\documentclass[12pt,notitlepage]{amsart} \usepackage[dvips]{graphicx} \usepackage{amsmath, amsfonts,amssymb} \usepackage{pstricks} \usepackage[myheadings]{fullpage} \usepackage{amsxtra} \setlength{\hfuzz}{5pt} \usepackage{amsthm} \usepackage{fullpage} \newtheorem{theorem}{Theorem}[section] \newtheorem{conjecture}[theorem]{Conjecture} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{definition}[theorem]{Definition} \newtheorem{solution}[theorem]{Solution} \newtheorem{example}[theorem]{Example} \newtheorem{remark}[theorem]{Remark} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{problem}[theorem]{Problem} \newtheorem{proposition}[theorem]{Proposition} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \setcounter{page}{1} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \newcommand{\R}{\mathbb R} \newcommand{\N}{\mathbb N} \newcommand{\Z}{\mathbb Z} \newcommand{\Q}{\mathbb Q} \newcommand{\Id}{\boldsymbol{1}} \newenvironment{enumroman} {\begin{enumerate}\renewcommand{\labelenumi}{\textnormal{(\roman{enumi})}}} {\end{enumerate}} \begin{document} \title{Math H104 Lecture 1, August 30, 2005} \author{Lecturer: Yuval Peres. \\ {} \\ Scribe: Thomson Nguyen} \address{University of California, Berkeley} \email{peres@stat.berkeley.edu} \date{\today} %\begin{abstract} %\end{abstract} \maketitle \markboth{Yuval Peres}{Math H104 Lecture 1} Let $\Z$ be the set of all integers $\{0,1,-1,2,-2,\ldots\}$ and let $\N$ be the positive integers $\{1,2,3,\ldots\}$. Denote the rational numbers by $\Q=\{\frac{a}{b}\, :\, a,b \in \Z \mbox{ \rm and } b\neq 0\}$. The ancient Greeks already discovered that rational numbers are not sufficient to describe certain natural geomtrical quantities, such as the diagonal in a square of side 1. \begin{proposition} \label{sqrt2} $\sqrt{2} \notin \Q$. That is, for every $a,b\in\Z$ with $b\neq 0$, we have $\displaystyle (a/b)^2\neq 2$. \end{proposition} \begin{proof} Suppose $(a/b)^2=2$ with $a,b \in \Z$. We may assume that $a,b>0$, otherwise we replace $a,b$ by their absolute values. We also may assume that we chose a solution with $a$ minimal. The equation $a^2=2b^2$ implies that $a$ is even, and therefore $a^2 \textrm{ is divisible by } 4.$ Consequently $b^2=a^2/2$ is even whence $b$ is even. Therefore we can replace $a$ and $b$ by $a/2$ and $b/2$, and obtain a smaller pair of integers where the ratio of their squares is $2$. This contradicts the minimality of $a$. \end{proof} The construction of the real numbers $\R$ can be done either via Dedekind cuts, or using Cauchy sequences. A {\bf Dedekind cut} $A|B$ consists of a pair of disjoint nonempty sets $A,B \subset \Q$, such that $A\cup B=\Q$ and $a0\textrm{ and } x^2>0\}. \end{equation} We will return to Dedekind cuts later. Recall that a sequence $\{x_n\}$ {\bf converges} to a limit $L$ (in symbols, $x_n \to L$ as $ n \to \infty$) if for any $\epsilon >0$ there is an $n_0 \in \N$ such that $|x_n-L| <\epsilon$ for all $n>n_0$. For now, focus on $x_n, L,\epsilon \in \Q$. This also applies to the next definition. However, these definitions will apply more generally later. We need a more sophisticated definition that describes when the members of a sequence are getting closer to each other without refering to any limit. \begin{definition} A sequence $\{x_n\}^\infty_{n=1}$ is a \textbf{Cauchy sequence} if for all (rational) $\epsilon>0$, there exists an $N$ such that $m,n>N\Rightarrow |x_m-x_n|<\epsilon$. \end{definition} For example, the sequence $\{ 3.1,3.14,3.141.3.1415,3.14159,...\}$ where each time we add another digit in the decimal expansion of $\pi$, is a Cauchy sequence. As we shall see later in the course, $\pi \notin \Q$, so this sequence does not converge in $\Q$. Similarly, if $x_n^2 \rightarrow 2$, then $\{x_n\}$ cannot converge to any rational $L$. \begin{problem}[Challenge] Find an explicit sequence $\{x_n\}\subset \Q$ such that $x_n^2 \rightarrow 2$ for all $x_n >0$. \end{problem} Following the preceding example, we can take $x_1=1.4$, and $x_n=x_{n-1}+\frac{a_n}{10^n}$ for $n>1$, where $a_n$ is the largest integer $a$ such that $(x_{n-1}+\frac{a}{10^n})^2<2$. Then $\{x_n\}$ is a Cauchy sequence, and $x_n^2 \to 2$ as $n \to \infty$. Here is an idea for a more insightful solution, motivated by a standard algorithm to approximate square roots. Let \begin{equation} \label{iter} x_1=2 \mbox{ \rm and } \, x_n=\frac{1}{2}\Bigl(x_{n-1}+\frac{2}{x_{n-1}}\Bigr) \mbox{ \rm for } n>1 \,. \end{equation} By induction, $x_n \in \Q$ for all $n$. \begin{problem}[Exercise] For the sequence in (\ref{iter}), check that $x_{n+1}0$ and that the Cauchy property holds. Hint: Consider $x_n^2-2$. \end{problem} To ensure that a sequence $\{y_n\}$ is Cauchy, it is {\bf not} enough to verify that $y_n-y_{n-1} \to 0$ as $n \to \infty$. \begin{example} Consider $H_n=1+\frac{1}{2}+\frac{1}{3}+\dots+\frac{1}{n}$, so that $H_n-H_{n-1}=\frac{1}{n}\rightarrow 0$. Nevertheless, $\{H_n\}$ is not a Cauchy sequence. To see this, take $\epsilon=1/3$, for instance. Given any $N$, we must find $m,n>N$ with $|H_n-H_m|\geq 1/3$. Let $m=N+1$ and $n=2m$. Then $$ H_{2m}-H_{m}=\frac{1}{m+1}+\frac{1}{m+2}+\ldots+\frac{1}{2m}\geq \frac{m}{2m}=\frac{1}{2}. $$ We are done. \end{example} In the preceding example, the sequence $H_n$ is not bounded. \begin{problem}[Exercise] \begin{itemize} \item Show that every Cauchy sequence is bounded. \item Show that every convergent sequence is a Cauchy sequence. \item Find an example of a bounded sequence $\{y_n\}$ such that $y_n-y_{n-1} \to 0$ yet $\{y_n\}$ is not a Cauchy sequence. Hint: Consider the distance from $H_n$ to the nearest integer. \end{itemize} \end{problem} To define real numbers via Cauchy sequences, we must deal with the fact that many different sequences might converge to the same limit. \begin{definition} Suppose $\{x_n\}$ and $\{y_n\}$ are Cauchy sequences of rational numbers. We say that $\{x_n\}$ is \textbf{equivalent to} $\{y_n\}$, and write $\{x_n\} \sim \{y_n\}$, if $x_n-y_n \rightarrow 0$. \end{definition} Given a Cauchy sequence $\{x_n\} \subset \Q$, consider its \textbf{equivalence class} $$ \overline{\{x_n\}}= \{\textrm{all sequences $\{y_n\}$ such that } \{x_n\} \sim \{y_n\}\} \, . $$ We can {\bf define a real number} as such an equivalence class. To do so, and still think of $\Q$ as a subset of $\R$, we identify every rational number with the equivalence class of (Cauchy) sequences converging to it. %\begin{thebibliography}{99} %\end{thebibliography} \end{document}