Add missing definitions to the glossary.
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\chapter{Glossary}%
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\label{chap:glossary}
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\section{\defined{Characteristic Function}}%
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\label{sec:def-characteristic-function}
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Let $S$ be a set of points on the real line.
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The \textbf{characteristic function} of $S$ is the function $\mathcal{X}_S$ such
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that $\mathcal{X}_S(x) = 1$ for every $x$ in $S$, and $\mathcal{X}_S(x) = 0$
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for those $x$ not in $S$.
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\begin{definition}
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\lean{Common/Set/Basic}{Set.characteristic}
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\end{definition}
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\section{\defined{Infimum}}%
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\label{sec:def-infimum}
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A number $B$ is called an \textbf{infimum} of a nonempty set $S$ if $B$ has
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the following two properties:
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\begin{enumerate}[(a)]
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\item $B$ is a lower bound for $S$.
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\item No number greater than $B$ is a lower bound for $S$.
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\end{enumerate}
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Such a number $B$ is also known as the \textbf{greatest lower bound}.
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\begin{definition}
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\lean{Mathlib/Order/Bounds/Basic}{IsGLB}
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\end{definition}
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\section{\defined{Partition}}%
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\label{sec:def-partition}
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@ -42,16 +73,26 @@ A collection of points satisfying \eqref{sec:partition-eq1} is called a
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\end{definition}
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\section{\partial{Refinement}}%
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\label{sec:def-refinement}
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Let $P$ be a \nameref{sec:def-partition} of closed interval $[a, b]$.
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A \textbf{refinement} $P'$ of $P$ is a partition formed by adjoining more
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subdivision points to those already in $P$.
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$P'$ is said to be \textbf{finer than} $P$. The union of two partitions $P_1$
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and $P_2$ is called the \textbf{common refinement} of $P_1$ and $P_2$.
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\section{\defined{Step Function}}%
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\label{sec:def-step-function}
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A function $s$, whose domain is a closed interval $[a, b]$, is called a step
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function if there is a \nameref{sec:def-partition}
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$P = \{x_0, x_1, \ldots, x_n\}$ of $[a b]$ such that $s$ is constant on each
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A function $s$, whose domain is a closed interval $[a, b]$, is called a
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\textbf{step function} if there is a \nameref{sec:def-partition}
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$P = \{x_0, x_1, \ldots, x_n\}$ of $[a, b]$ such that $s$ is constant on each
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open subinterval of $P$.
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That is to say, for each $k = 1, 2, \ldots, n$, there is a real number $s_k$
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such that $$s(x) = s_k \quad\text{if}\quad x_{k-1} < x < x_k.$$
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Step functions are sometimes called piecewise constant functions.
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Step functions are sometimes called \textbf{piecewise constant functions}.
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\vspace{8pt}
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\noindent
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@ -64,9 +105,38 @@ Step functions are sometimes called piecewise constant functions.
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\end{definition}
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\section{\defined{Supremum}}%
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\label{sec:def-supremum}
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A number $B$ is called a \textbf{supremum} of a nonempty set $S$ if $B$ has
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the following two properties:
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\begin{enumerate}[(a)]
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\item $B$ is an upper bound for $S$.
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\item No number less than $B$ is an upper bound for $S$.
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\end{enumerate}
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Such a number $B$ is also known as the \textbf{least upper bound}.
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\begin{definition}
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\lean{Mathlib/Order/Bounds/Basic}{IsLUB}
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\end{definition}
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\chapter{A Set of Axioms for the Real-Number System}%
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\label{chap:set-axioms-real-number-system}
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\section{\defined{Completeness Axiom}}%
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\label{sec:completeness-axiom}
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Every nonempty set $S$ of real numbers which is bounded above has a supremum;
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that is, there is a real number $B$ such that $B = \sup{S}$.
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\begin{axiom}
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\lean{Mathlib/Data/Real/Basic}{Real.exists\_isLUB}
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\end{axiom}
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\section{\verified{Lemma 1}}%
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\label{sec:lemma-1}
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@ -80,8 +150,8 @@ Nonempty set $S$ has supremum $L$ if and only if set $-S$ has infimum $-L$.
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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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By definition of the \nameref{sec:def-supremum}, $x \leq L$ and $L$ is the
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smallest value 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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@ -103,7 +173,8 @@ Every nonempty set $S$ that is bounded below has a greatest lower bound; that
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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 the \nameref{sec:completeness-axiom}, there exists a
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\nameref{sec:def-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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@ -130,8 +201,8 @@ For every real $x$ there exists a positive integer $n$ such that $n > x$.
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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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\section{\verified{Archimedean Property of the Reals}}%
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\label{sec:archimedean-property-reals}
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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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@ -182,8 +253,8 @@ If three real numbers $a$, $x$, and $y$ satisfy the inequalities
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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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By \nameref{sec:archimedean-property-reals}, there exists an integer $n > 0$
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such that $nc > y$.
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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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@ -220,8 +291,8 @@ If three real numbers $a$, $x$, and $y$ satisfy the inequalities
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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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By \nameref{sec:archimedean-property-reals}, there exists an integer $n > 0$
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such that $nc > y$.
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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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@ -257,7 +328,8 @@ If $S$ has a supremum, then for some $x$ in $S$ we have $x > \sup{S} - h$.
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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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By definition of a \nameref{sec:def-supremum}, $\sup{S}$ is the least upper
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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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@ -280,7 +352,8 @@ If $S$ has an infimum, then for some $x$ in $S$ we have $x < \inf{S} + h$.
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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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By definition of an \nameref{sec:def-infimum}, $\inf{S}$ is the greatest lower
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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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@ -321,7 +394,7 @@ If each of $A$ and $B$ has a supremum, then $C$ has a supremum, and
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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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By definition of a \nameref{sec:def-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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@ -390,7 +463,7 @@ If each of $A$ and $B$ has an infimum, then $C$ has an infimum, and
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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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By definition of an \nameref{sec:def-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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@ -577,7 +650,7 @@ A set consisting of a single point.
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\begin{proof}
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Let $S$ be a set consisting of a single point.
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By definition of a Point, $S$ is a rectangle in which all vertices coincide.
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By definition of a point, $S$ is a rectangle in which all vertices coincide.
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By \nameref{sec:choice-scale}, $S$ is measurable with area its width times
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its height.
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The width and height of $S$ is trivially zero.
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@ -658,7 +731,7 @@ The union of a finite collection of line segments in a plane.
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\paragraph{Base Case}%
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Consider a set $S$ consisting of a single line segment in a plane.
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By definition of a Line Segment, $S$ is a rectangle in which one side has
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By definition of a line segment, $S$ is a rectangle in which one side has
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dimension $0$.
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By \nameref{sec:choice-scale}, $S$ is measurable with area its width $w$
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times its height $h$.
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@ -1271,8 +1344,7 @@ $\floor{2x} = \floor{x} + \floor{x + \frac{1}{2}}.$
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\divider
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This is immediately proven by applying Hermite's Identity as shown in
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\nameref{sec:exercise-1.11.5}.
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This is immediately proven by applying \nameref{sec:hermites-identity}.
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\end{proof}
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@ -1288,20 +1360,17 @@ $\floor{3x} = \floor{x} + \floor{x + \frac{1}{3}} + \floor{x + \frac{2}{3}}.$
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\divider
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This is immediately proven by applying Hermite's Identity as shown in
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\nameref{sec:exercise-1.11.5}.
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This is immediately proven by applying \nameref{sec:hermites-identity}.
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\end{proof}
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\section{\partial{Exercise 1.11.5}}%
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\label{sec:exercise-1.11.5}
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\section{\partial{Hermite's Identity}}%
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\label{sec:hermites-identity}
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The formulas in Exercises 4(d) and 4(e) suggest a generalization for
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$\floor{nx}$.
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State and prove such a generalization.
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\note{The stated generalization is known as "Hermite's Identity."}
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\begin{proof}
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\lean{Bookshelf/Apostol/Chapter\_1\_11}
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@ -1568,11 +1637,10 @@ Now apply Exercises 4(a) and (b) to the bracket on the right.
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\label{sec:exercise-1.11.8}
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Let $S$ be a set of points on the real line.
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The \textit{characteristic function} of $S$ is, by definition, the function
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$\mathcal{X}_S$ such that $\mathcal{X}_S(x) = 1$ for every $x$ in $S$, and
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$\mathcal{X}_S(x) = 0$ for those $x$ not in $S$.
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Let $f$ be a step function which takes the constant value $c_k$ on the $k$th
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open subinterval $I_k$ of some partition of an interval $[a, b]$.
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Let $\mathcal{X}_S$ denote the \nameref{sec:def-characteristic-function} of $S$.
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Let $f$ be a \nameref{sec:def-step-function} which takes the constant value
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$c_k$ on the $k$th open subinterval $I_k$ of some partition of an interval
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$[a, b]$.
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Prove that for each $x$ in the union $I_1 \cup I_2 \cup \cdots \cup I_n$ we have
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$$f(x) = \sum_{k=1}^n c_k\mathcal{X}_{I_k}(x).$$
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This property is described by saying that every step function is a linear
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@ -1583,7 +1651,7 @@ This property is described by saying that every step function is a linear
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Let $x \in I_1 \cup I_2 \cup \cdots \cup I_n$ and $N = \{1, \ldots, n\}$.
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Let $k \in N$ such that $x \in I_k$.
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Consider an arbitrary $j \in N - \{k\}$.
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By definition of a partition, $I_j \cap I_k = \emptyset$.
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By definition of a nameref{sec:def-partition}, $I_j \cap I_k = \emptyset$.
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That is, $I_j$ and $I_k$ are disjoint for all $j \in N - \{k\}$.
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Therefore, by definition of the characteristic function,
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$\mathcal{X}_{I_k}(x) = 1$ and $\mathcal{X}_{I_j}(x) = 0$ for all
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