Real Analysis I (Workshop 2)
Author:
Philip Mak
Last Updated:
9年前
License:
Creative Commons CC BY 4.0
Abstract:
Real Analysis
Workshop 2
1.3.10
\begin
Discover why 18 million people worldwide trust Overleaf with their work.
Real Analysis
Workshop 2
1.3.10
\begin
Discover why 18 million people worldwide trust Overleaf with their work.
\documentclass[paper=a4, fontsize=11pt]{scrartcl} % A4 paper and 11pt font size
\usepackage[T1]{fontenc} % Use 8-bit encoding that has 256 glyphs
\usepackage{fourier} % Use the Adobe Utopia font for the document - comment this line to return to the LaTeX default
\usepackage[english]{babel} % English language/hyphenation
\usepackage{amsmath,amsfonts,amsthm} % Math packages
\usepackage{lipsum} % Used for inserting dummy 'Lorem ipsum' text into the template
\usepackage{sectsty} % Allows customizing section commands
\allsectionsfont{\centering \normalfont\scshape} % Make all sections centered, the default font and small caps
\usepackage{fancyhdr} % Custom headers and footers
\pagestyle{fancyplain} % Makes all pages in the document conform to the custom headers and footers
\fancyhead{} % No page header - if you want one, create it in the same way as the footers below
\fancyfoot[L]{} % Empty left footer
\fancyfoot[C]{} % Empty center footer
\fancyfoot[R]{\thepage} % Page numbering for right footer
\renewcommand{\headrulewidth}{0pt} % Remove header underlines
\renewcommand{\footrulewidth}{0pt} % Remove footer underlines
\setlength{\headheight}{13.6pt} % Customize the height of the header
\numberwithin{equation}{section} % Number equations within sections (i.e. 1.1, 1.2, 2.1, 2.2 instead of 1, 2, 3, 4)
\numberwithin{figure}{section} % Number figures within sections (i.e. 1.1, 1.2, 2.1, 2.2 instead of 1, 2, 3, 4)
\numberwithin{table}{section} % Number tables within sections (i.e. 1.1, 1.2, 2.1, 2.2 instead of 1, 2, 3, 4)
\setlength\parindent{0pt} % Removes all indentation from paragraphs - comment this line for an assignment with lots of text
%----------------------------------------------------------------------------------------
% TITLE SECTION
%----------------------------------------------------------------------------------------
\newcommand{\horrule}[1]{\rule{\linewidth}{#1}} % Create horizontal rule command with 1 argument of height
\title{
\normalfont \normalsize
\textsc{Rutgers University, Real Analysis I} \\ [25pt] % Your university, school and/or department name(s)
\horrule{0.5pt} \\[0.4cm] % Thin top horizontal rule
\huge Workshop 2 \\ % The assignment title
\horrule{2pt} \\[0.5cm] % Thick bottom horizontal rule
}
\author{Philip Mak} % Your name
\date{\normalsize\today} % Today's date or a custom date
\begin{document}
\maketitle % Print the title
%----------------------------------------------------------------------------------------
% PROBLEM 1
%----------------------------------------------------------------------------------------
\section{Problem 1.3.10}
The Cut Property of the real numbers is the following. If A and B are nonmempty, disjoint sets with A $\cup$ B = $\mathbb{R}$ and a < b for all $a \in A$ and $b \in B$, then there exists $c \in \mathbb{R}$ such that $x \leq c$ whenever $x \in A$ and $x \geq c$ whenever $x \in B$. $\newline$
%------------------------------------------------
(a) Use the Axiom of Completeness to prove the Cut Property $\newline$
Proof: Suppose sets A and B are nonempty, disjoint sets with $A \cup B = \mathbb{R}$ and $a < b$ for all $a \in A$ and $b \in B$. We want to show there exists $c \in \mathbb{R}$ such that $x \leq c$ whenever $x \in A$ and $x \geq c$ whenever $x \in B$. $\newline$
We know a<b $\forall a \in A$ and $\forall b \in B$. So, the set $B$ is the set of upper bounds of set $A$. Therefore, $A$ is bounded above and by the Axiom of Completeness, $A$ contains a supremum (least upper bound) the we call $s$.
Since $A \cup B = \mathbb{R}$ and $A \cap B = \emptyset$, either $s \in A$ and $s \notin B$ or $s \in B$ and $s \notin A$.
Case 1: $s \in A$ and $s \notin B$ $\newline$
$s \geq a \forall a \in A$ and $s < b \forall b\in B\newline$
Case 2: $s \notin A$ and $s \in B$ $\newline$
$s \leq b \forall b \in B$ and $s > a \forall a\in A\newline$
Therefore, $a \leq s \leq b \forall a \in A \forall b \in B$ and the Cut Property holds by the Axiom of Completeness. \newline
(b) Show that the implication goes the other way; that is, assume $\mathbb{R}$ possesses the Cut Property and let $E$ be a nonempty set that is bounded above. Prove sup $E$ exists. $\newline$
Assume that $\mathbb{R}$ possesses the Cut Property and let $E$ be a nonempty set that is bounded above. We want to show that sup($E$) exists. To do this, we have to show the two properties of a supremum. $\newline$
(i) s is an upper bound for $A$ $\newline$
(ii) if $b$ is any upper bound for $A$, then $s \leq b$.$\newline$
Since $E$ is bounded above and $\mathbb{R}$ has the Cut Property, then there exists a set $F$ such that $E \cup F = \mathbb{R}$ and $E \cap F = \emptyset$. Because $E$ is bounded above and $E \cup F = \mathbb{R}$, it implies that e < f $\forall e \in E \forall f \in F$. So $F$ is the set of upper bounds for $E$. Also by the definition of the Cut Property, we have some $g \in \mathbb{R}$ such that $e \leq g \leq f \forall e \in E \forall f \in F$. Since $g \leq f$, $g \in F$ and it is the smallest element in $F$. Therefore, $g$ is the least upper bound. $\newline$
(c) The punchline of parts (a) and (b) is that the cut property could be used in place of the Axiom f Completeness as the fundamental axiom that distinguishes the real numbers from the rational numbers. To drive this point home, give a concrete showing that the Cut Property is not a valid statement when $\mathbb{R}$ is replaced by $\mathbb{Q}$.
The easiest example of this would be to let $A = {a \in \mathbb{Q} : a^2 < 2}$ and $B = {b \in \mathbb{Q} : b^2 > 2}$. From this, it is easy to see that $A \cap B = \emptyset$ and $A \cup B = \mathbb{Q}$. To find the "Cut Value" $c$, some simple arithmetic will show that $c^2 = 2$. We want to show that $a \leq c \leq b \forall a \in A \forall b \in B$. However, the value of $c$ to solve this does not exist in $\mathbb{Q}$. Therefore, the Cut Property does not apply to the rational numbers.
\end{document}