% LaTeX Article Template - customizing header and footer \documentstyle[12pt, oldlfont]{article} \input{amssymb.def} \input{amssymb.tex} % Set left margin - The default is 1 inch, so the following % command sets a 1.25-inch left margin. %\setlength{\oddsidemargin}{0.25in} % Set width of the text - What is left will be the right margin. % In this case, right margin is 8.5in - 1.25in - 6in = 1.25in. %\setlength{\textwidth}{6in} % Set top margin - The default is 1 inch, so the following % command sets a 0.75-inch top margin. %\setlength{\topmargin}{-0.25in} % Set height of the header \setlength{\headheight}{0.3in} % Set vertical distance between the header and the text %\setlength{\headsep}{0.2in} % Set height of the text %\setlength{\textheight}{8.5in} % Set vertical distance between the text and the % bottom of footer %\setlength{\footskip}{0.3in} % Set the beginning of a LaTeX document \begin{document} \def\Ubf#1{{\baselineskip=0pt\vtop{\hbox{$#1$}\hbox{$\sim$}}}{}} \def\ubf#1{{\baselineskip=0pt\vtop{\hbox{$#1$}\hbox{$\scriptscriptstyle\sim$}}}{}} \def\R{{\Bbb R}} \def\V{{\Bbb V}} \def\N{{\Bbb N}} \def\Z{{\Bbb Z}} % Redefine "plain" pagestyle \makeatletter % `@' is now a normal "letter' for LaTeX \title{A new zero-dimensional Polish group} \author{Greg Hjorth} \date{\today} \maketitle % Set to use the "plain" pagestyle \pagestyle{plain} \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}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^n1-10^{-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 $-10^{-n}0$ 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