diff --git a/Bookshelf/Real/Sequence/Arithmetic_tex.tex b/Bookshelf/Real/Sequence/Arithmetic_tex.tex index 5e3c044..318d299 100644 --- a/Bookshelf/Real/Sequence/Arithmetic_tex.tex +++ b/Bookshelf/Real/Sequence/Arithmetic_tex.tex @@ -2,8 +2,11 @@ \input{preamble} -\newcommand{\link}[1]{\lean{../../..}{Bookshelf/Real/Sequence/Arithmetic} - {Real.Arithmetic.#1}} +\newcommand{\link}[1]{\lean{../../..} + {Bookshelf/Real/Sequence/Arithmetic} + {Real.Arithmetic.#1} + {Real.Arithmetic.#1} +} \begin{document} @@ -16,7 +19,7 @@ $$\sum_{i=0}^n a_i = \frac{(n + 1)(a_0 + a_n)}{2}.$$ \begin{proof} - \link{sum_recursive_closed} + \link{sum\_recursive\_closed} \end{proof} diff --git a/Bookshelf/Real/Sequence/Geometric_tex.tex b/Bookshelf/Real/Sequence/Geometric_tex.tex index e049c78..971177a 100644 --- a/Bookshelf/Real/Sequence/Geometric_tex.tex +++ b/Bookshelf/Real/Sequence/Geometric_tex.tex @@ -2,8 +2,11 @@ \input{preamble} -\newcommand{\link}[1]{\lean{../../..}{Bookshelf/Real/Sequence/Geometric} - {Real.Geometric.#1}} +\newcommand{\link}[1]{\lean{../../..} + {Bookshelf/Real/Sequence/Geometric} + {Real.Geometric.#1} + {Real.Geometric.#1} +} \begin{document} @@ -16,7 +19,7 @@ $$\sum_{i=0}^n a_i = \frac{a_0(1 - r^{n+1})}{1 - r}.$$ \begin{proof} - \link{sum_recursive_closed} + \link{sum\_recursive\_closed} \end{proof} diff --git a/Exercises/Apostol/Chapter_I_3_tex.tex b/Exercises/Apostol/Chapter_I_3_tex.tex index 23b3a5b..865a84b 100644 --- a/Exercises/Apostol/Chapter_I_3_tex.tex +++ b/Exercises/Apostol/Chapter_I_3_tex.tex @@ -3,8 +3,11 @@ \input{preamble} -\newcommand{\link}[1]{\lean{../..}{Exercises/Apostol/Chapter_I_3} - {Exercises.Apostol.Chapter\_I\_3.Real.#1}} +\newcommand{\link}[1]{\lean{../..} + {Exercises/Apostol/Chapter\_I\_3} + {Exercises.Apostol.Chapter\_I\_3.Real.#1} + {Chapter\_I\_3.Real.#1} +} \begin{document} @@ -16,7 +19,7 @@ is, there is a real number $L$ such that $L = \inf{S}$. \begin{proof} - \link{exists_isGLB} + \link{exists\_isGLB} \end{proof} @@ -27,7 +30,7 @@ For every real $x$ there exists a positive integer $n$ such that $n > x$. \begin{proof} - \link{exists_pnat_geq_self} + \link{exists\_pnat\_geq\_self} \end{proof} @@ -39,7 +42,7 @@ integer $n$ such that $nx > y$. \begin{proof} - \link{exists_pnat_mul_self_geq_of_pos} + \link{exists\_pnat\_mul\_self\_geq\_of\_pos} \end{proof} @@ -52,7 +55,7 @@ for every integer $n \geq 1$, then $x = a$. \begin{proof} - \link{forall_pnat_leq_self_leq_frac_imp_eq} + \link{forall\_pnat\_leq\_self\_leq\_frac\_imp\_eq} \end{proof} @@ -70,8 +73,8 @@ Let $h$ be a given positive number and let $S$ be a set of real numbers. \begin{proof} \begin{enumerate}[(a)] - \item \link{sup_imp_exists_gt_sup_sub_delta} - \item \link{inf_imp_exists_lt_inf_add_delta} + \item \link{sup\_imp\_exists\_gt\_sup\_sub\_delta} + \item \link{inf\_imp\_exists\_lt\_inf\_add\_delta} \end{enumerate} \end{proof} @@ -92,8 +95,8 @@ $$C = \{a + b : a \in A, b \in B\}.$$ \begin{proof} \begin{enumerate}[(a)] - \item \link{sup_minkowski_sum_eq_sup_add_sup} - \item \link{inf_minkowski_sum_eq_inf_add_inf} + \item \link{sup\_minkowski\_sum\_eq\_sup\_add\_sup} + \item \link{inf\_minkowski\_sum\_eq\_inf\_add\_inf} \end{enumerate} \end{proof} @@ -109,7 +112,7 @@ $$\sup{S} \leq \inf{T}.$$ \begin{proof} - \link{forall_mem_le_forall_mem_imp_sup_le_inf} + \link{forall\_mem\_le\_forall\_mem\_imp\_sup\_le\_inf} \end{proof} diff --git a/Exercises/Apostol/Exercises_1_7_tex.tex b/Exercises/Apostol/Exercises_1_7_tex.tex new file mode 100644 index 0000000..7a50d3d --- /dev/null +++ b/Exercises/Apostol/Exercises_1_7_tex.tex @@ -0,0 +1,177 @@ +\documentclass{article} +\usepackage{amsmath} + +\input{preamble} + +\newcommand{\larea}[2]{\lean{../..}{Bookshelf/Real/Geometry/Area}{#1}{#2}} +\newcommand{\lrect}[2]{\lean{../..}{Bookshelf/Real/Geometry/Rectangle}{#1}{#2}} + +\begin{document} + +The properties of area in this set of exercises are to be deduced from the +axioms for area stated in the foregoing section. + +\section{Exercise 1}% +\label{sec:exercise-1} + +Prove that each of the following sets is measurable and has zero area: + +\subsection{Exercise 1a}% +\label{sub:exercise-1a} + +A set consisting of a single point. + +\begin{proof} + + Let $S$ be a set consisting of a single point. + By definition of a \lrect{Real.Point}{Point}, $S$ is a rectangle in which all + vertices coincide. + By \larea{Choice-of-Scale}{Choice of Scale}, $S$ is measurable with area its + width times its height. + The width and height of $S$ is trivially zero. + Therefore $a(S) = (0)(0) = 0$. + +\end{proof} + +\subsection{Exercise 1b}% +\label{sub:exercise-1b} + +A set consisting of a finite number of points in a plane. + +\begin{proof} + + We show for all $k > 0$, a set consisting of $k$ points in a plane is + measurable with area $0$. + + \paragraph{Base Case}% + + Consider a set $S$ consisting of a single point in a plane. + By \eqref{sub:exercise-1a}, $S$ is measurable with area $0$. + + \paragraph{Induction Step}% + + Define our induction hypothesis as, "Let $k > 0$ and assume a set consisting + of $k$ points in a plane is measurable with area $0$." + + Consider a set $S_{k+1}$ consisting of $k + 1$ points in a plane. + Pick an arbitrary point of $S_{k+1}$. + Denote the set containing just this point as $T$. + Denote the remaining set of points as $S_k$. + By construction, $S_{k+1} = S_k \cup T$. + By the induction hypothesis, $S_k$ is measurable with area $0$. + By \eqref{sub:exercise-1a}, $T$ is measurable with area $0$. + By the \larea{Additive-Property}{Additive Property}, $S_k \cup T$ is + measurable, $S_k \cap T$ is measurable, and + \begin{align} + a(S_{k+1}) + & = a(S_k \cup T) \nonumber \\ + & = a(S_k) + a(T) - a(S_k \cap T) \nonumber \\ + & = 0 + 0 - a(S_k \cap T). \label{sub:exercise-1b-eq1} + \end{align} + + \noindent + There are two cases to consider: + + \subparagraph{Case 1}% + + $S_k \cap T = \emptyset$. + Then it trivially follows that $a(S_k \cap T) = 0$. + + \subparagraph{Case 2}% + + $S_k \cap T \neq \emptyset$. + Since $T$ consists of a single point, $S_k \cap T = T$. + By \eqref{sub:exercise-1a}, $a(S_k \cap T) = a(T) = 0$. + + \vspace{8pt} + \noindent + In both cases, \eqref{sub:exercise-1b-eq1} evaluates to $0$, implying + $a(S_{k+1}) = 0$ as expected. + + \paragraph{Conclusion}% + + By mathematical induction, it follows for all $n > 0$, a set consisting of + $n$ points in a plane is measurable with area $0$. + +\end{proof} + +\subsection{Exercise 1c}% +\label{sub:exercise-1c} + +The union of a finite collection of line segments in a plane. + +\begin{proof} + + We show for all $k > 0$, a set consisting of $k$ line segments in a plane is + measurable with area $0$. + + \paragraph{Base Case}% + + Consider a set $S$ consisting of a single line segment in a plane. + By definition of a \lrect{Real.LineSemgnet}{Line Segment}, $S$ is a + rectangle in which one side has dimension $0$. + By \larea{Choice-of-Scale}{Choice of Scale}, $S$ is measurable with area its + width $w$ times its height $h$. + Therefore $a(S) = wh = 0$. + + \paragraph{Induction Step}% + + Define our induction hypothesis as, "Let $k > 0$ and assume a set consisting + of $k$ line segments in a plane is measurable with area $0$." + + Consider a set $S_{k+1}$ consisting of $k + 1$ line segments in a plane. + Pick an arbitrary line segment of $S_{k+1}$. + Denote the set containing just this line segment as $T$. + Denote the remaining set of line segments as $S_k$. + By construction, $S_{k+1} = S_k \cup T$. + By the induction hypothesis, $S_k$ is measurable with area $0$. + By the base case, $T$ is measurable with area $0$. + By the \larea{Additive-Property}{Additive Property}, $S_k \cup T$ is + measurable, $S_k \cap T$ is measurable, and + \begin{align} + a(S_{k+1}) + & = a(S_k \cup T) \nonumber \\ + & = a(S_k) + a(T) - a(S_k \cap T) \nonumber \\ + & = 0 + 0 - a(S_k \cap T). \label{sub:exercise-1c-eq1} + \end{align} + + \noindent + There are two cases to consider: + + \subparagraph{Case 1}% + + $S_k \cap T = \emptyset$. + Then it trivially follows that $a(S_k \cap T) = 0$. + + \subparagraph{Case 2}% + + $S_k \cap T \neq \emptyset$. + Since $T$ consists of a single point, $S_k \cap T = T$. + By the base case, $a(S_k \cap T) = a(T) = 0$. + + \vspace{8pt} + \noindent + In both cases, \eqref{sub:exercise-1c-eq1} evaluates to $0$, implying + $a(S_{k+1}) = 0$ as expected. + + \paragraph{Conclusion}% + + By mathematical induction, it follows for all $n > 0$, a set consisting of + $n$ line segments in a plane is measurable with area $0$. + +\end{proof} + +\section{Exercise 2}% +\label{sec:exercise-2} + +Every right triangular region is measurable because it can be obtained as the +intersection of two rectangles. Prove that every triangular region is measurable +and that its area is one half the product of its base and altitude. + +\begin{proof} + + TODO + +\end{proof} + +\end{document} diff --git a/Exercises/Enderton/Chapter0_tex.tex b/Exercises/Enderton/Chapter0_tex.tex index e235dfd..916c711 100644 --- a/Exercises/Enderton/Chapter0_tex.tex +++ b/Exercises/Enderton/Chapter0_tex.tex @@ -2,8 +2,11 @@ \input{preamble} -\newcommand{\link}[1]{\lean{../..}{Exercises/Enderton/Chapter0} - {Exercises.Enderton.Chapter0.#1}} +\newcommand{\link}[1]{\lean{../..} + {Exercises/Enderton/Chapter0} + {Exercises.Enderton.Chapter0.#1} + {Chapter0.#1} +} \begin{document} @@ -15,7 +18,7 @@ Then $x_1 = \langle y_1, \ldots, y_{k+1} \rangle$. \begin{proof} - \link{lemma_0a} + \link{lemma\_0a} \end{proof} diff --git a/preamble.tex b/preamble.tex index 4a44c63..7f6a239 100644 --- a/preamble.tex +++ b/preamble.tex @@ -1,5 +1,4 @@ \usepackage{amsfonts, amsthm} -\usepackage[T1]{fontenc} \usepackage{hyperref} \newtheorem{theorem}{Theorem} @@ -11,11 +10,6 @@ \hypersetup{colorlinks=true, urlcolor=blue} -% https://tex.stackexchange.com/a/232188 -\newcommand{\startunderscoreletter}{\catcode`_ 12\relax} -\newcommand{\stopunderscoreletter}{\catcode`_ 8\relax} - % The first argument refers to a relative path upward from a current file to % the root of the workspace (i.e. where this `preamble.tex` file is located). -\newcommand{\lean}[3]{\href{#1/#2.html\##3}{ - \startunderscoreletter #3 \stopunderscoreletter}} +\newcommand{\lean}[4]{\href{#1/#2.html\##3}{#4}}