From 793cfbc8fd2df3f3e0d546295c3699088b87c68e Mon Sep 17 00:00:00 2001 From: Joshua Potter Date: Sat, 5 Aug 2023 11:18:53 -0600 Subject: [PATCH] Enderton. Finish drafting prompts of Natural Numbers section. --- Bookshelf/Enderton/Set.tex | 362 ++++++++++++++++++++++++++++++++++++- preamble.tex | 7 + 2 files changed, 364 insertions(+), 5 deletions(-) diff --git a/Bookshelf/Enderton/Set.tex b/Bookshelf/Enderton/Set.tex index b66d7dc..1e56092 100644 --- a/Bookshelf/Enderton/Set.tex +++ b/Bookshelf/Enderton/Set.tex @@ -7,6 +7,7 @@ \makeleancommands{../..} \newcommand{\pair}[1]{\left< #1 \right>} +\newcommand{\ineq}{\,\mathop{\underline{\in}}\,} \begin{document} @@ -446,7 +447,7 @@ Likewise, define $m$ to be \textbf{less than or equal to} $n$ if and only if That is, \begin{align*} m \leq n - & \iff m \mathop{\underline{\in}} n \\ + & \iff m \ineq n \\ & \iff m < n \lor m = n. \end{align*} @@ -548,6 +549,19 @@ For any set $a$, there is a set whose members are exactly the subsets of $a$: \end{axiom} +\section{\defined{Proper Subset}}% +\hyperlabel{ref:proper-subset} + +A set $A$ is said to be a \textbf{proper subset} of $B$ ($A \subset B$) if and + only if it is a subset of $B$ that is unequal to $B$. +$$A \subset B \iff A \subseteq B \land A \neq B.$$ + +\begin{definition} + + \lean*{Std/Classes/SetNotation}{HasSSubset} + +\end{definition} + \section{\defined{Quotient Set}}% \hyperlabel{ref:quotient-set} @@ -7212,7 +7226,7 @@ Show that $<_L$ is a linear ordering on $A \times B$. \end{proof} -\subsection{\pending{Lemma 4L(a)}}% +\subsection{\verified{Lemma 4L(a)}}% \hyperlabel{sub:lemma-4l-a} \begin{lemma}[4L(a)] @@ -7221,6 +7235,10 @@ Show that $<_L$ is a linear ordering on $A \times B$. \end{lemma} +\begin{note} + Here I referred to Enderton's proof in the forward direction. +\end{note} + \begin{proof} \lean{Std/Data/Nat/Lemmas}{Nat.succ\_lt\_succ\_iff} @@ -7229,13 +7247,50 @@ Show that $<_L$ is a linear ordering on $A \times B$. \paragraph{($\Rightarrow$)}% - Suppose $m \in n$. - TODO + Define $$S = \{n \in \omega \mid (\forall m \in n) m^+ \in n^+\}.$$ + We prove that (i) $0 \in S$ and (ii) if $n \in S$ then $n^+ \in S$. + Afterwards we show that (iii) the forward direction of the stated + biconditional holds. + + \subparagraph{(i)}% + \hyperlabel{spar:lemma-4l-a-i} + + $0 \in S$ vacuously. + That is, there are no members of $0 = \emptyset$ by definition. + + \subparagraph{(ii)}% + \hyperlabel{spar:lemma-4l-a-ii} + + Suppose $n \in S$. + We need to show for all $m \in n^+$, $m^+ \in n^{++}$. + Let $m \in n^+ = n \cup \{n\}$. + Then $m \in n$ or $m \in \{n\}$. + + If $m \in n$, then $n \in S$ implies $m^+ \in n^+ \in n^{++}$. + By \nameref{sub:theorem-4f}, every natural number is a + \nameref{ref:transitive-set}. + Therefore $m^+ \in n^{++}$. + On the other hand, if $m \in \{n\}$, then $m = n$. + Since $n^+ \in n^{++}$, it immediately follows $m^+ \in n^{++}$. + Hence $n^+ \in S$. + + \subparagraph{(iii)}% + + By \nameref{spar:lemma-4l-a-i} and \nameref{spar:lemma-4l-a-ii}, $S$ is an + \nameref{ref:inductive-set}. + Hence \nameref{sub:theorem-4b} implies $S = \omega$. + Thus for all $n \in \omega$, $m \in n \Rightarrow m^+ \in n^+$. \paragraph{($\Leftarrow$)}% Suppose $m^+ \in n^+$. - TODO + The definition of \nameref{ref:successor} immediately implies that + $m \in m^+$. + By \nameref{sec:ordering-natural-numbers}, $m^+ \in n^+$ implies $m^+ \in n$ + or $m^+ = n$. + If the latter, $m \in n$ immediately follows. + If the former, we note $n$ is a transitive set by \nameref{sub:theorem-4f}. + Thus $m \in m^+ \in n$ implies $m \in n$. \end{proof} @@ -7449,6 +7504,166 @@ Show that $<_L$ is a linear ordering on $A \times B$. \end{proof} +\subsection{\unverified{Corollary 4M}}% +\hyperlabel{sub:corollary-4m} + +\begin{corollary}[4M] + + For any natural numbers $m$ and $n$, + \begin{equation} + \hyperlabel{sub:corollary-4m-eq1} + m \in n \iff m \subset n + \end{equation} + and + \begin{equation} + \hyperlabel{sub:corollary-4m-eq2} + m \ineq n \iff m \subseteq n. + \end{equation} + +\end{corollary} + +\begin{proof} + + \paragraph{\eqref{sub:corollary-4m-eq1}}% + + We prove both directions of the biconditional specified in + \eqref{sub:corollary-4m-eq1}: + + \subparagraph{($\Rightarrow$)}% + + Suppose $m \in n$ and $t \in m$. + By \nameref{sub:theorem-4f}, $n$ is a \nameref{ref:transitive-set}. + Therefore $t \in n$. + Hence $m \subseteq n$. + Since $m \in n$, \nameref{sub:linear-ordering-natural-numbers} implies + $m \neq n$. + Thus, by definition of \nameref{ref:proper-subset}, $m \subset n$. + + \subparagraph{($\Leftarrow$)}% + + Suppose $m \subset n$. + By \nameref{sub:linear-ordering-natural-numbers}, exactly one of + $$m \in n, \quad m = n, \quad n \in m$$ holds. + By definition of \nameref{ref:proper-subset}, $m \subseteq n$ and + $m \neq n$. + Furthermore, it cannot be that $n \in m$ since otherwise $n \in n$, + contradicting \nameref{sub:lemma-4l-b}. + Thus $m \in n$ is the only possibility. + + \paragraph{\eqref{sub:corollary-4m-eq2}}% + + We prove both directions of the biconditional specified in + \eqref{sub:corollary-4m-eq2}: + + \subparagraph{($\Rightarrow$)}% + + Suppose $m \ineq n$. + By definition, $m \in n$ or $m = n$. + Let $p \in m$. + Then $p \in m \in n$ or $p \in m = n$. + By \nameref{sub:theorem-4f}, $n$ is a \nameref{ref:transitive-set}. + Thus $p \in n$ in either case. + Hence $m \subseteq n$. + + \subparagraph{($\Leftarrow$)}% + + Suppose $m \subseteq n$. + By \nameref{sub:linear-ordering-natural-numbers}, exactly one of + $$m \in n, \quad m = n, \quad n \in m$$ holds. + But it cannot be that $n \in m$ since that would imply $n \in n$, + contradicting \nameref{sub:lemma-4l-b}. + Therefore $m \in n$ or $m = n$. + Hence $m \ineq n$. + +\end{proof} + +\subsection{\sorry{Theorem 4N}}% +\hyperlabel{sub:theorem-4n} + +\begin{theorem}[4N] + + For any natural numbers $n$, $m$, and $p$, $$m \in n \iff m + p \in n + p.$$ + If, in addition, $p \neq 0$, then $$m \in n \iff m \cdot p \in n \cdot p.$$ + +\end{theorem} + +\begin{proof} + + TODO + +\end{proof} + +\subsection{\sorry{Corollary 4P}}% +\hyperlabel{sub:corollary-4p} + +\begin{corollary}[4P] + + The following cancellation laws hold for $m$, $n$, and $p$ in $\omega$: + \begin{align*} + m + p = n + p & \Rightarrow m = n, \\ + m \cdot p = n \cdot p & \Rightarrow m = n. + \end{align*} + +\end{corollary} + +\begin{proof} + + TODO + +\end{proof} + +\subsection{\sorry{% + Well Ordering of \texorpdfstring{$\omega$}{Natural Numbers}}}% +\hyperlabel{sub:well-ordering-natural-numbers} + +\begin{theorem} + + Let $A$ be a nonempty subset of $\omega$. + Then there is some $m \in A$ such that $m \ineq n$ for all $n \in A$. + +\end{theorem} + +\begin{proof} + + TODO + +\end{proof} + +\subsection{\sorry{Corollary 4Q}}% +\hyperlabel{sub:corollary-4q} + +\begin{corollary}[4Q] + + There is no function $f \colon \omega \rightarrow \omega$ such that + $f(n^+) \in f(n)$ for every natural number $n$. + +\end{corollary} + +\begin{proof} + + TODO + +\end{proof} + +\subsection{\sorry{% + Strong Induction Principle for \texorpdfstring{$\omega$}{Natural Numbers}}}% +\hyperlabel{sub:strong-induction-principle-natural-numbers} + +\begin{theorem} + + Let $A$ be a subset of $\omega$, and assume that for every $n \in \omega$, + $$\text{if every number less than } n \text{ is in } A, + \text{ then } n \in A.$$ + Then $A = \omega$. + +\end{theorem} + +\begin{proof} + + TODO + +\end{proof} + \section{Exercises 4}% \hyperlabel{sec:exercises-4} @@ -8027,4 +8242,141 @@ Prove that $m^{n+p} = m^n \cdot m^p$. \end{proof} +\subsection{\sorry{Exercise 4.18}}% +\hyperlabel{sub:exercise-4.18} + +Simplify $\img{\in_\omega^{-1}}{\{7, 8\}}$. + +\begin{proof} + + TODO + +\end{proof} + +\subsection{\sorry{Exercise 4.19}}% +\hyperlabel{sub:exercise-4.19} + +Prove that if $m$ is a natural number and $d$ is a nonzero number, then there + exist numbers $q$ and $r$ such that $m = (d \cdot q) + r$ and $r$ is less than + $d$. + +\begin{proof} + + TODO + +\end{proof} + +\subsection{\sorry{Exercise 4.20}}% +\hyperlabel{sub:exercise-4.20} + +Let $A$ be a nonempty subset of $\omega$ such that $\bigcup A = A$. +Show that $A = \omega$. + +\begin{proof} + + TODO + +\end{proof} + +\subsection{\sorry{Exercise 4.21}}% +\hyperlabel{sub:exercise-4.21} + +Show that no natural number is a subset of any of its elements. + +\begin{proof} + + TODO + +\end{proof} + +\subsection{\sorry{Exercise 4.22}}% +\hyperlabel{sub:exercise-4.22} + +Show that for any natural numbers $m$ and $p$ we have $m \in m + p^+$. + +\begin{proof} + + TODO + +\end{proof} + +\subsection{\sorry{Exercise 4.23}}% +\hyperlabel{sub:exercise-4.23} + +Assume that $m$ and $n$ are natural numbers with $m$ less than $n$. +Show that there is some $p$ in $\omega$ for which $m + p^+ = n$. +(It follows form this and the preceding exercise that $m$ is less than $n$ iff + $(\exists p \in \omega)m + p^+ = n$.) + +\begin{proof} + + TODO + +\end{proof} + +\subsection{\sorry{Exercise 4.24}}% +\hyperlabel{sub:exercise-4.24} + +Assume that $m + n = p + q$. +Show that $$m \in p \iff n \in q.$$ + +\begin{proof} + + TODO + +\end{proof} + +\subsection{\sorry{Exercise 4.25}}% +\hyperlabel{sub:exercise-4.25} + +Assume that $n \in m$ and $q \in p$. +Show that $$(m \cdot q) + (n \cdot p) \in (m \cdot p) + (n \cdot q).$$ +[\textit{Suggestion:} Use \nameref{sub:exercise-4.23}.] + +\begin{proof} + + TODO + +\end{proof} + +\subsection{\sorry{Exercise 4.26}}% +\hyperlabel{sub:exercise-4.26} + +Assume that $n \in \omega$ and $f \colon n^+ \rightarrow \omega$. +Show that $\ran{f}$ has a largest element. + +\begin{proof} + + TODO + +\end{proof} + +\subsection{\sorry{Exercise 4.27}}% +\hyperlabel{sub:exercise-4.27} + +Assume that $A$ is a set, $G$ is a function, and $f_1$ and $f_2$ map $\omega$ + into $A$. +Further assume that for each $n \in \omega$ both $f_1 \restriction n$ and + $f_2 \restriction n$ belong to $\dom{G}$ and + $$f_1(n) = G(f_1 \restriction n) \land f_2(n) = G(f_2 \restriction n).$$ +Show that $f_1 = f_2$. + +\begin{proof} + + TODO + +\end{proof} + +\subsection{\sorry{Exercise 4.28}}% +\hyperlabel{sub:exercise-4.28} + +Rewrite the proof of \nameref{sub:theorem-4g} using, in place of induction, the + well-ordering of $\omega$. + +\begin{proof} + + TODO + +\end{proof} + \end{document} diff --git a/preamble.tex b/preamble.tex index 3b4eec1..cb04f75 100644 --- a/preamble.tex +++ b/preamble.tex @@ -99,6 +99,13 @@ \newenvironment{definition}{\@statement{Definition}}{\hfill$\square$} \renewenvironment{proof}{\@statement{Proof}}{\hfill$\square$} +\newtheorem{corollaryinner}{Corollary} +\newenvironment{corollary}[1][]{% + \ifstrempty{#1} + {\corollaryinner} + {\renewcommand\thecorollaryinner{#1}\corollaryinner} +}{\endcorollaryinner} + \newtheorem{lemmainner}{Lemma} \newenvironment{lemma}[1][]{% \ifstrempty{#1}