bookshelf/Bookshelf/Apostol/Chapter_I_03.tex

406 lines
12 KiB
TeX
Raw Normal View History

2023-04-10 17:33:22 +00:00
\documentclass{article}
2023-05-07 21:57:40 +00:00
\input{../../preamble}
2023-04-10 17:33:22 +00:00
\newcommand{\lean}[1]{\leanref
{./Chapter\_I\_03.html\#Apostol.Chapter\_I\_03.#1}
{Apostol.Chapter\_I\_03.#1}}
2023-04-10 17:33:22 +00:00
\begin{document}
2023-04-10 17:33:22 +00:00
\header{A Set of Axioms for the Real-Number System}{Tom M. Apostol}
\section*{\verified{Lemma 1}}%
\label{sec:lemma-1}
Nonempty set $S$ has supremum $L$ if and only if set $-S$ has infimum $-L$.
\begin{proof}
\lean{is\_lub\_neg\_set\_iff\_is\_glb\_set\_neg}
\divider
Suppose $L = \sup{S}$ and fix $x \in S$.
By definition of the supremum, $x \leq L$ and $L$ is the smallest value
satisfying this inequality.
Negating both sides of the inequality yields $-x \geq -L$.
Furthermore, $-L$ must be the largest value satisfying this inequality.
Therefore $-L = \inf{-S}$.
\end{proof}
\section*{\verified{Theorem I.27}}%
\label{sec:theorem-i.27}
2023-04-10 17:33:22 +00:00
Every nonempty set $S$ that is bounded below has a greatest lower bound; that
is, there is a real number $L$ such that $L = \inf{S}$.
2023-04-10 17:33:22 +00:00
\begin{proof}
\lean{exists\_isGLB}
2023-04-10 17:33:22 +00:00
\divider
Let $S$ be a nonempty set bounded below by $x$.
Then $-S$ is nonempty and bounded above by $x$.
By the completeness axiom, there exists a supremum $L$ of $-S$.
By \nameref{sec:lemma-1}, $L$ is a supremum of $-S$ if and only if $-L$ is an
infimum of $S$.
2023-04-10 17:33:22 +00:00
\end{proof}
\section*{\verified{Theorem I.29}}%
\label{sec:theorem-i.29}
2023-04-10 17:33:22 +00:00
For every real $x$ there exists a positive integer $n$ such that $n > x$.
2023-04-10 17:33:22 +00:00
\begin{proof}
\lean{exists\_pnat\_geq\_self}
2023-04-10 17:33:22 +00:00
\divider
Let $n = \abs{\ceil{x}} + 1$.
It is trivial to see $n$ is a positive integer satisfying $n \geq 1$.
Thus all that remains to be shown is that $n > x$.
If $x$ is nonpositive, $n > x$ immediately follows from $n \geq 1$.
If $x$ is positive,
$$x = \abs{x} \leq \abs{\ceil{x}} < \abs{\ceil{x}} + 1 = n.$$
2023-04-10 17:33:22 +00:00
\end{proof}
\section*{\verified{Theorem I.30}}%
\label{sec:theorem-i.30}
2023-04-10 17:33:22 +00:00
If $x > 0$ and if $y$ is an arbitrary real number, there exists a positive
integer $n$ such that $nx > y$.
2023-04-10 17:33:22 +00:00
\note{This is known as the "Archimedean Property of the Reals."}
2023-04-10 17:33:22 +00:00
\begin{proof}
\lean{exists\_pnat\_mul\_self\_geq\_of\_pos}
2023-04-10 17:33:22 +00:00
\divider
Let $x > 0$ and $y$ be an arbitrary real number.
By \nameref{sec:theorem-i.29}, there exists a positive integer $n$ such that
$n > y / x$.
Multiplying both sides of the inequality yields $nx > y$ as expected.
2023-04-10 17:33:22 +00:00
\end{proof}
\section*{\verified{Theorem I.31}}%
\label{sec:theorem-i.31}
2023-04-10 17:33:22 +00:00
If three real numbers $a$, $x$, and $y$ satisfy the inequalities
$$a \leq x \leq a + \frac{y}{n}$$ for every integer $n \geq 1$, then $x = a$.
2023-04-10 17:33:22 +00:00
\begin{proof}
\lean{forall\_pnat\_leq\_self\_leq\_frac\_imp\_eq}
\divider
By the trichotomy of the reals, there are three cases to consider:
\paragraph{Case 1}%
Suppose $x = a$.
Then we are immediately finished.
\paragraph{Case 2}%
Suppose $x < a$.
But by hypothesis, $a \leq x$.
Thus $a < a$, a contradiction.
\paragraph{Case 3}%
Suppose $x > a$.
Then there exists some $c > 0$ such that $a + c = x$.
By \nameref{sec:theorem-i.30}, there exists an integer $n > 0$ such that
$nc > y$.
Rearranging terms, we see $y / n < c$.
Therefore $a + y / n < a + c = x$.
But by hypothesis, $x \leq a + y / n$.
Thus $a + y / n < a + y / n$, a contradiction.
\paragraph{Conclusion}%
Since these cases are exhaustive and both case 2 and 3 lead to
contradictions, $x = a$ is the only possibility.
\end{proof}
\section*{\verified{Lemma 2}}%
\label{sec:lemma-2}
If three real numbers $a$, $x$, and $y$ satisfy the inequalities
$$a - y / n \leq x \leq a$$ for every integer $n \geq 1$, then $x = a$.
\begin{proof}
\lean{forall\_pnat\_frac\_leq\_self\_leq\_imp\_eq}
\divider
By the trichotomy of the reals, there are three cases to consider:
\paragraph{Case 1}%
Suppose $x = a$.
Then we are immediately finished.
\paragraph{Case 2}%
Suppose $x < a$.
Then there exists some $c > 0$ such that $x = a - c$.
By \nameref{sec:theorem-i.30}, there exists an integer $n > 0$ such that
$nc > y$.
Rearranging terms, we see that $y / n < c$.
Therefore $a - y / n > a - c = x$.
But by hypothesis, $x \geq a - y / n$.
Thus $a - y / n < a - y / n$, a contradiction.
\paragraph{Case 3}%
Suppose $x > a$.
But by hypothesis $x \leq a$.
Thus $a < a$, a contradiction.
\paragraph{Conclusion}%
Since these cases are exhaustive and both case 2 and 3 lead to
contradictions, $x = a$ is the only possibility.
\end{proof}
\section*{Theorem I.32}%
\label{sec:theorem-i.32}
Let $h$ be a given positive number and let $S$ be a set of real numbers.
\subsection*{\verified{Theorem I.32a}}%
\label{sub:theorem-i.32a}
If $S$ has a supremum, then for some $x$ in $S$ we have $x > \sup{S} - h$.
\begin{proof}
\lean{sup\_imp\_exists\_gt\_sup\_sub\_delta}
2023-05-07 21:57:40 +00:00
\divider
By definition of a supremum, $\sup{S}$ is the least upper bound of $S$.
For the sake of contradiction, suppose for all $x \in S$,
$x \leq \sup{S} - h$.
This immediately implies $\sup{S} - h$ is an upper bound of $S$.
But $\sup{S} - h < \sup{S}$, contradicting $\sup{S}$ being the \textit{least}
upper bound.
Therefore our original hypothesis was wrong.
That is, there exists some $x \in S$ such that $x > \sup{S} - h$.
2023-04-10 17:33:22 +00:00
\end{proof}
\subsection*{\verified{Theorem I.32b}}%
\label{sub:theorem-i.32b}
If $S$ has an infimum, then for some $x$ in $S$ we have $x < \inf{S} + h$.
\begin{proof}
\lean{inf\_imp\_exists\_lt\_inf\_add\_delta}
\divider
By definition of an infimum, $\inf{S}$ is the greatest lower bound of $S$.
For the sake of contradiction, suppose for all $x \in S$,
$x \geq \inf{S} + h$.
This immediately implies $\inf{S} + h$ is a lower bound of $S$.
But $\inf{S} + h > \inf{S}$, contradicting $\inf{S}$ being the
\textit{greatest} lower bound.
Therefore our original hypothesis was wrong.
That is, there exists some $x \in S$ such that $x < \inf{S} + h$.
\end{proof}
\section*{Theorem I.33}%
\label{sec:theorem-i.33}
Given nonempty subsets $A$ and $B$ of $\mathbb{R}$, let $C$ denote the set
$$C = \{a + b : a \in A, b \in B\}.$$
\note{This is known as the "Additive Property."}
\subsection*{\verified{Theorem I.33a}}%
\label{sub:theorem-i.33a}
If each of $A$ and $B$ has a supremum, then $C$ has a supremum, and
$$\sup{C} = \sup{A} + \sup{B}.$$
\begin{proof}
\lean{sup\_minkowski\_sum\_eq\_sup\_add\_sup}
\divider
We prove (i) $\sup{A} + \sup{B}$ is an upper bound of $C$ and (ii)
$\sup{A} + \sup{B}$ is the \textit{least} upper bound of $C$.
\paragraph{(i)}%
\label{par:theorem-i.33a-i}
Let $x \in C$.
By definition of $C$, there exist elements $a' \in A$ and $b' \in B$ such
that $x = a' + b'$.
By definition of a supremum, $a' \leq \sup{A}$.
Likewise, $b' \leq \sup{B}$.
Therefore $a' + b' \leq \sup{A} + \sup{B}$.
Since $x = a' + b'$ was arbitrarily chosen, it follows $\sup{A} + \sup{B}$
is an upper bound of $C$.
\paragraph{(ii)}%
Since $A$ and $B$ have supremums, $C$ is nonempty.
By \nameref{par:theorem-i.33a-i}, $C$ is bounded above.
Therefore the completeness axiom tells us $C$ has a supremum.
Let $n > 0$ be an integer.
We now prove that
\begin{equation}
\label{par:theorem-i.33a-ii-eq1}
\sup{C} \leq \sup{A} + \sup{B} \leq \sup{C} + 1 / n.
\end{equation}
\subparagraph{Left-Hand Side}%
First consider the left-hand side of \eqref{par:theorem-i.33a-ii-eq1}.
By \nameref{par:theorem-i.33a-i}, $\sup{A} + \sup{B}$ is an upper bound of
$C$.
Since $\sup{C}$ is the \textit{least} upper bound of $C$, it follows
$\sup{C} \leq \sup{A} + \sup{B}$.
\subparagraph{Right-Hand Side}%
Next consider the right-hand side of \eqref{par:theorem-i.33a-ii-eq1}.
By \nameref{sub:theorem-i.32a}, there exists some $a' \in A$ such that
$\sup{A} < a' + 1 / (2n)$.
Likewise, there exists some $b' \in B$ such that
$\sup{B} < b' + 1 / (2n)$.
Adding these two inequalities together shows
\begin{align*}
\sup{A} + \sup{B}
& < a' + b' + 1 / n \\
& \leq \sup{C} + 1 / n.
\end{align*}
\subparagraph{Conclusion}%
Applying \nameref{sec:theorem-i.31} to \eqref{par:theorem-i.33a-ii-eq1}
proves $\sup{C} = \sup{A} + \sup{B}$ as expected.
\end{proof}
\subsection*{\verified{Theorem I.33b}}%
\label{sub:theorem-i.33b}
If each of $A$ and $B$ has an infimum, then $C$ has an infimum, and
$$\inf{C} = \inf{A} + \inf{B}.$$
\begin{proof}
\lean{inf\_minkowski\_sum\_eq\_inf\_add\_inf}
\divider
2023-05-07 21:57:40 +00:00
We prove (i) $\inf{A} + \inf{B}$ is a lower bound of $C$ and (ii)
$\inf{A} + \inf{B}$ is the \textit{greatest} lower bound of $C$.
\paragraph{(i)}%
\label{par:theorem-i.33b-i}
Let $x \in C$.
By definition of $C$, there exist elements $a' \in A$ and $b' \in B$ such
that $x = a' + b'$.
By definition of an infimum, $a' \geq \inf{A}$.
Likewise, $b' \geq \inf{B}$.
Therefore $a' + b' \geq \inf{A} + \inf{B}$.
Since $x = a' + b'$ was arbitrarily chosen, it follows $\inf{A} + \inf{B}$
is a lower bound of $C$.
\paragraph{(ii)}%
Since $A$ and $B$ have infimums, $C$ is nonempty.
By \nameref{par:theorem-i.33b-i}, $C$ is bounded below.
Therefore \nameref{sec:theorem-i.27} tells us $C$ has an infimum.
Let $n > 0$ be an integer.
We now prove that
\begin{equation}
\label{par:theorem-i.33b-ii-eq1}
\inf{C} - 1 / n \leq \inf{A} + \inf{B} \leq \inf{C}.
\end{equation}
\subparagraph{Right-Hand Side}%
First consider the right-hand side of \eqref{par:theorem-i.33b-ii-eq1}.
By \nameref{par:theorem-i.33b-i}, $\inf{A} + \inf{B}$ is a lower bound of
$C$.
Since $\inf{C}$ is the \textit{greatest} upper bound of $C$, it follows
$\inf{C} \geq \inf{A} + \inf{B}$.
\subparagraph{Left-Hand Side}%
Next consider the left-hand side of \eqref{par:theorem-i.33b-ii-eq1}.
By \nameref{sub:theorem-i.32b}, there exists some $a' \in A$ such that
$\inf{A} > a' - 1 / (2n)$.
Likewise, there exists some $b' \in B$ such that
$\inf{B} > b' - 1 / (2n)$.
Adding these two inequalities together shows
\begin{align*}
\inf{A} + \inf{B}
& > a' + b' - 1 / n \\
& \geq \inf{C} - 1 / n.
\end{align*}
\subparagraph{Conclusion}%
Applying \nameref{sec:lemma-2} to \eqref{par:theorem-i.33b-ii-eq1}
proves $\inf{C} = \inf{A} + \inf{B}$ as expected.
\end{proof}
\section*{\verified{Theorem I.34}}%
\label{sec:theorem-i.34}
Given two nonempty subsets $S$ and $T$ of $\mathbb{R}$ such that $$s \leq t$$
for every $s$ in $S$ and every $t$ in $T$. Then $S$ has a supremum, and $T$
has an infimum, and they satisfy the inequality $$\sup{S} \leq \inf{T}.$$
\begin{proof}
\lean{forall\_mem\_le\_forall\_mem\_imp\_sup\_le\_inf}
\divider
By hypothesis, $S$ and $T$ are nonempty sets.
Let $s \in S$ and $t \in T$.
Then $t$ is an upper bound of $S$ and $s$ is a lower bound of $T$.
By the completeness axiom, $S$ has a supremum.
By \nameref{sec:theorem-i.27}, $T$ has an infimum.
All that remains is showing $\sup{S} \leq \inf{T}$.
For the sake of contradiction, suppose $\sup{S} > \inf{T}$.
Then there exists some $c > 0$ such that $\sup{S} = \inf{T} + c$.
Therefore $\inf{T} < \sup{S} - c / 2$.
By \nameref{sub:theorem-i.32a}, there exists some $x \in S$ such that
$\sup{S} - c / 2 < x$.
Thus $$\inf{T} < \sup{S} - c / 2 < x.$$
But by hypothesis, $x \in S$ is a lower bound of $T$ meaning $x \leq \inf{T}$.
Therefore $x < x$, a contradiction.
Out original assumption is incorrect; that is, $\sup{S} \leq \inf{T}$.
\end{proof}
2023-04-10 17:33:22 +00:00
\end{document}