%  LaTex2e  11-9-00
%  Vitali sets and Hamel bases that are Marczewski measurable
%  by Arnold W. Miller and Strashimir G. Popvassilev
%  to appear in Fundamenta Mathematicae

\documentclass[12pt,leqno]{article}
\textwidth=30cc
\baselineskip=16pt

\usepackage{amssymb,latexsym}

\def\EE{E}
\def\FF{{^\omega\fair}}
\def\F{{\mathbb F}}
\def\K{{\mathbb K}}
\def\Z{{\mathbb Z}}
\def\al{\alpha}
\def\be{\beta}
\def\case{\par\bigskip\noindent Case }
\def\cc{{\mathfrak c}}
\def\cs{{^\om 2}}
\def\df#1{{\it #1}}
\def\empty{\emptyset}
\def\fair{{\cal F}}
\def\la{\langle}
\def\lex{\mathop{\rm{lex}}}
\def\newn{m}
\def\nwn{k}
\def\om{\omega}
\def\pf{\par\noindent Proof: }
\def\plane{\rr\times\rr}
\def\pr{^\prime}
\def\qed{\par$\Box$\par\bigskip}
\def\qq{{\mathbb Q}}
\def\ra{\rangle}
\def\res{{\upharpoonright}}
\def\rmand{\;\;\mbox{ and }\;\;}
\def\rmiff{\mbox{ iff }}
\def\rmif{\mbox{ if }}
\def\rmor{\mbox{ or }}
\def\rr{{\mathbb R}}
\def\sat#1{[#1]_{_\EE}}
\def\section#1{\bigskip\begin{center}{{\large #1}} \medskip \end{center}}
\def\shf(#1){{(\qq+#1)}}
\def\sig#1{\sigma(#1)}
\def\sm{\setminus}
\def\span{{\rm span}}
\def\su{\subseteq}
\def\zz{\underline{0}}

\newtheorem{theorem}{Theorem}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{conjecture}[theorem]{Conjecture}

\begin{document}

\bigskip
\begin{center}
{\Large  Vitali sets and Hamel bases that are Marczewski measurable}
\end{center}
\bigskip

\begin{center}
by Arnold W. Miller and Strashimir G. Popvassilev\footnote{The second
author was supported by the Bulgarian Foundation EVRIKA 
(I.3 IB-85/22.07.1997, 02.09.1997)
and by a GTA Fellowship from Auburn University. 
\par $\;\;$ Math Review Subject Classification:  03E15, 26A21, 28A05, 54H05
}
\end{center}

\begin{center} Abstract \end{center}

\begin{quote}
We give examples of a Vitali set and a Hamel basis which are
Marczewski measurable and perfectly dense. The Vitali set example answers a
question posed by Jack Brown. We also show there is a Marczewski null Hamel
basis for the reals, although a Vitali set cannot be Marczewski null. The
proof of the existence of a Marczewski null Hamel basis for the plane is
easier than for the reals and we give it first. We show that there is no
easy way to get a Marczewski null Hamel basis for the reals from one for the
plane by showing that there is no one-to-one additive Borel map from the
plane to the reals.
\end{quote}

\section{Basic definitions}

A subset $A$ of a complete separable
metric space $X$ is called \df{Marczewski measurable} 
if for every perfect set $P \subseteq X$ 
either $P\cap A$ or $P\sm A$
contains a perfect set. 
Recall that a \df{perfect set} is a non-empty closed set without isolated
points,  and a \df{Cantor set} is a homeomorphic copy of the Cantor
middle-third set. 
If every perfect set $P$ contains a perfect subset 
which misses $A$, then $A$ is called \df{Marczewski null}.  
The class of Marczewski measurable sets, denoted by $(s)$, 
and the class of Marczewski null sets, denoted by 
$(s^0)$, were defined by Marczewski \cite{M}, 
where it was shown that $(s)$ is a $\sigma$-algebra, i.e.
$X\in (s)$ and  $(s)$ is closed under complements and 
countable unions, and $(s^0)$ is a $\sigma$-ideal in  $(s)$, 
i.e. $(s^0)$ is closed under countable unions and
subsets.
Several equivalent definitions and important properties of $(s)$ and
$(s^0)$ were proved in \cite{M}, for example every analytic set is
Marczewski measurable, the properties $(s)$ and $(s^0)$ are preserved under
``generalized homeomorphisms'' (also called Borel
bijections),
i.e. one-to-one onto functions $f$ such that both
$f$ and $f^{-1}$ are Borel measurable (i.e. pre-images of 
open sets are Borel), a
countable product is in  $(s)$ if and only if each factor is in  $(s)$, and
a finite product is in $(s^0)$ if and only if each factor is in $(s^0)$.

The \df{perfect kernel} of a closed set $F$ is the set of all $a\in F$
such that $U\cap F$ is uncountable for every neighborhood $U$ of $a$.

A set is \df{totally imperfect} iff it contains no perfect subset. 
A totally imperfect set of reals cannot contain uncountable 
closed set, so it must have inner Lebesgue measure zero. 
A set $B$ is called
\df{Bernstein set} if every perfect set intersects
both $B$ and the complement of $B$, 
or, equivalently, both $B$ and its complement are totally
imperfect. 
Clearly, no Bernstein set can be Marczewski measurable.

A set $A$ is \df{perfectly dense} iff its intersection with
every nonempty open set contains a perfect set.


Let ${\rr}$ denote the set of all \df{real numbers} and  
${\qq}$ denote the set of all \df{rational numbers}. 
We use $\cc$ to denote the cardinality of the continuum. 

The \df{linear closure }(or span) over ${\qq}$
of a non-empty set $A\su{\rr}$ is the set
 $$\span(A)=\{q_1 a_1 + \ldots + q_n a_n : n <\om,\ q_j\in {\qq},\ a_j\in A \}
  $$
and $\span(\empty)=\{0\}$. 
 $A$ is
called \df{linearly independent} over ${\qq}$ if
$q_1 a_1+\ldots+q_n a_n\ne 0$
whenever $n<\om$,  $q_j\in {\qq}$ for $1\le j \le n$  with $q_j\ne 0$
for at least one $j$,  and $a_1,\ldots,a_n$   are different points from $A$.
 A linearly independent set $H$ such that ${\rr}=\span(H)$ is called a
\df{Hamel
basis}. Note a Hamel basis must have cardinality $\cc$. The inner Lebesgue
measure of any Hamel basis $H$ is zero (Sierpinski \cite{hamelzero} see also
Erdos \cite{erdos}). A Hamel basis can have Lebesgue measure $0$ 
(see Jones \cite{jones}, or Kuczma Chapter 11 \cite{kuczma}).

A Hamel basis $H$ which intersects every perfect set is called a
\df{Burstin set} \cite{burstin}. Every Burstin set $H$ is also a Bernstein
set,  otherwise if $P\su H$ for some perfect set $P$,  by the linear
independence of $H$ it follows that  $H\cap 2P = \emptyset$
(where $2P=\{2p : p \in P\}$), a contradiction
since $2P$ is a perfect set.

A Burstin set can be constructed as follows.  List all perfect subsets of
${\rr}$ as
 $$\{ P_\alpha : \alpha < \cc \},$$
pick a non-zero $p_0\in P_0$ and using that
 $$|\span(A)|\;\;\le\;\; |A|+ \om \;\; < \;\;\cc \;\; 
   \ \rmif \ |A| < \cc$$
and $|P_\alpha|= \cc $ for each $\alpha$, pick by induction
 $$p_\alpha \in P_\alpha\sm \span(\{p_\beta : \beta < \alpha \})$$
and let $H_{\cc }=\{p_\alpha : \alpha < \cc \}$. If $H$ is a maximal linearly
independent set with $H_{\cc }\su H$, then $H$ is a Burstin set. 

A set $V\su {\rr}$ is called a
\df{Vitali set} if  $V$ is a complete set of
representatives (or a transversal) for the equivalence 
relation defined by $x \sim y \rmiff x-y \in {\qq}$, 
i.e. for each $x\in {\rr}$ there exists a unique $v\in V$ 
such that $x-v\in {\qq}$.
 No Vitali set is Lebesgue measurable or,
has the Baire property.
One may construct a Vitali set which is a Bernstein set.




\section{Perfectly dense Marczewski measurable  Vitali set}


Recall that an equivalence relation on a space $X$ is called Borel if it is a
Borel subset of $X\times X$. The Vitali equivalence $\sim$ as defined above
is Borel. We first show that a Vitali set cannot be Marczewski null.

\begin{theorem} \label{countclass}
 Suppose $X$ is an uncountable separable completely metrizable
space with a Borel equivalence relation, $\equiv$, on it with
every equivalence class countable.  Then, if $V\su X$ meets each
equivalence class in exactly one element, $V$ cannot be
Marczewski null.
\end{theorem}
\pf
By a theorem of Feldman and Moore \cite{FM} every such Borel
equivalence relation is induced by a Borel action of a countable
group.  This implies that there are countably many Borel bijections
$f_n:X\to X$ for $n\in\om$ such that $x\equiv y$ iff
$f_n(x)=y$ for some $n$.  If $V$ were Marczewski null, then
$$X=\bigcup_{n<\om}f_n(V)$$
would be Marczewski null.
\qed

To obtain a Marczewski measurable Vitali set we will
use the following theorem:

\begin{theorem} (Silver \cite{silver})
If $\EE$ is a coanalytic equivalence relation on  the space of all real
numbers and $\EE$ has uncountably many  equivalence classes, then there is a
perfect set of mutually  $\EE$-inequivalent reals (in other words, an
$\EE$-independent perfect set). In the case of a Borel equivalence relation
$\EE$, one can drop the requirement that the field of the equivalence be the
whole set of reals.
\end{theorem}

If $\,\EE\,\su X\times X$ is a Borel
equivalence relation, where $X$ is an uncountable separable completely
metrizable space, and $B$ is a Borel subset of $X$, 
then the saturation of $B$, 
$\sat B = \bigcup_{x\in B} \sat x$, is analytic 
since it is the projection into the second coordinate
of the Borel set $(B\times X)\cap \EE$. 
The saturation need not be Borel, for example let $B$ be a Borel 
subset of $X=\rr^2$ whose projection $\pi_1(B)$ into the first 
coordinate is not Borel. Define $(x,y)\EE(u,v)$ iff $x=u$ (i.e. two
points are equivalent if they lie on the same vertical line). 
Then $\sat B = \pi_1(B)\times\rr$ is not Borel. 
On the other hand, if $\EE$ is a Borel equivalence with each equivalence class
countable, and $f_n$ are as in the proof of Theorem~\ref{countclass}, 
then the saturation $\sat B = \bigcup_{n<\om\,}f_n(B)$
of every Borel set $B$ is Borel.  

\begin{theorem} \label{transversal}
Suppose $X$ is an uncountable separable completely metrizable
space with a Borel equivalence relation $\EE$. 
Then there exists Marczewski measurable $V\su X$
which meets each equivalence class in
exactly one element.
\end{theorem}
\pf
Let $\{ P_\alpha : \alpha < \cc \}$ list all perfect 
subsets of $X$. We will describe how to construct disjoint 
$C_\alpha$~, each $C_\alpha$ either countable 
(possibly finite or empty) or a Cantor set such that the set 
$V_\alpha = \bigcup_{\beta<\alpha} C_\beta$ is $\EE$-independent. 
Then extend the set $V_\cc = \bigcup_{\alpha<\cc} C_\alpha$ to a 
maximal $\EE$-independent set $V$. 

\medskip
Case (a). If $P_\alpha\cap \sat{C_\beta}$ is uncountable for some 
$\beta<\alpha$, then let $C_\alpha = \emptyset$. 

\medskip
Subcase (a1). $|P_\alpha\cap C_\beta|>\om$. Then the perfect
kernel of $P_\alpha\cap C_\beta$ 
is contained in both $P_\alpha$ and $V_\alpha$ (and hence in $V$).

\medskip
Subcase (a2). $|P_\alpha\cap C_\beta|=\om$. Then, since 
$P_\alpha\cap\sat{C_\beta}\sm C_\beta$ is uncountable analytic, 
it contains a perfect set $Q$ which misses $V$. 

\medskip
Case (b). Not case (a). 
Then $|P_\al\cap\sat{V_\al}| = 
|P_\al\cap\bigcup_{\be<\al}\sat{C_\be}| 
\le |\al|\omega < \cc$~, 
and hence $P_\al\sm\sat{V_\al}$  contains a Cantor set $P$. 

\medskip
Subcase (b1). The restriction of $\EE$ to $P$ 
has only countably many classes. 
Let $C_\al$ be a countable $\EE$-independent subset of 
$P$ with $P\su\sat{C_\al}$. 
Then $P\sm C_\al$ contains a perfect set, which misses~$V$. 

\medskip
Subcase (b2). Case (b) but not case (b1). Then, by the above 
theorem of Silver, there is a perfect 
$\EE$-independent set $C_\al\su P$ (and $C_\al\su V$). 
\qed


\begin{remark}
The Vitali equivalence shows that a Borel equivalence need not
have a transversal that is Lebesgue measurable or has the Baire
property. See Kechris \cite{kechris} 18.D for more on transversals of
Borel equivalences.
\end{remark}

\begin{theorem}    \label{2.2}
There exists a Vitali set which is Marczewski measurable and 
its intersection with each non-empty
open set contains a perfect set.
\end{theorem}
\pf
By Theorem~\ref{transversal} there is a Marczewski measurable
Vitali set $V$, and by Theorem~\ref{countclass},
$V$ contains a perfect set $C$.
Split $C$ into countably many Cantor sets $C_0, C_1,\dots$,
fix a basis $\{B_n : n < \omega \}$ for the topology of
$\rr$ and pick rational numbers $q_n$ so that the set 
$q_n + C_n = \{q_n + c : c \in C_n\}$ intersects $B_n$ 
for each $n$. 
Then 
$$V'=(V\sm C)\cup\bigcup\{(q_n+C_n):n<\om\}$$ 
is a perfectly dense Marczewski measurable Vitali set.

\qed

\begin{remark}
 A Vitali set $V$ cannot have the stronger property that 
its intersection with every perfect set contains a perfect set.
This is because if $V$ contains the perfect set $P$,
then the perfect set
$$P' = P+1 = \{p+1 : p \in P \}$$ does not intersect $V$.
Similarly, if $H$ is a Hamel basis that contains the perfect set $P$,
then
$$2P= \{ 2p : p \in P \}$$ is a perfect set which misses $H$.
\end{remark}

\newpage
\section{Marczewski null Hamel bases}

\begin{remark}
(Erdos \cite{erdos}) Under CH there is a Hamel basis $H$ which
is a Lusin set (and hence Marczewski null). To see this,
note that by a
result of Sierpinski there is a Lusin set $X$ such that $X+X=\{x+y: x,y \in
X \} = {\rr}$ (see e.q. \cite{survey1}). Let $H$ be any maximal linearly
independent subset of $X$, then clearly $\span(H)=\span(X)= {\rr}$.
\end{remark}

Our construction (without CH) of a Marczewski null Hamel basis is slightly
simpler for the plane, so we do it first.

\begin{theorem} \label{plane}
There exists a Hamel basis, $H$, for $\plane$, i.e. a basis for the plane
as a vector space over $\qq$, which is a Marczewski null set,
i.e., every perfect set contains a perfect subset disjoint from $H$.
\end{theorem}

\begin{lemma} \label{planelemma}
Suppose $V$ with $|V|<\cc$ is a subspace of $\plane$
as a vector space over $\qq$ (not necessarily closed), 
$p\in \plane$, $y\in\rr$, 
and $$U\su U_y=(\{y\}\times \rr)\cup (\rr\times\{y\})$$
with $|U|<\cc$.  Then there exists a finite $F\su (U_y\sm U)$
with $p\in\span(F\cup V)$ and
such that $F$ is linearly independent over $\qq$ and independent over $V$,
i.e., $\span(F)$ meets $V$ only in the zero vector.
\end{lemma}

\pf \case 1. $p=(u,0)$.
\par Let $y_1$ and $y_2$ be so that
$$ y_2-y_1=u {\rm ,\ \ \ \ \ } 
  (y_1,y)\notin U
   {\rm  \ \ \ and \ \ \ } 
  (y_2,y)\notin U.$$
Clearly $p \in \span(\{(y_1,y),(y_2,y)\})$. Let 
$$ F \subseteq \{(y_1,y),(y_2,y)\} 
   \subseteq U_y\sm U $$ 
be minimal such that $p \in \span(V \cup F)$, then $F$
works. 

\case 2. $p=(0,v)$
\par Obviously this case is symmetric.
\case 3. $p=(u,v)$
\par Apply case 1 to $(u,0)$ obtaining $F_1$. Let
$$V\pr =\span(V\cup F_1)$$
and apply case 2 to $V\pr$ obtaining $F_2$ 
(and let $F=F_1\cup F_2$) 
so that $$(u,0),(0,v)\in\span(V\cup F_1\cup F_2).$$
\qed

The theorem is proved from the Lemma as follows.  Let
$\{B_\al:\al<\cc\}$ list all uncountable Borel subsets of
$\plane$ which have the property that for every
$y$ the set $B_\al\cap U_y$ is countable.  And let
$\{p_\al:\al<\cc\}=\plane$ and
$\{y_\al:\al<\cc\}=\rr$.
Construct an increasing sequence $H_\al\su \plane$ for
$\al<\cc$ so that
\begin{enumerate}
  \item $H_\al$ are linearly independent over the rationals,
%  \item $\al<\be$ implies $H_\al\su H_\be$,
  \item $\be<\al$ implies $H_\be\su H_\al$,
  \item $H_\lambda=\bigcup_{\al<\lambda}H_\al$ at limit ordinals $\lambda$,
  \item $(H_{\al+1}\sm H_\al)\su U_{y_\al}$ is finite,
  \item $p_\al \in \span(H_{\al+1})$
  \item $H_\al\cap B_\be\su H_{\be+1}$ whenever $\be<\al$.
  \item $H_\al\cap U_{y_\be}\su H_{\be+1}$ whenever $\be<\al$.
\end{enumerate}
At successor
ordinals $\al+1$ apply the lemma with $p=p_\al$, $V=\span(H_\al)$, and
$$U=\{p\in U_{y_\al}:\; \exists \be<\al \; (p\in B_\be
\mbox{ or } p\in U_{y_\be})\}.$$
Then let $H_{\al+1} = H_\al \cup F$. 

The set $H=\bigcup_{\al<\cc}H_\al$ is a Hamel basis and note that
for every $y_\al\in \rr$
we have that $H\cap U_{y_\al}\su H_{\al+1}$ and so
$$|H\cap U_{y_\al}|<\cc$$ and similarly for every $\al$
we have that $$|H\cap B_\al|<\cc.$$

To see that
$H$ is Marczewski null, suppose that $P$ is any perfect subset of
the plane. If for some $y\in \rr$ we have that $P\cap U_y$ is
uncountable and closed, then since $|H\cap U_y|<\cc$ and every perfect set
can be split into continuum many perfect subsets, there exists
a perfect set $P\pr\su P\cap U_y$ disjoint from $H$.

On the other hand
if there is no such $y$ then $P=B_\al$ for some $\al$ and so
$|P\cap H|<\cc$.  Thus again by splitting $P$
into continuum many pairwise disjoint perfect subsets,
there must be a perfect subset of $P$ disjoint from $H$.
\qed



\begin{theorem} \label{real}
There exists a Hamel basis, $H$, for the reals which is a Marczewski null set.
\end{theorem}

Obviously, this implies Theorem \ref{plane}, since
$$(H\times\{0\})\cup (\{0\}\times H)$$
is a Marczewski null Hamel basis for the plane.
But the proof is a little messier
so we chose to do the one for the plane first.

For $p,q\in\cs$ define
$$\sig{p,q}=\sum_{n=0}^\infty{p(n)\over 2^{2n+1}}+
            \sum_{n=0}^\infty{q(n)\over 2^{2n+2}}$$
So we are basically looking at the even and odd digits in the binary
expansion.
The function $\sig{p,q}$ maps $\cs\times\cs$ onto the
unit interval $[0,1]$.
For any $p\in \cs$ define
$$U_p=\{\sig{p,q}:\;\; q\in \cs\}$$
The following is the analogue of Lemma \ref{planelemma}.

\begin{lemma} \label{reallemma}
Suppose we have a subspace, $V\su\rr$, with
$|V|<\cc$ and $1\in V$, $p\in\cs$,
$U\su U_p$ with $|U|<\cc$, and $z\in\rr$.
Then there exists a finite $F\su U_p\sm U$ such that
%   $$z\in\span(V\cup F)\rmand \span(F)\cap V=\{0\}.$$
$$z\in\span(V\cup F)\rmand \span(F)\cap V 
\mbox{ is trivial}.$$
\end{lemma}
\pf
\case 1.  $z=\sig{\zz,q}$. ($\zz\in\cs$ is the constantly
zero function.)
\par
We may assume that there are infinitely many $n$ such that $q(n)=0$,
because otherwise $z\in\qq$ and so we may take $F$ to be empty.
Let $$A=\{n:\; q(n)=0\}.$$
For any $B\su A$ define the pair
$q_B,q_B\pr\in\cs$ as follows:
$$q_B(n)=\left\{\begin{array}{ll}
          q(n)  & \rmif n\notin B \\
           1    & \rmif n\in B\\
\end{array}\right. \;\;\;
q_B\pr(n)=\left\{\begin{array}{ll}
            0   & \rmif n\notin B \\
            1   & \rmif n\in B\\
\end{array}\right.$$
Since $q(n)=0$ for
each $n\in B$, it follows that
$q(n)=q_B(n)-q_B\pr(n)$ for every $n$.
Since we never do any ``borrowing'' we have that
$$z=\sig{\zz,q}=\sig{p,q_B}-\sig{p,q_B\pr}$$
Since $|U|<\cc$ there are
continuum many $B\su A$ such that neither $\sig{p,q_B}$
nor $\sig{p,q_B\pr}$ are in $U$.
Fix one of these $B$'s and let 
$$ F \su \{\sig{p,q_B},\sig{p,q_B\pr}\} \su U_p\sm U $$
be minimal, such that $ z \in \span(V \cup F)$. 

\case 2. $z=\sig{q,\zz}$
\par  Since
$${1\over 2}z={1\over 2}\sig{q,\zz}=\sig{\zz,q}$$
this follows easily from case 1.

To prove it for general $z\in\rr\sm\qq$ first we may assume that
$z=\sig{q_1,q_2}$
for some $q_1,q_2\in\cs$ since a rational multiple of $z$ is in
$[0,1]$.  Next we may apply case 1 to $\sig{\zz,q_2}$ and then
iteratively (as in the proof of Lemma \ref{planelemma}) to
$\sig{q_1,\zz}$.  Then since $z=\sig{q_1,\zz}+\sig{\zz,q_2}$
the Lemma is proved.
\qed
Note for any distinct $p_1,p_2\in\cs$ if neither is eventually one,
then $U_{p_1}$ and $U_{p_2}$ are disjoint.
The proof of Theorem \ref{real} is now similar to that of
Theorem \ref{plane}, using the family of $U_p$ for $p\in\cs$
which are not eventually one. 

\qed

\begin{remark}
Similar proofs can be given to produce Marczewski null Hamel
bases for $\rr^n$, $\qq^\om$, and $\rr^\om$.  
For $\rr^n$ one can either modify the proofs of
Theorem \ref{plane} and Lemma \ref{planelemma}, or else 
observe (for example when $n=3$) that if $H$ is a 
Marczewski null Hamel basis for $\rr$, then 
$$(H\times\{0\}\times\{0\})\cup 
(\{0\}\times H\times\{0\}) \cup
(\{0\}\times\{0\}\times H)$$ 
is a Marczewski null Hamel 
basis for $\rr^3$.
If $X=\qq^\om$ or $X=\rr^\om$ then $X$ is isomorphic to 
$X\times X$ and the proofs are similar to 
the proof for the plane.  
\end{remark}


\begin{conjecture}
Suppose $X$ is an uncountable completely metrizable separable metric
space which is also a vector space with respect to a 
% countable
field $\F$ and scalar multiplication and vector sum are Borel
maps.  Then there exists a basis $H$ for $X$ over $\F$ such
that $H$ is Marczewski null.
\end{conjecture}

Note that our conjecture reduces to the case that the field $\F$ is either
$\qq$ or $\Z_p$ for some prime $p$. This is because if $\K$ is a subfield of
$\F$ and $H$ is a Marczewski null basis for $X$ over $\K$, then some
maximal linearly independent over $\F$ subset of $H$ is a Marczewski null
basis for $X$ over $\F$.

F.B.~Jones \cite{jones} constructed a Hamel basis containing a
perfect set and attributed the construction of what might
be called Vitali-independent perfect set to R.L.~Swain.

\begin{theorem}\label{hamelperf}
There is a Hamel basis for $\rr$ which is Marczewski measurable and
perfectly dense.
\end{theorem}
\pf
Let $C$ be a linearly independent Cantor set and $H_0$ be a Marczewski null
Hamel basis. Split $C$ into countably many Cantor sets $C_0, C_1,\dots$,
fix a basis $\{B_n : n <\omega \}$ for the topology of the real
line and for each $n$ pick a non-zero rational $q_n$ such that
$q_n C_n$ intersects $B_n$. Note that
    $$C\pr = \bigcup \{ q_nC_n : n < \om \}$$
is still linearly independent
(though not a Cantor set) and for all open sets $U$ there exists a perfect
$P \su C' \cap U$. Let $H_1 \su H_0$ be maximal such that
    $$H =  C' \cup H_1$$
is linearly independent. It is easy to see that $H$ works.
\qed

\section{Borel Additive mappings}

We might hope to get Theorem \ref{real} as a corollary
to Theorem \ref{plane} getting
a Borel linear isomorphism between $\plane$ and $\rr$.  Since a
Borel bijection preserves the Marczewski null sets, we would
be able to obtain a Marczewski null Hamel basis for the reals
from one for the plane.

This will not work because of the following result.  A mapping
is called additive iff $f(x+y)=f(x)+f(y)$ for any $x$ and $y$.
Note that it $f$ is additive, then $f(rx)=rf(x)$ for any rational $r$.

\begin{theorem} \label{boreladd}
Any additive Borel map $f:\plane\to\rr$ fails to be one-to-one.
\end{theorem}

\begin{lemma}
Suppose $g:\rr\to \rr$ is an additive Borel map.  Then
there exists a comeager $G\su\rr$ and a real $a$ such that
$g(x)=ax$ for every $x\in G$.
\end{lemma}
\pf
This is due to F.Burton Jones \cite{jones}.  Since $g$ is additive it is not
hard to prove that for every rational $a\in\qq$ and real $x$ that
$g(ax)=ag(x)$.  Also since $g$ is Borel there exist a comeager $G$ such that
the restriction of $g$ to $G$ is continuous. Since $aG$ is comeager for any
nonzero $a$ we may without loss assume that $aG\su G$ for every nonzero
rational $a$.  Let $x_0$ be any fixed nonzero element of $G$. For any
$a\in\qq$ we have that $g(ax_0)=ag(x_0)$ and $ax_0\in G$. So by the
continuity of $g$ we have that $g(yx_0)=yg(x_0)$ for any $y$ with
$yx_0\in G$.  Now for any  $x\in G$
 $$g(x)=g({x\over{x_0}}x_0)= {x\over{x_0}} g(x_0)= x {g(x_0)\over{x_0}}$$
and so $a={g(x_0)\over{x_0}}$ works.
\qed


Assume that $f$ is an additive map.
% $f$ is one-to-one.
By the Lemma there exists comeager $G_i$ and reals $a_i$, 
$i=0,1$, such that
for every $x\in G_0$ we have $f(x,0)=a_0x$ and for every $y\in G_1$ we
have $f(0,y)=a_1y$.   Since $f$ is additive it follows that for
every $x,y\in G=G_0\cap G_1$ we have that
$$f(x,y)=a_0x+a_1y.$$
If either $a_i$ is zero, then of course $f$ is not one-to-one.  So
assume both are nonzero.  Let $x$ and $x\pr$ be arbitrary distinct
elements of $G$ and define
$$z= -{a_0\over a_1}(x-x\pr)$$
Since $G$ is comeager, so is $G+z$ and so we can choose $y$ in
both $G$ and $G+z$.
If we let $y\pr$ be so that $y=y\pr +z$,
then $y\pr=y-z\in G$ and
$$f(x,y)=a_0x+a_1y=a_0x+a_1y\pr-a_0(x-x\pr)=a_0x\pr+a_1y\pr=f(x\pr,y\pr)$$
and $f$ is not one-to-one.
\qed

We use similar Baire category arguments to prove the following theorem:

\begin{theorem}
There is no Borel (or even Baire) 1-1 additive
function $f$ of the following form for any $n=1,2,\ldots$
\begin{enumerate}
  \item $f:\rr^{n+1}\to \rr^n$
  \item $f:\rr^n\to \qq^\om$, or $f:\rr^n\to \Z^\om$ 
  ( even for \underline{any} 1-1 additive $f$ )
  \item $f:\qq^\om\to \rr^n$, or $f:\Z^\om\to \rr^n$.
\end{enumerate}
\end{theorem}
\pf

\par\bigskip\noindent (1) $f:\rr^{n+1}\to \rr^n$ \par
This argument is a generalization of Theorem \ref{boreladd}.
There exists a comeager $G\su\rr$ and a linear transformation
$L:\rr^{n+1}\to \rr^n$ with the property that
$$f(x_1,\ldots,x_{n+1})=L(x_1,\ldots,x_{n+1})
\mbox{ for any } x_1,\ldots,x_{n+1}\in G$$
Since $L$ cannot be 1-1 there must be distinct vectors
$u$ and $v$ with $L(u)=L(v)$.  Since $G$ is comeager
there exists a vector $w$ such that $u_i+w_i, v_i+w_i\in G$
for all coordinates $i=1,\ldots,n+1$ (choose $w_i\in(G-u_i)\cap(G-v_i)$).
But then
$$f(u+w)=L(u+w)=L(u)+L(w)=L(v)+L(w)=L(v+w)=f(v+w)$$
implies that $f$ is not 1-1.

\par\bigskip\noindent (2) $f:\rr^n\to \qq^\om,\ $ or
  $f:\rr^n\to \Z^\om$
  ( even for any 1-1 additive function f ). \par
It is enough to prove this for the case $f:\rr^1\to \qq^\om$,
since there are
such maps from $\rr^1$ into $\rr^n$ and
          from $\Z^\om$ into $\qq^\om$.
Let
$f(x)(\newn)\in\qq$ refer to the $\newn^{th}$
coordinate of $f(x)$.
If $f$ is 1-1 and additive, then for each non-zero $x\in\rr$
there is some $\newn$ such that $f(x)(\newn)\ne0$.
By Baire category there must exists some
$q_0\in\qq$ with $q_0\not=0$,
coordinate $\newn$, open interval $I$ and
$H\su I$ comeager in $I$ such that
$$f(x)(\newn)=q_0 \mbox { for every } x\in H.$$
But this is impossible because we can find $\epsilon\in\qq$ with
$\epsilon$ close to $1$ but different from $1$ and some $x$
we have $x,\epsilon x \in H$ but
$$f(x)+f(\epsilon x)=f(x+\epsilon x)=f((1+\epsilon)x) =(1+\epsilon)f(x)$$
Since both $x$ and $\epsilon x$ are in $H$ we have that 
$f(x)(\newn)=f(\epsilon x)(\newn)=q_0$,
contradicting $2 q_0\not= (1+\epsilon)q_0$.


\par\bigskip\noindent (3) $f:\qq^\om\to \rr^n,\ $ or $f:\Z^\om\to \rr^n$ \par
We show there is no such map $f:\Z^\om \to \rr^n$.
Since there is a 1-1 additive Borel map (inclusion) from $\Z^\om$ into
$\qq^\om$, this suffices. We start by giving the proof for $n=1$.  Assume for
contradiction that $G\su\Z^\om$ is a comeager $G_\delta$-set and $f\res G$
is continuous on $G$.

The topology on $\Z^\om$ is determined by the basic open sets
$$[s]=\{x\in \Z^\om : s\su x\}$$
where $s\in \Z^{<\om}$ --- the set of finite sequences from $\Z$.

\par\medskip\noindent Claim. For any $N\in\om$ for any
$s\in \Z^{<\om}$ there exists $t\in \Z^{<\om}$ with
$s\su t$ and for every $x\in G\cap [t]$ we have $f(x)>N$.
\par proof:
Let $m=|s|$ the length of $s$ 
(so $s=\la s(0),\dots,s(m-1) \ra$).
For each $\nwn\in Z$ let $x_\nwn\in \Z^\om$ be the sequence which
is all zeros except on the $m^{th}$ coordinate where it is
$\nwn$.  Since $f$ is additive and 1-1 it must be that either
$\lim_{\nwn\to\infty}f(x_\nwn)=\infty$ or 
$\lim_{\nwn\to-\infty}f(x_\nwn)=\infty$.
Since $G$ is comeager there exists $u\in [s]$ such that 
$u+x_\nwn\in G$ for every $\nwn\in Z$ (i.e, choose
$u\in \bigcap_{\nwn\in\Z}(-x_\nwn+G)$).  
Note that $(u+x_\nwn)\in [s]$ for every $\nwn$
and $f(u+x_\nwn)=f(u)+f(x_\nwn)$ and hence for some $\nwn\in \Z$
we have that $f(u+x_\nwn)>N$. Since $f$ is continuous on $G$ we can find
the $t$ as required.
\par This proves the Claim.
\par\medskip
According to the Claim for each $N$ there exists a dense open set
$D_N$ such that for every $x\in D_N\cap G$ we have $f(x)>N$.  But
this is a contradiction since it implies
$$G\cap \bigcap_{N\in\om}D_N=\emptyset$$

For the case that $f:\Z^\om \to \rr^n$ the argument is similar, 
we just prove a claim that says: For any $N\in\om$ for any
$s\in \Z^{<\om}$ there exists $t\in \Z^{<\om}$ with
$s\su t$ and for every $x\in G\cap [t]$ we have $f(x)(i)>N$ for
some coordinate $i<n$.

\qed

\begin{thebibliography}{99}

\bibitem{burstin}
C.Burstin, {\it Die Spaltung des Kontinuums in c in Lebesgueschem Sinne
nichtmessbare Mengen}, Sitzungsber. Akad. Wiss. Wien
Math. Nat. Klasse Abt. IIa 125(1916), 209-217.


\bibitem{erdos}
P. Erd\"os, {\it On some properties of Hamel basis},
Colloquium Mathematicum, 10(1963), 267-269.

\bibitem{FM}
J.Feldman and C.C.Moore,
{\it Ergodic equivalence relations, cohomology, and von Neumann
algebras I}, Trans. Amer. Math. Soc., 234(1977), 289-324.

\bibitem{jones}
F.B.Jones,
{\it Measure and other properties of a Hamel basis}, Bulletin of the
American Mathematical Society, 48(1942), 472-481.

\bibitem{kechris}
Alexander S. Kechris, {\bf Classical Descriptive Set Theory,} 
Graduate Texts in Mathematics 156, Springer-Verlag, 1995. 
ISBN:~0-387-94374-9, 3-540-94374-9

\bibitem{kuczma}
M.Kuczma, {\bf An introduction to the theory of functional equations and
inequalities. Cauchy's
equation and Jensen's inequality.} Prace Naukowe Uniwersytetu
\'Slpolhk
askiego w Katowicach
[Scientific Publications of the University of Silesia],
489. Uniwersytet \'Slpolhk aski, Katowice; Pa\'nstwowe Wydawnictwo
Naukowe (PWN), Warsaw, 1985. ISBN: 83-01-05508-1

\bibitem{survey1}
A.W.Miller,
{\it Special subsets of the real line}, 
in: {\bf Handbook of Set-Theoretic Topology}, 
ed. K. Kunen and J.E. Vaughan, Elsevier Sci. Publ., 1984.

\bibitem{hamelzero}
W.Sierpinski, {\it Sur la question de la mesurabilit\'e de la base
de Hamel}, Fundamenta Mathematica, 1(1920), 105-111.

\bibitem{silver}
Jack H. Silver, {\it Counting the number of equivalence classes of Borel
and coanalytic equivalence relations}, Annals of Mathematical Logic,
18(1980), 1-28.


\bibitem{M}
Edward Szpilrajn (Marczewski),
{\it Sur une classe de fonctions de M. Sierpinski
et la classe correspondante d'ensembles},
Fund. Math.,  24(1935), 17-34.

\end{thebibliography}

\bigskip\noindent
Arnold W. Miller\hfill
\break
University of Wisconsin-Madison\hfill
\break
Department of Mathematics, Van Vleck Hall\hfill
\break
480 Lincoln Drive\hfill
\break
Madison, Wisconsin 53706-1388\hfill\break
USA

\medskip\noindent
{\rm{E-mail: miller@math.wisc.edu}} \hfill

\medskip\noindent
{\rm{Homepage: http://www.math.wisc.edu/$\tilde{\ }$miller/}}

\bigskip\bigskip\bigskip\noindent
Strashimir G. Popvassilev\hfill
\break
Institute of Mathematics \hfill
\break
Bulgarian Academy of Sciences \hfill
\break
Acad. G. Bontchev street, bl.8 \hfill
\break
1113 Sofia, BULGARIA \hfill

\medskip\noindent
E{\rm{-mail: sgpopv@bgcict.acad.bg}} \hfill

\medskip\noindent
Department of Mathematics, 218 Parker Hall \hfill
\break
Auburn University \hfill
\break
Auburn, AL 36849-5310 \hfill\break
USA \hfill

\medskip\noindent
{\rm{E-mail: popvast@mail.auburn.edu}} \hfill

\medskip\noindent
{\rm{Homepage: http://www.auburn.edu/$\tilde{\ }$popvast/}}

\end{document}





