From f22712faf18298d0e8ceda440f71e70c60603ffa Mon Sep 17 00:00:00 2001 From: Joshua Potter Date: Wed, 23 Aug 2023 14:22:19 -0600 Subject: [PATCH] Enderton (set). Add cardinal arithmetic theorems and exercise prompts. --- Bookshelf/Enderton/Set.tex | 149 ++++++++++++++++++++++++++++++++++--- preamble.tex | 1 + 2 files changed, 141 insertions(+), 9 deletions(-) diff --git a/Bookshelf/Enderton/Set.tex b/Bookshelf/Enderton/Set.tex index 728cd53..0cd7e35 100644 --- a/Bookshelf/Enderton/Set.tex +++ b/Bookshelf/Enderton/Set.tex @@ -77,6 +77,19 @@ \lean{Mathlib/Data/Set/Prod}{Set.prod} +\section{\defined{Cardinal Arithmetic}}% +\hyperlabel{sec:cardinal-arithmetic} + + Let $\kappa$ and $\lambda$ be any cardinal numbers. + \begin{enumerate}[(a)] + \item $\kappa + \lambda = \card{(K \cup L)}$, where $K$ and $L$ are any + disjoint sets of cardinality $\kappa$ and $\lambda$, respectively. + \item $\kappa \cdot \lambda = \card{(K \times L)}$, where $K$ and $L$ are + any sets of cardinality $\kappa$ and $\lambda$, respectively. + \item $\kappa^\lambda = \card{^L{K}}$, where $K$ and $L$ are any sets of + cardinality $\kappa$ and $\lambda$, respectively. + \end{enumerate} + \section{\defined{Compatible}}% \hyperlabel{ref:compatible} @@ -8811,11 +8824,73 @@ Hence $S'$ is a finite set. \end{proof} +\subsection{\sorry{Theorem 6H}}% +\hyperlabel{sub:theorem-6h} + + Assume that $\equinumerous{K_1}{K_2}$ and $\equinumerous{L_1}{L_2}$. + \begin{enumerate}[(a)] + \item If $K_1 \cap L_1 = K_2 \cap L_2 = \emptyset$, then + $\equinumerous{K_1 \cup L_1}{K_2 \cup L_2}$. + \item $\equinumerous{K_1 \times L_1}{K_2 \times L_2}$. + \item $\equinumerous{^{(L_1)}{K_1}}{^{(L_2)}{K_2}}$. + \end{enumerate} + + \begin{proof} + TODO + \end{proof} + +\subsection{\sorry{Theorem 6I}}% +\hyperlabel{sub:theorem-6i} + + For any cardinal numbers $\kappa$, $\lambda$, and $\mu$: + \begin{enumerate} + \item $\kappa + \lambda = \lambda + \kappa$ and + $\kappa \cdot \lambda = \lambda \cdot \kappa$. + \item $\kappa + (\lambda + \mu) = (\kappa + \lambda) + \mu$ and + $\kappa \cdot (\lambda \cdot \mu) = (\kappa \cdot \lambda) \cdot \mu$. + \item $\kappa \cdot (\lambda + \mu) = + \kappa \cdot \lambda + \kappa \cdot \mu$. + \item $\kappa^{\lambda + \mu} = \kappa^\lambda \cdot \kappa^\mu$. + \item $(\kappa \cdot \lambda)^\mu = \kappa^\mu \cdot \lambda^\mu$. + \item $(\kappa^\lambda)^\mu = \kappa^{\lambda \cdot \mu}$. + \end{enumerate} + + \begin{proof} + TODO + \end{proof} + +\subsection{\sorry{Theorem 6J}}% +\hyperlabel{sub:theorem-6j} + + Let $m$ and $n$ be finite cardinals. + Then + \begin{align*} + m + n & = m +_\omega n, \\ + m \cdot n = m \cdot_\omega n, \\ + m^n = m^n, + \end{align*} + where on the right side we use the operations of $\omega$ defined via + recursion and on the left side we use the operations of cardinal arithmetic. + + \begin{proof} + TODO + \end{proof} + +\subsection{\sorry{Corollary 6K}}% +\hyperlabel{sub:corollary-6k} + + If $A$ and $B$ are finite, then $A \cup B$, $A \times B$, and $^B{A}$ are also + finite. + + \begin{proof} + TODO + \end{proof} + \section{Exercises 6}% \hyperlabel{sec:exercises-6} \subsection{\unverified{Exercise 6.1}}% -\hyperlabel{sub:exercise-6-1} +\hyperlabel{sub:exercise-6.1} Show that the equation $$f(m, n) = 2^m(2n + 1) - 1$$ defines a one-one-one correspondence between $\omega \times \omega$ and $\omega$. @@ -8894,7 +8969,7 @@ \end{proof} \subsection{\unverified{Exercise 6.2}}% -\hyperlabel{sub:exercise-6-2} +\hyperlabel{sub:exercise-6.2} Show that in Fig. 32 we have: \begin{align*} @@ -8933,7 +9008,7 @@ \end{proof} \subsection{\unverified{Exercise 6.3}}% -\hyperlabel{sub:exercise-6-3} +\hyperlabel{sub:exercise-6.3} Find a one-to-one correspondence between the open unit interval $\ioo{0}{1}$ and $\mathbb{R}$ that takes rationals to rationals and irrationals to @@ -9087,7 +9162,7 @@ \end{proof} \subsection{\sorry{Exercise 6.4}}% -\hyperlabel{sub:exercise-6-4} +\hyperlabel{sub:exercise-6.4} Construct a one-to-one correspondence between the closed unit interval $$\icc{0}{1} = \{x \in \mathbb{R} \mid 0 \leq x \leq 1\}$$ @@ -9098,7 +9173,7 @@ \end{proof} \subsection{\verified{Exercise 6.5}}% -\hyperlabel{sub:exercise-6-5} +\hyperlabel{sub:exercise-6.5} Prove \nameref{sub:theorem-6a}. @@ -9107,7 +9182,7 @@ \end{proof} \subsection{\sorry{Exercise 6.6}}% -\hyperlabel{sub:exercise-6-6} +\hyperlabel{sub:exercise-6.6} Let $\kappa$ be a nonzero cardinal number. Show there does not exist a set to which every set of cardinality $\kappa$ @@ -9118,7 +9193,7 @@ \end{proof} \subsection{\sorry{Exercise 6.7}}% -\hyperlabel{sub:exercise-6-7} +\hyperlabel{sub:exercise-6.7} Assume that $A$ is finite and $f \colon A \rightarrow A$. Show that $f$ is one-to-one iff $\ran{f} = A$. @@ -9128,7 +9203,7 @@ \end{proof} \subsection{\sorry{Exercise 6.8}}% -\hyperlabel{sub:exercise-6-8} +\hyperlabel{sub:exercise-6.8} Prove that the union of two finite sets is finite, without any use of arithmetic. @@ -9138,7 +9213,7 @@ \end{proof} \subsection{\sorry{Exercise 6.9}}% -\hyperlabel{sub:exercise-6-9} +\hyperlabel{sub:exercise-6.9} Prove that the Cartesian product of two finite sets is finite, without any use of arithmetic. @@ -9147,4 +9222,60 @@ TODO \end{proof} +\subsection{\sorry{Exercise 6.10}}% +\hyperlabel{sub:exercise-6.10} + + Prove part 4 of \nameref{sub:theorem-6i}. + + \begin{proof} + TODO + \end{proof} + +\subsection{\sorry{Exercise 6.11}}% +\hyperlabel{sub:exercise-6.11} + + Prove part 5 of \nameref{sub:theorem-6i}. + + \begin{proof} + TODO + \end{proof} + +\subsection{\sorry{Exercise 6.12}}% +\hyperlabel{sub:exercise-6.12} + + The proof to \nameref{sub:theorem-6i} involves eight instances of showing two + sets to be equinumerous. + (The eight are listed in the proof of the theorem as statements numbered 1-6.) + In which of these eight cases does equality actually hold? + + \begin{proof} + TODO + \end{proof} + +\subsection{\sorry{Exercise 6.13}}% +\hyperlabel{sub:exercise-6.13} + + Show that a finite union of finite sets is finite. + That is, show that if $B$ is a finite set whose members are themselves finite + sets, then $\bigcup{B}$ is finite. + + \begin{proof} + TODO + \end{proof} + +\subsection{\sorry{Exercise 6.14}}% +\hyperlabel{sub:exercise-6.14} + + Define a \textit{permutation} of $K$ to be any one-to-one function from $K$ + onto $K$. + We can the define the factorial operation on cardinal numbers by the equation + $$\kappa! = \card{\{f \mid f \text{ is a permutation of } K\}},$$ + where $K$ is any set of cardinality $\kappa$. + Show that $\kappa!$ is well defined, i.e. the value of $\kappa!$ is + independent of just which set $K$ is chosen. + + \begin{proof} + TODO + \end{proof} + \end{document} diff --git a/preamble.tex b/preamble.tex index f56500e..a5a6078 100644 --- a/preamble.tex +++ b/preamble.tex @@ -166,6 +166,7 @@ % ======================================== \newcommand{\abs}[1]{\left|#1\right|} +\newcommand{\card}[1]{\mathop{\text{card}}{#1}} \newcommand{\ceil}[1]{\left\lceil#1\right\rceil} \newcommand{\ctuple}[2]{\left< #1, \cdots, #2 \right>} \newcommand{\dom}[1]{\textop{dom}{#1}}