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Rename glossary to reference.

finite-set-exercises
Joshua Potter 2023-05-18 13:03:59 -06:00
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Bookshelf/Apostol.tex
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@ -15,13 +15,15 @@
\tableofcontents
\begingroup
\renewcommand\thechapter{G}
\renewcommand\thechapter{R}
\setcounter{chapter}{0}
\addtocounter{chapter}{-1}
\chapter{Glossary}%
\label{chap:glossary}
\chapter{Reference}%
\label{chap:reference}
\section{\defined{Characteristic Function}}%
\label{def:characteristic-function}
\label{ref: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
@ -35,7 +37,7 @@ The \textbf{characteristic function} of $S$ is the function $\mathcal{X}_S$ such
\end{definition}
\section{\defined{Infimum}}%
\label{def:infimum}
\label{ref:infimum}
A number $B$ is called an \textbf{infimum} of a nonempty set $S$ if $B$ has
the following two properties:
@ -52,36 +54,36 @@ Such a number $B$ is also known as the \textbf{greatest lower bound}.
\end{definition}
\section{\partial{Integrable}}%
\label{def:integrable}
\label{ref:integrable}
Let $f$ be a function defined and bounded on $[a, b]$.
$f$ is said to be \textbf{integrable} if there exists one and only one number
$I$ such that \eqref{def:integral-bounded-function-eq2} holds.
$I$ such that \eqref{ref:integral-bounded-function-eq2} holds.
If $f$ is integrable on $[a, b]$, we say that the integral
$\int_a^b f(x) \mathop{dx}$ \textbf{exists}.
\section{\partial{Integral of a Bounded Function}}%
\label{def:integral-bounded-function}
\label{ref:integral-bounded-function}
Let $f$ be a function defined and bounded on $[a, b]$.
Let $s$ and $t$ denote arbitrary step functions defined on $[a, b]$ such that
\begin{equation}
\label{def:integral-bounded-function-eq1}
\label{ref:integral-bounded-function-eq1}
s(x) \leq f(x) \leq t(x)
\end{equation}
for every $x$ in $[a, b]$.
If there is one and only one number $I$ such that
\begin{equation}
\label{def:integral-bounded-function-eq2}
\label{ref:integral-bounded-function-eq2}
\int_a^b s(x) \mathop{dx} \leq I \leq \int_a^b t(x) \mathop{dx}
\end{equation}
for every pair of step functions $s$ and $t$ satisfying
\eqref{def:integral-bounded-function-eq1}, then this number $I$ is called
\eqref{ref:integral-bounded-function-eq1}, then this number $I$ is called
the \textbf{integral of $f$ from $a$ to $b$}, and is denoted by the symbol
$\int_a^b f(x) \mathop{dx}$ or by $\int_a^b f$.
If $a < b$, we define $\int_b^a f(x) \mathop{dx} = -\int_a^b f(x) \mathop{dx}$,
provided $f$ is \nameref{def:integrable} on $[a, b]$.
provided $f$ is \nameref{ref:integrable} on $[a, b]$.
We also define $\int_a^a f(x) \mathop{dx} = 0$.
The function $f$ is called the \textbf{integrand}, the numbers $a$ and $b$ are
@ -89,10 +91,10 @@ The function $f$ is called the \textbf{integrand}, the numbers $a$ and $b$ are
\textbf{interval of integration}.
\section{\partial{Integral of a Step Function}}%
\label{def:integral-step-function}
\label{ref:integral-step-function}
Let $s$ be a \nameref{def:step-function} defined on $[a, b]$, and let
$P = \{x_0, x_1, \ldots, x_n\}$ be a \nameref{def:partition} of $[a, b]$
Let $s$ be a \nameref{ref:step-function} defined on $[a, b]$, and let
$P = \{x_0, x_1, \ldots, x_n\}$ be a \nameref{ref:partition} of $[a, b]$
such that $s$ is constant on the open subintervals of $P$.
Denote by $s_k$ the constant value that $s$ takes in the $k$th open subinterval
of $P$, so that
@ -105,17 +107,17 @@ If $a < b$, we define $\int_b^a s(x) \mathop{dx} = -\int_a^b s(x) \mathop{dx}$.
We also define $\int_a^a s(x) \mathop{dx} = 0$.
\section{\partial{Lower Integral}}%
\label{def:lower-integral}
\label{ref:lower-integral}
Let $f$ be a function bounded on $[a, b]$ and $S$ denote the set of numbers
$\int_a^b s(x) \mathop{dx}$ obtained as $s$ runs through all
\nameref{def:step-function}s below $f$.
\nameref{ref:step-function}s below $f$.
That is, let $$S = \left\{ \int_a^b s(x) \mathop{dx} : s \leq f \right\}.$$
The number $\sup{S}$ is called the \textbf{lower integral of $f$}.
It is denoted as $\ubar{I}(f)$.
\section{\partial{Monotonic}}%
\label{def:monotonic}
\label{ref:monotonic}
A function $f$ is called \textbf{monotonic} on set $S$ if it is increasing on
$S$ or if it is decreasing on $S$.
@ -125,11 +127,11 @@ $f$ is said to be \textbf{strictly monotonic} if it is strictly increasing on
A function $f$ is said to be \textbf{piecewise monotonic} on an interval if its
graph consists of a finite number of monotonic pieces.
In other words, $f$ is piecewise monotonic on $[a, b]$ if there is a
\nameref{def:partition} of $[a, b]$ such that $f$ is monotonic on each of
\nameref{ref:partition} of $[a, b]$ such that $f$ is monotonic on each of
the open subintervals of $P$.
\section{\defined{Partition}}%
\label{def:partition}
\label{ref:partition}
Let $[a, b]$ be a closed interval decomposed into $n$ subintervals by inserting
$n - 1$ points of subdivision, say $x_1$, $x_2$, $\ldots$, $x_{n-1}$, subject
@ -151,9 +153,9 @@ A collection of points satisfying \eqref{sec:partition-eq1} is called a
\end{definition}
\section{\partial{Refinement}}%
\label{def:refinement}
\label{ref:refinement}
Let $P$ be a \nameref{def:partition} of closed interval $[a, b]$.
Let $P$ be a \nameref{ref: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 $P$.
@ -161,10 +163,10 @@ $P'$ 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}}%
\label{def:step-function}
\label{ref:step-function}
A function $s$, whose domain is a closed interval $[a, b]$, is called a
\textbf{step function} if there is a \nameref{def:partition}
\textbf{step function} if there is a \nameref{ref:partition}
$P = \{x_0, x_1, \ldots, x_n\}$ of $[a, b]$ such that $s$ is constant on each
open subinterval of $P$.
That is to say, for each $k = 1, 2, \ldots, n$, there is a real number $s_k$
@ -183,7 +185,7 @@ That is to say, for each $k = 1, 2, \ldots, n$, there is a real number $s_k$
\end{definition}
\section{\defined{Supremum}}%
\label{def:supremum}
\label{ref:supremum}
A number $B$ is called a \textbf{supremum} of a nonempty set $S$ if $B$ has
the following two properties:
@ -200,11 +202,11 @@ Such a number $B$ is also known as the \textbf{least upper bound}.
\end{definition}
\section{\partial{Upper Integral}}%
\label{def:upper-integral}
\label{ref:upper-integral}
Let $f$ be a function bounded on $[a, b]$ and $T$ denote the set of numbers
$\int_a^b t(x) \mathop{dx}$ obtained as $t$ runs through all
\nameref{def:step-function}s above $f$.
\nameref{ref:step-function}s above $f$.
That is, let $$T = \left\{ \int_a^b t(x) \mathop{dx} : f \leq t \right\}.$$
The number $\inf{T}$ is called the \textbf{upper integral of $f$}.
It is denoted as $\bar{I}(f)$.
@ -241,7 +243,7 @@ Every nonempty set $S$ of real numbers which is bounded above has a supremum;
{Apostol.Chapter\_I\_03.is\_lub\_neg\_set\_iff\_is\_glb\_set\_neg}
Suppose $L = \sup{S}$ and fix $x \in S$.
By definition of the \nameref{def:supremum}, $x \leq L$ and $L$ is the
By definition of the \nameref{ref: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.
@ -268,7 +270,7 @@ Every nonempty set $S$ of real numbers which is bounded above has a supremum;
Let $S$ be a nonempty set bounded below by $x$.
Then $-S$ is nonempty and bounded above by $x$.
By the \nameref{sec:completeness-axiom}, there exists a
\nameref{def:supremum} $L$ of $-S$.
\nameref{ref:supremum} $L$ of $-S$.
By \nameref{sec:lemma-1}, $L$ is a supremum of $-S$ if and only if $-L$ is an
infimum of $S$.
@ -433,7 +435,7 @@ Let $h$ be a given positive number and let $S$ be a set of real numbers.
\lean{Bookshelf/Apostol/Chapter\_I\_03}
{Apostol.Chapter\_I\_03.sup\_imp\_exists\_gt\_sup\_sub\_delta}
By definition of a \nameref{def:supremum}, $\sup{S}$ is the least upper
By definition of a \nameref{ref:supremum}, $\sup{S}$ is the least upper
bound of $S$.
For the sake of contradiction, suppose for all $x \in S$,
$x \leq \sup{S} - h$.
@ -459,7 +461,7 @@ Let $h$ be a given positive number and let $S$ be a set of real numbers.
\lean{Bookshelf/Apostol/Chapter\_I\_03}
{Apostol.Chapter\_I\_03.inf\_imp\_exists\_lt\_inf\_add\_delta}
By definition of an \nameref{def:infimum}, $\inf{S}$ is the greatest lower
By definition of an \nameref{ref:infimum}, $\inf{S}$ is the greatest lower
bound of $S$.
For the sake of contradiction, suppose for all $x \in S$,
$x \geq \inf{S} + h$.
@ -504,7 +506,7 @@ Given nonempty subsets $A$ and $B$ of $\mathbb{R}$, let $C$ denote the set
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 \nameref{def:supremum}, $a' \leq \sup{A}$.
By definition of a \nameref{ref: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 $\sup{A} + \sup{B}$
@ -575,7 +577,7 @@ Given nonempty subsets $A$ and $B$ of $\mathbb{R}$, let $C$ denote the set
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 \nameref{def:infimum}, $a' \geq \inf{A}$.
By definition of an \nameref{ref: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 $\inf{A} + \inf{B}$
@ -1740,8 +1742,8 @@ Now apply Exercises 4(a) and (b) to the bracket on the right.
\label{sub:exercise-1.11.8}
Let $S$ be a set of points on the real line.
Let $\mathcal{X}_S$ denote the \nameref{def:characteristic-function} of $S$.
Let $f$ be a \nameref{def:step-function} which takes the constant value
Let $\mathcal{X}_S$ denote the \nameref{ref:characteristic-function} of $S$.
Let $f$ be a \nameref{ref:step-function} which takes the constant value
$c_k$ on the $k$th 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
@ -1754,7 +1756,7 @@ 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 $k \in N$ such that $x \in I_k$.
Consider an arbitrary $j \in N - \{k\}$.
By definition of a nameref{def:partition}, $I_j \cap I_k = \emptyset$.
By definition of a nameref{ref:partition}, $I_j \cap I_k = \emptyset$.
That is, $I_j$ and $I_k$ are disjoint for all $j \in N - \{k\}$.
Therefore, by definition of the characteristic function,
$\mathcal{X}_{I_k}(x) = 1$ and $\mathcal{X}_{I_j}(x) = 0$ for all
@ -1780,7 +1782,7 @@ This property is described by saying that every step function is a linear
\begin{theorem}[1.2]
Let $s$ and $t$ be \nameref{def:step-function}s on closed interval
Let $s$ and $t$ be \nameref{ref:step-function}s on closed interval
$[a, b]$.
Then
$$\int_a^b \left[ s(x) + t(x) \right] \mathop{dx} =
@ -1791,7 +1793,7 @@ This property is described by saying that every step function is a linear
\begin{proof}
Let $s$ and $t$ be step functions on closed interval $[a, b]$.
By definition of a step function, there exists a \nameref{def:partition}
By definition of a step function, there exists a \nameref{ref:partition}
$P_s$ such that $s$ is constant on each open subinterval of $P_s$.
Likewise, there exists a partition $P_t$ such that $t$ is constant on each
open subinterval of $P_t$.
@ -1804,7 +1806,7 @@ This property is described by saying that every step function is a linear
$P$.
Let $t_k$ denote the constant value of $t$ on the $k$th open subinterval of
$P$.
By definition of the \nameref{def:integral-step-function},
By definition of the \nameref{ref:integral-step-function},
\begin{align*}
\int_a^b \left[ s(x) + t(x) \right] \mathop{dx}
& = \sum_{k=1}^n (s_k + t_k) \cdot (x_k - x_{k-1}) \\
@ -1823,7 +1825,7 @@ This property is described by saying that every step function is a linear
\begin{theorem}[1.3]
Let $s$ be a \nameref{def:step-function} on closed interval $[a, b]$.
Let $s$ be a \nameref{ref:step-function} on closed interval $[a, b]$.
For every real number $c$, we have
$$\int_a^b c \cdot s(x) \mathop{dx} = c\int_a^b s(x) \mathop{dx}.$$
@ -1832,13 +1834,13 @@ This property is described by saying that every step function is a linear
\begin{proof}
Let $s$ be a step function on closed interval $[a, b]$.
By definition of a step function, there exists a \nameref{def:partition}
By definition of a step function, there exists a \nameref{ref:partition}
$P = \{x_0, x_1, \ldots, x_n\}$ such that $s$ is constant on each open
subinterval of $P$.
Let $s_k$ denote the constant value of $s$ on the $k$th open subinterval of
$P$.
Then $c \cdot s$ is a step function with step partition $P$.
By definition of the \nameref{def:integral-step-function},
By definition of the \nameref{ref:integral-step-function},
\begin{align*}
\int_a^b c \cdot s(x) \mathop{dx}
& = \sum_{k=1}^n c \cdot s_k \cdot (x_k - x_{k-1}) \\
@ -1854,7 +1856,7 @@ This property is described by saying that every step function is a linear
\begin{theorem}[1.4]
Let $s$ and $t$ be \nameref{def:step-function}s on closed interval
Let $s$ and $t$ be \nameref{ref:step-function}s on closed interval
$[a, b]$.
For every real $c_1$ and $c_2$, we have
$$\int_a^b \left[ c_1s(x) + c_2t(x) \right] \mathop{dx} =
@ -1884,7 +1886,7 @@ This property is described by saying that every step function is a linear
\begin{theorem}[1.5]
Let $s$ and $t$ be \nameref{def:step-function}s on closed interval
Let $s$ and $t$ be \nameref{ref:step-function}s on closed interval
$[a, b]$.
If $s(x) < t(x)$ for every $x$ in $[a, b]$, then
$$\int_a^b s(x) \mathop{dx} < \int_a^b t(x) \mathop{dx}.$$
@ -1894,7 +1896,7 @@ This property is described by saying that every step function is a linear
\begin{proof}
Let $s$ and $t$ be step functions on closed interval $[a, b]$.
By definition of a step function, there exists a \nameref{def:partition}
By definition of a step function, there exists a \nameref{ref:partition}
$P_s$ such that $s$ is constant on each open subinterval of $P_s$.
Likewise, there exists a partition $P_t$ such that $t$ is constant on each
open subinterval of $P_t$.
@ -1907,7 +1909,7 @@ This property is described by saying that every step function is a linear
$P$.
Let $t_k$ denote the constant value of $t$ on the $k$th open subinterval of
$P$.
By definition of the \nameref{def:integral-step-function},
By definition of the \nameref{ref:integral-step-function},
\begin{align*}
\int_a^b s(x) \mathop{dx}
& = \sum_{k=1}^n s_k \cdot (x_k - x_{k-1}) \\
@ -1923,7 +1925,7 @@ This property is described by saying that every step function is a linear
\begin{theorem}[1.6]
Let $a, b, c \in \mathbb{R}$ and $s$ a \nameref{def:step-function} on the
Let $a, b, c \in \mathbb{R}$ and $s$ a \nameref{ref:step-function} on the
smallest closed interval containing them.
Then
$$\int_a^c s(x) \mathop{dx} + \int_c^b s(x) \mathop{dx} +
@ -1935,7 +1937,7 @@ This property is described by saying that every step function is a linear
WLOG, suppose $a < c < b$ and $s$ be a step function on closed interval
$[a, b]$.
By definition of a step function, there exists a \nameref{def:partition}
By definition of a step function, there exists a \nameref{ref:partition}
$P$ such that $s$ is constant on each open subinterval of $P$.
Let $Q = \{x_0, x_1, \ldots, x_n\}$ be a refinement of $P$ that includes $c$
@ -1944,7 +1946,7 @@ This property is described by saying that every step function is a linear
that $x_i = c$.
Let $s_k$ denote the constant value of $s$ on the $k$th open subinterval of
$Q$.
By definition of the \nameref{def:integral-step-function},
By definition of the \nameref{ref:integral-step-function},
\begin{align*}
\int_a^b s(x) \mathop{dx}
& = \sum_{k=1}^n s_k \cdot (x_k - x_{k - 1}) \\
@ -1979,7 +1981,7 @@ This property is described by saying that every step function is a linear
\begin{proof}
Let $s$ be a step function on closed interval $[a, b]$.
By definition of a step function, there exists a \nameref{def:partition}
By definition of a step function, there exists a \nameref{ref:partition}
$P = \{x_0, x_1, \ldots, x_n\}$ such that $s$ is constant on each open
subinterval of $P$.
Let $s_k$ denote the constant value of $s$ on the $k$th open subinterval of
@ -1992,7 +1994,7 @@ This property is described by saying that every step function is a linear
Let $t_k$ denote the value of $t$ on the $k$th open subinterval of $Q$.
By construction, $t_k = s_k$.
By definition of the \nameref{def:integral-step-function},
By definition of the \nameref{ref:integral-step-function},
\begin{align*}
\int_{a+c}^{b+c} s(x - c) \mathop{dx}
& = \int_{a+c}^{b+c} t(x) \mathop{dx} \\
@ -2020,7 +2022,7 @@ This property is described by saying that every step function is a linear
\begin{proof}
Let $s$ be a step function on closed interval $[a, b]$.
By definition of a step function, there exists a \nameref{def:partition}
By definition of a step function, there exists a \nameref{ref:partition}
$P = \{x_0, x_1, \ldots, x_n\}$ such that $s$ is constant on each open
subinterval of $P$.
Let $s_i$ denote the value of $s$ on the $i$th open subinterval of $P$.
@ -2034,7 +2036,7 @@ This property is described by saying that every step function is a linear
Then $t(x) = s(x / k)$ is a step function on closed interval $[ka, kb]$
with partition $Q = \{kx_0, kx_1, \ldots, kx_n\}$.
Furthermore $t_i = s_i$.
By definition of the \nameref{def:integral-step-function},
By definition of the \nameref{ref:integral-step-function},
\begin{align*}
\int_{ka}^{kb} s(x / k) \mathop{dx}
& = \int_{ka}^{kb} t(x) \mathop{dx} \\
@ -2050,7 +2052,7 @@ This property is described by saying that every step function is a linear
Then $t(x) = s(x / k)$ is a step function on closed interval $[kb, ka]$
with partition $Q = \{kx_n, kx_{n-1}, \ldots, kx_0\}$.
Furthermore $t_i = s_i$.
By definition of the \nameref{def:integral-step-function},
By definition of the \nameref{ref:integral-step-function},
\begin{align*}
\int_{ka}^{kb} s(x / k) \mathop{dx}
& = -\int_{kb}^{ka} s(x / k) \mathop{dx} \\
@ -2102,7 +2104,7 @@ $\int_{-1}^3 \floor{x} \mathop{dx}$.
$P = \{-1, 0, 1, 2, 3\} = \{x_0, x_1, x_2, x_3, x_4\}$.
Let $s_k$ denote the constant value $s$ takes on the $k$th open subinterval of
$P$.
By definition of the \nameref{def:integral-step-function},
By definition of the \nameref{ref:integral-step-function},
\begin{align*}
\int_{-1}^3 \floor{x} \mathop{dx}
& = \sum_{k=1}^4 s_k \cdot (x_k - x_{k-1}) \\
@ -2129,7 +2131,7 @@ $\int_{-1}^3 \left(\floor{x} + \floor{x + \frac{1}{2}}\right) \mathop{dx}$.
\end{align*}
Let $s_k$ denote the constant value $s$ takes on the $k$th open subinterval of
$P$.
By definition of the \nameref{def:integral-step-function},
By definition of the \nameref{ref:integral-step-function},
\begin{align*}
\int_{-1}^3 \floor{x} + \floor{x + \frac{1}{2}}
& = \sum_{k=1}^8 s_k \cdot (x_k - x_{k-1}) \\
@ -2166,13 +2168,13 @@ Show that
Let $s(x) = \floor{x}$ and $t(x) = \floor{-x}$, both with domain $[a, b]$.
Let $x_1$, $\ldots$, $x_{n-1}$ denote the integers found in interval $(a, b)$.
Then $P = \{x_0, x_1, \ldots, x_n\}$, $x_0 = a$ and $x_n = b$, is a step
\nameref{def:partition} of both $s$ and $t$.
\nameref{ref:partition} of both $s$ and $t$.
Let $s_k$ and $t_k$ denote the constant values $s$ and $t$ take on the $k$th
open subinterval of $P$ respectively.
By \nameref{ssub:exercise-1.11.4b}, $\floor{-x} = -\floor{x} - 1$ for all $x$
in every open subinterval of $P$.
That is, $s_k = -t_k - 1$.
By definition of the \nameref{def:integral-step-function},
By definition of the \nameref{ref:integral-step-function},
\begin{align*}
\int_a^b \floor{x} \mathop{dx} + \int_a^b \floor{-x} \mathop{dx}
& = \sum_{k=1}^n s_k (x_k - x_{k-1}) +
@ -2197,11 +2199,11 @@ Prove that $\int_0^2 \floor{t^2} \mathop{dt} = 5 - \sqrt{2} - \sqrt{3}$.
\begin{proof}
Let $s(t) = \floor{t^2}$ with domain $[0, 2]$.
Then $s$ is a \nameref{def:step-function} with partition
Then $s$ is a \nameref{ref:step-function} with partition
$P = \{0, 1, \sqrt{2}, \sqrt{3}, 2\} = \{x_0, x_1, \ldots, x_4\}$.
Let $s_k$ denote the constant value that $s$ takes in the $k$th open
subinterval of $P$.
By the \nameref{def:integral-step-function},
By the \nameref{ref:integral-step-function},
\begin{align*}
\int_0^2 \floor{t^2} \mathop{dt}
& = \sum_{k=1}^4 s_k \cdot (x_k - x_{k-1}) \\
@ -2220,7 +2222,7 @@ Compute $\int_{-3}^3 \floor{t^2} \mathop{dt}$.
\begin{proof}
Let $s(t) = \floor{t^2}$ with domain $[0, 3]$.
Then $s$ is a \nameref{def:step-function} with \nameref{def:partition}
Then $s$ is a \nameref{ref:step-function} with \nameref{ref:partition}
\begin{align*}
P
& = \{\sqrt{0}, \sqrt{1}, \sqrt{2}, \sqrt{3}, \sqrt{4}, \sqrt{5},
@ -2229,7 +2231,7 @@ Compute $\int_{-3}^3 \floor{t^2} \mathop{dt}$.
\end{align*}
Let $s_k$ denote the constant value that $s$ takes in the $k$th open
subinterval of $P$.
By the \nameref{def:integral-step-function},
By the \nameref{ref:integral-step-function},
\begin{align}
\int_0^3 \floor{t^2} \mathop{dt}
& = \sum_{k=1}^9 s_k \cdot (x_k - x_{k-1})
@ -2267,11 +2269,11 @@ Compute $\int_0^9 \floor{\sqrt{t}} \mathop{dt}$.
\begin{proof}
Let $s(t) = \floor{\sqrt{t}}$ with domain $[0, 9]$.
Then $s$ is a \nameref{def:step-function} with \nameref{def:partition}
Then $s$ is a \nameref{ref:step-function} with \nameref{ref:partition}
$P = \{0, 1, 4, 9\} = \{x_0, x_1, x_2, x_3\}$.
Let $s_k$ denote the constant value that $s$ takes in the $k$th open
subinterval of $P$.
By the \nameref{def:integral-step-function},
By the \nameref{ref:integral-step-function},
\begin{align*}
\int_0^9 \floor{\sqrt{t}} \mathop{dt}
& = \sum_{k=1}^3 s_k \cdot (x_k - x_{k-1}) \\
@ -2300,11 +2302,11 @@ If $n$ is a positive integer, prove that
Let $n = 1$.
Define $s(t) = \floor{\sqrt{t}}$ with domain $[0, 1]$.
Then $s$ is a \nameref{def:step-function} with \nameref{def:partition}
Then $s$ is a \nameref{ref:step-function} with \nameref{ref:partition}
$P = \{0, 1\} = \{x_0, x_1\}$.
Let $s_k$ denote the constant value of $s$ on the $k$th open subinterval of
$P$.
By definition of the \nameref{def:integral-step-function}, the left-hand
By definition of the \nameref{ref:integral-step-function}, the left-hand
side of \eqref{sub:exercise-1.15.7b-eq1} evaluates to
\begin{align*}
\int_0^{n^2} \floor{\sqrt{t}} \mathop{dt}
@ -2320,7 +2322,7 @@ If $n$ is a positive integer, prove that
Let $n > 0$ be a positive integer and suppose $P(n)$ is true.
Define $s(t) = \floor{\sqrt{t}}$ with domain $[0, (n + 1)^2]$.
Then $s$ is a \nameref{def:step-function} with \nameref{def:partition}
Then $s$ is a \nameref{ref:step-function} with \nameref{ref:partition}
\begin{align*}
P
& = \{0, 1, 4, \ldots, n^2, (n + 1)^2\} \\
@ -2328,7 +2330,7 @@ If $n$ is a positive integer, prove that
\end{align*}
Let $s_k$ denote the constant value of $s$ on the $k$th open subinterval of
$P$.
By definition of the \nameref{def:integral-step-function}, it follows
By definition of the \nameref{ref:integral-step-function}, it follows
that
\begin{align*}
& \int_0^{(n + 1)^2} s(x) \mathop{dx} \\
@ -2405,8 +2407,8 @@ $\int_a^b s + \int_b^c s = \int_a^c s$.
WLOG, suppose $a < b < c$.
Let $s$ be a step function defined on closed interval $[a, c]$.
By definition of a \nameref{def:step-function}, there exists a
\nameref{def:partition} such that $s$ is constant on each open
By definition of a \nameref{ref:step-function}, there exists a
\nameref{ref:partition} such that $s$ is constant on each open
subinterval of $P$.
Let $Q = \{x_0, x_1, \ldots, x_n\}$ be a refinement of $P$ that includes $b$
as a subdivision point.
@ -2439,7 +2441,7 @@ $\int_a^b (s + t) = \int_a^b s + \int_a^b t$.
\vspace{6pt}
Let $s$ and $t$ be step functions on closed interval $[a, b]$.
By definition of a step function, there exists a \nameref{def:partition}
By definition of a step function, there exists a \nameref{ref:partition}
$P_s$ such that $s$ is constant on each open subinterval of $P_s$.
Likewise, there exists a partition $P_t$ such that $t$ is constant on each
open subinterval of $P_t$.
@ -2486,7 +2488,7 @@ $\int_a^b c \cdot s = c \int_a^b s$.
\vspace{6pt}
Let $s$ be a step function on closed interval $[a, b]$.
By definition of a step function, there exists a \nameref{def:partition}
By definition of a step function, there exists a \nameref{ref:partition}
$P = \{x_0, x_1, \ldots, x_n\}$ such that $s$ is constant on each open
subinterval of $P$.
Let $s_k$ denote the constant value of $s$ on the $k$th open subinterval of
@ -2518,8 +2520,8 @@ $\int_{a+c}^{b+c} s(x) \mathop{dx} = \int_a^b s(x + c) \mathop{dx}$.
\vspace{6pt}
Let $s$ be a step function on closed interval $[a + c, b + c]$.
By definition of a \nameref{def:step-function}, there exists a
\nameref{def:partition} $P = \{x_0, x_1, \ldots, x_n\}$ such that $s$ is
By definition of a \nameref{ref:step-function}, there exists a
\nameref{ref:partition} $P = \{x_0, x_1, \ldots, x_n\}$ such that $s$ is
constant on each open subinterval of $P$.
Let $s_k$ denote the constant value of $s$ on the $k$th open subinterval of
$P$.
@ -2556,8 +2558,8 @@ If $s(x) < t(x)$ for each $x$ in $[a, b]$, then $\int_a^b s < \int_a^b t$.
\vspace{6pt}
Let $s$ and $t$ be step functions on closed interval $[a, b]$.
By definition of a \nameref{def:step-function}, there exists a
\nameref{def:partition} $P_s$ such that $s$ is constant on each open
By definition of a \nameref{ref:step-function}, there exists a
\nameref{ref:partition} $P_s$ such that $s$ is constant on each open
subinterval of $P_s$.
Likewise, there exists a partition $P_t$ such that $t$ is constant on each
open subinterval of $P_t$.
@ -2596,8 +2598,8 @@ If $s(x) < t(x)$ for each $x$ in $[a, b]$, then $\int_a^b s < \int_a^b t$.
\int_a^b s(x) \mathop{dx} \leq \ubar{I}(f) \leq
\bar{I}(f) \leq \int_a^b t(x) \mathop{dx}
\end{equation}
for all \nameref{def:step-function}s $s$ and $t$ with $s \leq f \leq t$.
The function $f$ is \nameref{def:integrable} on $[a, b]$ if and only if
for all \nameref{ref:step-function}s $s$ and $t$ with $s \leq f \leq t$.
The function $f$ is \nameref{ref:integrable} on $[a, b]$ if and only if
its upper and lower integrals are equal, in which case we have
$$\int_a^b f(x) \mathop{dx} = \ubar{I}(f) = \bar{I}(f).$$
@ -2627,10 +2629,10 @@ If $s(x) < t(x)$ for each $x$ in $[a, b]$, then $\int_a^b s < \int_a^b t$.
By construction, $s \leq t$ for every $s$ in $S$ and $t$ in $T$.
Therefore \nameref{sec:theorem-i.34} tells us $S$ has a
\nameref{def:supremum}, $T$ has an \nameref{def:infimum}, and
\nameref{ref:supremum}, $T$ has an \nameref{ref:infimum}, and
$\sup{S} \leq \inf{T}$.
By definition of the \nameref{def:lower-integral}, $\ubar{I}(f) = \sup{S}$.
By definition of the \nameref{def:upper-integral}, $\bar{I}(f) = \inf{S}$.
By definition of the \nameref{ref:lower-integral}, $\ubar{I}(f) = \sup{S}$.
By definition of the \nameref{ref:upper-integral}, $\bar{I}(f) = \inf{S}$.
Thus \eqref{sub:theorem-1.9-eq1} holds.
\paragraph{(ii)}%
@ -2639,7 +2641,7 @@ If $s(x) < t(x)$ for each $x$ in $[a, b]$, then $\int_a^b s < \int_a^b t$.
there exists one and only one number $I$ such that
$$\int_a^b s(x) \mathop{dx} \leq I \leq \int_a^b t(x) \mathop{dx}$$
for every pair of step functions $s$ and $t$ satisfying
\eqref{def:integral-bounded-function-eq1}.
\eqref{ref:integral-bounded-function-eq1}.
By \eqref{sub:theorem-1.9-eq1} and the definition of the supremum/infimum,
this holds if and only if $\ubar{I}(f) = \bar{I}(f)$, concluding the
proof.
@ -2654,7 +2656,7 @@ If $s(x) < t(x)$ for each $x$ in $[a, b]$, then $\int_a^b s < \int_a^b t$.
\begin{theorem}[1.10]
Let $f$ be a nonnegative function, \nameref{def:integrable} on an interval
Let $f$ be a nonnegative function, \nameref{ref:integrable} on an interval
$[a, b]$, and let $Q$ denote the ordinate set of $f$ over $[a, b]$.
Then $Q$ is measurable and its area is equal to the integral
$\int_a^b f(x) \mathop{dx}$.
@ -2663,11 +2665,11 @@ If $s(x) < t(x)$ for each $x$ in $[a, b]$, then $\int_a^b s < \int_a^b t$.
\begin{proof}
Let $f$ be a nonnegative function, \nameref{def:integrable} on $[a, b]$.
Let $f$ be a nonnegative function, \nameref{ref:integrable} on $[a, b]$.
By definition of integrability, there exists one and only one number $I$ such
that $$\int_a^b s(x) \mathop{dx} \leq I \leq \int_a^b t(x) \mathop{dx}$$ for
every pair of step functions $s$ and $t$ satisfying
\eqref{def:integral-bounded-function-eq1}.
\eqref{ref:integral-bounded-function-eq1}.
In other words, $I$ is the one and only number that satisfies
$$a(S) \leq I \leq a(T)$$ for every pair of step regions
$S \subseteq Q \subseteq T$.
@ -2706,7 +2708,7 @@ If $s(x) < t(x)$ for each $x$ in $[a, b]$, then $\int_a^b s < \int_a^b t$.
such that
$$\int_a^b s(x) \mathop{dx} \leq I \leq \int_a^b t(x) \mathop{dx}$$
for every pair of step functions $s$ and $t$ satisfying
\eqref{def:integral-bounded-function-eq1}.
\eqref{ref:integral-bounded-function-eq1}.
In other words, $I$ is the one and only number that satisfies
$$a(S) \leq I \leq a(T)$$ for every pair of step regions
$S \subseteq Q' \subseteq T$.
@ -2738,8 +2740,8 @@ If $s(x) < t(x)$ for each $x$ in $[a, b]$, then $\int_a^b s < \int_a^b t$.
\begin{theorem}[1.12]
If $f$ is \nameref{def:monotonic} on a closed interval $[a, b]$, then $f$
is \nameref{def:integrable} on $[a, b]$.
If $f$ is \nameref{ref:monotonic} on a closed interval $[a, b]$, then $f$
is \nameref{ref:integrable} on $[a, b]$.
\end{theorem}
@ -2749,8 +2751,8 @@ If $s(x) < t(x)$ for each $x$ in $[a, b]$, then $\int_a^b s < \int_a^b t$.
That is to say, either $f$ is increasing on $[a, b]$ or $f$ is decreasing on
$[a, b]$.
Because $f$ is on a closed interval, it is bounded.
By \nameref{sub:theorem-1.9}, $f$ has a \nameref{def:lower-integral}
$\ubar{I}(f)$, $f$ has an \nameref{def:upper-integral} $\bar{I}(f)$,
By \nameref{sub:theorem-1.9}, $f$ has a \nameref{ref:lower-integral}
$\ubar{I}(f)$, $f$ has an \nameref{ref:upper-integral} $\bar{I}(f)$,
and $f$ is integrable if and only if $\ubar{I}(f) = \bar{I}(f)$.
Consider a partition $P = \{x_0, x_1, \ldots, x_n\}$ of $[a, b]$ in which
@ -2770,7 +2772,7 @@ If $s(x) < t(x)$ for each $x$ in $[a, b]$, then $\int_a^b s < \int_a^b t$.
\int_a^b s(x) \mathop{dx} \leq \ubar{I}(f)
\leq \bar{I}(f) \leq \int_a^b t(x) \mathop{dx}.
\end{equation}
By definition of the \nameref{def:integral-step-function},
By definition of the \nameref{ref:integral-step-function},
\begin{align*}
\int_a^b s(x) \mathop{dx}
& = \sum_{k=1}^n f(x_{k-1})\left[\frac{b - a}{n}\right] \\
@ -2809,7 +2811,7 @@ If $s(x) < t(x)$ for each $x$ in $[a, b]$, then $\int_a^b s < \int_a^b t$.
\int_a^b s(x) \mathop{dx} \leq \ubar{I}(f)
\leq \bar{I}(f) \leq \int_a^b t(x) \mathop{dx}.
\end{equation}
By definition of the \nameref{def:integral-step-function},
By definition of the \nameref{ref:integral-step-function},
\begin{align*}
\int_a^b s(x) \mathop{dx}
& = \sum_{k=1}^n f(x_k)\left[\frac{b - a}{n}\right] \\
@ -2863,7 +2865,7 @@ If $s(x) < t(x)$ for each $x$ in $[a, b]$, then $\int_a^b s < \int_a^b t$.
on every $k$th open subinterval of $P$.
Let $t$ be the step function above $f$ with constant value $f(x_k)$
on every $k$th open subinterval of $P$.
By definition of the \nameref{def:integral-step-function},
By definition of the \nameref{ref:integral-step-function},
\begin{align*}
\int_a^b s(x) \mathop{dx}
& = \sum_{k=1}^n f(x_{k-1})\left[\frac{b - a}{n}\right] \\
@ -2935,7 +2937,7 @@ If $s(x) < t(x)$ for each $x$ in $[a, b]$, then $\int_a^b s < \int_a^b t$.
on every $k$th open subinterval of $P$.
Let $t$ be the step function above $f$ with constant value $f(x_{k-1})$
on every $k$th open subinterval of $P$.
By definition of the \nameref{def:integral-step-function},
By definition of the \nameref{ref:integral-step-function},
\begin{align*}
\int_a^b s(x) \mathop{dx}
& = \sum_{k=1}^n f(x_k) \left[\frac{b - a}{n}\right] \\

6
Bookshelf/Enderton_Logic.tex
View File

@ -10,10 +10,10 @@
\tableofcontents
\begingroup
\renewcommand\thechapter{G}
\renewcommand\thechapter{R}
\chapter{Glossary}%
\label{chap:glossary}
\chapter{Reference}%
\label{chap:reference}
\endgroup