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Add missing definitions to the glossary.

finite-set-exercises
Joshua Potter 2023-05-13 10:39:02 -06:00
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Bookshelf/Apostol.tex
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@ -20,6 +20,37 @@
\chapter{Glossary}% \chapter{Glossary}%
\label{chap:glossary} \label{chap:glossary}
\section{\defined{Characteristic Function}}%
\label{sec:def-characteristic-function}
Let $S$ be a set of points on the real line.
The \textbf{characteristic function} of $S$ is the function $\mathcal{X}_S$ such
that $\mathcal{X}_S(x) = 1$ for every $x$ in $S$, and $\mathcal{X}_S(x) = 0$
for those $x$ not in $S$.
\begin{definition}
\lean{Common/Set/Basic}{Set.characteristic}
\end{definition}
\section{\defined{Infimum}}%
\label{sec:def-infimum}
A number $B$ is called an \textbf{infimum} of a nonempty set $S$ if $B$ has
the following two properties:
\begin{enumerate}[(a)]
\item $B$ is a lower bound for $S$.
\item No number greater than $B$ is a lower bound for $S$.
\end{enumerate}
Such a number $B$ is also known as the \textbf{greatest lower bound}.
\begin{definition}
\lean{Mathlib/Order/Bounds/Basic}{IsGLB}
\end{definition}
\section{\defined{Partition}}% \section{\defined{Partition}}%
\label{sec:def-partition} \label{sec:def-partition}
@ -42,16 +73,26 @@ A collection of points satisfying \eqref{sec:partition-eq1} is called a
\end{definition} \end{definition}
\section{\partial{Refinement}}%
\label{sec:def-refinement}
Let $P$ be a \nameref{sec:def-partition} of closed interval $[a, b]$.
A \textbf{refinement} $P'$ of $P$ is a partition formed by adjoining more
subdivision points to those already in $P$.
$P'$ is said to be \textbf{finer than} $P$. The union of two partitions $P_1$
and $P_2$ is called the \textbf{common refinement} of $P_1$ and $P_2$.
\section{\defined{Step Function}}% \section{\defined{Step Function}}%
\label{sec:def-step-function} \label{sec:def-step-function}
A function $s$, whose domain is a closed interval $[a, b]$, is called a step A function $s$, whose domain is a closed interval $[a, b]$, is called a
function if there is a \nameref{sec:def-partition} \textbf{step function} if there is a \nameref{sec:def-partition}
$P = \{x_0, x_1, \ldots, x_n\}$ of $[a b]$ such that $s$ is constant on each $P = \{x_0, x_1, \ldots, x_n\}$ of $[a, b]$ such that $s$ is constant on each
open subinterval of $P$. open subinterval of $P$.
That is to say, for each $k = 1, 2, \ldots, n$, there is a real number $s_k$ That is to say, for each $k = 1, 2, \ldots, n$, there is a real number $s_k$
such that $$s(x) = s_k \quad\text{if}\quad x_{k-1} < x < x_k.$$ such that $$s(x) = s_k \quad\text{if}\quad x_{k-1} < x < x_k.$$
Step functions are sometimes called piecewise constant functions. Step functions are sometimes called \textbf{piecewise constant functions}.
\vspace{8pt} \vspace{8pt}
\noindent \noindent
@ -64,9 +105,38 @@ Step functions are sometimes called piecewise constant functions.
\end{definition} \end{definition}
\section{\defined{Supremum}}%
\label{sec:def-supremum}
A number $B$ is called a \textbf{supremum} of a nonempty set $S$ if $B$ has
the following two properties:
\begin{enumerate}[(a)]
\item $B$ is an upper bound for $S$.
\item No number less than $B$ is an upper bound for $S$.
\end{enumerate}
Such a number $B$ is also known as the \textbf{least upper bound}.
\begin{definition}
\lean{Mathlib/Order/Bounds/Basic}{IsLUB}
\end{definition}
\chapter{A Set of Axioms for the Real-Number System}% \chapter{A Set of Axioms for the Real-Number System}%
\label{chap:set-axioms-real-number-system} \label{chap:set-axioms-real-number-system}
\section{\defined{Completeness Axiom}}%
\label{sec:completeness-axiom}
Every nonempty set $S$ of real numbers which is bounded above has a supremum;
that is, there is a real number $B$ such that $B = \sup{S}$.
\begin{axiom}
\lean{Mathlib/Data/Real/Basic}{Real.exists\_isLUB}
\end{axiom}
\section{\verified{Lemma 1}}% \section{\verified{Lemma 1}}%
\label{sec:lemma-1} \label{sec:lemma-1}
@ -80,8 +150,8 @@ Nonempty set $S$ has supremum $L$ if and only if set $-S$ has infimum $-L$.
\divider \divider
Suppose $L = \sup{S}$ and fix $x \in S$. Suppose $L = \sup{S}$ and fix $x \in S$.
By definition of the supremum, $x \leq L$ and $L$ is the smallest value By definition of the \nameref{sec:def-supremum}, $x \leq L$ and $L$ is the
satisfying this inequality. smallest value satisfying this inequality.
Negating both sides of the inequality yields $-x \geq -L$. Negating both sides of the inequality yields $-x \geq -L$.
Furthermore, $-L$ must be the largest value satisfying this inequality. Furthermore, $-L$ must be the largest value satisfying this inequality.
Therefore $-L = \inf{-S}$. Therefore $-L = \inf{-S}$.
@ -103,7 +173,8 @@ Every nonempty set $S$ that is bounded below has a greatest lower bound; that
Let $S$ be a nonempty set bounded below by $x$. Let $S$ be a nonempty set bounded below by $x$.
Then $-S$ is nonempty and bounded above by $x$. Then $-S$ is nonempty and bounded above by $x$.
By the completeness axiom, there exists a supremum $L$ of $-S$. By the \nameref{sec:completeness-axiom}, there exists a
\nameref{sec:def-supremum} $L$ of $-S$.
By \nameref{sec:lemma-1}, $L$ is a supremum of $-S$ if and only if $-L$ is an By \nameref{sec:lemma-1}, $L$ is a supremum of $-S$ if and only if $-L$ is an
infimum of $S$. infimum of $S$.
@ -130,8 +201,8 @@ For every real $x$ there exists a positive integer $n$ such that $n > x$.
\end{proof} \end{proof}
\section{\verified{Theorem I.30}}% \section{\verified{Archimedean Property of the Reals}}%
\label{sec:theorem-i.30} \label{sec:archimedean-property-reals}
If $x > 0$ and if $y$ is an arbitrary real number, there exists a positive If $x > 0$ and if $y$ is an arbitrary real number, there exists a positive
integer $n$ such that $nx > y$. integer $n$ such that $nx > y$.
@ -182,8 +253,8 @@ If three real numbers $a$, $x$, and $y$ satisfy the inequalities
Suppose $x > a$. Suppose $x > a$.
Then there exists some $c > 0$ such that $a + c = x$. 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 By \nameref{sec:archimedean-property-reals}, there exists an integer $n > 0$
$nc > y$. such that $nc > y$.
Rearranging terms, we see $y / n < c$. Rearranging terms, we see $y / n < c$.
Therefore $a + y / n < a + c = x$. Therefore $a + y / n < a + c = x$.
But by hypothesis, $x \leq a + y / n$. But by hypothesis, $x \leq a + y / n$.
@ -220,8 +291,8 @@ If three real numbers $a$, $x$, and $y$ satisfy the inequalities
Suppose $x < a$. Suppose $x < a$.
Then there exists some $c > 0$ such that $x = a - c$. 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 By \nameref{sec:archimedean-property-reals}, there exists an integer $n > 0$
$nc > y$. such that $nc > y$.
Rearranging terms, we see that $y / n < c$. Rearranging terms, we see that $y / n < c$.
Therefore $a - y / n > a - c = x$. Therefore $a - y / n > a - c = x$.
But by hypothesis, $x \geq a - y / n$. But by hypothesis, $x \geq a - y / n$.
@ -257,7 +328,8 @@ If $S$ has a supremum, then for some $x$ in $S$ we have $x > \sup{S} - h$.
\divider \divider
By definition of a supremum, $\sup{S}$ is the least upper bound of $S$. By definition of a \nameref{sec:def-supremum}, $\sup{S}$ is the least upper
bound of $S$.
For the sake of contradiction, suppose for all $x \in S$, For the sake of contradiction, suppose for all $x \in S$,
$x \leq \sup{S} - h$. $x \leq \sup{S} - h$.
This immediately implies $\sup{S} - h$ is an upper bound of $S$. This immediately implies $\sup{S} - h$ is an upper bound of $S$.
@ -280,7 +352,8 @@ If $S$ has an infimum, then for some $x$ in $S$ we have $x < \inf{S} + h$.
\divider \divider
By definition of an infimum, $\inf{S}$ is the greatest lower bound of $S$. By definition of an \nameref{sec:def-infimum}, $\inf{S}$ is the greatest lower
bound of $S$.
For the sake of contradiction, suppose for all $x \in S$, For the sake of contradiction, suppose for all $x \in S$,
$x \geq \inf{S} + h$. $x \geq \inf{S} + h$.
This immediately implies $\inf{S} + h$ is a lower bound of $S$. This immediately implies $\inf{S} + h$ is a lower bound of $S$.
@ -321,7 +394,7 @@ If each of $A$ and $B$ has a supremum, then $C$ has a supremum, and
Let $x \in C$. Let $x \in C$.
By definition of $C$, there exist elements $a' \in A$ and $b' \in B$ such By definition of $C$, there exist elements $a' \in A$ and $b' \in B$ such
that $x = a' + b'$. that $x = a' + b'$.
By definition of a supremum, $a' \leq \sup{A}$. By definition of a \nameref{sec:def-supremum}, $a' \leq \sup{A}$.
Likewise, $b' \leq \sup{B}$. Likewise, $b' \leq \sup{B}$.
Therefore $a' + b' \leq \sup{A} + \sup{B}$. Therefore $a' + b' \leq \sup{A} + \sup{B}$.
Since $x = a' + b'$ was arbitrarily chosen, it follows $\sup{A} + \sup{B}$ Since $x = a' + b'$ was arbitrarily chosen, it follows $\sup{A} + \sup{B}$
@ -390,7 +463,7 @@ If each of $A$ and $B$ has an infimum, then $C$ has an infimum, and
Let $x \in C$. Let $x \in C$.
By definition of $C$, there exist elements $a' \in A$ and $b' \in B$ such By definition of $C$, there exist elements $a' \in A$ and $b' \in B$ such
that $x = a' + b'$. that $x = a' + b'$.
By definition of an infimum, $a' \geq \inf{A}$. By definition of an \nameref{sec:def-infimum}, $a' \geq \inf{A}$.
Likewise, $b' \geq \inf{B}$. Likewise, $b' \geq \inf{B}$.
Therefore $a' + b' \geq \inf{A} + \inf{B}$. Therefore $a' + b' \geq \inf{A} + \inf{B}$.
Since $x = a' + b'$ was arbitrarily chosen, it follows $\inf{A} + \inf{B}$ Since $x = a' + b'$ was arbitrarily chosen, it follows $\inf{A} + \inf{B}$
@ -577,7 +650,7 @@ A set consisting of a single point.
\begin{proof} \begin{proof}
Let $S$ be a set consisting of a single point. Let $S$ be a set consisting of a single point.
By definition of a Point, $S$ is a rectangle in which all vertices coincide. By definition of a point, $S$ is a rectangle in which all vertices coincide.
By \nameref{sec:choice-scale}, $S$ is measurable with area its width times By \nameref{sec:choice-scale}, $S$ is measurable with area its width times
its height. its height.
The width and height of $S$ is trivially zero. The width and height of $S$ is trivially zero.
@ -658,7 +731,7 @@ The union of a finite collection of line segments in a plane.
\paragraph{Base Case}% \paragraph{Base Case}%
Consider a set $S$ consisting of a single line segment in a plane. Consider a set $S$ consisting of a single line segment in a plane.
By definition of a Line Segment, $S$ is a rectangle in which one side has By definition of a line segment, $S$ is a rectangle in which one side has
dimension $0$. dimension $0$.
By \nameref{sec:choice-scale}, $S$ is measurable with area its width $w$ By \nameref{sec:choice-scale}, $S$ is measurable with area its width $w$
times its height $h$. times its height $h$.
@ -1271,8 +1344,7 @@ $\floor{2x} = \floor{x} + \floor{x + \frac{1}{2}}.$
\divider \divider
This is immediately proven by applying Hermite's Identity as shown in This is immediately proven by applying \nameref{sec:hermites-identity}.
\nameref{sec:exercise-1.11.5}.
\end{proof} \end{proof}
@ -1288,20 +1360,17 @@ $\floor{3x} = \floor{x} + \floor{x + \frac{1}{3}} + \floor{x + \frac{2}{3}}.$
\divider \divider
This is immediately proven by applying Hermite's Identity as shown in This is immediately proven by applying \nameref{sec:hermites-identity}.
\nameref{sec:exercise-1.11.5}.
\end{proof} \end{proof}
\section{\partial{Exercise 1.11.5}}% \section{\partial{Hermite's Identity}}%
\label{sec:exercise-1.11.5} \label{sec:hermites-identity}
The formulas in Exercises 4(d) and 4(e) suggest a generalization for The formulas in Exercises 4(d) and 4(e) suggest a generalization for
$\floor{nx}$. $\floor{nx}$.
State and prove such a generalization. State and prove such a generalization.
\note{The stated generalization is known as "Hermite's Identity."}
\begin{proof} \begin{proof}
\lean{Bookshelf/Apostol/Chapter\_1\_11} \lean{Bookshelf/Apostol/Chapter\_1\_11}
@ -1568,11 +1637,10 @@ Now apply Exercises 4(a) and (b) to the bracket on the right.
\label{sec:exercise-1.11.8} \label{sec:exercise-1.11.8}
Let $S$ be a set of points on the real line. Let $S$ be a set of points on the real line.
The \textit{characteristic function} of $S$ is, by definition, the function Let $\mathcal{X}_S$ denote the \nameref{sec:def-characteristic-function} of $S$.
$\mathcal{X}_S$ such that $\mathcal{X}_S(x) = 1$ for every $x$ in $S$, and Let $f$ be a \nameref{sec:def-step-function} which takes the constant value
$\mathcal{X}_S(x) = 0$ for those $x$ not in $S$. $c_k$ on the $k$th open subinterval $I_k$ of some partition of an interval
Let $f$ be a step function which takes the constant value $c_k$ on the $k$th $[a, b]$.
open subinterval $I_k$ of some partition of an interval $[a, b]$.
Prove that for each $x$ in the union $I_1 \cup I_2 \cup \cdots \cup I_n$ we have Prove that for each $x$ in the union $I_1 \cup I_2 \cup \cdots \cup I_n$ we have
$$f(x) = \sum_{k=1}^n c_k\mathcal{X}_{I_k}(x).$$ $$f(x) = \sum_{k=1}^n c_k\mathcal{X}_{I_k}(x).$$
This property is described by saying that every step function is a linear This property is described by saying that every step function is a linear
@ -1583,7 +1651,7 @@ This property is described by saying that every step function is a linear
Let $x \in I_1 \cup I_2 \cup \cdots \cup I_n$ and $N = \{1, \ldots, n\}$. Let $x \in I_1 \cup I_2 \cup \cdots \cup I_n$ and $N = \{1, \ldots, n\}$.
Let $k \in N$ such that $x \in I_k$. Let $k \in N$ such that $x \in I_k$.
Consider an arbitrary $j \in N - \{k\}$. Consider an arbitrary $j \in N - \{k\}$.
By definition of a partition, $I_j \cap I_k = \emptyset$. By definition of a nameref{sec:def-partition}, $I_j \cap I_k = \emptyset$.
That is, $I_j$ and $I_k$ are disjoint for all $j \in N - \{k\}$. That is, $I_j$ and $I_k$ are disjoint for all $j \in N - \{k\}$.
Therefore, by definition of the characteristic function, Therefore, by definition of the characteristic function,
$\mathcal{X}_{I_k}(x) = 1$ and $\mathcal{X}_{I_j}(x) = 0$ for all $\mathcal{X}_{I_k}(x) = 1$ and $\mathcal{X}_{I_j}(x) = 0$ for all