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