% 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