b2)\cap (F-dM-r)$. Then $F_1$ is
meager and, $\forall x\in (\frac a2, \frac b2)\backslash F_1,$
we have $dM+r+x\subseteq I\backslash F$. Consequently, if we let
$J=(\frac a2, \frac b2)+r$, then $dM+j\subseteq I\backslash F$
for all $j$ in $J$ except for a meager
subset of $J$.\qed
\begin{lemma}
For some $j$ in $J$, we have $(dM+j)\subseteq
(I\backslash F)$ and
$$(dM+j)\cap \bigcup\limits_{i=1}^n (G_i\cup
E^{+}(H_i))=\emptyset$$
where
$d, M, F, I, J$ are defined in Lemma 4 and $G_i ,$
$E^{+}(H_i)$ are defined in Theorem 1.
\end{lemma}
\pf Assume that the conclusion of the lemma is
false. Then, according to Lemma 4,
$(dM+j)\cap \bigcup\limits_{i=1}^n (G_i\cup E^{+}(H_i))$
is nonempty for all $j$ in $J$ except for a meager subset of $J$. This implies
that, for some meager set $F_2,$
$$J\backslash F_2\subseteq \bigcup\limits_{i=1}^n (G_i\cup E^{+}(H_i))-dM.$$
Consequently, because
$\N (G_i\cup E^{+}(H_i))\subseteq (G_i\cup E^{+}(H_i)),$
we have
$$\N (J\backslash F_2)\subseteq
\bigcup\limits_{i=1}^n (G_i\cup E^{+}(H_i))-\N dM.$$
According to
Lemma 1, $\N (J\backslash F_2)$ contains a translated copy of
every countable bounded subset of $\R$. By applying Lemma(2) $n$ times, we
obtain that the set
$$\N (J\backslash F_2)\backslash \bigcup\limits_{i=1}^n (E^{+}(H_i)-\N dM)$$
contains a translated copy of every
countable bounded subset of $\R$. Note that for any
proper additive subgroup $G$ of $\R,$ the set $G-\N dM$
is contained in a proper additive subgroup of $\R.$ For, since $M$
is a subset of $\N$, the set $G-\N dM$ is contained in the subgroup
$G+Zd.$ If $G+Zd=\R$, since $d$ is rational, $G+Z=\R,$ which is
impossible. (For, if $\R=G+Z$, then $\frac 12=g+x$ for some $g\in G$ and
$x\in Z$. Consequently, $0\neq 2x-1\in G$. This implies
that the index of $G$ is
finite, which is a contradiction.) Now by applying Lemma 3 to
$\N (J\backslash
F_2)\backslash \bigcup\limits_{i=1}^n$ $(E^{+}(H_i)-\N dM),$ we obtain that
the set
$$(\N (J\backslash F_2)\backslash \bigcup\limits_{i=1}^n
(E^{+}(H_i)-\N dM))\backslash \bigcup \limits_{i=1}^n (G_i-\N dM)$$
contains a translated copy of every
finite subset of $\R$, but
$$(\N (J\backslash F_2)\backslash
\bigcup\limits_{i=1}^n (E^{+}(H_i)-\N dM))\backslash
\bigcup\limits_{i=1}^n (G_i-\N dM)=$$
$$\N (J\backslash F_2)\backslash\bigcup\limits_{i=1}^n
((G_i\cup E^{+}(H_i))-\N dM)$$ is an empty set.
This contradiction completes the proof of the lemma.\qed
\medskip
To conclude the proof of the theorem, suppose, to the contrary, that
$\R\backslash \bigcup\limits_{i=1}^n(G_i\cup A_i\cup E^{+}(H_i))$ is
meager in a nonempty open interval $I.$ Then, for some meager set $F$,
the set
$I\backslash F$ is contained in the set $\bigcup
\limits_{i=1}^n(G_i\cup A_i\cup E^{+}(H_i))$. Consequently, $(I\backslash
F)\backslash \bigcup\limits_{i=1}^n(G_i\cup E^{+}(H_i))\subseteq
\bigcup\limits_{i=1}^nA_i.$ Now according to Lemma 5, $dM+j\subseteq
\bigcup\limits_{i=1}^nA_i$ for some real numbers $d,j$ and $d\neq 0$.
Hence
$$M\subseteq \frac 1d(\bigcup\limits_{i=1}^nA_i-j).$$
By the definition of $M$ (see Lemma 4),
for some $i$, the set $\frac{1_{}}d(A_i-j)$ contains an
arithmetic progression of length three, which is impossible by the
definition of $A_i$. \qed
\begin{theorem}
Let $G$ be a subset
of $\R$ with cardinal smaller than the cardinality of the continuum and let
$P_1,...,P_n$ be subsets of $\R$ that have distinct
distances. (That is, if $\{x, y, z, w\} \subseteq P_n$ and $\mid
x-y\mid =\mid z-w\mid $, then $\{x,y\}=\{z, w\}.)$ Then
$(P_1+G)\cup \ldots\cup (P_n +G)$ cannot be residual in an interval.
\end{theorem}
First, we prove the following lemma.
\begin{lemma}
Let $\kappa $ be an infinite cardinal
number smaller than the cardinality of the continuum and let $G$ be a subset
of the reals such that $\mid G\mid <\mid\R\mid $. If $P$ is a set that has
distinct distances and $C$ is a set that contains a translated copy of every
set of size $\kappa $, so does $C\backslash (P+G)$.
\end{lemma}
\pf Assume that the conclusion of the lemma is false. Then,
for some subset $X$ of $\R$ of size $\kappa $, the set $(X+r)\cap (P+G)$ is
nonempty for every $r\in \{t\in\R:X+t\subseteq C\}$. Denote the set
$\{t\in\R:X+t\subseteq C\}$ by $Tr(X, C)$ and the set $G-X$ by
$Y$. Then $Tr(X, C)\subseteq P+G-X\subseteq P+Y$.
Let $a\in\R\backslash Q(Y-Y)$ and let $B=\{t\in\R:\{a, -a\}+t\subseteq
Tr(X, C) \}$. Then $\{a, -a\}+B\subseteq
Tr(X, C)\subseteq P+Y.$
\medskip
\noindent {\bf Claim.} $\mid B\mid =\mid\R\mid $
To justify the claim, assume that $\mid B\mid <\mid\R\mid $. Let
$S$ be the group generated by the set $B$. Then $\mid S\mid
<\mid\R\mid.$ Let $d\in\R\backslash S$. Since $\mid X+\{a, -a\}+\{0,
d\}\mid =\kappa $, by the definition of $C$, for some $r\in\R$, we have
$$X+\{a, -a\}+\{0, d\}+r\subseteq C.$$
This implies that $\{a, -a\}+\{0, d\}+r\subseteq Tr(X, C)$.
Now, by the definition of $B,$ $0+r$ and $d+r$ belong to the set
$B$ and hence $r$ and $d+r$ are in the additive subgroup $S.$ Consequently,
$d\in S$, which contradicts the choice that $d\in\R\backslash S.$ Thus
$\mid B\mid =\mid\R\mid .$ To conclude the proof of the lemma,
recall that
\medskip
\noindent {\bf (3)} $\{a, -a\}+B\subseteq P+Y,
\mid Y\mid <\mid\R\mid $ and $\mid B\mid =\mid\R\mid$.
\medskip
For $y$ in $Y$, let $B(y)=\{b\in B:a+b\in P+y\}$. Since $\bigcup\limits_{y
\in Y}B(y)=B$, by (3), there exists $y_1\in Y$ such that
$\mid B(y_1)\mid >\mid Y\mid $. Note that $a+B(y_1)\subseteq
P+y_1.$ Now, $-a+B(y_1)\subseteq P+Y$ and $\mid B(y_1)\mid >\mid
Y\mid $ imply that there exists $y_2\in Y$ and an infinite subset $E$ of
$B(y_1)$ such that $-a+E\subseteq P+y_2$. We have $a+E\subseteq P+y_1$ and
$-a+E\subseteq P+y_2$. For each $e\in E,$ $a+e-y_1$ and
$-a+e-y_2$ are distinct points in $P$ and
$$\mid(a+e-y_1)-(-a+e-y_2)\mid =\mid 2a-y_1+y_2\mid$$
is a constant, which contradicts the property of $P.$\qed
To conclude the proof of the theorem, assume that the conclusion of the
theorem is false. Then for some nonempty open interval $I$ and a
meager set $F,$ we have $I\backslash F\subseteq
\bigcup\limits_{i=1}^n(P_i+G_i)$. Now,
\medskip
\noindent {\bf (4)} $\R\backslash (F+\Q)=(I+\Q)\backslash
(F+\Q)\subseteq (I\backslash F)+\Q\subseteq
\bigcup\limits_{i=1}^n(P_i+G_i+\Q)$.
\medskip
Since $F+\Q$ is meager, it is easy to see that $\R\backslash (F+\Q)$
contains a translated copy of every countable subset of $\R.$ By
applying Lemma 6 $n$ times, we obtain that $(\R\backslash
(F+\Q))\backslash \bigcup\limits_{i=1}^n(P_i+G_i+\Q)$
contains a translated copy of every countable subset of $\R$, but
according to (4), $(\R\backslash (F+\Q))\backslash
\bigcup\limits_{i=1}^n(P_i+G_i+\Q)$ is an empty set. \qed
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\end{document}