From 1260c493bcae7094a2a770f99042554057f6c7f8 Mon Sep 17 00:00:00 2001 From: Joshua Potter Date: Mon, 24 Jul 2023 12:19:01 -0600 Subject: [PATCH] Enderton. Finish "Peano's Postulates" section. --- Bookshelf/Enderton/Set.tex | 116 ++++++++++++++++++++++++++++++++----- 1 file changed, 100 insertions(+), 16 deletions(-) diff --git a/Bookshelf/Enderton/Set.tex b/Bookshelf/Enderton/Set.tex index dcd5e05..17afa85 100644 --- a/Bookshelf/Enderton/Set.tex +++ b/Bookshelf/Enderton/Set.tex @@ -6170,7 +6170,7 @@ Show that $<_L$ is a linear ordering on $A \times B$. \end{proof} -\subsection{\sorry{Theorem 4G}}% +\subsection{\unverified{Theorem 4G}}% \label{sub:theorem-4g} \begin{theorem}[4G] @@ -6181,7 +6181,32 @@ Show that $<_L$ is a linear ordering on $A \times B$. \begin{proof} - TODO + Let $T = \{n \in \omega \mid \forall t \in n, t \in \omega\}$. + We prove that (i) $T$ is an \nameref{ref:inductive-set} and then (ii) every + member of a natural number is itself a natural number. + + \paragraph{(i)}% + \label{par:theorem-4g-i} + + First, it vacuously holds that $\emptyset \in T$. + Next, let $n \in T$. + We must prove that $n^+ \in T$ as well. + By definition of the \nameref{ref:successor}, $n^+ = n \cup \{n\}$. + That is, either $n^+ = n$ or $n^+ = \{n\}$. + If the former, then every member of $n^+$ must be a natural number since + this already holds for $n$. + If the latter, the only member of $n^+$ is $n$ which is, by definition of + $T$, a natural number. + Thus $n^+ \in T$. + We conclude that $T$ is an inductive set. + + \paragraph{(ii)}% + + Since $T \subseteq \omega$, \nameref{par:theorem-4g-i} and + \nameref{sub:theorem-4b} implies $T = \omega$. + Thus the member of every natural number is itself a natural number. + In other words, $\bigcup \omega \subseteq \omega$. + Therefore $\omega$ is indeed a \nameref{ref:transitive-set}. \end{proof} @@ -6208,18 +6233,23 @@ Show that $1 \neq 3$ i.e., that $\emptyset^+ \neq \emptyset^{+++}$. \end{proof} -\subsection{\sorry{Exercise 4.2}}% +\subsection{\unverified{Exercise 4.2}}% \hyperlabel{sub:exercise-4.2} Show that if $a$ is a transitive set, then $a^+$ is also a transitive set. \begin{proof} - TODO + Suppose $a$ is a \nameref{ref:transitive-set}. + By \nameref{sub:theorem-4e}, it follows $\bigcup \left(a^+\right) = a$. + By definition of a \nameref{ref:successor}, $a^+ = a \cup \{a\}$. + Thus it immediately follows + $$\bigcup \left(a^+\right) = a \subseteq a \cup \{a\} = a^+.$$ + Therefore $a^+$ is indeed a transitive set. \end{proof} -\subsection{\sorry{Exercise 4.3}}% +\subsection{\unverified{Exercise 4.3}}% \hyperlabel{sub:exercise-4.3} \begin{enumerate}[(a)] @@ -6231,47 +6261,101 @@ Show that if $a$ is a transitive set, then $a^+$ is also a transitive set. \begin{proof} - TODO + \paragraph{(a)}% + + Suppose $a$ is a \nameref{ref:transitive-set}. + We show that $\bigcup \powerset{a} \subseteq \powerset{a}$. + Let $t \in \bigcup \powerset{a}$. + By definition of the \nameref{ref:power-set}, there exists some + $X \subseteq a$ such that $t \in X$. + Thus $t \in a$. + Because $a$ is a transitive set, every member of $t$ is a member of $a$. + In other words, $t \subseteq a$. + Equivalently, $t \in \powerset{a}$. + + \paragraph{(b)}% + + Suppose $\powerset{a}$ is a transitive set. + We show that $\bigcup a \subseteq a$. + Let $t \in \bigcup a$. + Then there exists some $b \in a$ such that $t \in b$. + Since $\{b\}$ is a member of $\powerset{a}$ and $\powerset{a}$ is a + transitive set, $b \in \powerset{a}$. + That is, $b \subseteq a$. + Thus $t \in a$. \end{proof} -\subsection{\sorry{Exercise 4.4}}% +\subsection{\unverified{Exercise 4.4}}% \hyperlabel{sub:exercise-4.4} Show that if $a$ is a transitive set, then $\bigcup a$ is also a transitive set. \begin{proof} - TODO + Suppose $a$ is a transitive set. + We show that $\bigcup\bigcup{a} \subseteq \bigcup a$. + Let $t \in \bigcup\bigcup{a}$. + Then there exists some $b \in \bigcup{a}$ such that $t \in b$. + Since $a$ is transitive, $\bigcup{a} \subseteq a$. + Thus $b \in a$ and $t \in \bigcup a$. \end{proof} -\subsection{\sorry{Exercise 4.5}}% +\subsection{\unverified{Exercise 4.5}}% \hyperlabel{sub:exercise-4.5} Assume that every member of $\mathscr{A}$ is a transitive set. \begin{enumerate}[(a)] - \item Show that $\bigcup \mathscr{A}$ is a transitive set. - \item Show that $\bigcap \mathscr{A}$ is a transitive set (assume that + \item Show that $\bigcup\mathscr{A}$ is a transitive set. + \item Show that $\bigcap\mathscr{A}$ is a transitive set (assume that $\mathscr{A}$ is nonempty). \end{enumerate} \begin{proof} - TODO + \paragraph{(a)}% + + Suppose every member of $\mathscr{A}$ is a transitive set. + We show that $\bigcup\bigcup{\mathscr{A}} \subseteq \bigcup{\mathscr{A}}$. + Let $t \in \bigcup\bigcup{\mathscr{A}}$. + Then there exists some $b_1 \in \bigcup{\mathscr{A}}$ such that $t \in b_1$. + Likewise there exists some $b_2 \in \mathscr{A}$ such that $b_1 \in b_2$. + By hypothesis, $b_2$ is transitive meaning $t \in b_2$. + Thus $t \in \bigcup{\mathscr{A}}$. + + \paragraph{(b)}% + + Suppose every member of nonempty set $\mathscr{A}$ is a transitive set. + We show that $\bigcup\bigcap{\mathscr{A}} \subseteq \bigcap{\mathscr{A}}$. + Let $t \in \bigcup\bigcap{\mathscr{A}}$. + Then there exists some $b \in \bigcap{\mathscr{A}}$ such that $t \in b$. + Thus $b$ is a member of every member of $\mathscr{A}$. + By hypothesis, every member of $\mathscr{A}$ is a transitive set meaning + $t$ must be a member of every member of $\mathscr{A}$. + In other words, $t \in \bigcap{\mathscr{A}}$. \end{proof} -\subsection{\sorry{Exercise 4.6}}% +\subsection{\unverified{Exercise 4.6}}% \hyperlabel{sub:exercise-4.6} -Prove the converse to Theorem 4E: If $\bigcup \left(a^+\right) = a$, then $a$ is - a transitive set. +Prove the converse to \nameref{sub:theorem-4e}: If + $\bigcup \left(a^+\right) = a$, then $a$ is a transitive set. \begin{proof} - TODO + Let $a$ be a set such that $\bigcup \left(a^+\right) = a$. + Then + \begin{align*} + \bigcup \left(a^+\right) + & = \bigcup (a \cup \{a\}) & \textref{ref:successor} \\ + & = \bigcup a \cup \bigcup \{a\} & \textref{sub:exercise-2.21} \\ + & = a. & \text{by hypothesis} + \end{align*} + But $$\bigcup{a} \subseteq \bigcup a \cup \bigcup \{a\} = a.$$ + Thus $a$ is indeed a \nameref{ref:transitive-set}. \end{proof}