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\title{A new zero-dimensional Polish group}         
\author{Greg Hjorth}        
\date{\today}          
\maketitle

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\noindent{\large {\bf 0. Introduction}} 

Here we present an example of an abelian zero-dimensional Polish group $G$ that is generated by 
every open neighborhood of the identity -- that is to say, if ${\cal O}\subset G$ is 
open and $x\in G$ then there is some finite sequence $\epsilon_1, \epsilon_2, ...,\epsilon_n\in 
{\cal O}$ such that 
\[x=\epsilon_1+\epsilon_2+...\epsilon_n.\] 


The main point here is that any zero-dimensional {\it locally compact} Polish group has a neigborhood 
basis at the identity consisting of (necessarily clopen) subgroups.  

\bigskip 






\noindent{\large {\bf 1. The group}} 

\medskip 

\noindent{\bf 1.1 Definition} For $x\in\R$ we say that a sequence of 
integers $(a_i)_{i\in\N^+}$ is 
a {\it representation} of $x$ if: 

(i) at each $n$ we have $-10^{-n}<a_n<10^n$; 

(ii) there is some integer $k$ with 
\[k+\sum_{n\in\N^+}\frac{a_n}{10^1\cdot  10^2\cdot ...\cdot 10^n}=x.\] 

\noindent For $\vec a=(a_i)_{i\in\N^+}$ a representation of some real we let 
\[||\vec a||={\rm max}_{n\in\N^+}|\frac{a_n}{10^n}|.\] 
We let $||x||$ be the infinum of the set $\{||\vec a||: a$ is a 
representation of $x\}$. We then let $G$ be the set of $x\in\R$ for which there is some 
representation $\vec a$ of $x$ with 
\[\frac{a_n}{10^n}\rightarrow 0 \:\:\:\: {\rm as} \:\:\:\: n\rightarrow \infty.\] 
A representation $\vec a$ of a real $x$ is {\it legitimate} if we do indeed have that 
\[\frac{a_n}{10^n}\rightarrow 0 \] 
as $n\rightarrow \infty$. 

\medskip 


\noindent{\bf 1.2 Lemma} $G$ is a subgroup of $(\R, +)$. 

Proof. It is immediately clear from the structure of the definitions that 
$-x\in G$ if $x\in G$. For additive closure, suppose $x, y\in G$ with 
legitimate representations $\vec a, \vec b$. Then we may find a representation 
$\vec c$ of $x+y$ with 
\[|c_n|\leq |a_n|+|b_n|+1\] 
for each $n\in\N^+$. \hfill $\Box$ 

\medskip 

\noindent{\bf 1.3 Lemma} For all $x\in G$ we have 
\[||x||\leq\frac{6}{10}.\] 

Proof. Suppose $\vec a$ is a legitimate representation of $x$ and 
\[k+\sum_{n\in\N^+}\frac{a_n}{10^1\cdot  10^2 \cdot  ...\cdot  10^n}=x.\]  
Then we may successively produce integers $l$,  
$\hat{a}_1, \hat{a}_2,...$ and then $b_1, b_2, b_3, ...$ such that at each $M$ 
\[l+\sum_{n<M}\frac{b_n}{10^1 \cdot  10^2\cdot  ...\cdot  10^n}+ 
\hat{a}_M + \sum_{n>M}\frac{a_n}{10^1\cdot  10^2\cdot  ...\cdot  10^n} =x\] 
such that 
\[ |\frac{\hat{a}_n}{10^n}| \leq \frac{1}{2},\] 
\[|b_n|\leq |\hat{a}_n|+1.\] 

The first step of this process is the obvious one: If $|a_1|>5$ then we simply choose 
an integer $l\in \{k-1, k+1\}$ and $\hat{a}_1$ such that 
\[l+\frac{\hat{a}_1}{10^1} =k+\frac{a_1}{10^1},\] 
\[|\hat{a}_1|<5.\] 
If on the other hand $|a_1|\leq 5$ we have no work to do, and simply set $l=k$ and 
$\hat{a}_1=a_1$. 

At the next stage it is similar: If $|a_2|>50$ then we choose $b_1\in \{\hat{a}_1 -1, \hat{a}_1 +1\}$ 
and $\hat{a}_2$ such        that 
\[b_1+ \frac{\hat{a}_2}{10^2} =\hat{a}_1+\frac{a_2}{10^2},\] 
\[|\hat{a}_2|<50.\] 
Again if $|a_2|\leq 50$ then simply set $b_1=\hat{a}_1$ and 
$\hat{a}_2=a_2$. 

We continue in this manner. \hfill $\Box$ 

\medskip 

In general if $\vec a$ and $\vec b$ are two legitimate representations of a 
real $x$ then at each $N$ 
\[ |\sum_{n>N}\frac{a_n}{10^1\cdot  10^2\cdot  ...\cdot  10^n}-\sum_{n>N}\frac{b_n}{10^1\cdot  
10^2\cdot  ...\cdot  10^n}|\leq \frac{1}{10^1 \cdot 10^2\cdot ...\cdot 10^N}.\] 
Therefore for some $i\in\{0, 1, -1\}$ 
\[ |\sum_{n\leq N}\frac{a_n}{10^1\cdot  10^2\cdot  ...\cdot  10^n}-\sum_{n\leq N}\frac{b_n}{10^1\cdot  
10^2\cdot  ...\cdot  10^n}+1|\leq \frac{1}{10^1 \cdot 10^2\cdot ...\cdot 10^N},\] 
and thus $a_N$ must equal $b_N-1,$ $b_N,$ or $ b_N+1$ modulo $10^N$. 

\medskip 




\noindent{\bf 1.4 Lemma} If $x\in G$ then there is a legitimate representation 
$\vec a$ with 
\[||x||=||\vec a||.\] 

Proof. Let $\vec a$ be a legitimate representation of $x$. 
By the previous lemma we may assume 
\[||\vec a|| <\frac{8}{10}.\] 
Choose $N\in\N^+$ with 
\[\frac{|a_n|+1}{10^n}<10^{-1}\] 
all $n>N$. 
Let $\vec b$ be a competing legitimate representation of $x$ with 
$||\vec b||\leq ||\vec a||$. 

\noindent{\bf Claim}: For all $n>N$ 
\[a_n=b_n.\] 

Proof of claim: Assume for a contradiction that $a_n\neq b_n$ and 
$n>N$. There are two possibilities. 

\leftskip 0.4in 

\noindent Case(i) $a_n=b_n$ mod $10^n$. Then since $-10^n<a_n, b_n< 10^n$ 
we must have $|a_n|+|b_n|=10^n$. 
\[\therefore \frac{|b_n|}{10^n}=1-\frac{|a_n|}{10^n}>1-10^{-1},\] by the 
assumption on $n>N$, with a contradiction to $||\vec b||\leq ||\vec a||$. 

\noindent Case(ii)  $a_n=b_n-1$ mod $10^n$ or $a_n=b_n-1$ mod $10^n$. 
 
\noindent There is a 
further split. 

\leftskip 0.6in 

\noindent Case(iia) $a_n\neq b_n-1, b_n+1$. Then it is as case(i) with 
$|a_n|+|b_n|=10^n \pm 1$, and hence  
\[\frac{|b_n|}{10^n}=1-\frac{|a_n|\pm 1}{10^n}>1-\frac{2}{10}=\frac{8}{10},\] 
violating $||\vec b||\leq ||\vec a||$. 

\noindent Case(iib) $a_n=b_n-1$ or $a_n=b_n+1$.  
Thus for some $\eta$ with $\eta\in\{-1, 0, 1\}$. 
Then 
\[\frac{a_n}{10^n}+\frac{a_{n+1}}{10^n\cdot 10^{n+1}}= 
\frac{b_n}{10^n}+\frac{b_{n+1}}{10^n\cdot 10^{n+1}}+\frac{\eta}{10^n\cdot 10^{n+1}}.\] 
Then we have $a_{n+1}$ not equal to $b_{n+1}-1$, $b_{n+1}$, or $b_{n+1}+1$, and we are 
back into the contradiction of case(i). \hfill (Claim $\Box$) 

\leftskip 0in 

Thus in determining $||x||$ we are taking the infinum over a finite set. Thus this inf is 
attained. \hfill $\Box$ 

\medskip 

\noindent{\bf 1.5 Lemma} For $x, y\in G$, 
\[||x||=||-x||\] 
and 
\[||x+y||\leq ||x||+||y||.\] 

Proof. The  equality $||x||=||-x||$ is obvious from the very form of the definitions. 

For the inequality, we may as well assume $||x||+||y||<1$, or else we are done by lemma 
1.3. Fixing representations $\vec a$, $\vec b$ with 
\[||\vec a||=||x||\] 
and 
\[||\vec b||=||y||\] 
we have in particular at each $n$ that 
\[\frac{|a_n|}{10^n}+\frac{|b_n|}{10^n}<1.\] 
Therefore if we just let $c_n=a_n+b_n$ we obtain a representation for $x+y$ 
with 
\[|c_n|\leq|a_n|+|b_n|.\] 
\hfill $\Box$ 

\medskip 

\noindent{\bf 1.6 Corollary} If we set $\hat{d}(x, y)=||x-y||$, then 
$\hat{d}$ defines a pseudo-metric on $G$. 

\medskip 

\noindent{\bf 1.7 Definition} We let $d_{\R}$ be the usual euclidean metric on $\R$ and 
define $d_G$ on $G$ by 
\[d_G(x, y) =d_{\R}(x, y)+\hat{d}(x, y).\] 
Then $d_G$ is the sum of a pseudo-metric and metric, and therefore 
is a metric. 



\medskip 


\noindent{\bf 1.8 Lemma} $d_G$ is a complete metric on $G$. 

Proof. Let $(x_i)_{i\in\N}$ be a Cauchy sequence in $(G, d_G)$. 
Fix at each $i$ a representation ${\vec a}^i=(a_1^i, a_2^i, a_3^i,...)$ of $x_i$ with 
\[||{\vec a}^i||=||x_i||.\] 
After perhaps thinning out the original sequence of reals we may 
assume that for each $n$ the sequence $(a_n^0, a_n^1, a_n^2,...)$ is 
eventually constant. 
We may also assume that at each $i$ 
\[x_i=\sum_{n\in\N^+}\frac{a_n^i}{10^1\cdot  10^2\cdot ...\cdot 10^n};\] 
in other words, the $k$ is always zero.   Letting $a_n^{\infty}$ be the limit of the 
sequence $(a_n^0, a_n^1, a_n^2,...)$, 
we will show that ${\vec a}^{\infty}$ is a legitimate representation of  
some $x_{\infty}\in G$. 

\noindent{\bf Claim:} For all $\epsilon>0$ there is some $N_{\epsilon}$ and 
$I_{\epsilon}$ such that for all $m>N_{\epsilon}$ and $j>I_{\epsilon}$ 
\[\frac{a_m^j}{10^m}<{\epsilon}.\] 


Proof of claim: Fix $\epsilon>0$ and assume without loss of generality that 
$\epsilon<10^{-1}.$ Choose $I_{\epsilon}$ so that for all $i,j\geq I_{\epsilon}$ 
we have 
\[d_G(x_i, x_j)<\frac{\epsilon}{3}\] 
and 
\[\frac{\epsilon}{3}>10^{-I_{\epsilon}}.\] 
Let $N_{\epsilon}$ be such that for all $m>N_{\epsilon}$ 
\[\frac{a_m^{I_{\epsilon}}}{10^m}<\frac{\epsilon}{3}.\] 
Then for all $j>I_{\epsilon}$ we may obtain a legitimate representation $\vec c$ 
of $x_j$ such that at each $m>N_{\epsilon}$ 
\[\frac{c_m}{10^m} <\frac{\epsilon}{3}+\frac{\epsilon}{3}=\frac{2\epsilon}{3}.\] 
If 
\[\frac{a_m^j}{10^m}\geq \epsilon,\] 
then 
\[|a_m^j - c_m|>1,\] 
and thus we must be in the situation that there is some $\eta\in\{-1, 0, 1\}$ 
such that 
\[|a_m^j - c_m|=\eta\:{\rm mod}\: 10^j\] 
but 
\[|a_m^j - c_m|\neq\eta\] 
Thus 
\[\frac{|a_m^j|}{10^m}>1-\frac{|c_m|}{10^m}-\frac{1}{10^m},\] 
entailing the conclusion that 
\[ ||\vec a^j||>\frac{8}{10},\] 
which was already prohibited by lemma 1.3 and the assumption 
$||x_j||=||\vec a^j||$. \hfill (Claim$\Box$) 

Thus we may indeed obtain a real $x_{\infty}\in G$ with 
representation 
\[x_{\infty}=\sum_{n\in\N^+}\frac{a^{\infty}_n}{10^1\cdot ...10^n}.\] 
To see that 
\[x_i\rightarrow x_{\infty}\] 
in $(G, d_G)$ we appeal to the claim once more: Fix $\delta>0$ and 
assume that $\delta<1$.  
Then we may find $N_{\delta/2}$ and $I_{\delta/2}$ as in claim; 
in particular this gives 
\[\frac{a^{\infty}_n}{10^n}<\frac{\delta}{2}\] 
all $n>N_{\delta/2}$. 
By possibly 
enlarging $I_{\delta/2}$ we may assume that for all $i>I_{\delta/2}$ and 
$n\leq N_{\delta/2}$ 
\[a_n^i=a_n^{\infty}.\] 
Then we obtain that for all $ i>I_{\delta/2}$ and at {\it all} $n\in\N$ 
\[\frac{|a_n^i-a_n^{\infty}|}{10^n} <\delta,\] 
and thus $||x_n-x_{\infty}||$ has a representation $\vec c$ with 
\[||\vec c||<\delta.\] \hfill $\Box$ 

\medskip 

\noindent{\bf 1.9 Lemma} $(G, d_G)$ is separable. 

Proof. Let $A\subset G$ be the set of all reals in $G$ with a representation that 
has finite support -- that is to say, if $x\in A$ then there will be some 
$k\in\Z$ and $(a_i)_{i\in\N}$ such that at each $n$ we have $-10^{-n}<a_n<10^n$, 
$k+\sum_{n\in\N^+}\frac{a_n}{10^1\cdot  10^2\cdot ...\cdot 10^n}=x,$ 
and $a_n=0$ for sufficiently large $n$. 

Then given any $y\in G$ and $\epsilon>0$ we need to show 
\[\{x\in G: d_G(x, y)<\epsilon\}\cap A\neq 0.\] 
Suppose that 
\[l+\sum_{n\in\N^+}\frac{b_n}{10^1\cdot  10^2\cdot ...\cdot 10^n}=y\] 
with $l\in Z$ and $\vec b$ a legitimate representation. Then we may find $N$ 
such that for all $n>N$ 
\[\frac{|b_n|}{10^n}<\epsilon.\] 
Then if we let 
\[x=l+\sum_{n\leq N}\frac{b_n}{10^1\cdot  10^2\cdot ...\cdot 10^n}\] 
we have our member of $A$. \hfill $\Box$ 

\medskip 

\noindent{\bf 1.10 Lemma} The group operations in $(G, d_G)$ are continuous. 

Proof. Suppose that $x_0, y_0\in G$. Then for any $\epsilon>0$ and $x, y\in G$ with 
\[||x_0-x||, ||y_0-y||<\epsilon\] 
then by 1.5 we have 
\[||(-x_0)-(-x)||=||-(x_0-x)||=||x_0-x||<\epsilon\] 
and 
\[||(x_0+y_0)-(x+y)||=||(x_0-x)+(y_0-y)||\leq ||x_0-x||+||y_0-y||<2\epsilon.\] 
\hfill $\Box$ 

\medskip 







\noindent{\bf 1.11 Lemma} $(G, d_G)$ is zero-dimensional. 

Proof. It suffices by 1.10 to show that there is a neighborhood basis at the identity 
consisting of clopen sets. So fix $\epsilon>0$, and we will show that there is some 
clopen set ${\cal O}$ such that 
\[0\in {\cal O} \subset \{x\in G: d_G(x, 0)<\epsilon\}.\] 
Choose {\it irrational} $\delta>0$ with 
\[\delta<\frac{1}{10^2}, \epsilon.\] 
Let ${\cal O}=\{x\in G: d_{\R}(x, 0)<10\delta, ||x||<\delta\}$. 

Clearly ${\cal O}$ is open in $(G, d_G)$, so we just need to show it is closed. 
For this purpose, fix $x\in G\setminus {\cal O}$. 

If $||x||\geq\delta$ then since $||x||$ rational by 1.4, and we have 
$||x||>\delta$. Since $\{y\in G: ||y||>\delta\}$ is an open set, we 
have shown that there is an open neighborhood around $x$ disjoint from 
${\cal O}$. 

Alternatively if $||x||<\delta$ then we must have $d_{\R}(x, 0)\geq 10\delta$. 
Then it follows that $x$ has a representation $\vec a$ with $||\vec a||<\delta$ 
and 
\[k+\sum_{n\in\N^+}\frac{a_n}{10^1\cdot  10^2\cdot ...\cdot 10^n}=x,\] 
and necessarily $k\neq 0$. 
Thus in fact 
\[d_{\R}(x, 0)>\frac{1}{2}\]
and we have an open set containing $x$ and disjoint from ${\cal )}$. \hfill $\Box$ 

\medskip 

\noindent{\bf 1.12 Lemma} Every neighborhood of the identity in $(G, d_G)$ generates
the group. 

Proof. Fix $\epsilon>0$ and $x\in G$ with a legitimate representation $\vec a$ with 
\[k+\sum_{n\in\N^+}\frac{a_n}{10^1\cdot  10^2\cdot ...\cdot 10^n}=x.\] 
Choose $N\in \N$ with 
\[\frac{|a_n|}{10^n}, \frac{1}{10^n}<\epsilon\] 
all $n\geq N$. 

Then we may certainly obtain 
\[k+\sum_{n<N}\frac{a_n}{10^1\cdot  10^2\cdot ...\cdot 10^n}\] 
in the subgroup of $G$ generated by 
\[\frac{1}{10^N}.\]
Since 
\[d_G(0, \frac{1}{10^n}), \: d_G(0, \sum_{n\geq N}\frac{a_n}{10^1\cdot  10^2\cdot 
...\cdot 10^n})<\epsilon\] 
we have shown that $x$ is generated by $\{y\in G: d_G(0, y)<\epsilon\}$. \hfill $\Box$ 

\medskip 

\noindent{\bf 1.13 Theorem} In the topology generated by the metric $d_G$, 
$G$ is a zero-dimensional Polish group. 

Proof. This is exactly the content of the last five lemmas. \hfill $\Box$ 








\begin{thebibliography}{99}

\bibitem{scottish} R.D. Mauldin, {\bf The Scottish Book (Mathematics from the 
Scottish Caf\'{e})}, Birkh\"{a}user, Boston, 1981. 




\end{thebibliography}

6363 MSB

Mathematics

UCLA

CA90095-1555

greg@math.ucla.edu

www.math.ucla.edu/\~{}greg
























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